The theory of generalized thermoelasticity with fractional order strain for dipolar materials with double porosity

Abstract

This article continues the work on dipolar thermoelastic materials, which are a special case of multipolar continuum mechanics. This theory allows a double-porous structure: a macro-porosity related to pores in the material and a microporosity, which shows fissures in the porous skeleton. This paper constructs a mathematical model for dipolar materials, which have a double-porosity structure by considering a fractional order Duhamel–Neumann stress–strain relation. The heat conduction is described by Cattaneo’s equations. The results are the constitutive equations of the linear theory of thermoelasticity with fractional order strain. The equations are valid for anisotropic materials and are called the Duhamel–Neumann equations with fractional order. Finally, the isotropic case is considered under the conditions of plane strain in order to perform some numerical simulations for samples of porous copper.

Appendix

Proof of Lemma 1

Proof

The first equation from (1) is multiplied by ${\dot{u}}_i$ and is integrated over $\Omega$ to obtain

$$\begin{aligned} \int _{\Omega }\rho _0{\dot{u}}_i\ddot{u}_i \, \mathrm {d}V = \int _{\Omega } (t_{ji}+\eta _{ji})_{,j}{\dot{u}}_i \, \mathrm {d}V+\int _{\Omega }\rho _0 F_i{\dot{u}}_i \, \mathrm {d}V. \end{aligned}$$

(53)

The second equation from (1) is multiplied by ${\dot{\phi }}_{jk}$. Then, we sum up over j and k and integrate over $\Omega$ to obtain

$$\begin{aligned} \int _{\Omega } I_{ks}\ddot{\phi }_{js}\dot{\phi }_{jk} \, \mathrm {d}V = \int _{\Omega }(\mu _{ijk,i}\dot{\phi }_{jk}+\eta _{jk}\dot{\phi }_{jk}+\rho _0 M_{jk}\dot{\phi }_{jk}) \, \mathrm {d}V. \end{aligned}$$

(54)

The first equation from (2) is multiplied by $\dot{\varphi }$ and is integrated over $\Omega$ to obtain

$$\begin{aligned} \int _{\Omega }\kappa _1\ddot{\varphi }\dot{\varphi } \, \mathrm {d}V = \int _{\Omega }(\sigma _{i,i}\dot{\varphi }+\xi \dot{\varphi }+\rho _0 G\dot{\varphi } ) \, \mathrm {d}V. \end{aligned}$$

(55)

The second equation from (2) is multiplied by $\dot{\psi }$ and is integrated over $\Omega$ to obtain

$$\begin{aligned} \int _{\Omega }\kappa _2\ddot{\psi }\dot{\psi } \, \mathrm {d}V = \int _{\Omega }(\tau _{i,i}\dot{\psi }+\zeta \dot{\psi }+\rho _0 L\dot{\psi } ) \, \mathrm {d}V. \end{aligned}$$

(56)

We substitute relations (53), (54), (55) and (56) into the principle of conservation of energy (11) to obtain

$$\begin{aligned} \begin{aligned}&\int _{\Omega } (t_{ji}+\eta _{ji})_{,j}{\dot{u}}_i \, \mathrm {d}V + \int _{\Omega }\rho _0 F_i{\dot{u}}_i \, \mathrm {d}V \\&+ \int _{\Omega }( \mu _{ijk,i}\dot{\phi }_{jk}+\eta _{jk}\dot{\phi }_{jk}+\rho _0 M_{jk}{\dot{\phi }}_{jk}) \, \mathrm {d}V \\&+ \int _{\Omega }(\sigma _{i,i}\dot{\varphi }+\xi \dot{\varphi }+\rho _0 G\dot{\varphi })\, \mathrm {d}V+ \int _{\Omega }(\tau _{i,i}\dot{\psi }+\zeta \dot{\psi }+\rho _0 L\dot{\psi })\, \mathrm {d}V\\&+ \int _{\Omega }\rho _0{\dot{e}}\, \mathrm {d}V= \int _{\Omega }(\rho _0 F_i{\dot{u}}_i+\rho _0 M_{jk}\dot{\phi }_{jk}+\rho _0 G\dot{\varphi } + \rho _0 L\dot{\psi }+S) \, \mathrm {d}V \\&+\int _{\partial \Omega }( t_i{\dot{u}}_i+\mu _{jk}\dot{\phi }_{jk}+\sigma \dot{\varphi } +\tau \dot{\psi }+q)\, \mathrm {d}A \end{aligned} \end{aligned}$$

