CHAPTER 763
PROBLEM 6.1
Using the method of joints, determine the force in each member of the truss shown.
State whether each member is in tension or compression.
SOLUTION
AB = 32 + 1.252 = 3.25 m
BC = 32 + 42 = 5 m
Reactions:
SM A = 0 : (84 kN )(3 m) - C (5.25 m) = 0
+
C = 48 kN ¨
+
SFx = 0 : Ax - C = 0
A x = 48 kN Æ
+
SFy = 0 : Ay = 84 kN = 0
A y = 84 kN ≠
Joint A:
+
+
SFx = 0 : 48 kN -
SFy = 0 : 84 kN -
12
FAB = 0
13
FAB = +52 kN
FAB = 52 kN T 
5
(52 kN ) - FAC = 0
13
FAC = +64.0 kN FAC = 64.0 kN T 
Joint C:
FBC 48 kN
=
5
3
FBC = 80.0 kN C 
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3
PROBLEM 6.2
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
+
Free body: Entire truss
+
SFx = 0 : Bx = 0
SM B = 0 : C (5.25 m) - (5 kN )( 4 m) = 0
C=
+
SFy = 0 : By +
80
kN ≠
21
80
kN - 5 kN = 0
21
25
By =
kN ≠
21
Free body: Joint B:
FAB FBC 25 / 21
25
=
=
kN =
5
4
3
63
125
FAB =
kN = 1.9841 kN 68
FBC =
100
kN = 1.5873 kN 63
FAB = 1.984 kN C 
FBC = 1.587 kN T 
Free body: Joint C:
FAC FBC 80 / 21
80
=
=
kN =
kN
3.25 1.25
3
63
100
FAC =
kN = 1.5873 kN
63
FBC =
100
kN T
63
FAC = 1.587 kN C 
(Checks)
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4
PROBLEM 6.3
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss
+
+
SFx = 0 : C x = 0 C x = 0
SM B = 0 : (1.92 kN )(3 m) + C y ( 4.5 m) = 0
C y = -1.28 kN
+
C y = 1.28 kNØ
SFy = 0 : B - 1.92 kN - 1.28 kN = 0
B = 3.20 kN ≠
Free body: Joint B:
FAB FBC 3.20 kN
=
=
5
3
4
FAB = 4.00 kN C 
FBC = 2.40 kN C 
Free body: Joint C:
+
SFx = 0 : -
7.5
FAC + 2.40 kN = 0
8.5
FAC = +2.72 kN +
SFy =
FAC = 2.72 kN T 
4
(2.72 kN ) - 1.28 kN = 0 (Checks)
8.5
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5
PROBLEM 6.4
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.
SOLUTION
Reactions:
SM D = 0 : Fy (6) - (20 + 12)(3) - (5)(6) = 0
Fy = 21 kN ≠
+
SFx = 0 : Fx = 0
+
SFy = 0 : D - (5 + 20 + 5 + 12) + (21) = 0
D = 21 kN ≠
Joint A:
FAB = 0 
SFx = 0 : FAB = 0 +
SFy = 0 : - 5 - FAD = 0
FAD = -5 kN Joint D:
+
SFy = 0 : - 5 + 21 +
+
SFx = 0 :
8
FBD = 0
17
FBD = -34 kN FAD = 5.00 kN C 
FBD = 34.0 kN C 
15
( -34) + FDE = 0
17
FDE = +30 kN FDE = 30.0 kN T 
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6
PROBLEM 6.4 (Continued)
Joint E:
+
SFy = 0 : FBE - 12 = 0
FBE = +12 kN FBE = 12.00 kN T 
Truss and loading symmetrical about cL
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7
PROBLEM 6.5
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
Free body: Truss
+
+
SFx = 0 : A x = 0
SM A = 0 : D(8) - (50 kN )(8) - (50 kN )(20) = 0
D = 175 kN ≠
SFy = 0 : A y = 75 kNØ
Free body: Joint A:
FAB FAD 75 kN
=
=
2
1
5
FAB = 150.0 kN T 
FAD = 167.7 kN C 
Free body: Joint B:
SFx = 0:
FBC = 150.0 kN T 
SFy = 0:
FBD = 50 kN C 
Free body: Joint C:
FCD
10
=
FBC 50 kN
=
3
1
FBC = 150 kN T
FCD = 158.1 kN C 
(Checks)
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8
PROBLEM 6.6
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free Body: Truss
+
SM E = 0 : F (3 m) - (900 N )(2.25 m) - (900 N )( 4.5 m) = 0
F = 2025 N ≠
+
SFx = 0 : E x + 900 N + 900 N = 0
+
E x = -1800 N E x = 1800 N ¨
SFy = 0 : E y + 2025 N = 0
E y = -2025 N E y = 2025 N Ø
FAB = FBD = 0 
We note that AB and BD are zero-force members:
Free body: Joint A:
FAC FAD 900 N
=
=
2.25 3.75
3
FAC = 675 N T 
FAD = 1125 N C 
Free body: Joint D:
FCD FDE 1125 N
=
=
3
2.23
3.75
FCD = 900 N T 
FDF = 675 N C 
+
SFx = 0 : FEF - 1800 N = 0 FEF = 1800 N T 
+
Free body: Joint E:
SFy = 0 : FCE - 2025 N = 0
FCE = 2025 N T 
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9
PROBLEM 6.6 (Continued)
+
Free body: Joint F:
SFy = 0 :
2.25
FCF + 2025 N - 675 N = 0
3.75
FCF = -2250 N
+
SFx = -
FCF = 2250 N C 
3
( -2250 N ) - 1800 N = 0 (Checks)
3.75
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10
PROBLEM 6.7
Using the method of joints, determine the force in each member of the truss shown.
State whether each member is in tension or compression.
SOLUTION
+
Free body: Truss
+
SFy = 0 : B y = 0
SM B = 0 : D( 4.5 m) + (8.4 kN )( 4.5 m) = 0
D = -8.4 kN D = 8.4 kN ¨
+
SFx = 0 : Bx - 8.4 kN - 8.4 kN - 8.4 kN = 0
Bx = +25.2 kN B x = 25.2 kN Æ
Free body: Joint A:
FAB FAC 8.4 kN
=
=
5.3
4.5
2.8
FAB = 15.90 kN C 
FAC = 13.50 kN T 
+
Free body: Joint C:
SFy = 0 : 13.50 kN -
4.5
FCD = 0
5.3
FCD = +15.90 kN +
SFx = 0 : - FBC - 8.4 kN -
FCD = 15.90 kN T 
2.8
(15.90 kN ) = 0
5.3
FBC = -16.80 kN FBC = 16.80 kN C 
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11
PROBLEM 6.7 (Continued)
Free body: Joint D:
FBD 8.4 kN
=
4.5
2.8
FBD = 13.50 kN C 
We can also write the proportion
FBD 15.90 kN
=
4.5
5.3
FBD = 13.50 kN C (Checks)
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12
PROBLEM 6.8
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.
SOLUTION
AD = (2.5)2 + (6)2 = 6.5 m
BCD = (6)2 + (8)2 = 10 m
SFx = 0 : Dx = 0
Reactions:
SM E = 0 : D y (10.5 m) - (3 kN )(2.5 m) = 0 fi D y =
+
+
+
+
Joint D:
5
kN - 3 kN + E = 0 7
16
E=
kN
7
5
4
SFx = 0 :
FAD + FDC = 0 13
5
SFy = 0 :
SFy = 0 :
12
3
5
FAD + FDC + kN = 0
13
5
7
5
kN
7
D y = 0.714 kN ≠
E = 2.286 kN ≠
(1)
(2)
Solving (1) and (2), simultaneously:
260
kN 231
125
=+
kN 231
FAD = -
FAD = 1.126 kN C 
FDC
FDC = 0.541 kN T 
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13
PROBLEM 6.8 (Continued)
+
SFx = 0 :
5
4
FBE + FCE = 0 13
5
(3)
+
Joint E:
SFy = 0 :
12
3
16
FBE + FCE + kN = 0
13
5
7
(4)
Solving (3) and (4), simultaneously:
FBE = -
832
kN
231
FBE = 3.60 kN C 
FCE = +
400
kN 231
FCE = 1.732 kN T 
Joint C:
Force polygon is a parallelogram (see Fig. 6.11 p. 209)
FAC = 1.732 kN T 
FBC = 0.541 kN T 
Joint A:
+
SFx = 0 :
5 Ê 260
ˆ 4 Ê 400
ˆ
kN ˜ + Á
kN ˜ + FAB = 0
ÁË
¯ 5 Ë 231
¯
13 231
FAB = -
+
SFy = 0 :
20
kN FAB = 1.818 kN C 
11
12 Ê 260
ˆ 3 Ê 400
ˆ
kN ˜ - Á
kN ˜ = 0
ÁË
¯
Ë
¯
13 231
5 231
0 = 0 (Checks)
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14
PROBLEM 6.9
Determine the force in each member of the Pratt roof
truss shown. State whether each member is in tension
or compression.
SOLUTION
Free body: Truss
Due to symmetry of truss and load
SFx = 0 : Ax = 0
�� Ay = H =
1
total load = 21 kN ≠
2
Free body: Joint A:
FAB FAC 15.3 kN
=
=
37
35
12
FAB = 47.175 kN
FAC = 44.625 kN FAB = 47.2 kN C 
FAC = 44.6 kN T 
Free body: Joint B:
From force polygon:
FBD = 47.175 kN, FBC = 10.5 kN FBC = 10.50 kN C 
FBD = 47.2 kN
C
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15
PROBLEM 6.9 (Continued)
3
FCD - 10.5 = 0
5
SFy = 0 :
+
4
SFx = 0 : FCE + (17.50) - 44.625 = 0
5
FCE = 30.625 kN
Free body: Joint E:
FCD = 17.50 kN T 
+
Free body: Joint C:
DE is a zero-force member
FCE = 30.6 kN T 
FDE = 0 
Truss and loading symmetrical about cL
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16
PROBLEM 6.10
Determine the force in each member of the fan roof truss shown.
State whether each member is in tension or compression.
SOLUTION
Free body: Truss
SFx = 0 : A x = 0
From symmetry of truss and loading:
Ay = I =
1
2
Total load
A y = I = 6 kN ≠
Free body: Joint A:
F
FAB
5 kN
= AC =
9.849
9
4
FAB = 12.31 kN C 
FAC = 11.25 kN T 
Free body: Joint B:
SFx =
+
9
(12.31 kN + FBD + FDC ) = 0
9.849
FBD + FBC = -12.31 kN +
or
SFy =
(1)
4
(12.31 kN + FBD - FBC ) - 2 kN = 0
9.849
FBD - FBC = -7.386 kN (2)
Add (1) and (2):
2 FBD = -19.70 kN FBD = 9.85 kN C 
Subtract (2) from (1):
2 FBC = -4.924 kN FBC = 2.46 kN C 
or
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17
PROBLEM 6.10 (Continued)
Free body: Joint D:
From force polygon:
FCD = 2.00 kN C 
FDE = 9.85 kN C 
+
Free body: Joint C:
+
SFy =
4
4
FCE (2.46 kN ) - 2 kN = 0
5
9.849
FCE = 3.75 kN T 
3
9
SFx = 0 : FCG + (3.75 kN ) +
(2.46 kN ) - 11.25 kN = 0
5
9.849
FCG = +6.75 kN FCG = 6.75 kN T 
From the symmetry of the truss and loading:
FEF = FDE FEF = 9.85 kN C 
FEG = FCE FEG = 3.75 kN T 
FFG = FCD FFG = 2.00 kN C 
FFH = FBD FFH = 9.85 kN C 
FGH = FBC FGH = 2.46 kN C 
FGI = FAC FGI = 11.25 kN T 
FHI = FAB FHI = 12.31 kN C 
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18
PROBLEM 6.11
Determine the force in each member of the Howe roof
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss
SFx = 0 : H x = 0
Because of the symmetry of the truss and loading:
A = Hy =
1
Total load
2
A = H y = 6 kN ≠
Free body: Joint A:
FAB FAC 4.5 kN
=
=
5
4
3
FAB = 7.50 kN C 
FAC = 6.00 kN T 
Free body: Joint C:
BC is a zero-force member
FBC = 0 FCE = 6.00 kN T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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19
PROBLEM 6.11 (Continued)
Free body: Joint B:
SFx = 0 :
+
FBD + FBE = -7.5 kN +
or
4
4
4
FBD + FBC + (7.5 kN ) = 0
5
5
5
SFy = 0 :
(1)
3
3
3
FBD - FBE + (7.5 kN ) - 3 kN = 0
5
5
5
FBD - FBE = -2.5 kN (2)
Add Eqs. (1) and (2):
2 FBD = -10 kN FBD = 5.00 kN C 
Subtract (2) from (1):
2 FBE = -5 kN
FBE = 2.50 kN C 
or
Free Body: Joint D:
+
FDF
+
4
4
(5 kN ) + FDF = 0
5
5
= -5 kN FDF = 5.00 kN C 
3
3
(5 kN ) - ( -5 kN ) - 3 kN - FDE = 0
5
5
= + 3 kN FDE = 3.00 kN T 
SFx = 0 :
SFy = 0 :
FDE
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE FEF = 2.50 kN C 
FEG = FCE FEG = 6.00 kN T 
FFG = FBC FFG = 0 
FFH = FAB FFH = 7.50 kN C 
FGH = FAC FGH = 6.00 kN T 
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20
PROBLEM 6.12
Determine the force in each member of the Gambrel roof truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Truss
SFx = 0 : H x = 0
Because of the symmetry of the truss and loading
A = Hy =
1
2
Total load
A = H y = 6 kN ≠
Free body: Joint A:
FAB FAC 4.5 kN
=
=
5
4
3
FAB = 7.50 kN C 
FAC = 6.00 kN T 
Free body: Joint C:
BC is a zero-force member
FCE = 6.00 kN T 
FBC = 0 Free body: Joint B:
+
SFx = 0 :
0.9701FBD + 0.8FBE = -6 kN or
+
or
4
4
4
FBD + FBE + (7.5 kN ) = 0
5
5
17
SFy = 0 :
(1)
1
3
3
FBD - FBE + (7.5 kN ) - 3 kN = 0
5
5
17
0.2425FBD - 0.6 FBE = -1.5 kN
(2)
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21
PROBLEM 6.12 (Continued)
Solving (1) and (2)
FBD = -6.185 kN FBD = 6.19 kN C 
FBE = 1.9328 ¥ 10 -4 kN FBE ª 0 kN C 
Free body: Joint D:
+
SFx = 0 :
4
4
(6.185 kN ) +
FDF = 0
17
17
FDF = -6.185 kN FDF = 6.19 kN C 
+
SFy = 0 :
1
1
(6.185 kN ) ( -6.185 kN ) - 3 kN - FDE = 0
17
17
FDE = 1.6568 ¥ 10 -4 kN FDE ª 0 kN T 
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE FEF = 0 kN C 
FEG = FCE FEG = 6.00 kN T 
FFG = FBC FFG = 0 
FFH = FAB FFH = 7.50 kN C 
FGH = FAC FGH = 6.00 kN T 
Note: Compare results with those of Problem 6.9.
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22
PROBLEM 6.13
Determine the force in each member of the truss shown.
SOLUTION
12.5 kN FCD FDG
=
=
2.5
6
6.5
Joint D:
FCD = 30 kN T 
FDG = 32.5 kN C 
Joint G:
SF = 0 : FCG = 0

SF = 0 : FFG = 32.5 kN C 
BF
2
BF = (2.5 m) = 1.6667 m b = –BCF = tan -1
= 39.81∞
3
2
Joint C:
+
SFy = 0 : - 12.5 kN - FCF sin b = 0
-12.5 kN - FCF sin 39.81∞ = 0
FCF = -19.526 kN FCF = 19.53 kN C 
+
SFx = 0 : 30 kN - FBC - FCF cos b = 0
30 kN - FBC - ( -19.526 kN ) cos 39.81∞ = 0
FBC = +45.0 kN +
Joint F:
SFx = 0 : -
6
6
FEF (32.5 kN ) - FCF cos b = 0
6.5
6.5
Ê 6.5 ˆ
FEF = -32.5 kN - Á ˜ (19.526 kN ) cos 39.81∞
Ë 6 ¯
FEF = -48.75 kN +
FBC = 45.0 kN T 
SFy = 0 : FBF -
FEF = 48.8 kN C 
2.5
2.5
FEF (32.5 kN ) - (19.526 kN )sin 39.81∞ = 0
6.5
6.5
2.5
FBF ( -48.75 kN ) - 12.5 kN - 12.5 kN = 0
6.5
FBF = +6.25 kN
FBF = 6.25 kN T 
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23
PROBLEM 6.13 (Continued)
tan a =
+
Joint B:
2.5 m
; g = 51.34∞
2m
SFy = 0 : - 12.5 kN - 6.25 kN - FBE sin 51.34∞ = 0
FBE = -24.0 kN +
FBE = 24.0 kN C 
SFx = 0 : 45.0 kN - FAB + (24.0 kN ) cos 51.34∞ = 0
FAB = +60 kN FAB = 60.0 kN T 
Joint E:
g = 51.34∞
+
SFy = 0 : FAE - (24 kN )sin 51.34∞ - ( 48.75 kN )
FAE = +37.5 kN
2.5
=0
6.5
FAE = 37.5 kN T 
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24
PROBLEM 6.14
Determine the force in each member of the roof truss shown.
State whether each member is in tension or compression.
SOLUTION
Free body: Truss
SFx = 0 : A x = 0
From symmetry of loading:
Ay = E =
1
Total load
2
A y = E = 3.6 kN ≠
We note that DF is a zero-force member and that EF is aligned with the load. Thus
FDF = 0 
FEF = 1.2 kN C 
Free body: Joint A:
FAB FAC 2.4 kN
=
=
13
12
5
FAB = 6.24 kN C 
FAC = 2.76 kN T 
Free body: Joint B:
+
+
SFx = 0 :
3
12
12
FBC + FBD + (6.24 kN ) = 0
3.905
13
13
SFy = 0 :
-
2.5
5
5
FBC + FBD + (6.24 kN ) - 2.4 kN = 0
3.905
13
13
(1)
(2)
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25
PROBLEM 6.14 (Continued)
Multiply (1) by 2.5, (2) by 3, and add:
45
45
FBD + (6.24 kN ) - 7.2 kN = 0, FBD = -4.16 kN,
13
13
FBD = 4.16 kN C 
Multiply (1) by 5, (2) by –12, and add:
45
FBC + 28.8 kN = 0, FBC = -2.50 kN, 3.905
FBC = 2.50 kN C 
+
Free body: Joint C:
SFy = 0 :
5
2.5
FCD (2.50 kN ) = 0
5.831
3.905
FCD = 1.867 kN T 
+
SFx = 0 : FCE - 5.76 kN +
3
3
(2.50 kN ) +
(1.867 kN ) = 0
3.905
5.831
FCE = 2.88 kN T
+
Free body: Joint E:
+
SFy = 0 :
5
FDE + 3.6 kN - 1.2 kN = 0
7.81
FDE = -3.75 kN
SFx = 0 : - FCE -
FDE = 3.75 kN C 
6
( -3.75 kN ) = 0
7.81
FCE = +2.88 kN
FCE = 2.88 kN T
(Checks)
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26
PROBLEM 6.15
Determine the force in each member of the Warren bridge truss
shown. State whether each member is in tension or compression.
SOLUTION
SFx = 0 : Ax = 0
Free body: Truss
Due to symmetry of truss and loading
Ay = G =
1
Total load = 30 kN ≠
2
FAB FAC 30
=
=
kN
5
3
4
Free body: Joint A:
FBC FBD 37.5
=
=
kN
5
6
5
Free body: Joint B:
FAB = 37.5 kN C 
FAC = 22.5 kN T 
FBC = 37.5 kN T 
FBD = 45 kN C 
+
Free body: Joint C:
+
+
SFy = 0 :
4
4
(37.5) + FCD - 30 = 0
5
5
FCD = 0 
3
SFx = 0 : FCE - 22.5 - (37.5) = 0
5
FCE = +45 kN FCE = 45 kN T 
Truss and loading symmetrical about cL
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27
PROBLEM 6.16
Solve Problem 6.15 assuming that the load applied at E has
been removed.
PROBLEM 6.15 Determine the force in each member of the
Warren bridge truss shown. State whether each member is in
tension or compression.
SOLUTION
SFx = 0 : Ax = 0
Free body: Truss
SM G = 0 : 30(12) - Ay (18) = 0 A y = 20 kN ≠
+
+
SFy = 0 : 20 - 30 + G = 0 G = 10 kN ≠
FAB FAC 20 kN
=
=
5
3
4
Free body: Joint A:
FAB = 25.0 kN C 
FAC = 15.00 kN T 
FBC FBD 25 kN
=
=
5
6
5
Free body Joint B:
FBC = 25.0 kN T 
FBD = 30.0 kN C 
Free body Joint C:
4
4
(25) + FCD - 30 = 0 5
5
+
SFy = 0 :
+
3
3
SFx = 0 : FCE + (12.5) - (25) - 15 = 0 5
5
FCD = 12.50 kN T 
FCE = 22.5 kN T 
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28
PROBLEM 6.16 (Continued)
+
Free body: Joint D:
4
4
SFy = 0 : - (12.5) - FDE = 0
5
5
FDE = -12.5 kN FDE = 12.50 kN C 
+
Free body: Joint F:
3
3
SFx = 0 : FDF + 30 - (12.5) - (12.5) = 0
5
5
FDF = -15 kN FDF = 15.00 kN C 
FEF FFG 15 kN
=
=
5
5
6
FEF = 12.50 kN T 
FFG = 12.50 kN C 
Free body: Joint G:
FEG 10 kN
=
3
4
Also:
FFG 10 kN
=
5
4
FEG = 7.50 kN T 
FFG = 12.50 kN C (Checks)
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29
PROBLEM 6.17
Determine the force in member DE and in each of the members
located to the left of DE for the inverted Howe roof truss shown.
State whether each member is in tension or compression.
SOLUTION
Free body: Truss:
SFx = 0 : A x = 0
+
SM H = 0 : (2 kN )( 4d ) + ( 4 kN )(3d ) + ( 4 kN )(2d ) + ( 4 kN )d - Ay ( 4 d ) = 0
Ay = 8 kN ≠
Angles:
3
a = 36.87∞
4
3
tan b =
b = 18.43∞
9
tan a =
Free body: Joint A:
FAC
FAB
6 kN
=
=
sin 53.13∞ sin 108.44∞ sin 18.43∞
FAB = 15.2166 kN C
FAC = 18.004 kN T FAB = 15.22 kN C
FAC = 18.00 kN T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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30
PROBLEM 6.17 (Continued)
Free body: Joint B:
+
SFy = 0 : - FBC - ( 4 kN ) cos18.43∞ = 0
FBC = -3.7948 kN
+
FBC = 3.79 kN C 
SFx = 0 : FBD + 15.2166 kN + (4 kN) sin18.43∞ = 0
FBD = -16.4812 kN FBD = 16.48 kN C 
+
Free body: Joint C:
SFy = 0 : - FCE sin 36.87∞ + (18.004 kN )sin 36.87∞ - (3.7948 kN ) cos18.43∞ = 0
FCE = 12.0037 kN +
FCE = 12.00 kN T 
SFx = 0 : FCD - (18.004 kN) cos 36.87∞ + (12.0037 kN) cos 36.87∞ - (3.7948 kN)sin 18.43∞ = 0
FCD = 5.9999 kN FCD = 6.00 kN T 
Free body: Joint E:
12.0037 kN
FDE
=
sin 36.87∞
sin71.57∞
FDE = 7.59 kN C 
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31
PROBLEM 6.18
Determine the force in each of the members located to the right
of DE for the inverted Howe roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss
+
SM A = 0 : H ( 4d ) - ( 4 kN )d - ( 4 kN )(2d ) - ( 4 kN )(3d ) - (2 kN )( 4 d ) = 0
H = 8 kN ≠
Angles:
3
a = 36.87∞
4
3
tan b =
b = 18.43∞
9
tan a =
Free body: Joint H:
FGH = (6 kN ) cot 18.43∞
FGH = 18.0052 kN T FFH =
6 kN
= 18.9786 kN sin 18.43∞
FGH = 18.00 kN T 
FFH = 18.98 kN C 
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32
PROBLEM 6.18 (Continued)
Free body Joint F:
+
SFy = 0 : - FFG - ( 4 kN ) cos18.43∞ = 0
+
FFG = -3.7948 kN FFG = 3.79 kN C 
SFx = 0 : - FDF - 18.9786 kN + (4 kN) sin 18.43∞ = 0
FDF = -17.7140 kN
FDF = 17.71 kN C 
+
Free body: Joint G:
SFy = 0 : FDG sin 36.87∞ - (3.7948 kN ) cos18.43∞ = 0
FDG = +6.000 kN
+
FDG = 6.00 kN T 
SFx = 0 : - FEG + 18.0052 kN - (3.7948 kN )sin 18.43∞ - (6.00 kN ) cos 36.87∞ = 0
FEG = +12.0054 kN FEG = 12.01 kN T 
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33
PROBLEM 6.19
Determine the force in each of the members located to the left
of FG for the scissors roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free Body: Truss
SFx = 0 : A x = 0
+
SM L = 0 : (1 kN )(12 m) + (2 kN )(10 m) + (2 kN )(8 m) + (1 kN )(6 m) - Ay (12 m) = 0
A y = 4.50 kN ≠
FBC = 0 
We note that BC is a zero-force member:
FCE = FAC Free body: Joint A:
(1)
+
SFx = 0 :
1
2
FAB +
FAC = 0
2
5
(2)
+
Also:
SFy = 0 :
1
1
FAB +
FAC + 3.50 kN = 0 2
5
(3)
Multiply (3) by –2 and add (2):
-
1
FAB - 7 kN = 0 2
FAB = 9.90 kN C 
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34
PROBLEM 6.19 (Continued)
Subtract (3) from (2):
1
FAC - 3.50 kN = 0 FAC = 7.826 kN 5
FCE = FAC = 7.826 kN From (1):
FAC = 7.83 kN T 
FCE = 7.83 kN T 
+
Free body: Joint B:
SFy = 0 :
+
SFx = 0 : FBE +
Free body: Joint E:
+
+
1
1
FBD +
(9.90 kN ) - 2 kN = 0
2
2
FBD = -7.071 kN FBD = 7.07 kN C 
1
(9.90 - 7.071)kN = 0
2
FBE = -2.000 kN FBE = 2.00 kN C 
2
( FEG - 7.826 kN ) + 2.00 kN = 0
5
FEG = 5.590 kN FEG = 5.59 kN T 
1
SFy = 0 : FDE (7.826 - 5.590)kN = 0
5
SFx = 0 :
FDE = 1.000 kN FDE = 1.000 kN T 
Free body: Joint D:
+
SFx = 0 :
FDF + FDG = -5.590 kN +
or
or
2
1
( FDF + FDG ) +
(7.071 kN )
5
2
SFy = 0 :
(4)
1
1
( FDF - FDG ) +
(7.071 kN ) = 2 kN - 1 kN = 0
5
2
FDE - FDG = -4.472 (5)
Add (4) and (5):
2 FDF = -10.062 kN FDF = 5.03 kN C 
Subtract (5) from (4):
2 FDG = -1.1180 kN FDG = 0.559 kN C 
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35
PROBLEM 6.20
Determine the force in member FG and in each of the members
located to the right of FG for the scissors roof truss shown.
State whether each member is in tension or compression.
SOLUTION
Free body: Truss
+
SM A = 0 : L(12 m) - (2 kN )(2 m) - (2 kN )( 4 m) - (1 kN )(6 m) = 0
L = 1.500 kN ≠
Angles:
tan a = 1 a = 45∞
1
tan b =
b = 26.57∞
2
Zero-force members:
Examining successively joints K, J, and I, we note that the following members to the right of FG are zero-force
members: JK, IJ, and HI.
