This document discusses various metallic materials and processes used to manufacture and shape metals. It covers the following key topics in 3 sentences: Ferrous metals like steel and cast iron contain a large percentage of iron, while nonferrous metals do not or only contain a small percentage of iron. Common metal processing techniques include casting molten metal into molds, hot and cold rolling of sheets and plates, and extruding and forging metals into various shapes. The mechanical properties of metals like stress, strain, hardness, toughness and fracture are also described in detail.

1. PHY351
CHAPTER 6
1
Metals

2. METALLIC MATERIAL
Metals and Alloys
Ferrous
Eg: Steel,
Cast Iron


Nonferrous
Eg:Copper
Aluminum
Ferrous Metals and alloys
- Metals and alloys that contain a large percentage of iron such as steels and
cast irons
Nonferrous metals and alloys
- Metals and alloys that do not contain iron.
- If they do contain iron, it is only in a relatively small percentage.
2

3. QUESTION 1
1.
Define the following materials in terms of their
properties and give an example each of their
application.
a.
b.
c.
Stainless steel
Brass
Cast iron
3

4. PROCESSING OF METAL - CASTING
Most metals are first melted in a furnace.
 Alloying is done if required.
 Large ingots are then cast.
 Sheets and plates are then produced from ingots by
rolling (wrought alloy products).
 Channels and other shapes are produced by extrusion.
 Some small parts can be cast as final product.
Example :- Automobile Piston.

4

5. Casting Process
Casting mold
5

6. HOT ROLLING OF STEEL
Hot rolling : Greater reduction of thickneess in a single
pass.
 Rolling carried out at above recrystallization
temperature.
 Ingots preheated to about 12000C.
 Ingots reheated between passes if required.
 Usually, series of 4 high rolling mills are used.

6

7. COLD ROLLING OF METAL SHEET
Cold rolling is rolling performed below recrystallization
temperature.
 This results in strain hardening.
 Hot rolled slabs have to be annealed before cold rolling.
 Series of 4 high rolling mills are usually used.
 Less reduction of thickness.
 Needs high power.

7

8. % Cold work =
Initial metal thickness – Final metal thickness
x 100
Initial metal thickness
8

9. EXTRUSION
Metal under high pressure is forced through opening in a
die.
 Common Products are cylindrical bar, hollow
tubes from copper, aluminum etc.
 Normally done at high temperature.
 Indirect extrusion needs less power however has
limit on load applied

9

10. Direct
Extrusion
Die
Container
Metal
Indirect
Extrusion
Container
10
Metal
10

11. FORGING
Metal, usually hot, is hammered or pressed into desired
shape.
 Types:



Open die:
Dies are flat and simple in shape.
(Example products: Steel shafts)
Closed die:
Dies have upper and lower impresion.
(Example products: Automobile engine connection rod)
Forging increases structural properties, removes porosity
and increases homogeneity.
11

12. Direct
Forging
Indirect
Forging
Metal
Die
12

13. DRAWING

Wire drawing :- Starting rod or wire is drawn through
several drawing dies to reduce diameter.
% cold work =
Change in cross-sectional area
X 100
Original area
13

14. 
Deep drawing:- Used to shape cup like articles
from flats and sheets of metals
14
14

15. MECHANICAL PROPERTIES OF METAL
Stress
 Strain
 Hardness
 Impact Energy
 Fracture
 Toughness
 Fatigue
 Creep

15

16. STRESS

Metals undergo deformation under uniaxial tensile force.

Elastic deformation:
Metal returns to its original dimension after tensile
force is removed.

Plastic deformation:
The metal is deformed to such an extent such
that it cannot return to its original dimension
16

17. F
Engineering stress , σ
=
A0
Units of Stress are PSI or N/M2 (Pascals)
1 PSI = 6.89 x 103 Pa
17

18. Engineering strain , ε =

Change in length
Original length
  0
0



Units of strain are in/in or m/m.
18

19. SHEAR STRESS AND SHEAR STRAIN
τ = Shear stress =
S
A
Shear strain γ =
Amount of shear displacement
Distance ‘h’ over which shear acts.
19

20. 
Modulus of elasticity (E) or Young’s modulus : Stress
and strain are linearly related in elastic region. (Hooks law)
σ (Stress)
E=
ε (Strain)
Strain
Δσ
Δσ
E=
Δε
Δε
Stress
Linear portion of the stress strain curve
20

21. 
Higher the bonding strength, higher is the modulus of
elasticity.
Examples:
Modulus of Elasticity of steel is 207 Gpa.
Modulus of elasticity of Aluminum is 76 Gpa
21

22. YIELD STRENGTH

Yield strength is strength at
which metal or alloy show
significant amount of plastic
deformation.

