Project ForgeA rigid beam AD of length 3a=6 m is hinged at frictionless pin joint A and supported by two strings as shown in the figure. String BC passes over two small frictionless pulleys of negligible radius (running from B, up to a fixed pulley, horizontally to a second fixed pulley, then down to C, so it applies the same tension to both B and C). A separate vertical string of length a supports D directly, running up to a fixed anchor. All the strings are made of the same material and have equal cross-sectional area. A force F=9 kN is applied at C (at distance 2a from A) and the resulting stresses in the strings are within linear elastic limit. The self-weight of the beam is negligible with respect to the applied load. Assuming small deflections, the tension developed in the string at C is ________ kN (round off to 2 decimal places).
Explanation
This is a 1°-indeterminate elastic system: 2 unknown string tensions (T for string BC, T_D for the D-string) but only 1 independent equilibrium equation (moment about A), requiring a compatibility equation from the strings' elastic elongation matching the rigid beam's rotation θ. As the beam rotates, B, C, D move down by aθ, 2aθ, 3aθ respectively. String D (original length a) elongates by 3aθ, so T_D=(AE/a)(3aθ)=3AEθ. String BC (total original length 3a: two vertical a-segments plus one unchanged horizontal a-segment) elongates by aθ+2aθ=3aθ (only the vertical segments change), so T=(AE/3a)(3aθ)=AEθ. Thus T_D=3T. Moment equilibrium about A: T(a)+T(2a)+T_D(3a)=F(2a) → 3aT+3aT_D=2aF → 3T+9T=2F → T=F/6=9/6=1.5 kN.
Source: GATE ME 2022 Session 2 (ME-2) (2022)
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