Quantitative evaluation of adhesion of metallic coatings with an extended microbridge test

Abstract

An extended microbridge test (eMBT) was proposed to assess the adhesion of metallic coatings on metallic substrates. Through loading on the backside of narrow striped freestanding coatings, a two-dimensional stable interfacial delamination was introduced. A cross-sectional scanning electron microscope (SEM) was used to examine the interfacial fracture process. A large deflection solution for elastic deformation of the coating was derived, and an approximate model was established for the estimate of interfacial crack extension force G. The eMBT samples of electroplated Ni coatings on C45 carbon steel substrate were tested, and the measured interfacial fracture toughness was about 5.28 J/m². Cross-sectional SEM examination showed that the interface crack extended along the interface plane, and therefore the interfacial fracture proceeded by the debonding of Ni/steel interface.

Appendices

Appendix A: Large Deflection Solution of a Thin Plate

The schematic of the two-dimensional loading condition of the rectangular thin plate for the eMBT test is shown in Fig. A1.

FIG. A1

Schematic illustrations of a two-dimensional loading condition of the rectangular thin plate model for the eMBT test. Q_z denotes the transversal shearing force at the central line in the plate.

The boundary condition of the left part of the plate is given by

(A1)

where q is the load applied on the central point of the plate, w is the deflection of the thin plate, and Q_z is the transversal shearing force in the plate at the central point.

The equations for large deflection of plate27 are given by

$$ {D{\nabla ^4}w = q + {N_x}\frac{{{\partial ^2}w}}{{\partial {x^2}}} + {N_y}\frac{{{\partial ^2}w}}{{\partial {y^2}}} + {N_{xy}}\frac{{{\partial ^2}w}}{{\partial x\partial y}},} \\ {\frac{1}{E}{\nabla ^4} = \left[ {{{\left( {\frac{{{\partial ^2}w}}{{\partial x\partial y}}} \right)}^2} - \tfrac{{{\partial ^2}w}}{{\partial {x^2}}}\tfrac{{{\partial ^2}w}}{{\partial {y^2}}}} \right],} \\ $$

((A2))

where D is the flexural rigidity of the plate as defined in the main text, and N and F are the membrane force and stress function respectively. For the configuration of Fig. A1, the governing partial differential equations given above are simplified as follows:

$$D\frac{{{{\text{d}}^4}w}}{{{\text{d}}{y^4}}} = {N_y}\frac{{{{\text{d}}^2}w}}{{{\text{d}}{y^2}}}$$

((A3))

$$\int_0^{\bf{b}} {\left[ {\frac{{{\partial ^2}F}}{{\partial {x^2}}} - v\frac{{{\partial ^2}F}}{{\partial {y^2}}} - \frac{E}{2}{{\left( {\frac{{\partial w}}{{\partial y}}} \right)}^2}} \right]} \,{\text{d}}y = 0$$

((A4))

The general solution of Eq. (A3) is as follows (for the left part of the plate):

$$w = {A_1}sh{\lambda }y + {A_2}ch{\lambda }y + {A_3}y + {A_4}$$

((A5))

where λ² = hσ⁰_y/D, σ⁰_y is membrane stress in the y direction and defined as σ⁰_y = N_y/h. Substituting the boundary condition [Eq. (A1) ] into Eq. (A5) gives

$$ {{A_1} = \frac{{ - {A_3}}}{{\lambda }} = \frac{q}{{2D{{\lambda}^3}}}} \\ {{A_2} = - {A_4} = {A_1}\frac{{1 - ch\frac{{{\lambda }l}}{2}}}{{sh\frac{{{\lambda}l}}{2}}}} \\ $$

((A6))

Thus, the Eq. (A5) can be rewritten as

$$w = \frac{q}{{2D{{\lambda}^{\text{3}}}}}\left\{ {\frac{{sh{\lambda}\left( {y - \frac{1}{4}} \right) + sh\frac{{{\lambda }l}}{4}}}{{ch\frac{{{\lambda }l}}{4}}} - {\lambda y}} \right\}$$

