MATERIALS SCI + ENGR: INT W/ACCESS
MATERIALS SCI + ENGR: INT W/ACCESS
10th Edition
ISBN: 9781119808084
Author: Callister
Publisher: WILEY
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Chapter 7, Problem 1QAP
To determine

The distance of the chain length extended is to be determined and the chain length is to be determined.

Expert Solution & Answer
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Answer to Problem 1QAP

The chain extends up to 0.06462 mile , and the length of the chain is 0.6275 mile .

Explanation of Solution

Given:

Density of the dislocation, ρ=104 mm2

Volume of the dislocation, V=1000 mm3 .

The formula used for the length of the atomic chain without dislocation.

  Lw=ρV

Here, Lw is the length of the atomic chain without dislocation.

Substitute 104 mm2 for ρ , and 1000 mm3 for V in the above equation.

  Lw=104 mm2×1000 mm3=104000 mm×6.213×107 mile1 mm=0.0646 mile

For chain length:

Substitute 1010 mm2 for ρ , and 1000 mm3 for V in the above equation.

  Lw=1010 mm2×1000 mm3=1010000 mm×6.213×107 mile1 mm=0.6275 mile

Conclusion:

Thus, the required distance of the chain extended is 0.0646 mile , and the chain length is 0.6275 mile .

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