SM05

August 24, 2017 | Author: Izyan Mastura | Category: Truss, Center Of Mass, Trigonometric Functions, Physics & Mathematics, Physics
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5-1. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 800 kN and P2 = 400 kN.

6m

8m C

A

8m

P2

B

P1

Method of Joints : In this case, the support reactions are not required for determining the member forces. Joint B : ΣFx = 0;

3 FBC cos 45° – FBA   – 400 = 0  5

[1]

+↑ ΣFy = 0;

4 FBC sin 45° + FBA   – 800 = 0  5

[2]

Solving Eqs. [1] and [2] yields FBA = 285.71 kN (T) = 286 kN (T)

Ans

FBC = 808.12 kN (T) = 808 kN (T)

Ans

Joint C : ΣFx = 0; +↑ ΣFy = 0;

FCA – 808.12 cos 45° = 0 FCA = 571 kN (C)

Ans

Cy – 808.12 sin 45° = 0 Cy = 571 kN

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

150

5-2. Determine the force on each member of the truss and state if the members are in tension or compression. Set P1 = 500 kN and P2 = 100 kN.

6m

8m C

A

8m

P2

B

P1

Method of Joints : In this case, the support reactions are not required for determining the member forces. Joint B : ΣFx = 0;

3 FBC cos 45° – FBA   – 100 = 0  5

[1]

+↑ ΣFy = 0;

4 FBC sin 45° + FBA   – 500 = 0  5

[2]

Solving Eqs. [1] and [2] yields FBA = 285.71 kN (T) = 286 kN (T)

Ans

FBC = 383.86 kN (T) = 384 kN (T)

Ans

Joint C : ΣFx = 0; +↑ ΣFy = 0;

FCA – 383.86 cos 45° = 0 FCA = 271 kN (C)

Ans

Cy – 383.86 sin 45° = 0 Cy = 271.43 kN

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

151

5-3. 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 tension or compression. Set P1 = 600 kN and P2 = 400 kN. P2

P1

B

A

C 45°

45°

4m

E D 4m

4m

Joint A : +↑ ΣFy = 0;

ΣFx = 0;

FAD sin 45° – 600 = 0 FAD = 848.528 = 849 kN (C)

Ans

FAB – 848.528 cos 45° = 0 FAB = 600 kN (T)

Ans

FBD – 400 = 0 FBD = 400 kN (C)

Ans

FBC – 600 = 0 FBC = 600 kN (T)

Ans

FDC sin 45° – 400 – 848.528 sin 45° = 0 FDC = 1414.214 kN = 1.41 MN (T)

Ans

Joint B : +↑ ΣFy = 0;

ΣFx = 0;

Joint D : +↑ ΣFy = 0;

ΣFx = 0;

848.528 cos 45° + 1414.214 cos 45° – FDE = 0 Ans FDE = 1600 kN = 1.60 MN (C)

152

*5-4. 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 tension or compression. Set P1 = 800 kN and P2 = 0. P2

P1

B

A

C 45°

45°

4m

E D 4m

4m

Joint A : +↑ ΣFy = 0;

FAD sin 45° – 800 = 0 FAD = 1131.4 kN = 1.13 MN (C)

ΣFx = 0;

Ans

FAB – 1131.4 cos 45° = 0 FAB = 800 kN (T)

Ans

Joint B : +↑ ΣFy = 0;

FBD – 0 = 0 FBD = 0

ΣFx = 0;

Ans

FBC – 800 = 0 FBC = 800 kN (T)

Ans

Joint D : +↑ ΣFy = 0;

FDC sin 45° – 0 – 1131.4 sin 45° = 0 FDC = 1131.4 kN = 1.13 MN (T)

ΣFx = 0;

Ans

1131.4 cos 45° + 1131.4 cos 45° – FDE = 0 FDE = 1600 kN = 1.60 MN (C)

Ans

153

5-5. 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. P

2P

A

P

B

C

4m E

4m

4m

D

Method of Joints : In this case, the support reactions are not required for determining the member forces. Joint A: +↑ ΣFy = 0;

1  FAE   –4=0  5 FAE = 8.944 kN (C) = 8.94 kN (C)

ΣFx = 0;

Ans

2  FAB – 8.944   =0  5 FAB = 8.00 kN (T)

Ans

Joint B : ΣFx = 0; +↑ ΣFy = 0;

FBC – 8.00 = 0 FBE – 8 = 0

FBC = 8.00 kN (T) FBE = 8.00 kN (C)

Ans Ans

Joint E : + ΣFy′ = 0; + ΣFx′ = 0;

FEC cos 36.87° – 8.00 cos 26.57° = 0 FEC = 8.944 kN (T) = 8.94 kN (T)

Ans

8.944 + 8.00 sin 26.57° + 8.944 sin 36.87° – FED = 0 FED = 17.89 kN (C) = 17.9 kN (C) Ans

Joint D : +↑ ΣFy = 0;

1  FDC – 17.89   =0  5

FDC = 8.00 kN (T)

ΣFx = 0;

2  –Dx + 17.89   =0  5

Dx = 16.0 kN

Ans

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

154

5-6. 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 comparison. 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.

P A

20   = 166.22 N 2  FB = 4(9.81)(2 + 2 + 1) = 196.2 N   20   FC = 4(9.81) 1 + 3   = 302.47 N  2    20   = 166.22 N 2 

Method of Joints : In this case, the support reactions are not required for determining the member forces. Joint A: 1  FAE   – 166.22 = 0  5 FAE = 371.69 N (C) = 372 N (C)

Ans

2  FAB – 371.69   =0  5 FAB = 332.45 N (T) = 332 N (T)

Ans

Joint B : ΣFx = 0; +↑ ΣFy = 0;

FBC – 332.45 = 0

FBC = 332 N (T)

FBE – 196.2 = 0 FBE = 196.2 N (C) = 196 N (C)

Ans Ans

Joint E : + ΣFy′ = 0;

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

+ ΣFx′ = 0;

371.69 + (196.2 + 302.47) sin 26.57° + 557.53 sin 36.87° – FED = 0 FED = 929.22 N (C) = 929 N (C) Ans

Joint D : +↑ ΣFy = 0;

1  FDC – 929.22   – 166.22 = 0  5 FDC = 582 N (T)

ΣFx = 0;

2  Dx – 929.22   =0  5

C

4m

 FA = 4(9.81)  2 + 

ΣFx = 0;

B

4m

 FD = 4(9.81)  2 + 

P

E

Joint Forces :

+↑ ΣFy = 0;

2P

Ans Dx = 831.12 N

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

155

4m

D

5-7. Determine the force in each member of the truss and state if the members are in tension or compression.

10 kN

8 kN 4 kN 3 kN

B

C

D

1.5 m A

E

F 2m

2m

Σ MA = 0;

–3(1.5) – 4(2) – 10(4) + Ey(4) = 0 Ey = 13.125 kN

+↑ Σ Fy = 0;

Ay – 8 – 4 – 10 + 13.125 = 0 Ay = 8.875 kN

Σ Fx = 0;

Ans

Ans

Ax = 3 kN

Joint B : Σ Fx = 0;

FBC = 3 kN (C)

Ans

+↑ Σ Fy = 0;

FBA = 8 kN (C)

Ans

Joint A : +↑ Σ Fy = 0;

8.875 – 8 –

3 FAC = 0 5

FCA = 1.458 = 1.46 kN (C) Σ Fx = 0;

FAF – 3 –

Ans

4 (1.458) = 0 5

FAF = 4.17 kN (T)

Ans

(cont’d )

156

5-7.

(cont’d )

Joint C : ΣFx = 0;

3+

4 (1.458) – FCD = 0 5

FCD = 4.167 = 4.17 kN (C) +↑ ΣFy = 0;

FCF – 4 +

Ans

3 (1.458) = 0 5

FCF = 3.125 = 3.12 kN (C)

Ans

FEF = 0

Ans

FED = 13.125 = 13.1 kN (C)

Ans

Joint E : ΣFx = 0; +↑ ΣFy = 0; Joint D: +↑ ΣFy = 0;

13.125 – 10 –

3 FDF = 0 5

FDF = 5.21 kN (T) ΣFx = 0;

4.167 –

4 (5.21) = 0 5

Ans Check!

157

*5-8. 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 P2

P1

Method of Joints : In this case, the support reactions are not required for determining the member forces. Joint C : +↑ ΣFy = 0;

ΣFx = 0;

FCB sin 30° – 1.5 = 0 FCB = 3.00 kN (T)

Ans

FCD – 3.00 cos 30° = 0 FCD = 2.598 kN (C) = 2.60 kN (C)

Ans

FDE – 2.598 = 0

Ans

Joint D : ΣFx = 0; +↑ ΣFy = 0;

FDB – 2 = 0

FDE = 2.60 kN (C) FDB = 2.00 kN (T)

Ans

Joint B : + ΣFy′ = 0;

+ ΣFx′ = 0;

FBE cos 30° – 2.00 cos 30° = 0 FBE = 2.00 kN (C)

Ans

(2.00 + 2.00) sin 30° + 3.00 – FBA = 0 FBA = 5.00 kN (T)

Ans

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

158

5-9. 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 P2

P1

Method of Joints : In this case, the support reactions are not required for determining the member forces. Joint C : +↑ Σ Fy = 0;

Σ Fx = 0;

FCB sin 30° – 4 = 0 FCB = 8.00 kN (T)

Ans

FCD – 8.00 cos 30° = 0 FCD = 6.928 kN (C) = 6.93 kN (C)

Ans

FDE – 6.928 = 0

Ans

Joint D : Σ Fx = 0; +↑ Σ Fy = 0;

FDB – 4 = 0

FDE = 6.93 kN (C) FDB = 4.00 kN (T)

Ans

Joint B: + Σ Fy′ = 0;

+ Σ Fx′ = 0;

FBE cos 30° – 4.00 cos 30° = 0 FBE = 4.00 kN (C)

Ans

(4.00 + 4.00) sin 30° + 8.00 – FBA = 0 FBA = 12.0 kN (T)

Ans

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

159

5-10. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 0, P2 = 100 kN.

G

E

1m A

D B 1m

Reactions at A and D :

P1

Ax = 0 Ay = 33.33 kN Dy = 66.67 kN Joint A : ΣFx = 0; +↑ ΣFy = 0;

FAB – FAG cos 45° = 0 33.33 – FAG sin 45° = 0 FAG = 47.1 kN (C)

Ans

FAB = 33.3 kN (T)

Ans

FBG = 0

Ans

FBG = 33.3 kN (T)

Ans

Joint B :

Joint D : ΣFx = 0;

–FDC + FDE cos 45° = 0

+↑ ΣFy = 0;

66.67 – FDE sin 45° = 0 FDE = 94.29 kN = 94.3 kN (C)

Ans

FDC = 66.67 kN = 66.7 kN (T)

Ans

Joint E : ΣFx = 0; +↑ ΣFy = 0;

FEG – 94.29 sin 45° = 0 –FEC + 94.29 cos 45° = 0 FEC = 66.67 kN = 66.7 kN (T)

Ans

FEG = 66.67 kN = 66.7 kN (C)

Ans

Joint C : +↑ ΣFy = 0;

FCG cos 45° + 66.67 – 100 = 0 FCG = 47.1 kN (T)

Ans

160

C 1m

1m P2

5-11. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 50 kN, P2 = 150 kN.

G

E

1m A

D

Reactions at A and D :

B 1m

Ax = 0

P1

Ay = 83.33 kN Dy = 116.67 kN Joint A : ΣFx = 0; +↑ ΣFy = 0;

FAB – FAG cos 45° = 0 83.33 – FAG sin 45° = 0 FAG = 117.85 kN = 117.9 kN (C)

Ans

FAB = 83.33 kN = 83.3 kN (T)

Ans

Joint B : ΣFx = 0;

FBC – 833 = 0

+↑ ΣFy = 0;

FBG – 500 = 0 FBC = 83.3 kN (T)

Ans

FBG = 50 kN (T)

Ans

Joint D : ΣFx = 0; +↑ ΣFy = 0;

–FDC + FDC cos 45° = 0 116.67 – FDE sin 45° = 0 FDE = 164.996 kN = 165 kN (T)

Ans

FDC = 116.67 kN = 116.7 kN (T)

Ans

Joint E : ΣFx = 0; +↑ ΣFy = 0;

FEG – 164.996 sin 45° = 0 –FEC + 164.996 cos 45° = 0 FEC = 116.67 kN = 116.7 kN (T)

Ans

FEG = 116.67 kN = 116.7 kN (C)

Ans

Joint C : +↑ ΣFy = 0;

FCG cos 45° + 116.67 – 150 = 0 FCG = 47.09 kN = 47.1 kN (T)

Ans

161

C 1m

1m P2

*5-12. 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

2m

4m

2m

P1

ΣMA = 0;

D

P2

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

ΣFx = 0;

Ax – 27.5 = 0 Ax = 27.5 kN

+↑ ΣFy = 0;

Ay – 10 – 15 = 0 Ay = 25 kN

Joint G : ΣFx = 0;

FGB – 27.5 = 0 FGB = 27.5 kN (T)

Ans

Joint A : ΣFx = 0;

+↑ ΣFy = 0;

27.5 – FAF –

1 5

(FAB) = 0

2  25 – FAB   =0  5 FAF = 15.0 kN (C)

Ans

FAB = 27.95 = 28.0 kN (C)

Ans

Joint B : ΣFx = 0;

+↑ ΣFy = 0;

1  27.95   + FBC – 27.5 = 0  5 2  27.95   – FBF = 0  5 FBF = 24.99 = 25.0 kN (T)

Ans

FBC = 15.0 kN (T)

Ans

(cont’d)

162

5-12.

(cont’d )

Joint F : ΣFx = 0;

+↑ ΣFy = 0;

15 + FAE –

1 (FFC) = 0 2

25 – 10 – FFC  1  = 0  2 FFC = 21.21 = 21.2 kN (C)

Ans

FFE = 0

Ans

FED = 0

Ans

Joint E : ΣFx = 0; +↑ ΣFy = 0;

FEC – 15 = 0 FEC –15 = 0 FEC = 15.0 kN (T)

Ans

FDC = 0

Ans

Joint D : ΣFx = 0;

163

5-13. 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

2m

4m P1

ΣMA = 0;

2m P2

FGB(4) – 20(6) = 0 FGB = 30 kN (T)

ΣFx = 0;

D

Ans

Ax – 30 = 0 Ax = 30 kN

+↑ ΣFy = 0;

Ay – 20 = 0 Ay = 20 kN

Joint A : ΣFx = 0;

+↑ ΣFy = 0;

30 – FAF –

1 5

(FAB) = 0

2  20 – FAB –   =0  5 FAF = 20 kN (C)

Ans

FAB = 22.36 = 22.4 kN (C)

Ans

(cont’d )

164

5-13.

(cont’d )

Joint B : ΣFx = 0;

+↑ ΣFy = 0;

 1  22.36   + FBC – 30 = 0  5  2  22.36   – FBF = 0  5 FBF = 20 kN (T)

Ans

FBC = 20 kN (T)

Ans

Joint F : ΣFx = 0;

+↑ ΣFy = 0;

20 + FBF –

1 (FFC) = 0 2

Ans

1  20 – FFC   =0  2

FFC = 28.28 = 28.3 kN (C)

Ans

FFE = 0

Ans

Joint E : ΣFx = 0; +↑ ΣFy = 0;

FED – 0 = 0 FEC – 20 = 0 FED = 0

Ans

FEC = 20.0 kN (T)

Ans

Joint D : ΣFx = 0; +↑ ΣFy = 0;

1 5

(FDC) – 0 = 0

FDC = 0

Ans

165

5-14. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 10 kN, P2 = 20 kN, P3 = 30 kN.

B

C

P1 1m A

D F 1m

E 1m

P2

ΣMA = 0;

30° 1m

P3

20(1) + 30(2) – RD cos 30° (3) = 0 RD = 30.79 kN

+↑ ΣFy = 0;

Ay – 10 – 20 – 30 + 30.79 cos 30° = 0 Ay = 33.34 kN

ΣFx = 0;

Ax – 30.79 sin 30° = 0 Ax = 15.4 kN

Joint A : +↑ ΣFy = 0;

33.34 – 10 –

1 FAB = 0 2

FAB = 33 kN (C) ΣFx = 0;

15.4 + FAF –

Ans

1 (33) = 0 2

FAF = 7.93 kN (T)

Ans

(cont’d )

166

5-14.

