© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

This material is protected under all copyright laws as they currently

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3–5. A sign is subjected to a wind loading that exerts C

horizontal forces of 300 lb on joints B and C of one of the 300 lb
side supporting trusses. Determine the force in each
member of the truss and state if the members are in tension
or compression. 12 ft
13 ft
Joint C: Fig a.
D
5 ft
: a Fx = 0; 300 - FCD a b = 0 FCD = 780 lb (C)
+ 5
Ans.
13 300 lb

+ c a Fy = 0; 780 a b - FCB = 0 FCB = 720 lb (T)

12 B 12 ft
Ans.
13
13 ft
45
A E
Joint D: Fig. b.

+ Q a Fx = 0; FDB = 0 Ans.

+ a a Fy = 0; FDE - 780 = 0 FDE = 780 lb (C) Ans.

Joint B: Fig. c.
+
: a Fx = 0; 300 + FBE sin 45.24° - FBA cos 45° = 0

+ c a Fy = 0; 720 - FBE cos 45.24° - FBA sin 45° = 0

Solving

FBE = 296.99 lb = 297 lb (T) FBA = 722.49 lb (T) = 722 lb (T) Ans.

50
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3–6. Determine the force in each member of the truss. 1.5 k

Indicate if the members are in tension or compression.
Assume all members are pin connected. 2k F

2k G
8 ft

Support Reactions. Referring to the FBD of the entire truss, Fig. a H

E
a + a MD = 0; 2(8) + 2(16) - Ay(24) = 0 Ay = 2.0 k 4 ft
A

a Fx = 0;
+
: Ax = 0
B C D

Method of Joint.
8 ft 8 ft 8 ft
Joint A: Fig. b,

2.0 - FAH a b = 0
1
+ c a Fy = 0; FAH = 4.472 k (C) = 4.47k (C) Ans.
25

b = 0
2
a Fx = 0;
+
: FAB - 4.472a FAB = 4.00 k (T) Ans.
25
Joint B: Fig. c,

a Fx = 0;
+
: FBC - 4.00 = 0 FBC = 4.00 k (T) Ans.

+ c a Fy = 0; FBH = 0 Ans.

Joint H: Fig. d,
+ c a Fy = 0; FHC sin 53.13° - 2 sin 63.43° = 0 FHC = 2.236 k (C) = 2.24 k (C) Ans.

a Fx = 0;
+
: 4.472 - 2 cos 63.43° - 2.236 cos 53.13° - FHG = 0

FHG = 2.236 k (C) = 2.24 k (C) Ans.

Joint F: Fig. e,

a Fx = 0;
+
: FFG = 0 Ans.

+ c a Fy = 0; FFE - 1.5 = 0 FFE = 1.5 k (C) Ans.

Joint G: Fig. f,

2.236 a b - FGE = a b = 0
2 2
a Fx = 0;
+
: FGE = 2.236 k (C) = 2.24 k (C) Ans.
25 25

2.236 a b + 2.236 a b - 2 - FGC = 0

1 1
+ c a Fy = 0; FGC = 0 Ans.
25 25
Joint E: Fig. g,
2.236 a b - FEC a b = 0
2 2
a Fx = 0;
+
: FEC = 2.236 k (T) = 2.24 k (T) Ans.
25 25

FED = 2.236 a b - 2.236 a b - 1.5 = 0

1 1
+ c a Fy = 0; FED = 3.5 k (C) Ans.
25 25

Joint D: Fig. h,

a Fx = 0;
+
: FDC = 0 Ans.

51
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3–6. Continued

52
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3–7. Determine the force in each member of the truss. 4m

State whether the members are in tension or compression. B C
Set P = 8 kN.

Method of Joints: In this case, the support reactions are not required for 60 E 60
A D
determining the member forces.

4m 4m
Joint D:
P
+ c a Fy = 0; FDC sin 60° - 8 = 0

FDC = 9.238 kN (T) = 9.24 kN (T) Ans.

a Fx = 0;
+
: FDE - 9.238 cos 60° = 0

FDE = 4.619 kN (C) = 4.62 kN (C) Ans.

Joint C:

+ c a Fy = 0; FCE sin 60° - 9.328 sin 60° = 0

FCE = 9.238 kN (C) = 9.24 kN (C) Ans.

a Fx = 0;
+
: 2(9.238 cos 60°) - FCB = 0

FCB = 9.238 kN (T) = 9.24 kN (T) Ans.

