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Change in potential energy of elastic strip under deformation

  1. Mar 6, 2013 #1
    A linear elastic strip of natural length a and stiffness k lies between x = 0 and x = a. Each point on the strip is transformed by a differentiable, monotone increasing function f.

    a) Characterise the change in potential energy.

    b) Given the boundary conditions f(0) = 0 and f(a) = b, choose f such that the potential energy is minimised.

    My first thought was to find a piecewise linear approximation to the problem and then take the continuum limit.

    If we let [itex]x_0, x_1,..., x_i,...,x_n[/itex] denote an ordered set of points joined by springs then we have [itex]\Delta{E_i}=\int\limits_0^{e_i}\! kx + k(x - e_{i - 1})\, \mathrm{d}x=k(e_{i}^2-e_{i}e_{i-1})[/itex] where [itex]e_i=f(x_i) - x_i[/itex] and [itex]\Delta{E_i}[/itex] denotes the change in potential energy associated by the displacement of [itex]x_i[/itex] given that [itex]x_i[/itex] is displaced after [itex]x_{i-1}[/itex]. We then have (neglecting the endpoints) [itex]\Delta{E_{total}}\approx\!k\sum\limits_i\!(e_{i}^2-e_{i}e_{i-1})[/itex], but I am not sure where to go from there.

    Any help would be appreciated.

    Edit:

    If we factorise [itex]\!k\sum\limits_i\!(e_{i}^2-e_{i}e_{i-1})[/itex] to give [itex]\!k\sum\limits_i\!e_i(e_i - e_{i-1})[/itex] then in the limit we get [itex]\Delta{E_{total}} = \!k\int\limits_x\! e\mathrm{d}e=\frac{ke(x)^2}{2}\bigg|_{x_0}^{x_1}[/itex], but this lack of dependence of internal state runs counter to intuition; it seems to me that if you hold the ends of a rubber band fixed and pull the middle to one side it will snap back. Have I done something wrong? If so, what?
     
    Last edited: Mar 6, 2013
  2. jcsd
  3. Mar 7, 2013 #2

    TSny

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    Hello, eutectic. Welcome to PF!

    I'm not understanding your expression for ##\Delta E_i##. But, your idea of starting with a finite number of segments of the strip, treating each as a little spring, and then going to the continuum limit should work. You can let the length of each segment (before stretching) be the same amount ##\Delta x##. After the strip is stretched, each segment will have an energy ##\Delta E_i = \frac{1}{2} k_{seg} (\Delta L_i)^2##, where ##k_{seg}## is the effective spring constant of a segment and ##\Delta L_i## is the amount of stretch of the ith segment.

    You'll need to relate ##k_{seg}## to the overall force constant ##k## for the entire strip, and you'll need to find an expression for ##\Delta L_i## in terms of the function ##f(x)## and ##\Delta x##. See the attached figure.
     

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  4. Mar 7, 2013 #3

    rcgldr

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    If the natural length of the strip is a, and it lies between x = 0 and x = a, then the strip is not under tension, so it's not clear to me what f is supposed to represent. If the strip were under tension, then the tension would be the same at all points along the strip (Newton's third law) (assuming the strip is horizontal and not affected by gravity).
     
  5. Mar 7, 2013 #4
    That's before the transformation by f(x). After transformation, f(a) = b. So it represents a (potentially) non-uniform stretch.

    That's what the result of this exercise is supposed to demonstrate.
     
  6. Mar 7, 2013 #5

    rcgldr

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    The problem statement doesn't explain what the output of monotone increasing function f() or b are. So you're supposed to assume that f(a) means the strip is streched to length b? Is the student supposed allow for a strip where total stiffnes is k, but the stiffness varies across the length of the strip, which would require using calculus of variations? If the stiffness is k everywhere in the strip, then wouldn't f() have to be linearly increasing by definition of a constant stiffness?
     
    Last edited: Mar 7, 2013
  7. Mar 7, 2013 #6
    Yes, excepted that stiffness does not vary - it simply does not apply. It is a property of the entire object, and you need to replace it with something that could be used in infinitesimal analysis.
     
  8. Mar 7, 2013 #7

    TSny

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    Yes, that's how I interpret it.
    If you marked segments of equal length along the strip before it is stretched, then each segment would have the same stiffness , but the stiffness of the segments is not the same as the stiffness of the entire strip. Equivalently, http://en.wikipedia.org/wiki/Young's_modulus][/PLAIN] [Broken] Young's modulus would be the same throughout the strip.

