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\begin{center}
 SOME PROBLEMS OF ELASTICITY FOR THE
 TWO-COMPONENT  ISOTROPIC  MIXTURE

\vspace*{0.3cm}
{\it G. Khatiashvili}

\vspace*{0.3cm}
{\it N.I. Muskhelishvili Institute of Computational Mathematics\\
                                 of Georgian Academy of Sciences}
\end{center}

\vspace*{0.3cm}
\par
	The linearized theory of elasticity for the indicated medium was given by 
T.R. Steel [1]. The two-dimensional problems for the isotropic mixture are 
considered by T.R. Steel [2] and M.O. Basheleishvili [3]. Some three-dimensional basic 
problems for indicated medium are considered by D.G. Natroshvili, A.J. Jagmaidze 
and M.J.Svanadze [4].
\par We consider some problems of Saint-Venant's tyipe and antiplane deformation 
for the medium bounded by cylindrical surface.
\par Let us consider the rectangular system of Cartesian coordinates $0x_1x_2x_3$                      
and the body, made from the mixture, occupying the domain $\Omega$,
bounded by cylindrical surface $f(x_1,x_2)=0$ and  planes $x_3=0$ and                  
$x_3=l$ ($l > 0$).
\par	 It is asssumed that the components of  displacements, stress and strains
$u_j^{(\gamma)}$ $\tau_{ij}^{(\gamma)}$ and $e_{ij}^{(\gamma)}$ of the two different substances (with number 1          
and 2) constituent the mixture, obey the Steel's law [1] 
%(1)
\begin{equation}
\begin{array}{l}
\tau_{ij}^{(\gamma)}=[(-1)^\gamma \alpha_2+\lambda_{3\gamma-2}divu^{(1)}+
\lambda_{4-\gamma}divu^{(2)}]\delta_{jk}-\\
-2\mu_{2\gamma-1}e^{(1)}_{jk}+2\mu_{4-\gamma}e^{(2)}_{jk}+(-1)^\gamma
2\lambda_5h_{jk}\;\;(j,k=1,2,3;\;\;\gamma=1,2),
\end{array}
\end{equation}
where $\delta_{ij}$ is Kroneker's symbol; $\alpha_2$, $\lambda_m$ and $\mu_k$ are the constants of  
elasticity ($\alpha_2=$ $=\lambda_3-\lambda_4$) and the values $e_{jk}^{(\gamma)}$
and $h_{jk}$ are given by the equalities
%(2)
\begin{equation}
2e_{jk}^{(\gamma)}=D_ju_k^{(\gamma)}+D_ku_j^{(\gamma)},\;\;
2h_{jk}=D_j(u_k^{(1)}+u_k^{(2)})-D_k(u_j^{(1)}+u_j^{(2)}),
\end{equation}
where $D_j=\partial/\partial x_j$.
\par In [4] the inequalities 
%(3)
\begin{equation}
\begin{array}{l}
\displaystyle
\lambda_5<5;\;\;\mu_1,\;\mu_2>0,\;g=\mu_1\mu_2-\mu_3^2>0,\;
\lambda_1-\frac{\alpha_2}{\rho}\rho^{(2)}+\frac{2}{3}\mu_1>0,\\
\displaystyle
G=\left(\lambda_2+\frac{2}{3}\mu_2+\frac{\alpha_2}{\rho}\rho^{(2)}\right)
\left(\lambda_1+\frac{2}{3}\mu_1-\frac{\alpha_2}{\rho}\rho^{(1)}\right)>0,
\end{array}
\end{equation}
are established,
where $\rho^{(\gamma)}$ are the densities of substances with number 1 and 2 ($\gamma=1,2$).
\par It is assumed that $\gamma$ henceworth always will receive only meanings 1 and 2.      
