Mecânica GeralVínculos (Reações de Apoio)

Prof. Ettore Baldini-Netobaldini@uninove.br

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.

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.

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.

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.

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.

Tipos de apoio, reação e número de incógnitas

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.

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

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;

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.