The Antiplane Problem of a Lip-Shaped Orifice With 4 Edge Cracks in 1D Hexagonal Piezoelectric Quasicrystal
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摘要: 通过构造共形映射,利用Stroh型公式,研究了一维六方压电准晶体中唇形孔口次生四条裂纹的反平面问题,并对裂纹尖端处的应力强度因子及能量释放率进行了解析求解. 在数值算例中,分析了缺陷的几何参数和外部载荷对应力强度因子及能量释放率的影响规律. 结果表明:唇形孔口左右任意一侧裂纹长度或孔口长度的增长,对左右两侧裂纹的扩展有促进作用;上下两侧裂纹长度的增长,对左右两侧裂纹的扩展无明显的影响;唇形孔口的高度越高,对左右两侧裂纹扩展的抑制作用越显著;外部机械载荷和电载荷增大,对裂纹的扩展有促进作用. 一些特殊缺陷可由该缺陷的相关参数退化得来,如唇形孔口次生两条裂纹、唇型裂纹、Griffith裂纹等.Abstract: Through construction of the conformal mapping and with Stroh's formula, the antiplane problem of 4 secondary cracks at the lip-shaped orifice of 1D hexagonal piezoelectric quasicrystal, was studied. The effects of geometrical parameters and external loads on stress intensity factors and energy release rates were analyzed with numerical examples. The results show that, the crack length growth on either corner side of the orifice or the orifice length increase can promote the crack propagation. The crack length growth on the upper and lower sides has no obvious effect on the crack propagation on the left and right sides. The higher the orifice height is, the more significant the inhibition effect on the crack growth on both corner sides will be; the increase of the external mechanical load and electric load can promote the crack propagation. Some special defects can be derived from the degradation of the relevant parameters of the defects, such as the secondary 2 cracks in the orifice, the lip-shaped crack, and the Griffith crack, etc.
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Key words:
- conformal mapping /
- orifice edge crack /
- Stroh's formula /
- stress intensity factor
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0. 引言
准晶理论框架的建立来自于一些学者所做的物理实验及相关研究[1-2],实验结果表明: 准晶具有较强的脆性、耐磨性、抗氧化性及耐腐蚀性. 在使用和制备准晶的过程中,可能会出现孔洞、裂纹等缺陷,这些缺陷极大地影响了准晶材料的性能. 因此,研究压电准晶材料的裂纹及孔洞问题具有重要的理论意义和实际应用价值. Muskhelishvili[3]最早提出用复变函数法解决弹性孔口问题,为后人在准晶缺陷方面的研究提供了重要的参考价值. 在众多准晶中,一维六方准晶以其相对简单、具有代表性的研究方法成为研究准晶弹性问题的首选. Fan[4]提出了准晶弹性的数学理论,推导了一维六方准晶的弹性力学方程,对一些裂纹和孔洞问题进行了求解. 郭俊宏和刘官厅[5-6]推导了一维六方准晶中圆形孔口次生两条不对称裂纹和椭圆孔口次生两条不对称裂纹两种复杂缺陷的解析解. Li和Fan[7]得到了一维六方准晶条带中两个半无限共线裂纹的精确解. Shi[8]将边值问题简化为解析函数的Riemann-Hilbert问题,得到了一维六方准晶中反平面滑模共线周期裂纹和刚性包裹体的闭合解.
压电准晶体是指具有压电效应的准晶体,由于准晶体中存在相位子场,其压电性质比常规晶体复杂得多. Zhou和Li[9]研究了一维六方压电准晶条带在可渗透电边界条件下的两个共线裂纹,采用积分变换和积分方程的方法,得到了电弹性场的显式表达式. Altay等[10]通过研究得到了压电准晶体的基本方程. Li等[11]给出了三维压电准晶的基本方程及其精确解的算法. 张也等[12]研究了一维六方压电准晶体中Ⅲ型偏折裂纹的断裂问题,分析了裂纹尖端应力强度因子与缺陷几何尺寸的关系.
对于含唇形裂纹和孔口方面的研究,匡震邦[13]最先给出了含唇形裂纹的复合材料缺陷构型,推导了其应力及位移的解析表达式,并求得了裂纹尖端处的应力强度因子. 郭怀民等[14]研究了一维六方压电准晶中含唇形孔口次生两不对称裂纹的反平面剪切问题. 于静等[15]利用复变函数方法,研究了含唇形裂纹的压电复合材料受机械载荷和电载荷作用下的反平面问题. 受实际电站大型锻件断裂现象[16]的启发,本文提出了一种唇形孔口次生四条裂纹的缺陷模型,其意义在于这种缺陷较唇形裂纹更符合实际工程中存在的情况,且有更高的研究难度.
