Chapter 6.13, Problem 58AAP

a)

To determine

Constant for the hall- Petch equation has to be determined.

Answer to Problem 58AAP

Constants for the hall- Petch equation is $\underset{\_}{0.36\text{MPa}\cdot {\text{m}}^{1/2}}$ and $\underset{\_}{\text{89}\text{.1 MPa}}$.

Explanation of Solution

Hall-Petch Equation:

The Hall-Petch equation describes the strength of the metal as a combination of frictional stress and a factor ($k_y$) times the inverse of the square root of the grain size (d). It describes strengthening of material due to the reduction in grain size. Toughness of a metal also gets increased due to grain size reduction.

Write the Hall-Petch relation as given as follows:

${\sigma }_y={\sigma }_o+\frac{k_y}{\sqrt{d}}$ (I)

Here, the yield stress of the metal is ${\sigma }_y$, the material constant for starting stress is ${\sigma }_o$, the strengthening coefficient is $k_y$ and the grain diameter is $d.$

Conclusion:

Substitute 185 MPa for ${\sigma }_y$ and $14\text{μm}$ for $d$ in equation (I).

$185\text{ MPa}={\sigma }_0+\frac{k_y}{\sqrt{14\text{μm}}}$

${\sigma }_0=185-\frac{k_y}{\sqrt{14\text{μm}}}$ (II)

Substitute 140 MPa for ${\sigma }_y$ and $50\text{μm}$ for $d$ in equation (I).

$140\text{ MPa}={\sigma }_0+\frac{k_y}{\sqrt{50\text{μm}}}$ (III)

Substitute equation (II) in equation (III).

$\begin{array}{l}140\text{ MPa}=\left(185-\frac{k_y}{\sqrt{14\text{μm}}}\right)+\frac{k_y}{\sqrt{50\text{μm}}}\\ 140=185-k_y\left(\frac{1}{\sqrt{14}}-\frac{1}{\sqrt{50}}\right)\\ 45=126k_y\\ k_y=0.3576\\ k_y\simeq 0.36\text{MPa}\cdot {\text{m}}^{1/2}\end{array}$

Substitute $0.36\text{MPa}\cdot {\text{m}}^{1/2}$ for  $k_y$ in equation (III).

$\begin{array}{l}140\text{ MPa}={\sigma }_0+\frac{0.36\text{MPa}\cdot {\text{m}}^{1/2}}{\sqrt{50\text{μm}}}\\ 140\text{ MPa}={\sigma }_0+50.912\\ {\sigma }_0=89.1\text{MPa}\end{array}$

Thus, constants for the hall- Petch equation is $0.36\text{MPa}\cdot {\text{m}}^{1/2}$ and 89.1 MPa respectively

b)

To determine

Grain size for the strength of 220 MPa has to be determined.

Answer to Problem 58AAP

Grain size for the strength of 220 MPa is $7.55\text{ μm}$

Explanation of Solution

Hall-Petch Equation:

The Hall-Petch equation describes the strength of the metal as a combination of frictional stress and a factor ($k_y$) times the inverse of the square root of the grain size (d). It describes strengthening of material due to the reduction in grain size. Toughness of a metal also gets increased due to grain size reduction.

Write the Hall-Petch relation as given as follows:

${\sigma }_y={\sigma }_o+\frac{k_y}{\sqrt{d}}$ (I)

Here, the yield stress of the metal is ${\sigma }_y$, the material constant for starting stress is ${\sigma }_o$, the strengthening coefficient is $k_y$ and the grain diameter is $d.$

Conclusion:

Substitute 140 MPa for ${\sigma }_y$, 89 MPa for ${\sigma }_0$, and $0.36\text{MPa}\cdot {\text{m}}^{1/2}$ for $k_y$ equation (I).

$\begin{array}{l}220\text{ MPa}=89\text{MPa}+\frac{0.36\text{MPa}\cdot {\text{m}}^{1/2}}{\sqrt{d}}\\ 131\text{ MPa}=\frac{0.36\text{MPa}\cdot {\text{m}}^{1/2}}{\sqrt{d}}\\ \sqrt{d}=\frac{0.36\text{MPa}\cdot {\text{m}}^{1/2}}{131\text{ MPa}}\\ \sqrt{d}=2.7481\times {10}^{-3}\\ d=7.552\times {10}^{-6}\text{m}\times \frac{1\text{μm}}{{10}^{-6}\text{ m}}\\ d\simeq 7.55\text{ μm}\end{array}$

Thus, grain size diameter to obtain a 220 MPa strength is $7.55\text{ μm}$