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Re: Torsion on Wood Beam

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Torsion for a rectangular shape gets fun.  While the most significant aspect will be additional "shear stresses" due to torsion (remember "pure torsion" for a circular shape is ALL shear stresses), there will be some "warping" stresses that will produce some "normal" (aka axial/bending) stresses.

So, based upon what you got from the Wood Engineering Handbook, it appear they are just giving you the shear stresses from torsion and potentially ignoring any "warping" stresses (i.e. normal stresses).  If so, then the torsional shear stresses would have to be checked in combination with the "shear" shear stresses.  If you also got some value for the normal stresses due to "warping", then this would be checked in combination with bending and axial stresses.

The best analogy that is available that I can think of is to look at how concrete code (aka ACI 318) handles it.  Basically, torsion in concrete requires you to look at the "shear" reinforcement (aka "shear hoops") for both the shear and torsion...AND it requires to add additional longitudinal steel to what you have for bending because you will get the "warping" stresses due to most concrete beams being rectangular.

Regards,

Scott

On May 9, 2012, at 6:10 PM, Drew Morris wrote:

Wow, I was struggling with this earlier today.  I couldn't find anything in the NDS but found some information in the text book from Faherty and Williamson, "Wood Engineering and Construction Handbook" 2nd edition page 4.17.  Given a beam with depth 2a (in) and width 2b (in), the torsional stress is:

fs = Torque (in-lb) *(3*a+1.8*b)/(8*a*a*b*b)

The page refers to the "Wood Handbook" and gives the allowable torsional shear stress (Fs) as Fv for sawn lumber and the allowable radial tension stress for glulams (Fv/3?).
This reference shows you how to calculate the torsional stress but no guidance after that.

Having calculated the stresses, I guess you would need to use a form of the interaction equation:

   fs/Fs + fb/Fb + fv/Fv < 1.0  ????????????????

In my case, the archiitect wants to remove some columns that support trusses.  The bottom of the trusses is at 8'-9", so a GL6 3/4 x 18 centered under the support affects the height clearance.  I tried putting the bottom of the glulam flush with the bottom of the truss and side loading the glulam.  I was coming up with a much bigger glulam using the above approach.  As an alternative, I'm looking at using a W10 beam under the truss support or a laterally loaded HSS12x4.  I have to look at the building sections in more detail, I might be able to brace the top of the laterally loaded glulam up to the roof diaphragm.

On 5/9/2012 1:29 PM, Larry Hauer wrote:
Anyone know of some simple formulas, or reference, for designing an unbraced wood beam that has a torsional moment, (or twisting), at the center of the span as well as bending due to vertical loads on the beam?

Thanks in advance,

Larry Hauer, S.E.


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Drew Morris, PE | Project Engineer
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