Chapter 4, Problem 4.51P

Interpretation Introduction

(a)

Interpretation:

The Hall-Petch equation's constants needs to be calculated.

Concept Introduction:

The relation between the grain size to the yield strength is shown as follows:

  ${\sigma }_y={\sigma }_0+K{d}^{-1/2}\mathrm{..........}\left(1\right)$

Here, the yield strength is ${\sigma }_y$.

Answer to Problem 4.51P

The Hall-Petch equation constants are $K=6.77MPa.m{m}^{-1/2}\text{ and }{\sigma }_0=114.7MPa$

Explanation of Solution

The properties of a Cu-Zn alloy are as follows:

Grain Diameter (mm)	Strength (MPa)
$0.015$	$170MPa$   $\begin{array}{l}170MPa\\ \\ \\ \end{array}$
$0.025$	$158MPa$
$0.035$	$151MPa$
$0.050$	$145MPa$

S.N.	Grain Diameter (mm)	$d^{-1/2}$		Strength (MPa)
$1$	$0.015$	$8.165$  $$		$170MPa$   $\begin{array}{l}170MPa\\ \\ \\ \end{array}$
$2$	$0.025$	$6.325$		$158MPa$
$3$	$0.035$	$5.345$		$151MPa$
$4$	$0.050$	$4.472$		$145MPa$

In the above table, we have calculated $d^{-1/2}$ for each set of values as a part of hall Petch equation.

With the help of equation number $\left(1\right)$, we will set up two simultaneous equations from the given set of data in $S.N$

  $1$ and $4$, which are as follows.

For strength $170MPa$ and grain size is $0.015mm$.

Therefore,

  $170MPa={\sigma }_0+K\left(8.165\right)\mathrm{........}\left(2\right)$

For strength of titanium is $145MPa$ and grain size is $0.050mm$.

Therefore,

  $145MPa={\sigma }_0+K\left(4.472\right)\mathrm{........}\left(3\right)$

Solving equation $\left(2\right)\text{ and }\left(3\right)$, we get.

  $K=6.77MPa.m{m}^{-1/2}$

And

  ${\sigma }_0=114.7MPa$

Hence, the hall-Petch equation constants are $K=6.77MPa.m{m}^{-1/2}\text{ and }{\sigma }_0=114.7MPa$

Interpretation Introduction

(b)

Interpretation:

For a strength of 200MPa, the required grain size needs to be calculated.

Concept Introduction:

The Hall-Petch equation relates the grain size to the yield strength as follows:

  ${\sigma }_y={\sigma }_0+K{d}^{-1/2}\mathrm{..........}\left(1\right)$

Here,

The yield strength is ${\sigma }_y$.

Answer to Problem 4.51P

The grain size which is required is $0.006\text{3 mm}$

Explanation of Solution

  $\text{Strength=200MPa}$

Now, we have the Hall-Petch equation after we calculated the constants, then the equation $\left(1\right)$ will become as given below.

Here, the strength required, ${\sigma }_y\text{ as 200}MPa$.

  ${\sigma }_y={\sigma }_0+K{d}^{-1/2}$

  $\text{200}\text{MPa=114}\text{.7MPa}\text{+6}\text{.77MPa}{\text{.mm}}^{\text{-1/2}}{\left(\text{d}\right)}^{\text{-1/2}}$

  $\begin{array}{l}\text{85}\text{.3MPa=6}\text{.77MPa}{\text{.mm}}^{\text{-1/2}}{\left(\text{d}\right)}^{\text{-1/2}}\\ \text{d=0}\text{.0063 mm}\end{array}$

Hence, the grain size required is $\text{0}\text{.0063 mm}$.