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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 radil 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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The correct Answer is:
`2.786 xx 10^(-6) `
Load on one hollow cylinder `= 50,000 //4 = 12500 kg. a = PI ( 0.4^(2) -0.3^(@)) = 0.2198 m^(2) ` . Stain `= F//AY = 12500 xx9.8 //0.2198 xx 2 xx 10^(11) = 2.786 xx 10^(-6)`.
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