A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an 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, determine the critical resolved shear stress for aluminum.

A) The slip direction that is favored among the three given is;

slip direction that makes an angle of 62.4°

B) The critical resolved shear stress for aluminum is; 0.80 MPa

We are given;

Angle made by slip plane normal; Ф = 28.1°

Possible values angles of slip directions; λ = 62.4°, 72.0° and 81.1°.

The slip will occur in the direction where (cosФ × cosλ) is a maximum.

Since cosФ is constant, then the largest value of cosλ will make  (cosФ cosλ) to be a maximum. Thus;

The largest value for cosλ was for λ = 62.4°

Thus, the slip direction that is favored is at the angle of 62.4° with the tensile axis.

B) The critical resolved shear stress is gotten from the equation;

τ_crss = ( cosФ × cosλ)

We are given;

Tensile stress; = 1.95 MPa

Plugging in the relevant values gives;

τ_crss = (1.95)[cos(28.1) × cos(62.4)]

τ_crss = 0.80 MPa

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a) the slip direction is at an angle of 62.4° with the tensile axis

b) Critical  resolved shear stress is 0.8 MPa

Explanation:

Given that φ = 28.1°,  while possible values for λ are 62.4°, 72.0°, and 81.1°.

a) Slip would occur in a direction where cos Ф and cos λ is maximum. To get the slip direction we choose the highest cos λ

Therefore:

cos(62.4) = 0.46

cos(72) = 0.31

cos(81.1) = 0.15

Therefore, the slip direction is at an angle of 62.4° with the tensile axis

b)

Given σ = 1.95 MPa

Critical  resolved shear stress () is given as:

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