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Materials

Notes

No lecture next Friday.

Home work to be handed out next Thursday.

Hand in date 21st November. 4pm

Poisson's ratio

MECHANICS OF METALS AND ALLOYS 387

∆dd

∆L = L − L0σt

st

L0 L

Figure 5.5 Schematic illustration of tensile strain and corresponding lateral strain. FromZ. Jastrzebski, The Nature and Properties of Engineering Materials, 2nd ed. Copyright © 1976by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc.

number of three independent moduli, corresponding to the three forms of stress: tension,compression, and shear.

5.1.1.3 Moduli and Poisson’s Ratio. Let us examine the elastic deformation of amaterial on a more macroscopic level. Figure 5.5 shows a typical test material undertension on a scale somewhat larger than that represented in Figure 5.4. A stress, σt

is once again applied, resulting in a strain εt , which we define as #L/L0. For smalldeformations, the stress and strain are related by Eq. (5.5). Notice, however, that inorder for the material to (ideally) maintain a constant volume, the diameter, d , mustdecrease by an increment #d = d − d0. There is no requirement, of course, that volumebe maintained, but in most real systems, there is a decrease in the cross section withelongation. This response is characteristic of a material and is given by Poisson’s ratio,ν, which is defined as the ratio of the induced lateral strain to the axial strain:

ν = − #d/d0

#L/L0(5.9)

The minus sign in Eq. (5.9) is to account for the fact that #d as defined above isusually negative. Thus, Poisson’s ratio is normally a positive quantity, though there isnothing that prevents it from having a negative value. For constant volume deformations(such as in polymeric elastomers), ν = 0.5, but for most metals, Poisson’s ratio variesbetween 0.25 and 0.35. Values of Poisson’s ratio for selected materials are presentedin Appendix 7.

Poisson’s ratio provides a basis for relating the lateral response of a material tothe axial response. By definition, the forces that cause lateral deformation in a tensileexperiment must be shear forces, since they are normal to the direction of the applied

Material constant≈0.3

ExampleA tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 10 mm. Determine the magnitude of the load required to produce a 2.5 x 10-3mm change in diameter if the deformation is entirely elastic.

7.5 Elastic Properties of Materials ● 159

inorganic ceramic glasses are also isotropic. The remaining discussion of mechanicalbehavior assumes isotropy and polycrystallinity (formetals and crystalline ceramics)because such is the character of most engineering materials.

EXAMPLE PROBLEM 7.2

A tensile stress is to be applied along the long axis of a cylindrical brass rodthat has a diameter of 10 mm (0.4 in.). Determine the magnitude of the loadrequired to produce a 2.5 ! 10"3 mm (10"4 in.) change in diameter if thedeformation is entirely elastic.

S OLUT I ON

This deformation situation is represented in the accompanying drawing.

When the force F is applied, the specimen will elongate in the z direction andat the same time experience a reduction in diameter, #d, of 2.5 ! 10"3 mm inthe x direction. For the strain in the x direction,

$x %#dd0

%"2.5 ! 10"3 mm

10 mm% "2.5 ! 10"4

which is negative, since the diameter is reduced.It next becomes necessary to calculate the strain in the z direction using

Equation 7.8. The value for Poisson’s ratio for brass is 0.34 (Table 7.1), and thus

$z % "$x&

% "("2.5 ! 10"4)

0.34% 7.35 ! 10"4

The applied stress may now be computed using Equation 7.5 and the modulusof elasticity, given in Table 7.1 as 97 GPa (14 ! 106 psi), as

' % $zE % (7.35 ! 10"4)(97 ! 103 MPa) % 71.3 MPa

z

F

F

x

d0

di

l0li

$ z#ll0

li " l0l0

=

$x#dd0

di " d0d0

=

=

=

When the force F is applied, the specimen will elongate in the z direction and at the same time experience a reduction in diameter, Δd, of 2.5 x 10-3 mm in the x direction. For the strain in the x direction,



EXAMPLE PROBLEM 7.2


S OLUT I ON



$x %#dd0

%"2.5 ! 10"3 mm

10 mm% "2.5 ! 10"4



$z % "$x&

% "("2.5 ! 10"4)

0.34% 7.35 ! 10"4


' % $zE % (7.35 ! 10"4)(97 ! 103 MPa) % 71.3 MPa

z

F

F

x

d0

di

l0li

$ z#ll0

li " l0l0

=

$x#dd0

di " d0d0

=

=

=


The value for Poisson’s ratio for brass is 0.34, and thus



EXAMPLE PROBLEM 7.2


S OLUT I ON



$x %#dd0

%"2.5 ! 10"3 mm

10 mm% "2.5 ! 10"4



$z % "$x&

% "("2.5 ! 10"4)

0.34% 7.35 ! 10"4


' % $zE % (7.35 ! 10"4)(97 ! 103 MPa) % 71.3 MPa

z

F

F

x

d0

di

l0li

$ z#ll0

li " l0l0

=

$x#dd0

di " d0d0

=

=

=

ExampleA tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 10 mm. Determine the magnitude of the load required to produce a 2.5 x 10-3mm change in diameter if the deformation is entirely elastic.



