Calculate the maximum and minimum stress:
A cast iron column of 200 mm diameter carries a vertical load of 400 kN, at a distance of 50 mm from the centre. Calculate the maximum & minimum stress developed in the section, along with the diameter passing through the point of loading.
(a)
(b) Stress Distribution along the Diagonal
Figure
Solution
Vertical load, P = 400 kN = 400 × 10^{3} N
Diameter of the section, D = 200 mm
Area of the section, A = (π/4) (200)^{2 } = 31416 mm^{2}
Direct stress, f_{0} =(P/A) =4 × 10^{5}/31416 = 12.732 N/mm2
Eccentricity, e = 40 mm
Bending moment, M = P × e = (400 × 10^{3 }) 40 = 16 × 10^{6} N-mm
Section modulus, z = π D^{ 3}/32 = π (200)^{3} /32= 785.4 × 10^{3} mm^{3}
Bending stress, f =± Pe / Z =± 16 × 10^{6} /785.4 × 10^{3}=± 20.372 N/mm2
∴ Resultant stress at the edge, B = f_{0} + f_{b} = 12.732 - 20.372
= 33.104 N/mm^{2} (compressive)
∴ Resultant stress at the edge, A = f_{0} - f_{b} = 12.732 - 20.372
=- 7.640 N/mm^{2} (tensile)
The stress distribution along with the diameter is as illustrated in Figure (b).