# Solutions Simple Trusses

#### Short Description

Solution manual for simple trusses problems in Engineering Mechanics...

#### Description

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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.

600 N

D

2m E

900 N

C

2m A

B 2m

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6–2. The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the in tension tension or or compression. compression.Set SetPP1 1 == 600 3 kN, members are in lb, kN.lb. P2 ==2400

P2

P1

B

A

C 45⬚

45⬚

ft 14 m

E D 14 m ft

14 m ft

Joint A: +↑ΣFy = 0;

FAD sin 45° – 3 = 0 FAD = 4.2426 = 4.24 kN (C)

+ → ΣFx

= 0;

Ans

3 kN

FAB – 4.2426 cos 45° = 0 FAB = 3 kN (T)

Ans

Joint B: +↑ΣFy = 0;

FBD – 2 = 0 FBD = 2 kN (C)

+ → ΣFx

2 kN

Ans 3 kN

= 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)

+ → ΣFx

= 0;

Ans

4.2426 cos 45° + 7.0711 cos 45° – FDE = 0 FDE = 8.00 kN (C)

Ans

3 kN 4.2426 kN

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6–3. The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in in tension tension or or compression. compression. Set Set PP11 == 800 4 kN, lb, P22 ==0.0.

P2

P1

B

A

C 45⬚

45⬚

ft 14 m

E D 14 m ft

14 m ft

Joint A: +↑ΣFy = 0;

FAD sin 45° – 4 = 0 FAD = 5.657 kN = 5.66 kN (C)

+ → ΣFx

= 0;

Ans

4 kN

FAB – 5.657 cos 45° = 0 FAB = 4.00 kN (T)

Ans

Joint B: +↑ΣFy = 0;

FBD – 0 = 0 FBD = 0

+ → ΣFx

= 0;

Ans

FBC – 4 = 0 FBC = 4 kN (T)

4 kN

Ans

Joint D: +↑ΣFy = 0;

FDC sin 45° – 0 – 5.657 sin 45° = 0 FDC = 5.657 kN = 5.66 kN (T)

+ → ΣFx

= 0;

Ans

5.657 cos 45° + 5.657 cos 45° – FDE = 0 FDE = 8.00 kN (C)

5.657 kN

Ans

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

2P P

P C

B A

4m E

D 4m

4m

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•6–5. Assume that each member of the truss is made of steel having a mass per length of 4 kg/m. Set P = 0, determine the force in each member, and indicate if the members are in tension or compression. Neglect the weight of the gusset plates and assume each joint is a pin. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at the end of each member.

2P P

P C

B A

4m E

D 4m

4m

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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. Set P1 = 2 kN and P2 = 1.5 kN.

A

B

30⬚

30⬚ C

E

D 3m

3m P1

P2

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6–7. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = P2 = 4 kN.

A

B

30⬚

30⬚ C

E

D 3m

3m P1

P2

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*6–8. Determine the force in each member of the truss, and state if the members are in tension or compression. Set P ==4 800 kN. lb.

2.5 kN 500 lb ft 13 m

ft 13 m F

ft 13 m E

D

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

ft 13 m A

A ⇒ F ⇒ E ⇒ B ⇒ C.

B

Joint A: From the free – body diagram in Fig. a, +↑ΣFy = 0;

= 0;

P

FAF sin 45° – 4 = 0 FAF = 5.657 kN (T)

+ → ΣFx

Ans.

5.657 cos 45° – FAB = 0

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

FAB = 4 kN (C)

Ans.

+↑ΣFy = 0;

Joint F: From the free – body diagram in Fig. b, +↑ΣFy = 0;

= 0;

FFE – 9.192 sin 45° – 5.657 sin 45° = 0

= 0;

+↑ΣFy = 0;

Ans.

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, Ans.

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

= 0;

Ans.

FFE = 10.50 kN (T)

FBD sin 45° – 9.192 sin 45° = 0 FBD = 9.192 kN (T)

+ → ΣFx

FFB cos 45° – 5.657 cos 45° – 2.5 = 0 FFB = 9.192 kN (C)

+ → ΣFx

C

+↑ΣFy = 0;

FCD = 0

+ → ΣFx

17.0 – NC = 0

= 0;

NC = 17.0 kN

FED – 10.50 = 0 FED = 10.50 kN (T)

Ans.

