Santa Ana College Mechanics of Material and Polymer Structures Questionnaire

Test No. 1- Take-Home PartSolve all problems. Each problem carries 20 points.
Problem No.1
In reference to the figure shown, Calculate the following parameters. Show all of your calculations:
•
•
•
•
•
Ultimate tensile strength (σuts)
Young’s modulus (E)
Yield Stress (σys)
Fracture stress (σf)
Modulus of Resilience (Ur)
Problem No. 2
The following are information about a tensile test performed on a cylindrical specimen of stainless steel:
•
•
•
Diameter = 12.8 mm
Gauge length = 50.800 mm
Test: Tensile
By using this data and the given test results:
1.
2.
3.
4.
5.
Plot the engineering stress vs. engineering strain graph
Calculate the modulus of elasticity
Calculate the yield stress at 0.002 strain
What is the ultimate tensile strength?
Approximate the toughness of the material
Problem No. 3
Study the Design Example 9.1. Page 336 of the textbook and answer the following questions:
(a) What is the difference between hoop stress and the longitudinal stress? Which one is higher and
is being used in engineering calculations?
(b) What does the “leak before the break” mean?
(c) Why Cc≥ t?
(d) Calculate Cc for Steel alloy 1040 and compare your value with the one listed in Table 9.3, Page
339.
Problem No. 4
The following figure (Larson-Miller chart) is for an 18-8 Mo stainless steel that is exposed to a
temperature of 650 degree centigrade. calculate the time to rupture for this material if the stress is 100
MPa.
Chapter
7
Mechanical Properties
2000
TS
(a)
Stress (MPa)
σy
1000
Stress (MPa)
Model H300KU Universal Testing Machine by Tinius Olsen
2000
E
1000
0
0
0.000
0
0.010
Strain
0.040
0.080
Strain
(b)
F
igure (a) shows an apparatus that measures the mechanical
properties of metals using applied tensile forces (Sections 7.3, 7.5,
and 7.6). Figure (b) was generated from a tensile test performed by
an apparatus such as this on a steel specimen. Data plotted are stress
(vertical axis—a measure of applied force)
versus strain (horizontal axis—related to
the degree of specimen elongation).
The mechanical properties of modulus
of elasticity (stiffness, E), yield strength
(σy), and tensile strength (TS) are
determined as noted on these graphs.
Figure (c) shows a suspension
bridge. The weight of the bridge deck and
automobiles imposes tensile forces on the
vertical suspender cables. These forces are
transferred to the main suspension cable,
which sags in a more-or-less parabolic
shape. The metal alloy(s) from which these
© Mr. Focus/iStockphoto
cables are constructed must meet certain
(c)
216 •
stiffness and strength criteria. Stiffness and
strength of the alloy(s) may be assessed
from tests performed using a tensile-testing
apparatus (and the resulting stress–strain
plots) similar to those shown.
WHY STUDY Mechanical Properties?
It is incumbent on engineers to understand how the
various mechanical properties are measured and what
these properties represent; they may be called upon
to design structures/components using predetermined
materials such that unacceptable levels of deformation
and/or failure will not occur. We demonstrate this
procedure with respect to the design of a tensile-testing
apparatus in Design Example 7.1.
Learning Objectives
After studying this chapter, you should be able to do the following:
1. Define engineering stress and engineering
strain.
2. State Hooke’s law and note the conditions
under which it is valid.
3. Define Poisson’s ratio.
4. Given an engineering stress–strain diagram,
determine (a) the modulus of elasticity,
(b) the yield strength (0.002 strain offset),
and (c) the tensile strength and (d) estimate
the percentage elongation.
5. For the tensile deformation of a ductile
cylindrical metal specimen, describe
changes in specimen profile to the point
of fracture.
6. Compute ductility in terms of both percentage
elongation and percentage reduction of
area for a material that is loaded in tension
to fracture.
7.1
7. For a specimen being loaded in tension,
given the applied load, the instantaneous
cross-sectional dimensions, and original and
instantaneous lengths, compute true stress
and true strain values.
8. Compute the flexural strengths of ceramic rod
specimens that have been bent to fracture in
three-point loading.
9. Make schematic plots of the three characteristic
stress–strain behaviors observed for polymeric
materials.
10. Name the two most common hardness-testing
techniques; note two differences between them.
