Engineering mechanics statics 13th edition hibbeler solutions manual

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6–1.

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 800 lb and P2 = 400 lb.

6 ft

8 ft C

A

8 ft

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

P2

Joint B:

B P1

+ ©F = 0; : x

3 FBC cos 45° - FBA a b - 400 = 0 5

(1)

+ c ©Fy = 0;

4 FBC sin 45° + FBA a b - 800 = 0 5

(2)

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Solving Eqs. (1) and (2) yields FBA = 285.71 lb (T) = 286 lb (T)

Ans.

FBC = 808.12 lb (T) = 808 lb (T)

Ans.

Joint C: + ©F = 0; : x

FCA - 808.12 cos 45° = 0

FCA = 571 lb (C)

+ c ©Fy = 0;

Ans.

Cy - 808.12 sin 45° = 0 Cy = 571 lb

Note: The support reactions Ax and Ay can be determined by analyzing Joint A using the results obtained above.

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6–2.

Determine the force on each member of the truss and state if the members are in tension or compression. Set P1 = 500 lb and P2 = 100 lb.

6 ft

8 ft C

A

8 ft

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

P2

Joint B:

B P1

+ ©F = 0; : x

3 FBC cos 45° - FBA a b - 100 = 0 5

(1)

+ c ©Fy = 0;

4 FBC sin 45° + FBA a b - 500 = 0 5

(2)

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Solving Eqs. (1) and (2) yields FBA = 285.71 lb (T) = 286 lb (T)

Ans.

FBC = 383.86 lb (T) = 384 lb (T)

Ans.

Joint C: + ©F = 0; : x

FCA - 383.86 cos 45° = 0

FCA = 271 lb (C)

+ c ©Fy = 0;

Ans.

Cy - 383.86 sin 45° = 0 Cy = 271.43 lb

Note: The support reactions Ax and Ay can be determined by analyzing Joint A using the results obtained above.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–3.

Determine the force in each member of the truss, and state if the members are in tension or compression. Set u = 0°.

D

3 kN

1.5 m A

SOLUTION

B

Support Reactions: Applying the equations of equilibrium to the free-body diagram of the entire truss,Fig.a, we have a + ©MA = 0;

C

2m

2m 4 kN

NC (2 + 2) - 4(2) - 3(1.5) = 0 NC = 3.125 kN

+ ©F = 0; : x

3 - Ax = 0 A x = 3 kN

+ c ©Fy = 0;

A y + 3.125 - 4 = 0 A y = 0.875 kN

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

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 + c ©Fy = 0;

3 3.125 - FCD a b = 0 5

FCD = 5.208 kN = 5.21 kN (C)

+ ©F = 0; : x

Ans.

4 5.208 a b - FCB = 0 5

FCB = 4.167 kN = 4.17 kN (T)

Ans.

Joint A: From the free-body diagram in Fig. c, we can write + c ©Fy = 0;

3 0.875 - FAD a b = 0 5

FAD = 1.458 kN = 1.46 kN (C)

+ ©F = 0; : x

Ans.

4 FAB - 3 - 1.458a b = 0 5

FAB = 4.167 kN = 4.17 kN (T)

Ans.

Joint B: From the free-body diagram in Fig. d, we can write + c ©Fy = 0;

FBD - 4 = 0 FBD = 4 kN (T)

+ ©F = 0; : x

4.167 - 4.167 = 0

Ans. (check!)

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

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

u

*6–4.

Determine the force in each member of the truss, and state if the members are in tension or compression. Set u = 30°.

D

3 kN

1.5 m A

SOLUTION

B

Support Reactions: From the free-body diagram of the truss, Fig. a, and applying the equations of equilibrium, we have a + ©MA = 0;

C

2m

2m 4 kN

NC cos 30°(2 + 2) - 3(1.5) - 4(2) = 0 NC = 3.608 kN

+ ©F = 0; : x

3 - 3.608 sin 30° - A x = 0 A x = 1.196 kN

+ c ©Fy = 0;

A y + 3.608 cos 30° - 4 = 0 A y = 0.875 kN

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

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 + c ©Fy = 0;

3 3.608 cos 30° - FCD a b = 0 5

FCD = 5.208 kN = 5.21 kN (C)

+ ©F = 0; : x

Ans.

4 5.208 a b - 3.608 sin 30° - FCB = 0 5

FCB = 2.362 kN = 2.36 kN (T)

Ans.

Joint A: From the free-body diagram in Fig. c, we can write + c ©Fy = 0;

3 0.875 - FAD a b = 0 5

FAD = 1.458 kN = 1.46 kN (C)

+ ©F = 0; : x

Ans.

4 FAB - 1.458 a b - 1.196 = 0 5

FAB = 2.362 kN = 2.36 kN (T)

Ans.

Joint B: From the free-body diagram in Fig. d, we can write + c ©Fy = 0;

FBD - 4 = 0 FBD = 4 kN (T)

+ ©F = 0; : x

2.362 - 2.362 = 0

Ans. (check!)

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

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u

6–5.

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

300 N 400 N D

SOLUTION

C

2m

Method of Joints: Here, the support reactions A and C do not need to be determined.

We will first analyze the equilibrium of joints D and B, and then proceed to analyze joint C.

250 N

A B 2m

Joint D: From the free-body diagram in Fig. a, we can write + ©F = 0; : x

400 - FDC = 0 FDC = 400 N (C)

+ c ©Fy = 0;

200 N

Ans.

FDA - 300 = 0 FDA = 300 N (C)

Ans.

+ ©F = 0; : x

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Joint B: From the free-body diagram in Fig. b, we can write 250 - FBA = 0

FBA = 250 N (T)

+ c ©Fy = 0;

Ans.

FBC - 200 = 0

FBC = 200 N (T)

Ans.

Joint C: From the free-body diagram in Fig. c, we can write + c ©Fy = 0;

FCA sin 45° - 200 = 0

FCA = 282.84 N = 283 N (C)

+ ©F = 0; : x

Ans.

400 + 282.84 cos 45° - NC = 0

NC = 600 N

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

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6–6.

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

600 N

D

4m

SOLUTION

C

Method of Joints: We will begin by analyzing the equilibrium of joint D, and then proceed to analyze joints C and E. Joint D: From the free-body diagram in Fig. a, + ©F = 0; : x

4m B

A

3 FDE a b - 600 = 0 5 FDE = 1000 N = 1.00 kN (C)

+ c ©Fy = 0;

900 N

E

6m

Ans.