EXAMPLE PROBLEM 7.2


S OLUT I ON



$x %#dd0

%"2.5 ! 10"3 mm

10 mm% "2.5 ! 10"4



$z % "$x&

% "("2.5 ! 10"4)

0.34% 7.35 ! 10"4


' % $zE % (7.35 ! 10"4)(97 ! 103 MPa) % 71.3 MPa

z

F

F

x

d0

di

l0li

$ z#ll0

li " l0l0

=

$x#dd0

di " d0d0

=

=

=



EXAMPLE PROBLEM 7.2


S OLUT I ON



$x %#dd0

%"2.5 ! 10"3 mm

10 mm% "2.5 ! 10"4



$z % "$x&

% "("2.5 ! 10"4)

0.34% 7.35 ! 10"4


' % $zE % (7.35 ! 10"4)(97 ! 103 MPa) % 71.3 MPa

z

F

F

x

d0

di

l0li

$ z#ll0

li " l0l0

=

$x#dd0

di " d0d0

=

=

=

The applied stress may now be computed



EXAMPLE PROBLEM 7.2


S OLUT I ON



$x %#dd0

%"2.5 ! 10"3 mm

10 mm% "2.5 ! 10"4



$z % "$x&

% "("2.5 ! 10"4)

0.34% 7.35 ! 10"4


' % $zE % (7.35 ! 10"4)(97 ! 103 MPa) % 71.3 MPa

z

F

F

x

d0

di

l0li

$ z#ll0

li " l0l0

=

$x#dd0

di " d0d0

=

=

=

Finally, from Equation 7.1, the applied force may be determined as

F ! "A0 ! " !d0

2 "2

!

! (71.3 # 106 N/m2) !10 # 10$3 m2 "2

! ! 5600 N (1293 lbf)

MECHAN I CAL B EHAV IOR—METAL SFor most metallic materials, elastic deformation persists only to strains of about0.005. As the material is deformed beyond this point, the stress is no longerproportional to strain (Hooke’s law, Equation 7.5, ceases to be valid), andpermanent, nonrecoverable, or plastic deformation occurs. Figure 7.10a plotsschematically the tensile stress–strain behavior into the plastic region for atypical metal. The transition from elastic to plastic is a gradual one for mostmetals; some curvature results at the onset of plastic deformation, which increasesmore rapidly with rising stress.

From an atomic perspective, plastic deformation corresponds to the breakingof bonds with original atom neighbors and then reforming bonds with new neighborsas large numbers of atoms or molecules move relative to one another; upon removalof the stress they do not return to their original positions. This permanent deforma-tion for metals is accomplished by means of a process called slip, which involvesthe motion of dislocations as discussed in Section 8.3.

7.6 TENSILE PROPERTIES

YIELDING AND YIELD STRENGTHMost structures are designed to ensure that only elastic deformation will resultwhen a stress is applied. It is therefore desirable to know the stress level at which

160 ● Chapter 7 / Mechanical Properties

Stre

ss

"y

"y

Stre

ss

StrainStrain

PlasticElastic

0 .002

P

Upper yieldpoint

Lower yieldpoint

(b)(a)

FIGURE 7.10(a) Typical

stress–strain behaviorfor a metal showing

elastic and plasticdeformations, the

proportional limit P,and the yield strength "y ,

as determined usingthe 0.002 strain offset

method.(b) Representative

stress–strain behaviorfound for some steels

demonstrating the yieldpoint phenomenon.

THE END OF ELASTIC DEFORMATION


F ! "A0 ! " !d0

2 "2

!

! (71.3 # 106 N/m2) !10 # 10$3 m2 "2

! ! 5600 N (1293 lbf)






Stre

ss

"y

"y

Stre

ssStrainStrain

PlasticElastic

0 .002

P

Upper yieldpoint

Lower yieldpoint

(b)(a)









Nearly all engineeringApplications ensure that

the materials used under go elastic deformation only

Strain Offset Method

Resilience

Resilience is the capacity of a material to absorb energy under elastic deformation and release it upon reloading in such a way as to recover the energy. This is associated with the modulus of resilience Ur this is the strain energy per unit volume required to stress a material from an unloaded state up to the point of yielding.

values of yield strength, tensile strength, and ductility for several common metals(and also for a number of polymers and ceramics). These properties are sensitiveto any prior deformation, the presence of impurities, and/or any heat treatment towhich the metal has been subjected. The modulus of elasticity is one mechanicalparameter that is insensitive to these treatments. As with modulus of elasticity, themagnitudes of both yield and tensile strengths decline with increasing temperature;just the reverse holds for ductility—it usually increases with temperature. Figure7.14 shows how the stress–strain behavior of iron varies with temperature.

RESILIENCEResilience is the capacity of a material to absorb energy when it is deformedelastically and then, upon unloading, to have this energy recovered. The associatedproperty is the modulus of resilience, Ur , which is the strain energy per unit volumerequired to stress a material from an unloaded state up to the point of yielding.

