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MattusMaximus
24th September 2011, 06:19 PM
Howdy all,

I was asked by a friend the following question:

Say I shipped to you a very long steel beam and when off-loaded you only had two supports to keep it out of the dirt. Where would you place the supports to minimize the deflection of the beam?

This sounds to me like a job for the calculus of variations, but it also seems to me that this would probably be a pretty standard engineering question. So in the interest of being lazy, before I start working on the calculations myself, does anyone have a link to a reference on this? Thanks in advance!

Cheers - MM

the PC apeman
24th September 2011, 07:18 PM
Howdy all,

I was asked by a friend the following question:



This sounds to me like a job for the calculus of variations, but it also seems to me that this would probably be a pretty standard engineering question. So in the interest of being lazy, before I start working on the calculations myself, does anyone have a link to a reference on this? Thanks in advance!

Cheers - MM

If you only had one support you'd put it halfway. With two supports, you could consider the beam to be two beams that just happen to abut each other. In this case you would put the supports at the mid-point of each "half beam".

However, there would be some effect from the two "half beams" actually being one long beam. Here I would think the strength of the steel and the cross-section of the beam would be important factors. As the strength of the material increases, the mutual support coming from the resistance to bending at the "joint" would increase, allowing the supports to be moved toward the outer ends.

So I guess the answer would be 'It depends.' Pretty unsatisfying, eh? Just a wild guess though. It could all be crap.

casebro
24th September 2011, 07:19 PM
I'd say 25% in from each end.

Think of it as each 1/2 of the beam being balanced on a single point, the support.

Fabrication and machinist experience.

OnlyTellsTruths
24th September 2011, 07:24 PM
I'd say 25% in from each end.

Think of it as each 1/2 of the beam being balanced on a single point, the support.

Fabrication and machinist experience.

I think the two 25% from each end (half of half) posts are close, but I'm pretty sure you have to then move them a bit closer to center (towards each other) to account for it being one beam.

So my guess is somewhere between 25% and 30% from each end.... but it could very well be 25% and I'm wrong.

the PC apeman
24th September 2011, 07:26 PM
Hey, so far in three responses we have <=25%, =25%, and >=25% from each end. Take your pick. LOL

casebro
24th September 2011, 07:42 PM
"I'm trying to think but nuthin happens" [/Three Stooges]

In the fab shop, running a 12' press brake, we used to use thirds. So 1/6 of the way in from the ends. It made sense once the owner, a mechanical engineer out of Purdue, explained it. But I can't remember the reasoning. 35 years ago, hence the 3 stooges reference...

eta: Toilet bowl reference: every now and then something useful floats to the top- The thirds was to support the ends, push down at the 1/3 points together, bend the piece into a fairly true end to end arc. A 90% chance of no help today, I'll betcha. But there is only a 10% chance of that. ;)

Myriad
24th September 2011, 08:12 PM
I'd say you're better off putting it on the ground, minimizing the deflection to zero. Who cares if it gets a little dirt on it?

Respectfully,
Myriad

ellindsey
24th September 2011, 08:34 PM
Stand it vertically on end. If you balance it very carefully, it won't deflect at all.

temporalillusion
24th September 2011, 09:26 PM
Stand it vertically on end. If you balance it very carefully, it won't deflect at all.

Win

Roger Ramjets
25th September 2011, 12:47 AM
there would be some effect from the two "half beams" actually being one long beam.... As the strength of the material increases, the mutual support coming from the resistance to bending at the "joint" would increase, allowing the supports to be moved toward the outer ends.I don't think material strength will affect the required support positions, but you are correct that they should be a bit less than 25% from the ends. I tried an experiment with a 10" strip of paper supported on balsa triangles, and obtained minimal deflection at ~22%.

Here's an online calculator that may help...

Structural Beam Bending Equations / Calculation Supported on Both Ends With Overhanging Supports of equal Length and Uniform Loading (http://www.engineersedge.com/beam_bending/beam_bending5.htm)

the PC apeman
25th September 2011, 07:03 AM
I don't think material strength will affect the required support positions

Thanks but the deflection formulas from your link factor in these things...

Where:
E = Modulus of Elasticity
I = Moment of Inertia

My term "strength" isn't quite right for elasticity but "cross section" is pretty good for moment of inertia. I claim the "even a blind pig..." award on this one. Yay.

rwguinn
25th September 2011, 09:10 AM
Thanks but the deflection formulas from your link factor in these things...