    I think the problem is saying that there are external forces applied along the strip to stretch different small segments different amounts in a fairly arbitrary way. A point that was a distance x from the left end will end up at a distance f(x) from the left end. The resultant deformation will have a certain total elastic potential energy. The problem is to find the function f(x) that minimizes the energy and satisfies the boundary conditions. You can use calculus of variations on the resultant integral expression for the energy, but I think the answer can be found without using the formalism of calculus of variations.
     
    Last edited by a moderator: May 6, 2017
  9. Mar 7, 2013 #8

    rcgldr

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    I don't understand this. If all segments have the same stiffness, then why wouldn't the entire strip have the same stiffness?

    Correction to link:

    http://en.wikipedia.org/wiki/Youngs_modulus
     
  10. Mar 7, 2013 #9
    Stiffness depends on material properties and dimensions. You need something independent of dimensions - at least, of the length - here.
     
  11. Mar 7, 2013 #10
    If at first you don't succeed...

    OK, here goes round two, guided by TSny's advice.

    [itex]k_{seg}=\frac{ka}{\Delta\!x}[/itex]

    [itex]\Delta\!L_i=(f(x_i+\Delta\!x) - (x_i+\Delta\!x)) - (f(x_i) - x_i)=f(x_i+\Delta\!x)-f(x_i)-\Delta\!x=\Delta\!f-\Delta\!x[/itex]

    [itex]\Delta\!E_i
    =\frac{ka}{2\Delta\!x}(\Delta\!f-\Delta\!x)^2
    =\frac{ka}{2\Delta\!x}((\Delta\!f)^2+(\Delta\!x)^2 - 2 \Delta\!f \Delta\!x)
    =\frac{ka}{2}(\frac{(\Delta\!f)^2}{\Delta\!x}+ \Delta\!x - 2\Delta\!f)[/itex]

    [itex]\Delta\!E_{discrete}=\frac{ka}{2}\sum\!(\frac{ \Delta\!f}{\Delta\!x} \Delta\!f - 2\Delta\!f + \Delta\!x)[/itex]

    [itex]\Delta\!E_{continuous}=\frac{ka}{2}(\int\limits_{f(0)}^{f(a)}(\frac{\mathrm{d}f(x)}{\mathrm{d}x}-2)\mathrm{d}f(x)+ \int\limits_0^a\mathrm{d}x)
    =\frac{ka}{2}(\int\limits_0^a (f'(x))^2\mathrm{d}x+2(f(0)-f(a))+a)[/itex]

    Given the imposed boundary conditions the second part of the problem is obviously to be solved by minimising [itex]\int\limits_0^a (f'(x))^2\mathrm{d}x[/itex], subject to [itex]\int\limits_0^a f'(x)\mathrm{d}x=b[/itex]. Going back to the discrete case this gives us the minimisation of [itex]\sum\limits_i\!f'(x_i)^2\Delta\!x[/itex], subject to [itex]\sum\limits_i\!f'(x_i)\Delta\!x=b[/itex]. Substituting [itex]f'(x_n) = \frac{b}{\Delta\!x} - \sum_{i=0}^{n-1}\!f'(x_i)[/itex] gives [itex]\frac{\partial}{\partial\!(f'(x_j))}(\sum_{i=0}^{n-1}f'(x_i)^2 + (\frac{b}{\Delta\!x} - \sum_{i=0}^{n-1}\!f'(x_i))^2\Delta\!x) = 0 = 2\Delta\!x(f'(x_j) - (\frac{b}{\Delta\!x} - \sum_{i=0}^{n-1} f'(x_i)))[/itex].

    Hence [itex] f'(x_j) = \frac{b}{\Delta\!x} - \sum_{i=0}^{n-1} f'(x_i)=f'(x_n)=constant[/itex]
     
    Last edited: Mar 7, 2013
  12. Mar 7, 2013 #11

    haruspex

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    I find it a bit easier to think in terms of the displacement function, g(x) = f(x)-x.
    Over a segment δx, the extension is δg(x) and the tension is kδg(x)/δx. The energy stored in it is k(δg(x))2/2δx. In the limit, that's k(g')2δx/2. Integrating, ##E = \frac12\int_{x=0}^ag'^2.dx = \frac12\int_{x=0}^a(f'-1)^2.dx ##.
    Writing ##L(x, f, f') = \frac12 (f'-1)^2##, the Euler-Lagrange equation gives ##0 = \frac d{dx}\frac{∂L}{∂f'} = \frac d{dx}(f'-1)##. Hence f'-1 = constant.
     
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