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+// Problem no 7.2,Page no.184
+
+clc;clear;
+close;
+
+P=295 //KW //Power
+N=100 //R.p.m 
+sigma_s=80 //MPa //shear stress
+
+
+//Calculations
+
+T_mean=((P*60000)*(2*%pi*N)**-1) //N*m
+
+//T_max=T_mean=(%pi*D**3*sigma_s)*16**-1
+D=((T_mean*16)*(%pi*sigma_s*10**6)**-1)**0.333 //m //Diameter of solid shaft
+
+//For hollow shaft
+//I_p_h=%pi*32**-1*(D_1**4-d_1**4) (equation 1)
+
+//Now d_1=0.6*D_1
+//substituting above value in equation 1,we get,
+
+//I_p_h=0.0272*%pi*D_1**4
+
+//For solid shaft
+//I_p_s=%pi*32**-1*D**4
+
+//T and sigma_s being the same then I_p*R**-1 will be the same for the two shafts
+//Using relation  I_p_h*R_1**-1=I_p_s*R**-1
+
+//Substituting values and simplifying we get
+
+D_1=(D**3*0.8704**-1)**0.3333333 //m //External diameter of hollow shaft
+d_1=0.6*D_1 //cm //Internal diameter of hollow shaft
+
+A_s=%pi*4**-1*(D*10**2)**2 //cm**2 //Area of solid shaft
+A_h=%pi*4**-1*(((D_1*10**2)**2)-((d_1*10**2)**2))
+
+W=(A_s-A_h)*A_s**-1*100 //Percentage //Percentage saving in weight
+
+
+//Result
+printf("Diameter of solid shaft is %.5f m",D)
+printf("\n Percentage saving in weight is %.2f",W);printf(" %%")