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+// Problem 8.9,Page no.212
+
+clc;clear;
+close;
+
+L= 3 //m //Length of rod
+d_1=25*10**-3 //m //Diameter of rod
+n= 5 //no. of coils
+sigma=70*10**6 //MPa  //instantaneous stress
+E=70*10**9 //Pa 
+G=80*10**9 //Pa
+D=24*10**-2 //Spring diameter
+R=d_1*2**-1 //spring radius
+d=4*10**-2 //diameter of steel
+
+//Calculations
+
+dell_1=sigma*L*(E)**-1
+
+//Let P be the equivalent applied Load which will produce same stress of 70 MPa
+P=%pi*4**-1*(d_1)**2*E*10**-3 //KN
+
+//deflection of the spring is given by
+dell_2=P*64*R**3*n*(G*d**4)**-1 
+
+//Now Loss of Potential Energy of the weight=strain energy stored in the rod and the spring
+//Height measured from top of uncompressed  spring
+h=((P*dell_1*2**-1+P*dell_2*2**-1)*(5.5*10**3)**-1-(dell_1+dell_2))*10**2
+
+//Shear stress in the spring is given by
+sigma_s=16*P*R*(%pi*d**3)**-1*10**-6 //MPa 
+
+//Result
+printf("Height measured from top of uncompressed  spring %.2f",h);printf(" cm")
+printf("\n max shearing stress is %.2f",sigma_s);printf(" MPa")
+
+// Answer is wrong in the textbook.