Scaling Instability in Buckling of Axially Compressed Cylindrical Shells

Abstract

In this paper, we continue the development of mathematically rigorous theory of “near-flip” buckling of slender bodies of arbitrary geometry, based on hyperelasticity. In order to showcase the capabilities of this theory, we apply it to buckling of axially compressed circular cylindrical shells. The theory confirms the classical formula for the buckling load, whereby the perfect structure buckles at the stress that scales as the first power of shell’s thickness. However, in the case of imperfections of load, the theory predicts scaling instability of the buckling stress. Depending on the type of load imperfections, buckling may occur at stresses that scale as thickness to the power 1.5 or 1.25, corresponding to the lower and upper ends, respectively, of the historically accumulated experimental data.

Appendix: Nonlinear Trivial Branch for a Neo-Hookean Material

We note that in order to derive the asymptotics of the critical load in the presence of torsion, only the explicit linear trivial branch (4.14) is needed. However, for the result to be legitimate, we also need to know the existence of the regular trivial branch, and not its explicit representation. At the moment, we lack tools for establishing existence of regular trivial branches, nor can we exhibit one entirely explicitly for a material whose energy density satisfies our regularity assumptions. Here we derive an explicit form of the regular trivial branch for an incompressible neo-Hookean material in the way of providing “evidence” for our assumption of existence of the regular trivial branch for compressible materials.

The strain energy density function for a neo-Hookean solid has the form

$$\begin{aligned} W(\varvec{F})=\frac{E}{6}(|\varvec{F}|^{2}-3),\qquad \det \varvec{F}=1. \end{aligned}$$

(6.1)

We are looking for a trivial branch in a cylindrical shell, given in cylindrical coordinates by

$$\begin{aligned} y_{r}=\psi (r)\cos (\alpha z),\qquad y_{\theta }=\psi (r)\sin (\alpha z),\qquad y_{z}=(1-\lambda )z, \end{aligned}$$

(6.2)

where $\psi (r)$ also depends on $\alpha$, $\lambda$, and h. When $\alpha =0$, we expect that $\psi (r)$ will reduce to $(a(\lambda )+1)r$, as in (3.4). We remark that the ansatz (6.2) should also work for isotropic compressible materials with strain energy density

$$\begin{aligned} W(\varvec{F})=\displaystyle \frac{1}{2}|\varvec{F}|^{2}+H(\det \varvec{F}),\quad H'(1)=-1,\quad H''(1)>\displaystyle \frac{1}{3}, \end{aligned}$$

except the resulting nonlinear second-order boundary value problem for $\psi (r)$

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi ''+\dfrac{\psi '}{r}-\left( \alpha ^{2}+\displaystyle \frac{1}{r^{2}}\right) \psi =(1-\lambda )\dfrac{\psi }{r} \dfrac{d}{dr}\left[ H'\left( (1-\lambda )\dfrac{\psi \psi '}{r}\right) \right] ,&{}r\in I_{h},\\ \psi '+(1-\lambda )\dfrac{\psi }{r}H'\left( (1-\lambda )\dfrac{\psi \psi '}{r}\right) =0,&{}r=1 \pm \dfrac{h}{2} \end{array}\right. } \end{aligned}$$

(6.3)

cannot be solved explicitly.

Returning to the neo-Hookean solid (6.1), we must have

$$\begin{aligned} \det (\nabla \varvec{y})=(1-\lambda )\psi '(r)\frac{\psi (r)}{r}=1, \end{aligned}$$

and hence,

$$\begin{aligned} \psi (r)=\sqrt{\frac{r^{2}}{1-\lambda }+\beta } \end{aligned}$$

(6.4)

for some $\beta >-1$.

