Introduction to Engineering MaterialsENGR2000
Chapter 6: Mechanical Properties of Metals
Dr. Coates
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6.2 Concepts of Stress and Strain
• Principal ways in which a load may be applied– Tension– Compression– Shear– Torsional
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Fracture & Failure
• Fracture occurs when a structural componentseparates into two or more pieces.– fracture represents failure of the component
• Failure is the inability of a component to performits desired function.– failure may occur prior to fracture– e.g. a car axle that bends when you drive over a
pothole
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Tension Tests
• The specimen is deformed to fracture with agradually increasing tensile load…
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Engineering Stress & Engineering Strain
0
0
0
0
:strain gEngineerin
:stress gEngineerin
lll
lΔlε
AF
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Compression Tests
• Similar to the tensile tests• Force is compressive• Specimen contracts along direction of the stress
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Engineering Stress & Engineering Strain
0
:strain gEngineerin
:stress gEngineerin
0
0
0
0
lll
lΔlε
AF
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Shear and Torsional Tests
smallfor tan
:strainShear
:stressShear
0
hw
AF
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Shear and Torsional Tests
f
Tf
: torqueapplied toduestrain Shear
: torqueapplied todue stressShear
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Deformation
• All materials undergo changes in dimensions inresponse to mechanical forces. This phenomenonis called deformation.– If the material reverts back to its original size and
shape upon removal of the load, the deformation issaid to be elastic deformation.
– If application and removal of the load results in apermanent change in size or shape, the deformation issaid to be plastic deformation.
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Elastic Deformation6.3 Stress-Strain Behavior
εE
E
:elasticity of Modulus
:law sHooke'
G
:modulusShear
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Modulus of Elasticity
• Young’s modulus• Elastic modulus• Stiffness• Material’s resistance to elastic deformation
– greater the elastic modulus, the stiffer the material orsmaller the elastic strain that results from theapplication of a given stress.
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Elastic modulus of metals, ceramics & polymers
• Ceramics– high elastic modulus– diamond, graphite (covalent bonds)
• Metals– high elastic modulus
• Polymers– low elastic modulus– weak secondary bonds between chains
Increasing E
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Tangent or Secant Modulus
• Materials with non-linear elastic behavior– Gray cast iron,– concrete,– many polymers
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Elastic Modulus on an Atomic Scale
• Elastic strain– small changes in inter-atomic spacing– stretching of inter-atomic bonds
• Elastic modulus– resistance to separation of adjacent atoms– inter-atomic bonding forces
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Review2.5 Bonding Forces and Energies
length. bond theasknown is apart. are atoms two theof centers the-
0:mequilibriuAt
:forcenet The
0
0
rr
FFF
FFF
RAN
RAN
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Elastic Modulus on an Atomic Scale
??0
WhydrdFtorelatedbeEmightHow
r
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Effect of Temperature on the Elastic Modulus
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Example 6.1
• A piece of Cu originally 305 mm long is pulled intension with a stress of 276 Mpa. If thedeformation is entirely elastic, what will be theresultant elongation?
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Example 6.1
?110
:6.1 Table Using276305
:Given
0
lGpaE
MPamml
Cu
mmElσΔl
lΔlEEεσ
77.0
:law sHooke' Using
0
0
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6.4 Anelasticity- time dependent elastic behavior
• Time dependent elastic strain component.• Elastic deformation continues after the stress
application.• Upon load release, finite time is required for
complete recovery.
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Anelasticity of metals, ceramics & polymers
• Polymers– Visco-elastic behavior
• Metals
Increasinganelasticity
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6.5 Elastic Properties of Materials
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Poisson’s Ratio
35.025.0:metalsmost For
change net volume no50.0
25.041
0
:constant) material (a ratio sPoisson'
max
ltheoretica
z
y
z
x
09-13-10
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Elastic Properties of Materials
12
:)directions allin properties (same materials isotropicFor ,,
:)properties (material constants Elastic
EG
GE
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Example 6.2
• A tensile stress is to be applied along the long axisof a cylindrical brass rod that has a diameter of 10mm. Determine the magnitude of the loadrequired to produce a 2.5 x 10-3 mm change indiameter if the deformation is entirely elastic.
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?34.0
97:6.1 Table Using
105.2
10:Given
30
F
GpaE
mmdmmd
brass
brass
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NdF
MPaE
dd
z
xz
x
56004
:force theCompute3.71
:law sHooke' Using
1035.7
:ratio sPoisson' theUsing
105.2
:strain theCompute
20
4
4
0
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Plastic Deformation
• Elastic deformation– Up to strains of 0.005
• Plastic deformation– Hooke’s law is no longer valid– Non-recoverable or permanent deformation– Phenomenon of yielding
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6.6 Tensile Properties
fractureat strain fracture)at (stressstrength fracture
strain uniform stress gengineerin maximumstrength tensileultimate
strainpoint yieldlimit elastic the toingcorrespond stressstrength yield
f
f
u
uts
yp
ys
σ
σ
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Yielding and Yield Strength
• Gradual elastic-plastic transition• Proportional limit (P)
– Initial departure from linearity– Point of yielding
• Yield strength– Determined using a 0.002 strain
offset method
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Yielding and Yield Strength
• Steels & other materials• Yield point phenomenon• Yield strength
– Average stress associatedwith the lower yield point
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Elastic deformation, plastic deformation, necking & fracture
necking
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Example 6.3
• From the tensile stress-strain behavior for thebrass specimen shown, determine the following– The modulus of elasticity– The yield strength at a strain offset of 0.002– The maximum load that can be sustained by a
cylindrical specimen having an original diameter of12.8 mm
– The change in length of a specimen originally 250 mmlong that is subjected to a tensile stress of 345 Mpa.
