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Equilíbrio de um Corpo Rígido Prof. Ettore Baldini-Neto

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Page 1: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Equilíbrio de um Corpo Rígido

Prof. Ettore Baldini-Neto

Page 2: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Condições necessárias e suficientes para o equilíbrio de um corpo rígido.200 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

F1

M2

M1

F2

F3

F4

O

(a)

Fig. 5–1

R

W

2T

G

Fig. 5–2

FR ! 0

(MR)O ! 0

O

(b)

FR ! 0

(MR)O ! 0

O

A

r

(c)

Using the methods of the previous chapter, the force and couplemoment system acting on a body can be reduced to an equivalentresultant force and resultant couple moment at any arbitrary point O onor off the body, Fig. 5–1b. If this resultant force and couple moment areboth equal to zero, then the body is said to be in equilibrium.Mathematically, the equilibrium of a body is expressed as

(5–1)

The first of these equations states that the sum of the forces acting onthe body is equal to zero. The second equation states that the sum of themoments of all the forces in the system about point O, added to all thecouple moments, is equal to zero. These two equations are not onlynecessary for equilibrium, they are also sufficient. To show this, considersumming moments about some other point, such as point A in Fig. 5–1c.We require

Since , this equation is satisfied only if Eqs. 5–1 are satisfied,namely and .

When applying the equations of equilibrium, we will assume that thebody remains rigid. In reality, however, all bodies deform whensubjected to loads. Although this is the case, most engineering materialssuch as steel and concrete are very rigid and so their deformation isusually very small. Therefore, when applying the equations ofequilibrium, we can generally assume that the body will remain rigidand not deform under the applied load without introducing anysignificant error. This way the direction of the applied forces and theirmoment arms with respect to a fixed reference remain unchangedbefore and after the body is loaded.

EQUILIBRIUM IN TWO DIMENSIONS

In the first part of the chapter, we will consider the case where the forcesystem acting on a rigid body lies in or may be projected onto a singleplane and, furthermore, any couple moments acting on the body aredirected perpendicular to this plane.This type of force and couple systemis often referred to as a two-dimensional or coplanar force system. Forexample, the airplane in Fig. 5–2 has a plane of symmetry through itscenter axis, and so the loads acting on the airplane are symmetrical withrespect to this plane. Thus, each of the two wing tires will support thesame load T, which is represented on the side (two-dimensional) view ofthe plane as 2T.

(MR)O = 0FR = 0r Z 0

©MA = r * FR + (MR)O = 0

(MR)O = ©MO = 0

FR = ©F = 0

200 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

F1

M2

M1

F2

F3

F4

O

(a)

Fig. 5–1

R

W

2T

G

Fig. 5–2

FR ! 0

(MR)O ! 0

O

(b)

FR ! 0

(MR)O ! 0

O

A

r

(c)

Using the methods of the previous chapter, the force and couplemoment system acting on a body can be reduced to an equivalentresultant force and resultant couple moment at any arbitrary point O onor off the body, Fig. 5–1b. If this resultant force and couple moment areboth equal to zero, then the body is said to be in equilibrium.Mathematically, the equilibrium of a body is expressed as

(5–1)

The first of these equations states that the sum of the forces acting onthe body is equal to zero. The second equation states that the sum of themoments of all the forces in the system about point O, added to all thecouple moments, is equal to zero. These two equations are not onlynecessary for equilibrium, they are also sufficient. To show this, considersumming moments about some other point, such as point A in Fig. 5–1c.We require

Since , this equation is satisfied only if Eqs. 5–1 are satisfied,namely and .

When applying the equations of equilibrium, we will assume that thebody remains rigid. In reality, however, all bodies deform whensubjected to loads. Although this is the case, most engineering materialssuch as steel and concrete are very rigid and so their deformation isusually very small. Therefore, when applying the equations ofequilibrium, we can generally assume that the body will remain rigidand not deform under the applied load without introducing anysignificant error. This way the direction of the applied forces and theirmoment arms with respect to a fixed reference remain unchangedbefore and after the body is loaded.

EQUILIBRIUM IN TWO DIMENSIONS

In the first part of the chapter, we will consider the case where the forcesystem acting on a rigid body lies in or may be projected onto a singleplane and, furthermore, any couple moments acting on the body aredirected perpendicular to this plane.This type of force and couple systemis often referred to as a two-dimensional or coplanar force system. Forexample, the airplane in Fig. 5–2 has a plane of symmetry through itscenter axis, and so the loads acting on the airplane are symmetrical withrespect to this plane. Thus, each of the two wing tires will support thesame load T, which is represented on the side (two-dimensional) view ofthe plane as 2T.

(MR)O = 0FR = 0r Z 0

©MA = r * FR + (MR)O = 0

(MR)O = ©MO = 0

FR = ©F = 0

~FR =X

i

~Fi = 0

( ~MO)R =X

i

( ~Mi)O = 0

Quando aplicamos estas equações assumimos que o corpo permanece rígido. Embora hajam deformações, estas são consideradas pequenas e são desprezadas.

Page 3: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

200 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

F1

M2

M1

F2

F3

F4

O

(a)

Fig. 5–1

R

W

2T

G

Fig. 5–2

FR ! 0

(MR)O ! 0

O

(b)

FR ! 0

(MR)O ! 0

O

A

r

(c)

Using the methods of the previous chapter, the force and couplemoment system acting on a body can be reduced to an equivalentresultant force and resultant couple moment at any arbitrary point O onor off the body, Fig. 5–1b. If this resultant force and couple moment areboth equal to zero, then the body is said to be in equilibrium.Mathematically, the equilibrium of a body is expressed as

(5–1)

The first of these equations states that the sum of the forces acting onthe body is equal to zero. The second equation states that the sum of themoments of all the forces in the system about point O, added to all thecouple moments, is equal to zero. These two equations are not onlynecessary for equilibrium, they are also sufficient. To show this, considersumming moments about some other point, such as point A in Fig. 5–1c.We require

Since , this equation is satisfied only if Eqs. 5–1 are satisfied,namely and .

When applying the equations of equilibrium, we will assume that thebody remains rigid. In reality, however, all bodies deform whensubjected to loads. Although this is the case, most engineering materialssuch as steel and concrete are very rigid and so their deformation isusually very small. Therefore, when applying the equations ofequilibrium, we can generally assume that the body will remain rigidand not deform under the applied load without introducing anysignificant error. This way the direction of the applied forces and theirmoment arms with respect to a fixed reference remain unchangedbefore and after the body is loaded.

EQUILIBRIUM IN TWO DIMENSIONS

In the first part of the chapter, we will consider the case where the forcesystem acting on a rigid body lies in or may be projected onto a singleplane and, furthermore, any couple moments acting on the body aredirected perpendicular to this plane.This type of force and couple systemis often referred to as a two-dimensional or coplanar force system. Forexample, the airplane in Fig. 5–2 has a plane of symmetry through itscenter axis, and so the loads acting on the airplane are symmetrical withrespect to this plane. Thus, each of the two wing tires will support thesame load T, which is represented on the side (two-dimensional) view ofthe plane as 2T.

(MR)O = 0FR = 0r Z 0

©MA = r * FR + (MR)O = 0

(MR)O = ©MO = 0

FR = ©F = 0

Caso exista algum momento em relação a algum outro ponto, como o ponto A, devemos ter que:

X�MA = �r ⇥ �F + ( �MR)O = 0

Page 4: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Equilíbrio em duas dimensões

• O sistema de forças atua no mesmo plano.

• A força resultante encontra-se no mesmo plano das forças que a geram.

