An analytic derivation of the bifurcation conditions for localization in hyperelastic tubes and sheets

Abstract

We provide an analytic derivation of the bifurcation conditions for localized bulging in an inflated hyperelastic tube of arbitrary wall thickness and axisymmetric necking in a hyperelastic sheet under equibiaxial stretching. It has previously been shown numerically that the bifurcation condition for the former problem is equivalent to the vanishing of the Jacobian determinant of the internal pressure P and resultant axial force N, with each of them viewed as a function of the azimuthal stretch on the inner surface and the axial stretch. This equivalence is established here analytically. For the latter problem for which it has recently been shown that the bifurcation condition is not given by a Jacobian determinant equal to zero, we explain why this is the case and provide an alternative interpretation.

Appendix: A Proof of equality (2.58)

To prove (2.58), we start by analyzing the linearized forms of (2.57). Assuming that the perturbed configuration is also uniform, we show below that the linearized forms of the following two equations

$$\begin{aligned}&-P^*(\lambda _a,\lambda _z)+P=0,\end{aligned}$$

(A.1)

$$\begin{aligned}&\frac{1}{2}b^2 (P^*(\lambda _a,\lambda _z)-P)+\frac{1}{2\pi }(N^*(\lambda _a,\lambda _z)-N)=0 \end{aligned}$$

(A.2)

agree with (2.42) and (2.43).

Let us denote by $\overline{\varvec{\sigma }}$ and $\varvec{\sigma }$ the Cauchy stresses associated with the uniformly inflated configuration and perturbed configuration, respectively. Then it follows from the definition of the incremental stress tensor $\varvec{\chi }$ that

$$\begin{aligned} \varvec{\sigma }=(\varvec{I}+\varvec{\eta })(\overline{\varvec{\sigma }}+\varvec{\chi }). \end{aligned}$$

(A.3)

Note that the incremental deformation gradient $\varvec{\eta }$ is diagonal since the perturbed configuration is uniform.

Specifying (A.1) to the perturbed configuration, we see from the definition of $P^*$ that the resulted equation takes the form

$$\begin{aligned} \int _{{\tilde{a}}}^{{\tilde{b}}}\frac{\sigma _{33}-\sigma _{11}}{{\tilde{r}}}\,d{\tilde{r}}+P=0. \end{aligned}$$

(A.4)

where ${\tilde{r}}=r+u$ denotes the radius of the tube after perturbation, and ${\tilde{a}}={\tilde{r}}(A)$ and ${\tilde{b}}={\tilde{r}}(B)$. Substituting (A.3) into (A.4) and making a change of variables (integration by substitution) by applying the incompressibility equality

$$\begin{aligned} {\tilde{\lambda }}_z({\tilde{r}}^2-{\tilde{a}}^2)=\lambda _z (r^2-a^2), \end{aligned}$$

(A.5)

where ${\tilde{\lambda }}_z$ is the axial stretch of the tube in the perturbed configuration, we obtain

$$\begin{aligned} \int _{a}^b \frac{(1+\eta _{33})({\overline{\sigma }}_{33}+\chi _{33})-(1+\eta _{11})({\overline{\sigma }}_{11}+\chi _{11})}{r}\frac{\lambda _z r^2}{{\tilde{\lambda }}_z {\tilde{r}}^2}\,dr+P=0. \end{aligned}$$

(A.6)

When expanded to linear order, we have

$$\begin{aligned} \frac{\lambda _z r^2}{{\tilde{\lambda }}_z {\tilde{r}}^2}=1-2\frac{{\tilde{r}}-r}{r}-\frac{{\tilde{\lambda }}_z-\lambda _z}{\lambda _z}=1-2\eta _{11}-\eta _{22}. \end{aligned}$$

(A.7)

Thus by ignoring nonlinear terms, one can simplify (A.6) as

$$\begin{aligned} \int _a^b \frac{\chi _{33}-\chi _{11}}{r}\,dr+\int _a^b ({\overline{\sigma }}_{33}\frac{\eta _{33}-\eta _{11}}{r}+ \frac{{\overline{\sigma }}_{33}-{\overline{\sigma }}_{11}}{r}\eta _{33})\,dr=0, \end{aligned}$$