(57)

By doing integration by parts

$$\begin{aligned} \int _{\Omega }(t_{ji}+\eta _{ji})_{,j}{\dot{u}}_i \, \mathrm {d}V = &\,\,{} \int _{\partial \Omega } t_i{\dot{u}}_i \, \mathrm {d}A- \int _{\Omega }(t_{ji}+\eta _{ji}){\dot{u}}_{i,j}\, \mathrm {d}V \end{aligned}$$

(58)

$$\begin{aligned} \int _{\Omega }\mu _{ijk,i}{\dot{\phi }}_{jk}\, \mathrm {d}V = &\,\,{} \int _{\partial \Omega }\mu _{jk}\dot{\phi }_{jk} \, \mathrm {d}A- \int _{\Omega }\mu _{ijk}\dot{\phi }_{jk,i} \, \mathrm {d}V \end{aligned}$$

(59)

$$\begin{aligned} \int _{\Omega } \sigma _{i,i}\dot{\varphi } \, \mathrm {d}V = &\,\,{} \int _{\partial \Omega }\sigma \dot{\varphi } \, \mathrm {d}A- \int _{\Omega }\sigma _i\dot{\varphi }_{,i} \, \mathrm {d}V \end{aligned}$$

(60)

$$\begin{aligned} \int _{\Omega }\tau _{i,i}\dot{\psi }\, \mathrm {d}V = &\,\,{} \int _{\partial \Omega }\tau \dot{\psi } \, \mathrm {d}A- \int _{\Omega } \tau _i\dot{\psi }_{,i} \, \mathrm {d}V \end{aligned}$$

(61)

and by substituting (58), (59), (60) and (61) into (57), we obtain

$$\begin{aligned} \begin{aligned}&-\int _{\Omega }(t_{ji}+\eta _{ji}){\dot{u}}_{i,j} \, \mathrm {d}V-\int _{\Omega }\mu _{ijk}\dot{\phi }_{jk,i}\, \mathrm {d}V+ \int _{\Omega }\eta _{jk}\dot{\phi }_{jk}\, \mathrm {d}V \\&-\int _{\Omega }\sigma _i\dot{\varphi }_{,i}\, \mathrm {d}V+ \int _{\Omega }\xi \dot{\varphi }\, \mathrm {d}V-\int _{\Omega }\tau _i\dot{\psi }_{,i}\, \mathrm {d}V+ \int _{\Omega }\zeta \dot{\psi }\, \mathrm {d}V \\&+\int _{\Omega }\rho _0{\dot{e}} \, \mathrm {d}V= \int _{\Omega } S\, \mathrm {d}V+\int _{\partial \Omega }q \, \mathrm {d}A \end{aligned} \end{aligned}$$

(62)

which can be rewritten as

$$\begin{aligned} \begin{aligned}&\int _{\Omega } \left[\rho _0{\dot{e}}- (t_{ji}+\eta _{ji}){\dot{u}}_{i,j}-\mu _{ijk}\dot{\phi }_{jk,i}+\eta _{jk}{\dot{\phi }}_{jk}-\sigma _i \dot{\varphi }_{,i}+\xi \dot{\varphi }\right.\\& \quad \left.- \tau _i\dot{\psi }_{,i}+\zeta \dot{\psi }-q_{i,i}-\rho _0Q\right] \, \mathrm {d}V=0 \end{aligned} \end{aligned}$$

(63)

or in the pointwise form

$$\begin{aligned} \begin{aligned}&\rho _0{\dot{e}}= (t_{ji}+\eta _{ji}){\dot{u}}_{i,j}+\mu _{ijk}\dot{\phi }_{jk,i}-\eta _{jk}\dot{\phi }_{jk}+\sigma _i\dot{\varphi }_{,i}-\xi \dot{\varphi } \\& \qquad +\tau _i\dot{\psi }_{,i}-\zeta \dot{\psi }+q_{i,i}+\rho _0 Q \end{aligned} \end{aligned}$$

(64)