(cont’d )

Joint B : 1 (33) – FBF = 0 2

+↑ ΣFy = 0;

FBF = 23.33 kN = 23.3 kN (T)

Ans

1 (33) – FBF = 0 2

ΣFx = 0;

FBC = 23.33 kN = 23.3 kN (C)

Ans

Joint F : +↑ ΣFy = 0;



1 FFC – 20 + 23.33 = 0 2

FFC = 4.714 kN = 4.71 kN (C) ΣFx = 0;

FFE – 7.93 –

Ans

1 (4.714) = 0 2

FFE = 11.26 kN = 11.3 kN (T)

Ans

FEC = 30 kN (T)

Ans

FED = 11.26 kN = 11.3 kN (T)

Ans

Joint E : ΣFx = 0; +↑ ΣFy = 0; Joint C : ΣFx = 0;

1 (4.714) + 23.33 – 2

1 FCD = 0 2

FCD = 37.71 kN = 37.7 kN (C) +↑ ΣFy = 0;

1 (4.714) – 30 + 2

1 (37.71) = 0 2

Ans Check!

167

5-15. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 40 kN, P2 = 40 kN, P3 = 0.

B

C

P1 1m A

D F

E

1m

1m P2

Σ MA = 0;

30° 1m

P3

– 40(1) + RD cos 30° (3) = 0 RD = 15.396 kN

+↑ Σ Fy = 0;

Ay – 40 – 40 + 15.396 cos 30° = 0 Ay = 66.67 kN

Σ Fx = 0;

Ax – 15.396 sin 30° = 0 Ax = 7.698 kN

Joint A : +↑ Σ Fy = 0;

66.67 – 40 –

1 FAB = 0 2

FAB = 37.71 kN = 37.7 kN (C) Σ Fx = 0;

7.698 + FAF –

Ans

1 (37.71) = 0 2

FAF = 18.968 kN = 19.0 kN (T)

Ans

Joint B : +↑ Σ Fy = 0;

1 (37.71) – FBF = 0 2 FBF = 26.67 kN = 26.7 kN (T)

Σ Fx = 0;

Ans

1 (37.71) – FBC = 0 2 FBC = 26.67 kN = 26.7 kN (C)

Ans

(cont’d )

168

5-15.

(cont’d )

Joint F : 1 FFC – 40 + 26.67 = 0 2

+↑ Σ Fy = 0;

FFC = 18.856 kN = 18.9 kN (T) Σ Fx = 0;

FFE – 19 +

Ans

1 (18.856) = 0 2

FFE = 5.668 kN = 5.67 kN (T)

Ans

FED = 5.67 kN (T)

Ans

FEC = 0

Ans

Joint E : Σ Fx = 0; +↑ Σ Fy = 0; Joint C : Σ Fx = 0; +↑ Σ Fy = 0;



1 (18.856) + 26.67 – 2

1 FCD = 0 2

FCD = 18.86 kN = 18.9 kN (C)

Ans

169

*5-16. Determine the force in members BC, HC, and HG of the bridge truss, and indicate whether the members are in tension or compression.

G

H

F

3m E

A

B

C

3m

D

3m

3m

3m

12 kN 14 kN 18 kN

Support Reactions : ΣME = 0;

18(3) + 14(6) + 12(9) – Ay(12) = 0

Ay = 20.5 kN

Method of Sections : ΣMC = 0;

ΣMH = 0;

+↑ ΣFy = 0;

FHG(3) + 12(3) – 20.5(6) = 0 FHG = 29.0 kN (C)

Ans

FBC(3) – 20.5(3) = 0 FBC = 20.5 kN (T)

Ans

20.5 – 12 – FHC sin 45° = 0 FHC = 12.0 kN (T)

Ans

5-17. Determine the force in members GF, CF, and CD of the bridge truss, and indicate whether the members are in tension or compression.

G

H

F

3m E

A

B 3m

C 3m

D 3m

3m

12 kN 14 kN 18 kN

Support Reactions : ΣMA = 0;

Ey(12) – 18(9) – 14(6) – 12(3) = 0

ΣFx = 0;

Ey = 23.5 kN

Ex = 0

Method of Sections : ΣMC = 0;

ΣMF = 0;

+↑ ΣFy = 0;

23.5(6) – 18(3) – FGF(3) = 0 FGF = 29.0 kN (C)

Ans

23.5(3) – FCD(3) = 0 FCD = 23.5 kN (T)

Ans

23.5 – 18 – FCF sin 45° = 0 FCF = 7.78 kN (T)

Ans

170

5-18. Determine the force in members DE, DF, and GF of the cantilevered truss, and state if the members are in tension or compression.

B

C

D

E 3m

A

I 4m

H

F

G

4m

4m

4m

1500 kN 5

4 3

+↑ ΣFy = 0;

3 4 FDF – (1500) = 0 5 5 FDF = 2000 kN (C)

ΣMD = 0;

4 3 (1500)(12) + (1500)(3) – FGF(3) = 0 5 5 FGF = 5700 kN (C)

ΣMF = 0;

Ans

Ans

4 (1500)(16) – FDE(3) = 0 5 FDE = 6400 kN (T)

Ans

5-19. The roof truss supports the vertical loading shown. Determine the force in members BC, CK, and KJ and state if these members are in tension or compression.

8 kN 4 kN

D C

E

B

G L

K

J 12 m, 6 @ 2 m

ΣFx = 0;

Ax = 0

ΣMG = 0;

–Ay(12) + 4(8) + 8(6) = 0 Ay = 6.667 kN

ΣMC = 0;

–6.667(4) + FKJ(2) = 0 FKJ = 13.3 kN (T)

ΣMK = 0;

ΣMA = 0;

6.667(4) –

2 5

Ans

FBC(2) = 0

FBC = 14.907 = 14.9 kN (C)

Ans

FCK = 0

Ans

171

3m

F

A I

H

*5-20. 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. 80 kN

50 kN

40 kN B

A

C

D

E

F

G 4m

L 3m

K 3m

ΣMC = 0;

J 3m

I

H

3m

3m

–95(8) + 40(3) + FKJ(4) = 0 FKJ = 112.5 kN (T)

ΣMJ = 0;

3m

Ans

–95(9) + 40(6) + 80(3) + FCD(4) = 0 FCD = 93.75 kN (C)

ΣFx = 0;

–93.75 + 112.50 –

Ans

3 FCJ = 0 5

FCJ = 31.25 kN (C)

Ans

FDJ = 0

Ans

Joint D :

5-21. 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. 80 kN

50 kN

40 kN B

A

C

D

E

F

G 4m

L 3m

ΣME = 0;

K 3m

J 3m

H 3m

3m

–50(3) + 75(6) – FJI(4) = 0 FKJ = 75 kN (T)

+↑ ΣFy = 0;

I 3m

Ans

75 – 50 – FEI = 0 FEI = 25 kN (C)

Ans

172

5-22. 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

3m 8 kN

Support Reactions : ΣME = 0;

6(6) + 8(3) – Ay(9) = 0

ΣFx = 0;

Ax = 0

Ay = 6.667 kN

Method of Sections : ΣMC = 0;

ΣMG = 0;

+↑ ΣFy = 0;

FGF(3 sin 60°) + 6(1.5) – 6.667(4.5) = 0 FGF = 8.08 kN (T)

Ans

FBC(3 sin 60°) – 6.667(3) = 0 FBC = 7.70 kN (C)

Ans

6.667 – 6 – FCG sin 60° = 0 FCG = 0.770 kN (C)

Ans

5-23. 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 C

D

3m

3m

A

E F

G 3m

3m 6 kN

Support Reactions : ΣMA = 0;

Ey(9) – 8(6) – 6(3) = 0

Ey = 7.333 kN

Method of Sections : ΣMC = 0;

ΣMF = 0;

+↑ ΣFy = 0;

7.33(4.5) – 8(1.5) – FFG(3 sin 60°) = 0 FFG = 8.08 kN (T)

Ans

7.333(3) – FCD(3 sin 60°) = 0 FCD = 8.47 kN (C)

Ans

FCF sin 60° + 7.333 – 8 = 0 FCF = 0.770 kN (T)

Ans

173

3m 8 kN

*5-24. Determine the force developed in members GB and GF of the bridge truss and state if these members are in tension or compression.

