Meeg309 Chapter 5 Used Finally 2011
Transcript of Meeg309 Chapter 5 Used Finally 2011
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Chapter 6 - 1
ISSUES TO ADDRESS...
Stress and strain: What are they and why arethey used instead of load and deformation?
Elastic behavior: When loads are small, how muchdeformation occurs? What materials deform least?
Plastic behavior: At what point does permanentdeformation occur? What materials are most
resistant to permanent deformation? Toughness and ductility: What are they and how
do we measure them?
Chapter 5:Mechanical Properties of
Materials
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Chapter 6 -
Mechanical Properties of Metals
How do metals respond to external loads?
Stress and Strain
Tension
Compression
Shear
Torsion
Elastic deformation
Plastic Deformation
Yield Strength
Tensile Strength
Ductility
Toughness
Hardness
Chapter 6 Outline
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Chapter 6 - 3
Learning Objectives
Introduce the basic concepts associated with mechanical
properties of materials Review some of the basic testing procedures that engineers
use to evaluate many of these properties.
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Chapter 6 - 4
Often materials are subject to forces (loads) whenthey are used. It is important to know howmaterials deform (elongate, compress, twist) or
break as a function of applied load, time,temperature, and other conditions.
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Chapter 6 -5
(1) Important to understand capabilities andlimitations of materials: Failure is caused by a lack of fundamental
understanding of materials, their properties, andfailure modes.
Technological Importance:Why Study Properties and Failure of Materials ?
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Chapter 6 -
6
(2) An understanding of materials properties help us todesign better components, parts, devices, etc.Why and how do materials fail? Can we prevent failure? How do we make metals stronger?
Why do materials behave differently under dynamic loadscompared to static loads? How do we select the right material for the job?
(3) Its interesting and helps to make you a moreinformed person
Why study Properties and Failure of Materials ?
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Chapter 6 -
plastic (permanent) deformation of a
bridge
deformation led to eventual collapse
Tacoma Narrows suspension bridge,near Puget Sound, failed on at 11 amNov. 7, 1940, after only having beenopen for traffic a few months
Failure by Plastic Deformation
De Havilland Comet, first commercial jetaircraft, had five major crashes in 1952 -54 period
caused by fatigue cracks initiated at
square windows, driven by cabinpressurization and depressurization
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Chapter 6 -8
Catastrophic Failure - examples
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Chapter 6 -9
Catastrophic Failure - examples
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Chapter 6 -
10
Types of Loading
Tensile Compressive
Shear Torsion
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Chapter 6 - 11
Elastic means reversible!
Elastic Deformation
2. Small load
F
d
bondsstretch
1. Initial 3. Unload
return to
initial
F
d
Linear-
elastic
Non-Linear-elastic
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Chapter 6 - 12
Plastic means permanent!
Plastic Deformation (Metals)
F
d
linearelastic
linearelastic
dplastic
1. Initial 2. Small load 3. Unload
planesstillsheared
F
delastic + plastic
bondsstretch& planesshear
dplastic
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Chapter 6 - 13
Stress has units:N/m2 or lb
f/in2
Engineering Stress
Shear stress, t:
Area, Ao
Ft
Ft
Fs
F
F
Fs
t =Fs
Ao
Tensile stress, s:
original areabefore loading
s= FtAo
2f
2mNor
inlb=
Area, Ao
Ft
Ft
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Chapter 6 - 14
Simple tension: cable
Note: t = M/AcRhere.
Common States of Stress
os=
F
A
o
t =Fs
A
ss
M
M Ao
2R
FsAc
Torsion (a form of shear): drive shaftSki lift (photo courtesyP.M. Anderson)
Ao= cross sectional
area (when unloaded)
FF
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Chapter 6 - 15
(photo courtesy P.M. Anderson)Canyon Bridge, Los Alamos, NM
os= F
A
Simple compression:
Note: compressivestructure member(s < 0 here).(photo courtesy P.M. Anderson)
OTHER COMMON STRESS STATES (i)
Ao
Balanced Rock, ArchesNational Park
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Chapter 6 - 16
Tensile strain: Lateral strain:
Strain is alwaysdimensionless.
Engineering Strain
Shear strain:
q
90
90 - qy
x qg = x/y= tan
e =d
Lo
Adapted from Fig. 6.1(a) and (c), Callister & Rethwisch 8e.
d/2
Lowo
-deL =
L
wo
dL/2
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Chapter 6 - 17
Linear Elastic Properties
Modulus of Elasticity, E:(also known as Young's modulus)
Hooke's Law:
s = Ee s
Linear-elastic
Ee
F
Fsimpletensiontest
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Chapter 6 -
18
Nonlinear elastic behaviorIn some materials (many polymers, concrete...), elastic deformation is notlinear, but it is still reversible.