\par The equations of the elastic equilibrium of the isotropic mixture may be
represented in the form [1,4,2]
%(4)
\begin{equation}
\Delta(a_\gamma u^{(\gamma)}+Cu^{(3-\gamma)})+graddiv(bu^{(\gamma)}+du^{(3-\gamma)})=
-\rho^{(\gamma)}\Phi_\star^{(\gamma)},
\end{equation}
where $u^{(\gamma)}(u^{(\gamma)}_1,\;u^{(\gamma)}_2,\;u^{(\gamma)}_3)$ are vectors of the displacements,
$\Phi_\star^{(\gamma)}(\Phi_1^{(\gamma)},\;\Phi_2^{(\gamma)}\;\Phi_3^{(\gamma)})$
are vectors of the mass-forces and constants $a_\gamma,\;b_\gamma,\;c$ and $d$ are given  
by the equalities  
%(5)
\begin{equation}
\begin{array}{l}
\displaystyle
a_\gamma=\mu_\gamma-\lambda_5,\;\;b_\gamma=\mu_\gamma+\lambda_\gamma+\lambda_5+
(-1)^\gamma\frac{\alpha_2}{\rho}\rho^{(3-\gamma)},\\
\displaystyle
C=\mu_3+\lambda_5,\;\;d=\mu_3+\lambda_3-\lambda_5-
\frac{\alpha_2}{\rho}\rho^{(1)}=\mu_\gamma+\lambda_4-\lambda_5+
\frac{\alpha_2}{\rho}\rho^{(2)}.
\end{array}
\end{equation}
\par The components of the stress $\tau_{ij}^{(\gamma)}$.  determining the elastic equilibrium of the 
cylindrical beam due the exstension by  longitudonal force
$\tau_{33}^{(\gamma)}=\tau^{(\gamma)}$ (parallel to
$0x_3$ axis) are given on the form
$$
\begin{array}{l}
\tau_{jj}^{(\gamma)}=(-1)^\gamma\alpha_2+\lambda_{3\gamma-2}B_1+
\lambda_{4-\gamma}B_2+2\mu_{2\gamma-1}A_j^{(1)}
+2\mu_{4-\gamma}A_j^{(2)},\\
\\
\tau_{\beta k}^{(\gamma)}=0\;(\beta\ne k),\;(j,\beta,k=1,2,3),
\end{array}
$$
where
$$
\begin{array}{l}
\displaystyle
B_\gamma=\frac{1}{3G}\left[a^*_\gamma\left(\lambda_{3-\gamma}
+\frac{2}{3}\mu_{3-\gamma}-(-1)^\gamma\frac{\alpha_2}{\rho}\rho^{(\gamma)}\right)\right.-\\
\\\displaystyle
\left.
-a^*_{3-\gamma}\left(\lambda_{5-2\gamma}
+\frac{2}{3}\mu_{5-2\gamma}-\frac{\alpha_2}{\rho}\rho^{(\gamma)}\right)\right],\;\;
a^*_\gamma=\frac{1}{\omega_0}\tau^{(\gamma)},
\end{array}
$$
$\omega_0$ is the area of crossection of the beam  and the constants $A_j^{(\gamma)}$
are expressed by constants of elasticity and $B_k$, which easily will be easily obtained from the end 
conditions of the beam.
\par The components of  the stress in the problem of torsion due to torque      
(couple forces)  will be given in the form
$$
\tau_{13}^{(\gamma)}=(-1)^\gamma\tau_{\star}(\mu_3-\mu_\gamma)(D_1f_\star-x_2),\;
\tau_{23}^{(\gamma)}=(-1)^\gamma\tau_{\star}(\mu_3-\mu_\gamma)(D_2f_\star+x_1),
$$
where the function $f_\star(x_1,x_2)$ is the solution of the following boundary-value 
problem: $D_1^2f_\star+D_2^2f_\star=0$ in the domain $\omega$ of the crossection of beam,
$D_n^2f_\star\equiv$ $\equiv n_1D_1f_\star+n_2D_2f_\star=x_2n_1-x_1n_2$ in each point of the 
closed curve $S$ of the boundary of domain $\omega$ where $n(n_1,n_2)$ is the unit 
vector of the outward normal $S$ 
and the constants $\tau_\star$ is determined from the end conditions.
\par Other components of stress are equal to zero.
\par Now we consider the antiplane deformation of the medium,  made from 
the isotropic mixture, when the components of the displacements $u_1^{(\gamma)}$
 and $u_2^{(\gamma)}$ are  equal 
to zero and $u_3^{(\gamma)}=F^{(\gamma)}(x_1,x_2)$, where the functions 
$F^{(\gamma)}$ will be determined in the particular problem.
\par Introduce for the functions $T^{(\gamma)}(x_1,x_2)$ and
$V^{(\gamma)}(x_1,x_2)$ the linear algebraic operators
%(6)
\begin{equation}
\begin{array}{l}
M_\gamma T^{(1,2)}\equiv(\mu_{4-\gamma}+(-1)^\gamma\lambda_5)T^{(2)}+
(\mu_{2\gamma-1}+(-1)^\gamma\lambda_5)T^{(1)},\\
\\
N_\gamma V^{(1,2)}\equiv(-1)^\gamma g^{-1}[(\mu_{5-2\gamma}-\lambda_5)V^{(2)}-
(\mu_{\gamma+1}+\lambda_5)V^{(1)}].