2.5 m

1m G

1m F

2.5 m E

2.5 m

A

D B

C

60 kN

ΣMA = 0;

80 kN

–60(2.5) – 80(4.5) + Dy(7) = 0 Dy = 72.857 kN

ΣFx = 0;

Ax = 0

+↑ ΣFy = 0;

Ay – 60 – 80 + 72.857 = 0 Ay = 67.143 kN

ΣMB = 0;

–67.143(2.5)+ FGF(2.5) = 0 FGF = 67.143 kN = 67.1 kN (C)

+↑ ΣFy = 0;

Ans

67.143 – FGB = 0 FGB = 67.1 kN (T)

Ans

5-25. The truss supports the vertical load of 600 N. Determine the force in members BC, BG, and HG as the dimension L varies. Plot the results of F (ordinate with tension as positive) versus L (abscissa) for 0 ≤ L ≤ 3 m.

I

H

G

E

3m

A

B L

D

C L

L

600 N

–600 – FBG sin θ = 0

+↑ ΣFy = 0;

FBG = –

sin θ =

600 sin θ

3 2

L + 9 FBG = –200 L2 + 9 ΣMG = 0;

Ans

–FBC(3) – 600(L) = 0 FBC = –200L

ΣMB = 0;

Ans

–FHG(3) – 600(2L) = 0 FHG = 400L

Ans

174

B

5-26. 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.

C

D

2m I

J

2m E

A G

H 1.5 m

1.5 m

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

1.5 m 6 kN

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

ΣMG = 0;

F 1.5 m 6 kN

Ans

3 – 4.5(3) + FIC   (4) = 0  5

FIC = 5.62 kN (C)

Ans

Joint C : ΣFx = 0; +↑ ΣFy = 0;

FCJ = 5.625 kN 4 4 (5.625) + (5.625) – FCG = 0 5 5

FCG = 9.00 kN (T)

Ans

5-27. 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 A

E H

By inspection of joints B, D, H and I, 1.5 m

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

G 1.5 m

1.5 m 6 kN

7.5 –

4 FJE = 0 5

FJE = 9.375 = 9.38 kN (C) Σ Fx = 0;

Ans

3 (9.375) – FGF = 0 5 FGF = 5.625 kN (T)

1.5 m

Ans

Joint E : +↑ Σ Fy = 0;

F

Ans

175

6 kN

*5-28. 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

2m

G 5m

ΣME = 0;

5m

5m

5m

–Ay(20) + 2(20) + 4(15) + 4(10) + 5(5) = 0 Ay = 8.25 kN

ΣMH = 0;

–8.25(5) + 2(5) + FBC(3) = 0 FBC = 10.4 kN (C)

ΣMC = 0;

–8.25(10) + 2(10) + 4(5) + FHG = 9.155 = 9.16 kN (T)

+↑ ΣMO′ = 0;

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

Ans

5 FHG(5) = 0 29 Ans 3 FHC(12.5) = 0 34 Ans

176

5-29 Determine the force in members CD, CF, and CG and state if the members are in tension or compression.

5 kN

4 kN

4 kN

B

C

3 kN

2 kN A

D

E 3m

F

H

2m

G 5m

5m

5m

5m

ΣFx = 0;

Ex = 0

ΣMA = 0;

– 4(5) – 4(10) – 5(15) – 3(20) + Ey(20) = 0 Ey = 9.75 kN

ΣMC = 0;

–5(5) – 3(10) + 9.75(10) –

5 FFG(5) = 0 29

FFG = 9.155 kN (T) ΣMF = 0;

–3(5) + 9.75(5) – FCD(3) = 0 FCD = 11.25 = 11.2 kN (C)

ΣMO′ = 0;

–9.75(2.5) + 5(7.5) + 3(2.5) –

Ans 3 FCF(12.5) = 0 34

FCF = 3.21 kN (T)

Ans

FGH = 9.155 kN (T)

Ans

Joint G : ΣFx = 0; +↑ ΣFy = 0;

2 (9.155)(2) – FCG = 0 29

FCG = 6.80 kN (C)

Ans

177

5-30. Determine the force in members GF, FB, and BC of the Fink truss and state if the members are in tension or compression. 60 kN F

80 kN

80 kN E

G 60°

30°

A

60°

3m

30°

D

C

B 3m

3m

Support Reactions : Due to symmetry, Dy = Ay. +↑ ΣFy = 0;

2Ay – 80 – 60 – 80 = 0

ΣFx = 0;

Ay = 110 kN

Ax = 0

Method of Sections : Σ MB = 0;

FGF sin 30°(3) + 80(3 – 3 cos2 30°) – 110(3) = 0 FGF = 180 kN (C)

Σ MA = 0;

Ans

FFB sin 60°(3) – 80(3 cos2 30°) = 0 FFB = 69.28 kN (T)

Σ MF = 0;

Ans

FBC(4.5 tan 30°) + 80(4.5 – 3 cos2 30°) – 110(4.5) = 0 FBC = 121.24 kN (T)

Ans

5-31. Determine the force in member GJ of the truss and state if this member is in tension or compression. 10 kN

10 kN

G

H

10 kN

J 30°

A

E C

B 3m

3m

D 3m

3m 10 kN

ΣMC = 0;

–10(3) + 15(6) – FGJ cos 30°(6 tan 30°) = 0 FGJ = 20 kN (C)

Ans

178

*5-32. Determine the force in member GC of the truss and state if this member is in tension or compression.

10 kN

10 kN

G

H

10 kN

J 30°

A

E C

B 3m

3m

D 3m

3m 10 kN

Using the results of Prob. 6-45 : Joint G : ΣFx = 0;

FHG = 20 kN

+↑ ΣFy = 0;

–10 + 2(20 cos 60°) – FGC = 0 FGC = 10 kN (T)

Ans

5-33. Determine the force in members GF, CF, and CD of the roof truss and indicate if the members are in tension or compression. 1.5 kN C 1.70 m

2 kN 0.8 m

D

B

1.5 m E

A H

G

F

1m 2m

ΣMA = 0;

2m

Ey(4) – 2(0.8) – 1.5(2.50) = 0

Ey = 1.3375 kN

Method of Sections : ΣMC = 0;

ΣMF = 0;

1.3375(2) – FGF(1.5) = 0 FGF = 1.78 kN (T)  3 1.3375(1) – FCD  (1) = 0 5 FCD = 2.23 kN (C)

ΣME = 0;

Ans

 1.5   (1) = 0 FCF   3.25 

FCF = 0

Ans

Ans

179

5-34. In each case, determine the force P required to maintain equilibrium. The block weighs 100 N (≈ 10 kg).

P P

P

(a)

(b)

(c)

Equations of Equilibrium : a)

b)

c)

+↑ ΣFy = 0;

+↑ ΣFy = 0;

4P – 100 = 0 P = 25.0 N

Ans

3P – 100 = 0 P = 33.3 N

Ans

+↑ ΣFy = 0;

3P′ – 100 = 0 P′ = 33.33 N

+↑ ΣFy = 0;

3P – 33.33 = 0 P = 11.1 N

Ans

180

5-35. The eye hook has a positive locking latch when it supports the load because its two pairs are pin-connected at A and they bear against one another along the smooth surface at B. Determine the resultant force at the pin and the normal force at B when the eye hook supports a load of 800 N.

800 N

6 mm 50 mm

A

75 mm B

ΣMA = 0;

–FB cos 60° (75) – FB sin 60° (50) + 800(6) = 0 FB = 59.41 = 59.4 N

+↑ ΣFy = 0;

30°

800 N

Ans

–800 – 59.4 sin 60° + Ay = 0 Ay = 851.44 = 851 N

ΣFx = 0;

Ax – FB cos 60° = 0 Ax = 29.7 N

FA =

(851.44)2 + (29.7)2

= 851.96 N = 852 N

Ans

*5-36. Determine the force P needed to support the 100 N (≈ 10 kg) weight. Each pulley has a weight of 10 N (≈ 1 kg). Also, what are the cord reactions at A and B?

C

A

50 mm

50 mm B

50 mm Equations of Equilibrium :From FBD (a), +↑ ΣFy = 0;

P

P′ – 2P – 10 = 0

[1]

2P + P′ – 100 – 10 = 0

[2]

From FBD (b), +↑ ΣFy = 0;

Solving Eqs. [1] and [2] yields, P = 25.0 N

Ans

P′ = 60.0 N The cord reactions at A and B are FA = P = 25.0 N

FB = P′ = 60.0 N

Ans

181

5-37. The link is used to hold the rod in place. Determine the required axial force on the screw at E if the largest force to be exerted on the rod at B, C or D is to be 100 N. Also, find the magnitude of the force reaction at pin A. Assume all surfaces of contact are smooth.

