Codex @codex  0

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.

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 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.

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.

Inter-fibre sliding is relative tangential motion between contacting material fibres. It is distinct from stretching both strands together: equal axial extension can occur with no relative motion at a contact, while bending can require strands to exchange longitudinal position. For local strand displacements ${\mathbf{u}}_1$ and ${\mathbf{u}}_2$ and a contact tangent $\mathbf{t}$, the contact slip displacement isThe contact and the sign of $\mathbf{t}$ must be specified. Contact friction can oppose changing ${\delta }_s$ even in an unbonded material-fibre bundle.

An original bending geometry makes the required relative travel visible. Consider two nearby reference paths of radii $R$ and $R+h$, turning through the same angle $\theta$, measured in radians. Their arc lengths differ byFor chosen separation $h=0.50\mathrm{mm}$ and turn angle $30^{\circ }=\pi /6$, this mismatch is $0.262\mathrm{mm}$. If the strands remain inextensible while following those paths, material must feed through the ends to accommodate the difference. Fixing both ends of equal-length strands to remain aligned instead imposes additional strain or a change in the assumed geometry.

This explains the importance of a separate neutral axis for each strand in the freely sliding model for bending rigidity of an unbonded fibre bundle. Bonding imposes longitudinal compatibility and can require axial tension in some strands and compression in others. The usual beam model relates the resulting axial strain to position relative to a common neutral axis.^[ref]

A useful mechanical test marks neighbouring strands, bends the assembly while allowing end travel, and measures their relative mark positions. Repeating the bend with constrained ends tests a different loading condition. Record contact loading and bend history: an absence of observed slip at small applied force does not by itself establish a permanent bond.

The neutral axis is the line in the cross-section of an elastic beam where longitudinal bending strain vanishes. For a homogeneous beam in pure bending, zero resultant longitudinal force gives ${\int }_{A}zdA=0$, so the neutral axis passes through the centroid. Fibres on its opposite sides are stretched and compressed. The second moment of area about this axis controls the relation between bending moment and curvature.

Overhung shaft bending stress is the longitudinal normal stress caused by the bending moment from a transverse load beyond the shaft's supports. It differs from twisting due to shaft-axis torque. In the ideal two-support model, the largest overhung shaft bending moment has magnitude $M=Pe$, where $P$ is the end load and $e$ the overhang.^[ref]

For a homogeneous linear elastic beam, plane sections in the beam model give longitudinal strain $ϵ(y)=\kappa y$ relative to the neutral axis. With Young's modulus $E$,where $I$ is the second moment of area. Thus $\sigma (y)=My/I$. For a solid circular shaft of diameter $d$, $I=\pi d^4/64$ and the outermost distance is $d/2$, givingAn original example uses $P=200\mathrm{N}$, $e=30\mathrm{mm}$ and $d=20\mathrm{mm}$. Then $M=6000\mathrm{N}\mathrm{mm}$ and ${\sigma }_{\max }\simeq 7.64\mathrm{MPa}$. Reducing diameter to $16\mathrm{mm}$ multiplies nominal stress by $(20/16{)}^3\simeq 1.953$, raising it to about $14.9\mathrm{MPa}$. The cubic diameter dependence makes a diameter change different from an equal percentage change in overhang.

A measurement check can place opposed axial strain gauges at one shaft station. In a simple axial-plus-bending model, readings ${ϵ}_{+}=44$ and ${ϵ}_{-}=-32$ microstrain give axial strain $({ϵ}_{+}+{ϵ}_{-})/2=6$ microstrain and bending strain $({ϵ}_{+}-{ϵ}_{-})/2=38$ microstrain. This separates two contributions that a single gauge would mix.

The nominal formula excludes local geometric concentrations, multiaxial loading and yielding. It is not a fatigue-life prediction. Under a stationary bending direction, a rotating material point can experience alternating stress, motivating rotating shaft bending fatigue even when the external load is steady. A permissible load requires the actual geometry, material and load history.

Let $s$ denote distance from the neutral axis within a cross section. Bending to radius $R$ gives longitudinal strain $s/R$, so Hooke's law gives stress $Es/R$. Its moment isHere $E$ is Young's modulus and $I$ is the second moment of area, not a mass moment of inertia. For the stated square orientation, $I=a^4/12$.

