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With $k$ proportional to the square root of the compression force, an axially loaded (and otherwise unloaded) beam has a deflection following the DE

$$ \frac{\partial^4}{\partial x^4}z + k^2\frac{\partial^2}{\partial x^2}z = 0 $$

With solutions on the form

$$ z(x) = C_1 + C_2 k x + C_3 \sin(kx) + C_4\cos(kx) $$

Now, consider the following boundary conditions

$z(0) = 0, \left[\frac{\partial z}{\partial x}\right](0) = 0, z(L) = h, \left[\frac{\partial z}{\partial x}\right](L) = 0$

That is, the derivative is clamped to zero at both ends, and the function values are $0$ and $h$ respectively. Solving the corresponding system of equations for $C_1$, $C_2$, $C_3$, and $C_4$ gives $C_1 = - C_4$, and $C_2 = -C_3$, where

$$ C_1 = h\frac{\cos(kL) - 1}{\xi}\\ C_2 = h\frac{\sin(kL)}{\xi} $$

and

$$ \xi = kL \sin(kL)+2\cos(kL)-2 $$

If $kL = 2\pi n, n\in\mathbb{Z}$, $\xi$ becomes zero, and $C_1$ goes to infinity. Is this some kind of resonance, that occurs for certain wave numbers or compression forces?

user877329
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1 Answers1

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If $k$ becomes large enough, the beam buckles and collapses. The critical load is $$ F= \pi^2 Y I/(KL)^2 $$ where $Y$ is young's modulus, I is the Areal moment of inertia and for your boundary conditions $K=1/2$. The larger values values of $k$ that also give $\xi=0$ correspond to higher and higher bending modes becoming unstable (their energy goes down as the coeffecnient $C$ gets bigger. Of course the beam collapses as soon as any mode becomes unstable, and it seems that $k=2\pi /L$ is the first to go.

mike stone
  • 52,996