本文研究了一维六方压电准晶中唇形孔口次生四条裂纹的反平面问题,在电不可通的边界条件下求得了裂纹尖端处的应力强度因子和能量释放率,最后通过数值算例分析了孔口及裂纹的几何参数和外部载荷对裂纹扩展的影响. 此外,在退化一些参数后,可得到一些特殊缺陷,如唇形孔口次生两条裂纹、唇形裂纹、Griffith裂纹等.
1. 基本方程
取三维空间直角坐标系xi(i=1, 2, 3),设x3方向为一维六方压电准晶的准周期方向,各向同性平面x1Ox2与x3方向垂直,当缺陷沿准周期方向穿透时,研究问题变为二维平面问题,由文献[17]可知,一维六方压电准晶弹性问题的基本方程如下:
平衡方程
$$\sigma_{i j, i}=0, H_{i j, i}=0, D_{i, i}=0 ;$$ (1) 几何方程
$$\varepsilon_{i j}=\left(u_{i, j}+u_{j, i}\right) / 2, W_{i j}=v_{i, j}, E_i=-\phi_{, i} ;$$ (2) 本构方程
$$\left\{\begin{array}{l} \sigma_{i j}=C_{i j k l} \varepsilon_{k l}+R_{i j k l} W_{k l}-e_{k i j} E_k, \\ H_{i j}=R_{i j k l} \varepsilon_{k l}+K_{i j k l} W_{k l}-d_{k i j} E_k, \\ D_i=e_{k i j} \varepsilon_{j k}+d_{k i j} W_{j k}+\lambda_{i j} E_j, \end{array}\right.$$ (3) 其中i, j =1, 2, 3;εij, σij, ui分别表示声子场应变、应力和位移;Wij, Hij, vi分别表示相位子场应变、应力和位移;Ei, Di, φ分别表示电场、电位移和电势;Cijkl, Kijkl分别为声子场和相位子场的独立弹性常数;eijk, dijk分别为声子场和相位子场压电常数;λij, Rijkl分别为介电常数和声子场与相位子场的耦合弹性常数.
将式(3)代入式(1)后,将所得结果整理化简后可得[18]
$$\boldsymbol{B}_{0} \nabla^{2} \boldsymbol{u}={\mathbf 0} , $$ (4) 其中,$ \nabla^{2}$表示三维Laplace算子,u为广义位移向量,B0为压电准晶的材料系数矩阵,
$$ \boldsymbol{u}=\left[u_{3}, w_{3}, \varphi\right]^{\mathrm{T}}, \boldsymbol{B}_{0}=\left[\begin{array}{ccc} C_{44} & R_{3} & e_{15} \\ R_{3} & K_{2} & d_{15} \\ e_{15} & d_{15} & -\lambda_{11} \end{array}\right].$$ (5) 由文献[14]知,u可表示为
$$\boldsymbol{u}=\boldsymbol{A} \boldsymbol{f}(z)+\overline{\boldsymbol{A} \boldsymbol{f}(z)}, \quad z=x_{1}+\mathrm{i} x_{2}, $$ (6) 其中,A为三阶单位矩阵,f(z)为解析的复向量函数,$ \overline{\boldsymbol{f}(z)}$为f(z)的共轭函数.
引入广义函数向量Σ,使得
$$-\boldsymbol{\varSigma}_{, 2}=\left[\sigma_{31}, H_{31}, D_{1}\right]^{\mathrm{T}}, \boldsymbol{\varSigma}_{, 1}=\left[\sigma_{32}, H_{32}, D_{2}\right]^{\mathrm{T}} , $$ (7) 结合式(3)、式(6)和式(7)得
$$-\boldsymbol{\varSigma}_{, 2}=\boldsymbol{B}_{0} \frac{\partial \boldsymbol{f}}{\partial x_{1}}+\boldsymbol{B}_{0} \frac{\partial \overline{\boldsymbol{f}}}{\partial x_{1}}, \boldsymbol{\varSigma}_{, 1}=\boldsymbol{B}_{0} \frac{\partial \boldsymbol{f}}{\partial x_{2}}+\boldsymbol{B}_{0} \frac{\partial \overline{\boldsymbol{f}}}{\partial x_{2}}.$$ (8) 对式(8)积分得
$$\boldsymbol{\varSigma}=\boldsymbol{B} \boldsymbol{f}(z)+\overline{\boldsymbol{B} \boldsymbol{f}(z)}, $$ (9) 其中,B=iB0.
