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For independent sliding filaments in a filled bundle, bending energies add and $B\sim NE{r}^4\sim E{r}^2{R}^2$. A coherently bonded one-filament-thick tube has $B\sim Er{R}^3$, while a bonded filled rod has $B\sim E{R}^4$. The same torsional scalings apply with the shear modulus instead of Young's modulus if individual sliding filaments are forced to undergo a common material twist. Completely free filament spin/slip does not define that torsional response: collective helical bending alone begins at fourth order in the twist rate.

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.

The ideal rubber elasticity relation infers active strand number density from the shear modulus. If a quoted modulus is instead Young's modulus, nearly incompressible behavior gives ${\nu }_s\simeq E/(3k_{B}T)$. This is not automatically the junction density: for junction functionality $f$ and no dangling ends or loops, counting strand ends gives ${\nu }_s=f{\nu }_j/2$. Network fluctuations and inactive strands modify the modulus, so converting it into a chemical cross-link count requires a network model.

An adhesive sphere-on-flat contact approximation combining Hertzian contact deformation with an attractive force outside the contact area. It is most appropriate for sufficiently stiff, small-radius contacts with a small Tabor parameter; a fitted reduced modulus is not automatically the sample Young's modulus.

The effective elastic thickness is the thickness of a homogeneous elastic plate with the same effective bending stiffness as the loaded lithosphere. In an isotropic thin-plate model, $D=E{T}_e^3/[12(1-{\nu }^2)]$, with Young's modulus $E$ and Poisson's ratio $\nu$. It depends on thermal state, material creep, loading history and yielding, and is not generally the crust's thickness or the full thermal thickness.

A coefficient relating an appropriate stress and strain in the linear elastic regime. Young's modulus, shear modulus and bulk modulus describe different deformation modes.

An eigenstrain is the deformation a material would prefer locally without applied stress. In a one-dimensional linear elasticity model, let $E$ be Young's modulus, take tension as positive stress, and write $\sigma =E(\epsilon -{\epsilon }^{\mathrm{mag}})$. A free rod has $\sigma =0$ and follows $\epsilon ={\epsilon }^{\mathrm{mag}}$. A perfectly blocked rod has $\epsilon =0$ and $\sigma =-E{\epsilon }^{\mathrm{mag}}$. Thus positive preferred elongation produces compressive stress when blocked.

For stipulated $E=100\mathrm{GPa}$, cross-sectional area $A=1\mathrm{m}{\mathrm{m}}^2$, and ${\epsilon }^{\mathrm{mag}}=20\times 10^{-6}$, blocking gives $\sigma =-2\mathrm{MPa}$ and axial force $F=\sigma A=-2\mathrm{N}$. This idealization ignores changes of magnetization caused by the load itself. A real magnetostrictive actuator therefore needs both magnetic and mechanical constitutive information: a free-length measurement alone does not determine its loaded motion.

Under uniform compressive pressure $p$, isotropic Hooke's law gives each normal strain as $-p(1-2\nu )/E$. The volume strain is three times this, so the bulk modulus is $B=E/[3(1-2\nu )]$, where $E$ is Young's modulus and $\nu$ is Poisson's ratio. Positivity of $E$ and finite $B$ gives $\nu <1/2$. Positive shear modulus $G_s=E/[2(1+\nu )]$ adds the lower bound $\nu >-1$. Negative Poisson's ratio is allowed; the bulk relation alone does not supply the lower bound.

Linearity and reversibility answer different questions. The proportional limit ends the interval in which axial stress is proportional to axial strain. Elastic deformation means that deformation is recovered after unloading; a nonlinear response can still be elastic. Conversely, a short straight portion of a loading curve does not by itself demonstrate recovery.

To check recovery, unload and examine the local extensometer strain relative to its initial value, allowing for instrument resolution, drift and any time dependence. A residual extension can indicate plasticity in continuum mechanics, but grip slip can produce a residual in crosshead displacement without the same local residual. Compare the channels before assigning the effect to the specimen. Yield point concerns the onset of irreversible deformation; a specified offset proof stress is an operational convention rather than Young's modulus. For example, a $0.2\%$ offset proof stress is the intersection of the loading curve with $\sigma =E(\epsilon -0.002)$, using the correctly referenced strain. Report the fitted stress/strain window and the unloading evidence alongside the modulus, so that apparatus seating, departure from linearity and onset of plasticity in continuum mechanics remain distinguishable.

