Chapter 7: Mechanical Properties
Chapter 7: Mechanical Properties
Why mechanical properties?
ISSUES TO ADDRESS...
• Stress and strain: Normalized force and displacements.
• Need to design materials that will withstand
applied load and in-service uses for…
Bridges for autos and people
Engineering : σ e = Fi / A0
ε e = Δl / l 0
True : σ T = Fi / Ai
MEMS devices
εT = ln(l f / l 0 )
• Elastic behavior: When loads are small.
Young ' s Modulus : E
[GPa]
• Plastic behavior: dislocations and permanent deformation
Canyon Bridge, Los Alamos, NM
skyscrapers
Yield Strength : σ YS
Ulitmate Tensile Strength : σ TS
[MPa] (permanent deformation)
[MPa] (fracture)
• Toughness, ductility, resilience, toughness, and hardness:
Define and how do we measure?
Space elevator?
Space exploration
• Mechanical behavior of the various classes of materials.
1
MatSE 280: Introduction to Engineering Materials
2
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
Pure Tension
Stress and Strain
stress
Stress: Force per unit area arising from applied load.
Tension, compression, shear, torsion or any combination.
strain
σe =
εe =
©D.D. Johnson 2004/2006-2008
Pure Compression
Fnormal
Ao
l − lo
lo
Elastic σ e = Eε
response
Stress = σ = force/area
stress τ e = Fshear
Pure Shear
Ao
Strain: physical deformation response of a
material to stress, e.g., elongation.
strain
γ = tan θ
Elastic τ e = Gγ
response
Pure Torsional Shear
3
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
4
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
1
Engineering Stress
• Tensile stress, s:
Common States of Stress
• Shear stress, t:
• Simple tension: cable
σ=
Ski lift
• Simple shear: drive shaft
F
σ= t
Ao
original area
before loading
F
Ao
Fs
τ =
Ao
Stress has units: N/m 2 (or lb/in 2 )
Note: τ = M/AcR here.
5
MatSE 280: Introduction to Engineering Materials
(photo courtesy P.M. Anderson)
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
Common States of Stress
6
©D.D. Johnson 2004/2006-2008
Common States of Stress
• Simple compression:
• Bi-axial tension:
• Hydrostatic compression:
Ao
Canyon Bridge, Los Alamos, NM
(photo courtesy P.M. Anderson)
Balanced Rock, Arches
National Park
Note: compressive
structure member (σ < 0).
(photo courtesy P.M. Anderson)
Pressurized tank
(photo courtesy
P.M. Anderson)
Fish under water
(photo courtesy
P.M. Anderson)
σθ > 0
σz > 0
σ h< 0
7
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
8
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
2
Engineering Strain
• Tensile strain:
Elastic Deformation
• Lateral (width) strain:
δ/2
• Shear strain:
π/2
3. Unload
bonds
stretch
δ/2
δL/2
return to
initial
δ
θ/2
π/2 - θ
2. Small load
Lo
wo
δL/2
1. Initial
F
Strain is always
dimensionless.
γ = tan θ
F
Linearelastic
Elastic means reversible!
δ
θ/2
Non-Linearelastic
9
MatSE 280: Introduction to Engineering Materials
10
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
Plastic Deformation of Metals
1. Initial
2. Small load
bonds
stretch
& planes
shear
Strain Testing
3. Unload
• Tensile specimen
planes
still
sheared
Often 12.8 mm x 60 mm
Adapted from Fig. 7.2,
Callister & Rethwisch 3e.
extensometer
specimen
gauge
length
F
F
linear
elastic
linear
elastic
δplastic
MatSE 280: Introduction to Engineering Materials
• Tensile test machine
δplastic
δelastic + plastic
Plastic means permanent!
©D.D. Johnson 2004/2006-2008
δelastic
©D.D. Johnson 2004/2006-2008
δ
• Other types:
-compression: brittle materials (e.g., concrete)
-torsion: cylindrical tubes, shafts.
11
12
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
3
Linear Elasticity
• Modulus of Elasticity, E:
Example: Hooke’s Law
• Hooke's Law:
Units: E [GPa] or [psi]
(also known as Young's modulus)
σ=Eε
(linear elastic behavior)
Copper sample (305 mm long) is pulled in tension with stress of
276 MPa. If deformation is elastic, what is elongation?
• Hooke's Law: σ = E ε
σ
Axial strain
ε
Axial strain
 Δl 
σl
σ = Eε = E   ⇒ Δl = 0
E
 l0 
(276MPa)(305mm)
Δl =
= 0.77mm
110x103 MPa
E
Linearelastic
For Cu, E = 110 GPa.
