CHAPTER 6
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
Free body: Entire truss:
ΣFy = 0: By = 0
By = 0
ΣM C = 0: − Bx (3.2 m) − (48 kN)(7.2 m) = 0
Bx = −108 kN B x = 108 kN
ΣFx = 0: C − 108 kN + 48 kN = 0
C = 60 kN
C = 60 kN
Free body: Joint B:
FAB FBC 108 kN
=
=
5
4
3
FAB = 180.0 kN T 
FBC = 144.0 kN T 
Free body: Joint C:
FAC FBC 60 kN
=
=
13
12
5
FAC = 156.0 kN C 
FBC = 144 kN (checks)
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743
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
Reactions:
ΣM A = 0: C = 1260 lb
ΣFx = 0: A x = 0
ΣFy = 0: A y = 960 lb
Joint B:
FAB FBC 300 lb
=
=
12
13
5
FAB = 720 lb T 
FBC = 780 lb C 
Joint A:
ΣFy = 0: − 960 lb −
4
FAC = 0
5
FAC = 1200 lb
FAC = 1200 lb C 
3
ΣFx = 0: 720 lb − (1200 lb) = 0 (checks)
5
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744
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
AB = 32 + 1.252 = 3.25 m
BC = 32 + 42 = 5 m
Reactions:
ΣM A = 0: (84 kN)(3 m) − C (5.25 m) = 0
C = 48 kN
ΣFx = 0: Ax − C = 0
A x = 48 kN
ΣFy = 0: Ay = 84 kN = 0
A y = 84 kN
Joint A:
ΣFx = 0: 48 kN −
12
FAB = 0
13
FAB = +52 kN
ΣFy = 0: 84 kN −
5
(52 kN) − FAC = 0
13
FAC = +64.0 kN
FAB = 52.0 kN T 
FAC = 64.0 kN T 
Joint C:
FBC 48 kN
=
5
3
FBC = 80.0 kN C 
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745
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
Free body: Truss:
C = D = 600 lb
From the symmetry of the truss and loading, we find
Free body: Joint B:
FAB
5
=
FBC 300 lb
=
2
1
FAB = 671 lb T
FBC = 600 lb C 
Free body: Joint C:
ΣFy = 0:
3
FAC + 600 lb = 0
5
FAC = −1000 lb
ΣFx = 0:
From symmetry:
4
( −1000 lb) + 600 lb + FCD = 0
5
FAC = 1000 lb C 
FCD = 200 lb T 
FAD = FAC = 1000 lb C , FAE = FAB = 671 lb T , FDE = FBC = 600 lb C 
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746
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
Reactions:
ΣM D = 0: Fy (24) − (4 + 2.4)(12) − (1)(24) = 0
Fy = 4.2 kips
ΣFx = 0: Fx = 0
ΣFy = 0: D − (1 + 4 + 1 + 2.4) + 4.2 = 0
D = 4.2 kips
Joint A:
FAB = 0 
ΣFx = 0: FAB = 0
ΣFy = 0 : −1 − FAD = 0
FAD = −1 kip
Joint D:
ΣFy = 0: − 1 + 4.2 +
ΣFx = 0:
8
FBD = 0
17
FBD = −6.8 kips
15
(−6.8) + FDE = 0
17
FDE = +6 kips
FAD = 1.000 kip C 
FBD = 6.80 kips C 
FDE = 6.00 kips T 
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747
PROBLEM 6.5 (Continued)
Joint E:
ΣFy = 0 : FBE − 2.4 = 0
FBE = +2.4 kips
FBE = 2.40 kips T 
Truss and loading symmetrical about cL.
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748
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
AD = 52 + 122 = 13 ft
BCD = 122 + 162 = 20 ft
Reactions:
ΣFx = 0: Dx = 0
ΣM E = 0: D y (21 ft) − (693 lb)(5 ft) = 0
ΣFy = 0: 165 lb − 693 lb + E = 0
Joint D:
D y = 165 lb
E = 528 lb
ΣFx = 0:
5
4
FAD + FDC = 0
13
5
(1)
ΣFy = 0:
12
3
FAD + FDC + 165 lb = 0
13
5
(2)
Solving Eqs. (1) and (2) simultaneously,
FAD = −260 lb
FDC = +125 lb
FAD = 260 lb C 
FDC = 125 lb T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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749
PROBLEM 6.6 (Continued)
Joint E:
ΣFx = 0:
5
4
FBE + FCE = 0
13
5
(3)
ΣFy = 0:
12
3
FBE + FCE + 528 lb = 0
13
5
(4)
Solving Eqs. (3) and (4) simultaneously,
FBE = 832 lb C 
FBE = −832 lb
FCE = 400 lb T 
FCE = +400 lb
Joint C:
Force polygon is a parallelogram (see Fig. 6.11, p. 209).
FAC = 400 lb T 
FBC = 125.0 lb T 
Joint A:
ΣFx = 0:
5
4
(260 lb) + (400 lb) + FAB = 0
13
5
FAB = −420 lb
ΣFy = 0:
FAB = 420 lb C 
12
3
(260 lb) − (400 lb) = 0
13
5
0 = 0 (Checks)
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750
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: Entire truss:
ΣFx = 0: Cx + 2(5 kN) = 0
Cx = −10 kN

C x = 10 kN
ΣM C = 0: D(2 m) − (5 kN)(8 m) − (5 kN)(4 m) = 0

D = +30 kN
D = 30 kN

ΣFy = 0: C y + 30 kN = 0 C y = −30 kN C y = 30 kN
Free body: Joint A:
FAB FAD 5 kN
=
=
4
1
17


FAB = 20.0 kN T 
FAD = 20.6 kN C 
Free body: Joint B:
ΣFx = 0: 5 kN +
1
5
FBD = 0
FBD = −5 5 kN
ΣFy = 0: 20 kN − FBC −
2
5
(−5 5 kN) = 0
FBC = +30 kN

FBD = 11.18 kN C 
FBC = 30.0 kN T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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751
PROBLEM 6.7 (Continued)
Free body: Joint C:
ΣFx = 0: FCD − 10 kN = 0
FCD = +10 kN
FCD = 10.00 kN T 
ΣFy = 0: 30 kN − 30 kN = 0 (checks)
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752
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
Reactions:
ΣM C = 0: A x = 16 kN
ΣFy = 0: A y = 9 kN
ΣFx = 0:
C = 16 kN
Joint E:
FBE FDE 3 kN
=
=
5
4
3
FBE = 5.00 kN T 
FDE = 4.00 kN C 
Joint B:
ΣFx = 0:
4
(5 kN) − FAB = 0
5
FAB = +4 kN
3
ΣFy = 0: − 6 kN − (5 kN) − FBD = 0
5
FBD = −9 kN
FAB = 4.00 kN T 
FBD = 9.00 kN C 
Joint D:
ΣFy = 0: − 9 kN +
3
FAD = 0
5
FAD = +15 kN
4
ΣFx = 0: − 4 kN − (15 kN) − FCD = 0
5
FCD = −16 kN
FAD = 15.00 kN T 
FCD = 16.00 kN C 
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753
PROBLEM 6.9
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:
ΣFx = 0: H x = 0
Because of the symmetry of the truss and loading,
A = Hy =
1
total load
2
A = H y = 1200 lb
Free body: Joint A:
FAB = 1500 lb C 
FAB FAC 900 lb
=
=
5
4
3
FAC = 1200 lb T 
Free body: Joint C:
BC is a zero-force member.
FCE = 1200 lb T 
FBC = 0
Free body: Joint B:
ΣFx = 0:
24
4
4
FBD + FBE + (1500 lb) = 0
25
5
5
24 FBD + 20 FBE = −30, 000 lb
or
ΣFy = 0:
(1)
7
3
3
FBD − FBE + (1500) − 600 = 0
25
5
5
7 FBD − 15FBE = −7,500 lb
or
(2)
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754
PROBLEM 6.9 (Continued)
Multiply Eq. (1) by 3, Eq. (2) by 4, and add:
FBD = 1200 lb C 
100 FBD = −120, 000 lb
Multiply Eq. (1) by 7, Eq. (2) by –24, and add:
FBE = 60.0 lb C 
500 FBE = −30,000 lb
Free body: Joint D:
24
24
FDF = 0
(1200 lb) +
25
25
ΣFx = 0:
FDF = −1200 lb
FDF = 1200 lb C 
7
7
(1200 lb) − (−1200 lb) − 600 lb − FDE = 0
25
25
ΣFy = 0:
FDE = 72.0 lb
FDE = 72.0 lb T 
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE
FEF = 60.0 lb C 
FEG = FCE
FEG = 1200 lb T 
FFG = FBC
FFG = 0 
FFH = FAB
FFH = 1500 lb C 
FGH = FAC
FGH = 1200 lb T 
Note: Compare results with those of Problem 6.11.
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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755
PROBLEM 6.10
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:
ΣFx = 0: H x = 0
Because of the symmetry of the truss and loading,
A = Hy =
1
total load
2
A = H y = 1200 lb
Free body: Joint A:
FAB FAC 900 lb
=
=
5
4
3
FAB = 1500 lb C 
FAC = 1200 lb T 
Free body: Joint C:
BC is a zero-force member.
FCE = 1200 lb T 
FBC = 0
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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756
PROBLEM 6.10 (Continued)
Free body: Joint B:
ΣFx = 0:
or
FBD + FBE = −1500 lb
ΣFy = 0:
or
4
4
4
FBD + FBC + (1500 lb) = 0
5
5
5
(1)
3
3
3
FBD − FBE + (1500 lb) − 600 lb = 0
5
5
5
FBD − FBE = −500 lb
Add Eqs. (1) and (2):
2 FBD = −2000 lb
Subtract Eq. (2) from Eq. (1):
2 FBE = −1000 lb
(2)
FBD = 1000 lb C 
FBE = 500 lb C 
Free Body: Joint D:
4
4
(1000 lb) + FDF = 0
5
5
ΣFx = 0:
FDF = −1000 lb
FDF = 1000 lb C 
3
3
(1000 lb) − ( −1000 lb) − 600 lb − FDE = 0
5
5
ΣFy = 0:
FDE = +600 lb
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE
FEG = FCE
FFG = FBC
FFH = FAB
FGH = FAC
FDE = 600 lb T 
FEF = 500 lb C 
FEG = 1200 lb T 
FFG = 0 
FFH = 1500 lb C 
FGH = 1200 lb T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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757
PROBLEM 6.11
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:
ΣFx = 0: Ax = 0
Due to symmetry of truss and load,
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 
FBD = 47.175 kN, FBC = 10.5 kN
FBC = 10.50 kN C 
FAC = 44.6 kN T 
Free body: Joint B:
From force polygon:
FBD = 47.2 kN C 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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758
PROBLEM 6.11 (Continued)
Free body: Joint C:
ΣFy = 0:
FCD = 17.50 kN T 
3
FCD − 10.5 = 0
5
ΣFx = 0: FCE +
4
(17.50) − 44.625 = 0
5
FCE = 30.625 kN
Free body: Joint E:
DE is a zero-force member.
FCE = 30.6 kN T 
FDE = 0 
Truss and loading symmetrical about cL .
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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759
PROBLEM 6.12
Determine the force in each member of the Fink roof truss shown.
State whether each member is in tension or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: A x = 0
Because of the symmetry of the truss and loading,
Ay = G =
1
total load
2
A y = G = 6.00 kN
Free body: Joint A:
F
FAB
4.50 kN
= AC =
2.462 2.25
1
FAC = 10.125 kN
FAB = 11.08 kN C 
FAC = 10.13 kN T 
Free body: Joint B:
ΣFx = 0:
3
2.25
2.25
FBC +
FBD +
(11.08 kN) = 0
5
2.462
2.462
(1)
F
4
11.08 kN
FBC + BD +
− 3 kN = 0
5
2.462
2.462
(2)
ΣFy = 0: −
Multiply Eq. (2) by –2.25 and add to Eq. (1):
12
FBC + 6.75 kN = 0
5
FBC = −2.8125
FBC = 2.81 kN C 
Multiply Eq. (1) by 4, Eq. (2) by 3, and add:
12
12
FBD +
(11.08 kN) − 9 kN = 0
2.462
2.462
FBD = −9.2335 kN
FBD = 9.23 kN C 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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760
PROBLEM 6.12 (Continued)
Free body: Joint C:
ΣFy = 0:
4
4
FCD − (2.8125 kN) = 0
5
5
FCD = 2.8125 kN,
FCD = 2.81 kN T 
3
3
ΣFx = 0: FCE − 10.125 kN + (2.8125 kN) + (2.8125 kN) = 0
5
5
FCE = +6.7500 kN
Because of the symmetry of the truss and loading, we deduce that
FDE = FCD
FDF = FBD
FEF = FBC
FEG = FAC
FFG = FAB
FCE = 6.75 kN T 
FCD = 2.81 kN T 
FDF = 9.23 kN C 
FEF = 2.81 kN C 
FEG = 10.13 kN T 
FFG = 11.08 kN C 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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761
PROBLEM 6.13
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:
ΣM 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
ΣFx = 0: Ax = 0
ΣFy = 0: Ay + H − 9 = 0
Ay = 9 − 4.50
A y = 4.50 kN
Free body: Joint A:
FAB
FAC 3.50 kN
=
2
1
5
FAB = 7.8262 kN C
=
FAB = 7.83 kN C 
FAC = 7.00 kN T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
762
PROBLEM 6.13 (Continued)
Free body: Joint B:
ΣFx = 0:
2
5
FBD +
2
5
(7.8262 kN) +
1
2
FBC = 0
FBD + 0.79057 FBC = −7.8262 kN
or
ΣFy = 0:
1
5
FBD +
1
5
(7.8262 kN) −
1
2
(1)
FBC − 2 kN = 0
FBD − 1.58114 FBC = −3.3541
or
(2)
Multiply Eq. (1) by 2 and add Eq. (2):
3FBD = −19.0065
FBD = 6.34 kN C 
FBD = −6.3355 kN
Subtract Eq. (2) from Eq. (1):
2.37111FBC = −4.4721
FBC = 1.886 kN C 
FBC = −1.8861 kN
Free body: Joint C:
2
ΣFy = 0:
5
FCD −
1
2
(1.8861 kN) = 0
FCD = 1.491 kN T 
FCD = +1.4911 kN
ΣFx = 0: FCE − 7.00 kN +
FCE
1
(1.8861 kN) +
2
= +5.000 kN
1
5
(1.4911 kN) = 0
FCE = 5.00 kN T 
Free body: Joint D:
ΣFx = 0:
2
5
FDF +
1
2
FDE +
2
5
(6.3355 kN) −
1
5
(1.4911 kN) = 0
FDF + 0.79057 FDE = −5.5900 kN
or
ΣFy = 0:
or
1
5
FDF −
1
2
FDE +
1
5
(6.3355 kN) −
(1)
2
5
(1.4911 kN) − 2 kN = 0
FDF − 0.79057 FDE = −1.1188 kN
(2)
2 FDF = −6.7088 kN
FDF = 3.35 kN C 
Add Eqs. (1) and (2):
FDF = −3.3544 kN
Subtract Eq. (2) from Eq. (1):
1.58114 FDE = −4.4712 kN
FDE = −2.8278 kN
FDE = 2.83 kN C 
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763
PROBLEM 6.13 (Continued)
Free body: Joint F:
ΣFx = 0:
1
FFG +
2
2
(3.3544 kN) = 0
5
FFG = 4.24 kN C 
FFG = −4.243 kN
ΣFy = 0: −FEF − 1.75 kN +
1
5
(3.3544 kN) −
1
2
FEF = 2.750 kN
(−4.243 kN) = 0
FEF = 2.75 kN T 
Free body: Joint G:
ΣFx = 0:
1
FGH −
2
ΣFy = 0: −
1
2
1
2
(4.243 kN) = 0
FGH −
1
2
FEG −
1
2
(1)
(4.243 kN) − 1.5 kN = 0
FGH + FEG = −6.364 kN
or
2 FGH = −10.607
(2)
FGH = 5.30 kN C 
FGH = −5.303
Subtract Eq. (1) from Eq. (2):
2
FEG +
FGH − FEG = −4.243 kN
or
Add Eqs. (1) and (2):
1
2 FEG = −2.121 kN
FEG = −1.0605 kN
FEG = 1.061 kN C 
Free body: Joint H:
FEH = 3.75 kN T 
FEH 3.75 kN
=
1
1
We can also write
FGH
2
=
3.75 kN
1
FGH = 5.30 kN C (Checks)
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764
PROBLEM 6.14
The truss shown is one of several supporting an advertising panel. Determine
the force in each member of the truss for a wind load equivalent to the two
forces shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss:
ΣM F = 0: (800 N)(7.5 m) + (800 N)(3.75 m) − A(2 m) = 0
A = +2250 N
A = 2250 N
ΣFy = 0: 2250 N + Fy = 0
Fy = −2250 N Fy = 2250 N
ΣFx = 0: −800 N − 800 N + Fx = 0
Fx = +1600 N Fx = 1600 N
Joint D:
800 N FDE FBD
=
=
8
15
17
FBD = 1700 N C 
FDE = 1500 N T 
Joint A:
2250 N FAB FAC
=
=
15
17
8
FAB = 2250 N C 
FAC = 1200 N T 
Joint F:
ΣFx = 0: 1600 N − FCF = 0
FCF = +1600 N
ΣFy = 0: FEF − 2250 N = 0
FEF = +2250 N
FCF = 1600 N T 
FEF = 2250 N T 
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765
PROBLEM 6.14 (Continued)
Joint C:
ΣFx = 0:
8
FCE − 1200 N + 1600 N = 0
17
FCE = −850 N
ΣFy = 0: FBC +
15
FCE = 0
17
FBC = −
15
15
FCE = − ( −850 N)
17
17
FBC = +750 N
Joint E:
FCE = 850 N C 
ΣFx = 0: − FBE − 800 N +
8
(850 N) = 0
17
FBE = −400 N
ΣFy = 0: 1500 N − 2250 N +
FBC = 750 N T 
FBE = 400 N C 
15
(850 N) = 0
17
0 = 0 (checks)
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766
PROBLEM 6.15
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:
ΣFx = 0: A x = 0
Because of symmetry of loading,
Ay = L =
1
total load
2
A y = L = 1200 lb
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
(1)
Free body: Joint A:
FAB
FAC 1000 lb
=
2
1
5
FAB = 2236 lb C
=
FAB = 2240 lb C 
FAC = 2000 lb T 
FCE = 2000 lb T 
From Eq. (1):
Free body: Joint B:
ΣFx = 0:
2
5
FBD +
2
5
FBE +
2
5
(2236 lb) = 0
FBD + FBE = −2236 lb
or
ΣFy = 0:
1
5
FBD −
1
5
FBE +
1
FBD − FBE = −1342 lb
or
(2)
5
(2236 lb) − 400 lb = 0
(3)
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767
PROBLEM 6.15 (Continued)
FBD = 1789 lb C 
2 FBD = −3578 lb
Add Eqs. (2) and (3):
FBE = 447 lb C 
Subtract Eq. (3) from Eq. (1): 2 FBE = − 894 lb
Free body: Joint E:
ΣFx = 0:
2
5
FEG +
ΣFy = 0: FDE +
1
5
2
5
(447 lb) − 2000 lb = 0
FEG = 1789 lb T 
(1789 lb) −
1
5
FDE = − 600 lb
(447 lb) = 0
FDE = 600 lb C 
Free body: Joint D:
ΣFx = 0:
2
5
FDF +
2
5
FDG +
2
5
(1789 lb) = 0
FDF + FDG = −1789 lb
or
ΣFy = 0:
1
FDF −
1
FDG +
(4)
1
5
5
5
+ 600 lb − 400 lb = 0
FDF − FDG = −2236 lb
or
Add Eqs. (4) and (5):
(1789 lb)
(5)
2 FDF = − 4025 lb FDF = 2010 lb C 
Subtract Eq. (5) from Eq. (4): 2 FDG = 447 lb
FDG = 224 lb T 
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768
PROBLEM 6.16
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,
L=
1
total load,
2
L = 1200 lb
(See F.B. diagram to the left for more details.)
Free body: Joint L:
9
= 26.57°
18
3
β = tan −1 = 9.46°
18
FJL
FKL
1000 lb
=
=
sin 63.43° sin 99.46° sin17.11°
α = tan −1
FKL = 3352.7 lb C
FJL = 3040 lb T 
FKL = 3350 lb C 
Free body: Joint K:
ΣFx = 0: −
or
2
5
FIK −
2
5
2
FJK −
5
(3352.7 lb) = 0
FIK + FJK = − 3352.7 lb
ΣFy = 0:
or
1
5
FIK −
1
5
FJK +
(1)
1
5
(3352.7) − 400 = 0
FIK − FJK = − 2458.3 lb
2 FIK = − 5811.0
Add Eqs. (1) and (2):
FIK = − 2905.5 lb
Subtract Eq. (2) from Eq. (1): 2 FJK = − 894.4
FJK = − 447.2 lb
(2)
FIK = 2910 lb C 
FJK = 447 lb C 
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769
PROBLEM 6.16 (Continued)
Free body: Joint J:
2
ΣFx = 0: −
13
3
ΣFy = 0:
FIJ +
13
6
FIJ −
37
1
37
6
FGJ +
FGJ −
37
1
37
2
(3040 lb) −
(3040 lb) −
5
1
5
(447.2) = 0
(447.2) = 0
(3)
(4)
Multiply Eq. (4) by 6 and add to Eq. (3):
16
13
FIJ −
8
(447.2) = 0
5
FIJ = 361 lb T 
FIJ = 360.54 lb
Multiply Eq. (3) by 3, Eq. (4) by 2, and add:
−
16
37
( FGJ − 3040) −
8
(447.2) = 0
5
FGJ = 2430 lb T 
FGJ = 2431.7 lb
Free body: Joint I:
2
ΣFx = 0: −
5
2
FFI −
5
2
FGI −
5
2
(2905.5) +
13
(360.54) = 0
FFI + FGI = − 2681.9 lb
or
ΣFy = 0:
1
5
FFI −
1
5
FGI +
1
5
(5)
(2905.5) −
3
13
(360.54) − 400 = 0
FFI − FGI = −1340.3 lb
or
2 FFI = −4022.2
Add Eqs. (5) and (6):
FFI = −2011.1 lb
Subtract Eq. (6) from Eq. (5):
2 FGI = −1341.6 lb
(6)
FFI = 2010 lb C 
FGI = 671 lb C 
Free body: Joint F:
From
ΣFx = 0: FDF = FFI = 2011.1 lb C
 1

