Question

Four identical hollow cylindrical columns of steel support a big structure of mass $50,000kg$. The inner and outer radii of each column are $30cm$ and $40cm$ respectively, Assuming the load distribution to be uniform. Calculate the compressional strain 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 $2.78\times {10}^{-6}$

$\begin{array}{c}\text{Area}=\pi r_1^2-\pi r_2^2=\pi \left(\right(0.4{)}^2-\left(0.3{)}^2\right)m^2=0.22m^2\\ \text{mass}=50,000kg\\ Y=2\times {10}^{11}Pa\end{array}$

$\begin{array}{c}\text{Since there are}4\text{columumns, the tension due so the big structure}\\ \text{will divide among}4\text{columns}\to \end{array}$

$\begin{array}{cc}\frac{F}{A}=Y\frac{\Delta l}{l}& \\ \Rightarrow \frac{\left(50000/4\right)g}{0.22}=2\times {10}^{11}\times \frac{\Delta l}{l}& \\ \Rightarrow \frac{\Delta l}{l}=\frac{50000\times 9.8}{4\times 0.22\times 2\times {10}^{11}}& \end{array}$

$\begin{array}{c}\Rightarrow \frac{\Delta l}{l}=278409\times {10}^{-11}\\ \Rightarrow \frac{\Delta l}{l}=2\cdot {78}^{}\times {10}^{-6}\end{array}$