Respuesta :
Answer:
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 ([tex]\tau_{crss}[/tex]) is given as:
[tex]\tau_{crss} = \sigma_y*cos(\theta)*max[cos(\lambda)]\\\tau_{crss} = 1.95MPa*cos(28.1)*cos(62.4)\\\tau_{crss}=1.95MPa*0.88*0.46=0.80MPa = 114psi[/tex]
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°.
- A) We want to find the slip direction that is favored.
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;
- cos 62.4° = 0.46
- cos72.0° = 0.31
- cos81.1° = 0.15
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 = [tex]\sigma_{y}[/tex]( cosФ × cosλ)
We are given;
Tensile stress; [tex]\sigma_{y}[/tex] = 1.95 MPa
Plugging in the relevant values gives;
τ_crss = (1.95)[cos(28.1) × cos(62.4)]
τ_crss = 0.80 MPa
Read more about crystal structure at; https://brainly.com/question/14831455
