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/mm^{2}. Take I = 8.5 × 106 mm^{4}
Sol.: Given:
σ = 130 N/mm^{2}
I = 8.5 × 10^{6} mm^{4}
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
= WL_{2}/8
That is, M = WL_{2}/8 ...(i)
From bending equation M/I = σ/y_{max}
M = Ã.I/y_{max} = [(130) × (8.5 × 10_{6})]/ 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