Strain tensor selection and the elastic theory of incompatible thin sheets

The existing theory of incompatible elastic sheets uses the deviation of the surface metric from a reference metric to define the strain tensor [Efrati et al., J. Mech. Phys. Solids 57, 762 (2009)]. For a class of simple axisymmetric problems we examine an alternative formulation, defining the strain based on deviations of distances (rather than distances squared) from their rest values. While the two formulations converge in the limit of small slopes and in the limit of an incompressible sheet, for other cases they are found not to be equivalent. The alternative formulation offers several features which are absent in the existing theory. (a) In the case of planar deformations of flat incompatible sheets, it yields linear, exactly solvable, equations of equilibrium. (b) When reduced to uniaxial (one-dimensional) deformations, it coincides with the theory of extensible elastica; in particular, for a uniaxially bent sheet it yields an unstrained cylindrical configuration. (c) It gives a simple criterion determining whether an isometric immersion of an incompatible sheet is at mechanical equilibrium with respect to normal forces. For a reference metric of constant positive Gaussian curvature, a spherical cap is found to satisfy this criterion except in an arbitrarily narrow boundary layer.

Figure 1

(a) Deformation of an infinitesimal element in the radial direction [side view $\left(\widehat{\mathbf{r}},\widehat{\mathbf{z}}\right)$ plane]. The relaxed length of the element is $dr$ (solid line), and the deformed length is $|{\partial }_r\mathbf{f}|dr$. The radial strain component is ${\epsilon }_{rr}=\frac{|{\partial }_r\mathbf{f}|dr-dr}{dr}$, as given by Eq. (11a). The angle ${\varphi }^r$ satisfies $sin{\varphi }^r={\partial }_r\mathbf{f}·\widehat{\mathbf{z}}/\left|{\partial }_r\mathbf{f}\right|$. Substituting $\mathbf{f}(r,\theta )$ from Eq. (4) in the latter relation and using Eq. (13b) gives ${\varphi }_{\theta \theta }=sin{\varphi }^r/\mathrm{\Phi }$. In addition, by direct differentiation it can be verified that ${\varphi }_{rr}={\partial }_r{\varphi }^r$ as given by Eq. (13a). (b) Deformation of an infinitesimal sheet element in the azimuthal direction [top view $\left(\widehat{\mathbf{r}},\widehat{\theta }\right)$ plane]. The relaxed length in this direction is $\mathrm{\Phi }d\theta$ (solid line) and the deformed length is $|{\partial }_{\theta }\mathbf{f}|d\theta$ (dashed line). Thus, the azimuthal strain is ${\epsilon }_{\theta \theta }=\frac{|{\partial }_{\theta }\mathbf{f}|d\theta -\mathrm{\Phi }d\theta }{\mathrm{\Phi }d\theta }$, as given by Eq. (11b). (c) Deformation of an infinitesimal line element in the radial direction at height $x_3$ below the midsurface. By geometry, the shown angle $d\vartheta =(1+{\epsilon }_{rr})dr/R=(1+{\epsilon }_{rr}^{☆})dr/(R-x_3)$. Using $1/R={(1+{\epsilon }_{rr})}^{-1}d{\varphi }^r/dr$ and solving for ${\epsilon }_{rr}^{☆}$ gives Eq. (12a).

Figure 2

A flat thin sheet is deformed into a cylinder of constant radius without stretching of its midplane. This deformation is obtained for the extensible elastica by applying bending moments, $M_0$, on the sheet edges or by imposing $d\varphi /ds$ at the boundaries.

Figure 3

Radial force balance on an infinitesimal element of a flat sheet [59], p. 65]. At the point $P$ we have contributions from the two radial stresses, ${\left({\sigma }_{rr}\right)}_1\mathrm{\Phi }d\theta$ and $-{\left({\sigma }_{rr}\right)}_3\mathrm{\Phi }d\theta$, and from the two azimuthal stresses $-{\left({\sigma }_{\theta \theta }\right)}_{2}drsin(d\theta /2)$ and $-{\left({\sigma }_{\theta \theta }\right)}_{4}drsin(d\theta /2)$. Balancing these terms gives Eq. (31).