Joint B:

+ c a Fy = 0; FBE sin 60° - FBA sin 60° = 0

FBE = FBA = F

a Fx = 0;
+
: 9.238 - 2Fcos 60° = 0
F = 9.238 kN
Thus, FBE = 9.24 kN (C) FBA = 9.24 kN (T) Ans.

Joint E:

+ c a Fy = 0; Ey - 2(9.238 sin 60°) = 0 Ey = 16.0 kN

a Fx = 0;
+
: FBA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0
FEA = 4.62 kN (C) Ans.

Note: The support reactions Ax and Ay can be determined by analyzing Joint A

using the results obtained above.

53
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*3–8. If the maximum force that any member can support 4m

is 8 kN in tension and 6 kN in compression, determine the B C
maximum force P that can be supported at joint D.

60 E 60
A D

4m 4m
P

Method of Joints: In this case, the support reactions are not required for
determining the member forces.

Joint D:

+ c a Fy = 0; FDC sin 60° - P = 0 FDC = 1.1547P (T)

a Fx = 0;
+
: FDE - 1.1547P cos 60° = 0 FDE = 0.57735P (C)

Joint C:

+ c a Fy = 0; FCE sin 60° - 1.1547P sin 60° = 0

FCE = 1.1547P (C)

a Fx = 0;
+
: 2(1.1547P cos 60° - FCB = 0 FCB = 1.1547P (T)

Joint B:

+ c a Fy = 0; FBE sin 60° - FBE sin 60° = 0 FBE = FBA = F

a Fx = 0;
+
: 1.1547P - 2F cos 60° = 0 F = 1.1547P

Thus, FBE = 1.1547P (C) FBA = 1.1547P (T)

Joint E:

a Fx = 0;
+
: FEA + 1.1547P cos 60° - 1.1547P cos 60° - 0.57735P = 0
FEA = 0.57735P (C)

From the above analysis, the maximum compression and tension in the truss
members is 1.1547P. For this case, compression controls which requires

1.1547P = 6
P = 5.20 kN Ans.

54
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3–9. Determine the force in each member of the truss. 4k 4k

State if the members are in tension or compression.
F E

9 ft 2k
Reactions: D
By = 9.00 k, Dx = 0, Dy = 1.00 k A B C
30
12 ft 12 ft 12 ft
Joint A:
3
+ c a Fy = 0; (F ) - 2 = 0
5 AF
FAF = 3.333 k = 3.33 k (T) Ans.
4
a Fx = 0;
+
: -FAB + (3.333) = 0
5
FAB = 2.667 k = 2.67 k (C) Ans.

Joint B:

+ c a Fy = 0; 9.00 - (FBF) = 0

FBF = 9.00 k (C) Ans.

a Fx = 0;
+
: 2.667 - FBC = 0

FBC = 2.667 k = 2.67 k (C) Ans.

Joint F:
3 3
+ c a Fy = 0; - (F ) - 4 - (3.333) + 9 = 0
5 FC 5
FFC = 5.00 k (T) Ans.
4 4
a Fx = 0;
+
: -FFE - (3.333) + (5.00) = 0
5 5
FFE = 1.333 k = 1.33 k (C) Ans.

Joint C:
3
+ c a Fy = 0; -FCE + (5.00) = 0
5
FCE = 3.00 k (C) Ans.
4
a Fx = 0; FCD + (2.667) - 5 (5.00) = 0
+
:

FCD = 1.333 k = 1.33 k (T) Ans.

Joint D:
3
+ c a Fy = 0; - (F ) + 1 = 0
5 DE
FDE = 1.667 k = 1.67 k (C) Ans.
4
a Fx = 0;
+
: (1.667) - 1.333 = 0 (Check)
5

55
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3–10. Determine the force in each member of the truss. 3k

2k H
State if the members are in tension or comprehension.
3k
G
12 ft
F
Reactions:
A y = 1.65 k, Ex = 2.00 k, Ey = 4.35 k A
E
B C D
Joint E:
10 ft 10 ft 10 ft 10 ft
+ c a Fy = 0; -(FEF) sin 21.80° + 4.35 = 0
FEF = 11.71 k = 11.7 k (C) Ans.
a Fx = 0;
+
: -FED - 2 + 11.71 cos 21.80° = 0

FED = 8.875 k (T) Ans.

