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

(a) 9.924 mile

(b) 0.0994 mile

Step-by-step explanation:

We have given dislocation volume
`V=1600mm^3</p><p>(a) Density of the dislocation d = [tex]10^4mm^(-2)`

We know that dislocation density is given by

`d=(length\ of\ dislocation)/(volume)`

So
`{length\ of\ dislocation}={volume}* density`

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

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