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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.

The beam shear angle is the small-angle difference between centreline slope and material-section beam cross-section rotation:With transverse coordinate $z$ and kinematics $u_x=-z\theta (x)$, $u_z=w(x)$, the engineering shear strain is ${\partial }_z{u}_x+{\partial }_x{u}_z=\gamma$. The Euler-Bernoulli beam equation imposes $\gamma =0$; a shear-deformable model permits $w$ and $\theta$ to vary independently. The primary beam account distinguishes these kinematics and relates transverse force to the resulting shear measure.^[ref]

For an original local example, suppose shape measurements give $w_x=0.015$, while section markers give $\theta =0.010$ radians in the same sign convention. Then $\gamma =0.005$, about $0.286$ degrees. Setting the section rotation equal to the observed slope would erase this measured discrepancy by assumption.

In a simple linear shear model, the transverse force resultant is $V=S\gamma$, with effective beam shear rigidity $S=k_{s}GA$. Here $G$ is the shear modulus, $A$ the material area, and $k_s$ a section correction coefficient. Stipulating $G=1.0\mathrm{GPa}$, $A=20\mathrm{m}{\mathrm{m}}^2$ and $k_s=5/6$ gives $S=16667\mathrm{N}$ and $V\approx 83.3\mathrm{N}$. The correction coefficient accounts for using a reduced section model; it is not universally fixed at $5/6$.

A physical test must measure both the centreline shape and section orientation, for example by tracking markers during loading. Shape alone determines $w_x$ but cannot independently determine $\gamma$ without section-rotation information or an additional mechanical model. The comparison also requires compatible spatial resolution and angular measurement calibration.

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 Euler-Bernoulli beam equation models transverse linear bending waves in a slender beam. The normalized equation has $c_b=1$ and may be reduced by the Schrodinger factorization of the elastic beam equation. Prescribing displacement and curvature at an endpoint supplies two boundary traces for its fourth spatial derivative.

A flexural modulus is a stiffness inferred from a bending experiment using a specified stress–strain or force–deflection model. It has units of pressure. In a homogeneous, isotropic, linearly elastic slender beam with negligible shear and fixture deformation, it agrees with the Young's modulus entering $M=EI\kappa$. It is not the flexural strength or the load at fracture.

For a simply supported uniform beam in a symmetric four-point flexural test, let outer span be $L$, loading contacts lie at $a$ and $L-a$, and $P$ be their total force. Integration of the small-deflection Euler-Bernoulli beam equation gives midpoint deflection magnitudeEquivalently, with measured linear slope $S=dP/d\delta$,The formula uses midpoint displacement relative to the supports, not automatically the testing machine's crosshead displacement. The underlying moment-curvature law is the beam relation derived by Roylance; actual test conversion additionally requires the fixture spans and specimen geometry.^[ref][ref]

For an original rectangular example, choose $L=120\mathrm{mm}$, $a=30\mathrm{mm}$, $b=10\mathrm{mm}$ and $h=4\mathrm{mm}$, so $I=53.333\mathrm{m}{\mathrm{m}}^4$. A corrected midpoint slope $S=100\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$ gives $E_{\mathrm{flex}}\approx 46.4\mathrm{GPa}$. This is an inferred model value from stipulated inputs, not a tabulated property of a named material.

If measured displacement includes an extra compliance $C_{m}P$, the measured compliance is $d{\delta }_{\mathrm{meas}}/dP=C_{\mathrm{beam}}+C_m$. Ignoring positive fixture compliance reduces the apparent inferred modulus. Transverse shear, anisotropy, nonlinear response and time-dependent material behavior can also make flexural and tensile measurements differ. A reported value should therefore identify geometry, slope convention, displacement measurement and loading conditions.

The midspan compliance of a simply supported beam is the incremental central deflection per incremental central load, with the support geometry fixed. For a uniform, small-deflection Euler-Bernoulli beam equation model without axial preload,Its units are length per force. The simply supported beam contains the derivation; the experimental use is to infer beam flexural rigidity or, with known geometry, Young's modulus.^[ref]

An original two-span example separates beam compliance from an assumed constant machine contribution. Suppose the same uniform specimen is tested over spans $0.20$ and $0.30\mathrm{m}$. A machine-travel readout gives $0.060$ and $0.155mm/N$. Assumewhere $C_0$ is unchanged between the tests. Subtracting the readings givesHence $B=1/(48k)\approx 4.167\mathrm{N}{\mathrm{m}}^2$ and $C_0=0.020mm/N$. The inferred specimen compliances are $0.040$ and $0.135mm/N$, with ratio $3.375=(0.30/0.20{)}^3$. The raw readings have a different ratio because they include $C_0$.