FHI = FIJ = FJK = 0 
Thus:
We also note that
and
FGI = FIK = FKL (1)
FHJ = FJL (2)
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36
PROBLEM 6.20 (Continued)
Free body: Joint L:
FJL
1.500 kN
F
= KL =
sin 116.57∞ sin 45∞ sin18.43∞
FJL = 4.2436 kN FJL = 4.24 kN C 
FKL = 3.35 kN T 
From Eq. (1):
FGI = FIK = FKL FGI = FIK = 3.35 kN T 
From Eq. (2):
FHJ = FJL = 4.2436 kN FHJ = 4.24 kN T 
FGH
FFH
4.2436
=
=
sin 108.43∞ sin 18.43∞ sin 53.14∞
FFH = 5.03 kN C 
Free body: Joint H:
FGH = 1.677 kN T 
Free body: Joint F:
+
SFx = 0 : - FDF cos 26.57∞ - (5.03 kN ) cos 26.57∞ = 0
+
FDF = -5.03 kN
SFy = 0 : - FFG - 1 kN + (5.03 kN )sin 26.57∞ - ( -5.03 kN )sin 26.57∞ = 0
FFG = 3.500 kN FFG = 3.50 kN T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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37
PROBLEM 6.21
Determine the force in each of the members located to the left
of line FGH for the studio roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss
SFx = 0 : A x = 0
Because of symmetry of loading:
Ay = L =
1
Total load
2
A y = L = 6 kN ≠
Zero-Force Members. Examining joints C and H, we conclude that
BC, EH, and GH are zero-force members. Thus
FBC = FEH = 0
Also,
FCE = FAC Free body: Joint A
FAB

(1)
FAC 5 kN
=
2
1
5
FAB = 11.1803 kN C
=
FAB = 11.18 kN C 
FAC = 10.00 kN T 
From Eq. (1):
FCE = 10.00 kN T 
Free body: Joint B
+
SFx = 0 :
2
2
2
FBD +
FBE +
(11.1803 kN ) = 0
5
5
5
FBD + FBE = -11.1803 kN or
+
SFy = 0 :
1
1
1
FBD FBE +
(11.1803 kN) - 2 kN = 0
5
5
5
FBD - FBE = -6.7082 kN
or
(2)
(3)
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38
PROBLEM 6.21 (Continued)
2 FBD = -17.8885 kN
FBD = 8.94425 kN
Add (2) and (3):
Subtract (3) from (1):
2 FBE = - 4.4721 kN
FBE = -2.236 kN
FBD = 8.94 kN C 
FBE = 2.24 kN C 
Free body: Joint E
+
2
2
FEG +
(2.236 kN) - 10 kN = 0
5
5
SFx = 0 :
FEG = 8.94 kN T 
+
SFy = 0 : FDE +
1
1
(8.944 kN) (2.236 kN) = 0
5
5
FDE = -3.00 kN FDE = 3.00 kN C 
Free body: Joint D
+
SFx = 0 :
2
2
2
FDF +
FDG +
(8.944 kN) = 0
5
5
5
FDF + FDG = -8.944 kN or
+
SFy = 0 :
(4)
1
1
1
FDF FDG +
(8.944 kN)
5
5
5
+ 3 kN - 2 kN = 0
FDF - FDG = -11.1803 kN or
(5)
Add (4) and (5):
2 FDF = - 20.1243 kN FDF = 10.06 kN C 
Subtract (5) from (4):
2 FDG = 2.2363 kN FDG = 1.118 kN T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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39
PROBLEM 6.22
Determine the force in member FG and in each of the members
located to the right of FG for the studio roof truss shown. State
whether each member is in tension or compression.
SOLUTION
Reaction at L: Because of the symmetry of the loading,
1
L = Total load, L = 6 kN
2
(See F.B.D. to the left for more details)
Free body: Joint L
3
= 26.57∞
6
1
b = tan -1 = 9.46∞
6
FJL
FKL
5 kN
=
=
sin 63.43∞ sin 99.46∞ sin17.11∞
FJL = 15.20 kN C
FJL = 15.20 kN T 
FKL = 16.7637 kN C FKL = 16.76 kN C 
a = tan -1
Free body: Joint K
+
SFx = 0 : -
FIK + FJK = -16.7637 kN
or:
+
or:
2
2
2
FIK FJK (16.7637 kN ) = 0
5
5
5
SFy = 0 :
1
1
1
FIK FJK +
(16.7637) - 2 = 0
5
5
5
FIK - FJK = -12.2916 kN
Add (1) and (2):
(2)
2 FIK = - 29.0553
FIK = -14.5277 kN Subtract (2) from (1):
(1)
FIK = 14.53 kN C 
2 FJK = - 4.4721
FJK = - 2.236 kN
FJK = 2.24 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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40
PROBLEM 6.22 (Continued)
Free body: Joint J
+
2
6
6
2
FIJ FGJ +
(15.20 kN ) (2.236) = 0 13
37
37
5
SFx = 0 : -
3
1
1
1
FIJ +
FGJ (15.20 kN ) (2.236) = 0 13
37
37
5
+ SFy = 0 :
(3)
(4)
Solving (3) and (4) we obtain:
FIJ = 1.8027 kN FIJ = 1.803 kN T 
FGJ = 12.1587 kN FGJ = 12.16 kN T 
Free body: Joint I
+
2
2
2
2
FFI FGI (14.5277) +
(1.8027) = 0
5
5
5
13
SFx = 0 : -
FFI + FGI = -13.4097 kN or
+ SFy = 0 :
1
1
1
3
FFI FGI +
(14.5277) (1.8027) - 2 = 0
5
5
5
13
FFI - FGI = -6.7016 kN or
(5)
Add (5) and (6):
(6)
2 FFI = -20.1113
FFI = -10.0557 kN
Subtract (6) from (5):
2 FGI = -6.7081 kN FFI = 10.06 kN C 
FGI = 3.35 kN C 
Free body: Joint F
SFx = 0 : FDF = FFI = 10.0557 kN C
From
+
Ê 1 ˆ
SFy = 0 : FFG - 2 kN + 2 Á ˜ = 0
Ë 5¯
FFG = + 7.00 kN
FFG = 7.00 kN T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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41
PROBLEM 6.23
Determine the force in each of the members connecting joints A
through F of the vaulted roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss
SFx = 0 : A x = 0
+
SM K = 0 : (1.2 kN )6a + (2.4 kN )5a + (2.4 kN)4 a + (1.2 kN )3a
- Ay (6a) = 0 A y = 5.40 kN ≠
Free body: Joint A
FAC 4.20 kN
=
2
1
5
FAB = 9.3915 kN
FAB
=
FAB = 9.39 kN C 
FAC = 8.40 kN T 
Free body: Joint B
+
SFx = 0 :
+
SFy = 0 :
2
1
2
FBD +
FBC +
(9.3915) = 0 (1)
5
2
5
1
1
1
FBD FBC +
(9.3915) - 2.4 = 0 (2)
5
2
5
Add (1) and (2):
3
3
FBD +
(9.3915 kN ) - 2.4 kN = 0
5
5
FBD = -7.6026 kN
Multiply (2) by –2 and add (1):
3
FB + 4.8 kN = 0
2
FBC = -2.2627 kN
FBD = 7.60 kN C 
FBC = 2.26 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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42
PROBLEM 6.23 (Continued)
Free body: Joint C
+
SFx = 0 :
1
4
1
FCD +
FCE +
(2.2627) - 8.40 = 0 5
17
2
(3)
+
SFy = 0 :
2
1
1
FCD +
FCE (2.2627) = 0 5
17
2
(4)
Multiply (4) by - 4 and add (1):
-
7
5
FCD +
(2.2627) - 8.40 = 0
5
2
FCD = - 0.1278 kN FCD = 0.128 kN C 
Multiply (1) by 2 and subtract (2):
7
3
FCE +
(2.2627) - 2(8.40) = 0
17
2
FCE = 7.068 kN
FCE = 7.07 kN T 
Free body: Joint D
+
+
SFx = 0 :
SFy = 0 :
2
1
2
FDF +
FDE +
(7.6026)
1.524
5
5
1
+
(0.1278) = 0 5
(5)
1
1.15
1
FDF FDE +
(7.6026)
1.524
5
5
2
+
(0.1278) - 2.4 = 0 5
(6)
Multiply (5) by 1.15 and add (6):
3.30
3.30
3.15
FDF +
(7.6026) +
(0.1278) - 2.4 = 0
5
5
5
FDF = - 6.098 kN FDF = 6.10 kN C 
Multiply (6) by –2 and add (5):
3.30
3
FDE (0.1278) + 4.8 = 0
1.524
5
FDE = - 2.138 kN FDE = 2.14 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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43
PROBLEM 6.23 (Continued)
Free body: Joint E
+
SFx = 0 :
0.6
4
1
FEF +
( FEH - FCE ) +
(2.138) = 0 2.04
1.524
17
(7)
+
SFy = 0 :
1.95
1
1.15
FEF +
( FEH - FCE ) (2.138) = 0 2.04
1.524
17
(8)
Multiply (8) by 4 and subtract (7):
7.2
FEF - 7.856 kN = 0 2.04
FEF = 2.23 kN T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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44
PROBLEM 6.24
The portion of truss shown represents the upper part of a power
transmission line tower. For the given loading, determine the
force in each of the members located above HJ. State whether
each member is in tension or compression.
SOLUTION
Free body: Joint A
FAB FAC 1.2 kN
=
=
2.29 2.29
1.2
FAB = 2.29 kN T 
FAC = 2.29 kN C 
Free body: Joint F
FDF FEF 1.2 kN
=
=
2.29 2.29
2.1
FDF = 2.29 kN T 
FEF = 2.29 kN C 
Free body: Joint D
FBD FDE 2.29 kN
=
=
2.21 0.6
2.29
FBD = 2.21 kN T 
FDE = 0.600 kN C 
Free body: Joint B
+
SFx = 0 :
+
4
2.21
FBE + 2.21 kN (2.29 kN ) = 0
5
2.29
FBE = 0 
3
0.6
SFy = 0 : - FBC - (0) (2.29 kN ) = 0
5
2.29
FBC = - 0.600 kN FBC = 0.600 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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45
PROBLEM 6.24 (Continued)
Free body: Joint C
+
SFx = 0 : FCE +
2.21
(2.29 kN ) = 0
2.29
FCE = - 2.21 kN +
SFy = 0 : - FCH - 0.600 kN FCH = -1.200 kN FCE = 2.21 kN C 
0.6
(2.29 kN ) = 0
2.29
FCH = 1.200 kN C 
Free body: Joint E
+
SFx = 0 : 2.21 kN -
2.21
4
(2.29 kN ) - FEH = 0
2.29
5
FEH = 0 
+
0.6
(2.29 kN ) - 0 = 0
2.29
= -1.200 kN
FEJ = 1.200 kN C 
SFy = 0 : - FEJ - 0.600 kN FEJ
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46
PROBLEM 6.25
For the tower and loading of Problem 6.24 and knowing that
FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in member
HJ and in each of the members located between HJ and NO.
State whether each member is in tension or compression.
PROBLEM 6.24 The portion of truss shown represents the upper
part of a power transmission line tower. For the given loading,
determine the force in each of the members located above HJ.
State whether each member is in tension or compression.
SOLUTION
Free body: Joint G
FGH
F
1.2 kN
= GI =
3.03 3.03
1.2
FGH = 3.03 kN T 
FJL
1.2 kN
F
= KL =
3.03 3.03
1.2
FJL = 3.03 kN T 
Free body: Joint L
FGI = 3.03 kN C 
FKL = 3.03 kN C 
Free body: Joint J
+
SFx = 0 : - FHJ +
+
2.97
(3.03 kN ) = 0
3.03
FHJ = 2.97 kN T 
0.6
(3.03 kN ) = 0
3.03
= -1.800 kN FJK = 1.800 kN C 
Fy = 0 : - FJK - 1.2 kN FJK
Free body: Joint H
+
SFx = 0 :
+
4
2.97
FHK + 2.97 kN (3.03 kN ) = 0
5
3.03
FHK = 0 
0.6
3
(3.03) kN - (0) = 0
3.03
5
= -1.800 kN FHI = 1.800 kN C 
SFy = 0 : - FHI - 1.2 kN FHI
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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47
PROBLEM 6.25 (Continued)
Free body: Joint I
+
SFx = 0 : FIK +
2.97
(3.03 kN ) = 0
3.03
FIK = -2.97 kN +
FIK = 2.97 kN C 
0.6
(3.03 kN ) = 0
3.03
= -2.40 kN FIN = 2.40 kN C 
SFy = 0 : - FIN - 1.800 kN FIN
Free body: Joint K
+
SFx = 0 : -
4
2.97
FKN + 2.97 kN (3.03 kN ) = 0
5
3.03
FKN = 0 
+
SFy = 0 : - FKD -
0.6
3
(3.03 kN ) - 1.800 kN - (0) = 0
3.03
5
FKD = -2.40 kN
FKD = 2.40 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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48
PROBLEM 6.26
Solve Problem 6.24 assuming that the cables hanging from the
right side of the tower have fallen to the ground.
PROBLEM 6.24 The portion of truss shown represents the upper
part of a power transmission line tower. For the given loading,
determine the force in each of the members located above HJ.
State whether each member is in tension or compression.
SOLUTION
Zero-Force Members.
Considering joint F, we note that DF and EF are zero-force ­members:
FDF = FEF = 0 
Considering next joint D, we note that BD and DE are zero-force
­members:
FBD = FDE = 0 
Free body: Joint A
FAB FAC 1.2 kN
=
=
2.29 2.29
1.2
FAB = 2.29 kN T 
FAC = 2.29 kN C 
Free body: Joint B
+
SFx = 0 :
4
2.21
FBE (2.29 kN ) = 0
5
2.29
FBE = 2.7625 kN +
SFy = 0 : - FBC -
FBE = 2.76 kN T 
0.6
3
(2.29 kN ) - (2.7625 kN ) = 0
2.29
5
FBC = -2.2575 kN FBC = 2.26 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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49
PROBLEM 6.26 (Continued)
Free body: Joint C
+
SFx = 0 : FCE +
2.21
(2.29 kN ) = 0
2.29
FCE = 2.21 kN C 
+
0.6
(2.29 kN ) = 0
2.29
= -2.8575 kN FCH = 2.86 kN C 
SFy = 0 : - FCH - 2.2575 kN FCH
Free body: joint E
+
SFx = 0 : -
4
4
FEH - (2.7625 kN ) + 2.21 kN = 0
5
5
FEH = 0 
+
3
3
SFy = 0 : - FEJ + (2.7625 kN ) - (0) = 0
5
5
FEJ = +1.6575 kN FEJ = 1.658 kN T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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50
PROBLEM 6.27
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
Free body: Truss
SM F = 0 : G(10 m) - (75 kN )(8 m) - (200 kN )(7.5 m) = 0
+
G = 210 kN≠
+
SFx = 0 : Fx + 75 kN = 0
Fx = 75 kN ¨
+
SFy = 0 : Fy - 200 kN + 210 kN = 0
Fy = 10 kN Ø
Free body: Joint F
+
SFx = 0 :
1
FDF - 75 kN = 0
5
FDF = 167.705 kN +
SFy = 0 : FBF - 10 kN +
FDF = 167.7 kN T 
2
(167.705 kN ) = 0
5
FBF = -140.00 kN FBF = 140.0 kN C 
Free body: Joint B
+
+
SFx = 0 :
SFy = 0 :
5
5
FAB +
FBD + 75 kN = 0 29
61
2
6
FAB FBD + 140 kN = 0 29
61
(1)
(2)
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51
PROBLEM 6.27 (Continued)
Solving (1) and (2)
FAB = -154.823 kN FAB = 154.8 kN C 
FBD = 107.391 kN FBD = 107.4 kN T 
Free body: joint D
+
SFy = 0 :
2
2
6
FAD (167.705) +
(107.391) = 0
5
5
61
FAD = 75.467 kN T 
+
SFx = 0 : FDE +
1
5
(75.467 - 167.705) (107.391) = 0
5
61
FDE = 110.00 kN FDE = 110.0 kN T 
Free body: joint A
+
+
5
1
5
1
FAC +
FAE +
(154.823) (75.467) = 0 29
5
29
5
SFx = 0 :
SFy = 0 : -
2
2
2
2
FAC FAE +
(154.823) (75.467) = 0 29
5
29
5
(3)
(4)
Multiply (3) by 2 and add (4):
4
8
12
FAC +
(154.823) (75.467) = 0
29
29
5
FAC = -141.3602 kN,
FAC = 141.4 kN C 
Multiply (3) by 2 (4) by 5 and add:
-
8
20
12
(154.823) (75.467) = 0
FAE +
5
29
5
FAE = 47.5164 kN FAE = 47.5 kN T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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52
PROBLEM 6.27 (Continued)
Free body: Joint C
From force triangle
FCE
61
=
FCG 141.3602 kN
=
8
29
FCG = 210.0 kN FCE = 205 kN T 
FCG = 210 kN C 
Free body: Joint G
+
+
SFx = 0: FEG = 0 
SFy = 0 : 210 kN - 210 kN = 0
(Checks)
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53
PROBLEM 6.28
Determine the force in each member of the truss shown. State
whether each member is in tension or compression.
SOLUTION
Free body: Truss
SFx = 0 : H x = 0
+
SM H = 0 : 48(16) - G( 4) = 0 G = 192 kN ≠
+
SFy = 0 : 192 - 48 + H y = 0
H y = -144 kN H y = 144 kN Ø
Zero-Force Members:
Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC, BE,
DE, DG, FG
Thus,
FBC = FBE = FDE = FDG = FFG = 0
Also note:
FAB = FBD = FDF = FFH (1)
FAC = FCE = FEG (2)
Free body: Joint A:
FAB FAC 48 kN
=
=
8
3
73
FAB = 128 kN FAB = 128.0 kN T 
FAC = 136.704 kN FAC = 136.7 kN C 
From Eq. (1):
FBD = FDF = FFH = 128.0 kN T 
From Eq. (2):
FCE = FEG = 136.7 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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54
PROBLEM 6.28 (Continued)
Free body: Joint H
Also
FGH
145
=
144 kN
9
FFH 144 kN
=
8
9
FFH = 128.0 kN T
FGH = 192.7 kN C 
(Checks)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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55
PROBLEM 6.29
Determine whether the trusses of Problems 6.31a, 6.32a, and 6.33a
are simple trusses.
SOLUTION
Truss of Problem 6.31a
Starting with triangle ABC and adding two members at a time, we obtain
joints D, E, G, F, and H, but cannot go further thus, this truss
is not a simple truss 
Truss of Problem 6.32a
Starting with triangle HDI and adding two members at a time, we obtain
joints A, E, J, and B, but cannot go further. Thus, this truss
­successively
is not a simple truss 
Truss of Problem 6.33a
Starting with triangle EHK and adding two members at a time, we obtain
s­ uccessively joints D, J, C, G, I, B, F, and A, thus completing the truss.
Therefore, this truss
is a simple truss 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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56
PROBLEM 6.30
Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b
are simple trusses.
SOLUTION
Truss of Problem 6.31b
Starting with triangle ABC and adding two members at a time, we obtain
s­ uccessively joints E, D, F, G, and H, but cannot go further. Thus, this truss
is not a simple truss 
Truss of Problem 6.32b
Starting with triangle CGH and adding two members at a time, we obtain
­successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing
the truss.
Therefore, this truss is a simple truss 
Truss of Problem 6.33b
Starting with triangle GLM and adding two members at a time, we obtain
joints K and I but cannot continue, starting instead with triangle BCH, we obtain
joint D but cannot continue, thus, this truss
is not a simple truss 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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57
PROBLEM 6.31
For the given loading, determine the zero-force members in each of
the two trusses shown.
SOLUTION
Truss (a)
FB: Joint B: FBC = 0
FB: Joint C : FCD = 0
FB: Joint J : FIJ = 0
FB: Joint I : FIL = 0
FB: Joint N : FMN = 0
FB: Joint M : FLM = 0
The zero-force members, therefore, are
Truss (b)
BC , CD, IJ , IL, LM , MN 
FB: Joint C : FBC = 0
FB: Joint B: FBE = 0
FB: Joint G: FFG = 0
FB: Joint F : FEF = 0
FB: Joint E: FDE = 0
FB: Joint I : FIJ = 0
FB: Joint M : FMN = 0
FB: Joint N : FKN = 0
The zero-force members, therefore, are
BC , BE , DE , EF , FG, IJ , KN , MN 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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58
PROBLEM 6.32
For the given loading, determine the zero-force members in each of
the two trusses shown.
SOLUTION
Truss (a)
FB: Joint B: FBJ = 0
FB: Joint D: FDI = 0
FB: Joint E: FEI = 0
FB: Joint I : FAI = 0
FB: Joint F : FFK = 0
FB: Joint G: FGK = 0
FB: Joint K : FCK = 0
The zero-force members, therefore, are
Truss (b)
AI , BJ , CK , DI , EI , FK , GK 
FB: Joint K : FFK = 0
FB: Joint O: FIO = 0
The zero-force members, therefore, are
All other members are either in tension or compression.
FK and IO 
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59
PROBLEM 6.33
For the given loading, determine the zero-force members in each of the two trusses shown.
SOLUTION
Truss (a)
FB: Joint F : FBF = 0
FB: Joint B: FBG = 0
FB: Joint G: FGJ = 0
FB: Joint D: FDH = 0
FB: Joint J : FHJ = 0
FB: Joint H : FEH = 0
The zero-force members, therefore, are
Truss (b)
BF , BG, DH , EH , GJ , HJ 
FB: Joint A: FAB = FAF = 0
FB: Joint C : FCH = 0
FB: Joint E: FDE = FEJ = 0
FB: Joint L: FGL = 0
FB: Joint N : FIN = 0
The zero-force members, therefore, are
AB, AF , CH , DE , EJ , GL, IN 
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60
PROBLEM 6.34
Determine the zero-force members in the truss of (a) Problem 6.23, (b) Problem 6.28.
SOLUTION
(a)
Truss of Problem 6.23
FB : Joint I : FIJ = 0
FB : Joint J : FGJ = 0
FB : Joint G: FGH = 0
The zero-force members, therefore, are
(b)
GH , GJ , IJ 
Truss of Problem 6.28
FB : Joint C : FBC = 0
FB : Joint B: FBE = 0
FB : Joint E: FDE = 0
FB : Joint D: FDG = 0
FB : Joint F : FFG = 0
The zero-force members, therefore, are
BC , BE , DE , DG, FG 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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61
PROBLEM 6.35*
The truss shown consists of six members and is supported by a short
link at A, two short links at B, and a ball and socket at D. Determine
the force in each of the members for the given loading.
SOLUTION
Free body: Truss
From symmetry:
Dx = Bx
and
D y = By
SM z = 0 : - A(5 m) - (2 kN )(12 m) = 0
A = -4.8 kN
SFx = 0 : Bx + Dx + A = 0
2 Bx - 4.8 kN = 0, Bx = 2.4 kN
SFy = 0 : By + D y - 2 kN = 0
2 By = 2 kN
By = +1 kN
Thus
Free body: C
FCA
FCB
FCD
B = (2.40 kN )i + (1.000 kN ) j 
uuur
CA FAC
= FAC
=
( -12i + 5 j)
CA 13
uuur
CB FBC
= FBC
=
( -12i + 3.5k )
CB 12.5
uuur
CD FCD
( -12i - 3.5k )
= FCD
=
CD 12.5
SF = 0 : FCA + FCB + FCD - (2 kN ) j = 0
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62
PROBLEM 6.35* (Continued)
Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k :
-
i:
j:
k:
12
12
FAC ( FBC + FCD ) = 0 13
12.5
5
FAC - 2 kN = 0 13
3.5
( FBC - FCD ) = 0 FCD = FBC
12.5
(1)
FAC = 5.20 kN T 
Substitute for FAC and FCD in Eq. (1):
-
12
12
(5.20 kN ) (2 FBC ) = 0 FBC = -2.5 kN 13
12.5
Free body: B
FBC
FBA
FBD
FBC = FCD = 2.50 kN C 
uuur
CB
= (2.5 kN )
= -(2.4 kN )i + (0.7 kN )k
CB
uuur
F
BA
= FAB
= AB (5 j - 3.5k )
BA 6.1033
= - FBD k
SF = 0 : FBA + FBD + FBC + (2.4 kN )i + (1 kN ) j = 0
Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k:
j:
5
FAB + 1 kN = 0 FAB = -1.22066 kN 6.1033
k:
-
From symmetry:
3.5
FAB - FBD + 0.7 kN = 0
6.1033
3.5
FBD = ( -1.22066 kN ) + 0.7 kN = +1.400 kN 6.1033
FAD = FAB FAB = 1.221 kN C 
FBD = 1.400 kN T 
FAD = 1.221 kN C 
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63
PROBLEM 6.36*
The truss shown consists of six members and is supported by a ball
and socket at B, a short link at C, and two short links at D. Determine
the force in each of the members for P = (-2184 N)j and Q = 0.
SOLUTION
Free body: Truss
From symmetry:
Dx = Bx and D y = By
SFx = 0 : 2 Bx = 0
Bx = Dx = 0
SFz = 0 : Bz = 0
SM cz = 0 : - 2 By (2.8 m) + (2184 N )(2 m) = 0
By = 780 N
B = (780 N ) j 
Thus
Free body: A
uuur
AB FAB
( -0.8i - 4.8 j + 2.1k )
FAB = FAB
=
AB 5.30
uuur
AC FAC
(2i - 4.8 j)
FAC = FAC
=
AC 5.20
uuur
AD FAD
FAD = FAD
=
( +0.8i - 4.8 j - 2.1k )
AD 5.30
SF = 0 : FAB + FAC + FAD - (2184 N ) j = 0
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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64
PROBLEM 6.36* (Continued)
Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k :
i:
-
0.8
2
( FAB + FAD ) +
FAC = 0 5.30
5.20
(1)
j:
-
4.8
4.8
( FAB + FAD ) FAC - 2184 N = 0 5.30
5.20
(2)
k:
2.1
( FAB - FAD ) = 0
5.30
FAD = FAB
Multiply (1) by –6 and add (2):
Ê 16.8 ˆ
-Á
F - 2184 N = 0, FAC = -676 N Ë 5.20 ˜¯ AC
FAC = 676 N C 
Substitute for FAC and FAD in (1):
Ê 0.8 ˆ
Ê 2 ˆ
-Á
2 FAB + Á
( -676 N ) = 0, FAB = -861.25 N ˜
Ë 5.30 ¯
Ë 5.20 ˜¯
Free body: B
FAB = FAD = 861 N C 
uuur
AB
FAB = (861.25 N )
= -(130 N )i - (780 N ) j + (341.25 N )k
AB
Ê 2.8i - 2.1k ˆ
FBC = FBC Á
˜¯ = FBC (0.8i - 0.6k )
Ë
3.5
FBD = - FBD k
SF = 0 : FAB + FBC + FBD + (780 N ) j = 0
Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k :
i:
-130 N + 0.8 FBC = 0 FBC = +162.5 N k:
341.25 N - 0.6 FBC - FBD = 0
FBD = 341.25 - 0.6(162.5) = +243.75 N From symmetry:
FCD = FBC FBC = 162.5 N T 
FBD = 244 N T 
FCD = 162.5 N T 
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65
PROBLEM 6.37*
The truss shown consists of six members and is supported by a ball
and socket at B, a short link at C, and two short links at D. Determine
the force in each of the members for P = 0 and Q = (2968 N)i.
