Thermal cracks — COD scalar and resistivity contribution

Transposition of Crack opening displacement and compliance tensors to the 2nd-order (conductivity / diffusion / Darcy) problem, where the driving field is a vector rather than a symmetric 2-tensor.

Elasticity ↔ Conductivity — correspondence table

A scalar $b$ suffices because $[T]$ is a scalar and only $\mathbf q\cdot\hat{\mathbf n}$ produces a jump — there are no sliding or shear modes to resolve, unlike the 6-component $\mathbf B$. The direction is always $\hat{\mathbf n}$: the null space of $\mathbf K_0 - \mathbf K_0\mathbf P(0)\mathbf K_0$ is spanned by $\hat{\mathbf n}$ for any $\mathbf K_0$ (derivation below), so $b$ carries all the matrix anisotropy.

Crack geometry

Same geometric families as in the elasticity chapter:

• Elliptic cracks of aspect ratio $\eta = b/a \in (0, 1]$ (EllipticCrack), including the circular penny $\eta = 1$ (PennyCrack).
• Ribbon cracks (tunnel cracks) of half-width $b$ (RibbonCrack), infinite along $\hat{\boldsymbol\ell}$.

Hill tensor Taylor expansion and block-matrix limit

Mirror of the elasticity derivation of Barthélémy (2009). For a flat ellipsoidal inclusion of aspect ratio $\omega\to 0$, the 2nd-order Hill tensor is computed via the formula $\mathbf P(\mathbf A,\mathbf K_0) = \mathbf K_0^{-1/2}\cdot\mathbf I^{\mathbf A\cdot\mathbf K_0^{-1/2}} \cdot\mathbf K_0^{-1/2}$ (Giraud et al. 2019), where $\mathbf I^{\mathbf B}$ is assembled in the eigenbasis of $\mathbf B^T\mathbf B$ (right singular vectors of $\mathbf B = \mathbf A\cdot\mathbf K_0^{-1/2}$). As $\omega\to 0$:

• The null vector of $\mathbf B^T\mathbf B$ is $\mathbf v_3 = \mathbf K_0^{1/2}\hat{\mathbf n}/\sqrt{k_{nn}}$, $k_{nn}=\hat{\mathbf n}\cdot\mathbf K_0\hat{\mathbf n}$.
• The corresponding Newton potential $I_3\to 4\pi$ while $I_1, I_2\to 0$, so

\[\mathbf P(0) = \mathbf K_0^{-1/2}\mathbf v_3\otimes\mathbf v_3\mathbf K_0^{-1/2} = \frac{\hat{\mathbf n}\otimes\hat{\mathbf n}}{k_{nn}}\]

which is rank-1 along $\hat{\mathbf n}$. The acoustic block is then

\[\boldsymbol\Lambda(\omega) = \mathbf K_0 - \mathbf K_0\mathbf P(\omega)\mathbf K_0 = \boldsymbol\Lambda(0) + \omega\,\boldsymbol\Lambda_1 + o(\omega),\]

with $\boldsymbol\Lambda(0) = \mathbf K_0 - (\mathbf K_0\hat{\mathbf n})\otimes(\mathbf K_0\hat{\mathbf n})/k_{nn}$. Its null space is spanned by $\hat{\mathbf n}$ (one-dimensional in the 2-tensor case — to be contrasted with the 3-dimensional null space in the elasticity problem). The limit

\[\mathbf R = \lim_{\omega\to 0}\omega\,\boldsymbol\Lambda(\omega)^{-1} = \frac{1}{Y_{nn}}\,\hat{\mathbf n}\otimes\hat{\mathbf n},\]

is rank-1 along $\hat{\mathbf n}$, with $Y_{nn} = \hat{\mathbf n}\cdot\boldsymbol\Lambda_1\cdot\hat{\mathbf n}$.

Closed-form COD scalar $b$

Isotropic matrix

For $\mathbf K_0 = k_0\,\mathbf 1$ the square-root transform is trivial and $\hat{\mathbf w} = \hat{\mathbf n}$:

\[\boxed{\; b_{\text{ell}}^{\text{iso}} = \frac{\eta}{\pi\,k_0\,\mathcal E_\eta} \;}, \qquad \mathcal E_\eta = \mathcal E\!\bigl(\sqrt{1-\eta^{2}}\bigr),\]

with $\mathcal E$ the complete elliptic integral of the second kind. Penny limit $\eta = 1$: $b = 2/(\pi^{2} k_0)$. Ribbon: $b = 2/(\pi k_0)$ (direct 2-D computation, not the $\eta\to 0$ limit of the elliptic formula).

