A hollow cylindrical column of steel with inner and outer radii 20cm and 30cm supports a load of 12,000 kg. Assume the load distribution to be uniform. The value of compressional strain in terms of 10- 6 of the column is ... (Take `Y_("steel") = 2 xx 10^(11) Pa, pi ` = 3. 14, g = 9. 8 m/`s^(2)`

Answer

3.75

Explanation

Area of cross-section of the cylindrical column,
`a= pi (r_(2)^(2)- r_(1)^(2)) = pi (0.3^(2) - 0.2^(2)) =0.5 pi m^(2)`
Mass supported on the column, M = 12,000 kg Therefore, compressional force on the column,
F =Mg= 12,000 `xx` 9. 8 = 117600N
Compressional strain,
`(l)/(L) = (F)/(aY) = (117600)/(0.05pi xx 2xx 10^(11)) = (117600)/(3.14 xx 10^(10))`
`= 3.7452 xx 10^(-6)`
`= 3.75 xx 10^(-6) ` . (Rounding off to 2 decimal places.)