Question 8.5: Consider a copper sheet with a (011)[211] texture. Predict t......

Let x = [211] and z = [01\overline{1}] . The transverse direction must be y = [211] × [01\overline{1}] = [11\overline{1}] . Copper is fcc. The most favored slip systems would be (11\overline{1} )[101] and (11\overline{1} )[110]. Consider tension applied along [211]. For slip on (11\overline{1} )[101], the resulting strains can be found as

\varepsilon _{x}=\gamma \cos \lambda _{x} \cos \phi_{x}
\cos \lambda _{x}= [211]·[101] = \{(2·1 + 1·0 + 1·1)/[(2^2 +1^2 +1^2)(1^2+ 0^2+1^2)]^{1/2} \}
=3/√(6·2)
\cos \phi_{x}=[211]·[11\overline{1} ] = \{(2·1 + 1·1+ 1·\overline{1} )/[(2^2 +1^2 +1^2)(1^2+ 1^2+\overline{1} ^2)]^{1/2}
=3/√(6·3)
so
\varepsilon _{x}=\gamma (1/√\overline{6} ),
\varepsilon _{z}=\gamma \cos \lambda _{z} \cos \phi_{z};
\cos \lambda _{z}= [01\overline{1} ]·[101] = \{(0·1 + 1·0 + \overline{1} ·1)/[(0^2 +1^2 +\overline{1} ^2)(1^2+ 0^2+1^2)]^{1/2} \}
=\overline{1}/√(2·2)
\cos \phi_{z}=[01\overline{1} ]·[11\overline{1} ] = \{(0·1 + 1·1+ \overline{1} ·\overline{1} )/[(0^2 +1^2 +\overline{1} ^2)(1^2+ 1^2+\overline{1} ^2)]^{1/2} \}
=2/√(2·3)
so
\varepsilon _{z}=\gamma (-1/√\overline{6} ),
\varepsilon _{y}=\gamma \cos \lambda _{y} \cos \phi_{y};
\cos \lambda _{y}= [11\overline{1} ]·[101] = \{(1·1 + 1·0 + \overline{1} ·1)/[(1^2 +1^2 +\overline{1} ^2)(1^2+ 0^2+1^2)]^{1/2} \}
=0/√(3·2)
\cos \phi_{y}=[1\overline{11} ]·[11\overline{1} ] = \{(1·1 + \overline{1} ·1+ \overline{1· 1} )/[(1^2 +1^2 +\overline{1} ^2)(1^2+ 1^2+\overline{1} ^2)]^{1/2} \}
=1/√(3·3)
so
\varepsilon _{y}=0.
Dividing, R=\varepsilon_{y}/ \varepsilon_{z}=0.
Analysis of slip on the (1\overline{11} )[110] system would have resulted in exactly the same conclusion.