Steel Beam on Slab - Concentrated Moment

[CANPRO]

I'm looking for a little guidance on how to analyze this situation.

This will be applied to a job eventually, but for now I'm just trying to get the concept right.

For discussion sake I'll add some dimensions...

I have a 10' steel beam, 8' of the beam sits flat on a slab (I assume the slab is perfectly rigid) and is not fastened to the slab in any way, and there is a 2' cantilever. At the end of the canitlever is a concentrated moment. The cantilever supports gravity loads. The concentrated moment is applied so that the tip of the beam deflections upward. (see attached sketch)

I can picture how the beam will deflect...not all 8' of the beam will remain in contact with the slab, the remaining area of the beam in contact with the slab will resist the vertical load and the concentrated moment.

I would like to know how much of the beam remains in contact with the slab and what the pressure distribution will look like.

My approach right now is...

Take the vertical load, this is the resulant of the pressure between the beam and the slab.

Sum the moments at the far left of the beam (the end that is on the slab) and use this to get the location of my resultant.

And from there I'm guessing I have a triangular pressure distribution...

I'm just not sure if this is the right way to go. Does this sound about right?

 

[ztengguy]

How is the moment being induced into the beam?

 

[Ron]

Replace the induced moment with a force couple separated by the cantilever distance. This should tell you what you need to know once you resolve the forces. If the forces are not in equilibrium at that condition, move the couple separation to the cg of the beam to see how much imbalance there is. In this case, the self weight of the beam might be critical.

 

[msquared48]

If you put a shim plate at the edge of the cantilever under the beam, and at the other end of the beam, you will have a defined length of beam to determine your forces.

Otherwise, this is more like an inverse beam on an elastic foundation.

Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[BAretired]

Neglecting the self weight of the beam, if w is the load per foot on the 2' cantilever, then the applied vertical load is 2w. So the vertical reaction is also 2*w.

If the reaction is distance x from the c.g. of load, then:

2*w*x = M, so x = M/2w.

If 1'>x>9' equilibrium is not possible. If 1'<x<9' then the moment is zero at the reaction point and all points to the left.

Moments, curvatures and deformations can all be calculated in the usual manner.

Spreading the reaction over a nominal width will not make much difference but it could be included by treating the slab as an elastic foundation.

BA

 

[CANPRO]

I have looked at this similar to what you guys have said. Ron, I'm not sure I know what you mean by moving the force couple to the center of the beam to check unbalance...the geometry of the problem hasn't made equilibrium an issue so far, but I will come back to that if it comes up.

I guess at the end of the day as long as the reaction falls a reasonable distance from the edge of the slab its okay.

When I first looked at this I figured the slab would be so much stiffer than the beam that it wouldn't have much influence, but maybe it does...

If I were to assume that the slab was perfectly rigid, what do you think the pressure distribution would look like? I thought it would be triangular, with the point of zero pressure being where the beam lifted from the slab...but that would mean that I could get the length of bearing from just the equations of equilibrium, ignoring the stiffness of the beam...which can't be the case because if both beam and slab were infinitely stiff then the resultant would just be a point load at the far left of the beam.

I picture the pressure distribution being a trapezoid...a maximum pressure starting at the far left and staying constant for some distance, and then a linear drop to zero, at which point the beam lifts from the slab. I guess this is the part where I get a little lost running the numbers...

 

[BAretired]

If the slab is perfectly rigid, the beam reaction is a point load. There is no triangular distribution. To the left of the reaction point, the beam is perfecly straight. To the right, the moment increases linearly from zero at the reaction to the edge of the slab. Beyond the slab edge, moment increases at a lesser rate on a second degree curve, reaching a maximum of M at the end of the cantilever.

The slope of the beam is zero at the reaction point and increases by an amount equal to the area under the M/EI diagram.

The deflection of the beam at any point can be found using area-moment principles.

Theoretically, liftoff occurs at the reaction point.

BA