Chapter 6.13, Problem 92SEP

To determine

The tensile stress that must be applied to the $\left[1\overline{1}0\right]$ axis of a high-purity copper single crystal to cause slip on the $\left(1\overline{1}\overline{1}\right)\left[0\overline{1}1\right]$ slip system

Explanation of Solution

Write the expression to determine the angle $\left(\theta \right)$ between the two planes.

$\cos \theta =\left[\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)\cdot \left(a_2^2+b_2^2+c_2^2\right)}}\right]$ (I)

Here, $\left[a_1{b}_1{c}_1\right]$ and $\left[a_2{b}_2{c}_2\right]$ are the directional indices, and the angle between the indices is $\theta$.

Write the expression for the Schmidt law using the equation 6.17.

${\tau }_c=\sigma \cos \varphi \cos \lambda$ (II)

Here, the applied stress is $\sigma$, the angle between the normal to the slip plane and the applied stress direction is $\varphi$, and angle between the slip direction and the applied stress direction is $\lambda$.

Conclusion:

Substitute 1 for $a_1$, 1 for $a_2$, -1 for $b_1$, -1 for $b_2$, 0 for $c_1$, and -1 for $c_2$ in equation (I).

  $\begin{array}{c}\cos \varphi =\left[\frac{\left(1\times 1\right)+\left(-1\times -1\right)+\left(0\times -1\right)}{\sqrt{\left({\left(1\right)}^2+{\left(-1\right)}^2+{\left(0\right)}^2\right)\cdot \left({\left(1\right)}^2+{\left(-1\right)}^2+{\left(-1\right)}^2\right)}}\right]\\ =\frac{2}{\sqrt{2\times 3}}\\ =\frac{2}{\sqrt{6}}\end{array}$

Calculate the angle between $\left[1\overline{1}0\right]\text{ and }\left[0\overline{1}1\right]$ using equation (I).

Substitute 1 for $a_1$, 0 for $a_2$, -1 for $b_1$, -1 for $b_2$, 0 for $c_1$, and 1 for $c_2$ in equation (I).

  $\begin{array}{c}\cos \lambda =\left[\frac{\left(1\times 0\right)+\left(-1\times -1\right)+\left(0\times 1\right)}{\sqrt{\left({\left(1\right)}^2+{\left(-1\right)}^2+{\left(0\right)}^2\right)\cdot \left({\left(0\right)}^2+{\left(-1\right)}^2+{\left(1\right)}^2\right)}}\right]\\ =\frac{1}{\sqrt{2\times 2}}\\ =\frac{1}{2}\end{array}$

Substitute 0.85 MPa for ${\tau }_r$, $\frac{2}{\sqrt{6}}$ for $\cos \varphi$, $\frac{1}{2}$ for $\cos \lambda$ in equation (II).

  $\begin{array}{c}0.85\text{ MPa}=\sigma \times \frac{2}{\sqrt{6}}\times \frac{1}{2}\\ \sigma =\frac{0.85\times \sqrt{6}\times 2}{2}\\ =2.08\text{ MPa}\end{array}$

Thus, the resolved shear stress for the 0.85 MPa is 2.08 MPa.