Stretching Hookean ribbons part I: relative edge extension underlies transverse compression and buckling instability

Abstract

The wrinkle pattern exhibited upon stretching a rectangular sheet has attracted considerable interest in the “extreme mechanics” community. Nevertheless, key aspects of this notable phenomenon remain elusive. Specifically—what is the origin of the compressive stress underlying the instability of the planar state? what is the nature of the ensuing bifurcation? how does the shape evolve from a critical, near-threshold regime to a fully developed pattern of parallel wrinkles that permeate most of the sheet? In this paper we address some of these questions through numerical simulations and analytic study of the planar state in Hookean sheets. We show that transverse compression is a boundary effect, which originates from the relative extension of the clamped edges with respect to the transversely contracted, compression-free bulk of the sheet, and draw analogy between this edge-induced compression and Moffatt vortices in viscous, cavity-driven flow. Next, we address the instability of the planar state and show that it gives rise to a buckling pattern, localized near the clamped edges, which evolves—upon increasing the tensile load—to wrinkles that invade the uncompressed portion of the sheet. Crucially, we show that the key aspects of the process—from the formation of transversely compressed zones, to the consequent instability of the planar state and the emergence of a wrinkle pattern—can be understood within a Hookean framework, where the only origin of nonlinear response is geometric, rather than a non-Hookean stress–strain relation.

Graphic abstract

Solution of planar stress

In this Appendix, we describe the detailed solution of model B with BCs (6e–6h). In our analysis we follow closely the calculation of Benthem [22], who employed Laplace transformation to compute the stress at the clamped edge. Here we go beyond Benthem solution by performing an inverse Laplace transformation, which is required to evaluate the stress in the whole sheet.

1.1 The singularities in the corners

Fig. 8

A corner of a rectangular sheet. We set the corner as the origin of polar coordinates

Let us consider first the vicinity of one corner of the sheet in Fig. 1a, where one edge is clamped and the other one is free, and solve Eq. 4 in polar coordinates $(r, \theta )$, see Fig. 8, which depicts the left bottom corner.

$$\begin{aligned} \left( \frac{\partial ^2}{\partial r^2} + \frac{{1}}{r} \frac{\partial }{\partial r} + \frac{1}{r^2} \frac{\partial ^2}{\partial \theta ^2} \right) ^{ 2} \varPhi = 0 \end{aligned}$$

(16)

BCs Eqs. (5b, 5c) now become

$$\begin{aligned} \text {free boundary}(\theta = 0):&\sigma _{{ r}\theta } = {\sigma _{\theta \theta }}= 0 \ , \end{aligned}$$

(17a)

$$\begin{aligned} \text {clamped boundary}(\theta = \frac{\pi }{2}):&u_r = u_\theta = 0 \ . \end{aligned}$$

(17b)

The general solution of Eq. 16 is

$$\begin{aligned} \begin{aligned} \varPhi (r, \theta ) =&r^{\gamma + 1} ( D_1 \sin (\gamma +1)\theta + D_2 \cos (\gamma +1)\theta \\&+ D_3 \sin (\gamma - 1)\theta + D_4 \cos (\gamma - 1)\theta ) \end{aligned} \end{aligned}$$

(18)

Notice that Eq. (18) has 4 unknowns $D_i$, such that the four BCs (17) give rise to a nonzero solution only if the determinant of the corresponding $4 \times 4$ matrix vanishes, yielding:

$$\begin{aligned} \sin ^2 \left( \gamma \frac{\pi }{2} \right) = \frac{4 - \gamma ^2 (1+\nu )^2}{(3- \nu )(1+\nu )} \ . \end{aligned}$$

(19)

For a sheet with positive Poisson’s ratio, $\nu > 0$, the solution of Eq. (19) is $\gamma < 1$, such that:

$$\begin{aligned} \sigma \sim \frac{\partial \varPhi }{\partial r^2} \sim r^{\gamma - 1} \ , \end{aligned}$$