This beam compliance from two support spans relies on the unchanged-series-compliance assumption. Changing supports, contact indentation or shear contributions can invalidate it; two points fit its two coefficients without independently validating the model. A third span can test the cubic dependence. An incremental slope also differs from simply dividing a total displacement, including an initial offset, by the current force.

Work to first order in the slope $w^{\prime }$ and take $B$ constant. The elastic beam is slender, its cross-sections remain plane and normal to its centreline, transverse shear deformation is neglected, and the loading is quasi-static. There are no distributed couples. In a non-contact interval the distributed force is $-p{\mathbf{e}}_z$ and there is no axial distributed force. These are the assumptions of the small-deformation Euler-Bernoulli beam equation, with an optional prescribed tension retained in transverse force balance.

Choose the internal section resultant on the right face of a segment aswhere $N$ is the transverse shear force, and take the section bending moment as $M{\mathbf{e}}_y$. Longitudinal and transverse force balance on a segment of length $dx$ give $T^{\prime }=0$ and $(T{w}^{\prime }+N{)}^{\prime }=p$. Balancing the torques about its left end gives $M^{\prime }=N$: the torque from the transverse section resultant is $-Ndx{\mathbf{e}}_y$. ThereforeWith this positive-$y$ bending moment convention, the linear relation between bending moment and curvature is $M=-B{w}^{\prime \prime }$. Thus $N=-B{w}^{\prime \prime \prime }$, and substitution givesThe tension need not be small compared with the transverse restoring terms, but products arising solely from finite deformation, such as the change of axial extension proportional to $(w^{\prime }{)}^2$, are outside this linear model. A separate axial constitutive constraint would be required to determine an unknown tension from fixed end positions.

In an open contact interval, the upward distributed normal contact force is $R$. At this order the smooth obstacle supplies no axial force; its vertical resultant changes the downward load to $p-R$. HenceFor nonadhesive unilateral beam contact, the obstacle can push but cannot pull, so $R\ge 0$. The accompanying constraints are $w\ge f$ and $(w-f)R=0$. A negative $R$ would require adhesion and would invalidate the proposed contact region. If $f$ is four times differentiable, the distributed normal contact force inside an open contact interval is $R=B{f}^{\prime \prime \prime \prime }-T{f}^{\prime \prime }+p$.

The net upward force is the sum of the contact point forces, since the distributed normal contact force vanishes in every regime here. Before contact it is zero. During isolated contact it is $R_0=12B(\alpha L-2H)/L^3$. For an interval of contact it is $R_{-s}+R_s=12BH/d^3$, with $d=3H/\alpha$. ThereforeThe formulae agree at both transitions. At $\alpha =3H/L$ they give $12BH/L^3$, and their first derivatives also agree there. As a check on the sign and normalization, integrating the Euler-Bernoulli beam equation over the whole elastic beam gives $F_{\mathrm{up}}=B(w^{\prime \prime \prime }(L^{-})-w^{\prime \prime \prime }(-L^{+}))$, which yields the same values.

Splitting the free Schrodinger equation into real and imaginary parts givesDifferentiating the first relation in time and using the second yields the Euler-Bernoulli beam equation $u_{tt}+u_{xxxx}=0$. This is the Schrodinger factorization of the elastic beam equation.

Assume the initial velocity has an integrable first spatial moment, as allowed by sufficient decay, and defineThen $v_0^{\prime \prime }=-u_1$ and $v_0$ decays at infinity. To encode the second boundary datum, defineThe Dirichlet boundary condition for the resulting free Schrodinger equation is compatible at the corner, since $q_0(0)=g_0(0)$. Insert these explicit $q_0,g_0$ into the data-only complex integral in part (b), with ${\hat{q}}_0$ and $G_0$ defined as in part (a). The required displacement is the real part of that integral. Equivalently, the uniformly convergent lifted integral in part (b) may be used with the same complex data.

The Schrodinger factorization of the elastic beam equation verifies every condition: $u(x,0)=u_0(x)$, $u_t(x,0)=-v_0^{\prime \prime }(x)=u_1(x)$, $u(0,t)={\tilde{u}}_0(t)$, andThe corner requirements on $u_0^{\prime \prime }$ and ${\tilde{u}}_0^{\prime }$ ensure consistency of these derivative traces; the natural interpretation of the last printed compatibility is ${\tilde{u}}_0^{\prime }(0)=u_1(0)$. If its prime were instead imposed for every $t$, that would simply be an extra restriction on the data, and the same construction would still solve them.

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.

Elasticity describes recoverable deformation. For small displacements $\mathbf{u}(\mathbf{x})$, the infinitesimal strain tensor is the symmetric part of the displacement gradient,Its diagonal entries describe extension, while its off-diagonal entries describe shear. The engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$; the antisymmetric part of the displacement gradient is a small rigid rotation, not elastic strain.

