Chapter 4, Problem 4.28P

Interpretation Introduction

Interpretation:

The interplanar spacing, the length of the Burgers vector and the ratio between the shear stresses required for slip for the two systems should be determined.

Concept introduction:

The formula used is:

  $b=\left(\frac{1}{2}\right)\left(a_0\right)\left(\dot{A}\right)$

Here, $a_0\text{ and }\dot{A}$ are lattice parameters.

Answer to Problem 4.28P

The length of Burgers vector for slip system $\left(111\right)/\left[1\overline{1}0\right]$ in FCC aluminum is $2.863A$.

The inter planar spacing for slip system of plane $\left(111\right)$ in FCC aluminum is $2.338A$.

The length of Burgers vector for slip systems $\left(110\right)/\left[1\overline{1}1\right]$ in FCC aluminum is $7.014A$.

The inter planar spacing for slip system of plane $\left(110\right)$ in FCC aluminum is $2.863A$.

The ratio between the interplanar spacing and length of Burgers vector of the slip system

  $of\text{ 111/}\left[1\stackrel{-}{1}0\right]$ Is $0.8166$

The ratio between the interplanar spacing and length of Burgers vector of the slip system

  $of\text{ 110/}\left[1\stackrel{-}{1}1\right]$ Is $0.408$.

The ratio of shear stresses required for two slip systems is $0.44$.

Explanation of Solution

Plane: $\left(110\right)$ and direction $\left[1\stackrel{-}{1}1\right]$

The length of Burgers vector for slip system $\left(111\right)/\left[1\overline{1}0\right]$ in FCC aluminum is calculated as:

  $b=\left(\frac{1}{2}\right)\left(a_0\right)\left(\dot{A}\right)$

Here, $a_0\text{ and }\dot{A}$ are lattice parameter.

The lattice parameter $a_0\text{ and }\dot{A}$ of aluminum is $\sqrt{2}\text{ and 4}\text{.04958 }$.

Substitute $\sqrt{2}\text{ for }{a}_0\text{ and 4}\text{.04958 for }\dot{A}$

  $\begin{array}{c}b=\left(\frac{1}{2}\right)\left(\sqrt{2}\right)\left(4.04958\right)\\ =2.863\dot{A}\end{array}$

Hence, the length of Burgers vector for slip system $\left(111\right)/\left[1\overline{1}0\right]$ in FCC aluminum is $2.863A$.

The inter planar spacing for slip system of plane $\left(111\right)$ in FCC aluminum is calculated as:

  $d_{hkl}=\frac{\dot{A}}{\sqrt{h^2+k^2+l^2}}$

Here, Miller indices of adjacent planes are $h,k,\text{ and }l$.

Substituting $1\text{ for }h,1\text{ for }k,1\text{for }l,\text{ and 4}\text{.04958 for <mover accent="true"><mn>A</mn><mo>˙</mo></mover>}$

  $\begin{array}{c}{d}_{111}=\frac{4.04958\dot{A}}{\sqrt{1^2+1^2+1^2}}\\ =2.338\dot{A}\end{array}$

Hence, the inter planar spacing for slip system of plane $\left(111\right)$ in FCC aluminum is $2.338A$.

The length of Burgers vector for slip system $\left(110\right)/\left[1\overline{1}1\right]$ in FCC aluminum is calculated as:

  $b=\left(a_0\right)\left(\dot{A}\right)$

Here, $a_0\text{ and }\dot{A}$ are lattice parameter.

The lattice parameter $a_0\text{ and }\dot{A}$ of aluminum as $\sqrt{3}\text{ and 4}\text{.04958}$.

Substituting $\sqrt{3}\text{ for }{a}_0\text{and 4}\text{.04958}\text{for <mover accent="true"><mn>A</mn><mo>˙</mo></mover>}$.

  $\begin{array}{c}b=\left(\sqrt{3}\right)\left(4.04958\right)\\ =7.014\dot{A}\end{array}$

Hence, the length of Burgers vector for slip systems $\left(110\right)/\left[1\overline{1}1\right]$ in FCC aluminum is $7.014A$.

