*6–4.
Determine the force in each member of the truss, and state
if the members are in tension or compression. Set u = 30°.
D
3 kN
1.5 m
A
SOLUTION
B
Support Reactions: From the free-body diagram of the truss, Fig. a, and applying
the equations of equilibrium, we have
a + ©MA = 0;
NC cos 30°(2 + 2) - 3(1.5) - 4(2) = 0
NC = 3.608 kN
+ ©F = 0;
:
x
3 - 3.608 sin 30° - A x = 0
A x = 1.196 kN
+ c ©Fy = 0;
A y + 3.608 cos 30° - 4 = 0
A y = 0.875 kN
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
+ c ©Fy = 0;
3
3.608 cos 30° - FCD a b = 0
5
FCD = 5.208 kN = 5.21 kN (C)
+ ©F = 0;
:
x
Ans.
4
5.208 a b - 3.608 sin 30° - FCB = 0
5
FCB = 2.362 kN = 2.36 kN (T)
Ans.
Joint A: From the free-body diagram in Fig. c, we can write
+ c ©Fy = 0;
3
0.875 - FAD a b = 0
5
FAD = 1.458 kN = 1.46 kN (C)
+ ©F = 0;
:
x
Ans.
4
FAB - 1.458 a b - 1.196 = 0
5
FAB = 2.362 kN = 2.36 kN (T)
Ans.
Joint B: From the free-body diagram in Fig. d, we can write
+ c ©Fy = 0;
FBD - 4 = 0
FBD = 4 kN (T)
+ ©F = 0;
:
x
C
2.362 - 2.362 = 0
Ans.
(check!)
Note: The equilibrium analysis of joint D can be used to check the accuracy of the
solution obtained above.
2m
2m
4 kN
u