Definitions of E
s/e = tangent modulus at s2
s/e = secant modulus between originand s1
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Chapter 6 - 19
Linear Elastic Properties
Modulus of elasticity is defined as the ratio of stress to strain when
deformation is totally elastic
It is a measure of the stiffness of a material: the higher the modulus ofelasticity of a material, the stiffer the material
A plot of stress against strain for elastic deformation is linear and the
slope of this linear plot gives the modulus of elasticity For some materials such as gray cast iron, concrete and many
polymers, the initial elastic deformation is not linear.
Where the initial elastic modulus deformation is not linear, tangent orsecant modulus can be used to determine the modulus of elasticity
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Chapter 6 - 20
Anelasticity
The following assumptions are generally made with regard to elasticity of
materials:a. The deformation is time independent, i.e. an applied stress produces
an instantaneous elastic strain that remains constant as long as thestress is maintained
b. The strain is completely recovered after removal of the stress
However, elastic deformation will continue for a finite time and onremoving the applied stress, a finite time is required for a completerecovery of the strain
This time-dependent elastic behavior is referred to as an anelasticity
For metals, the anelastic components are negligible but for polymer its
magnitude is significant
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Chapter 6 -
Example:
A piece of metal originally 305 mm long is pulled in tension
with a stress of 276 MPa. If the deformation is entirely elasticwhat will be its elongation? E = 110 GPa
Solutions Lo = 305mm = 0.305 m,s = 270 x 106 Pa
L = ?E = 110 x 109 Pa
3
9
6
10x2.50910x110
10x276
E
-===
0.305mx10x2.509LLL
L
3
o
o
-===
= 0.765 mm
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Chapter 6 -
Tensile loads
Lf LO
s
ex
ey The ratio of lateral strain (in
direction perpendicular to theapplied stress) to axial strain (indirection parallel to the appliedstress) is called Poissons ratio.
For many metals and alloys,Poissons ratio lies between
0.25 and 0.35
For isotropic materials,
E = 2G(1 + )
y
x-ratio,sPoisson'e
e=
Poisson's ratio,
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Chapter 6 - 23
Poisson's ratio, Poisson's ratio, :
Units:E: [GPa] or [psi]: dimensionless
> 0.50 density increases
< 0.50 density decreases(voids form)
eL
e
-
e=- L
e
metals: ~ 0.33
ceramics: ~ 0.25
polymers: ~ 0.40
Example
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Chapter 6 -
Example
A tensile stress is to be applied along the long axis of a cylindrical brass rodthat has a diameter of 10 mm. Determine the magnitude of the load requiredto produce a 2.5 x 10-3 mm change in diameter if the deformation is entirely
elastic. Ebrass = 97 GPa and brass = 0.34
y
x
P
P
Do = 10 mm, D = 2.5 x 10-3 mm
E = 97 x 109 Pa u = 0.34
Solution
44
y
4
y
x 10x.353734.0
10x2.5
10x2.50.34
strainaxial
strainlateral
---
==-
-=-=-= ye
43
o
10x2.510
10x2.5DD -
-
-=-=-== xlateral ee
Pa10x71.32310xx7.35310x97 649 == -es E
NALoad o 56024
)01.0(x10x71.323
26 =
==
s
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Chapter 6 - 25
Mechanical Properties
Slope of stress strain plot (which is
proportional to the elastic modulus) dependson bond strength of metal
Adapted from Fig. 6.7,Callister & Rethwisch 8e.
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Chapter 6 - 26
Metals
Alloys
GraphiteCeramics
Semicond
PolymersComposites
/fibers
E(GPa)
Based on data in Table B.2,Callister & Rethwisch 8e.Composite data based onreinforced epoxy with 60 vol%of alignedcarbon (CFRE),aramid (AFRE), or
glass (GFRE)fibers.