SOLUTION
Free body: Truss
From symmetry:
Dx = Bx and D y = By
SFx = 0 : 2 Bx + 2968 N = 0
Bx = Dx = -1484 N
SM cz ¢ = 0 : - 2 By (2.8 m) - (2968 N )( 4.8 m) = 0
By = -2544 N
B = -(1484 N )i - (2544 N ) j 
Thus
Free body: A
FAB = FAB
uuur
AB
AB
FAB
( -0.8i - 4.8 j + 2.1k )
5.30
uuur
AC FAC
= FAC
=
(2i - 4.8 j)
AC 5.20
uuur
AD
= FAD
AD
F
= AD ( -0.8i - 4.8 j - 2.1k )
5.30
=
FAC
FAD
SF = 0 : FAB + FAC + FAD + (2968 N )i = 0
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66
PROBLEM 6.37* (Continued)
Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k :
i:
-
0.8
2
( FAB + FAD ) +
FAC + 2968 N = 0 5.30
5.20
(1)
j:
-
4.8
4.8
( FAB + FAD ) FAC = 0 5.30
5.20
(2)
k:
2.1
( FAB - FAD ) = 0
5.30
FAD = FAB
Multiply (1) by –6 and add (2):
Ê 16.8 ˆ
-Á
F - 6(2968 N) = 0, FAC = -5512 N
Ë 5.20 ˜¯ AC
FAC = 5510 N C 
Substitute for FAC and FAD in (2):
Ê 4.8 ˆ
Ê 4.8 ˆ
-Á
2 FAB - Á
( -5512 N ) = 0, FAB = +2809 N
˜
Ë 5.30 ¯
Ë 5.20 ˜¯
FAB = FAD = 2810 N T 
Free body: B
FAB
uuur
BA
= (2809 N )
= ( 424 N )i + (2544 N ) j - (1113 N )k
BA
Ê 2.8 i - 2.1k ˆ
FBC = FBC Á
˜¯ = FBC (0.8i - 0.6k )
Ë
3.5
FBD = - FBD k
SF = 0 : FAB + FBC + FBD - (1484 N )i - (2544 N ) j = 0
Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k:
From symmetry:
i:
+24 N + 0.8 FBC - 1484 N = 0, FBC = +1325 N k:
-1113 N - 0.6 FBC - FBD = 0
FBC = 1325 N T 
FBD = -1113 N - 0.6(1325 N ) = -1908 N, FBD = 1908 N C 
FCD = FBC FCD = 1325 N T 
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67
PROBLEM 6.38*
The truss shown consists of nine members and is supported
by a ball and socket at A, two short links at B, and a short
link at C. Determine the force in each of the members for
the given loading.
SOLUTION
Free body: Truss
From symmetry:
Az = Bz = 0
SFx = 0 : Ax = 0
SM BC = 0 : Ay (3 m) + (8 kN )(3.75 m) = 0
Ay = -10 kN
A = -(10.00 kN)j 
By = C
From symmetry:
SFy = 0 : 2 By - 10 kN - 8 kN = 0
By = 9 kN Free body: A
B = (9.00 kN ) j 
SF = 0 : FAB + FAC + FAD - (10 kN ) j = 0
FAB
i+k
i-k
+ FAC
+ FAD (0.6 i + 0.8 j) - (10 kN ) j = 0
2
2
Factoring i, j, k and equating their coefficient to zero:
1
1
FAB +
FAC + 0.6 FAD = 0 2
2
(1)
0.8FAD - 10 kN = 0 FAD = 12.5 kN T 
1
1
FAB FAC = 0
2
2
FAC = FAB
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68
PROBLEM 6.38* (Continued)
Substitute for FAD and FAC into (1):
Free body: B
2
FAB = -5.3033 kN, FAB = FAC = 5.30 kN C 
FAB + 0.6(12.5 kN ) = 0,
2
uuur
BA
i+k
FBA = FAB
= +(5.3033 kN )
= (3.75 kN )(i + k )
BA
2
FBC = - FBC k
FBD = FBD (0.8 j - 0.6k )
uuur
BE FBE
FBE = FBE
=
(3.75i + 4 j - 3k )
BE 6.25
SF = 0 : FBA + FBC + FBD + FBE + (9.00 kN ) j = 0
Substituting for FBA , FBC , FBD + FBE and equate to zero the coefficients of i, j, k :
i:
Ê 3.75 ˆ
3.75 kN + Á
F = 0, FBE = -6.25 kN Ë 6.25 ˜¯ BE
FBE = 6.25 kN C 
j:
Ê 4 ˆ
0.8 FBD + Á
( -6.25 kN ) + 9.0 kN = 0 Ë 6.25 ˜¯
FBD = 6.25 kN C 
k:
3.75 kN - FBC - 0.6( -6.25 kN ) -
3
( -6.25 kN ) = 0
6.25
FBC = 10.5 kN T 
FBD = FCD = 6.25 kN C 
From symmetry:
Free body: D
SF = 0: FDA + FDB + FDC + FDE i = 0
We now substitute for FDA , FDB , FDC and equate to zero the coefficient of i. Only FDA contains i and its
­coefficient is
-0.6 FAD = -0.6(12.5 kN ) = -7.5 kN
i:
-7.5 kN + FDE = 0 FDE = 7.5 kN T 
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69
PROBLEM 6.39*
The truss shown consists of nine members and is supported by a ball
and socket at B, a short link at C, and two short links at D. (a) Check
that this truss is a simple truss, that it is completely constrained,
and that the reactions at its supports are statically determinate.
(b) Determine the force in each member for P = (-1200 N)j and Q = 0.
SOLUTION
Free body: Truss
SM B = 0 : 1.8i ¥ Cj + (1.8i - 3k ) ¥ ( D y j + Dk )
+ (0.6i - 0.75k ) ¥ ( -1200 j) = 0
-1.8 Ck + 1.8 D y k - 1.8 Dz j
+ 3D y i - 720k - 900i = 0
Equate to zero the coefficients of i, j, k :
i:
3D y - 900 = 0, D y = 300 N
j:
Dz = 0, D = (300 N ) j 
k:
1.8C + 1.8(300) - 720 = 0 C = (100 N ) j 
SF = 0 : B + 300 j + 100 j - 1200 j = 0 B = (800 N ) j 
Free body: B
SF = 0 : FBA + FBC + FBE + (800 N ) j = 0, with
uuur
BA FAB
(0.6i + 3j - 0.75k )
FBA = FAB
=
BA 3.15
FBC = FBC i
FBE = - FBE k
Substitute and equate to zero the coefficient of j, i, k :
j:
Ê 3 ˆ
ÁË
˜ FAB + 800 N = 0, FAB = -840 N,
3.315 ¯
i:
Ê 0.6 ˆ
ÁË
˜ ( -840 N ) + FBC = 0 3.15 ¯
k:
Ê 0.75 ˆ
ÁË ˜ ( -840 N ) - FBE = 0 3.15 ¯
FAB = 840 N C 
FBC = 160.0 N T 
FBE = 200 N T 
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70
PROBLEM 6.39* (Continued)
Free body C:
SF = 0 : FCA + FCB + FCD + FCE + (100 N ) j = 0, with
uuur
CA FAC
FCA = FAC
( -1.2i + 3j - 0.75 k )
=
CA 3.317
FCB = -(160 N ) i
FCD = - FCD k FCE = FCE
uuur
F
CE
= CE ( -1.8i - 3k )
CE 3, 499
Substitute and equate to zero the coefficient of j, i, k :
Ê 3 ˆ
ÁË
˜ FAC + 100 N = 0, FAC = -110.57 N 3.317 ¯
j:
-
i:
k:
-
1.2
1.8
( -110.57) - 160 FCE = 0, FCE = -233.3 3.317
3.499
0.75
3
( -110.57) - FCD ( -233.3) = 0 3.317
3.499
FAC = 110.6 N C 
FCE = 233 N C 
FCD = 225 N T 
Free body: D
SF = 0 : FDA + FDC + FDE + (300 N ) j = 0, with
uuur
DA FAD
FDA = FAD
( -1.2i + 3j + 2.25k )
=
DA 3.937
FDC = FCD k = (225 N )k
FDE = - FDE i
Substitute and equate to zero the coefficient of j, i, k :
j:
Ê 3 ˆ
ÁË
˜ FAD + 300 N = 0,
3.937 ¯
i:
Ê 1.2 ˆ
ÁË ˜ ( -393.7 N ) - FDE = 0 3.937 ¯
k:
Ê 2.25 ˆ
ÁË
˜ ( -393.7 N ) + 225 N = 0 (Checks)
3.937 ¯
FAD = -393.7 N, FAD = 394 N C 
FDE = 1200 N T 
Free body: E
Member AE is the only member at E which does not lie in the xz plane.
Therefore, it is a zero-force member.
FAE = 0 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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71
PROBLEM 6.40*
Solve Problem 6.39 for P = 0 and Q = (-900 N)k.
PROBLEM 6.39* The truss shown consists of nine members
and is supported by a ball and socket at B, a short link at C, and
two short links at D. (a) Check that this truss is a simple truss,
that it is completely constrained, and that the reactions at its
supports are statically determinate. (b) Determine the force in
each member for P = (-1200 N)j and Q = 0.��
SOLUTION
Free body: Truss
SM B = 0 : 1.8i ¥ Cj + (1.8i - 3k ) ¥ ( D y j + Dz k )
+ (0.6i + 3j - 0.75k ) ¥ ( -900 N )k = 0
1.8 Ck + 1.8D y k - 1.8Dz j
+ 3D y i + 540 j - 2700i = 0
Equate to zero the coefficient of i, j, k :
3D y - 2700 = 0 D y = 900 N
-1.8 Dz + 540 = 0 Dz = 300 N
1.8C + 1.8D y = 0 C = - D y = -900 N
Thus:
C = -(900 N ) j D = (900 N ) j + (300 N )k 
SF = 0 : B - 900 j + 900 j + 300k - 900k = 0 B = (600 N )k 
Free body: B
SinceB is aligned with member BE:
FAB = FBC = 0, FBE = 600 N T 
Free body: C
SF = 0: FCA + FCD + FCE - (900 N ) j = 0, with
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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72
PROBLEM 6.40* (Continued)
uuur
CA FAC
FCA = FAC
( -1.2i + 3j - 0.75k )
=
CA 3.317
uuur
F
CE
FCD = - FCD k FCE = FCE
= CE ( -1.8i - 3k )
CE 3.499
Substitute and equate to zero the coefficient of j, i, k:
j:
Ê 3 ˆ
ÁË
˜ FAC - 900 N = 0, FAC = 995.1 N 3.317 ¯
i:
-
1.2
1.8
(995.1) FCE = 0, FCE = -699.8 N 3.317
3.499
FCE = 700 N C 
k:
-
0.75
3
(995.1) - FCD ( -699.8) = 0 3.317
3.499
FCD = 375 N T 
FAC = 995 N T 
Free body: D
SF = 0 : FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0
uuur
DA FAD
( -1.2i + 3j + 2.25k )
FDA = FAD
=
DA 3.937
with
FDE = - FDE i
and
Substitute and equate to zero the coefficient j, i, k:
j:
Ê 3 ˆ
ÁË
˜ FAD + 900 N = 0, FAD = -1181.1 N 3.937 ¯
FAD = 1181N C 
i:
Ê 1.2 ˆ
-Á
( -1181.1 N ) - FDE = 0 Ë 3.937 ˜¯
FDE = 360 N T 
k:
Ê 2.25 ˆ
ÁË
˜ ( -1181.1 N + 375 N + 300 N = 0)
3.937 ¯
(Checks)
Free body: E
Member AE is the only member at E which
does not lie in the XZ plane. Therefore, it is
a zero-force member.
FAE = 0 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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73
PROBLEM 6.41*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in each
of the six members joined at E.
SOLUTION
(a)
Check simple truss.
(1) start with tetrahedron BEFG
(2) Add members BD, ED, GD joining at D.
(3) Add members BA, DA, EA joining at A.
(4) Add members DH, EH, GH joining at H.
(5) Add members BC, DC, GC joining at C.
Truss has been completed: It is a simple truss
Free body: Truss
Check constraints and reactions:
Six unknown reactions-ok-however supports at A and B
­constrain truss to rotate about AB and support at G prevents
such a rotation. Thus,
Truss is completely constrained and reactions are statically
determinate
Determination of Reactions:
SM A = 0 : 5.5i ¥ ( By j + Bz k ) + (5.51i - 4.8k ) ¥ G j
+ (5.04 j - 4.8k ) ¥ (1375i + 1200k) = 0
5.5 B y k - 5.5 B y j + 5.5Gk + 1200Gi - (5.04)(1375)k
+ (5.04)(1200)i - ( 4.8)(1375)j = 0
Equate to zero the coefficient of i, j, k:
i: 4.8G + (5.04)(1200) = 0 G = -1260 N G = ( -1260 N ) j 
j: - 5.5 Bz - ( 4.8)(1375) = 0 Bz = -1200 N
k : 5.5 By + 5.5( -1260) - (5.04)(1375) = 0, B y = 2520 N B = (2520 N ) j - (1200 N )k 
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74
PROBLEM 6.41* (Continued)
SF = 0 : A + (2520 N ) j - (1200 N )k - (1260 N ) j
+ (1375 N )i + (1200 N )k = 0
A = -(1375 N )i - (1260 N ) j 
Zero-force members
The determination of these members will facilitate our solution.
FB: C. Writing
SFx = 0, SFy = 0, SFz = 0 Yields FBC = FCD = FCG = 0 
FB: F. Writing
SFx = 0, SFy = 0, SFz = 0 Yields FBF = FEF = FFG = 0 
FB: A: Since
FB: H: Writing
Az = 0, writing
Yields FAD = 0 
SFz = 0 SFy = 0 Yields FDH = 0 
FB: D: Since FAD = FCD = FDH = 0, we need
consider only members DB, DE, and DG.
Since FDE is the only force not contained in plane BDG, it must be zero.
Simple reasonings show that the other two forces are also zero.
FBD = FDE = FDG = 0 
The results obtained for the reactions at the supports and for the zero-force members are shown on the
figure below. Zero-force members are indicated by a zero (“0”).
(b)
Force in each of the members joined at E
FDE = FEF = 0 
We already found that
Free body: A
SFy = 0 Yields FAE = 1260 N T 
Free body: H
SFz = 0 Yields FEH = 1200 N C 
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75
PROBLEM 6.41* (Continued)
Free body: E
SF = 0 : FEB + FEG + (1200 N )k - (1260 N ) j = 0
F
FBE
(5.5i - 5.04 j) + EG (5.5i - 4.8k ) + 1200k - 1260 j = 0
7.46
7.3
Equate to zero the coefficient of y and k:
Ê 5.04 ˆ
F - 1260 = 0 j: - Á
Ë 7.46 ˜¯ BE
Ê 4.8 ˆ
F + 1200 = 0 k: -Á
Ë 7.3 ˜¯ EG
FBE = 1865 N C 
FEG = 1825 N T 
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76
PROBLEM 6.42*
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in each
of the six members joined at G.
SOLUTION
See solution to Problem 6.41 for Part (a) and for reactions and zero-force members.
(b)
Force in each of the members joined at G.
We already know that
FCG = FDG = FFG = 0 
Free body: H
SFx = 0 Yields FGH = 1375 N C 
Free body: G
SF = 0 : FGB + FGE + (1375 N )i - (1260 N ) j = 0
FBG
F
( -5.04 j + 4.8k ) + EG ( -5.5i + 4.8k ) + 1375i - 1260 j = 0
6.96
7.3
Equate to zero the coefficient of i, j, k:
i:
Ê 5.5 ˆ
- Á ˜ FEG + 1375 = 0 Ë 7.3 ¯
FEG = 1825 N T 
j:
Ê 5.04 ˆ
-Á
F - 1260 = 0 Ë 6.96 ˜¯ BG
FBG = 1740 N C 
k:
Ê 4.8 ˆ
Ê 4.8 ˆ
ÁË
˜¯ ( -1740) + ÁË
˜ (1825) = 0
6.96
7.3 ¯
(Checks)
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77
PROBLEM 6.43
A Warren bridge truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
Free body: Truss
SFx = 0 : K x = 0
+
SM A = 0 : K y (25 m) - (30 kN )(5 m)
+
- (30 kN )(10 m) = 0
K = K y = 18.00 kN ≠ 
SFy = 0 : A + 18 kN - 30 kN - 30 kN = 0
+
A = 42 kN ≠ 
We pass a section through members CE, DE, and DF and use the
free body shown.
SM D = 0 : FCE (6 m) - ( 42 kN )(7.5 m)
+
+ (30 kN )(2.5 m) = 0
FCE = +40 kN
SFy = 0 : 42 kN - 30 kN -
+
FDE = +13 kN +
FCE = 40.0 kN T 
6
FDE = 0
6.5
FDE = 13.00 kN T 
SM E = 0 : 30 kN (5 m) - ( 42 kN )(10 m)
- FDF (6 m) = 0
FDF = -45 kN FDF = 45.0 kN C 
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78
PROBLEM 6.44
A Warren bridge truss is loaded as shown. Determine the force in
members EG, FG, and FH.
SOLUTION
See solution of Problem 6.43 for free-body diagram of truss and determination of reactions:
A = 42 kN K = K y = 18 kN 
We pass a section through members EG, FG, and FH, and use the free body shown.
+
SM F = 0 : (18 kN )(12.5 m) - FEG (6 m) = 0
+ SFy = 0 :
+
FEG = +37.5 kN FEG = 37.5 kN T 
6
FFG + 18 kN = 0
6.5
FFG = -19.5 kN FFG = 19.50 kN C 
SM G = 0 : FFH (6 m) + (18 kN )(10 m) = 0
FFH = -30 kN FFH = 30.0 kN C 
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79
PROBLEM 6.45
Determine the force in members BD and DE of the truss shown.
SOLUTION
Member BD:
+
SM E = 0 : FBD ( 4.5 m) - (135 kN )( 4.8 m) - (135 kN )(2.4 m) = 0
FBD = +216 kN FBD = 216 kN T 
Member DE:
+
SFx = 0 : 135 kN + 135 kN - FDE = 0
FDE = +270 kN FDE = 270 kN T 
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80
PROBLEM 6.46
Determine the force in members DG and EG of the truss shown.
SOLUTION
Member DG:
SFx = 0 : 135 kN + 135 kN + 135 kN +
+
FDG = -459 kN 15
FDG = 0
17
FDG = 459 kN C 
Member EG:
+
SM D = 0 : (135 kN )( 4.8 m) + (135 kN)(2.4 m)
+ FEG ( 4.5 m) = 0
FEG = -216 kN FEG = 216 kN C 
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81
PROBLEM 6.47
A floor truss is loaded as shown. Determine the force in
members CF, EF, and EG.
SOLUTION
Free body: Truss
+
+
SFx = 0 : k x = 0
SM A = 0 : k y ( 4.8 m) - ( 4 kN )(0.8 m)
- ( 4 kN )(1.6 m) - (3 kN )(2.4 m)
- (2 kN )(3.2 m) - (2 kN )( 4 m) - (1 kN )( 4.8 m) = 0
k y = 7.5 kN
k = 7.5 kN≠ 
Thus:
+
SFy = 0 : A + 7.5 kN - 18 kN = 0
A = 10.5 kN A = 10.5 kN ≠ 
We pass a section through members CF, EF, and EG and use the free body shown.
+
SM E = 0 : FCF (0.4 m) - (10.5 kN )(1.6 m) + (2 kN )(1.6 m)
+ ( 4 kN )(0.8 m) = 0
FCF = +26.0 kN +
SFy = 0 : 10.5 kN - 2 kN - 4 kN - 4 kN FEF = +1.118 kN +
FCF = 26.0 kN T 
1
FEF = 0
5
FEF = 1.118 kN T 
SM F = 0 : (2 kN )(2.4 m) + ( 4 kN )(1.6 m) + ( 4 kN )(0.8 m)
- (10.5 kN )(2.4 m) - FEG (0.4 m) = 0
FEG = -27.0 kN FEG = 27.0 kN C 
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82
PROBLEM 6.48
A floor truss is loaded as shown. Determine the force in
members FI, HI, and HJ.
SOLUTION
See solution of Problem 6.47 for free-body diagram of truss and determination of reactions:
A = 10.5 kN ≠, k = 7.5 kN≠ 
We pass a section through members FI, HI, and HJ, and use the free body shown.
SM H = 0 : (7.5 kN )(1.6 m) - (2 kN )(0.8 m) - (1 kN )(1.6 m)
- FFI (0.4 m) = 0
FFI = +22.0 kN FFI = 22.0 kN T 
+
+
+
SFy = 0 :
1
FHI - 2 kN - 1 kN + 7.5 kN = 0
5
FHI = -10.06 kN FHI = 10.06 kN C 
SM I = 0 : FHJ (0.4 m) + (7.5 kN )(0.8 m) - (1 kN )(0.8 m) = 0
FHJ = -13.00 kN FHJ = 13.00 kN C 
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83
PROBLEM 6.49
A pitched flat roof truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
Reactions at supports: Because of the symmetry of the loading,
Ax = 0
Ay = I =
1
1
(Total load ) = (8 kN ) 2
2
A = I = 4 kN ≠ 
We pass a section through members CD, DE, and DF, and use the free body shown.
(We moved FDE to E and FDF to F)
Slope
BJ =
Slope
DE =
2.16 m 9
=
9.6 m 40
-1 m -5
=
2.4 m 12
0.46 m
0.46 m
=
= 2.0444 m
a=
9
Slope BJ
40
+
SM D = 0 : FCE (1 m) + (1 kN )(2.4 m) - ( 4 kN )(2.4 m) = 0
FCE = +7.20 kN +
FCE = 7.20 kN T 
SM K = 0 : ( 4 kN )(2.0444 m) - (1 kN )(2.0444 m)
ˆ
Ê5
- (2 kN )( 4.4444 m) - Á FDE ˜ (6.8444 m) = 0
¯
Ë 13
FDE = -1.047 kN +
FDE = 1.047 kN C 
SM E = 0 : (1 kN )( 4.8 m) + (2 kN )(2.4 m) - ( 4 kN )( 4.8 m)
ˆ
Ê 40
- Á FDF ˜ (1.54 m) = 0
¯
Ë 41
FDF = -6.39 kN FDF = 6.39 kN C 
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84
PROBLEM 6.50
A pitched flat roof truss is loaded as shown. Determine the force in
members EG, GH, and HJ.
SOLUTION
Reactions at supports: Because of the symmetry of the loading,
Ax = 0
1
1
(Total load ) = (8 kN ) 2
2
We pass a section through members EG, GH, and HJ, and use the free body shown.
Ay = I =
+
SM H = 0 : ( 4 kN )(2.4 m) - (1 kN )(2.4 m) - FEG (2.08 m) = 0
FEG = +3.4615 kN +
A = I = 4 kN ≠ 
FEG = 3.46 kN T 
SM J = 0 : - FGH (2.4 m) - FEG (2.62 m) = 0
2.62
(3.4615 kN )
2.4
= -3.7788 kN FGH = FGH
+
FGH = 3.78 kN C 
2.4
FHJ = 0
2.46
2.46
2.46
=FEG = (3.4615 kN )
2.4
2.4
= -3.548 kN
FHJ = 3.55 kN C 
SFx = 0 : - FEG FHJ
FHJ
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85
PROBLEM 6.51
A Howe scissors roof truss is loaded as shown. Determine
the force in members DF, DG, and EG.
SOLUTION
Reactions at supports.
Because of symmetry of loading.
Ax = 0,
1
1
Ay = L = (Total load) = ( 48 kN) = 24 kN
2
2
A = L = 24.0 kN ≠ 
We pass a section through members DF, DG, and EG, and use the free body shown.
We slide FDF to apply it at F:
SM G = 0 : (4 kN)(6 m) + (8 N)(4 m) + (8 kN)(2 m)
+
- (24 kN)(6 m) -
2 FDF
Ê 2.5 ˆ
2 +Á
Ë 3 ˜¯
2
2
(1.5 m) = 0
FDF = -52.0 kN, FDF = 52.0 kN C 
+
SM A = 0 : - (8 kN)(2 m) - (8 kN)(4 m)
-
2 / 3FDG
22 + (2 / 3)
2
( 4 m) -
2 FDG
22 + (2 / 3)
2
Ê5 ˆ
ÁË m˜¯ = 0
3
FDG = -16.8654 kN, FDG = 16.87 kN C 
+
SM D = 0 : ( 4 kN)( 4 m) + (8 kN)(2 m) - (24 kN)(4 m) -
2 FEG
Ê 1ˆ
2 +Á ˜
Ë 3¯
2
2
(1 m) = 0
FEG = -64.882 kN, FEG = 64.9 kN C 
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86
PROBLEM 6.52
A Howe scissors roof truss is loaded as shown. Determine
the force in members GI, HI, and HJ.
SOLUTION
Reactions at supports. Because of symmetry of loading:
1
Ax = 0, Ay = L = (Total load )
2
1
= ( 48 kN)
2
= 24 kN A = L = 24.0 kN ≠ 
We pass a section through members GI, HI, and HJ, and use the free body shown.
+
SM H = 0 :
4 FGI
Ê 2ˆ
42 + Á ˜
Ë 3¯
2
(1 m) + (24 kN )( 4 m) - ( 4 m)( 4 m) - (8 kN )(2 m) = 0
FGI = +64.883 kN +
FGI = 64.9 kN T 
SM L = 0 : (8 kN )(2 m) - FHI ( 4 m) = 0
FHI = +4.0 kN
FHI = 4.00 kN T 
We slide FHG to apply it at H.
+
SM I = 0 :
2 FHJ
Ê 2.5 ˆ
2 +Á ˜
Ë 3 ¯
2
2
(1 m) + (24 kN )( 4 m) - (8 kN )(2 m) - ( 4 kN )( 4 m)) = 0
FHJ = -69.333 kN FHJ = 69.3 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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87
PROBLEM 6.53
A Pratt roof truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
Free body: Entire truss
SFx = 0 : Ax = 0
Total load = 5(3 kN) + 2(1.5 kN)
= 18 kN
By symmetry:
1
Ay = L = (18 kN)
2
A = L = 9 kN 
Free body: Portion ACD
Note: Slope of ABDF is
6.75 3
=
9.00 4
Force in CE:
+
ˆ
Ê2
SM D = 0 : FCE Á ¥ 6.75 m˜ - (9 kN)(6 m) + (1.5 kN)(6 m) + (3 kN)(33 m) = 0
¯
Ë3
FCE ( 4.5 m) - 36 kN ◊ m = 0
FCE = +8 kN FCE = 8 kN T 
Force in DE:
+
SM A = 0 : FDE (6 m) + (3 kN)(6 m) + (3 kN)(3 m) = 0
FDE = -4.5 kN FDE = 4.5 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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88
PROBLEM 6.53 (Continued)
Force in DF:
Sum moments about E where FCE and FDE intersect.
+
SM E = 0 : (1.5 kN)(6 m) - (9 kN)(6 m) + (3 kN)(3 m) +
4
FCE ( 4.5 m) - 36 kN ◊ m
5
FCE = -10.00 kN 4
ˆ
Ê2
FCE Á ¥ 6.75 m˜ = 0
¯
Ë3
5
FCE = 10.00 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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89
PROBLEM 6.54
A Pratt roof truss is loaded as shown. Determine the force in
members FH, FI, and GI.
SOLUTION
Free body: Entire truss
SFx = 0 : Ax = 0
Total load = 5(3 kN) + 2(1.5 kN) = 18 kN
By symmetry:
1
A y = L = (18) = 9 kN ≠
2
Free body: Portion HIL
Slope of FHJL
6.75 3
=
9.00 4
tan a =
FG 6.75 m
=
a = 66.04∞
GI
3m
Force in FH:
+
SM I = 0 :
4
FFH
5
Ê2
ˆ
ÁË ¥ 6.75 m˜¯ + (9 kN)(6 m) - (1.5 kN)(6 m) - (3 kN)(3 m) = 0
3
4
FFH ( 4.5 m) + 36 kN ◊ m
5
FFH = -10.00 kN
FFH = 10.00 kN C 
Force in FI:
+
SM L = 0 : FFI sin a (6 m) - (3 kN)(6 m) - (3 kN)(3 m) = 0
FFI sin 66.04∞(6 m) = 27 kN ◊ m
FFI = +4.92 kN FFI = 4.92 kN T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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90
PROBLEM 6.54 (Continued)
Force in GI:
+
SM H = 0 : FGI (6.75 m) + (3 kN )(3 m) + (3 kN )(6 m) + (1.5 kN)(9 m) - (9 kN)(9 m) = 0
FGI (6.75 m) = +40.5 kN ◊ m
FGI = +6.00 kN FGI = 6.00 kN T 
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91
PROBLEM 6.55
Determine the force in members AD, CD, and CE of the
truss shown.
