Equilíbrio de um Corpo Rígido (Caso em 3 dimensões)estatica2015.weebly.com/uploads/3/7/5/7/37571293/aula11-parte2.pdf · Exemplo 6: Barra com três mancais radiais simples alinhados

Equilíbrio de um Corpo Rígido (Caso em 3 dimensões)

Prof. Ettore Baldini-Neto

Reações 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.


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.

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

EXAMPLE 5.14


SOLUTION45 N ! m

500 N


A

B

C

45 N ! m

500 N


Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C


Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy


C

200 lb ! ft


A

B300 lb

200 lb ! ft


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.


5


continued






F

F

F

Fz

FyFx


Fz

Fx

Mz

Mx

(1)

cable

(2)

(3)

roller

ball and socket

(4)

(5)




5


continued






F

F

F

Fz

FyFx


Fz

Fx

Mz

Mx

(1)

cable

(2)

(3)

roller

ball and socket

(4)

(5)



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


5

EXAMPLE 5.14


SOLUTION45 N ! m

500 N


A

B

C

45 N ! m

500 N


Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C


Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy


C

200 lb ! ft


A

B300 lb

200 lb ! ft


x

B300 lb

y

Az

z

MAz

MAx

Ax

Ay

T

Fig. 5–23


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.



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

EXAMPLE 5.14


SOLUTION45 N ! m

500 N


A

B

C

45 N ! m

500 N


Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C


Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy


C

200 lb ! ft


A

B300 lb

200 lb ! ft


x

B300 lb

y

Az

z

MAz

MAx

Ax

Ay

T

Fig. 5–23


5

EXAMPLE 5.14


SOLUTION45 N ! m

500 N


A

B

C

45 N ! m

500 N


Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C


Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy


C

200 lb ! ft


A

B300 lb

200 lb ! ft


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

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.




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

EXAMPLE 5.14


SOLUTION45 N ! m

500 N


A

B

C

45 N ! m

500 N


Bz

Bx

CxCy

xyAy

Az

z

400 lb

A

B

C


Ax

400 lb

Bz

z

yx

Az

Cx

Cz

Cy


C

200 lb ! ft


A

B300 lb

200 lb ! ft


x

B300 lb

y

Az

z

MAz

MAx

Ax

Ay

T

Fig. 5–23

Equações de Equilíbrio


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


5




(5–5)




(5–6a)

and

(5–6b)


©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


5




(5–5)




(5–6a)

and

(5–6b)


©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


5




(5–5)




(5–6a)

and

(5–6b)


©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

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.

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.

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.

Exemplo 8. Determine as componentes da reação que a junta esférica em A, o mancal radial liso em B e o rolete em C exercem sobre a montagem das barras na figura abaixo.


5

Determine the components of reaction that the ball-and-socket jointat A, the smooth journal bearing at B, and the roller support at Cexert on the rod assembly in Fig. 5–29a.

EXAMPLE 5.16

x

y

z

A

B

CD

0.4 m

0.4 m

(a)

0.6 m

900 N

0.4 m0.4 m

A

x

y

z

0.4 m

0.4 m

(b)

0.6 m0.4 m

0.4 mFC

Bz

Az Bx

Ax

Ay

900 N

Fig. 5–29

SOLUTIONFree-Body Diagram. As shown on the free-body diagram, Fig. 5–29b,the reactive forces of the supports will prevent the assembly fromrotating about each coordinate axis, and so the journal bearing at Bonly exerts reactive forces on the member.

Equations of Equilibrium. A direct solution for can be obtainedby summing forces along the y axis.

Ans.

The force can be determined directly by summing moments aboutthe y axis.

Ans.Using this result, can be determined by summing moments aboutthe x axis.

Ans.The negative sign indicates that acts downward. The force canbe found by summing moments about the z axis.

Ans.

Thus,

Ans.

Finally, using the results of and .

Ans.Az = 750 NAz + (-450 N) + 600 N - 900 N = 0©Fz = 0;

FCBz

Ax + 0 = 0 Ax = 0©Fx = 0;

-Bx(0.8 m) = 0 Bx = 0©Mz = 0;

BxBz

Bz = -450 NBz(0.8 m) + 600 N(1.2 m) - 900 N(0.4 m) = 0©Mx = 0;

Bz

FC = 600 NFC(0.6 m) - 900 N(0.4 m) = 0©My = 0;

FC

Ay = 0©Fy = 0;

Ay


5

Determine the components of reaction that the ball-and-socket jointat A, the smooth journal bearing at B, and the roller support at Cexert on the rod assembly in Fig. 5–29a.

