Chapter 2
Fundamentals of the Mechanical
Behavior of Materials
Outline
Introduction
Tensile test
True stress - true strain (flow curve)
mechanical properties:
- Resilience
- Ductility
- Toughness
- Hardness
1
Introduction
 One of the oldest and most important groups of
manufacturing process is plastic deformation (shaping
materials by applying forces by various means), also known
as deformation process, it includes bulk deformation
processes:
1. Forging
2. Rolling
3. Extrusion
4. Rod and wire drawing.
 Sheet metal-forming processes (cutting, bending, drawing
and general press working).
 This chapter deals with mechanical behavior of the material
during plastic deformation
2
Introduction
 In stretching a piece of metal to make an object such as
an automobile fender or a length of wire, the material is
subjected to tension.
 A solid cylindrical piece of metal is forged to make a
turbine disk, subjecting the material to compression.
 Sheet metal undergoes shearing stresses when for
example, a hole is punched through its cross section.
 In all these processes, the material is subjected to one
or more of the three basic modes of deformation,
namely, tension, compression and shear.
3
Introduction
 The degree of the deformation to which the material is
subjected is defined as strain.
 For tension or compression its called, the engineering strain
or nominal strain :
e=
 Shear strain is defined as:
4
Introduction
 In order to change the shape of an elements, or bodies,
forces must be applied to them as shown the previous
slide.
 The determination of these forces as a function of strain
is an important aspect in the study of manufacturing
process
 A knowledge of these forces is essential in order:
1. To design the proper equipment (weather these
equipment applies tension, compressive or shearing
forces) to be used in manufacturing
2. To select tool and die materials for proper strength
3. To determine weather a specific metalworking operation
can be accomplished on certain equipment (shearing
equipment differs from tension one)
5
Introduction
 Many materials, when in service, are subjected to forces
or loads.
 Airplane wing is constructed from aluminum alloy.
 An automobile axle from the steel.
• In such situations it is necessary to know the Service
conditions in order to determine characteristics of the
material and to design the member from which it is made
such that any resulting deformation will not be excessive
and fracture will not occur.
 The mechanical behavior of a material reflects the
response of the material to deformation from an applied
load or force.
 It is necessary to know the properties of materials.6
Properties of Materials
• Mechanical Properties: strength, toughness,
ductility, hardness, elasticity, fatigue, and creep
• Physical Properties: density, specific heat,
thermal expansion and conductivity, melting point,
and electrical and magnetic properties
• Chemical Properties: oxidation, corrosion.
• Manufacturing Properties: castability (is the
ease of forming a quality casting), formability,
machinability, weldability, …
7
Oxidation versus Corrosion.
They are essentially the same thing.
Oxidation is the process where electrons (which bind atoms
together to create materials) are drawn away by free oxygen
molecules which are relatively unstable and looking for
available electrons [rust forms on metal].
Corrosion is very similar, in that when material such as
steel is exposed to an environment that causes it to come
into contact with either a liquid, or dissimilar metal, a
galvanic reaction occurs where molecules seek to find a
balance between unequal numbers of electrons, the material
giving up more electrons tends to show a greater rate of
corrosion. iron's major drawback as a construction material
is that it reacts with moist air (in a process called corrosion)
8
Concepts of tests
 The mechanical behavior may be ascertained by a
simple stress–strain test.
 The test is conducted for metals at room temperature.
 Destructive test (tests are carried out to the specimen's
failure, in order to understand a specimen's structural
performance or material behavior under different loads)
 There are three principal ways in which a load may be
applied:
 Tension
 compression
 Shear
 In engineering practice many loads are torsional rather
9
than pure shear
Concepts of tests
10
Engineering Stress
• Tensile stress, s:
• Shear stress, t:
Ft
F
Area, A
Area, A
Ft
Ft
lb f
N
= 2 or
s=
2
in
m
Ao
original area
before loading
Ft
Fs
Fs
Fs
t=
Ao
Ft
F
 Stress has units:
N/m2 or lbf/in2
Chapter 6 - 11
Engineering Strain
• Tensile strain:
• Lateral strain:
d/2
e = d
Lo
wo
• Shear strain:
-dL
eL =
wo
Lo
dL /2
q
g = x/y = tan q
x
90º - q
y
90º
Strain is always
dimensionless.
Adapted from Fig. 6.1 (a) and (c), Callister 7e.
Chapter 6 - 12
Common States of Stress
• Simple tension: cable
F
F
A o = cross sectional
area (when unloaded)
F
s=
s
Ao
s
• Torsion (a form of shear): drive shaft
M
Ac
M
Fs
Ski lift
(photo courtesy
P.M. Anderson)
Ao
Fs
t =
Ao
2R
Note: t = M/AcR here.
Chapter 6 - 13
OTHER COMMON STRESS STATES (1)
• Simple compression:
Ao
Canyon Bridge, Los Alamos, NM
(photo courtesy P.M. Anderson)
Balanced Rock, Arches
National Park
(photo courtesy P.M. Anderson)
F
s=
Ao
Note: compressive
structure member
Chapter 6 - 14
OTHER COMMON STRESS STATES (2)
• Bi-axial tension:
Pressurized tank
(photo courtesy
P.M. Anderson)
• Hydrostatic compression:
Fish under water
sq > 0
sz > 0
(photo courtesy
P.M. Anderson)
sh< 0
Chapter 6 - 15
Tension
One of the most common mechanical stress–strain
tests is performed in tension.
