Equivalent circuit models of two-layer flexure beams with ... - JSSS

J. Sens. Sens. Syst., 3, 187–211, 2014 www.j-sens-sens-syst.net/3/187/2014/ doi:10.5194/jsss-3-187-2014 © Author(s) 2014. CC Attribution 3.0 License.

Equivalent circuit models of two-layer flexure beams with excitation by temperature, humidity, pressure, piezoelectric or piezomagnetic interactions U. Marschner1 , G. Gerlach2 , E. Starke3 , and A. Lenk4,* 1 Institute

of Semiconductors and Microsystems, Technische Universität Dresden, 01062 Dresden, Germany of Solid-State Electronics, Technische Universität Dresden, 01062 Dresden, Germany 3 Institute of Lightweight Engineering and Polymer Technology, Technische Universität Dresden, 01062 Dresden, Germany 4 Institute of Acoustics and Speech Communication, Technische Universität Dresden, 01062 Dresden, Germany * professor emeritus 2 Institute

Correspondence to: U. Marschner ([email protected]) Received: 17 March 2014 – Revised: 9 July 2014 – Accepted: 16 July 2014 – Published: 17 September 2014

Abstract. Two-layer flexure beams often serve as basic transducers in actuators and sensors. In this paper a

generalized description of their stimuli-influenced mechanical behavior is derived. For small deflection angles this description includes a multi-port circuit or network representation with lumped elements for a beam part of finite length. A number of coupled finite beam parts model the dynamic behavior including the first natural frequencies of the beam. For piezoelectric and piezomagnetic interactions, reversible transducer models are developed. The piezomagnetic two-layer beam model is extended to include solenoid and planar coils. Linear network theory is applied in order to determine network parameters and to simplify the circuit representation. The resulting circuit model is the basis for a fast simulation of the dynamic system behavior with advanced circuit simulators and, thus, the optimization of the system. It is also a useful tool for understanding and explaining this multi-domain system through basic principles of general system theory.

1

Introduction

Two-layer flexure beams have been present in engineering for more than 200 years. In the year 1766, a thermal bimetal strip was used for the first time in a practical application when it compensated environmental temperature influences in chronometers (Kašpar, 1960). Since then, a variety of two-layer problems that show the same effect have become known, but their operation is based on various physical causes (Fig. 1). In contrast with volume transducers, twolayer beams achieve significantly larger displacements, typically at the expense of a reduction in blocked force output. The mechanical transduction between deflection and stress is an essential property for sensor and energy-harvesting applications of two-layer beams, too. All these beams are described in the following in a unified representation as partially discussed by Gerlach and Lenk (1985). The large

spectrum of technically possible realizations is limited to the problem of the plate strip. The paper is organized as follows. In Sect. 2 the definition of the modeling of a unified one-dimensional two-layer beam element is narrowed down. The model is based on the material behavior state equations (Sect. 3) and ideal boundary conditions. Section 4 describes a differential beam element without shear forces and external pressure. From the beam element the general differential equation system of the actuating two-layer beam element for large angles and the differential equation system of the two-layer beam element for small angles are derived in Sects. 5 and 6, respectively. The latter is interpreted in Sect. 7 as a linear equivalent circuit of actuating two-layer beam elements, which is the main focus of the publication. The circuit description is extended in Sect. 8 to the compact low-frequency linear equivalent circuit of actuating two-layer beams. A number of coupled

Published by Copernicus Publications on behalf of the AMA Association for Sensor Technology.

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U. Marschner et al.: Equivalent circuit models of two-layer beams U. Marschner et al.: Equivalent circuit models of two-layer beams

ment is narrowed. The model is based on the Material bep0 ϑ;ψ(Sec. 3) and ideal boundary condihavior state equations tions. 1Sec. 4 describes a Differential beam element without 1 2 shear 2forces and external pressure. From the beam element the General differential equation system of the actuating p0 two-layer beam large angles and the Differϑ + ∆ϑ;element ψ + ∆ψ for b) a) ential equation system of the two-layer beam element for small anglesuare derived in Sec. 5 and 6, respectively. The u latter is interpreted in Sec. 7 as ai Linear equivalent circoil i cuit of actuating two-layer beam elements, which is the 1 1 circuit description is main intention of the publication. The 2 2 extended in Sec. 8 to the compact Low-frequency linear equivalent circuit of actuating two-layer beams. A number c) d) model the dynamic B; H of coupled finite beam elements behavior including the first natural frequency of a beam in Sec. 9 as Figure 1. Two-layer flexure beams with excitation by (a) tema demonstration of a High-frequency linear equivalent perature ϑ or humidity 9, (b) pressure p0 , (c) piezoelectric cirand cuitpiezomagnetic of an actuating two-layer beam. For piezoelectric and (d) interactions. piezomagnetic interactions Four-port models of reversible transducers are developed in Sec. 10. This includes both, piezoelectric and piezomagnetic unimorph transducrs. Linfinite beam elements model the dynamic behavior including ear network theory is applied in order to determine network the first natural frequency of a beam in Sect. 9 as a demonparameters and to simplify the circuit representation. Such stration of a high-frequency linear equivalent circuit of an a graphical representation of the system supports the underactuating two-layer beam. For piezoelectric and piezomagstanding of the involved physical phenomena. Furthermore, netic interactions multi-port models of reversible transducit enables the usage of powerful circuit simulators to calcuers are developed in Sect. 10. This includes both piezoeleclate the dynamic behavior of the system efficiently. tric and piezomagnetic unimorph transducers. Linear network theory is applied in order to determine network parameters and to of simplify the circuit representation. Such a graph2 Model a unified one-dimensional two-layer beam ical element representation of the system supports the understanding of the involved physical phenomena. Furthermore, it enables the usage of powerful circuit simulators to calculate the dyIn the following a one-dimensional two-layer beam element namic behavior of the system efficiently. is part of a plate strip or beam. It consists of two homogeneous layers with different elastomechanical properties (Fig. 2). The influence quantity Λ, e.g. temperature, humidity etc., 2which Model of a unified one-dimensional is constant with respect to the space coordinates, intwo-layer beam element duces a strain in the beam element. At the beam sections oriented in direction x1 a given disIn the following, two-layer beam eletributed stress T1 (xa3 )one-dimensional and a transverse force F3 act. The task ment is part of a plate strip or beam. It consists of two hois to determine the deflections ξ1 (x1 ) and ξ3 (x1 ) depending mogeneous layers with different elastomechanical properties on the load pressures and forces incorporating the boundary (Fig. 2). The influence quantity 3, e.g., temperature, humidconditions. ity In etc., which is constant with respect to the space coordiorder to solve the problem the following assumptions nates, induces a strain in the beam element. are made: At the beam sections oriented in direction x1 , a given dis- Thestress bondTlayer the laminae tributed and a transverse forceisF3infinitesimally act. The task 1 (x3 ) between small and there is no flaw or gap in the bond There is to determine the deflections ξ1 (x ) and ξ (x depending 1 3 1 ),layer. no shear deformation in the bond layer,the i.e.boundary the lamion theisload pressures and forces incorporating nae cannot slip relative to each other. conditions. In order to solve the problem the following assumptions The bond layer has infinite stiffness and hence the comare -made: posite beam behaves like a single-material beam elementbond with layer integrated properties. – The between the laminae is infinitesimally

small and there is no flaw or gap in the bond layer. There - Ideal elastic material behavior, i.e. validity of Hooke’s is no shear deformation in the bond layer, i.e., the lamilaw. nae cannot slip relative to each other. J. Sens. Sens. Syst., 3, 187–211, 2014

Λ ≠ f ( x1 x2 x3 ) ;

;

;

pa

c − h1 c

Material interface

1 2

0

F3

T1 ( x3 )

c + h2

pb

x3 p0

1

p

2

+

3 a)

p0 ξ1 ( x1 ; x3 ) 1

c

2

0

x1

ξ1 ( x1 ; x3 = 0 )

b)

x3

ξ3 ( x1 ; x3 = 0 )

ξ3 ( x1 ; x3 )

1 2

ϕ ( x1 )

Figure 2. (a) Dynamic and (b) kinematic conditions at the two-

Fig. 2.beam a) Dynamic b) 3 kinematic conditions at the two layer element, and where is the influence quantity; x1 , xlayer 2 , x3 beam element where Λ isdeflections; the influence x , x , x are are coordinates; ξ1 , ξ3 are h1 , hquantity; are the thicknesses of 1 2 3 2 coordinates; ξ1 ,2, ξ3respectively.; are deflections; h1 ,location h2 are the thicknesses of the layers 1 and c is the of the neutral axis; thep0layers 2,presp.; is the location neutral p, , pa =1pand pressures; F3 of is athe force; and Taxis; b = p0c are 0 + p, 1 is p, pnormal p0 + p, pb = p0 are pressures; F3 is a force; and T1 is the 0 , pa = stress. the normal stress

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95

– The bond layer has infinite stiffness and hence the composite beam behaves like a single-material beam ele- According to Bernoulli’s hypothesis the structure is ment with integrated properties. rigidly stiff. Rotational inertia is neglected and cross sections, originally perpendicular to the neutral plane – Ideal elastic behavior, i.e.,deformation. validity of Hooke’s or zero line, material remain planar during In conlaw. junction with Saint-Venant’s principle it is assumed that a distributed stress at the boundary T1 (x3 ) passes over – According Bernoulli’s structure is after a shorttodistance ∆ x1hypothesis, in a stress the which is deterrigidly stiff. Rotational inertia is ξneglected and cross mined by a displacement function (x ; x ). Mean and 1 1 3 sections, perpendicular to the neutral plane or moment originally of this displacement function are determined zero line, remain planar during In conjuncby the related quantities of the deformation. boundary distribution. tion with Saint-Venant’s principle, it is assumed that, after a short distance 1x1 , a distributed stress at the boundary T1 (x3 ) changes into a stress which is deter- The stress in x3 -direction is negligible (T3 = 0). mined by a displacement function. Mean and moment www.j-sens-sens-syst.net/3/187/2014/

U. Marschner et al.: Equivalent circuit models of two-layer beams

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- Contributions of shear deformations to the displacement of this displacement function are determined by the reξ3 (x1 ) can be neglected, i.e. the shear strains yield S4 = lated quantities of the boundary distribution. S5 = S6 = 0. –- The The coordinate stress in x3 origin direction 0).distance 3= x is = negligible 0 is placed(Tat the

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T3 = 0 T = 0 2

S2 = 0 w

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from the bond At this position thedisplacement translational – cContributions of layer. shear deformations to the system points are located at the beam ends. ξ3 (x1 ) can be neglected, i.e., the shear strains yield S4 = S 5 = S6 = 0. - Deflections of these system points at x3 = 0 are labeled with ξ1 and ξ3 or ξ1 (x1 ) and ξ3 (x1 ), respectively; de– The coordinate origin x3 = 0 is placed at the distance flections of arbitrary positions within the beam are lac from the bond layer. In this position the translational beled with (ξ1 (x1 ; x3 ), ξ3 (x1 ; x3 )). system points are located at the beam ends.

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a) p0 = 0; l

www.j-sens-sens-syst.net/3/187/2014/

w

b)

l

p0 = 0; w

S2 = αRΛ

≫l p0

p0 1

B

B

T3 = 0

(1a)

1 ν0 assume that isotropic material ν0 following considerations T1 + T2 − T3 + α0 Λ (1b) SAll 2=− E E behaviorEcan be described by 0 0 0 the following state equations: ν0 ν0 1 S3 = − 1 T1 − ν0 T2 + ν0 T3 + α0 Λ (1c) (1a) S1 = E0T1 − E0T2 − E0T3 + α0 3, E0 E0 E0 Reversibility is not considered at this stage but only actuation behavior. In order to be ν0 1 ν0 consistent with one-dimensional S2 = − theory, T1 + S2 ,TS23−and T32 ,+Tα3 0are 3, defined by boundary (1b) bending E E0 E0 conditions0 or viewed as strain-induced disturbance variables. In the νsimpliestν case of a1 very small beam element without 0 0 T1 − p0T2(Fig. + 3a) T3 + S3 = − pressure constant noαstress x2 0 3. components in (1c) E0 E0 E and x3 -direction occur (T2 0= T3 = 0) and Eq. (1a) simplifies to: Reversibility is not considered at this stage but only actuation behavior. In order to be consistent with one-dimensional 1 Sbending T1 + α0SΛ . and S and T and T are defined (2) 1= theory, by 2 3 2 3 E0 boundary conditions or viewed as strain-induced disturbance In case of a very wide beam, which is fixed at both ends variables. (w l) as depicted in Fig. 3b, the boundary condition S2 = In the simplest case of a very small beam element without 0 is acting internally. When, in addition, no constant pressure constant pressure p0 (Fig. 3a), no stress components in x2 p0 acts, for the stress T3 = 0 follows and for Eq. (1a) yields: and x3 direction occur (T2 = T3 = 0) and Eq. (1a) simplifies to 1 − ν02 T1 + (1 + ν0 ) α0 Λ . (3) S1 = 1E0 S1 = T1 + α0 3 . (2) E0 In some micro-mechanical constructions the beam ends are not blocked depicted in beam, Fig. 3bwhich but theisinfluence In the case of as a very wide fixed atquanboth 145 tity loads theasvery wide beam a straincondiS2 = endsΛ (w l) depicted in Fig.element 3b, thewith boundary α With no p0con=0 tion 0 is actingcoefficient internally. αWhen, in 3c). addition, R Λ Svia R (Fig. 2 = influence Eq. yields phere: stant(1a) pressure 0 acts, for the stress T3 = 0 follows, and for Eq. (1a) it yields 1 − ν02 S1 = T1 + [(1 + ν0 ) α0 − ν0 αR ] Λ . (4) E0ν 2 1− 0 S1 = T1 + (1 + ν0 ) α0 3 . (3) In the special case of αR = α0 , i.e. that the ends consist of E0 one of the layers, Eq. (1a) simplifies further to: In some micro-mechanical constructions the beam ends 150 1 −blocked ν02 are not as depicted in Fig. 3b, but the influence S1 = T1 + α0 Λ . (5) quantityE3 loads the very wide beam element with a strain 0

≫w w

Deflections of these system points at x3 = 0 are labeled 3 –Isotropic material behavior state equations either with ξ1 and ξ3 or ξ1 (x1 ) and ξ3 (x1 ); deflections of arbitrary positions assume within the are labeled All following conderations thatbeam isotropic materialwith be; x3described ), ξ3 (x1 ; xby 1 (x1be 3 )). havior(ξcan the state equations:

ν0 ν0 1 T1 − T2 −behavior T3 + αstate S31 =Isotropic 0 Λ equations material E0 E0 E0

T3 = 0

l

c) p0 = 0; w

2

Λ

l

≫l

1 2

3

p0

p0

d) T2 = T3 = −p0

Figure 3. Ideal boundary conditions at the beam. Fig. 3. Ideal boundary conditions at the beam

S = α 3 via influence coefficient α (Fig. 3c). With p = 0 Eq. (1a) here yields

2 R isotropic bimorph 0layers Table 1.RUniaxial stress conditions for real

1 − ν02 condition Boundary S1 = T1 + [(1 + ν0 ) α0 −Sν10=αR1] T 3 .+ αΛ inE0direction E 1

see(4) Fig.

