Prediction of In-Plane Shear Properties of a Composite with Debonded Interface

Abstract

An analytical approach is established to analyze the mechanical behavior of a unidirectional (UD) composite subjected to an in-plane shear. The effects of the interface debonding on the in-plane shear strength and plasticity of the composite are evaluated respectively. The volume-averaged internal stresses of the fiber and matrix are calculated through Bridging Model. Owing to the stress fluctuation in the matrix caused by other phases (such as the imbedded fiber and interface crack), the homogenized matrix stresses must be converted into true values based on a stress concentration factor (SCF). A new in-plane shear SCF of a composite with debonded interface is derived and applied to the predictions of strain and strength in this paper. Moreover, the contribution of the relative slippage between fiber and matrix to the non-linear shear deformation of the composite is analyzed. Finally, the efficiency and accuracy of our theory are verified with a series of examples. The approach is very efficient because the calculation can be achieved with explicit expressions mainly based on the constituent material properties. The comparisons between predicted strength and deformation with the experimental results show that our analytical approach can give out reliable underlying data for multi-scale analysis. Moreover, this theoretical method can tell if and how much the interface of a composite needs to be modified for a given load environment.

Appendices

Appendix A. Homogenized Internal Stresses by Bridging Model

For a UD composite with both fiber and matrix in an elastic deformation stage, the relationships between internal stresses of the fiber and matrix and the stresses applied on the composite are expressed as [42]

$$\left\{\sigma_i^f\right\}=\left(V_f\left[I\right]+V_m\left[A_{ij}\right]^{-1}\left\{\sigma_j\right\}\right)$$

(20)

$$\left\{\sigma_i^m\right\}=\left[A_{ij}\right]\left(V_f\left[I\right]+V_m\left[A_{ij}\right]\right)^{-1}\left\{\sigma_j\right\}$$

(21)

The first coordinate, x₁, is along the fiber axial direction. x₃ is along plate thickness directions. The arbitrary stress vector applied on the composite has six components as {σ_j} = {σ_11, σ_22, σ₃₃, σ₂₃, σ₁₃, σ₁₂}^T, and the vectors $\left\{\sigma_i^f\right\}$ and $\left\{\sigma_i^m\right\}$ have the same form. V_f and V_m (= 1-V_f) are the fiber and matrix volume fractions and [I] is a unit tensor. Explicit expressions for the bridging tensor [A_ij] in Eqs. (20) and (21) are given as [42]

$$\left[A_{ij}\right]=\begin{bmatrix}\alpha_{11}&\alpha_{12}&\alpha_{13}&0&0&0\\0&\alpha_{22}&0&0&0&0\\0&0&\alpha_{33}&0&0&0\\0&0&0&\alpha_{44}&0&0\\0&0&0&0&\alpha_{55}&0\\0&0&0&0&0&\alpha_{66}\end{bmatrix}$$

(22)

$$\alpha_{11}=\frac{E^m}{E_{11}^f}$$

(23)

$$\alpha_{22}=\alpha_{33}=\alpha_{44}=\beta+\left(1-\beta\right)\frac{E^m}{E_{22}^f}\;\left(0.3<\beta<0.6\right)$$

(24)

$$\alpha_{55}=\alpha_{66}=\alpha=\alpha+\left(1-\alpha\right)\frac{G^m}{G_{22}^f}\;\left(0.3<\alpha<0.6\right)$$

(25)

$$\alpha_{12}=\alpha_{13}=\frac{v_mE_{11}^f-E^mv_{12}^f}{E^m-E_{11}^f}\;\left(\alpha_{11}-\alpha_{22}\right)$$

(26)

In the equations above, $E_{11}^f$, $E_{22}^f$, $G_{12}^f$ and $v_{12}^f$ are, respectively, longitudinal, transverse and in-plane shear moduli, and longitudinal Poisson’s ratio of the fiber. E^m, G^m and ν^m are Young’s and shear moduli and Poisson’s ratio of the matrix. α and β are bridging parameters, which can be calibrated from comparison between the predicted in-plane shear and transverse Yong’s moduli of the composite, respectively. It is noted that

$$E_{22}=\frac{\left(V_f+V_m\alpha_{11}\right)\left(V_f+V_m\alpha_{22}\right)}{\left(V_f+V_m\alpha_{11}\right)\left(V_fS_{22}^f+\alpha_{22}V_mS_{22}^m\right)+V_fV_m\left(S_{21}^m-S_{21}^f\right)\alpha_{12}}$$

(27)

$$G_{12}=\frac{G_{12}^fG^m\left(V_f+V_m\alpha_{66}\right)}{V_fG^m+V_mG_{12}^f\alpha_{66}}$$

(28)

If no adjustment from an experiment is applicable, both α and β can be assumed to be 0.3 in most cases [42].

