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+clear
+//
+//
+
+//Initilization of Variables
+
+//Bar-A
+d1=30 //mm //Diameter of bar1
+L=600 //mm //length of bar1
+
+//Bar-B
+d2=30 //mm //Diameter of bar2
+d3=20 //mm //Diameter of bar2
+L2=600 //mm //length of bar2
+
+//Calculations
+
+//Area of bar-A
+A1=%pi*4**-1*d1**2
+
+//Area of bar-B
+A2=%pi*4**-1*d2**2
+A3=%pi*4**-1*d3**2
+
+//let SE be the Strain Energy
+//Strain Energy stored in Bar-A
+//SE=p**2*(2*E)**-1*V
+//After substituting values and simolifying further we get
+//SE=P**2*E**-1*0.4244
+
+//Strain Energy stored in Bar-B
+//SE2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1
+//After substituting values and simolifying further we get
+//SE2=0.6897*P**2*E**-1
+
+//Let X be the ratio of SE in Bar-B and SE in Bar-A
+X=0.6897*0.4244**-1
+
+//Part-2
+
+//When Max stress is produced is same:Let p be the max stress produced
+
+//Stress in bar A is p throughout
+//In bar B:stress in 20mm dia.portion=p2=p
+
+//Stress in 30 mm dia.portion
+//p1=P*A2*A3**-1
+//After substituting values and simolifying further we get
+//p1=4*9**-1*p
+
+//Strain Energy in bar A
+//SE_1=p**2*(2*E)**-1*A1*L1
+//After substituting values and simolifying further we get
+//SE_1=67500*p**2*%pi*E**-1
+
+//Strain Energy in bar B
+//SE_2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1
+//After substituting values and simolifying further we get
+//SE_2=21666.67*%pi*p**2*E**-1
+
+//Let Y be the Ratio of SE in bar B and SE in bar A
+Y=21666.67*67500**-1
+
+//result
+printf("\n Gradually applied Load is %0.2f ",X)
+printf("\n Gradually applied Load is %0.2f ",Y)