Question

Four identical hollow cylindrical columns of mild steel support a big structure of a 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.

Answer 1

Given data in the problem is:

Mass of the big structure, M = 50,000 kg
Total force exerted on the four columns= total weight of the structure=50000×9.8N
The compressional force on each column = Mg/4 = (50000×9.8)/4 N= 122500 N
Therefore, Stress = 122500 N
Young’s modulus of steel, Y=2×1011 Pa

Young’s modulus, Y= Stress/Strain

Strain = Young’s modulus/Stress

Strain = (F/A)/Y

Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Where,
Area, A=π(R2−r2)=π((0.6)2−(0.3)2) = 0.27 π m2

Strain =122500/[0.27 x 3.14×2×1011]=7.22×10−7

Hence, the compressional strain of each column is 7.22×10−7.

Compressional strain of each column is 7.22×10−7.