SOLUTION
Reactions:
SM k = 0 : 36(2.4) - B(13.5) + 20(9) + 20( 4.5) = 0
+
+
SFx = 0 : - 36 + K x = 0
+
+
K x = 36 kN ®
SFy = 0 : 26.4 - 20 - 20 + K y = 0 K y = 13.6 kN ≠
SM C = 0 : 36(1.2) - 26.4(2.25) - FAD (1.2) = 0
FAD = -13.5 kN
+
+
B = 26.4 kN ≠
Ê8
ˆ
SM A = 0 : Á FCD ˜ ( 4.5) = 0 Ë 17
¯
FAD = 13.5 kN C 
FCD = 0 
ˆ
Ê 15
SM D = 0 : Á FCE ˜ (2.4) - 26.4( 4.5) = 0
¯
Ë 17
FCE = +56.1 kN FCE = 56.1 kN T 
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92
PROBLEM 6.56
Determine the force in members DG, FG, and FH of the
truss shown.
SOLUTION
See the solution to Problem 6.55 for free-body diagram and analysis to determine the reactions at the supports
B and K.
B = 26.4 kN≠;
+
K x = 36.0 kN ®;
K y = 13.60 kN ≠
SM F = 0 : 36(1.2) - 26.4(6.75) + 20(2.25) - FDG (1.2) = 0
FDG = -75 kN +
Ê8
ˆ
SM D = 0 : - 26.4( 4.5) + Á FFG ˜ ( 4.5) = 0
Ë 17
¯
FFG = +56.1 kN
+
FDG = 75.0 kN C 
FFG = 56.1 kN T 
ˆ
Ê 15
SM G = 0 : 20(4.5) - 26.4(9) + Á FFH ˜ (2.4) = 0
¯
Ë 17
FFH = +69.7 kN FFH = 69.7 kN T 
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93
PROBLEM 6.57
A stadium roof truss is loaded as shown. Determine the force in
members AB, AG, and FG.
SOLUTION
We pass a section through members AB, AG, and FG, and use the free body shown.
ˆ Ê 15 ˆ
Ê 14
SM G = 0 : Á
F
m - (9 kN)(5 m) - ( 4.5 kN)(10 m) = 0
Ë 205 AB ˜¯ ÁË 7 ˜¯
+
FAB = +42.953 kN 3
Ê
ˆ
SM D = 0 : - Á FAG ˜ (10 m) + (9 kN)(10 m) + (9 kN)(5 m) = 0
Ë5
¯
+
FAG = +22.5 kN
+
FAB = 43.0 kN T 
FAG = 22.5 kN T 
M A = 0: - FFG (3 m) - (9 kN)(4 m) - (9 kN)(9 m) - ( 4.5 kN)(14 m) = 0
FFG = -60 kN FFG = 60.0 kN C 
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94
PROBLEM 6.58
A stadium roof truss is loaded as shown. Determine the force
in members AE, EF, and FJ.
SOLUTION
We pass a section through members AE, EF, and FJ, and use the free body shown.
+
Ê
ˆ
25
SM F = 0 : Á
FAE ˜ (3 m) - (9 kN)(4 m) - (9 kN)(9 m) - ( 4.5 kN)(14 m) = 0
Ë 2.52 + 32
¯
FAE = +71.4898 kN +
SM A = 0 : - FEF (3 m) - (9 kN)(4 m) - (9 kN)(9 m) - ( 4.5 kN)(14 m) = 0
FEF = -60.0 kN +
FAE = 71.5 kN T 
FEF = 60.0 kN C 
SM E = 0 : - FFJ (2.5 m) - ( 4.5 kN)(2.5 m) - (9 kN)(6.5 m) - (9 kN)(11.5 m) - ( 4.5 kN)(16.5 m) = 0
FFJ = -99.0 kN FFJ = 99.0 kN C 
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95
PROBLEM 6.59
A Polynesian, or duopitch, roof truss is loaded as
shown. Determine the force in members DF, EF,
and EG.
SOLUTION
Free body: Truss
SFx = 0 : N x = 0
+
SM N = 0 : (1 kN)(8a) + (2 kN)(7a + 6a + 5a) + (1.75 kN)(4a) + (1.5 kN)(3a + 2a + a) - A(8a) = 0
A = 7.50 kN ≠ 
+
SFy = 0 : 7.5 kN - 1 kN - 3(2 kN) - 1.75 kN - 3(1.5 kN) - 0.75 kN + N y = 0
N y = 6.50 kN N = 6.50 kN ≠ 
We pass a section through DF, EF, and EG, and use the free body shown.
(We apply FDF at F)
+
SM E = 0 : (1 kN)(6 m) + (2 kN)(4 m) + (2 kN)(2 m) - (7.5 kN)(6 m)
Ê
ˆ
6
-Á
FDF ˜ (1.5 m) = 0
Ë 62 + 1.52
¯
FDF = -18.554 kN
+
SM A = 0 : FEF (6 m) - (2 kN)(2 m) - (2 kN)(4 m) = 0
FEF = +2.00 kN +
FDF = 18.6 kN C 
FEF = 2.00 kN T 
SM F = 0 : FEG (1.5 m) - (7.5 kN)(6 m) + (1 kN)(6 m) + (2 kN)(4 m) + (2 kN)(2 m) = 0
FEG = +18.0 kN
FEG = 18.0 kN T 
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96
PROBLEM 6.60
A Polynesian, or duopitch, roof truss is loaded
as shown. Determine the force in members HI,
GI, and GJ.
SOLUTION
See solution of Problem 6.59 for reactions:
A = 7.5 kN ≠,
N = 6.5 kN ≠ 
We pass a section through HI, GI, and GJ, and use the free body shown.
(We apply FHI at H.)
+
Ê
ˆ
2
SM G = 0 : Á
FHI ˜ (275 m) + (6.5 kN)(8 m) - (1.5 kN)(2 m)
ÁË 22 + (1.25)2
˜¯
- (1.5 kN)(4 m) - (1.5 kN)(6 m) - (0.75 kN)(8 m) = 0
FHI = -12.007 kN
+
SM I = 0 : (6.5 kN)(6 m) - (1.5 kN)(2 m) - (1.5 kN)(4 m)
- (0.75 kN)(6 m) - FGJ (1.5 m) = 0
FGJ = +17.00 kN
+
FHI = 12.01 kN C 
SFx = 0 : -
FGJ = 17.00 kN T 
4
2
FGI ( -12.007 kN) - 17.00 kN = 0
5
22 + (1.25)2
FGI = -8.5226 kN
FGI = 8.52 kN C 
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97
PROBLEM 6.61
Determine the force in members AF and EJ of the truss shown when
P = Q = 1.2 kN. (Hint: Use section aa.)
SOLUTION
Free body: Entire truss
SM A = 0 : K (8 m) - (1.2 kN )(6 m) - (1.2 kN )(12 m) = 0
+
K = +2.70 kN K = 2.70 kNÆ 
Free body: Lower portion
SM F = 0 : FEJ (12 m) + (2.70 kN )( 4 m) - (1.2 kN )(6 m) - (1.2 kN )(12 m) = 0
+
FEJ = +0.900 kN +
FEJ = 0.900 kN T 
SFy = 0 : FAF + 0.9 kN - 1.2 kN - 1.2 kN = 0
FAF = +1.500 kN FAF = 1.500 kN T 
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98
PROBLEM 6.62
Determine the force in members AF and EJ of the truss shown when
P = 1.2 kN and Q = 0. (Hint: Use section aa.)
SOLUTION
Free body: Entire truss
SM A = 0 : K (8 m) - (1.2 kN )(6 m) = 0
+
K = +0.900 kN K = 0.900 kN ® 
Free body: Lower portion
SM F = 0 : FEJ (12 m) + (0.900 kN )( 4 m) - (1.2 kN )(6 m) = 0
+
FEJ = +0.300 kN +
FEJ = 0.300 kN T 
SFy = 0 : FAF + 0.300 kN - 1.2 kN = 0
FAF = +0.900 kN FAF = 1.900 kN T 
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99
PROBLEM 6.63
Determine the force in members EH and GI of the
truss shown. (Hint: Use section aa.)
SOLUTION
Reactions:
SFx = 0 : Ax = 0
SM P = 0 : 50(15) + 50(10) + 50(5) - Ay (30) = 0
+
A y = 50 kN ≠
+
+
SFy = 0 : 50 - 50 - 50 - 50 + P = 0
SM G = 0 : - (50 kN )(10 m) - FEH (6 m) = 0
FEH = -83.33 kN
+
P = 100 kN ≠
SFx = 0 : FGI - 83.33 kN = 0
FEH = 83.3 kN C 
FGI = 83.3 kN T 
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100
PROBLEM 6.64
Determine the force in members HJ and IL of the
truss shown. (Hint: Use section bb.)
SOLUTION
See the solution to Problem 6.63 for free body diagram and analysis to determine the reactions at supports A
and P.
A x = 0 ; A y = 50 kN≠;
+
P = 100 kN ≠
SM L = 0 : FHJ (6 m) - (50 kN )(5 m) + (100 kN )(10 m) = 0
+
FHJ = -125.0 kN FHJ = 125.0 kN C 
SFx = 0 : - 125 kN - FIL = 0
FIL = +125 kN FIL = 125.0 kN T 
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101
PROBLEM 6.65
The diagonal members in the center panels of
the power transmission line tower shown are
very slender and can act only in tension; such
members are known as counters. For the given
loading, determine (a) which of the two counters
listed below is acting, (b) the force in that counter.
Counters CJ and HE.
SOLUTION
Free body: Portion ABDFEC of tower
+
We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and EJ.
SFx = 0 :
4
FCJ - 2(1.2 kN )sin 20∞ = 0 FCJ = +1.026 kN
5
Since CJ is found to be in tension, our assumption was correct. Thus, the answers are
(a) CJ

(b) 1.026 kN T 
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102
PROBLEM 6.66
The diagonal members in the center panels of
the power transmission line tower shown are
very slender and can act only in tension; such
members are known as counters. For the given
loading, determine (a) which of the two counters
listed below is acting, (b) the force in that counter.
Counters IO and KN.
SOLUTION
Free body: Portion of tower shown
+
We assume that counter IO is acting and show the forces exerted by that counter and by members IN and KO.
SFx = 0 :
4
FIO - 4(1.2 kN )sin 20∞ = 0 FIO = +2.05 kN
5
Since IO is found to be in tension, our assumption was correct. Thus, the answers are
(a) IO

(b) 2.05 kN T 
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103
PROBLEM 6.67
The diagonal members in the center panels of the truss shown
are very slender and can act only in tension; such members are
known as counters. Determine the forces in the counters that are
acting under the given loading.
SOLUTION
Free body: Truss
SFx = 0 : Fx = 0
SM H = 0 : 24(3a) + 24(2a) + 24a - 12a - Fy (2a) = 0
+
Fy = +66 kN +
F = 66.0 kN ≠ 
SFy = 0 : H + 66 kN - 3(24 kN ) - 2(12 kN ) = 0
H = +30.0 kN H = 30.0 kN ≠ 
Free body: ABF
We assume that counter BG is acting.
+
SFy = 0 : -
2.4
FBG + 66 - 2(24) = 0
3.65
FBG = +27.375 kN FBG = 27.4 kN T 
Since BG is in tension, our assumption was correct
Free body: DEH
We assume that counter DG is acting.
+ SFy = 0 : -
2.4
FDG + 30 - 2(12) = 0
3.65
FDG = +9.125 kN FDG = 9.13 kN T 
Since DG is in tension, O.K.
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104
PROBLEM 6.68
The diagonal members in the center panels of the truss shown
are very slender and can act only in tension; such members are
known as counters. Determine the forces in the counters that
are acting under the given loading.
SOLUTION
Free body: Truss
SFx = 0 : Fx = 0
+
SM G = 0 : - Fy a + 24(2a) + 24a - 12a - 12(2a) = 0
Fy = 36.0 kN F = 36.0 kN≠ 
Free body: ABF
We assume that counter CF is acting.
+
SFy = 0 :
2.4
FCF + 36 - 2(24) = 0
3.65
FCF = 18.25 kN T 
FCF = +18.25 kN Since CF is in tension, O.K.
Free body: DEH
We assume that counter CH is acting.
+ SFy = 0 :
2.4
FCH - 2(12 kN ) = 0
3.65
FCH = +36.5 kN FCH = 36.5 kN T 
Since CH is in tension, O.K.
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105
PROBLEM 6.69
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a)
Number of members:
m = 16
Number of joints:
n = 10
Reaction components:
r=4
m + r = 20, 2n = 20
m + r = 2n 
Thus:
To determine whether the structure is actually completely constrained and determinate, we must try to find the
reactions at the supports. We divide the structure into two simple trusses and draw the free-body diagram of
each truss.
This is a properly supported simple truss – O.K.
This is an improperly supported simple truss.
Reaction at C passes through B. Thus,
Eq. SM B = 0 cannot be satisfied.)
Structure is improperly constrained 
Structure (b)
m = 16
n = 10
r=4
m + r = 20, 2n = 20
m + r = 2n 
Thus:
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106
PROBLEM 6.69 (Continued)
We must again try to find the reactions at the supports dividing the structure as shown.
Both portions are simply supported simple trusses.
Structure is completely constrained and determinate 
Structure (c)
m = 17
n = 10
r=4
m + r = 21, 2n = 20
m + r > 2n 
Thus:
This is a simple truss with an extra support which causes reactions (and forces in members) to be indeterminate.
Structure is completely constrained and indeterminate 
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107
PROBLEM 6.70
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Non-simple truss with r = 4, m = 16, n = 10
so m + r = 20 = 2n, but must examine further.
FBD Sections:
FBD
I:
SM A = 0
fi T1
II:
SFx = 0
fi T2
I:
SFx = 0
fi Ax
I:
SFy = 0
fi Ay
II:
SM E = 0
fi Cy
II:
SFy = 0
fi Ey
Since each section is a simple truss with reactions determined,
structure is completely constrained and determinate. 
Non-simple truss with r = 3, m = 16, n = 10
Structure (b):
so
m + r = 19 < 2n = 20
structure is partially constrained 
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108
PROBLEM 6.70 (Continued)
Structure (c):
Simple truss with r = 3, m = 17, n = 10 m + r = 20 = 2n, but the horizontal reaction forces Ax and E x are
collinear and no equilibrium equation will resolve them, so the structure is improperly constrained and
indeterminate

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109
PROBLEM 6.71
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Non-simple truss with r = 4, m = 12, n = 8 so r + m = 16 = 2n,
check for determinacy:
One can solve joint F for forces in EF, FG and then solve joint
E for E y and force in DE.
This leaves a simple truss ABCDGH with
r = 3, m = 9, n = 6 so r + m = 12 = 2n Structure is completely constrained and determinate 
Structure (b):
Simple truss (start with ABC and add joints alphabetically to complete truss) with r = 4, m = 13, n = 8
so
r + m = 17 > 2n = 16
Constrained but indeterminate 
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110
PROBLEM 6.71 (Continued)
Structure (c):
Non-simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n. To further examine, follow procedure in Part (a)
above to get truss at left.
Since F1 π 0 (from solution of joint F ),
SM A = aF1
π 0 and there is no equilibrium.
Structure is improperly constrained 
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111
PROBLEM 6.72
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a)
Number of members:
m = 12
Number of joints:
n=8
Reaction components:
r=3
m + r = 15,
Thus:
2n = 16
m + r 2n 
Structure is partially constrained 
Structure (b)
m = 13, n = 8
r=3
m + r = 16, 2n = 16
Thus:
m + r = 2n 
To verify that the structure is actually completely constrained and determinate, we observe that it is a simple
truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a roller. Thus:
Structure is completely constrained and determinate 
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112
PROBLEM 6.72 (Continued)
Structure (c)
m = 13, n = 8
r=4
m + r = 17, 2n = 16
Thus:
m + r 2n 
Structure is completely constrained and indeterminate 
This result can be verified by observing that the structure is a simple truss (follow lettering to check thus),
therefore rigid, and that its supports involve 4 unknowns.
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113
PROBLEM 6.73
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Rigid truss with r = 3, m = 14, n = 8
so
r + m = 17 2n = 16
so completely constrained but indeterminate 
Structure (b): Simple truss (start with ABC and add joints alphabetically), with
r = 3, m = 13, n = 8 so r + m = 16 = 2n
so completely constrained and determinate 
Structure (c):
Simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear so
cannot be resolved by any equilibrium equation.
structure is improperly constrained 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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114
PROBLEM 6.74
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
No. of members
m = 12
No. of joints
n = 8 m + r = 16 = 2n
No. of react. comps.
r = 4 unks = eqns
FBD of EH:
SM H = 0 Æ FDE ; SFx = 0 Æ FGH ; SFy = 0 Æ H y
Then ABCDGF is a simple truss and all forces can be determined.
This example is completely constrained and determinate. 
Structure (b):
No. of members
m = 12
No. of joints
n = 8 m + r = 15 2n = 16
No. of react. comps.
r = 3 unks eqns
partially constrained 
Note: Quadrilateral DEHG can collapse with joint D moving downward: in (a) the roller at F prevents this
action.
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115
PROBLEM 6.74 (Continued)
Structure (c):
No. of members
m = 13
No. of joints
n = 8 m + r = 17 2n = 16
No. of react. comps.
r = 4 unks eqns
completely constrained but indeterminate 
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116
PROBLEM 6.75
For the frame and loading shown, determine the force acting
on member ABC (a) at B, (b) at C.
SOLUTION
FBD ABC:
Note: BD is two-force member
(a)
Ê3
ˆ
SM C = 0 : (0.09 m)(200 N ) - (2.4 m) Á FBD ˜ = 0
Ë5
¯
FBD = 125.0 N
(b)
4
Æ SFx = 0 : 200 N - (125 N ) - C x = 0
5
≠ SFy = 0 :
3
FBD - C y = 0
5
36.9∞ �
C x = 100 N ¨
3
C y = (125 N ) = 75 N Ø
5
C = 125.0 N
36.9∞ 
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117
PROBLEM 6.76
Determine the force in member BD and the components of the reaction at C.
SOLUTION
We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along the BD.
Free body: ABC
Attaching FBD at D and resolving it into components, we write
ˆ
Ê 450
FBD ˜ (240 mm) = 0
SM C = 0: ( 400 N )(135 mm) + Á
¯
Ë 510
+
FBD = -255 N +
SFx = 0 : C x +
240
( -255 N ) = 0
510
C x = +120.0 N +
FBD = 255 N C 
SFy = 0 : C y - 400 N +
C x = 120.0 NÆ 
450
( -255 N ) = 0
510
C y = +625 N C y = 625 N≠ 
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118
PROBLEM 6.77
Rod CD is fitted with a collar at D that can be moved along rod AB, which is
bent in the shape of an arc of circle. For the position when q = 30∞, determine
(a) the force in rod CD, (b) the reaction at B.
SOLUTION
FBD:
(a)
SM C = 0 : ( 400 mm)(100 N - By ) = 0
≠ SFy = 0 : - 100 N + FCD sin 30∞ - 100 N = 0
(b)
B y = 100 N Ø
FCD = 400 N T 
Æ SFx = 0 : ( 400 N) cos 30∞ - Bx = 0
B x = 346.411 N ¨
B = Bx 2 + By 2 = 360.27 N so B = 360 N
16.10∞ 
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119
PROBLEM 6.78
Solve Problem 6.77 when q = 150∞.
PROBLEM 6.77 Rod CD is fitted with a collar at D that can be moved
along rod AB, which is bent in the shape of an arc of circle. For the position
when q = 30∞, determine (a) the force in rod CD, (b) the reaction at B.
SOLUTION
Note that CD is a two-force member, FCD must be directed along DC.
(a)
+
SM B = 0 : (100 N )(2 R) - ( FCD sin 30∞) R = 0
FCD = 400 N
(b)
+
SM C = 0 : (100 N ) R + ( By ) R = 0
By = 100 N
+
FCD = 400 N T 
B y = 100 N Ø
SFx = 0 : - FCD cos 30∞ + Bx = 0
-( 400 N ) cos 30∞ + Bx = 0
Bx = 346.41 N B x = 346.41 N Æ
B = 361 N
16.10° 
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120
PROBLEM 6.79
Determine the components of all forces acting on member
ABCD when q = 0.
SOLUTION
Free body: Entire assembly
+
SM B = 0 : A(200 mm) - (300 N )(500 mm) = 0
A = 750 N +
+
A = 750 N Æ 
B x = 750 N ¨ 
SFx = 0 : Bx + 750 N = 0 Bx = -750 N
SFy = 0 : By - 300 N = 0
B y = 300 N ≠ 
By = +300 N
We note that D is directed along DE, since DE is a two-force member.
Free body: Member ABCD
SM C = 0 : D(300) - (300 N )(100) + (750 N )(200) = 0
+
D = - 400 N D = 400 N Ø 
+
+
SFx = 0 : C x + 750 - 750 = 0
Cx = 0
SFy = 0 : C y + 300 - 400 = 0
C y = +100.0 N C = 100.0 N ≠ 
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121
PROBLEM 6.80
Determine the components of all forces acting on member
ABCD when q = 90∞.
SOLUTION
Free body: Entire assembly
+
SM B = 0 : A(200 mm) - (300 N )(200 mm) = 0
+
SFx = 0 : Bx + 300 N - 300 N = 0 Bx = 0
+
A = +300 N SFy = 0 : By = 0
A = 300 N Æ 
B=0 
Free body: Member ABCD We note that D is directed along DE, since DE is a two-force member.
+
SM C = 0 : D(300 mm) + (300 N )(200 mm) = 0
D = -200 N D = 200 N Ø 
+
SFx = 0 : C x + 300 N = 0
+
C x = -300 N C x = 300 N¨ 
SFy = 0 : C y - 200 N = 0
C y = +200 N C y = 200 N ≠ 
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122
3.6 m
A
PROBLEM 6.81
Determine the components of the forces acting on each member of the frame
shown.
2400 N
2.7 m
C
D
B
2.7 m
F
E
4.8 m
SOLUTION
Free body: Entire frame.
SM E = 0 : - (2400 N )(3.6 m) + F ( 4.8 m) = 0
F = +1800 N 3.6 m
A
+
SFy = 0 : - 2400 N + 1800 N + E y = 0
2400 N
E y = +600 N +
C
D
B
F = 1800 N ≠ 
E y = 600 N ≠ 
Ex = 0 
SFx = 0 : E x = 0 Members
F
E
Ey
Ex
Free body: Member BCD
F
4.8 m
+
SM B = 0 :
- (2400 N )(3.6 m) + C y (2.4 m) = 0
C y = +3600 N 
+
SM C = 0 :
- (2400 N )(1.2 m) + By (2.4 m) = 0
By = +1200 N 
+
2.4 m
1.2 m
Bx B
Ax
A
2.7 m
B
C
Cx
Free body: Member ABE
2400 N
D
+
SM A = 0 : Bx (2.7 m) = 0
Bx = 0 
SFx = 0: + Bx - Ax = 0
Ax = 0 
Ay
A
Ay
By
+
Ax
Free body: Member BCD.
Cx
+
2.7 m
Cy
E
600 N
SFy = 0 : - Ay + By + 600 N = 0
- Ay + 1200 N + 600 N = 0 2.4 m
C
B1
- Bx + C x = 0
+
Cy
By
SFx = 0 :
F
1800 N
SFx = 0 :
- Bx + C x = 0
Ay = +1800 N 
Returning now to member BCD, we write
0 + C x = 0
Cx = 0 
Free body: Member ACF (Check).
All unknown components have
now been found, to check the results, we verify that members ACF is in
equilibrium.
+
SM C = (1800 N )(2.4 m) + Ay (2.4 m) - Ax (2.7 m)
= (1800 N )(2.4 m) - (1800 N )(2.4 m) - 0 = 0
(checks)
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123
PROBLEM 6.82
Solve Sample Problem 6.5 assuming that the 18-kN load is replaced by a
clockwise couple of magnitude 72 kN · m applied to member CDEF at Point D.
SAMPLE PROBLEM 6.5 For the frame and loading shown, determine the
components of all forces acting on member ABC.
SOLUTION
Free body: Entire frame
+
SFx = 0 : Ax = 0
SM F = 0 : - 72 kN ◊ m - Ay (3.6 m) = 0
+
Ay = -20 kN A y = 20 kN Ø
A = 20.0 kN Ø 
Free body: Member ABC
Note: BE is a two-force member. Thus B is directed along line BE.
SM C = 0 : B( 4 m) + (20 kN )(3.6 m) = 0
+
B = -18 kN +
Fx = 0 : - 18 kN + C x = 0
C x = 18 kN +
B = 18.00 kN ¨ 
C x = 18.00 kN Æ 
SFy = 0 : C y - 20 kN = 0
C y = 20 kN C y = 20.0 kN ≠ 
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124
PROBLEM 6.83
Determine the components of the reactions at A and E if a 750-N force directed
vertically downward is applied (a) at B, (b) at D.
SOLUTION
Free-body: Entire Frame
The following analysis is valid for both (a) and (b) since position of load on its line of action is immaterial.
SM E = 0 : - (750 N )(240 mm) - Ax ( 400 mm) = 0
+
Ax = -450 N A x = 450 N ¨
+
+
(a)
SFx = 0 : E x - 450 N = 0 E x = 450 N E x = 450 N Æ
SFy = 0 : Ay + E y - 750 N = 0 (1)
Load applied at B.
Free body: Member CE
CE is a two-force member. Thus, the reaction at E must be directed along CE.
Ey
240 mm
=
; E y = 225 N ≠
450 N 480 mm
From Eq. (1):
Ay + 225 - 750 = 0; A y = 525 lb≠
Thus, reactions are:
(b)
A x = 450 N ¨,
A y = 525 lb≠ 
E x = 450NÆ,
E y = 225 lb≠ 
Load applied at D.
Free body: Member AC
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
160 mm
=
A y = 150.0 N ≠
450 N 480 mm
From Eq. (1):
Ay + E y - 750 N = 0
150 N + E y - 750 N = 0
E y = 600 N E y = 600 N ≠
A x = 450 N ¨,
Thus, reactions are: A y = 150.0 N ≠ 
E x = 450 N Æ, E y = 600 N ≠ 
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125
PROBLEM 6.84
Determine the components of the reactions at A and E if a 750-N force
directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
Free body: Entire frame
The following analysis is valid for both (a) and (b) since position of load on its line of action is immaterial.
SM E = 0 : - (750 N )(80 mm) - Ax (200 mm) = 0
+
Ax = - 300 N A x = 300 N ¨
+
+
(a)
SFx = 0 : E x - 300 N = 0 E x = 300 N E x = 300 N Æ
SFy = 0 : Ay + E y - 750 N = 0 (1)
Load applied at B.
Free body: Member CE
CE is a two-force member. Thus, the reaction at E must be directed along CE.
Ey
300 N
From Eq. (1):
=
75 mm
250 mm
Ay + 90 N - 750 N = 0
E y = 90 N ≠
Ay = 660 N ≠
Thus reactions are:
A x = 300 N¨, A y = 660 N ≠ 
E x = 300 N Æ, E y = 90.0 N ≠ 
(b)
Load applied at D.
Free body: Member AC
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
300 N
=
125 mm
250 mm
Ay = 150 N ≠
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126
PROBLEM 6.84 (Continued)
From Eq. (1):
Ay + E y - 750 N = 0
150 N + E y - 750 N = 0
E y = 600 N E y = 600 N ≠
Thus, reactions are:
A x = 300 N¨,
A y = 150.0 N ≠ 
E x = 300 N Æ,
E y = 600 N ≠ 
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127
PROBLEM 6.85
Determine the components of the reactions at A and E if the frame is loaded
by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.