Computationally, the modulus of resilience for a specimen subjected to a uni-axial tension test is just the area under the engineering stress–strain curve takento yielding (Figure 7.15), or

Ur ! !"y

0# d" (7.13a)


Stre

ss (

103 p

si)

800

600

400

200

0

Strain

Stre

ss (

MPa

)

120

100

80

60

40

20

00 0 .1 0 .2 0 .3 0 .4 0 .5

$200%C

$100%C

25%C

FIGURE 7.14Engineering stress–

strain behavior for ironat three temperatures.

FIGURE 7.15 Schematic representation showing howmodulus of resilience (corresponding to the shaded area) isdetermined from the tensile stress–strain behavior of amaterial.

Stre

ss

0 .002 Strain

σy

y




Ur ! !"y

0# d" (7.13a)


Stre

ss (

103 p

si)

800

600

400

200

0

Strain

Stre

ss (

MPa

)

120

100

80

60

40

20

00 0 .1 0 .2 0 .3 0 .4 0 .5

$200%C

$100%C

25%C




Stre

ss

0 .002 Strain

σy

y

This is the realm of non recoverable mechanical response, i.e permanent deformation

Breaking Stuff

applied, often a structure has experienced so much plastic deformation that it isuseless. Furthermore, fracture strengths are not normally specified for engineeringdesign purposes.

EXAMPLE PROBLEM 7.3

From the tensile stress–strain behavior for the brass specimen shown in Figure7.12, determine the following:

(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 havingan original diameter of 12.8 mm (0.505 in.).

(d) The change in length of a specimen originally 250 mm (10 in.) long that issubjected to a tensile stress of 345 MPa (50,000 psi).

S OLUT I ON

(a) The modulus of elasticity is the slope of the elastic or initial linear portionof the stress–strain curve. The strain axis has been expanded in the inset, Figure7.12, to facilitate this computation. The slope of this linear region is the riseover the run, or the change in stress divided by the corresponding change instrain; in mathematical terms,

E ! slope !"#"$

!#2 % #1

$2 % $1(7.10)

Inasmuch as the line segment passes through the origin, it is convenient to takeboth #1 and $1 as zero. If #2 is arbitrarily taken as 150 MPa, then $2 will have


Strain

M

F

TS

Stre

ssFIGURE 7.11

Typical engineeringstress–strain behavior tofracture, point F. Thetensile strength TS isindicated at point M.

The circular insetsrepresent the geometry

of the deformedspecimen at various

points along the curve.

Young Modulus

P

Plastics Limit

Yield Strength

Tensile Strength

Fracture

Question7.6 Tensile Properties ● 163

a value of 0.0016. Therefore,

E !(150 " 0) MPa

0.0016 " 0! 93.8 GPa (13.6 # 106 psi)

which is very close to the value of 97 GPa (14 # 106 psi) given for brass inTable 7.1.

(b) The 0.002 strain offset line is constructed as shown in the inset; its intersec-tion 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 byusing Equation 7.1, in which $ is taken to be the tensile strength, from Figure7.12, 450 MPa (65,000 psi). Solving for F, the maximum load, yields

F ! $A0 ! $ !d0

2 "2

!

! (450 # 106 N/m2) !12.8 # 10"3 m2 "2

! ! 57,900 N (13,000 lbf)

(d) To compute the change in length, %l, in Equation 7.2, it is first necessaryto determine the strain that is produced by a stress of 345 MPa. This is accom-plished by locating the stress point on the stress–strain curve, point A, andreading the corresponding strain from the strain axis, which is approximately0.06. Inasmuch as l0 ! 250 mm, we have

%l ! &l0 ! (0.06)(250 mm) ! 15 mm (0.6 in.)

Stre

ss (

103 p

si)

70

60

50

40

30

20

10

0

500

400

300

200

100

00 0 .10 0 .20 0 .30 0 .40

Strain

Stre

ss (

MPa

)

Tensile strength 450 MPa (65 ,000 psi)

A

Yield strength 250 MPa (36 ,000 psi)

MPa

200

100

0

40

30

20

10

00 0 .005

103 psi

FIGURE 7.12 The stress–strain behavior for the brass specimen discussed inExample Problem 7.3.

Calculate the Young’s Moduls

The Yield strength at offset 0.002

The maximum load that can be sus ta ined by a cy l indr ica l specimen diam. 12.8 mm

Change in length of 200 mm specimen under stress of 345 MPa

DefinitionsTensile Strength [ after yielding the stress necessary to continue plastic deformation increases at a maximum the TENSILE STRENGTH]. This is the maximum stress the system can take. This is not useful in engineering as you want to work in the plastic region. If your system is at its TENSILE STRENGTH it is well beyond its design

Yield Strength [ The stress at which the system has passed the plastic limit, this can be defined in a number of ways, generally it is defined as a strain offset.

