Solving boundary conditions for vibrating beam

[ajtgraves]

Hi there,

I'm solving the equation for the transverse vibrations of a Euler-Bernoulli beam fixed at both ends and subject to axial loading. It's a similar problem to that described by Rao on page 355 of his 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 $y(x) = C_1cosh(αx) + C_2sinh(αx) + C_3cos(βx) + C_4sin(βx)$ ,
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 $αC_2 + βC_4 = 0$, respectively, while the $y(L)$ and $y'(L)$ conditions give:

1) $C_1cosh(αL) + C_2sinh(αL) + C_3cos(βL) + C_4sin(βL) = 0$

2) $αC_1sinh(αL) + αC_2cosh(αL) – βC_3sin(βL) + βC_4cos(βL) = 0$

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

3) $C_1[cosh(αL) - cos(βL)] + C_2[sinh(αL) - (α/β)sin(βL)] = 0$

4) $C_1[αC_1sinh(αL) + βsin(βL)] + C_2[βcosh(αL) - αcos(βL)] = 0$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:

$C_2 = -C_1[cosh(αL) - cos(βL)] / [sinh(αL) - (α/β)sin(βL)]$

whereas 4) gives:

$C_2 = -C_1[αC_1sinh(αL) + βsin(βL)] / [βcosh(αL) - αcos(βL)]$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.

 

[SteamKing]

Hi there,

I'm solving the equation for the transverse vibrations of a Euler-Bernoulli beam fixed at both ends and subject to axial loading. It's a similar problem to that described by Rao on page 355 of his 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
$y(x) = C_1cosh(αx) + C_2sinh(αx) + C_3cos(βx) + C_4sin(βx)$,
where C₁ , C₂ , C₃ & C₄ 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 $αC_2 + βC_4 = 0$, respectively, while the $y(L)$ and $y'(L)$ conditions give:1) $C_1cosh(αL) + C_2sinh(αL) + C_3cos(βL) + C_4sin(βL) = 0$2) $αC_1sinh(αL) + αC_2cosh(αL) – βC_3sin(βL) + βC_4cos(βL) = 0$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(αL) - cos(βL)] + C_2[sinh(αL) - (α/β)sin(βL)] = 0$4) $C_1[αC_1sinh(αL) + βsin(βL)] + C_2[βcosh(αL) - αcos(βL)] = 0$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:

$C_2 = -C_1[cosh(αL) - cos(βL)] / [sinh(αL) - (α/β)sin(βL)]$

whereas 4) gives:$C_2 = -C_1[αC_1sinh(αL) + βsin(βL)] / [βcosh(αL) - αcos(βL)]$
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.

I've re-worked your Latex commands slightly to make your post more legible.