Compliance of a microfibril subjected to shear and normal loads

Abstract

Many synthetic bio-inspired adhesives consist of an array of microfibrils attached to an elastic backing layer, resulting in a tough and compliant structure. The surface region is usually subjected to large and nonlinear deformations during contact with an indenter, leading to a strongly nonlinear response. In order to understand the compliance of the fibrillar regions, we examine the nonlinear deformation of a single fibril subjected to a combination of shear and normal loads. An exact closed-form solution is obtained using elliptic functions. The prediction of our model compares well with the results of an indentation experiment.

1. Introduction

Recent interest in bio-inspired adhesives has motivated many researchers to fabricate microfibril arrays (Liu & Bhushan 2003; Peressadko & Gorb 2004; Chung & Chaudhury 2005; Crosby et al. 2005; Glassmaker et al. 2005, 2007; Huber et al. 2005; Northen & Turner 2005; Yurdumakan et al. 2005; Kim & Sitti 2006; Majidi et al. 2006; Aksak et al. 2007; Gorb et al. 2007; Greiner et al. 2007; Varenberg & Gorb 2007) and to study their contact mechanics and adhesion (Jagota & Bennison 2002; Gao et al. 2003, 2005; Persson & Gorb 2003; Hui et al. 2004; Persson et al. 2005; Spolenak et al. 2005a,b; Tang et al. 2005; Bhushan et al. 2006; Tian et al. 2006; Yao & Gao 2006; Chen & Gao 2007). Most of these studies focus on how the interface between the microfibrils and a smooth, hard substrate separates under a normal load. Of equal importance is how these fibrillar surfaces respond to a combination of normal and shear loads. For example, experiments have demonstrated that the maximum shear force a gecko seta can support is approximately six times greater than its normal pull-off force (Autumn et al. 2000), and direct measurements of how various species adhere to surfaces are conducted under shear (Irschick et al. 1996). However, these observations are often interpreted using theories based on the normal contact of surfaces. Therefore, there is a need to develop contact and adhesion models that take account of shear.

In the past year, there have been several experimental studies on the frictional behaviour of microfibril arrays against a flat substrate (Majidi et al. 2006; Ge et al. 2007). The fibril arrays fabricated by Majidi et al. (2006) and Ge et al. (2007) consist of very stiff fibrils, whereas those fabricated by P. R. Guduru (2007, personal communication) and Shen et al. (2008) are made of poly(dimethylsiloxane) (PDMS), a soft elastomer with a shear modulus of the order of 1 MPa. Despite the large differences in modulus, what emerges from these experiments is that the static friction of these arrays is much higher than that exhibited by flat unstructured controls made of the same material.

The mechanics of a flat elastic substrate indented by a smooth, soft elastic sphere under a normal load is well described by the Johnson–Kendall–Roberts theory (Johnson et al. 1971). A theory for the compliance of microfibril arrays under normal indentation has been developed (Noderer et al. 2007). While there is a strong quantitative influence of the fibrillar architecture on compliance, there are qualitative similarities between it and an unstructured flat control. For example, the load versus contact area curves have similar shapes and compliance generally decreases with increasing contact area (increasing load). The situation can be quite different with shear. Figure 1a shows schematically an experiment in which a film-terminated PDMS microfibril array is moved in shear relative to a fixed spherical indenter. Briefly (see Shen et al. (2008) for details), the microfibril array consists of micropillars oriented normal to an elastic PDMS backing layer. The micropillars are connected at their terminal ends by a thin, flexible film. This structure has been shown to significantly improve adhesion when compared with a flat unstructured control (Glassmaker et al. 2007; Noderer et al. 2007). The backing layer is bonded to a glass slide that is placed on an inverted optical microscope. Since PDMS is transparent, its deformation can be recorded by the microscope. A fixed normal load, F_N, is applied to press the indenter into contact with the sample surface (i.e. the thin film).

Figure 1.

(a) A glass spherical indenter is placed on the surface of a microfibril array under a fixed normal force (P) (applied via a mechanical balance). The sample is translated at a constant rate, u, and the shear force is measured by a load cell. Deformation near the contact region is recorded by means of an inverted optical microscope. The shear load versus shear displacement curve for a fibrillar sample is shown in (b). Three points are selected for comparison with the theory.

The shear force F_s is applied by translating the glass slide at a constant rate. A typical shear force versus shear displacement curve is shown in figure 1b. As the relative shear displacement between the indenter and the sample increases, the shear force increases to a peak value (stage 1). Beyond the peak, it decreases rapidly (stage 2) and then remains nearly constant (stage 3). Visual inspection of the contact region in stage 1 reveals that it changes in shape and size, but there is no macroscopic sliding between the indenter and the sample. In stage 3, the indenter slides steadily on the sample. More detailed explanations of the physics behind these different stages can be found in Shen et al. (2008). Briefly, in stage 1, the fibrils in contact with the indenter are loaded under shear. As shear increases, the elastic energy stored in these fibrils increases. This elastic energy is released suddenly in stage 2 due to the propagation of an interface crack. In stage 3, relative motion between the indenter and the sample appears to be accommodated by the propagation of Schallamach-like waves.

Owing to the applied shear displacement, the top of a typical fibril in a region enclosing the contact zone is displaced relative to the bottom. Figure 2 shows three optical micrographs of the contact region corresponding to the three points in figure 1b. In figure 2, the image of the top end of a fibril appears as a fuzzy grey circle, whereas the bottom end appears as a smaller dark square. This difference allows us to determine the relative deflection, Δ_T, of each fibril. The nearest-neighbour distance between fibrils is 87 μm, while their length is 67 μm. From figure 2c, it can be seen that the lateral displacement of a typical fibril can exceed its length. We show later that the usual small-deflection beam-column model (BCM; Timoshenko & Gere 1961) is not accurate enough to capture the deformation of microfibrils in our experiments. This motivates us to use a nonlinear large-deflection theory.

Figure 2.

