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+// Problem no 15.5,Page no.353
+
+clc;clear;
+close;
+
+t=0.015 //m //Thickness of plate
+sigma_t=120*10**6 //Pa //tensile stress
+sigma_l=120*10**6 //Pa //Longitudinal stress
+rho=0.7 //% //Efficiency of longitudinal joints
+rho_l=0.3 //% //Efficiency of circumferential joints
+p=2*10**6 //Pa //Internal pressure
+D=1.5 //m //shell diameter
+
+//Calculations (Part-1)
+
+D_1=sigma_t*2*t*rho*p**-1    //m 
+D_2=sigma_l*4*t*rho_l*p**-1  //m 
+
+//Thus max diameter of shell is min of above two cases
+
+//Calculations (Part-2)
+
+p_1=sigma_t*2*t*rho*D**-1*10**-6    //MPa
+p_2=sigma_l*4*t*rho_l*D**-1*10**-6  //MPa
+
+//Thus Internal pressure is min of above two cases
+
+//Result
+printf("Max Permissible Diameter of shell is %.2f",D_2);printf(" m")
+printf("\n Max Permissible Internal Pressure is %.2f",p_2);printf(" MPa")