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Force couple acting on rigid frame 2

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310toumad

Mechanical
May 12, 2016
63
How would I go about solving for the moment(s) at the base where this structure is fixed? From looking at examples of frames, a few assumptions are typically made to reduce the indeterminacy of the problem. For this case, can I assume the reactions acting opposite to the applied 1430 forces are also 1430? The problem I run into with this is that, if I assume that the horizontal member has a point of inflection at the midway point, how do I resolve the internal forces when "cutting it in half" and doing an force equilibrium analysis to find the moment at the base? If I already have assumed the applied force and the reaction are equal, then any internal force in the beam would have to be zero in a force balance, in which case your moment is just 1430*25.5. I don't think that is correct, the horizontal member connecting the two should 'stiffen' and lessen the bending moment logically. What am I assuming incorrectly? Wouldn't there also be a torsional component as well?
 
 http://files.engineering.com/getfile.aspx?folder=73d15ea1-73ed-48df-83c4-3dcebb261cac&file=Frame.jpg
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yes, i see the connection between normal flexure of the post and torsion of the beam ... a torsionally rigid beam would make the post double cantilever. if the beam is torsionally weak, then the posts are simple cantilevers

the big difference (I see) between your runs is the shear carried by the beam (instead of being reacted at the ground) ... this is flexure of the beam. And so the interconnection with torsion ... a torsionally strong beam allows the post to develop more top end moment, less ground shear reaction. similarly a torsionally strong post allows the beam react more fixed end moment, to behave more like a cantilever.

I don't know your sections well so I asked why would one section be so much stiffer in bending, and recongising the I beam has a strong and a weak axes, then I asked which way round is it in your run.



another day in paradise, or is paradise one day closer ?
 
I tried testing different stiffnesses for the HSS2.5x2.5, just because I like you so much.

I changed the load to 1 kip and all the member lengths to 10".

Everything normal, the beam takes 32% of the shear.



Now, leaving everything else the same, if I change the following:


Column torsional stiffness = ∞: Beam takes 82% of shear.

Beam flexural stiffness = ∞: Beam takes 34% of shear.


Column torsional stiffness = 0, Beam takes 0 shear.

Beam flexural stiffness = 0, Beam takes 0 shear.


Column torsional stiffness = ∞ and beam flexural stiffness = ∞, Beam takes 100% of shear.


The flexural strength has some part in attracting the shear, but it is primarily the torsional stiffness.










 
thx, I found that interesting.

a flexurally rigid up-right would react all the shear (zero beam shear),
a torsionally rigid beam would ... ?

a flexurally weak up-right would make the beam react all the shear, (may well collapse),
a torsionally weak beam would ... ?

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