0.2% offset yield strength is
that strength at which 0.2%
plastic deformation takes place.

Construction line, starting at
0.2% strain and parallel to
elastic region is drawn to 0.2%
offset yield strength.
22

23. ULTIMATE TENSILE STRENGTH


Ultimate tensile strength
(UTS) is the maximum
strength reached by the
engineering stress strain
curve.
Necking starts after UTS is
reached.
Al 2024-Tempered
S
T
R
E
S
S
Mpa
Necking Point
Al 2024-Annealed
Strain
Stress strain curves of
Al 2024 With two different
heat treatments. Ductile
annealed sample necks more
23

24. 
More ductile the metal is, more is the necking before
failure.

Stress increases till failure. Drop in stress strain curve is
due to stress calculation based on original area.
24

25. PERCENT ELONGATION

Percent elongation is a measure of ductility of a material.

It is the elongation of the metal before fracture expressed
as percentage of original length.
% Elongation =
Final length* – initial Length*
Initial Length
25

26. PERCENT REDUCTION IN AREA

Percent reduction area is also a measure of ductility.

The diameter of fractured end of specimen is measured
using caliper.
% Reduction
=
Area
Initial area – Final area
Initial area
26

27. 
Percent reduction in area in metals decreases in case of
presence of porosity.
27

28. TRUE STRESS – TRUE STRAIN

True stress and true strain are based upon instantaneous
cross-sectional area and length.

True stress is always greater than engineering stress.
28

29. 
True Stress = σt =
F
Ai (instantaneous area)
i

True Strain = εt =

0
d

 Ln
li
l0
 Ln
A0
Ai
29

30. QUESTION 2
1.
A 0.5cm diameter aluminium bar is subjected to a force of 500N.
Calculate the engineering stress in MPa on the bar.
(Answer: 25.5 MPa)
2.
A 1.25cm diameter bar is subjected to a load of 2500 kg. Calculate
the engineering stress on the bar in MPa.
(Answer: 200 MPa)
3.
A sample of commercially pure aluminium 1.27cm wide, 0.1cm
thick and 20.3cm long that has gage markings 5.1cm apart in the
middle of the sample is strained so that the gage markings are
6.7cm apart. Calculate the engineering strain and the percent
engineering strain elongation that the sample undergoes.
(Answer: 0.31, 31%)
30

31. 4.
A 12.7mm diameter round sample of a 1030 carbon steel is pulled
to failure in a tensile testing machine. The diameter of the sample
was 8.7mm at the fracture surface. Calculate the percent reduction
in area of the sample.
(Answer: 53%)
5.
A 70% Cu-30% Zn brass sheet is 0.12cm thick and is cold-rolled
with a 20 percent reduction in thickness. What must be the final
thickness of the sheet?
(Answer: 0.096cm)
6.
Calculate the percent cold reduction when an aluminium wire is
cold-drawn from a diameter of 6.5mm to a diameter of 4.25mm.
(Answer: 57.2%)
7.
A tensile specimen of cartridge brass sheet has a cross section of
10 mm x 4mm and a gage length of 51mm. Calculate the
engineering strain that occurred during a test if the distance
between gage markings is 63 mm after the test.
(Answer: 0.235)
31

32. 8.
Compare the engineering stress and strain with the true test and
strain for the tensile test of low-carbon steel that has the following
test values.
Load applied to specimen = 69 000N
Initial specimen diameter = 1.27 cm
Diameter specimen under 69 200 N load = 1.2 cm
(Answer: 544.6 MPa, 610 Mpa, 0.12, 0.117)
9.
A 20cm long rod with a diameter of 0.25cm is loaded on an FCC
single crystal. Calculate
a.
The engineering stress and strain at this load
(Answer: 1019 MPa, 0.147)
b.
The trues tress and strain at this load.
(Answer: 1443 MPa, 0.349)
32

33. HARDNESS
Hardness is a measure of the resistance of a metal to
permanent (plastic) deformation.
 General procedure:

Press the indenter that
is harder than the metal
Into metal surface.
Withdraw the indenter
Measure hardness by
measuring depth or
width of indentation.
33

34. Rockwell hardness tester
34

35. FRACTURE

Fracture results in separation of stressed solid into two or
more parts.

