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+// exa 3.6 Pg 69
+
+clc;clear;close;
+
+// Given Data
+N=200;// rpm
+P=200;// kW
+tau_d=42;// Mpa
+W=900;// N
+L=3;// m
+sigma_t=56;// MPa
+sigma_c=56;// MPa
+
+T=P*60*10**3/(2*%pi*N);// N.m
+M=W*L/4;// N.m
+Te=sqrt(M**2+T**2);// N.m
+//Te=(%pi/16)*d**3*tau_d
+d=(Te/((%pi/16)*tau_d)*1000)**(1/3);// mm
+printf('\n Using equivalent torque equation,\n shaft diameter d = %.f mm',d)
+
+Me=(1/2)*(M+sqrt(M**2+T**2));// N.m
+//Me=(%pi/32)*d**3*sigma_d
+d=(Me/((%pi/32)*sigma_c)*10**3)**(1/3);//mm
+printf('\n Using equivalent bending moment equation,\n shaft diameter d = %.2f mm or %.f mm',d, ceil(d))
+printf('\n Adopt d=105 mm.')