Mathematical modelling for multiple straight cracks in piezoelectric ceramics

Abstract

Piezoelectric materials are widely used in electronic devices like sensors. In this study, a strip saturation zone model is provided for a transversely isotropic piezoelectric plate which is cut along three equal collinear impermeable cracks. The developed saturation zones are assumed to be subjected to the linearly varying electric displacement conditions. The problem is solved using Stroh formalism and the complex variable approach. Analytic closed-form expressions are obtained for fracture parameters including crack opening displacement, crack opening potential, and saturation zone length. A numerical study also presented showing the effect of linearly varying electric displacement on the saturation zone length.

Keywords

Piezoelectric materials yield zones saturation zones crack opening displacement multiple cracks

Nomenclature

C _i, D _i, (i = 0,1,2,3)

Constants of the problem

F (𝜃, k), E (𝜃, k), 𝛱(𝜃, 𝛼², k)

Incomplete elliptic integral of first, second and third kind respectively

L _i (i = 1,2,3)

Crack lengths

P _n (z)

Polynomial of degree n

Δu ₂ (t)

Crack opening displacement

Δu ₄ (t)

Crack opening potential drop

± a, ±b, ±c

Crack tips

± a ₁, ±b ₁ ± c ₁

Tips of saturation zones

p (t), q (t)

Applied stresses on the yield zones

z = x ₁ + ix ₂

Complex variable

𝛤^′

$-\frac{1}{2}(N_{1}-N_{2})e^{-2i{\alpha}},N_{1}$ and N ₂ are the values of principal stresses at infinity, 𝛼 be the angle between N ₁ and the ox-axis

𝛶₂ (z), 𝛶₄(z)

Complex stress functions

𝜅

$=\frac{3-{\gamma}}{1+{\gamma}}$ for the plane-stress, =3 −4𝛾 for the plane-strain

𝜎_ij

Components of stress

𝜎₂₂ ^∞

Remotely applied stress at infinite boundary of the plate

D ₂ ^∞

Electric displacement at boundary of the plate

D _s

Electric displacement on electric saturation zones.

1. Introduction

Piezoelectric materials are extensively utilized in the modern electromechanical devices, construction of microbalance, actuators, sensors and many other devices [15]. There is a fair chance to develop a fatigue crack due to excessive use of the devices and brittle behaviour of the piezoelectric materials [18]. However, these type of materials can be damaged permanently owing to the influence of electromechanical load. Therefore, the study of the fracture problem of such materials is gaining more and more attention from researchers in the field of fracture mechanics.

Parton [13] provides an analysis of a single cracks in piezoelectric medium under mechanical stress. Pak [12] studied the stress intensity factor criterion and the energy-release rate criterion with the framework of linear elastic fracture mechanics and investigate that the electric fields will tried to slow down the crack growth. A plane crack problem at the interface of a bi-material piezoelectric was studied by Qin et al. [14]. A mode-III crack filled with a dielectric medium was studied by Zhang et al. [19]. Zhang and Tong [20] further studied an elliptical cylinder cavity using complex potential method. McMeeking [10] discussed the concept of energy release rate for Griffith crack propagation in an infinite body. Yang [21] developed the closed form solution of mode-I crack problem in piezoelectric materials for external load to open the crack. Yokobori et al. [22] discussed the interaction between crack and a slip band based on the concept of macro and micro fracture mechanics.

The analysis of two of more cracks is a very complicated procedure due their mathematical complexity. Multiple crack problem in a piezoelectric material under uniform remote loads was studied by Gao [5] and obtained the exact solutions of the problem using complex variable method. Considering the effect of permittivity of air, Hao [7] investigated the exact solution of the problem of multiple collinear cracks in piezoelectric materials. Li [8] analyzed an infinitely long piezoelectric strip of finite width containing two equal collinear cracks under different types of uniform anti-plane shear and uniform in-plane electric loading conditions. Using the Fourier transform method, basic solution of two collinear cracks in a piezoelectric material was obtained by Liang [9]. A two-dimensional mode-I strip electro-mechanical yielding model for two equal collinear cracks present in a transversely isotropic piezoelectric plate was studied by mean Stroh formalism and complex variable technique by Bhargava et al. [2] for both semi-permeable and impermeable cracks. They also studied cohesive saturation limit electric displacement [3]. Fourier series method together with integral equation method was used by Bhargava et al. [4] to solve the problem of two semi-permeable collinear cracks in a piezoelectric materials using Fourier transform method. Strip-saturation model for two cracks problem in piezoelectric media addressed by, his work is extended by Bhargava et al. [15], Singh et al. [16]. The problem of two cracks was discussed by various researchers for different loading conditions in piezoelectric materials. However, to study the effect of outer crack on inner crack and vice vera the modelling of cracks will be extended to more than two cracks under the different boundary conditions.

A strip-saturation model is suggested in the current study for a transversely isotropic piezoelectric plate that has three equal collinear cracks. An analysis has been made in the model to study the effect of linearly varying electric displacement on the growth of the three collinear straight cracks. Analytical expressions are obtained for saturation zone length, crack opening potential drop etc. A numerical study is also carried out to investigate the effect of linearly varying electric displacement on the saturation zone length.

2. Basic mathematical formulation

In a rectangular Cartesian coordinate system x _i(i = 1,2,3) the basic mathematical equations for anisotropic piezoelectric material in terms of stress components, 𝜎_ij and electric displacement components, D _i(i = 1,2,3) be expressed as:

Constitutive relations $\begin{eqnarray}\displaystyle {\sigma}_{ij}=C_{ijkl}{\gamma}_{kl}-e_{kij}E_{k},\quad D_{k}=e_{kij}{\gamma}_{ij}+{\varepsilon}_{kl}E_{l}. & & \displaystyle\end{eqnarray}$ (1)

Gradient equations $\begin{eqnarray}\displaystyle {\gamma}_{ij}={\displaystyle \frac{1}{2}}(u_{i,j}+u_{j,i}),\quad E_{i}=-{\varphi}_{,i}. & & \displaystyle\end{eqnarray}$ (2)

Mechanical and electrical equilibrium equations $\begin{eqnarray}\displaystyle {\sigma}_{ij,j}=0,\quad D_{i,i}=0, & & \displaystyle\end{eqnarray}$ (3) where 𝜎_ij is the stress tensor, D _i is the electric displacement vector, 𝜑 the electric potential, 𝛾_ij is the strain tensor, E _i the electric field intensity, C _ijkl the elastic stiffness tensor measured at a constant electric field, e _kij the piezoelectric tensor measured in possession of a spontaneous electric field, 𝜀_ik the dielectric tensor, (i, j) stress direction, k direction of electric displacement, l electric field direction and i, j, k, l = 1,2,3.

In case of two-dimensional problem all the field variables constitute a relation in x ₁, x ₂ and independent of x ₃. Hence, the generalized displacement vector u can be written as, using [1], $\begin{eqnarray}\displaystyle \mathbf{u}=[u_{1},u_{2},u_{3},{\varphi}]^{T}=\mathbf{a}f(x_{1}+px_{2}) & & \displaystyle\end{eqnarray}$ (4) where f (x ₁ + px ₂) is an analytic function, p is a constant, and a a four-element column, superscript T denotes the transpose.

For arbitrary function f (x ₁ + px ₂), Eqs (1)–(3) can be satisfied by Eqs (4) if $\begin{eqnarray}\displaystyle [\mathbf{Q}+p(\mathbf{R}+\mathbf{R}^{T})+p^{2}\mathbf{S}]\mathbf{a}=\mathbf{0}, & & \displaystyle\end{eqnarray}$ (5) where the matrices Q, R, S are given by $\begin{eqnarray}\displaystyle \mathbf{Q}=\left[\begin{array}{@{}cc@{}}C_{i1k1} & e_{11i}\\ e_{11i}^{T} & -{\varepsilon}_{11}\\ \end{array}\right],\quad \mathbf{R}=\left[\begin{array}{@{}cc@{}}C_{i1k2} & e_{21i}\\ e_{12i}^{T} & -{\varepsilon}_{12}\\ \end{array}\right],\quad \mathbf{S}=\left[\begin{array}{@{}cc@{}}C_{i2k2} & e_{22i}\\ e_{22i}^{T} & -{\varepsilon}22\\ \end{array}\right],\quad (i,k=1,2,3). & & \displaystyle\end{eqnarray}$ (6)

For a non-trivial solution of Eq. (5), $\begin{eqnarray}\displaystyle |\mathbf{Q}+p(\mathbf{R}+\mathbf{R}^{T})+p^{2}\mathbf{S}|=0. & & \displaystyle\end{eqnarray}$ (7)

The Eq. (7) has eight complex roots p _𝛼 and $\bar{p}_{{\alpha}}$ (𝛼 = 1,2,3,4) where p _𝛼 are all imaginary because of the positive definiteness of the strain energy and electric energy densities. One can obtain the values of p _𝛼 after solving the following eigenvalue problem. $\begin{eqnarray}\displaystyle N{\zeta}=p{\zeta} & & \displaystyle\end{eqnarray}$ (8) where $\begin{eqnarray} \displaystyle \text{} & \text{} & \displaystyle N=\left( \begin{array}{@{} cc@{}}N_{1} & N_{2}\\ N_{3} & N_{1}^{T}\\ \end{array} \right),\quad {\zeta}=\left( \begin{array}{@{} c@{}}\mathbf{a}\\ \mathbf{ b} \end{array} \right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle N_{1}=-S^{-1}R^{T},\quad N_{2}=S^{-1},\quad N_{3}=RS^{-1}R^{T}-Q,\quad N_{1}^{T}=-RS^{-1}.\nonumber \end{eqnarray}$