4 1000 a b - FDC = 0 5 Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FDC = 800 N (T) Joint C: From the free-body diagram in Fig. b, + ©F = 0; : x

FCE - 900 = 0

FCE = 900 N (C)

+ c ©Fy = 0;

Ans.

800 - FCB = 0

FCB = 800 N (T)

Ans.

Joint E: From the free-body diagram in Fig. c, R+ ©Fx ¿ = 0;

- 900 cos 36.87° + FEB sin 73.74° = 0 FEB = 750 N (T)

Q+ ©Fy ¿ = 0;

Ans.

FEA - 1000 - 900 sin 36.87° - 750 cos 73.74° = 0 FEA = 1750 N = 1.75 kN (C)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–7.

Determine the force in each member of the Pratt truss, and state if the members are in tension or compression.

J 2m

K

2m

SOLUTION

2m

Joint A: + c ©Fy = 0;

L

H

A B D C E F 2m 2m 2m 2m 2m 2m

20 - FAL sin 45° = 0

10 kN

FAL = 28.28 kN (C) + ©F = 0; : x

I

20 kN

10 kN

FAB - 28.28 cos 45° = 0 FAB = 20 kN (T)

Joint B: + ©F = 0; : x

FBC - 20 = 0

+ c ©Fy = 0;

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FBC = 20 kN (T) FBL = 0

Joint L: R+ ©Fx = 0;

FLC = 0

+Q©Fy = 0;

28.28 - FLK = 0

FLK = 28.28 kN (C) Joint C: + ©F = 0; : x

FCD - 20 = 0

FCD = 20 kN (T)

+ c ©Fy = 0;

FCK - 10 = 0

FCK = 10 kN (T)

Joint K: R+ ©Fx - 0;

10 sin 45° - FKD cos (45° - 26.57°) = 0 FKD = 7.454 kN (L)

+Q©Fy = 0;

28.28 - 10 cos 45° + 7.454 sin (45° - 26.57°) - FKJ = 0 FKJ = 23.57 kN (C)

Joint J: + ©F = 0; : x

23.57 sin 45° - FJI sin 45° = 0 FJI = 23.57 kN (L)

+ c ©Fy = 0;

2 (23.57 cos 45°) - FJD = 0 FJD = 33.3 kN (T)

Ans.

FAL = FGH = FLK = FHI = 28.3 kN (C)

Ans.

FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T)

Ans.

FBL = FFH = FLC = FHE = 0

Ans.

FCK = FEI = 10 kN (T)

Ans.

FKJ = FIJ = 23.6 kN (C)

Ans.

Due to Symmetry

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by = FID = 7.45 kN (C) FKD Copyright and written permission should be obtained from the publisher prior to anyAns. prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

G

*6–8.

Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The horizontal force component at A must be zero. Why?

800 lb

600 lb B

60

C

4 ft

SOLUTION Joint C: + ©F = 0; : x

FCB - 800 cos 60° = 0 FCB = 400 lb (C)

+ c ©Fy = 0;

Ans.

3 F - 400 = 0 5 BD

FBD = 666.7 = 667 lb (T) + c ©Fy = 0;

3 ft

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

+ ©F = 0; : x

D

FCD - 800 sin 60° = 0 FCD = 693 lb (C)

Joint B:

A

Ans.

FBA -

Ans.

4 (666.7) - 600 = 0 5

FBA = 1133 lb = 1.13 kip (C)

Ans.

Member AB is a two-force member and exerts only a vertical force along AB at A.

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6–9.

Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The vertical component of force at C must equal zero. Why?

B

C

2m

D

SOLUTION

A

Joint A: + c ©Fy = 0;

1.5 m

8 kN

Ans.

3 - FAE + 7.5 a b = 0 5 Ans.

+ ©F = 0; : x

FED = 4.5 kN(C)

Ans.

+ c ©Fy = 0;

FEB = 8 kN (T)

Ans.

Joint E:

Joint B: 1

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FAE = 4.5 kN (C)

+ c ©Fy = 0;

22

(FBD) - 8 -

4 (7.5) = 0 5

FBD = 19.8 kN (C) + ©F = 0; : x

2m

6 kN

4 FAB - 6 = 0 5 FAB = 7.5 kN (T)

+ ©F = 0; : x

E

FBC -

Ans.

1 3 (7.5) (19.8) = 0 5 22

FBC = 18.5 kN (T)

Ans.

Cy is zero because BC is a two-force member .

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6–10.

Each member of the truss is uniform and has a mass of 8 kg>m. Remove the external loads of 6 kN and 8 kN and determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member.

B

2m

D

SOLUTION

A 1.5 m

Joint A: + c ©Fy = 0;

Ans.

3 - FAE + 196.2a b = 0 5

+ ©F = 0; : x

FED = 117.7 = 118 N (C)

Ans.

+ c ©Fy = 0;

FEB = 215.8 = 216 N (T)

Ans.

Joint E:

Joint B: 1

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Ans.

22

(FBD) - 366.0 - 215.8 -

4 (196.2) = 0 5

FBD = 1045 = 1.04 kN (C) + ©F = 0; : x

2m

8 kN

FAE = 117.7 = 118 N (C)

+ c ©Fy = 0;

E

6 kN

4 - 157.0 = 0 F 5 AB FAB = 196.2 = 196 N (T)

+ ©F = 0; : y

C

FBC -

Ans.

1 3 (196.2) (1045) = 0 5 22

FBC = 857 N (T)

Ans.

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6–11.

Determine the force in each member of the truss and state if the members are in tension or compression.

4 kN 3m

3m

3m C

B

D 3m 5m

A F

SOLUTION

E

Support Reactions:

5 kN

a + ©MD = 0;

4162 + 5192 - Ey 132 = 0

+ c ©Fy = 0;

23.0 - 4 - 5 - Dy = 0

+ ©F = 0 : x

Ey = 23.0 kN Dy = 14.0 kN

Dx = 0

Joint D: + c ©Fy = 0;

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Method of Joints:

FDE ¢

5

234

≤ - 14.0 = 0

FDE = 16.33 kN 1C2 = 16.3 kN 1C2

+ ©F = 0; : x

16.33 ¢

3

234

≤ - FDC = 0

FDC = 8.40 kN 1T2

Joint E: + ©F = 0; : x

FEA ¢

3

210

≤ - 16.33 ¢

3

234

Ans.

≤ = 0

FEA = 8.854 kN 1C2 = 8.85 kN 1C2

+ c ©Fy = 0;

23.0 - 16.33 ¢

5

234

≤ - 8.854 ¢

1

210

+ c ©Fy = 0;

Ans.

