Question

Suppose that all the dislocations in 1600 mm3 of crystal were somehow removed and linked end to end. Given 1 m = 0.0006214 mile, how far (in miles) would this chain extend for dislocation densities of (a) 10^4 mm-2 (undeformed metal)? (b) 10^10 mm-2 (cold-worked metal)?

Answer 1

Answer:

(a) 9.924 mile

(b) 0.0994 mile

Explanation:

We have given dislocation volume $V=1600mm^3(a) Density of the dislocation d = [tex]10^4mm^{-2}$

We know that dislocation density is given by

$d=\frac{length\ of\ dislocation}{volume}$

So ${length\ of\ dislocation}={volume}\times density$

${length\ of\ dislocation}=1600\times 10^4mm=16000m =16000\times 0006214=9.924mile$

(b) ${length\ of\ dislocation}=1600\times 10^2mm=160m =160\times 0006214=0.0994mile$