Machine compliance is the extra load-path displacement per applied force attributed to the apparatus in a specified measurement model. Its units are $\mathrm{m}{\mathrm{N}}^{-1}$, the reciprocal units of a spring constant. Here $C_m$ includes the frame, load cell, adapters and grips insofar as their deformation contributes to the recorded crosshead displacement $x$. It is a property of that configuration and readout, rather than a universal constant of the testing machine.

Assume a uniform specimen with initial length $L$, constant $A_0$ and Young's modulus $E$, small axial elastic deformation, quasistatic loading, no slip, and a linear, length-independent apparatus response over the chosen force window. Equilibrium makes the same force act through specimen and apparatus, while their extensions add. The two springs in series therefore giveThe offset $x_0$ describes the displacement origin after seating; the slope $q$ describes total compliance. Fitting an intercept avoids forcing an arbitrary zero through the origin, but cannot repair slip or a curved response.

If all of $x-x_0$ is incorrectly attributed to specimen extension, the inferred modulus isFor positive $C_m$, this is too small. The fractional bias is most serious for a short, thick, stiff specimen because its own compliance $L/(E{A}_0)$ is small. A local extensometer instead has $d(\Delta L_g)/dF=L_g/(E{A}_0)$ in this ideal model, excluding deformation outside its contacts.

A shaped specimen requires more care. With varying initial cross-sectional area $A(s)$ and uniform $E$, its axial compliance is $\int ds/[EA(s)]$ in the one-dimensional approximation. Shoulder and end deformation must be included explicitly or remain identical between tests and be absorbed into an effective intercept. Changing only the extensometer gauge length on an unchanged specimen does not change its total compliance and cannot perform a two-length machine compliance separation.

Healthy cells use mechanical feedback to match growth, shape and movement to tissue demands; tumours can amplify or uncouple that feedback. Relevant inputs include extracellular resistance, tensile load, compression, fluid shear and confinement. The response depends on where a force is applied and how it is transmitted, rather than on a single scalar “stiffness” measurement.

A biological cell actively probes its surroundings through actomyosin contractility. myosin pulls on actin, and integrins couple the network to the extracellular matrix through focal adhesions. When the surroundings resist displacement, load can build across the linkage. cadherins similarly connect neighbouring cells, while the cytoskeleton and LINC complex transmit some loads to the cell nucleus. These coupled structures support mechanotransduction: conversion of a mechanical perturbation into altered molecular binding, signaling or gene expression.

One physical mechanism is force-dependent exposure of binding sites. In experiments combining single-molecule manipulation and imaging, stretching talin exposed otherwise inaccessible sites for vinculin.^[ref] Recruitment can strengthen the connection to actin, changing subsequent force transmission. A minimal two-state model illustrates why a small load can have a large effect. If opening lengthens a domain by $\Delta x$, constant tensile force $F$ changes its opening free energy to $\Delta G(F)=\Delta G(0)-F\Delta x$. Under equilibrium assumptions,The work $F\Delta x$ has units of energy; the model predicts an increased open-state population when pulling favors extension. This is an illustrative thermodynamic explanation, not a fit to a living adhesion or proof that every force-sensitive bond strengthens under load.

Mechanical context can alter cell fate. Micropatterned adhesive substrates controlling the spreading of human mesenchymal stem cells showed different differentiation outcomes, with RhoA and contractility perturbations linking cell geometry to the response.^[ref] Geometry, adhesive ligand density and substrate stiffness are related but distinct variables. To attribute a result to stiffness, keep ligand presentation and surface chemistry matched; to test spreading, constrain cell area independently. In healthy tissues, such feedback helps coordinate differentiation, wound closure, collective migration and adaptation to changing loads.

Nuclear transcriptional regulators integrate these inputs. YAP and TAZ changed localization and activity with matrix resistance, cell spreading and cytoskeletal organization in controlled culture experiments.^[ref] Nuclear enrichment is a useful readout, but does not by itself demonstrate the downstream program or a universal causal pathway. Measure target transcription and phenotype, then test loss and rescue of the regulator. hippo signaling and other biochemical pathways intersect with mechanical regulation; the balance is cell-type- and context-dependent. Force can additionally gate mechanosensitive ion channels, changing calcium entry, and alter nuclear shape or transport. Those mechanisms need direct channel or nuclear measurements rather than inference solely from a transcriptional endpoint.

