Calculating Horizontal Reaction Forces in L-Beam

[JHensley]

Hi everyone, I'm new here so I hope I'm posting this in the right place.

I'm struggling to duplicate the horizontal reaction forces being generated in an L shaped beam. Typically I like to do hand calculations and verify my answer using Solidworks, but in this case I can't seem to wrap my head around the results from Solidworks.

The beam is essentially pinned, or hinged, at each end with a moment applied at one end (see attached images). I realize this is considered an indeterminate situation and this is the root of my problem. I have some experience using Castigliano's theorem, but again that's for simple beams, or possibly superposition could be used in some way? I thought I was fairly well versed in statics and beam theory, but throughout my undergrad I had actually never encountered anything outside of the simple beam shape.

Any help, recommendation, references is much appreciated, thanks!

 

[jrmichler]

It's not clear from the figures exactly how you implemented your restraints and loads. I suggest making a new model where the beam is modeled as a solid, and the pinned connections as holes with elastic foundation restraints. You will need a high normal stiffness to correctly model your problem. Some iteration will be necessary. Then show us the deformed geometry plot.

 

[JHensley]

Thanks for the response jrmichler,

I drew a new free body diagram to show better how the beam is restrained and attached a deformation plot for reference.

I ran a new simulation with a "rigid" material and solid bars, the horizontal reaction force seems to vary between 430 N and 438 N depending on the beam geometry. I understand that the elastic interaction between the members has an impact, so is the best approach to this to use a strain energy method?

Out of curiosity, I managed to find a trial version of a frame analysis software just to validate the Solidworks results and it gives a slightly answer (see "L-beam 2nd Results").

 

[jrmichler]

Keep in mind that a pinned connection at each end enforces a mathematically exact constant distance between those two points. I find that a deformed geometry stress plot to be more useful. That's a deformed geometry plot with the stress contours shown in color.

1) Your horizontal and vertical forces sum to zero as they should. Do a sum of moments and confirm that sums to zero.

2) Try two different restraints on the left end - allow sliding in the horizontal direction, then allow sliding in the vertical direction. Then confirm that forces and moments sum to zero. I think that by the time you finish, the answer will appear.

 

[JHensley]

2) Try two different restraints on the left end - allow sliding in the horizontal direction, then allow sliding in the vertical direction. Then confirm that forces and moments sum to zero. I think that by the time you finish, the answer will appear.

So if I'm understanding correctly, if the Fy reaction is omitted from the left end the problem is now very basic, resulting in a 400 N horizontal force. But is this realistic, is it sufficient to ignore the variance between what Solidworks calculates and a simplified model? How do I know when or when not to omit certain reactions?

 

[jrmichler]

I can't seem to wrap my head around the results from Solidworks.

+1 for cross checking your own work. All too many people blindly accept the results.

The purpose is to better understand what SW is telling you. The SW FEA has some idiosyncrasies that can give you wrong answers. When you get answers that do not seem right, then you need to find out why. One technique is to break the problem down into simpler problems for the purpose of better understanding exactly what the software is doing. It might seem like wasted time, but I can assure you that it is not wasted.

I have extensive experience with SW FEA, mostly with analyses much more complex than yours. I created a document with the title SW Booby Traps and Tricks to keep track. It ran over ten pages.

 

[JHensley]

One technique is to break the problem down into simpler problems for the purpose of better understanding exactly what the software is doing. It might seem like wasted time, but I can assure you that it is not wasted.

Funny that you mention that because this actually is part of a larger system, so this was my attempt at trying to understand this specific section. It sounds like I need to do some research on some of the SW "idiosyncrasies", thanks for the help!