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892019 Reader Livro Muito Bom
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Aircraft Stress Analysis andStructural Design
Reader AE2-521N Version 102
Mostafa Abdalla
Roeland De Breuker
Zafer Gurdal
Jan Hol
Chair of Aerospace StructuresFaculty of Aerospace Engineering
Delft University of TechnologyThe Netherlands
January 2008
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ii
Copyright c 2007 by Mostafa Abdalla Roeland De Breuker Zafer
Gurdal and Jan HolldquoThe copyright of this manual rests with the authors No quotations from itshould be published without the authorrsquos prior written consent and informa-tion derived from it should be acknowledgedrdquo
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iii
Modifications
Date Version Modification19012008 101 Equation 39 changed19012008 102 Equations 48 and 42 changed
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iv
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Contents
List of Figures viii
1 Introduction 1
2 Stress Analysis and Design of Statically Determinate Trusses 5
21 Truss Properties 6
22 Static Determinacy of Trusses 6
23 Truss Analysis 7
231 Truss Definition 8232 Stress Analysis 9
24 Stress Design 10
25 Example 10
3 Stress Analysis and Design of Statically Determinate Beams 13
31 What is a beam 13
32 Equilibrium Equations 14
33 Stresses in Beams 18
34 Discrete Beam Analysis 19
35 Example 20
4 Stress Analysis and Design of Statically Determinate Plates 23
41 Uniform In-Plane Loading 23
42 Circle of Mohr 24
43 Stress Design 27
44 Example 28
5 Stress Concentrations and Multiple Loads 31
51 Stress Concentrations 31
52 Design for Multiple Loads 33
v
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CONTENTS vi
6 Displacement Analysis and Design of Statically Determinate
Trusses 37
61 Dummy Load Method 3862 Displacement Design 4163 Example 42
7 Displacement Analysis and Design of Statically Determinate
Beams 45
71 Example 46
8 Analysis of Statically Indeterminate Trusses 49
81 Displacement Compatibility 4982 Action Item List for Analysis of Indeterminate Trusses 5183 Example 52
9 Analysis of Statically Indeterminate Beams 55
91 Force Method of Analysis 5592 Flexibility Coefficients 5693 Example 57
10 Analysis of Statically Indeterminate Plates 61101 Biaxial Stress State 61102 Multimaterial Plates 62
11 Actuated Structures 65
111 Thermal Actuation 65112 Piezoelectric Actuation 66113 Response of an Actuated Member 67114 Dummy Load Method for Structures with Actuated Members 69115 Example 69
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List of Figures
21 Rib truss structure example [Courtesy of wwwsteenaerocom] 522 Typical simple truss structure [Courtesy of wwwcompuzonecokr] 623 Truss support types 724 Statically determinate and indeterminate trusses 725 Internal force distribution inside a truss 826 Three bar truss example 1127 Free-body diagram of each node 11
31 Aircraft beam model [Courtesy of wwwflightinternationalcom] 14
32 Engine beam model [Courtesy of wwwmilnetcom] 1533 Local beam axes 1534 Beam sign convenction 1635 Infinitesimal beam segment 1736 Force jump over an infinitesimal beam segment 18
41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom] 2442 Example of fully loaded stress element 2543 Free body diagram for the plate equilibrium equations 2544 Layout and definition of Mohrrsquos circle 2745 Special cases of Mohrrsquos circle 28
46 Example load case on a stress element 29
51 Illustration of the principle of De Saint-Venant 3252 Geometric definitions and K factor for a hole [Courtesy of
Beer PF Mechanics of Materials ] 3353 Geometric definitions and K factor for a cross-section change
[Courtesy of Beer PF Mechanics of Materials ] 3454 Combined airfoil load 3455 Flight envelope containing multiple load cases 3556 Plotted inequalities for the multiple load case 36
61 Principle of virtual work 38
vii
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LIST OF FIGURES viii
62 Principle of complementary energy 3963 Linear force-displacement diagram 4064 Truss structure with applied dummy load 4265 Nodal internal dummy loads 43
71 Statically determinate beam example 47
81 Force superposition principle 5082 Split-up of an indeterminate truss 50
83 Statically indeterminate six bar truss 5291 Illustration of the flexibility coefficient 5692 Solution of a statically indeterminate beam using flexibility
coefficients 5793 Analysis example of a statically indeterminate beam 58
101 Biaxial stress state plate 62102 Indeterminate multimaterial plates (left one statically inde-
terminate force right two statically indeterminate forces) 63
111 Thermal extension of a bar 66112 Piezoelectric thickness change of a bar 67113 Statically indeterminate six bar truss 70
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Chapter 1
Introduction
The primary purpose of this reader is to describe the use of analysis equationsand methodologies of structures for design purposes It is common these daysthat we hear from industry that the students graduating with engineeringdegrees do not know how to design whether it is structures or mechanicalsystems or systems from other engineering disciplines We therefore put theemphasis in this course on design
Anybody who has some understanding of the design process however re-alises that without a thorough understanding of the use of analysis methodsit will not be possible to design at least a reliable system It is thereforeimportant to establish a sound and firm analysis foundation before one canstart the design practice The approach used in this reader however is dif-ferent from the traditional one in which analysis and designs are taught indifferent portions of the course Instead we will use an approach in whichsmall portions (sections) of topics from analysis are first introduced immedi-ately followed by their design implementation A more detailed descriptionof the outline and the contents of the reader is provided in the following
The reader is divided into two primary sections because of a very impor-tant concept called statical determinacy that has a very strong influenceon the way structures are designed It is of course too early to completelydescribe the impact of statical indeterminacy on design It will be sufficientto state at this point that statical determinacy simplifies the design processof a structure made of multiple components by making it possible to designindividual components independent from one another Structural indeter-minacy on the other hand causes the internal load distribution in a givenstructural system to be dependent on the dimensional and material proper-ties of the individual component that are typically being designed That isas the design of the individual components change the loads acting on thosecomponents for which they are being designed also change Hence individ-
1
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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4
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
49
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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ii
Copyright c 2007 by Mostafa Abdalla Roeland De Breuker Zafer
Gurdal and Jan HolldquoThe copyright of this manual rests with the authors No quotations from itshould be published without the authorrsquos prior written consent and informa-tion derived from it should be acknowledgedrdquo
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iii
Modifications
Date Version Modification19012008 101 Equation 39 changed19012008 102 Equations 48 and 42 changed
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iv
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Contents
List of Figures viii
1 Introduction 1
2 Stress Analysis and Design of Statically Determinate Trusses 5
21 Truss Properties 6
22 Static Determinacy of Trusses 6
23 Truss Analysis 7
231 Truss Definition 8232 Stress Analysis 9
24 Stress Design 10
25 Example 10
3 Stress Analysis and Design of Statically Determinate Beams 13
31 What is a beam 13
32 Equilibrium Equations 14
33 Stresses in Beams 18
34 Discrete Beam Analysis 19
35 Example 20
4 Stress Analysis and Design of Statically Determinate Plates 23
41 Uniform In-Plane Loading 23
42 Circle of Mohr 24
43 Stress Design 27
44 Example 28
5 Stress Concentrations and Multiple Loads 31
51 Stress Concentrations 31
52 Design for Multiple Loads 33
v
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CONTENTS vi
6 Displacement Analysis and Design of Statically Determinate
Trusses 37
61 Dummy Load Method 3862 Displacement Design 4163 Example 42
7 Displacement Analysis and Design of Statically Determinate
Beams 45
71 Example 46
8 Analysis of Statically Indeterminate Trusses 49
81 Displacement Compatibility 4982 Action Item List for Analysis of Indeterminate Trusses 5183 Example 52
9 Analysis of Statically Indeterminate Beams 55
91 Force Method of Analysis 5592 Flexibility Coefficients 5693 Example 57
10 Analysis of Statically Indeterminate Plates 61101 Biaxial Stress State 61102 Multimaterial Plates 62
11 Actuated Structures 65
111 Thermal Actuation 65112 Piezoelectric Actuation 66113 Response of an Actuated Member 67114 Dummy Load Method for Structures with Actuated Members 69115 Example 69
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List of Figures
21 Rib truss structure example [Courtesy of wwwsteenaerocom] 522 Typical simple truss structure [Courtesy of wwwcompuzonecokr] 623 Truss support types 724 Statically determinate and indeterminate trusses 725 Internal force distribution inside a truss 826 Three bar truss example 1127 Free-body diagram of each node 11
31 Aircraft beam model [Courtesy of wwwflightinternationalcom] 14
32 Engine beam model [Courtesy of wwwmilnetcom] 1533 Local beam axes 1534 Beam sign convenction 1635 Infinitesimal beam segment 1736 Force jump over an infinitesimal beam segment 18
41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom] 2442 Example of fully loaded stress element 2543 Free body diagram for the plate equilibrium equations 2544 Layout and definition of Mohrrsquos circle 2745 Special cases of Mohrrsquos circle 28
46 Example load case on a stress element 29
51 Illustration of the principle of De Saint-Venant 3252 Geometric definitions and K factor for a hole [Courtesy of
Beer PF Mechanics of Materials ] 3353 Geometric definitions and K factor for a cross-section change
[Courtesy of Beer PF Mechanics of Materials ] 3454 Combined airfoil load 3455 Flight envelope containing multiple load cases 3556 Plotted inequalities for the multiple load case 36
61 Principle of virtual work 38
vii
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LIST OF FIGURES viii
62 Principle of complementary energy 3963 Linear force-displacement diagram 4064 Truss structure with applied dummy load 4265 Nodal internal dummy loads 43
71 Statically determinate beam example 47
81 Force superposition principle 5082 Split-up of an indeterminate truss 50
83 Statically indeterminate six bar truss 5291 Illustration of the flexibility coefficient 5692 Solution of a statically indeterminate beam using flexibility
coefficients 5793 Analysis example of a statically indeterminate beam 58
101 Biaxial stress state plate 62102 Indeterminate multimaterial plates (left one statically inde-
terminate force right two statically indeterminate forces) 63
111 Thermal extension of a bar 66112 Piezoelectric thickness change of a bar 67113 Statically indeterminate six bar truss 70
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Chapter 1
Introduction
The primary purpose of this reader is to describe the use of analysis equationsand methodologies of structures for design purposes It is common these daysthat we hear from industry that the students graduating with engineeringdegrees do not know how to design whether it is structures or mechanicalsystems or systems from other engineering disciplines We therefore put theemphasis in this course on design
Anybody who has some understanding of the design process however re-alises that without a thorough understanding of the use of analysis methodsit will not be possible to design at least a reliable system It is thereforeimportant to establish a sound and firm analysis foundation before one canstart the design practice The approach used in this reader however is dif-ferent from the traditional one in which analysis and designs are taught indifferent portions of the course Instead we will use an approach in whichsmall portions (sections) of topics from analysis are first introduced immedi-ately followed by their design implementation A more detailed descriptionof the outline and the contents of the reader is provided in the following
The reader is divided into two primary sections because of a very impor-tant concept called statical determinacy that has a very strong influenceon the way structures are designed It is of course too early to completelydescribe the impact of statical indeterminacy on design It will be sufficientto state at this point that statical determinacy simplifies the design processof a structure made of multiple components by making it possible to designindividual components independent from one another Structural indeter-minacy on the other hand causes the internal load distribution in a givenstructural system to be dependent on the dimensional and material proper-ties of the individual component that are typically being designed That isas the design of the individual components change the loads acting on thosecomponents for which they are being designed also change Hence individ-
1
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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4
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
49
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
55
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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iii
Modifications
Date Version Modification19012008 101 Equation 39 changed19012008 102 Equations 48 and 42 changed
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iv
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Contents
List of Figures viii
1 Introduction 1
2 Stress Analysis and Design of Statically Determinate Trusses 5
21 Truss Properties 6
22 Static Determinacy of Trusses 6
23 Truss Analysis 7
231 Truss Definition 8232 Stress Analysis 9
24 Stress Design 10
25 Example 10
3 Stress Analysis and Design of Statically Determinate Beams 13
31 What is a beam 13
32 Equilibrium Equations 14
33 Stresses in Beams 18
34 Discrete Beam Analysis 19
35 Example 20
4 Stress Analysis and Design of Statically Determinate Plates 23
41 Uniform In-Plane Loading 23
42 Circle of Mohr 24
43 Stress Design 27
44 Example 28
5 Stress Concentrations and Multiple Loads 31
51 Stress Concentrations 31
52 Design for Multiple Loads 33
v
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CONTENTS vi
6 Displacement Analysis and Design of Statically Determinate
Trusses 37
61 Dummy Load Method 3862 Displacement Design 4163 Example 42
7 Displacement Analysis and Design of Statically Determinate
Beams 45
71 Example 46
8 Analysis of Statically Indeterminate Trusses 49
81 Displacement Compatibility 4982 Action Item List for Analysis of Indeterminate Trusses 5183 Example 52
9 Analysis of Statically Indeterminate Beams 55
91 Force Method of Analysis 5592 Flexibility Coefficients 5693 Example 57
10 Analysis of Statically Indeterminate Plates 61101 Biaxial Stress State 61102 Multimaterial Plates 62
