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A circular elastic plate with bending stiffness $B$, constant excess pressure $p$ and clamped boundary conditions at radius $R$ obeys $B{\nabla }_r^{4}h=p$, where ${\nabla }_r^2=r^{-1}{\partial }_r(r{\partial }_r)$. Its regular solution is $h=p(R^2-r^2{)}^2/(64B)$. Consequently its enclosed volume is $V=\pi p{R}^6/(192B)$, and its edge curvature is $h_{rr}(R)=p{R}^2/(8B)=24V/(\pi R^4)$.

Beam flexural rigidity is the coefficient relating a slender beam's bending moment to its curvature in linear bending. For a homogeneous material with uniform Young's modulus $E$ over the material area,where $z$ is distance from the appropriate neutral axis. The magnitude relation is $|M|=B|\kappa |$; signs depend on the chosen moment and curvature conventions. Integrating the linear bending stress over the section gives this coefficient.^[ref]

For an original rectangular example, choose width $b=10\mathrm{mm}$, bending thickness $h=2\mathrm{mm}$ and $E=70\mathrm{GPa}$. The second moment of area is $I=b{h}^3/12=6.667\times 10^{-12}{\mathrm{m}}^4$, giving $B\approx 0.4667\mathrm{N}{\mathrm{m}}^2$. Doubling only the thickness multiplies $B$ by eight. Rotating the original rectangle by ninety degrees about its length instead exchanges width and thickness: $I$ increases by $(b/h{)}^2=25$ even though material area and modulus are unchanged.

Thus a bending experiment measures a combination of material response and geometry. Extracting $E$ requires the correct bending axis and second moment of area; the three-point flexural test is one route. This beam coefficient has units $\mathrm{N}{\mathrm{m}}^2$, distinct from the plate coefficient called bending stiffness. If modulus varies across a section, the relevant stiffness requires a material-weighted area integral and the appropriate neutral axis, rather than an arbitrary average modulus multiplied by $I$.

The part of an elastic rod's elastic energy that penalizes departure of its bending strain of a rod $u$ from the intrinsic value $\hat{u}$. Here $B$ is the bending stiffness and $S$ is reference arc length. For an inextensible rod, the same formula can be written using spatial curvature and current arc length.

A bending moment is an internal torque transmitted across a cut through a beam or elastic plate. For a slender linear elastic beam, $M=EI\kappa$ up to sign convention, where $E$ is Young's modulus, $I$ the second moment of area, and $\kappa$ the centreline curvature. The corresponding bending response of an elastic plate is described by its bending stiffness and curvature tensor.

Bending stiffness is the coefficient relating plate curvature to bending moment. In linear thin-plate theory a transverse displacement $\zeta$ produces a restoring load $B{\nabla }^4\zeta$.

Capillary origami uses surface tension to fold an initially flat elastic plate around a liquid droplet. The fold trades a reduction in surface energy for an increase in bending energy; the initial outline helps select the resulting enclosure.^[ref]

In the 2007 experiments of Charlotte Py and collaborators, 40–80 micrometre polydimethylsiloxane sheets rested on a water-repellent support. A water droplet reached the sheet's corners; time-lapse images followed its evaporation. Squares changed from four-corner bending to a tubular shape; triangles formed a tetrahedron. Smaller or thicker sheets could bend and then reopen without enclosing the droplet.^[ref]

The elastocapillary length makes the thickness dependence quantitative. For fixed Young's modulus, Poisson's ratio and surface tension, the bending stiffness obeys $B\propto h^3$ and ${\ell }_{\mathrm{ec}}\propto h^{3/2}$. Halving thickness therefore reduces ${\ell }_{\mathrm{ec}}$ by $2^{3/2}$, about $2.83$. At the same lateral size $L$, the sheet elastocapillary energy ratio $\eta =\gamma L^2/B$ increases eightfold. This calculation explains why making the sheet thinner can enable wrapping even without changing its outline.