2011.1 lb  = 0
ΣFy = 0: FFG − 400 lb + 2 
 5

FFG = + 1400 lb
FFG = 1400 lb T 
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770
PROBLEM 6.17
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:
ΣFx = 0: A x = 0
ΣM 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:
Also,
FCE = FAC
Free body: Joint A:
ΣFx = 0:
ΣFy = 0:
(1)
1
2
1
2
FAB +
FAB +
2
5
1
5
FAC = 0
(2)
FAC + 3.50 kN = 0
(3)
Multiply Eq. (3) by –2 and add Eq. (2):
−
1
2
FAB − 7 kN = 0
FAB = 9.90 kN C 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
771
PROBLEM 6.17 (Continued)
Subtract Eq. (3) from Eq. (2):
1
5
FAC − 3.50 kN = 0 FAC = 7.826 kN
FCE = 7.83 kN T 
FCE = FAC = 7.826 kN
From Eq. (1):
FAC = 7.83 kN T 
Free body: Joint B:
ΣFy = 0:
1
2
FBD +
1
2
(9.90 kN) − 2 kN = 0
FBD = −7.071 kN
1
ΣFx = 0: FBE +
2
(9.90 − 7.071) kN = 0
FBE = −2.000 kN
Free body: Joint E:
ΣFx = 0:
2
5
FBD = 7.07 kN C 
FBE = 2.00 kN C 
( FEG − 7.826 kN) + 2.00 kN = 0
FEG = 5.590 kN
ΣFy = 0: FDE −
1
5
FEG = 5.59 kN T 
(7.826 − 5.590) kN = 0
FDE = 1.000 kN
FDE = 1.000 kN T 
Free body: Joint D:
ΣFx = 0:
2
5
( FDF + FDG ) +
1
2
(7.071 kN)
FDF + FDG = −5.590 kN
or
ΣFy = 0:
or
1
5
( FDF − FDG ) +
1
2
(4)
(7.071 kN) = 2 kN − 1 kN = 0
FDE − FDG = −4.472
(5)
Add Eqs. (4) and (5):
2 FDF = −10.062 kN
FDF = 5.03 kN C 
Subtract Eq. (5) from Eq. (4):
2 FDG = −1.1180 kN
FDG = 0.559 kN C 
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772
PROBLEM 6.18
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:
ΣM A = 0: L(12 m) − (2 kN)(2 m) − (2 kN)(4 m) − (1 kN)(6 m) = 0
L = 1.500 kN
Angles:
tan α = 1
tan β =
1
2
α = 45°
β = 26.57°
Zero-force members:
Examining successively joints K, J, and I, we note that the following members to the right of FG are zeroforce 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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773
PROBLEM 6.18 (Continued)
Free body: Joint L:
FJL
F
1.500 kN
= KL =
sin116.57° sin 45° sin18.43°
FJL = 4.24 kN C 
FJL = 4.2436 kN
From Eq. (1):
FGI = FIK = FKL
From Eq. (2):
FHJ = FJL = 4.2436 kN
FKL = 3.35 kN T 
FGI = FIK = 3.35 kN T 
FHJ = 4.24 kN C 
Free body: Joint H:
FGH
FFH
4.2436
=
=
sin108.43° sin18.43° sin 53.14°
FFH = 5.03 kN C 
FGH = 1.677 kN T 
Free body: Joint F:
ΣFx = 0: − FDF cos 26.57° − (5.03 kN) cos 26.57° = 0
FDF = −5.03 kN
ΣFy = 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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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774
PROBLEM 6.19
Determine the force in each member of the Warren bridge truss
shown. State whether each member is in tension or compression.
SOLUTION
ΣFx = 0: Ax = 0
Free body: Truss:
Due to symmetry of truss and loading,
Ay = G =
Free body: Joint A:
1
total load = 6 kips
2
FAB = 7.50 kips C 
FAB FAC 6 kips
=
=
5
3
4

Free body: Joint B:
FBC FBD 7.5 kips
=
=
5
6
5
FAC = 4.50 kips T 
FBC = 7.50 kips T 
FBD = 9.00 kips C 
Free body: Joint C:
ΣFy = 0:
4
4
(7.5) + FCD − 6 = 0
5
5
3
ΣFx = 0: FCE − 4.5 − (7.5) = 0
5
FCE = +9 kips
FCD = 0 
FCE = 9.00 kips T 
Truss and loading is symmetrical about cL.
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775
PROBLEM 6.20
Solve Problem 6.19 assuming that the load applied at E has
been removed.
PROBLEM 6.19 Determine the force in each member of the
Warren bridge truss shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: Ax = 0
ΣM G = 0: 6(36) − Ay (54) = 0 A y = 4 kips
ΣFy = 0: 4 − 6 + G = 0 G = 2 kips
Free body: Joint A:
Free body Joint B:
FAB FAC 4 kips
=
=
5
3
4
FBC FBD 5 kips
=
=
5
6
5
FAB = 5.00 kips C 
FAC = 3.00 kips T 
FBC = 5.00 kips T 
FBD = 6.00 kips C 
Free body Joint C:
ΣM y = 0:
4
4
(5) + FCD − 6 = 0
5
5
3
3
ΣFx = 0: FCE + (2.5) − (5) − 3 = 0
5
5
FCD = 2.50 kips T 
FCE = 4.50 kips T 
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776
PROBLEM 6.20 (Continued)
Free body: Joint D:
4
4
ΣFy = 0: − (2.5) − FDE = 0
5
5
FDE = −2.5 kips
3
3
ΣFx = 0: FDF + 6 − (2.5) − (2.5) = 0
5
5
FDF = −3 kips
Free body: Joint F:
FEF FFG 3 kips
=
=
5
5
6

Free body: Joint G:
FEG 2 kips
=
3
4
Also,
FFG 2 kips
=
5
4
FDE = 2.50 kips C 
FDF = 3.00 kips C 
FEF = 2.50 kips T 
FFG = 2.50 kips C 
FEG = 1.500 kips T 
FFG = 2.50 kips C (Checks)
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777
PROBLEM 6.21
Determine the force in each member of the Pratt bridge truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Truss:
ΣFz = 0: A x = 0
ΣM A = 0 : H (36 ft) − (4 kips)(9 ft)
− (4 kips)(18 ft) − (4 kips)(27 ft) = 0
H = 6 kips
ΣFy = 0: Ay + 6 kips − 12 kips = 0
A y = 6 kips
Free body: Joint A:
FAB FAC 6 kips
=
=
5
3
4
FAB = 7.50 kips C 
FAC = 4.50 kips T 
Free body: Joint C:
ΣFx = 0:
ΣFy = 0:
FCE = 4.50 kips T 
FBC = 4.00 kips T 
Free body: Joint B:
ΣFy = 0: −
ΣFx = 0:
4
4
FBE + (7.50 kips) − 4.00 kips = 0 
5
5
8
3
(7.50 kips) + (2.50 kips) + FBD = 0 
5
5
FBD = −6.00 kips
FBE = 2.50 kips T 
FBD = 6.00 kips C 
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778
PROBLEM 6.21 (Continued)
Free body: Joint D:
We note that DE is a zero-force member:
Also,
From symmetry:
FFE = FBE
FEG = FCE
FFG = FBC
FFH = FAB
FGH = FAC
FDE = 0 
FDF = 6.00 kips C 
FEF = 2.50 kips T 
FEG = 4.50 kips T 
FFG = 4.00 kips T 
FFH = 7.50 kips C 
FGH = 4.50 kips T 
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779
PROBLEM 6.22
Solve Problem 6.21 assuming that the load applied at G has
been removed.
PROBLEM 6.21 Determine the force in each member of the
Pratt bridge truss shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: A x = 0
ΣM A = 0: H (36 ft) − (4 kips)(9 ft) − (4 kips)(18 ft) = 0
H = 3.00 kips
ΣFy = 0: A y + 5.00 kips
We note that DE and FG are zero-force members.
FDE = 0,
Therefore,
FFG = 0 
Also,
FBD = FDF
(1)
and
FEG = FGH
(2)
Free body: Joint A:
FAB FAC 5 kips
=
=
5
3
4
FAB = 6.25 kips C 
FAC = 3.75 kips T 

Free body: Joint C:
ΣFx = 0:
ΣFy = 0:
FCE = 3.75 kips T 
FBC = 4.00 kips T 
Free body: Joint B:
ΣFx = 0:
4
4
(6.25 kips) − 4.00 kips − FBE = 0
5
5
FBE = 1.250 kips T 
3
3
ΣFx = 0: FBD + (6.25 kips) + (1.250 kips) = 0 
5
5
FBD = −4.50 kips
FBD = 4.50 kips C 
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780
PROBLEM 6.22 (Continued)
Free body: Joint F:
We recall that FFG = 0, and from Eq. (1) that
FDF = FBD
FEF FFH 4.50 kips
=
=
5
5
6
FDF = 4.50 kips C 
FEF = 3.75 kips T 
FFH = 3.75 kips C 
Free body: Joint H:
FGH 3.00 kips
=
3
4
FGH = 2.25 kips T 
Also,
FFH 3.00 kips
=
5
4
FFH = 3.75 kips C (checks)
From Eq. (2):
FEG = FGH
FEG = 2.25 kips T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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781
PROBLEM 6.23
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:
F
FAB
1.2 kN
= AC =
2.29 2.29
1.2
FAB = 2.29 kN T 
FAC = 2.29 kN C 
Free body: Joint F:
FDF
F
1.2 kN
= EF =
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:
ΣFx = 0:
4
2.21
(2.29 kN) = 0
FBE + 2.21 kN −
5
2.29
FBE = 0 
3
0.6
(2.29 kN) = 0
ΣFy = 0: − FBC − (0) −
5
2.29
FBC = − 0.600 kN
FBC = 0.600 kN C 
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782
PROBLEM 6.23 (Continued)
Free body: Joint C:
ΣFx = 0: FCE +
2.21
(2.29 kN) = 0
2.29
FCE = − 2.21 kN
ΣFy = 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:
ΣFx = 0: 2.21 kN −
2.21
4
(2.29 kN) − FEH = 0
2.29
5
FEH = 0 
ΣFy = 0: − FEJ − 0.600 kN −
FEJ = −1.200 kN
0.6
(2.29 kN) − 0 = 0
2.29
FEJ = 1.200 kN C 
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783
PROBLEM 6.24
For the tower and loading of Problem 6.23 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 6.23 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 
FGI = 3.03 kN C 
Free body: Joint L:
FJL
F
1.2 kN
= KL =
3.03 3.03
1.2
FJL = 3.03 kN T 
FKL = 3.03 kN C 
Free body: Joint J:
ΣFx = 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:
ΣFx = 0:
4
2.97
(3.03 kN) = 0
FHK + 2.97 kN −
5
3.03
FHK = 0 
0.6
3
(3.03) kN − (0) = 0
3.03
5
= −1.800 kN
FHI = 1.800 kN C 
ΣFy = 0: − FHI − 1.2 kN −
FHI
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784
PROBLEM 6.24 (Continued)
Free body: Joint I:
ΣFx = 0: FIK +
2.97
(3.03 kN) = 0
3.03
FIK = −2.97 kN
ΣFy = 0: − FIN − 1.800 kN −
FIN = −2.40 kN
FIK = 2.97 kN C 
0.6
(3.03 kN) = 0
3.03
FIN = 2.40 kN C 
Free body: Joint K:
ΣFx = 0: −
4
2.97
(3.03 kN) = 0
FKN + 2.97 kN −
5
3.03
ΣFy = 0: − FKO −
FKN = 0 
0.6
3
(3.03 kN) − 1.800 kN − (0) = 0
3.03
5
FKO = −2.40 kN
FKO = 2.40 kN C 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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785
PROBLEM 6.25
Solve Problem 6.23 assuming that the cables hanging from the
right side of the tower have fallen to the ground.
PROBLEM 6.23 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:
F
FAB
1.2 kN
= AC =
2.29 2.29
1.2
FAB = 2.29 kN T 
FAC = 2.29 kN C 
Free body: Joint B:
ΣFx = 0:
4
2.21
(2.29 kN) = 0
FBE −
5
2.29
FBE = 2.7625 kN
ΣFy = 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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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786
PROBLEM 6.25 (Continued)
Free body: Joint C:
ΣFx = 0: FCE +
2.21
(2.29 kN) = 0
2.29
ΣFy = 0: − FCH − 2.2575 kN −
FCH = −2.8575 kN
FCE = 2.21 kN C 
0.6
(2.29 kN) = 0
2.29
FCH = 2.86 kN C 
Free body: Joint E:
ΣFx = 0: −
4
4
FEH − (2.7625 kN) + 2.21 kN = 0
5
5
FEH = 0 
3
3
ΣFy = 0: − FEJ + (2.7625 kN) − (0) = 0
5
5
FEJ = +1.6575 kN
FEJ = 1.658 kN T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
787
PROBLEM 6.26
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:
ΣFx = 0: A x = 0
ΣM K = 0: (1.2 kN)6a + (2.4 kN)5a + (2.4 kN)4a + (1.2 kN)3a
− Ay (6a) = 0 A y = 5.40 kN
Free body: Joint A:
FAB
FAC 4.20 kN
=
2
1
5
FAB = 9.3915 kN
=
FAB = 9.39 kN C 
FAC = 8.40 kN T 
Free body: Joint B:
2
ΣFx = 0:
5
1
ΣFy = 0:
5
FBD +
FBD −
1
2
1
2
FBC +
FBC +
2
5
1
5
(9.3915) = 0
(1)
(9.3915) − 2.4 = 0
(2)
Add Eqs. (1) and (2):
3
5
FBD +
3
5
(9.3915 kN) − 2.4 kN = 0
FBD = − 7.6026 kN
FBD = 7.60 kN C 
Multiply Eq. (2) by –2 and add Eq. (1):
3
2
FB + 4.8 kN = 0
FBC = −2.2627 kN
FBC = 2.26 kN C 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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788
PROBLEM 6.26 (Continued)
Free body: Joint C:
1
ΣFx = 0:
5
2
ΣFy = 0:
5
4
FCD +
17
1
FCD +
17
FCE +
FCE −
1
2
1
2
(2.2627) − 8.40 = 0
(3)
(2.2627) = 0
(4)
Multiply Eq. (4) by −4 and add Eq. (1):
−
7
5
FCD +
5
(2.2627) − 8.40 = 0
2
FCD = 0.128 kN C 
FCD = − 0.1278 kN
Multiply Eq. (1) by 2 and subtract Eq. (2):
7
17
FCE +
3
2
(2.2627) − 2(8.40) = 0
FCE = 7.07 kN T 
FCE = 7.068 kN
Free body: Joint D:
ΣFx = 0:
2
1
+
ΣFy = 0:
FDF +
5
5
1
5
+
(0.1278) = 0
FDF −
2
5
1
2
FDE +
(7.6026)
1.524
5
(5)
1.15
1
FDE +
(7.6026)
1.524
5
(0.1278) − 2.4 = 0
(6)
Multiply Eq. (5) by 1.15 and add Eq. (6):
3.30
5
FDF +
3.30
5
(7.6026) +
3.15
5
(0.1278) − 2.4 = 0
FDF = − 6.098 kN
FDF = 6.10 kN C 
Multiply Eq. (6) by –2 and add Eq. (5):
3.30
3
FDE −
(0.1278) + 4.8 = 0
1.524
5
FDE = − 2.138 kN
FDE = 2.14 kN C 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
789
PROBLEM 6.26 (Continued)
Free body: Joint E:
ΣFx = 0:
0.6
4
1
FEF +
( FEH − FCE ) +
(2.138) = 0
2.04
1.524
17
(7)
ΣFy = 0:
1.95
1
1.15
FEF +
( FEH − FCE ) −
(2.138) = 0
2.04
1.524
17
(8)
Multiply Eq. (8) by 4 and subtract Eq. (7):
7.2
FEF − 7.856 kN = 0
2.04
FEF = 2.23 kN T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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790
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:
ΣM F = 0: G (20 ft) − (15 kips)(16 ft) − (40 kips)(15 ft) = 0
G = 42 kips
ΣFx = 0: Fx + 15 kips = 0
Fx = 15 kips
ΣFy = 0: Fy − 40 kips + 42 kips = 0
Fy = 2 kips
Free body: Joint F:
ΣFx = 0:
1
5
FDF − 15 kips = 0
FDF = 33.54 kips
ΣFy = 0: FBF − 2 kips +
2
5
FDF = 33.5 kips T 
(33.54 kips) = 0
FBF = − 28.00 kips
FBF = 28.0 kips C 
Free body: Joint B:
ΣFx = 0:
ΣFy = 0:
5
29
2
29
FAB +
FAB −
5
61
6
61
FBD + 15 kips = 0
(1)
FBD + 28 kips = 0
(2)
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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791
PROBLEM 6.27 (Continued)
Multiply Eq. (1) by 6, Eq. (2) by 5, and add:
40
29
FAB + 230 kips = 0
FAB = 31.0 kips C 
FAB = −30.96 kips
Multiply Eq. (1) by 2, Eq. (2) by –5, and add:
40
61
FBD − 110 kips = 0
FBD = 21.5 kips T 
FBD = 21.48 kips
Free body: Joint D:
2
ΣFy = 0:
5
2
FAD −
1
ΣFx = 0: FDE +
5
5
(33.54) +
6
61
(15.09 − 33.54) −
(21.48) = 0
FAD = 15.09 kips T 
5
(21.48) = 0
61
FDE = 22.0 kips T 
Free body: Joint A:
5
ΣFx = 0:
29
ΣFy = 0: −
1
FAC +
2
29
5
FAC −
FAE +
2
5
5
FAE +
29
(30.36) −
2
29
1
5
(15.09) = 0
2
(30.96) −
5
(15.09) = 0
(3)
(4)
Multiply Eq. (3) by 2 and add Eq. (4):
8
29
FAC +
12
29
(30.96) −
4
5
FAC = −28.27 kips,
(15.09) = 0
FAC = 28.3 kips C 
Multiply Eq. (3) by 2, Eq. (4) by 5, and add:
−
8
5
FAE +
20
29
(30.96) −
12
5
(15.09) = 0
FAE = 9.50 kips T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
792
PROBLEM 6.27 (Continued)
Free body: Joint C:
From force triangle:
FCE
61
=
FCG 28.27 kips
=
8
29
FCE = 41.0 kips T 
FCG = 42.0 kips C 
Free body: Joint G:
ΣFx = 0:
ΣFy = 0: 42 kips − 42 kips = 0
(Checks) 
FEG = 0 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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793
PROBLEM 6.28
Determine the force in each member of the truss shown. State
whether each member is in tension or compression.
SOLUTION
Reactions:
ΣFx = 0:
Ex = 0
ΣM F = 0:
E y = 45 kips
ΣFy = 0:
F = 60 kips
Joint D:
FCD FDH 15 kips
=
=
12
13
5
FCD = 36.0 kips T 
FDH = 39.0 kips C 
Joint H:
ΣF = 0:
FCH = 0 
ΣF = 0:
FGH = 39.0 kips C 
ΣF = 0:
FCG = 0 
Joint C:
ΣF = 0:
FBC = 36.0 kips T 
ΣF = 0:
FBG = 0 
Joint G:
ΣF = 0:
FFG = 39.0 kips C 
ΣF = 0:
FBF = 0 
Joint B:
ΣF = 0:
FAB = 36.0 kips T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
794
PROBLEM 6.28 (Continued)
Joint A:
AE = 122 + 152 = 19.21 ft
tan 38.7° =
36 kips
FAF
sin 38.7° =
36 kips
FAE
FAF = 45.0 kips C 
FAE = 57.6 kips T 
Joint E:
ΣFx = 0: + (57.6 kips) sin 38.7° + FEF = 0
FEF = −36.0 kips
FEF = 36.0 kips C 
ΣFy = 0: (57.6 kips) cos 38.7° − 45 kips = 0 (Checks)
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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795
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 HDI and adding two members at a time, we obtain
successively joints A, E, J, and B, but cannot go further. Thus, this truss
is not a simple truss. 
Truss of Problem 6.32a:
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 6.33a:
Starting with triangle ABD and adding two members at a time, we obtain
successively joints H, G, F, E, I, C, and J, thus completing the truss.
Therefore, this is a simple truss. 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
796
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 CGM 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 6.32b:
Starting with triangle ABC and adding two members at a time, we obtain
successively joints E, D, F, G, and H, but cannot go further. Thus, this truss
is not a simple truss. 
Truss of Problem 6.33b:
Starting with triangle GFH and adding two members at a time, we obtain
successively joints D, E, C, A, and B, thus completing the truss.
Therefore, this is a simple truss. 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
797
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: FBJ = 0
FB: Joint D: FDI = 0
FB: Joint E: FEI = 0
FB: Joint I : FAI = 0
(a)
FB: Joint F : FFK = 0
FB: Joint G: FGK = 0
FB: Joint K : FCK = 0
AI , BJ , CK , DI , EI , FK , GK 
The zero-force members, therefore, are
Truss (b):
FB: Joint K : FFK = 0
FB: Joint O : FIO = 0
(b)
The zero-force members, therefore, are
FK and IO 
All other members are either in tension or compression.
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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798
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: 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):
(a)
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
(b)
FB: Joint M : FMN = 0
FB: Joint N : FKN = 0
The zero-force members, therefore, are
BC , BE , DE , EF , FG , IJ , KN , MN 
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799
PROBLEM 6.33
For the given loading, determine the zero-force members in each
of the two trusses shown.
SOLUTION
Truss (a):
Note: Reaction at F is vertical ( Fx = 0).
Joint G:
ΣF = 0,
Joint D:
ΣF = 0,
Joint F :
ΣF = 0,
Joint G:
ΣF = 0,
Joint J :
ΣF = 0,
Joint I :
ΣF = 0,
FDG = 0 
FDB = 0 
FFG = 0 
FGH = 0 
FIJ = 0 
FHI = 0 
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800
PROBLEM 6.33 (Continued)
Truss (b):
Joint A:
ΣF = 0,
Joint C:
ΣF = 0,
Joint E:
ΣF = 0,
Joint F :
ΣF = 0,
Joint G:
ΣF = 0,
FAC = 0 
FCE = 0 
FEF = 0 
FFG = 0 
FGH = 0 
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801
PROBLEM 6.34
Determine the zero-force members in the truss of (a) Problem 6.26, (b) Problem 6.28.
SOLUTION
(a)
Truss of Problem 6.26:
FB : Joint I : FIJ = 0
FB : Joint J : FGJ = 0
FB : Joint G: FGH = 0
GH , GJ , IJ 
The zero-force members, therefore, are
(b)
Truss of Problem 6.28:
FB : Joint B: FBF = 0
FB : Joint B: FBG = 0
FB : Joint C: FCG = 0
FB : Joint C: FCH = 0
The zero-force members, therefore, are
BF , BG, CG, CH 
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802
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
Dy = B y
ΣM z = 0: − A(10 ft) − (400 lb)(24 ft) = 0
A = −960 lb
ΣFx = 0: Bx + Dx + A = 0
2 Bx − 960 lb = 0, Bx = 480 lb
ΣFy = 0: B y + Dy − 400 lb = 0
2 By = 400 lb
By = +200 lb
Thus,
Free body: C:
FCA = FAC
FCB = FBC
FCD = FCD
B = (480 lb)i + (200 lb) j 

