Point load displacement 1

[4thorns]

I have a question regarding point load displacement thru multiple, stacked beams. Basically, if I have say, 4 stacked identical beams, how much of the load (eg. 1000 lbs. concentrated on the center of the beams. They are supported at each end) is transfered from the top beam to the one below it. Then from the second beam to the third, etc. This isn't common practice but it came up at work today and I didn't have an answer other than...Picture this, I put 200lbs. on your shoulder and it's the maximum you can carry without being crushed. That means you can carry 200 lbs. If I put 5oo lbs. on your shoulder, chances are you aren't going to stand ther and hold 200 of it up...You are on the ground. (What a mess!) I'm not sure how the math works in this situation and would appreciate any help

Doug

 

[apsix]

All the beams will have the same deflection, therefore each beam will be supporting 25% of the load.

 

[4thorns]

I understand what you are saying but where does it end? If I had 8 beams whould each of them be carrying 1/8 the deflection? It seems to me that eventually the beam at the bottom would not be affected by the load above, or at least the deflection would be proportionate to the support above. Not giving you a hard time, just trying to understand.

Doug

 

[desertfox]

Hi 4thorns

Think of it like this
If the beams are stacked together how can the bottom beam not deflect if all the beams above it have deflected? also the bottom beam carries the dead weight of the other beams.

regards

desertfox

 

[paddingtongreen]

The problem is that they would take equal load only if there was zero friction between the beams. As there is friction between the individual beams, they would, in part, act as a composite beam.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[msquared48]

A very small part...

Mike McCann
MMC Engineering

 

[apsix]

That's a good point, but I don't think it would affect the result much if the beams are steel and working hard.

 

[BAretired]

If the beams are separated from each other with a frictionless film, then each beam will carry precisely the same load. If there are eight beams, each will carry one eighth of the total.

If the beams are prevented from free deflection by friction forces acting between adjacent beams, they will tend to act a little bit differently. If they are welded together, they will all act together as a unit.

Stacked beams should be considered as individual members, sharing the load equally, neglecting the friction force between them because the friction force is not reliable in the absence of positive attachment.

BA

 

[Lion06]

I agree with Mike. If you have a frictionless surface between the beams and you have continuous contact, they will all deflect the same. They will only all carry the same load if they have identical section properties, otherwise they will carry load based on their relative stiffnesses.

In a design situation, you can't possibly count on friction to transfer any shear between the beam interfaces, so I would definitely design each beam for it's share of the total load. That being said, however, there really is friction between the beams, and if you stacked say (10) 10' long W8's on top of each other to take a 1kip point load at midspan, I think the bottom beam would end up taking almost no load in bending. I believe that (in actuality, not our idealized conditions), as you move from the top to the bottom each beam would have less bending stress and more tension stress than the one before it.

 

[4thorns]

Thanks for the replies. I wasn't actually intending for my question to travel in that direction but I'm glad it did. Your posts pointed out a couple of things that I hadn't thought of but almost seem like common sense. Thanks for that. Now let's assume I have 4 pine 6x8 beams stacked. Together they can handle the load and have zero deflection. Not sure how it would play out in real life but bear with me..I put that same load as above on 1 beam and it bends and breaks. I put the same on 2 of them(stacked) and it bends to near the breaking point. 3 of them bend less and adding the 4th results in no bending at all. My wording is quite vague but it's beacause I don't know how each beam is reacting to the load. I do know that eventually, with enough beams I'm going to stop the bending. Is it the same as before and once the the deflection is stopped all 4 beams are carrying equal loads. Or is the load distributed in proportion to the number of beams required to resit it? I've always said that it's hard to ask an intelligent question if you don't allready know the answer.(Explains why game show hosts appear to be so smart!!)

Thanks again,
Doug

 

[cessna98j]

In the case of the 6x8 beams scenario you mention, the same rules apply. The more beams you have sharing the load, the less the total deflection, and the load gets shared by all of the beams in the stack (theoretically, assuming zero friction between each beam). Granted, in reality there is friction between each beam which does make the load shared by bottom beam slightly less than that of the top beam.

Regardless, the principal is the same. In reality, deflection is a function of the applied load and no matter how many beams you have stacked together - when you apply a load there will be a deflection (no matter how small it may be), because nothing has infinite stiffness. This deflection is what allows the load to be shared by the entire stack of beams.

 

[desertfox]

Hi 4thorns

Well I agree with cessna98j in that nothings changed and that they will always be a deflection when a load is applied however small.
If we take out friction between the beams then if you stack two beams together taking your 6*8 then the second moment of area (I) for those beams would be:-

(I) = 2*6*8^3/12 = 256

however if those beams were rigidly connected then and worked as a composite beam then:-

(I) = 6*16^3/12 = 2048

assuming the beams and loading were identicle in every way then using the above (I)'s would reflect the changes to both stress and deflection.
Now you can add as many beams as you wish and compare them for the two extreme cases in terms of stress and deflection

desertfox

 

[4thorns]

Sorry for the late reply...Been crazy. Thank you all for your replies. You've been a great help. I understand what's going on now and picked up some info that I hadn't thought of.

Thanks again

Doug