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// Problem 8.2,Page no.207
clc;clear;
close;
L=15 //cm //Length of close coiled helical spring
U=50 //N*m //Strain energy
sigma_s=140 //MPa //Shear stress
D=10 //cm //Mean coil diameter
G=80 //GPa //Modulus of rigidity
R=D*2**-1 //cm //Mean coil Radius
//Calculations
//Let dell be the deflection of the spring when fully compressed
// 0.15-dell=n*d (Equation 1)
//U=(sigma_s)**2*V*(4*G)**-1 //Strain energy
//After substituting values in above equation and simolifying we get
V=50*4*80*10**9*((140*10**6)**2)**-1 //m**3 //Volume of spring
//But V=%pi*4**-1*d**2*2*%pi*R*n
//After substituting values in above equation and simolifying we get
//n=3.308*10**-3*(d**2)**-1 //Number of turns
//We know, T=P*R
//Now substituting values in T and simolifying we get
//P=549.7787*10**6*d**3 //Load
//U=P*dell*2**-1
//After substituting values in above equation and simolifying we get
//dell=0.18189*10**-6*d**3 //Deflection
//After substituting values in above equation and simolifying we get
//d**3-22.0533*10**-3*d**2-1.21261*10**-6=0
Coeff=[1 -22.0533*10**-3 0 -1.21261*10**-6]
d=roots(Coeff) //Diameter of steel wire
n=3.308*10**-3*((d(1)**2)**-1) //no.of coils
//Result
printf("Diameter of steel wire is %.5f m",d(1))
printf("\n number of coils = %d",ceil(n))
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