**For Aluminium; $E_{Al}= 70 GPa; α_{AL}= 23X10-6 /°C; A_{AL}=2000 mm^2\\ For Steel; E_{ST} = 190 GPa ; α_{ST} = 18X10-6 /°C; A_{ST}=8000 mm^2$

When the temperature reaches 140°C find i)Stresses in Al and ST ii) Exact length of the AL and ST rod.**

For parallel combination $\hspace{0.8cm}P_{Al} = P_{St}\\ \sigma_{Al}\hspace{0.05cm} .\hspace{0.05cm} A_{Al} = \sigma_{st}\hspace{0.05cm} .\hspace{0.05cm} A_{st}\\ \sigma_{Al}\hspace{0.05cm} \times\hspace{0.05cm} 2000 = \sigma_{st}\hspace{0.05cm} \times\hspace{0.05cm} 8000\\ \hspace{1cm} \sigma_{Al} = 4\hspace{0.05cm} \sigma_{st}\\$ (i) Stresses in AL and ST $T = 140 - 20 = 120\hspace{0.05cm} ^\circ{C}\ (\alpha L t)_{Al} + (\alpha L t)_{st} - (\frac{PL}{AE})_{Al} - (\frac{PL}{AE})_{st} = 0.5\ (23\hspace{0.05cm} \times\hspace{0.05cm} 10^{-6}\hspace{0.05cm} \times\hspace{0.05cm} 300\hspace{0.05cm} \times\hspace{0.05cm} 120) + (18\hspace{0.05cm} \times\hspace{0.05cm} 10^{-6}\hspace{0.05cm} \times\hspace{0.05cm} 250\hspace{0.05cm} \times\hspace{0.05cm} 120) - (\frac{4\hspace{0.05cm} \sigma_{st}\hspace{0.05cm} \times\hspace{0.05cm} 300}{70\hspace{0.05cm} \times\hspace{0.05cm} 10^3}) - (\frac{\sigma_{st}\hspace{0.05cm} \times\hspace{0.05cm} 250}{190\hspace{0.05cm}\times\hspace{0.05cm} 10^3}) = 0.5\ \sigma_{st} = 53\hspace{0.05cm} MPa\ \sigma_{Al} = 214\hspace{0.05cm} MPa$ (ii) Exact lengths of AL and ST Exact length in AL = $L + \delta L\ =300 + (\frac{PL}{AE})_{Al}\ =300 + (\frac{\sigma_{Al} + L_{Al}}{E_{Al}})\ =300.91\hspace{0.05cm} mm$