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Course Title : metallic condensed matters
Lecture 4 Mechanical properties
condensed matters
Lecture 4. Mechanical properties of metallic materials and
mechanical test techniques
Lect e Plan
mechanical test techniques
Lecture Plan:1. Concepts of stress and strain. Mechanical tests
2. Elastic deformation3. Plastic deformation4. Strain & stress
diagram5 Design safety factors5. Design safety factors
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The fist problem is to ensure the safe service of mechanisms
andconstructions those are usually under load and prevent their
fracture and thesecond is to fabricate new structural materials
with desired mechanicalsecond is to fabricate new structural
materials with desired mechanicalproperties and microstructure. At
that the key mechanical design properties arestiffness, strength,
hardness, ductility, and toughness.
The mechanical properties of materials are investigated by
performingp p g y p gcarefully designed standardized laboratory
experiments that replicate as nearlyas possible the service
conditions. Factors to be considered include the nature ofthe
applied load and duration of test, as well as the environmental
conditions.
Important Terms and ConceptsSample - design stress -
proportional limit - ( resilience
ductility - elastic deformation
safe stress - shear - stiffness - tensile strength -
elastic recovery - engineering strain
tensile strength toughness ,
() engineering stress () hardness -
true strain true stress yielding - hardness
modulus of elasticity Poissons ratio
yielding yield strength -
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CONCEPTS OF STRESS AND STRAIN
A load is static when all changes are relatively slow with time
and the load isA load is static when all changes are relatively
slow with time and the load isapplied uniformly over a cross
section or surface of a sample. In this case themechanical behavior
may be defined by a simple stressstrain test; these aremost
commonly conducted for metals at room temperature. There are
threeprincipal ways in which a load may be applied: namely,
tension, compression,shear. Consider each kind of mechanical test
in more detail.
Tensile TestsTensile Tests
(c) The specimen is elongated by
(a) (b) (c)is elongated by the moving crosshead; load cell
and
t t extensometer measure, respectively, the magnitude of the
magnitude of the applied load and the elongation.
Fig 1 (a) Schematic illustration of how a tensile load produces
an Fig.1 (a) Schematic illustration of how a tensile load produces
an elongation and positive linear strain. Dashed lines represent
the shape
before deformation; solid lines, after deformation. (b)
tensile-testing machine.
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Fig.2 A standard tensile specimens with circular and rectangular
Fig.2 A standard tensile specimens with circular and rectangular
cross section (cylindrical shape and plate shape,
respectively).
One of the most common mechanical stressstrain tests is tensile
test A One of the most common mechanical stress strain tests is
tensile test. Aspecimen is deformed, usually to fracture, with a
gradually increasing tensileload that is applied uniaxially along
the long axis of a specimen. Tensile specimens can be with circular
or rectangular cross sections. The
d b d h h ld f h lspecimen is mounted by its ends into the
holding grips of the tensile testingmachine. This machine is
designed to elongate the specimen at a constant rateand to
continuously and simultaneously measure the instantaneous
appliedload (with a load cell) and the resulting elongations (using
an extensometer).load (with a load cell) and the resulting
elongations (using an extensometer). A stressstrain test typically
takes several minutes to perform and isdestructive; that is, the
test specimen is permanently deformed and usuallyfractured. The
output of such a tensile test is recorded (usually on a
computer)
l d f l ias load or force versus elongation.4
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Load and elongation are expressed in terms of respective
parameters of
F
Load and elongation are expressed in terms of respective
parameters ofengineering stress and engineering strain. Engineering
stress is defined bythe relationship
(1) 0SF=
in which F is the instantaneous load applied perpendicular to
the specimen crossin which F is the instantaneous load applied
perpendicular to the specimen crosssection, in units of [N], and S0
is the original cross-sectional area before any loadis applied
[m2]. The unit of engineering stress (referred to subsequently as
juststress) is [MPa] (where 1 MPa = 106 N/m2).) [ ] ( / )
(2)0 lll iEngineering strain is defined according to
(2) 00
0
ll ==i
in which l0 is the original length before any load is applied
and li is the currentin which l0 is the original length before any
load is applied and li is the currentvalue of length. The
difference li - l0 is denoted as l and is the deformationelongation
or change in length at some instant, as referenced to the
originallength. Engineering strain (or deformation) is unitless or
can be expressed in %,f hi h i l i l i li d b 100for this the
strain value is multiplied by 100.
