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+// Problem 3.15,Page no.63
+
+clc;clear;
+close;
+
+A_s=0.003848 //m**2 //Area of steel bar
+A_al=0.003436 //m**2 //Area of Aluminium tube
+E=220*10*9 //N //Young's modulus of steel
+E=70*10*9 //N //Young's modulus of aluminium
+P=600*10**3 //N //Load applied to the bar
+//dell_L_al-dell_L_s=0.00015 //mm //difference between strain in aluminium bar and steel bar
+
+//Calculations
+
+
+//Let the aluminium tube be compressed by dell_L_al and steel bar by by dellL_s
+//dell_L_al=sigma_al*E_al**-1*L_al
+//dell_L_s=sigma_s*E_s**-1*L_s
+
+//After substituting and simplifying above equation we get,
+//((sigma_al*70**-1)-(sigma_s*220**-1))=300000 //(equation 1)
+
+//After simplifying above equation we get,
+//sigma_al=17462.165*10**4-1.1199*sigma_s //(equation 2)
+
+//Now substituting sigma_al in equation(1)
+//((17462.165*10**4-1.1199*sigma_s)*(70)**-1)-(sigma_s*220**-1)=300000
+
+//After simplifying above equation we get,
+
+sigma_s=-((300000-249.4594*10**4)*0.0205444**-1)*10**-6 //MN/m**2 //stress developed in steel bar
+//sigma_al=17462.165*10**4-1.1199*sigma_s
+sigma_al=(17462.165*10**4-1.1199*106822005.02)*10**-6
+
+
+//Result
+printf("stress developed in steel bar is %.2f MN/m^2",sigma_s)
+printf("\n stress developed in aluminium bar is %.2f MN/M^2",sigma_al)