Fracture path in brittle thin sheets: a unifying review on tearing

Abstract

We review several studies of crack path in brittle thin sheets where large out of plane bending is involved. Fracture path are observed to be very reproducible. We present a unifying framework based on an energetic point of view. A simplified description, where the sheet is considered to behave as an inextensible fabric, captures important features of experiments: the fact that fracture path seems to obey geometry. We quantify the possible effects of additional bending and stretching terms, and estimate the validity of the model.

Appendix

1.1 Peeling an elastica

We consider an inextensible rod, with a tangent having an angle $\theta$ at the point of curvilinear abscissa $s$, subject to a horizontal force $F$ (see Fig. 23). The torque balance on an element with size $ds$ reads $dM/ds + F \sin \theta =0$, where the constitutive relation $M=Bd\theta /ds$ can be used. Here the bending rigidity is $B=Et^3/12(1-\nu ^{2})$, where $t$ is the thickness of the sheet, $E$ its Young modulus, and $\nu$ Poisson’s ratio.

Fig. 23

A portion of the shape of a peeled elastica with an angle $\phi =\pi$ (left) leads to an angle $\phi$ when rotated (right)

Finally, the elastica equation is Landau and Lifshitz (1967), Love (1944)

$$\begin{aligned} B \ddot{\theta }+\frac{F}{w} \sin \theta =0 \end{aligned}$$

with the boundary conditions $\theta (0)=0$, and a force $F$, but no torque applied at $s=L$, $\dot{\theta }(L)=0$. We can expect the flap to be curved only on a localized region near the clamped condition $s=0$. What is the size of this region? Dimensional analysis directly shows that the only length-scale left in the problem is $\sqrt{Bw/F},$ so that the flap shapes for different loading and rigidity will all be similar, up to a simple scaling factor, as long as they are long compared to this radius of curvature, $L\gg \sqrt{Bw/F}.$ The elastic energy per unit width which only depends on $B$ and $F/w$ can only be written as $E_{el}/w = a \sqrt{FB/w}$.

These results are also found by estimating the radius of curvature of the fold $R$ from a torque balance. The torque $Bw/R \sim FR$ is produced by force $F$ with a lever arm of the order of $R$. Because $1/R \sim \sqrt{ F/Bw}$, we also find that $E_{el} \sim Bw/R \sim \sqrt{BFw}.$ We also note that the bending energy density scales like $F/w.$

In fact these quick arguments can be made exactly because an explicit solution is available in the case where $L=\infty$: we first normalize all distances by the typical length $\sqrt{Bw/F}$ and find $\ddot{\theta }+ \sin \theta =0$. Here we look for the solution where with the condition $\theta (0)=0, \theta (\infty )=\pi , \dot{\theta }(\infty )=0.$ These solutions are the same as the 2D meniscus of a liquid under gravity and surface tension (Roman et al.).

A first integral of this equation gives $\dot{\theta }^2 /2 = 1+\cos \theta$, using the boundary conditions at $s=\infty$. If we keep $\dot{\theta }>0$, this can be rewritten into $\dot{\theta }= 2\cos \theta /2,$ which can be integrated into

$$\begin{aligned} \sin (\theta /2) = \tanh (s). \end{aligned}$$

This implicit solution with $s \in [0,\infty ]$ corresponds to a peeling angle $\phi =\pi$. But for a different peeling angle $\phi$, the solution is simply a rotated portion of the same solution $s \in [s_0, \infty ]$, where $\tanh (s_0)= \sin (\pi /2-\phi /2)=\cos (\phi /2)$, as seen in Fig. 23.

We compute the nondimensional elastic energy using these solutions:

$$\begin{aligned} E_{el}/\sqrt{FBw}&= \int \limits _{s_0}^\infty \dot{\theta }^2/2 ds = 2 \int \limits _{s_0}^\infty \cos ^2(\theta /2)\\ ds&= 2 [\tanh (s)]_{s_0}^\infty = 2[1-\cos (\phi /2)]. \end{aligned}$$

Finally we obtain

$$\begin{aligned} E_{el} = 2 \sqrt{FBw}[1-\cos (\phi /2)] \end{aligned}$$

(25)

