I'm interested in how to work out whether torsional effects can be ignored or not, in calculating the adequacy of a given steel beam in the following idealised eccentric scenario.

Or, equivalently, the criteria which must apply, for it to be appropriate to treat the beam and its UDLs as a simple question about vertical plane bending, needing only to consider the actual vs design bending moment, deflection, and bearing support strength.

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Description of problem

The problem is idealised and static, so the available parameters are the beam's inherent properties, the UDLs Load 1 and Load 2, and the length (clear span) of the beam. For simplicity, the beam can be deemed horizontal and both positionally and rotationally fixed at both ends. (It might be the top beam in a moment resisting frame, or both ends concreted into adjoining walls.)

Load 2 is one side of a massive and stiff horizontal rectangular object, such as a thick steel sheet or precast concrete floor, that's simply resting (unattached) on the bottom flange. (The other side of the sheet is simply resting unattached on a [relatively distant] parallel wall or second beam to the left). There are adequate stiffeners in the flange itself


  • If Load 2 is small enough, it will be contextually trivial and we can totally ignore the torsional element, and just treat the beam as subject to vertical bending only from Load 1.

  • If Load 2 is large enough, there will be a torsional dimension to the problem, requiring more complex calculations or numerical methods.

What method, criteria, or formula would be used to determine whether or not the torsional aspects of Load 2 can be ignored, for the purposes of calculating whether the beam is adequately strong for the setup and loads?

Or is it the case that enough stiffeners and the problem can be ignored?


1 Answer 1


In your sketch, there seems to be no off-center loading so no torque.

But if we had a slab sitting on say the lip of the lower flange then we had torque, $$t=w*x (\text{w is slab weight}/ ft \ , \ x= offset),$$
This torque would rotate the fix- fix (fix for rotation )beam by this equation: $$ \theta_{max-at- center}= \frac{t*L^2}{8*G*j} $$ G = Shear modulus

J = section polar moment of inertia or St Venant's section modulus.

Note at the similarity to M of a simply supported beam.

And the diagram of the torque stress would be similar to that of the shear stress of a simply supported beam with uniformly distributed load: a straight line from, $-tL/2 \ $ max at left support to zero at the mid-span to $+tL/2 \ $max at the right support, depending on sign convention.

Then we would have to add the torque stress to the shear stress with sign and direction, a bit complicated. $\tau+v < allowable \ shear-stress .\ $ And depending on load combination multiply by appropriate factors.



AISC webinar on designing beams for torsion has condensed useful material and real-life examples and links. Webinar

  • $\begingroup$ Thanks! I'm surprised you've said there isn't torque. Because Load 2 is supported presumably by the width of the flange, no reason to assume its entire load supported by the infinitely thin "corner" at the centre and rest of flange carries zero load. So I placed Load 2 at the flange midpoint (total beam width /4) or even at the lip for max conservative safety - why wouldn't this be right? $\endgroup$
    – Stilez
    Nov 24, 2019 at 17:45
  • $\begingroup$ And also can you expand a bit about moving from max rotation at centre, and the torque stress graph (understandable), to (tau + v) and where allowable shear stress is obtained/calculated? Does the zero for stress graph at centre mean maximal torque+shear stresses both occur at the fixed ends? $\endgroup$
    – Stilez
    Nov 24, 2019 at 17:47
  • $\begingroup$ @Stilez, 1- I think the code says up to 20% of incidental section torque needs no treatment for WF, there is code in the link. 2- yes, torque and shear max both happen at supports. but of course, torque stress is directing to different radians in XY plane versus sheer pointing vertical. 3-Torque in open section members is a bit complicated. Two mechanisms act, torque which causes tensile and compression stress on the flanges and shear on the flanges due to twisting of the beam. read the link it is brief and helpful. $\endgroup$
    – kamran
    Nov 24, 2019 at 20:57

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