microstructure*proper-es.pajarito.materials.cmu.edu/rollett/27301/l11-anisotropy-elasticity... ·...

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27-‐301 Microstructure-‐Proper-es

L11: Tensors and Anisotropy, Part 4 Profs. A. D. Rolle?, M. De Graef

Microstructure Properties

Processing Performance

Last modi.ied: 25th Oct. ‘15

Please acknowledge Carnegie Mellon if you make public use of these slides

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Objective •  The objecGve of this lecture is to provide a mathemaGcal framework for the descripGon of properGes, especially when they vary with direcGon.

•  A basic property that occurs in almost applicaGons is elas%city. Although elasGc response is linear for all pracGcal purposes, it is oNen anisotropic (composites, textured polycrystals etc.).


Questions & Answers 1.  Why is it useful to rotate/transform the compliance tensor or matrix? ONen we

need to compute the elasGc modulus in some parGcular direcGon that is not [100] or [111]. Why do we compute the compliance rather than the sGffness in the 1-‐direcGon? This is subtle: we use compliance because one can impose a stress state that has only one non-‐zero component, from which we only need the strain component parallel to it. Poisson’s raGo tells us that imposing a strain in one direcGon automaGcally results in lateral strains (unless ν=0), which means that it is not possible to have one and only one strain component contribuGng to a parGcular stress component.

2.  How are the quanGGes in the equaGon for the rotated/transformed s11 related to the same equaGon with the Young’s moduli in the <100> and <111> direcGons? Comparison of the two formulae shows how to relate the three S values to the Youngs’ moduli in the two direcGons.

3.  What is Zener’s anisotropy raGo? C' = (C11 -‐ C12)/2; Zener’s raGo = C44/C’. 4.  Which materials are most nearly isotropic? W at room temperature is almost

isotropic and Al is not quite so close to being isotropic. 5.  How do we apply the equaGons to calculate the variaGon in Young’s modulus

between [100] and [110] in a cubic metals such as Cu? DirecGon cosines are the quanGGes that are needed to define the direcGon in relaGon to the crystal axes.

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Q&A -‐ 2 6.  What are the Lamé constants? These are the two constants λ and G that are

needed for isotropic elasGcity. What do they have to do with isotropic elasGcity? G has its usual meaning of shear modulus, or C44: see the notes for how they relate to C11 and C12. How do they relate to Young’s modulus, bulk modulus and Poisson’s raGo? See the notes for the formulae.

7.  How do we write the piezoelectric matrix for quartz? 6x3 matrix. What sGmuli and responses do each coefficient in the “d” matrix relate? SGmulus is the electric field and the response is the strain. What are the “BT” and “AT” cuts of a quartz crystal? These are cuts that maximize the usefulness of the thickness shear mode of oscillaGon.

8.  What equaGon describes the resonant frequency? See the notes. Why does temperature ma?er here? Temperature ma?ers because one prefers to have a crystal that does not change its resonant frequency with temperature. Why does the density vary as the sum of 2α11+α33? This sum is the trace of the matrix for the coefficient of thermal expansion, i..e the variaGon in volume with change in T.

9.  How does the angle θ relate to the AT and BT cuts already described? This angle is a rotaGon of the normal to the surface of the crystal in the y-‐z plane (i.e. rotaGon about x). How do we set up the equaGon that tells us the variaGon in d66 with angle of cut? The Eq we need is that which describes the rate of change of resonant frequency with temperature.

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Rotated compliance (matrix) Given an orientaGon aij, we transform the compliance tensor, using cubic point group symmetry:

! S 11 = S11 a114 + a12

4 + a134( )

+ 2S12 a122 a13

2 + a112 a12

2 + a112 a13

2( )+ S44 a12

2 a132 + a11

2 a122 + a11

2 a132( )

WriGng this out in full for the 1111 component:

Re-‐wriGng this with vector-‐matrix notaGon gives:


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Rotated compliance (matrix) •  This can be further simplified with the aid of the standard

relaGons between the direcGon cosines, aikajk = 1 for i=j; aikajk = 0 for i≠j, (aikajk = δij) to read as follows:

•  By definiGon, the Young’s modulus in any direcGon is given by the reciprocal of the compliance, E = 1/S’11.

•  Thus the second term on the RHS is zero for <100> direcGons and, for C44/C'>1, a maximum in <111> direcGons (conversely a minimum for C44/C'<1).

