Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
PROBLEM 6.30
Determine the force in members EG and EF of the truss
shown when P = 20 kips.
SOLUTION
Reactions:
2.5
7.5
tan 6
51.34
P
a
a
= =
=
= °
CK
Member EG:
0: (18 ft) 2.5 (12 ft) (6 ft) (7.5 ft) 0
F EG
M P P PFΣ= − − + =
0.8 ;
EG
FP= +
For
20 kips,P=
0.8(20) 16 kips
EG
F= = +
16.00 kips
EG
FT=
Member EF:
0: 2.5 (6 ft) (12 ft) sin51.34 (12 ft) 0
A EF
M PP FΣ= − + ° =
0.320 ;
EF
FP= −
For
20 kips,P=
0.320(20) 6.4 kips
EF
F=−=−
6.40 kips
EF
FC=
consent of McGraw-Hill Education.
PROBLEM 6.31
Determine the force in members DF and DE of the truss shown.
SOLUTION
1
1
tan 7.1
8
a
−
= =
Member CE:
0: (1.75 m) (30 kN)(4 m) (20 kN)(2 m) 0
F DF
MFΣ=− + + =
91.4 kN
DF
F= +
91.4 kN
DF
FT=
Member EF:
0: (14 m) (30 kN)(10 m) 20 kN)(12 m) 0
O DE
MFΣ= − − − =
38.6 kN
DE
F= −
38.6 kN
DE
FC=
consent of McGraw-Hill Education.
PROBLEM 6.32
Determine the force in members CD and CE of the truss shown.
SOLUTION
1
1
1
tan 7.1
8
1.5
tan 36.9
2
a
β
−
−
= =
= =
Member CE:
( )
0: sin 36.9 (14 m) (30 kN)(10 m) (20 kN)(12 m)0
O CD
MFΣ= − − =
64.2 kN
CD
F= +
64.2 kN
CD
FT
=
Member EF:
( )
0: cos 7.1 (1.75 m) (30 kN)(4 m) (20 kN)(2 m) 0
O CE
MFΣ= + + =
92.1 kN
CE
F= −
92.1 kN
CE
FC=
consent of McGraw-Hill Education.
PROBLEM 6.33
A monosloped roof truss is loaded as shown. Determine the
force in members CE, DE, and DF.
SOLUTION
PROBLEM 6.34
A monosloped roof truss is loaded as shown. Determine the
force in members EG, GH, and HJ.
SOLUTION
Reactions at supports: Because of the symmetry of the loading,
0
11
(Total load) (8 kN)
22
x
y
A
AI
=
= = =
4 kN= =AI
We pass a section through members EG, GH, and HJ, and use the free body shown.
0: (4 kN)(2.4 m) (1kN)(2.4 m) (2.08 m) 0
H EG
MFΣ= − − =
3.4615 kN
EG
F= +
3.46 kN
EG
FT=
0: (2.4 m) (2.62 m) 0
J GH EG
MF F
Σ= − − =
2.62 (3.4615 kN)
2.4
GH
F= −
3.7788 kN
GH
F= −
3.78 kN
GH
FC
=
2.4
0: 0
2.46
x EG HJ
FF FΣ= − − =
2.46 2.46 (3.4615 kN)
2.4 2.4
HJ EG
FF=−=−
3.548 kN
HJ
F= −
3.55 kN
HJ
FC=
consent of McGraw-Hill Education.
PROBLEM 6.35
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.
40
0: (6.3 ft) (1.8 kips)(14 ft) (0.9 kips)(28 ft) 0
41
G AB
MF
Σ= − − =
8.20 kips
AB
F= +
8.20 kips
AB
FT=
3
0: (28 ft) (1.8 kips)(28 ft) (1.8 kips)(14 ft) 0
5
D AG
MF
Σ=− + + =
4.50 kips
AG
F= +
4.50 kips
AG
FT=
0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0
A FG
MF=−−−− =
11.60 kips
FG
F= −
11.60 kips
FG
FC=
consent of McGraw-Hill Education.
PROBLEM 6.36
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.