SOLUTION
Free body: Entire frame
The following analysis is valid for both (a) and (b) since the point of application of the couple is immaterial.
SM E = 0 : -36 N ◊ m - Ax (0.4 m) = 0
+
Ax = - 90 N A x = 90.0 N ¨
+
SFx = 0 : - 90 +E x = 0
E x = 90 N E x = 90.0 N Æ
+
(a)
SFy = 0 : Ay + E y = 0 (1)
Couple applied at B.
Free body: Member CE
AC is a two-force member. Thus, the reaction at E must be directed along EC.
Ey
90 N
From Eq. (1):
=
0.24 m
; E y = 45.0 N ≠
0.48 m
Ay + 45 N = 0
Ay = - 45 N A y = 45.0 N Ø
Thus, reactions are
(b)
A x = 90.0 N ¨,
A y = 45.0 N Ø 
E x = 90.0 N Æ,
E y = 45.0 N ≠ 
Couple applied at D.
Free body: Member AC
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
90 N
=
0.16 m
; A y = 30 N ≠
0.48 m
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128
PROBLEM 6.85 (Continued)
From Eq. (1):
Ay + E y = 0
30 N + E y = 0
E y = - 30 N E y = 30 N Ø
Thus, reactions are:
A x = 90.0 N ¨,
A y = 30.0 N ≠ 
E x = - 90.0 N Æ,
E y = 30.0 N Ø 
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129
PROBLEM 6.86
Determine the components of the reactions at A and E if the frame is loaded
by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.
SOLUTION
Free body: Entire frame
The following analysis is valid for both (a) and (b) since the point of application of the couple is immaterial.
+
SM E = 0 : -36 N ◊ m - Ax (0.2 m) = 0
Ax = - 180 N A x = 180 N ¨
+
SFx = 0 : - 180 N + E x = 0
E x = 180 N E x = 180 N Æ
+
(a)
SFy = 0 : Ay + E y = 0 (1)
Couple applied at B.
Free body: Member CE
AC is a two-force member. Thus, the reaction at E must be directed along EC.
Ey
180 N
From Eq. (1):
=
0.075 m
0.25 m
E y = 54 N ≠
Ay + 54 N = 0
Ay = - 54 N A y = 54.0 N Ø
Thus reactions are
(b)
A x = 180.0 N¨,
A y = 54.0 N Ø 
E x = 180.0 N Æ,
E y = 54.0 N ≠ 
Couple applied at D.
Free body: Member AC
AC is a two-force member. Thus, the reaction at A must be
directed along EC.
Ay
0.125 m
=
Ay = 90 N ≠
180 N 0.25 m
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130
PROBLEM 6.86 (Continued)
From Eq. (1):
Ay + E y = 0
90 N + E y = 0
E y = - 90 N E y = 90 N Ø
Thus, reactions are
A x = 180.0 N¨,
A y = 90.0 N ≠ 
E x = -180.0 N Æ,
E y = 90.0 N Ø 
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131
PROBLEM 6.87
Determine the components of the reactions at A and B, (a) if the 500-N load is
applied as shown, (b) if the 500-N load is moved along its line of action and is
applied at Point F.
SOLUTION
Free body: Entire frame
Analysis is valid for either (a) or (b), since position of 500 N load on its line of action is immaterial.
SM A = 0 : By (250) - (500 N )(150) = 0 By = + 300 N
+
+
+
(a)
SFy = 0 : Ay + 300 - 500 = 0
Ay = + 200 N
SFx = 0 : Ax + Bx = 0 (1)
Load applied at E.
Free body: Member AC
Since AC is a two-force member, the reaction at A must be directed along CA. We have
Ax
200 N
=
250 mm 125 mm
From Eq. (1):
A x = 400 N ¨,
A y = 200 N ≠ 
B x = 400 N Æ,
B y = 300 N ≠ 
- 400 + Bx = 0 Bx = + 400 N
Thus,
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132
PROBLEM 6.87 (Continued)
(b)
Load applied at F.
Free body: Member BCD
Since BCD is a two-force member (with forces applied at B and C only),
the reaction at B must be directed along CB. We have therefore
Bx = 0
The reaction at B is
From Eq. (1):
The reaction at A is
B x = 0
Ax + 0 = 0
B y = 300 N ≠ 
Ax = 0
A x = 0
A y = 200 N ≠ 
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133
PROBLEM 6.88
The 250 N load can be moved along the line of action shown and applied at A,
D, or E. Determine the components of the reactions at B and F if the 250 N load
is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b) and (c) since the position of the load
along its line of action is immaterial.
SM F = 0 : (250 N )(200 mm) - Bx (300 mm) = 0
+
500
N = 166.67 N B x = 166.7 N Æ
3
SFx = 0 : 166.67 N + Fx = 0
Bx =
+
Fx = -166.67 N Fx = 166.67 N ¨
+
(a)
SFy = 0 : By + Fy - 250 N = 0
(1)
Load applied at A.
Free body: Member CDB
CDB is a two-force member. Thus, the reaction at B must be directed along BC.
By
500 / 3 N
From Eq. (1):
=
125 mm
400 mm
By =
625
N = 52.083 N
12
B y = 52.1 N ≠
625
N + Fy - 250 N = 0
12
Fy = 197.92 N Fy = 197.9 N ≠
Thus reactions are:
B x = 166.7 N Æ,
B y = 52.1 N≠ 
Fx = 166.7 N ¨,
Fy = 197.9 N ≠ 
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134
PROBLEM 6.88 (Continued)
(b)
Load applied at D.
Free body: Member ACF.
ACF is a two-force member. Thus, the reaction at F must be
directed along CF.
Fy
500 / 3 N
From Eq. (1):
By +
=
175 mm
400 mm
Fy =
875
N = 72.92 N ≠
12
875
N - 250 N = 0
12
By = 177.083 N
By = 177.1 N ≠
Thus, reactions are:
(c)
B x = 166.7 N ¨,
B y = 177.1 N ≠ 
Fx = 166.7 N Æ,
Fy = 72.9 N≠ 
Load applied at E.
Free body: Member CDB
This is the same free body as in Part (a).
Reactions are same as (a) 
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135
PROBLEM 6.89
The 250-N load is removed and a 40 N · m. clockwise couple is applied
successively at A, D, and E. Determine the components of the reactions at B
and F if the couple is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b), and (c), since the point of
application of the couple is immaterial.
SM F = 0 : - 40 N ◊ m - Bx (0.3 m) = 0
+
Bx = +
SFx = 0 : - 400 / 3 N + Fx = 0
Fx =
+
(a)
400
N B x = 133.33 N ¨
3
SFy = 0 : By + Fy = 0 400
N Fx = 133.33 N Æ
3
(1)
Couple applied at A.
Free body: Member CDB
CDB is a two-force member. Thus, reaction at B must be directed along BC.
By
From Eq. (1):
=
125 mm
400 mm
By =
125
N = 41.667 N Ø
3
400 / 3 N
125
N + Fy = 0
3
Fy = 125 / 3 N Fy = 41.667 N ≠
Thus, reactions are:
B x = 133.3 N ¨,
B y = 41.7 N Ø 
Fx = 133.3 N Æ,
Fy = 41.7 N ≠ 
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136
PROBLEM 6.89 (Continued)
(b)
Couple applied at D.
Free body: Member ACF.
ACF is a two-force member. Thus, the reaction at F must be directed along CF.
Fy
175 mm
=
400
N 400 mm
3
From Eq. (1):
By -
Fy =
175
N = 58.333 N Ø
3
175
N= 0
3
By = +
175
N
3
B y = 58.333 N ≠
Thus, reactions are:
(c)
B x = 133.3 N ¨,
B y = 58.30 N ≠ 
Fx = 133.3 N Æ,
Fy = 58.3 N Ø 
Couple applied at E.
Free body: Member CDB
This is the same free body as in Part (a).
Reactions are same as in (a) 
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137
PROBLEM 6.90
(a) Show that when a frame supports a pulley at A, an equivalent loading of the frame and of each of its
component parts can be obtained by removing the pulley and applying at A two forces equal and parallel to the
forces that the cable exerted on the pulley. (b) Show that if one end of the cable is attached to the frame at a
point B, a force of magnitude equal to the tension in the cable should also be applied at B.
SOLUTION
First note that, when a cable or cord passes over a frictionless, motionless pulley,
the tension is unchanged.
SM C = 0 : rT1 - rT2 = 0 T1 = T2
(a)
Replace each force with an equivalent force-couple.
(b)
Cut cable and replace forces on pulley with equivalent pair of forces at A as above.
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138
PROBLEM 6.91
Knowing that the pulley has a radius of 50 mm, determine the components
of the reactions at B and E.
SOLUTION
Free body: Entire assembly
SM E = 0 : - (300 N )(350 mm) - Bx (150 mm) = 0
+
Bx = -700 N
+
SFx = 0 : - 700 N + E x = 0
E x = 700 N
+
B x = 700 N ¨
E x = 700 NÆ
SFy = 0 : By + E y - 300 N = 0 (1)
Free body: Member ACE
+
SM C = 0 : (700 N )(150 mm) - (300 N )(50 mm) - E y (180 mm) = 0
E y = 500 N
From Eq. (1):
E y = 500 N ≠
By + 500 N - 300 N = 0
By = -200 N
B y = 200 N Ø
Thus, reactions are:
B x = 700 N ¨,
B y = 200 N Ø 
E x = 700 NÆ,
E y = 500 N ≠ 
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139
PROBLEM 6.92
Knowing that each pulley has a radius of 250 mm, determine
the components of the reactions at D and E.
SOLUTION
Free body: Entire assembly
SM E = 0 : ( 4.8 kN )( 4.25 m) - Dx (1.5 m) = 0
+
Dx = +13.60 kN +
SFx = 0 : E x + 13.60 kN = 0
E x = 13.60 kN¨ 
E x = -13.60 kN +
D x = 13.60 kN Æ 
SFy = 0 : D y + E y - 4.8 kN = 0 (1)
Free body: Member ACE
+
SM A = 0 : ( 4.8 kN )(2.25 m) + E y ( 4 m) = 0
E y = -2.70 kN From Eq. (1):
E y = 2.70 kNØ 
D y - 2.70 kN - 4.80 kN = 0
D y = +7.50 kN D y = 7.50 kN ≠ 
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140
PROBLEM 6.93
Two 225-mm diameter pipes (pipe 1 and pipe 2) are supported every
2.5 m by a small frame like that shown. Knowing that the combined
weight of each pipe and its contents is 450 N/m and assuming
frictionless surfaces, determine the components of the reactions at
A and G.
SOLUTION
425
Free-body: Pipe 2
375
200
|
17
18
15
W = ( 450 N/m)(2.5 m) = 1125 N
F D 1125 N
=
=
8 17
15
F = 600 NÆ
D = 1275 N
r = 112.5 mm
Geometry of pipe 2
CF = CD By symmetry:
(1)
Equate horizontal distance:
r+
8
Ê 15 ˆ
r = CD Á ˜
Ë 17 ¯
17
25
Ê 15 ˆ
r = CD Á ˜
Ë 17 ¯
17
25
5
CD = r = r
15
3
5
5
CF = r = (112.5 mm)
3
3
CF = 187.5 mm
From Eq. (1):
Free-body: Member CFG
+
SM C = 0 : (600 N )(187.5 mm) - Gx ( 400 mm) = 0
Gx = 281.25 N G x = 281 N Æ 
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141
PROBLEM 6.93 (Continued)
Free body: Frame and pipes
Note: Pipe 2 is similar to pipe 1.
AE = CF = 187.5 mm
E = F = 600 N
+
SM A = 0 : G y (375 mm) - (281.25 N )(600 mm) - (1125 N )(112.5 mm)
-(1125 N )( 487.5 mm) - (600 N )(187.5 mm) = 0
G y = 2550 N +
SFx = 0 : Ax + 600 N + 281.25 N = 0
Ax = -881.25 N +
G y = 2550 N ≠ 
A x = 881 N ¨ 
SFy = 0 : Ay + 2550 N - 1125 N - 1125 N = 0
Ay = -300 N A y = 300 N Ø 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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142
PROBLEM 6.94
Solve Problem 6.93 assuming that pipe 1 is removed and that only pipe
2 is supported by the frames.
PROBLEM 6.93 Two 225-mm diameter pipes (pipe 1 and pipe 2)
are supported every 2.5 m by a small frame like that shown. Knowing
that the combined weight of each pipe and its contents is 450 N/m
and assuming frictionless surfaces, determine the components of the
reactions at A and G.
SOLUTION
425
Free-body: Pipe 2
W = ( 450 N/m)(2.5 m) = 1125 N
375
200
|
17
18
15
F D 1125 N
=
=
8 17
15
F = 600 NÆ
D = 1275 N
r = 112.5 mm
Geometry of pipe 2
CF = CD By symmetry:
(1)
Equate horizontal distance:
r+
8
Ê 15 ˆ
r = CD Á ˜
Ë 17 ¯
17
25
Ê 15 ˆ
r = CD Á ˜
Ë 17 ¯
17
25
5
CD = r = r
15
3
5
5
CF = r = (112.5 mm)
3
3
CF = 187.5 mm
From Eq. (1):
Free-body: Member CFG
+
SM C = 0 : (600 N )(187.5 mm) - Gx ( 400 mm) = 0
Gx = 281.25 N G x = 281 N Æ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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143
PROBLEM 6.94 (Continued)
Free body: Frame and pipe 2
G x = 281 N Æ 
+
SM A = 0 : G y (375 mm) - (281.25 N )(600 mm) - (1125 N )( 487.5 mm) = 0
G y = 1912.5 N G y = 1913 N ≠ 
+
SFx = 0 : Ax + 281.25 N = 0
A x = 281 N ¨ 
Ax = -281.25 N +
SFy = 0 : Ay + 1912.5 N - 1125 N = 0
Ay = -787.5 N A y = 788 N Ø 
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144
PROBLEM 6.95
A trailer weighing 12 kN is attached to a 14.5 kN pickup truck by a ball-and-socket truck hitch at D. Determine
(a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each
of the truck wheels due to the trailer.
SOLUTION
(a)
Free body: Trailer
(We shall denote by A, B, C the reaction at one wheel)
SM A = 0 : - (12 kN )(0.6 m) + D(3.3 m) = 0
+
D = 2.1818 kN
+
SFy = 0 : 2 A - 12 kN + 2.1818 kN = 0
A = 4.9091 kN A = 4.91 kN ≠ 
Free body: Truck
SM B = 0 : (2.1818 kN )(0.9 m) - (14.5 kN )(1.5 m) + 2C (2.7 m) = 0
+
C = 3.6641 kN
+
SFy = 0 : 2 B - 2.1818 kN - 14.5 kN + 2(3.6641 kN ) = 0
B = 4.6768 kN (b)
C = 3.66 kN ≠ 
B = 4.68 kN ≠ 
Additional load on truck wheels
Use free body diagram of truck without 14.5 kN
+
SM B = 0 : (2.1818 kN )(0.9 m) + 2C (2.7 m) = 0
C = -0.3636 kN +
DC = -0.364 kN 
SFy = 0 : 2 B - 2.1818 N - 2(0.3636 kN ) = 0
B = 1.4545 kN DB = +1.455 kN 
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145
PROBLEM 6.96
In order to obtain a better weight distribution over the four
wheels of the pickup truck of Problem 6.95, a compensating
hitch of the type shown is used to attach the trailer to the truck.
The hitch consists of two bar springs (only one is shown in the
figure) that fit into bearings inside a support rigidly attached to
the truck. The springs are also connected by chains to the trailer
frame, and specially designed hooks make it possible to place
both chains in tension. (a) Determine the tension T required in
each of the two chains if the additional load due to the trailer
is to be evenly distributed over the four wheels of the truck.
(b) What are the resulting reactions at each of the six wheels of
the trailer-truck combination?
PROBLEM 6.95 A trailer weighing 12 kN is attached to a
14.5 kN pickup truck by a ball-and-socket truck hitch at D.
Determine (a) the reactions at each of the six wheels when the
truck and trailer are at rest, (b) the additional load on each of
the truck wheels due to the trailer.
SOLUTION
(a)We shall first find the additional reaction D at each wheel due the
trailer.
Free body diagram (Same D at each truck wheel)
+
SM A = 0 : - (12 kN )(0.6 m) + 2 D( 4.2 m) + 2 D(6.9 m) = 0
D = 0.32432 kN
SFY = 0 : 2 A - 12 kN + 4(0.32432 kN ) = 0;
A = 5.3514 kN;
Free body: Truck
(Trailer loading only)
+
+
SM D = 0 : 2 D(3.6 m) + 2 D(0.9 m) - 2T (0.51 m) = 0
T = 8.824 D
= 8.824(0.32432 kN )
T = 2.8616 kN
A = 5.35 kN ≠
T = 2.86 kN 
Free body: Truck
(Truck weight only)
+
SM B = 0 : - (14.5 kN )(1.5 m) + 2C ¢(2.7 m) = 0
C ¢ = 4.02778 kN
C¢ = 4.03 kN ≠
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146
PROBLEM 6.96 (Continued)
+
SFy = 0 : 2 B ¢ - 14.5 kN + 2( 4.02778 kN ) = 0
B ¢ = 3.2222 kN B ¢ = 3.22 kN ≠
Actual reactions
B = B ¢ + D = 3.2222 kN + 0.32432 kN = 3.5465 kN B = 3.55 kN≠ 
C = C ¢ + D = 4.02778 kN + 0.32432 kN = 4.3521 kN
C = 4.35 kN ≠ 
(From Part a):
A = 5.35 kN≠ 
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147
PROBLEM 6.97
The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the
cab wheels. The distance from C to D is 1 m. The center of gravity of the 300 kN motor unit is located at Gm,
while the centers of gravity of the 100 kN cab and 75 kN load are located, respectively, at Gc and Gl. Knowing
that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels,
(b) the forces exerted on the motor unit at C and D.
SOLUTION
(a)
Free body: Entire machine
A =Reaction at each front wheel
B =Reaction at each rear wheel
+
SM A = 0 : 75(3.2 m) - 100(1.2 m) + 2 B( 4.8 m) - 300(5.6 m) = 0
2 B = 325 kN +
B = 162.5 kN ≠ 
SFy = 0 : 2 A + 325 - 75 - 100 - 300 = 0
2 A = 150 kN A = 75.0 kN ≠ 
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148
PROBLEM 6.97 (Continued)
(b)
Free body: Motor unit
+
SM D = 0 : C (1 m) + 2 B(2.8 m) - 300(3.6 m) = 0
C = 1080 - 5.6 B (1)
B = 162.5 kN, C = 1080 - 5.6(162.5) = 170 kN
Recalling
C = 170.0 kN ¨ 
+
SFx = 0 : Dx - 170 = 0 +
SFy = 0 : 2(162.5) - D y - 300 = 0 D x = 170.0 kN Æ 
D y = 25.0 kN Ø 
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149
PROBLEM 6.98
Solve Problem 6.97 assuming that the 75 kN load has been removed.
PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin
located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor
unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively,
at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each
of the four wheels, (b) the forces exerted on the motor unit at C and D.
SOLUTION
(a)
Free body: Entire machine
A =Reaction at each front wheel
B =Reaction at each rear wheel
+
SM A = 0 : 2 B( 4.8 m) - 100(1.2 m) - 300(5.6 m) = 0
2 B = 375 kN +
(b)
SFy = 0 : 2 A + 375 - 100 - 300 = 0
2 A = 25 kN B = 187.5 kN ≠ 
A = 12.50 kN ≠ 
Free body: Motor unit
See solution of Problem 6.97 for free body diagram and derivation of Eq. (1). With B = 187.5 kN,
we have
C = 1080 - 5.6(187.5) = 30 kN
C = 30.0 kN ¨ 
+
+
SFx = 0 : Dx - 30 = 0 SFy = 0 : 2(187.5) - D y - 300 = 0 D x = 30.0 kN Æ 
D y = 75.0 kN Ø 
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150
PROBLEM 6.99
For the frame and loading shown, determine the components of the forces
acting on member CFE at C and F.
SOLUTION
Free body: Entire frame
+
SM D = 0 : (200 N )(325 mm) + Ax (250 mm) = 0
Ax = -260 N,
A x = 260 N Æ
Free body: Member ABF
+
SM B = 0 : - (260 N )(150 mm) + Fx (100 mm) = 0
Fx = +390 N
Free body: Member CFE
Fx = 390 N¨ 
From above:
SM C = 0 : (200 N )(225 mm) - (390 N )(100 mm) - Fy (100 mm) = 0
+
Fy = +60 N +
+
Fy = 60.0 N≠ 
SFx = 0 : C x - 390 N = 0
C x = +390 N C x = 390 NÆ 
SFy = 0 : - 200 N + 60 N + C y = 0; C y = +140 N C y = 140.0 N ≠ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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151
PROBLEM 6.100
For the frame and loading shown, determine the components of the forces acting
on member CDE at C and D.
SOLUTION
Free body: Entire frame
+ SM y = 0 : Ay - 125 N = 0
Ay = 125 N
A y = 125.0 N ≠
SM F = 0 : Ax (173.2 + 2 ¥ 86.6) - (125 N )(300 mm) = 0
+
Ax = 108.256 N
A x = 108.3 N ¨
+ SFx = 0 : F - 108.256 N = 0
F = 108.256 N
F = 108.3 N Æ
Free body: Member CDE
+
SM C = 0 : D y (100 mm) - (125 N )(250 mm) = 0
D y = +312.5 N +
D y = 313 N ≠ 
SFy = 0 : -C y + 312.5 N - 125 N = 0
C y = +187.5 N C y = 187.5 NØ 
Free body: Member ABD
+
SM B = 0 : Dx (86.66 mm) + (108.256 N )(173.2 mm)
-(125 N )(100 mm) - (312.5 N )(50 mm)
Dx = +108.182 N
Return to free body: Member CDE
From above
+
Dx = +108.182 N D x = 108.2 N ¨ 
SFx = 0 : C x - 108.182 N
Dx = 108.2 N¬ 
C x = +108.182 N C x = 108.2 N Æ 
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152
PROBLEM 6.101
For the frame and loading shown, determine the components of all forces
­acting on member ABE.
SOLUTION
FBD Frame:
SM E = 0 : (1.8 m) Fy - (2.1 m)(12 kN ) = 0
Fy = 14.00 kN ≠
≠SFy = 0 : - E y + 14.00 kN - 12 kN = 0
� E y = 2 kN Ø
E y = 2.00 kNØ 
FBD member BCD:
SM B = 0 : (1.2 m)C y - (12 kN )(1.8 m) = 0 C y = 18.00 kN ≠
But C is ^ ACF, so C x = 2C y ; C x = 36.0 kN Æ
ÆSFx = 0 : - Bx + C x = 0 Bx = C x = 36.0 kN
Bx = 36.0 kN ¨ on BCD
≠SFy = 0 : - By + 18.00 kN - 12 kN = 0 By = 6.00 kN Ø on BCD
B x = 36.0 kN Æ 
on ABE:
B y = 6.00 kN≠ 
FBD member ABE:
SM A = 0 : (1.2 m)(36.0 kN ) - (0.6 m)(6.00 kN )
+ (0.9 m)(2.00 kN ) - (1.8 m)( E x ) = 0
E x = 23.0 kN ¨ 
Æ SFx = 0 : - 23.0 kN + 36.0 kN - Ax = 0 A x = 13.00 kN ¨ 
≠SFy = 0 : - 2.00 kN + 6.00 kN - Ay = 0 A y = 4.00 kN Ø 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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153
PROBLEM 6.102
For the frame and loading shown, determine the components of all forces
acting on member ABE.
PROBLEM 6.101 For the frame and loading shown, determine the components
of all forces acting on member ABE.
SOLUTION
FBD Frame:
SM F = 0 : (1.2 m)(2400 N ) - ( 4.8 m) E y = 0 E y = 600 N ≠ 
FBD member BC:
Cy =
4.8
8
Cx = Cx
5.4
9
SM C = 0 : (2.4 m) By - (1.2 m)(2400 N ) = 0 By = 1200 N Ø
B y = 1200 N ≠ 
on ABE:
so
≠ SFy = 0 : - 1200 N + C y - 2400 N = 0 C y = 3600 N ≠
9
C x = C y C x = 4050 N Æ
8
ÆSFx = 0 : - Bx + C x = 0 Bx = 4050 N ¨ on BC
B x = 4050 N Æ 
on ABE:
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154
PROBLEM 6.102 (Continued)
FBD member AB0E:
SM A = 0 : a( 4050 N ) - 2aE x = 0
E x = 2025 N ÆSFx = 0 : - Ax + ( 4050 - 2025) N = 0 ≠SFy = 0 : 600 N + 1200 N - Ay = 0 E x = 2025 N ¨ 
A x = 2025 N¨ 
A y = 1800 N Ø 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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155
PROBLEM 6.103
Knowing that P = 75 N and Q = 325 N, determine the components of the
forces exerted (a) on member BCDF at C and D, (b) on member ACEG at E.
SOLUTION
Free body: Entire frame
+
SM E = 0 : ( P + Q )(625 mm) - ( P + Q )(375 mm) - Dx (200 mm) = 0
5
4
5
D x = ( P + Q) Æ
4
+ SF = 0 : - E + ( P + Q ) 5 = 0
x
x
4
5
5
Ex = ( P + Q) ; Ex = ( P + Q) ¨
4
4
Free body: Member BCDF
Dx = ( P + Q )
From above:
5
D x = ( P + Q) Æ 
4
+
+
5
Cx = ( P + Q) ¨ 
4
SFx = 0 : -C x + Dx = 0 5
SM D = 0 : - ( P + Q ) (100 mm) - P (375 mm) + Q(625 mm) - C y (250 mm)
4
+
C y = 2Q - 2 P C y = 2Q - 2 P ≠ 
SFy = 0 : D y + (2Q - 2 P ) = P + Q = 0
D y = - Q + 3P D y = - Q + 3P ≠ 
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156
PROBLEM 6.103 (Continued)
Free body: Member ACEG
From above
5
E x = ( P + Q) ¨ 
4
C y = 2 Q - 2P
+
and
SFy = 0 : E y - C y - P - Q = 0
E y - ( 2Q - 2 P ) - P - Q = 0
E y = 3Q - P E y = 3Q - P ≠ 
P = 75 N and Q = 325 N
Cx = ( P + Q)
5
5
= ( 400) = +500 N 4
4
C y = 2 Q - 2 P = 2(325) - 2(75) = +500 N Dx = ( P + Q )
5
5
= ( 400) = +500 N 4
4
D y = - Q + 3P = - 325 + 3(75) = -100 N Ex = ( P + Q)
5
5
= ( 400) = +500 N
4
4
E y = 3Q - P = 3(325) - 75 = + 900 N
C x = 500 N¨ 
C y = 500 N ≠ 
D x = 500 N Æ 
D y = 100.0 N Ø 
E x = 500 N ¨ 
E y = 900 N ≠ 
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157
PROBLEM 6.104
Knowing that P = 125 N and Q = 275 N, determine the components of
the forces exerted (a) on member BCDF at C and D, (b) on member
ACEG at E.
