Question

Find the displacement ( in $\text{mm}$) of the midpoint of a steel rod of length $2\text{m}$ and mass $10\text{kg}$ due to its own weight

[Assume, modulus of elasticity $Y=2\times {10}^{11}{\text{N/m}}^2$, area of cross-section $A={10}^{-8}{\text{m}}^2$]

Answer 1

Let B be the mid point of the rod, $W$ be the weight of the rod , $L$ be the total length of the rod from the point of suspension and $A$ be the cross-sectional area of the rod.
Extension in the rod at point $B$ is due to weight of the part of the rod from the point of suspension to $B$ and weight of the rod below the point $B$.


From the definition of Young's modulus of elasticity, we know that $Y=\frac{F}{A}\times \frac{L}{\Delta L}$
$\Rightarrow \Delta L=\frac{FL}{AY}$
From figure:
Weight of a small element of the rod $dx$ at distance $x$ from B will be $\frac{W}{L}dx$
So, extension at point B due to weight of that small element is given by $\frac{Wdx}{L}\times \frac{x}{AY}$
Also, the weight acting at $B$ due to the portion of rod present below it is $\frac{W}{2}$
Extension due to this weight $=\frac{\frac{W}{2}\times \frac{L}{2}}{AY}$

Hence, total extension at point $B$ is given by
$(\Delta L{)}_B=\underset{0}{\overset{L/2}{\int }}\frac{Wx}{ALY}dx+\frac{(\frac{W}{2}\times \frac{L}{2})}{AY}$
$(\Delta L{)}_B=\frac{WL}{8AY}+\frac{WL}{4AY}$
$\Rightarrow (\Delta L{)}_B=\frac{3WL}{8AY}$
Given,
Area of cross-section of rod $\left(A\right)={10}^{-8}{\text{m}}^2$
Young's modulus of elasticity $\left(Y\right)=2\times {10}^{11}{\text{N/m}}^2$
Length of the rod $\left(L\right)=2\text{m}$
Mass of the rod $\left(m\right)=10\text{kg}$
Thus,
$(\Delta L{)}_B=\frac{3\left(mg\right)L}{8YA}=\frac{3\times 10\times 10\times 2}{8\times {10}^{-8}\times (2\times {10}^{11})}$
$\Rightarrow (\Delta L{)}_B=37.5\text{mm}$