Joint D:

+ c a Fy = 0; FDF = 0 Ans.

a Fx = 0;
+
: -FDC + 8.875 = 0
FDC = 8.875 k (T) Ans.

Joint A:

+ c a Fy = 0; -FAH sin 50.19° + 1.65 = 0

FAH = 2.148 k = 2.15 k (C) Ans.

a Fx = 0; FAB - 2.148 (cos 50.19°) = 0

+
:
FAB = 1.375 k (T) Ans.

Joint B:
+ c a Fy = 0; FBH = 0 Ans.

a Fx = 0;
+
: FBC - 1.375 = 0
FBC = 1.375 k (T) Ans.

Joint F:
+ Q a Fy = 0; FFC cos 46.40° - 3 cos 21.80° = 0
FFC = 4.039 k = 4.04 k (C) Ans.
+ R a Fx = 0; FFG + 3 sin 21.80° + 4.039 sin 46.40° - 11.71 = 0
FFG = 7.671 k = 7.67 k (C) Ans.

Joint G:
+ Q a Fy = 0; FGC cos 21.80° - 3 cos 21.80° = 0 FGC = 3.00 k (C) Ans.

+ R a Fx = 0; FGH + 3 sin 21.80° - 3 sin 21.80° - 7.671 = 0;

FGH = 7.671 k = 7.67 k (C) Ans.

Joint C:
+ c a Fy = 0; FCH sin 50.19° - 3.00 - 4.039 sin 21.80° = 0
FCH = 5.858 k = 5.86 k (T) Ans.

a Fx = 0;
+
: -4.039 cos 21.80° - 5.858 cos 51.9° - 1.375 + 8.875 = 0 (Check)

56
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3–11. Determine the force in each member of the truss.

State if the members are in tension or compression. Assume
all members are pin connected.
G
F

E
3m

A B C D
2m 2m 2m

5 kN 5 kN 5 kN

Joint D:

FED a b - 5 = 0;
3
+ c a Fy = 0; FED = 8.33 kN (T) Ans.
5
4
a Fx = 0;
+
: FCD - (8.33) = 0;
5
FCD = 6.67 kN (C) Ans.

Joint C:

a Fx = 0;
+
: FBC - 6.67 = 0;
FBC = 6.67 kN (C) Ans.
+ c a Fy = 0; FCE - 5 = 0;
FCE = 5 kN (T) Ans.

Joint G:

a Fx = 0;
+
: FGF - 20 = 0; FGF = 20 kN (T) Ans.

+ c a Fy = 0; 15 - FGA = 0; FGA = 15 kN (T) Ans.

Joint A:
+ c a Fy = 0; 15 - FAF (sin 56.3°) = 0;
FAF = 18.0 kN (C) Ans.

a Fx = 0;
+
: -FAB - 18.0(cos 56.3°) + 20 = 0;
FAB = 10.0 kN (C) Ans.

Joint B:
-FBE a b + 10.0 - 6.67 = 0;
4
a Fx = 0;
+
:
5
FBE = 4.17 kN (C) Ans.
FFB - 5 - 4.17 a b = 0;
3
+ c a Fy = 0;
5
FFB = 7.50 kN (T) Ans.

Joint F:
18(sin 56.3°) - 7.5 - FFE a b = 0;
3
+ c a Fy = 0;
5
FFE = 12.5 kN (T) Ans.

57
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*3–12. Determine the force in each member of the truss. 8 kN

State if the members are in tension or compression. Assume
all members are pin connected. AG = GF = FE = ED. 8 kN 8 kN
F

4 kN G E 4 kN
2m
A
D
B C

4m 4m

Reactions:

A x = 0, A y = 16.0 kN

Joint A:

+ c a Fy = 0; 16 - 4 - FAG sin 26.565° = 0

FAG = 26.83 kN = 26.8 kN (C) Ans.

a Fx = 0;
+
: -26.83 cos 26.565° + FAB = 0

FAB = 24.0 kN (T) Ans.

Joint G:

+ a a Fy = 0; -8 cos 26.565° + FGB = 0

FGB = 7.155 kN = 7.16 kN (C) Ans.

+ Q a Fx = 0; 26.83 - FGF - 8 sin 26.56° = 0

FGF = 23.36 kN = 23.3 kN (C) Ans.

