FEIT EDUCATION  
'6��.Q:�  
Week 12 Quiz4梳理 – CIVL1110 
by 
Winnie老师 

1  
CIVL1110 Quiz 4 Tips  
- Phase Diagram 重点，理解图，易错点  
- Isothermal transformation diagram 重点，理解图 计算较少  
- Alloys (Concepts 居多)  
- Welding: CE formula重点；其他 concept居多  
PART 1 Phase Diagram  
1. Concepts  

Phase diagram (相图) is used to address combining two elements.  
• The development of microstructure of an alloy is related to the characteristics of its phase  
diagram, and microstructure is related to mechamical properties  
• Phase diagrams provide valuable information about melting, casting, crystallization, etc  
a) Phase: Physically and chemically distinct material region  
eg Chemical 成分不同；physical：liquid and solid不同  
- Homogeneous: a single-phase system  
- Heterogeneous (mixture): two or more phases  

b) Solutions – solid solutions,(or liquid solution) single phase  
– Mixtures – more than one phase  

c) Solubility Limit: Max concentration for which only a single phase solution occurs.  
- The boundary between single and multi-phase regions represents the Solubility limit.  
- Isomorphous (同晶的) = complete solubility of one component in anohter  
- The copper–nickel system is termed isomorphous because of this complete liquid and solid  
solubility of the two components  
- Binary - two components  


2  
2.Composition of phases  
a) Rules: If we know T and Co, then we know: -the number and types of phases present; the  
composition of each phase; the amount of each phase (given in wt%).  

b) The Lever Rule  

= 

+  
= 

+  
= 
− 0 
−  

= 

+  
= 
0 −  
−  


- Tie line – connects the phases in equilibrium with each other 

Example 1 :  
The above figure gives 35wt% Ni - 65wt% Cu alloy at 1250o. (Point B)  
a) Phase/phases?  

b) Phase composition is : (Tie Line)  


c) Phase composition using Lever Rule:  



3  
Example 2  
For a 40 wt% Sn–60 wt% Pb alloy at 150oC,  
(a) What phase(s) is (are) present?  
(b) What is (are) the composition(s) of the phase(s)?  
calculate the relative amount of each phase present in terms of (c) mass fraction and (d) volume  
fraction. At 150oC take the densities of Pb and Sn to be 11.23 and 7.24 g/cm3, respectively.  



4  
Example 3.  
A 50 wt% Ni–50 wt% Cu alloy is slowly cooled from 1400oC to 1200oC.  


(a) At what temperature does the first solid phase form?  



(b) What is the composition of this solid phase?  



(c) At what temperature does the liquid solidify?  




(d) What is the composition of this last remaining liquid phase?  



5  
Example 4  

Using the lead-sin phase diagram above, answer the following questions:  
a) What is the maximum solubility of lead in tin in wt%?  


b) What is the melting temperature of lead in degrees Celsius?  


c) For a 40wt% Tin and 60 wt% Lead allloy at 100oC.  
c1) What is the phase(s) present?  



c2) What is the composition(s) of the phase(s)?  




c3) What is the fraction of each phase present? Explain how you obtain the data.  










6  
PART 2 Iron, Steel, Aluminum  
1. Iron - Carbon Phase Diagram  

(1) Ferrite (铁酸盐；铁素体)  
- At room temperature the stable form called ferrite, or iron,  
- has a BCC crystal structure  

(2) Austenite （奥氏体）  
- Up to 2.14wt%C  
- At 912oC, Ferrite experiences a polymorphic transformation to FCC austenite, or iron  

(2) Cementite（碳化铁）  
- intermediate compound iron carbide, or cementite (Fe3C)  
- represented by a vertical line on the phase diagram  
- The composition axis extends only to 6.7wt%C.  
- The composition is expressed in "wt%C" rather than "wt%Fe3C" .  
eg, 6.7wt%C = _____wt%Fe3C  

2 Eutectic, Eutectoid & Peritectic  
a) Eutectic 共熔的 - early melted  
- liquid in equilibrium with two solids L cool α+β  
⇋ 1 + 2  

* b) Eutectoid 共析的 - solid phase in equation with two solid phases  
2 ⇋ 1 + 3  
eg. At 7270C: γ ⇋ α + Fe3C (intermetallic compound - cementite)  

7  

c) Peritectic 包晶的- liquid + solid 1 → solid 2  
1 + ⇋ 2  
eg. At 1493oC： δ + L ⇋ γ  
3 Microstructure for eutectoid composition  
In metal alloys, microstructure is characterized by the number of phases present, their proportions, and the manner in  
which they are distributed or arranged. （可以说 microstructure是部分取决于 phase的）  
a) Formation of Pearlite  
- Resulted from eutectoid reaction  
- Consists of alternating layers of ferrite and cementite 

b) Hypoeutectoid Steel (在左) Left, less than eutectoid,  
→ + → + 3  
c) Hypereutectoid Steel （在右）Right, more than eutectoid  
→ 3 + → + 3  


