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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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Mass of big structure = 50,000 kg

Net force = (5 × 104 × 9.8)N

= 4.9 × 105 N = FB

Since it is supported by four cylinders the weight is equally distributed.

Force (F) on each cylinder is

F = \(\frac {FB}{4}\) = 122500 N

Young’s Modulus = Y

∴ Compressional strain of each column is given by, 

Here, F = 122500 N

A = π(0.6)2 – π(0.3)2 = π(0.36 – 0.09)

= 0.27π = 8.48 × 10-1 m2

Y = 2 × 1011 N m-2 (Standard Values)

∴ Strain = \(\frac {122500}{(0.27π) \, (2 \times 10^{11})}\)

= 7.22 × 10-7

Thus the compressional strain per cylinder is 7.22 × 10-7.

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