Complex stress potential for antiplane shear (${\sigma }_{23}+i{\sigma }_{13}=\mu F^{\prime }(z)$)

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.