6.20 - FCF sin 45° = 0

FCF = 8.768 kN 1T2 = 8.77 kN 1T2 + ©F = 0; : x

Ans.

≤ - FEC = 0

FEC = 6.20 kN 1C2

Joint C:

Ans.

Ans.

8.40 - 8.768 cos 45° - FCB = 0 FCB = 2.20 kN T

Ans.

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6–11. (continued) Joint B: + ©F = 0; : x

2.20 - FBA cos 45° = 0 FBA = 3.111 kN 1T2 = 3.11 kN 1T2

+ c ©Fy = 0;

Ans.

FBF - 4 - 3.111 sin 45° = 0 FBF = 6.20 kN 1C2

Ans.

+ c ©Fy = 0;

8.768 sin 45° - 6.20 = 0

(Check!)

+ ©F = 0; : x

8.768 cos 45° - FFA = 0

Joint F:

Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FFA = 6.20 kN 1T2

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*6–12.

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 10 kN, P2 = 15 kN.

G

B

C

4m

A

F

E

D

SOLUTION a + ©MA = 0;

2m

Gx (4) - 10(2) - 15(6) = 0 Gx = 27.5 kN

+ ©F = 0; : x

4m P1

2m P2

Ax - 27.5 = 0 Ax = 27.5 kN

+ c ©Fy = 0;

Ay - 10 - 15 = 0

Joint G: + ©F = 0; : x

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Ay = 25 kN

FGB - 27.5 = 0

FGB = 27.5 kN (T) Joint A: + ©F = 0; : x

27.5 - FAF -

+ c ©Fy = 0;

25 - FAB a

1

(FAB) = 0

25

2

25

Ans.

b = 0

FAF = 15.0 kN (C)

Ans.

FAB = 27.95 = 28.0 kN (C)

Ans.

Joint B: + ©F = 0; : x

27.95 a

+ c ©Fy = 0;

27.95 a

1

25 2

25

b + FBC - 27.5 = 0

b - FBF = 0

FBF = 25.0 kN (T)

Ans.

FBC = 15.0 kN (T)

Ans.

Joint F: 1

+ ©F = 0; : x

15 + FFE -

+ c ©Fy = 0;

25 - 10 - FFC a

22

(FFC) = 0 1 22

b = 0

FFC = 21.21 = 21.2 kN (C)

Ans.

FFE = 0

Ans.

Joint E: + ©F = 0; :

FED = 0

Ans.

+ c ©Fy = 0;

FEC - 15 = 0

x

FEC = 15.0 kN (T)

Ans.

Joint D: + ©©F 2013=Pearson Education, is protected by 0; F = 0Inc., Upper Saddle River, NJ. All rights reserved. This publication : Ans. x

DC

Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–13.

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 0, P2 = 20 kN.

G

B

C

4m

A

F

E

D

SOLUTION a + ©MA = 0;

FGB = 30 kN (T) + ©F = 0; : x

2m

-FGB (4) + 20(6) = 0

4m P1

2m P2

Ans.

Ax - 30 = 0 Ax = 30 kN

+ c ©Fy = 0;

Ay - 20 = 0 Ay = 20 kN

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Joint A:

1

+ ©F = 0; : x

30 - FAF -

+ c ©Fy = 0;

20 - FAB a

25 2

25

(FAB) = 0

b = 0

FAF = 20 kN (C)

Ans.

FAB = 22.36 = 22.4 kN (C)

Ans.

Joint B: + ©F = 0; : x

22.36 a

+ c ©Fy = 0;

22.36 a

1

25 2

25

b + FBC - 30 = 0

b - FBF = 0

FBF = 20 kN (T)

Ans.

FBC = 20 kN (T)

Ans.

Joint F: + ©F = 0; : x

20 + FFE -

+ c ©Fy = 0;

20 - FFC a

1

22

1

22

(FFC) = 0

b = 0

FFC = 28.28 = 28.3 kN (C)

Ans.

FFE = 0

Ans.

Joint E: + ©F = 0; : x

FED - 0 = 0

+ c ©Fy = 0;

FEC - 20 = 0 FED = 0

Ans.

FEC = 20.0 kN (T)

Ans.

Joint D: + ©F = 0; : x

1 25

(FDC) - 0 = 0

0 Upper Saddle River, NJ. All rights reserved. ThisAns. FDC = Inc., © 2013 Pearson Education, publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–14.

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 100 lb, P2 = 200 lb, P3 = 300 lb.

B

C

P1 10 ft A

D F

c + ©MA = 0;

10 ft

10 ft

SOLUTION

P2

200(10) + 300(20) - RD cos 30°(30) = 0

E

30 10 ft

P3

RD = 307.9 lb + c ©Fy = 0;

Ay - 100 - 200 - 300 + 307.9 cos 30° = 0 Ay = 333.3 lb

+ ©F = 0; : x

Ax - 307.9 sin 30° = 0

Joint A: + c ©Fy = 0;

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Ax = 154.0 lb

333.3 - 100 -

1

22

FAB = 0

FAB = 330 lb (C) + ©F = 0; : x

154.0 + FAF -

Ans.

1

22

(330) = 0

FAF = 79.37 = 79.4 lb (T) Joint B: 1

+ c ©Fy = 0;

22

(330) - FBF = 0

FBF = 233.3 = 233 lb (T) 1

+ ©F = 0; : x

22

+ c ©Fy = 0;

-

1

22

Ans.

FFC - 200 + 233.3 = 0

FFC = 47.14 = 47.1 lb (C) + ©F = 0; : x

Ans.

(330) - FBC = 0

FBC = 233.3 = 233 lb (C) Joint F:

Ans.

FFE - 79.37 -

1 22

Ans.

(47.14) = 0

FFE = 112.7 = 113 lb (T)

Ans.

+ ©F = 0; : x

FEC = 300 lb (T)

Ans.

+ c ©Fy = 0;

FED = 112.7 = 113 lb (T)

Ans.

Joint E:

Joint C: + ©F = 0; : x

1 22

(47.14) + 233.3 -

1 22

FCD = 0

FCD = 377.1 = 377 lb (C) + c ©Fy = 0;

1

(47.14) - 300 +

1

(377.1) = 0

Ans. Check!

© 2013 Pearson Education, Inc., Upper Saddle 22 22 River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–15.

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 400 lb, P2 = 400 lb, P3 = 0.

