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:
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$$   \displaystyle \boldsymbol \nabla \cdot \boldsymbol \sigma = \mathbf 0 $$   (1)
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we follow the procedures of Kochmann,, express the Cauchy' equations in the polar coordinates:
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$$   \displaystyle (\lambda+\mu)(r^2 u_{r,rr}+ r u_{r,r} - u_r + r u_{\theta,r\theta} - u_{\theta,\theta}) + \mu(r^2 u_{r,rr}+ r u_{r,r} - u_r + u_{r,\theta\theta} - 2u_{\theta,\theta}) = 0 $$   (2)
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$$   \displaystyle (\lambda+\mu)(ru_{r,r\theta}+ u_{r,\theta} + u_{\theta,\theta\theta}) + \mu(r^2 u_{\theta,rr} + r u_{\theta,r} + 2u_{r,\theta} + u_{\theta,\theta\theta} - u_\theta) = 0 $$   (3)
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and assume the separable form of the solutions
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$$   \displaystyle u_r=f(r)e^{im\theta}, \qquad u_{\theta}=g(r)e^{im\theta} $$   (4)
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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:


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$$   \displaystyle u_r=\sum_{m=-\infty}^{\infty}f_m(r)e^{im\theta}, u_\theta=\sum_{m=-\infty}^{\infty}g_m(r)e^{im\theta} $$   (5)
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On the other hand, displacements must be real, i.e $$\displaystyle \overline u_r=u_r, \quad \overline u_\theta=u_\theta \quad$$ or


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$$   \displaystyle \sum_{m=-\infty}^{\infty}f_m(r)e^{im\theta}=\sum_{m=-\infty}^{\infty}\overline {f_m(r)}e^{-im\theta}, \quad \sum_{m=-\infty}^{\infty}g_m(r)e^{im\theta}=\sum_{m=-\infty}^{\infty}\overline{g_m(r)}e^{-im\theta} $$   (6)
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It is easy to see that


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$$   \displaystyle \sum_{m=-\infty}^{\infty}\overline {f_m(r)}e^{-im\theta}=\sum_{m=-\infty}^{\infty}\overline {f_{-m}(r)}e^{im\theta}, \quad \sum_{m=-\infty}^{\infty}\overline{g_m(r)}e^{-im\theta}=\sum_{m=-\infty}^{\infty}\overline {g_{-m}(r)}e^{im\theta} $$   (7)
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Comparing equation ... and ... to get


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$$   \displaystyle f_m(r)=\overline{f_{-m}(r)} \quad \text{and} \quad g_m(r)=\overline{g_{-m}(r)} $$   (8)
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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:


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$$   \displaystyle \begin{align} u_r & =\sum_{m=-\infty}^{\infty}{f_m(r)}e^{im\theta}=f_0(r)+\sum_{m=1}^{\infty}f_m(r)e^{im\theta} + \sum_{m=-\infty}^{-1}f_m(r)e^{im\theta}=f_0(r)+ \sum_{m=1}^{\infty}f_m(r)e^{im\theta}           +\sum_{m=1}^{\infty}f_{-m}(r)e^{-im\theta} \\ & =f_0(r)+ \sum_{m=1}^{\infty}f_m(r)e^{im\theta}+\sum_{m=1}^{\infty}\overline {f_{m}(r)}e^{-im\theta}=f_0(r)+ \sum_{m=1}^{\infty}f_m(r)e^{im\theta}+\sum_{m=1}^{\infty}\overline {f_{m}(r)e^{im\theta}} \\ & =f_0(r)+ \sum_{m=1}^{\infty} \left( f_m(r)e^{im\theta}+\overline {f_{m}(r)e^{im\theta}}\right )=f_0(r)+ 2\sum_{m=1}^{\infty} \mathbb{R}\left( f_m(r)e^{im\theta}\right )\equiv \mathbb{R}\sum_{m=0}^{\infty} F_m(r)e^{im\theta}\\ \end{align} $$   (9)
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We conclude by writing the general representation of solutions as follows:


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$$  \displaystyle u_r=\mathbb{R}\sum_{m=0}^{\infty} f_m(r)e^{im\theta}, \qquad u_{\theta}=\mathbb{R}\sum_{m=0}^{\infty} g_m(r)e^{im\theta} $$  (10)
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it means that m is a nonnegative integer. Now let find the solution of the Navier's equation. Subtitute ... to ...