FEB = 0

Ans.

Ans.

Ans.

2.5 kN

FFE = 10.50 kN

FAF = 5.657 kN 4 kN

FFB = 9.192 kN

FBC = 17.0 kN FAB = 4 kN

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•6–9. Remove the 2.5-kN force and then determine the 500-lb force greatest force P that can be applied to the truss so that none of the members are subjected to a force exceeding either 4800 kNlbinintension kN lb in in compression. tensionoror3 600 compression.

2.5 kN 500 lb 13 m ft

13 m ft F

13 m ft E

D 13 m ft

A B

C

P

2P = 4 kN 3P = 3 kN

P = 2 kN P = 1 kN

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6–10. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 ==4800 kN,lb P,2P=20.= 0.

m 41 ft

m 41 ft

14 m ft

14 m ft

P1 P2

C B

D

6 ftm 1.5 E

A

G

F

4=0 3.33 kN (C)

4 kN

3.33 kN (C) 3.33 kN (C)

Ans 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. Set 400 lb. P1 ==3600 kN,lb P,2P =22=kN.

14m ft

14m ft

14mft

14mft

P1 P2

C B

A

D

1.5 6 ftm E

G

F

Joint B: FBG = 0

ΣFy = 0;

FBC = FBA

+

ΣFy = 0;

+

Ans

Joint G: +↑ΣFy = 0;

+ → ΣFx

= 0;

FGC sin ␪ = 0 FGC = 0

Ans

FGA = 0

Ans

2 kN

Joint D:

+

ΣFx = 0;

FDE – FDC = 0 FDE = FDC

+

ΣFy = 0;

FDF – 2 = 0 FDF = 2 kN (C)

Ans FDF = 2 kN

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:

3 kN

+↑ΣFx = 0;

FBC cos 36.87° – FDC cos 36.87° + 1.6667 cos 73.74° = 0

+ → ΣFy

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

= 0;

FBC = FBA = 3.54 kN (C) Ans FDC = FDE = 4.13 kN (C) Ans

FFC = 1.67 kN

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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 N,, P P22 ==500 P1 ==1200 240 lb 100N.lb.

D

C

P1

P2

1.55mft

A

B 3.6 m 12 ft

Joint D: +↑ΣFy = 0;

⎛ 5⎞ FBD ⎜ ⎟ – 500 = 0 ⎝ 13 ⎠ 1200 N

FBD = 1300 N (C) Ans + → ΣFx

= 0;

⎛ 12 ⎞ 1200 – FCD + 1300 ⎜ ⎟ = 0 ⎝ 13 ⎠ FCD = 2400 N (T) Ans

500 N

Joint A: +↑ΣFy = 0;

FAC = 0

Ans

+ → ΣFy

FAB = 0

Ans

= 0;

Joint B: +↑ΣFy = 0;

⎛ 5⎞ FBC – 1300 ⎜ ⎟ = 0 ⎝ 13 ⎠ FBC = 500 N (T) Ans FBD = 1300 N

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•6–13. Determine the largest load P2 that can be applied to the truss so that the force in any member does not exceed 2.5 kN (T)or or350 1.75lbkN (C). Take 500 lb (T) (C). Take P1P=1 =0.0.

D

C

P1

P2

1.55m ft

A

B 12 3.6ftm

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

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6–14. Determine the force in each member of the truss, and state if the members are in tension or compression. Set P ==12.5 2500kN. lb.

1200 6 kNlb 14 m ft

+ ΣMA = 0;

14 m ft

E

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

1200 6 kNlb

P 14m ft D

14 m ft C

14 m ft

NB (2 + 2) – 6 (2 + 2) – 12.5 (2) = 0

F

NB = 12.25 kN

30⬚

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

A

G 30⬚ B

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

= 0;

FBG = 0

+↑ΣFy = 0;

Ans.

12.25 – FBC = 0 FBC = 12.25 kN (C)

Ans.

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)

+ → 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)

+ → ΣFx

= 0;

Ans.

