What methods have you been taught so far? That would be helpful info because there are several ways to approach the problem.

Your structure looks to be singly redundant, you can solve it statically if you remove one of the X-direction constraints (so that both x-direction reactions disappear (yes?).

To figure this out, you need to solve an an indeterminate structure method. A method is the unit force method (read up about this). An outline of the steps ...
1) remove the redundant constraint (the RH x-reaction), solve the structure.
2) calculate the x-displacement to the RH end (where you removed the x-constraint, the end of the RH leg, yes?); call this X. This is going to be tricky as all the elements are bending.
3) then load your determinate structure (what you considered in 1) above) with 1 lbf (1N if you like) in the x-direction, and solve this structure
4) calculate the x-displacement of the RH end under this unit load, call this x.
5) so the RH x-reaction is a load that will return the determinate displacement (from 3) above) to zero, a multiple (= X/x) of your unit load

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

Why would there be horizontal forces at 2 and 4. The loads applied would be the same as a moment applied to the frame and the reactions at 2 and 4 would be vertical and opposite with a magnitude of 2057 kN. (8100 * 261.6 / 1030)

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

I assume he constrained the model this way (x- and y- at both legs) ... otherwise the LH leg would lift up

The more "normal" constraint would put a roller under one leg.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

With fixed supports, this cannot be solved without axial and bending stiffness values for the members and moment restraint values at nodes 1 & 3.

I see that i wrote in the situation that the 2 forces are 8100kN, but this should say 8100N! Sorry

Guys i think got it!
I changed the support at 4 to a roller and it all makes sense now.
The reaction should indeed be (I think) around 2,057kN! The simulation also confirms this after i changed the support at 4.

Thank you all so much for the quick response! Im really sorry that i didnt think of this in the first place.

have a good one!

As posted the applied moment is 8100*0.2616 = 2119, the pure reaction couple should be 2119/1.03 = 2057

as posted reactions ... 1735*1.03 +510-180 = 2132 (I figured your reactions moment's directions)

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.