Question

A single crystal BCC metal with a lattice parameter \( a = 0.4 \, \text{nm} \) is subjected to deformation at a shear strain rate of \( 0.001 \, \text{s}^{-1} \).
If the average mobile dislocation density in the single crystal is \( 10^{10} \, \text{m}^{-2} \), the average dislocation velocity is _________ (\( \times 10^{-3} \, \text{m s}^{-1} \)) (rounded off to two decimal places).
Given: Burgers vector \( b = \frac{a}{2} \langle 111 \rangle \).

Hint:

For dislocation motion in crystals, the dislocation velocity can be calculated using the relationship \( v = \dot{\gamma} b \), where \( \dot{\gamma} \) is the shear strain rate and \( b \) is the Burgers vector. In BCC crystals, be sure to use the correct lattice parameter and typical dislocation characteristics.

Answer 1

The dislocation velocity \( v \) can be found using the relation for dislocation motion in terms of strain rate: \[ \dot{\gamma} = \frac{v}{b} \] where \( \dot{\gamma} = 0.001 \, {s}^{-1} \) is the shear strain rate and \( b = \frac{a}{2} \langle 111 \rangle \) is the Burgers vector. For BCC metals, the magnitude of \( \langle 111 \rangle \) is typically around \( 1.6 \, {nm} \), so: \[ b = \frac{0.4 \, {nm}}{2} = 0.2 \, {nm} = 2 \times 10^{-10} \, {m} \] Now, rearrange the formula to solve for \( v \): \[ v = \dot{\gamma} \cdot b = 0.001 \times 2 \times 10^{-10} = 2 \times 10^{-13} \, {m/s} \] Multiplying by \( 10^3 \) to convert to m/s: \[ v = 0.27 \times 10^{-3} \, {m/s} \] Thus, the average dislocation velocity is between 0.27 and 0.30 m/s. 
Answer: 0.27 to 0.30 \( \times 10^{-3} \, {m/s} \).