Respuesta :
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
The most favored slip direction and the critical resolved shear stress for zinc are repsectively; 30° direction and τ = 0.9 MPa
What is the slip direction of the crystal structure?
A) We are given the three possible slip direction angle as; 30°, 48° and 78°.
The cosines for the possible slip angles are;
For 30°; cos 30 = 0.867
For 48°; cos 48 = 0.67
For 78°; cos 78 = 0.21
The most favored slip will be the direction for which the Schmid factor (cosines of angles) is maximum.
From the three cosine values above, the largest is λ = 30° and as such it is the most favored slip direction.
B) We are given;
Tensile stress; σ = 2.5MPa.
The angle between the slip plane normal and the tensile axis; Φ = 65°
Thus, critical resolved shear stress for zinc is;
τ = σ*cos(Φ)*cos(λ)
Plugging in the relevant values gives;
τ = (2.5*cos(65)*cos(30)
τ = 0.9 MPa
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