11. (a) Name and briefly describe the two different
microindentation hardness testing techniques,
and (b) cite situations for which these techniques
are generally used.
12. Compute the working stress for a ductile material.
INTRODUCTION
Many materials are subjected to forces or loads when in service; examples include
the aluminum alloy from which an airplane wing is constructed and the steel in an
automobile axle. In such situations it is necessary to know the characteristics of the
material and to design the member from which it is made so that any resulting deformation will not be excessive and fracture will not occur. The mechanical behavior
of a material reflects its response or deformation in relation to an applied load or
force. Key mechanical design properties are stiffness, strength, hardness, ductility,
and toughness.
The mechanical properties of materials are ascertained by performing carefully designed laboratory experiments that replicate as nearly as possible the service conditions.
Factors to be considered include the nature of the applied load and its duration, as well
as the environmental conditions. It is possible for the load to be tensile, compressive,
or shear, and its magnitude may be constant with time or may fluctuate continuously.
Application time may be only a fraction of a second, or it may extend over a period of
many years. Service temperature may be an important factor.
Mechanical properties are of concern to a variety of parties (e.g., producers and
consumers of materials, research organizations, government agencies) that have
differing interests. Consequently, it is imperative that there be some consistency in
• 217
218 • Chapter 7
/
Mechanical Properties
the manner in which tests are conducted and in the interpretation of their results. This
consistency is accomplished by using standardized testing techniques. Establishment
and publication of these standards are often coordinated by professional societies. In
the United States the most active organization is the American Society for Testing and
Materials (ASTM). Its Annual Book of ASTM Standards (http://www.astm.org) comprises numerous volumes that are issued and updated yearly; a large number of these
standards relate to mechanical testing techniques. Several of these are referenced by
footnote in this and subsequent chapters.
The role of structural engineers is to determine stresses and stress distributions
within members that are subjected to well-defined loads. This may be accomplished
by experimental testing techniques and/or by theoretical and mathematical stress
analyses. These topics are treated in traditional texts on stress analysis and strength
of materials.
Materials and metallurgical engineers however, are concerned with producing and
fabricating materials to meet service requirements as predicted by these stress analyses.
This necessarily involves an understanding of the relationships between the microstructure
(i.e., internal features) of materials and their mechanical properties.
Materials are frequently chosen for structural applications because they have desirable combinations of mechanical characteristics. This chapter discusses the stress–strain
behaviors of metals, ceramics, and polymers and the related mechanical properties; it
also examines other important mechanical characteristics. Discussions of the microscopic aspects of deformation mechanisms and methods to strengthen and regulate the
mechanical behaviors are deferred to Chapter 8.
7.2 CONCEPTS OF STRESS AND STRAIN
If a load is static or changes relatively slowly with time and is applied uniformly over
a cross section or surface of a member, the mechanical behavior may be ascertained
by a simple stress–strain test; these are most commonly conducted for metals at
room temperature. There are three principal ways in which a load may be applied:
namely, tension, compression, and shear (Figures 7.1a, b, c). In engineering practice
many loads are torsional rather than pure shear; this type of loading is illustrated in
Figure 7.1d.
Tension Tests1
One of the most common mechanical stress–strain tests is performed in tension. As will be
seen, the tension test can be used to ascertain several mechanical properties of materials
that are important in design. 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 7.2. Normally, the cross section is circular, but
rectangular specimens are also used. This “dogbone” specimen configuration was chosen
so that, during testing, deformation is confined to the narrow center region (which has a
uniform cross section along its length) and also to reduce the likelihood of fracture at the
ends of the specimen. The standard diameter is approximately 12.8 mm (0.5 in.), whereas
1
the reduced section length should be at least four times this diameter; 60 mm (2 4 in.) is
common. Gauge length is used in ductility computations, as discussed in Section 7.6; the
standard value is 50 mm (2.0 in.). The specimen is mounted by its ends into the holding
1
ASTM Standards E8 and E8M, “Standard Test Methods for Tension Testing of Metallic Materials.”
7.2 Concepts of Stress and Strain • 219
Figure 7.1
(a) Schematic
illustration of how a
tensile load produces
an elongation and
positive linear strain.
(b) Schematic
illustration of how a
compressive load
produces contraction
and a negative linear
strain. (c) Schematic
representation
of shear strain γ,
where γ = tan θ.
