Appendices
Derivation in the local frame
The local basis is given by \(\varvec{e}_r,\varvec{e}_{\theta },\varvec{e}_{s}\) for this frame we have the relations
$$\begin{aligned} \dfrac{\partial \varvec{e}_r}{\partial \theta }= & {} \varvec{e}_{\theta };\quad \dfrac{\partial \varvec{e}_r}{\partial s}= \varGamma \cos \theta \varvec{e}_{s}\\ \dfrac{\partial \varvec{e}_{\theta }}{\partial \theta }= & {} -\varvec{e}_r;\quad \dfrac{\partial \varvec{e}_{\theta }}{\partial s}=-\varGamma \sin \theta \varvec{e}_{s}\\ \dfrac{\partial \varvec{e}_{s}}{\partial s}= & {} \varGamma \varvec{N}=-\varGamma (\cos \theta \varvec{e}_r-\sin \theta \varvec{e}_{\theta }) \end{aligned}$$
We can replace the derivative with respect to \(\,\mathrm{d}s\) by \(\mathrm{d}\phi \) taking account of \(\dfrac{\mathrm{d}\phi }{\,\mathrm{d}s} = \varGamma \).
Gradient of a vector
For a vector \(\varvec{u}=u\varvec{e}_r+ v\varvec{e}_{\theta }+ w\varvec{e}_{s}\) the gradient is
$$\begin{aligned} \nabla \varvec{u}= & {} \dfrac{\partial u}{\partial r}\varvec{e}_r\otimes \varvec{e}_r+\left( \dfrac{1}{r}\dfrac{\partial u}{\partial \theta }-\dfrac{v}{r}\right) \varvec{e}_r\otimes \varvec{e}_{\theta }\\&+(\dfrac{1}{\gamma }\dfrac{\partial u}{\partial s}-w\dfrac{\varGamma }{\gamma } \,\cos \theta )\varvec{e}_r\otimes \varvec{e}_{s} \\&+ \dfrac{\partial v}{\partial r} \varvec{e}_{\theta }\otimes \varvec{e}_r +\left( \dfrac{1}{\gamma }\dfrac{\partial v}{\partial s}+w\dfrac{\varGamma }{\gamma }\,\sin \theta \right) \varvec{e}_{\theta }\otimes \varvec{e}_{s}\\&+\left( \dfrac{1}{r}\dfrac{\partial v}{\partial \theta } + \dfrac{u}{r}\right) \varvec{e}_{\theta }\otimes \varvec{e}_{\theta }\\&+ \dfrac{\partial w}{\partial r} \varvec{e}_{s}\otimes \varvec{e}_r + \dfrac{1}{r}\dfrac{\partial w}{\partial \theta } \varvec{e}_{s}\otimes \varvec{e}_{\theta }\\&+ \left( \dfrac{1}{\gamma }\dfrac{\partial w}{\partial s} +u \dfrac{ \varGamma }{\gamma }\cos \theta -v\dfrac{\varGamma }{\gamma }\,\sin \theta \right) \varvec{e}_{s}\otimes \varvec{e}_{s} \end{aligned}$$
Divergence of a second order tensor
$$\begin{aligned}&P= P^{rr}\varvec{e}_r\otimes \varvec{e}_r+ P^{r\theta }\varvec{e}_r\otimes \varvec{e}_{\theta }+ P^{rs}\varvec{e}_r\otimes \varvec{e}_{s}\\&\quad + P^{\theta r}\varvec{e}_{\theta }\otimes \varvec{e}_r+ P^{\theta \theta }\varvec{e}_{\theta }\otimes \varvec{e}_{\theta }+ P^{\theta s}\varvec{e}_{\theta }\otimes \varvec{e}_{s}\\&\quad + P^{sr}\varvec{e}_{s}\otimes \varvec{e}_r+ P^{\theta s}\varvec{e}_{s}\otimes \varvec{e}_{\theta }+ P^{ss}\varvec{e}_{s}\otimes \varvec{e}_{s} \\&{\mathrm{div}}( P) = \left( \dfrac{\partial P^{rr}}{\partial r}+ \dfrac{1}{r}\dfrac{\partial P^{r\theta }}{\partial \theta } +\dfrac{P^{rr}-P^{\theta \theta }}{r} \right) \varvec{e}_r\\&\quad + \left( \dfrac{\partial P^{\theta r}}{\partial r}+\dfrac{1}{r}\dfrac{\partial P^{\theta \theta }}{\partial \theta } + (P^{r\theta }+P^{\theta r}) \dfrac{1}{r} \right) \varvec{e}_{\theta }\\&\quad +\left( \dfrac{1}{\gamma }\dfrac{\partial P^{rs}}{\partial s} + (P^{rr}-P^{ss}) \cos \theta \dfrac{\varGamma }{\gamma } - P^{r\theta }\sin \theta \dfrac{\varGamma }{\gamma } \right) \varvec{e}_r\\&\quad + \left( \dfrac{1}{\gamma }\dfrac{\partial P^{\theta s}}{\partial s}+P^{\theta r}\cos \theta \dfrac{\varGamma }{\gamma } +(P^{ss}-P^{\theta \theta }) \sin \theta \dfrac{\varGamma }{\gamma }\right) \varvec{e}_{\theta }\\&\quad +\left( \dfrac{\partial P^{sr}}{\partial r} +\dfrac{1}{r}\dfrac{\partial P^{\theta s}}{\partial \theta }+P^{sr}\dfrac{1}{r}+ \dfrac{1}{\gamma }\dfrac{\partial P^{ss}}{\partial s}\right) \varvec{e}_{s}\\&\quad +\left( (P^{rs}+P^{sr})\cos \theta - (P^{\theta s}+P^{\theta s}) \sin \theta \right) \dfrac{\varGamma }{\gamma } \varvec{e}_{s}\end{aligned}$$
Consider, that P is a symmetric tensor \(\sigma \), with components independent of s, we obtain
$$\begin{aligned} {\mathrm{div}}( \sigma )= & {} \left( \dfrac{\partial \sigma ^{rr}}{\partial r}+ \dfrac{1}{r}\dfrac{\partial \sigma ^{r\theta }}{\partial \theta } +\dfrac{\sigma ^{rr}-\sigma ^{\theta \theta }}{r} \right) \varvec{e}_r\\&+ \left( \dfrac{\partial \sigma ^{\theta r}}{\partial r}+\dfrac{1}{r}\dfrac{\partial \sigma ^{\theta \theta }}{\partial \theta } + 2\dfrac{\sigma ^{r\theta }}{r} \right) \varvec{e}_{\theta }\\&+ \left( \dfrac{\partial \sigma ^{sr}}{\partial r} +\dfrac{1}{r}\dfrac{\partial \sigma ^{\theta s}}{\partial \theta }+\sigma ^{sr}\dfrac{1}{r}\right) \varvec{e}_{s}\\&+\dfrac{\varGamma }{\gamma } \left( (\sigma ^{rr}-\sigma ^{ss}) \cos \theta - \sigma ^{r\theta }\sin \theta \right) \varvec{e}_r\\&+\dfrac{\varGamma }{\gamma } \left( \sigma ^{\theta r}\cos \theta +(\sigma ^{ss}-\sigma ^{\theta \theta }) \sin \theta \right) \varvec{e}_{\theta }\\&+ 2\dfrac{\varGamma }{\gamma } (\sigma ^{rs}\cos \theta - \sigma ^{\theta s}) \sin \theta ) \varvec{e}_{s}\end{aligned}$$
The first three terms correspond to the plane and anti-plane classical equation.
$$\begin{aligned} {{\mathrm{div}}\sigma = {\mathrm{div}}_{\pi } \sigma + \dfrac{\varGamma }{\gamma } {\mathrm{div}}_{\perp } \sigma } \end{aligned}$$
(77)
For stress tensor with components \(\sigma ^{ij}= r^\alpha \varGamma ^\beta \Sigma ^{ij}(\theta )\) we have
$$\begin{aligned} {{\mathrm{div}}\sigma = \alpha r^{\alpha -1} T_{\pi }(\theta ) \varGamma ^\beta + \dfrac{\varGamma ^{\beta +1}}{\gamma } r^\alpha T_{\perp }(\theta )} \end{aligned}$$
(78)
This remark is used in the process of construction of more consistent field.
