Monorail beam design to AS1418.18 1

[apsix]

The equations for both flange and web thickness checks include the dynamically factored wheel load Nw.

As most trolleys have 4 wheels I typically use;
(dyn. fac. hoisted load + hoist + trolley)/4.

What I don't understand is that the wheel longitudinal spacing is not accounted for. I would expect interaction between the wheel load effects if the spacing is small enough.

Does anyone have any comment on my procedure and/or dilemma?

Thanks.

 

[sdz]

Good question. Unfortunately I can't answer it.

 

[apsix]

This is the response I received elsewhere.

I believe the reason that only one wheel is used is because for most typical arrangements, the c/c wheel spacing of the trolley would be significantly greater than the half-flange width, so only one wheel would contribute in any significant way to the calculated stresses in the flange or web near the point of application of the wheel load. (e.g. Assume a 45° spread of load from each wheel contact patch to the root of the flange – if the zones of influence of each of the wheels do not overlap, you can consider them to be independent load effects.)

If you had an unusually wide-flanged monorail, or a trolley with unusually small diameter, closely-spaced wheels compared to the beam size, you might need to consider both wheels as contributing to the critical stress state, but this would be a very unusual configuration.

My response;
"Your explanation is most probably the answer, but seems slightly at odds with the effect that a free end has on flange thickness. (where KL = 1.3 if the wheel is within (depth + width) of beam free end)"

Yes, I agree that the referenced expressions are a bit philosophically inconsistent – but then again, they are also pretty much “black box” expression anyway. (e.g. where do the 2400 and 600 terms come from?) I suspect the expressions just embody some fairly conservative “rules of thumb”.

If you were to try to work out an appropriate flange thickness from first principles, you would start by estimating an effective tributary width of flange (e.g. assume a 45° angle of spread from the point of wheel contact, and then calculate the effective bending stress on that tributary width). If you try this, you end up with thicknesses pretty consistent with the code expressions. It doesn’t make a lot of sense that the required flange thickness instantly increases by 30% when you cross the magic line 2BF+D from the end of the beam, but having some sort of linearly varying expression for KL as you move from the end of the beam (KL = 1.0 at 2BF+D ), increasing to KL = 1.3 at the very end of the beam is probably unjustifiable complexity.