((A7))

wherein, the value of λ can be obtained by the following equation:

$${\sigma }_{\text{y}}^0 = \frac{E}{{1 - {v^2}}}\frac{1}{{2b}}{\int_0^l {\left( {\frac{{{\text{d}}w}}{{{\text{d}}y}}} \right)} ^2}{\text{d}}y$$

((A8))

Appendix B: Load–Deflection Relationship

The strain energy of the plate is defined as

$$U = \frac{1}{2}{\smallint _v}{\sigma _{ij}}{ \in _{ij}}{\text{d}}V$$

((B1))

For the elastic plate the above equation can be expressed as

$$U = \frac{1}{2}\smallint \smallint \smallint \left( {{{\sigma }_x}{ \in _x} + \,{{\sigma }_y}{ \in _y} + \,\,{{\tau }_{{\text{xy}}}}{{\gamma }_{xy}}} \right){\text{d}}x{\text{d}}y{\text{d}}z = {U_{\text{m}}} + {U_{\text{b}}}\,,$$

((B2))

where U_m and U_b are the membrane strain energy and bending strain energy, respectively, defined as27

$$ {{U_{\text{m}}} = } {\frac{h}{{2E}}\smallint \smallint \left\{ {{{\left( {{V^2}F} \right)}^2} - 2\left( {1 + v} \right)\left[ {\frac{{{\partial ^2}F}}{{\partial {x^2}}}\frac{{{\partial ^2}F}}{{{\sigma}{y^2}}}} \right.} \right.} \\ {} {\left. {\left. { - {{\left( {\frac{{{\partial ^2}F}}{{\partial x\partial y}}} \right)}^2}} \right]} \right\}{\text{d}}x{\text{d}}y\,,} \\ $$

((B3))

and

$$ {{U_{\text{m}}} = } {\frac{h}{{2E}}\smallint \smallint {{\left\{ {\left( {{\nabla _2}F} \right)} \right.}^2} - 2\left( {1\,|v} \right)\left[ {\frac{{{\partial ^{\text{2}}}F}}{{\partial {x^2}}}} \right.\frac{{{\partial ^{\text{2}}}F}}{{{\sigma }{y^2}}}} \\ {} {\left. {\left. { - {{\left( {\frac{{{\partial ^{\text{2}}}F}}{{{\text{d}}x{\text{d}}y}}} \right)}^2}} \right]} \right\}{\text{d}}x{\text{d}}y} \\ $$

((B4))

For the following deflection function:

$$w = {\delta }\,{\text{si}}{{\text{n}}^{\text{2}}}\frac{{{\pi y}}}{l}$$

((B5))

the membrane stress σ⁰_y Eq. (A8), membrane strain energy U_m Eq. (B3), and bending strain energy U_b Eq. (B4) are expressed as

$${\sigma }_{\text{y}}^0 = \frac{{{{\pi }2}E{{\delta }^{\text{2}}}}}{{4\left( {1 - {v^2}} \right){l^2}}}$$

((B6))

$${U_{\text{m}}} = b{U_{{\text{my}}}} = b\frac{{{{\pi }^4}hE{{\delta }^4}}}{{32\left( {1 - {v^2}} \right){l^3}}}$$

((B7))

and

$${U_{\text{b}}} = b{U_{{\text{sc}}}} = b\frac{{{{\pi }^4}{\text{D}}{{\delta }^2}}}{{{l^3}}}$$

((B8))

The maximum stress in the bending plate, which is located at the point y = 0 and z = −h/2, is expressed as Eq. (B9):

$${\left( {{{\sigma }_y}} \right)_{\max }} = {\sigma }_{\text{y}}^{\text{o}} + \frac{{{{\pi }^2}Eh{{\delta }^{}}}}{{\left( {1 - {v^2}} \right){l^2}}}$$

((B9))

Energy minimization method is used to derivate the deflection δ at the center, which is

$$\frac{{\partial I}}{{\partial f}} = 0$$

((B10))

where I = U_my + U_by − Λ, Λ = w₀q. The following equation then gives:

$$12.18{\left( {\frac{{\delta }}{h}} \right)^3} + 16.23\left( {\frac{{\delta }}{h}} \right) = \frac{{1 - {v^2}}}{{Eh}}{\left( {\frac{l}{h}} \right)^3}q$$

((B11))