CA FAC
=
( −24i + 10 j)
CA 26

CB FBC
=
(−24i + 7k )
CB
25

CD FCD
=
(−24i − 7k )
CD
25
ΣF = 0: FCA + FCB + FCD − (400 lb) j = 0
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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803
PROBLEM 6.35* (Continued)
Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k :
24
24
FAC − ( FBC + FCD ) = 0
26
25
i:
−
j:
10
FAC − 400 lb = 0
26
k:
7
( FBC − FCD ) = 0 FCD = FBC
25
(1)
FAC = 1040 lb T 
Substitute for FAC and FCD in Eq. (1):
−
24
24
(10.40 lb) − (2 FBC ) = 0 FBC = −500 lb
26
25
Free body: B:
FBC
FBA
FBD
FBC = FCD = 500 lb C 

CB
= (500 lb)
= −(480 lb)i + (140 lb)k
CB

BA FAB
= FAB
=
(10 j − 7k )
BA 12.21
= − FBD k
ΣF = 0: FBA + FBD + FBC + (480 lb)i + (200 lb) j = 0
Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k:
j:
10
FAB + 200 lb = 0 FAB = −244.2 lb
12.21
k: −
7
FAB − FBD + 140 lb = 0
12.21
FBD = −
From symmetry:
7
(−244.2 lb) + 140 lb = +280 lb
12.21
FAD = FAB
FAB = 244 lb C 
FBD = 280 lb T 
FAD = 244 lb C 
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804
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 Dy = By
ΣFx = 0: 2 Bx = 0
Bx = Dx = 0
ΣFz = 0: Bz = 0
ΣM c z = 0: − 2 By (2.8 m) + (2184 N)(2 m) = 0
By = 780 N
Thus,
Free body: A:
B = (780 N) j 

AB FAB
FAB = FAB
=
( −0.8i − 4.8 j + 2.1k )
AB 5.30

AC FAC
=
FAC = FAC
(2i − 4.8 j)
AC 5.20

AD FAD
=
FAD = FAD
(+0.8i − 4.8 j − 2.1k )
AD 5.30
ΣF = 0: FAB + FAC + FAD − (2184 N) j = 0
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805
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)
2.1
( FAB − FAD ) = 0
5.30
k:
FAD = FAB
Multiply Eq. (1) by –6 and add Eq. (2):
 16.8 
−
 FAC − 2184 N = 0, FAC = −676 N
 5.20 
FAC = 676 N C 
Substitute for FAC and FAD in Eq. (1):
 0.8 
 2 
−
 2 FAB + 
 (−676 N) = 0, FAB = −861.25 N
 5.30 
 5.20 
Free body: B:
FAB = FAD = 861 N C 

AB
= −(130 N)i − (780 N) j + (341.25 N)k
FAB = (861.25 N)
AB
 2.8i − 2.1k 
FBC = FBC 
 = FBC (0.8i − 0.6k )
3.5


FBD = − FBD k
ΣF = 0: FAB + FBC + FBD + (780 N) j = 0
Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k ,
i:
k:
−130 N + 0.8FBC = 0 FBC = +162.5 N
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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806
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 Dy = By
ΣFx = 0: 2 Bx + 2968 N = 0
Bx = Dx = −1484 N
ΣM cz ′ = 0: − 2 By (2.8 m) − (2968 N)(4.8 m) = 0
By = −2544 N
Thus,
Free body: A:
FAB = FAB
B = −(1484 N)i − (2544 N) j 

AB
AB
FAB
(−0.8i − 4.8 j + 2.1k )
5.30

AC FAC
= FAC
=
(2i − 4.8 j)
AC 5.20

AD
= FAD
AD
FAD
=
(−0.8i − 4.8 j − 2.1k )
5.30
=
FAC
FAD
ΣF = 0: FAB + FAC + FAD + (2968 N)i = 0
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807
PROBLEM 6.37* (Continued)
Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k ,
0.8
2
FAC + 2968 N = 0
( FAB + FAD ) +
5.30
5.20
(1)
4.8
4.8
FAC = 0
( FAB + FAD ) −
5.30
5.20
(2)
i:
−
j:
−
k:
2.1
( FAB − FAD ) = 0
5.30
FAD = FAB
Multiply Eq. (1) by –6 and add Eq. (2):
 16.8 
−
 FAC − 6(2968 N) = 0, FAC = −5512 N
 5.20 
FAC = 5510 N C 
Substitute for FAC and FAD in Eq. (2):
 4.8 
 4.8 
−
 2 FAB − 
 (−5512 N) = 0, FAB = +2809 N
 5.30 
 5.20 
Free body: B:
FAB
FBC
FBD
FAB = FAD = 2810 N T 

BA
= (2809 N)
= (424 N)i + (2544 N) j − (1113 N)k
BA
 2.8 i − 2.1k 
= FBC 
 = FBC (0.8i − 0.6k )
3.5


= − FBD k
ΣF = 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 ,
i:
+24 N + 0.8FBC − 1484 N = 0, FBC = +1325 N
k:
−1113 N − 0.6 FBC − FBD = 0
FBD = −1113 N − 0.6(1325 N) = −1908 N,
From symmetry:
FCD = FBC
FBC = 1325 N T 
FBD = 1908 N C 
FCD = 1325 N T 
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808
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
ΣFx = 0: Ax = 0
ΣM BC = 0: Ay (6 ft) + (1600 lb)(7.5 ft) = 0
Ay = −2000 lb
A = −(2000 lb)j 
By = C
From symmetry:
ΣFy = 0: 2 By − 2000 lb − 1600 lb = 0
By = 1800 lb
B = (1800 lb) j 
ΣF = 0: FAB + FAC + FAD − (2000 lb) j = 0
Free body: A:
FAB
i+k
+ FAC
2
i−k
2
+ FAD (0.6 i + 0.8 j) − (2000 lb) j = 0
Factoring i, j, k and equating their coefficient to zero,
1
2
FAB +
1
FAC + 0.6 FAD = 0
(1)
0.8FAD − 2000 lb = 0
FAD = 2500 lb T 
2
1
2
FAB −
1
2
FAC = 0
FAC = FAB
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809
PROBLEM 6.38* (Continued)
Substitute for FAD and FAC into Eq. (1):
2
2
Free body: B:
FAB + 0.6(2500 lb) = 0, FAB = −1060.7 lb,
FBA
FBC
FAB = FAC = 1061 lb C 

BA
i+k
= FAB
= + (1060.7 lb)
= (750 lb)(i + k )
BA
2
= − FBC k
FBD = FBD (0.8 j − 0.6k )

BE FBE
(7.5i + 8 j − 6k )
FBE = FBE
=
BE 12.5
ΣF = 0: FBA + FBC + FBD + FBE + (1800 lb) j = 0
Substituting for FBA , FBC , FBD , and FBE and equating to zero the coefficients of i, j, k ,
i:
 7.5 
750 lb + 
 FBE = 0, FBE = −1250 lb
 12.5 
j:
 8 
0.8 FBD + 
 (−1250 lb) + 1800 lb = 0
 12.5 
k:
750 lb − FBC − 0.6( −1250 lb) −
FBE = 1250 lb C 
FBD = 1250 lb C 
6
( −1250 lb) = 0
12.5
FBC = 2100 lb T 
FBD = FCD = 1250 lb C 
From symmetry:
Free body: D:
ΣF = 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(2500 lb) = −1500 lb
i:
−1500 lb + FDE = 0
FDE = 1500 lb T 
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810
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:
ΣM 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:
3Dy − 900 = 0, D y = 300 N
j:
Dz = 0,
k:
1.8C + 1.8(300) − 720 = 0
D = (300 N) j 
C = (100 N) j 
ΣF = 0: B + 300 j + 100 j − 1200 j = 0
B = (800 N) j 
Free body: B:
ΣF = 0: FBA + FBC + FBE + (800 N) j = 0, with

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:
i:
k:
 3 

 FAB + 800 N = 0, FAB = −840 N,
 3.315 
 0.6 

 (−840 N) + FBC = 0
 3.15 
 0.75 
−
 (−840 N) − FBE = 0
 3.15 
FAB = 840 N C 
FBC = 160.0 N T 
FBE = 200 N T 
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811
PROBLEM 6.39* (Continued)
Free body: C:
ΣF = 0: FCA + FCB + FCD + FCE + (100 N) j = 0, with

F
CA
FCA = FAC
= AC (−1.2i + 3j − 0.75 k )
CA 3.317
FCB = −(160 N) i
FCD = − FCD k FCE = FCE

F
CE
= CE (−1.8i − 3k )
CE 3, 499
Substitute and equate to zero the coefficient of j, i, k :
 3 

 FAC + 100 N = 0, FAC = −110.57 N
 3.317 
j:
−
i:
k:
−
FAC = 110.6 N C 
FCE = 233 N C 
1.2
1.8
FCE = 0, FCE = −233.3
(−110.57) − 160 −
3.317
3.499
FCD = 225 N T 
0.75
3
(−110.57) − FCD −
(−233.3) = 0
3.317
3.499
Free body: D:
ΣF = 0: FDA + FDC + FDE + (300 N) j = 0, with

F
DA
FDA = FAD
= AD (−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:
i:
k:
 3 

 FAD + 300 N = 0,
 3.937 
FAD = −393.7 N,
 1.2 
−
 ( −393.7 N) − FDE = 0
 3.937 
FAD = 394 N C 
FDE = 120.0 N T 
 2.25 

 (−393.7 N) + 225 N = 0 (Checks)
 3.937 
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 
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812
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:
ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dz k )
+ (0.6i + 3j − 0.75k ) × (−900N)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 :
3Dy − 2700 = 0
Dy = 900 N
−1.8Dz + 540 = 0
Dz = 300 N
1.8C + 1.8 Dy = 0
Thus,
C = − D y = −900 N
C = −(900 N) j D = (900 N) j + (300 N)k
ΣF = 0: B − 900 j + 900 j + 300k − 900k = 0

B = (600 N)k 
Free body: B:
Since B is aligned with member BE,
FAB = FBC = 0,
Free body: C:

FBE = 600 N T 
ΣF = 0: FCA + FCD + FCE − (900 N) j = 0, with
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813
PROBLEM 6.40* (Continued)

FCA
FCD

F
CA
= FAC
= AC ( −1.2i + 3j − 0.75k ) 
CA 3.317

F
CE
= − FCD k FCE = FCE
= CE (−1.8i − 3k )
CE 3.499
Substitute and equate to zero the coefficient of j, i, k:
j:
 3 

 FAC − 900 N = 0,
 3.317 
FAC = 995.1 N
FAC = 995 N T 
FCE = −699.8 N
FCE = 700 N C 
i:
−
1.2
1.8
FCE = 0,
(995.1) −
3.317
3.499
k:
−
0.75
3
(995.1) − FCD −
( −699.8) = 0
3.317
3.499
FCD = 375 N T 
Free body: D:
ΣF = 0: FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0

F
DA
FDA = FAD
= AD (−1.2i + 3j + 2.25k )
DA 3.937
with
FDE = − FDE i
and
Substitute and equate to zero the coefficient j, i, k:
j:
i:
k:
 3 

 FAD + 900 N = 0, FAD = −1181.1 N
 3.937 
FAD = 1181 N C 
 1.2 
−
 (−1181.1 N) − FDE = 0
 3.937 
FDE = 360 N T 
 2.25 

 (−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 
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814
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:
ΣM A = 0: 11i × ( By j + Bz k ) + (11i − 9.6k ) × G j
+ (10.08 j − 9.6k ) × (275i + 240k ) = 0
11By k − 11By j + 11Gk + 9.6Gi − (10.08)(275)k
+ (10.08)(240)i − (9.6)(275) j = 0
Equate to zero the coefficient of i, j, k:
i : 9.6G + (10.08)(240) = 0 G = −252 lb
j: − 11Bz − (9.6)(275) = 0 Bz = −240 lb
k : 11By + 11(−252) − (10.08)(275) = 0, By = 504 lb
G = (−252 lb) j 
B = (504 lb) j − (240 lb)k 
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815
PROBLEM 6.41* (Continued)
ΣF = 0: A + (504 lb) j − (240 lb)k − (252 lb) j
+ (275 lb)i + (240 lb)k = 0
A = −(275 lb)i − (252 lb) j 
Zero-force members.
The determination of these members will facilitate our solution.
FB: C: Writing
ΣFx = 0, ΣFy = 0, ΣFz = 0
FB: F: Writing
ΣFx = 0, ΣFy = 0, ΣFz = 0
FB: A: Since
FB: H: Writing
Az = 0, writing ΣFz = 0
ΣFy = 0
yields FBC = FCD = FCG = 0 
yields FBF = FEF = FFG = 0 
yields FAD = 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:
ΣFy = 0
Free body: H:
ΣFz = 0
yields FAE = 252 lb T 
yields FEH = 240 lb C 
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816
PROBLEM 6.41* (Continued)
ΣF = 0: FEB + FEG + (240 lb)k − (252 lb) j = 0
Free body: E:
F
FBE
(11i − 10.08 j) + EG (11i − 9.6k ) + 240k − 252 j = 0
14.92
14.6
Equate to zero the coefficient of y and k:
 10.08 
j: − 
 FBE − 252 = 0
 14.92 
k:
FBE = 373 lb C 
 9.6 
−
 FEG + 240 = 0
 14.6 
FEG = 365 lb T 
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817
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
Free body: H:
Free body: G:
FCG = FDG = FFG = 0 
yields FGH = 275 lb C 
ΣFx = 0
ΣF = 0: FGB + FGE + (275 lb)i − (252 lb) j = 0
FBG
F
(−10.08 j + 9.6k ) + EG (−11i + 9.6k ) + 275i − 252 j = 0
13.92
14.6
Equate to zero the coefficient of i, j, k:
 11 
i: − 
 FEG + 275 = 0
 14.6 
FEG = 365 lb T 
 10.08 
j: − 
 FBG − 252 = 0
 13.92 
 9.6 
 9.6 
k: 
 (−348) + 
 (365) = 0
 13.92 
 14.6 
FBG = 348 lb C 
(Checks)
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818
PROBLEM 6.43
Determine the force in members CD and DF of the truss shown.
SOLUTION
Reactions:
ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B(9.6 m) = 0
B = 9.00 kN
ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0
J = 15.00 kN
Member CD:
ΣFy = 0: 9.00 kN + FCD = 0
Member DF:
ΣM C = 0: FDF (1.8 m) − (9.00 kN)(2.4 m) = 0
FCD = 9.00 kN C 
FDF = 12.00 kN T 
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819
PROBLEM 6.44
Determine the force in members FG and FH of the truss shown.
SOLUTION
Reactions:
ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B (9.6 m) = 0
B = 9.00 kN
ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0
J = 15.00 kN
Member FG:
3
ΣFy = 0: − FFG − 12.00 kN + 15.00 kN = 0
5
FFG = 5.00 kN T 
Member FH:
ΣM G = 0: (15.00 kN)(2.4 m) − FFH (1.8 m) = 0
FFH = 20.0 kN T 
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820
PROBLEM 6.45
A Warren bridge truss is loaded as shown. Determine the force in
members CE, DE, and DF.
SOLUTION
Free body: Truss:
ΣFx = 0: k x = 0
ΣM A = 0: k y (62.5 ft) − (6000 lb)(12.5 ft)
− (6000 lb)(25 ft) = 0
k = k y = 3600 lb 
ΣFy = 0: A + 3600 lb − 6000 lb − 6000 lb = 0
A = 8400 lb 
We pass a section through members CE, DE, and DF and use the
free body shown.
ΣM D = 0: FCE (15 ft) − (8400 lb)(18.75 ft)
+ (6000 lb)(6.25 ft) = 0
FCE = +8000 lb
ΣFy = 0: 8400 lb − 6000 lb −
FCE = 8000 lb T 
15
FDE = 0
16.25
FDE = +2600 lb
FDE = 2600 lb T 
ΣM E = 0: 6000 lb (12.5 ft) − (8400 lb)(25 ft)
− FDF (15 ft) = 0
FDF = −9000 lb
FDF = 9000 lb C 
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821
PROBLEM 6.46
A Warren bridge truss is loaded as shown. Determine the force in
members EG, FG, and FH.
SOLUTION
See solution of Problem 6.45 for free-body diagram of truss and determination of reactions:
A = 8400 lb
k = 3600 lb 
We pass a section through members EG, FG, and FH, and use the free body shown.
ΣM F = 0: (3600 lb)(31.25 ft) − FEG (15 ft) = 0
FEG = +7500 lb
ΣFy = 0:
15
FFG + 3600 lb = 0
16.25
FFG = −3900 lb
ΣM G = 0: FFH (15 ft) + (3600 lb)(25 ft) = 0
FFH = −6000 lb
FEG = 7500 lb T 
FFG = 3900 lb C 
FFH = 6000 lb C 
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822
PROBLEM 6.47
Determine the force in members DF, EF, and EG of the
truss shown.
SOLUTION
tan β =
3
4
Reactions:
A=N=0
Member DF:
3
ΣM E = 0: + (16 kN)(6 m) − FDF (4 m) = 0
5
FDF = + 40 kN
Member EF:
+ ΣF = 0: (16 kN) sin β − FEF cos β = 0
FEF = 16 tan β = 16(0.75) = 12 kN
Member EG:
ΣM F = 0: (16 kN)(9 m) +
4
FEG (3 m) = 0
5
FEG = −60 kN
FDF = 40.0 kN T 
FEF = 12.00 kN T 
FEG = 60.0 kN C 
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823
PROBLEM 6.48
Determine the force in members GI, GJ, and HI of the
truss shown.
SOLUTION
Reactions:
A=N=0
Member GI:
+ ΣF = 0: (16 kN)sin β + FGI sin β = 0
FGI = −16 kN
Member GJ:
ΣM I = 0: − (16 kN)(9 m) −
4
FGJ (3 m) = 0
5
FGJ = −60 kN
Member HI:
ΣM G = 0: − (16 kN)(9 m) +
3
FHI (4 m) = 0
5
FHI = +60 kN
FGI = 16.00 kN C 
FGJ = 60.0 kN C 
FHI = 60.0 kN T 
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824
PROBLEM 6.49
Determine the force in members AD, CD, and CE of the
truss shown.
SOLUTION
Reactions:
ΣM k = 0: 36(2.4) − B(13.5) + 20(9) + 20(4.5) = 0
ΣFx = 0: −36 + K x = 0
B = 26.4 kN
K x = 36 kN
ΣFy = 0: 26.4 − 20 − 20 + K y = 0 K y = 13.6 kN
ΣM C = 0: 36(1.2) − 26.4(2.25) − FAD (1.2) = 0
 8

ΣM A = 0:  FCD  (4.5) = 0
17


FAD = −13.5 kN
FAD = 13.5 kN C 
FCD = 0 
 15

ΣM D = 0:  FCE  (2.4) − 26.4(4.5) = 0
 17

FCE = +56.1 kN
FCE = 56.1 kN T 
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825
PROBLEM 6.50
Determine the force in members DG, FG, and FH of the
truss shown.
SOLUTION
See the solution to Problem 6.49 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
ΣM F = 0: 36(1.2) − 26.4(6.75) + 20(2.25) − FDG (1.2) = 0
 8

ΣM D = 0: − 26.4(4.5) +  FFG  (4.5) = 0
17


FDG = −75 kN
FFG = +56.1 kN


 15
ΣM G = 0: 20(4.5) − 26.4(9) +  FFH
 17

 (2.4) = 0 

FFH = +69.7 kN 
FDG = 75.0 kN C 
FFG = 56.1 kN T 
FFH = 69.7 kN T 
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826
PROBLEM 6.51
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.
ΣM G = 0:
 40

 41 FAB  (6.3 ft) − (1.8 kips)(14 ft)


− (0.9 kips)(28 ft) = 0
FAB = +8.20 kips
FAB = 8.20 kips T 
FAG = +4.50 kips
FAG = 4.50 kips T 
3

ΣM D = 0: −  FAG  (28 ft) + (1.8 kips)(28 ft)
5

+ (1.8 kips)(14 ft) = 0
ΣM A = 0: − FFG (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft)
− (0.9 kips)(40 ft) = 0
FFG = 11.60 kips C 
FFG = −11.60 kips
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827
PROBLEM 6.52
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.


