Linear elasticity

Linear elasticity approximates strain and stress to first order in a small displacement field.

For a infinitesimal strain tensor $e=(\nabla u+\nabla u^T)/2$, commuting displacement derivatives gives the symmetric incompatibility tensor $C=0$. Its six component equations are the Saint-Venant compatibility equations. They satisfy three differential identities because ${\partial }_i{\partial }_k$ is symmetric while ${ϵ}_{ik\ell }$ is antisymmetric. Thus there are three independent differential constraints, although keeping all six equations is a coordinate-symmetric way to state them. At a nonzero Fourier wavevector along the third axis the constrained components are $e_{11},e_{22},e_{12}$, while $e_{13},e_{23},e_{33}$ arise from the three displacement components. On a simply connected region compatibility gives local integrability up to a rigid motion; on a multiply connected region circulation conditions must also be imposed.

For uniform twist $\Omega$ and displacement $(-\Omega yz,\Omega xz,w(x,y))$, linear elasticity gives ${\tau }_{xz}=\mu (w_x-\Omega y)$ and ${\tau }_{yz}=\mu (w_y+\Omega x)$. Their planar divergence vanishes by axial equilibrium. On a simply connected section there is therefore a stress function $\varphi$ with ${\tau }_{xz}={\varphi }_y$ and ${\tau }_{yz}=-{\varphi }_x$.

The mixed derivatives of $w$ imply $\Delta \varphi =-2\mu \Omega$. The lateral axial traction is the tangential derivative of $\varphi$, so a traction-free connected boundary makes $\varphi$ constant; choose it to be zero. Finally the torque is $M={\int }_D(x{\tau }_{yz}-y{\tau }_{xz})dA=-{\int }_D(x{\varphi }_x+y{\varphi }_y)dA=2{\int }_D\varphi dA$, since its boundary term vanishes. Thus one scalar Poisson equation determines both the shear field and torsional stiffness. Multiply connected sections require additional constants and compatibility conditions on the separate boundary components.

In uniform torsion about the $z$-axis, the axial displacement $w(x,y)$ describes departure of a cross-section from a plane. It is determined up to an additive axial translation by $w_x=\Omega y+{\varphi }_y/\mu$, $w_y=-\Omega x-{\varphi }_x/\mu$, where $\varphi$ is the torsion stress function for a simply connected bar. For a circular section these derivatives vanish. For an ellipse they give $w=\Omega (b^2-a^2)xy/(a^2+b^2)+\text{constant}$, so a noncircular section generally warps. The stress-function Poisson equation is exactly the mixed-derivative compatibility condition for this axial displacement.

On $x^2/a^2+y^2/b^2<1$, the torsion stress function for a simply connected bar is $\varphi =\mu \Omega a^2{b}^2(1-x^2/a^2-y^2/b^2)/(a^2+b^2)$. Its Laplacian is $-2\mu \Omega$ and its boundary value is zero, proving the formula by uniqueness of the Poisson equation with Dirichlet data. Integration gives $M=\pi \mu \Omega a^3{b}^3/(a^2+b^2)$, hence the displayed stiffness $R=M/\Omega$.

The three principal stresses are $0,\pm |\nabla \varphi |$. With ${\tau }_Y$ defined as the shear yield stress, the Tresca yield criterion is $|\nabla \varphi |\le {\tau }_Y$. For $a>b$ its maximum is $2|C|/b$ at $(0,\pm b)$, where $C=\mu \Omega a^2{b}^2/(a^2+b^2)$. Equating that maximum to ${\tau }_Y$ gives the displayed first-yield twist. The ends of the minor axis yield first; geometric distance from the bar axis alone does not determine stress in a noncircular section.

On a connected isotropic linear elastic body, assume $\mu >0$ and $3\lambda +2\mu >0$. Then $W(e)=\mu |e^{\mathrm{dev}}{|}^2+(3\lambda +2\mu )(\mathrm{tr}e{)}^2/6$ is positive for every nonzero symmetric strain tensor. If two equilibria have the same body force and boundary traction, their difference $v$ satisfies zero loading. Multiplication of equilibrium by $v$ and the divergence theorem yield ${\int }_{B}2W(e(v))dV=0$. Hence $e(v)=0$, and the kernel of the symmetric gradient consists of infinitesimal translations and rotations: $v=a+\omega \times x$.

