Materials Science and Engineering A 420 (2006) 220–227

Transformation characteristics of shape memory alloys in a thermal cycle

L. An a, W.M. Huang b,∗a College of Transportation Engineering, Southeast University, Nanjing, PR China

b School of Mechanical and Aerospace Engineering, Nanyang Technological University, Nanyang Avenue, Singapore

Received 1 February 2005; received in revised form 4 January 2006; accepted 17 January 2006

Abstract

This paper presents a systematic study on the transformation characteristics of shape memory alloys in a thermal cycle with/without a load. Themethods to determine some key material properties are proposed. The expressions for estimating the hysteresis and transformation interval underdifferent types of loads are obtained. The influential factors in real experiments are identified and discussed.© 2006 Elsevier B.V. All rights reserved.

Keywords: Shape memory alloy; Phase transformation; Hysteresis; Transformation interval; Thermal cycling

1

httet

apptrt(trhuc

rcci

0d

. Introduction

As a promising smart material, shape memory alloy (SMA)as attracted a lot of attention from the scientific and indus-rial communities because of its two unique features, namely,he recovery to the original shape upon heating (shape memoryffect, or in short term, SME) and super-elasticity at a tempera-ure above the austenite finish temperature [1,2].

Since the properties of SMAs, in particular NiTi based alloys,re very sensitive to, for instance, the exact composition, androcessing procedure, etc. [3], from the engineering applicationoint of view, it is essential to know the exact properties of a par-icular SMA, so that the design can be efficient in handling theequired task. While some of the properties, e.g. the transforma-ion temperatures in the thermally induced phase transformationload-free), can be directly determined from a standard test, in ahermal cycle, the transformation characteristics may be alteredemarkably if a load is applied. For instance, Huang and Xu [4]ave investigated whether the hysteresis is a constant in two sit-ations: thermal cycling under a constant load and mechanicalycling of a super-elastic SMA.

thermal cycling under a variable load, our particular concern ison the thermal cycling of a SMA film deposited atop an elasticsubstrate, which is largely motivated by the current intensiveinterest in the NiTi based SMA thin films for MEMS applica-tions.

The layout of this paper is as follows. Section 2 summariesthe framework. Section 3 presents the expressions of some keymaterial properties in terms of the easily measurable param-eters in some typical thermal cyclic tests, and the simplifiedformulas for transformation temperatures under different loadconditions. Subsequently, two important transformation charac-teristics, namely, hysteresis and transformation interval, underdifferent loading conditions are compared quantitatively. Sec-tion 4 addresses the influential factors that may happen in a realtest. Section 5 is conclusion.

2. Framework

A piece of SMA is assumed to be homogenous. Therefore,the complementary free energy (Ψ ) of a unit volume may bewritten as [7]

While Ortin and Plane [5,6] have previously investigated theelationships among the material properties and transformationharacteristics in the load-free case, this paper aims to provide aomplete picture on the transformation characteristics of SMAsn a thermal cycle under different types of loads. In the case of

Ψ (Σij, T, ξ) = −[Φ(Σij, T, ξ) − ΣijEij] (1)

where Σij is the macroscopic stress tensor, Eij is the macroscopicstrain tensor, T is the temperature and ξ is the fraction of phasetifd

∗ Corresponding author. Tel.: +65 67904859; fax: +65 67911859.E-mail address: mwmhuang@ntu.edu.sg (W.M. Huang).

921-5093/$ – see front matter © 2006 Elsevier B.V. All rights reserved.oi:10.1016/j.msea.2006.01.062

ransformation. Although the strain in the phase transformations not small (the maximum is about 6% in NiTi SMAs [3]),or simplicity, it is still reasonable to assume that Eij can beivided into two parts, namely, the elastic strain Ee

ij and the

L. An, W.M. Huang / Materials Science and Engineering A 420 (2006) 220–227 221

transformation strain Etij . In Eq. (1), the Helmoltz free energy

Φ is defined as

Φ(Σij, T, ξ) = �Gch(T ) + Wmech + Wsurf (2)

where �Gch is the change of the chemical free energy due tothe phase transformation. Wsurf is the surface energy at the inter-phase (between the austenite and martensite phases) and inter-face (between martensite variants), and Wmech is the mechanicalenergy. As a conventional practice, Wsurf is small and can beignored for simplicity [1].