The normal feedback is bidirectional. Cells deposit and remodel extracellular matrix, and their newly produced matrix changes later mechanical input. During repair, that loop can transiently stabilize a wound; persistent activation can instead produce fibrosis. An external substrate's Young's modulus is not the same as the force within a particular adhesion: cell contractility, geometry, matrix architecture and loading rate determine the force actually transmitted. traction force microscopy, calibrated molecular tension sensors and local mechanical measurements constrain different parts of this relationship.

In tumours, excessive matrix deposition and collagen cross-linking can raise tissue resistance, promote sustained integrin signaling and favor invasive behavior. In mammary tumour models, manipulation of lysyl oxidase-dependent matrix cross-linking linked matrix mechanics to integrin signaling and progression.^[ref] This goes beyond comparing a stiff tumour with softer normal tissue, but interventions can still change matrix chemistry and growth-factor presentation along with mechanics. Reconstitution with defined matrices and independent control of signaling are needed to distinguish those contributions. Increased stiffness can then stimulate further contractility and remodeling, creating a positive feedback loop.

Growing tumours also generate solid stress and elevated interstitial pressure. Compression can distort neighbouring tissue and impair perfusion, producing heterogeneous oxygen and nutrient delivery.^[ref] A stiff matrix and high compressive stress can coexist; they are not interchangeable. Spatial differences matter: an invasive edge may experience tension and aligned matrix fibres while the interior is compressed. Loss of normal polarity, cell–cell adhesion or growth inhibition allows cells to respond to these environments differently from their healthy counterparts. These consequences do not imply that every isolated tumour cell is intrinsically stiffer; tissue mechanics can be dominated by matrix and collective organization.

Cancer-associated changes in RhoA, adhesion signaling or transcription can also alter how cells generate or interpret force. A mechanically responsive program that normally supports regeneration can become sustained inappropriately, while weakened junctional feedback can permit growth despite crowding. To test causality, vary a mechanical input before a response occurs, perturb a candidate sensor, and ask whether downstream rescue restores the response. Monitor proliferation, apoptosis and cell-cycle state because each can change both mechanics and the measured phenotype. Three-dimensional matrices should be examined alongside flat substrates: confinement, viscoelastic relaxation and degradability can change the outcome even at similar elastic moduli.

The imbalance is a failure of regulated mechanical homeostasis: altered matrix, force generation and signaling reinforce one another, while normal spatial and growth constraints lose control. The clearest experiments separate applied mechanics from ligand chemistry, identify a force-transmitting molecular step and link it to a physiological output.

With $w$ fixed, bending strength requires $h=\sqrt{6FL/(w{\sigma }_f)}$. Substitution into the cantilever mass givesThe material performance index to maximize is thereforeChanging the free geometric variable changes the strength exponent. A stiffness-limited problem would instead involve Young's modulus; the supplied modulus does not enter this strength-only index.

Check stiffness and tip deflection using Young's modulus, failure from existing cracks using fracture toughness or the supplied critical energy-release rate, local crushing at supports, shear failure, buckling, fatigue and joint design. In a simple linear-elastic crack model, $K_c^2\simeq E^{\prime }{G}_c$, so toughness cannot be inferred from strength alone. Wood also has grain-direction dependence, moisture sensitivity, natural defects, property scatter and time-dependent creep; the design must control grain alignment and compression-side stability. Metals raise issues of corrosion, mass, joining, machining, residual stress and heat-treatment state. Cost, availability, reproducibility, service temperature and end-of-life treatment can decide between otherwise feasible designs. A high merit index is a screening result, not a complete safety specification.

The second moment of area about an axis weights each element of cross-sectional area by its squared perpendicular distance from that axis. The ellipse extends farther along ${\mathbf{d}}_2$ than along ${\mathbf{d}}_1$, so its material lies, on balance, farther from the ${\mathbf{d}}_1$ axis. ThereforeWith zero intrinsic curvature of an elastic filament, the respective bending moments required by the constitutive equation are $E{I}_1\kappa {\mathbf{d}}_1$ and $E{I}_2\kappa {\mathbf{d}}_2$. Thus their magnitudes satisfyfor $\kappa \ne 0$: bending about the minor axis is harder. Here $E$ is Young's modulus, and $E{I}_i$ is the corresponding bending stiffness.

