SPRINGBACK [Archive] - The WoodenBoat Forum

Ross M

10-04-2005, 09:16 AM

This is listed in the WBF B&R Faq but does not exist in the database:

sailordave
posted 01-02-2001 04:28 PM
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Okay, I'm laminating frames and beams on a 1955 Lightning that I'm restoring. Some time back I saw a Gougeon Bros. formula for calculating springback that is an okay approximation but not the greatest. Springback in inches= x/n^2 where "x" is the height of the bend (dist from the middle of the chord of the endpts to the bottom lam.) and "n" is the number of laminate layers, obviously squared. In other words take a yardstick and place it on a 4" block of wood at the 18" mark. Now bend the ends down to touch the table. x= 4". But if you use a 6' long pc. of wood and do the same thing x still equals 4" but the tension in the beam isn't going to be the same. As you can see by the square function in hte denominator, the more laminates the faster the beam gets stiff. The problem w/ this equation is it doesn't take the LENGTH of the beam in to account nor the modulus of Elasticity "E" of the material. I'm sure there is a big difference in bending mahog. v. oak v. teak v. pine etc.
So does anyone know of an equation that factors this in? Thanks, David

JimConlin
posted 01-02-2001 05:40 PM
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I believe that the formula is correct and doesn't depend on the length of the piece or the modulus of the material.
Here's how to visualise the problem. The individual laminations of the part are pushing the part to be straight. The assemblage of laminations is pushing the part to be bent. The laminated part will be in equilibrium when the springback is such that the total force toward bending the several laminations balances the force needed to straighten the laminated part.
The algebra representing this is something for a smarter day than I'm having, but it's based on the stress in a bent thing being proportional to the deflection and the cube of the thickness. So, think of N planks of thickness T opposing one plank of thickness N*T and solving for the deflection.
It turns out that the formula:
springback = (formed bend) / (n^2)
is about right.

Notice that for N=1, springback is complete and for n=large, it approaches zero.

sailordave
posted 01-03-2001 11:16 AM
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I hear what you're saying but it doesn't compute nor is it what one experiences in the real world. Use my yardstick example; the more you bend the yardstick the stiffer it gets- it fights you until it fails. You might be able to bend it over a 20" high block by pushing down on the ends. The amount of force opposing that effort is certainly going to be more than if you took a 10' long pc of the same material and dimension and pushed it down over the same 20" high block. (The percentage displacement of the whole pc. is different)

To illustrate further take a pc of wood and bend it in your hands. Now cut the pc in half and try to displace the shorter pc the same amount. NOTE this is not the same arc curve! Not trying to argue but I think the formula fails to account for the different ARC

htom
posted 01-03-2001 11:51 AM
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You're comparing different things, or the same things differently.

Thought experiment: we make up different "beams". One is 1" spruce, one two layers of 1/2" spruce, one three layers of 1/3" spruce, one four layers of 1/4" spruce ...

All of these are bent to the same curve, a distance D, and the glue is allowed to set.
The beams are all removed from the mold.

Now we measure the springback (that is, the distance the end of the beam moved from the mold.)
We look for a relationship between the beams, and find it (the approximation formula you've quoted.)

The formula says that the more layers there are in the lamination, the less the springback will be, and provides an estimate of what that springback will be.

The formula does not explain the results. The results explain the formula.

Emerson
posted 01-03-2001 01:37 PM
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Dave,
Seems to me sizing laminations is determining the number of layers and how thick each one is based on the the radius of curvature and the ultimate size of your frames or beams. There is a formula which the woodenboat series (re?)published in their Frame, Stem, and Keel Repair book which would help you (I don't have it here, and will likely forget to find it tonight); this formula will enable you to determine the appropriate the thickness of each layer so your frames will have adequate strength when in place (it takes into account the radius of curvature - arc length based - and thickness of laminated layer). I think everyone here is trying to tell you to size the laminated frames first based upon the number of and thickness of layers. Your template if you insist, could account for springback. The formula you quote must make a basic assumption regarding the arc length (ie. "x" is the height of the bend over a 1' chord) and material's ME.
One last thought - if the damn thing springs back after its dried - so what. Whenit goes in the boat, you fasten it in place and remove the springback. If its properly sized, it will work.

I like to obsess over minute details, this just doesn't seem so complex.

