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+// Problem 8.16,Page no.217
+
+clc;clear;
+close;
+
+n=10 //Number of coils
+sigma=100 //MPa //Bending stress
+sigma_s=110 //MPa //Twisting stress
+//D=8*d
+dell=1.8 //cm //Max extension of of wire
+E=200 //GPa //Modulus of Elasticity
+G=80 //GPa //Modulus od Rigidity
+
+//Calculation
+
+//M=W*R*sin_alpha=%pi*d**3*sigma_1*32**-1 //(Equation 1) //Bending moment
+//As D=8*d 
+//then R=D*2**-1
+//Therefore, R=4*d
+
+//Now substituting values in equation 1 we get
+//W*sin_alpha=2454369.3*d**2  (Equation 2)
+
+//T=W*R*cos_alpha=%pi*d**3*sigma_s //Torque  (Equation 3)  
+//Now substituting values in equation 3 we get
+//W*cos_alpha=5399612.4*d**2    (Equation 4)
+
+//Dividing Equation 2 by Equation 4 we get,
+//tan_alpha=0.4545
+alpha=atan(0.4545)*180*%pi**-1
+
+//From Equation 2 we get
+//W=2454369.3*d**2*(sin24.443)**-1
+//W=5931241.1*d**2   (Equation 5)
+
+//dell=64*W*R**3*n*sec_alpha*(d**4)**-1*((cos_alpha)**2*G**-1+2*sin_alpha**2*E**-1)
+//Now substituting values in above equation  we get
+//W=33140.016*d  (Equation 6)
+
+//From Equation 5 and Equation 6 we get
+//5931241.1*d**2=33140.016*d
+//After simplifying above equation we get
+d=33140.016*5931241.1**-1 //m //Diameter of wire
+W=33140.016*d //N //MAx Permissible Load
+
+//Result
+printf("The Max Permissible Load is %.2f",W);printf(" N")
+printf("\n The Wire Diameter is %.6f m",d)