Write $B$ for the filament bending modulus and $C$ for the filament torsional stiffness, so that the energy per unit length is $(B{\kappa }^2+C{\Omega }^2)/2$. For one circular filament,where $I_1$ is the second moment of area, $J_1$ the polar moment, $G_s$ the shear modulus, and $\nu$ Poisson's ratio. Numerical twisting constants cannot be fixed by Young's modulus alone without specifying $\nu$.

For a filled sliding bundle, $N\sim (R/r{)}^2$. Filaments share the centerline curvature but slide instead of accumulating the additional axial strains of a bonded cross-section. Their bending energies add: $B\sim N{B}_1\sim E{r}^2{R}^2$. If end constraints impose the same material twist on every filament, their torsional energies also add: $C\sim G_s{r}^2{R}^2$.

For a bonded, one-filament-thick circular tube, the wall thickness $h$ is of order $r$. Its moments are $I\sim \pi R^{3}h$ and $J_p\sim 2\pi R^{3}h$, giving $B\sim Er{R}^3$ and $C\sim G_{s}r{R}^3$. For a bonded filled rod, $I\sim \pi R^4/4$ and $J_p\sim \pi R^4/2$, giving $B\sim E{R}^4$ and $C\sim G_s{R}^4$. Packing fractions and the convention for tube radius affect constants, not these powers.

There is an important qualification to bending and torsion of filament bundles: free axial sliding does not specify whether each filament can also spin freely. If both spin and slip are unconstrained, the common-material-twist calculation is not forced by the geometry. A helix of small twist rate has centerline curvature of order $\rho {\Omega }^2$, so its bending energy starts at ${\Omega }^4$, not ${\Omega }^2$. A nonzero linear collective torsional stiffness then requires additional end, contact, or spin constraints.

For a three-dimensional worm-like chain, the tangent correlation length is $l_p=B/(k_{B}T)$, where $B$ is the filament bending modulus. The measured persistence length therefore givesRepresent the actin filament as an isotropic solid circular cylinder with diameter $d=10^{-9}\mathrm{m}$. Its second moment of area is $I=\pi d^4/64=4.909\times 10^{-38}{\mathrm{m}}^4$. Since $B=EI$,This is an effective Young's modulus within the given continuum cross-sectional model; the numerical diameter and persistence length must not be replaced by unrelated typical biological values.

The same measurement corrects the independent-link approximation: the Kuhn length is $b_K=2l_p=40\mathrm{nm}$, with $N_K=L/b_K=2.5\times 10^9$. Thus the long-chain entropic stiffness is $3k_{B}T/(2l_{p}L)=3.105\times 10^{-15}\mathrm{N}{\mathrm{m}}^{-1}$, and the root-mean-square end-to-end distance is $2\mathrm{mm}$. For comparison, the material axial stiffness is $EA/L=1.325\times 10^{-11}\mathrm{N}{\mathrm{m}}^{-1}$, using $A=\pi d^2/4$. The corrected coil response remains much softer than axial backbone stretching.

Assume small-amplitude motion, a thin uniform elastic plate, no prestress, and inviscid, irrotational deep water. Write the upward displacement as $\eta (x,t)$, the ice mass per unit area as $m={\rho }_{i}h$, and its flexural rigidity aswhere $E$ is Young's modulus and $\nu$ is Poisson's ratio. The plate equation is $m{\eta }_{tt}+D{\eta }_{xxxx}=p$, where $p$ is the perturbation of upward water pressure.

Let the water velocity potential be $\varphi =A{e}^{kz}{e}^{i(kx-\omega t)}$ below the mean surface $z=0$, with $k>0$. This solves the Laplace equation and decays at depth. For $\eta ={\eta }_0{e}^{i(kx-\omega t)}$, the kinematic boundary condition ${\eta }_t={\varphi }_z$ gives $A=-i\omega {\eta }_0/k$. The linear Bernoulli equation givesEliminating $p$ proves the dispersion relation of a floating elastic plate:Thus the desired wavenumber is the positive root ofFor positive $D$ and $\omega$, this root is unique: the polynomial is negative at zero and positive eventually; if its derivative starts negative, that derivative crosses zero only once, at a negative-valued minimum before the positive root. If a finite water depth $d$ is required, the corresponding relation is ${\omega }^2=(gk+D{k}^5/{\rho }_w)\tanh (kd)/[1+({\rho }_{i}h/{\rho }_w)k\tanh (kd)]$.