Optical micrographs of the contact region. The direction of shear displacement u is indicated by the arrow on the right. The image of the top end of a fibril appears as a fuzzy grey circle, whereas the bottom end appears as a smaller dark square. This difference allows us to determine the relative deflection Δ_T of each fibril. (a–c) correspond to the points A–C in figure 1b, respectively. The contact region Ω_c is the white polygon.

It should be noted that, when the same indentation test is performed on a flat unstructured control PDMS substrate, the contact area is found to decrease monotonically with increasing shear, consistent with a theory proposed by Savkoor & Briggs (1977). In contrast, with increasing shear, the contact area for fibrillar samples initially increases. This difference in behaviour can be explained by the fact that, when a microfibril under shear bends, it significantly increases the normal compliance of the array. Since the normal load is fixed in these experiments, increase in normal compliance initially results in an increase in the contact area.

In order to develop a quantitative understanding of the load-bearing capacity of a fibrillar array under combined normal and shear loads, we study the nonlinear deformation of individual microfibrils. The model for fibril deflection and theoretical results based on it are presented in §2. In §3 we compare model predictions with the experiments.

2. Elastica model of a stretchable beam

We model the deformation of a typical microfibril in stage 1. Since the length of a typical microfibril is significantly greater than its lateral dimensions, it will be modelled as a stretchable elastica. Inside the contact zone, which is denoted by Ω_c, the thin film is well adhered to the indenter. Therefore, a fibril in Ω_c cannot rotate at this end, implying a clamped boundary condition. For a sufficiently long fibril, it is reasonable to assume that its bottom end is also clamped. We further assume that fibrils do not twist; this assumption is consistent with the loading conditions in our experiments. Finally, we neglect the compliance of the half-space to which the fibril is attached; this is again reasonable since fibrils are slender.

The problem of interest is an initially straight elastic beam with an initial length, L₀, that is clamped at both ends. One end of the beam is fixed to a rigid wall (the backing layer) while the other end is subjected to vertical and tangential displacements Δ_N and Δ_T, respectively. Let N₀, T₀, and M₀ denote the unknown normal, shear and moment applied at this end, respectively. Let s denote the arc length of the deformed beam and x(s), y(s) denote the deformed coordinates of the beam at a point ‘p’ on the bar (figure 3).

Figure 3.

The coordinate system used in our analysis. The undeformed beam is straight and has length L₀. Both the ends of the beam are clamped. The right end is subjected to a vertical and tangential displacement Δ_N and Δ_T, respectively. The coordinates of the point ‘p’ are (x(s), y(s)).

Following Frisch-Fay (1962), we assume a linear relation between curvature and moment, i.e.

$$EI\frac{\text{d}\psi }{\text{d}s}=M(s),$$ (2.1a)

where E is the Young modulus; I is the moment of inertia; ψ is the rotation of the deformed bar relative to the x-axis; and M(s) is the moment acting at a generic point ‘p’. (The free-body diagram of the bar is shown in figure 4.)

Figure 4.

The free-body diagram of a section of the deformed bar. The right end of this section is clamped (ψ=0). Δ_T and Δ_N are the shear and normal displacements, respectively, at the right end of the beam. N₀ and T₀ denote the reaction forces, whereas the reaction moment is denoted by M₀.

The relevant boundary conditions are as follows:

$$\psi (s=0)=\psi (s=L)=0,$$ (2.1b)

$$x(s=0)=0y(s=0)=0,$$ (2.1c)

$$x(s=L)=L_0+{\Delta }_{\text{N}},y(s=L)={\Delta }_{\text{T}},$$ (2.1d)

where L denotes the deformed length of the beam. The free-body diagram in figure 4 shows that

$$M(s)=M_0-N_0({\Delta }_{\text{T}}-y(s))+T_0(L_0+{\Delta }_{\text{N}}-x(s)).$$ (2.2)

Substituting (2.2) into (2.1a) gives

$$EI\frac{\text{d}\psi }{\text{d}s}=M(s)=M_0-N_0({\Delta }_{\text{T}}-y(s))+T_0(L_0+{\Delta }_{\text{N}}-x(s)).$$ (2.3)

Let N(s) and T(s) denote the normal (normal to the cross section of the deformed bar) and shear forces along the deformed bar, respectively. Force balance requires that

$$\begin{array}{l}N\text{cos}\psi -T\text{sin}\psi =N_0,\hfill \\ N\text{sin}\psi +T\text{cos}\psi =T_0.\hfill \end{array}\}$$ (2.4)

Equation (2.4) implies that

$$N=N_0\text{cos}\psi +T_0\text{sin}\psi .$$ (2.5)

Note that, by definition,

$$\frac{\text{d}x}{\text{d}s}=\text{cos}\psi ,$$ (2.6a)

$$\frac{\text{d}y}{\text{d}s}=\text{sin}\psi .$$ (2.6b)

Equations (2.6a) and (2.6b) imply that

$$x={\int }_0^s\text{cos}\psi (s^{\prime })\text{d}{s}^{\prime },$$ (2.7a)

$$y={\int }_0^s\text{sin}\psi (s^{\prime })\text{d}{s}^{\prime }.$$ (2.7b)

Combining (2.1d), (2.7a) and (2.7b), we have

$$L_0+{\Delta }_{\text{N}}={\int }_0^L\text{cos}\psi (s^{\prime })\text{d}{s}^{\prime },$$ (2.8a)

$${\Delta }_{\text{T}}={\int }_0^L\text{sin}\psi (s^{\prime })\text{d}{s}^{\prime }.$$ (2.8b)

To eliminate x and y from equation (2.3), we differentiate (2.3) by s and use (2.6a) and (2.6b),

$$EI\frac{{\text{d}}^2\psi }{\text{d}{s}^2}=N_0\text{sin}\psi (s)-T_0\text{cos}\psi (s).$$ (2.9)

Multiplying both sides of (2.9) by dψ/ds and integrating the resulting expression gives

$${\left(\frac{\text{d}\psi }{\text{d}s}\right)}^2=-\frac{2}{EI}(N_0\text{cos}\psi (s)+T_0\text{sin}\psi (s))+D.$$ (2.10)

