5–1. Determine the force in each member of the truss, D E Cfaculty.kfupm.edu.sa/ce/saghamdi/hmpage/CE202/S_M… · · 2014-12-21Determine the force in each member of the truss,

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281401

•5–1. Determine the force in each member of the truss,and state if the members are in tension or compression.

© 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.

600 N

900 N

2 m

2 m

2 m

A

CE

D

B

06a Ch06a 401-445.indd 401 6/12/09 8:48:31 AM

Method of Joints: We will begin by analyzing the equilibrium of joint D, and then proceed toanalyze joints C and D.

Joint D: From geometry, u = tan–1 = 26.57°. Thus, from the free-body diagram in Fig. a,

Joint C: From the free - body diagram in Fig. a,

Joint E: From the free - body diagram in Fig. c,

12

FDC = 1341.64 N = 1.34 kN (C) Ans

FDE = 1200 N = 1.20 kN (T) Ans

FFB = 1272.79 N = 1.27 kN (C) Ans

+ c ©Fy = 0; 1341.64 cos 26.57° – FDE = 0

FEA = 2100 N = 2.10 kN (T) Ans+ c ©Fy = 0; 1200 + 1272.79 cos° – FEA = 0

+S ©Fx = 0; 600 – FDC sin 26.57° = 0

+S ©Fx = 0; 900 – FEB sin 45° = 0

©Fx¿ = 0; FCE cos 26.57° = 0 FCE = 0 Ans

+

↑©Fy¿ = 0; FCB 1341.64 = 0

FCB = 1341.64 N = 1.34 kN (C) Ans

+

↑

Note: The equilibrium analysis of joint A can be used to determine the components ofsupport reaction at A.

SM_CH05.indd 281 4/8/11 11:49:44 AM


282402

6–2. The truss, used to support a balcony, is subjected tothe loading shown. Approximate each joint as a pin anddetermine the force in each member. State whether themembers are in tension or compression. Set P2 = 400 lb.

P1 = 600 lb,


45

4 ft 4 ft

45

D

E

CB

P2

A

4 ft

P1

1 m

1 m 1 m

3 kN

2 kN

3 kN

3 kN

4.2426 kN

members are in tension or compression. Set P1 = 3 kN, P2 = 2 kN.

Joint A:

+↑ΣFy = 0; FAD sin 45° – 3 = 0

FAD = 4.2426 = 4.24 kN (C) Ans

+→ ΣFx = 0; FAB – 4.2426 cos 45° = 0

FAB = 3 kN (T) Ans

Joint B:

+↑ΣFy = 0; FBD – 2 = 0

FBD = 2 kN (C) Ans

+→ ΣFx = 0; FBC – 3 = 0

FBC = 3 kN (T) Ans

Joint D:

+↑ΣFy = 0; FDC sin 45° – 2 – 4.2426 sin 45° = 0

FDC = 7.0711 kN = 7.07 kN (T) Ans

+→ ΣFx = 0; 4.2426 cos 45° + 7.0711 cos 45° – FDE = 0

FDE = 8.00 kN (C) Ans

06a Ch06a 401-445.indd 402 6/12/09 8:48:33 AM

5–2.

SM_CH05.indd 282 4/8/11 11:49:44 AM


283403

6–3. The truss, used to support a balcony, is subjected tothe loading shown. Approximate each joint as a pin anddetermine the force in each member. State whether themembers are in tension or compression. Set P2 = 0.

P1 = 800 lb,


45

4 ft 4 ft

45

D

E

CB

P2

A

4 ft

P1

1 m

1 m 1 m

4 kN

4 kN

5.657 kN

members are in tension or compression. Set P1 = 4 kN, P2 = 0.

Joint A:

+↑ΣFy = 0; FAD sin 45° – 4 = 0

FAD = 5.657 kN = 5.66 kN (C) Ans

+→ ΣFx = 0; FAB – 5.657 cos 45° = 0

FAB = 4.00 kN (T) Ans

Joint B:

+↑ΣFy = 0; FBD – 0 = 0

FBD = 0 Ans

+→ ΣFx = 0; FBC – 4 = 0

FBC = 4 kN (T) Ans

Joint D:

+↑ΣFy = 0; FDC sin 45° – 0 – 5.657 sin 45° = 0

FDC = 5.657 kN = 5.66 kN (T) Ans

+→ ΣFx = 0; 5.657 cos 45° + 5.657 cos 45° – FDE = 0


06a Ch06a 401-445.indd 403 6/12/09 8:48:36 AM

5–3.

SM_CH05.indd 283 4/8/11 11:49:44 AM


284404

*6–4. Determine the force in each member of the trussand state if the members are in tension or compression.Assume each joint as a pin. Set P = 4 kN.


A

E

D

CB

PP

2P

4 m

4 m 4 m

06a Ch06a 401-445.indd 404 6/12/09 8:48:38 AM

Joint A

Joint B

Joint E

Joint D

FAE = 8.944 kN (C) = 8.94 kN (C) Ans

+ c ©Fy = 0; FAE – 4 = 0

©Fx¿ = 0; 8.944 + 8.00 sin 26.57° + 8.944 sin 36.87° – FED = 0 + ↑

©Fy¿ = 0; FEC cos 36.87° – 8.00 cos 26.57° = 0 FEC = 8.944 kN (T) = 8.94 kN (T) Ans

FED = 17.89 kN (C) = 17.9 kN (C) Ans

+↑

Note: The support reactions, Cx and Cy can be determined by analysing Joint C using the results obtained above.

Method of Joints: In this case, the support reactions arenot required for determining the member forces.

15

FAB = 8.00 kN (T) Ans

S ©Fx = 0; FAB – 8.944 = 0 +

S ©Fx = 0; FBC – 8.00 = 0 FBC = 8.00 kN (T) Ans

+ c ©Fy = 0; FBE – 8 = 0 FBE = 8.00 kN (C) Ans

+

25

+ c ©Fy = 0; FDC – 17.89 = 0 FDC = 8.00 kN (T) Ans15

S ©Fx = 0; Dx – 17.89 = 0 Dx = 16.0 kN25

+

*5–4.

SM_CH05.indd 284 4/8/11 11:49:45 AM


285405

•6–5. Assume that each member of the truss is made of steelhaving a mass per length of 4 kg/m. Set , determine theforce in each member, and indicate if the members are intension or compression. Neglect the weight of the gusset platesand assume each joint is a pin. Solve the problem by assumingthe weight of each member can be represented as a verticalforce, half of which is applied at the end of each member.

P = 0


A

E

D

CB

PP

2P

4 m

4 m 4 m

06a Ch06a 401-445.indd 405 6/12/09 8:48:39 AM


Joint Forces:

FA = 4(9.81) = 166.22 N

FB = 4(9.81)(2 + 2 + 1) = 196.2 N

202 +2

FD = 4(9.81) = 166.22 N202 +2

FE = 4(9.81) = 302.47 N201 + 32

� �

FAE = 371.69N (C) = 372 N (C) Ans

+ c ©Fy = 0; FAE – 166.22 = 015

FAB = 332.45 N (T) = 322 N (T) Ans

S ©Fx = 0; FAB – 371.69 = 0 + 25

Joint A

Joint B

S ©Fx = 0; FBC – 332.45 = 0 FBC = 332 N (T) Ans

+ c ©Fy = 0; FBE – 196.2 = 0

FBE = 196.2 N (C) = 196 N (C) Ans

+

Joint E

©Fx¿ = 0; 371.69 + (196.2 + 302.47) sin 26.57° +

↑

©Fy¿ = 0; FEC cos 36.87° – (196.2 + 302.47) cos 26.57° = 0 FEC = 557.53 N (T) = 558 N (T) Ans

FED = 929.22 N (C) = 929 N (C) Ans

FDC = 582 N (T) Ans

+↑

+ 557.53 sin 36.87° – FED = 0

Joint D

+ c ©Fy = 0; – 166.22 – 929.22 + FDC = 015

•5–5.

SM_CH05.indd 285 4/8/11 11:49:45 AM


286406

6–6. Determine the force in each member of the truss andstate if the members are in tension or compression. Set

and .P2 = 1.5 kNP1 = 2 kN


A

DE

30 30

B

C

3 m 3 m

P2P1

06a Ch06a 401-445.indd 406 6/12/09 8:48:40 AM


Joint D

Joint C

S ©Fx = 0; FDE – 2.598 = 0 FDE = 2.60 kN (C) Ans

+ c ©Fy = 0; FDB – 2 = 0 FDB = 2.00 kN (T) Ans

+ c ©Fy = 0; FCB sin 30° – 1.5 = 0

FCB = 3.00 kN (T) Ans

+

S ©Fx = 0; FCD – 3.00 cos 30° = 0 Ans

FCD = 2.598 kN (C) = 2.60 kN (C) Ans

+

Joint B

©Fx¿ = 0; (2.00 + 2.00) sin 30° + 3.00 – FBA = 0 +↑

©Fy¿ = 0; FBE cos 30° – 2.00 cos 30° = 0 FBE = 2.00 kN (C) Ans

FBA = 5.00 kN (T) Ans

+↑

Note: The support reactions at support A and E can be determined by analysing Joints A and E respectively using the results obtained above.

5–6.

SM_CH05.indd 286 4/8/11 11:49:45 AM


287407

6–7. Determine the force in each member of the truss andstate if the members are in tension or compression. Set

.P1 = P2 = 4 kN


A

DE

30 30

B

C

3 m 3 m

P2P1

06a Ch06a 401-445.indd 407 6/12/09 8:48:42 AM

Method of Joints: In this case, the support reactions are not required for determining the member forces.

Joint C

Joint D

Joint B

Note: The support reactions at support A and E can be determined by analyzing Joints A and E respectively using the results obtained above.

+c©Fy = 0; FCB sin 30° – 4 = 0 FCB = 8.00 kN (T) Ans

S+ ©Fx = 0; FCD – 8.00 cos 30° = 0 FCD = 6.928 kN (C) = 6.93 kN Ans

S+ ©Fx = 0; FDE – 6.928 = 0 FDE = 6.93 kN (C) Ans

+c©Fy = 0; FDB – 4 = 0 FDB = 4.00 kN (T) Ans

Q+©Fy¿ = 0; FBE cos 30° – 4.00 cos 30° = 0 FBE = 4.00 kN (C) Ans

R+©Fx¿ = 0; (4.00 + 4.00) sin 30° + 8.00 – FBA = 0 FBA = 12.0 kN (T) Ans

5–7.

SM_CH05.indd 287 4/8/11 11:49:46 AM


288408

*6–8. Determine the force in each member of the truss,and state if the members are in tension or compression. Set

.P = 800 lb


3 ft

3 ft

3 ft

P

3 ft

500 lb

ACB

DF E1 m 1 m 1 m

1 m

2.5 kN

2.5 kN

FFE = 10.50 kN

FAF = 5.657 kN

FBC = 17.0 kN

FAB = 4 kN

FFB = 9.192 kN

4 kN

P = 4 kN.

Method of Joints: We will analyze the equilibrium of the joints in the following sequence:

A ⇒ F ⇒ E ⇒ B ⇒ C.

Joint A: From the free – body diagram in Fig. a,

+↑ΣFy = 0; FAF sin 45° – 4 = 0

FAF = 5.657 kN (T) Ans

+→ ΣFx = 0; 5.657 cos 45° – FAB = 0

FAB = 4 kN (C) Ans

Joint F: From the free – body diagram in Fig. b,

+↑ΣFy = 0; FFB cos 45° – 5.657 cos 45° – 2.5 = 0

FFB = 9.192 kN (C) Ans

+→ ΣFx = 0; FFE – 9.192 sin 45° – 5.657 sin 45° = 0

FFE = 10.50 kN (T) Ans

Joint E: From the free – body diagram in Fig. c,

+→ ΣFx = 0; FED – 10.50 = 0

FED = 10.50 kN (T) Ans

+↑ΣFy = 0; FEB = 0 Ans

Joint B: From the free – body diagram in Fig. d,

+↑ΣFy = 0; FBD sin 45° – 9.192 sin 45° = 0

FBD = 9.192 kN (T) Ans

+→ ΣFx = 0; 4 + 9.192 cos 45° + 9.192 cos 45° – FBC = 0

FBC = 17.0 kN (C) Ans

Joint C: From the free – body diagram in Fig. e,

+↑ΣFy = 0; FCD = 0 Ans

+→ ΣFx = 0; 17.0 – NC = 0

NC = 17.0 kN Ans

06a Ch06a 401-445.indd 408 6/12/09 8:48:42 AM

*5–8.

SM_CH05.indd 288 4/8/11 11:49:46 AM


289409


•6–9. Remove the 500-lb force and then determine thegreatest force P that can be applied to the truss so that noneof the members are subjected to a force exceeding either800 lb in tension or in compression.600 lb

3 ft

3 ft

3 ft

P

3 ft

500 lb

ACB

DF E1 m 1 m 1 m

1 m

2.5 kN•6–9. Remove the 2.5-kN force and then determine the greatest force P that can be applied to the truss so that none of the members are subjected to a force exceeding either 4 kN in tension or 3 kN in compression.

2P = 4 kN P = 2 kN3P = 3 kN P = 1 kN (controls) Ans

06a Ch06a 401-445.indd 409 6/12/09 8:48:44 AM

Method of Joints : We will analyze the equilibrium of the joints in the following sequence: A S F S E S B S CJoint A: From the free-body diagram in Fig. a,

+c©Fy = 0; FAF sin 45° – P = 0

FAF = 1.4142P (T)

S+ ©Fx = 0; 1.4142P cos 45° – FAB = 0

FAB = P (C)

Joint F: From the free-body diagram in Fig. b,

+c©Fy = 0; FFB cos 45° – 1.4142P cos 45° = 0

FFB = 1.4142P (C)

S+ ©Fx = 0; FFE – 1.4142P sin 45° – 1.4142P sin 45° = 0

FFE = 2P (T)

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

+c©Fy = 0; FBD sin 45° – 1.4142P sin 45° = 0

FBD = 1.4142P (T)

S+ ©Fx = 0; P + 1.4142P cos 45° + 1.4142P cos 45° – FBC = 0

FBC = 3P (C)

Joint E: From the free-body diagram in Fig. c,

S+ ©Fx = 0; FED – 2P = 0

FED = 2P (T)

+c©Fy = 0; FEB = 0

Joint C: From the free-body diagram in Fig. e,

S+ ©Fx = 0; 3P – NC = 0

NC = 3P

+c©Fy = 0; FCD = 0

From the above results, the greatest compressive and tensile forces developed in the member are 3Pand 2P, respectively.

•5–9.