Eshelby’s momentum tensor
$$\begin{aligned} {\mathsf {\varPsi }}^{rr}&=W- \sigma ^{rr}\dfrac{\partial u}{\partial r}-\sigma ^{r\theta }\dfrac{\partial v}{\partial r}- \sigma ^{rs}\dfrac{\partial w}{\partial r}\\ {\mathsf {\varPsi }}^{\theta \theta }&=W- \sigma ^{\theta \theta }\left( \dfrac{1}{r}\dfrac{\partial v}{\partial \theta }+\dfrac{u}{r}\right) -\sigma ^{\theta r}\left( \dfrac{1}{r}\dfrac{\partial u}{\partial \theta }-\dfrac{v}{r}\right) -\sigma ^{\theta s}\dfrac{1}{r}\dfrac{\partial w}{\partial \theta }\\ {\mathsf {\varPsi }}^{ss}&= W-\sigma ^{ss}\left( \dfrac{\varGamma }{\gamma }(u\cos \theta -v\sin \theta )+\dfrac{1}{\gamma }\dfrac{\partial w}{\partial s}\right) -\sigma ^{sr}\\&\quad \left( \dfrac{1}{\gamma }\dfrac{\partial u}{\partial s}-\dfrac{w\varGamma }{\gamma }\cos \theta \right) -\sigma ^{\theta s}\left( \dfrac{1}{\gamma }\dfrac{\partial v}{\partial s}+\dfrac{w\varGamma }{\gamma }\sin \theta \right) \\ {\mathsf {\varPsi }}^{r\theta }&=-\sigma ^{rr}\left( \dfrac{1}{r}\dfrac{\partial u}{\partial \theta }-\dfrac{v}{r}\right) -\sigma ^{r\theta }\left( \dfrac{1}{r}\dfrac{\partial v}{\partial \theta }+\dfrac{u}{r}\right) -\sigma ^{rs}\dfrac{1}{\gamma }\dfrac{\partial w}{\partial s}\\ {\mathsf {\varPsi }}^{\theta r}&= -\sigma ^{\theta r}\dfrac{\partial u}{\partial r}-\sigma ^{\theta \theta }\dfrac{\partial v}{\partial r}-\sigma ^{\theta s}\dfrac{\partial w}{\partial r}\\ {\mathsf {\varPsi }}^{rs}&= -\sigma ^{rr}\left( \dfrac{1}{\gamma }\dfrac{\partial u}{\partial s}-\dfrac{w\varGamma }{\gamma }\cos \theta \right) -\sigma ^{r\theta }\left( \dfrac{1}{\gamma }\dfrac{\partial v}{\partial s}+\dfrac{w\varGamma }{\gamma }\sin \theta \right) \\&-\sigma ^{rs}\left( \dfrac{\varGamma }{\gamma }(u\cos \theta -v\sin \theta )+\dfrac{1}{\gamma }\dfrac{\partial w}{\partial s}\right) \\ {\mathsf {\varPsi }}^{sr}&= -\sigma ^{ss}\dfrac{\partial w}{\partial r}-\sigma ^{sr}\dfrac{\partial u}{\partial r}-\sigma ^{\theta s}\dfrac{\partial v}{\partial r}\\ {\mathsf {\varPsi }}^{\theta s}&= -\sigma ^{\theta \theta }\left( \dfrac{1}{\gamma }\dfrac{\partial v}{\partial s}+\dfrac{w\varGamma }{\gamma }\sin \theta \right) -\sigma ^{\theta r}\left( \dfrac{1}{\gamma }\dfrac{\partial u}{\partial s}-\dfrac{w\varGamma }{\gamma }\cos \theta \right) \\&-\sigma ^{\theta s}\left( \dfrac{\varGamma }{\gamma }(u\cos \theta -v\sin \theta )+\dfrac{1}{\gamma }\dfrac{\partial w}{\partial s}\right) \\ {\mathsf {\varPsi }}^{s\theta }&= -\sigma ^{sr}\left( \dfrac{1}{r}\dfrac{\partial u}{\partial \theta }-\dfrac{v}{r}\right) -\sigma ^{\theta s}\left( \dfrac{1}{r}\dfrac{\partial v}{\partial \theta }+\dfrac{u}{r}\right) -\sigma ^{ss}\dfrac{1}{r}\dfrac{\partial w}{\partial \theta } \end{aligned}$$
General process : case of mode I
Consider the singular part of the displacement: \(\varvec{u}_o=\sqrt{r} \dfrac{\mathrm{K}_\mathrm{I}}{2\mu } \varvec{u}_o^\mathrm{I}\) we have
$$\begin{aligned} \begin{aligned} \varvec{u}_o^\mathrm{I}(\theta )&= U_o(\theta ) \varvec{e}_r+V_o(\theta )\varvec{e}_{\theta }\\&=\frac{1}{2}\left( -\,\cos \dfrac{3\theta }{2}+(5-8 \nu )\,\cos \dfrac{\theta }{2}\right) \varvec{e}_r\\&\quad + \frac{1}{2} \left( \,\sin \dfrac{3\theta }{2}+(8\nu -7)\,\sin \dfrac{\theta }{2}\right) \varvec{e}_{\theta }\end{aligned} \end{aligned}$$
(79)
