Calculating Torsion...

[DMWWEngr]

Real quick question (I hope)...

Say you have a beam that that is simply supported by HSS sections on each end.

How does one determine the torsion that the beam places on the HSS sections??

Sorry for this simple question...but I can't find any good examples of this in my books.

TIA! ---
Andrew

 

[austim]

Hi, DMWWEngr,

Quick answer coming up.. It all depends on what you mean by 'simply supported'.

If your beam is genuinely simply supported, with bearings at each end permitting free end rotations, then your answer is simply that the total torque applied to the HSS will be the end reaction multiplied by any eccentricity between the bearing and the HSS centreline. The total should be proportioned between the adjacent sections of HSS according to their lengths. (eg if you have 2 feet one side, 4 feet the other, the torque in the shorter length is twice the longer).

If, as I suspect, you really mean that you have assumed the beam to be simply supported for analysis purposes, but it will be rigidly connected to the HSS, that is another matter.

For that case, (short of running a simple 3D frame analysis, which could be quicker than digesting my response), you will get a slightly conservative value for your torques thus:
(a) calculate the end slopes of your simply supported beam.
(b) Apply your calculated slopes to the HSS elements as imposed rotations, and determine the resulting torques from the standard formulae. (eg Timoshenko and Goodier, Theory of Elasticity, Chapter 11).

Good luck.

 

[ishvaaag]

If you let fall a beam on 2 parallel HSS sections of same deformation, the conjoint rotation with such beams at both ends is the same, and, except you are counting the created eccentricity of the load (P·e) there's no torsion, that requires relative torsional distortion between sections.

However, as austim says, making an analysis taking unto account what the case requires is the general approach.

 

[DMWWEngr]

Thanks,

Yep austim, it is the second approach. Guess it wasn't as simple as I hoped but it's still do-able

Thanks for the responses. ---
Andrew