Joint B:
+ c a Fy = 0; FBF sin 53.13° - 7.155 sin 63.43° = 0

FBF = 8.00 kN (T) Ans.

a Fx = 0;
+
: FBC - 24.0 + 7.155 cos 63.43° + 8.00 cos 53.13° = 0
FBC = 16.0 kN (T) Ans.

Due to symmetrical loading and geometry:

FCD = FAB = 24.0 kN (T) Ans.
FEF = FGF = 23.3 kN (C) Ans.
FDE = FAG = 26.8 kN (C) Ans.
FEC = FGB = 7.16 kN (C) Ans.
FCF = FBF = 8.00 kN (T) Ans.

58
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3–13. Determine the force in each member of the truss 4 kN

and state if the members are in tension or compression. 3m 3m 3m
B C
Support Reactions:
D
a + a MD = 0; 4(6) + 5(9) - Ey (3) = 0 Ey = 23.0 kN
3m
+ c a Fy = 0; 23.0 - 4 - 5 - Dy = 0 Dy = 14.0 kN A 5m
F
a Fx = 0;
+
: Dx = 0
E
5 kN
Method of Joints:

Joint D:

FDE a b - 14.0 = 0
5
+ c a Fy = 0;
234
FDE = 16.33 kN (C) = 16.3 kN (C) Ans.

16.33 a b - FDC = 0
+ 3
: a Fx = 0;
234
FDC = 8.40 kN (T) Ans.

Joint E:
FEA a b - 16.33 a b = 0
+ 3 3
: a Fx = 0;
210 234
FEA = 8.854 kN (C) = 8.85 kN (C) Ans.

23.0 - 16.33 a b - 8.854 a b - FEC = 0

5 1
+ c a Fy = 0;
234 210
FEC = 6.20 kN (C) Ans.

Joint C:

+ c a Fy = 0; 6.20 - FCF sin 45° = 0

FCF = 8.768 kN (T) = 8.77 kN (T) Ans.

+
: a Fx = 0; 8.40 - 8.768 cos 45° - FCB = 0

FCB = 2.20 kN (T) Ans.

Joint B:
+
: a Fx = 0; 2.20 - FBA cos 45° = 0

FBA = 3.111 kN (T) = 3.11 kN (T) Ans.

+ c a Fy = 0; FBF - 4 - 3.111 sin 45° = 0

FBF = 6.20 kN (C) Ans.

Joint F:
+ c a Fy = 0; 8.768 sin 45° - 6.20 = 0 (Check!)

+
: a Fx = 0; 8.768 cos 45° - FFA = 0

FFA = 6.20 kN (T) Ans.

59
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3–14. Determine the force in each member of the roof 8 kN

truss. State if the members are in tension or compression. 4 kN 4 kN
4 kN 4 kN
I
4 kN J 4 kN
Reactions: K H
G 3.5 m
A y = 16.0 kN, A x = 0, Fy = 16.0 kN A
B C D E F
Joint A:
+ c a Fy = 0; -FAK sin 16.26° - 4 + 16 = 0 6 @ 4 m  24 m

FAK = 42.86 kN = 42.9 kN (C) Ans.

a Fx = 0;
+
: FAB - 42.86 cos 16.26° = 0

FAB = 41.14 kN = 41.1 kN (T) Ans.

Joint K:
+ a a Fy = 0; -4 cos 16.26° + FKB cos 16.26° = 0

FKB = 4.00 kN (C) Ans.

+ Q a Fx = 0; 42.86 + 4.00 sin 16.26° - 4.00 sin 16.26° - FKJ = 0

FKJ = 42.86 kN = 42.9 kN (C) Ans.

Joint B:
+ c a Fy = 0; FBJ sin 30.26° - 4 = 0

FBJ = 7.938 kN = 7.94 kN (T) Ans.

a Fx = 0;
+
: FBC + 7.938 cos 30.26° - 41.14 = 0

FBC = 34.29 kN = 34.3 kN (T) Ans.

Joint J:

a Fx = 0;
+
: -FJI cos 16.26° - 7.939 sin 59.74° + 42.86 cos 16.26° = 0

FJI = 35.71 kN = 35.7 kN (C) Ans.

+ c a Fy = 0; FJC + 42.86 sin 16.26° - 7.939 cos 59.74° - 4 - 35.71 sin 16.26° = 0

FJC = 6.00 kN (C) Ans.