22
8
0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0
89
F AE
MF
Σ= −−− =
+
17.46 kips
AE
F= +
17.46 kips
AE
FT=
0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0
A EF
MFΣ=−−−− =
11.60 kips
EF
F= −
11.60 kips
EF
FC
=
0: (8 ft) (0.9 kips)(8 ft) (1.8 kips)(20 ft) (1.8 kips)(34 ft) (0.9 kips)(48 ft) 0
E FJ
MFΣ=−−−−− =
18.45 kips
FJ
F= −
18.45 kips
FJ
FC=
consent of McGraw-Hill Education.
PROBLEM 6.37
Determine the force in members DF, EF, and EG of the
truss shown.
SOLUTION
3
tan 4
β
=
Reactions:
0= =AN
Member DF:
3
0: (16 kN)(6 m) (4 m) 0
5
E DF
MFΣ= + − =
40 kN
DF
F= +
40.0 kN
DF
FT=
Member EF: +
0: (16 kN)sin cos 0
EF
FF
ββ
Σ= − =
16tan 16(0.75) 12 kN
EF
F
β
= = =
12.00 kN
EF
FT=
Member EG:
4
0: (16 kN)(9 m) (3 m) 0
5
F EG
MFΣ= + =
60 kN
EG
F= −
60.0 kN
EG
FC=
consent of McGraw-Hill Education.
PROBLEM 6.38
Determine the force in members GI, GJ, and HI of the
truss shown.
SOLUTION
Reactions:
0= =AN
Member GI: +
0: (16 kN)sin sin 0
GI
FF
ββ
Σ= + =
16 kN
GI
F= −
16.00 kN
GI
FC
=
Member GJ:
4
0: (16 kN)(9 m) (3 m) 0
5
I GJ
MFΣ= − − =
60 kN
GJ
F= −
60.0 kN
GJ
FC=
Member HI:
3
0: (16 kN)(9 m) (4 m) 0
5
G HI
MFΣ=− + =
60 kN
HI
F= +
60.0 kN
HI
FT=
consent of McGraw-Hill Education.
PROBLEM 6.30
Determine the force in members EG and EF of the truss
shown when P = 20 kips.
SOLUTION
Reactions:
2.5
7.5
tan 6
51.34
P
a
a
= =
=
= °
CK
Member EG:
0: (18 ft) 2.5 (12 ft) (6 ft) (7.5 ft) 0
F EG
M P P PFΣ= − − + =
0.8 ;
EG
FP= +
For
20 kips,P=
0.8(20) 16 kips
EG
F= = +
16.00 kips
EG
FT=
Member EF:
0: 2.5 (6 ft) (12 ft) sin51.34 (12 ft) 0
A EF
M PP FΣ= − + ° =
0.320 ;
EF
FP= −
For
20 kips,P=
0.320(20) 6.4 kips
EF
F=−=−
6.40 kips
EF
FC=
consent of McGraw-Hill Education.
PROBLEM 6.31
Determine the force in members DF and DE of the truss shown.
SOLUTION
1
1
tan 7.1
8
a
−
= =
Member CE:
0: (1.75 m) (30 kN)(4 m) (20 kN)(2 m) 0
F DF
MFΣ=− + + =
91.4 kN
DF
F= +
91.4 kN
DF
FT=
Member EF:
0: (14 m) (30 kN)(10 m) 20 kN)(12 m) 0
O DE
MFΣ= − − − =
38.6 kN
DE
F= −
38.6 kN
DE
FC=
consent of McGraw-Hill Education.
PROBLEM 6.32
Determine the force in members CD and CE of the truss shown.
SOLUTION
1
1
1
tan 7.1
8
1.5
tan 36.9
2
a
β
−
−
= =
= =
Member CE:
( )
0: sin 36.9 (14 m) (30 kN)(10 m) (20 kN)(12 m)0
O CD
MFΣ= − − =
64.2 kN
CD
F= +
64.2 kN
CD
FT
=
Member EF:
( )
0: cos 7.1 (1.75 m) (30 kN)(4 m) (20 kN)(2 m) 0
O CE
MFΣ= + + =
92.1 kN
CE
F= −
92.1 kN
CE
FC=
consent of McGraw-Hill Education.
