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Numerical solution for the stress near a hole with corners in an infinite plate under biaxial loading

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We consider the elastic stress near a hole with corners in an infinite plate under biaxial stress. The elasticity problem is formulated using complex Goursat functions, resulting in a set of singular integro-differential equations on the boundary. The resulting boundary integral equations are solved numerically using a Chebyshev collocation method which is augmented by a fractional power term, derived by asymptotic analysis of the corner region, to resolve stress singularities at corners of the hole. We apply our numerical method to the test case of the hole formed by two partially overlapping circles, which can include either a corner pointing into the solid or a corner pointing out of the solid. Our numerical results recover the exact stress on the boundary to within relative error \(10^{-3}\) for modest computational effort.

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We thank Jeremy Hoskins for his helpful discussion on numerical aspects of this work.

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Correspondence to Brian J. Spencer.

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This work was supported by a grant from the Simons Foundation (Award #354717, BJS)

Appendix A: Asymptotic analysis near the corner for overlapping circles case.

Appendix A: Asymptotic analysis near the corner for overlapping circles case.

Here we derive the asymptotic behavior of the stress near the corner for the overlapping circle case of Sect. 4.5. The trace of the stress tensor for the overlapping circles case is given by Eq. (63) which we write as

$$\begin{aligned} \sigma _{x}+\sigma _{y}=H(\xi ,\alpha )\int _{0}^{\infty }F(s,\alpha )\cos s\xi \,\mathrm {d}s, \end{aligned}$$


$$\begin{aligned}&H(\xi ,\alpha )=4\left( \cosh \xi -\cos \alpha \right) \sin \alpha , \end{aligned}$$
$$\begin{aligned}&F(s,\alpha )=\frac{2K-\left( N_{1}-N_{2}\right) s\left( s-\cot \alpha \coth s\alpha \right) }{\sinh 2s\alpha +s\sin 2\alpha }\cdot \sinh s\alpha \end{aligned}$$


$$\begin{aligned} \cosh \xi =\frac{1+\cos \alpha \cos \gamma }{\cos \alpha +\cos \gamma } \end{aligned}$$


$$\begin{aligned} \gamma =\theta +\arcsin \left( \sin \theta \cos \alpha \right) . \end{aligned}$$

Here, the polar angle from the center of the circle \(\gamma \), polar angle \(\theta \), amount of overlap between the circles \(\alpha \), tensions \(N_{1}\) and \(N_{2}\) are defined as in Sect. 4.4. We use \(\alpha =2\pi /3\) as an example in this Appendix. The result can be generalized to other \(\alpha \) in \(\pi /2<\alpha <\pi \) using a similar approach as described here.

The corner location is at \(\theta =\pi /2\). As \(\theta \rightarrow \pi /2\), \(\gamma \rightarrow \pi /2+\arcsin (\cos \alpha )\). Thus,

$$\begin{aligned} \cos \gamma \rightarrow \cos \left[ \pi /2+\arcsin (\cos \alpha )\right] =-\cos \alpha . \end{aligned}$$

It follows from (70) that

$$\begin{aligned} \cosh \xi \rightarrow \infty , \end{aligned}$$

which gives \(\xi \rightarrow \infty \) as \(\theta \rightarrow \pi /2\). Thus the \(\cos (s\xi )\) term in integral (67) is highly oscillatory near the corner which may cause inaccuracy of the numerical integration.

We determine the behavior of the integral near the corner by applying asymptotic analysis. Let \(\theta =\pi /2-\varepsilon \) where \(0<\varepsilon \ll 1\), then \(\cos \theta =\sin \varepsilon =\varepsilon -\varepsilon ^{3}/6+{\mathcal {O}}(\varepsilon ^{5})\) and \(\sin \theta =\cos \varepsilon =1-\varepsilon ^{2}/2+{\mathcal {O}}(\varepsilon ^{2})\). By Eq. (71), we have

$$\begin{aligned} \cos \gamma =\cos \theta \sqrt{1-\sin ^{2}\theta \cos ^{2}\alpha }-\sin ^{2}\theta \cos \alpha . \end{aligned}$$

