Analysis of a cylindrically orthotropic disk using a regular perturbation method

Abstract

We employ a regular perturbation method to construct approximate solutions for a class of boundary value problems in classical linear elasticity. A problem in this class consists of a cylindrically orthotropic disk in equilibrium with no body force and subjected to a prescribed radial displacement along its boundary. The problem has an exact solution that is singular at the centre of the disk. The singularity is dictated by the ratio between the tangential and the radial components of the elasticity tensor. The approximate solutions are given in terms of truncated power series of a small parameter $\varepsilon$, which is equal to one minus the above ratio. In the particular case of an isotropic material, this ratio is one, yielding $\varepsilon = 0$. The approximate solutions can be cast in a very concise form and tend to the exact solution as the number of terms in the truncated series increases, even for $\varepsilon$ not small. On the other hand, convergence rates of these series depend strongly on $\varepsilon$. Since the zeroth-order term, corresponding to $\varepsilon = 0$, yields the solution of the disk problem for an isotropic material, we can study the stresses in a neighbourhood of the centre of the disk as if the disk was isotropic with an error of order $\varepsilon$. For comparison purposes, we have also considered finite element approximations of the exact solutions obtained with uniform meshes. We have found that, differently from the perturbation method approximations, the finite element approximations are not very sensitive to changes in $\varepsilon$. The approach presented in this work can be used to validate numerical solutions and to obtain insight on the solutions of complex problems that are not known in closed form.

Appendices

Appendix A

We show details of the derivation of the expressions of $A_{(i,j)}$ in (22). It follows from (21) that the derivatives of ${\hat{u}}_i(\rho )$ are given by:

$$\begin{aligned} \begin{aligned}&\frac{d}{d\rho }{\hat{u}}_i(\rho ) = \frac{1}{2^i\,i!}\, \beta \, \sum _{j=1}^i \left[ (-1)^j\,A_{(i,j)}\, (\log ^j\rho + j\,\log ^{j-1}\rho )\right] , \\&\frac{d^2}{d\rho ^2}{\hat{u}}_i(\rho ) = \frac{1}{2^i\,i!}\, \frac{\beta }{\rho }\, \sum _{j=1}^i \left[ (-1)^j\,A_{(i,j)}\, \left( j\,\log ^{j-1}\rho + j\,(j-1)\,\log ^{j-2}\rho \right) \right] . \end{aligned} \end{aligned}$$

(34)

By substituting (21), (34), and (19) in (16.b) for $i = 1$, we obtain

$$\begin{aligned} \frac{1}{2}\,\frac{\beta }{\rho }\,(-2\,A_{(1,1)} + 2)\,\log ^0\rho = 0. \end{aligned}$$

(35)

For (35) to hold for all $\rho \in (0,1)$, we must have $A_{(1,1)} = 1$.

Next, for $i \ge 2$, we substitute (21) and (34) into (16.b) to obtain

$$\begin{aligned} \begin{aligned} 0 =&\;(-1)^i\, i\,(i-1)\, A_{(i,i)}\,\log ^{i-2}\rho + (-1)^i\, 2\,i\, A_{(i,i)}\, \log ^{i-1}\rho \\&\;+\sum _{j=1}^{i-1}\,(-1)^j \left[ j\,(j-1)\, A_{(i,j)}\,\log ^{j-2}\rho + 2\,j\, A_{(i,j)}\,\log ^{j-1}\rho + 2\,i\, A_{(i-1,\,j)}\,\log ^j\rho \right] , \end{aligned} \end{aligned}$$

(36)

which can be rewritten as

$$\begin{aligned} \begin{aligned} 0 =&\left( -2\,A_{(i,1)} + 2\,A_{(i,2)}\right) \log ^0\rho \\&+ \left( -2\,i\,A_{(i-1,\,1)} + 4\,A_{(i,2)} - 6\,A_{(i,3)}\right) \log ^1\rho \\&+ \left( 2\,i\,A_{(i-1,2)} - 6\,A_{(i,3)} + 12\,A_{(i,4)}\right) \log ^2\rho \\&+ \ldots \\&+ \left[ (-1)^{l}\,2\,i\,A_{(i-1,\,l)} + (-1)^{l + 1}\,2\,(l + 1)\,A_{(i,\,l + 1)} + (-1)^{l + 2}\, (l + 2)\, (l + 1)\, A_{(i,\,l + 2)}\right] \log ^{l}\rho \\&+ \ldots \\&+ \left[ (-1)^{i-2}\,2\,i\,A_{(i-1,\,i-2)} + (-1)^{i-1}\,2\,(i-1)\,A_{(i,\,i-1)} + (-1)^{i}\, i\, (i-1)\, A_{(i,i)}\right] \log ^{i-2}\rho \\&+ \left[ (-1)^{i-1}\,2\,i\,A_{(i-1,\,i-1)} + (-1)^{i}\, 2\,i\, A_{(i,i)}\right] \log ^{i-1}\rho , \end{aligned} \end{aligned}$$

