GATE 2025 CE (CE2) – Question 52

Structural Analysis · Statically indeterminate structures: force and displacement methods · 2 marks · Numerical answer

A steel beam supported by three parallel pin-jointed steel rods is shown in the figure. The moment of inertia of the beam is $8 \times 10^7$ mm$^4$. Take modulus of elasticity of steel as 210 GPa. The beam is subjected to uniformly distributed load of 6.25 kN/m, including its self-weight.
The axial force (in kN) in the centre rod CD is ______ (round off to one decimal place).

a horizontal beam B-D-F of total length 4 m (2 m + 2 m) hung from the ceiling by three vertical rods, each 1 m long: AB and EF of 12 mm diameter at the ends and CD of 30 mm diameter in the centre. The UDL is 6.25 kN/m.

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Correct answer: 16.42 to 16.58

Explanation

The load on the beam is $6.25 \times 4 = 25$ kN. Let $R$ be the force in the centre rod. By symmetry each end rod carries $\frac{25 - R}{2}$. The stiffness of a rod is $\frac{EA}{L}$: for the 12 mm rods $k_1 = \frac{210 \times 10^6 \times 1.131 \times 10^{-4}}{1} = 23750$ kN/m and for the 30 mm rod $k_2 = 148440$ kN/m. The beam stiffness is $EI = 210 \times 10^6 \times 8 \times 10^{-5} = 16800$ kN m². The centre of the beam has to move down by the extension of the centre rod: end extension $\frac{25 - R}{2k_1}$ plus the deflection of the beam relative to its ends, $\frac{5wL^4}{384EI} - \frac{RL^3}{48EI}$ (with $L = 4$ m), equals $\frac{R}{k_2}$. That is $\frac{25 - R}{47500} + 0.0012401 - 7.9365 \times 10^{-5}R = \frac{R}{148440}$, which gives $R = 16.5$ kN.