Joint C:
+ c a Fy = 0; FCI sin 41.19° - 6.00 = 0

FCI = 9.111 kN = 9.11 kN (T) Ans.

a Fx = 0;
+
: FCD + 9.111 cos 41.19° - 34.29 = 0
FCD = 27.4 kN (T) Ans.

Due to symmetrical loading and geometry

FIH = 35.7 kN (C) Ans.
FHD = 6.00 kN (C) Ans.
FHE = 7.94 kN (T) Ans.
FHG = 42.9 kN (C) Ans.
FED = 34.3 kN (T) Ans.
FID = 9.11 kN (T) Ans.
FFG = 42.9 kN (C) Ans.
FGE = 4.00 kN (C) Ans.
FFE = 41.1 kN (T) Ans.

60
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3–15. Determine the force in each member of the roof G

truss. State if the members are in tension or compression.
Assume all members are pin connected. 3m
H F

3m
A E
B C D

10 kN 10 kN 10 kN
4m 4m 4m 4m

Joint A:
3
a Fy = 0; - F
5 AH
+ 15 kN = 0

FAH = 25 kN (C) Ans.

4
a Fx = 0; - (25 kN) + FAB = 0
5
FAB = 20 kN (T) Ans.

Joint B:

a Fx = 0; FBC = 20 kN (T) Ans.

a Fy = 0; FBH = 10 kN (T) Ans.

Joint H:
3 3 3
a Fy = 0; 5
(25 kN) - 10 kN + FHC - FHG = 0
5 5
4 4 4
a Fx = 0; 5
(25 kN) - FHC - FHG = 0
5 5
FHG = 16.7 kN (C) Ans.
FHC = 8.33 kN (C) Ans.

Joint G:
4 4
a Fx = 0; 5
(16.67 kN) - FGF = 0
5
FGF = 16.7 kN (C) Ans.
3 3
a Fx = 0; 5
(16.67 kN) + (16.67 kN) - FGC = 0
5
FGC = 20 kN (C) Ans.

The other members are determined from symmetry.

61
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*3–16. Determine the force in each member of the truss. F E

State if the members are in tension or compression.

B
30 30

45 45
A C
Joint E: 30 30

+ c a Fy = 0; FEA = FEC 2m 2m

a Fx = 0;
+
: 2.31 - 2 FEA sin 30° = 0 2 kN

FEA = 2.31 kN (C) Ans.

FEC = 2.31 kN (T) Ans.

Joint A:

a Fx = 0;
+
: 2.31 - 2.31 sin 30° - FAB cos 30° + FAD cos 45° = 0

+ c a Fy = 0; 2 - 2.31 cos 30° + FAD sin 45° - FAB sin 30° = 0

FAD = 2.24 kN (T) Ans.
FAB = 3.16 kN (C) Ans.

Joint B:

a Fx = 0;
+
: FBC = 3.16 kN (C) Ans.

+ c a Fy = 0; 2(3.16) sin 30° - FBD = 0

FBD = 3.16 kN (C) Ans.

Joint D:
FDC = 2.24 kN (T) Ans.

62
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3–17. Determine the force in each member of the roof truss. F

State if the members are in tension or compression. Assume
B is a pin and C is a roller support.
G E
60
30 60 60 60
A D
Support Reactions. Referring to the FBD of the entire truss, Fig. a, B C
2m 2m
a + a MC = 0; 2(4) - 2(2) - NB (2) = 0 NB = 2.00 kN
Method of joint. 2 kN 2 kN

Joint A: Fig. b,

+ c a Fy = 0; FAG sin 30° - 2 = 0 FAG = 4.00 kN (T) Ans.

a Fx = 0;
+
: 4.00 cos 30° - FAB = 0 FAB = 3.464 kN (C) = 3.46 kN (C) Ans.

Joint G: Fig. c,

+Q a Fx = 0; FGF - 4.00 = 0 FGF = 4.00 kN (T) Ans.

+a a Fy = 0; FGB = 0 Ans.

Joint B: Fig. d,

+ c a Fy = 0; 2 - FBF sin 60° = 0 FBF = 2.309 kN (C) = 2.31 kN (C) Ans.

a Fx = 0;
+
: 3.464 - 2.309 cos 60° - FBC = 0 FBC = 2.309 kN (C) - 2.31 kN (C) Ans.