Some more definitions

Hooke’s Law

Hooke’s Law for shear

Shear stress Shear Modulus

Shear strain

The shear and elastic modulus can be related by the Poissons ratio

A few moreFor larger strains the Compressive (K) modulus and the Young’s

modulus can be related

What happens at the Yield point

Atoms rearrange them selves in to lower energy configurations

Two methods are outlined here

Slip and twiningMECHANICS OF METALS AND ALLOYS 391

Slip plane

(a) (b)

Twinningplane

Figure 5.9 Schematic illustration of plastic deformation in single crystals by (a) slip and(b) twinning. From Z. Jastrzebski, The Nature and Properties of Engineering Materials, 2nded. Copyright © 1976 by John Wiley & Sons, Inc. This material is used by permission of JohnWiley & Sons, Inc.

A

B

a

She

ar s

tres

s

Displacement, x

(a)

(b)

Figure 5.10 Schematic illustration of (a) relative shear of two planes of atoms in a strainedmaterial and (b) shear stress as a function of relative displacement of the planes from their equi-librium positions. Reprinted, by permission, from C. Kittel, Introduction to Solid State Physics,2nd ed., p. 517. Copyright © 1957 by John Wiley & Sons, Inc.

Based on the concepts of intermolecular forces and shear modulus introduced in theprevious section, it is relatively easy to estimate the theoretical stress required to causeslip in a single crystal. We call this the critical shear stress, σcr . Refer to Figure 5.10a,and consider the force required to shear two planes of atoms past each other. In theregion of small elastic strains, the stress, τ , is related to the displacement, x, relativeto the initial interplanar spacing, d , according to a modified form of Eq. (5.10) for theshear modulus, G:

τ = Gx

d(5.14)

As a first approximation, we may represent the stress-displacement relation by a sinefunction (Figure 5.10b), since when atom A in Figure 5.10a is displaced to the pointwhere it is directly over atom B in the plane below, an unstable equilibrium exists and

Twining occurs when a whole planerearranges to reduce stress concentration

SlippingMECHANICS OF METALS AND ALLOYS 391

Slip plane

(a) (b)

Twinningplane


A

B

a

She

ar s

tres

s

Displacement, x

(a)

(b)



τ = Gx

d(5.14)


MECHANICS OF METALS AND ALLOYS 391

Slip plane

(a) (b)

Twinningplane


A

B

a

She

ar s

tres

s

Displacement, x

(a)

(b)



τ = Gx

d(5.14)



F ! "A0 ! " !d0

2 "2

!

! (71.3 # 106 N/m2) !10 # 10$3 m2 "2

! ! 5600 N (1293 lbf)






Stre

ss

"y

"y

Stre

ss

StrainStrain

PlasticElastic

0 .002

P

Upper yieldpoint

Lower yieldpoint

(b)(a)









This simple explanationDoes not give us a particularly

good answer

DefectsMECHANICS OF METALS AND ALLOYS 393

Shear stress

Slip plane

A B C D

Edgedislocation

line

Shear stress

A B C D

(a) (b)

Shear stress

Unit stepof slip

A B C D

(c)

Figure 5.11 Schematic illustration edge dislocation motion in response to an applied shearstress, where (a) the extra half-plane is labeled as A (cf. Figure 1.28), (b) the dislocation movesone atomic distance to the right, and (c) a step forms on the crystal surface as the extra half-planeexits the crystal. Reprinted, by permission, from W. Callister, Materials Science and Engineer-ing: An Introduction, 5th ed., p. 155. Copyright © 2000 by John Wiley & Sons, Inc.

possible slip systems for BCC and HCP metals are listed in Table 5.1. Body-centeredcubic structures have a high number of slip systems and, as a result, undergo exten-sive plastic deformation; that is, they are highly ductile. The opposite is true of HCPstructures, which have relatively few slip systems. For reference, the correspondingtwinning systems are listed in Table 5.2.

The final factor in determining the critical shear stress for plastic deformation ishow the preferred slip planes are oriented relative to the applied stress. Figure 5.12illustrates how a test sample may be oriented such that the preferred slip plane is notnecessarily along the direction of the applied force, F . Shear forces will still exist,and their magnitude will depend not only on the applied stress, but on orientation. Theangle φ in Figure 5.12 represents the angle between the normal to the slip plane andthe applied stress direction. The angle λ represents the angle between the slip directionand the applied stress direction. Remember that the slip plane and the slip directionare not the same. It is also important to note that the three vectors in Figure 5.12 are


F ! "A0 ! " !d0

2 "2

!

! (71.3 # 106 N/m2) !10 # 10$3 m2 "2

! ! 5600 N (1293 lbf)






Stre

ss

"y

"y

Stre

ss

StrainStrain

PlasticElastic

0 .002

P

Upper yieldpoint

Lower yieldpoint

(b)(a)









Ductility (AKA plastic flow)

DUCTILITYDuctility is another important mechanical property. It is a measure of the degreeof plastic deformation that has been sustained at fracture. A material that experi-ences very little or no plastic deformation upon fracture is termed brittle. Thetensile stress–strain behaviors for both ductile and brittlematerials are schematicallyillustrated in Figure 7.13.

Ductility may be expressed quantitatively as either percent elongation or percentreduction in area. The percent elongation %EL is the percentage of plastic strainat fracture, or

%EL ! !lf " l0l0

"# 100 (7.11)

where lf is the fracture length10 and l0 is the original gauge length as above. Inasmuchas a significant proportion of the plastic deformation at fracture is confined to theneck region, the magnitude of %EL will depend on specimen gauge length. Theshorter l0 , the greater is the fraction of total elongation from the neck and, conse-quently, the higher the value of%EL. Therefore, l0 should be specified when percentelongation values are cited; it is commonly 50 mm (2 in.).

