Question 8.2: A cubic crystal is subjected to a stress state, σx = 15 kPa,......
A cubic crystal is subjected to a stress state, \sigma_{x} = 15 kPa, \sigma_{y} = 0, \sigma_{z} = 7.5 kPa, \tau _{yz} = \tau _{zx}=\tau _{xy} =0, where x = [100], y = [010], and z = [001]. What is the shear stress on the (11\overline{1} )[101] slip system? See Figure 8.3.
\tau _{nd}=l_{nx} l_{dx} \sigma_{x}+ l_{nz} l_{dz} \sigma_{z} , where d = [101] and n = [11\overline{1} ]. Taking dot products,
l_{nx}= [11\overline{1} ]⋅ [100] =(1·1+1⋅ 0+\overline{1}⋅ 0 )/√[(1^2+ 1^2 +\overline{1}^2 )(1^2 +0^2 +0^2)]
=1/ √3,
l_{dx}= [101 ]⋅ [100] =(1·1+0⋅ 0+1⋅ 0 )/√[(1^2+ 0^2 +1^2 )(1^2 +0^2 +0^2)]
=1/ √2,
l_{nz}= [11\overline{1} ]⋅ [001] =(1·0+1⋅ 0+\overline{1} ⋅ 1 )/√[(1^2+ 1^2 +\overline{1} ^2 )(0^2 +0^2 +1^2)]
=-1/ √3,
l_{dz}= [101 ]⋅ [001] =(1·0+0⋅ 0+1 ⋅ 1 )/√[(1^2+ 0^2 +1^2 )(0^2 +0^2 +1^2)]
=1/ √2.
Substituting,
\tau _{nd}=(1/√3)(1/√2)15 \ kPa + (-1/√3)(1/√2)7.5 \ kPa
=7.5/ √6 = 3.06 \ kPa.