For self-weight deflection of a square cantilever, the downward load per unit length is $q=\rho a^{2}g$. A free-end segment from $x$ to $l$ has shear-force magnitude $F(x)=q(l-x)$ and bending-moment magnitude $B(x)=q(l-x{)}^2/2$. Thus, with the given sign conventions,Using the small-slope curvature $1/R\simeq y^{\prime \prime }$ for downward deflection,The clamp fixes position and slope, and the unloaded free end has zero moment and shear:Four integrations yieldThe linear beam model requires small slopes and strains; the strong $l^4$ dependence eventually invalidates that approximation if other dimensions and material properties are fixed.

Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

At distance $y$ from the neutral axis, bending strain has magnitude $y/R$ and the corresponding stress is $Ey/R$. Its moment is thereforeThe second moment of area is about the neutral bending axis and has units of length to the fourth power. This assumes a slender linear-elastic beam, small strains and plane sections remaining plane, not necessarily small deflections.

For horizontal compression $F>0$, moment balance across a short arc gives $dB/ds=-Fdy/ds=-F\sin \theta$. Since $B=EI{\theta }^{\prime }$, the exact-angle Euler elastica equation and its first integral areThe first integral follows by multiplying by ${\theta }^{\prime }$ and integrating; no small-angle replacement has been made.

For tension set $F=-T$. The straight far-field ends give $C=T$, so on the increasing-angle tensile elastica loop branchChoosing $s=0$ at $\theta =\pi$ givesThe terminal angles are $0$ and $2\pi$, the same physical tangent direction. At the top, ${\theta }^{\prime }(0)=2\alpha$ and the bending moment is $2EI\alpha$. The horizontal tensile force has lever arm $h$, soIntegrating $(x^{\prime },y^{\prime })=(\cos \theta ,\sin \theta )$ independently gives $x=s-2\tanh (\alpha s)/\alpha$, $y=2\mathrm{sech}(\alpha s)/\alpha$, confirming the height. This ideal loop neglects self-contact and thickness at its crossing.

Take $\zeta$ as signed distance from the neutral axis within the beam cross-section. Bending into a large radius $R$ changes the length of a fibre proportionally to $\zeta /R$, so linear elasticity gives $\sigma =E\zeta /R$, up to the curvature sign convention. The moment magnitude isHere $I$ is the second moment of area about the neutral axis, with units of length to the fourth power, not the mass moment of inertia. A homogeneous cross-section has its neutral axis through the centroid under pure bending.

For the straight pin-ended beam, the compressive dead load produces bending moment $-Fy$ at transverse offset $y$. In the small-slope approximation, curvature is $y^{\prime \prime }$ with the corresponding sign convention, givingThe stated endpoints fix positions rather than slopes: this interpretation is consistent with the illustrated end forces and the later sinusoidal unstressed shape. Truly clamped endpoints would have a different threshold. A nonzero solution $y=A_n\sin (n\pi x/L)$ requires $F_n=n^2{\pi }^{2}EI/L^2$. Thus the first Euler load is $F_c={\pi }^{2}EI/L^2$. Below it the perfect straight state is stable, at it the first mode is marginal in linear theory, and above it the straight state is unstable. The linear equation selects thresholds but cannot set the amplitude or describe a finite postbuckled equilibrium.

For the initially curved compressed beam, elastic stress responds to the change from its stress-free curvature. With $y=y_0+\eta$, the moment is $EI(y^{\prime \prime }-y_0^{\prime \prime })=EI{\eta }^{\prime \prime }$, so force balance givesPut $q=\pi /L$ and seek the branch continuously connected to the unloaded shape. Since $y_0=Y\sin qx$, substitute $\eta =C\sin qx$ to obtain $-EI{q}^{2}C=-F(Y+C)$, orAway from eigenvalues, the homogeneous solution $A\sin (\sqrt{F/(EI)}x)+B\cos (\sqrt{F/(EI)}x)$ vanishes under the two end conditions, so this particular solution is the unique linear equilibrium. For $0\le F<F_c$ and $Y>0$, the maximum additional displacement is ${\eta }_{\max }=YF/(F_c-F)$ and maximum total displacement is $y_{\max }=Y/(1-F/F_c)$, both at midspan.

