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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.

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Here, outer radius of each column

r0 = 60 cm = 60 x 10-2 m

Inner radius, ri = 30 cm = 30 x 10-2 m

Area of cross-section of each column,

A = πr02 -πri2 = π(r02 - ri2)

= 3.14 x [(60 x 10-2)2 - (30 x 10-2)2]

= 0.85 m2

Total mass supported by four columns,

= 50000 kg

Mass supported by each column,

= 50000/4 = 12500 kg

Stress on each column = mg/A = 12500 x 9.8/0.85

= 1.44 x 105 Nm-2

Now, Young's modulus, Y = 2.0 x 1011 Nm-2

Using the relation, Stress/Strain = Y

We have, Strain = Stress/Y = 1.44 x 105/2 x 1011 = 7.2 x 10-7

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