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RE: Torsion in Wide Flange Beam

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A torsionally pinned condition means that the supports cannot develop axial stresses in the beam flanges, compression on one side of the flange and tension on the other side, due to warping torsion. Simple support connections for gravity loads are, I think, necessarily also torsionaly simply supported.  For a beam to be torsionally fixed, there needs to be some means of developing axial stresses in the beam flanges.  A fixed end, moment connection, can do this but the stresses due to torsion are additive/deductive to the stresses due to flexure.
 
HTH,
Warren Foy
-----Original Message-----
From: Rich Lewis [mailto:seaint02(--nospam--at)lewisengineering.com]
Sent: Wednesday, May 25, 2005 11:31 AM
To: seaint(--nospam--at)seaint.org
Subject: Torsion in Wide Flange Beam

This is probably more a strength of materials type question rather than a steel question.  I have searched through my Strength and Statics book but haven’t found this described or illustrated.  Torsion is always shown with a fixed end.

 

I am investigating a wide flange beam that is supporting a uniform eccentric load.  It is an existing condition, not a new design.  I have AISC Design Guide 9 on torsion in steel.  I am trying to understand what a torsionally pinned connection is.  I believe this is what I have since I do not see any torsional restraints at the ends of the beam.

 

If I have a uniform torsional moment along the beam (caused by a uniform eccentric load) with pinned ends, is the maximum torsional moment at the center of the beam?  Is it zero at the ends? If it’s pinned I would think so.   I guess I would picture the torsion loading diagram like a uniform load diagram.  I would picture the torsion diagram like a moment diagram for a simply supported beam with a peak in the center and zero at the ends.  I’m having a hard time picturing an equivalent shear diagram to the torsion.  This would have a maximum value at the ends and a zero value at the loads.  If there is no torsional restraint how could I have a force at the ends?  I was thinking that there should be diagrams like load, shear and moment where you can get the value of one by integrating the area under the curve for the previous one.

 

If there is a torsional force on the beam, and it has pinned ends, how is it in equilibrium?  What produces the equal and opposite force of the torsion?

 

I guess if I answer these questions I can better understand how to apply Design Guide 9.

 

Thanks for any insight you may give.

 

Rich Lewis