The general solution of the problem discussed in Eqs (1)–(3) can be written in the form of the following equations $\begin{eqnarray}\displaystyle \mathbf{u} & = & \displaystyle \mathbf{A}\mathbf{f}(z)+\overline{\mathbf{A}}\overline{\mathbf{f}(z)},\end{eqnarray}$ (9) $\begin{eqnarray}\displaystyle \boldsymbol{{\phi}} & = & \displaystyle \mathbf{B}\mathbf{f}(z)+\overline{\mathbf{B}}\overline{\mathbf{f}(z)}\end{eqnarray}$ (10) where $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \mathbf{A}=[\mathbf{a}_{1},\mathbf{a}_{2},\mathbf{a}_{3},\mathbf{a}_{4}],\quad \mathbf{B}=[\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3},\mathbf{b}_{4}],\quad \mathbf{b}_{k}=(\mathbf{R}^{T}+p_{k}\mathbf{S})\mathbf{a}_{k},\quad (k=1,2,3,4)\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \mathbf{f}(z_{{\alpha}})=[f_{1}(z_{1}),f_{2}(z_{2}),f_{3}(z_{3}),f_{4}(z_{4})]^{T},\quad z_{{\alpha}}=x_{1}+p_{{\alpha}}x_{2},\nonumber\end{eqnarray}$ and 𝜙 is the generalized stress function such that $\begin{eqnarray}\displaystyle \boldsymbol{{\sigma}}_{2}=[{\sigma}_{2j},D_{2}]^{T}=\boldsymbol{{\phi}}_{,1},\quad \boldsymbol{{\sigma}}_{1}=[{\sigma}_{1j},D_{1}]^{T}=\boldsymbol{{\phi}}_{,2}. & & \displaystyle\end{eqnarray}$ (11)

In case of two-dimensional in-plane problem, the stress components 𝜎₁₁, 𝜎₂₂ and 𝜎₁₂, may be expressed in terms of two complex potential functions 𝛷(z) and 𝛹(z) as, using Muskhelishvili [11] $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\sigma}_{11}+{\sigma}_{22}=2[{\Phi}(z)+\overline{{\Phi}(z)}],\end{eqnarray}$ (12) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\sigma}_{22}-i{\sigma}_{12}={\Phi}(z)+{\Psi}(\bar{z})+(z-\bar{z})\overline{{\Phi}^{\prime }(z)}.\end{eqnarray}$ (13)

Consider a plate weakened by n straight cracks L _i(i = 1,2, …, n) which are lying on the real axis. The upper and lower edges of these cracks are subjected to the uniform stress distribution ${\sigma}_{22}^{\pm }$ , ${\sigma}_{12}^{\pm }$ . The Eqs (12) and (13) may be expressed in terms of two Hilbert problems, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Phi}^{+}(t)+{\Psi}^{-}(t)={\sigma}_{22}^{+}-i{\sigma}_{12}^{+},\end{eqnarray}$ (14) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Phi}^{-}(t)+{\Psi}^{+}(t)={\sigma}_{22}^{-}-i{\sigma}_{12}^{-},\quad \text{on }\mathop{\bigcup }_{i=1}^{n}L_{i}\end{eqnarray}$ (15) under the assumption lim_x
₂→0 x ₂𝛷^′(x ₁ + ix ₂) = 0.

The general solution of the Eqs (14) and (15) can be written as $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Phi}(z)={\Phi}_{0}(z)+{\displaystyle \frac{P_{n}(z)}{X(z)}}-{\displaystyle \frac{1}{2}}\overline{{\Gamma}}^{\prime },\end{eqnarray}$ (16) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Psi}(z)={\Psi}_{0}(z)+{\displaystyle \frac{P_{n}(z)}{X(z)}}+{\displaystyle \frac{1}{2}}\overline{{\Gamma}}^{\prime },\end{eqnarray}$ (17) where $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Phi}_{0}(z)={\displaystyle \frac{1}{2{\pi}iX(z)}}\displaystyle \int _{\mathop{\bigcup }_{i=1}^{n}L_{i}}{\displaystyle \frac{X^{+}(t)p(t)}{t-z}}dt+{\displaystyle \frac{1}{2{\pi}i}}\displaystyle \int _{\mathop{\bigcup }_{i=1}^{n}L_{i}}{\displaystyle \frac{q(t)}{t-z}}dt,\end{eqnarray}$ (18) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Psi}_{0}(z)={\displaystyle \frac{1}{2{\pi}iX(z)}}\displaystyle \int _{\mathop{\bigcup }_{i=1}^{n}L_{i}}{\displaystyle \frac{X^{+}(t)p(t)}{t-z}}dt+{\displaystyle \frac{1}{2{\pi}i}}\displaystyle \int _{\mathop{\bigcup }_{i=1}^{n}L_{i}}{\displaystyle \frac{q(t)}{t-z}}dt,\end{eqnarray}$ (19) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle p(t)={\displaystyle \frac{1}{2}}({\sigma}_{22}^{+}+{\sigma}_{22}^{-})-{\displaystyle \frac{i}{2}}({\sigma}_{12}^{+}+{\sigma}_{12}^{-}),q(t)={\displaystyle \frac{1}{2}}({\sigma}_{22}^{-}-{\sigma}_{22}^{-})-{\displaystyle \frac{i}{2}}({\sigma}_{12}^{-}-{\sigma}_{12}^{-}),\end{eqnarray}$ (20) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle X(z)=\mathop{\prod }_{k=1}^{n}\sqrt{z-a_{k}}\sqrt{z-b_{k}},P_{n}(z)=C_{0}z^{n}+C_{1}z^{n-1}+C_{2}z^{n-2}+\cdots +C_{n}.\end{eqnarray}$ (21)

Constant C ₀ is determined using the loading condition at the boundary of the plate and remaining constants C _i(i =1,2, …, n) are determined by using condition of single-valued displacements, $\begin{eqnarray}\displaystyle 2({\kappa}+1)\displaystyle \int _{L_{i}}{\displaystyle \frac{P_{n}(z)}{X(z)}}dz+{\kappa}\displaystyle \int _{L_{i}}[{\Phi}_{0}^{+}(z)-{\Phi}_{0}^{+}(z)]dz+\displaystyle \int _{L_{i}}[{\Psi}_{0}^{+}(z)-{\Psi}_{0}^{-}(z)]dz=0. & & \displaystyle\end{eqnarray}$ (22)

The mathematical formulation given above is taken from the work of Wang [17], Gao [6] and Muskhelishvili [11] in order to make the paper self-sufficient.

3. Statement of the problem

Consider an infinite transversely isotropic piezoelectric plate containing three straight collinear cracks as shown in Fig. 1. The poling direction is normal to the faces of the cracks. Assume that the cracks are lying symmetrically along the real axis and occupying the intervals (−a, −b), (−c, c) and (b, a) respectively. Saturation zones develop ahead of each crack tip due to the application of constant uniform stress ${\sigma}_{22}^{\infty }$ and in-plane electric displacement loading $D_{2}^{\infty }$ at the infinite boundary of the plate. These saturation zones occupy the intervals (−a ₁, −a), (−b, −b ₁), (−c ₁, −c), (c, c ₁), (b ₁, b) and (a, a ₁) ahead the respective crack tips. To stop the further opening of the cracks these developed saturation zones are subjected to a linearly varying electric displacement value $\frac{t}{a_{1}}D_{s}$ , (say D _a) where t is any point on the rims of the crack and D _s is saturation-limit electric displacement.

Fig. 1.

Schematic representation of the problem.

4. Solution of the problem

The problem discussed in the Section-3 is solved under the following boundary conditions, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\sigma}_{2j}^{+}={\sigma}_{2j}^{-}=0,\quad \text{on }L=\mathop{\bigcup }_{i=1}^{3}L_{i},\end{eqnarray}$ (23) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\sigma}_{22}={\sigma}_{22}^{\infty },\quad D_{2}=D_{2}^{\infty },\quad \text{for }|x_{2}|\rightarrow \infty ,\end{eqnarray}$ (24) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle D_{2}^{+}=D_{2}^{-}={\displaystyle \frac{t}{a_{1}}}D_{s},\quad \text{for }c\leq |x_{1}|\leq c_{1},b_{1}\leq |x_{1}|\leq b,a\leq |x_{1}|\leq a_{1},\end{eqnarray}$ (25) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\phi}_{,1}^{+}(x_{1})={\phi}_{,1}^{-}(x_{1})=-V,\quad \text{for }|x_{1}|\lt c,b\lt |x_{1}|\lt a,\end{eqnarray}$ (26) where, $V=[0,{\sigma}_{22}^{\infty },0,D_{2}^{\infty }]^{T}$ .

Assuming the continuity of 𝜙(x ₁) on the entire ox ₁ axis and using the Eq. (10) along with the boundary condition (23) the following Hilbert problem be formulated as, $\begin{eqnarray}\displaystyle [\mathbf{B}\mathbf{F}(x_{1})-\bar{\mathbf{B}\mathbf{F}(x_{1})}]^{+}-[\mathbf{B}\mathbf{F}(x_{1})-\bar{\mathbf{B}\mathbf{F}(x_{1})}]^{-}=0. & & \displaystyle\end{eqnarray}$ (27)

The Hilbert problem given in Eq. (27) is solved using the methodology given in Section-2 and can be written as $\begin{eqnarray}\displaystyle \mathbf{B}\mathbf{F}(\mathbf{z})-\bar{\mathbf{B}}\bar{\mathbf{F}(z)}=\mathbf{h}(\mathbf{z}). & & \displaystyle\end{eqnarray}$ (28)

Further, the boundary condition (26) along with Eq. (28) yields $\begin{eqnarray}\displaystyle \mathbf{h}^{+}(x_{1})+\mathbf{h}^{-}(x_{1})=-\mathbf{V },\quad |x_{1}|<c,b<|x_{1}|<a. & & \displaystyle\end{eqnarray}$ (29)

To solve the problem, complex function 𝛶(z) = [𝛶₁(z), 𝛶₂(z), 𝛶₃(z), 𝛶₄(z)]^T is introduced in such a way that $\begin{eqnarray}\displaystyle \boldsymbol{{\Upsilon}}(\mathbf{z})=H^{R}BF(z). & & \displaystyle\end{eqnarray}$ (30)

Consequently, the solution of Eqs (28) and (30) may be expressed as $\begin{eqnarray}\displaystyle h(z)=\boldsymbol{{\Theta}}\boldsymbol{{\Upsilon}}(z), & & \displaystyle\end{eqnarray}$ (31) where 𝛩 = [H ^R]⁻¹, H ^R = 2ReY , Y = iAB ⁻¹.