Percent reduction in area %RA is defined as

%RA ! !A0 " Af

A0"# 100 (7.12)

where A0 is the original cross-sectional area and Af is the cross-sectional area atthe point of fracture.10 Percent reduction in area values are independent of bothl0 and A0 . Furthermore, for a given material the magnitudes of %EL and %RAwill, in general, be different. Most metals possess at least a moderate degree ofductility at room temperature; however, some become brittle as the temperatureis lowered (Section 9.8).

A knowledge of the ductility of materials is important for at least two reasons.First, it indicates to a designer the degree to which a structure will deform plastically


A

B

C

B'

C'

BrittleDuctile

Strain

Stre

ss

FIGURE 7.13 Schematic representations oftensile stress–strain behavior for brittle andductile materials loaded to fracture.

10 Both lf and Af are measured subsequent to fracture, and after the two broken ends havebeen repositioned back together.



%EL ! !lf " l0l0

"# 100 (7.11)



%RA ! !A0 " Af

A0"# 100 (7.12)




A

B

C

B'

C'

BrittleDuctile

Strain

Str

ess





%EL ! !lf " l0l0

"# 100 (7.11)



%RA ! !A0 " Af

A0"# 100 (7.12)




A

B

C

B'

C'

BrittleDuctile

Strain

Str

ess



All materials undergo elastic extension to some point, what happens after that is defined by the materials atomic structure.

Elastic recovery7.10 Flexural Strength ● 171

the plastic region will be similar to the tensile counterpart (Figure 7.10a: yieldingand the associated curvature). However, for compression, there will be nomaximum,since necking does not occur; furthermore, the mode of fracture will be differentfrom that for tension.

MECHAN I CAL B EHAV IOR— CERAM IC SCeramic materials are somewhat limited in applicability by theirmechanical proper-ties, which in many respects are inferior to those of metals. The principal drawbackis a disposition to catastrophic fracture in a brittle manner with very little energyabsorption. In this section we explore the salient mechanical characteristics of thesematerials and how these properties are measured.

7.10 FLEXURAL STRENGTHThe stress–strain behavior of brittle ceramics is not usually ascertained by a tensiletest as outlined in Section 7.2, for three reasons. First, it is difficult to prepare andtest specimens having the required geometry. Second, it is difficult to grip brittlematerials without fracturing them; and third, ceramics fail after only about 0.1%strain, which necessitates that tensile specimens be perfectly aligned in order toavoid the presence of bending stresses, which are not easily calculated. Therefore,a more suitable transverse bending test is most frequently employed, in which arod specimen having either a circular or rectangular cross section is bent untilfracture using a three- or four-point loading technique;11 the three-point loadingscheme is illustrated in Figure 7.18. At the point of loading, the top surface of thespecimen is placed in a state of compression, whereas the bottom surface is intension. Stress is computed from the specimen thickness, the bending moment, and

11 ASTM Standard C 1161, ‘‘Standard Test Method for Flexural Strength of AdvancedCeramics at Ambient Temperature.’’

Stre

ss

Strain

Unload

Reapplyload

Elastic strainrecovery

!y0

!yiD

FIGURE 7.17 Schematic tensilestress–strain diagram showing thephenomena of elastic strain recoveryand strain hardening. The initial yieldstrength is designated as !y0

; !yiis the

yield strength after releasing the loadat point D, and then upon reloading.




Ur ! !"y

0# d" (7.13a)


Stre

ss (

103 p

si)

800

600

400

200

0

Strain

Stre

ss (

MPa

)

120

100

80

60

40

20

00 0 .1 0 .2 0 .3 0 .4 0 .5

$200%C

$100%C

25%C




Stre

ss

0 .002 Strain

σy

y

C h a p t e r 9 / Failure

An oil tanker that fractured in a brittle manner by crack propagation around its girth. (Pho-

tography by Neal Boenzi. Reprinted with permission from The New York Times.)

Why Study Failure?

The design of a component or structure often calls

upon the engineer to minimize the possibility of fail-

ure. Thus, it is important to understand the mechan-

ics of the various failure modes—i.e., fracture, fa-

tigue, and creep—and, in addition, be familiar with

appropriate design principles that may be employed

to prevent in-service failures. !For example, we dis-

cuss in Section 20.5 material selection and pro-

cessing issues relating to the fatigue of an automo-

bile valve spring."

234

Flexural Strength

the moment of inertia of the cross section; these parameters are noted in Figure7.18 for rectangular and circular cross sections. The maximum tensile stress (asdetermined using these stress expressions) exists at the bottom specimen surfacedirectly below the point of load application. Since the tensile strengths of ceramicsare about one-tenth of their compressive strengths, and since fracture occurs on thetensile specimen face, the flexure test is a reasonable substitute for the tensile test.