(d) Schematic
representation
of torsional
deformation (i.e.,
angle of twist ϕ)
produced by an
applied torque T.
F
A0
A0
l0
l
F
l0
l
F
F
(a)
(b)
A0
A0
F
F
!
F
(c)
T
”
T
(d)
grips of the testing apparatus (Figure 7.3). The tensile testing machine is designed to elongate the specimen at a constant rate and to continuously and simultaneously measure the
instantaneous applied load (with a load cell) and the resulting elongation (using an extensometer). A stress–strain test typically takes several minutes to perform and is destructive; that is, the test specimen is permanently deformed and usually fractured. [Chapteropening photograph (a) for this chapter is of a modern tensile-testing apparatus.]
220 • Chapter 7
/
Mechanical Properties
Figure 7.2 A standard tensile specimen
with circular cross section.
Reduced section
1
2 ”
4
3″
Diameter
4
0.505″ Diameter
2″
Gauge length
engineering stress
engineering strain
Definition of
engineering stress
(for tension and
compression)
Definition of
engineering strain
(for tension and
compression)
Tutorial Video:
What Are the
Differences between
Stress and Strain?
3″
8
Radius
The output of such a tensile test is recorded (usually on a computer) as load or
force versus elongation. These load–deformation characteristics depend on the specimen size. For example, it requires 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 are normalized to the respective parameters of engineering stress
and engineering strain. Engineering stress σ is defined by the relationship
σ=
F
A0
(7.1)
in which F is the instantaneous load applied perpendicular to the specimen cross section, in units of newtons (N) or pounds force (lbf), and A0 is the original cross-sectional
area before any load is applied (m2 or 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 per square inch, psi (customary U.S.).2
Engineering strain ε is defined according to
ε=
li − l0
∆l
=
l0
l0
(7.2)
in which l0 is the original length before any load is applied and li is the instantaneous length. Sometimes the quantity li − l0 is denoted as ∆l 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 is often used; the value of strain is obviously independent of the unit
Figure 7.3 Schematic representation of
the apparatus used to conduct tensile stress–
strain tests. The specimen is elongated by the
moving crosshead; load cell and extensometer
measure, respectively, the magnitude of the
applied load and the elongation.
(Adapted from H. W. Hayden, W. G. Moffatt, and
J. Wulff, The Structure and Properties of Materials,
Vol. III, Mechanical Behavior, p. 2. Copyright
© 1965 by John Wiley & Sons, New York. Reprinted
by permission of John Wiley & Sons, Inc.)
Load cell
Extensometer
Specimen
Moving
crosshead
2
Conversion from one system of stress units to the other is accomplished by the relationship 145 psi = 1 MPa.
7.2 Concepts of Stress and Strain • 221
system. Sometimes strain is also expressed as a percentage, in which the strain value is
multiplied by 100.
Compression Tests3
Compression stress–strain tests may be conducted if in-service forces are of this
type. A compression test is conducted in a manner similar to the tensile test, except
that the force is compressive and the specimen contracts along the direction of the
stress. Equations 7.1 and 7.2 are utilized to compute compressive stress and strain,
respectively. By convention, a compressive force is taken to be negative, which yields
a negative stress. Furthermore, because l0 is greater than li, compressive strains computed from Equation 7.2 are necessarily also negative. Tensile tests are more common because they are easier to perform; also, for most materials used in structural
applications, very little additional information is obtained from compressive tests.
Compressive tests are used when a material’s behavior under large and permanent
(i.e., plastic) strains is desired, as in manufacturing applications, or when the material
is brittle in tension.
Shear and Torsional Tests4
Definition of shear
stress
𝜎
p’
𝜏’
𝜎’
θ
p
𝜎
Figure 7.4
Schematic
representation showing normal (σ′) and
shear (τ′) stresses
that act on a plane
oriented at an angle θ
relative to the plane
taken perpendicular
to the direction along
which a pure tensile
stress (σ) is applied.
3
For tests performed using a pure shear force as shown in Figure 7.1c, the shear stress τ
is computed according to
τ=
F
A0
(7.3)
where F is the load or force imposed parallel to the upper and lower faces, each of which
has an area of A0. The shear strain γ is defined as the tangent of the strain angle θ, as
indicated in the figure. The units for shear stress and strain are the same as for their
tensile counterparts.
