# Internal forces in arches

In this guide, we will deal with the calculation of internal forces in the  arch. The greatest difficulty we encounter while calculating the internal forces is the curvilinear diagram of the rod, but with the help of trigonometry and already derived transformation patterns, which I will present and explain how to use them. It’s best to learn through practice, so as part of the arches guide, I will solve the whole task from beginning to end.

Below is a diagram of the arched structure that we will calculate.

Before starting any calculations, you should take the coordinate system in which we will work.

Calculation of support reactions.
Distribution of force P at an angle to the components X and Y.
Before starting to calculate the reaction, I will divide the force at an angle into Px (horizontal) and Py (vertical) components.

$P_{y}&space;=&space;8&space;*&space;\sin&space;(24º)&space;=&space;3,254kN$
$P_{x}&space;=&space;8&space;*&space;cos&space;(24º)&space;=&space;7,308&space;kN$

Calculation of the reactions.
Calculation of support reactions in curves, nothing different from calculating supporting reactions in an beams, an frames or an trusses. Let me just remind you that the shape of the diagram is not important, but vertical and horizontal distances when writing actions on the sum of moments of forces to the adopted point, and for the sum of projections of forces on a given axis, we do not even need distances. If anyone has problems with this, please go back to the appropriate guides.

Let’s now calculate the support reactions.

$\Sigma&space;M_{A}&space;=&space;0$
$-R_{B}&space;*&space;24&space;+&space;3,254&space;*&space;18&space;-&space;7,308&space;*&space;10,392&space;+&space;7&space;*&space;12&space;*&space;6&space;=&space;0$
$-24R_{B}&space;=&space;-486,627&space;/:&space;(-&space;24,00)$
$R_{B}&space;=&space;20,27&space;kN$

$\Sigma&space;Y&space;=&space;0$
$VA&space;-&space;7&space;*&space;12&space;-&space;3,254&space;+&space;20,27&space;=&space;0$
$VA&space;=&space;66,977&space;kN$

$\Sigma&space;X&space;=&space;0$
$HA&space;=&space;7,308&space;kN$

Calculation of internal forces.
Before starting calculations, it should be noted that internal forces (cutting and normal) that arise after cutting the bar, which are parallel and perpendicular to the axis of the bar, will change their position along with the curved line of the arc. The bending moment does not apply because it rotates.
Having this knowledge, we need to write the equation of the arc in polar coordinates, they depend on the adopted coordinate system. For the above arrangement, they will look as follows.

$\left.\begin{matrix}&space;x(\alpha&space;)=R*\cos&space;(\alpha&space;)&space;&&space;\\&space;y(\alpha&space;)=R*\sin&space;(\alpha&space;)&&space;\end{matrix}\right\}$

The radius of the arc is constant, so the independent variable in the arc equation, as well as in the equation of the cross-sectional forces will be the angle $\alpha$.

Let’s start calculating internal forces.

Section 1-1; it is in the range of 0º <α <90º

We cut our 1-1 section, from 0 ° to 90 °, so in theory we should check SW in places every 1º. In practice, it is enough to check SW at 10 ° -15 °, depending on the size of the range.
In the above drawing, you can also notice a little too much internal forces, I explain.

1. The resulting internal force from the bending moment (Mα) is already the proper one, so arranging a function for it (equation), remembering the polar course of forces, we can calculate the acting bending forces.
2. Calculation of internal and normal cutting forces is slightly more complicated. At the beginning, we arrange the equations by projecting forces on the vertical and horizontal axis, obtaining the functions of the forces H and V. Next, using transformational formulas, we go to the proper functions that will allow us to calculate cutting forces (Qα) and normal (Nα).

Let’s put together the functions needed to calculate internal forces.

$\Sigma&space;M&space;=&space;0$

$-M(\alpha&space;)+66,977*(12-12cos&space;(\alpha&space;))-7,308*12&space;*sin(\alpha&space;)-7*(12-12cos&space;(\alpha&space;))^{2}*0,50&space;=&space;0$

$M&space;(\alpha&space;)&space;=&space;803,724&space;-&space;803,724&space;*&space;cos&space;(\alpha&space;)&space;-&space;87,696&space;*&space;sin&space;(\alpha&space;)&space;-&space;3,50&space;*&space;(144-288&space;*&space;cos&space;(\alpha&space;)&space;+&space;144&space;*&space;cos&space;(\alpha&space;)^{2})&space;=&space;0$

$M&space;(\alpha&space;)&space;=&space;803,724&space;-&space;803,724&space;*&space;cos&space;(\alpha&space;)&space;-&space;87,696&space;*&space;sin&space;(\alpha&space;)&space;-&space;504&space;+&space;1008&space;*&space;cos&space;(\alpha&space;)&space;-&space;504&space;*&space;cos&space;(\alpha&space;)^{2}&space;=&space;0$

$M&space;(\alpha&space;)&space;=&space;299,724&space;+&space;204,276&space;*&space;cos&space;(\alpha&space;)&space;-&space;87,696&space;*&space;sin&space;(\alpha&space;)&space;-&space;504&space;*&space;cos&space;(\alpha&space;)^{2}$

We have put together the function needed to calculate the bending moments. Now, by putting the angles under α, we will obtain the internal forces of the bending moment in the places of interest to us.

Now, to calculate the cutting and normal forces, we will have to use transformational formulas, they look like this.

