Microstructure of a lead-tin alloy - Course...

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Microstructure of a lead-tin alloy

Example: Building Structures

Burj Dubai under construction 2007; completed 2010

Single-Crystal Titanium panels

F/A22 Raptor

Example: Aerospace Industry

Example: Aerospace Industry

Pratt and Whitney F119 Engine

Why study Phase Diagrams?

• Provide valuable information about: – Melting

– Casting

– Microstructure

– Crystallization

– Temperature

– Pressure

– Composition

– Phase transformations

Phase Diagrams

• Component: a chemical constituent of an alloy, which may be used to specify its composition – Example: In a copper-zinc brass, the

components are Cu and Zn

• Solid solution (Chapter 4): – Consists of atoms of at least two different types

– Solute atoms occupy either substitutional or interstitial positions

– Crystal structure of solvent is maintained

Solubility Limit

• Maximum concentration of solute atoms that may dissolve in the solvent, at a specific temperature, to form a solid solution

• Adding solute past the solubility limit yields another solid solution (different crystal structure) or compound that has a different composition

Phases • Phase: a homogeneous portion of a system that

has uniform physical and chemical characteristics • The following are considered phases:

– Solid solution – Liquid solution – Gaseous solution

• If more than one phase is present in a system, each will have its own distinct properties

• Phases exist over a range of compositions, temperatures and pressures

• Example: Water and Ice two phases (physically dissimilar, but chemically the same)

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Stability of Phase

• Phase stability depends on temperature, pressure and composition,

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Phases

For each of the following situations how many phases are present?

1. Ice cubes floating in a glass of water.

2. Water at room temperature containing 20 wt% sugar.

3. A bottle of 7-Up.

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Components vs. Phases • Components:

– The elements (or compounds) that are mixed to make the alloy

• Al and Cu

• MgO and Al2O3

• styrene and butadiene

– The composition is the concentration of each component present in an alloy

• Phase: – Physically and/or chemically

distinct regions in the material

Phase β Phase α

Aluminum atom

Copper atom

Al-Cu alloy

Sugar- Water Demo

• Initially, sugar is added to water and there is a sugar- water solution or syrup

• Solution becomes more concentrated with sugar until the solubility limit is reached (saturated)

• System cannot dissolve any more sugar (at the specified temperature) and further additions settle to the bottom of the container

• System now contains two phases; syrup and solid (undissolved) sugar crystals

Solid Solubility

• We define solubility as the maximum equilibrium concentration of a solute in a host material.

Answer: 65 wt% sugar

ALS: What is the solubility of sugar in water at 20oC?

ALS: If you have 1 L (1000 g) of water, how much sugar can you dissolve in it?

a) 650 g of sugar

b) 2000 g of sugar

c) 1350 g of sugar

Solid Solubility

• We define solubility as the maximum equilibrium concentration of a solute in a host material Previous Answer: 65 wt% sugar

ALS: If you have 1 L (1000 g) of water, how much sugar can you dissolve in it?

Solid Solubility

• We define solubility as the maximum equilibrium concentration of a solute in a host material Previous Answer: 65 wt% sugar

Answer: The solution is 65% (about 2/3) sugar and 1/3 water. The weight of the sugar is twice that of the water, so 2000 g is the answer.

If you leave a saturated water/sugar solution open to the air and the water starts to evaporate, what will happen?

Solubility Limits

Answer: At some point there won’t be enough water to hold all of the sugar. The solubility for sugar will be exceeded, and some of it will come out of the solution in crystal form (i.e. precipitate) to compensate.

Phase Diagrams

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One-Component (Or Unary) Phase Diagrams

• 3 externally controllable parameters that affect phase structure:

– Temperature

– Pressure (usually scaled logarithmically)

– Composition

• Unary phase diagrams only deal with pressure-temperature graphs composition remains constant

• Along the phase boundaries, the phases on either side are in equilibrium

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(H20) Each of the phases will exist in equilibrium

conditions over the corresponding temperature

• Triple point: all three states are in equilibrium (at 273.16 K at a pressure of 6.04x10-3 atm in the example)

Binary Phase Diagrams

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Binary Phase Diagrams

• Temperature and composition are variable parameters

• Binary alloys – contain two components

• Pressure held constant normally ≈ 1 atm

• Maps that represent the relationship between temperatures, compositions and quantities of phases at equilibrium

• Accurately predict phase transformations and resulting microstructures

Binary Isomorphous Systems

• Easiest binary phase diagram to understand – Example: copper-nickel system

• Temperature is plotted along the vertical axis

• Composition of the alloy is plotted along the horizontal axis – Bottom- weight %

– Top – atom %

– Composition ranges from 0 wt% Ni (100 wt% Cu) on the left of the horizontal axis to 100 wt% Ni (0 wt% Cu) on the right of the horizontal axis