We obtain from Eqs. (64) and (14)

$$\begin{aligned} \begin{aligned}&\rho _0{\dot{\phi }}= (t_{ji}+\eta _{ji}){\dot{u}}_{i,j}+\mu _{ijk}\dot{\phi }_{jk,i}- \eta _{jk}\dot{\phi }_{jk}+\sigma _i\dot{\varphi }_{,i}-\xi \dot{\varphi } \\& \qquad +\tau _i\dot{\psi }_{,i}-\zeta \dot{\psi }+q_{i,i}+\rho _0 Q-\rho _0\eta {\dot{T}}- \rho _0 T\dot{\eta } \end{aligned} \end{aligned}$$

(65)

$\square$

Proof of Theorem 1

Proof

We obtain from the first relation in (15) and the chain rule

$$\begin{aligned} \begin{aligned}&\rho _0\dot{\phi }=\rho _0 \frac{\partial \phi }{\partial \tilde{\varepsilon }_{ij}}\tilde{\dot{\varepsilon }}_{ij}+ \rho _0 \frac{\partial \phi }{\partial \kappa _{ij}} \dot{\kappa }_{ij}+\rho _0\frac{\partial \phi }{\partial \chi _{ijk}}\dot{\chi }_{ijk}+ \rho _0\frac{\partial \phi }{\partial \varphi }\dot{\varphi }+\rho _0 \frac{\partial \phi }{\partial \varphi _{,i}}\dot{\varphi }_{,i} \\& \qquad + \rho _0\frac{\partial \phi }{\partial \psi }{\dot{\psi }}+\rho _0 \frac{\partial \phi }{\partial \psi _{,i}}{\dot{\psi }}_{,i}+ \rho _0\frac{\partial \phi }{\partial T}{\dot{T}}+\rho _0\frac{\partial \phi }{\partial T_{,i} }{\dot{T}}_{,i} \end{aligned} \end{aligned}$$

(66)

By comparing formulae (19) $\rho _0\dot{\phi }=t_{ij}\dot{\tilde{\varepsilon }}_{ij}+\eta _{ij}\dot{\kappa }_{ij}+\mu _{ijk}\dot{\chi }_{ijk}+ \sigma _i\dot{\varphi }_{,i}-\xi \dot{\varphi }+\tau _i\dot{\psi }_{,i}-\zeta \dot{\psi }+q_{i,i}+\rho _0Q- \rho _0\eta {\dot{T}}-\rho _0 T\dot{\eta }$ and (66), we obtain

$$\begin{aligned} \begin{aligned}&t_{ij}=\rho _0\frac{\partial \phi }{\partial {\tilde{\varepsilon }}_{ij}}\quad \eta _{ij}= \rho _0 \frac{\partial \phi }{\partial \kappa _{ij}} \quad \mu _{ijk}=\rho _0 \frac{\partial \phi }{\partial \chi _{ijk}} \\&\xi =-\rho _0 \frac{\partial \phi }{\partial \varphi } \quad \zeta =- \rho _0 \frac{\partial \phi }{\partial \psi } \\&\sigma _i=\rho _0 \frac{\partial \phi }{\partial \varphi _{,i}} \quad \tau _i=\rho _0 \frac{\partial \phi }{\partial \psi _{,i}} \\&\eta =-\frac{\partial \phi }{\partial T} \\&\frac{\partial \phi }{\partial T_{,i}}=0 \end{aligned} \end{aligned}$$

(67)

and

$$\begin{aligned} q_{i,i}+\rho _0 Q-\rho _0 T\dot{\eta } =0. \end{aligned}$$

(68)

The proof is complete if we compute the derivatives above by formula (17). $\square$