\end{array}
\end{equation}
%(7)
\begin{equation}
\begin{array}{l}
{\rm If}\;\; T^{(\gamma)}=N_\gamma V^{(1,2)},\;\; {\rm then}\;\;
M_jT^{(1,2)}=V^{(j)}\;\; (j=1,2)               .
\end{array}
\end{equation}
\par Conforming stress to displacements $u_3^{(\gamma)}=F^{(\gamma)}(x_1,x_2)$ will be given in 
the form
%(8)
\begin{equation}
\begin{array}{l}
\tau_{jj}^{(\gamma)}=(-1)^\gamma\alpha_2,\;\;\tau_{m3}^{(\gamma)}=M_\gamma(D_mF^{(1,2)}),\\
\tau_{12}^{(\gamma)}=\tau_{21}^{(\gamma)}=0,\;\;(m=1,2;\;j=1,2,3).
\end{array}
\end{equation}
\par Consider the medium  (finite or infinite), bounded by  cylindrical  surface
$f(x_1,x_2)=0$ and  propose  that  in  its each point external forces 
are acting uniformly (along 
the axis) $\tau^{(\gamma)}(s)$, where $s$ is the arc of the
boundary $S$ of  the domain $\omega$,
 of the crossection of cylindrical surface.
\par In this case in the medium  the displacements $u_3^{(\gamma)}=F^{(\gamma)}(x_1,x_2)$ 
must satisfay the  equations $D_1^2F^{(\gamma)}+D_2^2F^{(\gamma)}=0$
in the domain $\omega$ and   the boundary conditions  $D_nF^{(\gamma)}=N_\gamma\tau^{(1,\epsilon)}$
on the $S$. Then, the indicated       
displacements $u_j^{(\gamma)}$ will satisfy    the   equations   (4)   
and    corresponding  components  of  the  stress  on  the  cylindrical  surface  will  satisfy  the  
boundary conditions 
%(9)
\begin{equation}
\tau_{jn}^{(\gamma)}=\tau_{1j}^{(\gamma)}n_1+\tau_{2j}^{(\gamma)}n_2=
\frac{1}{2}(j-1)(j-2)\tau^{(\gamma)}\;\;(j=1,2,3).
\end{equation}
\par Let us consider some particular cases exercises
\par 1. We consider the infinite elliptic cylinder, the crossection of which by plane
$0x_1x_2$ is the elliptic domain, bounded by the ellipse
%(10)
\begin{equation}
x_1=acos\theta,\;\;x_2=bsin\theta\;\;(0\le\theta\le 2\pi;\;a>b,\;a^2-b^2=c^2).
\end{equation}
\par The displacements $u_3^{(\gamma)}$,  determining the elastic state will be represented 
as the series with respect to Faber's polynomials
$$
u_3^{(\gamma)}=\frac{1}{\pi}Re{\mathop\sum\limits^\infty_{k=1}}\frac{1}{k}
\frac{t_1^k+t_2^k}{1-\lambda^{2k}}\int\limits_0^{2\pi}\varphi_k(\theta)N_\gamma
\tau^{(1,2)}d\theta,\;\;(\lambda=\frac{a-b}{a+b}<1),
$$
where
$$
\begin{array}{l}
t_1=z+\sqrt{z^2-c^2},\;\;t_2=z-\sqrt{z^2-c^2},\;\;z=x_1+ix_2,\;\;(i^2=-1),\\
\\
\varphi_k(\theta)=(a^2sin^2\theta+b^2cos^2\theta)^{1/2}[(1+\lambda^k)cosk\theta-
i(1-\lambda^k)sink\theta].
\end{array}
$$
\par 2. Now we consider the problem of the antiplane deformation of the infinite 
medium with the 
elliptic hole (10), in each point of which the exterior forces (9) are acting 
uniformly (along the axis $0x_3$). 
\par Coresponding displacements, determining the solution of this problem  will 
be given in the form
$$
\begin{array}{c}
\displaystyle
u_3^{(\gamma)}=\frac{1}{\pi}Re\left[{\mathop\sum\limits^\infty_{k=1}}(kt_1^k)^{-1}
\int\limits_0^{2\pi}(a^2sin^2\theta+b^2cos^2\theta)^{1/2}e^{ik\theta}N_\gamma\tau^{(1,2)}
d\theta+\right.\\
\\ \displaystyle+
\left.\left(\int\limits_sN_\gamma\tau^{(1,2)}ds\right)lnt_1\right].