Binary Isomorphous Systems

• Three phase regions or fields: – Alpha(α)

• Substitutional solid solution consisting of Cu and Ni and both have an FCC crystal structure

– Liquid (L) • Homogeneous liquid solution

composed of both copper and nickel

– Two phase α + L

• The copper-nickel system is termed isomorphous due to the complete liquid and solid solubility of the two components

Phase Boundary

Interpretation of Phase Diagrams

• For a binary system of known composition and temperature that is at equilibrium, you can identify:

1. Phases present 2. Compositions of phases 3. Percentages or fractions of phases

• Phases Present:

– Locate the temperature-composition point on the diagram and one can identify the phase(s) within the phase field that are/ is present


• Phase compositions: – One phase present:

• composition of phase is the same as the overall composition of the alloy

– Two phases present: • A Tie line is constructed across two-phase region at

a specified temperature of the alloy

• Intersections of tie lines and phase boundaries on both sides are noted

• The compositions corresponding to these points are read from the x-axis.


• Determining Phase Amounts: – One phase present:

• Alloy is composed entirely of that phase; phase fraction is 1.0 or percentage is 100%

– Two phases present: • Use the lever rule; tie line is constructed across the two-phase

region at a specified temperature of the alloy • Overall alloy composition is located on the tie line • Fraction of one phase is determined by taking the length of the tie

line from the overall alloy composition to the phase boundary for the other phase, and dividing by the total lie line length

• The fraction of the other phase is determined in the same manner • Segment lengths can be determined either by direct measurement

from the phase diagram using ruler or by subtracting compositions from the composition axis

• Liquid mass fraction:

• α-Phase mass fraction:

Alternate Approach - Phase diagrams

L

LCC

CC

SR

SW

0

L

L

CC

CC

SR

RW

0

Developing Microstructures – Isomorphous Alloys

• Equilibrium cooling:

– Cooling occurs very slowly

Mechanical Properties of Isomorphous Alloys

• Solute Strengthening

TS for pure Cu

TS for pure Ni

%EL for pure Cu

%EL for pure Ni

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Binary Eutectic Systems

One common type of alloy system is called the eutectic system. Binary eutectic systems occur when 2 elements completely dissolve into each other in the liquid state, but have only limited solubility in the solid state.

More than one solid phase exists in these systems. There are multiphase regions between single-phase regions. Which phases are present depends, as before, on the temperature and the composition of the material.

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Binary Eutectic Systems

The copper-silver alloy system is binary eutectic.

• Consider Cu-Ag alloys:

− How many phases are there in the system?

− What are the two solid phases?

− Where is the eutectic point?

3 (L, , b)

: mostly Cu

b: mostly Ag

eutectic temperature: TE eutectic composition: CE

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ALS: Pb-Sn Eutectic System

• Consider an alloy which contains 40 wt% Sn at 150oC.

• What phases are present? What are their compositions?

• Answer:

There are 2 phases: a and b.

Ca = 11 wt% Sn

Cb = 99 wt% Sn

The lead-tin system is also binary eutectic

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ALS: Pb-Sn Eutectic System

• How much of each phase is present?

%6788

59

W

1 33%W Wb

• Answer: From the lever rule:

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Adapted from Fig. 9.9,

Callister 6e.

Microstructures in Eutectic Systems

• Consider a eutectic system (e.g. Pb-Sn) with an initial concentration (Co) less than 6 wt% Sn.

• The result is... – … a single phase

polycrystal with grains.

10% 30% 20% 0%

100

300

400

200

%Sn

Te

mp

era

ture

oC

Co

L

L

b

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Pb-Sn

system


Callister 6e.


• Consider a eutectic system with an initial concentration (Co) between 2 and 18.3 wt% Sn.

• The result is... –… a 2-phase

polycrystal with... • … and β grains. • … a fine grained

structure.

: C o wt%Sn L +

2 0 0

T(°C)

C o , wt% Sn 1 0

18.3

2 0 0 C o

3 00

1 00

L

3 0

L: C o wt%Sn

+ b

4 00

(sol. limit at TE)

T E

2 (sol. limit at Troom)

L

b

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Pb-Sn

system

Adapted from Fig. 9.11, Callister 6e.

Adapted from Fig. 9.12, Callister 6e. (Fig. 9.12 from Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)

Microstructures in Eutectic Systems • When the initial concentration (Co) equals the eutectic

composition (CE) the result is...

– … alternating layers of and β crystals

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L +

2 0 0

T(°C)

wt% Sn

2 0 4 0 0

3 00

1 00

L

6 0

L: C o wt%Sn

+ b

T E b

0 8 0 100

L + b

C o 18.3 61.9

L

L

primary

97.8

S

S

R

R

eutectic eutectic b

Pb-Sn

system


Callister 6e.