SM_CH05.indd 289 4/8/11 11:49:46 AM


290410

6–10. Determine the force in each member of the trussand state if the members are in tension or compression. Set

, .P2 = 0P1 = 800 lb


6 ft

AG

B

C

F

D

E

P1

P2

4 ft 4 ft 4 ft 4 ft1 m 1 m 1 m

1.5 m

1 m

4 kN

P1 = 4 kN, P2 = 0.

06a Ch06a 401-445.indd 410 6/12/09 8:48:45 AM

Joint B:

R+©Fy = 0; FBG = 0

+b©Fx = 0; FBA = FBC

Joint G:

+c©Fy = 0; FCG sin u = 0

FCG = 0

S+ ©Fx = 0; FAG = 0

Joint C:

S+ ©Fx = 0; FBC – FCD = 0 45

45

+c©Fy = 0; (FBC) + (FCD) – 4 = 0

FBC = FCD = 3.33 kN (C)

35

35

Due to symmetry:

FDF = FBG = 0 Ans

FCF = FCG = 0 Ans

FgF = FAG = 0 Ans

FAB = FDg = 3.33 kN (C) Ans

FBC = FCD = 3.33 kN (C) Ans

5–10.

SM_CH05.indd 290 4/8/11 11:49:47 AM


291411


, .P2 = 400 lbP1 = 600 lb


6 ft

AG

B

C

F

D

E

P1

P2

4 ft 4 ft 4 ft 4 ft1 m 1 m 1 m

1.5 m

1 m

3 kN

2 kN

FDF = 2 kN

FFC = 1.67 kN

P1 = 3 kN, P2 = 2 kN.

Joint B:

+↑

ΣFy = 0; FBG = 0 Ans

+↑ΣFy = 0; FBC = FBA

Joint G:

+↑ΣFy = 0; FGC sin = 0

FGC = 0 Ans

+→ ΣFx = 0; FGA = 0 Ans

Joint D:

+ ↑ΣFx = 0; FDE – FDC = 0

FDE = FDC

+↑ ΣFy = 0; FDF – 2 = 0

FDF = 2 kN (C) Ans

Joint F:

+↑

ΣFx = 0; FFE sin 53.13° – FFC sin 53.13° = 0

FFE = FFC

+↑ΣFy = 0; 2 F cos 53.13° – 2 = 0

FPC = FFE = 1.6667 = 1.67 kN (T) Ans

Joint C:

+↑ΣFx = 0; FBC cos 36.87° – FDC cos 36.87° + 1.6667 cos 73.74° = 0

+→ ΣFy = 0; FBC sin 36.87° + FDC sin 36.87° – 3 – 1.6667 sin 73.74° = 0

FBC = FBA = 3.54 kN (C) Ans

FDC = FDE = 4.13 kN (C) Ans

06a Ch06a 401-445.indd 411 6/12/09 8:48:47 AM

5–11.

SM_CH05.indd 291 4/8/11 11:49:47 AM


292412

*6–12. Determine the force in each member of the trussand state if the members are in tension or compression. Set

, .P2 = 100 lbP1 = 240 lb


B

C D

A

12 ft

5 ft

P1

P2

1200 N

500 N

FBD = 1300 N

1.5 m

3.6 m

P1 = 1200 N, P2 = 500 N.

Joint D:

+↑ΣFy = 0; FBD 5

13

– 500 = 0

FBD = 1300 N (C) Ans

+→ ΣFx = 0; 1200 – FCD + 1300

12

13

= 0

FCD = 2400 N (T) Ans

Joint A:

+↑ΣFy = 0; FAC = 0 Ans

+→ ΣFy = 0; FAB = 0 Ans

Joint B:

+↑ΣFy = 0; FBC – 1300 5

13

= 0

FBC = 500 N (T) Ans

06a Ch06a 401-445.indd 412 6/12/09 8:48:48 AM

*5–12.

SM_CH05.indd 292 4/8/11 11:49:47 AM


293413

•6–13. Determine the largest load that can be appliedto the truss so that the force in any member does not exceed500 lb (T) or 350 lb (C). Take .P1 = 0

P2


B

C D

A

12 ft

5 ft

P1

P21.5 m

3.6 m

2.5 kN (T) or 1.75 kN (C). Take P1 = 0.

Maximum tension member is DC :

2.5 = 2.40 P2

P2 = 1.04 kN

Maximum compression member is DB :

1.75 = 2.60 P2

P2 = 0.673 kN

The member DB reaches the critical value first.

P2 = 0.673 kN Ans

06a Ch06a 401-445.indd 413 6/12/09 8:48:50 AM

Joint A:

+c©Fy = 0; FAC sin u = 0

FAC = 0

S+ ©Fx = 0; FAB = 0

Joint D:

+c©Fy = 0; –P2 + FDB sin 22.62° = 0 FDB = 2.60 P2 (C)

S+ ©Fx = 0; 2.60 P2 cos 22.62° – FDC = 0 FDC = 2.40 P2 (T)

Joint B:

+c©Fy = 0; FBC – 2.60 P2 sin 22.62° = 0 FBC = P2 (T)

•5–13.

SM_CH05.indd 293 4/8/11 11:49:47 AM


294414

6–14. Determine the force in each member of the truss,and state if the members are in tension or compression. Set

.P = 2500 lb


4 ft

4 ft

bl 0021bl 0021 P

4 ft 4 ft 4 ft

A B

F

E D C

G

30 30

1 m

1 m 1 m 1 m 1 m

6 kN 6 kN

NB = 12.25 kN FBC = 12.25 kN

FCG = 8.8389 kN

6 kN6 kN 6 kN12.5 kN

2 m 2 m

P = 12.5 kN.

Support Reactions: Applying the moment equation of equilibrium about point A to the free – body diagram of the truss, Fig. a,

+ ΣMA = 0; NB (2 + 2) – 6 (2 + 2) – 12.5 (2) = 0

NB = 12.25 kN

Method of Joints: We will begin by analyzing the equilibrium of joint B, and then that of joints C and G.

Joint B: From the free – body diagram in Fig. b,

+→ ΣFx = 0; FBG = 0 Ans

+↑ΣFy = 0; 12.25 – FBC = 0


Joint C: From the free – body diagram in Fig. c,

+↑ΣFy = 0; 12.25 – 6 – FCG sin 45° = 0

FCG = 8.8389 kN = 8.84 kN (T) Ans

+→ ΣFx = 0; FCD – 8.8389 cos 45° = 0

FCD = 6.25 kN (C) Ans

Joint G: From the free – body diagram in Fig. d,

+↑ΣFy = 0; 8.8389 cos 45° – FGD cos 45° = 0

FGD = 8.8389 kN = 8.84 kN (C) Ans

+→ ΣFx = 0; 8.8389 sin 45° + 8.8389 sin 45° – FGF = 0

FGF = 12.5 kN (T) Ans

Due to the symmetry of the system and the loading,

FAE = FBC = 12.25 kN Ans

FAF = FBG = 0 Ans

FED = FCD = 6.25 kN (C) Ans

FEF = FCG = 8.8389 kN = 8.84 kN (T) Ans

FFD = FGD = 8.8389 kN = 8.84 kN (C) Ans

06a Ch06a 401-445.indd 414 6/12/09 8:48:50 AM

5–14.

SM_CH05.indd 294 4/8/11 11:49:48 AM


295415


6–15. Remove the 1200-lb forces and determine thegreatest force P that can be applied to the truss so that noneof the members are subjected to a force exceeding either2000 lb in tension or 1500 lb in compression.

4 ft

4 ft

bl 0021bl 0021 P

4 ft 4 ft 4 ft

A B

F

E D C

G

30 30

1 m

1 m 1 m 1 m 1 m

6 kN 6 kN

2 m 2 m

6–15. Remove the 6-kN forces and determine the greatest force P that can be applied to the truss so that none of the members are subjected to a force exceeding either 10 kN in tension or 7.5 kN in compression.

Support Reactions: Applying the moment equation of equilibrium about point A to the free – body diagram of the truss, Fig. a,

+ ΣMA = 0; NB (2 + 2) – P (2) = 0

NB = 0.5 P

0.7071P = 7.5 P = 10.6067 kNP = 10 kN (controls) Ans

06a Ch06a 401-445.indd 415 6/12/09 8:48:53 AM

Method of Joints: We will begin by analyzing the equilibrium of joint B, and then that of joints C and G.

Joint B: From the free-body diagram in Fig. b,

S+ ©Fx = 0; FBG = 0

+c©Fy = 0; 0.5P – FBC = 0 FBC = 0.5P (C)

Joint C: From the free-body diagram in Fig. c,

+c©Fy = 0; 0.5P – FCG sin 45° = 0 FCG = 0.7071P (T)

S+ ©Fx = 0; FCD = 0.7071P cos 45° = 0 FCD = 0.5P (C)

Joint G: From the free-body diagram in Fig. d,

+c©Fy = 0; 0.7071P cos 45° – FGD cos 45° = 0 FGD = 0.7071P (C)

S+ ©Fx = 0; 0.7071P sin 45° + 0.7071P sin 45° – FGF = 0 FGF = P (T)

Due to the symmetry of the system and the loading,FAE = FBC = 0.5P (C) FAF = FBG = 0 FED = FCD = 0.5P (C) FEF = FCG = 0.7071P (T) FFD = FGD = 0.7071P (C)

From the above results, the greatest tensile and compressive forces developed in the member of the truss are P and 0.7071P, respectively. Thus,

5–15.

SM_CH05.indd 295 4/8/11 11:49:49 AM


296424



.P = 4 kN

P

3 m

A CB

E

D

F

P

3 m 3 m3 m

06a Ch06a 401-445.indd 424 6/12/09 8:49:07 AM

Method of Joints: We will analyze the equilibrium of the joints in the following sequence: A S D S F S E S CJoint A: From the free-body diagram in Fig. a,

+c©Fy = 0; FAF sin 45° – 4 = 0 FAF = 5.657 kN = 5.66 kN (T) Ans

S+ ©Fx = 0; 5.657 cos 45° – FAB = 0 FAB = 4 kN (C) AnsJoint D: From the free-body diagram in Fig. b,+c©Fy = 0; FDE sin 45° – 4 = 0 FDE = 5.657 kN = 5.66 kN (T) Ans

S+ ©Fx = 0; FDC – 5.657 cos 45° = 0 FDC = 4 kN (C) Ans

Joint F: From the free-body diagram in Fig. c,+c©Fy = 0; FFB – 5.657 cos 45° = 0 FFB = 4 kN (C) Ans

S+ ©Fx = 0; FFE – 5.657 sin 45° = 0 FFE = 4 kN (T) AnsJoint E: From the free-body diagram in Fig. d,

S+ ©Fx = 0; 5.657 sin 45° – 4 – FEB sin 45° = 0 FEB = 0 Ans+c©Fy = 0; FBC – 5.657 cos 45° = 0 FBC = 4 kN (C) Ans

Joint C: From the free-body diagram in Fig. e,

S+ ©Fx = 0; FCB – 4 = 0 FCB = 4 kN (C) Ans+c©Fy = 0; NC – 4 = 0 NC = 4 kN Ans

*5–16.

SM_CH05.indd 296 4/8/11 11:49:49 AM


297425


•6–25. Determine the greatest force P that can be appliedto the truss so that none of the members are subjected to aforce exceeding either in tension or incompression.

1 kN1.5 kN

P

3 m

A CB

E

D

F

P

3 m 3 m3 m

06a Ch06a 401-445.indd 425 6/12/09 8:49:09 AM

Method of Joints: We will analyze the equilibrium of the joints in the following sequence: A S D S F S E S CJoint A: From the free-body diagram in Fig. a,

+c©Fy = 0; FAF sin 45° – P = 0 FAF = 1.4142P (T)

S+ ©Fx = 0; 1.4142P cos 45° – FAB = 0 FAB = P (C)

Joint D: From the free-body diagram in Fig. b,+c©Fy = 0; FDE sin 45° – P = 0 FDE = 1.4142P (T)

S+ ©Fx = 0; FDC – 1.4142P cos 45° = 0 FDC = P (C)

Joint F: From the free-body diagram in Fig. c,

S+ ©Fx = 0; FFE – 1.4142P sin 45° = 0 FFE = P (T)+c©Fy = 0; FFB – 1.4142P cos 45° = 0 FFB = P (C)

Joint E: From the free-body diagram in Fig. d,

S+ ©Fx = 0; 1.4142P sin 45° – P – FEB sin 45° = 0 FEB = 0+c©Fy = 0; FBC – 1.4142P cos 45° = 0 FBC = P (C)

Joint C: From the free-body diagram in Fig. e, FCB = P (C)

From the above results, the greatest compressive and tensile forces developed in the member are P and 1.4142P, respectively. Thus,P = 1 kN (controls) Ans1.4142P = 1.5 P = 1.06 kN

•5–17.

SM_CH05.indd 297 4/8/11 11:49:49 AM


298426


6–26. A sign is subjected to a wind loading that exertshorizontal forces of 300 lb on joints B and C of one of theside supporting trusses. Determine the force in eachmember of the truss and state if the members are in tensionor compression.

A

C

B

D

E

13 ft

13 ft

12 ft

5 ft

300 lb

300 lb

12 ft

45

3.9 m

3.6 m

1.5 kN

1.5 kN

1.5 m

3.6 m

3.9 m

1.5 kN

1.5 kN

FCD = 3.90 kN

FCB = 3.60 kN

horizontal forces 1.5 kN on joints B and C of one of the

Joint C:

+→ ΣFx = 0; 1.5 – FCD sin 22.62° = 0


+↑ΣFy = 0; –FCB + 3.90 cos 22.62° = 0


Joint D:

+ ↓ΣFx = 0; FDB = 0 Ans

+ ↓ ΣFy = 0; 3.90 – FDE = 0


Joint B:

+→ ΣFx = 0; 1.5 – FBA cos 45° + FBE sin 45.24° = 0

+↑ΣFy = 0; 3.60 – FBA sin 45° – FBE cos 45.24° = 0

FBE = 1.485 kN (T) Ans


06a Ch06a 401-445.indd 426 6/12/09 8:49:10 AM

5–18.

SM_CH05.indd 298 4/8/11 11:49:50 AM


299426


6–26. A sign is subjected to a wind loading that exertshorizontal forces of 300 lb on joints B and C of one of theside supporting trusses. Determine the force in eachmember of the truss and state if the members are in tensionor compression.

