(8 marks) May-18

Area of steel rod $A_s=\frac{\pi}{4}*15^2=56.25 \pi mm^2$

Area of coppper rod $A-c=\frac{\pi}{4}*(50-40)^2=225\pi mm^2$

Free expansion:

$\delta_{steel}=\alpha_s TL $

$\delta_{copper}=\alpha_c TL $

Let the actual expansion of each component be \delta

$\alpha_c TL \gt\delta\gt \alpha_s TL $

Steel is in tension and copper is in compression. Let $P_s$ and $P_c$ be the stresses in steel and copper. For the equilibrium system,

Tension in steel=Compression in copper

$P_sA_s=P_cA_c$

$P_s=\frac{A_c}{A_s}P_c$

$P_s=\frac{225\pi}{56.25\pi}P_c$

$P_s=4P_c$

Actual expansion of steel=actual expansion of copper

$\alpha_s T_l+\frac{P_s}{E_s}l=\alpha_c T_l-\frac{P_c}{E_c}l$

ie.$ \alpha_sT+\frac{P_s}{E_s}=\alpha_cT-\frac{P_c}{E_c}$

Substituting values of $P_s, \alpha_s, \alpha_c, E_s, E_c$

$12*10^{-6}*60+\frac{4P_c}{2.1*10^5}=17.5*10^{-6}*60-\frac{P_c}{1.05*10^5}$

$P_c=11.55 N/mm^2$ (Compressive)

$P_s=4P_c=4*11.55 N/mm^2=46.20 N/mm^2$ (Tensile)