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$$   \displaystyle (\lambda+2\mu)\left( f''+\frac{f'}{r} \right ) - \left(\frac{\lambda+2\mu+m^2\mu}{r^2} \right )f = im \left[(\lambda+3\mu)\frac{g}{r^2}-(\lambda+\mu)\frac{g'}{r}\right] $$   (11)
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$$   \displaystyle \mu\left( g''+\frac{g'}{r} \right ) - \left(\frac{(\lambda+2\mu)m^2+\mu}{r^2} \right )g = -im \left[(\lambda+3\mu)\frac{f}{r^2}+(\lambda+\mu)\frac{f'}{r}\right] $$   (12)
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Let consider possible situations.


 * $$\displaystyle \lambda+2\mu=0$$


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$$   \displaystyle - \frac{m^2\mu}{r^2}f = im\mu \left[\frac{g}{r^2} + \frac{g'}{r}\right] $$   (13)
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$$   \displaystyle \mu\left( g''+\frac{g'}{r} \right ) - \frac{\mu}{r^2}g = -im\mu \left[\frac{f}{r^2}-\frac{f'}{r}\right] $$   (14)
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 * 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. $$\displaystyle \mu \neq 0$$


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$$   \displaystyle - \frac{m^2}{r^2}f = im \left(\frac{g}{r^2}+\frac{g'}{r}\right) $$   (15)
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$$   \displaystyle \left(g''+\frac{g'}{r}\right) - \frac{g}{r^2} = -im \left(\frac{f}{r^2}-\frac{f'}{r}\right) $$   (16)
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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.


 * $$\displaystyle \mu=0$$


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$$   \displaystyle \lambda\left( f''+\frac{f'}{r} \right ) - \frac{\lambda f}{r^2} = im \lambda\left[\frac{g}{r^2}-\frac{g'}{r}\right] $$   (17)
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$$   \displaystyle -\lambda\frac{m^2 g}{r^2} = -im\lambda \left[\frac{f}{r^2}+\frac{f'}{r}\right] $$   (18)
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 * A. $$\displaystyle \lambda = 0 $$

The same conclusion as 1.a above.


 * B. $$\displaystyle \lambda \neq 0 $$


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$$   \displaystyle \left( f''+\frac{f'}{r} \right ) - \frac{f}{r^2} = im \left[\frac{g}{r^2}-\frac{g'}{r}\right] $$   (19)
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$$   \displaystyle \frac{m^2 g}{r^2} = im \left[\frac{f}{r^2}+\frac{f'}{r}\right] $$   (20)
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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.


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


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$$   \displaystyle f''+\frac{f'}{r} - \left(1+\frac{m^2\mu}{\lambda+2\mu} \right )\frac{f}{r^2} = \frac{im}{\lambda+2\mu} \left[(\lambda+3\mu)\frac{g}{r^2}-(\lambda+\mu)\frac{g'}{r}\right] $$   (21)
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$$   \displaystyle g''+\frac{g'}{r} - \left(1+\frac{(\lambda+2\mu)m^2}{\mu} \right )\frac{g}{r^2} = -\frac{im}{\mu} \left[(\lambda+3\mu)\frac{f}{r^2}+(\lambda+\mu)\frac{f'}{r}\right] $$   (22)
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 * A. $$\displaystyle m = 0 $$


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$$   \displaystyle f_0(r)= A_0r+\frac{A_1}{r}, \qquad g_0(r)=B_0r + \frac{B_1}{r} $$   (23)
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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


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$$   \displaystyle (k^2-\alpha^2)A-\frac{im}{\lambda+2\mu} \left[(\lambda+3\mu)-(\lambda+\mu)k\right]B=0 $$   (24)
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$$   \displaystyle (k^2-\beta^2)B+\frac{im}{\mu} \left[(\lambda+3\mu)+(\lambda+\mu)k\right]A=0 $$   (25)
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where


