Chapter 6.13, Problem 53AAP

a)

To determine

Resolve shear stress acting on the $\left(101\right)\left[\overline{1}11\right]$ system has to be determined.

Answer to Problem 53AAP

Resolve shear stress acting on the $\left(101\right)\left[\overline{1}11\right]$ system is $\underset{\_}{22.45\text{ MPa}}$.

Explanation of Solution

Write the expression for the Schmidt’s law:

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

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

Write the expression for the angle between two systems $\left[a_1{b}_1{c}_1\right]\text{ and}\left[a_2{b}_2{c}_2\right]$.

$\cos \varphi =\left[\frac{\left(a_1\times a_2\right)+\left(b_1\times b_2\right)+\left(c_1\times c_2\right)}{\left[\sqrt{{\left(a_1\right)}^2+{\left(b_1\right)}^2+{\left(c_1\right)}^2}\right]\left[\sqrt{{\left(a_2\right)}^2+{\left(b_2\right)}^2+{\left(c_2\right)}^2}\right]}\right]$ (II)

Here, Directions of BCC crystal are $\left[a_1,b_1,c_1\right]$ and directions are $\left[a_2,b_2,c_2\right]$.

Conclusion:

Substitute 75 MPa for $\sigma$, $45°$ for $\lambda$, and $54.74°$ 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 resolve shear stress acting on the (111) $\left[\overline{1}01\right]$ is $\underset{\_}{30.6\text{ MPa}}$.

Calculate the angle between the uniaxial force and normal to the slip plane system.

Write the expression for the angle between two systems $\left[a_1{b}_1{c}_1\right]\text{ and}\left[a_2{b}_2{c}_2\right]$.

$\cos \varphi =\left[\frac{\left(a_1\times a_2\right)+\left(b_1\times b_2\right)+\left(c_1\times c_2\right)}{\left[\sqrt{{\left(a_1\right)}^2+{\left(b_1\right)}^2+{\left(c_1\right)}^2}\right]\left[\sqrt{{\left(a_2\right)}^2+{\left(b_2\right)}^2+{\left(c_2\right)}^2}\right]}\right]$ (II)

Here, Directions of BCC crystal are $\left[a_1,b_1,c_1\right]$ and directions are $\left[a_2,b_2,c_2\right]$.

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

$\begin{array}{l}\cos \varphi =\left[\frac{\left(0\times 1\right)+\left(0\times 0\right)+\left(1\times 1\right)}{\left[\sqrt{{\left(0\right)}^2+{\left(0\right)}^2+{\left(1\right)}^2}\right]\left[\sqrt{{\left(1\right)}^2+{\left(0\right)}^2+{\left(1\right)}^2}\right]}\right]\\ \cos \varphi =\left[\frac{1}{\sqrt{\left(1\right)\left(2\right)}}\right]\\ \varphi ={\cos }^{-1}\left(0.707\right)\\ \varphi =45°\end{array}$

Calculate the angle between the axial forces and slip direction planes $\left[\text{001}\right]\text{and}\left[\overline{\text{1}}\text{11}\right]$

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

$\begin{array}{l}\cos \lambda =\left[\frac{\left(0\times 1\right)+\left(0\times 0\right)+\left(1\times 1\right)}{\left[\sqrt{{\left(0\right)}^2+{\left(0\right)}^2+{\left(1\right)}^2}\right]\left[\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}\right]}\right]\\ \cos \lambda =\left[\frac{1}{\sqrt{\left(1\right)\left(3\right)}}\right]\\ \lambda ={\cos }^{-1}\left(0.5773\right)\\ \lambda =54.73°\end{array}$

Calculate the resolved shear stress using equation (I).

Substitute 55 MPa for $\sigma$, $45°$ for $\lambda$, and $54.73°$ for $\varphi$ in equation (I).

$\begin{array}{c}{\tau }_r=55\text{ MPa}\times \cos \left(45°\right)\cos \left(54.73°\right)\\ =22.45\text{ MPa}\end{array}$

Thus, the resolved shear stress is $\underset{\_}{22.45\text{ MPa}}$.

b)

To determine

Resolve shear stress acting on the $\left(110\right)\left[\overline{1}11\right]$ system has to be determined.

Answer to Problem 53AAP

Resolve shear stress acting on the $\left(110\right)\left[\overline{1}11\right]$ system is zero.

Explanation of Solution

Write the expression for the angle between two systems $\left[a_1{b}_1{c}_1\right]\text{ and}\left[a_2{b}_2{c}_2\right]$.

$\cos \varphi =\left[\frac{\left(a_1\times a_2\right)+\left(b_1\times b_2\right)+\left(c_1\times c_2\right)}{\left[\sqrt{{\left(a_1\right)}^2+{\left(b_1\right)}^2+{\left(c_1\right)}^2}\right]\left[\sqrt{{\left(a_2\right)}^2+{\left(b_2\right)}^2+{\left(c_2\right)}^2}\right]}\right]$ (II)

Here, Directions of BCC crystal are $\left[a_1,b_1,c_1\right]$ and directions are $\left[a_2,b_2,c_2\right]$.

Conclusion:

Calculate the angle between the uniaxial force and normal to the slip planes.

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

$\begin{array}{l}\cos \lambda =\left[\frac{\left(0\times 1\right)+\left(0\times 0\right)+\left(1\times 1\right)}{\left[\sqrt{{\left(0\right)}^2+{\left(0\right)}^2+{\left(1\right)}^2}\right]\left[\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}\right]}\right]\\ \cos \lambda =\left[\frac{1}{\sqrt{\left(1\right)\left(3\right)}}\right]\\ \lambda ={\cos }^{-1}\left(0.5773\right)\\ \lambda =54.73°\end{array}$

Substitute 55 MPa for $\sigma$, $90°$ for $\lambda$, and $54.73°$ for $\varphi$ in equation (I).

$\begin{array}{c}{\tau }_r=55\text{ MPa}\times \cos \left(90°\right)\cos \left(54.73°\right)\\ =0\text{ MPa}\end{array}$

Thus, the resolved shear stress is zero , it indicates that this plane is shear less planes.

c)

To determine

Check the slip is occur or not in the two given system.

Answer to Problem 53AAP

Slip is occurring in $\left(111\right)\left[\overline{1}01\right]$ system.

Explanation of Solution

Refer the table 6.4,” Room-temperature slips systems and critical resolved shear stress for metal single crystals”, the $\left(111\right)\left[\overline{1}01\right]$ slip system value of resolved shear stress is much more higher than the table values in the FCC crystal structure column, so the slip is occur in the $\left(111\right)\left[\overline{1}01\right]$ slip system.