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. Code: #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; }

So a+.002 should be what I have listed below in red? If this is the case. I am still getting garbage answers. Code: #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;[B][COLOR=red]a+=.002[/COLOR][/B]) { Sigma = (1500 * .020)/ ((1/12)*(.060)*pow(.040-2*a,3)); cout<<Sigma<<endl; }; system("pause"); return 0; }

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

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 : Code: 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