The tension test can be used to ascertain several
mechanical properties of materials that are important
in design.
1. Strength: Yielding and yield strength.
2. Ductility,
3. Toughness,
4. Elastic modulus or Young’s modulus (E) (the
stiffness of the material)
5. Modulus of resilience (U) : (the specific energy that
the material can store elastically).
16
Tensile-Test Specimen and Machine
• A specimen is deformed, usually to fracture, with a
gradually increasing tensile load that is applied uniaxially
along the long axis of a specimen.
• Normally, the cross section is circular, but rectangular
specimens are also used.
• The standard diameter is approximately 12.8 mm (0.5 in.)
• The standard length of the specimen is 50 mm
• Test is done according to American Society for Testing and
Materials (ASTM).
17
Stress-Strain Testing
• Typical tensile test
machine
extensometer
• Typical tensile
specimen
specimen
Adapted from
Fig. 6.2,
Callister 7e.
gauge
length
Adapted from Fig. 6.3, Callister 7e. (Fig. 6.3 is taken from H.W.
Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of
Materials, Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons,
New York, 1965.)
Chapter 6 - 18
Tensile-Test Specimen and Machine
Figure 2.1 (a) A standard tensile-test specimen before and after pulling, showing original and final
gage lengths. (b) A typical tensile-testing machine.
19
Tensile Test
 When the load is applied, the
specimen elongates
proportionately to the load up
to the proportional limit ( this is
the range of linear elastic
behavior).
 Continuing applying the load,
the material will continue to
deform elastically, although
not linearly, up to yielding
point Y.
 When the load is removed
before the yield point is
reached, the specimen will
return to its original length.
20
Elastic Deformation
1. Initial
2. Small load
3. Unload
bonds
stretch
return to
initial
d
F
F
Linearelastic
Elastic means reversible!
d
Non-Linearelastic
Chapter 6 - 21
Elastic Properties
• Macroscopic elastic strain is manifested as small
changes in the interatomic spacing and the stretching of
interatomic bonds. Thus, the magnitude of E is a
measure of the resistance to atoms separation, that is,
the interatomic bonding forces.
Chapter 6 - 22
Tensile Test
 When increasing the load beyond
the yield point, the specimen
begins to yield; that is, it begins
to undergo plastic (permanent)
deformation, and the relationship
between the stress and strain is
no longer linear.
 For most materials the rate of
change of the slope of the stressstrain curve from linear to nonlinear up to yield point is very
small, thus the determination of
the yield point, Y, can be
difficult
 Yield point determination will be
discussed later
23
Plastic Deformation (Metals)
1. Initial
2. Small load
bonds
stretch
& planes
shear
delastic + plastic
3. Unload
planes
still
sheared
dplastic
F
F
Plastic means permanent!
linear
elastic
linear
elastic
dplastic
d
Chapter 6 - 24
Tensile Test
 Its important to note that yielding dose not necessarily
mean failure.
 In the design of structures and load-bearing members,
yielding is not acceptable since it leads to permanent
deformation [ it will not do its functionality].
 However, yielding is necessary in metalworking
process, such as forging, rolling and sheet-metal
forming operations, where materials have to be
subjected to permanent deformation to develop the
desired part shape
25
Tensile Test
As the specimen
continuous to elongate
under increasing the
load beyond Y, its crosssectional area decreases
permanently and
uniformly throughout its
gage length.
26
Tensile Test
 If the specimen is unloaded from a
stress level higher than Y, the curve
follows a straight line downward and
parallel to the original elastic slope as
shown.
 Some fraction of the total deformation
is recovered as elastic strain.
 This elastic strain, which is regained
during unloading, corresponds to the
strain recovery.
 If the load is reapplied yielding will
again occur at the unloading stress
level where the unloading began.
27
Plastic (Permanent) Deformation
(at lower temperatures, i.e. T < Tmelt/3)
• Simple tension test:
Elastic+Plastic
at larger stress
engineering stress, s
Elastic
initially
permanent (plastic)
after load is removed
ep
engineering strain, e
plastic strain
Adapted from Fig. 6.10 (a),
Callister 7e.
Chapter 6 - 28
Loading and Unloading of Tensile-Test Specimen
slope
E = ……..
Note that, during unloading,
the curve follows a path
parallel to the original elastic
slope.
spring back
……………...
in bending
Schematic illustration of the loading and
the unloading of a tensile- test specimen.
Tensile Test
 As the load is further increased,
the curve eventually reaches a
maximum point and then begins
to decrease.
 This maximum point is known as
the tensile strength or ultimate
tensile strength (UTS) of the
material.
 UTS is thus a simple and practical
measure of the overall strength
of the material (strength of the
material that could withstand
without failure)
Tensile Test
 When the specimen is loaded beyond
its UTS, it begins to neck and
elongation is no longer uniform.
 That is, the change in the crosssectional area of the specimen is no
longer uniform but is concentrated
locally in a neck formed in the
specimen (called necking).
 As the test progress, the engineering
stress drops further and specimen
finally fractures within the necked
region.
 The final stress level (marked by an x
in the figure) at fracture is known as
breaking or fracture stress.
Mechanical Behaviour
As plastic deformation
proceeds, the force
increases due to work-hardening
……………….
As more of the stress becomes
concentrated in the neck,
formation of voids
……… occur
These voids result in even
higher stress concentrations
and eventual fracture
Tensile Strength, TS
• Maximum stress on engineering stress-strain curve.