According to the second law of thermodynamics, one has

dΨ = dEfric ≥ 0 (3)

for any transformation process. Here, Efric, the energy dissipa-tion, is largely due to the movement of interface and/or inter-phase. A precise mathematic expression for Efric is difficult toobtain unless there is only one interface/inter-phase. For sim-plicity, we assume it is a linear function of the transformationfraction. For instance, in the martensitic transformation fromaustenite (A) to martensite (M), one has

dEfric = EA→Mfric dξA→M (4)

where ξA→M is the martensite fraction.In a uniform and homogenous material, the chemical free

ecf

�

If(

�

Tb

W

wnt

W

Fl

3c

Tofi

3.1. Without load

Upon cooling a piece of austenite SMA to a critical tem-perature (Ms), the thermally induced martensitic transformationstarts. The resultant martensite, i.e., thermally induced marten-site (Mtim), is twinned, and one has

dΨ = −�Gch(T − Tequ)dξA→Mtim

− WA→Mtim

int (ξA→Mtim)dξA→Mtim

= dEfric = EA→Mtim

fric dξA→Mtim(9)

in the transformation process. Solving for T yields

T = Tequ − EA→Mtim

fric + WA→Mtim

int (ξA→Mtim)

�Gch(10)

Substituting the martensite start temperature (Ms) and themartensite finish temperature (Mf), and the corresponding val-ues of ξA→Mtim

(i.e., ξA→Mtim = 0, Wint = 0, and ξA→Mtim = 1,Wint = WA→Mtim

int (1)) into Eq. (10) yields

Ms = Tequ − EA→Mtim

fric

�Gch(11a)

Mf = Tequ − EA→Mtim

fric + WA→Mtim

int (1)(11b)

I

T

ffi

A

A

a

•

•

•

•

T

nergy is a function of temperature. At a temperature T, thehange of the chemical energy due to the phase transformationrom austenite to martensite in SMAs can be written as

Gch(T ) = [GMch(T ) − GA

ch(T )]ξA→M = �GA→Mch ξA→M (5)

t should be reasonably accurate to use a linear approximationor �Gch(T) around the equilibrium temperature Tequ. Thus, Eq.5) can be rewritten as

Gch(T ) = �Gch(T − Tequ)ξA→M (6)

he mechanical energy Wmech (elastic energy) in Eq. (2) is giveny

mech = 12ΣijCijklΣkl + Wint (7)

here Cijkl is the elastic compliance tensor and Wint is the inter-al energy, which is a function of ξA→M in the martensiticransformation, i.e.

int = Wint(ξA→M) (8)

or an estimation of the internal energy in SMAs from theirattice structures, one may refer to, for instance, [8].

. Material properties and transformationharacteristics

Consider a SMA under a thermal cycle with/without a load.he applied load can be either a constant (e.g., a dead weight)r a variable (e.g., thermal cycling a sputter deposited SMA thinlm atop a silicon wafer).

�Gch

n a similar way, one has

= Tequ + EMtim→Afric − WA→Mtim

int (ξA→Mtim)

�Gch(12)

or the reverse transformation. Hence, the austenite start andnish temperatures are

s = Tequ + EMtim→Afric − WA→Mtim

int (1)

�Gch(13a)

f = Tequ + EMtim→Afric

�Gch(13b)

Fig. 1 shows the evolution of ξA→Mtimof four different situ-

tions in a thermal cycle:

Fig. 1(a): Energy dissipation Efric = 0, internal energy Wint =0. Hence, Ms = Mf = As = Af = Tequ.Fig. 1(b): Energy dissipation Efric �= 0, internal energy Wint =0. Hence, Ms = Mf < Tequ < As = Af.Fig. 1(c): Energy dissipation Efric = 0, internal energy Wint �=0. Hence, Ms = Af < Mf = As = Tequ.Fig. 1(d): Energy dissipation Efric �= 0, internal energy Wint �=0. Depending on the magnitude of WA→Mtim

int (1), it might beMf < Ms < As < Af or Mf < As < Ms < Af.

