Unveiling the Mechanism Behind the Asymmetric Bending Compliance of Thin-Walled U-Shaped Strips: A Study Inspired by Plant Leaves

Abstract

Inspired by the shape of some plant leaves, we find that the thin-walled U-shaped strips exhibit different compliances under bending with opposite orientations. The asymmetric bending compliance is attributed to the buckling of sidewalls of strips caused by the bending-induced compression. Integrating the Euler–Bernoulli beam theory with the Kirchhoff–Love thin plate theory, a theoretical model is derived for the in-depth understanding of the sidewall buckling. For pure bending, the critical moment applied to the strip for the sidewall buckling is found to be insensitive to the height, width and length of strip, which is the result of the compromise between the opposite geometric effects on the buckling behavior of sidewalls and the characteristics of cross sections. Then the critical moment can be approximated as a linear function of flexural rigidity $D = Et^{3} /12\left( {1 - {\nu}^{2} } \right)$, where t is the wall thickness of strip, E is Young’s modulus, and v is Poisson’s ratio. These predictions by our model agree well with the results obtained by finite element analysis. We also investigate the buckling behavior of sidewalls for bending under transverse loads, considering the loading conditions of concentrated force and distributed force. Our study unveils the mechanism behind the asymmetric bending compliance of thin-walled U-shaped strips. These results would offer convenient guidance for the promising engineering applications related to this structure, such as the design of soft robots with enhanced locomotion performance.

Appendix: Theoretical derivations of effective elastic coefficients \({\varvec{k}}_{1}\) and \({\varvec{k}}_{2}\)

A thin-walled U-shaped strip under positive bending can be divided into two parts by the neutral surface. The upper part is regarded as thin plates under compression with lower sides connected to the remaining bottom part. The bottom part is also a U-shaped strip with a cross-section in the size of $h_{2} \times b$, as depicted in Fig. 

Fig. 9

Schematics of the derivations of elastic coefficients ${k}_{1}$ and ${k}_{2}$. For a U-shaped strip, the part below the neutral surface shown in a is modeled as a frame system $\overline{ABC}$ shown in b

9a. Take the unit length for analysis. Since the wall thickness t is much smaller than b and $h_{2}$, the bottom part can be approximated as a two-dimensional frame system $\overline{ABC}$, which is illustrated in Fig. 9b. The bars $\overline{AB}$ and $\overline{BC}$ are firmly jointed at point B. At point A, the displacement along the z-direction and the rotation are prohibited to represent the symmetry of the bottom structure. At point C, the displacement along the y-direction is fixed to hold the static determinacy of the frame system, while no constraints are applied to the movement along the z-direction or the rotation, allowing the deformation of the sidewalls in the upper part.

Then $k_{1}$ can be obtained from the relationship between the applied force F and the resulting displacement $\delta$ at point C along the z-direction, which is written as $k_{1} = F/\delta$. Many methods can be found in the textbooks of structural mechanics for the solution of $k_{1}$. Using the unit load method based on the principle of virtual work, we have

$$ \begin{array}{*{20}c} {\delta = \frac{1}{{ EI^{\prime } }}\left( {\mathop \smallint \limits_{0}^{{h_{2} }} Fy^{2} {\text{d}}y + \mathop \smallint \limits_{0}^{b/2} Fh_{2}^{2} {\text{d}}z} \right)} \\ \end{array} $$

(22)

Substituting $I^{\prime } = t^{3} /12$, $k_{1}$ is calculated as

$$ \begin{aligned} {k_{1} = \frac{{Et^{3} }}{{4h_{2}^{3} + 6h_{2}^{2} b}}} \\ \end{aligned} $$

(23)

Similarly, $k_{2}$ can be obtained as $k_{2} = M/\theta$, where M is the applied moment and $\theta$ is the resulting rotation angle at point C. Using the same method, we have

$$ \begin{array}{*{20}c} {\theta = \frac{1}{{ EI^{\prime } }}\left( {\mathop \smallint \limits_{0}^{{h_{2} }} M{\text{d}}y + \mathop \smallint \limits_{0}^{b/2} M{\text{d}}z} \right)} \\ \end{array} $$

(24)

Then $k_{2}$ is given by

$$ \begin{array}{*{20}c} {k_{2} = \frac{{Et^{3} }}{{12h_{2} + 6b}}} \\ \end{array} $$

(25)