When calculating a resolved shear stress within a single unit cell, the angle between the applied stress direction and the slip direction must be determined. Using a BCC unit cell, if a applied stress is in the [110] direction, but slip occurs in the [111] direction, what is value of the angle between the applied direction and the slip direction?

Body-centered cubic lattice (bcc or cubic-I), just like all lattices, has lattice points at the eight corners of the unit cell with an additional points at the center of the cell. It has unit cell vectors a = b = c and interaxial angles α=β=γ=90°.

The simplest crystal structures are those that have present only a single atom at each lattice point.

body-centered cubic unit cell has atoms at each of the eight corners of a cube (like the cubic unit cell) plus one atom in the center of the cube. Each of the corner atoms is the corner of another cube so the corner atoms are shared between eight unit cells. It is said to have a coordination number of 8. The bcc unit cell consists of a net total of two atoms; one in the center and eight eighths from corners atoms

With the use of BCC unit cell, if a applied stress is in [110] direction, but slip applies in [111] direction, the angle between applied direction and slip direction is given as:

[1 1 0] [1 1 1]

λ = Cos^-1 ( 1×1 + 1×1 + 0×1 ÷ (1^2 + 1^2 +0^2) (1^2 + 1^2+ 1^2))

Cos^-1 2/ sqrt 6

= 35.386°

AFOKE88 AFOKE88 · Answer 2 · 2022-03-24T14:34:54+01:00

The value of the angle between the applied direction and the slip direction is; θ = 35.26°

What is the angle for the crystal structure?

Body centered cubic lattice (BCC) is a unit cell that has lattice points at its' eight corners with additional points being at the center of the cell. It also has unit cell vectors a = b = c and interaxial angles α = β = γ=90°.

BCC unit cell has atoms at each of the eight lattice points at the 8 corners with an additional one atom at the center of the cube. The corner atoms are shared between the eight unit cells and as such will have a coordination number of 8.

Finally when using BCC unit cell, if an applied stress is in the [110] direction, but slip occurs in the [111] direction, it means that the angle between applied direction and slip direction is;

θ = Cos⁻¹ ([(1 × 1) + (1 × 1) + (0 × 1)] ÷ √((1² + 1² + 0²)(1² + 1² + 1²))

θ = Cos⁻¹ (2/√6)

θ = 35.26°

Read more about BCC structures at; https://brainly.com/question/14831455