SM05
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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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