Question

A steel rod of length $5\text{m}$ and diameter $30\text{mm}$ is fixed between two rigid supports. Determine the thermal stress and change in diameter of the rod, when the temperature increases by $50{}^{\circ }\text{C}$. Take Young's modulus of elasticity $\left(Y\right)=2.0\times {10}^6{\text{kg/cm}}^2$, coefficient of thermal expansion $\left(\alpha \right)=12\times {10}^{-6}{/}^{\circ }\text{C}$ and Poisson's ratio $\left(\nu \right)=0.3$.

• A. $120{\text{kg/cm}}^2,0.0234\text{mm}$
• B. $1200{\text{kg/cm}}^2,0.018\text{mm}$
• C. $12000{\text{kg/cm}}^2,0.0054\text{mm}$
• D. $1200{\text{kg/cm}}^2,0.0234\text{mm}$

Answer 1

The correct option is D $1200{\text{kg/cm}}^2,0.0234\text{mm}$

Let ${\sigma }_t$ be the thermal stress produced in steel rod due to change in temperature.

And, we know that
${\sigma }_t=\alpha \times \Delta T\times Y$
$=12\times {10}^{-6}\times 50\times 2\times {10}^6$
$=1200kg/{cm}^2$
$\therefore {\sigma }_t=1200{kg/cm}^2$

So, compressive thermal stress produced in the rod is $1200{\text{kg/cm}}^2$.
Let $\Delta d$ be the total change in the diameter.
Here rod is fixed at both the ends, so diameter will change due to :
(i) thermal expansion
(ii) compressive thermal stress

Change in diameter $\left(\Delta d_1\right)$ due to thermal expension
$\Delta d_1=\alpha \Delta Td$
$=12\times {10}^{-6}\times 50\times 30$
$=0.018mm$

We know that,
$\frac{\Delta L}{L}=\frac{\Delta d}{d}=\alpha \Delta T$
$L=$ length of rod
$d=$ diameter of rod
and Poisson ratio $\nu =-\frac{\frac{\Delta d}{d}}{-\frac{\Delta L}{L}}$
H​ere $\Delta d_2=\Delta d=$ change in diameter due to thermal stress
$\Rightarrow \frac{\Delta d_2}{d}=\nu \times \frac{\Delta L}{L}=\nu \alpha \Delta T$
$\Rightarrow \Delta d_2=\nu \alpha \Delta Td$
$\Rightarrow \Delta d_2=\nu \alpha \Delta Td$
$=0.3\times 12\times {10}^{-6}\times 50\times 30$
$=0.0054mm$
Hence, total change in diameter
$\left(\Delta d\right)=\Delta d_1+\Delta d_2$
$=0.018+0.0054=0.0234\text{mm}$