8  
Hypoeutectoid (< 0.76wt%C) Hypereutectoid (>0.76wt%C)  
4. Isothermal Transformation Diagrams  
a) Iron-iron carbide eutectoid reaction:  
- Undercool: γ (0.76 wt% C) ⇄ α (0.022 wt% C) + Fe3C (6.7 wt% C)  
- Temperature plays an important role in the rate of the austenite-to-pearlite transformation  
- Different microstructures that may be produced in iron–carbon alloys depending on heat treatment.  

b) Isothermal transformation diagram (bottom) is generated from percentage transformation VS  
logarithm of time measurements  
- Only valid for an iron-carbon alloy of eutectoid composition  
- Only accurate for transformations in which the temperature of the alloy is held constant throughout  
the duration of the reaction  


c) Microstructures  
Summary of Processing Options  



9  
1) Spheroidite 球状渗碳体  
-α grains with spherical Fe3C -  
- diffusion dependent. --heat - -  

2) Bainite: 贝氏体 (B) --α lathes  
(strips) with long  
rods of Fe3C --diffusion controlled.  

3) Martensite: 马氏体  
-γ(FCC) to Martensite (BCT) (involves single atom jumps)  
- M = martensite is body centered tetragonal (BCT)  
- γ to M transformation -- is rapid!  
- percentage transformation depends on T only.  


4) Tempering Martensite 回火马氏体  
• reduces brittleness of martensite,  
• reduces internal stress caused by quenching  
• produces extremely small Fe3C particles surrounded by α.  
• decreases TS, YS but increases %RA  

5) Hardness:  
fine pearlite << martensite.  
fine pearlite > coarse pearlite > spheroidite  

6) Strength and Ductility  


10  

Example 1: Rank the following iron-carbon alloys and associated microstructures from the highest to  
the lowest tensile strength:  
a)0.25 wt% C with spheroidite  
b)0.25 wt% C with coarse pearlite  
c)0.6 wt% C with fine pearlite  
d)0.6 wt% C with coarse pearlite  
e)0.6 wt% C with bainite  



11  
Example 2.  
a) What is the structure of pearlite? Draw a schematic of all the appropriate phases.  
b) Distinguish between the following three types of plain-carbon steels: i) eutctoid; ii) hypoeutectoid  
and iii) hypereutectoid.  
c) Consider 6.0 Kg of austenite containing 0.5 wt% C, cooled to below 725°C. Using the iron-  
carbon phase diagram given on the figure below, answer the following questions:  
c1) What is the proeutectoid phase of this alloy?  
c2) How many kilograms each of total ferrite and cementite form?  
c3) How many kilograms each of pearlite and proeutectoid phase form?  





12  
Example 3:  
Using the diagram above, answer the following questions:  

a) What is the melting temperature of pure iron in degree Celsius?  



b) At 726oC, for a composition of 0.5wt% carbon, what is the phase present?  




c) For the same conditions as question b), what is the composition of each phase?  




d) For the same conditions as for question b), what is the pro-eutectoid phase?  











13  


Example 4.  
Using the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition,  
specify the nature of the final microstructure (in terms of micro-constituents present and approximate  
percentages) of a specimen that has been subjected to the following time-temperature treatments. In  
each case assume that the specimen begins at 760°C and that it has been held at temperature for long  
enough to have achieved a homogenous austenitic structure.  

a) Rapidly cool to 650°C, hold for 20 seconds, and quench to room temperature.  

b) Rapidly cool to 500°C, hold for 1000 seconds, and quench to room temperature.  

c) Rapidly cool to 580°C, hold for 3 seconds, rapidly cool to 500°C, hold for 4 seconds and quench  
to room temperature.  








14  
Example 5.  






15  

Example 6.  
Use the isothermal transforma- tion diagram for an iron–carbon alloy of eutectoid composition and  
then sketch and label time–temperature paths on this diagram to produce the following  
microstructures  
(a) 100% coarse pearlite  
(b) 50% martensite and 50% austenite  
(c) 50% coarse pearlite, 25% bainite, and 25% martensite  



16  
5. Alloys  
1) Classification  



2) High Strength Steels  
– Composition: Carbon: 0.08% Manganese: 1% Nickel: 1.25% Chromium: 0.45% Molybdenum:  
0.4% Copper: 0.2%  
– Property of interest: Yield strength = 800 MPa  

3) Stainless Steel Alloy  
Advantages:  
- Resistance to corrosion in an oxidizing environment  
- Increased strength at elevated temperature  
- 5 categories: martensitic, ferretic, austenitic, precipitation, duplex  




