[chemr2]

I have taken c++ but it has been about a year ago. I am taking mechanics of materials now and need to solve a bending moment problem. I am trying to get values of Sigma when a is between 0 and 20mm in 2mm increments. I have wrote a little bit of code and need to figure out if this would the correct approach? Also why is this loop infinite?
I do not have a copy of my text anymore so any help would be appreciated.

The equation I am solving for is Sigma = MC / I
Remember I need 16 values for Sigma when a is varying.

#include<iostream>
#include<cmath>
using namespace std;
 
int main()
{
 
    double n,M,C,I,Sigma,b,h,a,x,i;
 
    n = 2.66;
    C = .020;
    M = 1500;
 
    for(a=0;a<=.020;a+.002)
    {
 
           Sigma = (1500 * .020)/ ((1/12)*(.060)*pow(.040-2*a,3));
           cout<<Sigma<<endl;
     };
 
 
 
 
    system("pause");
    return 0;
 
}

[xpi0t0s]

The loop does not modify a. Look carefully at the 3rd clause in the for statement.

[chemr2]

Quote:

Originally Posted by xpi0t0s

The loop does not modify a. Look carefully at the 3rd clause in the for statement.

So a+.002 should be what I have listed below in red? If this is the case. I am still getting garbage answers.

 
#include<iostream>
#include<cmath>
using namespace std;
 
int main()
{
 
    double n,M,C,I,Sigma,b,h,a,x,i;
 
    n = 2.66;
    C = .020;
    M = 1500;
 
    for(a=0;a<=.020;a+=.002)
    {
 
           Sigma = (1500 * .020)/ ((1/12)*(.060)*pow(.040-2*a,3));
           cout<<Sigma<<endl;
     };
 
 
 
 
    system("pause");
    return 0;
 
}

[xpi0t0s]

What answers do you get and what answers do you expect?
Do you get different answers if you add ".0" to the end of each integer in the expression?

Sigma = (1500.0 * .020)/ ((1.0/12.0)*(.060)*pow(.040-2.0*a,3.0));

[SaswatPadhi]

Probably, he was getting all values as "Infinity".

@chemr2 :
The problem is because of the 1/12 in the denominator.

Both 1 and 12 are integers. Change at least one of them to float to get proper results.
This should be sufficient :

Sigma = (1500.0 * .020)/ ((1.0/12)*(.060)*pow(.040-2.0*a,3));

BTW, this is NOT what we call an infinite loop.
Your loop is finite, but the output values are infinite