The stress at fracture using this flexure test is known as the flexural strength,modulus of rupture, fracture strength, or the bend strength, an important mechanicalparameter for brittle ceramics. For a rectangular cross section, the flexural strength!fs is equal to

!fs "3FfL2bd 2 (7.20a)

where Ff is the load at fracture, L is the distance between support points, andthe other parameters are as indicated in Figure 7.18. When the cross section iscircular, then

!fs "FfL!R3 (7.20b)

R being the specimen radius.Characteristic flexural strength values for several ceramic materials are given

in Table 7.2. Since, during bending, a specimen is subjected to both compressiveand tensile stresses, the magnitude of its flexural strength is greater than the tensilefracture strength. Furthermore, !fs will depend on specimen size; as explained inSection 9.6, with increasing specimen volume (under stress) there is an increase inthe probability of the existence of a crack-producing flaw and, consequently, adecrease in flexural strength.


Support

= stress =!

= maximum bending momentwhere M

M c I !

= distance from center of specimen to outer fibers

c

= moment of inertia of cross sectionI= applied load

Rectangular

Circular

F

Possible cross sections

Rectangular

Circular

F

R

d

b

2LL2

McI

FL4

d2

bd3

123FL2bd2

FL4

R #R4

4FL

#R3

FIGURE 7.18 A three-pointloading scheme for measuring thestress–strain behavior and flexuralstrength of brittle ceramics,including expressions forcomputing stress for rectangularand circular cross sections.

Used to study materials that are very brittle or not at all m a l l e a b l e , g e n e r a l l y ceramics.

7.13 Stress–Strain Behavior ● 173

7.11 ELASTIC BEHAVIOR

The elastic stress–strain behavior for ceramic materials using these flexure tests issimilar to the tensile test results for metals: a linear relationship exists betweenstress and strain. Figure 7.19 compares the stress–strain behavior to fracture foraluminum oxide (alumina) and glass. Again, the slope in the elastic region is themodulus of elasticity; also, the moduli of elasticity for ceramic materials are slightlyhigher than for metals (Table 7.2 and Table B.2, Appendix B). From Figure 7.19it may be noted that neither of the materials experiences plastic deformation priorto fracture.

7.12 INFLUENCE OF POROSITY ON THE MECHANICAL

PROPERTIES OF CERAMICS (CD-ROM)

MECHAN I CAL B EHAV IOR— POLYMER S7.13 STRESS–STRAIN BEHAVIOR

The mechanical properties of polymers are specified with many of the same parame-ters that are used for metals, that is, modulus of elasticity, and yield and tensilestrengths. For many polymeric materials, the simple stress–strain test is employedfor the characterization of some of these mechanical parameters.12 The mechanicalcharacteristics of polymers, for the most part, are highly sensitive to the rate ofdeformation (strain rate), the temperature, and the chemical nature of the environ-ment (the presence of water, oxygen, organic solvents, etc.). Some modificationsof the testing techniques and specimen configurations used for metals are necessarywith polymers, especially for the highly elastic materials, such as rubbers.

40

30

20

10

00

250

200

150

100

50

0 0 .0004 0 .0008 0 .0012

Strain

Glass

Aluminum oxide

Stre

ss (

103

psi)

Stre

ss (

MPa

)

FIGURE 7.19 Typical stress–strainbehavior to fracture for aluminumoxide and glass.

12 ASTM Standard D 638, ‘‘Standard Test Method for Tensile Properties of Plastics.’’

Toughness



%EL ! !lf " l0l0

"# 100 (7.11)



%RA ! !A0 " Af

A0"# 100 (7.12)




A

B

C

B'

C'

BrittleDuctile

Strain

Stre

ss



Toughness is a mechanical term that is used in several contexts; loosely speaking, it is a measure of the ability of a material to absorb energy up to fracture. Specimen geometry as well as the manner of load application are i m p o r t a n t i n t o u g h n e s s determinations. For dynamic (high strain rate) loading conditions and when a notch (or point of stress concentration) is present, notch toughness is assessed by using an impact test, as discussed in Section 9.8. Furthermore, fracture toughness is a property indicative of a material’s resistance to fracture when a crack is present (Section 9.5).

applied, often a structure has experienced so much plastic deformation that it isuseless. Furthermore, fracture strengths are not normally specified for engineeringdesign purposes.

EXAMPLE PROBLEM 7.3

From the tensile stress–strain behavior for the brass specimen shown in Figure7.12, determine the following:

(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 havingan original diameter of 12.8 mm (0.505 in.).

(d) The change in length of a specimen originally 250 mm (10 in.) long that issubjected to a tensile stress of 345 MPa (50,000 psi).

S OLUT I ON

(a) The modulus of elasticity is the slope of the elastic or initial linear portionof the stress–strain curve. The strain axis has been expanded in the inset, Figure7.12, to facilitate this computation. The slope of this linear region is the riseover the run, or the change in stress divided by the corresponding change instrain; in mathematical terms,

E ! slope !"#"$

!#2 % #1

$2 % $1(7.10)

Inasmuch as the line segment passes through the origin, it is convenient to takeboth #1 and $1 as zero. If #2 is arbitrarily taken as 150 MPa, then $2 will have


Strain

M

F

TS

Stre

ss

FIGURE 7.11Typical engineering

stress–strain behavior tofracture, point F. Thetensile strength TS isindicated at point M.