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.

8.8389 sin 45° + 8.8389 sin 45° – FGF = 0 FGF = 12.5 kN (T)

6 kN

6 kN

12.5 kN 2m

Ans.

6 kN

2m

NB = 12.25 kN

FBC = 12.25 kN

FCG = 8.8389 kN

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forcesforces and determine the greatest 6–15. Remove the the6-kN 1200-lb and determine the force P that be applied to the truss thatso none the greatest forcecan P that can be applied to thesotruss thatof none members are subjected to a force either 10 either kN in of the members are subjected to exceeding a force exceeding tension or tension 7.5 kN in 2000 lb in orcompression. 1500 lb in compression.

6 kNlb 1200 14 m ft

+ ΣMA = 0;

F

NB = 0.5 P

30⬚

A

2m

14m ft D

14 m ft C

14 m ft

NB (2 + 2) – P (2) = 0

0.7071P = 7.5 P = 10 kN

14 m ft

E

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

6 kNlb 1200

P

G 30⬚ B

P = 10.6067 kN

2m

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*6–16. Determine the force in each member of the truss, and state if the members are in tension or compression. Set P = 5 kN.

E 1.5 m D 1.5 m C

A

1.5 m P

B 2m

2m

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•6–17. 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 2.5 kN in tension or 2 kN in compression.

E 1.5 m D 1.5 m C

A

1.5 m P

B 2m

2m

417

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6–18. Determine the force in each member of the truss, and state if the members are in tension or compression.

4.5 900kN lb

3 kN 600 lb 4 ftm 1.2

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

+ ΣMC = 0;

3 (1.2) + 4.5 (1.2 + 1.2 + 1.2) – NA (1.2 + 1.2) = 0

4 ftm 1.2

4 ftm 1.2 E

F

D

0.9 3 ftm C

NA = 8.25 kN A

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

B

F → E → A → B → D. Joint F: From the free – body diagram in Fig. b, ⎛ 3⎞ +↑ΣFy = 0; FFA ⎜ ⎟ – 4.5 = 0 ⎝ 5⎠ FFA = 7.5 kN (C) + → ΣFx

= 0;

Joint B: From the free – body diagram in Fig. e, + → ΣFx

Ans.

⎛ 4⎞ FFE – 7.5 ⎜ ⎟ = 0 ⎝ 5⎠

+↑ΣFy = 0;

FFE = 6.0 kN (T)

= 0;

+ → ΣFx

Ans.

FEA = 0

+↑ΣFy = 0;

Ans.

= 0;

Ans.

FBD = 0

Ans.

⎛ 4⎞ ⎛ 4⎞ FDC ⎜ ⎟ + 6.25 ⎜ ⎟ – 6 = 0 ⎝ 5⎠ ⎝ 5⎠ Ans.

⎛ 3⎞ ⎛ 3⎞ 6.25 ⎜ ⎟ – 1.25 ⎜ ⎟ – 3 = 0 ⎝ 5⎠ ⎝ 5⎠

(check)

Note The equilibrium analysis of joint C must be used to determine the components of the support reaction at C.

⎛ 3⎞ ⎛ 4⎞ 8.25 – 7.50 ⎜ ⎟ – FAD ⎜ ⎟ = 0 ⎝ 5⎠ ⎝ 5⎠ FAD = 6.25 kN (C)

+ → ΣFx

FBC = 1.0 kN (C)

FDC = 1.25 kN (T)

Joint A: From the free – body diagram in Fig. d, +↑ΣFy = 0;

= 0;

FED – 6.0 = 0 FED = 6.0 kN (T)

+↑ΣFy = 0;

1.0 – FBC = 0

Joint D: From the free – body diagram in Fig. f,

Ans.

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

= 0;

Ans.

