1
`
`UTC 2018
`General Electric v. United Technologies
`IPR2016-01289
`
`

`

`This book is in the
`
`Addison-Wesley Series in Metallurgy and Materials Engineering
`
`Consulting Editors
`Morris Cohen
`
`Merton C. Flemings
`
`Sponsoring Editor: Don Fowley
`Production Supervisor: Bette J. Aaronson
`Copy Editor: Lyn Dupré
`Text Designer: Deborah Schneck
`Technical Art Consultant: Joseph Vetere
`Illustrators: Dick Morton and C & C Associates, Inc.
`Cover Designer: Marshal] Henrichs
`Production Coordinators: Helen Wythe and Sheila Bendikian
`Manufacturing Supervisor: Hugh Crawford
`
`Library of Congress Cataloging-in—Publication Data
`
`Van Vlack, Lawrence H.
`Elements of materials science and engineering I by Lamenoe H. Van
`Vlack.—6th ed.
`p. cm.
`
`Includes index.
`ISBN 0-201-09314-6
`1. Materials.
`2. Solids.
`TA403.V35 1989
`620.1’1—dc19
`
`I. Title.
`
`Reprinted with corrections June, 1990
`
`88-15383
`CIP
`
`Copyright © 1989, 1985, 1980, 1975, 1964, 1959 by Addison-Wesley Publishing
`Company, Inc.
`All nghts reserved. No part of this publication may be reproduced, stored in a retrieval
`system, or transmitted, in any form or by any means, electronic, mechanical, photocopy-
`ing, recordin or otherwise, without the prior written permission ofthe publisher. Printed
`in the Unit
`States of America. Published simultaneously in Canada.
`
`10 DO 9594
`
`
`
`2
`
`

`

` To Today’s Students, who will be Tomorrow’s Engineers
`
`
`
`mtsm
`
`f" 2:
`
`(33mg;
`
`
`
`3
`
`

`

`
`
`
`
`Elements of Materials Science .
`_ and Engineering
`
`SIXTH EDITION
`
`
`
`Lawrence H. Van Vlack
`
`
`
`
`4
`
`

`

`STRUCTURE
`
`
`_
`ELEMENTS OF
`MATERIALS SCIENCE
`
`AND ENGINEERING
`
`I
`
`‘. Q
`_
`|
`
`I
`
`'
`|
`
`5
`i
`
`'
`
`l
`I '
`
`Lawrence H. Van Vlack
`UNIVERSITY OF MICHIGAN
`AmArbor,Mfchigan
`
`This Book Belongs
`To Prokassor D. R. Clarke.
`Please Refurn So omen
`Mayflomow It.
`_
`
`3””?
`
`m
`
`Addison-Wesley Publishing Company
`
`Reading, Massachusetts 0 Menlo Park, California I New York
`Don Mills, Ontario 0 Wokingham, England 0 Amsterdam
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`Bonn II Sydney 0 Singapore 0 Tokyo 0 Madrid 0 San'J'uan
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`5
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`

`

`
`
`WORLD STUDENT SERIES EDITION
`
`This book is in the
`
`Addison-Wesley Series in Metallurgy and Materials Engineering
`
`Consufling Editors
`Morris Cohen
`
`Merton C. Flemings
`
`Sponsoring Editor: Don Fowley
`Production Supervisor: Bette J. Aaronson
`Copy Editor: Lyn Dupré
`Text Designer: Deborah Schneck
`Technical Art Consultant: Joseph Vetere
`Illustrators: Dick Morton and C & C Associates, Inc.
`Cover Designer: Marshall Henrichs
`Production Coordinators: Helen Wythe and Sheila Bendikian
`Manufacturing Supervisor: Hugh Crawford
`
`
`
`Copyright © 1989, 1985, 1980, 1975, 1964, 1959 by Addison-Wesley Publishing
`Company, Inc.
`All rights reserved. No part of this publication may be reproduced, stored in a retrieval
`system, or transmitted, in any form or by any means, electronic, mechanical, photocopy-
`ing, recording, or otherwise, without the prior written permission of the publisher. Printed
`in the United States of America.
`
`ISBN 0-201-52822—3
`
`45.5?89 DO 9321
`
`6
`
`

`

`
`__ MArION
`URE
`
`
`
`
`.
`
`
`
`8—1 ELASTIC DEFORMATION
`Elastic Moduli
`
`Elastic Moduli Versus Temperature
`Elastic Moduli Versus Crystal Direction
`
`8'2 PLASTIC DEFORMATION
`
`Strengths
`Hardness
`
`Ductility
`True Stress and Strain (Optional)
`
`8-3 DEFORMATION MECHANISMS
`Slip Systems
`Mechanism of Slip
`Dislocation Movements in Solid Solutions
`
`Plastic Deformation of Compounds
`Deformation by Twinning (Optional)
`
`8—4 FRACI'URE
`
`Toughness Tests
`motility-Transition Temperature
`Fracture Toughness K1,,
`Design Considerations
`
`Most materials are subjected to stresses and the accompanying deformation
`during processing and use. Plastic deformation is required in processing most
`metals. Furthermore, the engineer must provide for elastic deformation of I
`structural parts in designs. Even electrical materials, such as a wire, undergo
`deformation during manufacture and use. Normally, the originally intended role
`of any material is terminated if there is fracture, whether it is from an overload,
`from a sudden impact, or from sharp thermal gradients.
`In this chapter, we shall consider the nature and mechanism of deformation—
`both elastic and plastic (recoverable and permanent). Then, we shall be in a better
`position to anticipate the rigidity and deformation strength of materials for
`processing and design. When we design to avoid fracture, we must consider stress
`concentrations and toughness.
`-
`
`251
`
`
`
`7
`
`