After expressing the Eq. (29) in component form by putting the value of h (z) from Eq. (31), we get the following Hilbert problems $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Theta}_{22}[{\Upsilon}_{2}^{+}(x_{1})+{\Upsilon}_{2}^{-}(x_{1})]+{\Theta}_{24}[{\Upsilon}_{4}^{+}(x_{1})+{\Upsilon}_{4}^{-}(x_{1})]=-{\sigma}_{22}^{\infty },\quad |x_{1}|<c,b<|x_{1}|<a,\end{eqnarray}$ (32) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Theta}_{42}[{\Upsilon}_{2}^{+}(x_{1})+{\Upsilon}_{2}^{-}(x_{1})]+{\Theta}_{44}[{\Upsilon}_{4}^{+}(x_{1})+{\Upsilon}_{4}^{-}(x_{1})]=-D_{2}^{\infty },\quad |x_{1}|<c,b<|x_{1}|<a.\end{eqnarray}$ (33)

Eliminating the term ${\Upsilon}_{4}^{+}(x_{1})+{\Upsilon}_{4}^{-}(x_{1})$ from Eqs (32) and (33), one can obtain $\begin{eqnarray}\displaystyle {\Upsilon}_{2}^{+}(x_{1})+{\Upsilon}_{2}^{-}(x_{1})=-{\displaystyle \frac{{\Theta}_{44}{\sigma}_{22}^{\infty }-{\Theta}_{44}D_{2}^{\infty }}{{\Delta}}},\quad |x_{1}|<c,\quad b<|x_{1}|<a. & & \displaystyle\end{eqnarray}$ (34)

Solution of the Eq. (34) is obtained using the methodology discussed in Section 2. The solution can be written as follows $\begin{eqnarray}\displaystyle {\Upsilon}_{2}(z)=-{\displaystyle \frac{({\Theta}_{44}{\sigma}_{22}^{\infty }-{\Theta}_{44}D_{2}^{\infty })}{{\Delta}}}+{\displaystyle \frac{P_{1}(z)}{2X_{1}(z)}} & & \displaystyle\end{eqnarray}$ (35) where $\begin{eqnarray}\displaystyle \hspace{-10.0pt}P_{1}(z)=C_{0}z^{3}+C_{1}z^{2}+C_{2}z+C_{3},\quad {\Delta}={\Theta}_{22}{\Theta}_{44}-{\Theta}_{24}{\Theta}_{42},\quad X_{1}(z)=\sqrt{(z^{2}-a^{2})(z^{2}-b^{2})(z^{2}-c^{2})}. & & \displaystyle \nonumber\end{eqnarray}$

The constants C ₁ = 0, C ₃ = 0 due to the symmetry of the geometry and the applied loading conditions. Further, the constants C ₀, C ₂ are determined using the loading conditions at the infinite boundary lim_z→∞𝛶₂(z) = 0 and single valuedness condition around cracks rims respectively. Therefore, the constants C ₀, C ₂ are $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle C_{0}={\displaystyle \frac{{\Theta}_{44}{\sigma}_{22}^{\infty }-{\Theta}_{44}D_{2}^{\infty }}{{\Delta}}},\end{eqnarray}$ (36) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle C_{2}=-{\displaystyle \frac{({\Theta}_{44}{\sigma}_{22}^{\infty }-{\Theta}_{44}D_{2}^{\infty })}{{\Delta}}}\{c^{2}+{\lambda}^{2}(a^{2}-c^{2})\}.\end{eqnarray}$ (37)

After substituting the value of constants in Eq. (35), the final expression for potential function 𝛶₂(z) is $\begin{eqnarray}\displaystyle {\Upsilon}_{2}(z)={\displaystyle \frac{({\Theta}_{44}{\sigma}_{22}^{\infty }-{\Theta}_{24}D_{2}^{\infty })}{2{\Delta}}}\left[{\displaystyle \frac{z^{3}-z((1-{\lambda}_{1}^{2})c^{2}+a^{2}{\lambda}_{1}^{2})}{X_{1}(z)}}-1\right] & & \displaystyle\end{eqnarray}$ (38) where $\begin{eqnarray}\displaystyle {\lambda}_{1}^{2}={\displaystyle \frac{E(k_{1})}{F(k_{1})}},\quad k_{1}^{2}={\displaystyle \frac{(a^{2}-b^{2})}{(a^{2}-c^{2})}}. & & \displaystyle \nonumber\end{eqnarray}$

Moreover, in case of linearly varying electric displacement acting on the saturation zones, the Eq. (33) together with the boundary condition (25) yields $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Theta}_{42}[{\Upsilon}_{2}^{+}(x_{1})+{\Upsilon}_{2}^{-}(x_{1})]+{\Theta}_{44}[{\Upsilon}_{4}^{+}(x_{1})+{\Upsilon}_{4}^{-}(x_{1})]={\displaystyle \frac{|t|}{a_{1}}}D_{s}-D_{2}^{\infty },\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad c\leq |x_{1}|\leq c_{1},b_{1}\leq |x_{1}|\leq b,a\leq |x_{1}|\leq a_{1}.\end{eqnarray}$ (39)

Following the methodology given by Muskhelishvili [11], the solution of the Eq. (39) can be expressed as $\begin{eqnarray}\displaystyle {\Upsilon}_{4}(z)=-{\displaystyle \frac{{\Theta}_{42}}{{\Theta}_{44}}}{\Upsilon}_{2}(z)-{\displaystyle \frac{D_{2}^{\infty }}{2{\Theta}_{44}}}+{\displaystyle \frac{P_{2}(z)}{2X_{1}(z)}}+{\displaystyle \frac{D_{s}}{2{\pi}ia_{1}{\Theta}_{44}X_{2}(z)}}\displaystyle \int _{{\Gamma}}{\displaystyle \frac{tX_{2}(t)}{(t-z)}}dt, & & \displaystyle\end{eqnarray}$ (40) where $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle P_{2}(z)=D_{0}z^{3}+D_{1}z^{2}+D_{2}z+D_{3},\quad X_{2}(z)=\sqrt{(z^{2}-a_{1}^{2})(z^{2}-b_{1}^{2})(z^{2}-c_{1}^{2})},\end{eqnarray}$ (41) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\Gamma}=[-a_{1},-a]\cup [-b,-b_{1}]\cup [-c_{1},-c]\cup [c,c_{1}]\cup [b_{1},b]\cup [a,a_{1}].\end{eqnarray}$ (42)

