Four identical hollow cylindrical cloumns of steel support a big structure of mass 50.000 kg. the inner and outer radii of each column are 30 cm and 60 cm respectively. Assume the load distribution to be uniform , calculate the compressional strain of each column. the Young's modulus of steel is `2.0 xx 10^(11) Pa`.

Explanation

Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y = 2`xx``10^(11)` Pa
Total force exerted, F = Mg = 50000`xx`9.8 N
Stress = Force exerted on a single column = `(50000xx9.8)/(4)=122500` N
Young’s modulus, Y = `("Stress")/("Strain")`
Strain = `((F)/(A))/(Y)`
Where, Area, A = `pi(R^(2)-r^(2))=pi((0.6)^(2)-(0.3)^(2))`
Strain = `(122500)/(pi[(0.6)^(2)-(0.3)^(2)]xx2xx10^(11))=7.22xx10^(-7)`
Hence, the compressional strain of each column is 7.22`xx``10^(-7)`.