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+// Problem no 15.12,Page no.359
+
+clc;clear;
+close;
+
+D=0.038 //m //External Diameter
+d=0.035 //m //Internal Diameter
+d_1=0.0008 //m //Steel wire diameter
+p=2*10**6 //pa //Pa //Internal Pressure
+sigma_t_1=7*10**6 //Pa //Circumferential stress
+//E_s=1.6*E_s
+m=0.3
+
+//Calculation
+
+t=(D-d)*2**-1 //m Thickness 
+
+//sigma_t*2*t=%pi*d*2**-1*sigma_w
+//From Above equation we get
+
+//sigma_t=0.419*sigma_w   (Equation 1)
+
+sigma_w_1=(p*d-sigma_t_1*2*t)*(2*d_1**-1*%pi*4**-1*d_1**2)**-1 //stress in wire
+sigma_l=p*d*(4*t)**-1 //Longitudinal stress in tube
+
+//Now Equating equations of strain in tube and wire we get
+sigma_w=-(1.6*(sigma_t_1-sigma_l*m)-sigma_w_1)*1.67**-1*10**-6
+
+//Result
+printf("The Tension at which wire must have been wound is %.2f",sigma_w);printf(" MPa")