A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65∘ with the tensile axis. Three possible slip directions make angles of 30∘, 65∘, and 78∘ 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 2.5 MPa (355 psi), determine the critical resolved shear stress for zinc.

(a) 30°      (b)0.9MPa

Explanation:

Ф = 65°

(a) First, you find the cosines for the possible λ values given

λ ∈ ( 30°, 65°, 78°)

30° ; cos 30 = 0.87

65° ; cos 65 = 0.42

78° ; cos 78  = 0.21

(make sure when using your calculator to calculate for cos, input the values as degrees. e.g cos 30 in deg form is 0.87, but in rad form is 0.15).

among these calculated cosine values, slip will occur along the direction for which (cosФ cos λ) is maximum

The maximum value of Schmid factor is 0.87. Thus, the most favored slip direction is 30° with the tensile axis.

(b)The plastic deformation begins at a tensile stress of 2.5MPa. Also, the value of the angle between the slip plane normal and the tensile axis is mentioned as 65°

From the expression for Schmid’s law:

τ = σ*cos(Φ)*cos(λ)

Substituting 2.5MPa for σ, 30° for λ and 65° for Φ

τ = 2.5*cos(65°)*cos(30°)

=   2.5 × 0.42 × 0.87 = 0.91

The critical resolved shear stress for zinc, τ = 0.9 MPa

layer a is the exosphere and layer b is the thermosphere.   is the answer i think but i hope this .

i think the answer is a, hope this is right, that would be my best guess.