60 kN, P2 = 40 kN.
m
m
m
60 kN
60 = 0
84.853 = 84.9 kN
Ans.
84.853
.
60 kN(T)
4 0 kN
4 0=0
40 kN(C )
60 kN
.
60 = 0
60 kN(T)
.
40 − 84.853sin 45 = 0
141.42 kN
141 kN
.
84.853cos 45 + 141.42cos 45 − FOE = 0
160 kN (C )
4 0 kN
84.853 kN
.
8 0 kN, P2 = 0.
8 0=0
113.14 kN
.
113 kN
80 kN
113.14
.
80 kN (T)
.
80 kN
80 = 0
.
80 kN (T)
113.14
113.14 kN
113.14
113 kN
.
113.14
160 kN (C)
.
113.14 kN
6–3.
Determine the force in each member of the truss, and state
if the members are in tension or compression. Set u = 0°.
D
3 kN
1.5 m
A
SOLUTION
B
Support Reactions: Applying the equations of equilibrium to the free-body diagram
of the entire truss,Fig.a, we have
a + ©MA = 0;
NC (2 + 2) - 4(2) - 3(1.5) = 0
NC = 3.125 kN
+ ©F = 0;
:
x
3 - Ax = 0
A x = 3 kN
+ c ©Fy = 0;
A y + 3.125 - 4 = 0
A y = 0.875 kN
Method of Joints: We will use the above result to analyze the equilibrium of
joints C and A, and then proceed to analyze of joint B.
Joint C: From the free-body diagram in Fig. b, we can write
+ c ©Fy = 0;
3
3.125 - FCD a b = 0
5
FCD = 5.208 kN = 5.21 kN (C)
+ ©F = 0;
:
x
Ans.
4
5.208 a b - FCB = 0
5
FCB = 4.167 kN = 4.17 kN (T)
Ans.
Joint A: From the free-body diagram in Fig. c, we can write
+ c ©Fy = 0;
3
0.875 - FAD a b = 0
5
FAD = 1.458 kN = 1.46 kN (C)
+ ©F = 0;
:
x
Ans.
4
FAB - 3 - 1.458 a b = 0
5
FAB = 4.167 kN = 4.17 kN (T)
Ans.
Joint B: From the free-body diagram in Fig. d, we can write
+ c ©Fy = 0;
FBD - 4 = 0
FBD = 4 kN (T)
+ ©F = 0;
:
x
C
4.167 - 4.167 = 0
Ans.
(check!)
Note: The equilibrium analysis of joint D can be used to check the accuracy of the
solution obtained above.
2m
2m
4 kN
u
6–4.
Determine the force in each member of the truss, and state
if the members are in tension or compression. Set u = 30°.
D
3 kN
1.5 m
A
SOLUTION
B
Support Reactions: From the free-body diagram of the truss, Fig. a, and applying
the equations of equilibrium, we have
a + ©MA = 0;
NC cos 30°(2 + 2) - 3(1.5) - 4(2) = 0
NC = 3.608 kN
+ ©F = 0;
:
x
3 - 3.608 sin 30° - A x = 0
A x = 1.196 kN
+ c ©Fy = 0;
A y + 3.608 cos 30° - 4 = 0
A y = 0.875 kN
Method of Joints: We will use the above result to analyze the equilibrium of
joints C and A, and then proceed to analyze of joint B.
Joint C: From the free-body diagram in Fig. b, we can write
+ c ©Fy = 0;
3
3.608 cos 30° - FCD a b = 0
5
FCD = 5.208 kN = 5.21 kN (C)
+ ©F = 0;
:
x
Ans.
4
5.208 a b - 3.608 sin 30° - FCB = 0
5
FCB = 2.362 kN = 2.36 kN (T)
Ans.
Joint A: From the free-body diagram in Fig. c, we can write
+ c ©Fy = 0;
3
0.875 - FAD a b = 0
5
FAD = 1.458 kN = 1.46 kN (C)
+ ©F = 0;
:
x
Ans.
4
FAB - 1.458a b - 1.196 = 0
5
FAB = 2.362 kN = 2.36 kN (T)
Ans.
Joint B: From the free-body diagram in Fig. d, we can write
+ c ©Fy = 0;
FBD - 4 = 0
FBD = 4 kN (T)
+ ©F = 0;
:
x
C
2.362 - 2.362 = 0
Ans.
(check!)
Note: The equilibrium analysis of joint D can be used to check the accuracy of the
solution obtained above.
2m
2m
4 kN
u
6–5.
Determine the force in each member of the truss, and state
if the members are in tension or compression.
300 N
400 N D
SOLUTION
C
2m
Method of Joints: Here, the support reactions A and C do not need to be determined.
We will first analyze the equilibrium of joints D and B, and then proceed to analyze
joint C.
250 N
A
B
2m
Joint D: From the free-body diagram in Fig. a, we can write
+ ©F = 0;
:
x
400 - FDC = 0
FDC = 400 N (C)
+ c ©Fy = 0;
200 N
Ans.
FDA - 300 = 0
FDA = 300 N (C)
Ans.
Joint B: From the free-body diagram in Fig. b, we can write
+ ©F = 0;
:
x
250 - FBA = 0
FBA = 250 N (T)
+ c ©Fy = 0;
Ans.
FBC - 200 = 0
FBC = 200 N (T)
Ans.
Joint C: From the free-body diagram in Fig. c, we can write
+ c ©Fy = 0;
FCA sin 45° - 200 = 0
FCA = 282.84 N = 283 N (C)
+ ©F = 0;
:
x
Ans.
400 + 282.84 cos 45° - NC = 0
NC = 600 N
Note: The equilibrium analysis of joint A can be used to determine the components
of support reaction at A.
6–6.
Determine the force in each member of the truss, and state
if the members are in tension or compression.
600 N
D
4m
SOLUTION
C
Method of Joints: We will begin by analyzing the equilibrium of joint D, and then
proceed to analyze joints C and E.
Joint D: From the free-body diagram in Fig. a,
+ ©F = 0;
:
x
Ans.
Ans.
Joint C: From the free-body diagram in Fig. b,
+ ©F = 0;
:
x
FCE - 900 = 0
FCE = 900 N (C)
+ c ©Fy = 0;
Ans.
800 - FCB = 0
FCB = 800 N (T)
Ans.
Joint E: From the free-body diagram in Fig. c,
R+ ©Fx ¿ = 0;
- 900 cos 36.87° + FEB sin 73.74° = 0
FEB = 750 N (T)
Q+ ©Fy ¿ = 0;
Ans.
FEA - 1000 - 900 sin 36.87° - 750 cos 73.74° = 0
FEA = 1750 N = 1.75 kN (C)
B
6m
4
1000 a b - FDC = 0
5
FDC = 800 N (T)
4m
A
3
FDE a b - 600 = 0
5
FDE = 1000 N = 1.00 kN (C)
+ c ©Fy = 0;
900 N
E
Ans.
6–7.
Determine the force in each member of the Pratt truss, and
state if the members are in tension or compression.
J
2m
K
2m
SOLUTION
2m
Joint A:
+ c ©Fy = 0;
20 - FAL sin 45° = 0
FAB - 28.28 cos 45° = 0
Joint B:
FBC - 20 = 0
FBC = 20 kN (T)
+ c ©Fy = 0;
FBL = 0
Joint L:
R+ ©Fx = 0;
FLC = 0
+Q©Fy = 0;
28.28 - FLK = 0
FLK = 28.28 kN (C)
Joint C:
+ ©F = 0;
:
x
FCD - 20 = 0
FCD = 20 kN (T)
+ c ©Fy = 0;
FCK - 10 = 0
FCK = 10 kN (T)
Joint K:
R+ ©Fx - 0;
10 sin 45° - FKD cos (45° - 26.57°) = 0
FKD = 7.454 kN (L)
+Q©Fy = 0;
28.28 - 10 cos 45° + 7.454 sin (45° - 26.57°) - FKJ = 0
FKJ = 23.57 kN (C)
Joint J:
+ ©F = 0;
:
x
23.57 sin 45° - FJI sin 45° = 0
FJI = 23.57 kN (L)
+ c ©Fy = 0;
H
A
10 kN
FAB = 20 kN (T)
+ ©F = 0;
:
x
L
B
D
C
E
F
2m 2m
2m 2m 2m 2m
FAL = 28.28 kN (C)
+ ©F = 0;
:
x
I
2 (23.57 cos 45°) - FJD = 0
FJD = 33.3 kN (T)
Ans.
FAL = FGH = FLK = FHI = 28.3 kN (C)
Ans.
FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T)
Ans.
FBL = FFH = FLC = FHE = 0
Ans.
FCK = FEI = 10 kN (T)
Ans.
FKJ = FIJ = 23.6 kN (C)
Ans.
FKD = FID = 7.45 kN (C)
Ans.
Due to Symmetry
20 kN
10 kN
G
– .
Determine the force in each member of the truss and state if the members are in tension or
compression. Hint: The horizontal force component at A must be zero. Why?
8QLWV8VHG
3
N1 10 1
*LYHQ
) 600N1
) 800N1
D 4P
E 3P
T 60GHJ
6ROXWLRQ
Initial Guesses
)%$ 1N1
)%' 1N1
)&% 1N1
)&' 1N1
*LYHQ
Joint C
Joint B
)&% )˜ FRV ( T ) = 0
)&% )%'˜
E
2
D E
2
)&' ) ˜ VLQ ( T ) = 0
= 0
§ )%$ ·
¨
¸
¨ )%' ¸
¨
¸ )LQG )%$ )%' )&% )&'
¨ )&% ¸
¨) ¸
© &' ¹
)%$ )%'˜
§ )%$ ·
¨
¸
¨ )%' ¸
¨
¸
¨ )&% ¸
¨) ¸
© &' ¹
D
2
D E
§ 1133 ·
¨
¸
¨ 667 ¸ N1
¨ 400 ¸
¨
¸
© 693 ¹
2
) = 0
Positive means Tension
Negative means
Compression
Ans.
6–9.
Determine the force in each member of the truss and state if
the members are in tension or compression. Hint: The
vertical component of force at C must equal zero. Why?
B
C
2m
D
SOLUTION
A
Joint A:
+ c ©Fy = 0;
1.5 m
6 kN
4
FAB - 6 = 0
5
8 kN
FAB = 7.5 kN (T)
+ ©F = 0;
:
x
Ans.
3
-FAE + 7.5 a b = 0
5
FAE = 4.5 kN (C)
Ans.
+ ©F = 0;
:
x
FED = 4.5 kN(C)
Ans.
+ c ©Fy = 0;
FEB = 8 kN (T)
Ans.
Joint E:
Joint B:
+ c ©Fy = 0;
1
22
(FBD) - 8 -
4
(7.5) = 0
5
FBD = 19.8 kN (C)
+ ©F = 0;
:
x
FBC -
E
Ans.
1
3
(7.5) (19.8) = 0
5
22
FBC = 18.5 kN (T)
Cy is zero because BC is a two-force member .
Ans.
2m
6–10.
Each member of the truss is uniform and has a mass of
8 kg>m. Remove the external loads of 6 kN and 8 kN and
determine the approximate force in each member due to
the weight of the truss. State if the members are in tension
or compression. Solve the problem by assuming the weight
of each member can be represented as a vertical force, half
of which is applied at each end of the member.
B
2m
D
SOLUTION
A
1.5 m
Joint A:
+ c ©Fy = 0;
8 kN
Ans.
3
- FAE + 196.2 a b = 0
5
FAE = 117.7 = 118 N (C)
Ans.
+ ©F = 0;
:
x
FED = 117.7 = 118 N (C)
Ans.
+ c ©Fy = 0;
FEB = 215.8 = 216 N (T)
Ans.
Joint E:
Joint B:
+ c ©Fy = 0;
1
22
(FBD) - 366.0 - 215.8 -
4
(196.2) = 0
5
FBD = 1045 = 1.04 kN (C)
+ ©F = 0;
:
x
FBC -
Ans.
1
3
(196.2) (1045) = 0
5
22
FBC = 857 N (T)
E
6 kN
4
- 157.0 = 0
F
5 AB
FAB = 196.2 = 196 N (T)
+ ©F = 0;
:
y
C
Ans.
2m
6–11.
Determine the force in each member of the truss and state
if the members are in tension or compression.
4 kN
3m
3m
3m
C
B
D
3m
5m
A
F
SOLUTION
E
Support Reactions:
5 kN
a + ©MD = 0;
4162 + 5192 - Ey 132 = 0
+ c ©Fy = 0;
23.0 - 4 - 5 - Dy = 0
+ ©F = 0
:
x
Ey = 23.0 kN
Dy = 14.0 kN
Dx = 0
Method of Joints:
Joint D:
+ c ©Fy = 0;
FDE ¢
5
234
≤ - 14.0 = 0
FDE = 16.33 kN 1C2 = 16.3 kN 1C2
+ ©F = 0;
:
x
16.33 ¢
3
234
Ans.
≤ - FDC = 0
FDC = 8.40 kN 1T2
Ans.
Joint E:
+ ©F = 0;
:
x
FEA ¢
3
210
≤ - 16.33 ¢
3
234
≤ = 0
FEA = 8.854 kN 1C2 = 8.85 kN 1C2
+ c ©Fy = 0;
23.0 - 16.33 ¢
5
234
≤ - 8.854 ¢
1
210
Ans.
≤ - FEC = 0
FEC = 6.20 kN 1C2
Ans.
Joint C:
+ c ©Fy = 0;
6.20 - FCF sin 45° = 0
FCF = 8.768 kN 1T2 = 8.77 kN 1T2
+ ©F = 0;
:
x
Ans.
8.40 - 8.768 cos 45° - FCB = 0
FCB = 2.20 kN T
Ans.
6–11. (continued)
Joint B:
+ ©F = 0;
:
x
2.20 - FBA cos 45° = 0
FBA = 3.111 kN 1T2 = 3.11 kN 1T2
+ c ©Fy = 0;
Ans.
FBF - 4 - 3.111 sin 45° = 0
FBF = 6.20 kN 1C2
Ans.
Joint F:
+ c ©Fy = 0;
8.768 sin 45° - 6.20 = 0
+ ©F = 0;
:
x
8.768 cos 45° - FFA = 0
FFA = 6.20 kN 1T2
(Check!)
Ans.
6–12.
Determine the force in each member of the truss and state if
the members are in tension or compression. Set P1 = 10 kN,
P2 = 15 kN.
G
B
C
4m
A
F
E
D
SOLUTION
a + ©MA = 0;
2m
Gx (4) - 10(2) - 15(6) = 0
Gx = 27.5 kN
+ ©F = 0;
:
x
Ax - 27.5 = 0
Ax = 27.5 kN
+ c ©Fy = 0;
Ay - 10 - 15 = 0
Ay = 25 kN
Joint G:
+ ©F = 0;
:
x
FGB - 27.5 = 0
FGB = 27.5 kN (T)
Ans.
Joint A:
+ ©F = 0;
:
x
27.5 - FAF -
+ c ©Fy = 0;
25 - FAB a
1
(FAB) = 0
25
2
25
b = 0
FAF = 15.0 kN (C)
Ans.
