Plastic Em

8/21/2019 Plastic Em

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“Gh. Asachi” Technical University of Iaşi Hyperstatic structures

Department of Structural Mechanics Cezar Aanicăi, Dr. Eng.

1

POSTELASTIC STRUCTURAL ANALYSIS

1. Basics of postelastic analysis

In plastic analysis and design of a structure, the ultimate load of the structure as a whole is

regarded as the design criterion. The term plastic has occurred due to the fact that the ultimate load is

found from the strength of steel in the plastic range. This method is rapid and provides a rational

approach for the analysis of the structure. It also provides striking economy as regards the weight of

steel since the sections required by this method are smaller in size than those required by the method

of elastic analysis. Plastic analysis and design has its main application in the analysis and design of

statically indeterminate framed structures.

Fig, 1 – Mathematical models for steel

Many mathematical models are considered in nonlinear structural analysis. Choosing the

appropriate mathematical model for the material from which a structure is made is the main key for a

successful structural response evaluation (Fig. 1). The plastic method is applicable to structures

constructed with an ideal elastic-plastic material that exhibits the stress-strain relationship shown in

Figure 2. The moment-curvature relationship for any section of the structure is assumed to have the

ideal form shown in Figure 3. Thus, on applying a uniform sagging moment to a member the moment-

curvature relationship is linear until the applied moment reaches the value of M p the plastic moment

of resistance of the section. At this stage, all material above the zero-strain axis of the section has

yielded in compression and all material below has yielded in tension, and a plastic hinge has formed.Then the section can offer no additional resistance to deformation, and increase in curvature continues

at a constant applied moment. In addition, in determining the collapse load of a structure, it is

assumed that elastic deformations are negligible and do not affect the geometry of the structure. It


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means that the width-thickness ratio of plate elements is small so that local buckling does not occur –

in other words, the sections will classify as plastic. Thus, the structure behaves in a rigid-plastic

manner with zero deformation until the formation of sufficient plastic hinges to produce a mechanism.

With these assumptions, it can be said that the section will reach its plastic moment capacity and then

undergo considerable rotation at this moment. With these assumptions, we will now look at the

behaviour of a beam up to collapse.

Fig. 2 Fig. 3

2. Formation of plastic hinges

Consider a simply supported beam subjected to a uniformly distributed working load of total

magnitude w, as shown in Fig. 4. The elastic bending moment at the ends is wl 2 /12, and at mid-span is

wl 2

/24, where l is the span. The stress distribution across any cross section is linear (Fig. 5.a). As w isincreased gradually, the bending moment at every section increases and the stresses also increase. At a

section close to the support where the maximum bending moment is, the stresses in the extreme fibers

reach the yield stress. The moment corresponding to this state is called the first yield moment M y, of

the cross section. But this does not imply failure as the beam can continue to take additional load. As

the load continues to increase, more and more fibers reach the yield stress and the stress distribution is

as shown in Fig 5.b. Eventually, the whole of the cross section reaches the yield stress and the

corresponding stress distribution is as shown in Fig. 5.c. The moment corresponding to this state is

known as the plastic moment of the cross section and is denoted by M p. In order to find out the fully

plastic moment of a yielded section of a beam, we employ the force equilibrium equation, namely the

total force in compression and the total force in tension over that section are equal.

w

plastic hinges

collapse mechanism

M /2 p

M pwl /24

2

wl /122

wl /122

M p

wu

M p M p

l l

Fig. 4 – Formation of a collapse mechanism in a fully fixed beam


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a) Elastic limit b) Elastic-plastic c) Quasi-fully plasticized

cross-section

Fig. 5

Finally, as the applied load is increased still further, a plastic hinge forms in the center of the

beam, and the distribution of bending moment is as shown. The beam has now been converted to theunstable collapse mechanism shown above, and collapse is imminent under the ultimate load wu. The

ratio of the collapse load to the working load is:

uw

=w

λ

where is the collapse the load factor . Since the structure is statically determinate at the point of

collapse, the collapse load is readily determined as:

16 p

uw

l =

and this value is unaffected by settlement of the supports or elastically restrained end connections.

Fig. 6 – Elastic-plastic response Fig. 7 – Mathematical model

of beams with different cross-sections elastic-perfectly plastic

Fig. 8 - Ideal plastic hinge formationi i

p p p- M M + M ≤ ≤

1/R

Mp

M

o

formation of plastic hinge


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Theoretically, the plastic hinges are assumed to form at points at which plastic rotations occur.

