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+clear
+//
+//
+
+//Initilization of Variables
+
+D=25 //mm //Diameter of Brass
+De=50 //mm //External Diameter of steel tube
+Di=25 //mm //Internal Diameter of steel tube
+L=1.5 //m //Length of both bars
+t1=30 //degree celsius //Initial Temperature
+t2=100 //degree celsius //final Temperature
+E_s=2*10**5 //N/mm**2 //Modulus of ELasticity of steel bar
+E_b=1*10**5 //N/mm**2 //Modulus of Elasticity of brass bar
+alpha_s=11.6*10**-6 //Temperature Coeff of steel
+alpha_b=18.7*10**-6 //Temperature coeff of brass bar
+d=20 //mm //diameter of pins
+
+//Calculations
+
+t=t2-t1 //Temperature Difference
+A_s=%pi*4**-1*(De**2-Di**2) //mm**2 //Area of steel
+A_b=%pi*4**-1*D**2 //mm**2 //Area of brass
+
+//Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar
+//But from Equilibrium of Forces 
+//P_b=P_s=P
+
+//Let dell=dell_s+dell_b
+dell=(alpha_b-alpha_s)*t*L*1000
+
+P=dell*(1*(A_s*E_s)**-1+1*(A_b*E_b)**-1)**-1*(L*1000)**-1
+P_b=P
+P_s=P
+//Stress in steel
+sigma_s=P*A_s**-1
+
+//Stress in Brass
+sigma_b=P_b*A_b**-1
+
+//Area of Pins
+A_p=%pi*4**-1*d**2
+
+//Since,the force is resisted by two cross section of pins
+tou=P*(2*A_p)**-1
+    
+//Result
+printf("\n Stress in steel bar is %0.2f  N/mm**2",sigma_s)
+printf("\n Stress in Brass bar is %0.2f  N/mm**2",sigma_b)
+printf("\n Shear Stresss induced in pins is %0.2f  N/mm**2",tou)