A

C

B

D

E

13 ft

13 ft

12 ft

5 ft

300 lb

300 lb

12 ft

45

3.9 m

3.6 m

1.5 kN

1.5 kN

1.5 m

3.6 m

3.9 m

1.5 kN

1.5 kN

FCD = 3.90 kN

FCB = 3.60 kN

horizontal forces 1.5 kN on joints B and C of one of the

Joint C:

+→ ΣFx = 0; 1.5 – FCD sin 22.62° = 0


+↑ΣFy = 0; –FCB + 3.90 cos 22.62° = 0


Joint D:

+ ↓ΣFx = 0; FDB = 0 Ans

+ ↓ ΣFy = 0; 3.90 – FDE = 0


Joint B:

+→ ΣFx = 0; 1.5 – FBA cos 45° + FBE sin 45.24° = 0

+↑ΣFy = 0; 3.60 – FBA sin 45° – FBE cos 45.24° = 0

FBE = 1.485 kN (T) Ans


06a Ch06a 401-445.indd 426 6/12/09 8:49:10 AM

427


6–27. Determine the force in each member of the doublescissors truss in terms of the load P and state if the membersare in tension or compression.

ADFE

P P

B C

L/3

L/3L/3L/3

06a Ch06a 401-445.indd 427 6/12/09 8:49:12 AM

+©MA = 0; P + P – (Dy)(L) = 0

Dy = P

+c©Fy = 0; Ay = P

3L

3

2L

Joint F:

Joint A:

+c©Fy = 0; FFB – P = 0

FFB = 2P = 1.41P (T)

2

1

S+ ©Fx = 0; FCA – 2P – FCD = 05

2

2

1

2

1

FCA – FCD = P5

2

2

1

S+ ©Fx = 0; FFD – FFE – FFB = 0

FFD – FFE = P (1)

2

1

FCA = P = 1.4907P = 1.49P (C)52

3

FCD = P = 0.4714P = 0.471P (C)2

3

S+ ©Fx = 0; FAE – P – P = 02

3 2

1

5

2

52

3

FAE = P = 1.67 P (T)53

From Eqs. (1) and (2) :

FFE = 0.667 P (T) Ans

FFD = 1.67 P (T) Ans

FAB = 0.471 P (C) Ans

FAE = 1.67 P (T) Ans

FAC = 1.49 P (C) Ans

FBF = 1.41 P (T) Ans

FBD = 1.49 P (C) Ans

FEC = 1.41 P (T) Ans

FCD = 0.471 P (C) Ans

5–19.

SM_CH05.indd 299 4/8/11 11:49:50 AM


300428


*6–28. Determine the force in each member of the truss interms of the load P, and indicate whether the members arein tension or compression.

A

B

C

DF

E

P

d

d

d d/2 d/2 d

06a Ch06a 401-445.indd 428 6/12/09 8:49:13 AM

32

d 43

Support Reactions:

+©Mg = 0; P (2d) – Ay = 0 Ay = P

+c©Fy = 0; P – Ey = 0 Ey = P

S+ ©Fx = 0; Ex – P = 0 Ex = P

43

43

Method of Joints: By inspection of joint C, members CB and CD are zero force member. Hence

FCB = FCD = 0 Ans

Joint A

Joint B

Joint F

Joint D

Solving Eqs. [1] and [2] yield,

43

+c©Fy = 0; FAB – P = 0

FAB = 2.404P (C) = 2.40P (C) Ans

3.25

1

S+ ©Fx = 0; 2.404P – P

–FBF – FBD = 0

1.00P – 0.4472 FBF – 0.4472 FBD = 0 [1]

3.25

1.5

1.25

0.5

1.25

0.5

+c©Fy = 0; 2.404P + FBD – FBF = 0

1.333P + 0.8944 FBD – 0.8944 FBF = 0 [2]

3.25

1

1.25

1

1.25

1

+c©Fy = 0; 1.863P – FFE = 0

FFE = 1.863P (T) = 1.86P (T) Ans

1.25

1

1.25

1

+c©Fy = 0; FDE – 0.3727P = 0

FDE = 0.3727P (C) = 0.373P (C) Ans

1.25

1

1.25

1

FBF = 1.863P (T) = 1.86P (T) Ans

FBD = 0.3727P (C) = 0.373P (C) Ans

S+ ©Fx = 0; FAF – 2.404P = 0

FAF = 2.00P (T) Ans

3.25

1.5

S+ ©Fx = 0; FFD + 2 1.863P – 2.00P = 0

FFD = 0.3333P (T) = 0.333P (T) Ans

1.25

0.5

S+ ©Fy = 0; 2 0.3727P – 0.3333P = 0 (check!)1.25

0.5

*5–20.

SM_CH05.indd 300 4/8/11 11:49:51 AM


301428


*6–28. Determine the force in each member of the truss interms of the load P, and indicate whether the members arein tension or compression.

A

B

C

DF

E

P

d

d

d d/2 d/2 d

06a Ch06a 401-445.indd 428 6/12/09 8:49:13 AM

32

d 43

Support Reactions:

+©Mg = 0; P (2d) – Ay = 0 Ay = P

+c©Fy = 0; P – Ey = 0 Ey = P

S+ ©Fx = 0; Ex – P = 0 Ex = P

43

43


FCB = FCD = 0 Ans

Joint A

Joint B

Joint F

Joint D


43

+c©Fy = 0; FAB – P = 0

FAB = 2.404P (C) = 2.40P (C) Ans

3.25

1

S+ ©Fx = 0; 2.404P – P

–FBF – FBD = 0

1.00P – 0.4472 FBF – 0.4472 FBD = 0 [1]

3.25

1.5

1.25

0.5

1.25

0.5

+c©Fy = 0; 2.404P + FBD – FBF = 0

1.333P + 0.8944 FBD – 0.8944 FBF = 0 [2]

3.25

1

1.25

1

1.25

1

+c©Fy = 0; 1.863P – FFE = 0

FFE = 1.863P (T) = 1.86P (T) Ans

1.25

1

1.25

1

+c©Fy = 0; FDE – 0.3727P = 0

FDE = 0.3727P (C) = 0.373P (C) Ans

1.25

1

1.25

1

FBF = 1.863P (T) = 1.86P (T) Ans

FBD = 0.3727P (C) = 0.373P (C) Ans

S+ ©Fx = 0; FAF – 2.404P = 0

FAF = 2.00P (T) Ans

3.25

1.5

S+ ©Fx = 0; FFD + 2 1.863P – 2.00P = 0

FFD = 0.3333P (T) = 0.333P (T) Ans

1.25

0.5

S+ ©Fy = 0; 2 0.3727P – 0.3333P = 0 (check!)1.25

0.5

429


•6–29. If the maximum force that any member can supportis 4 kN in tension and 3 kN in compression, determine themaximum force P that can be applied at joint B. Take

.d = 1 m

A

B

C

DF

E

P

d

d

d d/2 d/2 d

06a Ch06a 401-445.indd 429 6/12/09 8:49:14 AM

32

d 43

Support Reactions:

+©ME = 0; P (2d) – Ay = 0 Ay = P

+c©Fy = 0; P – Ey = 0 Ey = P

S+ ©Fx = 0; Ex – P = 0 Ex = P

43

43


FCB = FCD = 0

Joint A

Joint B

Joint F

Joint D

S+ ©Fx = 0; FAF – 2.404P = 0 FAF = 2.00P (T)3.25

1.5

+c©Fy = 0; –FAB + P = 0 FAB = 2.404P (C)3.25

1

43

S+ ©Fx = 0; 2.404P – P

–FBF – FBD = 0

1.00P – 0.4472 FBF – 0.4472 FBD = 0 [1]

3.25

1.5

1.25

0.5

1.25

0.5

+c©Fy = 0; 2.404P + FBD – FBF = 0

1.333P + 0.8944 FBD – 0.8944 FBF = 0 [2]

3.25

1

1.25

1

1.25

1


FBF = 1.863P (T) FBD = 0.3727P (C)

+c©Fy = 0; 1.863P – FFE = 0

FFE = 1.863P (T)

1.25

1

1.25

1

S+ ©Fx = 0; FFD + 2 1.863P – 2.00P = 0

FFD = 0.333P (T)

1.25

0.5

+c©Fy = 0; FDE – 0.3727P = 0

FDE = 0.3727P (C)

1.25

1

1.25

1

S+ ©Fy = 0; 2 0.3727P – 0.3333P = 0 (check!)1.25

0.5

From the above analysis, the maximum compression and tension in the truss members are 2.404P and 2.00P, respectively. For this case, compression controls which requires

2.404P = 3P = 1.25 kN

•5–21.

SM_CH05.indd 301 4/8/11 11:49:51 AM


302430


6–30. The two-member truss is subjected to the force of300 lb. Determine the range of for application of the load sothat the force in either member does not exceed 400 lb (T) or200 lb (C).

uB

CA

4 ft

3 ft

300 lb

u

0.9 m

1.5 kN

1.2 m

1.5 kN

1.5 kN

6–30. The two-member truss is subjected to the force of 1.5 kN. Determine the range of for application of the load so that the force in either member does not exceed 2 kN (T) or 1 kN (C).

Joint A:

+→ ΣFx = 0; 1.5 cos + FAC + FAB

4

5

= 0

+↑ΣFy = 0; –1.5 sin + FAB 3

5

= 0

Thus,

FAB = 2.5 sin

FAC = –1.5 cos – 2 sin

For AB required:

–1 ≤ 2.5 sin ≤ 2

–2 ≤ 5 sin ≤ )1( 4

For AC required:

–1 ≤ –1.5 cos – 2 sin ≤ 2

–4 ≤ 3 cos + 4 sin ≤ 2 (2)

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

127° ≤ – 196° Ans

336° ≤ – 347° Ans

Possible hand solution:

2 = 1 + tan–1 3

4

= 1 + 36.870

Then

FAB = 2.5 sin 1

FAC = –1.5 cos ( 2 – 36.870°) – 2 sin ( 2 – 36.870°)

= –1.5 [cos 2 cos 36.870° + sin 2 sin 36.870°]

– 2 [sin 2 cos 36.870° – cos 2 sin 36.870°]

= –1.20 cos 2 – 0.90 sin 2 – 1.60 sin 2 + 1.20 cos 2

= –2.50 sin 2

Thus, we require

–2 ≤ 5 sin 1 ≤ 4 or –0.4 ≤ sin 1 ≤ 0.8 (1)

–4 ≤ 5 sin 2 ≤ 2 or –0.8 ≤ sin 2 ≤ 0.4 (2)

Since 1 = 2 – 36.870°, the range of acceptable values for = 1 is

127° ≤ ≤ 196° Ans

336° ≤ ≤ 347° Ans

06a Ch06a 401-445.indd 430 6/12/09 8:49:15 AM

5–22.

SM_CH05.indd 302 4/8/11 11:49:52 AM


303430


6–30. The two-member truss is subjected to the force of300 lb. Determine the range of for application of the load sothat the force in either member does not exceed 400 lb (T) or200 lb (C).

uB

CA

4 ft

3 ft

300 lb

u

0.9 m

1.5 kN

1.2 m

1.5 kN

1.5 kN

6–30. The two-member truss is subjected to the force of 1.5 kN. Determine the range of for application of the load so that the force in either member does not exceed 2 kN (T) or 1 kN (C).

Joint A:

+→ ΣFx = 0; 1.5 cos + FAC + FAB

4

5

= 0

+↑ΣFy = 0; –1.5 sin + FAB 3

5

= 0

Thus,

FAB = 2.5 sin

FAC = –1.5 cos – 2 sin

For AB required:

–1 ≤ 2.5 sin ≤ 2

–2 ≤ 5 sin ≤ )1( 4

For AC required:

–1 ≤ –1.5 cos – 2 sin ≤ 2

–4 ≤ 3 cos + 4 sin ≤ 2 (2)

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

127° ≤ – 196° Ans

336° ≤ – 347° Ans

Possible hand solution:

2 = 1 + tan–1 3

4

= 1 + 36.870

Then

FAB = 2.5 sin 1

FAC = –1.5 cos ( 2 – 36.870°) – 2 sin ( 2 – 36.870°)

= –1.5 [cos 2 cos 36.870° + sin 2 sin 36.870°]

– 2 [sin 2 cos 36.870° – cos 2 sin 36.870°]

= –1.20 cos 2 – 0.90 sin 2 – 1.60 sin 2 + 1.20 cos 2

= –2.50 sin 2

Thus, we require

–2 ≤ 5 sin 1 ≤ 4 or –0.4 ≤ sin 1 ≤ 0.8 (1)

–4 ≤ 5 sin 2 ≤ 2 or –0.8 ≤ sin 2 ≤ 0.4 (2)

Since 1 = 2 – 36.870°, the range of acceptable values for = 1 is

127° ≤ ≤ 196° Ans

336° ≤ ≤ 347° Ans

06a Ch06a 401-445.indd 430 6/12/09 8:49:15 AM

431


6–31. The internal drag truss for the wing of a lightairplane is subjected to the forces shown. Determine theforce in members BC, BH, and HC, and state if themembers are in tension or compression.

2 ft

A B C D

J I H G

E

F

2 ft 2 ft 2 ft 1.5 ft

80 lb 80 lb60 lb

40 lb

0.8 m

0.8 m 0.8 m 0.8 m 0.6 m

0.8 m

0.8 m 0.8 m 0.6 m

0.8 m

0.8 m 0.6 m

400 N 400 N300 N

200 N

400 N 300 N 200 N

400 N 300 N 200 N

+↑ΣFy = 0; 900 – FBH sin 45° = 0

FBH = 1273 N (T) Ans

+ ΣMB = 0; –FBC (0.8) + 300 (0.8) + 200 (1.4) = 0

FBC = 650 N (T) Ans

Section 2:

+↑ΣFy = 0; 400 + 300 + 200 – FHC = 0

FHC = 900 N (C) Ans

06a Ch06a 401-445.indd 431 6/12/09 8:49:16 AM

5–23.

SM_CH05.indd 303 4/8/11 11:49:52 AM


304432


*6–32. The Howe bridge truss is subjected to the loadingshown. Determine the force in members HD, CD, and GD,and state if the members are in tension or compression.

EAB C D

IJ

30 kN20 kN 20 kN

40 kN

H G F

4 m

16 m, 4@4m

06a Ch06a 401-445.indd 432 6/12/09 8:49:17 AM

Support Reactions:

+©MA = 0; Ey (16) – 40(12) – 20(8) – 20(4) = 0

Ey = 45.0 kN

Method of Sections:

+©MH = 0; 45.0(8) – 40(4) – FCD (4) = 0

FCD = 50.0 kN (T) Ans

+c©Fy = 0; 45.0 – 40 – FHD sin 45° = 0

FHD = 7.071 kN (C) = 7.07 kN (C) Ans

Method of Joints: Analysing joint D, we have

+c©Fy = 0; FGD – 7.071 sin 45° = 0

FGD = 5.00 kN (T) Ans

*5–24.

SM_CH05.indd 304 4/8/11 11:49:52 AM


305433


•6–33. The Howe bridge truss is subjected to the loadingshown. Determine the force in members HI, HB, and BC,and state if the members are in tension or compression.