The integration constant D is determined using the boundary conditions $EI\text{d}\psi (s=L)/\text{d}s=M_0$ and ψ(s=L)=0, and is found to be

$$D=\frac{2}{EI}{N}_0+{\left[\frac{M_0}{EI}\right]}^2.$$ (2.11)

Integrating (2.10) with respect to ψ, we obtain

$$s={\int }_0^{\psi }\frac{\text{d}{\psi }^{\prime }}{\sqrt{-\frac{2}{EI}(N_0\text{cos}{\psi }^{\prime }+T_0\text{sin}{\psi }^{\prime })+D}},0\le \psi \le {\psi }_{\text{max}},$$ (2.12)

where we have retained only the positive root. By symmetry (figure 3), the angle ψ increases with arc length, s, from either end and reaches a maximum value of ψ_max at the midpoint. The value of ψ_max is found using dψ/ds=0,

$$-\frac{2}{EI}(N_0\text{cos}{\psi }_{\text{max}}+T_0\text{sin}{\psi }_{\text{max}})+D=0.$$ (2.13)

2.1 Extensibility

Let λ denote the stretch ratio of a material point ‘p’ on the beam. It can be labelled by its coordinates, X, on the initially straight beam, which coincides with the horizontal axis, i.e. $X\in (0,L_0)$. Point ‘p’ is displaced to coordinates (x(s), y(s)) on the deformed beam. Assuming that the stretch is proportional to the local normal force,

$$\lambda =\frac{\text{d}s}{\text{d}X}\iff \lambda -1=\frac{\text{d}s-\text{d}X}{\text{d}X}=cN(x(s),y(s))\iff \frac{\text{d}s}{\text{d}X}=1+cN(x(s),y(s)),$$ (2.14)

where c=1/EA and A is the cross-sectional area of the beam. Equation (2.14) can be integrated to give

$${\int }_0^L\frac{\text{d}{s}^{\prime }}{1+cN(x(s^{\prime }),y(s^{\prime }))}={\int }_0^{L_0}\text{d}X=L_0.$$ (2.15)

Using (2.5), equation (2.15) becomes

$${\int }_0^L\frac{\text{d}{s}^{\prime }}{1+c[N_0\text{cos}\psi (s^{\prime })+T_0\text{sin}\psi (s^{\prime })]}=L_0.$$ (2.16)

For a given T₀ and N₀, one can solve equations (2.8a), (2.8b), (2.12) and (2.16) for Δ_N, Δ_T, M₀ and L, with

$$\psi \left(s=\frac{L}{2}\right)={\psi }_{\text{max}}.$$ (2.17)

2.2 An equivalent problem

The deflection in figure 3 can also be obtained by moving the right (left) end of the beam up (down) by ±Δ_T/2 and outwards by Δ_N/2, with the midpoint of the beam fixed. If we measure s from the left end of the beam, then it is easily seen that ψ_max is attained at s=L/2. In addition, we can solve the problem for s∈(0, L/2). The boundary conditions are

$$\psi (s=0)=0,$$ (2.18a)

$${\psi }^{\prime }(s=L/2)=0,$$ (2.18b)

$$\psi \left(s=\frac{L}{2}\right)={\psi }_{\text{max}},$$ (2.18c)

$$x(s=0)=y(s=0)=0,$$ (2.18d)

$$x(s=L/2)=(L_0+{\Delta }_{\text{N}})/2,$$ (2.18e)

$$y(s=L/2)={\Delta }_{\text{T}}/2.$$ (2.18f)

The known quantities are N₀, T₀, L₀; the unknowns are Δ_N, Δ_T, L and the constant D. The constant D can be expressed in terms of the unknowns Δ_N, Δ_T, using (2.11) and noting that the moment at the centre of the deflected beam is zero, i.e.

$$M_0+\frac{(L_0+{\Delta }_{\text{N}})T_0}{2}-N_0\frac{{\Delta }_{\text{T}}}{2}=0.$$ (2.19)

The extensibility condition, equation (2.16), becomes

$${\int }_0^{L/2}\frac{\text{d}{s}^{\prime }}{1+c[N_0\text{cos}\psi (s^{\prime })+T_0\text{sin}\psi (s^{\prime })]}=L_0/2.$$ (2.20)

Likewise, equations (2.8a) and (2.8b) become

$$\frac{L_0+{\Delta }_{\text{N}}}{2}={\int }_0^{L/2}\text{cos}\psi (s^{\prime })\text{d}{s}^{\prime },$$ (2.21a)

$$\frac{{\Delta }_{\text{T}}}{2}={\int }_0^{L/2}\text{sin}\psi (s^{\prime })\text{d}{s}^{\prime }.$$ (2.21b)

Given N₀, T₀, L₀, equations (2.12), (2.20), (2.21a) and (2.21b) can be solved to find the unknowns Δ_N, Δ_T, L.

2.3 Results

There is a simple way to solve the above problem and to reduce the solution to elliptic integrals (see equations (A 20), (A 22) and (A 23b) in appendix A). Details are given in appendix A; here we state the main results.

The problem can be reduced to the solution of the following three decoupled equations:

$$\sqrt{\frac{EI}{2\Lambda }}{\int }_{{\theta }_0}^{q_{\text{max}}}\frac{\text{d}q}{(1+c\Lambda \text{sin}q)\sqrt{\text{sin}{q}_{\text{max}}-\text{sin}q}}=L_0/2,$$ (2.22)

$$\frac{{\Delta }_{\text{T}}}{2}=\sqrt{\frac{EI}{2\Lambda }}{\int }_{{\theta }_0}^{q_{\text{max}}}\text{sin}(q-{\theta }_0)\frac{\text{d}q}{\sqrt{\text{sin}{q}_{\text{max}}-\text{sin}q}},$$ (2.23)

$$\frac{(L_0+{\Delta }_{\text{N}})}{2}=\sqrt{\frac{EI}{2\Lambda }}{\int }_{{\theta }_0}^{q_{\text{max}}}\text{cos}(q-{\theta }_0)\frac{\text{d}q}{\sqrt{\text{sin}{q}_{\text{max}}-\text{sin}q}},$$ (2.24)

where Λ is the magnitude of the total applied force, i.e.

