Question

A steel rod of length 2000cm and uniform cross sectional area 3×10^3cm2 at 28°C. Find the change in length and the force due to exerted by the rod

(Linear expansivity of steel=1.2×10^-5/K

Young's modulus of steel = 2.0×10^7/N/cm2

Answer 1

We use the definition for the tensile stress $F/A$ required to keep the rod’s length constant, then we find the force exerted with the Young's modulus (Y), the linear expansivity of steel ($\alpha$), the change of temperature as $\Delta T =T-T_0=(28-20)°C=8°C=8\,K$, and the cross-section area (A):

$\frac{F}{A} = -Y\alpha \Delta T \implies F =-YA\alpha \Delta T \\ F =-(2\times10^{7}\frac{N}{cm^2})(3\times 10^3\,cm^2)(1.2\times10^{-5}\,K^{-1})(8\,K) \\ F =-5.76\times10^{6}\,N$

The change in length of the rod can be found with the relations for the thermal and tension fractional changes:

$(\frac{\Delta L }{L_0})_{thermal} = \alpha \Delta T \\ \implies \Delta L_{thermal} =\alpha \Delta T L_0=... \\ ...= (1.2\times10^{-5}K^{-1})(8K)(2000\,cm) \\ \therefore \Delta L_{thermal} =0.192\,cm \\ (\frac{\Delta L }{L_0})_{tension} = \frac{F}{AY}=-(\frac{\Delta L }{L_0})_{thermal} \\ \therefore \Delta L_{tension} =-\,0.192\,cm$

In conclusion, we find that the change in length is $\Delta L_{tension}=-0.192\,cm$, and the force exerted by the rod as $F=-5.76\times 10^{6}\,N$ (F<0 because it is a compressive force on the steel rod).

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.