In thex3special case α0 , i.e., that theα ends consist of x 2 of αR = E simplifies furtherα to one of the layers, Eq. (1a) T =0 E 3a 2

0

1 − ν2 S = 0 S1 T =3 = 0 0 T1 +2 α0 3 . E0 S = α Λ

E0

0

( 1 + ν 0 ) α0

3b(5)

( 1 + ν 0 ) α0 − ν 0 α R 2 R 1 − ν02 3c In the caseSof =a αconstant pressure p0 , which α0 also acts on 2 0Λ the surfaces perpendicular to axis x3 , T2 = T3 = −p0 yields T3 = − p0 such T2 that = − pEq. 2 ν0 E0 3d 0 (1a)Esimplifies 0 further to conditions 1 2ν0 S1 = T1 + p0 . (6) E0 E0 In case of a constant pressure p0 , which also acts on the All considered casesto can bexdescribed linear surfaces perpendicular axis −punified con3 , T2 = Tby 3 =the 0 yields relation ditions such that Eq. (1a) simplifies further to 1 2ν0 (7) S1 = 1 T1 + α3. S1 = E T1 + p0 . (6) E0 E0 The meaning of 1/E and α in each case is summarized in Table 1. Even in the case of anisotropic material properties, All considered cases can be described by the unified linear generalized Eq. (7) applies. relation 4

Differential beam element without shear forces 1 S1 and = external T1 + αΛ pressure . (7) E

4.1 meaning Stressesofand strains The 1/E and α in each case is summarized in Table 1. Even in the case anisotropic materialdifferential properties The object to be studied ofnext is an arbitrary generalized Eq. 7 applies. two-layer beam element of the plate strip. The layers are J. Sens. Sens. Syst., 3, 187–211, 2014

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4

Differential beam element without shear forces and external pressure layers.

Table 1. Uniaxial stress conditions for real isotropic bimorph

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4.1

Stresses and strains

Boundary condition in direction

S = 1 T + α3

E 1 The object to be studied next is an1arbitrary differential see twox3 x2 of the plate E strip. The layers α Fig. layer beam element are characterized by their Tmaterial properties well E0 E1 , α1 and α0 E2 , α2 as 3a 2=0 as their thicknesses h and h . From Bernoulli’s hypothesis 2 S2 = 0 1 3b (1 + ν0 ) α0 follows (x T3 = 0that the deflection ξ1E 0 1 ; x3 ) at defined position is a S = αR 3 (1 + ν0 ) α0 − ν0 αR 1−ν linear function2of coordinate x302: 3c

S2 = α0 3

ξ1 (x 3 )0 = C 1 )0· x3 + CE b 0(x1 ) . T3 1=; x −p T2a=(x−p

α0

2ν0 /E0

3d(8)

∆ϕ 2

c − h1

E1, α1

c

0 c + h2

E2 , α2

F1

165

ξ (x ; x ) = Cd ϕ (x ) · x3 + Cb (x1 ) . S0 S11 (x11 ; x33 ) = a 1· x3 + d x1

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(8) (10)

The same conclusion where S0 is the strain applies at x3 =to0:the strain: ∂ ξ1 (x1 ; x3 ) ∆ ξ1,0 dξ1,0 S0 = = . (11) dCa = ∆dC dξ1 b ∂x x (9) S1 (x1 ; x3 ) = 1 = x3 =0· x3 + 1 . dx1 dx1 dx1 dx1 Considering the material state equations which are valid in each region, the stress in the beam element yields: The differential beam element in Fig. 4, which is bent about 180 ! the angle ϕ(x 1 ), illustrates the meaning of dCa /dx1 and  d ϕ d ξ 1,0   dCb /dx1 : x3 + − α1 Λ  E1   d x1 d x1     for c − h1 6 x3 6 c − 0  T (x11;;xx33))= = dϕ · x3 + S0 , (12) S11 (x (10) !  dx  1  d ξ d ϕ 1,0   E2 x3 + − α2 Λ    d x1 d x1   where S0 is thestrain at x3for = 0: c + 0 6 x3 6 c + h 2 dξ1,0 ∂ ξ1 (x1 ; x3 ) 1 ξ1,0 S = = = F1 and . Moment(11) 0 4.2 Introduction of internal force M ∂x1 1 x dx1 1 x3 =0 as coordinates

From Eqn. (12)the thematerial balancing normal force which F1 andare balancing Considering state equations valid in moment M for section Fig. yields 5: each region, thethe stress in thefollow beamwith element Z 2 c+h d ξ1,0 dϕ  x + − α 3 E1 T1dx 3 1 F1 (x1 ) = −w (x ) d x 3 3 d x  1 1   c−h for c − h1 6 x3 6 c − 0 1 . (12) T1 (x1 ; x3 ) = dϕ 1 d 2ϕ  d 2ξ1,0   EE x3E+2 hdx −(E α213 = −w − h1 + E2 h2 ) c 2 1h  21+ dx11 −  d x1 185 2 for c + 0 6 x3 6 c + h2 d ξ1,0 + (E1 h1 + E2 h2 ) − (α1 E1 h1 + α2 E2 h2 ) Λ d x1 J. Sens. Sens. Syst., 3, 187–211, 2014

F1

x1

∆x1 x3

a)

∆ξ1 ( x1; x3 = 0 ) = ∆ξ1 0 ,

The same conclusion applies to the strain: d Ca d Cb d ξ1 = material · x3 + . E1 , α1 and E2 ,(9) S 1 (x1 ; x3 ) = by their characterized properties α2 d x1 d x1 d x1 as well as their thicknesses h1 and h2 . From Bernoulli’s hyThe differential beam in ξFig. is rotated pothesis follows that theelement deflection x3which ) at a defined po1 (x1 ;4, about thea linear angle function ϕ(x1 ), illustrates the meaning of dCa /dx1 sition is of coordinate x3 : and dCb /dx1 :

∆ϕ 2 Ca 1 ⋅ x3 = ∆ϕ ⋅ x3 2 2

1 ∆ϕ ⋅ x3 2

SB

c − h1 c

α1Λ SΛ α2Λ

0 c + h2

b)

dξ1 0 dx1 ,

S0 x3

x3

 ∂ ξ1 ( x1; x3 )    = Cb = ∆x1 ⋅  ∂x1    S0  x = 0 3

S1

S1 − SΛ x3

T1 x3

Figure 4. Strains and stresses in a two-layer beam: (a) bimorph

Fig. 4. Strains and stressesand in a deformed two-layer state; beam:(b) a) bimorph element element in non-deformed strain and stress in non-deformed deformed distributions at theand cross section.state; b) strain and stress distributions at the cross section

4.2

Introduction of internal force F1 and moment M as (13) coordinates

From Eq. (12) the balancing normal force F1 and balancing c+h Z 2 moment M for the section follow with Fig. 5: M (x1 ) = −w T1 (x3 ) · x3 dx3 c+h 2 Z c−h 1 F1 (x1 ) = −w T1 (x3 ) dx3 (13) 1 3 3 2 E h + E h = −w + (E h + E h ) c 1 2 1 2 1 2 1 2 c−h1 3 dϕ 1 2 d ϕ 2 2 2 = −w − E h − E h + E h + E h c 22 2 c 1 1 2 2 − 2 E1 h11 − 1 E2 h dx1 d x1 dξ 1,0 1 +E − (α1 E1 h1 + α2 E2 hd2ξ)1,0 3 , + 1 1 h2 2 h2 2 h2 − E1 hE 1−E 2 − (E1 h1 + E2 h2 ) c dx 1 2 d x1 1 2 2 − Λ − α1 Ec+h Z1 h21 − α2 E2 h2 + c (α1 E1 h1 + α2 E2 h2 ) 2 M (x1 ) = −w T1 (x3 ) · x3 dx3 (14) (14) c−h 1

c ischosen at When 1 2 3 + (E 2 1 h1 + E2 h2 ) c2 h = −w 1 EEh12h31−+EE 2 1 h2 2 h 2 χζ − c = cS = 3 1 1 2 = · , (15) 2 E1 h1+ Edϕ 1 + χζ 2 2 h2 − E1 h21 − E2 h22 c E1 h1 dx with and ζ = , χ= dξ1,0 E22 1 2 h2 − E1 h1 − E2 h2 − E1 h1 + E2 h2 c then 2F1 depends not on dϕ/dx1 and M not ondxd1ξ1 /dx1 . This the material interface distance 1from is the well-known loca2 2 − 3of−the neutral α1 E1 h1layer − α2in E2case h2 + α2 E2 h2 ) This . (αexternal 1 E1 h1 +moment. tion ofcan 2 www.j-sens-sens-syst.net/3/187/2014/


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case is incthe following When is chosen at referred to as sensing case. With cS , normal force and moment yield: h21 − E2 h22 1 E1 χζ 2 − 1 h2 c = cS = = (15) d ξ1,0 1 + χζ · 2 , E1 h1αΛ +E F1 (x1 ) =2EA −2 h2 (16) d x1h1 E1 with χ = and ζ = , E2 d ϕ h2 + MΛ (17) M (x1 ) = −EI x1 then F1 does notddepend on dϕ/dx1 nor M on dξ1 /dx1 . This distance from the material interface coefficient is the well-known locawith the equivalent linear expansion tion of the neutral layer in the case of an external moment. h1in+the α2following E2 h2 This α case referred to as sensing case. With 1 E1is α= , (18) cS , normal E1 hforce 1 + Eand 2 h2moment yield dξ1,0 the length-related translational compliance or extensional , (16) F1 (x1 ) = EA α3 − stiffness of a homogeneous beam dx1 dϕE h ) , EA w (E1 h1 + (19) 2 M 2 3 , M (x= + (17) 1 ) = −EI dx1 the compliance or bending stiffness withlength-related the equivalentrotational linear expansion coefficient of a homogeneous beam α1 E1 h1 + α2 E2 h2 α = w E 2 h4 + E E ,h h 4 h2 + 6 h h + 4 h2 + h(18) 4 2 1 2 1 2 1 2 1 2 2 E2 EI = E1 h11 +1 E2 h2 12 E1 h1 + E2 h2 the length-related translational compliance or extensional stiffness a homogeneous w of 1+ χ2 ζ 4 + 4χζ 3 +beam 6χζ 2 + 4χζ 3 (20) · h2 E2 = +1 EA =12 w (E1 h1 + E2 hχζ (19) 2) ,

and the moment source the length-related rotational compliance or bending stiffness w 2 2 of Λa = homogeneous beam M E1 α1 −h 1 + 2cS h1 + E2 α2 h2 + 2cS h2 Λ 2 w E12 h41 + E1 E2 h1 h2 4 h21 + 6 h1 h2 + 4 h22 + h42 E22 EI = w E1 E2 h1 h2 (h1 + h2 ) (21) 12 E1 h1(α +1E−2 hα22 ) Λ =− 2 E1 h1 + E2 h2 225 w 1 + χ 2 ζ 4 + 4χζ 3 + 6χζ 2 + 4χζ 3 (20) · h2 E2 , = by Λ. driven 12 χζ + 1 When a mean stress T 1 = F1 /A is related to force F1 then and the moment sourcethe strain S0 = d ξ1 /d x1 . The differthis stress determines T (x ) = T (x ; x )represents iniaddience T1w(xh1 ; x3 ) − 1 1 B 1 3 M3 = E1 α1 −h21 + 2cS h1 + E2 α2 h22 + 2cS h2 3 tion the2mean-free part of the stress distribution, whose moment determines change h1 hangle h2 ) dϕ/dx1 (Fig. 5). w E1 E2the 2 (h1 + (21) =− (α1 − α2 ) 3 2 E1 h1 + E2 h2 4.3 Curvature and neutral layer driven by 3. When a mean T1 = F1 /A F1 from then 230 When there is nostress external force, i.e.is Frelated it force follows 1 = 0, to this (13) stressthat determines strain d ξ1,0by: /d x1 . In addition, Eq. the strainthe dξ1,0 /dxS 1 0is=given the difference T1 (x1 ; x3 ) − T 1 (x1 ) = TB (x1 ; x3 ) represents 2 the mean-free whose dξ1,0 1 E1part h21 −ofEthe dϕ α1 E1 h E2 h2 2 h2stress distribution, 1 + α2moment = − c + Λ determines the angle change dϕ/dx (Fig. 5). dx1 2 E1 h1 + E2 h2 dx11 E1 h1 (22) 4.3

220

Curvature and neutral layer

which is substituted in Eq. (14) to get the curvature: When there is no external force, i.e., F1 = 0, it follows from 235 Eq. strain dξ21,0 dϕ (13) that the6E h2/dx (h11+ishgiven − α1 )Λ 1 h1 E 2 )(α2by = 4 2 2 ! dx1 h2 E2 + 2E E h h (2h + 3h h + 2h22 ) + E12 h41 2 1 2 1 1 2 1 dξ1,0 1 E1 h21 − E2 h22 dϕ = −c 2 (22) dx1 6 2 E1 h1 + E2 h(1 2 + ζ) dx1 = · (α2 − α1 )Λ (23) (h1 +αh12E ) 1 h11+ α2 E2 h2 2 ) + χζ 3 + 6ζ + + 4 (1 + ζ3, E1 h1 χζ www.j-sens-sens-syst.net/3/187/2014/

5

191 ∆ϕ 2

∆ϕ 2

T1

TB with TB = 0

∆ ξ1

M

M F1

F1

F1

M = 0; ∆ϕ = 0

F1 = 0; ∆ξ1 = 0 ∆ϕ 2

F1

∆ϕ 2

M

M

c − h1 c

0

∆x1

∆ξ1

T1 ( x3 )

c + h2

x1

x3

Figure 5. Forces and moments at the differential beam element Fig. 5. Forces moments at the differential without shear and forces and difference pressure. beam element without shear forces and difference pressure

which is substituted in Eq. (14) to get the curvature as shown, e.g., by Guerrero and Wetherhold (2003) for magnetostrictive unimorphs. For a given total bimorph thickness , the results dϕ h = h1 +h26E h2 )(α2 − αcurvature 1 hwell-known 1 E2 h2 (h1 +maximum 1 )3 = 4 22 . Geometry and transduction 2 2 2 4 for coefficients have dx1χ =h1/ζ 2 E2 + 2E2 E1 h2 h1 2h1 + 3h1 h2 + 2h2 + E1 h1 a larger effect on the deflection than a variation of Young’s modulus (Gerlach and D¨otzel, 2008). Backsubstitution of Eq. 2 Eq. (22) gives the location 6 + ζ )in (1(23) = the neutral· layer of in the actuation case: (α2 − α1 )3 (23) (h1 + h2 ) χ1ζ + 4 1!+ ζ 2 + χ ζ 3 + 6ζ 1 2 3 4ζby +Guerrero A+ 3ζWetherhold (A + 1) + ζ(2003) χ + 4 for magand as shown, e.g., ζχ h2 unimorphs. For a given total bimorph thickness cnetostrictive (24) A=− (ζ + 1)maximum (A − 1) curvature results for h = h1 +6h2 , the well-known χ = 1/ζ 2 . αGeometry and transduction coefficients have a 2 with = .onThe location follows for theofminimum largerAeffect thesame deflection than a variation Young’s α1 modulus (Gerlachenergy and Dötzel, of the potential of the2008). deflected beam w.r.t. c for of Eq. (23) in Eq. (22) gives the location dξ1Back /dx1 substitution = 0: of the layer in the actuation case:  neutral  ∆x Z  2 d 1 MΛ  . 3ζ (A + 1) + ζ 3 χ + 4 (25) 4ζ 2dx + 1ζ1χ =A0 + dc 2 h2 EI cA = − 0 , (24) 6 (ζ + 1) (A − 1) In case of only layer 1 being active (α2 = 0), Eq. (24) simwith Ato= αα21 . The same location follows for the minimum of plifies the potential energy of the deflected beam with regard to c for dξ1 /dx1 =10:h31 E2 + 3E1 h1 h22 + 4E1 h32 cA |α2 =0 = − 6  1x  h2 E1 (h1 + h2 ) (26) Z 1 2 M3h 3ζ + 4 + χζ 3 d 1  2 dx = 0 . (25) 1 − dc 2 = EI 6 1+ζ 0

J. Sens. Sens. Syst., 3, 187–211, 2014

6

192

U. Marschner e


In the case of only layer 1 being active (α2 = 0), Eq. (24) simplifies to 1 h3 E2 + 3E1 h1 h22 + 4E1 h32 cA |α2 =0 = − 1 6 h2 E1 (h1 + h2 ) h2 3ζ + 4 + χζ 3 . =− 6 1+ζ