The Piola–Kirchhoff stress function is

$$\begin{aligned} \varvec{P}(\varvec{F})=\frac{E}{3}\left( \varvec{F}-\frac{3\hat{p}}{E}\mathrm {cof}(\varvec{F})\right) , \end{aligned}$$

where the Lagrange multiplier $\hat{p}$ plays the role of pressure. For $\varvec{y}$, given by (6.2) and $\varvec{F}=\nabla \varvec{y}$, we compute

$$\begin{aligned} \varvec{F}^{T}\varvec{F}= \begin{bmatrix} (\psi '(r))^{2}&0&0\\ 0&\frac{\psi (r)^{2}}{r^{2}}&\frac{\alpha \psi (r)^{2}}{r}\\ 0&\frac{\alpha \psi (r)^{2}}{r}&\alpha ^{2}\psi (r)^{2} + (1-\lambda )^{2} \end{bmatrix}. \end{aligned}$$

The traction-free condition $\varvec{P}\varvec{e}_{r}=\varvec{0}$ on $r=1\pm h/2$ can be written as

$$\begin{aligned} \varvec{F}^{T}\varvec{F}\varvec{e}_{r}=p\varvec{e}_{r},\quad r=1\pm \frac{h}{2},\qquad p=3\hat{p}/E. \end{aligned}$$

The formula for $\varvec{F}^{T}\varvec{F}$, together with $\det \varvec{F}=1$, implies that

$$\begin{aligned} p(r,\theta ,z)=(\psi '(r))^{2},\quad r=1\pm \frac{h}{2}. \end{aligned}$$

(6.5)

This suggests that it is reasonable to look for the trivial branch for which the function $p(r,\theta ,z)$ depends only on r. Under this assumption, we compute

$$\begin{aligned} \frac{3}{E}\varvec{P}= \begin{bmatrix} s_{1}(r)\cos (\alpha z)&\quad -s_{2}(r)\sin (\alpha z)&\quad -s_{3}(r)\sin (\alpha z)\\ s_{1}(r)\sin (\alpha z)&\quad s_{2}(r)\cos (\alpha z)&\quad s_{3}(r)\cos (\alpha z)\\ 0&\quad q_{1}(r)&\quad q_{2}(r) \end{bmatrix}, \end{aligned}$$

where

$$\begin{aligned} s_{1}=\psi '-\frac{p}{\psi '},\quad s_{2}=\frac{\psi }{r}-\frac{rp}{\psi },\quad s_{3}=\alpha \psi ,\quad q_{1}=\frac{\alpha rp}{1-\lambda },\quad q_{2}=1-\lambda -\frac{p}{1-\lambda }. \end{aligned}$$

It follows that $\nabla \cdot \varvec{P}=\varvec{0}$ results in a single ODE for p(r):

$$\begin{aligned} (rs_{1})'=s_{2}+\alpha rs_{3}. \end{aligned}$$

(6.6)

Substituting (6.4) for $\psi (r)$ into (6.6) and solving for p(r), we obtain

$$\begin{aligned} p(r)=\displaystyle \frac{1}{2(1-\lambda )}\left( \ln \left( \frac{1}{1-\lambda }+\frac{\beta }{r^{2}}\right) -r^{2}\alpha ^{2} -\frac{\beta (1-\lambda )}{r^{2}+\beta (1-\lambda )}+\gamma \right) \end{aligned}$$

for some constant of integration $\gamma$. The traction-free boundary conditions (6.5) become

$$\begin{aligned} \frac{r^{2}}{r^{2}+\beta (1-\lambda )}=\ln \left( \frac{1}{1-\lambda }+\frac{\beta }{r^{2}}\right) -r^{2}\alpha ^{2} +\gamma -1,\quad r=1\pm \frac{h}{2}. \end{aligned}$$

(6.7)

To simplify notation, we denote

$$\begin{aligned} \Phi (r;\lambda ,\beta )=\ln \left( \frac{1}{1-\lambda }+\frac{\beta }{r^{2}}\right) - \frac{r^{2}}{r^{2}+\beta (1-\lambda )}. \end{aligned}$$

Then (6.7) can be written as

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha ^{2}\left( 1+\frac{h}{2}\right) ^{2}=\Phi \left( 1+\frac{h}{2};\lambda ,\beta \right) +\gamma -1,\\ \alpha ^{2}\left( 1-\frac{h}{2}\right) ^{2}=\Phi \left( 1-\frac{h}{2};\lambda ,\beta \right) +\gamma -1 \end{array}\right. } \end{aligned}$$

(6.8)

