Edge Effect on Crack Patterns in Thermally Sprayed Ceramic Splats

Abstract

To explore the edge effect on intrasplat cracking of thermally sprayed ceramic splats, crack patterns of splats were experimentally observed and investigated through mechanical analysis. Both the polycrystalline splats and single-crystal splats showed obvious edge effects, i.e., preferential cracking orientation and differences in domain size between center fragments and edge fragments. In addition, substrate/interface delamination on the periphery was clearly observed for single-crystal splats. Mechanical analysis of edge effect was also carried out, and it was found that both singular normal stress in the substrate and huge peeling stress and shear stress at the interface were induced. Moreover, effective relief of tensile stress in splats is discussed. The good correspondence between experimental observations and mechanical analysis is elaborated. The edge effect can be used to tailor the pattern morphology and shed further light on coating structure design and optimization.

Appendix

The constitutive law in regard to thermal strain without consideration of bending problem reads

$$\varepsilon_{\text{f}}^{0} = \frac{{\sigma_{\text{f}}^{\text{i}} }}{{E_{\text{f}} }} + \alpha_{\text{f}} \Delta T = \varepsilon_{\text{f}}^{\text{i}} + \alpha_{\text{f}} \Delta T$$

$$\varepsilon_{\text{s}}^{0} = \frac{{\sigma_{\text{s}}^{\text{i}} }}{{E_{\text{s}} }} + \alpha_{\text{s}} \Delta T = \varepsilon_{\text{s}}^{\text{i}} + \alpha_{\text{s}} \Delta T$$

(27)

where ɛ ⁰ _f and ɛ ⁰ _s are the uniform strain of the splat and substrate without consideration of bending, respectively. Moreover, σ ⁱ _f and σ ⁱ _s are the uniform stress of the splat and substrate, respectively. Using the condition of the strain compatibility at the film/substrate interface gives

$$\varepsilon_{\text{f}}^{\text{i}} - \varepsilon_{\text{s}}^{\text{i}} = \left( {\alpha_{\text{s}} - \alpha_{\text{f}} } \right)\Delta T = \Delta \alpha \Delta T$$

(28)

The force balance of the whole bimaterial beams dictates

$$E_{\text{f}} \varepsilon_{\text{f}}^{\text{i}} h_{\text{f}} + E_{\text{s}} \varepsilon_{\text{s}}^{\text{i}} h_{\text{s}} = 0$$

(29)

Combination of Eq 28 and 29 gives

$$\varepsilon_{\text{f}}^{\text{i}} = \frac{{E_{\text{s}} h_{\text{s}} \Delta \alpha \Delta T}}{{E_{\text{f}} h_{\text{f}} + E_{\text{s}} h_{\text{s}} }}$$

$$\varepsilon_{s}^{\text{i}} = \frac{{ - E_{\text{f}} h_{\text{f}} \Delta \alpha \Delta T}}{{E_{\text{f}} h_{\text{f}} + E_{\text{s}} h_{\text{s}} }}$$

(30)

The moments due to uniform strain component, $M^{\text{i}}$, can thus be expressed as

$$M^{\text{i}} = \mathop \smallint \limits_{0}^{{h_{f} }} E_{\text{f}} \varepsilon_{\text{f}}^{\text{i}} (y - \delta ){\text{d}}y + \mathop \smallint \limits_{{ - h_{\text{s}} }}^{0} E_{\text{s}} \varepsilon_{\text{s}}^{\text{i}} (y - \delta ){\text{d}}y$$

(31)

where δ is the location of the neutral axis composite beam. The quantity δ was calculated on the assumption of equivalent cross section (Ref 25) as shown in Fig. 10, which was defined as

$$\delta = \frac{{y_{\text{f}} A_{\text{f}} + y_{\text{s}} A_{\text{s}} }}{{A_{\text{f}} + A_{\text{s}} }}$$

(32)

where y _f and y _s are the location of the neutral axis of film and substrate, respectively. The quantities A _f and A _s are the equivalent cross-sectional area of film and substrate, as shown in Fig. 10(b), respectively. Therefore, δ can further be reduced to

$$\delta = \frac{{E_{\text{f}} h_{\text{f}}^{2} - E_{\text{s}} h_{\text{s}}^{2} }}{{2(E_{\text{f}} h_{\text{f}} + E_{\text{s}} h_{\text{s}} )}}$$

(33)

Fig. 10

Procedure for finding the location of the neutral axis of a composite beam. (a) Actual cross section of bimaterial beam and (b) equivalent cross section of bimaterial beam on the assumption of equal bending stiffness

To balance the moment $M^{\text{i}}$, an extra moment $M^{\text{b}}$ is required, which in return gives rise to stress redistribution in bimaterial beam.

