Question 8.1: A tensile stress of σ = 5 kPa is applied parallel to the [43......
A tensile stress of σ = 5 kPa is applied parallel to the [432] direction of a cubic crystal. Find the shear stress, \tau , on the (11\overline{1} ) plane in the [011] direction.
First find m= \cos \lambda \cos \phi for (11\overline{1} ) [011] slip. In cubic crystals, the normal to a plane has the same indices as the plane, so the normal to (11\overline{1} ) is [11\overline{1}] . Also, in cubic crystals, the cosine of the angle between two directions is given by the dot product of unit vectors in those directions.
Therefore
\cos \phi =(4\cdot 1+3\cdot 1+2\cdot \overline{1} )/[(4^2 +3^2 +2^2)^{1/2} (1^2 +1^2 +\overline{1}^2 )^{1/2} ]
=5/(√29 √3),
\cos \lambda =(4\cdot 0+3\cdot 1+2\cdot 1)/[(4^2 +3^2 +2^2)^{1/2} (0^2 +1^2 +1^2 )^{1/2} ]
=5/( √29 √2),
m= \cos \lambda \cos \phi =25/(29 √6)=0.352,
\tau =m \ \sigma =0.352\times 5 \ kPa = 1.76 \ kPa.