Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ = σ
( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23