User:Tmhoang81/Instability analysis of composite materials

The general solution of the Navier's equations in the circular domain for a homogeneous body

Starting from the the Cauchy's equations:

$\displaystyle {\boldsymbol {\nabla }}\cdot {\boldsymbol {\sigma }}=\mathbf {0}$

(1)

we follow the procedures of Kochmann, , express the Cauchy' equations in the polar coordinates:

$\displaystyle (\lambda +\mu )(r^{2}u_{r,rr}+ru_{r,r}-u_{r}+ru_{\theta ,r\theta }-u_{\theta ,\theta })+\mu (r^{2}u_{r,rr}+ru_{r,r}-u_{r}+u_{r,\theta \theta }-2u_{\theta ,\theta })=0$

(2)

$\displaystyle (\lambda +\mu )(ru_{r,r\theta }+u_{r,\theta }+u_{\theta ,\theta \theta })+\mu (r^{2}u_{\theta ,rr}+ru_{\theta ,r}+2u_{r,\theta }+u_{\theta ,\theta \theta }-u_{\theta })=0$

(3)

and assume the separable form of the solutions

$\displaystyle u_{r}=f(r)e^{im\theta },\qquad u_{\theta }=g(r)e^{im\theta }$

(4)

For each value of $\displaystyle m$, we have a set of solutions $\displaystyle f(r)=f_{m}(r),g(r)=g_{m}(r)$. Since solutions $\displaystyle u_{r},\quad u_{\theta }$ are $\displaystyle 2\pi$ periodic of $\displaystyle \theta$, $\displaystyle m$ must be an integer and hence $\displaystyle m$ ranges from $\displaystyle -\infty$ to $\displaystyle \infty$. The general solutions would be:

$\displaystyle u_{r}=\sum _{m=-\infty }^{\infty }f_{m}(r)e^{im\theta },u_{\theta }=\sum _{m=-\infty }^{\infty }g_{m}(r)e^{im\theta }$

(5)

On the other hand, displacements must be real, i.e $\displaystyle {\overline {u}}_{r}=u_{r},\quad {\overline {u}}_{\theta }=u_{\theta }\quad$ or

$\displaystyle \sum _{m=-\infty }^{\infty }f_{m}(r)e^{im\theta }=\sum _{m=-\infty }^{\infty }{\overline {f_{m}(r)}}e^{-im\theta },\quad \sum _{m=-\infty }^{\infty }g_{m}(r)e^{im\theta }=\sum _{m=-\infty }^{\infty }{\overline {g_{m}(r)}}e^{-im\theta }$

(6)

It is easy to see that

$\displaystyle \sum _{m=-\infty }^{\infty }{\overline {f_{m}(r)}}e^{-im\theta }=\sum _{m=-\infty }^{\infty }{\overline {f_{-m}(r)}}e^{im\theta },\quad \sum _{m=-\infty }^{\infty }{\overline {g_{m}(r)}}e^{-im\theta }=\sum _{m=-\infty }^{\infty }{\overline {g_{-m}(r)}}e^{im\theta }$

(7)

Comparing equation ... and ... to get

$\displaystyle f_{m}(r)={\overline {f_{-m}(r)}}\quad {\text{and}}\quad g_{m}(r)={\overline {g_{-m}(r)}}$

(8)

However, we can go further to get simplier (in this context) representation of solutions by doing some operations. For example, from the above conditions, one can wrire:

$\displaystyle {\begin{aligned}u_{r}&=\sum _{m=-\infty }^{\infty }{f_{m}(r)}e^{im\theta }=f_{0}(r)+\sum _{m=1}^{\infty }f_{m}(r)e^{im\theta }+\sum _{m=-\infty }^{-1}f_{m}(r)e^{im\theta }=f_{0}(r)+\sum _{m=1}^{\infty }f_{m}(r)e^{im\theta }+\sum _{m=1}^{\infty }f_{-m}(r)e^{-im\theta }\\&=f_{0}(r)+\sum _{m=1}^{\infty }f_{m}(r)e^{im\theta }+\sum _{m=1}^{\infty }{\overline {f_{m}(r)}}e^{-im\theta }=f_{0}(r)+\sum _{m=1}^{\infty }f_{m}(r)e^{im\theta }+\sum _{m=1}^{\infty }{\overline {f_{m}(r)e^{im\theta }}}\\&=f_{0}(r)+\sum _{m=1}^{\infty }\left(f_{m}(r)e^{im\theta }+{\overline {f_{m}(r)e^{im\theta }}}\right)=f_{0}(r)+2\sum _{m=1}^{\infty }\mathbb {R} \left(f_{m}(r)e^{im\theta }\right)\equiv \mathbb {R} \sum _{m=0}^{\infty }F_{m}(r)e^{im\theta }\\\end{aligned}}$

