Chapter 4, Problem 4.29P

Interpretation Introduction

Interpretation:

The inter planar spacing, the length of the Burgers vector for slip 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 4.29P

The length of Burgers vector for slip system $\left(110\right)/\left[1\overline{1}1\right]$ in BCC tantalum is $2.860\stackrel{0}{A}$

The inter planar spacing for lip system of plane $\left(110\right)$ in BCC tantalum is $2.335\dot{A}$.

The length of Burgers vector for slip system $\left(111\right)/\left[1\overline{1}0\right]$ in BCC tantalum is $4.671\stackrel{0}{A}$.

The inter planar spacing for slip system of plane $\left(111\right)$ in BCC tantalum is $1.907\dot{A}$.

  ${\left(d/b\right)}_{\left(111\right)/\left[1\overline{1}0\right]}=0.408$

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 $\left(110\right)/\left[1\overline{1}1\right]$ in BCC tantalum 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 parameters.

The lattice parameter $a_0\text{ and }\dot{A}$ of tantalum are $\sqrt{3}\text{ and 3}\text{.3026}$.

Substituting as $\sqrt{3}\text{ for }{a}_0\text{ and 3}\text{.3026 for }\dot{A}$

  $\begin{array}{c}b=\left(\frac{1}{2}\right)\left(\sqrt{3}\right)\left(3.3026\right)\\ =2.860\dot{A}\end{array}$

Hence, the length of Burgers vector for slip system $\left(110\right)/\left[1\overline{1}1\right]$ in BCC tantalum is $2.860\dot{A}$

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

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

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

Substitute $1\text{for }h,1\text{ for }k,0\text{ for }l\text{ and 3}\text{.3026 for }\dot{A}$

  $\begin{array}{c}{d}_{110}=\frac{3.3026\dot{A}}{\sqrt{1^2+1^2+0^2}}\\ =2.335\dot{A}\end{array}$

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

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

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

The lattice parameter $a_0\text{ and}\dot{A}$ of tantalum are $\sqrt{2}\text{ and 3}\text{.3026}$.

Substitute $\sqrt{2}\text{ for }{a}_0\text{and 3}\text{.3026 for }\dot{A}$.

  $\begin{array}{c}b=\left(\sqrt{2}\right)\left(3.3026\right)\\ =4.671\dot{A}\end{array}$

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

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

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

Substitute $1\text{for }h,1\text{ for }k,1\text{ for }l\text{ and 3}\text{.3026 for }\dot{A}$

  $\begin{array}{c}{d}_{\left(111\right)}=\frac{3.3026\dot{A}}{\sqrt{1^2+1^2+1^2}}\\ =1.907\dot{A}\end{array}$

Hence, the inter planar spacing for slip system of plane $\left(111\right)$ in BCC tantalum Is $1.907\dot{A}$.

The ratio between the inter planar spacing and length of Burgers vector for slip system of $\left(110\right)/\left[1\overline{1}1\right]$ is calculated as:

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

Substitute $2.335\dot{A}\text{ for }{d}_{110}\text{ and 2}\text{.860}\dot{A}$ for $b$.

  $\begin{array}{c}{\left(d/b\right)}_{\left(110\right)/\left[1\overline{1}1\right]}=\frac{2.335\dot{A}}{2.860\dot{A}}\\ =0.816\end{array}$

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:

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

Substitute $1.907\dot{A}\text{ for }{d}_{111}\text{ and 4}\text{.671}\dot{A}\text{for }b.$

  $\begin{array}{c}{\left(d/b\right)}_{\left(111\right)/\left[1\overline{1}0\right]}=\frac{1.907\dot{A}}{4.671\dot{A}}\\ =0.408\end{array}$

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

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

Substitute $2\text{ for }k,\text{0}\text{.816}\text{for }{\left(d/b\right)}_{\left(110\right)/\left[1\overline{1}1\right]}$, and $0.408\text{ for }{\left(d/b\right)}_{\left(111\right)/\left[1\overline{1}0\right]}$

  $\begin{array}{c}\frac{{\tau }_{\left(111\right)/\left[1\overline{1}0\right]}}{{\tau }_{\left(110\right)/\left[1\overline{1}1\right]}}=\frac{\exp \left[-2\left(0.816\right)\right]}{\exp \left[-2\left(0.408\right)\right]}\\ =0.44\end{array}$

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

Conclusion

The length of Burgers vector for slip system $\left(110\right)/\left[1\overline{1}1\right]$ in BCC tantalum is $2.860\stackrel{0}{A}$

The inter planar spacing for lip system of plane $\left(110\right)$ in BCC tantalum is $2.335\dot{A}$.

The length of Burgers vector for slip system $\left(111\right)/\left[1\overline{1}0\right]$ in BCC tantalum is $4.671\stackrel{0}{A}$.

The inter planar spacing for slip system of plane $\left(111\right)$ in BCC tantalum is $1.907\dot{A}$.

  ${\left(d/b\right)}_{\left(111\right)/\left[1\overline{1}0\right]}=0.408$

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