The conclusion is uniqueness modulo these modes, not existence for arbitrary tractions. Existence additionally requires zero total resultant force and torque. On a disconnected body each connected component has its own rigid modes; fixing enough displacements removes those modes and can produce full uniqueness.

A linear map from symmetric three-dimensional infinitesimal strain tensors to symmetric stress tensors acts between two six-dimensional spaces. Choosing the minor index symmetries $C_{ijkl}=C_{jikl}=C_{ijlk}$ leaves $6\times 6=36$ independent constitutive coefficients. Symmetry of the two second-rank tensors alone does not force pair interchange. If a strain-energy density exists and stress is its derivative, equality of mixed second derivatives gives the major symmetry $C_{ijkl}=C_{klij}$. The resulting symmetric six-by-six coefficient matrix has $6\cdot 7/2=21$ independent coefficients. This distinguishes minor tensor symmetry from the extra reciprocity supplied by an energy potential.

Write a quadratic small-strain energy as $\tilde{W}(F)=Q(\mathrm{sym}(F-I))$ with $Q$ homogeneous of degree two. If it were fully frame indifferent, the rotation $R=\mathrm{diag}(1,-1,-1)$ would have zero energy. Quadratic homogeneity would then make every $F_s=I+s(R-I)$ have zero energy. At $s=1/4$, $F_s=\mathrm{diag}(1,1/2,1/2)$ has determinant $1/4$, yet zero energy. Thus a nondegenerate linear elasticity energy cannot also be used as a fully objective finite-deformation energy. Its invariance concerns infinitesimal rotations around the reference state.

Fix a positive shear modulus $\mu$ and let the other Lamé parameter tend to infinity in $S_{\lambda }=\lambda \mathrm{tr}(e_{\lambda })I+2\mu e_{\lambda }$. If strain and stress converge to finite limits and differentiation commutes with the limit, taking the trace shows $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$. Thus $\mathrm{div}u=0$ and $S=-pI+2\mu e$ for $p=-\lim \lambda \mathrm{tr}{e}_{\lambda }$. Force balance becomes the displayed Stokes equation, with an elastic shear modulus in place of fluid dynamic viscosity. The vanishing volumetric strain is the linearization of elastic incompressibility $\det Dy=1$.

In a homogeneous isotropic solid without body force, a radial displacement field satisfies $(\lambda +2\mu )(u^{\prime }+2u/r{)}^{\prime }=0$. For nonzero $\lambda +2\mu$, its volumetric strain is constant. Since Hooke's law gives $\mathrm{tr}\tau =(3\lambda +2\mu )(u^{\prime }+2u/r)$, its stress tensor has constant trace. After plasticity in continuum mechanics has occurred, this conclusion applies to purely elastic increments during elastic unloading, and need not apply to the total residual stress.

The elasticity tensor relates stress to infinitesimal strain by ${\sigma }_{ij}=C_{ijkl}{ϵ}_{kl}$. In conservative linear elasticity, the minor symmetries reflect symmetric stress and strain, while the major symmetry $C_{ijkl}=C_{klij}$ expresses the existence of an elastic energy density.

The uniaxial strain modulus relates longitudinal normal stress to longitudinal strain when both transverse normal strains vanish. For isotropic linear elasticity, solving the transverse compliance equations gives the transverse stresses $\nu {\tau }_1/(1-\nu )$ and hence the displayed modulus in terms of Young's modulus and Poisson's ratio. It differs from Young's modulus, which applies when the transverse stresses vanish. Their difference is crucial for thin elastic layers bonded to rigid faces.

For a thin sheet of initial face area $A$, compressed normally by force $F$ between frictionless plates, each lateral strain is $\nu F/(EA)$. To first order the area of either broad face increases by $2\nu F/E$; total two-face area changes twice as much.

