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Chapter 3: Mechanical Properties of Materials
Why mechanical properties?Why mechanical properties?Why mechanical properties?Why mechanical properties?
Need to design materials that will withstand
applied load and in-service uses for
Bridges for autos and people MEMS devices
1
Space elevator?
skyscrapers
Space exploration
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Chapter 7: Mechanical Properties
ISSUES TO ADDRESS...
Stress and strain: Normalized force and displacements.
Elastic behavior: When loads are small.
Engineering: e= F
i/A
0e= l / l
0
True: T= F
i/A
iT= ln(l
f/ l
0)
'
Chapter 3: Mechanical Properties of Materials
2
Plastic behavior: dislocations andpermanent deformation
Toughness, ductility, resilience, toughness, and hardness:
Define and how do we measure?
Mechanical behavior of the various classes of materials.
Yield Strength: YS
[MPa] (permanent deformation)
UlitmateTensile Strength: TS
[MPa] (fracture)
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Tensile specimen Tensile test machine
Stress & strain Testing
Often 12.8 mm x 60 mm
specimenextensometerAdapted from Fig. 7.2,
Callister & Rethwisch 3e.
4
Other types:-compression: brittle materials (e.g., concrete)
-torsion: cylindrical tubes, shafts.
gauge
length
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Tensile stress, s: Shear stress, t:
Area, A
Ft
Area, A
Ft
Fs
F
Engineering Stress
5
Ft =
FtA
ooriginal areabefore loading
FtF
Fs
=Fs
Ao
Stress has units: N/m2 (or lb/in2 )
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Tensile strain: Lateral (width) strain:
/2
/2
Lowo =
Lo L =L
wo
Engineering Strain
6
Shear strain:/2
/2
/2 -
/2
= tan Strain is always
dimensionless.
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Simple tension: cable
=F
Ao = cross sectionalArea (when unloaded)
FF
Common States of Stress
7
o Simple shear: drive shaft
o =
Fs
A
Note: = M/AcR here.
Ski lift (photo courtesy P.M. Anderson)
M
M Ao
2R
Fs
Ac
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Simple compression:
Ao
Common States of Stress
8
Canyon Bridge, Los Alamos, NM
Balanced Rock, ArchesNational Park o =
F
A
Note: compressive
structural member ( < 0).
(photo courtesy P.M. Anderson)
(photo courtesy P.M. Anderson)
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Bi-axial tension: Hydrostatic compression:
Common States of Stress
9
Fish under waterPressurized tank
z > 0
> 0
< 0h
(photo courtesy
P.M. Anderson)
(photo courtesy
P.M. Anderson)
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Pure Tension Pure Compression
e=Fnormal
Ao
e=
l lo
lo
stress
strain
e=EElastic
response
10
Pure Shear
Pure Torsional Shear
e=Fshear
Ao
= tan
stress
strain
e=GElastic
response
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Elastic Deformation
2. Small load
bonds
stretch
1. Initial 3. Unload
return to
initial
11
Elastic means reversible!
F
Linear-
elastic
Non-Linear-
elastic
F
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Plastic Deformation of Metals
1. Initial 2. Small load 3. Unload
planes
still
sheared
bonds
stretch
& planesshear
12
Plastic means permanent!
F
elastic + plastic plastic
elastic
F
linear
elasticlinear
elastic
plastic
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Modulus of Elasticity, E:(also known as Young's modulus)
Hooke's Law: = E
Units: E [GPa] or [psi]
Linear Elasticity
FAxial strain
13
Linear-elastic
E
FWidth strain
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Tangent Modulus is experienced in service.
Secant Modulus is effective modulus at 2% strain.- grey cast iron is also an example
Modulus of polymer changes with time and strain-rate.- must report strain-rate d/dt for polymers.
- must report fracture strain fbefore fracture.
Polymers: Tangent and Secant Modulus
14
%strain
Stress (MPa)initial E
secant E
1 2 3 4 5 ..
tangent E
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Hooke's Law: = E (linear elastic behavior)
Copper sample (305 mm long) is pulled in tension with stress of276
MPa. If deformation is elastic, what is elongation?