In addition to the previous considerations, the influence of a shear force F3 and of a difference pressure p on the twolayer beam element is now taken into account. In order to obtain a general differential equation system model of the two-layer beam element, the deformed element isconsidered in the transformed coordinate system x1∗ , x2 , x3∗ according to Fig. 7, where (x1 , x2 , x3 ) serves as reference system for the deflections ξ1 and ξ3 . This leads to the following relationships: – As illustrated by Fig. 7b, the length of the deformed – originally 1x1 -long – bimorph element becomes ∗ 1 x1 · 1 + S0∗ (x1 ) = 1 x1 + 1 ξ1,0 (x1 ) 1 = 1x1 + 1 ξ1,0 (x1 ) cos ϕ 1 1 x1 · (1 + S0 (x1 )) . cos ϕ

For the strain S0 (x1 ) = dξ1,0 /dx1 , S0 (x1 ) = cos ϕ · 1 + S0∗ (x1 ) − 1

(27)

J. Sens. Sens. Syst., 3, 187–211, 2014

0.6 c

250

cS

0.5

h2

0.45

α1 Λ κ ⋅

1.25

gin. B an indu thickn to the angle, S0 . Th mome prereq 5

1.3

1.35

1.4

1.5 1.45 10 -2

1.55 255

Figure 6. Solution of Eq. (14) with d ξ1,0 /d x1 = 0 and α2 = 0

to the neutral-layer position normalized h2 0exemplarily Fig. 6. Solution of Eq. (14)c with dξ1 /dxto1 = and α2 = for 0 to = 1 mm,layer E1 =position 23 GPa, hc2 normalized = 5 mm, E2 to = 71 w = 5 mm,for theh1 neutral h2GPa, exemplarily = 0.1 m and h1l = 1 mm, E1α= 230.GPa, h2 = 5 mm, E2 = 71 GPa, w = 5 mm, 2= l = 0.1 m and α2 = 0 260

F1∗ (x1 ),

– The force which acts at the cross section of the element, consists of two components in the x1 - and When xthe active layer is thin (h1 ≈ 0) the neutral layer is 3 directions:

Ge tw

In add of a sh two-la obtain two-la in the to Fig. the de tionshi

- A o

located at −h2 · 2/3 and not at −h2 /2 where it is located in ϕ +an F3external sin ϕ . moment. This location (30) F1∗ (x1case, ) = Fi.e. the sensing 1 cosfor matches that found by Stoney (1909) for thin films. With resultpinacting Eq. (24), theside two-layer beamelement actually 265 – A the pressure on one of the beam unveilscauses pure bending Mforce 6= f (S locations of xthe oritwo acting components the x1 and 0 ) for two in 3 di-

∆

rections, too (Fig. 7). For the deformed beam element in coordinate system p ⋅ w ⋅which ∆x1 ⋅ 1is+ consistent S0∗ ⋅ cos ϕ with x1∗ , x2 , x3∗ , the following holds, Eqs. (16) and (17) from Sect. 4.2:

(

)

c − h1 ∗ dξ1,0 ∗ c F1 (x1 ) = EA α3 − , dx1

0 dϕ + M3 . M (x1 ) = −EI dx1

c + h2 x3∗

(31)

ϕ ( x1 )

∆F = p ⋅

p ⋅ w ⋅ ∆x1

w

x1∗

(32)

The coefficients α, EA, EI and M3 are identical with the coefficients in Eqs. (16) and (17). From Fig. 7 the following ∆ξ1 relations can also be derived: – horizontal balance of forces: w

F

F3

F1∗

c − h1 (28)

and for the relationship between the deflections ξ1,0 (x1 ) ∗ (x ) in the x ∗ direction in the x1 direction and ξ1,0 1 1 ∗ dξ1,0 dξ1,0 = cos ϕ · 1 + − 1. dx1 dx1

245

cA

0.65 0.55

General differential equation system of the actuating two-layer beam element for large angles 240

1 x1 · 1 + S0∗ (x1 ) =

0.7

(26)

When the active layer is thin (h1 ≈ 0), the neutral layer is located at −h2 · 2/3 and not at −h2 /2, where it is located in the sensing case, i.e., for an external moment. This location matches that found by Stoney (1909) for thin films. With the result in Eq. (24), the two-layer beam actually unveils pure bending M 6 = f (S0 ) for two locations of the origin. Both locations, cA and cS , give the same curvature for an induced stress. Figure 6 depicts c normalized to the substrate thickness as a function of a given excitation α1 3, normalized to the curvature. Therefore, the curvature or the deflection angle, respectively, can be calculated with cS countenancing S0 . Then the source moment can be treated as an external moment acting on the unimorph with the compliance nR as a prerequisite for a reversible transducer model in Sect. 10. 5

Solution

(29)

ξ1 ( x1 + ∆x1 ) =(33) ξ1 ( x1 ) + ∆ξ1 ( x1 ) F1 c(x1 + 1x1 ) − F1 (x1 ) ∗ + p · w · 1x1 · 1 + S0 · sin ϕ = 0; dξ

0

x1

–c +vertical h 2 balance of forces:

x3 ∆x F3 (x1 + 1x1 ) − F3 (x1 1 ) a) − p · w · 1x1 1 + S0∗ · cos ϕ = 0;

(34)

Fig. 7. Definitions at thewww.j-sens-sens-syst.net/3/187/2014/ two-layer beam element with respect to large angles

b)


p ⋅ w ⋅ ∆x1 ⋅ 1 + S0∗

(

Mx

( 1)

F1∗ ( x1 )

F1 ( x1 )

) cos ϕ ⋅

∆F = p ⋅ w ⋅ ∆x1 ⋅ ( 1 + S0∗ )

c − h1

F3 ( x1 )

c

0

ϕ ( x1 )

c + h2

193

p ⋅ w ⋅ ∆x1 ⋅ 1 + S

w

x1∗

∗ 0

(

c − h1

) sin ϕ

c

⋅

F1 ( x1 + ∆x1 )

x3∗

ϕ ( x1 + ∆x1 )

c + h2

M x + ∆x (

ξ1,0 ( x1 )

c − h1

F3

w

,

c

ξ1 0 ( x1 ) ,

0 c + h2

a)

∆ + ∆ξ1

ξ1 0 ( x1 + x1 ) = (

F3 ( x1 + ∆x1 ) sin ϕ F1 ( x1 + ∆x1 ) cos ϕ

x3

S0 =

∆x1

1

S0∗ ( x1 )

x3∗

1)

∆ξ1∗ 0 ,

∆x1

d ξ3 ≈ ∆ ξ3

x1 )

x1

∗

S1∗ ( x1 ) = dξ1 dx1

0

∆ ξ1 0 ∆ x1

ϕ

( x3 = 0 ) ∆ξ1 0 ,

,

b)

∆x1

∆x1 ( 1 + S0

x

( 1)

)

Figure 7. Definitions at the two-layer beam element with respect to large angles.

– balance of moment: (35)

M (x1 + 1x1 ) − M (x1 ) ∗

− F1 (x1 + 1x1 ) · 1x1 · 1 + S0 · sin ϕ + F3 (x1 + 1x1 ) · 1x1 · 1 + S0∗ · cos ϕ 1x1 · 1 + S0∗ ∗ ≈ p · w · 1x1 · 1 + S0 · ≈ 0; 2 – kinematics (Fig. 7b): ξ3 (x1 + 1x1 ) − ξ3 (x1 ) = 1x1 (1 + S0 ) · tan ϕ .

(36)

Equations (33)–(36) together with Eqs. (27) and (32) constitute the differential equation set of a general plate strip element in set of equations (SOE) (37). Its validity is restricted by – the assumption of small deflections d2 ξ1 /dx12 ; – the limited validity of Bernoulli’s hypothesis; – nonlinear stress–strain relations that occur in reality and – the idealized formulation of the boundary conditions for S2 and T2 .


dF1 = − p · w · sin ϕ · 1 + α3 dx1 1 − (F1 cos ϕ + F3 sin ϕ) EA dξ1,0 =S0 = cos ϕ · 1 + α3 dx1 1 − (F1 cos ϕ + F3 sin ϕ) − 1 EA dF3 =p · w · cos ϕ · 1 + α3 dx1 1 − (F1 cos ϕ + F3 sin ϕ) EA dξ3 = tan ϕ · 1 + α3 dx1 1 − (F1 cos ϕ + F3 sin ϕ) EA dM = (F1 sin ϕ − F3 cos ϕ) · 1 + α3 dx1 1 − (F1 cos ϕ + F3 sin ϕ) EA dϕ 1 =− (M + M3 ) dx1 EI

(37a)

(37b)

(37c)

(37d)

(37e)

(37f)

For F1 (ξ1 (x1 ) = 0) > 0, i.e., if the beam is axially compressed, courses of the functions F3 , ξ3 , M, ϕ = f (x1 ) can result which do not fulfill the demand for biuniqueness. J. Sens. Sens. Syst., 3, 187–211, 2014

8 U. Marschner et al U. Marschner et al.: Equivalent circuit models of two-layer beams

194

6

∆F = p ⋅ w ⋅ ∆x1 w F1 ( x1 + ∆x1 ) M ( x + ∆x ) 1 1 F3 ( x1 + ∆x1 ) ϕ ( x1 )

ϕ( x1 )

This problem must be solved with stability theory methods (Pflüger, 1975).

M( x ) 1

F1 ( x1 )

ξ3 ( x1 )

Differential equation system of the two-layer beam element for small angles

ξ3(x1

+∆x1 )

Contrary to the previous considerations, Fig. 8 depicts the definitions of the involved quantities and directions when small angles and deflections are assumed. With these definitions the following relations can be derived:

ξ1,0 ( x1 )

F3 ( x1 ) ξ3 ( x1 )

b)

∆ξ1 0 ( x1 ) w

ξ1 0 ( x1 ) ,

c

F1 (x1 + 1x1 ) − F1 (x1 ) = 0,

0

(38)

ϕ( x1 + ∆x1 ) 335

∆

dM d x1

2. The f sourc

,

340

a)

(39)

x3 x1 + ξ1 0 ( x1 ) ,

– balance of moment:

x1

+ F3 (x1 + 1x1 ) · 1x1 = pw1x1 ·

1x1 ≈ 0; 2

(41)

(42a) (42b) (42c) (42d) (42e) (42f)

dM ξ3 (x1 + ∆x1 ) − ξ3 (x1 ) = ϕ (x1 ) · ∆x1 . (41) = −F3 . (43) dx1 From Eqs. (16) to (41) SOE (42) can be derived. With this differential all problems can be 355 2. set Theof force F1 isequations treated almost as a timeand motionsolved where |ϕ (x ) | 1 and F 6 0 hold. For the case 1 independent source quantity. 1 F1 (ξ1 (x1 )) > 0, the same remarks as in Sec. 5 apply. In order to obtain a circuit representation of the linearized set of Eq. (42) the time-dependent complex coordinates angular d F1 (42a) velocity = 0 d x1 d ξ1,0 1 dϕ (42b) = d x=1 jω=ϕS0 = αΛ − EA F1 (44) dt d F3 = p·w (42c) and velocity d x1 ddξ ξ3 =ϕ (42d) 1,0;3 υ 1;3 =d x1 = jω ξ 1;3 (45)360 d Mdt = F1 ϕ − F3 (42e) d x1 are introduced. Here, ω is the angular frequency and j the d ϕ unit. Integration 1 imaginary of Eq. (42) over x1 then gives the =− (M + MΛ ) (42f) following d x1solution: EI F 1 = F 1 (x1 ) ,

dϕ = dt

and veloci

345

325

J. Sens. Sens. Syst., 3, 187–211, 2014

x1

Linear equivalent circuits of actuating two-layer - Balance of moment: beam elements M (x1 + ∆x1 ) − M (x1 ) The differential system (42) is nonlinear due to the + F1 (x1 +equation ∆x1 ) · ∆x 1 · ϕ (x1 ) (40) coupling F1 ·ϕ. The nonlinearity can be avoided ∆x1 in two ways: ≈0 + F3 (x1 + ∆x1 ) · ∆x1 = pw∆x1 · 350 1. For small forces F1 the condition |F12· ϕ| F3 is asThis will change Eq. (42e) to - sumed. Kinematics:

From Eqs. (16) to (41) SOE (42) can be derived. With this set of differential equations almost all problems can be solved where |ϕ (x1 ) | 1 and F1 6 0 hold. For the case F1 (ξ1 (x1 )) > 0, the same remarks as in Sect. 5 apply. dF1 =0 dx1 dξ1,0 1 = S0 = α3 − F1 dx1 EA dF3 = p·w dx1 dξ3 =ϕ dx1 dM = F1 ϕ − F3 dx1 dϕ 1 =− (M + M3 ) dx1 EI

x1 + ∆x1

In order to of equatio angular ve Ω=

,

7

– kinematics: ξ3 (x1 + 1x1 ) − ξ3 (x1 ) = ϕ (x1 ) · 1x1 .

x1 + ∆ ( x1 ) + ξ1 0 ( x1 + ∆ ( x1 ) )

Figure 8. Definitions at the two-layer beam element for small anFig. 8. at theand two-layer beam element gles: (a)Definitions non-deformed (b) deformed bimorph.for small angles: a) non-deformed and b) deformed bimorph

(40)

M (x1 + 1x1 ) − M (x1 ) + F1 (x1 + 1x1 ) · 1x1 · ϕ (x1 )

1. For s sume

ξ1,0 ( x1 + x1 ) ξ3 ( x1 + ∆x1 )

∆ ξ1 0 ( x1 ) + ∆ξ1 ( x1 )

c + h2

F3 (x1 + 1x1 ) − F3 (x1 ) − pw1x1 = 0;

Linea beam

The differ coupling F

ξ1,0 ( x1 + x1 ) =

0

ξ3

7

,

c − h1

– balances of forces:

ϕ ( x1 + ∆x1 )

330

(46a) www.j-sens-sens-syst.net/3/187/2014/

dξ υ 1,3 =

d

are introd imaginary lution:

F1 = F1(

υ 1 = jω α

F3 = p·w M=

F1 ∆ jω F 3 (x

Ω=

jω EI

υ3 =

jω EI

Applicatio υ 1,3 (x1 ), ditions F Ω (x1 + ∆ using the a

~1 : F 1 (x

U. Marschner Marschneretetal.: al.:Equivalent Equivalentcircuit circuit models two-layer beams U. models ofof two-layer beams 365

370

x1 −jω ∆n·F 1 1(x : υjω 1+ 1 ) = ∆υ 1 (x υ 11= α3 x1∆x − jω FΛ ) ,1 )+υ 1 (x1 ) 1 (x1 ) + υ 1 (x EA ~2 : F 3 (x1 + ∆x1 ) = ∆F + F 3 (x1 ) F 3 = p · w · x1 + F 3 (x1 ) , Ω (x1 + ∆x1 ) ~3 : M (x1 + ∆x1 ) = M (x1 ) − jω ∆nF1 2 x F1 1 x1 ∆x w 1 ∆x1 M = − (∆F (x 1 1x)) 1F+ (x 1 ) − p1· − + F 3 jω 23 (x1 ) 2 2 F 3(x(x1+ ) x∆x (x1 ) , 1 +M 2−: Ω 1 1 ) ≈ Ω (x1 ) − ∆Ω Λ −   3 2 x1 ∆x1 jω ∆x1  x1 = jω ∆npR· ·wM + 1F)3−(xF1 )3 (x (x ) M (x1+) x∆F  1 1− 2 6} 6 2 EI | {z

(47b) (46b) (47c) (46c) (47d) (46d)

a) F1 ( x1 )

v1 ( x1 )

(47e) (46e)

Ω ( x1 )