`

`
`
`8 — 1
`
`ELASTIC DEFORMATION
`
`Elastic deformation is a reversible strain. If a stress is applied in tension, the
`material becomes slightly longer; removal of the load permits the material to
`return to its original dimension. Conversely, when it is under compression, the
`material becomes slightly shorter. The dimensions ofthe-unit cells change when
`the material undergoes elastic strain (Fig. 8 — 1.1).
`
`Elastic Moduli
`
`When only elastic deformation exists, the strain is proportional to the applied
`stress (Fig. 8 w 1 .2). The ratio ofstress to strain is the modulus ofelasticity (Young’s
`modulus), and is a property ofthe material (Eq. 1 — 2. 3). The greater the forces of
`attraction between atoms in a. material, the higher the modulus of elasticity (Fig.
`2 — 6. 1 ).
`Any lengthening or compression ofthe crystal structure in one direction, due to
`a uniaxial force, produces an adjustment in the dimensions at right angles to the
`force. In Fig. 8 — l . 1(a), for example, a small contraction is indicated at right angles
`to the tensile force. The negative ratio between the lateral strain a, and the direct
`tensile strain e, is called Poisson ’s ratio, v:
`e
`,, = __r
`ex
`
`(3—1.1)
`
`Engineering materials may be loaded in shear as well as in tension (and com-
`pression). In shear loading, the two forces are parallel but are not aligned (Fig.
`8 — 1 .3b). As a result, the shear stress, I, is the shear force, F, divided by the sheared
`area, A,:
`
`t = FJA,
`
`(3—1.2)
`
`FIG. 8— 1.1
`
`Elastic Normal Strain
`(Greatly Exaggerated). Atoms
`are not permanently displaced
`from their original neighbors.
`(a) Tension {+). (b) No strain.
`(c) Compression (—).
`
`252
`
` (b)
`
`
`
`(C)
`
`
`
`
`
`8
`
`

`

`3_1 ELASTIC DEFORMATION
`
`253
`
`FIG. 8—4.2
`
`Stress Versus Strain (Elastic). With only
`elastic strain, the two are proportional, and Eq.
`(1-2.3) applies. The slope of the curve is the
`modulus of elasticity {Young’s modulus},
`
`E = sfe.
`
`Strain. e
`
`A shear stress produces an angular displacement, a. We define shear strain, 3), as
`the tangent ofthat angle; that is, as 36/3) in Fig. 8 — 1.3(b). The recoverable or elastic
`shear strain is preportional to the shear stress:
`
`G = If?
`
`(8~l.3)
`
`where G is the shear modulus. Also called the modulus of rigidity, the shear
`modulus is diflerent from the modulus ofelasticity, E; however, the two are related
`at small strains by
`
`'
`
`E=2G(1+v)
`
`(3—1.4)
`
`Since Poisson’s ratio v is normally between 0.25 and 0. 5, the value ofG is approxi-
`mately 35 percent of E.
`A third elastic modulus is the bqu modulus, K. It is the reciprocal of the
`compressibility [1’ ofthe material and is equal to the hydrostatic pressure P], per unit
`
` (a)
`
`FIG. 8—1.3
`
`Elastic Shear Strain. Shear couples produce a relative displacement of one plane of
`atoms past the next. This strain is elastic as long as atoms keep their original neighbors.
`(a) No strain. (13) Shear strain.
`
`
`
`
`
`I
`
`
`
`9
`
`

