1. CHAPTER 9: Definitions A. Solid Solution
Consists of atoms of at least 2 different types
Solute (minority) and Solvent (majority)
Substitutional vs. Interstitial
Components vs. System
Components – pure metals or compounds that compose an alloy
System – series of possible alloys of different composition

2. B. Solubility Limit
• Maximum concentration of solute atoms that may dissolve
• Ex: Phase Diagram:
Water-Sugar System
Question: What is the
solubility limit at 20C?
Answer: 65wt% sugar.
If Co < 65wt% sugar: solution
If Co > 65wt% sugar: syrup + sugar.
• Solubility limit increases with T:
e.g., if T = 100C, solubility limit = 80wt% sugar.

3. C. Phase
• A homogeneous portion with uniform physical and chemical characteristics
Every pure metal, every liquid, solid and gaseous solution
Note: Different phases may be different physically, chemically or both. For polymorphic substance, each structure is a phase.
Aluminum-
Copper
Alloy

4. D. Equilibrium E. Phase Equilibrium
At a specified condition (temperature, pressure and composition), free energy (internal energy & entropy) is minimum.
E. Phase Equilibrium
Constant phase characteristics
In solids, rate of approach to equilibrium is extremely slow; sometimes take metastable state.

5. Equilibrium Phase Diagram
Represent relationship between temperature, and composition and quantities of phases at equilibrium.
Assume:
P is constant
Variables are T and composition
Only binary alloys (2 components)

6. Isomorphous System • Copper-Nickel System
isomorphous because of complete solid solution solubility
horizontal – composition; vertical – temperature
3 Regions:
Alpha () phase – FCC solid solution
below 1080C, complete solubility
Liquid (L) phase – liquid solution
Two-phase ( + L) region
Liquidus line – separates L and  + L
Solidus line – separates  and  + L
Intersection of lines – melting point of pure metal (one temperature)
Melting of solution: happens over a range of T’s.

7. A. Finding phases present
Given know T and Co, locate T-Co point on the graph.
• Examples:
A(60,1100):
1 phase: 
B(35,1250):
2 phases: L +
Cu-Ni
phase
diagram

8. B. Finding composition of phases
i) one phase – same as overall composition
ii) two-phase :
- Construct a tie line (isotherm, horizontal)
- Intersection of tie line with phase boundaries
• Examples:
Cu-Ni system

9. C. Phase amounts (mass/weight fraction of each phase)
i) single phase – 100%
ii) 2-phase: Lever Rule
Cu-Ni
system
• Examples:
CO = 35 wt% Ni
At TA: L only
WL=1.0 or 100%
At TD:  only
W=1.0 or 100%
At TB:  + L
Use Lever Rule
= 0.27

10. Lever Rule
• A geometric interpretation:

11. Microstructural Development: Isomorphous
• Phase diagram:
Cu-Ni system.
Cu-Ni
system
• System is:
--binary
--isomorphous
i.e., complete solubility of one component in another; a phase field extends from 0 to 100wt% Ni.
• Consider
Co = 35wt%Ni.
Start at A and cool slowly (equilibrium cooling)
Follow development of microstructures

12. Non-equilibrium solidification
• Ca changes as we solidify.
• Cu-Ni case:
First a to solidify has Ca = 46wt%Ni.
Last a to solidify has Ca = 35wt%Ni.
• Fast rate of cooling:
Cored structure
• Slow rate of cooling:
Equilibrium structure

13. MECHANICAL PROPERTIES: Cu-Ni System
• Effect of solid solution strengthening on:
--Tensile strength (TS)
--Ductility (%EL,%AR)
Adapted from Fig. 9.5(a), Callister 6e.
Adapted from Fig. 9.5(b), Callister 6e.
--Peak as a function of Co
--Min. as a function of Co

14. BINARY-EUTECTIC SYSTEMS
limited solid solution solubility
has a special composition with min. melting temp.
Cu-Ag
system
Solvus line – separates + from , and + from 
Eutectic Point, E – an invariant pt (3 phases in equilibriium):
CE = 71.9wt% Ag, TE = 779ºC

15. EX: Pb-Sn (Eutectic)
TE = 183ºC
CE = 61.9 wt% Sn

16. EX: Pb-Sn Eutectic • For a 40wt%Sn-60wt%Pb alloy at 150C, find...
a) the phases present:
a + b
b) the compositions of
the phases:
Ca = 11wt%Sn
Cb = 99wt%Sn
c) the relative amounts
of each phase:
Pb-Sn
system
W= (99-40)/(99-11)
= 59/88=0.67
W= (40-11)/(99-11)
= 29/88=0.33

17. Microstructures: Eutectic
A. Co < 2wt%Sn
• Result:
- polycrystal of a grains.
Adapted from Fig. 9.9, Callister 6e.

18. Microstructures: Eutectic
B. 2wt%Sn < Co < 18.3wt%Sn
• Result:
--grains of a crystal with fine b crystals.
Pb-Sn
system
Adapted from Fig. 9.10, Callister 6e.