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Compression TestsFig.3. Schematic illustration of how a
compressive load Fig.3. Schematic illustration of how a compressive
load produces contraction and a negative linear strain.
A compression test is conducted in a manner similar to the
tensile test except that the force is compressive and the
specimen
By convention, a compressive force is taken to be negative,
which yields a
test, except that the force is compressive and the specimen
contracts along the direction of the stress. Equations (1-2) are
utilized to compute compressive stress and strain,
respectively.
By convention, a compressive force is taken to be negative,
which yields anegative stress. Furthermore, because l0 is greater
than li , compressive strainscomputed from (2) are necessarily also
negative. Compressive tests are usedwhen a materials behavior under
large and permanent (i.e., plastic) strains isd i d i f i li i h h
i l i b i l idesired, as in manufacturing applications, or when the
material is brittle intension.
Shear and Torsional Tests
Fig.4 (c) Schematic representation of shear strain (d) Schematic
shear strain . (d) Schematic representation of torsional
deformation (i.e., angle of twist) produced by an applied torque
T.
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Th h t t i h i Fi 4 H th l f f th b i fi d The shear test is
shown in Fig. 4c. Here the lower face of the cube is fixed rigidly
and the force is imposed parallel to the upper face of the cube.
The resulting shear stress is calculated according to
, 0SF= )3( tan =
0
where F is the load or force imposed, S0 an area of upper or
bottom face.
Torsion is a variation of pure shear, wherein a sample is
twisted: torsionalforces produce a rotational motion about the
longitudinal axis of one end of thesample relative to the other
end. Examples of torsion are found for machine
l d d i h ft d l f t i t d ill T i l t t llaxles and drive
shafts, and also for twist drills. Torsional tests are
normallyperformed on cylindrical solid shafts or tubes. A shear
stress is a function ofthe applied torque T, whereas shear strain
is related to the angle of twist.
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Geometric Considerations of the Stress State
Fig.5 Schematic representation showing normal and shear ()
stresses that act on a plane oriented at an ( ) pangle relative to
the plane taken perpendicular to the direction along which a pure
tensile stress is applied.
The stress value depends on the orientation of the plane upon
which the stress is taken to act. For example, consider the
cylindrical tensile specimen that is subjected to a tensile stress
applied parallel to its axis (Figure 4) Consider also the plane
p-applied parallel to its axis (Figure 4). Consider also the plane
pp that is oriented at some arbitrary angle relative to the plane
of the specimen cross-section. The external stress can be expressed
as a vector sum of a tensile stress normal to the p-p plane and a
shear stress that acts parallel to this plane. and in terms of and
, take the form
, 2
2cos1cos2
+== (4) 22sincossin
==
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STRESSSTRAIN BEHAVIOR : Elastic deformation
At l l l f t f t t l th t l i ti l t t iAt low levels of stress
for most metals the stress value is proportional to strain
(5) E= This is Hookes law, where E (GPa) is the modulus of
elasticity, or Youngs modulus. For most typical metals the
magnitude of this modulus ranges between 45 GPa for Mg and 407 GPa
for Wthis modulus ranges between 45 GPa for Mg, and 407 GPa for
W.
Fig.6. Schematic stressstrain diagram showing linear (a) and
nonlinear elastic deformation for
The slope of this linear segment is the modulus ofelasticity E -
an indicator of materials resistance to
( )loading and unloading cycles.
yelastic deformation or stiffness. The greater themodulus, the
stiffer the material, or the smaller theelastic strain that results
from the application of a givenstress Elastic deformation is
reversiblestress. Elastic deformation is reversible.
For some materials (gray cast iron, concrete, polymers)the
elastic portion of the stressstrain curve is not linear.For this
nonlinear behavior, either tangent ()or secant modulus ( ) is
normally used.Tangent modulus is taken as the slope of the
stressstraincurve at some point of stress whereas secant
moduluscurve at some point of stress, whereas secant
modulusrepresents the slope of a secant drawn from the origin
tosome given point of the curve. 9
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Table 1 Room-Temperature Elastic and Shear Moduli and Poissons
Ratio for Various M t l All
Plot of modulus of elasticity versus temperature for tungsten,
steel, and
aluminum.Metal Alloys
aluminum.
