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+// Problem 8.1,Page no.206
+
+clc;clear;
+close;
+
+k=1 //KN/m //stiffness of spring
+P=45 //N //Maximum Load
+sigma_s=126 //MPa //Max shear stress
+L=4.5 //cm //Lenght of spring
+G=42 //GPa //Modulus of rigidity
+
+//Calculations
+
+//sigma_s_max=16*P*R*(%pi*d**3)**-1 //Max shear stress
+
+//After substituting values in above equation and simolifying we get
+//1000=42*10**9*d**4*(64*R**3*n)**-1   (//Equation 1)
+
+//R=0.175*10**6*%pi*d**3 //Radius of spring (Equation 2)
+
+//L=n*d //solid length of spring
+//Thus simplifying above equation, n=L*d**-1
+
+//substituting value of n and R in (equation 1) we get,
+
+d=(42*10**9*(1000*64*4.5*10**-2*(0.175*%pi)**3*(10**6)**3)**-1)**0.25*10**2 //cm //diameter of helical spring
+
+//substituting value d in (equation 2) we get,
+R=0.175*10**6*%pi*(d)**3*10**-6*100 //cm //Radius of coil
+D=2*R //cm //Mean diameter of coil
+n=0.045*0.00306**-1 //Number of turns
+
+
+//Result
+printf("The Diameter of wire is %.3f cm",d)
+printf("\n The Mean Diameter of coil is %.2f",D);printf(" cm")