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Is the top beam helping? 2

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4thorns

Structural
Jan 22, 2009
152
In the attached, the load is resting on a smaller beam. Does the smaller beam actually assist the bigger beam by resisting the 500# to keep the deflection within the allowable 1/2"

If so then can it be said that at the 1/2" deflection only 1000# plus the wall weight is being transferred to the bottom beam?
 
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All I will say is that load is attracted to stiffness.
 
The upper beam will help. The upper beam will take Iu/(Iu+Il) of the load and the lower beam will take the remainder of the load. Iu = I of the upper beam, Il = I of the lower beam.

 
So the whole load will push the bigger beam beyond the allowed 1/2" deflection?
 
If the I's work out to exactly what you have (i.e. that I of the upper beam is 1/2 of I of the lower beam and they both deflect 1/2" under the assumed loading), then they will both deflect 1/2".
 
Thanks Lion,

Is Iu/(Iu+Il) a constant regardless of the proportionate size of the two beams?
 
if there is OSB or Plywood sheathing on the wall, everything changes.
 
Understood Toad.:) This is a hypothetical situation as shown...no other variables.
 
This is assuming the beams are of the same material, so "E" is constant. It also assumes that lateral buckling of either beam is not a governing factor.

Bottom line here is that your problem cannot be answered with any degree of certainty as it is not specific enough.

Mike McCann
MMC Engineering

 
Mike-
I don't think the capacity of the beams matters for the load distribution. That's a capacity problem. The beam may, or may not, be able to take the load, but it will attract it based on the relative stifnesses. Point taken, though, that if one of the beams fails prematurely that the other beam is doing all the work.


Toad-
I considered the diapgragm behavior of the wall, but discounted it when I saw the load and aspect ratio of the wall. The wall is longer than it is tall so the plf of the diaphragm would be greater than the load plf, which is significant for a wood diaphragm.
 
I put the wall studs in this scenario merely to transfer the load to the lower beam. Let's remove the wall and put the smaller beam on top of the bigger one with the same loading conditions and deflection limits. If the beams are deflected 1/2" is the upper(smaller) beam carrying 500# and the lower (bigger) beam carrying 1000#?
 
Relative stiffness, I agree Toad.

Lion:

I would have to disagree as the detail is dimensionless, only subjective in nature. I do get your drift however.

It is interesting to note though that if the studs transfer the vertical force with no plywood, then the size of the top and bottom beams (speaking of the larger and smaller context of the OP) will not matter. However, with the plywood, and the aspect of a deep plywood beam between the two solid beams, then the shear stress in the end of the plywood wall will be greater if the smaller beam is placed on the top since a greater portion of the load will have to be transferred through the plywood wall to the larger beam below. However, in the end, regardless of the placement of the beams, all three will deflect the same.

In essence, with the plywood, there will be three beams of different relative stiffnesses.

Mike McCann
MMC Engineering

 
Toad-
Which part do you not follow? I did say this was a relative stiffness problem, no? Iu/(Iu+Il).
 
As 4thorns defined the problem, I would say the answer is YES. I don't know why you guys have made it so confusing. He did say the little beam deflects 1/2" under 500#, and the big beam deflects the same amount under 1000#, so the sum of the load taken by the two beams is 1500# when deflecting 1/2".
 
Agreed hokie, I just wasn't sure if that was purposeful or not. I guessed not, because if he understood the problem well enough to make it purposeful, there's no need to ask the question.
 
One thing to note here is that IF either of the beams fail or go into the inelastic range beyond a 1/2" deflection, then load will get shed over to the non-inelastic or non-failed beam.

You might flesh out this problem for us by telling us why you are asking.

 
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