Thermal Stress

Illustration :

A brass rod of length \(1 \mathrm{~m}\) is fixed to a vertical wall, at one end, with the other end free to expand. When the temperature of the rod is increased by \(120^{\circ} \mathrm{C}\), the length increases by \(2 \mathrm{~cm}\). What is the strain?

Sol. After the rod expands, to the new length there are no elastic forces developed internally in it. So, strain \(=0\).

top view

A change in shape/size i.e., dimensions need not necessarily imply a strain. For example, if a body is heated to expand, its volume change, as it acquires a new size, due to expansion. However, the strain remains zero. Unless and until, internal elastic forces operate, to bring the body to the original state, no strain exists. When a body is heated, the total energy of molecule increase, owing to an increase in the kinetic energy of the molecules. This results in a shift (increment) of the "equilibrium distance" of molecules and the body acquires a new shape and size, in the expanded form, whereby the molecules are in "zero force" state. Hence, there is no strain. However, if the body is resistricted to expand, during the process of heating, then the molecules become "strained", and even if there is no apparent change in dimensions of the body, there is strain. In such cases, strain is measured as the ratio. In dimension that would have occured, and the change in dimension that would have occured, had the body been free to expand or contract, to the original dimension.

When a metal rod is heated or cooled it tends to expand or contract. If it is left free to expand or contract, no temperature stresses will be induced. However, if the rod be restricted to change its length, then temperature stresses are generated within it. Stress induced due to temperature change can be understood as follows:

Consider a uniform rod AB fixed rigidly between two supports. (fig.) If L be its length, \(\alpha\) the coefficient of linear expansion, then a change in temperature of \(\Delta \theta\), would tend to bring a change in its length by \(l=L \alpha \Delta \theta\). Had the rod been free (say one of its ends) its length would have changed by l. Now, let a force be gradually applied so as to restore the natural length. Since the rod, tends to remain in the new state, due to a change in temperature, so when a force \(F\) is applied, thermal stress is induced. In equilibrium.

\( \frac{\mathrm{F}}{\mathrm{A}}=\frac{l}{(\mathrm{~L} \pm l)} Y \quad[\because \text { stress }=\operatorname{strain} \times Y]\)

Neglecting l in comparison to \(L\), 

\(\mathrm{F}=\frac{l \mathrm{~A}}{\mathrm{~L}} \mathrm{Y}=\mathrm{AY} \alpha \Delta \theta\)

Now, if the two ends remain fixed, then this external force is provided from the support.

Clearly strain \(=\frac{\ell}{\mathrm{L}}=\alpha \Delta \theta\)

Illustration :

The stress on a steel beam

A steel beam is used in the roadbed of a bridge. The beam is mounted between two concrete supports when the temperature is \(22^{\circ} \mathrm{C}\), with no room provided for thermal expansion.

What compressional stress must the concrete supports apply to each end of the beam, if they are to keep the beam from expanding when the temperature rises to \(42^{\circ} \mathrm{C}\) ?

Sol. Reasoning Recall from that the stress (force per unit cross-sectional area or \(F / A\) ) required to change the length \(L_{0}\) of an object by an amount \(\Delta L\) is

Stress \(=Y \frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}\)

where \(Y\) is Young's modulus. If the steel beam were free to expand because of the change in temperature, the length would change by \(\Delta L=\alpha L_{0} \Delta T\). Because the concrete supports do not permit any expansion, they must supply a stress to compress the beam by an amount \(\triangle L\). Thus,

\(\text { Stress }=Y \frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}=Y \frac{\alpha \mathrm{L}_{0} \Delta \mathrm{T}}{\mathrm{L}_{0}}=Y \alpha \Delta T\)

Young's modulus and the coefficient of linear expansion for steel are \(Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\) and \(\alpha=12 \times 10^{-6}\left(C^{\circ}\right)^{-1}\), respectively. The change in temperature from 22 to \(42^{\circ} \mathrm{C}\) is \(\Delta T=20^{\circ} \mathrm{C}\). The thermal stress is

Stress \(\left.=Y \alpha \Delta T=\left(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\left[12 \times 10^{-6}\left(C^{\circ}\right)^{-1}\right]\left(20 C^{\circ}\right)=4.8 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}\right]\)

Illustration :

A rod of length \(2 m\) is at a temperature of \(20^{\circ} \mathrm{C}\). Find the free expansion of the rod, if the temperature is increased to \(50^{\circ} \mathrm{C}\). Find the temperature stresses produced when the rod is (i) fully prevented to expand, (ii) permitted to expand by \(0.4 \mathrm{~mm} . Y=2 \times 10^{11} \mathrm{Nm}^{-2} ; \alpha=15 \times 10^{-6}\) per \({ }^{\circ} \mathrm{C}\).

Sol. Free expansion of the rod \(=\alpha \Delta \theta\)

\(=\left(15 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right) \times(2 \mathrm{~m})\) 

\(\times\left(50^{\circ}-20^{\circ}\right) \mathrm{C} \)

\( =9 \times 10^{-4} \mathrm{~m}=0.9 \mathrm{~mm}\)

(i) If the expansion is fully prevented

then strain \(=\frac{9 \times 10^{-4}}{2} \Rightarrow 4.5 \times 10^{-4}\) \(\therefore \quad\) temperature stress \(=\) strain \(\times Y\) \(=4.5 \times 10^{-4} \times 2 \times 10^{11}=9 \times 10^{7} \mathrm{Nm}^{-2}\)

(ii) If \(0.4 \mathrm{~mm}\) expansion is allowed, then length restricted to expand \(=0.9-0.4=0.5 \mathrm{~mm}\)

\(\therefore \quad\) Strain \(=\frac{5 \times 10^{-4}}{2}=2.5 \times 10^{-4}\)

\(\therefore \quad\) Temperature stress \(=\operatorname{strain} \times Y\)

\(=2.5 \times 10^{-4} \times 2 \times 10^{11}=5 \times 10^{7} \mathrm{Nm}^{-2}\)