`

`—.——_——————
`
`254
`
`_ DEFQRMAT'ION AND FRAC'I'URE
`
`of volume compression, AV] V:
`
`K — E r i
`AV 1;
`
`The bulk modulus is related to the modulus of elasticity as follows:
`E
`K: 3(1—2v)
`
`You will derive this equation in Problem 817.
`
`(s 1 5)
`' ‘
`
`(8—1.6)
`
`Elastic Moduli Versus Temperature
`Elastic moduli decrease as temperature increases, as shown in Fig. 8 — 1.4 for four
`common metals; In terms of Fig. 2 — 5 2(a), a thermal expansion reduces the value
`of dF/da, and therefore decreases the modulus of elasticity. The discontinuity in
`the curve for iron in Fig. 8— 1.4 is due to the change from bcc to fee at 912° C
`(1673” F). Not surprisingly, the more densely packed fcc polymorph requires
`greater stresses for a given strain; that is, the elastic modulus is greater for fee. Also
`note from Fig. 8 — 1.4 that higher-melting—temperature metals have greater elastic
`moduli.
`
`Elastic Moduli Versus Crystal Direction
`Elastic moduli are anisotropic within materials; that is, they vary with crystallo-
`graphic direction. As an example, iron has an average modulus of elasticity of
`about 205 GPa (30,000,000 psi); however, the actual modulus of a crystal of iron
`varies from 280 GPa (41,000,000 psi) in the [111] direction to only 125 _GPa
`(18,000,000 psi) in the [100] direction (Table 8— 1-1). The consequence of any
`such anisotropy becomes significant in polycrystalline materials. Assume, for
`example, that Fig. 8 fl 1 5(a) represents the cross-section ofa steel wire in which the
`average stress is 205 MPa (30,000 psi). If the grains are randomly oriented, the
`
`
`
`Temperature, “F
`500
`1000
`1500
`30“ r—l—'_‘l—_—l—‘
`— 40
`e
`
`— 30 a?
`
`s:
`‘3.
`3? zoo
`
`‘
`
`
`
`
`
`___——
`
`FIG. 8—1.4
`Modulus ofElasticity Versus
`Temperature. (Adapted from A. G. Guy and
`J. J. Hren, Elements ofPhysical Metallurgy.
`Addison-Wesley.)
`
`100
`
`o
`
`Z“:
`_
`Fe
`— 20
`O
`g
`g N —10 ‘2‘
`E 1% 3
`0 _———__—_ 0 E
`0
`500
`1070
`Temperature, °C
`
`U—l
`
`0
`
`:l
`
`10
`
`
`
`10
`
`

`

`8—1 ELASTIC DEFORMATION
`
`255
`
`TABLE 8— 1.1 Moduli of Elasticity (Young’s Modulus)*
`
`RANDOM
`MINIMUM
`MAXIMUM
`
`METAL
`GPa
`10‘5 psi
`GPa
`10“ psi
`GPa
`lO‘psi
`Aluminum
`75
`11
`60
`9
`I0
`10
`Gold
`1 10
`16
`40
`6
`80
`12
`Copper
`195
`28
`70
`10
`110
`16
`
`30
`205
`18
`125
`41
`280
`Iron (boo)
`
`
`
`
`
`
`345 50 345 50 345Tungsten 50
`
`* Adapted from E. Schmid and W. Boas, Plasticity in Crystals. English translation, London: Chapman
`Hall.
`
`
`elastic strain is 0001, because the average modulus of elasticity is 205 GPa
`(30,000,000 psi). However, in reality, the stress varies from 125 MPa (18,000 psi)
`to 280 MPa (41 ,000 psi) as shown in Fig. 8 —1.S(b), because grains have difierent
`orientations, but each is strained equally (0.001). Ofcourse, this means that some
`grains will exceed their yield strength before other grains reach their yield strength.
`
`.
`
`
`
`
`
`
`
`
`
`
`Elastic__..,
`
`Stress
`
`Average
`1___
`
`(b)
`
`11
`
`'
`
`I
`1
`
`
`
`'
`
`I
`
`FIG. 8—1.5
`
`Stress Heterogeneities (Sche-
`matic). Elastic stresses vary with
`grain orientation, because the
`moduli of elasticity are not
`isotropic.
`
`11
`
`

`

`256
`
`Example 8 — 1 .1
`
`direction (perpendicularto opposite sides ofthe square) with a200-MPa (29,000-psi)stress.
`(a) What are the dimensions ofthe scribed area? (Poisson’s ratio of steel = 0.29).
`Withoutthe initial stressbeingremoved, asecondtensionstress of410 MPa(60,000 psi)
`is applied at right angles to the firstethatis, perpendicularto the other edges ofthesquare.
`(b) What are the new dimensions of the scribed area?
`Procedure Since no preferred orientation is indicated, we shall assume the random grain
`value of the modulus of elasticity {Table 8 — 1.1). The strains are additive when the two
`stresses are applied.
`
`Calculation
`e, = 200 MPaj205,000 MPa = 0.000975
`(a) From Eq. (1 —2.3):
`e}, = -0.29(0.000975) = —0.00028
`From Eq. (8— 1.1):
`1000 mm (1 + 0.000975) X 1000 mm (1 — 0.00023) = 1001.0 mm X 999.7 mm
`(b) 6’, = —0.00028 + 410 MPaj'205,000 MPa = 0.00172
`92 = 0.000975 - 0.29[410f205,000) = 0.00040
`1000 mm (1 + 0.0004) X 1000 mm (1 + 0.00172) = 1000.4 mm X 1001.? mm
`Additional information We can write a general equation for elastic deformation in three
`dimensions from Eqs. (1 —2.3 and 8— 1.1):
`
`VS
`VS
`— l —— ._._z
`
`8x:
`
`hilt“
`
`_
`
`______.————-
`Example 8 — 1 .2
`What is the percentage volume change in iron ifit is hydrostatically compressedwith 1400
`MPa {200,000 psi)? (Poisson’s ratio = 0.29.)
`Procedure We need the bulk modulus (=th(11Kill-7)) which is obtainable from E and v
`(Eq. 8— 1.6). An alternativesolution makes use oqu. (8—1.7). Since c}( = e}, = ez (=ALIL),
`and with AVIV z 3(ALIL), we get the same answer.
`
`Solution
`
`K = (205,000 MPaysu — 0.53))
`= 162,700 MPa.
`MW: — 1400 MPaflfiZflOfl MPa = -—0.86 vie
`
`Alternative solution
`ex = (1 — 2v)(- 1400 MPa)](205,000 MPa) = -0.0023T
`Al’sz 3.2 = —0.86 we
`
`12
`
`
`
`
`
`12
`
`