Appendix B. Expressions of Matrix SCFs

Definition of a SCF of the matrix in a composite is generally given by [23]

$$\begin{aligned}K_{ij}=\frac1{\left|\overrightarrow R^b-\overrightarrow R^\alpha\right|}\int_{\left|\overrightarrow R^\alpha\right|}^{\left|\overrightarrow R^b\right|}\frac{\widetilde\sigma_{ij}^m}{{\left(\widetilde\sigma_{ij}^m\right)}_{BM}}d\left|\overrightarrow R\right|\end{aligned}$$

(29)

$K_{ij}$ is the SCF related to $\sigma_{ij}^m$ or $d\sigma_{ij}^m\cdot{\left(\sigma_{ij}^m\right)}_{BM}$ is the matrix stress calculated from Bridging Model $\widetilde\sigma_{ij}^m$ and is a point-wise matrix stress generally obtained through an elasticity on a CCA (coaxial cylinder assemblage) model. $\overrightarrow R$ is a vector along a line perpendicular to the fracture surface of the composite under the given load. $\overrightarrow R^{a}$ and $\overrightarrow R^{b}$ are the vectors of ending at the surfaces of the fiber and matrix cylinders, respectively. a is the radius of the fiber. b and a are correlated as $b=\frac\alpha{\sqrt{V_f}}$. Explicit formulae for the SCFs of matrix in a composite are listed as follows [22,23,24].

The transverse tension SCF of the matrix having a perfect interface bonding with the fiber, $K_{22}^{t}$ is given by

$$K_{22}^t=\left[1+\frac{\sqrt{V_f}}2A+\frac{\sqrt{V_f}}2\left(3-V_f-\sqrt{V_f}\right)B\right]\frac{\left(V_f+0.3V_m\right)E_{22}^f+0.7V_mE^m}{0.3E_{22}^f+0.7E^m}$$

(30)

The expression of transverse compressive SCF of the matrix, $K_{22}^{c}$, is

$$\begin{aligned}K_{22}^c&=\Bigg\{\Bigg[1-A\sqrt{V_f}\frac{\sigma_{u,c}^m-\sigma_{u,t}^m}{4\sigma_{u,c}^m}+\frac B{2\left(1-\sqrt{V_f}\right)}\Bigg[-V_f^2\Bigg[1-2\left(\frac{\sigma_{u,c}^m-\sigma_{u,t}^m}{2\sigma_{u,c}^m}\right)^2\Bigg]{+} \\& \frac{\left(\sigma_{u,c}^m-\sigma_{u,t}^m\right)}{\sigma_{u,c}^m}\left(1+\frac{\sigma_{u,c}^m-\sigma_{u,t}^m}{\sigma_{u,c}^m}\right)-\sqrt{V_f}\Bigg[\frac{\sigma_{u,c}^m-\sigma_{u,t}^m}{\sigma_{u,c}^m}+1-2\left(\frac{\sigma_{u,c}^m-\sigma_{u,t}^m}{2\sigma_{u,c}^m}\right)^2\Bigg]\Bigg]\Bigg]\Bigg\} \\ &\frac{\left(V_f+0.3V_m\right)E_{22}^f+0.7V_mE^m}{0.3E_{22}^f+0.7E^m}\end{aligned}$$

(31)

$$A=\frac{2E_{22}^fE^m\left(V_{12}^f\right)^2+E_{11}^f\left\{E^m\left(V_{23}^f-1\right)-E_{12}^f\left[2\left(v^m\right)^2+v^m-1\right]\right\}}{E_{11}^f\left[E_{22}^f+E^m\left(1-v_{23}^f\right)+E_{22}^fv^m\right]-2E_{22}^fE^m\left(v_{12}^f\right)^2}$$

(32)

$$B=\frac{E^m\left(1+v_{23}^f\right)-E_{22}^f\left(1+v^m\right)}{E_{22}^f\left[v^m+4\left(v^m\right)^2-3\right]-E^m\left(1+v_{23}^f\right)}$$

(33)

$E_{11}^f$, $E_{22}^f$, and $G_{12}^f$ are, respectively, longitudinal, transverse, and in-plane shear moduli of the fiber, $v_{23}^f$is its transversal Poisson’s ratio, $v_{12}^f$ and is its longitudinal Poisson’s ratio. E^m and G^m are Young’s and shear moduli of the matrix, respectively, and ν^m is its Poisson’s ratio.

