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.

Explanation

Here , `Y = 2.0 xx 10^11 Pa` , Mass (M) = 51,000 kg.
inner radius, `r_1`= 30cm = 0.3 m
Outer radius, `r_2` = 60 cm = 0.6 m
Force acting on each columns,
`F = (Mg)/(4) = (50,000 xx 9.8)/(4) = 122500N`
Area of cross-section of each column, i.e.,
`A = pi r_2^2 -pi r_1^2`
` =pi (r_2^2 - r_1^2) = 3.14 (0.36 - 0.09) = 0.85 m^2`
Compressional stress on each column,
`F/A = (122500)/(0.85) = 1.44 xx 10^5 Pa ` As Y = stress/strain
stress (compressional) = Strain (compressional)/Y
strain = `(1.44 xx 10^5)/(2.0 xx 10^11) = 7.2 xx 10^(-7)`