Proof of Theorem 2

Proof

Using the equality for $\eta$ from (67) and (68), we obtain

$$\begin{aligned} \begin{aligned}&-q_{i,i} = \rho _0 Q+\rho _0 T \left( \frac{\partial ^2 \phi }{\partial \tilde{\varepsilon }_{ij}\partial T} \dot{\tilde{\varepsilon }}_{ij}+ \frac{\partial ^2\phi }{\partial \kappa _{ij}\partial T }\dot{\kappa }_{ij}+ \frac{\partial ^2 \phi }{\partial \chi _{ijk}\partial T}\dot{\chi }_{ijk}\right. \\& \qquad \quad +\left. \frac{\partial ^2 \phi }{\partial \varphi \partial T}\dot{\varphi }+ \frac{\partial ^2\phi }{\partial \varphi _{,i}\partial T}\dot{\varphi }_{,i}+ \frac{\partial ^2 \phi }{\partial \psi \partial T}\dot{\psi }+\frac{\partial ^2 \phi }{\partial \psi _{,i}\partial T}{\dot{\psi }}_{,i}+ \frac{\partial ^2\phi }{\partial T^2}{\dot{T}} \right) \end{aligned} \end{aligned}$$

(69)

By using Eqs. (69) and (17), we obtain the gradient of the heat flux in the form

$$\begin{aligned} - q_{i,i}=\rho _0 Q-T \alpha _{ij}\dot{\tilde{\varepsilon }}_{ij} -T\beta _{ij}\dot{\kappa }_{ij}-T\gamma _{ijk}\dot{\chi }_{ijk}-T\gamma _1\dot{\varphi }-T\gamma _2\dot{\psi } -aT{\dot{T}} \end{aligned}$$

(70)

Let $T\approx T_0$ for linearity, where $T_0$ is the constant absolute temperature of the body in its reference state. Therefore, we get

$$\begin{aligned} \begin{aligned}&-q_{i,i}=\rho _0 Q- T_0 \alpha _{ij} \dot{\tilde{\varepsilon }}_{ij} -T_0\beta _{ij}\dot{\kappa }_{ij}-T_0 \gamma _{ijk}\dot{\chi }_{ijk} \\& \qquad \quad -T_0\gamma _1\dot{\varphi }- T_0\gamma _2\dot{\psi } -a T_0{\dot{T}} \end{aligned} \end{aligned}$$

(71)

The non-Fourier heat equations give

$$\begin{aligned} \begin{aligned}&(K_{ij} T_{,j})_{,i}=-q_{i,i}-\tau _0{\dot{q}}_{i,i}\\& \qquad \quad \,\,=\rho _0 Q-T_0 \alpha _{ij} \dot{\tilde{\varepsilon }}_{ij}-T_0 \beta _{ij}\dot{\kappa }_{ij}-T_0\gamma _{ijk}\dot{\chi }_{ijk}- T_0\gamma _1\dot{\varphi } \\& \qquad \qquad -T_0\gamma _2\dot{\psi }-a T_0{\dot{T}} +\tau _0\left( \rho _0\dot{Q}-T_0 \alpha _{ij}\ddot{\tilde{\varepsilon }}_{ij}- T_0\beta _{ij}\ddot{\kappa }_{ij} \right. \\& \qquad \qquad \left. -T_0\gamma _{ijk}\ddot{\chi }_{ijk}-T_0\gamma _1\ddot{\varphi }-T_0\gamma _2\ddot{\psi }-aT_0\ddot{T} \right) \end{aligned} \end{aligned}$$

(72)

$\square$

Closed-form solution in the isotropic case

We consider the problem of plane strain biaxial deformation of a rectangular specimen as in [37] for isotropic dipolar elasticity with double porosity in the stationary case. We consider that $\tau =0$, $\theta =0$ and we assume that $u_1(x_1)=c_1 x_1$, $u_2(x_2)=c_2 x_2$, $\phi _{11}=c_3$, $\phi _{22}=c_4$, $\phi _{12}=\phi _{21}=0$, $\varphi =c_5$ and $\psi =c_6$, where $c_i$ are constants. These satisfy the displacement boundary conditions $u_1(x_1=0)=0$ and $u_2(x_2=0)=0$. Following [37], we show in the sequel that they also satisfy the governing equations and the remaining boundary conditions, so that they must be the unique solution of the problem.

