What is the correct way of setting up this problem ?

[Milos Stancic]

Hello everyone,

This is my first time posting, so forgive me if I did it in the wrong sub thread or wrong forum.

I have a very specific problem that I though I have solution to, but then after talking to my colleague and understanding his point of view I am left a bit puzzled now.

Here's the setup (I also uploaded a screenshot of it for more clarity):

I have a square tube that is fixed on one end (left end on the picture). The square tube is being supported by I beam piles that have bearings on top. So this allows the square tubing to rotate around it's axis.

The square tubing gets a moment loading at the end of the tube. We know what that moment is and we know that it creates a 30 degree deflection at the end of the tube (torsional deflection).

For design purposes and interference with some other parts of design we don't want this tube to rotate that much. We want to allow it to rotate 15 degrees only. So we designed a hard stop at the end. It allows the torque tube to rotate 15 degrees and when the moment gets too high it hits the hard stop that doesn't allow it to rotate more than 15 degrees at the end.

NOW this is the question. How much load does that hard stop takes ?

For the sake of argument let's assume that it takes 1000 KNm to torsionally deflect the tube to 30 degrees at the end of it. At 500 KNm it gets to 15 degrees and hits the hard stop. The torque keeps being applied to 1000 KNm. Does hard stop sees full load divided /2 (half of the full load on left fixed point and half of the load on the new fixed point (the hard stop)) or is load divided 75% for the fixed point to the left and 25% for the new fixed point ton the right (the hard stop) ? My colleague says that new hard stop doesn't see any of the first 500KNm load because the tube is not yet engaged, and once it gets engaged that's when it starts seeing the load, but only other 500KNm are being shared between the fixed points, not total load. Hence 75% to 25% distribution.

Thanks for helping out and sorry if this sounds confusing. I will do my best to explain it better if needed.

 

[BAretired]

BAretired[/color]]Unless the fixed end is a spring, it's not a spring...it is fixed. we rotate the free end by a torque then release it, the tube won't rotate back by itself, it will unless it is held against rotation or, alternatively, unless it is strained beyond the elastic limit what is the stress at the free end? depends on the size of the HSS, but it must resist a torsion of 500kN-m Then we place a stop, we already did that and apply additional torque, which end will engage to hold the free end (now with a stop) from rotation? the end stop has become a fixed end which will engage the free end of the HSS, preventing rotation, and the free end is no longer free I think the superposition method answers this question - the shear diagram of a cantilever plus a beam with both ends restrained for rotation. Can you provide a sketch showing this? It's not a cantilever if both ends are restrained against rotation. We are talking about torsional moments.

BA

 

[BAretired]

I am wondering if chris3eb and le99 are thinking of a different problem, i.e. bending of a beam. The OP is talking about torsional moments, not bending moments. My comments are addressing torsion, not bending.

BA

 

[le99]

The total torque at each end is the sum of the cases below.

Correction: Both T2 shall pointing left side (the right side reaction was drawn in the wrong direction)

 

[Milos Stancic]

The system sees torsion only. And the reason for even asking this is designing the hard stop. I want to know if the hard stop sees full torque OR half torque OR 25%.
I was under impression that it should be 50% 50%, but the way @BAretired explains it makes a lot sense on the paper.

So to clarify again, there is a square tube that is fixed at one end. There is torsion being applied throughout the whole tube evenly (let's say clockwise) twisting the square tube. The tube is free at the other end. With the torsion applied to the tube it can rotate 30 degrees. But I want to avoid the tube being twisted at the end that much. So I am designing a hard stop that will allow the tube to twist 15 degrees and then it will hit the hard stop. At that moment the torsional load will continue to be applied evenly to the tube. So the question is how much load will the hard stop see.

A) Since there is total 1000KNm at the end of the loading it will be spread 500 KNm between fixed points OR
B) Fixed end will see 500KNm before the free end engages the hard stop and then additional load will be divided between the fixed end and a hard stop resulting in original fixed end seeing 750 KNm and a new hard stop seeing 250 KNm. Hence the 75% 25 % distribution

 

[Milos Stancic]

Also to add to this, @le99 asks a good question. Should we consider this torsion at the end as a spring ? Because if that's the case I understand how new hard stop doesn't see any load until it gets touched and then it starts sharing the new load. In that case I can see how distribution is 75 25.

 

[le99]

@Milos,

If you bother to go back a few threads, you will find I've suggested a shear friction-type connection utilizing slotted holes to allow limited rotation, once the torque break through the resistance, the connection will be forced into bearing, which is similar to a hard stop. This mechanism requires design for the attached end only - 1) set shear friction limit, 2) determine bearing strength.

 

[Milos Stancic]

Hey @le99 I saw your suggestion, but it's much more complicated than that. I simplified this problem a lot to post it here without creating too much confusion. I was trying to not involve anyone in the design part but more in understanding the issue and how to treat it. Thank you for the suggestion tho

 

[BAretired]

My bad. I goofed. I agree with 75-25 now. But it's late and I will think about it overnight.

Okay, thought about it!
T1 is applied at the right end, not uniformly over the span. T1 at the fixed end is in the opposite direction. When the end stop is fixed, T1 does not change and the beam remains rotated at 15 degrees.

T2 = t2*L/2 where L is length of beam. T2 is the reaction to a uniform torsion and is in the same direction at each end.

Maximum torque reaction at fixed end is 750kN-m after uniformly distributed torque is added.
Maximum torque reaction at end stop is 500kN-m before uniformly distributed torque is added.
Torque reaction at end stop after uniformly distributed torque is added is 250kN-m.

BA