`

`
`
`8—2 PLASTIC DEFORMATION
`
`257
`
`approximation A V] V: 3(ALiL) originates
`The
`Comment
`(l + ALKLP, and it is valid when the As are small.
`
`from 1 + AW V =
`
`8 — 2
`
`PLASTIC DEFORMATION
`
`Figure 8 — l .2 shows only elastic strain. This situation is typical ofbrittle (nonduc-
`tile) materials such as cast iron, glass, and phenol—formaldehyde polymers. Duc-
`tit'e materials undergo some plastic (permanent) strain before fracture. For exam-
`ple, ifa steel beam is loaded, it will first deflect elasticly. The deflection disappears
`when the load is removed. An overload will permanently bend the beam in the
`locations where the stresses exceed the yield strength ofthe steel. In this case, the
`bent beam has failed, but it has not fractured. In contrast, on a production line, the
`yield strength of a sheet of steel may be intentionally exceeded to bend the sheet
`into the shape of a car fender. At this stage, the metal has yielded, but it has not
`failed because production requires considerable plastic strain. It is necessary in
`both production and service to know (1) the critical stress requirements to initiate
`permanent deformation, and (2) the amount of plastic strain available before
`eventual fracture of a ductile material.
`
`Strengths
`
`The laboratory test bar of Fig. 8 ~2. 1(a) was used to produce the accompanying
`stress—strain diagram. The stress, 5, is the force per unit area expressed in Nirn2
`(or lbfllinF), but we more commonly expressed it in MPa (or psi). The strain, 6, is
`AL/L, and therefore is dimensionless. The increasing stress produces only elastic
`strain in the linear part of the test. Beyond the preportional limit, plastic strain
`accompanies and generally exceeds the elastic strain:
`
`9 = 3,, + an,
`
`(8—2.1)
`
`The yield strength, 8,, is the critical stress required to introduce yielding, or
`plastic strain. The point of deviation from the elastic slope is gradual in some
`materials, such as aluminum and zinc. It also may be abrupt, as it is in structural
`steels. In either case, continued loading produces both elastic and plastic strain'“
`Inasmuch as the first increment ofplastic strain that is detected is dependent 0n
`the sensitivity of the testing equipment, an unstandardized test procedure could
`
`“ In many steels, where initial yielding is pronounced, the engineer sometimes speaks ofafieldpoint,
`or an elastic limit, rather than of the yield strength. Although used, the latter term is not fully
`appropriate, because continued loading produces additional elastic strain.
`
`
`
`-WW##_._.___._.—_
`
`13
`
`13
`
`

`

`258
`
`DEFORMATION AND FRACTURE
`
`
`
`FIG. 8-4.1
`
` change
`
`0
`
`0.002
`
`in scale
`
`Stresss
`
`5
`
`[
`BI
`
`Strain.e
`(b)
`
`Tensile Test. (a) Standardized test bar after ductile fracture. (Courtesy of U.S. Steel
`Corp.) (in) Stress—strain diagram (ductile material). The yield strength, S3,, is the stress
`that initiates plastic deformation (commonly defined by 0.2 percent permanent offset).
`The ultimate strength, 8“, is the maximum stress based on the original, or nominal, area.
`
`lead to considerable data variability from test bar to test bar, and from laboratory
`to laboratory. Pragmatically, the engineer standardizes the testing by defining the
`yield strength as the stress required to produce a tolerable amount of strain —
`commonly 0.2 percent. This strength is obtained by plotting the 0.2 percent afiirei
`from the linear portion of the stress—strain curve, then reading the intercept (Fig.
`8 — 2. 1 b).
`Almost all ductile materials become stronger when they are deformed plasticly.
`We call this process Strain hardening. It is used advantageously in the design of
`engineering materials (Section 9 — 3). This added strength appears in Fig. 8 — 2. l (b)
`as the concurrent increase in stress and strain beyond the yield strength.
`As presented in the Fig. 8 —2. 1(b), the stress—strain curve reaches a maximum
`and then falls off. The maximum is called the ultimate tensile strength, or more
`simply the ultimate strength, Sn. The ultimate strength is the limiting stress used
`by the engineer. Beyond this point, the test bar shows areduction in cross-sectional
`area, called necking. As a result, the load-carrying capability ofthe bar is reduced
`until the final fracture occurs. A breaking strength, SB, can be calculated, but has
`little engineering significance because the nominal test-bar area is not pertinent at
`this stage.
`
`
`
`14
`
`14
`
`