Width strain
Width strain
Hooke’s law involves axial (parallel to applied tensile load) elastic deformation.
13
MatSE 280: Introduction to Engineering Materials
Elastic Deformation
1. Initial
14
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
©D.D. Johnson 2004/2006-2008
Mechanical Properties
2. Small load
3. Unload
• Recall: Bonding Energy vs distance plots
bonds
stretch
return to
initial
δ
F
F
Linearelastic
Elastic means reversible!
δ
tension
Non-Linearelastic
compression
Adapted from Fig. 2.8
Callister & Rethwisch 3e.
15
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
16
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
4
Elasticity of Ceramics
Mechanical Properties
• Recall: Slope of stress strain plot (proportional to the E)
depends on bond strength of metal
And Effects of Porosity
• Elastic Behavior
E= E0(1 - 1.9P + 0.9 P 2)
E larger
E smaller
Al2O3
Adapted from Fig. 7.7,
Callister & Rethwisch 3e.
Neither Glass or Alumina experience plastic deformation before fracture!
17
MatSE 280: Introduction to Engineering Materials
18
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
©D.D. Johnson 2004/2006-2008
Polymers: Tangent and Secant Modulus
Comparison of Elastic Moduli
• 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 dε/dt for polymers.
- must report fracture strain εf before fracture.
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
400
Carbon Nanotubes
~1000
Normalize by density, 20x steel wire.
initial E
Stress (MPa)
secant E
tangent E
strength normalized by density is 56x wire.
%strain
1
2
3
4
5 …..
19
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
20
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
5
Poisson's ratio, ν
Young’s Modulus, E
Metals
Alloys
1200
1000
800
600
400
E(GPa)
200
100
80
60
40
109 Pa
Graphite
Composites
Ceramics Polymers
/fibers
Semicond
• Poisson's ratio, ν:
Units: ν dimensionless
Diamond
Tungsten
Molybdenum
Steel, Ni
Tantalum
Platinum
Cu alloys
Zinc, Ti
Silver, Gold
Aluminum
Si carbide
Al oxide
Si nitride
ν =−
Carbon fibers only
CFRE(|| fibers)*
<111>
Si crystal
width strain
Δw / w
ε
=−
=− L
axial strain
Δl / l
ε
Aramid fibers only
<100>
AFRE(|| fibers)*
Glass-soda
Glass fibers only
Magnesium,
Tin
GFRE(|| fibers)*
Axial strain
Concrete
GFRE*
20
CFRE*
GFRE( fibers)*
Graphite
10
8
6
4
εL
Polyester
PET
PS
PC
2
CFRE( fibers)*
AFRE( fibers)*
Epoxy only
PP
HDPE
1
0.8
0.6
0.4
PTFE
Wood(
ε
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.
metals: ν ~ 0.33
ceramics: ν ~ 0.25
polymers: ν ~ 0.40
Width strain
grain)
-ν
Why does ν have minus sign?
LDPE
0.2
21
MatSE 280: Introduction to Engineering Materials
22
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
Limits of the Poisson Ratio
Poisson Ratio: materials specific
• Poisson Ratio has a range –1 ≤ ν ≤ 1/2
Metals:
Look at extremes
• No change in aspect ratio:
Ir
0.26
Solid Argon:
0.25
ν =−
Δw / w
= −1
Δl / l
Covalent Solids:
• Volume (V = AL) remains constant:
Hence, ΔV = (L ΔA+A ΔL) = 0.
ΔV =0.
So,
ΔA / A = −ΔL /L
In terms of width, A = w 2, then ΔA/A = 2 w Δw/w2 = 2Δw/w = –ΔL/L.
Hence,
ν =−
Δw / w
(− Δl / l)
=− €
= 1/ 2
Δl / l
Δl / l
1
2
W
0.29
Ni
0.31
Cu
0.34
Al
0.34
Ag
0.38
Incompressible solid.
Water (almost).