11 Actuated Structures 65
111 Thermal Actuation 65112 Piezoelectric Actuation 66113 Response of an Actuated Member 67114 Dummy Load Method for Structures with Actuated Members 69115 Example 69
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List of Figures
21 Rib truss structure example [Courtesy of wwwsteenaerocom] 522 Typical simple truss structure [Courtesy of wwwcompuzonecokr] 623 Truss support types 724 Statically determinate and indeterminate trusses 725 Internal force distribution inside a truss 826 Three bar truss example 1127 Free-body diagram of each node 11
31 Aircraft beam model [Courtesy of wwwflightinternationalcom] 14
32 Engine beam model [Courtesy of wwwmilnetcom] 1533 Local beam axes 1534 Beam sign convenction 1635 Infinitesimal beam segment 1736 Force jump over an infinitesimal beam segment 18
41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom] 2442 Example of fully loaded stress element 2543 Free body diagram for the plate equilibrium equations 2544 Layout and definition of Mohrrsquos circle 2745 Special cases of Mohrrsquos circle 28
46 Example load case on a stress element 29
51 Illustration of the principle of De Saint-Venant 3252 Geometric definitions and K factor for a hole [Courtesy of
Beer PF Mechanics of Materials ] 3353 Geometric definitions and K factor for a cross-section change
[Courtesy of Beer PF Mechanics of Materials ] 3454 Combined airfoil load 3455 Flight envelope containing multiple load cases 3556 Plotted inequalities for the multiple load case 36
61 Principle of virtual work 38
vii
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LIST OF FIGURES viii
62 Principle of complementary energy 3963 Linear force-displacement diagram 4064 Truss structure with applied dummy load 4265 Nodal internal dummy loads 43
71 Statically determinate beam example 47
81 Force superposition principle 5082 Split-up of an indeterminate truss 50
83 Statically indeterminate six bar truss 5291 Illustration of the flexibility coefficient 5692 Solution of a statically indeterminate beam using flexibility
coefficients 5793 Analysis example of a statically indeterminate beam 58
101 Biaxial stress state plate 62102 Indeterminate multimaterial plates (left one statically inde-
terminate force right two statically indeterminate forces) 63
111 Thermal extension of a bar 66112 Piezoelectric thickness change of a bar 67113 Statically indeterminate six bar truss 70
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Chapter 1
Introduction
The primary purpose of this reader is to describe the use of analysis equationsand methodologies of structures for design purposes It is common these daysthat we hear from industry that the students graduating with engineeringdegrees do not know how to design whether it is structures or mechanicalsystems or systems from other engineering disciplines We therefore put theemphasis in this course on design
Anybody who has some understanding of the design process however re-alises that without a thorough understanding of the use of analysis methodsit will not be possible to design at least a reliable system It is thereforeimportant to establish a sound and firm analysis foundation before one canstart the design practice The approach used in this reader however is dif-ferent from the traditional one in which analysis and designs are taught indifferent portions of the course Instead we will use an approach in whichsmall portions (sections) of topics from analysis are first introduced immedi-ately followed by their design implementation A more detailed descriptionof the outline and the contents of the reader is provided in the following
The reader is divided into two primary sections because of a very impor-tant concept called statical determinacy that has a very strong influenceon the way structures are designed It is of course too early to completelydescribe the impact of statical indeterminacy on design It will be sufficientto state at this point that statical determinacy simplifies the design processof a structure made of multiple components by making it possible to designindividual components independent from one another Structural indeter-minacy on the other hand causes the internal load distribution in a givenstructural system to be dependent on the dimensional and material proper-ties of the individual component that are typically being designed That isas the design of the individual components change the loads acting on thosecomponents for which they are being designed also change Hence individ-
1
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
55
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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Contents
List of Figures viii
1 Introduction 1
2 Stress Analysis and Design of Statically Determinate Trusses 5
21 Truss Properties 6
22 Static Determinacy of Trusses 6
23 Truss Analysis 7
231 Truss Definition 8232 Stress Analysis 9
24 Stress Design 10
25 Example 10
3 Stress Analysis and Design of Statically Determinate Beams 13
31 What is a beam 13
32 Equilibrium Equations 14
33 Stresses in Beams 18
34 Discrete Beam Analysis 19
35 Example 20
4 Stress Analysis and Design of Statically Determinate Plates 23
41 Uniform In-Plane Loading 23
42 Circle of Mohr 24
43 Stress Design 27
44 Example 28
5 Stress Concentrations and Multiple Loads 31
51 Stress Concentrations 31
52 Design for Multiple Loads 33
v
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CONTENTS vi
6 Displacement Analysis and Design of Statically Determinate
Trusses 37
61 Dummy Load Method 3862 Displacement Design 4163 Example 42
7 Displacement Analysis and Design of Statically Determinate
Beams 45
71 Example 46
8 Analysis of Statically Indeterminate Trusses 49
81 Displacement Compatibility 4982 Action Item List for Analysis of Indeterminate Trusses 5183 Example 52
9 Analysis of Statically Indeterminate Beams 55
91 Force Method of Analysis 5592 Flexibility Coefficients 5693 Example 57
10 Analysis of Statically Indeterminate Plates 61101 Biaxial Stress State 61102 Multimaterial Plates 62
11 Actuated Structures 65
111 Thermal Actuation 65112 Piezoelectric Actuation 66113 Response of an Actuated Member 67114 Dummy Load Method for Structures with Actuated Members 69115 Example 69
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List of Figures
21 Rib truss structure example [Courtesy of wwwsteenaerocom] 522 Typical simple truss structure [Courtesy of wwwcompuzonecokr] 623 Truss support types 724 Statically determinate and indeterminate trusses 725 Internal force distribution inside a truss 826 Three bar truss example 1127 Free-body diagram of each node 11
31 Aircraft beam model [Courtesy of wwwflightinternationalcom] 14
32 Engine beam model [Courtesy of wwwmilnetcom] 1533 Local beam axes 1534 Beam sign convenction 1635 Infinitesimal beam segment 1736 Force jump over an infinitesimal beam segment 18
41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom] 2442 Example of fully loaded stress element 2543 Free body diagram for the plate equilibrium equations 2544 Layout and definition of Mohrrsquos circle 2745 Special cases of Mohrrsquos circle 28
46 Example load case on a stress element 29
51 Illustration of the principle of De Saint-Venant 3252 Geometric definitions and K factor for a hole [Courtesy of
Beer PF Mechanics of Materials ] 3353 Geometric definitions and K factor for a cross-section change
[Courtesy of Beer PF Mechanics of Materials ] 3454 Combined airfoil load 3455 Flight envelope containing multiple load cases 3556 Plotted inequalities for the multiple load case 36
61 Principle of virtual work 38
vii
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LIST OF FIGURES viii
62 Principle of complementary energy 3963 Linear force-displacement diagram 4064 Truss structure with applied dummy load 4265 Nodal internal dummy loads 43
71 Statically determinate beam example 47
81 Force superposition principle 5082 Split-up of an indeterminate truss 50
83 Statically indeterminate six bar truss 5291 Illustration of the flexibility coefficient 5692 Solution of a statically indeterminate beam using flexibility
coefficients 5793 Analysis example of a statically indeterminate beam 58
101 Biaxial stress state plate 62102 Indeterminate multimaterial plates (left one statically inde-
terminate force right two statically indeterminate forces) 63
111 Thermal extension of a bar 66112 Piezoelectric thickness change of a bar 67113 Statically indeterminate six bar truss 70
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Chapter 1
Introduction
The primary purpose of this reader is to describe the use of analysis equationsand methodologies of structures for design purposes It is common these daysthat we hear from industry that the students graduating with engineeringdegrees do not know how to design whether it is structures or mechanicalsystems or systems from other engineering disciplines We therefore put theemphasis in this course on design
Anybody who has some understanding of the design process however re-alises that without a thorough understanding of the use of analysis methodsit will not be possible to design at least a reliable system It is thereforeimportant to establish a sound and firm analysis foundation before one canstart the design practice The approach used in this reader however is dif-ferent from the traditional one in which analysis and designs are taught indifferent portions of the course Instead we will use an approach in whichsmall portions (sections) of topics from analysis are first introduced immedi-ately followed by their design implementation A more detailed descriptionof the outline and the contents of the reader is provided in the following
The reader is divided into two primary sections because of a very impor-tant concept called statical determinacy that has a very strong influenceon the way structures are designed It is of course too early to completelydescribe the impact of statical indeterminacy on design It will be sufficientto state at this point that statical determinacy simplifies the design processof a structure made of multiple components by making it possible to designindividual components independent from one another Structural indeter-minacy on the other hand causes the internal load distribution in a givenstructural system to be dependent on the dimensional and material proper-ties of the individual component that are typically being designed That isas the design of the individual components change the loads acting on thosecomponents for which they are being designed also change Hence individ-
1
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
49
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
55
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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Contents
List of Figures viii
1 Introduction 1
2 Stress Analysis and Design of Statically Determinate Trusses 5
21 Truss Properties 6
22 Static Determinacy of Trusses 6
23 Truss Analysis 7
231 Truss Definition 8232 Stress Analysis 9
24 Stress Design 10
25 Example 10
3 Stress Analysis and Design of Statically Determinate Beams 13
31 What is a beam 13
32 Equilibrium Equations 14
33 Stresses in Beams 18
34 Discrete Beam Analysis 19
35 Example 20
4 Stress Analysis and Design of Statically Determinate Plates 23
41 Uniform In-Plane Loading 23
42 Circle of Mohr 24
43 Stress Design 27
44 Example 28
5 Stress Concentrations and Multiple Loads 31
51 Stress Concentrations 31
52 Design for Multiple Loads 33
v
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CONTENTS vi
6 Displacement Analysis and Design of Statically Determinate
Trusses 37
61 Dummy Load Method 3862 Displacement Design 4163 Example 42
7 Displacement Analysis and Design of Statically Determinate
Beams 45
71 Example 46
8 Analysis of Statically Indeterminate Trusses 49
81 Displacement Compatibility 4982 Action Item List for Analysis of Indeterminate Trusses 5183 Example 52
9 Analysis of Statically Indeterminate Beams 55
91 Force Method of Analysis 5592 Flexibility Coefficients 5693 Example 57
10 Analysis of Statically Indeterminate Plates 61101 Biaxial Stress State 61102 Multimaterial Plates 62
11 Actuated Structures 65
111 Thermal Actuation 65112 Piezoelectric Actuation 66113 Response of an Actuated Member 67114 Dummy Load Method for Structures with Actuated Members 69115 Example 69
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List of Figures
21 Rib truss structure example [Courtesy of wwwsteenaerocom] 522 Typical simple truss structure [Courtesy of wwwcompuzonecokr] 623 Truss support types 724 Statically determinate and indeterminate trusses 725 Internal force distribution inside a truss 826 Three bar truss example 1127 Free-body diagram of each node 11
31 Aircraft beam model [Courtesy of wwwflightinternationalcom] 14
32 Engine beam model [Courtesy of wwwmilnetcom] 1533 Local beam axes 1534 Beam sign convenction 1635 Infinitesimal beam segment 1736 Force jump over an infinitesimal beam segment 18
41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom] 2442 Example of fully loaded stress element 2543 Free body diagram for the plate equilibrium equations 2544 Layout and definition of Mohrrsquos circle 2745 Special cases of Mohrrsquos circle 28
46 Example load case on a stress element 29
51 Illustration of the principle of De Saint-Venant 3252 Geometric definitions and K factor for a hole [Courtesy of
Beer PF Mechanics of Materials ] 3353 Geometric definitions and K factor for a cross-section change
[Courtesy of Beer PF Mechanics of Materials ] 3454 Combined airfoil load 3455 Flight envelope containing multiple load cases 3556 Plotted inequalities for the multiple load case 36
61 Principle of virtual work 38
vii
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LIST OF FIGURES viii
62 Principle of complementary energy 3963 Linear force-displacement diagram 4064 Truss structure with applied dummy load 4265 Nodal internal dummy loads 43
71 Statically determinate beam example 47
81 Force superposition principle 5082 Split-up of an indeterminate truss 50
83 Statically indeterminate six bar truss 5291 Illustration of the flexibility coefficient 5692 Solution of a statically indeterminate beam using flexibility
coefficients 5793 Analysis example of a statically indeterminate beam 58
101 Biaxial stress state plate 62102 Indeterminate multimaterial plates (left one statically inde-
terminate force right two statically indeterminate forces) 63
111 Thermal extension of a bar 66112 Piezoelectric thickness change of a bar 67113 Statically indeterminate six bar truss 70
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Chapter 1
Introduction
The primary purpose of this reader is to describe the use of analysis equationsand methodologies of structures for design purposes It is common these daysthat we hear from industry that the students graduating with engineeringdegrees do not know how to design whether it is structures or mechanicalsystems or systems from other engineering disciplines We therefore put theemphasis in this course on design
Anybody who has some understanding of the design process however re-alises that without a thorough understanding of the use of analysis methodsit will not be possible to design at least a reliable system It is thereforeimportant to establish a sound and firm analysis foundation before one canstart the design practice The approach used in this reader however is dif-ferent from the traditional one in which analysis and designs are taught indifferent portions of the course Instead we will use an approach in whichsmall portions (sections) of topics from analysis are first introduced immedi-ately followed by their design implementation A more detailed descriptionof the outline and the contents of the reader is provided in the following
