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+clear
+//
+//
+
+//Initilization of Variables
+
+d=250 //mm //Diameter iron pipe
+t=10 //mm //Thickness
+d2=6 //mm //Diameter of steel
+p=80 //N/mm**2 //stress
+P=3 //N/mm**2 //Pressure
+E_c=1*10**5 //N/mm**2
+mu=0.3 //Poissons ratio
+E_s=2*10**5 //N/mm**2
+n=1 //No.of wires
+
+//Calculations
+
+L=6 //mm //Length of cyclinder
+
+//Force Exerted by steel wire at diameterical section
+F=p*2*%pi*d2**2*1*4**-1 //N
+
+//Initial stress in cyclinder
+f_c=F*(2*t*d2)**-1 //N/mm**2
+
+//LEt due to fluid pressure alone stresses developed in steel wire be F_w and in cyclinder f1 and f2
+f2=P*d*(4*t)**-1 //N/mm**2
+
+//Considering the equilibrium of half the cyclinder, 6mm long we get
+//F_w*2*%pi*4**-1*d2**2*n+f1*2*t*d2=P*d*d2
+//After further simplifying we get
+//F_w+2.122*f1=79.58 . ......................................(1)
+
+//Equating strain in wire to circumferential strain in cyclinder
+//F_w=(f1-mu*f2)*E_s*E_c**-1 //N/mm**2
+//After further simplifying we get
+//F_w=2*f1-11.25 ....................................(2)
+
+//Sub in equation in1 we get
+f1=(79.58+11.25)*(4.122)**-1 //N/mm**2
+F_w=2*f1-11.25 //N/mm**2
+
+//Final stresses
+//1) In steel Wir
+sigma=F_w+p //N/mm**2
+
+//2) In Cyclinde
+sigma2=f1-f_c
+
+//Result
+printf("\n Final Stresses developed in:cyclinder is %0.2f N/mm**2",sigma)
+printf("\n :Steel is %0.2f N/mm**2",sigma2)