Force to Straighten Curved Beam 4

[ryldbl]

I have a 25' round steel rod with a diameter of 1.125". This rod is not straight, but rather has a radius of curvature of "p". Does anyone know how to calculate the axial force necessary to straighten the rod? I don't know a formula to determine lateral deflection in a curved beam due to axial tension.

When I say "straighten" I mean elastically deform the rod into a straight beam, not permanently yield the rod to a straight condition.

 

[Allen]

Correct me if I'm wrong, but you sould thread this problem as a spring not a beam!

Somethig simmilar I can see on "Spotts Design of machine elements" 4-th edition on page 197 as "Belleville Spring"
may be you can use that theory, check here with other proffesionals.

 

[jamesjm]

What is the application for that you are describing?

 

[EnglishMuffin]

I believe it requires an infinite amount of axial force to make such a beam absolutely straight, so you would first have to define some tolerance within which you considered it to be "straight".

 

[JStephen]

Ditto to English's answer. As long as you don't plastically deform it, it will just come closer and closer to straight as force is increased, but never perfectly straight.

 

[prex]

I'll propose the following argument.
Assume your beam of length L may be considered as bent to a constant radius of curvature R (as you suggest): if that is not exactly true the following will give anyway a good approximation.
If the deflection at center with respect to a straight line joining the two ends is f then R=L²/8f and the energy that must be spent to go elastically from the bent shape to the straight one is (the same that would be required to bend it from straight to the deformed shape) W=EJL/2R²=32EJf²/L³.
In going from the deformed shape to the straight one the distance between the end points raises by [Δ]L=L³/24R²=8f²/3L.
Now if you call F the average value of the force by which you pull axially the beam to straighten it one has
F=W/[Δ]L=12EJ/L²
Hoping that all the above contains no mistake, if you take the double of that force I guess you'll be quite close to a straight shape.

prex

http://www.xcalcs.com

Online tools for structural design

 

[desertfox]

Hi ryldbl

I would use a strain energy method or Castigliano's theorem
which can be found in most mechanics books.
If you assume one end is fixed and you apply your load at the other end parallel to that end face you should be able to work out the force required.
It would help if you could describe what angle the radius 'p' goes through and also the actual value of 'p'


regards desertfox

 

[rb1957]

i don't know if this approach turns out the same as prex's but ...

i did like prex's observation that this is a two force member, so the bar looks like a shallow arch being pulled in tension. it's also symetrical so you can consider just 1/2, cutting at the crest of the arch. this end has a moment (due to the off-set). two questions come to mind ...

the off-set is reduced as the arch flattens, reducing the moment which is the only thing trying to straighten the arch. i suspect that the arch will want to form a plastic hinge, and will adopt a double arch shape as it is stretched.

also, this is a large displacement problem because as the displacements (that flatten the arch) have a significant impact on the internal loads.

possibly you could analyze the starting geometry, a cantilevered semi-arch. radial sections will be sufficient to determine the internal loads along the arch. apply the load in increments, determine the deformed shape, at some stage you'll need to reform the internal loads (depending on how you've formed the expressions (maybe you can iterate and converge on a solution for eah load step).

alternatively use some non-linear FE code (MARC).

whatever happens, i doubt that you can elastically straighten an arch (depending on your definition of straight!)

good luck

 

[prex]

Going one step further in my reasoning, that is of course based onto energy conservation principles, one can state that the axial force that will be required to go from a deflection f to a deflection f-df is:
F=(dW/df)/(d[Δ]L/df)
Now dW/df=64EJfdf/L³
and d[Δ]L/df=16fdf/3L
Now making the ratio as above the term fdf cancels out and one finds that the pulling force F=12EJ/L² is constant throughout the straightening process. Hence, contrary to what almost everyone above was expecting, the required load does not tend to infinity, it is on the contrary constant!
So I must come back on my conclusion in the preceding post: you don't need to double the value given by the equation above, just take it as F=12EJ/L².

prex

http://www.xcalcs.com

Online tools for structural design

 

[rb1957]

sorry, don't buy that, mostly because the curvature of the arc (ie the offset in the ends of the semi-arch) doesn't enter into it ... and it has to (no?), are we dealing with a semi-circle or a shallow arc ?

but also the problem is non-linear (each increment in load will not have the same transverse displacement) and the displacement is large (particularly in its effect on the internal loads). i believe that your analysis includes these assumptions, which are violated.

i did a quick linear run (a cantilevered semi-arch). i had 1" offset between the ends, radius of beam = 50", a steel bar 1/2" radius, applied 1000 lbs and got 0.03" transverse displacement with a max stress of 12ksi; so the linear (wrong) answer is 30,000 lbs with a stress of 360 ksi (a bit above yield).

looking at your equation F = 12EI/L^2, I = A^2/(4pi),
so axial stress = F/A = 12EA/(4piL^2), about EA/L^2
then stress/E = A/L^2, or E/stress*A = L^2,
E=30Mpsi, stress=100ksi, A=1in2, then L=sqrt(30000)=173"
but you previous comments still apply (i don't think this is right)

 

[GregLocock]

Hang on, is anyone claiming that there is a linear elastic solution leading to a perfectly straight bar? You can disprove that in three lines or less using superposition.





Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[JStephen]

Draw a straight line between the endpoints.

At any point in the rod, the moment is equal to tension at the ends times the deflection away from that line.

The curvature of the rod due to applied forces is proportional to the moment at each point.

In order to get the rod straight, the curvature has to be zero, which requires some finite moment to offset the inital curvature. But if rod is straight, there IS no moment. Or alternatively, as the centerline rod approaches a straight line, the required end force increases without limit. And note that the moments at the ends of the rod are always zero, so the ends of the rod are never straight.

One problem I see with the energy method is that it assumes the rod will take a certain shape, only the assumed shape is not one that can be formed by applying end tension. If the problem was restated as "What shape will be formed by applying end tension?", I think this would be more evident. You could assume that rod deflected into a pretzel shape and calculate the tension required to do that using energy methods, but I don't think you'd have a meaningful result.

 

[wes616]

Energy methods.... shouldn't the beam be modeles as a continuoum.

Wes C.

 

[prex]

As ryldbl provided all the required data, I can calculate the required load with my formula: F=1390 Newtons (sorry for not using american units). If ryldbl is able to make the measurement (hope this is not a school assignement), he could enlighten us on the result. Note that this same force, as I stated above, will be required to start straightening and to arrive to the straight beam.
For those that question this result (that surprises me the first) without providing a different route, I'll try to detail here the assumptions on which it is based.
The first assumption is that no account is made for the axial strain of the beam: this one is quite correct till the beam approaches the straight shape, after that any increase in the axial load will of course go into an axial strain.
The second assumption is that we may use the common formula for the energy of deformation in bending: W=([∫]EJy''²dx)/2 the integral being extended over beam length and y being the deflection. I think no one can question this formula, that is known to work well for quite high deflections (though we are not dealing with a semi circle of course) and is valid for any deformed shape.
The third assumption is that the beam bent shape may be approximated with an arc of a circle: this one will be fairly good for most beam deflections that resemble an arc of a circle (not an S shape for instance).
The fourth assumption is that the arc is quite shallow. In this case we have the following approximations (the symbols are as in my first post above):
y''=1/R
R=L²/8f : this one comes from simple geometry and includes the approximations f<<L<<R that are of course acceptable for a shallow arc
[&Delta;]L=L³/24R² : this one is obtained by making the difference [&alpha;]-sin[&alpha;], 2[&alpha;] being the angle subtended by beam length from the center of curvature, and using the first two terms of the Taylor expansion for sin[&alpha;] (which is again good for a shallow arc).
From all the above, based on the relationship dW=Fd[&Delta;]L (in this differential form it is of course one of the least questionable principles of the classical physics), one obtains the formula F=12EJ/L².
In all this:
- to rb1957: you are correct that this is a problem in the so called large deformation field (that is where the stress depends on the strain), but nevertheless all the above stays correct; the proof? The large deformation theory is not linear and indeed we have (surprisingly) a constant load for a changing deformation!
- to JStephen: I too was expecting a force tending to infinite when the shape approaches the straightness, but, as you certainly know, one has to be careful when dividing two quantities that both approach zero. It is true that the lever arm of the axial force tends to zero, but the same does the resistance of the beam to bending (or in other words the energy required to deform it), and the surprising result is that the force stays constant over the whole deformation (assuming as stated above a shallow arc)


prex

http://www.xcalcs.com

Online tools for structural design

 

[rb1957]

prex,
"As ryldbl provided all the required data, I can calculate the required load with my formula: F=1390 Newtons (sorry for not using american units)."

um, i didn't see a radius posted (only a length (25') at some radius, p).

um, this is a solid steel (which flavour; never mind, doesn't matter much) bar 1.125" diameter ... and you're going to straighten it with a force of 1400N (= 300 lbs) ...
i don't think so ... 300psi ??? you'd get this load just by hanging the bar from one end ! (25' = 300"*1in2 = 300in3*0.3 = 90 lbf ... ok, you get only 1/3 the force !!

again i don't think you're equations account for large displacement in a non-linear problem, and i think a quick sanity check on your result bears this out.