! s 11 = s11 −

2 s11 − s12 − 12s44( ) α12α22 +α 22α3

2 +α32α1

2{ }


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Anisotropy in terms of moduli •  Another way to write the above equaGon is to insert the values for the Young's modulus in the soN and hard direcGons, assuming that the <100> are the most compliant direcGon(s). (Courtney uses α, β, and γ in place of my α1, α2, and α3.) The advantage of this formula is that moduli in specific direcGons can be used directly.

1Euvw

=1E100

− 3 1E100

−1E111

" # $

% & ' α12α 2

2 + α22α 3

2 +α32α1

2( )


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Cubic crystals: anisotropy factor •  If one applies the symmetry elements of the cubic system, it turns out that only three independent coefficients remain: C11, C12 and C44, (similar set for compliance). From these three, a useful combinaGon of the first two is

C' = (C11 -‐ C12)/2

•  See Nye, Physical Proper%es of Crystals


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Zener’s anisotropy factor •  C' = (C11 -‐ C12)/2 turns out to be the sGffness associated with a shear in a <110> direcGon on a {110} plane. In certain martensiGc transformaGons, this modulus can approach zero which corresponds to a structural instability. Zener proposed a measure of elasGc anisotropy based on the raGo C44/C'. This turns out to be a useful criterion for idenGfying materials that are elasGcally anisotropic.


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Anisotropy in cubic materials

•  The following table shows that most cubic metals have posiGve values of Zener's coefficient so that <100> is most compliant and <111> is most sGff, with the excepGons of V, Nb and NaCl.

Material C44/C' E111/E100

Cu 3.21 2.87Ni 2.45 2.18A1 1.22 1.19Fe 2.41 2.15Ta 1.57 1.50

W (2000K) 1.23 1.35W (R.T.) 1.01 1.01

V 0.78 0.72Nb 0.55 0.57

β-CuZn 18.68 8.21spinel 2.43 2.13MgO 1.49 1.37NaC1 0.69 0.74


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Stiffness coef.icients, cubics

Nb (niobium): beta1=17:60 (TPa)-‐1 , Bcub= 0.50. s11 = 6.56, s44 = 35.20, s12 = -‐2.29 (in (TPa)-‐1 ). Emin = 0.081, Emax = 0.152 GPa.

Units:1010 Pa

or10 GPa

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Example Problem

Should be E<111>= 18.89

[Courtney]


For an elastically isotropic body, there are only 2 elastic moduli, known as the Lamé constants.

Lamé constants (isotropic elasticity)


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Young’s, Bulk moduli, Poisson’s ratio


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Engineering with the Piezoelectric Effect

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•  [Newnham, secGons 12.8 and 13.10] The use of quartz as a resonant crystal for oscillator circuits with highly stable frequency depends strongly on the details of its properGes.

•  Although quartz is only weakly piezoelectric, other aspects of its properGes provide the key, namely thermal stability.

•  Most elasGc sGffness coefficients have nega%ve temperature coefficients, meaning that materials become less sGff with rising temperature. The c66 coefficient of quartz, however, is posiGve; Table 13.7. This offsets the effect of thermal expansion, which increases dimensions and decreases density. This is what makes it possible to have an oscillator that is insensiGve to temperature changes.

(�1, �2, �3, �4, �5, �6) =

0

BBBB@

d11 0 0�d11 0 0

0 0 0d14 0 00 �2d11 0

1

CCCCA

0

@E1

E2

E3

1

A

d11 = 2.27; d14 =-‐0.67 pC/N http://en.wikipedia.org/wiki/Electromagnetic_acoustic_transducer


Quartz Oscillator Crystals, contd. 16

•  Resonant frequency, f, for thickness (t) shear mode, as a funcGon of the rotaGon of axes to get c’66, where ρ is the density:

•  AT and BT cut modes are thickness shear modes driven by the piezoelectric coefficient d’26: ε’6 = d’26 E’2

f =12t

sc066

�

A parGcular angle must be determined for the ideal cut to minimize the temperature dependence of the resonant frequency.



•  Temperature dependence of the resonant frequency, f, for thickness (t) shear mode, as a funcGon of the rotaGon of axes to get c’66, where ρ is the density:

1f

df

dT= �1

t

dt

dT+

12c0

66

dc066

dT� 1

2�

d�

dT

•  Temperature derivaGve of the density: 1⇥

d⇥

dT= �(2�11 + �33) = �36.4⇥ 10�6K�1

•  Temperature derivaGve of the thickness in the Z’2 (Y’) direcGon: 1t

dt

dT= �0

2 = �11cos2⇥ + �33sin

2⇥

•  Transformed (rotated) sGffness coefficient: c066 = c1313sin

2� + c1312sin�cos� + c1213sin�cos� + c1212cos2�



•  Quartz belongs to point group 23. Therefore c1313 = c55 = c44 and c1213 = c65 = c56 = c14.