(20)

and hence the components of stress tensor diverge at the corner (i.e., $r \rightarrow 0$). For convenience, we choose to focus on a specific value, $\gamma = 3/ 4$, which corresponds to $\nu = \frac{17 + 4 \sqrt{2} - 8 \sqrt{7}}{4 \sqrt{2} - 1} \approx 0.32$. For this value, the boundary values of ${\sigma _{\theta \theta }}$ and ${\sigma _{r\theta }}$ are given by [22]:

$$\begin{aligned}&{\sigma _{\theta \theta }}\left( \theta =\frac{\pi }{2}\right) = c_1 r^{-\frac{1}{4}}, \qquad {\sigma _{r\theta }}\left( \theta =\frac{\pi }{2}\right) = -c_2 r^{-\frac{1}{4}} \nonumber \\&c_1 = \sqrt{8 + 3 \sqrt{7}} c_2 \end{aligned}$$

(21)

1.2 Laplace transformation

Let us address now the Airy potential in a semi-infinite sheet (Fig. 9). For convenience, we set the width $W = 2$ (i.e., $y \rightarrow {\tfrac{y}{2W}}, x \rightarrow {\tfrac{x}{2W}}$), and transform the origin of the coordinate system to the middle of left edge, such that the left clamped boundary is $x=0$ and the two free boundaries are $y = \pm 1$.

Fig. 9

The configuration of a semi-infinite sheet ($L\rightarrow \infty$). We set the width $W = 2$ for convenience, the left boundary $x = 0$ is clamped and the two boundaries $y = \pm 1$ are free

To solve Eq. (4), we employ Laplace transformation,

$$\begin{aligned} f(p, y) = \int _0^{\infty } \varPhi (x,y) e^{-s x} dx \ , \end{aligned}$$

(22)

such that the transformed Eq. (4) becomes

$$\begin{aligned} s^4 f + 2 s^2 \frac{\partial ^2 f}{\partial y^2} + \frac{\partial ^4 f}{\partial y^4}= & {} s^3 \varPhi (0,y) + s^2 \frac{\partial \varPhi }{\partial x} (0, y) \nonumber \\&+ s \left( \frac{\partial ^2 \varPhi }{\partial x^2}(0,y) + 2 \frac{\partial ^2 \varPhi }{\partial y^2} (0,y) \right) \nonumber \\&+ \frac{\partial ^2 \varPhi }{\partial x^3}(0,y) + 2 \frac{\partial ^3 \varPhi }{\partial x \partial y^2}(0,y)\qquad \end{aligned}$$

(23)

With the up-down symmetry $\varPhi (x,y) = \varPhi (x,-y)$, the singularities at corners, Eq. (21), and BC (6e), we can assume [22]

$$\begin{aligned} \frac{\partial ^2\varPhi }{\partial y^2} = {{\sigma _{xx}}}(0, y)= & {} \ c_1 \{ (1-y)^{-\frac{1}{4}} + (1+y)^{-\frac{1}{4}} - 2^{-1/4} \} \nonumber \\&+ \sum _{n=1} a_n \cos (q_n y) \nonumber \\ \frac{\partial ^2\varPhi }{\partial x \partial y} = -{{\sigma _{xy}}}(0, y)= & {} c_2 \{ (1-y)^{-\frac{1}{4}} - (1+y)^{-\frac{1}{4}} + y 2^{-1/4} \} \nonumber \\&+ \sum _{n=1} b_n \sin (k_n y) \end{aligned}$$

(24)

where $q_n = (2n-1) \frac{\pi }{2}$, $k_n = n \pi$ and $a_n,\ b_n$ are constants that must be computed. Equation  (24) and the BCs (6g, 6h) enable us to express all boundary terms in the right side of Eq. (23) in terms of the unknowns $c_1,c_2$ and $a_n,b_n$, whereas the BCs (6e, 6f) become:

$$\begin{aligned} f(s, y=\pm 1)&= 0 \end{aligned}$$

(25a)

$$\begin{aligned} \frac{\partial f}{\partial y}(s, y=\pm 1)&= 0 \ . \end{aligned}$$

(25b)

