Critical buckling force of a thin plate
1. Assumptions and notation
- Plate is thin (classical Kirchhoff–Love plate theory applies): transverse shear is neglected.
- Linear elastic, isotropic material with Young’s modulus $E$ and Poisson’s ratio $\nu$.
- Plate midsurface coordinates: $x\in[0,a]$, $y\in[0,b]$. Transverse deflection $w(x,y)$ (positive out of plane).
- In-plane compressive load is uniform along $y$ and equal to $N_x$ (force per unit length in N/m) directed along $-x$ (i.e. compression). We assume no in-plane $y$ compression $N_y=0$ and no twisting in-plane coupling (pure uniaxial).
- Bending rigidity (flexural rigidity) of the plate [[Thin plate bending]]: \(D=\frac{E t^3}{12(1-\nu^2)}.\)
We seek the critical value $N_{x,\mathrm{cr}}$ at which the trivial flat solution $w\equiv0$ becomes unstable (buckling).
2. Governing linearized buckling equation
\[D \left( \frac{\partial^4 w}{\partial x^4} + 2 \frac{\partial^4 w}{\partial x^2 \partial y^2} + \frac{\partial^4 w}{\partial y^4} \right) + N_x \frac{\partial^2 w}{\partial x^2} + N_y \frac{\partial^2 w}{\partial y^2} + 2N_{xy} \frac{\partial^2 w}{\partial x \partial y} = 0\]From linear plate theory (small deflections, linear kinematics) the static buckling equation under an in-plane compressive stress resultant $N_x$ (per unit length) is: (derived in section 5)
\[D\nabla^4 w + N_x \frac{\partial^2 w}{\partial x^2} = 0,\]where $\nabla^4$ is the biharmonic operator $\nabla^4 = (\partial^2/\partial x^2 + \partial^2/\partial y^2)^2$.
This equation arises from equating bending moments (through $D\nabla^4 w$) and the destabilizing effect of in-plane compression (term with $N_x$) after linearization.
3. Choose admissible buckling mode (simply supported on all edges)
A convenient set of admissible functions for a plate simply supported on all four edges $x=0,a$ and $y=0,b$ is the double sine series. The fundamental separable eigenfunctions are
\[w(x,y) = W \sin\!\bigg(\frac{m\pi x}{a}\bigg)\sin\!\bigg(\frac{n\pi y}{b}\bigg),\]with integers $m,n=1,2,\dots$ and amplitude $W$ (nonzero for buckling modes). These satisfy
- $w=0$ at $x=0,a$ and $y=0,b$,
- and the bending-moment/slope compatibility conditions for simply supported edges (these sine forms are standard eigenfunctions).
Compute the necessary derivatives of $w$:
\[\frac{\partial^2 w}{\partial x^2} = -\left(\frac{m\pi}{a}\right)^2 w, \qquad \frac{\partial^2 w}{\partial y^2} = -\left(\frac{n\pi}{b}\right)^2 w.\]Then
\[\nabla^2 w = -\left[\left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2\right] w,\]and consequently
\[\nabla^4 w = \bigg(\left(\tfrac{m\pi}{a}\right)^2 + \left(\tfrac{n\pi}{b}\right)^2\bigg)^2 w.\]4. Substitute into governing equation → eigenvalue for $N_x$
Substitute the above into
\[D\nabla^4 w + N_x \frac{\partial^2 w}{\partial x^2} = 0.\]Since $w$ is not identically zero for a buckling mode, factor $w$ out and set the coefficient to zero:
\[D\bigg(\Big(\tfrac{m\pi}{a}\Big)^2 + \Big(\tfrac{n\pi}{b}\Big)^2\bigg)^2 \;-\; N_x \Big(\tfrac{m\pi}{a}\Big)^2 \;=\; 0.\]Solve for the critical in-plane compressive resultant $N_{x,\mathrm{cr}}$ for mode $(m,n)$:
\[\boxed{ \; N_{x,\mathrm{cr}}(m,n) \;=\; D \; \frac{\Big(\big(\tfrac{m\pi}{a}\big)^2 + \big(\tfrac{n\pi}{b}\big)^2\Big)^2} {\big(\tfrac{m\pi}{a}\big)^2} \; =\; D \; \frac{\big(k_x^2+k_y^2\big)^2}{k_x^2} \; }\]where I used $k_x=\dfrac{m\pi}{a}$ and $k_y=\dfrac{n\pi}{b}$ for compactness.