• Co between 18.3 and 61.9 wt% Sn results in... – … a combination of crystals and eutectic regions

Just above TE:

W L = (1- W ) = 50 wt%

C = 18.3 wt% Sn

C L = 61.9 wt% Sn

S

R + S W = = 50 wt%

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Pb-Sn

system


Callister 6e.


• Co between 18.3 and 61.9 wt% Sn results in... – … a combination of crystals and eutectic regions

Just below TE:

C = 18.3 wt%Sn

C b = 97.8 wt%Sn

S

R + S W = = 73 wt%

W b = 27 wt%

Eutectoid and Peritectic Reactions

• Eutectoid reaction: a reaction wherein upon cooling, one solid phase transforms isothermally and reversibly into two new solid phases that are intimately mixed

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cooling

heating

cooling

heating

L

• Peritectic reaction: a reaction wherein upon cooling, a solid and a liquid phases transform isothermally and reversibly to a solid phase having a different composition

Copper-zinc phase diagram

The Iron-Carbon System

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The Iron-Iron Carbide (Fe-Fe3C) Phase Diagram

• When heated, pure iron experiences two changes in its crystal structure before it reaches its melting temperature:

– At room temperature, exists as ferrite (α iron) and has a BCC structure

– Ferrite experiences a polymorphic transformation to FCC austenite (γ iron) at 912°C

– Experiences a second transformation to BCC phase known as δ ferrite at 1394°C, which finally melts at 1538°C


• Cementite, (Iron Carbide - Fe3C). It is 25at% C or 6.7 wt% C.


Euctectic Point

Eutectoid Point

Eutectic reaction for the iron-iron carbide

system:

Eutectoid reaction for the iron-iron carbide system:

)%7.6()%022.0()%76.0( 3 CwtCFeCwtCwtcooling

heating

CFeLcooling

heating3

Development of Microstructure in Iron-Carbon Alloys

• The microstructure depends on: – Concentration of carbon – Heat treatment

• Pearlite: a two-phase microstucture results from the transformation of austenite and consists of alternating layers ( or lamellae) of -ferrite and cementite

43 Pearlite

Hypoeutectoid Alloys

• Hypoeutectoid alloy : (less than eutectoid) between 0.022 and 0.76wt% C

• Proeutectoid ferrite: that formed above the eutectoid temperature

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Hypereutectoid Alloys • Hypereutectoid alloys:

containing between 0.76 and 2.14 wt% C

• Proeutectoid cementite: that which forms before the eutectoid reaction cementite composition remains constant as temperature changes

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Non-Equilibrium Cooling

• In most situations, the cooling rates are impractically slow and unnecessary

• Nonequilibrium effects of practical importance: – Phase changes or transformations at temperatures other than those

predicted in phase diagrams – The existence of non-equilibrium phases that do not appear on the phase

diagram

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Chapter 9 Review

What did you learn?

•

•

•

•

•

•

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Chapter 9

Practice Problems

Practice Problems

1. Salt is often spread on roads during the winter in order to depress the freezing point of water. A binary phase diagram for water and salt (NaCl) is shown below.

Liquid (Brine)

Field “Z”

Practice Problems

a) What are the two phases that exist in equilibrium in the field “Z” region.

b) At -10°C determine the overall composition for which

equal amounts of these phases exist in equilibrium. c) What are the compositions of each phase under these

conditions? d) What is the lowest temperature at which adding salt

can prevent water from freezing completely?

Practice Problems

Answer:

a)Liquid (Brine) + Ice

b)8 wt% NaCl

c)Ice – 0.5 wt% NaCl

Salt + Brine – 16 wt% NaCl

d) -21°C

Practice Problems

2. Using the Fe-Fe3C phase diagram, compute the weight fractions of proeutectoid ferrite and pearlite, respectively, that form in an iron-carbon alloy containing 0.25 wt% C if the alloy is slowly cooled to a temperature just below the eutectoid temperature.

Practice Problems

a) Wα = 0.31, Wp = 0.69 b) Wα = 0.69, Wp = 0.31 c) Wα = 0.56, Wp = 0.44 d) Wα = 0.44, Wp = 0.56

Practice Problems

Answer:

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b) isanswer theTherefore,

31.0022.076.0

022.025.0

69.0022.076.0

25.076.0

pW

W

Practice Problems

c)

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% 12.11or 1211.07317.08528.0

:reaction eutectic during formed

8528.03.188.97

308.97

:reaction eutecticAfter

7317.03.189.61

309.61

:reaction eutectic Before

W

W

W

Microstructure of a lead-tin alloy - Course...

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