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+clear all; clc;
+
+disp("Scilab Code Ex 5.13 : ")
+
+//Given:
+ l = 1.2; //m
+ a = 40; //mm
+ tou_allow = 56; //MPa
+ phi_allow = 0.02; //rad
+ G = 26; //GPa
+ alpha = (60*%pi)/180; //degrees
+ 
+ //Calculations:
+ T_shear1 = (tou_allow*a^3)/(20*1000); // allowable shear stress = (20T)/(a^3)
+ T_twist1 = (phi_allow*a^4*G*10^3)/(46*l*10^6); //angle of twist =(46TL)/(a^4*G)
+ 
+ T1 = min(T_shear1, T_twist1);
+ 
+//Circular Cross Section:
+c_ = (a*a*sin(alpha))/(%pi*2);
+c = sqrt(c_);
+
+J = (%pi/2)*(c^4);
+T_shear2 = (tou_allow*J)/(c*1000);
+T_twist2 = (phi_allow*J*G*10^3)/(l*10^6);
+
+ T2 = min(T_shear2, T_twist2);
+
+
+//Display:
+
+printf('\n\nThe largest torque that can be applied at the end of the triangular shaft    = %1.2f Nm',T1);
+printf('\nThe largest torque that can be applied at the end of the circular shaft      = %1.2f Nm',T2);
+
+
+//------------------------------------------------------------------------------END------------------------------------------------------------------