TS
F = fracture or
ultimate
strength
engineering
stress
sy
Typical response of a metal
Neck – acts
as stress
concentrator
strain
engineering strain
• Metals: occurs when noticeable necking starts.
• Polymers: occurs when polymer backbone chains are
aligned and about to break.
Chapter 6 - 33
Tensile Test
34
Linear Elastic Properties
• Modulus of Elasticity, E (also known as Young's modulus):
E can be thought as the stiffness, or the resistance to elastic
deformation.
• Hooke's Law:
s=Ee
s
F
E
e
Linearelastic
F
simple
tension
test
Chapter 6 - 35
Stress-Strain Curves
 Modulus of Elasticity or
Young’s modulus (E) = Slope
 The elongation of the specimen
is accomplished by a
contraction of its lateral
dimension.
 The absolute value of the ratio
of lateral strain to longitudinal
strain is know as Poisson’s
ratio, v.
• E in Mpa
Y
36
Poisson’s Ratio
When pulled in tension (Z), a sample gets
longer and thinner, i.e., a contraction in
the width (X) and breadth (z)
if compressed gets fatter
Poisson’s ratio defines how much
strain occurs in the lateral directions
(x & y) when strained in the (z)
direction:
lateral strain
 =longitudin al strain
ex
ey
 ==ez
ez
Typical values = 0.2 to 0.5
Stress-Strain Curves
 The area under stress-stain
curve up to the yield point, Y,
of the material is known as the
modulus of resilience
 (U)= Area under stress – strain
curve up to the yield point (Y)
U : has the units of Energy per unit volume (N-m/ m ^3), and indicates
38
the specific energy that the material can store elastically.
Resilience
Ability of material to absorb energy during elastic deformation and
then to give it back when unloaded.
Measured with Modulus of Resilience, Ur
Ur , is area under s - e curve up to yielding:
ey

U r = s de
0
Assuming a linear elastic region:
2
s
s
 y
y
 =
U r = 12 s ye y = 12 s y 
 E  2E
Units are J/m3
Stress-Strain Curve
• Engineering stress
s = P/Ao
• Engineering strain
e = (l- lo) / lo
• Measures of ductility
% elongation
(lf - lo) / lo x 100
A typical stress- strain curve obtained
from a tension test, showing various
features.
% Reduction area
(Af - Ao) / Ao x 100
Engineering Stress-Strain
Engineering stress --- the ratio of the applied
load P to the original cross-sectional area A0
 Engineering Stress (σ) =
 Engineering Strain
(e) =
P
A0
(l - l0 )
l0
41
Engineering Stress-Strain
 Engineering Strain =
(l - l0 )
e=
l0
42
Mechanical Properties of Various Materials
at Room Temperature
TABLE 2.2 Mechanical Properties of Various Materials at Room Temperature
Metals (Wrought)
E (GPa)
Y (MPa)
UTS (MPa)
Elongation
in 50 mm
(%)
Aluminum and its alloys
Copper and its alloys
Lead and its alloys
Magnesium and its alloys
Molybdenum and its alloys
Nickel and its alloys
Steels
Titanium and its alloys
Tungsten and its alloys
69–79
105–150
14
41–45
330–360
180–214
190–200
80–130
350–400
35–550
76–1100
14
130–305
80–2070
105–1200
205–1725
344–1380
550–690
90–600
140–1310
20–55
240–380
90–2340
345–1450
415–1750
415–1450
620–760
45–4
65–3
50–9
21–5
40–30
60–5
65–2
25–7
0
Nonmetallic materials
Ceramics
70–1000
—
140–2600
0
Diamond
820–1050
—
—
—
Glass and porcelain
70-80
—
140
—
Rubbers
0.01–0.1
—
—
—
Thermoplastics
1.4–3.4
—
7–80
1000–5
Thermoplastics, reinforced
2–50
—
20–120
10–1
Thermosets
3.5–17
—
35–170
0
Boron fibers
380
—
3500
0
Carbon fibers
275–415
—
2000–3000
0
Glass fibers
73–85
—
3500–4600
0
Kevlar fibers
62–117
—
2800
0
Note: In the upper table the lowest values for E, Y, and UTS and the highest values for elongation are for pure metals.
Multiply gigapascals (GPa) by 145,000 to obtain pounds per square in. (psi), megapascals (MPa) by 145 to obtain psi.
43
True Stress and True Strain
True stress --- is the ratio of the load (P) to the
actual (hence true) or instantaneous cross-section
area (A)
True Stress (s) =
P
A
Where (A) is the instantaneous (true or actual)
cross-sectional area
44
True Stress and True Strain
In tension test, there will be a series of incremental
tension where, for each increment, the specimen is a
little longer than at the preceding stage. Thus, the true
strain, ϵ, can be defined as
• True strain --- the elongation of the specimen in increments of
instantaneous change in length
True Strain (e) =

l
l0
dl
l
= ln( )
l
l0
45
True Stress and True Strain
 For small values of engineering strain, we have e = ϵ
 For larger values of strain, however, the values rapidly
diverge as seen the table above.
 As we have seen that, at small stains, the engineering and
true strains are very close to each other and, therefore,
either one can be used in calculations.
 However, for large strains encountered in metalworking,
the true strain should be used because it is the true 46
measure of the strain.