Values of the modulus E for ceramic materials are about the same
as for metals; Values of the modulus E for ceramic materials are
about the same as for metals; for polymers they are lower. With
increasing temperature, the modulus of elasticity diminishes.
,kxl
lESFo
o == , o
o
lESk = lx =
1where
2
21)/( EdxFlSU ooe == Elastic energy 10
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ELASTIC PROPERTIES OF MATERIALS
Fig. 7 Axial (z) elongation (positive strain) and lateral (x and
y) contractions (negative strains) in response to an imposed
tensile stress Solid lines represent dimensions after
YX stress. Solid lines represent dimensions after stress
application; dashed lines, before.
Z
Y
Z
Xv ==Theoretically Poissons ratio for isotropic materials should
be 1/4 Theoretically, Poissons ratio for isotropic materials should
be 1/4 ; the maximum value (or that value for which there is no net
volume change) is 0.50. For many metals and other alloys, values of
Poissons ratio range between 0.25 and 0.35. g
For isotropic materials, shear or bulk and elastic moduli are
connected with Poissons ratio by the relationwith Poisson s ratio
by the relation
( ) )21(312 vBvGE =+=11
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Elastic constants for homogeneous materials
Shear stress and strain are proportional to each other
/ =Gwhere G is the shear modulus, the slope of the linear
elastic region of the shear stressstrain curve.
The bulk modulus (B) of a substance measures the substance's
resistance to uniform compression.
dP;
dVdPVB =
EEG)21(3)3(3 v
EEG
EGB == ( ) )21(312 vBvGE =+=
)1(2)21(3
93
vvB
EBBEG +
== )3(2231
263
GBGB
GE
BEBv +
===
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Plastic DeformationFor most metallic materials elastic
deformation persists only to strains of For most metallic
materials, elastic deformation persists only to strains of
about 0.005. As the material is deformed beyond this point, the
stress is no longer proportional to strain, and permanent,
nonrecoverable, i.e. plastic deformation occurs. From an atomic
perspective, plastic deformation corresponds to the p p p pbreaking
of bonds with original atom neighbors and then re-forming bonds
with new neighbors as large numbers of atoms or molecules move
relative to one another; upon removal of the stress they do not
return to their original positions.
For crystalline solids deformation is accomplished by means of a
process For crystalline solids, deformation is accomplished by
means of a process called slip, which involves the motion of
dislocations. Plastic deformation in noncrystalline solids (as well
as liquids) occurs by a viscous flow mechanism.
Fig. 9 (a) Typical stressstrain behavior for a metal showing
elastic and plastic d f i h i l li i P d deformations, the
proportional limit P, and the yield strength as determined using
the 0.002 strain offset method. (b) Representative stressstrain
behavior (b) Representative stress strain behavior found for some
steels demonstrating the yield point phenomenon. ( )
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TENSILE PROPERTIES Yielding and Yield Strength
Figure 10. Typical yield behavior for non-ferrous alloys.
1: True elastic limit2: Proportionality limit3: Elastic limit3:
Elastic limit4: Yield strength
True elastic limit : The lowest stress at which dislocations
move. Thisdefinition is rarely used, since dislocations move at
very low stresses, andde t o s a e y used, s ce d s ocat o s o e at
e y o st esses, a ddetecting such movement is very difficult.
Proportionality limit : Point P (Fig. 9a). Up to this point,
stress isti l t t i (H k l ) th t t i h i t i htproportional to
strain (Hookes law), so the stress-strain graph is a straight
line, and the gradient will be equal to the elastic modulus of
the material.
Elastic limit : Beyond the elastic limit, permanent deformation
will occur. They , plowest stress at which permanent deformation
can be measured.
