# Conservation Integrals in Nonhomogeneous Materials with Flexoelectricity
## 5 A Nonhomogeneous Bernoulli-Euler Beam Problem with Flexoelectricity
At the crack tip, \(x=0\), one has  
\[v(0)=0,\ \ \ \ \ v^{\prime}(0)=0 \tag{67}\]  
At the right end, \(x=-l\), the boundary conditions of the upper arm read  
\[\left(M+P\right)_{x=-l}=M_{0},\qquad\qquad\frac{d}{dx}\left(M+P\right)_{x=-l}=0 \tag{68}\]  
Under these conditions, the general solution of the governing Equation (63) can be derived as  
\[v=B_{2}\big{(}e^{-\alpha x}-1+\alpha x\big{)}+B_{3}x^{2} \tag{69}\]  
where let \(B_{2}=\frac{1}{a^{2}}\frac{M_{0}}{A_{1}!+B_{1}A},\ B_{3}=\frac{1}{4}\frac{\mu _{0}}{A_{1}!+B_{1}A}\frac{VA}{h}\) for convenience. Using geometric relation (53), the strain and strain gradient are  
\[ \begin{array}{l}\varepsilon_{xx}=-z\frac{d^{2}y}{dx^{2}}=-z\big{(}B_{2}\alpha^{2 }e^{-ax}+2B_{3}\big{)}\\ w_{xxx}=-\frac{d^{2}y}{dx^{3}}=zB_{2}\alpha^{3}e^{-ax}\\ w_{xxz}=-\frac{d^{2}z}{dx^{2}}=-\big{(}B_{2}\alpha^{2}e^{-ax}+2B_{3}\big{)} \end{array}  \tag{70}\]  
Then, according to the constitutive equations, the stress, high-order stress, the electric displacement, and high-order electric quadrupole can be obtained  
\[ \begin{array}{l}\sigma_{xx}=A_{1}z\big{(}B_{2}\alpha^{2}+2B_{3}e^{ax}\big{)} +A_{2}\big{(}B_{2}\alpha^{2}+2B_{3}e^{ax}\big{)}-d_{0}e^{ax}\frac{V}{\hbar}\\ \tau_{xxx}=A_{2}z\big{(}B_{2}\alpha^{2}+2B_{3}e^{ax}\big{)}+B_{1}\big{(}B_{2} \alpha^{2}+2B_{3}e^{ax}\big{)}-\frac{1}{2}\mu_{0}e^{xx}\frac{V}{\hbar}\\ D_{z}=a_{0}e^{ax}\frac{V}{\hbar}-(2 zd_{0}+\mu_{0})\big{(}B_{2}\alpha^{2}+2B_{3 }e^{ax}\big{)}\\ Q_{zz}=\frac{1}{2}\mu_{0}z\big{(}B_{2}\alpha^{2}+2B_{3}e^{ax}\big{)}\end{array}  \tag{71}\]  
The \(x-\) direction path-independent integral is  
\[ \begin{array}{l}\mathrm{G}_{1}^{J}=\int_{1}^{G}-\int_{V}A_{1}^{G}dV\\ =\int_{V}\partial_{i}\big{(}g^{E}\delta_{i1}-\sigma_{ij}u_{j,1}-\tau_{mj}u_{m, j1}+\tau_{jim,m}u_{j,1}-D_{i}\varphi_{,1}-Q_{ji}\varphi_{,j1}+Q_{ijij}\varphi_{,1} \big{)}dV\\ \quad+\int_{V}\alpha g^{E}dV\end{array}  \tag{72}\]  
Using the governing equation, the electric Gibbs free energy can be rewritten as  
\[ \begin{array}{l}g^{E}=\frac{1}{2}\sigma_{ij}u_{i,j}+\frac{1}{2}\tau_{ijm}u_{ i,jm}-\frac{1}{2}D_{i}E_{i}-\frac{1}{2}Q_{ij}V_{ij}\\ =\frac{1}{2}\Big{[}\big{(}\sigma_{ij}u_{i}\big{)}_{,j}+\big{(}\tau_{ijm}u_{i, j}\big{)}_{,m}-\big{(}\tau_{ijm,m}u_{i}\big{)}_{,j}\Big{]}\\ +\frac{1}{2}\Big{[}\big{(}D_{i}\varphi\big{)}_{,i}+\big{(}Q_{ij}\varphi_{,i} \big{)}_{,j}-\big{(}Q_{ij,ij}\varphi\big{)}_{,i}\Big{]}\end{array}  \tag{73}\]  
Substituting Equation (73) into Equation (72), the path-independent integral can be rewritten as  
\[ \begin{array}{l}G_{1}^{J}\\ =\int_{S}\big{[}g^{E}n_{1}-n_{i}\big{(}\sigma_{ij}u_{j,1}+\tau_{mj}u_{m,j1}- \tau_{jim,m}u_{j,1}+D_{i}\varphi_{,1}+Q_{ji}\varphi_{,j1}-Q_{ijij}\varphi_{,1 }\big{)}\big{]}dS\\ +\int_{S}\frac{\alpha}{2}\big{(}\sigma_{ij}u_{i}n_{j}+\tau_{ijm}u_{i,j}n_{m}- \tau_{jim,m}u_{i}n_{j}+D_{i}\varphi n_{i}+Q_{ij}\varphi_{,i}n_{j}-Q_{ijij} \varphi n_{i}\big{)}dS\end{array}  \tag{74}\]  
If the gradient term is omitted, the conservation integral \(G_{1}^{J}\) will reduce to the result in [33]. Due to symmetry, we just need integrate along half of the contour, which is depicted by the dotted curve in Figure 1. It is straightforward to see that \(n_{1}=1,\ \ n_{3}=0\) on the left curve, \(n_{1}=0,\ \ n_{3}=1\) on the top curve, \(n_{1}=-1,\ \ n_{3}=0\) on the right curve. The whole integral is the sum of three parts:  
\[G_{1}^{J}=2\Big{(}G_{1}^{right}+G_{1}^{left}+G_{1}^{up}\Big{)} \tag{75}\]  
with  
\[ \begin{array}{l}G_{1}^{up}=\int_{-I}^{L-l}(-\tau_{113}u_{1,11}-D_{3}\varphi _{,1}-Q_{33}\varphi_{,31}+Q_{33,3}\varphi_{,1})dx\\ \quad+\int_{-I}^{L-l}\frac{g}{2}(\tau_{113}u_{1,1}+D_{3}\varphi+Q_{33}\varphi _{,3}-Q_{33,3}\varphi)dx\end{array}  \tag{76}\]  
\[ \begin{array}{l}G_{1}^{right}=\int_{0}^{\frac{1}{2}h}\Big{(}-g^{E}+\sigma_{11 }u_{1,1}-\tau_{113,3}u_{1,1}\Big{)}dz+\int_{0}^{\frac{1}{2}h}\frac{\alpha}{2}( -\sigma_{11}u_{1}+\tau_{113,3}u_{1})dz\end{array}  \tag{77}\]  
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