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// Problem 11.11.6,Page no.278
clc;clear;
close;
//d=Diameter of rod
P=500 //KN
e=0.75 //cm //eccentricity
//calculation
//A=%pi*d**2*4**-1 //cm**2 //Area of rod
//sigma_d=P*A**-1 //KN/cm**2 //stress due to direct load
//After substituting value and simplifying we get,
//sigma_d=2000*(%pi*d**2)**-1 //KN/cm**2
M=P*e //Kn*cm //Moment
//Z=%pi*d**3*32**-1 //cm**3 //section modulus
//sigma_b=M*Z**-1 //KN/cm**2 //Stress due to moment
//After substituting value and simplifying we get,
//sigma_b=12000*(%pi*d**3)**-1 //KN/cm**2
//Max stress
//sigma=sigma_d+sigma_b
//After substituting value and simplifying we get,
//2000*(%pi*d**2)**-1+12000*(%pi*d**3)**-1=12.5
//After simplifying we get,
//d**3-53.05*d-318.3=0
//From Synthetic Division we get d**2+4.73*d-42.918
a=1
b=-4.73
c=-42.918
X=b**2-(4*a*c)
d_1=(-b+X**0.5)*(2*a)**-1
d_2=(-b-X**0.5)*(2*a)**-1
//Result
printf("The minimum diameter of the rod is %.2f",d_1);printf(" cm")
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