Note that if EA→Mtim

fric = EMtim→Afric ,

equ = Ms + Af

2(14)

222 L. An, W.M. Huang / Materials Science and Engineering A 420 (2006) 220–227

Fig. 1. Thermally induced phase transformation. (a) Efric = 0, Wint = 0; (b) Efric �= 0, Wint = 0; (c) Efric = 0, Wint �= 0; and (d) Efric �= 0, Wint �= 0 [9].

Eq. (14) is the formula used by many researchers to determineTequ (e.g., [1,5,6]). However as revealed above, it is a precise

solution only if EA→Mtim

fric = EMtim→Afric . It is well known that in

some NiTi SMAs, there is an intermediate R-phase upon coolingfrom austenite to martensite. But in the reverse transformation,the transformation is from marteniste to austenite directly. It isa clear evidence indicating EA→Mtim

fric �= EMtim→Afric .

The hysteresis, H(=Af − Ms or As − Mf), in a thermal cycleis given by

H = EA→Mtim

fric + EMtim→Afric

�Gch(15)

and the transformation interval �T(=Af − As or Ms − Mf) is

�T = WA→Mtim

int (1)

�Gch(16)

In a differential scanning calorimeter (DSC) test (Fig. 2), thetotal energy dissipation in a full thermal cycle (EA→Mtim

fric +EMtim→A

fric ) is the difference between the areas of the peak (Ah)and the trough (Ac). Since all transformation temperatures canbe estimated from a DSC curve, H and �T can be estimated.

Subsequently, �Gch and WA→Mtim

int (1) can be calculated fromEqs. (15) and (16), respectively.

3.2. Under a constant load

If a constant load is applied in a thermal cycle, the resultantmartensite variant(s) is detwinned, and the terms of the externalload Σc

ij (hereinafter, superscript c means constant load) shouldbe included. Hence, one has

− 12Σc

ij(CAijkl − CMi

ijkl)Σckl + Σc

ijEA→Mi

ij

= EA→Mi

fric + �Gch(Mcs − Tequ) (17a)

for the start of martensitic transformation, and

− 12Σc

ij(CAijkl − CMi

ijkl)Σckl + Σc

ijEA→Mi

ij

= EA→Mi

fric + WA→Mi

int (1) + �Gch(Mcf − Tequ) (17b)

for the finish of martensitic transformation. For simplicity, weassume the resultant martensite variant(s) is the same in thewhole cycle.

Here, CAijkl and CMi

ijkl are the elastic compliance tensors of

austenite and martensite Mi, respectively. As a rough estimation,o

[

[

wot

Fig. 2. Schematic illustration of DSC result.

ne may take [10]

CAijkl] = 1

DA

⎡⎢⎣

1 −ν −ν

−ν 1 −ν

−ν −ν 1

⎤⎥⎦ (18a)

CMi

ijkl] = 1

DM

⎡⎢⎣

1 −ν −ν

−ν 1 −ν

−ν −ν 1

⎤⎥⎦ (18b)

here ν is the Poisson’s ratio (assuming that the Poisson’s ratiosf martensite and austenite are the same), and DA and DM arehe Young’s moduli of austenite and martensite, respectively.

L. An, W.M. Huang / Materials Science and Engineering A 420 (2006) 220–227 223

Similarly, the conditions for the start and finish of the reversetransformation are

12Σc

ij(CAijkl − CMi

ijkl)Σckl + Σc

ijEMi→Aij

= EMi→Afric − WA→Mi

int (1) + �Gch(Tequ − Acs) (19a)

and

12Σc

ij(CAijkl − CMi

ijkl)Σckl + Σc

ijEMi→Aij

= EMi→Afric + �Gch(Tequ − Ac

f ) (19b)

Subsequently, the four transformation temperature can beobtained from Eqs. (17) and (19). Since EA→Mi

ij = −EMi→Aij ,

the hysteresis, Hc(= Acf − Mc

s or Acs − Mc

f ), in a thermal cycleis

Hc = EA→Mi

fric + EMi→Afric

�Gch(20)

and the transformation interval �T c(= Acf − Ac

s or Mcs − Mc

f )is

�T c = WA→Mi

int (1)

�Gch(21)

