What uniformly distributed load per meter beam may carry?, Mechanical Engineering

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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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