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An active swimming beam has a total internal bending moment $G$ containing passive elasticity and muscular actuation. Neglecting cross-sectional rotary inertia, slice angular momentum balance identifies the internal transverse force with $G_x$. Slice linear momentum balance then gives $G_{xx}=m_b{h}_{tt}-F_{\mathrm{body}}$. In Lighthill elongated-body theory, $F_{\mathrm{body}}=-\mathcal{D}(m\mathcal{D}h)$, where $\mathcal{D}={\partial }_t+U{\partial }_x$. The body and its hydrodynamic added mass therefore contribute with the same sign to the acceleration coefficient. The positive quantity $\mathcal{D}(m\mathcal{D}h)$ is the reaction on the fluid.

At leading order in slope, the hydrodynamic torque on the distal filament is $-\mu K_N{\int }_0^{L}s(Y_t-V)ds$. The displayed opposite torque is required at its base. It is the total internal bending moment; identifying it with the active moment assumes no separate passive elastic moment. If a passive moment is specified, subtract that contribution to determine the required active moment.

Beam cross-section rotation is the signed change in orientation of a material section in the bending plane. The Euler-Bernoulli beam equation assumes that sections remain plane and perpendicular to the deformed centreline. With a consistent angle convention, the section rotation therefore agrees with the centreline tangent angle; for a graph $w(x)$,The last step is a small-angle approximation, with the angle measured in radians. This kinematic relation underlies the axial bending strain, rather than being an extra condition imposed by every support.^[ref]

For an original geometric profile, take $w(x)=a{x}^2$ with $a=0.10{\mathrm{m}}^{-1}$. At $x=0.50\mathrm{m}$, the slope is $0.10$, while the exact tangent angle is $\arctan (0.10)\approx 0.09967$ radians. Replacing it by $0.10$ overestimates the angle by about 0.33 percent. This is a chosen shape illustrating the approximation, not a measured equilibrium profile.

The beam shear angle records a possible difference between centreline slope and section rotation in a model that permits transverse shear strain. The Euler-Bernoulli kinematics set that difference to zero; a short or shear-compliant specimen can require a different model. A simply supported beam leaves its end section rotation free, whereas a clamped boundary condition prescribes it. Neither zero transverse displacement alone nor zero endpoint bending moment requires zero rotation.

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 $I$.

Bending changes the curvature of an elastic beam, rod or sheet. In the elementary beam model, material on opposite sides of the neutral axis stretches and shortens. Their longitudinal normal stresses can balance to give zero resultant force while still transmitting a nonzero bending moment. A bent ruler illustrates why resistance to changing shape cannot be described by axial stretching alone.

For a homogeneous slender elastic beam in small-deflection pure bending, take $x$ along the beam and $y$ perpendicular to its centroidal neutral axis. Let $\kappa$ be the signed centreline curvature. The plane-section kinematics and Hooke's law giveHere $E$ is Young's modulus, taken constant over the section. Choosing the positive bending moment as $M=-{\int }_{A}y{\sigma }_{xx}dA$ yieldsThus the second moment of area $I$, rather than just the cross-sectional area, controls bending stress. The beam flexural rigidity is $EI$. Zero axial resultant requires ${\int }_{A}ydA=0$, locating the neutral axis at the section centroid under these assumptions.^[ref]

A four-point bending test makes the constant-moment region experimentally accessible. Two outer supports carry a specimen while two inner contacts apply the load. The Illinois teaching laboratory uses adjustable inner and outer spans on an Instron 4400 frame and records load and deflection electronically; it also tests notched specimens to examine stress concentration.^[ref]

For an original symmetric loading example, place the outer supports at $x=0,L$ and apply total downward force $P$ equally at $x=a,L-a$, with $0<a<L/2$. Each support reaction is $P/2$. Between the inner contacts the transverse shear force is zero and the bending moment is constant, $M=Pa/2$. This central interval realizes the ideal beam conditions for pure bending away from contact disturbances. A three-point bending test instead has its maximum moment at the single central loading contact.