8
ΣM F = 0: 
FAE  (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0
 82 + 92



FAE = +17.46 kips
FAE = 17.46 kips T 
ΣM A = 0: − FEF (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0
FEF = −11.60 kips
FEF = 11.60 kips C 
ΣM E = 0: − FFJ (8 ft) − (0.9 kips)(8 ft) − (1.8 kips)(20 ft) − (1.8 kips)(34 ft) − (0.9 kips)(48 ft) = 0
FFJ = −18.45 kips
FFJ = 18.45 kips C 
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828
PROBLEM 6.53
Determine the force in members CD and DF of the truss shown.
SOLUTION
tan α =
5
α = 22.62°
12
sin α =
5
12
cos α =
13
13
Member CD:
ΣM I = 0: FCD (9 m) + (10 kN)(9 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0
FCD = 20.0 kN C 
FCD = −20 kN
Member DF:
ΣM C = 0: ( FDF cos α )(3.75 m) + (10 kN)(3 m) + (10 kN)(6 m) + (10 kN)(9 m) = 0
FDF cos α = −48 kN
 12 
FDF   = −48 kN
 13 
FDF = −52.0 kN
FDF = 52.0 kN C 
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829
PROBLEM 6.54
Determine the force in members CE and EF of the truss shown.
SOLUTION
Member CE:
ΣM F = 0: FCE (2.5 m) − (10 kN)(3 m) − (10 kN)(6 m) = 0
FCE = +36 kN
FCE = 36.0 kN T 
Member EF:
ΣM I = 0: FEF (6 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0
FEF = −15 kN
FEF = 15.00 kN C 
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830
PROBLEM 6.55
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,
Ay = O =
1
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
ΣM E = 0: (0.6 kN)(7.44 m) + (1.24 kN)(3.84 m)
− (4.48 kN)(7.44 m)
 6

−
FFG  (4.80 m) = 0
6.25


FFG = −5.231 kN
FFG = 5.23 kN C 
ΣM 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
ΣFy = 0:
FEH = 5.08 kN T 
5.50
1.75
(−5.231 kN) + 4.48 kN − 0.6 kN − 1.24 kN − 1.04 kN = 0
FEG +
6.001
6.25
FEG = −0.1476 kN
FEG = 0.1476 kN C 
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831
PROBLEM 6.56
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
Because of symmetry of loading,
O=
O = 4.48 kN 
1
load
2
We pass a section through KM, LM, LN, and use free body shown.
 3.84

ΣM M = 0: 
FLN  (3.68 m)
 4

+ (4.48 kN − 0.6 kN)(3.6 m) = 0
FLN = −3.954 kN
FLN = 3.95 kN C 
ΣM 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
ΣFy = 0:
FKM = 5.02 kN T 
4.80
1.12
(−3.954 kN) − 1.24 kN − 0.6 kN + 4.48 kN = 0
FLM +
6.147
4
FLM = −1.963 kN
FLM = 1.963 kN C 
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832
PROBLEM 6.57
A Polynesian, or duopitch, roof truss is loaded as
shown. Determine the force in members DF, EF,
and EG.
SOLUTION
Free body: Truss:
ΣFx = 0: N x = 0
ΣM N = 0: (200 lb)(8a ) + (400 lb)(7a + 6a + 5a )+(350 lb)(4a ) + (300 lb)(3a + 2a + a ) − A(8a ) = 0
A = 1500 lb 
ΣFy = 0: 1500 lb − 200 lb − 3(400 lb) − 350 lb − 3(300 lb) − 150 lb + N y = 0
N y = 1300 lb N = 1300 lb 
We pass a section through DF, EF, and EG, and use the free body shown.
(We apply FDF at F.)
ΣM E = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) − (1500 lb)(18 ft)


18
FDF  (4.5 ft) = 0
−
 182 + 4.52



FDF = −3711 lb
ΣM A = 0: FEF (18 ft) − (400 lb)(6 ft) − (400 lb)(12 ft) = 0
FDF = 3710 lb C 
FEF = 400 lb T 
FEF = +400 lb
ΣM F = 0: FEG (4.5 ft) − (1500 lb)(18 ft)+(200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) = 0
FEG = +3600 lb
FEG = 3600 lb T 
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833
PROBLEM 6.58
A Polynesian, or duopitch, roof truss is loaded as
shown. Determine the force in members HI, GI,
and GJ.
SOLUTION
See solution of Problem 6.57 for reactions:
A = 1500 lb ,
N = 1300 lb 
We pass a section through HI, GI, and GJ, and use the free body shown.
(We apply FHI at H.)


6
ΣM G = 0: 
FHI  (8.5 ft) + (1300 lb)(24 ft) − (300 lb)(6 ft)
 2

2
 6 +4

− (300 lb)(12 ft) − (300 lb)(18 ft) − (150 lb)(24 ft) = 0
FHI = −2375.4 lb
FHI = 2375 lb C 
ΣM I = 0: (1300 lb)(18 ft) − (300 lb)(6 ft) − (300 lb)(12 ft)
− (150 lb)(18 ft) − FGJ (4.5 ft) = 0
FGJ = +3400 lb
ΣFx = 0: −
4
6
FGI −
(−2375.4 lb) − 3400 lb = 0
2
5
6 + 42
FGI = −1779.4 lb
FGJ = 3400 lb T 
FGI = 1779 lb C 
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834
PROBLEM 6.59
Determine the force in members DE and DF of the truss shown
when P = 20 kips.
SOLUTION
Reactions:
C = K = 2.5P
7.5
18
β = 22.62°
tan β =
Member DE:
ΣM A = 0: (2.5 P )(6 ft) − FDE (12 ft) = 0
FDE = +1.25 P
For P = 20 kips,
FDE = +1.25(20) = +25 kips
FDE = 25.0 kips T 
Member DF:
ΣM E = 0: P (12 ft) − (2.5 P )(6 ft) − FDF cos β (5 ft) = 0
12 P − 15 P − FDF cos 22.62°(5 ft) = 0
FDF = −0.65 P
For P = 20 kips,
FDF = −0.65(20) = −13 kips
FDF = 13.00 kips C 
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835
PROBLEM 6.60
Determine the force in members EG and EF of the truss shown
when P = 20 kips.
SOLUTION
Reactions:
C = K = 2.5P
7.5
6
α = 51.34°
tan α =
Member EG:
ΣM F = 0: P (18 ft) − 2.5 P (12 ft) − P (6 ft) + FEG (7.5 ft) = 0
FEG = +0.8 P;
For P = 20 kips,
FEG = 16.00 kips T 
FEG = 0.8(20) = +16 kips
Member EF:
ΣM A = 0: 2.5 P (6 ft) − P (12 ft) + FEF sin 51.34°(12 ft) = 0
FEF = −0.320 P;
For P = 20 kips,
FEF = −0.320(20) = −6.4 kips
FEF = 6.40 kips C 
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836
PROBLEM 6.61
Determine the force in members EH and GI of the
truss shown. (Hint: Use section aa.)
SOLUTION
Reactions:
ΣFx = 0: Ax = 0
ΣM P = 0: 12(45) + 12(30) + 12(15) − Ay (90) = 0
A y = 12 kips
ΣFy = 0: 12 − 12 − 12 − 12 + P = 0
P = 24 kips
ΣM G = 0: − (12 kips)(30 ft) − FEH (16 ft) = 0
FEH = −22.5 kips
ΣFx = 0: FGI − 22.5 kips = 0
FEH = 22.5 kips C 
FGI = 22.5 kips T 
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837
PROBLEM 6.62
Determine the force in members HJ and IL of the truss
shown. (Hint: Use section bb.)
SOLUTION
See the solution to Problem 6.61 for free body diagram and analysis to determine the reactions at supports A
and P.
A x = 0; A y = 12.00 kips ;
P = 24.0 kips
ΣM L = 0: FHJ (16 ft) − (12 kips)(15 ft) + (24 kips)(30 ft) = 0
FHJ = −33.75 kips
ΣFx = 0: 33.75 kips − FIL = 0
FIL = +33.75 kips
FHJ = 33.8 kips C 
FIL = 33.8 kips T 
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838
PROBLEM 6.63
Determine the force in members DG and FI of the truss shown. (Hint: Use
section aa.)
SOLUTION
ΣM F = 0: FDG (4 m) − (5 kN)(3 m) = 0
FDG = +3.75 kN
ΣFy = 0: − 3.75 kN − FFI = 0
FFI = −3.75 kN
FDG = 3.75 kN T 
FFI = 3.75 kN C 
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839
PROBLEM 6.64
Determine the force in members GJ and IK of the truss shown. (Hint: Use
section bb.)
SOLUTION
ΣM I = 0: FGJ (4 m) − (5 kN)(6 m) − (5 kN)(3 m) = 0
FGJ = +11.25 kN
ΣFy = 0: − 11.25 kN − FIK = 0
FIK = −11.25 kN
FGJ = 11.25 kN T 
FIK = 11.25 kN C 
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840
PROBLEM 6.65
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:
ΣFx = 0: Fx = 0
ΣM H = 0: 4.8(3a) + 4.8(2a) + 4.8a − 2.4a − Fy (2a) = 0
F = 13.20 kips 
Fy = +13.20 kips
ΣFy = 0: H + 13.20 kips − 3(4.8 kips) − 2(2.4 kips) = 0
H = +6.00 kips
H = 6.00 kips 
Free body: ABF:
We assume that counter BG is acting.
ΣFy = 0: −
9.6
FBG + 13.20 − 2(4.8) = 0
14.6
FBG = +5.475
FBG = 5.48 kips T 
Since BG is in tension, our assumption was correct.
Free body: DEH:
We assume that counter DG is acting.
ΣFy = 0: −
9.6
FDG + 6.00 − 2(2.4) = 0
14.6
FDG = +1.825
FDG = 1.825 kips T 
Since DG is in tension, O.K.
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841
PROBLEM 6.66
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:
ΣFx = 0: Fx = 0
ΣM G = 0: − Fy a + 4.8(2a) + 4.8a − 2.4a − 2.4(2a) = 0
Fy = 7.20
F = 7.20 kips 
Free body: ABF:
We assume that counter CF is acting.
ΣFy = 0:
9.6
FCF + 7.20 − 2(4.8) = 0
14.6
FCF = +3.65
FCF = 3.65 kips T 
Since CF is in tension, O.K.
Free body: DEH:
We assume that counter CH is acting.
ΣFy = 0:
9.6
FCH − 2(2.4 kips) = 0
14.6
FCH = +7.30
FCH = 7.30 kips T 
Since CH is in tension, O.K.
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842
PROBLEM 6.67
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.
ΣFx = 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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843
PROBLEM 6.68
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.
ΣFx = 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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844
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, 2 n = 20
(a)
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. ΣM B = 0 cannot be satisfied.)
Structure is improperly constrained. 
Structure (b)
m = 16
n = 10
r=4
m + r = 20, 2n = 20
(b)
m + r = 2n 
Thus,
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845
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
(c)
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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846
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):
Nonsimple truss with r = 4, m = 16, n = 10,
so m + r = 20 = 2n, but we must examine further.
FBD Sections:
FBD
I:
ΣM A = 0 
T1
II:
ΣFx = 0 
I:
ΣFx = 0

T2
I:
ΣFy = 0

Ax
ΣM E = 0 
Ay
ΣFy = 0 
Cy
II:
II:
Ey
Since each section is a simple truss with reactions determined,
structure is completely constrained and determinate. 
Structure (b):
Nonsimple truss with r = 3, m = 16, n = 10,
so
Structure is partially constrained. 
m + r = 19 < 2n = 20
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847
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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848
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):
Nonsimple 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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849
PROBLEM 6.71 (Continued)
Structure (c):
Nonsimple 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 ≠ 0 (from solution of joint F),
ΣM A = aF1
≠ 0 and there is no equilibrium.
Structure is improperly constrained. 
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850
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,
2 n = 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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851
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 this),
therefore it is rigid, and that its supports involve four unknowns.
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852
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
Structure (b):
r + m = 17 > 2n = 16
so completely constrained but indeterminate 
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. 
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853
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):
m = 12
No. of members
No. of joints
n = 8 m + r = 16 = 2n
No. of reaction components
r = 4 unknows = equations
FBD of EH:
ΣM H = 0
FDE ; ΣFx = 0
FGH ; ΣFy = 0
Hy
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 reaction components
r = 3 unknows < equations
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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854
PROBLEM 6.74 (Continued)
Structure (c):
No. of members
m = 13
No. of joints
n = 8 m + r = 17 > 2n = 16
No. of reaction components
r = 4 unknows > equations
completely constrained but indeterminate 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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855
PROBLEM 6.75
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 line BD.
Free body: ABC:
BD = (24) 2 + (10) 2 = 26 in.
 10

ΣM C = 0: (160 lb)(30 in.) − 
FBD  (16 in.) = 0
 26

FBD = +780 lb
ΣM x = 0: C x +
FBD = 780 lb T 
24
(780 lb) = 0
26
C x = −720 lb
ΣFy = 0: C y − 160 lb +
10
(780 lb) = 0
26
C y = −140.0 lb
C x = 720 lb

C y = 140.0 lb 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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856
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 line BD.
Free body: ABC:
Attaching FBD at D and resolving it into components, we write
ΣM C = 0:
 450

FBD  (240 mm) = 0
(400 N)(135 mm) + 
 510

FBD = −255 N
ΣFx = 0: C x +
FBD = 255 N C 
240
(−255 N) = 0
510
C x = +120.0 N
ΣFy = 0: C y − 400 N +
450
(−255 N) = 0
510
C y = +625 N
C x = 120.0 N

C y = 625 N 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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857
PROBLEM 6.77
Determine the components of all forces acting on member ABCD of the
assembly shown.
SOLUTION
Free body: Entire assembly:
ΣM B = 0: D (120 mm) − (480 N)(80 mm) = 0
D = 320 N 
ΣFx = 0: Bx + 480 N = 0
B x = 480 N
ΣFy = 0: By + 320 N = 0

B y = 320 N 
Free body: Member ABCD:
ΣM A = 0: (320 N)(200 mm) − C (160 mm) − (320 N)(80 mm)
− (480 N)(40 mm) = 0
C = 120.0 N 
ΣFx = 0: Ax − 480 N = 0
A x = 480 N
ΣFy = 0: Ay − 320 N − 120 N + 320 N = 0

A y = 120.0 N 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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858
PROBLEM 6.78
Determine the components of all forces acting on member ABD
of the frame shown.
SOLUTION
Free body: Entire frame:
ΣM D = 0: (300 lb) − (12 ft) + (450 lb)(4 ft) − E (6 ft) = 0
E = +900 lb
E = 900 lb

D x = 900 lb

ΣFx = 0: Dx − 900 lb = 0
ΣFy = 0: Dy − 300 lb − 450 lb = 0
D y = 750 lb 
Free body: Member ABD:
We note that BC is a two-force member and that B is directed along BC.
ΣM A = 0: (750 lb)(16 ft) − (900 lb)(6 ft) − B (8 ft) = 0
B = +825 lb
B = 825 lb 
ΣFx = 0: Ax + 900 lb = 0
Ax = −900 lb
ΣFy = 0: Ay + 750 lb − 825 lb = 0
Ay = +75 lb
A x = 900 lb

A y = 75.0 lb 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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859
PROBLEM 6.79
For the frame and loading shown, determine the components of all
forces acting on member ABC.
SOLUTION
Free body: Entire frame:
ΣM E = 0: − Ax (4) − (20 kips)(5) = 0
Ax = −25 kips,
A x = 25.0 kips

ΣFy = 0: Ay − 20 kips = 0
A y = 20.0 kips 
Ay = 20 kips
Free body: Member ABC:
Note: BE is a two-force member, thus B is directed along line BE and By =
2
Bx .
5
ΣM C = 0: (25 kips)(4 ft) − (20 kips)(10 ft) + Bx (2 ft) + By (5 ft) = 0
−100 kip ⋅ ft + Bx (2 ft) +
2
Bx (5 ft) = 0
5
Bx = 25 kips
By =
B x = 25.0 kips
2
2
( Bx ) = (25) = 10 kips
5
5
B y = 10.00 kips 
ΣFx = 0: C x − 25 kips − 25 kips = 0
C x = 50 kips
ΣFy = 0: C y + 20 kips − 10 kips = 0
C y = −10 kips

C x = 50.0 kips

C y = 10.00 kips 
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860
PROBLEM 6.80
Solve Problem 6.79 assuming that the 20-kip load is replaced by a
clockwise couple of magnitude 100 kip ⋅ ft applied to member EDC
at Point D.
PROBLEM 6.79 For the frame and loading shown, determine the
components of all forces acting on member ABC.
SOLUTION
Free body: Entire frame:
ΣFy = 0: Ay = 0
ΣM E = 0: − Ax (4 ft) − 100 kip ⋅ ft = 0
Ax = −25 kips
A x = 25.0 kips
A = 25.0 kips

B x = 25.0 kips

Free body: Member ABC:
Note: BE is a two-force member, thus B is directed along line BE and By =
2
Bx .
5
ΣM C = 0: (25 kips)(4 ft) + Bx (2 ft) + By (5 ft) = 0
100 kip ⋅ ft + Bx (2 ft) +
2
Bx (5 ft) = 0
5
Bx = −25 kips
By =
B y = 10.00 kips 
2
2
Bx = ( −25) = −10 kips;
5
5
ΣFx = 0: − 25 kips + 25 kips + C x = 0
Cx = 0
ΣFy = 0: +10 kips + C y = 0
C y = −10 kips
C y = 10 kips
C = 10.00 kips 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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861
PROBLEM 6.81
Determine the components of all forces acting on member
ABCD when θ = 0.
SOLUTION
Free body: Entire assembly:
ΣM B = 0: A(200) − (150 N)(500) = 0
A = +375 N
A = 375 N

B x = 375 N

ΣFx = 0: Bx + 375 N = 0
Bx = −375 N
ΣFy = 0: By − 150 N = 0
By = +150 N
B y = 150 N 
Free body: Member ABCD:
We note that D is directed along DE, since DE is a two-force member.
ΣM C = 0: D (300) − (150 N)(100) + (375 N)(200) = 0
D = −200 N
D = 200 N 
ΣFx = 0: C x + 375 − 375 = 0
Cx = 0
ΣFy = 0: C y + 150 − 200 = 0
C y = +50.0 N
C = 50.0 N 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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862
PROBLEM 6.82
Determine the components of all forces acting on member
ABCD when θ = 90°.
SOLUTION
Free body: Entire assembly:
ΣM B = 0: A(200) − (150 N)(200) = 0
A = +150.0 N
A = 150.0 N

ΣFx = 0: Bx + 150 − 150 = 0
Bx = 0
B = 0 
ΣFy = 0: By = 0
Free body: Member ABCD:
We note that D is directed along DE, since DE is a two-force member.
ΣM C = 0: D (300) + (150 N)(200) = 0
D = −100.0 N
D = 100.0 N 
ΣFx = 0: C x + 150 N = 0
C x = −150 N
ΣFy = 0: C y − 100 N = 0
C y = +100.0 N
Cx = 150.0 N

C y = 100.0 N 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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863
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 parts (a) and (b) since position of load on its line of action is immaterial.
ΣM E = 0: − (750 N)(80 mm) − Ax (200 mm) = 0
Ax = − 300 N
A x = 300 N
ΣFx = 0: E x − 300 N = 0 E x = 300 N E x = 300 N
ΣFy = 0: Ay + E y − 750 N = 0
(a)
(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
E y = 90 N
Ay + 90 N − 750 N = 0
Ay = 660 N
Thus, reactions are
(b)
A x = 300 N
,
E x = 300 N
,
A y = 660 N 
E y = 90.0 N 
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
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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864
PROBLEM 6.83 (Continued)
Ay + E y − 750 N = 0
From Eq. (1):
150 N + E y − 750 N = 0
E y = 600 N E y = 600 N
Thus, reactions are
A x = 300 N
,
E x = 300 N
,
A y = 150.0 N 
E y = 600 N 
PROPRIETARY MATERIAL. © 2013 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
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865
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 parts (a) and (b) since position of load on its line of action is immaterial.
ΣM E = 0: − (750 N)(240 mm) − Ax (400 mm) = 0
Ax = −450 N
A x = 450 N
ΣFx = 0: E x − 450 N = 0 E x = 450 N E x = 450 N
ΣFy = 0: Ay + E y − 750 N = 0
(a)
(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
450 N
From Eq. (1):
=
240 mm
; E y = 225 N
480 mm
Ay + 225 − 750 = 0; A y = 525 N
Thus, reactions are
A x = 450 N
,
,
E y = 225 N 
A x = 450 N
,
A y = 150.0 N 
E x = 450 N
,
E x = 450N
(b)
A y = 525 N 
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
450 N
From Eq. (1):
=
160 mm
480 mm
A y = 150.0 N
Ay + E y − 750 N = 0
150 N + E y − 750 N = 0
E y = 600 N E y = 600 N
Thus, reactions are
E y = 600 N 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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866
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 parts (a) and (b) since the point of application of the couple is
immaterial.
ΣM E = 0: −36 N ⋅ m − Ax (0.2 m) = 0
Ax = − 180 N
A x = 180 N
ΣFx = 0: − 180 N + E x = 0
E x = 180 N E x = 180 N
ΣFy = 0: Ay + E y = 0
(a)
(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
Thus, reactions are
(b)
A x = 180.0 N
,
E x = 180.0 N
,
A y = 54.0 N 
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
180 N
=
0.125 m
0.25 m
Ay = 90 N
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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867
PROBLEM 6.85 (Continued)
From Eq. (1):
Ay + E y = 0
90 N + E y = 0
E y = − 90 N E y = 90 N
Thus, reactions are
A x = 180.0 N
E x = −180.0 N
A y = 90.0 N 
,
,
E y = 90.0 N 
PROPRIETARY MATERIAL. © 2013 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
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868
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 parts (a) and (b) since the point of application of the couple is
immaterial.
ΣM E = 0: −36 N ⋅ m − Ax (0.4 m) = 0
Ax = − 90 N
A x = 90.0 N
ΣFx = 0: −90 +E x = 0
E x = 90 N
E x = 90.0 N
ΣFy = 0: Ay + E y = 0
(a)
(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
Thus, reactions are
(b)
A x = 90.0 N
,
E x = 90.0 N
,
A y = 45.0 N 
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
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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869
PROBLEM 6.86 (Continued)
From Eq. (1):
Ay + E y = 0
30 N + E y = 0
E y = − 30 N E y = 30 N
A x = 90.0 N
Thus, reactions are
E x = − 90.0 N
A y = 30.0 N 
,
,
E y = 30.0 N 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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870
PROBLEM 6.87
Determine all the forces exerted on member AI if the frame is loaded
by a clockwise couple of magnitude 1200 lb · in. applied (a) at Point D,
(b) at Point E.
SOLUTION
Free body: Entire frame:
Location of couple is immaterial.
ΣM H = 0: I (48 in.) − 1200 lb ⋅ in. = 0
I = +25.0 lb
(a) and (b)
I = 25.0 lb 
We note that AB, BC, and FG are two-force members.
Free body: Member AI:
tan α =
(a)
20 5
=
48 12
α = 22.6°
Couple applied at D.
ΣFy = 0: −
ΣM G = 0:
5
A + 25 lb = 0
13
A = +65.0 lb
A = 65.0 lb
12
(65 lb)(40 in.) − C (20 in.) = 0
13
C = +120 lb
ΣFx = 0: −
12
(65 lb) + 120 lb + G = 0
13
G = −60.0 lb
22.6° 
C = 120 lb
G = 60 lb