FAB = 27.95 = 28.0 kN (C)
Ans.
Joint B:
+ ©F = 0;
:
x
27.95 a
+ c ©Fy = 0;
27.95 a
1
25
2
25
b + FBC - 27.5 = 0
b - FBF = 0
FBF = 25.0 kN (T)
Ans.
FBC = 15.0 kN (T)
Ans.
Joint F:
1
+ ©F = 0;
:
x
15 + FFE -
+ c ©Fy = 0;
25 - 10 - FFC a
22
(FFC) = 0
1
22
b = 0
FFC = 21.21 = 21.2 kN (C)
Ans.
FFE = 0
Ans.
Joint E:
+ ©F = 0;
:
x
FED = 0
Ans.
+ c ©Fy = 0;
FEC - 15 = 0
Joint D:
+ ©F = 0;
:
x
4m
P1
FEC = 15.0 kN (T)
Ans.
FDC = 0
Ans.
2m
P2
6–13.
Determine the force in each member of the truss and state
if the members are in tension or compression. Set P1 = 0,
P2 = 20 kN.
G
B
C
4m
A
F
E
D
SOLUTION
a + ©MA = 0;
FGB = 30 kN (T)
+ ©F = 0;
:
x
2m
-FGB (4) + 20(6) = 0
Ans.
Ax - 30 = 0
Ax = 30 kN
+ c ©Fy = 0;
Ay - 20 = 0
Ay = 20 kN
Joint A:
1
+ ©F = 0;
:
x
30 - FAF -
+ c ©Fy = 0;
20 - FAB a
25
2
25
(FAB) = 0
b = 0
FAF = 20 kN (C)
Ans.
FAB = 22.36 = 22.4 kN (C)
Ans.
Joint B:
+ ©F = 0;
:
x
22.36 a
+ c ©Fy = 0;
22.36 a
1
25
2
25
b + FBC - 30 = 0
b - FBF = 0
FBF = 20 kN (T)
Ans.
FBC = 20 kN (T)
Ans.
Joint F:
+ ©F = 0;
:
x
20 + FFE -
+ c ©Fy = 0;
20 - FFC a
1
22
1
22
(FFC) = 0
b = 0
FFC = 28.28 = 28.3 kN (C)
Ans.
FFE = 0
Ans.
Joint E:
+ ©F = 0;
:
x
FED - 0 = 0
+ c ©Fy = 0;
FEC - 20 = 0
FED = 0
Ans.
FEC = 20.0 kN (T)
Ans.
Joint D:
+ ©F = 0;
:
x
1
25
(FDC) - 0 = 0
FDC = 0
4m
P1
Ans.
2m
P2
.
.
.
.
.
.
.
.
.
.
.
.
6–16.
Determine the force in each member of the truss. State
whether the members are in tension or compression. Set
P = 8 kN.
4m
B
A
SOLUTION
Joint D:
FDC sin 60° - 8 = 0
FDC = 9.238 kN 1T2 = 9.24 kN 1T2
+ ©F = 0;
:
x
Ans.
FDE - 9.238 cos 60° = 0
FDE = 4.619 kN 1C2 = 4.62 kN 1C2
Ans.
Joint C:
+ c ©Fy = 0;
FCE sin 60° - 9.238 sin 60° = 0
FCE = 9.238 kN 1C2 = 9.24 kN 1C2
+ ©F = 0;
:
x
Ans.
219.238 cos 60°2 - FCB = 0
FCB = 9.238 kN 1T2 = 9.24 kN 1T2
Ans.
Joint B:
+ c ©Fy = 0;
FBE sin 60° - FBA sin 60° = 0
FBE = FBA = F
+ ©F = 0;
:
x
9.238 - 2F cos 60° = 0
F = 9.238 kN
Thus,
FBE = 9.24 kN 1C2
FBA = 9.24 kN 1T2
Ans.
Joint E:
+ c ©Fy = 0;
Ey - 219.238 sin 60°2 = 0
+ ©F = 0;
:
x
FEA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0
FEA = 4.62 kN 1C2
Ey = 16.0 kN
Ans.
Note: The support reactions Ax and Ay can be determinedd by analyzing Joint A
using the results obtained above.
60°
E
4m
Method of Joints: In this case, the support reactions are not required for
determining the member forces.
+ c ©Fy = 0;
60°
C
D
4m
P
6–17.
If the maximum force that any member can support is 8 kN
in tension and 6 kN in compression, determine the
maximum force P that can be supported at joint D.
4m
B
A
SOLUTION
Joint D:
+ c ©Fy = 0;
FDC sin 60° - P = 0
+ ©F = 0;
:
x
FDE - 1.1547P cos 60° = 0
FDC = 1.1547P 1T2
FDE = 0.57735P 1C2
Joint C:
FCE sin 60° - 1.1547P sin 60° = 0
FCE = 1.1547P 1C2
+ ©F = 0;
:
x
211.1547P cos 60°2 - FCB = 0
FCB = 1.1547P 1T2
+ c ©Fy = 0;
FBE sin 60° - FBA sin 60° = 0
FBE = FBA = F
+ ©F = 0;
:
x
1.1547P - 2F cos 60° = 0
Joint B:
F = 1.1547P
Thus,
FBE = 1.1547P 1C2
FBA = 1.1547P 1T2
Joint E:
+ ©F = 0;
:
x
FEA + 1.1547P cos 60° - 1.1547P cos 60°
- 0.57735P = 0
FEA = 0.57735P 1C2
From the above analysis, the maximum compression and tension in the truss
member is 1.1547P. For this case, compression controls which requires
1.1547P = 6
P = 5.20 kN
Ans.
60°
E
4m
Method of Joints: In this case, the support reactions are not required for
determining the member forces.
+ c ©Fy = 0;
60°
C
D
4m
P
6–18.
Determine the force in each member of the truss and state if
the members are in tension or compression. Hint: The
resultant force at the pin E acts along member ED. Why?
3 kN
3m
A
4m
Joint C:
E
+ c ©Fy = 0;
2
213
FCD - 2 = 0
FCD = 3.606 = 3.61 kN (C)
+ ©F = 0;
:
x
-FCD + 3.606a
3
213
Ans.
b = 0
FCB = 3 kN (T)
Ans.
+ ©F = 0;
:
x
FBA = 3 kN (T)
Ans.
+ c ©Fy = 0;
FBD = 3 kN (C)
Ans.
Joint B:
Joint D:
+ ©F = 0;
:
x
+ c ©Fy = 0;
3
213
2
213
FDE -
(FDE) -
3
213
(3.606) +
2
213
(FDA) -
3
213
2
213
FDA = 0
(3.606) - 3 = 0
FDA = 2.70 kN (T)
Ans.
FDE = 6.31 kN (C)
Ans.
B
D
SOLUTION
2 kN
3m
C
6–19.
Each member of the truss is uniform and has a mass of
8 kg>m. Remove the external loads of 3 kN and 2 kN and
determine the approximate force in each member due to
the weight of the truss. State if the members are in tension
or compression. Solve the problem by assuming the weight
of each member can be represented as a vertical force, half
of which is applied at each end of the member.
3 kN
3m
A
4m
Joint C:
+ c ©Fy = 0;
2
213
E
FCD - 259.2 = 0
FCD = 467.3 = 467 N (C)
+ ©F = 0;
:
x
- FCB + 467.3 a
3
213
Ans.
b = 0
FCB = 388.8 = 389 N (T)
Ans.
+ ©F = 0;
:
x
FBA = 388.8 = 389 N (T)
Ans.
+ c ©Fy = 0;
FBD = 313.9 = 314 N (C)
Ans.
Joint B:
Joint D:
+ ©F = 0;
:
x
+ c ©Fy = 0;
3
213
2
213
FDE -
3
213
(FDE) +
(467.3) -
2
213
(FDA) -
3
213
FDA = 0
2
213
(467.3) - 313.9 - 502.9 = 0
FDE = 1204 = 1.20 kN (C)
Ans.
FDA = 736 N (T)
Ans.
B
D
SOLUTION
2 kN
3m
C
6–20.
Determine the force in each member of the truss in terms of
the load P, and indicate whether the members are in tension
or compression.
B
P
d
C
A
D
F
d
SOLUTION
Support Reactions:
E
a + ©ME = 0;
3
P(2d) - Ay a d b = 0
2
+ c ©Fy = 0;
4
P - Ey = 0
3
+ ©F = 0;
:
x
Ex - P = 0
Ey =
4
Ay = P
3
d
4
P
3
Ex = P
Method of Joints: By inspection of joint C, members CB and CD are zero force
members. Hence
Ans.
FCB = FCD = 0
Joint A:
+ c ©Fy = 0;
FAB ¢
1
23.25
≤ -
4
P = 0
3
FAB = 2.40P (C) = 2.40P (C)
+ ©F = 0;
:
x
FAF - 2.404P ¢
1.5
23.25
Ans.
≤ = 0
FAF = 2.00P (T)
Ans.
Joint B:
+ ©F = 0;
:
x
2.404P ¢
1.5
23.25
≤ - P - FBF ¢
0.5
21.25
≤ - FBD ¢
0.5
21.25
≤ = 0
1.00P - 0.4472FBF - 0.4472FBD = 0
+ c ©Fy = 0;
2.404P ¢
1
23.25
≤ + FBD ¢
1
21.25
≤ - FBF ¢
(1)
1
21.25
1.333P + 0.8944FBD - 0.8944FBF = 0
≤ = 0
(2)
Solving Eqs. (1) and (2) yield,
FBF = 1.863P (T) = 1.86P (T)
Ans.
FBD = 0.3727P (C) = 0.373P (C)
Ans.
Joint F:
+ c ©Fy = 0;
FFE
+ ©F = 0;
:
x
1
1
≤ - FFE ¢
≤ = 0
21.25
21.25
= 1.863P (T) = 1.86P (T)
1.863P ¢
FFD + 2 B 1.863P ¢
0.5
21.25
Ans.
≤ R - 2.00P = 0
FFD = 0.3333P (T) = 0.333P (T)
Ans.
Joint D:
+ c ©Fy = 0;
FDE ¢
1
21.25
≤ - 0.3727P ¢
1
21.25
≤ = 0
FDE = 0.3727P (C) = 0.373P (C)
+ ©F = 0;
:
y
2 B 0.3727P ¢
0.5
21.25
≤ R - 0.3333P = 0 (Check!)
Ans.
d/2
d/2
d
6–21.
If the maximum force that any member can support is 4 kN
in tension and 3 kN in compression, determine the maximum
force P that can be applied at joint B. Take d = 1 m.
B
P
d
SOLUTION
a + ©ME = 0;
+ c ©Fy = 0;
4
P - Ey = 0
3
Ey =
+ ©F = 0;
:
x
Ex - P = 0
Ex = P
4
Ay = P
3
d
4
P
3
E
d
Method of Joints: By inspection of joint C, members CB and CD are zero force
members. Hence
FCB = FCD = 0
Joint A:
1
+ c ©Fy = 0;
FAB ¢
+ ©F = 0;
:
x
FAF - 2.404P ¢
4
P = 0
3
≤ -
23.25
1.5
23.25
FAB = 2.404P (C)
≤ = 0
FAF = 2.00P (T)
Joint B:
2.404P ¢
1.5
23.25
≤ - P - FBF ¢
0.5
21.25
≤ - FBD ¢
0.5
21.25
≤ = 0
1.00P - 0.4472FBF - 0.4472FBD = 0
+ c ©Fy = 0;
2.404P ¢
1
23.25
≤ + FBD ¢
1
21.25
≤ - FBF ¢
(1)
1
21.25
1.333P + 0.8944FBD - 0.8944FBF = 0
≤ = 0
(2)
Solving Eqs. (1) and (2) yield,
FBF = 1.863P (T)
FBD = 0.3727P (C)
Joint F:
+ c ©Fy = 0;
1.863P ¢
1
21.25
≤ - FFE ¢
1
21.25
≤ = 0
FFE = 1.863P (T)
+ ©F = 0;
:
x
FFD + 2 B 1.863P ¢
0.5
21.25
≤ R - 2.00P = 0
FFD = 0.3333P (T)
Joint D:
+ c ©Fy = 0;
FDE ¢
1
21.25
≤ - 0.3727P ¢
1
21.25
≤ = 0
FDE = 0.3727P (C)
+ ©F = 0;
:
y
D
F
3
P(2d) - Ay a d b = 0
2
+ ©F = 0;
:
x
C
A
Support Reactions:
2 B 0.3727P ¢
0.5
21.25
≤ R - 0.3333P = 0 (Check!)
From the above analysis, the maximum compression and tension in the truss members
are 2.404P and 2.00P, respectively. For this case, compression controls which requires
2.404P = 3
P = 1.25 kN
d/2
d/2
d
6–22.
B
Determine the force in each member of the double scissors
truss in terms of the load P and state if the members are in
tension or compression.
C
L/3
SOLUTION
c + ©MA = 0;
L
2L
Pa b + P a
b - (Dy)(L) = 0
3
3
Dy = P
+ c ©Fy = 0;
Ay = P
L/3
FFD - FFE - FFB a
1
22
(1)
b = 0
FFD - FFE = P
+ ©F = 0;
:
y
1
FFB a
22
b -P = 0
FFB = 22P = 1.41 P (T)
Similarly,
FEC = 22P
Joint C:
+ ©F = 0;
:
x
2
FCA a
2
25
+c ©Fy = 0;
FCA
25
b - 22P a
FCA 1
25
1
22
1
22
b - FCD a
1
22
b = 0
FCD = P
- 22P
1
22
+ FCD
1
22
=0
FCA =
2 25
P = 1.4907P = 1.49P (C)
3
FCD =
22
P = 0.4714P = 0.471P (C)
3
FAE -
1
2
22
2 25
Pa
Pa
b b = 0
3
3
22
25
FAE =
5
P = 1.67 P (T)
3
Joint A:
+ ©F = 0;
:
x
Similarly,
FFD=1.67 P (T)
From Eq.(1), and Symmetry,
FFE = 0.667 P (T)
Ans.
FFD = 1.67 P (T)
Ans.
FAB = 0.471 P (C)
Ans.
FAE = 1.67 P (T)
Ans.
FAC = 1.49 P (C)
Ans.
FBF = 1.41 P (T)
Ans.
FBD = 1.49 P (C)
Ans.
FEC = 1.41 P (T)
Ans.
FCD = 0.471 P (C)
Ans.
F
L/3
P
Joint F:
+ ©F = 0;
:
x
E
A
D
L/3
P
6–23.
Determine the force in each member of the truss in terms of
the load P and state if the members are in tension or
compression.
L
B
L
L
C
L
L
A
D
E
SOLUTION
L
P
Entire truss:
a + ©MA = 0;
- P(L) + Dy (2 L) = 0
Dy =
P
2
P
- P + Ay = 0
2
+ c ©Fy = 0;
Ay =
+ ©F = 0;
:
x
P
2
Ax = 0
Joint D:
+ c ©Fy = 0;
-FCD sin 60° +
P
= 0
2
FCD = 0.577 P (C)
+ ©F = 0;
;
x
Ans.
FDB - 0.577P cos 60° = 0
FDB = 0.289 P(T)
Ans.