• Os momentos são perpendiculares ao plano onde as forças atuam.

Diagrama de corpo livre

• Todas as forças externas e momentos que atuam no corpo devem ser desenhados.

• As equações de equilíbrio devem ser aplicadas.

• O completo entendimento de como se desenhar o diagrama de corpo livre é fundamental na resolução dos problemas.

Page 5: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Suportes, pontos de contato, etc

Consideramos os vários tipos de reações que ocorrem em suportes e pontos de contato entre corpos rígidos quando temos um sistema de forças coplanares atuando sobre eles.

Regra Geral:

a) Se um suporte evita que ocorra translação de um corpo em uma dada direção, uma força aparece no corpo nesta direção.

b) Se uma rotação é evitada, um momento de um binário aparece no corpo.

Page 6: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

5.2 FREE-BODY DIAGRAMS 201

5.2 Free-Body Diagrams

Successful application of the equations of equilibrium requires a completespecification of all the known and unknown external forces that act onthe body. The best way to account for these forces is to draw a free-bodydiagram.This diagram is a sketch of the outlined shape of the body, whichrepresents it as being isolated or “free” from its surroundings, i.e., a “freebody.” On this sketch it is necessary to show all the forces and couplemoments that the surroundings exert on the body so that these effects canbe accounted for when the equations of equilibrium are applied. Athorough understanding of how to draw a free-body diagram is of primaryimportance for solving problems in mechanics.

Support Reactions. Before presenting a formal procedure as tohow to draw a free-body diagram, we will first consider the various typesof reactions that occur at supports and points of contact between bodiessubjected to coplanar force systems. As a general rule,

• If a support prevents the translation of a body in a given direction,then a force is developed on the body in that direction.

• If rotation is prevented, a couple moment is exerted on the body.

For example, let us consider three ways in which a horizontal member,such as a beam, is supported at its end. One method consists of a roller orcylinder, Fig. 5–3a. Since this support only prevents the beam fromtranslating in the vertical direction, the roller will only exert a force onthe beam in this direction, Fig. 5–3b.

The beam can be supported in a more restrictive manner by using a pin,Fig. 5–3c. The pin passes through a hole in the beam and two leaves whichare fixed to the ground. Here the pin can prevent translation of the beamin any direction Fig. 5–3d, and so the pin must exert a force F on thebeam in this direction. For purposes of analysis, it is generally easier torepresent this resultant force F by its two rectangular components and

Fig. 5–3e. If and are known, then F and can be calculated.The most restrictive way to support the beam would be to use a fixed

support as shown in Fig. 5–3f. This support will prevent both translationand rotation of the beam. To do this a force and couple moment must bedeveloped on the beam at its point of connection, Fig. 5–3g. As in thecase of the pin, the force is usually represented by its rectangularcomponents and

Table 5–1 lists other common types of supports for bodies subjected tocoplanar force systems. (In all cases the angle is assumed to be known.)Carefully study each of the symbols used to represent these supports andthe types of reactions they exert on their contacting members.

u

Fy.Fx

fFyFxFy,Fx

f,

5

(a)roller

(b)F

(c)

pin

or

Fy

Fx

F

(e)(d)

f

(f)

fixed support

Fy

Fx

M

(g)

Fig. 5–3

5.2 FREE-BODY DIAGRAMS 201

5.2 Free-Body Diagrams

Successful application of the equations of equilibrium requires a completespecification of all the known and unknown external forces that act onthe body. The best way to account for these forces is to draw a free-bodydiagram.This diagram is a sketch of the outlined shape of the body, whichrepresents it as being isolated or “free” from its surroundings, i.e., a “freebody.” On this sketch it is necessary to show all the forces and couplemoments that the surroundings exert on the body so that these effects canbe accounted for when the equations of equilibrium are applied. Athorough understanding of how to draw a free-body diagram is of primaryimportance for solving problems in mechanics.

Support Reactions. Before presenting a formal procedure as tohow to draw a free-body diagram, we will first consider the various typesof reactions that occur at supports and points of contact between bodiessubjected to coplanar force systems. As a general rule,

• If a support prevents the translation of a body in a given direction,then a force is developed on the body in that direction.

• If rotation is prevented, a couple moment is exerted on the body.

For example, let us consider three ways in which a horizontal member,such as a beam, is supported at its end. One method consists of a roller orcylinder, Fig. 5–3a. Since this support only prevents the beam fromtranslating in the vertical direction, the roller will only exert a force onthe beam in this direction, Fig. 5–3b.

The beam can be supported in a more restrictive manner by using a pin,Fig. 5–3c. The pin passes through a hole in the beam and two leaves whichare fixed to the ground. Here the pin can prevent translation of the beamin any direction Fig. 5–3d, and so the pin must exert a force F on thebeam in this direction. For purposes of analysis, it is generally easier torepresent this resultant force F by its two rectangular components and

Fig. 5–3e. If and are known, then F and can be calculated.The most restrictive way to support the beam would be to use a fixed

support as shown in Fig. 5–3f. This support will prevent both translationand rotation of the beam. To do this a force and couple moment must bedeveloped on the beam at its point of connection, Fig. 5–3g. As in thecase of the pin, the force is usually represented by its rectangularcomponents and

Table 5–1 lists other common types of supports for bodies subjected tocoplanar force systems. (In all cases the angle is assumed to be known.)Carefully study each of the symbols used to represent these supports andthe types of reactions they exert on their contacting members.

u

Fy.Fx

fFyFxFy,Fx

f,

5

(a)roller

(b)F

(c)

pin

or

Fy

Fx

F

(e)(d)

f

(f)

fixed support

Fy

Fx

M

(g)

Fig. 5–3

(rolete)

(pino)

O rolete impede que a viga translade na vertical.

O pino impede que a viga translade em qualquer direção.

Page 7: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

5.2 FREE-BODY DIAGRAMS 201

5.2 Free-Body Diagrams

Successful application of the equations of equilibrium requires a completespecification of all the known and unknown external forces that act onthe body. The best way to account for these forces is to draw a free-bodydiagram.This diagram is a sketch of the outlined shape of the body, whichrepresents it as being isolated or “free” from its surroundings, i.e., a “freebody.” On this sketch it is necessary to show all the forces and couplemoments that the surroundings exert on the body so that these effects canbe accounted for when the equations of equilibrium are applied. Athorough understanding of how to draw a free-body diagram is of primaryimportance for solving problems in mechanics.

Support Reactions. Before presenting a formal procedure as tohow to draw a free-body diagram, we will first consider the various typesof reactions that occur at supports and points of contact between bodiessubjected to coplanar force systems. As a general rule,

• If a support prevents the translation of a body in a given direction,then a force is developed on the body in that direction.

• If rotation is prevented, a couple moment is exerted on the body.

For example, let us consider three ways in which a horizontal member,such as a beam, is supported at its end. One method consists of a roller orcylinder, Fig. 5–3a. Since this support only prevents the beam fromtranslating in the vertical direction, the roller will only exert a force onthe beam in this direction, Fig. 5–3b.

