Design of Ferromagnetic Shape Memory Alloy...

Design of Ferromagnetic ShapeMemory Alloy Composites

MASAHIRO KUSAKA* AND MINORU TAYA

Department of Mechanical Engineering

Center for Intelligent Materials and Systems

University of Washington

Box 352600, Seattle, WA, 98195-2600, USA

(Received April 17, 2003)(Revised September 12, 2003)

ABSTRACT: Ferromagnetic shape memory alloy (FSMA) composites composed ofa ferromagnetic material and a shape memory alloy (SMA) are key material systemsfor fast-responsive and compact actuators.The function of ferromagnetic material is to induce magnetic force which is then

used to induce the stress in the SMA, resulting in the stress-induced martensitetransformation (SIM), i.e. change in the Young’s modulus, stiff (austenite) to soft(martensite). This SIM-induced phase change causes larger deformation in the SMA,which is often termed as ‘‘superelastic’’.This paper discusses a simple model by which the stress and strain field in the

FSMA composites subjected to bending and torsion loading are computed withthe aim of identifying the optimum geometry of FSMA composites. The results ofthe present analytical study are utilized to design torque actuator (bending of FSMAcomposite plate) and spring actuator (torsion of helical FSMA composite spring).

KEY WORDS: shape memory alloy, ferromagnetic material, stress-inducedmartensite transformation, superelasticity, bending plate, coil spring.

INTRODUCTION

RECENTLY, FERROMAGNETIC SHAPE memory alloys (FSMAs) attract strong attentionas a fast responsive actuator material. There are three actuation mechanisms

identified in FSMAs, (1) magnetic field-reduced phase transformation, (2) martensitevariant rearrangement and (3) hybrid mechanism by magnetic field gradient [1,2]. The firstmechanism often requires large magnetic field, thus necessitating the design of largeelectromagnetic driving unit, not suited for compact actuators, while the secondmechanism can provide large strain but at low stress level. Therefore, use of the hybridmechanism is the most effective to design high-force actuators, yet at fast speed.

*Author to whom correspondence should be addressed. E-mail: [email protected]

Journal of COMPOSITE MATERIALS, Vol. 00, No. 00/200? 1

0021-9983/0?/00 0001–25 $10.00/0 DOI: 10.1177/0021998302040565� 200? Sage Publications

+ [8.11.2003–1:25pm] [1–26] [Page No. 1] FIRST PROOFS {Sage}Jcm/JCM-40565.3d (JCM) Paper: JCM-40565 Keyword

binsumol

The hybrid mechanism is a sequence of chain reaction events, applied magnetic fieldgradient, magnetic force in a FSMA, stress-induced phase transformation from stiffaustenite to softer martensite, resulting in a large deformation, yet large stress can berealized due to superelastic plateau in the stress–strain curve of a FSMA. In the hybridmechanism, the magnetic force F is given by

F ¼ �0VM@H

@xð1Þ

where �0 is the magnetic permeability in vacuum, V is the volume of a ferromagneticmaterial, M is the magnetization vector and H is the magnetic field, thus Fis proportional to both magnetization vector and magnetic field gradient. It is notedthat the magnetic force F influences the internal stress field within a FSMA, i.e., thelarger F is, the larger stress field is induced in FSMA. It is also here that use of aportable electromagnet or permanent magnet can provide large magnetic field gradient,resulting in larger magnetic force, thus larger stress-induced martensite phasetransformation.

The cost of processing FSMAs such as FePd [2] is usually very expensive. Superelasticshape memory alloys (SMAs) have high mechanical performances, large transformationstrain and stress capability. But, the speed of superelastic SMAs by changing temperatureis usually slow. If a FSMA composite composed of a ferromagnetic material and asuperelastic SMA can be developed, cost-effective and high-speed actuators can bedesigned. In the design of this composite, the requirements are: no plastic deformationof the ferromagnetic material and large transformation strain in superelastic SMA.It is necessary to design the optimum microstructure (cross-section) of composite withhigh performance (high load capacity and large deformability) while satisfying theserequirements. In order to obtain the optimum microstructure of FSMA composites withhigh performance, one needs to use either numerical models such as finite element method(FEM), or analytical model. There have been a number of works on finite element analysis(FEA) of SMA structures [3–7]. The FEA which uses commercial FEM is time consumingin the preliminary design. It would be easier for a designer to use a simple analyticalmodel to obtain the optimal microstructure of FSMA composite, if the simple analyticalmodel provides closed form solutions. We made a preliminary model for FSMAcomposites [8]. The analytical model in this paper is a further extension of our preliminarymodel, and it is aimed at detailed modeling of the superelastic behavior of a SMA in aFSMA composite.

In this study, two cases of loading, bending and the twist modes of the compositesare considered with emphasis on how the geometry and the mechanical properties ofthe components influence the superelastic SMA behavior of the composite. First, thebending deformation of the composite plate with application to torque actuators [9] istheoretically analyzed. That is, the relation between the curvature and the bendingmoment for the composite plate. Next, the spring of the composite wire with therectangular section form is designed in consideration of application to spring actuators[10], and the deformation characteristic of the spring is examined. For both models ofbending and torsion of FSMA composites, the optimized microstructures of thecomposites are identified with the aim of maximizing force and deformation of FSMAcomposite actuators.

2 M. KUSAKA AND M. TAYA


SUPERELASTIC BEHAVIOR OF BENDING COMPOSITE PLATE

Analytical Model

For bending type actuation, the laminated composite plate composed of a ferromagneticmaterial layer and superelastic SMA layer as shown in Figure 1(a), is examined. Thecomposite plate is subject to bending moment M induced by the magnetic force generatedby the ferromagnetic material. After the maximum bending stresses on the plate surface ofSMA layer reach the transformation stress (onset of superelastic plateau in the upper loopof the stress–strain curve, Figure 2(b)), the phase transformation proceeds from the platesurface as shown in Figure 1(b). The stress in the transformed region remains constant dueto the superelastic behavior of SMA. It is assumed throughout in this paper to facilitatethe analysis that the superelastic loop of SMA is ‘‘flat’’ i.e. no working–hardening typeslope allowed, and the Young’s modulus of the austenite is the same as that of themartensite. These assumptions allow us to obtain simple closed form solutions in thepresent model, although the predictions are still to the first order approximation. The aimof using this simple model is to identify the best thickness ratio of a ferromagnetic layerand SMA layer in the composite plate.

