A simply supported beam 5 m span of cross - sectional area of 40 mm² with second area of moment is equal to 400 × 10⁶ mm⁴ and Young's modulus of elasticity is equal to 1.2 × 10⁶ N/mm². The centroid of bending moment lies at 2 m. It carries an UDL of 5 KN/m over the entire span as shown in the figure. Calculate the maximum deflection by area moment method.

Concept

The maximum deflection at centre of simply supported beam with uniformly distributed load (UDL) is given by

\(\delta= \frac{5wL^4}{384EI}\)

where 

w = U.D.L. load in beam

E = Modulus of elasticity

I = Moment of inertia

L = length of beam

Calculation

Given data

w = 5 kN/m = 5 N/mm

L = 5 m = 5000 mm

E =  1.2 × 106 N/mm2

I = 400 × 106 mm4 

So the deflection on beam 

\(\delta= \frac{5wL^4}{384EI}\)

\(\delta= \frac{5\times5\times5000^4}{384\times 1.2\times10^6\times400\times10^6}\)

\(\delta = 0.0847 mm\)