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How would I figure out the Cartesian graph that describes a bar clamped flat for a length on one end with downward force being applied to the other?

I have an idea that the bar will try to average out the stress over the free length. Does it in fact evenly distribute stress over the whole free length? If so, can I use that fact to calculate curvature at each point to derive a Cartesian graph?

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I saw this very briefly on a course of mathematics applied to engineering. I remember that the differential equation for the shape of the bar had derivatives of up to fourth order, but that's all I've got. :/ –  Bruce Connor Jan 13 '11 at 4:18
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@Bruce That's right. try page 38 here: pma.caltech.edu/Courses/ph136/yr2008/0810.3.K.pdf –  Mark Eichenlaub Jan 13 '11 at 11:32
    
@Mark Nice! Thanks. –  Bruce Connor Jan 13 '11 at 14:09

2 Answers 2

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I know this question is a little old, but so far there is no correct answer posted. It happens that I used to argue this question with Civil Engineering students years ago. They had trouble with it because they learned about "bending moment", which is basically the same thing as curvature only no one told them that. The curvature, or the bending moment if you will, goes to zero at the free end and increases uniformly (linearly) up to the point of constraint. You can tell that this makes sense because if the bar is over-stressed, that's where it will break. The question asks us to plot the graph of the bar, which you basically get by integrating twice. (In the limit of small displacements, the curvature of course is equal to the second derivative of displacement.) If you differentiate twice in the other direction, you get the distribution of forces. You can see that the step function at the point of constraint gives you a couple when you differentiate twice.

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Thank you, that is very helpful. Can you explain why it must increase linearly and not as the square, for instance? –  jnm2 Jun 4 '11 at 14:32
    
I can't exactly explain it, but I can point out that in the beam equation, for a uniform cross-section the force distribution is the second derivative of the curvature. So wherever there are no forces the chance in curvature can be at most linear. –  Marty Green Jun 4 '11 at 14:55

I think if the boundary conditions were perfectly symmetric, you'd get a simple circular shape. In your case you have somewhat different conditions on both ends. The inside of the curve will be in compression, and the outside in tension. But I think your clamped end restricts some of the material degrees of freedom, so there is probably a transition region near that end where the curvature differs somewhat from the rest of the bar.

Real materials can be nonlinear. If the material weakens with greater stress -especially if it can plastically deform there can be a tendency for the strain to accumulate in a small region (like bending a coat hanger, and getting most of the bending in one small area).

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