$$\Lambda =\sqrt{N_0^2+T_0^2}$$ (2.25a)

and θ₀ is the phase angle of the applied force, i.e.

$$N_0/\Lambda =\text{sin}{\theta }_0,T_0/\Lambda =\text{cos}{\theta }_0.$$ (2.25b)

Finally,

$$q_{\text{max}}\equiv {\theta }_0+{\psi }_{\max },$$ (2.26)

in (2.22)–(2.24). To find Δ_N and Δ_T given T₀ and N₀, we solve equation (2.22) for q_max given T₀ and N₀. We then substitute q_max into (2.23) and (2.24) to find Δ_N and Δ_T.

2.4 Normalization

To expedite the analysis, we define the following normalized variables:

$$\begin{array}{c}{\overline{T}}_0=\frac{T_0{L}_0^2}{2EI},\\ {\overline{N}}_0=\frac{N_0{L}_0^2}{2EI}\overline{\Lambda }=\frac{\Lambda L_0^2}{2EI}{\overline{\Delta }}_{\mathrm{T},\mathrm{N}}={\Delta }_{\mathrm{T},\mathrm{N}}/L_0.\end{array}\}$$ (2.27)

After normalization, equations (2.22)–(2.24) become

$${\int }_{{\theta }_0}^{q_{\text{max}}}\frac{\text{d}q}{\left(1+c\frac{2EI}{L_0^2}\overline{\Lambda }\text{sin}q\right)\sqrt{\text{sin}{q}_{\text{max}}-\text{sin}q}}=\sqrt{\overline{\Lambda }},$$ (2.28)

$${\overline{\Delta }}_{\text{T}}=\frac{1}{\sqrt{\overline{\Lambda }}}{\int }_{{\theta }_0}^{q_{\text{max}}}\frac{\text{sin}(q-{\theta }_0)}{\sqrt{\text{sin}{q}_{\text{max}}-\text{sin}q}}\text{d}q,$$ (2.29)

$$1+{\overline{\Delta }}_{\text{N}}=\frac{1}{\sqrt{\overline{\Lambda }}}{\int }_{{\theta }_0}^{q_{\text{max}}}\text{cos}(q-{\theta }_0)\frac{\text{d}q}{\sqrt{\text{sin}{q}_{\text{max}}-\text{sin}q}}.$$ (2.30)

It is interesting to note that the normalized equations (2.28)–(2.30) depend on two dimensionless parameters: θ₀, which is the phase angle of loading, and $2cEI/L_0^2=2I/A{L}_0^2$, which is a purely geometric quantity. For a beam with a square cross section with side length b, $2I/A{L}_0^2$ is proportional to (b/L₀)², the aspect ratio squared. The numerical results presented in §2.5 are for b=14 μm and length L₀=67 μm, which are the dimensions of fibrils used in our experiments.

2.5 Numerical results

The normalized shear displacement ${\overline{\Delta }}_{\text{T}}$ is plotted against ${\overline{T}}_0$ for ${\overline{N}}_0=0$ in figure 5. The prediction of the BCM (derived in appendix B for our boundary condition) for small $\overline{\Lambda }$ is

$${\overline{\Delta }}_{\text{T}}=\frac{1}{6}{\overline{T}}_0,$$ (2.31)

$${\overline{\Delta }}_{\text{N}}={\overline{C}}_{\text{N}0}{\overline{N}}_0+\frac{1}{6}\frac{{\overline{C}}_{\text{N}0}}{(1-{\overline{C}}_{\text{N}0}{\overline{N}}_0)}{\overline{T}}_0^2,$$ (2.32)

where ${\overline{C}}_{\text{N}0}\equiv 2I/b^2{L}_0^2$ is the normalized compliance for a bar under a pure normal load. Note that the shear compliance for small $\overline{\Lambda }$ is independent of the normal load.

Figure 5.

Normalized shear displacement, ${\overline{\Delta }}_{\text{T}}$, versus normalized shear force, ${\overline{T}}_0$, for ${\overline{N}}_0=0$ (or θ₀=0) and b/L₀=14/67. Solid line, the solution of the nonlinear theory (2.28)–(2.30); dashed line, the prediction of BCM (2.31).

Figure 5 shows that the small-deflection BCM result (equations (2.31) and (2.32)) for ${\overline{N}}_0=0$ is accurate as long as ${\overline{T}}_0\le 1.5$ or when the shear displacement is less than 30% of the original length of the beam.

Figure 6 plots the normalized shear displacement ${\overline{\Delta }}_{\text{T}}$ versus the normalized shear load ${\overline{T}}_0$ for different normalized normal forces ${\overline{N}}_0$ or θ₀. The normalized compressive load in figure 6 is chosen to be less than the buckling load. Owing to adhesion, a fibril (e.g. those at the edge of the contact zone) can be subjected to tension (i.e. N₀>0). The incremental shear compliance at a fixed normal load is $\partial {\overline{\Delta }}_{\text{T}}/\partial {\overline{T}}_0$. The prediction of the BCM is indicated by the dashed line. In the BCM, the shear compliance is independent of the normal load (see appendix B). Figure 6 shows that the BCM is valid only in the limit when both normal and shear forces are small. For example, even for small deflections, the shear compliance is not a constant but increases with increasing compression (more negative ${\overline{N}}_0$). In general, when the beam is under compression, the BCM theory underestimates the shear compliance for small shear and it overestimates it for large shear. It is also clear from figure 6a that, for the range of deflections of interest in our experiments $({\overline{\Delta }}_{\text{T}}\approx 1)$, the shear force predicted by the small-deflection theory can be much smaller than that predicted by the nonlinear theory. The normalized incremental shear compliance ${\overline{C}}_{\text{T}}={\partial {\overline{\Delta }}_{\text{T}}/\partial {\overline{T}}_0|}_{N_0=0}$ versus Δ_T for different three normal forces is shown in figure 6b. The dashed line is the prediction of the BCM. Note that, for ${\overline{\Delta }}_{\text{T}}\ge 0.75$, the incremental shear compliance for different normal loads becomes nearly identical. In this regime, the shear response is governed almost entirely by stretching of the fibril.