B

C

P1 10 ft A

D F 10 ft

SOLUTION a + ©MA = 0;

10 ft P2

- 400(10) + RD cos 30°(30) = 0

E

30 10 ft

P3

RD = 153.96 lb + c ©Fy = 0;

Ay - 400 - 400 + 153.96 cos 30° = 0 Ay = 666.67 lb

+ ©F = 0; : x

Ax - 153.96 sin 30° = 0

Joint A: + c ©Fy = 0;

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Ax = 76.98 lb

666.67 - 400 -

1

22

FAB = 0

FAB = 377.12 = 377 lb (C) + ©F = 0; : x

76.98 + FAF -

1

22

Ans.

(377.12) = 0

FAF = 189.69 = 190 lb (T) Joint B: + c ©Fy = 0;

1 22

Ans.

(377.12) - FBF = 0

FBF = 266.67 = 267 lb (T) + ©F = 0; : x

1 22

Ans.

(377.12) - FBC = 0

FBC = 266.67 = 267 lb (C) Joint F: + c ©Fy = 0;

1 22

Ans.

FFC - 400 + 266.67 = 0

FFC = 188.56 = 189 lb (T) + ©F = 0; : x

FFE - 189.69 +

1 22

Ans.

(188.56) = 0

FFE = 56.35 = 56.4 lb (T)

Ans.

+ ©F = 0; : x

FED = 56.4 lb (T)

Ans.

+ c ©Fy = 0;

FEC = 0

Ans.

Joint E:

Joint C: + ©F = 0; : x

-

1 22

(188.56) + 266.67 -

FCD = 188.56 = 189 lb (C)

1 22

FCD = 0 Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*6–16.

Determine the force in each member of the truss. State whether the members are in tension or compression. Set P = 8 kN.

4m B

A

60°

SOLUTION

C

60°

E

4m

4m

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

Joint D: + c ©Fy = 0;

FDC sin 60° - 8 = 0 FDC = 9.238 kN 1T2 = 9.24 kN 1T2 FDE - 9.238 cos 60° = 0 FDE = 4.619 kN 1C2 = 4.62 kN 1C2

Joint C: + c ©Fy = 0;

+ c ©Fy = 0;

Ans.

219.238 cos 60°2 - FCB = 0

FCB = 9.238 kN 1T2 = 9.24 kN 1T2 Joint B:

Ans.

FCE sin 60° - 9.238 sin 60° = 0

FCE = 9.238 kN 1C2 = 9.24 kN 1C2 + ©F = 0; : x

Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

+ ©F = 0; : x

Ans.

FBE sin 60° - FBA sin 60° = 0 FBE = FBA = F

+ ©F = 0; : x

9.238 - 2F cos 60° = 0

F = 9.238 kN

Thus,

FBE = 9.24 kN 1C2

FBA = 9.24 kN 1T2

Ans.

Joint E: + c ©Fy = 0;

Ey - 219.238 sin 60°2 = 0

+ ©F = 0; : x

FEA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0 FEA = 4.62 kN 1C2

D

Ey = 16.0 kN

Ans.

Note: The support reactions Ax and Ay can be determinedd by analyzing Joint A using the results obtained above.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

P

6–17.

If the maximum force that any member can support is 8 kN in tension and 6 kN in compression, determine the maximum force P that can be supported at joint D.

4m B

A

SOLUTION

60°

C

60°

E

4m

4m

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

Joint D: + c ©Fy = 0;

FDC sin 60° - P = 0

+ ©F = 0; : x

FDE - 1.1547P cos 60° = 0

FDC = 1.1547P 1T2 FDE = 0.57735P 1C2

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Joint C: + c ©Fy = 0;

FCE sin 60° - 1.1547P sin 60° = 0 FCE = 1.1547P 1C2

+ ©F = 0; : x

211.1547P cos 60°2 - FCB = 0

FCB = 1.1547P 1T2

+ c ©Fy = 0;

FBE sin 60° - FBA sin 60° = 0

FBE = FBA = F

+ ©F = 0; : x

1.1547P - 2F cos 60° = 0

Joint B:

Thus,

FBE = 1.1547P 1C2 Joint E: + ©F = 0; : x

F = 1.1547P

FBA = 1.1547P 1T2

FEA + 1.1547P cos 60° - 1.1547P cos 60°

- 0.57735P = 0

FEA = 0.57735P 1C2

From the above analysis, the maximum compression and tension in the truss member is 1.1547P. For this case, compression controls which requires 1.1547P = 6 P = 5.20 kN

D

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

P

6–18.

Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The resultant force at the pin E acts along member ED. Why?

3 kN 3m A

4m

2 kN 3m

B

D

SOLUTION Joint C: E

+ c ©Fy = 0;

2 213

FCD - 2 = 0

FCD = 3.606 = 3.61 kN (C) + ©F = 0; : x

-FCD + 3.606a

3 213

Ans.

b = 0 Ans.

+ ©F = 0; : x

FBA = 3 kN (T)

Ans.

+ c ©Fy = 0;

FBD = 3 kN (C)

Ans.

Joint B:

Joint D: + ©F = 0; : x + c ©Fy = 0;

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FCB = 3 kN (T)

3

213 2

213

FDE -

(FDE) -

3

213

(3.606) +

2

213

(FDA) -

3

213 2

213

FDA = 0

(3.606) - 3 = 0

FDA = 2.70 kN (T)

Ans.

FDE = 6.31 kN (C)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

C

6–19.

Each member of the truss is uniform and has a mass of 8 kg>m. Remove the external loads of 3 kN and 2 kN and determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member.

3 kN 3m A

4m

2 kN 3m

B

D

SOLUTION Joint C: + c ©Fy = 0;

E

2 213

FCD - 259.2 = 0

FCD = 467.3 = 467 N (C) + ©F = 0; : x

- FCB + 467.3a

3 213

Ans.

b = 0 Ans.

+ ©F = 0; : x

FBA = 388.8 = 389 N (T)

Ans.

+ c ©Fy = 0;

FBD = 313.9 = 314 N (C)

Ans.

Joint B:

Joint D: + ©F = 0; : x

+ c ©Fy = 0;

3

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FCB = 388.8 = 389 N (T)

213 2

213

FDE -

3

213

(FDE) +

(467.3) -

2

213

(FDA) -

3

213

FDA = 0

2

213

(467.3) - 313.9 - 502.9 = 0

FDE = 1204 = 1.20 kN (C)

Ans.

FDA = 736 N (T)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

C

*6–20.

Determine the force in each member of the truss in terms of the load P, and indicate whether the members are in tension or compression.