SOLUTION
Free body: Entire frame
SM E = 0 : ( P + Q )(625 mm) - ( P + Q )(375 mm) - Dx (200 mm) = 0
+
5
4
5
D x = ( P + Q) Æ
4
Dx = ( P + Q )
+
SFx = 0 : - E x + ( P + Q )
5
=0
4
5
5
Ex = ( P + Q) ; Ex = ( P + Q) ¨
4
4
Free body: Member BCDF
From above:
5
D x = ( P + Q) Æ 
4
+
+
SFx = 0 : -C x + Dx = 0 5
Cx = ( P + Q) ¨ 
4
5
SM D = 0 : - ( P + Q ) (100 mm) - P (375 mm) + Q(625 mm) - C y (250 mm)
4
+
C y = 2Q - 2 P C y = 2Q - 2 P ≠ 
SFy = 0 : D y + (2Q - 2 P ) = P + Q = 0
D y = - Q + 3P D y = - Q + 3P ≠ 
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158
PROBLEM 6.104 (Continued)
Free body: Member ACEG
From above
5
E x = ( P + Q) ¨ 
4
C y = 2 Q - 2P
+
and
SFy = 0 : E y - C y - P - Q = 0
E y - ( 2Q - 2 P ) - P - Q = 0
E y = 3Q - P E y = 3Q - P ≠ 
P = 125 N and Q = 275 N
Cx = ( P + Q)
5
5
= ( 400) = +500 N 4
4
C y = 2 Q - 2 P = 2(275) - 2(125) = +300 N Dx = ( P + Q )
5
5
= ( 400) = +500 N 4
4
D y = - Q + 3P = - 275 + 3(125) = +100 N
Ex = ( P + Q)
5
5
= ( 400) = +500 N
4
4
E y = 3Q - P = 3(275) - 125 = + 700 N C x = 500 N¨ 
C y = 300 N ≠ 
D x = 500 N Æ 
D y = 100.0 N ≠ 
E x = 500 N ¨ 
E y = 700 N ≠ 
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159
PROBLEM 6.105
For the frame and loading shown, determine the components of the
forces acting on member DABC at B and D.
SOLUTION
Free body: Entire frame
+
SM G = 0 : H (0.6 m) - (12 kN )(1 m) - (6 kN )(0.5 m) = 0
H = 25 kN H = 25 kN ≠
Free body: Member BEH
+
SM F = 0 : Bx (0.5 m) - (25 kN )(0.2 m) = 0
Bx = +10 kN
Free body: Member DABC
B x = 10.00 kN Æ 
From above:
SM D = 0 : - By (0.8 m) + (10 kN + 12 kN )(0.5 m) = 0
+
By = +13.75 kN +
SFx = 0 : - Dx + 10 kN + 12 kN = 0
Dx = + 22 kN +
B y = 13.75 kN ≠ 
D x = 22.0 kN ¨ 
SFy = 0 : - D y + 13.75 kN = 0
D y = +13.75 kN D y = 13.75 kN Ø 
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160
PROBLEM 6.106
Solve Problem 6.105 assuming that the 6-kN load has been removed.
PROBLEM 6.105 For the frame and loading shown, determine the
components of the forces acting on member DABC at B and D.
SOLUTION
Free body: Entire frame
+
SM G = 0 : H (0.6 m) - (12 kN )(1 m) = 0
H = 20 kN H = 20 kN ≠
Free body: Member BEH
+
SM E = 0 : Bx (0.5 m) - (20 kN )(0.2 m) = 0
Bx = +8 kN
Free body: Member DABC
B x = -8.00 kN Æ 
From above:
+
SM D = 0 : - By (0.8 m) + (8 kN + 12 kN )(0.5 m) = 0
By = +12.5 kN +
SFx = 0 : - Dx + 8 kN + 12 kN = 0
Dx = + 20 kN +
B y = 12.50 kN ≠ 
SFy = 0 : - D y + 12.5 kN = 0; D y = + 12.5 kN D x = 20.0 kN ¨ 
D y = 12.50 kNØ 
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161
PROBLEM 6.107
The axis of the three-hinge arch ABC is a ­parabola with
vertex at B. Knowing that P = 112 kN and Q = 140 kN,
determine (a) the components of the ­reaction at A, (b) the
components of the force exerted at B on segment AB.
SOLUTION
Free body: Segment AB:
+
SM A = 0 : Bx (3.2 m) - By (8 m) - P (5 m) = 0 (1)
Bx (2.4 m) - By (6 m) - P (3.75 m) = 0 (2)
SM C = 0 : Bx (1.8 m) + By (6 m) - Q(3 m) = 0 (3)
0.75 (Eq. 1)
Free body: Segment BC:
+
4.2 Bx - 3.75P - 3Q = 0
Add (2) and (3):
Bx = (3.75 P + 3Q )/4.2 3.2
(3.75P + 3Q )
- 8B y - 5P = 0
4.2
By = ( - 9P + 9.6Q ) / 33.6 Eq. (1):
(4)
(5)
Given that P = 112 kN and Q = 140 kN
(a)
Reaction at A:
Considering again AB as a free body
+
+
SFx = 0 : Ax - Bx = 0;
Ax = Bx = 200 kN A x = 200 kN Æ 
SFy = 0 : Ay - P - By = 0
Ay - 112 kN - 10 kN = 0
Ay = +122 kN A y = 122.0 kN ≠ 
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162
PROBLEM 6.107 (Continued)
(b)
Force exerted at B on AB
Eq. (4):
Bx = (3.75 ¥ 112 + 3 ¥ 140)/4.2 = 200 kN
B x = 200 kN ¨ 
Eq. (5):
By = ( - 9 ¥ 112 + 9.6 ¥ 140) / 33.6 = + 10 kN
B y = 10.00 kN Ø 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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163
PROBLEM 6.108
The axis of the three-hinge arch ABC is a ­parabola
with vertex at B. Knowing that P = 140 kN and
Q = 112 kN, determine (a) the components of the
­reaction at A, (b) the components of the force exerted
at B on segment AB.
SOLUTION
Free body: Segment AB:
+
SM A = 0 : Bx (3.2 m) - By (8 m) - P (5 m) = 0 (1)
Bx (2.4 m) - By (6 m) - P (3.75 m) = 0 (2)
0.75 (Eq. 1)
Free body: Segment BC:
+
SM C = 0 : Bx (1.8 m) + By (6 m) - Q(3 m) = 0 Add (2) and (3):
4.2 Bx - 3.75P - 3Q = 0
Bx = (3.75 P + 3Q )/4.2 (3.75P + 3Q )
Eq. (1):
(3)
3.2
- 8B y - 5P = 0
4.2
By = ( - 9P + 9.6Q ) / 33.6 (4)
(5)
Given that P = 140 kN and Q = 112 kN
(a)
Reaction at A:
SFx = 0 : Ax - Bx = 0;
+
+
Ax = Bx = 205 kN A x = 205 kN Æ 
SFy = 0 : Ay - P - By = 0
Ay - 140 kN - ( - 5.5 kN ) = 0
Ay = 134.5 kN A y = 134.5 kN ≠ 
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164
PROBLEM 6.108 (Continued)
(b)
Force exerted at B on AB
Eq. (4):
Bx = (3.75 ¥ 140 + 3 ¥ 112) /4.2 = 205 kN
B x = 205 kN ¨ 
Eq. (5):
By = ( - 9 ¥ 140 + 9.6 ¥ 112) / 33.6 = - 5.5 kN
B y = 5.5 kN ≠ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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165
PROBLEM 6.109
Knowing that the surfaces at A and D are frictionless, determine
the forces exerted at B and C on member BCE.
SOLUTION
Free body of Member ACD
SM H = 0 : C x (50 mm) - C y ( 450 mm) = 0 C x = 9C y +
(1)
Free body of Member BCE
+
SM B = 0 : C x (150 mm) + C y (150 mm) - (250 N )(300 mm) = 0
9C y (150) + C y (150) - 75000 = 0
Substitute from (1):
C y = + 50 N; C x = 9C y = 9(50 N ) = + 450 N
+
SFx = 0 : Bx - 450 N = 0
Bx = 450 N
+
SFy = 0 : By + 50 N - 250 N = 0
By = 200 N
C = 453 N
6.34° 
B = 492 N
24.0° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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166
PROBLEM 6.110
For the frame and loading shown, determine (a) the reaction at C, (b) the
force in member AD.
SOLUTION
Free body: Member ABC
+
SM C = 0 : + (500 N )(1125 mm) +
4
4
FAD (1125 mm) + FBE (750 mm) = 0
5
5
562500 + 900 FAD + 600 FBF = 0 (1)
fi 3FAD + 2 FBE = - 1875 N
Free Body: Member DEF
+
SM F = 0 :
4
4
FAD (750 mm) + FBE (375 mm) = 0
5
5
FBE = -2 FAD (a)
(2)
Substitute from (2) into (1)
3FAD + 2( - 2 FAD ) = - 1875 N
FAD = +1875 N (2)
FAD = 1875 N 
FBE = - 2 FAD = - 2(1875 N )
FBE = -3750 N FBE = 3750 N comp.
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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167
PROBLEM 6.110 (Continued)
(b)
Return to free body of member ABC
+
SFx = 0 : C x + 500 N +
4
4
FAD + FBE = 0
5
5
4
4
C x + 500 + (1875) + ( -3750) = 0
5
5
+
C x = + 1000 N C x = 1000 N Æ
3
3
SFy = 0 : C y - FAD - FBF = 0
5
5
3
3
C y - (1875) - ( - 3750) = 0
5
5
C y = - 1125 N C y = 1125 N Ø
a = 48.37∞
C = 1505.2 N C = 1505 N
48.4° 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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168
PROBLEM 6.111
Members ABC and CDE are pin-connected at C and supported by four
links. For the loading shown, determine the force in each link.
SOLUTION
Member FBDs:
I
II
From FBD I:
SM J = 0 :
a
3a
a
C x + C y - P = 0 C x + 3C y = P
2
2
2
FBD II:
SM K = 0 :
a
3a
C x - C y = 0 C x - 3C y = 0
2
2
Solving:
Cx =
FBD I:
P
P
; C y = as drawn
2
6
SM B = 0 : aC y - a
ÆSFx = 0 : -
1
2
FAG = 0 FAG = 2C y =
P
6
2
FBF =
SM D = 0 : a
2
P C 
6
1
1
2
2
FAG +
FBF - C x = 0 FBF = FAG + C x 2 =
P+
P
6
2
2
2
FBD II:
FAG =
1
2
FEH + aC y = 0 FEH = - 2C y = P
6
2
ÆSFx = 0 : C x -
2 2
P C
3
2
P T 
6
FEH =
1
2
2
1
FDI +
FEH = 0 FDI = FEH + C x 2 = P+
P
6
2
2
2
FDI =
2
P C 
3
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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169
PROBLEM 6.112
Members ABC and CDE are pin-connected at C and supported by four
links. For the loading shown, determine the force in each link.
SOLUTION
Member FBDs:
I
FBD I:
FBD II:
FBDs combined:
II
aC y - a
1
FAF = 0 FAF = 2C y
2
M D = 0 : aC y - a
1
FEH = 0 FEH = 2C y
2
SM B = 0 :
SM G = 0 :
aP - a
1
1
1
1
FAF - a
FEH = 0 P =
2C y +
2C y
2
2
2
2
Cy =
P
2
so FAF =
2
P C
2
FEH =
2
P T 
2
FBD I:
≠SFy = 0 :
1
1
FAF +
FBG - P + C y = 0
2
2
P
1
P
+
FBG - P + = 0
2
2
2
FBG = 0 
FBD II:
≠ SFy = 0 : - C y +
1
1
P
1
P
FDG FEH = 0 - +
FDG - = 0
2
2
2
2
2
FDG = 2 P C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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170
PROBLEM 6.113
Members ABC and CDE are pin-connected at C and supported by four
links. For the loading shown, determine the force in each link.
SOLUTION
Member FBDs:
I
II
FBD I:
SM I = 0 : 2aC y + aC x - aP = 0 2C y + C x = P
FBD II:
SM J = 0 : 2aC y - aC x = 0 2C y - C x = 0
Solving:
FBD I:
P
P
; C y = as shown
2
4
1
FBG + C x = 0 FBG = C x 2 ÆSFx = 0 : 2
Cx =
≠SFy = 0 : FAF - P +
FBD II:
1 Ê 2 ˆ P
P + = 0
2 ÁË 2 ˜¯ 4
ÆSFx = 0 : - C x +
1
FDG = 0 FDG = C x 2 2
≠SFy = 0 : - C y +
1 Ê 2 ˆ
P P
P
P ˜ + FEH = 0 FEH = - = - Á
4 2
4
2Ë 2 ¯
P
2
C 
P
4
C 
P
2
C 
P
4
T 
FBG =
FAF =
FDG =
FEH =
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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171
PROBLEM 6.114
Members ABC and CDE are pin-connected at C and supported by the
four links AF, BG, DG, and EH. For the loading shown, determine the
force in each link.
SOLUTION
Free body: Member ABC
+
SM H = 0 : C x ( a) - C y (2a) = 0
C x = 2C y
Note: This checks that for 3-force member the forces are concurrent.
Free body: Member CDE
+
SM F = 0 : C x (2a) - C y ( a) + P ( a) = 0
2C x - C y + P = 0
1
Cy = - P 
3
2(2C y ) - C y + P = 0 2
Cx = - P 
3
C x = 2C y :
+
SF = 0 : C x -
E
2
E
= 0; - P =0
3
2
2
E=-
+
2 2
P
3
SM E = 0 : D(2a) + C y (3a) - P (3a) = 0
P
D(2a) - (3a) - P (3a) = 0
3
D = + 2P FEH =
2 2
P comp.
3
FDG = 2 P T 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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172
PROBLEM 6.114 (Continued)
+
Return to free body of ABC
SFy = 0 :
A
+ Cy = 0
2
A P
- =0
2 3
A= +
+
2
P
3
FAF =
2
P T 
3
SM A = 0 : B(2a) - C x (3a) = 0
B( 2a) +
2
P (3a) = 0
3
B = -P FBG = P C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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173
PROBLEM 6.115
Solve Problem 6.113 assuming that the force P is replaced by a
­clockwise couple of moment M0 applied to member CDE at D.
PROBLEM 6.113 Members ABC and CDE are pin-connected at C
and supported by four links. For the loading shown, determine the
force in each link.
SOLUTION
Free body: Member ABC
SM J = 0 : C y (2a) + C x ( a) = 0
+
C x = - 2C y
Free body: Member CDE
+
SM K = 0 : C y (2a) - C x ( a) - M 0 = 0
C y (2a) - ( - 2C y )( a) - M 0 = 0 C x = - 2C y :
+
M0
D=
+
D M0
=0
2 2a
D
+ Cx = 0 ;
2
SFx = 0 :
2a
M0

4a
M
Cx = - 0 
2a
Cy =
FDG =
M0
2a
T 
SM D = 0 : E ( a) - C y ( a) + M 0 = 0
ÊM ˆ
E ( a) - Á 0 ˜ ( a) + M 0 = 0
Ë 4a ¯
E=-
3 M0
4 a
FEH =
3 M0
4 a
C 
Return to Free Body of ABC
+
SFx = 0 :
B
+ C x = 0;
2
B=
+
M0
2a
SM B = 0 : A( a) + C y ( a);
B M0
=0
2 2a
FBG =
A( a) +
M0
2a
T 
M0
( a) = 0
4a
A= -
M0
4a
FAF =
M0
4a
C 
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174
PROBLEM 6.116
Solve Problem 6.114 assuming that the force P is replaced by a
­clockwise couple of moment M0 applied to member CDE at D.
PROBLEM 6.114 Members ABC and CDE are pin-connected at C
and supported by the four links AF, BG, DG, and EH. For the loading
shown, determine the force in each link.
SOLUTION
Free body: Member ABC
+
SM H = 0 : C x ( a) - C y (2a) = 0
C x = 2C y
Free body: Member CDE
+
SM F = 0 : C x (2a) - C y ( a) - M 0 = 0
(2C y )(2a) - C y ( a) - M 0 = 0 C x = 2C y :
+
SFx = 0 : C x -
+
D+
Cx =
M0

3a
2M0

3a
2M0
E
E
= 0;
=0
3
a
2
2
E=
SFy = 0 : D +
Cy =
2 2 M0
3 a
FEH =
2 2 M0

3 a
E
+ Cy = 0
2
M
2 2 M0 1
+ 0 =0
3 a 2 3a
D=-
M0
a
FDG =
M0
a
C 
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175
PROBLEM 6.116 (Continued)
+
Return to free body of ABC
SFy = 0 :
A
+ C y = 0;
2
A= -
+
A M0
+
=0
2 3a
2 M0
3 a
FAF =
2 M0
3 a
C 
M0
a
T 
SM A = 0 : B(2a) - C x (3a) = 0
Ê 2 M0 ˆ
(3a) = 0
B( 2a) - Á
Ë 3 a ˜¯
B=+
M0
a
FBG =
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176
PROBLEM 6.117
Four beams, each of length 3a, are held together by single nails at A, B, C, and D. Each beam is attached to a
support located at a distance a from an end of the beam as shown. Assuming that only vertical forces are exerted
at the connections, determine the vertical reactions at E, F, G, and H.
SOLUTION
We shall draw the free body of each member. Force P will be applied to member AFB. Starting with member
AED, we shall express all forces in terms of reaction E.
Member AFB:
+
SM D = 0 : A(3a) + E ( a) = 0
E
3
SM A = 0 : - D(3a) - E (2a) = 0
A= -
+
D=Member DHC:
+
+
2E
3
Ê 2E ˆ
(3a) - H ( a) = 0
SM C = 0 : Á Ë 3 ˜¯
H = - 2E Ê 2E ˆ
( 2a) + C ( a) = 0
SM H = 0 : Á Ë 3 ˜¯
C=+
(1)
4E
3
Member CGB:
+
Ê 4E ˆ
(3a) - G( a) = 0
SM B = 0 : + Á
Ë 3 ˜¯
G = + 4E (2)
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177
PROBLEM 6.117 (Continued)
+
Ê 4E ˆ
( 2a) + B( a) = 0
SM G = 0 : + Á
Ë 3 ˜¯
8E
B=3
+
Member AFB:
SFy = 0 : F - A - B - P = 0
Ê E ˆ Ê 8E ˆ
F - Á- ˜ - Á- ˜ - P = 0
Ë 3¯ Ë 3 ¯
F = P - 3E +
(3)
SM A = 0 : F ( a) - B(3a) = 0
Ê 8E ˆ
( P - 3E )( a) - Á - ˜ (3a) = 0
Ë 3¯
P - 3E + 8E = 0; E = -
P
5
E=
P
Ø
5
Substitute E = - P5 into Eqs. (1), (2), and (3).
Ê Pˆ
H = -2 E = - 2 Á - ˜
Ë 5¯
H =+
2P
5
H=
2P
≠
5
Ê Pˆ
G = + 4E = 4 Á - ˜
Ë 5¯
G=-
4P
5
G=
4P
Ø
5
Ê Pˆ
F = P - 3E = P - 3 Á - ˜
Ë 5¯
F =+
8P
5
F=
8P
Ø
5
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178
PROBLEM 6.118
Four beams, each of length 2a, are nailed together at their midpoints to form the support system shown.
­Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at A, D, E,
and H.
SOLUTION
Note that, if we assume P is applied to EG, each individual member FBD looks like
so
2 Fleft = 2 Fright = Fmiddle
Labeling each interaction force with the letter corresponding to the joint of its application, we see that
B = 2 A = 2F
C = 2 B = 2D
G = 2C = 2 H
P + F = 2G( = 4C = 8 B = 16 F ) = 2 E
From
P + F = 16 F , F =
P
15
so A =
P
≠
15
D=
2P
≠
15
H=
4P
≠
15
E=
8P
≠
15
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179
PROBLEM 6.119
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame,
determine the reactions at the supports and indicate whether the frame is rigid.
SOLUTION
Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the third
force, at B, must be parallel to the link forces.
(a)
FBD whole:
SM A = 0 : - 2aP -
a 4
1
B + 5a
B = 0 B = 2.06 P 4 17
17
4
B=0
A x = 2P ¨
17
1
P
Ay - P +
B = 0 A y = ≠
2
17
B = 2.06 P
14.04∞ 
A = 2.06 P
14.04∞ 
ÆSFx = 0 : Ax ≠SFy = 0 :
(b)
rigid 
FBD whole:
Since B passes through A, SM A = 2aP = 0 only if P = 0
no equilibrium if P π 0
not rigid 
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180
PROBLEM 6.119 (Continued)
(c)
FBD whole:
SM A = 0 : 5a
1
3a 4
17
B+
B - 2aP = 0 B =
P
4 17
4
17
ÆSFx = 0 : Ax +
4
B=0
17
≠SFy = 0 : Ay - P +
B = 1.031P
14.04∞ 
A = 1.250 P
36.9∞ 
Ax = - P
1
B=0
17
Ay = P -
P 3P
=
4
4
System is rigid 
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181
PROBLEM 6.120
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame,
determine the reactions at the supports and indicate whether the frame is rigid.
SOLUTION
(a)
Member FBDs:
I
II
FBD I:
SM A = 0 : aF1 - 2aP = 0 F1 = 2 P; ≠SFy = 0 :
FBD II:
SM B = 0 : - aF2 = 0 F2 = 0
ÆSFx = 0 : Bx + F1 = 0, Bx = - F1 = -2 P
Ay - P = 0 A y = P ≠
B x = 2P Æ
≠SFy = 0 : By = 0 FBD I:
ÆSFx = 0 : - Ax - F1 + F2 = 0
so B = 2P Æ 
Ax = F2 - F1 = 0 - 2 P
(b)
A x = 2P Æ
so A = 2.24 P
26.6∞ 
frame is rigid 
FBD left:
FBD whole:
I
FBD I:
SM E = 0 :
FBD II:
SM B = 0 :
II
a
a
5a
P + Ax Ay = 0 Ax - 5 Ay = - P
2
2
2
3aP + aAx - 5aAy = 0 Ax - 5 Ay = -3P
This is impossible unless P = 0
not rigid 
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182
PROBLEM 6.120 (Continued)
(c)
Member FBDs:
I
FBD I:
II
SFy = 0 : A - P = 0 A = P≠ 
SM D = 0 : aF1 - 2aA = 0 F1 = 2 P
FBD II:
ÆSFx = 0 : F2 - F1 = 0 F2 = 2 P
F
SM B = 0 : 2aC - aF1 = 0 C = 1 = P 2
ÆSFx = 0 : F1 - F2 + Bx = 0 Bx = P - P = 0
≠SFx = 0 : By + C = 0 By = - C = - P C = P≠ 
B=P 
Frame is rigid 
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183
PROBLEM 6.121
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame,
determine the reactions at the supports and indicate whether the frame is rigid.
SOLUTION
(a)
Member FBDs:
I
FBD II:
FBD I:
≠SFy = 0 : By = 0
II
SM B = 0 : aF2 = 0
SM A = 0 : aF2 - 2aP = 0 but F2 = 0
so P = 0
(b)
F2 = 0
not rigid for P π 0 
Member FBDs:
Note: 7 Unknowns ( Ax , Ay , Bx , By , F1 , F2 , C ) but only 6 independent equations.
System is statically indeterminate 
System is, however, rigid 
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184
PROBLEM 6.121 (Continued)
(c)
FBD whole:
FBD right:
I
FBD I:
SM A = 0 : 5aBy - 2aP = 0
≠SFy = 0 : Ay - P +
FBD II:
FBD I:
II
2
By = P ≠
5
SM c = 0 :
2
P=0
5
Ay =
a
5a
Bx By = 0 Bx = 5 By
2
2
ÆSFx = 0 : Ax + Bx = 0
Ax = - Bx
3
P≠
5
B x = 2P Æ
A x = 2P ¨
A = 2.09 P
16.70∞ 
B = 2.04 P
11.31∞ 
System is rigid 
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185
PROBLEM 6.122
A 400-N force is applied to the toggle vise at C. Knowing that q = 90∞,
­
determine (a) the vertical force exerted on the block at D, (b) the force exerted
on member ABC at B.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC
BD = (175)2 + (600)2 = 625 mm
We have
( FBD ) x =
+
24
7
FBD , ( FBD ) y =
FBD
25
25
SM A = 0 : ( FBD ) x (600) + ( FBD ) y (175) - 400( 400) = 0
ˆ
Ê 24
ˆ
Ê 7
ÁË FBD ˜¯ (600) + ÁË FBD ˜¯ (175) = 400( 400)
25
25
336 FBD = 160, 000
FBD = 476.19 N
tan a =
600
a = 73.7∞
175
FBD = 476 N
(b)
Force exerted at B:
(a)
Vertical force exerted on block
( FBD ) y =
73.7° 
24
24
FBD = ( 476.19 N ) = 457.14 N
25
25
( FBD ) y = 457 N Ø 
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186
PROBLEM 6.123
Solve Problem 6.122 when q = 0.
PROBLEM 6.122 A 400-N force is applied to the toggle vise at C. Knowing
that q = 90∞, determine (a) the vertical force exerted on the block at D, (b) the
force exerted on member ABC at B.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC
BD = (175)2 + (600)2 = 625 mm
We have
( FBD ) x =
+
24
7
FBD , ( FBD ) y =
FBD
25
25
SM A = 0 : ( FBD ) x (600) + ( FBD ) y (175) - 400(1000) = 0
ˆ
Ê 24
ˆ
Ê 7
ÁË FBD ˜¯ (600) + ÁË FBD ˜¯ (175) = 400(1000)
25
25
336 FBD = 400, 000
FBD = 1190.48 N
tan a =
24
a = 73.7∞
7
FBD = 1190 N
(b)
Force exerted at B:
(a)
Vertical force exerted on block
( FBD ) y =
73.7° 
24
24
FBD = (1190.48 N ) = 1142.86 N
25
25
( FBD ) y = 1143 N Ø 
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187
PROBLEM 6.124
The control rod CE passes through a horizontal hole in the body
of the toggle system shown. Knowing that link BD is 250 mm
long, determine the force Q required to hold the system in
­equilibrium when b = 20∞.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC
Dimensions in mm
33.404
; q = 7.679∞
250
SM C = 0 : ( FBD sin q )187.94 - ( FBD cos q )68.404 + (100 N )328.89 = 0
BD = 250, q = sin -1
Since
+
FBD [187.94 sin 7.679∞ - 68.404 cos 7.679∞] = 32889
FBD = 770.6 N
+
SFx = 0 : (770.6 N ) cos 7.679∞ = C x = 0
C x = + 763.7 N
Member CQ:
SFx = 0 : Q = C x = 763.7 N
Q = 764 N ¨ 
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188
PROBLEM 6.125
Solve Problem 6.124 when (a) b = 0, (b) b = 6∞.
PROBLEM 6.124 The control rod CE passes
through a horizontal hole in the body of the toggle
system shown. Knowing that link BD is 250 mm
long, determine the force Q required to hold the
system in equilibrium when b = 20∞.
SOLUTION
We note that BD is a two-force member.
(a)
b = 0:
Free body: Member ABC
BD = 250 mm, sinq =
Since
+
35 mm
; q = 8.048∞
250 mm
SM C = 0 : (100 N )(350 mm) - FBD sin q (200 mm) = 0
FBD = 1250 N
+
SFx = 0 : FBD cosq - C x = 0
(1250 N )(cos 8.048∞) - C x = 0
C x = 1237.7 N
Member CE:
+
SFx = 0 : (1237.7 N ) - Q = 0
Q = 1237.7 N (b)
b = 6∞
Q = 1238 N ¨ 
Free body: Member ABC
Dimensions in mm
14.094 mm
250 mm
q = 3.232∞
BD = 250 mm, q = sin -1
Since
+
SM C = 0 : ( FBD sin q )198.90 + ( FBD cos q )20.906 - (100 N )348.08 = 0
FBD [198.90 sin 3.232∞ + 20.906 cos 3.232∞] = 34808
FBD = 1084.8 N
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189
PROBLEM 6.125 (Continued)
+
SFx = 0 : FBD cosq - C x = 0
(1084.8 N ) cos 3.232∞ - C x = 0
C x = +1083.1 N
Member DE:
SFx = 0: Q = C x
Q = 1083.1 N Q = 1083 N ¨ 
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190
PROBLEM 6.126
The press shown is used to emboss a small seal at E. Knowing
that P = 250 N, determine (a) the vertical ­component of the force
exerted on the seal, (b) the reaction at A.