Example: Hookes Law
For Cu (polycrystalline), E = 110 GPa. FAxial strain
15
=E=E ll
0
l =
l 0E
l =(276MPa)(305mm)
110x10
3
MPa
= 0.77mm
Hookes law involves axial (parallel to applied tensile load) elastic deformation.
FWidth strain
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Mechanical Properties
Recall: Bonding Energy vs distance plots
16
Adapted from Fig. 2.8
Callister & Rethwisch 3e.
tension
compression
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Elastic Behavior
Elasticity of Ceramics
Al O
And Effects of PorosityE= E0(1 - 1.9P + 0.9 P
2)
17
Neither Glass or Alumina experience plastic deformation before fracture!
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Comparison of Elastic Moduli
18
Silicon (single xtal) 120-190 (depends on crystallographic direction)
Glass (pyrex) 70
SiC (fused or sintered) 207-483
Graphite (molded) ~12
High modulus C-fiber 400Carbon Nanotubes ~1000
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Magnesium,
Aluminum
Platinum
Silver, Gold
Tantalum
Zinc, Ti
Steel, Ni
Molybdenum
Si crystal
Glass-soda
Si nitrideAl oxide
Glass fibers only
Carbon fibers only
Aramid fibers only
6080
100
200
600800
10001200
400
Cu alloys
Tungsten
Si carbide
Diamond
*
AFRE(|| fibers)*
CFRE(|| fibers)*
Metals
Alloys
GraphiteCeramics
Semicond
PolymersComposites
/fibers
E(GPa)
Eceramics
> Emetals>> Epolymers
Youngs Modulus, E
19
0.2
8
0.6
1
Graphite
Concrete
PC
Wood( grain)
AFRE( fibers)*
CFRE*
GFRE*
Epoxy only
0.4
0.8
2
4
6
10
20
40 n
PTFE
HDPE
LDPE
PP
Polyester
PSPET
CFRE( fibers)*
GFRE( fibers)*
109 Pa
Based on data in Table B2, Callister 6e.
Composite data based on
reinforced epoxy with 60 vol%
of aligned carbon (CFRE),aramid (AFRE), or glass (GFRE) fibers.
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Poisson's ratio, :
Poisson's ratio,
=
width strain
axial strain =
w /w
l /l =
L
Units: dimensionless
F
L
20
Why does have minus sign?
metals: ~ 0.33
ceramics: ~ 0.25polymers: ~ 0.40
F
Axial strain
Width strain
-
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Limits of the Poisson Ratio
= w/w
l / l= 1
Poisson Ratio has a range 1 1/2
Look at extremes
No change in aspect ratio: w/w = l/l
21
Volume (V = AL) remains constant: V =0.
Hence, V = (L A+A L) = 0. So,
In terms of width, A = w2, then A/A = 2 w w/w2 = 2w/w = L/L.
Hence,
A/A = L/L
= w/w
l / l=
(1
2l / l )
l / l=
1
2
Incompressible solid.
Water (almost).
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Tensile stress is applied along cylindrical brass rod (10 mm diameter).
Poisson ratio is = 0.34 and E = 97 GPa.
Determine load needed for 2.5x103 mm change in diameter if the
deformation is entirely elastic?
F
Example: Poisson Effect
Width strain: (note reduction in diameter)
23
simpletensiontest
x= = . x mm mm = . x
Axial strain: Given Poisson ratio
z= x/ = (2.5x104)/0.34 = +7.35x104
Axial Stress: z = Ez = (97x103 MPa)(7.35x104) = 71.3 MPa.
Required Load: F = zA0 = (71.3 MPa) (5 mm)2 = 5600 N.
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Elastic Shear
modulus, G:
1
G
= G
M
M
simple
Torsion test
Other Elastic Properties
24
modulus, K:
P= -KVVo
V
1-K
Vo
P P
Pressure test:
Init. vol = Vo.
Vol chg. = V
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Simple tension test:
engineering stress,
Elastic+Plastic
at larger stress
Elastic
(at lower temperatures, i.e. T< Tmelt/3)
Plastic (Permanent) Deformation
25
engineering strain, p
plastic strain
initially
Adapted from Fig. 7.10 (a),
Callister & Rethwisch 3e.
permanent (plastic)
after load is removed
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Stress where noticeable plastic deformation occurs.