EXAMPLE 5.16

x

y

z

A

B

CD

0.4 m

0.4 m

(a)

0.6 m

900 N

0.4 m0.4 m

A

x

y

z

0.4 m

0.4 m

(b)

0.6 m0.4 m

0.4 mFC

Bz

Az Bx

Ax

Ay

900 N

Fig. 5–29

SOLUTIONFree-Body Diagram. As shown on the free-body diagram, Fig. 5–29b,the reactive forces of the supports will prevent the assembly fromrotating about each coordinate axis, and so the journal bearing at Bonly exerts reactive forces on the member.

Equations of Equilibrium. A direct solution for can be obtainedby summing forces along the y axis.

Ans.

The force can be determined directly by summing moments aboutthe y axis.

Ans.Using this result, can be determined by summing moments aboutthe x axis.

Ans.The negative sign indicates that acts downward. The force canbe found by summing moments about the z axis.

Ans.

Thus,

Ans.

Finally, using the results of and .

Ans.Az = 750 NAz + (-450 N) + 600 N - 900 N = 0©Fz = 0;

FCBz

Ax + 0 = 0 Ax = 0©Fx = 0;

-Bx(0.8 m) = 0 Bx = 0©Mz = 0;

BxBz

Bz = -450 NBz(0.8 m) + 600 N(1.2 m) - 900 N(0.4 m) = 0©Mx = 0;

Bz

FC = 600 NFC(0.6 m) - 900 N(0.4 m) = 0©My = 0;

FC

Ay = 0©Fy = 0;

Ay

Exemplo 9. A barra AB mostrada está sujeita a uma força de 200N. Determine as reações na junta esférica A e a tração nos cabos BD e BE.


5

EXAMPLE 5.18

1.5 m

2 m

200 N

1.5 m

2 m

E

A

B

D

C

(a)

1 m

200 N

y

B

C

x

z

rB

rC

TD

TE

Az

A AyAx

(b)

Fig. 5–31

Rod AB shown in Fig. 5–31a is subjected to the 200-N force.Determine the reactions at the ball-and-socket joint A and thetension in the cables BD and BE.

SOLUTION (VECTOR ANALYSIS)

Free-Body Diagram. Fig. 5–31b.

Equations of Equilibrium. Representing each force on the free-bodydiagram in Cartesian vector form, we have

Applying the force equation of equilibrium.

(1)

(2)(3)

Summing moments about point A yields

Since then

Expanding and rearranging terms gives

(4)

(5)(6)

Solving Eqs. 1 through 5, we getAns.

Ans.Ans.Ans.Ans.

NOTE: The negative sign indicates that and have a sense whichis opposite to that shown on the free-body diagram, Fig. 5–31b.

AyAx

Az = 200 NAy = -100 NAx = -50 NTE = 50 N

TD = 100 N

TD - 2TE = 0©Mz = 0;-2TE + 100 = 0©My = 0;

2TD - 200 = 0©Mx = 0;

12TD - 2002i + 1-2TE + 1002j + 1TD - 2TE2k = 0

10.5i + 1j - 1k2 * 1-200k2 + 11i + 2j - 2k2 * 1TEi + TDj2 = 0

rC = 12 rB,

rC * F + rB * 1TE + TD2 = 0©MA = 0;

Az - 200 = 0©Fz = 0;Ay + TD = 0©Fy = 0;

Ax + TE = 0©Fx = 0;

1Ax + TE2i + 1Ay + TD2j + 1Az - 2002k = 0

FA + TE + TD + F = 0©F = 0;

F = 5-200k6 NTD = TDjTE = TEi

FA = Axi + Ayj + Azk


5

EXAMPLE 5.18

1.5 m

2 m

200 N

1.5 m

2 m

E

A

B

D

C

(a)

1 m

200 N

y

B

C

x

z

rB

rC

TD

TE

Az

A AyAx

(b)

Fig. 5–31

Rod AB shown in Fig. 5–31a is subjected to the 200-N force.Determine the reactions at the ball-and-socket joint A and thetension in the cables BD and BE.

SOLUTION (VECTOR ANALYSIS)

Free-Body Diagram. Fig. 5–31b.

Equations of Equilibrium. Representing each force on the free-bodydiagram in Cartesian vector form, we have

Applying the force equation of equilibrium.

(1)

(2)(3)

Summing moments about point A yields

Since then

Expanding and rearranging terms gives

(4)

(5)(6)

Solving Eqs. 1 through 5, we getAns.

Ans.Ans.Ans.Ans.

NOTE: The negative sign indicates that and have a sense whichis opposite to that shown on the free-body diagram, Fig. 5–31b.