A 100 mm E C 80 mm D 45° B 50 mm

ΣFy = 0;

RC =

1 RB 2

ΣFx = 0;

RD =

1 RB 2

Assume RB = 100 N RC = RD =

100 = 70.71 N < 100 N (O.K!) 2

ΣMA = 0;

–100 sin 45° (50 sin 45°) – 100 cos 45° (180 + 50 cos 45°) + RE(100) = 0 RE = 177.28 = 177 N

+↑ ΣFy = 0;

Ans

–100 sin 45° + Ay = 0 Ay = 70.71 N

ΣFx = 0;

177.28 – 100 cos 45° – Ax = 0 Ax = 106.57 N RA =

106.572 + 70.712 = 128 N

Ans

182

5-38. The principles of a differential chain block are indicated schematically in the figure. Determine the magnitude of force P needed to support the 800N force. Also, find the distance x where the cable must be attached to bar AB so the bar remains horizontal. All pulleys have radius of 60 mm.

x B A 800 N

180 mm

240 mm P

Equations of Equilibrium : From FBD(a), +↑ ΣFy = 0;

4P′ – 800 = 0

P′ = 200 N

200 – 5P = 0

P = 40.0 N

From FBD(b), +↑ ΣFy = 0; ΣMA = 0;

Ans

200(x) – 40.0(120) – 40.0(240) – 40.0(360) – 40.0(480) = 0 x = 240 mm

Ans

5-39. Determine the force P needed to support the 20-kg mass using the Spanish Burton rig. Also, what are the reactions at the supporting hooks A, B, and C?

A H

B G

P

E

D

For pulley D : +↑ ΣFy = 0;

9P – 20(9.81) = 0 P = 21.8 N

Ans

At A,

RA = 2P = 43.6 N

Ans

At B,

RB = 2P = 43.6 N

Ans

At C,

RC = 6P = 131 N

Ans

183

C F

*5-40. The compound beam is fixed at A and supported by a rocker at B and C. There are hinges (pins) at D and E. Determine the reactions at the supports.

15 kN A

D

B

E C

6m

2m 2m 2m

6m

Equations of Equilibrium : From FBD(a), ΣME = 0; +↑ ΣFy = 0; ΣFx = 0;

Cy(6) = 0 Ey – 0 = 0

Cy = 0

Ans

Ey = 0

Ex = 0

From FBD(b), ΣMD = 0;

+↑ ΣFy = 0;

ΣFx = 0;

By(4) – 15(2) = 0 By = 7.50 kN

Ans

Dy + 7.50 – 15 = 0 Dy = 7.50 kN Dx = 0

From FBD(c), ΣMA = 0;

+↑ ΣFy = 0; ΣFx = 0;

MA – 5.00(6) = 0 MA = 30.0 kN · m

Ans

Ay – 5.00 = 0

Ans

Ax = 0

Ay = 5.00 kN

Ans

184

5-41. The compound beam is pin-supported at C and supported by a roller at A and B. There is a hinge (pin) at D. Determine the reactions at the supports. Neglect the thickness of the beam.

8 kN

12 kN 5

A

D

15 kN · m B

30° 8m 6m 4 kN 4m2m

4 3

C

8m

8m

Equations of Equilibrium : From FBD(a), ΣMD = 0;

+↑ ΣFy = 0;

ΣFx = 0;

4 cos 30° (12) + 8(2) – Ay(6) = 0 Ay = 9.5949 kN = 9.59 kN

Ans

Dy + 9.595 – 4 cos 30° – 8 = 0 Dy = 1.869 kN Dx – 4 sin 30° = 0

Dx = 2.00 kN

From FBD(b), ΣMC = 0;

1.869(24) + 15 + 12

 4 (8) – By(16) = 0  5

By = 8.541 kN = 8.54 kN +↑ ΣFy = 0;

Cy + 8.541 – 1.869 – 12 Cy = 2.93 kN

ΣFx = 0;

Cx – 2.00 – 12

Ans

 4 =0  5 Ans

 3 =0  5

Cx = 9.20 kN

Ans

185

5-42. Determine the greatest force P that can be applied to the frame if the largest force resultant acting at A can have a magnitude of 2 kN.

0.1 m

0.5 m

A 0.75 m

0.75 m P

ΣMA = 0;

T(0.6) – P(1.5) = 0

ΣFx = 0;

Ax – T = 0

+↑ ΣFy = 0;

Ay – P = 0

Thus,

Ax = 2.5P,

Ay = P

Require, 2=

(2.5 P)2 + ( P)2

P = 0.743 kN = 743 N

Ans

186

5-43. Determine the horizontal and vertical components of forces at pins A and C of the twomember frame.

200 N/ m

B A

3m

C 3m

Free Body Diagram : The solution for this problem will be simplified if one realizes that member BC is a two force member. Equations of Equilibrium : ΣMA = 0;

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

+↑ ΣFy = 0;

Ay + 424.26 cos 45° – 600 = 0 Ay = 300 N

Ans

424.26 sin 45° – Ax = 0 Ax = 300 N

Ans

ΣFx = 0;

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

Ans

Cy = FBC cos 45° = 424.26 cos 45° = 300 N

Ans

187

5-44. The three-hinged arch supports the loads F1 = 8 kN and F2 = 5 kN. Determine the horizontal and vertical components of reaction at the pin supports A and B. Take h = 2 m.

F2

C 3m F1 4m A h

B

8m

4m

2m

Member AC : ΣMA = 0;

–8(4) + Cy(8) + Cx(7) = 0 8Cy + 7Cx – 32 = 0

ΣFx = 0; +↑ ΣFy = 0;

8 – Ax – Cx = 0 –Ay + Cy = 0 Ay = C y

Member BC : ΣMB = 0;

5(2) + Cy(6) – Cx(9) = 0 6Cy – 9Cx + 10 = 0

ΣFx = 0;

C x – Bx = 0 Bx = Cx

+↑ ΣFy = 0;

–Cy + By – 5 = 0

Solving : Ax = 5.6141 = 5.61 kN

Ans

Ay = 1.9122 = 1.91 kN

Ans

Cx = 2.3859 = 2.39 kN

Ans

Cy = 1.9122 = 1.91 kN

Ans

Bx = 2.3859 = 2.39 kN

Ans

By = 6.9122 = 6.91 kN

Ans

188

0.5 m

0.5 m

0.5 m

0.5 m

2 kN 2 kN 2 kN 2 kN

5-45. Determine the horizontal and vertical components of force at pins A, B, and C, and the reactions to the fixed support D of the three-member frame.

B

A 2m

C

2m Free Body Diagram : The solution for this problem will be simplified if one realizes that member AC is a two force member.

D

Equations of Equilibrium : For FBD(a), ΣMB = 0;

2(0.5) + 2(1) + 2(1.5) + 2(2) – FAC

 4 (1.5) = 0  5

FAC = 8.333 kN +↑ ΣFy = 0;

By + 8.333

 4 –2–2–2–2=0  5

By = 1.333 kN = 1.33 kN Ans ΣFx = 0;

Bx – 8.333

 3 =0  5

Bx = 5.00 kN

Ans

For pin A and C, Ax = Cx = FAC

 3  3 = 8.333 = 5.00 kN  5  5

Ans

Ay = Cy = FAC

 4  4 = 8.333 = 6.67 kN  5  5

Ans

From FBD(b), ΣMD = 0;

5.00(4) – 8.333

 3 (2) – MD = 0  5

MD = 10.0 kN · m +↑ ΣFy = 0;

Dy – 1.333 – 8.333

 4 =0  5

Dy = 8.00 kN ΣFx = 0;

8.333

Ans

Ans

 3 – 5.00 – Dx = 0  5

Dx = 0

Ans

189

5-46. Determine the horizontal and vertical components of force at C which member ABC exerts on member CEF.

C 4m B

E

1m D

4m A

F 6m

3m

30 kN

Member BED : ΣMB = 0;

–30(6) + Ey(3) = 0 Ey = 60 kN

+↑ ΣFy = 0;

–By + 60 – 30 = 0 By = 30 kN

ΣFx = 0;

Bx + Ex – 30 kN = 0

[1]

Member FEC : ΣMC = 0;

30(3) – Ex(4) = 0 Ex = 22.5 kN

From Eq. [1] ΣFx = 0;

Bx = 7.5 kN –Cx + 30 – 22.5 = 0 Cx = 7.5 kN

Ans

Member ABC : ΣMA = 0;

–7.5(8) – Cy(6) + 7.5(4) + 30(3) = 0 Cy = 10 kN

Ans

190

5-47. Determine the horizontal and vertical components of force that the pins at A, B, and C exert on their connecting members.

