solutions simple trusses

401

•6–1. Determine the force in each member of the truss,and state if the members are in tension or compression.

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600 N

900 N

2 m

2 m

2 m

A

CE

D

B

06a Ch06a 401-445.indd 40106a Ch06a 401-445.indd 401 6/12/09 8:48:31 AM6/12/09 8:48:31 AM

402

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

P1 = 600 lb,


45�

4 ft 4 ft

45�

D

E

CB

P2

A

4 ft

P1

1 m

1 m 1 m

3 kN

2 kN

3 kN

3 kN

4.2426 kN

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

Joint A:

+↑ΣFy = 0; FAD sin 45° – 3 = 0

FAD = 4.2426 = 4.24 kN (C) Ans

+→ ΣFx = 0; FAB – 4.2426 cos 45° = 0

FAB = 3 kN (T) Ans

Joint B:

+↑ΣFy = 0; FBD – 2 = 0

FBD = 2 kN (C) Ans

+→ ΣFx = 0; FBC – 3 = 0

FBC = 3 kN (T) Ans

Joint D:

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

FDC = 7.0711 kN = 7.07 kN (T) Ans

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

FDE = 8.00 kN (C) Ans


403

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

P1 = 800 lb,


45�

4 ft 4 ft

45�

D

E

CB

P2

A

4 ft

P1

1 m

1 m 1 m

4 kN

4 kN

5.657 kN

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

Joint A:

+↑ΣFy = 0; FAD sin 45° – 4 = 0

FAD = 5.657 kN = 5.66 kN (C) Ans

+→ ΣFx = 0; FAB – 5.657 cos 45° = 0

FAB = 4.00 kN (T) Ans

Joint B:

+↑ΣFy = 0; FBD – 0 = 0

FBD = 0 Ans

+→ ΣFx = 0; FBC – 4 = 0

FBC = 4 kN (T) Ans

Joint D:

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

FDC = 5.657 kN = 5.66 kN (T) Ans

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



404

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


A

E

D

CB

PP

2P

4 m

4 m 4 m


405

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

P = 0


A

E

D

CB

PP

2P

4 m

4 m 4 m


406

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

and .P2 = 1.5 kNP1 = 2 kN


A

E D

30� 30�

B

C

3 m 3 m

P2P1


407

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

.P1 = P2 = 4 kN


A

E D

30� 30�

B

C

3 m 3 m

P2P1


408

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

.P = 800 lb


3 ft

3 ft

3 ft

P

3 ft

500 lb

ACB

DF E1 m 1 m 1 m

1 m

2.5 kN

2.5 kN

FFE = 10.50 kN

FAF = 5.657 kN

FBC = 17.0 kN

FAB = 4 kN

FFB = 9.192 kN

4 kN

P = 4 kN.

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

A ⇒ F ⇒ E ⇒ B ⇒ C.

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

+↑ΣFy = 0; FAF sin 45° – 4 = 0

FAF = 5.657 kN (T) Ans.

+→ ΣFx = 0; 5.657 cos 45° – FAB = 0

FAB = 4 kN (C) Ans.

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

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

FFB = 9.192 kN (C) Ans.

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

FFE = 10.50 kN (T) Ans.

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

+→ ΣFx = 0; FED – 10.50 = 0

FED = 10.50 kN (T) Ans.

+↑ΣFy = 0; FEB = 0 Ans.

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

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

FBD = 9.192 kN (T) Ans.

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

FBC = 17.0 kN (C) Ans.

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

+↑ΣFy = 0; FCD = 0 Ans.

+→ ΣFx = 0; 17.0 – NC = 0

NC = 17.0 kN Ans.


409


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

3 ft

3 ft

3 ft

P

3 ft

500 lb

ACB

DF E1 m 1 m 1 m

1 m

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

2P = 4 kN P = 2 kN3P = 3 kN P = 1 kN


410

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

, .P2 = 0P1 = 800 lb


6 ft

AG

B

C

F

D

E

P1

P2

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

1.5 m

1 m

4 kN

P1 = 4 kN, P2 = 0.

