Question

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).
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Hint:

When solving problems related to beam deflection and axial force, carefully apply symmetry principles, calculate the elongation and deflection using the respective formulas, and ensure that equilibrium conditions are satisfied for the structure.

Answer 1

Correct Answer: 16.3

Given: Moment of inertia of beam, \( I = 8 \times 10^7 \, {mm}^4 \)
Modulus of elasticity of steel, \( E = 210 \times 10^3 \, {N/mm}^2 \)
UDL, \( w = 6.25 \, {kN/m} = 6.25 \, {N/mm} \) (including self-weight)
Axial force in CD, \( P_1 = ? \)
Flexural rigidity, \( EI = 210 \times 10^3 \times 8 \times 10^7 = 1.68 \times 10^{13} \, {N-mm}^2 = 16800 \, {kNm}^2 \) Due to symmetry, we have: \[ \Delta_{EF} = \Delta_{AB} \quad {and} \quad \Delta_{CD} \neq \Delta_{EF} \] From equilibrium: \[ 2P_2 + P_1 = 6.25 \times 40 \] \[ 2P_2 + P_1 = 25 \, {kN} \] Solving for \( P_2 \): \[ P_2 = \frac{25 - P_1}{2} = 12.5 - 0.5 P_1 \] The net elongation of rod CD is given by: \[ \Delta_{CD \, {net}} = \frac{P_1 \times 1}{\frac{\pi}{4} (0.03)^2 \times 210 \times 10^6} + \frac{0.5 \times 4 \times 12.5}{\frac{\pi}{4} (0.012)^2 \times 210 \times 10^6} \] Substitute the values: \[ \Delta_{CD \, {net}} = 2.77 \times 10^{-5} P_1 - 5.26 \times 10^{-4} \] Now, for the deflection of the beam at point D: \[ \Delta_{{beam}} = \frac{5 w l^4}{384 EI} - \frac{P_1 l^3}{48 EI} \] Substitute the values for \( w = 6.25 \, {kN/m} \) and \( l = 2 \, {m} \): \[ \Delta_{{beam}} = \frac{4^3}{48 \times 16800} \left( 5 \times 6.25 \times 4 - P_1 \right) \] Finally, solving for the axial force in Rod CD: \[ \Delta_{{CD \, net}} = \Delta_{{beam}} = \frac{4^3}{48 \times 16800} \left( 5 \times 6.25 \times 4 - P_1 \right) \] After solving, we find: \[ P_1 = 16.48 \, {kN} \] Thus, the axial force in rod CD is \( P_1 = 16.48 \, {kN} \).