The reader is divided into two primary sections because of a very impor-tant concept called statical determinacy that has a very strong influenceon the way structures are designed It is of course too early to completelydescribe the impact of statical indeterminacy on design It will be sufficientto state at this point that statical determinacy simplifies the design processof a structure made of multiple components by making it possible to designindividual components independent from one another Structural indeter-minacy on the other hand causes the internal load distribution in a givenstructural system to be dependent on the dimensional and material proper-ties of the individual component that are typically being designed That isas the design of the individual components change the loads acting on thosecomponents for which they are being designed also change Hence individ-
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
49
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
55
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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CONTENTS vi
6 Displacement Analysis and Design of Statically Determinate
Trusses 37
61 Dummy Load Method 3862 Displacement Design 4163 Example 42
7 Displacement Analysis and Design of Statically Determinate
Beams 45
71 Example 46
8 Analysis of Statically Indeterminate Trusses 49
81 Displacement Compatibility 4982 Action Item List for Analysis of Indeterminate Trusses 5183 Example 52
9 Analysis of Statically Indeterminate Beams 55
91 Force Method of Analysis 5592 Flexibility Coefficients 5693 Example 57
10 Analysis of Statically Indeterminate Plates 61101 Biaxial Stress State 61102 Multimaterial Plates 62
11 Actuated Structures 65
111 Thermal Actuation 65112 Piezoelectric Actuation 66113 Response of an Actuated Member 67114 Dummy Load Method for Structures with Actuated Members 69115 Example 69
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List of Figures
21 Rib truss structure example [Courtesy of wwwsteenaerocom] 522 Typical simple truss structure [Courtesy of wwwcompuzonecokr] 623 Truss support types 724 Statically determinate and indeterminate trusses 725 Internal force distribution inside a truss 826 Three bar truss example 1127 Free-body diagram of each node 11
31 Aircraft beam model [Courtesy of wwwflightinternationalcom] 14
32 Engine beam model [Courtesy of wwwmilnetcom] 1533 Local beam axes 1534 Beam sign convenction 1635 Infinitesimal beam segment 1736 Force jump over an infinitesimal beam segment 18
41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom] 2442 Example of fully loaded stress element 2543 Free body diagram for the plate equilibrium equations 2544 Layout and definition of Mohrrsquos circle 2745 Special cases of Mohrrsquos circle 28
46 Example load case on a stress element 29
51 Illustration of the principle of De Saint-Venant 3252 Geometric definitions and K factor for a hole [Courtesy of
Beer PF Mechanics of Materials ] 3353 Geometric definitions and K factor for a cross-section change
[Courtesy of Beer PF Mechanics of Materials ] 3454 Combined airfoil load 3455 Flight envelope containing multiple load cases 3556 Plotted inequalities for the multiple load case 36
61 Principle of virtual work 38
vii
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LIST OF FIGURES viii
62 Principle of complementary energy 3963 Linear force-displacement diagram 4064 Truss structure with applied dummy load 4265 Nodal internal dummy loads 43
71 Statically determinate beam example 47
81 Force superposition principle 5082 Split-up of an indeterminate truss 50
83 Statically indeterminate six bar truss 5291 Illustration of the flexibility coefficient 5692 Solution of a statically indeterminate beam using flexibility
coefficients 5793 Analysis example of a statically indeterminate beam 58
101 Biaxial stress state plate 62102 Indeterminate multimaterial plates (left one statically inde-
terminate force right two statically indeterminate forces) 63
111 Thermal extension of a bar 66112 Piezoelectric thickness change of a bar 67113 Statically indeterminate six bar truss 70
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Chapter 1
Introduction
The primary purpose of this reader is to describe the use of analysis equationsand methodologies of structures for design purposes It is common these daysthat we hear from industry that the students graduating with engineeringdegrees do not know how to design whether it is structures or mechanicalsystems or systems from other engineering disciplines We therefore put theemphasis in this course on design
Anybody who has some understanding of the design process however re-alises that without a thorough understanding of the use of analysis methodsit will not be possible to design at least a reliable system It is thereforeimportant to establish a sound and firm analysis foundation before one canstart the design practice The approach used in this reader however is dif-ferent from the traditional one in which analysis and designs are taught indifferent portions of the course Instead we will use an approach in whichsmall portions (sections) of topics from analysis are first introduced immedi-ately followed by their design implementation A more detailed descriptionof the outline and the contents of the reader is provided in the following
The reader is divided into two primary sections because of a very impor-tant concept called statical determinacy that has a very strong influenceon the way structures are designed It is of course too early to completelydescribe the impact of statical indeterminacy on design It will be sufficientto state at this point that statical determinacy simplifies the design processof a structure made of multiple components by making it possible to designindividual components independent from one another Structural indeter-minacy on the other hand causes the internal load distribution in a givenstructural system to be dependent on the dimensional and material proper-ties of the individual component that are typically being designed That isas the design of the individual components change the loads acting on thosecomponents for which they are being designed also change Hence individ-
1
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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4
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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List of Figures
21 Rib truss structure example [Courtesy of wwwsteenaerocom] 522 Typical simple truss structure [Courtesy of wwwcompuzonecokr] 623 Truss support types 724 Statically determinate and indeterminate trusses 725 Internal force distribution inside a truss 826 Three bar truss example 1127 Free-body diagram of each node 11
31 Aircraft beam model [Courtesy of wwwflightinternationalcom] 14
32 Engine beam model [Courtesy of wwwmilnetcom] 1533 Local beam axes 1534 Beam sign convenction 1635 Infinitesimal beam segment 1736 Force jump over an infinitesimal beam segment 18
41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom] 2442 Example of fully loaded stress element 2543 Free body diagram for the plate equilibrium equations 2544 Layout and definition of Mohrrsquos circle 2745 Special cases of Mohrrsquos circle 28
46 Example load case on a stress element 29
51 Illustration of the principle of De Saint-Venant 3252 Geometric definitions and K factor for a hole [Courtesy of
Beer PF Mechanics of Materials ] 3353 Geometric definitions and K factor for a cross-section change
[Courtesy of Beer PF Mechanics of Materials ] 3454 Combined airfoil load 3455 Flight envelope containing multiple load cases 3556 Plotted inequalities for the multiple load case 36
61 Principle of virtual work 38
vii
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LIST OF FIGURES viii
62 Principle of complementary energy 3963 Linear force-displacement diagram 4064 Truss structure with applied dummy load 4265 Nodal internal dummy loads 43
71 Statically determinate beam example 47
81 Force superposition principle 5082 Split-up of an indeterminate truss 50
83 Statically indeterminate six bar truss 5291 Illustration of the flexibility coefficient 5692 Solution of a statically indeterminate beam using flexibility
coefficients 5793 Analysis example of a statically indeterminate beam 58
101 Biaxial stress state plate 62102 Indeterminate multimaterial plates (left one statically inde-
terminate force right two statically indeterminate forces) 63
111 Thermal extension of a bar 66112 Piezoelectric thickness change of a bar 67113 Statically indeterminate six bar truss 70
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Chapter 1
Introduction
The primary purpose of this reader is to describe the use of analysis equationsand methodologies of structures for design purposes It is common these daysthat we hear from industry that the students graduating with engineeringdegrees do not know how to design whether it is structures or mechanicalsystems or systems from other engineering disciplines We therefore put theemphasis in this course on design
Anybody who has some understanding of the design process however re-alises that without a thorough understanding of the use of analysis methodsit will not be possible to design at least a reliable system It is thereforeimportant to establish a sound and firm analysis foundation before one canstart the design practice The approach used in this reader however is dif-ferent from the traditional one in which analysis and designs are taught indifferent portions of the course Instead we will use an approach in whichsmall portions (sections) of topics from analysis are first introduced immedi-ately followed by their design implementation A more detailed descriptionof the outline and the contents of the reader is provided in the following
The reader is divided into two primary sections because of a very impor-tant concept called statical determinacy that has a very strong influenceon the way structures are designed It is of course too early to completelydescribe the impact of statical indeterminacy on design It will be sufficientto state at this point that statical determinacy simplifies the design processof a structure made of multiple components by making it possible to designindividual components independent from one another Structural indeter-minacy on the other hand causes the internal load distribution in a givenstructural system to be dependent on the dimensional and material proper-ties of the individual component that are typically being designed That isas the design of the individual components change the loads acting on thosecomponents for which they are being designed also change Hence individ-
1
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
49
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
55
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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LIST OF FIGURES viii
62 Principle of complementary energy 3963 Linear force-displacement diagram 4064 Truss structure with applied dummy load 4265 Nodal internal dummy loads 43
71 Statically determinate beam example 47
81 Force superposition principle 5082 Split-up of an indeterminate truss 50
83 Statically indeterminate six bar truss 5291 Illustration of the flexibility coefficient 5692 Solution of a statically indeterminate beam using flexibility
coefficients 5793 Analysis example of a statically indeterminate beam 58
101 Biaxial stress state plate 62102 Indeterminate multimaterial plates (left one statically inde-
terminate force right two statically indeterminate forces) 63
111 Thermal extension of a bar 66112 Piezoelectric thickness change of a bar 67113 Statically indeterminate six bar truss 70
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Chapter 1
Introduction
The primary purpose of this reader is to describe the use of analysis equationsand methodologies of structures for design purposes It is common these daysthat we hear from industry that the students graduating with engineeringdegrees do not know how to design whether it is structures or mechanicalsystems or systems from other engineering disciplines We therefore put theemphasis in this course on design
Anybody who has some understanding of the design process however re-alises that without a thorough understanding of the use of analysis methodsit will not be possible to design at least a reliable system It is thereforeimportant to establish a sound and firm analysis foundation before one canstart the design practice The approach used in this reader however is dif-ferent from the traditional one in which analysis and designs are taught indifferent portions of the course Instead we will use an approach in whichsmall portions (sections) of topics from analysis are first introduced immedi-ately followed by their design implementation A more detailed descriptionof the outline and the contents of the reader is provided in the following
The reader is divided into two primary sections because of a very impor-tant concept called statical determinacy that has a very strong influenceon the way structures are designed It is of course too early to completelydescribe the impact of statical indeterminacy on design It will be sufficientto state at this point that statical determinacy simplifies the design processof a structure made of multiple components by making it possible to designindividual components independent from one another Structural indeter-minacy on the other hand causes the internal load distribution in a givenstructural system to be dependent on the dimensional and material proper-ties of the individual component that are typically being designed That isas the design of the individual components change the loads acting on thosecomponents for which they are being designed also change Hence individ-
1
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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4
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
49
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
55
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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Chapter 1
Introduction
The primary purpose of this reader is to describe the use of analysis equationsand methodologies of structures for design purposes It is common these daysthat we hear from industry that the students graduating with engineeringdegrees do not know how to design whether it is structures or mechanicalsystems or systems from other engineering disciplines We therefore put theemphasis in this course on design
Anybody who has some understanding of the design process however re-alises that without a thorough understanding of the use of analysis methodsit will not be possible to design at least a reliable system It is thereforeimportant to establish a sound and firm analysis foundation before one canstart the design practice The approach used in this reader however is dif-ferent from the traditional one in which analysis and designs are taught indifferent portions of the course Instead we will use an approach in whichsmall portions (sections) of topics from analysis are first introduced immedi-ately followed by their design implementation A more detailed descriptionof the outline and the contents of the reader is provided in the following
