# Internal forces in the frame

Calculation of internal forces in frame constructions is analogous to beam structures, with the difference that we will also have vertical bars, which does not make a big problem. The frame that I will calculate looks as follows.

Before calculating the internal forces, calculate the support reactions and forces in the joint. The next step will be to plan how we will divide the construction that we count. Everything is clearly visible in the picture below.

I divided the frame into 5 sections. The forces in the joint will be visible when I calculate parts C and D. Now we can start calculating the internal forces.

Section A-A, distance: 0 ≤ x ≤ 2

$\Sigma&space;X&space;=&space;0$
$T&space;(x)&space;-&space;4,50kN&space;=&space;0$
$T&space;(x)&space;=&space;4,50kN$

$\Sigma&space;Y&space;=&space;0$
$N&space;(x)&space;+&space;1kN&space;=&space;0$
$N&space;(x)&space;=&space;-1kN$

$\Sigma&space;M&space;=&space;0$
$-M&space;(x)&space;+&space;4,50x&space;=&space;0$
$M&space;(x)&space;=&space;4,50kN&space;*&space;x$
$x&space;=0&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(0)&space;=&space;0kNm$
$x&space;=&space;2\:&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(2)&space;=&space;9kNm$

Section B-B, distance: 2 ≤ x ≤ 4

$\Sigma&space;X&space;=&space;0$
$T&space;(x)&space;+&space;5kN&space;-&space;4,50kN&space;=&space;0$
$T&space;(x)&space;=&space;-0,50kN$

$\Sigma&space;Y&space;=&space;0$
$N&space;(x)&space;+&space;1kN&space;=&space;0$
$N&space;(x)&space;=&space;-1kN$

$\Sigma&space;M&space;=&space;0$
$-M&space;(x)&space;-&space;5kN&space;*&space;(2-x)&space;+&space;4,50kN&space;*&space;x&space;=&space;0$
$M&space;(x)&space;=&space;-5kN&space;*&space;(2-x)&space;+&space;4,50kN&space;*&space;x$
$x&space;=&space;2\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(2)&space;=&space;9kNm$
$x&space;=&space;4\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(4)&space;=&space;8kNm$

Section C-C, distance: 0 ≤ x ≤ 2

$\Sigma&space;X&space;=&space;0$
$-N&space;(x)&space;-&space;0,50kN&space;=&space;0$
$N&space;(x)&space;=&space;-0,50kN$

$\Sigma&space;Y&space;=&space;0$
$T&space;(x)-1kN&space;=&space;0$
$T&space;(x)&space;=&space;1kN$

$\Sigma&space;M&space;=&space;0$
$M&space;(x)&space;+&space;1kN&space;*&space;x&space;=&space;0$
$M&space;(x)&space;=&space;-1kN&space;*&space;x$
$x&space;=&space;0\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(0)&space;=&space;0kNm$
$x&space;=&space;2\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(2)&space;=&space;-2kNm$

Section D-D, distance: 0 ≤ x ≤ 2

$\Sigma&space;X&space;=&space;0$
$N&space;(x)&space;+&space;0,50kN&space;=&space;0$
$N&space;(x)&space;=&space;-&space;0,50kN$

$\Sigma&space;Y&space;=&space;0$
$-T&space;(x)&space;+&space;1kN&space;-&space;2kN&space;*&space;x&space;=&space;0$
$T&space;(x)&space;=&space;1kN&space;-&space;2kN&space;*&space;x$
$x&space;=&space;0\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;T&space;(0)&space;=&space;1,00kN$
$x&space;=&space;2\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;T(2)&space;=&space;-3,00kN$

$\Sigma&space;M&space;=&space;0$
$-M&space;(x)&space;+&space;1kN&space;*&space;x&space;-&space;2kN&space;*&space;x&space;*&space;x&space;*&space;0.5$
$M&space;(x)&space;=&space;1kN&space;*&space;x&space;-&space;2kN&space;*&space;x&space;*&space;x&space;*&space;0.5$
$x&space;=&space;0\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(0)&space;=&space;0kNm$
$x&space;=&space;2\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;\;&space;M&space;(2)&space;=&space;-2kNm$

Section E-E, distance: 0 ≤ x ≤ 4

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

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

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

We already know the values of all internal forces, we can make their graphs.

Bending moment

Shear forces

Normal forces

We can see from the cutting force diagram that there will be some extreme value in part D. Let’s calculate it, we equate the equation of the cutting force from section D to zero.

$-2&space;*&space;x&space;+&space;1&space;=&space;0$
$x&space;=&space;0,50m$

We substitute the distance to the bending moment equation from section D.

$M&space;(0,50)&space;=&space;1&space;*&space;0,50&space;-&space;2&space;*&space;0,50&space;*&space;0,50&space;*&space;0,50$
$M&space;(0,50)&space;=&space;0,25&space;kNm$

The extreme value is $0,25kNm.$

In this way, we have reached the end of the task, I hope that everything is understandably explained and as usual I encourage you to take a few examples to consolidate knowledge. If everything is clear, we can move on!