Si
0.27
Ionic Solids:
MgO
Silica Glass:
0.20
Ge
0.28
Al2O3
0.23
TiC
0.19
generic value ~ 1/4
0.19
Polymers:
Network (Bakelite) 0.49
Chain (PE) 0.40 ~generic value
Elastomer:
Hard Rubber (Ebonite) 0.39
(Natural) 0.49
23
MatSE 280: Introduction to Engineering Materials
Au
0.42
generic value ~ 1/3
Δw /w = Δl /l
€
©D.D. Johnson 2004/2006-2008
©D.D. Johnson 2004/2006-2008
24
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
6
Other Elastic Properties
Example: Poisson Effect
Tensile stress is applied along cylindrical brass rod (10 mm
diameter). Poisson ratio is ν = 0.34 and E = 97 GPa.
• Determine load needed for 2.5x10 –3 mm change in diameter if
the deformation is entirely elastic?
M
τ
• Elastic Shear
modulus, G:
G
1
τ=Gγ
simple
Torsion test
γ
M
Width strain: (note reduction in diameter)
Axial strain:
mm)/(10 mm) =
P
• Elastic Bulk
modulus, K:
–2.5x10 –4
P
Given Poisson ratio
P
Pressure test:
Init. vol = Vo.
Vol chg. = ΔV
εz = –εx/ν = –(–2.5x10 –4)/0.34 = +7.35x10 –4
Axial Stress:
• Special relations for isotropic materials:
E
E
So, only 2 independent elastic
G=
K=
2(1+ ν)
3(1− 2ν) constants for isotropic media
σz = Eεz = (97x10 3 MPa)(7.35x10 –4) = 71.3 MPa.
Required Load: F = σzA 0 = (71.3 MPa)π(5 mm)2 = 5600 N.
25
MatSE 280: Introduction to Engineering Materials
δ = FL o δw = −ν Fw o
EA o
EA o
F
Ao
wo
• For linear elastic, isotropic case, use “linear superposition”.
α = 2ML o
πr o4 G
• Strain || to load by Hooke’s Law: εi=σi/E,
i=1,2,3 (maybe x,y,z).
• Strain ⊥ to load governed by Poisson effect: εwidth = –νεaxial .
M = moment
α = angle of twist
Lo
δw /2
• There are 3 principal components of stress and (small) strain.
• Simple torsion:
δ/2
©D.D. Johnson 2004/2006-2008
Complex States of Stress in 3D
Useful Linear Elastic Relationships
• Simple tension:
26
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
stress
strain
ε1
ε2
ε3
Lo
2ro
σ1
σ2
σ3
σ1/E
-νσ1/E
-νσ1/E
-νσ2/E
σ2/E
-νσ2/E
-νσ3/E
-νσ3/E
σ3/E
In x-direction, total linear strain is:
• Material, geometric, and loading parameters all contribute to deflection.
• Larger elastic moduli minimize elastic deflection.
Total Strain
in x
in y
in z
1
{σ − ν (σ 2 + σ 3 )}
E 1
1
= {(1+ ν )σ 1 − ν (σ 1 + σ 2 + σ 3 )}
E
ε1 =
or
27
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
Poisson
εx = Δd/d =
–(2.5x10 –3
εz , σz
28
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
7
Complex State of Stress and Strain in 3-D Solid
Plastic (Permanent) Deformation
• Hooke’s Law and Poisson effect gives total linear strain:
ε1 =
1
{σ − ν (σ 2 + σ 3 )}
E 1
(at lower temperatures, i.e. T < Tmelt/3)
1
{(1+ ν )σ 1 − ν (σ 1 + σ 2 + σ 3 )}
E
or
• For uniaxial tension test σ1= σ2 =0,
• Simple tension test:
so ε3= σ3/E and ε1=ε2= –νε3.
• Hydrostatic Pressure:
P = σ Hyd =
σ 1 + σ 2 + σ 3 Tr σ
=
3
3
ε1 =
Elastic+Plastic
at larger stress
engineering stress, σ
Elastic
initially
1
{(1+ ν )σ 1 − 3ν P}
E
permanent (plastic)
after load is removed
• For volume (V=l1l2l3) strain, ΔV/V = ε1+ ε 2+ ε3 = (1-2ν)σ3/E
εp
ΔV
P
= 3(1− 2ν )
V
E
Bulk Modulus, B or K:
P = –K ΔV/V so K = E/3(1-2ν)
engineering strain, ε
plastic strain
(sec. 7.5)
Adapted from Fig. 7.10 (a),
Callister & Rethwisch 3e.