[This message has been edited by Emerson (edited 01-03-2001).]

casem
posted 01-03-2001 08:33 PM
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I would guess that the formula assumes "small deflections", so that the deflection y is related to the radius of curvature r by d^2y/dx^2=1/r (and x is the coordinate along the beam). So you don't really have to account for the fact that the radius of curvature changes along the length of the beam. Releasing that assumption is the only way I can see why the length would come into play, but that makes the problem non-linear and probably very difficult/impossible to solve. So the formula might not work when you laminate things with alot of curvature. I know that when I laminated my tiller (two laminations and a very subtle curve) the equation predicted the spring back very well. But when I laminated my stems (5 laminations and a whole lot of curvature) it didn't.
As for E, I wouldn't expect that to be in the final equations because both the forces that cause the springback and the forces which oppose it are both proportional to E, so it should cancel out.

sailordave
posted 01-04-2001 12:10 AM
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I hear what most are saying and some folks have heard what I am saying. I realize more thinner lams will reduce springback. And experience has given me the ability to build in a little extra curve (LAR Looks About Right) in the curve. I just want to find something that will refine the basic y=x/n^2 equation to account for the fact that x can vary enough to skew the results. Let's face it if I try to laminate 4 layers of 1/4" half way around around a 55gal drum thats a lot different than laminating 4 layers of 1/4" that are 20 feet long and using the same drum as the midpoint and anchoring the ends. In the second example I've got twice as much "x" value but I can guarantee you I won't get as much springback. An extreme example but that's what I think people are overlooking. "x" is a variable dependent on an outside, unconsidered value, not a constant. Don't mean to sound arguementative but when I exlain this to my non boatbuilding friends that know higher math than I, they immediately saw the point I was driving at. Any engineers lurking here? Regardless, I love the feedback this question has provoked. Thanks all. D

sailordave
posted 01-04-2001 12:15 AM
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Emerson: You're the closest I think. HTOM is right when you compare the same curve using more thinner lams. That is a given; As you said you need to compare the rise over the chord to a certain length. Apples and Oranges if the chord length is different.
Thanks, David

casem
posted 01-04-2001 11:44 AM
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Sorry Dave, I guess I got the impression that you were an engineer. Anyway, the point I was trying to make is this: That equation only works if your curve is pretty small relative to the original straight beam. For example, it would work fairly well if you laminated a 20' beam over a 55 gallon drum at the midpoint (ends fixed to the ground). That's because the final beam is still kind-of straight, and it is clear what the initial deflection is (the x would be equal to the diameter of the barrel).
But when you wrap the same laminations around the barrel, you no longer have "small deflections". In fact, you really can't easily define what x is because the thing curves so much over its entire length. That's what that equation I wrote up there is saying (kind of). If you have alot of curvature, you can't easily relate your deflection x to the curvature (and I think its the curvature, not the x, that really matters).

Think about this. If you wrap your 20' laminations around the barrel, you can't use that equation. This is primarily because you can't easily specify what x is. Now use 6" pieces instead of 20' pieces and just press them down on the barrel. The CURVATURE is the same, but now your x is a little more obvious, when referenced to the ends of the 6" pieces. Depending on the shape of the barrel, it might be about 1/4". Since 1/4" is small compared to 6", you can use that equation, even though you really haven't done anything different with the laminations. You could think of your 20' piece wrapped around the barrel as a series of 6" pieces and predict the springback.

From a practical stanpoint, what Emerson said is probably the best advice - use enough laminations and you'll be able to fasten it into the right shape when it goes back in the boat. That's what I did with my stems and it worked out okay (of course the boat hasn't seen any sea-trials yet).

marc
posted 01-04-2001 11:48 AM
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HTOM: Your example is correct but I don't think that is what Dave is asking. Yes for a given curve more thinner laminations will be stiffer. The part of the equation that he is trying to refine is the X part. That value X will be the same regardless of the RADIUS of the curve as long as the height above the chord is the same! His yardstick example is a perfect demo. Try this: take 3 wooden yardsticks, stack them together and bend them over a 1" pipe so that the ends touch the table and the pipe is in the middle of the sticks. Feel the amount of resistance in your hand. Now cut off 12" from the same 3 sticks, and repeat, bending over the 1" pipe and pushing the ends down to touch the table. Feel how much more resistance you encounter. (natural springback!) In BOTH cases the variable X=1 and N=3 but you know that the amount of springback is going to be more in the second example. That is what Dave is trying to account for. Marc

casem
posted 01-04-2001 12:01 PM
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Marc - I would argue that the springback would be the same in both cases, just that the residual stress in the shorter beam would be higher. In terms of the beam trying to relieve its curvature, the short beam will springback more, but when you relate that to a deflection x, it works out to be the same. I think.