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871
PROBLEM 6.87 (Continued)
(b)
Couple applied at E.
ΣFy = 0: −
ΣM G = 0:
5
A + 25 lb = 0
13
A = +65.0 lb
A = 65.0 lb
22.6° 
12
(65 lb) + (40 in.) − C (20 in.) − 1200 lb ⋅ in. = 0
13
C = +60.0 lb
C = 60.0 lb
ΣFx = 0: −
12
(65 lb) + 60 lb + G = 0
13

G=0 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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872
PROBLEM 6.88
Determine all the forces exerted on member AI if the frame is
loaded by a 40-lb force directed horizontally to the right and applied
(a) at Point D, (b) at Point E.
SOLUTION
Free body: Entire frame:
Location of 40-lb force on its line of action DE is immaterial.
ΣM H = 0: I (48 in.) − (40 lb)(30 in.) = 0
I = +25.0 lb
(a) and (b)
I = 25.0 lb 
We note that AB, BC, and FG are two-force members.
Free body: Member AI:
tan α =
(a)
20 5
=
48 12
α = 22.6°
Force applied at D.
ΣFy = 0: −
ΣM G = 0:
5
A + 25 lb = 0
13
A = +65.0 lb
12
(65 lb)(40 in.) − C (20 in.) = 0
13
C = +120.0 lb
ΣFx = 0: −
12
(65 lb) + 120 lb + G = 0
13
G = −60.0 lb
A = 65.0 lb
22.6° 
C = 120.0 lb

G = 60.0 lb

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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873
PROBLEM 6.88 (Continued)
(b)
Force applied at E.
ΣFy = 0: −
ΣM G = 0:
5
A + 25 lb = 0
13
A = +65.0 lb
A = 65.0 lb
22.6° 
12
(65 lb)(40 in.) − C (20 in.) − (40 lb)(10 in.) = 0
13
C = +100.0 lb
C = 100.0 lb
ΣFx = 0: −
12
(65 lb) + 100 lb + 40 lb + G = 0
13
G = −80.0 lb
G = 80.0 lb


PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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874
PROBLEM 6.89
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 parts (a) or (b), since position of 100-lb load on its line of action is immaterial.
ΣM A = 0: By (10) − (100 lb)(6) = 0 By = + 60 lb
ΣFy = 0: Ay + 60 − 100 = 0
Ay = + 40 lb
ΣFx = 0: Ax + Bx = 0
(a)
(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
40 lb
=
10 in. 5 in.
From Eq. (1):
A x = 80.0 lb
,
A y = 40.0 lb 
B x = 80.0 lb
,
B y = 60.0 lb 
− 80 + Bx = 0 Bx = + 80 lb
Thus,
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875
PROBLEM 6.89 (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 y = 60.0 lb 
Bx = 0
Ax + 0 = 0
Ax = 0
Ax = 0
A y = 40.0 lb 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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876
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.
ΣM C = 0: rT1 − rT2 = 0 T1 = T2
(a)
Replace each force with an equivalent force-couple.
(b)
Cut the cable and replace the forces on pulley with equivalent pair of forces at A as above.

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877
PROBLEM 6.91
A 3-ft-diameter pipe is supported every 16 ft by a small frame like
that shown. Knowing that the combined weight of the pipe and its
contents is 500 lb/ft and assuming frictionless surfaces, determine the
components (a) of the reaction at E, (b) of the force exerted at C on
member CDE.
SOLUTION
Free body: 16-ft length of pipe:
W = (500 lb/ft)(16 ft) = 8 kips
Force Triangle
B D 8 kips
= =
3 5
4
B = 6 kips
D = 10 kips
Determination of CB = CD.
 

We note that horizontal projection of BO + OD = horizontal projection of CD
Thus,
3
4
r + r = (CD )
5
5
8
CB = CD = r = 2(1.5 ft) = 3 ft
4
Free body: Member ABC:
ΣM A = 0: Cx h − (6 kips)(h − 3)
Cx =
h−3
(6 kips)
h
Cx =
9−3
(6 kips) = 4 kips
9
(1)
For h = 9 ft,
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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878
PROBLEM 6.91 (Continued)
Free body: Member CDE:
From above, we have
Cx = 4.00 kips

ΣM E = 0: (10 kips)(7 ft) − (4 kips)(6 ft) − C y (8 ft) = 0
C y = +5.75 kips,
3
ΣFx = 0: − 4 kips + (10 kips) + Ex = 0
5
Ex = −2 kips,
4
ΣFy = 0: 5.75 kips − (10 kips) + E y = 0,
5
C y = 5.75 kips 
E x = 2.00 kips

E y = 2.25 kips 
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879
PROBLEM 6.92
Solve Problem 6.91 for a frame where h = 6 ft.
PROBLEM 6.91 A 3-ft-diameter pipe is supported every 16 ft by a
small frame like that shown. Knowing that the combined weight of the
pipe and its contents is 500 lb/ft and assuming frictionless surfaces,
determine the components (a) of the reaction at E, (b) of the force
exerted at C on member CDE.
SOLUTION
See solution of Problem 6.91 for derivation of Eq. (1).
For h = 6 ft, C x =
h−3
6−3
= 3 kips
(6 kips) =
h
6
Free body: Member CDE:
From above, we have
Cx = 3.00 kips

ΣM E = 0: (10 kips)(7 ft) − (3 kips)(6 ft) − C y (8 ft) = 0
C y = +6.50 kips,
3
ΣFx = 0: −3 kips + (10 kips) + Ex = 0
5
Ex = −3.00 kips,
4
ΣFy = 0: 6.5 kips − (10 kips) + E y = 0
5
E y = 1.500 kips
C y = 6.50 kips 
E x = 3.00 kips

E y = 1.500 kips 
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880
PROBLEM 6.93
Knowing that the pulley has a radius of 0.5 m, determine the
components of the reactions at A and E.
SOLUTION
FBD Frame:
ΣM A = 0: (7 m) E y − (4.5 m)(700 N) = 0
ΣFy = 0: Ay − 700 N + 450 N = 0
ΣFx = 0: Ax − E x = 0
E y = 450 N 
A y = 250 N 
Ax = E x
Dimensions in m
FBD Member ABC:
ΣM 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



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881
PROBLEM 6.94
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:
ΣM E = 0 : − (300 N)(350 mm) − Bx (150 mm) = 0
Bx = −700 N
B x = 700 N
ΣFx = 0: − 700 N + E x = 0
E x = 700 N
E x = 700 N
ΣFy = 0: By + E y − 300 N = 0
(1)
Free body: Member ACE:
ΣM 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
Thus, reactions are
B x = 700 N
,
E x = 700 N
,
B y = 200 N 
E y = 500 N 
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882
PROBLEM 6.95
A trailer weighing 2400 lb is attached to a 2900-lb 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.)
ΣM A = 0: − (2400 lb)(2 ft) + D (11 ft) = 0
D = 436.36 lb
ΣFy = 0: 2 A − 2400 lb + 436.36 lb = 0
A = 982 lb 
A = 981.82 lb
Free body: Truck.
ΣM B = 0: (436.36 lb)(3 ft) − (2900 lb)(5 ft) + 2C (9 ft) = 0
C = 732.83 lb
C = 733 lb 
ΣFy = 0: 2B − 436.36 lb − 2900 lb + 2(732.83 lb) = 0
B = 935.35 lb
(b)
B = 935 lb 
Additional load on truck wheels.
Use free body diagram of truck without 2900 lb.
ΣM B = 0: (436.36 lb)(3 ft) + 2C (9 ft) = 0
C = −72.73 lb
ΣFy = 0: 2B − 436.36 lb − 2(72.73 lb) = 0
B = 290.9 lb
ΔC = −72.7 lb 
ΔB = +291 lb 
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883
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 2400 lb is attached
to a 2900-lb 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 small first find the additional reaction Δ at each wheel due to the trailer.
Free body diagram: (Same Δ at each truck wheel)
ΣM A = 0: − (2400 lb)(2 ft) + 2 Δ (14 ft) + 2 Δ (23 ft) = 0
Δ = 64.86 lb
ΣFy = 0: 2 A − 2400 lb + 4(64.86 lb) = 0;
A = 1070 lb;
A = 1070 lb
Free body: Truck:
(Trailer loading only)
ΣM D = 0: 2 Δ (12 ft) + 2 Δ (3 ft) − 2T (1.7 ft) = 0
T = 8.824Δ
= 8.824(64.86 lb)
T = 572.3 lb
T = 572 lb 
Free body: Truck:
(Truck weight only)
ΣM B = 0: − (2900 lb)(5 ft) + 2C ′(9 ft) = 0
C ′ = 805.6 lb
C′ = 805.6 lb
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884
PROBLEM 6.96 (Continued)
ΣFy = 0: 2B′ − 2900 lb + 2(805.6 lb) = 0
B′ = 644.4 lb
Actual reactions:
B = B′ + Δ = 644.4 lb + 64.86 = 709.2 lb
C = C ′ + Δ = 805.6 lb + 64.86 = 870.46 lb
From part a:
B′ = 644.4 lb
B = 709 lb 
C = 870 lb 
A = 1070 lb 
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885
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
ΣM 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 
ΣFy = 0: 2 A + 325 − 75 − 100 − 300 = 0
2 A = 150 kN
A = 75.0 kN 
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886
PROBLEM 6.97 (Continued)
(b)
Free body: Motor unit:
ΣM D = 0: C (1 m) + 2 B (2.8 m) − 300(3.6 m) = 0
C = 1080 − 5.6 B
Recalling
(1)
B = 162.5 kN, C = 1080 − 5.6(162.5) = 170 kN
C = 170.0 kN



ΣFx = 0: Dx − 170 = 0
D x = 170.0 kN
ΣFy = 0: 2(162.5) − Dy − 300 = 0


D y = 25.0 kN 
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887
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
ΣM A = 0: 2 B (4.8 m) − 100(1.2 m) − 300(5.6 m) = 0
2 B = 375 kN
B = 187.5 kN 
ΣFy = 0: 2 A + 375 − 100 − 300 = 0
2 A = 25 kN
(b)
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
ΣFx = 0: Dx − 30 = 0
D x = 30.0 kN
ΣFy = 0: 2(187.5) − Dy − 300 = 0


D y = 75.0 kN 
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888
PROBLEM 6.99
For the frame and loading shown, determine the components of all forces
acting on member ABE.
SOLUTION
FBD Frame:
ΣM E = 0: (1.8 m) Fy − (2.1 m)(12 kN) = 0
Fy = 14.00 kN
ΣFy = 0: − E y + 14.00 kN − 12 kN = 0
E y = 2 kN
E y = 2.00 kN 
FBD member BCD:
ΣM B = 0: (1.2 m)C y − (12 kN)(1.8 m) = 0 C y = 18.00 kN
But C is ⊥ ACF, so Cx = 2C y ; Cx = 36.0 kN
ΣFx = 0: − Bx + C x = 0
Bx = C x = 36.0 kN
Bx = 36.0 kN
on BCD
ΣFy = 0: − By + 18.00 kN − 12 kN = 0 By = 6.00 kN on BCD
On ABE:
B x = 36.0 kN

E x = 23.0 kN

B y = 6.00 kN 
FBD member ABE:
ΣM 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
ΣFx = 0: − 23.0 kN + 36.0 kN − Ax = 0
ΣFy = 0: − 2.00 kN + 6.00 kN − Ay = 0
A x = 13.00 kN

A y = 4.00 kN 
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889
PROBLEM 6.100
For the frame and loading shown, determine the components of all
forces acting on member ABE.
PROBLEM 6.99 For the frame and loading shown, determine the
components of all forces acting on member ABE.
SOLUTION
FBD Frame:
ΣM 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
ΣM C = 0: (2.4 m) By − (1.2 m)(2400 N) = 0 By = 1200 N
B y = 1200 N 
On ABE:
ΣFy = 0: −1200 N + C y − 2400 N = 0 C y = 3600 N
9
Cx = C y
8
so
ΣFx = 0: − Bx + C x = 0
Bx = 4050 N
C x = 4050 N
on BC
B x = 4050 N
On ABE:

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890
PROBLEM 6.100 (Continued)
FBD member AB0E:
ΣM A = 0: a (4050 N) − 2 aE x = 0
E x = 2025 N
ΣFx = 0 : − Ax + (4050 − 2025) N = 0
ΣFy = 0: 600 N + 1200 N − Ay = 0
E x = 2025 N
A x = 2025 N


A y = 1800 N 
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891
PROBLEM 6.101
For the frame and loading shown, determine the components of all forces
acting on member ABD.
SOLUTION
Free body: Entire frame:
Σ M A = 0: E (12 in.) − (360 lb)(15 in.) − (240 lb)(33 in.) = 0
E = +1110 lb
E = +1110 lb
ΣFx = 0: Ax + 1110 lb = 0
Ax = −1110 lb
A x = 1110 lb
ΣFy = 0: Ay − 360 lb − 240 lb = 0

A y = 600 lb 
Ay = +600 lb
Free body: Member CDE:
Σ M C = 0: (1110 lb)(24 in.) − Dx (12 in.) = 0
Dx = +2220 lb
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892
PROBLEM 6.101 (Continued)
Free body: Member ABD:
D x = 2220 lb
From above:
ΣM B = 0: Dy (18 in.) − (600 lb)(6 in.) = 0
Dy = +200 lb
D y = 200 lb 
ΣFx = 0: Bx + 2220 lb − 1110 lb = 0
Bx = −1110 lb
ΣFy = 0: By + 200 lb + 600 lb = 0
By = −800 lb

B x = 1110 lb

B y = 800 lb 
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893
PROBLEM 6.102
Solve Problem 6.101 assuming that the 360-lb load has been removed.
PROBLEM 6.101 For the frame and loading shown, determine the
components of all forces acting on member ABD.
SOLUTION
Free body diagram of entire frame.
ΣM A = 0: E (12 in.) − (240 lb)(33 in.) = 0
E = +660 lb
E = 660 lb
ΣFx = 0: Ax + 660 lb = 0
Ax = −660 lb
A x = 660 lb
ΣFy = 0: Ay − 240 lb = 0

A y = 240 lb 
Ay = +240 lb
Free body: Member CDE:
ΣM C = 0: (660 lb)(24 in.) − Dx (12 in.) = 0
Dx = +1320 lb
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894
PROBLEM 6.102 (Continued)
Free body: Member ABD:
D x = 1320 lb
From above:
ΣM B = 0: Dy (18 in.) − (240 lb)(6 in.) = 0
Dy = +80 lb
D y = 80.0 lb 
ΣFx = 0: Bx + 1320 lb − 660 lb = 0
Bx = −660 lb
ΣFy = 0: By + 80 lb + 240 lb = 0
By = −320 lb

B x = 660 lb

B y = 320 lb 
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895
PROBLEM 6.103
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:
ΣM y = 0: Ay − 25 lb = 0
A y = 25 lb 
Ay = 25 lb
ΣM F = 0: Ax (6.928 + 2 × 3.464) − (25 lb)(12 in.) = 0
Ax = 21.651 lb
A y = 21.65 lb
ΣFx = 0: F − 21.651 lb = 0
F = 21.651 lb
F = 21.65 lb
Free body: Member CDE:
ΣM C = 0: Dy (4 in.) − (25 lb)(10 in.) = 0
Dy = +62.5 lb
ΣFy = 0: −C y + 62.5 lb − 25 lb = 0
C y = +37.5 lb
D y = 62.5 lb 
C y = 37.5 lb 
Free body: Member ABD:
ΣM B = 0: Dx (3.464 in.) + (21.65 lb)(6.928 in.)
−(25 lb)(4 in.) − (62.5 lb)(2 in.) = 0
Dx = +21.65 lb
Return to free body: Member CDE:
From above:
Dx = +21.65 lb
D x = 21.7 lb

C x = 21.7 lb

ΣFx = 0: C x − 21.65 lb
C x = +21.65 lb
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896
PROBLEM 6.104
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:
ΣM D = 0: (40 lb)(13 in.) + Ax (10 in.) = 0
Ax = −52 lb,
A x = 52 lb

Free body: Member ABF:
ΣM B = 0: − (52 lb)(6 in.) + Fx (4 in.) = 0
Fx = +78 lb
Free body: Member CFE:
Fx = 78.0 lb
From above:
ΣM C = 0: (40 lb)(9 in.) − (78 lb)(4 in.) − Fy (4 in.) = 0
Fy = +12 lb
Fy = 12.00 lb 
ΣFx = 0: C x − 78 lb = 0
C x = +78 lb
ΣFy = 0: − 40 lb + 12 lb + C y = 0; C y = +28 lb

C x = 78.0 lb

C y = 28.0 lb 
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897
PROBLEM 6.105
For the frame and loading shown, determine the components of all
forces acting on member ABD.
SOLUTION
Free body: Entire frame:
Dimensions in mm
ΣM A = 0: F (500) − (3 kN)(600) − (2 kN)(1000) = 0
F = +7.60 kN
F = 7.60 kN
ΣFx = 0: Ax + 7.60 N = 0,
Ax = −7.60 N
ΣFy = 0: Ay − 3 kN − 2 kN = 0
Ay = +5 kN
A x = 7.60 kN

A y = 5.00 kN 
Free body: Member CDE:
ΣM C = 0: D y (400) − (3 kN)(200) − (2 kN)(600) = 0
Dy = +4.50 kN
Free body: Member ABD:
D y = 4.50 kN 
From above:
ΣM B = 0: Dx (200) − (4.50 kN)(400) − (5 kN)(400) = 0
Dx = +19 kN
D x = 19.00 kN

B x = 11.40 kN

ΣFx = 0: Bx + 19 kN − 7.60 kN = 0
Bx = −11.40 kN
ΣFy = 0: By + 5 kN − 4.50 kN = 0
By = −0.50 kN
B y = 0.500 kN 
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898
PROBLEM 6.106
Solve Problem 6.105 assuming that the 3-kN load has been removed.
PROBLEM 6.105 For the frame and loading shown, determine the
components of all forces acting on member ABD.
SOLUTION
Free body: Entire frame:
Dimensions in mm
ΣM A = 0: F (500) − (2 kN)(1000) = 0
F = +4 kN
F = 4.00 kN
ΣFx = 0: Ax + 4 kN = 0,
Ax = − 4 kN
A x = 4.00 kN
ΣFy = 0: Ay − 2 kN = 0,

A y = 2.00 kN 
Ay = +2 kN
Free body: Member CDE:
ΣM C = 0: D y (400) − (2 kN)(600) = 0
Dy = +3.00 kN
Free body: Member ABD:
D y = 3.00 kN 
From above:
ΣM B = 0: Dx (200) − (3 kN)(400) − (2 kN)(400) = 0
Dx = +10.00 kN
D x = 10.00 kN
ΣFx = 0: Bx + 10 kN − 4 kN = 0
Bx = −6 kN
ΣFy = 0: By + 2 kN − 3 kN = 0
By = +1 kN
B x = 6.00 kN


B y = 1.000 kN 
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899
PROBLEM 6.107
Determine the reaction at F and the force in members AE and BD.
SOLUTION
Free body: Entire frame:
ΣM C = 0: Fy (9 in.) − (450 lb)(24 in.) = 0
Fy = 1200 lb
Fy = 1200 lb
Free body: Member DEF:
ΣM J = 0: (1200 lb)(4.5 in.) − Fx (18 in.) = 0
Fx = 300 lb
3

ΣM D = 0: − Fx (24 in.) −  FAE  (12 in.) = 0
5

10
10
FAE = − Fx = − (300 lb)
3
3
FAE = −1000 lb
ΣFy = 0: 1200 lb +
FBD =
4
4
(−1000 lb) − FBD = 0
5
5
5
(1200 lb) − 1000 lb = +500 lb
4
Fx = 300 lb

FAE = 1000 lb C 
FBD = 500 lb T 
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900
PROBLEM 6.108
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:
ΣFx = 0:
A − B + (1000 lb)sin 30° = 0
A − B + 500 = 0
(1)
ΣFy = 0: D + E − (1000 lb) cos 30° = 0
D + E − 866.03 = 0
(2)
Free body: Member ACE:
ΣM C = 0: − A(6 in.) + E (8 in.) = 0
E=
3
A
4
(3)
Free body: Member BCD:
ΣM C = 0: − D(8 in.) + B (6 in.) = 0
D=
3
B
4
(4)
Substitute E and D from Eqs. (3) and (4) into Eq. (2):
−
3
3
A + B − 866.06 = 0
4
4
A + B − 1154.71 = 0
(5)
From Eq. (1): A − B + 500 = 0
(6)
Eqs. (5) + (6): 2 A − 654.71 = 0
A = 327.4 lb
A = 327 lb
Eqs. (5) − (6): 2 B − 1654.71 = 0
B = 827.4 lb
B = 827 lb