Choose $L=160\mathrm{mm}$, $a=40\mathrm{mm}$, $P=40\mathrm{N}$, rectangular width $b=10\mathrm{mm}$ and thickness $h=4\mathrm{mm}$. Then $I=b{h}^3/12=53.333\mathrm{m}{\mathrm{m}}^4$ and $M=0.800\mathrm{N}\mathrm{m}$. With stipulated $E=70\mathrm{GPa}$, the model givesThese independently chosen numbers describe an assumed elastic specimen, rather than measurements of a particular material.

The same relation can infer a flexural modulus from a measured bending moment and curvature, provided the geometry and beam assumptions are adequate. Added axial force contributes a uniform longitudinal normal stress and can shift the zero-stress line. Short beams require attention to transverse shear strain; large rotations require a geometrically nonlinear description. The small-deflection Euler-Bernoulli beam equation and the more general Euler elastica cover different levels of that modeling.

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 bending rigidity of an unbonded fibre bundle relates total bending moment to curvature in a stated bending model. If slender material fibres slide longitudinally without appreciable friction, each bends about its own neutral axis. With the same imposed small curvature $\kappa$ and no additional axial tension, strand $i$ contributes $M_i=E_i{I}_i\kappa$, givingHere $I_i$ is that strand's second moment of area about its own bending axis. The underlying linear elasticity relation is the beam moment-curvature law.^[ref] Adding offset-area terms as though all strands shared one neutral axis would impose a different mechanical constraint.

An original two-strand comparison makes the difference quantitative. Take two identical circular strands of radius $r$, with centres vertically separated by $2r$, bending in the plane containing their centres. Each has $A=\pi r^2$ and $I=\pi r^4/4$. In the freely sliding model,For a separate idealization, bond the strands perfectly along their contact so they share a common neutral axis halfway between the centres. Neglect the bond's own area and compliance. Integrating squared distance from that common axis givesFor stipulated $E=2\mathrm{GPa}$ and $r=1\mathrm{mm}$, these are approximately $3142$ and $15708\mathrm{N}\mathrm{m}{\mathrm{m}}^2$. At $\kappa =0.001\mathrm{m}{\mathrm{m}}^{-1}$, the corresponding moments are $3.142$ and $15.708\mathrm{N}\mathrm{mm}$. Both specimens contain the same strand material; their ability to exchange longitudinal force differs.

A physical comparison would impose end rotations, permit the relative end travel required for inter-fibre sliding, and measure bending moment and curvature. Clamping every strand to identical end displacements, adding axial tension or allowing friction can change the response. The factor five above belongs to this two-strand geometry and ideal bond, not to every bundle. The equal-material subdivision model in material fibre makes a different comparison with a solid circular rod.

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

Bending stress is the longitudinal normal stress associated with a bending moment in a beam. In a homogeneous slender elastic beam with small deformation, plane sections remaining plane give longitudinal strain ${\epsilon }_{xx}=-\kappa y$ relative to the neutral axis. Applying Hooke's law gives ${\sigma }_{xx}=-E\kappa y$.^[ref]

Choose the moment convention $M=-{\int }_{A}y{\sigma }_{xx}dA$. ThenHere $I$ is the second moment of area, and the neutral axis passes through the section centroid when axial resultant is zero and $E$ is uniform. The sign tells which side is in tension; the magnitude grows linearly with distance from that axis.

For an original rectangular section with width $b=12\mathrm{mm}$ and thickness $h=6\mathrm{mm}$, $I=b{h}^3/12=216\mathrm{m}{\mathrm{m}}^4$. A moment $M=1.2\mathrm{N}\mathrm{m}$ produces maximum surface magnitudeThe moment is $1200\mathrm{N}\mathrm{mm}$ when using millimetre section dimensions. At fixed $M$ and $b$, doubling $h$ reduces this nominal surface stress fourfold, not merely twofold, because $I$ grows as $h^3$.

These are beam-model stresses away from local load contacts and discontinuities. Added axial force contributes $N/A$ to the longitudinal stress; a notch can produce a stress concentration beyond the nominal distribution. For inhomogeneous sections the neutral-axis position must be obtained from the actual stiffness distribution rather than assumed from geometry alone.

The anisotropic-drag contribution to axial force from a small-slope transverse motion is $F=(c_{\perp }-c_{\parallel })\int y_x{y}_{t}dx$. Using the small-slope elastohydrodynamic filament equation and integration by parts givesThus this propulsive contribution is determined by endpoint slope, bending moment, and shear. A separately imposed axial translation adds ordinary longitudinal drag. Exact second-order longitudinal motion required by inextensibility also contributes to instantaneous total drag, but its time-derivative contribution averages to zero over a deformation cycle.

