solution 2 m a - florida international university

481

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SolutionMethod of Joints. Start at joint C and then proceed to join D.

Joint C. Fig. a

S+ ΣFx = 0; FCB = 0 Ans.

+ cΣFy = 0; FCD - 20 = 0 FCD = 20.0 kN (C) Ans.

Joint D. Fig. b

+ cΣFy = 0; FDB a35b - 20.0 = 0 FDB = 33.33 kN (T) = 33.3 kN (T) Ans.

S+ ΣFx = 0; 10 + 33.33 a45b - FDA = 0

FDA = 36.67 kN (C) = 36.7 kN (C) Ans.

6–1.

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 20 kN, P2 = 10 kN.

C B

A

D

1.5 m

2 m

P1

P2

Ans: FCB = 0 FCD = 20.0 kN (C) FDB = 33.3 kN (T) FDA = 36.7 kN (C)

486


6–5.

Determine the force in each member of the truss, and stateif the members are in tension or compression. Set .

SOLUTION

Support Reactions: Applying the equations of equilibrium to the free-body diagram of the entire truss,Fig.a, we have

a

Method of Joints: We will use the above result to analyze the equilibrium of joints C and A, and then proceed to analyze of joint B.

Joint C: From the free-body diagram in Fig. b, we can write

Ans.

Ans.

Joint A: From the free-body diagram in Fig. c, we can write

Ans.

Ans.

Joint B: From the free-body diagram in Fig. d, we can write

Ans.

Note: The equilibrium analysis of joint D can be used to check the accuracy of thesolution obtained above.

4.167 - 4.167 = 0 (check!)©Fx = 0;:+

FBD = 4 kN (T)

FBD - 4 = 0+ c ©Fy = 0;

FAB = 4.167 kN = 4.17 kN (T)

FAB - 3 - 1.458a45b = 0©Fx = 0;:+

FAD = 1.458 kN = 1.46 kN (C)

0.875 - FAD a35b = 0+ c ©Fy = 0;

FCB = 4.167 kN = 4.17 kN (T)

5.208 a45b - FCB = 0©Fx = 0;:+

FCD = 5.208 kN = 5.21 kN (C)

3.125 - FCD a35b = 0+ c ©Fy = 0;

Ay = 0.875 kN

Ay + 3.125 - 4 = 0+ c ©Fy = 0;

Ax = 3 kN

3 - Ax = 0©Fx = 0;:+

NC = 3.125 kN

NC (2 + 2) - 4(2) - 3(1.5) = 0+ ©MA = 0;

u = 0°

A C

B

D

2 m

4 kN

3 kN

2 m

1.5 m

u

Ans:FCD = 5.21 kN (C)FCB = 4.17 kN (T)FAD = 1.46 kN (C)FAB = 4.17 kN (T)FBD = 4 kN (T)

487


6–5.


SOLUTION

Support Reactions: Applying the equations of equilibrium to the free-body diagram of the entire truss,Fig.a, we have

a



Ans.

Ans.


Ans.

Ans.


Ans.


4.167 - 4.167 = 0 (check!)©Fx = 0;:+

FBD = 4 kN (T)

FBD - 4 = 0+ c ©Fy = 0;

FAB = 4.167 kN = 4.17 kN (T)

FAB - 3 - 1.458a45b = 0©Fx = 0;:+

FAD = 1.458 kN = 1.46 kN (C)

0.875 - FAD a35b = 0+ c ©Fy = 0;

FCB = 4.167 kN = 4.17 kN (T)

5.208 a45b - FCB = 0©Fx = 0;:+

FCD = 5.208 kN = 5.21 kN (C)

3.125 - FCD a35b = 0+ c ©Fy = 0;

Ay = 0.875 kN

Ay + 3.125 - 4 = 0+ c ©Fy = 0;

Ax = 3 kN

3 - Ax = 0©Fx = 0;:+

NC = 3.125 kN

NC (2 + 2) - 4(2) - 3(1.5) = 0+ ©MA = 0;

u = 0°

A C

B

D

2 m

4 kN

3 kN

2 m

1.5 m

u

6–6.


SOLUTION

Support Reactions: From the free-body diagram of the truss, Fig. a, and applyingthe equations of equilibrium, we have

a



Ans.

Ans.


Ans.

Ans.


Ans.


