Chapter 4, Problem 4.42P

Interpretation Introduction

Interpretation:

The resolved shear stress acting on the (111) slip plane in the $\left[\overline{1}10\right]$ and $\left[10\overline{1}\right]$ slip directions needs to be calculated. The slip system(s) that become active first needs to be identified.

Concept Introduction:

The resolved shear stress operating on a slip direction/plane is given by $\pi =\sigma \mathrm{Cos}\lambda \cos \varphi$

The expression for resolved shear stress is as follows:

  ${\tau }_r=\sigma \cos \lambda \cos \phi$

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

Answer to Problem 4.42P

The resolved shear stress acting on the slip plane $\left(111\right)$ in the $\left[\stackrel{-}{1}10\right]$ direction is zero.

Explanation of Solution

Considering the slip occurring on the $\left(111\right)$ plane in a direction of $\left[\overline{1}10\right]$.

The angle between the applied stress direction $\left[001\right]$ and normal to the $\left(111\right)$ plane can be calculated as follows:

  $\phi ={\cos }^{-1}\left[\frac{u_1{u}_2+v_1{v}_2+w_1{w}_2}{\sqrt{\left(u_1^2+v_1^2+w_1^2\right)\left(u_2^2+v_2^2+w_2^2\right)}}\right]$

Here, $u_1,v_1,w_1$ are the components of direction of applied stress and $u_2,v_2,w_2$ are the components of slip plane.

Substitute $0\text{ for }{u}_1,0\text{ for }{v}_1,1\text{ for }{w}_1,1\text{for }{u}_2,1\text{ for }{v}_2,\text{and}\text{1 for }{w}_2$.

  $\begin{array}{c}\phi ={\cos }^{-1}\left[\frac{u_1{u}_2+v_1{v}_2+w_1{w}_2}{\sqrt{\left(u_1^2+v_1^2+w_1^2\right)\left(u_2^2+v_2^2+w_2^2\right)}}\right]\\ ={\cos }^{-1}\left[\frac{0+0+1}{\sqrt{\left(0^2+0^2+1^2\right)\left(1^2+1^2+1^2\right)}}\right]\\ ={54.73}^{\text{o}}\end{array}$

Find the angle between the direction of the applied stress $\left[001\right]$ and normal to the $\left[\overline{1}10\right]$ plane.

  $\lambda ={\cos }^{-1}\left[\frac{u_1{u}_2+v_1{v}_2+w_1{w}_2}{\sqrt{\left(u_1^2+v_1^2+w_1^2\right)\left(u_2^2+v_2^2+w_2^2\right)}}\right]\mathrm{..............}\left(1\right)$

Here, $u_1,v_1,w_1$ are the components of direction of applied stress and $u_2,v_2,w_2$ are the components of slip plane.

Substitute $0\text{ for }{u}_1,0\text{ for }{v}_1,1\text{ for }{w}_1,-1\text{for }{u}_2,1\text{ for }{v}_2,\text{and}0\text{ for }{w}_2$.

  $\begin{array}{c}\lambda ={\cos }^{-1}\left[\frac{u_1{u}_2+v_1{v}_2+w_1{w}_2}{\sqrt{\left(u_1^2+v_1^2+w_1^2\right)\left(u_2^2+v_2^2+w_2^2\right)}}\right]\\ ={\cos }^{-1}\left[\frac{0}{\sqrt{\left(0^2+0^2+1^2\right)\left({\left(-1\right)}^2+1^2+0^2\right)}}\right]\\ ={90}^{\text{o}}\end{array}$

Find the resolved shear stress acting on the $\left(111\right)$ plane.

  ${\tau }_r=\sigma \cos \lambda \cos \phi \mathrm{.......}\left(2\right)$

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

Substitute $5000psi$ for $\sigma ,{90}^{\text{o}}\text{ for }\lambda \text{and 54}{\text{.73}}^{\text{o}}\text{for }\phi$.

  $\begin{array}{c}{\tau }_r=\sigma \cos \lambda \cos \phi \\ =5000\times \cos 90\times \cos 54.73\\ =0\end{array}$

Therefore, the resolved shear stress acting on the slip plane $\left(111\right)\text{ in the }\left[\overline{1}10\right]$ direction is $0$.

Find the Schmidt factor.

  $\begin{array}{c}\cos \lambda \cos \phi =\cos 90\cos 54.73\\ =0\end{array}$

Now, considering the slip occurring on the $\left(111\right)$ plane in a direction $\left[10\overline{1}\right]$.

The angle between the applied stress $\left[001\right]$ direction and the slip direction $\left[10\overline{1}\right]$ can be calculated as follows:

  $\lambda ={\cos }^{-1}\left[\frac{u_1{u}_2+v_1{v}_2+w_1{w}_2}{\sqrt{\left(u_1^2+v_1^2+w_1^2\right)\left(u_2^2+v_2^2+w_2^2\right)}}\right]$

Substitute $0\text{ for }{u}_1,0\text{ for }{v}_1,1\text{ for }{w}_1,1\text{for }{u}_2,0\text{ for }{v}_2,\text{and}-\text{1 for }{w}_2$ in equation $\left(1\right)$.

  $\begin{array}{c}\lambda ={\cos }^{-1}\left[\frac{u_1{u}_2+v_1{v}_2+w_1{w}_2}{\sqrt{\left(u_1^2+v_1^2+w_1^2\right)\left(u_2^2+v_2^2+w_2^2\right)}}\right]\\ ={\cos }^{-1}\left[\frac{0+0-1}{\sqrt{\left(0^2+0^2+1^2\right)\left({\left(1\right)}^2+0^2+{\left(-1\right)}^2\right)}}\right]\\ ={135}^{\text{o}}\end{array}$

Substitute $5000psi\text{ for }\sigma {\text{, 135}}^{\text{o}}\text{ for }\lambda \text{and 54}{\text{.73}}^{\text{o}}\text{for }\phi$ in equation $\left(2\right)$.

  $\begin{array}{c}{\tau }_r=\sigma \cos \lambda \cos \phi \\ =5000\times \cos 135\times \cos 54.73\\ =-2041.5psi\end{array}$ ]

Therefore, the resolved shear stress acting on the slip plane $\left(111\right)$ in the $\left[10\overline{1}\right]$ direction is $-2041.5psi$.

Find the Schmidt factor.

  $\begin{array}{c}\cos \lambda \cos \phi =\cos 135\cos 54.73\\ =-0.408\end{array}$

The Schmidt factor is more for $\left(111\right)$ plane in a $\left[10\overline{1}\right]$ direction. Therefore, the slip system $\left(111\right),\left[10\overline{1}\right]$ will become active first.

Conclusion

The resolved shear stress acting on the slip plane $\left(111\right)$ in the $\left[\stackrel{-}{1}10\right]$ direction is zero.