Due to symmetry,
FDE = FAG = 4.00 kN (T) FDC = FAB = 3.46 kN (C) Ans.
FEF = FGF = 4.00 kN (T) FEC = FGB = 0 Ans.
FCF = FBF = 2.31 kN (C)

63
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3–18. Determine the force in members GF, FC, and CD of G

the bridge truss. State if the members are in tension of
H F 15 ft
compression. Assume all members are pin connected.

30 ft
A E
B C D
40 ft 40 ft 40 ft 40 ft

15 k 10 k

a + a MF = 0; -FDC(30) + 8.75 (40) = 0

FDC = 11.7 k (T) Ans.

-FFC a b(45) + 8.75 (80) = 0

8
a + a MC = 0;
273
FFG = 16.6 k (C) Ans.

8.75 - 16.6 a b # FFC a b = 0

3 3
+ c a Fy = 0;
273 5
FFC = 4.86 k (T) Ans.

3–19. Determine the force in members JK, JN, and CD. J

State if the members are in tension of compression. Identify
all the zero-force members. K I
L 20 ft
H
M N O G
F
A
B C D E
20 ft 30 ft 20 ft

2k 2k

Reactions:
A x = 0, A y = 2.0 k, Fy = 2.0 k
a + a MJ = 0; FCD(20) + 2(15) - 2(35) = 0
FCD = 2.00 k (T) Ans.

+ c a Fy = 0; Jy = 0

a Fx = 0;
+
: -Jx + 2.00 = 0; Jx = 2.00 k

Joint J:

a+ a Fy = 0; -FJN sin 23.39° + 2 sin 29.74° = 0

FJN = 2.50 k (T) Ans.

+Q a Fx = 0; FJK cos 29.74° - 2.50 cos 23.39°

FJK = 4.03 k (C) Ans.

Members KN, NL, MB, BL, CL, IO, OH, GE, EH, HD
are zero force members. Ans.

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*3–20. Determine the force in members GF, FC, and CD 12 kN

of the cantilever truss. State if the members are in tension of E
compression. Assume all members are pin connected. 12 kN
F
12 kN 3m
G

A
B C D
2m 2m 2m
a + a MC = 0; 12 kN (cos 26.57°) (4 m) + 12 kN (cos 26.57°)(2m)
-12 kN (sin 26.57°) (1 m) - FGF sin 26.57° (4 m) = 0

FGF = 33.0 kN (T) Ans.

a + a MA = 0; -12 kN (2.236 m) + FFC (4 m) = 0

FFC = 6.71 kN (T) Ans.

a + a MF = 0; 12 kN (2.236 m) + 12 kN (2)(2.236 m) - FCD (2 m) = 0

FCD = 40.2 kN (C) Ans.

3–21. The Howe truss is subjected to the loading shown. 5 kN

Determine the forces in members GF, CD, and GC. State if
the members are in tension or compression. Assume all G
members are pin connected. 5 kN 5 kN

3m
H F
2 kN 2 kN

E
A
B C D

2m 2m 2m 2m
c + a MG = 0; FCD(3) - 9.5(4) + 5(2) + 2(4) = 0

FCD = 6.67 kN (T) Ans.

4
c + a MD = 0; -9.5(2) + 2(2) + (1.5) FGF = 0
5
FGF = 12.5 kN (C) Ans.

Joint C:
FGC = 0 Ans.

65
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3–22. Determine the force in members BG, HG, and BC G

of the truss and state if the members are in tension or
H F
compression.
4.5 m
3m
A E
B C D

6 kN 7 kN 4 kN
12 m, 4 @ 3 m

a + a ME = 0; 6(9) + 7(6) + 4(3) - Ay (12) = 0 Ay = 9.00 kN

a Fx = 0;
+
: Ax = 0

Method of Sections:

a + a MG = 0; FBC(4.5) + 6(3) - 9(6) = 0

FBC = 8.00 kN (T) Ans.

FHG a b(6) - 9(3) = 0

1
a + a MB = 0;
25
FHG = 10.1 kN (C) Ans.

FBG a b(6) + 9(3) - 6(6) = 0

1.5
a + a MO = 0;
23.25
FBG = 1.80 kN (T) Ans.