Then, the relation between the bending moment and the curvature is theoreticallycalculated by using stress–strain curves of the constituent materials. Figure 2(a) shows the

M

hf

h

ρ

Superelastic SMA layer

Ferromagnetic layer

Plate width; b

Stre

ss

Strain

Ferromagnetic layer Young's modulus; Ef

Yield stress;σf

Superelastic SMA layer Young's modulus; ESMA

f

Onset stess for SIM;σ0

Onset stess for Reverse Transformation;σ1

(a) (b)

σ

Figure 2. Material properties and model for the theoretical examination: (a) plate bending model; (b) stress–strain relations for ferromagnetic material and superelastic SMA.

TransformationMM

Ferromagnetic layer


(a) (b)

Figure 1. Composite plate for bending mode actuation: (a) material composition; (b) stress distribution incross section.

Design of Ferromagnetic Shape Memory Alloy Composites 3


analytical model. Radius of curvature of the composite plate subject to bending momentM is �, the thickness of the composite plate is h, the thickness of the ferromagnetic layer ishf, and the plate width is b. Figure 2(b) shows the stress–strain curves of the ferromagneticmaterial and the superelastic SMA, where the Young’s modulus of the ferromagneticmaterial is Ef, that of the SMA are ESMA, the yield stress of the ferromagnetic material is�f, and only elastic portion of the ferromagnetic material is shown. The onset stress forphase transformation of superelastic SMA is �0, the onset stress for reverse transformationis �1 in the superelastic loop portion of SMA. As a result, the relation between the bendingmoment and the curvature of the composite plate also is expected to exhibit thesuperelastic loop if properly designed. This superelastic loop of the FSMA composites isindeed desired.

The curvature which reaches yield stress �f in a ferromagnetic layer and the curvaturewhich reaches transformation stress �0 in superelastic SMA layer are strongly influencedby the mechanical properties and the thickness of both materials. Stress distribution isclassified into the following three cases because of the relation between the transformationstress in the SMA layer and the yield stress of a ferromagnetic layer.

Case 1 The stress in a ferromagnetic layer reaches the yield stress �f, before reachingthe transformation stress �0 in the superelastic SMA layer.

The stress distribution of this case upon loading and unloading is shown in Figure 3,where the bending stress by elastic deformation is illustrated in each material.

Case 2 The stress in a ferromagnetic layer reaches the yield stress, after SMA layerreaching the transformation stress in some part.

The stress distribution of Case 2 upon loading and unloading is shown in Figure 4.Under increasing bending moment first elastic stress distribution (a), then the stress in theSMA layer reaches the transformation stress �0 at the position of y1 (b), and when thetransformation domain advances to y1¼Y1, a ferromagnetic layer reaches the yield stress�f (c). It is noted in (b)–(e) that Y1 remains constant until y3 reaches Y1. During unloading,the stress decreases first elastically in all domains (d), next, the stress becomes constantfrom the upper part of the SMA layer to the position of y3 where the stress reached reversetransformation stress �1 (e). In addition, after the stress at location y3¼Y1 reaches �1, thestress inside portion (y<y2) decreases elastically ( f ). Finally, the stress in the entire SMAlayer decreases elastically when the stress in the SMA on the top surface becomes smallthan �1 (g).

Case 3 The stress in a ferromagnetic layer reaches the yield stress, after the entiredomain of the superelastic SMA layer reaching the transformation stress �0.

σf

σ

σf


Ferromagnetic layer

MM

Loading

−σ0

−σ0

(a) (b)

Unloading

y

x

σ

Figure 3. Changes in stress distribution in cross section according to load (Case 1).



The stress distribution of Case 3 upon the loading and the unloading is shown inFigure 5. In early stage of loading, the stress in a ferromagnetic layer does not reach theyield stress yet even after the stress in all domains of the SMA layer reaches thetransformation stress �0 (c). A neutral axis position changes with an increase in the load,and the stress reaches the yield stress �f finally in a ferromagnetic layer (d). The process ofunloading is shown in Figure 5 (e)–(h).

For each stress distribution �x( y) of the three cases, the following equations are valid,i.e. the equilibrium of force and moment.

Z h

0

�xðyÞbdy ¼ 0 ð2aÞ

M ¼ �

Z h

0

�xðyÞybdy ð2bÞ

σf

σf σ f

σ f

σ fσ

fσ

f

−σ0−σ

1−σ

0−σ1

−σ0−σ

1−σ

0−σ1


Ferromagnetic layer

MM

Loading

Unloading

−σ0

−σ0

−σ0

(a) (b) (c)

(d)(e)(f)(g)

y1

2Y1

y

x

Y1Y1

y2

y3


y

σ σ

yσ σ σ σy

−σ0

σ

y

σ

yyyσ σ σ σ y

σ σ σ σ

Loading

Unloading

(a) (b ) (c ) (d )

(e)(f )(g )(h )

y

x

−σ0

−σ0

−σ0

−σ0

−σ1

−σ0 −σ

1−σ

0 −σ1

−σ0 −σ

1




The neutral axis position and the relation between the moment and the curvature areobtained by solving these equations. Let us focus on the case 2, particularly the stress stateof Figure 4(b). When a neutral axis position is �2, and the transformation stress position isy1, the stress distribution in each domain becomes

in ferromagnetic layer (0<y<hf)

�ð yÞ ¼ Ef�2 � y

�ð3Þ

in SMA layer below the transformation stress �0 (hf< y<y1)

�ð yÞ ¼ ESMA�2 � y

�ð4Þ

in the transformation domain of SAM ( y1<y<h)

�ð yÞ ¼ ��0 ð5Þ

By substituting Equations (3), (4), and (5) to Equations (2), unknown �2 and y1 are solvedand they are given by