The circular insetsrepresent the geometry

of the deformedspecimen at various

points along the curve.

Ai over which deformation is occurring (i.e., the neck, past the tensile point), or

!T "FAi

(7.15)

Furthermore, it is occasionally more convenient to represent strain as truestrain #T , defined by

#T " lnlil0

(7.16)

If no volume change occurs during deformation, that is, if

Aili " A0l0 (7.17)

true and engineering stress and strain are related according to

!T " !(1 $ #) (7.18a)

#T " ln(1 $ #) (7.18b)

Equations 7.18a and 7.18b are valid only to the onset of necking; beyond this pointtrue stress and strain should be computed from actual load, cross-sectional area,and gauge length measurements.

A schematic comparison of engineering and true stress–strain behavior is madein Figure 7.16. It is worth noting that the true stress necessary to sustain increasingstrain continues to rise past the tensile point M%.

Coincident with the formation of a neck is the introduction of a complex stressstate within the neck region (i.e., the existence of other stress components in additionto the axial stress). As a consequence, the correct stress (axial) within the neck isslightly lower than the stress computed from the applied load and neck cross-sectional area. This leads to the ‘‘corrected’’ curve in Figure 7.16.

For some metals and alloys the region of the true stress-strain curve fromthe onset of plastic deformation to the point at which necking begins may beapproximated by

!T " K#nT (7.19)


Strain

Engineering

Corrected

True

M%

M

Stre

ss

FIGURE 7.16 A comparison of typicaltensile engineering stress–strain andtrue stress–strain behaviors. Neckingbegins at point M on the engineeringcurve, which corresponds to M% on thetrue curve. The ‘‘corrected’’ truestress–strain curve takes into accountthe complex stress state within theneck region.

Modeling True Stress7.7 True Stress and Strain ● 169

In this expression, K and n are constants, which values will vary from alloy to alloy,and will also depend on the condition of the material (i.e., whether it has beenplastically deformed, heat treated, etc.). The parameter n is often termed the strain-hardening exponent and has a value less than unity. Values of n and K for severalalloys are contained in Table 7.3.

EXAMPLE PROBLEM 7.4

A cylindrical specimen of steel having an original diameter of 12.8 mm (0.505in.) is tensile tested to fracture and found to have an engineering fracturestrength !f of 460 MPa (67,000 psi). If its cross-sectional diameter at fractureis 10.7 mm (0.422 in.), determine:

(a) The ductility in terms of percent reduction in area.

(b) The true stress at fracture.

S OLUT I ON

(a) Ductility is computed using Equation 7.12, as

%RA "!12.8 mm

2 "2

! # !10.7 mm2 "2

!

!12.8 mm2 "2

!

$ 100

"128.7 mm2 # 89.9 mm2

128.7 mm2 $ 100 " 30%

(b) True stress is defined by Equation 7.15, where in this case the area is takenas the fracture area Af . However, the load at fracture must first be computed

Table 7.3 Tabulation of n and K Values (Equation 7.19) forSeveral Alloys

K

Material n MPa psi

Low-carbon steel 530 77,0000.26(annealed)

Alloy steel 93,0006400.15(Type 4340, annealed)

Stainless steel 185,00012750.45(Type 304, annealed)

Aluminum (annealed) 26,0001800.20100,000Aluminum alloy 6900.16

(Type 2024, heat treated)46,000315Copper (annealed) 0.54

Brass 895 130,0000.49(70Cu–30Zn, annealed)

Source: From Manufacturing Processes for Engineering Materials bySeope Kalpakjian, 1997. Reprinted by permission of Prentice-Hall,Inc., Upper Saddle River, NJ.

Ai over which deformation is occurring (i.e., the neck, past the tensile point), or

!T "FAi

(7.15)

Furthermore, it is occasionally more convenient to represent strain as truestrain #T , defined by

#T " lnlil0

(7.16)

If no volume change occurs during deformation, that is, if

Aili " A0l0 (7.17)

true and engineering stress and strain are related according to

!T " !(1 $ #) (7.18a)

#T " ln(1 $ #) (7.18b)

Equations 7.18a and 7.18b are valid only to the onset of necking; beyond this pointtrue stress and strain should be computed from actual load, cross-sectional area,and gauge length measurements.

A schematic comparison of engineering and true stress–strain behavior is madein Figure 7.16. It is worth noting that the true stress necessary to sustain increasingstrain continues to rise past the tensile point M%.

Coincident with the formation of a neck is the introduction of a complex stressstate within the neck region (i.e., the existence of other stress components in additionto the axial stress). As a consequence, the correct stress (axial) within the neck isslightly lower than the stress computed from the applied load and neck cross-sectional area. This leads to the ‘‘corrected’’ curve in Figure 7.16.