Figure 6.

(a) The normalized shear displacement versus shear force for different applied normal loads. The dashed line is the BCM (2.31) for ${\overline{N}}_0=0$ (circles, ${\overline{N}}_0=0$; crosses, ${\overline{N}}_0=-2$; solid line, ${\overline{N}}_0=2$). (b) The incremental shear compliance versus normalized shear displacement for different applied normal loads. The dashed line is the BCM (2.31) for ${\overline{N}}_0=0$ (squares, ${\overline{N}}_0=0$; filled circles, ${\overline{N}}_0=-2$; inverted triangles, ${\overline{N}}_0=2$). The symbols are the numerical solution of (2.28)–(2.30).

Figure 7 shows that a fibril with no applied normal load can still have normal displacement due to shear. Figure 7 also shows that the small-deflection theory considerably underestimates the normal displacement for large ${\overline{T}}_0$. To study the incremental normal compliance of the fibril ${\overline{C}}_{\text{N}}={\partial {\overline{\Delta }}_{\text{N}}/\partial {\overline{N}}_0|}_{{\overline{T}}_0}$, the dependence of ${\overline{\Delta }}_{\text{N}}$ on ${\overline{N}}_0$ for different ${\overline{T}}_0$ is shown in figure 8a. For small ${\overline{T}}_0$, the compliance is small. For a fixed ${\overline{N}}_0$, the incremental normal compliance increases significantly with increasing applied shear load ${\overline{T}}_0$; for a fixed ${\overline{T}}_0$, it also increases slightly with increasing compressive normal load, as shown in figure 8b.

Figure 7.

The normalized normal displacement, ${\overline{\Delta }}_{\text{N}}$, versus normalized shear force ${\overline{T}}_0$, for ${\overline{N}}_0=0$ (or θ₀=0). It shows that a significant normal deflection at the fibril end can occur with no applied normal force. Note that the small-deflection theory underestimates the normal displacement. Circles, large-deflection theory; crosses, small-deflection BCM.

Figure 8.

(a) The dependence of ${\overline{\Delta }}_{\text{N}}$ on ${\overline{N}}_0$ for different ${\overline{T}}_0$. (b) The normalized normal compliance of a fibril against ${\overline{N}}_0$ for different ${\overline{T}}_0$. It shows that, for small ${\overline{T}}_0$, the incremental compliance is small. For a fixed ${\overline{N}}_0$, the incremental normal compliance increases significantly with increasing applied shear load ${\overline{T}}_0$; for a fixed ${\overline{T}}_0$, it also increases slightly with increasing normal load. The dashed lines are the predictions of the BCM with ${\overline{T}}_0=0$. Circles, ${\overline{T}}_0=0.5$; squares, ${\overline{T}}_0=1.5$; asterisks, ${\overline{T}}_0=3.5$.

3. Comparison with the experiments

We give an example to illustrate how our model can be used to interpret the experiments. As mentioned in §1, the relative deflection of the fibrils (i.e. Δ_T's) can be measured in our experiments. Since fibrils far away from the contact zone Ω_c do not carry load, we need only to measure deflections inside a sufficiently large region Ω that contains Ω_c. In our experiments, the normal force applied on the indenter, F_N, is much smaller than the typical shear force F_s on the sphere. Therefore, we assume that the normal force N₀ acting on each fibril is approximately zero. With this assumption, Δ_T and N₀ are known for each fibril (N₀=0), and the shear force, T₀, acting on these fibrils can be computed using our model. To expedite the computation, the result of figure 9 is represented as a relationship between ${\overline{T}}_0$ and ${\overline{\Delta }}_{\text{T}}$ using a seventh-degree polynomial, i.e.

$${\overline{T}}_0=261.65{\overline{\Delta }}_{\text{T}}^7-1296.2{\overline{\Delta }}_{\text{T}}^6+2373.5{\overline{\Delta }}_{\text{T}}^5-1950.9{\overline{\Delta }}_{\text{T}}^4+765.36{\overline{\Delta }}_{\text{T}}^3-117{\overline{\Delta }}_{\text{T}}^2+6{\overline{\Delta }}_{\text{T}}.$$ (3.1)

Figure 9 shows that equation (3.1) is very accurate and converges to (2.31) for very small ${\overline{\Delta }}_{\text{T}}$.

Figure 9.

Normalized shear force versus normalized shear displacement for N₀=0. The dashed line is the small-deflection BCM (2.31). The seventh-order polynomial fit (3.1) (solid line) captures the numerical result (diamonds) obtained by solving (2.28)–(2.30). Normal compression is assumed to be zero.

The total shear force acting on the indenter, F_s, is given by

$$F_{\text{s}}=\sum _{i=1}^n{T}_0(i),$$ (3.2)

where T₀(i) denotes the shear force acting on the ith fibril and n is the total number of fibrils in Ω. It should be noted that, while fibrils inside Ω_c obey the clamped–clamped boundary condition, the top of the fibrils in Ω−Ω_c can rotate freely. However, there is no difficulty in computing T₀ for these fibrils. Indeed, the deformation of these fibrils can be obtained by replacing L₀/2 in (2.22) by L₀, Δ_T/2 in (2.23) by Δ_T and (L₀+Δ_N)/2 in (2.24) by L₀+Δ_N.