There are two types of fracture:
 Ductile fracture
 Brittle Fracture
35

36. DUCTILE FRACTURE

Ductile fracture :
High plastic deformation & slow crack propagation (fracture
due to slow crack propagation).

Three steps :
 Specimen forms neck and cavities within neck.
 Cavities form crack and crack propagates towards
surface, perpendicular to stress.
 Direction of crack changes to 450 resulting in cup-cone
fracture.
36

37. Ductile fracture
37

38. BRITTLE FRACTURE

No significant plastic deformation before fracture (fracture
due to rapid crack propagation).

Common at high strain rates and low temperature.

Three stages:
 Plastic deformation concentrates dislocation along slip
planes.
 Microcracks nucleate due to shear stress where
dislocations are blocked.
 Crack propagates to fracture.
38

39. Example:
HCP Zinc ingle crystal under high stress along {0001}
plane undergoes brittle fracture.
SEM of ductile fracture
SEM of brittle fracture
39

40. Brittle Fracture
40

41. 
Brittle fractures are due to defects like
 Folds
 Undesirable grain flow
 Porosity
 Tears and Cracks
 Corrosion damage
 Embrittlement due to atomic hydrogen

At low operating temperature, ductile to brittle transition
takes place
41

42. TOUGHNESS
Toughness is a measure of energy absorbed before
failure.
 Impact test measures the ability of metal to absorb
impact.
 Toughness is measured using impact testing
machine

42

43. FRACTURE TOUGHNESS

Cracks and flaws cause stress concentration.
K1  Y a
where
K1 = Stress intensity factor.
σ = Applied stress.
a = edge crack length
Y = geometric constant.
43

44. 
KIc = critical value of stress intensity factor (fracture
toughness)
Example:
Al 2024 T851
26.2MPam1/2
4340 alloy steel 60.4MPam1/2
44

45. QUESTION 3
1.
A structural plate component of an engineering design must
support 207 MPa in tension. If aluminium alloy 2024-T851 is
used for this application, what is the largest internal flaw size
that this material can support? (Use Y = 1, KIc =26.4 MPa)
(Answer: 5.18 mm)
45

46. FATIGUE

The phenomenon leading to fracture under repealed
stresses having a maximum value less than the ultimate
strength of the material.

Metals often fail at much lower stress at cyclic loading
compared to static loading.

Crack nucleates at region of stress concentration and
propagates due to cyclic loading.

Failure occurs when cross sectional area of the metal too
small to withstand applied load.
46

47. Fracture started here
Final rupture
Fatigue fractured surface of keyed shaft
47

48. 
Factors affecting fatigue strength:




Stress concentration: Fatigue strength is reduced by stress
concentration.
Surface roughness: Smoother surface increases the fatigue
strength.
Surface condition: Surface treatments like carburizing and
nitriding increases fatigue life.
Environment: Chemically reactive environment, which
might result in corrosion, decreases fatigue life.
48

49. CREEP

Creep is a time-dependent plastic deformation when
subjected to a constant stress or load.

Important in high temperature applications:
i.
Primary creep: creep rate decreases with time due
to strain hardening.
ii.
Secondary creep: Creep rate is constant due to
simultaneous strain hard- ening and recovery
process.
iii. Tertiary creep: Creep rate increases with time
leading to necking and fracture.
49

50. Creep curve.
The slope of the linear part of the curve is the steady-state creep rate.
50

51. Creep test determines the effect of temperature and
stress on creep rate.
 Metals are tested at constant stress at different
temperature & constant temperature with different stress.

High temperature
or stress
Medium temperature
or stress
Low temperature
or stress
51

52. 
Creep strength: Stress to produce minimum creep rate
of 10-5%/h at a given temperature.
52

53. 
Creep rupture test is same as creep test but aimed at
failing the specimen.

Plotted as log stress versus log rupture time.
53

54. 
Time for stress rupture decreases with increased
and temperature
stress
54

55. REFERENCES




A.G. Guy (1972) Introduction to Material Science,
McGraw Hill.
J.F. Shackelford (2000). Introduction to Material Science
for Engineers, (5th Edition), Prentice Hall.
W.F. Smith (1996). Principle to Material Science and
Engineering, (3rd Edition), McGraw Hill.
W.D. Callister Jr. (1997) Material Science and
Engineering: An Introduction, (4th Edition) John Wiley.
55