GATE 2022 ME (ME1) – Question 49

Engineering Mechanics · Plane trusses and frames · 2 marks · Numerical answer

A structure, along with the loads applied on it, is shown in the figure. Self-weight of all the members is negligible and all the pin joints are friction-less. AE is a single member that contains pin C. Likewise, BE is a single member that contains pin D. Members GI and FH are overlapping rigid members. The magnitude of the force carried by member CI is ________ kN (in integer).

pin supports at A (top left) and B (bottom left). Member A-C-E and member B-D-E. A vertical link CD, a horizontal top chord C-I-H with CI = 3.5 m and IH = 3 m, links EI, IF, HG, EF, FG and the overlapping members GI and FH. The bottom row has B-D 2 m, D-E 2 m, E-F 3 m and F-G 3 m, and the height of the structure is 3 m. A 2 kN force acts to the right at H and a 4 kN force acts downwards at G.

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Correct answer: 18

Explanation

Take the whole structure first. Moments about B: $A_x \times 3 + (\text{loads}) = 0$, with the 2 kN load at height 1.5 m and the 4 kN load at 10 m, gives $-3A_x - 3 - 40 = 0$, so $A_x = -14.33$ kN. Then $B_x = 12.33$ kN, and $A_y + B_y = 4$ kN. For the member BE (pins B, D, E) take moments about E: the link CD is vertical, so $-4B_y - 2T_{CD} = 0$. Cut the structure vertically between E and F through CI, EI and EF and consider the left part. The vertical balance gives $A_y + B_y + 0.707T_{EI} = 0$, so $T_{EI} = -5.657$ kN. Moments about B for the left part: $43 - 1.5T_{CI} + 2.828T_{EI} = 0$, so $T_{CI} = \frac{43 - 16}{1.5} = 18$ kN (tension).