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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.
Solution
Method of Joints. Start at joint C and then proceed to join D.
Joint C****. Fig. a
S+^ Σ Fx =^ 0;^ FCB =^0 Ans.
- c Σ Fy = 0; FCD - 20 = 0 FCD = 20.0 kN (C) Ans.
Joint D****. Fig. b
b - 20.0 = 0 FDB = 33.33 kN (T) = 33.3 kN (T) Ans.
S+^ Σ Fx =^ 0;^10 +^ 33.33^ a
b - FDA = 0
FDA = 36.67 kN (C) = 36.7 kN (C) Ans.
Determine the force in each member of the truss and state if the members are in tension or compression. Set P 1 = 20 kN, P 2 = 10 kN. C B
A
D
1.5 m
2 m
P 1
P 2
Ans: FCB = 0 FCD = 20.0 kN (C) FDB = 33.3 kN (T) FDA = 36.7 kN (C)
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Solution
Method of Joints. Start at joint C and then proceed to joint D.
Joint C****. Fig. a
S+^ Σ Fx =^ 0;^ FCB =^0 Ans.
- c Σ Fy = 0; FCD - 45 = 0 FCD = 45.0 kN ( C ) Ans.
Joint D****. Fig. b
b - 45.0 = 0 FDB = 75.0 kN ( T ) Ans.
S+^ Σ Fx =^ 0;^30 +^ 75.0^ a
b - FDA = 0 FDA = 90.0 kN ( C ) Ans.
Determine the force in each member of the truss and state if the members are in tension or compression. Set P 1 = 45 kN, P 2 = 30 kN. C B
A
D
1.5 m
2 m
P 1
P 2
Ans: FCB = 0 FCD = 45.0 kN (C) FDB = 75.0 kN (T) FDA = 90.0 kN (C)
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
SOLUTION
c
Joint A :
Ans.
Ans.
Joint E :
Ans.
Ans.
Joint B :
Ans.
Ans.
Joint D :
Ans.
Ans.
Joint F :
Ans.
Ans.
Joint H :
Ans.
- c ©F (^) y = 0; 2.154 sin 21.80° - FHI = 0 F (^) HI = 0.8 kip (T) Ans.
F (^) HG = 2.154 kip = 2.15 kip (C)
:+^ ©F (^) x = 0; 2 - F (^) HG cos 21.80° = 0
FFG = 8.078 kip = 8.08 kip (C)
+R©F (^) x¿ = 0; FFG + 3 sin 21.80° + 4.039 sin 46.40° - 12.12 = 0
FFC = 4.039 kip = 4.04 kip (C)
+Q©F (^) y¿ = 0; F (^) FC cos 46.40° - 3 cos 21.80° = 0
:+ ©F (^) x = 0; - F (^) DC + 7.75 = 0 FDC = 7.75 kip (T)
- c ©F (^) y = 0; F (^) DF = 0
:+ ©F (^) x = 0; FBC - 3.75 = 0 FBC = 3.75 kip (T)
FED = 7.75 kip (T)
:+ ©F (^) x = 0; - F (^) ED - 3.5 + 12.12 cos 21.80° = 0
FEF = 12.12 kip = 12.1 kip (C)
- c ©F (^) y = 0; 4.5 - FEF sin 21.80° = 0
:+ ©F (^) x = 0; FAB - 4.039 cos 21.80° = 0 FAB = 3.75 kip (T)
F (^) Al = 4.039 kip = 4.04 kip (C)
©F (^) x = 0; 1.5 + 2 - Ex = 0 Ex = 3.5 kip
- ©M (^) E = 0; A (^) y (40) + 1.5(4) + 2(12) - 3(10) - 3(20) = 0 A (^) y = 1.5 kip
Determine the force in each member of the truss and state if the members are in tension or compression. (^) 2 kip
1.5 kip 4 ft
10 ft 10 ft 10 ft
3 kip 3 kip
10 ft
A (^) B
I
H
G F
C D
E
8 ft
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*6–4. Continued
Joint C :
Ans.
Ans.
Joint G :
Ans.