�2h¼ �

Ef

ESMA� 1

� �hf

hþ

�0ESMA

�

h

� �

þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiEf

ESMA

Ef

ESMA� 1

� �hf

h

� �2

þ2�0

ESMA

�

h1þ

Ef

ESMA� 1

� �hf

h

� �sð6Þ

y1

h¼

�2hþ

�0ESMA

�

hð7Þ

Moreover, by substituting Equations (3)–(7) to Equation (2b), the relation between thenormalized bending moment and curvature is obtained as

M

ESMAbh2¼

h

�

Ef

ESMA

1

3

hf

h

� �3

�1

2

�2h

hf

h

� �2( )

þ1

3

y1

h

� �3�

hf

h

� �3( )"

�1

2

�2h

y1

h

� �2�

hf

h

� �2( )#

þ1

2

�0ESMA

1�y1

h

� �2� �ð8Þ

Equation (8) is valid for the range of curvature, i.e. from the curvature withtransformation stress �0 in top ( y¼ h) of SMA layer to the curvature with yield stress�f at bottom ( y¼ 0) of ferromagnetic layer. This range of the curvature is given by

�0ESMA

2 1þ ðEf=ESMAÞ � 1ð Þ hf=hð Þ�

1þ ðEf=ESMAÞ � 1ð Þ 2� ðhf=hÞð Þ hf=hð Þ<

h

��

h

�1ð9Þ



where

h

�1¼

�fEf

þ�0

ESMA

� �2( ),(

��fEf

Ef

ESMA� 1

� �hf

h�

�0ESMA

� �

þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�fEf

Ef

ESMA� 1

� �hf

h�

�0ESMA

� �2

þ�fEf

þ�0

ESMA

� �2Ef

ESMA� 1

� �hf

h

� �2s ) ð10Þ

Similarly, the relations between the bending moment and the curvature for the three casesof Figures 3–5 can be calculated. The results for Cases 1, 2 and 3 are shown in Appendix.The conditions under which three cases are valid, are obtained as

Case 1

�f�0

<Ef

ESMA

1þ ðEf=ESMAÞ � 1ð Þ hf=hð Þ2

1þ ðEf=ESMAÞ � 1ð Þ 2� ðhf=hÞð Þ hf=hð Þð11Þ

Case 2

2h

hf

�0ESMA

þ�0Ef

hf

h� 1

� �� >

h

�1ð12Þ

Case 3

2h

hf

�0ESMA

þ�0Ef

hf

h� 1

� ��

h

�1ð13Þ

The maximum normalized curvatures in these cases are given byCase 1 Case 2 Case 3

h

�¼

�fEf

2 1þ ðEf=ESMAÞ � 1ð Þ hf=hð Þ� 1þ ðEf=ESMAÞ � 1ð Þ hf=hð Þ

2,

h

�¼

h

�1,

h

�¼ 2

h

hf

�fEf

��0Ef

h

hf� 1

� �� ð14Þ

The maximum deformability of the composite plate can be analyzed for a given set of themechanical properties and the thickness ratio of materials by using Equation (14).

Analytical Results and Discussion

The relation between the bending moment and the curvature is predicted by the presentmodel for two types of the composite, i.e. Fe/CuAlMn and FeCoV/CuAlMn. Figure 6(a)is the idealized stress–strain curves of Fe and CuAlMn. The results of the predictedrelation between the normalized bending moment and the normalized curvature forthickness ratio hf/h¼ 0.5 are shown in Figure 6(b). The state of the stress for this casecorresponds to Case 1, Figure 3, i.e. the stress in SMA layer is not superelastic plateau,and thus, the superelastic loop is not observed as evidenced in Figure 6(b). Therefore, the



composite plate of Fe and CuAlMn is undesirable as effective bending actuatorcomponent.

Next, the FeCoV/CuAlMn composite plate was analyzed by using the mechanicalproperty data shown in Figure 7(a). Figure 7(b) shows the analytical results for hf/h¼ 0.5,exhibiting clearly superelastic behavior. By using FeCoV whose yield stress is larger thanFe, yet its soft magnetic property is better than Fe, we can achieve now the state wheremost of the CuAlMn layer becomes a transformation domain, corresponding to almostthe state of Case 3. Moreover, the maximum curvature was 2.22 times larger and thebending moment was 1.60 times larger than those of the composite with Fe. Therefore, theFSMA composite so identified is promising as an effective bending actuator component.

Next, we performed a set of parametric studies to examine the effects of materialparameters (�f, Ef, �0, �1, ESMA) and geometrical parameter, i.e., thickness ratio (hf/h).The predicted results are shown in Figure 8, where (a)–(f) denote the case of changingparameters, yield stress of ferromagnetic material (�f), the upper plateau stress (�0) andlower plateau stress (�1) of CuAlMn superelastic loop, and ratio of ferromagnetic plate(hf) to the composite (h), hf/h, Young’s modulus of ferromagnetic material (Ef) and that ofSMA (ESMA), respectively.

0

100

200

300

400

500

0 0.002 0.004 0.006 0.008 0.01 0.012

Fe (Ef=200GPa)

CuAlMn (ESMA

=60GPa)

Str

ess

: σ (

MPa

)

Strain : ε

0=100MPa

1=50MPa

σ

σ

σ

f=200MPa

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7

Nor

mal

ized

mom

ent,

M/E

SM

Abh

2 x10

-4

Nolmalized curvature, h/ρ x10-3

hf/h=0.5

(a) (b)

Figure 6. (a) Stress–strain curve for Fe and CuAlMn; (b) relation between normalized bending moment andnormalized curvature.

0

100

200

300

400

500

0 0.002 0.004 0.006 0.008 0.01 0.012

FeCoV (Ef=200GPa)

CuAlMn (ESMA

=60GPa)

Str

ess

: σ (M

Pa)

Strain : ε

σ 0=100MPa

σ 1=50MPa

f=400MPa

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7

Nor

mal

ized

mom

ent,

M/E

SMAbh

2 x10

-4

Nolmalized curvature, h/ρ x10-3

hf/h=0.5

Figure 7. (a) Stress–strain curve for FeCoV and CuAlMn; (b) relation between normalized bending momentand normalized curvature.