An axisymmetric small radial displacement $R(r)$ gives radial strain $R^{\prime }$ and hoop strain $R/r$. Radial force balance on an annular wedge gives ${\tau }_r^{\prime }+({\tau }_r-{\tau }_{\varphi })/r=0$. Under plane stress, isotropic Hooke's law gives ${\tau }_r=E(R^{\prime }+\nu R/r)/(1-{\nu }^2)$ and ${\tau }_{\varphi }=E(R/r+\nu R^{\prime })/(1-{\nu }^2)$. Substitution into equilibrium gives the displayed displacement equation. Its power-law exponents are $+1$ and $-1$, so $R=c_{1}r+c_2/r$. The hoop-stress solution is not based on treating a thick wall as a membrane.

For inner and outer radii $a,b$, with internal and external pressures $p_1,p_0$, the boundary tractions are ${\tau }_r(a)=-p_1$ and ${\tau }_r(b)=-p_0$. The radial equilibrium of a thick elastic cylinder gives $P=(p_1{a}^2-p_0{b}^2)/(b^2-a^2)$ and $Q=(p_1-p_0)a^2{b}^2/(b^2-a^2)$. The radial compressive pressure is $p(r)=-P+Q/r^2$. Since hoop stress is monotone in $r$ when $Q\ne 0$, its maximum absolute value is the larger of $|P+Q/a^2|$ and $|P+Q/b^2|$. With negligible external pressure, its maximum is the tensile inner-wall value $p_1(a^2+b^2)/(b^2-a^2)$. The stress distribution itself is independent of elastic moduli in this linear homogeneous axisymmetric model; its associated displacement depends on them.

A stress tensor state with all traction components involving one normal direction zero. In an isotropic solid, its two remaining normal stresses obey ${\tau }_x=E(e_x+\nu e_y)/(1-{\nu }^2)$ and ${\tau }_y=E(e_y+\nu e_x)/(1-{\nu }^2)$. The out-of-plane strain need not vanish: $e_z=-\nu ({\tau }_x+{\tau }_y)/E$. Constraining the out-of-plane deformation instead would be a different boundary condition.

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

The proportional limit bounds the loading interval over which stress is proportional to strain. It does not by itself establish the last load that permits complete recovery of elastic deformation.

Apply the divergence theorem to ${\partial }_k(x_i{\sigma }_{jk})$ and use momentum balance ${\sigma }_{jk,k}=\rho (a_j-b_j)$ and stress symmetry to obtain the displayed mean stress identity. Applying the same theorem to the infinitesimal strain tensor gives $V{\overline{e}}_{ij}=\frac{1}{2}{\int }_S(u_i{n}_j+u_j{n}_i)dS$. The acceleration term cannot be dropped during a transient.

For a uniform traction-free sphere, first-order self-gravity is $b=-gx/a$. Static averaging gives $\overline{\sigma }=-\rho gaI/5$. Homogeneous linear elasticity then gives $\overline{e}=-\rho gaI/(15\kappa )$. The boundary strain identity identifies $\overline{e}=u(a)I/a$, proving the displayed positive radius decrease. The local strain is not uniform, and the cancellation of the shear modulus concerns only the mean contraction.

For a homogeneous comparison elastic stiffness tensor $C^0$, the elastic strain Green operator maps an elastic polarization to minus the induced infinitesimal strain tensor. In a bounded domain the induced displacement field has zero boundary data; in an infinite medium it is the decaying perturbation. The defining static elastic equilibrium equation is $\nabla \cdot (C^0\mathrm{sym}\nabla u+\tau )=0$. The bounded-domain operator is self-adjoint, annihilates constant elastic polarizations, has zero mean output, and satisfies $\Gamma C^0\Gamma =\Gamma$.

For a three-dimensional isotropic comparison material, the contraction of its elastic strain Green operator is the displayed distributional derivative. Its double trace is ${\Gamma }_{iikk}=\delta /L$, because $\Delta |x{|}^{-1}=-4\pi \delta$. Away from zero the contraction is $({\delta }_{ij}|x{|}^2-3x_i{x}_j)/(4\pi L|x{|}^5)$; the full distribution also includes ${\delta }_{ij}\delta /(3L)$. This contact term must not be discarded when computing hydrostatic strain.