2. 复势函数及边界条件
考虑一维六方压电准晶体含唇形孔口次生四条裂纹的缺陷模型,如图 1所示. 假设唇口沿x1方向的裂纹长度分别为L1和L2,沿x2方向的上下两侧裂纹长度均为L3,a为孔口长度,h为孔口高度,z1为孔口右侧裂纹尖端处的点,准晶基体在无穷远处受声子场、相位子场反平面剪切应力和电场的共同作用.
引入复向量函数[6]:
$$\boldsymbol{f}(z)=\boldsymbol{c}^{\infty} z+\boldsymbol{f}_{0}(z) , $$ (10) 其中,c∞为与远场载荷相关的复常数向量,f0(z)为待定的复向量,且f0(∞)=0.
对式(6)和式(9)两端关于x1求导得
$$ \left\{\begin{array}{l} \boldsymbol{u}_{, 1}=\boldsymbol{A} \boldsymbol{F}(z)+\overline{\boldsymbol{A} \boldsymbol{F}(z)} , \\ \boldsymbol{\varSigma}_{, 1}=\boldsymbol{B F}(z)+\overline{\boldsymbol{B} \boldsymbol{F}(z)}, \end{array}\right.$$ (11) 其中,F(z)=df(z)/dz. 将式(10)代入式(11),并令z→∞可得
$$\boldsymbol{u}_{, 1}^{\infty}=\boldsymbol{A} \boldsymbol{c}^{\infty}+\overline{\boldsymbol{A} \boldsymbol{c}^{\infty}}, \boldsymbol{\varSigma}_{, 1}^{\infty}=\boldsymbol{B} \boldsymbol{c}^{\infty}+\overline{\boldsymbol{B} \boldsymbol{c}^{\infty}}, $$ (12) 其中,Σ, 1∞=(σ32∞, H32∞, D2∞)T, u, 1∞=(ε31∞, w31∞, E1∞)T.
在力和电载荷的作用下,唇形孔口和裂纹面上的边界条件为
$$\boldsymbol{B} \boldsymbol{f}(z)+\overline{\boldsymbol{B} \boldsymbol{f}(z)}=\int_{s} \boldsymbol{t}_{s} \mathrm{~d} s, \quad \boldsymbol{t}_{s}=\left(t_{3}, h_{3}, -D_{\mathrm{n}}\right)^{\mathrm{T}}, $$ (13) 其中,t3, h3分别为唇形孔口和裂纹面上的声子场和相位子场的反平面剪切力,Dn为沿着边界的法向电位移.
在实际工程中,由于唇形孔口的内部电场微小,可以采用电不可通的边界条件,假设在唇形孔口周围及裂纹面上不受载荷作用,则有
$$\boldsymbol{B} \boldsymbol{f}(z)+\overline{\boldsymbol{B} \boldsymbol{f}(z)}={\mathbf 0} .$$ (14) 将式(10)代入式(14)得
$$\boldsymbol{B} \boldsymbol{f}_{0}(z)+\overline{\boldsymbol{B} \boldsymbol{f}_{0}(z)}=-\left(\boldsymbol{B} \boldsymbol{c}^{\infty} z+\overline{\boldsymbol{B} \boldsymbol{c}^{\infty} z}\right).$$ (15) 3. 共形映射及问题求解
为求得f0(z),构造共形映射函数如下(构造过程见附录A):
$$z=\omega(\zeta)=\frac{a \rho}{2}\left\{H(\zeta)-\frac{n}{H(\zeta)}+\frac{H(\zeta)}{\rho^{2}\left[H^{2}(\zeta)-n\right]}\right\} .$$ (16) 该共形映射函数将z平面上唇形孔口次生四条裂纹的外部区域映射到ζ平面单位圆的内部区域,且
$$ \omega(1)=\frac{a}{2}\left(1+l_{1}+\frac{1}{1+l_{1}}\right)=a+L_{1}, \omega(-1)=-\frac{a}{2}\left(1+l_{2}+\frac{1}{1+l_{2}}\right)=-\left(a+L_{2}\right) . $$ 将式(16)代入式(15),在ζ平面上有
