Mecânica Geral Momento de Inércia de áreasmecanicageral2015.weebly.com/uploads/3/7/5/7/37571293/aula04-mec... · Em relação ao eixo z, ou pólo, o momento de inércia é chamado

Mecânica Geral

Momento de Inércia de áreasMomento de uma força

Prof. Ettore [email protected]

mailto:[email protected]

• Nas aulas anteriores aprendemos:

• A calcular os Centros de Massa

• de um sistema com poucas partículas

• de sistemas geométricos simples

• de sistemas onde a massa é continuamente distribuída

• Calcular o Momento de Inércia

• de um sistema com poucas partículas

• de sistemas geométricos simples

• de sistemas onde a massa é continuamente distribuída

• Na aula de hoje continuaremos com discussão sobre momentos de inércia desta vez voltando nossa atenção para o cálculo destas grandezas para áreas.

• Motivação:

• Sempre que uma carga distribuída, que varia linearmente, atuar perpendicularmente a uma área, o momento de inércia desta área será importante no cálculo do momento de distribuição de carga em relação a um dado eixo.

Momentos de Inércia de áreas512 CH A P T E R 10 MO M E N T S O F IN E RT I A

10

Moment of Inertia. By definition, the moments of inertia of adifferential area dA about the x and y axes are and

respectively, Fig. 10–2. For the entire area A the momentsof inertia are determined by integration; i.e.,

(10–1)

We can also formulate this quantity for dA about the “pole” O orz axis, Fig. 10–2. This is referred to as the polar moment of inertia. It isdefined as where r is the perpendicular distance from thepole (z axis) to the element dA. For the entire area the polar moment ofinertia is

(10–2)

This relation between and is possible since Fig. 10–2.

From the above formulations it is seen that and will alwaysbe positive since they involve the product of distance squared and area.Furthermore, the units for moment of inertia involve length raised to thefourth power, e.g., or

10.2 Parallel-Axis Theorem for an Area

The parallel-axis theorem can be used to find the moment of inertia of anarea about any axis that is parallel to an axis passing through the centroidand about which the moment of inertia is known.To develop this theorem,we will consider finding the moment of inertia of the shaded area shownin Fig. 10–3 about the x axis.To start, we choose a differential element dAlocated at an arbitrary distance from the centroidal axis. If thedistance between the parallel x and axes is then the moment ofinertia of dA about the x axis is . For the entire area,

= LAy¿2 dA + 2dyLA

y¿ dA + dy2LA

dA

Ix = LA1y¿ + dy22 dA

dIx = 1y¿ + dy22 dAdy,x¿

x¿y¿

in4.ft4,mm4,m4,

JOIy,Ix,

r2 = x2 + y2,IyIx,JO

JO = LAr2 dA = Ix + Iy

dJO = r2 dA,

Ix = LAy2 dA

Iy = LAx2 dA

dIy = x2 dA,dIx = y2 dA

Ox

y

y

x

r

dA

A

Fig. 10–2

Ox

y

d

dx

dy

x¿

y!

x¿

y¿dA

C

Fig. 10–3

Por definição os momentos de inércia de uma área em relação aos eixos x e y são dados, respectivamente por

Ix

=

Zy2dA

Iy =

Zx

2dA

Em relação ao eixo z, ou pólo, o momento de inércia é chamado de Momento de Inércia Polar, sendo calculado através de:

J0 =

Zr2dA

como r

2 = x

2 + y

2

J0 = Ix

+ Iy

Note que r é a distância perpendicular do polo (eixo z) até o elemento dA.

512 CH A P T E R 10 MO M E N T S O F IN E RT I A

10



(10–1)


(10–2)





= LAy¿2 dA + 2dyLA

y¿ dA + dy2LA

dA



x¿y¿

in4.ft4,mm4,m4,

JOIy,Ix,



dJO = r2 dA,

Ix = LAy2 dA

Iy = LAx2 dA


Ox

y

y

x

r

dA

A

Fig. 10–2

Ox

y

d

dx

dy

x¿

y!

x¿

y¿dA

C

Fig. 10–3

Teorema dos eixos paralelos para uma área

512 CH A P T E R 10 MO M E N T S O F IN E RT I A

10



(10–1)


(10–2)





