Given

$E_{st} = 2.1\hspace{0.05cm}\times\hspace{0.05cm}10^5,\hspace{0.25cm}E_{Al} = 0.7\hspace{0.05cm}\times\hspace{0.05cm}10^5\\ T = 200\hspace{0.05cm}^\circ{C},\hspace{0.9cm}\delta = 0.1\\ \alpha_{st} = 12\hspace{0.05cm}\times\hspace{0.05cm}10^{-6}/\hspace{0.05cm}^\circ{C}\\ \alpha_{Al} = 23\hspace{0.05cm}\times\hspace{0.05cm}10^{-6}/\hspace{0.05cm}^\circ{C}$

Solution

$A_{St} = \frac{\pi}{4}\hspace{0.05cm}\times\hspace{0.05cm}10^2 = 78.54\hspace{0.05cm}mm^2\\ A_{Al} = \frac{\pi}{4}\hspace{0.05cm}\times\hspace{0.05cm}20^2 = 314.16\hspace{0.05cm}mm^2\\ (\alpha L t)_{Al} + (\alpha L t)_{st} = (\frac{PL}{AE})_{Al} + (\frac{PL}{AE})_{St} + 0.1\\ (\alpha L t)_{Al} + (\alpha L t)_{st} - (\frac{\sigma L}{E})_{Al} - (\frac{\sigma L}{E})_{St} = 0.1\\ (\alpha_{Al}. L_{Al}.t) + (\alpha_{St}. L_{St}.t) - \frac{\sigma_{Al}.L_{Al}}{E_{al}} - \frac{\sigma_{st}.L_{st}}{E_{st}} = 0.1$

For parallel combination,

$\hspace{0.6cm}P_{Al} = P_{St}\\ \sigma_{Al}\hspace{0.05cm}.\hspace{0.05cm}A_{Al} = \sigma_{st}\hspace{0.05cm}.\hspace{0.05cm}A_{st}\\ \sigma_{Al}\hspace{0.05cm}\times\hspace{0.05cm}314.16 = \sigma_{st}\hspace{0.05cm}\times\hspace{0.05cm}78.5\\ \sigma_{Al} = 0.24\hspace{0.05cm}\sigma_{st}$

Therefore,

$(23\hspace{0.05cm}\times\hspace{0.05cm}10^{-6}\hspace{0.05cm}\times\hspace{0.05cm}300\hspace{0.05cm}\times\hspace{0.05cm}200) + (12\hspace{0.05cm}\times\hspace{0.05cm}10^{-6}\hspace{0.05cm}\times\hspace{0.05cm}600\hspace{0.05cm}\times\hspace{0.05cm}200) - (\frac{0.24\sigma_{st}\hspace{0.05cm}\times\hspace{0.05cm}300}{0.7\hspace{0.05cm}\times\hspace{0.05cm}10^5}) - (\frac{\sigma_{st}\hspace{0.05cm}\times\hspace{0.05cm}600}{2.1\hspace{0.05cm}\times\hspace{0.05cm}10^5}) = 0.1\\ \hspace{05cm}\sigma_{st} = 700\hspace{0.05cm}N/mm^2\\ \hspace{05cm}\sigma_{Al} = 0.24\hspace{0.05cm}\sigma_{st} = 168\hspace{0.05cm}N/mm^2$