When p = 0.002
Yield Stress, Y
For metals agreed upon 0.2%
P is theproportional limitwhere deviation
from linear behavior occurs.
-
tensile stress,
Y
26
Note: for 2 in. sample
= 0.002 = z/z
z = 0.004 in
Start at 0.2% strain (for most metals). Draw line parallel to elastic curve (slope of E).
Y is value of stress where dotted line crosses
stress-strain curve (dashed line).
Eng. strain, p = 0.002
Elastic
recovery
P
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Yield-point phenomenon occurs when elastic plastic
transition is abrupt.
Yield Points and YS
No offset method required.
In steels, this effect is seen when
dislocations start to move and unbind
27
For steels, take the avg. stress
of lower yield point since less
sensitive to testing methods.
.
Lower yield point taken as Y.
Jagged curve at lower yield point occurs
when solute binds dislocation and
dislocation unbinding again, until work-hardening begins to occur.
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3 different types of behavior
Stress-Strain in Polymers
Brittle
plastic
For plastic polymers: YS at maximum stress just
after elastic region.
TS is stress at fracture!
28
Highly elastic
Highly elastic polymers:
Elongate to as much as 1000% (e.g. silly putty).
7 MPa < E < 4 GPa 3 order of magnitude!
TS(max) ~ 100 MPa some metal alloys up to 4 GPa
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Graphite/
Ceramics/Semicond
Metals/
Alloys
Composites/
fibersPolymers
,y(MP
a)
ure,
occursbefore
yield.
300
400
500600700
1000
2000
Al (6061)ag
hr
Steel (1020)cdSteel (4140)a
Steel (4140)qt
Ti (5Al-2.5Sn)aW (pure)
Mo (pure)Cu (71500)cw
ure,
trixcomposite
s,since
ccursbeforey
ield.
y(ceramics)
>>y(metals)
>> y(polymers)
Compare Yield Stress, YS
29
Yield
strength
PVC
Hardtome
as
sinceinte
nsion,fractureusuall
Nylon 6,6
LDPE
70
20
40
6050
100
10
30
Tin (pure)
Al (6061)a
Cu (71500)hrTa (pure)
Ti (pure)a
Hardtom
ea
inceramicmatrixandepoxym
intension,fractureusua
lly
HDPEPP
humid
dry
PC
PET
oom va ues
Based on data in Table B4,
Callister 6e.
a = annealed
hr = hot rolled
ag = aged
cd = cold drawncw = cold worked
qt = quenched & tempered
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Maximum possible engineering stress in tension.
(Ultimate) Tensile Strength, TS
y F= fracture orultimate
strength
eering
TS
ress
30
Metals: occurs when necking starts.
Ceramics: occurs when crack propagation starts.
Polymers: occurs when polymer backbones are
aligned and about to break.
strain
Typical response of a metalNeck acts
as stress
concentratorengi
s
engineering strain
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Metals: Tensile Strength, vTS
For Metals: max. stress in tension when necking starts, which
is the metals work-hardening tendencies vis--vis those that
initiate instabilities.
dF= 0 Maximum eng. Stress (at necking)
dF= 0 = dA +Ad
dT =
dAi
31
decreased force due to
decrease in gage diameter
Increased force due to
increase in applied stress
At the point where these two competing changes in forceequal, there is permanent neck.
Determined by slope of true stress - true strain curve
T iFractionalIncrease in
Flow stress
fractionaldecrease
in load-
bearing
area
dT
T
= dA
i
Ai
=dl
i
li
dT
d
T
dT
=T
IfT=K(
T)n, then n=
T
n = strain-hardening coefficient
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Room T values
TS(ceram)
~TS(met)
~ TS(comp)
>> TS(poly)
Compare Tensile Strength, TSGraphite/
Ceramics/Semicond
Metals/
Alloys
Composites/
fibersPolymers
TS
(M
Pa)
200
300
1000
Al (6061) ag
Cu (71500) hr
Ta (pure)Ti (pure) a
Steel (1020)
Steel (4140) a
Steel (4140) qt
Ti (5Al-2.5Sn) a
W (pure)
Cu (71500) cw
2000
3000
5000
Al oxide
Diamond
Si nitride
GFRE (|| fiber)
C FRE (|| fiber)
A FRE (|| fiber)
E-glass fib
C fibersAramid fib
32
Based on data in Table B4,
Callister & Rethwisch 3e.