The solution f(s, y) to Eq. (23) can now be expressed in the following form [22]:

$$\begin{aligned}&f(s, y) = g(s,y)\nonumber \\&\quad - 2 g(s,1) \frac{(s \cos s + \sin s) \cos s y + s (\sin s) y \sin s y}{\sin 2s + 2 s} \nonumber \\&\quad + 2 \frac{\partial g}{\partial y}(s, 1) \frac{(\sin s) \cos s y - (\cos s) y \sin s y}{\sin 2 s + 2s} \end{aligned}$$

(26)

where g(s, y) is the particular solution of the non-homogeneous ODE (23), while the left two terms correspond to solutions of the homogeneous equation (i.e., replacing the right side with 0), which enable satisfying the BCs (25). The function g(s, y) is relatively complicated, and in order to express it compactly we define an “auxiliary” function:

$$\begin{aligned} \begin{aligned} R(q; s, y)&= \frac{1}{4 s^{4+2 q}} \Big ( e^{isy} (-i s)^q ( (-1 + i s y) \varGamma (1+q; is) \\&\quad \, + (1-isy) \varGamma (1+q; isy) \\&\quad \, - \varGamma (2+q; is) + \varGamma (2+q; is y) ) \\&\quad \, + e^{-is y} (is)^q ( (-1 - isy) \varGamma (1+q; -is) \\&\quad \, + (1+isy)\varGamma (1+q; -is y) \\&\quad \, - \varGamma (2+q; -is) + \varGamma (2+q; -is y) ) \Big ) \end{aligned} \end{aligned}$$

(27)

where $\varGamma (x; y)$ is the incomplete gamma function. Then g(s, y) can be written as

$$\begin{aligned} \begin{aligned} g(s,y) =&\frac{s^3 c_1}{1.3125} \left( R\left( \frac{7}{4};s, 1-y\right) + R(\frac{7}{4}; s, 1+y) \right) \\&-\frac{4}{3} s^2 c_2 \left( R\left( \frac{3}{4}; s, 1-y\right) + R\left( \frac{3}{4}; s, 1+y\right) \right) \\&+s(2+\nu ) c_1 \left( R\left( -\frac{1}{4}; s, 1-y\right) + R\left( -\frac{1}{4}; s, 1+y\right) \right) \\&-\frac{1}{4} \nu c_2 \left( R\left( -\frac{5}{4}; s, 1-y\right) + R\left( -\frac{5}{4}; s, 1+y\right) \right) \\&+ \left( -s^3 c_1 2^{-\frac{5}{4}} + s^2 c_2 2^{-\frac{5}{4}} \right) \frac{s^2 y^2 - 4}{s^6} \\&+ \frac{c_1}{s} \left( 2^{-\frac{5}{4}} - \frac{2^{\frac{7}{4}}}{1.3125} \right) - \frac{2 + \nu }{s^3} c_1 2^{-\frac{1}{4}} \\&- \frac{\nu }{s^4} c_2 2^{-\frac{1}{4}} + \frac{\alpha }{s^3} + \frac{b_0}{s^2} \\&{+ \sum _{n=1} \frac{- \frac{s^3}{q_n^2} + s (2+\nu )}{(s^2 - q_n^2)^2} a_n \cos q_n y -\frac{ \frac{s^2}{k_n} + \nu k_n}{(s^2 - k_n^2)^2} b_n \cos k_n y }\\ \end{aligned} \end{aligned}$$

(28)

where $b_0$ is an integration constant:

$$\begin{aligned} \frac{\partial \varPhi }{\partial x}(0,y) = \int \frac{\partial ^2 \varPhi }{\partial x \partial y}(0,y) \hbox {d}y + b_0 \ . \end{aligned}$$

(29)

1.3 Solution of Laplace transform and its inversion

Next we must solve all unknowns $a_n, b_n, c_1, b_0$ that define the function g(s, y), Eq. (28), substitute it in Eq. (26) to obtain the Laplace transform f(s, y), and perform the inverse transform

$$\begin{aligned} \varPhi (x,y) = \frac{1}{2 i} \int _{c-i \infty }^{c+ i\infty } f(s,y) e^{sx} ds \qquad (c>0) \end{aligned}$$

(30)

to obtain the Airy potential $\varPhi (x,y)$.