The sheet elastocapillary energy ratio compares available energy scales; it does not determine a unique final shape. The contact angle, initial liquid volume and outline also matter. A cut chosen to permit several panels to meet is a mechanical design choice, while the surface tension supplies the driving force. Predicting the complete folding path requires more than the scalar energy comparison. The same group reported cross-shaped templates forming a cube and flower-shaped templates forming nearly spherical enclosures.^[ref]

For a unit-length elastic beam with unit bending stiffness, the Green function of $D_x^4$ with $G=G_x=0$ at both endpoints is the displayed kernel. It is piecewise cubic and continuous through its second derivative, while $G_{xxx}(\xi +,\xi )-G_{xxx}(\xi -,\xi )=1$. A constructive form is $G=a{x}^2+b{x}^3+(x-\xi {)}_{+}^3/6$, with $a=\xi (1-\xi {)}^2/2$ and $b=-(1-\xi {)}^2(1+2\xi )/6$. The right endpoint conditions determine these coefficients. Thus $y(x)={\int }_0^{1}G(x,\xi )f(\xi )d\xi$ solves the clamped static beam problem.

Positive compression $\alpha$ contributes negative quadratic potential energy $-\alpha {\varphi }_x^2/2$, while positive bending stiffness $\beta$ contributes $\beta {\varphi }_{xx}^2/2$. A plane wave has ${\omega }^2=(\beta k^4-\alpha k^2)/\rho$, so sufficiently long waves grow exponentially. For displacement-fixed, moment-free endpoints at $0,l$, the natural boundary conditions for a free endpoint on the slope give ${\varphi }_{xx}=0$ as well as $\varphi =0$. The modes are $\sin (n\pi x/l)$, and the first Euler buckling of an elastic filament threshold is ${\alpha }_c=\beta {\pi }^2/l^2$. For truly clamped ends, $\varphi ={\varphi }_x=0$, the threshold is instead $4{\pi }^2\beta /l^2$, with first static mode $1-\cos (2\pi x/l)$. More generally,with the infimum taken over the specified endpoint class. At threshold there is a static zero mode and a marginal linear-in-time partner in the linear dynamics; above it the straight configuration is unstable.

For bending stiffness $D$ and mass per area $m$, a plane wave of surface displacement $\eta$ satisfies $m{\eta }_{tt}+D{\nabla }_H^4\eta =p$. An inviscid deep-water velocity potential has vertical dependence $e^{kz}$; the kinematic boundary condition and linear Bernoulli equation give $p=({\rho }_w{\omega }^2/k-{\rho }_{w}g)\eta$. Eliminating $p$ gives the displayed dispersion relation. With water depth $d$, replace it by ${\omega }^2=(gk+D{k}^5/{\rho }_w)\tanh (kd)/[1+(m/{\rho }_w)k\tanh (kd)]$.

In the specified thin-sheet deep-water model, bending stiffness adds restoring force while sheet mass adds inertia. Their ratio to an uncovered gravity-capillary wave gives the displayed expression. Added inertia first lowers frequency at small wavenumber; at formally large wavenumber bending produces a frequency ratio proportional to $k^{1/2}$. That formal asymptote must not be extrapolated past the range where a thin-sheet continuum description is valid.

An elastic plate is a body whose thickness is small compared with its in-plane dimensions, allowing a three-dimensional elasticity problem to be reduced to its middle surface. It can carry both in-plane stretching loads and transverse bending loads. A sheet supported around its edges and pressed sideways is a simple bending experiment; its response depends strongly on thickness and on the boundary conditions.

For an isotropic plate of thickness $h$, Young's modulus $E$ and Poisson's ratio $\nu$, small-slope bending has bending stiffnessWith constant $B$, no applied in-plane tension and transverse load per area $p$, static deflection $w$ satisfies $B{\nabla }^{4}w=p$, using a consistent sign for load and deflection. The fourth-order biharmonic equation needs both displacement-related and moment-related edge conditions. Doubling thickness multiplies $B$ by eight, illustrating why a geometrical change can greatly stiffen a sheet without changing its material.