When the yield stress of the ferromagnetic material increases, it is clear from Figure 8(a)that both bending moment and the curvature increase. When transformation stress �0 ofSMA increases, it is found from Figure 8(b) that the bending moment increases and thecurvature decreases. It can be seen from Figure 8(c), the lower limit of the superelastic loopdecreases if the reverse transformation stress �1 decreases. When the thickness of theferromagnetic layer increases, it is clear from Figure 8(d) that the bending moment increases

0

2

4

6

8

1010

0 20 2 4 6 8 10

No

Nor

mal

izeded

m mom

ent,

M/

M/E

SM

SM

Abhbh

2 x1

x10

−4

Normrmalizedlized cu curvature, e, h/ρ x1 x10−3

303004040050500

σ σ f(M(MPa)

0

2

4

6

8

1010

0 20 2 4 6 8 10

No

Nor

mal

izeded

m mom

ent,

M/

M/E

SMSMAbhbh

2 x10

x10

−4

Normrmalizlizeded cu curvature,e, h h/ρ x10 x10−3

70 70100100150150

σ σ 0(M(MPa)

(a) (b)

0

2

4

6

8

1010

0 20 2 4 6 84 6 8 10

Nor

Nor

mal

izedzed

m mom

ent,

M/

M/E

SM

SM

Abhbh

2 x10

x10

−4

NormNormalizedlized cu curvature, he, h/ρ x10 x10−3

303050507070

σ σ 1(M(MPa)

0

2

4

6

8

1010

0 20 2 4 6 8 10

Nor

Nor

mal

izlizeded

m mom

ent,

M/

M/E

SMSMAbhbh

2 x10

x10

−4

Normrmalizeded cu curvature, h/ρ x10 x10−3

0.30.30.50.50.70.7

hf/h/h

(c) (d)

0

2

4

6

8

1010

0 20 2 4 6 84 6 8 10

Nor

Nor

mal

ize

lized

m mom

ent,

M/

M/E

SM

SM

Ab

hb

h2 x

10 x

10−4

NormNormalizlizeded cu curvature, h/ρ x10 x10−3

150150200200250250

Ef(G(GPa)

0

2

4

6

8

1010

0 20 2 4 6 8 10

Nor

Nor

mal

izizeded

m mom

ent,

M/

M/E

SM

SM

Ab

hb

h2 x

10 x

10−4

NorNormalizedzed cu curvature,e, h/ρ x10 x10−3

404060608080

ESMSMA

(G(GPa)

(e) (f)

Figure 8. Change in superelastic behavior of bending plate influenced by various parameters: (a) yield stressof Fe; (b) upper transformation stress of SMA; (c) lower transformation stress of SMA; (d) thickness ratio of Feto FSMA composite; (e) Young’s modulus of Fe; and (f) Young’s modulus of SMA. (a) �f ; (b) �0 ; (c) �1 ; (d) hf/h ;(e) Ef ; (f ) ESMA.



though the curvature decreases. Oppositely, because the thickness of superelastic SMAlayer increases when the thickness of a ferromagnetic layer decreases, the superelasticitybehavior increases. Therefore, the bending moment decreases, and the curvature increases.From Figure 8(e), the maximum curvature decreases though the bending moment does notchange when the Young’s modulus of the ferromagnetic material increases. Therefore, anincrease in the Young’s modulus of the ferromagnetic material is undesirable as thecomposite. From Figure 8(f), the bending moment decreases when the Young’s modulus ofSMA increases. The design of a more high performance FSMA composites becomespossible by the materials design based on the above analysis.

SUPERELASTIC BEHAVIOR OF COIL SPRING MODE

OF A COMPOSITE WIRE WITH RECTANGULAR CROSS SECTION

Analytical Model

With the aim of designing a high-speed linear actuator, the superelastic characteristic ofa coiled spring of the ferromagnetic shape memory composite wire with rectangularsection is analyzed. Figure 9 shows the analytical model. The magnetic force is generatedin the ferromagnetic material by the magnetic field gradient, and displacement is generatedin the spring by the hybrid mechanism described in Introduction. The relation between thisspring force and displacement is analyzed.

When axial force P is given to the spring, the wire of the ferromagnetic shape memorycomposite is subjected to torque T. The relation between spring force P and torque T isgiven by

T ¼ PR cos � ð15Þ

For a twist angle per unit length of the rectangular section wire of !, the total twist angle �is 2n�R!sec� as the total length of the wire is 2n�R!sec�. Therefore, the displacement ofthe spring is calculated by the next equation.

� ¼ �torsion þ �shear ffi �torsion

¼ R� ¼ 2n�R2! sec �ð16Þ

Figure 9. Analytical model of coil spring with rectangular cross section. D: the diameter of spring (D¼ 2R),d: the diameter of wire, p: the pitch of one cycle, n: the number of turns, L: the length of spring without load(L¼np), �: the inclined angle of the wire to the x–y plane.



It is assumed in the present model that the displacement due to direct shear, �shear isneglected. This is justified for large ratio of D to a or b. Then, the relation between thespring force, P and displacement, � can be calculated if the relation between the twist angleper unit length ! and the torque T of the rectangular section wire is known, which will beobtained in the following.

Analytical Model for Torsion of Composite Wire with Rectangular Section

To generate large magnetic force by the hybrid mechanism, it is necessary to increase thearea of a ferromagnetic material in the rectangular section, while meeting the requirementthat the ferromagnetic material should not reach its yield stress. The stress field in therectangular section can be calculated from the shear strain distribution of the rectangularsection for a given twist angle.