Put $\beta =D/{\rho }_w$ and $\alpha ={\rho }_{i}h/{\rho }_w$. Differentiation gives the group velocityFor long periods, the surface gravity wave limit gives $c_g\sim gT/(4\pi )$. At sufficiently short periods, plate inertia and bending dominate, $\omega \sim \sqrt{D/m}{k}^2$ and $c_g\sim 2\sqrt{D/m}k$. Consequently the curve has a group-velocity minimum of a flexural-gravity wave between branches that rise toward both long and short periods, within the formal deep-water plate model. Very short waves eventually fall outside the thin-plate approximation.

The plot is an original model illustration with $h=1\mathrm{m}$, $E=5\mathrm{GPa}$, $\nu =0.3$, ${\rho }_i=917\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_w=1025\mathrm{kg}{\mathrm{m}}^{-3}$, and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. Its two minima occur at different periods.

A wave packet near the group-speed minimum carries energy away relatively slowly, so sustained forcing can concentrate energy in that period band. However, wind generation cannot be decided from $c_g$ alone. A steadily translating pressure pattern resonates through its phase velocity $c_p=\omega /k$; it must reach the minimum phase speed of a flexural-gravity wave. Actual wind-wave growth also requires suitable pressure coupling and enough energy input to overcome damping. A coherent cover suppresses ordinary free-surface wave generation, while incoming ocean swell can still excite bending and break the cover.

For disconnected pancake ice or frazil ice, remove the assumption of coherent elastic bending and retain the effective mass loading as a first conservative approximation. Setting $D=0$ gives ${\omega }^2=gk/[1+({\rho }_{i}h/{\rho }_w)k]$. To describe damping as well, replace the elastic layer by a viscous suspension and match velocity and stress to the water beneath: the resulting viscous layer model of frazil-pancake ice has complex $k$, whose real part gives dispersion and imaginary part gives attenuation. Keller's 1998 two-layer model implements this replacement. Effective viscosity and connectedness must reflect the particular cover; zero rigidity alone does not model attenuation.

Write $w(x,t)$ for the vertical displacement of the sea ice, and take the undisturbed water surface as $z=0$, with water occupying $z<0$. The elastic plate has areal mass $m={\rho }_{i}h$ and bending stiffnessHere $E$ is Young's modulus and $\eta$ is Poisson's ratio. We neglect in-plane prestress, viscosity and plate shear deformation, and linearize about hydrostatic equilibrium. These are important assumptions: perfect elasticity alone does not specify every term in a floating-plate model.

For a plane wave $w=\hat{w}{e}^{i(kx-\omega t)}$, $k>0$, potential flow in deep water has velocity potential $\varphi =\hat{\varphi }{e}^{kz}{e}^{i(kx-\omega t)}$. This solves Laplace's equation and decays downwards. The kinematic boundary condition gives $k\hat{\varphi }=-i\omega \hat{w}$. Linearizing the water pressure at the displaced interface gives an upward excess loadThe elastic plate equation is $m{w}_{tt}+D{w}_{xxxx}=p$. Substitution and multiplication by $k$ therefore give the flexural-gravity wave dispersion relationEquivalently, with $\alpha ={\rho }_{i}h/{\rho }_w$ and $\beta =D/{\rho }_w$,This determines the positive wavenumber implicitly for prescribed positive angular frequency. It is unambiguous: the derivative of ${\omega }^2$ is $(g+5\beta k^4+4\alpha \beta k^5)/(1+\alpha k{)}^2>0$, while ${\omega }^2$ runs from zero to infinity.

The phase velocity and group velocity areFor open-water deep-water gravity waves, ${\omega }^2=gk$, and hence $c_p=gT/(2\pi )$ and $c_g=gT/(4\pi )$. At large period the ice-covered curves approach these straight lines. At shorter period, plate bending raises the speeds, so both ice-covered curves turn upward as period decreases. In the bending regime with negligible plate inertia, $\omega \propto k^{5/2}$ and $c_g\simeq (5/2)c_p$; in the formal plate-inertia-dominated limit, $\omega \propto k^2$ and $c_g\simeq 2c_p$. The latter extrapolation eventually leaves thin-plate validity and should not be read as a prediction at arbitrarily small wavelength.