My term "strength" isn't quite right for elasticity but "cross section" is pretty good for moment of inertia. I claim the "even a blind pig..." award on this one. Yay.
Only because it is irrelevant.
The assumption is that the beam is homogenous and of constant cross section. W (load) E, I, and L are the same at any section of the beam (They cancel out), and thus the equation is merely a function of l and c (from the link nomenclature)

the PC apeman
25th September 2011, 11:16 AM
Only because it is irrelevant.
The assumption is that the beam is homogenous and of constant cross section. W (load) E, I, and L are the same at any section of the beam (They cancel out), and thus the equation is merely a function of l and c (from the link nomenclature)
I guess I don't understand then. Aren't the deflections just at the midpoint and at the endpoints the important values? Using these two formulas to calculate just those values, where does the cancelling out occur?
Deflection at the ends
http://www.engineersedge.com/beam_bending/images/beam_b59.gif

Deflection at center
http://www.engineersedge.com/beam_bending/images/beam_b59_1.gif

the PC apeman
25th September 2011, 11:30 AM
Wait, I think I get it now...

If we look for the support placement value (c) where the midpoint and endpoint deflections are the same we're solving...

http://www.engineersedge.com/beam_bending/images/beam_b59.gif = http://www.engineersedge.com/beam_bending/images/beam_b59_1.gif

rwguinn
25th September 2011, 11:36 AM
I guess I don't understand then. Aren't the deflections just at the midpoint and at the endpoints the important values? Using these two formulas to calculate just those values, where does the cancelling out occur?
Sorry--experience shortcut was taken, and I didn't make it clear
For a homogenous, constant section beam, Minimum deflection is going to occur when the 2 ends and the center are at the same deflection (Y value).

Thus, (from the link)
http://www.engineersedge.com/beam_bending/images/beam_b59_1.gif=http://www.engineersedge.com/beam_bending/images/beam_b59.gif



and W, E, and I cancell out

sol invictus
25th September 2011, 11:55 AM
This sounds to me like a job for the calculus of variations

If you do it that way I'd be curious to see the derivation. For a hanging rope or chain you can ignore the energy cost in bends, and that makes it pretty easy (you get a catenary) - but clearly that's not true for a steel beam.

the PC apeman
25th September 2011, 12:01 PM
Yep, rwguinn. One very small adjustment to this approach just hit me in the shower though...

The deflection formulas give a measurement down to the top of the mid- and endpoints. The more relevant measurement is from the bottom of the beam to the ground. For the midpoint this calculation is trivial: Height of the supports minus the deflection. However the endpoints are a different matter. The end faces of the beam will be bent over facing downward, causing the distance from the bottom of the end faces to be slightly higher off the ground than the bottom of the midpoint cross-section. Perhaps trivial but fun to think about.

rwguinn
25th September 2011, 12:06 PM
Yep, rwguinn. One very small adjustment to this approach just hit me in the shower though...

The deflection formulas give a measurement down to the top of the mid- and endpoints. The more relevant measurement is from the bottom of the beam to the ground. For the midpoint this calculation is trivial: Height of the supports minus the deflection. However the endpoints are a different matter. The end faces of the beam will be bent over facing downward, causing the distance from the bottom of the end faces to be slightly higher off the ground than the bottom of the midpoint cross-section. Perhaps trivial but fun to think about.
The calculations are for the Neutral Axis of the beam, so yeah, assuming a square cut on the ends, there will be a small difference, simply due to geometry...

Dave Rogers
26th September 2011, 02:29 AM
I'd say you're better off putting it on the ground, minimizing the deflection to zero the amount required to conform to a circular arc of 6371km radius.

FTFY.

Dave

Evilgiraffe
26th September 2011, 05:52 AM
i'd say you're better off putting it on the ground, minimizing the deflection to zero the amount required to conform to a circular arc of 6371km radius whatever is required to conform to the local topography.

Respectfully,
myriad

ftfy.

Dave

ftftfy.

:D

quadraginta
26th September 2011, 06:36 AM
Howdy all,

I was asked by a friend the following question:

Say I shipped to you a very long steel beam and when off-loaded you only had two supports to keep it out of the dirt. Where would you place the supports to minimize the deflection of the beam?


This sounds to me like a job for the calculus of variations, but it also seems to me that this would probably be a pretty standard engineering question. So in the interest of being lazy, before I start working on the calculations myself, does anyone have a link to a reference on this? Thanks in advance!

Cheers - MM


I can't offer any equation-backed engineering solutions, but I can offer the rather mundane perspective of having off-loaded hundreds of truckloads of steel, all types, in the field.

The applied physics answer is, "It depends." and, "Not enough data.".


