Chapter 6:
Mechanical Properties (of Metals)
ISSUES TO ADDRESS...
• Stress and strain
• Elastic behavior
• Plastic behavior
• Toughness and ductility
Chapter 6 -
1
Chapter 6:
Mechanical Properties (of Metals)
Callister study guide.
• 6.1 – 6.9: Study all parts
• 6.10: Understand terms and concepts – conversion
between different hardness scales is not required.
• 6.11 – 6.12: Study all parts
Chapter 6 -
2
The most important properties of structural materials, including metals and alloys,
ceramics, polymers, and composites, are their mechanical properties.
Most researches (research publications) related to the development of new
structural materials include mechanical property testing.
Example:
Hardness test
Tensile test
M. N. Hasan, et al., International Journal of Plasticity 123 (2019) 178–195
Chapter 6 -
3
A Simple Question
Why do we care about mechanical properties of a material?
https://www.youtube.com/watch?v=KeTizNY0zDA
Mechanical properties reflect how materials respond to applied load.
How do we know the mechanical properties of materials?
Apply load and check how materials respond.
Chapter 6 -
4
Elastic Deformation
1. Initial
Elastic means reversible!
3. Unload
bonds
stretch
Atomic bonds are
stretched during an elastic
deformation process and
will return to their initial
state once external load is
released.
Elastic deformation is
reversible but is not
necessarily linear!
(e.g., a rubber band)
2. Small load
return to
initial
d
F
(load)
F
Linearelastic
Non-Linearelastic
d (displacement)
Chapter 6 -
5
Plastic Deformation (Metals)
Plastic means permanent!
1. Initial
Plastic deformation is not
reversible and does not
present a linear force–
displacement relationship.
Plastic deformation in
metals can occur via plane
shearing
2. Large load
bonds
stretch
& planes
shear
delastic + plastic
F
3. Unload
planes
still
sheared
dplastic
F
linear
elastic
linear
elastic
dplastic
d
Elastic deformation may continue after plastic
deformation has initiated. This will be further
discussed later.
Chapter 6 -
6
Various kinds of applied forces
Tensile
Compressive
Tension test is one of the most
common mechanical tests
Compression test is
conducted when the inservice force is compressive
or when materials are brittle in
tension.
Shear
Torsional
Torsion is a variation
of pure shear
Chapter 6 -
7
Engineering Stress
• Tensile stress, s:
It is always easier to break a
thin chopstick than a thick one
made from a same material.
We introduce the concept of
stress and strain to exclude
the specimen size effect on
mechanical properties.
• Shear stress, t:
Ft
Ft
Area, A
Area, A
Tensile stress (shear stress )
applied to a plane equals to the
external load component
perpendicular to (parallel with)
the plane divided by the area of
the plane.
F
Ft
F
s = t = lb2f or N2
m
A o in
original area
before loading
Pa or pascal
Fs
Fs
F
t= s
Ao
F
Ft
\ Stress has units:
N/m2 or lbf/in2
Chapter 6 -
8
Pay Special Attention to Units in Calculations
The unit of Pascal is too small.
We usually use MPa and GPa
MPa = 106 Pa
GPa = 109 Pa
Example: Tensile force of 1 KN is
applied to a rod sample with a cross
sectional diameter of 10 mm. What is
the engineering tensile stress applied
to the sample?
#!
#!
!=
= #
$" %&
1×10$*
& 45
=
≅
12.7×10
3.14×(5×10%$)#
= 12.7 645
Chapter 6 -
9
For a sample under tensile stress, is there any shear stress
component applied to the sample?
Plastic deformation occurs via
shearing of atomic planes. Later
we will know that shearing is
possible only along some specific
atomic planes. This makes it
important for us to calculate shear
stress on these specific atomic
planes.