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Yield point or yield strength (proof stress ) : The position of
this point P is difficult to measure precisely.
h b bl h d h hAs a consequence, a convention has been
established wherein a straight line is constructed parallel to the
elastic portion of the stressstrain curve at some specified strain
offset, usually 0.002. The stress corresponding to the intersection
of this line and the stressstrain curve as it bends over in the
intersection of this line and the stress strain curve as it bends
over in the plastic region is defined as the yield strength Y
[MPa]. Upper yield point and lower yield point :Some metals, such
as mild steel ( ), demonstrate the onset of plastic deformation as
shown in Fig 9b. At the upper yield point, plastic deformation is
initiated with an apparent decrease in engineering stress Continued
deformation fluctuates slightly about some in engineering stress.
Continued deformation fluctuates slightly about some constant
stress value, termed the lower yield point; stress subsequently
rises with increasing strain. The material response is linear up
until the upper yield point, but the lower yield point is used in
structural engineering as a conservative value.
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Stress-strain diagram. Tensile Strength
Fig. 11 Typical engineering stressstrain behavior to fracture,
point F. The tensile strength TS is fracture, point F. The tensile
strength TS is indicated at point M. The circular insets represent
the geometry of the deformed specimen at various points along the
curve.
The tensile strength TS is the stress at the maximum on the
engineering stressstrain curve.
All deformation up to this point is uniform. At the maximum
stress, a small constriction or neck begins to form at some point,
and all subsequent deformation is confined at this neck. If this
stress continues to be applied, fracture will result in the
neck.
Tensile strengths ( ) may vary anywhere from 50 g ( ) y y yMPa
for an aluminum to as high as 3000 MPa for the high-strength
steels. Ordinarily, when the strength of a metal is cited for
design purposes, the yield strength is used.
The fracture strength corresponds to the stress at fracture.
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Ductility is a measure of the degree of plastic deformation that
has beensustained at fracture.Ductility may be expressed
quantitatively as either percent elongation orpercent reduction in
area. The percent elongation is the percentage of plasticstrain at
fracture he e l is the f act e strain at fracture
%1000
0
= l
ll fwhere lf is the fracture length and l0 is the original gauge
length as given earlier0
Percent reduction in area is defined as
%1000
=
SSS f
0 Swhere S0 is the original cross-sectional area and Sf is the
cross-sectional area at the point
Fig. 12 Schematic representations of tensile stressstrain
behavior for brittle and
of fracture.
strain behavior for brittle and ductile metals loaded to
fracture. 17
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Table 2 Typical Mechanical Properties of Several Metals and
Alloys in an Annealed Statean Annealed State
Fig. 13Engineering stressstrain behavior for iron at three
temperatures iron at three temperatures.
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Resilience is the capacity of a material to absorb energy when
it is deformedelastically and then, upon unloading, to have this
energy recovered (J/m3). They , p g, gy ( / )associated property is
the modulus of resilience, Ur, which is the strain energyper unit
volume required to stress a material from an unloaded state up to
thepoint of yielding. Computationally, the modulus of resilience
for a specimensubjected to a uniaxial tension test is just the area
under the engineering stresssubjected to a uniaxial tension test is
just the area under the engineering stressstrain curve taken to
yielding (Fig.14) = Y dU r 0
Assuming a linear elastic region,
in which Y is the strain at yielding. U YYYr1 2 == Y y gEYYr
22
Thus this area under the stressstrain curve represents energy
absorption per unit volume of material The energy absorption per
unit volume of material. The resilient materials are those having
high yield strengths and low moduli of elasticity; such alloys can
be used in spring applications.
Fig.14 Schematic representation showing how modulus f l ( d h h
d d )of resilience (corresponding to the shaded area) is
determined from the tensile stressstrain behavior of a
material.
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ELASTIC RECOVERY AFTER PLASTIC DEFORMATION
Fig.15 Schematic tensile stressstrain diagram showing the
phenomena of elastic strain recovery and strain hardening. The
initial yield strength is designated as is the yield strength after
releasing the designated as is the yield strength after releasing
the load at point D, and then upon reloading.
During the unloading cycle, the curve traces a near t i ht li th
f th i t f l di ( i t D) straight-line path from the point of
unloading (point D),
and its slope is virtually identical to the modulus of
elasticity, or parallel to the initial elastic portion of the
curve. The magnitude of this elastic strain, which is
Toughness
cu e e ag tude o t s e ast c st a , c sregained during
unloading, corresponds to the elastic strain recovery.