4 = 0

3.33 kN (C)

3.33 kN (C) Ans

3.33 kN (C) Ans


411

6–11. Determine the force in each member of the trussand state if the members are in tension or compression. Set

, .P2 = 400 lbP1 = 600 lb


6 ft

AG

B

C

F

D

E

P1

P2

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

1.5 m

1 m

3 kN

2 kN

FDF = 2 kN

FFC = 1.67 kN

P1 = 3 kN, P2 = 2 kN.

Joint B:

+↑

ΣFy = 0; FBG = 0 Ans

+↑ΣFy = 0; FBC = FBA

Joint G:

+↑ΣFy = 0; FGC sin � = 0

FGC = 0 Ans

+→ ΣFx = 0; FGA = 0 Ans

Joint D:

+ ↑ΣFx = 0; FDE – FDC = 0

FDE = FDC

+↑ ΣFy = 0; FDF – 2 = 0

FDF = 2 kN (C) Ans

Joint F:

+↑

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

FFE = FFC

+↑ΣFy = 0; 2 F cos 53.13° – 2 = 0

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

Joint C:

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

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

FBC = FBA = 3.54 kN (C) Ans

FDC = FDE = 4.13 kN (C) Ans


412

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

, .P2 = 100 lbP1 = 240 lb


B

C D

A

12 ft

5 ft

P1

P2

1200 N

500 N

FBD = 1300 N

1.5 m

3.6 m

P1 = 1200 N, P2 = 500 N.

Joint D:

+↑ΣFy = 0; FBD 5

13

⎛⎝⎜

⎞⎠⎟

– 500 = 0

FBD = 1300 N (C) Ans

+→ ΣFx = 0; 1200 – FCD + 1300

12

13

⎛⎝⎜

⎞⎠⎟

= 0

FCD = 2400 N (T) Ans

Joint A:

+↑ΣFy = 0; FAC = 0 Ans

+→ ΣFy = 0; FAB = 0 Ans

Joint B:

+↑ΣFy = 0; FBC – 1300 5

13

⎛⎝⎜

⎞⎠⎟

= 0

FBC = 500 N (T) Ans


413

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

P2


B

C D

A

12 ft

5 ft

P1

P21.5 m

3.6 m

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

Maximum tension member is DC :

2.5 = 2.40 P2

P2 = 1.04 kN

Maximum compression member is DB :

1.75 = 2.60 P2

P2 = 0.673 kN

The member DB reaches the critical value first.

P2 = 0.673 kN Ans


414

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

.P = 2500 lb


4 ft

4 ft

1200 lb 1200 lbP

4 ft 4 ft 4 ft

A B

F

E D C

G

30� 30�

1 m

1 m 1 m 1 m 1 m

6 kN 6 kN

NB = 12.25 kN FBC = 12.25 kN

FCG = 8.8389 kN

6 kN6 kN 6 kN12.5 kN

2 m 2 m

P = 12.5 kN.

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

�

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

NB = 12.25 kN

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

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

+→ ΣFx = 0; FBG = 0 Ans.

+↑ΣFy = 0; 12.25 – FBC = 0

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) Ans.

+→ ΣFx = 0; FCD – 8.8389 cos 45° = 0

FCD = 6.25 kN (C) Ans.

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

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

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

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

FGF = 12.5 kN (T) Ans.

Due to the symmetry of the system and the loading,

FAE = FBC = 12.25 kN Ans.

FAF = FBG = 0 Ans.

FED = FCD = 6.25 kN (C) Ans.