EAB C D

IJ

30 kN20 kN 20 kN

40 kN

H G F

4 m

16 m, 4@4m

06a Ch06a 401-445.indd 433 6/12/09 8:49:18 AM

Support Reactions:

+©ME = 0; 30(16) + 20(12) + 20(8) + 40(4) – Ay (16) = 0

Ay = 65.0 kN

S+ ©Fx = 0; Ax = 0

Method of Sections:

+©MH = 0; FBC (4) + 20(4) + 30(8) – 65.0(8) = 0

FBC = 50.0 kN (T) Ans

+©MB = 0; FHI (4) + 30(4) – 65.0(4) = 0

FHI = 35.0 kN (C) Ans

+c©Fy = 0; 65.0 – 30 – 20 – FHB sin 45° = 0

FHB = 21.2 kN (C) Ans

•5–25.

SM_CH05.indd 305 4/8/11 11:49:53 AM


306434


6–34. Determine the force in members JK, CJ, and CD ofthe truss, and state if the members are in tension orcompression.

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

06a Ch06a 401-445.indd 434 6/12/09 8:49:19 AM

Method of Joints: Applying the equations of equilibrium to the free-body diagram of the truss, Fig. a,

S+ ©Fx = 0; Ax = 0

+©MG = 0; 6(2) + 8(4) + 5(8) + 4(10) – Ay (12) = 0

Ay = 10.33 kN

Method of Sections: Using the left portion of the free-body diagram, Fig. a,

+©MC = 0; FJK (3) + 4(2) – 10.33(4) = 0

FJK = 11.111 kN = 11.1 kN (C) Ans

+©MJ = 0; FCD (3) + 5(2) + 4(4) – 10.33(6) = 0

FCD = 12 kN (T) Ans

+c©Fy = 0; 10.33 – 4 – 5 – FCJ sin 56.31° = 0

FCJ = 1.602 kN = 1.60 kN (C) Ans

5–26.

SM_CH05.indd 306 4/8/11 11:49:53 AM


307435


6–35. Determine the force in members HI, FI, and EF ofthe truss, and state if the members are in tension orcompression.

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

06a Ch06a 401-445.indd 435 6/12/09 8:49:21 AM

Support Reactions: Applying the moment equations of equilibrium about point A to the free-body diagram of the truss, Fig. a,

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

NG = 12.67 kN

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

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

FEF = 12.89 kN = 12.9 kN (T) Ans

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

FFI = 7.211 kN = 7.21 kN (T) Ans

+©MF = 0; 12.67(2) – FHI (2) = 0

FHI = 21.11 kN = 21.1 kN (C) Ans

35

5–27.

SM_CH05.indd 307 4/8/11 11:49:54 AM


308436


*6–36. Determine the force in members BC, CG, and GFof the Warren truss. Indicate if the members are in tensionor compression.

A E

B C D

6 kN8 kN

G F

3 m

3 m 3 m 3 m

3 m

3 m3 m

•6–37. Determine the force in members CD, CF, and FGof the Warren truss. Indicate if the members are in tensionor compression.

A E

B C D

6 kN8 kN

G F

3 m

3 m 3 m 3 m

3 m

3 m3 m

06a Ch06a 401-445.indd 436 6/12/09 8:49:23 AM

Support Reactions:

+©ME = 0; 6(6) + 8(3) – Ay (9) = 0 Ay = 6.667 kN

S+ ©Fx = 0; Ax = 0

Method of Sections:

+©MC = 0; FGF (3 sin 60°) + 6(1.5) – 6.667(4.5) = 0


+©MG = 0; FBC (3 sin 60°) – 6.667(3) = 0


+c©Fy = 0; 6.667 – 6 – FCG sin 60° = 0

FCG = 0.770 kN (C) Ans

Support Reactions:

+©MA = 0; Ey (9) – 8(6) – 6(3) = 0 Ey = 7.333 kN

Method of Sections:

+©MC = 0; 7.333(4.5) – 8(1.5) – FFG (3 sin 60°) = 0

FFG = 8.08 kN (T) Ans

+©MG = 0; 7.333(3) – FCD (3 sin 60°) = 0


+c©Fy = 0; FCF sin 60° + 7.333 – 8 = 0

FCF = 0.770 kN (T) Ans

*5–28.

•5–29.

SM_CH05.indd 308 4/8/11 11:49:54 AM


309437


6–38. Determine the force in members DC, HC, and HI ofthe truss, and state if the members are in tension orcompression.

A

C

G

E D

HF

IB

2 m 2 m 2 m

1.5 m

50 kN40 kN

40 kN

30 kN

1.5 m

1.5 m

06a Ch06a 401-445.indd 437 6/12/09 8:49:25 AM

Support Reactions: Applying the moment equation of equilibrium about point A to the free-bodydiagram of the truss, Fig. a,

+©MA = 0; 40(1.5) + 30(3) + 40(2) – Fy (4) = 0

Fy = 57.5 kN

S+ ©Fx = 0; Ax – 30 – 40 = 0; Ax = 70 kN

+c©Fy = 0; 57.5 – 40 – 50 + Ay = 0; Ay = 32.5 kN

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

+©MC = 0; 70(3) – 32.5(2) – 40(1.5) – FHI (2) = 0

FHI = 42.5 kN (T) Ans

+©MD = 0; 70(4.5) – 40(3) – 30(1.5) – FHC (1.5) = 0

FHC = 100 kN (T) Ans

+c©Fy = 0; 32.5 + 42.5 – FDC = 0

FDC = 125 kN (C) Ans

35

5–30.

SM_CH05.indd 309 4/8/11 11:49:55 AM


310438


6–39. Determine the force in members ED, EH, and GHof the truss, and state if the members are in tension orcompression.

A

C

G

E D

HF

IB

2 m 2 m 2 m

1.5 m

50 kN40 kN

40 kN

30 kN

1.5 m

1.5 m

06a Ch06a 401-445.indd 438 6/12/09 8:49:26 AM

Support Reactions: Applying the moment equation of equilibrium about point A to the free-bodydiagram of the truss, Fig. a,

+©MA = 0; 40(1.5) + 30(3) + 40(2) – Fy (4) = 0

Fy = 57.5 kN

S+ ©Fx = 0; Ax – 30 – 40 = 0; Ax = 70 kN

+c©Fy = 0; 57.5 – 40 – 50 + Ay = 0; Ay = 32.5 kN

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

+©ME = 0; –57.5(2) + FGH (1.5) = 0

FGH = 76.7 kN (T) Ans

+©MH = 0; –57.5(4) + FED (1.5) + 40(2) = 0

FED = 100 kN (C) Ans

+c©Fy = 0; 57.5 – FEH – 40 = 0

FEH = 29.2 kN (T) Ans

35

5–31.

SM_CH05.indd 310 4/8/11 11:49:55 AM


311439


*6–40. Determine the force in members GF, GD, and CDof the truss and state if the members are in tension orcompression. 260 lb

4 ft 4 ft

4 ft

3 ft 3 ft

4 ft 4 ft

5

1213

AH G F

B

C

D

E

•6–41. Determine the force in members BG, BC, and HGof the truss and state if the members are in tension orcompression. 260 lb

4 ft 4 ft

4 ft

3 ft 3 ft

4 ft 4 ft

5

1213

AH G F

B

C

D

E

1.2 m 1.2 m 1.2 m 1.2 m

1300 N1.2 m

0.9 m0.9 m

1.2 m 1.2 m 1.2 m 1.2 m

1300 N1.2 m

0.9 m0.9 m

1300 N

1300 N

Ay = 1200 N

1.2 m 1.2 m 2.4 m

0.9 m1.2 m

2.4 m 2.4 m

2.4 m 1.2 m 1.2 m

1.2 m0.9 m

Ax = 500 N

+ ΣMO = 0; 12

13

1300 (2.4) – FGD sin 36.87° (4.8) = 0

FGD = 1000 N (C) Ans

+ ΣMD = 0; FGF (0.9) – 12

13

1300 (1.2)

– 5

13

(1300) (0.9) = 0

FGF = 2100 N (C) Ans

+ ΣMG = 0; FCD cos 14.04° (1.2) – 12

13

1300 (2.4) = 0

FCD = 2473.9 N (T) Ans

Entire truss:

+ ΣMG = 0; Ay (2.4) – 12

13

(1300) (2.4) = 0

Ay = 1200 N

+→ ΣFx = 0; Ax –

5

13

(1300) = 0

Ax = 500 N

Section:

+ ΣMO = 0; 1200 (2.4) – FBC cos 14.04° (1.2) = 0

FBC = 2473.9 N (T) Ans

+ ΣMB = 0; 1200 (1.2) + 500 (0.9) – FHG (0.9) = 0

FHG = 2100 N (C) Ans

+ ΣMG = 0; –1200 (2.4) + FBG sin 36.87° (4.8) = 0

FBG = 1000 N (C) Ans

06a Ch06a 401-445.indd 439 6/12/09 8:49:28 AM

*5–32.

•5–33.

SM_CH05.indd 311 4/8/11 11:49:56 AM


312440


6–42. Determine the force in members IC and CG of thetruss and state if these members are in tension orcompression. Also, indicate all zero-force members.

1.5 m 1.5 m 1.5 m

AH

B C D

JI

G FE

1.5 m

2 m

2 m

6 kN 6 kN

6–43. Determine the force in members JE and GF of thetruss and state if these members are in tension orcompression. Also, indicate all zero-force members.

1.5 m 1.5 m 1.5 m

AH

B C D

JI

G FE

1.5 m

2 m

2 m

6 kN 6 kN

06a Ch06a 401-445.indd 440 6/12/09 8:49:29 AM

By inspection of joints B, D, H and I,

AB, BC, CD, DE, HI, and GI are all zero-force members. Ans

+©MG = 0; – 4.5(3) + FIC (4) = 0

FIC = 5.62 kN (C) Ans

35

By inspection of joints B, D, H and I,

AB, BC, CD, DE, HI, and GI are all zero-force members. Ans

Joint C:

S+ ©Fx = 0; FCJ = 5.625 kN

+c©Fy = 0; (5.625) + (5.625) – FCG = 0

FCG = 9.00 kN (T) Ans

45

45

Joint E:

+c©Fy = 0; 7.5 – FJE = 0

FJE = 9.375 = 9.38 kN (C) Ans

45

S+ ©Fx = 0; (9.375) – FGF = 0;


35

5–34.

5–35.

SM_CH05.indd 312 4/8/11 11:49:56 AM


313441


*6–44. Determine the force in members JI, EF, EI, and JEof the truss, and state if the members are in tension orcompression.

8 ft

8 ft

8 ft

900 lb

1500 lb1000 lb1000 lb

A G

N BH

F

M

C D EI

JL K

8 ft 8 ft 8 ft 8 ft 8 ft 8 ft2.5 m 2.5 m 2.5 m 2.5 m 2.5 m 2.5 m

2.5 m

2.5 m

2.5 m

7.5 kN5 kN5 kN

4.5 kN

7.5 kN5 kN5 kN

4.5 kN

2.5 m 2.5 m 5 m5 m

7.5 m 2.5 m

2.5 m

Gx = 4.5 kN

2.5 m 2.5 m

Gy = 11 kN

FEF = 2.828 kN

FEI = 6.5 kN

Support Reactions: Applying the equations of equilibrium to the free – body diagram of the truss, Fig. a,

+→ ΣFx = 0; 4.5 – Gx = 0

Gx = 4.5 kN

+ ΣMA = 0; 5 (5) + 7.5 (7.5) + 5 (10) + 4.5 (7.5) – Gy (15) = 0

Gy = 11 kN

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

+ ΣME = 0; 11 (5) – 4.5 (5) – FJI sin 45° (2.15) = 0

FJI = 18.385 kN (C) Ans

+ ΣMI = 0; 11 (2.5) – 4.5 (5) – FEF cos 45° (2.5) = 0

FEF = 2.828 kN (T) Ans

Using the above results and writing the force equation of equilibrium along the x axis,

+→ ΣFx = 0; 18.385 cos 45° – 2.828 cos 45° – 4.5 – FEI = 0

FEI = 6.5 kN (T) Ans

Method of Joints: From the free – body diagram of joint E, Fig. c,

+↑ΣFy = 0; FJE – 2.828 sin 45° = 0

FHE = 2.0 kN (T) Ans

06a Ch06a 401-445.indd 441 6/12/09 8:49:31 AM

*5–36.

SM_CH05.indd 313 4/8/11 11:49:57 AM


314442


•6–45. Determine the force in members CD, LD, and KLof the truss, and state if the members are in tension orcompression.

8 ft

8 ft

8 ft

900 lb

1500 lb1000 lb1000 lb

A G

N BH

F

M

C D EI

JL K

8 ft 8 ft 8 ft 8 ft 8 ft 8 ft2.5 m 2.5 m 2.5 m 2.5 m 2.5 m 2.5 m

2.5 m

2.5 m

2.5 m

7.5 kN5 kN5 kN

4.5 kN

7.5 kN5 kN5 kN

4.5 kN

2.5 m 2.5 m 5 m5 m

7.5 m

5 kN

4.5 kN

2.5 m

2.5 m5 m

NA = 6.5 kN

Support Reactions: Applying the equation of equilibrium about point G to the free – body diagram of the truss, Fig. a,

+ ΣMG = 0; 5 (5.0) + 7.5 (7.5) + 5 (10.0) – 4.5 (7.5) – NA (15) = 0

NA = 6.5 kN

Method of Sections: Using the left portion of the free – body diagram, Fig. b.

+ ΣMD = 0; FKL (2.5) + 5 (2.5) – 4.5 (2.5) – 6.5 (7.5) = 0

FKL = 19 kN (C) Ans

+ ΣML = 0; FCD (2.5) – 6.5 (5) = 0

FCD = 13 kN (T) Ans

+↑ΣFy = 0; 6.5 – 5 – FLD sin 45° = 0

FLD = 2.121 kN (T) Ans

06a Ch06a 401-445.indd 442 6/12/09 8:49:33 AM

•5–37.

SM_CH05.indd 314 4/8/11 11:49:57 AM


315443


6–46. Determine the force developed in members BC andCH of the roof truss and state if the members are in tensionor compression.

1.5 m

2 m2 m

m 1m 1

0.8 m

2 kN

1.5 kN

AH

BD

G

C

F E

06a Ch06a 401-445.indd 443 6/12/09 8:49:35 AM

+©MA = 0; Ey (4) – 2(0.8) – 1.5(2) = 0 Ey = 1.15 kN

+©MH = 0; 1.15(3) – 1.5(1) – FBC (1) = 0


+©MA = 0; 1.15(4) – 1.5(2) – FCH sin 56.31° (1) = 0

FCH = 1.92 kN (T) Ans

35

5–38.

SM_CH05.indd 315 4/8/11 11:49:58 AM


316444


6–47. Determine the force in members CD and GF of thetruss and state if the members are in tension orcompression. Also indicate all zero-force members.

