# Eccentricity forces

One of the issues that you will deal with in Strength of Materials is to determine stresses from the eccentric action of forces. It may be eccentric stretching or compression. I would like to note that the knowledge contained in the General Mechanics manual, the flat figures chapter, is needed to calculate this problem, as we will need, for example, the moment of inertia of this figure. At the end of the introduction, I would like to assure you that there is nothing to be afraid of because this topic is very simple. It’s all about substituting for the formula, but I recommend that you understand what you are taking to avoid making mistakes on the exam.
Drawings in this guide will only be previewed to see what it should look like.

Let’s begin with the fact that the eccentric action of force is the same as the simultaneous operation of the normal force and the bending moment.

Figures that you can get on projects can be different, starting from connected geometric figures, for example: two squares, ending with constructions composed of sections, for example: I-sections, channel sections, etc.
Here I put a I-beam 140, with the following appearance.

In the first step, find the neutral axis, to do this use the following formula.

$\\&space;X_{0}=-\frac{i_{y}^{2}}{x_{N}}&space;\\&space;\\&space;\\&space;Y_{0}=-\frac{i_{x}^{2}}{y_{N}}$

where:

$i_{x}^{2}\;&space;\;&space;and\;&space;\;&space;i_{y}^{2}$  – these are the rays of inertia

$x_{N}&space;\;&space;\;&space;and\;&space;\;&space;y_{N}$ – these are the coordinates of the force P applied to our system.

After calculating the co-ordinates of the neutral axis, we apply them to the drawing and combine them so that we have a straight line. We can also draw a straight perpendicular (at an angle of 90) to the neutral axis, on which we draw a stress plot. At the intersection point of the neutral axis with the straight perpendicular drawn, the stress value is zero.

Now you should calculate the value of the force effect at individual points. These are usually characteristic points of the outline of the figure. I will select them in the drawing with a ready graph. To calculate these values, use the following formula.

$\sigma&space;_{x}=\frac{P}{A}*(1+\frac{y_{p}}{i_{x}^{2}}*y+\frac{x_{p}}{i_{y}^{2}}*x)$

where:

$P$ – value of the force applied to our scheme

$A$ – the surface area of our schematic

$x_{p}&space;\;&space;\;&space;and&space;\;&space;\;&space;y_{p}$  – coordinates of the point at which force is applied

$y&space;\;&space;\;&space;and\;&space;\;&space;x$  – coordinates of the point for which we count the value of stresses

$i_{x}^{2}&space;\;&space;\;&space;and&space;\;&space;\;&space;i_{y}^{2}$  – the radius of inertia of the figure

If we are to calculate the stress value at four points, then the above action should be performed four times with the appropriate data. If in five points, then five times, etc … After calculating the stress values, you should draw their graph. An example of the appearance of the stress chart.

I remind you that all drawings are preview. I have not calculated an example here because you can see that there is nothing complicated here. Of course, you have to sit on this issue for a long time so as not to make mistakes. Please remember that you can always return to this guide and seek information.

I hope that everything is clearly described and the topic is understandable!