− M3 x1 + (x1 ) , ∆x1 3 : υ 3 (x1 + ∆x1 ) ≈ υ 3 (x1 ) + Ω (x1 ) + 2 2 (47f) x13 R ∆x1 · F x12(x1 ) x1 )14 ∆x1 − jω ∆n jω (x + ∆x Ω 1 p·w +2F 3 (x1 ) −M (46f) υ3 = (x1 ) 3 12 | 6 {z 24 2 } EI ≈0 x2 with − M3 1 + (x1 ) x1 + υ 3 (x1 ) . 2 ∆υ Λ = jω αΛ∆x Application of the1boundary conditions at the left side of(48) the ∆x 1bending element F 1,3 (x1 ), υ 1,3 (x1 ), M (x1 ) and differential (49) ∆n = (x1 ) and EAat the right side F 1,3 (x1 + 1x1 ), υ 1,3 (x1 + 1x1 ), 1 (x1 + 1x1 ) acting at a beam element of M (x1 + 1x1 ) and ∆nF1 = − (50) length 1x1 (F and x1 ) the approximation of Eq. (40) yields 1 ∆using the ∆Ffollowing: = p · w · ∆x (51)

F 3 ( x1 )

380

385

390

395

400

∆x+ ~1 : F 1 (x (47a) 1 1x1 ) = F 1 (x1 ) , ∆nR = 1 (52) EI ∆ Ω Λ = jω ∆ nR · M Λ . (53) 1 : υ 1 (x1 + 1x1 ) = 1υ 3 −jω 1n·F 1 (x1 )+υ 1 (x1 ) , (47b) Following Kichhoff’s laws, sums of flow quantities constitute nodes — marked with ~ and sums of across quantities constitute meshes — marked with . Figs. 9 and 10 depict ~2 : F 3 (x1 + 1x1 ) = 1F + F 3 (x1 ) , (47c) the resulting circuit representation of the two-layer beam element for the two linearization cases of the first paragraph in Sec. 7. The angular velocity source Ω Λ) can also be in1x (x1 +∆ 1 ~ +moment 1x1 ) = M − Λ acting on ∆nR . These(47d) (x1 ) M terpreted source net3 : M (x1as jω 1nF1 works include for Λ = 0 and h1 or h2 = 0 the case of a homo 1x1 1xsubject 1 geneous monomorph-plate strip is to bending. − Fwhich , − 1F + F 3 (x1 ) 3 (x1 ) 2 (47e) and (47f) 2 do not apply for The approximations in Eqs. a large ∆x. This case is treated in Sec. 8. ≈ (x1 ) circuits − 13assume that |F1 ·ϕ| (47e) 2 All : (x 1 + 1x1 )equivalent following F3   and that no pressure load occurs.  1x1 1x1   − jω 1nR ·  M (x1 ) − F 3 (x1 ) 2 + 1F 6  , | {zof }actuating 8 Low frequency linear equivalent circuit ≈0 two-layer beams

∆ x1 EA

∆n =

M ( x1 )

1

F1 ( x1 + ∆ x1 )

∗1 1

v1 ( x1 + ∆ x1 )

Translational domain (x1)

∆vΛ

b)

∗3

∆nR

∆Ω Λ

Ω ( x1 + ∆ x1 )

Rotational domain

( F 3 ( x1 ) + ∆ F ) ∆2x1 2 ∆ x1

v3 ( x1 ) Ω ( x1 )

2 ∆ x1 F 3 ( x1 + ∆ x1 )

∗2 ∆F

∆ x1 2

3

Ω ( x1 + ∆ x1 )

Ideal rod

∆ x1 2

v3 ( x1 + ∆ x1 )

Translational domain (x3)

Ideal rod

∆x ∆ nR = 1 EI

c)

M ( x1 + ∆ x1 )

∗3

2

∆x F 3 ( x1 ) 1 2

≈0

375

9 195

Source

∆Ω Λ

Bending compliance

Figure 9. Circuit representation of a two-layer beam element when

Fig. 9. Circuit representation two-layerx3beam element when |F F3 in direction x1 (a)of anda direction as well as of the dy1 ·ϕ| |F1 · ϕ|rotational F3 inbehavior direction(b), x1 and (a) system and direction x3 asideal wellrods as as of namic model with the dynamic rotational behavior (b),(introduced and systembymodel with1968) ideal rotational–translational transducers Schroth, rodsThe as rotational-translational (introduced by Schroth to the location cS (c). axial velocity source transducers 1υ 3 is related is related to the location (1968)) (c). The axial velocity source ∆υ (1υ = 0 at the neutral axis c ). Λ A 3 cS (∆υ Λ = 0 at the neutral axis cA ).

with

M ( x1 )

∗3

∆Ω Λ

∗3

∗3

M ( x1 + ∆ x1 )

1υ 3 = jω α31x1 , (48) ∆nR 1x1 ( x1 ) ,∆ x 1nΩ= Ω ( x1 + ∆(49) x1 ) ∗ ∆ nF M 1 1 FEA 2 3 ( x1 ) 21 , (50) 1nF1 = − (F 1 1 x1 ) F 3 ( x1 ) 2 ∆ x1 ∗ 2 2 ∆ x1 F 3 ( x1 + ∆ x1 ) 1F = p · w · 1 x1 , (51) ∆ F ∆ x ∆ x 1 1x1 Ω ( x1 + ∆ x1 ) 1 v3 ( x1 + ∆(52) x1 ) 1nv3R( x=1 ) Ω ( x,1 ) 2 2 EI 3 1 3 = jω 1 nR · M 3 . (53)

≈0

∆x 1 ∆ x1 consti∗ Following quantities ∆ nR = 1 Kirchhoff’s , ∆ nF = − laws,, sums M =of Fflow ( 3 ( x1 ) + ∆ F ) 1 EI – marked F 1 ⋅ ∆ x1~ and sums of across2quantities tute nodes with constitute meshes – marked with . Figures 9 and 10 depict the circuit representation of the two-layer beam elFig.resulting 10. Circuit representation of a two-layer beam element when ement for the two linearization cases of the first paragraph F1 6= F1 (t) (source quantity) in Sect. 7. The angular velocity source 1 3 can also be interpreted as moment source M 3 acting on 1nR . For 3 = 0 and h1 or h2 = 0, these networks include the case of a homogeneous monomorph-plate strip which is subject to bending.


J. Sens. Sens. Syst., 3, 187–211, 2014

1xof 1 finite length ∆x In :the previous section a beam element (47f) 3 υ 3 (x1 + 1x1 ) ≈ υ 3 (x1 ) + (x1 ) 2 beam of length l was derived. In order to describe a long for quasistatic processes 1x12 load (p = 0) and 1x1 without pressure + + 1x − jω 1n F 3 (x1elements (x ) ), 1 1 R with the assumption 2|F1 · ϕ| F3 , 12 the ·beam in | {z } Fig. 9 can be connected chain-like. This chain circuit can

(51)

Fig. 9. Circuit representation of a two-layer beam element when |F1 · ϕ| F3 in direction x1 (a) and direction x3 as well as of be (52) simplified to a much smaller 4-port circuit as long as the the dynamic rotational behavior (b), and system model with ideal ΩI = 0 F II = 0 links between the coordinates at thetransducers beam ends are ofbyin-Schroth rods as rotational-translational (introduced (53) vEquivalent =0 I 10 U. Marschner et al.: circuit models beams = F (0), υ = υ (0), terest. These coordinates become F is related to the location (1968)) (c). The axial velocity source ∆υ I 3 Λ I 3 196 U. Marschner et al.: Equivalent circuit models of of two-layer two-layer beams l = 0 at the neutral axis c ). c (∆υ A S M = M (0), Ω = Ω (0) at the left side and F = F (l), Λ es constiI

I

II

3

to aΩmuch smaller circuit (l) at 4-port the right sideasaslong de- as the υ II = υ 3 (l),beMsimplified quantities II = M (l), II = Ω ∆Ω Λ links between the coordinates at the beam ends are of in10picted depict in Fig. 11. Integration of SOE (42) over beam length ∗3 ∗ 3 ∗ 3 M ( x1 + ∆ x1 ) M (x ) 405 υI = terest. 1These F I =atF beam 3I(0), l andel-application of thecoordinates boundary become conditions and II υ 3 (0), M = M Ω (0) at the left side and F = F 3 (l), 3 (0), Ω I =∆n paragraph I II R gives with n0 = l /EI: ( x1υ)3 (l),∆M = M (l),∆ΩnII = Ω (l)∗at the rightΩside υ IIΩ= lso be inx +as ∆ xde( II 1 1) x1 F1 M (42) over beam length pictedFin 11. Integration of SOE hese net2 3 ( xFig. 1) 2 F II = F I l and application of the boundary conditions at(54a)

I and II f a homo3 410 gives with n = l /EI: F 3 ( x1 ) bending. 0 2 ∆ x1 ∗ 2 2 ∆ x1 F 3 ( x1 + ∆ x1 ) apply M II for = FI − lMI (54b) ∆F F II = F I (54a) ∆ x1 ∆ x1 Ω ( x1 + ∆ x1 ) v x Ω ( x1 ) v3 ( x1 + ∆ x1 ) ·ϕ| F3 n0 3 ( 1 ) n0 2 2 3 +Ω Ω II = −jω 2 (M I + M Λ ) + jω F I M (54b) l II = F I − l M I 2l I

n0 ∆ x 1 ∆x ∗ 1 n = −jω ∆2nR =M I − , 0l∆F nFI +M , jω M n=0( F x1 )Ω + ∆ F) 1 Λ )+ Ω II = |−jω F3I(+ }1 =I −+FM 2l actuating Λx I 2 ⋅ ∆ EI {zl2 (M 1 1 2l M II (54c) n0 n0 = −jω Figure 10. Circuit Mrepresentation − l F I +M Λof a two-layer beam element 2 )|+I Ω −jω (M + {zIquantity). } of a two-layer beam element when 2l I6 = F M Λ(source Fig. 10. Circuit representation 2 when F (t) ength ∆x 1 1 2l M} II | {z (54c) length l 415 F1 6= F1 (t) (source quantity)

= 0) and ments in rcuit can

n0 −jω 2 (M I + M Λ ) + Ω I The approximations | 2l {zin Eqs. (47e) } and (47f) do not apply for ΩT

M

M

Ω =0 v =0

F = 0 ξ II

I

M

l3 n0 = E⋅I n M0 = I

II

I

I

I

II

I

I

1 Ω T = ⋅vT l 1l

MT =n l ⋅0 F2lI

2

M T = l ⋅ FI

II

i

M

II

M

II

Ω Ω

II

II

F =0 II

−n0 6 vT = v − vI 1 l v

F =F

v =0

II

II

n0 2l 2

MΛ

n01 2l Ω T = ⋅vT l

I

I

ξ −ϕ 1

2

I

Ω =0

F =F

II

MΛ

l3 I 2 n 2l E⋅I 0

Ω =M0

M

−ϕ 1 l

F =0 II

v

i

II

−n0 6

Ω T is treated in Sect. 8. v v = 0 vT = v − vI a large 1x. This case v All following equivalent circuits assume that |F ·ϕ| F n0 n0 1 3 υ II = −jω and(M + M Λ ) + jω F I + lΩ I + υ I load6occurs. 2l thatI no pressure Fig. 11. Low-frequency linear equivalent circuit of an actuating n0 n0 n0 Figure 11. Low-frequency lineardirection equivalent1 circuit of an actuating II = I = l · (Ω Iυ− Ω −jω ) +2l jω (M IF+I M + Λυ)I + jω 6 F I + lΩ I + υ(54d) fixed-free two-layer beam without T 8{z Low-frequency } | 6 linear Fig. 11. Low-frequency equivalent of an actuating fixed–free two-layer beamlinear without direction circuit 1. n0 equivalent circuit of (54d) υ T actuating F I + υI = l · (Ω I −two-layer Ω T ) + jω beams fixed-free two-layer beam without direction 1 {z } | 6 I

i

i

II

υT n0 2 M Λ l The equivalent is section shown ainbeam Fig. element 11 for the example In thecircuit previous of finite length 1x n0 2 M Λ l of a left-side fixed-free prevented movement and was derived. Inbeam. order toisdescribe ainlong beam length l for The equivalent circuitThe shown Fig. 11 forofthe example processes without load quantities. (pmovement = 0) and with of aleft left-side fixed-free beam.pressure The across prevented and rotation at quasi-static the side are short-circuited n0 2 the assumption |Fincludes F3the , thenegative beam elements in quantities. Fig. 9 can 420 rotation at the left sideare short-circuited across 1 ·ϕ| The translational domain compliance n0 2 be connected in a chain-like manner. Thisnegative chain circuit can be The translational domain includes the compliance −n0 /6 which can be handled by circuit simulators. However, v n simplified to much smaller four-port circuit as long as the −n0 /6 which can be handled by circuit simulators. However, v the total compliance of athe bending beam is positive. − 0 n0 vT links between the coordinates at thebeam beamisends (I) and (II) the total compliance of the bending positive. vT 6− When only the translational behavior in direction 3 is of 6 areWhen of interest. These coordinatesbehavior becomein F Idirection = F 3 (0),3 υisI = only the translational of interest,425then the network can be simplified. Fig. 12 shows υinterest, and I can = be at the left side F II = (0)simplified. the(0)network Fig.and 12 shows I=M 3 (0), Mthen Figure 12. Translational linear equivalent circuit of an actuating the result of the ofMthe Fthe M II = and II rotational the right υtransformation = elements (l) rotational (l) at elements result of υthe transformation of the 3 (l), II = 3 (l), fixed–free two-layer beam. 2 linear equivalent fixed2 Fig. 12. into the translational domain F 3With = MnF/l and n/l=and nRn·l= .nR ·lof Fig.Translational 12. Translational linear equivalentcircuit circuitof ofan an actuating actuating fixedside, as depicted in with Fig. 11. l 3M /EI , integration into the translational domain with . 03== free two-layer beam free two-layer beam The force source linked to the transformed moment generates SOE (42) over beam length l and application of the boundary The force source linked to the transformed moment generates at υ(I) and (II) as gives the , which appears as υ IIno when noFforce the velocityconditions υ Wvelocity , which appears υ II when force W II is F II is present:430 present: equivalent n0 n0 9 High-frequency linear circuit of of an actuatcircuit F II = F I , (54a)9 High-frequency υ II = −jω M Ilinear + M 3 equivalent + jω F I + lI + υ I an actuat(54d) ing two-layer beam 2l 6 ing two-layer beam n M 0 Λ n0 0=M υ T = −υ II = −jω ξn −jω (55) Λ II = −jω ξ II−=l M υ T = −υ II M −jω .2 l . (55) (54b) = l · I − T + jω F I + υ I . If no pressure difference is considered but the average mass = F , 2 l | {z } II I I If no pressure difference is 6considered but the average mass υfor 440 is assumed each beam element, then Eqs. (47a, c and d) T With 440 is assumed for each beam element, change in the case of |F · ϕ| F3then to Eqs. (47a, c and d) With 1 n0 n0 The equivalent circuit is shown in II = −jω 2 Mjω + jω F I + I (54c)change in the case of |F1 · ϕ| F3 to Fig. 11 for the examI +nM 0 3 2l l = Ω II = jω jω ϕ − · M . (56) ple ro~1 of : Fa1 fixed–free (x1 + ∆x1beam. ) = F 1The (x1 )prevented + ∆F m,1 movement and (58a) Λ nII0 2 Ω II = jω ϕII = = −−jω2 n0· MMΛ l.− l F +M (56) ~1 : tation F 1 (xat + ∆x ) = F (x ) + ∆F (58a) the left side are short-circuited across quantities. 1 1 1 1 m,1 I I 3 l 2 1 : translational υ 1 (x1 + ∆xdomain (58b) 1 ) = ∆υincludes 1) Λ − jω ∆n 1 + υ 1 (x The the· Fnegative compliance follows 2l | {z } 445 M II 1 : −n υ210:(x + ∆xcan jω · Fsimulators. + υ 1 (x1However, ) (58c) (58b) 1) = /6 be ∆υ handled by ∆n circuit Λ− ~ F1which 3 (x1 + ∆x1 ) = ∆F m,3 + F 3 (x11) follows 445 n0 l the total compliance of the bending beam is positive. −jωϕ 2. M I + M 3 + I , (x1 + ∆x∆x F 3 (x1 ) (58c) 435 ξ II = − (57)~2 : F 1) = ~When (x ) ∆F = Mm,3 (x1+ ) behavior 33 : M 1 +the 1 II only translational in direction 3 is of | 2 2l {z } l (58d) ∆x ∆x1 ξ II = − ϕII . (57) T then ~3 :interest, M (x1 + = M (xcan 1 )network 1 ) be1 simplified. − ∆x (Fthe − F 3 (x1 ) Figure 12 shows 2 m + F 3 (x1 )) 2 the result of the transformation elements (58d) ∆x12 of the rotational ∆x 1 − (F m + F 3 (x1 )) − F 3 (x1 ) 2 2 II II

J. Sens. Sens. Syst., 3, 187–211, 2014



11


M ( x1 )

Ω ( x1 )

F 3 ( x1 )

v3 ( x1 )

MΛ

MΛ

MΛ

∆Ω Λ

∗3

M =0

∗3

∆nR F 3 ( x1 ) ⋅

MΛ

∆Ω Λ

M =0

M ( x1 + ∆ x1 )

…

F 3 ( x1 + ∆ x1 )

∆m

… v3 ( x1 + ∆ x1 )

MΛ

∆x 2

ΩΛ

M ( x1 + N∆ x1 )

M

I

Ω ( x1 + N∆ x1 )

Ω

I

∆nR

Ω ( x1 + ∆ x1 )

∆x 2

∆x 2 ∗ 2 ∆x 2 ∆ F m,3

197

FI

F 3 ( x1 + N∆ x1 )

∆x 2

Ω Bending waveguide

II

II

F II vII

vI

v3 ( x1 + N∆ x1 )

∆m

M

Figure 13. High-frequency linear equivalent circuit of direction 3 and dynamic rotational behavior of an actuating two-layer beam.

