1. No category

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.
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
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
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.
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,
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.
833