# chapter6 ```Chapter 6:
Mechanical Properties
• Stress and strain: What are they and stress-strain curve.
• Elastic behavior: When loads are small, how much
deformation occurs? What materials deform least?
• Plastic behavior: At what point does permanent
deformation occur? What materials are most
resistant to permanent deformation?
• Ductility, Toughness, Resilience, Hardness
Chapter 6 - 1
CONCEPTS OF STRESS AND STRAIN
•
If a load is static or changes relatively slowly with time and is applied
uniformly over a cross section or surface of a member, the mechanical
behavior may be ascertained by a simple stress–strain test;
•
There are four principal ways in which a load may be applied: namely,
tension, compression, shear, and torsion
Chapter 6 - 2
Chapter 6 - 3
Elastic Deformation
1. Initial
bonds
stretch
initial
d
F
Elastic means reversible
Time independent!
F
Linearelastic
d
Non-Linearelastic
Chapter 6 - 4
Plastic Deformation (Metals)
1. Initial
bonds
stretch
&amp; planes
shear
delastic + plastic
planes
still
sheared
dplastic
F
F
Plastic means permanent!
linear
elastic
linear
elastic
dplastic
d
Chapter 6 - 5
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
Ft
Fs
Fs
Fs
t=
Ao
Ft
F
 Stress has units:
N/m2 or lbf/in2
Chapter 6 - 6
Common States of Stress
• Simple tension: cable
F
F
A o = cross sectional
F
s=
s
Ao
s
• Torsion (a form of shear): drive shaft
M
Ac
M
Fs
Ski lift
Ao
MR
t =
Io
2R
Chapter 6 - 7
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
(s &lt; 0 here).
Chapter 6 - 8
OTHER COMMON STRESS STATES (2)
• Bi-axial tension:
Pressurized tank
(photo courtesy
P.M. Anderson)
• Hydrostatic compression:
Fish under water
sq &gt; 0
sz &gt; 0
(photo courtesy
P.M. Anderson)
sh&lt; 0
Chapter 6 - 9
Engineering Strain
• Tensile strain:
• Lateral strain:
d/2
e = d
Lo
wo
d = L=L-L0
• Shear strain:
eL =
Lo
dL
wo
d L = W  W0
dL /2
q
g = x/y = tan q
x
90&ordm; - q
y
90&ordm;
Strain is always
dimensionless.
Chapter 6 - 10
Stress-Strain Testing
• Typical tensile test
machine
• Typical tensile
specimen
extensometer
specimen
Strain
measured
gauge
length
We can obtained many important properties of
materials from stress-strain diagram
Chapter 6 - 11
(linear)Elastic Properties of materials
• Modulus of Elasticity, E :
(also known as Young's modulus)
• Hooke's Law:
s=Ee
s
F
E
e
Linearelastic
F
simple
tension
test
Chapter 6 - 12
Poisson's ratio, n
• Poisson's ratio, n:
eL
n= e
metals: n ~ 0.33
ceramics: n ~ 0.25
polymers: n ~ 0.40
Units:
E: [GPa] or [psi]
n: dimensionless
Chapter 6 - 13
Mechanical Properties
• Slope of stress strain plot (which is
proportional to the elastic modulus) depends
on bond strength of metal
The magnitude of E is measured of the
atoms, that is, the interatomic bonding
forces
So E α (dF/dr)
Chapter 6 - 14
T – E diagram
Plot of
modulus of
elasticity
versus
temperature
for tungsten,
steel, and
aluminum.
Chapter 6 - 15
Other Elastic Properties
• Elastic Shear
modulus, G:
stress
t
M
G
t=Gg
g
strain
• Elastic Bulk
modulus, K:
V
P = -K
Vo
M
P
P
K
V P
Vo
• Special relations for isotropic materials:
E
G=
2(1 + n)
simple
torsion
test
E
K=
3(1  2n)
P
pressure
test: Init.
vol =Vo.
Vol chg.
= V
Chapter 6 - 16
Young’s Moduli: Comparison
Metals
Alloys
1200
1000
800
600
400
E(GPa)
200
100
80
60
40
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)*
&lt;111&gt;
Si crystal
Aramid fibers only
&lt;100&gt;
AFRE(|| fibers)*
Glass -soda
Glass fibers only
GFRE(|| fibers)*
Concrete
109 Pa
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 - 17
Useful Linear Elastic Relationships
• Simple tension:
d = FL o d = n Fw o
L
EA o
EA o
F
a=
wo
2ML o
pr o4 G
M = moment
a = angle of twist
d/2
Ao
dL /2
• Simple torsion:
Lo
Lo
2ro
contribute to deflection.
• Larger elastic moduli minimize elastic deflection.