When only the bulk modulus differs from an isotropic matrix, the elastic polarization is spherical: $\tau =(\kappa -{\kappa }^0)\mathrm{tr}(\epsilon )I$. The double trace of the isotropic elastic strain Green kernel makes the trace equation local, giving the displayed formula. Individual strain components still depend on a nonlocal potential of the spherical polarization, so the complete infinitesimal strain tensor is not generally a local multiple of its reference value.

Let $B_{n}a=\mathrm{sym}(a\otimes n)$; its adjoint on symmetric stress tensors is $B_n^T\tau =\tau n$. Then $B_n^{T}C{B}_n$ is the acoustic tensor. The directional elastic strain Green operator satisfies the resolvent identity ${\tilde{\Gamma }}_r(C^r-C^s){\tilde{\Gamma }}_s={\tilde{\Gamma }}_s-{\tilde{\Gamma }}_r$, by the elementary inverse identity $K_r^{-1}(K_r-K_s)K_s^{-1}=K_s^{-1}-K_r^{-1}$.

The elastic polarization is the excess stress relative to a homogeneous comparison elastic stiffness tensor. In heterogeneous linear elasticity, $\tau =(C-C^0)\epsilon$. The comparison static elastic equilibrium equation gives the integral equation $\epsilon ={\epsilon }^0-\Gamma \tau$, where ${\epsilon }^0$ is the reference strain under the same external loading.

The bulk modulus relates hydrostatic compressive pressure to fractional volume compression. In isotropic three-dimensional linear elasticity, the spherical part of the stress tensor is $\kappa \mathrm{tr}(\epsilon )I$, so uniform dilation has strain energy density $\kappa (\mathrm{tr}\epsilon {)}^2/2$. Positive bulk modulus and positive shear modulus make the isotropic elastic stiffness tensor positive on nonzero symmetric strains.

For an isotropic linearly elastic solid, equal compressive principal stresses $-p$ produce three equal normal strains $-(1-2\nu )p/E$. Adding them gives the displayed volume strain, or bulk modulus $K=E/[3(1-2\nu )]$. It is a small-strain relation using the pressure change, not an arbitrary absolute pressure reference.

The pressure response to compression at fixed entropy, $K_S=\rho (\partial P/\partial \rho {)}_S$. Seismic compressions are commonly approximately adiabatic; a background density gradient follows the corresponding derivative only when composition and entropy assumptions also apply.

Antiplane shear has displacement normal to a plane and independent of that normal coordinate. In an isotropic linear elastic body, ${\sigma }_{13}=\mu w_{,1}$ and ${\sigma }_{23}=\mu w_{,2}$, together with their symmetric partners, are the only nonzero stress tensor entries. Static balance without body force reduces to the Laplace equation for $w$.

On an antiplane shear $F=I+e_3\otimes (p,q,0)$, the first strain invariant satisfy $I_1=I_2=3+p^2+q^2$, $I_3=1$. Restricting an isotropic objective strain energy density to this family determines only derivatives tangent to the family. It fixes $S_{31}=2w^{\prime }(p^2+q^2)p$ and $S_{32}=2w^{\prime }(p^2+q^2)q$, but does not specify the whole Cauchy stress tensor. For example the neo-Hookean solid energy $W=\mu (|F{|}^2-3)/2-\mu \log \det F$ gives $S=\mu (B-I)$ on this family. Even though the reference state is stress free, this is not generally a scalar multiple of $B$.

For subsonic motion $|V|<c$, an antiplane shear displacement $w(x_1-Vt,x_2)$ satisfies ${\beta }^2{w}_{xx}+w_{x_2{x}_2}=0$. The displayed rescaling makes it a harmonic function of $(x,y)$, so locally $w=\mathrm{Re}H(x+iy)$ for a holomorphic function $H$. Set $F=\beta H$. The Cauchy-Riemann equations then give $\mu F^{\prime }=\beta {\sigma }_{13}-i{\sigma }_{23}$. Keeping this potential normalization explicit avoids losing a factor of $\beta$ in the dynamic stress field. At sonic or supersonic speeds the elliptic harmonic reduction does not apply.