Thus we have

$$\begin{aligned} \cos \gamma =-\cos \alpha +\varepsilon \sin \alpha +{\mathcal {O}}(\varepsilon ^{2}). \end{aligned}$$

Substitute Eq. (75) into Eq. (70), the asymptotic approximation of \(\cosh \xi \) is

$$\begin{aligned} \cosh \xi =\frac{\text {e}^{\xi }+\text {e}^{-\xi }}{2}=\frac{1}{\varepsilon }\sin \alpha +\cos \alpha +{\mathcal {O}}(\varepsilon ). \end{aligned}$$


$$\begin{aligned} \xi =\ln \left[ \frac{2}{\varepsilon }\sin \alpha +2\cos \alpha +{\mathcal {O}}(\varepsilon )\right] . \end{aligned}$$

Now consider the integral in (67). F has properties \(F(-s)=F(s)\), \(F(s)\rightarrow 0\) as \(s\rightarrow \pm \infty \) and \(F(s,\alpha )\) is bounded for all s (\(s=0\) is a removable singularity), so we can rewrite the integral on \(-\infty<s<+\infty \) as

$$\begin{aligned} I=\int _{0}^{\infty }F(s,\alpha )\cos s\xi \,\mathrm {d}s=\frac{1}{2}\int _{-\infty }^{\infty }F(s,\alpha )\cos s\xi \,\mathrm {d}s. \end{aligned}$$

We evaluate I by considering the contour integral on complex plane \(z=s+\text {i}t\) as follows:

$$\begin{aligned} {\tilde{I}}=\underset{R\rightarrow \infty }{\lim }\int _{C}F(z,\alpha )\text {e}^{\text {i}\xi z} \mathrm {d}z, \end{aligned}$$


$$\begin{aligned} I=\frac{1}{2}\text{ Re }\left\{ {\tilde{I}}\right\} , \end{aligned}$$

where C is the line along s axis from \(-R\) to \(-R\). Note \(F(z,\alpha )\) is analytic in upper half-plane \(\text{ Im }(z)>0\) except at zeros of the denominator located by the roots of

$$\begin{aligned} \sinh 2z\alpha +z\sin 2\alpha =0. \end{aligned}$$
Fig. 21
figure 21

Contour plot of \(\left| F\right| \) on upper half-plane for \(\alpha =2\pi /3\)

Fig. 22
figure 22

Complex contour integral

There are infinitely many singularities of \(F(z,\alpha )\) on the upper half-plane (see Fig. 21). We claim that since \({\tilde{I}}\) contains \(\text {e}^{\text {i}\xi z}=\text {e}^{-\xi t}\text {e}^{\text {i}\xi s}\) and \(\xi \gg 1\), the integral \({\tilde{I}}\) can be approximated by the contribution from the residual of the singularity that occurs at the location with smallest imaginary part t in the upper half-plane. For \(\alpha =2\pi /3\), as shown in Fig. 21, this first singularity lies on the imaginary axis, and so satisfies \(s=0\) and

$$\begin{aligned} \sin 2\alpha t+t\sin 2\alpha =0. \end{aligned}$$

For \(\alpha =2\pi /3\) the solution of Eq. (82) is \(t_1^*\approx 0.6157\) and the first singularity is at \(z=\text {i}t_1^*\). We construct a rectangular contour on the complex plane with the first singularity inside as in Fig. 22 to evaluate the integral (79), where C is the line segment from \(z=-R\) to \(z=R\), \(\varGamma _1\) is the line segment from \(z=R\) to \(z=R+\text {i}t_2\), \(\varGamma _2\) is the line segment from \(z=R+\text {i}t_2\) to \(z=-R+\text {i}t_2\), and \(\varGamma _3\) is the line segment from \(z=-R+\text {i}t_2\) to \(z=-R\). The height of the rectangle is defined by \(t_2=\pi /(2\alpha )>t_1^*\) such that only the first singularity is inside the contour. By the residue theorem and taking the limit \(R\rightarrow \infty \),