⎛ 4⎞ ⎛ 4⎞ 7.50 ⎜ ⎟ – 6.25 ⎜ ⎟ – FAB = 0 ⎝ 5⎠ ⎝ 5⎠ FAB = 1.0 kN (C)

Ans. 4.5 kN

4.5 kN 1.2 m

3 kN

1.2 m

6 kN

1.2 m

7.50 kN

3 kN

1 kN 8.25 kN

6.25 kN

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4.5 900kN lb

6–19. The truss is fabricated using members having a 10 lb>ft weight of 0.2 . Remove kN/m. Removethe the external external forces forces from from the truss, and determine the force in each member due to the weight of the members. State whether the members are in tension or compression. Assume that the total force acting on a joint is the sum of half of the weight of every member connected to the joint.

3 kNlb 600

4 ftm 1.2

4 ftm 1.2

1.2 m 4 ft E

F

D

0.9 3 ftm C A

Joint Loadings: ⎛ 1.2 + 1.5 ⎞ FC = FF = 0.2 ⎜ ⎟⎠ = 0.27 kN 2 ⎝

B

FF = 0.27 kN FE = 0.33 kN FD = 0.51 kN 1.2 m 1.2 m 1.2 m

⎛ 1.2 + 1.2 + 0.9 ⎞ FE = FB = 0.2 ⎜ ⎟⎠ = 0.33 kN 2 ⎝

FC = 0.27 kN

0.9 m

⎛ 1.5 + 1.5 + 1.2 + 0.9 ⎞ FA = FD = 0.2 ⎜ ⎟⎠ = 0.51 kN 2 ⎝

FA = 0.51 kN FB = 0.33 kN

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

+ ΣMC = 0;

0.27 (1.2 + 1.2 + 1.2) + 0.33 (1.2 + 1.2) + 0.51 (1.2 + 1.2) + 0.51 (1.2) + 0.33 (1.2) – NA (1.2 + 1.2) = 0 NA = 1.665 kN

0.27 kN

Method of Joints: We will analyze the equilibrium of the joints in the following sequence: F → E → A → B → D. Joint F: From the free – body diagram in Fig. b, ⎛ 3⎞ +↑ΣFy = 0; FFA ⎜ ⎟ – 0.27 = 0 ⎝ 5⎠ + → ΣFx

= 0;

0.33 kN

FFA = 0.45 kN (C)

Ans.

⎛ 4⎞ FFE – 0.45 ⎜ ⎟ = 0 ⎝ 5⎠ FFE = 0.36 kN (T)

Ans.

FFE = 0.36 kN

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

= 0;

FED – 0.36 = 0 FED = 0.36 kN (T)

+↑ΣFy = 0;

FEA – 0.33 = 0 FEA = 0.33 kN (C)

Ans.

FEA = 0.33 kN FFA = 0.45 kN

0.51 kN

Ans.

Joint A: From the free – body diagram in Fig. d, ⎛ 3⎞ ⎛ 3⎞ +↑ΣFy = 0; 1.665 – 0.33 – 0.51 – 0.45 ⎜ ⎟ – FAD ⎜ ⎟ = 0 ⎝ 5⎠ ⎝ 5⎠ FAD = 0.925 kN = 0.925 kN (C) Ans. + → ΣFx

= 0;

⎛ 4⎞ ⎛ 4⎞ FAB + 0.45 ⎜ ⎟ – 0.925 ⎜ ⎟ = 0 ⎝ 5⎠ ⎝ 5⎠ FAB = 0.38 kN (T)

NA = 1.665 kN

Ans.

Joint B: From the free – body diagram in Fig. e, + → ΣFx

= 0;

+↑ΣFy = 0;

FAB = 0.38 kN

FBC – 0.38 = 0 FBC = 0.38 kN = 0.38 kN

Ans.

FBD – 0.33 = 0 FBD = 0.33 kN (T)

Ans.

0.33 kN

Joint D: From the free – body diagram in Fig. f, ⎛ 4⎞ ⎛ 4⎞ + 0.925 ⎜ ⎟ – 0.36 – FDC ⎜ ⎟ = 0 → ΣFx = 0; ⎝ 5⎠ ⎝ 5⎠ FDC = 0.475 kN = 0.475 kN (C) +↑ΣFy = 0;

0.51 kN FED = 0.36 kN

Ans.

