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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.

By vertical force balance and reflection symmetry, each support exerts an upward force $W/2$. Use the paper's convention in which the sagging bending moment is negative. For a cut at $0<x<L/2$, the left support's moment about that cut givesFor $L/2<x<L$, include the central downward force when taking moments of all loads to the left:ThusThe bending moment is continuous at the load, with $B(L/2)=-WL/4$, and vanishes at both freely rotating supports. The jump is in its first derivative, reflecting the concentrated transverse force.

The uniform weight per unit length is $q=\rho abg$, where $\rho$ is the mass density and $g$ the gravitational acceleration. The total weight is $qL$, so each support provides an upward force $qL/2$. The weight on the portion from $0$ to $x$ is $qx$ acting at that portion's midpoint, a distance $x/2$ from the cut. In the same bending moment convention as part (c),The units are $(\mathrm{N}{\mathrm{m}}^{-1}){\mathrm{m}}^2=\mathrm{N}\mathrm{m}$. Its greatest magnitude occurs at $x=L/2$, where $B=-q{L}^2/8$.

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.

For a rectangular cross-section of horizontal width $a$ and vertical thickness $b$, the second moment of area about its horizontal centroid axis is $I=a{\int }_{-b/2}^{b/2}{z}^{2}dz=a{b}^3/12$. Thus increasing vertical thickness is especially effective at reducing the curvature caused by a given bending moment.

A simply supported beam is prevented from moving transversely at its two supports while its ends remain free to rotate. In the small-deflection Euler-Bernoulli beam equation, a uniform span $0\le x\le L$ with transverse displacement $w(x)$ and positive beam flexural rigidity $B=EI$ has the support boundary conditionsHere $E$ is Young's modulus and $I$ is the second moment of area about the bending axis. The second pair expresses zero applied endpoint bending moment, since $M=-B{w}^{\prime \prime }$ in the convention used here. It does not set the endpoint beam cross-section rotation or slope to 0. A clamped boundary condition instead constrains both displacement and slope.

For an original derivation, apply a downward central point force $P$ and take $w$ positive downward. Assume small elastic deflections, constant $B$ and no axial preload. Symmetry and force balance give an upward reaction $P/2$ at each support. On the left half, the bending moment is $M(x)=Px/2$, so $w^{\prime \prime }=-Px/(2B)$. Integrate twice, using $w(0)=0$ and the symmetry condition $w^{\prime }(L/2)=0$, to obtainThus the midspan compliance of a simply supported beam isThe endpoint slopes are $w^{\prime }(0)=P{L}^2/(16EI)$ and $w^{\prime }(L)=-P{L}^2/(16EI)$: they are generally nonzero. This support geometry and central-load formula are used to infer material stiffness from a three-point bending test.^[ref]

The moment-free natural boundary condition for a beam also follows from the principle of virtual work. For a central load, the total potential energy is $\Pi [w]=\frac{1}{2}{\int }_0^{L}B(w^{\prime \prime }{)}^{2}dx-Pw(L/2)$. An admissible virtual displacement $\eta$ satisfies $\eta (0)=\eta (L)=0$, while its endpoint first derivatives remain free. Two integrations by parts produce the boundary contributionThe terms multiplying $\eta$ vanish because the endpoint displacements are prescribed. Arbitrary endpoint values of ${\eta }^{\prime }$ then require $B{w}^{\prime \prime }=0$ there. The free end rotation and zero end moment are conjugate parts of the same variational boundary condition.

Clamping both ends changes the central-load deflection to $P{L}^3/(192EI)$, one quarter of the simply supported value for the same $P,L,E,I$. It introduces endpoint reaction moments; it does not merely rename the same support. The clamped-beam Green function provides the corresponding fourth-order boundary-value solution. The support conditions likewise determine the simply supported beam normal modes, so a static stiffness measurement and a vibration measurement must use the same mechanical support model when they are compared.

A three-point bending test places a specimen on two separated supports and applies a central force. A load cell measures force, and a midspan displacement readout measures response. A direct plunger and extensometer can measure specimen deflection separately from machine crosshead displacement, as in the published Instron fixture arrangement.^[ref] Fitting a proportional small-load segment estimates $E=L^3/(48IC)$, where $C$ is the measured incremental midspan compliance. Record support spacing, load position and cross-section dimensions with that estimate.

For stipulated $L=0.50\mathrm{m}$, rectangular width $b=20\mathrm{mm}$, thickness $h=3\mathrm{mm}$ and $E=200\mathrm{GPa}$, the second moment of area is $I=b{h}^3/12=4.5\times 10^{-11}{\mathrm{m}}^4$ and $EI=9\mathrm{N}{\mathrm{m}}^2$. An incremental $1\mathrm{N}$ central load gives $0.289\mathrm{mm}$ incremental deflection under the simply supported model, versus $0.0723\mathrm{mm}$ for the clamped model. Incremental readings remove an unchanged baseline deflection in this linear model; support settling, contact indentation and large rotations require separate interpretation.

