PROBLEM 6.1
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 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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737
PROBLEM 6.2
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: Entire truss
ΣFx = 0: Bx = 0
ΣM B = 0: C (15.75 ft) − (945 lb)(12 ft) = 0
C = 720 lb
ΣFy = 0: By + 720 lb − 945 lb = 0
B y = 225 lb
Free body: Joint B:
FAB FBC 225 lb
=
=
5
4
3
FAB = 375 lb C
FBC = 300 lb T
Free body: Joint C:
FAC FBC 720 lb
=
=
9.75 3.75
9
FBC = 300 lb T
FAC = 780 lb C
(Checks)
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
738
!
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
Free body: Entire truss
ΣFx = 0: C x = 0 C x = 0
ΣM B = 0: (1.92 kN)(3 m) + C y (4.5 m) = 0
C y = −1.28 kN C y = 1.28 kN
ΣFy = 0: B − 1.92 kN − 1.28 kN = 0
B = 3.20 kN
Free body: Joint B:
FAB FBC 3.20 kN
=
=
5
3
4
FAB = 4.00 kN C
FBC = 2.40 kN C
Free body: Joint C:
ΣFx = 0: −
7.5
FAC + 2.40 kN = 0
8.5
FAC = +2.72 kN
!
! ΣFy =
FAC = 2.72 kN T
!
4
(2.72 kN) − 1.28 kN = 0 (Checks)
8.5
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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you are using it without permission.
739
PROBLEM 6.4
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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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740
PROBLEM 6.4 (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. © 2010 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.
741
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
Free body: Truss
ΣFx = 0: A x = 0
ΣM A = 0: D (22.5) − (10.8 kips)(22.5) − (10.8 kips)(57.5) = 0
D = 38.4 kips
ΣFy = 0: A y = 16.8 kips
Free body: Joint A:
FAB
F
16.8 kips
= AD =
22.5 25.5
12
FAB = 31.5 kips T
FAD = 35.7 kips C
Free body: Joint B:
ΣFx = 0:
FBC = 31.5 kips T
ΣFy = 0:
FBD = 10.80 kips C
Free body: Joint C:
FCD FBC 10.8 kips
=
=
37
35
12
FBC = 31.5 kips T
FCD = 33.3 kips C
(Checks)
PROPRIETARY MATERIAL. © 2010 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.
742
!
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
Free Body: Truss
ΣM E = 0: F (3 m) − (900 N)(2.25 m) − (900 N)(4.5 m) = 0
F = 2025 N
ΣFx = 0: Ex + 900 N + 900 N = 0
Ex = −1800 N E x = 1800 N
ΣFy = 0: E y + 2025 N = 0
E y = −2025 N E y = 2025 N
We note that AB and BD are zero-force members:
FAB = FBD = 0
Free body: Joint A:
FAC FAD 900 N
=
=
2.25 3.75
3
FAC = 675 N T
FAD = 1125 N C
Free body: Joint D:
FCD FDE 1125 N
=
=
3
2.23
3.75
FCD = 900 N T
FDF = 675 N C
Free body: Joint E:
ΣFx = 0: FEF − 1800 N = 0
FEF = 1800 N T
ΣFy = 0: FCE − 2025 N = 0
FCE = 2025 N T
PROPRIETARY MATERIAL. © 2010 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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743
PROBLEM 6.6 (Continued)
Free body: Joint F:
ΣFy = 0:
2.25
FCF + 2025 N − 675 N = 0
3.75
FCF = −2250 N
ΣFx = −
FCF = 2250 N C
3
( −2250 N) − 1800 N = 0 (Checks)
3.75
PROPRIETARY MATERIAL. © 2010 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.
744
PROBLEM 6.7
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
ΣFy = 0: B y = 0
ΣM B = 0: D(4.5 m) + (8.4 kN)(4.5 m) = 0
D = −8.4 kN
D = 8.4 kN
ΣFx = 0: Bx − 8.4 kN − 8.4 kN − 8.4 kN = 0
Bx = +25.2 kN B x = 25.2 kN
Free body: Joint A:
FAB FAC 8.4 kN
=
=
5.3
4.5
2.8
FAB = 15.90 kN C
FAC = 13.50 kN T
Free body: Joint C:
ΣFy = 0: 13.50 kN −
4.5
FCD = 0
5.3
FCD = +15.90 kN
ΣFx = 0: − FBC − 8.4 kN −
FCD = 15.90 kN T
2.8
(15.90 kN) = 0
5.3
FBC = −16.80 kN
FBC = 16.80 kN C
PROPRIETARY MATERIAL. © 2010 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.
745
PROBLEM 6.7 (Continued)
Free body: Joint D:
FBD 8.4 kN
=
4.5
2.8
FBD = 13.50 kN C
We can also write the proportion
FBD 15.90 kN
=
4.5
5.3
FBD = 13.50 kN C (Checks)
PROPRIETARY MATERIAL. © 2010 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.
746