Chapter 6 -
10
Common States of Stress
• Simple tension: cable
F
F
Ao = cross sectional
area (when unloaded)
F
s
s=
Ao
s
• Torsion (a form of shear): drive shaft
M
Ac
M
2R
Fs
Ao
Ski lift
(photo
courtesy P.M. Anderson)
Fs
t =
Ao
Note: t = M/AcR here.
Chapter 6 -
11
OTHER COMMON STRESS STATES (1)
• Simple compression:
Ao
Canyon Bridge, Los Alamos, NM
Balanced Rock,
Arches National Park
s=
F
Ao
Note: compressive
structure member
(s < 0 here).
Chapter 6 -
12
OTHER COMMON STRESS STATES (2)
• Bi-axial tension:
Pressurized tank
sq > 0
sz > 0
• Hydrostatic compression:
Fish in water
sh< 0
Chapter 6 -
13
Geometric Consideration of Stress State
Tensile and shear stresses applied to the yellow plane
s’ = s cos2q
t’ = s sinq cosq
s=
F
Ao
Fs
t =
Ao
!'
<'
#(
# 789:
=
=
= ! 789 #:
$)( $"⁄789:
#*
# 9=>:
=
=
= ! 9=>: 789:
$)( $"⁄789:
Chapter 6 -
14
Angles for Max. Stresses
s’ = s cos2q
t’ = s sinq cosq
The maximum normal stress at q = 0°.
The maximum shear stress at q = 45°.
Chapter 6 -
15
Engineering Strain
• Tensile strain:
• Lateral strain:
d/2
e = d
Lo
• Shear strain:
wo
-dL
eL =
wo
Lo
dL /2
q
g = Dx/y = tan q
Dx
90º - q
y
90º
When a tensile load is
applied to two chopsticks
made from the same material
with different lengths but the
same diameter, the longer
one deforms more.
To exclude this dimensional
effect, the concept of strain is
introduced.
Strain is always
dimensionless.
Chapter 6 -
16
Stress–Strain Testing
• Typical tensile test
machine
extensometer
• Typical tensile
specimen
specimen
We have introduced
stress and strain, will the
dimensions of specimens
still affect the measured
mechanical properties?
Yes, ductility will be affected!
=
(elongation at tensile failure)
gauge
length
Chapter 6 -
17
Effect of Specimen Dimension on Measured Ductility
We will learn ductility later.
ΔA
?=
A"
This example shows that fractured
areas of superplastically deformed
samples are very fine tips.
Imaging the impact of
(1) sample thickness on ∆l and
(2) A" on the value of ∆,D,!
J. Mater. Res. 32, 4541–4553 (2017)
Chapter 6 -
18
Linear Elastic Properties
• Modulus of Elasticity, E (GPa or psi):
(also known as Young's modulus)
F
s
• Hooke's Law:
s=Ee
E is the slope of the linear part of
stress–strain curves. It shows
material’s resistance to elastic
deformation (stiffness).
Is it a good idea to write Hooke’s Law as:
# = E∆A
where F is the applied load, K is a
constant, ∆A is the change in length
E
Linearelastic
e
F
simple
tension
test
For the same material, E is a constant,
but K is affected by sample dimensions
Chapter 6 -
19
Non-Linear Elastic Properties
(grey cast ion, concrete, and many polymers)
Tangent modulus:
The slope of the stress–strain curve at
a specific stress value
Secant modulus:
The slope of a secant from the origin
to a specific point of the stress–strain
curve
Chapter 6 -
20
Bonding Forces
There is a repulsive force
(caused by the overlapping of
electron clouds of two atoms)
and an attractive force
(depending on type of bonding)
between neighbouring atoms.
The net force is the sum of the
repulsive and attractive forces.