We obtain T₀(i) using the following procedure. First, we select three points on the shear force versus shear displacement curve in figure 1b. The coordinates of these points are A(0.0155 mm, 3.80 mN), B(0.0456 mm, 18.68 mN) and C(0.106 mm, 40.03 mN), respectively. We select Ω with the condition that fibrils outside of Ω do not have measurable deflections. Once this is done, we measure the deflection for each fibril (Δ_T(i)) inside Ω. ${\overline{\Delta }}_{\text{T}}(i)$ in (3.1) is computed from Δ_T(i) using L=67 μm. We then use (3.1) to obtain ${\overline{T}}_0(i)$. To compute T₀(i) from ${\overline{T}}_0(i)$, we need the Young modulus and the moment of inertia I. We use a Young modulus of 3 MPa for PDMS, the fibril material. This modulus has been measured independently using an indentation test (see Noderer et al. 2007). The moment of inertia I is 3.2×10³ (μm)⁴ since the fibrils in our experiments have a square cross section of 14 μm. The number of fibrils in Ω is 324. We then compute the indenter shear forces using (3.2). They are 5.23 mN for point A, 11.22 mN for point B and 31.54 mN for point C. We also use the small-deflection theory (2.31) to compute these shear forces. They are 3.24 mN for point A, 3.75 mN for point B and 4.02 mN for point C. These results are shown in figure 10, which compares the experimental data (stage 1) with the large- and small-deflection results. As expected, the small-deflection theory works well for small shear displacements but strongly underestimates the shear force for large shear displacements. For example, near the peak load (e.g. point C), the use of the small-deflection theory underestimates the shear force by approximately an order of magnitude. Given the fact that we have used no adjustable parameters, the agreement between our large-deflection theory and the experimental data is quite good.

Figure 10.

Comparison of the experimental data (stage 1, filled circles) with the large (asterisks) and small (squares) deflection theoretical results. The small-deflection theory (BCM) works well for very small shear displacements but considerably underestimates the shear force for large shear displacements.

4. Discussion and conclusion

The mechanical behaviour of fibrils under combined normal and shear loads underlies the response of biomimetic fibrillar arrays. Deformations are typically large compared with fibril dimensions. We have developed a nonlinear model to compute the deflection of fibrils in microfibril arrays subjected to normal and shear loads. To simplify the analysis, we have assumed that the fibrils do not twist. Also, we assume that the beam can undergo very large deflection but material behaviour is still linear.

Our model isolates a single fibril in the array to study its behaviour, whereas, in practice, the entire array is subjected to normal and shear loads. To illustrate how our model can be applied in this situation, we use our model to predict the shear force acting on a glass indenter in contact with an array of film-terminated fibrils. The computed shear forces are then compared with those obtained from the experiments. In the simulations, we have made the approximation that the normal force acting on all the fibrils is zero, which is strictly valid only for fibrils that are outside the contact zone. For fibrils inside the contact zone, some of the fibrils can be under tension (e.g. those close to the contact edge), whereas others can be under compression, hence our assumption is only approximately valid if the normal indentation force is very small in comparison with the shear forces, which is the case in our experiments. Nevertheless, the prediction of our nonlinear model is in reasonably good agreement with the experimental data. We should point out that there is some error in measuring the relative displacement of each fibril. Since the nonlinear theory is very sensitive to the relative shear displacement, it is not surprising that the nonlinear theory did worst for small deflections, where the relative error of the measurements can be large. Finally, there is no fitting parameter in our calculation.

Although our analysis is valid for any loading phase angle as well as for arbitrary aspect ratios, explicit results are presented for a particular aspect ratio. Also, we focus on the case where the normal load is zero. In general, there is no difficulty in generating results for other aspect ratios and different phase angles. Also, our model can be easily modified to describe the deformation of preoriented fibrils.

The model of fibril deformation studied in this work is quite general and can be used to study fibrillar structures other than our own. For example, it is applicable to a similar fibrillar structure (with angled fibrils) fabricated recently by Yao et al. (2007). It can also be used to analyse the shear deformation of fibrils with spatulated tips, such as those fabricated by Kim et al. (2007).

The analysis will be more complicated for the general case where the normal indentation force is significant. In this case, most of the fibrils inside the contact zone will be under compression except those near the edge. These edge fibrils will be under tension owing to adhesion. For this case, the normal and shear loads on each fibril in the array must be determined using the contact condition. This will be studied in a future work.

Appendix A. Derivation of (2.22)–(2.24)

Substituting (2.13) into (2.12) gives

$$s={\int }_0^{\psi }\frac{\text{d}{\psi }^{\prime }}{\sqrt{\begin{array}{l}\frac{2}{EI}(N_0\text{cos}{\psi }_{\text{max}}+T_0\text{sin}{\psi }_{\text{max}})\\ -\frac{2}{EI}(N_0\text{cos}{\psi }^{\prime }+T_0\text{sin}{\psi }^{\prime })\end{array}}}.$$ (A1)

Using (2.25a) and (2.25b), the terms inside the square root in (A 1) are

$$\frac{2}{EI}(N_0\text{cos}{\psi }_{\text{max}}+T_0\text{sin}{\psi }_{\text{max}})-\frac{2}{EI}(N_0\text{cos}{\psi }^{\prime }+T_0\text{sin}{\psi }^{\prime })=\frac{2\Lambda }{EI}(\text{sin}{q}_{\text{max}}-\text{sin}q),$$ (A2)

where

$$q\equiv ({\theta }_0+{\psi }^{\prime }),q_{\text{max}}\equiv ({\theta }_0+{\psi }_{\text{max}}).$$ (A3)

Since

$$\frac{2\Lambda }{EI}(\text{sin}{q}_{\text{max}}-\text{sin}q)=\frac{2\Lambda }{EI}(1+\text{sin}{q}_{\text{max}}-1-\text{sin}q),$$ (A4)

we can define a constant p by

$$2p^2\equiv 1+\text{sin}{q}_{\text{max}}=1+\text{sin}\left({\theta }_0+{\psi }_{\text{max}}\right).$$ (A5)

Note that 0≤p≤1. Also, introduce a new variable ϕ by

$$1+\text{sin}q\equiv 2p^2{\text{sin}}^2\varphi ,$$ (A6)

so that (A 4) is

$$\sqrt{\frac{2\Lambda }{EI}(\text{sin}{q}_{\text{max}}-\text{sin}q)}=2\sqrt{\frac{\Lambda }{EI}}p\text{cos}\varphi .$$ (A7)