AyAx

Az = 200 NAy = -100 NAx = -50 NTE = 50 N

TD = 100 N

TD - 2TE = 0©Mz = 0;-2TE + 100 = 0©My = 0;

2TD - 200 = 0©Mx = 0;

12TD - 2002i + 1-2TE + 1002j + 1TD - 2TE2k = 0

10.5i + 1j - 1k2 * 1-200k2 + 11i + 2j - 2k2 * 1TEi + TDj2 = 0

rC = 12 rB,

rC * F + rB * 1TE + TD2 = 0©MA = 0;

Az - 200 = 0©Fz = 0;Ay + TD = 0©Fy = 0;

Ax + TE = 0©Fx = 0;

1Ax + TE2i + 1Ay + TD2j + 1Az - 2002k = 0

FA + TE + TD + F = 0©F = 0;

F = 5-200k6 NTD = TDjTE = TEi

FA = Axi + Ayj + Azk

Exemplo 5.84. Determine o maior peso do barril de óleo que a grua pode sustentar sem tombar. Além disso, quais são as reações verticais nas rodas lisas A, B e C para este caso? A grua tem um peso de 1,5kN com seu centro de gravidade localizado em G.

389

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*5–84. Determine the largest weight of the oil drum thatthe floor crane can support without overturning. Also, whatare the vertical reactions at the smooth wheels A, B, and Cfor this case. The floor crane has a weight of 300 lb, with itscenter of gravity located at G.

x

C

G

B

Ay

0.9 m

0.45 m

3 m

1.2 m0.6 m

0.75 m0.75 m

0.3 m

30

3 m

1.5 kN

0.3 m

0.6 m

1.2 m

0.75 m

0.75 m

for this case. The floor crane has a weight of 1.5 kN, with its center of gravity located at G.

Equations of Equilibrium: The floor crane has a tendency to overturn about the y′ axis, as shown on the free-body diagram in Fig. a. When the crane is about to overturn, the wheel at C loses contact with the ground. Thus

NC = 0 Ans

Applying the moment equation of equilibrium about the y′ axis,

ΣMy′ = 0; W(3 cos 30° – 0.6 – 1.2) – 1.5 (1.2) = 0

W = 2.255 kN Ans

Using this result and writing the moment equation of equilibrium about the x axis and the force equation of equilibrium along the z axis,

ΣMx = 0; NB (0.75) – NA (0.75) = 0 (1)

ΣFz = 0; NA + NB – 1.5 – 2.255 = 0 (2)

Solving Eqs. (1) and (2), yields

NA = NB = 1.878 kN Ans

05b Ch05b 361-400.indd 38905b Ch05b 361-400.indd 389 6/12/09 8:46:36 AM6/12/09 8:46:36 AM

389

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*5–84. Determine the largest weight of the oil drum thatthe floor crane can support without overturning. Also, whatare the vertical reactions at the smooth wheels A, B, and Cfor this case. The floor crane has a weight of 300 lb, with itscenter of gravity located at G.

x

C

G

B

Ay

0.9 m

0.45 m

3 m

1.2 m0.6 m

0.75 m0.75 m

0.3 m

30

3 m

1.5 kN

0.3 m

0.6 m

1.2 m

0.75 m

0.75 m

for this case. The floor crane has a weight of 1.5 kN, with its center of gravity located at G.

Equations of Equilibrium: The floor crane has a tendency to overturn about the y′ axis, as shown on the free-body diagram in Fig. a. When the crane is about to overturn, the wheel at C loses contact with the ground. Thus

NC = 0 Ans

Applying the moment equation of equilibrium about the y′ axis,

ΣMy′ = 0; W(3 cos 30° – 0.6 – 1.2) – 1.5 (1.2) = 0

W = 2.255 kN Ans

Using this result and writing the moment equation of equilibrium about the x axis and the force equation of equilibrium along the z axis,

ΣMx = 0; NB (0.75) – NA (0.75) = 0 (1)

ΣFz = 0; NA + NB – 1.5 – 2.255 = 0 (2)

Solving Eqs. (1) and (2), yields

NA = NB = 1.878 kN Ans

05b Ch05b 361-400.indd 38905b Ch05b 361-400.indd 389 6/12/09 8:46:36 AM6/12/09 8:46:36 AM

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Equilíbrio de um Corpo Rígido (Caso em 3 dimensões)estatica2015.weebly.com/uploads/3/7/5/7/37571293/aula11-parte2.pdf · Exemplo 6: Barra com três mancais radiais simples alinhados