The formula of in-plane shear SCF of the matrix, is shown as

$$K_{12}=\left\{1-V_f\frac{G_{12}^f-G^m}{G_{12}^f+G^m}\left(W\left(V_f\right)-\frac13\right)\right\}\frac{V_f+A_{66}V_m}{A_{66}}$$

(34)

$$W\left(V_f\right)=\int\limits_0^a\frac1\alpha\sqrt{1-\frac{x_3^2}{\alpha^2}}\sqrt{\frac1{V_f}-\frac{x_3^2}{\alpha^2}}dx_3\approx\pi\sqrt{V_f}\left[\frac1{4V_f}-\frac4{128}-\frac2{512}V_f-\frac5{4096}V_f^2\right]$$

(35)

$G_{12}^f$ and G^m are the longitudinal shear moduli of the fiber and matrix, respectively.

Besides, the transverse tension SCF of the matrix after the interface debonding, is given by

$$\begin{aligned}\hat K_t^{22}=\hat K_t^{22}\left(\psi\right)=Re\Big\{e^{-2i\psi}M\left(be^{i\psi}\right)\left(\alpha^2/b-b\right)-e^{-2i\psi}\left(N_2-N_1\left(\frac{\alpha^2}be^{-i\psi}\right)\right)\\+e^{-i\psi}\left(2+e^{-2i\psi}\right)\left[N\left(be^{i\psi}\right)-N_3\right]\Big\}\frac{\left(V_f+0.3V_m\right)E_{22}^f+0.7V_mE^m}{2\left(b-a\right)\left(0.3E_{22}^f+0.7E^m\right)}\end{aligned}$$

(36)

The functions in Eq. (35), N, N₁, N₂ and N₃, are given by

$$N\left(z\right)=Fz+\frac{a^2k}z-\left(z-\alpha e^{i\psi}\right)^{0.5+i\lambda}\left(z-\alpha e^{i\psi}\right)^{0.5-i\lambda}\left[\left(F-0.5\right)-\frac D{\alpha^2z}\right]$$

(37)

$$N_1\left(z\right)=Fz+\frac{a^2k}z+\frac1\xi\left(z-\alpha e^{i\psi}\right)^{0.5+i\lambda}\left(z-\alpha e^{i\psi}\right)^{0.5-i\lambda}\left[\left(F-0.5\right)-\frac D{\alpha^2z}\right]$$

(38)

$$N_2=\alpha Fe^{-i\psi}+\alpha ke^{i\psi},\;N_3=Fae^{i\psi}+e^{-i\psi}\alpha k$$

(39)

$$M\left(z\right)=F\frac{\alpha^2k}{z^2}-\left[\left(F-0.5\right)z+H+\frac Cz+\frac D{z^2}\right]\chi\left(z\right)$$

(40)

$$F=\frac{1-\left[\cos\left(\psi\right)+2\lambda\sin\left(\psi\right)\right]\mathrm{exp}\left[2\lambda\left(\pi-\psi\right)\right]+\left(1-k\right)\left(1+4\lambda^2\right)\sin\left(\psi\right)}{{\displaystyle\frac4k}-2-2\left[\cos\left(\psi\right)+2\lambda\sin\left(\psi\right)\right]\mathrm{exp}\left[2\lambda\left(\pi-\psi\right)\right]}$$

(41)

$$H=\alpha\left(0.5-F\right)\left(\cos\left(\psi\right)+2\lambda\sin\left(\psi\right)\right)$$

(42)

$$C=\alpha^2\left(k-1\right)\left[\cos\left(\psi\right)-2\lambda\sin\left(\psi\right)\right]exp\left[2\lambda\left(\psi-\pi\right)\right]$$

(43)

$$D=\left(1-k\right)\alpha^3\mathrm{exp}\left[2\lambda\left(\psi-\pi\right)\right]$$

(44)

$$\chi\left(z\right)=\left(z-\alpha e^\psi\right)^{-0.5+i\lambda}\left(z-\alpha e^\psi\right)^{-0.5-i\lambda}$$