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

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


415


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

4 ft

4 ft

1200 lb 1200 lbP

4 ft 4 ft 4 ft

A B

F

E D C

G

30� 30�

1 m

1 m 1 m 1 m 1 m

6 kN 6 kN

2 m 2 m

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

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

�

+ ΣMA = 0; NB (2 + 2) – P (2) = 0

NB = 0.5 P

0.7071P = 7.5 P = 10.6067 kNP = 10 kN


416


.P = 5 kN


AC

B

D

E

P

1.5 m

1.5 m

2 m2 m

1.5 m


417

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

2 kN2.5 kN


AC

B

D

E

P

1.5 m

1.5 m

2 m2 m

1.5 m


418

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


3 ft

4 ft

900 lb600 lb

4 ft 4 ft

A B

C

DEF

0.9 m

1.2 m 1.2 m 1.2 m

4.5 kN3 kN

4.5 kN6 kN3 kN

4.5 kN1.2 m

1.2 m 1.2 m

7.50 kN

8.25 kN

1 kN

6.25 kN

3 kN

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

NA = 8.25 kN

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

F → E → A → B → D.


+↑ΣFy = 0; FFA 3

5

⎛⎝⎜

⎞⎠⎟

– 4.5 = 0

FFA = 7.5 kN (C) Ans.

+→ ΣFx = 0; FFE – 7.5

4

5

⎛⎝⎜

⎞⎠⎟

= 0

FFE = 6.0 kN (T) Ans.

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

+→ ΣFx = 0; FED – 6.0 = 0

FED = 6.0 kN (T) Ans.

+↑ΣFy = 0; FEA = 0 Ans.

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

+↑ΣFy = 0; 8.25 – 7.50 3

5

⎛⎝⎜

⎞⎠⎟

– FAD 4

5

⎛⎝⎜

⎞⎠⎟

= 0

FAD = 6.25 kN (C) Ans.

+→ ΣFx = 0; 7.50

4

5

⎛⎝⎜

⎞⎠⎟

– 6.25 4

5

⎛⎝⎜

⎞⎠⎟

– FAB = 0

FAB = 1.0 kN (C) Ans.

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

+→ ΣFx = 0; 1.0 – FBC = 0

FBC = 1.0 kN (C) Ans.

+↑ΣFy = 0; FBD = 0 Ans.

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

+→ ΣFx = 0; FDC

4

5

⎛⎝⎜

⎞⎠⎟

+ 6.25 4

5

⎛⎝⎜

⎞⎠⎟

– 6 = 0

FDC = 1.25 kN (T) Ans.

+↑ΣFy = 0; 6.25 3

5

⎛⎝⎜

⎞⎠⎟

– 1.25 3

5

⎛⎝⎜

⎞⎠⎟

– 3 = 0 (check)

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


419


6–19. The truss is fabricated using members having aweight of . Remove the external forces from thetruss, and determine the force in each member due to theweight of the members. State whether the members are intension or compression. Assume that the total force actingon a joint is the sum of half of the weight of every memberconnected to the joint.

10 lb>ft

3 ft

4 ft

900 lb600 lb

4 ft 4 ft

A B

C

DEF

0.27 kN

0.33 kN

0.33 kN

0.51 kN

0.51 kN

FBD = 0.33 kNFAD = 0.925 kN

FED = 0.36 kN

FAB = 0.38 kN

NA = 1.665 kN

FEA = 0.33 kN

FFA = 0.45 kN

FFE = 0.36 kN

FC = 0.27 kN

FB = 0.33 kN

FA = 0.51 kN

FF = 0.27 kN FE = 0.33 kN FD = 0.51 kN1.2 m 1.2 m 1.2 m

0.9 m

0.9 m

1.2 m 1.2 m 1.2 m

4.5 kN3 kN

weight of 0.2 kN/m. Remove the external forces from the

Joint Loadings:

FC = FF = 0.21.2 + 1.5

2

⎛⎝⎜

⎞⎠⎟

= 0.27 kN

FE = FB = 0.21.2 + 1.2 + 0.9

2

⎛⎝⎜

⎞⎠⎟

= 0.33 kN

FA = FD = 0.21.5 + 1.5 + 1.2 + 0.9

2

⎛⎝⎜

⎞⎠⎟

= 0.51 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

Method of Joints: We will analyze the equilibrium of the joints in the following sequence:F → E → A → B → D.