Joint C:
+ c ©Fy = 0;
0.577 P sin 60° - FCE sin 60° = 0
FCE = 0.577 P (T)
+ ©F = 0;
:
x
L
Ans.
FBC - 0.577 P cos 60° - 0.577P cos 60° = 0
FBC = 0.577 P (C)
Ans.
Due to symmetry:
FBE = FCE = 0.577 P (T)
Ans.
FAB = FCD = 0.577 P (C)
Ans.
FAE = FDE = 0.577 P (T)
Ans.
6–24.
Each member of the truss is uniform and has a weight W.
Remove the external force P and determine the approximate
force in each member due to the weight of the truss. State if
the members are in tension or compression. Solve the
problem by assuming the weight of each member can be
represented as a vertical force, half of which is applied at
each end of the member.
L
B
L
L
C
L
L
A
D
E
SOLUTION
L
P
Entire truss:
a + ©MA = 0;
3
L
3
3
- W a b - 2 W(L) - W a L b - W(2 L) + Dy (2 L) = 0
2
2
2
2
Dy =
7
W
2
Joint D:
+ c ©Fy = 0;
7
W - W - FCD sin 60° = 0
2
FCD = 2.887W = 2.89 W (C)
+ ©F = 0;
:
x
Ans.
2.887W cos 60° - FDE = 0
FDE = 1.44 W (T)
Ans.
Joint C:
+ c ©Fy = 0;
2.887W sin 60° -
3
W - FCE sin 60° = 0
2
FCE = 1.1547W = 1.15 W (T)
+ ©F = 0;
:
x
L
Ans.
FBC - 1.1547W cos 60° - 2.887W cos 60° = 0
FBC = 2.02 W (C)
Ans.
Due to symmetry:
FBE = FCE = 1.15 W (T)
Ans.
FAB = FCD = 2.89 W (C)
Ans.
FAE = FDE = 1.44 W (T)
Ans.
6–25.
Determine the force in each member of the truss in terms of
the external loading and state if the members are in tension
or compression.
P
B
L
C
u
L
L
SOLUTION
A
Joint B:
+ c ©Fy = 0;
L
FBA sin 2u - P = 0
FBA = P csc 2u (C)
+ ©F = 0;
:
x
Ans.
P csc 2u(cos 2u) - FBC = 0
FBC = P cot 2 u (C)
Ans.
Joint C:
+ ©F = 0;
:
x
P cot 2 u + P + FCD cos 2 u - FCA cos u = 0
+ c ©Fy = 0;
FCD sin 2 u - FCA sin u = 0
FCA =
cot 2 u + 1
P
cos u - sin u cot 2 u
FCA = (cot u cos u - sin u + 2 cos u) P (T)
Ans.
FCD = (cot 2 u + 1) P
Ans.
(C)
Joint D:
+ ©F = 0;
:
x
FDA - (cot 2 u + 1)(cos 2 u) P = 0
FDA = (cot 2 u + 1)(cos 2 u) (P)
(C)
Ans.
D
P
6–26.
The maximum allowable tensile force in the members of the
truss is 1Ft2max = 2 kN, and the maximum allowable
compressive force is 1Fc2max = 1.2 kN. Determine the
maximum magnitude P of the two loads that can be applied
to the truss. Take L = 2 m and u = 30°.
P
B
L
u
L
L
SOLUTION
A
(Tt)max = 2 kN
L
(FC)max = 1.2 kN
Joint B:
+ c ©Fy = 0;
FBA cos 30° - P = 0
FBA =
+ ©F = 0;
:
x
P
= 1.1547 P (C)
cos 30°
FAB sin 30° - FBC = 0
FBC = P tan 30° = 0.57735 P (C)
Joint C:
+ c ©Fy = 0;
-FCA sin 30° + FCD sin 60° = 0
FCA = FCD a
+ ©F = 0;
:
x
sin 60°
b = 1.732 FCD
sin 30°
P tan 30° + P + FCD cos 60° - FCA cos 30° = 0
FCD = a
tan 30° + 1
23 cos 30° - cos 60°
b P = 1.577 P (C)
FCA = 2.732 P (T)
Joint D:
+ ©F = 0;
:
x
FDA - 1.577 P sin 30° = 0
FDA = 0.7887 P (C)
1) Assume FCA = 2 kN = 2.732 P
P = 732.05 N
FCD = 1.577(732.05) = 1154.7 N 6 (Fc)max = 1200 N
Thus,
Pmax = 732 N
C
(O.K.!)
Ans.
D
P
.
.
.
.
.
.
6 –29.
The Pratt bridge truss is subjected to the loading shown. Determine the force in members LD,
LK, CD, and KD, and state if the members are in tension or compression.
Units Used:
3
kN
10 N
Given:
F1
50 kN
F2
50 kN
F3
50 kN
a
4m
b
3m
Solution:
Ax
0
Ay
3F 3 4F2 5F 1
6
Guesses
F LD
1 kN
F LK
1 kN
F CD
1 kN
F KD
1 kN
Given
F 2 b F 1( 2b) Ay( 3b) F LK a
F CD a F1 b Ay( 2b)
Ay F1 F2 F 3 F KD
§ FLD ·
¨
¸
¨ FLK ¸
¨
¸
¨ FCD ¸
¨F ¸
© KD ¹
0
0
a
·
§
¨ 2 2 ¸ FLD
© a b ¹
0
0
Find F LD FLK F CD F KD
§ FLD ·
¨
¸
¨ FLK ¸
¨
¸
¨ FCD ¸
¨F ¸
© KD ¹
§ 0 ·
¨
¸
¨ 112.5 ¸ kN
¨ 112.5 ¸
¨
¸
© 50 ¹
Positive (T)
Negative (C)
Ans.
6–30.
The Pratt bridge truss is subjected to the loading shown. Determine the force in members JI,
JE, and DE, and state if the members are in tension or compression.
Units Used:
3
kN
10 N
Given:
F1
50 kN
F2
50 kN
F3
50 kN
a
4m
b
3m
Solution:
Initial Guesses
Gy 1 kN
F JI
F JE
1 kN F DE
1 kN
1 kN
Given
Entire Truss
F 1 b F2( 2b) F3( 3b) Gy( 6b)
0
Section
F DE F JI
Gy( 2b) F DE a
§ Gy ·
¨
¸
¨ FJI ¸
¨
¸
¨ FJE ¸
¨F ¸
© DE ¹
F JE Gy
0
0
0
Find Gy F JI FJE FDE
Gy
50 kN
§ FJI ·
¨
¸
¨ FJE ¸
¨F ¸
© DE ¹
§ 75 ·
¨ 50 ¸ kN
¨
¸
© 75 ¹
Positive means Tension,
Negative means Compression
Ans.
6 –31.
Determine the force in members BE, EF, and CB, and state if the members are in tension or
compression.
Units Used:
kN
3
10 N
Given:
F1
5 kN
F4
F2
10 kN
a
4m
F3
5 kN
b
4m
T
Solution:
10 kN
§ a·
¸
© b¹
atan ¨
Inital Guesses
F CB
1 kN F BE
1 kN
F EF
1 kN
Given
F 1 F 2 F BE cos T
0
F CB FEF FBE sin T F3
F 1 a F CB b
0
0
§ FCB ·
¨
¸
¨ FBE ¸
¨F ¸
© EF ¹
Find FCB F BE F EF
§ FCB ·
¨
¸
¨ FBE ¸
¨F ¸
© EF ¹
§ 5 ·
¨ 21.2 ¸ kN
¨
¸
© 25 ¹
Positive (T)
Negative (C)
Ans.
6–32.
Determine the force in members BF, BG, and AB, and state if the members are in tension or
compression.
Units Used:
kN
3
10 N
Given:
F1
5 kN
F4
F2
10 kN
a
4m
F3
5 kN
b
4m
T
Solution:
10 kN
atan §¨
a·
¸
© b¹
Inital Guesses
F AB
1 kN F BG
1 kN
F BF
1 kN
Given
F 1 F 2 F 4 F BG cos T
F 1 3a F 22a F 4 a FAB b
F BF
0
0
0
§ FAB ·
¨
¸
¨ FBG ¸
¨F ¸
© BF ¹
Find F AB F BG F BF
§ FAB ·
¨
¸
¨ FBG ¸
¨F ¸
© BF ¹
§ 45 ·
¨ 35.4 ¸ kN Positive (T)
¨
¸
Negative (C)
© 0 ¹
Ans.
6–33.
Determine the force developed in members BC and CH of the roof truss and state if the
members are in tension or compression.
Units Used:
3
kN
10 N
Given:
F1
1.5 kN
F2
2 kN
a
1.5 m
b
1m
c
2m
d
0.8 m
Solution:
§ a·
¸
©c¹
atan ¨
T
I
§ a ·
¸
© c b¹
atan ¨
Initial Guesses:
Ey
1 kN
F BC
F CH
1 kN
1 kN
Given
F 2( d) F1( c) Ey( 2c)
0
F BC sin T ( c) FCH sin I ( c b) E y( c)
F BC sin T FCH sin I F1 Ey
§ Ey ·
¨
¸
¨ FBC ¸
¨F ¸
© CH ¹
Find Ey F BC F CH
Ey
0
0
1.15 kN
§¨ FBC ·¸
¨ FCH ¸
©
¹
§ 3.25 ·
¨
¸ kN
© 1.923 ¹
Positive (T)
Negative (C)
Ans.
6–34.
Determine the force in members CD
and GF of the truss and state if the
members are in tension or
compression. Also indicate all
zero-force members.
Units Used:
3
kN
10 N
Given:
F1
1.5 kN
F2
2 kN
a
1.5 m
b
1m
c
2m
d
0.8 m
Solution:
atan §¨
a·
¸
©c¹
T
·
¸
© c b¹
atan §¨
a
1 kN
F GF
I
Initial Guesses:
Ey
1 kN
F CD
1 kN
Given
F 2( d) F1( c) Ey( 2c)
E y( b) F CD sin T ( b)
E y( c) FGF( a)
§ Ey ·
¨
¸
¨ FCD ¸
¨F ¸
© GF ¹
0
0
0
Find E y FCD FGF
Ey
1.15 kN
§¨ FCD ·¸
¨ FGF ¸
©
¹
§ 1.917 ·
Positive (T)
¨
¸ kN Negative (C)
1.533
©
¹
DF and CF are zero force members.
Ans.
6–35.
Determine the force in members BC, HC, and HG. After
the truss is sectioned use a single equation of equilibrium
for the calculation of each force. State if these members are
in tension or compression.
5 kN
4 kN
4 kN
B
C
3 kN
2 kN
A
D
E
3m
F
H
SOLUTION
a + ©ME = 0;
G
5m
- Ay (20) + 2(20) + 4(15) + 4(10) + 5(5) = 0
Ay = 8.25 kN
a + ©MH = 0;
-8.25(5) + 2(5) + FBC(3) = 0
FBC = 10.4 kN (C)
a + ©MC = 0;
-8.25(10) + 2(10) + 4(5) +
Ans.
5
229
FHG(5) = 0
FHG = 9.1548 = 9.15 kN (T)
a + ©MO¿ = 0;
2m
-2(2.5) + 8.25(2.5) - 4(7.5) +
FHC = 2.24 kN (T)
Ans.
3
234
FHC(12.5) = 0
Ans.
5m
5m
5m
6–36.
Determine the force in members CD, CF, and CG and state
if these members are in tension or compression.
5 kN
4 kN
4 kN
B
C
3 kN
2 kN
A
D
E
3m
F
H
SOLUTION
G
5m
+ ©F = 0;
:
x
Ex = 0
a + ©MA = 0;
-4(5) - 4(10) - 5(15) - 3(20) + Ey (20) = 0
Ey = 9.75 kN
a + ©MC = 0;
-5(5) - 3(10) + 9.75(10) -
5
229
FFG(5) = 0
FFG = 9.155 kN (T)
a + ©MF = 0;
-3(5) + 9.75(5) - FCD(3) = 0
FCD = 11.25 = 11.2 kN (C)
a + ©MO¿ = 0;
-9.75(2.5) + 5(7.5) + 3(2.5) FCF = 3.21 kN (T)
Ans.
3
234
FCF(12.5) = 0
Ans.
Joint G:
+ ©F = 0;
:
x
+ c ©Fy = 0;
2m
FGH = 9.155 kN (T)
2
229
(9.155)(2) - FCG = 0
FCG = 6.80 kN (C)
Ans.
5m
5m
5m
1.5 m
1.5 m
N
2m
N
2m
2m
N
2m
2m
2m
1.20(10) + 1.50(8) + 1.80(6) − Ay (12) = 0
kN
2 m 2 m
m
2 m
kN
kN
kN
(3) + 1.20(2) − 2.90(4) = 0
kN
kN
.
3 m
(3) + 1.20(2) − 2.90(4) = 0
kN
kN
.
2 m
kN
2 m
kN
kN
kN
3 m
kN
kN
2 m
.
kN
kN
2 m
kN
kN
1.5 m
1.5 m
N
N
2m
2m
2m
N
2m
2m
(6) – 1.50(4) – 1.20(2) = 0
kN
1.60(4)
1.60(4)
(3) = 0
kN
.
3 m
(3) = 0
kN
.
4m
kN
2m
6–39.
Determine the force in members IC and CG of the truss
and state if these members are in tension or compression.
Also, indicate all zero-force members.
B
C
D
2m
I
J
2m
SOLUTION
A
By inspection of joints B, D, H and I,
AB, BC, CD, DE, HI, and GI are all zero-force members.
a + ©MG = 0;
Ans.
3
-4.5(3) + FIC a b (4) = 0
5
FIC = 5.625 = 5.62 kN (C)
+ c ©Fy = 0;
Ans.
FCJ = 5.625 kN
4
4
(5.625) + (5.625) - FCG = 0
5
5
FCG = 9.00 kN (T)
1.5 m
G
1.5 m
Ans.
F
1.5 m
6 kN
Joint C:
+ ©F = 0;
:
x
E
H
1.5 m
6 kN
6–40.
Determine the force in members JE and GF of the truss
and state if these members are in tension or compression.
Also, indicate all zero-force members.
B
C
D
2m
I
J
2m
SOLUTION
A
By inspection of joints B, D, H and I,
AB, BC, CD, DE, HI, and GI are zero-force members.
Ans.
7.5 -
4
F = 0
5 JE
FJE = 9.375 = 9.38 kN (C)
+ ©F = 0;
:
x
1.5 m
G
1.5 m
Ans.
3
(9.375) - FGF = 0
5
FGF = 5.62 kN (T)
Ans.
F
1.5 m
6 kN
Joint E:
+ c ©Fy = 0;
E
H
1.5 m
6 kN
6–41.
Determine the force in members FG, GC and CB of the truss
used to support the sign, and state if the members are in
tension or compression.
1.5 m
1.5 m
1.5 m
E
900 N
3m
1800 N
F
D
SOLUTION
3m
Method of Sections: The forces in members FG, GC, and CB are exposed by cutting
the truss into two portions through section a–a on the upper portion of the free-body
diagram, Fig. a. From this free-body diagram, FCB, FGC, and FFG can be obtained by
writing the moment equations of equilibrium about points G, E, and C, respectively.
a + ©MG = 0;
FCB = 3600 N = 3.60 kN (T)
a + ©ME = 0;
a + ©MC = 0;
Ans.
