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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.
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