`

`259
`
`
`
`3—2 PLASTIC DEFORMATION
`
`Hardness
`
`As we indicated in Section 1 —2, hardness is the resistance to penetration. Various
`procedures are used to measure hardness. These depend on the material, its thick~
`ness, the indentor used, and the load applied. More common hardness indices are
`the Brinell Hardness Number (BI-IN), and the Rockwell hardnesses (R). The latter
`has several someszc, Rb, Rf, and so on—which have appropriate loads and
`indentors for harder steels, softer brasses, and thin sheet metal. Although hardness
`is not a basic property of a material, we shall consider hardness data as useful
`indices of strength. For example, with steels, a rule of thumb is that the ultimate
`strength in psi is 500 times the BI-IN value. Since the hardness can be measured in
`rim, and does not require the machining of a test bar, its index is useful in quality
`control and service checking.
`-
`
`Ductility
`
`We use the term ductility to define the permanent strain that is realized before the
`test bar fractures. As she does with strength, the engineer has more than one way to
`define this strain. Elongation is the linear plastic strain accompanying fracture:
`
`Elgagelength = (Lf — Lea/Lo
`
`It is imperative that for the engineer to identify the gage length, because the plastic
`strain is almost invariably localized (Fig. 8 —-2.2).
`A second measure of ductility is the reduction ofarea, R of A, at the point of
`fracture:
`
`R of A = (A, — Aa/A,
`
`(3—2.3)
`
`FIG. 8—2.2
`
`Gage length
`200 mm
`50 mm (2 in.)
`
`Elongation
`20%
`50%
`
`15
`
`Elongation Versus Gage Length. Since final
`deformation is localized, an elongation value is
`meaningless unless the gage length is indicated.
`For routine testing, a 50-min (2-in.) gage length
`_
`13 common.
`
`15
`
`

`

`
`
`260
`
`. DEFORMATION AND FRACTURE
`
`Both measures ofductility normally are expressed in percent. Values for these two
`measures ofductility parallel each other: both are high in ductile materials, and are
`nil in a brittle material. However, there is no established mathematical relation-
`ship between the two, because the final plastic deformation is highly localized.
`
`True Stress and Strain
`
`The stress — strain curve as commonly presented (Fig. 8 — 2. lb) is based on force per
`nominal, or original, area. As previously noted, however, necking occurs before a
`ductile material fractures. After necking starts, the true stress, a, is higher than is
`the nominal stress, 5. Likewise, the true strain, 6, differs from the nominal strain, 8.
`We must modify the nominal stress — strain curve ofFig. 3 — 2. 1(b) ifwe want a true
`stress—strain curve (Fig. 8 — 2.3).
`The design engineer almost always uses the nominal s/e data, rather than the
`true ole data. The nominal data are used partly because design calculations for
`engineering products are based on original dimensions. More important, it would
`be impractical to reduce the load as the material plasticly deforms during the last
`moments before complete failure.*
`Although products are never designed on the basis ofthe true fracture strength,
`of, (Fig. 8—2.3), knowledge of this stress is useful in the design of deformation
`processes. We shall observe in the next chapter that it is possible to make materials
`in which the ultimate strength, 5“, is raised toward that limit.
`
`
`FIG. 8~2.3
`
`or I:
`
`
`
`_/L,_
`
`Strain
`
`l
`9'
`
`|
`fl
`
`
`
`3
`
`ga
`
`“ _ I
`1
`Sb - I
`l
`5 _ l
`3'
`
`True Stress and Strain. The true
`stress, a, is based on the actual area
`rather than on the nominal (original)
`area. Therefore, the true fracture
`stress, or, exceeds the nominal
`breaking strength, Sb. Also, because
`the strain is localized, the true strain
`at the point of fracture, Er, exceeds
`the nominal strain for fracture, 2;.
`(See Example 8—2.2.)
`
`* In a facetious but illustrative vein, you as the design engineer will probably not volunteer to crawl out
`on the bridge to measure the diameter ofa tie barjust before it breaks and the bridge falls into the river
`so that the true stress and strain are known. Sometimes, it does not pay to know the truth!
`
`16
`
`
`
`16
`
`