•  Taking the temperature derivaGve for c’66 and subsGtuGng all the relevant values into the equaGon, one obtains the following. Here, “T(c66)” denotes the temperature coefficient of the relevant sGffness coefficient (Table 13.7). The derivaGve of the resonant frequency, f, can be set equal to zero in the standard fashion in order to find the minima.

•  Applying the soluGon procedure yields two values with theta = -‐35° and +49°, corresponding to the two cuts illustrated.

•  Further discussion is provided by Newnham on how to make AC and BC cuts that are useful for transducers for transversely-‐polarized acousGc waves.

1f

df

dT= ��11cos

2⇥ � �33sin2⇥ +

12(2�11 + �33) +

12

c44T (c44)sin2⇥ + 2c14T (c14)sin⇥cos⇥ + c66T (c66)cos2⇥

c44sin2⇥ + 2c14sin⇥cos⇥ + c66cos2⇥


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Summary

•  We have covered the following topics: –  Examples of elasGc property values –  Anisotropy coefficients (Zener) –  Dependence of Young’s modulus on direcGon (in a crystal)

–  Worked example –  Quartz oscillator crystals


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Supplemental Slides •  The following slides contain some useful material for those who are not familiar with all the detailed mathemaGcal methods of matrices, transformaGon of axes etc.


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Notation F SGmulus (field) R Response P Property j electric current E electric field D electric polarizaGon ε  Strain σ  Stress (or conducGvity) ρ  ResisGvity d piezoelectric tensor

C elasGc sGffness S elasGc compliance a rotaGon matrix W work done (energy) I idenGty matrix O symmetry operator (matrix)

Y Young’s modulus δ  Kronecker delta e axis (unit) vector T tensor, or temperature α direcGon cosine


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Bibliography •  R.E. Newnham, Proper'es of Materials: Anisotropy, Symmetry, Structure, Oxford

University Press, 2004, 620.112 N55P. •  De Graef, M., lecture notes for 27-‐201. •  Nye, J. F. (1957). Physical Proper%es of Crystals. Oxford, Clarendon Press. •  Kocks, U. F., C. Tomé & R. Wenk, Eds. (1998). Texture and Anisotropy, Cambridge

University Press, Cambridge, UK. •  T. Courtney, Mechanical Behavior of Materials, McGraw-‐Hill, 0-‐07-‐013265-‐8, 620.11292

C86M. •  Landolt, H.H., Börnstein, R., 1992. Numerical Data and Func%onal Rela%onships in

Science and Technology, III/29/a. Second and Higher Order ElasGc Constants. Springer-‐Verlag, Berlin.

•  Zener, C., 1960. Elas%city And Anelas%city Of Metals, The University of Chicago Press. •  GurGn, M.E., 1972. The linear theory of elasGcity. Handbuch der Physik, Vol. VIa/2.

Springer-‐Verlag, Berlin, pp. 1–295. •  HunGngton, H.B., 1958. The elasGc constants of crystals. Solid State Physics 7, 213–351. •  Love, A.E.H., 1944. A Trea%se on the Mathema%cal Theory of Elas%city, 4th Ed., Dover,

New York. •  Newey, C. and G. Weaver (1991). Materials Principles and Prac%ce. Oxford, England,

Bu?erworth-‐Heinemann. •  Reid, C. N. (1973). Deforma%on Geometry for Materials Scien%sts. Oxford, UK,

Pergamon.


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Transformations of Stress & Strain Vectors •  It is useful to be able to transform the axes of

stress tensors when wri?en in vector form (equaGon on the leN). The table (right) is taken from Newnham’s book. In vector-‐matrix form, the transformaGons are:

€

" σ 1" σ 2" σ 3" σ 4" σ 5" σ 6

$

%

& & & & & & &

'

(

) ) ) ) ) ) )

=

α11 α12 α13 α14 α15 α16α21 α22 α23 α24 α25 α26

α31 α32 α33 α34 α35 α36

α41 α42 α43 α44 α45 α46

α51 α52 α53 α54 α55 α56

α61 α62 α63 α64 α65 α66

+

,

- - - - - - -

.

/

0 0 0 0 0 0 0

σ1σ 2

σ 3

σ 4

σ 5

σ 6

$

%

& & & & & & &

'

(

) ) ) ) ) ) )

€

" σ i =α ijσ j

σ i =α ij−1 " σ j

" ε i =α ij−1Tε j

εi =α ijT " ε j


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