= LAy¿2 dA + 2dyLA

y¿ dA + dy2LA

dA



x¿y¿

in4.ft4,mm4,m4,

JOIy,Ix,



dJO = r2 dA,

Ix = LAy2 dA

Iy = LAx2 dA


Ox

y

y

x

r

dA

A

Fig. 10–2

Ox

y

d

dx

dy

x¿

y!

x¿

y¿dA

C

Fig. 10–3

Conforme vimos, o teorema de Steiner pode ser usado para determinar o momento de inércia de uma área em relação a qualquer eixo que seja paralelo a um eixo passando pelo CM

Ix

=

Z

A

(y0 + dy

)2dA

Ix

=

Z

A

y02dA+ 2dy

Z

A

y0dA+ d2y

Z

A

dA

Ix

= Ix

0 +Ad2y

Iy

= Iy

0 +Ad2x

Analogamente,

E também

J0 = JC +Ad2

Concluindo: O momento de inércia para uma área ao redor de um eixo é igual ao seu momento de inércia em volta de um eixo paralelo que passa pelo CM desta área mais o produto da área pelo quadrado da distância perpendicular entre os eixos.

Exemplo 1: Determine o momento de inércia de uma área retangular em relação ao eixo x', em relação ao eixo xb e em relação ao polo z que é perpendicular ao plano x’y’e passa pelo centróide C.

10.3 RADIUS OF GYRATION OF AN AREA 515

10

EXAMPLE 10.1

Determine the moment of inertia for the rectangular area shown inFig. 10–5 with respect to (a) the centroidal axis, (b) the axis passing through the base of the rectangle, and (c) the pole or axisperpendicular to the plane and passing through the centroid C.

SOLUTION (CASE 1)

Part (a). The differential element shown in Fig. 10–5 is chosen forintegration. Because of its location and orientation, the entire elementis at a distance from the axis. Here it is necessary to integratefrom to Since then

Ans.

Part (b). The moment of inertia about an axis passing through thebase of the rectangle can be obtained by using the above result of part(a) and applying the parallel-axis theorem, Eq. 10–3.

Ans.

Part (c). To obtain the polar moment of inertia about point C, wemust first obtain which may be found by interchanging thedimensions b and h in the result of part (a), i.e.,

Using Eq. 10–2, the polar moment of inertia about C is therefore

Ans.JC = Ix¿ + Iy¿ = 112

bh1h2 + b22

Iy¿ = 112

hb3

Iy¿,

= 112

bh3 + bhah2b2

= 13

bh3

Ixb= Ix¿ + Ady

2

Ix¿ = 112

bh3

Ix¿ = LAyœ2 dA = L

h>2-h>2yœ21b dy¿2 = bL

h>2-h>2yœ2 dyœ

dA = b dy¿,y¿ = h>2.y¿ = -h>2 x¿y¿

x¿–y¿z¿

xbx¿

x¿

y¿

y¿

xb

C

dy¿

b2

b2

h2

h2

Fig. 10–5

Exemplo 2: Determine o momento de inércia em relação ao eixo x da área hachurada na figura abaixo.516 CH A P T E R 10 MO M E N T S O F IN E RT I A

10

x

y

200 mm

100 mm

y

xdy

y2 ! 400x

(a)

(100 – x)

x

y

200 mm

x

y

100 mm

dx

x¿

y2 ! 400x

(b)

y !~ y––2

Fig. 10–6

Determine the moment of inertia for the shaded area shown in Fig. 10–6a about the x axis.

SOLUTION I (CASE 1)A differential element of area that is parallel to the x axis, as shown inFig. 10–6a, is chosen for integration. Since this element has a thicknessdy and intersects the curve at the arbitrary point (x, y), its area is

Furthermore, the element lies at the samedistance y from the x axis. Hence, integrating with respect to y, from

to yieldsy = 200 mm,y = 0

dA = 1100 - x2 dy.