The place where the net force
is 0 is the equilibrium
interatomic spacing
Chapter 6 -
21
Mechanical Properties
• Slope of stress strain plot (which is
proportional to the elastic modulus) depends
on the bond strength of materials
Eµ
E is a measure of the resistance to change the interatomic distance
Chapter 6 -
22
Young’s Moduli: Comparison
General trend on E for
different materials
(caused by atomic
bonding):
Metals
Alloys
1200
1000
800
600
400
E(GPa)
Metals/alloys
Polymer-based
composites
polymers
Increasing E
Ceramics
200
100
80
60
40
109 Pa
Graphite
Composites
Ceramics Polymers
/fibers
Semicond
Diamond
Tungsten
Molybdenum
Steel, Ni
Tantalum
Platinum
Cu alloys
Zinc, Ti
Silver, Gold
Aluminum
Magnesium,
Tin
Si carbide
Al oxide
Si nitride
Carbon fibers only
CFRE(|| fibers)*
<111>
Si crystal
Aramid fibers only
<100>
AFRE(|| fibers)*
Glass -soda
Glass fibers only
GFRE(|| fibers)*
Concrete
GFRE*
20
10
8
6
4
2
1
0.8
0.6
0.4
0.2
CFRE*
GFRE( fibers)*
Graphite
Polyester
PET
PS
PC
CFRE( fibers) *
AFRE( fibers) *
Composite data
based on
reinforced epoxy with
60 vol%
of aligned carbon
(CFRE),
aramid (AFRE), or
glass (GFRE) fibers.
Epoxy only
PP
HDPE
PTFE
LDPE
Wood(
grain)
Chapter 6 -
23
Stress–strain sample question (1)
A copper specimen having a rectangular cross section of
15.2 mm x 19.1 mm is pulled in tension with 44,500 N force,
producing only elastic deformation. Calculate the resulting
strain. If the original specimen length is 100 mm, calculate
the elongation. E = 110 GPa.
Chapter 6 -
24
Stress–strain sample question (1)
A copper specimen having a rectangular cross section of
15.2 mm x 19.1 mm is pulled in tension with 44,500 N force,
producing only elastic deformation. Calculate the resulting
strain. If the original specimen length is 100 mm, calculate
the elongation. E = 110 GPa.
! = F?
à
? = !DF
! = #D$ -----(2)
"
-----(1)
Combining (1) and (2):
#
44500
%$
?=
=
=
1.39×10
$"F (15.2×10%$)×(19.1×10%$)×(110×10∆A
?=
A"
à
∆A = ?A" = 1.39×10%$×100 HH = 0.139 HH
Chapter 6 -
25
What We Have Learnt So Far
Tensile Stress
Pa (Pascal) = N/m2
or MPa (M = 106)
Shear Stress
Young’s Modulus
GPa (G = 109)
Tensile Strain
Shear Strain
L
Dimensionless
Lateral Strain
Hooke's Law:
s=Ee
# = E∆A
K is determined by both materials and dimensions
E is a function of materials
Chapter 6 -
26
E vs Temperature
Young’s modulus decreases
monotonically, but not linearly,
with increasing temperature.
Chapter 6 -
27
Anelasticity
• Time dependent elastic behaviour
What are the primary differences among
elastic, anelastic, and plastic deformation
behaviours?
DOI: 10.1021/nl401175t
Elastic deformation is reversible. It
occurs by interatomic bond stretching.
Anelastic deformation is a special kind of
elastic deformation. It takes time for a
material to return to its original shape.
Plastic deformation is permanent
deformation. It occurs via atomic plane
shearing.
Applications in damping systems to reduce/eliminate noise for a broad
range of industrial, electronic, and structural applications
Chapter 6 -
28
Poisson's ratio, n
• Poisson's ratio, n:
eL
eL
n=- e
e
-n
metals: n ~ 0.33
ceramics: n ~ 0.25
polymers: n ~ 0.40
Units:
E: [GPa] or [psi]
n: dimensionless
Volumetric
change
∆V
≈ (1 − 2M)?
J
When tensile (compressive)
load is applied to a sample,
the sample is stretched
(compressed) that changes
the dimension along the
loading direction. At the
same time, the lateral
dimension of the sample
also changes. The ratio of
(lateral strain)/(strain along
loading direction) is called
Poisson’s ratio.