It is common to express the result as
\[N_{x,\mathrm{cr}} = k\,\frac{\pi^2 D}{a^2},\]where $k$ is a buckling coefficient depending on aspect ratio $a/b$ and boundary conditions. Comparing with the explicit expression above yields for simply supported edges
\[k(m,n) \;=\; \frac{a^2}{\pi^2}\, \frac{\Big(\big(\tfrac{m\pi}{a}\big)^2 + \big(\tfrac{n\pi}{b}\big)^2\Big)^2}{\big(\tfrac{m\pi}{a}\big)^2} \;=\; \frac{\big(m^2 + (a^2/b^2)\,n^2\big)^2}{m^2}.\]For given aspect ratio $a/b$ you choose integers $m,n$ to get the smallest $N_{x,\mathrm{cr}}$ (the first buckling mode). The smallest $k$ gives the critical load.
4.1. Simple special cases
(a) Square plate $a=b$, lowest mode usually $m=n=1$
Take $a=b$ and $m=n=1$:
\[N_{x,\mathrm{cr}} = D \frac{\big((\pi/a)^2 + (\pi/a)^2\big)^2}{(\pi/a)^2} = D \frac{(2\pi^2/a^2)^2}{\pi^2/a^2} = D \cdot \frac{4\pi^4/a^4}{\pi^2/a^2} = 4\pi^2 \frac{D}{a^2}.\]So
\[P_{\mathrm{cr}} = b\,N_{x,\mathrm{cr}} = a \cdot 4\pi^2 \frac{D}{a^2} = 4\pi^2 \frac{D}{a} \quad\text{(for }a=b\text{)}.\](b) Very long strip ( $a \gg b$ ) — plate behaves like a column across length with many half-waves in $x$
If $a$ is very long, continuous $m$ modes can be approximated and often the minimum occurs with $n=1$ and $m$ chosen to minimize expression. In the continuous wavelength limit you can minimize
\[N_{x,\mathrm{cr}}(k_x,k_y)=D\frac{(k_x^2+k_y^2)^2}{k_x^2}\]w.r.t $k_x$ for fixed $k_y$. Doing $\partial N/\partial k_x =0$ yields $k_x^2 = k_y^2$, so the optimal buckling cell is square $k_x=k_y$. For discrete $m,n$ you pick the nearest integers.
5. Governing differential equation
To derive the governing differential equation for plate buckling, we will use the Equilibrium Method.1 This involves analyzing a tiny differential element of the plate after it has slightly buckled.
We are looking for the point where the flat equilibrium becomes unstable and a new, bent equilibrium state is possible.
5.1. The Differential Element
Imagine a tiny rectangular piece of the plate with dimensions $dx$ and $dy$.
- $w$: The vertical deflection of the plate (downward in the $z$-direction).
- $N_x$: The compressive force per unit length acting on the $x$-face.
- $Q_x, Q_y$: Transverse shear forces (acting vertically).
- $M_x, M_y, M_{xy}$: Bending and twisting moments.
5.2. Equilibrium of Forces in the Z-Direction
We sum all forces acting vertically ($z$-direction) on this element. For the plate to be in equilibrium in a buckled shape, these forces must sum to zero.
The forces come from two sources:
- Vertical Shear Forces ($Q$): The standard resistance of the plate to bending.
- In-Plane Compressive Forces ($N_x$): These are horizontal forces, but because the plate is bent, they now have a small vertical component.
5.2.1. Step 2a: The Net Shear Force
On the left face, we have shear force $-Q_x dy$. On the right face, we have $(Q_x + \frac{\partial Q_x}{\partial x}dx) dy$.
The net force from $Q_x$ is:
\[\left(Q_x + \frac{\partial Q_x}{\partial x}dx\right)dy - Q_x dy = \frac{\partial Q_x}{\partial x} dx dy\]Similarly for the $y$-direction shear ($Q_y$):
\[\frac{\partial Q_y}{\partial y} dy dx\]Total force from shear:
\[\left( \frac{\partial Q_x}{\partial x} + \frac{\partial Q_y}{\partial y} \right) dx dy\]5.2.2. Step 2b: The “P-Delta” Effect (Vertical Component of $N_x$)
This is the critical step for buckling.
The compressive force $N_x dy$ acts on the edges. If the plate is bent, the two ends of the element are not perfectly horizontal. They are tilted by the slope $\frac{\partial w}{\partial x}$.
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Slope on the left edge: $\frac{\partial w}{\partial x}$
-
Slope on the right edge: $\frac{\partial w}{\partial x} + \frac{\partial^2 w}{\partial x^2}dx$
The vertical component of the force $N_x$ is $N_x \cdot \sin(\theta) \approx N_x \cdot \tan(\theta) \approx N_x \frac{\partial w}{\partial x}$ (for small angles).