Youngs Moduli: Comparison
109 Pa
0.2
8
0.6
1
Magnesium,
Aluminum
Platinum
Silver, Gold
Tantalum
Zinc, Ti
Steel, Ni
Molybdenum
Graphite
Si crystal
Glass-soda
Concrete
Si nitrideAl oxide
PC
Wood( grain)
AFRE( fibers) *
CFRE*
GFRE*
Glass fibers only
Carbon fibers only
Aramid fibers only
Epoxy only
0.4
0.8
2
4
6
10
2 0
4 0
6 08 010 0
200
600800
10 001200
400
Tin
Cu alloys
Tungsten
Si carbide
Diamond
PTFE
HDPE
LDPE
PP
Polyester
PSPET
CFRE( fibers) *
GFRE( fibers)*
GFRE(|| fibers)*
AFRE(|| fibers)*
CFRE(|| fibers)*
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Chapter 6 -
27
Stress-Strain Behavior(Tension)
Elastic Plastic
Stre
ss
Strain
Elastic deformation
Reversible:
( For small strains)
Stress removed
material returns tooriginal size
Plastic deformation
Irreversible:
Stress removed material does notreturn to original dimensions.
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Chapter 6 - 28
Stress at which noticeableplastic deformation hasoccurred.
when ep = 0.002
Yield Strength, sy
sy= yield strength
Note: for 2 inch sample
e = 0.002 = z/z
z= 0.004 in
Adapted from Fig. 6.10(a),Callister & Rethwisch 8e.
tensile stress,s
engineering strain, e
sy
ep= 0.002
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Chapter 6 -
March 07, 2008
29
1:
Example 1: Convert the change in length data in Table 1 to engineering stress
and strain and plot a stress-strain curve.
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Chapter 6 -30
Example 1: SOLUTION
below
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Chapter 6 -31
(c)2003 Brooks/Co le, a division of Thomson Learning, Inc. Thomson Learning
is a trademark used herein under license.
The stress-strain curve for an aluminum alloy fromTable 1
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Chapter 6 -32
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Chapter 6 - 33
Stress-Strain Testing
Typical tensile test
machine
Adapted from Fig. 6.3, Callister & Rethwisch 8e. (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.)
specimenextensometer
Typical tensile
specimen
Adapted fromFig. 6.2,
Callister &Rethwisch 8e.
gauge
length
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Chapter 6 -34
A unidirectional force is
applied to a specimen in
the tensile test by means
of the moveable
crosshead. The cross-
head movement can be
performed using screws
or a hydraulic
mechanism
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Chapter 6 - 35
Experimental Tests:
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Chapter 6 -
36
Experimental Tests:Test Specimen Geometry, Facilities
& Set-Up
Tension Torsion Load Frame
Optical Strain Measurement System
Specimen with Random Patterns
D = 25.38
r= 6.35RAD
+
2
11
3
D = 25.382
1
1
3
90.00o0.50
r= 3.00 0.03
Mild Notch
Sharp Notch
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Chapter 6 -37
The Tensile Test: Use of the Stress-Strain
Diagram
Load - The force applied to a material during testing.
Strain gage or Extensometer - A device used for measuring
change in length and hence strain. Engineering stress - The applied load, or force, divided by
the original cross-sectional area of the material.
Engineering strain - The amount that a material deforms per
unit length in a tensile test.
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Chapter 6 - 38
Tensile Strength, TS
Metals: occurs when noticeable necking starts. Polymers: occurs when polymer backbonechains are
aligned and about to break.
Adapted from Fig. 6.11,Callister & Rethwisch 8e.
sy
strain
Typical response of a metal
F= fracture or
ultimate
strength
Neck actsas stressconcentrator
enginee
ring
TS
stres
s
engineering strain
Maximum stress on engineering stress-strain curve.
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Chapter 6 -
The maximum stress on the stress-strain curve of a materialrepresents its tensile strength
Deformation up to the maximum stress value is uniform
throughout the gauge length of the tensile specimen
At the maximum stress, necking occurs at some point andall subsequent deformation is confined to that point untilfracture occurs
The fracture strength corresponds to the stress level, atwhich failure occurs
Yield strength is normally used for design purpose instead oftensile strength because a material will undergo a large
amount of undesirable plastic deformation at maximumstress
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Chapter 6 -40
Stress-Strain Curve
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Chapter 6 -41
Important Design Considerations
D tilit
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Chapter 6 - 42
Plastic tensile strain at failure:
Ductility
Another ductility measure: 100xA
AARA%
o
fo-
=
x 100
LLL
EL%o
of -=
LfAo AfLoAdapted from Fig. 6.13,Callister & Rethwisch 8e.
Engineering tensile strain, e
Engineeringtensilestress, s
smaller %EL
larger %EL
D tilit
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Chapter 6 -
100xEL%
o
of
L
LL -=
Ductility is a measure of plastic deformation that a materialcan sustain before fracture
It is expressed either as percent elongation to fracture or aspercent reduction in area
When the fracture strain is low and less than 5 %, thematerial is said to be brittle
Ductility
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Chapter 6 -
Toughness is a measure of materials ability to absorb
energy up to fracture It is a measure of a materials resistance to fracture when a
crack is present
For low strain rate condition, toughness is given by the areaunder the stress-strain curve measured up to fracture point
For high strain condition and when a notch is present, notchtoughness is assessed using impact test
A tough material must have high strength and ductility
Toughness
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Chapter 6 - 45
Energy to break a unit volume of material
Approximate by the area under the stress-strain curve.