True Stress and True Strain
Since the volume of the
material specimen remains
constant in the plastic
region, thus the true strain
can be expressed as
Ao
Do 2
Do
l
e = ln( ) = ln( ) = ln( ) = 2 ln( )
lo
A
D
D
47
Engineering Stress vs. True Stress
Since the actual cross-sectional area is reduced, use of the initial
area gives a lower value than the actual one (the ratio is Ao/Ac).
True stress
…….
True stress, s = P/Ac
– P: load
– Ac: current area
True strain, e = ln (lc/lo)
- lc: current length
- lo: original length
Engineering stress
…….………
s
e
Even though the true stress-strain curve gives a more accurate picture of
the breaking strength of a material, it is difficult to obtain measurements
of the actual area in real-time.
Usually, the reported values are the engineering stress.
True fracture strength > tensile strength
 but the engineering s - e diagram does not show this
True Stress- Strain Curves
The relationship between engineering and true
values for stress and strain can now be used to
construct true stress-true strain curves.
How???????
Construction of True Stress-True Strain Curve
(a) Load-elongation curve in tension
testing of a stainless steel specimen.
(b) Engineering stress-engineering strain
curve, drawn from the data in Fig. a.
(c) True stress-true strain curve, drawn
from the data in Fig. b.
Note that this curve has a
positive slope, indicating that
the material is becoming
stronger as it is strained.
(d) True stress-true strain curve
plotted on log-log scale, drawn
from data in fig. c and based on
the corrected curve in Fig. c. The
correction is due to the triaxial
state of stress that exists in the
necked region of a specimen
True Stress- Strain Curves
 A typical true stress-true strain curve is
shown
 Such curve is typically approximated by
the equation
 The above equation indicates neither
the elastic region nor the yield point Y
 The strains at the yield point are very small ϵ
= e , the differences between the true yield
stress and engineering stress is negligible.
 The reason is that, at yielding, the differences
in the cross-sectional areas Ao and A is
negligible
True Stress- Strain Curves
 If we plot the true stress-true strain curve on a log-log scale, we
obtain the figure shown to the right.
 The slope of this figure is n : is known as the strain-hardening
exponent.
 It is similar to solve this
equation by taking the log for
each side
log s = log K  n log e
Strength of coefficient K
K is known as the
strength of
coefficient.
Note that K is the
true stress at a true
strain of unity
Note from the figure that the elastic strain is much
smaller than the plastic strain.
Consequently, and although both effects exist, we will
ignore elastic strains in our calculations for forming
process, thus the plastic strain will be the total strain that
the material undergoes
Flow stress Yf
 From the figure, Yf, is known
as the flow stress.
 The flow stress is defined as
the trues stress required to
continue plastic deformation
at a particular stain, ϵ1.
Flow stress is defined as the
instantaneous value
of stress required to continue
plastically deforming the material to keep the metal flowing. Flow
stress can also be defined as the
stress required to sustain plastic
deformation at a particular strain.,
Flow stress Yf
 Flow stress is the stress required to sustain a certain plastic
strain on the material.
 In forming of materials, we are concerned with flow stress of
material being formed, as this affects the ability of material to
undergo deformation
 Factors such as strain rate, temperature, affect the flow
stress of materials.
 A simple power law expression for flow stress of a
material which does not show anisotropy can be
expressed as:
 where n is known as strain hardening exponent.
Chapter 6 -
Flow stress Yf
 Higher strain hardening exponent values enhance the
flow stress. Similarly, flow stress is enhanced with
increase in strain rate during a plastic deformation
process.
 Effect of strain rate on flow stress becomes more
pronounced at higher temperatures.
 AT higher temperatures [hot working], strain hardening
may not have effect on flow stress. However, during
cold working effect of strain on flow stress cannot be
neglected as the material gets work hardened .
 In such case, average flow stress can be determined
between two given strains.
Chapter 6 - 56
Flow stress Yf
 Flow curve is the stress-strain curve for a material in the
plastic range. It describes material behavior in metal
forming. From flow curve, we can determine the flow
stress as
( )
s =K e
n
 In forming processes, such as forging, the
instantaneous flow stress can be found from the flow
curve, as the stress required to cause a given strain or
deformation.
Chapter 6 - 57
Flow stress Yf
In Rolling, for example, the flow stress
considerably changes during the forming
process as the material gets work hardened
considerably. In such case, an average flow
stress is determined from the flow curve. The
average flow stress is given as:
Chapter 6 - 58
True Stress & Strain
• Curve fit to the stress-strain response:
( )
sT = K eT
“true” stress (F/A)
n
hardening exponent:
n = 0.15 (some steels)
to n = 0.5 (some coppers)
“true” strain: ln(L/Lo)
Chapter 6 - 59
Hardening
• An increase in sy due to plastic deformation.
s
sy
1
sy
0
large hardening
small hardening
e
Chapter 6 - 60
True Stress-True Strain
Curves
Figure 2.6 True stress-true
strain curves in tension at
room temperature for
various metals. The
curves start at a finite level
of stress: The elastic
regions have too steep a
slope to be shown in this
figure, and so each curve
starts at the yield stress, Y,
of the material.
61
Typical Values for K and n at
Room Temperature
TABLE 2.3
Aluminum
1100–O
2024–T4
6061–O
6061–T6
7075–O
Brass
70–30, annealed
85–15, cold-rolled
Cobalt-base alloy, heat-treated
Copper, annealed
Steel
Low-C annealed
4135 annealed
4135 cold-rolled
4340 annealed
304 stainless, annealed
410 stainless, annealed
K (MPa)
n
180
690
205
410
400
0.20
0.16
0.20
0.05
0.17
900
580
2070
315
0.49
0.34
0.50
0.54
530
1015
1100
640
1275
960
0.26
0.17
0.14
0.15
0.45
0.10
62
Instability in Tension
 We have observed that once ultimate tensile strength is
reached, the specimen will begin to neck and thus
deformation is no longer uniform.