(A.8)

where we have used incompressibility constraint $\eta _{11}+\eta _{22}+\eta _{33}=0.$ The radial equilibrium for the unperturbed deformation implies that

$$\begin{aligned} \frac{d{\overline{\sigma }}_{33}}{dr}+\frac{{\overline{\sigma }}_{33}-{\overline{\sigma }}_{11}}{r}=0. \end{aligned}$$

(A.9)

Using integration by parts, one can write the second integral in (A.8) as

$$\begin{aligned} \begin{aligned} \int _a^b \left( {\overline{\sigma }}_{33}\frac{\eta _{33}-\eta _{11}}{r}+ \frac{{\overline{\sigma }}_{33}-{\overline{\sigma }}_{11}}{r}\eta _{33}\right) \,dr&=\int _a^b \left( {\overline{\sigma }}_{33}\frac{\eta _{33}-\eta _{11}}{r}-\frac{d\overline{\sigma }_{33}}{dr}\eta _{33}\right) \,dr\\ {}&=-\overline{\sigma } _{33}\eta _{33}|_{r=a}^{r=b}+\int _a^b {\overline{\sigma }}_{33}\left( \frac{\eta _{33}-\eta _{11}}{r}+\frac{d\eta _{33}}{dr}\right) \,dr\\ {}&=-Pu_{r}|_{r=a}+\int _a^b {\overline{\sigma }}_{33}\left( \frac{\eta _{33}-\eta _{11}}{r}+\frac{d\eta _{33}}{dr}\right) \,dr, \end{aligned} \end{aligned}$$

(A.10)

where use has been made of the boundary conditions ${\overline{\sigma }}_{33}|_{r=a}=-P$ and ${\overline{\sigma }}_{33}|_{r=b}=0$. In view of incompressibility constraint and the fact that $\eta _{22}$ is constant since the perturbed configuration is uniform, we deduce that

$$\begin{aligned} \begin{aligned} \frac{\eta _{33}-\eta _{11}}{r}+\frac{d\eta _{33}}{dr}&=\frac{\eta _{33}-\eta _{11}}{r}-\frac{d\eta _{11}}{dr}=\frac{u_r-u/r}{r}-\frac{d}{dr}\left( \frac{u}{r}\right) =0. \end{aligned} \end{aligned}$$

(A.11)

Putting these together, we see that the linearized form of (A.1) can be written as

$$\begin{aligned} \int _a^b \frac{\chi _{33}-\chi _{11}}{r}\,dr-Pu_r|_{r=a}=0, \end{aligned}$$

(A.12)

which agrees with (2.42) (note that $\chi _{32}=0$ for uniform inflations).

Equation (A.2) applied to the perturbed configuration can be written as

$$\begin{aligned} \int _{{\tilde{a}}}^{{\tilde{b}}}\sigma _{22}{\tilde{r}}\,d{\tilde{r}}-\frac{1}{2}{\tilde{a}}^2P-\frac{N}{2\pi }=0. \end{aligned}$$

(A.13)

Using (A.3) and (A.5), we can rewrite the above equation as

$$\begin{aligned} \int _a^b ({\overline{\sigma }}_{22}+\chi _{22})r\,dr-\frac{1}{2}{\tilde{a}}^2P-\frac{N}{2\pi }=0. \end{aligned}$$

(A.14)

Its linearized form is

$$\begin{aligned} \int _a^b \chi _{22}r\,dr-Pau|_{r=a}=0, \end{aligned}$$

(A.15)

which is the same as (2.43).

