Using the method of joints, determine the force in each member of the truss
ic
s
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
at
Free body: Joint B:
FAB FBC 108 kN
=
=
5
4
3
FAB = 180.0 kN T 
FBC = 144.0 kN T 
st
Free body: Joint C:
FAC FBC 60 kN
=
=
13
12
5
FAC = 156.0 kN C 
FBC = 144 kN (checks)
Using the method of joints, determine the force in each member of the
ic
s
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 
at
FBC = 780 lb C 
Joint A:
ΣFy = 0: − 960 lb −
4
FAC = 0
5
FAC = 1200 lb
FAC = 1200 lb C 
st
3
ΣFx = 0: 720 lb − (1200 lb) = 0 (checks)
5
Using the method of joints, determine the force in each member of the truss
ic
s
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
at
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
st
ΣFy = 0: 84 kN −
FAB = 52.0 kN T 
5
(52 kN) − FAC = 0
13
FAC = +64.0 kN
FAC = 64.0 kN T 
Joint C:
FBC 48 kN
=
5
3
FBC = 80.0 kN C 
ic
s
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:
3
FAC + 600 lb = 0
5
at
ΣFy = 0:
FAC = −1000 lb
ΣFx = 0:
4
( −1000 lb) + 600 lb + FCD = 0
5
FAC = 1000 lb C 
FCD = 200 lb T 
From symmetry:
st
FAD = FAC = 1000 lb C , FAE = FAB = 671 lb T , FDE = FBC = 600 lb C 
ic
s
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:
at
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:
ΣFx = 0: FAB = 0
FAB = 0 
ΣFy = 0 : −1 − FAD = 0
st
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 
PROBLEM (Continued)
ic
s
Joint E:
ΣFy = 0 : FBE − 2.4 = 0
FBE = +2.4 kips
FBE = 2.40 kips T 
st
at
Truss and loading symmetrical about cL.
ic
s
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
at
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
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)
st
Joint D:
D y = 165 lb
Solving Eqs. (1) and (2) simultaneously,
FAD = −260 lb
FAD = 260 lb C 
FDC = +125 lb
FDC = 125 lb T 
PROBLEM (Continued)
Joint E:
5
4
FBE + FCE = 0
13
5
(3)
ic
s
ΣFx = 0:
ΣFy = 0:
12
3
FBE + FCE + 528 lb = 0
13
5
(4)
Solving Eqs. (3) and (4) simultaneously,
Joint C:
FBE = −832 lb
FBE = 832 lb C 
FCE = +400 lb
FCE = 400 lb T 
Force polygon is a parallelogram (see Fig. 6.11, p. 209).
FAC = 400 lb T 
at
FBC = 125.0 lb T 
ΣFx = 0:
5
4
(260 lb) + (400 lb) + FAB = 0
13
5
FAB = −420 lb
ΣFy = 0:
12
3
(260 lb) − (400 lb) = 0
13
5
0 = 0 (Checks)
st
Joint A:
FAB = 420 lb C 
Using the method of joints, determine the force in each member of the truss shown.
ic
s
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
at
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:
st
ΣFx = 0: 5 kN +
1
5
FBD = 0
FBD = −5 5 kN
ΣFy = 0: 20 kN − FBC −

2
5
FBD = 11.18 kN C 
(−5 5 kN) = 0
FBC = +30 kN
FBC = 30.0 kN T 
PROBLEM(Continued)
ic
s
Free body: Joint C:
ΣFx = 0: FCD − 10 kN = 0
FCD = +10 kN
FCD = 10.00 kN T 
st
at
ΣFy = 0: 30 kN − 30 kN = 0 (checks)
PROBLEM 6.8
ic
s
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 
at
FDE = 4.00 kN C 
Joint B:
ΣFx = 0:
4
(5 kN) − FAB = 0
5
FAB = +4 kN
FAB = 4.00 kN T 
3
ΣFy = 0: − 6 kN − (5 kN) − FBD = 0
5
FBD = −9 kN
FBD = 9.00 kN C 
Joint D:
st
ΣFy = 0: − 9 kN +
3
FAD = 0
5
FAD = +15 kN
FAD = 15.00 kN T 
4
ΣFx = 0: − 4 kN − (15 kN) − FCD = 0
5
FCD = −16 kN
FCD = 16.00 kN C 
ic
s
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 FAC 900 lb
=
=
5
4
3
FAB = 1500 lb C 
at
FAC = 1200 lb T 
Free body: Joint C:
BC is a zero-force member.
