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+// Problem 8.5,Page no.210
+
+clc;clear;
+close;
+
+d=6 //mm //Diameter of steel wire
+n=50 //number of turns
+D=5 //cm //Mean Diameter
+R=D*2**-1 //cm //Radius of coil
+G=80 //GPa //Modulus of Rigidity
+P=150 //KN //Load
+
+//Calculation
+
+//Dell=64*P*R**3*n*(G*d**4)**-1 //Deflection
+//After substituting values in above equation and simplifying we get
+//P=2073.6*dell //Gradually applied equivalent Load
+
+//loss of potential Energy of the weight=Gain of strain Energy of the spring
+//150*(0.05+dell)=P*dell*2**-1
+//After substituting values in above equation we get
+
+//dell**2-0.1446*dell-0.00723=0
+//Above Equation is in the form of ax^2+bx+c=0
+
+a=1
+b=-0.1446
+c=-0.00723
+
+//First computing value of  b^2-4ac and store it in a variable say X
+X=b**2-(4*a*c)
+//now roots are given as x=(-b+X**0.5)/(2*a) and second root is negative sign before X
+
+
+dell_1=(-b+X**0.5)*(2*a)**-1*10**2
+dell_2=(-b-X**0.5)*(2*a)**-1*10**2
+
+P=2073.6*dell_1*10**-2 //N 
+
+sigma=16*P*R*10**-2*(%pi*(d*10**-3)**3)**-1 //N/m**2 //Max stress
+
+//Result
+printf("The Max Extension of the Spring is %.2f",dell_1);printf(" cm")
+printf("\n The Max stress is %.3e N/m^2",sigma)