Question

A hollow cylindrical column of steel supports a load of $12,000kg$. The inner and outer radii of the column are $20cm$ and $30cm$ respectively. Assuming the load distribution to be uniform, calculate the value of compressional strain in terms of $10^{-6}$ of the column. (Take Young's modulus of steel $=2\times 10^{11}Pa,\pi =3.14,g=9.8m/s^2$ )

Answer 1

Answer: 3.75

Area of cross-section of the cylindrical column,
$a=\pi \left(r_2^2-r_1^2\right)=\pi \left(0.3^2-0.2^2\right)=0.05\pi m^2$
Mass supported on the column, $M=12,000kg$
Therefore, compressional force on the column,
$F=Mg=12,000\times 9.8=117600N$
Compressional strain,
$\frac{l}{L}=\frac{F}{aY}$
$=\frac{117600}{0.05\pi \times 2\times 10^{11}}$
$=\frac{117600}{3.14\times 10^{10}}$
$=3.7452\times 10^{-6}$
$=3.75\times 10^{-6}$
....(Rounding off to $2$ decimal places.)