Question

Four identical hollow cylindrical columns of steel support a big structure of mass $50,000$ kg. The inner and outer radii of each column are $30$cm and $40$cm respectively.Assuming the load distribution to be uniform. Calculate the compressional strains of each column, the young's modulus of steel is $2\times {10}^{11}Pa$

• A. $2.78\times {10}^{-6}$
• B. $3.78\times {10}^{-6}$
• C. $2.78\times {10}^{-4}$
• D. $3.78\times {10}^{-4}$

Answer 1

The correct option is A $2.78\times {10}^{-6}$

$\begin{array}{c}\text{Inner radius of each column,}{r}_1=30cm=0.3m\\ \text{Outer radius of each column,}{r}_2=40cm=0.4m\\ \text{Area of each column}=\pi (r_2^2-r_1^2)\\ \begin{array}{cc}& =\frac{22}{7}(0.16-0.09)\\ & =22\times {10}^{-2}{m}^2\end{array}\end{array}$

$\begin{array}{c}\text{Young modulus}=2\times {10}^{11}{Nm}^{-2}\\ \text{As load is distributed equally among the four columns, hence}\\ \text{load on each column}m=\frac{50,000}{4}kg=12,500kg\\ \begin{array}{c}F=mg=12,500\times 9.8\\ \end{array}\end{array}$

$\begin{array}{cc}Y=& \frac{\text{stress}}{\text{Strain.}}\\ \text{Strain}=& \frac{\text{Stress}}{Y}=\frac{F/A}{Y}\\ =& \frac{12,500\times 9.8}{0.22\times 2\times {10}^{11}}\\ =& 2.78\times {10}^{-6}\end{array}$