CHAPTER 6
6.2)
Given: Cu rod heated from 0 1300 K shows increase in length of 2.1%.
Data: , Cu is FCC (n = 4)
Fractional increase in length = , where subscript 0 refers to the 0K state.
There are two contributions to this increase in length (): (i) from thermal expansion () and (ii) from increase in fraction of vacancies due to heating (). The vacancies are created by atoms migrating to the surface leading to an increase in volume of the material. The vacancies are incorporated in the crystal due to the entropic stabilization that it provides (which more than offsets the increase in enthalpy caused by broken bonds).
V = L3 Þ dV = 3L2 dL
Þ (6.2-1)
The fraction required to be calculated is ®
(= x)
1 unit cell gives a volume ↑ of a3 Þ 4 vacancies give a volume of a3
Þ nv vacancies give a volume of
Equation (6.2-1) Þ . Where V0 is given by:
Þ = x ,
6.6)
Given: SC crystal with , positive edge dislocation () l = 1mm climbs down 1µm (= c).
[l = 1mm = 10-3m, c = 1µ = 10-6m].
Number of atoms along l , Number of atoms along c
Area created
Atoms required for climb down by c
These atoms come from the lattice no. of vacancies created = x
6.7)
Given: dislocation density Þ length of dislocation line
Data: , , .
Core saturation with Interstitial implies that there is for each Fe atom along the dislocation line one C atom is present.
In a volume of a3 (Å3) (1 unit cell) there are 2 atoms of Fe.
in 1m3 the number of Fe atoms (nfe) =
For 1 C atom is present.
In of dislocation line the number of atoms of C present is: .
Atom% = = 4.755 × 10-7 %
This implies that a small concentration of carbon can saturate the core of all the edge dislocations.
6.10)
Given: ,
The problem refers to a general case but the illustrative figures are drawn assuming a cubic case for simplicity. refers to the direction perpendicular to and unit vectors are shown by hats.
, ,
, , . (, ).
Slip plane contains both & . Let slip plane (hkl).
Applying Wiess zone law:
On → h + k = 0
On → h + k + 2l = 0 Þ l = 0, h = -k Þ the slip plane
For the screw segment of a dislocation: ||
For the edge segment of the dislocation:
Looking at the figure: , ,
Let the direction to be . This direction lies on plane and is ^ to .
Applying Weiss zone law for these conditions:
u - v = 0, u + v + 2w = 0, u = -w, u = v Þ
Þ ,
,
,
6.14)
Analytically:
Graphically:
It is to be noted that is the Burgers vector in an FCC crystal and the energetically favorable reaction is the opposite of the one given (i.e. splitting of a full dislocation with into two partial dislocations with Burgers vector belong to the family .
The energy of a dislocation is proportional to .
I.e. the energetically preferred reaction is: ® + , with ()
Additionally the following may be noted:
The red vector connects B-B (or A-A), blue connects B-C and green C-B
6.16)
Data: aCu = 3.61 Å,
6.17)
Given: Au wire, R=0.08mm = 8×10-5m, F = 0.736×10-3 N
γ → surface tension
At the melting point we assume that all the weight is borne by the surface tension forces.
Þ F = γ × 2pr
γ = =1.46 N/m
6.28)
Given: ,
do = 3×10-5 m. Let A0 be the surface area of an average grain and let its volume be V0.
No. of grains in unit volume (n0)=
Starting with an initial state of n0 grains the grain growth process leads finally to 1 grain.
Initial surface area = S0 = n × Ao × (the multiplier half is each surface is shared between two grains® becomes an interface).
S0 = (7.07×1015)(2.82×10-19) = 9.99×104 ~ 105
Initial Energy of grain boundaries (which can be reduced by grain growth)
= So × EGB = 105 × 0.5 = 5×104
7