$$\boldsymbol{B} \boldsymbol{f}_{0}(\sigma)+\overline{\boldsymbol{B} \boldsymbol{f}_{0}(\sigma)}=-\left[\boldsymbol{B} \boldsymbol{c}^{\infty} \omega(\sigma)+\overline{\boldsymbol{B} \boldsymbol{c}^{\infty} \omega(\sigma)}\right], $$ (17) 其中,σ=eiθ为单位圆上的点,f0(σ)为f0(ζ)在单位圆上的取值,且f0(σ)=f0[ω(σ)]. 通过式(16)发现ω(σ)=$ \overline{\omega(\sigma)}$,由式(12)和式(17)得
$${\boldsymbol{B f}}_{0}(\sigma)+\overline{{\boldsymbol{B f}}_{0}(\sigma)}=-{\boldsymbol{\varSigma}}_{, 1}^{\infty} \omega(\sigma).$$ (18) 将式(18)两边乘以$ \frac{1}{2 {\rm{ \mathsf{ π}}} \mathrm{i}} \int_{\tau} \frac{\mathrm{d} \sigma}{\sigma-\zeta}$,由Cauchy积分公式得
$$ {\boldsymbol{B f}}_{0}(\zeta)=-{\boldsymbol{\varSigma}}_{, 1}^{\infty} \frac{1}{2 {\rm{ \mathsf{ π}}} \mathrm{i}} \int_{\tau} \frac{\omega(\sigma)}{\sigma-\zeta} \mathrm{d} \sigma, \quad|\zeta|<1 .$$ (19) 由式(16)可知,ω(ζ)在单位圆内除一个极点ζ=0外处处解析,由留数定理得
$$\frac{1}{2 {\rm{ \mathsf{ π}}} \mathrm{i}} \int_\tau \frac{\omega(\sigma)}{\sigma-\zeta} \mathrm{d} \sigma=\omega(\zeta)-\frac{a \rho\left(\sqrt{g_1^2+g_3^2}+\sqrt{g_2^2+g_3^2}\right)}{4 \zeta} .$$ (20) 将式(20)代入式(19),并对等式两边求导得
$$\boldsymbol{B} \boldsymbol{F}_0(\zeta)=-\boldsymbol{\varSigma}_{, 1}^{\infty}\left(\omega^{\prime}(\zeta)+\frac{a \boldsymbol{\rho}\left(\sqrt{g_1^2+g_3^2}+\sqrt{g_2^2+g_3^2}\right)}{4 \zeta^2}\right), $$ (21) 其中,F0(ζ)=df(ζ)/dζ.
对式(16)两边求导得
$$\omega^{\prime}(\zeta)=\frac{a \rho}{2}\left\{1+\frac{n}{H^2(\zeta)}-\frac{H^2(\zeta)+n}{\rho^2\left[H^2(\zeta)-n\right]^2}\right\} H^{\prime}(\zeta), $$ (22) 其中
$$ H^{\prime}(\zeta)=\frac{\left[\sqrt{\mu(\zeta)-16 \zeta^2}+\sqrt{\mu(\zeta)}\right]\left[-2 \mu(\zeta)+\zeta \mu^{\prime}(\zeta)\right]}{8 \zeta^2 \sqrt{\mu(\zeta)\left[\mu(\zeta)-16 \zeta^2\right]}} . $$ 4. 应力强度因子的精确解
$$\boldsymbol{\kappa}=\left[K_{\text {III }}^\sigma, K_{\text {III }}^H, K_{\text {III }}^D\right]^{\mathrm{T}}=\lim\limits_{z \rightarrow z_1} \sqrt{2 {\rm{ \mathsf{ π}}}\left(z-z_1\right)} \boldsymbol{\varSigma}_{, 1}, $$ (23) 其中,KⅢσ, KⅢH, KⅢD分别为声子场应力强度因子、相位子场应力强度因子和电场强度因子.