The stress tensor specifies the force per unit area on each oriented plane: $\mathbf{t}(\hat{\mathbf{n}})=\mathbf{\sigma }\hat{\mathbf{n}}$. Positive normal stress is tensile in the convention used here; tangential traction gives shear stress. In the absence of internal couples, angular-momentum balance makes the stress tensor symmetric. Force balance gives $\nabla \cdot \mathbf{\sigma }+\mathbf{f}=0$ in static equilibrium, where $\mathbf{f}$ is body force per unit volume.

Linear elasticity relates stress and strain. A homogeneous isotropic solid needs two independent elastic constants:Here $G$ is the shear modulus and $\lambda$ is the other Lamé parameter. For uniaxial loading, Young's modulus is $E={\sigma }_{xx}/{ϵ}_{xx}$, with lateral contraction described by Poisson's ratio $\nu =-{ϵ}_{yy}/{ϵ}_{xx}$. The bulk modulus is $K=-p/(\Delta V/V)$ for a small hydrostatic compression. The relations between these constants areStable isotropic elastic strain energy density requires $G>0$ and $K>0$, implying $-1<\nu <1/2$. The incompressible limit is approached as $\nu \to 1/2$. Anisotropic crystals generally need more constants; the isotropic relations should not be imposed on arbitrary directions in such a crystal.

The elastic energy density is $w=\mathbf{\sigma }:\mathbf{ϵ}/2$. In the isotropic linear model this becomes $\lambda (\mathrm{tr}\mathbf{ϵ}{)}^2/2+G\mathbf{ϵ}:\mathbf{ϵ}$. The quadratic form explains why reversible loading stores energy and unloading can return it, unlike plastic deformation or a dissipative hysteresis cycle.

For a slender beam described by the Euler-Bernoulli beam equation, plane sections approximately remain plane. A fibre a distance $y$ from the neutral axis has axial strain $-y\kappa$, where $\kappa$ is curvature, and stress $-Ey\kappa$. Integrating its moment about the neutral axis gives, with consistent signs,$I_A$ is a second moment of area, with units of length to the fourth power; it is not a mass moment of inertia. For a rectangular section of width $b$ and thickness $h$, $I_A=b{h}^3/12$, explaining the strong dependence of bending stiffness on thickness. For small transverse deflection $w(x)$, curvature is approximately $w^{\prime \prime }$, and the Euler-Bernoulli beam equation is $E{I}_A{w}^{\prime \prime \prime \prime }=q(x)$ with a suitable load-sign convention. A cantilever of length $\ell$ under end load $F$ has tip deflection $F{\ell }^3/(3E{I}_A)$ and largest bending stress of magnitude $F\ell c/I_A$, where $c$ is the furthest distance from the neutral axis.

In torsion of a solid circular shaft, a cross-section rotated by angle $\theta (x)$ has shear strain $r{\theta }^{\prime }(x)$. ThereforeHere $J$ is the polar second moment of area. A uniform shaft of length $\ell$ under constant torque twists by $\mathcal{T}\ell /(GJ)$, and its largest shear stress is $\mathcal{T}a/J$. Noncircular sections can warp, so replacing their torsional stiffness by $G$ times their polar area moment is not generally valid. Both bending and torsion formulas presume small, elastic deformations; exceeding the elastic limit can produce permanent deformation or fracture.

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.

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.

Writing $q=u+iv$ in the free Schrodinger equation gives $u_t=-v_{xx}$ and $v_t=u_{xx}$, hence $u_{tt}+u_{xxxx}=0$. To encode initial velocity $u_1$, choose the decaying primitive $v_0(x)=-{\int }_x^{\infty }(s-x)u_1(s)ds$. A prescribed endpoint curvature becomes the time derivative of the imaginary boundary trace. This turns the normalized Euler-Bernoulli beam equation into a complex second-order boundary problem.

Engineering shear strain is the first-order change of the right angle between material lines initially directed along two perpendicular axes. For a displacement $\mathbf{u}$, ${\gamma }_{xy}={\partial }_y{u}_x+{\partial }_x{u}_y=2{\epsilon }_{xy}$, where ${\epsilon }_{xy}$ is the tensor strain component. Slender Euler-Bernoulli beam equation models neglect the transverse shear strain contribution to deflection.

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.

An elastic beam above a smooth obstacle cannot penetrate it, and a nonadhesive obstacle can only push. The displayed constraints express this when $R$ is an upward distributed normal contact force; concentrated point forces must also be nonnegative. In the small-slope Euler-Bernoulli beam equation,Integrating across a contact point with no applied couple gives jump conditions $[w]=[w^{\prime }]=[B{w}^{\prime \prime }]=0$ and $[B{w}^{\prime \prime \prime }]=R_i$, provided the tension is continuous. For a flat obstacle, no distributed load and zero tension, an open contact interval has $R=0$; its entire normal contact force can therefore be concentrated at its endpoints.

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.