FGC(6) - 900(6) - 1800(3) = 0
FGC = 1800 N = 1.80 kN (C)
Ans.
900(6) + 1800(3) - FFG sin 26.57°(6) = 0
FFG = 4024.92 N = 4.02 kN (C)
C
3m
A
900(6) + 1800(3) - FCB(3) = 0
Ans.
900 N
G
B
6–42.
Determine the force in members DE, DL,
and ML of the roof truss and state if the
members are in tension or compression.
Units Used:
3
kN
10 N
Given:
F1
6 kN
F2
12 kN
F3
12 kN
F4
12 kN
a
4m
b
3m
c
6m
Solution:
T
I
atan §¨
c b·
¸
© 3a ¹
ªb «
atan «
¬
2
º
( c b) »
3
a
»
¼
Initial Guesses:
Ay
1 kN
F ML
1 kN
F DL
1 kN
F DE
1 kN
Given
F 2( a) F3( 2a) F4( 3a) F3( 4a) F2( 5a) F1( 6a) A y( 6a)
2
F 1( 2a) F 2( a) A y( 2a) FMLª«b ( c b)º»
¬ 3
¼
Ay F 1 F 2 F 3 F DE sin T FDL sin I
F ML F DL cos I FDE cos T
0
0
0
0
§ Ay ·
¨
¸
¨ FML ¸
¨
¸
¨ FDE ¸
¨F ¸
© DL ¹
Find Ay F ML F DE F DL
Ay
36 kN
§ FML ·
¨
¸
¨ FDE ¸
¨F ¸
© DL ¹
§ 38.4 ·
¨ 37.1 ¸ kN
¨
¸
© 3.8 ¹
Positive (T), Negative (C)
Ans.
6–43.
Determine the force in members EF and EL of the roof truss and state if the members are in
tension or compression.
Units Used:
3
kN
10 N
Given:
F1
6 kN
F2
12 kN
F3
12 kN
F4
12 kN
a
4m
b
3m
c
6m
Solution:
T
atan §¨
c b·
¸
© 3a ¹
Initial Guesses:
Iy
F EF
1 kN
F EL
1 kN
1 kN
Given
F 2( a) F3( 2a) F4( 3a) F3( 4a) F2( 5a) F1( 6a) Iy( 6a)
F 3( a) F2( 2a) F1( 3a) Iy( 3a) F EF cos T ( c)
F 4 F EL 2F EF sin T
§ Iy ·
¨
¸
¨ FEF ¸
¨F ¸
© EL ¹
Find Iy FEF FEL
0
0
0
Iy
36 kN
§¨ FEF ·¸
¨ FEL ¸
©
¹
§ 37.108 ·
¨
¸ kN
© 6 ¹
Positive (T)
Negative (C)
Ans.
6–44.
The skewed truss carries the load shown. Determine the
force in members CB, BE, and EF and state if these
members are in tension or compression. Assume that all
joints are pinned.
d/ 2
C
d
D
P
d
P
SOLUTION
a + ©MB = 0;
B
E
-P(d) + FEF(d) = 0
d
FEF = P (C)
a + ©ME = 0;
- P(d) + B
Ans.
d
2(d) +
2
A B
d 2
2
FCB = 1.118 P (T) = 1.12 P (T)
+ ©F = 0;
:
x
P -
0.5
21.25
Ans.
(1.118 P) - FBE = 0
FBE = 0.5P (T)
F
A
R FCB (d) = 0
Ans.
d
d/ 2
6–45.
The skewed truss carries the load shown. Determine the
force in members AB, BF, and EF and state if these
members are in tension or compression. Assume that all
joints are pinned.
d/ 2
C
d
D
P
d
P
SOLUTION
a + ©MF = 0;
B
E
- P(2d) + P(d) + FAB (d) = 0
d
FAB = P (T)
a + ©MB = 0;
Ans.
- P(d) + FEF(d) = 0
FEF = P (C)
+ ©F = 0;
:
x
P - FBF a
1
22
Ans.
b = 0
FBF = 1.41P (C)
F
A
Ans.
d
d/ 2
6–46.
Determine the force in members CD and CM of the
Baltimore bridge truss and state if the members are in
tension or compression. Also, indicate all zero-force
members.
M
N
O
SOLUTION
C
21122 + 5182 + 3162 + 2142 - Ay 1162 = 0
Ay = 5.625 kN
Ax = 0
Method of Joints: By inspection, members BN, NC, DO, OC, HJ
LE and JG are zero force members.
Ans.
Method of Sections:
a + ©MM = 0;
FCD142 - 5.625142 = 0
FCD = 5.625 kN 1T2
a + ©MA = 0;
D
E
F
5 kN 3 kN
16 m, 8 @ 2 m
Support Reactions:
Ans.
FCM 142 - 2142 = 0
FCM = 2.00 kN T
Ans.
2m
J
P
I
2 kN
+ ©F = 0;
:
x
K
A
B
a + ©MI = 0;
L
G
2 kN
H
2m
6–47.
Determine the force in members EF, EP, and LK of the
Baltimore bridge truss and state if the members are in
tension or compression. Also, indicate all zero-force
members.
M
N
O
SOLUTION
C
Iy 1162 - 21122 - 31102 - 5182 - 2142 = 0
Iy = 6.375 kN
Method of Joints: By inspection, members BN, NC, DO, OC, HJ
LE and JG are zero force members.
Ans.
Method of Sections:
3122 + 6.375142 - FEF142 = 0
FEF = 7.875 = 7.88 kN 1T2
a + ©ME = 0;
Ans.
6.375182 - 2142 - 3122 - FLK 142 = 0
FLK = 9.25 kN 1C2
+ c ©Fy = 0;
D
E
F
5 kN 3 kN
16 m, 8 @ 2 m
Support Reactions:
Ans.
6.375 - 3 - 2 - FED sin 45° = 0
FED = 1.94 kN T
Ans.
2m
J
P
I
2 kN
a + ©MK = 0;
K
A
B
a + ©MA = 0;
L
G
2 kN
H
2m
6–48.
The truss supports the vertical load of 600 N. If L = 2 m,
determine the force on members HG and HB of the truss
and state if the members are in tension or compression.
I
H
G
E
3m
SOLUTION
A
Method of Section: Consider the FBD of the right portion of the truss cut through
sec. a–a, Fig. a, we notice that FHB and FHG can be obtained directly by writing the
force equation of equilibrium along vertical and moment equation of equilibrium
about joint B, respectively.
+ c ©Fy = 0;
FHB - 600 = 0
FHB = 600 N (T)
Ans.
a + ©MB = 0;
FHG (3) - 600(4) = 0 FHG = 800 N (T)
Ans.
B
L
D
C
L
L
600 N
6–49.
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
+ c ©Fy = 0;
FBG
3
2L2 + 9
FBG = - 200 2L2 + 9
- FBC132 - 6001L2 = 0
FBC = - 200L
a + ©MB = 0;
FHG132 - 60012L2 = 0
FHG = 400L
D
C
L
L
600 N
600
= sin u
sin u =
a + ©MG = 0;
B
L
- 600 - FBG sin u = 0
E
3m
A
SOLUTION
G
Ans.
6–5 0.
Two space trusses are used to equally support the uniform sign of mass M. Determine the force
developed in members AB, AC, and BC of truss ABCD and state if the members are in tension
or compression. Horizontal short links support the truss at joints B and D and there is a ball-andsocket joint at C.
Given:
M
50 kg
a
0.25 m
b
0.5 m
c
2m
Solution:
h
AB
§ a ·
¨ c ¸
¨ ¸
©h¹
BD
§ 2a ·
¨0¸
¨ ¸
©0¹
g
9.81
m
2
s
2
2
b a
AD
§a·
¨ c ¸
¨ ¸
©h¹
BC
§a·
¨0¸
¨ ¸
© h ¹
AC
§0·
¨ c ¸
¨ ¸
©0¹
1N
Guesses
F AB
1N
F AD
1N
F AC
F BC
1N
F BD
1N
By
1N
Given
§¨ 0 ·¸
0 ¸
AB
AD
AC
F AB
FAD
FAC
¨
¨ M g ¸
AB
AD
AC
¨ 2 ¸
©
¹
§¨ 0 ·¸
F AB
FBD
F BC
¨ B y ¸
AB
BD
BC
¨ 0 ¸
©
¹
AB
BD
BC
0
0
§ FAB ·
¨
¸
¨ FAD ¸
¨F ¸
¨ AC ¸
¨ FBC ¸
¨
¸
¨ FBD ¸
¨
¸
© By ¹
Find F AB F AD F AC F BC F BD B y
§ By ·
¨
¸
¨ FAD ¸
¨F ¸
© BD ¹
§ 566 ·
¨ 584 ¸ N
¨
¸
© 0 ¹
§ FAB ·
¨
¸
¨ FAC ¸
¨F ¸
© BC ¹
§ 584 ·
¨ 1133 ¸ N
¨
¸
© 142 ¹
Positive (T), Negative (C)
Ans.
6–51.
Determine the force in each member of the space truss and
state if the members are in tension or compression. Hint:
The support reaction at E acts along member EB. Why?
z
2m
E
SOLUTION
B
5m
3m
C
Method of Joints: In this case, the support reactions are not required for
determining the member forces.
3m
Joint A:
©Fz = 0;
A
FAB ¢
5
229
≤ - 6 = 0
x
FAB = 6.462 kN 1T2 = 6.46 kN 1T2
Ans.
©Fx = 0;
3
3
FAC a b - FAD a b = 0
5
5
©Fy = 0;
2
4
4
FAC a b + FAD a b - 6.462 ¢
≤ = 0
5
5
229
FAC = FAD
(1)
FAC + FAD = 3.00
(2)
Solving Eqs. (1) and (2) yields
FAC = FAD = 1.50 kN 1C2
Ans.
Joint B:
©Fx = 0;
FBC ¢
©Fz = 0;
FBC ¢
3
238
5
238
≤ - FBD ¢
≤ + FBD ¢
3
238
5
238
≤ = 0
FBC = FBD
≤ - 6.462 ¢
5
229
(1)
≤ = 0
FBC + FBD = 7.397
(2)
Solving Eqs. (1) and (2) yields
FBC = FBD = 3.699 kN 1C2 = 3.70 kN 1C2
©Fy = 0;
2 B 3.699 ¢
2
238
y
D
≤ R + 6.462 ¢
2
229
FBE = 4.80 kN 1T2
Ans.
≤ - FBE = 0
Ans.
Note: The support reactions at supports C and D can be determined by analyzing
joints C and D, respectively using the results obtained above.
4m
6 kN
6–52.
Determine the force in each member of the space truss and
state if the members are in tension or compression. The
truss is supported by rollers at A, B, and C.
8 kN
z
6m
SOLUTION
©Fx = 0;
3m
3
3
F
- FDA = 0
7 DC
7
x
2
2
2.5
F
+ FDA F
= 0
7 DC
7
6.5 DB
6
6
F
= 0
- 8 + 2a b FDC +
7
6.5 DB
FDC = FDA = 2.59 kN 1C2
Ans.
FDB = 3.85 kN 1C2
Ans.
©Fx = 0;
FBC = FBA
©Fy = 0;
3.85 a
4.5
2.5
b - 2¢
≤ FBC = 0
6.5
229.25
FBC = FBA = 0.890 kN 1T2
©Fx = 0;
Ans.
3
3
2.59a b - 0.890 ¢
≤ - FAC = 0
7
229.25
FAC = 0.617 kN 1T2
y
B
A
FDB = 1.486 FDC
©Fz = 0;
C
3m
FDC = FDA
©Fy = 0;
D
Ans.
2m
2.5 m
6–53.
The space truss supports a force F = [300i + 400j - 500k] N.
Determine the force in each member, and state if the members
are in tension or compression.
z
F
D
SOLUTION
Method of Joints: In this case, there is no need to compute the support reactions. We
1.5 m
will begin by analyzing the equilibrium of joint D, and then that of joints A and C.
(1)
1
1
1
b - FDA a
b - FDC a
b + 400 = 0
3.5
3.5
110
(2)
FDA a
©Fy = 0;
FDB a
©Fz = 0;
- FDA a
3
3
3
b - FDC a
b - FDB a
b - 500 = 0
3.5
3.5
110
(3)
Solving Eqs. (1) through (3) yields
FDB = - 895.98 N = 896 N (C)
Ans.
FDC = 554.17 N = 554 N (T)
Ans.
FDA = - 145.83 N = 146 N (C)
Ans.
Joint A: From the free-body diagram, Fig. b,
1
2
b - 145.83 a
b = 0
2.5
3.5
©Fy = 0;
FAB a
©Fx = 0;
1.5
1.5
b - 52.08 a
b - FAC = 0
145.83 a
3.5
2.5
FAB = 52.08 N = 52.1 N (T)
FAC = 31.25 N (T)
©Fz = 0;
Ans.
Ans.
3
b = 0
A z - 145.83 a
3.5
A z = 125 N
Ans.
Joint C: From the free-body diagram, Fig. c,
1.5
1.5
b - FCB a
b = 0
3.5
2.5
©Fx = 0;
31.25 + 554.17 a
©Fy = 0;
2
1
b - 447.92 a
b + Cy = 0
554.17 a
3.5
2.5
FCB = 447.92 N = 448 N (C)
Ans.
Cy = 200 N
©Fz = 0;
554.17 a
B
y
1.5
1.5
b - FDC a
b + 300 = 0
3.5
3.5
©Fx = 0;
C
A
Joint D: From the free-body diagram, Fig. a, we can write
3m
1.5 m
3
b - Cz = 0
3.5
Cz = 475 N
Note: The equilibrium analysis of joint B can be used to determine the components
of support reaction of the ball and socket support at B.
1m
x
1m
z
6–54.
The space truss supports a force F = [⫺400i ⫹ 500j ⫹ 600k] N.
Determine the force in each member, and state if the members
are in tension or compression.
F
D
C
1.5 m
Method of Joints: In this case, there is no need to compute the support reactions.
We will begin by analyzing the equilibrium of joint D, and then that of joints
A and C.
©Fy = 0;
©Fz = 0;
1.5
1.5
b - FDC a
b - 400 = 0
3.5
3.5
1
1
1
b - FDA a
b - FDC a
b + 500 = 0
FDB - a
3.5
3.5
110
3
3
3
600 - FDA a
b - FDC a
b - FDB a
b = 0
3.5
3.5
110
FDA a
(1)
(2)
(3)
FDB = - 474.34 N = 474 N (C)
Ans.
FDC = 145.83 N = 146 N (T)
Ans.
FDA = 1079.17 N = 1.08 kN (T)
Ans.
Joint A: From the free-body diagram, Fig. b,
1079.17 a
1
2
b - FAB a
b = 0
3.5
2.5
FAB = 385.42 N = 385 N (C)
Ans.