The plotted parameters are illustrative rather than measured at the observation site. They give a group-velocity minimum of a flexural-gravity wave near $15.1\mathrm{m}{\mathrm{s}}^{-1}$ at period $17.6\mathrm{s}$. There is no arbitrarily slow wave-energy branch under a continuous elastic sheet. Energy put into a localized disturbance travels away at at least this minimum group velocity; a slowly moving wind system cannot retain a wave packet indefinitely beneath itself. This reduces the opportunity for sustained local growth compared with slow, short open-water waves, and the continuous cover also prevents direct wind forcing of an exposed water surface. Incoming long swell can still propagate.

A group-velocity minimum of a flexural-gravity wave is not by itself a universal minimum wind speed for wave generation. A steadily translating forcing pattern requires a phase velocity matching its translation speed, so the minimum of $c_p$, not $c_g$, supplies the corresponding resonance threshold. Random wind forcing, dissipation and aerodynamic coupling must be specified before making an absolute generation claim.

Take $h$ positive downwards and write $B$ for the filament bending modulus. Under the material-modulus interpretation of the printed $A$ as Young's modulus, the circular cross-section gives the second moment of area and bending modulusIndeed a small curvature $\kappa$ gives axial strain $-y\kappa$, so integrating $A{y}^2{\kappa }^2/2$ over the section gives bending energy $B{\kappa }^2/2$ per unit length. Keeping this distinction is essential to the powers of radius in the answer.

After allowing for buoyancy, the effective downward force per unit length isThe filament loses potential energy as it moves down; the displaced fluid's hydrostatic contribution is included through the density difference. For an inextensible filament with arc-length coordinate $s$, the geometric bending and gravitational energy is ${\int }_0^L[B\kappa (s{)}^2/2-wh(s)]ds$, up to a constant. For the small-slope Monge representation this reduces toThe upward-positive convention reverses the sign of $h$ and the gravitational term together. There is no stretching contribution at this order when bending a slender filament without imposed axial tension. The wording does not explicitly distinguish a material elastic modulus from a flexural modulus: if $A$ instead denotes the filament bending modulus in the intended convention, set $B=A$ throughout; then the $\pi a^4/4$ factor is already included in $A$.

Use the downward-positive displacement and the filament bending modulus $B$ and load density $w$ defined above. Two applications of integration by parts give the first variationThus the elastic-plus-gravitational force density is $-\delta E/\delta h=-B{h}_{xxxx}+w$. In resistive-force theory, transverse motion has drag per unit length $-{\zeta }_{\perp }{h}_t$. Neglecting filament and fluid inertia, instantaneous force balance gives the forced small-slope elastohydrodynamic filament equationClamping fixes position and tangent, while the free end has zero bending moment and zero transverse shear force. ThereforeThe last two conditions also follow as natural boundary conditions from the variation, since the free-end values of $\delta h$ and $\delta h_x$ are arbitrary. Distributed weight does not add a concentrated force or moment at the tip.

At steady state, $B{h}_s^{\prime \prime \prime \prime }=w$. Integrating first from the free end gives $h_s^{\prime \prime \prime }=(w/B)(x-L)$ and $h_s^{\prime \prime }=(w/2B)(x-L{)}^2$. Integrating twice more with the clamp conditions gives the uniform-load bending of a clamped filament:In particular the free-end displacement is $\delta =h_s(L)=w{L}^4/(8B)$ and the total buoyancy-corrected gravitational load is $F=wL$. Thus the tip stiffness of a uniformly loaded cantilever isThe last expression uses $A$ as Young's modulus; if $A$ itself is flexural rigidity, the same result reads $k_{\mathrm{eff}}=8A/L^3$. This is Hooke's law for the specified loading pattern. It is not the stiffness for a point force at the tip, which would be $3B/L^3$; replacing a distributed force by its resultant does not preserve its bending moment distribution.

For dimensional analysis, $[A]=$ force/length squared, $[I]=$ length to the fourth power, hence $[B]=$ force times length squared. Consequently $[B/L^3]=$ force/length, as required for a spring constant. Also $[{\zeta }_{\perp }]=$ force times time/length squared makes ${\zeta }_{\perp }{h}_t$, $B{h}_{xxxx}$ and $w$ all forces per unit length. Neither viscosity nor density enters the stiffness, although viscous drag controls the rate of approach to the steady shape. The small-slope approximation requires $w{L}^3/B\ll 1$, since $h_s^{\prime }(L)=w{L}^3/(6B)$.