The reader is divided into two primary sections because of a very impor-tant concept called statical determinacy that has a very strong influenceon the way structures are designed It is of course too early to completelydescribe the impact of statical indeterminacy on design It will be sufficientto state at this point that statical determinacy simplifies the design processof a structure made of multiple components by making it possible to designindividual components independent from one another Structural indeter-minacy on the other hand causes the internal load distribution in a givenstructural system to be dependent on the dimensional and material proper-ties of the individual component that are typically being designed That isas the design of the individual components change the loads acting on thosecomponents for which they are being designed also change Hence individ-
1
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
49
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
55
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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2
ual components cannot be designed independently as the design changes inthese components and the other components around them alter the loadsthat they are being designed for The resulting process is an iterative onerequiring design of all the individual components to be repeated again andagain until the internal load redistribution stabilises reaching an equilibriumstate with the prescribed external loading
For the reason briefly described above the first section of the reader isdedicated to statically determinate structural systems An important impli-cation of the independence of internal loads to design changes is the fact
that stresses in the individual components depend only on the sizing vari-ables which determine the cross-sectional geometry of the components Thatis since internal stresses are determined by dividing the internal load by aproperty of the cross section of the component (be it area or thickness orarea moment of inertia) a designer can make sure that by proper selection of the sizing variables safe levels of stresses compared to stress allowables can bemaintained Furthermore material selection of the components can be madeindependent of the rest of the components The first section therefore concen-trates on the use of stress analysis of structural components to design crosssectional properties based on stress considerations only There are of course
other considerations that influence the design decision besides stresses evenif we limit ourselves to statically determinate systems For example struc-tural responses such as buckling and vibration are typically very importantfor many aircraft and spacecraft design problems Such considerations willbe discussed later in the reader
The second section of the reader is dedicated to statically indeterminate
structural systems An important element of the stress analysis of indeter-minate systems is the need to compute displacements and deformations of the members As stated earlier internal load distribution in indeterminatesystems is influenced by the cross-sectional properties of the individual com-
ponents as well as their material properties This dependency is in fact stemsfrom the dependence of the internal loads to the stiffnesses of the individ-ual component which strongly influence their deformation characteristicsStructural stiffness is a product of the material properties and the dimen-sions As such the design effort requires evaluation of the member stiffnessesas a function of the designed quantities determining the effect of the mem-ber stiffnesses on member deformations and in turn internal forces Hencethe second section concentrates on design of indeterminate structures andstructures with displacement constraints
Finally the material in the reader is organised depending on the type of structural component to be analysed and designed Most complex structuralsystems are composed of three different types of components namely truss
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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4
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to master the skills of analysing staticallydeterminate structures composed of beams or a combination of beam andtruss members using both the continuous and discrete approaches The dis-crete approach is based on splitting the structure into short beam segments
over which the loading can be assumed constant In this case analysing thestructure is reduced to well-structured calculation which can be carried outeither by hand or using a computer
31 What is a beam
A beam is any structure or structural member which has one dimensionmuch larger than the other two This is the simplest possible definition andthe one mostly used in engineering practice The long direction is usually
called the beam axis The beam axis is not necessarily a straight line Theplanes normal to the beam axis (in the undeformed configuration) intersectthe structure in cross-sections The point at which the beam axis intersectsthe cross-section need not be the centroid although under certain conditionsthis choice results in considerable simplification
Beam analysis is common in preliminary design of aerospace structuresAs shown in figure 31 the complete structure of an aircraft can be idealisedas a collection of beams In other situations structural members rather thanbuilt-up structures are modelled as beam Examples of such use are tail orwing spars and shafts of a turbojet or turbofan engine as shown in figure 32
Apart from its use in preliminary design beam models are important forthe insight they provide into the behaviour of the structure Terminology
13
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32 Equilibrium Equations 14
Figure 31 Aircraft beam model [Courtesy of wwwflightinternationalcom]
associated with beam analysis is an essential part of the vocabulary of struc-tural engineers A thorough understanding of and familiarity with beamtheory is one of the cornerstones of the engineering intuition of structuralengineers
32 Equilibrium Equations
A beam has cross-section dimensions which are much smaller than the beamlength For this reason a beam is usually sketched as lines representing thebeam axis If we imagine the beam to be separated into two parts at acertain (see figure 33) cross-section the internal forces at that cross-sectionwill have a static resultant force and moment When resolved along the localbeam axes the static resultant is composed of three force components andthree moment components The force component along the beam axis xis termed the normal or axial force The force components along the twocross-section axes y and z are called shear forces The moment component
along the beam axis is called the twisting moment or torque The momentcomponents along the two cross-section axes are called bending moments
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15 3 Stress Analysis and Design of Statically Determinate Beams
Figure 32 Engine beam model [Courtesy of wwwmilnetcom]
x
y
z
Figure 33 Local beam axes
It is important to note that while the names of the different cross-sectionforce and moment components are almost universal the sign conventions
symbols and how the variation of these quantities is plotted along the beamaxis can vary between different books and computer software It is essential
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32 Equilibrium Equations 16
to review the conventions adopted by specific books or software packagesbefore using them
In the following we are going to use a sign convention that is commonin aerospace applications The normal force is denoted by N The signconvention for the normal force is a positive sign for a tensile force and anegative sign for a compressive force The torque is denoted by M x and ispositive when rotating counter clockwise out of the cross-section The bend-ing moments about the two cross-section axes are denoted by M y and M zA bending moment is positive if it causes compression in the first quadrant
Finally the shear forces are denoted by V y and V z The sign convention of all cross-sectional resultants is shown in figure 34
+
+
++
+
Figure 34 Beam sign convenction
Consider an arbitrary infinitesimal segment of the beam as shown in fig-ure 35 This segment will be in equilibrium under the action of internaland external forces and moments For simplicity we derive the vertical equi-librium equations Similar arguments can be used to derive the rest of theequilibrium equations
Let pz(x) define the vertical loading per unit length If the beam seg-ment is short enough the load can be assumed uniform The vertical forceequilibrium of the beam segment requires
pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (31)
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17 3 Stress Analysis and Design of Statically Determinate Beams
∆xVz(x) Vz(x+∆x)
pz
My(x) My(x+∆x)
Figure 35 Infinitesimal beam segment
Dividing by ∆x and taking the limit as ∆x rarr 0 we find the continuousform of the vertical force equilibrium as
dV zdx
= pz (32)
The student is encouraged to derive the moment equilibrium equation aroundthe y axis The complete set of equilibrium equations are listed below
dN
dx = minus px
dV ydx
= py dV z
dx = pz (33a)
M xdx
= minustx M ydx
= V z dM zdx
= V y (33b)
In these equations px(x) is the distributed thrust load per unit length py
and pz are the lateral distributed loads per unit length in y and z directionrespectively and tx(x) is the distributid twisting moment per unit length
The equilibrium equations 33 are valid when the applied distributed loadsvary smoothly This however is frequently not the case In many practicalsituations load introduction is abrupt and the loading can be idealised asconcentrated forces andor moments In such cases the distribution of the
cross-sectional forces and moments is not continuous Special jump condi-tions have to applied at such locations
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33 Stresses in Beams 18
The jump condition when a vertical concentrated force is applied arederived Referring to figure 36 the vertical force equilibrium equation 31 ismodified to read
F z + pz(x)∆x + V z(x) minus V z(x + ∆x) = 0 (34)
Taking the limit as ∆x rarr 0 we obtain the jump condition
V +z minus V minusz = F z (35)
where V minusz and V +z are the shear force right before and right after the appliedconcentrated load F z
Similar jump conditions can be derived for all forces and moments Thestudent should be able to carry out the derivation without difficulty
∆x
Vz(x) Vz(x+∆x)
Fz
pz
Figure 36 Force jump over an infinitesimal beam segment
33 Stresses in Beams
Once the distribution of the internal forces and moments is obtained throughthe integration of the equilibrium equations 33 the stress state at any pointof the cross section can be calculated This might at first seem strange sincedifferent loading can lead to the same internal force and moment resultant at
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19 3 Stress Analysis and Design of Statically Determinate Beams
a given cross-section The question arises whether the stress distribution dueto two statically equivalent loads are the same The answer to this questionis that the stress distributions due to statically equivalent systems of loadsare not identically the same but are generally close away from the pointof application of load This is a statement of the de St Venant principleStresses calculated based only on the resultant internal force and moment areusually accurate enough away from points of load introduction supports anddiscontinuity in the cross-sections
If for the sake of simplicity we assume that the beam is loaded in the xz
plane only the formulae for the stress distribution are particularly simpleThe distribution of the normal stress σx is given by
σx = N
A minus M y
I yz (36)
where A is the total area of the cross-section I y is the second moment of area about the y axis and the z coordinate is measured from the centroid of the cross section The distribution of the shear stress is given by
τ = minusV zQ
I yt (37)
where Q is the first moment of area of the part of the cross section up to thepoint of calculation
34 Discrete Beam Analysis
While the equations of equilibrium are straightforward to integrate for beamswith simple loading the integration can become very cumbersome for more
complex loading (eg lift distribution over a wing) For such cases it ispossible to compute approximate distribution of cross-sectional forces andmoments by discretising the beam into a finite number of segments n andassuming the loading to be constant over each segment
For simplicity we consider only lateral loads in the xz plane The tech-nique can be easily extended for the general case The beam is divided into nsegments The ith segment having a length (i) and uniform applied load p
(i)z
It is assumed that any concentrated forces are applied between the segmentsand not inside the segment Integration of the equilibrium equations givesthe following relations
V (i)z
L
= V (iminus1)z
R
+ F (i)z V (i)z
R
= V (i)z
L
+ p(i)z (i) (38)
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35 Example 20
and
M (i)y
L
= M (iminus1)y
R
M (i)y
R
= M (i)y
L
+
V (i)z
L
+ p(i)z (i)2
(i) (39)
where the L and R subscripts denote the left end and right end of the segmentrespectively
The above equations are very easy to program using Maple Matlab oreven using a simple excel sheet The way the equations are used can be bestillustrated by an example
35 Example
A representative example will be analysed using both continuous integrationand a discrete beam model Consider an aircraft wing modelled as a can-tilever beam and loaded with a vertical load distribution (lift distribution)approximated as
pz(x) = p0
3
x
Lminus 3
x
L2
+
x
L3
(310)
where for numerical computations p0 = 1000Nm and L = 5m are usedThe shear force at the leftmost edge is zero since it is a free end This
allows us to evaluate the integration constant when integrating equation 310Thus we can obtain the shear force distribution as
V z(x) = p0L
4
6x
L
2 minus 4x
L
3+x
L
4 (311)
Similarly equation 311 can be integrated and the integration constant de-termined by noting that at x = 0 the bending moment vanishes The finalexpression for the bending moment is
M y(x) = p0L2
20
10x
L
3 minus 5x
L
4+x
L
5 (312)
To solve this example using the discrete approach we divide the beaminto ten segments The results of the computation is summarised in Table35 Sample Maple codes for discrete beam analysis are available online
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21 3 Stress Analysis and Design of Statically Determinate Beams
Table 31 Discrete Shear Force and Bending Moment Distribution
Segment pz V z|L V z|R M y|L M y|R1 05 1426 00 713 00 1782 05 3859 713 2643 178 10173 05 5781 2643 5533 1017 30614 05 7254 5533 9160 3061 67345 05 8 336 9160 13328 6734 123566 05 9089 13328 17873 12356 201577 05 9571 17873 22658 20157 302898 05 9844 22658 27580 30289 428499 05 9966 27580 32563 42849 57885
10 05 9999 32563 37563 57885 75416
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35 Example 22
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Chapter 4
Stress Analysis and Design of Statically Determinate Plates