Ross Faneuf
posted 01-04-2001 11:12 AM
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When I did the laminated frames for 'Ceol Mor', I braced each frame with pieces of 1x4 pine screwed to the face of the frame, which I removed once the hull was complete. 'Ceol Mor' is bigger than a Lightning, but you could probably do the same thing with lighter pieces - say 1x2 or 1x3. If you're doing individual frames a single pieces connecting the ends (like the string on a bow) would probably reduce springback to an amount you couldn't even see.

sailordave
posted 01-04-2001 11:16 AM
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Casem: Marc got it! That's exactly what I'm trying to say. And as you pointed out the example of the 55 gal drum is extreme; I was just trying to make the point that X is the same. And you last comment about the two different length yardsticks having the same springback is in error. I'm currently bending beams (to a radius of 20') and Frames (to a radius of 10'4") and when I laminate pieces of wood around the tighter radius there is indeed more springback compared to a longer piece on the bigger curve (X remains the same) Also, EMERSON said he gets rid of the sprinback by fastening it in the boat and that pulls it back into shape. That isn't a good way and it WON'T work when you are building from scratch or are replacing deck beams w/ no shape to force it against. Thus it becomes imperative that you get the shape correct right from the start. (or at least pretty damn close: I try to stay less than 1/8" sprinback at one end which when both ends are fastened means I'm out 1/16", well w/in 45 y.o. wooden boat tolerances! ) Thanks, David

Emerson
posted 01-04-2001 12:43 PM
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Feels like I'm taking a pop quiz and didn't know it.
Casem - what are you study at um? That old glue account sounds nostalgic - familiar.

Dave - have you considered that although the 4' and 20' strips may resist more over a 55 gallon drum that there should be a laminate glue between layers in shear. If you think about the resistance in the glue layer, you may consider that one bent more severely and then glued stiff may actually springback (undoing initial deflection) less than you think. Yes, it seems that a harder bend would spring more, but would it - case in point, a post-tensioned concrete beam.

Try experimenting a bit with scrap - if you're a wood boat lover you should have plenty - and see if it works out ok. For deck beams the springback equations should be adequate. Have you researched the equation yet? I hate when people use formulas not knowing what asusmptions are behind it. Find these out and you will know its limitations. Otherwise we're all shooting in the dark and hypothesizing our selves to a delusionally happier day.

Good luck.
-Matt

casem
posted 01-04-2001 02:03 PM
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I don't know Dave, I still think I'm right, at least within the realm of application of that equation (small deflection, linear elastic material, presence of transverse shear not affecting deflection, and probably some other stuff). I'd bet a six-pack on it. If that equation isn't working, it's probably due to some violation of those assumptions, which very well may be the case. The equation you want to use (and rightfully so) releases all those assumptions but there probably isn't anyone around smart enough to solve it. The value of the equation is to illustrate that if you're getting an unacceptable amount of springback, you should be able to reduce it by using more laminations (perhaps to within your 1/8" tolerance). I'm not saying you should use three laminations and force the ends in 3 inches (although thats just about what happened with my stems!).
Actually my stems are a pretty good example. I tried to use that equation to anticipate the amount of springback and build my jig so that the final result would be perfect. But I ended up with about an extra inch of springback. My tiller, on the other hand, was about 4 feet long with about 4" of x, and the equation was very accurate (about 1/8" off, I think). It's definately worth investigating though, since your Lightning is a much more serious boat than the one I'm building.

Emerson - I actually just graduated from UM with a degree in mech. engineering. I guess I should change my user profile to read "unemployed" instead of student.

[This message has been edited by casem (edited 01-04-2001).]

htom
posted 01-05-2001 12:45 PM
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Dave, you might be happier with the equation in this form:

S = springback

n = number of layers
Di = initial displacement

Df = final displacement

S = Di - Df

S = Di / n^2

(Di - Df) = Di / n^2

n^2 * (Di - Df) = Di

(n^2 * Di) - (n^2 * Df) = Di

(n^2 * Di) - Di = n^2 * Df

(n^2 - 1) * Di = (n^2) * Df

The energy put into bending the layers is still there after the glue sets; I've used what I thought was water-proof glue, left something out in the wet, and it came apart. Lots of energy there for a brief instant.