All steel beams are not created equal. Assuming a typical "I" beam, the cross section (flange width and chord height) will make a big difference. Box beam (tubular) cross sections are affected similarly. Also, how long is "long?". All of these will be relevant to the deflection of the piece.

Another factor is the purpose for which the member has been fabricated. Although the term 'steel beam' is used fairly casually in common parlance it can have a very specific meaning structurally. As opposed to column, for example. 'Beams' are generally fabricated with a camber (an arc) calculated to be relieved (deflected) by both the dead-load weight of the member itself as well as any further load for which it may be designed (e.g.; concrete slab).

In practice this means that often the best places to crib up a 'beam' are closer to the ends, where the beam will actually be supported when it is in place. Too far in toward the middle and the unloaded camber of the beam can have the ends touching the ground even with no deflection at all. (ETA: this would be a very long beam with a lot of camber. :p)

So most of the time you don't want to minimize the deflection. You want to take advantage of it.

If the piece in question is a 'column' this won't be the case. Of course, columns are often shorter and generally much more rigid.

None of this addresses the dimensions of the dunnage being used to crib up the piece. 4x4 is standard. Sometimes bigger. Sometimes chunks of old railroad ties. If it isn't thick enough for whatever piece you are trying to keep off the ground it doesn't matter where you put it.

So. In real life. "It depends." "Not enough data."

And "Oh, hell. That ain't gonna work."

It's "more better" to have plenty of dunnage handy and not get into such a fix. :)

Or just let it hit the ground. The important thing is to be able to get your chokers (rigging) under the piece where you need to when you want to hook it back up again and install it in the structure. Until then it doesn't make any real difference if part of it is touching the ground. It looks prettier, but that's about all.

Crossbow
26th September 2011, 07:33 AM
Howdy all,

I was asked by a friend the following question:



This sounds to me like a job for the calculus of variations, but it also seems to me that this would probably be a pretty standard engineering question. So in the interest of being lazy, before I start working on the calculations myself, does anyone have a link to a reference on this? Thanks in advance!

Cheers - MM

It has been a while since I studied this, so if anyone has a correction, then please post it ...

However, for beams that are uniformly loaded (such as the one in this case), I think that in order to minimize the amount of deflection, then the two supports should be located at a distance which is one-fourth of the total length from the ends.

Because placing the supports at this location means that the sum of the moments about both of the supports will be zero.

Wangler
26th September 2011, 12:47 PM
It has been a while since I studied this, so if anyone has a correction, then please post it ...

However, for beams that are uniformly loaded (such as the one in this case), I think that in order to minimize the amount of deflection, then the two supports should be located at a distance which is one-fourth of the total length from the ends.

Because placing the supports at this location means that the sum of the moments about both of the supports will be zero.

You are correct. This website (http://www.engineersedge.com/beam_bending/beam_bending5.htm) has just the information needed to solve this problem.

If L is the total length of the beam, and c is the distance of each support from the beam end, deflection is minimum for c = 0.25L.

I think we used COV to solve problems like this in my graduate finite elements class, but don't ask me how after all this time.

ben m
26th September 2011, 01:20 PM
It has been a while since I studied this, so if anyone has a correction, then please post it ...

However, for beams that are uniformly loaded (such as the one in this case), I think that in order to minimize the amount of deflection, then the two supports should be located at a distance which is one-fourth of the total length from the ends.

Because placing the supports at this location means that the sum of the moments about both of the supports will be zero.

I disagree. Let's start with your arrangement and show that it's not optimal.

Under your arrangement, the bending moments would be equal, and this would result in equal sag to either side of either support, even if the beam were sawn through at the center.

The real beam is not sawn through, and has additional strength at the center. Compared to the sawn-through beam, the strong, non-sawn beam can take more load. Therefore, you can safely move the supports *outwards* from 25% and 75%.

I'm pretty sure this is what rgwinn and PC Apeman are doing above mut for some reason I can't see their equations.

sol invictus
26th September 2011, 03:15 PM
You are correct. This website (http://www.engineersedge.com/beam_bending/beam_bending5.htm) has just the information needed to solve this problem.

If L is the total length of the beam, and c is the distance of each support from the beam end, deflection is minimum for c = 0.25L.

I think we used COV to solve problems like this in my graduate finite elements class, but don't ask me how after all this time.

That's not what that website says. If you set "deflection at the center" equal to "deflection at the end", you get a cubic equation. The solution is 21.9%, not 25%.

Wangler
26th September 2011, 03:22 PM
If you set "deflection at the center" equal to "deflection at the end", you get a cubic equation. The solution is 21.9%, not 25%.