By substituting the expressions above into the equations from Theorem 4, we are only left with Eqs. (46), (49), (51) and (52), which become

$$\begin{aligned} c_1(b_2+b_3+2 g_2)-c_3(b_2+b_3) = &\,\,{} 0 \end{aligned}$$

(73)

$$\begin{aligned} c_2(b_2+b_3+2 g_2)-c_4(b_2+b_3) = &\,\,{} 0 \end{aligned}$$

(74)

$$\begin{aligned} d_1(c_1+c_2)+\alpha _1 c_5+\alpha _3 c_6 = &\,\,{} 0 \end{aligned}$$

(75)

$$\begin{aligned} d_2(c_1+c_2)+\alpha _3 c_5+\alpha _2 c_6 = &\,\,{} 0 \end{aligned}$$

(76)

The boundary conditions are $t_{22}+\eta _{22}=\tilde{t}_2$ at the top and $t_{11}+\eta _{11}=0$ at the sides. Hence, we obtain by virtue of the expressions for $t_{ij}$ and $\eta _{ij}$

$$\begin{aligned}&c_1\lambda +c_2(\lambda +2\mu +4 g_2+b_2+b_3)-c_4(2g_2+b_2+b_3)+c_5d_1+c_6 d_2=\tilde{t}_2 \end{aligned}$$

(77)

$$\begin{aligned}&c_1(\lambda +2\mu +4g_2+b_2+b_3)+c_2\lambda -c_3(2g_2+b_2+b_3)+c_5d_1+c_6d_2=0 \end{aligned}$$

(78)

Then, we solve the above linear system of six equations for $c_i, i=1,6$, and we substitute into the definitions of the functions, which yields

$$\begin{aligned}&\begin{aligned}&\Delta _1 = - 2\alpha _3(b_2+b_3)d_1d_2+\alpha _1(b_2+b_3)d_2^2 \\& \qquad +\alpha _3^2(-2g_2^2+(b_2+b_3)(\lambda +\mu )) \\& \qquad +\alpha _2(2\alpha _1 g_2^2 +b_2 (d_1^2-\alpha _1(\lambda +\mu ))+b_3(d_1^2-\alpha _1(\lambda +\mu ))) \end{aligned} \end{aligned}$$

(79)

$$\Delta = 4(2g_2^2-(b_2+b_3)\mu )\Delta _1 $$

(80)

$$\begin{aligned}&u_1 = \frac{1}{\Delta }(b_2+b_3)^2(-2\alpha _3d_1d_2+\alpha _1d_2^2+\alpha _3^2\lambda +\alpha _2(d_1^2-\alpha _1\lambda ))\tilde{t}_2 x_1 \end{aligned}$$

(81)

$$\begin{aligned}&\begin{aligned}&u_2 = - \frac{1}{\Delta }(b_2+b_3)(-2\alpha _3(b_2+b_3)d_1d_2+\alpha _1(b_2+b_3)d_2^2\\& \qquad +\alpha _3^2(-4g_2^2+(b_2+b_3)(\lambda +2\mu )) \\& \qquad +\alpha _2(4\alpha _1 g_2^2+(d_1^2-\alpha _1 (\lambda +2\mu ))(b_2+b_3)\tilde{t}_2 x_2 \end{aligned} \end{aligned}$$

(82)

$$\begin{aligned}&\phi _{11}=\frac{1}{\Delta }(b_2+b_3)(b_2+b_3+2 g_2) (-2\alpha _3 d_1 d_2+\alpha _1 d_2^2+ \alpha _3^2\lambda +\alpha _2(d_1^2-\alpha _1\lambda ))\tilde{t}_2 \end{aligned}$$

(83)

$$\begin{aligned}&\begin{aligned}&\phi _{22}=-\frac{1}{\Delta }(b_2+b_3+2 g_2)(-2\alpha _3(b_2+b_3)d_1 d_2+\alpha _1(b_2+b_3)d_2^2\\&\qquad \,\,+ \alpha _3^2(-4g_2^2+(b_2+b_3)(\lambda +2\mu ))\\& \qquad \,\, +\alpha _2(4\alpha _1 g_2^2+ (d_1^2-\alpha _1(\lambda +2\mu ))(b_2+b_3)))\tilde{t}_2 \end{aligned} \end{aligned}$$

(84)

$$\begin{aligned}&\varphi =\frac{1}{2\Delta _1}(b_2+b_3)(\alpha _2 d_1-\alpha _3 d_2) \tilde{t}_2 \end{aligned}$$

(85)

$$\begin{aligned}&\psi =-\frac{1}{2\Delta _1}(b_2+b_3)(\alpha _3 d_1-\alpha _1 d_2)\tilde{t}_2. \end{aligned}$$

(86)