I suspect that the formula is increasingly inaccurate after a bend of 10-20 degrees; for some reason I think it's using the sin x ~= x approximation somewhere.

casem
posted 01-05-2001 01:21 PM
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It took me all morning to come up with this so hopefully it will help. Note it will only work if you're bending things into TRUE CIRCULAR ARCS (or have a very small deflection, in which case you're probably better off using the x/n^2 equation):
R=r*(n^2)/(n^2-1)

where

R = final radius (the curve you get after the glue dries), to the center of the finished beam

r = initial radius (the curve you make the mold to), to the center of the finished beam

There's still some assumptions in it but I bet it will get you closer. So if you mold 4 layers on an arc with a radius of curvature of 20', you end up with a 21.3' radius. Its a little more awkward because you have to think of things in terms of radii of curvature instead of deflection (e.g., the flatter the board, the larger its radius of curvature) but it avoids the sin(x)=x /small deflection assumption htom is talking about. You can relate it to the length of the beam and the midpoint deflection by geometry but it gets a little ugly. Let me know if you really want to try it and I'll type it in (I'm dying to see if it works).

casem
posted 01-05-2001 01:28 PM
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P.S. You're still probably better off with LAR!

sailordave
posted 01-05-2001 03:27 PM
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Casem: Your formula sounds about right (SAR!) I think that when the snow clears that we just got I shall get some pine lattice and make some test runs. I'm dying of curiousity now. I've got most of the replacement beams laminated and they came off the jig pretty close to the line.
One thing that threw me for 0.1 sec. In your form. you need to add a set of parenthese in the denominator.
R=r*(n^2)/((n^2)-1)

Thanks to all for the thought this has provoked. It's been an interesting, passionately opinionated, but civil discussion! David

sailordave
posted 04-10-2001 12:29 PM
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Well three months later and I've run the test strips for laminate springback. First I laminated 3 layers of lattice 31.5 inches long, bent over a 2.75" can and secured at the ends. Then I did the same w/ three strips 47" long. NOTE this yields diff. radius curves even though the deflection is the same at 2.75"
Results: the short pc. sprung 13/16"
the long pc. sprung 9/16"
My interpretation of the results would be that for large radius curves (small displacement) the equation def= X/Y^2
where X is the displacement and Y is the number of laminates. For tighter curves this is not quite accurate. Any thoughts?

casem
posted 04-12-2001 11:10 PM
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I think that makes sense because the springback formula assumes the length of the beam is "much" greater than the deflection. So for the 2.75" deflection, the equation should be more accurate for the longer laminate. I calculated 2.75/3^2 to be about 5/16" so it's still off by a little, even for the longer lattice. I wonder how long the pieces have to be before the equation becomes perfectly accurate (to within a few percent or so)?

sailordave
posted 04-13-2001 08:31 PM
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Casem: That's kinda the whole point of the original thread. The LENGTH of the laminates in relation to the displacement has to be considered. IE the radius of the curve needs to somehow be factored into the equation. For the deck beams on my Lightning which were about a 20 foot radius (240") the equation worked almost perfectly. The frames at 10'5" radius I tightened up a bit from 125" to about 115". Still I wish I could calculate it a bit more accurately instead of running test lams for each layout.

gashmore
posted 04-13-2001 09:03 PM
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You guys can argue theory all you want but I have been laminating trim every night for the last three weeks. Everything from long 24" radius sweeps with 4 plys to 3" radius corners with 12 plys. The angles had to line up to mate to the solid fiddles and door jams so I used the formula S = D/p^2 for all the jigs. I know it is a rule of thumb but the springback came out just as predicted.
Engineers tend to think a problem to death. I am guilty of it myself 'cause I was born and bred to be one. But sometimes you just have to trust the tables and rules of thumb.<g>

casem
posted 04-16-2001 10:01 AM
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That equation with the radius of curvature should eliminate the length variable (somewhat). The "springback=deflection/n^2" equation is a special case of that equation which is only valid when the length of the beam is much much greater than the deflection. I suppose the only problem with the radius of curvature equation is that you need be able to measure radius of curvature.
Gashmore, you ain't kidding. I can't help but go overboard on this stuff. That's why I'm limiting my posts to this thread to one every three days. Talk to you on the 19th.