Question

A simply supported beam with a rectangular cross-section is subjected to a central concentrated load. If the width and depth of the beam are doubled, while retaining the same elastic properties, then the deflection at the center of the beam w.r.t. the original deflection will be reduced to
1. 50%
2. 25%
3. 75%
4. 12.50%
5. 6.25%

Answer 1

Correct Answer - Option 5 : 6.25%

Concept:

Deflection at the center due to central concentrated load on a simply supported beam, \(δ = \frac{{W{L^3}}}{{48EI}}\)

For rectangular beam Moment of inertia, \(I = \frac{{B{D^3}}}{{12}}\)

Calculation:

Given,

width = B and Depth = D

Deflection at centre = δ₁ 

When B´ = 2B and D´ = 2D 

E, L remains the same.

Since, \(\delta = \frac{{W{L^3}}}{{48E\left( {\frac{{B{D^3}}}{{12}}} \right)}}\)

\(\delta \propto \frac{1}{{B{D^3}}}\)

\(\frac{{{\delta _1}}}{{{\delta _2}}} = \frac{{B'{{\left( {D'} \right)}^3}}}{{B{D^3}}} = \frac{{\left( {2B} \right){{\left( {2D} \right)}^3}}}{{B{D^3}}} = 16\)

\(\frac{{{\delta _2}}}{{{\delta _1}}} = \frac{1}{{16}}\)

\(\frac{{{\delta _2}}}{{{\delta _1}}} \times 100 = 6.25\% \)

The deflection at the center of the beam w.r.t. the original deflection will be reduced to 6.25 %.