Materials Engineering – Day 6

• Quiz• Review crystal structure• Review crystal defects• Review dislocations• Review the golden rule for strengthening

metal.• Things that can be used to make metals

strong.

You need to know/be able to

• For the following processes, determine (from graphs and/or calculations) the strength/ductility and describe the governing microstructural mechanism – Solid Solution Strengthening– Grain Size Refinement– Cold Work and Annealing

• Name the two types of solid solutions (interstitial and substitutional) and explain how they differ.

Crystallinity in Metals

• Three types of unit cells. List in order of slip systems.

• Name a point defect, a line defect, and an area defect.

• What is the relationship between slip and plastic deformation?

• What is the relationship between dislocation motion and slip?

4

Dislocation MotionDislocations & plastic deformation• Cubic & hexagonal metals - plastic deformation by plastic

shear or slip where one plane of atoms slides over adjacent plane by defect motion (dislocations).

• If dislocations don't move, deformation doesn't occur!

Adapted from Fig. 7.1, Callister 7e.

BLOCK THAT DISLOCATION

• We will tell the story of one of the earliest attempts to “block that dislocation” discovered by humans. Our story starts around 700 BC…

• Chinese bronze from Spring and Autumn period

Bronze sword from Troy

What’s Happening?• The tin atoms dissolve in the matrix of copper. There

are many, many substitutional solute atoms.• These atoms interact with dislocations, impeding their

motion.

1. The solute atoms are not quite the right size. This produces stress and strain in the lattice.

2. The solute atoms’s stess field attracts or repels the stress field around the dislocation.

3. The result is that the dislocation is pinned or blocked – It’s motion is impeded!

“Rules” for substitutional and Interstitial sold solutions

• Hume Rothery Rules for substitutional– Similar size– Similar crystal structure– Similar electronegativity

• For Interstitial– Solute atom must be small relative to solvent

atom so it can fit in the spaces between atoms

8

Two outcomes if impurity (B) added to host (A):

• Solid solution of B in A (i.e., random dist. of point defects)

• Solid solution of B in A plus particles of a new phase (usually for a larger amount of B)

OR

Substitutional solid soln.(e.g., Cu in Ni)

Interstitial solid soln.(e.g., C in Fe)

Second phase particle--different composition--often different structure.

Point Defects in Alloys

9

Imperfections in SolidsConditions for substitutional solid solution (S.S.)• W. Hume – Rothery rule

– 1. r (atomic radius) < 15%– 2. Proximity in periodic table

• i.e., similar electronegativities

– 3. Same crystal structure for pure metals– 4. Valency

• All else being equal, a metal will have a greater tendency to dissolve a metal of higher valency than one of lower valency

10

Imperfections in SolidsApplication of Hume–Rothery rules – Solid Solutions

1. Would you predictmore Al or Ag to dissolve in Zn?

2. More Zn or Al in Cu?

Table on p. 106, Callister 7e.

Element Atomic Crystal Electro- ValenceRadius Structure nega-

(nm) tivity

Cu 0.1278 FCC 1.9 +2C 0.071H 0.046O 0.060Ag 0.1445 FCC 1.9 +1Al 0.1431 FCC 1.5 +3Co 0.1253 HCP 1.8 +2Cr 0.1249 BCC 1.6 +3Fe 0.1241 BCC 1.8 +2Ni 0.1246 FCC 1.8 +2Pd 0.1376 FCC 2.2 +2Zn 0.1332 HCP 1.6 +2

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Strengthening by Alloying• small impurities tend to concentrate at dislocations• reduce mobility of dislocation increase strength

Adapted from Fig. 7.17, Callister 7e.

12

Strengthening by alloying• large impurities concentrate at dislocations on low density

side

Adapted from Fig. 7.18, Callister 7e.

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Ex: Solid SolutionStrengthening in Copper

• Tensile strength & yield strength increase with wt% Ni.

