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+// exa 3.10 Pg 71
+
+clc;clear;close;
+
+// Given Data
+d=4;// cm
+M=15000;// N.cm
+Syt=20000;// N/cm.sq.
+
+printf('\n (i) Maximum Principal Stress Theory-')
+z=%pi*d**3/32;// cm.cube.
+sigma_b=M/z;// N/cm.sq.
+T=poly(0,'T')
+tau=16*T/(%pi*d**3);// N/cm.sq.
+//sigma1=(1/2)*(sigma_b+sqrt(sigma_b**2+4*tau**2)) // Maximum principal stress
+//sigma1=(sigma_b/2+sqrt(sigma_b**2/4+tau**2)) // on solving
+//tau=sqrt((sigma1-sigma_b/2)**2-sigma_b**2/4)
+sigma1=Syt;// N/cm.sq.
+T=sqrt((sigma1-sigma_b/2)**2-sigma_b**2/4)*(%pi*d**3)/16;// N.cm.
+printf('\n Maximum value of torque, T = %.f N.cm.',T)
+
+printf('\n (ii) Maximum Shear Stress Theory')
+tau_d=0.5*Syt;// N.cm.
+//Te=sqrt(M**2+T**2)=(%pi/16)*d**3*tau_d
+T=sqrt(((%pi/16)*d**3*tau_d)**2-M**2);// N.cm.
+printf('\n Maximum value of torque, T = %.f N.cm.',T)
+// Answer in the textbook is not accurate.