-
Force on left edge (pushing up): $-N_x dy \left( \frac{\partial w}{\partial x} \right)$
-
Force on right edge (pushing down/up): $+N_x dy \left( \frac{\partial w}{\partial x} + \frac{\partial^2 w}{\partial x^2}dx \right)$
Summing these gives the net vertical push caused by compression:
\[N_x dy \frac{\partial^2 w}{\partial x^2} dx = N_x \frac{\partial^2 w}{\partial x^2} dx dy\]5.2.3. Step 2c: The Combined Equation
Combine terms and divide by the area $dx dy$:
\[\frac{\partial Q_x}{\partial x} + \frac{\partial Q_y}{\partial y} + N_x \frac{\partial^2 w}{\partial x^2} = 0\](Note: If we had compression $N_y$ or shear $N{xy}$, they would add similar terms here, but you asked to simplify for x-compression only.)_
5.3. Relate Shear ($Q$) to Moments ($M$)
We cannot solve the equation above yet because it uses Shear ($Q$), but plate deformation is driven by Moments ($M$). We need to use moment equilibrium to bridge the gap.
Summing moments about the y-axis (for the element):
\[Q_x - \frac{\partial M_x}{\partial x} - \frac{\partial M_{xy}}{\partial y} = 0 \implies Q_x = \frac{\partial M_x}{\partial x} + \frac{\partial M_{xy}}{\partial y}\]Summing moments about the x-axis:
\[Q_y - \frac{\partial M_y}{\partial y} - \frac{\partial M_{xy}}{\partial x} = 0 \implies Q_y = \frac{\partial M_y}{\partial y} + \frac{\partial M_{xy}}{\partial x}\]Substitute these expressions for $Q_x$ and $Q_y$ back into our Force equation from Phase 2c:
\[\frac{\partial}{\partial x} \left( \frac{\partial M_x}{\partial x} + \frac{\partial M_{xy}}{\partial y} \right) + \frac{\partial}{\partial y} \left( \frac{\partial M_y}{\partial y} + \frac{\partial M_{xy}}{\partial x} \right) + N_x \frac{\partial^2 w}{\partial x^2} = 0\]Expanding this gives:
\[\frac{\partial^2 M_x}{\partial x^2} + 2\frac{\partial^2 M_{xy}}{\partial x \partial y} + \frac{\partial^2 M_y}{\partial y^2} + N_x \frac{\partial^2 w}{\partial x^2} = 0\]5.4. Relate Moments ($M$) to Deflection ($w$)
From standard plate bending theory (Kirchhoff-Love theory), moments are related to the curvature (second derivatives of $w$) via the plate stiffness $D$.
\[M_x = -D \left( \frac{\partial^2 w}{\partial x^2} + \nu \frac{\partial^2 w}{\partial y^2} \right)\] \[M_y = -D \left( \frac{\partial^2 w}{\partial y^2} + \nu \frac{\partial^2 w}{\partial x^2} \right)\] \[M_{xy} = -D (1-\nu) \frac{\partial^2 w}{\partial x \partial y}\]Where 2$D = \frac{Et^3}{12(1-\nu^2)}$ is the flexural rigidity.3
5.5. The Final Substitution
Substitute the Moment-Deflection equations (Phase 4) into the Moment Equilibrium equation (Phase 3).
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Term 1 ($\frac{\partial^2 M_x}{\partial x^2}$):
\[-D \left( \frac{\partial^4 w}{\partial x^4} + \nu \frac{\partial^4 w}{\partial x^2 \partial y^2} \right)\] -
Term 2 ($2\frac{\partial^2 M_{xy}}{\partial x \partial y}$):
\[-2D (1-\nu) \frac{\partial^4 w}{\partial x^2 \partial y^2}\] -
Term 3 ($\frac{\partial^2 M_y}{\partial y^2}$):
\[-D \left( \frac{\partial^4 w}{\partial y^4} + \nu \frac{\partial^4 w}{\partial x^2 \partial y^2} \right)\]
Sum them up:
Notice the cross terms $\frac{\partial^4 w}{\partial x^2 \partial y^2}$. We have $-\nu$, $-2(1-\nu)$, and $-\nu$.
Sum = $-\nu - 2 + 2\nu - \nu = -2$.
So the sum of the moment terms is:
\[-D \left( \frac{\partial^4 w}{\partial x^4} + 2 \frac{\partial^4 w}{\partial x^2 \partial y^2} + \frac{\partial^4 w}{\partial y^4} \right)\]Or more compactly using the biharmonic operator: $-D \nabla^4 w$.
5.6. The Final Governing Equation
Combine the result from Phase 5 with the $N_x$ term from Phase 2c:
\[-D \nabla^4 w + N_x \frac{\partial^2 w}{\partial x^2} = 0\]Rearranging to the standard form:
\[D \left( \frac{\partial^4 w}{\partial x^4} + 2 \frac{\partial^4 w}{\partial x^2 \partial y^2} + \frac{\partial^4 w}{\partial y^4} \right) + N_x \frac{\partial^2 w}{\partial x^2} = 0\]This is the governing differential equation for the buckling of a thin plate under uniaxial compression $N_x$.