# Internal forces in the beams

We will now calculate the internal forces in the beams. This is a very important topic. You have to devote a long time to practicing this issue. You can not learn this, you must understand it!

In this guide, I will show how such forces count in the continuous beam. In the previous guide you got to know some rules in the diagrams, now we will deal with the topic from the practical side. I will calculate internal forces on the example of the beam visible below.

This is exactly the same beam as in the previous guide. Everyone who has become familiar with the principle of division into parts should know how many will be able to count them. I divide the entire beam into 6 sections. From the right:

1. Roller support- Bending moment
2. Bending moment – Joint
3. Joint – Roller support
4. Roller support – Joint
5. Joint – Consolidated power
6. Consolidated power – Fixed support

The first step is to calculate support forces and force values in joints, because we will need them during the “cutting”. I will not elaborate the calculations of support reactions here, you must already know how to do it. Below the beam with the values of support reactions marked.

There is nothing else to do but to start calculating internal forces in particular cross-sections. The following figure shows the designations of the individual sections.

We see that we have 6 sections from A to F. I will solve them and translate them in alphabetical order.

Section A-A, for distance: 0 ≤ x ≤ 2, that is from zero to two meters.

A piece of beam from the roller support to the bending moment. Note the lack of bending moment in the drawing. The element to which it is cut is not taken into account during calculations. Which is good for us because we have fewer things to consider.
Once we have cut, we create internal forces N, T and M. We assume their returns arbitrarily, but I recommend taking the same as in this example. We only have two options because we can go from the left or the right side of the beam.

Calculation of internal forces.

$\Sigma&space;X&space;=&space;0$
$N&space;(x)&space;+&space;3kN&space;=&space;0$
$N&space;(x)&space;=&space;-3kN$

$\Sigma&space;Y&space;=&space;0$
$T&space;(x)&space;+&space;2.5kN&space;=&space;0$
$T&space;(x)&space;=&space;-2,5kN$
$\Sigma&space;M&space;=&space;0$
$M&space;(x)&space;-&space;2.5kN&space;*&space;x&space;=&space;0$, where x is the distance.
$M&space;(x)&space;=&space;2.5kN&space;*&space;x$
$x&space;=&space;0$
$M&space;(0)&space;=&space;0kNm$
$x&space;=&space;2$
$M&space;(2)&space;=&space;2.5&space;*&space;2&space;=&space;5kNm$

We already have counted internal forces for part A of our beam. I do not explain how I made calculations because we already learned this when calculating support reactions. These are exactly the same equations of equilibrium.

Section B-B, distance: 0 ≤ x ≤ 2

In this case, I had two options. I could continue to cut the beam from the end of the sliding support, but then I would have more forces to take into account, and my distance would be between 2 and 4 meters. I chose a different way. Well, we can cut the beam even in the middle, but ONLY when there is a joint, which can greatly facilitate counting. Of course, we must know the forces values in the joint. In part F, I will use the first way to show what it will look like.

Calculation of internal forces.

$\Sigma&space;X&space;=&space;0$
$N&space;(x)&space;+&space;3&space;kN&space;=&space;0$
$N&space;(x)&space;=&space;-3kN$

$\Sigma&space;Y&space;=&space;0$
$-T&space;(x)&space;-&space;2.5kN&space;=&space;0$
$T&space;(x)&space;=&space;-2,5kN$

$\Sigma&space;M&space;=&space;0$
$-M&space;(x)&space;-&space;2.5&space;*&space;x&space;=&space;0$
$M&space;(x)&space;=&space;-2.5&space;*&space;x$
$x&space;=&space;0$
$M&space;(0)&space;=&space;0kNm$
$x&space;=&space;2$
$M&space;(2)&space;=&space;-5kNm$

Section C-C, distance: 0 ≤ x ≤ 4

Note that this is the same wrist as in part B, but on the left. This is what I wrote above, if we have a joint in the center of the beam, we can really divide it easily and count our beam.

Calculation of internal forces.

$\Sigma&space;X&space;=&space;0$
$-N&space;(x)&space;-&space;3kN&space;=&space;0$
$N&space;(x)&space;=&space;-3kN$

$\Sigma&space;Y=0$
$T&space;(x)&space;+&space;2,5&space;kN&space;=&space;0$
$T&space;(x)&space;=&space;-2,5kN$

$\Sigma&space;M&space;=&space;0$
$M&space;(x)&space;-&space;2,5&space;*&space;x&space;=&space;0$
$M&space;(x)&space;=&space;2,5&space;*&space;x$
$x&space;=&space;0$
$M&space;(0)&space;=&space;0kNm$
$x&space;=&space;4$
$M&space;(4)&space;=&space;2,5&space;kN&space;*&space;4m&space;=&space;10kNm$

Section D-D, distance: 0 ≤ x ≤ 4

In this part we have a case with a distributed force, very often here problems arise, how to take into account this distance when calculating the bending moment. It should be remembered that taking into account the rectangular force, we take the distance twice. The first time it will be the value of the distributed force operating on one meter multiplied by the entire distance on which this force operates, i.e. we can say that we already have a focused force. We have been to multiply this strength by the distance to the point we are counting on. It should be remembered that the resultant force in the case of a rectangular distributed force lies exactly in the middle, what is the conclusion? This second distance, in this case, will be half of the first. IN THIS CASE!