EXAMPLE 10.2

Ans.= 10711062 mm4

= L200 mm

0y2a100 -

y2

400b dy = L

200 mm

0a100y2 -

y4

400bdy

Ix = LAy2 dA = L

200 mm

0y21100 - x2 dy

SOLUTION II (CASE 2)A differential element parallel to the y axis, as shown in Fig. 10–6b, ischosen for integration. It intersects the curve at the arbitrary point (x, y).In this case, all points of the element do not lie at the same distancefrom the x axis, and therefore the parallel-axis theorem must be usedto determine the moment of inertia of the element with respect to thisaxis. For a rectangle having a base b and height h, the moment ofinertia about its centroidal axis has been determined in part (a) ofExample 10.1.There it was found that For the differentialelement shown in Fig. 10–6b, and and thus

Since the centroid of the element is from thex axis, the moment of inertia of the element about this axis is

(This result can also be concluded from part (b) of Example 10.1.)Integrating with respect to x, from to yields

Ans.= 10711062 mm4

Ix = LdIx = L100 mm

0

13

y3 dx = L100 mm

0

131400x23>2 dx

x = 100 mm,x = 0

dIx = dIx¿ + dA y'2 = 1

12dx y3 + y dxay

2b2

= 13

y3 dx

y' = y>2dIx¿ = 1

12 dx y3.h = y,b = dx

Ix¿ = 112 bh3.

Momento de inércia de áreas compostas

• Desde que os momentos de inércia de cada área sejam conhecidos ou possam ser determinados em relação a um eixo comum, o momento de inércia da área composta em relação a este eixo é igual à soma algébrica dos momentos de inércia de suas partes.

• Para o cálculo:

• Divida a área em suas partes compostas indicando a distância perpendicular a partir do CM de cada parte até o eixo de interesse.

• Use o Teorema dos Eixos Paralelos quando necessário.

• Momento de inércia de buracos devem ser subtraídos ao invés de somados.

10.4 MOMENTS OF INERTIA FOR COMPOSITE AREAS 523

10

x

100 mm

75 mm

75 mm

25 mm

(a)x

100 mm

75 mm

75 mm

25 mm

–

(b)

Fig. 10–8

EXAMPLE 10.4

Determine the moment of inertia of the area shown in Fig. 10–8aabout the x axis.

SOLUTION

Composite Parts. The area can be obtained by subtracting thecircle from the rectangle shown in Fig. 10–8b. The centroid of eacharea is located in the figure.

Parallel-Axis Theorem. The moments of inertia about the x axisare determined using the parallel-axis theorem and the data in thetable on the inside back cover.

Circle

Rectangle

Summation. The moment of inertia for the area is therefore

Ans.= 10111062 mm4

Ix = -11.411062 + 112.511062= 1

1211002115023 + 110021150217522 = 112.511062 mm4

Ix = Ixœ + Ady2

= 14p12524 + p1252217522 = 11.411062 mm4

Ix = Ixœ + Ady2

Exemplo 3: Determine o momento de inércia da área mostrada em relação ao eixo x

Momento de uma força (Torque)

• Quando uma força é aplicada sobre um corpo ela vai produzir uma tendência de rotação em um ponto que não está em sua linha de ação.

• Esta tendência de rotação é conhecida como torque ou momento de uma força.

• O módulo do momento da força é proporcional `a intensidade da força e à distância perpendicular `a aplicação da força, ou braço do momento.

Force SystemResultants

CHAPTER OBJECTIVES

• To discuss the concept of the moment of a force and show how tocalculate it in two and three dimensions.

• To provide a method for finding the moment of a force about aspecified axis.

• To define the moment of a couple.

• To present methods for determining the resultants of nonconcurrentforce systems.

• To indicate how to reduce a simple distributed loading to a resultantforce having a specified location.

4.1 Moment of a Force—Scalar Formulation

When a force is applied to a body it will produce a tendency for the bodyto rotate about a point that is not on the line of action of the force. Thistendency to rotate is sometimes called a torque, but most often it is calledthe moment of a force or simply the moment. For example, consider awrench used to unscrew the bolt in Fig. 4–1a. If a force is applied to thehandle of the wrench it will tend to turn the bolt about point O (or the zaxis). The magnitude of the moment is directly proportional to themagnitude of F and the perpendicular distance or moment arm d. Thelarger the force or the longer the moment arm, the greater the moment orturning effect. Note that if the force F is applied at an angle ,Fig. 4–1b, then it will be more difficult to turn the bolt since the momentarm will be smaller than d. If F is applied along the wrench,Fig. 4–1c, its moment arm will be zero since the line of action of F willintersect point O (the z axis).As a result, the moment of F about O is alsozero and no turning can occur.

d¿ = d sinu

u Z 90°

4z

O d

F

(a)z

O

F

d¿ ! d sin u

(b)

u

d

z

O

(c)

F

Fig. 4–1

117

Note que: Se o ângulo entre F e d for diferente de 90o, a força aplicada para rodar o parafuso será maior.