1.5 m

2 m2 m

m 1m 1

0.8 m

2 kN

1.5 kN

AH

BD

G

C

F E

06a Ch06a 401-445.indd 444 6/12/09 8:49:37 AM

Entire truss:

+©MA = 0; –2(0.8) – 1.5(2) + Ey (4) = 0

Ey = 1.15 kN

Section:

+©MF = 0; 1.15(1) – FCD sin 36.87° (1) = 0


+©MC = 0; –FGF (1.5) + 1.15(2) = 0


Joint D:

+b©Fy = 0; FFD = 0 Ans

Joint F:

+c©Fy = 0; FFC cos u = 0

FFC = 0 Ans

5–39.

SM_CH05.indd 316 4/8/11 11:49:58 AM


317445


*6–48. Determine the force in members IJ, EJ, and CD ofthe Howe truss, and state if the members are in tension orcompression.

2 kN3 kN

4 kN5 kN

4 kN

6 kN

5 kN

BA

C D E FG

H

I

J

K

L

2 m

4 m

2 m 2 m 2 m 2 m 2 m

06a Ch06a 401-445.indd 445 6/12/09 8:49:38 AM

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

+©MA = 0; NG (12) – 2(12) – 4(10) – 4(8) – 6(6) – 5(4) – 5(2) = 0

NG = 13.5 kN

Method of Sections: By inspecting joint D, we find that member DJ is a zero-force member, thus FDJ = 0. Using the right portion of the free-body diagram, Fig. b,

+©MJ = 0; 13.5(6) – 4(2) – 4(4) – 2(6) – FCD (4) = 0


+©ME = 0; 13.5(4) – 2(4) – 4(2) – FIJ sin 33.69°(4) = 0

FIJ = 17.13 kN = 17.1 kN (C) Ans

+©MG = 0; 4(2) + 4(4) – FEJ sin 63.43°(4) = 0

FEJ = 6.708 kN = 6.71 kN (T) Ans

*5–40.

SM_CH05.indd 317 4/8/11 11:49:59 AM


318446


•6–49. Determine the force in members KJ, KC, and BCof the Howe truss, and state if the members are in tension orcompression.

2 kN3 kN

4 kN5 kN

4 kN

6 kN

5 kN

BA

C D E FG

H

I

J

K

L

2 m

4 m

2 m 2 m 2 m 2 m 2 m

06b Ch06b 446-490.indd 446 6/12/09 8:51:41 AM

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

Method of Sections: Using the left portion of the free–body diagram, Fig. a,

S+ ©Fx = 0; Ax = 0

+©MG = 0 3(12) + 5(10) + 5(8) + 6(6) + 4(4) + 4(2) – Ay(12) = 0

Ay = 15.5 kN

+©MC = 0; FKJ sin 33.69° (4) + 5(2) + 3(4) – 15.5(4) = 0

FKJ = 18.03 kN = 18.0 kN (C) Ans

+©MA = 0; FKC (4) – 5(4) – 5(2) = 0

FKC = 7.50 kN (C) Ans

+©MK = 0; FBC (2.667) + 5(2) + 3(4) – 15.5(4) = 0

FBC = 15 kN (T) Ans

•5–41.

SM_CH05.indd 318 4/8/11 11:49:59 AM


319447



, .P2 = 10 kNP1 = 20 kN

06b Ch06b 446-490.indd 447 6/12/09 8:51:43 AM

AG F

E

DCB

P2P1

1.5 m 1.5 m 1.5 m 1.5 m

2 m

+©MA = 0; –20(1.5) – 10(4.5) + Ey(6) = 0

Ey = 12.5 kN

+↑©Fy = 0; Ay – 20 – 10 + 12.5 = 0

Ay = 17.5 kN

S+ ©Fx = 0; Ax = 0

Entire truss:

Joint A:

Joint B:

Joint G:

Joint C:

45

+↑©Fy = 0; 17.5 – FAB = 0

FAB = 21.875 = 21.9 kN (C) Ans

+↑©Fy = 0; (21.875) – FBG = 0

FBG = 17.5 kN (T) Ans

45

+↑©Fy = 0; FCF – (3.125) = 0

FCF = 3.125 = 3.12 kN (C) Ans

45

45

+↑©Fy = 0; 17.5 – 20 + FCG = 0

FCG = 3.125 = 3.12 kN (T) Ans

45

S+ ©Fx = 0; FAG – (21.875) = 0

FAG = 13.125 = 13.1 kN (T) Ans

35

S+ ©Fx = 0; (21.875) – FBC = 0

FBC = 13.125 = 13.1 kN (C) Ans

35

S+ ©Fx = 0; (3.125) + FFG – 13.125 = 0

FFG = 11.25 = 11.2 kN (T) Ans

35

S+ ©Fx = 0; 13.125 – (3.125) – (3.125) – FCD = 0

FCD = 9.375 = 9.38 kN (C) Ans

35

35

5–42.

SM_CH05.indd 319 4/8/11 11:49:59 AM


320448


06b Ch06b 446-490.indd 448 6/12/09 8:51:44 AM

Joint D:

Joint F:

35S+ ©Fx = 0; 9.375 – FDE = 0

FDE = 15.625 = 15.6 kN (C) Ans

+↑©Fy = 0; (15.625) – FDF = 0

FDF = 12.5 kN (T) Ans

45

S+ ©Fx = 0; (3.125) – 11.25 + FEF = 0

FEF = 9.38 kN (T) Ans

35

+↑©Fy = 0; 12.5 – 10 – (3.125) = 0 Check!45

SM_CH05.indd 320 4/8/11 11:50:00 AM


321449



, .P2 = 20 kNP1 = 40 kN

AG F

E

DCB

P2P1

1.5 m 1.5 m 1.5 m 1.5 m

2 m

06b Ch06b 446-490.indd 449 6/12/09 8:51:45 AM

+©MA = 0; –40(1.5) – 20(4.5) + Ey(6) = 0

Ey = 25 kN

+↑©Fy = 0; Ay – 40 – 20 + 25 = 0

Ay = 35 kN

S+ ©Fx = 0; Ax = 0

Entire truss:

Joint A:

45

+↑©Fy = 0; 35 – FAB = 0

FAB = 43.75 = 43.8 kN (C) Ans

S+ ©Fx = 0; FAG – (43.75) = 0

FAG = 26.25 = 26.2 kN (T) Ans

35

Joint B:

+↑©Fy = 0; (43.75) – FBG = 0

FBG = 35.0 kN (T) Ans

45

S+ ©Fx = 0; (43.75) – FBC = 0

FBC = 26.25 = 26.2 kN (C) Ans

35

5–43.

SM_CH05.indd 321 4/8/11 11:50:00 AM


322450


06b Ch06b 446-490.indd 450 6/12/09 8:51:45 AM

Joint G:

Joint E:

+↑©Fy = 0; –40 + 35 + FGC = 0

FGC = 6.25 kN (T) Ans

45

S+ ©Fx = 0; (6.25) + FGF – 26.25 = 0


35

Joint D:

Joint F:

35S+ ©Fx = 0; FDC – (31.25) = 0

FDC = 18.75 = 18.8 kN (C) Ans

+↑©Fy = 0; (31.25) – FDF = 0

FDF = 25.0 kN (T) Ans

45

+↑©Fy = 0; 25 – (FFC) – 20 = 0

FFC = 6.25 kN (C) Ans

45

S+ ©Fx = 0; –22.5 + 18.75 + (6.25) = 0 Check!35

45

+↑©Fy = 0; 25 – FED = 0

FED = 31.25 = 31.2 kN (C) Ans

S+ ©Fx = 0; –FEF + (31.25) = 0

FEF = 18.75 = 18.8 kN (T) Ans

35

SM_CH05.indd 322 4/8/11 11:50:00 AM


323451


*6–52. Determine the force in members KJ, NJ, ND, andCD of the K truss. Indicate if the members are in tension orcompression. Hint: Use sections aa and bb.

1800 lb

15 ft

15 ft

20 ft 20 ft 20 ft 20 ft 20 ft

A B

I H

20 ft

L

M N O P

GFEDC

1500 lb1200 lb

a b

JKa b

6 kN7.5 kN 9 kN

4.5 m

4.5 m

6 m 6 m 6 m 6 m 6 m 6 m

6 m 6 m 6 m 18 m

9 m

6 m 6 m

9 m

6 m 6 m

7.5 kN

7.5 kN

7.5 kN

6 kN 9 kN

6 kN

6 kN

Ay = 14.5 kN

Ay = 14.5 kN

Support Reactions :

+ ΣMG = 0; 6 (30) + 7.5 (24) + 9 (18) – Ay (36) = 0

Ay = 14.5 kN

+→ ΣFx = 0; Ax = 0

Method of Sections : From section a – a, FKJ and FCD can be obtained directly by summing moment about points C and K respectively.

+ ΣMC = 0; FKI (9) + 6 (6) – 14.5 (12) = 0

FKJ = 15.33 kN (C) Ans

+ ΣMK = 0; FCD (9) + 6 (6) – 14.5 (12) = 0


From sec b – b, summing forces along x and y axes yields

+→ ΣFx = 0; FND

4

5

– FNJ 4

5

+ 15.33 – 15.33 = 0

FND = FNJ ]1[

+↑ΣFy = 0; 14.5 – 6 – 7.5 – FND 3

5

– FNJ 3

5

= 0

FND + FNJ = 1.6667 kN [2]

Solving Eqs. [1] and [2] yields

FND = 0.833 kN (T)

FNJ = 0.833 kN (C) Ans

06b Ch06b 446-490.indd 451 6/12/09 8:51:46 AM

*5–44.

SM_CH05.indd 323 4/8/11 11:50:01 AM


324452


•6–53. Determine the force in members JI and DE ofthe K truss. Indicate if the members are in tension orcompression.

1800 lb

15 ft

15 ft

20 ft 20 ft 20 ft 20 ft 20 ft

A B

I H

20 ft

L

M N O P

GFEDC

1500 lb1200 lb

a b

JKa b

6 kN7.5 kN 9 kN

4.5 m

4.5 m

6 m 6 m 6 m 6 m 6 m 6 m

6 m 6 m 6 m 18 m

7.5 kN6 kN 9 kN

12 m

9 m

Gy = 8 kN

Support Reactions :

+ ΣMA = 0; Gy (36) – 9 (18) – 7.5 (12) – 6 (6) = 0

Gy = 8 kN

Method of Sections :

+ ΣMg = 0; 8 (12) – FJI (9) = 0

FJI = 10.67 kN (C) Ans

+ ΣMl = 0; 8 (12) – FDE (9) = 0

FDg = 10.67 kN (T) Ans

06b Ch06b 446-490.indd 452 6/12/09 8:51:47 AM

•5–45.

SM_CH05.indd 324 4/8/11 11:50:01 AM


325471


6–71. Determine the force needed to support the 100-lbweight. Each pulley has a weight of 10 lb. Also, what are thecord reactions at A and B?

P

P

2 in.

2 in.

2 in.

CA

B

50 (9.81) N

50 N50 N

50 mm

50 mm

50 mm

6–71. Determine the force P needed to support the 50-kg weight. Each pulley has a weight of 50 N. Also, what are the cord reactions at A and B?

Equations of Equilibrium : From FBD (a),

+↑ΣFy = 0; P′ – 2P – 50 = 0 [1]

From FBD (b),

+↑ΣFy = 0; 2P + P′ – 50 (9.81) – 50 = 0 [2]

Solving Eqs. [1] and [2] yields,

P = 122.625 N Ans

P′ = 295.25 N

The cord reactions at A and B are

FA = P = 122.625 N

FB = P′ = 295.25 N Ans

06b Ch06b 446-490.indd 471 6/12/09 8:52:26 AM

5–46.

SM_CH05.indd 325 4/8/11 11:50:01 AM


326472


*6–72. The cable and pulleys are used to lift the 600-lbstone. Determine the force that must be exerted on the cableat A and the corresponding magnitude of the resultant forcethe pulley at C exerts on pin B when the cables are in theposition shown.

P

A

C

B

D

30

300 (9.81) N

T = 1471.5 N

T = 1471.5 N

*6–72. The cable and pulleys are used to lift the 300-kg

Pulley D :

+↑ΣFy = 0; 2T – 300 (9.81) = 0

T = 1471.5 N = 1.47 kN Ans

Pulley B :

+↑ΣFx = 0; Bx – 1471.5 sin 30° = 0

Bx = 735.75 N = 0.74 kN

+↑ΣFy = 0; By – 1471.5 – 1471.5 cos 30° = 0

By = 2745.86 N

FB = (735.75) + (2745.86)2 2

= 2842.7 N = 2.84 kN Ans

06b Ch06b 446-490.indd 472 6/12/09 8:52:28 AM

5–47.

SM_CH05.indd 326 4/8/11 11:50:02 AM


327473


•6–73. If the peg at B is smooth, determine thecomponents of reaction at the pin A and fixed support C.

A

B C

600 mm

800 mm

900 N m

600 mm

500 N

45

06b Ch06b 446-490.indd 473 6/12/09 8:52:30 AM

ΣMA = 0; NB (0.8) – 900 = 0 NB = 1125 N+Equations of Equilibrium: From the free – body diagram of member AB, Fig. a,

+

S ©Fx= 0; Ax – 1125 cos 45° = 0 Ax = 795.50 N = 795 N Ans

+c©Fy = 0; 1125 sin 45° – Ay = 0 Ay = 795.50 N = 795 N Ans

+

Applying the equations of equilibrium to the free-body diagram of member BC, Fig. b,

S ©Fx= 0; 1125 cos 45° – Cx = 0 Cx = 795.50 N = 795 N Ans

+c©Fy = 0; Cy 1125 sin 45° – 500 = 0 Cy = 1295.50 N = 130 kN Ans

+

MC = 1254.59 N · m = 1.25 kN · m AnsΣMC = 0; 1125 sin 45° (1.2) + 500 (0.6) = MC = 0+

*5–48.

SM_CH05.indd 327 4/8/11 11:50:02 AM


328474


6–74. Determine the horizontal and vertical componentsof reaction at pins A and C.

BA

C

2 ft3 ft

150 lb

100 lb

2 ft

45

By = 450 N

500 N0.6 m

0.6 m0.9 m

750 N

0.6 m0.9 m

750 N

500 N

0.6 m

Equations of Equilibrium: From the free – body diagram of member AB in Fig. a, we have

+ ΣMA = 0; By (1.5) – 750 (0.9) = 0 By = 450 N

+ ΣMB = 0; 750 (0.6) – Ay (1.5) = 0 Ay = 300 N Ans

+↑ΣFx = 0; Ax – Bx )1( 0 =

From the free – body diagram of the member BC in Fig. b and using the result for By, we can write

+ ΣMC = 0; 450 (0.6) + 500 sin 45° (0.6) – Bx (0.6) = 0

Bx = 803.55 N

+↑ΣFy = 0; Cy – 450 – 500 sin 45° = 0

Cy = 803.55 N Ans

+↑ΣFx = 0; 803.55 – 500 cos 45° – Cx = 0

Cx N 054 = Ans

Substituting Bx = 803.55 N into Eq. (1) yields

Ax = 803.55 N Ans

06b Ch06b 446-490.indd 474 6/12/09 8:52:32 AM

5–49.