It is easy to verify that

$$\text{d}{\psi }^{\prime }=\text{d}q=\frac{2p\text{cos}\varphi \text{d}\varphi }{\sqrt{1-p^2{\text{sin}}^2\varphi }}.$$ (A8)

Substituting (A 7) and (A 8) into (A 1), we have

$$s=\sqrt{\frac{EI}{\Lambda }}{\int }_{{\varphi }_0}^{\varphi }\frac{\text{d}{\varphi }^{\prime }}{\sqrt{1-p^2{\text{sin}}^2{\varphi }^{\prime }}},$$ (A9)

where

$${\varphi }_0={\text{sin}}^{-1}\left[\sqrt{\frac{1+\text{sin}{\theta }_0}{2p^2}}\right].$$ (A10)

Equations (A 5) and (A 6) imply that

$$\psi ={\psi }_{\text{max}}\iff q=q_{\text{max}}\iff \varphi =\frac{\pi }{2}.$$ (A11)

Thus, setting ϕ=π/2 in (A 9) gives

$$L/2=\sqrt{\frac{EI}{\Lambda }}{\int }_{{\varphi }_0}^{\pi /2}\frac{\text{d}{\varphi }^{\prime }}{\sqrt{1-p^2{\text{sin}}^2{\varphi }^{\prime }}}.$$ (A12)

Equations (A 3) and (A 9) imply that

$$\text{d}s=\sqrt{\frac{EI}{2\Lambda }}\frac{\text{d}\psi }{\sqrt{(\text{sin}{q}_{\text{max}}-\text{sin}q)}}=\sqrt{\frac{EI}{2\Lambda }}\frac{\text{d}q}{\sqrt{(\text{sin}{q}_{\text{max}}-\text{sin}q)}},$$ (A13)

so (2.21a) is

$$\frac{(L_0+{\Delta }_{\text{N}})}{2}=\sqrt{\frac{EI}{2\Lambda }}{\int }_{{\theta }_0}^{{\theta }_0+{\psi }_{\text{max}}}\frac{\text{cos}(q-{\theta }_0)\text{d}q}{\sqrt{(\text{sin}{q}_{\text{max}}-\text{sin}q)}},$$ (A14)

which is (2.24). Equation (A 14) can be rewritten as

$$\frac{(L_0+{\Delta }_{\text{N}})}{2}=\sqrt{\frac{EI}{2\Lambda }}\left[\text{cos}{\theta }_0{\int }_{{\theta }_0}^{{\theta }_0+{\psi }_{\text{max}}}\frac{\text{cos}q\text{d}q}{\sqrt{(\text{sin}{q}_{\text{max}}-\text{sin}q)}}+\text{sin}{\theta }_0{\int }_{{\theta }_0}^{{\theta }_0+{\psi }_{\text{max}}}\frac{\text{sin}q\text{d}q}{\sqrt{(\text{sin}{q}_{\text{max}}-\text{sin}q)}}\right].$$ (A15)

The first integral in (A 15) can be integrated exactly, i.e.

$${\int }_{{\theta }_0}^{{\theta }_0+{\psi }_{\text{max}}}\frac{\text{cos}q\text{d}q}{\sqrt{(\text{sin}{q}_{\text{max}}-\text{sin}q)}}=2\sqrt{\text{sin}({\theta }_0+{\psi }_{\text{max}})-\text{sin}{\theta }_0}.$$ (A16)

The second integral can be expressed in terms of elliptic functions using (A 5)–(A 7)

$${\int }_{{\theta }_0}^{{\theta }_0+{\psi }_{\text{max}}}\frac{\text{sin}q\text{d}q}{\sqrt{(\text{sin}{q}_{\text{max}}-\text{sin}q)}}=\sqrt{2}[-2E(p)+2E({\varphi }_0,p)+K(p)-F({\varphi }_0,p)],$$ (A17)

where

$$F({\varphi }_0,p)\equiv {\int }_0^{{\varphi }_0}\frac{1}{\sqrt{1-p^2{\text{sin}}^2\varphi }}\text{d}\varphi$$ (A18)

is the incomplete elliptic integral of the first kind; $K(p)=F(\pi /2,p)$ is the complete elliptic integral of the first kind;

$$E({\varphi }_0,p)={\int }_0^{{\varphi }_0}\sqrt{1-p^2{\text{sin}}^2\varphi }\text{d}\varphi$$ (A19)

is the incomplete elliptic integral of the second kind; and $E(p)=E(\pi /2,p)$ is the complete elliptic integral of the second kind. Using (A 16) and (A 17), (A 14) becomes

$$\frac{(L_0+{\Delta }_{\text{N}})}{2}=\sqrt{\frac{EI}{\Lambda }}\left[\sqrt{2}\text{cos}{\theta }_0\sqrt{\text{sin}({\theta }_0+{\psi }_{\text{max}})-\text{sin}{\theta }_0}+\text{sin}{\theta }_0[-2E(p)+2E(p,{\varphi }_0)+K(p)-F(p,{\varphi }_0)]\right].$$ (A20)

Likewise, (2.23) can be obtained by substituting (A 13) into (2.21b),

$${\Delta }_{\text{T}}/2={\int }_0^{L/2}\text{sin}\psi (s^{\prime })\text{d}{s}^{\prime }=\sqrt{\frac{EI}{2\Lambda }}{\int }_{{\theta }_0}^{q_{\text{max}}}\text{sin}(q-{\theta }_0)\frac{\text{d}q}{\sqrt{\text{sin}{q}_{\text{max}}-\text{sin}q}}.$$ (A21)

In exactly the same way, (A 21) becomes

$${\Delta }_{\text{T}}/2=\sqrt{\frac{EI}{\Lambda }}\left[-\sqrt{2}\text{sin}{\theta }_0\sqrt{\text{sin}({\theta }_0+{\psi }_{\text{max}})-\text{sin}{\theta }_0}+\text{cos}{\theta }_0\left[-2E(p)+2E(p,{\varphi }_0)+K(p)-F(p,{\varphi }_0)\right]\right].$$ (A22)