- 1.40 sin 21.80° - 8.081 = 0 (Check)
+R©Fx¿ = 0; 2.154 + 3 sin 21.80° + 5.924 sin 46.40°
FGI = 5.924 kip = 5.92 kip (C)
+Q©Fy¿ = 0; FGI cos 46.40° - 3 cos 21.80° - 1.40 cos 21.80° = 0
FCG = 1.40 kip (T)
- c ©Fy = 0; FCG + 0.2692 sin 21.80° - 4.039 sin 21.80° = 0
F (^) CI = 0.2692 kip = 0.269 kip (T)
:+ ©Fx = 0; - FCI cos 21.80° - 4.039 cos 21.80° - 3.75 + 7.75 = 0
Ans: FAl = 4.04 kip (C) FAB = 3.75 kip (T) FEF = 12.1 kip (C) FED = 7.75 kip (T) FBI = 0 FBC = 3.75 kip (T) FDF = 0 FDC = 7.75 kip (T) FFC = 4.04 kip (C) FFG = 8.08 kip (C) FHG = 2.15 kip (C) FHI = 0.8 kip (T) FCI = 0.269 kip (T) FCG = 1.40 kip (T) FGI = 5.92 kip (C)
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Determine the force in each member of the truss, and state if the members are in tension or compression. Set.
SOLUTION
Support Reactions: From the free-body diagram of the truss, Fig. a , and applying the equations of equilibrium, we have
a
Method of Joints: We will use the above result to analyze the equilibrium of joints C and A , and then proceed to analyze of joint B.
Joint C : From the free-body diagram in Fig. b, we can write
Ans.
Ans.
Joint A : From the free-body diagram in Fig. c, we can write
Ans.
Ans.
Joint B : From the free-body diagram in Fig. d, we can write
Ans.
Note: The equilibrium analysis of joint D can be used to check the accuracy of the solution obtained above.
:^ +©Fx =^ 0; 2.362^ -^ 2.362^ =^0 (check!)
FBD = 4 kN (T)
FAB = 2.362 kN = 2.36 kN (T)
FAB - 1.458a
:^ +©Fx =^ 0; b^ -^ 1.196^ =^0
FAD = 1.458 kN = 1.46 kN (C)
0.875 - FAD a
FCB = 2.362 kN = 2.36 kN (T)
5.208 a
:+©Fx =^ 0; b^ -^ 3.608 sin 30°^ -^ FCB =^0
FCD = 5.208 kN = 5.21 kN (C)
3.608 cos 30° - FCD a
A (^) y = 0.875 kN
- c ©Fy = 0; A (^) y + 3.608 cos 30° - 4 = 0
A (^) x = 1.196 kN
:^ +©Fx =^ 0;^3 -^ 3.608 sin 30°^ -^ A^ x =^0
NC = 3.608 kN
- ©MA = 0; NC cos 30°(2 + 2) - 3(1.5) - 4(2) = 0
u = 30°
A (^) C
B
D
2 m 4 kN
3 kN
2 m
1.5 m
u
Ans: FCD = 5.21 kN (C) FCB = 2.36 kN (T) FAD = 1.46 kN (C) FAB = 2.36 kN (T) FBD = 4 kN (T)
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
SOLUTION
Support Reactions:
a
Method of Joints:
Joint D :
Ans.
Ans.
Joint E :
Ans.
Ans.
Joint C :
Ans.
F (^) CB = 2.20 kN T Ans.
:+ ©Fx = 0 ; 8.40 - 8.768 cos 45° - F (^) CB = 0
F (^) CF = 8.768 kN 1 T 2 = 8.77 kN 1 T 2
- c ©Fy = 0 ; 6.20 - F (^) CF sin 45° = 0
F (^) EC = 6.20 kN 1 C 2
≤ -^ 8.854^ ¢
+ c ©Fy = 0 ; ≤ - F EC = 0
F (^) EA = 8.854 kN 1 C 2 = 8.85 kN 1 C 2
FEA ¢
:+ ©Fx = 0 ; ≤ = 0
F (^) DC =^ 8.40 kN^1 T^2
:+ ©Fx = 0 ; ≤ - F DC = 0
F (^) DE = 16.33 kN 1 C 2 = 16.3 kN 1 C 2
FDE ¢
+ c ©Fy = 0 ; ≤ - 14.0 = 0
:+ ©Fx = 0 D (^) x = 0
Determine the force in each member of the truss and state if the members are in tension or compression.
E
D
B C
F
A 5 m
3 m
5 kN
4 kN 3 m 3 m 3 m
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Determine the force in each member of the truss, and state if the members are in tension or compression.
SOLUTION
Method of Joints: We will begin by analyzing the equilibrium of joint D , and then proceed to analyze joints C and E.
Joint D : From the free-body diagram in Fig. a,
Ans.
Ans.
Joint C : From the free-body diagram in Fig. b,
Ans.
Ans.
Joint E : From the free-body diagram in Fig. c,
Ans.