–n > 0.50 density increases
–n < 0.50 density decreases
(voids form)
Chapter 6 -
29
Stress–strain sample question (2)
A cylindrical copper specimen having a diameter of 10 mm is pulled in
tension. If diameter reduction is 0.0025 mm and the deformation is
entirely elastic. Determine the applied load. E = 110 GPa.
#
!=
$"
à # = !$"
----(1)
! = F? ----(2)
∆O
−?.
?. =
----(4)
N=
----(3)
L
O
eL
eL =
"
?
n%∆2
= ⁄-2!e
%/o"
w
# = !$" = F? $" = F( 0 )$" = F( 0 )$"
s=Ee
metals: n ~ 0.33
ceramics: n ~ 0.25
polymers: n ~ 0.40
F %$)×3.14×(5×10%$)#
−F∆O$" (110×10-)×(0.0025×10
s=
=
=
%$ )×0.33
O"N
(10×10A
o
= 6.53×10$*
Anything wrong here?
Number of significant figures
Chapter 6 -
30
Relationship among moduli and Poisson ratio
• Elastic Shear
modulus, G:
t=Gg
t
G
g
Same properties
along all directions
• Special relations for isotropic materials:
E
G=
2(1 + n)
E
K=
3(1 - 2n)
Poisson ratio
Chapter 6 -
31
Plastic (Permanent) Deformation
Unloading line // elastic loading
line
• Simple tension test:
Elastic+Plastic
at larger stress
engineering
stress, s
Elastic
initially
ep
plastic strain
The recovered part is larger than
the initial elastic deform.
Further elastic def. occurs
during plastic def.
permanent (plastic)
after load is removed
engineering strain, e
Elastic strain
Note: shearing occurs only
on some specific planes. The
shear planes are closely
related to crystal structure.
Chapter 6 -
32
Yield Strength, sy
• Stress at which noticeable plastic deformation has
occurred.
when ep = 0.002
tensile stress, s
sy
Yield strength
Draw a straight line
starting from 0.2% strain
point and parallel to the
elastic stress–strain (SS)
sy = yield strength line. The straight line
intercepts with the SS
curve. The stress value at
Note: for 2 inch sample the intercept is the yield
strength.
e = 0.002 = Dz/z
Proportional limit
\ Dz = 0.004 in
Behavior of some steels
engineering strain, e
ep = 0.002
0.2%
Adapted from Fig. 6.10 (a),
Callister 7e.
Question: Does plastic deformation occur
before the stress reaches the yield strength?
Chapter 6 -
33
Yield Strength : Comparison
Metals/
Alloys
2000
Graphite/
Ceramics/
Semicond
Polymers
Composites/
fibers
!3 can be easily manipulated by changing
the microstructures of materials
300
200
Al (6061) ag
Steel (1020) hr
Ti (pure) a
Ta (pure)
Cu (71500) hr
100
70
60
50
40
Al (6061) a
30
20
10
Tin (pure)
¨
dry
PC
Nylon 6,6
PET
PVC humid
PP
HDPE
LDPE
in ceramic matrix and epoxy matrix composites, since
in tension, fracture usually occurs before yield.
700
600
500
400
Ti (5Al-2.5Sn) a
W (pure)
Cu (71500) cw
Mo (pure)
Steel (4140) a
Steel (1020) cd
Hard to measure,
1000
Hard to measure ,
since in tension, fracture usually occurs before yield.
Yield strength, sy (MPa)
Steel (4140) qt
Room temperature
values
Based on data in Table B.4,
Callister & Rethwisch 8e.
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched & tempered
Al alloy with σy > 1 GPa
DOI: 10.1038/ncomms1062
Chapter 6 -
34
Tensile Strength, TS
• Maximum stress on engineering stress–strain curve.