 This phenomena has important significance because
nonuniform deformation will cause part thickness
variation and localization in processing of materials,
particularly in sheet forming operation where the
material is subjected to tension.
 Numerically, it was found that the true strain at the
onset of necking is equal to the strain-hardening
exponent (n).
 ϵT = n at the UTS
63
Instability in Tension
 Note in figure, that the slope of the
load-elongation curve at UTS is
zero ( meaning that dp = 0).
 It is here that the instability begins;
that is, the specimen begins to neck
and can not support the load
because the cross-sectional area of
the necked region is becoming
smaller as the test progresses
Proof that ϵ = n ?????????????????
proof that the value of true strain
at the onset of necking is equal to
strain hardening exponent. Using
the relationships
True Stress and True Strain
σT = σE(1+ ϵE )
ϵT = ln (1+ ϵE)
Not that ϵE = e
65
Yield point determination
 Yielding point is determined as the
initial departure from linearity of
stress – strain curve.
 In such cases the position of this
point may not be determined
precisely.
 As a consequence, a convention has
been established wherein a straight
line is constructed parallel to the
elastic portion of the stress–strain
curve at some specified strain offset,
usually 0.002. The stress
corresponding to the intersection of
this line is defined as the yield
strength σy
 The units of yield strength are MPa
or psi
66
Yield point determination
 Some steels (low carbon steel) and other
materials exhibit the tensile stress–strain
behavior as shown in Figure.
 The elastic–plastic transition is very well
defined and occurs in what is termed a
yield point phenomenon.
 At the upper yield point, plastic deformation
is initiated with an actual decrease in
stress.
 Continued deformation fluctuates slightly
about some constant stress value, termed
the lower yield point.
 For metals that display this effect, the yield
strength is taken as the average stress that
is associated with the lower yield point.
67
Yielding –Yield strength
• There are some materials (e.g., gray
cast iron, concrete, and many
polymers) for which this initial elastic
portion of the stress–strain curve is not
linear
• For this nonlinear behavior, either
tangent or secant modulus is
normally used. Tangent modulus is
taken as the slope of the stress–
strain curve at some specified
level of stress, while secant
modulus represents the slope of a
secant drawn from the origin to
some given point of the – curve.
The determination of these moduli
is illustrated in Figure
68
Temperature Effects on Stress - Strain Curve
 Various factors have an influence on the shape of the stressstrain curves.
 The first factor is temperature. Temperature usually
1. Lowers the modulus
of elasticity
2. Lowers yield stress
3. Lower ultimate
tensile strength
4. Increases ductility
and toughness
5. Temperatures also affects the strain-hardening exponent, n,
of most metals, in that n decreases with increasing
temperature
69
Temperature Effects on Stress - Strain Curve
Increasing temperature
 ductility and toughness
 yield stress and
the modulus of elasticity
n (strain hardening exponent)
 Work hardening, also known as
strain hardening or cold
working, is the strengthening
of a metal by plastic
deformation
70
Strain rate
 Depending on the particular manufacturing operation, a
piece of material may be formed at speed ranges from
low to high.
 Some machines, such as hydraulic presses, form
materials at low speeds; others, such as mechanical
presses, form at high speeds.
 To simulate such differences, the test specimen can be
strained at a rate corresponding to that which the metal
will experience in the actual manufacturing process.
 Deformation rate (v) is typically defined as the speed at
which a tension test is being carried out such as m/s
(ex: student and balls)
 The strain rate ( such as 102 s-1, 104 s-1) is a function
of
71
the specimen length
Strain rate
 For example, let’s take two rubber bands, one of 20
mm and the other of 100 mm in length, respectively,
and elongate them both by 10 mm within a period of
1 second.
 The engineering strain in the shorter specimen is
10/20 = 0.5 and in the longer is 10/100 = 0.1. thus the
strain rates are 0.5 s -1 and 0.1 s -1, respectively,
with the short band being subjected to strain rate five
times higher than that for long band ( although they
are both being stretched at the same deformation
rate)
 There is a typical deformation rates and strain rates
in various metal working process
72
Typical Ranges of Strain and
Deformation Rate in Manufacturing
Processes
TABLE 2.4
Process
Cold working
Forging, rolling
Wire and tube drawing
Explosive forming
Hot working and warm working
Forging, rolling
Extrusion
Machining
Sheet-metal forming
Superplastic forming
True strain
Deformation rate
(m/s)
0.1–0.5
0.05–0.5
0.05–0.2
0.1–100
0.1–100
10–100
0.1–0.5
2–5
1–10
0.1–0.5
0.2–3
0.1–30
0.1–1
0.1–100
0.05–2
-4
-2
10 -10
73
Strain rate
 In order to simulate the actual metalworking process, the specimen in a
tension test can be strained at different rates
 Where v is the rate of deformation (Deformation speed),
for example, the speed of the jaws of testing machine in
which the specimen is clamped
The effect of temperature and strain rate on the ultimate tensile
strength of aluminum.
 Typical effects of temperature and
strain rate on the ultimate tensile
strength of metals are shown the
figure.
