GATE 2021 CE, Set 1, Solution of Q.22. A truss EFGH is shown in the figure

Q.22 (GATE 2021, CE, Set1)

A truss EFGH is shown in the figure, in which all the members have the same axial rigidity R. In the figure, P is the magnitude of external horizontal forces acting at joints F and G.

If R = 500×10³ kN, P = 150 kN and L = 3 m, the magnitude of the horizontal displacement of joint G (in mm, round off to one decimal place)
is _______________

Solution:

Also Check: Solution of GATE 2021, CE, Set1 (Q.1 to Q.100)

Please note that HG is a zero-force-member hence, F_HG = 0

Hence force in FG is F_FG = P (tensile)

Now, at joint F, using ∑Fx = 0, F_FH = 0, which in turn implies that F_EF = 0 , Similarly F_EH = 0 at Joint H.

Hence, only member EG is carrying a force, all others are zero-force members.

Apply Hooke’s Law,

Elongation of member EG, d = PL/AE,

note AE = R = 500×10³ kN,  P = 150 kN, L = 3 m

Hence, d = 150*3/(500×10³) = 0.0009 m

or d = 0.9 mm

Also Check: Solution of GATE 2021, CE, Set1 (Q.1 to Q.100)