Thus the length of a plastic hinge is considered as zero. However, the values of moment, at the

adjacent section of the yield zone are more than the yield moment up to a certain length Δ L, of the

structural member. This length Δ L, is known as the hinged length. The hinged length depends upon

the type of loading and the geometry of the cross-section of the structural member. The region of

hinged length is known as region of yield or plasticity.

3. Plastic moment of resistance

After the formation of a plastic

hinge in the section shown above, the

rectangular stress distribution shown at is

produced. Equating horizontal

compressive and tensile forces:

c t y c y t

P P f A f A= ⇒ ='

where f y and f y′ are the yield stresses in

tension and compression (which may be

assumed to be equal) and At and Ac are

the cross-sectional areas in tension and

compression. Thus, the plastic moment of

resistance is:

p p yW f = ⋅

where M p is the plastic bending moment, W pl is the plastic section modulus, and f y the yield limit;

For the cross sections subjected to pure axial efforts:

p y N A f = ⋅

where N p is the plastic axial force, and A the cross section area.

5. Shape factor

The shape factor is defined by the following relationship:

p

s

el

W =

W

α

where W el is the elastic section modulus. The shape factor value strongly depends by the cross-

section shape:

- for hot rolled elements:

I cross section: ,(y) (z) s s= 1.85 = 1.15 1.17;α α ÷

U (or C) cross section:(z)(y)

s s= 1.20;= 2.10 , α α

- rectangular hollow cross section: s = 1.27 ;α

- full rectangular cross section: s = 1.50 ;α

- circular cross section: s = 1.70 ;α

- rhombic cross section: s = 2.00 ;α

- triangular cross section: s = 2.37 α .

Thus, for a rectangular section the plastic moment M p is about 1.5 times greater than the

elastic moment capacity.


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If the section is thin walled, due to local buckling, it may not be able to sustain the moment for

additional rotations and may collapse either before or soon after attaining the plastic moment. It may

be noted that formation of a single plastic hinge gives a collapse mechanism for a simply supported

beam. The ratio of the ultimate rotation to the yield rotation is called the rotation capacity of the

section. The yield and the plastic moments together with the rotation capacity of the cross section are

used to classify the sections.

6. Application

Let find the shape factor of the cross section shown in Figure 9:

8 4 6

8 4 12

z G = 1 0 . 4

3

8 4 6

8 4 12

z p = 9

2 2 - y p =

1 3

1 1 . 5

7

AN

AN

Elastic behaviour Fully plasticized cross-section

z 1 = 6 . 0

2

z 2 = 9 . 3

7+

G2

[ cm] [ cm]

+G1

4

1 6

2

4

1 6

2

+

G+

G

Fig. 9

Solve for the centroid position and for the elastic section modulus:

G

3 3 32 2 2

3

18 4 2+16 4 12+24 2 21z = 10 43

18 4+16 4+24 2

18 4 4 16 24 218 4 8 43 4 16 1 57 24 2 10 57

12 12 12 1047 0711 57

el

cm

W cm

⋅ ⋅ ⋅ ⋅ ⋅ ⋅=

⋅ ⋅ ⋅

⋅ ⋅ ⋅+ ⋅ ⋅ + + ⋅ ⋅ + + ⋅ ⋅

= =

.

. . .

..

Now, considering that the cross section is fully plasticized, in order to find the plastic

section modulus the new position of the neutral axis must be founded:

p p (8+4+12) 2+(22-z -2) 4=(8+4+6) 4+(z -4) 4 A A

− += ⇒ ⋅ ⋅ ⋅ ⋅

( ) ( )

p p p48+80-4 z =72+4 z -16 z = 9 cm⇒ ⋅ ⋅ ⇒

z1

8 4+ 6+( ) 4⋅ 7⋅ 9 4−( ) 4⋅ 2.5⋅+

18 4⋅ 5 4⋅+6.02=:= z

2

8 4+ 12+( ) 2⋅ 12⋅ 13 2−( ) 4⋅ 6.5⋅+

24 2⋅ 11 4⋅+9.37=:=

cm3

Wpl 18 4⋅ 5 4⋅+( ) 6.02⋅ 24 2⋅ 11 4⋅+( ) 9.37⋅+ 1415.88=:=


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The shape factor is:

1451 881 352

1047 07

pl

s

el

W =

W α = =

..

.

7. Principles of plastic analysis

Fundamental conditions for plastic analysis

- Mechanism condition: The ultimate or collapse load is reached when a mechanism is

formed. The number of plastic hinges developed should be just sufficient to form a mechanism.