SM_CH05.indd 328 4/8/11 11:50:03 AM


329475


6–75. The compound beam is fixed at A and supported byrockers at B and C. There are hinges (pins) at D and E.Determine the components of reaction at the supports.

6 m2 m

6 m

30 kN m

2 m 2 m

15 kN

A D B EC

06b Ch06b 446-490.indd 475 6/12/09 8:52:33 AM

Equations of Equilibrium : From FBD(a),

ΣME = 0; 30 – Cy (6) = 0 Cy = 5.00 kN Ans+

+ ↑ΣFy = 0; Ey – 5.00 = 0 Ey = 5.00 kN

S ©Fx = 0; Ey = 0+

From FBD(b),

ΣMD = 0; By (4) – 15(2) – 5.00(6) = 0 By = 15.0 kN Ans

+

+ ↑ΣFy = 0; Dy + 15.0 – 15 – 5.00 = 0

Dy = 5.00 kN

S ©Fx = 0; Dx = 0+

From FBD(b),

ΣMD = 0; MA – 5.00(6) = 0 MA = 30.0 kN · m Ans

+

+ ↑ΣFy = 0; Ay + 5.00 = 0 Ay 5.00 kN Ans

S ©Fx = 0; Ax = 0 Ans+

5–50.

SM_CH05.indd 329 4/8/11 11:50:03 AM


330476


*6–76. The compound beam is pin-supported at C andsupported by rollers at A and B. There is a hinge (pin) at D.Determine the components of reaction at the supports.Neglect the thickness of the beam. A D B C

8 ft

3

45

8 ft

12 kip

15 kip ft

4 kip30

8 kip

8 ft4 ft 2 ft

6 ft3 m 4 m 4 m 4 m2 m 1 m

40 kN 60 kN

25 kN · m

4 m 4 m 4 m

Dy = 9.35 kN

60 kN

25 kN · mDx = 10 kN

40 kN

20 kN

3 m 2 m1 m

20 kN

Equations of Equilibrium : From FBD (a),

+ ΣMD = 0; 20 cos 30° (6) + 40 (1) – Ay (3) = 0

Ay = 47.97 kN Ans

+↑ΣFy = 0; Dy + 47.97 – 20 cos 30° – 40 = 0

Dy = 9.35 kN

+→ ΣFx = 0; Dx – 20 sin 30° = 0

Dx = 10 kN

From FBD (b),

+ ΣMC = 0; 9.35 (12) + 25 + 60 4

5

(4) – By (8) = 0

By = 41.15 kN Ans

+↑ΣFy = 0; Cy + 41.15 – 9.35 – 60 4

5

= 0

Cy = 16.2 kN Ans

+→ ΣFx = 0; Cx – 10 – 60

3

5

= 0

Cx = 46.0 kN Ans

06b Ch06b 446-490.indd 476 6/12/09 8:52:34 AM

5–51.

SM_CH05.indd 330 4/8/11 11:50:03 AM


331477


•6–77. The compound beam is supported by a rocker at Band is fixed to the wall at A. If it is hinged (pinned) togetherat C, determine the components of reaction at the supports.Neglect the thickness of the beam.

4 ft 4 ft

500 lb200 lb

4000 lb ft

4 ft8 ft

A C B

1213

5 60

1 m

2.5 kN1 kN

6 kN · m

1 m 2 m 1 m

1 kN

6 kN · m

2 m 1 m

1 m 1 m

Cx = 0.5 kN

Cy = 1.711 kN

2.5 kN

Member CB :

+→ ΣFx = 0; –Cx + 1 cos 60° = 0

Cx = 0.5 kN

+ ΣMC = 0; –1 sin 60° (2) + By (3) – 6 = 0

By = 2.577 kN Ans

+↑ΣFy = 0; Cy – 1 sin 60° + 2.577 = 0

Cy = –1.711 kN

Member AC :

+→ ΣFx = 0; Ax – 2.5

5

13

+ 0.5 = 0

Ax = 0.462 kN Ans

+↑ΣFy = 0; Ay – 2.5 12

13

+ 1.711 = 0

Ay = 0.597 kN Ans

+ ΣMA = 0; –MA – 2.5 12

13

(1) + 1.711 (2) = 0

MA = 1.114 kN · m Ans

06b Ch06b 446-490.indd 477 6/12/09 8:52:36 AM

*5–52.

SM_CH05.indd 331 4/8/11 11:50:04 AM


332478


6–78. Determine the horizontal and vertical componentsof reaction at pins A and C of the two-member frame.

3 m

3 m

200 N/m

A

B

C

06b Ch06b 446-490.indd 478 6/12/09 8:52:36 AM

Equations of Equilibrium:

ΣMA = 0; FBC cos 45° (3) – 600(1.5) = 0FBC = 0 424.26 N

+

+ ↑©Fy = 0; Ay + 424.26 cos 45° – 600 = 0 Ay = 300 N Ans

+ ↑©Fx = 0; 424.26 sin 45° – Ax = 0 Ax = 300 N Ans

For pin C, Cx = FBC sin 45° = 424.26 sin 45° = 300 N Ans Cy = FBC cos 45°= 424.26 cos 45° = 300 N Ans

5–53.

SM_CH05.indd 332 4/8/11 11:50:04 AM


333479


6–79. If a force of acts on the rope, determinethe cutting force on the smooth tree limb at D and thehorizontal and vertical components of force acting on pin A.The rope passes through a small pulley at C and a smoothring at E.

F = 50 N

F 50 N

BC

E

30 mm

100 mm

A

D

06b Ch06b 446-490.indd 479 6/12/09 8:52:38 AM

Equations of Equilibrium: From the free-body diagram of pulley C in Fig. a,

+c©Fy = 0; FBC – 50 – 50 = 0 FBC = 100 N

From the free-body diagram of segment BAD in Fig. b and using the result FBC = 100 N,

ND (30) – 100(100) = 0 ND = 333.33 N = 333 N Ans

S+ ©Fx = 0; 333.33 – Ax= 0 Ax = 333.33 N = 333 N Ans

+c©Fy = 0; Ay – 100 = 0 Ax = 100 N

+ΣMA = 0;

5–54.

SM_CH05.indd 333 4/8/11 11:50:04 AM


334480


*6–80. Two beams are connected together by the shortlink BC. Determine the components of reaction at the fixedsupport A and at pin D.

A B

C

D

10 kN12 kN

3 m1.5 m1 m 1.5 m

06b Ch06b 446-490.indd 480 6/12/09 8:52:38 AM

Equations of Equilibrium: First, we will consider the free-body diagram of member BD in Fig. a,+c©MD = 0; 10(1.5) – FBC (3) = 0

FBC = 5 kN

S+ ©Fx = 0; Dx = 0 Ans

Dy (3) – 10(1.5) = 0

+ΣMB = 0;

Dy = 5 kN Ans

S+ ©Fx = 0; Ax = 0 Ans

Subsequently, the free-body diagram of member AC in Fig. b will be considered using the result FBC = 5 kN.

+c©Fy = 0; Ay – 12 – 5 = 0

Ay – 17 kN Ans

+ΣMA = 0; MA – 12(1) – 5(4) = 0

MA – 32 kN · m Ans

5–55.

SM_CH05.indd 334 4/8/11 11:50:05 AM


335481


•6–81. The bridge frame consists of three segments whichcan be considered pinned at A, D, and E, rocker supportedat C and F, and roller supported at B. Determine thehorizontal and vertical components of reaction at all thesesupports due to the loading shown. 15 ft

20 ft

5 ft 5 ft

15 ft

2 kip/ft

30 ft

A

B

FC

DE

30 kN/m

9 m

6 m4.5 m

1.5 m 1.5 m

4.5 m

4.5 m4.5 m

1.5 m

2.25 m 2.25 m

1.5 m

2.25 m2.25 m

30 (4.5) kN

30 (4.5) kN By = 135 kN

Dy = 135 kN

30 (9) kN

For segment BD :

+ ΣMD = 0; 30 (9) (4.5) – By (9) = 0

By = 135 kN Ans

+→ ΣFx = 0; Dx = 0 Ans

+↑ΣFy = 0; Dy + 135 – 30 (9) = 0

Dy = 135 kN Ans

For segment ABC :

+ ΣMA = 0; Cy (1.5) – 30 (4.5) (2.25) – 135 (4.5) = 0

Cy = 607.5 kN Ans

+→ ΣFx = 0; Ax = 0 Ans

+↑ΣFy = 0; –Ay + 607.5 – 30 (4.5) – 135 = 0

Ay = 337.5 kN Ans

For segment DEF :

+ ΣMg = 0; –Fy (1.5) – 30 (4.5) (2.25) + 135 (4.5) = 0

Fy = 607.5 kN Ans

+→ ΣFx = 0; Ex = 0 Ans

+↑ΣFy = 0; –Ey + 607.5 – 30 (4.5) – 135 = 0

Ey = 337.5 kN Ans

06b Ch06b 446-490.indd 481 6/12/09 8:52:40 AM

*5–56.

SM_CH05.indd 335 4/8/11 11:50:05 AM


336482


6–82. If the 300-kg drum has a center of mass at point G,determine the horizontal and vertical components of forceacting at pin A and the reactions on the smooth pads Cand D. The grip at B on member DAB resists bothhorizontal and vertical components of force at the rim ofthe drum.

P

390 mm

100 mm

60 mm60 mm

600 mm

30

B

A

C

D G

E

06b Ch06b 446-490.indd 482 6/12/09 8:52:41 AM

+ΣMA = 0; 300(9.81) (600 cos 30°) – NC (120) = 0 NC = 12 743.56 N = 12.7 kN Ans

S+ ©Fx = 0; Ax = 12 743.56 = 0 Ans

Ax = 12 743.56 N = 12.7 kN Ans

+c©Fy = 0; 300(9.81) – Ay = 0

Ay = 2943 N = 2.94 kN Ans

+ΣMB = 0; 12 743.56(60) – 2943(100) – ND (450) = 0 ND = 1045.14 N = 1.05 kN Ans

Equations of Equilibrium: From the free-body diagram of segment CAE Fig. a,

Using the results for Ay and Ay obtained above and applying the moment equation of equilibrium

about point B on the free-body diagram of segment BAD, Fig. b,

5–57.

SM_CH05.indd 336 4/8/11 11:50:05 AM


337483


6–83. Determine the horizontal and vertical componentsof reaction that pins A and C exert on the two-member arch.

1 m1.5 m

2 kN

1.5 kN

0.5 m

A

B

C

06b Ch06b 446-490.indd 483 6/12/09 8:52:42 AM

+ΣMA = 0; –2(0.5) + By (1.5) – Bx (1.5) = 0

Member BC:

+ΣMC = 0; By (1.5) + Bx (1.5) = 1.5(1) = 0

Solving:

By = 0.8333 kN = 833 N

Bx = 0.1667 kN = 167 N

Member AB:

S+ ©Fx = 0; –Ax + 167 = 0

Ax = 167 N Ans

+c©Fy = 0; Ay – 2000 + 833 = 0

Ay = 1.17 kN Ans

Member BC:

S+ ©Fx = 0; –Cx + 1500 – 167 = 0

Cx = 1.33 kN Ans

+c©Fy = 0; Cy – 833 = 0

Cy = 833 kN Ans

Member AB:

5–58.

SM_CH05.indd 337 4/8/11 11:50:06 AM


338484


*6–84. The truck and the tanker have weights of 8000 lband 20 000 lb respectively. Their respective centers ofgravity are located at points and . If the truck is atrest, determine the reactions on both wheels at A, at B, andat C. The tanker is connected to the truck at the turntableD which acts as a pin.

G2G1G1

15 ft 10 ft 9 ft5 ft

A B

D

C

G2

Dy = 60 kN100 kN 40 kN

1.5 m

2.7 m4.5 m 3 m

4.5 m 3 m 2.7 m1.5 m

*6–84. The truck and the tanker have weights of 40 kg and 100 kN respectively. Their respective centers of gravity are located at points G1 and G2 . If the truck is at rest, determine the reactions on both wheels at A, at B, and at C. The tanker is connected to the truck at the turntable D which acts as a pin.

Equations of Equilibrium: First, we will consider the free – body diagram of the tanker in Fig. a.

+ ΣMD = 0; 100 (3) – NA (7.5) = 0

NA = 40 kN Ans

+→ ΣFx = 0; Dx = 0

+↑ΣFy = 0; Dy + 40 – 100 = 0

Dy = 60 kN

Using the results of Dx and Dy obtained above and considering the free – body diagram of the truck in Fig. b,

+ ΣMD = 0; NC (4.2) – 40 (2.7) = 0

NC = 25.71 kN Ans

+→ ΣFy = 0; NB + 25.71 – 40 – 60 = 0

NB = 74.29 kN Ans

06b Ch06b 446-490.indd 484 6/12/09 8:52:43 AM

5–59.

SM_CH05.indd 338 4/8/11 11:50:06 AM


339485


•6–85. The platform scale consists of a combination ofthird and first class levers so that the load on one leverbecomes the effort that moves the next lever. Through thisarrangement, a small weight can balance a massive object.If , determine the required mass of thecounterweight S required to balance a 90-kg load, L.x = 450 mm

350 mm150 mm

150 mm100 mm250 mm

BA

C D

E F

H

G

x

L

S

06b Ch06b 446-490.indd 485 6/12/09 8:52:46 AM

Equations of Equilibrium: Applying the moment equation of equilibrium about point A to the free-body diagram of member AB in Fig. a,

+ΣMF = 0; FED (250) – 264.87(150) = 0 FED = 158.922 N

+ΣMA = 0; FBG (500) – 90(9.81)(150) = 0 FBG = 264.87 N

Using the result of FBG and writing the moment equation of equilibrium about point F to the free-body diagram of member EFG in Fig. b,

Using the result of FED and writing the moment equation of equilibrium about point C to the free-body diagram of member CDI in Fig. c,

+ΣMC = 0; 158.922(100) – ms (9.81)(950) = 0 ms = 1.705 kg = 1.71 kg Ans

*5–60.