SOLUTION
FBD Stamp D:
≠SFy = 0 : E - FBD cos 20∞ = 0, E = FBD cos 20∞
FBD ABC:
SM A = 0 : (0.2 m)(sin 30∞)( FBD cos 20∞) + (0.2 m)(cos 30∞)( FBD sin 20∞)
- [(0.2 m)sin 30∞ + (0.4 m) cos15∞](250 N ) = 0
FBD = 793.64 N C
and, from above,
E = (793.64 N ) cos 20∞
(a)
E = 746 N Ø 
ÆSFx = 0 : Ax - (793.64 N )sin 20∞ = 0
A x = 271.44 N Æ
≠SFy = 0 : Ay + (793.64 N ) cos 20∞ - 250 N = 0
A y = 495.78 N Ø
so (b)
A = 565 N
61.3° 
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191
PROBLEM 6.127
The press shown is used to emboss a small seal at E. Knowing
that the vertical component of the force exerted on the seal must
be 900 N, determine (a) the required vertical force P, (b) the
­corresponding reaction at A.
SOLUTION
FBD Stamp D:
≠SFy = 0 : 900 N - FBD cos 20∞ = 0, FBD = 957.76 N C
(a)
FBD ABC:
SM A = 0 : [(0.2 m)(sin 30∞)](957.76 N ) cos 20∞ + [(0.2 m)(cos 30∞)](957.76 N )sin 20∞
- [(0.2 m)sin 30∞ + (0.4 m) cos15∞]P = 0
P = 301.70 N,
(b)
P = 302 N Ø 
ÆSFx = 0 : Ax - (957.76 N )sin 20∞ = 0
A x = 327.57 N Æ
≠SFy = 0 : - Ay + (957.76 N ) cos 20∞ - 301.70 N = 0
A y = 598.30 N Ø
so
A = 682 N
61.3° 
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192
PROBLEM 6.128
Water pressure in the supply system exerts a downward force of 135 N on the
vertical plug at A. Determine the tension in the fusible link DE and the force
exerted on member BCE at B.
SOLUTION
Free body: Entire linkage
+ SFy = 0 : C - 135 = 0
C = +135 N
Free body: Member BCE
+
+
SFx = 0 : Bx = 0
SM B = 0 : (135 N )(6 mm) - TDE (10 mm) = 0
TDE = 81.0 N 
+
SFy = 0 : 135 + 81 - By = 0
By = + 216 N B = 216 N Ø 
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193
PROBLEM 6.129
A couple M of magnitude 1.5 kN ◊ m is applied to the crank of the engine system shown. For each of the two
positions shown, determine the force P required to hold the system in equilibrium.
SOLUTION
(a)
FBDs:
Dimensions in mm
50 mm
175 mm
2
=
7
Note:
tanq =
FBD whole:
SM A = 0 : (0.250 m)C y - 1.5 kN ◊ m = 0 C y = 6.00 kN
FBD piston:
≠SFy = 0 : C y - FBC sin q = 0 FBC =
Cy
sin q
=
6.00 kN
sinq
ÆSFx = 0 : FBC cosq - P = 0
P = FBC cos q =
6.00 kN
= 7 kips
tan q
P = 21.0 kN ¨ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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194
PROBLEM 6.129 (Continued)
(b)
FBDs:
Dimensions in mm
2
as above
7
Note:
tanq =
FBD whole:
SM A = 0 : (0.100 m)C y - 1.5 kN ◊ m = 0 C y = 15 kN
SFy = 0 : C y - FBC sin q = 0 FBC =
ÆSFx = 0 :
Cy
sin q
FBC cosq - P = 0
P = FBC cos q =
Cy
tan q
=
15 kN
2/7
P = 52.5 kN ¨ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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195
PROBLEM 6.130
A force P of magnitude 16 kN is applied to the piston of the engine system shown. For each of the two positions
shown, determine the couple M required to hold the system in equilibrium.
SOLUTION
(a)
FBDs:
Dimensions in mm
Note:
FBD piston:
50 mm
tanq =
175 mm
2
=
7
ÆSFx = 0 : FBC cos q - P = 0 FBC =
P
cos q
≠SFy = 0 : C y - FBC sin q = 0 C y = FBC sin q = P tan q =
FBD whole:
2
P
7
SM A = 0 : (0.250 m)C y - M = 0
Ê 2ˆ
M = (0.250 m) Á ˜ (16 kN )
Ë 7¯
= 1.14286 kN ◊ m
M = 1143 N ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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196
PROBLEM 6.130 (Continued)
(b)
FBDs:
Dimensions in mm
2
as above
7
2
FBD piston: as above
C y = P tanq = P
7
2
FBD whole:
SM A = 0 : (0.100 m)C y - M = 0 M = (0.100 m) (16 kN )
7
M = 0.45714 kN ◊ m
Note:
tanq =
M = 457 N ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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197
PROBLEM 6.131
The pin at B is attached to member ABC and can slide freely along the
slot cut in the fixed plate. Neglecting the effect of friction, determine the
couple M required to hold the system in equilibrium when q = 30∞.
SOLUTION
Free body: Member ABC
+
SM C = 0 : (125 N )(346.41 mm) - B(75 mm) = 0
+
B = +577.35 N
SFy = 0 : - 125 N + C y = 0
C y = +125 N
+
SFx = 0 : 577.35 N - C x = 0
C x = +577.35 N
Free body: Member CD
129.9
; b = 40.5∞
200
CD cos b = (200) cos 40.50∞ = 152.08 mm
b = sin -1
+
SM D = 0 : M - (125 N )(129.9 mm) - (577.35 N )(152.08 mm) = 0
M = +104040.89 N ◊ mm
= 104.04 N.m
M = 104.0 N ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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198
PROBLEM 6.132
The pin at B is attached to member ABC and can slide freely along the
slot cut in the fixed plate. Neglecting the effect of friction, determine the
couple M required to hold the system in equilibrium when q = 60∞.
SOLUTION
Free body: Member ABC
+
SM C = 0 : (125 N )(200 mm) - B(129.9 mm) = 0
B = +192.456 N
+
SFx = 0 : 192.456 N - C x = 0
+
C x = +192.456 N
SFy = 0 : - 125 N + C y = 0
C y = +125 N
Free body: Member CD
b = sin -1
75
; b = 22.024∞
200
CD cos b = (200 mm) cos 22.024∞ = 185.405 mm
+
SM D = 0 : M - (125 N )(75 mm) - (192.456 N )(185.405 mm) = 0
M = +45057.3 N ◊ mm
M = 45.1 N ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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199
PROBLEM 6.133
Arm ABC is connected by pins to a collar at B and to crank CD at C.
Neglecting the effect of friction, determine the couple M required to
hold the system in equilibrium when q = 0.
SOLUTION
Free body: Member ABC
+
SFx = 0 : C x - 240 N = 0
C x = +240 N
+
SM C = 0 : (240 N )(500 mm) - B(160 mm) = 0
+
B = +750 N
SFy = 0 : C y - 750 N = 0
C y = +750 N
Free body: Member CD
+
SM D = 0 : M + (750 N )(300 mm) - (240 N )(125 mm) = 0
M = -195 ¥ 103 N ◊ mm
M = 195.0 kN ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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200
PROBLEM 6.134
Arm ABC is connected by pins to a collar at B and to crank CD at C.
Neglecting the effect of friction, determine the couple M required to
hold the system in equilibrium when q = 90∞.
SOLUTION
Free body: Member ABC
+
+
SFx = 0 : C x = 0
SM B = 0 : C y (160 mm) - (240 N )(90 mm) = 0
C y = +135 N
Free body: Member CD
+
SM D = 0 : M - (135 N )(300 mm) = 0
M = +40.5 ¥ 103 N ◊ mm
M = 40.5 kN ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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201
PROBLEM 6.135
Two rods are connected by a slider block as shown. Neglecting the
effect of friction, determine the couple MA required to hold the system
in equilibrium.
SOLUTION
D ^ AC
Note:
Member FBDs:
SM B = 0 :
lD sin 115∞ - 30, 000 N ◊ mm = 0
D=
30000 N ◊ mm
2(375 mm) cos 25∞ sin 115∞
= 48.6977 N
SM A = 0 :
M A - (375 mm)( 48.6977 N ) = 0
M A = 18261.64 N ◊ mm
= 18.26 N.m
M A = 18.26 N ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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202
PROBLEM 6.136
Two rods are connected by a slider block as shown. Neglecting the
effect of friction, determine the couple MA required to hold the system
in equilibrium.
SOLUTION
C ^ BD
Note:
Member FBD’s:
SM B = 0 : lC - 30, 000 N ◊ mm = 0
C=
30, 000 N ◊ mm
2(375 mm) cos 25∞
C = 44.135 N
SM A = 0 : M A - (375 mm)C sin 65∞ = 0
M A = (375 mm)( 44.135 N )sin 65∞
M A = 14999.96 N ◊ mm
= 15.00 N.m
M A = 15.00 N ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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203
PROBLEM 6.137
Rod CD is attached to the collar D and passes through a
collar welded to end B of lever AB. Neglecting the effect of
friction, determine the couple M required to hold the system in
equilibrium when q = 30∞.
SOLUTION
B ^ CD
Note:
FBD DC:
SFx ¢ = 0 : D y sin 30∞ - (300 N ) cos 30∞ = 0
Dy =
300 N
= 519.62 N
tan 30∞
FBD machine:
SM A = 0 :
0.200 m
519.62 N - M = 0
sin 30∞
M = 207.85 N ◊ m
M = 208 N ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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204
PROBLEM 6.138
Rod CD is attached to the collar D and passes through a collar welded
to end B of lever AB. Neglecting the effect of friction, determine the
couple M required to hold the system in equilibrium when q = 30∞.
SOLUTION
FBD DC:
SFx ¢ = 0 :
D y sin 30∞ - (150 N ) cos 30∞ = 0
D y = (150 N ) ctn 30∞ = 259.81 N
FBD machine:
SM A = 0 : (0.100 m)(150 N ) + d (259.81 N ) - M = 0
d = b - 0.040 m
b=
so
0.030718 m
tan 30
b = 0.053210 m
d = 0.0132100 m
M = 18.4321 N ◊ m
M = 18.43 N ◊ m 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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205
PROBLEM 6.139
Two hydraulic cylinders control the position of
the robotic arm ABC. Knowing that in the position
shown the cylinders are parallel, determine the
force exerted by each cylinder when P = 160 N
and Q = 80 N.
SOLUTION
Free body: Member ABC
+
4
FAE (150 mm) - (160 N )(600 mm) = 0
5
FAE = +800 N 3
SFx = 0 : - (800 N ) + Bx - 80 N = 0
5
SM B = 0 :
+
FAE = 800 N T 
+
Bx = +560 N
4
SFy = 0 : - (800 N ) + By - 160 N = 0
5
By = +800 N
Free body: Member BDF
+
SM F = 0 : (560 N )( 400 mm) - (800 N )(300 mm) FDG = -100 N 4
FDG (200 mm) = 0
5
FDG = 100.0 N C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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206
PROBLEM 6.140
Two hydraulic cylinders control the position of the robotic arm ABC. In the position shown, the cylinders are
parallel and both are in tension. Knowing the FAE = 600 N and FDG = 50 N, determine the forces P and Q
applied at C to arm ABC.
SOLUTION
Free body: Member ABC
+
SM B = 0 :
4
(600 N )(150 mm) - P (600 mm) = 0
5
P = +120 N +
SM C = 0 :
P = 120.0 NØ 
4
(600)(750 mm) - By (600 mm) = 0
5
By = +600 N
Free body: Member BDF
+
SM F = 0 : Bx ( 400 mm) - (600 N )(300 mm)
4
- (50 N )(200 mm) = 0
5
Bx = +470 N
Return to free body: Member ABC
+ SF = 0 : - 3 (600 N ) + 470 N - Q = 0
x
5
Q = +110 N Q = 110.0 N ¨ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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207
PROBLEM 6.141
A log weighing 4 kN is lifted by a pair of tongs as shown. Determine the
forces exerted at E and F on tong DEF.
SOLUTION
FBD AB:
Ay = By = 2 kN
By symmetry:
Ax = Bx
and
6
= (2 kN )
5
= 2.4 kN
D = -B
Note:
FBD DEF:
Dx = 2.4 kN
so
D y = 2 kN
SM F = (210 mm)(2 kN ) + (310 mm)(2.4 kN ) - (240 mm) E x = 0
E x = 4.85 kN E = 4.85 kN Æ 
Æ SFx = 0 : - 2.4 kN + 4.85 kN - Fx = 0
Fx = 2.45 kN
≠SFy = 0 : 2 kN - Fy = 0
Fy = 2 kN
F = 3.16 kN
39.2∞ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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208
PROBLEM 6.142
A 12 m length of railroad rail of weight 750 N/m is lifted by the tongs shown.
Determine the forces exerted at D and F on tong BDF.
SOLUTION
W = (12 m)(750 N/m) = 9000 N
Free body: Rail
1
E y = Fy = W = 4500 N
2
1
By symmetry,
( FAB ) y = ( FAC ) y = W = 4500 N
2
Since AB is a two-force member,
By symmetry
Free body: Upper link
( FAB ) x ( FAB ) y
=
192
120
+
( FAB ) x =
192
( 4500) = 7200 N
120
SM D = 0: ( Attach FAB at A)
Free Body: Tong BDF
Fx (160) - ( FAB ) x (360) - Fy (16) = 0
Fx (160) - (7200 N )(360) - ( 4500 N )(16) = 0
Fx = +16650 N
+
F = 17250 N
15.12∞ 
SFx = 0 : - Dx + ( FAB ) x + Fx = 0
+
Dx = ( FAB ) x + Fx = 7200 + 16650 = 23850 N
SFy = 0 : D y + ( FAB ) y - Fy = 0
Dy = 0 D = 23900 N ¨ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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209
PROBLEM 6.143
The tongs shown are used to apply a total upward force of 45 kN on a pipe cap.
Determine the forces exerted at D and F on tong ADF.
SOLUTION
FBD whole:
By symmetry�A = B = 22.5 kN ≠
SM F = 0 : (75 mm)CD - (100 mm)(22.5 kN ) = 0
FBD ADF:
CD = 30.0 kN ¨ 
ÆSFx = 0 : Fx - CD = 0
Fx = CD = 30 kN
≠SFy = 0 : 22.5 kN - Fy = 0
Fy = 22.5 kN
F = 37.5 kN
so
36.9∞ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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210
PROBLEM 6.144
If the toggle shown is added to the tongs of Problem 6.143 and a single vertical
force is applied at G, determine the forces exerted at D and F on tong ADF.
SOLUTION
Free body: Toggle
1
Ay = ( 45 kN ) = 22.5 kN
2
By symmetry
AG is a two-force member
Ax
22.5 kN
=
22 mm 55 mm
Ax = 56.25 kN
+
Free body: Tong ADF
SFy = 0 : 22.5 kN - Fy = 0
Fy = +22.5 kN
+
SM F = 0 : D(75 mm) - (22.5 kN )(100 mm) - (56.25 kN )(160 mm) = 0
D = +150 kN +
D = 150.0 kN ¨
SFx = 0 : 56.25 kN - 150 kN + Fx = 0
Fx = 93.75 kN
F = 96.4 kN
13.50∞ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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211
PROBLEM 6.145
The pliers shown are used to grip a 6 mm-diameter rod. Knowing
that two 300 N forces are applied to the handles, determine (a) the
magnitude of the forces exerted on the rod, (b) the force exerted by
the pin at A on portion AB of the pliers.
SOLUTION
Free body: Portion AB
(a)
+
SM A = 0 : Q(24 mm) - (300 N )(190 mm) = 0 fi Q = 2375 N
Q = 2380 N 
(b)
+
SFx = 0 : Q(sin 30∞) + Ax = 0
(2375 N )(sin 30∞) + Ax = 0
+
Ax = -1187.5 N
A y = 1188 N ¨
SFy = 0 : - Q(cos 30∞) + Ay - 300 N = 0
-(2375 N )(cos 30∞) + Ay - 300 N = 0
Ay = +2356.81 N
A y = 2360 N ≠
A = 2640 N
63.3∞ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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212
PROBLEM 6.146
In using the bolt cutter shown, a worker applies two 300-N forces
to the handles. Determine the magnitude of the forces exerted by
the cutter on the bolt.
SOLUTION
FBD cutter AB:
I
FBD I: Æ SFx = 0 : Bx = 0
FBD handle BC:
Dimensions in mm
II
FBD II: SM C = 0 : (12 mm)By - ( 448 mm)300 N = 0
By = 11, 200.0 N
Then
FBD I: SM A = 0 : (96 mm) By - (24 mm) F = 0
F = 4 By
F = 44, 800 N = 44.8 kN 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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213
PROBLEM 6.147
Determine the magnitude of the gripping forces exerted along
line aa on the nut when two 250 N forces are applied to the
handles as shown. Assume that pins A and D slide freely in
slots cut in the jaws.
SOLUTION
FBD jaw AB:
ÆSFx = 0 : Bx = 0
SM B = 0 : (10 mm)Q - (30 mm) A = 0
A=
Q
3
≠SFy = 0 : A + Q - By = 0
By = A + Q =
FBD handle ACE:
4Q
3
By symmetry and FBD jaw DE:
Q
3
E x = Bx = 0
D = A=
E y = By =
4Q
3
SM C = 0 : (105 mm)(250 N ) + (15 mm)
Q
4Q
- (15 mm)
=0
3
3
Q = 1750 N 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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214
PROBLEM 6.148
Determine the magnitude of the gripping forces produced when
two 300 N forces are applied as shown.
SOLUTION
We note that AC is a two-force member
FBD handle CD:
SM D = 0 : - (126 mm)(300 N ) - (6 mm)
Ê 1
+ (30 mm) Á
Ë 8.84
2.8
A
8.84
ˆ
A˜ = 0
¯
A = 2863.6 8.84 N
Dimensions in mm
FBD handle AB:
SM B = 0 : (132 mm)(300 N ) - (120 mm)
1
(2863.6 8.84 N )
8.84
+ (36 mm) F = 0
F = 8.45 kN 
Dimensions in mm
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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215
PROBLEM 6.149
Knowing that the frame shown has a sag at B of a = 25 mm, determine
the force P required to maintain equilibrium in the position shown.
SOLUTION
We note that AB and BC are two-force members
Free body: Toggle
By symmetry:
Cy =
P
2
Cy
Cx
=
250 mm
a
250
250 P 125P
Cx =
Cy =
◊ =
a
a 2
a
Free body: Member CDE
+
SM E = 0 : C x (150 mm) - C y (500 mm) - (250 N )(250 mm) = 0
125P
P
(150) - (500) = 62500
a
2
Ê 75 ˆ
P Á - 1˜ = 250 ¯
Ë a
For
(1)
a = 25 mm
2 P = 250
P = 125 N
P = 125 N Ø 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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216
PROBLEM 6.150
Knowing that the frame shown has a sag at B of a = 12.5 mm,
determine the force P required to maintain equilibrium in the position
shown.
SOLUTION
We note that AB and BC are two-force members
Free body: Toggle
By symmetry:
Cy =
P
2
Cy
Cx
=
250 mm
a
250
250 P 125P
Cx =
Cy =
◊ =
a
a 2
a
Free body: Member CDE
+
SM E = 0 : C x (150 mm) - C y (500 mm) - (250 N )(250 mm) = 0
125P
P
(150) - (500) = 62500
2
a
Ê 75 ˆ
P Á - 1˜ = 250
¯
Ë a
For
a = 12.5 mm
5P = 250 N
P = 50 N
(1)
P = 50.0 N Ø 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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217
PROBLEM 6.151
The garden shears shown consist of two blades and two handles. The two handles are connected by pin C and
the two blades are connected by pin D. The left blade and the right handle are connected by pin A; the right
blade and the left handle are connected by pin B. Determine the magnitude of the forces exerted on the small
branch at E when two 80-N forces are applied to the handles as shown.
SOLUTION
By symmetry vertical components Cy, Dy, Ey are 0. Then by considering SFy = 0 on the blades or handles, we
find that Ay and By are 0.
Thus forces at A, B, C, D, and E are horizontal.
Free body: Right handle
SM C = 0 : A(30 mm) - (80 N )(270 mm) = 0
+
A = +720 N
+
SFx = 0 : C - 720 N - 80 N = 0
C = +800 N
Free body: Left blade
+
 M D = 0 : E (180 mm) - (720 N)(60 mm) = 0
E = 240 N 
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218
PROBLEM 6.152
The telescoping arm ABC is used to provide an elevated platform for
construction workers. The workers and the platform together have a
mass of 200 kg and have a combined center of gravity located directly
above C. For the position when q = 20∞, determine (a) the force exerted
at B by the single hydraulic cylinder BD, (b) the force exerted on the
supporting carriage at A.
SOLUTION
a = (5 m) cos 20∞ = 4.6985 m
b = (2.4 m) cos 20∞ = 2.2553 m
c = (2.4 m)sin 20∞ = 0.8208 m
d = b - 0.5 = 1.7553 m
e = c + 0.9 = 1.7208 m
Geometry:
tan b =
e 1.7208
=
; b = 44.43∞
d 1.7553
Free body: Arm ABC
We note that BD is a two-force member
W = (200 kg)(9.81 m/s2 ) = 1.962 kN
(a)
+
SM A = 0 : (1.962 kN )( 4.6985 m) - FBD sin 44.43∞(2.2553 m) + FBD cos 44.43(0.8208 m) = 0
9.2185 - FBD (0.9927) = 0 : FBD = 9.2867 kN
FBD = 9.29 kN
(b)
+
SFx = 0 : Ax - FBD cos b = 0
Ax = (9.2867 kN ) cos 44.43∞ = 6.632 kN
+
44.4∞ 
A x = 6.632 kN Æ
SFy = 0 : Ay - 1.962 kN + FBD sin b = 0
Ay = 1.962 kN - (9.2867 kN )sin 44.43∞ = -4.539 kN
A y = 4.539 kN Ø
A = 8.04 kN
34.4∞ 
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219
PROBLEM 6.153
The telescoping arm ABC can be lowered until end C is close to
the ground, so that workers can easily board the platform. For the
position when q = - 20∞, determine (a) the force exerted at B by the
single hydraulic cylinder BD, (b) the force exerted on the supporting
carriage at A.
SOLUTION
a = (5 m) cos 20∞ = 4.6985 m
b = (2.4 m) cos 20∞ = 2.2552 m
c = (2.4 m)sin 20∞ = 0.8208 m
d = b - 0.5 = 1.7553 m
e = 0.9 - c = 0.0792 m
Geometry:
tan b =
e 0.0792
=
; b = 2.584∞
d 1.7552
Free body: Arm ABC
We note that BD is a two-force member
W = (200 kg)(9.81 m/s2 )
W = 1962 N = 1.962 kN
(a)
+
SM A = 0 : (1.962 kN )( 4.6985 m) - FBD sin 2.584∞(2.2553 m) - FBD cos 2.584∞(0.8208 m) = 0
9.2185 - FBD (0.9216) = 0 FBD = 10.003 kN
FBD = 10.00 kN
(b)
+
SFx = 0 : Ax - FBD cos b = 0
Ax = (10.003 kN ) cos 2.583∞ = 9.993 kN
+
2.58∞ 
A x = 9.993 kN Æ
SFy = 0 : Ay - 1.962 kN + FBD sin b = 0
Ay = 1.962 kN - (10.003 kN )sin 2.583∞ = -1.5112 kN
A y = 1.5112 kNØ
A = 10.11 kN
8.60∞ 
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220
PROBLEM 6.154
The position of member ABC is controlled by the hydraulic
cylinder CD. Knowing that q = 30∞, determine for the loading
shown (a) the force exerted by the hydraulic cylinder on pin C,
(b) the reaction at B.
SOLUTION
Geometry: In DBCD
(CD )2 = (0.5)2 + (1.5)2 - 2(0.5)(1.5) cos 60∞
CD = 1.3229 m
Law of cosines
sin b
sin 60∞
=
0.5 m 1.3229 m
b = 19.107∞
sin b = 0.3273
Law of sines
Free body: Entire system
Move force FCD along its line of action so it acts at D.
(a)
+
SM B = 0 : (10 kN )(0.69282 m) - FCD sin b (1.5 m) = 0
6.9282 kN ◊ m - FCD sin 19.107∞(1.5 m) = 0
FCD = 14.111 kN
FCD = 14.11 kN
(b)
+
19.11° 
SFx = 0 : Bx + FCD cos b = 0
Bx + (14.111 kN ) cos 19.107∞ = 0
+
Bx = -13.333 kN
B x = 13.333 kN ¨
SFy = 0 : By - 10 kN - FCD sin 19.107∞ = 0
By - 10 kN - (14.111 kN )sin 19.107∞ = 0
By = +14.619 kN
B y = 14.619 kN≠
B = 19.79 kN
47.6° 
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221
PROBLEM 6.155
The motion of the bucket of the front-end loader
shown is controlled by two arms and a linkage that are
pin-connected at D. The arms are located symmetrically
with respect to the central, vertical, and longitudinal plane
of the loader; one arm AFJ and its control cylinder EF are
shown. The single linkage GHDB and its control cylinder
BC are located in the plane of symmetry. For the position
and loading shown, determine the force exerted (a) by
cylinder BC, (b) by cylinder EF.
SOLUTION
Free body: Bucket
+
SM J = 0 : (20 kN )(500 mm) - FGH (550 mm) = 0
FGH =
Free body: Arm BDH
+
200
kN = 18.1818 kN
11
Ê 200 ˆ
SM D = 0 : - Á
kN ˜ (600 mm) - FBC (500 mm) = 0
¯
Ë 11
FBC =
Free body: Entire mechanism
-240
= 21.8181 kN 11
FBC = 21.8 kN C 
(Two arms and cylinders AFJE)
Note: Two arms thus 2FEF
450 mm
1625 mm
b = 15.48∞
tan b =
+
SM A = 0 : (20 kN)(3075 mm) + FBC (300 mm) + 2 FEF cos b (600 mm) = 0
Ê 240 ˆ
(20 kN)(3075 mm) - Á
(300 mm) + 2 FEF cos (15.48∞)(600 mm) = 0
Ë 11 ˜¯
FEF = -47.5193 kN FEF = 47.5 kN C 
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222
PROBLEM 6.156
The bucket of the front-end loader shown carries a 15 kN
load. The motion of the bucket is controlled by two identical
mechanisms, only one of which is shown. Knowing that
the mechanism shown supports one-half of the 15 kN load,
determine the force exerted (a) by cylinder CD, (b) by
cylinder FH.
SOLUTION
Free body: Bucket (One mechanism)
+
SM D = 0 : (7.5 kN )(375 mm) - FAB ( 400 mm) = 0
FAB = 7.03125 kN
Note: There are 2 identical support mechanisms.
Free body: One arm BCE
200
500
b = 21.8∞
tan b =
+
SM E = 0 : (7.03125 kN)(575 mm) + FCD cos 21.8∞(375 mm) - FCD sin 21.8∞(125 mm) = 0
FCD = -13.3979 kN FCD = 13.40 kN C 
Free body: Arm DFG
+
SM G = 0 : (7.5 kN)(1875 mm) + FFH sin 45∞(600 mm) - FFH cos 45∞(150 mm) = 0
FFH = - 44.1941 kN FFH = 44.19 kN C 
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223
PROBLEM 6.157
The motion of the backhoe bucket shown is controlled
by the hydraulic cylinders AD, CG, and EF. As a
result of an attempt to dislodge a portion of a slab,
a 10 kN force P is exerted on the bucket teeth at J.
Knowing that q = 45∞, determine the force exerted by
each cylinder.