- Equilibrium condition: Σ Fx = 0, Σ Fy = 0, Σ Mxy = 0

- Plastic moment condition: The bending moment at any section of the structure should not

be more than the fully plastic moment of the section.

Collapse mechanisms

When a system of loads is applied to an elastic body, it will deform and will show a resistance

against deformation. Such a body is known as a structure. On the other hand if no resistance is set up

against deformation in the body, then it is known as a mechanism. Various types of independent

mechanisms are identified to enable prediction of possible failure modes of a structure.

a) Beam mechanism

Fig. 4 (right) shows a fully fixed beam and the corresponding mechanism.

b) Panel or Sway mechanism

The Figure 10.a shows a panel or sway mechanism for a portal frame fixed at both ends.

a) Panel (sway) mechanism b) Gable mechanism c) Joint mechanism

Fig. 10

c) Gable mechanism

Fig. 10.b shows the gable mechanism for a gable structure fixed at both the supports.

d) Joint mechanism

Fig. 10.c shows a joint mechanism. It occurs at a joint where more than two structural

members meet.


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Combined mechanism

Various combinations of independent mechanisms can be made depending upon whether the

frame is made of strong beam and weak column combination or strong column and weak beam

combination. The one shown in Fig. 11 is a combination of a beam and sway mechanism. Failure istriggered by formation of hinges at the bases of the columns and the weak beam developing two

hinges. This is illustrated by the right hinge being shown on the beam, in a position slightly away

from the joint. From the above examples, it is seen that the number of hinges needed to form a

mechanism equals the static redundancy of the structure plus one.

Fig. 11

8. Plastic load factor and theorems of plastic collapse

The plastic load factor at rigid plastic collapse ( λ p ) is defined as the lowest multiple of the

design loads which will cause the whole structure, or any part of it to become a mechanism. In a limit

state approach, the designer is seeking to ensure that at the appropriate factored loads the structurewill not fail. Thus the rigid plastic load factor λ p must not be less than unity. The number of

independent mechanisms ( N ) is related to the number of possible plastic hinge locations ( X ) and the

number of degree of redundancy or the static indeterminacy ( ns ) of the frame by the equation.

N = X - ns

The three theorems of plastic collapse are given below.

(I) Lower bound or Static theorem

A load factor ( λs ) computed on the basis of an arbitrarily assumed bending moment diagram

which is in equilibrium with the applied loads and where the fully plastic moment of resistance is

nowhere exceeded will always be less than or at best equal to the load factor at rigid plasticcollapse, ( λ p ). In other words, λ p is the highest value of λ s which can be found.

(II) Upper bound or Kinematic theorem

A load factor ( λk ) computed on the basis of an arbitrarily assumed mechanism will always be

greater than, or at best equal to the load factor at rigid plastic collapse ( λ p ). In other words, λ p is the

lowest value of λk which can be found.

(III) Uniqueness theorem

If both the above criteria are satisfied, then the resulting load factor corresponds to its

value at rigid plastic collapse ( λ p ).


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9. Mechanism method

In the mechanism or kinematics method of plastic analysis, various plastic failure mechanisms

are evaluated. The plastic collapse loads corresponding to various failure mechanisms are obtained by

equating the internal work at the plastic hinges to the external work by loads during the virtual

displacement. This requires evaluation of displacements and plastic hinge rotations. As the plastic

deformations at collapse are considerably larger than elastic ones, it is assumed that the frame remains

rigid between supports and hinge positions i.e. all plastic rotation occurs at the plastic hinges.

Considering a simply supported beam subjected to a point load at midspan or a uniform

distributed working (Fig. 4), the maximum strain will take place at the centre of the span where a

plastic hinge will be formed at yield of full section. The remainder of the beam will remain straight,

thus the entire energy will be absorbed by the rotation of the plastic hinge.

10. Stability

For plastically designed frames three stability criteria have to be considered for ensuring thesafety of the frame. These are:

1. general frame stability,

2. local buckling criterion,

3. restraints.