Let us look at the rectangular section of a composite with width 2a and height 2b asshown in Figure 9. We introduce the assumption that the spring deformation is uniformalong the wire direction (z-axis) and plane displacements u and v are in proportion to z, asfollows;

u ¼ �!yz, v ¼ !xz, w ¼ !’jðx, yÞ ð17Þ

where the function ’(x, y) is the Saint-Venant’s function [11] that satisfies the equilibriumequation and 2D compatibility equation of strain. For the spring with rectangular crosssection, the shear strain components are expressed as

zx!a

¼ �16

�2

X1n¼1

�1ð Þn�1

2n� 1ð Þ2

sinh 2n� 1ð Þ�y=2a½ �

cosh 2n� 1ð Þ�b=2a½ �cos 2n� 1ð Þ�x=2a½ � ð18Þ

zy!a

¼16

�2

X1n¼1

�1ð Þn�1

2n� 1ð Þ2

1�cosh 2n� 1ð Þ�y=2a½ �

cosh 2n� 1ð Þ�b=2a½ �

� �sin 2n� 1ð Þ�x=2a½ � ð19Þ

Therefore, the effective shear strain acting on the rectangular cross section, iscalculated by

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2zx þ 2zy

qð20Þ

For a¼ 2 and b¼ 1, the contour line distributions of shear strain component zx, zyand effective shear strain divided by a! are shown in Figure 10(a), (b) and (c)respectively. zx becomes 0 at x¼�a and a, and it reaches to the minimum value at y¼ bon the y axis, and becomes the maximum at y¼�b on the y axis. zy reaches to theminimum value at x¼�2, y¼ 0, and becomes the maximum at x¼ 2, y¼ 0. Thenormalized effective shear strain, /a! reaches the maximum value 0.930 at the center oflong side edges, and reduces toward the center.

The effective shear stress induced in the ferromagnetic material is calculated bymultiplying by the shear modulus Gf of the ferromagnetic material. The effective shearstress distribution of the ferromagnetic material in the rectangular section is calculated for



a given set of twist angle per unit length !, size a and b. Then, the optimum shape of theferromagnetic material can be determined from its domain under the condition that theeffective shear stress does not exceed the yield stress in shear f of the ferromagneticmaterial.

(a)

(b)

(c)

Figure 10. Contour line distributions of shear strain in rectangular section: (a) zx/a!; (b) zy/a! ; and (c) /a!,where a is the length of longer side of a rectangular cross section of a FSMA composite and ! is the twistangle per unit length.



If FeCoV (Gf¼ 70GPa, f¼ 231MPa) is used as a ferromagnetic material, andCuAlMn is used as a superelastic SMA, then for !¼ 0.003, a¼ 2, and b¼ 1, /!a<0.55 isobtained from the requirement of Gf <f. Figure 11 shows the optimized rectangularsection of the composite obtained by this design, where the dark area of FeCoV satisfies /!a<0.55.

Next, we examine the relation between the twist angle per unit length ! and the torque Tof the composite wire with rectangular section. The torque is calculated by

T ¼ Mz ¼ PR cos� ¼

Z b

�b

Z a

�a

xzy � yzx �

dxdy

¼

Z b

�b

Z a

�a

G xzy � yzx �

dxdy

ð21Þ

Here, we can define three domains in the composite during loading.

Domain 1 Domain of ferromagnetic materialDomain 2 Domain with effective shear stress less than the forward transformation

shear stress of SMA, 0Domain 3 Transformation domain of SMA

The effective stress in the ferromagnetic material is obtained by multiplying shearmodulus Gf by the corresponding effective shear strain for the elastic deformation. InSMA, it is necessary to judge if the effective shear stress is below the forwardtransformation shear stress 0. is obtained by multiplying shear modulus GSMA by ifthe effective shear stress of domain 2 is below the forward transformation shear stress 0.In domain 3 where the effective shear stress reaches the upper transformation shearstress 0, then ¼ 0.

Because a shear strain component proportionally increases with an increase in !, bymultiplying the corresponding shear strain component by the modified shear modulusG¼ 0/, the shear stress component for which becomes 0 is calculated. That is,Equation (21) is applicable to domain 3 by using Equation (24). Then, torque T

Figure 11. Optimized rectangular section of the composite made of ferromagnetic, FeCoV and superelasticSMA, CuAlMn.



corresponding to the twist angle per unit length ! is calculated by Equation (21) by usingthe modified shear modulus in each domain according to the following equations.

Domain 1:

<f!

Gf!f

G ¼ Gf

ð22Þ

Domain 2:

�f!

Gf!fand <

0GSMA

G ¼ GSMA

ð23Þ

Domain 3:

�f!

Gf!fand �

0GSMA

G ¼0

ð24Þ

where, !f is input data, and it is the maximum twist angle per unit length when the crosssection is optimized, !f¼ 0.0015 for cross section shape of Figure 11.

Next, the case of unloading is considered. The stress in each domain decreases duringthe unloading, but the superelasticity in SMA was generated in Domain 3 where theeffective shear stress reached the transformation stress during the preceding loading, it isnecessary to divide Domain 3 into three sub-domains.

Domain 3-1 above the reverse transformation stress 1Domain 3-2 equal to the reverse transformation stress 1Domain 3-3 below the reverse transformation stress 1

For Domain 3-1, the effective shear stress is larger than the reverse transformationstress 1. The shear stress component of ¼ 0 is calculated by multiplying modified shearmodulus of Equation (24) by the shear strain component, and it decreases from this stressstate elastically in proportion to GSMA in Domain 3. That is, the shear stress component iscalculated by multiplying the modified shear modulus of Equation (26) by the shear straincomponent in the range of the effective shear strain of Equation (25).

Domain 3-1

>1

GSMAand �

0 � 1ð Þ!

GSMA !f � !ð Þð25Þ

G ¼0� GSMA ð26Þ

For Domain 3-2, because the effective shear stress reaches the reverse transformationstress 1, the shear stress remains constant, i.e. ¼ 1. That is, the shear stress is calculated



by multiplying the modified shear modulus of Equation (28) by the shear strain in therange of effective shear strain of Equation (27).

Domain 3-2

>1

GSMAand >

0 � 1ð Þ!

GSMA !f � !ð Þð27Þ

G ¼1

ð28Þ

For Domain 3-3, the superelasticity disappears because the effective shear stress lowersmore than 1. The range of effective shear strain and modified shear modulus are given by

Domain 3-3

�1

GSMAð29Þ

G ¼ GSMA ð30Þ

The torque T corresponding to ! can be analyzed from Equation (21) by calculatingeffective shear strain of each area using the modified shear modulus corresponding to eachdomain defined by Equations (22)–(24), (26), (28) and (30). The relation between the forceand displacement of a spring can be calculated by using Equations (15) and (16).