Plates are extensively used structural elements in aerospace constructionsSkin material ribs aircraft floors their application possibilities areinfinite An example of the use of plate material as aircraft skin is given infigure 41 At the very dawn of aviation aircraft wing and fuselage skins
were applied to preserve the required aerodynamic shape and all the loadsacting on the vehicle were carried mainly by truss structures However withthe introduction of aluminium in the aircraft industry this changed radiacllyThe so-called fully stressed skin was invented This means that the platesforming the skin take their part of the loads
A plate can be loaded mainly in two ways in the plane of the plate andperpendicular to its plane The stress analysis of bending of plates has provento be too complicated to deal with in this course and as such will not betreated Also the loading in the plane is not exactly trivial either howeverthere are some particular cases that will be investigated in this course These
simplified cases already provide initial insight in the essentials of stresses inplates
41 Uniform In-Plane Loading
The loading condition considered is a uniform in-plane loading This means aconstant stress field that is distributed uniformly over the entire plate Sucha stress field is the resultant of an applied stress in one direction of the plateThe plate can also be loaded uniformly in the other direction This results ina uniform stress field in that direction as well A combination of two uniformstress fields in either direction of the plate is called a bi-axial uniform stress
23
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42 Circle of Mohr 24
Figure 41 Example of an aircraft riveted plate skin [Courtesy of wwwromeolimacom]
state Finally a plate can in combination with the previously mentionedapplied normal stresses also be loaded in uniform shear This results in auniformly distributed shear stress state in the plate The total combinationis shown in figure Please note the particular directions of the applied shearstress components The combined load cases can be seen in figure 42
When these three components act on a plate a stress analysis can be
carried out Even in case of a plate that has a complex stress distributiondue to a certain applied load combination locally the stress distribution isuniform Such a small local plate element is called a stress element
42 Circle of Mohr
If a bi-axial stress state and a uniform shear stress are present the questionis what the occurring normal and shear stress would be in any arbitrarydirection of the plate On top of that one might wonder what the maximum
normal and shear stress would be in the plate and in which direction thatone would act This issue is solved by looking at the force equilibrium in an
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25 4 Stress Analysis and Design of Statically Determinate Plates
σx
σy
τxy
σy
σx
τxy
x
y
Figure 42 Example of fully loaded stress element
arbitrary direction of a stress element as shown in figure 43
τθ
σy
σx
τxy
σθ
983105 983088
θ
983160
983161
983160θ
983161θ
θ
Figure 43 Free body diagram for the plate equilibrium equations
Keeping in mind that a force is the applied stress times the area over
which the stress is distributed the force equilibrium in x and y-direction canbe written as
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42 Circle of Mohr 26
F xθ 0 = σθA0 minus σx(A0 cos θ)cos θ minus τ xy(A0 sin θ)cos θ minus σy(A0 sin θ)sin θ
minusτ xy(A0 cos θ)sin θF yθ 0 = τ θA0 + σx(A0 cos θ)sin θ + τ xy(A0 sin θ)sin θ minus σy(A0 sin θ)cos θ
minusτ xy(A0 cos θ)cos θ(41)
Assembling corresponding terms of the above equation yields
σθ = σx cos2 θ + σy sin2 θ + 2τ xy sin θ cos θτ θ = minus(σx minus σy)sin θ cos θ + τ xy(cos2 θ minus sin2 θ)
(42)
Using standard trigonometric relations the above expressions can be sim-
plified even further σθ = σx+σy
2 + σxminusσy
2 cos2θ + τ xy sin 2θ
τ θ = minusσxminusσy2
sin2θ + τ xy cos 2θ (43)
If the second of the above equation is substituted into the first one one
obtains an expression of a circleσθ minus σx + σy
2
2
+ τ 2θ =
σx minus σy
2
2
+ τ 2xy (44)
This circle is called Mohrrsquos circle It is obvious that this is a circle in theσθ minus τ θ - plane and has as centre point σx+σy
2 which is the average normal
stress σav and radius R =
σxminusσy2
2+ τ 2xy Assume the applied stresses σx
σy and τ xy are known the stresses σθ and τ θ can be determined using theequation of the circle of Mohr The layout of the circle is given in figure 44
It is clear that the maximum (σ1) and minimum (σ2) normal stresses can
be read off Mohrrsquos circle directly (where the circle intersects with the σθ-axis) as well as the maximum shear stress In some cases it can also occurthat σ2 becomes larger than σ1 in absolute value Obviously σ2 will be themaximum normal stress in that case Also the double of the direction ϑ inwhich the normal stresses act can be read from the circle It is the anglebetween the σθ-axis and the line connecting the circlersquos centre point and the(σx τ xy) point The reason why it is a double angle can be found in equation43 The equation for the direction angle ϑ is
ϑ = 1
2
arctan τ xy
(σx minus σy)2 (45)
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27 4 Stress Analysis and Design of Statically Determinate Plates
σθ
τθ
2
2
2
x y
xy R
σ σ τ
minus = +
σ1 = σθ max
σ2= σ
θ min 0
2
x yσ σ +
τθ max
( ) x xyσ τ
( ) y xyσ τ minus
tan 2
2
xy
x y
τ ϑ
σ σ =
minus
2ϑ
Figure 44 Layout and definition of Mohrrsquos circle
43 Stress Design
The stress design of a statically determinate plate is very similar to that of a statically determinate truss or beam In this case also the fully stressed
design philosophy applies as already explained in section 24 The majordifference is that in the case of a plate we are not looking to update thecross sectional area but the thickness since the other geometric quantitiesare usually prescribed The plate thickness can be updated according totwo design criteria maximum normal stress or maximum shear stress Themaximum normal stress is the maximum value of the two stresses σ1 and σ2 -in absolute value - obtained using Mohrrsquos circle The maximum shear stress
is obtained directly from using the circle of Mohr The new plate thicknessis calculated according to the following equations which are expressions formaximum normal and shear stress respectively
tnew = σ
σall
told (46)
tnew = τ
τ alltold (47)
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44 Example 28
44 Example
Special cases of Mohrrsquos circle
σx or σy is zeroτxy is zero
σx and σy are equalτxy is zero
σx and σy are zeroτxy is finite
σx = -σy andτxy is zero
σx = σy andτxy = σx = σy
σ
τ
σ
τ
σ
τ
Figure 45 Special cases of Mohrrsquos circle
Bi-axially Loaded Stress Element
A stress element is loaded with two normal stresses and one shear stress asindicated in figure 46
In orde to construct Mohrrsquos circle we need the circle midpoint and thecircle radius The midpoint located on the σθ-axis is the following
σav = σx + σy
2 =
100 + 50
2 = 75MP a (48)
The circle radius is
R =
σx minus σy
2
2
+ τ 2xy =
100 minus 50
2
2
+ 252 = 3536MP a (49)
Using this information the maximum normal and shear stress can befound
σ1 = σav + R = 11036MP a (410)
τ max = R = 3536MP a (411)
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29 4 Stress Analysis and Design of Statically Determinate Plates
σ983160 983101 983089983088983088 983117983120983137
σ983161 983101 983093983088 983117983120983137
τ983160983161
983101 983090983093 983117983120983137
x
y
Figure 46 Example load case on a stress element
The value for σ2 in this case is 3964 MPa The direction ϑ in which themaximum normal stress acts is calculated as follows
ϑ = 1
2 arctan
25
(100minus 50)2
= 225 (412)
Assuming the initial thickness of the plate is 2 mm the allowable normalstress is 100 MPa and the allowable shear stress is 50 MPa the fully stressedthickness both for maximum normal and shear stress becomes respectively
tnew = 11036
100 2 = 22mm (413)
tnew = 3536
50 2 = 14mm (414)
It is evident that one can get totally different answers based on the stresscriterion one uses
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44 Example 30
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Chapter 5
Stress Concentrations andMultiple Loads
51 Stress Concentrations
We have seen in the stress analysis methods of trusses beams and plates that
one deals with average stresses uniformly distributed over the cross-sectionof the structure In most cases that is a fair assumption and leads to goodanalysis results However particular load introductions holes in a structureor sudden changes in cross-section can cause stress concentrations to occurThese stresses are larger than the average stresses calculated far from thosecritical locations and as such can cause the occurring stress to become largerthan the allowable stress at certain locations in the structure Therefore it isimportant to take these stress concentrations into account and involve theminto the stress analysis
Stress concentrations happen at places where there are discontinuities in
the structure such as holes cracks cross-section changes They are alsopresent at load introduction locations It can be stated generally that brittlematerials are more susceptible to stress concentrations than ductile onesThe reason for this is twofold First of all cracks and cognates influence thefatigue life of a structure and it is generally known that brittle materialsare more sensitive to fatigue Secondly the elevated stresses might causeplasticity to emerge in the material Also in this case ductile materials aremore favourable than brittle ones
The stress level at the stress concentration location σmax is higher thanthe average stress in the structure σav The ratio between these two stressesis a measure for the intensity of the stress concentration and indicated withthe stress concentration factor K
31
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51 Stress Concentrations 32
K = σmax
σav
(51)
De Saint-Venantrsquos Principle
The question is now how big the influence of such local phenomena is on theremainder of the structure De Saint-Venant postulated a theorem on thismatter which is still accepted today He claims that such local phenomena
have only a local influence Far from the influence the occurring stresses areindependent of how the loads are applied or whether there are holes presentor not This is depicted in figure 51 As a rule of thumb one may assumefor load introduction problems that the effect of the applied axial load isrelevant up to a distance equal to the width of the plate
σ983149983137983160
σ983137983158
σ983149983137983160
σ983137983158σ
983137983158
σ983149983137983160
P
Figure 51 Illustration of the principle of De Saint-Venant
Circular holes and Cross-Section Changes
As mentioned earlier stress concentrations can occur near circular holes orcross-section changes The question is how to determine the stress concentra-tion factor K based on the geometric properties of the holes or area changeThe following geometric quantities are important when calculating K
bull Circular hole the ratio of hole radius r and the distance from the holeto the plate edge d
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33 5 Stress Concentrations and Multiple Loads
bull Cross-section change
1 the ratio of the fillet radius r and the length of the smallest cross-section d
2 the ratio of the smallest cross-section length to the largest cross-section length
Symbols r and d are used multiple times however these are the con-ventions for hole and cross-section change stress concentration factors so be
always mindful of what K factor you are dealing withThe geometric definitions and K factor relations for the hole are given in
figure 52 while those for the cross-section change are given in figure 53
Figure 52 Geometric definitions and K factor for a hole [Courtesy of BeerPF Mechanics of Materials ]
52 Design for Multiple Loads
The terms combined load and multiple loads are often used in design In thefirst case a structure is submitted to a combination of loads simultaneouslyA good example is an aircraft wing this structure is submitted to a lift forcea drag force and an aerodynamic moment at the same time during flight (seefigure 54)
Calculating the stresses due to such a combined load can be approached
by calculating the internal normal and shear forces and the bending andtorsional moments resulting from the combined load at the location where
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52 Design for Multiple Loads 34
Figure 53 Geometric definitions and K factor for a cross-section change[Courtesy of Beer PF Mechanics of Materials ]
L
DM
Figure 54 Combined airfoil load
one desires to know the stresses The stresses resulting from each of theseinternal forces can added together since we are dealing with linear structuresthroughout the entire course As such desinging for combined loads is notfundamentally different from designing for a single load only
When one designs for multiple loads on the other hand the objective isto make sure that the structure is able to withstand several load cases whichdo not occur simultaneously To clarify the difference between multiple loadand combined load each of the individual load cases of a multiple loadingcondition can be a combined load As an example for a multiple load caseserves a flight envelope (see figure 55) An aircraft wing needs to be designed
for each of the points inside the boundaries of the flight envelopeThe challenge is that one is often confronted with completely different
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35 5 Stress Concentrations and Multiple Loads
Figure 55 Flight envelope containing multiple load cases
requirements Take as an example a simply supported beam which needs tobe designed for tension and compression simultaneously In the tensile casethe only important factor is the total cross-section area of the beam On theother hand in the compressive case the beam is prone to buckling For asimply supported beam the buckling load P cr is equal to π2EI
L2 where I is thesecond moment of area of the beam and as such not only the area of the beambecomes important but also the shape Let us assume a rectangular thin-walled shape with wall thickness 1 mm width b and height h The appliedtensilecompressive load is 10000 N and the material Youngrsquos modulus is 70
GPa and its allowable stress 310 MPa This leads to two requirements
Amin ge P
σall
(52)
I min ge P L2
π2E (53)
If we keep the formulas for A and I in mind (A = 2t(b + h) and I =
th2(1
6h+
1
2b)) and plugging in the values given above we obtain the following
inequalities
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52 Design for Multiple Loads 36
323 le 2(b + h) (54)
1450 le h2(1
6h +
1
2b) (55)
Plotting both curves gives a graph as depicted in figure 56
Figure 56 Plotted inequalities for the multiple load case
The inequalities given above indicate that all combinations of b and h that
are above both curve are good candidates for the design problem Obviouslyboth values should be positive so b = 0-line forms a boundary as well If bothtensile failure and buckling should occur simultaneously then the intersectionof both graphs in the positive b-plane is the only option
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Chapter 6
Displacement Analysis and
Design of Statically
Determinate Trusses