(9)

We conclude by writing the general representation of solutions as follows:

$\displaystyle u_{r}=\mathbb {R} \sum _{m=0}^{\infty }f_{m}(r)e^{im\theta },\qquad u_{\theta }=\mathbb {R} \sum _{m=0}^{\infty }g_{m}(r)e^{im\theta }$

(10)

it means that m is a nonnegative integer. Now let find the solution of the Navier's equation. Subtitute ... to ...

$\displaystyle (\lambda +2\mu )\left(f''+{\frac {f'}{r}}\right)-\left({\frac {\lambda +2\mu +m^{2}\mu }{r^{2}}}\right)f=im\left[(\lambda +3\mu ){\frac {g}{r^{2}}}-(\lambda +\mu ){\frac {g'}{r}}\right]$

(11)

$\displaystyle \mu \left(g''+{\frac {g'}{r}}\right)-\left({\frac {(\lambda +2\mu )m^{2}+\mu }{r^{2}}}\right)g=-im\left[(\lambda +3\mu ){\frac {f}{r^{2}}}+(\lambda +\mu ){\frac {f'}{r}}\right]$

(12)

Let consider possible situations.

• $\displaystyle \lambda +2\mu =0$

$\displaystyle -{\frac {m^{2}\mu }{r^{2}}}f=im\mu \left[{\frac {g}{r^{2}}}+{\frac {g'}{r}}\right]$

(13)

$\displaystyle \mu \left(g''+{\frac {g'}{r}}\right)-{\frac {\mu }{r^{2}}}g=-im\mu \left[{\frac {f}{r^{2}}}-{\frac {f'}{r}}\right]$

(14)

A. $\displaystyle \mu =0\quad$

The Navier's equations auto-satisfied or any arbitratry functions $\displaystyle u_{r},u_{\theta }\quad$ can be solutions. In this case the trivial solutions $\displaystyle \mathbf {u} =0\quad$ is unstable.

B. $\displaystyle \mu \neq 0$

$\displaystyle -{\frac {m^{2}}{r^{2}}}f=im\left({\frac {g}{r^{2}}}+{\frac {g'}{r}}\right)$

(15)

$\displaystyle \left(g''+{\frac {g'}{r}}\right)-{\frac {g}{r^{2}}}=-im\left({\frac {f}{r^{2}}}-{\frac {f'}{r}}\right)$

(16)

It is obvious that the system, in this case, always admits nontrivial axisysmetric solutions (w.r.t $\displaystyle m=0$) $\displaystyle \quad u_{r}=f_{0}(r)$ arbitrary and hence the instability of trivial solution $\displaystyle \mathbf {u} =0\quad$ occurs.

• $\displaystyle \mu =0$

$\displaystyle \lambda \left(f''+{\frac {f'}{r}}\right)-{\frac {\lambda f}{r^{2}}}=im\lambda \left[{\frac {g}{r^{2}}}-{\frac {g'}{r}}\right]$

(17)

$\displaystyle -\lambda {\frac {m^{2}g}{r^{2}}}=-im\lambda \left[{\frac {f}{r^{2}}}+{\frac {f'}{r}}\right]$

(18)

A. $\displaystyle \lambda =0$

The same conclusion as 1.a above.

B. $\displaystyle \lambda \neq 0$

$\displaystyle \left(f''+{\frac {f'}{r}}\right)-{\frac {f}{r^{2}}}=im\left[{\frac {g}{r^{2}}}-{\frac {g'}{r}}\right]$

(19)

$\displaystyle {\frac {m^{2}g}{r^{2}}}=im\left[{\frac {f}{r^{2}}}+{\frac {f'}{r}}\right]$

(20)

Same conclusion as 1.b except that in this case nontrivial,arbitrary axisysmetric solutions (w.r.t $\quad m=0$) is $\quad u_{\theta }=g_{0}(r)$ arbitrary.