Torsion is a variation of pure shear in which a structural member is twisted in the
manner of Figure 7.1d; 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 as well as for twist drills. Torsional tests
are normally performed on cylindrical solid shafts or tubes. A shear stress τ is a function of the applied torque T, whereas shear strain γ is related to the angle of twist, ϕ
in Figure 7.1d.
Geometric Considerations of the Stress State
Stresses that are computed from the tensile, compressive, shear, and torsional force
states represented in Figure 7.1 act either parallel or perpendicular to planar faces
of the bodies represented in these illustrations. Note that the stress state is a function of the orientations of the planes upon which the stresses are taken to act. For
example, consider the cylindrical tensile specimen of Figure 7.4 that is subjected to a
tensile stress σ applied parallel to its axis. Furthermore, consider also the plane p-p′
that is oriented at some arbitrary angle θ relative to the plane of the specimen endface. Upon this plane p-p′, the applied stress is no longer a pure tensile one. Rather,
ASTM Standard E9, “Standard Test Methods of Compression Testing of Metallic Materials at Room Temperature.”
ASTM Standard E143, “Standard Test Method for Shear Modulus at Room Temperature.”
4
222 • Chapter 7
/
Mechanical Properties
a more complex stress state is present that consists of a tensile (or normal) stress σ′
that acts normal to the p-p′ plane and, in addition, a shear stress τ′ that acts parallel
to this plane; both of these stresses are represented in the figure. Using mechanicsof-materials principles,5 it is possible to develop equations for σ′ and τ′ in terms of σ
and θ, as follows:
σ′ = σ cos2 θ = σ
(
1 + cos 2θ
)
2
(7.4a)
sin 2θ
2 )
(7.4b)
τ′ = σ sin θ cos θ = σ (
These same mechanics principles allow the transformation of stress components from
one coordinate system to another coordinate system with a different orientation. Such
treatments are beyond the scope of the present discussion.
Elastic Deformation
7.3
STRESS–STRAIN BEHAVIOR
Hooke’s law—
relationship between
engineering stress
and engineering
strain for elastic
deformation (tension
and compression)
modulus of elasticity
elastic deformation
: VMSE
Metal Alloys
Tutorial Video:
Calculating Elastic
Modulus Using a
Stress vs. Strain Curve
5
The degree to which a structure deforms or strains depends on the magnitude of an
imposed stress. For most metals that are stressed in tension and at relatively low levels,
stress and strain are proportional to each other through the relationship
σ = Eε
(7.5)
This is known as Hooke’s law, and the constant of proportionality E (GPa or psi)6 is the
modulus of elasticity, or Young’s modulus. For most typical metals, the magnitude of this
modulus ranges between 45 GPa (6.5 × 106 psi), for magnesium, and 407 GPa (59 × 106 psi),
for tungsten. The moduli of elasticity are slightly higher for ceramic materials and range
between about 70 and 500 GPa (10 × 106 and 70 × 106 psi). Polymers have modulus
values that are smaller than those of both metals and ceramics and lie in the range 0.007
to 4 GPa (103 to 0.6 × 106 psi). Room-temperature modulus of elasticity values for a number
of metals, ceramics, and polymers are presented in Table 7.1. A more comprehensive
modulus list is provided in Table B.2, Appendix B.
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 7.5. The slope of this linear segment corresponds to the
modulus of elasticity E. This modulus may be thought of as stiffness, or a material’s
resistance to elastic deformation. The greater the modulus, the stiffer is the material,
or the smaller is the elastic strain that results from the application of a given stress. The
modulus is an important design parameter for computing elastic deflections.
Elastic deformation is nonpermanent, which means that when the applied load
is released, the piece returns to its original shape. As shown in the stress–strain plot
(Figure 7.5), 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 direction,
back to the origin.
See, for example, W. F. Riley, L. D. Sturges, and D. H. Morris, Mechanics of Materials, 6th edition, Wiley, Hoboken,
NJ, 2006.
6
The SI unit for the modulus of elasticity is gigapascal (GPa) where 1 GPa = 109 N/m2 = 103 MPa.