B

P

d C

A D

F

d

SOLUTION Support Reactions:

E

a + ©ME = 0;

3 P(2d) - Ay a d b = 0 2

+ c ©Fy = 0;

4 P - Ey = 0 3

Ey =

+ ©F = 0; : x

Ex - P = 0

Ex = P

4 Ay = P 3

d

d/2

d/2

d

4 P 3

Joint A: + c ©Fy = 0;

FAB ¢

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Method of Joints: By inspection of joint C, members CB and CD are zero force members. Hence Ans. FCB = FCD = 0 1

23.25

≤ -

4 P = 0 3

FAB = 2.40P (C) = 2.40P (C) + ©F = 0; : x

FAF - 2.404P ¢

1.5

23.25

Ans.

≤ = 0

FAF = 2.00P (T) Joint B: + ©F = 0; : x

2.404P ¢

1.5

23.25

Ans.

≤ - P - FBF ¢

0.5

21.25

≤ - FBD ¢

0.5

21.25

≤ = 0

1.00P - 0.4472FBF - 0.4472FBD = 0 + c ©Fy = 0;

2.404P ¢

1

23.25

≤ + FBD ¢

1

21.25

≤ - FBF ¢

(1)

1

21.25

1.333P + 0.8944FBD - 0.8944FBF = 0

≤ = 0

(2)

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

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

Ans.

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

Ans.

1 1 ≤ - FFE ¢ ≤ = 0 21.25 21.25 FFE = 1.863P (T) = 1.86P (T)

Ans.

Joint F: + c ©Fy = 0;

+ ©F = 0; : x

1.863P ¢

FFD + 2 B 1.863P ¢

0.5 21.25

≤ R - 2.00P = 0

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

Ans.

Joint D: + c ©Fy = 0;

FDE ¢

1 21.25

≤ - 0.3727P ¢

1 21.25

≤ = 0

FDE = 0.3727P (C) = 0.373P (C) + ©F = 0; : y

2 B 0.3727P ¢

0.5 21.25

Ans.

≤ R - 0.3333P = 0 (Check!)

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–21.

If the maximum force that any member can support is 4 kN in tension and 3 kN in compression, determine the maximum force P that can be applied at joint B. Take d = 1 m.

B

P

d

SOLUTION

C

A

Support Reactions:

D

F

a + ©ME = 0;

3 P(2d) - Ay a d b = 0 2

4 Ay = P 3

+ c ©Fy = 0;

4 P - Ey = 0 3

Ey =

+ ©F = 0; : x

Ex - P = 0

Ex = P

d

4 P 3

E d

d/2

d/2

d

Method of Joints: By inspection of joint C, members CB and CD are zero force members. Hence FCB = FCD = 0

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Joint A: 1

+ c ©Fy = 0;

FAB ¢

+ ©F = 0; : x

FAF - 2.404P ¢

23.25

Joint B: + ©F = 0; : x

2.404P ¢

4 P = 0 3

FAB = 2.404P (C)

1.5

FAF = 2.00P (T)

≤ -

1.5

23.25

23.25

≤ = 0

≤ - P - FBF ¢

0.5

21.25

≤ - FBD ¢

0.5

21.25

≤ = 0

1.00P - 0.4472FBF - 0.4472FBD = 0 + c ©Fy = 0;

2.404P ¢

1

23.25

≤ + FBD ¢

1

21.25

≤ - FBF ¢

(1)

1

21.25

1.333P + 0.8944FBD - 0.8944FBF = 0

≤ = 0

(2)

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

FBF = 1.863P (T) Joint F: + c ©Fy = 0;

1.863P ¢

1

21.25

FBD = 0.3727P (C)

≤ - FFE ¢

1

21.25

≤ = 0

FFE = 1.863P (T) + ©F = 0; : x

FFD + 2 B 1.863P ¢

0.5 21.25

≤ R - 2.00P = 0

FFD = 0.3333P (T) Joint D: + c ©Fy = 0;

FDE ¢

1 21.25

≤ - 0.3727P ¢

1 21.25

≤ = 0

FDE = 0.3727P (C) + ©F = 0; : y

2 B 0.3727P ¢

0.5 21.25

≤ R - 0.3333P = 0 (Check!)

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 = 3 © 2013 Pearson Education, Inc., Saddle River, NJ. All rights reserved. This publication is protected by P =Upper 1.25 kN Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–22.

B

Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression.

C

L/3

SOLUTION c + ©MA = 0;

L 2L Pa b + P a b - (Dy)(L) = 0 3 3 Dy = P

+ c ©Fy = 0;

Ay = P

E

A L/3

F L/3

P

D L/3

P

Joint F: + ©F = 0; : x

FFD - FFE - FFB a

1 22

(1)

b = 0

FFD - FFE = P 1

FFB a

22

b -P = 0

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

+ ©F = 0; : y

FFB = 22P = 1.41 P (T) Similarly,

FEC = 22P Joint C: + ©F = 0; : x

2

FCA a 2 25

+c ©Fy = 0;

FCA

25

FCA 1

25

1

22

1

22

b - FCD a

1

22

- 22P

1

22

+ FCD

1

22

=0

2 25 P = 1.4907P = 1.49P (C) 3

FCD =

22 P = 0.4714P = 0.471P (C) 3

FAE FAE =

b = 0

FCD = P

FCA =

Joint A: + ©F = 0; : x

b - 22P a

1 2 22 2 25 Pa Pa b b = 0 3 3 22 25

5 P = 1.67 P (T) 3

Similarly, FFD=1.67 P (T) From Eq.(1), and Symmetry, 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.

© 2013 Pearson is protected by Ans. FEC = 1.41 P (T) Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, FCD Ans. = 0.471 P (C) photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–23.

Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

L

B

L

L

C

L

L

A

D

E

SOLUTION

L P

Entire truss: a + ©MA = 0;

- P(L) + Dy (2 L) = 0 Dy =

P 2

P - P + Ay = 0 2 Ay =

+ ©F = 0; : x

Ax = 0

Joint D:

P 2

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

+ c ©Fy = 0;

+ c ©Fy = 0;

- FCD sin 60° +

P = 0 2

FCD = 0.577 P (C) + ©F = 0; ; x

Ans.

FDB - 0.577P cos 60° = 0 FDB = 0.289 P(T)

Joint C: + c ©Fy = 0;

Ans.

0.577 P sin 60° - FCE sin 60° = 0 FCE = 0.577 P (T)

+ ©F = 0; : x

L

Ans.