TS
F = fracture
strength
sy
engineering
stress
A key point: “on engineering
stress–strain curve”
Typical response of a metal
strain
engineering strain
Neck – acts
as stress
concentrator
A question: how will the S–S
curve be affected if we use a
thicker specimen?
• Metals: occurs when noticeable necking starts.
• Polymers: occurs when polymer backbone chains are
aligned and about to break.
Chapter 6 -
35
Mechanical properties from stress-strain plot
•
•
•
•
Young’s modulus
Yield strength at a strain offset of 0.002 (0.2%)
Tensile strength, maximum load
Change in length (strain) under a specific stress
Chapter 6 -
36
Tensile Strength : Comparison
Metals/
Alloys
Tensile strength, TS (MPa)
5000
3000
2000
1000
300
200
100
40
30
20
Graphite/
Ceramics/
Semicond
Polymers
Composites/
fibers
C fibers
Aramid fib
E-glass fib
Steel (4140) qt
Diamond
W (pure)
Ti (5Al-2.5Sn)aa
Steel (4140)cw
Si nitride
Cu (71500)
Cu (71500) hr
Al oxide
Steel (1020)
ag
Al (6061) a
Ti (pure)
Ta (pure)
Al (6061) a
Si crystal
<100>
Glass-soda
Concrete
Graphite
AFRE(|| fiber)
GFRE(|| fiber)
CFRE(|| fiber)
We will learn the strengthening
mechanisms later.
Room Temp. values
Nylon 6,6
PC PET
PVC
PP
HDPE
wood(|| fiber)
GFRE( fiber)
CFRE( fiber)
AFRE( fiber)
LDPE
10
wood (
1
Let’s compare:
a vs ag, pure vs alloying, hr vs cw
fiber)
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched & tempered
AFRE, GFRE, & CFRE =
aramid, glass, & carbon fiber-reinforced epoxy
composites, with 60 vol% fibers.
Chapter 6 -
37
Ductility
• Plastic tensile strain at failure:
Engineering
tensile
stress, s
Lf - Lo
x 100
%EL =
Lo
Key points:
• Plastic
• Tensile
smaller %EL (brittle)
larger %EL
(ductile)
Lo
Ao
Af
Lf
X
Engineering tensile strain, e
• Another ductility measure:
(used less frequently)
Ao - Af
%RA =
x 100
Ao
Chapter 6 -
38
Toughness
• Energy to break a unit volume of material
• Approximate by the area under the stress-strain
curve.
small toughness (ceramics)
Engineering
tensile
large toughness (metals)
stress, s
very small toughness
(unreinforced polymers)
Engineering tensile strain, e
Brittle fracture: elastic energy
Ductile fracture: elastic + plastic energy
Chapter 6 -
39
Resilience, Ur
• Ability of a material to store energy for elastic deformation
– Modulus of resilience – the strain energy per unit volume for stress up to the
point of yielding
Modulus of resilience =
Ur = ò
ey
0
sde
If we assume a linear
stress-strain curve, this
simplifies to
1
Ur @ sy e y =(sy)2/2E
2
Chapter 6 -
40
Concept check
• (a) Which will experience the greatest percent reduction in area?
(largest deformation)
• (b) Which is the strongest? (largest strengths)
• (c) Which is the stiffest? (largest E)
Chapter 6 -
41
Mechanical properties from s-e plot
•
•
•
•
•
•
•
•
Young’s modulus
Yield strength at a strain offset of 0.002 (0.2%)
Tensile strength, maximum load
Change in length under a specific stress
Ductility
Elastic strain
Toughness
Modulus of resilience
Chapter 6 -
42
True Stress & Strain
Note: The instantaneous dimensions of a sample change when the
sample is stretched
sT = s(1 + e )
• True tensile stress sT = F Ai
?
eT = ln(1 + e )
∆A5
A5
• True tensile strain ?4 = Q A = ln(A ) ?
5
5
"
Valid before necking
& assuming same
volume, i.e. A0l0 = Aili
M’ < M Why?
