0

```
> Hello Can Someone please help me with this problem.
This is what i have sofar but i am not sure whats missing or how to end it please help.
A solid steel column of circular cross section is needed to support a load of 65,000 lb.
The ends of the column are not rigidly fixed but are able to rotate, and the point of
application of the load may be as much as 1 in. from the axis of the column.
The yield point of steel is 50,000 lb./in2, and the desired safety factor is 3.
The column is 45 in. long. Determine the required diameter (up to four decimal places)
to safely carry the load.
where
s = (F *P) / A * ( L + ( ( e * c ) / ( i * i ) * ( 1.0 / cos ( ( L / ( 2.0 * i ) ) * sqrt (( F * P ) / ( A * E ) ) ) ) ))
s = maximum stress = 50,000 lb./ in2
F = factor of safety = 3
P = applied load = 65,000 lb.
A = cross section area in2 = for a circular column
e = eccentricity of applied load = 1 in.
c = distance from column axis to extreme edge of cross section
= d/2 for circular column
i = radius of gyration = d/4 for circular column
E = modulus of elasticity = 30,000,000 for steel
L = column length = 45 in.
*
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main ()
{
int F, P, e, E, L;
double pi = 3.14159;
double d , c , i , A, s;
cout << "Enter Diameter" <<endl;
cin >> d;
do
{
F = 3;
P = 65000;
e = 1;
E = 30000000;
L = 45;
d = d + 0.01;
c = d / 2.0;
i = d / 4.0;
A = ( pi * d * d ) / 4.0;
s = (F *P) / A * ( L + ( ( e * c ) / ( i * i ) * ( 1.0 / cos ( ( L / ( 2.0 * i ) ) * sqrt (( F * P ) / ( A * E ) ) ) ) ));
} while ( fabs ( 50000 - s) > 0.001 );
cout << " Solution of diameter" << endl;
cin >> s;
} // end main
```