Toughness
Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy
Adapted from Fig. 6.13,Callister & Rethwisch 8e.
very small toughness(unreinforced polymers)
Engineering tensile strain, e
Engineeringtensile
stress, s
small toughness (ceramics)
large toughness (metals)
R ili U
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Chapter 6 -
Resilience of a material is a measure of its capacity to absorb energywhen it elastically deformed and have this energy recovered on
removing the applied load. Modulus of resilience, Ur is the strain energy per unit volume required to
stress a material from an unloaded state up to the yield point
Modulus of resilience, Ur, is the area under the stress-strain curve takento the yield point
=y
0
r dU
e
es
Assuming a linear elastic regionyesy2
1r U =
sy and ey are yield stress and yield strain respectively
Substituting for strain
2E
E
U
2
yy
21
y21
r =
== yy ses
Thus resilient materials are those having high yieldstrength and low moduli of elasticity
Resilience, Ur
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Chapter 6 - 47
Resilience, Ur
Ability of a material to store energy
Energy stored best in elastic region
If we assume a linearstress-strain curve thissimplifies to
Adapted from Fig. 6.15,Callister & Rethwisch 8e.
yyr2
1U es@
e
es= y dUr0
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Chapter 6 - 48
Elastic Strain Recovery
Adapted from Fig. 6.17,Callister & Rethwisch 8e.
Stress
Strain
3. Reapplyload
2. Unload
D
Elastic strainrecovery
1. Load
syo
syi
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Chapter 6 - 49
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 measure size
of indent afterremoving load
dDSmaller indentsmean largerhardness.
increasing hardness
mostplastics
brassesAl alloys
easy to machinesteels file hard
cuttingtools
nitridedsteels diamond
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Chapter 6 - 50
Hardness: Measurement
Rockwell
No major sample damage
Each scale runs to 130 but only useful in range20-100.
Minor load 10 kg Major load 60 (A), 100 (B) & 150 (C) kg
A = diamond, B = 1/16 in. ball, C = diamond
HB = Brinell Hardness TS(psia) = 500 x HB
TS(MPa) = 3.45 x HB
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Chapter 6 - 51
Hardness: MeasurementTable 6.5
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Chapter 6 - 52
Hardening
Curve fit to the stress-strain response:
sT =K eT( )n
true stress (F/A) true strain: ln(L/Lo)
hardening exponent:n= 0.15 (some steels)to n= 0.5 (some coppers)
An increase in sydue to plastic deformation.s
e
large hardening
small hardeningsy0
sy1
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Chapter 6 - 53
Variability in Material Properties
Elastic modulus is material property
Critical properties depend largely on sample flaws(defects, etc.). Large sample to sample variability.
Statistics
Mean
Standard Deviation s=
n
xi -
x
( )
2
n-1
1
2
nxx n
n
=
where nis the number of data points
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Chapter 6 - 54
Design uncertainties mean we do not push the limit.
Factor of safety, N
N
y
working
s=s
Often Nisbetween1.2 and 4
Example: Calculate a diameter, d, to ensure that yield does
not occur in the 1045 carbon steel rod below. Use afactor of safety of 5.
Design or Safety Factors
220,000N
d2/ 4( )
5N
y
working
s=s 1045 plain
carbon steel:
sy= 310 MPaTS= 565 MPa
F= 220,000N
d
Lo
d= 0.067 m = 6.7 cm
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Chapter 6 - 55
Stress and strain: These are size-independentmeasures of load and displacement, respectively.
Elastic behavior: This reversible behavior oftenshows a linear relation between stress and strain.To minimize deformation, select a material with a
large elastic modulus (Eor G).
Toughness: The energy needed to break a unitvolume of material.
Ductility: The plastic strain at failure.
Summary
Plastic behavior: This permanent deformationbehavior occurs when the tensile (or compressive)
uniaxial stress reaches sy.
S
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56
Summary
Anelasticity
Ductility Elastic deformation
Elastic recovery
Engineering strain
Engineering stress
Hardness Modulus of elasticity
Plastic deformation
Poissons ratio
Proportional limit
Shear Tensile strength
Toughness
Yielding
Yield strength
Make sure you understand language and concepts: