Question

A steel bar $2.5\text{cm}$ in diameter is rigidly attached to two parallel supports which are $5\text{m}$ apart. Find the stress and the change in the diameter of the bar when the temperature is increased by ${100}^{\circ }C$.
Take $\alpha =12\times {10}^{-6}/{}^{\circ }C,Y=210\text{GPa}$.

• A. $252\text{MPa}$, $0.03\text{mm}$
• B. $150\text{MPa}$, $0.13\text{mm}$
• C. $59\text{MPa}$, $0.33\text{mm}$
• D. $212\text{MPa}$, $0.13\text{mm}$

Answer 1

The correct option is A $252\text{MPa}$, $0.03\text{mm}$


Let us assume $\sigma$ be the thermal stress induced by the temperature increment.
Since supports are rigid , there is zero net deflection in bar $AB$
$\therefore {\delta }_{AB}=0$
$\Rightarrow {\delta }_{fe}+{\delta }_{th}=0......\left(1\right)$
where,
${\delta }_{fe}=$ Thermal free expansion in bar, in the absence of rigid support.
${\delta }_{th}=$ contraction due to thermal stress.
We know that, when the temperature of a rod is changed by $\Delta T$ , From the concept of linear expansion ,the Change in length of the rod is given by
$\Delta L={\delta }_{fe}=L\alpha \Delta T$ and contraction due to thermal stress is given by ${\delta }_{th}=-\frac{\sigma L}{Y}\because Y=\frac{\text{stress}}{\text{strain}}$
Using the above formulae in $\left(1\right)$ we get,
$\alpha \Delta TL-\frac{\sigma L}{Y}=0\Rightarrow \sigma =Y\alpha \Delta T$
From the data given in the question ,
$\sigma =210\times {10}^9\times 12\times {10}^{-6}\times (100-0)\Rightarrow \sigma =252\text{MPa}$
Change in the diameter $\left(d\right)$ of the bar is given by
$\Delta d=d\alpha \Delta T$ $=(0.025\times 12\times {10}^{-6}\times (100-0\left)\right)\times {10}^3\text{mm}$
$\Rightarrow \Delta d=0.03\text{mm}$
Hence, option $\left(a\right)$ is the correct answer.