 It clearly indicates that increasing
strain rate increases ultimate tensile
strength [ strain-rate hardening]
 And that the sensitivity of the
strength to strain rate increases with
temperature. [Note that as
temperature increases, the slope
increases. Thus, tensile strength
becomes more and more sensitive to
strain rate as temperature increases].
 Note also that the sensitivity of the
strength to strain is relatively small at
room temperature.
 The slope of these curves
is called Strain-rate
sensitivity exponent, m.
The effect of temperature and strain rate on the ultimate tensile
strength of aluminum.





The relationship is give by the equation above
Where C is the strength coefficient, similar to K, m is strain-rate sensitivity exponent
For cold working m values up to (0.05)
For hot working (0.05-0.4)
For superplastic material (0.3-0.85). Refers to capability of some material to under go76large
uniform elongation prior to failure
The effect of temperature and strain rate on the ultimate tensile
strength of aluminum.
 The value of m decrease with metals with increasing strength.
 Experimental observations have shown that with higher m values,
the material stretches to a greater length before it fails, an indication
that necking is delayed with increasing m.
 Explain how ????????????????????
1. When necking is about to begin, the region’s strength with respect to
the rest of the specimen increases because of strain hardening.
2. However, the strain rate in the neck region is also higher than the
rest of the specimen because the material is elongating faster there.
3. Since the material in the necked region is becoming stronger as it is
strained at a higher rate, this region exhibits a higher resistance to
necking.
4. Thus, the increase in the resistance to necking depends on the
magnitude of m.
77
Superplasticity
 Superplasticity: the capability of some materials to
undergo large, uniform elongation prior necking and
fracturing in tension.
 The elongation may be on the order of few hundred
percent to as much as 200%.
 Common non-metallic materials exhibiting superplastic
behavior are : bubble gum, thermoplastics, and glass (at
elevated temperatures). As a result, glass and
thermoplastics can be formed successful into complex
shapes, such as beverage bottles
 Among metals exhibiting superplastic behavior are very
fine grain [10-15µm] titanium alloys and alloys of zincaluminum; when heated, they can elongate to many
78
times their original length.
Values for C and m
79
Ductility
 The strain in the specimen at
fracture is a measure of ductility,
that is, how large a strain the material
withstands before fracture
 Ductility is the extent of plastic
deformation that the material
undergoes before fracture.
 The strain up to the UTS is called
uniform strain.
 The elongation at fracture is known as
the total elongation.
 The total elongation is measured
between the original gage marks after
the two pieces of the broken specimen
are placed together.
80
Ductility
• Plastic tensile strain at failure:
Lf - Lo
x 100
%EL =
Lo
smaller %EL
Engineering
tensile
stress, s
larger %EL
Lo
Ao
Af
Lf
Engineering tensile strain, e
• Another ductility measure:
Ao - Af
%RA =
x 100
Ao
Chapter 6 - 81
Ductility
Two quantities that are commonly used to define
ductility
1. Elongation
2. Reduction of area
Elongation =
(l f - l0 )
l0
Reduction of Area =
100
( A0 - A f )
A0
Elongation ranges
approximately (10%-60%)
100
Reduction of Area
ranges (20%-90%)
82
Elongation versus % Area
Reduction
Figure 2.4
Approximate
relationship
between elongation
and tensile
reduction of area
for various groups
of metals.
83
TOUGHNESS
Toughness is the area under s - e curve up to fracture.
ef
Toughness =  sde
0
Where ϵf is the true strain at fracture.
Note that toughness is the energy per unit
volume that has been dissipated up to the
point of fracture
Toughness is the material’s ability
to absorb energy before fracture
- Similar to Resilience (same units J/m3).
- Larger area  tougher material.
ductility such as
strength And …………
So tough materials have a combination of ………...
ruber
Can be measured by an impact test.
Toughness
• Energy to break a unit volume of material
• Approximate by the area under the stress-strain curve.
Engineering
tensile
stress, s
small toughness (ceramics)
large toughness (metals)
very small toughness
(unreinforced polymers)
Engineering tensile strain,
e
Brittle fracture: elastic energy
Ductile fracture: elastic + plastic energy
Chapter 6 - 85
Example 1
• A cable is made of material of (K =
60,000 psi, n = 0.5) Calculate the true
and engineering UTS for this cable prior
to necking.
• σ= Kεn
86
Example 2
• Assume that a material has true stress
– true strain curve given by:
σ = 100,000 ε0.5 psi. Calculate the true
ultimate tensile strength and
engineering UTS of this material.
87
Example 3
• A cable is made of two strands of different
materials, A and B, and cross sections as
follows:
For material A: K=70,000 psi, n=0.5, A0=0.6 in2.
For material B: K=25,000 psi, n=0.5, A0=0.3 in2
Calculate the maximum tensile force that this
cable can withstand.
88
Example 4
89
Compression Test
 Many operation in metalworking, such as
forging, rolling, and extrusion, are
performed with the workpiece under
externally applied compressive forces.
 The compression test, in which the
specimen is subjected to a compressive
load, can give useful information for these
process, such as
1. stresses required and
2. the behavior of the material under
compression.
 The deformation shown in figure is ideal.
 The test is usually carried out by
compressing (upsetting) a solid cylindrical
specimen between two flat dies
90
Compression Test
 The friction between the
specimen and the die is an
important factor, in that, it causes
barreling because friction
prevents the top and bottom
surfaces from expanding freely.
