1. Heat Treatment (Annealing) of Cold-Worked Metals
Annealing: heat treatment whereby ________________________________
______________________________________________________________
i. Cold working: increase y, TS; decrease ductility (see next slide)
ii. Annealing: decrease y, TS; increase ductility
3 Annealing Stages
tensile strength (MPa)
ductility (%EL)
tensile strength
ductility
Recovery
Recrystallization
Grain Growth
600
300
400
500 60 50 40 30 20
annealing temperature (ºC)
200
100
700
Three Stages of Anealling with
Increasing Temp.: -
(Note:Annealling stages may also occur by holding T above Trecrys)

2. Cold working (strain-hardening, work hardening) Plastic deformation 
Recall….
Cold working (strain-hardening, work hardening)
Plastic deformation 
Increase # of or dislocation density (now, very high dislocation density) 
___________________________ (many atoms out of place) 
Dislocation motion _______________________________: _____________
___________________________________ 
__________________________________________

3. As temp. increases, ______ _______________ increases
3 Stages of Annealing:
1. Recovery:
As temp. increases, ______
_______________ increases
Some dislocations move to relieve strain
Slight _________________ in # of dislocations
Minor changes in mechanical properties
Brass alloy

4. ________________________
2. Recrystallization:
_______________________
________________________
New set of ___________ grains:
Have ______ dislocation density
Gradually consume cold-worked grains
Occurs when T increased above Trecrys or held at T > Trecrys
Trecrys = min temp required for complete recrystallization in 1 h
Trecrys = ½ to 1/3 of Tmelt
T in Kelvin
[see photomicrographs; magnification 75X]
Brass alloy
crystals

5. Initial recrystallization
Cold-worked grains
3 580 C
Initial recrystallization
Brass
Trecrys = 450 C
4 580 C
8 580 C
Complete recrystallization
(Annealling stages occur by holding T above Trecrys)

6. Next Stage : grain growthC C
New set of strain-free grains w/ low dislocation density

7. If left at elevated temperature following recrystallization:
3. Grain Growth:
If left at elevated temperature following recrystallization:
_________________________ (coarsening)
some grow while others shrink
[see photomicrographs; magnification 75X]
Brass alloy
Cold-worked Grains:
_______ dislocation density
_______ strain
 stronger
After Annealing, Grains:
 more ductile

8. ______ dislocation density: 109 – 1010 / mm2 (highly deformed)
Cold-worked Metals:
Grains:
______ dislocation density:
109 – 1010 / mm2 (highly deformed)
______ strain state
 stronger
After Annealing, Metals:
105 – 106 / mm2
______strain state
 more ductile
Brass alloy

9. Hot-Working “metal forming operation performed at T > Trecryst”
Metal remains ductile or (if previously cold-worked) becomes more ductile
-Forging A o d
force
die
blank
-Rolling
roll A o d pp
-Drawing
tensile
force A o d
die
-Extrusion
ram
billet
container
force
die holder
die A o d
extrusion
Steel 1020
y (MPa)
TS (MPa)
%EL
Cold-drawn
350
420 15
Hot-rolled
210
380 25

10. Mechanical Properties of Ceramics (12.8-12.11, 13.11)

11. Elastic Modulus Most Ceramics: E = ___________ GPa
Most Metals: E = _____________ GPa
Most polymers: E < ____ GPa

12. Density

13. Most ceramics fail/fracture __________________________________ (see below)
Very __________ (not tough, not ductile)
Exhibits ___________________________________ (which parallels?)
______________________ (ionic) prevents dislocation motion (i.e. “slip”)
Recall:
Ionic bonds: kJ/mol
Metallic bonds: <850 kJ/mol
Covalent bonds: 346 kJ/mol (for C-C)

14. Al2O3 and Soda-lime glass

15. Why Flexural Strength (not Tensile Strength) for Ceramics?
Difficult to prepare and test specimens with the required geometry
Difficult to grip brittle materials into tensile tester clamps w/o fracture
Ceramics fail after ~ 0.1% strain
Material s fs
(MPa) E(GPa)
Si nitride
Si carbide
Al2O3
Zirconia (ZrO2)
glass (soda) 69
304
345
380
205 69

16. Measuring Flexural Strength (fs)
• _____________________________ test to measure strength. F
L/2
d = midpoint
deflection
cross section R b d
rect.
circ.
location of max tension
Adapted from Fig , Callister 7e.
For rectangular specimen:
Specimen:
- rod with circular or rectangular cross-section
Apply force:
- “top” face [ in compression]
- “bottom” face [in tension]
3. Fracture occurs on “tension face”
Why? For ceramics: TS = 1/10 of compressive strength
So, flexural test good substitute for tensile test
fs = (3Ff L)/(2bd2)
Ff = load at fracture
L = distance between
support pts
For rectangular specimen:
fs = (Ff L)/(R2)
R = radius of specimen

17. Influence of Pores in Ceramics
Figure 13.16
powder
With
pressure
Figure 13.14
pore
With
sintering
Pores introduced during “powder processing”

18. fs = o exp (-nP) E = Eo(1-1.9P + 0.9P2) Fig. 12.35 Fig. 12.36
E versus P
fs versus P
fs = o exp (-nP)
E = Eo(1-1.9P + 0.9P2)
0, n: experimental constants
Increase P  _______________ fs
10% porosity  decrease fs 50% (vs non-porous)
Eo = elastic modulus of non-porous material
P = volume fraction porosity
Increase P  __________________ E