2.362 - 2.362 = 0 (check!)©Fx = 0;:+

FBD = 4 kN (T)

FBD - 4 = 0+ c ©Fy = 0;

FAB = 2.362 kN = 2.36 kN (T)

FAB - 1.458a45b - 1.196 = 0©Fx = 0;:+

FAD = 1.458 kN = 1.46 kN (C)

0.875 - FAD a35b = 0+ c ©Fy = 0;

FCB = 2.362 kN = 2.36 kN (T)

5.208a45b - 3.608 sin 30° - FCB = 0©Fx = 0;:+

FCD = 5.208 kN = 5.21 kN (C)

3.608 cos 30° - FCD a35b = 0+ c ©Fy = 0;

Ay = 0.875 kN

Ay + 3.608 cos 30° - 4 = 0+ c ©Fy = 0;

Ax = 1.196 kN

3 - 3.608 sin 30° - Ax = 0©Fx = 0;:+

NC = 3.608 kN

NC cos 30°(2 + 2) - 3(1.5) - 4(2) = 0+ ©MA = 0;

u = 30°

A C

B

D

2 m

4 kN

3 kN

2 m

1.5 m

u

Ans:FCD = 5.21 kN (C)FCB = 2.36 kN (T)FAD = 1.46 kN (C)FAB = 2.36 kN (T)FBD = 4 kN (T)

500


*6–16.

SOLUTIONMethod of Joints: In this case, the support reactions are not required fordetermining the member forces.

Joint D:

Ans.

Ans.

Joint C:

Ans.

Ans.

Joint B:

Thus,

Ans.

Joint E:

Ans.

Note: The support reactions and can be determinedd by analyzing Joint Ausing the results obtained above.

AyAx

FEA = 4.62 kN 1C2FEA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0:+ ©Fx = 0;

Ey - 219.238 sin 60°2 = 0 Ey = 16.0 kN+ c ©Fy = 0;

FBE = 9.24 kN 1C2 FBA = 9.24 kN 1T2

F = 9.238 kN

9.238 - 2F cos 60° = 0:+ ©Fx = 0;

FBE = FBA = F

FBE sin 60° - FBA sin 60° = 0+ c ©Fy = 0;

FCB = 9.238 kN 1T2 = 9.24 kN 1T2219.238 cos 60°2 - FCB = 0:+ ©Fx = 0;

FCE = 9.238 kN 1C2 = 9.24 kN 1C2FCE sin 60° - 9.238 sin 60° = 0+ c ©Fy = 0;

FDE = 4.619 kN 1C2 = 4.62 kN 1C2FDE - 9.238 cos 60° = 0:+ ©Fx = 0;

FDC = 9.238 kN 1T2 = 9.24 kN 1T2FDC sin 60° - 8 = 0+ c ©Fy = 0;

Determine the force in each member of the truss. Statewhether the members are in tension or compression. SetP = 8 kN.

60••60••

4 m 4 m

B

E D

C

A

4 m

P

Ans:FDC = 9.24 kN (T)FDE = 4.62 kN (C)FCE = 9.24 kN (C)FCB = 9.24 kN (T)FBA = 9.24 kN (T)FEA = 4.62 kN (C)

518


6–30.

Determine the force in members CD, HI, and CJ of the truss,and state if the members are in tension or compression.

SOLUTION

Method of Sections: The forces in members HI, CH, and CD are exposed by cuttingthe truss into two portions through section b–b on the right portion of the free-bodydiagram, Fig. a. From this free-body diagram, and can be obtained by writingthe moment equations of equilibrium about points H and C, respectively. can beobtained by writing the force equation of equilibrium along the y axis.

a

Ans.

a

Ans.

Ans.FCH = 5625 lb (C)

FCHa45b - 1500 - 1500 = 0 + c ©Fy = 0;

FHI = 6750 lb (T)

FHI(4) - 1500(3) - 1500(6) - 1500(9) = 0 + ©Mc = 0;

FCD = 3375 lb (C)

FCD(4) - 1500(6) - 1500(3) = 0 + ©MH = 0;

FCH

FHIFCD

AB C D E F

GHIJK

4 ft

3 ft 3 ft3 ft3 ft3 ft

1500 lb1500 lb1500 lb1500 lb1500 lb

Ans:FCD = 3375 lb (C)FHI = 6750 lb (T)FCH = 5625 lb (C)

529


6–41.

Determine the force developed in members FE, EB, and BC of the truss and state if these members are in tension or compression.

11 kN

B

A D

C

F E

22 kN

2 m 1.5 m

2 m

2 m

SolutionSupport Reactions. Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A.

a+ΣMA = 0; ND(5.5) - 11(2) - 22(3.5) = 0 ND = 18.0 kN

Method of Sections. Referring to the FBD of the right portion of the truss sectioned through a–a shown in Fig. b, FBC and FFE can be determined directly by writing the moment equations of equilibrium about point E and B, respectively.

a+ΣME = 0; 18.0(2) - FBC(2) = 0 FBC = 18.0 kN (T) Ans.

a+ΣMB = 0; 18.0(3.5) - 22(1.5) - FFE(2) = 0 FFE = 15.0 kN (C) Ans.

Also, FEB can be obtained directly by writing force equation of equilibrium along the y axis

+ cΣFy = 0; FEB a45b + 18.0 - 22 = 0 FEB = 5.00 kN (C) Ans.

Ans:FBC = 18.0 kN (T)FFE = 15.0 kN (C)FEB = 5.00 kN (C)

537


6–49.