The pure bending strain due to $M^{\text{b}}$ dictates the function form

$$\varepsilon^{\text{b}} = \frac{y - \delta }{r}$$

(34)

where r is the radius of curvature for neutral axis of composite beam.

Fig. 11

The SEM (a) and EDS (b, c) morphologyof single-crystal LZ splats deposited on single-crystal YSZ substrate. The element shown in (b) and (c) was lanthanum (La) and yttrium (Y) corresponding to LZ splats and YSZ substrate, respectively

The bending moment, $M^{\text{b}}$, related to the pure bending strains $\varepsilon^{\text{b}}$, can be expressed as

$$M^{\text{b}} = \mathop \smallint \limits_{{ - h_{\text{s}} }}^{0} E_{\text{s}} \frac{{\left( {y - \delta } \right)^{2} }}{r}{\text{d}}y + \mathop \smallint \limits_{0}^{{h_{\text{f}} }} E_{\text{f}} \frac{{\left( {y - \delta } \right)^{2} }}{r}{\text{d}}y$$

(35)

The moment balance of the whole bimaterial beam follows $M^{\text{i}} + M^{\text{b}} = 0$, which in conjunction with Eq 31, 34, and 35 dictates

$$\frac{1}{r} = \frac{3}{2} \cdot \frac{{E_{\text{s}} \varepsilon_{\text{s}}^{\text{i}} h_{\text{s}} \left( {h_{\text{s}} + 2\delta } \right) - E_{\text{f}} \varepsilon_{\text{f}}^{\text{i}} h_{\text{f}} \left( {h_{\text{f}} - 2\delta } \right)}}{{E_{\text{s}} h_{\text{s}} \left( {h_{\text{s}}^{2} + 3h_{\text{s}} \delta + 3\delta^{2} } \right) + E_{\text{f}} h_{\text{f}} \left( {h_{\text{f}}^{2} - 3h_{\text{f}} \delta + 3\delta^{2} } \right)}}$$

(36)

The total strain in the film (the upper beam) can thus be expressed as

$$\varepsilon_{\text{f}}^{\text{t}} = \alpha_{\text{f}} \Delta T + \varepsilon_{\text{f}}^{\text{i}} + \varepsilon_{\text{f}}^{\text{b}} = \alpha_{\text{f}} \Delta T + \frac{{E_{\text{s}} h_{\text{s}} \Delta \alpha \Delta T}}{{E_{\text{f}} h_{\text{f}} + E_{\text{s}} h_{\text{s}} }} + \frac{y - \delta }{r}$$

(37)

In the present study,

$$\eta = h_{\text{f}} /h_{\text{s}} = 1/500 \to 0$$

$$\mathop {\lim }\limits_{\eta \to 0} \varepsilon_{\text{f}}^{\text{i}} = (\alpha_{\text{s}} - \alpha_{\text{f}} )\Delta T$$

$$\mathop {\lim }\limits_{\eta \to 0} \delta = - \frac{{h_{s} }}{2}$$

$$\mathop {\lim }\limits_{\eta \to 0} \frac{1}{r} = \frac{{ - 6E_{\text{f}} h_{\text{f}} }}{{E_{\text{s}} h_{\text{s}}^{2} }}\Delta \alpha \Delta T = 0$$

(38)

Therefore, the strain and stress in the film can be reduced to

$$\varepsilon_{\text{f}}^{\text{t}} = - \frac{{6E_{\text{f}} h_{\text{f}} }}{{E_{\text{s}} h_{\text{s}}^{2} }}\Delta \alpha \Delta T \cdot \left( {y + \frac{{h_{\text{s}} }}{2}} \right)$$

$$\sigma_{\text{f}}^{\text{t}} = E_{\text{f}} \left[ { - \frac{{6E_{\text{f}} h_{\text{f}} }}{{E_{\text{s}} h_{\text{s}}^{2} }}\Delta \alpha \Delta T \cdot \left( {y + \frac{{h_{\text{s}} }}{2}} \right) - \alpha_{\text{f}} \Delta T} \right]$$

(39)