 This phenomena makes it difficult
to obtain relevant data and
construct a compressive stressstrain curve because:
1. The cross sectional area of the
specimen changes along its
With effective lubrication, it is
height (barreling effect)
2. Friction dissipated energy and this possible to minimize friction, and
hence barreling to obtain
energy is supplied through an
increased in compressive force. reasonable constant cross sectional
91
area.
Compression Test
by convention, stress and strain are negative
 Engineering strain rate
Where v is the speed of the die
used for measuring
ho is the original length of the specimen strength of brittle
materials and for
calculating forces
required in
True strain rate
manufacturing
processing which
Where v is the speed of the die
involve
h is the current length of the specimen
compressive
deformation
92
Shear Test
Shear stress is t = F/Ao and g (shear strain) is
tangent of shear angle, q
G = t /g, G is shear modulus
Shear tests are often used to measure
adhesive bonding, riveted joints etc
Torsion-Test Specimen
 Another method of determining material properties is torsion test.
 This test is generally carried out on tubular specimen with a reduced
mid section
 A typical torsion-test specimen. It is mounted between the two heads of a machine
and is twisted. Note the shear deformation of an element in the reduced section.
T
Shear stress, t =
2
2r t
Shear strain, g =
r
l
T: Torque applied, t: thickness of the reduced section, r: is mean
radius, Ф: twist angle, l: the length of reduced section
Shear modulus or modulus of rigidity (G)=(shear stress/shear strain) in
94
the elastic range.
Testing of Brittle Materials
A perfect elastic material
displays linear behavior with
slope E
This is the behavior of brittle
material, such as most
ceramics, common glass,
and some cast iron.
These brittle materials may
be represented by such
curve.
Perfectly elastic
No plastic flow
95
Testing of Brittle Materials
 Recall: Hard brittle materials (e.g., ceramics) possess elasticity but little
or no plasticity.
 Ceramics are not normally tested in tension because:
1. It is difficult to machine brittle material to the required geometry
(Shaping and machining them to proper dimensions can be challenging)
2. it is difficult to grip brittle materials without inducing fracture
(Clamping brittle test specimen for testing can be difficult)
3. Brittle materials are sensitive to surface defects and scratches. Why
4. ceramics typically fail after only ~ 0.1% strain
5. Improper alignment of the test specimen may result in nonuniform
stress distribution along cross section of the specimen
For these reasons, the mechanical properties are determined using a
different approach, the Three point bend test : or Four point bend test

Specimen geometry is either circular or
rectangular cross section
 During the test, the top surface is under
compression while the bottom surface is
under tension
Brittle materials are sensitive to
surface defects and scratches
Why brittle materials are so weak in tension
compared to their strength in compression.
This because of the presence of defects in
brittle material (brittle materials are so
sensitive to defects) explain why????
Under tension, the tip of a crack is subjected
to high tensile stresses which propagate the
crack rapidly, because brittle materials has
little capacity to absorb or dissipate energy.
Bending
 A common test method for brittle materials is the bend test
 Usually involves a specimen with
rectangular cross section and
supported at both ends.
 The load is applied vertically, either at
one or two points (three point or four
point bending test).
 The basic difference between the two
loading conditions:
 In three- point bending test, the
maximum stress is at the center of
the beam
 In four-point test, the maximum
stress is constant between the two
loading points
 The stress magnitude is the
same in both situations when
all other parameters are
maintained.
98
Hardness
One of the most common tests for assessing
the mechanical properties of materials is the
hardness test.
Hardness of material is generally defined
as its resistance to permanent indentation; it
can also be defined as its resistance to
scratching or wear
99
Hardness
• Resistance to permanently indenting the surface.
• Large hardness means:
--resistance to plastic deformation or cracking in
compression.
--better wear properties.
apply known force
measure size
of indent after
removing load
e.g.,
10 mm sphere
D
most
plastics
brasses
Al alloys
Smaller indents
mean larger
hardness.
d
easy to machine
steels
file hard
cutting
tools
nitrided
steels
diamond
increasing hardness
Chapter 6 - 100
Hardness Tests
π
101
Brinell Testing
 In the Brinell test, a steel or tungsten carbide ball
10mm in diameter is pressed against a surface with
a load of 500, 1500, or 3000kg.
 The Brinell hardness number is defined as the ratio
of the load P to the curved area of indentation
π
 D is the diameter of the ball
 d is the diameter of the impression
102
Brinell Testing
 Depending on the condition of the material tested, different
type of impressions are obtained on the surface after a
Brinell hardness test has been performed.
 Annealed materials, generally have a round profile,
whereas cold work materials have a sharp profile. The
correct method of measuring the indentation diameter d for
both cases is shown below:
(a) annealed metal
(b) work-hardened metal
103
Brinell Testing
 Because the steel ball indenter has a finite elastic
modulus, it undergoes elastic deformation under the
applied load, P; thus hardness measurements may not be
as accurate as expected.
 A common method of minimizing this effect is to use
tungsten carbide balls, because of their high modulus of
elasticity, deform less than steel ball.
104
RocwellTesting
In Rockwell test, the indenter is pressed on surface,
first with minor load and then with a major load.
The differences in the depth of penetration is a
measure of the hardness.
Rockwell hardness test uses different scales that
employ different loads.
55 HRC is read as : the hardness number is 55
using the C scale
105
The effects of grain size on yield strength
A change in grain size affects the yield strength due to
the dislocations interacting with the grain boundary as
they move.
