The length of the bar is increased by 1.5mm. Determine the original length and find the stresses in the bar. Take $E_S= 2.1X10^5 MPa; α_S= 12X10^{-6} /°C; E_C= 1X10^5 MPa; α_S= 17.5X10^{-6} /°C$

For parallel combination

$2\hspace{0.05cm}\times\hspace{0.05cm}P_{cu} = P_{st}\\ 2\hspace{0.05cm}\times\hspace{0.05cm}\sigma_cA_c = \sigma_sA_s\\ 2\hspace{0.05cm}\times\hspace{0.05cm}\sigma_c\hspace{0.05cm}\times\hspace{0.05cm}1 = \sigma_s\hspace{0.05cm}\times\hspace{0.05cm}1\\ \hspace{01.5cm} \sigma_s = 2\hspace{0.05cm}\sigma_c$

$(\alpha L t)_{cu} + (\alpha L t)_{st} - (\frac{PL}{AE})_{cu} - (\frac{PL}{AE})_{st} = 1.5\\ (17.5\hspace{0.05cm}\times\hspace{0.05cm}10^{-6}\hspace{0.05cm}\times\hspace{0.05cm}L\hspace{0.05cm}\times\hspace{0.05cm}300) + (12\hspace{0.05cm}\times\hspace{0.05cm}10^{-6}\hspace{0.05cm}\times\hspace{0.05cm}L\hspace{0.05cm}\times\hspace{0.05cm}300) - (\frac{\sigma_c\hspace{0.05cm}\times\hspace{0.05cm}L}{1\hspace{0.05cm}\times\hspace{0.05cm}10^5})_{cu} - (\frac{2\sigma_{cu}\hspace{0.05cm}\times\hspace{0.05cm}L}{2.1\hspace{0.05cm}\times\hspace{0.05cm}10^5})_{st} = 1.5\\ \epsilon_{cu} + \epsilon_{st} = (\alpha_{cu} - \alpha_{st}).t\\ (\frac{\sigma}{E})_{cu} + (\frac{\sigma}{E})_{st} = (\alpha_{cu} - \alpha_{st})t\\ \frac{0.5\hspace{0.05cm}\sigma_{st}}{1\hspace{0.05cm}\times\hspace{0.05cm}10^5} + \frac{\sigma_{st}}{2.1\hspace{0.05cm}\times\hspace{0.05cm}10^5} = (17.5\hspace{0.05cm}\times\hspace{0.05cm}10^{-6} - 12\hspace{0.05cm}\times\hspace{0.05cm}10^{-6})\hspace{0.05cm}\times\hspace{0.05cm}300\\ \sigma_{st} = 169.024\hspace{0.05cm}N/mm^2\\ \sigma_{cu} = 84.51\hspace{0.05cm}N/mm^2\\ L = 2.857\hspace{0.05cm}mm$