For the static deformation, a measure of toughness in metals
(derived from plastic deformation) may be ascertained from the
results of a tensile stressstrain test. It is the area under the -
curve up to the point of fracture. The units are [energy per unit
volume of material]. For a metal to be tough, it must display both
strength and ductility. Hence, even though the brittle metal has
higher yield and tensile strengths, it has a lower toughness than
the ductile one as can be seen by comparing the has a lower
toughness than the ductile one, as can be seen by comparing the
areas ABC and ABC (Fig.12).
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TRUE STRESS AND STRAIN
True stress is defined as the load F divided by the
instantaneous cross-sectional area Si over which deformation is
occurring. True strain - T .
F l
If no volume change occurs during deformation
iT S
F=0l
ll iT n=
00ll SS ii =( ) ( )true and engineering stress and strain are
related as
If no volume change occurs during deformation
( ) += 1T ( ) += 1nT lEquations are valid only to the onset of
necking; the true stress necessary to
t i i i t i ti t i t th t il i t M C i id t ith sustain
increasing strain continues to rise past the tensile point M.
Coincident with the formation of a neck is the introduction of a
complex stress state within the neck region (i.e., the existence of
other stress components in addition to the axial stress). So the
correct stress (axial) within the neck is slightly lower than the )
( ) g ystress computed from the applied load and neck
cross-sectional area. For some metals and alloys the region of the
true stressstrain curve from the onset of plastic deformation to
the point at which necking begins may be approximated by
Th t i ft t d th t i h d i nTT K = The parameter n is often
termed the strain hardening exponent and has a value less than
unity. 21
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HARDNESS is a measure of a materials resistance to localized
plastic deformation (e.g., a small dent or a scratch).
Hardness tests are performed more frequently than any other
mechanical test for several reasons:1. They are simple and
inexpensiveordinarily no special specimen need beprepared, and the
testing apparatus is relatively inexpensive.2. The test is
nondestructivethe specimen is neither fractured nor
excessivelydeformed; a small indentation is the only deformation.3
Other mechanical properties often may be estimated from hardness
data3. Other mechanical properties often may be estimated from
hardness data,such as tensile strength.
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Correlation between Hardness and Tensile Strength
Both tensile strength and hardness are indicators of a metals
resistance to plastic deformation. Consequently, they are roughly
proportional, as shown in Figure 19, for tensile strength as a
function of the HB for cast iron, steel, and b Th ti lit l ti hi d
t h ld f ll t l Fi brass. The same proportionality relationship
does not hold for all metals, as Figure 19 indicates.
As a rule of thumb for most steels, the HB and the tensile
strength are related according to ( ) HBMPTS 453according to
Figure 17 Relationships between hardness and
( ) HBMPaTS = 45.3p
tensile strength for steel, brass, and cast iron
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DESIGN/SAFETY FACTORS
For less critical static situations and when tough materials are
used a design For less critical static situations and when tough
materials are used, a design stress, d, is taken as the calculated
stress level c (on the basis of the estimated maximum load)
multiplied by a design factor, N
where N is greater than unity. Thus, the material to be used for
Cd N =
where N is greater than unity. Thus, the material to be used for
the particular application is chosen so as to have a yield strength
at least as high as this value of d.
Alternatively a safe stress or working stress is used instead of
design stress
N/
Alternatively, a safe stress or working stress, w, is used
instead of design stress. This safe stress is based on the yield
strength of the material and is defined as the yield strength
divided by a factor of safety, N, or
NYw / =The choice of an appropriate value of N is necessary. If
N is too large, then
t d i ill lt th t i ith t h t i l llcomponent overdesign will
result; that is, either too much material or an alloyhaving a
higher-than-necessary strength will be used. Values normally
rangebetween 1.2 and 4.0. Selection of N will depend on a number of
factors,including economics, previous experience, the accuracy with
which mechanicalincluding economics, previous experience, the
accuracy with which mechanicalforces and material properties may be
determined, and, most important, theconsequences of failure in
terms of loss of life and/or property damage. Becauselarge N values
lead to increased material cost and weight, structural
designers
i t d i t h t i l ith d d t ( d i t bl )are moving toward using
tougher materials with redundant (and inspectable)designs, where
economically feasible. 24