A harmonic conjugate $\varphi$ satisfying ${\varphi }_{,1}=w_{,2}$ and ${\varphi }_{,2}=-w_{,1}$ makes $F=\varphi +iw$ a holomorphic function. Its derivative packages the two antiplane shear stresses into the displayed complex field. On multiply connected domains a global potential additionally requires the appropriate periods to vanish; prescribed displacement jumps can instead be represented by branches of the potential.

For a crack on the real axis under applied shear traction $p$, write $G=\mu F^{\prime }$ for the additional complex stress potential for antiplane shear derivative. Reflection of the induced displacement gives $G^{-}=\overline{G^{+}}$. Traction-free crack faces therefore require $G^{+}+G^{-}=-2p$; intact portions have $G^{+}=G^{-}$. This is a scalar Riemann-Hilbert problem. Endpoint regularity, single-valued displacement and the absence of additional remote loads fix its homogeneous terms.

For the moving semi-infinite crack, take the branch cut of $\sqrt{z}$ on the negative real axis and $\sqrt{x}>0$ ahead of the tip. With steadily moving antiplane shear, face tractions ${\sigma }_{23}^{\pm }=-f(x)$ imply $\mathrm{Im}{F}^{\prime \pm }=f/\mu$. The antisymmetric displacement response gives $F^{\prime -}=-\overline{F^{\prime +}}$. Hence $G=\sqrt{z}{F}^{\prime }$ has jump $G^{+}-G^{-}=-2\sqrt{-x}f(x)/\mu$. The Sokhotski–Plemelj theorem gives the displayed Cauchy integral, assuming convergence and no extra remote homogeneous loading. For $x>0$ it is purely imaginary, so ${\sigma }_{23}={\int }_{-\infty }^0\sqrt{-s}f(s)/(x-s)ds/(\pi \sqrt{x})$. At fixed co-moving loading this ahead-of-tip stress is independent of speed; the displacement still has the factor $1/\beta$.

For real $C$, this analytic homogeneous field has real boundary values on both faces of the negative-axis crack, so it adds no face shear traction. Ahead of the tip it adds stress $-\mu C/\sqrt{x}$. Its stress tends to zero at infinity, but its primitive grows like $\sqrt{z}$, producing unbounded remote displacement and infinite total elastic energy. Thus merely requiring vanishing far stress does not uniquely fix the face-loaded crack solution. Excluding an added remote stress-intensity field, for example through the localized-load bounded-displacement far condition, removes this homogeneous ambiguity. Ordinary inverse-square-root tip stress remains allowed and has finite energy in any bounded neighbourhood of the tip.

For two semi-infinite cracks separated by a ligament, choose $\chi (z)=\sqrt{a^2-z^2}>0$ on the ligament. Its upper crack values are $-i\sqrt{t^2-a^2}$ on the right and $+i\sqrt{t^2-a^2}$ on the left. Multiplication by $\chi$ converts the Hilbert problem for an antiplane crack to an additive jump; the Sokhotski–Plemelj theorem gives the displayed solution, assuming convergence. A homogeneous term $C/\chi$ transfers an extra resultant $\pi C$ through the ligament and has logarithmic displacement at infinity. The usual correction with no additional remote resultant excludes it. The far-field normalization is essential to uniqueness.

Choose $\chi (z)=\sqrt{z^2-a^2}\sim z$ at infinity, cut on the finite crack. Its upper crack value is $i\sqrt{a^2-t^2}$, so $H=\chi G$ has jump $-2i\sqrt{a^2-t^2}p(t)$. The Sokhotski–Plemelj theorem gives the displayed Cauchy integral solution. It decays as $z^{-2}$ for integrable weighted traction. A possible homogeneous term $C/\chi$ would give a logarithmic displacement period; it is removed for a dislocation-free finite crack with no extra remote loading.