(45)

$$k=\frac{\mu_1\left(1+\kappa_2\right)}{\left(1+\xi\right)\left(\mu_1+\kappa_1\mu_2\right)}$$

(46)

$$\lambda=-\left(\ln\xi\right)/\left(2\pi\right),\;\xi=\left(\mu_2+\kappa_2\mu_1\right)/\left(\mu_1+\kappa_1\mu_2\right)$$

(47)

$$\kappa_1=3-4v^m,\;\kappa_2=\frac{3-v_{23}^f-4v_{12}^fv_{21}^f}{1+v_{23}^f}$$

(48)

$$\mu_1=\frac{E^m}{2\left(1+v^m\right)},\;\mu_2=\frac{E_{22}^f}{2\left(1+v_{23}^f\right)}$$

(49)

$$b=\alpha/\sqrt{V_f}$$

(50)

The debonding half angle, $\psi$ is the solution to the following equation [44]:

$$\mathrm{Re}{\left\{\left(G_0-\frac1k-\frac{2\left(1-k\right)}{k\;\mathrm{exp}\left(i\varphi\right)}\mathrm{exp}\left[2\lambda\left(\psi-\pi\right)\right]\right)R\left(e^{i\varphi}\right)\right\}}_{\varphi=\psi-\gamma}=0$$

(51)

$$R\left(exp\left(i\varphi\right)\right)=\left[exp\left(i\left(\varphi\right)\right)-e^{i\psi}\right]^{0.5+i\lambda}\left[exp\left(i\left(\varphi\right)\right)-e^{i\psi}\right]^{0.5-i\lambda}exp\left(-i\left(\varphi\right)\right)$$

(52)

$$G_0=\frac{1-\left[\cos\left(\psi\right)+2\lambda\sin\left(\psi\right)\right]\mathrm{exp}\left[2\lambda\left(\pi-\psi\right)\right]+\left(1-k\right)\left(1+4\lambda^2\right)\sin\left(\psi\right)}{2-k-k\left[\cos\left(\psi\right)+2\lambda\sin\left(\psi\right)\right]\mathrm{exp}\left[2\lambda\left(\pi-\psi\right)\right]}$$

(53)

$$\gamma=\left\{\begin{array}{l}\frac{2\lambda\left(J_1^2+J_2^2\right)}{J_1^2+J_2^2-2J_2J_3},\;if\;\xi<1\\-\frac{2\lambda\left(J_1^2+J_2^2\right)}{J_1^2+J_2^2-2J_2J_3},\;if\;\xi>1\end{array}\right.$$

(54)

$$J_1=kG_0-1-2\left(1-k\right)\xi\;\mathrm{exp}\left(2\lambda\psi\right)\cos\left(\psi\right)$$

(55)

$$J_2=2\left(1-k\right)\xi\;exp\left(2\lambda\psi\right)\;\sin\left(\psi\right)$$

(56)

$$J_3=2\left(1-k\right)\xi\;exp\left(2\lambda\psi\right)\left[J_1\;\cos\left(\psi\right)-J_2\;\sin\left(\psi\right)\right]/J_2$$

(57)

If = 1, no solution for $\psi$ is available, and the corresponding interface crack is singular. But, one can always adjust a fiber or matrix property so that 1, as deviation exists in measurement of it $v_{23}^f$. is the transverse Poisson’s ratio of the fiber. $\sigma_{u,\;t}^m,\;\sigma_{u,\;c}^m$, and $\sigma_{u,\;s}^m$ are, respectively, the matrix tensile, compressive and shear strengths.

Appendix C. Integration After Interface Debonding

For a certain longitudinal section located at $x_3=x_3^0,\alpha\leq x_3^0\leq b$, the length of the integration line between the starting and the ending points is $\sqrt{2\left[b^2-\left(x_3^0\right)^2\right]}$. Thus, the matrix shear SCF along this line is given by

$${\hat{K}}_{12\vert\psi=0.5\pi,a\leq x_3^0\leq b}=\frac{1}{\sqrt{2(b^2-{(x_3^0)}^2)}}\int\limits_0^{\sqrt{2(b^2-{(x_3^0)}^2)}}\frac{\sigma_{12}^m}{{(\sigma_{12}^m)}_{BM}}dS$$

(58)

