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**Strength of a material can be increase by reducing the average crystallite (grain) size of the material.**

During nucleation and growth, each crystal or grain grow separately as solidification occures, and they have different orientations while growing. These crystals/grains having different orientations meet without a disruption in the continuity of the material, and the interface formed between these crystal/grains is called as grain boundary. As the crystal orientation changes at a grain boundary, so the slip planes does not continue after grain boundary. Hence, the dislocations, gliding in a crystal can not cross the grain boundary and end up getting pile up at the grain boundary.

Inside a large grain, the number of dislocations getting piled-up at the grain boundary is more and hence the stress concentration produced at the leading edge of the pile up will be higher compare to smaller grains.

\[ \boxed{ \text{Stress concentration at the grain boundary} \propto \sqrt{\text{Grain Diameter}} } \]
When the stress concentration at the edge of grain boundary is more then less stress will be required to move dislocation. Which means it becomes easier to move dislocation across the grain boundary in large grain, hence strength of the material decreases.

Hall - Petch Equation :

The Hall-Petch equation expresses the inverse-square-root dependence of grain size on yield or flow strength of polycrystalline metals...

\[ \boxed{ \sigma_{y} = \sigma_{i} + \frac{K}{\sqrt{d}} } \] Here,

\( \sigma_{y} \) : Yield strength of the polycrystalline metal.

\( \sigma_{i} \) : Yield strength of same polycrystalline metal having infinite grain size.

K : Hall - Petch Constant / Locking parameter

\( d \) : Avg grain diameter

For grain size measurement there are several approach such as -

- Counting the number of grains within a given area on a polished surface (ex. circle or rectangle)
- Determining the number of grains (or grain boundary) that intersect a given length of random line.
- Comparing with standard grain-size chart from ASTM.

\[ \boxed{ N^* = 2^{n-1} } \]

GATE 2012 : A material with grain size of ASTM No. 6 has a lattice frictional stress 100 MN/m

Q.1 Grain size of the material is approximately

(A) 45 \(\mu\)m (B) 35 \(\mu\)m (C) 4.5 \(\mu\)m (D) 3.5 \(\mu\)m

Q.2 Yield strength of the material is approximately

(A) 100 MPa (B) 115 MPa (C) 165 MPa (D) 215 MPa

Show Answer

Answer : (A) 45 \( \mu \)m , (B) 115 MPa

**Solution : **

As given,

ASTM No. = 6 , \( \sigma_{i} = 100\) MN/m^{2}, \( k = 0.10 \) MN/m^{3/2}

No. of grains per square inch at 100X = N^{*} = 2^{6-1} = 32

No. of grain per square meter at 100X \[ N = \frac{32}{(2.54 \times 10^{-2})^2} \tag{As 1 inch = 0.0254 m }\]
No. of grain per square meter at 1X \[ N = \frac{32}{(2.54 \times 10^{-2})^2} \times (100)^2\]
\[ N = 496 \times 10^6 \frac{\text{grains}}{m^2} \text{ At 1 X} \]
\[ \textrm{As } d = \sqrt{\frac{1}{N}} \]
\[ \implies d = \sqrt{ \frac{1}{496 \times 10^6} } \]
\[ \implies d= 44.9 \mu m \]
Now using hall petch equation..

\[ \sigma_{y} = 100 + \frac{0.10}{ \sqrt{44.9 \times 10^{-6}} } \]
\[ \sigma_{y} = 115 \text{ } MN/m^2 \textrm{ or MPa} \]