The beam can be supported in a more restrictive manner by using a pin,Fig. 5–3c. The pin passes through a hole in the beam and two leaves whichare fixed to the ground. Here the pin can prevent translation of the beamin any direction Fig. 5–3d, and so the pin must exert a force F on thebeam in this direction. For purposes of analysis, it is generally easier torepresent this resultant force F by its two rectangular components and

Fig. 5–3e. If and are known, then F and can be calculated.The most restrictive way to support the beam would be to use a fixed

support as shown in Fig. 5–3f. This support will prevent both translationand rotation of the beam. To do this a force and couple moment must bedeveloped on the beam at its point of connection, Fig. 5–3g. As in thecase of the pin, the force is usually represented by its rectangularcomponents and

Table 5–1 lists other common types of supports for bodies subjected tocoplanar force systems. (In all cases the angle is assumed to be known.)Carefully study each of the symbols used to represent these supports andthe types of reactions they exert on their contacting members.

u

Fy.Fx

fFyFxFy,Fx

f,

5

(a)roller

(b)F

(c)

pin

or

Fy

Fx

F

(e)(d)

f

(f)

fixed support

Fy

Fx

M

(g)

Fig. 5–3Restringe totalmente a translação e rotação. Por isso, aparecem uma força e um momento de binário no ponto de conexão.

Page 8: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Tipos de apoio, reação e número de incógnitas

Page 9: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)
Page 10: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Receita para análise

• Esboçe o sistema físico isolado de suas restrições e conexões

• Identifique todas as forças externas e momentos dos binários conhecidos e desconhecidos.

• As forças são decorrentes de: carregamentos, reações que ocorrem em apoios ou pontos de contato com outros corpos e o peso do corpo.

• Identifique cada carga e as dimensões dadas

Page 11: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Importante• Nenhum problema deve ser resolvido sem antes fazer o desenho do

DCL.

• Se um apoio impede a translação de um corpo em uma dada direção, então o apoio exerce uma força sobre este corpo nessa direção.

• Se a rotação é impedida, então o apoio exerce um momento de binário sobre o corpo.

• Forças internas são desprezadas

• O peso de um corpo é uma força externa que atua sobre o centro de gravidade deste corpo.

• Momentos de binário podem ser colocados em qualquer lugar do DCL pois são vetores livres. As forças podem agir em qualquer ponto em sua linha de ação pois são vetores deslizantes.

Page 12: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Exemplo 1: Desenhe o diagrama de corpo livre da estrutura abaixo.

5.2 FREE-BODY DIAGRAMS 207

5

EXAMPLE 5.1

Draw the free-body diagram of the uniform beam shown in Fig. 5–7a.The beam has a mass of 100 kg.

SOLUTION

The free-body diagram of the beam is shown in Fig. 5–7b. Since thesupport at A is fixed, the wall exerts three reactions on the beam,denoted as and . The magnitudes of these reactions areunknown, and their sense has been assumed. The weight of the beam,

acts through the beam’s center of gravityG, which is 3 m from A since the beam is uniform.W = 10019.812 N = 981 N,

MAAy,Ax,

(a)

2 m 1200 N

6 m

A

Ay

Ax

2 m1200 N

3 m

A

981 N

MA

G

Effect of appliedforce acting on beam

Effect of gravity (weight)acting on beam

Effect of fixedsupport actingon beam

(b)

y

x

Fig. 5–7

Solução:

5.2 FREE-BODY DIAGRAMS 207

5

EXAMPLE 5.1

Draw the free-body diagram of the uniform beam shown in Fig. 5–7a.The beam has a mass of 100 kg.

SOLUTION

The free-body diagram of the beam is shown in Fig. 5–7b. Since thesupport at A is fixed, the wall exerts three reactions on the beam,denoted as and . The magnitudes of these reactions areunknown, and their sense has been assumed. The weight of the beam,

acts through the beam’s center of gravityG, which is 3 m from A since the beam is uniform.W = 10019.812 N = 981 N,

MAAy,Ax,

(a)

2 m 1200 N

6 m

A

Ay

Ax

2 m1200 N

3 m

A

981 N

MA

G

Effect of appliedforce acting on beam

Effect of gravity (weight)acting on beam

Effect of fixedsupport actingon beam

(b)

y

x

Fig. 5–7

Page 13: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Exemplo 2: Desenhe o diagrama de corpo livre do pedal mostrado na figura (b). O operador aplica uma força vertical no pedal de modo que a mola é estendida de 1.5 in e a força aplicada no elo curto em B vale 20lb.

208 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

Draw the free-body diagram of the foot lever shown in Fig. 5–8a. Theoperator applies a vertical force to the pedal so that the spring isstretched 1.5 in. and the force in the short link at B is 20 lb.

EXAMPLE 5.2

A

B

(a)

F

5 in.

1.5 in.

1 in.

A

B

k ! 20 lb/in.

(b)

F

30 lb

5 in.

1.5 in.

1 in.

A

B

20 lb

Ay

Ax

(c)

Fig. 5–8

SOLUTIONBy inspection of the photo the lever is loosely bolted to the frameat A. The rod at B is pinned at its ends and acts as a “short link.”After making the proper measurements, the idealized model of thelever is shown in Fig. 5–8b. From this, the free-body diagram isshown in Fig. 5–8c. The pin support at A exerts force components

and on the lever. The link at B exerts a force of 20 lb, actingin the direction of the link. In addition the spring also exerts ahorizontal force on the lever. If the stiffness is measured and foundto be then since the stretch using Eq. 3–2,

Finally, the operator’s shoeapplies a vertical force of F on the pedal. The dimensions of thelever are also shown on the free-body diagram, since thisinformation will be useful when computing the moments of theforces. As usual, the senses of the unknown forces at A have beenassumed. The correct senses will become apparent after solving theequilibrium equations.

Fs = ks = 20 lb>in. 11.5 in.2 = 30 lb.s = 1.5 in.,k = 20 lb>in.,

AyAx

Solução:

208 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

Draw the free-body diagram of the foot lever shown in Fig. 5–8a. Theoperator applies a vertical force to the pedal so that the spring isstretched 1.5 in. and the force in the short link at B is 20 lb.

EXAMPLE 5.2

A

B

(a)

F

5 in.

1.5 in.

1 in.

A

B

k ! 20 lb/in.

(b)

F

30 lb

5 in.

1.5 in.

1 in.

A

B

20 lb

Ay

Ax

(c)

Fig. 5–8

SOLUTIONBy inspection of the photo the lever is loosely bolted to the frameat A. The rod at B is pinned at its ends and acts as a “short link.”After making the proper measurements, the idealized model of thelever is shown in Fig. 5–8b. From this, the free-body diagram isshown in Fig. 5–8c. The pin support at A exerts force components

and on the lever. The link at B exerts a force of 20 lb, actingin the direction of the link. In addition the spring also exerts ahorizontal force on the lever. If the stiffness is measured and foundto be then since the stretch using Eq. 3–2,

Finally, the operator’s shoeapplies a vertical force of F on the pedal. The dimensions of thelever are also shown on the free-body diagram, since thisinformation will be useful when computing the moments of theforces. As usual, the senses of the unknown forces at A have beenassumed. The correct senses will become apparent after solving theequilibrium equations.

Fs = ks = 20 lb>in. 11.5 in.2 = 30 lb.s = 1.5 in.,k = 20 lb>in.,

AyAx

Como k=20 lb/in, então com s=1,5in, então a força na mola é dada por F=ks=20.1,5=30lb

Page 14: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Equações de Equilíbrio

No plano, quando um corpo está sujeito a um sistema de forças e torques temos que:

X

x

Fx

= 0

X

y

Fy

= 0

XM0 = 0

214 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

5.3 Equations of Equilibrium

In Sec. 5.1 we developed the two equations which are both necessary andsufficient for the equilibrium of a rigid body, namely, and

When the body is subjected to a system of forces, which all liein the x–y plane, then the forces can be resolved into their x and ycomponents. Consequently, the conditions for equilibrium in twodimensions are

(5–2)

Here and represent, respectively, the algebraic sums of the xand y components of all the forces acting on the body, and represents the algebraic sum of the couple moments and the moments ofall the force components about the z axis, which is perpendicular to thex–y plane and passes through the arbitrary point O.