29
MatSE 280: Introduction to Engineering Materials
Yield Stress, σY
©D.D. Johnson 2004/2006-2008
Yield Points and σYS
• Yield-point phenomenon occurs when elastic
plastic transition is abrupt.
• Stress where noticeable plastic deformation occurs.
when εp = 0.002
For metals agreed upon 0.2%
30
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
No offset method required.
tensile stress, σ
• P is the proportional limit where
deviation from linear behavior occurs.
• In steels, this effect is seen when
dislocations start to move and unbind
for interstitial solute.
σy
P
Elastic
recovery
Strain off-set method for Yield Stress
• Start at 0.2% strain (for most metals).
• Draw line parallel to elastic curve (slope of E).
• σY is value of stress where dotted line
crosses stress-strain curve (dashed line).
Eng. strain, ε
εp = 0.002
Note: for 2 in. sample
• Lower yield point taken as σY .
For steels, take the avg.
stress of lower yield point
since less sensitive to
testing methods.
• Jagged curve at lower yield point
occurs when solute binds dislocation
and dislocation unbinding again, until
work-hardening begins to occur.
ε = 0.002 = Δz/z
∴ Δz = 0.004 in
31
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
32
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
8
Compare Yield Stress, σYS
Stress-Strain in Polymers
• 3 different types of behavior
σy(ceramics)
>>σy(metals)
>> σy(polymers)
For plastic polymers:
• YS at maximum stress just
after elastic region.
• TS is stress at fracture!
Brittle
plastic
Room T values
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
MatSE 280: Introduction to Engineering Materials
Based on data in Table B4,
Callister 6e .
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched & tempered
33
(Ultimate) Tensile Strength, σTS
For Metals: max. stress in tension when necking starts,
which is the metals work-hardening tendencies vis-à-vis
those that initiate instabilities.
TS
F = fracture or
ultimate
strength
engineering
stress
Typical response of a metal
strain
engineering strain
©D.D. Johnson 2004/2006-2008
Metals: Tensile Strength, σTS
• Maximum possible engineering stress in tension.
σy
34
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
Neck – acts
as stress
concentrator
dF = 0
Maximum eng. Stress (at necking)
dσ T
dA
=− i
σT
Ai
dF = 0 = σ T dAi + Ai dσ T
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 force equal, there is permanent neck.
•
Metals: occurs when necking starts.
• Ceramics: occurs when crack propagation starts.
• Polymers: occurs when polymer backbones are
aligned and about to break.
Determined by slope of “true stress” - “true strain” curve
Fractional
Increase in
Flow stress
dσ T
dA dl
= − i = i ≡ dεT
σT
Ai
li
©D.D. Johnson 2004/2006-2008
dσ T
= σT
dεT
⇒
If σ T = K(εT )n , then n = εT
n = strain-hardening coefficient
35
MatSE 280: Introduction to Engineering Materials
fractional
decrease
in loadbearing
area
36
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
9
Compare Tensile Strength, σTS
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
W (pure)
Ti (5Al-2.5Sn) a
Steel (4140)cwa
Cu (71500)
Cu (71500) hr
Steel (1020)
Al (6061) ag
Ti (pure) a
Ta (pure)
Al (6061) a
A FRE(|| fiber)
GFRE(|| fiber)
CFRE(|| fiber)
Diamond
Si nitride
Al oxide
Si crystal
<100>
Glass-soda
Concrete
Graphite
Nylon 6,6
PC PET
PVC
PP
HDPE
Example for Metals: Determine E, YS, and TS
Room T values
Stress-Strain for Brass
• Young’s Modulus, E
(bond stretch)
σ − σ 1 (150 − 0)MPa
E= 2
=
= 93.8GPa
ε2 − ε1
0.0016 − 0
TS(ceram)
~TS(met)
~ TS(comp)
>> TS(poly)
• 0ffset Yield-Stress, YS
€
(plastic deformation)
YS = 250 MPa
wood(|| fiber)
• Max. Load from Tensile Strength TS
GFRE( fiber)
CFRE( fiber)
A FRE( fiber)
€
d 2
Fmax = σTS A0 = σTS π  0 
2
LDPE
Based on data in Table B4,
Callister & Rethwisch 3e.
