Lecture 1B: Physical Metallurgy of Aluminum part 2

Question 1

Atomic number of aluminum

Answer 1

13

Question 2

Valency of aluminum

Answer 2

3

Question 3

What are some physical properties of aluminum?

Answer 3

Low melting point
Low density
High electrical and thermal conductivity
High latent heat of fusion
Alloying will change the properties

Question 4

Which properties are related to aluminum’s binding energy (U)?

Answer 4

• Melting point (bond breaking)
• Coefficient of thermal expansion (bond stretching with thermal energy)
• Young’s modulus (stress stretching of bonds)

Question 5

Where does the binding energy occur?

Answer 5

At the typical neighbor bonding energy (r0)

Question 6

Which metals are FCC?

Answer 6

Atoms with 3 valence electrons
Ex: aluminum

Question 7

Which metals are BCC?

Answer 7

Atoms with 1 valence electron

Question 8

Which metals are HCP?

Answer 8

Atoms with 2 valence electrons

Question 9

Unit cell

Answer 9

Smallest repeating unit that produces the periodic arrangement of atoms in a crystalline solid

Question 10

What is the packing structure for FCC lattices?

Answer 10

ABC repeated

Question 11

What are the closed-packed planes and directions in Aluminum?

Answer 11

The FCC unit cell of aluminum has 4{111} planes and three <110> directions per plant

Question 12

Shear Modulus of Al

Answer 12

26 GPa

Question 13

Young’s Modulus of Al

Answer 13

70 GPa

Question 14

Poisson’s ratio of Al

Answer 14

0.34

Question 15

Critical Resolved Shear Stress (tCRSS)

Answer 15

The threshold stress to start dislocation motion (slip) on a slip system
This is a material property

Question 16

Critical Resolved Shear Stress for Aluminum

Answer 16

148 psi (~1 MPa) on {111}<110>
FCC metals have relatively low tCRSS

Question 17

When does plastic yielding start?

Answer 17

Plastic yielding starts when the resolved shear stress is equal to the critical resolved shear stress

Question 18

Stage 1 of plastic yielding

Answer 18

Single slip (easy glide) on the primary slip system (interaction between dislocations)

Question 19

Stage 2 of plastic yielding

Answer 19

Dislocation pile-up
(absent if cross-slip is easy)

Question 20

Stage 3 of plastic yielding

Answer 20

Dislocation intersection

Question 21

What is the width of a stacking fault determined by?

Answer 21

It is determined by the stacking fault energy (SFE)

Question 22

What happens due to aluminum having a high stacking fault energy?

Answer 22

Only one atom row exists between partial dislocations

Question 23

Why is the activation energy for cross-slip low in aluminum (can more easily cross-slip?)

Answer 23

Because of the short separation distance between partial dislocations, these can more easily form a constriction in the stacking fault and cross slip

Question 24

Which metal experiences all 3 stages of plastic deformation?

Answer 24

HCP is the only one that experiences easy glide

Question 25

Which metal experiences stages 2 and 3 of plastic deformation?

Answer 25

FCC crystals with low stacking fault energy, like Cu, experience stages 2 and 3

Question 26

Which metal experiences stage 3 only of plastic deformation?

Answer 26

FCC crystals with high stacking fault energy, like Al, and BCC crystals experience only stage 3 of plastic deformation

Question 27

Why is stage 1 (easy glide) not present in BCC and FCC?

Answer 27

Due to the existence of a large number of slip systems

Question 28

Why do FCC crystals with low stacking fault energy experience stage 2 of plastic deformation?

Answer 28

Because of low stacking fault energy, FCC crystals have wide stacking faults.
Since it cannot cross-slip its partials, it shows stage 2, dislocation pile-up

Question 29

Why do BCC crystals not experience stage 2 of plastic deformation?

Answer 29

Because BCC has 24 independent cross-slip systems, it cross-slips really easily, and has less tendency to form partial dislocations

Question 30

Why do FCC crystals with HIGH stacking fault energy experience only stage 3 of plastic deformation?

Answer 30

Because of cross-slip of full dislocations (5 independent slip systems) and cross-slip of partials (high SFE narrow stacking faults) is easy, dislocation pile-up does not exist

Question 31

What is the proportional limit?

Answer 31

The proportional limit is the end of the linear portion of the stress vs. strain graph, where the elastic region ends and the plastic region begins
In Al, since the transition is gradual, this proportional limit is difficult to find

Question 32

Proof stress

Answer 32

Maximum elastic strain in most materials is 0.001-0.002
As a result, offset/proof stress is taken at 0.001-0.002 offset from strain

Question 33

How can pure aluminum be strengthened?

Answer 33

strain hardening
solid solution strengthening
second phase and age hardening
grain refinement and grain refinement design

Question 34

How can you increase stress, and therefore strength in aluminum?

Answer 34

By adding obstacles to dislocation motion
tCRSS increases, and strength increases

Question 35

What are examples of obstacles to dislocation motion?

Answer 35

Other dislocations
Solute atoms
Second phase particles
Grain boundaries

Question 36

Why does the number of valence electrons determine the crystal structure of a metal?

Answer 36

Because of efficiency (energy minimization) in locating electrons in the interstitial spaces in the crystal structure
(note only sp electrons are considered, since d electrons are tightly held to nucleus)