Note that $b=a/ \sqrt{V_f}$, where V_f is the fiber volume fraction and S is along the integration line. In Eq. (57), ${\left(\sigma_{12}^m\right)}_{BM}$ is the matrix in-plane shear stress calculated through Bridging Model. is a point-wise stress in the matrix given by Eq. (7). Since dS equals (-$\sqrt2$  dx₂) on the integration line (the value of x₃ is a constant), the line integration of Eq.(57) is changed to (noticing that  $x_3^0=x_3$).

$${\hat K}_{12\vert\psi=0.5\pi,\;\alpha\leq x_3^0\leq b}=\frac1{\sqrt{b^2-\chi_3^2}}\int_{-\sqrt{b^2-\chi_3^2}}^0\frac{\sigma_{12}^m}{{\left(\sigma_{12}^m\right)}_{BM}}dx_2$$

(59)

Substituting ψ = 0.5π into Eq. (7), one has

$$\sigma_{12}^m=\sigma_{12}^0\;Re\left\{\frac{G^m}{G^m+G_{12}^f}\left(1-\frac{a^2}{z^2}\right)+\frac{G_{12}^f}{G^m+G_{12}^f}\left(z-ai\right)^{-0.5}\left(z+ai\right)^{-0.5}\left(z-\frac{a^3}{z^2}\right)\right\}$$

(60)

Replacing z in Eq. (59) with x₂ + ix₃ and taking the integration along the x₂ axis gives

$$\begin{aligned}&\int_{-\sqrt{b^2-x_3^2}}^0\;\sigma_{12}^mdx_2\\&=\sigma_{12}^0\;\int_{-\sqrt{b^2-x_3^2}}^0\;Re\left\{\frac{G^m}{G^m+G_{12}^f}\left[1-\frac{\alpha^2}{\left(x_2+x_3i\right)^2}\right]+\frac{G_{12}^f}{G^m+G_{12}^f}N\left(x_2+x_3i\right)\right\}dx_2\end{aligned}$$

(61)

$$N\left(x_2+x_3i\right)=\chi\left(x_2+x_3i\right)\left(x_2+x_3i-\frac{\alpha^3}{\left(x_2+x_3i\right)^2}\right)$$

(62)

$$\chi\left(x_2+x_3i\right)=\chi\left(x_2+x_3i+\alpha i\right)^{-0.5}\left(x_2+x_3i-\alpha i\right)^{-0.5}$$

(63)

Now let us derive the integration on the function N with respect to $x_{2}$ 

Setting $R_0\left(x_2\right)=\left(x_2+x_3i-\alpha i\right)^{0.5}\left(x_2+x_3i-\alpha i\right)^{0.5}$, its derivative with respect to $x_{2}$ is

$$\begin{aligned}&\frac{dR_0}{dx_2}=\frac{d\left[\left(x_2+x_3i-\alpha i\right)^{0.5}\left(x_2+x_3i+\alpha i\right)^{0.5}\right]}{dx_2}=0.5\times\frac{\left(x_2+x_3i+\alpha i\right)^{0.5}}{\left(x_2+x_3i-\alpha i\right)^{0.5}}+0.5\frac{\left(x_2+x_3i-\alpha i\right)^{0.5}}{\left(x_2+x_3i+\alpha i\right)^{0.5}}\\&=\frac{x_2+x_3i}{\left(x_2+x_3i-\alpha i\right)^{0.5}\left(x_2+x_3i+\alpha i\right)^{0.5}}=\left(x_2+x_3i\right)\chi\left(x_2+x_3i\right)\end{aligned}$$

(64)

Similarly, setting $R_1\left(x_2\right)=\frac{\alpha\left(x_2+x_3i-\alpha i\right)^{0.5}\left(x_2+x_3i+\alpha i\right)}{x_2+x_3i}$ and taking a derivative with respect to $x_{2}$ results in

$$\begin{aligned}&\frac{dR_1}{dx_2}=d\left[\frac{\left(x_2+x_3i-\alpha i\right)^{0.5}\left(x_2+x_3i+\alpha i\right)^{0.5}}{x_2+x_3i}\right]/dx_2\\&=\frac{\alpha\left(x_2+x_3i-\alpha i\right)^{0.5}\left(x_2+x_3i+\alpha i\right)^{0.5}}{\left(x_2+x_3i\right)^2}+\frac\alpha{x_2+x_3i}\frac{d\left[\left(x_2+x_3i-\alpha i\right)^{0.5}\left(x_2+x_3i+\alpha i\right)^{0.5}\right]}{dx_2}\\&=\frac{\alpha\left(x_2+x_3i-\alpha i\right)^{0.5}\left(x_2+x_3i+\alpha i\right)^{0.5}}{\left(x_2+x_3i\right)^2}+\frac\alpha{\left(x_2+x_3i-\alpha i\right)^{0.5}\left(x_2+x_3i+\alpha i\right)^{0.5}}\\&=\frac\alpha{\left(x_2+x_3i\right)^2}\chi\left(x_2+x_3i\right)\end{aligned}$$