Figure 4

Layouts of the three considered reference metrics. (a) Flat metric, Eq. (34). When the two radii (dash-dotted red lines) are held together, the rest length of concentric circles on the closed disk becomes $2\pi \alpha r<2\pi r$. (b) Elliptic metric, Eq. (40). Gluing together the two curved dash-dotted red lines creates a frustrated disk, where concentric circles have rest length of $2\pi \mathrm{\Phi }\left(r\right)<2\pi r$. (c) Hyperbolic metric, Eq. (47). In this panel dashing represents unseen lines; concentric circles have rest length $2\pi \mathrm{\Phi }\left(r\right)>2\pi r$, causing pieces of the disk to be placed in the relaxed configuration one over the other (marked in blue). Attaching together the lower (hidden) red-dashed line with the upper solid red line results in a disk with a hyperbolic metric.

Figure 5

Comparison between the exact solution for the radial displacement [Eq. (39a); black, solid line] and the numerical solution of Eq. (10) in Ref. [13] (dashed, blue line) for a flat reference metric. We consider an annulus with inner and outer radii $R_i=0.1$ and $R_o=1.1$. In accordance with the example in Ref. [13], we use $\nu =0$.

Figure 6

Comparison between the exact plane-stress solutions [Eqs. (39), black solid line] and the numerical solution of Eq. (10) in Ref. [13] (dashed blue line) for a flat reference metric. Parameters are as in Fig. 5.

Figure 7

Exact solution for the radial displacement [Eqs. (42) and (45); black solid lines], plotted alongside the numerical solution of Eq. (10) in Ref. [13] (dashed blue lines) for an elliptic reference metric. Displacement and radial position are both normalized by the reference Gaussian curvature. Results are presented for an annulus of normalized inner radius $\sqrt{K}{R}_i=0.1, \nu =0$, and two different values of $\sqrt{K}{R}_o$ as indicated.

Figure 8

Comparison between the exact plane-stress solutions, Eqs. (46) (black solid lines), and numerical results based on Ref. [13] (dashed blue lines) for an elliptic reference metric. Parameters are as in Fig. 7.

Figure 9

Exact solution for the radial displacement (black solid lines), plotted alongside the numerical solution of Eq. (10) in Ref. [13] (dashed blue lines) for a hyperbolic reference metric. Displacement and radial position are both normalized by the reference Gaussian curvature. Results are presented for an annulus of normalized inner radius $\sqrt{K}{R}_i=0.1, \nu =0$, and two different values of $\sqrt{K}{R}_o$ as indicated.

Figure 10

Exact radial and azimuthal plane-stress solutions for a flat annulus with a hyperbolic reference metric (solid black lines) are compared with the numerical solution of Eq. (10) in Ref. [13] (dashed blue lines). Parameters are as in Fig. 9.

Figure 11

(a) Schematic force balance on a finite sheet segment. A bending moment $M_o$, applied at the boundary, is balanced by the reaction forces ${\sigma }_{ss}$ and ${\sigma }_{sn}$ and the reaction bending moment $M_{ss}$. Under these conditions ${\sigma }_{ss}={\sigma }_{sn}=0$ and $M_{ss}=M_o$, consistent with Eqs. (A10). (b) Schematic force balance on an infinitesimal sheet segment of length $d\widehat{s}$. Balance of forces in the tangential direction $\widehat{\mathbf{t}}\left(\widehat{s}\right)$ is given by $-{\sigma }_{ss}\left(\widehat{s}\right)+{\sigma }_{ss}(\widehat{s}+d\widehat{s})\widehat{\mathbf{t}}(\widehat{s}+d\widehat{s})·\widehat{\mathbf{t}}\left(\widehat{s}\right)+{\sigma }_{sn}(\widehat{s}+d\widehat{s})\widehat{\mathbf{n}}(\widehat{s}+d\widehat{s})·\widehat{\mathbf{t}}\left(\widehat{s}\right)=0$. Expanding this equation to leading order in the differential $d\widehat{s}$ [using Eqs. (A3)] we obtain $d{\sigma }_{ss}/d\widehat{s}-\kappa {\sigma }_{sn}=0$. Similarly, force balance in the normal direction and balance of bending moments gives $d{\sigma }_{sn}/d\widehat{s}+\kappa {\sigma }_{ss}=0$ and $d{M}_{ss}/d\widehat{s}+{\sigma }_{ns}=0$, consistent with Eqs. (A12).