Ch6

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
MECHANICALMECHANICAL
PROPERTIES
OF METALS
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
MECHANICAL PROPERTIES OF METALS
WHY STUDY The Mechanical Properties of Metals?
It is incumbent on engineers to understand howIt is incumbent on engineers to understand how
the various mechanical properties are
measured and what these properties represent.measured and what these properties represent.
They may be called upon to designThey may be called upon to design
structures/components using predetermined
materials such that unacceptable levels ofa e a s suc a u accep ab e e e s o
deformation and/or failure will not occur.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
INTRODUCTION
The mechanical behavior of a material reflects the relationship between
its response or deformation to an applied load or force.ts espo se o de o at o to a app ed oad o o ce
Important mechanical properties are strength, hardness, ductility, and
stiffness.
Factors to be considered desiging:
The nature of the applied load and its duration: Load may be tensile,
i h d it it d b t t ith ti itcompressive, or shear, and its magnitude may be constant with time, or it
may fluctuate continuously.
The environmental conditionsThe environmental conditions.
The American Society for Testing and Materials (ASTM),
www.alquds.edu
(http://www.astm.org).

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
CONCEPTS OF STRESS AND STRAIN
There are three principal ways in which a load may be applied, Figure 6.1:
Tension
CompressionCompression
Shear and Torsional deformation.
Tension TestsTension Tests
A specimen is deformed, usually to fracture, with a gradually increasing
tensile load that is applied uniaxially along the long axis of a specimen.
A standard tensile specimen is shown in Figure 6.2. The specimen is
mounted by its ends into the holding grips of the testing apparatus (Figure
6.3).
This “dogbone” specimen configuration was chosen :
During testing, deformation is confined to the narrow center region
(which has a uniform cross section along its length)
www.alquds.edu
(which has a uniform cross section along its length).
To reduce the likelihood of fracture at the ends of the specimen.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.1
(a) Schematic
illustration of how a
tensile load producestensile load produces
an elongation and
positive linear
strain.
Dashed lines
represent the shape
before deformation;
solid lines aftersolid lines, after
deformation.
(b) Schematic
ill t ti f hillustration of how a
compressive load
produces contraction
and a negative linear
www.alquds.edu
a d a egat e ea
strain.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.1 (c) Schematic representation of shear strain , where . (d)
Schematic Representation of torsional deformation (i.e., angle of twist Ф) produced
www.alquds.edu
Schematic Representation of torsional deformation (i.e., angle of twist Ф) produced
by an applied torque T.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.2 A standard tensile specimenFigure 6.2 A standard tensile specimen
with circular cross section.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.3
The apparatus used toThe apparatus used to
conduct tensile
stress–strain tests. The
specimen is elongatedp g
by the moving crosshead; load
cell and extensometer
measure, respectively, the
magnitude of the applied load
and the elongation.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Th t t f h t il t t i d d ( ll t )
The output of such a tensile test is recorded (usually on a computer) as
load or force versus elongation.
These load deformation characteristics are dependent on the specimenThese load–deformation characteristics are dependent on the specimen
size.
Example it will require twice the load to produce the same elongation if theExample, it will require twice the load to produce the same elongation if the
cross-sectional area of the specimen is doubled.
To minimize these geometrical factors, load and elongation areg , g
normalized to the respective parameters of engineering stress and
engineering strain.
Engineering stress σ is defined by the relationship
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
i hi h F i th i t t l d li d di l t th
in which F is the instantaneous load applied perpendicular to the
specimen cross section, in units of newtons (N) or pounds force (lbf), and
Ao is the original cross sectional area before any load is applied (m2 or
in 2)in.2).
The units of engineering stress (referred to subsequently as just stress) are
megapascals MPa (SI) (where 1 MPa 106 N/m2) and pounds force permegapascals, MPa (SI) (where 1 MPa 10 N/m ), and pounds force per
square inch, psi.
145 psi = 1 MPa.p
Engineering strain ε is defined according to
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
in lo which is the original length before any load is applied, and li is the
instantaneous length.
S ti th tit (l l ) i d t d Δl d i th d f tiSometimes the quantity (li - lo) is denoted as Δl and is the deformation
elongation or change in length at some instant, as referenced to the original
length.
Engineering strain (subsequently called just strain) is unitless, but meters
per meter or inches per inch are often used; the value of strain is obviously
independent of the unit system.independent of the unit system.
Sometimes strain is also expressed as a percentage, in which the strain
value is multiplied by 100.y
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Compression Tests
A compression test is conducted in a manner similar to the tensile test,
except that the force is compressive and the specimen contracts alongexcept that the force is compressive and the specimen contracts along
the direction of the stress.
Above equations are utilized to compute compressive stress and strain.q p p
By convention, a compressive force is taken to be negative, which yields a
negative stress.
Furthermore, since lo is greater than li , compressive strains computed
above equation are necessarily also negative.
Tensile tests are more common because:
They are easier to perform.
Very little additional information is obtained from compressive tests
www.alquds.edu