Now consider the integrals on \(\varGamma _1\), \(\varGamma _2\), and \(\varGamma _3\). First, on \(\varGamma _1\)


We can bound |F| in (84) using following inequalities

$$\begin{aligned}&|\sinh \alpha z|=|\sinh \alpha R\cos \alpha t+\text {i}\cosh \alpha R\sin \alpha t| \le 2|\cosh \alpha R|=|\text {e}^{\alpha R}+\text {e}^{-\alpha R}|, \end{aligned}$$
$$\begin{aligned}&|\cosh \alpha z|=|\cosh \alpha R\cos \alpha t+\text {i}\sinh \alpha R\sin \alpha t| \le 2|\cosh \alpha R|=|\text {e}^{\alpha R}+\text {e}^{-\alpha R}|, \end{aligned}$$
$$\begin{aligned}&|z|\le \sqrt{R^2+t_2^2}, \end{aligned}$$
$$\begin{aligned}&|\sinh 2\alpha z|=|\cosh 2\alpha R\cos 2\alpha t+\text {i}\sinh 2\alpha R\sin 2\alpha t|\nonumber \\&\quad \quad \quad \quad \quad \quad \ge |\sinh 2\alpha R\cos 2\alpha t+\text {i}\sinh 2\alpha R\sin 2\alpha t|\ge |\sinh 2\alpha R|. \end{aligned}$$


$$\begin{aligned} \underset{R\rightarrow \infty }{\lim }\left| F\right|\le & {} \underset{R\rightarrow \infty }{\lim }\left[ \frac{\left[ 2\left| K\right| +\left| N_1-N_2\right| \left( R^2+t_2^2\right) \right] \left| 2\cosh (\alpha R)\right| }{|\sinh (2\alpha R)|-\sqrt{R^2+t_2^2}\left| \sin {2\alpha }\right| }\right. \nonumber \\&+ \left. \frac{\left| N_1-N_2\right| \left| \cot \alpha \right| \sqrt{R^2+t_2^2}\left| 2\cosh (\alpha R)\right| }{|\sinh (2\alpha R)|-\sqrt{R^2+t_2^2}\left| \sin {2\alpha }\right| }\right] =0. \end{aligned}$$


$$\begin{aligned} \underset{R\rightarrow \infty }{\lim }\int _{\varGamma _1}F(z,\alpha )\text {e}^{\text {i}\xi z} \mathrm {d}z=0. \end{aligned}$$

For the same reason,

$$\begin{aligned} \underset{R\rightarrow \infty }{\lim }\int _{\varGamma _3}F(z,\alpha )\text {e}^{\text {i}\xi z} \mathrm {d}z=0. \end{aligned}$$

For the \(\varGamma _2\) contour

$$\begin{aligned}&\underset{R\rightarrow \infty }{\lim }\left| \int _{\varGamma _2}F(z,\alpha )\text {e}^{\text {i}\xi z} \mathrm {d}z\right| \le \underset{R\rightarrow \infty }{\lim }\int _{-R}^{R}\left| F(s+\text {i}{t_2},\alpha )\text {e}^{-\xi t_2+\text {i}\xi s}\right| \mathrm {d}s\nonumber \\&\quad \le \left\{ 2\left| K\right| \left| \int _{-\infty }^{\infty }\frac{\sinh (\alpha (s+\text {i}t_2))}{\sinh (2\alpha (s+\text {i}t_2))+(s+\text {i}t_2)\sin (2\alpha )} \mathrm {d}s\right| \right. \nonumber \\&\qquad +\left[ \, \left| \int _{-\infty }^{\infty }\frac{(s+\text {i}t_2)^2\sinh (\alpha (s+\text {i}t_2))}{\sinh (2\alpha (s+\text {i}t_2))+(s+\text {i}t_2)\sin (2\alpha )} \mathrm {d}s\right| \right. \nonumber \\&\qquad +\left. \left. \left| \int _{-\infty }^{\infty }\frac{(s+\text {i}t_2)\cot {\alpha }\cosh (\alpha (s+\text {i}t_2))}{\sinh (2\alpha (s+\text {i}t_2))+(s+\text {i}t_2)\sin (2\alpha )} \mathrm {d}s \right| \, \right] \cdot \left| N_1-N_2\right| \right\} \times \text {e}^{-\xi t_2}. \end{aligned}$$

