## Bar Expansion During Its Cooling

### Task number: 2220

We heat up a steel bar with cross section diameter of 2 cm^{2} from temperature 0 °C to 50 °C. Then we cool it down back to its original temperature.

Calculate the smallest force acting on the axis of the bar which we would need to use to prevent the shortening of the bar. Young’s module is for simplicity considered a constant.

*Note*: Coefficients of thermal expansion of steel and iron are approximately the same.

#### Hint

To prevent the shortening of the bar, we need to act on it with the same force as if we wanted to prolong it at 0 °C to the length it will have at 50 °C.

#### Analysis

The prolonging of the heated bar is directly dependent on the temperature difference. The bar will shorten back to its original length during the cooling. If we want to prevent it, we have to act on it with the same force as if we wanted to prolong it from its original length to the length it will have at 50 °C. This prolonging follows Hooke’s law for deformation by stretching.

#### List of Known Information

*S*= 2 cm^{2}= 2·10^{−4}m^{2}bar’s cross section diameter *t*_{0}= 0 °Cbar’s original temperature *t*_{1}= 50 °Cbar’s temperature after heating up *F*= ?force that we have to act with *From tables:**α*= 12·10^{−6}°C^{−1}coefficient of thermal expansion of steel *E*= 21·10^{10}PaYoung’s module of steel #### Solution

The bar’s length increases when being heated up. This increasement Δ

\[ \mathrm{\Delta} l=l_0 \alpha\mathrm{\Delta}t =l_0\alpha(t_1-t_0), \]*l*is directly dependent on the temperature difference Δ*t*=*t*_{1}−*t*_{0}according to equationwhere

*l*_{0}is the original length of the bar and*α*is the coefficient of thermal expansion.The bar’s length should decrease back to the original length

*l*0 after cooling down to the original temperature. If we want to prevent this, we have to strech the bar with a force equal to the force we would need to stretch it with if we wanted to lengthen it from length*l*_{0}to*l*_{0}+Δ*l*.Force

\[ \sigma=E\varepsilon, \]*F*required to increase the relative length of the bar by*ε*is directly dependent on this elongation according to Hooke’s law:where the constant of proportion is Young’s module of elasticity

\[\sigma=\frac{F}{S}\]*E*and*σ*is normal tension caused by force*F*defined asand relative elongation

\[\varepsilon=\frac{\mathrm{\Delta}l}{l_0}.\]*ε*is defined by relationWe gain after substitution for

\[\frac{F}{S}=E\,\varepsilon\qquad\Rightarrow\qquad F=E\,S\,\varepsilon=E\,S\,\frac{\mathrm{\Delta}l}{l_0}.\]*σ*and*ε*into Hooke’s law:Now we can substitute for Δ

\[F=E\,S\,\frac{\mathrm{\Delta}l}{l_0}=E\,S\,\frac{l_0\,\alpha\,\left(t_1-t_0\right)}{l_0}=E\,S\,\alpha\,\left(t_1-t_0\right).\]*l*and get the final relation:Note that the required force is independent of the bar’s length so it doesn’t matter that it wasn’t given.

**Numerical substitution:**#### Answer

We need to act on the bar with force of 25.2 kN to prevent it from shortening.