In comparison with Eqs. (15) and (16), one can see that if theetcaetl

3

it

elastic substrate) as an example. By measuring the curvature(K) verse temperature relationship of the bilayer structure, onecan work out the evolution of stress and strain in the film againtemperature [11]. If the film is very thin so that the stress (σf)and strain (εf) inside of it are more or less uniform [12], one has

σf = − KEst3s

6tf(tf + ts)(22a)

εf = 16K[4ts + tf

(5 + tf

tf+ts

)]+ αsT (22b)

where t is thickness, andα is the coefficient of thermal expansion.Here, subscripts f and s stand for the SMA film and substrate,respectively. εf has two components, namely, thermal strain andmechanical strain. Given the coefficient of thermal expansionof SMA film, one can work out the mechanical strain of thefilm. After deducting elastic strain from mechanical strain, theremaining is the strain resulted by the phase transformation,reorientation of martensite variant(s), or a mixture of them.

Eq. (22a) is for the uniaxial stress state. In the case of biaxial

stress state, E should be replaced by E(= E

1−v

), the biaxial

modulus [13]. A simplified typical film stress verse temperaturerelationship in a thermal cycle is illustrated in Fig. 3 (dark lines).Upon cooling from point o to point b, the film is austenite. Thecurvature versus temperature slope in this range can be expresseda

wfi

w

n

ture re

nergy dissipation and internal energy are constants, the hys-eresis and transformation interval in a thermal cycle under aonstant load are identical to that in a thermal cycle without anypplied load. Otherwise Eqs. (20) and (21) can be applied tostimate the real energy dissipation and internal energy, respec-ively. It is possible that the values of them are stress-state andoad level dependent.

.3. Under a variable load

Now we consider the case that the applied load Σvij (here-

nafter, superscript v means a variable load) is a variable. Takehermal cycling a bilayer structure (a SMA thin film atop an

Fig. 3. Film stress vs. tempera

s [12]

dK

dT= 6EA

f Estfts(tf + ts)(αAf − αs)

(EAf )

2t4f + E2

s E4s + 2EA

f Estfts(2t2f + 3tfts + 2t2

s )(23)

here superscript A stands for austenite. In the case of very thinlm, the slope of ob is

dσf

dT= EA

f (αAf − αs) (24)

hich has been widely used in thin film analysis.At point b, the martensitic transformation starts. At point c,

o austenite exists but martensite only. Hence, the condition for

lationship in thermal cycling.

224 L. An, W.M. Huang / Materials Science and Engineering A 420 (2006) 220–227

the start of martensitic transformation is

− 12Σvb

ij (CAijkl − CMj

ijkl)Σvb

kl + Σvb

ij EA→Mj

ij

= EA→Mj

fric + �Gch(Mvs − Tequ) (25a)

while that for the finish of martensitic transformation is

− 12Σvc

ij (CAijkl − CMj′

ijkl )Σvc

kl + Σvc

ij EA→Mj′ij

= EA→Mj′fric + WA→Mj′

int (1) + �Gch(Mvf − Tequ) (25b)

Practically, the resultant martensite variant(s) can be differentfrom that in thermal cycling under a variable or constant load,and may even vary during thermal cycling with the variation ofload.

Upon heating from point c to point d, the deformation of thefilm is roughly elastic if the V-shape effect in the temperatureversus transformation start stress relationship is ignored [10].Thus, similar to Eq. (24), the slope of cd is given by

dσf

dT= EM

f (αMf − αs) (26)

Here, superscript M is for martensite. It is obvious that in gen-eral, the slopes of ab and cd are not the same. It should be pointedout that Eqs. (24) and (26) are valid only if the deformation inSMA thin film is elastic in the referred temperature range. Ingeneral, this condition can be satisfied for oab line as the mate-rrodE

ps

a

It is clear that the hysteresis in this case is (Fig. 3)

Hv ={

Avs − Mv

f

Avf − Mv

s(28)

On the other hand, the transformation interval in thermal cyclingthe SMA film atop an elastic substrate is

�T v ={

Avf − Av

s

Mvs − Mv

f(29)