An elastic stiffness tensor maps a small displacement gradient tensor to the stress tensor. Minor symmetries in each index pair ensure symmetric stress and dependence only on symmetric strain; major symmetry $c_{ijpq}=c_{pqij}$ follows from an elastic energy and underlies reciprocity. Stability requires positive elastic energy on nonzero strains. In an isotropic material, $c_{ijpq}=\lambda {\delta }_{ij}{\delta }_{pq}+\mu ({\delta }_{ip}{\delta }_{jq}+{\delta }_{iq}{\delta }_{jp})$, with Lamé parameters $\lambda ,\mu$.

The two minor symmetries of the elastic stiffness tensor make it a map between six-dimensional spaces of symmetric strain and stress. An elastic energy gives major symmetry, so the matrix of this map is symmetric and has 21 independent entries. For a symmetry operation acting on strains with parity spaces of dimensions $m$ and $6-m$, invariance eliminates the mixed block and leaves $m(m+1)/2+(6-m)(7-m)/2$ independent coefficients.

A rotation through $\pi$ about the third coordinate axis preserves the strain components $e_{11},e_{22},e_{33},e_{12}$ and reverses $e_{13},e_{23}$. A symmetry-invariant elastic stiffness tensor cannot couple the two parity spaces. Its independent symmetric blocks therefore contain $4\cdot 5/2+2\cdot 3/2=13$ coefficients. A twofold axis alone does not impose invariance under every rotation about that axis.

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 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.

An elastic membrane is a thin deformable sheet whose transverse restoring force is supplied by in-plane tension. For uniform tension $T$ and mass per area $m$, a two-dimensional small-displacement model has $m{\eta }_{tt}=T{\eta }_{xx}-[p]$, where the pressure jump is defined from the lower fluid to the upper fluid. Fluid loading modifies its traveling modes through the acoustic wave on a tensioned massive membrane.

Two equal complex amplitudes with opposite transverse wavevector components sum to a cosine across a strip with edges $y=\pm a/2$. Identical phases satisfy the edge condition only for the odd transverse indices displayed. Equal magnitudes with opposite phase give the sine family; together the families form all Dirichlet strip modes $n\pi /a$, $n\ge 1$. The lowest axial phase velocity exceeds the unrestricted membrane wave speed, while its group velocity is smaller; neither should be substituted for the other.

Below the lowest strip cutoff, the axial wavevector is imaginary. A localized harmonic drive produces a displacement decaying as $e^{-\kappa |z-z_0|}$ away from the source in the dominant lowest mode. Cross-sections still oscillate in time, but there is no far-field travelling axial wave. Several more rapidly decaying transverse modes can contribute near the source. For a symmetric central point drive, transverse modes vanishing at the centre are not excited.

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 uniform elastic beam with endpoint conditions $y=y_{xx}=0$ has normal modes $\sin (n\pi x/L)$ for positive integers $n$. Substitute $y=X{e}^{-i\omega t}$ into $EI{y}_{xxxx}+{\rho }_{\ell }{y}_{tt}=0$ and impose the four endpoint conditions to obtain the displayed spectrum. Here ${\rho }_{\ell }$ is mass per unit length, $I$ is a second moment of area, and $EI$ is bending stiffness. frequencies proportional to $n^2$ distinguish flexural waves from the nondispersive ideal stretched string.

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 unit-length elastic beam with unit bending stiffness, the Green function of $D_x^4$ with $G=G_x=0$ at both endpoints is the displayed kernel. It is piecewise cubic and continuous through its second derivative, while $G_{xxx}(\xi +,\xi )-G_{xxx}(\xi -,\xi )=1$. A constructive form is $G=a{x}^2+b{x}^3+(x-\xi {)}_{+}^3/6$, with $a=\xi (1-\xi {)}^2/2$ and $b=-(1-\xi {)}^2(1+2\xi )/6$. The right endpoint conditions determine these coefficients. Thus $y(x)={\int }_0^{1}G(x,\xi )f(\xi )d\xi$ solves the clamped static beam problem.