Now let ${\tilde{\lambda }}_a$ and ${\tilde{\lambda }}_z$ be the two principal stretches of the perturbed configuration, thus

$$\begin{aligned} {\tilde{\lambda }}_a&=\lambda _a+d\lambda _a=\lambda _a+\alpha ^2\left( c_1\lambda _a+\frac{c_2}{\lambda _a A^2}\right) ,\end{aligned}$$

(A.16)

$$\begin{aligned} {\tilde{\lambda }}_z&=\lambda _z+d\lambda _z=\lambda _z-2\alpha ^2 \lambda _z c_1. \end{aligned}$$

(A.17)

Then equation (A.1) applied to the unperturbed and perturbed configurations takes the form $-P^*(\lambda _a,\lambda _z)+P=0$ and $-P^*({\tilde{\lambda }}_a,{\tilde{\lambda }}_z)+P=0$, respectively. Subtraction of these two equalities yields

$$\begin{aligned} -P^*({\tilde{\lambda }}_a,{\tilde{\lambda }}_z)+P^*(\lambda _a,\lambda _z)=0. \end{aligned}$$

(A.18)

In a similar way, we can deduce from (A.2) that

$$\begin{aligned} \frac{1}{2}{\tilde{b}}^2(P^*({\tilde{\lambda }}_a,{\tilde{\lambda }}_z)-P^*(\lambda _a,\lambda _z))+\frac{1}{2\pi }(N^*({\tilde{\lambda }}_a,{\tilde{\lambda }}_z)-N^*(\lambda _a,\lambda _z))=0, \end{aligned}$$

(A.19)

where ${\tilde{b}}={\tilde{r}}(B)$ is the outer radius of the tube after perturbation. Linearizing the above two equations at $(\lambda _a,\lambda _z)$ followed by the use of (A.16) and (A.17), we obtain

$$\begin{aligned}&\left( 2\lambda _z\frac{\partial P^*}{\partial \lambda _z}-\lambda _a\frac{\partial P^*}{\partial \lambda _a}\right) c_1-\frac{1}{\lambda _aA^2}\frac{\partial P^*}{\partial \lambda _a}c_2=0,\end{aligned}$$

(A.20)

$$\begin{aligned}&\Big (\frac{\lambda _a}{2\pi }\frac{\partial N^*}{\partial \lambda _a}+\frac{\lambda _ab^2}{2}\frac{\partial P^*}{\partial \lambda _a}-\frac{\lambda _z}{\pi }\frac{\partial N^*}{\partial \lambda _z}-\lambda _zb^2\frac{\partial P^*}{\partial \lambda _z}\Big )c_1+\Big (\frac{1}{2\pi \lambda _a A^2}\frac{\partial N^*}{\partial \lambda _a}+\frac{b^2}{2\lambda _a A^2}\frac{\partial P^*}{\partial \lambda _a}\Big )c_2=0, \end{aligned}$$

(A.21)

Comparing them with (2.39) and (2.44), we conclude that

$$\begin{aligned} \begin{aligned} m_{11}&=2\lambda _z\frac{\partial P^*}{\partial \lambda _z}-\lambda _a\frac{\partial P^*}{\partial \lambda _a},\\ m_{12}&=-\frac{1}{\lambda _aA^2}\frac{\partial P^*}{\partial \lambda _a},\\ m_{21}&=\frac{\lambda _a}{2\pi }\frac{\partial N^*}{\partial \lambda _a}+\frac{\lambda _ab^2}{2}\frac{\partial P^*}{\partial \lambda _a}-\frac{\lambda _z}{\pi }\frac{\partial N^*}{\partial \lambda _z}-\lambda _zb^2\frac{\partial P^*}{\partial \lambda _z},\\ m_{22}&=\frac{1}{2\pi \lambda _a A^2}\frac{\partial N^*}{\partial \lambda _a}+\frac{b^2}{2\lambda _a A^2}\frac{\partial P^*}{\partial \lambda _a}. \end{aligned} \end{aligned}$$

(A.22)

In view of (2.49), it follows from (A.22) that $\Omega (\lambda _a,\lambda _z)$ can be expressed as

$$\begin{aligned} \Omega (\lambda _a,\lambda _z)=-\frac{\lambda _z}{\pi \lambda _a A^2}\left( \frac{\partial P^*}{\partial \lambda _a}\frac{\partial N^*}{\partial \lambda _z}-\frac{\partial P^*}{\partial \lambda _z}\frac{\partial N^*}{\partial \lambda _a}\right) , \end{aligned}$$

(A.23)

which completes the proof.