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$$   \displaystyle \alpha^2=1+\frac{m^2\mu}{\lambda+2\mu},\qquad \beta^2=1+\frac{m^2(\lambda+2\mu)}{\mu} $$   (26)
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Rewrite the system of equation in the matrix form
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$$   \displaystyle \begin{bmatrix} \displaystyle (k^2-\alpha^2) & \displaystyle -\frac{im}{\lambda+2\mu} \left[(\lambda+3\mu)-(\lambda+\mu)k\right]\\ \displaystyle \frac{im}{\mu} \left[(\lambda+3\mu)+(\lambda+\mu)k\right] & (k^2-\beta^2) \end{bmatrix} \begin{Bmatrix} A\\ B \end{Bmatrix} = \begin{Bmatrix} 0\\ 	0	\end{Bmatrix} $$   (27)
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and realize that the system has non-trivial solutions if the determinant equal zero or


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$$   \displaystyle (k^2-\alpha^2)(k^2-\beta^2) +\frac{(im)^2}{\mu(\lambda+2\mu)} \left[(\lambda+3\mu)-(\lambda+\mu)k\right]\left[(\lambda+3\mu)+(\lambda+\mu)k\right] = 0 $$   (28)
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Solving the above equation to obtain


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$$   \displaystyle k^4-2(m^2+1)k^2+(m^2-1)^2=0 \to k=\pm m \pm 1 $$   (29)
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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


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$$   \displaystyle f_1(r)=A^1_1r^2 + A^2_1 + A^3_1lnr + A^4_1r^{-2}, \qquad g_1(r)=B^1_1r^2 + B^2_1 + B^3_1lnr + B^4_1r^{-2} $$   (30)
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where


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$$   \displaystyle B^1_1=\frac{i(3\lambda+5\mu)}{\lambda-\mu}A^1_1, \quad B^2_1=iA^2_1+i \frac{(\lambda+\mu)}{\lambda+3\mu}A^3_1, \quad B^3_1= iA^3_1, \quad B^4_1=-iA^4_1$$ || <p style="text-align:right"> (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:
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$$   \displaystyle f_m(r)=A^1_mr^{k_1} + A^2_mr^{k_2} + A^3_m r^{k_3} + A^4_mr^{k_2} = A^1_mr^{m+1} + A^2_mr^{m-1} + A^3_m r^{-m+1} + A^4_mr^{-m-1} $$   (32)
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$$   \displaystyle g_m(r)=B^1_mr^{k_1} + B^2_mr^{k_2} + B^3_m r^{k_3} + B^4_mr^{k_2} = B^1_mr^{m+1} + B^2_mr^{m-1} + B^3_m r^{-m+1} + B^4_mr^{-m-1} $$   (33)
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with relations


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$$   \displaystyle B^s_m=-\frac{im}{\mu((k_s)^2-\beta^2)}\left[(\lambda+3\mu)+(\lambda+\mu)k \right], \qquad s=\overline{1,4} $$   (34)
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Finally, the form of the general solutions after superimposing all modes are:


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$$  \displaystyle u_r= A_0r+\frac{A_1}{r} + \mathbb{R}\left(A^1_1r^2 + A^2_1 + A^3_1lnr + A^4_1r^{-2} \right)e^{i\theta} + \mathbb{R}\sum^{\infty}_{m=2}\left(A^1_mr^{m+1} + A^2_mr^{m-1} + A^3_m r^{-m+1} + A^4_mr^{-m-1} \right )e^{im\theta} $$  (35)
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$$  \displaystyle u_\theta= B_0r+\frac{B_1}{r} + \mathbb{R}\left(B^1_1r^2 + B^2_1 + B^3_1lnr + B^4_1r^{-2} \right)e^{i\theta} + \mathbb{R}\sum^{\infty}_{m=2}\left(B^1_mr^{m+1} + B^2_mr^{m-1} + B^3_m r^{-m+1} + B^4_mr^{-m-1} \right )e^{im\theta} $$  (36)
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When no displacements at the outer boundary
= The general solution of the Navier's equations in the circular domain for a composite body =

When no displcaments at the outer boundary
= Appendix =

Effective bulk modulus for a composite cylinder (2D case)
From the definition,i.e, the effective bulk modulus is defined as:


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$$   \displaystyle \overline K = \frac{\text{tr} \boldsymbol {\overline \sigma}}{2\text{tr}\boldsymbol {\overline  \epsilon}} $$   (N)
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$$   \displaystyle \boldsymbol {\overline \sigma}\equiv \frac{1}{V} \int_{V}\boldsymbol \sigma dV $$ (N)
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$$   \displaystyle \boldsymbol {\overline \epsilon}\equiv \frac{1}{V} \int_{V}\boldsymbol \epsilon dV $$ (N)
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$$   \displaystyle \int_{S}\mathbf u\otimes (\mathbf T^T\cdot \mathbf n)dS = \int_{V} \begin{bmatrix} \mathbf u \otimes (\nabla \cdot \mathbf T) + (\mathbf u {\nabla})\cdot \mathbf T \end{bmatrix} dV $$ (N)
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$$   \displaystyle \mathbf u\equiv \mathbf x, \quad \mathbf T \equiv \boldsymbol \sigma $$   (N)
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$$   \displaystyle \nabla\cdot \mathbf T = \nabla \cdot \boldsymbol \sigma = \mathbf 0 $$   (N)
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$$   \displaystyle \int_{S}\mathbf x\otimes \underbrace {(\boldsymbol \sigma\cdot \mathbf n)}_{\mathbf t}dS = \int_{V} \begin{bmatrix} \mathbf x \otimes \underbrace{(\nabla \cdot \boldsymbol \sigma)}_{\mathbf 0} + \underbrace{(\mathbf x {\nabla})}_{\mathbf I}\cdot \boldsymbol \sigma dV \end{bmatrix} = \int_{V}\boldsymbol \sigma dV $$ (N)
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$$  \displaystyle \boldsymbol {\overline \sigma}\equiv \frac{1}{V} \int_{V}\boldsymbol \sigma dV = \frac{1}{V} \int_{S}\mathbf x\otimes \mathbf t dS $$ (N)
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$$   \displaystyle \mathbf T \equiv \mathbf I $$ (N)
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$$   \displaystyle \int_{S}\mathbf u\otimes \underbrace {(\boldsymbol I\cdot \mathbf n)}_{\mathbf n}dS = \int_{V} \begin{bmatrix} \mathbf u \otimes \underbrace{(\nabla \cdot \mathbf I)}_{\mathbf 0} + \underbrace{(\mathbf u {\nabla})\cdot \mathbf I}_{\mathbf u {\nabla}} \end{bmatrix} dV $$ (N)
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$$   \displaystyle \frac{1}{2}\int_{S}\left( \mathbf u\otimes \mathbf n + \mathbf n\otimes \mathbf u\right ) dS = \int_{V} \frac {1}{2} \left[ \mathbf u {\nabla} + (\mathbf u {\nabla})^T\right] dV = \int_{V} \boldsymbol \epsilon dV $$ (N)
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$$  \displaystyle \boldsymbol {\overline \epsilon} = \frac{1}{V}\int_{V} \boldsymbol \epsilon dV = \frac{1}{2V}\int_{S}\left( \mathbf u\otimes \mathbf n + \mathbf n \otimes \mathbf u\right ) dS $$ (N)
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$$   \displaystyle \mathbf t = \sigma \mathbf e_r, \,\mathbf x = b\mathbf e_r, \, \mathbf n = \mathbf e_r, \, \mathbf u = u_r\mathbf e_r $$   (N)
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$$   \displaystyle \boldsymbol {\overline \sigma}\equiv \frac{1}{V} b\sigma \int_{S}\mathbf e_r \mathbf e_r dS = \frac{1}{V} b\sigma \mathbf e_r \mathbf e_r\int_{S} dS = \frac{Sb}{V} \sigma \mathbf e_r \mathbf e_r = \sigma \mathbf e_r \mathbf e_r $$   (N)
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$$   \displaystyle \boldsymbol {\overline \epsilon} = \frac{u_r(b)}{2V}\int_{S}\left( \mathbf e_r \otimes \mathbf e_r + \mathbf e_r \otimes \mathbf e_r\right ) dS = \frac{Su_r(b)}{V} \mathbf e_r \otimes \mathbf e_r = \frac{u_r(b)}{b} \mathbf e_r \otimes \mathbf e_r $$   (N)
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$$  \displaystyle \overline K = \frac{\text{tr} \boldsymbol {\overline \sigma}}{2\text{tr}\boldsymbol {\overline  \epsilon}} = \frac{b\sigma}{2u_r(b)} $$  (N)
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Effective bulk modulus for a composite sphere (3D case)
=References =