Effect of dislocation absorption by surfaces on strain hardening of single crystalline thin films

Abstract

In the paper, by taking advantage of a strain gradient crystal plasticity theory with consideration of dislocation absorption by surfaces, plastic behaviors of thin films with two active slip systems under constrained shear is analytically studied. It is found that the critical loads for the onset of dislocations absorption by surfaces for the two slip systems are size dependent and are greatly affected by the latent hardening in the grain interior, and dislocations absorption by surfaces can significantly change the distributions of the plastic deformation and dislocation density and hence the strain-hardening behaviors.

Appendix

The factors $a_i$ and $b_i (i=1,2,3,4,5,6,7)$ in Eqs. (35) and (36) are expressed as

$$\begin{aligned} a_1= & {} \eta R^{2}\sin ^{2}\theta _1 , a_2 =\eta R^{2}\sin \theta _2 \sin (2\theta _1 -\theta _2 ), \quad a_3 =-\mu (1-\kappa )\sin ^{2}2\theta _1 ,\\ a_4= & {} -\mu (1-\kappa )\sin 2\theta _1 \sin 2\theta _2 , \quad a_5 =-\mu (\cos ^{2}2\theta _1 +\kappa \sin ^{2}2\theta _1 ),\\ a_6= & {} -\mu (\cos 2\theta _1 \cos 2\theta _2 +\kappa \sin 2\theta _1 \sin 2\theta _2 ), \quad a_7 =\mu \cos 2\theta _1 ,\\ b_1= & {} \eta R^{2}\sin \theta _1 \sin (2\theta _2 -\theta _1 ), \quad b_2 =\eta R^{2}\sin ^{2}\theta _2 ,\\ b_3= & {} -\mu (1-\kappa )\sin 2\theta _1 \sin 2\theta _2 , \quad b_4 =-\mu (1-\kappa )\sin ^{2}2\theta _2 ,\\ b_5= & {} -\mu (\cos 2\theta _1 \cos 2\theta _2 +\kappa \sin 2\theta _1 \sin 2\theta _2 ), \quad b_6 =-\mu (\cos ^{2}2\theta _2 +\kappa \sin ^{2}2\theta _2 ),\\ b_7= & {} \mu \cos 2\theta _2. \end{aligned}$$

The constants $D_{i1}$,$D_{i2}$ and $D_{i3} (i=3,4,5)$ associated with $D_3 ,D_4$ and $D_5$ in Eqs. (37) and (38) are

$$\begin{aligned} D_{31}= & {} \frac{-\left( a_4 b_5 -a_5 b_4 \right) }{2\left( a_2 b_3 -a_3 b_2 -a_1 b_4 +a_4 b_1 \right) }, D_{32} =\frac{-\left( a_4 b_6 -a_6 b_4 \right) }{2\left( a_2 b_3 -a_3 b_2 -a_1 b_4 +a_4 b_1 \right) },\\ D_{33}= & {} \frac{-\left( a_4 b_7 -a_7 b_4 \right) }{2\left( a_2 b_3 -a_3 b_2 -a_1 b_4 +a_4 b_1 \right) }, D_{41} =\frac{\left( a_3 b_2 b_3 -a_3 b_1 b_4 \right) a_5 -\left( a_2 a_3 b_3 -a_1 a_3 b_4 \right) b_5 }{a_3 b_3 \left( a_2 b_3 -a_3 b_2 -a_1 b_4 +a_4 b_1 \right) },\\ D_{42}= & {} \frac{\left( a_3 b_2 b_3 -a_3 b_1 b_4 \right) a_6 -\left( a_2 a_3 b_3 -a_1 a_3 b_4 \right) b_6 }{a_3 b_3 \left( a_2 b_3 -a_3 b_2 -a_1 b_4 +a_4 b_1 \right) },\\ D_{43}= & {} \frac{\left( a_3 b_2 b_3 -a_3 b_1 b_4 \right) a_7 -\left( a_2 a_3 b_3 -a_1 a_3 b_4 \right) b_7 }{a_3 b_3 \left( a_2 b_3 -a_3 b_2 -a_1 b_4 +a_4 b_1 \right) }, \quad D_{51} =\frac{\left( a_3 b_5 -a_5 b_3 \right) }{2\left( a_2 b_3 -a_3 b_2 -a_1 b_4 +a_4 b_1 \right) },\\ D_{52}= & {} \frac{\left( a_3 b_6 -a_6 b_3 \right) }{2\left( a_2 b_3 -a_3 b_2 -a_1 b_4 +a_4 b_1 \right) }, D_{53} =\frac{\left( a_3 b_7 -a_7 b_3 \right) }{2\left( a_2 b_3 -a_3 b_2 -a_1 b_4 +a_4 b_1 \right) }. \end{aligned}$$