and the associated strain
$$\begin{aligned} {\varepsilon (\varvec{u}_o^\mathrm{I})= \dfrac{\mathrm{K}_\mathrm{I}}{2\mu }\dfrac{\varvec{E}^o}{\sqrt{r}} + \dfrac{\varGamma }{\gamma }\dfrac{\mathrm{K}_\mathrm{I}}{2\mu } \sqrt{r}\;E^1_{ss}\;\varvec{e}_{s}\otimes \varvec{e}_{s}} \end{aligned}$$
(80)
with
$$\begin{aligned} \begin{aligned} E^o_{rr}&= \dfrac{1}{2}U_o, \quad E^o_{\theta \theta }=U_o + \dfrac{\partial V_o}{\partial \theta } \\ E^o_{r\theta }&= \dfrac{\partial U_o}{\partial \theta }- V_o, \quad E^1_{ss}= U_o \,\cos \theta -V_o\,\sin \theta \end{aligned} \end{aligned}$$
(81)
The Cauchy stress \(\sigma \) admits the same decomposition
$$\begin{aligned} {\sigma (\varvec{u}_o)= \dfrac{\mathrm{K}_\mathrm{I}}{\sqrt{r}}\Sigma ^\mathrm{I}_{o}(\theta ) +\dfrac{\mathrm{K}_\mathrm{I}\varGamma }{\gamma }\;\sqrt{r}\; \Sigma _{o1}} \end{aligned}$$
(82)
with
$$\begin{aligned} {\Sigma _{o1} = \lambda \varvec{E}_{o1}^\mathrm{I}\mathbb {I}+2 \mu \varvec{E}_{o1}^\mathrm{I}\; \varvec{e}_{s}\otimes \varvec{e}_{s}} \end{aligned}$$
(83)
The contributions of \(\varGamma \) is not reduced to order 1, because \(1/\gamma \) has contributions to higher order:
$$\begin{aligned} {\dfrac{1}{\gamma }= 1 -r\varGamma \,\cos \theta + (r\varGamma )^2 \cos ^2\theta +...} \end{aligned}$$
(84)
The same thing occurs for \(\varvec{\varPsi }\), the order 0 is the contribution of the singular part, and the other are contributions due to the expansion of \(1/\gamma \).
For pure mode I, when \(\mathrm{K}_\mathrm{I}\) is constant, there is no contribution along \(\varvec{e}_{s}\) : \(\dfrac{1}{\gamma }\dfrac{\partial f}{\partial s}=0\) . The expressions are simpler and in particular J-integral is reduced to
$$\begin{aligned} { J= \int _{\partial A} \varvec{e}_r.\varvec{\varPsi }.\varvec{N}\gamma (R) R\mathrm{d}{\theta }+\int _A \varPsi ^{ss} \varGamma r \mathrm{d}r\mathrm{d}{\theta }} \end{aligned}$$
(85)
on the disc A with radius R, and \(\varvec{N}= -\,\cos \theta \varvec{e}_r+\,\sin \theta \varvec{e}_{\theta }\). We have finally
$$\begin{aligned} J&=\int _{\partial A} (-\varPsi ^{rr}\,\cos \theta +\varPsi ^{rs}\,\sin \theta )(1+R \varGamma ) R\mathrm{d}{\theta }\nonumber \\&\quad +\int _A \varPsi ^{ss}\varGamma r \mathrm{d}{\theta }\end{aligned}$$
(86)
It can be noticed that,
$$\begin{aligned} {\varvec{\varPsi }(\varvec{u}_o)= \varvec{\varPsi }_o + \varGamma \varvec{\varPsi }_{o1}(r,\theta )+ \varGamma ^2 \varvec{\varPsi }_{o2}(r,\theta )+...} \end{aligned}$$
(87)
The evaluation of J-integral is obtained using its definition considering \({\mathrm{div}}\varvec{\varPsi }^T\) or applying 85.
On the torus, the divergence of \(\varvec{\varPsi }(\varvec{u}_o)\) is not vanishing, because the stress \(\sigma (\varvec{u}_o)\) is not in equilibrium (17), then
$$\begin{aligned} {\mathrm{div}}\varvec{\varPsi }^T(\varvec{u}_o)&= A \left( 4-\cos {3\theta }+4\cos {2\theta }\right. \\&\qquad \left. +\cos {\theta }(9-64\nu (1-\nu ))\right) \varvec{e}_r\\&\quad A \left( \sin {3\theta }-\sin {2\theta }(-24+64\nu )\right. \\&\qquad \left. -\sin {\theta }(11-64\nu (1-\nu ))\right) \varvec{e}_{\theta }\\&+ \varGamma ^2 P_2(r,\theta ) +... \end{aligned}$$
where \(A= \dfrac{1-\nu }{16(2\nu -1)}\dfrac{\varGamma }{r}\mathrm{K}_\mathrm{I}\)