MIT's Spring 2004 laboratory assignment Properties of Heat Treated Steel gives a concrete tensile testing protocol. Record transducer channels and calibration factors; measure specimen diameter and length, and the extensometer gauge length with its locking pin inserted. Zero the load cell, pin the threaded grips into their adapters, and attach the extensometer centrally. Apply the specified seating load, record actuator position, then acquire readings every second while moving at about $0.05\mathrm{cm}\mathrm{mi}{\mathrm{n}}^{-1}$. The handout derives engineering stress from load cell force, engineering strain from extensometer deformation, and Young's modulus by linear regression below the proportional limit; it warns that actuator-derived strain includes apparatus deformation.^[ref]

For a modulus-focused measurement, select a load window after seating, hold temperature and loading rate consistent, and record force, local extension and crosshead displacement together. Fit $F=F_0+k_g\Delta L_g$ within that window and calculate $E=k_g{L}_g/A_0$. Inspect the regression residuals for curvature; repeat at a lower maximum load and include unloading to check elastic deformation. These are diagnostic additions to the cited protocol, whose full test continues through failure. The handout's seating load is specific to its specimens and apparatus, rather than a universal preload.

An extensometer measures the change between its contact points; crosshead displacement measures travel of the machine's moving member. The manufacturer Instron explicitly distinguishes local strain from travel that includes machine deformation and machine compliance.^[ref] The gauge length $L_g$ can be shorter than the uniform length $L$ between grips. Thus even a perfectly rigid machine generally has $\Delta L_g\ne x$: the local instrument sees only its own segment. Do not divide one channel's extension by the other channel's reference length.

Grips must transmit axial force without slip. Misalignment adds bending moment and makes opposite sides experience different axial strains; an instrument on just one side may then report a biased modulus. A changing extension offset, hysteresis between loading and unloading, or load-dependent fitted slope is evidence to investigate seating, slip or material response. A straight line over a short interval alone does not identify which is responsible.

The component of the internal resultant force perpendicular to an elastic beam's tangent. For a planar elastic beam, one consistent sign convention has bending moment $M=-B{w}^{\prime \prime }$ and $M^{\prime }=N$, so $N=-B{w}^{\prime \prime \prime }$. Reversing the section or bending moment convention reverses the corresponding signs.

The vibroscope area estimator converts the fundamental frequency of a tensioned material fibre into a cross-sectional area through a mechanical model. For a uniform flexible strand of length $L$, positive tension $T$, linear mass density $\mu$ and mass density $\rho$, the fixed-end string relation givesThe second equation uses $\mu =\rho A$: the frequency first identifies mass per length under the stated tension, and a chosen mass density then converts it into area. Circular geometry is needed only to infer a diameter from that area.^[ref]

The experimental study by Bo Madsen, Rajnish Kumar and Hans Lilholt used a Favimat+ fibre testing machine. Acoustic waves excited the strand, and a frequency sweep plus a polynomial fit located the resonance peak. A load cell measured tension, while grip separation supplied $L$. The reported frequency range was 500–4000 Hz, with an estimated resolution of 0.1 Hz; the stated force and length readout resolutions were $1\mu \mathrm{N}$ and $0.1\mu \mathrm{m}$. Those readout resolutions do not themselves specify the estimator's full measurement uncertainty.^[ref]

The area and mass density must refer to the same loaded or reference configuration. The companion theory derives volume and area corrections for recovering an untensioned reference area from the tensioned measurement.^[ref]

The vibroscope area uncertainty follows locally by logarithmic differentiation:Treat the small input errors as independent random variables, and let each $u$ denote the corresponding standard uncertainty. First-order uncertainty propagation then givesCorrelated inputs require their covariances as well. Four independent 1% relative standard uncertainties give $\sqrt{10}\%\approx 3.16\%$ for the area. A systematic 1% overestimate of frequency alone instead multiplies the area estimate by $(1.01{)}^{-2}\approx 0.9803$: it produces about a 1.97% underestimate.

Bending resistance can also increase the observed frequency. The companion theory paper models fully clamped supports and derives the corresponding frequency equation and deformation corrections.^[ref] An independently derived, simply supported example isolates the bending-stiffness bias in the vibroscope area estimator. Let $B=EI$ be the filament bending modulus and impose zero deflection and zero bending moment at each endpoint. Substituting $y=\sin (n\pi x/L)\cos (2\pi f_{n}t)$ into the tensioned Euler-Bernoulli beam equation $\mu y_{tt}+B{y}_{xxxx}-T{y}_{xx}=0$ givesWith $r=b/a={\pi }^{2}B/(T{L}^2)$, the uncorrected flexible-string estimate satisfies $\hat{A}/A=1/(1+r)$. At $r=0.10$, it is 9.09% too small. The same model yields two-mode separation of tension and bending stiffness:For $T>0$ and $B\ge 0$, the predicted mode ratio is $f_2/f_1=2\sqrt{(1+4r)/(1+r)}$, ranging from 2 toward 4. These identities belong to the simply supported model; clamped grips require their own boundary conditions. Neither a single resonance peak nor an assumed diameter establishes the appropriate support, density or bending correction.