l11 single xtal aniso1 1oct09

8/8/2019 L11 Single Xtal Aniso1 1Oct09

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Single Xtal Plasticity, A.D.Rollett, Carnegie Mellon U., 2009

Plastic DeformationPlastic Deformation

of Single Crystals (L11)of Single Crystals (L11)

27-750, Fall 2009

Texture, Microstructure & Anisotropy,Fall 2009

A.D. Rollett, P. Kalu

CarnegieMellon

MRSEC

With thanks to H. Garmestani (GaTech),G. Branco (FAMU/FSU)

Last revised: 4th Oct. 09


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ObjectiveObjective

The objective of this lecture is to explain howsingle crystals deform plastically.

Subsidiary objectives include:

S

chmid Law Critical Resolved ShearStress

Reorientation during deformation

Note that this development assumes that we

load each crystal under stress boundaryconditions. That is, we impose a stress andlook for a resulting strain (rate).


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BibliographyBibliography

Kocks, U. F., C. Tom, H.-R. Wenk, Eds. (1998).Texture and Anisotropy, Cambridge UniversityPress, Cambridge, UK.

Reid, C. N. (1973). Deformation Geometry forMaterials Scientists. Oxford, UK, Pergamon.

Khan and Huang (1999), Continuum Theory ofPlasticity, ISBN: 0-471-31043-3, Wiley.

W. Hosford, (1993), The Mechanics of Crystals andTextured Polycrystals, Oxford Univ. Press.

Nye, J. F. (1957). Physical Properties of Crystals.Oxford, Clarendon Press. T. Courtney, Mechanical Behavior of Materials,

McGraw-Hill, 0-07-013265-8, 620.11292 C86M.


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NotationNotation Strain (tensor), local: Elocal; global: Eglobal

Slip direction (unit vector): b (orm) Slip plane (unit vector) normal: n (ors) Stress (tensor): W Shear stress (scalar, usually on a slip system): X Angle between tensile axis and slip direction: P Angle between tensile axis and slip plane normal: U

Schmid factor (scalar): m Slip system geometry matrix (3x3): m

Taylor factor (scalar): M Shear strain (scalar, usually on a slip system): K Stress deviator (tensor): S Rate sensitivity exponent: n

Slip system index: s (orE)


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Historical DevelopmentHistorical Development

Physics of single-crystal plasticity

Established by Ewing and Rosenhaim(1900),Polanyi(1922), Taylor and others(1923/25/34/38),

Schimd(1924), Bragg(1933)Mathematical representation

Initially proposed by Taylor in 1938, followed by Bishop& Hill (1951)

Further developments by Hill(1966), Kocks (1970), Hilland Rice(1972) , Asaro and Rice(1977), Hill andHavner(1983)


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Experimental measurements showed that

At room temperature the major source for plasticdeformation is the dislocation motion through the crystallattice Dislocation motion occurs on fixed crystal planes (slip

planes) in fixed crystallographic directions (correspondingto the Burgers vector of the dislocation that carries the slip) The crystal structure of metals is notaltered by the plasticflowVolume changes during plastic flow are negligible

Physics of SlipPhysics of Slip

Experimental techniqueUniaxial Tension or Compression


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Shear StressShear Stress Shear Strain CurvesShear Strain Curves

A typical flow curve (stress-strain) for a single crystal showsthree stages of work hardening:- Stage I = easy glide with lowhardening rates;

- Stage II with high, constanthardening rate, nearlyindependent of temperature orstrain rate;- Stage III with decreasinghardening rate and verysensitive to temperature andstrain rate.Hardening behavior will bediscussed in another lecture.


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Burgers vector:Burgers vector: bb

Screw position:

line direction//b

Edge position:

line directionBb

A dislocation is a line defect in the crystal lattice. The defect has adefinite magnitude and direction determined by the closure failurein the lattice found by performing a circuit around the dislocationline; this vector is known as the Burgers vector of the dislocation.It is everywhere the same regardless of the line direction of thedislocation. Dislocations act as carriers of strain in a crystalbecause they are able to change position by purely local

exchange in atom positions (conservative motion) without anylong range atom motion (i.e. no mass transport required).


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Slip stepsSlip steps

Slip steps (wheredislocations exitfrom the crystal) on the

surface of compressedsingle crystal ofNb.

[Reid]


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Dislocation glideDislocation glide

The effect of dislocation motion in a crystal:passage causes one half of the crystal to bedisplaced relative to the other. This is a sheardisplacement, giving rise to a shear strain.

[Dieter]


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Single Crystal DeformationSingle Crystal Deformation

To make the connection between dislocationbehavior and yield strength as measured in tension,consider the deformation of a single crystal.

Given an orientation for single slip, i.e. the resolved

shear stress reaches the critical value on one systemahead of all others, then one obtains a pack-of-cards straining.

[Dieter]


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Resolved Shear StressResolved Shear Stress

Geometry of slip: how big anapplied stress is required for slip?