10
wood (
fiber)
• 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)

2
12.8x 10−3 m 
= 450MPa
π = 57,900N


2


• Change in length at Point A, Δl = εl0
€
1
Δl = εl0 = (0.06)250 mm = 15 mm
37
MatSE 280: Introduction to Engineering Materials
38
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
©D.D. Johnson 2004/2006-2008
Ductility (%EL and %RA)
Temperature matters (see Failure)
Most metals are ductile at RT and above, but can become brittle at low T
• Plastic tensile strain at failure:
%EL =
Lf − Lo
bcc Fe
Lo
x100
Adapted from Fig. 7.13,
Callister & Rethwisch 3e.
• Another ductility measure:
cup-and-cone fracture in Al
%RA =
Ao − Af
Ao
• Note: %RA and %EL are often comparable.
- Reason: crystal slip does not change material volume.
- %RA > %EL possible if internal voids form in neck.
brittle fracture in mild steel
39
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
x100
40
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
10
Toughness
Resilience, Ur
• Energy to break a unit volume of material,
or absorb energy to fracture.
• Approximate as area under the stress-strain curve.
• Resilience is capacity to absorb energy when deformed elastically
and recover all energy when unloaded (=σ2YS/2E).
• Approximate as area under the elastic stress-strain curve.
small toughness (ceramics)
Engineering
tensile
stress, σ
large toughness (metals)
very small toughness
(unreinforced polymers)
ε
UT = ∫of σ dε
ε
Ur = ∫oY σ dε
ε2 σ ε
σ2
ε
= ∫oY Eε dε ~ E Y = Y Y = Y
2
2
2E
Area up to 0.2% strain
If linear elastic
Engineering tensile strain, ε
Brittle fracture: elastic energy
Ductile fracture: elastic + plastic energy
41
MatSE 280: Introduction to Engineering Materials
Elastic Strain Recovery
D
Stress
Ceramic materials are more brittle than metals.
Why?
• Consider mechanism of deformation
– In crystalline materials, by dislocation motion
– In highly ionic solids, dislocation motion is difficult
• few slip systems
• resistance to motion of ions of like
charge (e.g., anions) past one another.
2. Unload
1. Load
©D.D. Johnson 2004/2006-2008
Ceramics Mechanical Properties
• Unloading in step 2 allows elastic strain to be recovered from bonds.
• Reloading leads to higher YS, due to work-hardening already done.
σyi
σyo
42
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
3. Reapply
load
Strain
Adapted from Fig. 7.17,
Callister & Rethwisch 3e.
Elastic strain
recovery
43
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
44
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
11
Strength of Ceramics - Elastic Modulus
Strength of Ceramics - Flexural Strength
• RT behavior is usually elastic with brittle failure.
• 3-point bend test employed (tensile test not best for brittle materials).
• 3-point bend test employed for RT Flexural strength.
Al2O3
cross section
d
b
rect.
R
δ = midpoint
deflection
circ.
• Determine elastic modulus according to:
F
x
F
slope =
δ
δ
linear-elastic behavior
E=
F L3
δ 4bd 3
Rectangular cross-section
d
(rect. cross section)
b
σ fs =
3Ff L
• Typical values:
2bd 2
Material
Si nitride
250-1000 304
Si carbide
100-820 345
Al oxide
275-700 393
glass (soda-lime) 69
69
Circular cross-section
F L3
(circ. cross section)
E=
δ 12πR 4
R
σ fs =
8Ff L
πd 3
L= length between
load pts
b = width
d = height or
diameter
σ fs (MPa) E(GPa)
Data from Table 7.2, Callister & Rethwisch 3e.
45
MatSE 280: Introduction to Engineering Materials
46
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
Stress-Strain in Polymers
Influence of T and Strain Rate on Thermoplastics
brittle polymer
• Decreasing T...
-- increases E
-- increases TS
-- decreases %EL
plastic
elastomer
elastic moduli
– less than for metals
©D.D. Johnson 2004/2006-2008
Adapted from Fig. 7.22,
Callister & Rethwisch 3e.
• Increasing
strain rate...
-- same effects
as decreasing T.
σ(MPa)
80 4°C
60
Plots for
semicrystalline
PMMA (Plexiglas)
20°C
40
40°C
20
0
60°C
0
0.1
0.2
ε
to 1.3
0.3
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.)
• Fracture strengths of polymers ~ 10% of those for metals.
• Deformation strains for polymers > 1000%.
– for most metals, deformation strains < 10%.
