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Chapter 7: Problem 9

Equations 7.1a and 7.1b, expressions for Burgers vectors for \(\mathrm{FCC}\) and BCC crystal structures, are of the form $$ \mathbf{b}=\frac{a}{2}\langle u v w\rangle $$ where \(a\) is the unit cell edge length. Also, because the magnitudes of these Burgers vectors may be determined from the following equation: $$ |\mathbf{b}|=\frac{a}{2}\left(u^{2}+v^{2}+w^{2}\right)^{1 / 2} $$ determine values of \(|\mathbf{b}|\) for aluminum and chromium. You may want to consult Table 3.1.

Short Answer

Expert verified
The magnitudes of the Burgers vectors for aluminum (Al) and chromium (Cr) are: - For aluminum (Al), \(|\mathbf{b}|_{Al} = 2.864\) Å - For chromium (Cr), \(|\mathbf{b}|_{Cr} = 2.523\) Å

Step by step solution

01

Find the unit cell edge length for aluminum and chromium

We consult Table 3.1 to find the unit cell edge lengths for aluminum (Al) and chromium (Cr). For aluminum (Al), which has an FCC structure, the unit cell edge length \(a_{Al} = 4.05\) Å. For chromium (Cr), which has a BCC structure, the unit cell edge length \(a_{Cr} = 2.91\) Å.
02

Use the Burgers vector magnitude equation for aluminum

We will determine the Burgers vector magnitude for aluminum (Al) using the given equation: $$ |\mathbf{b}|_{Al}=\frac{a_{Al}}{2}\left(u^{2}+v^{2}+w^{2}\right)^{1 / 2} $$ For an FCC structure, the Burgers vector is given in the form of \(\frac{a}{2}\langle u v w\rangle\). It is known that, in an FCC crystal, the Burgers vector is \(\frac{a}{2}\langle 1 1 0 \rangle\). Now, substitute \(a_{Al}\) and the values of \(u\), \(v\), and \(w\) into the equation: $$ |\mathbf{b}|_{Al}=\frac{4.05}{2}\sqrt{(1)^{2}+(1)^{2}+(0)^{2}} $$
03

Calculate the Burgers vector magnitude for aluminum

Now, calculate the Burgers vector magnitude for aluminum: $$ |\mathbf{b}|_{Al}=\frac{4.05}{2}\sqrt{1+1}=\frac{4.05}{2}\sqrt{2} $$ $$ |\mathbf{b}|_{Al}=2.864 \ Å $$
04

Use the Burgers vector magnitude equation for chromium

We will determine the Burgers vector magnitude for chromium (Cr) using the given equation: $$ |\mathbf{b}|_{Cr}=\frac{a_{Cr}}{2}\left(u^{2}+v^{2}+w^{2}\right)^{1 / 2} $$ For a BCC structure, the Burgers vector is given in the form of \(\frac{a}{2}\langle u v w\rangle\). It is known that, in a BCC crystal, the Burgers vector is \(\frac{a}{2}\langle 1 1 1 \rangle\). Now, substitute \(a_{Cr}\) and the values of \(u\), \(v\), and \(w\) into the equation: $$ |\mathbf{b}|_{Cr}=\frac{2.91}{2}\sqrt{(1)^{2}+(1)^{2}+(1)^{2}} $$
05

Calculate the Burgers vector magnitude for chromium

Now, calculate the Burgers vector magnitude for chromium: $$ |\mathbf{b}|_{Cr}=\frac{2.91}{2}\sqrt{1+1+1}=\frac{2.91}{2}\sqrt{3} $$ $$ |\mathbf{b}|_{Cr}=2.523 \ Å $$
06

State the results

We found the following magnitudes of Burgers vectors for aluminum (Al) and chromium (Cr): - For aluminum (Al), \(|\mathbf{b}|_{Al} = 2.864\) Å - For chromium (Cr), \(|\mathbf{b}|_{Cr} = 2.523\) Å

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Most popular questions from this chapter

Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows: $$ \begin{array}{lcr} \hline & \begin{array}{c} \text { Circular } \\ \text { (diameter, } \mathbf{m m}) \end{array} & \begin{array}{c} \text { Rectangular } \\ (\mathbf{m m}) \end{array} \\ \hline \text { Original dimensions } & 15.2 & 125 \times 175 \\ \text { Deformed dimensions } & 11.4 & 75 \times 200 \\ \hline \end{array} $$ Which of these specimens will be the hardest after plastic deformation, and why?

A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an angle of \(28.1^{\circ}\) with the tensile axis. Three possible slip directions make angles of \(62.4^{\circ}, 72.0^{\circ}\), and \(81.1^{\circ}\) with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of \(1.95 \mathrm{MPa}(280 \mathrm{psi})\), determine the critical resolved shear stress for aluminum.

(a) Show, for a tensile test, that $$ \% \mathrm{CW}=\left(\frac{\epsilon}{\epsilon+1}\right) \times 100 $$ if there is no change in specimen volume during the deformation process (i.e., \(A_{0} l_{0}=A_{d} l_{d}\) ). (b) Using the result of part (a), compute the percent cold work experienced by naval brass (the stress-strain behavior of which is shown in Figure 6.12) when a stress of 400 MPa \((58,000\) psi) is applied.

Consider a single crystal of silver oriented such that a tensile stress is applied along a \([001]\) direction. If slip occurs on a (111) plane and in a [ \(\overline{101}\) ] direction, and is initiated at an applied tensile stress of \(1.1 \mathrm{MPa}\) (160 psi), compute the critical resolved shear stress.

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the \([110]\) direction. If the critical resolved shear stress for this material is \(1.75 \mathrm{MPa}\), calculate the magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the [1\overline{110], [10\overline{1} ] } \text { and } [ 0 1 \overline { 1 } ] \text { directions. }
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