(65)

Based on Eqs. (63) and (64), one can obtain that

$$\int_{-\sqrt{b^2-x_3^2}}^0\;N\left(x_2+x_3i\right)dx_2=\mathrm{Re}\left[R_0\left(x_2\right)+R_1\left(x_2\right)\right]\;\left|\begin{array}{c}0\\-\sqrt{b^2-x_3^2}\end{array}\right.$$

(66)

Therefore, the integration on Eq. (60) is found to be

$$\begin{array}{l}\int_{-\sqrt{b^2-x_3^2}}^0\sigma_{12}^mdx_2=\sigma_{12}^0\left\{\frac{G^m}{G^m+G_{12}^f}\left(x_2+\frac{x_2\alpha^2}{x_2^2+x_3^2}\right)+\frac{G_{12}^f}{G^m+G_{12}^f}\mathrm{Re}\left[R_0\left(x_2\right)+R_1\left(x_2\right)\right]\right\}\vert_{-\sqrt{b^2-x_3^2}}^0\\=\sigma_{12}^0\left\{\frac{G^m\left(1+V_f\right)}{G^mG_{12}^f}\sqrt{b^2-x_3^2}+\frac{G_{12}^f}{G^mG_{12}^f}\mathrm{Re}\lbrack\frac{\alpha+x_3i}{x_3i}\left(x_3i-\alpha i\right)^{0.5}\left(x_3i-\alpha i\right)\right. \\ \left.-(1+\frac\alpha{-\sqrt{b^2-x_3^2+x_3i}})\left(-\sqrt{b^2-x_3^2}+x_3i-\alpha i\right)^{0.5}\rbrack\right\}\end{array}$$

(67)

Substituting Eq. (66) into Eq. (58), one obtains

$$\begin{array}{l}{\hat K}_{12\vert\psi=0.5\pi,\;\alpha\leq x_3^0\leq b}=\frac{\sigma_{12}^0}{{\left(\sigma_{12}^m\right)}_{BM}\sqrt{b^2-x_3^2}}\{\frac{G^m\left(1+V_f\right)}{G^m+G_{12}^f}\sqrt{b^2-x_3^2}\\+\frac{G_{12}^f}{G^m+G_{12}^f}Re\lbrack\left(1+\frac\alpha{x_3i}\right)\left(x_3i-\alpha i\right)^{0.5}\left(x_3i+\alpha i\right)^{0.5}\\-(1+\frac\alpha{-\sqrt{b^2-x_3^2+x_3i}})\left(-\sqrt{b^2-x_3^2}+x_3i-\alpha i\right)^{0.5}\rbrack\}\end{array}$$

(68)

It should be noted that Eq. (67) is obtained only for a particular longitudinal cross-section. As done in Ref. [44], the matrix in-plane shear SCF must be determined based on an average along the thickness direction from to. This gives rise to

$${\hat K}_{12}=\frac{\int_\alpha^b{\hat K}_{12\vert\psi=0.5\pi,\;\alpha\leq x_3^0\leq b}dx_3^0}{b-\alpha}=\frac{\int_\alpha^b\lbrack\int_{-\sqrt{b^2-x_3^2}}^0\sigma_{12}^m\left(x_2,x_3\right)dx_2/\sqrt{b^2-x_3^2}\rbrack dx_3}{\left(b-\alpha\right){\left(\sigma_{12}^m\right)}_{BM}}$$

(69)

Letting $m=-\sqrt{b^2-x_3^2}+x_3i$ and $R_2=-\left(\sqrt{b^2-x_3^2}+x_3i-\alpha i\right)^{0.5}\left(-\sqrt{b^2-x_3^2}+x_3i-\alpha i\right)^{0.5}=\left(m-\alpha i\right)^{0.5}\left(m+\alpha i\right)^{0.5}$