From Eq. (4): D =
3
(827.4)
4
D = 620.5 lb
D = 621 lb 
From Eq. (3): E =
3
(327.4)
4
E = 245.5 lb
E = 246 lb 
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901
PROBLEM 6.109
The axis of the three-hinge arch ABC is a parabola
with the 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:
ΣM A = 0: Bx (3.2 m) − B y (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:
ΣM C = 0: Bx (1.8 m) + B y (6 m) − Q(3 m) = 0
Add Eqs. (2) and (3):
(3)
4.2 Bx − 3.75P − 3Q = 0
Bx = (3.75 P + 3Q)/4.2
From Eq. (1):
(3.75P + 3Q)
(4)
3.2
− 8B y − 5 P = 0
4.2
B y = (− 9P + 9.6Q )/33.6
(5)
given that P = 112 kN and Q = 140 kN.
(a)
Reaction at A.
Considering again AB as a free body,
ΣFx = 0: Ax − Bx = 0;
Ax = Bx = 200 kN
A x = 200 kN

ΣFy = 0: Ay − P − B y = 0
Ay − 112 kN − 10 kN = 0
Ay = + 122 kN
A y = 122.0 kN 
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902
PROBLEM 6.109 (Continued)
(b)
Force exerted at B on AB.
From Eq. (4):
Bx = (3.75 × 112 + 3 × 140)/4.2 = 200 kN
B x = 200 kN
From Eq. (5):
B y = (− 9 × 112 + 9.6 × 140)/33.6 = + 10 kN

B y = 10.00 kN 
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903
PROBLEM 6.110
The axis of the three-hinge arch ABC is a parabola
with the 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:
ΣM A = 0: Bx (3.2 m) − B y (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:
ΣM C = 0: Bx (1.8 m) + B y (6 m) − Q(3 m) = 0
Add Eqs. (2) and (3):
(3)
4.2 Bx − 3.75P − 3Q = 0
Bx = (3.75 P + 3Q)/4.2
(3.75P + 3Q)
From Eq. (1):
(4)
3.2
− 8B y − 5 P = 0
4.2
B y = (− 9P + 9.6Q )/33.6
(5)
given that P = 140 kN and Q = 112 kN.
(a)
Reaction at A.
ΣFx = 0: Ax − Bx = 0;
Ax = Bx = 205 kN
A x = 205 kN

ΣFy = 0: Ay − P − B y = 0
Ay − 140 kN − (− 5.5 kN) = 0
A y = 134.5 kN 
Ay = 134.5 kN
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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904
PROBLEM 6.110 (Continued)
(b)
Force exerted at B on AB.
From Eq. (4):
Bx = (3.75 × 140 + 3 × 112)/4.2 = 205 kN
B x = 205 kN
From Eq. (5):

B y = (− 9 × 140 + 9.6 × 112)/33.6 = − 5.5 kN
B y = 5.50 kN 
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905
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
FBD I:
FBD II:
FBDs combined:
ΣM B = 0:
II
1
aC y − a
1
M D = 0: aC y − a
ΣM G = 0:
aP − a
FAF = 0 FAF = 2C y
2
2
1
2
FEH = 0 FEH = 2C y
1
FAF − a
2
FEH = 0 P =
Cy =
P
2
1
2
2C y +
1
2C y
2
so FAF =
2
P C 
2
FEH =
2
P T 
2

FBD I:
FBD II:
ΣFy = 0:
1
2
FAF +
ΣFy = 0: − C y +
1
2
1
2
FBG − P + C y = 0
FDG −
1
2
FEH = 0 −
1
P
P
+
FBG − P + = 0
2
2
2
FBG = 0 
1
P
P
+
FDG − = 0
2
2
2
FDG = 2 P C 
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906
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
II
FBD I:
ΣM I = 0: 2aC y + aC x − aP = 0 2C y + C x = P
FBD II:
ΣM J = 0: 2aC y − aC x = 0 2C y − C x = 0
Solving,
FBD I:
Cx =
P
P
; C y = as shown.
2
4
ΣFx = 0: −
1
2
FBG + C x = 0 FBG = C x 2
ΣFy = 0: FAF − P +
FBD II:
ΣFx = 0: − Cx +
ΣFy = 0: − C y +
1
2
1  2  P
P + = 0

2  2  4
FDG = 0 FDG = Cx 2
1  2 
P P
P
P  + FEH = 0 FEH = − = −


4 2
4
2 2 
FBG =
P
2
P
4
FAF =
FDG =
P
FEH =
2
C 
C 
C 
P
T 
4
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907
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
From FBD I:
ΣM J = 0:
a
a
3a
Cx + C y − P = 0 Cx + 3C y = P
2
2
2
FBD II:
ΣM K = 0:
a
3a
C x − C y = 0 Cx − 3C y = 0
2
2
Solving,
Cx =
FBD I:
P
P
; C y = as drawn.
2
6
ΣM B = 0: aC y − a
ΣFx = 0: −
1
2
FAG = 0 FAG = 2C y =
P
6
2
FAG =
1
1
2
2
FAG +
FBF − Cx = 0 FBF = FAG + Cx 2 =
P+
P
6
2
2
2
FBF =
FBD II:
ΣM D = 0: a
2
P C 
6
1
2
FEH + aC y = 0 FEH = − 2C y = −
P
6
2
ΣFx = 0: C x −
2 2
P C 
3
FEH =
2
P T 
6
1
1
2
2
FDI +
FEH = 0 FDI = FEH + C x 2 = −
P+
P
6
2
2
2
FDI =
2
P C 
3
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908
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
We consider members ABC and CDE:
Free body: CDE:
ΣM J = 0: C x (4a) + C y (2a) = 0
C y = −2Cx
(1)
Free body: ABC:
ΣM K = 0: Cx (2a) + C y (4a) − P(3a) = 0
Substituting for Cy from Eq. (1):
Cx (2a ) − 2Cx (4a) − P(3a ) = 0
1
Cx = − P
2
ΣFx = 0:
1
2
 1 
C y = −2  − P  = + P
 2 
FBG + C x = 0
P
 1 
,
FBG = − 2Cx = − 2  − P  = +
2
 2 
FBG =
P
2
T 
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909
PROBLEM 6.114 (Continued)
ΣFy = 0: − FAF −
1
2
FBG − P + C y = 0
FAF = −
1
2
P
2
−P+P=−
P
2
FAF =
P
2
C 
Free body: CDE:
ΣFy = 0:
1
2
FDG − C y = 0
FDG = 2C y = + 2 P
ΣFx = 0: − FEH − Cx −
1
2
FDG = 2 P T 
FDG = 0
P
 1  1
2P = −
FEH = −  − P  −
2
2
 2 
FEH =
P
2
C 
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910
PROBLEM 6.115
Solve Problem 6.112 assuming that the force P is replaced by a
clockwise couple of moment M0 applied to member CDE at D.
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
Free body: Member ABC:
ΣM J = 0: C y (2a) + Cx ( a) = 0
Cx = − 2C y
Free body: Member CDE:
ΣM K = 0: C y (2a ) − Cx (a ) − M 0 = 0
C y (2a) − (− 2C y )(a) − M 0 = 0
Cx = − 2C y :
D
ΣFx = 0:
2
D
+ Cx = 0;
2
−
M0
D=
M0
䉰
4a
M
Cx = − 0 䉰
2a
Cy =
M0
=0
2a
FDG =
2a
M0
2a
T 
ΣM D = 0: E (a) − C y ( a) + M 0 = 0
M
E (a) −  0
 4a

 (a) + M 0 = 0

E=−
3 M0
4 a
FEH =
3 M0
4 a
C 
Return to free body of ABC:
ΣFx = 0:
B
2
+ Cx = 0;
B=
B
2
−
M0
=0
2a
M0
FBG =
2a
ΣM B = 0: A( a) + C y (a); A(a) +
M0
2a
M0
(a ) = 0
4a
A=−
M0
4a
FAF =
M0
4a
T 
C 
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911
PROBLEM 6.116
Solve Problem 6.114 assuming that the force P is replaced by a clockwise
couple of moment M0 applied at the same point.
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: CDE: (same as for Problem 6.114)
ΣM J = 0: C x (4a) + C y (2a) = 0
C y = −2Cx
(1)
Free body: ABC:
ΣM K = 0: C x (2a) + C y (4a) − M 0 = 0
Substituting for Cy from Eq. (1): Cx (2a ) − 2Cx (4a) − M 0 = 0
M0
6a
 M
C y = −2  − 0
 6a
Cx = −
ΣFx = 0:
1
2
M0

 = + 3a

FBG + C x = 0
2M 0
6a
FBG = − 2C x = +
ΣFy = 0: − FAF −
1
2
FBG =
2M 0
6a
T 
FBG + C y = 0
FAF = −
1
2
M
2M 0 M 0
+
=+ 0
6a
3a
6a
FAF =
M0
6a
T 
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912
PROBLEM 6.116 (Continued)
Free body: CDE:
(Use F.B. diagram of Problem 6.114.)
ΣFy = 0:
1
2
FDG − C y = 0
FDG = 2C y = +
ΣFx = 0: − FEH − C x −
1
2
2M 0
3a
FDG =
2M 0
3a
T 
FDG = 0
 M
FEH = −  − 0
 6a
  1   2M 0

−
  2   3a

M
 = − 0 ,
6a

FEH =
M0
6a
C 
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913
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:
ΣM D = 0: A(3a) + E (a ) = 0
E
3
ΣM A = 0: − D (3a ) − E (2a ) = 0
A=−
D=−
Member DHC:
2E
3
 2E 
ΣM C = 0:  −
 (3a) − H (a) = 0
 3 
H = − 2E
 2E 
ΣM H = 0:  −
 (2a) + C (a ) = 0
 3 
C=+
Member CGB:
4E
3
 4E 
ΣM B = 0: + 
 (3a ) − G (a ) = 0
 3 
G = + 4E
 4E 
ΣM G = 0: + 
 (2a) + B(a ) = 0
 3 
B=−
(1)
(2)
8E
3
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914
PROBLEM 6.117 (Continued)
Member AFB:
ΣFy = 0: F − A − B − P = 0
 E   8E 
F − −  − −
−P =0
 3  3 
F = P − 3E
(3)
ΣM A = 0: F ( a) − B(3a) = 0
 8E 
( P − 3E )(a) −  −
 (3a ) = 0
 3 
P − 3E + 8 E = 0; E = −
P
5
E=
P

5
Substitute E = − P5 into Eqs. (1), (2), and (3).
 P
H = −2 E = − 2  − 
 5
 P
G = + 4E = 4  − 
 5
 P
F = P − 3E = P − 3  − 
 5
H =+
2P
5
H=
2P

5
G=−
4P
5
G=
4P

5
F =+
8P
5
F=
8P

5

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915
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 = 2B = 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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916
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
(a)
Member FBDs:
I
II
FBD I:
ΣM A = 0: aF1 − 2aP = 0 F1 = 2 P;
FBD II:
ΣM B = 0: − aF2 = 0 F2 = 0
ΣFy = 0:
Ay − P = 0 A y = P
ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2 P
B x = 2P
ΣFy = 0: B y = 0
FBD I:
ΣFx = 0: − Ax − F1 + F2 = 0
so B = 2P
Ax = F2 − F1 = 0 − 2 P
A x = 2P
so A = 2.24P

26.6° 
Frame is rigid. 
(b)
FBD left:
FBD whole:
I
FBD I:
ΣM E = 0:
FBD II:
ΣM 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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917
PROBLEM 6.119 (Continued)
(c)
Member FBDs:
I
FBD I:
II
ΣFy = 0: A − P = 0
A = P 
ΣM D = 0: aF1 − 2aA = 0 F1 = 2 P
ΣFx = 0: F2 − F1 = 0 F2 = 2 P 
FBD II:
ΣM B = 0: 2aC − aF1 = 0 C =
F1
=P
2
ΣFx = 0: F1 − F2 + Bx = 0 Bx = P − P = 0 
ΣFx = 0: By + C = 0 B y = − C = − P
C = P 
B=P 
Frame is rigid. 
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918
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
FBD II:
FBD I:
ΣFy = 0: B y = 0
II
ΣM B = 0: aF2 = 0
F2 = 0
ΣM A = 0: aF2 − 2aP = 0, but F2 = 0
so P = 0
(b)
not rigid for P ≠ 0 
Member FBDs:
Note: Seven unknowns ( Ax , Ay , Bx , B y , F1 , F2 , C ), but only six independent equations
System is statically indeterminate. 

System is, however, rigid. 
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919
PROBLEM 6.120 (Continued)
(c)
FBD whole:
FBD right:
I
FBD I:
ΣM A = 0: 5aBy − 2aP = 0
ΣFy = 0: Ay − P +
FBD II:
FBD I:
II
ΣM C = 0:
2
P=0
5
a
5a
Bx −
B y = 0 Bx = 5 B y
2
2
ΣFx = 0: Ax + Bx = 0
Ax = − Bx
By =
2
P 
5
Ay =
3
P
5
B x = 2P
A x = 2P
A = 2.09P
B = 2.04P
16.70° 
11.31° 
System is rigid. 
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920
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
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:
ΣM A = 0: − 2aP −
ΣFx = 0: Ax −
ΣFy = 0:
(b)
a 4
1
B + 5a
B = 0 B = 2.06 P
4 17
17
4
17
Ay − P +
B=0
1
17
B = 2.06P
14.04° 
A = 2.06P
14.04° 
A x = 2P
B = 0 Ay =
P
2
rigid 
FBD whole:
Since B passes through A,
ΣM A = 2aP = 0 only if P = 0.
no equilibrium if P ≠ 0
not rigid 
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921
PROBLEM 6.121 (Continued)
(c)
FBD whole:
ΣM A = 0: 5a
1
17
ΣFx = 0: Ax +
B+
4
17
ΣFy = 0: Ay − P +
3a 4
17
B − 2aP = 0 B =
P
4 17
4
B=0
1
17
B = 1.031P
14.04° 
A = 1.250P
36.9° 
Ax = − P
B=0
Ay = P −
P 3P
=
4
4
System is rigid. 
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922
PROBLEM 6.122
The shear shown is used to cut and trim electronic circuit board
laminates. For the position shown, determine (a) the vertical component
of the force exerted on the shearing blade at D, (b) the reaction at C.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC: We have the components:
Px = (400 N) sin 30° = 200 N
Py = (400 N) cos 30° = 346.41 N
(FBD ) x =
25
FBD
65
(FBD ) y =
60
FBD
65
ΣM C = 0: ( FBD ) x (45) + ( FBD ) y (30) − Px (45 + 300sin 30°)
− Py (30 + 300cos 30°) = 0
 25

 60

 FBD  (45) +  FBD  (30) = (200)(195) + (346.41)(289.81)
 65

 65

45 FBD = 139.39 × 103
FBD = 3097.6 N
(a)
Vertical component of force exerted on shearing blade at D.
( FBD ) y =
60
60
(3097.6 N) = 2859.3 N
FBD =
65
65
(FBD ) y = 2860 N 
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923
PROBLEM 6.122 (Continued)
(b)
Returning to FB diagram of member ABC,
ΣFx = 0: ( FBD ) x − Px − C x = 0
Cx = ( FBD ) x − Px =
25
FBD − Px
65
25
(3097.6) − 200
65
Cx = +991.39
=
C x = 991.39 N
ΣFy = 0: ( FBD ) y − Py − C y = 0
C y = ( FBD ) y − Py =
C y = +2512.9 N
C = 2295 N
60
60
(3097.6) − 346.41
FBD − Py =
65
65
C y = 2512.9
C = 2700 N
68.5° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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924
PROBLEM 6.123
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:
ΣFy = 0: E − FBD cos 20° = 0, E = FBD cos 20°
FBD ABC:
ΣM 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 
ΣFx = 0: Ax − (793.64 N)sin 20° = 0
A x = 271.44 N
ΣFy = 0: Ay + (793.64 N) cos 20° − 250 N = 0
A y = 495.78 N
so
(b)
A = 565 N
61.3° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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925
PROBLEM 6.124
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:
ΣFy = 0: 900 N − FBD cos 20° = 0, FBD = 957.76 N C
(a)
FBD ABC:
ΣM 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 
ΣFx = 0: Ax − (957.76 N)sin 20° = 0
A x = 327.57 N
ΣFy = 0: −Ay + (957.76 N) cos 20° − 301.70 N = 0
A y = 598.30 N
so
A = 682 N
61.3° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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926
PROBLEM 6.125
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:
+ ΣFy = 0: C − 135 = 0
C = + 135 N
Free body: Member BCE:
ΣFx = 0: Bx = 0
ΣM B = 0: (135 N)(6 mm) − TDE (10 mm) = 0
TDE = 81.0 N 
ΣFy = 0: 135 + 81 − By = 0
By = + 216 N
B = 216 N 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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927
PROBLEM 6.126
An 84-lb force is applied to the toggle vise at C. Knowing that θ = 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 = (7) 2 + (24)2 = 25 in.
We have
( FBD ) x =
7
24
FBD , ( FBD ) y =
FBD
25
25
ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(16) = 0
 7

 24

 FBD  (24) +  FBD  (7) = 84(16)
25
25




336
FBD = 1344
25
FBD = 100 lb
tan α =
24
α = 73.7°
7
FBD = 100.0 lb
(b)
Force exerted at B.
(a)
Vertical force exerted on block.
( FBD ) y =
24
24
FBD =
(100 lb) = 96 lb
25
25
73.7° 
(FBD ) y = 96.0 lb 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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928
PROBLEM 6.127
Solve Problem 6.126 when θ = 0.
PROBLEM 6.126 An 84-lb force is applied to the toggle vise at C. Knowing
that θ = 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 = (7) 2 + (24)2 = 25 in.
We have
( FBD ) x =
7
24
FBD , ( FBD ) y =
FBD
25
25
ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(40) = 0
 7

 24

 FBD  (24) +  FBD  (7) = 84(40)
25
25




336
FBD = 3360
25
FBD = 250 lb
tan α =
24
α = 73.7°
7
FBD = 250.0 lb
(b)
Force exerted at B.
(a)
Vertical force exerted on block.
( FBD ) y =
24
24
FBD =
(250 lb) = 240 lb
25
25
73.7° 
(FBD ) y = 240 lb 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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929
PROBLEM 6.128
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:
ΣM C = 0: R( FBD sin 30°) − [ R(1 − cos 30°)](100 N) − R(50 N) = 0
FBD = 126.795 N
ΣFx = 0: −C x + (126.795 N) cos 30° = 0
(b) FBD = 126.8 N T 
C x = 109.808 N
ΣFy = 0: C y + (126.795 N) sin 30° − 100 N − 50 N = 0
C y = 86.603 N
II:
so (c) C = 139.8 N
38.3° 
FBD II:
ΣM A = 0: aP − a[(126.795 N) cos 30°] = 0
(a ) P = 109.8 N

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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930
PROBLEM 6.129
The Whitworth mechanism shown is used to produce a quick-return motion
of Point D. The block at B is pinned to the crank AB and is free to slide in a
slot cut in member CD. Determine the couple M that must be applied to the
crank AB to hold the mechanism in equilibrium when (a) α = 0, (b) α = 30°.
SOLUTION
(a)
Free body: Member CD:
ΣM C = 0: B(0.5 m) − (1200 N)(0.7 m) = 0
B = 1680 N
Free body: Crank AB:
ΣM A = 0: M − (1680 N)(0.1 m) = 0
M = 168.0 N ⋅ m
(b)
M = 168.0 N ⋅ m

Geometry:
AB = 100 mm, α = 30°
tan θ =
50
θ = 5.87°
486.6
BC = (50)2 + (486.6) 2
= 489.16 mm
Free body: Member CD:
ΣM C = 0: B (0.48916) − (1200 N)(0.7) cos 5.87° = 0
B = 1708.2 N
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931
PROBLEM 6.129 (Continued)
Free body: Crank AB:
ΣM A = 0: M − ( B sin 65.87°)(0.1 m) = 0
M = (1708.2 N)(0.1 m)sin 65.87°
M = 155.90 N ⋅ m
M = 155.9 N ⋅ m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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932
PROBLEM 6.130
Solve Problem 6.129 when (a) α = 60°, (b) α = 90°.
PROBLEM 6.129 The Whitworth mechanism shown is used to produce a
quick-return motion of Point D. The block at B is pinned to the crank AB and is
free to slide in a slot cut in member CD. Determine the couple M that must be
applied to the crank AB to hold the mechanism in equilibrium when (a) α = 0,
(b) α = 30°.
SOLUTION
(a)
Geometry:
AB = 100 mm, α = 60°
BE = 100 cos 60° = 86.60
CE = 400 + 100sin 60° = 450
86.60
tan θ =
θ = 10.89°
450
BC = (86.60)2 + (450) 2 = 458.26 mm
Free body: Member CD:
ΣM C = 0: B(0.45826 m) − (1200 N)(0.7 m) cos10.89° = 0
B = 1800.0 N
Free body: Crank AB:
ΣM A = 0: M − ( B sin 40.89°)(0.1 m) = 0
M = (1800.0 N)(0.1 m)sin 40.89°
M = 117.83 N ⋅ m
M = 117.8 N ⋅ m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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933
PROBLEM 6.130 (Continued)
(b)
Free body: Member CD:
0.1 m
0.4 m
θ = 14.04°
tan θ =
BC = (0.1) 2 + (0.4) 2 = 0.41231 m
ΣM C = 0: B(0.41231) − (1200 N)(0.7 cos14.04°) = 0
B = 1976.4 N
Free body: Crank AB:
ΣM A = 0: M − ( B sin14.04°)(0.1 m) = 0
M = (1976.4 N)(0.1 m) sin14.04°
M = 47.948 N ⋅ m
M = 47.9 N ⋅ m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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934
PROBLEM 6.131
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:
50 mm
175 mm
2
=
7
Dimensions in mm
Note:
tan θ =
FBD whole:
ΣM A = 0: (0.250 m)C y − 1.5 kN ⋅ m = 0 C y = 6.00 kN
FBD piston:
ΣFy = 0: C y − FBC sin θ = 0 FBC =
Cy
sin θ
=
6.00 kN
sinθ
ΣFx = 0: FBC cos θ − P = 0
P = FBC cos θ =
6.00 kN
= 7 kips
tan θ
P = 21.0 kN