⎛ 3⎞ ⎛ 3⎞ 0.925 ⎜ ⎟ + 0.475 ⎜ ⎟ – 0.51 – 0.33 = 0 ⎝ 5⎠ ⎝ 5⎠

(check)

FBD = 0.33 kN

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*6–20. Determine the force in each member of the truss and state if the members are in tension or compression. The load has a mass of 40 kg.

0.1 m D E 3.5 m

F C G B

2.5 m

A 6m

420

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•6–21. Determine the largest mass m of the suspended block so that the force in any member does not exceed 30 kN (T) or 25 kN (C).

0.1 m D E 3.5 m

F C G B

2.5 m

A 6m

421

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6–22. Determine the force in each member of the truss, and state if the members are in tension or compression.

600 N

400 N

D

E

A

45⬚

45⬚

45⬚

45⬚

C

B 2m

2m

422

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6–23. The truss is fabricated using uniform members having a mass of 5 kg>m. Remove the external forces from the truss, and determine the force in each member due to the weight of the truss. State whether the members are in tension or compression. Assume that the total force acting on a joint is the sum of half of the weight of every member connected to the joint.

600 N

400 N

D

E

A

45⬚

45⬚

45⬚

45⬚

C

B 2m

2m

423

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*6–24. Determine the force in each member of the truss, and state if the members are in tension or compression. Set P = 4 kN.

3m

3m

3m F

E

B

C

3m A P

D P

424

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•6–25. 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 1.5 kN in tension or 1 kN in compression.

3m

3m

3m F

E 3m

B

A P

C

D P

425

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6–26. A sign is subjected to a wind loading that exerts forces of 1.5300 kNlbon horizontal forces onjoints jointsBBand andCC of of one one of the side supporting trusses. Determine the force in each member of the truss and state if the members are in tension or compression.

C

300kN lb 1.5

3.6 12 m ft 13 ft 3.9 m 1.5 5 ftm

D

1.5 300kN lb B

12 ft 3.6 m 3.913mft

45⬚ A

E

1.5 kN

Joint C: + → ΣFx

= 0;

1.5 – FCD sin 22.62° = 0 FCD = 3.90 kN (C) Ans

+↑ΣFy = 0;

–FCB + 3.90 cos 22.62° = 0 FCB = 3.60 kN (T)

FCD = 3.90 kN

Ans

Joint D:

ΣFx = 0;

FDB = 0

+

ΣFy = 0;

3.90 – FDE = 0

+

Ans

FDE = 3.90 kN (C) Ans Joint B: + → ΣFx

= 0;

+↑ΣFy = 0;

1.5 – FBA cos 45° + FBE sin 45.24° = 0 FCB = 3.60 kN

3.60 – FBA sin 45° – FBE cos 45.24° = 0 FBE = 1.485 kN (T)

Ans

FBA = 3.612 kN (T)

Ans 1.5 kN

426

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6–27. 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.

B

C

L/3 E

A L/3

F L/3

P

D L/3

P

427

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*6–28. 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

E d

d/2

d/2

d

428

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•6–29. 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 C

A D

F

d

E d

d/2

d/2

d

429

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6–30. The two-member truss is subjected to the force of 1.5 kN. Determinethe therange rangeofofu for ␪ for application load 300 lb. Determine application of of thethe load so so that force in either member does exceed that thethe force in either member does not not exceed 400 2lbkN (T)(T) or or 1 lb kN(C). (C). 200

B

0.9 3 ftm

Joint A: + → ΣFx

= 0;

+↑ΣFy = 0;

⎛ 4⎞ 1.5 cos ␪ + FAC + FAB ⎜ ⎟ = 0 ⎝ 5⎠

C

A

⎛ 3⎞ –1.5 sin ␪ + FAB ⎜ ⎟ = 0 ⎝ 5⎠

u 1.2 4 ftm 300kN lb 1.5

Thus, FAB = 2.5 sin ␪ FAC = –1.5 cos ␪ – 2 sin ␪ For AB required:

1.5 kN

–1 ≤ 2.5 sin ␪ ≤ 2 –2 ≤ 5 sin ␪ ≤ 4

(1)

For AC required: –1 ≤ –1.5 cos ␪ – 2 sin ␪ ≤ 2 – 4 ≤ 3 cos ␪ + 4 sin ␪ ≤ 2

(2)