Using (A 2) and (A 13), (2.20) becomes

$$\sqrt{\frac{EI}{2\Lambda }}{\int }_{{\theta }_0}^{q_{\text{max}}}\frac{\text{d}q}{(1+c\Lambda \text{sin}q)\sqrt{\text{sin}{q}_{\text{max}}-\text{sin}q}}=L_0/2.$$ (A23a)

Equation (A 23a) can be expressed in terms of incomplete elliptic integral of the third kind, $\Pi (n;\varphi ,k)$, i.e.

$$-\frac{1}{(1+c\Lambda \text{sin}{q}_{\text{max}})\sqrt{1+\text{sin}{q}_{\text{max}}}}\Pi \left(\sqrt{\frac{\text{sin}{\theta }_0-\text{sin}{q}_{\text{max}}}{1-\text{sin}{q}_{\text{max}}}};\frac{(\text{sin}{q}_{\text{max}}-1)c\Lambda }{1+c\Lambda \text{sin}{q}_{\text{max}}},\text{i}\frac{\left|\text{cos}{q}_{\text{max}}\right|}{1+\text{sin}{q}_{\text{max}}}\right)=\frac{1}{2}\sqrt{\frac{\Lambda L_0^2}{2EI}},$$

where

$$\Pi (n;\varphi ,k)={\int }_0^{\varphi }\frac{\text{d}\theta }{(1-n{\text{sin}}^2\theta )\sqrt{1-k^2{\text{sin}}^2\theta }}.$$ (A24)

Appendix B. Small-angle approximation (small-deflection approximation)

Assuming ψ is small, (2.10) can be approximated at

$$\frac{\text{d}\psi }{\text{d}s}\approx \sqrt{-\frac{2}{EI}(N_0+T_0\psi (s))+D},0<s<L/2.$$ (B1)

Substituting $D=2/EI(N_0\text{cos}{\psi }_{\text{max}}+T_0\text{sin}{\psi }_{\text{max}})\approx 2/EI(N_0+T_0{\psi }_{\text{max}})$ into (B 1) and integrating, we have

$$s\approx \sqrt{\frac{EI}{2T_0}}{\int }_0^{\psi }\frac{\text{d}{\psi }^{\prime }}{\sqrt{{\psi }_{\text{max}}-{\psi }^{\prime }}}=\sqrt{\frac{2EI}{T_0}}\left[\sqrt{{\psi }_{\text{max}}}-\sqrt{{\psi }_{\text{max}}-\psi }\right],{\psi }_{\text{max}}>\psi ,$$ (B2)

or

$$\psi ={\psi }_{\text{max}}-{\left[\sqrt{\frac{T_0}{2EI}}s-\sqrt{{\psi }_{\text{max}}}\right]}^2.$$ (B3)

Using (2.10) and (2.11) and the fact that (dψ/ds)=0 at ψ=ψ_max, we have

$$-\frac{2}{EI}(N_0+T_0{\psi }_{\text{max}})+D=0\iff -\frac{2}{EI}{T}_0{\psi }_{\text{max}}+{\left[\frac{M_0}{EI}\right]}^2=0\Rightarrow {\psi }_{\text{max}}=\frac{1}{2T_0}\left[\frac{M_0^2}{EI}\right].$$ (B4)

Since ψ=ψ_max when s=L/2, (B 3) implies that

$${\psi }_{\text{max}}=\frac{T_0{L}^2}{8EI}.$$ (B5)

Using (B 4), we have

$$M_0=-\frac{T_{0}L}{2}.$$ (B6)

Combining (B 3) and (B 5) gives

$$\psi =\frac{T_0}{2EI}s(L-s).$$ (B7)

Using (B 5), the small-deflection version of (2.21b) is

$${\Delta }_{\text{T}}/2\approx {\int }_0^{L/2}\psi (s^{\prime })\text{d}{s}^{\prime }=\frac{T_0{L}^3}{24EI}.$$ (B8)

The normalized form of (B 8) is (2.31) using L≈L₀,

$${\overline{\Delta }}_{\text{T}}=\frac{1}{6}{\overline{T}}_0.$$ (B9)

To determine the relationship between normal displacement and normal load for small deflections, we approximate (2.20) by assuming that the slope is small, so that ψ≈y′, ds≈dx, i.e.

$${\int }_0^{L/2}\frac{\text{d}x}{1+c[N_0+T_0{y}^{\prime }]}=L_0/2.$$ (B10)

Consistent with small deflections, we assume

$$c{T}_0{y}^{\prime }/(1+c{N}_0)\ll 1.$$ (B11)

The integral in (B 10) can be written as

$${\int }_0^{L/2}\frac{\text{d}x}{(1+c{N}_0)\left[1+\frac{c{T}_0{y}^{\prime }}{(1+c{N}_0)}\right]}\approx \frac{1}{(1+c{N}_0)}{\int }_0^{L/2}\left[1-\frac{c{T}_0{y}^{\prime }}{(1+c{N}_0)}\right]\text{d}x=\frac{1}{(1+c{N}_0)}\left[\frac{L}{2}-\frac{c{T}_0{\Delta }_T}{2(1+c{N}_0)}\right],$$ (B12)

where we have used

$${\int }_0^{L/2}{y}^{\prime }\text{d}x=y(L/2)={\Delta }_{\text{T}}/2.$$ (B13)

Using (B 12) and (B 13), (B 10) becomes

$${\Delta }_n-\frac{c{T}_0{\Delta }_{\text{T}}}{(1+c{N}_0)}=L_{0}c{N}_0,$$ (B14)

where L≈L₀+Δ_N. Equation (2.32) is obtained by substituting (B 9) into (B 14), i.e.

$${\Delta }_n=L_{0}c{N}_0+\frac{12cEI{\Delta }_{\text{T}}^2}{L^3(1+c{N}_0)}.$$ (B15)