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
935
PROBLEM 6.131 (Continued)
(b)
FBDs:
Dimensions in mm
2
as above
7
Note:
tan θ =
FBD whole:
ΣM A = 0: (0.100 m)C y − 1.5 kN ⋅ m = 0 C y = 15 kN
ΣFy = 0: C y − FBC sin θ = 0 FBC =
ΣFx = 0:
Cy
sin θ
FBC cos θ − P = 0
P = FBC cos θ =
Cy
tan θ
=
15 kN
2/7
P = 52.5 kN

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
936
PROBLEM 6.132
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:
Note:
FBD piston:
50 mm
175 mm
2
=
7
Dimensions in mm
tan θ =
ΣFx = 0: FBC cos θ − P = 0 FBC =
P
cos θ
ΣFy = 0: C y − FBC sin θ = 0 C y = FBC sin θ = P tan θ =
FBD whole:
2
P
7
ΣM A = 0: (0.250 m)C y − M = 0
2
M = (0.250 m)   (16 kN)
7
= 1.14286 kN ⋅ m
M = 1143 N ⋅ m

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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937
PROBLEM 6.132 (Continued)
(b)
FBDs:
Dimensions in mm
tan θ =
Note:
FBD piston, as above:
FBD whole:
2
as above
7
C y = P tan θ =
2
P
7
2
ΣM A = 0: (0.100 m)C y − M = 0 M = (0.100 m) (16 kN)
7
M = 0.45714 kN ⋅ m
M = 457 N ⋅ m

PROPRIETARY MATERIAL. © 2013 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
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you are using it without permission.
938
PROBLEM 6.133
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 θ = 30°.
SOLUTION
Free body: Member ABC:
ΣM C = 0: (25 lb)(13.856 in.) − B(3 in.) = 0
B = +115.47 lb
ΣFy = 0: − 25 lb + C y = 0
C y = +25 lb
ΣFx = 0: 115.47 lb − Cx = 0
Cx = +115.47 lb
Free body: Member CD:
β = sin −1
5.196
; β = 40.505°
8
CD cos β = (8 in.) cos 40.505° = 6.083 in.
ΣM D = 0: M − (25 lb)(5.196 in.) − (115.47 lb)(6.083 in.) = 0
M = +832.3 lb ⋅ in.
M = 832 lb ⋅ in. 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
939
PROBLEM 6.134
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 θ = 60°.
SOLUTION
Free body: Member ABC:
ΣM C = 0: (25 lb)(8 in.) − B(5.196 in.) = 0
B = +38.49 lb
ΣFx = 0: 38.49 lb − C x = 0
Cx = +38.49 lb
ΣFy = 0: − 25 lb + C y = 0
C y = +25 lb
Free body: Member CD:
3
8
β = sin −1 ; β = 22.024°
CD cos β = (8 in.) cos 22.024° = 7.416 in.
ΣM D = 0: M − (25 lb)(3 in.) − (38.49 lb)(7.416 in.) = 0
M = +360.4 lb ⋅ in.
M = 360 lb ⋅ in.

PROPRIETARY MATERIAL. © 2013 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
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940
PROBLEM 6.135
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 θ = 30°.
SOLUTION
FBD DC:
ΣFx′ = 0:
D y sin 30° − (150 N) cos 30° = 0
Dy = (150 N) ctn 30° = 259.81 N
FBD machine:
ΣM 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. © 2013 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
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you are using it without permission.
941
PROBLEM 6.136
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 θ = 30°.
SOLUTION
B ⊥ CD
Note:
FBD DC:
ΣFx′ = 0: D y sin 30° − (300 N) cos 30° = 0
Dy =
300 N
= 519.62 N
tan 30°
FBD machine:
ΣM A = 0:
0.200 m
519.62 N − M = 0
sin 30°
M = 207.85 N ⋅ m
M = 208 N ⋅ m

PROPRIETARY MATERIAL. © 2013 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
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you are using it without permission.
942
PROBLEM 6.137
Two rods are connected by a frictionless collar B. Knowing that the magnitude
of the couple MA is 500 lb · in., determine (a) the couple MC required for
equilibrium, (b) the corresponding components of the reaction at C.
SOLUTION
(a)
Free body: Rod AB & collar:
ΣM A = 0: ( B cos α )(6 in.) + ( B sin α )(8 in.) − M A = 0
B = (6cos 21.8° + 8sin 21.8°) − 500 = 0
B = 58.535 lb
Free body: Rod BC:
+ΣM C = 0: M C − B = 0
M C = B = (58.535 lb)(21.541 in.) = 1260.9 lb ⋅ in. M C = 1261 lb ⋅ in.
(b)

ΣFx = 0: Cx + B cos α = 0
Cx = − B cos α = −(58.535 lb) cos 21.8° = −54.3 lb
C x = 54.3 lb

ΣFy = 0: C y − B sin α = 0
C y = B sin α = (58.535 lb) sin 21.8° = +21.7 lb
C y = 21.7 lb 
PROPRIETARY MATERIAL. © 2013 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
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you are using it without permission.
943
PROBLEM 6.138
Two rods are connected by a frictionless collar B. Knowing that the magnitude
of the couple MA is 500 lb · in., determine (a) the couple MC required for
equilibrium, (b) the corresponding components of the reaction at C.
SOLUTION
(a)
Free body: Rod AB:
ΣM A = 0: B(10 in.) − 500 lb ⋅ in. = 0
B = 50.0
Free body: Rod BC & collar:
ΣM C = 0: M C − (0.6 B)(20 in.) − (0.8 B)(8 in.) = 0
M C − (30 lb)(20 in.) − (40 lb)(8 in.) = 0
M C = 920 lb ⋅ in.
(b)
M C = 920 lb ⋅ in.

C x = 30.0 lb

ΣFx = 0: Cx + 0.6 B = 0
Cx = −0.6 B = −0.6(50.0 lb) = −30.0 lb
ΣFy = 0: C y − 0.8B = 0
C y = 0.8 B = 0.8(50.0 lb) = +40.0 lb
C y = 40.0 lb 
PROPRIETARY MATERIAL. © 2013 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
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you are using it without permission.
944
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:
ΣM B = 0:
4
FAE (150 mm) − (160 N)(600 mm) = 0
5
FAE = +800 N
FAE = 800 N T 
3
ΣFx = 0: − (800 N) + Bx − 80 N = 0
5
Bx = +560 N
4
ΣFy = 0: − (800 N) + By − 160 N = 0
5
By = +800 N
Free body: Member BDF:
ΣM 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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
945
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:
ΣM B = 0:
4
(600 N)(150 mm) − P(600 mm) = 0
5
P = 120.0 N 
P = +120 N
ΣM C = 0:
4
(600)(750 mm) − By (600 mm) = 0
5
By = +600 N
Free body: Member BDF:
ΣM 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:
3
ΣFx = 0: − (600 N) + 470 N − Q = 0
5
Q = +110 N
Q = 110.0 N

PROPRIETARY MATERIAL. © 2013 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
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you are using it without permission.
946
PROBLEM 6.141
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,
FBD ADF:
A = B = 22.5 kN
ΣM F = 0: (75 mm) D − (100 mm)(22.5 kN) = 0
D = 30.0 kN

ΣFx = 0: Fx − D = 0
Fx = D = 30 kN
ΣFy = 0: 22.5 kN − Fy = 0
Fy = 22.5 kN
F = 37.5 kN
so
36.9° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
947
PROBLEM 6.142
If the toggle shown is added to the tongs of Problem 6.141 and a single vertical
force is applied at G, determine the forces exerted at D and F on tong ADF.
SOLUTION
Free body: Toggle:
By symmetry,
Ay =
1
(45 kN) = 22.5 kN
2
AG is a two-force member.
Ax
22.5 kN
=
22 mm 55 mm
Ax = 56.25 kN
Free body: Tong ADF:
ΣFy = 0: 22.5 kN − Fy = 0
Fy = +22.5 kN
ΣM F = 0 : D (75 mm) − (22.5 kN)(100 mm) − (56.25 kN)(160 mm) = 0
D = +150 kN
D = 150.0 kN

ΣFx = 0: 56.25 kN − 150 kN + Fx = 0
Fx = 93.75 kN
F = 96.4 kN
13.50° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
948
PROBLEM 6.143
A small barrel weighing 60 lb is lifted by a pair of tongs as shown. Knowing
that a = 5 in., determine the forces exerted at B and D on tong ABD.
SOLUTION
We note that BC is a two-force member.
Free body: Tong ABD:
Bx By
=
15
5
Bx = 3By
ΣM D = 0: By (3 in.) + 3By (5 in.) − (60 lb)(9 in.) = 0
By = 30 lb
Bx = 3By : Bx = 90 lb
ΣFx = 0: −90 lb + Dx = 0
D x = 90 lb
ΣFy = 0: 60 lb − 30 lb − D y = 0
D y = 30 lb
B = 94.9 lb
D = 94.9 lb
18.43° 
18.43° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
949
PROBLEM 6.144
A 39-ft length of railroad rail of weight 44 lb/ft is lifted by the tongs
shown. Determine the forces exerted at D and F on tong BDF.
SOLUTION
Free body: Rail:
W = (39 ft)(44 lb/ft) = 1716 lb
Free body: Upper link:
By symmetry,
1
E y = Fy = W = 858 lb
2
By symmetry,
1
( FAB ) y = ( FAC ) y = W = 858 lb
2
Since AB is a two-force member,
( FAB ) x ( FAB ) y
=
9.6
6
Free Body: Tong BDF:
( FAB ) x =
9.6
(858) = 1372.8 lb
6
ΣM D = 0: (Attach FAB at A.)
Fx (8) − ( FAB ) x (18) − Fy (0.8) = 0
Fx (8) − (1372.8 lb)(18) − (858 lb)(0.8) = 0
Fx = +3174.6 lb
F = 3290 lb
15.12° 
ΣFx = 0: − Dx + ( FAB ) x + Fx = 0
Dx = ( FAB ) x + Fx = 1372.8 + 3174.6 = 4547.4 lb
ΣFy = 0:
Dy + ( FAB ) y − Fy = 0
Dy = 0
D = 4550 lb

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
950
PROBLEM 6.145
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:
ΣM D = 0: − (126 mm)(300 N) − (6 mm)
 1

+ (30 mm) 
A = 0
 8.84 
2.8
8.84
A
A = 2863.6 8.84 N
Dimensions in mm
FBD handle AB:
ΣM B = 0: (132 mm)(300 N) − (120 mm)
+ (36 mm) F = 0
1
8.84
(2863.6 8.84 N)
F = 8.45 kN 
Dimensions in mm
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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951
PROBLEM 6.146
The compound-lever pruning shears shown can be adjusted
by placing pin A at various ratchet positions on blade ACE.
Knowing that 300-lb vertical forces are required to complete
the pruning of a small branch, determine the magnitude P of
the forces that must be applied to the handles when the shears
are adjusted as shown.
SOLUTION
We note that AB is a two-force member.
( FAB ) x ( FAB ) y
=
0.65 in. 0.55 in.
11
( FAB ) y = ( FAB ) x
13
(1)
Free body: Blade ACE:
ΣM C = 0: (300 lb)(1.6 in.) − ( FAB ) x (0.5 in.) − ( FAB ) y (1.4 in.) = 0
Use Eq. (1):
( FAB ) x (0.5 in.) +
11
( FAB ) x (1.4 in.) = 480 lb ⋅ in.
13
1.6846( FAB ) x = 480
( FAB ) y =
11
(284.9 lb)
13
( FAB ) x = 284.9 lb
( FAB ) y = 241.1 lb
Free body: Lower handle:
ΣM D = 0: (241.1 lb)(0.75 in.) − (284.9 lb)(0.25 in.) − P(3.5 in.) = 0
P = 31.3 lb 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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952
PROBLEM 6.147
The pliers shown are used to grip a 0.3-in.-diameter rod.
Knowing that two 60-lb 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)
(b)
ΣM A = 0: Q (1.2 in.) − (60 lb)(9.5 in.) = 0
Q = 475 lb 
ΣFx = 0: Q(sin 30°) + Ax = 0
(475 lb)(sin 30°) + Ax = 0
Ax = −237.5 lb
A x = 237.5 lb
ΣFy = 0: − Q (cos 30°) + Ay − 60 lb = 0
−(475 lb)(cos 30°) + Ay − 60 lb = 0
Ay = +471.4 lb
A y = 471.4 lb
A = 528 lb
63.3° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
953
PROBLEM 6.148
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:
FBD handle BC:
II
Dimensions in mm
ΣFx = 0: Bx = 0
FBD II:
ΣM C = 0: (12 mm)By − (448 mm)300 N = 0
By = 11, 200.0 N
Then
FBD I:
ΣM A = 0: (96 mm) By − (24 mm) F = 0
F = 4 By
F = 44,800 N = 44.8 kN 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
954
PROBLEM 6.149
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 135 lb ⋅ in., 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
1.5 in.
Free body: Nut:
=
Cx
C y = 2.4 C x
in.
5
8
ΣFx = 0: Bx = C x
(1)
ΣFy = 0: By = C y = 2.4 C x
(2)
ΣFx = 0: C x = Dx
ΣM = 135 lb ⋅ in.
C x (1.125 in.) = 135 lb ⋅ in.
C x = 120 lb
(a)
Eq. (1):
Bx = C x = 120 lb
Eq. (2):
By = C y = 2.4(120 lb) = 288 lb
(
B = Bx2 + By2
(b)
1/ 2
)
B = 312 lb 
= (1202 + 2882 )1/2
Free body: Rod:
ΣM D = 0: − M 0 + By (0.625 in.) − Bx (0.375 in.) = 0
− M 0 + (288)(0.625) − (120)(0.375) = 0
M 0 = 135.0 lb ⋅ in.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
955
PROBLEM 6.150
Determine the force P that must be applied to the toggle CDE to maintain
bracket ABC in the position shown.
SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:
( FCD ) x ( FCD ) y
=
30
150
Similarly,
( FCD ) y = 5( FCD ) x
Dimensions in mm
( FDE ) y = 5( FDE ) x
ΣFy = 0: ( FDE ) y = ( FCD ) y
It follows that
( FDE ) x = ( FCD ) x
ΣFx = 0: P − ( FDE ) x − ( FCD ) x = 0
( FDE ) x = ( FCD ) x =
Also,
1
P
2
1 
( FDE ) y = ( FCD ) y = 5  P  = 2.5P
2 
Free body: Member ABC:
1 
ΣM A = 0: (2.5 P )(300) +  P  (450) − (910 N)(150) = 0
2 
(750 + 225) P = (910 N)(150)
P = 140.0 N 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
956
PROBLEM 6.151
Determine the force P that must be applied to the toggle CDE to maintain
bracket ABC in the position shown.
SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:
( FCD ) x ( FCD ) y
=
30
150
Similarly,
( FCD ) y = 5( FCD ) x
( FDE ) y = 5( FDE ) x
Dimensions
in mm
ΣFy = 0: ( FDE ) y = ( FCD ) y
It follows that
( FDE ) x = ( FCD ) x
Σ Fx = 0: ( FDE ) x + ( FCD ) x − P = 0
( FDE ) x = ( FCD ) x =
Also,
1
P
2
1 
( FDE ) y = ( FCD ) y = 5  P  = 2.5P
2 
Free body: Member ABC:
1 
ΣM A = 0: (2.5P)(300) −  P  (450) − (910 N)(150) = 0
2 
(750 − 225) P = (910 N)(150)
P = 260 N 
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957
PROBLEM 6.152
A 45-lb 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:
ΣM A = 0: By (10 in.) − (45 lb)(6.25 in.) = 0
By = 28.125 lb
Free body: Portion ECB:
ΣM E = 0: − Bx (7.5 in.) − P(7.5 in.) − (28.125 lb)(10 in.) = 0
Bx = −37.5 − P
Free body: Portion CDB:
ΣM C = 0: − (28.125 lb)(4.6 in.) − Bx (2.2 in.) = 0
−(28.125 lb)(4.6 in.) − ( −37.5 − P)(2.2 in.) = 0
P = 21.3 lb P = 21.3 lb