Analytical Results and Discussion

Based on the above model, we made predictions of the torque (T ) – twist angle (!)relation, and also of the spring force (P) – displacement (�) relation where the idealizedstress–strain relations of ferromagnetic FeCoV and superelastic CuAlMn shown inFigure 12 are used.

Figure 13 shows the analytical results for the case of maximum twist angle per unitlength !¼ 0.003 of a composite plate wire with a¼ 2mm (width is 4mm), and b¼ 1mm(height is 2mm). Figure 13(a) shows the relation between the torque and the normalized

0

5050

10100

15150

20200

25250

0 0.002002 0.0.004004 0.0060.006 0.00.008 0.01 0.012.012

FeFeCoCoV ( (Gf=70G=70GPa)

CuCuAlMnMn ( (GSMSMA

=25G=25GPa)

Shea

ring

str

ess:

Shea

ring

str

ess:

τ (M

Pa)

(M

Pa)

Shearing strain : Shearing strain : γ

τ0=57.57.7M7MPa

τ1=28.28.9M9MPa

τf=230.9M=230.9MPa

Figure 12. Idealized stress–strain curves of FeCoV and CuAlMn.



twist angle, indicating that the torque rises proportionally as the twist angle increases, andthe transformation of SMA begins at !a¼ 0.0025, reaching the transformation stress with!a¼ 0.0042 in all domain of SMA. After !a reaches 0.006, the superelastic loop exhibitsthe reverse transformation corresponding to the unloading.

Figure 13(b) shows the relation between the spring force and the displacement of the coilspring of length L¼ 100mm, diameter D¼ 25mm, pitch p¼ 5mm and number of turnsn¼ 20. The maximum displacement of this coiled spring was 59.2mm, the spring forcebecame 78.4N.

We made a parametric study to examine the effects of each parameter on the P–�relation. Figure 14 shows the analytical results of the P–� relations influenced by variousparameters, (a) GSMA, (b) 0, (c) Gf, (d) f and (e) 1. From Figure 14(a), it is clear thatshear modulus of superelasticity SMA does not influence the maximum displacement andthe maximum spring force. It is noted from Figure 14(b), that the spring force increaseswith an increase in forward transformation shear stress 0. It is clear from Figure 14(c),that the spring force does not change and only the maximum displacement increases if theshear modulus of the ferromagnetic material becomes small resulting in largerdisplacement of the spring. It can be seen from Figure 14(d), that both the spring forceand displacement increase the superelastic behavior when the yield stress of theferromagnetic material increases. It is noted from Figure 14(e), that the lower limit ofsuperelastic loop decreases if the reverse transformation stress 1 decreases.

In summary, larger f of the ferromagnetic material and softer ferromagnetic materialwill provide a spring actuator with larger displacement. And, to obtain large force of thespring, use of SMA of larger 0 is desired.

We are examining two kinds [Fe(Gf¼ 70GPa, f¼ 116MPa) and FeCoV(Gf¼ 70GPa,f¼ 231MPa)] as a ferromagnetic material from the view point of low cost and easiness ofprocessing. It follows from Figure 14(d) that Fe does not show the superelastic behaviorand the spring force and displacement are small. Therefore, we considered that FeCoVwhose f is large was suitable as the ferromagnetic material.

Next, we shall compare the mechanical performance (P–� relation) of a spring between‘‘rectangular’’ and ‘‘square’’ cross section. To this end, the cross section area of the squareis made equal to that of the rectangular studied earlier (Figure 12). The analytical resultsof the optimum square cross section of FeCoV/CuAlMn composite are shown in Figure15(a), while the P–� relation of the FSMA spring with this square cross section is given in

0

200200

400400

600600

800800

1000000

0 0.00202 0.00004 0.00606 0.00008

(a)(a) (b)(b)

T (

T (

Nm

)

ωa

0

1010

2020

3030

4040

5050

6060

7070

8080

0 20 40 40 60 80

Forc

e: P

(N)

Forc

e: P

(N)

DisDisplalacemcemenent : : δ ( (mm)mm)

Figure 13. Superelastic behavior of Fe/CuAlMn composite spring: (a) relation between torque and normalizedtwist angle; (b) spring force (P)–displacement (�) curve.



Figure 15(b) as a dashed line where the results of the rectangular cross section are alsoshown by solid line. A comparison between the square cross section of Figure 15(a) andthe rectangular cross section of Figure 11 reveals that the FSMA composite spring withsquare cross section provides larger force capability than that with the rectangular crosssection for the same cross section area. However, the effectiveness of using the spring with

0

2020

4040

6060

8080

100100

0 20 40 60 60 80 80 100

Fo

For

ce:

P P(N

)

DiDispsplacemcemenent : : δ (m (mm)

GSMSMA

(G (GPa)

152550

0

2020

4040

6060

8080

100100

0 20 40 40 60 80 80 100

Fo

For

ce:

P

P (

N)

DiDispsplacement : : δ (m (mm)

0 (M (MPa)

40. 40.4 57. 57.7 86. 86.6

(a) (b)

0

2020

4040

6060

8080

100100

0 20 40 40 60 80 80 100

Fo

For

ce:

P

P (

N)

DispDisplalacemcemenent : : δ (m (mm)

G Gf (G (GPa)

505070709090

0

2020

4040

6060

8080

10100

0 20 40 40 60 60 80 100

Fo

For

ce:

P

P (

N)

DispDisplacecement : : δ ( (mm)mm)

f (M (MPa)

173173231231289289

(c) (d)

0

2020

4040

6060

8080

100100

0 20 40 60 60 80 80 100

Fo

For

ce:

P P(N

)

DispDisplalacemcemenent : : δ (m (mm)

1 (M (MPa)

17.328.940.4

(e)

Figure 14. Effects of various parameters on P–� relation of FMSA composite springs: (a) SMA shear modulus,GSMA ; (b) forward transformation shear stress, 0 ; (c) shear modulus of a ferromagnetic material, Gf ; (d) theyield stress in shear of a ferromagnetic material, f ; and (e) reverse transformation shear stress, 1.



the square cross section remains to be determined after its effectiveness of inducing largemagnetic force between the neighboring turns of the spring.

