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+// exa 6.1 Pg 168
+clc;clear;close;
+
+// Given Data
+Sut=650;// MPa
+Syt=380;// MPa
+F1BYF2 = 2.5;// ratio of tensions
+Fmax=2.5;// kN
+da=200;// mm
+db=400;// mm
+L=1*1000;//mm
+Km=1.5;// fatigue factor
+Kt=1;// shock factor
+
+
+tau_d1=0.30*Syt;// MPa
+tau_d2=0.18*Sut;// MPa
+tau_d=min(tau_d1, tau_d2);// MPa (taking minimum value)
+tau_d=0.75*tau_d;//MPa (Accounting keyway effect)
+
+// Pulley A
+F1=2500;// N
+F2=1000;// N
+T=(F1-F2)*da/2;// N.mm
+Fa=F1+F2;// N (resultant pull Downwards)
+
+// Pulley B
+// F3 & F4 are tension in belt (assumed)
+//T=(F3-F4)*db/2
+SUB_F3F4 = 2*T/db;// N (where SUB_F3F4 = F3-F4) --eqn(1)
+F3BYF4=F1BYF2;// ratio of tensions --eqn(2)
+F4 = SUB_F3F4/(F3BYF4-1);// N (using above 2 equations)
+F3=F3BYF4*F4;// N
+Fb=F3+F4;// N (resultant pull right side( -->))
+
+// BENDING MOMENTS -
+Mav=Fa*L/4;// N.mm (vertical force)
+Mc=Fb*da;// N.mm
+Mah=Mc/2;// N.mm (vertical force)
+M = sqrt(Mav**2+Mah**2);// N.mm (resultant bending moment at A)
+d=((16/%pi/tau_d)*sqrt((Km*M)**2+(Kt*T)**2))**(1/3);// mm
+
+printf('shaft diameter = %.2f mm. Use diameter = 45 mm.',d)