# Trusses

We are almost at the finish of our first General Mechanics course. Here we will deal with the calculation of support reactions, but also determine the axial forces in the rods, unfortunately it is not enough to count only the support reactions, your professors will also require you to calculate the values of forces in the rods. With the concept of axial forces you can meet for the first time, carefully read it in the next guide, Strength of Materials. In short, this is based on the determination of internal forces: M – moments, T – cutting and N – normal that work in a given structure. We will be interested in the latter at the moment because the moments and cutting forces are zero in the trusses. Before starting calculations, there must be a dose of theory.

When calculating the values in bars, we can use two basic calculation and graphic methods. Let’s deal with computational methods, in trusses to calculate the cross-sectional forces we will use the following methods:

1. Ritter method (intersections)
2. The method of balancing nodes
3. The method of balancing a part of a truss

Very often, when calculating truss bars, we will use these methods for a change. We must also learn to recognize “zero bars”, the following drawings show in which cases zero bars are present. This is very useful because we have less counting.

Cases of zero force member in truss.
We have three cases that are really worth remembering!

1. Two rods, unloaded node – both zero rods.
2. Two bars, a node loaded with a force parallel to one of the bars  – the second zero bar
3. Unloaded node, three bars, including 2 lying on one straight line – third zero bar

Three very nice pictures, and they make life easier!
Well, let’s start learning through practice.
We have such a truss.

Let’s see what the support reactions look like.

Calculation of support reactions.
The horizontal reactions Hb are calculated from the projection of forces on the X axis

$\Sigma&space;_{X}&space;=&space;0$
$-H_{b}&space;+&space;10,00kN&space;=0$
$H_{b}&space;=&space;10,00kN$

Turn for vertical reactions, the equation of moments to point A.

$\Sigma&space;M_{a}&space;=&space;0$
$10,00kN&space;*&space;4,00m&space;+&space;20,00kN&space;*&space;4,00m&space;-&space;V_{b}&space;*&space;8,00m&space;=0,00$
$-8,00V_{b}&space;=&space;-120,00&space;kN&space;/&space;:(-8&space;,00)$
$V_{b}&space;=&space;15,00kN$

Now Ra reaction.

$\Sigma&space;_{Y}&space;=&space;0$
$R_{a}&space;-&space;20,00kN&space;+&space;15,00kN&space;=0$
$R_{a}&space;=&space;5,00kN$

Calculation of cross-sectional forces
The Ritter method (intersections).

Using this method to calculate the axial forces in the bars of the truss, we “imagine” intersect three bars or more, so that it is possible to create Ritter points.

What is the Ritter point? This is the point where all the cut bars intersect except for one. Having a Ritter point, we can, using the sum of moments, calculate a bar whose operation does not intersect the Ritter point.

In the figure below, I will present 4 sections that I created to calculate individual bars, here they are.

I created 4 cross-sections, namely: A-A, B-B, C-C and D-D. When creating the cross-section, we divide our truss into two parts, so when calculating, select which part of the truss you choose left or right. This choice of course depends only on us, because the results will always come out the same, regardless of the section we choose. However, let us choose the side of the section where there is less counting, so there are fewer forces, support reactions, etc…

After cutting the truss, in the cut bars, we create tensile forces and we have to calculate these forces.

Section A-A (left side of the truss).

We see that after cutting the bars No. 2, 3 and 4, we created forces stretching these rods. To calculate the force in bar no. 2, we create the equation of the sum of moments to the Ritter point No. 1, because this point intersects the bars number 4 and 3, so we do not take them into account in the calculations. With the capital letter P with the number in the index, I will mark bars, eg P2 – rod number 2.

Remember that vertical rods have a dimension of 4 m, horizontal 2 m, and a rod at an angle of 4.47 m.
Let’s start calculating.

$\Sigma&space;M_{pR1}&space;=&space;0$
$P_{2}&space;*&space;4,00&space;+&space;10,00kN&space;*&space;4,00&space;+&space;5,00kN&space;*&space;2,00&space;=0$
$P_{2}&space;=&space;-12.5&space;kN$ – we have the axial force of the rod number 2.

$\Sigma&space;M_{pR2}&space;=&space;0$
$P_{4}&space;*&space;4&space;=&space;0$
$P_{4}&space;=&space;0,00kN$ – we know the axial force of the rod number 4.

$\Sigma&space;_{Y}&space;=&space;O$
$5,00kN&space;-&space;P_{3}&space;*&space;sin&space;=0$
The angle between the bars 3 and 2.
Calculated from Pythagoras.
$sin=\frac{4}{4,472}=0,894$

$5,00kN&space;-&space;P_{3}&space;*&space;0,89&space;=&space;0$
$P_{3}&space;=&space;5,60kN$ – the value of the axial force in the rod No. 3.

