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+// Problem no 7.1,Page no.183
+
+clc;clear;
+close;
+
+G=84 //Gpa //Modulus of Rigidity
+N=110 //no. of revolution
+//d*D**-1=0.6 //Ratio of inner diameter to outer diameter
+sigma_s=63 //MPa //shear stress
+L=3 //m //Length of shaft
+P=590 //KW //Power
+
+//Calculation
+
+//P=2*%pi*N*T_mean*60000**-1 //KW //Power
+T_mean=P*60000*(2*%pi*N)**-1 //N*m //Mean Torque
+
+//I_p=p*32**-1*(D**4-d**4)
+
+//After substituting value of d in above equation we get
+//I_p=0.0272*%pi*D**4 //m**4 //Polar moment of Inertia
+
+T_max=1.2*T_mean //N*m //Max torque
+
+//Using Relation
+//T_max*T_p**-1=sigma_s*R**-1=G*theta*L**-1 
+
+//After substituting values and simplifying we get
+
+D=(5.7085*10**-3)**0.3333 //m //Diameter of shaft
+
+theta=1.4*%pi*180**-1 //radians
+
+//theta=((T_max*L)*(G*10**9*I_p)) //radians
+
+//After substituting values and simplifying we get
+D_1=(1.0513*10**-3)**0.25
+
+//Result
+printf("The Minimum external diameter is %.2f",D_1);printf(" m")