The integrals appeared in Eq. (40) are evaluated using the following condition $\begin{eqnarray} \displaystyle \text{} & \text{} & \displaystyle \hspace{-20.0pt}X_{2}(t)=X_{2}(-t)=i\sqrt{(a_{1}^{2}-t^{2})(t^{2}-b_{1}^{2})(t^{2}-c_{1}^{2})},\quad \text{when }-a_{1}\lt -b_{1}\lt -c_{1}\lt c_{1}\lt b_{1}\lt t\lt a_{1}, \end{eqnarray}$ (43) $\begin{eqnarray} \displaystyle \text{} & \text{} & \displaystyle \hspace{-20.0pt}X_{2}(t)=X_{2}(-t)=-i\sqrt{(a_{1}^{2}-t^{2})(b_{1}^{2}-t^{2})(c_{1}^{2}-t^{2})},\quad \text{when }-a_{1}\lt -b_{1}\lt -c_{1}\lt t\lt c_{1}\lt b_{1}\lt a_{1}, \end{eqnarray}$ (44) and written in terms of elliptic integrals as $\begin{eqnarray} \displaystyle {\displaystyle \frac{1}{2za_{1}}}\displaystyle \int _{{\Gamma}}{\displaystyle \frac{tX_{2}(t)}{(t-z)}}dt\text{} & =\text{} & \displaystyle \displaystyle \int _{a}^{a_{1}}{\displaystyle \frac{X_{1}(t)}{(t^{2}-z^{2})}}dt+\displaystyle \int _{b_{1}}^{b}{\displaystyle \frac{X_{1}(t)}{(t^{2}-z^{2})}}dt+\displaystyle \int _{c}^{c_{1}}{\displaystyle \frac{X_{1}(t)}{(t^{2}-z^{2})}}dt,\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle ia_{1}^{3}\left[{\displaystyle \frac{m_{2}^{3}}{k_{2}}}\{H_{1}+(n_{2}^{2}(z)-1)H_{2}\}+{\displaystyle \frac{m_{3}^{2}}{k_{3}}}\{J_{1}+(n_{3}^{2}(z)-1)J_{2}\}\right] \end{eqnarray}$ (45) where, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle H_{1}=h_{1}({\phi}_{2}(a)k_{2})+h_{1}\left({\displaystyle \frac{{\pi}}{2}},k_{2}\right)-h_{1}({\phi}_{2}(b),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle H_{2}=h_{2}({\phi}_{2}(a),n_{2}(z),k_{2})+h_{2}\left({\displaystyle \frac{{\pi}}{2}},n_{2}(z),k_{2}\right)-h_{2}({\phi}_{2}(b),n_{2}(z),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle J_{1}=j_{1}\left({\displaystyle \frac{{\pi}}{2}},m_{3},k_{3}\right)-j_{1}({\phi}_{4}(c),m_{3},k_{3}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle J_{2}=j_{2}\left({\displaystyle \frac{{\pi}}{2}},m_{3},n_{3},k_{3}\right)-j_{2}({\phi}_{4}(c),m_{3},n_{3},k_{3}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle h_{1}(z,k)={\displaystyle \frac{1}{3k^{2}}}\{(1-k^{2})F(z,k)+(2k^{2}-1)E(z,k)-k^{2}\sin z\cos z.p_{0}(z,k)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle h_{2}(z,n(z),k)={\displaystyle \frac{1}{n^{2}}}E(z,k)+{\displaystyle \frac{k^{2}}{n^{4}}}F(z,k)+{\displaystyle \frac{(n^{2}-k^{2})}{n^{4}}}{\Pi}(z,n(z),k),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle j_{1}(z,m,k)=-{\displaystyle \frac{2\sqrt{1-m^{2}}(1-k^{2})}{3k}}F({\psi}_{2}(z,k),r_{0})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{36.0pt}\qquad +∼{\displaystyle \frac{\sqrt{1-m^{2}}(m^{2}-k^{2}-2m^{2}k^{2})}{3m^{2}k}}E({\psi}_{2}(z,k),r_{0})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{36.0pt}\qquad -∼{\displaystyle \frac{(2m^{2}(1-k^{2})+k^{2}-m^{2}k^{2}\sin ^{2}z)}{3m^{2}}}{\displaystyle \frac{p_{0}(z,m)\cos z}{p_{0}(z,k)}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle j_{2}(z,m,n(z),k)=-{\displaystyle \frac{m^{2}}{n^{2}}}{\displaystyle \frac{p_{0}(z,k)\cos z}{p_{0}(z,m)}}-{\displaystyle \frac{k\sqrt{1-m^{2}}}{n^{2}}}E({\psi}_{1}(z,m),r_{0})+{\displaystyle \frac{k\sqrt{1-m^{2}}}{n^{2}}}F({\psi}_{1}(z,m),r_{0})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{60.0pt}\qquad +∼{\displaystyle \frac{m^{2}(n^{2}-k^{2})}{kn^{4}\sqrt{1-m^{2}}}}{\Pi}\left({\psi}_{1}(z,m),{\displaystyle \frac{(n^{2}-m^{2})}{n^{2}(1-m^{2})}}r_{0}\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle k_{1}^{2}={\displaystyle \frac{(a^{2}-b^{2})}{(a^{2}-c^{2})}},\quad k_{2}^{2}={\displaystyle \frac{(a_{1}^{2}-b_{1}^{2})}{(a_{1}^{2}-c_{1}^{2})}},\quad k_{3}^{2}={\displaystyle \frac{c_{1}^{2}}{b_{1}^{2}}},\quad k_{4}^{2}=k_{3}^{3}k_{2}^{2},\quad k_{2}^{^{\prime }2}={\displaystyle \frac{(b_{1}^{2}-c_{1}^{2})}{(a_{1}^{2}-c_{1}^{2})}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{0}^{2}={\displaystyle \frac{(k^{2}-m^{2})}{k^{2}(1-m^{2})}},\quad m_{1}^{2}={\displaystyle \frac{(a^{2}-b^{2})}{a^{2}}},\quad m_{2}^{2}={\displaystyle \frac{(a_{1}^{2}-b_{1}^{2})}{a_{1}^{2}}},\quad m_{3}^{2}={\displaystyle \frac{c_{1}^{2}}{a_{1}^{2}}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle n_{2}^{2}(z)={\displaystyle \frac{(a_{1}^{2}-b_{1}^{2})}{(a_{1}^{2}-z^{2})}},\quad n_{3}^{2}(z)={\displaystyle \frac{c_{1}^{2}}{z^{2}}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle p_{0}(z,k)=\sqrt{1-k^{2}\sin ^{2}z},\quad p_{1}(z,m,k)=\tan ^{-1}\left({\displaystyle \frac{mk\sin z\cos z}{p_{0}(z,k).p_{0}(z,m)}}\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle p_{2}(z,k)=E(z,k)-\tan z.p_{0}(z,k),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle p_{3}(z,k)=E(z,k)-{\displaystyle \frac{k^{2}\sin z\cos z}{p_{0}(z,k)}},\quad p_{4}(z,m,k)={\displaystyle \frac{\sin 2z.p_{0}(z,k)}{4.p_{0}(z,m)}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\psi}_{1}(z,m)=\tan ^{-1}\left({\displaystyle \frac{\sqrt{1-m^{2}}}{m\cos z}}\right),\quad {\psi}_{2}(z,k)=\tan ^{-1}\left({\displaystyle \frac{k\cos z}{\sqrt{1-k^{2}}}}\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \sin ^{2}{\phi}_{1}(z)={\displaystyle \frac{(a^{2}-z^{2})}{(a^{2}-b^{2})}},\quad \sin ^{2}{\phi}_{2}(z)={\displaystyle \frac{(a_{1}^{2}-z^{2})}{(a_{1}^{2}-b_{1}^{2})}},\quad \sin ^{2}{\phi}_{3}(z)={\displaystyle \frac{(z^{2}-c_{1}^{2})}{(z^{2}-b_{1}^{2})}},\quad \sin ^{2}{\phi}_{4}(z)={\displaystyle \frac{z^{2}}{c_{1}^{2}}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \sin ^{2}{\phi}_{5}(z)={\displaystyle \frac{(z^{2}-a^{2})}{(z^{2}-b^{2})}},\quad \sin ^{2}{\phi}_{6}(z)={\displaystyle \frac{1}{k_{2}^{2}}}{\displaystyle \frac{(z^{2}-b_{1}^{2})}{(z^{2}-c_{1}^{2})}},\quad \sin ^{2}{\phi}_{7}(z)={\displaystyle \frac{1}{k_{3}^{2}}}{\displaystyle \frac{(z^{2}-c_{1}^{2})}{(z^{2}-b_{1}^{2})}}.\nonumber\end{eqnarray}$

Further, the constant D ₀ is determine using the condition lim_z→∞𝛶₄(z) =0 and D ₂ is determined using single value of displacement components $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle D_{0}={\displaystyle \frac{D_{2}^{\infty }}{{\Theta}_{44}}},\end{eqnarray}$ (46) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle D_{2}={\displaystyle \frac{D_{2}^{\infty }}{{\Theta}_{44}}}\{c_{1}^{2}+{\lambda}_{2}^{2}(a_{1}^{2}-c_{1}^{2})\}-{\displaystyle \frac{2TD_{s}\sqrt{a_{1}^{2}-c_{1}^{2}}}{a_{1}{\pi}{\Theta}_{44}}}\end{eqnarray}$ (47) where, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle T=[(a_{1}^{2}-b_{1}^{2})\{I_{1}-{\lambda}_{2}^{2}I_{2}\}+(b_{1}^{2}-c_{1}^{2})\{I_{3}-I_{4}\}],\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle I_{1}=i_{1}({\phi}_{2}(a),k_{2})+i_{1}\left({\displaystyle \frac{{\pi}}{2}},k_{2}\right)-i_{1}({\phi}_{2}(b),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle I_{2}=i_{2}({\phi}_{2}(a),k_{2})+i_{2}\left({\displaystyle \frac{{\pi}}{2}},k_{2}\right)-i_{2}({\phi}_{2}(b),k_{2}),\quad I_{4}=i_{3}({\phi}_{3}(c),k_{2})-{\lambda}_{2}^{2}i_{4}({\phi}_{3}(c),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle I_{3}={\displaystyle \frac{1}{3k_{2}^{^{\prime }2}}}\left\{\vphantom{{\displaystyle \frac{p_{0}({\phi}_{3}(c),k_{2})\tan {\phi}_{3}(c)}{\cos ^{2}{\phi}_{3}(c)}}}(1-2k_{2}^{2})E({\phi}_{3}(c),k_{2})-k_{2}^{^{\prime }2}F({\phi}_{3}(c),k_{2})\right.\nonumber\\ \displaystyle & \text{} & \displaystyle \text{}\qquad +\left.{\displaystyle \frac{p_{0}({\phi}_{3}(c),k_{2})\tan {\phi}_{3}(c)}{\cos ^{2}{\phi}_{3}(c)}}(k_{2}^{^{\prime }2}+(1-2k_{2}^{^{\prime }2})\cos ^{2}{\phi}_{3}(c))\right\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{1}(z,k)={\displaystyle \frac{p_{0}(z,k)\sin z\cos z}{6}}-{\displaystyle \frac{(1-k^{2})}{6k^{2}}}F(z,k)+\left({\displaystyle \frac{(2-k^{2})}{3k^{2}}}-{\displaystyle \frac{(1-k^{2}\sin ^{2}z)}{2k^{2}}}\right)E(z,k),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{2}(z,k)={\displaystyle \frac{E(z,k)}{2k^{2}}}-{\displaystyle \frac{(1-k^{2}\sin ^{2}z)}{2k^{2}}}F(z,k),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{3}(z,k)={\displaystyle \frac{(1+\sec ^{2}z)}{2}}E(z,k)-{\displaystyle \frac{F(z,k)}{2}}-{\displaystyle \frac{p_{0}(z,k)\tan z}{2}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{4}(z,k)={\displaystyle \frac{(-1+\sec ^{2}z)}{2}}F(z,k)+{\displaystyle \frac{E(z,k)}{2(1-k^{2})}}-{\displaystyle \frac{p_{0}(z,k)\tan z}{2(1-k^{2})}}.\nonumber\end{eqnarray}$

The desired complex potential function 𝛶₄(z) can be written as $\begin{eqnarray}\displaystyle {\Upsilon}_{4}(z)\text{} & =\text{} & \displaystyle -{\displaystyle \frac{{\Theta}_{42}}{{\Theta}_{44}}}{\Upsilon}_{2}(z)+{\displaystyle \frac{D_{2}^{\infty }}{2{\Theta}_{44}}}\left\{{\displaystyle \frac{z^{3}-z(c_{1}^{2}+{\lambda}_{2}^{2}(a_{1}^{2}-c_{1}^{2}))}{X_{2}(z)}}-1\right\}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle -∼{\displaystyle \frac{zD_{s}}{{\pi}a_{1}{\Theta}_{44}X_{2}(z)}}\left\{T\sqrt{(a_{2}^{2}-c_{2}^{2})}-S(z)\right\},\end{eqnarray}$ (48) where $\begin{eqnarray}\displaystyle S(z)=a_{1}^{3}\left[{\displaystyle \frac{m_{2}^{3}}{k_{2}}}\{H_{1}+(n_{2}^{2}(z)-1)H_{2}\}+{\displaystyle \frac{m_{3}^{2}}{k_{3}}}\{J_{1}+(n_{3}^{2}(z)-1)J_{2}\}\right]. & & \displaystyle \nonumber\end{eqnarray}$

Having solved the problem as posed, it is necessary to discuss the applications of the complex potential function given in Eq. (48). The following sections deal with the applicability of the results obtained.

5. Applications

In this section, analytical expressions of important fracture parameters are determined. These parameters are used to describe the ability of the materials used against the growth of the saturation zone.

5.1. Stress intensity factor

Stress intensity factors (SIF) at each crack tip a, b, c is obtained using the well known formulae, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle K(a)=\displaystyle \lim _{x_{1}\rightarrow a^{+}}\sqrt{2{\pi}(x_{1}-a)}{\sigma}_{22}(x_{1}),\end{eqnarray}$ (49) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle K(b)=\displaystyle \lim _{x_{1}\rightarrow b^{-}}\sqrt{2{\pi}(b-x_{1})}{\sigma}_{22}(x_{1}),\end{eqnarray}$ (50) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle K(c)=\displaystyle \lim _{x_{1}\rightarrow c^{+}}\sqrt{2{\pi}(x_{1}-c)}{\sigma}_{22}(x_{1}).\end{eqnarray}$ (51)

Stress component 𝜎₂₂(x ₁) is determined using Eqs (10), (30) and (35) and substitute it into Eqs (49)–(51). Hence, the mathematical expressions for SIF at crack tips a, b and c are $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle K(a)=\left({\sigma}_{22}^{\infty }-{\displaystyle \frac{{\Theta}_{24}}{{\Theta}_{44}}}D_{2}^{\infty }\right){\displaystyle \frac{(1-{\lambda}_{1}^{2})}{k_{1}}}\sqrt{a{\pi}},\end{eqnarray}$ (52) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle K(b)=\left({\sigma}_{22}^{\infty }-{\displaystyle \frac{{\Theta}_{24}}{{\Theta}_{44}}}D_{2}^{\infty }\right){\displaystyle \frac{({\lambda}_{1}^{2}+k_{1}^{2}-1)}{k_{1}\sqrt{(1-k_{1}^{2})}}}\sqrt{b{\pi}},\end{eqnarray}$ (53) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle K(c)=\left({\sigma}_{22}^{\infty }-{\displaystyle \frac{{\Theta}_{24}}{{\Theta}_{44}}}D_{2}^{\infty }\right){\displaystyle \frac{{\lambda}_{1}^{2}}{k_{1}^{\prime }}}\sqrt{c{\pi}}.\end{eqnarray}$ (54)

5.2. Electric saturation zone length

Using the Eqs (10) and (31), we can write $\begin{eqnarray}\displaystyle {\phi}_{,1}(x_{1})=BF^{+}(x_{1})+BF^{-}(x_{1})={\Theta}[{\Upsilon}^{+}(x_{1})+{\Upsilon}^{-}(x_{1})],\quad |x_{1}|>a_{2}. & & \displaystyle\end{eqnarray}$ (55)

The electric-displacement component is obtained by taking the fourth component of the Eq. (55) $\begin{eqnarray}\displaystyle D_{2}(x_{1})={\Theta}_{42}[{\Upsilon}_{2}^{+}(x_{1})+{\Upsilon}_{2}^{-}(x_{1})]+{\Theta}_{44}[{\Upsilon}_{4}^{+}(x_{1})+{\Upsilon}_{4}^{-}(x_{1})]. & & \displaystyle\end{eqnarray}$ (56)

After subsisting values of 𝛶₂(z) from Eq. (38) and 𝛶₄(z) from Eq. (48) into Eq. (56), we get $\begin{eqnarray}\displaystyle D_{2}(x_{1})=D_{2}^{\infty }\left\{{\displaystyle \frac{x_{1}^{3}-x_{1}(c_{1}^{2}+{\lambda}_{2}^{2}(a_{1}^{2}-c_{1}^{2}))}{X_{2}(x_{1})}}-1\right\}-{\displaystyle \frac{2x_{1}D_{s}}{a_{1}{\pi}X_{2}(x_{1})}}\left[T\sqrt{(a_{1}^{2}-c_{1}^{2})}-S(x_{1})\right]. & & \displaystyle\end{eqnarray}$ (57)

Using the hypothesis that the electric-displacement remains finite at each tip of the cracks, we obtained three non-linear equations at cracks tips a ₁, b ₁ and c ₁, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}{\displaystyle \frac{m_{2}^{2}}{k_{2}^{2}}}(1-{\lambda}_{2}^{2}){\displaystyle \frac{D_{2}^{\infty }}{D_{s}}}-{\displaystyle \frac{2}{{\pi}}}{\displaystyle \frac{1}{a_{1}}}\left\{{\displaystyle \frac{m_{2}}{a_{1}k_{2}}}T-a_{1}\left\{{\displaystyle \frac{m_{2}^{3}}{k_{2}}}\{H_{1}+{\tau}_{1}\}+{\displaystyle \frac{m_{3}^{2}}{k_{3}}}\left\{J_{1}+\left(1-{\displaystyle \frac{1}{m_{3}^{2}}}\right)J_{1}^{^{\prime }}\right\}\right\}\right\}=0,\end{eqnarray}$ (58) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \hspace{-18.0pt}{\displaystyle \frac{m_{2}^{2}}{k_{2}^{2}}}(1-k_{2}^{2}-{\lambda}_{2}^{2}){\displaystyle \frac{D_{2}^{\infty }}{D_{s}}}-{\displaystyle \frac{2}{{\pi}}}{\displaystyle \frac{1}{a_{1}}}\left\{{\displaystyle \frac{m_{2}}{a_{1}k_{2}}}T-a_{1}\left\{{\displaystyle \frac{m_{2}^{3}}{k_{2}}}H_{1}+{\displaystyle \frac{m_{3}^{2}}{k_{3}}}\left\{J_{1}+\left(1-{\displaystyle \frac{1}{k_{3}^{2}}}\right){\tau}_{2}\right\}\right\}\right\}=0,\end{eqnarray}$ (59) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\displaystyle \frac{m_{2}^{2}}{k_{2}^{2}}}{\lambda}_{2}^{2}{\displaystyle \frac{D_{2}^{\infty }}{D_{s}}}+{\displaystyle \frac{2}{{\pi}}}{\displaystyle \frac{1}{a_{1}}}\left\{{\displaystyle \frac{m_{2}}{a_{1}k_{2}}}T-a_{1}\left\{{\displaystyle \frac{m_{2}^{3}}{k_{2}}}\left\{H_{1}+\left(1-{\displaystyle \frac{1}{k_{2}^{2}}}\right)H_{1}^{^{\prime }}\right\}+{\displaystyle \frac{m_{3}^{2}}{k_{3}}}J_{1}\right\}\right\}=0.\end{eqnarray}$ (60) where, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle l(z,m,k)={\displaystyle \frac{m^{2}p_{0}(z,k)\cos z}{p_{0}(z,m)}}-k\sqrt{1-m^{2}}E({\psi}_{1}(z,m),r_{0})+k\sqrt{1-m^{2}}F({\psi}_{1}(z,m),r_{0}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\tau}_{1}=E({\phi}_{1}(a),k_{2})+E\left({\displaystyle \frac{{\pi}}{2}},k_{2}\right)-E({\phi}_{1}(b),k_{2}),\quad {\tau}_{2}=l\left({\displaystyle \frac{{\pi}}{2}},m_{3},k_{3}\right)-l({\phi}_{4}(c),m_{3},k_{3}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle H_{1}^{^{\prime }}={\tau}_{1}+F({\phi}_{1}(a),k_{2})+F\left({\displaystyle \frac{{\pi}}{2}},k_{2}\right)-F({\phi}_{1}(b),k_{2})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle J_{1}^{^{\prime }}={\tau}_{2}-k_{3}m_{2}k_{2}\left(F({\psi}_{1}({\phi}_{4}(c),k_{3}),k_{2})-F\left({\displaystyle \frac{{\pi}}{2}},k_{2}\right)\right),\quad {\lambda}_{2}^{2}={\displaystyle \frac{E(k_{2})}{F(k_{2})}}.\nonumber\end{eqnarray}$

The Eqs (58), (59) and (60) enables to investigate the saturation zone length |a ₁ − a|, |b − b ₁| and |c ₁ − c| at the crack tips a, b and c respectively.

5.3. Crack opening displacement

As defined by Bhargava [2] the jump displacement vector is $\begin{eqnarray}\displaystyle i{\delta}\mathbf{u}_{,1}(x_{1})=i[u_{1,1}^{+}-u_{1,1}^{-},u_{2,1}^{+}-u_{2,1}^{-},u_{3,1}^{+}-u_{3,1}^{-},u_{4,1}^{+}-u_{4,1}^{-}]^{T}={\Upsilon}^{+}(x_{1})-{\Upsilon}^{-}(x_{1}). & & \displaystyle\end{eqnarray}$ (61)

Taking the second component of the Eq. (61) one can write $\begin{eqnarray}\displaystyle i{\delta}u_{2,1}={\Upsilon}_{2}^{+}(x_{1})-{\Upsilon}_{2}^{-}(x_{1}). & & \displaystyle\end{eqnarray}$ (62)

On substituting value of 𝛶₂(x ₁) from Eq. (38) to Eq. (62) $\begin{eqnarray}\displaystyle {\delta}u_{2}(x_{1})=2\sqrt{a_{1}^{2}-c_{1}^{2}}\left({\displaystyle \frac{{\sigma}_{22}^{\infty }{\Theta}_{44}-D_{2}^{\infty }{\Theta}_{24}}{{\Theta}_{22}{\Theta}_{44}-{\Theta}_{24}{\Theta}_{42}}}\right)\{E({\phi}_{1},k_{1})-{\lambda}_{1}^{2}F({\phi}_{1},k_{1})\}, & & \displaystyle\end{eqnarray}$ (63) where, $\begin{eqnarray}\displaystyle \sin ^{2}{\phi}_{1}={\displaystyle \frac{(a^{2}-b^{2})}{(a^{2}-c^{2})}}. & & \displaystyle \nonumber\end{eqnarray}$

5.4. Crack opening potential drop

By taking the fourth component of Eq. (61), we get $\begin{eqnarray}\displaystyle i{\delta}u_{4,1}={\Upsilon}_{4}^{+}(x_{1})-{\Upsilon}_{4}^{-}(x_{1}). & & \displaystyle\end{eqnarray}$ (64)