Se, por sua vez a força for aplicada junto ao eixo da chave o momento será zero e não haverá rotação.


CHAPTER OBJECTIVES








d¿ = d sinu

u Z 90°

4z

O d

F

(a)z

O

F

d¿ ! d sin u

(b)

u

d

z

O

(c)

F

Fig. 4–1

117


CHAPTER OBJECTIVES








d¿ = d sinu

u Z 90°

4z

O d

F

(a)z

O

F

d¿ ! d sin u

(b)

u

d

z

O

(c)

F

Fig. 4–1

117

Generalizando a discussão

118 CHAPTER 4 FORCE SYSTEM RESULTANTS

4

We can generalize the above discussion and consider the force F andpoint O which lie in the shaded plane as shown in Fig. 4–2a. The moment

about point O, or about an axis passing through O and perpendicularto the plane, is a vector quantity since it has a specified magnitude anddirection.

Magnitude. The magnitude of is

(4–1)

where d is the moment arm or perpendicular distance from the axis atpoint O to the line of action of the force. Units of moment magnitudeconsist of force times distance, e.g., or

Direction. The direction of is defined by its moment axis, whichis perpendicular to the plane that contains the force F and its momentarm d. The right-hand rule is used to establish the sense of direction of

. According to this rule, the natural curl of the fingers of the righthand, as they are drawn towards the palm, represent the tendency forrotation caused by the moment. As this action is performed, the thumbof the right hand will give the directional sense of , Fig. 4–2a. Noticethat the moment vector is represented three-dimensionally by a curlaround an arrow. In two dimensions this vector is represented only bythe curl as in Fig. 4–2b. Since in this case the moment will tend to cause acounterclockwise rotation, the moment vector is actually directed out ofthe page.

Resultant Moment. For two-dimensional problems, where all theforces lie within the x–y plane, Fig. 4–3, the resultant momentabout point O (the z axis) can be determined by finding the algebraic sumof the moments caused by all the forces in the system. As a convention,we will generally consider positive moments as counterclockwise sincethey are directed along the positive z axis (out of the page). Clockwisemoments will be negative. Doing this, the directional sense of eachmoment can be represented by a plus or minus sign. Using this signconvention, the resultant moment in Fig. 4–3 is therefore

a

If the numerical result of this sum is a positive scalar, will be acounterclockwise moment (out of the page); and if the result is negative,

will be a clockwise moment (into the page).(MR)O

(MR)O

+ (MR)O

= ©Fd; (MR)O

= F1d1 - F2d2 + F3d3

(MR)O

MO

MO

MO

lb # ft.N # m

MO = Fd

MO

MO

Sense of rotation

O

Moment axis

dF

MO

MO

F

d

O

(a)

(b)

Fig. 4–2

y

xO

F3

F2

F1

M3

M2 M1

d3

d2d1

Fig. 4–3


4




(4–1)





a



(MR)O

+ (MR)O

= ©Fd; (MR)O

= F1d1 - F2d2 + F3d3

(MR)O

MO

MO

MO

lb # ft.N # m

MO = Fd

MO

MO

Sense of rotation

O

Moment axis

dF

MO

MO

F

d

O

(a)

(b)

Fig. 4–2

y

xO

F3

F2

F1

M3

M2 M1

d3

d2d1

Fig. 4–3

M0 = Fd [N.m]

O Momento de uma força é uma grandeza vetorial

Direção: Regra da mão direita

Intensidade:

Momento Resultante


4




(4–1)





a



(MR)O

+ (MR)O

= ©Fd; (MR)O

= F1d1 - F2d2 + F3d3

(MR)O

MO

MO

MO

lb # ft.N # m

MO = Fd

MO

MO

Sense of rotation

O

Moment axis

dF

MO

MO

F

d

O

(a)

(b)

Fig. 4–2

y

xO

F3

F2

F1

M3

M2 M1

d3

d2d1

Fig. 4–3Em problemas bi-dimensionais, o momento resultante das forças que atuam no plano xy é simplesmente a soma algébrica dos vários momentos com a seguinte convenção: Quando as forças atuam em sentido anti-horário, o momento é positivo. Quando atuam no sentido horário o momento é negativo.

MR0 = F1d1 � F2d2 + F3d3

Exemplo 4: Para cada caso determine o momento da força em volta do ponto O.

4.1 MOMENT OF A FORCE—SCALAR FORMULATION 119

4

EXAMPLE 4.1

For each case illustrated in Fig. 4–4, determine the moment of theforce about point O.

SOLUTION (SCALAR ANALYSIS)The line of action of each force is extended as a dashed line in order toestablish the moment arm d.Also illustrated is the tendency of rotationof the member as caused by the force. Furthermore, the orbit of theforce about O is shown as a colored curl. Thus,

Fig. 4–4a b Ans.

Fig. 4–4b b Ans.

Fig. 4–4c b Ans.

Fig. 4–4d d Ans.

Fig. 4–4e d Ans.MO = 17 kN214 m - 1 m2 = 21.0 kN # mMO = 160 lb211 sin 45° ft2 = 42.4 lb # ftMO = 140 lb214 ft + 2 cos 30° ft2 = 229 lb # ftMO = 150 N210.75 m2 = 37.5 N # mMO = 1100 N212 m2 = 200 N # m

2 m

O

(a)

100 N

Fig. 4–4

2 m

O

(b)

50 N

0.75 m

(d)

O1 sin 45! ft

60 lb

3 ft

45!1 ft

2 m

O (e)

4 m

1 m7 kN

2 ft

(c)

O

4 ft2 cos 30! ft

40 lb30!


4

EXAMPLE 4.1



Fig. 4–4a b Ans.

Fig. 4–4b b Ans.

Fig. 4–4c b Ans.

Fig. 4–4d d Ans.


2 m

O

(a)

100 N

Fig. 4–4

2 m

O

(b)

50 N

0.75 m

(d)

O1 sin 45! ft

60 lb

3 ft

45!1 ft

2 m

O (e)

4 m

1 m7 kN

2 ft

(c)

O

4 ft2 cos 30! ft

40 lb30!


4

EXAMPLE 4.1



Fig. 4–4a b Ans.

Fig. 4–4b b Ans.

Fig. 4–4c b Ans.

Fig. 4–4d d Ans.


2 m

O

(a)

100 N

Fig. 4–4

2 m

O

(b)

50 N

0.75 m

(d)

O1 sin 45! ft

60 lb

3 ft

45!1 ft

2 m

O (e)

4 m

1 m7 kN

2 ft

(c)

O

4 ft2 cos 30! ft

40 lb30!


4

EXAMPLE 4.1



Fig. 4–4a b Ans.

Fig. 4–4b b Ans.

Fig. 4–4c b Ans.

Fig. 4–4d d Ans.


2 m

O

(a)

100 N

Fig. 4–4

2 m

O

(b)

50 N

0.75 m

(d)

O1 sin 45! ft

60 lb

3 ft

45!1 ft

2 m

O (e)

4 m

1 m7 kN

2 ft

(c)

O

4 ft2 cos 30! ft

40 lb30!

Exemplo 5: Determine o momento resultante em relação ao ponto O das forças atuando no bastão. Assuma que os momentos positivos atuam na direção k (sentido anti-horário).


4

EXAMPLE 4.2

50 N

40 N

20 N3 m

2 m 2 m

Ox

y

60 N

30!

Fig. 4–5

As illustrated by the example problems, the moment of aforce does not always cause a rotation. For example, the forceF tends to rotate the beam clockwise about its support at Awith a moment The actual rotation would occurif the support at B were removed.

MA = FdA.

O

FN

FH

The ability to remove the nail will require the momentof about point O to be larger than the moment ofthe force about O that is needed to pull the nail out.FN

FH

MA " FdA

dA

F

A B

Determine the resultant moment of the four forces acting on the rodshown in Fig. 4–5 about point O.

SOLUTIONAssuming that positive moments act in the direction, i.e.,counterclockwise, we have

a

b Ans.

