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.

Question

Engineering

Paris Dawson

6 January, 18:25

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.

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Answers (1)

Know the Answer?

Shaniya Schmidt · Answer 1

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