将式(11)代入式(23)得
$$\boldsymbol{\kappa}=\left[K_{\text {III }}^\sigma, K_{\text {III }}^H, K_{\text {III }}^D\right]^{\mathrm{T}}=\lim\limits_{z \rightarrow z_1} \sqrt{2 {\rm{ \mathsf{ π}}}\left(z-z_1\right)} \boldsymbol{B} \boldsymbol{F}_0(z) .$$ (24) 将式(16)代入式(24)得
$$\boldsymbol{\kappa}=2 \sqrt{2 {\rm{ \mathsf{ π}}}} \lim\limits_{\zeta \rightarrow 1} \sqrt{\omega(\zeta)-\omega(1)} \frac{\boldsymbol{B} \boldsymbol{F}_0(\zeta)}{\omega^{\prime}(\zeta)}=2 \sqrt{{\rm{ \mathsf{ π}}}} \lim\limits_{\zeta \rightarrow 1} \frac{\boldsymbol{B} \boldsymbol{F}_0(\zeta)}{\sqrt{\omega^{\prime \prime}(\zeta)}} .$$ (25) 将式(16)和式(21)代入式(25)得
$$\boldsymbol{\kappa}=K \sqrt{{\rm{ \mathsf{ π}}}}\left(\begin{array}{l} \sigma_{32}^{\infty} \\ H_{32}^{\infty} \\ D_{2}^{\infty} \end{array}\right), $$ (26) K为无量纲应力强度因子,
$$K=\frac{a \rho\left(\sqrt{g_{2}^{2}+g_{3}^{2}}+\sqrt{g_{1}^{2}+g_{3}^{2}}\right)}{2} \cdot \frac{1}{\sqrt{\omega^{\prime \prime}(1)}}, $$ (27) 其中
$$ \begin{gather*} \omega^{\prime \prime}(1)=\frac{a \rho}{4}\left\{1+\frac{n}{\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)^{2}}-\frac{\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)^{2}+n}{\rho^{2}\left[\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)^{2}-n\right]^{2}}\right\} \times \\ \frac{\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)\left(g_{1}^{2}+g_{3}^{2}+\sqrt{g_{2}^{2}+g_{3}^{2}} \sqrt{g_{1}^{2}+g_{3}^{2}}\right)}{g_{1} \sqrt{g_{1}^{2}-1}}. \end{gather*}$$ (28) 5. 能量释放率的精确解
$$G=\frac{K_{{\rm{III}}}^{\sigma} K_{{\rm{II}}}^{S}+K_{{\rm{III}}}^{H} K_{\perp}^{S}-K_{{\rm{III}}}^{D} K_{E}}{2}, $$ (29) 其中,KⅡS和K⊥S分别为声子场应变强度因子和相位子场应变强度因子,KE为电场强度因子. 式(29)中强度因子关系如下:
$$ \left\{\begin{array}{l} K_{\text {III }}^{s}=C_{44} K_{\text {II }}^{S}+R_{3} K_{\perp}^{S}-e_{15} K_{E} , \\ K_{\text {III }}^{H}=R_{3} K_{\text {II }}^{S}+K_{2} K_{\perp}^{S}-d_{15} K_{E}, \\ K_{\text {III }}^{D}=e_{15} K_{\text {II }}^{S}+d_{15} K_{\perp}^{S}+\lambda_{11} K_{E}. \end{array}\right.$$ (30) 将式(30)代入式(29)可得
$$ G=\frac{1}{2}\left(K_{\text {III }}^{\sigma}, K_{\text {III }}^{H}, K_{\text {III }}^{D}\right)\left(\begin{array}{ccc} C_{44} & R_{3} & e_{15} \\ R_{3} & K_{2} & d_{15} \\ e_{15} & d_{15} & -\lambda_{11} \end{array}\right)^{-1}\left(\begin{array}{l} K_{\text {III }}^{\sigma} \\ K_{\text {III }}^{H} \\ K_{\text {III }}^{D} \end{array}\right).$$ (31) 将式(5)和式(30)代入式(31)得
$$G=\frac{{\rm{ \mathsf{ π}}}}{2 \operatorname{det} \boldsymbol{B}_{0}} \Pi K^{2}, $$ (32) 其中
$$ \left\{\begin{array}{l} \operatorname{det} \boldsymbol{B}_{0}=-C_{44} K_{2} \lambda_{11}+2 R_{3} e_{15} d_{15}-K_{2} \mathrm{e}_{15}^{2}+R_{3}^{2} \lambda_{11}-C_{44} d_{15}^{2} , \\ \Pi=\left(K_{2} \lambda_{11}+d_{15}^{2}\right)\left(\sigma_{32}^{\infty}\right)^{2}+\left(C_{44} \lambda_{11}+\mathrm{e}_{15}^{2}\right)\left(H_{32}^{\infty}\right)^{2}-\left(C_{44} K_{2}-R_{3}^{2}\right)\left(D_{2}^{\infty}\right)^{2}- \\ \ \ \quad 2\left(R_{3} \lambda_{11}+e_{15} d_{15}\right) \sigma_{32}^{\infty} H_{32}^{\infty}-2\left(R_{3} d_{15}-K_{2} e_{15}\right) \sigma_{32}^{\infty} D_{2}^{\infty}+2\left(C_{44} d_{15}-R_{3} e_{15}\right) H_{32}^{\infty} D_{2}^{\infty}, \end{array}\right.$$ (33) K由式(27)决定. 由上述结果可见,若不考虑相位子场的影响,可得到压电复合材料中唇形孔口次生四条裂纹的特殊情况;若忽略电场,则可得到一维六方准晶中含唇形孔口次生四条裂纹的特殊情况.