• $\displaystyle \lambda +2\mu \neq 0,\mu \neq 0$

$\displaystyle f''+{\frac {f'}{r}}-\left(1+{\frac {m^{2}\mu }{\lambda +2\mu }}\right){\frac {f}{r^{2}}}={\frac {im}{\lambda +2\mu }}\left[(\lambda +3\mu ){\frac {g}{r^{2}}}-(\lambda +\mu ){\frac {g'}{r}}\right]$

(21)

$\displaystyle g''+{\frac {g'}{r}}-\left(1+{\frac {(\lambda +2\mu )m^{2}}{\mu }}\right){\frac {g}{r^{2}}}=-{\frac {im}{\mu }}\left[(\lambda +3\mu ){\frac {f}{r^{2}}}+(\lambda +\mu ){\frac {f'}{r}}\right]$

(22)

A. $\displaystyle m=0$

$\displaystyle f_{0}(r)=A_{0}r+{\frac {A_{1}}{r}},\qquad g_{0}(r)=B_{0}r+{\frac {B_{1}}{r}}$

(23)

where $\displaystyle \quad A_{0},A_{1},B_{0}{\text{and}}B_{1}\quad$ are arbitrary real constants.

B. $\displaystyle m\geq 0\quad$

Assume $\displaystyle f(r)=Ar^{k},g(r)=Br^{k}$ are solutions of ... Plug them to

$\displaystyle (k^{2}-\alpha ^{2})A-{\frac {im}{\lambda +2\mu }}\left[(\lambda +3\mu )-(\lambda +\mu )k\right]B=0$

(24)

$\displaystyle (k^{2}-\beta ^{2})B+{\frac {im}{\mu }}\left[(\lambda +3\mu )+(\lambda +\mu )k\right]A=0$

(25)

where

$\displaystyle \alpha ^{2}=1+{\frac {m^{2}\mu }{\lambda +2\mu }},\qquad \beta ^{2}=1+{\frac {m^{2}(\lambda +2\mu )}{\mu }}$

(26)

Rewrite the system of equation in the matrix form

$\displaystyle {\begin{bmatrix}\displaystyle (k^{2}-\alpha ^{2})&\displaystyle -{\frac {im}{\lambda +2\mu }}\left[(\lambda +3\mu )-(\lambda +\mu )k\right]\\\displaystyle {\frac {im}{\mu }}\left[(\lambda +3\mu )+(\lambda +\mu )k\right]&(k^{2}-\beta ^{2})\end{bmatrix}}{\begin{Bmatrix}A\\B\end{Bmatrix}}={\begin{Bmatrix}0\\0\end{Bmatrix}}$

(27)

and realize that the system has non-trivial solutions if the determinant equal zero or

$\displaystyle (k^{2}-\alpha ^{2})(k^{2}-\beta ^{2})+{\frac {(im)^{2}}{\mu (\lambda +2\mu )}}\left[(\lambda +3\mu )-(\lambda +\mu )k\right]\left[(\lambda +3\mu )+(\lambda +\mu )k\right]=0$

(28)

Solving the above equation to obtain

$\displaystyle k^{4}-2(m^{2}+1)k^{2}+(m^{2}-1)^{2}=0\to k=\pm m\pm 1$

(29)

Recognize that when $\displaystyle m=1\quad$ then $\displaystyle k_{1}=2,k_{2}=k_{3}=0,k_{4}=-2$. We have three independent solutions in this case, $\displaystyle r^{2},\quad r^{0}=const,\quad r^{-2}$

Since the solution $\displaystyle r^{0}=const$ is the one with respect to the double root of the characteristic equation, we can find another independent solution, which is $\displaystyle ln(r)$. So for this case the general solution would be

$\displaystyle f_{1}(r)=A_{1}^{1}r^{2}+A_{1}^{2}+A_{1}^{3}lnr+A_{1}^{4}r^{-2},\qquad g_{1}(r)=B_{1}^{1}r^{2}+B_{1}^{2}+B_{1}^{3}lnr+B_{1}^{4}r^{-2}$