Why are you setting the "deflection at the center" equal to the "deflection at the end"?

sol invictus
26th September 2011, 03:32 PM
Why are you setting the "deflection at the center" equal to the "deflection at the end"?

Because the task was to minimize the deflection of the beam. If the deflection at the center is either greater than or less than the deflection at the ends, you can reduce one by increasing the other. The minimum deflection is when the two are equal.

Dinwar
26th September 2011, 04:09 PM
y I shipped to you a very long steel beam and when off-loaded you only had two supports to keep it out of the dirt. Where would you place the supports to minimize the deflection of the beam?
I'd see what else was lying around and get my forklift operator to move it to support the beam (there's always a few pallets lying around the jobsite). Or I'd have them go to Home Depot and pick up some 4x4s and cut them into supports. No foreman is going to do long math when yelling at someone and using what's available will work just as well. ;)

MattusMaximus
26th September 2011, 04:35 PM
If you do it that way I'd be curious to see the derivation. For a hanging rope or chain you can ignore the energy cost in bends, and that makes it pretty easy (you get a catenary) - but clearly that's not true for a steel beam.

Yup, that's exactly what I was thinking as well. I'm glad for the online calculator, and I'll probably use that for my friend's question, but now that I'm thinking about calculus of variations I think I'll try that as well and see how the two compare :)

ben m
26th September 2011, 04:54 PM
There's one subtlety which is not captured in the cited equations. The online calculator is assuming that the beam is horizontal at the support point. This would be the correct treatment if you were asking "I'm going to clamp this beam to two flat supports, where should I put them?" But you're actually going to rest the beam on the supports, atop which it's free to tilt, and this is not reflected in the equations. If my mental picture is correct, the minimum-deflect solution will be such that the highest point on the beam is not at the support itself, but actually slightly endwards of the support, and the beam is aimed upwards as it crosses the pivot.

Wangler
26th September 2011, 04:58 PM
Because the task was to minimize the deflection of the beam. If the deflection at the center is either greater than or less than the deflection at the ends, you can reduce one by increasing the other. The minimum deflection is when the two are equal.

Conceeded! Of course Sol was right on this, please see deflection graph based upon the equations from that website.

Equal deflections at edge and center are the minimum, with supports in from the edge at approximately 22.5% of the total beam length.

http://forums.randi.org/imagehosting/thum_235364e81119e4ced9.jpg (http://forums.randi.org/vbimghost.php?do=displayimg&imgid=24288)

Wangler
26th September 2011, 05:01 PM
There's one subtlety which is not captured in the cited equations. The online calculator is assuming that the beam is horizontal at the support point. This would be the correct treatment if you were asking "I'm going to clamp this beam to two flat supports, where should I put them?" But you're actually going to rest the beam on the supports, atop which it's free to tilt, and this is not reflected in the equations. If my mental picture is correct, the minimum-deflect solution will be such that the highest point on the beam is not at the support itself, but actually slightly endwards of the support, and the beam is aimed upwards as it crosses the pivot.

I don't think that the online calculator assumes a clamp at the support points. It only assumes that the beam is fixed in the vertical, or "Y" direction at the support points. The beam is free to move around the "Z" axis (this axis is out of the paper) at the support point.

The beam is simply supported at these points, not clamped.

rwguinn
26th September 2011, 05:28 PM
I'd see what else was lying around and get my forklift operator to move it to support the beam (there's always a few pallets lying around the jobsite). Or I'd have them go to Home Depot and pick up some 4x4s and cut them into supports. No foreman is going to do long math when yelling at someone and using what's available will work just as well. ;)
And there is the reality of the situation!
The thought experiment of where to place the supports is a wonderful exercise, and get the student familiar with the way it works, but nobody is actually going to take deliveries that way. they're going to throw a couple of 4x4's on the ground and dump the load on that.
It is important, however, if the beam is going to support a distributed load, and you need to minimize deflection, as in a floor joist, or a pipe carrying a fluid.
carry on...

ben m
26th September 2011, 05:33 PM
I don't think that the online calculator assumes a clamp at the support points. It only assumes that the beam is fixed in the vertical, or "Y" direction at the support points. The beam is free to move around the "Z" axis (this axis is out of the paper) at the support point.

The beam is simply supported at these points, not clamped.

You're right, my mistake. I was looking at a different calculator than the one you linked to.

Crossbow
27th September 2011, 06:10 AM
That's not what that website says. If you set "deflection at the center" equal to "deflection at the end", you get a cubic equation. The solution is 21.9%, not 25%.

Yes, you are correct!

Thanks much, it has been a while since I have studied this stuff.