HSS statics problem

[310toumad]

If I have a piece of rectangular HSS, with another formed channel resting on top with spacer blocks on the inside, how do I find the reactions of the blocks on the sides that resist the moment shown? I'm not sure if my convention of the angle of the reactions is correct. I believe for the block on the right, its corner follows the path traced by the circle with the center at the pivot point of the moment. For the block on the left, the corner that contacts the HSS does so only after the one on the right makes contact, and starts pivoting at that location.

So that leaves me with the question, how to find the reactionary forces if I know the magnitude of the moment and all the dimensions? Its not a force couple because they aren't parallel (although their x-y components would be I suppose), unless I have their direction screwed up then its a simple M/d = F situation, but I'm not sure if that is the case.

 

[rb1957]

it looks like you're trying to say you have tangential forces at two of the three contact points. why ?

1) why two of three ?
2) why different arcs/centers of rotation ? (wouldn't the outer part rotate about a center ?)
3) why tangential ?

this looks to be a highly redundant reaction set. Whilst you know the three forces form a closed polygon(triangle), I don't think you can readily set the direction of magnitude of any of them. Sure you can make assumptions, but ...
The stiffness of the channel will also affect the solution.

another day in paradise, or is paradise one day closer ?

 

[jlnsol]

Well...think of the above U shape as a wrench that tries to rotate a nut (the rect tube)then it is not necessary that there is a contact location where the torque arrow is indicated, so suppose that force=0N
And since immediately flattening of the contact locations L(eft) en R(ight) will occur we can say that these forces are equal and opposit working forming a moment since then are at a vertical parallel distance y being the width of the spacers.
So the forces can be calculated.

Note with a real wench there the hand force on the wrench will plays its role in the horizontal Equilibrium (when the hand force is hor)

 

[310toumad]

So essentially you are saying to make the assumption the forces act perpendicular to the HSS like this? And then use the spread to find the forces using M/1.16 = F?

If that is what you are suggesting that is how I initially approached the problem, but then I thought about an eccentrically loaded bolt pattern and how the reaction forces are perpendicular to a line drawn from the pivot point to the bolt, and figured this must be similar.

 

[desertfox]

This problem won't be easy to solve, yes you can assume that the top block is a pivot and work out the reactions as in your sketch.
That said with tolerance build up in the components how are you going to guarantee that hen the top block pivots the wo side plates will touchdown on the rectangular section simultaneosly?
If the above doesn't happen then the u section is taking the load all on one side.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[jlnsol]

310toumad: yes that is what I am saying and as your second pdf shows. Not sure what you mean by bolt pattern..
desertfox: the OP has not mentioned any vertical forces up till now, so we can assume the friction in the 'pivot' point to be low, therefore as soon as one side touches its reaction force will overcome that friction and slide the top U until the other side also touches.
Actually it isn't a real pivot but more a slidable contact which may not even be there as with the wrench example.
Ofcoarse with a vertical weight in the game it would be different.

 

[handleman]

Golly jeepers folks. The solution is in the freaking post title: STATICS. Quit being lazy or overthinking. Write the FBD, sum forces and moments. Reaction at the top block is zero? Really? Yep (given the problem statment).

 

[rb1957]

really ? how to draw the FBD when the reactions are not clear ? and dependent (to some extent) of the rigidity of the structure, and certainly to the (undefined) gaps and clearances ??

another day in paradise, or is paradise one day closer ?

 

[desertfox]

Well judging by the angle of the reactions shown in the diagrams there is very little force available to slide the surface of the U section over the surface of the rectangular section.
Imagine that we push the U section either left or right so that one of the side plates contacts the rectangular section the other side having all the clearence between the two faces, now apply a moment to the U section does the U section rotate? Or is it prevented from rotating by the side wall in full contact with the side plate on the U section? Now move the U section so that the clearence is equal on both sides and apply the moment just as you have in your diagram, now it may well rotate till one or both side plates come into contact with box section but the reactions depending on what actually happens in practice will be different in each case.
You can make simplifying assumptions and calculate theoretical reactions but they may be far from the truth.


“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Shu Jiang PEng SE]

You may ignore the spacer on the long side of HSS, the torsion will be resisted the reactions of the two spacers on HSS short sides. Per your second attachment 1.16xr=Torsion