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Bending of a long, thin elastic rod or wire (finding shape & height)

  1. Jan 27, 2014 #1
    Hi all,

    I'm trying to do something that might be impossible, but seems to me should have a solution: Finding the height (or displacement) of a long, thin rod or wire under horizontal compression forces at the ends (pinned, not clamped) causing it to bend into a stable arch-like shape given only the length of the rod, the distance between the pinned ends ("width" in the images, always < length. imagine a rope being tied between the endpoints holding the arch in a stable shape), and possibly the parameters unique to the rod's material & cross-section (Young’s modulus, E, and the area moment of inertia, I). Ideally, I'd like to find a way to describe parametrically the shape the rod makes, from which I could easily find the max height (displacement). I have yet to find a formula for this though. The closest I've come to finding a solution is these resources:

    http://www.scielo.org.mx/pdf/rmfe/v53n2/v53n2a8.pdf (Figure 1 on pg 189 is exactly what I'm trying to find a formula to describe)
    http://www.princeton.edu/mae/people/faculty/faccvroyce/homepage/publications/DomHRoy-JNLS97.pdf
    http://solidmechanics.org/Text/Chapter10_4/Chapter10_4.php
    http://levien.com/phd/elastica_hist.pdf
    http://thegeometryofbending.blogspot.com/

    I unfortunately don't understand Euler's work well enough to derive this on my own. I've seen references to the "Jacobi elliptic functions" and "complete elliptic integral of the first (or second) kind", but don't understand how to find solutions for those either.

    Also, if it's a matter of finding a solution numerically, that would work great as I'm trying to write an algorithm (in C or python) to solve this anyway. I would ideally take any 2 of the 3 parameters (length, height, and width between the endpoints) and output the unknown variable and a formula for the shape of the curve.

    Thanks for any help or insight you can provide!


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  2. jcsd
  3. Jan 29, 2014 #2

    Baluncore

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    Am I correct in thinking you want to take gravity into consideration, (hence the height), or are you considering a rod that is laying flat on the floor. Does it have a mass per unit length?
    Is the section always constant throughout the length?
     
  4. Jan 29, 2014 #3

    SteamKing

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    What you are trying to accomplish is I think related to trying to find the elastica of a beam. This is the shape that a beam takes when it is loaded beyond its elastic limit. It is an advanced topic in the strength of materials and most of the solutions involve complicated elliptic functions and such, so that numerical solutions are the only practical ones which can be found.

    http://en.wikipedia.org/wiki/Elastica_theory

    Regular beam theory is not applicable because the deflections and slopes of the beams are not small. Essentially, you are looking for the solutions to a system of differential equations.
     
  5. Jan 30, 2014 #4
    I did look into elastica, but again, I haven't been able to find a straightforward way to derive or numerically solve for height yet, let alone plot the actual shape of the curve. I did find a few more relevant resources though:

    http://myweb.lsbu.ac.uk/~gossga/thesisFinal.pdf
    http://www.levien.com/phd/thesis.pdf
    http://www.scribd.com/doc/50402462/Timoshenko-Theory-of-Elastic-Stability [Broken]

    I'm probably just overlooking a key formula, but as I haven't done much engineering in a while, I'm a little rusty.

    As for Baluncore's questions, no on gravity (it could be lying flat on a floor). It would have mass per unit length, as well as a specific area moment of inertia and Young's modulus (like for aluminum). And yes, the section would be constant throughout the length.
     
    Last edited by a moderator: May 6, 2017
  6. Jan 30, 2014 #5

    SteamKing

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    Just glancing over the thesis by Levien, it appears this problem has occupied minds having quite a bit of mathematical knowledge (particularly Carl deBoor). It seems that the analogy of the spline curve is the most useful in grappling with the problem of the elastica.

    There are various formulations of mathematical splines which are used to interpolate discrete data points (e.g., cubic splines, tension splines, B-splines, NURBS, etc.) all of which have their particular formulations and advantages for certain applications. Probably Section 5.15 of Levien's thesis points the way to further research on calculating the spline analog to an elastica. If you go to Levien's web site, he mentions an interactive software called Spiro which is used to design typefaces. It seems that some of the example typeface designs contain elements similar to the examples you showed in the earlier posts in this thread.

    In any event, even the basic cubic spline formulation uses several routines to calculate the spline parameters from the data points and to evaluate the resulting spline, so I don't think you will find a simple formula into which you plug a few parameters and get out an elastica.
     
  7. Jan 30, 2014 #6

    Baluncore

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    Splines tend to convert orientation using polynomial derivatives. The length of a polynomial curve can become very involved. Because the length of the rod is reasonably well controlled, IMO simple splines will be at a disadvantage.

    I suspect there is an orderly FEA solution that can be applied to an array of elements. In many situations there will be multiple solutions, so there will need to be some constraint, maybe in the form of an initial crude approximation such as a circular arc of the rod length passing through the defined rod end points.