PROBLEM 6.33
A monosloped roof truss is loaded as shown. Determine the
force in members CE, DE, and DF.
SOLUTION
PROBLEM 6.34
A monosloped roof truss is loaded as shown. Determine the
force in members EG, GH, and HJ.
SOLUTION
Reactions at supports: Because of the symmetry of the loading,
0
11
(Total load) (8 kN)
22
x
y
A
AI
=
= = =
4 kN= =AI
We pass a section through members EG, GH, and HJ, and use the free body shown.
0: (4 kN)(2.4 m) (1kN)(2.4 m) (2.08 m) 0
H EG
MFΣ= − − =
3.4615 kN
EG
F= +
3.46 kN
EG
FT=
0: (2.4 m) (2.62 m) 0
J GH EG
MF F
Σ= − − =
2.62 (3.4615 kN)
2.4
GH
F= −
3.7788 kN
GH
F= −
3.78 kN
GH
FC
=
2.4
0: 0
2.46
x EG HJ
FF FΣ= − − =
2.46 2.46 (3.4615 kN)
2.4 2.4
HJ EG
FF=−=−
3.548 kN
HJ
F= −
3.55 kN
HJ
FC=
consent of McGraw-Hill Education.
PROBLEM 6.35
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.
40
0: (6.3 ft) (1.8 kips)(14 ft) (0.9 kips)(28 ft) 0
41
G AB
MF
Σ= − − =
8.20 kips
AB
F= +
8.20 kips
AB
FT=
3
0: (28 ft) (1.8 kips)(28 ft) (1.8 kips)(14 ft) 0
5
D AG
MF
Σ=− + + =
4.50 kips
AG
F= +
4.50 kips
AG
FT=
0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0
A FG
MF=−−−− =
11.60 kips
FG
F= −
11.60 kips
FG
FC=
consent of McGraw-Hill Education.
PROBLEM 6.36
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.
22
8
0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0
89
F AE
MF
Σ= −−− =
+
17.46 kips
AE
F= +
17.46 kips
AE
FT=
0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0
A EF
MFΣ=−−−− =
11.60 kips
EF
F= −
11.60 kips
EF
FC
=
0: (8 ft) (0.9 kips)(8 ft) (1.8 kips)(20 ft) (1.8 kips)(34 ft) (0.9 kips)(48 ft) 0
E FJ
MFΣ=−−−−− =
18.45 kips
FJ
F= −
18.45 kips
FJ
FC=
consent of McGraw-Hill Education.
PROBLEM 6.37
Determine the force in members DF, EF, and EG of the
truss shown.
SOLUTION
3
tan 4
β
=
Reactions:
0= =AN
Member DF:
3
0: (16 kN)(6 m) (4 m) 0
5
E DF
MFΣ= + − =
40 kN
DF
F= +
40.0 kN
DF
FT=
Member EF: +
0: (16 kN)sin cos 0
EF
FF
ββ
Σ= − =
16tan 16(0.75) 12 kN
EF
F
β
= = =
12.00 kN
EF
FT=
Member EG:
4
0: (16 kN)(9 m) (3 m) 0
5
F EG
MFΣ= + =
60 kN
EG
F= −
60.0 kN
EG
FC=
consent of McGraw-Hill Education.
PROBLEM 6.38
Determine the force in members GI, GJ, and HI of the
truss shown.
SOLUTION
Reactions:
0= =AN
Member GI: +
0: (16 kN)sin sin 0
GI
FF
ββ
Σ= + =
16 kN
GI
F= −
16.00 kN
GI
FC
=
Member GJ:
4
0: (16 kN)(9 m) (3 m) 0
5
I GJ
MFΣ= − − =
60 kN
GJ
F= −
60.0 kN
GJ
FC=
Member HI:
3
0: (16 kN)(9 m) (4 m) 0
5
G HI
MFΣ=− + =
60 kN
HI
F= +
60.0 kN
HI
FT=
consent of McGraw-Hill Education.