Alternative Sets of Equilibrium Equations. AlthoughEqs. 5–2 are most often used for solving coplanar equilibrium problems,two alternative sets of three independent equilibrium equations may alsobe used. One such set is

(5–3)

When using these equations it is required that a line passing throughpoints A and B is not parallel to the y axis.To prove that Eqs. 5–3 providethe conditions for equilibrium, consider the free-body diagram of theplate shown in Fig. 5–11a. Using the methods of Sec. 4.8, all the forces onthe free-body diagram may be replaced by an equivalent resultant force

acting at point A, and a resultant couple momentFig. 5–11b. If is satisfied, it is necessary that

Furthermore, in order that satisfy it must have nocomponent along the x axis, and therefore must be parallel to the yaxis, Fig. 5–11c. Finally, if it is required that where B does notlie on the line of action of then Since Eqs. 5–3 show that bothof these resultants are zero, indeed the body in Fig. 5–11a must be inequilibrium.

FR = 0.FR,©MB = 0,FR

©Fx = 0,FRMRA= 0.

©MA = 0MRA= ©MA,

FR = ©F,

©Fx = 0©MA = 0©MB = 0

©MO

©Fy©Fx

©Fx = 0©Fy = 0©MO = 0

©MO = 0.©F = 0

B

A

C

(a)

F4

F3

F1

F2

x

y

A

MRA

FR

(b)

B C

x

y

(c)

A

FR

B C

x

y

Fig. 5–11

Note que o último termo representa a soma algébrica dos momentos de binário e das componentes das forças em volta do eixo z que é perpendicular ao plano e que passa por um ponto qualquer O no plano.

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Exemplo 2: Determine as componentes vertical e horizontal da reação sobre a viga causada pelo pino em B e o pelo apoio oscilante em A. Despreze o peso da viga.

216 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

(a)

600 N

D

100 N

A B

200 N

2 m 3 m 2 m

0.2 m

By

2 m

600 sin 45! N

3 m 2 m

AB

200 N

600 cos 45! N

Ay

Bxx

y

(b)100 N

0.2 m

D

Fig. 5–12

Determine the horizontal and vertical components of reaction on thebeam caused by the pin at B and the rocker at as shown in Fig. 5–12a.Neglect the weight of the beam.

A

EXAMPLE 5.5

SOLUTIONFree-Body Diagram. Identify each of the forces shown on the free-body diagram of the beam, Fig. 5–12b. (See Example 5.1.) Forsimplicity, the 600-N force is represented by its x and y components asshown in Fig. 5–12b.

Equations of Equilibrium. Summing forces in the x direction yields

Ans.

A direct solution for can be obtained by applying the momentequation about point B.

a

Ans.

Summing forces in the y direction, using this result, gives

Ans.

NOTE: We can check this result by summing moments about point A.

a

Ans.By = 405 N

-1100 N215 m2 - 1200 N217 m2 + By17 m2 = 0

-1600 sin 45° N212 m2 - 1600 cos 45° N210.2 m2+©MA = 0;

By = 405 N

319 N - 600 sin 45° N - 100 N - 200 N + By = 0+ c©Fy = 0;

Ay = 319 N

- 1600 cos 45° N210.2 m2 - Ay17 m2 = 0

100 N12 m2 + 1600 sin 45° N215 m2+©MB = 0;

©MB = 0Ay

Bx = 424 N

600 cos 45° N - Bx = 0:+ ©Fx = 0;

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Exemplo 3: O braço mostrado abaixo está conectado por um pino em A e apoia-se em um suporte liso em B. Determine as componentes horizontal e vertical da reação ao ponto A.

218 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

EXAMPLE 5.7

0.75 m

30!

1 m0.5 m

60 N

90 N " m

A

B

(a)NB

30!

0.75 m1 m

60 N

A

Ax

Ay

30!

(b)

x

y90 N " m

Fig. 5–14

SOLUTION

Free-Body Diagram. As shown in Fig. 5–14b, the reaction isperpendicular to the member at B. Also, horizontal and verticalcomponents of reaction are represented at A.

Equations of Equilibrium. Summing moments about A, we obtain adirect solution for

a

Using this result,

Ans.

Ans.Ay = 233 N

Ay - 200 cos 30° N - 60 N = 0+ c©Fy = 0;

Ax = 100 N

Ax - 200 sin 30° N = 0:+ ©Fx = 0;

NB = 200 N

-90 N # m - 60 N11 m2 + NB10.75 m2 = 0+©MA = 0;

NB,

NB

The member shown in Fig. 5–14a is pin-connected at A and restsagainst a smooth support at B. Determine the horizontal and verticalcomponents of reaction at the pin A.

218 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

EXAMPLE 5.7

0.75 m

30!

1 m0.5 m

60 N

90 N " m

A

B

(a)NB

30!

0.75 m1 m

60 N

A

Ax

Ay

30!

(b)

x

y90 N " m

Fig. 5–14

SOLUTION

Free-Body Diagram. As shown in Fig. 5–14b, the reaction isperpendicular to the member at B. Also, horizontal and verticalcomponents of reaction are represented at A.

Equations of Equilibrium. Summing moments about A, we obtain adirect solution for

a

Using this result,

Ans.

Ans.Ay = 233 N

Ay - 200 cos 30° N - 60 N = 0+ c©Fy = 0;

Ax = 100 N

Ax - 200 sin 30° N = 0:+ ©Fx = 0;

NB = 200 N

-90 N # m - 60 N11 m2 + NB10.75 m2 = 0+©MA = 0;

NB,

NB

The member shown in Fig. 5–14a is pin-connected at A and restsagainst a smooth support at B. Determine the horizontal and verticalcomponents of reaction at the pin A.

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220 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

Determine the horizontal and vertical components of reaction on themember at the pin A, and the normal reaction at the roller B inFig. 5–16a.

SOLUTION

Free-Body Diagram. The free-body diagram is shown in Fig. 5–16b.The pin at A exerts two components of reaction on the member,and .Ay

Ax

EXAMPLE 5.9

3 ft

A

B

3 ft

2 ft

(a)

30!

750 lb

A

B

2 ft

3 ft 3 ft

750 lb

Ax

Ay

NB30!

y

x

(b)

Fig. 5–16

Equations of Equilibrium. The reaction NB can be obtained directlyby summing moments about point A since and produce nomoment about A.

a

Ans.

Using this result,

Ans.

Ans.Ay = 286 lb

Ay + (536.2 lb) cos 30° - 750 lb = 0+ c©Fy = 0;

Ax = 268 lb

Ax - (536.2 lb ) sin 30° = 0:+ © Fx = 0;

NB = 536.2 lb = 536 lb

[NB cos 30°](6 ft) - [NB sin 30°](2 ft) - 750 lb(3 ft) = 0

+©MA = 0;

AyAx

Exemplo 4: Determine as componentes vertical e horizontal da reação sobre o braço no pino A e a reação normal no rolamento(rolete) B da figura abaixo.

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Exemplo 5: O bastão liso e uniforme da figura está sujeito a uma força e um momento de binário. Se o bastão é apoiado em A por uma parede lisa e em B e C, tanto em cima quanto embaixo, por roletes, determine as reações nestes suportes. Ignore o peso do bastão.

5.3 EQUATIONS OF EQUILIBRIUM 221

5

EXAMPLE 5.10

The uniform smooth rod shown in Fig. 5–17a is subjected to a forceand couple moment. If the rod is supported at A by a smooth wall andat B and C either at the top or bottom by rollers, determine thereactions at these supports. Neglect the weight of the rod.