(37)

where $1 \le l \le i-2$. For (37) to be satisfied for all $\rho \in (0,1)$, the coefficients that multiply each power of $\log \rho$ must be null. Therefore, for $\log ^0\rho$, $\log ^{l}\rho$ and $\log ^{i-1}\rho$, it yields, respectively,

$$\begin{aligned} A_{(i,1)}= & {} A_{(i,2)}, \end{aligned}$$

(38)

$$\begin{aligned} A_{(i,\,l+1)}= & {} \frac{i}{l+1} \, A_{(i-1,\,l)} +\frac{l+2}{2}\,A_{(i,\,l+2)}, \end{aligned}$$

(39)

$$\begin{aligned} A_{(i,i)}= & {} A_{(i-1,\,i-1)}. \end{aligned}$$

(40)

By defining $k {\mathop {=}\limits ^{\text {def}}}l + 1$, we rewrite (39) as

$$\begin{aligned} A_{(i,k)} = \frac{i}{k} \, A_{(i-1,\,k-1)} + \frac{k+1}{2}\,A_{(i,\,k+1)}. \end{aligned}$$

(41)

Finally, it follows from $A_{(1,1)} = 1$, (38), (40), and (41) that $A_{(i,j)}$ is given by (22).

Appendix B

Using proof by induction in i, we derive the expression (25.a) and show that the coefficients $B_{(i,j)}$ in this expression are given by:

$$\begin{aligned} \begin{aligned}&B_{(1,1)} = 1,\\&B_{(i+1,\,1)} = (2\,i-1) \, B_{(i,1)},\\&B_{(i+1,\,i+1)} = B_{(i,i)},\\&B_{(i+1,\,j)} = B_{(i,\,j-1)} + (2\,i-j)\,B_{(i,j)}, \quad \text {for}\; 2\le j \le i, \end{aligned} \end{aligned}$$

(42)

which can be rewritten as (26). First, we use (13) together with both (6.b) and (10) to write

$$\begin{aligned} \begin{aligned} \frac{\partial {\hat{u}}}{\partial \varepsilon }(\rho ,\varepsilon )&= \frac{1}{2}\, \beta \,\rho ^\alpha \,\left[ -\frac{\log \rho }{\alpha }\right] , \qquad \alpha = \sqrt{1-\varepsilon }, \\ \frac{\partial ^2 {\hat{u}}}{\partial \varepsilon ^2}(\rho ,\varepsilon )&= \frac{1}{4}\, \beta \,\rho ^\alpha \,\left[ - \frac{\log \rho }{\alpha ^3} + \frac{\log ^2\rho }{\alpha ^2}\right] , \\ \frac{\partial ^3 {\hat{u}}}{\partial \varepsilon ^3}(\rho ,\varepsilon )&= \frac{1}{8}\, \beta \,\rho ^\alpha \,\left[ - 3\,\frac{\log \rho }{\alpha ^5} + 3\,\frac{\log ^2\rho }{\alpha ^4} -\frac{\log ^3\rho }{\alpha ^3}\right] , \\ \frac{\partial ^4 {\hat{u}}}{\partial \varepsilon ^4}(\rho ,\varepsilon )&= \frac{1}{16}\, \beta \,\rho ^\alpha \,\left[ - 15\,\frac{\log \rho }{\alpha ^7} + 15\,\frac{\log ^2\rho }{\alpha ^6} -6\,\frac{\log ^3\rho }{\alpha ^5} + \frac{\log ^4\rho }{\alpha ^4}\right] . \end{aligned} \end{aligned}$$

(43)

Comparing (43) with (25), we obtain the relations in (42) for $i = 1,2,3,4$. This ends the proof of the base case.

To prove the induction step, we first apply the chain rule to (13) and write

$$\begin{aligned} \frac{\partial ^{i+1} {\hat{u}}}{\partial \varepsilon ^{i+1}}(\rho ,\varepsilon ) = \frac{\partial }{\partial \varepsilon }\left[ \frac{\partial ^{i} {\hat{u}}}{\partial \varepsilon ^{i}}(\rho ,\varepsilon )\right] = \frac{\partial }{\partial \alpha }\left[ \frac{\partial ^{i} {\hat{u}}}{\partial \varepsilon ^{i}}(\rho ,\varepsilon )\right] \frac{\partial \alpha }{\partial \varepsilon } = -\frac{1}{2\,\alpha }\,\frac{\partial }{\partial \alpha }\left[ \frac{\partial ^{i} {\hat{u}}}{\partial \varepsilon ^{i}}(\rho ,\varepsilon )\right] . \end{aligned}$$

(44)