FBC = 0
FCE = 1200 lb T 
Free body: Joint B:
24
4
4
FBD + FBE + (1500 lb) = 0
25
5
5
st
ΣFx = 0:
or
ΣFy = 0:
or
24 FBD + 20 FBE = −30, 000 lb
(1)
7
3
3
FBD − FBE + (1500) − 600 = 0
25
5
5
7 FBD − 15FBE = −7,500 lb
(2)
PROBLEM 6.9 (Continued)
Multiply Eq. (1) by 3, Eq. (2) by 4, and add:
FBD = 1200 lb C 
ic
s
100 FBD = −120, 000 lb
Multiply Eq. (1) by 7, Eq. (2) by –24, and add:
500 FBE = −30,000 lb
FBE = 60.0 lb C 
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
at
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 
st
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,
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.
755
ic
s
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:
at
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 
st
FAC = 1200 lb T 
Free body: Joint C:
BC is a zero-force member.
FBC = 0
FCE = 1200 lb T 
PROBLEM 6.10 (Continued)
Free body: Joint B:
4
4
4
FBD + FBC + (1500 lb) = 0
5
5
5
ic
s
ΣFx = 0:
or
FBD + FBE = −1500 lb
ΣFy = 0:
or
(1)
3
3
3
FBD − FBE + (1500 lb) − 600 lb = 0
5
5
5
FBD − FBE = −500 lb
(2)
Add Eqs. (1) and (2):
2 FBD = −2000 lb
FBD = 1000 lb C 
Subtract Eq. (2) from Eq. (1):
2 FBE = −1000 lb
FBE = 500 lb C 
Free Body: Joint D:
ΣFx = 0:
4
4
(1000 lb) + FDF = 0
5
5
FDF = −1000 lb
3
3
(1000 lb) − ( −1000 lb) − 600 lb − FDE = 0
5
5
at
ΣFy = 0:
FDF = 1000 lb C 
FDE = +600 lb
FDE = 600 lb T 
st
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE
FEF = 500 lb C 
FEG = FCE
FEG = 1200 lb T 
FFG = FBC
FFG = 0 
FFH = FAB
FFH = 1500 lb C 
FGH = FAC
FGH = 1200 lb T 
ic
s
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,
1
total load = 21 kN
2
at
Ay = H =
Free body: Joint A:
FAB FAC 15.3 kN
=
=
37
35
12
FAB = 47.175 kN FAC = 44.625 kN
FAB = 47.2 kN C 
FAC = 44.6 kN T 
st
Free body: Joint B:
From force polygon:
FBD = 47.175 kN, FBC = 10.5 kN
FBC = 10.50 kN C 
FBD = 47.2 kN C 
PROBLEM 6.11 (Continued)
Free body: Joint C:
ΣFy = 0:
3
FCD − 10.5 = 0
5
4
(17.50) − 44.625 = 0
5
ic
s
ΣFx = 0: FCE +
FCD = 17.50 kN T 
FCE = 30.625 kN
Free body: Joint E:
DE is a zero-force member.
FCE = 30.6 kN T 
FDE = 0 
st
at
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,
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.
759
Determine the force in each member of the Fink roof truss shown.