Si crystal
Tensile
strength
,
PVC
Nylon 6,6
10
100Al (6061) a
LDPE
PP
PC PET
20
3040
Graphite
Concrete
Glass-soda
HDPE
wood ( fiber)
wood(|| fiber)
1
GFRE ( fiber)C FRE ( fiber)A FRE( fiber)
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Example for Metals: Determine E, YS, and TS
Stress-Strain for Brass Youngs Modulus, E (bond stretch)
GPaMPa
E 8.9300016.0
)0150(
12
12 =
=
=
0ffset Yield-Stress, YS (plastic deformation)
YS= 250 MPa
33
Max. Load from Tensile Strength TS
Fmax =TSA0 =TSd0
2
2
= 450MPa12.8x103m
2
2
= 57,900N
Change in length atPoint A, l = l0
Gage is 250 mm (10 in) in length and 12.8 mm
(0.505 in) in diameter.
Subject to tensile stress of 345 MPa (50 ksi)
l = l0 = (0.06)250 mm = 15 mm
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Most metals are ductile at RT and above, but can become brittle at low T
bcc Fe
Temperature matters (see Failure)
34
cup-and-cone fracture in Al brittle fracture in mild steel
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Energy to break a unit volume of material,
or absorb energy to fracture.
Approximate as area under the stress-strain curve.
Toughness
E ngineering
tensile
small toughness (ceramics)
large toughness (metals)
36
very small toughness
(unreinforced polymers)
Engineering tensile strain,
,
UT= do
f
Brittle fracture: elastic energy
Ductile fracture: elastic + plastic energy
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Stress-Strain in Polymers
brittle polymer
plastic
elastomer
37
elastic moduli
less than for metals Adapted from Fig. 7.22,Callister & Rethwisch 3e.
Fracture strengths of polymers ~ 10% of those for metals.
Deformation strains for polymers > 1000%.
for most metals, deformation strains < 10%.
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Decreasing T...
-- increases E
-- increases TS
-- decreases %EL
Increasing
Influence ofTand Strain Rate on Thermoplastics
40
60
80 4C
20C
(MPa)
Plots for
semicrystalline
PMMA (Plexiglas)
38
strain rate...-- same effectsas decreasing T.
Adapted from Fig. 7.24, Callister & Rethwisch 3e. (Fig. 7.24 is from T.S. Carswell
and J.K. Nason, 'Effect of Environmental Conditions on the Mechanical Properties
of Organic Plastics", Symposium on Plastics, American Society for Testing and
Materials, Philadelphia, PA, 1944.)
20
00 0.1 0.2 0.3
60Cto 1.3
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Necking appears along entire
sample after YS!
Stress-Strain in Polymers
Mechanism unlike metals, necking due
to alignment of crystallites.
Load vertical
39
Align crystalline sections by
straightening chains in the
amorphous sections
After YS, necking
proceeds by
unraveling; hence,
neck propagates,
unlike in metals!See Chpt 8
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Time-dependent deformation in Polymers
Stress relaxation test:
-strain in tension to and hold.
- observe decrease instress with time.
tensile test
Large decrease in Erfor T> Tg.
(amorphous
polystyrene)
Fig. 7.28, Callister &
Rethwisch 3e.
(Fig. 7.28 from A.V.Tobolsky, Properties
and Structures of
Polymers, Wiley and
Sons, Inc., 1960.)
103
101
10-1
105
rigid solid
(small relax)
transitionregion
Er(10 s)
in MPa
viscous liquid
40
Representative Tg values (in C):
PE (low density)
PE (high density)PVC
PS
PC
-110
- 90+ 87
+100
+150Selected values from Table 11.3, Callister & Rethwisch 3e.
o
r
ttE
=
)()(
Relaxation modulus:
time
straino
(t)
10-
60 100 140 180 T(C)Tg
(large relax)
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Wh T S i ?