FBC - 0.577 P cos 60° - 0.577P cos 60° = 0 FBC = 0.577 P (C)

Ans.

Due to symmetry: FBE = FCE = 0.577 P (T)

Ans.

FAB = FCD = 0.577 P (C)

Ans.

FAE = FDE = 0.577 P (T)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*6–24.

Each member of the truss is uniform and has a weight W. Remove the external force P and determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member.

L

B

L

L

C

L

L

A

D

E

SOLUTION

L P

Entire truss: a + ©MA = 0;

L

3 L 3 3 - W a b - 2 W(L) - W a L b - W(2 L) + Dy (2 L) = 0 2 2 2 2 Dy =

7 W 2

Joint D: 7 W - W - FCD sin 60° = 0 2

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

+ c ©Fy = 0;

FCD = 2.887W = 2.89 W (C) + ©F = 0; : x

2.887W cos 60° - FDE = 0 FDE = 1.44 W (T)

Joint C: + c ©Fy = 0;

2.887W sin 60° -

Ans.

FBC - 1.1547W cos 60° - 2.887W cos 60° = 0 FBC = 2.02 W (C)

Due to symmetry:

Ans.

3 W - FCE sin 60° = 0 2

FCE = 1.1547W = 1.15 W (T) + ©F = 0; : x

Ans.

Ans.

FBE = FCE = 1.15 W (T)

Ans.

FAB = FCD = 2.89 W (C)

Ans.

FAE = FDE = 1.44 W (T)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–25.

Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

P B

L

C u

L

L

SOLUTION A

Joint B: + c ©Fy = 0;

L

FBA sin 2u - P = 0 FBA = P csc 2u (C)

+ ©F = 0; : x

Ans.

P csc 2u(cos 2u) - FBC = 0 FBC = P cot 2 u (C)

Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Joint C: + ©F = 0; : x

P cot 2 u + P + FCD cos 2 u - FCA cos u = 0

+ c ©Fy = 0;

FCD sin 2 u - FCA sin u = 0 FCA =

Joint D: + ©F = 0; : x

D

cot 2 u + 1 P cos u - sin u cot 2 u

FCA = (cot u cos u - sin u + 2 cos u) P (T)

Ans.

FCD = (cot 2 u + 1) P

Ans.

(C)

FDA - (cot 2 u + 1)(cos 2 u) P = 0 FDA = (cot 2 u + 1)(cos 2 u) (P)

(C)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

P

6–26.

The maximum allowable tensile force in the members of the truss is 1Ft2max = 2 kN, and the maximum allowable compressive force is 1Fc2max = 1.2 kN. Determine the maximum magnitude P of the two loads that can be applied to the truss. Take L = 2 m and u = 30°.

P B

L

C u

L

L

SOLUTION A

(Tt)max = 2 kN

L

D

(FC)max = 1.2 kN Joint B: + c ©Fy = 0;

FBA cos 30° - P = 0

+ ©F = 0; : x

P = 1.1547 P (C) cos 30°

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FBA =

FAB sin 30° - FBC = 0

FBC = P tan 30° = 0.57735 P (C) Joint C: + c ©Fy = 0;

-FCA sin 30° + FCD sin 60° = 0 FCA = FCD a

+ ©F = 0; : x

sin 60° b = 1.732 FCD sin 30°

P tan 30° + P + FCD cos 60° - FCA cos 30° = 0 FCD = a

tan 30° + 1

23 cos 30° - cos 60°

b P = 1.577 P (C)

FCA = 2.732 P (T) Joint D: + ©F = 0; : x

FDA - 1.577 P sin 30° = 0 FDA = 0.7887 P (C)

1) Assume FCA = 2 kN = 2.732 P P = 732.05 N

FCD = 1.577(732.05) = 1154.7 N 6 (Fc)max = 1200 N Thus,

Pmax = 732 N

(O.K.!)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

P

6–27.

Determine the force in members HG, HE, and DE of the truss, and state if the members are in tension or compression.

K

J

I

H

G

B

C

D

E

4 ft A

SOLUTION

3 ft

Method of Sections: The forces in members HG, HE, and DE are exposed by cutting the truss into two portions through section a–a and using the upper portion of the free-body diagram, Fig. a. From this free-body diagram, FHG and FDE can be obtained by writing the moment equations of equilibrium about points E and H, respectively. FHE can be obtained by writing the force equation of equilibrium along the y axis.

3 ft

3 ft

3 ft

F 3 ft

1500 lb 1500 lb 1500 lb 1500 lb 1500 lb

Joint D: From the free-body diagram in Fig. a, a + ©ME = 0;

FHG(4) - 1500(3) = 0 FHG = 1125 lb (T) FDE(4) - 1500(6) - 1500(3) = 0

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

a + ©MH = 0;

FDE = 3375 lb (C)

+ c ©Fy = 0;

Ans.

Ans.

4 FHE a b - 1500 - 1500 = 0 5

FEH = 3750 lb (T)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*6–28.

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

K

J

I

H

G

B

C

D

E

4 ft A

SOLUTION

3 ft

Method of Sections: The forces in members HI, CH, and CD are exposed by cutting the truss into two portions through section b –b on the right portion of the free-body diagram, Fig. a. From this free-body diagram, FCD and FHI can be obtained by writing the moment equations of equilibrium about points H and C, respectively. FCH can be obtained by writing the force equation of equilibrium along the y axis. a + ©MH = 0;

3 ft

3 ft

1500 lb 1500 lb 1500 lb 1500 lb 1500 lb

Ans.

FHI(4) - 1500(3) - 1500(6) - 1500(9) = 0 FHI = 6750 lb (T)

Ans.

4 FCH a b - 1500 - 1500 = 0 5

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

+ c ©Fy = 0;

3 ft

FCD(4) - 1500(6) - 1500(3) = 0 FCD = 3375 lb (C)

a + ©Mc = 0;

3 ft

F

FCH = 5625 lb (C)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–29.

Determine the force developed in members GB and GF of the bridge truss and state if these members are in tension or compression.

10 ft

4 ft G

4 ft F

10 ft E

10 ft A

SOLUTION a + ©MA = 0;

D B 600 lb

-600(10) - 800(18) + Dy (28) = 0

C 800 lb

Dy = 728.571 lb + ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

Ay - 600 - 800 + 728.571 = 0 Ay = 671.429 lb - 671.429(10) + FGF(10) = 0

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

a + ©MB = 0;

FGF = 671.429 lb = 671 lb (C)

+ c ©Fy = 0;

Ans.