(30)

where

$\displaystyle B_{1}^{1}={\frac {i(3\lambda +5\mu )}{\lambda -\mu }}A_{1}^{1},\quad B_{1}^{2}=iA_{1}^{2}+i{\frac {(\lambda +\mu )}{\lambda +3\mu }}A_{1}^{3},\quad B_{1}^{3}=iA_{1}^{3},\quad B_{1}^{4}=-iA_{1}^{4}$

(31)

When $\displaystyle m\geq 2$ we have four different values of $\displaystyle k,\quad k_{1}=m+1,\quad k_{2}=m-1,\quad k_{3}=-m+1,\quad k_{4}=-m-1$, giving the solutions:

$\displaystyle f_{m}(r)=A_{m}^{1}r^{k_{1}}+A_{m}^{2}r^{k_{2}}+A_{m}^{3}r^{k_{3}}+A_{m}^{4}r^{k_{2}}=A_{m}^{1}r^{m+1}+A_{m}^{2}r^{m-1}+A_{m}^{3}r^{-m+1}+A_{m}^{4}r^{-m-1}$

(32)

$\displaystyle g_{m}(r)=B_{m}^{1}r^{k_{1}}+B_{m}^{2}r^{k_{2}}+B_{m}^{3}r^{k_{3}}+B_{m}^{4}r^{k_{2}}=B_{m}^{1}r^{m+1}+B_{m}^{2}r^{m-1}+B_{m}^{3}r^{-m+1}+B_{m}^{4}r^{-m-1}$

(33)

with relations

$\displaystyle B_{m}^{s}=-{\frac {im}{\mu ((k_{s})^{2}-\beta ^{2})}}\left[(\lambda +3\mu )+(\lambda +\mu )k\right],\qquad s={\overline {1,4}}$

(34)

Finally, the form of the general solutions after superimposing all modes are:

$\displaystyle u_{r}=A_{0}r+{\frac {A_{1}}{r}}+\mathbb {R} \left(A_{1}^{1}r^{2}+A_{1}^{2}+A_{1}^{3}lnr+A_{1}^{4}r^{-2}\right)e^{i\theta }+\mathbb {R} \sum _{m=2}^{\infty }\left(A_{m}^{1}r^{m+1}+A_{m}^{2}r^{m-1}+A_{m}^{3}r^{-m+1}+A_{m}^{4}r^{-m-1}\right)e^{im\theta }$

(35)

$\displaystyle u_{\theta }=B_{0}r+{\frac {B_{1}}{r}}+\mathbb {R} \left(B_{1}^{1}r^{2}+B_{1}^{2}+B_{1}^{3}lnr+B_{1}^{4}r^{-2}\right)e^{i\theta }+\mathbb {R} \sum _{m=2}^{\infty }\left(B_{m}^{1}r^{m+1}+B_{m}^{2}r^{m-1}+B_{m}^{3}r^{-m+1}+B_{m}^{4}r^{-m-1}\right)e^{im\theta }$

(36)

The general solution of the Navier's equations in the circular domain for a composite body

Appendix

Effective bulk modulus for a composite cylinder (2D case)

From the definition,i.e, the effective bulk modulus is defined as:

$\displaystyle {\overline {K}}={\frac {{\text{tr}}{\boldsymbol {\overline {\sigma }}}}{2{\text{tr}}{\boldsymbol {\overline {\epsilon }}}}}$

(N)

$\displaystyle {\boldsymbol {\overline {\sigma }}}\equiv {\frac {1}{V}}\int _{V}{\boldsymbol {\sigma }}dV$

(N)

$\displaystyle {\boldsymbol {\overline {\epsilon }}}\equiv {\frac {1}{V}}\int _{V}{\boldsymbol {\epsilon }}dV$

(N)

$\displaystyle \int _{S}\mathbf {u} \otimes (\mathbf {T} ^{T}\cdot \mathbf {n} )dS=\int _{V}{\begin{bmatrix}\mathbf {u} \otimes (\nabla \cdot \mathbf {T} )+(\mathbf {u} {\nabla })\cdot \mathbf {T} \end{bmatrix}}dV$