1) 如果L3=0,则可得到唇形孔口次生横向两条裂纹的解析解,即
$$ K=\frac{\sqrt{a \rho\left(g_{1}+g_{2}\right)}}{\sqrt{\left[1+\frac{n}{\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)^{2}}-\frac{\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)^{2}+n}{\rho^{2}\left[\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)^{2}-n\right]^{2}}\right] \cdot \frac{g_{1}+\sqrt{g_{1}^{2}-1}}{\sqrt{g_{1}^{2}-1}}}}, $$ 此结果与文献[14]相同,从而验证了式(27)的正确性.
2) 若L2=0,L3=0,则可得到唇形孔口次生右侧裂纹的解析解:
$$ K=\frac{\sqrt{a \rho\left(g_{1}+1\right)}}{\sqrt{\left[1+\frac{n}{\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)^{2}}-\frac{\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)^{2}+n}{\rho^{2}\left[\left(g_{1}+\sqrt{g_{1}^{2}-1}\right)^{2}-n\right]^{2}}\right] \cdot \frac{g_{1}+\sqrt{g_{1}^{2}-1}}{\sqrt{g_{1}^{2}-1}}}}. $$ 3) 若L1=0,L2=0,L3=0,g1=g2=1,ω(-1)=-a,ω(1)=a,$ \omega(\mathrm{i})=-\frac{2 n a}{1-n^{2}} \mathrm{i}, \omega(-\mathrm{i})=\frac{2 n a}{1-n^{2}} \mathrm{i}$,则可得到唇形裂纹的解析解为
$$ K=\frac{\sqrt{a}\left(1-\beta+\sqrt{\beta^{2}+1}\right)}{2}, $$ 其中$ \beta=\frac{h}{a}=\frac{2 n}{1-n^{2}}$, 该结果与文献[22]结果一致.
4) 若L1→0,L2→0,L3→0,h→0,m→0,有K=$ \sqrt{a}$,该结果为经典Griffith裂纹的解析解[23].
6. 数值算例
由式(27)和式(32)可知,裂纹尖端的应力强度因子和能量释放率与孔口高度、孔口长度、裂纹长度及外部载荷有着密切的关系. 下面分析缺陷的几何参数和外部载荷对应力强度因子及能量释放率的影响,选取材料参数如表 1所示[11].
图 2为应力强度因子K随唇形孔口长度a的变化曲线,其中h=0.005 m,L2=0.006 m,L3=0.006 m. 结果表明,当右侧裂纹长度L1固定时,应力强度因子随孔口长度的增加而增大,即孔口长度的增加对裂纹的扩展有促进作用. 该结果与文献[15]所得结果相吻合.
图 3讨论了L1/a与K的关系,其中a=0.01 m,h=0.005 m,L3=0.005 m. 由图可知,当左侧裂纹长度L2固定时,应力强度因子K随孔口右侧裂纹长度L1的增加而增大;当L1固定时,K随孔口左侧裂纹长度L2的增加而增大. 由于缺陷的对称性及边界条件的对称性,左侧裂纹尖端处的应力强度因子随右侧裂纹长度的增加而增大. 由此可知,左右任意一侧裂纹长度的增加对两侧裂纹的扩展有促进作用.
图 4讨论了h/a与K的关系,其中a=0.01 m,L1=0.005 m,L3=0.005 m. 由图可知:当左侧裂纹长度L2=0 m,孔口高度h增加时,K先增大后减小而后趋近常数;当左侧裂纹长度L2不为零,孔口高度h增加时,K逐渐减小,最后趋近常数. 该结果与文献[15]所得结果一致,从而验证了该结论的正确性. 两侧裂纹长度均不为零时,孔口高度的增加对裂纹的扩展有抑制作用,左右两侧裂纹长度的增加,对裂纹的扩展有促进作用.
图 5给出了当a=0.01 m,L1=0 m,L2=0 m,L3=0 m时,K随h/a的变化图像. 图像表明,K随h的增加逐渐减小最后趋于常数,该结果与文献[22]所得结论一致.