Very little additional information is obtained from compressive tests.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Shear and Torsional Tests
The shear stress т is,
where F is the load or force imposed parallel to the upper and lower faceswhere F is the load or force imposed parallel to the upper and lower faces,
each of which has an area of Ao.
The shear strain γ is defined as the tangent of the strain angle θ, asγ g g ,
indicated in the figure.
The units for shear stress and strain are the same as for their tensile
counterparts.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Torsion is a variation of pure shear, wherein a structural member is
twisted in the manner of Figure 6.1d.
Torsional forces produce a rotational motion about the longitudinal axis
of one end of the member relative to the other end.
Examples of torsion are found for machine axles and drive shafts, and also
for twist drillsfor twist drills.
Torsional tests are normally performed on cylindrical solid shafts orTorsional tests are normally performed on cylindrical solid shafts or
tubes.
A shear stress т is a function of the applied torque T, whereas shear
www.alquds.edu
pp q ,
strain γ is related to the angle of twist, in Figure 6.1d.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Geometric Considerations of the Stress State
The stress state is a function of the orientations of the
planes upon which the stresses are taken to actplanes upon which the stresses are taken to act.
Example: cylindrical tensile specimen that is subjected to
a tensile stress σ applied parallel to its axis.a tensile stress σ applied parallel to its axis.
Consider also the plane p-p’ that is oriented at some
arbitrary angle θ relative to the plane of the specimen
end-face.
Upon this plane p-p’ , the applied stress is no longer a
ilpure tensile one.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Rather, a more complex stress state is present that consists of a tensile
(or normal) stress σ’ that acts normal to the plane.
In addition a shear stress ’ that acts parallel to this plane both of theseIn addition, a shear stress т’ that acts parallel to this plane; both of these
stresses are represented in the figure.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
STRESS–STRAIN BEHAVIOR
Stress and strain are proportional to each other through the relationship:
This is known as Hooke’s law, and the constant of proportionality E (GPa
or psi) is the modulus of elasticity, or Young’s modulus.
Modulus of elasticity values for several metals at room temperature are
presented in Table 6.1.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Deformation in which stress and strain are proportional is called elastic
deformation.
a plot of stress (ordinate) versus strain (abscissa) results in a linear
relationship, as shown in Figure 6.5.
The slope of this linear segment corresponds to the modulus of elasticity
E.
Thi d l b th ht f tiff t i l’ i t tThis modulus may be thought of as stiffness, or a material’s resistance to
elastic deformation.
The greater the modulus the stiffer the material or the smaller the elasticThe greater the modulus, the stiffer the material, or the smaller the elastic
strain that results from the application of a given stress.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6 5Figure 6.5
Schematic stress–
strain diagramstrain diagram
showing linear
elastic deformation for
loading and unloadingg g
cycles.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Elastic deformation is nonpermanent, which means that when the
applied load is released, the piece returns to its original shape.
A h i th t t i l t (Fi 6 5) li ti f th l dAs shown in the stress–strain plot (Figure 6.5), application of the load
corresponds to moving from the origin up and along the straight line.
Upon release of the load the line is traversed in the opposite directionUpon release of the load, the line is traversed in the opposite direction,
back to the origin.
There are some materials (e g gray cast iron concrete and manyThere are some materials (e.g., gray cast iron, concrete, and many
polymers) for which this elastic portion of the stress–strain curve is not
linear (Figure 6.6);
Hence, it is not possible to determine a modulus of elasticity as
described above.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.6
Schematic
stress–strain diagram
showing
non-linear elastic
behavior, and
how secant and tangent
moduli
d t i dare determined.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
For this nonlinear behavior, either tangent or secant modulus is
normally used, Figure 6.6..
Tangent modulus is taken as the slope of the stress–strain curve at
some specified level of stress.
Secant modulus represents the slope of a secant drawn from the originSecant modulus represents the slope of a secant drawn from the origin
to some given point of the σ-ε curve.
On an atomic scale the magnitude of the modulus of elasticity is aOn an atomic scale, the magnitude of the modulus of elasticity is a
measure of the resistance to separation of adjacent atoms, that is, the
interatomic bonding forces.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.7 Force versus interatomic separation for weakly and strongly
www.alquds.edu
Figure 6.7 Force versus interatomic separation for weakly and strongly
bonded atoms. The magnitude of the modulus of elasticity is proportional to
the slope of each curve at the equilibrium interatomic separation r0.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
With increasing temperature, the modulus of elasticity diminishes, as is
shown for several metals in Figure 6.8.
The imposition of compressi e shear or torsional stresses also e okesThe imposition of compressive, shear, or torsional stresses also evokes
elastic behavior.
The stress–strain characteristics at low stress levels are virtually theThe stress–strain characteristics at low stress levels are virtually the
same for both tensile and compressive situations, to include the
magnitude of the modulus of elasticity.
Shear stress and strain are proportional to each other through the
expression,
where G is the shear modulus, the slope of the linear elastic region of the
h i T bl 6 1
www.alquds.edu
shear stress–strain curve, Table 6.1.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.8
