Question

For the truss shown in Figure 16, find the forces in members DE, BD and CB.

Answer 1

ΣFx = 0

H_A = 0

ΣFy =0
VA – 2 -2 -2 + R_B = 0

V_A + R_B = 6 ……..(1)

`ΣM_(A^F) = 0`

(2x4) + (2x8) + (2x12) – (R_Bx8) = 0

R_B = 6 kN

Now,
V_A + R_B = 6
V_A = 6 – 6 = 0 kN

In ΔABE,
Tan(α) = `(EB)/(AB) = 5/8`

`alpha = tan^(-1) 0.625 = 32°`

Taking section DD’,

ΣFx = 0
F_DEcos(32) + F_BDcos(32) + F_CB = 0 ……(1)

ΣFy = 0

- 2 + F_DEsin(32) - F_BDsin(32) = 0

F_DEsin(32) - F_BDsin(32) = 2 …….(2)

ΣMD
F = 0
F_CB x perpendicular distance of F_CB from D = 0
F_CB = 0 kN…..(3)

Solving (1), (2) and (3),
`F_(DE) = 1.887 kN`
`F_(BD )= - 1.887 kN`
`F_(CB) = 0 kN`

The forces in the members DE, BD and CB are 1.887 kN (compression), 1.887 kN
(tension) and 0 kN respectivel .