图 6描述了应力强度因子K与唇形孔口上下两侧裂纹长度L3的关系. 由图可知,当h固定时,随着L3的增加,K无明显变化趋势,即上下两侧裂纹长度的变化对左右两侧裂纹的扩展无明显影响. 此结果与文献[24]结论一致.
图 7讨论了不同L1/a条件下,能量释放率G/Gcr与唇形孔口长度a的关系,其中h=0.005 m,L2=0.006 m,L3=0.006 m,Gcr=5.0 N/m,Gcr为标准能量释放率. 结果表明:当左侧裂纹长度L1固定时,G/Gcr随a的增加而增大,孔口长度的增加促进了裂纹的扩展;当a固定时,G/Gcr随L1的增加而增大,右侧裂纹长度的增加促进了裂纹的扩展.
图 8给出了当a=0.01 m,h=0.005 m,L1=0.006 m,L2=0.01 m,L3=0.005 m时,能量释放率G/Gcr随声子场应力σ32∞变化的曲线. 由图像可以看出,当电位移D2∞固定时,G/Gcr随σ32∞的增加而增大,可见声子场应力的增大对裂纹的扩展有促进作用,此结果与文献[25]所得结果一致.
当D2∞=0 C/m2时,本文材料退化为一维六方准晶材料,图 9给出了当a=0.01 m,h=0.005 m,D2∞=0 C/m2,H32∞=0.2 MPa,L3=0.005 m时,能量释放率G/Gcr和声子场应力σ32∞的关系曲线. 由图可知,当左侧裂纹长度L2固定时,能量释放率随声子场应力的增加而增大;当σ32∞固定时,G/Gcr随L2的增加而增大.
图 10为能量释放率G/Gcr随相位子场应力H32∞的变化曲线. 结果表明,G/Gcr随H32∞的增加而增大,相位子场应力的增大对裂纹扩展有促进作用. 当H32∞≤1 MPa时,左侧裂纹长度对能量释放率的影响可以忽略不计;当H32∞≥1 MPa时,随着H32∞不断增加,左侧裂纹长度对能量释放率的影响越来越显著.
图 11描述了准晶耦合系数R3对能量释放率G/Gcr的影响规律. 由图可知:G/Gcr随R3的增加而增大,当0 GPa<R3<4 GPa时,能量释放率随耦合系数的增加而缓慢增加;当R3>4 GPa时,能量释放率随耦合系数的增加而迅速增加. 结果表明,耦合系数越小含缺陷的材料越稳定,该结论与文献[14]所得结论一致.
由图 12可知,当相位子场弹性常数K2不变时,能量释放率G/Gcr随左侧裂纹长度L2的增加而增大. 当L2不变时,G/Gcr随K2的增加而减小,即相位子场弹性常数越大,裂纹越不易扩展,材料稳定性越高.
图 13描述了能量释放率G/Gcr随右侧裂纹长度L1的变化规律,当外部载荷给定时,左侧裂纹长度L2不变时,G/Gcr随L1的增加而增大,当L1不变时,G/Gcr随L2的增加而增大. 由此可知,左右任意一侧裂纹长度的增加都会促进裂纹的扩展.
图 14表明了孔口高度h对能量释放率G/Gcr的影响. 当左侧裂纹长度L2=0 m时,G/Gcr随h的增加先增大而后减小;当L2>0 m时,G/Gcr随h的增加而不断减小. 结果表明,左右两侧裂纹长度均不为零时,孔口高度的增加会抑制裂纹的扩展.
图 15揭示了孔口上下两侧裂纹长度L3对能量释放率G/Gcr的影响. 由图可见,能量释放率随上下两侧裂纹长度的增加无明显变化.
7. 结论
本文采用复变函数方法,结合断裂力学相关知识,讨论了一维六方压电准晶中唇形孔口次生四条裂纹的反平面剪切问题,得到了裂纹尖端处应力强度因子及机械能释放率的精确解. 通过改变孔口尺寸和裂纹长度,本文的结果可用于模拟实际工程中许多缺陷构型,其中有唇形孔口次生两条裂纹、唇形孔口次生一条裂纹、唇形裂纹、Griffith裂纹等特殊情况. 数值算例中分析了缺陷的几何参数和外部载荷对应力强度因子及能量释放率的影响,得到了如下结论:
1) 唇形孔口右侧裂纹尖端处的应力强度因子及能量释放率随左右任意一侧裂纹长度的增加而增大,由于外部载荷及边界条件的对称性,左右任意一侧裂纹的增长对两侧裂纹的扩展都有促进作用. 当上下两侧裂纹长度增加时,应力强度因子及能量释放率无明显变化,说明上下两侧裂纹长度对左右两侧裂纹的扩展无明显影响.
