Question

A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the [100] direction. If the magnitude of this stress is 2.47 MPa, compute the resolved shear stress in direction and on the (110) plane.

Answer 1

For [1 1 0] and [1 0 1] plane, σₓ = 6.05 MPa

For [0 1 1] plane, σ = 0; slip will not occur

Step-by-step explanation:

compute the resolved shear stress in [111] direction on each of the [110], [011] and on the [101] plane.

Given;

Stress direction: [1 0 0] ⇒ A

Slip direction: [1 1 1]

Normal to slip direction: [1 1 1] ⇒ B

∅ is the angle between A & B

Step 1: cos∅ = A·B/|A| |B| =
`([100][111])/(√(1).√(3) )` ⇒ cos∅ = 1/
`√(3)`

σₓ = τ/cos ∅·cosλ

where τ is the critical resolved shear stress given as 2.47MPa

Step 2: Solve for the slip along each plane

(a) [1 1 0]

cosλ = [1 1 0]·[1 0 0]/(
`√(2)`·
`√(1)`)

note: cosλ = slip D·stress D/|slip D||stress D|

cosλ = 1/
`√(2)`

∵ σₓ = τ/
`(1)/(√(2) )` ·
`(1)/(√(3) )` =
`√(6)` * 2.47MPa = 6.05MPa

Hence, stress necessary to cause slip on [1 1 0] is 6.05MPa

(b) [0 1 1]

cosλ = [0 1 1]·[1 0 0]/(
`√(2)`·
`√(1)`) = 0

∵ σₓ = 2.47MPa/0, which is not defined

Hence, for stress along [1 0 0], slip will not occur along [0 1 1]

(c) [1 0 1]

cosλ = [0 1 1]·[1 0 0]/(
`√(2)`·
`√(1)`)

cosλ = 1/
`√(2)`

∵ σₓ = τ/
`(1)/(√(2) )` ·
`(1)/(√(3) )` =
`√(6)` * 2.47MPa = 6.05MPa

See attachment for the space diagram