(a)

A

2 m

300 N

4000 N ! m

4 m

2 m

C

B

30"2 m

(b)

2 m

300 N

4000 N ! m4 m

2 m30"

30"

Cy¿

By¿

30" 30"

Ax

y y¿

x

x¿

30"

SOLUTION

Free-Body Diagram. As shown in Fig. 5–17b, all the supportreactions act normal to the surfaces of contact since these surfaces aresmooth. The reactions at B and C are shown acting in the positive direction. This assumes that only the rollers located on the bottom ofthe rod are used for support.

Equations of Equilibrium. Using the x, y coordinate system inFig. 5–17b, we have

(1)

(2)

a

(3)

When writing the moment equation, it should be noted that the line ofaction of the force component 300 sin 30° N passes through point A,and therefore this force is not included in the moment equation.

Solving Eqs. 2 and 3 simultaneously, we obtain

Ans.

Ans.

Since is a negative scalar, the sense of is opposite to that shownon the free-body diagram in Fig. 5–17b. Therefore, the top roller at Bserves as the support rather than the bottom one. Retaining the negativesign for (Why?) and substituting the results into Eq. 1, we obtain

Ans.Ax = 173 N

1346.4 sin 30° N + (-1000.0 sin 30° N) - Ax = 0

By¿

By¿By¿

Cy¿ = 1346.4 N = 1.35 kN

By¿ = -1000.0 N = -1 kN

+ 1300 cos 30° N218 m2 = 0

-By¿12 m2 + 4000 N # m - Cy¿16 m2+©MA = 0;

-300 N + Cy¿ cos 30° + By¿ cos 30° = 0+ c©Fy = 0;

Cy¿ sin 30° + By¿ sin 30° - Ax = 0:+ ©Fx = 0;

y¿ Fig. 5–17

5.3 EQUATIONS OF EQUILIBRIUM 221

5

EXAMPLE 5.10

The uniform smooth rod shown in Fig. 5–17a is subjected to a forceand couple moment. If the rod is supported at A by a smooth wall andat B and C either at the top or bottom by rollers, determine thereactions at these supports. Neglect the weight of the rod.

(a)

A

2 m

300 N

4000 N ! m

4 m

2 m

C

B

30"2 m

(b)

2 m

300 N

4000 N ! m4 m

2 m30"

30"

Cy¿

By¿

30" 30"

Ax

y y¿

x

x¿

30"

SOLUTION

Free-Body Diagram. As shown in Fig. 5–17b, all the supportreactions act normal to the surfaces of contact since these surfaces aresmooth. The reactions at B and C are shown acting in the positive direction. This assumes that only the rollers located on the bottom ofthe rod are used for support.

Equations of Equilibrium. Using the x, y coordinate system inFig. 5–17b, we have

(1)

(2)

a

(3)

When writing the moment equation, it should be noted that the line ofaction of the force component 300 sin 30° N passes through point A,and therefore this force is not included in the moment equation.

Solving Eqs. 2 and 3 simultaneously, we obtain

Ans.

Ans.

Since is a negative scalar, the sense of is opposite to that shownon the free-body diagram in Fig. 5–17b. Therefore, the top roller at Bserves as the support rather than the bottom one. Retaining the negativesign for (Why?) and substituting the results into Eq. 1, we obtain

Ans.Ax = 173 N

1346.4 sin 30° N + (-1000.0 sin 30° N) - Ax = 0

By¿

By¿By¿

Cy¿ = 1346.4 N = 1.35 kN

By¿ = -1000.0 N = -1 kN

+ 1300 cos 30° N218 m2 = 0

-By¿12 m2 + 4000 N # m - Cy¿16 m2+©MA = 0;

-300 N + Cy¿ cos 30° + By¿ cos 30° = 0+ c©Fy = 0;

Cy¿ sin 30° + By¿ sin 30° - Ax = 0:+ ©Fx = 0;

y¿ Fig. 5–17

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Membros de duas e três forças

• A idéia é simplificar a solução de alguns problemas se reconhecermos membros que estão sujeitos a duas/três forças apenas.

• Membro de duas forças: As forças são aplicadas somente em dois pontos e para que estejam em equilíbrio, elas devem ser iguais e opostas e devem estar na mesma linha de ação e junto à linha que une os dois pontos onde as forças atuam. Assim, ∑MA=0 ou ∑MB=0 (figura c)

A

B

224 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

5.4 Two- and Three-Force MembersThe solutions to some equilibrium problems can be simplified byrecognizing members that are subjected to only two or three forces.

Two-Force Members As the name implies, a two-force member hasforces applied at only two points on the member. An example of a two-force member is shown in Fig. 5–20a. To satisfy force equilibrium, and

must be equal in magnitude, , but opposite in direction, Fig. 5–20b. Furthermore, moment equilibrium requires that

and share the same line of action, which can only happen if they aredirected along the line joining points A and B ( or ),Fig. 5–20c. Therefore, for any two-force member to be in equilibrium, thetwo forces acting on the member must have the same magnitude, act inopposite directions, and have the same line of action, directed along the linejoining the two points where these forces act.

©MB = 0©MA = 0FB

FA(©F = 0)FA = FB = FFB

FA

Three-Force Members If a member is subjected to only threeforces, it is called a three-force member. Moment equilibrium can besatisfied only if the three forces form a concurrent or parallel forcesystem. To illustrate, consider the member subjected to the three forces

, , and , shown in Fig. 5–21a. If the lines of action of and intersect at point O, then the line of action of must also pass throughpoint O so that the forces satisfy .As a special case, if the threeforces are all parallel, Fig. 5–21b, the location of the point of intersection,O, will approach infinity.

©MO = 0F3

F2F1F3F2F1

B

FB(a)

A FA

(b)

Two-force member

A FA ! F

FB ! F

A

FB ! F(c)

B

FA ! F

Fig. 5–20

F3F1

O

F1

F3

Three-force member

F2 F2

(b)(a)

Fig. 5–21

The bucket link AB on the back-hoeis a typical example of a two-forcemember since it is pin connected atits ends and, provided its weight isneglected, no other force acts on thismember.

The link used for this railroad car brakeis a three-force member. Since the force

in the tie rod at B and from thelink at C are parallel, then forequilibrium the resultant force at thepin A must also be parallel with thesetwo forces.

FA

FCFB

FB

FAFC

BA C

The boom on this lift is a three-forcemember, provided its weight is neglected.Here the lines of action of the weight of theworker, W, and the force of the two-forcemember (hydraulic cylinder) at B, ,intersect at O.For moment equilibrium,theresultant force at the pin A, , must alsobe directed towards O.

FA

FB

FAB

W

O

AFB

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• Membro de três forças: Neste caso, o equilíbrio só pode ser alcançado se as três forças formarem um sistema de forças concorrentes ou paralelas. Note que se as linhas de ação das três forças passam por O, o momento total deve ser anular, figura (a).

• Se as forças forem paralelas, como na figura (b), o ponto de intersecção entre elas ocorre no infinito.

A

B

224 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

5.4 Two- and Three-Force MembersThe solutions to some equilibrium problems can be simplified byrecognizing members that are subjected to only two or three forces.

Two-Force Members As the name implies, a two-force member hasforces applied at only two points on the member. An example of a two-force member is shown in Fig. 5–20a. To satisfy force equilibrium, and

must be equal in magnitude, , but opposite in direction, Fig. 5–20b. Furthermore, moment equilibrium requires that

and share the same line of action, which can only happen if they aredirected along the line joining points A and B ( or ),Fig. 5–20c. Therefore, for any two-force member to be in equilibrium, thetwo forces acting on the member must have the same magnitude, act inopposite directions, and have the same line of action, directed along the linejoining the two points where these forces act.

©MB = 0©MA = 0FB

FA(©F = 0)FA = FB = FFB

FA

Three-Force Members If a member is subjected to only threeforces, it is called a three-force member. Moment equilibrium can besatisfied only if the three forces form a concurrent or parallel forcesystem. To illustrate, consider the member subjected to the three forces

, , and , shown in Fig. 5–21a. If the lines of action of and intersect at point O, then the line of action of must also pass throughpoint O so that the forces satisfy .As a special case, if the threeforces are all parallel, Fig. 5–21b, the location of the point of intersection,O, will approach infinity.

©MO = 0F3

F2F1F3F2F1

B

FB(a)

A FA

(b)

Two-force member

A FA ! F

FB ! F

A

FB ! F(c)

B

FA ! F

Fig. 5–20

F3F1

O

F1

F3

Three-force member

F2 F2

(b)(a)

Fig. 5–21

The bucket link AB on the back-hoeis a typical example of a two-forcemember since it is pin connected atits ends and, provided its weight isneglected, no other force acts on thismember.

The link used for this railroad car brakeis a three-force member. Since the force

in the tie rod at B and from thelink at C are parallel, then forequilibrium the resultant force at thepin A must also be parallel with thesetwo forces.

FA

FCFB

FB

FAFC

BA C

The boom on this lift is a three-forcemember, provided its weight is neglected.Here the lines of action of the weight of theworker, W, and the force of the two-forcemember (hydraulic cylinder) at B, ,intersect at O.For moment equilibrium,theresultant force at the pin A, , must alsobe directed towards O.

FA

FB

FAB

W

O

AFB

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Equilíbrio em 3DReações de Apoio:

Uma força é desenvolvida por um suporte na direção em que o suporte previne um movimento de translação do membro atachado a ele.

Um momento de binário é desenvolvido por um suporte quando uma rotação do membro atachado a ele é evitada.

238 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

Types of Connection Reaction Number of Unknowns

continued

One unknown. The reaction is a force which acts away from the member in the known direction of the cable.

One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact.

One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact.

Three unknowns. The reactions are three rectangular force components.

Four unknowns. The reactions are two force and two couple-moment components which act perpendicular to the shaft. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

F

F

F

Fz

FyFx

single journal bearing

Fz

Fx

Mz

Mx

(1)

cable

(2)

(3)

roller

ball and socket

(4)

(5)

smooth surface support

TABLE 5–2 Supports for Rigid Bodies Subjected to Three-Dimensional Force Systems

A junta esférica previne três translações e por isso aparecem três componentes da força no suporte. Como o membro permite a rotação livre do membro, não há momentos de binário.

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Exemplo 6: Barra com três mancais radiais simples alinhados em A, B e C.

5.5 FREE-BODY DIAGRAMS 241

5

EXAMPLE 5.14

Consider the two rods and plate, along with their associated free-bodydiagrams shown in Fig. 5–23. The x, y, z axes are established on thediagram and the unknown reaction components are indicated in thepositive sense. The weight is neglected.

SOLUTION45 N ! m

500 N

Properly aligned journalbearings at A, B, C.

A

B

C

45 N ! m

500 N

The force reactions developed bythe bearings are sufficient for equilibrium since they prevent the shaft from rotating abouteach of the coordinate axes.

Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C

Properly aligned journal bearingat A and hinge at C. Roller at B.

Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy

Only force reactions are developed bythe bearing and hinge on the plate to prevent rotation about each coordinate axis.No moments at the hinge are developed.

C

200 lb ! ft

Pin at A and cable BC.

A

B300 lb

200 lb ! ft

Moment components are developedby the pin on the rod to preventrotation about the x and z axes.

x

B300 lb

y

Az

z

MAz

MAx

Ax

Ay

T

Fig. 5–23

5.5 FREE-BODY DIAGRAMS 239

5

Reaction Number of Unknowns

Five unknowns. The reactions are two force and three couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Six unknowns. The reactions are three force and three couple-moment components.

Fz

Fx

Mz

Mx

Fy

Fz

Fx

Mz

MxMy

Fz

Mz

FxFy My

Mz

Fx

Fy

Mx

Fz

Mz

Fx MyMxFy

Fz

Types of Connection

TABLE 5–2 Continued

single hinge

fixed support

single thrust bearing

single journal bearingwith square shaft

single smooth pin

(7)

(6)

(8)

(10)

(9)

5.5 FREE-BODY DIAGRAMS 239

5

Reaction Number of Unknowns

Five unknowns. The reactions are two force and three couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Six unknowns. The reactions are three force and three couple-moment components.

Fz

Fx

Mz

Mx

Fy

Fz

Fx

Mz

MxMy

Fz

Mz

FxFy My

Mz

Fx

Fy

Mx

Fz

Mz

Fx MyMxFy

Fz

Types of Connection

TABLE 5–2 Continued

single hinge

fixed support

single thrust bearing

single journal bearingwith square shaft

single smooth pin

(7)

(6)

(8)

(10)

(9)

5.5 FREE-BODY DIAGRAMS 241

5

EXAMPLE 5.14

Consider the two rods and plate, along with their associated free-bodydiagrams shown in Fig. 5–23. The x, y, z axes are established on thediagram and the unknown reaction components are indicated in thepositive sense. The weight is neglected.

SOLUTION45 N ! m

500 N

Properly aligned journalbearings at A, B, C.

A

B

C

45 N ! m

500 N

The force reactions developed bythe bearings are sufficient for equilibrium since they prevent the shaft from rotating abouteach of the coordinate axes.

Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C

Properly aligned journal bearingat A and hinge at C. Roller at B.

Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy

Only force reactions are developed bythe bearing and hinge on the plate to prevent rotation about each coordinate axis.No moments at the hinge are developed.

C

200 lb ! ft

Pin at A and cable BC.

A

B300 lb

200 lb ! ft

Moment components are developedby the pin on the rod to preventrotation about the x and z axes.

x

B300 lb

y

Az

z

MAz

MAx

Ax

Ay

T

Fig. 5–23

Deve se notar que mancais, pinos e dobradiças simples, se usados com outros componentes no apoio ao corpo rígido e adequadamente alinhados quando conectados ao corpo, não necessitam dos momentos de binário, ou seja, as forças sozinhas, são suficientes para apoiar o corpo.

Page 25: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Exemplo 7: Barra com pino em A e cabo em BC

5.5 FREE-BODY DIAGRAMS 241

5

EXAMPLE 5.14

Consider the two rods and plate, along with their associated free-bodydiagrams shown in Fig. 5–23. The x, y, z axes are established on thediagram and the unknown reaction components are indicated in thepositive sense. The weight is neglected.

SOLUTION45 N ! m

500 N

Properly aligned journalbearings at A, B, C.

A

B

C

45 N ! m

500 N

The force reactions developed bythe bearings are sufficient for equilibrium since they prevent the shaft from rotating abouteach of the coordinate axes.

Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C

Properly aligned journal bearingat A and hinge at C. Roller at B.

Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy

Only force reactions are developed bythe bearing and hinge on the plate to prevent rotation about each coordinate axis.No moments at the hinge are developed.

C

200 lb ! ft

Pin at A and cable BC.