To obtain the resolved shearstress based on an applied tensilestress, P, take the component ofthe stress along the slip directionwhich is given by FcosP, and divideby the area over which the (shear)force is applied, A/cosJ. Note thatthe two angles are not complementary unless the slip

direction, slip plane normal and tensile directionhappen to be co-planar.

X= (F/A) cosPcosJ !WcosPcosJ!W* m

Schmid factor :=m

In tensor (index) form:

X=bi Wij nj

= b

= n


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Schmids LawSchmids Law

Initial yield stress varies from sample to sampledepending on, among several factors, the position of the

crystal lattice relative to the loading axis. It is the shear stress resolved along the slip direction onthe slip plane that initiates plastic deformation.

Yield will begin on a slip system when the shear stresson this system first reaches a critical value (criticalresolved shear stress, crss), independent of the tensilestress or any other normal stress on the lattice plane.

Schmid postulated that:

E.S

chmid & W.B

oas (1950), Plasticity of Crystals, Hughes & Co., London.


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CriticalResolved Shear StressCriticalResolved Shear Stress

The experimental evidence ofSchmids Law is that there is acritical resolved shear stress. This is verified by measuring theyield stress of single crystals as a function of orientation. Theexample below is for Mg which is hexagonal and slips mostreadily on the basal plane (all otherXcrss are much larger).

Soft orientation,

with slip plane at

45rto tensile axis

Hard orientation,

with slip plane at

~90rto tensile axis

W! XcosPcosJ

Exercise:draw a series of

diagrams that illustratewhere the tensile axispoints in relation to thebasal plane normal fordifferent points alongthis curve


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Schmids LawSchmids Law

Using chmids law


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Rotation of the Crystal LatticeRotation of the Crystal Lattice

The slip direction rotates towards the tensile axis

[Khan]


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FCCFCC

Geometry ofGeometry ofSlip SystemsSlip Systems

[Reid]

In fcccrystals, the slipsystems are combinationsof slip directions(the Burgers vectors) and

{111} slip planes.


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Slip Systems

in fcc materialsFor FCC materials there are12 slip systems (with + and -shear directions:

Four {111} planes, each withthree directions

[Khan]

The combination of slip plane {a,b,c,d} andslip direction {1,2,3} that operates withineach unit triangle is shown in the figure

Note correction to system b2

101? A


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Geometry of Single SlipGeometry of Single Slip

For tensile stress applied in the[100]-[110]-[111] unit triangle,the most highly stressed slipsystem (highest Schmid factor)has a (11-1) slip plane and a[101] slip direction (the indicesof both plane and direction arethe negative of those shown onthe previous page).

Caution: this diagram places[100] in the center, not [001]. [Hosford]


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Schmid factorsSchmid factors

The Schmid factors, m, vary markedly within the unittriangle (a). One can also (b) locate the position ofthe maximum (=0.5) as being equidistant betweenthe slip plane and slip direction.

(a)(b)

[Hosford]


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Names of Slip SystemsNames of Slip Systems

In addition to the primary slipsystem in a given triangle,there are systems withsmaller resolved shearstresses. Particular names

are given to some of these.For example the system thatshares the same Burgersvector allows for cross-slip ofscrews and so is known as

the cross slip system. Thesystem in the triangle acrossthe [100]-[111] boundary isthe conjugate slip system.

[Hosford]


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Rotation of the Crystal Lattice in

Tensile Test of an fcc Single Crystal

The tensile axis rotates in tension towards the [100]-[111] line. If

the tensile axis is in the conjugate triangle, then it rotates to thesame line so there is convergence on this symmetry line. Onceon the line, the tensile axis will rotate towards [211] which is astable orientation. Note: the behavior in multiple slip is similar butthere are significant differences.

[Hosford]


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Rotation of the Crystal Lattice in CompressionRotation of the Crystal Lattice in Compression

Test of an fcc Single CrystalTest of an fcc Single Crystal

The slip plane normal rotates towards the compression axis

[Khan]


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Useful EquationsUseful Equations

Following the notation in Reid: n:=slip plane normal (unit vector), b:=slip direction (unit vector).

To find a new direction, D, based on

an initial direction, d, after slip, use(and remember that crystaldirections do not change):

To find a new plane normal, P, based on an initialplane normal, p, after slip, use the following:

[Reid]


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Taylor rateTaylor rate--sensitive modelsensitive model

We will find out later that the classicalS

chmid Lawpicture of elastic-perfectly plastic behavior is notsufficient.

In fact, there is a smooth transition from elastic to

plastic behavior that can be described by a power-law behavior.

The shear strain rate on each slip system is given bythe following (for a specified stress state),

where*m

ij=

binj:

K s) ! Im

(s) :Wc

X(s)

n(s )

sgn m(s) :Wc * m is a slip tensor, formed as the outer product of the slip direction and slip plane normal


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Schmid Law ProblemsSchmid Law Problems

How to solve problems using the Schmid Law Check that single slip is the appropriate model to use (as opposed to, say,

multiple slip and the Taylor model) Make a list of all 12 slip systems with slip plane normals (as unit vectors)

and slip directions (as unit vectors that are perpendicular to their associatedslip planes)

Convert whatever information you have on the orientation of the single

crystal into an orientation matrix, g Apply the inverse of the orientation to all planes and directions so that they

are in specimen coordinates If the tensile stress is applied along the z-axis, for example, compute the

Schmid factor as the product of the third components of the transformedplane and direction

Inspect the list of absolute values of the Schmid factors: the slip system with

the largest absolute value is the one that will begin to slip before the others Alternatively (for a general multi-axial state of stress), compute the following

quantity, which projects the stress, W, onto the kth slip system:

X(k) ! gimbi(k)W

mngjnn j

(k)


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SummarySummary

The Schmid Law is well established forthe dependence of onset of plastic slipas well as the geometry of slip.