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MPa
GPa
E
ys 250:strength Yield
8.9300016.0
0150
:modulus Elastic
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NdF
MPa
Fmmd
uts
uts
900,574
450:strength tensile theUsing
?8.12
:Given
20
max
max
0
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mmll
lMPamml
15:elongation theCompute
06.0:A)(point strain theDetermine
?345250
:Given
0
0
09-12-11
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Ductility
• Measure of the degree of plastic deformation thathas been sustained at fracture.
• Brittle materials– Little or no plastic deformation
prior to fracture• Ductile materials
– Significant plastic deformationprior to fracture
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Ductility
region. necked theof area sectional-cross final theis and area sectional-cross original theis where
100%
:areain reduction Percent fracture.at length theis and
length original theis where
100100%
:elongationPercent
0
0
0
0
0
0
f
f
f
ff
AA
AAA
RA
ll
lll
EL
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Ductility
• Why do we need to know the ductility ofmaterials?– The degree to which a structure will deform plastically
prior to fracture (design & safety measures)– The degree of allowable deformation during fabrication
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Mechanical properties of metals
• Depend on– prior deformation,– presence of impurities &– heat treatments
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Engineering stress-strain behavior for Fe at different temperatures
brittle
ductile
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Resilience
• Capacity of a material to absorb energy when it isdeformed elastically and then, upon unloading, tohave this energy recovered.
• Modulus of resilience– Strain energy per unit volume required to stress a
material from an unloaded state up to the point ofyielding.
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Resilience
EE
U
dU
yyy
yyr
r
y
22121
:region elasticlinear a Assuming
:resilience of Modulus
2
0
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Resilience
• Materials used in spring applications– High resilience– High yield strength– Low elastic modulus
09-17-10
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Toughness
• Measure of the ability of a material to absorbenergy up to fracture.
• Area under the stress-strain curve up to the pointof fracture.
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Toughness
f
dU
0
:Toughness
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Toughness
• Ductile materials• Brittle materials
Increasingtoughness
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Example
• Which material has the highest elastic modulus?• Which material has the highest ductility?• Which material has the highest toughness?• Which material does not exhibit
any significant plastic deformationprior to fracture?
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6.7 True Stress and Strain
• Decline in stress necessary to continuedeformation past the maximum point M.
• Is the material getting weaker?
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6.7 True Stress and Strain
• Short-coming in the engineering stress-straincurve– Decreasing cross-sectional area not considered
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6.7 True Stress and Strain
0
lln
:strain True area. ousinstantane theis where
:stress True
0 ll
ldlε
AAF
lt
i
it
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6.7 True Stress and Strain
11ln
:necking Prior to
:change volumeno Assuming
00
t
t
ii lAlA
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6.7 True Stress and Strain
• Necking introduces a complex stress state…
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Strain-hardening exponent
t.coefficienstrength theis andexponent hardening-strain theis where
:necking ofpoint thetondeformatio plastic ofonset theFrom
Kn
K ntt
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Example 6.4
• A cylindrical specimen of steel having an originaldiameter of 12.8 mm is tensile tested to fractureand found to have an engineering fracture stress of460 Mpa. If its cross-sectional diameter atfracture is 10.7 mm, determine– The ductility in terms of percent reduction in area– The true stress at fracture
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??%
7.10
4608.12
:Given
0
tf
f
f
RAmmdMPammd
%30100
100%
:Ductility
20
220
0
0
ddd
AAA
RA
f
f
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MPaAF
NAF
AFMPa
ftf
f
f
660
:fractureat stress True
200,59:load applied The
460
:fractureat stress gEngineerin
0
0
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Example 6.5
• Compute the strain-hardening exponent n for analloy in which a true stress of 415 Mpa produces atrue strain of 0.10. Assume a value of 1035 Mpafor K.
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Example 6.5
?1035
10.0415
:Given
nMPaK
MPa
t
t
40.0
:exponent hardening-Strain
n
K ntt
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6.10 Hardness
• Measure of a materials resistance to localizedplastic deformation (small dent or scratch)
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Rockwell Hardness Tests
• A hardness number is determined– Difference in depth of penetration
from an initial minor loadfollowed by a major load
• Combinations of indenters &loads are used
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Hardness Testing Techniques
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Minor load=10 kg- used for thick specimens
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Minor load = 3 kg- thin specimens
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Correlations between hardness & tensile strength
• Both are measures ofresistance to plasticdeformation
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Property Variability & Design/Safety Factors
• Reading assignment– Section 6.11 on variability of material properties– Example 6.6
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6.12 Design/Safety Factors
dys
c
cd
N
N
:criteriaselection Materialfactor.design theis 1'
load)max estimated theof basis (on the stress calculated theis '
:stressDesign
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6.12 Design/Safety Factors
0.42.1:safety ofFactor
safety. offactor theis 1and material theofstrength yield theis
:stress or working stress Safe
N
N
Nys
ysw
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Design Example 6.1
• A tensile-testing apparatus is to be constructedthat must withstand a maximum load of 220 kN.The design calls for two cylindrical support posts,each of which is to support half of the max load.Furthermore, plain carbon steel ground andpolished shafting rounds are to be used; theminimum yield and tensile strengths of this alloyare 310 Mpa and 565 Mpa respectively. Specify asuitable diameter for these support posts.
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5:safety offactor a Assume
?565
310
11022021:Given
max
N
dMPaMPa
kNkNF
uts
ys
mmd
dF
MPaN
w
ysw
5.474/
:requireddiameter Minimum
62
:stress Working
0
20
max
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Design problem 6.D1
• A large tower is to be supported by a series ofsteel wires; it is estimated that the load on eachwire will be 13,300 N. Determine the minimumrequired wire diameter, assuming a factor of safetyof 2 and a yield strength of 860 Mpa for the steel.
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