• Empirical relation:

• Alloying increases y and TS.

21 /y C~

Adapted from Fig. 7.16 (a) and (b), Callister 7e.

Ten

sile

str

engt

h (M

Pa)

wt.% Ni, (Concentration C)

200

300

400

0 10 20 30 40 50 Yie

ld s

tren

gth

(MP

a)wt.%Ni, (Concentration C)

60

120

180

0 10 20 30 40 50

Result

Here is the plot in the notes.

S o lid S o lutio n S tre ng the ning

Effect dependsupon alloying element

S treng th

% alloy

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Strategies for Strengthening: Reduce Grain Size

• Grain boundaries are barriers to slip.• Barrier "strength" increases with Increasing angle of misorientation.• Smaller grain size:

more barriers to slip.

• Hall-Petch Equation:

21 /yoyield dk

Adapted from Fig. 7.14, Callister 7e.(Fig. 7.14 is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.)

Another Dislocation blocker: The grain boundary

A dislocation coming up on the grain boundary will not be able to cross easily into the adjacent grain.

It will probably stop waiting for more stress to be applied. Other dislocations will pile up behind it.

The Hall-Petch Relationship

E f fe c t o f G rain S ize R e d uc tio n

s ys =s o+kyd

Yie ld S treng th

-1/2

d -1/2

Another Blocker: Other Dislocations

• Recall that as plastic deformation proceeds the density of dislocations increases by several orders of magnitude.

• So dislocations block themselves! This accounts for the strengthening that occurs during plastic deformation. (Done on purpose, we call it cold work.

E f fe c t o f P las tic D e fo rm atio n

Degree of strengtheningdepends onmaterial

Yie ld S treng th

%area reduction

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Strategies for Strengthening: Cold Work (%CW)

• Room temperature deformation.• Common forming operations change the cross sectional area:

Adapted from Fig. 11.8, Callister 7e.

-Forging

Ao Ad

force

dieblank

force-Drawing

tensile force

AoAddie

die

-Extrusion

ram billet

container

containerforce

die holder

die

Ao

Adextrusion

100 x %o

do

A

AACW

-Rolling

roll

AoAd

roll

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• Ti alloy after cold working:

• Dislocations entangle with one another during cold work.• Dislocation motion becomes more difficult.

Adapted from Fig. 4.6, Callister 7e. (Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.)

Dislocations During Cold Work

0.9 m

21

Result of Cold WorkDislocation density =

– Carefully grown single crystal ca. 103 mm-2

– Deforming sample increases density 109-1010 mm-2

– Heat treatment reduces density 105-106 mm-2

• Yield stress increases

as d increases:

total dislocation length

unit volume

large hardening

small hardening

y0 y1

22

Impact of Cold Work

Adapted from Fig. 7.20, Callister 7e.

• Yield strength (y) increases.

• Tensile strength (TS) increases.

• Ductility (%EL or %AR) decreases.

As cold work is increased

Mechanism:Grain boundaries block dislocation motion. More grains (smaller grains) means more boundaries and more blocking of dislocations

Process:Cold work to add internal energy, anneal to recrystallize and form new small grains.Note: Strength increases without loss of toughness

Mechanism:Solute atoms are too big or too small and cause distortion ins the crystal lattice

Process:Add other elements to the melt e.g. add Al and V to Ti to get Ti6Al4V.

Mechanism:Number of dislocations increases by orders of magnitude, distorting lattice and impeding dislocations

Process:Mechanically deform plastically. (e.g. cold roll, wire draw)

E f fe c t o f G rain S ize R e d uc tio n

s ys =s o+kyd

Yie ld S treng th

-1/2

d -1/2

S o lid S o lutio n S tre ng the ning

Effect dependsupon alloying element

S treng th

% alloy

E f fe c t o f P las tic D e fo rm atio n

Degree of strengtheningdepends onmaterial

Yie ld S treng th

%area reduction