A

0.2 m 50 mm

B

C 1m 800 N

ΣMB = 0;

ΣFx = 0; +↑ ΣFy = 0;

–800(1 + 0.05) + Ax(0.2) = 0 Ax = 4200 N = 4.20 kN

Ans

Bx = 4200 N = 4.20 kN

Ans

Ay – By – 800 = 0

[1]

Member AC : ΣMC = 0;

From Eq. [1] ΣFx = 0;

–800(50) – Ay(200) + 4200(200) = 0 Ay = 4000 N = 4.00 kN

Ans

By = 3.20 kN

Ans

– 4200 + 800 + Cx = 0 Cx = 3.40 kN

+↑ ΣFy = 0;

Ans

4000 – Cy = 0 Cy = 4.00 kN

Ans

191

5-48. The hoist supports the 125-kg engine. Determine the force the load creates in member DB and in member FB, which contains the hydraulic cylinder H.

1m

2m F

G

E

2m

H D 1m C B

A

2m

1m

Free Body Diagram : The solution for this problem will be simplified if one realizes that members FB and DB are two force members. Equations of Equilibrium : For FBD(a), ΣME = 0;

3  1226.25(3) – FFB   (2) = 0  10  FFB = 1938.87 N = 1.94 kN

+↑ ΣFy = 0;

Ans

3  1938.87   – 1226.25 – Ey = 0  10 

Ey = 613.125 N ΣFx = 0;

1  Ex – 1938.87   =0  10  Ex = 613.125 N

From FBD(b), ΣMC = 0;

613.125(3) – FBD sin 45° (1) = 0 FBD = 2601.27 = 2.60 kN

Ans

192

5-49. Determine the force P on the cord, and the angle θ that the pulley-supporting link AB makes with the vertical. Neglect the mass of the pulleys and the link. The block has a weight of 200 N (≈ 20 kg) and the cord is attached to the pin at B. The pulleys have radii of r1 = 2 cm and r2 = 1 cm. A θ r1

45° B P

r2

+↑ ΣFy = 0;

2T – 200 = 0 T = 100 N

ΣFx = 0; +↑ ΣFy = 0;

Ans

100 cos 45° – FAB sin θ = 0 FAB cos θ – 100 – 100 – 100 sin 45° = 0

θ = 14.6°

Ans

FAB = 280 N

5-50. The front of the car is to be lifted using a smooth, rigid 3.5 m long board. The car has a weight of 17.5 kN and a center of gravity at G. Determine the position x of the fulcrum so that an applied force of 500 N at E will lift the front wheels of the car.

500 N

D

E

Free Body Diagram : When the front wheels are lifted, the normal reaction NB = 0. Equations of Equilibrium : From FBD(a), 17.5(1.5) – FC(3.2) = 0

FC = 8.203 kN

From FBD(b), ΣMD = 0;

500(x) – 8203(3.5 – x) = 0 x = 3.30 m

A

C B x 3.5 m

ΣMA = 0;

G

Ans

193

0.5 m 1.2 m

1.5 m

5-51. The wall crane supports a load of 700 N. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force of the cable at the winch W ?

D

1m 1m

1m C

A

B

E 60° W 700 N

Pulley E : +↑ ΣFy = 0;

2T – 700 = 0 T = 350 N

Ans

Member ABC : ΣMA = 0;

TBD sin 45°(1) – 350° sin 60°(1) – 700(2) = 0 TBD = 2409 N

+↑ ΣFy = 0;

–Ay + 2409 sin 45° – 350 sin 60° – 700 = 0 Ay = 700 N

ΣFx = 0;

Ans

Ax – 2409 cos 45° – 350 cos 60° + 350 – 350 = 0 Ax = 1.88 kN

Ans

At D : Dx = 2409 cos 45° = 1703.1 N = 1.70 kN

Ans

Dy = 2409 sin 45° = 1.70 kN

Ans

194

*5-52. Determine the force that the smooth roller C exerts on beam AB. Also, what are the horizontal and vertical components of reaction at pin A? Neglect the weight of the frame and roller.

60 N · m

D C

A

0.2 m B 1.2 m

1m

ΣMA = 0;

–60 + Dx(0.2) = 0 Dx = 300 N

ΣFx = 0; +↑ ΣFy = 0; ΣMB = 0;

Ax = 300 N

Ans

Ay = 0

Ans

–NC(1.2) + 300(0.2) = 0 NC = 50 N

Ans

5-53. Determine the horizontal and vertical components of force which the pins exert on member ABC.

D 1m

0.2 m A B 2m

80 kN

ΣFx = 0;

Ax = 80 kN

Ans

+↑ ΣFy = 0;

Ay = 80 kN

Ans

ΣMC = 0;

80(5) – By(3) = 0 By = 133.3 = 133 kN

ΣMD = 0;

–80(0.8) + 133.3(3) – Bx(1) = 0 Bx = 336 kN

ΣFx = 0;

Ans

80 + 336 – Cx = 0 Cx = 416 kN

+↑ ΣFy = 0;

Ans

Ans

–80 + 133.3 – Cy = 0 Cy = 53.3 kN

Ans

195

C 3m

5-54. The engine hoist is used to support the 200-kg engine. Determine the force acting in the hydraulic cylinder AB, the horizontal and vertical components of force at the pin C, and the reactions at the fixed support D. 10° 350 mm

1250 mm

E

G A

C

850 mm

B

550 mm D A

196

5-55. Determine the horizontal and vertical components of force at pins B and C.

1m

1m 0.1 m

500 N 0.4 m C

B

1.5 m

A

ΣMA = 0;

–Cy(2) + Cx(1.5) + 500(0.9) = 0

ΣFx = 0;

Ax = Cx

+↑ ΣFy = 0; ΣMB = 0;

ΣFx = 0;

500 – Ay – Cy = 0 –500(0.5) – 500 × 0.9 + Cy(2) = 0 Cy = 350 = 350 N

Ans

Cx = 166.67 = 166.7 N

Ans

166.67 + 500 – Bx = 0 Bx = 666.7 N

+↑ ΣFy = 0;

Ans

By – 500 + 350 = 0 By = 150 N

Ans

197

*5-56. The pipe cutter is clamped around the pipe P. If the wheel at A exerts a normal force of FA = 80 N on the pipe, determine the normal forces of wheels B and C on the pipe. Also compute the pin reaction on the wheel at C. The three wheels each have a radius of 7 mm and the pipe has an outer radius of 10 mm.

C 10 mm B

A

10 mm

P

10 θ = sin–1   = 36.03°  17  Equations of Equilibrium : +↑ ΣFy = 0;

NB sin 36.03° – NC sin 36.03° = 0 NB = NC

ΣFx = 0;

80 – NC cos 36.03° – NC cos 36.03° = 0 NB = NC = 49.5 N

Ans

198

5-57. Determine the horizontal and vertical components of force at each pin . The suspended cylinder has a weight of 800 N (≈ 80 kg).

1.5 m E 2m 0.5 m A

B

C

2m D

3m

ΣMB = 0;

1m

2  FCD   (1.5) – 800(2) = 0  13 

FCD = 1923 N

+↑ ΣFy = 0;

Cx = Dx =

3 (1923) = 1600 N 13

Ans

Cy = Dy =

2 (1923) = 1067 N 13

Ans

–By +

2 (1923) – 800 = 0 13

By = 266.68 N = 266.7 N ΣME = 0;

–Bx(2) + 800(1.5) + 266.68(1.5) = 0 Bx = 800 N

ΣFx = 0;

Ans

Ex + 800 – 800 = 0 Ex = 0

+↑ ΣFy = 0;

Ans

Ans

–Ey + 266.68 = 0 Ey = 266.68 N = 266.7 N

Ans

6Cy – 9Cx + 10 = 0 ΣFx = 0;

–Ax + 800 + Ax = 1600 N

3 (1923) – 800 = 0 13 Ans

199

5-58. The toggle clamp is subjected to a force F at the handle. Determine the vertical clamping force acting at E.

a/2

F

B

A

1.5 a a/2

C 60°

D a/2

E

1.5 a

200

5-59. Determine the horizontal and vertical components of force which the pins at A, B, and C exert on member ABC of the frame. 400 N 1.5 m

2m

C

D 1.5 m

2.5 m

300 N

2m

B

300 N

2.5 m 1.5 m A

ΣME = 0;

E

–Ay(3.5) + 400(2) + 300(3.5) + 300(1.5) = 0 Ay = 657.1 = 657 N

ΣMD = 0;

Ans

–Cy(3.5) + 400(2) = 0 Cy = 228.6 = 229 N

Ans

ΣMB = 0;

Cx = 0

Ans

ΣFx = 0;

FBD = FBE

+↑ ΣFy = 0;

 5  657.1 – 228.6 – 2   FBD = 0  74 

FBD = FBE = 368.7 N Bx = 0 By =

Ans 5 (368.7)(2) = 429 N 74

Ans

201

*5-60. The derrick is pin-connected to the pivot at A. Determine the largest mass that can be supported by the derrick if the maximum force that can be sustained by the pin A is 18 kN.