Effect of axial load and shear

If a member is subjected to the combined action of bending moment and axial force, the plastic

moment capacity will be reduced. The presence of an axial load implies that the sum of the tension

and compression forces in the section is not zero. This means that the neutral axis moves away from

the equal area axis providing an additional area in tension or compression depending on the type ofaxial load. The interaction equation is:

2

1 x

p y

M N

N = −

The presence of shear forces will also reduce the moment capacity.

b f y

f y

h

h/2

z1

f y

f y

f y

= +

Total stresses = Bending Axial compression+

Fig. 12


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Plastic analysis for more than one condition of loading

When more than one condition of loading can be applied to a beam or structure, it may not

always be obvious which is critical. It is necessary then to perform separate calculations, one for each

loading condition, the section being determined by the solution requiring the largest plastic moment.Unlike the elastic method of design in which moments produced by different loading systems can be

added together, plastic moments obtained by different loading systems cannot be combined, i.e. the

plastic moment calculated for a given set of loads is only valid for that loading condition. This is

because the “ principle of superposition” becomes invalid when parts of the structure have yielded.

11. Application

Let find the collapse mechanism of the following frame system using the combined

mechanism method.

The number of elementary mechanism is:

N= X – n = 10 – 6 = 4

two beam mechanisms (fig. 13.a, b), one sway mechanism (fig. 13.c) and one joint mechanism

(fig. 13.d).

71 10

2M p 4M p

M p

4m

λ

2λ 4λ

32 4 5

6

8 9

2λ4λ

b)a)

M p M p

2M p

2M p2M p

M p 4M p

M p

4M p4M p

4m 8m 8m

Fig. 13


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2M p2M p

2M p

M p

λ

2λ

λ

c) d)

e)

M p

M p

M p

M p

M p

M p

M pM p

M p

M p

Fig. 13 (cont.)

For the elementary mechanisms the following relationships could be written:

- for the beam mechanism shown in Figure 13.a:

(a)72 2 2 0.8758

p p p p p p2 4 = M + M M M M M λ θ θ θ θ θ λ ⋅ ⋅ ⋅ + ⋅ + ⋅ ⇒ = =

- for the beam mechanism shown in Figure 13.b:

(b)

134 8 3 4 + 0.406

32 p p p p = M M M M λ θ θ θ λ ⋅ ⋅ ⋅ ⋅ ⇒ = =

- for the floor (or displacement) mechanism shown in Figure 13.c:

(c)

64 6 1.5

4 p p p = M M M λ θ θ λ ⋅ ⋅ ⇒ = =

The combined mechanisms are obtained as follows:

- Combining the elementary mechanisms (a) and (b), another possible mechanism results

(Fig. 12.e); the joint mechanism (d) is not useful, because rotating the joint high values for the

internal work are obtained (8M pθ - Fig. 12.g – clockwise rotation) and, 6M pθ , respectively, (Fig.

12.h – counterclockwise rotation), higher than the value of the internal work when the joint is

not rotated (Fig. 12.f).

M p

2M p

4M p

2M p2M p

M p

M p

4M p

f) La = 3M p g) La = 8M p > 3M p h) La = 6M p > 3M p

Fig. 13 (cont.)


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For the mechanism shown in Fig. 12.e, the equilibrium relationship could be written

in two forms:

a)( ) ( ) ( ) ( )a c a c articulatii inchise

F F a a a L L L L L+ = + −

(e) 114 2 4 7 6 2 0.91712

p p p p p = M M M M M λ θ λ θ θ θ θ λ ⋅ + ⋅ ⋅ + ⋅ − ⋅ ⇒ = =

b) or directly on the combined mechanism:

(e)

114 2 4 5 3 2 0.917

12 p p p p

= M M M M λ θ λ θ θ θ λ ⋅ + ⋅ ⋅ + ⋅ ⋅ ⇒ = =

- Let now combine the mechanisms (12.a), (12.b) (12.i), and (12.d) – clockwise joint

rotation (Fig. 12.l, comparing with Fig. 12.k – fixed joint, and Fig. 12.m – counterclockwise joint

rotation). Thus, on the mechanism shown in Fig. 12.j, the following equilibrium equations could

be written:

M p

M p

M p

M p M p

M p

4M p

4M p 4M p

M p4λ

4λ2M p

M p

M p

M p

M p

M p

4M p 4M p

M p

i)

j)

λ

λ

k) La = 5M p l) La = 7M p > 5Mp m) La = 12M p > 5M p

4M p

M p

2M p

2M p

4M p

M p

M p

4M p

M p

4M p

Fig. 13 (cont.)