Eq. (68) is changed to

$$\begin{array}{l}{\hat K}_{12}=\frac{\delta_{12}^0}{\left(b-\alpha\right){\left(\sigma_{12}^m\right)}_{BM}}\int\limits_a^b\{\frac{G^m}{G^m+G_{12}^f}\left(1+V_f\right)+\frac{G_{12}^f}{G^m+G_{12}^f}\mathrm{Re}\lbrack\frac{\alpha+x_3^2}{x_3i\sqrt{b^2-x_3^2}}\left(x_3i-\alpha i\right)^{0.5}\left(x_3i+\alpha i\right)^{0.5}\\-(\frac{R_2}{\sqrt{b^2-x_3^2}}+\frac{\alpha R_2}{\sqrt{b^2-x_3^2}(-\sqrt{b^2-x_3^2}+x_3i)})\rbrack\}dx_3\end{array}$$

(70)

The expression in Eq. (70) containing R₂ is hard to integrate, and must be converted into a function of m. However, $(m-\alpha{i})^{0.5}(m+\alpha{i})^{0.5}$ might equal to $(m^{2}+\alpha^{2})^{0.5}$ or $-(m^{2}+\alpha^{2})^{0.5}$. We need to choose the correct form within the integration range. Because x₃ ∈ [a, b], the phase angles of $(m-\alpha{i})$ and $(m+\alpha{i})$ are within [0.5π, π]. Thus, the phase angles of both and are in between [0.25π, 0.5π]. Hence, the phase angle of is in the second quadrant, indicating that its imaginary part is positive.

As $(m^2+\alpha^2)^{0.5}=({(b^2-2x_3^2+\alpha^2)-2x_3i\sqrt{b^2-x_3^2})}^{0.5}$ and $\mathrm{Im}(-2x_3i\sqrt{b^2-x_3^2})<0$, the phase angle of $(b^2-x_3^2+\alpha^2)-2x_3i\sqrt{b^2-x_3^2}$ is in between -π and 0. The corresponding square root is within [-0.5π,0], and the imaginary part of $(m^2+\alpha^2)^{0.5}$ is negative. Overall, for x₃ ∈ [a, b] and $m=-\sqrt{b^2-x_3^2}+x_3i$, one has $(m-\alpha{i})^{0.5}(m+\alpha{i})^{0.5}=-(m^2-\alpha{2})^{0.5}$ 

Next, from $\frac{dm}{dx_3}=\frac{x_3}{\sqrt{b^2-x_3^2}}+i=\frac{x_3+i\sqrt{b^2-x_3^2}}{\sqrt{b^2-x_3^2}}=\frac{x_3-i\sqrt{b^2-x_3^2}}{\sqrt{b^2-x_3^2}}=\frac m{i\sqrt{b^2-x_3^2}}$, one derives

$$\begin{array}{l}\int_\alpha^b\mathrm{Re}\left\{\frac{R_2\left(x_3\right)}{\sqrt{b^2-x_3^2}}\right\}dx_3=\mathrm{Re}\left\{\int_\alpha^b\frac{-\left(m^2+\alpha^2\right)^{0.5}}{\sqrt{b^2-x_3^2}}dx_3\right\}=\mathrm{Re}\left\{\int_{-\sqrt{b^2-\alpha^2+\alpha i}}^{bi}\frac{-i\left(m^2+\alpha^2\right)^{0.5}}mdm\right\}\\=-\mathrm{Re}\left\{i\left(m^2+\alpha^2\right)^{0.5}+\alpha\;\ln\frac{\left(m^2+\alpha^2\right)^{0.5}-\alpha}{\left|m\right|}\left|\begin{array}{c}bi\\-\sqrt{b^2-\alpha^2}+\alpha i\end{array}\right.\right\}=-Re\left\{i\lbrack\left(b^2-\alpha^2-2\alpha i\sqrt{b^2-\alpha^2}\right)^{0.5} \right.\\ \left. +\alpha\;\ln\frac{\left(b^2-\alpha^2-2\alpha i\sqrt{b^2-\alpha^2}\right)^{0.5}-\alpha}b-i\left(\alpha^2-b^2\right)^{0.5}-\alpha\;\ln\frac{\left(\alpha^2-b^2\right)^{0.5}-\alpha}b\rbrack\right\}\end{array}$$

(71)