An elastic sheet is a body whose thickness $h$ is small relative to its lateral dimensions and which responds elastically to deformation. Its geometry separates in-plane stretching from out-of-plane bending. A thin elastic plate is a useful model when the assumptions of linear elasticity and small through-thickness strain apply.

The different thickness dependences explain much of a sheet's behaviour. For uniform uniaxial loading with free lateral contraction, Hooke's law gives $\sigma =E\epsilon$. Multiplying by thickness gives the tensile force per unit width $N=Eh\epsilon$: the uniaxial sheet stretching stiffness is $Eh$. For a homogeneous, isotropic thin elastic plate, its bending stiffness iswhere $E$ is Young's modulus and $\nu$ is Poisson's ratio.^[ref]

For an original scaling example, halve $h$ while keeping material and lateral dimensions fixed. The uniaxial tensile stiffness halves, but the bending stiffness falls to one eighth. Thin sheets can therefore become much easier to bend without becoming equally easy to stretch. In simple bending to radius $R$, a layer a distance $z$ from the middle surface has strain approximately $z/R$; choosing $h=0.1\mathrm{mm}$ and $R=10\mathrm{mm}$ gives outer-layer strain about $0.005$, or $0.5\%$.

The 2007 capillary origami study made this competition observable by placing a water droplet on cut polydimethylsiloxane sheets and following their folding during evaporation. Whether a sheet enclosed the drop depended on size, thickness and outline.^[ref] The sheet elastocapillary energy ratio compares the resulting capillary and bending energy scales.

Elastocapillarity is the coupling between surface tension and elastic deformation. Changing an interface can reduce surface energy while bending a compliant solid increases bending energy. Thin structures can deform strongly when the available capillary energy competes with their elastic resistance.

For an isotropic thin elastic plate, its bending stiffness iswhere $E$ is Young's modulus, $h$ the thickness and $\nu$ Poisson's ratio. A sheet bent over a size $L$ has curvature of order $1/L$. Its bending energy scales as $B(1/L^2)L^2\sim B$, while the capillary energy scale is $\gamma L^2$, where $\gamma$ is surface tension. Balancing these scales gives the elastocapillary lengthHere $B$ has units of energy and $\gamma$ energy per area, so the ratio has units of length squared. At fixed material properties, ${\ell }_{\mathrm{ec}}\propto h^{3/2}$; doubling thickness multiplies this length by $2\sqrt{2}$. Geometry and contact angle enter the more detailed wrapping criterion.^[ref]

The 2007 capillary origami experiments of Charlotte Py and collaborators provide a direct demonstration. They cut polydimethylsiloxane membranes, 40–80 micrometres thick, into planar shapes, placed them on a water-repellent support, added a water droplet and recorded the deformation during evaporation. Suitable triangular sheets folded around the shrinking drop into a tetrahedral enclosure. Smaller or stiffer sheets failed to close completely. The observations connected a controlled material thickness and initial shape to the competition between surface energy and bending energy.^[ref]

For a thin elastic plate interacting with a liquid interface, the elastocapillary length isHere $B$ is the plate's bending stiffness, $\gamma$ the surface tension, $E$ Young's modulus, $h$ the thickness and $\nu$ Poisson's ratio. In this plate convention $B$ has units $\mathrm{N}\mathrm{m}$, while $\gamma$ has units $N/m$; their ratio has units of length squared. The quantity is different from the gravity-dependent capillary length.

