On Poisson’s functions of compressible elastomeric materials under compression tests in the framework of linear elasticity theory

Abstract

We analyze the compression of a right cylinder made of an elastomeric material, sliding on a rigid plate in the framework of linear elasticity theory. Lubrication seems to reduce the heterogeneous effects of lateral bulging so that the deformations can be considered as “quasi-homogeneous.” As a result, we show that the ratio between the transversal deformation and axial strain is not a Poisson’s ratio but a Poisson’s function.

Appendix A

We consider the following boundary value problem:

$$\begin{aligned}&\bar{{m}}^{4}S^{4}\bar{{\phi }}-2 \bar{{m}}^{2}S^{2}\left( {{\bar{{\phi }}}''+\frac{{\bar{{\phi }}}'}{\bar{{r}}}} \right) +\frac{1}{\bar{{r}}}\left[ {\bar{{r}} \left( {{\bar{{\phi }}}''+\frac{{\bar{{\phi }}}'}{\bar{{r}}}} \right) ^{\prime }} \right] ^{\prime }=0 \end{aligned}$$

(A.1)

where

$$\begin{aligned}&\bar{{\phi }}\left( 0 \right) =\bar{{\phi }}\left( 1 \right) = 1\hbox { and }{\bar{{\phi }}}'\left( 0 \right) =0. \end{aligned}$$

(A.2)

We pose: $\forall r \in ] {-R, R} [ , \bar{{\phi }}\left( {\bar{{r}}} \right) =\sum \nolimits _{n\ge 0} {a_n } \bar{{r}}^{n}$.

So, we have then

$$\begin{aligned} \bar{{\phi }}\left( 0 \right) =1\hbox { and }{\bar{{\phi }}}'\left( 0 \right) =0\Rightarrow a_0 =1\hbox { and }a_1 =0. \end{aligned}$$

(A.3.1-4)

We substitute Eq. (A.3.1) in (A.1), and by a changing of the indices, we get the following equation:

$$\begin{aligned} \left\{ {\begin{array}{ll} a_{n+1} =- \frac{2 \bar{{m}}^{2}S^{2}\left( {n-1} \right) ^{2}a_{n-1} -\bar{{m}}^{4}S^{4}a_{n-3} }{\left( {n+1} \right) ^{2}\left( {n-1} \right) ^{2}}, &{}\quad \hbox {if }n\ge 3; \\ a_n =0, &{}\quad \hbox {if }n=2p+1, \forall p\ge 0 \\ \end{array}} \right. \end{aligned}$$

In order to obtain the coefficients $a_{n}$, we vary the value of $a_{2}$ until to satisfy the boundary condition $\bar{{\phi }}\left( 1 \right) =1$.