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+clear
+//
+//
+
+//Initilization of Variables
+
+d1=100 //mm //External Diameter
+d2=50 //mm //Internal Diameter
+N=500 //mm //r.p.m
+P=60*10**6 //N-mm/sec //Power
+p=100 //N/mm**2 //principal stress
+
+//Calculations
+
+//M.I
+I=%pi*(d1**4-d2**4)*64**-1 //mm**4
+
+//Bending Stress
+//f=M*I*d1*2**-1 //N/mm**2
+
+//Principal Planes
+//p_x=32*M*(%pi*(d1**4-d2**4))*d1
+//p_y=0
+
+//Shear stress
+//q=T*J**-1*(d1*2**-1)
+//After sub values and further simplifying we get
+//q=16*T*d1*(%pi*(d1**4-d2**4))*d1
+
+//Principal stresses
+//P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2
+//After sub values and further simplifying we get
+//P1=16*(%pi*(d1**4-d2**4))*d1*(M+(M**2+t**2)**0.5) ...............(1)
+
+//P=2*%pi*N*T*60**-1
+//After sub values and further simplifying we get
+T=P*60*(2*%pi*N)**-1*10**-6 //N-mm
+
+//Again Sub values and further simplifying Equation 1 we get
+M=(337.533)*(36.84)**-1 //KN-m
+
+//Min Principal stress
+//P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 //N/mm**2
+//Sub values and further simplifying we get
+P2=16*(%pi*(d1**4-d2**4))*d1*(M-(M**2+T**2)**0.5)*10**-11
+
+//Result
+printf("\n Bending Moment safely applied to shaft is %0.2f KN-m",M)
+printf("\n Min Principal Stress is %0.3f N/mm**2",P2)