`

`8—2 PLASTIC DEFORMATION
`
`Example 8-2.]
`
`A copper test bar has a 2.00-in. (51-min) gage length. After testing, the gage marks are
`2.82 in. (71.6 mm) apart. What is the ductility?
`
`Calculation
`
`elongationzmfim = (2.82 in. — 2.00 in.)f2.00 in.
`= {141
`
`{or 41%)
`
`Comments The reduction of area is expected to be high, but it cannot be calculated from
`the elongation data. A calculation in SI units will give the same ductility.
`
`
`
`Example 8 — 2.2
`
`
`
`A copper wire has a nominal breaking strength of 300 MPa (43,000 psi). Its ductility is 7?
`percent reduction of area. Calculate the true stress or for fracture.
`
`Solution Based on the original area, An,
`
`Ai = 300 MPa
`
`F = (300 x 106 mezplo
`
`H i =
`‘7‘ Atr
`
`F
`(1 — 0.77%,
`
`_ (300 x 106 sigma/1‘, _
`0 23Ao
`_ 1300 MP3
`
`01'
`
`i _ (43,000 psiJAc _
`Alf ——0_23Ao
`
`.
`187,000 p51
`
`Comment We can determine the true strain E from the cross~sectional dimensions. If we
`define the true strain a as
`
`‘ in:
`
`<1)
`“ 1.,
`
`and assume constant volume, A! = A010, then
`
`6 = 1n
`
`= 1n
`
`(8—4.4)
`
`This equation gives a definition of true strain that holds for all strains and is independent of
`gage length.
`
`Example 8 — 2.3
`
`A 212-crn copper wire is 0.76 mm (0.03 in.) in diameter. Plastic deformafiou started when
`the load was 3.7 kg. (a) What was the force prmided in N and 1b;? {13) When loaded to
`15.2 kg (33 .9 113;), the total strain was 0.01 1. The wire was then unloaded. Whatis the length
`of the wire after unloading? (c) What is the yield strength of the copper?
`
`
`
`17
`
`17
`
`

`

`262
`
`DEFORMATION AND FRACTURE
`
`Procedure The yield strength is the critical stress to initiate plastic strain. Therefore, we
`must calculate the stress with the 8.7-kg load for (c). After yielding is initiated, any addi-
`tional stress produces both elastic and plastic deformation. Thus, both are included in the
`0.011 strain; we can calculate the elastic strain from the modulus of elasticity and the
`15 .Z-kg load.
`
`Calculation
`
`(8.7 kg}(2.2 lbfiikg) = 19.1 lbr
`(a) {8.7 kg){9.8 111152): 85.25 N
`area = 11 0.76 X 10—3 m)2,t4 = 0.45 X 10—6 m2
`
`(b) Stress at 15.2 kg:
`
`(15.2 kg)(9.8 mls2)i(0.45 X 10—5 m2) = 331 MPa
`
`Elastic strain at 15.2 kg:
`
`Plastic strain at 15.2 kg:
`
`(331 MPa),i(l 10,000 MPa) = 0.003
`0.011 # 0.003 = 0.008 permanent
`
`Wire length after unloading:
`
`1.008(212 cm)= 213.7 cm
`
`(c) Yield strength:
`
`8,, = (85.2 N)f(0.45 X [0’5 m2) = 190 MPa
`
`Comment Elastic strain continues to increase after plastic deformation starts, because the
`bonds of all atoms experience increased forces, even if they do not establish new neighbors
`by plastic deformation.
`
`H D
`
`EFORMATION MECHANISMS
`
`Cubic metals and their nonordered alloys deform predominantly by plastic shear,
`or slip, in which one plane of atoms slides over the next adjacent plane. Plastic
`shear also is one ofthe methods ofdeformation in hexagonal metals, and in a few
`ceramic materials. Shear deformation even occurs when compression or tensicm
`forces are applied, because the stresses may be resolved into shear stresses.
`
`Slip Systems
`
`Slip occurs more readily along certain crystal directions and planes than along
`others. This is illustrated in Fig. 8 — 3.1, where a single crystal of an hep metal was
`deformed plastically. The shear stress required to produce slip on a crystal plane is
`called the critical shear stress, T9.
`The predominant sets ofslip systems in several familiar metals are summarized -
`in Table 8 — 3.1. A slip system includes the slip plane (hki),* and a slip direction
`[new]. There are a number of slip systems, because of the multiple planes in a
`family ofplanes and the multiple directions in a family ofdirections (Sections 3 - 6
`and 3—7). Two facts stand out in Table 8—3. 1.
`
`* See Section 3—? for (hkil) indices of hexagonal crystals.
`
`18
`
`18
`
`