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958
PROBLEM 6.153
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 θ = 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
Geometry:
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
tan β =
e 1.7208
=
; β = 44.43°
d 1.7553
Free body: Arm ABC:
We note that BD is a two-force member.
W = (200 kg)(9.81 m/s 2 ) = 1.962 kN
(a)
ΣM 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)
44.4° 
ΣFx = 0: Ax − FBD cos β = 0
Ax = (9.2867 kN) cos 44.43° = 6.632 kN
A x = 6.632 kN
ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0
Ay = 1.962 kN − (9.2867 kN)sin 44.43° = −4.539 kN
A y = 4.539 kN
A = 8.04 kN
34.4° 
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959
PROBLEM 6.154
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 θ = −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
Geometry:
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
tan β =
e 0.0792
=
; β = 2.584°
d 1.7552
Free body: Arm ABC:
We note that BD is a two-force member.
W = (200 kg)(9.81 m/s 2 )
W = 1962 N = 1.962 kN
(a)
ΣM 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)
2.58° 
ΣFx = 0: Ax − FBD cos β = 0
Ax = (10.003 kN)cos 2.583° = 9.993 kN
A x = 9.993 kN
ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0
Ay = 1.962 kN − (10.003 kN)sin 2.583° = −1.5112 kN
A y = 1.5112 kN
A = 10.11 kN
8.60° 
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960
PROBLEM 6.155
The bucket of the front-end loader shown carries a 3200-lb
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 3200-lb load, determine the force exerted (a) by
cylinder CD, (b) by cylinder FH.
SOLUTION
Free body: Bucket: (one mechanism)
ΣM D = 0: (1600 lb)(15 in.) − FAB (16 in.) = 0
FAB = 1500 lb
Note: There are two identical support mechanisms.
Free body: One arm BCE:
8
20
β = 21.8°
tan β =
ΣM E = 0: (1500 lb)(23 in.) + FCD cos 21.8°(15 in.) − FCD sin 21.8°(5 in.) = 0
FCD = −2858 lb
FCD = 2.86 kips C 
Free body: Arm DFG:
ΣM G = 0: (1600 lb)(75 in.) + FFH sin 45°(24 in.) − FFH cos 45°(6 in.) = 0
FFH = −9.428 kips
FFH = 9.43 kips C 
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961
PROBLEM 6.156
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
ΣM J = 0: (4500 lb)(20 in.) − FGH (22 in.) = 0
FGH = 4091 lb
Free body: Arm BDH
ΣM D = 0: − (4091 lb)(24 in.) − FBC (20 in.) = 0
FBC = −4909 lb
FBC = 4.91 kips C 
Free body: Entire mechanism
(Two arms and cylinders AFJE)
Note: Two arms thus 2 FEF
18 in.
65 in.
β = 15.48°
tan β =
ΣM A = 0: (4500 lb)(123 in.) + FBC (12 in.) + 2 FEF cos β (24 in.) = 0
(4500 lb)(123 in.) − (4909 lb)(12 in.) + 2 FEF cos 15.48°(24 in.) = 0
FEF = 10.69 kips C 
FEF = −10.690 lb
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962
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 2-kip force P is exerted on the bucket
teeth at J. Knowing that θ = 45°, determine the
force exerted by each cylinder.
SOLUTION
Free body: Bucket:
ΣM H = 0:
(Dimensions in inches)
4
3
FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0
5
5
FCG = −
P
(16 cos θ + 8 sin θ )
14
(1)
Free body: Arm ABH and bucket:
(Dimensions in inches)
ΣM B = 0:
4
3
FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0
5
5
FAD = −
P
(86 cos θ − 42 sin θ )
15.6
(2)
Free body: Bucket and arms IEB + ABH:
Geometry of cylinder EF:
16 in.
40 in.
β = 21.801°
tan β =
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963
PROBLEM 6.157 (Continued)
ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0
P(120 sin θ − 28 cos θ )
cos 21.8°(18)
P
=
(120 sin θ − 28 cos θ )
16.7126
FEF =
For P = 2 kips,
(3)
θ = 45°
From Eq. (1):
FCG = −
2
(16 cos 45° + 8 sin 45°) = −2.42 kips
14
From Eq. (2):
FAD = −
2
(86 cos 45° − 42 sin 45°) = −3.99 kips
15.6
From Eq. (3):
FEF =
2
(120 sin 45° − 28 cos 45°) = +7.79 kips
16.7126
FCG = 2.42 kips C 
FAD = 3.99 kips C 
FEF = 7.79 kips T 
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964
PROBLEM 6.158
Solve Problem 6.157 assuming that the 2-kip force P
acts horizontally to the right (θ = 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 2-kip force P is exerted on the
bucket teeth at J. Knowing that θ = 45°, determine the
force exerted by each cylinder.
SOLUTION
Free body: Bucket:
ΣM H = 0:
(Dimensions in inches)
4
3
FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0
5
5
FCG = −
P
(16 cos θ + 8 sin θ )
14
(1)
Free body: Arm ABH and bucket:
(Dimensions in inches)
ΣM B = 0:
4
3
FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0
5
5
FAD = −
P
(86 cos θ − 42 sin θ )
15.6
(2)
Free body: Bucket and arms IEB + ABH:
Geometry of cylinder EF:
16 in.
40 in.
β = 21.801°
tan β =
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965
PROBLEM 6.158 (Continued)
ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0
P(120 sin θ − 28 cos θ )
cos 21.8°(18)
P
(120 sin θ − 28 cos θ )
=
16.7126
FEF =
For P = 2 kips,
(3)
θ =0
From Eq. (1):
FCG = −
2
(16 cos 0 + 8 sin 0) = −2.29 kips
14
From Eq. (2):
FAD = −
2
(86 cos 0 − 42 sin 0) = −11.03 kips
15.6
From Eq. (3):
FEF =
2
(120 sin 0 − 28 cos 0) = −3.35 kips
16.7126
FCG = 2.29 kips C 
FAD = 11.03 kips C 
FEF = 3.35 kips C 
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966
PROBLEM 6.159
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
ΣM A = 0: 3(rA F ) − 10 N ⋅ m = 0,
ΣM C = 0: rB ( F − F4 ) = 0,
FBD Gear C:
F=
10
N⋅m
3rA
F4 = F
ΣFx′ = 0: Cx′ = 0
ΣFy′ = 0: C y′ − 2 F = 0,
C y′ = 2 F
Gears B and D are analogous, each having a central force of 2F.
ΣM A = 0: 50 N ⋅ m − 3(rA + rB )2F = 0
FBD Spider:
50 N ⋅ m − 3(rA + rB )
20
N⋅m = 0
rA
rA + rB
r
= 2.5 = 1 + B ,
rA
rA
Since rA = 18 mm,
rB = 1.5rA
rB = 27.0 mm 
FBD Outer Gear:
(a)
ΣM 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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967
PROBLEM 6.160
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:
ΣM G = 0: 30 N ⋅ m − J (0.03 m) = 0; J = 1000 N
Free body: Gear D:
ΣM D = 0: M 0 − (1000 N)(0.09 m) = 0
M 0 = (90.0 N ⋅ m)i 
M 0 = 90 N ⋅ m
(b)
Gear G and axle FH:
ΣM F = 0: H (0.3 m) − (1000 N)(0.18 m) = 0
H = 600 N
ΣFy = 0: F + 600 − 1000 = 0
F = 400 N
Gear D and axle CE:
ΣM C = 0: (1000 N)(0.18 m) − E (0.3 m) = 0
E = 600 N
ΣFy = 0: 1000 − C − 600 = 0
C = 400 N
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968
PROBLEM 6.160 (Continued)
Free body: Bracket AE:
ΣFy = 0: A − 400 + 400 = 0
A = 0 
ΣM A = 0: M A + (400 N)(0.32 m) − (400 N)(0.2 m) = 0
M A = −48 N ⋅ m
Free body: Bracket BH:
ΣFy = 0: B − 600 + 600 = 0
M A = −(48.0 N ⋅ m)i 
B=0 
ΣM 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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969
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 500 lb ⋅ in. (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:
ΣM x = 0: M C cos30° − 500 lb ⋅ in. = 0
M C = 577.35 lb ⋅ in.
Free body: Shaft BC:
We use here x′, y ′, z with x′ along BC .
ΣM C = 0: − M R i′ − (577.35 lb ⋅ in.)i′ + (−5 in.)i′ × ( By j ′ + Bz k ) = 0
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970
PROBLEM 6.161* (Continued)
Equate coefficients of unit vectors to zero:
i:
M A − 577.35 lb ⋅ in. = 0
j:
Bz = 0
k:
By = 0
M A = 577.35 lb ⋅ in.
M A = 577 lb ⋅ in. 
B=0
ΣF = 0:
B + C = 0,
B=0
since B = 0,
C=0
Return to free body of shaft DF.
ΣM D = 0
(Note that C = 0 and M C = 577.35 lb ⋅ in. )
(577.35 lb ⋅ in.)(cos 30°i + sin 30° j) − (500 lb ⋅ in.)i
+(6 in.)i × ( Ex i + E y j + Ez k ) = 0
(500 lb ⋅ in.)i + (288.68 lb ⋅ in.) j − (500 lb ⋅ in.)i
+ (6 in.) E y k − (6 in.) Ez j = 0
Equate coefficients of unit vectors to zero:
j:
288.68 lb ⋅ in. − (6 in.)Ez = 0 Ez = 48.1 lb
k:
Ey = 0
ΣF = 0: C + D + E = 0
0 + Dy j + Dz k + Ex i + (48.1 lb)k = 0
i:
Ex = 0
j:
Dy = 0
k:
Dz + 48.1 lb = 0
Dz = −48.1 lb
Reactions are:
B=0 
D = −(48.1 lb)k 
E = (48.1 lb)k 
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971
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 500 lb · in. (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.
Σ M x = 0: M C − 500 lb ⋅ in. = 0
M C = 500 lb ⋅ in.
Free body: Shaft BC:
We resolve −(520 lb ⋅ in.)i into components along x′ and y ′ axes:
−MC = −(500 lb ⋅ in.)(cos30°i′ + sin 30° j′)
ΣMC = 0: M Ai′ − (500 lb ⋅ in.)(cos30°i′ + sin 30° j′) + (5 in.)i′ × ( By′ j′ + Bz k ) = 0
M A i′ − (433 lb ⋅ in.)i ′ − (250 lb ⋅ in.) j + (5 in.) By′k − (5 in.) Bz j′ = 0
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972
PROBLEM 6.162* (Continued)
Equate to zero coefficients of unit vectors:
M A = 433 lb ⋅ in. 
i′: M A − 433 lb ⋅ in. = 0
j′: − 250 lb ⋅ in. − (5 in.) Bz = 0
Bz = −50 lb
k : B y′ = 0
B = −(50 lb)k
Reactions at B.
ΣF = 0: B − C = 0
−(50 lb)k − C = 0
C = −(50 lb)k
Return to free body of shaft DF.
ΣM D = 0: (6 in.)i × ( Ex i + E y j + Ez k ) − (4 in.)i × (−50 lb)k
− (500 lb ⋅ in.)i + (500 lb ⋅ in.)i = 0
(6 in.) E y k − (6 in.) Ez j − (200 lb ⋅ in.) j = 0
k : Ey = 0
j: −(6 in.) Ez − 200 lb ⋅ in. = 0
Ez = −33.3 lb
ΣF = 0: C + D + E = 0
−(50 lb)k + Dy j + Dz k + Ex i − (33.3 lb)k = 0
i : Ex = 0
k : −50 lb − 33.3 lb + Dz = 0
Dz = 83.3 lb
Reactions are
B = −(50 lb)k 
D = (83.3 lb)k 
E = −(33.3 lb)k 
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973
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 DGJ.)
SOLUTION
Free body: Pin A:
T = W = mg = (7500 kg)(9.81 m/s 2 ) = 73.575 kN
ΣFx = 0: ( FAB ) x = ( FAC ) x
1
ΣFy = 0: ( FAB ) y = ( FAC ) y = W
2
Also,
( FAC ) x = 2( FAC ) y = W
Free body: Member CDF:
1
Σ M D = 0: W (0.3) + W (2.3) − Fx (1.8) − Fy (0.5 m) = 0
2
or
1.8Fx + 0.5Fy = 1.45W
(1)
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974
PROBLEM 6.163* (Continued)
ΣFx = 0: Dx − Fx − W = 0
Ex − Fx = W
or
(2)
1
ΣFy = 0: Fy − Dy + W = 0
2
1
E y − Fy = W
2
or
(3)
Free body: Member EFH:
1
ΣM 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)
ΣFx = 0: Ex + Fx − H x = 0
Ex + Fx = H x
or
2 Fx = H x − W
Subtract Eq. (2) from Eq. (5):
Subtract Eq. (4) from 3 × (1):
3.6Fx = 5.25W − 2.3H x
Add Eq. (7) to 2.3 × Eq. (6):
8.2Fx = 2.95W
Fx = 0.35976W
(5)
(6)
(7)
(8)
Substitute from Eq. (8) into Eq. (1):
(1.8)(0.35976W ) + 0.5Fy = 1.45W
0.5Fy = 1.45W − 0.64756W = 0.80244W
Fy = 1.6049W
(9)
Substitute from Eq. (8) into Eq. (2):
Ex − 0.35976W = W ; Ex = 1.35976W
Substitute from Eq. (9) into Eq. (3):
1
E y − 1.6049W = W
2
From Eq. (5):
H x = Ex + Fx = 1.35976W + 0.35976W = 1.71952W
Recall that
1
Hy = W
2
E y = 2.1049W
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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975
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 y = 154.9 kN 
E x = 100.0 kN
Fx = 26.5 kN

Fy = 118.1 kN 
H y = 36.8 kN 
H x = 126.5 kN
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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976
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:
ΣM E = 0: F (3 m) − (900 N)(2.25 m) − (900 N)(4.5 m) = 0
F = 2025 N
ΣFx = 0: Ex + 900 N + 900 N = 0
Ex = −1800 N E x = 1800 N
ΣFy = 0: E y + 2025 N = 0
E y = −2025 N E y = 2025 N
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
Free body: Joint E:
ΣFx = 0: FEF − 1800 N = 0
ΣFy = 0: FCE − 2025 N = 0
FCD = 900 N T 
FDF = 675 N C 
FEF = 1800 N T 
FCE = 2025 N T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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977
PROBLEM 6.164 (Continued)
Free body: Joint F:
ΣFy = 0:
2.25
FCF + 2025 N − 675 N = 0
3.75
FCF = −2250 N
ΣFx = −
FCF = 2250 N C 
3
(−2250 N) − 1800 N = 0 (Checks)
3.75
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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978
PROBLEM 6.165
Using the method of joints, determine the force in each member
of the roof truss shown. State whether each member is in tension
or compression.
SOLUTION
Free body: Truss:
ΣFx = 0: A x = 0
From symmetry of loading:
Ay = E =
1
2
total load
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:
ΣFx = 0:
Σ Fy = 0:
3
12
12
FBC + FBD + (6.24 kN) = 0
3.905
13
13
−
2.5
5
5
FBC + FBD + (6.24 kN) − 2.4 kN = 0
3.905
13
13
(1)
(2)
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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979
PROBLEM 6.165 (Continued)
Multiply Eq. (1) by 2.5, Eq. (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 Eq. (1) by 5, Eq. (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:
ΣFy = 0:
5
2.5
(2.50 kN) = 0
FCD −
5.831
3.905
ΣFx = 0: FCE − 5.76 kN +
FCD = 1.867 kN T 
3
3
(2.50 kN) +
(1.867 kN) = 0
3.905
5.831
FCE = 2.88 kN T
Free body: Joint E:
ΣFy = 0:
5
FDE + 3.6 kN − 1.2 kN = 0
7.81
FDE = −3.75 kN
ΣFx = 0: − FCE −
FDE = 3.75 kN C 
6
(−3.75 kN) = 0
7.81
FCE = +2.88 kN FCE = 2.88 kN T
(Checks)
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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980
PROBLEM 6.166
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,
Ay = L =
1
1
(total load) = (9.60 kips) = 4.80 kips
2
2
A = L = 4.80 kips 
We pass a section through members DF, DG, and EG, and use the free body shown.
We slide FDF to apply it at F:
ΣM G = 0: (0.8 kip)(24 ft) + (1.6 kips)(16 ft) + (1.6 kips)(8 ft)
8 FDF
− (4.8 kips)(24 ft) −
82 + 3.52
(6 ft) = 0
FDF = −10.48 kips, FDF = 10.48 kips C 
ΣM A = 0: − (1.6 kips)(8 ft) − (1.6 kips)(16 ft)
−
2.5FDG
2
8 + 2.5
2
(16 ft) −
8FDG
82 + 2.52
FDG = −3.35 kips, FDG = 3.35 kips C 
ΣM D = 0: (0.8 kips)(16 ft) + (1.6 kips)(8 ft) − (4.8 kips)(16 ft) −

(7 ft) = 0
8 FEG
82 + 1.52
(4 ft) = 0
FEG = +13.02 kips, FEG = 13.02 kips T 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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981
PROBLEM 6.167
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
(Total load)
2
1
= (9.60 kips)
2
Ax = 0,
Ay = L =
= 4.80 kips
A = L = 4.80 kips 
We pass a section through members GI, HI, and HJ, and use the free body shown.
ΣM H = 0: −
16 FGI
162 + 32
(4 ft) + (4.8 kips)(16 ft) − (0.8 kip)(16 ft) − (1.6 kips)(8 ft) = 0
FGI = +13.02 kips
FGI = 13.02 kips T 
ΣM L = 0: (1.6 kips)(8 ft) − FHI (16 ft) = 0
FHI = +0.800 kips
FHI = 0.800 kips T 
We slide FHG to apply it at H.
ΣM I = 0:
8FHJ
82 + 3.52
(4 ft) + (4.8 kips)(16 ft) − (1.6 kips)(8 ft) − (0.8 kip)(16 ft) = 0
FHJ = −13.97 kips
FHJ = 13.97 kips C 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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982
PROBLEM 6.168
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 θ = 30°,
determine (a) the force in rod CD, (b) the reaction at B.
SOLUTION
FBD:
(a)
ΣM C = 0: (15 in.)(20 lb − By ) = 0
ΣFy = 0: − 20 lb + FCD sin 30° − 20 lb = 0
(b)
B y = 20 lb
FCD = 80.0 lb T 
ΣFx = 0: (80 lb) cos30° − Bx = 0
B x = 69.282 lb
so B = 72.1 lb
16.10° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
983
PROBLEM 6.169
For the frame and loading shown, determine the components of all forces acting
on member ABC.
SOLUTION
Free body: Entire frame:
ΣFx = 0: Ax + 18 kN = 0
Ax = −18 kN
A x = 18.00 kN

ΣM E = 0: −(18 kN)(4 m) − Ay (3.6 m) = 0
Ay = −20 kN
A y = 20.0 kN 
ΣFy = 0: − 20 kN + F = 0
F = +20 kN
F = 20 kN
Free body: Member ABC
Note: BE is a two-force member, thus B is directed along line BE.
ΣM C = 0: B(4 m) − (18 kN)(6 m) + (20 kN)(3.6 m) = 0
B = 9 kN
B = 9.00 kN

Cx = 9.00 kN

ΣFx = 0: Cx − 18 kN + 9 kN = 0
Cx = 9 kN
ΣFy = 0: C y − 20 kN = 0
C y = 20 kN
C y = 20.0 kN 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
984
PROBLEM 6.170
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:
ΣM E = 0: (4.8 kN)(4.25 m) − Dx (1.5 m) = 0
Dx = +13.60 kN
D x = 13.60 kN

E x = 13.60 kN

ΣFx = 0: Ex + 13.60 kN = 0
Ex = −13.60 kN
Σ Fy = 0: Dy + E y − 4.8 kN = 0
(1)
Free body: Member ACE:
ΣM 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 kN 
Dy − 2.70 kN − 4.80 kN = 0
Dy = +7.50 kN
D y = 7.50 kN 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
985
PROBLEM 6.171
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:
ΣM 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:
ΣM F = 0: Bx (0.5 m) − (25 kN)(0.2 m) = 0
Bx = +10 kN
Free body: Member DABC:
B x = 10.00 kN
From above:

ΣM D = 0: − By (0.8 m) + (10 kN + 12 kN)(0.5 m) = 0
By = +13.75 kN
B y = 13.75 kN 
ΣFx = 0: − Dx + 10 kN + 12 kN = 0
Dx = + 22 kN
D x = 22.0 kN

ΣFy = 0: − Dy + 13.75 kN = 0
Dy = +13.75 kN
D y = 13.75 kN 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
986
PROBLEM 6.172
For the frame and loading shown, determine (a) the reaction at C, (b) the force
in member AD.
SOLUTION
Free body: Member ABC:
ΣM C = 0: + (100 lb)(45 in.) +
4
4
FAD (45 in.) + FBE (30 in.) = 0
5
5
3FAD + 2 FBF = − 375 lb
(1)
Free Body: Member DEF:
ΣM F = 0:
4
4
FAD (30 in.) + FBF (15 in.) = 0
5
5
FBE = −2FAD
(b)
Substitute from Eq. (2) into Eq. (1):
3FAD + 2(− 2FAD ) = − 375 lb
FAD = + 375 lb
From Eq. (2):
FAD = 375 lb T 
FBE = − 2 FAD = − 2(375 lb)
FBE = −750 lb
(a)
(2)
FBE = 750 lb C.
Return to free body of member ABC.
4
4
ΣFx = 0: C x + 100 lb + FAD + FBE = 0
5
5
C x + 100 +
4
4
(375) + (−750) = 0
5
5
Cx = + 200 lb Cx = 200 lb
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
987
PROBLEM 6.172 (Continued)
3
3
ΣFy = 0: C y − FAD − FBF = 0
5
5
3
3
C y − (375) − ( − 750) = 0
5
5
C y = − 225 lb C y = 225 lb
α = 48.37°
C = 301.0 lb
C = 301 lb
48.4° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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988
PROBLEM 6.173
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 β = 20°.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC:
Dimensions in mm
Since BD = 250,
θ = sin −1
33.404
; θ = 7.679°
250
Σ M C = 0: ( FBD sin θ )187.94 − ( FBD cos θ )68.404 + (100 N)328.89 = 0
FBD [187.94sin 7.679° − 68.404 cos 7.679°] = 32,889
FBD = 770.6 N
ΣFx = 0: (770.6 N) cos 7.679° = Cx = 0
Cx = + 763.7 N
Member CQ:
ΣFx = 0: Q = Cx = 763.7 N
Q = 764 N

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
989
PROBLEM 6.174
Determine the magnitude of the gripping forces exerted along
line aa on the nut when two 50-lb 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:
ΣFx = 0: Bx = 0
ΣM B = 0: (0.5 in.)Q − (1.5 in.) A = 0
A=
Q
3
ΣFy = 0: A + Q − By = 0
By = A + Q =
FBD handle ACE:
4Q
3
By symmetry and FBD jaw DE,
Q
3
Ex = Bx = 0
D= A=
E y = By =
4Q
3
ΣM C = 0: (5.25 in.)(50 lb) + (0.75 in.)
Q
4Q
− (0.75 in.)
=0
3
3
Q = 350 lb 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
990
PROBLEM 6.175
Knowing that the frame shown has a sag at B of a = 1 in., 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:
Cy =
By symmetry,
P
2
Cy
Cx
=
10 in. a
Cx =
10
10 P 5 P
Cy = ⋅ =
a
a 2
a
Free body: Member CDE:
ΣM E = 0: Cx (6 in.) − C y (20 in.) − (50 lb)(10 in.) = 0
5P
P
(b) − (20) = 500
2
a
30


P  − 10  = 500
 a

For
(1)
a = 1.0 in.
 30

P  − 10  = 500
1


20 P = 500
P = 25.0 lb 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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991
PROBLEM 6.F1
For the frame and loading shown, draw the free-body diagram(s)
needed to determine the forces acting on member ABC at B and C.
SOLUTION
We note that BD is a two-force member.
Free body: ABC


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992
PROBLEM 6.F2
For the frame and loading shown, draw the free-body diagram(s) needed to
determine all forces acting on member GBEH.
SOLUTION
We note that AG, DEF, and DH are two-force members.
Free body: ABC
Note:
Sum moments about A to determine By.
Note:
Sum moments about E to determine FDH
(which becomes zero).
Free body: DEF
Sum forces in y to determine Ey
(which also becomes zero).
Free body: GBEH
Note:
With By, Ey, and FDH previously
determined, apply three equilibrium
equations to determine three remaining
forces on GBEH.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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993
PROBLEM 6.F3
For the frame and loading shown, draw the free-body diagram(s) needed to
determine the reactions at B and F.
SOLUTION
We note that BC is two-force member.
Free body: ACF

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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994
PROBLEM 6.F4
Knowing that the surfaces at A and D are frictionless, draw the
free-body diagram(s) needed to determine the forces exerted at
B and C on member BCE.
SOLUTION
Free body: ACD
Note:
Sum moments about H to
relate Cx and Cy.
Free body: BCE
Note:
Sum moments about B to relate Cx and Cy ;
combine with relation from free body ACD
to determine Cx and Cy ; finally, sum forces
in x and y to determine Bx and By.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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995
PROBLEM 6.F5
The position of member ABC is controlled by the hydraulic
cylinder CD. Knowing that θ = 30°, draw the free-body
diagram(s) needed to determine the force exerted by the
hydraulic cylinder on pin C, and the reaction at B.
SOLUTION
We note that CD is a two-force member.
Free body: ABC
Note: To find θ, consider geometry of triangle BCD:
Law of cosines:
(CD)2 = (0.5)2 + (1.5)2 − 2(0.5)(1.5) cos 60°
CD = 1.32288 m
Law of sines:
sin q
sin 60°
=
0.5 m 1.32288 m
θ = 19.1066° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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996
PROBLEM 6.F6
Arm ABC is connected by pins to a collar at B and to crank CD at C.
Neglecting the effect of friction, draw the free-body diagram(s)
needed to determine the couple M to hold the system in equilibrium
when θ = 30°.
SOLUTION
Free body: ABC
Note:
Sum forces in x to determine Cx ; sum moments
about B to determine Cy.
Free body: CD
Note: Sum moments about D to determine M.

PROPRIETARY MATERIAL. © 2013 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
997
PROBLEM 6.F7
Since the brace shown must remain in position even when the magnitude of P is very
small, a single safety spring is attached at D and E. The spring DE has a constant of
50 lb/in. and an unstretched length of 7 in. Knowing that l = 10 in. and that the
magnitude of P is 800 lb, draw the free-body diagram(s) needed to determine the
force Q required to release the brace.
SOLUTION
Free body: ABC
Note:
Sum moments about C to relate Q to A.
Spring force.
Unstretched length:
Stretched length:
 0 = 7 in.
 = 10 in.
k = 50 lb/in.
F = kx = k ( −  0 )
= 50(10 − 7) = 150 lb
Free body: ADB
Note:
Sum moments about B to determine A; use relation from
free body ABC to determine Q.

PROPRIETARY MATERIAL. © 2013 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
998
PROBLEM 6.F8
A log weighing 800 lb is lifted by a pair of tongs as shown. Draw the
free-body diagram(s) needed to determine the forces exerted at E and
F on tong DEF.
SOLUTION
We note that AC and BD are two-force members.
Free body: AB
Note:
Free body: LOG
Sum moments about A to determine FBD.
Free body: DEF
Note: Sum moments
about E to find
Fx; sum forces in
x and y to find Ex
and Ey.
Note: Sum moments about G to
determine Fy.

PROPRIETARY MATERIAL. © 2013 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 teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
999