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

=

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