Superimpose the transformation line versus temperaturerelationships, which indicate the variation of transformationtemperatures in thermal cycling under a constant load orthe transformation stresses at a constant temperature (referto, for instance, [14,7], for details about the relationships),into Fig. 3 (grey lines). Points c, b, d, and a should passthe transformation lines of Mf, Ms, As and Af, respectively.On the one hand, since ha(= Hc) > ga(= Hv), Hc > Hv. Onthe other hand, as cf (= �T v) > ce(= �T c), �Tv > �Tc.For a simple comparison, we ignore the difference in elasticconstants between austenite and martensite. Therefore, onehas

Mvs = Tequ

(Σvb

ij EA→Mj

ij − EA→Mj

fric

�G

)(30a)

M

A

A

T

vs −

vf −

w

(Σvij

(Σ

ial is austenite, which is hard. However, particular attention isequired in applying Eq. (26), since martensite reorientation mayccur under an almost constant load in a small strain range. If thisoes happen, the slope can only be obtained by differentiatingq. (22a) with respect to temperature.

Upon further heating, the reverse transformation starts atoint d and finishes at point a. The critical conditions corre-ponding to these two points are

12Σvd

ij (CAijkl − CMj′

ijkl )Σvd

kl + Σvd

ij EMj′→Aij

= EMj′→Afric − WA→Mj′

int (1) + �Gch(Tequ − Avs ) (27a)

nd

12Σva

ij (CAijkl − CMj

ijkl)Σva

kl + Σva

ij EMj→Aij

= EMj→Afric + �Gch(Tequ − Av

f ) (27b)

Hv =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

A

A

�T v =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

Avf − Av

s = WA→Mj′int (1) +

Mvs − Mv

f = WA→Mj′int (1) +

ch

vf = Tequ +

⎛⎝Σvc

ij EA→Mj′ij − EA→Mj′

fric − WA→Mj′int (1)

�Gch

⎞⎠

(30b)

vs = Tequ −

⎛⎝Σvd

ij EMj′→Aij − EMj′→A

fric + WA→Mj′int (1)

�Gch

⎞⎠

(30c)

vf = Tequ −

(Σva

ij EMj→Aij − EMj→A

fric

�Gch

)(30d)

hus, the hysteresis can be expressed as

Mvf = EMj′→A

fric + EA→Mj′fric + (Σvd

ij − Σvc

ij )EA→Mj′ij

∆Gch

Mvs = EMj→A

fric + EA→Mj

fric + (Σva

ij − Σvb

ij )EA→Mj

ij

�Gch

(31)

hile the transformation interval can be expressed as

aEMj→A

ij − Σvd

ij EMj′→Aij ) − (EMj→A

fric − EMj′→Afric )

∆Gch

vb

ij EA→Mj

ij − Σvc

ij EA→Mj′ij ) − (EA→Mj

fric − EA→Mj′fric )

�Gch

(32)

L. An, W.M. Huang / Materials Science and Engineering A 420 (2006) 220–227 225

In comparison with the hysteresis and transformation intervalin thermal cycling without a load or under a constant load, thevariable load causes a slightly reduced hysteresis (the reductionis roughly proportional to (Σvd

ij − Σvc

ij ) or (Σva

ij − Σvb

ij )) anda significantly widened transformation interval (roughly, theincrement is proportional to (Σvb

ij − Σvc

ij ) or (Σva

ij − Σvd

ij )). If

EMj→Afric ≈ EMj′→A

fric , the slopes of bc and ad are the same, andcan be expressed as

dσf

dT= �Gch

EMj→Aij

(33)

Hence, by measuring the slope of bc or ad, with �Gch obtainedelsewhere (e.g., DSC test, mentioned in Section 3.1), onecan estimate EMj→A

ij from Eq. (33). Subsequently, from Eqs.

(31) and (32), (EMj′→Afric + EA→Mj′

fric ) and WA→Mj′int (1) can be

determined.

In the case of recovery under a full constraint, i.e., thermalcycling with the deformation of the SMA fixed, one can referto the above-mentioned procedure and follow a similar pro-cess as in [15] for a detailed investigation. A much prolongedtransformation interval and a further shortened hysteresis areexpected.

4. Influential factors

Above-discussion provides a full picture of the transforma-tion characteristics of SMAs in a thermal cycle. Apart from theassumption that the results are ideal, some other assumptionsare also made so that simple expressions can be derived.