CONCLUSION

The predicted results of the bending moment–curvature of a FSMA composite plateexhibit superelastic behavior of the composite beam while those of the FSMA compositespring with rectangular cross section show also similar superelastic behavior. The abovesuperelastic behavior is the performance required for FSMA composite actuators withhigh force and displacement capability. The results of the simple model were usedeffectively for optimization of the cross section geometry of two types of FSMAcomposite, bending and torsion types.

ACKNOWLEDGMENT

This study was supported by a Grant from AFOSR to University of Washington(F49620-02-1-0028) where Dr. Les Lee is the Program Manager.

APPENDIX

Relation Between Bending Moment and Curvature

The relation between the normalized bending moment and the normalized curvatureof the FMSA composite plate is classified into the following eight patterns as shownFigure A.1.

0

2020

4040

6060

8080

10100

0 20 40 40 60 60 80 100

Fo

For

ce:

P

P (

N)

DispDisplalacemcemenent : : (m (mm)

ReRectangularSquarquare

(a) (b)

Figure 15. Superelastic behavior of Fe/CuAlMn composites: (a) shape of cross section; (b) spring force–displacement curve.



Case 1 is constructed with only Pattern 1. (Figure A.1(a))Case 2 is constructed with Patterns 1 and 2 for the loading, and Patterns 1, 4, 5, and 6

for the unloading. (Figure A.1(b))Case 3 is constructed with Patterns 1, 2 and 3 for the loading, and Patterns 1, 4, 7, and 8

for the unloading. (Figure A.1(c))Equations of each pattern are shown as follows.

Pattern 1 (Cases 1–3)

M

ESMAbh2¼

h

�

Ef

ESMA

1

3

hf

h

� �3

�1

2

�1h

hf

h

� �2( )

þ1

31�

hf

h

� �3( )

�1

2

�1h

1�hf

h

� �2( )" #

where, �1 is the distance of the neutral axis.

�1h¼

ðEf=ESMAÞ � 1ð Þ hf=hð Þ2þ1

2 ðEf=ESMAÞ � 1ð Þ hf=hð Þ þ 1�

Pattern 2 (Cases 2,3)

M

ESMAbh2¼

h

�

Ef

ESMA

1

3

hf

h

� �3

�1

2

�2h

hf

h

� �2( )

þ1

3

y1

h

� �3�

hf

h

� �3( )

�1

2

�2h

y1

h

� �2�

hf

h

� �2( )" #

þ1

2

�0ESMA

1�y1

h

� �2� �

where, �2 is the distance of the neutral axis, and y1 is the position for � ¼ �0.

�2h¼ �

Ef

ESMA� 1

� �hf

hþ

�0ESMA

�

h

� �

þ


ESMA

Ef

ESMA� 1

� �hf

h

� �2

þ2�0

ESMA

�

h1þ

Ef

ESMA� 1

� �hf

h

� �s

y1

h¼

�2hþ

�0ESMA

�

h

Pattern 1

Loading and

Unloading

M/E

Abh

2

h/ρ

Pattern 1 Pattern 5

Pattern 6

Pattern 2

Pattern 4

Loading

Unloading

M/E

Abh

2

h/ρ

Pattern 1Pattern 7

Pattern 8

Pattern 3

Pattern 2

Pattern 4

Loading

Unloading

M/E

Abh

2

h/ρ

(a) (b) (c)

Figure A1. Relation between normalized bending moment and normalized curvature: (a) Case 1; (b) Case 2;(c) Case 3.



Pattern 3 (Case 3)

M

ESMAbh2¼

h

�

Ef

ESMA

1

3

hf

h

� �3

�1

2

�3h

hf

h

� �2( )

þ1

2

�0ESMA

1�hf

h

� �2( )


�3h¼

�0Ef

�

h

h

hf� 1

� �þ1

2

hf

h

Pattern 4 (Cases 2,3)

M

ESMAbh2¼

h

�

Ef

ESMA

1

3

hf

h

� �3

�1

2

�4h

hf

h

� �2( )

þ1

3

y2

h

� �3�

hf

h

� �3( )

�1

2

�4h

y1

h

� �2�

hf

h

� �2( )" #

þ1

2

�1ESMA

1�y2

h

� �2� �

where, �4 is the distance of the neutral axis, and y2 is the position for � ¼ �0.

�4h¼ �

Ef

ESMA� 1

� �hf

hþ

�1ESMA

�

h

� �

þ


ESMA

Ef

ESMA� 1

� �hf

h

� �2

þ2�1

ESMA

�

h1þ

Ef

ESMA� 1

� �hf

h

� �s

y2

h¼

�4hþ

�1ESMA

�

h

Pattern 5 (Case 2)

M

ESMAbh2¼

h

�

1

3

Ef

ESMA� 1

� �hf

h

� �3

þy3

h

� �3( )�1

2

�5h

Ef

ESMA� 1

� �hf

h

� �2

þy3

h

� �2( )" #

�h

�1

1

3

y3

h

� �3�

Y1

h

� �3( )

�1

2

�5h

y3

h

� �2�

Y1

h

� �2( )" #

þ1

2

�0ESMA

y3

h

� �2�

Y1

h

� �2( )

þ1

2

�1ESMA

1�y3

h

� �2� �




�5h¼

�B3 þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiB23 �A3C3

qA3

A3 ¼ 1�h

�1

�

h

B3 ¼Ef

ESMA� 1

� �hf

hþ

h

�1

Y1

h��0 � �1ESMA

� ��

h

C3 ¼ �Ef

ESMA� 1

� �hf

h

� �2

�h

�1

Y1

h� 2

�0ESMA

� �Y1

hþ 2

�1ESMA

��0 � �1ESMA

� �2 �1�1 � �

�

h

( )�

h

y3

h¼

�5h��0 � �1ESMA

�1�1 � �

�

h

Y1

h¼

�fEf

þ�0

ESMA

� ��1h

h

�1¼

�fEf

þ�0

ESMA

� �2( ),(

��fEf

Ef

ESMA� 1

� �hf

h�

�0ESMA

� �

þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�fEf

Ef

ESMA� 1

� �hf

h�

�0ESMA

� �2

þ�fEf

þ�0

ESMA

� �2Ef

ESMA� 1

� �hf

h

� �2s )

Pattern 6 (Case 2)