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Why use True Strain?
Up to YS, there is volume change due to Poisson Effect!
In a metal, from YS and TS, there is plastic deformation, as dislocations move
atoms by slip, but V=0 (volume is constant).
A0l 0 = Ail i
t= ln
li
l0
lnli l
0+ l
0
l0
= ln(1+)
42
Sum of incremental strain does
NOT equal total strain!
Test length Eng. Eng.
0-1-2-3 0-3
0 2.00
1 2.20 0.1
2 2.42 0.1
3 2.662 0.1 0.662/2.0
TOTAL 0.3 0.331
t= ln
2.2
2.0+ ln
2.42
2.20+ ln
2.662
2.42= ln
2.662
2.00
Eng.
Strain
True
StrainSum of incremental strain does
equal total strain.
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Example:
A cylindrical specimen of steel having an original diameter of 12.8
mm (0.505 in.) is tensile tested to fracture and found to have an
engineering fracture strength of 460 MPa (67,000 psi). If its cross-
sectional diameter at fracture is 10.7 mm (0.422 in.), determine:
(a) The ductility in terms of percent reduction in area(b) The true stress at fracture
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Hardness
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Resistance to permanently indenting the surface.
Large hardness means:--resistance to plastic deformation or cracking in compression.
--better wear properties.
e.g.,apply known force(1 to 1000g)
measure sizeof indent after
Hardness
45
mm sp ere
removing load
dDSmaller indentsmean largerhardness.
increasing hardness
mostplastics
brassesAl alloys
easy to machinesteels file hard
cuttingtools
nitridedsteels diamond
Adapted from Fig. 7.18.
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Hardness: Measurement
Rockwell
No major sample damage
Each scale runs to 130 (useful in range 20-100). Minor load 10 kg
Major load 60 (A), 100 (B) & 150 (C) kg
46
A = diamond, B = 1/16 in. ball, C = diamond
HB = Brinell Hardness
TS (psia) = 500 x HB
TS (MPa) = 3.45 x HB
H d M t
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Hardness: Measurement
47
F ti
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Under fluctuating / cyclic stresses, failure can occur atconsiderably lower loads than tensile or yield strengths of
material under a static load.Causes 90% of all failures of metallic structures (bridges,aircraft, machine components, etc.)
Fatigue
Failure under fluctuating stress
48
Fatigue failure is brittle-like (relatively little plastic deformation)
- even in normally ductile materials. Thus sudden and
catastrophic!
Fatigue failure has three stages:
crack initiation in areas of stress concentration
(near stress raisers)
incremental crack propagation
catastrophic failure
F i C li S
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Fatigue: Cyclic StressesCyclic stresses characterized by maximum, minimum and mean stress, the range of stress, the
stress amplitude, and the stress ratio
Mean stress m = (max + min) / 2
Range of stress r = (max - min)
Stress amplitude a = r/2 = (max - min) / 2
Stress ratio R = min / max
49
Convention: tensile stresses positive
compressive stresses negative
Creep
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pTime-dependent and permanent
deformation of materials subjected to a constant load at hightemperature (> 0.4 Tm).
Examples: turbine blades, steam generators.
Creep testing:
50 Creep testing
Furnace
Stages of creep
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S ages o c eep
51
1. Instantaneous deformation, mainly elastic.
2. Primary/transient creep. Slope of strain vs. time decreases with time:
work-hardening
3. Secondary/steady-state creep. Rate of straining constant: work-hardening
and recovery.
4. Tertiary. Rapidly accelerating strain rate up to failure: formation of internal
cracks, voids, grain boundary separation, necking, etc.
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Mechanisms of Creep
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Mechanisms of Creep
Different mechanisms act in different materials and underdifferent loading and temperature conditions:
Stress-assisted vacancy diffusion
Grain boundary diffusion
Grain boundary sliding
Dislocation motion
53
Different mechanisms different n, Qc.
Grain boundary diffusion Dislocation glide and climb
Typical Forming Operations
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Typical Forming Operations
rollingforging
54
Wire drawing extrusion Deep drawing
die
Stretch formingbending
shearing
Summary
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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).
Summary
55
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.