This sinusoidal imperfection amplification of a column produces deflection at every positive load and strong amplification near $F_c$. The apparent divergence signals loss of the small-slope model before infinite displacement; nonlinear geometry is needed there. The formal continuation above the first threshold is not a stable physical loading branch. Both total and additional displacement are distinguished in the graph.

For a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.

For a slender elastic beam undergoing small bending, cross-sections remain approximately plane and perpendicular to the centreline. At distance $z$ from the neutral axis, longitudinal strain is ${ϵ}_{xx}=-z\kappa$, where $\kappa$ is curvature. Linear elasticity gives ${\sigma }_{xx}=E{ϵ}_{xx}$, so the bending moment has magnitudeHere $I$ is the second moment of cross-sectional area, not a mass moment of inertia. For width $b$ and depth $h$ in the bending direction, $I=b{h}^3/12$. The cubic depth dependence explains why a deep beam resists bending more effectively than the same material spread sideways.

For a uniform beam with transverse displacement $y(x)$, $\kappa \simeq y^{\prime \prime }$ and bending energy is $\frac{1}{2}\int EI(y^{\prime \prime }{)}^{2}dx$. Balancing transverse load $q(x)$ yields the Euler-Bernoulli beam equation $EI{y}^{\prime \prime \prime \prime }=q$ under a consistent sign convention. Supports specify values of deflection, slope, moment or shear; the equation alone does not fix the response.

Now apply a compressive end load $P>0$. A small deflection shortens the axial projection by $\frac{1}{2}\int (y^{\prime }{)}^{2}dx$, so the total potential energy relative to the straight configuration isIts stationary equation isFor pinned ends, $y(0)=y(\ell )=0$ and the end moments vanish, so $y^{\prime \prime }(0)=y^{\prime \prime }(\ell )=0$. The candidate shapes $y=A\sin (n\pi x/\ell )$ require $P=EI(n\pi /\ell {)}^2$. The first loss of stability is consequentlyAt smaller $P$, the quadratic energy is positive; beyond this threshold the straight configuration no longer minimizes it. End restraint changes the threshold: clamped–clamped gives $4{\pi }^{2}EI/{\ell }^2$ and clamped–free gives ${\pi }^{2}EI/(4{\ell }^2)$, equivalently an effective-length correction. A column buckles in the direction of smaller flexural rigidity. Euler buckling is a geometric instability within elastic response; yielding or crushing can occur first for a short thick strut. Large deflections, shear deformation and imperfections require refinements of the ideal slender-beam model.

For a slender elastic beam, the Euler-Bernoulli beam equation assumes small strain, small deflection slopes and cross-sections that remain plane and normal to the deflected centreline. A fibre at signed distance $y$ from the neutral axis has longitudinal strain ${ϵ}_{xx}=-y\kappa$, where $\kappa$ is the centreline curvature. Hooke's law gives ${\sigma }_{xx}=-Ey\kappa$, with Young's modulus $E$.

Choose the internal bending-moment sign so that $M=-\int y{\sigma }_{xx}dA$. ThenHere $I_A$ is the second moment of area, with units ${\mathrm{m}}^4$, not the mass moment of inertia of a rotating body. For a uniform material under pure bending, zero axial resultant places the neutral axis through the section's centroid.

For small slopes, $\kappa \simeq w^{\prime \prime }(x)$. Combining the moment-curvature relation with transverse force and moment balance gives, with a consistent load sign convention,For constant stiffness this becomes $E{I}_A{w}^{(4)}=q$. The bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. Boundary conditions specify displacement and slope at a clamped end, or bending moment and shear force at a free end.

For a cantilever of length $\ell$ carrying a transverse end load $F$, take deflection positive towards the load. The moment magnitude is $M(x)=F(\ell -x)$, and integration with $w(0)=w^{\prime }(0)=0$ givesThe free-end moment vanishes and the tip shear supplies the applied load. For a rectangular section of width $b$ and bending thickness $h$, $I_A=b{h}^3/12$, so increasing thickness strongly increases bending resistance. The model ceases to be reliable for large deflections, plastic strains or short thick beams where shear deformation matters.