Fig. 13. High-frequency linear equivalent circuit of direction 3 and dynamic rotational behavior of an actuating two-layer beam

450

into the translational domain with F 3 = M/ l and n = nR ·l 2 . with force source linked to the transformed moment generThe ates the velocity υ W , which appears as υ II when no force F II ∆F (59) is present: m,1 = jω ∆ m υ 1 (x1 + ∆x1 ) n0 M 3 υand . T = −υ II = −jω ξ II = −jω 2 l

455

460

With ∆F m,3 = jω ∆ m υ 3 (x1 + ∆x1 /2) . (60) jω n0 II = jω ϕ II = − 2 · M 3 , (56) The beam elementl describing direction 3 includes its mass instead of that the pressure-related force source in Fig. 9. Fig. 13 it follows shows the resulting mechanical network with distributed pal direction 1 when the beam is discretized ξrameters = − without ϕ . (57) II 2 over x3 intoII N finite beam elements. A finite beam element in direction 1 is represented by a circuit similar to Fig. 9a 9withHigh-frequency equivalent of is conthe difference oflinear the included mass circuit ∆m, which an actuating two-layer beam nected to the mechanical reference (ground). A though experiment leads to the conclusion that the If no pressure but the mass source momentdifference is coupledisatconsidered the beam ends andaverage that it is not is assumed for each beam element, then Eqs. (47a), (47c) and branched into the inner beam, i.e. (47d) change in the case of |F1 · ϕ| F3 to ∆1M =M ~ :F F 1. (x1 ) + 1F m,1 , (xx Λ 1(∆ 1 1+) 1x 1 ) Λ=(l)

465

470

475

(55)

(61) (58a)

Therefore the unimorph can be grouped to a bending (58b) wave 1 : υ 1 (x1 + 1x1 ) = 1υ 3 − jω 1n · F 1 + υ 1 (x1 ) , guide with the moment source as an additional boundary condition, as shown in Fig. 13. 480 ~2The : F 3structures + F 3 (x1 ) , wave guide in (58c) (x1 + 1x1constitute ) = 1F m,3 a longitudinal direction 1 and the bending wave guide in Fig. 13. The behavior of 1 this can ~ + 1xnetwork (58d) 1 ) = M (x 1 ) be investigated with standard 3 : M (x circuit simulators, like pSpice. is presented in 1x1 An example 1x1 − FWhen − , Sec. 10.2.3. areFconnected chain-like then 1) m + Fthe 3 (xelements 3 (x1 ) 2 2 two connected translational-rotational transducers of length 485 with ∆ x1 /2 can be combined to one single transducer of length ∆ x as demonstrated in Fig. 32. 1F 1m,1 = jω 1 m υ 1 (x1 + 1x1 ) (59) www.j-sens-sens-syst.net/3/187/2014/

and

Rotational

Interacting

X υ 3 (x1 + 1xdomain 1F m,3 =domain jω 1 m 1 /2) .

M

Translational domain (60)

F

j ω element I 3 3 mass The beam describing direction 3 includes its instead of the pressure-related force Rotationalsource in Fig. 9. Figure V13 shows theX-rotational resulting mechanical network withv3distranslational Transducer tributed parameters without direction 1transducer when the beam is discretized over x3 into N finite beam elements. A finite beam element in direction 1 is represented by a circuit similar to ω Idifference F1 which Fig. 9a with jthe of the included mass 1m, 1 is connected to the mechanical reference (ground). A thought experimentx3leads to X-translational the conclusion that v1 the Transducer source moment is coupled at xthe beam ends and that it is not 1 branched into the inner beam, i.e.,

Ω

1 M 3 (1 x1 ) = M 3 (l) .

(61)

Fig. 14. Difference and flow coordinates in into the circuit description Therefore, the unimorph can be grouped a bending wave involving the interacting domain X guide with the moment source as an additional boundary condition, as shown in Fig. 13. The structures constitute a longitudinal wave guide in di10 Four-port reversible rection 1 and themodels bendingofwave guide transducers in Fig. 13. The behavior of this network can be investigated with standard circuit Eq. (7) describes the relation between the mechanical field simulators, like SPICE (Simulation Program with Integrated pair (S, T ) and a transduction relation between the two physCircuit Emphasis). An example is presented in Sect. 10.2.3. ical domains. Reversibility entails byina second set of material When the elements are connected a chain-like manner state equations, which relates field pair components (Λi ,with Ki ) then two connected rotational–translational transducers of the interacting domain X to each other in addition to the a length of 1 x1 /2 can be combined to one single transducer transduction mechanism. From this second set only the relawith a length of 1 x1 as demonstrated in Fig. 32. tion between the influence quantity Λ, stress T 1 or strain S 1 , obtained with Eq. (7), and quantity K is considered:

10

Multi-port models of reversible transducers

T K = α T 1(7) + βdescribes Λ = αEthe S 1 relation + β T −between α2 E Λ the mechanical (62) Equation | {z } between the two field pair (S, T ) and a transduction relation βS physical domains. Reversibility is entailed by a second set of J. Sens. Sens. Syst., 3, 187–211, 2014

3 and dynamic rotational behavior of an actuating two-layer beam 198


(59)

j

ωI

M

3

X-rotational Transducer

V

Ω

(60)

mass g. 13 d patized ment g. 9a con-

j

ωI

Translational domain

Rotational domain

Interacting domain X

F

Quantity/parameter 3

Rotationaltranslational transducer

x3 x1

X-translational Transducer

Time-integrated flow I

Electric R u = xE E RR Q = AD

Magnetic R V m = xH H RR 8 = AB

Flow coordinate dI/dt

i

Im

Across coordinate V

v3

F1

1

Table 2. Electrical and magnetic quantities and parameters

10.1

v1

Piezoelectric bimorph

Using Einstein notation, piezoelectric state equations are written as T D n = εnm E m + dnj T j

n = 1 . . . 3, m = 1 . . . 3,

(63)

j = 1 . . . 6, Figure 14. Across and flow coordinates in the circuit description

involving the interacting domain X.

Fig. 14. Difference and flow coordinates in the circuit description S = d E + s E T mi m ij j i involving the interacting domain X

the s not

(61)

wave con480

n die bedard ed in then ngth 485 ngth

material state equations, which relates field pair components (3i , Ki ) of the interacting domain X to each other in addition to the transduction mechanism. From this second set only the 10 Four-port models of reversible transducers relation between the influence quantity 3, stress T 1 or strain S 1 , obtained with Eq. (7), and quantity K is considered: Eq. (7) describes the relation between the mechanical field pairK(S, T transduction phys= α T) 1and + βaT 3 = αE S 1 + relation β T − α 2between E 3, the two(62) ical domains. Reversibility entails by set of material | {z a second } S state equations, which relates fieldβpair components (Λi , Ki )

i = 1 . . . 6, m = 1 . . . 3,

(64)

j = 1 . . . 6, with displacement current D, electrical field strength E, piezoelectric charge constant d and permittivity ε. Two typical electrode arrangements are considered next. 10.1.1

Transverse piezoelectric coupling

An important technical configuration is transverse coupling of two piezoelectric layers as shown in Fig. 15. For a piezoof the interacting domain X to each other in addition to the electric longitudinally acting bimorph with small beam width where β is mechanism. an additional From constant material With transduction this secondcoefficient. set only the rela- (E 2 , E 1 = 0 and T 2 . . . T6 = 0), Eqs. (63) and (64) simplify this second set a multi-domain port description including theS , to tion between the influence quantity Λ, stress T 1 or strain physical domain X can be derived. This domain controls the 1 obtained with Eq. (7), and quantity K is considered: T moment source, velocity or rotational velocity source in the D 3 = d31 T 1 + ε33 E3 , (65) previously developed models. Figure 14 shows the extended T port Each direction α2 E Λinvolves a sep-(62) K= α Tdescription. αEtranslational S1 + βT − 1+β Λ = E | these {z sources. } In the inter(66a) S 1 = s11 · T 1 + d13 · E 3 , arate transducer which includes S acting domain X, the flow rate jωIβ and across quantity V serve as network coordinates, which are integrals of the field E S 2 = s21 (66b) · T 1 + d23 · E 3 . quantities. The product V · jωI is a power. Reversible transducer models for piezoelectric and piezoThe indices match the directional conventions in Fig. 2. The magnetic two-layer bimorphs are derived next. Table 2 lists deformation of the small beam in direction 2 is not further the coordinates used in the electrical and magnetic domain. considered. In the case of a large beam width this deformaDue to Eq. (61), no distinction is made between a beam of tion is suppressed (S 2 = 0), thus causing T 2 6= 0, such that length l and a beam element of length 1x1 . In accordance with terms used in material science, the indices used in material equations relate direction 3 to the acE E S 1 = s11 · T 1 + s12 · T 2 + d13 · E 3 , (67a) tive direction. The convention will be clarified in the context of the analyzed examples. E E S 2 = s21 · T 1 + s22 · T 2 + d23 · E 3 = 0, (67b) T D 3 = d31 · T 1 + d32 · T 2 + ε33 · E3

(67c)

E = s E and d = d , or, including s12 23 32 21

J. Sens. Sens. Syst., 3, 187–211, 2014


U. Marschner et al.: Equivalent circuit models two-layer beams beams U. Marschner et al.: Equivalent circuitofmodels of two-layer U. Marschner et al.: Equivalent circuit models of two-layer beams

rs

where A = w · l is the through-flow-area:

ic

1 i E1 1 EI M =− Ω− α1 −h21 + 2chl1 i1 h2 jω |{z} l u jω 2l } | P E{z 3 , D3 1 1/nR,o S ,T −1/X R⊥1

H

A

13

199 c = cS

h

H B

P

2

E3 , D3

F1

M, Ω (72)

E2 3α2 h22 + 2ch2 i2w − jω 2l {z } | Figure 15. Transversely coupled piezoelectric two-layer beam. −1/X R⊥2

ent. With uding the ntrols the rce in the extended 525 ves a sep-555 the interuantity V f the field

nd piezoble 2 lists 530 c domain.560 beam of

ce, the into the ace context565 535

tions are

(63)

ength E,

ed next.

545

coupling a piezoam width ) simplify

(65) 570

(66a) 550

Ω = Ω −Ω I

− i1

u R,1

i2

F1

2

I

Cb,1

i

Fig. 15. Transversely coupled piezoelectric two-layer beam

Ω M II

II

1 ⋅Ω uR,1 = X R⊥1 i1 = X R⊥1 ⋅ M Λ,1

−M Λ

,1

1 EA E1 h1 E2 h2 α1 i1 + α2 i2 (73) υ1 + jω |{z} l jω l jω l | {z ! } 2 ! | .{z } S 2 = sE (66b) 1/n E oEd23 · E 21 · T 1+ 3 t⊥1 1/X −1/X s d 23 s t⊥2 E S 1 = s11 T 1 + d13 − E 12 ET3 , (68a) − 12E Thewith indices directionalαconventions in Fig. 2. The s22 theparameters and the match material =s−d 22 31 /ε33 and | | {z } } 2 is not further deformation of the small beam in {z direction ∗ E∗ 1 d s considered. In width this deformation is 11 case. of a large beam 13 (74) E = E 2 ) s11 (1 − = 0), thus causing T 2 6= 0, such that suppressed (Sk231 ! ! 2 E d32 d32 sE21 Eqs. (72) describe a moment node a force (68b) node T E and (73) D = s11d31 Eand S 13 = · T−1 + sE12 · T T2 + d13ε·33 E− (67a) 1+ 3 3 E rotational including the complex translational and transducs22 s22 {z XE }and X | , from {z which } the mechanical | tion R⊥· E = 0 sE · T 2 + d23 (67b) S 2 =coefficients ∗ s22t⊥ 21 · T 1d + 3T ∗ ε33The setup related direcdomain in Fig.3116 can be deduced. D3 =ofd31 + dand εT33 ·reversed E3 (67c) tions i1 ·, TM1 Λ,1 were to match the direc32 · Tυ21 + and (68a) differ only in the can be written. Equations (66a) tions of the arrows E in Fig. E 14. or, encounting s12 =and s21Eandare d23 = d32 , of x while the discoefficients. functions 3 over x3 in layer — named with subIntegratingTE 1, S 3 each 1 D23!and placement current does⊥2not onEcoordinate be—depend script and considering Eqs.x3(44), ,1 and ,2 or E ⊥1 s120. When these equations d23 s12are rearranged to E cause of div D = (45), gives s11 and − (10) + voltage d13 − E (68a) S 1 =(69b) T 1the E sE s22352) 3 (T , E = f (S, D)) 22(Lenk et al., 2011, pp. | {z } | Zc {z } c+h Z 2 ∗ ! d E∗ 13 s 2 11 d x = 1 3,1 1 E 3,2 kd31 u=− E x3 ·3S 1 − ·D , T1 = E (69a) 2 2 3 s11 11 − kd31 sE cd31 1 − k31 c−h 2 d 32 21 32 D3 =Zc d (68b) T + εT33 − E E 31 − ξ1 ! 1 sdE d ϕ s22 S 3 22 2 = − | 1 E1 {z α1k31 +} E1 α1|x3 {z1+ β1}D3,1 dx3 (69b) E3 = − d∗31 d x21 · S 1 + Td xε1T33∗ 2 · D 3 , c−h1d 1−k ε 1−k

Figure 16. Transversely coupled piezoelectric transducer model Fig. coupled piezoelectric transducer model given given16. byTransversely Eqs. (75), (72) and (73). by Eqs. (75), (72) and (73)