\end{array}
$$
\par 3. Now we consider the case,when along the axis $0x_3$ of the infinite 
medium of the space $0x_1x_2x_3$ are acting the concentrated forces $P^{(\gamma)}=const$, 
parallel to axis $0x_3$ and concentrated couple forces by moments 
$m_1^{(\gamma)}$ and $m_2^{(\gamma)}$ where
the vectors $\vec{m}_1^{(\gamma)}$ and $\vec{m}_2^{(\gamma)}$ are parallel to
$0x_1$ and $0x_2$ respectively.
\par The displacements coresponding to indicated loading may be represented 
in the form ($m_j^{(\gamma)}=const$)
$$
\displaystyle
(u_3^{(\gamma)})_p=\frac{1}{2\pi}(N_\gamma P^{(1,2)})lnr,\;\;(x_1^2+x_2^2=r^2),\;\;
(u_3^{(\gamma)})_m=\frac{(-1)^\gamma}{\pi r^2g}\left[(\mu_{(5-2\gamma)}-\right.
$$
$$\left.
-\lambda_5)
(m_1^{(2)}x_2-m_2^{(2)}x_1)-(\mu_{\gamma+1}+\lambda_5)(m_1^{(1)}x_2-m_1^{(1)}x_1)
\right].
$$
\par 4. The stress state of the infinite medium with 
circular cylindrical hole $x_1^2+x_2^2=R^2$ is considered when in its each point 
jumping forces of the following kind are acting 
uniformly (along the $0x_3$ axis)                              
$$
\begin{array}{c}
(\tau^{(\gamma)})_1=\left\{
\begin{array}{l}
\displaystyle
p_1^{(\gamma)},\;\;0\le\theta\le\pi ;\\\displaystyle
-p_1^{(\gamma)},\;\;\pi\le\theta\le 2\pi .
\end{array}\right.
(\tau^{(\gamma)})_2=\left\{
\begin{array}{l}
\displaystyle
-p_2^{(\gamma)},\;\;-\frac{\pi}{2}\le\theta\le\frac{\pi}{2};\\\displaystyle
p_2^{(\gamma)},\;\;\frac{\pi}{2}\le\theta\le \frac{3}{2}\pi .
\end{array}\right.\\
(p_j^{(\gamma)}\hspace{-0.1cm}=const).
\end{array}
$$
\par We represent  the corresponding displacements in the form $u_3^{(\gamma)}=ReH_\gamma(z)$,
where $H_\gamma(z)$ are  the analytical functions of the complex variable $z=x_1+ix_2$
in the exterior (with respect to circular hole) domain $\omega$. If the $H'_\gamma(z)$
vanish at the infinity for the indicated jumping forces, the introduced complex
potential obtains the following form
$$
\displaystyle
iz[H_\gamma(z)]_j=\frac{2R}{\pi}(N_\gamma P_j^{(1,2)})ln\frac{z+z_0^{(j)}}{z-z_0^{(j)}},\;\;
(j=1,2;\;z_0^{(1)}=R,\;z_0^{(2)}=Ri)
$$
\par From these expressions we may obtain the  values (meanings) of the complex 
potentials for the hydrodynamic dipole  of the plane, incompressible , stationary 
and irrotational movement of  fluids.

\vspace*{0.4cm}
\footnotesize
\begin{center}
{\bf R e f e r e n c e s}

\vspace*{0.2cm}
\end{center}
\par
[1] Steel T.R. Applicatioins of a Theory of Interacting Continua. Quart.J.
       Mech. and Appl. Math., 1967, vol..XX, part. 1, pp. 57-72.
\par
[2] Steel T.R. Linarized Theory of Plane Strain of Mixture of Two Solids.Jnt.J.
Eng. Scienc., 1967, v.5, n.10, pp.775-790.
\par
[3] Basheleishvili M.O. Two-Dimensional Boundary Value Problems of Statics of
the Elastic Mixtures. Memories on Diff.Eq.and Math.Physics, Tbilisi, v.6, 1995,
pp.59-105.
\par
[4] Natroshvili D.G., Djagmaidze A.J., Svanadze M.J. Some Problems of the
Linear Theory of the Elastic Mixture. Publish. House of Tbilisi State
University, 1986, 210 p. (in Russian).
\end{document}