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

1.5 kN

127° ≤ ␪ – 196°

Ans

336° ≤ ␪ – 347°

Ans

Possible hand solution: ⎛ 3⎞ ␪2 = ␪1 + tan–1 ⎜ ⎟ = ␪1 + 36.870 ⎝ 4⎠ 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

430

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

J

I

H

G

F 2 ftm 0.8

A

B 0.8 2 ftm

C 0.8 2 ftm

80 lb 400 N

D 0.8 m 2 ft

80 lb 400 N

E

0.6 1.5 m ft 40 lb 200 N 60 lb 300 N

0.8 m

+↑ΣFy = 0;

900 – FBH sin 45° = 0 FBH = 1273 N (T)

+ ΣMB = 0;

0.8 m

Ans

0.8 m

0.6 m 300 N

400 N

–FBC (0.8) + 300 (0.8) + 200 (1.4) = 0 FBC = 650 N (T)

200 N

Ans

Section 2: +↑ΣFy = 0;

400 + 300 + 200 – FHC = 0 FHC = 900 N (C)

Ans. 0.8 m

0.8 m 400 N

0.6 m 300 N

200 N

431

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

40 kN

30 kN J

20 kN

20 kN

I

H

G

F 4m

A

E B

C 16 m, [email protected]

D

432

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

40 kN

30 kN J

20 kN

20 kN

I

H

G

F 4m

A

E B

C 16 m, [email protected]

D

433

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6–34. Determine the force in members JK, CJ, and CD of the truss, and state if the members are in tension or compression.

J

K 3m

I H

L

G

A B 2m

C

2m

4 kN

2m

5 kN

D 2m

F

E 2m

8 kN

2m

6 kN

434

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6–35. Determine the force in members HI, FI, and EF of the truss, and state if the members are in tension or compression.

J

K 3m

I H

L

G

A B 2m

C

2m

4 kN

2m

5 kN

D 2m

F

E 2m

8 kN

2m

6 kN

435

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*6–36. Determine the force in members BC, CG, and GF of the Warren truss. Indicate if the members are in tension or compression.

3m B

3m C

D

3m

3m

A

E F

G 3m

3m 6 kN

•6–37. Determine the force in members CD, CF, and FG of the Warren truss. Indicate if the members are in tension or compression.

3m B

3m 8 kN

3m C

D

3m

3m

A

E F

G 3m

3m 6 kN

3m 8 kN

436

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6–38. Determine the force in members DC, HC, and HI of the truss, and state if the members are in tension or compression.

50 kN

40 kN 2m

2m

2m D

E

1.5 m C

F G

H I

B

30 kN 1.5 m 40 kN 1.5 m

A

437

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6–39. Determine the force in members ED, EH, and GH of the truss, and state if the members are in tension or compression.

50 kN

40 kN 2m

2m

2m D

E

1.5 m C

F G

H I

B

30 kN 1.5 m 40 kN 1.5 m

A

438

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*6–40. Determine the force in members GF, GD, and CD of the truss and state if the members are in tension or compression.

C B

D 260 lbN 1300

41.2 ft m

13

+ ΣMO = 0;

⎛ 12 ⎞ ⎜⎝ 13 ⎟⎠ 1300 (2.4) – FGD sin 36.87° (4.8) = 0

A

30.9 ft m G

H

FGD = 1000 N (C) Ans

+ ΣMD = 0;

30.9 ft m

1.2 4 ftm

1.2 4 ftm

⎛ 12 ⎞ FGF (0.9) – ⎜ ⎟ 1300 (1.2) ⎝ 13 ⎠

E

F

1.2 4 ftm

12 5

1.2 m 4 ft

1300 N

⎛ 5⎞ – ⎜ ⎟ (1300) (0.9) = 0 ⎝ 13 ⎠ 1.2 m

FGF = 2100 N (C) Ans

+ ΣMG = 0;

0.9 m

⎛ 12 ⎞ FCD cos 14.04° (1.2) – ⎜ ⎟ 1300 (2.4) = 0 ⎝ 13 ⎠ FCD = 2473.9 N (T)

1.2 m

Ans

1.2 m

•6–41. Determine the force in members BG, BC, and HG of the truss and state if the members are in tension or compression.