Zbe 2written. canc+h Eqs. (66a) and (68a) differ onlyin the coefd ξ1 dϕ S x while the diswith ficients. TE12, α S21 and = +EE α2 xfunctions + βof 3 2are 3 2 D33,2 dx3 d x d x 1 placement current 1D3 does not depend on coordinate x3 2 c dof 31 div D = 0. When these Eqs. are rearranged 2 because to k31 E =1 hT1 E , E1 1 h1 β1S (70) 2 ε33fαs(S, (T D)) (Lenk et al., 2011, pp. 352): = ,E = α −h + 2ch υ + Ω − i 1 1 1 1 1 jω l 11 1 jω 2l jω | wl {z } | {z } | {z 2 } moment M(x F 1(x1 ) follow from Eqs. 11 ) and force k 1/C(13) b,1 and 575 t⊥1 T 1 1/X =with · S 1−1/X − 1 R⊥1 31 2 · D3 (69a) E 2 (14) s11h(1 − k31 ) E d31 1 − k31 S 1 h β E 2 2 2 2 2 α2 υ 1 − α2 h22 + 2ch2 Ω + i2 =− jωD l1 3,1 jω 2 jωi2l= jωA D i 1 = jωA and , k31 | wl {z }(71) 3,2 1 } 2 | {z } | {z E3 = − · S1 + T · D3 1/C (69b) 2 2 ) b,2 d31 t⊥21 − k31 −1/X −1/X R⊥2 ε33 (1 − k31 where A = w · l is the through-flow area (75) 580 with 1 EI E1 with M2 = − d231 − α1 −h21 + 2ch1 i 1 (72) k31 = jω (70) l, jω 2l |{z} 2 | εT33εsTE {z } β S = 1/ 1 − k . (76) 11 33 31 1/nR,o −1/XR⊥1 moment M ) follow from Eq. and 585 The resulting meshes in 1Fig. 16 include the(13) electrivoltage F 1 (x 1 ) and force E2 (x α2 h22 + 2chand , b,2 of the two layers. The caEq.−capacitances (14) with cal Cb,1 2 i 2C jω 2l pacitance Cb {z = Cb,1 + C}b.2 can be measured in case of the | i1 = jωAbeam D3,1(υ and i20). = jωA D3,2 , (71) blocked =Ω= −1/X

1 EA E1 h1 E2 h2 F1 = − υ1 + α1 i 1 + α2 i 2 (73) jω l jω l jω l Compliance |{z}no and |rotational {z } compliance | {z }nR,o are labelled with the subscript for “open” since could be mea1/no “o”1/X −1/Xthey t⊥1 t⊥2 sured in case of no current (i1 = i2 = 0). If the two layers are T non-symmetric, then this boundary be realand with the material parameters α condition = −d31 /εcannot 33 and ized. The open electrical circuit connects Cb,1 and Cb,2 in series which 1 allows a current flow. This connection provides . = between (74) aElink 2rotation and translation. The transfer function s E 1 − k31 υ ∗1 /Ω 011 |i=0 for this case can be determined analytically for the circuit in(72) Fig.and 16 which is simplified for this pupose. Thea Equations (73) describe a moment node and first step is to explore the effect of an assumed rotational veforce node including the complex translational and rotational locity source Ω in the electrical domain. The effect can be 0 transduction coefficients X t⊥ and X R⊥ , from which the mefound by a source transformation as demonstrated in Fig. 17. chanical domain in Fig. 16 can be deduced. The setup-related The electrorotational be ignored afterwards directions of i 1 , M 3,1transducers and υ 1 werecan reversed to match the disince they do not influence the ideal sorces and current i1 rections of the arrows in Fig. 14. which flows through transducers. The Integrating E overthe x3electrotranslational in each layer – designated by subresult of this reduction step is depicted in Fig. 18a. The cirscript and or and – and considering Eqs. (44),

F1 = −

31

(64) 540

1

31

33

R⊥2


31

Cb,2 1

uR,2 =

⋅Ω

X R⊥2 i2 = X R⊥2 ⋅ M Λ,2

u R,2

u

− i1

i2

u

2

nR,o

M Ω

F1,1 1 ⋅v X⊥ i = X ⊥ ⋅F u = t ,1

t

t ,1

1

u

M Λ,

t

1

1

1,1

1 u ,2 = ⋅v X ⊥2 i2 = X ⊥2 ⋅ F1,2 t

t

t ,2

F1,2

F1

no

v1∗

t

v1∗ = −v1

,1

,2

⊥1

⊥2

J. Sens. Sens. Syst., 3, 187–211, 2014

14 200

590


cuit can be simplified further, when the combined sources (45), (69b) and (10) gives the voltage c 1 1 c+h Z ·2 Ω (77) uΩ = Z − 0 E 3,1 d xX3 R⊥2 u = − X R⊥1 = E 3,2 d x3 (75) c c−h1capacitance and total c Z dξ 1 dϕ S C · C E1 α1b,2 + E1 α1 x3 =ser − = b,1 + β1 D 3,1 dx3 C dx1 dx1 Cb,1 + Cb,2

Cb,1 − i 1 u R,1

i1

600

605

610

615

2 Ztransformed are c+h the translational mechanical domain. dξinto dϕ S 1 The a transition to a = complex E2 α2transduction + E2 α2coefficient x3 + βcauses D dx 3 3,2 2 dx1 or inductance dx1 symbol, compliant behavior respectively. Then c to each other. They theEtwo transducers connected E1are solely 1 h1 2 = be combined α1 υ 1 + to a αlever 2chratio can with 1 −h 1 or mechanical ad1 +the jω l jω 2l vantage {z } | {z } | 1/Xt⊥1 −1/XR⊥1 X (79) t = 1t⊥1h1 β1S −X t⊥2 i1 jω | wl {z } as depicted Fig. 18b. The sought transfer function 1/Cin b,1 ! E2 h2 E21 2 1 =− α υ − α h + 2ch 2 2 2 1 2 jω l X t⊥1 · jωX2l − X | } R⊥1 {z R⊥2 υ ∗1 | {z } = (80) −1/X −1/X t⊥2 R⊥2 Ω 0 i=0 nC t − 1 + 1 h2 β2S no · (t − 1) i2, + jω | wl {z } with 1/Cb,2 X2 with nC = − 2 t⊥1 . (81) ω Cser S T 2 β = 1/ ε33 1 − k31 . (76) is found by applying of Kirchhoff’s laws: The resulting voltage meshes in Fig. 16 include the electrical i1Cb,2 of the two∗layers. capacitances and −υ +t·υ = 0 (82) : −υ ∗1 + jωCnb,1 C· The capacitance X Cbt⊥1 = Cb,1Ω+ Cb,2 1can be measured in the case of the blocked ∗beam (υ = = 0). υ 1 rotational compliance n are labeled R,o ~ Compliance : t · i∗1 − i∗1 −no and =0 (83) jω n0 for “open” since they could be meawith the subscript “o” sured with in the case of no current (i 1 = i 2 = 0). If the two layers are nonsymmetric, then this boundary condition cannot be rei1 open electrical circuit connects C and C in alized. b,1 b,2 . (84) i∗1 = The X t⊥1 allows a current flow. This connection provides series, which a link between rotation and translation. The transfer function υ ∗1 / |i=0 for thisoccurs case can be X determined analytically for No0dependency when . R⊥1 = X R⊥2 the circuit in Fig. 16, which is simplified for this purpose. 10.1.2 Symmetric laminate The first step is to explore the effect of an assumed rotational velocity source 0 in the electrical domain. The effect can be A special case oftransformation transverse coupling is the laminate of found by a source as demonstrated in Fig. 17. two equal piezoelectric layers as shown in Fig. 15 where The electrorotational transducers can be ignored afterwards i1 = −i i/2 Suchthe a symmetric laminate is char2 =do since they notholds. influence ideal sources and current i1 = X and X = X , i.e. t= 1. acterized by X R⊥1 the electrotranslational R⊥2 t⊥1 t⊥2 which flows through transducers. The Simplified network parameters — as derived by Lenk and result of this reduction step is depicted in Fig. 18a. The cirIrrgang — result from when Eqs. (75), (72) and (73): cuit can (1977) be simplified further, the combined source

J. Sens. Sens. Syst., 3, 187–211, 2014

(85a) (77)

1

uR,2 =

2

M

nR,o

Ω

M Λ,

⋅Ω

X R⊥2 i1 = X R⊥2 ⋅ M Λ,2

u R,2

0

− i1

i1

Transformation

Cb,1 −i 1

−M Λ

1 uR,1 = ⋅Ω X R⊥1 i1 = X R⊥1 ⋅ M Λ,1

Ω0

X R⊥1

i1

Cb,2

,1

M Λ, M 2

⋅Ω

X R⊥2 i1 = X R⊥2 ⋅ M Λ,2

X R⊥2 i1

1

uR,2 =

Ω0

Ω

nR,o

0

− i1

Figure 17. Transformation of a rotational velocity source. Fig. 17. Transformation of a rotational velocity source

and total capacitance

1  Cb,1 · C1b,2 Cser =uΩ =  X − X  ⋅ 0 R⊥b,2 1 R⊥2  Cb,1+ C

Ω

vΩ = uΩ ⋅ X (78) ⊥1 t

∗

i

1 are transformed into the mechanical domain. i1 = 1 translational X ⊥1 The complex transduction coefficient causes a transition2 to X − X ⊥1 a compliant behavior or⊥1 inductance symbol. Then the2 two ∗ ω Cserbe F t ⋅ i1They can Cser are solely connected 1 transducers to each other. combined into a lever with the ratio or mechanical advantage t

t

Cb,1 Cb,2 Cb,1 Cb,2 ⋅

=

t

v1∗

v1∗

v∗

t⋅ 1

no no X t = t⊥1 (79) Xt⊥2 1: X a) b) t = X ⊥1 : X ⊥2 ⊥2 as depicted in Fig. 18b. The sought transfer function: Fig. 18. Length variation coupled piezoelectric 1 1 of a transversally ∗ X · − t⊥1 υ Xvelocity unimorph caused by aXrotational R ⊥1 R ⊥2 1 (80) = t − 1 + nC +

t

0 i=0

620

1 1 M =− 1 ·Ω + ·i 1 X R⊥ u = jω nR,o− · 0 X R⊥1 X R⊥2

,1

Cb,2

(78)

c−h1

595

−M Λ

1 uR,1 = ⋅Ω X R⊥1 i1 = X R⊥1 ⋅ M Λ,1

t

t

no ·(t−1)

with 1 F 1 = − X 2t⊥1· υ nC = − jω2 n , ω Cser

(85b) (81) www.j-sens-sens-syst.net/3/187/2014/

− i1

i1

(81)


Fig. 17. Transformation of a rotational velocity source


201

(82)

uΩ

(83)

 1 = −  X R⊥1

1

 ⋅ X R⊥2 

Ω0 i1

=

C Cb,1 Cb,2 Cb,1 Cb,2 ser

⋅

=

X t ⊥1

no

+

1: X ⊥2

a)

aminate of 15 where te is chari.e. t = 1. Lenk and (73):

t

− X ⊥1

t

= X ⊥1 : X ⊥2 t

t

υ ∗1 = 0, jω n0

625

(82) (85b)

(83) 630

with i1 . X t⊥1

(84)

No dependency occurs when X R⊥1 = X R⊥2 . 635

Symmetric laminate

A special case of transverse coupling is the laminate of two equal piezoelectric layers as shown in Fig. 15 where i 1 = −i 2 = i/2 holds. Such a symmetric laminate is characterized by X R⊥1 = X R⊥2 and Xt⊥1 = Xt⊥2 , i.e., t = 1. Simplified 640 network parameters – as derived by Lenk and Irrgang (1977) – result from Eqs. (75), (72) and (73): 1 1 M =− ·+ · i, jω nR,o X R⊥

F1 = −

(85a)

1 · υ, jω n

1 1 1 ·υ + ·+ X t⊥1 XR⊥1 jω Cb,1 1 1 1 =− ·υ + ·+ X t⊥2 XR⊥2 jω Cb,2

u=−

nR,s

Ω

n

v1

(85b)

i 2 i · , 2 ·

(85c) 645

with EI nR,o = , l www.j-sens-sens-syst.net/3/187/2014/

1

no

i1 ∗ 1 − υ + t · υ ∗1 = 0, : C· F 1−υ = 1−+ jω n· υ X jω n t⊥1

10.1.2

1 ⋅Ω YR⊥

F1

Figure 19. Model of a transversely coupled symmetric piezoelecFig. 19. Modelbeam. of a transversely coupled symmetric piezoelectric tric two-layer two-layer beam

ser

is found by applying Kirchhoff’s laws:

i ∗1 =

iR =

M

2

v1∗

v∗

t⋅ 1

b)

ωC

Figure 18. Length variation of a transversally coupled piezoelectric Fig. 18. caused Lengthby variation of a velocity. transversally coupled piezoelectric bimorph a rotational unimorph caused by a rotational velocity

~ : t · i ∗1 − i ∗1 −

u = YR⊥ ⋅ M T

MT

t

∗ t ⋅ i1

620

(85a)

Cb

2

F1 v1∗

u

i1

t

(84)

iR

t

∗

1

X ⊥1

i

vΩ = uΩ ⋅ X ⊥1

(86)

1 i w(2h11 )3 · υ + 1 · Ω + u=− · (87) I = X t⊥1 , X R⊥1 jω Cb,1 2 12 (85c) 1 1 i 1 =− ·υ+ ·Ω + · X t⊥2 X R⊥2 jω Cb,2 2 650 3 nR,o α 1 with = , (88) XR⊥ EI 4 wh1 jω nR,o = (86) l 3 w(2h EA 1 ) In = (87) (89) 655 = l 12 1 3 nR,o α (88) = and X R⊥ 4 wh1 jω EA A= = 2wh1 . (90) n (89) l and The neutral layer is located at the material interface, i.e., 660 cA==2wh cS . Therefore, no translational electromechanical effect A (90) 1. occurs (F 1 6= f (i)). In the sensing case, opposite voltages The neutral layer is located at the material interface, i.e. are caused by υ 1 which cancel each other out (u 6= f (υ 1 )). cA = cS . Therefore, no translational electromechanical effect Following Lenk and Irrgang (1977), a circuit description occurs (F 1 6= f (i)). In the sensing case, opposite voltages with real transduction coefficients are caused by υ which cancel each other out (u 6= f (υ)). 1Following jω CbLenk d31and Irrgang (1977) a circuit description (91) 665 = transduction = E wh 1 with real YR⊥ XR⊥ s11 coefficients jω Cb d31 1 = wh1 SOE (85) to (91) is obtained by = rearranging YR⊥ XR⊥ sE 11 1 SOE (85) to: is obtained1by rearranging M =− ·+ ·u (92) jω nR,s YR⊥ 1 1 M =− 1 ·Ω + ·u · + jωYCR⊥ i= jω nR,s b · u, 670 YR⊥ (92) 1 i = n is the · Ω + jω Cb · u where R,s YR⊥ short-circuit compliance (subscript “s”) l where nR,s nR,s = EI l nR,s = EI and and

is the1short-circuit compliance (subscript “s”): , (93) 2 1 −143 k31 (93) 2 1 − 34 k31

lw lw TT 2 Cbb = 1 − k 231 (94) = Cb,1 +C Cb,2 = 22εε33 b,1 + b,2 = 33 1 − k31 hh11 is is the the total total blocked blocked capacitance. capacitance. SOE SOE (92) (92) can can be be interpreted interpreted as the circuit in Fig. 19 (Lenk et al., 2011). 675 as the circuit in Fig. 19 (Lenk et al., 2011). J. Sens. Sens. Syst., 3, 187–211, 2014

Fig. 20. L

10.1.3

L

Longitud interdigit separate e ers are po but with ing thin l mately un rections 1 nate syste Longit T3 =

1

sE 33

D3 = d∗33

matching β T = εT33∗ sE 32 d23 /s Calcul Eqs. (13) tromecha equation

Mλ = − |

and the t cient in th

1 F = 1 Yt u

As in Sec electric la

202


P

Ne = 2

E 3 , D3

i

i

l

h1

u

h2

u

iR

u = YR ⋅ M Λ iR =

Cb,1 Cb,2

M

MΛ

nR,s

1 ⋅Ω YR

Ω

S ,T

3

2

it

u = Y ⋅ FT

w

1

F3

FT

t

1 i = ⋅v3 Y

Figure 20. Longitudinally coupled piezoelectric two-layer beam.

t

n

s

v3

t

10.1.3

Longitudinal piezoelectric coupling Figure 21. Model of a longitudinally coupled piezoelectric beam.