19. Microstructures: Eutectic
Co = CE
As we cross TE,
Eutectic reaction: L (61.9%)   (18.3%) + (97.8%)
• Result: Eutectic microstructure
-alternating layers of a and b crystals (short dist. to diffuse).
Pb-Sn
system
Adapted from Fig. 9.11, Callister 6e.

20. Microstructures: Eutectic
• 18.3wt%Sn < Co < 61.9wt%Sn
• Result: primary a and a eutectic microstructure
Pb-Sn
system
Question: What are the relative amounts of primary and eutectic ?

21. HYPOEUTECTIC & HYPEREUTECTIC
Adapted from Fig. 9.7,
Callister 6e. (Fig. 9.7 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)
(Figs and 9.15 from Metals Handbook, 9th ed.,
Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)
Adapted from
Fig. 9.15, Callister 6e.
Adapted from Fig. 9.15, Callister 6e. (Illustration only)
Adapted from Fig. 9.12, Callister 6e.
Differentiate: components vs. phases vs. microconstituents

22. Other Invariant Points:
Eutectoid – (“eutectic like”): solid1  solid2 +solid3
Peritectic: solid1 + L  solid2

23. The Iron-Carbon System

24. The Iron-Carbon System
Iron – iron carbide phase diagram: iron rich (0-6.7 wt% C)
i. pure iron:
ferrite, -iron: BCC
austenite, -iron: FCC
-ferrte: BCC
ii. 6.7 wt% C: compound – iron carbide, or cementite (Fe3C)
Cementite – Fe3C
Very hard and brittle
Metastable only but will remain indefinitely at room T
iii. Singe-phase solid solutions – C is an interstitial impurity
 and : BCC – very limited solubility (max solubility = wt%C)
: FCC – higher solubility limit (2.11 wt%C)
Why does BCC have lower solubility than FCC?
iv. Two-Phase Regions:
Eutectic Point at 4.3 wt% C and 1147C
Eutectoid Point at 0.76 wt%C and 727C

25. IRON-CARBON (Fe-C) PHASE DIAGRAM
(Adapted from Fig. 9.24, Callister 6e. (Fig from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)
Adapted from Fig. 9.21,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed.,
Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)

26. HYPOEUTECTOID STEEL
Adapted from Figs and 9.26,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)
Adapted from
Fig. 9.27,Callister
6e. (Fig courtesy Republic Steel Corporation.)

27. HYPEREUTECTOID STEEL
Adapted from Figs and 9.29,Callister 6e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)
Adapted from
Fig. 9.30,Callister
6e. (Fig. 9.30
copyright 1971 by United States Steel Corporation.)

28. ALLOYING STEEL WITH MORE ELEMENTS
• Teutectoid changes:
• Ceutectoid changes:
Adapted from Fig. 9.31,Callister 6e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)
Adapted from Fig. 9.32,Callister 6e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)

29. Example. For a 0.35 wt%C steel at just below 727 ºC, determine:
Fraction of total cementite and ferrite
Fraction of proeutectoid ferrite and pearlite
Fraction of eutectoid ferrite
Answers: a) W=0.95, Wcementite=0.05, b) W’=0.56, Wp=0.44 c) We=0.39

30. It’s Clicker Time!
A reaction wherein, upon cooling, a solid phase and a liquid phase transform isothermally and reversibly to another solid phase having a different composition is
a) eutectic
b) eutectoid
c) peritectic
d) proeutectoid

31. 2. Proeutectoid -ferrite forms during the transformation of austenite
a) above the eutectoid temperature in a hypoeutectoid steel
b) above the eutectoid temperature in a hypereutectoid steel
c) below the eutectoid temperature in a hypoeutectoid steel
d) below the eutectoid temperature in a hypereutectoid steel

32. 3. The line separating a one-phase solid region from a two-solid region is called
a) isomorphous
b) solidus
c) solvus
d) liquidus

33. 4. Which of the following is/are true about an isomorphous system?
I. The is only one solid phase.
II. The phase diagram has no solvus line.
III. There is at least two two-phase regions.
IV. The phase diagram has a euctectic point.
a) I and II
b) I only
c) I and III
d) II and IV

34. 5. For the equilibrium phase diagram of the iron-iron carbide system, which of the following solid solutions has the highest solubility limit?
a) BCC delta ferrite
b) FCC gamma austenite
c) BCC alpha ferrite
d) cementite

35. 6. A Cu-Ag alloy of composition 80 wt% Ag is cooled from a temperature of 1200C. What is the approximate composition of the first solid that forms?
a) 8.0 wt% Ag
b) 71.9 wt% Ag
c) 91.2 wt% Ag
d) 93.0 wt% Ag

36. 7. A Cu-Ag alloy of composition 40 wt% Ag undergoes equilibrium cooling from a temperature of 1200C. What is the approximate composition of the last liquid to solidify?
a) 8.0 wt% Ag
b) 71.9 wt% Ag
c) 91.2 wt% Ag
d) 93.0 wt% Ag

37. 8. A hypoeutectic Cu-Ag alloy of wt% Ag undergoes equilibrium cooling from a temperature of 1200C. Which of the following microconstituents will be present at room temperature.
a) only alternating layers
of alpha and beta.
b) primary alpha and
primary beta
c) primary alpha plus
eutectic alpha and beta
d) primary beta plus