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+clear
+//
+//
+
+//Initilization of Variables
+
+M=40*10**6 //N-mm //Bending moment
+T=10*10**6 //N-mm //TOrque
+mu=0.25 //Poissons ratio
+P_e=200 //N/mm**2 //Stress at Elastic Limit
+FOS=2
+
+//Calculations
+
+//Let d be the diameter of the shaft
+
+//Principal stresses are given by
+
+//P1=16*(%pi*d**3)**-1*(M+(M**2+T**2)**0.5)
+//After substituting values and further simplifying we get
+//P1=4.13706*10**8*(d**3)**-1     ............................(1)
+
+//P2=16*(%pi*d**3)**-1*(M-(M**2+T**2)**0.5)
+//After substituting values and further simplifying we get
+//P2=-6269718*(%pi*d**3)**-1       ..............................(2)
+
+//q_max=(P1-P2)*2**-1
+//After substituting values and further simplifying we get
+//q_max=2.09988*10**8*(d**3)**-1
+
+//Max Principal stress theory
+//P1=P_e*(FOS)**-1
+//After substituting values and further simplifying we get
+d=(4.13706*10**8*2*200**-1)**0.33333 //mm 
+
+//Max shear stress theory
+//q_max=shear stress at elastic limit*(FOS)**-1
+//After substituting values and further simplifying we get
+d2=(2.09988*10**8*4*200**-1)**0.33333
+
+//Max strain energy theory
+//P_3=0
+//P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1
+//After substituting values and further simplifying we get
+d3=(8.62444*10**12)**0.166666
+
+//Result
+printf("\n Diameter of shaft according to:MAx Principal stress theory %0.2f  mm",d)
+printf("\n                               :Max shear stress theory %0.2f  mm",d2)
+printf("\n                               :Max strain energy theory %0.2f  mm",d3)