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a) ( ) ( ) ( ) ( )b c b c articulatii inchise articulatii deschise F F a a a a L L L L L L+ = + − +

(j)

164 4 8 13 6 ( 4 ) 2 0.444

36 p p p p p p p

= M M M M M M M λ θ λ θ θ θ θ θ θ λ ⋅ + ⋅ ⋅ + ⋅ − ⋅ + ⋅ + ⋅ ⇒ = =

b) or directly on the combined mechanism:

(j)

164 4 8 6 2 2 4 0.444

36 p p p p p

= M M M M M λ θ λ θ θ θ θ λ ⋅ + ⋅ ⋅ + ⋅ + ⋅ ⋅ ⇒ = =

Consequently, the collapse load factor is:

min (b)= = 0.406

u im pλ λ λ =

corresponding to a partial collapse mechanism (beam mechanism) in a second bay of the frame.

12. Biographical method

In what follows, a computer program is presented for evaluating the collapse mechanism of

plane frames applying the biographical method, or, in other words, finding the order of the plastic

hinge formation. The frames must be loaded only with concentrated forces applied strictly in the

joints of the frames, and not on the beams. It means that a beam which is loaded with a concentrated

force somewhere along its span must be divided into separate beams having joint connections where

the forces are located. If a uniform distributed load is applied on to a member, it must be replaced by

its resultant load.

First, an elastic analysis will be done for

the given external loads Poi. The load factor of a

cross section is defined by the ratio between the

plastic moment and the corresponding elastic

one produced by the load combination Poi. This

load factor is calculated for all possible critical

cross-sections, thus for all beam ends. (The

bending moment diagrams have linear variations

on all the beams because the frame is loaded

only with concentrated forces applied in the

joints.) The smallest load factor will identify thefirst plastic hinge – the first cross-section which

will be fully plasticized.

Then, different parameters will be

computed (displacements, reactive forces etc.) corresponding to this first plastic hinge formation,

multiplying the load factor with all the values already founded from the load combination Poi.

In that cross-section a hinge is now introduced, and the stiffness matrix of the structure is re-

assembled. This plastic hinge is considered loaded with its corresponding plastic bending moment,

and the modified structure will be used for the next step, to find the second plastic hinge, applying the

same linear structural analysis.

When increasing the magnitudes of the external loads applied on the structure, the variation ofthe plastic bending moment is null, but a free cross-section rotation being allowed). All the other

beams of the frame have an elastic structural behaviour.

(1) (2)

NO

YES

321

321

321

(1) (2)

(1) (2)


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INPUT DATA- joints coordinates

- beam mechanical properties

- boundary conditions

Assembling

the stiffness matrix

LOADS INPUT

HINGE=HINGE + 1

Inverting the N-th stiffness

matrix of the auxiliarystructure

IMPOSSIBLE STRUCTURALCOLLAPSE!

POSSIBLE

RESULTS

IN THE N-th AUXILIARY STRUCTURE

- joint displacements

- beam internal efforts

- hinges rotations

THE PLASTIC HINGE ISCLOSING?

YES

NO

THE NEXT PLASTIC HINGE IS FORMED

- finding the new plastic hinge- internal efforts in the auxiliary structure

- load factor

- joint displacements- beam internal efforts

- plastic hinge rotations

Modifying the stiffness matrix of the N-thauxiliary structure and updating and reassembling

it for the (N+1)-th structure


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The load factors between the plastic hinges formations are obtained by multiplying the load

factor already founded in the earlier step in the attached (auxiliary) structure, subjected to the load

combination Po

i , with the next load factor. This step-by-step technique will continue till the plastic

hinges will form a collapse mechanism, meaning that on the main diagonal of the assembled structural

stiffness matrix a null or negative coefficient will be identified (a singular matrix could not beinverted).

Closing the plastic hinges

During a proportional load combination, a plastic hinge could be unloaded. The plastic

bending moment of the corresponding cross section could decrease, having a linear elastic behaviour,

but with a remanent (residual) plastic rotation. Such situations are quite frequently possible because

the new plastic hinges developed into the structure modify the internal efforts distribution decreasing

their magnitudes, or having opposite senses

in the auxiliary (attached) structure, but only

if the structure has a linear elastic behaviour between two consecutive steps. When this

case is happening in the auxiliary structure

(the k plastic hinge formation) the hinge

rotation is in the opposite sense of the (full)

hinge rotation. Thus, the auxiliary structure

must be modified (being unacceptable) by

replacing it with another structure with a

removed plastic hinge in the corresponding

cross section, and the stiffness matrix of the

structure will be updated and reassembled.

Fig. 13 – Closing the plastic hinge

Mp

M

1/Ro

plastic hinge

formation

closing

the plastic hinge

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