2) 应力强度因子及能量释放率随孔口长度的增加而增大,所以孔口长度的增加对裂纹的扩展有促进作用. 当唇形孔口左侧裂纹长度为零时,随着孔口高度的增加,右侧裂纹尖端的应力强度因子及能量释放率先增加后减小,最后趋于稳定,当左侧裂纹长度不为零时,应力强度因子随着孔口高度的增加不断减小最后趋于稳定. 由此可知:当一侧裂纹长度不为零时,另一侧裂纹尖端处的应力强度因子随孔口高度增加而减小,从而起到抑制裂纹扩展的作用;当孔口足够高时,应力强度因子及能量释放率几乎不受两侧裂纹长度的影响.
3) 能量释放率随声子场应力、相位子场应力、外加电场载荷和机械载荷的增加而增大,说明外部载荷对裂纹的扩展有促进作用. 当准晶耦合系数增大时,能量释放率也随之增大,即声子场和相位子场耦合程度越大,材料的稳定性越差.
4) 由文中结果可知:如果不考虑相位子场的影响,即相位子场弹性常数退化为零,则可得到压电复合材料中唇形孔口次生四条裂纹的反平面剪切问题精确解;如果不考虑电场的影响,则可得到一维六方准晶中唇形孔口次生四条裂纹的反平面剪切问题精确解.
本文研究结果可为一维六方压电准晶材料结构设计(例如电站大型锻件生产过程)和断裂力学研究提供一定的理论参考,为工程材料的制备和应用提供了良好的理论依据.
附录A. 共形映射函数的推导
图A1介绍了唇形孔口次生四条裂纹的外部区域到ζ平面单位圆的内部的共形映射详细步骤. 式(A1)为唇形孔口次生四条裂纹的外部区域到椭圆孔口次生四条裂纹的外部区域的共形映射; 受文献[26]启发, 式(A2)为椭圆孔口次生四条裂纹的外部区域到单位圆内的共形映射:
$$z=\frac{a}{2}\left(t+\frac{1}{t}\right) \text {. }$$ (A1) 受文献[26]启发,椭圆孔口次生四条裂纹的外部区域到单位圆内的共形映射为
$$t=\frac{1}{1-n}\left(H(\zeta)-\frac{n}{H(\zeta)}\right) , $$ (A2) 其中
$$ \begin{aligned} & H(\zeta)=\left[\sqrt{\mu(\zeta)}+\sqrt{\mu(\zeta)-16 \zeta^{2}}\right] /(4 \zeta) , \\ & \mu(\zeta)=-16 g_{3}^{2} \zeta^{2}+\left[\sqrt{g_{2}^{2}+g_{3}^{2}}(\zeta-1)^{2}+\sqrt{g_{1}^{2}+g_{3}^{2}}(\zeta+1)^{2}\right]^{2} , \\ & g_{1}=\frac{e_{1}^{2}+1}{2 e_{1}}, g_{2}=\frac{e_{2}^{2}+1}{2 e_{2}}, g_{3}=\frac{e_{3}^{2}-1}{2 e_{3}} , \\ & e_{1}=\frac{c_{1}+\sqrt{c_{1}^{2}-1+b^{2}}}{1+b}, e_{2}=\frac{c_{2}+\sqrt{c_{2}^{2}-1+b^{2}}}{1+b}, e_{3}=\frac{c_{3}+\sqrt{c_{3}^{2}-1+b^{2}}}{1+b}, \\ & c_{1}=1+l_{1}, c_{2}=1+l_{2}, c_{3}=b+l_{3}, \\ & b=\frac{h}{a}+\sqrt{\frac{h^{2}}{a^{2}}+1}, n=\frac{b-1}{b+1}, \rho=\frac{1}{1-n}, \quad 0 \leqslant n \leqslant 1 , \\ & l_{m}=L_{m}+\sqrt{L_{m}^{2}+2 a L_{m}}, \quad m=1, 2, l_{3}=\frac{L_{3}+\sqrt{a^{2}+\left(h+L_{3}\right)^{2}}+\sqrt{a^{2}+h^{2}}}{a}. \end{aligned} $$ -
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