1.5
1.5
b - 1079.17 a
b + FAC = 0
2.5
3.5
©Fx = 0;
385.42 a
©Fz = 0;
1
b - Az = 0
1079.17 a
3.5
FAC = 231.25 N = 231 N (C)
A z = 925 N
Ans.
Ans.
Joint C: From the free-body diagram, Fig. c,
©Fx = 0;
FCB a
1.5
1.5
b - 231.25 + 145.83 a
b = 0
2.5
3.5
FCB = 281.25 N = 281 N (T)
©Fy = 0;
©Fz = 0;
281.25 a
Ans.
2
1
b + 145.83 a
b - Cy = 0
2.5
3.5
Cy = 266.67 N
3
b - Cz = 0
145.83 a
3.5
Cz = 125 N
Note: The equilibrium analysis of joint B can be used to determine the components
of support reaction of the ball and socket support at B.
B
y
x
Solving Eqs. (1) through (3) yields
©Fy = 0;
A
1m
Joint D: From the free-body diagram, Fig. a, we can write
©Fx = 0;
3m
1.5 m
SOLUTION
1m
6–55.
z
Determine the force in each member of the space truss and
state if the members are in tension or compression. The
truss is supported by ball-and-socket joints at C, D, E, and G.
1m
1m
G
E
SOLUTION
©(MEG)x = 0;
2
25
FBC (2) +
2
25
FBD (2) -
FBC = FBD = 1.342 = 1.34 kN (C)
FAB -
x
©Fx = 0;
FAG = FAE
©Fy = 0;
2
3
2
(3) FAE FAG = 0
5
25
25
FAG = FAE = 1.01 kN (T)
©Fy = 0;
©Fz = 0;
25
2
25
(1.342) +
1
1
1
FBE (1.342) - FBG = 0
3
3
25
(1.342) -
2
2
2
F +
(1.342) - FBG = 0
3 BE
3
25
B
Ans.
1.5 m
Ans.
Ans.
Joint B:
1
3 kN
C
2m
4
(3) = 0
5
FAB = 2.4 kN (C)
©Fx = 0;
F
4
(3)(2) = 0
5
Joint A:
©Fz = 0;
D
2m
FBc + FBD = 2.683 kN
Due to symmetry:
A
2
2
F + FBG - 2.4 = 0
3 BE
3
FB G = 1.80 kN (T)
Ans.
FBE = 1.80 kN (T)
Ans.
y
6–56.
The space truss is used to support vertical forces at joints B,
C, and D. Determine the force in each member and state if
the members are in tension or compression. There is a roller
at E, and A and F are ball-and-socket joints.
0.75 m
6 kN
B
8 kN
C
9 kN
90
D
1.5 m
SOLUTION
1m
F
A
Joint C:
E
1.25 m
©Fx = 0;
FBC = 0
Ans.
©Fy = 0;
FCD = 0
Ans.
©Fz = 0;
FCF = 8 kN (C)
Ans.
©Fy = 0;
FBD = 0
Ans.
©Fz = 0;
FBA = 6 kN (C)
Ans.
©Fy = 0;
FAD = 0
Ans.
©Fx = 0;
FDF = 0
Ans.
©Fz = 0;
FDE = 9 kN (C)
Ans.
©Fx = 0;
FEF = 0
Ans.
©Fy = 0;
FEA = 0
Ans.
Joint B:
Joint D:
Joint E:
6–57.
Determine the force in members BE, BC, BF, and CE of the
space truss, and state if the members are in tension or
compression.
z
F
SOLUTION
D
1.5 m
A
Method of Joints: In this case, there is no need to compute the support reactions.We will
begin by analyzing the equilibrium of joint C, and then that of joints E and B.
x
Joint C: From the free-body diagram, Fig. a, we can write
©Fz = 0;
©Fx = 0;
1.5
b - 600 = 0
FCE a
13.25
FCE = 721.11 N = 721 N (T)
1
b - FBC = 0
721.11 a
13.25
FBC = 400 N (C)
Ans.
Ans.
FBE = cos u = 0
FBE = 0
Ans.
Joint B: From the free-body diagram, Fig. c,
©Fz = 0;
FBF a
1.5
b - 900 = 0
3.5
FBF = 2100 N = 2.10 kN (T)
3m
B
Joint E: From the free-body diagram, Fig, b, notice that FEF, FED, and FEC lie in the
same plane (shown shaded), and FBE is the only force that acts outside of this plane.
If the x⬘ axis is perpendicular to this plane and the force equation of equilibrium is
written along this axis, we have
©Fx¿ = 0;
E
Ans.
1m
900 N
C
1m
600 N
y
6–58.
Determine the force in members AF, AB, AD, ED, FD, and
BD of the space truss, and state if the members are in tension
or compression.
z
F
SOLUTION
D
1.5 m
Support Reactions: In this case, it will be easier to compute the support reactions
first. From the free-body diagram of the truss, Fig. a, and writing the equations of
x
equilibrium, we have
©Mx = 0; Fy(1.5) - 900(3) - 600(3) = 0
Fy = 3000 N
©My = 0; 900(2) - A z(2) = 0
A z = 900 N
©Mz = 0;
A y(2) - 3000(1) = 0
A y = 1500 N
©Fx = 0;
Ax = 0
©Fy = 0;
Dy + 1500 - 3000 = 0
Dy = 1500 N
©Fz = 0;
Dz + 900 - 900 - 600 = 0
Dz = 600 N
©Fz = 0;
1500 - FAB = 0
900 - FAF a
©Fx = 0;
1081.67 a
Ans.
1.5
b = 0
13.25
FAF = 1081.67 N = 1.08 kN (C)
Ans.
1
b - FAD = 0
13.25
FAD = 600 N (T)
Ans.
Joint C: From the free-body diagram of the joint in Fig. c, notice that FCE, FCB, and
the 600-N force lie in the x–z plane (shown shaded). Thus, if we write the force
equation of equilibrium along the y axis, we have
©Fy = 0;
FDC = 0
Joint D: From the free-body diagram, Fig. d,
©Fx = 0;
©Fy = 0;
©Fz = 0;
2
1
1
b + FFD a
b + FFD a
b + 600 = 0
3.5
113
13.25
3
3
b + FED a
b + 1500 = 0
FBD a
3.5
113
1.5
1.5
b + FED a
b + 600 = 0
FFD a
3.5
113
FBC - a
(1)
(2)
(3)
Solving Eqs. (1) through (3) yields
FFD = 0
FED = - 1400N = 1.40 kN (C)
FBD = - 360.56 N = 361 N (C)
3m
1m
900 N
Joint A: From the free-body diagram, Fig. b, we can write
FAB = 1500 N = 1.50 kN (C)
A
B
Method of Joints: Using the above results, we will begin by analyzing the equilibrium
of joint A, and then that of joints C and D.
©Fy = 0;
E
Ans.
Ans.
C
1m
600 N
y
6–59. If the truss supports a force of F = 200 N,
determine the force in each member and state if the
members are in tension or compression.
z
200 mm
200 mm
D
C
200 mm
200 mm
E
B
x
y
500 mm
A
300 mm
F
6–60. If each member of the space truss can support a
maximum force of 600 N in compression and 800 N in
tension, determine the greatest force F the truss can
support.
z
200 mm
200 mm
D
C
200 mm
200 mm
E
B
x
y
500 mm
A
300 mm
F
6–61. Determine the force P required to hold the 50-kg
mass in equilibrium.
C
B
A
P
6–62. Determine the force P required to hold the
150-kg crate in equilibrium. The two cables are connected
to the bottom of the hanger.
B
A
C
P
6–63.
The principles of a differential chain block are indicated
schematically in the figure. Determine the magnitude of
force P needed to support the 800-N force. Also, find the
distance x where the cable must be attached to bar AB so
the bar remains horizontal. All pulleys have a radius of
60 mm.
x
B
A
SOLUTION
800 N
Equations of Equilibrium: From FBD(a),
+ c ©Fy = 0;
4P¿ - 800 = 0
From FBD(b),
200 - 5P = 0
a + ©MA = 0;
2001x2 - 40.011202 - 40.012402
P = 40.0 N
Ans.
- 40.013602 - 40.014802 = 0
x = 240 mm
240 mm
P
P¿ = 200 N
+ c ©Fy = 0;
180 mm
Ans.
6–64.
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
SOLUTION
For pulley D:
+ c ©Fy = 0;
D
9P - 2019.812 = 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.
C
F
5
0.7 m
0.8 m
0.8 m
8
6
0.8 m
2 P,
1.6 m
5
(2 P)2 + (P)2
2.24 kN
.
6–66.
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
SOLUTION
a + ©MB = 0;
800 N
- 800(1 + 0.05) + Ax (0.2) = 0
Ax = 4200 N = 4.20 kN
Ans.
+ ©F = 0;
:
x
Bx = 4200 N = 4.20 kN
Ans.
+ c ©Fy = 0;
Ay - By - 800 = 0
(1)
Member AC:
a + ©MC = 0;
- 800(50) - Ay(200) + 4200(200) = 0
Ay = 4000 N = 4.00 kN
Ans.
From Eq. (1)
By = 3.20 kN
Ans.
+ ©F = 0;
:
x
- 4200 + 800 + Cx = 0
Cx = 3.40 kN
+ c ©Fy = 0;
Ans.
4000 - Cy = 0
Cy = 4.00 kN
Ans.
.
.
.
.
6–68.
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
SOLUTION
a + ©MA = 0;
0.75 m
P
T(0.6) - P(1.5) = 0
+ ©F = 0;
:
x
Ax - T = 0
+ c ©Fy = 0;
Ay - P = 0
Thus, Ax = 2.5 P, Ay = P
Require,
2 = 2(2.5P)2 + (P)2
P = 0.743 kN = 743 N
Ans.
6– 69.
Determine the horizontal and vertical components of force that pins A and C exert on the frame.
Given:
F
500 N
a
0.8 m
d
0.4 m
b
0.9 m
e
1.2 m
c
0.5 m
T
45deg
Solution:
BC is a two-force member
Member AB :
6MA = 0;
e
F c FBC
2
2
a e
b F BC
2
F BC
Fc
a
2
2
( c d)
0
a e
2
a e
eba ca d
F BC
200.3 N
Thus,
6F x = 0;
6F y = 0;
Cx
F BC
Cy
F BC
Ax F BC
e
2
2
a e
a
2
2
a e
e
2
Cx
167 N
Ans.
Cy
111 N
Ans.
0
2
a e
Ay F FBC
a
2
2
a e
0
e
Ax
FBC
Ay
F F BC
2
2
a e
a
2
2
a e
Ax
167 N
Ans.
Ay
389 N
Ans.
.
6–71.
Determine the support reactions at A, C, and E on the
compound beam which is pin connected at B and D.
10 kN
9 kN
10 kN m
B C
E
D
SOLUTION
A
Equations of Equilibrium: First, we will consider the free-body diagram of
segment DE in Fig. c.
1.5 m 1.5 m 1.5 m 1.5 m 1.5 m 1.5 m
+ ©MD = 0;
NE(3) - 10(1.5) = 0
NE = 5 kN
+ ©ME = 0;
Ans.
10(1.5) - Dy(3) = 0
Dy = 5 kN
+ ©F = 0;
:
x
Dx = 0
Ans.
Subsequently, the free-body diagram of segment BD in Fig. b will be considered
using the results of Dx and Dy obtained above.
+ ©MB = 0;
NC(1.5) - 5(3) - 10 = 0
NC = 16.67 kN = 16.7 kN
+ ©MC = 0;
Ans.
By(1.5) - 5(1.5) - 10 = 0
By = 11.67 kN
+ ©F = 0;
:
x
By = 0
Finally, the free-body diagram of segment AB in Fig. a will be considered using the
results of Bx and By obtained above.
+ ©F = 0;
:
x
Ax = 0
+ c ©Fy = 0;
11.67 - 9 - A y = 0
A y = 2.67 kN
+ ©MA = 0;
Ans.
Ans.
11.67(3) - 9(1.5) - MA = 0
MA = 21.5 kN # m
Ans.
6–72.
Determine the horizontal and vertical components of force
at pins A,B, and C,and the reactions at the fixed support D
of the three-member frame.
0.5 m
0.5 m
0.5 m
0.5 m
2 kN 2 kN 2 kN 2 kN
B
A
2m
SOLUTION
Free Body Diagram: The solution for this problem will be simplified if one realizes
that member AC is a two force member.
C
2m
Equations of Equilibrium: For FBD(a),
a + ©MB = 0;
4
210.52 + 2112 + 211.52 + 2122 - FAC a b11.52 = 0
5
D
FAC = 8.333 kN
4
By + 8.333a b - 2 - 2 - 2 - 2 = 0
5
+ c ©Fy = 0;
By = 1.333 kN = 1.33 kN
Ans.
3
Bx - 8.333a b = 0
5
+ ©F = 0;
:
x
Bx = 5.00 kN
Ans.
3
3
Ax = Cx = FAC a b = 8.333 a b = 5.00 kN
5
5
Ans.
4
4
Ay = Cy = FAC a b = 8.333 a b = 6.67 kN
5
5
Ans.
For pin A and C,
From FBD (b),
a + ©MD = 0;
3
5.00142 - 8.333 a b 122 - MD = 0
5
MD = 10.0 kN # m
+ c ©Fy = 0;
4
Dy - 1.333 - 8.333a b = 0
5
Dy = 8.00 kN
+ ©F = 0;
:
x
Ans.
Ans.
3
8.333 a b - 5.00 - Dx = 0
5
Dx = 0
Ans.
6–73.
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
SOLUTION
Equations of Equilibrium: From FBD(a),
a + ©ME = 0;
Cy 162 = 0
+ c ©Fy = 0;
Ey - 0 = 0
+ ©F = 0;
:
x
Ex = 0
Cy = 0
Ans.
Ey = 0
From FBD(b),
a + ©MD = 0;
By 142 - 15122 = 0
By = 7.50 kN
+ c ©Fy = 0;
Ans.
Dy + 7.50 - 15 = 0
Dy = 7.50 kN
+ ©F = 0;
:
x
Dx = 0
From FBD(c),
a + ©MA = 0;
MA - 7.50162 = 0
MA = 45.0 kN # m
Ans.
+ c ©Fy = 0;
Ay - 7.5 = 0
Ans.
+ ©F = 0;
:
x
Ax = 0
Ay = 7.5 kN
Ans.
2m 2m 2m
6m
6–74. The platform scale consists of a combination of
third and first class levers so that the load on one lever
becomes the effort that moves the next lever. Through this
arrangement, a small weight can balance a massive object.
If x = 450 mm, determine the required mass of the
counterweight S required to balance a 90-kg load, L.
100 mm
250 mm
150 mm
H
E
C
F
G
D
150 mm
S
350 mm
B
A
x
L
6–75. The platform scale consists of a combination of
third and first class levers so that the load on one lever
becomes the effort that moves the next lever. Through this
arrangement, a small weight can balance a massive object. If
x = 450 mm and, the mass of the counterweight S is 2 kg,
determine the mass of the load L required to maintain the
balance.
100 mm
250 mm
150 mm
H
E
C
F
G
D
150 mm
S
350 mm
B
A
x
L
mL
6–76.
