If we clamp the ends of a rod rigidly to prevent expansion or contraction and then
change the temperature, tensile or compressive stresses called thermal stresses
develop.
The rod would like to expand or contract, but the clamps won't let it.
The resulting stresses may become large enough to strain the rod irreversibly or
even break it.
Engineers must account for thermal stress when designing structures. Concrete highways and bridge decks usually have gaps between sections, filled with
a flexible material or bridged by interlocking teeth , to permit expansion and contraction of the concrete.
To calculate the thermal stress in a clamped rod, we compute the amount the rod would expand (or contract) if not held and then find the stress needed to compress (or stretch) it back to its original length. Suppose that a rod with length $L$
and cross-sectional area $A$ is held at constant length while the temperature is reduced (negative $\Delta T$), causing a tensile stress. The fractional change in length if the rod were free to contract would be
$\left ( \frac{\Delta L}{L} \right )_{thermal}=\alpha \Delta T$
Here both $\Delta L$ and $\Delta T$ are negative.The tension must increase by an amount F that is just enough to produce an equal and opposite fractional change in length $\left ( \frac{\Delta L}{L} \right )_{tension}$. From the definition of Young's modulus,
$Y=\frac{F/A}{\Delta L /L}$
or,
$\left ( \frac{\Delta L}{L} \right )_{tension}=\frac{F}{AY}$
If the length is to be constant, the total fractional change in length must be zero. This means that
$\left ( \frac{\Delta L}{L} \right )_{thermal}+\left ( \frac{\Delta L}{L} \right )_{tension} = \alpha \Delta T + \frac{F}{AY} = 0$
Solving for the tensile stress $F/A$ required to keep the rod's length constant, we find
\begin{equation}
\frac{F}{A}=-Y\alpha \Delta T
\tag{7}
\end{equation}
For a decrease in temperature, $\Delta T$ is negative, so $F$ and $F/A$ are positive; this
means that a tensile force and stress are needed to maintain the length. If $\Delta T$ is
positive, $F$ and $F/A$ are negative, and the required force and stress are compressive.
Solved Example
Question 1
An aluminium bar has some length at 20°C. How much stress is required to keep it at the same length if the temperature is increase to 40 °C?
($\alpha=25 \times 10^{-6} /^0 C$, $Y=70 \times 10^9 N/m^2$) Solution
Thermal stress is given by
$\frac{F}{A}=Y\alpha \Delta T$
$=(70 \times 10^9) \times (25 \times 10^{-6}) \times 20 = 3.5 \times 10^7 N/m^2$
Question 2
A rod of length L having coefficient of Linear expansion $\alpha$ is lying freely on the floor.it is heated so that temperature changes by $\Delta T$ .Find the longitudinal strain developed in the rod
a. 0
b. $\alpha \Delta T$
c. $-\alpha \Delta T$
d. none of the above Solution
There was no restriction for it expansion.So no tensile or compressive force developed.Longitudinal strains happens only when tensile or compressive force developed in the rod.So answer a Question 3
Calculate the stress developed inside a tooth cavity filled with copper, when hot tea at temperature of 57°C is drunk. You can take body (tooth) temperature to be 37°C and $\alpha = 1.7 \times 10^{-5} /K$ and bulk modulus for copper = $140 \times 10^9 N/m^2$? Solution
Change in temperature ($\Delta T$) = 57 - 37 = 20°C
Linear expansion $\alpha$ of (tooth) body = $1.7 \times 10^{-5} /K$
Cubical expansion $\alpha _V = 3\alpha = 3 \times 1.7 \times 10^{-5} = 5.1 \times 10^{-5} K^{-1}$
Let the volume of the cavity be V and its volume increased by $\Delta V$ due to increase in temperature $\Delta T$.
Volumetric Strain=$\frac {\Delta V}{V} = \alpha _V \Delta T$
Thermal stress produced = B × Volumetric strain
$= B \alpha _V \Delta T$
Thermal stress =$ 140 \times 10^9 \times 5.1 \times 10^{-5} \times 20 = 14280 \times 10^4 $
$= 1.428 \times 10^8 N/m^2$