After putting the mathematical expression of 𝛶₄(x ₁) from Eq. (48) to equation (64), one can write the 𝛿u ₄(x ₁) at crack tip x ₁ = a as $\begin{eqnarray}\displaystyle {\delta}u_{4}(a)\text{} & =\text{} & \displaystyle {\displaystyle \frac{D_{s}}{{\pi}{\Theta}_{44}}}\left\{{\pi}\sqrt{a_{1}^{2}-c_{1}^{2}}\left({\displaystyle \frac{D_{2}^{\infty }}{D_{s}}}\right)_{a}(E({\phi}_{2}(a),k_{2})-{\lambda}_{2}^{2}F({\phi}_{2}(a),k_{2}))\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle -\left.\vphantom{\left({\displaystyle \frac{D_{2}^{\infty }}{D_{s}}}\right)_{a}}{\displaystyle \frac{2}{a_{1}}}(I_{5}-F({\phi}_{2}(a),k_{2})T)\right\}.\end{eqnarray}$ (65)

Also, at the crack tip x ₁ = b $\begin{eqnarray}\displaystyle {\delta}u_{4}(b)\text{} & =\text{} & \displaystyle -{\displaystyle \frac{D_{s}}{{\pi}{\Theta}_{44}}}\left\{{\pi}\sqrt{a_{1}^{2}-c_{1}^{2}}\left({\displaystyle \frac{D_{2}^{\infty }}{D_{s}}}\right)_{b}(p_{3}({\phi}_{6}(b),k_{2})-{\lambda}_{2}^{2}F({\phi}_{6}(b),k_{2}))\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \left.\vphantom{\left({\displaystyle \frac{D_{2}^{\infty }}{D_{s}}}\right)_{a}}-I_{6}-{\displaystyle \frac{2T}{a_{1}}}F({\phi}_{6}(b),k_{2})\right\}\end{eqnarray}$ (66) and at the crack tip x ₁ = c $\begin{eqnarray}\displaystyle {\delta}u_{4}(c)\text{} & =\text{} & \displaystyle -{\displaystyle \frac{D_{s}}{{\pi}{\Theta}_{44}}}\left\{{\pi}\sqrt{a_{1}^{2}-c_{1}^{2}}\left({\displaystyle \frac{D_{2}^{\infty }}{D_{s}}}\right)_{c}(p_{2}({\phi}_{3}(c),k_{2})-{\lambda}_{2}^{2}F({\phi}_{3}(c),k_{2}))\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \left.\vphantom{\left({\displaystyle \frac{D_{2}^{\infty }}{D_{s}}}\right)_{a}}+I_{7}+{\displaystyle \frac{2T}{a_{1}}}F({\phi}_{3}(c),k_{2})\right\}\end{eqnarray}$ (67) where, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\theta}(t)=\tan ^{-1}{\displaystyle \frac{\sqrt{a_{1}^{2}-c_{1}^{2}}}{\sqrt{c_{1}^{2}-t^{2}}}},\quad K=F\left({\displaystyle \frac{{\pi}}{2}},k_{2}\right),\quad {\alpha}_{1}^{2}(t)=k_{2}^{2}{\displaystyle \frac{(t^{2}-c_{1}^{2})}{(t^{2}-b_{1}^{2})}},\quad {\alpha}_{2}^{2}(t)={\displaystyle \frac{t^{2}(a_{1}^{2}-b_{1}^{2})}{b_{1}^{2}(a_{1}^{2}-t^{2})}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\alpha}_{3}^{2}(t)={\displaystyle \frac{c_{1}^{2}(a_{1}^{2}-t^{2})}{t^{2}(a_{1}^{2}-c_{1}^{2})}},\quad {\alpha}_{4}^{2}={\displaystyle \frac{(b_{1}^{2}-a_{1}^{2})}{b_{1}^{2}}},\quad {\alpha}_{5}^{2}(t)={\displaystyle \frac{(t^{2}-b_{1}^{2})}{(t^{2}-c_{1}^{2})}},\quad {\alpha}_{6}^{2}(t)={\displaystyle \frac{(a_{1}^{2}-t^{2})}{(a_{1}^{2}-c_{1}^{2})}},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle I_{5}={\textstyle \frac{1}{2}}(-s_{1}-s_{2}-T_{1}+T_{2}+T_{3}),\quad I_{6}=s_{3}+s_{4}+s_{5}+T_{4}+T_{5}+T_{6}+T_{7},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle I_{7}=-s_{6}-s_{7}+s_{8}+T_{8}+T_{9}+T_{10}+T_{11},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\tau}_{1}=E({\phi}_{2}(a),k_{2})+E\left({\displaystyle \frac{{\pi}}{2}},k_{2}\right)-E({\phi}_{2}(b),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\tau}_{2}=l\left({\displaystyle \frac{{\pi}}{2}},m_{3},k_{3}\right)-l({\phi}_{4}(c),m_{3},k_{3}),\quad {\tau}_{1}^{^{\prime }}=F({\phi}_{2}(a),k_{2})+F\left({\displaystyle \frac{{\pi}}{2}},k_{2}\right)-F({\phi}_{2}(b),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle s_{1}={\displaystyle \frac{b^{2}\sqrt{b^{2}-b_{1}^{2}}\sqrt{b^{2}-c_{1}^{2}}}{\sqrt{a_{1}^{2}-c_{1}^{2}}\sqrt{a_{1}^{2}-b_{1}^{2}}}}{\Pi}({\phi}_{2}(a),n_{2}^{2}(b),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle s_{2}={\displaystyle \frac{c^{2}\sqrt{b_{1}^{2}-c^{2}}\sqrt{c_{1}^{2}-c^{2}}}{\sqrt{a_{1}^{2}-c_{1}^{2}}\sqrt{c^{2}-a_{1}^{2}}}}{\Pi}({\phi}_{2}(a),n_{2}^{2}(c),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle s_{3}={\displaystyle \frac{b^{2}\sqrt{(b^{2}-b_{1}^{2})(b^{2}-c_{1}^{2})}}{a_{1}\sqrt{a_{1}^{2}-c_{1}^{2}}\sqrt{a_{1}^{2}-b^{2}}}}\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \qquad \times ∼\left({\Pi}({\phi}_{2}(a),n_{2}^{2}(b),k_{2})-{\Pi}({\theta}(c),{\alpha}_{6}^{2}(b),k_{2})-F({\phi}_{2}(b),k_{2})+{\Pi}\left({\phi}_{2}(b),{\displaystyle \frac{k_{2}^{2}}{n_{2}^{2}(b)}},k_{2}\right)\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle s_{4}={\displaystyle \frac{b^{2}(b_{1}^{2}-c_{1}^{2})\sqrt{a_{1}^{2}-b^{2}}}{a_{1}\sqrt{a_{1}^{2}-c_{1}^{2}}\sqrt{(b^{2}-b_{1}^{2})(b^{2}-c_{1}^{2})}}}\left(F({\phi}_{6}(b),k_{2})-{\Pi}\left({\phi}_{6}(b),{\displaystyle \frac{k_{2}^{2}}{{\alpha}_{1}^{2}(b)}},k_{2}\right)\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle s_{5}={\displaystyle \frac{c_{1}^{2}\sqrt{(a_{1}^{2}-b^{2})(b^{2}-b_{1}^{2})}}{a_{1}\sqrt{a_{1}^{2}-c_{1}^{2}}\sqrt{b^{2}-c_{1}^{2}}}}F({\phi}_{6}(b),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle s_{6}={\displaystyle \frac{c^{2}\sqrt{(b_{1}^{2}-c^{2})(c_{1}^{2}-c^{2})}}{a_{1}\sqrt{a_{1}^{2}-c_{1}^{2}}\sqrt{a_{1}^{2}-c^{2}}}}({\Pi}({\phi}_{2}(a),n_{2}^{2}(c),k_{2})-{\Pi}({\phi}_{2}(b),n_{2}^{2}(c),k_{2})),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle s_{7}={\displaystyle \frac{c^{2}(b_{1}^{2}-c_{1}^{2})\sqrt{a_{1}^{2}-c^{2}}}{a_{1}\sqrt{a_{1}^{2}-c_{1}^{2}}\sqrt{(b_{1}^{2}-c^{2})(c_{1}^{2}-c^{2})}}}\left(F({\phi}_{3}(c),k_{2})-{\Pi}\left({\phi}_{3}(c),{\displaystyle \frac{k_{2}^{2}}{{\alpha}_{5}^{2}(c)}},k_{2}\right)\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle s_{8}={\displaystyle \frac{c^{2}\sqrt{b_{1}^{2}-c^{2}}\sqrt{c_{1}^{2}-c^{2}}}{a_{1}\sqrt{a_{1}^{2}-c_{1}^{2}}\sqrt{(a_{1}^{2}-c^{2})(b_{1}^{2}-c^{2})}}}\left(F({\theta}(c),k_{2})-{\Pi}\left({\theta}(c),{\displaystyle \frac{k_{2}^{2}}{{\alpha}_{6}^{2}(c)}},k_{2}\right)\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{1}=E({\phi}_{2}(a),k_{2})(r_{1}+r_{2}-r_{3}),\quad T_{2}={\displaystyle \frac{1}{\sqrt{a_{1}^{2}-c_{1}^{2}}}}F({\phi}_{2}(a),k_{2})(r_{4}+r_{5}+r_{6}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{3}={\displaystyle \frac{\sqrt{(a_{1}^{2}-a^{2})(a^{2}-b_{1}^{2})(a^{2}-c_{1}^{2})}}{\sqrt{a_{1}^{2}-c_{1}^{2}}}}(-F({\phi}_{2}(a),k_{2})+r_{7}-r_{8}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{1}=a_{1}^{2}i_{5}({\phi}_{2}(a),m_{2},k_{2}),\quad r_{2}=b_{1}^{2}i_{6}({\phi}_{6}(b),k_{2},k_{4},k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{3}=c_{1}^{2}\left(i_{7}({\phi}_{3}(c),k_{2})-{\displaystyle \frac{b_{1}^{2}}{c_{1}^{2}}}i_{8}({\phi}_{3}(c),k_{2})\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{4}=\sqrt{a_{1}^{2}-c_{1}^{2}}(a_{1}^{2}E({\phi}_{2}(a),k_{2})-(a_{1}^{2}-b_{1}^{2})i_{9}({\phi}_{2}(a),k_{2})),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{5}={\displaystyle \frac{k_{2}^{^{\prime }2}m_{2}a_{1}}{k_{2}}}(a_{1}^{2}i_{10}({\phi}_{6}(b),k_{2})-(a_{1}^{2}-b_{1}^{2})i_{11}({\phi}_{6}(b),k_{2})),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{6}={\displaystyle \frac{k_{2}^{^{\prime }2}m_{2}a_{1}}{k_{2}}}(a_{1}^{2}i_{8}({\phi}_{3}(c),k_{2})-(a_{1}^{2}-c_{1}^{2})i_{12}({\phi}_{3}(c),k_{2})),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{7}={\displaystyle \frac{b_{1}^{2}}{(a^{2}-b_{1}^{2})}}i_{6}({\phi}_{6}(b),{\alpha}_{1}(a),k_{4},k_{2}),\quad r_{8}={\displaystyle \frac{c_{1}^{2}}{(a^{2}-c_{1}^{2})}}i_{6}\left({\phi}_{3}(c),{\alpha}_{5}(a),{\displaystyle \frac{1}{k_{3}}},k_{2}\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{4}=2{\displaystyle \frac{a_{1}^{2}}{\sqrt{a_{1}^{2}-c_{1}^{2}}}}\left({\displaystyle \frac{m_{2}^{3}}{k_{2}}}H_{1}+{\displaystyle \frac{m_{3}^{2}}{k_{3}}}J_{1}\right)F({\phi}_{6}(b),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{5}=-r_{9}+r_{10}+\{E({\phi}_{2}(a),k_{2})-E({\phi}_{2}(b),k_{2})+E({\theta}(c),k_{2})\}r_{11},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{6}=\{F({\phi}_{2}(a),k_{2})-F({\phi}_{2}(b),k_{2})\}r_{12}+r_{13}-r_{14}+r_{15},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{7}=F({\theta}(c),k_{2})r_{16}-{\displaystyle \frac{(a_{1}^{2}-c^{2})}{a_{1}\sqrt{a_{1}^{2}-c_{1}^{2}}}}2p_{4}({\theta}(c),0,k_{2})r_{17},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{9}=2b_{1}(1-k_{3}^{2}){\displaystyle \frac{m_{3}}{k_{3}}}k_{2}^{2}{\tau}_{1}i_{5}({\phi}_{6}(b),k_{2}),\quad r_{10}=2a_{1}{\tau}_{1}^{^{\prime }}m_{2}^{2}k_{2}^{^{\prime }2}k_{2}^{2}i_{6}({\phi}_{6}(b),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{11}={\displaystyle \frac{b_{1}m_{3}}{k_{3}}}i_{6}({\phi}_{6}(b),k_{2},k_{4},k_{2}),\quad r_{12}={\displaystyle \frac{k_{2}^{^{\prime }2}}{a_{1}}}\{(b_{1}^{2}-c_{1}^{2})k_{2}^{2}i_{15}({\phi}_{6}(b),k_{2})+b_{1}^{2}i_{16}({\phi}_{6}(b),k_{2})\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{13}={\displaystyle \frac{2b_{1}}{c_{1}}}\sqrt{b_{1}^{2}-c_{1}^{2}}k_{2}^{^{\prime }2}{\tau}_{2}i_{16}({\phi}_{6}(b),k_{2}),\quad r_{15}=2\left({\displaystyle \frac{1}{k_{3}^{2}}}-1\right)Km_{3}k_{2}^{^{\prime }2}k_{2}^{2}i_{15}({\phi}_{6}(b),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{14}=2p_{4}({\phi}_{2}(a),0,k_{2}){\displaystyle \frac{(a_{1}^{2}-b_{1}^{2})}{a_{1}(c_{1}^{2}-a^{2})}}\left\{{\displaystyle \frac{a^{2}(b_{1}^{2}-c_{1}^{2})}{(a^{2}-b_{1}^{2})}}{\Pi}({\phi}_{6}(b),{\alpha}_{1}^{2}(a),k_{2})+c_{1}^{2}F({\phi}_{6}(b),k_{2})\right\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{16}={\displaystyle \frac{1}{a_{1}}}\{a_{1}^{2}k_{2}^{^{\prime }2}i_{8}({\phi}_{6}(b),k_{2})-(b_{1}^{2}-c_{1}^{2})k_{2}^{2}i_{13}({\phi}_{6}(b),k_{2})\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{17}=a_{1}k_{2}m_{2}\left\{i_{17}({\phi}_{6}(b),k_{2})+{\displaystyle \frac{c^{2}}{(b_{1}^{2}-c^{2})}}i_{18}({\phi}_{6}(b),k_{2})\right\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{8}={\displaystyle \frac{2a_{1}^{2}}{\sqrt{a_{1}^{2}-c_{1}^{2}}}}\left({\displaystyle \frac{m_{2}^{3}}{k_{2}}}H_{1}+{\displaystyle \frac{m_{3}^{2}}{k_{3}}}J_{1}\right)F({\phi}_{3}(c),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{9}=r_{18}-r_{19}+\{E({\phi}_{2}(a),k_{2})-E({\phi}_{2}(b),k_{2})+E({\theta}(c),k_{2})\}r_{20},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{10}=\{F({\phi}_{2}(a),k_{2})-F({\phi}_{2}(b),k_{2})\}r_{21}+2p_{4}({\phi}_{2}(a),0,k_{2})r_{22}(a)-2p_{4}({\phi}_{2}(b),0,k_{2})r_{22}(b),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle T_{11}=r_{23}+r_{24}-F({\theta}(c),k_{2})r_{25}+r_{26},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{18}=2{\displaystyle \frac{(b_{1}^{2}-c_{1}^{2})}{a_{1}}}{\tau}_{1}i_{19}({\phi}_{3}(c),k_{2}),\quad r_{19}=2{\displaystyle \frac{(b_{1}^{2}-c_{1}^{2})}{a_{1}}}{\tau}_{1}^{^{\prime }}i_{20}({\phi}_{3}(c),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{20}={\displaystyle \frac{b_{1}^{2}}{a_{1}}}\{F({\phi}_{3}(c),k_{2})-(1-k_{3}^{2})i_{19}({\phi}_{3}(c),k_{2})\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{21}=-{\displaystyle \frac{b_{1}^{2}}{a_{1}}}k_{2}^{^{\prime }2}\{(1-k_{3}^{2})i_{21}({\phi}_{3}(c),k_{2})-i_{8}({\phi}_{3}(c),k_{2})\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{22}(t)={\displaystyle \frac{a_{1}c_{1}^{2}m_{2}^{2}}{(t^{2}-c_{1}^{2})}}i_{6}\left({\phi}_{3}(c),{\alpha}_{5}(t),{\displaystyle \frac{1}{k_{3}}},k_{2}\right),\quad r_{23}=2b_{1}^{2}{\displaystyle \frac{m_{3}}{k_{3}}}(1-k_{3}^{2})k_{2}^{^{\prime }}{\tau}_{2}i_{8}({\phi}_{3}(c),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{24}=2c_{1}{\displaystyle \frac{m_{3}}{k_{3}^{2}}}(1-k_{3}^{2})k_{2}^{^{\prime }2}F({\pi}/2,k_{2})i_{21}({\phi}_{3}(c),k_{2}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{25}=-{\displaystyle \frac{1}{a_{1}}}\{(b_{1}^{2}-c_{1}^{2})i_{19}({\phi}_{3}(c),k_{2})-a_{1}^{2}k_{2}^{^{\prime }2}i_{16}({\phi}_{3}(c),k_{2})\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle r_{26}=-{\displaystyle \frac{2(a_{1}^{2}-c^{2})}{a_{1}}}p_{4}({\theta}(c),0,k_{2})\left\{i_{16}({\phi}_{3}(c),k_{2})-{\displaystyle \frac{c^{2}}{(c^{2}-b_{1}^{2})}}k_{2}^{2}F({\phi}_{3}(c),k_{2})\right\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{5}(z,m,k)={\displaystyle \frac{1}{k^{2}}}\{(k^{2}-m^{2})F(z,k)+m^{2}E(z,k)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{6}(z,m,n,k)={\displaystyle \frac{1}{m^{2}}}((m^{2}-n^{2}){\Pi}(z,m,k)+n^{2}F(z,k)),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{7}(z,k)={\displaystyle \frac{1}{k^{^{\prime }2}}}\{k^{^{\prime }2}F(z,k)-E(z,k)+p_{0}(z,k)\tan (z)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{8}(z,k)={\displaystyle \frac{1}{k^{^{\prime }2}}}\{p_{0}(z,k)\tan (z)-E(z,k)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{9}(z,k)={\displaystyle \frac{1}{3k^{^{\prime }2}}}\{(2k^{2}-1)E(z,k)+k^{^{\prime }2}F(z,k)-2k^{2}p_{4}(z,0,k)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{10}(z,k)={\displaystyle \frac{1}{k^{^{\prime }2}}}\{E(z,k)-2k^{2}p_{4}(z,0,k)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{11}(z,k)={\displaystyle \frac{1}{3k^{2}k^{^{\prime }2}}}\left\{\vphantom{{\displaystyle \frac{k^{2}k^{^{\prime }2}}{(1-k^{2}\sin ^{2}z)}}}(2k^{2}-1)E(z,k)+k^{^{\prime }2}F(z,k)+2p_{4}(z,0,k)\left(k^{2}-2k^{4}\right.\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{30.0pt}\qquad +\left.\left.{\displaystyle \frac{k^{2}k^{^{\prime }2}}{(1-k^{2}\sin ^{2}z)}}\right)∼\right\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{12}(z,k)={\displaystyle \frac{1}{3k^{^{\prime }2}}}\left\{\vphantom{{\displaystyle \frac{1}{\cos ^{2}z}}}(1-2k^{2})E(z,k)-k^{^{\prime }2}F(z,k)+p_{0}(z,k)\tan (z)\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{30.0pt}\qquad \times \left.{\displaystyle \frac{1}{\cos ^{2}z}}(k^{^{\prime }2}+\cos ^{2}z-2k^{^{\prime }2}\cos ^{2}z)\right\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{13}(z,k)={\displaystyle \frac{1}{k^{2}k^{^{\prime }2}}}\left(E(z,k)-k^{^{\prime }2}F(z,k)-{\displaystyle \frac{k^{2}\sin z\cos z}{p_{0}(z,k)}}\right),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{14}(z,k)={\displaystyle \frac{1}{3k^{4}k^{^{\prime }2}}}\left\{\vphantom{\left({\displaystyle \frac{k^{^{\prime }2}}{(1-k^{2}\sin ^{2}z)}}-1-k^{^{\prime }2}\right)}(1+k^{^{\prime }2})E(z,k)-2k^{^{\prime }2}F(z,k)+2k^{2}p_{4}(z,0,k)\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{30.0pt}\qquad \times \left.\left({\displaystyle \frac{k^{^{\prime }2}}{(1-k^{2}\sin ^{2}z)}}-1-k^{^{\prime }2}\right)\right\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{15}(z,k)={\displaystyle \frac{1}{3k^{2}k^{^{\prime }4}}}\left\{\vphantom{\left({\displaystyle \frac{k^{^{\prime }2}}{(1-k^{2}\sin ^{2}z)}}-1-k^{^{\prime }2}\right)}(1+k^{2})E(z,k)-k^{^{\prime }2}F(z,k)+2k^{2}p_{4}(z,0,k)\right.\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \hspace{30.0pt}\qquad \times \left.\left(k^{^{\prime }2}-{\displaystyle \frac{k^{^{\prime }2}}{(1-k^{2}\sin ^{2}z)}}-2\right)\right\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{16}(z,k)={\displaystyle \frac{1}{k^{^{\prime }2}}}\{E(z,k)-2k^{2}p_{4}(z,0,k)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{17}(z,k)={\displaystyle \frac{1}{k^{2}}}\{F(z,k)-E(z,k)+2k^{2}p_{4}(z,0,k)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{18}(z,k)={\displaystyle \frac{1}{m^{2}}}\{(m^{2}-1){\Pi}(z,m,k)+F(z,k)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{19}(z,k)={\displaystyle \frac{1}{k^{^{\prime }2}}}\{k^{^{\prime }2}F(z,k)-E(z,k)+p_{0}(z,k)\tan (z)\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{20}(z,k)={\displaystyle \frac{1}{3k^{^{\prime }2}}}\left\{2k^{^{\prime }2}F(z,k)+(k^{2}-2)E(z,k)+p_{0}(z,k)\tan (z)\left(2-k^{2}+{\displaystyle \frac{k^{^{\prime }2}}{\cos ^{2}z}}\right)\right\},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle i_{21}(z,k)={\displaystyle \frac{1}{3k^{^{\prime }4}}}\left\{-k^{^{\prime }2}F(z,k)+(1+k^{2})E(z,k)+p_{0}(z,k)\tan (z)\left({\displaystyle \frac{k^{^{\prime }2}}{\cos ^{2}z}}-1-k^{2}\right)\right\}\nonumber\end{eqnarray}$