To obtain more accurate value of J-integral, we add a correction to field \(\varvec{u}_o\) in order to satisfy the equilibrium state at order 1, the additional term but be in \(r\sqrt{r}\)
$$\begin{aligned} {\varvec{u}_1= \sqrt{r} \dfrac{\mathrm{K}_\mathrm{I}}{2\mu } (\varvec{u}^\mathrm{I}_o + r \varGamma \varvec{u}_1^\mathrm{I})} \end{aligned}$$
(88)
The value of \({\mathrm{div}}\sigma (\varvec{u}_o)\) contains trigonometric functions of \(\theta /2\) and \(3\theta /2\), because \(\sigma \) is a linear function of the displacement, the correction is necessary of the form
$$\begin{aligned} \begin{aligned} \varvec{u}_1^\mathrm{I}(\theta )&= U^\mathrm{I}_1(\theta ) \varvec{e}_r+ V^\mathrm{I}(\theta ) \varvec{e}_{\theta }\\ U^\mathrm{I}_1&= U_{ct}^\mathrm{I}\,\cos \dfrac{3\theta }{2}+ U_{st}^\mathrm{I}\,\sin \dfrac{3\theta }{2}+ U_{cu}^\mathrm{I}\,\cos \dfrac{\theta }{2}+ U_{su}^\mathrm{I}\,\sin \dfrac{\theta }{2}\\ V^\mathrm{I}_1&= V_{ct}^\mathrm{I}\,\cos \dfrac{3\theta }{2}+ V_{st}^\mathrm{I}\,\sin \dfrac{3\theta }{2}+ V_{cu}^\mathrm{I}\,\cos \dfrac{\theta }{2}+ V_{su}^\mathrm{I}\,\sin \dfrac{\theta }{2}\end{aligned} \end{aligned}$$
To respect the equilibrium state \({\mathrm{div}}\sigma (\varvec{u}_1)=o(\varGamma ^2)\), we must solve a system of linear equations for the constants \(U_{ct}^\mathrm{I}, U_{st}^\mathrm{I}, U_{cu}^\mathrm{I},U_{su}^\mathrm{I}, V_{ct}^\mathrm{I},V_{st}^\mathrm{I}, V_{cu}^\mathrm{I},V_{su}^\mathrm{I}\). When we consider the boundary conditions on the crack lips, due to symmetry of the field, the system is reduced to a system of four constant, which is easily solved and we obtain (45). The state of stress \(\sigma (\varvec{u}_1)\) satisfies (\(A=\mathrm{K}_\mathrm{I}\dfrac{\varGamma ^2 \sqrt{r}}{16(2\nu -1)}\))
$$\begin{aligned} \begin{aligned}&{\mathrm{div}}\sigma (\varvec{u}_1)=\\&A\Big ((24\nu -3)\,\cos \dfrac{5\theta }{2}-12\,\cos \dfrac{3\theta }{2}\\&\qquad +(128\nu ^2-29)\,\cos \dfrac{\theta }{2}\Big )\varvec{e}_r\\&\qquad +A \Big ((24\nu -3)\,\sin \dfrac{5\theta }{2}-12 \,\sin \dfrac{3\theta }{2}\\&\qquad +(128\nu ^2-32\nu -15)\,\sin \dfrac{\theta }{2}\Big )\varvec{e}_{\theta }\end{aligned} \end{aligned}$$
For order 2, the added term has the form
$$\begin{aligned} {\varvec{u}_2= \varvec{u}_1+ \dfrac{\mathrm{K}_\mathrm{I}}{2\mu }\sqrt{r} (r\varGamma )^2 \varvec{u}_2^\mathrm{I}(\theta )} \end{aligned}$$
(89)
where \(\varvec{u}_1\) has being determined previously The additional term must be proportional to \(\sqrt{r} (r\varGamma )^2\) in order to cancel \(\sqrt{r} \) in \({\mathrm{div}}\sigma (\varvec{u}_1)\) and \(\varvec{u}_2^\mathrm{I}(\theta )\) depends now upon \(\theta /2,3\theta /2,5\theta /2\). Due to symmetry, the linear system of equations issued from \({\mathrm{div}}(\sigma (\varvec{u}_2)=o(\varGamma ^3)\) is reduced to 6 constants, and the solution is easily solved and we obtain (50).
William’s asymptotic expansion for the displacement in plane strain
The displacement takes the formal form
$$\begin{aligned} \varvec{u}= \sqrt{r} \sum _{p\alpha } k_{p\alpha } r^p \left( u^\alpha _p(\theta ) \varvec{e}_r+ v^\alpha _p(\theta ) \varvec{e}_{\theta }+w^\alpha _p(\theta ) \varvec{e}_{s}\right) \end{aligned}$$
where the components \((u^\alpha _p,v^\alpha _p,w^\alpha _p)\) depend on the mode of fracture \(\alpha \).
