Question

A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65 degree with the tensile axis. Three possible slip directions make angles of 30 degree , 48 degree , and 78 degree 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.

Answer 1

Answer:

a. 30°

b. 0.9MPa

Explanation:

The slip will occur along that direction for which the Schmid factor is maximum. The three possible slip directions are mentioned as 30°, 48°, 78°

The cosines for the possible λ values are given as

For 30°, cos 30 = 0.867

For 48°, cos 48 = 0.67

For 78°, cos 78 = 0.21

Among the three-calculated cosine values, the largest cos(λ) gives the favored slip direction

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

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°

Thus, calculate the value of critical resolved shear stress for zinc:

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

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