No structure is rigid The floor you were just walking on deformed underyour feet the chair you are sitting on shortened just a little bit Howevernot evident from every day life practice everything deforms under a certainloading Especially in the aerospace industry deformations are importantas a consequence of the design philosophy Weight is the driving factor inevery aerospace design and this constant striving for lightness renders struc-tures to become flexible The danger of such structures is that the occurringdisplacements might jeopardise the structural functionality Imagine thatbecause of weight savings the torsional rigidity of a wing would be reducedThis would result in large local angles-of-attack altering the aerodynamic
performance considerably
Therefore a structure should always be designed such that it is able tocarry the applied loads without violating certain displacement constraintsBecause of that displacement analysis of a structure is very important More-over as we will see later the internal stress distribution in a statically in-determinate structure depends entirely on the displacement field This is asecond example the importance of displacement calculation
In this chapter we will deal with the displacement of trusses But first weintroduce a tool with which we can calculate displacements of trusses andbeams using energy principles This tool is called the Dummy Load Method
37
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61 Dummy Load Method 38
61 Dummy Load Method
Energy Principles
Work W exerted on a structure is equal to a force F times the displacementδ the force creates
W = F δ (61)
Doing work requires energy U which is stored in the structure If allenergy is released when the applied forces are removed from the structuresuch a structure is called conservative or elastic If some energy remains inthe structure it is called nonconservative or plastic The externally appliedwork resulting in an external energy U e needs to be in equilibrium with theinternal structural strain energy U i This principle is called conservation of
energy Another important concept is the principle of virtual work This type
of work can be interpreted as an applied force effectuating an infinitessimalsmall displacement in the direction of the force It can be explained bylooking at figure 61
F
δ
U
∆U
∆δ δ δ δ
Figure 61 Principle of virtual work
It is clear that the energy stored in the structure due to a finite displace-ment ∆δ is equal to ∆U = F ∆δ If the latter displacement is going to zeroin the limit - in order to obtain the infinitessimally small displacement - theprinciple of virtual work becomes
dU = F dδ (62)
This means that the rate of change of the internal strain energy with thedisplacement is equal to the applied force Associated with virtual work is
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39
6 Displacement Analysis and Design of Statically Determinate
Trusses
the principle of complementary energy U lowast This can be inspected in figure62
F
δ
U
U
∆ F
∆U
Figure 62 Principle of complementary energy
Instead of an infinitessimal displacement we are now looking at a smallforce ∆F Analoguous to the method described above for the virtual workwe obtain
dU lowast = dF δ (63)
This formula indicates that the rate of change of the complementaryenergy with the force is equal to the resulting displacement
Castiglianorsquos Theorem
For linear structures which we use thorugout this course the relation be-tween an applied force and the resulting displacement is linear F = kδ wherek is the structural stiffness It is evident from figure 63 that in the linearcase the internal energy is equal to the complementary energy U = U lowast
Therefore the following important relation can be derived from equations62 and 63
dU
dF = δ (64)
This means that we can find the displacement δ in the direction of aforce F by differentiating the strain energy with respect to that force Thisequation is called the theorem of Castigliano If multiple forces act on thestructure each displacement in the direction of a particular force F i is thenexpressed as
dU dF i= δ i (65)
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61 Dummy Load Method 40
F
δ
U
∆U
∆δ δ δ δ
U
∆ F
∆U
Figure 63 Linear force-displacement diagram
Note that the strain energy U is a result from all applied forces F whilethe displacement in the direction of a particular force F i is the derivativewith that force only
If we apply the above explanation to a truss member in particular weend up with the following expression for the energy
U =
F 0
δdF = F 2L2EA
(66)
keeping in mind that F = kδ and that the stiffness k of an axial truss memberis equal to k = EA
L For an entire truss structure the energy becomes the
summation over the individual contributions of each member
U =n
i=1
F 2i Li
2E iAi
(67)
where n is the number of truss members in the truss structure This equationassumes the properties like cross-section length and Youngrsquos modulus to beconstant over the truss member
If we apply Castiglianorsquos theorem to the above expressions we obtain anexpression for the displacement in the direction of a certain applied force P (do not confuse with the internal forces F )
δ = dU
dP =
ni=1
F idF idP
Li
E iAi
(68)
The quantity dF
idP is how the internal force in truss member i changes if the external loading P changes This is also denoted with the symbol f i
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41
6 Displacement Analysis and Design of Statically Determinate
Trusses
δ = dU
dP =
ni=1
F if iLi
E iAi
(69)
Notice that f i is dimensionless
Dummy Load Method
In practice the above derived theory works as follows The most difficult and
important part of the expression is to determine the unity force distributionf i This quantity indicates how the internal forces change with changingexternal force Keeping this in mind and also thinking of the fact thatCastiglianorsquos theorem says that the displacement is obtained by derivingthe total strain energy U by a force in that direction one can conclude thefollowing If one wants to know the displacement of a structure in a certaindirection place a unit load (remind that f i is dimensionless) at that particularpoint in the direction in which one desires to know the displacement Dothis regardless whether an external load P is already applied or not Usingthis applied unit load one can calculate f i Use the following approach to
calculate the displacements of a truss
bull Calculate the internal force F i in each truss member due to the exter-nally applied forces P i
bull Remove all externally applied forces
bull Apply a unit load at the point where you want to know the displacementin the direction you want to calculate the displacement
bull Calculate the reaction forces at the supports due to the dummy load
bull Calculate the internal unity forces f i due to the dummy load
bull Apply equation 68 to obtain the magintude of the displacement in thedirection of the dummy load
62 Displacement Design
If the geometry of a structure is given including the loading conditions theconstruction displaces according to the given parameters However it mightbe desirable to alter the geometry of the construction such that a certainpoint of the truss meets a prescribed displacement δ 0 Usually the truss
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63 Example 42
layout in terms of member lenghts is fixed as well as the material propertiesThis leaves the cross-sectional areas as variables Assume that only the cross-sectional area of member n An is variable we get the following equation forthe displacement (analogue to equation 68)
nminus1i=1
F if iLi
E iAi
+ F nf nLn
E nAn
= δ 0 (610)
The only unknown in this equation is the unknown cross-section area An
which can be determined easily The same procedure holds if a certain lengthLi or Youngrsquos modulus E i is unknown
63 Example
As an example we take the same truss as in the example of chapter 2 seefigure 26 We already know the internal force distribution due to the appliedload P which is given in section 25 so we know all the unknown quantitiesF i Now it we want to know the displacement of node B in positive y-direction we first remove the load P and apply a dummy load as given infigure 64
x
y
A
BC
1
Figure 64 Truss structure with applied dummy load
By inspection it is evident that the reaction dummy force cy is equal to1 Furthermore ax = LBC
LAC and as such cx = minusLBC
LAC Note that each of these
dummy reaction forces are dimensionless as well The internal dummy forcesf i can be calculated by looking at nodal equilibrium as given in figure 65
The equilibrium equations for node A are
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43
6 Displacement Analysis and Design of Statically Determinate
Trusses
ax
fABfAC
AfAB
fBC
B
cx fBC
C
1
cy
fAC
Figure 65 Nodal internal dummy loads
F rarr+x 0 = ax + f AB
LBC
LAB
(611)
F uarr+y 0 = minusf AC minus f AB
LAC
LAB
(612)
The equilibrium equations for node B are
F rarr+x 0 = minusf BC minus f AB
LBC
LAB
(613)
F uarr+y 0 = 1 + f AB
LAC
LAB
(614)
The equilibrium equations for node C are
F rarr+x 0 = cx + f BC (615)
F uarr+y 0 = cy + f AC (616)
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
ax =
LBC
LAC
cx = minusLBC
LAC
cy = minus1 f AB = minusLAB
LAC
f AC = 1 f BC = LBC
LAC
(617)
Note that the reaction forces are indeed the ones we predicted which givesconfidence that our solution is correct indeed Note furthermore that thedummy forces f i are minusdF i
dP indeed The minus sign comes from the opposite
signs of P and the dummy loadIf then equation 68 is applied and assuming lengths LAC and LBC are
equal to L (to simplify the resulting expression) we obtain for the vertical
displacement
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63 Example 44
δ = minusP 2(1 +radic
2)L
EA (618)
Since we chose the dummy load to be positive upwards the resultingdisplacement is negative since node B will displace downwards under theapplied load P Verify this by inspection
Assume we prescribe the vertical displacement positive upwards accord-ing to the dummy load direction to be minusδ 0 All geometric values are knownexcept the cross-sectional area AAC The other cross-sectional areas are all
equal to A Determine that value in order the displacement to be δ 0
δ 0 = minus
F AC L
EAAC
+ F BC L
EA +
F AB
radic 2L
EA
(619)
The unknown can now be solved as
AAC = P L A
A E δ 0 minus (2radic
2 + 1)P L(620)
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Chapter 7
Displacement Analysis andDesign of Statically
Determinate Beams
In this chapter we consider the calculation of beam displacements and ro-tations at selected points This is done as explained before for trusses by
applying the unit load method Again we restrict ourselves to beams in thex-z plane subject to lateral loading
If it is desired to calculate the displacement at a certain point p weapply a unit later force at p The direction of the unit load is arbitrary (itcan be pointing up or down) It must be kept in mind that the displacementcalculated will be in the direction of the load Thus a negative displacementmeans that the actual sense of the displacement is opposite to that of theunit load Similarly if it is desired to calculate the beam rotation at a certainpoint we apply a unit moment at this point
The bending moment distribution due to the application of the unit load
(force or moment) is denoted by my(x) Then the displacement (or rotation)at the given point in the direction of the unit load is given by
δ =
beam
my(x)M y(x)
EI (x) dx (71)
where M (x) is the bending moment distribution due to actual appliedloads
The integration in 71 can be evaluated analytically only in special casesFor general load distribution and variable cross-sectional geometry along thebeam (eg due to tapering) it is best to evaluate the integral 71 in con- junction with the discrete method described in the previous chapter The
45
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71 Example 46
discrete version of the equation reads
δ =n
i=1
m(i)y
M (i)y
EI (i) dx (72)
where
M (i) =
M (i)y
L
+ M (i)y
R
2 (73)
and similarly for m(i)y
71 Example
We take a beam with length L Youngrsquos modulus E and moment of inertiaI which is loaded with a constant distributed load q 0 The objective is tocalculate the tip rotation of the beam For that purpose we are going to usethe integral equation as given in equation 71 We first calculate the momentdistribution due to the distributed loading q 0
M (x) = minusq 02 (Lminus x)
2
(74)
Since we are interested in the tip rotation we apply a unit moment at thetip of the beam as depicted in figure 71 The resulting moment distributionis hence
m(x) = 1 (75)
Applying the integral equation you end up with a tip rotation θ
θ =minus
q 0
2EI
L
0
(Lminus
x2)dx =minus
q 0L3
6EI (76)
Check this expression with the basic mechanics expression for a tip rota-tion of a beam loaded under a distributed load
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47
7 Displacement Analysis and Design of Statically Determinate
Beams
A B
q0
A B
1
Figure 71 Statically determinate beam example
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71 Example 48
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Chapter 8
Analysis of StaticallyIndeterminate Trusses
In previous chapters statically determinate trusses have been discussed Alsothe way to determine whether a structure is determinate or indeterminateand the degree of indeterminacy is explained This basically breaks down tochecking whether the number of available equilibrium equations is equal to
the number of unknown forces in the system If that is the case then thestructure is statically determinate If not the number of missing equationsindicates the degree of indeterminacy A structure can be indeterminatefor several reasons Trusses are often indeterminate for redundancy reasonsmeaning that additional truss member are added to the structure in caseanother member would fail This makes the stress analysis of trusses morecomplex since additional equations need to be formulated in order to solvefor all unknowns Such equations can be formulated as displacement com-patibility equations as will be explained next
81 Displacement Compatibility
Essential in the technique of displacement compatibility is the fact that thestructural response is linear This allows the application of the principle of superposition Resultant displacements of a set of loadings acting simulta-neously is equal to the sum of the effect of individual loadings acting aloneThis is illustrated in figure 81
So if this is applied to a redundant or statically indeterminate truss onecan split up the indeterminate structure into a statically determinate partand a redundant force as is shown in the following figure
Figure 82 shows that the internal forces can be calculated as function of
49
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81 Displacement Compatibility 50
P
Pi
Pi
P
+ =
Figure 81 Force superposition principle
= +
P P
F1
F2
F3
F1cut F
2cut
f 1cut f
2cut unit
loads
F3 F
3F
3
Figure 82 Split-up of an indeterminate truss
the statically indeterminate force F 3 So this particular force remains to besolved This can be done by introducing a displacement compatibility at thelocation where the structure has been cut to create the statically determinateversion The relative displacement of the point where the structure is cutneeds to be compatible for both superposed configurations This can be illus-trated easily if we take figure 82 as an example The relative displacementof the point where force P is applied needs to be zero for the two superposed
