GATE 2025 CE (CE1) – Question 51

Solid Mechanics · Simple stress and strain relationships · 2 marks · Numerical answer

Consider the rigid bar ABC supported by the pin-jointed links BD and CE and subjected to a load $P$ at the end A, as shown in the figure. The axial rigidities of BD and CE are 22500 kN and 15000 kN, respectively. If CE elongates by 5 mm due to the load $P$, the magnitude of the downward deflection (in mm) of the end A would be ________________________ (*rounded off to the nearest integer*).

A horizontal rigid bar ABC with A at the left end, B 200 mm from A and C 200 mm further right. Link BD is vertical, 200 mm long, joining B to a support D above. Link CE is vertical, 300 mm long, joining C to a support E below. A downward load P acts at A.

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: 14

Explanation

Moments about B give $P \times 200 + F_C \times 200 = 0$, so the force in CE is $P$ (a pull downward on C). Then the vertical balance gives $F_B = 2P$, a tension in BD. CE elongates by $\frac{P \times 300}{15000} = 5$ mm, so $P = 250$ kN and the force in BD is 500 kN, which stretches it by $\frac{500 \times 200}{22500} = 4.44$ mm, so B moves down 4.44 mm. C moves up 5 mm because CE lengthens away from E. The rigid bar is straight, so the slope is $\frac{5 + 4.44}{200} = 0.0472$, and A, 200 mm to the other side of B, moves down by $4.44 + 200 \times 0.0472 = 13.9$ mm, which is 14 mm.