Question

A simply supported beam with rectangular cross-section section is subjected to central concentrated load. If width and depth of the beam are doubled, then the deflection at the centre of the beam will be reduced to ............

• A. 50%
• B. 25%
• C. 12.5%
• D. 6.25%

Hint:

When both the width and depth of a beam's cross-section are doubled, the moment of inertia increases by a factor of 8, resulting in a deflection that is reduced by a factor of 8.

Answer 1

The Correct Option is D

The deflection \( \delta \) at the centre of a simply supported beam subjected to a central concentrated load is given by the formula: \[ \delta = \frac{W L^3}{48 E I} \] where:
- \( W \) is the applied load,
- \( L \) is the span of the beam,
- \( E \) is the Young's Modulus of the material,
- \( I \) is the moment of inertia of the beam's cross-section.
For a rectangular cross-section, the moment of inertia \( I \) is given by: \[ I = \frac{b d^3}{12} \] where:
- \( b \) is the width of the beam,
- \( d \) is the depth of the beam.
If both the width and depth of the beam are doubled, the new moment of inertia becomes: \[ I_{\text{new}} = \frac{(2b) (2d)^3}{12} = \frac{2b \cdot 8d^3}{12} = 8 \cdot \frac{b d^3}{12} = 8 I \] Since the deflection is inversely proportional to the moment of inertia, the new deflection will be: \[ \delta_{\text{new}} = \frac{\delta}{8} \] Thus, the deflection at the centre of the beam will be reduced to \( \frac{1}{8} \) of the original deflection, which is 12.5% of the original deflection. Therefore, the correct answer is 6.25%.