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+// Problem no 15.11,Page no.358
+
+clc;
+clear;
+close;
+
+
+d=0.3 //m //Diameter
+D=0.003 //m //Diameter of steel wire
+t=0.006 //m //thickness
+sigma_w=8*10**6 //Pa //Stress
+p=1*10**6 //Pa //Internal pressure
+E_s=200*10**9 //Pa //Modulus of Elasticity for steel
+E_c=100*10**9 //Pa //Modulus of Elasticity for cast iron
+m=1*0.3**-1
+
+//Calculations
+
+sigma_p=(sigma_w*%pi*2**-1*d)*(2*t)**-1 //compressive hoop stress
+sigma_l=p*d*(4*t)**-1 //Longitudinal stress
+
+//when internal presure is apllied Let sigma_w_1=Tensile in wire and sigma_p_1=tensile hoop in wire
+//sigma_p_1*2*t+sigma_w_1*2*d**-1*%pi*4**-1*d**2=p*D
+
+//After substituting values and further simplifying we get
+//1.2*sigma_p_1+0.471*sigma_w_1=3000    Equation 1
+
+//1*E_c**-1(sigma_p_1-sigma_1*m**-1+sigma_p)=1*E_s**-1(sigma_w_1-sigma_w)
+
+//After substituting values and further simplifying we get
+//sigma_p_1-0.5*sigma_w_1=1.36*10**6    
+//sigma_p_1=0.5*sigma_w_1-3.39*10**6   Equation 2
+
+//From Equation 2 substituting value of sigma_p_1 in Equation 1
+
+
+sigma_w_1=(40.68*10**3+0.3*10**6)*(10.71238*10**-3)**-1
+sigma_p_1=0.5*sigma_w_1-3.39*10**6
+
+//Let X=sigma_p_1 and Y=sigma_w_1
+X=sigma_p_1*10**-6 //MPa //Stresses in %pipe
+Y=sigma_w_1*10**-6 //MPa //Stresses in wire
+
+//Result
+printf("Stress in the pipe is %.2f",X);printf(" MN/m**2")
+printf("\n Stress in the wire is %.2f",Y);printf(" MN/m**2")