Anisotropic matrix (K⁻¹ᐟ² transform)

Applying the square-root change of variable $\tilde{\mathbf x} = \mathbf K_0^{-1/2}\mathbf x$ reduces the problem to the isotropic one with a transformed crack shape. Let $\tilde{\mathbf A} = \mathbf A\cdot\mathbf K_0^{-1/2}$ with $\mathbf A = \mathbf R_c\,\mathrm{diag}(a,b,0)\,\mathbf R_c^{T}$, and denote its singular values $\sigma_1 \ge \sigma_2 \ge \sigma_3 = 0$. The transformed in-plane aspect ratio is $\eta_t = \sigma_2/\sigma_1 \in (0,1]$ and the transformed ellipse first-kind integral parameter is $k_t^{2} = 1 - \eta_t^{2}$. Then

\[\boxed{\; b_{\text{ell}}^{\text{aniso}} = \frac{\sigma_2} {\pi\,a_\text{max}\,\sqrt{\hat{\mathbf n}\cdot\mathbf K_0\hat{\mathbf n}}\,\mathcal E_{\eta_t}} \;}, \qquad a_\text{max} = \max(a,b).\]

It reduces to the isotropic formula above when $\mathbf K_0 = k_0\,\mathbf 1$ (then $\sigma_2 = \min(a,b)/\sqrt{k_0}$ and $\hat{\mathbf n}\cdot\mathbf K_0\hat{\mathbf n} = k_0$). For a TI matrix aligned with $\hat{\mathbf n}$ ($\mathbf K_0 = \mathrm{diag}(k_t, k_t, k_n)$ in the crack frame, penny):

\[b_{\text{penny}}^{\text{aligned TI}} = \frac{2}{\pi^{2}\,\sqrt{k_t k_n}} \qquad\text{(geometric mean of }k_t\text{ and }k_n\text{)}.\]

Ribbon crack — 2D formula

Only the $(\hat{\mathbf m}, \hat{\mathbf n})$ transverse block of $\mathbf K_0$ enters the formula:

\[\boxed{\; b_{\text{ribbon}}^{\text{aniso}} = \frac{2}{\pi\,\sqrt{\det\bigl(\mathbf K_0|_{(\hat{\mathbf m},\hat{\mathbf n})}\bigr)}} \;}\]

which reduces to $b = 2/(\pi k_0)$ for an isotropic matrix.

Resistivity contribution $\mathbf R$ and dilute correction $\Delta\mathbf R$

The size-independent crack resistivity contribution tensor is assembled from the scalar $b$ and the effective direction $\hat{\mathbf w}$:

\[\mathbf R^{\mathcal E} = \tfrac{3}{4}\,b\,\hat{\mathbf n}\otimes\hat{\mathbf n} \qquad\text{(elliptic)}, \qquad \mathbf R^{\mathcal R} = \tfrac{2}{\pi}\,b\,\hat{\mathbf n}\otimes\hat{\mathbf n} \qquad\text{(ribbon)}.\]

The geometric prefactors $3/4$ and $2/\pi$ are the same as in the elasticity case (they come from $cS/V$ evaluated on the ellipsoidal and ribbon geometries — see Crack compliance and COD tensor). The rank-1 direction is always the crack normal $\hat{\mathbf n}$, for any conductivity tensor $\mathbf K_0$.

compliance_contribution(crack, K₀) returns $\mathbf R$ directly. The dilute resistivity correction to the effective resistivity of the cracked conductor is obtained via delta_resistivity(crack, R, ε):

\[\Delta\mathbf R = \tfrac{4\pi}{3}\,\varepsilon^{3\mathrm d}\,\mathbf R \qquad\text{(elliptic, }\varepsilon^{3\mathrm d} = Nab^{2}\text{)}, \qquad \Delta\mathbf R = \pi\,\varepsilon^{2\mathrm d}\,\mathbf R \qquad\text{(ribbon, }\varepsilon^{2\mathrm d} = Nb^{2}\text{)}.\]

These reduce to the Sevostianov–Kachanov expressions (see Sevostianov & Kachanov (2002), Kachanov (2018)).

Intensity factors

Thermal analogs of the elastic stress / displacement intensity factors:

• Heat-flux intensity factor $K_T$ (mode I analog, scalar): the singular crack-tip field scales as $\sim K_T/\sqrt{r}$. For a ribbon crack of half-width $b$ and remote flux $\mathbf q^\infty$, $K_T = \sqrt{\pi b}\,(\hat{\mathbf n}\cdot\mathbf q^{\infty})$. For an elliptic crack the formula involves the tangent-ribbon COD ratio exactly as in the elasticity case ($b^\mathcal E/b^\mathcal R$ replaces $\mathbf B^\mathcal E(\mathbf B^\mathcal R)^{-1}$).
• Temperature intensity factor (scalar): $[T]_\text{avg} = b\,(\hat{\mathbf n}\cdot\mathbf q^{\infty})$.

See sif and dif for the full signatures (dispatched on $\mathbf K_0::\texttt{AbstractTens\{2,3\}}$).