Question

A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensile axis. Three possible slip directions make angles of 62.4 72.0 , and 81.1 with the same tensile axis.(a) Which of these three slip directions is most favored?(b) If plastic deformation begins at a tensile stress of 1.95 MPa (280 psi), determine the critical resolved shear stress for aluminum.

Answer 1

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$_{crss}$ = σ$_{Y}$ ( cosФ cosλ)$_{max}$

now by putting values,

     = (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23