6. Numerical study

During the process of solving the above problems, we found that it is almost impossible to obtain the mathematical expressions of saturation zone length in terms of applied electric displacement. In view of this, a numerical study is carried out to discuss the growth of the saturation zone on the increasing value of applied stress and electric displacement. The analysis has been made for different types of piezoelectric materials given in the Table 1.

Table 1
Constants for different piezoelectric materials

Unit PZT-4 PZT-5H PZT-7A PZT-7 BaTiO₃

c ₁₁ 10⁹ Nm⁻² 139 126 148 130 150

c ₁₃ 10⁹ Nm⁻² 74.3 53.0 74.2 83.0 66.0

c ₁₂ 10⁹ Nm⁻² 77.8 55.0 76.2 83.0 66.0

c ₃₃ 10⁹ Nm⁻² 113 117 131 119 146

c ₄₄ 10⁹ Nm⁻² 25.6 35.3 25.4 25.0 44

e ₃₁ Cm⁻² −6.98 −6.50 −2.10 −10.3 −4.35

e ₃₃ Cm⁻² 13.8 23.3 9.50 14.7 17.5

e ₁₅ Cm⁻² 13.4 17.0 9.70 13.5 11.4

𝜀₁₁ 10⁻⁹ C(Vm)⁻¹ 6.00 15.1 8.11 17.1 9.87

𝜀₃₃ 10⁻⁹ C(Vm)⁻¹ 5.47 13.0 7.35 18.6 11.2

	Unit	PZT-4	PZT-5H	PZT-7A	PZT-7	BaTiO₃
c ₁₁	10⁹ Nm⁻²	139	126	148	130	150
c ₁₃	10⁹ Nm⁻²	74.3	53.0	74.2	83.0	66.0
c ₁₂	10⁹ Nm⁻²	77.8	55.0	76.2	83.0	66.0
c ₃₃	10⁹ Nm⁻²	113	117	131	119	146
c ₄₄	10⁹ Nm⁻²	25.6	35.3	25.4	25.0	44
e ₃₁	Cm⁻²	−6.98	−6.50	−2.10	−10.3	−4.35
e ₃₃	Cm⁻²	13.8	23.3	9.50	14.7	17.5
e ₁₅	Cm⁻²	13.4	17.0	9.70	13.5	11.4
𝜀₁₁	10⁻⁹ C(Vm)⁻¹	6.00	15.1	8.11	17.1	9.87
𝜀₃₃	10⁻⁹ C(Vm)⁻¹	5.47	13.0	7.35	18.6	11.2

Fig. 2.

Electric load ratio versus normalized saturation zone length at tip a.

Fig. 3.

Electric load ratio versus normalized saturation zone length at tip b.

The analysis has been made to discuss the effect of the applied electric displacement ratio on the growth of saturation zones for three collinear straight cracks. Figure 2 shows the variation between electric displacement ratio, $\frac{D_{2}^{\infty }}{D_{s}}$ and normalized saturation zone length $\frac{(a_{1}-a)}{(a-b)}$ at the crack tip z = a using the Eq. (58). The parameter ${\rho}=\frac{2c}{b+c}$ shows the inter crack distance. The higher value of 𝜌 = 0.9 demonstrates that the cracks are situated close to each other while the value 𝜌 = 0.1 shows the cracks are situated far away from each other.

It is observed from the results shown in Fig. 2 that the saturation zone length increases as applied electrical displacement to the normal saturation limit electric displacement, $\frac{D_{2}^{\infty }}{D_{s}}$ increases at crack tip x = a. It is observed that the saturation zone length is almost unaffected by the linear nature of electric displacement. Saturation zone length is bigger in case of closely located cracks (𝜌 = 0.9) in comparison to far away located cracks (𝜌 = 0.1).

Figure 3 shows the variation between normalized saturation zone length $\frac{(b-b_{1})}{(a-b)}$ and electric displacement ratio $\frac{D_{2}^{\infty }}{D_{s}}$ at the crack tip x = b. It is observed that the zone developed at the interior tip x = b is bigger than that at the exterior tip x = a for the same applied load $\frac{D_{2}^{\infty }}{D_{s}}$ . And as the prescribed electric load increases the developed saturation zone size also increases at both the interior and exterior tips, as expected.

Fig. 4.

Electric load ratio versus normalized saturation zone length at tip c.

Fig. 5.

Normalized COP versus electric load ratio at tip a.

The reverse effect is seen at the crack tip x = c due to small value of $\frac{t}{a_{1}}D_{s}$ for three cracks located far away from each other as shown in Fig. 4. However, for closely located cracks it shows same behavior as crack tip x = b. This means that the crack tip x = c is highly effected by the linearly varying electric displacement.

Fig. 6.

Normalized COP versus electric load ratio at tip b.

Fig. 7.

Normalized COP versus electric load ratio at tip c.

COP is calculated numerically at the crack tips x = a, b, c using Eqs (65)–(67). Figure 5 depicts the variation of COP over crack rims against the increasing value of $\frac{D_{2}^{\infty }}{D_{s}}$ . It is observed that COP remains negative. COP is almost same for PZT-4 and PZT-7 and less as compared to other type of materials. COP is higher for the material type PZT-7A as compared to others.

At crack tip x = b, COP is decreases faster as compared to crack tip x = a as shown in Fig. 6 at a smaller value of $\frac{D_{2}^{\infty }}{D_{s}}$ for different types of piezoelectric materials. Almost negligible potential drop is observed for the closely located cracks. As far COP at the crack tip x = c is concern, the tip is highly effected by the linear nature of electric displacement. A sharp decline is seen in COP at the crack tip at x = c due to very small value of electric displacement.

SIF given in Eqs (52)–(54) are evaluated numerically. The effect of piezoelectric properties on SIF at the crack tips x = a, x = b and x = c are plotted against the increasing value of applied electric-displacement $D_{2}^{\infty }$ for different piezoelectric materials. Figure 8 represents the variation between SIF at the crack tip x = a. It is seen from the results that the values of SIFs are changing by using different piezoelectric materials. It is observed as the $D_{2}^{\infty }$ increases the SIF increases linearly and the value is higher for PZT-4 and lower for PZT-7A and never takes zero value.

Fig. 8.

$D_{2}^{\infty }$ versus K _I(a).

Fig. 9.

$D_{2}^{\infty }$ versus K _I(b).

Fig. 10.

$D_{2}^{\infty }$ versus K _I(c).

Figure 9 shows the variation between $D_{2}^{\infty }$ and SIF at the crack tip b. Big difference is seen between the values of SIF for 𝜌 = 0.3 and 𝜌 = 0.7 means closely located cracks shows high value of SIF at small value of $D_{2}^{\infty }$ . Piezoelectric material PZT-4 shows the high value of SIF in comparison to others. The effect of the applied electric field on the SIF at the crack tip c is plotted and depicted in Fig. 10. It can be seen that the SIF is higher at the cracks tip c at lesser value of applied electric field $D_{2}^{\infty }$ (or electric displacement) as compared to other crack tips. This shows that the middle crack is highly sensitive about applied electric field when 𝜌 = 0.7.

7. Conclusion

A mathematical model is proposed for piezoelectric plate which is cut along three equal collinear hairline straights cracks under linearly varying electric field. Mathematical expressions for different fracture parameters are derived in terms of elliptic integrals. Numerical study is presented for different types of piezoelectric ceramics. A comparative study is also presented for different types of piezoelectric materials.

Footnotes

Acknowledgements

The authors are grateful to the referees and the editors for their valuable suggestions, which improved the understandability of the paper.

Conflict of interest

None to report.

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