 

[prex]

Well rb1957 I just can't see why that load should be so inadequate: under its own weight (1/4 of that load) that same bar, when suspended horizontally at the ends, would deflect by about half a meter (being close but not beyond yield by the way).
Also what my formula says is that the load doesn't depend on the radius of curvature (provided this one is large with respect to length).
As of course I can't personally be 100% sure of a result obtained on the fly (no literature references that I know of), why don't you try to discuss the merit of my arguments or to obtain an independent result?

prex

http://www.xcalcs.com

Online tools for structural design

 

[rb1957]

prex,

i've run a couple of quick models. curved beams, about 30" long, 1" diameter, steel, loaded with 1,000lbs (about 3* your load).

one model had about 1" rise, and the load produced about 0.05" lateral displacement.

the 2nd model had 0.05" rise, and produced 0.0015" lateral displacement.

first, the lateral displacement is non-linear, the same force produced different effects.

2nd, the result is proportionally constant (which i thought was quite intereting). what i mean is that in both cases the load produced a lateral displacement = 3% of the rise.

maybe i should try my models with 300" length; but basically, i think that trying to straighten something by teniosn isn't particularly effective.

i did think your fR = L^2/8 is a neat observation, and a way to avoid f and R, so long as you're always using (fR) and not f or R. but think about the effect of 300psi (=2MPa, =1400N/625mm2) on a steel bar.

 

[EnglishMuffin]

For a hanging cable, for which the formulae are readily available and discussed on other threads, there is no question that the cable cannot be made absolutely straight regardless of the tension. Although this case is slightly different, it is highly unlikely that you could straighten a beam under similar conditions, since the transverse resisting force increases as the beam becomes straighter, whereas the cable weight is constant. Although I could be wrong, without wading through the whole of Prex's derivation, it seems that the most likely error is that he has confused the y deflection in the energy integral with the y in the equation y" = 1/R, since in that equation, y and the y deflection are the same thing only for an initially straight beam.

 

[rb1957]

i liked prex's approach that the work done to straighten the bar is the same as the work done to bend it.

so the external work done is applied force P, displaced by L-2Rsinx; note we can't say sinx =x (even tho' x is small) because L=2Rx.

if x = 0.01c (L=300", R=15000", f=0.75")
then delta = 0.005" (well, 0.004999975")
so the external work is 0.005P

if x = 0.1c (L=300", R=1500", f=7.49375")
delta = 0.49975006"

then the external work is about 50*x^2*P

and the internal work is ...

 

[prex]

I'm assuming we are not speaking about a hanging cable or beam of course: that is a completely different problem. What I'm assuming is the bent bar is fully supported by a flat horizontal table, so that beam weight has no effect: I see now that this was unspecified in the original formulation, but assume this was the intention of ryldbl, as the focus was on recovering straightness against a previously bent shape (of course in a cable this would have no meaning, as there is no bending).
And EnglishMuffin I see no confusion: y''=1/R is of course the second derivative with respect to x of the deflection y.
rb1957 I agree of course that the relationship between loads and deflection is not linear in a beam beam bent by transverse loads and with an axial one, if that is what you mean: this is a well known result (a cubic equation is found for the deflection if I recall correctly) and you can see calculation sheets for this condition in the site below.
Coming to your models, I can easily confirm your results with my formulae (though .0015" is perhaps .0025"?): for a 30" long beam your load of 1000 lbs is about 1/20th of the straightening load I calculate, and in fact your calculated lateral displacement is about 1/20th (or close to that) of the initial rise. You should apply a load 20 times higher (or use a length some 5 times higher) and you should see in both models that the residual deflection goes close to zero.
I can check your results also in a more detailed way, using only the first model for brevity.
In the initial configuration the radius of curvature is (excuse me if I will use the SI) R₁=2857.5 mm and after applying the load R₂=3007.9 mm.
This requires a work of deformation given by W=EJL(R₂-R₁)²/R₁⁴/2=544 Joules
The displacement of the end under the load (you could certainly confirm that in your calculation) is [&Delta;]L=L³(1/R₁²-1/R₂²)/24=0.22 mm
and the force to do that is F=W/[&Delta;]L=2473 N = 556 lbs
This is not exactly what you find, but it must be recognized that the radius of curvature in your first model is not really big, so there is quite a big error in differentiating 1/R². Suppose that the second model would give a better result, can you check and confirm the data?
And thanks to you rb1957 I think I discovered an error in my second post above (not in the first one however), but I'll leave to find it as an exercise...

prex

http://www.xcalcs.com

Online tools for structural design