structures First of all the displacement of the aforementioned point of thestatically determinate part is as follows
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51 8 Analysis of Statically Indeterminate Trusses
δ det = F cuti f cuti Li
E iAi
(81)
The displacement due to the indeterminate force is now calculated asfollows
δ F 3 = (F 3f cuti ) f cuti Li
E iAi
(82)
Now the relative displacement needs to be zero enforced by the followingdisplacement compatibility equation
δ det + δ F 3 = 0 (83)
Note that in the above equation the only unknown is the indeterminateforce F 3 If the value for that force is known the internal force distribution isdefined It can be seen from equation 83 that the force F 3 is a function of thegeometric parameters such as cross-sectional areas and Youngrsquos moduli Thismeans that if the geometry changes the internal force distribution is alteredas well It is important to keep this in mind when designing an indeterminate
structure
82 Action Item List for Analysis of Indeter-
minate Trusses
This following list of actions should be followed when analysing a staticallyindeterminate truss
bull Make the structure statically determinate remove truss members or
supports that render the truss to be indeterminate Often this canbe done in multiple ways though the different individual approachesshould yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement compatibility at the locations where the stat-ically indeterminate forces have been removed At those locations cal-culate the displacements in the direction of the released forces usingthe dummy load method
bull Next calculate the displacements again in the aforementioned direc-tions due to the released forces
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53 8 Analysis of Statically Indeterminate Trusses
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (84)
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structure
For simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f f LAB 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200times 103 15 1414 1061 1414 3536 times 10minus4 0707BD 200times 103 15 1000 750 0 2500 times 10minus4 0CD 200times 103 15 1000 750 0 2500 times 10minus4 0
1207
Table 82 Calculation of displacement under mechanical load
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83 Example 54
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Chapter 9
Analysis of StaticallyIndeterminate Beams
In the previous chapter statically indeterminate trusses have been discussedObviously beams can be indeterminate as well The reason why beams areindeterminate is because of redundancy reasons as well or often supportingtruss members are connected to beams to restrict displacements or redis-
tribute stresses yielding lighter designs for the beam Just as for trusses thismakes the stress analysis of the beam more complex since the number of available equilibrium equations is smaller than the number of internal forcesand moments that need to be solved for Therefore displacement compatibil-ity equations need to be formulated which will make the internal force andmoment distribution dependent on the structurersquos geometry
The principle of static indeterminacy and analysis is basically the sameas for trusses This is discussed in the previous chapter However thereare certain differences as already highlighted in the chapters on staticallydeterminate trusses and beams The main issue is that beams apart fromaxial forces can also take bending moments This extends the principle of displacement compatibility of beams to displacement and rotation compati-bility
91 Force Method of Analysis
This following list of actions should be followed when analysing a staticallyindeterminate beam
bull Make the structure statically determinate remove additional forcesand moments that render the beam to be indeterminate Often this canbe done in multiple ways though the different individual approaches
55
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92 Flexibility Coefficients 56
should yield the same result This can be a convenient way of checkingthe correctness of the solution
bull Identify the displacement and rotation compatibility at the locationswhere the statically indeterminate forces and moments have been re-moved At those locations calculate the displacements and rotationsin the direction of the released forces using the dummy load method
bull Next calculate the displacements and rotations again in the aforemen-
tioned directions due to the released forces and momentsbull Sum up all individual displacements and rotation and equate them
to zero Obtain an equation for each displacement and rotation thatneeds to be compatible As such you will get as many equations asunknowns which should be a solvable system of equations
In case of beams it is convenient to use the principle of flexibility coeffi-cients This technique is highlighted in the following section
92 Flexibility CoefficientsThe flexibility coefficient f BA is defined as the displacement in point B dueto a unit force in point A This definition is illustrated in figure 91
A
1
B
f BA
f AA
Figure 91 Illustration of the flexibility coefficient
The use of such coefficients is very effective since if one wants to knowthe displacement in letrsquos say point X due to a force in point Y one takesthe flexibility coefficient f XY and one multiplies that coefficient with theforce in point Y and you obtain the displacement in point X Moreover theflexibility coefficient f XY is equal to f Y X according to Maxwellrsquos theoremThe convenience of flexibility coefficients becomes clear when looking at thefollowing figure
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57 9 Analysis of Statically Indeterminate Beams
A
P1
D
P2
B C
A
P1
D
P2
B C
∆B ∆C
A DB C
1xR B
f BBR B
f CBR B
A DB C
1xR C
f BCR C f
CCR C
||
+
+
Figure 92 Solution of a statically indeterminate beam using flexibility co-efficients
It is easy to see that this beam is statically indeterminate to the seconddegree This structure can be solved by enforcing the displacement compat-ibility at points B and C The displacement at point B is calculated byadding the displacements due to the external loads P 1 and P 2 and due tothe statically indeterminate forces RB and RC As such the displacementcompatibility equations for points B and C are calculated as
0 = ∆B + f BBRB + f BC RC (91)
0 = ∆C + f BC RB + f CC RC (92)
The flexibility coefficients can be calculated easily by using the dummyload method
93 Example
Let us consider a beam of length L which is cantilevered on the left handside in point A and supported by a roller on the right hand side in point
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93 Example 58
B The beam has Youngrsquos modulus E and a moment of inertia I It can beseen easily that the beam is statically indeterminate In order to analyse thestructure the first step is to make it statically determinate and superposethe statically indeterminate force This is shown in figure 93
A
B
A
B
=
+
A
B
RB
q0
q0
q0
Figure 93 Analysis example of a statically indeterminate beam
In this example the use of flexibility coefficients comes in handy In thiscase the flexibility coefficient f BB is needed The reason for this is that weare going to apply compatibility at the beam tip It can be seen easily that
the tip deflection in point B needs to be equal to zero For this particularbeam this influence coefficient is equal to
f BB = L3
3EI (93)
Try to figure out by using the dummy load method Furthermore fromstandard mechanics it is evident that the tip deflection due to a distributedload is equal to
δ B = q 0L4
8EI (94)
The compatibility equation now reads
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59 9 Analysis of Statically Indeterminate Beams
0 = q 0L4
8EI + f BBRB (95)
By adding the sign we assume the force RB to be acting in the downwarddirection Try to figure out yourself why The solution of this equation givesthe expression for the unknown force RB
RB = minus3
8q 0L (96)
Note the minus sign indicating that the force RB acts in the oppositethan defined meaning the force points upwards which is consistent with thephysical problem
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93 Example 60
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Chapter 10
Analysis of StaticallyIndeterminate Plates
The analysis of plates is rather intricate as already indicated in the chapteron statically determinate plates Only in exceptional cases of uniaxial orbiaxial stress states analytical methods can offer a way out In this chapterit is explained how internal stresses in plates can be calculated for indeter-
minate plates Again just as in the case of trusses and beams displacementcompatibility is the key to solving the stress state
As a start we consider a statically determinate plate If a square plate of length L and thickness t is loaded in x-direction with a force P the resultingstress in that direction is σx = P
Lt and according to Hookersquos law the strain
is x = P ELt
Due to the Poisson effect which accounts for the fact that aplate shrinks in a direction perpendicular to the loading direction the strainin the transverse direction is y = νx ν is the Poisson ratio with has atypical value of around 03 for metals Note that the stress in the transversedirection remains zero
101 Biaxial Stress State
Let us first consider a simple biaxial stress state plate which is given in figure101
Obviously there is a stress in x-direction due to the applied force P but because of both side restraints on the plate there is normal stress iny-direction as well due to the constraint force Ry The displacement in x-direction is defined as
δ x = P LEwt
minus ν Ry
Et (101)
61
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102 Multimaterial Plates 62
y
P P
R y R y+=
x
Figure 101 Biaxial stress state plate
while the transverse displacement is equal to
δ y = Ryw
ELt minus ν
P
Et (102)
Now the unknown force Ry is calculated by requiring the displacement iny-direction is equal to zero (due to the restraints)
102 Multimaterial Plates
Plates consisting of multiple materials are inherently indeterminate even if the geometric boundary conditions do not seem to indicate so For instancethe left plate of figure 102 contains two different materials each having theirown internal force In this particular case there are two unknown forcesand one force equilibrium equation Therefore a displacement compatibilityequation needs to be formulated If we take a closer look at the problemitself it is evident that the vertical displacements of the three parts of theplate need to extend equally under the given load Therefore an additional
equation can be formulated requiring the displacement of the Aluminiumbeing equal to the displacement of the Brass Note that this condition onlyapplies because of the fact that the two Aluminium part have exactly thesame dimensions otherwise the force in both Aluminium plates would bedifferent
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63 10 Analysis of Statically Indeterminate Plates
P
Al Al Brass
waw
a w
b
L
Al Al Brass
waw
a w
b
L
W
P
y
x
Figure 102 Indeterminate multimaterial plates (left one statically indeter-minate force right two statically indeterminate forces)
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102 Multimaterial Plates 64
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Chapter 11
Actuated Structures
Linear actuation can be achieved through various means A simple exampleis the change in length of bars under thermal actuation (sometimes referredto as thermal load) Hydraulic and pneumatic cylinders which work withpressurised liquids and gases respectively are the most common commerciallyavailable linear actuators A more modern version of similar devices are theelectrically actuated active material systems such as piezoelectric devices
While we consider temperature as an actuation the methods presentedhere are also applicable for the practically important case of thermal loading For many aerospace structures such as launchers reentry vehicles and en-gines structural components are subject to elevated temperatures that leadto considerable stresses and deformations The stress and deflection analy-sis of thermally loaded structures follows the same methodology as thermalactuation
111 Thermal Actuation
A bar with a uniform cross section A will change its length due to a uniformapplied change in the temperature ∆ T The change in length is linearlyproportional to the temperature change as shown in figure 111 Typicallythis relation is represented in terms of thermal strain and is given by therelation
T = α∆ T (111)
where α is the coefficient of thermal expansion (CTE) This is a special onedimensional case of a more general three dimensional description of thermalexpansion and is sometimes called linear coefficient of thermal expansion Ingeneral application of the thermal field would also generate strain in other
65
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112 Piezoelectric Actuation 66
directions (like in the cross-sectional expansion of a bar) but for the timebeing we will ignore that for one-dimensional bars
Figure 111 Thermal extension of a bar
Following are the CTEs for three popular materials namely copper alu-minium and steel respectively all in the units of cmcmC (strainC)αcu = 18times 10minus5 αAl = 25times 10minus5 and αst = 19times 10minus5
Note that there is a limit on the maximum thermal strain that can beachieved for a given material dictated by the maximum temperature that thematerial is able to withstand without loosing its elastic properties We willcall this strain max
a Also the temperature difference may be either positiveor negative with respect to the room (or operating) temperature
112 Piezoelectric Actuation
For a piezoelectric material an electric field E 3 produces strains representedby S 33 as shown in figure 112
S 33 = d33E 3 (112)
In this representation the subscript 33 refers to the thickness direction inthe piezoelectric material The use of such subscripts should not be confus-ing as the basic equation above 112 is same as thermal actuation of a barpresented earlier in equation 111 it is just different notation
In an actual piezoelectric material (like in the case of thermal loads) thesame electric field across the thickness also produces strains in the transversedirections but these will not be needed until we study bending actuatorsconstructed using piezoelectric material
For a typical piezoelectric material d33 = 500 times 10minus12
mV
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67 11 Actuated Structures
Figure 112 Piezoelectric thickness change of a bar
In contrast with the thermal case we rarely use the applied electric fielddirectly The applied actuation is customarily specified as applied voltageThe electric field E 3 generated by applying voltage difference V across apiezoelectric substrate of thickness t is
E 3 = V
t (113)
Similar to the thermal case in which there is a maximum temperaturethat can be applied to the material there is a maximum electric field thatcan be applied called the coercive electric field E c Strain correspondingto this coercive field is
S max33 = d33E 3 (114)
For a typical value of the coercive field E c = 04 times 106V m the maximumpossible strain is S max
33 = 02 times 10minus3Typically in order to increase the stroke (displacement) capability the
piezoceramic material is arranged in a stacks Lets say we put 200 layers on