Similarly, 

$$\begin{aligned}&\int_\alpha^b\frac{\alpha R_2\left(x_3\right)}{\left(-\sqrt{b^2-x_3^2+x_3i}\right)\sqrt{b^2-x_3^2}}dx_3=\mathrm{Re}\;\int_{-\sqrt{b^2-\alpha^2+\alpha i}}^{bi}\;\frac{-\alpha\left(m^2+\alpha^2\right)^{0.5}}{m^2}dm\\&=\mathrm{Re}\left\{-\alpha i\lbrack-\left(1+\frac{\alpha^2}{m^2}\right)^{0.5}+\ln({(m^2+\alpha^2)}^{0.5}+m)\rbrack\; \Bigg|\begin{array}{c}bi\\-\sqrt{b^2-\alpha^2+\alpha i}\end{array}\right\}\\&=\mathrm{Re}\left\{\alpha i\lbrack\ln\left(\left(b^2-\alpha^2-2\alpha i\sqrt{b^2-\alpha^2}\right)^{0.5}-\sqrt{b^2-\alpha^2}+\alpha i\right)\right. \\ &\left. -(1+\frac\alpha{-\sqrt{b^2-x_3^2+x_3i}})\left(-\sqrt{b^2-x_3^2}+x_3i-\alpha i\right)^{0.5}\rbrack\right\}\end{aligned}$$

(72)

Now, let us derive the integration for the other part of Eq. (68), which contains

$$Re\left\{\left(\frac{\alpha+x_3i}{x_3i\sqrt{b^2-x_3^2}}\right)^{0.5}\left(x_3i-\alpha i\right)^{0.5}\left(x_3i+\alpha i\right)^{0.5}\right\}$$

(73)

As $\left(x_3i-\alpha i\right)^{0.5}\left(x_3i+\alpha i\right)^{0.5}=i\sqrt{x_3^2-\alpha^2}$, Eq. (73) equals to $\frac{\alpha\sqrt{x_3^2-}\alpha^2}{x_3\sqrt{b^2}-x_3^2}$ 

The integration on Eq. (73) becomes

$$\int\limits_a^b\frac{a\sqrt{x_3^2-a^2}}{x_3\sqrt{b^2-x_3^2}}dx_3=-\int\limits_{\sqrt{b^2-a^2}}^{0}\frac{a\sqrt{x_3^2-a^2}}{x_3^2}d(\sqrt{b^2-x_3^2})=\int\limits_0^{\sqrt{b^2-a^2}}\frac{a\sqrt{b^2-a^2-x^2}}{b^2-x^2}dx$$

(74)

In order to get the explicit integral expression, the parameter $x(0\leq x\leq\sqrt{b^2-\alpha^2})$ in Eq. (74) is replaced with $(\sqrt{b^2-\alpha^2}\sin\left(\theta\right),\;0\leq\theta\leq0.5\pi)$. It becomes

$$\begin{aligned}&\int\limits_0^{\sqrt{(b^2-a^2)}}\frac{a\sqrt{(b^2-a^2)(1-\sin^2(\theta))}}{b^2-(b^2-a^2)\sin^2(\theta)} d{\sqrt{(b^2-a^2)}\sin(\theta)}=a\int\limits_0^{\frac{\pi}{2}}\frac{(b^2-a^2)\cos^2(\theta)}{a^2+(b^2-a^2)\cos^2(\theta)} \\ &=\int\limits_0^{\frac{\pi}{2}}a\lbrack1-\frac{2a^2}{2a^2+(b^2-a^2)(1+\cos(2\theta))}\rbrack d\theta\\ &=\frac{a\pi}2-\int\limits_0^{\frac{\pi}{2}}\lbrack\frac{a^2}{(b^2-a^2)\cos(\theta)+(b^2+a^2)}\rbrack d\theta\end{aligned}$$

(75)

Making use of the integral formula

$\int\frac1{\alpha+b\;\cos\left(x\right)}d\left(x\right)=\frac2{\alpha+b}\sqrt{\frac{\alpha+b}{\alpha-b}}\;{\mathrm{arc}\tan}\left(\sqrt{\frac{\alpha-b}{\alpha+b}}\tan\frac x2\right)$,

Eq. (75) is simplified to $\frac{\alpha\pi}2-\alpha\left[\sqrt{V_f}\;arc\tan\left(\sqrt{V_f}\;\tan\frac\theta2\right)\right]\left|\begin{array}{c}\pi\\0\end{array}=\right.\frac{\alpha\pi}2\left(1-\sqrt{V_f}\right).$ 

Combining Eqs. (67), (70), (71), (72) and (75), one finally obtains the expression of the in-plane shear SCF with debonded interface as Eq. (8).