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The Kirsch equations describe the stress tensor around a traction-free circular hole in an infinite linear elastic plate under uniform remote tension. For remote stress ${\sigma }_{\infty }$ along $x$, hole radius $a$ and polar coordinates $(r,\theta )$ measured from that direction, put $q=a^2/r^2$. In the classical isotropic plane stress model,The fields satisfy equilibrium and the prescribed boundary tractions. They can be obtained using an Airy stress function satisfying the biharmonic compatibility equation.^[ref]

At $r=a$, $q=1$, so ${\sigma }_{rr}={\sigma }_{r\theta }=0$ whileThe tangential stress is $3{\sigma }_{\infty }$ at the top and bottom of the hole and $-{\sigma }_{\infty }$ at its sides. A traction-free boundary does not require every stress component to vanish: the outward boundary normal is radial, so radial normal and shear tractions must vanish.

For an original distance comparison along $\theta =\pi /2$, the tangential stress is ${\sigma }_{\infty }(1+q/2+3q^2/2)$. At $r=2a$ it is $1.21875{\sigma }_{\infty }$, already far below the edge value three. At large distance the polar components approach those of the original uniform Cartesian tension.

A finite-width plate, internal hole pressure, plasticity or out-of-plane bending changes the boundary-value problem. The stress concentration factor three is therefore specific to the stated remote-tension model. The MIT laboratory uses local strain measurements and photoelasticity to examine how actual geometries compare with ideal concentration fields.^[ref]

The Airy stress function determines the in-plane stress through its second derivatives. If two functions give the same stress, their difference $B$ has $B_{xx}=B_{xy}=B_{yy}=0$. On a connected domain this implies that its gradient is constant. Thereforeis exactly the local affine gauge freedom. It also preserves biharmonicity. An imposed decay condition for $A$ can fix this freedom; adding an arbitrary biharmonic function would usually change the physical stress and is not a gauge transformation.

The voltage is a mechanically mediated effect: nickel magnetostriction strains the bonded central layer, and the direct piezoelectric effect converts stress into electrical displacement. An estimate of $|d_{31}|$ requires a strain-transfer and electrical-boundary assumption.

Let $x$ run along the applied field and $z$ through the layers. Assume saturated, approximately volume-preserving isotropic nickel magnetostriction, perfect bonding, mechanically dominant nickel faces and a high-impedance voltage measurement. Relative to the random-domain state,Treat the thin central layer as in plane stress, with these imposed in-plane strains, negligible converse-effect backreaction and isotropic elastic constants. ThensoFor a transversely polarized layer,The open-circuit increment has $\Delta D_3\simeq 0$, and $|E_3|=|V|/t=800\mathrm{V}{\mathrm{m}}^{-1}$. HenceThe observed voltage magnitude does not specify the sign without electrode polarity and poling direction. A one-dimensional approximation retaining only axial stress $E\Lambda$ instead gives $4.78\mathrm{pC}{\mathrm{N}}^{-1}$. That is a different mechanical model. Nickel stiffness and thickness, interface compliance, central-layer tensor coefficients and electrical loading are not specified, so the experiment does not determine a unique exact intrinsic coefficient without these assumptions.

Constitutive law and Poisson ratio. Use positive stress for tension. In an isotropic linearly elastic medium, Hooke's law givesThe analogous equations follow by cyclic permutation. Under uniaxial tension in direction 1, $e_1={\tau }_1/E$ and $e_2=e_3=-\sigma {\tau }_1/E$. Thus Poisson's ratio $\sigma =-e_2/e_1$ measures transverse contraction relative to longitudinal extension when the transverse stresses are zero. It is dimensionless; it is not the ratio of transverse stress to longitudinal stress.

Radial and hoop strains. Let a material point originally at radius $r$ move to $r+R(r)$, with small axisymmetric displacement. Two neighbouring radii separated by $dr$ move apart by $dr+dR$, so the radial fractional extension is $dR/dr$. A circle's circumference changes from $2\pi r$ to $2\pi (r+R)$, so its fractional extension is $R/r$. ConsequentlyThe absence of longitudinal stress is a plane stress condition, not a claim of zero longitudinal strain.

Equilibrium of an annular wedge. Consider a slice of unit axial length, radial thickness $dr$ and small angular width $d\varphi$. The difference of the outward and inward radial face forces is $(r{\tau }_r{)}^{\prime }drd\varphi$ to leading order. The two tangential face tensions have inward radial resultant ${\tau }_{\varphi }drd\varphi$: each contributes ${\tau }_{\varphi }dr\sin (d\varphi /2)$. With no body force, their balance gives the radial equilibrium of a thick elastic cylinder,This hoop contribution is why the radial stress need not be spatially constant in a thick wall.

Displacement equation. With ${\tau }_z=0$, the two Hooke's law equations readInvert them:Substitute in ${\tau }_{\varphi }=(r{\tau }_r{)}^{\prime }$. After cancelling the common factor, the right-hand side is $r{R}^{\prime \prime }+(1+\sigma )R^{\prime }$, soThe cancellation explains why this displacement equation does not depend on the value of Poisson's ratio.

Power laws and wall pressure. Try $R\propto r^n$. The indicial equation is $n(n-1)+n-1=n^2-1=0$, soThe resulting Lamé stress distribution in a pressurized cylindrical wall can be writtenThe pressure boundary tractions are ${\tau }_r(a)=-p_1$ and ${\tau }_r(b)=-p_0$. Solving them yieldsHence the positive compressive radial pressure in the wall isIt equals $p_1$ at $a$ and $p_0$ at $b$. This directional radial pressure is not an isotropic fluid pressure: radial and tangential stresses differ. The associated displacement is $R=[(1-\sigma )Pr+(1+\sigma )Q/r]/E$. Also $e_z=-\sigma ({\tau }_r+{\tau }_{\varphi })/E=-2\sigma P/E$, generally nonzero even though ${\tau }_z=0$.