(N)

$\displaystyle \mathbf {u} \equiv \mathbf {x} ,\quad \mathbf {T} \equiv {\boldsymbol {\sigma }}$

(N)

$\displaystyle \nabla \cdot \mathbf {T} =\nabla \cdot {\boldsymbol {\sigma }}=\mathbf {0}$

(N)

$\displaystyle \int _{S}\mathbf {x} \otimes \underbrace {({\boldsymbol {\sigma }}\cdot \mathbf {n} )} _{\mathbf {t} }dS=\int _{V}{\begin{bmatrix}\mathbf {x} \otimes \underbrace {(\nabla \cdot {\boldsymbol {\sigma }})} _{\mathbf {0} }+\underbrace {(\mathbf {x} {\nabla })} _{\mathbf {I} }\cdot {\boldsymbol {\sigma }}dV\end{bmatrix}}=\int _{V}{\boldsymbol {\sigma }}dV$

(N)

$\displaystyle {\boldsymbol {\overline {\sigma }}}\equiv {\frac {1}{V}}\int _{V}{\boldsymbol {\sigma }}dV={\frac {1}{V}}\int _{S}\mathbf {x} \otimes \mathbf {t} dS$

(N)

$\displaystyle \mathbf {T} \equiv \mathbf {I}$

(N)

$\displaystyle \int _{S}\mathbf {u} \otimes \underbrace {({\boldsymbol {I}}\cdot \mathbf {n} )} _{\mathbf {n} }dS=\int _{V}{\begin{bmatrix}\mathbf {u} \otimes \underbrace {(\nabla \cdot \mathbf {I} )} _{\mathbf {0} }+\underbrace {(\mathbf {u} {\nabla })\cdot \mathbf {I} } _{\mathbf {u} {\nabla }}\end{bmatrix}}dV$

(N)

$\displaystyle {\frac {1}{2}}\int _{S}\left(\mathbf {u} \otimes \mathbf {n} +\mathbf {n} \otimes \mathbf {u} \right)dS=\int _{V}{\frac {1}{2}}\left[\mathbf {u} {\nabla }+(\mathbf {u} {\nabla })^{T}\right]dV=\int _{V}{\boldsymbol {\epsilon }}dV$

(N)

$\displaystyle {\boldsymbol {\overline {\epsilon }}}={\frac {1}{V}}\int _{V}{\boldsymbol {\epsilon }}dV={\frac {1}{2V}}\int _{S}\left(\mathbf {u} \otimes \mathbf {n} +\mathbf {n} \otimes \mathbf {u} \right)dS$

(N)

$\displaystyle \mathbf {t} =\sigma \mathbf {e} _{r},\,\mathbf {x} =b\mathbf {e} _{r},\,\mathbf {n} =\mathbf {e} _{r},\,\mathbf {u} =u_{r}\mathbf {e} _{r}$

(N)

$\displaystyle {\boldsymbol {\overline {\sigma }}}\equiv {\frac {1}{V}}b\sigma \int _{S}\mathbf {e} _{r}\mathbf {e} _{r}dS={\frac {1}{V}}b\sigma \mathbf {e} _{r}\mathbf {e} _{r}\int _{S}dS={\frac {Sb}{V}}\sigma \mathbf {e} _{r}\mathbf {e} _{r}=\sigma \mathbf {e} _{r}\mathbf {e} _{r}$

(N)

$\displaystyle {\boldsymbol {\overline {\epsilon }}}={\frac {u_{r}(b)}{2V}}\int _{S}\left(\mathbf {e} _{r}\otimes \mathbf {e} _{r}+\mathbf {e} _{r}\otimes \mathbf {e} _{r}\right)dS={\frac {Su_{r}(b)}{V}}\mathbf {e} _{r}\otimes \mathbf {e} _{r}={\frac {u_{r}(b)}{b}}\mathbf {e} _{r}\otimes \mathbf {e} _{r}$

(N)

$\displaystyle {\overline {K}}={\frac {{\text{tr}}{\boldsymbol {\overline {\sigma }}}}{2{\text{tr}}{\boldsymbol {\overline {\epsilon }}}}}={\frac {b\sigma }{2u_{r}(b)}}$

(N)