# Frames

In this guide you will learn how to calculate support reactions within a framework. The frame is nothing but a few connected beams. It is not difficult, if you have learned well and practiced calculating the reactions in beams, the frames will be easy too. When calculating the beams, we took only the length dimension when calculating the moments, because the beams were horizontal lines, and here we will have only two dimensions, about which we must remember and know which one to choose. This does not mean that we will take two dimensions at once, we will just have to know which dimension to choose, although it seems to me that it is obvious and I am sure that we will not have any problem with it.

Why two dimensions? Because the frame elements are horizontal and vertical. In the drawings it will be clearly visible. We will solve the example framework that I will try to describe step by step, so that it is clear to you.

We have such a frame, let’s see what reactions we have here and how I have accepted their returns.

We already have all the necessary data for calculations, we know the turns of the arrowheads and we are sure about the static determination and geometrical constancy of our design, of course if someone is not sure, please check. Now we are proceeding to calculate the reaction.

Let’s start with the horizontal Ha reaction. We make a force projection on the X axis, so.

$\Sigma&space;_{X}=0$
$H_{a}&space;+&space;8,00kN&space;=&space;0$
$H_{a}=-8,00kN$

We see that the course of action is exactly the same as in ordinary beams. Let us now calculate the horizontal reactions Va and Rb. First, we throw a force on the Y axis.

$\Sigma&space;_{Y}=0$
$V_{a}&space;-&space;10,00kN&space;-&space;5,00kN&space;+&space;R_{b}&space;=&space;0$

We see that we have two unknown Va and Rb, what should we do? Of course, the equation of moments, we will do them for point A.

$\Sigma&space;M_{a}&space;=&space;0$

And one by one we gather strength, here is what I wrote at the beginning. We need to know which dimension to use vertically or horizontally, so we have.

$8,00kN&space;*&space;2,00m&space;(vertical)&space;+&space;(2,00kN/m&space;*&space;5,00m)&space;*&space;2,50m(horizontal)&space;+&space;5,00kN&space;*&space;8,00m&space;(horizontal)&space;-&space;R_{b}&space;*&space;8,00m&space;=&space;0$

Comment.

1. The power of 8kN to point A has a horizontal distance equal to Om, and a vertical distance of 2m.
2. Spreading force 2kN / m, to make it easier we determine Net force. The force of 2 kN / m works on the length of 5 m, therefore (2 kN / m * 5m) and is located in the middle of the spreading force, or 5m / 2. Having the resultant force, let’s set the distance to point A, the vertical distance to this force is Om, and in the level of 2.5 m.
3. Last concentrated force 5 kN, its distance to points A are vertical Om, and horizontally 8m.
4. The Rb response lies on one line with a force of 5 kN, therefore its distance to point A is the same. Vertical Om, and horizontally 8m.

Warning.
Forces do not lie on one straight line with point A, but remember that forces can move freely forward and backward, that’s why they would be level with this punt. In this case, only one horizontal or vertical dimension remains, depending on the direction of force.

Well, let’s finish counting our vertical reactions, our equation looks like this.

$8,00kN&space;*&space;2,00m&space;(vertical)&space;+&space;(2,00kN/m&space;*&space;5,00m)&space;*&space;2,50m(horizontal)&space;+&space;5,00kN&space;*&space;8,00m&space;(horizontal)&space;-&space;R_{b}&space;*&space;8,00m&space;=&space;0$, so.
$8R_{b}&space;=&space;81,00kN&space;/&space;:8,00$
$R_{b}&space;=&space;10,125kN$

$\Sigma&space;_{Y}&space;=&space;0$
$V_{a}&space;-&space;10,00kN&space;-&space;5,00kN&space;+&space;R_{b}&space;=&space;0$ – instead of Rb we substitute the calculated value of 10,12 kN
$V_{a}&space;-&space;10,00kN&space;-&space;5,00kN&space;+&space;10,125kN&space;=&space;0$
$V_{a}&space;=&space;4,875&space;kN$

In this way, we calculated the reactions in a simple articulated frame, which amounts to

$H_{a}&space;=&space;-8,00kN$
$V_{a}&space;=&space;4,875&space;kN$
$R_{b}&space;=&space;10,125&space;kN$

I invite you to the next stage of our guide.
I hope that everything is explained in a simple and understandable way.