Q. A rectangular beam which is 300 mm deep is simply supported over span of 5 m. What uniformly distributed load per meter beam may carry? If bending stress is not to exceed 130N/mm². Take I = 8.5 × 106 mm⁴

Sol.: Given:

σ = 130 N/mm²

I = 8.5 × 10⁶ mm⁴

y =  d/2 = 300/2 = 150mm

L = 5m = 5000mm

Let UDL = W N/m
Maximum bending moment for simply supported beam with UDL on the entire span can be given by

= WL₂/8

That is, M = WL₂/8 ...(i)

From bending equation M/I = σ/y_max

M = Ã.I/y_max = [(130) × (8.5 × 10₆)]/ 150 = 7366666.67 Nmm    ...(ii)

Bi putting this value in equation (i); we get

7366666.67 = W(5000)

2/8

W = 2.357 N/mm = 2357.3 N/m .......ANS