Q. A supported rectangular beam with the symmetrical section 200mm in depth is having moment of inertia of 2.26 x 10^{-5 }m^{4} about its neutral axis. Find out longest span over which beam would carry a uniformly distributed load of 4KN/m run such that the stress because of bending does not exceed 125 MN/m^{2}.
Sol.: Given:
Depth d = 200mm = 0.2m
I = The moment of inertia = 2.26 × 10^{-5 }m^{4}
UDL = 4KN/m
Bending stress σ = 125 MN/m^{2} = 125 × 106 N/m^{2}
Span = ?
As we know that Maximum bending moment for supported beam with UDL on its entire span can be given by = WL^{2}/8
That is; M = WL^{2/}8 ...(i)
y_{max} = d/2 = 0.2/2 = 0.1m
M = σ.I/y_{max} = [(125 × 10^{6}) × (2.26 × 10^{-5})]/ 0.1 = 28250 Nm ...(ii)
Substituting this value in the equation (i);
28250 = (4 × 103)L^{2}/8
L = 7.52m .......ANS