+↑ΣFy = 0; FFA 3

5

⎛⎝⎜

⎞⎠⎟

– 0.27 = 0

FFA = 0.45 kN (C) Ans.

+→ ΣFx = 0; FFE – 0.45

4

5

⎛⎝⎜

⎞⎠⎟

= 0

FFE = 0.36 kN (T) Ans.

Joint E: From the free – body diagram in Fig. c,+→ ΣFx = 0; FED – 0.36 = 0 FED = 0.36 kN (T) Ans.

+↑ΣFy = 0; FEA – 0.33 = 0 FEA = 0.33 kN (C) Ans.

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

+↑ΣFy = 0; 1.665 – 0.33 – 0.51 – 0.45 3

5

⎛⎝⎜

⎞⎠⎟

– FAD 3

5

⎛⎝⎜

⎞⎠⎟

= 0

FAD = 0.925 kN = 0.925 kN (C) Ans.

+→ ΣFx = 0; FAB + 0.45

4

5

⎛⎝⎜

⎞⎠⎟

– 0.925 4

5

⎛⎝⎜

⎞⎠⎟

= 0

FAB = 0.38 kN (T) Ans.

Joint B: From the free – body diagram in Fig. e,+→ ΣFx = 0; FBC – 0.38 = 0 FBC = 0.38 kN = 0.38 kN Ans.

+↑ΣFy = 0; FBD – 0.33 = 0 FBD = 0.33 kN (T) Ans.

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

+→ ΣFx = 0; 0.925

4

5

⎛⎝⎜

⎞⎠⎟

– 0.36 – FDC 4

5

⎛⎝⎜

⎞⎠⎟

= 0

FDC = 0.475 kN = 0.475 kN (C) Ans.

+↑ΣFy = 0; 0.925 3

5

⎛⎝⎜

⎞⎠⎟

+ 0.475 3

5

⎛⎝⎜

⎞⎠⎟

– 0.51 – 0.33 = 0 (check)


420


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

G

A

B

FC

ED

0.1 m

6 m

2.5 m

3.5 m


421


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

G

A

B

FC

ED

0.1 m

6 m

2.5 m

3.5 m


422


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

A

E D

BC

2 m

400 N

45� 45�45� 45�

2 m

600 N


423


6–23. The truss is fabricated using uniform membershaving a mass of . Remove the external forces fromthe truss, and determine the force in each member due tothe weight of the truss. State whether the members are intension or compression. Assume that the total force actingon a joint is the sum of half of the weight of every memberconnected to the joint.

5 kg>m

A

E D

BC

2 m

400 N

45� 45�45� 45�

2 m

600 N


424



.P = 4 kN

P

3 m

A CB

E

D

F

P

3 m 3 m3 m


425


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

1 kN1.5 kN

P

3 m

A CB

E

D

F

P

3 m 3 m3 m


426


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

A

C

B

D

E

13 ft

13 ft

12 ft

5 ft

300 lb

300 lb

12 ft

45�

3.9 m

3.6 m

1.5 kN

1.5 kN

1.5 m

3.6 m

3.9 m

1.5 kN

1.5 kN

FCD = 3.90 kN

FCB = 3.60 kN

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

Joint C:

+→ ΣFx = 0; 1.5 – FCD sin 22.62° = 0

FCD = 3.90 kN (C) Ans

+↑ΣFy = 0; –FCB + 3.90 cos 22.62° = 0

FCB = 3.60 kN (T) Ans

Joint D:

+ ↓ΣFx = 0; FDB = 0 Ans

+ ↓ ΣFy = 0; 3.90 – FDE = 0


Joint B:

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

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

FBE = 1.485 kN (T) Ans

FBA = 3.612 kN (T) Ans


427


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

ADFE

P P

B C

L/3

L/3L/3L/3


428


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

A

B

C

DF

E

P

d

d

d d/2 d/2 d


429


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

.d = 1 m

A

B

C

DF

E

P

d

d

d d/2 d/2 d


430


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

uB

CA

4 ft

3 ft

300 lb

u

0.9 m

1.5 kN

1.2 m

1.5 kN

1.5 kN

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

Joint A:

+→ ΣFx = 0; 1.5 cos � + FAC + FAB

4

5

⎛⎝⎜

⎞⎠⎟

= 0

+↑ΣFy = 0; –1.5 sin � + FAB 3

5

⎛⎝⎜

⎞⎠⎟

= 0

Thus,

FAB = 2.5 sin �

FAC = –1.5 cos � – 2 sin �

For AB required:

–1 ≤ 2.5 sin � ≤ 2

–2 ≤ 5 sin � ≤ 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,

127° ≤ � – 196° Ans

336° ≤ � – 347° Ans

Possible hand solution:

�2 = �1 + tan–1 3

4

⎛⎝⎜

⎞⎠⎟

= �1 + 36.870

Then

FAB = 2.5 sin �1

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

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

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

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

= –2.50 sin �2

Thus, we require

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

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

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

127° ≤ � ≤ 196° Ans

336° ≤ � ≤ 347° Ans


431


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

2 ft

A B C D

J I H G

E

F

2 ft 2 ft 2 ft 1.5 ft

80 lb 80 lb60 lb

40 lb

0.8 m

0.8 m 0.8 m 0.8 m 0.6 m

0.8 m

0.8 m 0.8 m 0.6 m

0.8 m

0.8 m 0.6 m

400 N 400 N300 N

200 N

400 N 300 N 200 N

400 N 300 N 200 N

+↑ΣFy = 0; 900 – FBH sin 45° = 0

FBH = 1273 N (T) Ans

�

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

FBC = 650 N (T) Ans

Section 2:

+↑ΣFy = 0; 400 + 300 + 200 – FHC = 0

FHC = 900 N (C) Ans.


432


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

A EB C D

IJ

30 kN20 kN 20 kN

40 kN

H G F

4 m

16 m, 4@4m


433


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

A EB C D

IJ

30 kN20 kN 20 kN

40 kN

H G F

4 m

16 m, 4@4m


434


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

AB C D FE

G

H

IJ

L

K

6 kN8 kN5 kN4 kN

3 m

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


435


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

AB C D FE

G

H

IJ

L

K

6 kN8 kN5 kN4 kN

3 m

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


436


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

A E

B C D

6 kN8 kN

G F

3 m

3 m 3 m 3 m

3 m

3 m3 m

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

A E

B C D

6 kN8 kN

G F

3 m

3 m 3 m 3 m

3 m

3 m3 m


437


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

A

C

G

E D

HF

IB

2 m 2 m 2 m

1.5 m

50 kN40 kN

40 kN

30 kN

1.5 m

1.5 m


438


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

A

C

G

E D

HF

IB

2 m 2 m 2 m

1.5 m

50 kN40 kN

40 kN

30 kN

1.5 m

1.5 m


439


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

4 ft 4 ft

4 ft

3 ft 3 ft

4 ft 4 ft

5

1213

AH G F

B

C

D

E

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

4 ft 4 ft

4 ft

3 ft 3 ft

4 ft 4 ft

5

1213

AH G F

B

C

D

E

1.2 m 1.2 m 1.2 m 1.2 m

1300 N1.2 m

0.9 m0.9 m

1.2 m 1.2 m 1.2 m 1.2 m

1300 N1.2 m

0.9 m0.9 m

1300 N

1300 N

Ay = 1200 N

1.2 m 1.2 m 2.4 m

0.9 m1.2 m

2.4 m 2.4 m

2.4 m 1.2 m 1.2 m

1.2 m0.9 m

Ax = 500 N

�

+ ΣMO = 0; 12

13

⎛⎝⎜

⎞⎠⎟

1300 (2.4) – FGD sin 36.87° (4.8) = 0

FGD = 1000 N (C) Ans

�

+ ΣMD = 0; FGF (0.9) – 12

13

⎛⎝⎜

⎞⎠⎟

1300 (1.2)