`

`8—3 DEFORMATION MECHANISMS
`
`.
`
`263
`
`FIG. 8—3.1
`
`Slip in a Single Crystal
`(hep). Slip paralleled the
`(000]) plane, which contains
`the shortest slip vector. (See
`Section 3 — 7 {or (hkii ) indioes
`_ of hexagonal crystals.)
`(Constance Elam, Distortion
`ofMetai Crystals, Oxford:
`Clarendon Press.)
`
`
`
`
`
`(a)
`
`(b)
`
`(c)
`
`1 . The slip direction in each metal crystal is the direction with the highest linear
`density of lattice points, or the shortest distance between lattice points
`2. The slip planes are planes that have wide interplanar spacings and, therefore,
`high planar densities (Section 3 — 8)
`
`Mechanism of Slip
`
`Figure 8—3.2 shows a simplified mechanism of slip. If we were to calculate the
`strength ofmetals on this basis, the result would indicate that the strength ofmetals
`should be approximately 6/6, where G is the shear modulus. Since metals have
`only a fraction of that strength, a difl‘erent slip mechanism 'must be operative. All
`experimental evidence supports a mechanism involving dislocation movements.
`
`TABLE 8—3.1 Predominant Slip Systems in Metals
`
`NUMBER OF
`INDEPENDENT
`
`STRUCTURE
`EXAMPLES
`SLIP DIRECTIONS
`SLIP PLANES
`SLIP SYSTEMS
`
`12
`{101}
`(T11)
`ctr-Fe, Mo, Na, w
`bcc
`12
`{211}
`{111)
`ot-Fe, Mo, Na, W
`bcc
`12**
`{111}
`(110)
`Ag, Al, Cu, y—Fe, Ni, Pb
`foc
`3
`{0001}*
`(11:20)
`Cd, Mg, ot-Ti, Zn
`hep
`
`
`a—Tihcp 3 ( 1 1:0) {10i0}*
`
`
`
`* See Section 3 H”! for{hk1'1'} indices of hexagonal crystals. ** See Example 8—3.[ and Fig. 8—3. 10.
`
`
`
`19
`
`19
`
`

`

`
`
`
`
`264
`
`DEFORlliATION AND FRACTURE
`
`
`
`FIG. 8—3.2
`
`An Assumed Mechanism of Slip (Simplified). Metals actually deform with less shear
`stress than this mechanism would require.
`
`If we use Fig. 8 — 3.3 as a model of a dislocation and place a shear stress along the
`horizontal direction, the dislocation can be moved (Fig. 8 #3.4) with a shearing
`displacement within the crystal. (See also Fig. 4 — l .5.) The shear stress required for
`this type of deformation is a small fraction ofthe previously cited value of G/6, and
`it matches observed shear strengths.
`The mechanism ofslip requires the growth and movement ofa dislocation line;
`therefore, energy is required. The energy, E, of a dislocation line is proportional to
`the length of the dislocation line, I, the product of the shear modulus, G, and the
`square of the unit slip vector II (Section 4~ l):
`
`escrow
`
`(ti—3.1)
`
`Thus, the easiest dislocations to generate and expand-for plastic deformation are
`those with the shattest unit slip vector, 11 (Fig. 4 — 1 .2), particularly since the b term
`
`
`
`
`
`
`
`
`
`
`
`
`FIG. 8—3.3
`
`Edge Dislocation. “Bubble-
`raft” model of an imperfection
`in a crystal structure. Note the
`extra row of atoms. (Bragg
`and Nye, Proc. Roy. Soc.
`(London).)
`
`
`
`20
`
`20
`
`

`

`
`
`8—3 DEFORMATION MECHANISMS
`
`265
`
`
`
`FIG. 8—3.4
`
`Slip by Dislocations. In this model, only a few atoms at a time are moved from their
`low-energy positions. Less stress is therefore required to produce slip than would be
`needed if all the atoms moved at once as proposed in Fig. 8—3.2.
`
`is squared. The directions in a metal with the shortest slip vector will be the
`directions with the greatest linear density of atoms. The lowest value for the shear
`modulus, G, accompanies the planes that are farthest apart and hence have the
`greatest planar density of atoms. Thus, we can develop the rule of thumb that
`predicts that the lowest critical shear stresses will occur on the most denserpacked
`planes and in the most densely packed directions.
`
`Dislocation Movements in Solid Solutions
`
`The energy associated with an edge dislocation (Fig. 4— 1.3) is the same whether
`the dislocation is located at (b) or at (c) in Fig. 8— 3.4. Therefore, no net energy is
`required for the movement between the two.* Such is not the case when solute
`atoms are present. As shown in Fig. 8 — 3.5, when an impurity atom is present, the
`energy associated with a dislocation is less than it is in a pure metal. Thus, when a
`dislocation encounters foreign atoms, its movement is restrained because energy
`must be supplied to release it for further slip. As a result, solid-solution alloys
`always have higher strengths than do pure metals (Fig. 8 — 3.6). We call this process
`solution hardening (Section 9 — 2).
`
`* This statement does not apply if( I) the movement includes an increase in the length ofthe disloca-
`tion, or (2) there is a pile-up of dislocations.
`
`
`
`
`
`21
`
`21
`
`