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
C
2m
D
1.5 m
2.5 m
300 N
2m
B
300 N
2.5 m
1.5 m
A
SOLUTION
a + ©ME = 0;
-Ay(3.5) + 400(2) + 300(3.5) + 300(1.5) = 0
Ay = 657.1 = 657 N
a + ©MD = 0;
Ans.
- Cy (3.5) + 400(2) = 0
Cy = 228.6 = 229 N
Ans.
a + ©MB = 0;
Cx = 0
Ans.
+ ©F = 0;
:
x
FBD = FBE
+ c ©Fy = 0;
657.1 - 228.6 - 2 a
5
274
b FBD = 0
FBD = FBE = 368.7 N
Bx = 0
By =
Ans.
5
274
(368.7)(2) = 429 N
Ans.
E
6–77.
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
2 ft
A
SOLUTION
a + ©MA = 0;
-4(2 cos 30°) + W cos 45°(2 cos 30°) + Wsin 45°(2 sin 30°) = 0
W = 3.586 kN
m = 3.586(1000)/9.81 = 366 kg
+ ©F = 0;
:
x
Ans.
4 - 3.586 cos 45° - Ax = 0
Ax = 1.464 kN
+ c ©Fy = 0;
3.586 sin 45° - Ay = 0
Ay = 2.536 kN
FA = 2(1.464)2 + (2.536)2 = 2.93 kN
Ans.
6–78.
Determine the reactions on the collar at A and the pin at C.
The collar fits over a smooth rod, and rod AB is fixed
connected to the collar.
600 N
750 N
1.25 m
1.25 m
A
45
SOLUTION
Equations of Equilibrium: From the force equation of equilibrium of member AB,
Fig. a, we can write
+ ©MA = 0;
MA - 750(1.25) - By(2.5) = 0
(1)
+ ©F = 0;
:
x
NA cos 45° - Bx = 0
(2)
+ c ©Fy = 0;
NA sin 45° - 750 - By = 0
(3)
From the free-body diagram of member BC in Fig. b,
+ ©MC = 0;
Bx(2 sin 30°) - By(2 cos 30°) + 600(1) = 0
(4)
+
: ©Fx = 0;
Bx + 600 sin 30° - Cx = 0
(5)
+ c ©Fy = 0;
By - Cy - 600 cos 30° = 0
(6)
Solving Eqs. (2), (3), and (4) yields
By = 1844.13
N = 1.84 kN
NA = 3668.66
N = 3.67 kN
Bx = 2594.13 N
Ans.
Substituting the results of Bx and By into Eqs. (1), (5), and (6) yields
MA = 5547.84 N # m = 5.55 kN # m
Ans.
Cx = 2894.13
N = 2.89 kN
Ans.
Cy = 1324.52
N = 1.32 kN
Ans.
C
30
B
1m
1m
6–79.
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
60°
1.2 a
Equations of Equilibrium: From FBD (a),
a
a
FCD cos 30° a b - FCD sin 30°a b - F12a2 = 0
2
2
FCD = 10.93F
+ ©F = 0;
:
x
Bx - 10.93 sin 30° = 0
Bx = 5.464F
From (b),
a + ©MA = 0;
5.464F1a2 - FE 11.2a2 = 0
FE = 4.55F
D
a/2
Free Body Diagram: The solution for this problem will be simplified if one realizes
that member CD is a two force member.
a + ©MB = 0;
a/2
C
E
SOLUTION
1.5 a
Ans.
1.2
6–80.
.
6–81.
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 connection D.
10
350 mm
1250 mm
C
G
A
850 mm
SOLUTION
Free-Body Diagram: The solution for this problem will be simplified if one realizes
that member AB is a two force member. From the geometry,
B
lAB = 23502 + 8502 - 2(350)(850) cos 80° = 861.21 mm
sin u
sin 80°
=
850
861.24
550 mm
D
u = 76.41°
Equations of Equilibrium: From FBD (a),
a + ©MC = 0;
1962(1.60) - FAB sin 76.41°(0.35) = 0
FAB = 9227.60 N = 9.23 kN
+ ©F = 0;
:
x
Cx - 9227.60 cos 76.41° = 0
Cx = 2168.65 N = 2.17 kN
+ c ©Fy = 0;
Ans.
Ans.
9227.60 sin 76.41° - 1962 - Cy = 0
Cy = 7007.14 N = 7.01 kN
Ans.
+ ©F = 0;
:
x
Dx = 0
Ans.
+ c ©Fy = 0;
Dy - 1962 = 0
From FBD (b),
Dy = 1962 N = 1.96 kN
a + ©MD = 0;
Ans.
1962(1.60 - 1.40 sin 10°) - MD = 0
MD = 2662.22 N # m = 2.66 kN # m
Ans.
.
.
.
.
.
.
6–83.
Determine the horizontal and vertical components of force that pins A and C exert on the
frame.
Units Used:
kN
3
10 N
Given:
F1
1 kN
F2
500 N
T
45 deg
a
0.2 m
b
0.2 m
c
0.4 m
d
0.4 m
Solution:
Guesses
Ax
1N
Ay
1N
Cx
1N
Cy
1N
Given
Ax Cx
0
Ay Cy F1 F2
F1 a Ay 2 a Ax d
§ Ax ·
¨ ¸
¨ Ay ¸
¨ ¸
¨ Cx ¸
¨C ¸
© y¹
0
Find Ax Ay Cx Cy
0
F 2 b Cy( b c) Cx d
§ Ax ·
¨ ¸
¨ Ay ¸
¨ ¸
¨ Cx ¸
¨C ¸
© y¹
§ 500 ·
¨
¸
¨ 1000 ¸ N
¨ 500 ¸
¨
¸
© 500 ¹
0
Ans.
6–84.
The compound beam is fixed supported at A and supported by rockers at B and C. If there are
hinges (pins) at D and E, determine the reactions at the supports A, B, and C.
Units Used:
3
kN
10 N
Given:
a
2m M
b
4m
c
2m
d
6m
e
3m
48 kN˜ m
w1
8
w2
6
kN
m
kN
m
Solution:
Guesses
Ax
1N
Ay
1N
MA
Dx
1N
Dy
1N
By
1N
Ey
1N
Ex
1N
Cy
1N
1 Nm
Given
Ay w2 a Dy
a
MA w2 a Dy a
2
w1
( b c)
w1§¨
de
2
B y b E y( b c)
Cy
d e·§ d e·
0
Dx E x
E x
0
¸¨
¸ Cy d M
© 2 ¹© 3 ¹
0
Dy w1( b c) By E y
0
2
2
E y w1
Ax Dx
0
0
0
0
0
§ Ax ·
¨
¸
¨ Ay ¸
¨M ¸
¨ A¸
¨ Dx ¸
¨
¸
¨ Dy ¸
¨B ¸
¨ y¸
¨ Ey ¸
¨
¸
¨ Ex ¸
¨
¸
© Cy ¹
Find A x A y MA Dx Dy B y Ey E x Cy
§¨ Ax ·¸
¨ Ay ¸
© ¹
MA
§0·
¨ ¸ kN
© 19 ¹
26 kN m
Ans.
Ans.
By
51 kN
Ans.
Cy
26 kN
Ans.
6–85.
The pruner multiplies blade-cutting power with the compound
leverage mechanism. If a 20-N force is applied to the handles,
determine the cutting force generated at A. Assume that the
contact surface at A is smooth.
20 N
60 mm
150 mm
F
E
B
10 mm
A
SOLUTION
Equations of Equilibrium: Applying the moment equation of equilibrium about
point C to the free-body diagram of handle CDG in Fig. a, we have
C
D
45
25 mm 30 mm
G
+ ©MC = 0;
20(150) - FDE sin 45°(25) = 0
FDE = 169.71 N
20 N
Using the result of FDE and applying the moment equation of equilibrium about
point B on the free-body diagram of the cutter in Fig. b, we obtain
+ ©MB = 0;
169.71 sin 45°(55) + 169.71 cos 45°(10) - NA (60) = 0
FA = 130 N
Ans.
6–86.
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. 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
P
SOLUTION
u = sin -1 a
10
b = 36.03°
17
Equations of Equilibrium:
+ c ©Fy = 0;
NB sin 36.03° - NC sin 36.03° = 0
NB = NC
+ ©F = 0;
:
x
80 - NC cos 36.03° - NC cos 36.03° = 0
NB = NC = 49.5 N
Ans.
10 mm
N.
N
N
N
N
.
N
N
N
N
.
6–88.
Show that the weight W1 of the counterweight at H required
for equilibrium is W1 = (b>a)W, and so it is independent of
the placement of the load W on the platform.
D
c
4
A
W(x) - NB a 3b +
NB =
+ c ©Fy = 0;
FEF +
3
cb = 0
4
Wx
3
a 3b + c b
4
Wx
- W = 0
3
a 3b + c b
4
FEF = W §1 -
x
3b +
3
c
4
¥
Using the result of NB and applying the moment equation of equilibrium about
point A on the free-body diagram in Fig. b, we obtain
+ ©MA = 0;
Wx
1
a cb = 0
3
4
3b + c
4
Wx
=
12b + 3c
FCD (c) -
NCD
Writing the moment equation of equilibrium about point G on the free-body
diagram in Fig. c, we have
+ ©MG = 0;
E
C
c
Equations of Equilibrium: First, we will consider the free-body diagram of
member BE in Fig. a,
Wx
(4b) + W § 1 12b + 3c
W1 =
b
W
a
x
3b +
3
c
4
¥(b) - W1(a) = 0
Ans.
This result shows that the required weight W1 of the counterweight is independent
of the position x of the load on the platform.
F
W
B
SOLUTION
+ ©ME = 0;
b
3b
a
G
H
6–89.
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 at A
is 18 kN.
B
C
5m
SOLUTION
AB is a two-force member.
D
Pin B
A
Require FAB = 18 kN
+ c ©Fy = 0;
18 sin 60° -
W
sin 60° - W = 0
2
W = 10.878 kN
m =
10.878
= 1.11 Mg
9.81
Ans.
60
6–90.
20°
Determine the force that the jaws J of the metal cutters
exert on the smooth cable C if 200-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.
400 mm
20°
A
20 mm
J
E
C
B
D
30 mm 80 mm
SOLUTION
Free Body Diagram: The solution for this problem will be simplified if one realizes
that member ED is a two force member.
F
20°
20°
20 mm
400 mm
Equations of Equilibrium: From FBD (b),
+ ©F = 0;
:
x
20°
Ax = 0
20°
From (a),
a + ©MF = 0;
Ay sin 20°1202 + 200 sin 20°1202
- 200 cos 20°14002 = 0
20°
Ay = 10790 N
From FBD (b),
a + ©ME = 0;
107901802 - FC 1302 = 0
FC = 28773 N = 28.8 kN
200 N
Ans.
2 00 N
200 N
6–91.
The compound beam is pin supported at
B and supported by rockers at A and C.
There is a hinge (pin) at D. Determine
the reactions at the supports.
Units Used:
3
kN
10 N
Given:
F1
7 kN
a
4m
F2
6 kN
b
2m
F3
16 kN
c
3m
T
60 deg
Solution:
Member DC :
6MD = 0;
F 1 sin T ( a c) Cy a
Cy
6F y = 0;
ac
a
Dy F 1 sin T Cy
Dy
6F x = 0;
F 1 sin T
Cy
1.52 kN
Dy
4.55 kN
Dx
3.5 kN
By
23.5 kN
Ans.
Ay
3.09 kN
Ans.
Bx
3.5 kN
Ans.
Ans.
0
F1 sin T Cy
Dx F 1 cos T
Dx
0
0
F1 cos T
Member ABD :
6MA = 0;
F 3 a F2( 2 a b) Dy( 3 a b) By2 a
By
6F y = 0;
2a
Ay F 3 B y F2 Dy
Ay
6F x = 0;
F 3 a F2( 2a b) Dy ( 3a b)
Dy F3 By F 2
B x F1 cos T
Bx
F 1 cos T
0
0
0
6–92.
The scissors lift consists of two sets of cross members and
two hydraulic cylinders, DE, symmetrically located on
each side of the platform. The platform has a uniform mass
of 60 kg, with a center of gravity at G1. The load of 85 kg,
with center of gravity at G2, is centrally located between
each side of the platform. Determine the force in each of
the hydraulic cylinders for equilibrium. Rollers are located
at B and D.
0.8 m
1.2 m
2m
G2
G1
A
B
1m
C
SOLUTION
Free Body Diagram: The solution for this problem will be simplified if one realizes
that the hydraulic cyclinder DE is a two force member.
Equations of Equilibrium: From FBD (a),
a + ©MA = 0;
2NB 132 - 833.8510.82 - 588.6122 = 0
2NB = 614.76 N
+ ©F = 0;
:
x
+ c ©Fy = 0;
Ax = 0
2Ay + 614.76 - 833.85 - 588.6 = 0
2Ay = 807.69 N
From FBD (b),
a + ©MD = 0;
807.69132 - 2Cy 11.52 - 2Cx 112 = 0
2Cx + 3Cy = 2423.07
(1)
From FBD (c),
a + ©MF = 0;
2Cx 112 - 2Cy 11.52 - 614.76132 = 0
2Cx - 3Cy = 1844.28
(2)
Solving Eqs. (1) and (2) yields
Cx = 1066.84 N
Cy = 96.465 N
From FBD (b),
+ ©F = 0;
:
x
D
F
211066.842 - 2FDE = 0
FDE = 1066.84 N = 1.07 kN
Ans.
1.5 m
1.5 m
1m
E
6–93.
The two disks each have a mass of 20 kg and are attached at
their centers by an elastic cord that has a stiffness of
k = 2 kN>m. Determine the stretch of the cord when the
system is in equilibrium, and the angle u of the cord.
r
A
5
3
4
SOLUTION
Entire system:
+ ©F = 0;
:
x
3
NB - NA a b = 0
5
+ c ©Fy = 0;
4
NA a b - 2 (196.2) = 0
5
a + ©MO = 0;
NB (l sin u) - 196.2 l cos u = 0
Solving,
NA = 490.5 N
NB = 294.3 N
u = 33.69° = 33.7°
Ans.
Disk B:
+ ©F = 0;
:
x
- T cos 33.69° + 294.3 = 0
T = 353.70 N
Fx = kx;
353.70 = 2000 x
x = 0.177 m = 177 mm
Ans.
θ
l
r
B
6–107. A man
of of
175875
lb attempts
to hold
man having
havinga aweight
weight
N (⬇ 87.5
kg)
6–94.
himself
one ofhimself
the twousing
methods
the
attemptsusing
to hold
one shown.
of theDetermine
two methods
total
force
he mustthe
exert
bar he
ABmust
in each
shown.
Determine
totalonforce
exertcase
on and
bar
the
exerts
on the
platform
at C. Neglect
AB normal
in eachreaction
case andhethe
normal
reaction
he exerts
on the
the
weightatof
platform.
platform
C.the
Neglect
the weight of the platform.
A
A
(a)
Bar:
437.5 N
+↑ΣFy = 0;
437.5 N
2(F/2) – 2(437.5) = 0
F = 875 N
Ans.
Man:
+↑ΣFy = 0;
NC – 875 – 2(437.5) = 0
NC = 1750 N
875 N
437.5 N
437.5 N
Ans.