Maximum absolute tangential stress. For $p_1>p_0$, $Q>0$ and ${\tau }_{\varphi }^{\prime }=-2Q/r^3<0$. Therefore the maximum of its absolute value must be at one of the two surfaces. Keeping the specified external pressure gives the fully general answerFor the usual high internal pressure with negligible external pressure, the largest magnitude is the tensile stress on the inner wall:For finite $p_0$, the inner-wall expression is $[p_1(a^2+b^2)-2p_0{b}^2]/(b^2-a^2)$, but its absolute value is not always the larger endpoint value. For example, nearly equal internal and external pressures can leave the outer surface more compressive. With $p_1=p_0=p$, both stresses correctly reduce to $-p$ everywhere. These results use a homogeneous linear elastic wall without axial loading; a closed-ended pipe would require a different longitudinal-stress condition.

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.

A stress concentration is a localized increase of stress caused by geometry or another disturbance of the load path. Holes, notches and abrupt section changes can make the largest local normal stress much greater than a stress calculated from the total force divided by a reference cross-sectional area. A stress concentration factor $K_t={\sigma }_{\max }/{\sigma }_{\mathrm{ref}}$ is meaningful only after both the stress component and its reference have been specified.

The Kirsch equations give an instructive exact result for a circular hole of radius $a$ in an infinite, isotropic linear elastic elastic plate under remote tensile stress ${\sigma }_{\infty }$ in one direction, using a plane stress model. Measure the polar angle $\theta$ from the loading direction. At the hole boundary,At $\theta =\pi /2,3\pi /2$, the tangential normal stress reaches $3{\sigma }_{\infty }$, giving $K_t=3$ relative to remote stress. At $\theta =0,\pi$, it is $-{\sigma }_{\infty }$. A traction-free hole therefore does not require every component of its stress tensor to vanish: it requires zero traction on the radial boundary.^[ref]

For a finite plate, distinguish gross and net reference areas. A plate of width $W$, thickness $t$ and central hole diameter $d$ transmits force $P$ through gross area $A_g=tW$ and minimum net area $A_n=t(W-d)$. Defining ${\sigma }_g=P/A_g$ and ${\sigma }_n=P/A_n$ gives the exact conversionThis is a change of normalization, not a solution for the finite plate's stress field. For an original numerical example, take $W=40\mathrm{mm}$, $t=2\mathrm{mm}$, $d=10\mathrm{mm}$ and $P=800\mathrm{N}$. The reference stresses are $10$ and $13.333\mathrm{MPa}$. If the local maximum is stipulated as $32\mathrm{MPa}$, the same physical maximum gives $K_{tg}=3.2$ and $K_{tn}=2.4$. Neither number changes the local stress by choosing a different denominator.

The distinction appears in real historical measurements. In 1939, C. Dumont loaded a perforated aluminum-alloy plate in out-of-plane bending and measured strain and deflection at two loads. The reported peak tangential stress was 1.59 times the computed net-section stress, but 1.85 times the computed stress for an unperforated plate under the same bending forces. These are bending results for that geometry, rather than the infinite-plate tensile factor three.^[ref]

Modern demonstrations can combine a metal plate instrumented with strain gauges and a transparent polymer viewed in a polariscope. The latter uses stress-induced birefringence; its fringes respond to differences of principal stresses. The MIT laboratory compares local and remote strain measurements with qualitative photoelasticity observations around holes and a slot.^[ref]

The elastic stress concentration factor is dimensionless. It differs from the stress intensity factor used to describe a sharp crack's singular field. Local yielding or a crack changes the mechanical problem; multiplying a remote stress by the elastic $K_t$ does not, by itself, establish a failure load.

The uniaxial sheet stretching stiffness is the coefficient relating axial force per unit width to axial engineering strain in a homogeneous thin elastic sheet under uniaxial stress with free lateral contraction. With Young's modulus $E$ and thickness $h$,Here $N=F/w$ for tensile force $F$ and sheet width $w$. Since Hooke's law gives $\sigma =E\epsilon$ and $F=\sigma wh$, division by $w$ gives the result. $Y_{2D}$ has units $N/m$, while $E$ has units $N/{\mathrm{m}}^2$.^[ref]

For an original example, choose $E=2\mathrm{GPa}$, $h=50\mu \mathrm{m}$, $w=10\mathrm{mm}$ and gauge length $L=50\mathrm{mm}$. Then $Y_{2D}=100000N/m$. At strain $\epsilon =0.001$, $N=100N/m$, $F=1\mathrm{N}$ and extension $\delta =\epsilon L=50\mu \mathrm{m}$. The specimen's axial stiffness is $F/\delta =Y_{2D}w/L=20000N/m$, which depends on its width and gauge length.

A tensile measurement can infer $Y_{2D}$ from the slope of $F/w$ against $\delta /L$. It needs known dimensions, reasonably uniform loading and freedom for lateral contraction; unloading tests elastic recovery. If lateral strain is instead constrained to zero while through-thickness stress remains zero, the plane stress equations give coefficient $Eh/(1-{\nu }^2)$, with $\nu$ Poisson's ratio.^[ref] The loading condition is part of the meaning of the reported stiffness.