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+// Problem 3.9,Page no.59
+
+clc;clear;
+close;
+
+//P is the force acting on the bar BC compressive in nature and force on AB is (100-P) Tensile in nature
+E=200*10*9 //N
+A_1=3*10**-4 //cm**2 //Area of AB
+A_2=4*10**-4 //cm**2 //Area of BC
+L=1.5 //cm //Length of bar
+
+//Calculations
+
+//The total elongation of bar
+//(((100-P)*10**3*1.5)*(3*10**-4*E)**-1)-((P*10**3*1.5)*(4*10**-4*E)**-1)=0 
+
+//The total elongation of bar is limited to 1
+//(25-0.4375*P)*10**-4=1*10**-3
+
+//After simplifying above equation we get,
+P=-(10-25)*0.4375**-1 //KN //Total elongation of bar
+F_AB=100-P //KN //force in AB
+F_BC=P //KN //Force in BC
+sigma_AB=(((F_AB)*(3*10**-4)**-1)*10**-3) //KN //Stress in AB
+sigma_BC=((F_BC)*(4*10**-4)**-1*10**-3) //KN //Stress in Bc
+
+
+//Result
+printf("F_AB = %.2f kN",F_AB)
+printf("\n F_BC = %.2f kN",F_BC)
+printf("\n sigma_AB = %.2f kN",sigma_AB)
+printf("\n sigma_BC = %.2f kN",sigma_BC)