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Formula for X-Y-Theta bend test

  1. Aug 2, 2010 #1
    Being a 4th year mech. engineering student, you would think I could figure this out. But, for some reason, I can't get the right values. I am designing a bend test fixture for a company I am co-oping for, and although the idea isn't final yet, I figured I'd get a head start on this particular idea. I am looking to setup MATLAB code to give the X- and Y-coordinates for the bend test idea pictured below. The idea came from an article that gave some formulas, but they didn't really help me in my MATLAB code. Here are the formulas given, and the schematic of the fixture:

    θ = L0/r
    X = (L0/θ)*sin(θ)
    Y = (L0/θ)*[1–cos(θ)]

    where (inputs):
    r = bend radius
    L0 = specimen length

    [PLAIN]http://a.imageshack.us/img251/2532/idea2schem.png [Broken]

    Can anyone give me a clue as to how I would be able to calculate the X- and Y-coordinates for this setup? We will have two linear motors driving these directions, and would like to go from straight to bent to straight to bent the other direction. Let me know what you can come up with... Thanks.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 3, 2010 #2


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    Sorry I don't quite understand what the bend radius is. Can you give a little more insight as to how this works and such.
  4. Aug 3, 2010 #3


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    It looks to me like the curves are given by the formulas for x and y that you posted, as long as you treat [itex]L_0[/itex] not as the specimen length but as the length along the specimen.

    If you treat [itex]L_0[/itex] as the total length, a parametric plot of [itex](x(\theta),y(\theta))[/itex] will give you a curve that represents the endpoints of the beam as theta varies.

    If you rewrite things as

    [tex]x(s) = r \sin\left(\frac{s}{r}\right)[/tex]
    [tex]y(s) = r\left(1-\cos\left(\frac{s}{r}\right)\right)[/tex]

    where s is the arc length along the beam from the origin, and plot from s = 0 to the total specimen length L_0 for a given value of r (chosen such that the final angle is [itex]\theta = L_0/r[/itex]) it should give the curves in the image you posted. (perhaps rotated)
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