Correct Answer - Option 4 : 3/8 times
Concept:
Deflection at centre of a simply supported beam due to point load (W):
\(δ_1=\frac{WL^3}{48EI}\)
Deflection at centre of a simply supported beam due to uniformly distribute load (W = wL):
\(δ_2=\frac{5wL^4}{384EI}=\frac{5WL^3}{384EI}\;\;\;(\because w=\frac{W}{L})\)
Calculation:
Given:
\(\frac{\delta_2}{\delta_1}=\frac{5WL^3}{384EI}\times \frac{48EI}{WL^3}=\frac{5}{8}\)
Hence,
The deflection will be reduced by
\(={\delta _1} - {\delta _2} = \left( {1 - \frac{5}{8}} \right){\delta _1} = \frac{3}{8}{\delta _1}\)