Q.  A supported rectangular beam with the symmetrical section 200mm in depth is having moment of inertia of 2.26 x 10^-5 m⁴ 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/m². 

Sol.: Given:

Depth d = 200mm = 0.2m

I = The moment of inertia = 2.26 × 10^-5 m⁴

UDL = 4KN/m

Bending stress σ = 125 MN/m² = 125 × 106 N/m²

Span = ?

As we know that Maximum bending moment for supported beam with UDL on its entire span can be given by = WL²/8

That is; M = WL^2/8 ...(i)

y_max =  d/2 = 0.2/2 = 0.1m

M = σ.I/y_max = [(125 × 10⁶) × (2.26 × 10^-5)]/ 0.1 = 28250 Nm ...(ii)

Substituting this value in the equation (i);

28250 = (4 × 103)L²/8

L = 7.52m .......ANS