(b)
Bar:
+↑ΣFy = 0;
2(218.75) – 2(F/2) = 0
F = 437.5 N
Ans.
218.75 N
218.75 N
Man:
+↑ΣFy = 0;
NC – 875 + 2(218.75) = 0
NC = 437.5 N
Ans.
B
875 N
218.75 N
218.75 N
B
C
C
(a)
(b)
6–95.
6–108.
A man
of of
175875
lb attempts
to hold
man having
havinga aweight
weight
N (⬇ 87.5
kg)
himself
one ofhimself
the twousing
methods
the
attemptsusing
to hold
one shown.
of theDetermine
two methods
total
force
he mustthe
exert
barhe
AB
in each
and
shown.
Determine
totalon
force
must
exert case
on bar
ABthe
in
normal
reaction
exerts reaction
on the platform
at on
C.The
each case
and thehenormal
he exerts
the platform
has
weight
of 30 lb.
at C.a The
platform
has a weight of 150 N (⬇ 15 kg).
A
A
512.5 N
512.5 N
512.5 N
512.5 N
150 N
875 N
(a)
Bar:
+↑ΣFy = 0;
2(F/2) – 512.5 – 512.5 = 0
F = 1025 N
Ans.
875 N
512.5 N 512.5 N
Man:
+↑ΣFy = 0;
NC – 875 – 512.5 – 512.5 = 0
NC = 1900 N
Ans.
(b)
Bar:
512.5 N
+↑ΣFy = 0;
2(F/2) – 256.25 – 256.25 = 0
F = 512.5 N
Ans.
256.25 N 256.25 N
Man:
+↑ΣFy = 0;
NC – 875 + 256.25 + 256.25 = 0
NC = 362.5 N
256.25 N
B
256.25 N
Ans.
875 N
256.25 N
256.25 N
B
C
C
(a)
(b)
6–96.
The double link grip is used to lift the beam. If the beam
weighs 8 kN, determine the horizontal and vertical
components of force acting on the pin at A and the
horizontal and vertical components of force that the flange
of the beam exerts on the jaw at B.
8 kN
160 mm
160 mm
D
45°
E
SOLUTION
C
140 mm
A
Free Body Diagram: The solution for this problem will be simplified if one realizes
that members ED and CD are two force members.
300 mm
B
F
Equations of Equilibrium: Using method of joint, [FBD (a)],
+ c ©Fy = 0;
8 - 2F sin 45° = 0
300 mm 300 mm
F = 5.657 kN
From FBD (b),
+ c ©Fy = 0;
2By - 8 = 0
By = 4.00 kN
Ans.
From FBD (c),
a + ©MA = 0;
8 kN
Bx 13002 - 4.0013002 - 5.657 cos 45°11402
- 5.657 sin 45°11602 = 0
Bx = 8.00 kN
+ c ©Fy = 0;
Ay + 5.657 sin 45° - 4.00 = 0
Ay = 0
+ ©F = 0;
:
x
Ans.
Ans.
5.657 kN
8.00 + 5.657 cos 45° - Ax = 0
Ax = 12.00 kN
140 mm
Ans.
4.00
300 mm
300 mm
8
300 mm
300 mm
6–97.
Operation of exhaust and intake valves in an automobile engine consists of the cam C, push rod
DE, rocker arm EFG which is pinned at F, and a spring and valve, V. If the spring is
compressed a distance G when the valve is open as shown, determine the normal force acting on
the cam lobe at C. Assume the cam and bearings at H, I, and J are smooth.The spring has a
stiffness k.
Given:
a
25 mm
b
40 mm
G
20 mm
k
300
N
m
Solution:
Fs
kG
Fs
6N
6F y = 0;
F G F s
FG
0
Fs
FG
6N
6MF = 0;
FG b T a
T
FG
b
a
0
T
9.60 N
Ans.
6–98.
Determine the horizontal and vertical components of force that the pins at A, B, and C exert on their
connecting members.
Units Used:
3
kN 10 N
Given:
F 9 00N
a 0.9 m
r 50mm
b 0.2m
Solution:
F ˜ ( a r) Ax˜ b = 0
ar
Ax F˜
b
Ax
4.28 kN Ans.
Bx
4.28 kN Ans.
Ax Bx = 0
Bx Ax
F˜ r Ay˜ b Ax˜ b = 0
Ay F˜ r Ax˜ b
b
Ay
4.05 kN Ans.
By
3.15 kN Ans.
Cx
3.38 kN Ans.
Cy
4.05 kN Ans.
Ay By F = 0
By Ay F
Ax F Cx = 0
Cx Ax F
Ay Cy = 0
Cy Ay
6–99.
If a clamping force of 300 N is required at A, determine the
amount of force F that must be applied to the handle of the
toggle clamp.
F
70 mm
235 mm
30 mm
C
30 mm
A
SOLUTION
Equations of Equilibrium: First, we will consider the free-body diagram of the clamp in
Fig. a. Writing the moment equation of equilibrium about point D,
a + ©MD = 0;
Cx (60) - 300(235) = 0
Cx = 1175 N
Subsequently, the free - body diagram of the handle in Fig. b will be considered.
a + ©MC = 0;
FBE cos 30°(70) - FBE sin 30°(30) - F cos 30°(275 cos 30° + 70)
-F sin 30°(275 sin 30°) = 0
45.62FBE - 335.62F = 0
+ ©F = 0;
:
x
(1)
1175 + F sin 30° - FBE sin 30° = 0
0.5FBE - 0.5F = 1175
(2)
Solving Eqs. (1) and (2) yields
F = 369.69 N = 370 N
FBE = 2719.69N
Ans.
30
B
E
D
30
275 mm
6–100.
If a force of F = 350 N is applied to the handle of the toggle
clamp, determine the resulting clamping force at A.
F
70 mm
235 mm
30 mm
C
30 mm
A
SOLUTION
Equations of Equilibrium: First, we will consider the free-body diagram of the handle
in Fig. a.
a + ©MC = 0;
FBE cos 30°(70) - FBE sin 30°(30) - 350 cos 30°(275 cos 30° + 70)
-350 sin 30°(275 sin 30°) = 0
FBE = 2574.81 N
+ ©F = 0;
:
x
Cx - 2574.81 sin 30° + 350 sin 30° = 0
Cx = 1112.41 N
Subsequently,, the free-body diagram of the clamp in Fig. b will be considered.
Using the result of Cx and writing the moment equation of equilibrium about
point D,
a + ©MD = 0;
1112.41(60)- NA (235) = 0
NA = 284.01 N = 284 N
Ans.
30
B
E
D
30
275 mm
6 –101.
The skid steer loader has a mass M1, and in the position shown the center of mass is at G1. If
there is a stone of mass M2 in the bucket, with center of mass at G2 determine the reactions
of each pair of wheels A and B on the ground and the force in the hydraulic cylinder CD and
at the pin E. There is a similar linkage on each side of the loader.
Units Used:
3
Mg
10 kg
kN
10 N
3
Given:
M1
1.18 Mg
M2
300 kg
a
1.25 m
d
0.15 m
b
1.5 m
e
0.5 m
c
0.75 m
T
30 deg
Solution:
Entire System:
6MA = 0;
M2 g b M1 g( c d) NB c
NB
6F y = 0;
M1 g( c d) M2 g b
c
NB M2 g M1 g NA
NA
0
3.37 kN
(Both wheels)
Ans.
NA
11.1 kN
(Both wheels)
Ans.
0
NB M2 g M1 g
Upper member:
6ME = 0;
NB
M2 g( a b) 2 F CD sin T a
0
M 2 g( a b)
F CD
6F x = 0;
Ex
6F y = 0;
Ey Ey
FR
F CD
2 sin T a
F CD cos T
M2 g
FCD sin T
2
M2 g
2
2
F CD sin T
2
Ex Ey
6.5 kN
Ex
5607 N
Ey
1766 N
FR
5.879 kN
Ans.
0
Ans.
6–102.
The tractor boom supports the uniform mass of 600 kg in
the bucket which has a center of mass at G. Determine the
force in each hydraulic cylinder AB and CD and the
resultant force at pins E and F. The load is supported
equally on each side of the tractor by a similar mechanism.
G
B
A
0.25 m
E
C
1.5 m
0.3 m
0.1 m
SOLUTION
a + ©ME = 0;
1.25 m
0.2 m
294310.12 - FAB10.252 = 0
FAB = 1177.2 N = 1.18 kN
+ ©F = 0;
:
x
- Ex + 1177.2 = 0;
Ex = 1177.2 N
+ c ©Fy = 0;
Ey - 2943 = 0;
Ey = 2943 N
F
Ans.
0.6 m
D
0.4 m 0.3 m
FE = 211177.222 + 1294322
a + ©MF = 0;
= 3.17 kN
294312.802 - FCD1cos 12.2°210.72 + FCD1sin 12.2°211.252 = 0
FCD = 19 618 N = 19.6 kN
+ ©F = 0;
:
x
Ans.
2943 N
Ans.
Fx - 19 618 sin 12.2° = 0
Fx = 4145.8 N
+ c ©Fy = 0;
- Fy - 2943 + 19 618 cos 12.2° = 0
Fy = 16 232 N
FF = 214145.822 + 116 23222 = 16.8 kN
Ans.
2943 N
–.
The nail cutter consists of the handle and the two cutting blades. Assuming the blades are pin
connected at B and the surface at D is smooth, determine the normal force on the fingernail
when a force ) is applied to the handles as shown.The pin AC slides through a smooth hole at
A and is attached to the bottom member at C.
*LYHQ
) 51
D 6PP
E 36PP
6ROXWLRQ
Handle :
6MD = 0;
) $˜ D )˜ E = 0
)$ )˜
6Fy = 0;
E
D
)$
30 1
1'
35 1
1' ) $ ) = 0
1' )$ )
Top blade :
6MB = 0;
1'˜ E )1˜ ( 2D E) = 0
) 1 1'˜
E
2˜ D E
)1
26.25 1
Ans.
6–104.
Determine the force created in
the hydraulic cylinders EF and
AD in order to hold the shovel
in equilibrium. The shovel load
has a mass W and a center of
gravity at G. All joints are pin
connected.
Units Used:
3
Mg
10 kg
kN
10 N
3
Given:
a
0.25 m T 1
30 deg
b
0.25 m T 2
10 deg
c
1.5 m
T3
60 deg
d
2m
W
1.25 Mg
e
0.5 m
Solution:
Assembly FHG :
6MH = 0; [ W g( e) ] FEF c sin T 1
F EF
Wg
§ e · F
¨ c sin T ¸ EF
1 ¹
©
0
(T)
8.175 kN
Ans.
Assembly CEFHG:
6MC = 0; F AD cos T 1 T 2 b W gª( a b c)cos T 2 eº
¬
¼
0
§ cos T 2 a cos T 2 b cos T 2 c e ·
¨
¸
cos T 1 T 2 b
©
¹
F AD
Wg
F AD
158 kN (C)
Ans.
6–105.
The hoist supports the engine of
mass M. Determine the force in
member DB and in the hydraulic
cylinder H of member FB.
Units Used:
3
kN
10 N
Given:
M
125 kg
d
1m
a
1m
e
1m
b
2m
f
2m
c
2m
g
9.81
m
2
s
Solution:
Member GFE:
cd
6ME = 0; F FBª
«
º b M g ( a b)
2»
2
0
¬ ( c d) ( b e) ¼
ª a b º ( c d) 2 ( b e) 2
«
»
¬b( c d) ¼
F FB
Mg
F FB
1.94 kN
6F x = 0;
Ans.
be
ª
E x FFB«
¬
Ex
º
2
( c d) ( b e)
be
ª
F FB«
0
2»
¼
º
2»
2
¬ ( c d) ( b e) ¼
Member EDC:
60c = 0;
§
E x( c d) F DB¨
©
F DB
Ex
e
2
·d
2¸
e d
0
¹
§ c d · e2 d2
¨
¸
© ed ¹
F DB
2.601 kN
Ans.
6–106. If P = 75 N, determine the force F that the toggle
clamp exerts on the wooden block.
140 mm
85 mm
P
140 mm
50 mm
A
D
50 mm
C
B
20 mm
E
F
P
6–107. If the wooden block exerts a force of F = 600 N
on the toggle clamp, determine the force P applied to the
handle.
140 mm
85 mm
P
140 mm
50 mm
A
D
50 mm
C
B
20 mm
E
F
P
6–108.
The pillar crane is subjected to the load having a mass M. Determine the force developed in the
tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tied
in the position shown.
Units Used:
3
kN
10 N
Given:
M
500 kg
a
1.8 m
b
2.4 m
T1
10 deg
T2
20 deg
g
9.81
m
2
s
Solution:
initial guesses: F CB
10 kN
F AB
10 kN
Given
M
g cos T 1 F AB cos T 2 FCB
2
M
2
g sin T 1 FAB sin T 2 FCB
§¨ FAB ·¸
¨ FCB ¸
©
¹
Find FAB FCB
§¨ Cx ·¸
¨ Cy ¸
© ¹
b
2
0
2
a b
a
2
2
Mg
0
a b
FCB
2
a b
§b·
¨ ¸
2©a¹
§ FAB ·
¨
¸
¨ Cx ¸
¨ C ¸
© y ¹
§ 9.7 ·
¨ 11.53 ¸ kN
¨
¸
© 8.65 ¹
Ans.
6–109.
The symmetric coil tong supports the coil which has a mass
of 800 kg and center of mass at G. Determine the horizontal
and vertical components of force the linkage exerts on
plate DEIJH at points D and E. The coil exerts only vertical
reactions at K and L.
H
300 mm
D
J
E
I
400 mm
SOLUTION
100 mm
Free-Body Diagram: The solution for this problem will be simplified if one realizes
that links BD and CF are two-force members.
Equations of Equilibrium : From FBD (a),
78481x2 - FK12x2 = 0
a + ©ML = 0;
FK = 3924 N
FBD cos 45°11002 + FBD sin 45°11002 - 39241502 = 0
FBD = 1387.34 N
+ ©F = 0;
:
x
A x - 1387.34 cos 45° = 0
+ c ©Fy = 0;
A y - 3924 - 1387.34 sin 45° = 0
A x = 981 N
A y = 4905 N
From FBD (c),
a + ©ME = 0;
4905 sin 45°17002 - 981 sin 45°17002
- FCF cos 15°13002 = 0
FCF = 6702.66 N
+ ©F = 0;
:
x
Ex - 981 - 6702.66 cos 30° = 0
Ex = 6785.67 N = 6.79 kN
+ c ©Fy = 0;
A
Ans.
Ey + 6702.66 sin 30° - 4905 = 0
Ey = 1553.67 N = 1.55 kN
Ans.
Dx = FBD cos 45° = 1387.34 cos 45° = 981 N
Ans.
Dy = FBD sin 45° = 1387.34 sin 45° = 981 N
Ans.
At point D,
C
30°
45°
30°
F
50 mm
100 mm
K
From FBD (b),
a + ©MA = 0;
45°
B
G
L
6–110. A 300-kg counterweight, with center of mass at G, is
mounted on the pitman crank AB of the oil-pumping unit.