To see the balance, bend a portion of sheet of lateral size $L$ through an angle of order one. Its curvature is of order $1/L$, so its bending energy scales as $B(1/L{)}^2{L}^2\sim B$. The competing surface energy scale is $\gamma L^2$. Their dimensionless quantity is the sheet elastocapillary energy ratioLarge $\eta$ means that the capillary energy scale is large relative to the bending energy scale. This dimensional analysis does not make $L={\ell }_{\mathrm{ec}}$ a universal threshold for complete closure in capillary origami.^[ref]

For chosen illustrative values $B=10^{-9}\mathrm{N}\mathrm{m}$ and $\gamma =0.05N/m$, the elastocapillary length is $\sqrt{2\times 10^{-8}}\mathrm{m}\simeq 141\mu \mathrm{m}$. A lateral size of $0.1\mathrm{mm}$ gives $\eta =0.5$, whereas $1\mathrm{mm}$ gives $\eta =50$. These are calculations with assumed parameters, not measurements of a particular material.

Two ways of changing size give opposite trends. Keep Young's modulus, Poisson's ratio and surface tension fixed, and let $s=L/L_0$ be a lateral scale factor. At fixed thickness, $B$ stays fixed and $\eta /{\eta }_0=s^2$. If all sheet dimensions shrink in the same proportion, then $h/h_0=s$, $B/B_0=s^3$ and $\eta /{\eta }_0=s^{-1}$. Halving lateral size alone decreases the sheet elastocapillary energy ratio fourfold; halving lateral size and thickness together doubles it. The phrase "smaller sheets" therefore needs a specified thickness rule.

The length can also be calibrated directly: the 2007 study fitted an elastic-loop model to a long polydimethylsiloxane strip held by an ethanol meniscus. For their complete-wetting configuration, loop width was $0.89{\ell }_{\mathrm{ec}}$. The coefficient belongs to that model and geometry.^[ref]

An Euler strut is an initially straight slender elastic column loaded in axial compression. Its lateral stability depends on bending stiffness, length and endpoint constraints. Euler buckling predicts a loss of straight-configuration stability before material failure only in the appropriate slender elastic regime; it is distinct from crushing or plastic yielding of a short column.

A flexural-gravity wave bends a floating elastic plate while displacing the underlying water. For a uniform plate with areal mass $m$ and bending stiffness $D$ above inviscid deep water of mass density ${\rho }_w$, put $\alpha =m/{\rho }_w$ and $\beta =D/{\rho }_w$. Combining the plate equation with the water's kinematic boundary condition gives the displayed dispersion relation. Both gravitational acceleration and plate bending restore displacement; water and plate inertia resist acceleration. It assumes small amplitude, no prestress and negligible shear deformation of the plate.

For a thin plate with areal mass $m$ and bending stiffness $B$, small free transverse disturbances satisfy $m{\eta }_{tt}+B{\nabla }^4\eta =0$. A plane wave of tangential wavenumber $k$ therefore satisfiesThe frequency is quadratic in wavenumber, so these waves are dispersive. Matching this free-wave relation cancels the plate's inertial and bending response in an acoustic transmission problem.

The group velocity of a flexural-gravity wave diverges on the long-wave gravity branch and on the short-wave bending branch, and therefore has a positive minimum between them when the bending stiffness is positive. This minimum bounds the propagation speed of narrow wave packets within the ideal model. It does not by itself determine a wind-generation threshold: steady forcing requires matching phase velocity, while dissipation and coupling determine growth.

For a homogeneous isotropic plate of thickness $h$, Young's modulus $E$ and Poisson's ratio $\nu$, the bending stiffness is $D=E{h}^3/[12(1-{\nu }^2)]$. Its quadratic bending energy isIt follows by integrating the strain energy density $\frac{1}{2}{\tau }_{ij}{e}_{ij}$ through the thickness: linear bending strains are proportional to $z$, and ${\int }_{-h/2}^{h/2}{z}^{2}dz=h^3/12$. Taking a first variation and integrating by parts twice gives $D{\Delta }^{2}w=0$ without applied transverse load. Clamped boundary variations have both zero value and zero normal derivative.

A slender flexible body modeled by a moving curve that transmits tension while its bending stiffness is neglected. For a uniform stretched string with constant tension $T$ and linear mass density $\mu$, small transverse oscillations obey $\mu y_{tt}=T{y}_{xx}$; see the wave equation on a string.