3–23. Determine the force in members GF, CF, and CD of 1.5 kN

the roof truss and indicate if the members are in tension or
compression. C
2 kN 1.70 m

a + a MA = 0; Ey (4) - 2(0.8) - 1.5(2.50) = 0 Ey = 1.3375 kN 1.5 m

0.8 m D
B
Method of Sections: E
A
a + a MC = 0; 1.3375(2) - FGF (1.5) = 0 H G F
1m
FGF = 1.78 kN(T) Ans. 2m 2m

1.3375(1) - FCD a b(1) = 0

3
a + a MF = 0;
5
FCD = 2.23 kN (C) Ans.

FCF a b(1) = 0
1.5
a + a ME = 0 FCF = 0 Ans.
23.25

66
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*3–24. Determine the force in members GF, FB, and BC 600 lb

of the Fink truss and state if the members are in tension or
compression. 800 lb F 800 lb

G E

30 60 60 30

A D
B C
10 ft 10 ft 10 ft

Support Reactions: Due to symmetry. Dy = A y.

+ c a Fy = 0; 2Ay - 800 - 600 - 800 = 0 Ay = 1100 lb

a Fx = 0;
+
: Ax = 0

Method of Sections:
a + a MB = 0; FGF sin 30° (10) + 800(10 - 10 cos2 30°) - 1100(10) = 0
FGF = 1800 lb (C) = 1.80 k (C) Ans.

a + a MA = 0; FFB sin 60° (10) - 800(10 cos2 30°) = 0

FFB = 692.82 lb (T) = 693 lb (T) Ans.

a + a MF = 0; FBC (15 tan 30°) + 800(15 - 10 cos2 30°) - 1100(15) = 0

FBC = 1212.43 lb (T) = 1.21 k (T) Ans.

3–25. Determine the force in members IH, ID, and CD of 10 m, 5 @ 2 m

the truss. State if the members are in tension or compression.
Assume all members are pin connected. 3 kN 3 kN 3 kN 3 kN
1.5 kN
K J I H G
F

E
5m D
C
B
A

Referring to the FBD of the right segment of the truss sectioned

through a–a, Fig. a,

a + a MD = 0; FIH(2) - 3(2) - 1.5(4) = 0

FIH = 6.00 kN (T) Ans.

3(2) + 3(4) - FID a b (6) = 0

1
a + a MF = 0;
22
FID = 4.243 kN (T) = 4.24 kN (T) Ans.

FCD a b(6) - 3(2) - 3(4) - 1.5(6) = 0

1
a + a MI = 0;
25
FCD = 10.06 kN = 10.1 kN (C) Ans.

67
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3–26. Determine the force in members JI, IC, and CD of 10 m, 5 @ 2 m

the truss. State if the members are in tension or compression.
Assume all members are pin connected. 3 kN 3 kN 3 kN 3 kN
1.5 kN
K J I H G
F

E
5m D
C
B
A

Consider the FBD of the right segment of the truss

sectioned through a–a, Fig. a,

a + a MC = 0; FJI(3) - 3(2) - 3(4) - 1.5(6) = 0

FJI = 9.00 kN (T) Ans.

a + a MF = 0; 3(6) + 3(4) + 3(2) - FIC (6) = 0

FIC = 6.00 kN (C) Ans.

FCD a b(6) - 1.5(6) - 3(4) - 3(2) = 0

1
a + a MI = 0;
25
FCD = 10.06 kN (C) = 10.1 kN (C) Ans.

3–27. Determine the forces in members KJ, CD, and CJ of 30 kN 20 kN

the truss. State if the members are in tension or compression. 15 kN 15 kN 10 kN
5 kN 5 kN
L K J I H G
A
F 3@1m3m
B
C E
Entire truss: D
a Fx = 0;
:+ Ax = 0 6 @ 3 m  18 m

a + a MA = 0; -15(3) - 15(6) - 30(9) - 20(12) - 10(15) - 5(18) + Gy(18) = 0

Gy = 49.17 kN

+ c a Fy = 0; Ay - 5 - 15 - 15 - 30 - 20 - 10 - 5 + 49.167 = 0

A y = 50.83 kN

Section:
a + a MC = 0; 15(3) + 5(6) - 50.83(6) + FKJ(2) = 0

FKJ = 115 kN (C) Ans.

a + a MA = 0; -15(3) - 15(6) + FCJ sin 33.69° (9) = 0

FCJ = 27.0 kN (T) Ans.

a + a MJ = 0; -50.83(9) + 5 (9) + 15(6) + 15(3) + FCD cos 18.43° (3) = 0

FCD = 97.5 kN (T) Ans.

68