Assignment 21

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 −
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
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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:
ΣFx = 0: FAB = 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
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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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.
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.
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
FAD = 260 lb C
FDC = +125 lb
FDC = 125 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.
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
FBE = 832 lb C
FCE = +400 lb
FCE = 400 lb T
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)
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.
750

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