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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/ymax
M = Ã.I/ymax = [(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
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