Longitudinal piezoelectric coupling is achieved with (Ne +1) interdigital electrodes on the piezoelectric layers creating Ne separate electric field regions as sketched in Fig. 20. The layers are polarized longitudinally and match the comb structure but with opposite direction in layers 1 and 2. In the following, thin layers are assumed which accommodate an approximately uniform electric field in direction 1. The material directions 1 and 3 are now exchanged compared to the coordinate system in Fig. 2. Longitudinal piezoelectric coupling is described by T3 =

1 E∗ s33

· S3 −

∗ d33 E∗ s33

· E3 ,

∗ T∗ · T 3 + ε33 · E3 , D 3 = d33

(95a)

(95b)

E∗ = s E − s E 2 /s E , matching the abbreviations 1/E = s33 33 23 22 T ∗ = ε T − d 2 /s E ∗ = β T = ε33 and α1 = −α2 = d33 33 23 22 E d /s E , as well as 3 = E . d33 − s32 23 22 3 Calculation of moment M and longitudinal force F 3 with Eqs. (13) and (14) and u = E 3 · l/Ne give the rotational electromechanical transduction coefficient YR for the actuation Eq. (21) w E1 E2 h1 h2 (h1 + h2 ) (α1 − α2 ) Mλ = − u, 2 (E1 h1 + E2 h2 ) l/Ne | {z } 1/YR

(96)

and the translational electromechanical transduction coefficient in the combination of Eq. (42b) and Eqs. (18) and (19): 1 F 1 EA · α = = Yt u dξ3 =0 l/Ne

(97)

dx3

=

w (α1 E1 h1 + α2 E2 h2 ) . l/Ne

As in Sect. 10.1.2 this coefficient is 0 when the two piezoelectric layers are equal and cA = cS , too. J. Sens. Sens. Syst., 3, 187–211, 2014

The time-integrated flow coordinate charge Q is gained from the density quantity D 3 by c+h Z 2

Q = n·w

αE

dξ 3 dx3

+ αEx1

dϕ dx3

+ β − α 2 E 3 dx1

c−h1

dϕ 1 α2 E2 h22 − α1 E1 h21 + (α1 E1 h1 + α2 E2 h2 ) c 2 dx3 dξ 1 + α1 E1 h1 + α2 E2 h2 dx3 h i (98) + β1 − α12 E1 h1 + β2 − α22 E2 h2 3 =w

in the case of a uniform electrical field in direction 2. When curvature, strain along x3 and the 3 field coordinate are all constant, then the flow coordinate – current – i = jωQ w E1 E2 h1 h2 (h1 + h2 ) (α1 − α2 ) i=− 2 (E1 h1 + E2 h2 ) l/Ne | {z } 1/YR w + α1 E1 h1 + α2 E2 h2 υ l/Ne 3 | {z } 1/Yt 

(99)



 wh wh   T 1 2 2 T 2 u + jω  β − α E + β − α E 1 1 2 2 2  1 l/Ne l/Ne  | {z } | {z } Cb,1 Cb,2 includes a bending-induced, a x3 -translationally induced and a voltage-induced component with the transduction coefficients YR and Yt and the layer capacitancies Cb,1 and Cb,2 . Together with Eqs. (42f) and (56), a circuit interpretation of Eqs. (99) and (96) leads to the reversible transducer model in Fig. 21. Short-circuiting the electrical port decouples translational and rotational domain. Then the compliance ns and rotational compliance nR,s can be measured. www.j-sens-sens-syst.net/3/187/2014/

i1 =

(98)

Yt

⋅v

s

U. Marschner et al.: Equivalent circuit models of two-layer beams and in case of a large beam width to i iR H H S 3 = s33 · T 3 + s31 · T 1 + d33u·=HY3R ⋅ M Λ H n + d31 · H 1= 0 Su1 = sH 13 · T 3 + s11 · T 1R,s 3 Cd · TYt 2+ µTi R·=HY ⋅ Ω B 3 = d33 C · Tb,13 + b,2 31 33 1 3R

en curonstant,

720

MΛ

(99)

u

ced and 725 coeffid Cb,2 . eads to- 705 nsducer decou- 730 compliasured. e comformed 735 e trans- 710 n addi-

740

715

745

750

755

Ω nR,s (102)

The equations and linearized in the caseconstitutive of a large beam widthfor to transverse coupling of mechanical and magnetic field quantities for a small beam H H can S 3 be = sformulated · T 1 + d33 · H 3 (102) 33 · T 3 + s31as:

33

3

1


m

dt

11 | {z 11 } | {z } ∗ When only one mechanical domain is d 33 of interest, the comsH∗ 31 (103) pliances of the other mechanical domain can be transformed is givenby: d31 sH d231 13 T 22, the transinto the electrical domain. As depicted in Fig. B 3 = d33 − H T 3 + µ33 − H H 3 s11 for example, s11 lationalTcompliance, is recognized as an addi| nm H m {z+ dnj T} j n |= 1 . .{z . 3, m }= 1 . . . 3, Btional n=µ capacitance d∗33 for F 1 = 0. µT33∗ j = 1...6 (100) H 10.2 bimorph S dPiezomagnetic H m + sdomain i = 1 . . . 6, m = V1m. .and . 3, the time mi ij T j iIn=the magnetic magnetic voltage derivative flux fluxpiezorate In the caseofofthea magnetic piezomagnetic – analog to j = dΦ/dt, 1bimorph . . . 6 or magnetic Ielectric as coordinates listed in Table 2. An ideal materials – the linear as magnetomechanical coupling m , are chosen solenoid coil then acts as electromechanical transducer with matrixd (= for the quantities stressconstant T , strain S,µmagnetic where α) isfield the piezomagnetic and (= β) the coefficient N , which is coupled with the piezomagnetic beam flux density B(= K ) and magnetic field strength H (= 3) is i permeability. given by et al., (Marschner 2010). With magnetic coils voltage andplanar flux rate, Technically important are solenoid and coils a magnetic reluctance R is described by the capacitor symm T asBelectromagnetic transducers. = µto H manalogy + dnj T jto then capacitance = 1 . . . 3, m u == 1 .1/(jωC) . . 3, (100) nmthe boln due · i as explained in Fig. 23. j = 1 . . . 6 10.2.1 Longitudinal coupling with an ideal solenoid With the abbreviations E = 1/sE , α = d and β T = µT33 S i = dmi H m + sijH T j i = 1 . 33 . . 6, m =33 1 . . . 3, the transduction coefficients are obtained with Eqs. (96) and Longitudinal coupling is achieved when a solenoid coil with 1 . . . 6 to the material (97) when Ne is set to 1, andjin=addition coefN turns around the unimorph serves as an electromagnetic ficients, u is substituted by V m and Λ by H 3 . Replacing also where d (=asα)shown is the piezomagnetic and µ (=coil β) the transducer in Fig. 1d. Anconstant ideal solenoid conQ by the magnetic flux Φ, the magnetic flux rate I m instead permeability. centrates a uniform H-field in the unimorph and covers the of i follows with Eq. (99) where the reluctances of the layTechnically important electromagnetic transducers are unimorph gap.beDemagnetization are not ers Rm,b,1without and R an aircan measured when effects the bimorph solenoid coils andm,b,2 planar coils. considered. is mechanically blocked. Fig. 24 shows the resulting elecFor longitudinal the magnetomechanical coutromechanical circuitcoupling, representation. Compliance n and ro10.2.1 Longitudinal coupling with an ideal solenoid pling matrix of SOE (100) simplifies for a small beam width tational compliance nR are separated in case of a suppressed to: magnetic voltage variation (V m =when 0) which is achieved by Longitudinal coupling is achieved a solenoid coil with an electrical circuit (i = 0). N open turns around a considered unimorph – with one piezo= sH ·T3 + d33 · Hfor S 3 The 33 model variations of the magnetic magnetic layeris–valid serves3assmall an electromagnetic transducer as (101) T and mechanical system quantities. In order to model nonlinB = d · T + µ · H 1d. An ideal solenoid coil concentrates a unishown in Fig. 33 33 3 3 3 ear sensing and behaviors of piezomagnetic form H field in actuating the unimorph and covers the unimorphmatewithrials, models comprise and out anFinite-Element-based air gap. Demagnetization effects are notmagnetic considered. elastic value problems that are bidirecFor boundary longitudinal coupling, the (BVPs) magnetomechanical cou- 760 tionally coupled through stress and field-dependent coupling simplifies for a small beam width pling matrix of SOE (100) variables — magnetostriction and magnetization (Mudivarthi to et al., 2008). H S 3 = s33 · T 3 + d33 · H 3 (101) 10.2.2 Transverse coupling 765 B 3 = d33 · T 3 + µT33 · H 3

3

Magnetic flux rate d Φ = I 1 Rm

m

MMF = ∫ H ⋅ ds Magnetomotive Force

or

H = ds13 T3+ s HH· ·TT1 + d31 · H 3 = 0 13 ·· H SS11 = 11 3 + s11 1 T T B = d · T + d · T B 3 = µ33 · H3 + d3131 · T1 + µ33 · H 3

203

Magnetic flux Φ Rm

M

Figure 22. Model of coupled piezoelectric beam a!longitudinally H 2 with22. transformed compliance. Fig. Model ofsatranslational coupled beam with d31piezoelectric sH 31longitudinally 31 H T 3 + d33 − H S 3 = s33translational H3 − H transformed compliance s s

piezooupling agnetic H (= Λ)

17

Fig. 21. Model ofeta al.: longitudinally beam U. Marschner Equivalentcoupled circuit piezoelectric models of two-layer beams

(104) 770

Vm = ∫ H ⋅ ds

Magnetic Voltage

MMF =Rm ⋅Φm

Vm =

Rm ⋅ I ω

j

m

Figure 23. Magnetic coordinates on a reluctance. Fig. 23. Magnetic coordinates on a reluctance

or

i

MΛ M

I m,R

Im

H= Y ⋅ M u =HN ⋅ I m s31 dV31 Λ m,ms31 R T 3 + d33 − H3 (103) S 3u= s33 − H H V Ω 1 1 m s11 = ⋅ Ω I i| = ⋅V{zm s11 } n m,R | {z } R YR N ∗ H∗ d33 s31 1 ! 1 ! H 2 d31m,b,1 s13 d m,b,2 F T F1 B 3 = d33 − H T 3 + µT33 − 31 H 3. H s11 s11 V = Y ⋅ F T | |I m,1 {z m,m } {z } v1 ∗ T ∗ 1 d33 µ33I = ⋅ v

! H 2

R

!

R

t

m,1

Y

t

1

n

In the magnetic domain, magnetic voltage V m and the time derivative of the magnetic flux d8/dt, or magnetic flux rate are chosen as listed in Table 2. An ideal I m ,Solenoid coil as coordinates Piezomagnetic unimorph solenoid coil then acts as electromechanical transducer with coefficient N, which is coupled with the piezomagnetic beam (Marschner et al., 2010). magnetic voltageelectromagnetic and flux rate, Fig. 24. Circuit model forWith uniform longitudinal coupling achieved by applying ideal solenoid a magnetic reluctance Rm isan described by thecoil capacitor symbol due to the analogy with the capacitance u = 1/(jωC) · i as explained in Fig. 23. or With the abbreviations E = 1/s E , α = d and β T = µT , 33 33 33 the transduction coefficients are obtained by Eqs. (96) and d13 1 T 1 = − H · H 3 + H · S1 (97) when s Ne is set to s111, and in addition to the material coef u11is substituted (105) ficients, by Vdm and 3 by H 3 . Also replacing d213 13 T B 3by = theµmagnetic − · H + · S Q flux 8, the magnetic flux rate I instead 33 3 1 m sH sH 11 the reluctances of the layof i follows with11Eq. (99), where ers Rm,b,1 and Rm,b,2 for can abelarge measured when the Similar considerations beam width holdbimorph as disis mechanically blocked. Figure 24 shows the resulting eleccussed for longitudinal coupling. tromechanical circuit representation. Compliance n and A technical solution to realize transverse coupling arerotaplational compliance are coil separated in case of and a suppressed R the nar coils. In a part nof the mechanical magnetic magnetic voltage variation (V m = 0)towhich is achieved by quantities are directed perpendicular each other as visualan open electrical circuit (i = 0). ized by Fig. 25. When a uniform magnetic field strength in model is valid small variations of theparameters magnetic theThe piezomagnetic layerfor is assumed, then network and mechanical system quantities. In order to model nonlinsimilar to longitudinal coupling are found except exchanged ear sensing and actuating behaviors of piezomagnetic length l and width w of the beam as listed in Table 4. materials, finite-element-based models comprise magnetic and 10.2.3 Example:value Transverse coupling coil elastic boundary problems (BVPs)with thatplanar are bidirectionally coupled through stress and field-dependent coupling Next, the technically interesting of a thin piezomagnetic variables – magnetostriction andcase magnetization (Mudivarthi layer a non-magnetic carrier in combination with a planar et al.,on 2008). J. Sens. Sens. Syst., 3, 187–211, 2014

MMF =Rm ⋅Φm

Vm =

Rm ⋅ I ω

j

m

(103)

the time flux rate An ideal ucer with etic beam flux rate, itor symωC) · i as

β T = µT33 (96) and rial coefcing also m instead f the laybimorph ing elecn and roppressed hieved by

