Chapter 6.13, Problem 52AAP

a)

To determine

The resolved shear stress acting on the $\left(111\right)\left[\overline{1}01\right]$ slips system has to be determined.

Answer to Problem 52AAP

The resolved shear stress acting on the $\left(111\right)\left[\overline{1}01\right]$ slips system is $\underset{\_}{30.6\text{ MPa}}$.

Explanation of Solution

Write the expression for the Schmidt’s law:

  ${\tau }_r=\sigma \cos \lambda \cos \varphi$                                                                               …… (I)

Here, ${\tau }_r$ is resolved shear stress , $\sigma$ is uniaxial stress, $\lambda$ is the angle between the axial force and slip direction, and $\varphi$ is the angle between the uniaxial force and normal to the slip plane.

Conclusion:

Draw the schematic diagram of slip plane and direction of the (111) $[\overline{1}01]$ slip system as in Figure 1.

Draw the geometry of slip plane ($\varphi$ on (110) plane) as shown in Figure 2.

Draw the geometry of slip plane ($\lambda$ on (001) plane) as shown in Figure 3.

The miller indices of the crystal plane and the direction indices of the normal to a crystal plane are same in the cubic system.

Use the figure (3), if the line AE is bisected the the [001] plane then calculate the angle of EAD.

  $\lambda =\angle EAD=45°$

Calculate the angle between the uniaxial force and normal to the slip plane using the Figure 2.

  $\begin{array}{c}\cos \varphi =\frac{a}{\sqrt{3}a}\\ =\frac{1}{\sqrt{3}}\\ ={\cos }^{-1}\left(0.5773\right)\\ \varphi =54.74°\end{array}$

Substitute $45°$ for $\lambda$, 75 MPa for $\sigma$, $54.74°$ for $\varphi$ in Equation (I).

  $\begin{array}{c}{\tau }_r=75\text{ MPa}\times \cos \left(45°\right)\cos \left(54.74°\right)\\ =30.6\text{ MPa}\end{array}$

Thus, the resolved shear stress acting on the (111) $\left[\overline{1}01\right]$ is $\underset{\_}{30.6\text{ MPa}}$.

b)

To determine

The resolved shear stress acting on the $\left(111\right)\left[\overline{1}10\right]$ system has to be determined.

Answer to Problem 52AAP

The resolved shear stress acting on the $\left(111\right)\left[\overline{1}10\right]$ system is zero.

Explanation of Solution

Draw the schematic diagram of slip plane and direction of the (111) $[\overline{1}10]$ slip system as in Figure 4.

Draw the geometry of slip plane ($\varphi$ on (110) plane) as shown in Figure 5.

Refer the Figure 5, the angle between the (111) and [111] are $\left(\gamma \right)$ is $90°$.

Calculate the angle between the axial plane [001] and normal slip plane [111] $\left(\varphi \right)$.

  $\cos \varphi =\frac{AD}{BD}$

Substitute $a$ for $AD$ and $\sqrt{3}a$ for $BD$  in equation (II).

  $\begin{array}{c}\cos \varphi =\frac{a}{\sqrt{3}a}\\ =\frac{1}{\sqrt{3}}\\ ={\cos }^{-1}\left(0.5773\right)\\ \varphi =54.74°\end{array}$

Substitute $90°$ for $\lambda$, 75 MPa for $\sigma$, $54.74°$ for $\varphi$ in equation (I).

  $\begin{array}{c}{\tau }_r=75\text{ MPa}\times \cos \left(90°\right)\cos \left(54.74°\right)\\ =0\end{array}$

Thus, the resolve shear stress acting on the (111) $\left[\overline{1}10\right]$ is $\underset{\_}{0}$.