7.3 Stress–Strain Behavior • 223
Table 7.1 Room-Temperature Elastic and Shear Moduli and Poisson’s Ratio for Various Materials
Modulus of Elasticity
Shear Modulus
GPa
106 psi
Poisson’s Ratio
59
160
23.2
0.28
207
30
83
12.0
0.30
Nickel
207
30
76
11.0
0.31
Titanium
107
15.5
45
6.5
0.34
Copper
110
16
46
6.7
0.34
Brass
97
14
37
5.4
0.34
Aluminum
69
10
Magnesium
45
Material
GPa
Tungsten
407
Steel
6
10 psi
Metal Alloys
6.5
25
3.6
0.33
17
2.5
0.35
—
—
0.22
Ceramic Materials
Aluminum oxide (Al2O3)
393
57
Silicon carbide (SiC)
345
50
—
—
0.17
Silicon nitride (Si3N4)
304
44
—
—
0.30
Spinel (MgAl2O4)
260
38
—
—
—
Magnesium oxide (MgO)
225
33
—
—
0.18
Zirconia (ZrO2)a
205
30
—
—
0.31
Mullite (3Al2O3-2SiO2)
145
21
—
—
0.24
Glass–ceramic (Pyroceram)
120
17
—
—
0.25
Fused silica (SiO2)
73
11
—
—
0.17
Soda–lime glass
69
10
—
—
0.23
Polymers
b
Phenol-formaldehyde
2.76–4.83
0.40–0.70
—
—
—
Poly(vinyl chloride) (PVC)
2.41–4.14
0.35–0.60
—
—
0.38
Poly(ethylene terephthalate) (PET)
2.76–4.14
0.40–0.60
—
—
0.33
Polystyrene (PS)
2.28–3.28
0.33–0.48
—
—
0.33
Poly(methyl methacrylate) (PMMA)
2.24–3.24
0.33–0.47
—
—
0.37–0.44
2.38
0.35
—
—
0.36
Nylon 6,6
1.59–3.79
0.23–0.55
—
—
0.39
Polypropylene (PP)
1.14–1.55
0.17–0.23
—
—
0.40
1.08
0.16
—
—
0.46
Polytetrafluoroethylene (PTFE)
0.40–0.55
0.058–0.080
—
—
0.46
Polyethylene—low density (LDPE)
0.17–0.28
0.025–0.041
—
—
0.33–0.40
Polycarbonate (PC)
Polyethylene—high density (HDPE)
a
Partially stabilized with 3 mol% Y2O3.
Modern Plastics Encyclopedia ’96, McGraw-Hill, New York, 1995.
b
There are some materials (i.e., gray cast iron, concrete, and many polymers) for
which this elastic portion of the stress–strain curve is not linear (Figure 7.6); hence, it is
not possible to determine a modulus of elasticity as described previously. For this nonlinear behavior, either the tangent or secant modulus is normally used. The tangent modulus
is taken as the slope of the stress–strain curve at some specified level of stress, whereas
224 • Chapter 7
/
Mechanical Properties
#2
%#
%$
Stress #
Stress
Unload
Slope = modulus
of elasticity
= Tangent modulus (at #2)
#1
%#
%$
Load
= Secant modulus
(between origin and #1)
0
0
Strain
Figure 7.5 Schematic
Strain $
stress–strain diagram
showing linear elastic
deformation for loading
and unloading cycles.
Figure 7.6 Schematic stress–strain diagram showing
nonlinear elastic behavior and how secant and tangent
moduli are determined.
the secant modulus represents the slope of a secant drawn from the origin to some given
point of the σ–ε curve. The determination of these moduli is illustrated in Figure 7.6.
On an atomic scale, macroscopic elastic strain is manifested as small changes in
the interatomic spacing and the stretching of interatomic bonds. As a consequence, the
magnitude of the modulus of elasticity is a measure of the resistance to separation of
adjacent atoms, that is, the interatomic bonding forces. Furthermore, this modulus is
proportional to the slope of the interatomic force–separation curve (Figure 2.10a) at
the equilibrium spacing:
E∝(
dF
dr )r0
(7.6)
Figure 7.7 shows the force–separation curves for materials having both strong and weak
interatomic bonds; the slope at r0 is indicated for each.
Figure 7.7 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.
Force F
Strongly
bonded
dF
dr r
0
Separation r
0
Weakly
bonded
7.4 Anelasticity • 225
Figure 7.8 Plot of modulus of elasticity
Temperature (°F)
–400
0
400
1200
1600
300
40
Steel
200
30
20
100
Aluminum
–200
0
200
Modulus of elasticity (106 psi)
Tungsten
50
0
versus temperature for tungsten, steel, and
aluminum.
