Question

A girder of uniform section and constant depth is freely supported over a span of 3 metres. The point load at the midpoint is 30 kN and Moment of inertia = 15 × 10^-6 m⁴ and youngs modulus = 200 GN / m². The deflection at centre will be
1. 6.6 mm
2. 8.6 mm
3. 5.6 mm
4. 6.6 m

Answer 1

Correct Answer - Option 3 : 5.6 mm

Concept:

When in the simply supported beam the point load act at the center and there is symmetric loading then,

∴ Deflection at center will be given \( = \frac{{P{L^3}}}{{48EI}}\)

Where, P = Load, L = Length, E = modulus of elasticity, I = Moment of inertia.

Calculation:

Given:

P = 30 × 10³ N, L = 3 m, I = 15 × 10^-6 m⁴, E = 200 × 10⁹ N/m²

The deflection at the center will be given by the formula,

\( = \frac{{P{L^3}}}{{48EI}} = \frac{{30 \times {{10}^3} \times {3^3}}}{{48 \times 200 \times {{10}^9} \times 15 \times {{10}^{ - 6}}}} = 5.6 \times {10^{ - 3}}m = 5.6\;mm\)

The deflection at the center will be 5.6 mm.