For this calculation, note how the moment-arm distances for the 20-Nand 40-N forces are established from the extended (dashed) lines ofaction of each of these forces.

MRO= -334 N # m = 334 N # m

-40 N14 m + 3 cos 30° m2MRO= -50 N12 m2 + 60 N102 + 20 N13 sin 30° m2+MRO= ©Fd;

+k

O momento de uma força em relação a um dado ponto é igual à soma dos momentos das componentes da força em relação ao mesmo ponto.


4

Important Points

● The moment of a force creates the tendency of a body to turnabout an axis passing through a specific point O.

● Using the right-hand rule, the sense of rotation is indicated by thecurl of the fingers, and the thumb is directed along the momentaxis, or line of action of the moment.

● The magnitude of the moment is determined from where d is called the moment arm, which represents theperpendicular or shortest distance from point O to the line ofaction of the force.

● In three dimensions the vector cross product is used to determinethe moment, i.e., Remember that r is directed frompoint O to any point on the line of action of F.

● The principle of moments states that the moment of a forceabout a point is equal to the sum of the moments of the force’scomponents about the point. This is a very convenient method touse in two dimensions.

MO = r * F.

MO = Fd,

F2

O

r

F1F

Fig 4–16

MO

Fx

FFy

O

d

x

y

Fig. 4–17

FFyFy

FxFx

F

d

O

The moment of the applied force F aboutpoint O is easy to determine if we use theprinciple of moments. It is simply

.MO = Fxd

4.4 Principle of Moments

A concept often used in mechanics is the principle of moments, which issometimes referred to as Varignon’s theorem since it was originallydeveloped by the French mathematician Varignon (1654–1722). It statesthat the moment of a force about a point is equal to the sum of the momentsof the components of the force about the point.This theorem can be proveneasily using the vector cross product since the cross product obeys thedistributive law. For example, consider the moments of the force andtwo of its components about point O. Fig. 4–16. Since we have

For two-dimensional problems, Fig. 4–17, we can use the principle ofmoments by resolving the force into its rectangular components andthen determine the moment using a scalar analysis. Thus,

This method is generally easier than finding the same moment using.MO = Fd

MO = Fxy - Fyx

MO = r * F = r * 1F1 + F22 = r * F1 + r * F2

F = F1 + F2

F

~M0 = ~r ⇥ ~F = ~r ⇥ (~F1 + ~F2)

= ~r ⇥ ~F1 + ~r ⇥ ~F2

Princípios dos Momentos -Teorema de Varignon-

Princípios dos Momentos Problemas bidimensionais


4

Important Points

● The moment of a force creates the tendency of a body to turnabout an axis passing through a specific point O.

● Using the right-hand rule, the sense of rotation is indicated by thecurl of the fingers, and the thumb is directed along the momentaxis, or line of action of the moment.

● The magnitude of the moment is determined from where d is called the moment arm, which represents theperpendicular or shortest distance from point O to the line ofaction of the force.

● In three dimensions the vector cross product is used to determinethe moment, i.e., Remember that r is directed frompoint O to any point on the line of action of F.

● The principle of moments states that the moment of a forceabout a point is equal to the sum of the moments of the force’scomponents about the point. This is a very convenient method touse in two dimensions.

MO = r * F.

MO = Fd,

F2

O

r

F1F

Fig 4–16

MO

Fx

FFy

O

d

x

y

Fig. 4–17

FFyFy

FxFx

F

d

O

The moment of the applied force F aboutpoint O is easy to determine if we use theprinciple of moments. It is simply

.MO = Fxd

4.4 Principle of Moments

A concept often used in mechanics is the principle of moments, which issometimes referred to as Varignon’s theorem since it was originallydeveloped by the French mathematician Varignon (1654–1722). It statesthat the moment of a force about a point is equal to the sum of the momentsof the components of the force about the point.This theorem can be proveneasily using the vector cross product since the cross product obeys thedistributive law. For example, consider the moments of the force andtwo of its components about point O. Fig. 4–16. Since we have

For two-dimensional problems, Fig. 4–17, we can use the principle ofmoments by resolving the force into its rectangular components andthen determine the moment using a scalar analysis. Thus,

This method is generally easier than finding the same moment using.MO = Fd

MO = Fxy - Fyx

MO = r * F = r * 1F1 + F22 = r * F1 + r * F2

F = F1 + F2

F

Neste caso a decomposição da força resultante em suas componentes resolve o problema.