Mode I.
$$\begin{aligned} u_p(\theta )= & {} -\dfrac{1}{2}[ (1-2p)\cos \big ((2p+3) \dfrac{\theta }{2}\big ) \\&+ (8\nu -5+2p)\cos \big ((2p-1)\dfrac{\theta }{2}\big )]\\ v_p(\theta )= & {} -\dfrac{1}{2}[ (2p-1)\sin \big ((2p+3) \dfrac{\theta }{2}\big )\\&+ (8\nu -7-2p)\sin \big ((2p-1)\dfrac{\theta }{2}\big )] \end{aligned}$$
Mode II.
$$\begin{aligned} u_p(\theta )= & {} -\dfrac{1}{2}[ (3+2p)\sin \big ((2p+3) \dfrac{\theta }{2}\big ) \\&+ (5-8\nu -2p)\sin \big ((2p-1)\dfrac{\theta }{2}\big )]\\ v_p(\theta )= & {} -\dfrac{1}{2}[ (2p+3)\cos \big ((2p+3) \dfrac{\theta }{2}\big )\\&+ (8\nu -7-2p)\cos \big ((2p-1)\dfrac{\theta }{2}\big )] \end{aligned}$$
Mode III.
$$\begin{aligned} w_p = \sin (2p+1) \dfrac{\theta }{2}\end{aligned}$$
Property of the field \(\varvec{u}\). Consider the p term
$$\begin{aligned} {\varvec{u}^\alpha = \sqrt{r}\;\; r^p \varvec{u}^\alpha _p(\theta )} \end{aligned}$$
(90)
the associated strain is decomposed in two terms
$$\begin{aligned} {\varepsilon (\varvec{u}^\alpha )= \varepsilon _o^\alpha + \varGamma \varepsilon ^\alpha _1} \end{aligned}$$
(91)
where \(\varepsilon _o^\alpha \) is the plane and anti-plane strain, then we have
$$\begin{aligned} { \sigma (\varvec{u}^\alpha )= \sigma ^\alpha _o + \varGamma \sigma ^\alpha _1, \quad {\mathrm{div}}_\pi \sigma ^\alpha _o =0} \end{aligned}$$
(92)
the divergence of \(\sigma (\varvec{u}^\alpha )\) is proportional to the curvature.
Fabrikant solution
1.1 Circular crack under uniform pressure
In the frame (\(\varvec{e}_{x},\varvec{e}_{z}\)) the displacement is given as Fabrikant (1988)
$$\begin{aligned} u_x&= \dfrac{p \rho }{2\pi \mu } \Bigg \{ (1-2\nu ) \Big [\dfrac{a\sqrt{l^2_2-a^2}}{l^2_2}- \arcsin {\dfrac{a}{l_2}}\Big ]\\&+\dfrac{2 a^2 |z|\sqrt{a^2-l_1^2}}{l^2_2(l^2_2-l^2_1)} \Bigg \} \\ u_z&=\dfrac{2p}{\pi \mu }\Bigg \{ 2 (1-\nu ) \Big [ \dfrac{z}{|z|} \sqrt{a^2-l_1^2}-z \arcsin {\dfrac{a}{l_2}}\Big ] \\&+z \big [ \arcsin {\dfrac{a}{l_2}}- \dfrac{a \sqrt{l^2_2-a^2}}{l^2_2-l^2_1}\big ] \Bigg \} \end{aligned}$$
with
$$\begin{aligned} \rho= & {} 1 +\eta \cos \theta ,\quad z = \eta \sin \theta , \quad \eta =r \varGamma \\ l_1= & {} \sqrt{1+\eta \cos \theta + \dfrac{\eta ^2}{4}}-\dfrac{\eta }{2}\\ l_2= & {} \sqrt{1+\eta \cos \theta + \dfrac{\eta ^2}{4}}+\dfrac{\eta }{2} \end{aligned}$$
The solution is developed up to order 2 in \(\varGamma \)
$$\begin{aligned}&1-l_1^2 = \eta (2-\eta ) \sin ^2 \dfrac{\theta }{2}\\&l^2_2-1 = \eta (2+\eta ) \cos ^2\dfrac{\theta }{2}\\&A = \sqrt{l^2_2-1}{l^2_2}= \sqrt{\eta }\;\sqrt{1+\cos \theta }\;\\&\qquad (1-\dfrac{\eta }{4}(3+4\cos \theta ))\\&\arcsin {\dfrac{1}{l_2}}= \dfrac{\pi }{2} +\sqrt{\eta }\sqrt{\cos \theta +1}\\&\qquad (-1+\dfrac{\eta }{12}(1+4\cos \theta ))\\&\sqrt{A}-\arcsin {\dfrac{1}{l_2}}= -\dfrac{\pi }{2}+\sqrt{\eta }\;\sqrt{\cos \theta +1} \;\\&\qquad (2-\dfrac{\eta }{6}(5+8\cos \theta ))\\&\dfrac{2z}{l_2^2}= \eta \sin \theta (2-2\eta (1+\cos \theta )+...) \end{aligned}$$
with these quantities we obtain
$$\begin{aligned} U_x= & {} \dfrac{\pi }{2} (1+\eta \cos \theta ) (2\nu -1)+ \sqrt{\eta } U^0_x+ \eta \sqrt{\eta } U^1_x+ ...\\ U_z= & {} \pi \eta \sin \theta (2\nu -1) +\sqrt{\eta } U^0_z+ \eta \sqrt{\eta } U^1_z+ .. \end{aligned}$$