Chapter 6 - 18
Chapter 6 - 19
Chapter 6 - 20
Plastic (Permanent) Deformation
(at lower temperatures, i.e. T &lt; Tmelt/3)
• Simple tension test:
Elastic+Plastic
at larger stress
engineering stress, s
Yielding stress y
Elastic
initially
permanent (plastic)
ep
engineering strain, e
plastic strain
Chapter 6 - 21
Plastic properties of materials
•
•
•
•
•
Yield Strength, sy
Tensile Strength, TS
Ductility
Toughness
Resilience, Ur
Chapter 6 - 22
Yield Strength, sy
• Stress at which noticeable plastic deformation has
occurred.
when ep = 0.002
tensile stress, s
sy
sy = yield strength
Note: for 2 inch sample
e = 0.002 = z/z
 z = 0.004 in
engineering strain, e
ep = 0.002
Chapter 6 - 23
Yield Strength : Comparison
Metals/
Alloys
Graphite/
Ceramics/
Semicond
Polymers
Composites/
fibers
2000
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)
&uml;
dry
PC
Nylon 6,6
PET
PVC humid
PP
HDPE
LDPE
Hard to measure,
300
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
since in tension, fracture usually occurs before yield.
1000
Hard to measure ,
Yield strength, sy (MPa)
Steel (4140) qt
Room T values
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched &amp; tempered
Chapter 6 - 24
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
Chapter 6 - 25
Tensile Strength : Comparison
Metals/
Alloys
Tensile strength, TS (MPa)
5000
3000
2000
1000
300
200
100
40
30
Graphite/
Ceramics/
Semicond
Polymers
C fibers
Aramid fib
E-glass fib
Steel (4140) qt
AFRE(|| fiber)
GFRE(|| fiber)
CFRE(|| fiber)
Diamond
W (pure)
Ti (5Al-2.5Sn)aa
Steel (4140)
Si nitride
Cu (71500) cw
Cu (71500) hr
Al oxide
Steel (1020)
ag
Al (6061) a
Ti (pure)
Ta (pure)
Al (6061) a
Si crystal
&lt;100&gt;
Glass-soda
Concrete
Room Temp. values
Nylon 6,6
PC PET
PVC
PP
HDPE
20
Composites/
fibers
Graphite
wood(|| fiber)
GFRE( fiber)
CFRE( fiber)
AFRE( fiber)
LDPE
10
wood (
1
fiber)
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched &amp; tempered
AFRE, GFRE, &amp; CFRE =
aramid, glass, &amp; carbon
fiber-reinforced epoxy
composites, with 60 vol%
fibers.
Chapter 6 - 26
Ductility
Ductility can be expressed quantita
• Plastic tensile strain at failure:
Lf  Lo
x 100
%EL =
Percent elongation
Lo
smaller %EL
Engineering
tensile
stress, s
brittle
larger %EL
Lo
ductile
Ao
Af
Lf
Engineering tensile strain, e
• Another ductility measure:
Percent reduction of area
%RA =
Ao - Af
x 100
Ao
Chapter 6 - 27
Ductility
• Brittle materials are approximately
considered to be those having a fracture
strain of less than about 5%.
• Brittle materials have little or no plastic
deformation upon fracture
Chapter 6 - 28
Toughness
• Energy to break a unit volume of material
or it is a measure of the ability of a material to absorb
energy up to fracture( or impact resistance)
• 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( no apparent plastic
deformation takes place fracture)
Ductile fracture: elastic + plastic energy(extensive
plastic deformation takes place before fracture)
Chapter 6 - 29
Fracture toughness is a property indicative of a material’s
resistance to fracture when a crack is present
For a material to be tough, it must display both strength and
ductility; and often, ductile materials are tougher than brittle ones
Chapter 6 - 30
Resilience, Ur
• Ability of a material to store energy when it is
deformed elastically, and then, upon
– Energy stored best in elastic region
Ur = 
ey
0
sde
If we assume a linear
stress-strain curve this
simplifies to
1
Ur @ sy e y
2
Area under s- e curve
taken to yielding(the
Chapter 6 -
31
1
1 s y  s y
U r = s ye y = s y   =
2
2  E  2E
2
Modulus of resilience
( is the area under curve)
Thus, resilient materials are those having high yield
strengths and low module of elasticity; such alloys would be
used in spring applications.
Chapter 6 - 32
Elastic Strain Recovery
Chapter 6 - 33
Hardness
• Resistance to permanently indenting the surface, or
is a measure of a materials resistance to deformation by
surface indentation or by abrasion.
• Large hardness means:
--resistance to plastic deformation or cracking in
compression.
--better wear properties.
e.g.,
10 mm sphere
D
most
plastics
brasses
Al alloys
measure size
of indent after
apply known force
Smaller indents
mean larger
hardness.
d
easy to machine
steels
file hard
cutting
tools
nitrided
steels
diamond
increasing hardness
Chapter 6 - 34
Hardness: Measurement
• Rockwell
– No major sample damage
– Each scale runs to 130 but only useful in range
20-100.