Another estimate gives in $\int \dot{\theta }^2 /2 ds = \int (1+\cos \theta ) ds = l-\delta$, where $l$ and $\delta$ are the distances on Fig. 23. In dimensional terms, we find

$$\begin{aligned} E_{el} = F(l-\delta ), \end{aligned}$$

(26)

which shows that

$$\begin{aligned} l-\delta = 2\sqrt{Bw/F} [1-\cos (\phi /2)] \end{aligned}$$

(27)

and

$$\begin{aligned} E_{el}= \frac{4 Bw}{(l-\delta )}[1-\cos (\phi /2) ]^2 \end{aligned}$$

(28)

Yet another interesting quantity is based on direct integration, which shows that $h= \int \sin \theta = -[\dot{\theta }]_{s_0}^\infty = 2\cos (\theta (0))=2\sin (\phi /2)$. In dimensional form, this means that

$$\begin{aligned} h = 2 \sqrt{Bw/F}\sin (\phi /2) = (l-\delta )\frac{\sin (\phi /2)}{1-\cos (\phi /2)} \end{aligned}$$

(29)

and to the elastic energy

$$\begin{aligned} E_{el} = \frac{ 4Bw}{h} [1-\cos (\phi /2)] \sin (\phi /2) \end{aligned}$$

(30)

1.2 Why does the crack loose memory (almost) instantaneously?

In the pulling configuration of pulling on an adhering sheet (Fig. 16), the past history of the crack only enters the problem through the shape of the flap. We consider that the flap continues to have a cylindrical shape invariant in the $z$ direction. The elastic energy reads

$$\begin{aligned} E_{el}= \frac{B}{2}\int \limits _{0}^{\infty } w(u) \kappa ^2(u) du \end{aligned}$$

where $u$ is the curvilinear abscissa along the fold, and the function $\kappa (.)$ is the curvature of the fold, an universal function that depends on $(l-\delta )^{-1}$.  As the cracks propagate by $\delta s$, this energy varies for two reasons: the profile $w(l)$ is modified because the origin of the fold has advanced by $\delta l$, and the curvature profile is modified (because $l-\delta$ has changed).

$$\begin{aligned} \delta E_{el}&= \frac{B}{2}\int \limits _{0}^{\infty } [ w(u+\delta l)-w(u)]\kappa ^2(u) du \\&\quad + \frac{B}{2}\int \limits _{0}^{\infty } w(u) \delta [\kappa ^2(u)] du \end{aligned}$$

The key point is that the curvature profile is localized on a small region with size $r$ comparable to $l-\delta .$ If we assume that on this small lengthscale, $w(u+\delta l)-w(u)$ can be replaced by $\delta l (dw/du)_{u=0}$ and $w(u)\sim w(0)$, we get

$$\begin{aligned} \delta E_{el}&= \frac{dw}{du}\Bigg )_{u=0} \!\! \!\! \delta l \frac{B}{2} \int \limits _{0}^{\infty } \kappa ^2(u) du\\&\quad + w(0) \frac{B}{2}\delta \left[ \int \limits _{0}^{\infty } \kappa ^2(u)du \right] \end{aligned}$$

In the first term we recognize the elastic energy of a slice of fold with unit width, multiplied by the the variation $\delta w$. Because of the invariance of the fold in direction $z$, this is exactly

$$\begin{aligned} d w \frac{\partial E_{el} }{\partial w } \Bigg )_{(l-\delta )} \end{aligned}$$

whereas the second term is in fact a derivative where the width $w=w(0)$ is held constant:

$$\begin{aligned} \frac{\partial E_{el} }{\partial (l-\delta )} \Bigg )_{w} d (l-\delta ) \end{aligned}$$

so that we recover the equations of Sect. 4.1

$$\begin{aligned} dE_{el}= \frac{\partial E_{el} }{\partial (l-\delta )} \Bigg )_{w} d (l-\delta ) + d w \frac{\partial E_{el} }{\partial w } \Bigg )_{(l-\delta )}. \end{aligned}$$

When inserted in Griffith’s criterion, all the quantities depend on $w$ and $dw/ds$, so that finally the equation of evolution of the width can only be a first order equation of the type $dw/ds = \mathcal{F} (w)$: the evolution of the inter-crack distance $w$ only depends on its actual value, not on the past.