SM_CH05.indd 339 4/8/11 11:50:07 AM


340486


6–86. The platform scale consists of a combination ofthird and first class levers so that the load on one leverbecomes the effort that moves the next lever. Through thisarrangement, a small weight can balance a massive object. If

and, the mass of the counterweight S is 2 kg,determine the mass of the load L required to maintain thebalance.

x = 450 mm

350 mm150 mm

150 mm100 mm250 mm

BA

C D

E F

H

G

x

L

S

06b Ch06b 446-490.indd 486 6/12/09 8:52:47 AM

Equations of Equilibrium: Applying the moment equation of equilibrium about point A to the free-body diagram of member AB in Fig. a,

FBG (500) – ML (9.81)(150) = 0 FBG = 2.943 lb+ΣMA = 0;

Using the result of FBG and writing the moment equation of equilibrium about point F on the free-body diagram of member EFG in Fig. b,

+ΣMF = 0; FED (250) – 2.943mL (150) = 0 FED = 1.7658mL

Using the result of FED and writing the moment equation of equilibrium about point F on the free-body diagram of member CDI in Fig. c,

+ΣMC = 0; 1.7658mL (100) – 2(9.81)(950) = 0 mL = 105.56 kg = 106 kg Ans

5–61.

SM_CH05.indd 340 4/8/11 11:50:07 AM


341487


6–87. The hoist supports the 125-kg engine. Determinethe force the load creates in member DB and in memberFB, which contains the hydraulic cylinder H.

C

D

EFG

H

2 m

1 m

1 m

2 m1 m

2 m

A B

06b Ch06b 446-490.indd 487 6/12/09 8:52:49 AM

Force Vectors: The solution for this problem will be simplified ifone realizes that members FB and DB are two-force members.

Equations of Equilibrium: For FBD(a).

1226.25(3) – FFB (2) = 0+ΣMg = 0;

F FB = 1938.87 N = 1.94 kN Ans

Ey = 613.125 N

ΣMC = 0; 613.125(3) – FBD sin 45° (1) = 0FBD = 2601.27 N = 2.60 kN Ans

+

310

310

1938.87 – 1226.25 –Ey = 0+c©Fy = 0;

Ey = 613.125 N

S+ ©Fx = 0;110

Ex –1938.87 = 0

From FBD(b),

5–62.

SM_CH05.indd 341 4/8/11 11:50:08 AM


342488


*6–88. The frame is used to support the 100-kg cylinder E.Determine the horizontal and vertical components ofreaction at A and D.

A

CD

E

0.6 m

1.2 m

r 0.1 m

06b Ch06b 446-490.indd 488 6/12/09 8:52:51 AM

Consider the free-body diagram of member AC in Fig. (a).

+ΣMA = 0; Dx (0.6) – 981(1.2) + 981(0.6) = 0

Dx – 981 N Ans

S+ ©Fx = 0; Ax – 981 – 981 = 0

Ax = 1962 N Ans

+c©Fy = 0; Ay – 981 = 0

Ay = 981 N Ans

Equations of Equilibrium: Member DC is a two-force member.

Dy = 0 Ans

*5–63.

SM_CH05.indd 342 4/8/11 11:50:08 AM


343489


•6–89. Determine the horizontal and vertical componentsof reaction which the pins exert on member AB of the frame.

A

E

B

CD

500 lb

300 lb

3 ft 3 ft

4 ft

60

1.5 kN

0.9 m0.9 m

0.9 m 0.9 m

2.5 kN

1.5 kN

1.2 m

Member AB :

+ ΣMA = 0; –1.5 sin 60° (0.9) + 4

5 FBD (1.8) = 0

FBD = 0.8119 kN

Thus,

Bx = 3

5 (0.8119) = 0.4871 kN Ans

By = 4

5 (0.8119) = 0.6495 kN Ans

+→ ΣFx = 0; –1.5 cos 60° +

3

5 (0.8119) + Ax = 0

Ax = 0.2629 kN Ans

+↑ΣFy = 0; Ay – 1.5 sin 60° + 4

5 (0.8119) = 0

Ay = 0.6495 kN Ans

06b Ch06b 446-490.indd 489 6/12/09 8:52:52 AM

*5–64.

SM_CH05.indd 343 4/8/11 11:50:09 AM


344490


6–90. Determine the horizontal and vertical components ofreaction which the pins exert on member EDC of the frame.

A

E

B

CD

500 lb

300 lb

3 ft 3 ft

4 ft

60

1.5 kN

0.9 m0.9 m

2.5 kN

1.2 m

0.9 m 0.9 m

1.5 kN

0.9 m 0.9 m

2.5 kN

FBD = 0.8119 kN

FBD = 0.8119 kNFAD = 7.0619 kN

Member AB :

+ ΣMA = 0; –1.5 sin 60° (0.9) + 4

5 FBD (1.8) = 0

FBD = 0.8119 kN

Member EDC :

+ ΣMg = 0; –2.5 (1.8) – 4

5 (0.8119) (0.9)

+ 4

5 FAD (0.9) = 0

FAD = 7.0619 kN

+→ ΣFx = 0; Ex – 0.8119

3

5

– 7.0619 3

5

= 0

Ex = 4.7243 kN Ans

+↑ΣFy = 0; –Ey + 7.0619 4

5

– 0.8119 4

5

– 2.5 = 0

Ey = 2.5 kN Ans

Pin D :

+→ ΣFx = 0; Dx –

3

5 (0.8119) –

3

5 (7.0619) = 0

Dx = 4.7243 kN Ans

+↑ΣFy = 0; –Dy – 4

5 (0.8119) +

4

5 (7.0619) = 0

Dy = 5 kN Ans

06b Ch06b 446-490.indd 490 6/12/09 8:52:53 AM

5–65.

SM_CH05.indd 344 4/8/11 11:50:09 AM


345491


6–91. The clamping hooks are used to lift the uniformsmooth 500-kg plate. Determine the resultant compressiveforce that the hook exerts on the plate at A and B, and thepin reaction at C.

A

B

80 mm

P

P P

150 mm

C

06c Ch06c 491-535.indd 491 6/12/09 8:53:46 AM

+©MC = 0; NA (80) – 2452.5 (150) = 0

NA = 4598.4 N = 4.60 kN Ans

+↑©Fy = 0; 2452.5 + 4598.4 – Cy = 0

Cy = 7050.9 N = 7.05 kN Ans

CB is a two-force member.

NB = Cy = 7.05 Ans

5–66.

SM_CH05.indd 345 4/8/11 11:50:09 AM


346492


*6–92. The wall crane supports a load of 700 lb. Determinethe horizontal and vertical components of reaction at the pinsA and D.Also, what is the force in the cable at the winch W?

4 ft

D

A B

C

E

W

4 ft

700 lb

60

4 ft1.2 m

1.2 m 1.2 m

3.5 kN

1.2 m 1.2 m

3.5 kN

3.5 kN

1.75 kN1.75 kN

1.75 kN

TBD = 12.04 kN

*6–92. The wall crane supports a load of 3.5 kN. Determine

Pulley E :

+↑ΣFy = 0; 2 T – 3.5 = 0

T = 1.75 kN Ans

Member ABC :

+ ΣMA = 0; TBD sin 45° (1.2) – 1.75 sin 60° (1.2) – 3.5 (2.4) = 0

TBD = 12.04 kN

+↑ΣFy = 0; –Ay + 12.04 sin 45° – 1.75 sin 60° – 3.5 = 0

Ay = 3.50 kN Ans

+→ ΣFx = 0; Ax – 12.04 cos 45° – 1.75 cos 60° + 1.75 – 1.75 = 0

Ax = 9.39 kN Ans

A1D :

Dx = 12.04 cos 45° = 8.51 kN Ans

Dy = 12.04 sin 45° = 8.51 kN Ans

06c Ch06c 491-535.indd 492 6/12/09 8:53:48 AM

5–67.

SM_CH05.indd 346 4/8/11 11:50:10 AM


347493


•6–93. The wall crane supports a load of 700 lb.Determine the horizontal and vertical components ofreaction at the pins A and D. Also, what is the force in thecable at the winch W? The jib ABC has a weight of 100 lband member BD has a weight of 40 lb. Each member isuniform and has a center of gravity at its center. 4 ft

D

A B

C

E

W

4 ft

700 lb

60

4 ft1.2 m

1.2 m 1.2 m

3.5 kN

1.2 m 1.2 m

3.5 kN

1.75 kN1.75 kN

3.5 kN1.75 kN 500 N

200 N

1.2 m

0.6 m 0.6 m

By = 9.016 kN

*6–93. The wall crane supports a load of 3.5 kN. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force in the cable at the winch W? The jib ABC has a weight of 500 N and member BD has a weight of 200 N. Each member is uniform and has a center of gravity at its center.

Pulley E :

+↑ΣFy = 0; 2 T – 3.5 = 0

T = 1.75 kN Ans

Member ABC :

+ ΣMA = 0; By (1.2) – 3.5 (2.4) – 0.5 (1.2) – 1.75 sin 60° (1.2) = 0

By = 9.016 kN

+↑ΣFy = 0; –Ay – 1.75 sin 60° – 0.5 – 3.5 + 9.016 = 0

Ay = 3.5 kN Ans

+→ ΣFx = 0; Ax – 1.75 cos 60° – Bx + 1.75 – 1.75 = 0

Ax = Bx + 0.85 (1)

Member DB :

+ ΣMD = 0; –0.2 (0.6) – 9.016 (1.2) + Bx (1.2) = 0

Bx = 9.116 kN

+→ ΣFx = 0; –Dx + 9.116 = 0

Dx = 9.116 kN Ans

+↑ΣFy = 0; Dy – 0.2 – 9.016 = 0

Dy = 9.216 kN Ans

From Eq. (1)

Ax = 9.966 Ans

06c Ch06c 491-535.indd 493 6/12/09 8:53:50 AM

*5–68.

SM_CH05.indd 347 4/8/11 11:50:10 AM


348494


6–94. The lever-actuated scale consists of a series ofcompound levers. If a load of weight is placedon the platform, determine the required weight of thecounterweight S to balance the load. Is it necessary to placethe load symmetrically on the platform? Explain.

W = 150 lb

BA

CDE

FG H

I

JKS

M

W

L

1.5 in.

1.5 in.

7.5 in. 7.5 in.4.5 in.

4 in.1.25 in.100 mm 31.25 mm

112.5 mm

37.5 mm

187.5 mm 187.5 mm37.5 mm

37.5 mm187.5 mm

Dy = 750 – 2x

750 N

375 mm

31.25 mm

FIJ = 53.57 N

100 mm

37.5 mm187.5 mm 300 mm

FEH = 125 – 0.3333x

compound levers. If a load of weight W = 750 N is placed

Equations of Equilibrium: First, we will consider the free – body diagram of the platform in Fig. a.

+ ΣMD = 0; 750 (x) – Gy (375) = 0

Gy = 2x

+↑ΣFy = 0; Dy + 2x – 750 = 0

Dy = 750 – 2x

From the free – body diagram of member CDE in Fig. b,

+ ΣMC = 0; (750 – 2x) (37.5) – FEH (225) = 0

FEH = 125 – 0.3333x

From the free – body diagram of member FGHI in Fig. c,

+ ΣMF = 0; FIJ (525) – (125 – 0.3333x) (225) – 2x (37.5) = 0

FIJ = 53.57 N

Finally, from the free – body diagram of member JKL in Fig. d,

+ ΣMK = 0; WS (100) – 53.57 (31.25) = 0

WS = 16.74 N Ans

This result shows that WS is independent of the position x of the load on the platform. Thus, the load can be placed at any position on the platform.

06c Ch06c 491-535.indd 494 6/12/09 8:53:52 AM

5–69.

SM_CH05.indd 348 4/8/11 11:50:11 AM


349506


6–106. The bucket of the backhoe and its contents have aweight of 1200 lb and a center of gravity at G. Determinethe forces of the hydraulic cylinder AB and in links AC andAD in order to hold the load in the position shown. Thebucket is pinned at E.

120

45

1.5 ft

1 ft

B

G

EA D

C

0.25 ft

0.3 m

0.45 m0.075 m

6 kN

0.3 m

0.45 m0.075 m

FAC = 12.561 kN

weight of 6 kN and a center of gravity at G. Determine

Free Body Diagram : The solution for this problem will be simplified if one realizes that the hydranlic cylinder AB, links AD and AC are two-force members.

Equations of Equilibrium : From FBD (a).

+ ΣMg = 0; FAC cos 60° (0.3) + FAC sin 60° (0.075) – 6 (0.45) = 0

FAC = 12.561 kN Ans

Using method of joint [FBD (b)],

+↑ΣFy = 0; 12.561 sin 60° – FAB cos 45° = 0

FAB = 15.384 kN Ans

+→ ΣFx = 0; FAD – 15.384 sin 45° – 12.561 cos 60° = 0

FAD = 17.159 kN Ans

06c Ch06c 491-535.indd 506 6/12/09 8:54:25 AM

5–70.

SM_CH05.indd 349 4/8/11 11:50:11 AM


350507


6–107. A man having a weight of 175 lb attempts to holdhimself using one of the two methods shown. Determine thetotal force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglectthe weight of the platform.

CC

A BA B

)b()a(

6–107. A man having a weight of 875 N ( 87.5 kg) attempts to hold himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglect the weight of the platform.

218.75 N218.75 N875 N

218.75 N218.75 N

437.5 N437.5 N875 N

437.5 N437.5 N

(a)

Bar:

+↑ΣFy = 0; 2(F/2) – 2(437.5) = 0

F = 875 N Ans

Man:

+↑ΣFy = 0; NC – 875 – 2(437.5) = 0

NC = 1750 N Ans(b)

Bar:

+↑ΣFy = 0; 2(218.75) – 2(F/2) = 0

F = 437.5 N Ans

Man:

+↑ΣFy = 0; NC – 875 + 2(218.75) = 0

NC = 437.5 N Ans

06c Ch06c 491-535.indd 507 6/12/09 8:54:28 AM

5–71.

SM_CH05.indd 350 4/8/11 11:50:11 AM


351508


*6–108. A man having a weight of 175 lb attempts to holdhimself using one of the two methods shown. Determine thetotal force he must exert on bar AB in each case and thenormal reaction he exerts on the platform at C.The platformhas a weight of 30 lb.

CC

A BA B

)b()a(

*6–108. A man having a weight of 875 N ( 87.5 kg) attempts to hold himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has a weight of 150 N ( 15 kg).

512.5 N512.5 N875 N

512.5 N512.5 N

512.5 N 512.5 N

150 N

875 N

512.5 N

256.25 N256.25 N

256.25 N 256.25 N

256.25 N256.25 N875 N

(a)

Bar:

+↑ΣFy = 0; 2(F/2) – 512.5 – 512.5 = 0

F = 1025 N Ans

Man:

+↑ΣFy = 0; NC – 875 – 512.5 – 512.5 = 0

NC = 1900 N Ans(b)

Bar:

+↑ΣFy = 0; 2(F/2) – 256.25 – 256.25 = 0

F = 512.5 N Ans

Man:

+↑ΣFy = 0; NC – 875 + 256.25 + 256.25 = 0

NC = 362.5 N Ans

06c Ch06c 491-535.indd 508 6/12/09 8:54:29 AM

*5–72.

