Respuesta :
Answer:
Shear stress in the [1’11] direction on the plane [110] is 1.124 MPa, Shear stress in the [1’11] direction on the plane [101] is 0MPa, b. Slip system [110] – [1’11] is most favorably orientated.
Explanation:
a. Calculate the angle between [1’11] and [010] using the following relation
λ = cos⁻¹ [u₁u₂ +v₁v₂ +w₁w₂/√{(u²₁+ v²₁ + w²₁)(u²₂ + v²₂ + w²₂)}]
Where [u1v1w1] = [1’11] and [u2v2w2] = [010]
Putting the values for u, v, w
λ = cos⁻¹[(-1 x 0) + (1 x 1) + (1 x 0)/√{(-1² + 1² + 1²)(0² + 1² + 0²)}]
λ = cos⁻¹[1/√3] = 54.7deg
Calculate the angle φ₁ between the direction [110] and [010] using the following relation
φ₁ = cos⁻¹ [u₁u₂ +v₁v₂ +w₁w₂/√{(u²₁+ v²₁ + w²₁)(u²₂ + v²₂ + w²₂)}]
Here [u1v1w1] = [110] and [u2v2w2] = [010]
φ₁ = cos⁻¹[(1 x 0) + (1 x 1) + (0 x 0)/√(1² + 1² + 0²)(0² + 1² + 0²)]
φ₁ = cos⁻¹[1/√2] = 45°
Determine the angle φ₂ between the directions [101] and [010] using the following relation
φ₂ = cos⁻¹ [u₁u₂ +v₁v₂ +w₁w₂/√{(u²₁+ v²₁ + w²₁)(u²₂ + v²₂ + w²₂)}]
Here [u1v1w1] = [101] and [u2v2w2] = [010]
φ₂ = cos-1[(1 x 0) + (0 x 1) + (1 x 0)/sq(1² + 0² + 1²)(0² + 1² + 0²)]
φ₂ = cos-1[0] = 90°
Calculate the resolved shear stress in the [1’11] direction on the plane [110] by using the following equation
τ(r) = σcosφ1cosλ, where σ is the applied tensile stress
Substitute 2.75 MPa for σ, 45° for φ1 and 54.7° for λ
τ(r) = (2.75 x cos45cos54.7) = 1.124 MPa
Therefore, the shear stress in the [1’11] direction on the plane [110] is 1.124 MPa
Calculate the resolved shear stress in the [1’11] direction on the plane [101] by using the following equation
τ(r) = σcosφ2cosλ
Substitute 2.75 MPa for σ, 90° for φ2 and 54.7° for λ
τ(r) = (2.75 x cos90cos54.7) = 0MPa
Therefore, the shear stress in the [1’11] direction on the plane [101] is 0MPa
b. The most favorable slip system is one that has the largest resolved shear stress value. Therefore, the slip system [110] – [1’11] is most favorably orientated.