SOLUTIONSupport Reactions: Applying the moment equation of equilibrium about point A tothe free - body diagram of the truss, Fig. a,

a

Method of Sections: Using the right portion of the free - body diagram, Fig. b.

a

Ans.

a

Ans.

a

Ans.FHI = 21.11 kN = 21.1 kN (C)

+©MF = 0; 12.67(2) - FHIa35b(2) = 0

FFI = 7.211 kN = 7.21 kN (T)

+©MG = 0; -FFI sin 56.31°(2) + 6(2) = 0

FEF = 12.89 kN = 12.9 kN (T)

+©MI = 0; 12.67(4) - 6(2) - FEF(3) = 0

NG = 12.67 kN

+©MA = 0; NG(2) - 4(2) - 5(4) - 8(8) - 6(10) = 0

Determine the force in members HI, FI, and EF of the truss,and state if the members are in tension or compression.

AB C D FE

G

H

IJ

L

K

6 kN8 kN5 kN4 kN

3 m

2 m 2 m 2 m 2 m 2 m 2 m

Ans:FEF = 12.9 kN (T)FFI = 7.21 kN (T)FHI = 21.1 kN (C)

558


SolutionFree Body Diagram. The assembly will be dismembered into member AC, BD and pulley E. The solution will be very much simplified if one recognizes that member BD is a two force member. The FBD of pulley E and member AC are shown in Fig. a and b respectively.

Equations of Equilibrium. Consider the equilibrium of pulley E, Fig. a,

+ cΣFy = 0; 2T - 12 = 0 T = 6.00 kN

Then, the equilibrium of member AC gives

a+ΣMA = 0; FBD a45b(1.5) + 6(0.3) - 6(3) - 6(3.3) = 0

FBD = 30.0 kN

S+ ΣFx = 0; Ax - 30.0 a35b - 6 = 0 Ax = 24.0 kN Ans.

+ cΣFy = 0; 30.0 a45b - 6 - 6 - Ay = 0 Ay = 12.0 kN Ans.

Thus,

FA = 2Ax2 + Ay

2 = 224.02 + 12.02 = 26.83 kN = 26.8 kN

FB = FBD = 30.0 kN

Dx =35

(30) = 18.0 kN Ans.

Dy =45

(30) = 24.0 kN Ans.

6–66.

Determine the horizontal and vertical components of force at pins A and D.

1.5 m

D

A B

C

E

1.5 m

0.3 m

12 kN

2 m

Ans:Ax = 24.0 kNAy = 12.0 kNDx = 18.0 kNDy = 24.0 kN

562


SolutionFree Body Diagram. The solution will be very much simplified if one realizes that member AB is a two force member. Also, the tension in the cable is equal to the weight of the cylinder and is constant throughout the cable.

Equations of Equilibrium. Consider the equilibrium of member BC by referring to its FBD, Fig. a,

a+ΣMC = 0; FAB a35b(2) + 75(9.81)(0.3) - 75(9.81)(2.8) = 0

FAB = 1532.81 N

a+ΣMB = 0; Cy (2) + 75(9.81)(0.3) - 75(9.81)(0.8) = 0

Cy = 183.94 N = 184 N Ans.

S+ ΣFx = 0; 1532.81a45b - 75(9.81) - Cx = 0

Cx = 490.5 N Ans.

Thus,

FB = FAB = 1532.81 N

Bx =45

(1532.81) = 1226.25 N = 1.23 kN Ans.

By =35

(1532.81) = 919.69 N = 920 kN Ans.

6–70.

Determine the horizontal and vertical components of force at pins B and C. The suspended cylinder has a mass of 75 kg.

A

BC

1.5 m

0.3 m

2 m0.5 m

Ans:Cy = 184 NCx = 490.5 NBx = 1.23 kNBy = 920 kN

569


6–77.

SOLUTIONMember AC:

a

Ans.

Ans.

Member BDE:

a

Ans.

Ans.

Ans.Ey = 1608 lb = 1.61 kip

+ c ©Fy = 0; -500 - 120 a35b + 2180 - Ey = 0

Ex = 96 lb

:+ ©Fx = 0; -Ex + 120a45b = 0

Dy = 2180 lb = 2.18 kip

+ ©Mg = 0; 500 (8) + 120a35b (5) - Dy (2) = 0

Ay = 72 lb

+ c ©Fy = 0; -Ay + 120a35b = 0

Ax = 96 lb

:+ ©Fx = 0; Ax - 120a45b = 0

NC = 120 lb

+ ©MA = 0; NC (5) - 600 = 0

The two-member structure is connected at C by a pin, whichis fixed to BDE and passes through the smooth slot inmember AC. Determine the horizontal and verticalcomponents of reaction at the supports.

3 ft 3 ft 2 ft

4 ft

A

B

C DE

600 lb ft

500 lb

Ans:Ax = 96 lbAy = 72 lbDy = 2.18 kipEx = 96.0 lbEy = 1.61 kip

solution 2 m a - florida international university

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