671.429 - FGB = 0 FGB = 671 lb (T)

Ans.

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6–30.

10 ft

Determine the force in members EC, EF, and FC of the bridge truss and state if these members are in tension or compression.

4 ft G

4 ft F

10 ft E

10 ft

A

D B

SOLUTION

600 lb

C 800 lb

Support Reactions: Applying the moment equation of equilibrium about point A by referring to the FBD of the entire truss shown in Fig. a, a + ©MA = 0;

ND (28) - 600(10) - 800(18) = 0 ND = 728.57 lb

Ans.

a + ©MC = 0;

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Method of Sections: Consider the FBD of the right portion of the truss cut through sec. a–a, Fig. b, we notice that FEF and FFC can be obtained directly by writing moment equation of equilibrium about joint C and force equation of equilibrium along y-axis, respectively.

728.57(10) - FEF(10) = 0

FEF = 728.57 lb (C) = 729 lb (C) + c ©Fy = 0;

FFC a

b + 728.57 - 800 = 0 229 FFC = 76.93 lb (T) = 76.916 (T)

Ans.

5

Ans.

Method of joints: Using the result FEF to consider joint E, Fig. c, + ©F = 0; : x

728.57 - FDE cos 45° = 0 FDE = 1030.36 lb (C)

+ c ©Fy = 0;

1030.36 sin 45° - FEC = 0

FEC = 728.57 lb (T) = 729 lb (T)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–31.

Determine the force in members CD, CJ, KJ, and DJ of the truss which serves to support the deck of a bridge. State if these members are in tension or compression.

8000 lb

5000 lb

4000 lb B

A

C

D

E

F

G 12 ft

SOLUTION a + ©MC = 0;

L 9 ft

H 9 ft

9 ft

Ans.

-9375 + 11 250 -

Ans.

3 FCJ = 0 5

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

Joint D:

I 9 ft

-9500(27) + 4000(18) + 8000(9) + FCD(12) = 0 FCD = 9375 lb = 9.38 kip (C)

+ ©F = 0; : x

J 9 ft

- 9500(18) + 4000(9) + FKJ(12) = 0 FKJ = 11 250 lb = 11.2 kip (T)

a + ©MJ = 0;

K 9 ft

FCJ = 3125 lb = 3.12 kip (C)

Ans.

FDJ = 0

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*6–32.

Determine the force in members EI and JI of the truss which serves to support the deck of a bridge. State if these members are in tension or compression.

8000 lb

5000 lb

4000 lb B

A

C

D

E

F

G 12 ft

SOLUTION a + ©ME = 0;

L 9 ft

J 9 ft

I 9 ft

H 9 ft

9 ft

- 5000(9) + 7500(18) - FJI(12) = 0 FJI = 7500 lb = 7.50 kip (T)

+ c ©Fy = 0;

K 9 ft

Ans.

7500 - 5000 - FEI = 0 Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FEI = 2500 lb = 2.50 kip (C)

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6–33.

Determine the force in member GJ of the truss and state if this member is in tension or compression.

1000 lb

1000 lb H

1000 lb

J

30

SOLUTION a + ©MC = 0;

G

A

E C

B

-1000(10) + 1500(20) - FGJ cos 30°(20 tan 30°) = 0

10 ft

Ans.

10 ft

10 ft 1000 lb

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FGJ = 2.00 kip (C)

10 ft

D

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6–34.

Determine the force in member GC of the truss and state if this member is in tension or compression.

1000 lb

1000 lb H

1000 lb

J

30

SOLUTION a + ©MC = 0;

G

A

E B

-1000(10) + 1500(20) - FGJ cos 30°(20 tan 30°) = 0 FGJ = 2.00 kip (C)

10 ft

C 10 ft

Ans.

D 10 ft

10 ft 1000 lb

Joint G: + ©F = 0; : x

FHG = 2000 lb

+ c ©Fy = 0;

-1000 + 2(2000 cos 60°) - FGC = 0 Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FGC = 1.00 kip (T)

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6–35.

Determine the force in members BC, HC, and HG. After the truss is sectioned use a single equation of equilibrium for the calculation of each force. State if these members are in tension or compression.

5 kN

4 kN

4 kN

B

C

3 kN

2 kN A

D

E 3m

F

H

SOLUTION a + ©ME = 0;

2m

G 5m

- Ay (20) + 2(20) + 4(15) + 4(10) + 5(5) = 0

5m

5m

5m

Ay = 8.25 kN a + ©MH = 0;

-8.25(5) + 2(5) + FBC(3) = 0 FBC = 10.4 kN (C) -8.25(10) + 2(10) + 4(5) +

5 229

FHG(5) = 0

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

a + ©MC = 0;

Ans.

FHG = 9.1548 = 9.15 kN (T) a + ©MO¿ = 0;

-2(2.5) + 8.25(2.5) - 4(7.5) + FHC = 2.24 kN (T)

Ans.

3

234

FHC(12.5) = 0

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*6–36.

Determine the force in members CD, CF, and CG and state if these members are in tension or compression.

5 kN

4 kN

4 kN

B

C

3 kN

2 kN A

D

E 3m

F

H

SOLUTION

2m

G 5m

+ ©F = 0; : x

Ex = 0

a + ©MA = 0;

-4(5) - 4(10) - 5(15) - 3(20) + Ey (20) = 0

5m

5m

5m

Ey = 9.75 kN a + ©MC = 0;

-5(5) - 3(10) + 9.75(10) -

5 229

FFG(5) = 0

a + ©MF = 0;

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FFG = 9.155 kN (T) -3(5) + 9.75(5) - FCD(3) = 0 FCD = 11.25 = 11.2 kN (C)

a + ©MO¿ = 0;

-9.75(2.5) + 5(7.5) + 3(2.5) -

FCF = 3.21 kN (T)

Joint G: + ©F = 0; : x + c ©Fy = 0;

Ans.

3

234

FCF(12.5) = 0

Ans.

FGH = 9.155 kN (T) 2

229

(9.155)(2) - FCG = 0

FCG = 6.80 kN (C)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–37.

Determine the force in members GF, FB, and BC of the Fink truss and state if the members are in tension or compression.