Plot of
modulus of
elasticityelasticity
versus
temperature
for
tungsten,
steel,
and
aluminum.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
ANELASTICITY
Elastic deformation is time independent: that an applied stress produces
an instantaneous elastic strain that remains constant over the period ofan instantaneous elastic strain that remains constant over the period of
time the stress is maintained.
It has also been assumed that upon release of the load the strain is totallyy
recovered: that the strain immediately returns to zero.
Anelasticity is time-dependent elastic deformation will continue after the
stress application, and upon load release some finite time is required for
complete recovery.
For metals the anelastic component is normally small and is oftenFor metals the anelastic component is normally small and is often
neglected.
For some polymeric materials its magnitude is significant; in this case it
www.alquds.edu
For some polymeric materials its magnitude is significant; in this case it
is termed viscoelastic behavior.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Ex: A piece of copper originally 305 mm (12 in.) long is pulled in tension with
a stress of 276 MPa (40,000 psi). If the deformation is entirely elastic, what
will be the resultant elongation?
Th l f d l i 276 MP d 305 ti l dThe values of σ and lo are given as 276 MPa and 305 mm, respectively, and
the magnitude of E for copper from Table 6.1 is 110 GPa (16x106 psi).
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
ELASTIC PROPERTIES OF MATERIALS
When a tensile stress is imposed on a metal specimen, an elastic
elongation and accompanying strain, εz result in the direction of thee o gat o a d acco pa y g st a , εz esu e d ect o o e
applied stress, Figure 6.9.
As a result of this elongation, there will be constrictions in the lateral (x
and y) directions perpendicular to the applied stress; from these
contractions, the compressive strains εx and εy may be determined.
If th li d t i i i l ( l i th di ti ) d th t i l iIf the applied stress is uniaxial (only in the z direction), and the material is
isotropic,
Poisson’s ratio (v) is defined as the ratio of the lateral and axial strains,
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.9
Axial (z) elongation
(positive strain) and lateral (x
and y) contractions (negative
strains) in response to anstrains) in response to an
imposed tensile stress.
Solid lines represent
dimensions afterdimensions after
stress application; dashed
lines, before.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
The negative sign is included in the expression so that Poisson’s ratio will
always be positive, since εx and εy will always be of opposite sign.
Theoretically Poisson’s ratio for isotropic materials should be 1/4Theoretically, Poisson’s ratio for isotropic materials should be 1/4.
The maximum value for Poisson’s ratio (or that value for which there is no
net volume change) is 0 50net volume change) is 0.50.
For many metals and other alloys, values of Poisson’s ratio range
between 0.25 and 0.35, Table 6.1.,
For isotropic materials, shear modulus G and elastic modulus E are
related to each other and to Poisson’s ratio according to
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
In most metals, G is about 0.4E; thus, if the value of one modulus is known,
the other may be approximated.
Ex: A tensile stress is to be applied
along the long axis of a cylindrical
brass rod that has a diameter of 10
mm (0.4 in.).
Determine the magnitude of the load
required to produce a 2.5x10-3 mm (
10-4 in.) change in diameter if the
deformation is entirely elastic.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
The value for Poisson’s ratio for brass is 0.34 (Table 6.1), and thus
The modulus of elasticity E, given in Table 6.1 as 97 GPa
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Plastic Deformation
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 (Hooke’s law, ceases to be valid), and permanent,
nonrecoverable, or plastic deformation occurs, Figure 6.10a.
From an atomic perspective, plastic deformation corresponds to the
b ki f b d ith i i l t i hb d th f ibreaking of bonds with original atom neighbors and then reforming
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 positionstheir original positions.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.10
Figure 6.10
(a) Typical stress–
strain behavior for
a metal showing
l ti d l tielastic and plastic
deformations, the
proportional limit P,
and the yieldy
strength
as determined
using the 0.002
strain offset methodstrain offset method.
(b) Representative
stress–strain
behavior
found for some
steels demonstrating
the yield point
www.alquds.edu
the yield point
phenomenon.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
TENSILE PROPERTIES
1) Yielding and Yield Strength
Most structures are designed to ensure that only elastic deformation will
result when a stress is applied.
A structure or component that has plastically deformed, or experienced a
permanent change in shape, may not be capable of functioning as intended.
It i th f d i bl t k th t l l t hi h l tiIt is therefore desirable to know the stress level at which plastic
deformation begins, or where the phenomenon of yielding occurs.
For metals the point of yielding may be determined as the initialFor metals, the point of yielding may be determined as the initial
departure from linearity of the stress–strain curve; this is sometimes called
the proportional limit, as indicated by point P in Figure 6.10a.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
The position of this point may not be determined precisely.
So, a convention has been established wherein a straight line is
t t d ll l t th l ti ti f th t t i tconstructed parallel to the elastic portion of the stress–strain curve at
some specified strain offset, usually 0.002.
The stress corresponding to the intersection of this line and the stressThe stress corresponding to the intersection of this line and the stress–
strain curve as it bends over in the plastic region is defined as the yield
strength (σy) and its unit is MPa or psi.
The magnitude of the yield strength for a metal is a measure of its
resistance to plastic deformation.