$Q&space;(\alpha&space;)&space;=&space;-H&space;*&space;cos&space;(\alpha&space;)&space;-V&space;*&space;sin&space;(\alpha&space;)$$N&space;(\alpha&space;)&space;=&space;H&space;*&space;sin&space;(\alpha&space;)&space;-&space;V&space;*&space;cos&space;(\alpha&space;)$

It seems to me that everything is clear. We just need to arrange the functions for the substitute internal forces H and V, and then substituting them for the above formulas, we get the appropriate internal forces Q (α) and N (α).

$\Sigma&space;Y&space;=&space;0$
$V&space;=&space;66,977&space;-&space;7&space;*&space;(12-12&space;*&space;cos&space;(\alpha&space;))$
$V&space;=&space;66,977&space;-&space;84&space;+&space;84&space;*&space;cos&space;(\alpha&space;)$
$V&space;=&space;-17,023&space;+&space;84&space;*&space;cos&space;(\alpha&space;)$

$\Sigma&space;X&space;=&space;0$
$H&space;=&space;-7,308$

Now it is enough to substitute the calculated V and H forces for the transformation patterns. We will have the following functions (equations).

$Q&space;(\alpha&space;)&space;=&space;-7,308&space;*&space;cos&space;(\alpha&space;)&space;-&space;17,023&space;*&space;sin&space;(\alpha&space;)&space;+&space;84&space;*&space;cos&space;(\alpha&space;)&space;*&space;sin&space;(\alpha&space;)$$N&space;(\alpha&space;)&space;=&space;-7,308&space;*&space;sin&space;(\alpha&space;)&space;+&space;17,023&space;*&space;cos&space;(\alpha&space;)&space;-&space;84&space;*&space;cos&space;(\alpha&space;)^{2}$

We already have all the equations arranged, i.e. for the cutting forces, normal forces and for the bending moment that looks like this.

$M&space;(\alpha&space;)&space;=&space;299,724&space;+&space;204,276&space;*&space;cos&space;(\alpha&space;)&space;-&space;87,696&space;*&space;sin&space;(\alpha&space;)&space;-&space;504&space;*&space;cos&space;(\alpha&space;)^{2}$

Now, I suggest you calculate internal forces using the Excel program, due to the large amount of calculations, and present the results in tabular form. The internal forces for section 1-1 are as follows.

 α M(α) [kNm] Q(α) [kN] N(α) [kN] 0º 0 -7,308 -66,977 15º 4,103 9,54 -63,82 30º 54,78 21,53 -51,91 45º 130,15 24,79 -35,13 60º 199,91 17,98 -18,82 75º 234,12 2,67 -8,28 90º 212,02 -17,02 -7,308

We already know the internal forces in the middle of our bow. Now we calculate the remaining half. Due to the change in the direction of the SW forces, transformation patterns are also changing, now they look as follows.

$Q&space;(\alpha&space;)&space;=&space;H&space;(\alpha&space;)&space;*&space;cos&space;(\alpha&space;)&space;+&space;V&space;*&space;sin&space;(\alpha&space;)$$N&space;(\alpha&space;)&space;=&space;H&space;(\alpha&space;)&space;*&space;sin&space;(\alpha&space;)&space;-&space;V&space;*&space;cos&space;(\alpha&space;)$

Section 3 -3              0º <α <60º

$\Sigma&space;M&space;=&space;0$
$M&space;(\alpha&space;)&space;=&space;20,27&space;*&space;(12-12&space;*&space;cos&space;(\alpha&space;))$
$M&space;(\alpha&space;)&space;=&space;243,24&space;-&space;243,24&space;*&space;cos&space;(\alpha&space;)$

$\Sigma&space;Y&space;=&space;0$
$V&space;=&space;20,27$

$\Sigma&space;X&space;=&space;0$
$H&space;=&space;0$        $Q&space;(\alpha&space;)&space;=&space;-20,27&space;*&space;sin&space;(\alpha&space;)$$N&space;(\alpha&space;)&space;=&space;-20,27&space;*&space;cos&space;(\alpha&space;)$

 α M(α) [kNm] Q(α) [kN] N(α) [kN] 0º 0 0 -20,27 15º 8,29 -5,25 -19,58 30º 32,59 -10,14 -17,55 45º 71,24 -14,33 -14,33 60º 121,62 -17,55 -10,14

Section 2 – 2             60º <α <90º

$\Sigma&space;M&space;=&space;0$
$M&space;(\alpha&space;)&space;=&space;299,661&space;-&space;204,192&space;*&space;cos&space;(\alpha&space;)&space;-&space;87,696&space;*&space;sin&space;(\alpha&space;)$

$\Sigma&space;Y&space;=&space;0$
$V&space;=&space;20,27&space;-&space;3,254$
$V&space;=&space;17,023$

$\Sigma&space;X&space;=&space;0$
$H&space;=&space;-7,308$

$Q&space;(\alpha&space;)&space;=&space;7,308&space;*&space;cos&space;(\alpha&space;)&space;-&space;17,023&space;*&space;sin&space;(\alpha&space;)$$N&space;(\alpha&space;)&space;=&space;-7,308&space;*&space;sin&space;(\alpha&space;)&space;-&space;17,023&space;*&space;cos&space;(\alpha&space;)$

 α M(α) [kNm] Q(α) [kN] N(α) [kN] 60º 121,62 -11,09 -14,84 75º 162,10 -14,55 -11,46 90º 212,02 -17,02 -7,308

The last point is to draw graphs of internal forces in the arch.
I hope that this guide will bring you closer to the correct solution of your projects and to pass all exams. If you still have any questions, please contact me and I will be happy to help you! For my part, I encourage you to take a few examples for training and invite you to the other guides.