All three integrals in the square brackets of Eq. (92) are finite, so the bound on the integral on contour \(\varGamma _2\) is of order \(\text {e}^{-\xi t_2}\) for \(\xi \gg 1\).

Finally, consider the residue from the singularity:

$$\begin{aligned}&2\pi \text {i}\cdot \;\text {residue of}\;F(z,\alpha )\text {e}^{\text {i}\xi z}\;\text {at}\;{it_{1}}^{*} =2\pi \text {i}\underset{z\rightarrow z_{1}^{*}}{\lim }(z-z_{1}^{*})F(z,\alpha )\text {e}^{\text {i}\xi z}\nonumber \\&\quad =-2\pi \text {e}^{-\xi {t_{1}^{*}}}\cdot \frac{2K+\left( N_{1}-N_{2}\right) {t_{1}^{*}}\left( {t_{1}^{*}}-\cot \alpha \coth \alpha {t_{1}^{*}}\right) }{2\alpha \cos 2\alpha t_{1}^{*}+\sin 2\alpha } \times \sin \alpha {t_{1}^{*}}\sim {\mathcal {O}}\left( \text {e}^{-\xi t_1^*}\right) . \end{aligned}$$

From Eq. (77), \(\xi \gg 1\) near the corner. Then the integral on \(\varGamma _2\) is asymptotically smaller than the residue at \(z=\text {i}t_1^*\) because \(\text {e}^{-\xi t_2}\ll \text {e}^{-\xi t_1^*}\). Thus the dominant asymptotic contribution to the integral is

$$\begin{aligned} I\sim \frac{1}{2}\text{ Re }\left\{ 2\pi \text {i}\cdot \text{ residue } \text{ of } F(z,\alpha )\text {e}^{\text {i}\xi z} \text{ at } z=\text {i}t_1^*\right\} . \end{aligned}$$

Using this result in Eq. (67) we obtain

$$\begin{aligned}&\sigma _{x}+\sigma _{y}=H(\xi ,\alpha )\cdot I =4\left( \cosh \xi -\cos \alpha \right) \sin \alpha \cdot I\nonumber \\&\quad \sim -\frac{2K+\left( N_{1}-N_{2}\right) t_{1}^{*}\left( t_{1}^{*}-\cot \alpha \coth \alpha t_{1}^{*}\right) }{2\alpha \cos 2\alpha t_{1}^{*}+\sin 2\alpha } 2\pi \sin \alpha \sin \alpha t_{1}^{*}\cdot \left( \frac{2}{\varepsilon }\sin \alpha \right) ^{1-t_{1}^{*}}, \end{aligned}$$

where \(t_{1}^{*}\) is the smallest non-zero root of Eq. (82). Recalling that \(\varepsilon =\pi /2-\theta \) is the proximity to the corner, since \(t_{1}^{*}<1\), \(\sigma _{x}+\sigma _{y}\) has an integrable singularity at the corner. Note the exponent \(1-t_1^*\) matches the exponent for the singular solutions for an infinite wedge geometry [12], as Eq. (82) is equivalent to Eq. (30). More generally, for \(\pi /2<\alpha <\pi \) (cases like Fig. 13) there is an integrable singularity with \(1<t_1^*<3/2\), and for \(0<\alpha <\pi /2\) (cases like Fig. 9) there is no singularity because \(0<t_1^*<1\).

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Wang, W., Spencer, B.J. Numerical solution for the stress near a hole with corners in an infinite plate under biaxial loading. J Eng Math 127, 13 (2021).

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