top of one another In this way the displacement output is amplified 200times Piezoelectric material arranged in a stack is rather stiff on the otherhand They produce small deformations but they can produce very largeactuation forces This will be demonstrated in the next section However wefirst learn how to handle strainsdeformations created under the combinedaction of applied forces and applied actuation
113 Response of an Actuated Member
More than often the actuators described above will have to operate under
mechanical loads acting on the members In this section we will assume that
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113 Response of an Actuated Member 68
the entire bar is made of a material that will be actuated the cross sectionalarea of the bar is uniform A the modulus is E and the bar is under theaction of an axial load F Also for simplicity we will consider the thermalcase only the piezoelectric case being similar
Under the action of combined thermal and mechanical load the strainthat the active member will experience is the superposition of the strainsdue to the thermal load and the mechanical loads
= T + M (115)
The mechanical part of the strain M is the one responsible for thegeneration of mechanical stress thus
σ = EM (116)
where E is Youngrsquos modulus of the material The stress in the member canbe expressed as
σ = F
A (117)
The strain in the member is
= α∆ T + F
EA (118)
Clearly when the applied external force is zero then the total strain of theactuator is force free (M = 0) and is equal to the T
The other extreme case is when the expansion of the bar is completelyrestrained ( = 0) In this case the corresponding force is called the blocked
force F b which can be computed by setting = 0 in 118 to obtain
F b = αE ∆ T A (119)
with a corresponding (compressive) stress of
σb = minusαE ∆ T (1110)
In realistic applications when an actuated member is part of a structurethat is performing a function the members is neither fully blocked nor com-pletely free to extend It will be under the action of combined mechanicaland thermal (actuation) loads
In the following section we will use the dummy-load method to determinethe actuation temperature of an axial member
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69 11 Actuated Structures
114 Dummy Load Method for Structures with
Actuated Members
We certainly do not need the dummy-load method to determine the responseof a single member under a mechanical load and actuation The equationderived in the previous section based on superposition would have beenenough to compute anythingeverything we need However as we will seelater with examples the dummy-load method provides a powerful tool to
compute response efficiently in more complicated structural geometries es-pecially the ones that are statically indeterminate Therefore although wewill still use the method of superposition the argument provided for thecalculation of the displacement under the thermal load may be useful (eventhough it is implementation is ridiculously simple)
The basic equation for the unit load method is not changed The dis-placement at a given point and in a given direction is calculated by applyingthe corresponding unit load and calculating
δ =
f iiLi (1111)
where f i are the member forces under the unit load The only additionalconsideration is that the strain in the i-th member is calculated according to118
115 Example
The same truss is considered as in chapter 8 The picture is shown again infigure 113 The length of the truss is 1000 mm the cross-sectional areas of the members are 15 mm2 and the applied force P is 1500 N The Youngrsquos
modulus is 200 GP aAs usual we use superposition Moreover we will compute the displace-ment due to mechanical and thermal loads separately One important pointto remember is that for statically indeterminate structures under thermalloads member forces are not zero
First we calculate all the necessary ingredients for the displacement com-patibility equation used to calculate the redundant force F AD under theloading P
The value of the redundant reaction in member AD is calculated fromthe condition
1707 + 1609 times 10minus3 RAD = 0 =rArr RAD = minus1061N (1112)
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115 Example 70
A
B
C
D
A
B
C
D
L
L
L
L
P
FAD
FAD
Figure 113 Statically indeterminate six bar truss
member E A L F sd f sd f L f 2LEAAB 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AC 200 times 103 15 1000 -1500 -07071 minus50000times 10minus4 0354 1667 times 10minus4
AD 200 times 103 15 1414 0 10 0 00 4714 times 10minus4
BC 200 times 103 15 1414 2121 10 70710times 10minus4 1 4714 times 10minus4
BD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
CD 200 times 103 15 1000 0 -07071 0 0 1667 times 10minus4
1707 1609times
10minus3
Table 111 Calculation of redundant force under mechanical load
For the calculation of displacements we need the total member forces F which is obtained by superposing the statically determinate values with thevalues due to the redundant reaction The unit load at point C needed tocalculate the displacement is applied to a statically determinate structureFor simplicity we use the same statically determinate structure as was usedfor the calculation of the redundant
member E A L F f fLAB 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AC 200 times 103 15 1000 -750 -1 minus2500 times 10minus4 0250AD 200 times 103 15 1414 -1061 0 minus3536 times 10minus4 00BC 200 times 103 15 1414 1061 1414 3536times 10minus4 0707BD 200 times 103 15 1000 750 0 2500times 10minus4 0CD 200 times 103 15 1000 750 0 2500times 10minus4 0
1207
Table 112 Calculation of displacement under mechanical load
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3 1 Introduction
and beam members and plate segments The primary difference betweenthe three types of components is the type of loads that are carried by themTruss members are two-load members which can only carry co-linear loads(either tensile or compressive) acting along a line that passes through thetwo pin joints connecting the member to other truss members or structuralcomponents Furthermore the member is assumed to be straight between thetwo pins hence no bending of the member is allowed The beam membersare primarily used for carrying bending loads which are typically induced bymore than two forces or two force members with non co-linear forces Finally
plate segments are the more general form of a load carrying componentreacting to in plane and out of plane loads A more formal definition of thesecomponents will be provided in appropriate chapters later on For the timebeing we only consider some examples to them in aircraft and spacecraftstructures
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Chapter 2
Stress Analysis and Design of Statically Determinate Trusses
A lot of aerospace structures can be idealised as truss structures This isclearly illustrated in picture 21 where it can be seen that the ribs in thewing are built up as a truss structure Also in space applications trussesare widely used because of their simplicity and light-weightness Now the
question rises what is a truss
Figure 21 Rib truss structure example [Courtesy of wwwsteenaerocom]
5
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21 Truss Properties 6
21 Truss Properties
A truss is a structure built up from truss members which are slender barswith a cross-sectional area A and having a Youngrsquos modulus E
A generic picture is given in figure 22
Figure 22 Typical simple truss structure [Courtesy of wwwcompuzonecokr]
The following assumptions apply when analysing a truss structure
bull Bar elements can only transfer loads axially These forces can be eithertensile tending to elongate the bar or compressive tending to shortenthe bar
bull The bar elements are pin-joined together This has as a consequencethat the joints only transfer forces from one bar element to the otherand no moments If the bar elements were welded together or attached
with a plate also moments would have been transferred which is incontradiction with the earlier mentioning that bar elements are onlysuited to take axial loads and no moments
bull Loadings can only be applied at the joints of the truss This inherentlymeans that the weight of the bar which would act at the midpoint of a uniform bar is neglected
22 Static Determinacy of Trusses
A truss is statically determinate if the numbers of unknowns is equal to thenumbers of equations (which is twice the number of nodes n because at each
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7 2 Stress Analysis and Design of Statically Determinate Trusses
node two forces in x and y direction act) that can be constructed for theproblem The unknowns are the truss member forces and the reaction forcesat the truss supports There are m truss member forces and r reaction forces
So a necessary condition for static determinacy of a truss structure is
m + r = 2n (21)
If m + r gt 2n the truss structure is indeterminate but that will bediscussed in later chapters
The reaction forces are due to the supports of the truss structure Thetwo support types that we use for truss structures are the pinned joint andthe roller support Both supports can be inspected in figure 23
Pinned node
Roller support
Rx
Ry
Ry
Support Type Reactions
Figure 23 Truss support types
Two trusses are given in figure 24 One of them is statically determinateand one is indeterminate Figure out why they are determinate or indeter-minate
Figure 24 Statically determinate and indeterminate trusses
23 Truss Analysis
The internal forces created by an external loading to the truss can be solvedby using the method of joints This method implies that the summation of all
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23 Truss Analysis 8
forces in either of the two directions at each node is equal to zero (
F x = 0and
F y = 0) The method of joints is highlighted in figure 25
Figure 25 Internal force distribution inside a truss
231 Truss Definition
Defining and analysing a truss structure is not only a matter of stress analysisbut also a matter of good book keeping Therefore we define the nodallocations the members along with their begin and end nodes and materialproperties in an orderly fashion The ith node has the following location pi
pi = xi yi i = 1 n (22)
where n is the number of nodes in the truss The jth member or element e jhas a start point ps and an end point pe
e j = ps pe j = 1 m (23)
where m is the number of members Each member has its Youngrsquos modulusE j and cross-sectional area A j The length l j of the jth member is thendefined as
l j =
(xs minus xe)2 + (ys minus ye)2 (24)
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9 2 Stress Analysis and Design of Statically Determinate Trusses
Finally each node has its own applied force P j which can be zero in caseof the absence of the force
232 Stress Analysis
The procedure for analysing a truss is as follows
bull Define the truss as described in the previous section
bull Draw each node separately including the member forces and if relevantthe applied forces and reaction forces on the node By convention wedraw the direction of the member forces away from the node (see alsofigure 25) This does not mean that all forces are tensile Compressiveforces will come out negative The reaction forces should be drawn inthe positive axis directions and applied forces should remain in theiroriginal direction
bull Sum all the forces per node and per direction and equate them to zero
for equilibrium
bull Solve the obtained equations from the previous item to get the unknownforces (notice that in case of a statically indeterminate truss you wouldhave too few equations) both member and reaction forces
bull Check global equilibrium to see whether the reaction forces come outright This gives additional confidence in the solution
Notice that if the convention of drawing the unknown member forces away
from the node is used the compressive member forces come out negativeautomatically Now that the member forces are known the member stressesσ j can be retrieved using
σ j = P jA j
(25)
Notice that only the cross-sectional area itself is needed for the calculationof the member stress so the shape of the cross section albeit a square ora circle is irrelevant Also notice that since a truss member can only takenormal forces only normal stress is calculated so no shear stress is presentin a truss member
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24 Stress Design 10
24 Stress Design
The stresses in the truss members can be calculated according to equation25 This means that if the loading on the member and its geometry areknown the resulting stress can be derived straightly Now the questionoccurs whether the resulting stress is larger than the stress the material cantake without failure The latter stress is the allowable stress σall
If the occurring stress in a truss member is lower than the allowable stressit means that there is too much material present Obviously an aerospace
structure should be as light as possible So therefore in a stress design oneshould always strive for equating the occurring stress and the allowable stressSuch a desing philosophy is called a fully stressed design In a staticallydeterminate truss the member forces are solely dependent on the externallyapplied loads As such one should adapt the cross sectional area to a new
one as
σall = F
Anew
(26)
Keeping in mind that the original expression for the stress is σ = F
Aold
we get as full stressed design update criterion for the cross-sectional area thefollowing expression
Anew =
σ
σall
Aold (27)
25 Example
Given is a three bar truss shown in figure 26 with all element having equal
length A force P is applied at node B in negative y direction First free-body diagrams are drawn for each node separately They can be viewed infigure 27
The equilibrium equations for node A are
F rarr+x 0 = Ax + F AB
LBC
LAB
(28)
F uarr+y 0 = minusF AC minus F AB
LAC
LAB
(29)
The equilibrium equations for node B are
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112 Stress Analysis and Design of Statically Determinate Trusses
P
A
B
C x
y
Figure 26 Three bar truss example
A
FABFAC
Ax
BFAB
P
FBC
C
FBC
Cy
Cx
FAC
Figure 27 Free-body diagram of each node
F rarr+x 0 = minusF BC minus F AB
LBC
LAB
(210)
F uarr+y 0 = minusP + F AB
LAC
LAB
(211)
The equilibrium equations for node C are
F rarr+x 0 = C x + F BC (212)
F uarr+y 0 = C y + F AC (213)
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25 Example 12
Solution of the six equations simultaneously yields the following expres-sions for the unknowns
Ax = minusP
LBC
LAC
C x = P LBC
LAC
C y = P F AB = P LAB
LAC
F AC = minusP F BC = minusP LBC
LAC
(214)
By inspection it is evident that the reaction forces equilibrate each otherand the externally applied force P
Now that the three member forces are known their cross-sectional areascan be changed such that each of the members becomes fully stressed usingequation 27 Assume the material has an allowable stress σall the new crosssectional areas become
AABnew =
P LABLAC
σall
AACnew = P
σall
ABCnew =P LBC
LAC
σall
(215)
Note that although some of the member forces were negative their ab-solute value needs to be taken to calculate the cross-sectional areas becauseobviously an area cannot be negative
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Chapter 3
Stress Analysis and Design of Statically Determinate Beams
The objective of this chapter is to