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+// Problem 8.19,Page no.219
+
+clc;clear;
+close;
+
+d=6 //mm //Diameter of steel wire
+n=1 //number of turns
+D=6.5 //cm //Mean of diameter
+G=80 //GPa //modulus of rigidity
+P_1=150 //Load
+p=1.5 //cm //%pitch of coil
+
+//Calculation
+
+R=D*2**-1
+//For one turn deflection is
+dell=p-d*10**-1 //cm
+
+//dell=64*P*R**3*n*(G*d**4)**-1
+//Now, after simplifying further we get,
+P=dell*10**-2*G*10**9*(d*10**-3)**4*(64*(R*10**-2)**3*n)**-1 //N //Axial Load
+
+dell_2=dell*8 //cm //Total Displacement //Notification has been changed
+U=P*dell_2*10**-2*2**-1 //N-m //Strain Energy
+
+//Potential Energy given by 150N Load is
+//U=150*(h+0.072)
+
+//After simplifying above equation we get
+h=(U*P_1**-1-0.072)*10**2 //cm //Height from which 150 N load falls
+
+//Result
+printf("Axial Load is %.2f",P);printf(" N")
+printf("\n Height from which 150 N load falls is %.2f",h);printf(" cm" )