Yield strengths may range from 35 MPa (5000 psi) for a low strength
aluminum to over 1400 MPa (200,000 psi) for high-strength steels.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
2) Tensile Strength
After yielding, the stress necessary to continue plastic deformation in
metals increases to a maximum point M (Figure 6 11) and then decreasesmetals increases to a maximum, point M (Figure 6.11) and then decreases
to the eventual fracture, point F.
The tensile strength (TS) (MPa or psi) is the stress at the maximum ong ( ) ( p )
the engineering stress–strain curve (Figure 6.11).
This corresponds to the maximum stress that can be sustained by a
structure in tension; if this stress is applied and maintained, fracture will
result.
All d f ti t thi i t i if th h t th i fAll deformation up to this point is uniform throughout the narrow region of
the tensile specimen.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.11
Typical
engineering
stress–strain
behavior to
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.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
However, at this maximum stress, a small constriction or neck begins to
form at some point, and all subsequent deformation is confined at this neck,
Figure 6.11.
This phenomenon is termed “necking” and fracture ultimately occurs at the
neck.
The fracture strength corresponds to the stress at fracture.
Tensile strengths may vary anywhere from 50 MPa (7000 psi) for anTensile strengths may vary anywhere from 50 MPa (7000 psi) for an
aluminum to as high as 3000 MPa (450,000 psi) for the high-strength steels.
When the strength of a metal is cited for design purposes, the yieldWhen the strength of a metal is cited for design purposes, the yield
strength is used. This is because by the time a stress corresponding to the
tensile strength has been applied, often a structure has experienced so
much plastic deformation that it is useless. Furthermore, fracture
www.alquds.edu
p
strengths are not normally specified for engineering design purposes.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Ex: From the tensile stress–strain behavior for the brass specimen shown in
Figure 6.12, determine the following:
(a) The modulus of elasticity(a) The modulus of elasticity.
(b) The yield strength at a strain offset of 0.002.
(c) The maximum load that can be sustained by a cylindrical specimen
having an original diameter of 12 8 mm (0 505 in )having an original diameter of 12.8 mm (0.505 in.).
(d) The change in length of a specimen originally 250 mm (10 in.) long that
is subjected to a tensile stress of 345 MPa (50,000 psi).
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
(b) The 0.002 strain offset line is constructed as shown in the inset; its
intersection with the stress–strain curve is at approximately 250 MPa
(36,000 psi), which is the yield strength of the brass.
( c) The maximum load that can be sustained by the specimen is calculated
in which tensile strength from the figure is 450 MPa (65,000 psi). Solving for
F th i l d i ldF, the maximum load, yields
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
(d) T t th h i l th Δl it i fi t t d t i th
(d) To compute the change in length Δl, it is first necessary to determine the
strain that is produced by a stress of 345 MPa.
This is accomplished by locating the stress point on the stress strain curveThis is accomplished by locating the stress point on the stress–strain curve,
point A, and reading the corresponding strain from the strain axis, which is
approximately 0.06. Inasmuch as lo= 250 mm, we have
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
3 D tilit
3. Ductility
It is a measure of the degree of plastic deformation that has been
sustained at fracture.
A material that experiences very little or no plastic deformation upon
fracture is termed brittle, Figure 6.13.
Ductility may be expressed quantitatively as either percent elongation or
percent reduction in area.
1) The percent elongation %EL is the percentage of plastic strain at
fracture, or
www.alquds.edu
lf : fracture length, lo : the original gauge length.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.13
SchematicSchematic
representations
of tensile stress–
strain behavior
for brittle and
ductile materials
loaded to
ffracture.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
I h i ifi t ti f th l ti d f ti t f t i
Inasmuch as a significant proportion of the plastic deformation at fracture is
confined to the neck region, the magnitude of %EL will depend on
specimen gauge length.
The shorter lo , the greater is the fraction of total elongation from the neck
and, consequently, the higher the value of %EL.
Therefore, lo should be specified when percent elongation values are cited;
it is commonly 50 mm (2 in.).
2) Percent reduction in area %RA is defined as
Ao: the original cross-sectional area.
A th ti l t th i t f f t
www.alquds.edu
Af: the cross-sectional area at the point of fracture.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
A knowledge of the ductility of materials is important for at least two
reasons:
1) It indicates to a designer the degree to which a structure will deform
plastically before fracture.
2) It ifi th d f ll bl d f ti d i f b i ti2) It specifies the degree of allowable deformation during fabrication
operations.
Brittle materials are approximately considered to be those having aBrittle materials are approximately considered to be those having a
fracture strain of less than about 5%.
Table 6.2: Yield strength tensile strength and ductility are sensitive toTable 6.2: Yield strength, tensile strength, and ductility are sensitive to
any prior deformation, the presence of impurities, and/or any heat
treatment to which the metal has been subjected.
www.alquds.edu
Where modulus of elasticity is insensitive to these treatments.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Th it d f b th i ld d t il t th d li ith
The magnitudes of both yield and tensile strengths decline with
increasing temperature; just the reverse holds for ductility—it usually
increases with temperature, Figure 6.14.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.14 Engineering stress–strain behavior for iron at three
www.alquds.edu
g g g
temperatures

