# Calculation of displacements – addition

After analyzing the previous guide, calculating displacements in flat systems you should already know what the displacement method is all about. Of course, not be able to do it at 100%, but at least be more or less aware of it.
This guide will complement the previous one. It will contain parameters of geometric figures, drawn and calculated examples of multiplication of graphs (functions) from the previous guide and invented, I will also take into account curvilinear functions and the last thing, an exemplary division of the complex figure and the method of its multiplication.
I still skip the supports and temperature, these elements will use the example in the next guide.

Parameters of basic geometric figures.

Below I will present the parameters of the four basic figures that will be needed when calculating displacements. There are many more, but these are used the most. During the classes, the instructors will present to you the remaining figures, if needed. Here we will focus on the most important ones, here they are.

Now I will show you how to do multiplication of graphs (functions), I will only consider graphs of bending moments, they look like this.

Multiplying the graphs by yourself, divide the entire frame into elements, just as you would when calculating internal forces. The only difference is that the forces that were applied are disappearing. In this case it will be a division from the beginning of the horizontal bar, to joining with the vertical bar (AB) and from joining to the end of the vertical bar (BC). The pattern in this case is as follows.

We multiply graphs of the horizontal bar AB. The length of the rod is 6 meters. The charts are on the same side, so the result will be positive. The shapes of the function are triangles.
$EJ&space;=&space;const$.
The diagrams are rectilinear, so there is no difference whether we take the triangle area from the virtual force of 4kNm or from the real forces of 72kNm.

$4&space;kNm&space;*&space;1&space;'*&space;6m&space;*&space;\frac{1}{2}&space;(surface\:&space;area)&space;*\frac{2}{3}&space;*&space;72&space;kNm&space;(center\:&space;of&space;\:&space;gravity)$

Now the BC (vertical) rod. The length of the rod is 4 meters. The charts are also on the same side. EJ = const.

$4&space;kNm&space;*&space;1&space;'*&space;4m&space;*&space;\frac{1}{2}&space;(surface\:&space;area)&space;*&space;\frac{2}{3}&space;*&space;72&space;kNm&space;(center\:&space;of\:&space;gravity)$

We have just analyzed the multiplication of charts from the previous guide. Now we will look at examples where the graphs are on different sides and the functions are curvilinear. It should be remembered that having the graphs on the same side of the bar, we always get a positive value from the multiplication, so I did not even take into account the minus points visible on the charts. Let’s continue with sample multiplication of charts.
I will skip the part with the formula and that it should be divided by EJ, GA, EA.

Example 1.
The charts are located on opposite sides of the rod.

The red line is a rod.

$5kNm&space;*&space;1&space;'*&space;3m&space;(surface\:&space;area)&space;*&space;\frac{1}{2}&space;*&space;(-10kNm)&space;(center&space;\:&space;of\:&space;gravity)&space;=&space;...$

Note that we project the center of gravity of the figure, which we counted the surface area for the second figure.
In this case,  $\frac{1}{2}$ , because we count the area of the rectangle.
Now I will multiply the graphs vice versa, I will take the triangle field and the center of gravity in the rectangle. The operation must go the same.

$(-10kNm)&space;*&space;3&space;*&space;\frac{1}{2}&space;(surface\:&space;area)&space;*&space;5kNm&space;*&space;1&space;'(center\:&space;of&space;\:&space;gravity)&space;=&space;...$

If we “projected” the center of gravity on a triangle, not a rectangle, then in action would appear:  $\frac{2}{3}&space;\:&space;or\:&space;\frac{1}{3},$ but here it does
not exist, because the rectangle has the same value along its entire length. In example number 2 this is shown.

The fraction in this operation appears because we calculate the area of the triangle.

Example 2.
The graphs are on the same side of the bar, but the maximum values are reversed.

The diagram from the virtual force is on the triangle with the value of 3kNm.
In this case, it is also indifferent, from which figure we take into account the surface area and from which the center of gravity.

$6,00kNm&space;*&space;5,00m&space;*&space;\frac{1}{2}&space;(surface\;&space;area)&space;*&space;\frac{1}{3}&space;*&space;3,00kNm&space;*&space;1&space;'(center&space;\;&space;of\;&space;gravity)&space;=$

Here we see what I wrote in the previous example. By projecting the center of gravity of the triangle with a value of 6kNm, we will hit the second triangle in place of  $\frac{1}{3}$  of the triangle’s value. If the value of 6kNm was at the top of the bar, then we would multiply by  $\frac{&space;2}{3}$, an example of such multiplication is at the top of this guide or in the previous guide.

Example 3.
Charts on the same side, but one is a parabola (curved). We must remember about the order. We multiply the area of the function higher, by the center of gravity of the lower function.

Immediately on this example, I will take care of the next and last thing in this guide. I am talking about the distribution of charts so that it is easy to multiply them. A parabola and a triangle will be created from a curved plot of 5kNm.

Multiplication will look as follows.

$5,00kNm&space;*&space;10,00m&space;*&space;\frac{1}{2}&space;*&space;\frac{2}{3}&space;*&space;2,00kNm&space;*&space;1&space;'+&space;(\frac{-&space;(q*l^{2})}{8})&space;*&space;10,00m&space;*&space;\frac{2}{3}&space;*&space;\frac{1}{2}&space;*&space;2,00kNm&space;*&space;1')&space;=&space;...$

The first is the multiplication of the triangle by the triangle, the second is the multiplication of the parabola by the triangle.

Let’s look at the division of various curvilinear functions into components. The first example will divide the curvilinear function into two triangles and a parabola. Please remember that the red line is the axis of the rod. It looks like this.

From which side the parabola will be located depends on what sign will appear between the figures. This is clearly seen in the picture above.

The last thing will be the section of the frame with the rod at an angle on which the curvilinear function is located.
We will divide this function into a triangle and a parabola. We must pay attention to the distances that we use when calculating the maximum value of the parabola, and then multiplying the charts. For the maximum value of the parabola we take the length of force on which it works, while when we perform multiplication, we use the length of the rod at an angle. This is clearly seen in the picture below.

To calculate the maximum value of the parabola, we use the distance L (the force is horizontal), unless the force is distributed at an angle (like a bar), then we use the length x.
To multiply the graphs we use the distance x (bar length).

In this way we have reached the end of the next guide. It seems to me that the information in it will make it easier for you to learn. Traditionally, I encourage you to exercise, exercise and exercise again. In case you had any questions, I’ll be glad to help!