Then, we use the induction hypothesis and (44) to write

$$\begin{aligned} \begin{aligned} \frac{\partial ^{i+1} {\hat{u}}}{\partial \varepsilon ^{i+1}}(\rho ,\varepsilon )&= -\frac{1}{2\,\alpha }\,\frac{\partial }{\partial \alpha }\left[ \frac{1}{2^i}\,\beta \,\rho ^\alpha \sum _{j=1}^i \frac{(-\log \rho )^j}{\alpha ^{2\,i-j}}\,B_{(i,j)}\right] \\ {}&= -\frac{\beta }{2^{i+1}\,\alpha } \sum _{j=1}^i \left[ (-\log \rho )^j\,\,B_{(i,j)}\, \frac{\partial }{\partial \alpha }\!\!\left( \frac{\rho ^\alpha }{\alpha ^{2\,i-j}}\right) \right] \ \\ {}&= -\frac{\beta }{2^{i+1}\,\alpha }\, \sum _{j=1}^i \left[ (-\log \rho )^j\,\,B_{(i,j)}\, \left( \frac{\rho ^\alpha \,\log \rho }{\alpha ^{2\,i-j}} - \frac{(2\,i-j)\,\rho ^\alpha }{\alpha ^{2\,i-j+1}} \right) \right] \\ {}&= \frac{\beta }{2^{i+1}}\,\rho ^\alpha \sum _{j=1}^i \left[ \frac{(-\log \rho )^{j+1}}{\alpha ^{2\,i-j+1}}\,B_{(i,j)} + \frac{(-\log \rho )^j}{\alpha ^{2\,i-j+2}}\,(2\,i-j)\,B_{(i,j)} \right] . \end{aligned} \end{aligned}$$

(45)

We define $k{\mathop {=}\limits ^{\text {def}}}j+1$ and, after some algebraic manipulation, we obtain

$$\begin{aligned} \begin{aligned} \frac{\partial ^{i+1} {\hat{u}}}{\partial \varepsilon ^{i+1}}(\rho ,\varepsilon )&= \frac{\beta }{2^{i+1}}\,\rho ^\alpha \left[ \sum _{k=2}^{i+1} \frac{(-\log \rho )^{k}}{\alpha ^{2\,i-k+2}}\,B_{(i,\,k-1)} + \sum _{j=1}^i \frac{(-\log \rho )^j}{\alpha ^{2\,i-j+2}}\,(2\,i-j)\,B_{(i,j)} \right] \\ {}&= \frac{\beta }{2^{i+1}}\,\rho ^\alpha \,\Bigg [ \sum _{k=2}^{i} \frac{(-\log \rho )^{k}}{\alpha ^{2\,i-k+2}}\,B_{(i,\,k-1)} + \sum _{j=2}^i \frac{(-\log \rho )^j}{\alpha ^{2\,i-j+2}}\,(2\,i-j)\,B_{(i,j)} \\ {}&\qquad \qquad \quad \;+ \frac{(-\log \rho )^{i+1}}{\alpha ^{i+1}}\,B_{(i,i)} + \frac{(-\log \rho )}{\alpha ^{2\,i+1}}\,(2\,i-1)\,B_{(i,1)} \Bigg ] \\ {}&= \frac{\beta }{2^{i+1}}\,\rho ^\alpha \,\Bigg [ \sum _{j=2}^{i} \frac{(-\log \rho )^{j}}{\alpha ^{2\,(i+1)-j}}\, \left[ B_{(i,\,j-1)} + (2\,i-j)\,B_{(i,j)}\right] \\ {}&\qquad \qquad \quad \;+ \frac{(-\log \rho )^{i+1}}{\alpha ^{2\,(i+1)-(i+1)}}\,B_{(i,i)} + \frac{(-\log \rho )}{\alpha ^{2\,(i+1)-1}}\,(2\,i-1)\,B_{(i,1)} \Bigg ]. \end{aligned} \end{aligned}$$

(46)

Finally, we use (42) to identify the terms $B_{(i+1,\,1)}$, $B_{(i+1,\,i+1)}$, and $B_{(i+1,\,j)}$, yielding

$$\begin{aligned} \begin{aligned} \frac{\partial ^{i+1} {\hat{u}}}{\partial \varepsilon ^{i+1}}(\rho ,\varepsilon )&= \frac{\beta }{2^{i+1}}\,\rho ^\alpha \,\Bigg [ \sum _{j=2}^{i} \frac{(-\log \rho )^{j}}{\alpha ^{2\,(i+1)-j}}\, B_{(i+1,\,j)} \\ {}&\quad + \frac{(-\log \rho )^{i+1}}{\alpha ^{2\,(i+1)-(i+1)}}\,B_{(i+1,\,i+1)} + \frac{(-\log \rho )^1}{\alpha ^{2\,(i+1)-1}}\,B_{(i+1,\,1)} \Bigg ] \\ {}&= \frac{1}{2^{i+1}}\,\beta \,\rho ^\alpha \sum _{j=1}^{i+1} \frac{(-\log \rho )^j}{\alpha ^{2\,(i+1)-j}}\,B_{(i+1,\,j)}, \end{aligned} \end{aligned}$$

(47)

and this completes the proof.

Appendix C