600 lb F

800 lb

800 lb E

G A

60°

30°

60°

10 ft

SOLUTION

30°

D

C

B 10 ft

10 ft

Support Reactions: Due to symmetry, + c ©Fy = 0;

2Ay - 800 - 600 - 800 = 0

+ ©F = 0; : x

Ax = 0

Ay = 1100 lb

Method of Sections: a + ©MB = 0;

FGF sin 30°1102 + 800110 - 10 cos2 30°2 - 11001102 = 0

a + ©MA = 0;

FFB sin 60°1102 - 800110 cos2 30°2 = 0

FFB = 692.82 lb 1T2 = 693 lb 1T2

a + ©MF = 0;

Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FGF = 1800 lb 1C2 = 1.80 kip 1C2

Ans.

FBC115 tan 30°2 + 800115 - 10 cos2 30°2 - 11001152 = 0 FBC = 1212.43 lb 1T2 = 1.21 kip 1T2

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–38.

Determine the force in members FE and EC of the Fink truss and state if the members are in tension or compression.

600 lb F

800 lb

800 lb E

G

SOLUTION

A

60

10 ft

30 C

B

Support Reactions: Due to symmetry, + c ©Fy = 0;

60

30

10 ft

10 ft

2By - 800 - 600 - 800 = 0; By = 1100 lb

Method of Sections: a + ©MC = 0;

1100(10) - 800(10 - 7.5) - (FFE sin 30°)(10) = 0 FFE = 1.80 kip (C)

Ans.

Joint E: + c ©Fy = 0;

FEC - 800 cos 30° = 0 Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FEC = 693 lb (C)

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

D

6–39.

Determine the force in members IC and CG of the truss and state if these members are in tension or compression.

Also, indicate all zero-force members.

B

C

D

2m I

J

2m

SOLUTION A

By inspection of joints B, D, H and I, AB, BC, CD, DE, HI, and GI are all zero-force members. a + ©MG = 0;

E H

Ans

1.5 m

G 1.5 m

3 - 4.5(3) + FIC a b (4) = 0 5 FIC = 5.625 = 5.62 kN (C)

F 1.5 m

6 kN

1.5 m 6 kN

Ans.

Joint C: FCJ = 5.625 kN

+ c ©Fy = 0;

4 4 (5.625) + (5.625) - FCG = 0 5 5

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

+ ©F = 0; : x

FCG = 9.00 kN (T)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

*6–40.

Determine the force in members JE and GF of the truss and state if these members are in tension or compression.

Also, indicate all zero-force members.

B

C

D

2m I

J

2m

SOLUTION A

By inspection of joints B, D, H and I, AB, BC, CD, DE, HI, and GI are zero-force members.

E H

Ans.

1.5 m

G 1.5 m

+ c ©Fy = 0;

7.5 -

1.5 m 6 kN

4 F = 0 5 JE Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FJE = 9.375 = 9.38 kN (C) + ©F = 0; : x

1.5 m 6 kN

Joint E:

F

3 (9.375) - FGF = 0 5 FGF = 5.62 kN (T)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–41.

Determine the force in members FG, GC and CB of the truss used to support the sign, and state if the members are in tension or compression.

1.5 m

1.5 m

1.5 m

E

900 N

3m 1800 N

F D

SOLUTION

3m

Method of Sections: The forces in members FG, GC, and CB are exposed by cutting the truss into two portions through section a–a on the upper portion of the free-body diagram, Fig. a. From this free-body diagram, FCB, FGC, and FFG can be obtained by writing the moment equations of equilibrium about points G, E, and C, respectively. a + ©MG = 0;

FCB = 3600 N = 3.60 kN (T) a + ©ME = 0;

3m B

Ans.

FGC(6) - 900(6) - 1800(3) = 0 Ans.

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

FGC = 1800 N = 1.80 kN (C) a + ©MC = 0;

C

A

900(6) + 1800(3) - FCB(3) = 0

900 N

G

900(6) + 1800(3) - FFG sin 26.57°(6) = 0

FFG = 4024.92 N = 4.02 kN (C)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–42.

K

Determine the force in members LK, LC, and BC of the truss, and state if the members are in tension or compression.

3 ft 3 ft

I

L

H

A

G B 4 ft

D C 4 ft 4 ft

SOLUTION

4 ft

Support Reactions: Applying the moment equation of equilibrium about point G by referring to the FBD of the entire truss shown in Fig. a,

3000 lb 3000 lb

a + ©MG = 0;

J

E 4 ft

F 4 ft

2000 lb 2000 lb 4000 lb

2000(4) + 2000(8) + 4000(12) + 3000(16) = 3000(20) - A y (24) = 0 A y = 7500 lb

Method of Section: Consider the FBD of the left portion of the truss cut through sec a–a, Fig. b, we notice that FLK, FLC and FBC can be obtained directly by writing moment equation of equilibrium about joint C, A, and L, respectively.

3 FLK a b (8) + 3000(4) - 7500(8) = 0 5

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

a + ©MC = 0;

FLK = 10 000 lb (C) = 10.0 kip (C)

a + ©MA = 0;

3 FLC a b (8) - 3000(4) = 0 5

FLC = 2500 lb (C) = 2.50 kip (C)

a + ©ML = 0;

Ans.

Ans.

FBC(3) - 7500(4) = 0

FBC = 10 000 lb (T) = 10.0 kip (T)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

6–43.

Determine the force in members JI, JE, and DE of the truss, and state if the members are in tension or compression.

K 3 ft 3 ft

H

A

G 4 ft

+ ©MA = 0;

I

L

SOLUTION Support Reactions: Applying the equations of equilibrium about point A to the freebody diagram of the truss, Fig. a, we have

J

B 4 ft

D C 4 ft 4 ft

3000 lb 3000 lb

E 4 ft

F 4 ft

2000 lb 2000 lb 4000 lb

3000(4) + 3000(8) + 4000(12) + 2000(16) + 2000(20) - NG(24) = 0 NG = 6500 lb

Method of Sections: The force in members JI, JE, and DE are exposed by cutting the truss into two portions through section b–b on the right portion of the free-body diagram, Fig. a. From this free-body diagram, FJI and FDE can be obtained by writing the moment equations of equilibrium about points E and J, respectively. FJE can be obtained by writing the force equation of equilibrium along the y axis.

6500(8) - 2000(4) - FJI (6) = 0

T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b)

+ ©ME = 0;

FJI = 7333.33 lb = 7333 lb (C)

+ ©MJ = 0;

6500(12) - 2000(8) - 2000(4) - FDE(6) = 0

FDE = 9000 lb (T)

+ c ©Fy = 0;

Ans.

Ans.

6500 - 2000 - 2000 - FJE sin 56.31° = 0

FJE = 3004.63 lb = 3005 lb (C)

Ans.

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.