M0 = xF

y

� yF

x

Resumo até aqui...

• O momento de uma força cria a tendência de um corpo girar ao redor de um eixo que passa por um ponto específico O.

• A regra da mão direita diz que o sentido de rotação segue a curva dos dedos e o polegar fornece a direção do momento da força (torque).

• O módulo do momento da força (torque) é M0=F.d onde d é chamado de braço do momento e representa a distância perpendicular do ponto O até a linha de ação da força.

• Em 3D, o torque é dado por M0= r x F, e r está direcionado de O até qualquer ponto sobre a linha de ação de F.

Exemplo 1: Determine o momento da força em relação ao ponto O da figura.

4.4 PRINCIPLE OF MOMENTS 129

4

EXAMPLE 4.5

Determine the moment of the force in Fig. 4–18a about point O.

Fig. 4–18

x

y

(c)

45!

30!

30!3 m

O

Fx " (5 kN) sin 75!

Fy " (5 kN) sin 75!

SOLUTION IThe moment arm d in Fig. 4–18a can be found from trigonometry.

Thus,

b Ans.

Since the force tends to rotate or orbit clockwise about point O, themoment is directed into the page.

SOLUTION IIThe x and y components of the force are indicated in Fig. 4–18b.Considering counterclockwise moments as positive, and applying theprinciple of moments, we have

MO = Fd = (5kN)(2.898 m2 = 14.5 kN # m

d = (3 m) sin 75° = 2.898 m

a

b Ans.= -14.5 kN # m = 14.5 kN # m= -15 cos 45° kN213 sin 30° m2 - 15 sin 45° kN213 cos 30° m2+ MO = - Fxdy - Fydx

SOLUTION IIIThe x and y axes can be set parallel and perpendicular to the rod’s axisas shown in Fig. 4-18c. Here produces no moment about point Osince its line of action passes through this point. Therefore,

a

b Ans.= -14.5 kN # m = 14.5 kN # m= -(5 sin 75° kN)(3 m)

+ MO = -Fy dx

Fx

30!

(a)

45!

F " 5 kN3 m

O

d75!

y

x

(b)

30!

45!

O

dy " 3 sin 30! m

dx " 3 cos 30! mFx " (5 kN) cos 45!

Fy " (5 kN) sin 45!

4.4 PRINCIPLE OF MOMENTS 129

4

EXAMPLE 4.5

Determine the moment of the force in Fig. 4–18a about point O.

Fig. 4–18

x

y

(c)

45!

30!

30!3 m

O

Fx " (5 kN) sin 75!

Fy " (5 kN) sin 75!

SOLUTION IThe moment arm d in Fig. 4–18a can be found from trigonometry.

Thus,

b Ans.

Since the force tends to rotate or orbit clockwise about point O, themoment is directed into the page.

SOLUTION IIThe x and y components of the force are indicated in Fig. 4–18b.Considering counterclockwise moments as positive, and applying theprinciple of moments, we have

MO = Fd = (5kN)(2.898 m2 = 14.5 kN # m

d = (3 m) sin 75° = 2.898 m

a

b Ans.= -14.5 kN # m = 14.5 kN # m= -15 cos 45° kN213 sin 30° m2 - 15 sin 45° kN213 cos 30° m2+ MO = - Fxdy - Fydx

SOLUTION IIIThe x and y axes can be set parallel and perpendicular to the rod’s axisas shown in Fig. 4-18c. Here produces no moment about point Osince its line of action passes through this point. Therefore,

a

b Ans.= -14.5 kN # m = 14.5 kN # m= -(5 sin 75° kN)(3 m)

+ MO = -Fy dx

Fx

30!

(a)

45!

F " 5 kN3 m

O

d75!

y

x

(b)

30!

45!

O

dy " 3 sin 30! m

dx " 3 cos 30! mFx " (5 kN) cos 45!

Fy " (5 kN) sin 45!

Exemplo 2: A força F age na extremidade da cantoneira mostrada na figura abaixo. Determine o momento da força em relação ao ponto O.