Cubic metals have a limited set of slipsystems: {111} forfcc, and{110} forbcc(neglecting pencil

glide for now).


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Supplemental SlidesSupplemental Slides


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Dislocation MotionDislocation Motion

Dislocations control most aspects of strength andductility in structural (crystalline) materials.

The strength of a material is controlled by the

density of defects (dislocations, secondphase particles, boundaries).

For a polycrystal:

WWyieldyield

== XXcrsscrss

= EE Gb Gb VV


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Dislocations & YieldDislocations & Yield

Straight lines are not a good approximation for theshape of dislocations, however: dislocations reallymove as expanding loops.

The essential feature of yield strength is the densityof obstacles that dislocations encounter as they moveacross the slip plane. Higher obstacle density higher strength.

[Dieter]


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Orowan bowingOrowan bowing

stressstress

1

2

3

2r

P

If you consider the three consecutivepositions of the dislocation loop, it isnot hard to see that the shear stress

required to support the line tension ofthe dislocation is roughly equal for positions 1 and 3, buthigher for position 2. Moreover, the largest shear stressrequired is at position 2, because this has the smallest radiusof curvature. A simple force balance (ignoring edge-screwdifferences) between the force on the dislocation versus the

line tension force on each obstacle then givesXmaxbP = (b2/2),where P is the separation between the obstacles (strictlyspeaking one subtracts their diameter), b is the Burgers vectorand G is the shear modulus (Gb2/2 is the approximatedislocation line tension).


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Orowan Bowing Stress, contd.Orowan Bowing Stress, contd.

To see how the force balance applies,consider the relationship between the shapeof the dislocation loop and the force on thedislocation.

Line tension = Gb2/2Force resolved in the vertical direction =2cosJGb2/2

Force exerted on the dislocation per unitlength (Peach-Koehler Eq.) = XbForce on dislocation per obstacle (only thelength perpendicular to the shear stressmatters) = PXb

At each position of the dislocation, theforces balance, soX=cosJGb2/Pb

The maximum force occurs when the angleJ= 90r, which is when the dislocation isbowed out into a complete semicirclebetween the obstacle pair.

X

P

Gb2/2 Gb2/2

J

X

P

Gb2/2 Gb2/2

J!r

MOVIES: http://www.gpm2.inpg.fr/axes/plast/MicroPlast/ddd/


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Critical stressCritical stress

It should now be apparent that dislocations will onlymove short distances if the stress on the crystal is lessthan the Orowan bowing stress. Once the stress risesabove this value then any dislocation can move past

all obstacles and will travel across the crystal or grain. This analysis is correct for all types of obstacles fromprecipitates to dislocations (that intersect the slipplane). For weak obstacles, the shape of the criticalconfiguration is not the semi-circle shown above (to be

discussed later) - the dislocation does not bow out sofar before it breaks through.


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Stereology: NearestNeighbor DistanceStereology: NearestNeighbor Distance

The nearest neighbor distance(in a plane), 2, can beobtained from the point densityin a plane, P

A.

The probability density, P(r), is

given by consideringsuccessive shells of radius, r:the density is the shell area,multiplied by the point density ,PA, multiplied by the remainingfraction of the cumulative

probability. For strictly 1D objects such as

dislocations, 2 may be usedas the mean free distancebetween intersection points on

a plane.

P(r)dr! 1 P(r)dr0

r

? APA2TrdrP(r)dr! 2TrP

AeT r

2P

A

(2 !

rP(r)dr

0

g

( 2 !

1

2 PA

r

dr

Ref: Underwood, pp 84,85,185.


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Dislocations as obstaclesDislocations as obstacles Dislocations can be considered either as a set of

randomly oriented lines within a crystal, or as aset of parallel, straight lines. The latter is easierto work with whereas the former is more realistic.

Dislocation density, V, is defined as either linelength per unit volume, L

V. It can also be defined

by the areal density of intersections of

dislocations with a plane, PA. Randomly oriented dislocations: use

V =LV= 2PA; 2=(2PA)

-1/2; thusP=(2{L

V/2})-1| (2{V/2})-1.

P is the obstacle spacing in any plane.

Straight, parallel dislocations: use V = LV= P

Awhere PA applies to the plane orthogonal to thedislocation lines only; 2=(PA)-1/2;thus P= 1/LV| 1/V where P is the obstacle spacing in theplane orthogonal to the dislocation lines only.

Thus, we can write Xcrss = EbV

U

l11 single xtal aniso1 1oct09

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