The inter planar spacing for slip system of plane $\left(110\right)$ in FCC aluminum is calculated as:

  $d_{hkl}=\frac{\dot{A}}{\sqrt{h^2+k^2+l^2}}$

Here, Miller indices of adjacent planes are $h,k\text{ and }l$.

Substituting $1\text{ for }h,1\text{ for }k,0\text{for }l,\text{and}\text{4}\text{.04958 for <mover accent="true"><mn>A</mn><mo>˙</mo></mover>}$.

  $\begin{array}{c}{d}_{110}=\frac{4.04958\dot{A}}{\sqrt{1^2+1^2+0^2}}\\ =2.863\dot{A}\end{array}$

Hence, the inter planar spacing for slip system of plane $\left(110\right)$ in FCC aluminum is $2.863\dot{A}$.

The ratio between the inter planar spacing and length of Burgers vector for slip system of $\left(111\right)/\left[1\overline{1}0\right]$ is calculated as:

  ${\left(d/b\right)}_{\left(111\right)/\left[1\overline{1}0\right]}=\frac{d_{111}}{b}$

Substituting $2.338\dot{A}\text{ for }{d}_{111}\text{ and 2}\text{.863}\dot{A}\text{ for }b$

  $\begin{array}{c}{\left(d/b\right)}_{\left(111\right)/\left[1\overline{1}0\right]}=\frac{2.338\dot{A}}{2.863\dot{A}}\\ =0.8166\end{array}$

The ratio between the inter planar spacing and length of Burgers vector for slip system of $\left(111\right)/\left[1\overline{1}0\right]$ is calculated as:

  ${\left(d/b\right)}_{\left(110\right)/\left[1\overline{1}1\right]}=\frac{d_{111}}{b}$

Substitute $2.338\dot{A}\text{ for }{d}_{111}\text{ and 2}\text{.863}\dot{A}\text{ for }b$

  $\begin{array}{c}{\left(d/b\right)}_{\left(110\right)/\left[1\overline{1}1\right]}=\frac{2.863\dot{A}}{7.014\dot{A}}\\ =0.408\end{array}$

The ratio of shear stresses required for two slip system is calculated as:

  $\frac{{\tau }_{\left(110\right)/\left[1\overline{1}1\right]}}{{\tau }_{\left(110\right)/\left[1\overline{1}1\right]}}=\frac{\exp \left[-k{\left(d/b\right)}_{\left(111\right)/\left[1\overline{1}0\right]}\right]}{\exp \left[-k{\left(d/b\right)}_{\left(110\right)/\left[1\overline{1}1\right]}\right]}$

Substituting $2\text{ for }k,\text{0}\text{.8166}\text{for }{\left(d/b\right)}_{\left(111\right)/\left[1\overline{1}0\right]}$, and $0.408\text{ for }{\left(d/b\right)}_{\left(110\right)/\left[1\overline{1}1\right]}$

  $\begin{array}{c}\frac{{\tau }_{\left(111\right)/\left[1\overline{1}0\right]}}{{\tau }_{\left(110\right)/\left[1\overline{1}1\right]}}=\frac{\exp \left[-2\left(0.8166\right)\right]}{\exp \left[-2\left(0.408\right)\right]}\\ =0.44\end{array}$

Hence, the ratio of shear stresses required for two slip systems is $0.44$.

Conclusion

The length of Burgers vector for slip system $\left(111\right)/\left[1\overline{1}0\right]$ in FCC aluminum is $2.863A$.

The inter planar spacing for slip system of plane $\left(111\right)$ in FCC aluminum is $2.338A$.

The length of Burgers vector for slip systems $\left(110\right)/\left[1\overline{1}1\right]$ in FCC aluminum is $7.014A$.

The inter planar spacing for slip system of plane $\left(110\right)$ in FCC aluminum is $2.863A$.

The ratio between the interplanar spacing and length of Burgers vector of the slip system

  $of\text{ 111/}\left[1\stackrel{-}{1}0\right]$ Is $0.8166$

The ratio between the interplanar spacing and length of Burgers vector of the slip system

  $of\text{ 110/}\left[1\stackrel{-}{1}1\right]$ Is $0.408$.

The ratio of shear stresses required for two slip systems is $0.44$.