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# hand calculation simple frame

## hand calculation simple frame

(OP)
Hello guys,

Im really struggling to do a proper hand calculation for some reason on this really simple frame. none of my methodes seem to solve this one. i need to know what the reaction forces are in the 2 supports at the bottom. Does anyone know something on how to solve this situation? the reaction forces/moments are as you can see already calculated by a simulation.

I hope someone can me help out!

Greetings from the netherlands!
Replies continue below

### RE: hand calculation simple frame

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

### RE: hand calculation simple frame

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.

### RE: hand calculation simple frame

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

### RE: hand calculation simple frame

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.

### RE: hand calculation simple frame

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

### RE: hand calculation simple frame

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

### RE: hand calculation simple frame

(OP)
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!

### RE: hand calculation simple frame

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.

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