– Major load 60 (A), 100 (B) &amp; 150 (C) kg
• A = diamond, B = 1.588 mm ball of steel, C = diamond
• the hardness can be taken directly from the machine, so
it is a quick test
• HB = Brinell Hardness
- P= 500, 1500, 3000 kg
– TS (psia) = 500 x HB
– TS (MPa) = 3.45 x HB
(The HB and tensile strength
relationship)
Chapter 6 - 35
Hardness: Measurement
Table 6.5
Chapter 6 - 36
problem
Material
Yielding
strength
( MPa)
Tensile
strength
( MPa)
Strain at
fracture
Fracture
strength
(MPa)
Elastic
modulus
(GPa)
A
310
340
0.23
265
210
B
100
120
0.4
105
150
C
415
550
0.15
500
310
D
700
850
0.14
720
210
E
Fracture before yielding
650
350
a) Which experience the greatest percent reduction in area?
Why?
b) Which is the strongest? Why?
c) Which is the stiffest? Why?
d) Which is the hardest? Why?
Chapter 6 - 37
a) Material B will experience the greatest percent area
reduction since it has the highest strain at fracture, and,
therefore is most ductile.
b) Material D is the strongest because it has the highest
yield and tensile strengths.
c) Material E is the stiffest because it has the highest elastic
modulus.
stiffness=E=σ/ε, the higher E the material more stiffest
d) Material D is the hardest because it has the highest
tensile strength.
Chapter 6 - 38
Problem:
Chapter 6 - 39
True Stress &amp; Strain
Note: Surf.Are. changes when sample stretched
• True stress
• True Strain
sT = F Ai
eT = ln i  o 
sT = s1 + e 
eT = ln1 + e 
Chapter 6 - 40
Hardening
• An increase in sy due to plastic deformation.
s
large hardening
sy
1
sy
small hardening
0
e
• Curve fit to the stress-strain response:
The region of the true
stress-strain curve from
the onset of plastic
deformation to the point
at which necking begins
may be approximated by
 
sT = K eT
“true” stress (F/Ai )
n
hardening exponent:
n = 0.15 (some steels)
to n = 0.5 (some coppers)
“true” strain: ln(L/Lo)
(n and K constants)
Chapter 6 - 41
Design or Safety Factors
• Design uncertainties mean we do not push the limit.
• Factor of safety, N
Often N is
Safe stress, or
sworking =
sy
N
between
1.2 and 4
The choice of an appropriate value of N is necessary. If N
is too large, then component overdesign will result, that is,
either too much material or a material having a higherthan-necessary strength will be used. Values normally
range between1.2 and 4.0.
Selection of N will depend on a number of factors,
including economics, previous experience, the accuracy
with which mechanical forces and material properties may
be determined, and, most important, the consequences of
Chapter 6 - 42
failure in terms of loss of life and/or property damage.
Example: Calculate a diameter, d, to ensure that yield
does not occur in the 1045 carbon steel rod below.
Use a factor of safety of 5.
sworking =
220,000N
p d2 / 4


5
sy
N
d
1045 plain
carbon steel:
sy = 310 MPa
TS = 565 MPa
Lo
F = 220,000N
d = 0.067 m = 6.7 cm
Chapter 6 - 43
problem
Chapter 6 - 44
Chapter 6 - 45
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 amount of plastic strain that has occurred
at fracture.
Chapter 6 - 46
problem
A brass alloy is known to have a yield strength of 240
MPa (35,000 psi), a tensile strength of 310 MPa (45,000
psi), and an elastic modulus of 110 GPa (16.0 10 6 psi). A
cylindrical specimen of this alloy 15.2 mm (0.60 in.) in
diameter and 380 mm (15.0 in.) long is stressed in tension
and found to elongate 1.9 mm (0.075 in.). On the basis of
the information given, is it possible to compute the
magnitude of the load that is necessary to produce this
change in length? If so, calculate the load. If not, explain
why.
Chapter 6 - 47
We are asked to ascertain whether or not it is possible to
compute, for brass, the magnitude of the load necessary to
produce an elongation of 1.9 mm (0.075 in.). It is first necessary to
compute the strain at yielding from the yield strength and the
elastic modulus, and then the strain experienced by the test
specimen. Then, if
ε(test) &lt; ε(yield)
deformation is elastic, and the load may be computed using
Equations 6.1 and 6.5. However, if
F
(s =
A
, s = Ee )
ε(test) &gt; ε(yield)
computation of the load is not possible inasmuch as deformation
is plastic and we have neither a stress-strain plot nor a
mathematical expression relating plastic stress and strain. We
compute these two strain values as
Chapter 6 - 48
Therefore, computation of the load is not possible since
ε(test) &gt; ε(yield).
Chapter 6 - 49
problem
A cylindrical metal specimen having an original
diameter of 12.8 mm(0.505 in.) and gauge length of
50.80 mm (2.000 in.) is pulled in tension until
fracture occurs. The diameter at the point of
fracture is 8.13 mm (0.320 in.), and the fractured
gauge length is 74.17 mm (2.920 in.). Calculate the
ductility in terms of percent reduction in area and
percent elongation.
Chapter 6 - 50
This problem calls for the computation of ductility in both percent
reduction in area and percent elongation. Percent reduction in
area is computed using Equation 6.12 as
in which d0 and df are, respectively, the original and fracture cross-sectional areas. Thus,
While, for percent elongation, we use Equation 6.11 as
Chapter 6 - 51
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