Calculation of internal forces.

$\Sigma&space;X&space;=&space;0$
$N&space;(x)&space;+&space;3,00kN&space;=&space;0$
$N&space;(x)&space;=&space;-3,00kN$

$\Sigma&space;Y&space;=&space;0$
$-T&space;(x)&space;-&space;2kN&space;*&space;x&space;+&space;6,50kN&space;=&space;0$
$T&space;(x)&space;=&space;-&space;2kN&space;*&space;x&space;+&space;6,500kN$
$x=0\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;T&space;(0)&space;=&space;6,50kN$
$x&space;=&space;4\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;T&space;(4)&space;=&space;-8kN&space;+&space;6,50kN$            $T&space;(4)&space;=&space;-1,50kN$

$\Sigma&space;M&space;=&space;0$
$-M&space;(x)&space;+&space;6,50kN&space;*&space;x-2kN&space;*&space;x&space;*&space;x&space;*&space;0,50&space;=&space;0$
$M(x)&space;=&space;6,50&space;*&space;x&space;-&space;2kN&space;*&space;x&space;*&space;x&space;*&space;0,50$
$x&space;=&space;0\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(0)&space;=&space;0kNm$
$x&space;=&space;4\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(4)&space;=&space;6,50kN&space;*&space;4m&space;-&space;2kN&space;*&space;2m&space;*&space;2m&space;*&space;0,50$              $M&space;(4)&space;=&space;10,00kNm$

Section E-E, distance: 0 ≤ x ≤ 2

Calculation of internal forces.

$\Sigma&space;X=0$
$-N(x)-3kN&space;=&space;0$
$N(x)&space;=&space;-3kN$

$\Sigma&space;Y=0$
$T(x)-6,50kN=0$
$T(x)&space;=&space;6,50kN$

$\Sigma&space;M=0$
$M(x)&space;+&space;6,50kN&space;*&space;x&space;=&space;0$
$M(x)&space;=&space;6,50kN$
$x=0\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M(0)&space;=&space;0kNm$
$x=2\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M(2)&space;=&space;-13kNm$

Section F-F, distance 2 ≤ x ≤ 4

In this case, I could divide the bar from restraint to internal forces that I would not have seen. To show that dividing the beam greatly affects the ease of calculation, I will make a division where I will see section E and F.

Calculation of internal forces.

$\Sigma&space;X&space;=&space;0$
$-N&space;(x)&space;-&space;3kN&space;=&space;0$
$N&space;(x)&space;=&space;-3kN$

$\Sigma&space;Y&space;=&space;0$
$T&space;(x)&space;-5kN&space;-&space;6,50kN&space;=&space;0$
$T&space;(x)&space;=&space;11,50kN$

$\Sigma&space;M&space;=&space;0$
$M&space;(x)&space;+&space;5kN&space;*&space;(2-x)&space;+&space;6,50kN&space;*&space;x&space;=&space;0$
$M&space;(x)&space;=&space;-5&space;*&space;(2-x)&space;-&space;6,50kN&space;*&space;x$
$x&space;=&space;2\:&space;\:&space;\:&space;\:&space;\:\:\:&space;M&space;(2)&space;=&space;-13kNm$
$x&space;=&space;4\;&space;\:&space;\:\:\:\:\:&space;M&space;(4)&space;=&space;-36kNm$

And that’s it for the calculations, I encourage you to solve the same beam yourself, while checking my calculations. Now we have to plot graphs of internal forces N, T, M. They will look like this.

We see that in section D the cutting forces diagram cuts the axis, so we already know that there will be an extreme in the bending moment. We can even see it in the chart above, but how to calculate it? I already explain.
First, find the place where the axis is cut from the cutting force equation in part D.

$T&space;(x)&space;=&space;-2kN&space;*&space;x&space;+&space;6,50&space;kN$
$-2&space;*&space;x&space;+&space;6,5&space;=&space;0$
$2,00&space;*&space;x&space;=&space;6,50/:&space;2$
$x&space;=&space;3,25m$

We already know the place, now we will know the value. To calculate the value, one should substitute 3,25 m for the moment in the D section of the equation.

$M(x)&space;=&space;6.5kN&space;*&space;x-2kN&space;*&space;x&space;*&space;x&space;*&space;0.5$
$x&space;=&space;3,25m$
$M(3,25)=6,5&space;*&space;3,25&space;-&space;2&space;*&space;3,25&space;*&space;3,25&space;*&space;0,50$
$M&space;(3,25)&space;=&space;10,56kNm$

This way, we have finished the practical part of calculating internal forces in beams. I hope that the guide lightly explained to you what and how. I will add that you should practice, practice and practice again so that you do not have problems with it.