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 FAC 900 lb
=
=
5
4
3
FAB = 1500 lb C
FAC = 1200 lb T
Free body: Joint C:
BC is a zero-force member.
FBC = 0
FCE = 1200 lb T
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:
100 FBD = −120, 000 lb
FBD = 1200 lb C
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
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.
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755
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 kN
FCD = 20.0 kN C
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
FDF
12
= −48 kN
13
FDF = −52.0 kN
FDF = 52.0 kN C
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829
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
182 + 4.52
FDF (4.5 ft) = 0
FDF = −3711 lb
FDF = 3710 lb C
ΣM A = 0: FEF (18 ft) − (400 lb)(6 ft) − (400 lb)(12 ft) = 0
FEF = +400 lb
FEF = 400 lb T
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
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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833