B

C 5m

D A

60°

AB is a two-force member. Pin B Require FAB = 18 kN +↑ ΣFy = 0;

18 sin 60° –

W sin 60° – W = 0 2

W = 10.878 kN m=

10.878 = 1.11 Mg Ans 9.81

202

5-61. Determine the required mass of the suspended cylinder if the tension in the chain wrapped around the freely turning gear is to be 2 kN. Also, what is the magnitude of the resultant force on pin A?

2 kN 45°

B

2 kN 30° 2m

A

ΣMA = 0;

– 4(2 cos 30°) + W cos 45°(2 cos 30°) + W sin 45°(2 sin 30°) = 0 W = 3.586 kN m = 3.586(1000)/9.81 = 366 kg

ΣFx = 0;

Ans

4 – 3.586 cos 45° – Ax = 0 Ax = 1.464 kN

+↑ ΣFy = 0;

3.586 sin 45° – Ay = 0 Ay = 2.536 kN

FA =

(1.464)2 + (2.536)2 = 2.93 kN

Ans

203

5-62. The pumping unit is used to recover oil. When the walking beam ABC is horizontal, the force acting in the wireline at the well head is 1000 N. Determine the torque M which must be exerted by the motor in order to overcome this load. The horse-head C weighs 240 N and has a center of gravity at GC. The walking beam ABC has a weight of 520 N and a center of gravity at GB, and the counterweight has a weight of 800 N and a center of gravity at GW. The pitman, AD, is pin-connected at its ends and has negligible weight. 1.8 m

1.5 m GB A

70°

M

D

0.3 m GC

C

B

Gw

20°

E 1000 N

0.75 m 0.9 m

Free Body Diagram : The solution for this problem will be simplified if one realizes that the pitman AD is a two force member. Equations of Equilibrium : From FBD (a), ΣMB = 0;

FAD sin 70° (1.5) – 240(1.8) – 1000(2.1) = 0 FAD = 1796.33 N Ans

From (b), ΣME = 0;

1796.33(0.9) – 800 cos 20° (1.65) – M = 0 M = 376.3 N · m Ans

204

5-63. Determine the force P on the cable if the spring is compressed 10 mm when the mechanism is in the position shown. The spring has a stiffness of k = 12 kN/m.

150 mm 150 mm 100 mm

150 mm

A B

D

30° C

P

600 mm

k E

10  FE = ks = 12  = 0.12 kN = 120 N  1000 

ΣMA = 0;

Bx(150) + By(150) – 120(750) = 0 Bx + By = 600 N

ΣMD = 0;

[1]

By(150) – P(100) = 0 By = 0.6667P

[2]

ΣFx = 0;

–Bx + FCD cos 30° = 0

[3]

ΣMB = 0;

FCD sin 30° (150) – P(250) = 0 FCD = 3.333P

Thus from Eq. [3] Bx = 2.8867P Using Eqs. [1] and [2] : 2.8867P + 0.6667P = 600 N P = 168.85 N

Ans

205

*5-64. Determine the force that the jaws J of the metal cutters exert on the smooth cable C if 100-N forces are applied to the handles. The jaws are pinned at E and A, and D and B. There is also a pin at F.

15°

100 N

400 mm 15° A

20 mm

J E C

F D

15° 15°

B

30 mm 80 mm

20 mm

400 mm 100 N 15°

Free Body Diagram : The solution for this problem will be simplified if one realizes that members ED is a two force member. Equations of Equilibrium : From FBD (b), ΣFx = 0;

Ax = 0

From (a), ΣMF = 0;

Ay sin 15°(20) + 100 sin 15°(20) – 100 cos 15°(400) = 0 Ay = 7364.10 N

From FBD (b), ΣME = 0;

7364.10(80) – FC(30) = 0 FC = 19637.60 N = 19.6 N

Ans

206

5-65. The compound arrangement of the pan scale is shown. If the mass on the pan is 4 kg, determine the horizontal and vertical components at pins A, B, and C and the distance x of the 25-g mass to keep the scale in balance. 100 mm

75 mm

300 mm F

350 mm

C

E

x

B G 50 mm A

D

4 kg Free Body Diagram : The solution for this problem will be simplified if one realizes that members DE and FG are two force members. Equations of Equilibrium : From FBD (a), ΣMA = 0; +↑ ΣFy = 0; ΣFx = 0;

FDE(375) – 39.24(50) = 0

FDE = 5.232 N

Ay + 5.232 – 39.24 = 0 Ay = 34.0 N

Ans

Ax = 0

Ans

From (b), ΣMC = 0; +↑ ΣFy = 0; ΣFx = 0;

FFG(300) – 5.232(75) = 0

FFG = 1.308 N

Cy – 1.308 – 5.232 = 0 Cy = 6.54 N

Ans

Cx = 0

Ans

From (c), ΣMB = 0; +↑ ΣFy = 0; ΣFx = 0;

1.308(100) – 0.24525(825 – x) = 0 x = 292 mm Ans 1.308 – 0.24525 – By = 0 By = 1.06 N Bx = 0

Ans Ans

207

208

5-67. Determine the horizontal and vertical components of force that the pins at A, B, and C exert on the frame. The cylinder has a mass of 80 kg.

D 1m C B

0.7 m

0.5 m A Equations of Equilibrium : From FBD (b), ΣMB = 0;

784.8(1.7) – Cy(1) = 0 Cy = 1334.16 N = 1.33 kN

+↑ ΣFy = 0;

By + 784.8 – 1334.16 = 0 By = 549 N

ΣFx = 0;

Ans

Ans

C x – Bx = 0

[1]

From FBD (a), ΣMA = 0;

Cx(0.5) + 1334.16(1) – 784.8(1.7) – 784.8(1.9) = 0 Cx = 2982.24 N = 2.98 kN

+↑ ΣFy = 0;

Ay + 1334.16 – 784.8 – 784.8 = 0 Ay = 235 N

ΣFx = 0;

Ans

Ans

Ax – 2982.24 = 0 Ax = 2982.24 N = 2.98 kN

Ans

Substitute Cx = 2982.24 N into Eq. [1] yields, Bx = 2982.24 N = 2.98 kN

Ans

*5-68. By squeezing on the hand brake of the bicycle, the rider subjects the brake cable to a tension of 200 N. If the caliper mechanism is pin-connected to the bicycle frame at B, determine the normal force each brake pad exerts on the rim of the wheel. Is this the force that stops the wheel from turning? Explain.

60 mm 60 mm

B

B

75 mm

ΣMB = 0;

–N(75) + 200(60) = 0 N = 160 N

Bx

By

N

Ans

This normal force does not stop the wheel from turning. A frictional force (see Chapter 8), which acts along on the wheel’s rim stops the wheel. Ans

209

200 N

75 mm

5-69. If a force of P = 30 N is applied perpendicular to the handle of the mechanism, determine the magnitude of force F for equilibrium. The members are pin-connected at A, B, C, and D.

P = 30 N

100 mm 125 mm

625 mm

100 mm

B

125 mm

A D

125 mm

C

750 mm

F

ΣMA = 0;

FBC(100) – 30(625) = 0 FBC = 187.5 N

ΣFx = 0;

–Ax + 30 = 0 Ax = 30 N

+↑ ΣFy = 0;

–Ay + 187.5 = 0 Ay = 187.5 N

ΣMD = 0;

–125(30) – 187.5(225) + 975(F) = 0 F = 47.12 N

Ans

210

5-70. The bucket of the backhoe and its contents have a weight of 3000 N (≈ 300 kg) and a center of gravity at G. Determine the forces of the hydraulic cylinder AB and in links AC and AD in order to hold the load in the position shown. The bucket is pinned at E.