17  
4) Aluminium Alloys  
Ore 矿: Bauxite (Al2O3:3H2O, Al2O3:H2O) (铝土矿)  
• Advantages of Aluminium over Steels  
o Light: 1/3 of weight of steels  
o Better corrosion resistance than most steels in most environments  
o Better fabricability加工性: more freedom of shapes  
o Weldability 可焊接性  
o Better machinability than steels 机械加工性  
o No brittle fracture at low temperatures  

• Bad Points for Aluminium  
o Cost: metal more expensive than steels (but can be compensated by lower costs of  
machining and maintenance)  
o Buckling: failure by buckling occur at lower load in aluminium than in steels  
o Lower strength at high temperature  
o More chance of failure by fatigue  
o More deflection due to lower elastic modulus  

5) Wrought Aluminium alloys 锻造铝合金  
• Alloys that are so brittle that forming or shaping by appreciable deformation is not possible  
ordinarily are cast; those that are subject to mechanical deformation are termed wrought  
alloys.  
• Ingots are homogenized and rolled  
• Classification: According to major alloying elements.  


18  
Multiple Choice (Past Quiz)  

1) What is tempered martensite?  
a) Martensite that has been heat treated at a temperature below the eutectoid temperature.  
b) Martensite that has been heat treated at a temperature above the eutectoid temperature.  
c) Martensite that has been heat treated at a temperature above the ferrite-to-austensite phase  
transition temperature.  
d) Martensite that has been heat treated at a temperature above the melting temperature.  



2) Which of these is not a type of a stainless steel?  
a) pearlitic  
b) austenitic  
c) martensitic  
d) ferritic  


3) Which of these statements is incorrect?  
a) Aluminum alloys are less susceptible to fatigue failure than steels.  
b) Aluminum alloys have a better corrosion resistance than most steels.  
c) Aluminum alloys have a better machinability than steels.  
d) Aluminum alloys are lighter than steels.  


4) What are the external parameters that affect the phase structure?  
a) Temperature, Composition  
b) Pressure, Composition  
c) Temperature, Composition, Pressure  
d) Temperature, Pressure  


19  
PART 3 Welding  
1 Categories  
- 50 different types of welding operations classified in two  
categories:  
Fusion welding (熔焊接)  
- uses heat to melt the base metals  
- most of the time a filler metal is used  
- most welding processes are fusion welding  

Solid-state welding (固态焊)  
- pressure alone or combination of pressure and heat  
- if heat involves, below melting temperature of base metals.  
- no filler metal used  
2 Advantages & Drawbacks of Welding  
a) Pros  
– It’s Permanent joint  
– Welded joint can be stronger than the parent material  
– Welding is usually the most economical way to join components in terms of material usage and  
fabrication costs.  
– Welding is not restricted to the factory environment (can be on-site)  

b) Cons  
– Most welding operations are performed manually and are expensive in terms of labor costs  
(skilled trades)  
– Most welding processes are dangerous because they involve the use of high energy  
– Permanent joint = no disassembly （无法拆解的）  
– Defects can be created in the welds that are difficult to detect and can  
cause lost of material properties.  













20  
3 Parts of Welding  
A: Fusion zone : Mixture of filler material and base materials that  
have completely melted.  

B: Heat-Affected Zone (HAZ) : The metal in this zone has  
experienced temperatures that are below the melting temperature of  
the metal, yet high enough to create micro-structural changes.  

C: Unaffected Base Metal : has not seen enough heat to change its  
properties; but there can be high residual stresses due to expansion  
and shrinkage in the fusion zone  
4 Residual Stresses  
Factors affecting distortion and residual stress:  
• Coefficient of thermal expansion  
• Melting point  
• Phase changes  
• Difference of thermal expansion between two metals  
• Energy input  
• Heat treatment history  
• External restraints during welding  

2. Method to reduce:  
Peening （锤）  
Heat treatment: controllable and more effective than peening  

*5 Carbon Equivalent  
- Definition of Weldability: Capacity of a material to be welded under imposed fabrication  
conditions into a specific suitably designed structure and to perform satisfactorily in the intended  
service.  
- Carbon Equivalent (CE) formula  
If CE ≤ 0.4: no adjustment necessary, the alloy is weldable  
If 0.4 < CE ≤ 0.6: need to use low-hydrogen electrodes  
* If CE > 0.6: Difficult to weld: need to use low-hydrogen electrodes, increase welding heat input,  
pre-heat, post-heat or impose slower cooling rates (to prevent formation of martensite and  
hydrogen embrittlement)  

21  
Example:  
The following table gives the composition of four steel alloys. Rank these 4 alloys from the easiest  
to the most difficult to weld. Explain your reasoning.