`

`DEFORMATION AND FRACTURE
`
`
`
`(a) Larger impurity atom
`
`(b) Smaller impurity atom
`
`(c) Same size atom
`
`FIG. 8 — 3.5
`
`Solid Solution and Dislocations. An oddwsized atom decreases the stress around a
`dislocation. As a result, energy must be supplied and additional stress applied to detach
`the dislocation from the solute atom. This process accounts for solution hardening.
`
`300
`
`a
`E 200
`—='
`E0
`g
`g 100
`5:
`
`
`
`70 .fi
`CL
`
`8
`60 E
`5'
`a“
`350 g
`g
`—40 g
`5
`
`500
`
`mi
`
`s
`45' 400
`a
`a
`g
`g 390
`5
`
`
`
`
`
`
`
`
`—40
`
`._
`
`
`a
`—30 g
`r
`a”
`'-
`— 20 g”
`E
`"—"10 a
`>-‘
`
`
`
`2
`000
`
`10
`
`30
`
`20
`Zinc, %
`(a)
`
`— 30
`4o
`
`.
`
`00
`
`10
`
`——I 0
`20
`30
`40
`Zinc, %
`(b).
`
`'
`
`FIG. 8—3.6
`
`Solution Hardening (Annealed arc-Brass). The addition of zinc to copper increases both
`the yield and the ultimate strengths. (Thirty-five percent zinc is the practical limit for
`commercial brasses. See Fig. 5~6. 1.)
`
`Plastic Deformation of Compounds
`
`An atom in a single-component metal always has like atoms for neighbors, even .
`during deformation. A compound, however, has two or more types ofatoms with a
`preference for unlike neighbors (see, for example, Fig. 3 — l. 1). On many potential
`slip planes, defamation brings together like atoms and separates a fraction of the
`unlike atoms (Fig. 8 — 3.7). Within a comporrnd, this means higher energy, which
`shows up as a resistance to shear. Consequently, the critical shear stresses on some
`planes are sufficiently high that slip is essentially impossible. In effect, the number
`
`22
`
`
`
`ll!
`
`22
`
`

`

`267
`
`
`
`8—3 DEFORMATION MECHANISMS
`
`Ni Ni Ni Ni Ni
`
`Ni“ 02— Ni“ 02_ Ni2+ 02—
`
`Ni Ni Ni Ni Ni
`
`02— Ni2+ 02— Ni2+ 03— Ni“
`
`-—rNi Ni Ni Ni Ni
`
`Ni Ni Ni Ni Ni (-—
`
`——>Ni2+ 02" Ni2+ 02— Ni2+ 02—
`
`02" Ni“ 02+,\1‘\li2+ 02' Nil+ *—
`
`(a)
`
`FIG. 8—3.7
`
`(b)
`
`Comparison of Slip Processes (Metallic Nickel and Nickel Oxide). More force is
`required to displace the ions in NiO than is needed for the atoms in nickel. The strong
`rcpulsive forces between like ions become siginificant. Nickel also has more slip systems -
`than does nickel oxide.
`
`of possible slip systems is reduced, and ductility decreases. This result is revealed in
`Table 8 ~32, which cites the slip systems for several relatively simple compounds.
`For example, only six sets are operative in MgO. In more complex compounds—
`such as in Ni8 13616024, used in magnet ceramics, and in PbZrO3 , used in piezoe-
`lectric transducers—the possibility of slip is negligible at normal temperatures;
`hence, the materials behave in a brittle manner.
`As revealed in Fig. 3 — 1. I, the shortest repeating distance (i.e., the slip vector) in
`an NaC1-type crystal is along the several ( 1 10) directions. Quite expectedly, there-
`fore, the ( 1 10) directions have been found by experiment to be the slip directions.
`The most common slip plane for NaCl-type crystals is one of those in the {110}
`form. This is particularly true in those compounds, such as LiF and MgO, that
`have small, “nondeformable” ions. Other NaCI—type compounds with large ions,
`such as PbS and MnSe, possess other slip planes, but still have the ( l 10} slip
`directions (Table 8 — 3.2).
`
`TABLE 8—3.2 Slip Systems in Simple Compounds
`
`NUMBER OF
`STRUCTURE
`EXAMPLES
`SLIP DIRECTIONS
`SLIP PLANES
`COMBINATIONS
`
`
`6
`{110}
`{110)
`LiF, MgO, MnS, TiC
`NaCIl
`6
`{001}
`(1 10}
`P138
`NaCl
`12
`{l1 1}
`(110)
`MnSe
`NaCl
`6
`{001 }
`{100)
`CsCl
`CsCl
`A1203A1203 3 { 1 150') {0001}
`
`
`
`
`
`
`
`23
`
`23
`
`

`

`
`
`
`
`268
`
`DEEORMATTON AND FRACIURE
`
`We conclude that both intermetallic and ceramic compounds are inherently
`less deformable than are pure metals and their solid solution phases, because
`critical shear stresses are high and the number of slip planes is small. Conse-
`quently, fracture stresses are commonly exceeded before plastic deformation is
`initiated (Section 8 —4).
`
`fllf’lihli%l%l’.lfiili
`
`Deformation by Twinning
`
`A crystal is twinnedifit possesses a mirror boundary. In Fig. 8 — 3.8, the (1 10) plane
`of a bcc metal is sketched in two dimensions in the plane of the paper. Vertical to
`this plane is a twin boundary, across which the atoms are arranged in the same, but
`reflected, pattern.
`'
`Twinning can be induced by shear stresses, as indicated in Fig. 8-6.9. This
`twinning produces a permanent displacement, but is li