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