I'm solving the equation for the transverse vibrations of a Euler-Bernoulli beam fixed at both ends and subject to axial loading (as per this diagram). It's a similar problem to that described by Rao on page 355 of his excellent book "Vibration of Continuous Systems" (Google books link), except the example he uses is for a simply supported beam.

The general solution takes the form of $y(x) = C_1\cosh(\alpha x) + C_2\sinh(\alpha x) + C_3\cos(\beta x) + C_4\sin(\beta x)$, where $C_1$, $C_2$, $C_3$ & $C_4$ are the constants I need to find. The BCs are standard:

  • $y(0)=y(L) = 0$ (zero displacement at ends)
  • $y'(0)=y'(L) = 0$ (zero gradient at ends)

When I substitute these in the $y(0)$ and $y'(0)$ conditions give $C_1 + C_3 = 0$ and $\alpha C_2 + \beta C_4 = 0$, respectively, while the $y(L)$ and $y'(L)$ conditions give:

1) $C_1\cosh(\alpha L) + C_2\sinh(\alpha L) + C_3\cos(\beta L) + C_4\sin(\beta L) = 0$

2) $\alpha C_1\sinh(\alpha L) + \alpha C_2\cosh(\alpha L) – \beta C_3\sin(\beta L) + \beta C_4\cos(\beta L) = 0$

Clearly the first 2 conditions can be used to reduce these last two equations into functions of $C_1$ and $C_2$ only:

3) $C_1[\cosh(\alpha L) - \cos(\beta L)] + C_2\left[\sinh(\alpha L) - \frac{\alpha}{ \beta}\sin(\beta L)\right] = 0$

4) $C_1[\alpha C_1\sinh(\alpha L) + \beta \sin(\beta L)] + C_2[\beta \cosh(\alpha L) - \alpha \cos(\beta L)] = 0$

We can now solve for $C_1$ (or $C_2$) and use this to write all the terms of the original governing equation in terms of it alone. However, there are two possible expressions for $C_1$ (and $C_2$), depending on which equation is used. 3) gives:

$C_2 = -C_1\frac{[\cosh(\alpha L) - \cos(\beta L)]}{[\sinh(\alpha L) - (\alpha /\beta )\sin(\beta L)]}$

whereas 4) gives:

$C_2 = -C_1\frac{[\alpha C_1\sinh(\alpha L) + \beta \sin(\beta L)]}{[\beta \cosh(\alpha L) - \alpha \cos(\beta L)]}$

These are clearly different, but are they both correct? Which one should be used?

Many thanks in advance for your help, it would be much appreciated.

  • $\begingroup$ See this page for tips on how to use mathjax/latex to typeset math here. In particular add \ in front of standard functions like \sin(x) to get proper formatting and fractions can be done as \frac{ a } { b }. $\endgroup$ – Winther Jan 17 '16 at 16:36
  • $\begingroup$ Thanks very much for your comments and apologies for my typos! It should all be correct now, but I don't suppose you could suggest a solution?! $\endgroup$ – AJG Jan 17 '16 at 17:17
  • $\begingroup$ Please give a scan of the page ( Google preview is limited). Is it a pure buckling problem ? Or is there a transverse load also ? The principle of superposition or combining between axial loading and lateral loading is not always valid. $\endgroup$ – Narasimham Jan 17 '16 at 17:46
  • $\begingroup$ I don't think the general solution that you proposed is the same for simply supported beam. Beam fixed with both ends are more complicated and in my opinion should have a different form. Plus, it is also symmetric. So it won't be one sinusoidal + one hyperbolic as you thought... just my 2 cents. $\endgroup$ – kensaii Jan 17 '16 at 17:59
  • $\begingroup$ Your comment was addressed to OP, right? $\endgroup$ – Narasimham Jan 17 '16 at 20:31

If the beam is fixed at both ends and subjected to axial loads, there would be no deformation !

  • $\begingroup$ Look at Euler's problem of buckling to see a counterexample... $\endgroup$ – vonbrand Jan 18 '16 at 1:20
  • $\begingroup$ In the classical Euler buckling axial translation is permitted, leading to the buckling load $ \pi^2 EI/l^2$ or $ 4 \pi^2 EI/l^2$ depending on whether end rotation is allowed or suppressed. In the present case all degrees of freedom are arrested. $\endgroup$ – Narasimham Jan 18 '16 at 1:43
  • $\begingroup$ Exactly, although the ends of the beam are fixed into the supports, the tension is applied to the supports, i.e they are pulled apart $\endgroup$ – AJG Jan 18 '16 at 10:46

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