For some metals and alloys the region of the true stress-strain curve fromthe onset of plastic deformation to the point at which necking begins may beapproximated by

!T " K#nT (7.19)


Strain

Engineering

Corrected

True

M%

M

Stre

ss

FIGURE 7.16 A comparison of typicaltensile engineering stress–strain andtrue stress–strain behaviors. Neckingbegins at point M on the engineeringcurve, which corresponds to M% on thetrue curve. The ‘‘corrected’’ truestress–strain curve takes into accountthe complex stress state within theneck region.

Polymers

Three typically different types of stress–strain behavior are found for polymericmaterials, as represented in Figure 7.22. Curve A illustrates the stress–strain charac-ter for a brittle polymer, inasmuch as it fractures while deforming elastically. Thebehavior for the plastic material, curve B, is similar to that found for many metallicmaterials; the initial deformation is elastic, which is followed by yielding and aregion of plastic deformation. Finally, the deformation displayed by curve C istotally elastic; this rubberlike elasticity (large recoverable strains produced at lowstress levels) is displayed by a class of polymers termed the elastomers.

Modulus of elasticity (termed tensile modulus or sometimes just modulus forpolymers) and ductility in percent elongation are determined for polymers in thesame manner as for metals (Section 7.6). For plastic polymers (curve B, Figure7.22), the yield point is taken as a maximum on the curve, which occurs just beyondthe termination of the linear-elastic region (Figure 7.23); the stress at this maximumis the yield strength (!y). Furthermore, tensile strength (TS) corresponds to thestress at which fracture occurs (Figure 7.23); TS may be greater than or less than!y . Strength, for these plastic polymers, is normally taken as tensile strength. Table7.2 and Tables B.2, B.3, and B.4 in Appendix B give these mechanical propertiesfor a number of polymeric materials.


Stre

ss (

MPa

)

Stre

ss (

103

psi)

10

8

6

4

2

00

60

50

40

30

20

10

10 2 3 4 5 6 7 8

Strain

10

8

6

4

2

00

60

50

40

30

20

10

B

A

C

10 2 3 4 5 6 7 8

Strain

FIGURE 7.22 Thestress–strain behavior forbrittle (curve A), plastic(curve B), and highlyelastic (elastomeric) (curveC) polymers.

FIGURE 7.23 Schematicstress–strain curve for a plasticpolymer showing how yield andtensile strengths are determined.

Stre

ss

Strain

TS

y!

Effect on temp (polymer)7.14 Macroscopic Deformation ● 175

Polymers are, in many respects, mechanically dissimilar to metals (and ceramicmaterials). For example, the modulus for highly elastic polymeric materials maybe as low as 7 MPa (103 psi), but may run as high as 4 GPa (0.6 ! 106 psi) forsome of the very stiff polymers; modulus values for metals are much larger (Table7.1). Maximum tensile strengths for polymers are on the order of 100 MPa (15,000psi)—for some metal alloys 4100 MPa (600,000 psi). And, whereas metals rarelyelongate plastically tomore than 100%, some highly elastic polymers may experienceelongations to as much as 1000%.

In addition, the mechanical characteristics of polymers are much more sensitiveto temperature changes within the vicinity of room temperature. Consider thestress–strain behavior for polymethyl methacrylate (Plexiglas) at several tempera-tures between 4 and 60"C (40 and 140"F) (Figure 7.24). Several features of thisfigure are worth noting, as follows: increasing the temperature produces (1) adecrease in elastic modulus, (2) a reduction in tensile strength, and (3) an enhance-ment of ductility—at 4"C (40"F) the material is totally brittle, whereas considerableplastic deformation is realized at both 50 and 60"C (122 and 140"F).

The influence of strain rate on the mechanical behavior may also be important.In general, decreasing the rate of deformation has the same influence on the stress–strain characteristics as increasing the temperature; that is, the material becomessofter and more ductile.

7.14 MACROSCOPIC DEFORMATION

Some aspects of the macroscopic deformation of semicrystalline polymers deserveour attention. The tensile stress–strain curve for a semicrystalline material, whichwas initially unoriented, is shown in Figure 7.25; also included in the figure are

Strain

Stre

ss (

MPa

)

Stre

ss (

103 p

si)

00

0 .1 0 .2 0 .3

10

20

30

40

50

60

70

80

0

2

4

6

8

10

12

4°C (40°F)

20°C (68°F)30°C (86°F)

40°C (104°F)

50°C (122°F)

60°C (140°F)

To 1 .30

FIGURE 7.24 The influence of temperature on thestress–strain characteristics of polymethyl methacrylate.(From T. S. Carswell and H. K. Nason, ‘‘Effect ofEnvironmental Conditions on the Mechanical Properties ofOrganic Plastics,’’ Symposium on Plastics, AmericanSociety for Testing and Materials, Philadelphia, 1944.Copyright, ASTM. Reprinted with permission.)

HardnessHardness tests are performed more frequently than any other mechanical test for several reasons:

1. They are simple and inexpensive—ordinarily no special specimen need be prepared, and the testing apparatus is relatively inexpensive.

2. The test is nondestructive—the specimen is neither fractured nor excessively deformed; a small indentation is the only deformation.

3. Other mechanical properties often may be estimated from hardness data, such as tensile strength

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