– 5

13

⎛⎝⎜

⎞⎠⎟

(1300) (0.9) = 0

FGF = 2100 N (C) Ans

�

+ ΣMG = 0; FCD cos 14.04° (1.2) – 12

13

⎛⎝⎜

⎞⎠⎟

1300 (2.4) = 0

FCD = 2473.9 N (T) Ans

Entire truss:

�

+ ΣMG = 0; Ay (2.4) – 12

13

⎛⎝⎜

⎞⎠⎟

(1300) (2.4) = 0

Ay = 1200 N

+→ ΣFx = 0; Ax –

5

13

⎛⎝⎜

⎞⎠⎟

(1300) = 0

Ax = 500 N

Section:

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

FBC = 2473.9 N (T) Ans

�

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

FHG = 2100 N (C) Ans

�

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

FBG = 1000 N (C) Ans


440


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

1.5 m 1.5 m 1.5 m

AH

B C D

JI

G FE

1.5 m

2 m

2 m

6 kN 6 kN

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

1.5 m 1.5 m 1.5 m

AH

B C D

JI

G FE

1.5 m

2 m

2 m

6 kN 6 kN


441


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

8 ft

8 ft

8 ft

900 lb

1500 lb1000 lb1000 lb

A G

N BH

F

M

C D EI

JL K

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

2.5 m

2.5 m

2.5 m

7.5 kN5 kN5 kN

4.5 kN

7.5 kN5 kN5 kN

4.5 kN

2.5 m 2.5 m 5 m5 m

7.5 m 2.5 m

2.5 m

Gx = 4.5 kN

2.5 m 2.5 m

Gy = 11 kN

FEF = 2.828 kN

FEI = 6.5 kN

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

+→ ΣFx = 0; 4.5 – Gx = 0

Gx = 4.5 kN

�

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

Gy = 11 kN

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

�

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

FJI = 18.385 kN (C) Ans.

�

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

FEF = 2.828 kN (T) Ans.

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

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

FEI = 6.5 kN (T) Ans.

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

+↑ΣFy = 0; FJE – 2.828 sin 45° = 0

FHE = 2.0 kN (T) Ans.


442


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

8 ft

8 ft

8 ft

900 lb

1500 lb1000 lb1000 lb

A G

N BH

F

M

C D EI

JL K

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

2.5 m

2.5 m

2.5 m

7.5 kN5 kN5 kN

4.5 kN

7.5 kN5 kN5 kN

4.5 kN

2.5 m 2.5 m 5 m5 m

7.5 m

5 kN

4.5 kN

2.5 m

2.5 m5 m

NA = 6.5 kN

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

�

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

NA = 6.5 kN

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

�

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

FKL = 19 kN (C) Ans.

�

+ ΣML = 0; FCD (2.5) – 6.5 (5) = 0

FCD = 13 kN (T) Ans.

+↑ΣFy = 0; 6.5 – 5 – FLD sin 45° = 0

FLD = 2.121 kN (T) Ans.


443


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

1.5 m

2 m2 m

1 m 1 m

0.8 m

2 kN

1.5 kN

AH

BD

G

C

F E


444


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

1.5 m

2 m2 m

1 m 1 m

0.8 m

2 kN

1.5 kN

AH

BD

G

C

F E


445


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

2 kN3 kN

4 kN5 kN

4 kN

6 kN

5 kN

BA

C D E FG

H

I

J

K

L

2 m

4 m

2 m 2 m 2 m 2 m 2 m


solutions simple trusses

Documents

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