2.4 m

C B

D 260 lbN 1300

4 ft m 1.2

13

30.9 ft m

A

30.9 ft m G

H

Entire truss:

+ ΣMG = 0;

1.2 4 ftm

⎛ 12 ⎞ Ay (2.4) – ⎜ ⎟ (1300) (2.4) = 0 ⎝ 13 ⎠

1.2 4 ftm

E

F

1.2 m 4 ft

12 5

1.2 m 4 ft

1300 N

Ay = 1200 N + → ΣFx

= 0;

⎛ 5⎞ Ax – ⎜ ⎟ (1300) = 0 ⎝ 13 ⎠ Ax = 500 N

2.4 m

2.4 m

Section:

+ ΣMO = 0;

1200 (2.4) – FBC cos 14.04° (1.2) = 0 FBC = 2473.9 N (T)

+ ΣMB = 0;

Ans

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

0.9 m

1.2 m

FBG = 1000 N (C) Ans Ax = 500 N 1.2 m

2.4 m

1.2 m

Ay = 1200 N

439

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6–42. 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 A

E H 1.5 m

G 1.5 m

F 1.5 m

6 kN

6–43. 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

1.5 m 6 kN

C

D

2m I

J

2m A

E H 1.5 m

G 1.5 m

F 1.5 m

6 kN

1.5 m 6 kN

440

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*6–44. Determine the force in members JI, EF, EI, and JE of the truss, and state if the members are in tension or compression.

7.5 kN 1500 lb 5 kN lb 51000 kN lb 1000 4.5 900kN lb L

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

= 0;

+ ΣMA = 0;

2.5 8 ftm

4.5 – Gx = 0

8 ftm 2.5

Gx = 4.5 kN

8 ftm 2.5

K

J

M

I C

N

B

D E H

F

A

G

5 (5) + 7.5 (7.5) + 5 (10) + 4.5 (7.5) – Gy (15) = 0 8 ft 8 ftm 2.5 8 ftm 2.5 8 ftm 2.5 8 ftm 2.5 8 ftm 2.5 m 2.5

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)

+ ΣMI = 0;

Ans.

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)

5 kN

Ans.

7.5 kN 5 kN

4.5 kN

7.5 m

2.5 m

2.5 m 5m

2.5 m 2.5 m

5m Gx = 4.5 kN

2.5 m

2.5 m Gy = 11 kN

FEI = 6.5 kN

FEF = 2.828 kN

441

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•6–45. Determine the force in members CD, LD, and KL of the truss, and state if the members are in tension or compression.

7.5 kN 1500 lb 5 kNlb 5 kN lb 1000 1000 4.5 900kN lb L 8 ftm 2.5

J

M

I C

8 ftm 2.5 N 2.5 8 ftm A

K

D E

B

H

F

G 8 ftm 2.5 8 ftm 2.5 8 ftm 2.5 8 ftm 2.5 8 ftm 2.5 8 ftm 2.5

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)

+ ΣML = 0;

Ans.

FCD (2.5) – 6.5 (5) = 0 FCD = 13 kN (T)

+↑ΣFy = 0;

Ans.

6.5 – 5 – FLD sin 45° = 0 FLD = 2.121 kN (T)

Ans.

5 kN 7.5 kN 5 kN

5 kN 4.5 kN

4.5 kN

2.5 m 7.5 m

5m

2.5 m 2.5 m

5m

5m

2.5 m

NA = 6.5 kN

442

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6–46. Determine the force developed in members BC and CH of the roof truss and state if the members are in tension or compression.

C 2 kN 0.8 m

D

B

A H 1m

G

E

F 1m

1.5 kN 2m

1.5 m

2m

443

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

C 2 kN 0.8 m

D

B

A H 1m

G

E

F 1m

1.5 kN 2m

1.5 m

2m

444

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

6 kN 4 kN

5 kN J 5 kN 3 kN 4m

K

4 kN I H 2 kN

L

A

G B 2m

2m

C 2m

D 2m

E 2m

F 2m

445

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