In real practice, a number of factors have to be consid-ered in conducting the tests and interpreting the measuredresults. Probably, the most influential factors are: transformationfront propagation, heating/cooling rate in thermal cycling, andpreload.

F

ig. 4. Observation of phase transformation front movement in a NiTi wire at differe nt strain rate of (a) 1.3 × 10−3/s, (b) 6.5 × 10−4/s, and (c) 2.6 × 10−4/s [9].

226 L. An, W.M. Huang / Materials Science and Engineering A 420 (2006) 220–227

4.1. Transformation front

It has been well documented in the literature (e.g., [7,16]) thatupon unixaxial tension one can often observe the so-called phasetransformation front movement phenomenon. In Fig. 4, a stripof temperature sensitive liquid crystal film was stuck to a NiTiSMA wire, that is 1 mm in diameter, to catch the temperaturedistribution in it in real time during uniaxial tension at threedifferent speeds. It is based on the principle that upon loading,the heat generated during the phase transformation will causethe temperature sensitive film to change colour. A closer-lookreveals that the front may propagate at a fixed angle with thedirection of the applied load [17,18].

This phenomenon is more likely to occur in long SMA sam-ples in uniaxial tension but not in short samples and under someother stress-states. For instance, in uniaxial compression of asingle crystalline CuAlNi SMA, there is no transformation front[19], partially due to the constraint from the boundaries as thetested sample has to be short enough to avoid buckling. In thecase of a deposited SMA thin film atop a wafer, upon ther-mal cycling, the deformation should be more or less uniformeverywhere and thus, theoretically there should not be any phasetransformation front.

With or without a phase transformation front may directlyaffect the resultant martensite variant(s), the exact transforma-tion progress, and the stress versus strain relationship [20].

4

pps[

SfiSd(apahD

hpa

4

cp[

Fig. 5. DSC results of a NiTi thin film at different heating/cooling rates [25].

Fig. 6. Heating/cooling rate vs. peak temperatures upon heating and cooling.

in the following thermal cycle, the reverse transformation startsat a temperature somewhere between the ones that correspondto the full preload and the later reduced load upon heating [30].There is not enough experimental result reported in the literature.Further investigation is required for a better understanding of thiseffect.

5. Conclusions

This paper presents a systematic study on the transformationcharacteristics of SMAs in a thermal cycle with/without a load.Simple methods to estimate some key material properties areproposed. So that, by applying them, one can quantitativelyinvestigate, for instance, the evolution of properties of SMAsupon thermal cycling. The expressions for estimating thehysteresis and transformation interval under different typesof loads, namely, load free, under a constant load, and undera variable load, are obtained. In addition, the influentialfactors in testing and interpreting the results are identified anddiscussed.

.2. Heating/cooling rate

Heating/cooling rate is another concern in testing SMAs. It isarticularly important in bulk SMAs due to a normally prolongederiod of heat transfer [21]. On the other hand, with a largerurface to volume ratio, SMA thin films can react much faster22,23].

Apart from the influence of heat transfer, even in smallMA samples, one can get remarkably different result at dif-erent cyclic rate. Shield [24] has shown the influence of heat-ng/cooling rate on the resultant DSC curve of a single crystallineMA. Fu [25] has conducted a series of DSC tests on a sputtereposited NiTi thin film at some different heating/cooling speedsFig. 5). Taking the peak temperature in the phase transformations an example, Fig. 6 plots heating/cooling rate against peak tem-erature relationship based on the data plotted in Fig. 5. It revealskind of linear relationship between the peak temperature andeating/cooling rate. The measured energy dissipation from theSC tests is also strongly affected by the heating/cooling rate.In thermal cycling a SMA under a load, additional attention

as to be paid on the temperature uniformity in the specimen, inarticular the non-uniformity caused by the boundary conditionsnd phase transformation front phenomenon [9].

.3. Preload

After a loading/unloading cycle, in the subsequent thermalyclic DSC test, one may see the shift of the transformationeak in the heating process towards a higher temperature (e.g.,26–29]). If the applied load is reduced instead of fully removed,

L. An, W.M. Huang / Materials Science and Engineering A 420 (2006) 220–227 227

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