A

B300 lb

200 lb ! ft

Moment components are developedby the pin on the rod to preventrotation about the x and z axes.

x

B300 lb

y

Az

z

MAz

MAx

Ax

Ay

T

Fig. 5–23

5.5 FREE-BODY DIAGRAMS 239

5

Reaction Number of Unknowns

Five unknowns. The reactions are two force and three couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Six unknowns. The reactions are three force and three couple-moment components.

Fz

Fx

Mz

Mx

Fy

Fz

Fx

Mz

MxMy

Fz

Mz

FxFy My

Mz

Fx

Fy

Mx

Fz

Mz

Fx MyMxFy

Fz

Types of Connection

TABLE 5–2 Continued

single hinge

fixed support

single thrust bearing

single journal bearingwith square shaft

single smooth pin

(7)

(6)

(8)

(10)

(9)

5.5 FREE-BODY DIAGRAMS 241

5

EXAMPLE 5.14

Consider the two rods and plate, along with their associated free-bodydiagrams shown in Fig. 5–23. The x, y, z axes are established on thediagram and the unknown reaction components are indicated in thepositive sense. The weight is neglected.

SOLUTION45 N ! m

500 N

Properly aligned journalbearings at A, B, C.

A

B

C

45 N ! m

500 N

The force reactions developed bythe bearings are sufficient for equilibrium since they prevent the shaft from rotating abouteach of the coordinate axes.

Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C

Properly aligned journal bearingat A and hinge at C. Roller at B.

Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy

Only force reactions are developed bythe bearing and hinge on the plate to prevent rotation about each coordinate axis.No moments at the hinge are developed.

C

200 lb ! ft

Pin at A and cable BC.

A

B300 lb

200 lb ! ft

Moment components are developedby the pin on the rod to preventrotation about the x and z axes.

x

B300 lb

y

Az

z

MAz

MAx

Ax

Ay

T

Fig. 5–23

Page 26: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

5.5 FREE-BODY DIAGRAMS 241

5

EXAMPLE 5.14

Consider the two rods and plate, along with their associated free-bodydiagrams shown in Fig. 5–23. The x, y, z axes are established on thediagram and the unknown reaction components are indicated in thepositive sense. The weight is neglected.

SOLUTION45 N ! m

500 N

Properly aligned journalbearings at A, B, C.

A

B

C

45 N ! m

500 N

The force reactions developed bythe bearings are sufficient for equilibrium since they prevent the shaft from rotating abouteach of the coordinate axes.

Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C

Properly aligned journal bearingat A and hinge at C. Roller at B.

Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy

Only force reactions are developed bythe bearing and hinge on the plate to prevent rotation about each coordinate axis.No moments at the hinge are developed.

C

200 lb ! ft

Pin at A and cable BC.

A

B300 lb

200 lb ! ft

Moment components are developedby the pin on the rod to preventrotation about the x and z axes.

x

B300 lb

y

Az

z

MAz

MAx

Ax

Ay

T

Fig. 5–23

Exemplo 8: Chapa com mancal alinhado em A e dobradiça em C

5.5 FREE-BODY DIAGRAMS 239

5

Reaction Number of Unknowns

Five unknowns. The reactions are two force and three couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Five unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the examples.

Six unknowns. The reactions are three force and three couple-moment components.

Fz

Fx

Mz

Mx

Fy

Fz

Fx

Mz

MxMy

Fz

Mz

FxFy My

Mz

Fx

Fy

Mx

Fz

Mz

Fx MyMxFy

Fz

Types of Connection

TABLE 5–2 Continued

single hinge

fixed support

single thrust bearing

single journal bearingwith square shaft

single smooth pin

(7)

(6)

(8)

(10)

(9)

5.5 FREE-BODY DIAGRAMS 241

5

EXAMPLE 5.14

Consider the two rods and plate, along with their associated free-bodydiagrams shown in Fig. 5–23. The x, y, z axes are established on thediagram and the unknown reaction components are indicated in thepositive sense. The weight is neglected.

SOLUTION45 N ! m

500 N

Properly aligned journalbearings at A, B, C.

A

B

C

45 N ! m

500 N

The force reactions developed bythe bearings are sufficient for equilibrium since they prevent the shaft from rotating abouteach of the coordinate axes.

Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C

Properly aligned journal bearingat A and hinge at C. Roller at B.

Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy

Only force reactions are developed bythe bearing and hinge on the plate to prevent rotation about each coordinate axis.No moments at the hinge are developed.

C

200 lb ! ft

Pin at A and cable BC.

A

B300 lb

200 lb ! ft

Moment components are developedby the pin on the rod to preventrotation about the x and z axes.

x

B300 lb

y

Az

z

MAz

MAx

Ax

Ay

T

Fig. 5–23

Page 27: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Equações de Equilíbrio

242 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

5.6 Equations of Equilibrium

As stated in Sec. 5.1, the conditions for equilibrium of a rigid bodysubjected to a three-dimensional force system require that both theresultant force and resultant couple moment acting on the body be equalto zero.

Vector Equations of Equilibrium. The two conditions forequilibrium of a rigid body may be expressed mathematically in vectorform as

(5–5)

where is the vector sum of all the external forces acting on the bodyand is the sum of the couple moments and the moments of all theforces about any point O located either on or off the body.

Scalar Equations of Equilibrium. If all the external forces andcouple moments are expressed in Cartesian vector form and substitutedinto Eqs. 5–5, we have

Since the i, j, and k components are independent from one another, theabove equations are satisfied provided

(5–6a)

and

(5–6b)

These six scalar equilibrium equations may be used to solve for at mostsix unknowns shown on the free-body diagram. Equations 5–6a requirethe sum of the external force components acting in the x, y, and zdirections to be zero, and Eqs. 5–6b require the sum of the momentcomponents about the x, y, and z axes to be zero.

©Mx = 0©My = 0©Mz = 0

©Fx = 0©Fy = 0©Fz = 0

©MO = ©Mxi + ©Myj + ©Mzk = 0

©F = ©Fxi + ©Fy j + ©Fzk = 0

©MO

©F

©F = 0©MO = 0

242 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

5.6 Equations of Equilibrium

As stated in Sec. 5.1, the conditions for equilibrium of a rigid bodysubjected to a three-dimensional force system require that both theresultant force and resultant couple moment acting on the body be equalto zero.

Vector Equations of Equilibrium. The two conditions forequilibrium of a rigid body may be expressed mathematically in vectorform as

(5–5)

where is the vector sum of all the external forces acting on the bodyand is the sum of the couple moments and the moments of all theforces about any point O located either on or off the body.

Scalar Equations of Equilibrium. If all the external forces andcouple moments are expressed in Cartesian vector form and substitutedinto Eqs. 5–5, we have

Since the i, j, and k components are independent from one another, theabove equations are satisfied provided

(5–6a)

and

(5–6b)

These six scalar equilibrium equations may be used to solve for at mostsix unknowns shown on the free-body diagram. Equations 5–6a requirethe sum of the external force components acting in the x, y, and zdirections to be zero, and Eqs. 5–6b require the sum of the momentcomponents about the x, y, and z axes to be zero.

©Mx = 0©My = 0©Mz = 0

©Fx = 0©Fy = 0©Fz = 0

©MO = ©Mxi + ©Myj + ©Mzk = 0

©F = ©Fxi + ©Fy j + ©Fzk = 0

©MO

©F

©F = 0©MO = 0

242 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

5.6 Equations of Equilibrium

As stated in Sec. 5.1, the conditions for equilibrium of a rigid bodysubjected to a three-dimensional force system require that both theresultant force and resultant couple moment acting on the body be equalto zero.

Vector Equations of Equilibrium. The two conditions forequilibrium of a rigid body may be expressed mathematically in vectorform as

(5–5)

where is the vector sum of all the external forces acting on the bodyand is the sum of the couple moments and the moments of all theforces about any point O located either on or off the body.