The clamped-beam Green function is symmetric. Its explicit formula depends only on $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. More generally, two integrations by parts give $\int u{v}^{\prime \prime \prime \prime }dx=\int u^{\prime \prime }{v}^{\prime \prime }dx=\int u^{\prime \prime \prime \prime }vdx$ when $u=u^{\prime }=v=v^{\prime }=0$ at both endpoints. Applying this identity to two point-load responses shows that the displacement at one point caused by a unit force at a second equals the displacement at the second caused by a unit force at the first.

A straight slender elastic beam under axial compression loses stability when bending can decrease its total potential energy. With pinned ends a distance $L$ apart, its quadratic potential is $\frac{1}{2}{\int }_0^L[E{I}_A(y^{\prime \prime }{)}^2-P(y^{\prime }{)}^2]dx$. Stationarity gives $E{I}_A{y}^{\prime \prime \prime \prime }+P{y}^{\prime \prime }=0$, with $y=y^{\prime \prime }=0$ at each end. The first nontrivial mode is $\sin (\pi x/L)$, at $P_{\mathrm{cr}}={\pi }^{2}E{I}_A/L^2$. More generally $P_{\mathrm{cr}}={\pi }^{2}E{I}_A/(KL{)}^2$, with an effective-length factor $K$ determined by endpoint constraints: $K=1$ for pinned-pinned, $2$ for fixed-free and $1/2$ for fixed-fixed. Real imperfections remove the perfectly sharp onset, and short columns can yield before this elastic instability.

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.

With $y_0=Y\sin (\pi x/L)$ and $F_c={\pi }^{2}EI/L^2$, the equilibrium branch connected to zero load has $y=Y\sin (\pi x/L)/(1-F/F_c)$. The pole as $F$ approaches $F_c$ marks breakdown of the small-deflection model, not a physically infinite beam deflection.

Positive compression $\alpha$ contributes negative quadratic potential energy $-\alpha {\varphi }_x^2/2$, while positive bending stiffness $\beta$ contributes $\beta {\varphi }_{xx}^2/2$. A plane wave has ${\omega }^2=(\beta k^4-\alpha k^2)/\rho$, so sufficiently long waves grow exponentially. For displacement-fixed, moment-free endpoints at $0,l$, the natural boundary conditions for a free endpoint on the slope give ${\varphi }_{xx}=0$ as well as $\varphi =0$. The modes are $\sin (n\pi x/l)$, and the first Euler buckling of an elastic filament threshold is ${\alpha }_c=\beta {\pi }^2/l^2$. For truly clamped ends, $\varphi ={\varphi }_x=0$, the threshold is instead $4{\pi }^2\beta /l^2$, with first static mode $1-\cos (2\pi x/l)$. More generally,with the infimum taken over the specified endpoint class. At threshold there is a static zero mode and a marginal linear-in-time partner in the linear dynamics; above it the straight configuration is unstable.

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.

The second moment of area of a cross-section about an axis is $I=\int y^{2}dA$, where $y$ is the perpendicular distance from that axis. Its dimensions are length to the fourth power. For a circular section of radius $a$, polar integration gives $I={\int }_0^a{r}^{3}dr{\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi a^4/4$. A slender rod with Young's modulus $E$ has filament bending modulus $B=EI$: the axial strain caused by curvature $\kappa$ is $-y\kappa$, and integrating the linear elasticity energy density $E{y}^2{\kappa }^2/2$ over the section gives $B{\kappa }^2/2$ per unit length.

The polar second moment of a planar area about a perpendicular axis is the integral of squared distance from that axis over area. It has units ${\mathrm{m}}^4$ and equals the sum of second moments about two perpendicular in-plane axes through the same origin. For a disk of radius $a$, $J_p={\int }_0^{a}2\pi r^{3}dr=\pi a^4/2$. It is geometric, not a mass moment of inertia; a uniform cylinder has axial mass moment $\rho L{J}_p$.

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.

Poisson's ratio is minus the transverse strain divided by longitudinal strain under uniaxial loading in isotropic linear elasticity.

Young's modulus is the ratio of axial stress to axial strain in uniaxial linear elasticity. For an isotropic plate, the bending rigidity is $E{t}^3/[12(1-{\nu }^2)]$, with thickness $t$ and Poisson's ratio $\nu$.