ic
s
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
at
FAB = 11.08 kN C 
FAC = 10.125 kN
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):
st
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 
PROBLEM (Continued)
Free body: Joint C:
ΣFy = 0:
4
4
FCD − (2.8125 kN) = 0
5
5
FCD = 2.81 kN T 
ic
s
FCD = 2.8125 kN,
3
3
ΣFx = 0: FCE − 10.125 kN + (2.8125 kN) + (2.8125 kN) = 0
5
5
FCE = +6.7500 kN
FCE = 6.75 kN T 
FCD = 2.81 kN T 
FDF = FBD
FDF = 9.23 kN C 
FEF = FBC
FEF = 2.81 kN C 
FEG = FAC
FEG = 10.13 kN T 
FFG = FAB
FFG = 11.08 kN C 
st
at
Because of the symmetry of the truss and loading, we deduce that
FDE = FCD
PROBLEM 6.13
ic
s
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
at
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
st
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 
PROBLEM 6.13 (Continued)
Free body: Joint B:
2
5
FBD +
2
5
(7.8262 kN) +
1
2
FBC = 0
ic
s
ΣFx = 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 = −6.3355 kN
FBD = 6.34 kN C 
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 = +1.4911 kN
at
1
FCD = 1.491 kN T 
ΣFx = 0: FCE − 7.00 kN +
FCE
(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
1
5
FDF −
1
2
st
ΣFy = 0:
or
FDE +
1
5
(6.3355 kN) −
FDF − 0.79057 FDE = −1.1188 kN
Add Eqs. (1) and (2):
Subtract Eq. (2) from Eq. (1):
2
5
(1)
(1.4911 kN) − 2 kN = 0
(2)
2 FDF = −6.7088 kN
FDF = −3.3544 kN
FDF = 3.35 kN C 
1.58114 FDE = −4.4712 kN
FDE = −2.8278 kN
FDE = 2.83 kN C 
PROBLEM 6.13 (Continued)
Free body: Joint F:
1
FFG +
2
2
(3.3544 kN) = 0
5
ic
s
ΣFx = 0:
FFG = −4.243 kN
ΣFy = 0: −FEF − 1.75 kN +
1
5
FFG = 4.24 kN C 
(3.3544 kN) −
1
2
FEF = 2.750 kN
(−4.243 kN) = 0
FEF = 2.75 kN T 
Free body: Joint G:
ΣFx = 0:
or
1
2
1
2
FEG +
1
2
(4.243 kN) = 0
FGH − FEG = −4.243 kN
ΣFy = 0: −
1
2
FGH −
1
2
FEG −
1
2
(1)
(4.243 kN) − 1.5 kN = 0
FGH + FEG = −6.364 kN
(2)
at
or
FGH −
Add Eqs. (1) and (2):
2 FGH = −10.607
FGH = −5.303
Subtract Eq. (1) from Eq. (2):
FGH = 5.30 kN C 
2 FEG = −2.121 kN
FEG = −1.0605 kN
FEG = 1.061 kN C 
st
Free body: Joint H:
FEH 3.75 kN
=
1
1
We can also write
FGH
2
=
3.75 kN
1
FEH = 3.75 kN T 
FGH = 5.30 kN C (Checks)
PROBLEM 6.14
ic
s
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
at
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:
st
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
FCF = 1600 N T 
ΣFy = 0: FEF − 2250 N = 0
FEF = +2250 N
FEF = 2250 N T 
PROBLEM 6.14 (Continued)
Joint C:
ΣFx = 0:
8
FCE − 1200 N + 1600 N = 0
17
ΣFy = 0: FBC +
15
FCE = 0
17
FBC = −
15
15
FCE = − ( −850 N)
17
17
FBC = +750 N
ΣFx = 0: − FBE − 800 N +
FBC = 750 N T 
8
(850 N) = 0
17
FBE = −400 N
ΣFy = 0: 1500 N − 2250 N +
FBE = 400 N C 
15
(850 N) = 0
17
0 = 0 (checks)
st
at
Joint E:
FCE = 850 N C 
ic
s
FCE = −850 N
PROBLEM 6.23
ic
s
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 
at
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 
st
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 
PROBLEM 6.23 (Continued)
Free body: Joint C:
2.21
(2.29 kN) = 0
2.29
ic
s
ΣFx = 0: FCE +
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 C 
st
at
FEJ = −1.200 kN
0.6
(2.29 kN) − 0 = 0
2.29
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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783
PROBLEM 6.24
ic
s
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:
at
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
st
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
PROBLEM 6.24 (Continued)
Free body: Joint I:
2.97
(3.03 kN) = 0
3.03
ic
s
ΣFx = 0: FIK +
FIK = −2.97 kN
ΣFy = 0: − FIN − 1.800 kN −
FIN = −2.40 kN
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
FKN = 0 
ΣFy = 0: − FKO −
0.6
3
(3.03 kN) − 1.800 kN − (0) = 0
3.03
5
FKO = −2.40 kN
at
st
FIK = 2.97 kN C 
FKO = 2.40 kN C 
PROBLEM 6.25
Solve Problem 6.23 assuming that the cables hanging from the
right side of the tower have fallen to the ground.
ic
s
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 
at
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
st
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 
PROBLEM 6.25 (Continued)
Free body: Joint C:
2.21
(2.29 kN) = 0
2.29
ic
s
ΣFx = 0: FCE +
FCE = 2.21 kN C 
ΣFy = 0: − FCH − 2.2575 kN −
FCH = −2.8575 kN
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
st
at
FEJ = +1.6575 kN
FEJ = 1.658 kN T 