the components of the expansion are obtained successively,
These components are decomposed into two contributions, the correction for the order 0
$$\begin{aligned} u^1_x= & {} \dfrac{1}{24}\Big (3\,\cos \dfrac{5\theta }{2}+(128\nu ^2-144\nu +34)\,\cos \dfrac{3\theta }{2}\\&\qquad +(72\nu -33)\,\cos \dfrac{\theta }{2}\Big )\\ u^1_z= & {} \dfrac{1}{24}\Big (3\,\sin \dfrac{5\theta }{2}+(128\nu ^2-144\nu +34)\,\sin \dfrac{3\theta }{2}\\&\qquad 4+(24\nu -9)\,\sin \dfrac{\theta }{2}\Big ) \end{aligned}$$
and a plane strain term as Williams had proposed for \(r\sqrt{r}\):
$$\begin{aligned} W^1_x= & {} (7-8\nu ) \,\cos \dfrac{3\theta }{2}-3 \,\cos \dfrac{\theta }{2}\\ W^1_y= & {} (5-8\nu )\,\sin \dfrac{3\theta }{2}-3 \,\sin \dfrac{\theta }{2}\\ \end{aligned}$$
Then the order one displacement is
$$\begin{aligned} U^1= u^1+\dfrac{1-8\nu }{12} W^1 \end{aligned}$$
(93)
1.2 Circular crack under uniform shear
For uniform shear, the decomposition is similar . On the crack an uniform shear is imposed
$$\begin{aligned} u_x= & {} 2\dfrac{1-\nu }{2-\nu } \dfrac{\pi }{\mu } \tau _x \sqrt{a^2 -\rho ^2},\nonumber \\ u_y= & {} 2\dfrac{1-\nu }{2-\nu } \dfrac{\pi }{\mu } \tau _y \sqrt{a^2 -\rho ^2}, \end{aligned}$$
(94)
\(\rho \) is the distance to the axis Oz. Then we introduce the stress intensity factors proposed by Williams (\(\rho =a-r\))
$$\begin{aligned} K_{II}= & {} 2\dfrac{\sqrt{2a}}{\pi } \dfrac{\tau _x}{2-\nu } \end{aligned}$$
(95)
$$\begin{aligned} K_{III}= & {} 2\dfrac{\sqrt{2a}}{\pi } \dfrac{1-\nu }{2-\nu }\tau _y \end{aligned}$$
(96)
These quantities varies along the front of the crack, a pure mode II along axis \(\varvec{e}_{x}\) becomes a pure anti-plane shear along \(\varvec{e}_{y}\). This is conformed to the relations given in (Bui 1975, 1977). The Fabrikant solution Fabrikant (1989) is written as \((A=1/(\pi \mu (2-\nu ))\)
$$\begin{aligned}&F(r,\theta ) = (-5+4\nu ) z\arcsin \dfrac{a}{l_2} + 4(1-\nu ) \sqrt{a^2-l_1^2}\\&G(r,\theta ) =\dfrac{z \sqrt{l_2^2-a^2}}{l_2^2-l_1^2} \\&\dfrac{u_x}{A}= F(r,\theta ) \tau _x + a G(r,\theta )\\&\qquad \big (\tau _x+\dfrac{l_1^2}{l_2^2} (\tau _x\cos 2\phi +\tau _y\sin 2\phi )\big )\\&\dfrac{u_y}{A}=F(r,\theta ) \tau _y +a G(r,\theta )\\&\qquad \big (\tau _y++ \frac{l_1^2}{l_2^2} (\tau _x\sin 2\phi -\tau _y\cos 2\phi )\big ) \\&\dfrac{u_z}{A} = \rho (\tau _x\cos \phi +\tau _y\sin \phi )\Big ( \frac{1-2\nu }{2}\\&\qquad (\arcsin \frac{a}{l_2} - a \frac{\sqrt{l_2^2-a^2}}{l_2^2} ) +\frac{a^2}{l_2^2} G(r,\theta ) \Big ) \end{aligned}$$
Now, we consider the normal plane to the crack front, with direction \(\varvec{T}(\phi )\). In the frame (\(-\varvec{N},\varvec{e}_{z}\)) , in this plane we have \((U,V,W)= u \dfrac{\sin \phi }{1-\nu }, v\dfrac{ \sin \phi }{1-\nu }, w\; \cos \phi ) \))