Recalling that our analysis here assumes a semi-infinite sheet, for which we expect the stress potential $\varPhi (x,y)$ vanishes exponentially away from the clamped edge at $x =0$, we note that the Laplace transform f(s, y) must not have any poles at the right half of complex plane $\text {Re}\,(s) > 0$. Inspection of Eqs. (26, 28) reveals that the only poles originate from the contribution of the homogeneous solutions of Eq. (23), namely, complex numbers s which solve the nonlinear equation:

$$\begin{aligned} \sin 2s + 2 s = 0 \ . \end{aligned}$$

(31)

This equation has an infinite sequence of solutions in the right half plane, $\{ s_i \ and \ {\bar{s}}_i; i\in {\mathbb {N}}, \text {Re}\, (s_i)> 0, \text {Im}\, (s_i) > 0\}$, which can be arranged in increasing order, $|s_{i+1}| > |s_i|$. Requiring the residues of f(s, y) at $s_i$ and ${\bar{s}}_i$ to vanish, yields an infinite sequences of algebraic equations

$$\begin{aligned} \text {Res}\, f(s_i, y) = 2 \left( \frac{\partial g}{\partial y}(s_i,1) - g(s_i, 1) \sin ^2 \! s_i \right) G(s_i,y) = 0,\nonumber \\ \end{aligned}$$

(32)

where we used Eqs. (26, 28) and defined

$$\begin{aligned} G(s_i,y) = \frac{(\sin s_i)\cos s_i y - (\cos s_i) y \sin s_i y}{4 \cos ^2 s_i} \ . \end{aligned}$$

(33)

Equations (32, 28), together with the complementary set of conjugate equations for the residues $\text {Res}\, f({\bar{s}}_i, y) = \overline{\text {Res}\, f(s_i, y)}$, determine the unknowns $a_n, b_n, c_1, b_0$ in Eq.( 28), and thereby the Laplace transform f(s, y) through Eq. (26). In order to obtain a numerical solution of this system of equations, we truncate the sequence of unknowns, and keep only the 1000 leading unknowns (ordered by increasing magnitude of the corresponding poles $|s_i|$).

Furthermore, in order to carry out the integration in Eq. (30), we can close the contour in the left half complex plane $\text {Re}\, (s) < 0$ and compute it through the sum of residues of f(s, y). Noticing that the poles of f(s, y) in the complex s plane are solutions of Eq. (31), one readily realizes that the poles in the left half of the complex plane are $\{-s_i,- {\bar{s}}_i \}$, and consequently:

$$\begin{aligned} \begin{aligned} \varPhi (x,y)&=\sum _{i=1} \text {Res} \left( \,(f(s,y)e^{sx}; -s_i) + \text {Res}\,(f(s,y)e^{s x}; -{\bar{s}}_i) \right) \\&= 4 \text {Re} \sum _{i=1} \, \left( g(-s_i, 1) \sin ^2 s_i - \frac{\partial g}{\partial y}(-s_i,1) \right) G(s_i, y)e^{-s_i x} \end{aligned}\nonumber \\ \end{aligned}$$

(34)

Finally, recovering our transformation $y\rightarrow \tfrac{2y}{W},\ x\rightarrow \tfrac{2x}{W}$ and define $p_i = 2 s_i$, we can get Eq. (9b)

$$\begin{aligned} \varPhi (x,y)= & {} \text {Re}\! \left\{ \! \sum _i C_i \!\cdot \! \left( \varPhi ^{i,c}(x,y) + A_i \varPhi ^{i,s}(x,y) \right) \! \right\} , \nonumber \\ C_i= & {} \frac{\sin \frac{p_i}{2}}{\cos ^2 \frac{p_i}{2}} \left( g(-\frac{p_i}{2}, 1) \sin ^2 p_i - \frac{\partial g}{\partial y}(-\frac{p_i}{2},1) \right) \qquad \end{aligned}$$

(35)