Eliminating $\gamma$ from (6.8), we obtain

$$\begin{aligned} \alpha ^{2}=\displaystyle \frac{1}{2h}\left( \Phi \left( 1+\frac{h}{2};\lambda ,\beta \right) -\Phi \left( 1-\frac{h}{2};\lambda ,\beta \right) \right) . \end{aligned}$$

when h is small

$$\begin{aligned} \alpha ^{2}\approx \displaystyle \frac{1}{2}\Phi '(1;\lambda ,\beta )=-\frac{\beta (1-\lambda )(2+\beta (1-\lambda ))}{(1+\beta (1-\lambda ))^{2}}. \end{aligned}$$

Thus, when $(h,\lambda )\rightarrow (0,0)$, $\beta \approx -\alpha ^{2}/2$. We conclude that $\alpha$, and, therefore, $\beta$ must go to zero, as $\lambda \rightarrow 0$, since otherwise the trivial branch $\varvec{y}(\varvec{x};h,\lambda )$, given by (6.2), (6.4) will not emanate from the undeformed configuration. The regularity of the trivial branch in $\lambda$ demands that $\alpha (h,\lambda )\sim \alpha _{0}(h)\lambda$, as $\lambda \rightarrow 0$. Thus, for an arbitrary fixed parameter $\beta _{0}>0$, we set¹¹ $\beta =-\beta _{0}^{2}\lambda ^{2}/2$, resulting in the explicit expression for the parameter $\alpha$:

$$\begin{aligned} \alpha (\lambda ,h)=\sqrt{\frac{\Phi (1+h/2;\lambda ,-\beta _{0}^{2}\lambda ^{2}/2) -\Phi (1-h/2;\lambda ,-\beta _{0}^{2}\lambda ^{2}/2)}{2h}}. \end{aligned}$$

(6.9)

Hence, the nonlinear trivial branch has the form

$$\begin{aligned} y_{r}= & {} \sqrt{\frac{r^{2}}{1-\lambda }-\frac{\beta _{0}^{2}\lambda ^{2}}{2}}\cos (\alpha (\lambda ,h)z),\nonumber \\ y_{\theta }= & {} \sqrt{\frac{r^{2}}{1-\lambda }-\frac{\beta _{0}^{2}\lambda ^{2}}{2}}\sin (\alpha (\lambda ,h)z),\quad y_{z}=(1-\lambda )z, \end{aligned}$$

(6.10)

where $\beta _{0}>0$ is a constant and $\alpha (\lambda ,h)$ is given by (6.9). We compute

$$\begin{aligned} \left. \displaystyle \frac{\partial \alpha }{\partial \lambda }\right| _{\lambda =0}=\frac{4\beta _{0}}{4-h^{2}},\qquad \left. \displaystyle \frac{\partial \psi }{\partial \lambda }\right| _{\lambda =0}=\frac{r}{2}. \end{aligned}$$

Therefore, the linearized trivial branch displacement $\varvec{u}^{h}$ is given by

$$\begin{aligned} u^{h}_{r}=\left. \displaystyle \frac{\partial y_{r}}{\partial \lambda }\right| _{\lambda =0}=\frac{r}{2},\qquad u^{h}_{\theta }=\left. \displaystyle \frac{\partial y_{\theta }}{\partial \lambda }\right| _{\lambda =0}=\frac{4\beta _{0}rz}{4-h^{2}},\qquad u^{h}_{z}=\left. \displaystyle \frac{\partial y_{z}}{\partial \lambda }\right| _{\lambda =0}=-z. \end{aligned}$$

The corresponding linear stress and its $h\rightarrow 0$ limit are

$$\begin{aligned} \varvec{\sigma }_{h}=E \begin{bmatrix} 0&0&0\\ 0&0&\frac{4\beta _{0}r}{3(4-h^{2})}\\ 0&\frac{4\beta _{0}r}{3(4-h^{2})}&-1 \end{bmatrix},\qquad \varvec{\sigma }^{0}=E \begin{bmatrix} 0&0&0\\ 0&0&\frac{\beta _{0}}{3}\\ 0&\frac{\beta _{0}}{3}&-1 \end{bmatrix}. \end{aligned}$$

These agree with formulas (4.14), (4.16) for $\nu =1/2$.