Steel rod inside copper tube

Section areas

$A_s = \frac{\pi}{4}(30)^2 = 706.9\ \text{mm}^2$, $A_c = \frac{\pi}{4}(50^2 - 30^2) = 1256.6\ \text{mm}^2$.

(a) Load sharing

Equilibrium: $P_s + P_c = 150\times10^3$ N. Compatibility (same shortening, same length): $\varepsilon_s = \varepsilon_c$, so

$$\frac{\sigma_s}{E_s} = \frac{\sigma_c}{E_c} \;\Rightarrow\; \sigma_c = \sigma_s\frac{E_c}{E_s} = 0.5\,\sigma_s$$ $$\sigma_s A_s + \sigma_c A_c = 150000 \;\Rightarrow\; \sigma_s\,(706.9 + 0.5\times1256.6) = 150000$$ $$\sigma_s = 112.3\ \text{MPa}, \qquad \sigma_c = 56.2\ \text{MPa}$$ $$P_s = 112.3\times706.9 = 79.4\ \text{kN}, \qquad P_c = 56.2\times1256.6 = 70.6\ \text{kN}$$

(check: $79.4 + 70.6 = 150$ kN)

$$\delta = \frac{\sigma_s L}{E_s} = \frac{112.3\times600}{200000} = 0.337\ \text{mm (shortening)}$$

(b) Temperature rise of 40 degrees C

Copper tends to expand more than steel, but they are held together. Copper is held back (extra compression) and steel is pulled along (extra tension). The internal forces are equal and opposite, $P_t$:

$$\alpha_c\Delta T L - \frac{P_t L}{A_cE_c} = \alpha_s\Delta T L + \frac{P_t L}{A_sE_s}$$ $$P_t = \frac{(\alpha_c - \alpha_s)\Delta T}{\dfrac{1}{A_cE_c} + \dfrac{1}{A_sE_s}} = \frac{5\times10^{-6}\times 40}{\dfrac{1}{1256.6\times10^5} + \dfrac{1}{706.9\times2\times10^5}} = 13.31\ \text{kN}$$

Thermal stresses: steel $= +13310/706.9 = +18.8$ MPa (tension), copper $= -13310/1256.6 = -10.6$ MPa (compression).

Final stresses (add to the load stresses, compression negative):

Check: $-93.5\times706.9 - 66.8\times1256.6 = -150$ kN.

Answer: (a) $\sigma_s = 112.3$ MPa, $\sigma_c = 56.2$ MPa (both compressive), $P_s = 79.4$ kN, $P_c = 70.6$ kN, shortening $0.337$ mm; (b) final $\sigma_s = -93.5$ MPa, $\sigma_c = -66.8$ MPa.