SOLUTION
Free body: Bucket
+
SM H = 0
(Dimensions in mm)
4
3
FCG (250) + FCG (250) + P cos q ( 400) + P sin q (200) = 0
5
5
- 4P
FCG =
(2 cos q + sin q ) 7
Free body: Arm ABH and bucket
(1)
(Dimensions in mm)
+
SM B = 0 :
4
3
FAD (300) + FAD (250) + P cos q (2150) - P sin q (1050) = 0
5
5
FAD = -
5P
( 43 cos q - 21 sin q ) 39
(2)
Free body: Bucket and arms IEB + ABH
Geometry of cylinder EF
400 mm
1000 mm
b = 21.801∞
tan b =
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224
PROBLEM 6.157 (Continued)
+
SM I = 0 : FEF cos b ( 450 mm) + P cos q (700 mm) - P sin q (3000 mm) = 0
FEF = 0.23934 P (30 sin q - 7 cos q ) (3)
For P = 10 kN, q = 45∞
Eq. (1):
FCG = -
4
¥ 10 (2 cos 45∞ + sin 45∞) = -12.122 kN 7
FCG = 12.12 kN C 
Eq. (2):
FAD = -
5
¥ 10 ( 43 cos 45∞ - 21 sin 45∞) = -19.944 kN 39
FAD = 19.94 kN C 
Eq. (3):
FEF = 0.23934 ¥ 10 (30 sin 45∞ - 7 cos 45∞) = 38.9249 kN FEF = 38.9 kN T 
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225
PROBLEM 6.158
Solve Problem 6.157 assuming that the 10 kN force
P acts horizontally to the right (q = 0).
PROBLEM 6.157 The motion of the backhoe bucket
shown is controlled by the hydraulic cylinders AD,
CG, and EF. As a result of an attempt to dislodge
a portion of a slab, a 10-kN force P is exerted on the
bucket teeth at J. Knowing that q = 45∞, determine
the force exerted by each cylinder.
SOLUTION
Free body: Bucket
+
SM H = 0
(Dimensions in mm)
3
4
FCG (250) + FCG (250) + P cos q ( 400) + P sin q (200) = 0
5
5
FCG =
- 4P
(2 cos q + sin q ) 7
(1)
Free body: Arm ABH and bucket
(Dimensions in mm)
4
3
FAD (300) + FAD (250) + P cos q (2150) - P sin q (1050) = 0
5
5
-5P
FAD =
( 43 cos q - 21 sin q ) 39
Free body: Bucket and arms IEB + ABH
400 mm
Geometry of cylinder EF
tan b =
1000 mm
b = 21.801∞
+
SM B = 0 :
(2)
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226
PROBLEM 6.158 (Continued)
+
SM I = 0 : FEF cos b ( 450 mm) + P cos q (700 mm) - P sin q (3000 mm) = 0
FEF = 0.23934 P (30 sin q - 7 cos q ) (3)
For P = 10 kN , q = 0
Eq. (1):
FCG =
-4 ¥ 10
(2 cos 0∞ + sin 0∞) = -11.4286 kN 7
Eq. (2):
FAD =
-5 ¥ 10
( 43 cos 0∞ - 21 sin 0∞) = -55.128 kN 39
Eq. (3):
FEF = 0.2394 ¥ 10 (30 sin 0∞ - 7 cos 0∞) = -16.758 kN FCG = 11.43 kN C 
FAD = 55.1 kN C 
FEF = 16.76 kN C 
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227
PROBLEM 6.159
The gears D and G are rigidly attached to shafts that
are held by frictionless bearings. If rD = 90 mm and
rG = 30 mm, determine (a) the couple M0 that must be
applied for equilibrium, (b) the reactions at A and B.
SOLUTION
(a)
Projections on yz plane
Free body: Gear G
+
SM G = 0 : 30 N ◊ m - J (0.03 m) = 0; J = 1000 N
Free body: Gear D
+
SM D = 0 : M 0 - (1000 N )(0.09 m) = 0
M 0 = 90 N ◊ m (b)
M 0 = (90.0 N ◊ m)i 
Gear G and axle FH
+
SM F = 0 : H (0.3 m) - (1000 N )(0.18 m) = 0
+
H = 600 N
SFy = 0 : F + 600 - 1000 = 0
F = 400 N
Gear D and axle CE
+
SM C = 0 : (1000 N )(0.18 m) - E (0.3 m) = 0
+
E = 600 N
SFy = 0 : 1000 - C - 600 = 0
C = 400 N
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228
PROBLEM 6.159 (Continued)
+
Free body: Bracket AE
+
SFy = 0 : A - 400 +400 = 0 A=0 
SM A = 0 : M A + ( 400 N )(0.32 m) - ( 400 N )(0.2 m) = 0
M A = -48 N ◊ m M A = -( 48.0 N ◊ m)i 
+
Free body: Bracket BH
+
SFy = 0 : B - 600 + 600 = 0 B=0
SM B = 0 : M B + (600 N )(0.32 m) - (600 N )(0.2 m) = 0
M B = -72 N ◊ m
M B = -(72.0 N ◊ m)i 
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229
PROBLEM 6.160
In the planetary gear system shown, the radius of the central gear A is a = 18 mm,
the radius of each planetary gear is b, and the radius of the outer gear E is
(a + 2b). A clockwise couple of magnitude MA = 10 N ◊ m is applied to the
central gear A and a counterclockwise couple of magnitude MS = 50 N ◊ m is
applied to the spider BCD. If the system is to be in equilibrium, determine
(a) the required radius b of the planetary gears, (b) the magnitude ME of the
couple that must be applied to the outer gear E.
SOLUTION
FBD Central Gear:
By symmetry:
F1 = F2 = F3 = F
SM A = 0 : 3( rA F ) - 10 N ◊ m = 0,
F=
10
N◊m
3rA
SM C = 0 : rB ( F - F4 ) = 0, F4 = F
FBD Gear C:
SFx ¢ = 0 : C x ¢ = 0
SFy ¢ = 0 : C y ¢ - 2 F = 0,
C y¢ = 2F
Gears B and D are analogous, each having a central force of 2F
SM A = 0 : 50 N◊ m - 3( rA + rB )2 F = 0
FBD Spider:
50 N◊ m - 3( rA + rB )
20
N◊ m = 0
rA
rA + rB
r
= 2.5 = 1 + B , rA
rA
rB = 1.5rA
Since rA = 18 mm,
FBD Outer Gear:
rB = 27.0 mm 
(a)
SM A = 0 : 3( rA + 2rB ) F - M E = 0
3(18 mm + 54 mm)
10 N◊ m
- ME = 0
54 mm
M E = 40.0 N◊ m 
(b)
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230
PROBLEM 6.161*
Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B
and D do not exert any axial force. A couple of magnitude 60 N ◊ m (clockwise when viewed from the positive x axis)
is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine
(a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B,
D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.)
SOLUTION
We recall from Figure 4.10, that a universal joint exerts on members it connects a force of unknown direction
and a couple about an axis perpendicular to the crosspiece.
Free body: Shaft DF
SM x = 0 : M C cos 30∞ - 60 N◊ m = 0
M C = 69.282 N ◊ m
Free body: Shaft BC
We use here x ¢, y ¢, z with x ¢ along BC
SM C = 0 : - M Ai ¢ - (69.282 N ◊ m)i ¢ + ( -0.125 m)i ¢ ¥ ( By j ¢ + Bz k ) = 0
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231
PROBLEM 6.161* (Continued)
Equate coefficients of unit vectors to zero:
M A - 69.282 N◊ m = 0
i:
j:
Bz = 0
k:
By = 0
}
SF = 0 :
B + C = 0,
M A = 69.282 N◊ m
M A = 69.3 N◊ m 
B=0
B=0
since B = 0,
C=0
Return to free body of shaft DF
(Note that C = 0 and M C = 69.282 N ◊ m )
SM D = 0
(69.282 N ◊ m)(cos 30∞i + sin 30∞ j) - (60 N ◊ m)i
+ (0.15 m)i ¥ ( E x i + E y j + E z k ) = 0
(60 N ◊ m)i + (34.641 N◊ m) j - (60 N ◊ m)i
+ (0.15 m) E y k - (0.15 m) E z j = 0
Equate coefficients of unit vectors to zero:
j:
34.641 N ◊ m - (0.15 m)E z = 0
k:
Ey = 0
E z = 230.94 N
SF = 0 : C + D + E = 0
0 + D y j + Dz k + E x i + (230.94 N)k = 0
i:
Ex = 0
j:
Dy = 0
k:
Dz + 230.94 N = 0
Dz = -230.94 N
B=0 
Reactions are:
D = -(231 N)k 
E = (231 N)k 
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232
PROBLEM 6.162*
Solve Problem 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical.
PROBLEM 6.161 Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at
C. The bearings at B and D do not exert any axial force. A couple of magnitude 60 N ◊ m (clockwise when viewed
from the positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is
horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium,
(b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.)
SOLUTION
Free body: Shaft DF
SM x = 0 : M C - 60 N◊ m = 0
M C = 60 N◊ m
Free body: Shaft BC
We resolve -(60 N ◊ m)i into components along x ¢ and y ¢ axes:
- MC = -(60 N ◊ m)(cos 30∞i ¢ + sin 30∞ j¢ )
SMC = 0 : M Ai ¢ - (60 N ◊ m)(cos 30∞i ¢ + sin 30∞ j¢ ) + (0.125 m)i ¢ ¥ ( By ¢ j¢ + Bz k ) = 0
M Ai ¢ - (51.9615 N ◊ m)i ¢ - (30 N ◊ m) j + (0.125 m) By ¢ k - (0.125 m) Bz j¢ = 0
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233
PROBLEM 6.162* (Continued)
Equate to zero coefficients of unit vectors:
i ¢ : M A - 51.9615 N ◊ m = 0 j¢ : - 30 N ◊ m - (0.125 m) Bz = 0
M A = 52.0 N ◊ m 
Bz = -240 N
k : B y¢ = 0
B = -240 N
Reactions at B:
SF = 0 : B - C = 0
-(240 N)k - C = 0
C = -(240 N) k
Return to free body of shaft DF:
SM D = 0 : (0.15 m)i ¥ ( E x i + E y j + E z k ) - (0.1 m)i ¥ ( -240 N)k
- (60 N ◊ m)i + (60 N ◊ m)i = 0
(0.15 m) E y k - (0.15 m) E z j - (24 N ◊ m) j = 0
k : Ey = 0
j: - (0.15 m) E z - 24 N ◊ m = 0
E z = -160.0 N
SF = 0 : C + D + E = 0
i : Ex = 0
-(240 N)k + D y j + Dz k + E x i - (160 N)k = 0
k : - 240 N - 160 N + Dz = 0
Dz = 400 N
B = -(240 N)k 
Reactions are:
D = ( 400 N)k 
E = -(160.0 N)k 
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234
PROBLEM 6.163*
The large mechanical tongs shown are used to grab and lift a thick 7500 kg steel
slab HJ. Knowing that slipping does not occur between the tong grips and the
slab at H and J, determine the components of all forces acting on member EFH.
(Hint: Consider the symmetry of the tongs to establish relationships between
the components of the force acting at E on EFH and the components of the
force acting at D on CDF.)
SOLUTION
T = W = mg = (7500 kg)(9.81 m/s2 ) = 73.575 kN
Free body: Pin A
SFx = 0 : ( FAB ) x = ( FAC ) x
1
SFy = 0 : ( FAB ) y = ( FAC ) y = W
2
Also:
( FAC ) x = 2( FAC ) y = W
Free body: Member CDF
+
or
1
SM D = 0 : W (0.3) + W (2.3) - Fx (1.8) - Fy (0.5 m) = 0
2
1.8Fx + 0.5Fy = 1.45W (1)
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235
PROBLEM 6.163* (Continued)
+
SFx = 0 : Dx - Fx - W = 0
E x - Fx = W +
or
or
1
SFy = 0 : Fy - D y + W = 0
2
1
E y - Fy = W 2
(2)
(3)
Free body: Member EFH
+
1
SM E = 0 : Fx (1.8) + Fy (1.5) - H x (2.3) + W (1.8 m) = 0
2
1.8Fx + 1.5Fy = 2.3H x - 0.9W or:
+
(4)
SFx = 0 : E x + Fx - H x = 0
E x + Fx = H x or
2Fx = H x - W Subtract (2) from (5):
Subtract (4) from 3 ¥ (1):
3.6 Fx = 5.25W - 2.3H x Add (7) to 2.3 ¥ (6):
8.2 Fx = 2.95W
Fx = 0.35976W (5)
(6)
(7)
(8)
Substitute from (8) into (1):
(1.8)(0.35976W ) + 0.5Fy = 1.45W
0.5Fy = 1.45W - 0.64756W = 0.80244W
Fy = 1.6049W Substitute from (8) into (2):
Substitute from (9) into (3):
From (5):
Recall that:
(9)
E x - 0.35976W = W ; E x = 1.35976W
1
E y - 1.6049W = W E y = 2.1049W
2
H x = E x + Fx = 1.35976W + 0.35976W = 1.71952W
1
Hy = W
2
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236
PROBLEM 6.163* (Continued)
Since all expressions obtained are positive, all forces are directed as shown on the free-body diagrams.
Substitute
W = 73.575 kN
E x = 100.0 kN Æ
E y = 154.9 kN ≠ 
Fx = 26.5 kN Æ
Fy = 118.1 kN Ø 
H x = 126.5 kN ¨
H y = 36.8 kN Ø 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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237
PROBLEM 6.164
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.
SOLUTION
Free body: Truss.
C = D = 3 kN ≠
From the symmetry of the truss and loading, we find
Free body: Joint B
FAB
5
=
FBC 1.5 kN
=
2
1
FAB = 3.3541 N T FBC = 3.00 kN C 
+
Free body: Joint C
+
SFy = 0 :
SFx = 0 :
3
FAC + 3 kN = 0
5
FAC = -5 kN 4
( -5 kN) + 3 kN + FCD = 0 5
FAC = 5.00 kN C 
FCD = 1.000 kN T 
From symmetry:
FAD = FAC = 5.00 kN C , FAE = FAB = 3.35 kN T , FDE = FBC = 3.00 kN C 
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238
PROBLEM 6.165
Using the method of joints, determine the force in each member of
the double-pitch roof truss shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Truss
+
SM A = 0 : H (18 m) - (2 kN )( 4 m) - (2 kN )(8 m) - (1.75 kN )(12 m)
- (1.5 kN )(15 m)) - (0.75 kN )(18 m) = 0
H = 4.50 kN ≠
SFx = 0 : Ax = 0
SFy = 0 : Ay + H - 9 = 0
Ay = 9 - 4.50,
Ay = 4.50 kN ≠
Free body: Joint A
FAC 3.50 kN
=
2
1
5
FAB = 7.8262 kN C FAB
=
FAB = 7.83 kN C 
FAC = 7.00 kN T 
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239
PROBLEM 6.165 (Continued)
Free body: Joint B
+
SFx = 0 :
2
2
1
FBD +
(7.8262 kN ) +
FBC = 0
5
5
2
FBD + 0.79057 FBC = -7.8262 kN (1)
1
1
1
FBD +
(7.8262 kN ) FBC - 2 kN = 0
5
5
2
FBD - 1.58114 FBC = -3.3541 (2)
+
or
SFy = 0 :
or
Multiply (1) by 2 and add (2):
3FBD = -19.0065
FBD = -6.3355 kN FBD = 6.34 kN C 
Subtract (2) from (1):
2.37111FBC = -4.4721
FBC = -1.8861 kN FBC = 1.886 kN C 
+
Free body: Joint C
SFy = 0 :
2
1
FCD (1.8861 kN ) = 0
5
2
FCD = +1.4911 kN
+
SFx = 0 : FCE - 7.00 kN +
FCD = 1.491 kN T 
1
1
(1.8861 kN ) +
(1.4911 kN ) = 0
2
5
FCE = 5.000 kN FCE = 5.00 kN T 
Free body: Joint D
+
SFx = 0 :
2
1
1
2
FDF +
FDE +
(6.3355 kN ) (1.4911 kN ) = 0
5
2
5
5
FDF + 0.79057 FDE = -5.5900 kN +
or
SFy = 0 :
or
Add (1) and (2):
1
2
1
1
FDF FDE +
(6.3355 kN ) (1.4911 kN ) - 2 kN = 0
5
2
5
5
FDF - 0.79057 FDE = -1.1188 kN
(2)
2 FDF = -6.7088 kN
FDF = -3.3544 kN Subtract (2) from (1):
(1)
FDF = 3.35 kN C 
1.58114 FDE = -4.4712 kN
FDE = -2.8278 kN
FDE = 2.83 kN C 
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240
PROBLEM 6.165 (Continued)
Free body: Joint F
+
SFx = 0 :
1
2
FFG +
(3.3544 kN ) = 0
2
5
FFG = -4.243 kN +
SFy = 0 : -FEF - 1.75 kN +
FFG = 4.24 kN C 
1
1
(3.3544 kN ) ( -4.243 kN ) = 0
5
2
FEF = 2.750 kN FEF = 2.75 kN T 
Free body: Joint G
+
SFx = 0 :
1
1
1
FGH FEG +
( 4.243 kN ) = 0
2
2
2
FGH - FEG = -4.243 kN +
or:
SFy = 0 : -
1
1
1
FGH FEG ( 4.243 kN ) - 1.5 kN = 0
2
2
2
FGH + FEG = -6.364 kN or:
Add (1) and (2):
(2)
2 FGH = -10.607
FGH = -5.303 Subtract (1) from (2):
(1)
FGH = 5.30 kN C 
2 FEG = -2.121 kN
FEG = -1.0605 kN FEG = 1.061 kN C 
Free body: Joint H
FEH 3.75 kN
=
1
1
We can also write:
FGH
2
=
3.75 kN
1
FEH = 3.75 kN T 
FGH = 5.30 kN C (Checks)
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241
PROBLEM 6.166
The truss shown was designed to support the roof of a
food market. For the given loading, determine the force in
members FG, EG, and EH.
SOLUTION
Reactions at supports. Because of the symmetry of the loading
Ax = 0,
1
Ay = O = (Total load)
2
A = O = 4.48 kN ≠ 
We pass a section through members FG, EG, and EH, and use the Free body shown.
Slope FG = Slope FI =
Slope EG =
1.75 m
6m
5.50 m
2.4 m
SM E = 0 : (0.6 kN )(7.44 m) + (1.24 kN )(3.84 m)
+
- ( 4.48 kN )(7.44 m)
ˆ
Ê 6
FFG ˜ ( 4.80 m) = 0
-Á
¯
Ë 6.25
FFG = -5.231 kN FFG = 5.23 kN C 
SM G = 0 : FEH (5.50 m) + (0.6 kN )(9.84 m)
+
+(1.24 kN )(6.24 m) + (1.04 kN )(2.4 m)
-( 4.48 kN )(9.84 m) = 0
+
SFy = 0 :
FEH = 5.08 kN T 
5.50
1.75
FEG +
( -5.231 kN ) + 4.48 kN - 0.6 kN - 1.24 kN - 1.04 kN = 0
6.001
6.25
FEG = -0.1476 kN FEG = 0.1476 kN C 
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242
PROBLEM 6.167
The truss shown was designed to support the roof of
a food market. For the given loading, determine the
force in members KM, LM, and LN.
SOLUTION
1
O = ( Load ) 2
We pass a section through KM, LM, LN, and use free body shown
Because of symmetry of loading,
+
O = 4.48 kN ≠ 
ˆ
Ê 3.84
(3.68 m)
SM M = 0 : Á
F
Ë 4 LN ˜¯
+ ( 4.48 kN - 0.6 kN )(3.6 m) = 0
FLN = -3.954 kN +
FLN = 3.95 kN C 
SM L = 0 : - FKM ( 4.80 m) - (1.24 kN )(3.84 m)
+ ( 4.48 kN - 0.6 kN )(7.44 m) = 0
FKM = +5.022 kN
+ SFy = 0 :
FKM = 5.02 kN T 
4.80
1.12
FLM +
( -3.954 kN ) - 1.24 kN - 0.6 kN + 4.48 kN
N=0
6.147
4
FLM = -1.963 kN FLM = 1.963 kN C 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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243
PROBLEM 6.168
For the frame and loading shown, determine the components of all
forces acting on member ABC.
SOLUTION
Free body: Entire frame
+
SM E = 0 : - Ax (2) - (100 kN)(2.5) = 0
Ax = -125 kN , +
A x = 125.0 kN ¨ 
SFy = 0 : Ay - 100 kN = 0
Ay = 100 kN A y = 100.0 kN ≠ 
Free body: Member ABC
2
Bx
5
SM C = 0 : (125 kN)(2 m) - (100 kN)(5 m) + Bx (1 m) + By (2.5 m) = 0
Note: BE is a two-force member, thus B is directed along line BE and By =
+
-250 kN◊ m + Bx (1 m) +
Bx = 125 kN
2
Bx (2.5 m) = 0
5
B x = 125.0 kN ¨ 
2
2
By = ( Bx ) = (125) = 50 kN 5
5
+
SFx = 0 : C x - 125 kN - 125 kN = 0
C x = 250 kN +
B y = 50.0 kNØ 
C x = 250 kN Æ 
SFy = 0 : C y + 100 kN - 50 kN = 0
C y = -50 kN C y = 50.0 kNØ 
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244
PROBLEM 6.169
Solve Problem 6.168 assuming that the 100 kN load is replaced by a
clockwise couple of magnitude 150 kN ◊ m applied to member EDC at
Point D.
PROBLEM 6.168 For the frame and loading shown, determine the
components of all forces acting on member ABC.
SOLUTION
Free body: Entire frame
+
SFy = 0 : Ay = 0
SM E = 0 : - Ax (2 m) - 150 kN ◊ m = 0
+
Ax = -75 kN
A x = 75 kN ¨
A = 75.0 kN ¨ 
Free Body: Member ABC
2
Bx
5
SM C = 0 : (75 kN)(2 m) + Bx (1 m) + By (2.5 m) = 0
Note: BE is a two-force member, thus B is directed along line BE and By =
+
150 kN + Bx (1 m) +
2
Bx (2.5 m) = 0
5
Bx = -75 kN 2
2
Bx = ( -75) = -30 kN 5
5
SFx = 0 : - 75 kN + 75 kN + C x = 0
Cx = 0
By =
+
+
B x = 75.0 kN Æ 
B y = 30.0 kN≠ 
SFy = 0 : + 30 kN + C y = 0
C y = -30 kN
C y = 30 kN Ø
C = 30.0 kN Ø 
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245
PROBLEM 6.170
Knowing that the pulley has a radius of 0.5 m, determine the components
of the reactions at A and E.
SOLUTION
FBD Frame:
SM A = 0 : (7 m) E y - ( 4.5 m)(700 N ) = 0 E y = 450 N ≠ 
≠SFy = 0 : Ay - 700 N + 450 N = 0 A y = 250 N ≠ 
ÆSFx = 0 : Ax - E x = 0
Ax = E x
Dimensions in m
FBD Member ABC:
SM C = 0 : (1 m)(700 N ) - (1 m)(250 N ) - (3 m) Ax = 0
A x = 150.0 N ¨ 
so E x = 150.0 N Æ 
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246
PROBLEM 6.171
For the frame and loading shown, determine the reactions at A, B, D, and E.
Assume that the surface at each support is frictionless.
SOLUTION
Free body: Entire frame
+
SFx = 0 : A - B + (5 kN)sin 30∞ = 0
A - B + 2.5 kN = 0 +
(1)
SFy = 0 : D + E - (5 kN) cos 30∞ = 0
D + E - 4.3301 kN = 0 (2)
Free body: Member ACE
+
SM C = 0 : - A(150 mm) + E (200 mm) = 0
E=
3
A
4
(3)
Free body: Member BCD
+
SM C = 0 : - D(200 mm) + B(150 mm) = 0
D=
3
B
4
Substitute E and D from (3) and (4) into (2)
3
3
A + B - 4.3301 = 0
4
4
A + B - 5.7735 = 0
(1)
A - B + 2.5 = 0 (4)
(5)
(6)
(5) + (6)
2 A - 3.2735 = 0
A = 1.6368 kN A = 1.637 kN Æ 
(5) - (6)
2 B - 8.2735 = 0
B = 4.1368 kN B = 4.14 kN ¨ 
(4)
D=
D = 3.1026 N D = 3.10 kN ≠ 
E = 1.2276 kN E = 1.228 kN ≠ 
(3)
3
( 4.1368)
4
3
E = (1.6368)
4
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247
PROBLEM 6.172
For the system and loading shown, determine (a) the force P required for
equilibrium, (b) the corresponding force in member BD, (c) the corresponding
reaction at C.
SOLUTION
Member FBDs:
FBD I:
I:
SM C = 0 : R( FBD sin 30∞) - [ R(1 - cos 30∞)](100 N ) - R(50 N ) = 0
FBD = 126.795 N (b) FBD = 126.8 N T 
ÆSFx = 0 : -C x + (126.795 N ) cos 30∞ = 0
C x = 109.808 N ¨
≠SFy = 0 : C y + (126.795 N )sin 30∞ - 100 N - 50 N = 0
C y = 86.603 N ≠
II:
(c) so C = 139.8 N
38.3∞ 
FBD II:
SM A = 0 : aP - a[(126.795 N ) cos 30∞] = 0 ( a) P = 109.8 N Æ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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248
PROBLEM 6.173
A small barrel weighing 30 kg is lifted by a pair of tongs as shown.
Knowing that a = 125 mm, determine the forces exerted at B and D on
tong ABD.
SOLUTION
We note that BC is a two-force member.
P = Weight of barrel = 30 ¥ 9.81 N
= 294.3 N
Free body: Tong ABD
By
Bx
=
375 125
Bx = 3By
SM D = 0 : By (75 mm) + 3By (125 mm) - (294.3 N )(225 mm) = 0
+
By = 147.15 N Ø
Bx = 3By : Bx = 441.45 N ¨
+
+
SFx = 0 : - 441.45 N + Dx = 0
SFy = 0 : 294.3 N - 147.15 N - D y = 0
D x = 441 N Æ
D y = 147.2 N Ø
B = 465 N
18.43∞ 
D = 465 N
18.43∞ 
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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249
PROBLEM 6.174
A 20-kg shelf is held horizontally by a self-locking brace that consists
of two parts EDC and CDB hinged at C and bearing against each other
at D. Determine the force P required to release the brace.
SOLUTION
Free body: Shelf
W = (20 kg)(9.81 m/s2 ) = 196.2 N
+
SM A = 0 : By (200 mm) - (196.2 N )(125 mm) = 0
By = 122.63 N
Free body: Portion ACB
+
SM A = 0 : - Bx (150 mm) - P (150 mm) - (122.63 N )(200 mm) = 0
Bx = -163.5 - P +
(1)
SM C = 0 : + (122.63 N )(92 mm) + Bx (94 mm) = 0
+ (122.63 N )(92 mm) + ( -163.5 - P )( 44 mm) = 0
P = 92.9 N P = 92.9 N

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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250
PROBLEM 6.175
The specialized plumbing wrench shown is used in confined areas (e.g., under
a basin or sink). It consists essentially of a jaw BC pinned at B to a long rod.
Knowing that the forces exerted on the nut are equivalent to a clockwise (when
viewed from above) couple of magnitude 16 N ◊ m, determine (a) the magnitude
of the force exerted by pin B on jaw BC, (b) the couple M0 that is applied to the
wrench.
SOLUTION
Free body: Jaw BC
This is a two-force member
Cy
Cx
=
C y = 2.4 C x
30 mm 12.5 mm
SFx = 0: Bx = C x (1)
SFy = 0 : By = C y = 2.4 C x (2)
Free body: Nut SFx = 0: C x = Dx
SM = 16 N ◊ m = 16000 N ◊ mm
C x (22.5 mm) = 16000 N ◊ mm
C x = 711.111 N
(a)
Eq. (1):
Bx = C x = 711.111 N
Eq. (2):
By = C y = 2.4(711.111 N ) = 1706.667 N
(
B = Bx2 + By2
)
1/ 2
= 711.1112 + 1706.6672
= 1848.89 N (b)
B = 1850 N 
Free body: Rod
+
SM D = 0 : - M 0 + By (12.5 mm) - Bx (7.5 mm) = 0
- M 0 + (1706.667)(12.5) - (711.111)(7.5) = 0
M 0 = -16000 N ◊ mm
= -16 N ◊ m
M 0 = 16.00 N ◊ m

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to
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251