# Conservation Integrals in Nonhomogeneous Materials with Flexoelectricity
## 5 A Nonhomogeneous Bernoulli-Euler Beam Problem with Flexoelectricity  
In this section, a nonhomogeneous Bernoulli-Euler beam problem will be solved to illustrate the use of the J-integral derived in the previous section. A series of work has been carried out on the Bernoulli-Euler beam [54; 55; 56; 57]. Based on their work, a double cantilever beam problem with a crack loaded by a couple \(M_{0}\) at the free end and the open circuit condition will be solved. Using the coordinate system \((x,y,z)\) shown in Figure 1, the displacement components can be obtained as follows [58; 59],  
\[u_{x}=-z\frac{\partial v}{\partial x},\quad\quad u_{y}=0,\quad\quad u_{z}=v \tag{52}\]  
The elastic strain and the strain gradient can be obtained from Equation (52) as  
\[\varepsilon_{xx}=-z\frac{d^{2}v}{dx^{2}}\quad\quad\quad w_{xxx}=-z\frac{d^{3} v}{dx^{3}}\quad\quad\quad w_{xxz}=-\frac{d^{2}v}{dx^{2}} \tag{53}\]  
The one-dimensional constitutive equations can be expressed as follows:  
\[ \begin{array}{l}\sigma_{xx}=\frac{\partial x^{E}}{\partial x_{ij}}=C_{11} \varepsilon_{xx}-d_{31}E_{z}+\frac{1}{2}\mu_{31}E_{z,z}\\ \tau_{xxz}=\frac{\partial x^{E}}{\partial y_{ijm}}=-\frac{1}{2}\mu_{31}E_{z} \\ D_{z}=a_{33}E_{z}+d_{31}\varepsilon_{xx}+\frac{1}{2}\mu_{31}\varepsilon_{xx,z }\\ Q_{zz}=-\frac{1}{2}\mu_{31}\varepsilon_{xx}\end{array}  \tag{54}\]  
where \(\mu_{31}=f_{31}-e_{31}\). It should be stressed that all the material coefficients depend on space position. In this example, for the convenience of mathematical calculations, all the material coefficients take the form:  
\[\Omega=\Omega_{0}\varrho^{xx} \tag{55}\]  
where \(\Omega_{0}\) and \(\alpha\) are constant. This form can also be found in previous works [51; 60]. The electric governing equation is:  
\[D_{i,i}-Q_{ij,ji}=0 \tag{56}\]  
Figure 1: Schematic of a beam with crack. The electric boundary conditions are assumed as:  
\[\varphi\bigg{(}\frac{1}{2}h\bigg{)}=V,\qquad\qquad\varphi\bigg{(}-\frac{1}{2}h \bigg{)}=0,\qquad\qquad Q_{33}\bigg{(}\pm\frac{1}{2}h\bigg{)}=0 \tag{57}\]  
Thus, combining Equations (54), (56), and (57), the electric potential, the electric field strength, and electric field gradient can be obtained:  
\[\varphi(z)=\frac{V}{2}+\frac{Vz}{h}+\frac{d_{31}}{2a_{33}}\bigg{(}\frac{h^{2}} {4}-z^{2}\bigg{)}\frac{d^{2}v}{dx^{2}}\ -\frac{\mu_{31}z}{2a_{33}}\frac{d^{2}v}{dx^{2}} \tag{58}\]  
\[E_{z}=\frac{V}{h}-z\frac{d_{31}}{a_{33}}\frac{d^{2}v}{dx^{2}}-\frac{\mu_{31}}{ 2a_{33}}\frac{d^{2}v}{dx^{2}} \tag{59}\]  
\[E_{z,z}=-\frac{d_{31}}{a_{33}}\frac{d^{2}v}{dx^{2}} \tag{60}\]  
Substituting Equations (58)-(60) into Equation (54), the stress and higher-order stress are:  
\[ \begin{array}{l}\sigma_{xx}=A_{1}e^{ax}z\frac{d^{2}v}{dx^{2}}+A_{2}e^{ax} \frac{d^{2}v}{dx^{2}}-d_{0}e^{ax}\frac{V}{h}\\ \tau_{xxz}=A_{2}e^{ax}z\frac{d^{2}v}{dx^{2}}+B_{1}e^{ax}\frac{d^{2}v}{dx^{2}} -\frac{1}{2}\mu_{0}e^{ax}\frac{V}{h}\end{array}  \tag{61}\]  
where  
\[A_{1}=\frac{d_{0}^{2}-C_{0}a_{0}}{a_{0}},\qquad\qquad A_{2}=\frac{\mu_{0}d_{0 }}{2a_{0}},\qquad\qquad B_{1}=\frac{\mu_{0}^{2}}{4a_{0}} \tag{62}\]  
The mechanical governing equation is [55]  
\[\frac{d^{2}}{dx^{2}}(M+P)=0 \tag{63}\]  
where \(M\) is the internal bending moment and \(P\) is the higher-order axial couple, respectively, as  
\[M=\int_{A}\sigma_{11}zdA=A_{1}Ie^{ax}\frac{d^{2}v}{dx^{2}} \tag{64}\]  
\[P=\int_{A}\tau_{113}dA=B_{1}Ae^{ax}\frac{d^{2}v}{dx^{2}}-\frac{1}{2}\mu_{0}e^ {ax}\frac{VA}{h} \tag{65}\]  
where \(A\) is the area of the cross section and \(I\) is the usual second moment of the cross-sectional area defined as  
\[A=\int_{A}dA\qquad\qquad I=\int_{A}z^{2}dA \tag{66}\]  
The governing equation must be supplemented by boundary conditions.