Section B-B (left side of the truss).
In this section, only the axial force of the number 5 bar is calculated because I want to show two possibilities of calculating these forces. First of all, we can use the Ritter method, it will look like this.

The second method is the method of balancing nodes, it simply consists in cutting out the node we are interested in from the truss, it will look like this.

And then we make a force projection on the Y axis to calculate the rod number 5.
It will use the Ritter method.

$\Sigma&space;_{Y}&space;=&space;0$
$P_{5}&space;+&space;5,00kN&space;=&space;0$
$P_{5}&space;=&space;-5,00kN$ – value of the axial force of rod no. 5

Section C-C (left side of the truss)

Remember that vertical rods have a dimension of 4 m, horizontal 2 m, and rods at an angle of 4.47 m.

$\Sigma&space;M_{pR4}&space;=&space;0$
$P_{6}&space;*&space;4,00&space;+&space;10,00kN&space;*&space;4,00&space;+&space;5,00kN&space;*&space;4,00&space;=&space;0$
$P_{6}&space;=&space;-15,00kN$

$\Sigma&space;M_{pR3}&space;=&space;0$
$-P_{8}&space;*&space;4,00&space;+&space;5,00kN&space;*&space;2,00&space;=&space;0$
$P_{8}&space;=&space;2,50kN$

$\Sigma&space;_{Y}&space;=&space;0$
$5,00kN&space;-&space;P_{7}&space;*&space;sin&space;=&space;0$
$5,00kN&space;-&space;P_{7}&space;*&space;0,8944&space;=&space;0$
$P_{7}&space;=&space;5,60kN$

Cross-section D-D (left side of the truss)

Remember that vertical rods have a dimension of 4 m, horizontal 2 m, and rods at an angle of 4.47 m.

$\Sigma&space;M_{pR6}&space;=&space;0$
$-P_{10}&space;*&space;4,00&space;-&space;15,00kN&space;*&space;4,00&space;=0$
$P_{10}&space;=&space;-15,00kN$

$\Sigma&space;M_{pR5}&space;=&space;0$
$P_{12}&space;*&space;4,00&space;-&space;15,00kN&space;*&space;2,00&space;+&space;10,00kN&space;*&space;4,00&space;=0$
$P_{12}&space;=&space;-2,50kN$

$\Sigma&space;_{Y}&space;=&space;0$
$-P_{11}&space;*&space;sin&space;+&space;15,00&space;kN&space;=&space;0$
$P_{11}&space;=&space;16,80&space;kN$

The method of balancing nodes and the method of balancing a part of a truss.
We were left to count the force in the bars number: 1, 9, 14 and 15. In these methods we will calculate the forces in the bars only by throwing forces on the given axis Y or X.

Balancing the node with bars 1 and 4.

We are only interested in rod number 1, because 4 we have already counted.

$\Sigma&space;_{Y}&space;=&space;0$
$5,00kN&space;+&space;P_{1}&space;=&space;0$
$P_{1}&space;=&space;-5,00kN$

Balancing the node with bars No. 6, 9 and 10

$\Sigma&space;_{Y}&space;=&space;0$
$-20,00kN-P_{9}&space;=&space;0$
$P_{9}&space;=&space;-20,00kN$

Balancing a part of a truss with bars number 14 and 15.

Remember that vertical rods have a dimension of 4 m, horizontal 2 m, and rods at an angle of 4.47 m.
First, count the rod at an angle to get rid of the two unknowns.

$\Sigma_{Y}&space;=0$

$15,00kN&space;+&space;P_{14}&space;*&space;sin&space;=&space;0$
Calculated from Pythagoras.
$sin=\frac{4}{4,472}=0,894$

$15,00kN&space;+&space;P_{14}&space;*&space;0,894&space;=&space;0$
$0,894P_{14}=&space;-15,00kN&space;/:0,894$
$P_{14}&space;=&space;-16,85kN$

Now with only one unknown, the last rod P15, we can cast a force on the X axis.

$\Sigma&space;_{X}&space;=0$

$-P_{15}&space;-&space;P_{14}&space;*&space;cos&space;-&space;10,00kN&space;=0$
Calculated from Pythagoras.
$cos=\frac{2}{4,472}=0,447$
$-P_{15}-&space;(-16,85kN)&space;*&space;0,447&space;-&space;10,00kN&space;=&space;0$
$P_{15}&space;=&space;-2,46kN\approx&space;-2,50kN$

With these methods, we counted all the axial forces in the bars of the given truss. It’s always worth checking if we did it right. In order to perform the check, we can cut nodes one by one, make a force projection on a given axis and check whether we will get zero, this method is described in previous lessons. This is good for the colloquium, but when it comes to performing projects at home, I recommend programs like RM-Win or Soldis. Below is our truss made in the RM-Win program, let’s check if the results have come out well.

Everything is well counted.