Scalar Equations of Equilibrium. If all the external forces andcouple moments are expressed in Cartesian vector form and substitutedinto Eqs. 5–5, we have

Since the i, j, and k components are independent from one another, theabove equations are satisfied provided

(5–6a)

and

(5–6b)

These six scalar equilibrium equations may be used to solve for at mostsix unknowns shown on the free-body diagram. Equations 5–6a requirethe sum of the external force components acting in the x, y, and zdirections to be zero, and Eqs. 5–6b require the sum of the momentcomponents about the x, y, and z axes to be zero.

©Mx = 0©My = 0©Mz = 0

©Fx = 0©Fy = 0©Fz = 0

©MO = ©Mxi + ©Myj + ©Mzk = 0

©F = ©Fxi + ©Fy j + ©Fzk = 0

©MO

©F

©F = 0©MO = 0

242 CH A P T E R 5 EQ U I L I B R I U M O F A RI G I D BO D Y

5

5.6 Equations of Equilibrium

As stated in Sec. 5.1, the conditions for equilibrium of a rigid bodysubjected to a three-dimensional force system require that both theresultant force and resultant couple moment acting on the body be equalto zero.

Vector Equations of Equilibrium. The two conditions forequilibrium of a rigid body may be expressed mathematically in vectorform as

(5–5)

where is the vector sum of all the external forces acting on the bodyand is the sum of the couple moments and the moments of all theforces about any point O located either on or off the body.

Scalar Equations of Equilibrium. If all the external forces andcouple moments are expressed in Cartesian vector form and substitutedinto Eqs. 5–5, we have

Since the i, j, and k components are independent from one another, theabove equations are satisfied provided

(5–6a)

and

(5–6b)

These six scalar equilibrium equations may be used to solve for at mostsix unknowns shown on the free-body diagram. Equations 5–6a requirethe sum of the external force components acting in the x, y, and zdirections to be zero, and Eqs. 5–6b require the sum of the momentcomponents about the x, y, and z axes to be zero.

©Mx = 0©My = 0©Mz = 0

©Fx = 0©Fy = 0©Fz = 0

©MO = ©Mxi + ©Myj + ©Mzk = 0

©F = ©Fxi + ©Fy j + ©Fzk = 0

©MO

©F

©F = 0©MO = 0

Page 28: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

Problemas de indeterminação estática

• Para garantir o equilíbrio do corpo rígido, além das equações de equilíbrio serem satisfeitas, é necessário que não haja indeterminação estática.

• Isto pode ocorrer sempre que o número de reações de apoio seja redundante de tal modo a fazer com que o número de incógnitas seja maior do que o número de equações disponíveis.

• A resolução deste problema está além do escopo deste curso e será discutida no curso de Mecânica dos Sólidos (Resistência dos Materiais) através da consideração das deformações que um corpo rígido sofre.

Page 29: Equilíbrio de um Corpo Rígido · Condições necessárias e suficientes para o equilíbrio de um corpo rígido. 200 CHAPTER 5EQUILIBRIUM OF A RIGID BODY 5 F1 M2 M1 F2 F3 F4 O (a)

5.7 CONSTRAINTS AND STATICAL DETERMINACY 243

5

5.7 Constraints and Statical Determinacy

To ensure the equilibrium of a rigid body, it is not only necessary to satisfythe equations of equilibrium, but the body must also be properly held orconstrained by its supports. Some bodies may have more supports than arenecessary for equilibrium, whereas others may not have enough or thesupports may be arranged in a particular manner that could cause thebody to move. Each of these cases will now be discussed.

Redundant Constraints. When a body has redundant supports,that is, more supports than are necessary to hold it in equilibrium, itbecomes statically indeterminate. Statically indeterminate means thatthere will be more unknown loadings on the body than equations ofequilibrium available for their solution. For example, the beam in Fig. 5–24a and the pipe assembly in Fig. 5–24b, shown together withtheir free-body diagrams, are both statically indeterminate because ofadditional (or redundant) support reactions. For the beam there are fiveunknowns, and for which only three equilibriumequations can be written ( and Eqs. 5–2).The pipe assembly has eight unknowns, for which only six equilibriumequations can be written, Eqs. 5–6.

The additional equations needed to solve statically indeterminateproblems of the type shown in Fig. 5–24 are generally obtained from thedeformation conditions at the points of support.These equations involvethe physical properties of the body which are studied in subjects dealingwith the mechanics of deformation, such as “mechanics of materials.”*

©MO = 0,©Fy = 0,©Fx = 0,Cy,By,Ay,Ax,MA,

500 N

B C

A

2 kN ! m

500 N

2 kN ! m

Ax

Ay

MABy Cy

(a)

x

y

B

A

400 N

200 N

400 N

200 N

Ay

Az

ByBx

Mx My

Bz

Mz

(b)

y

z

x

Fig. 5–24

* See R. C. Hibbeler, Mechanics of Materials, 7th edition, Pearson Education/PrenticeHall, Inc.

5.7 CONSTRAINTS AND STATICAL DETERMINACY 243

5

5.7 Constraints and Statical Determinacy

To ensure the equilibrium of a rigid body, it is not only necessary to satisfythe equations of equilibrium, but the body must also be properly held orconstrained by its supports. Some bodies may have more supports than arenecessary for equilibrium, whereas others may not have enough or thesupports may be arranged in a particular manner that could cause thebody to move. Each of these cases will now be discussed.

Redundant Constraints. When a body has redundant supports,that is, more supports than are necessary to hold it in equilibrium, itbecomes statically indeterminate. Statically indeterminate means thatthere will be more unknown loadings on the body than equations ofequilibrium available for their solution. For example, the beam in Fig. 5–24a and the pipe assembly in Fig. 5–24b, shown together withtheir free-body diagrams, are both statically indeterminate because ofadditional (or redundant) support reactions. For the beam there are fiveunknowns, and for which only three equilibriumequations can be written ( and Eqs. 5–2).The pipe assembly has eight unknowns, for which only six equilibriumequations can be written, Eqs. 5–6.

The additional equations needed to solve statically indeterminateproblems of the type shown in Fig. 5–24 are generally obtained from thedeformation conditions at the points of support.These equations involvethe physical properties of the body which are studied in subjects dealingwith the mechanics of deformation, such as “mechanics of materials.”*

©MO = 0,©Fy = 0,©Fx = 0,Cy,By,Ay,Ax,MA,

500 N

B C

A

2 kN ! m

500 N

2 kN ! m

Ax

Ay

MABy Cy

(a)

x

y

B

A

400 N

200 N

400 N

200 N

Ay

Az

ByBx

Mx My

Bz

Mz

(b)

y

z

x

Fig. 5–24

* See R. C. Hibbeler, Mechanics of Materials, 7th edition, Pearson Education/PrenticeHall, Inc.

Este problema apresenta 5 incógnitas e 3 equações de equilíbrio.

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Pontos importantes

1. Sempre desenhe o DCL

2. Se um apoio evita a translação de um corpo em uma dada direção, então o apoio exerce uma força sobre o corpo nesta direção.

3. Se um apoio evita uma rotação em volta de um eixo, então o suporte exerce um momento de binário sobre o corpo, em volta deste eixo.

4. Se um corpo é submetido a mais incógnitas (apoios) do que as equações disponíveis, então o problema é estaticamente indeterminado.

5. Um corpo rígido estável requer que as linhas de ação das forças de reação de apoio não cruzem um eixo comum e que também não sejam paralelas.