    The positions, vector forces and torques applied at the two ends of the rod will be equal and opposite. Each element will be accelerated towards a position and orientation where it's end forces are balanced.
     
  8. Jan 30, 2014 #7

    SteamKing

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    Not all splines are based on polynomials. The tension splines, IIRC, wind up using hyperbolics in their formulation.

    Certainly, discretization and minimization of some constraint could be applied to solving the problem, as is discussed in some of the papers cited. But, I believe the OP was looking for a simple plug-and-chug formula for calculating the elastica given a few parameters, which, given the mathematical complexity of the problem, is not indicated by perusing the papers provided. By looking at some of the references in the Levien thesis, there may be an approach which lies somewhere between a spline calculation and a FEM calculation, which is why I suggested further study of the Levien thesis, Sec. 5.15.
     
  9. Jan 30, 2014 #8

    AlephZero

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    The basic principle is the same as beam theory, so long as the strains are small (even though the deflections are large) and the stress-strain relationship is linear (e.g. no plastiicity).

    As for a beam, the basic equation relates the curvature at any point to the bending moment. The "only" problem is that the math gets messy when you can't assume the displacements are small.

    There are some simple special cases, for example when the applied loads at the ends are pure bending (not shear), in which case the curvature is constant and the shape is just an arc of a circle. (That is a nice test for computer software that claims to solve this type of problem - see if a straight beam can be bent round into a complete circle.)
     
  10. Jan 30, 2014 #9

    Baluncore

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    I agree with all comments to date. Historically, the elastica is a problem that has a few elegant solutions in special cases but many real world problems appear to be intractable.

    My initial approach is from the practical FEM direction because I know I am not up to the mathematics that has challenged the very greatest minds. FEM appears to guarantee a solution to any particular class of elastica.

    The OP problem has pinned ends, the length of the rod and the distance between the pins is specified. That can be a scaled ratiometric solution, it requires only solution at a number of points along one axis, namely the ratio of rod length to pin separation.

    For one particular class of problem, such as that presented in the OP, a numerical solution would generate a curve from which a general relationship could be identified numerically. I suspect that the solution to a sub-class of problems will be reasonable and once identified numerically can be shown to have a simple theoretical basis. The advantage of FEM here is that it can handle other 2D and 3D cases as they arise in engineering. I see the numerical code as a good long term investment.

    This is a really elegant and interesting corner of engineering, I would like to explore some of the hidden valleys. If I can find an elegant solution without seeing the footprints of others, then it has to beat the Himalayas as a destination. If when I find a solution the initials of Euler or Bernoulli are on the peak equation, I will see it as a great holiday, if not a pilgrimage.
     
  11. Jan 30, 2014 #10
    Oh wow: http://thegeometryofbending.blogspot.se/2014/01/these-saw-blade-curves-seen-in-two.html

    I had e-mailed the guy, Mårten, about my inquiry, and he wrote me back with the above formula, [itex]h=\sqrt{(\frac{2L}{5})^2-(\frac{d-\frac{L}{5}}{2})^2}[/itex] with d being the distance between the two endpoints and L the length of the rod/wire/sawblade. Exactly what I was looking for, and so far it seems to hold up to the experimental data I have. Incredible. Now if only I could get the shape of the curve!
     
  12. Jan 30, 2014 #11

    Baluncore

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    “Oh wow” is a very appropriate expression for a find like that.
     
    Last edited by a moderator: May 6, 2017
  13. Jan 30, 2014 #12
    Seriously...I told him it was surely worthy of publishing somewhere, no?
     
  14. Jan 30, 2014 #13

    SteamKing

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    It is published. Look at the RHS of the webpage: "This blog is also a book." with details about price and purchase.
     
  15. Jan 30, 2014 #14
    I don't believe that formula was in his book...he only posted it to his blog today after I had posed the question to him (the book came out last June). He had figured out the circle relation earlier though (here), so that was likely in his book.
     
  16. Jan 31, 2014 #15

    Baluncore

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  17. Mar 6, 2014 #16
    Solved!

    After further research, I've discovered a more accurate answer to this problem, and I also put together some VB.NET code that will find the form of this curve (well, a bunch of {x, y} points along the curve). I found a new paper that was one of the keys to figuring this out: "An experiment in nonlinear beam theory" by A. Valiente. That one plus these were the most helpful:

    "The elastic rod" by M.E. Pacheco Q. & E. Pina (linked above)
    "Snap buckling, writhing and Loop formation In twisted rods" by V.G.A. GOSS (linked above)
    "Theory of Elastic Stability" by Stephen Timoshenko (start on p. 76, linked above)