FEA = 1750 N = 1.75 kN (C) Ans.
Q+ ©Fy¿ = 0; FEA - 1000 - 900 sin 36.87° - 750 cos 73.74° = 0
FEB = 750 N (T)
R+ ©Fx¿ = 0; - 900 cos 36.87° + FEB sin 73.74° = 0
FCB = 800 N (T)
FCE = 900 N (C)
:+©Fx =^ 0;^ FCE -^900 =^0
FDC = 800 N (T)
1000 a
FDE = 1000 N = 1.00 kN (C)
FDE a
:^ +©Fx =^ 0; b^ -^600 =^0
B
E
D
A
C
600 N
900 N
4 m
4 m
6 m
Ans: FDE = 1.00 kN (C) FDC = 800 N (T) FCE = 900 N (C) FCB = 800 N (T) FEB = 750 N (T) FEA = 1.75 kN (C)
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Solution
Support Reactions. Referring to the FBD of the entire truss shown in Fig. a ,
a+Σ MA = 0; ND (12) - 3(4) - 6(8) = 0 ND = 5.00 kN
a+Σ MD = 0; 6(4) + 3(8) - Ay (12) = 0 Ay = 4.00 kN
S+^ Σ Fx = 0; Ax = 0
Method of Joints. We will carry out the analysis of joint equilibrium according to the sequence of joints A , D , B and C.
Joint A****. Fig. b
b = 0
FAE = 4 2 2 kN (C) = 5.66 kN (C) Ans.
S+^ Σ Fx =^ 0;^ FAB -^4 22 a^
b = 0 FAB = 4.00 kN (T) Ans.
Determine the force in each member of the truss and state if the members are in tension or compression. Set P 1 = 3 kN, P 2 = 6 kN.
A
D
E
B C
P 1 P 2
4 m 4 m 4 m
6 m
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Solution
Support Reactions. Referring to the FBD of the entire truss shown in Fig. a ,
a+Σ MA = 0; ND (12) - 6(4) - 9(8) = 0 ND = 8.00 kN
a+Σ MD = 0; 9(4) + 6(8) - Ay (12) = 0 Ay = 7.00 kN
S+^ Σ Fx = 0; Ax = 0
Method of Joints. We will carry out the analysis of joint equilibrium according to the sequence of joints A , D , B and C.
Joint A****. Fig. a
b = 0 FAE = 7 2 2 kN (C) = 9.90 kN (C) Ans.
S+^ Σ Fx = 0; FAB - 722 a
b = 0 FAB = 7.00 kN (T) Ans.
Joint D****. Fig. c
b = 0 FDE = 8 2 2 kN (C) = 11.3 kN (C) Ans.
S+^ Σ Fx =^ 0;^822 a^
b - FDC = 0 FDC = 8.00 kN (T) Ans.
Determine the force in each member of the truss and state if the members are in tension or compression. Set P 1 = 6 kN, P 2 = 9 kN.
A
D
E
B C
P 1 P 2
4 m 4 m 4 m
6 m
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Joint B****. Fig. d
b - 6 = 0 FBE = 2 2 10 kN (T) = 6.32 kN (T) Ans.
S+^ Σ Fx =^ 0;^ FBC -^ 7.00^ +^ a^2210 b^ a^
b = 0 FBC = 5.00 kN (T) Ans.
Joint C. Fig. e
b - 9 = 0 FCE = 32 10 kN = 9.49 kN (T) Ans.
S+^ Σ Fx =^ 0;^ 8.00^ -^ 5.00^ -^ a^3210 ba^
b = 0 (Check!!)
6–10. Continued
Ans: FAE = 9.90 kN (C) FAB = 7.00 kN (T) FDE = 11.3 kN (C) FDC = 8.00 kN (T) FBE = 6.32 kN (T) FBC = 5.00 kN (T) FCE = 9.49 kN (T)
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
SOLUTION
Joint D :
Ans.
Ans.
Joint C :
Ans.
Ans.
Joint E :
Ans.
Ans.
Joint B :
Ans.
Ans.
Joint B :
Ans.
Joint D :
+b©Fy = 0; FDF = 0 Ans.
