(a) a single crystal of a metal that has the bcc crystal structure is oriented such that a tensile stress is applied in the [010] direction. if the magnitude of this stress is 2.75 mpa, compute the resolved shear stress in the direction on each of the (110) and (101) planes. (b) on the basis of these resolved shear stress values, which slip system is more favorably oriented

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.

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explanation: