Elastic–plastic analysis of the peel test for ductile thin film presenting a saturation of the yield stress

Abstract

The paper investigates the peel test of an elastic–plastic film on an elastic substrate. The case of a film material presenting a saturation of the yield stress is considered. Based on earlier approaches of the literature, see for instance Kim and Aravas (Int J Solids Struct 24:417–435, 1988), a semi-analytical expression for the work done by bending plasticity is proposed. The validity of the present expression is established based on finite element calculations. It is shown that for the interpretation of the results of peel test at 90$^{\circ }$ when the peel force and the curvature are measured, the present approach can provide a precise value of the interface fracture energy.

Appendix A: Derivation of the expression for the work done by bending plasticity

The present appendix provides some details concerning the derivation of the expression for the work done by bending plasticity ${\varPsi }$ along the loading path (OABCD). From the elastic response of the film along path (OA), as depicted in Fig. 1b, one obtains based on Eq. (6):

$$\begin{aligned} \int _{0}^{K_e} MdK = \frac{\sigma _o^2 t}{6 E}. \end{aligned}$$

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Since for a thin copper film, the curvature $K_B$ is larger than $K_e$, one needs to evaluate the contribution to ${\varPsi }$ along (AB). The relationship between the moment M and the curvature K is given by Eq. (10). With the definition $F\left( -\frac{2 \gamma K}{\sqrt{3}}, \frac{t}{2}\right) {=} -\exp \left( {-}\frac{\gamma K t}{\sqrt{3}}\right) \left( \frac{t\sqrt{3}}{4\gamma K}+\frac{3}{4 \gamma ^2 K^2}\right)$, and with a change of variable $u=\frac{\gamma K t}{\sqrt{3}}$, one easily gets:

$$\begin{aligned} \begin{aligned}&\int _{K_e}^{K_B} F\left( -\frac{2 \gamma K}{\sqrt{3}}, \frac{t}{2}\right) dK \\&\quad =\frac{3}{4 \gamma ^2}\left[ G\left( -\frac{\gamma t}{\sqrt{3}}, K_B\right) - G\left( -\frac{\gamma t}{\sqrt{3}}, K_e\right) \right] , \end{aligned} \end{aligned}$$

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where $G(x,y) =\frac{\exp (xy)}{y}$. The contribution linked to the term $F\left( -\frac{2 \gamma K}{\sqrt{3}}, h\right)$ with $h=\frac{\sigma _o \sqrt{3}}{2 E K}$ is evaluated in a straightforward manner, so the work done by bending plasticity along the path (AB) is:

$$\begin{aligned} \begin{aligned}&\int _{K_e}^{K_B} MdK \\&\quad = \frac{\alpha t^2}{2 \sqrt{3}}(K_B-K_e) \\&\qquad +\left( \frac{\sigma _o}{\sqrt{3}}- \frac{\sqrt{3} \alpha }{2}\right) \frac{\sigma _o^2}{E^2} \left( \frac{1}{K_e}-\frac{1}{K_B}\right) \\&\qquad -\frac{\sqrt{3} \beta }{\gamma ^2} \left[ G\left( -\frac{\gamma t}{\sqrt{3}}, K_B\right) - G\left( -\frac{\gamma t}{\sqrt{3}}, K_e\right) \right] \\&\qquad +\frac{\sqrt{3}\beta }{\gamma }\exp \left( -\frac{\gamma \sigma _o}{E}\right) \left( \frac{\sigma _o}{E}+\frac{1}{\gamma }\right) \left( \frac{1}{K_B}-\frac{1}{K_e}\right) \end{aligned}. \end{aligned}$$

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For the unloading path (BC), the contribution to ${\varPsi }$ is readily obtained:

$$\begin{aligned} \int _{K_B}^{K_B-K_B^*} M dK = - M_B K_B^* + \frac{E t^3}{18} (K_B^*)^2. \end{aligned}$$

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The last term corresponding to the reverse plastic loading phase (CD) is the most complex contribution. Indeed, the expression of M is given by Eq. (15) where an integral term is already present. In addition, it is related to the position of the elastic–plastic transition surface $h^{'}$ defined by an implicit relation (13). Therefore one can simply notice that:

$$\begin{aligned} \begin{aligned} \int _{K_B-K_B^*}^{0} M dK&= \left[ \frac{8EK_Bh_B^3}{9} -\frac{2 \alpha }{\sqrt{3}}\left( h_B^2+\frac{t^2}{4}\right) \right. \\&\qquad + \left. \frac{4 \beta }{\sqrt{3}} F\left( -\frac{2\gamma K_B}{\sqrt{3}},h_B\right) \right] \\&\quad \times (K_B^*-K_B) + I, \end{aligned}\nonumber \\ \end{aligned}$$

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where the term I is defined by Eq. (22). As mentioned in Sect. 2, for a metallic film with large values of $K_B$ and $\gamma$, the accumulated strain Eq. (14) (resp. $h^{'}$) can be approximated as proposed in Eq. (16) (resp. by $h^{'}= \frac{\sqrt{3} \alpha }{E (K_B-K)}$). After a standard mathematical development, the integral term I is evaluated and the approximate closed form expression Eq. (23) is found.