SM_CH05.indd 351 4/8/11 11:50:12 AM


352509


•6–109. If a clamping force of is required at A,determine the amount of force F that must be applied to thehandle of the toggle clamp.

300 N

275 mm30

30

235 mm

30 mm

30 mm

70 mm

F

C

E

B

D

A

06c Ch06c 491-535.indd 509 6/12/09 8:54:33 AM

Equations of Equilibrium: First, we will consider the free-body diagram of the clampin Fiq. a. Writing the moment equation of equilibrium about point D,

+©MD = 0; Cx (60) – 300(235) = 0

+©MC = 0; FBE cos 30° (70) – FBE sin 30° (30) – F cos 30° (275 cos 30° + 70)

Cx = 1175 N

Subsequently, the free-body diagram of the handle in Fig. b will be considered.

– F sin 30° (275 sin 30°) = 0

©Fx = 0; 1175 + F sin 30° – FBE sin 30° = 00.5FBE – 0.5F = 1175 (2)

45.62FBE – 335.62F = 0 (1)+→

Solving Eqs. (1) and (2) yieldsF = 369.69 N = 370 N AnsFBE = 2719.69 N

•5–73.

SM_CH05.indd 352 4/8/11 11:50:12 AM


353510


6–110. If a force of is applied to the handle ofthe toggle clamp, determine the resulting clamping force at A.

F = 350 N

275 mm30

30

235 mm

30 mm

30 mm

70 mm

F

C

E

B

D

A

06c Ch06c 491-535.indd 510 6/12/09 8:54:34 AM

Equations of Equilibrium: First, we will consider the free-body diagram of the handlein Fiq. a,

+©MC = 0; FBE cos 30° (70) – FBE sin 30° (30) – 350 cos 30° (275 cos 30° + 70)

– 350 sin 30° (275 sin 30°) = 0

+©MD = 0; 1112.41 (60) – NA (235) = 0

NA = 284.01 N = 284 N Ans

©Fx = 0; Cx – 2574.81 sin 30° + 350 sin 30° = 0Cx = 1112.41 N

+→

FBE = 2574.81 N

Subsequently, the free-body diagram of the clamp in Fig. b will be considered. Using theresult of Cx and writing the moment equation of equilibrium about point D,

5–74.

SM_CH05.indd 353 4/8/11 11:50:13 AM


354511


6–111. Two smooth tubes A and B, each having the sameweight, W, are suspended from a common point O by meansof equal-length cords. A third tube, C, is placed between Aand B. Determine the greatest weight of C withoutupsetting equilibrium.

r/2

rB

C

3r 3r

O

rA

06c Ch06c 491-535.indd 511 6/12/09 8:54:37 AM

Free Body Diagram: When the equilibrium is about to be upset, the

reaction at B must be zero (NB = 0). From the geometry. f= cos –1

r

r23

r

4r= 48.19° and u = cos –1 = 75.52°

Equations of Equilibrium: From FBD (a),

©Fx = 0; T cos 75.52° – NC cos 48.19° = 0 [1]

+↑©Fy = 0; T sin 75.52° – Nc sin 48.19° – W = 0 [2]

+↑©Fy = 0; 2(0.5445W sin 48.19°) – Wc = 0

+→

Solving Eq. [1] and [2] yields,

T = 1.452W Nc = 0.5445W

From FBD (b),

Wc = 0.812W Ans

5–75.

SM_CH05.indd 354 4/8/11 11:50:13 AM


355526


6–127. Determine the clamping force exerted on thesmooth pipe at B if a force of 20 lb is applied to the handlesof the pliers. The pliers are pinned together at A.

A

20 lb

20 lb

10 in. 401.5 in.

0.5 in.

B

100 N

100 N

100 N

12 mm

40 mm

250 mm

250 mm40 mm

smooth pipe at B if a force of 100 N is applied to the handles

+ ΣMA = 0; 100 (250) – 40 (FB) = 0

FB = 625 N Ans

06c Ch06c 491-535.indd 526 6/12/09 8:55:27 AM

5–76.

SM_CH05.indd 355 4/8/11 11:50:13 AM


356527


*6–128. Determine the forces which the pins at A andB exert on the two-member frame which supports the100-kg crate.

A

C

B

D

0.6 m

0.8 m 0.6 m

0.4 m

06c Ch06c 491-535.indd 527 6/12/09 8:55:29 AM

AC and BC are two-force members.

Pin C:

+↑ΣFy = 0; FAC + FBC – 100(9.81) = 03

5

13

2

+→ ΣFx = 0; FAC – FBC – 50(9.81) = 0

4

5

13

3

FAC = 1154 N = 1.15 kN Ans

FBC = 520 N Ans

5–77.

SM_CH05.indd 356 4/8/11 11:50:14 AM


357528


•6–129. Determine the force in each member of the trussand state if the members are in tension or compression. D

A

E

3 m 3 m

3 m

8 kN

B

0.1 m

C

06c Ch06c 491-535.indd 528 6/12/09 8:55:30 AM

Method of Joint: In this case, support reactions are not required fordetermining the member forces. By inspection, members DB and BE are zero force members. Hence

FDB = FBE = 0 Ans

Joint C

Joint D

Joint B

Joint A

+↑©Fy = 0; FCB – 8 = 0

+↑©Fx¿ = 0; FBA – 17.89 = 0

15

FCB = 17.89 kN (C) = 17.9 kN (C) Ans

©Fx = 0; 17.89 – 8 – FCD = 025

+→

©Fx = 0; Ax – 17.89 = 0 Ax = 16.0 kN+→

©Fx = 0; 8.00 – FDE = 0 FDE = 8.00 kN (T) Ans+→


FBA = 17.89 kN (C) = 17.9 kN (C) Ans

+↑©Fy = 0; FAE – 17.89 = 015

FAE = 8.00 kN (T) Ans

25

Note: The support reactions Ex and Ey can be determined by analyzing Joint E using the results obtained above.

•5–78.

SM_CH05.indd 357 4/8/11 11:50:14 AM


358417

•6–17. Determine the greatest force P that can be appliedto the truss so that none of the members are subjected to aforce exceeding either in tension or incompression.

2 kN2.5 kN


AC

B

D

E

P

1.5 m

1.5 m

2 m2 m

1.5 m

06a Ch06a 401-445.indd 417 6/12/09 8:48:56 AM

Joint A: From the free-body diagram in Fig. a,+Q©Fy¿ = 0; FAD sin 73.74° – P sin 53.13° = 0 FAD = 0.8333P (T)+R©Fx¿ = 0; 0.8333P cos 73.74° + P cos 53.13° – FAB = 0 FAB = 0.8333P (C)

Joint D: From the free-body diagram in Fig. c,+a©Fy¿ = 0; FDC sin 73.74° – P sin 53.13° = 0 FDC = 0.8333P (C)+Q©Fx¿ = 0; FDE – 0.8333P – P cos 53.13° – 0.8333P cos 73.74° = 0 FDE = 1.6667P (T)

Method of Joints: We will begin by analyzing the equilibrium of joint A, and then proceed toanalyzing that of joints B and D.


S+ ©Fx = 0; 0.8333P – FBC = 0

FBC = 0.8333P (C)

45

45

+c©Fy = 0; FBD – 0.8333P – 0.8333P = 0

FBD = P (T)

35

35

From the above results, the greatest compressive and tensile forces developed in the member are 0.8333P and 1.6667P, respectively. Thus,

0.8333P = 2 P = 2.40 kN

1.6667P = 2.5 P = 1.50 kN (controls) Ans

•5–79.

SM_CH05.indd 358 4/8/11 11:50:15 AM


359416


.P = 5 kN


AC

B

D

E

P

1.5 m

1.5 m

2 m2 m

1.5 m

06a Ch06a 401-445.indd 416 6/12/09 8:48:54 AM

Method of Joints: We will begin by analyzing the equilibrium of joint A, and then proceed to analyzing that of joints B and D.

Joint A: From the free-body diagram in Fig. a,©Fy¿ = 0; FAD sin 73.74° – 5 sin 53.13° = 0 FAD = 4.167 kN = 4.17 kN (T) Ans©Fx¿ = 0; 4.167 cos 73.74° + 5 cos 53.13° – FAB = 0 FAB = 4.167 kN = 4.17 kN (C) Ans

Joint D: From the free-body diagram in Fig. c,+a©Fy¿ = 0; FDC sin 73.74° – 5 sin 53.13° = 0 FDC = 4.167 kN = 4.17 kN (C) Ans+Q©Fx¿ = 0; FDE – 4.167 – 5 cos 53.13° – 4.167 cos 73.74° = 0 FDE = 8.333 kN = 8.33 kN (T) Ans

Note: The equilibrium analysis of joints E and C can be used to determine the components of the support reaction at supports E and C, respectively.


S+ ©Fx = 0; 4.167 – FBC = 0

FBC = 4.167 kN = 4.17 kN (C) Ans

45

45

+c©Fy = 0; FBD – 4.167 – 4.167 = 0

FBD = 5 kN (T) Ans

35

35

*5–80.

SM_CH05.indd 359 4/8/11 11:50:15 AM


360531


*6–132. Determine the horizontal and vertical componentsof reaction that the pins A and B exert on the two-memberframe. Set .F = 0

1.5 m

400 N/m

60

1 m

1 m

B

C

A

F

06c Ch06c 491-535.indd 531 6/12/09 8:55:37 AM

CB is a two-force member.

Member AC:

+©MA = 0; – 600(0.75) + 1.5 (FCB sin 75°) = 0

FCB = 310.6

Thus,

Bx = By = 310.6 = 220 N Ans12

+→ ©Fx = 0; – Az + 600 sin 60° – 310.6 cos 45° = 0

Az = 300 N Ans

+↑©Fy = 0; Ay – 600 cos 60° + 310.6 sin 45° = 0

Ay = 80.4 N Ans

5–81.

SM_CH05.indd 360 4/8/11 11:50:16 AM


361532


•6–133. Determine the horizontal and vertical componentsof reaction that pins A and B exert on the two-memberframe. Set .F = 500 N

1.5 m

400 N/m

60

1 m

1 m

B

C

A

F

06c Ch06c 491-535.indd 532 6/12/09 8:55:38 AM

Member AC:

Member CB:

Member AC:

Member CB:

Solving,

+©MB = 0; – Cx (1) – Cy (1) + 500(1) = 0

Cx = 402.6 N

Cy = 97.4 N

Ax = 117 N Ans

Ay = 397 N Ans

Bx = 97.4 N Ans

By = 97.4 N Ans

©Fx = 0; – Ax + 600 sin 60° – 402.6 = 0+→

©Fx = 0; 402.6 – 500 + Bx = 0+→

+↑©Fy = 0; Ay – 600 cos 60° – 97.4 = 0

+↑©Fy = 0; – By + 97.4 = 0

+©MA = 0; – 600(0.75) – Cy (1.5 cos 60°) + Cx (1.5 sins 60°) = 0

•5–82.

SM_CH05.indd 361 4/8/11 11:50:16 AM


362533


6–134. The two-bar mechanism consists of a lever arm ABand smooth link CD, which has a fixed smooth collar at itsend C and a roller at the other end D. Determine the force Pneeded to hold the lever in the position . The spring has astiffness k and unstretched length 2L. The roller contactseither the top or bottom portion of the horizontal guide.

u

2 L

L

k

C

A

B

D

P

u

06c Ch06c 491-535.indd 533 6/12/09 8:55:41 AM

Free Body Diagram: The spring compresses x = 2L – . Then, Lsin u

the spring force developed is Fsp = kx = kL .1

sin u2 –

Equations of Equilibrium: From FBD (a),

From FBD (b),

©Fx = 0; kL – FCD sin u = 0+→

1sin u

2 –

FCD =

P = (2 – cos u) Ans

kLsin u

1sin u

2 –

kLsin u

1sin u

2 –

+©MD = 0; MC = 0

+©MA = 0; P(2L) – (L cos u) = 0

kL2 tan u sin u

5–83.

SM_CH05.indd 362 4/8/11 11:50:16 AM


363534


6–135. Determine the horizontal and vertical componentsof reaction at the pin supports A and E of the compoundbeam assembly. 2 ft

2 kip/ft

1 ft

3 ft 6 ft2 ft1 ft

A

C

EDB

0.9 m 0.6 m 1.8 m0.3 m

0.3 m0.6 m

50 kN/m

0.3 m

1.5 m

0.9 m 1.2 m 0.6 m

12

(50) (1.8) = 45 kN

Member BDE :

+ ΣMg = 0; 45 (0.6) + T2

5

(1.8) – R

1

5

(2.7) = 0

Member AC :

+ ΣMA = 0; T1

5

(0.3) + T

2

5

(1.5) – R

1

5

(1.5) = 0

Solving,

T = 25.80 kN, R = 56.76 kN

Member AC :

+→ ΣFx = 0; Ax – 56.76

2

5

– 25.80

1

5

= 0

Ax = 62.31 kN Ans

+↑ΣFy = 0; Ay – 56.761

5

+ 25.80

2

5

= 0

Ay = 2.31 kN Ans

Member BDE :

+→ ΣFx = 0; 56.76

2

5

+ 25.80

1

5

– Ex = 0

Ex = 62.31 kN Ans

+↑ΣFy = 0; 56.761

5

– 25.80

2

5

– 4.5 + Ey = 0

Ey = 42.69 kN Ans

06c Ch06c 491-535.indd 534 6/12/09 8:55:42 AM

*5–84.

SM_CH05.indd 363 4/8/11 11:50:17 AM


364469


•6–69. Determine the force required to hold the 50-kgmass in equilibrium.

P

P

A

B

C

06b Ch06b 446-490.indd 469 6/12/09 8:52:20 AM

Equations of Equilibrium: Applying the force equation of equilibrium along the y axis of each pulley.

+c©Fy = 0; R – 3P = 0; R = 3P

+c©Fy = 0; R – 3R = 0; T = 3P = 9P

+c©Fy = 0; 2P + 2R + 2T – 50(9.81) = 0 Ans

2P + 2(3P) + 2(9P) = 50(9.81)

P = 18.9 N Ans

Substituting Eqs.(1) and (2) into Eq.(3) and solving for P,

•5–85.

SM_CH05.indd 364 4/8/11 11:50:20 AM

Documents

5–1. Determine the force in each member of the truss, D E Cfaculty.kfupm.edu.sa/ce/saghamdi/hmpage/CE202/S_M… · · 2014-12-21Determine the force in each member of the truss,