F (^) BF = 155 lb (C)
+a©Fy = 0; FBF sin 75° - 150 = 0
+Q©F (^) x¿ = 0; FBA - 515.39 = 0 FBA = 515 lb (C)
+a©Fy¿ = 0; FBF cos u = 0 FBF = 0
+R©Fx¿ = 0; 515.39 - FEF = 0 FEF = 515 lb (C)
+Q©Fy¿ = 0; FEB cos u = 0 FEB = 0
FCB = 515.39 lb = 515 lb (C)
+Q©Fx¿ = 0; FCB - 500 sin 75.96° - 125 sin 14.04° = 0
FCE = 0
+a©Fy¿ = 0; FCE cos 39.09° + 125 cos 14.04° - 500 cos 75.96° = 0
:+ ©Fx = 0; FCD - 515.39 cos 75.96° = 0 FCD = 125 lb (C)
FDE = 515.39 lb = 515 lb (C)
- c ©Fy = 0; FDE sin 75.96° - 500 = 0
Determine the force in each member of the truss and state if the members are in tension or compression.
500 lb 3 ft
500 lb
C
B
A F
E
D
9 ft
6 ft
6 ft
3 ft 3 ft
Ans: FDE = 515 lb (C) FCD = 125 lb (C) FCE = 0 FCB = 515 lb (C) FEB = 0 FEF = 515 lb (C) FBF = 0 FBA = 515 lb (C) FBF = 155 lb (C) FDF = 0
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
SOLUTION
Joint A :
Ans.
Ans.
Joint D :
FDB = 1.33 P (T) Ans.
+ c ©Fy = 0; FDB - 0.687 P ¢
(0.687 P) - P = 0
FCB = FAB = 0.943 P (C)
FCD = FAD = 0.687 P (T)
P
(FAB) +
(FAD) = 0
:+ ©Fx = 0;
(FAD) -
FAB = 0
Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.
B
D
P
A C a a
a
a
—^3 4
—^1 4
Ans: FCD = FAD = 0.687 P (T) FCB = FAB = 0.943 P (C) FDB = 1.33 P (T)
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Members AB and BC can each support a maximum compressive force of 800 lb, and members AD , DC , and BD can support a maximum tensile force of 2000 lb. If a = 6 ft, determine the greatest load P the truss can support.
Ans: P max = 849 lb
B
D
3
A C a a
a
a
—^3 4
—^1 4
Solution
- Assume FAB = 800 lb (C)
Joint A :
S+^ Σ Fx =^ 0;^ -^800 a
b +^ FAD a
b =^0
FAD = 583.0952 lb 6 2000 lb OK
P
P = 848.5297 lb OK
Joint D :
- c Σ Fy = 0; - 848.5297 - 583.0952(2) (^) a
b +^ FDB =^0
FBD = 1131.3724 lb 6 2000 lb OK
Thus, Pmax = 849 lb Ans.
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
SOLUTION
Method of Joints: In this case, the support reactions are not required for determining the member forces.
Joint D :
Ans.
Ans.
Joint C :
Ans.
Ans.
Joint B :
Thus,
Ans.
Joint E :
Ans.
Note: The support reactions and can be determinedd by analyzing Joint A using the results obtained above.
Ax^ Ay
FEA = 4.62 kN 1 C 2
:+ ©F (^) x = 0; F (^) EA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0
- c ©F (^) y = 0; Ey - 21 9.238 sin 60° 2 = 0 Ey = 16.0 kN
FBE = 9.24 kN 1 C 2 FBA = 9.24 kN 1 T 2
F = 9.238 kN
:+ ©F (^) x = 0; 9.238 - 2 F cos 60° = 0
FBE = FBA = F
- c ©F (^) y = 0; F (^) BE sin 60° - FBA sin 60° = 0
FCB = 9.238 kN 1 T 2 = 9.24 kN 1 T 2
:+ ©F (^) x = 0; 21 9.238 cos 60° 2 - FCB = 0
FCE = 9.238 kN 1 C 2 = 9.24 kN 1 C 2
- c ©F (^) y = 0; F (^) CE sin 60° - 9.238 sin 60° = 0
FDE = 4.619 kN 1 C 2 = 4.62 kN 1 C 2
:+ ©F (^) x = 0; F (^) DE - 9.238 cos 60° = 0
FDC = 9.238 kN 1 T 2 = 9.24 kN 1 T 2
- c ©F (^) y = 0; F (^) DC sin 60° - 8 = 0
Determine the force in each member of the truss. State whether the members are in tension or compression. Set P = 8 kN.
60 •• 60 ••
4 m 4 m
B
E (^) D
C
A
4 m
P
Ans: FDC = 9.24 kN (T) FDE = 4.62 kN (C) FCE = 9.24 kN (C) FCB = 9.24 kN (T) FBA = 9.24 kN (T) FEA = 4.62 kN (C)
