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+clear
+//
+//
+
+dell=0.25 //mm //Instantaneous Extension
+
+//Bar-A
+b1=25 //mm //width of bar
+D1=500 //mm //Depth of bar
+
+//Bar-B
+b2_1=25 //mm //width of upper bar
+b2_2=15 //mm //Width of Lower Bar
+L2=200 //mm //Length of upper bar
+L1=300 //mm //Length of Lower bar
+
+E=2*10**5 //N/mm**2 //Youngs Modulus of bar
+
+//Calculations
+
+//Strain
+e=dell*D1**-1
+
+//Load
+p=e*E
+
+//Area of bar-A
+A=%pi*4**-1*25**2
+
+//Volume of bar-A
+V=A*D1
+
+//Let E1 be the Energy of Blow
+//Energy of Blow
+E1=p**2*(E)**-1*V
+
+//Let p2 be the Max stress in bar B When this blow is applied.
+//the max stress occurs in the 15mm dia. portion,Hence, the stress in 25 mm dia.portion is
+//p2*%pi*4**-1*b2_2**2*(%pi*4**-1*b2_2**2=0.36*p
+
+//Strain Energy of bar B
+//E2=p**2*(2*E)**-1*v1+1*(2*E)**-1*(0.36*p2)**2*v2
+//After substituting values and Further substituting values we get
+//E2=0.1643445*p2**2
+
+//Equating it to Energy of applied blow,we get
+p2=(12271.846*0.1643445**-1)**0.5
+
+//Stress in top portion
+sigma=0.36*p2
+
+//Extension in Bar-1
+dell_1=p2*E**-1*L1
+
+//Extension in Bar-2
+dell_2=0.36*p2*E**-1*L2
+
+//Extension of bar
+dell_3=dell_1+dell_2
+
+//Result
+printf("\n Instantaneous Max stress is %0.2f N/mm**2",sigma)
+printf("\n extension in Bar is %0.2f mm",dell_3)