M

ESMAbh2¼

h

�

Ef

ESMA

1

3

hf

h

� �3

�1

2

�6h

hf

h

� �2( )

þ1

31�

hf

h

� �3( )

�1

2

�6h

1�hf

h

� �2( )" #

�h

�1

1

31�

Y1

h

� �3( )

�1

2

�6h

1�Y1

h

� �2( )" #

þ1

2

�0ESMA

1�Y1

h

� �2( )


�6h¼

ðEf=ESMAÞ � 1ð Þ hf=hð Þ2þ1� ðh=�1Þ 1þ ðY1=hÞð Þ � 2ð�0=ESMAÞ

� 1� ðY1=hÞð Þð�=hÞ

2 ðEf=ESMAÞ � 1ð Þ hf=hð Þ þ 1� ðh=�1Þ 1� ðY1=hÞð Þð�=hÞ�

Pattern 7 (Case 3)

M

ESMAbh2¼

h

�

1

3

Ef

ESMA� 1

� �hf

h

� �3

þy4

h

� �3( )�1

2

�7h

Ef

ESMA� 1

� �hf

h

� �2

þy4

h

� �2( )" #

�h

�2

1

3

y4

h

� �3�

hf

h

� �3( )

�1

2

�7h

y4

h

� �2�

hf

h

� �2( )" #

þ1

2

�0ESMA

y4

h

� �2�

hf

h

� �2( )

þ1

2

�1ESMA

1�y4

h

� �2� �




�7h¼

�B4 þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiB24 � A4C4

qA4

A4 ¼ 1�h

�2

�

h

B4 ¼Ef

ESMA� 1

� �hf

hþ

h

�2

hf

h��0 � �1ESMA

� ��

h

C4 ¼ �Ef

ESMA� 1

� �hf

h

� �2

�h

�2

hf

h� 2

�0ESMA

� �hf

hþ 2

�1ESMA

��0 � �1ESMA

� �2 �2�2 � �

�

h

( )�

h

y4

h¼

�7h��0 � �1ESMA

�2�2 � �

�

h

h

�2¼ 2

h

hf

�fEf

��0Ef

h

hf� 1

� ��

Pattern 8 (Case 3)

M

ESMAbh2¼

h

�

Ef

ESMA

1

3

hf

h

� �3

�1

2

�8h

hf

h

� �2( )

þ1

31�

hf

h

� �3( )

�1

2

�8h

1�hf

h

� �2( )" #

�h

�2

1

31�

hf

h

� �3( )

�1

2

�8h

1�hf

h

� �2( )" #

þ1

2

�0ESMA

1�hf

h

� �2( )


�8h¼

ðEf=ESMAÞ � 1ð Þ hf=hð Þ2þ1� ðh=�2Þ 1þ ðhf=hÞð Þ � 2ð�0=ESMAÞ

� 1� ðhf=hÞð Þð�=hÞ

2 ðEf=ESMAÞ � 1ð Þ hf=hð Þ þ 1� ðh=�2Þ 1� ðhf=hÞð Þð�=hÞ�

USEFUL RANGE

The useful range of the curvature of each pattern is shown as follows.

Case 1

Pattern 1 (Loading and Unloading)

0 �h

��

�fEf

2 1þ ðEf=ESMAÞ � 1ð Þ hf=hð Þ� 1þ ðEf=ESMAÞ � 1ð Þ hf=hð Þ

2



Case 2

Pattern 1 (Loading)

0 �h

��

�0ESMA


1þ ðEf=ESMAÞ � 1ð Þ 2� ðhf=hÞð Þ hf=hð Þ

Pattern 2 (Loading)

�0ESMA



h

��

h

�1

Pattern 1 (Unloading)

0 �h

��

�1ESMA




�1ESMA



<h

��

2ð�1=ESMAÞ 1þ ðEf=ESMAÞ � 1ð Þ hf=hð Þ�

Y1=hð Þ2þ ðEf=ESMAÞ � 1ð Þ 2ðY1=hÞ � ðhf=hÞð Þ hf=hð Þ



Y1=hð Þ2þ ðEf=ESMAÞ � 1ð Þ 2ðY=h1Þ � ðhf=hÞð Þ hf=hð Þ

<h

��

2A1

�B1 �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiB21 � 4A1C1

q

where,

A1 ¼ 2�1

ESMAþ 1�

Y1

h

� �h

�1

� �1�

Y1

h

� �

B1 ¼ 2Ef

ESMA� 1

� �hf

hþ 1

� ��0 � �1ESMA

�1h� 1

� �þ

Ef

ESMA� 1

� �hf

h

� �2

þ 1� 1�Y1

h

� �1�

Y1

hþ 2

�0ESMA

�1h

� �

C1 ¼�1h

Ef

ESMA� 1

� �2�

hf

h

� �hf

hþ 1

� �




2A1

�B1 �


q <h

��

h

�1

Case 3

Pattern 1 (Loading)

0 �h

�� ð�0=ESMAÞ



Pattern 2 (Loading)

�0ESMA



h

�� 2

h

hf

�0ESMA

þ�0Ef

hf

h� 1

� ��

Pattern 3 (Loading)

2h

hf

�0ESMA

þ�0Ef

hf

h� 1

� �� <

h

��

h

�2


0 �h

��

�1ESMA




�1ESMA



<h

��






<h

��

2A2

�B2 �


q



where,

A2 ¼ 2�1

ESMAþ 1�

hf

h

� �h

�2

� �1�

hf

h

� �

B2 ¼ 2Ef

ESMA� 1

� �hf

hþ 1

� ��0 � �1ESMA

�2h� 1

� �þ

Ef

ESMA� 1

� �hf

h

� �2

þ 1� 1�hf

h

� �1�

hf

hþ 2

�0ESMA

�2h

� �

C2 ¼�2h

Ef

ESMA� 1

� �2�

hf

h

� �hf

hþ 1

� �


2A2

�B2 �


q <h

��

h

�2

REFERENCES

1. Kato, H., Wada, T., Tagawa, T., Liang, Y. and Taya, M. (2001). Proc. of 50th Anniversary ofJapan Society of Materials Sciences, A: 296–305.

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