For tensile deformation, A5 > A".
∆6
∆6
Therefore, ∑ 6 # < 6
#
!
(how do we know this is tensile deformation?)
Chapter 6 -
43
A578 − A5
∆A5
?4 = Q
=Q
A5
A5
5
5
(by making delta(li) infinitesimal)
=
6# 26
∫6 6
!
After necking, deformation is no longer uniform.
True strain varies from one place to another. The
calculated value then becomes meanless
= ln(6#D6! )
Assuming constant volume and no necking
#
!4 =
$5
$ 5 A5 = $ " A"
$ " A"
à $5 =
A5
Valid only before necking.
What if necking has occurred?
#
# A5
A5 − A" + A"
!4 =
=
=!
= !(? + 1)
A" $ " A"
A"
$"
A5
(1)
(2)
"!
(3)
""
!!
!"
Chapter 6 -
44
Work Hardening & Elastic Strain Recovery
Work hardening is an increase in sy due to plastic deformation
Releasing load recovers the elastic strain. The
unloading curve is therefore a straight line with a
slope the same that of original linear line. When
reloading, yielding occurs at the last unloading
stress, i.e., the yield point increases from !!! to
!!" . The phenomenon is called work hardening
ab < cd, implying that elastic deformation
continues after yielding.
a b
c
d
Later we will learn that plastic deformation
occurs via dislocation motion. Work hardening
simply means dislocation motion becomes
more difficult.
Work hardening is a very important method to
strengthen materials
Chapter 6 -
45
Work Hardening
• An increase in sy due to plastic deformation.
s
large hardening
sy
1
sy
small hardening
0
Work hardening capability
affects significantly the
ductility of materials
e
• Curve fit to the stress-strain response:
( )
sT = K eT
“true” stress (F/A)
n
strain hardening exponent:
n = 0.15 (some steels)
to n = 0.5 (some coppers)
“true” strain: ln(L/Lo)
Chapter 6 -
46
Work Hardening Capability, Necking and Ductility
TS
engineering
stress
sy
Typical response of a metal
strain
engineering strain
Macroscopically uniform plastic
deformation is not “uniform” at the
microscale. Deformation occurs
F = fracture on individual atomic planes. This
strength leads to local work hardening and
stops the local deformation. The
Neck – acts subsequent deformation then
as stress
occurs elsewhere. However, if the
concentrator material loses its hardening
capability, the deformation will
continue because of higher local
stress (smaller cross-sectional
area caused by necking) and
fracture will finally occur.
You will have better understanding
of this issue later.
Chapter 6 47
Concept check
. A schematic tensile stress-strain curve of a
metallic sample is shown below.
The elastic strain at point A is:
1)
2)
3)
4)
OB
OC
CD
OD
The plastic strain at point A is:
1)
2)
3)
4)
OC
OD
BC
OB
Chapter 6 -
48
Problem 6.35
Prove that true strain can also be expressed as
eT = ln(A0 /Ai)
Original definition:
eT = ln(! i ! o )
Assuming no volume change à Aili = A0l0
Therefore li/l0 = A0/Ai
Put this to the original definition
eT = ln(A0 /Ai)
This expression is more valid during necking
because Ai is taken as the area of the neck.
Chapter 6 -
49
Concept check
Make a schematic plot showing the tensile engineering stress–strain behavior
for a typical metal alloy to the point of fracture. Now superimpose on this plot a
schematic compressive engineering stress-strain curve for the same alloy.
Explain any differences between the two curves.
Hint: pay attention to the evolution of cross-sectional area during deformation
Reasons:
• The true cross-sectional area in
compression is larger than the original
one.
• There is no necking in compression.
Therefore, the engineering stress keeps
increasing.
• Larger strain in compression because of
no necking.
Chapter 6 -
50
#
#
!" =
!=
$#
$!
Tension
F
Compression
F
$!
!!