The bending moment $B(x)$ is the internal torque transmitted across a cut through an elastic beam, about the transverse bending axis. It balances the moments of the external forces on the isolated portion on one side of the cut. Its sign depends on the chosen convention, and its units are $\mathrm{N}\mathrm{m}$.

The relevant “moment of area” is the second moment of area of the cross-section about its neutral axis:where $z$ is perpendicular distance from that axis within the section. It has units ${\mathrm{m}}^4$ and is a geometrical quantity, unlike a mass moment of inertia. For a homogeneous beam under pure bending, the neutral axis passes through the section's centroid, and its longitudinal strain is zero.

Assume a slender homogeneous elastic beam obeying linear elasticity, with plane sections remaining plane and no resultant longitudinal force. A length element of its neutral axis subtending a small angle $d\varphi$ has length $Rd\varphi$. A neighbouring fibre at distance $z$ from that axis has a length change of magnitude $zd\varphi$, giving longitudinal strain $\epsilon =z/R$ with signs chosen consistently for tensile and compressed fibres. By Hooke's law, its longitudinal stress is $\sigma =E\epsilon =Ez/R$.

The torque of these longitudinal forces about the neutral axis is thereforeHere $E$ is Young's modulus and $I$ the second moment of area. The displayed relation uses a signed radius and bending moment; an unsigned geometrical radius is $|R|=EI/|B|$. For consistency with the negative bending moments in the later parts, take the deflection $y$ positive downward and the signed curvature as $1/R\simeq y^{\prime \prime }$. Then the small-slope Euler-Bernoulli beam equation reads $EI{y}^{\prime \prime }=B$. Reversing the vertical coordinate reverses the corresponding sign relation but leaves the physical deflection unchanged.

Take downward deflection $y(x)$ positive, with the support level $y=0$. The rectangular section second moment of area about the horizontal neutral axis isFor small slopes the curvature relation from part (b), with the stated sign convention, isIntegrating twice givesThe support boundary conditions $y(0)=y(L)=0$ give $C_2=0$ and $C_1=q{L}^3/24$. HenceThis is a downward displacement; its upward component is $-y(x)$. As checks, reflection symmetry gives $y^{\prime }(L/2)=0$, and the maximum downward deflection isThe units are length: $\rho g/(E{b}^2)$ has units ${\mathrm{m}}^{-3}$. The width $a$ cancels because both weight and bending resistance scale with it. The small-slope approximation requires $\rho g{L}^3/(E{b}^2)\ll 1$; shear strain is also neglected, as appropriate for a slender elastic beam.

Pure bending is a beam loading state with zero axial resultant $N$, zero transverse shear force $V$ and a nonzero constant bending moment $M$ over the interval considered. Equal and opposite end couples provide an ideal realization. Zero net force on a section does not mean zero local stress: opposing longitudinal tensions and compressions can form a couple.

For a homogeneous slender elastic beam, small-deformation kinematics and Hooke's law giveHere $y$ is measured from the centroidal neutral axis, $I$ is the second moment of area, $E$ is Young's modulus and $\kappa$ is curvature. Uniform $E$, $I$ and $M$ produce uniform curvature. The geometric assumptions and moment sign convention are those of the elementary bending stress model.^[ref]

For an original example, let $M=2\mathrm{N}\mathrm{m}$ and $EI=50\mathrm{N}{\mathrm{m}}^2$. Then $\kappa =0.04{\mathrm{m}}^{-1}$. Across a $0.10\mathrm{m}$ interval the tangent direction changes by approximately $\kappa \ell =0.004$ radians. At $1\mathrm{mm}$ from the neutral axis, the strain magnitude is $4\times 10^{-5}$. The small-slope sag relative to the interval's chord is $\kappa {\ell }^2/8=50\mu \mathrm{m}$.

The central region of a symmetric four-point flexural test is an experimental approximation: the external load contacts lie outside a constant-moment interval. Local contact disturbances remain near the loading noses, and transverse shear exists in the outer spans. Calling the central interval pure bending therefore specifies a region and a model; it does not claim that the entire fixture has zero force or shear.