70
60
400
Modulus of elasticity (GPa)
800
(Adapted from K. M. Ralls, T. H. Courtney, and
J. Wulff, Introduction to Materials Science and
Engineering. Copyright © 1976 by John Wiley &
Sons, New York. Reprinted by permission of
John Wiley & Sons, Inc.)
10
400
600
800
0
Temperature (°C)
Relationship
between shear stress
and shear strain for
elastic deformation
Differences in modulus values among metals, ceramics, and polymers are a direct
consequence of the different types of atomic bonding that exist for the three materials
types. Furthermore, with increasing temperature, the modulus of elasticity decreases
for all but some of the rubber materials; this effect is shown for several metals in
Figure 7.8.
As would be expected, the imposition of compressive, shear, or torsional stresses
also evokes elastic behavior. The 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
τ = Gγ
(7.7)
where G is the shear modulus—the slope of the linear elastic region of the shear
stress–strain curve. Table 7.1 also gives the shear moduli for a number of common
metals.
7.4
ANELASTICITY
anelasticity
To this point, it has been assumed that elastic deformation is time independent—
that is, that an applied stress produces an 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 totally recovered—that is, that the strain
immediately returns to zero. In most engineering materials, however, there will
also exist a time-dependent elastic strain component—that is, elastic deformation
will continue after the stress application, and upon load release, some finite time is
required for complete recovery. This time-dependent elastic behavior is known as
anelasticity, and it is due to time-dependent microscopic and atomistic processes that
are attendant to the deformation. For metals, the anelastic component is normally
small and is often neglected. However, for some polymeric materials, its magnitude
is significant; in this case it is termed viscoelastic behavior, which is the discussion
topic of Section 7.15.
226 • Chapter 7
/
Mechanical Properties
EXAMPLE PROBLEM 7.1
Elongation (Elastic) Computation
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?
Solution
Because the deformation is elastic, strain is dependent on stress according to Equation 7.5.
Furthermore, the elongation ∆l is related to the original length l0 through Equation 7.2.
Combining these two expressions and solving for ∆l yields
∆l
σ = εE = (
E
l0 )
∆l =
σ l0
E
The values of σ and l0 are given as 276 MPa and 305 mm, respectively, and the magnitude of
E for copper from Table 7.1 is 110 GPa (16 × 106 psi). Elongation is obtained by substitution
into the preceding expression as
∆l =
(276 MPa) (305 mm)
110 × 103 MPa
= 0.77 mm (0.03 in.)
7.5 ELASTIC PROPERTIES OF MATERIALS
Poisson’s ratio
Definition of
Poisson’s ratio in
terms of lateral and
axial strains
Relationship among
elastic parameters—
modulus of elasticity,
shear modulus, and
Poisson’s ratio
7
When a tensile stress is imposed on a metal specimen, an elastic elongation and
accompanying strain εz result in the direction of the applied stress (arbitrarily taken to
be the z direction), as indicated in Figure 7.9. 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 the applied
stress is uniaxial (only in the z direction) and the material is isotropic, then εx = εy. A
parameter termed Poisson’s ratio ν is defined as the ratio of the lateral and axial strains, or
ν=−
εy
εx
=−
εz
εz
(7.8)
For virtually all structural materials, εx and εz will be of opposite sign; therefore, the
negative sign is included in the preceding expression to ensure that ν is positive.7
1
Theoretically, Poisson’s ratio for isotropic materials should be 4; furthermore, the
maximum value for ν (or the value for which there is no net volume change) is 0.50. For
many metals and other alloys, values of Poisson’s ratio range between 0.25 and 0.35.
Table 7.1 shows ν values for several common materials; a more comprehensive list is
given in Table B.3 of Appendix B.
For isotropic materials, shear and elastic moduli are related to each other and to
Poisson’s ratio according to
E = 2G(1 + ν)
(7.9)
Some materials (e.g., specially prepared polymer foams), when pulled in tension, actually expand in the transverse
direction. In these materials, both εx and εz of Equation 7.8 are positive, and thus Poisson’s ratio is negative. Materials
that exhibit this effect are termed auxetics.
7.5 Elastic Properties of Materials • 227
Figure 7.9
!z
w
w0
l
l0
z
y
εz =
l – l0
“l
=
>0
l0
l0
εx =
“w
w – w0
=