!#
Chapter 6 -
51
Hardness
• Resistance to permanently indenting the surface.
• Large hardness means:
--resistance to plastic deformation or cracking in
compression.
--better wear properties.
e.g.,
10 mm sphere
apply known force
D
most
plastics
brasses
Al alloys
measure size
of indent after
removing load
Smaller indents
mean larger
hardness.
d
easy to machine
steels
file hard
cutting
tools
nitrided
steels
diamond
increasing hardness
Chapter 6 -
52
Hardness
Hardness tests more frequently than other mechanical properties because:
•Simple & inexpensive
•Nondestructive – only small indentation
•Can be approximately converted to other properties
(e.g., tensile strength)
•High spatial resolution (especially for nanoindentation), making the connection
between local microstructure and mechanical properties possible.
Chapter 6 -
53
Hardness: Measurement
• Rockwell
– Each scale runs to 130 but only useful in
range 20-100.
– Minor load 10 kg
– Major load 60 (A), 100 (B) & 150 (C) kg
• A = diamond, B = 1/16 in. ball, C = diamond
Hardness specification includes:
1. hardness number
2. scale symbol
ex: 80 HRB à 80 Rockwell hardness on the B scale
Chapter 6 -
54
Hardness: Measurement
Chapter 6 -
55
Comparison of different hardness scales
•Hardness is not a well-defined
property
•Comprehensive conversion
scheme not available
•Conversion data depend on
materials
•Most reliable data exist for
steels
Artificial materials harder than natural diamond:
• Nanotwinned diamond. DOI: 10.1038/nature13381
• Amorphous carbon. DOI: 10.1093/nsr/nwab140
Fig 6.18
Chapter 6 -
56
Hardness-Tensile Strength relationship
HB = Brinell Hardness
For most steels:
TS (psia) = 500 x HB
TS (MPa) = 3.45 x HB
Chapter 6 -
57
Variability in Material Properties
• Elastic modulus is material property
• Critical properties depend largely on sample flaws
(defects, etc.). Large sample to sample variability.
• Statistics
n
– Mean
– Standard Deviation
S xn
x=
n
2ù
én
S(x i - x ) ú
ê
s=
ê n -1 ú
ë
û
1
2
where n is the number of data points
Chapter 6 -
58
Scatter
•Test method
•Inhomogeneity
•Slight structural difference
•Operator bias
•…
in experimental data
Typical strength-ductility trade-off
Curves with the same color
come from specimens made
from the same material
Chapter 6 -
59
Design/Safety Factors
Uncertainties:
• Uncertain stress level for in-service applications
• Variability in materials properties
• Imperfections introduced during manufacture
• Sustained damage during service
Need to introduce a design safety factor so that the applied stress is reduced
Design Stress Y;
!2 = *′!9
*′ > 1 is a design factor, !9 is calculated stress level
Safe stress (work stress) Y<
!3
!: =
*
N is a factor of safety, !3 is the yield strength
Chapter 6 -
60
Summary
• Stress and strain: These are size-independent
measures of load and displacement, respectively.
• Elastic behavior: This reversible behavior often
shows a linear relation between stress and strain.
To minimize deformation, select a material with a
large elastic modulus (E or G).
• Plastic behavior: This permanent deformation
behavior occurs when the tensile (or compressive)
uniaxial stress reaches sy.
• Toughness: The energy needed to break a unit
volume of material.
• Ductility: The plastic strain at failure.
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61
Concepts And Equations To Be Fully Understood
Engineer stress (6.1), engineering strain (6.2), shear stress
(6.3), Hooke’s law (6.5), Young’s modulus, elastic and
plastic deformation, anelasticity, yield strength, tensile
strength, ductility (6.11 & 6.12), resilience, modulus of
resilience (6.13 & 6.14), toughness, true stress (6.15), true
strain (6.16), hardness, variability of materials properties
(6.21 & 6.22), and design safety factors (6.23 & 6.24).
Chapter 6 -
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