Q7  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.

Answers (1)
S Sayak

Inner radii of each column, r1 = 30 cm = 0.3 m

Outer radii of each colum, r2 = 60 cm = 0.6 m

Mass of the structure, m = 50000 kg

Stress on each column is P

\\P=\frac{50000\times 9.8}{4\times \pi \times (0.6^{2}-0.3^{2})}\\ P=1.444\times 10^{5}Nm^{-2}

Youngs Modulus of steel is Y=2\times 10^{11}Nm^{-2}

\\Compressional\ strain=\frac{P}{Y} \\=\frac{1.444\times 10^{5}}{2\times 10^{11}}\\ =7.22\times 10^{-7}