The boundaries act as obstacles, hindering the
dislocation glide along the slip planes. As subsequent
dislocations move along the same slip plane so that the
dislocations pile-up at the grain boundaries..
The dislocations repel each other, so as the number of
dislocations in the pile-up increases the stress on the
grain boundary increases.
In fact, if there are n dislocations in the pile-up, the
stress at the grain boundary will be n times the applied
106
stress.
The effects of grain size on yield strength
If the grain boundary in a sample gives way at a
stress τ, there needs to be a stress of τ/n applied to
the sample in order to cause the boundary to
collapse.
In a larger grain there will be more dislocations within
the grain, so there will be more dislocations in the
pile-up. Therefore a lower applied stress is required to
produce a local stress great enough to cause the
grain boundary to collapse.
107
The effects of grain size on yield strength
Accurate modelling is difficult, but it is found that the
tensile yield stress, σy, is related to grain diameter, d,
by the Hall-Petch :
where σi is the 'intrinsic' yield stress, and k is a
constant for a particular material.
108
Grain Size and Material Strength
It was noted that most low-temperature, permanent
deformation of metal comes from the movement of
crystalline imperfections, known as dislocations,
through the grains in the metal..
Given enough stress and thermal energy, dislocations
will easily move throughout the crystalline grains,
resulting in permanent distortion of the grain itself.
 However, once a dislocation reaches a grain
boundary, it has nowhere to go. In other words, grain
boundaries stop dislocations.
109
Grain Size and Material Strength
Representation of a Dislocation Stopped by a Grain
Boundary (Red Line)..
110
Grain Size and Material Strength
Thus, an easy way to improve the strength of a
material is to make the grains as small as possible,
increasing the amount of grain boundary.
Smaller grains have greater ratios of surface area to
volume, which means a greater ratio of grain
boundary to dislocations.
The more grain boundaries that exist, the higher the
strength becomes.
111
Grain Size and Material Strength
The following example illustrates this principle. Figure
shows a crude representation of two grains. For the
sake of simplicity, the grains are illustrated as perfect
rectangular prisms. Each prism is made up of several
cubic units.
112
Grain Size and Material Strength
 For the sake of this analysis, each unit contains exactly six dislocations.
 For the larger grain, there are 2 x 3 x 4 = 24 cubic units, and the
smaller grain is one cubic unit.
 The larger grain will have 24 x 6 = 144 dislocations, and the smaller
grain has six.
 The larger grain has a total surface area of 2 x (2 x 4) + 2 x (2 x 3)
+ 2 x (3 x 4) = 52 square units.
113
 The smaller grain has a surface area of 6 square units.
Grain Size and Material Strength
 For every dislocation in the large grain, there is 0.36 square units of
grain boundary.
 In the smaller grain, there is one square unit of grain boundary
for each dislocation.
 There is a much greater chance for a dislocation to be stopped at a
grain boundary in the smaller grain. Therefore, the smaller grain
114 is
stronger.
Grain Size and Material Strength
 In the larger grain, a dislocation can travel up to 4 units without being
stopped by a grain boundary,indicating the potential for extensive
plastic flow.
 In the small grain, no dislocation can travel more than 1unit of
distance. This type of strengthening is known as Hall-Petch
strengthening.
115
Isotropy versus anisotropy
Within each grain, there are preferred planes where
the atoms (in the form of dislocations) are free to
move across each other. These are known as slip
planes.
If the applied stress coincides with a slip plane, the
dislocations can move easily.
If the applied stress is perpendicular to the slip plane,
it would be extremely difficult for the dislocations to
move. Therefore, each grain is weaker in certain
directions than in others.
116
Isotropy versus anisotropy
If all the grains in the base metal are oriented the
same way, the metal would certainly show signs of
weakness in a particular direction. The same would
be true if there were just one or two grains across a
critical dimension.
 However, with many grains oriented in random
directions, the microscopic directionality of strength
would tend to be averaged out. This would provide
equal strength in all directions.
117
Single crystal
 A single crystal is a material in which the crystal lattice
of the entire sample is continuous, with no grain
boundaries . The absence of the defects associated with
grain boundaries can give unique properties, particularly
mechanical, which can also be anisotropic.
 Anisotropy is the property of being directionally
dependent, as opposed to isotropy, which implies
identical properties in all directions. It can be defined as
a difference, when measured along different axes, in a
material's physical or mechanical properties
(absorbance, refractive index, conductivity, tensile
strength, etc.) An example of anisotropy is the wood,
which is easier to split along its grain than against it.
118
Grain boundary
 A grain boundary is the interface between two grains, or
crystallites , in a polycrystalline material.
 Grain boundaries are defects in the crystal structure, and
tend to decrease the electrical and thermal conductivity of
the material.
 Most grain boundaries are preferred sites for the onset of
corrosion and for the precipitation of new phases from the
solid.
 They are also important to many of the mechanisms of creep.
 On the other hand, grain boundaries disrupt the motion
of Dislocations through a material, so reducing crystallite size
is a common way to improve strength, as described by
the Hall-Petch relationship.
119
Relationship between Hardness and Strength
120
Relationship between Hardness and Strength
 Note the smaller values
of C for cold worked
metals
 the higher value of C
for annealed materials
is explained by the fact
that, due to strain
hardening
These results are
not reasonable
 Hardness of cold worked metals=
3*Y (close to being perfectly plastic
121
in their behavior).
 Hardness of annealed metals= 5*Y.