Answer:
grain diameter is ![d=4.34*10^-^3\\d =0.00434 mm](https://tex.z-dn.net/?f=d%3D4.34%2A10%5E-%5E3%5C%5Cd%20%3D0.00434%20mm)
Explanation:
i) Lower yield point =230MPa when grain diameter is
mm
ii) Lower yield point =275MPa when grain diameter is
mm
The relation between yield stress and grain size is
-----1
This is the Hall–Petch equation
where Yp is the yield stress,
a is a materials constant
b is the strengthening coefficient (material specific)
and d is the average grain diameter
therefore (i) can be represented as
-----2
And ii) represented as
------3
Further simplifying since
= 10
And
= 12. 91
We have that eqn 2 and 3 becomes respectively
![230 = a + 10b -----4\\275 = a + 12.91b - - - - - -5\\](https://tex.z-dn.net/?f=230%20%3D%20a%20%2B%2010b%20-----4%5C%5C275%20%3D%20a%20%2B%2012.91b%20-%20-%20-%20-%20-%20-5%5C%5C)
Solving the simultaneous eqns 4 and 5
Eqn 5 – eqn 4 gives
![230 = a + 10b -----4\\-\\275 = a + 12.91b - - - - - -5\\](https://tex.z-dn.net/?f=230%20%3D%20a%20%2B%2010b%20-----4%5C%5C-%5C%5C275%20%3D%20a%20%2B%2012.91b%20-%20-%20-%20-%20-%20-5%5C%5C)
![45 =2.91b](https://tex.z-dn.net/?f=45%20%3D2.91b)
Therefore b = 15.46
Substituting the value of b in eqn 4 we have that
![230 =a + (10*15.46) = a +154.6](https://tex.z-dn.net/?f=230%20%3Da%20%2B%20%2810%2A15.46%29%20%3D%20a%20%2B154.6)
Therefore a becomes 230 -154.6 = 75.4
a=75.4
Eqn 1 now becones
-----6
Therefore for a yield point of 310, eqn 6 becomes
![310=75.4 + 15.46/\sqrt{d}\\234.6 = 15.46/\sqrt{d}\\ Cross multiplying,\\\sqrt{d} = 15.46/234.6\\d =(15.46/234.6)^2\\d=4.34*10^-^3\\d =0.00434 mm in 3 significant figures](https://tex.z-dn.net/?f=310%3D75.4%20%2B%2015.46%2F%5Csqrt%7Bd%7D%5C%5C234.6%20%3D%2015.46%2F%5Csqrt%7Bd%7D%5C%5C%20Cross%20multiplying%2C%5C%5C%5Csqrt%7Bd%7D%20%20%3D%2015.46%2F234.6%5C%5Cd%20%3D%2815.46%2F234.6%29%5E2%5C%5Cd%3D4.34%2A10%5E-%5E3%5C%5Cd%20%3D0.00434%20mm%20in%203%20significant%20figures)