Chapter 8.19, Problem 1QP

To determine

The extension or expansion in the chain (in miles) and the chain length of dislocations:

Answer to Problem 1QP

The extension or expansion in the chain (in miles) is $\text{62}\text{mi}$ and the chain length of dislocations is $6.2\left({10}^5\right)\text{mi}$.

Explanation of Solution

Given:

The dislocation density of a metal specimen, ${10}^5{\text{mm}}^{-2}$.

The volume of metal specimen, $\text{1000}{\text{mm}}^{\text{3}}$.

The increased density of metal specimen after cold working, ${10}^9{\text{mm}}^{-2}$.

Explanation:

The dislocation density is defined as the total dislocation length per unit volume.

Express the extension or expansion in the chain.

$l_{\text{chain}}=V_{\text{metal}}\left({\rho }_{\text{dislocation}}\right)$ (I)

Here, expansion in the chain is $l_{\text{chain}}$, volume of metal specimen is $V_{\text{metal}}$ and dislocation density is ${\rho }_{\text{dislocation}}$.

Express the chain length of dislocations.

$l_{\text{chain}\text{dislocations}}=V_{\text{metal}}\left({\rho }_{\text{metal}}\right)$ (II)

Here, chain length of dislocation is $l_{\text{chain}\text{dislocations}}$, and metal density is ${\rho }_{\text{metal}}$.

Conclusion:

Substitute ${10}^5{\text{mm}}^{-2}$ for ${\rho }_{\text{dislocation}}$ and $\text{1000}{\text{mm}}^{\text{3}}$ for $V_{\text{metal}}$ in Equation (I).

$\begin{array}{c}{l}_{\text{chain}}=\text{1000}{\text{mm}}^{\text{3}}\left({10}^5{\text{mm}}^{-2}\right)\\ ={10}^8\text{mm}\left[\frac{{10}^{-3}\text{m}}{1\text{mm}}\right]\\ ={10}^5\text{m}\left[\frac{0.00062\text{mi}}{1\text{m}}\right]\\ =62\text{mi}\end{array}$

Hence, the extension or expansion in the chain (in miles) is $\text{62}\text{mi}$.

Substitute ${10}^9{\text{mm}}^{-2}$ for ${\rho }_{\text{metal}}$ and $\text{1000}{\text{mm}}^{\text{3}}$ for $V_{\text{metal}}$ in Equation (II).

$\begin{array}{c}{l}_{\text{chain}\text{dislocations}}=\text{1000}{\text{mm}}^{\text{3}}\left({10}^9{\text{mm}}^{-2}\right)\\ ={10}^{12}\text{mm}\left[\frac{{10}^{-3}\text{m}}{1\text{mm}}\right]\\ ={10}^9\text{m}\left[\frac{0.00062\text{mi}}{1\text{m}}\right]\\ =6.2\times {10}^5\text{mi}\end{array}$

Hence, the chain length of dislocations is $\underset{\_}{6.2\left({10}^5\right)\text{mi}}$.