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
4. Resilience
Is the capacity of a material to absorb energy when it is deformed
elastically and then, upon unloading, to have this energy recovered.
The associated property is the modulus of resilience Ur, which is the strain
energy per unit volume required to stress a material from an unloaded state
t th i t f i ldi Fi 6 15up to the point of yielding, Figure 6.15.
Assuming a linear elastic regionAssuming a linear elastic region,
www.alquds.edu
in which εy is the strain at yielding.

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
Figure 6.15
Schematic
representation
showing how
modulus of
iliresilience
(corresponding to
the shaded area) is
determineddetermined
from the tensile
stress–strain
behavior of abehavior of a
material.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
5. Toughness
It is a measure of the ability of a material to absorb energy up to fracture.
It is the area under the stress–strain curve up to the point of fracture.p p
The units for toughness are the same as for resilience (i.e., energy per unit
volume of material).
For a material to be tough, it must display both strength and ductility;
often, ductile materials are tougher than brittle ones, Figure 6.13.
Hence, even though the brittle material has higher yield and tensile
strengths, it has a lower toughness than the ductile one, Figure 6.13.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
HARDNESS
is a measure of a material’s resistance to localized plastic deformation
(e.g., a small dent or a scratch).( g )
It is measured by Mohs scale, which ranged from 1 on the soft end for talc
to 10 for diamond.
A small indenter is forced into the surface of a material to be tested, under
controlled conditions of load and rate of application.
The depth or size of the resulting indentation is measured, which in turn is
related to a hardness number; the softer the material, the larger and
deeper is the indentation and the lower the hardness index numberdeeper is the indentation, and the lower the hardness index number.
www.alquds.edu

ALQUDS
UNIVERSITY
CHEMISTRY OF
MATERIALS
(a) Briefly explain the concept of a driving force.
(b) What is the driving force for steady-state
diffusion?diffusion?
(a) The driving force is that which compels a reaction to
occur.
(b) The driving force for steady-state diffusion is the
www.alquds.edu
( ) g y
concentration gradient.

Ch6

More Related Content

What's hot (20)

Similar to Ch6 (20)

Recently uploaded (20)

Ch6