Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Given Data,
Mass of the big structure, M = 50,000 kg.
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Ys, from the standard chart is 2×1011 Pa
Total force exerted on all the columns, F = Mg = 5000kg × 9.8 N
Amount of force exerted on each column, F1 = (5000Kg× 9.8N)/4
⇒ F1 = 122500 N
Young’s modulus, Y = Stress/ Strain
⇒ Y = (
/Strain
⇒ Strain = 
…………………………… (1)
Where, A = area of each column = π (R2-r2)
⇒ A = π ((0.6)2 - (0.3)2)
Substituting value of Area in equation (1), we get,
Strain = 122500 N/ π ((0.6)2-(0.3)2) × 2×1011 Pa
⇒ Strain = 7.22× 10-7
Hence, the compressional strain on each column is 7.22 × 10-7.