A beam fixed at one end and free at the other. A transverse end force $F$ gives bending moment $M(x)=F(L-x)$ and, for a rectangular section, maximum bending stress $6FL/(w{h}^2)$ at the fixed end. Longitudinal tension and compression are greatest on opposite outer fibres.

An elastic beam is a slender body that supports bending moments and transverse loads. In the small-slope Euler-Bernoulli approximation its bending moment is $M=EI{y}_{xx}$, and transverse momentum balance gives $\rho A{y}_{tt}+EI{y}_{xxxx}=q$ for a uniform beam under distributed force per length $q$. Here $I$ is the second moment of area; boundary conditions specify the end constraints and applied moments and forces.

A material fibre is a long, slender physical strand: its length is large compared with its transverse dimensions. The geometry is shared by materials with different roles. Carbon fibres and glass fibres can reinforce a composite material, while an optical fibre guides light. The material's properties, the strand dimensions and the way strands are assembled all affect its response.

A single-fibre tensile test grips one strand, measures its axial force and extension over a specified gauge length, and uses its cross-sectional area to obtain stress and strain. A 2020 study used a semi-automated instrument to characterize glass fibres and carbon fibres through their stress-strain curves and inferred moduli.^[ref] Its public dataset supplies individual tensile curves at several gauge lengths, summary sheets and calculations of test-setup machine compliance.^[ref] This matters because crosshead displacement can include deformation of the apparatus as well as extension of the material fibre; its slope alone need not give the specimen's Young's modulus.

An original ideal model shows why making thinner strands changes some responses more strongly than others. Keep total material cross-sectional area $A$, length $L$ and Young's modulus $E$ fixed. Compare one solid circular rod with $N$ identical circular fibres, each of radiusFor axial loading, grip every strand so that all undergo the same extension $\Delta L$. Each supplies force $E(A/N)\Delta L/L$, and adding the forces gives the axial stiffnessindependent of $N$ under these assumptions. This is the uniform-bar linear elasticity relation applied to parallel strands.

Now make a separate pure-bending comparison of an unbonded material-fibre bundle. Allow longitudinal sliding, neglect friction, and let each strand bend about its own neutral axis with the same curvature $\kappa$. A circular strand has second moment of area $I_N=\pi r_N^4/4$ and filament bending modulus $E{I}_N$. Its bending moment is $E{I}_N\kappa$ in the small-deformation model.^[ref] Adding those moments gives the bending rigidity of an unbonded fibre bundle:Each material fibre has its own neutral axis, which is crucial to this calculation. Bonding strands together transfers shear stress and couples their axial deformation; the independent-strand sum then ceases to describe the assembly. In a reinforced composite material, that coupling is part of fibre-matrix load transfer.

For chosen $A=1{\mathrm{mm}}^2$, $L=100\mathrm{mm}$ and $E=70\mathrm{GPa}$, both axial specimens have $K_{\mathrm{ax}}=700\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$. At $N=100$, each radius is approximately $0.0564\mathrm{mm}$, and the independent-bending value is $55.7\mathrm{N}\mathrm{m}{\mathrm{m}}^2$ compared with $5570.4\mathrm{N}\mathrm{m}{\mathrm{m}}^2$ for the solid rod. The parameters are stipulated for this model; $A$ counts material and excludes spaces between strands.

A useful mechanical comparison therefore records which material fibres are gripped, whether they can slide, their arrangement and the loading direction. Equal material quantity alone does not determine an assembly's stiffness, and small filament bending modulus does not imply small axial Young's modulus.

A four-point flexural test bends a specimen using two outer supports and two inner loading contacts. For symmetric loading, the interval between the inner contacts has constant bending moment and zero transverse shear force, approximating pure bending away from contact disturbances. The Illinois laboratory uses adjustable spans and records force and deflection with a mechanical testing frame.^[ref]

Let the support span be $L$, the distance from each support to its nearest loading contact be $a$, and the total downward applied force be $P$. Each loading contact and each support carries force magnitude $P/2$. Static equilibrium givesThe central constant-moment region has length $L-2a$. Specifying whether a force symbol means total force or force at each nose prevents a factor-of-two error.