    As stated in the above papers, the solution to my problem lies with elliptic integrals, specifically [itex]K(m)[/itex], the complete elliptic integral of the first kind, and [itex]E(m)[/itex], the complete elliptic integral of the second kind. There was a lot of confusion over the [itex]m[/itex] and [itex]k[/itex] parameters for these functions, as some people use them interchangeably, but they are not the same. [itex]m=k^{2}[/itex] (thus [itex]k=\sqrt{m}[/itex]). I try to use the [itex]m[/itex] parameter exclusively to avoid this confusion. There is a unique [itex]m[/itex] parameter for every configuration/shape of this curve, and to solve for any of the unknowns, we first must find [itex]m[/itex]. Here are the key formulas:

    [itex]\frac{width}{length}=2\frac{E(m)}{K(m)}-1[/itex] (note that [itex]width[/itex] is the distance between the two endpoints)

    [itex]\frac{height}{length}=\frac{\sqrt{m}}{K(m)}[/itex] (it's actually possible to have 2 values for [itex]m[/itex] here for certain height/length ratios)

    [itex]\frac{width}{height}=\frac{2 E(m) - K(m)}{\sqrt{m}}[/itex]


    The easiest way to solve for [itex]m[/itex] (for individual calculations) is to use Wolfram|Alpha. Here's an example for width=45.2 and length=67.1: "solve for m: 45.2/67.1=2*E(m)/K(m)-1". It should output [itex]m \approx 0.311837[/itex]. With [itex]m[/itex] found, we can now calculate the unknown [itex]height[/itex]:

    [itex]height=length \frac{\sqrt{m}}{K(m)}=67.1 \frac{\sqrt{0.311837}}{K(0.311837)}=67.1 \frac{0.55842}{1.720878} \approx 21.7739[/itex]

    Another useful formula is for the angle (in radians) of the tangent at the start (left side) of the curve:

    [itex]\Theta=\arccos (1-2m)[/itex] (or to find [itex]m[/itex] from [itex]\Theta[/itex]: [itex]m=\frac{1-\cos \Theta}{2}[/itex])

    It seems like there are a couple of ways to find the form or shape of the curve, but the one I found the easiest to follow was from "An experiment in nonlinear beam theory". The setup is a little different than what I drew in my first post here, as it's rotated 90° and only considers half of the curve, but it is easily transposed & mirrored after the fact. Their process involves finding [itex]x[/itex] and [itex]y[/itex] values for each tangent angle [itex]\theta[/itex] along the curve, starting at the initial tangent angle [itex]\Theta[/itex] (found via the [itex]m[/itex] parameter, though rotated -90° before plugging into the following formulas) and ending at the top (or in their case, the side) of the curve where the tangent angle is [itex]\frac{-\pi}{2}[/itex]. Here are the formulas to find each [itex]x[/itex] and [itex]y[/itex] value:

    [itex]x=\frac{length \sqrt{2}\sqrt{\sin \Theta-\sin \theta}}{K(m)}[/itex]

    [itex]y=\frac{length}{\sqrt{2}K(m)}\int_\theta^\frac{-\pi}{2} \frac{\sin \omega}{\sin \Theta-\sin \omega}\,\mathrm{d}\omega[/itex] (note: there's a typo in the paper...the limit for the integral should start at [itex]\theta[/itex] instead of [itex]\Theta[/itex])

    I wrote a VB script (attached) for a Rhino plugin called Grasshopper that will find the {x, y} points along this curve. Even though it uses Rhino- and Grasshopper-specific code, it should be relatively straightforward to port the code to say Excel (VBA) or a standalone application (I tried to comment the code pretty well). The key function here would be FindBendForm.

    As a side note, some interesting values for [itex]m[/itex]:

    [itex]m=0.826114765984970336[/itex] (value of [itex]m[/itex] when [itex]width=0[/itex])
    [itex]m=0.701327460663101223[/itex] (value of [itex]m[/itex] at maximum possible height of the curve)
    [itex]m=0.180254422335013983[/itex] (minimum value of [itex]m[/itex] when two width values are possible for a given height and length)

    Hope folks find this useful!
     

    Attached Files:

  18. Jul 4, 2014 #17
    Hi willmac8,
    Am just getting into the game as a newby member.
    If I may suggest, there is a book Advanced Mathematics For Engineers by H.W. Reddick and F.H. Miller, second edition, sixth printing, 1949.
    On page 138 Art. 33 is an article called The Elastica in which a uniform elastic spring, originally straight, in which two equal and opposite forces are applied to the ends and resulting in the curves you are interested in. The derivation does in fact use elliptic integrals in the solution. The math is a bit sticky but with a bit of work can be worked through. My interest was the application of this work to a flat thin blade, bent into a U or C shape. The equations derived in the book lend themselves nicely for use with Excel. I took the results and compared nicely with an equation posted previously by another. Consider the analysis by Bisshopp & Drucker on Fly Fishing. This is a bit late but hope it helps. Regards
     
    Last edited: Jul 4, 2014
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