$$\begin{aligned} u= & {} -\dfrac{\varGamma }{12} r\sqrt{r} \Big ((16\nu -26)\,\sin \dfrac{5\theta }{2}+(24\nu -9)\,\sin \dfrac{3\theta }{2}\\&+(32\nu -7)\,\sin \dfrac{\theta }{2}\Big )\\&+\sqrt{r}\Big (3 \,\sin \dfrac{3\theta }{2}+(8\nu -5)\,\sin \dfrac{\theta }{2}\Big ) \\&- \dfrac{1}{2}\pi \;r (2-\nu ) \sqrt{2\varGamma } \sin 2\theta +(1-2\nu ) \dfrac{\pi }{\sqrt{2\varGamma }} \sin \theta \\ v= & {} -\dfrac{\varGamma }{12} r\sqrt{r} \Big (2(8\nu -13)\,\cos \dfrac{5\theta }{2}+3(8\nu -5)\,\cos \dfrac{3\theta }{2}\\&-(32\nu -37)\,\cos \dfrac{\theta }{2}\Big )\\&- \sqrt{r}\; \Big ( 3 \,\cos \dfrac{3\theta }{2}+(8\nu -7)\,\cos \dfrac{\theta }{2}\Big ) \\&\dfrac{1}{2}\pi \; r\; \sqrt{2\varGamma }((\nu -2) \cos 2\theta -3(1-\nu ) \cos \theta )\\&+(1-2\nu ) \dfrac{\pi }{\sqrt{2\varGamma }} \cos \theta \\ w= & {} 8 \sqrt{r} \,\sin \dfrac{\theta }{2}+ r \sqrt{r} \;\dfrac{2\varGamma }{1-\nu } ((3-2\nu ) \,\sin \dfrac{3\theta }{2}\\&+(2-\nu )\,\sin \dfrac{\theta }{2})\dfrac{\pi (5-4\nu )}{\sqrt{2}(1-\nu )} r \sqrt{\varGamma } \sin \theta \end{aligned}$$
Consider the first singular terms (\(\sqrt{r}\)), we recognize the mode II for (u, v) and mode III for w. The pure mode III valid for \(\phi =0\) is changed in pure mode II at \(\phi =\pi /2\).
The displacement is decomposed into three contributions at order 1 in \(\varGamma \) :
-
a plane strain correction
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an anti-plane strain correction,
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a correction with coupling into plane and anti-plane solutions,
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and additional Williams contribution proportional to \(r\sqrt{r}\)
The plane and anti-plane corrections have the form given in section 4.
For the third contribution, the anti-plane shear gives a contribution in plane strain, this is due to the fact that the stress intensity factors are non uniform and depend on \(\phi \). For given \(w_o\), the equilibrium equation in direction \(\varvec{e}_r,\varvec{e}_{\theta }\) must be satisfied, with the boundary conditions on the lips. Solving this system we obtain with ( \(K_2=\sin \phi /(1-\nu ), K_3=\cos \phi \))
$$\begin{aligned} u_1= & {} -K_2\; r \sqrt{r}\;\varGamma \dfrac{64\nu ^2-48\nu -16}{15}\,\sin \dfrac{\theta }{2}\\ v_1= & {} K_2\; r\sqrt{r}\;\varGamma \dfrac{64\nu ^2-96\nu +32}{15}\,\cos \dfrac{\theta }{2}\end{aligned}$$
In a similar way, the plane strain gives a contribution normal to the plane (equilibrium equation on direction (\(\varvec{e}_{z}\)))
$$\begin{aligned} w_1 =\dfrac{K_3}{1-\nu }\; r\sqrt{r}\;\varGamma \,\sin \dfrac{\theta }{2}\end{aligned}$$
The terms of Williams are given for \(r\sqrt{r}\) with \(n=3, p=1\) :
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plane strain mode II
$$\begin{aligned} u_w^{II}= & {} r \sqrt{r}\; (5 \,\sin \dfrac{5\theta }{2}+ (8\nu -3) \,\sin \dfrac{\theta }{2})\\ v_w^{II}= & {} r \sqrt{r}\; (5\,\cos \dfrac{5\theta }{2}+ (8\nu -9) \,\cos \dfrac{\theta }{2}) \end{aligned}$$
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anti-plane mode III
$$\begin{aligned} w_w= r\sqrt{r} \,\sin \dfrac{3\theta }{2}\end{aligned}$$
$$\begin{aligned} u^w_1= & {} (8\nu -13) u_w^{II} /120 \end{aligned}$$
(97)
$$\begin{aligned} v^w_1= & {} (8\nu -13) v_w^{II}/120 \end{aligned}$$
(98)
$$\begin{aligned} w^w_1= & {} \dfrac{(2\nu -3)}{1-\nu } w_w \end{aligned}$$
(99)
It can be noticed that this type of decomposition is conformed to the asymptotic expansion proposed in Leblond and Torlai (1992).