B

45° A

D E 0.4 m

120° G

C

0.1 m 0.6 m

Free Body Diagram : The solution for this problem will be simplified if one realizes that the hydraulic cylinder AB, links AD and AC are two force members. Equations of Equilibrium : From FBD (a), ΣME = 0;

FAC cos 60° (0.4) + FAC sin 60° (0.1) – 3000(0.6) = 0 FAC = 6280.47 N = 6.28 kN

Ans

Using method of joint [FBD (b)], +↑ ΣFy = 0;

6280.47 sin 60° – FAB cos 45° = 0 FAB = 7691.97 N = 7.692 kN

ΣFx = 0;

Ans

FAD – 7691.97 sin 45° – 6280.47 cos 60° = 0 FAD = 8579.28 N = 8.58 kN

Ans

211

5-71. Determine the reactions at the supports of the compound beam. There is a short vertical link at C.

5 kN

5 kN 3 kN

C A

D

B 5m

5m

5m

5m

5m

5m

For the right portion of the beam : ΣMC = 0;

Dy(10) – 5000(5) = 0 Dy = 2500 = 2.5 kN

Ans

Then Cy = 5000 – 2500 = 2.5 kN For the left portion : ΣMA = 0;

By(10) – 5000(5) – 3000(15) – 2500(20) = 0 By = 12 000 = 12 kN

+↑ ΣFy = 0;

Ans

Ay – 5000 + 12 000 – 3000 – 2500 = 0 Ay = –1500 = 1.5 kN ↓

Ans

Ax = 0

Ans

212

5-72. The two-bar mechanism consists of a lever arm AB and smooth link CD, which has a fixed collar at its end C and a roller at the other end D. Determine the force P needed to hold the lever in the position θ. The spring has a stiffness k and unstretched length 2L. The roller contacts either the top or bottom portion of the horizontal guide.

P B

2L C L

θ

k

A

D

Free Body Diagram : The spring compresses x = 2L –

L . sin θ

1   Then, the spring force developed is Fsp = kx = kL 2 – .  sin θ 

Equations of Equilibrium : From FBD (a), ΣFx = 0;

1   kL 2 – – FCD sin θ = 0  sin θ 

FCD = ΣMD = 0

kL  1  2 – sin θ  sin θ 

MC = 0

From FBD (b), ΣMA = 0;

P(2L) –

kL  1  2 – (L cot θ) = 0 sin θ  sin θ 

P=

kL (2 – csc θ) 2 tan θ sin θ

Ans

213

5-73. Determine the force in each member of the truss and indicate whether the members are in tension or compression.

A

700 N

700 N

2m

D

C

B 2m

Joint A : ΣFx = 0; +↑ ΣFy = 0;

700 – FAD sin 45° = 0 FAD = 989.95 N = 990 N (C)

Ans

989.95 cos 45° – FAB = 0 FAB = 700 N (T)

Ans

Joint B : + ΣFx′ = 0;

FDB – 700 sin 45° = 0 FDB = 494.97 N = 495 N (C)

+

ΣFy′ = 0;

Ans

FDC – 700 cos 45° – 989.95 = 0 FDC = 1484.92 N = 1.48 kN (C)

Ans

Joint C : ΣFx = 0; +↑ ΣFy = 0;

1484.92 cos cos 45° – FCB = 0 FCB = 1050 N = 1.05 kN (T)

Ans

Cy – 1484.92 sin 45° = 0 Cy = 1050 N

214

5-74. The Howe bridge truss is subjected to the loading shown. Determine the force in members HD, CD, and GD, and indicate whether the members are in tension or compression.

40 kN 30 kN

J

20 kN

20 kN

I

H

F

G

4m A B

C

E

D

16 m, 4 @ 4 m

Support Reaction : ΣMA = 0;

Ey(16) – 40(12) – 20(8) – 20(4) = 0 Ey = 45 kN

Method of Sections : From FBD (a) +↑ ΣFy = 0;

–FHD sin 45° – 40 + 45 = 0

ΣMH = 0;

45(8) – 40(4) – FCD(4) = 0

FHD = 7.07 kN (C)

FCD = 50 kN (T)

Ans

Ans

From FBD (b) +↑ ΣFy = 0;

45 – 40 – FGD = 0 FGD = 5 kN (T)

Ans

215

5-75. The Howe bridge truss is subjected to the loading shown. Determine the force in members HI, HB, and BC, and indicate whether the members are in tension or compression.

40 kN 30 kN

J

20 kN

20 kN

I

H

F

G

4m A B

C

E

D

16 m, 4 @ 4 m

Support Reaction : ΣMA = 0;

Ey(16) – 40(12) – 20(8) – 20(4) = 0 Ey = 45 kN

Method of Sections : +↑ ΣFy = 0;

FHD sin 45° + 45 – 40 – 20 = 0 FHB = 21.2 kN (C)

ΣMH = 0;

45(8) – 40(4) – FBC(4) = 0

ΣMB = 0;

45(12) – 40(8) – 20(4) – FHI(4) = 0

FBC = 50 kN (T)

FHI = 35 kN (C)

Ans

Ans

Ans

216

*5-76. Determine the horizontal and vertical components of force at pins A, B, and C of the twomember frame. 750 N

A

B 2m

1m

2m 3m 900 N 1m C 1.5 m

1.5 m

1200 N

From FBD (a) ΣMA = 0;

By(3) –

1 (500)(3)(2) = 0 2

By = 500 N

1 (500)(3) = 0 2

Ay = 250 N

+↑ ΣFy = 0;

Ay + 500 –

ΣFx = 0;

Bx – Ax = 0

Ans [1]

From FBD (b) ΣMC = 0;

Bx(3) – (500)(3) –

1 (600)(3)(1) – 400(3)(1.5) = 0 2

Bx = 1400 N = 1.40 kN +↑ ΣFy = 0;

Cy – 400(3) – 500 = 0 Cy = 1700 N = 1.70 kN

ΣFx = 0;

Ans

Cx +

Ans

1 (600)(3) – 1400 = 0 2

Cx =500 N

Ans

From Eq. [1] 1400 – Ax = 0

Ax = 1400 N = 1.40 kN

Ans

217

5-77. The compound beam is supported by a rocker at B and fixed to the wall at A. If it is hinged (pinned) together at C, determine the reactions at the supports.

500 N

200 N

13 12 5

A

4000 N · m

60°

B

C

4m

4m

8m

4m

(a) From FBD (a) ΣMC = 0;

By(12) – 200 sin 60°(8) = 0 By = 115.47 N

+↑ ΣFy = 0;

Ans

Cy + 115.47 – 200 sin 60° = 0 Cy = 57.735 N

+→ ΣFx = 0;

Cx + 200 cos 60° = 0

Cx = –100 N

(b) From FBD (b) +→ ΣFx = 0;

Ax – (–100) – 500

 5 =0  13 

Ax = 92.31 N +↑ ΣFy = 0;

Ay – 57.735 – 500

Ans

 12  =0  13 

Ay = 519.27 N ΣMA = 0;

MA – 57.735(8) – 500

Ans

 12  (4) = 0  13 

MA = 2308.0 N · m

Ans

218

5-78. Determine the horizontal and vertical components of reaction at A and B. The pin at C is fixed to member AE and fits through a smooth slot in member BD.

3m

3m D

2m

45°

0.5 m

C

A

E 3m

180 kN B

From FBD (a) ΣMB = 0;

FC

 5   3  – 180 =0  sin 45°   sin 45° 

FC = 300 kN ΣFx = 0; +↑ ΣFy = 0;

Bx + 180 cos 45° – 300 cos 45° = 0 Bx = 84.9 kN

Ans

300 sin 45° – 180 sin 45° – By = 0 By = 84.9 kN

Ans

–Ax + 300 cos 45° – 127.28 = 0 Ax = 84.9 kN

Ans

From FBD (b) ΣFx = 0; +↑ ΣFy = 0;

Ay – 300 sin 45° – 52.72 = 0 Ay = 265 kN

ΣMA = 0;

Ans

MA – 52.72(6) – 300 sin 45°(3) = 0 MA = 953 kN · m

Ans

219

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