47
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
48
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
12
Stress-Strain in Polymers
• Necking appears along
entire sample after YS!
Time-dependent deformation in Polymers
• Mechanism unlike metals, necking
due to alignment of crystallites.
Load vertical
• Stress relaxation test:
-strain in tension to εο
and hold.
- observe decrease in
stress with time.
• Large decrease in Er for T > Tg.
105
Er (10 s) 3
in MPa 10
10 -1
• Representative Tg values (°C):
time
See Chpt 8
E r (t ) =
Fig. 7.28, Callister &
Rethwisch 3e.
(Fig. 7.28 from A.V.
Tobolsky, Properties
and Structures of
Polymers, Wiley and
Sons, Inc., 1960.)
60 100 140 180 T(°C)
Tg
PE (low density)
PE (high density)
PVC
PS
PC
• Relaxation modulus:
•After YS, necking
proceeds by
unraveling; hence,
neck propagates,
unlike in metals!
viscous liquid
(amorphous
polystyrene)
10 -3 (large relax)
strain
σ(t)
•Align crystalline sections
by straightening chains in
the amorphous sections
transition
region
101
tensile test
εo
rigid solid
(small relax)
σ(t )
εo
- 110
- 90
+ 87
+100
+150
Selected values from Table 11.3, Callister & Rethwisch 3e.
49
MatSE 280: Introduction to Engineering Materials
50
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
True Stress and Strain
Engineering stress σ =
€
True strain
Why use True Strain?
F
A0
Initial area always
Relation before necking
F
Ai
instantaneous area
σ T = σ (1+ ε )
True stress σt =
l
εt = ln i
l0
Relative change
€
©D.D. Johnson 2004/2006-2008
• 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 ΔV=0 (volume is constant).
εT = ln (1+ ε )
l
l − l0 + l0
ε t = ln i  
→ ln i
= ln(1+ ε )
l0
l0
Necking: 3D state of stress!
Eng.
Strain
€
Test
length
0
1
2
3
TOTAL
2.00
2.20
2.42
2.662
Eng.
0-1-2-3
Eng.
0-3
0.1
0.1
0.1
0.3
0.662/2.0
0.331
True
2.2
2.42
2.662
2.662
Strain ε t = ln 2.0 + ln 2.20 + ln 2.42 = ln 2.00
A0 l 0 = Ai l i
Sum of incremental strain
does NOT equal total strain!
Sum of incremental strain
does equal total strain.
51
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
52
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
13
Hardening
Using Work-Hardening
Influence of “cold working” on low-carbon steel.
• An increase in σy due to plastic deformation.
σ
2nd drawn
1st drawn
large hardening
σy
1
σy
small hardening
0
Undrawn wire
ε
• Curve fit to the stress-strain response after YS:
Processing: Forging, Rolling, Extrusion, Drawing,…
• Each draw of the wire decreases ductility, increases YS.
• Use drawing to strengthen and thin “aluminum” soda can.
53
MatSE 280: Introduction to Engineering Materials
54
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
Hardness
©D.D. Johnson 2004/2006-2008
Hardness: Measurement
• Resistance to permanently indenting the surface.
• Large hardness means:
• 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
--resistance to plastic deformation or cracking in compression.
--better wear properties.
• A = diamond, B = 1/16 in. ball, C = diamond
• HB = Brinell Hardness
– TS (psia) = 500 x HB
– TS (MPa) = 3.45 x HB
Adapted from Fig. 7.18.
55
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
56
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
14
Hardness: Measurement
Account for 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
Σ xn
x=
n
– Mean
– Standard Deviation
n
 Σ xi − x
s =
 n −1

(
)
1
2 2





where n is the number of data points
57
MatSE 280: Introduction to Engineering Materials
• Design uncertainties mean we do not push the limit.
• Factor of safety, N (sometime given as S)
Often N is between 1.2 and 4
σy
σ working =
N
• Ex: Calculate diameter, d, to ensure that no yielding occurs
in the 1045 carbon steel rod. Use safety factor of 5.
220,000N


π  d2 / 4 


©D.D. Johnson 2004/2006-2008
Summary
Design Safety Factors
σ working =
58
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
σy
N
5
• 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 s y.
• Toughness: The energy needed to break a unit
volume of material.
• Ductility: The plastic strain at failure.
d = 0.067 m = 6.7 cm
59
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
60
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004/2006-2008
15