If a force of F = 5 kN is to be developed in the fixed cable
attached to the end of the walking beam DEF, determine
the torque M that must be supplied by the motor.
1.75 m
2.50 m
D
30⬚
E
M
A
F
B
G
30⬚
0.5 m
0.65 m
F
6–111. A 300-kg counterweight, with center of mass at G, is
mounted on the pitman crank AB of the oil-pumping unit.
If the motor supplies a torque of M = 2500 N # m, determine
the force F developed in the fixed cable attached to the end
of the walking beam DEF.
1.75 m
2.50 m
D
30⬚
E
M
A
F
B
G
30⬚
0.5 m
0.65 m
F
6–112.
.
6–113. Determine the horizontal and vertical components
of reaction which the pins exert on member AB of the frame.
1.5
300kN
lb
60⬚
A
B
1.2
4 ftm
E
C
D
0.9
3 ftm
0.9
3 ftm
500kN
lb
2.5
Member AB :
哭
+ ΣMA = 0;
–1.5 sin 60° (0.9) +
4
FBD (1.8) = 0
5
FBD = 0.8119 kN
1.5 kN
Thus,
+
→ ΣFx
= 0;
Bx =
3
(0.8119) = 0.4871 kN Ans.
5
By =
4
(0.8119) = 0.6495 kN Ans.
5
–1.5 cos 60° +
3
(0.8119) + Ax = 0
5
Ax = 0.2629 kN
+↑ΣFy = 0;
Ay – 1.5 sin 60° +
Ay = 0.6495 kN
Ans.
4
(0.8119) = 0
5
Ans.
0.9 m
0.9 m
6–114. Determine the horizontal and vertical components of
reaction which the pins exert on member EDC of the frame.
1.5
300kN
lb
60⬚
A
B
1.2
4 ftm
E
C
D
0.9
3 ftm
0.9
3 ftm
500kN
lb
2.5
1.5 kN
Member AB :
哭
+ ΣMA = 0;
–1.5 sin 60° (0.9) +
4
FBD (1.8) = 0
5
FBD = 0.8119 kN
0.9 m
0.9 m
Member EDC :
哭
+ ΣMg = 0;
–2.5 (1.8) –
+
4
(0.8119) (0.9)
5
4
FAD (0.9) = 0
5
FBD = 0.8119 kN
FAD = 7.0619 kN
+
→ ΣFx
= 0;
⎛ 3⎞
⎛ 3⎞
Ex – 0.8119 ⎜ ⎟ – 7.0619 ⎜ ⎟ = 0
⎝ 5⎠
⎝ 5⎠
Ex = 4.7243 kN
+↑ΣFy = 0;
Ans.
⎛ 4⎞
⎛ 4⎞
–Ey + 7.0619 ⎜ ⎟ – 0.8119 ⎜ ⎟ – 2.5 = 0
⎝ 5⎠
⎝ 5⎠
Ey = 2.5 kN
= 0;
3
3
(0.8119) –
(7.0619) = 0
5
5
Dx = 4.7243 kN
+↑ΣFy = 0;
2.5 kN
FAD = 7.0619 kN
Dx –
–Dy –
Ans.
4
4
(0.8119) +
(7.0619) = 0
5
5
Dy = 5 kN
0.9 m
Ans.
Pin D :
+
→ ΣFx
0.9 m
Ans.
FBD = 0.8119 kN
6–115.
The handle of the sector press is fixed to gear G, which in turn is in mesh with the sector gear
C. Note that AB is pinned at its ends to gear C and the underside of the table EF, which is
allowed to move vertically due to the smooth guides at E and F. If the gears exert tangential
forces between them, determine the compressive force developed on the cylinder S when a
vertical force F is applied to the handle of the press.
Given:
F
40 N
a
0.5 m
b
0.2 m
c
1.2 m
d
0.35 m
e
0.65 m
Solution:
Member GD:
6MG = 0;
F a F CG b
a
F CG F
b
0
F CG
100 N
Sector gear :
6MH = 0; F CG( d e) FAB§
¨
·d
2
2¸
© c d ¹
F AB
FCG ( d e)
c
0
§ c2 d2 ·
¨
¸ F
© c d ¹ AB
297.62 N
Table:
6F y = 0;
F AB§
c
·
¨ 2 2 ¸ Fs
© c d ¹
Fs
FAB
c
§
·
¨ 2 2¸
© c d ¹
0
Fs
286 N
Ans.
6–116.
The structure is subjected to the loading shown. Member AD
is supported by a cable AB and roller at C and fits through
a smooth circular hole at D. Member ED is supported by a
roller at D and a pole that fits in a smooth snug circular hole
at E. Determine the x, y, z components of reaction at E and
the tension in cable AB.
z
B
E
0.8 m
SOLUTION
©My = 0;
4
- FAB (0.6) + 2.5(0.3) = 0
5
FAB = 1.5625 = 1.56 kN
©Fz = 0;
D
C
0.5 m
x
Ans.
Ans.
Dz = 1.25 kN
Dy = 0
©Fx = 0;
Dx + Cx -
©Mx = 0;
MDx +
3
(1.5625) = 0
5
(1)
4
(1.5625)(0.4) - 2.5(0.4) = 0
5
MDx = 0.5 kN # m
3
(1.5625)(0.4) - Cx (0.4) = 0
5
©Mz = 0;
MDz +
©Fz = 0;
Dz¿ = 1.25 kN
©Mx = 0;
MEx = 0.5 kN # m
Ans.
©My = 0;
MEy = 0
Ans.
©Fy = 0;
Ey = 0
Ans.
©Mz = 0;
Dx (0.5) - MDz = 0
Solving Eqs. (1), (2) and (3):
Cx = 0.938 kN
MDz = 0
Dx = 0
0.4 m
{ 2.5 } kN
4
(1.5625) - 2.5 + Dz = 0
5
©Fy = 0;
0.3 m
A
(2)
(3)
0.3 m
y
6–117.
The four-member "A" frame is supported at A and E by smooth collars and at G by a pin. All
the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is
F max, determine the largest vertical force P that can be supported by the frame. Also, what
are the x, y, z force components which member BD exerts on members EDC and ABC? The
collars at A and E and the pon at G only exert force components on the frame.
Given:
F max
800 N
a
300 mm
b
600 mm
c
600 mm
Solution:
6Mx = 0;
b
P 2 c 2
2
b c
Fmax c
F max b
P
2
2
P
2
c
2
2
2
b c
b
2
2
2
Bz
2
b c
Dy
Bz
Bz
283 N
Ans.
Dz
283 N
Ans.
0
Dy
By
By
By
283 N
Bx
Dx
b c
F max b
2
Dz
2
B y Dy Fmax
By
Ans.
Dz
0
b c
F max c
2
282.843 N
b c
B z Dz Fmax
Bz
0
0
Dy
Ans.
283 N
Ans.
6 –118.
The structure is subjected to the loadings shown.Member AB is supported by a ball-and-socket
at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z
components of reaction at A and C.
Given:
a
2m
M
800 N˜ m
b
1.5 m
F
250 N
c
3m
T1
60 deg
d
4m
T2
45 deg
T3
60 deg
Solution:
Guesses
Bx
1N
By
1N
Ax
1N
Ay
1N
Az
1N
Cx
1N
Cy
1N
Cz
1N
MBx
1 N˜ m
MBy
1 N˜ m
MCy
1 N˜ m
MCz
1 N˜ m
Given
§ Ax · § Bx ·
¨ ¸ ¨ ¸
¨ Ay ¸ ¨ By ¸
¨A ¸ ¨ ¸
© z¹ © 0 ¹
0
§ c · §¨ Bx ·¸ § M · §¨ MBx ·¸
¨ a ¸ u B ¨ 0 ¸ M
¨ ¸ ¨ y ¸ ¨ ¸ ¨ By ¸
© 0 ¹ ¨© 0 ¸¹ © 0 ¹ ¨© 0 ¸¹
§ cos
¨
F ¨ cos
¨
© cos
0
T1 ·
T2
T3
§ Bx · § Cx ·
¸ ¨
¸ ¨ ¸
¸ ¨ By ¸ ¨ Cy ¸
¸ ¨
¸ ¨ ¸
¹ © 0 ¹ © Cz ¹
ª § cos
§ 0 · « ¨
¨ 0 ¸ u F cos
¸ « ¨
¨
© b d ¹ « ¨ cos
¬ ©
0
T 1 ·º
T2
T3
§ Bx · § MBx · § 0 ·
0
¸» §¨ ·¸ ¨
¸ ¨
¸ ¨
¸
¸» ¨ 0 ¸ u ¨ By ¸ ¨ MBy ¸ ¨ MCy ¸
¸» © b ¹ ¨
¸ ¨
¸ ¨
¸
© 0 ¹ © 0 ¹ © MCz ¹
¹¼
0
§¨ Ax ·¸
¨ Ay ¸
¨
¸
¨ Az ¸
¨ C ¸
¨ x ¸
¨ Cy ¸
¨
¸
¨ Cz ¸
¨
¸
¨ Bx ¸
¨ B ¸
¨ y ¸
¨ MBx ¸
¨
¸
¨ MBy ¸
¨M ¸
¨ Cy ¸
¨ MCz ¸
©
¹
§ Ax ·
¨ ¸
¨ Ay ¸
¨A ¸
¨ z¸
¨ Cx ¸
¨ ¸
¨ Cy ¸
¨ ¸
© Cz ¹
Find Ax Ay Az Cx Cy Cz Bx B y MBx MBy MCy MCz
§¨ 172.3 ¸·
¨ 114.8 ¸
¨ 0 ¸
¨
¸N
¨ 47.3 ¸
¨ 61.9 ¸
¨ 125 ¸
©
¹
§¨ MCy ·¸
¨ MCz ¸
©
¹
§ 429 ·
¨
¸ N˜ m
© 0 ¹
Ans.
.
.
.
.
6–120.
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
SOLUTION
a + ©MA = 0;
A
2m
Ey = 13.125 kN
+ c ©Fy = 0;
Ay - 8 - 4 - 10 + 13.125 = 0
Ay = 8.875 kN
+ c ©Fx = 0;
Ax = 3 kN
Joint B:
+ ©F = 0;
:
x
FBC = 3 kN (C)
Ans.
+ c ©Fy = 0;
FBA = 8 kN (C)
Ans.
Joint A:
+ c ©Fy = 0;
8.875 - 8 -
3
F
= 0
5 AC
FAC = 1.458 = 1.46 kN (C)
+ ©F = 0;
:
x
FAF - 3 -
Ans.
4
(1.458) = 0
5
FAF = 4.17 kN (T)
Ans.
Joint C:
+ ©F = 0;
:
x
3 +
4
(1.458) - FCD = 0
5
FCD = 4.167 = 4.17 kN (C)
+ c ©Fy = 0;
FCF - 4 +
Ans.
3
(1.458) = 0
5
FCF = 3.125 = 3.12 kN (C)
Ans.
+ ©F = 0;
:
x
FEF = 0
Ans.
+ c ©Fy = 0;
FED = 13.125 = 13.1 kN (C)
Ans.
Joint E:
Joint D:
+ c ©Fy = 0;
13.125 - 10 -
3
F
= 0
5 DF
FDF = 5.21 kN (T)
+ ©F = 0;
:
x
4.167 -
4
(5.21) = 0
5
E
F
- 3(1.5) - 4(2) - 10(4) + Ey (4) = 0
Ans.
Check!
2m
6–121.
Determine the horizontal and vertical components of force
at pins A and C of the two-member frame.
500 N/m
A
B
3m
3m
SOLUTION
Member AB:
a + ©MA = 0;
-750 (2) + By (3) = 0
C
By = 500 N
600 N/m
Member BC:
a + ©MC = 0;
400 N/m
- 1200 (1.5) - 900 (1) + Bx(3) - 500 (3) = 0
Bx = 1400 N
+ c ©Fy = 0;
Ay - 750 + 500 = 0
Ay = 250 N
Ans.
Member AB:
+ ©F = 0;
:
x
- Ax + 1400 = 0
Ax = 1400 N = 1.40 kN
Ans.
Member BC:
+ ©F = 0;
:
x
Cx + 900 - 1400 = 0
Cx = 500 N
+ c ©Fy = 0;
Ans.
- 500 - 1200 + Cy = 0
Cy = 1700 N = 1.70 kN
Ans.
6–122. The clamping hooks are used to lift the uniform
smooth 500-kg plate. Determine the resultant compressive
force that the hook exerts on the plate at A and B, and the
pin reaction at C.
P
P
80 mm
150 mm
C
A
B
.
.
.
P
6–123.
The spring has an unstretched length of 0.3 m. Determine
the mass m of each uniform link if the angle u = 20° for
equilibrium.
0.1 m
B
0.6 m
A
k
SOLUTION
y
= sin 20°
2(0.6)
D
E
y = 1.2 sin 20°
Fs = (1.2 sin 20° - 0.3)(400) = 44.1697 N
a + ©MA = 0;
Ex (1.4 sin 20°) - 2(mg)(0.35 cos 20°) = 0
Ex = 1.37374(mg)
a + ©MC = 0;
1.37374mg(0.7 sin 20°) + mg(0.35 cos 20°) - 44.1697(0.6 cos 20°) = 0
mg = 37.860
m = 37.860/9.81 = 3.86 kg
Ans.
400 N/ m
u
u
C
6–124.
Determine the horizontal and vertical components of force
that the pins A and B exert on the two-member frame. Set
F = 0.
1m
F
C
1.5 m
1m
SOLUTION
B
CB is a two-force member.
60
400 N/ m
Member AC:
a + ©MA = 0;
- 600 (0.75) + 1.5 (FCB sin 75°) = 0
FCB = 310.6
Thus,
Bx = By = 310.6 a
+ ©F = 0;
:
x
1
22
b = 220 N
- Ax + 600 sin 60° - 310.6 cos 45° = 0
Ax = 300 N
+ c ©Fy = 0;
Ans.
Ans.
Ay - 600 cos 60° + 310.6 sin 45° = 0
Ay = 80.4 N
Ans.
A
6–125.
Determine the horizontal and vertical components of force
that pins A and B exert on the two-member frame. Set
F = 500 N.
1m
F
C
1.5 m
1m
B
SOLUTION
60
400 N/ m
Member AC:
a + ©MA = 0;
- 600 (0.75) - Cy (1.5 cos 60°) + Cx (1.5 sin 60°) = 0
Member CB:
a + ©MB = 0;
- Cx (1) - Cy (1) + 500 (1) = 0
Solving,
Cx = 402.6 N
Cy = 97.4 N
Member AC:
+ ©F = 0;
:
x
- Ax + 600 sin 60° - 402.6 = 0
Ax = 117 N
+ c ©Fy = 0;
Ans.
Ay - 600 cos 60° - 97.4 = 0
Ay = 397 N
Ans.
Member CB:
+ ©F = 0;
:
x
402.6 - 500 + Bx = 0
Bx = 97.4 N
+ c ©Fy = 0;
Ans.
- By + 97.4 = 0
By = 97.4 N
Ans.
A
.
.
.
.
Cont’d
.
.
.
.
.
.
.