4

EXAMPLE 4.6

Force F acts at the end of the angle bracket shown in Fig. 4–19a.Determine the moment of the force about point O.

SOLUTION I (SCALAR ANALYSIS)The force is resolved into its x and y components as shown inFig. 4–19b, then

a

b

or

Ans.

SOLUTION II (VECTOR ANALYSIS)Using a Cartesian vector approach, the force and position vectorsshown in Fig. 4–19c are

The moment is therefore

Ans.

NOTE: It is seen that the scalar analysis (Solution I) provides amore convenient method for analysis than Solution II since thedirection of the moment and the moment arm for each componentforce are easy to establish. Hence, this method is generallyrecommended for solving problems displayed in two dimensions,whereas a Cartesian vector analysis is generally recommended onlyfor solving three-dimensional problems.

= 5-98.6k6 N # m= 0i - 0j + [0.41-346.42 - 1-0.221200.02]k

MO = r * F = 3 i j k0.4 -0.2 0

200.0 -346.4 0

3= 5200.0i - 346.4j6 NF = 5400 sin 30°i - 400 cos 30°j6 Nr = 50.4i - 0.2j6 m

MO = 5-98.6k6 N # m

= -98.6 N # m = 98.6 N # m+MO = 400 sin 30° N10.2 m2 - 400 cos 30° N10.4 m2

0.4 m

0.2 m

30!

O

F = 400 N(a)

Fig. 4–19

0.4 m

0.2 m

(b)

x

400 cos 30! N

400 sin 30! N

O

y

y

x

0.4 m

0.2 m

30!

O

F(c)

r


4

EXAMPLE 4.6

Force F acts at the end of the angle bracket shown in Fig. 4–19a.Determine the moment of the force about point O.

SOLUTION I (SCALAR ANALYSIS)The force is resolved into its x and y components as shown inFig. 4–19b, then

a

b

or

Ans.

SOLUTION II (VECTOR ANALYSIS)Using a Cartesian vector approach, the force and position vectorsshown in Fig. 4–19c are

The moment is therefore

Ans.

NOTE: It is seen that the scalar analysis (Solution I) provides amore convenient method for analysis than Solution II since thedirection of the moment and the moment arm for each componentforce are easy to establish. Hence, this method is generallyrecommended for solving problems displayed in two dimensions,whereas a Cartesian vector analysis is generally recommended onlyfor solving three-dimensional problems.

= 5-98.6k6 N # m= 0i - 0j + [0.41-346.42 - 1-0.221200.02]k

MO = r * F = 3 i j k0.4 -0.2 0

200.0 -346.4 0

3= 5200.0i - 346.4j6 NF = 5400 sin 30°i - 400 cos 30°j6 Nr = 50.4i - 0.2j6 m

MO = 5-98.6k6 N # m

= -98.6 N # m = 98.6 N # m+MO = 400 sin 30° N10.2 m2 - 400 cos 30° N10.4 m2

0.4 m

0.2 m

30!

O

F = 400 N(a)

Fig. 4–19

0.4 m

0.2 m

(b)

x

400 cos 30! N

400 sin 30! N

O

y

y

x

0.4 m

0.2 m

30!

O

F(c)

r

Apêndice: Integração de funções constantes e funções polinomiais

• Fazer a operação de integração é a grosso modo efetuar uma soma no contínuo, ou seja, quando os elementos a serem somados são tantos e tão pequenos que não podemos contá-los

• Historicamente, o cálculo de integrais está relacionado com a obtenção de áreas

• Existem uma infinidade de funções integráveis, mas neste momento queremos só utilizar uma regra que diz respeito à integração de funções constantes e de funções que são polinômios.

Z b

aCx

n = C

x

n+1

(n+ 1)

�b

a

n 6= �1

Z b

aCdx = C

Z b

adx = Cx|ba = C(b� a)

Exercícios

Calcule as seguintes integrais

Z 1

0x

2dx

Z 2

1(x+ 2x3)dx

Z 3

12dx

Z 2

0(2x+ 4)dx

Documents

Mecânica Geral Momento de Inércia de áreasmecanicageral2015.weebly.com/uploads/3/7/5/7/37571293/aula04-mec... · Em relação ao eixo z, ou pólo, o momento de inércia é chamado