For an original example, $L=180\mathrm{mm}$, $a=60\mathrm{mm}$ and $P=80\mathrm{N}$ give reactions $40\mathrm{N}$ and central moment $2.4\mathrm{N}\mathrm{m}$. A rectangular specimen with width $b=20\mathrm{mm}$ and thickness $h=5\mathrm{mm}$ has $I=b{h}^3/12=208.33\mathrm{m}{\mathrm{m}}^4$. The nominal maximum bending stress magnitude is $M(h/2)/I=28.8\mathrm{MPa}$.

The apparatus measures force and a chosen displacement; the beam model converts them into stress, strain or flexural modulus. Midpoint displacement and displacement at the inner loading contacts are generally different. Fixture motion can additionally include machine compliance and contact indentation. Record the displacement location and both spans along with specimen dimensions.^[ref]

Unlike a central three-point load, this geometry provides a finite interval of maximum nominal moment. It tests a larger highly stressed region, so flaw distribution and where failure begins can matter when comparing results between fixtures.

Variation of the small-slope bending energy of an elastic filament gives endpoint work $A[h^{(2)}\delta h^{(1)}-h^{(3)}\delta h]$. Arbitrary endpoint displacement and slope variations force the bending moment and transverse force to vanish. These are $h^{(2)}=h^{(3)}=0$. They define a self-adjoint nonnegative fourth-order bending operator on a finite interval; its kernel contains rigid translation and tilt.

For $G_{xx}=R$, zero bending moment and transverse force at the leading end give $G(x)={\int }_0^x(x-s)R(s)ds$. The two trailing-end free-end bending boundary conditions hold exactly when the displayed two integrals vanish. They express total transverse force and torque balance. Adding unknown rigid translation and yaw to a prescribed deformation supplies the two degrees of freedom of the swimming recoil correction. Its principal inertial matrix is the weighted moment matrix of $1$ and $x-x_c$ with weight $m_b+m$, hence is positive definite for a nondegenerate positive mass distribution.

A prescribed initial beam profile may vanish at a pivot while having nonzero endpoint curvature. If the post-release pivot is moment-free, its bending moment condition is $EI{y}_{xx}=0$. The data then cannot have a classical solution with continuous curvature at the initial boundary corners. A finite-energy weak solution can still represent release from preparing end moments. For example $y_0=ax(x-L)$ has odd sine coefficients proportional to $n^{-3}$; $\sum n^4|A_n{|}^2$ converges, giving finite bending energy. The displacement series takes the prescribed initial shape, but termwise endpoint second derivatives cannot be identified with its interior curvature limit. This conservative beam evolution does not have the immediate smoothing of a heat equation.

For a pin-ended slender beam with unstressed profile $y_0$, bending moment depends on curvature change: $M=EI(y-y_0)$. An axial dead load gives $EI(y-y_0)=-Fy$ in the small-slope model.

For a quasistatic cantilever of Kelvin-Voigt material, the bending moment is $M=EI{y}_{xx}+{\eta }_{s}I{y}_{xxt}$. A tip load gives $M=F(t)(\ell -x)$, so the free-tip displacement obeys the displayed equation with $\tau ={\eta }_s/E$. A step load gives a finite asymptotic deflection. If inertia is retained, the interior equation is $\rho A{y}_{tt}+EI{y}_{xxxx}+{\eta }_{s}I{y}_{xxxxt}=0$, with the corresponding moment and shear conditions at the tip.

For a functional $J[y]=\int F(x,y,y^{\prime },y^{\prime \prime })dx$, two applications of integration by parts leave the boundary termIf the endpoint displacement is fixed but its slope is free, $\delta y=0$ while $\delta y^{\prime }$ is arbitrary. The first variation therefore forces $F_{y^{\prime \prime }}=0$ at that endpoint. For the beam energy density $F=\beta (y^{\prime \prime }{)}^2/2-\alpha (y^{\prime }{)}^2/2$ with $\beta >0$, this gives $y^{\prime \prime }=0$, the zero-bending moment condition. A clamped endpoint instead prescribes the slope and has $\delta y^{\prime }=0$.