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+// Problem 8.2,Page no.207
+
+
+clc;clear;
+close;
+
+L=15 //cm //Length of close coiled helical spring
+U=50 //N*m //Strain energy
+sigma_s=140 //MPa //Shear stress
+D=10 //cm //Mean coil diameter
+G=80 //GPa //Modulus of rigidity
+
+R=D*2**-1 //cm //Mean coil Radius
+
+//Calculations
+
+//Let dell be the deflection of the spring when fully compressed
+// 0.15-dell=n*d (Equation 1)
+ 
+//U=(sigma_s)**2*V*(4*G)**-1 //Strain energy
+
+//After substituting values in above equation and simolifying we get
+V=50*4*80*10**9*((140*10**6)**2)**-1 //m**3 //Volume of spring
+
+//But V=%pi*4**-1*d**2*2*%pi*R*n
+//After substituting values in above equation and simolifying we get
+//n=3.308*10**-3*(d**2)**-1 //Number of turns
+
+//We know, T=P*R 
+//Now substituting values in T and simolifying we get
+//P=549.7787*10**6*d**3  //Load
+
+//U=P*dell*2**-1
+//After substituting values in above equation and simolifying we get
+//dell=0.18189*10**-6*d**3 //Deflection 
+
+//After substituting values in above equation and simolifying we get
+
+//d**3-22.0533*10**-3*d**2-1.21261*10**-6=0
+
+Coeff=[1 -22.0533*10**-3 0 -1.21261*10**-6]
+d=roots(Coeff)  //Diameter of steel wire
+n=3.308*10**-3*((d(1)**2)**-1) //no.of coils
+
+//Result
+printf("Diameter of steel wire is %.5f m",d(1))
+printf("\n number of coils  = %d",ceil(n))