18

204 23. Magnetic coordinates on a reluctance Fig. I m,R

Im

i

u = N ⋅Im u 1 i = ⋅V m N

Vm 1

V m,m = YR ⋅ M Λ I m,R =

1

1

YR

⋅Ω

Rm,b,1 Rm,b,2 I m,1

18

l

Coil pin µ 33 coil Solenoid

V m,m = Y ⋅ F T

MΛ M nR

1

I m,1 = ⋅v1 Y t

µ 33

n

1

1

T1,S1 H3,B3 µ 31 Transverse coupling

Longitudinal coupling

l = 100

w H1 B1

U. Marschner et al.: Equivalent circuit models of two-layer beams Figure 25. Rectangular planar coil coupling. v1

w=2

Fig. 25. Rectangular planar coil coupling

Longitudinal coupling T1,S1 Piezomagnetic unimorph

1 Figure 24. Circuit model for uniform longitudinal electromagnetic

10.2.2

i

3

F T F1

t

l

Coil pin

Ω

w H B1 coupling by applying an ideallongitudinal solenoid coil. Fig. 24.3 achieved Circuit model for uniform electromagnetic H3,Ban3 idealµ solenoid 31 coupling achieved by applying coil

i

U. Marschner et a


775

Transverse coupling

Transverse coupling

or Fig. 25. Rectangular planar coil couplingfor transverse coupling The linearized constitutive equations d13 1 of magnetic field quantities for a small beam 780 T 1mechanical = − H · Hand 3 + H · S1 s11 s can beformulated as 11 (105) set of inductors 2 will be discussed. The planar set of inductors, magnetic d d13 13 T H B µ − · H · S = + as shown in Fig. 26, serves as electromagnetic transducer. el nonlin1 · H 3 s+Hs11 · T 13 sH S31 = d1333 (104)It 11 11rectangular planar coil. In this can be viewed as section of a tic mateB 3specific = µT33 ·example H 3 + d31a ·thin T for 1 isolation 0.2 mm Similar considerations a large layer beamofwidth holdthickness as disnetic and 785 775 is located between conductor layer and magnetic layer. for longitudinal coupling. bidirec- 760 cussed or A a) technical solution to realize transverse coupling are placoupling Coil inductance dplacement 1 the nar coils. In a part of coil the mechanical and magnetic udivarthi 13 The the the · S 1 ferromagnetic material beside T 1 = − H · H 3 + of (105) H perpendicular quantities are directed toon each other as visualplanar sturns has much less influence the coil inductance s 11 ! 11properties ized Fig. 25.2When a uniforminmagnetic field astrength in or by electromagnetic general than ferromagd d13 13 layer 765 780 the piezomagnetic is+assumed, then network parameters core a· solenoid coil. While the inductance of a H · S . B 3netic = µT33 −inside 3 1 H H s11 similar to longitudinal coupling are material found except exchanged solenoid coils11 with isotropic core depends linearly coupling length l and width w ofthe theinductance beam as listed Tablecoil 4. L∞ is790 on the permeability, of a in planar mall beam Similar considerations for a large beam width hold as dis-of doubled at the maximum by an isotropic permeable layer cussed for longitudinal coupling. 10.2.3 Example: Transverse coupling with planar coilair infinite thickness underneath the turns compared to the A technical solution to realizing transverse coupling are 785 coil inductance L0 . For an infinite layer thickness Roshen Next, the technically interesting case of a thin piezomagnetic planar coils. In a part of the coil the mechanical and magnetic and Turcotte (1988) showed that: (104) 770 layer on a are non-magnetic carrier in combination withasavisualplanar quantities directed perpendicular to each other 795 2µr ized a uniform magnetic field strength in L∞by=Fig. 25.LWhen . (106) 0 µr + 1 the piezomagnetic layer is assumed, then network parameters found except with It similar should to be longitudinal noticed, thatcoupling already are a relative permeability theµrlength l and width w of the abeam exchanged as = 20 increases L0 by about factorbeing of 1.9. The network in Table 790 listed model of the 4. coil arrangement, which is derived next, relates800 the magnetic reluctances to the inductance. This will give an explanation for thetransverse inductancecoupling peculiarity. 10.2.3 Example: with planar coil

setl of inductors = 100 mm will be discussed. The planar set of inductors, Air as shown in Fig. 26, serves as electromagnetic transducer. It can be viewed as section of a rectangular planar (0.5 coil.mm) In this specific example a thin isolation layer of 0.2 mm thickness Magnetic layer is located between conductor layer and magnetic (h1 = layer. 1 mm) Source current a) Coil inductance Aluminum The material beside the w =placement 23 mm of2,the ferromagnetic (h2 = 5 mm) 1, x1 planar turns has much x3 less influence on the coil inductance x2 general than a ferromagor electromagnetic properties3, in netic core inside a solenoid coil. While the inductance of a Figure 26. Section of a planar coil on top of a thin piezomagnetic solenoid coil with isotropic core material depends linearly Fig. 26. Section of a planar coil on top of a thin piezomagnetic layer layer. on the permeability, the inductance of a planar coil L∞ is doubled at the maximum by an isotropic permeable layer of infinite underneath the turns compared to the air (a) Coil thickness inductance coil inductance L0 . For an infinite layer thickness Roshen The Turcotte placement of theshowed ferromagnetic and (1988) that: material beside the planar turns has much less influence on the coil inductance 2µr 0.598 A or electromagnetic L L0 . properties in general than a ferromag(106) ∞= µr + 1 a solenoid coil. While the inductance of a netic core inside 0.11 A linearly solenoid isotropic core material depends 0.16 Acoil It should be with noticed, that already a relative permeability A on the permeability, the 0.129 inductance of a planar coil L∞ is µrPiezo= 20 increases L0 by about a factor of 1.9. The network doubled at the maximum by an isotropic permeable layer of magnetic model the coil arrangement, which is derived next, relates layer ofthickness infinite underneath Athe turns compared to the air the magnetic reluctances 0.129 to the inductance. This will give an coil inductance L0 . For an infinite layer thickness, Roshen explanation for the inductance peculiarity. and Turcotte (1988) showed that b) Combined Simulation Fig. 27. Absolute value of the magnetic field strength H in the right 2µr The electromagnetic system is studied for the setup in Lhalf L0 voltages . (106) ∞= and magnetic across each path µr = 20 Fig. 26µusing r + 1 Finite-Element (FE) simulations to determine the structures and parameters of the magnetic network. The It should be noted, that already a relative permeability µr = method of deriving network parameters from FE-simulations 20 increases LSymmetry of 1.9. The network model 0 by about axisa factor is called Combined Simulation (Starke, 2009), (Starke et al., of the coil arrangement, which is derived next, relates the 2011). The network elements of the circuit in Fig. 29 can be magnetic reluctances to the inductance. This will give an exAir Inner Turnand separately by applying Outer Turn determined individually particu- 810 planation for the inductance peculiarity. lar boundary conditions (e.g. short- or open-circuits, excitation quantities). The same conditions are used as boundary (b) |B| Combined conditions for Simulation the FE-simulations to determine the element values from the complex geometries distributions. The electromagnetic system is studiedand for field the setup in Fig. 26 A magnetostatic FE-analysis, not considering anystruceddy using finite-element (FE) simulations to determine the current losses, reveals the actuation situation when only the tures andConcentration parameters of network. method of 815 Fig. 28. ofthe the magnetic magnetic flux in the The magnetic layer low-frequency properties of the unimorph are of interest. The deriving network parameters from FE simulations is called near the turns generated field strength is depicted The Combinedmagnetic Simulation (Starke, 2009), (StarkeinetFig. al.,27. 2011). magnetic flux is concentrated near the turns with a slight peak The network elements of the circuit in Fig. 29 can be near the middle of the turn area while the flux in x3 -direction is nearly uniform (Fig. 28).

Combined Simulation Next,b)the technically interesting case of a thin piezomagnetic The electromagnetic system is studied forwith the asetup layer on a nonmagnetic carrier in combination planarin 795 Fig. 26 using Finite-Element (FE) simulations to determine set of inductors will be discussed. The planar set of inductors, 805 structures of the magnetictransducer. network. The asthe shown in Fig.and 26, parameters serves as electromagnetic It method of deriving network parameters from FE-simulations can be viewed as a section of a rectangular planar coil. In this is called Combined Simulation (Starke, 2009), (Starke et al., specific example a thin isolation layer of 0.2 mm thickness is 2011). The network elements of the circuit in Fig. 29 can be located between the conductor layer and the magnetic layer. 800 determined individually and separately by applying particu- 810 lar boundary conditions (e.g. short- or open-circuits, excitaquantities). The 3, same conditions J. tion Sens. Sens. Syst., 187–211, 2014are used as boundary www.j-sens-sens-syst.net/3/187/2014/ conditions for the FE-simulations to determine the element c) Equivalent circuit values from the complex geometries and field distributions. When each partial magnetic voltage along a path around the conductors is related to a reluctance, then a magnetic volt805 A magnetostatic FE-analysis, not considering any eddy in T

Fig. 26. S

0.16 Piezomagnetic layer

Fig. 27. A half and m

|B| in T

Fig. 28. C near the tu

near the m is nearly

c) Equ When the condu age divid portional netic vol magnetic

w = 23 mm

inductors, nsducer. It inductors, oil. In this nsducer. It thickness oil. In this ayer. thickness ayer. beside the nductance beside the ferromagnductance tance of a ferromagds linearly tance of a oil L∞ is ds linearly le layer of oil L∞ is to the air le layer of ss Roshen to the air ss Roshen (106)

1, 3, x1 x2 3, x2

2,

x3

Aluminum (h2 = 5 mm)

Fig. 26. Section of a planar coil on top of a thin piezomagnetic layer


U. Marschner Equivalent models of two-layer Fig. 26. Section et of aal.: planar coil on topcircuit of a thin piezomagnetic layer beams

205

R

0.16 A Piezo-0.16 A magnetic Piezolayer magnetic layer

0.598 A

0.11 A

0.129 A

0.11 A

I m,R

Im

i

u = N ⋅Im u 1 i = ⋅V m,0 N

0.598 A

Vm 1

I m,1

Fig. 27. Absolute value of the magnetic field strength H in the right Figure Absolute value across of the each magnetic field strength H in the half and27. magnetic voltages path µ r = 20 Fig. 27. value voltages of the magnetic field strength in the right right halfAbsolute and magnetic across each path µr H = 20. half and magnetic voltages across each path µr = 20

1

YR⊥

⋅Ω

V m = Y ⊥ ⋅ FT

nR

d33 = sH 33 =

Ω

sH 12 = sH 44 =

F T F1

t

I m,1 =

1

Y⊥

⋅ v1

t

n

v1

i

u

Planar coil

Symmetry axis

Piezomagnetic unimorph

u=

i=

N=

Figure 29. Circuit model for uniform transversal electromagnetic

Symmetry axis Air Inner Turn

Outer Turn

Air

Outer Turn

Fig. 29. Circuit model for uniform transversal electromagnetic coucoupling applying an ideal planar conductor setup on a single piezopling applying an shown ideal planar setup on a single piezomagnetic layer as in Fig.conductor 26. magnetic layer as shown in Fig. 26

Z=

ϕ

Table 3. Material parameters of the magnetic layer.

|B| in T |B| in T

d33 = 10 · 10−9 m A−1 i

Figure 28. Concentration of the magnetic flux in the magnetic layer Fig. the 28.turns. Concentration of the magnetic flux in the magnetic layer near near the turns Fig. 28. Concentration of the magnetic flux in the magnetic layer near the turns

Im

H = 3.8 · 10−11 m2 N−1 s33

determined individually and separately by applying particnearboundary the middleconditions of the turn(e.g., area while flux incircuits, x3 -direction ular short the or open exciis nearly uniform (Fig. 28). tation quantities). The same conditions are used as boundary near the middle of the turn area while the flux in x3 -direction conditions for the (Fig. FE simulations to determine the element is nearly uniform c) Equivalent circuit 28). values from the complex geometries and field distributions. When each partial magnetic voltage along a path around Equivalent circuit Ac) magnetostatic FE analysis, not considering any eddy the conductors is related to a reluctance, then a magnetic voltWhenlosses, each partial voltage along when a pathonly around current revealsmagnetic the actuation situation the age divider is found where the magnetic voltages are prothe conductors properties is related toofathe reluctance, then magnetic low-frequency unimorph areaof interest.voltThe portional to the reluctances. Due to the ideally equal magage divider is foundfield where the magnetic voltages 27.proThe 820 generated magnetic strength is depicted in Fig.are netic voltage in a thin piezomagnetic layer and in the nonportional flux to the reluctances. near Due the to the ideally equal magmagnetic is concentrated turns with a slight peak magnetic substrate and air below, as depicted in Fig. 27, the neticthe voltage thinturn piezomagnetic layer the nonnear middleinofa the area while the fluxand in xin3 direction magnetic substrate and air below, as depicted in Fig. 27, the is nearly uniform (Fig. 28). 825

When each partial magnetic voltage along a path around the conductors is related to a reluctance, then a magnetic voltage divider is found where the magnetic voltages are proportional to the reluctances. Due to the ideally equal magnetic voltage in a thin piezomagnetic layer and in the nonmagnetic 830 substrate and air below, as depicted in Fig. 27, the related reluctances Rm,m and Rm,A,b are connected in parallel. The equivalent circuit of Fig. 24 can thus be extended by the magnetic voltage divider including reluctance Rm,A,a of the air above and beside the conductors, as shown in Fig. 29. Con- 835 trary to the ideal solenoid coupling, the magnetic voltage V m across the magnetic layer cannot be suppressed by setting i = 0. Therefore, nR and n cannot be measured separately, www.j-sens-sens-syst.net/3/187/2014/

I m,R =

MΛ M

1

0.129 A

(c) Equivalent circuit

V m = YR⊥ ⋅ M Λ

Rm,m Rm,A,b

0.129 A 0.129 A

Inner Turn

Table 3. M

V m,A,a

1 m,A,a

(106) rmeability e network rmeability ext, relates e network ill give an ext, relates ill give an

e setup in determine e setup in work. The determine mulations work. The arke et al., mulations 29 can be arke et al., g particu- 810 29 can be ts, excitang particu- 810 boundary its, excitae element boundary ibutions. e element any eddy ributions. n only the 815 any eddy erest. The n only the 815 g. 27. The terest. The light peak g. 27. The slight peak

x3

i

1 V m,0 L d31 ==−5 · 10⋅ −92 m A−1 m,1 V m,1 N H = 4.4 · 10−11 m2 N−1 s11

R

u = N ⋅Im V −11 L2 −1 H = −1.1 · 10−11 m2 N−1 H = −1.65 1 · 10 u s12 V m,1s13 1 = m,0 ⋅ m 2 N V m,m V m,0 i = −11 ⋅V m,0 V m2 NN−1 H = 24 · 10 H = 11 ·m,2 s44 s66 10−11 m,2 N m2 N−1

R

Fig. 31. N planar con Fig. calculation based on the inductance. magand 30. the Reluctance rotational and translational domain cannot beThe decomnetomechanical is switched off by blockingreluctances the unimormph. posed this waysystem but only when the magnetic are

taken into account (Marschner et al., 2014). 840

related reluctances Rm,m and Rm,A,b are connected in paral(d) Reluctances lel. The equivalent circuit of Fig. 24 can thus be extended by the voltage including of m,A,a inThemagnetic reluctances can bedivider determined on thereluctance basis of theRtotal the air above and beside the conductors, as shown 29. ductance of the setup as described by Marschner et in al. Fig. (2010). Contrary to the are ideal solenoidbycoupling, magnetic voltThe reluctances separated a disabledthe magnetomechanacrossis the magnetic age change V mwhich ical interaction, achieved eitherlayer by cannot = 0 andbeυ sup=0 pressed = 0. nR and n cannot meaor I m,R by = 0setting and Iim,1 = Therefore, 0. With a pure magnetic FEbeanaly845 sured seperately and rotational and translational domain sis these ideal boundary conditions are matched. Oncecanthe not be decomposed way but only when the magnetic reinductance L of thethis arrangement is determined by FE simluctances are total takenreluctance into account al., individual 2014). ulations, the can(Marschner be related toet the reluctances by applying the magnetic voltage divider rule as d) Reluctances shown in Fig. 30. can The be total reluctanceonRthe Rm,1of+the Rm,2 m= The reluctances determined basis toand total inductance L are related to each other by tal inductance of the setup as described by Marschner et al. 850

(2010). The reluctances are separated by a disabled magneN2 and tomechanical interaction, which is achieved by Ω = 0(107) = L. m 0 or I υR= = 0 and I = 0, respectively. With a pure m,R m,1 magnetic these ideal conditionssysare A total FE-analysis inductance of 1.1 µH of boundary the electromagnetic matched. Once the inductance L of the arrangement is detertem is calculated with FE analysis from the static magmined by FE-simulations, the totalvoltages reluctance can R bem,1 related netic field energy. The magnetic across and to the individual reluctances by applying the magnetic voltRm,2 reach 0.871 A and 0.129 A, as noted in Fig. 27. Back J. Sens. Sens. Syst., 3, 187–211, 2014

age divid Rm = R each othe

N2 =L Rm

A total in is calcula energy. T 0.871 A a tion gives 3 MA/Wb layer its d Rm,m =

where wa tion unde from eac

Rm,A,b =

V m = YR⊥ ⋅ M Λ u = N ⋅Im u V Ωcou1 1 m Fig. i29. = Circuit ⋅V m,0 model for uniform transversal I m,R = electromagnetic ⋅ n R pling applying an ideal planar conductor setup piezoN YR⊥ on a single

Ω

magnetic layer as shown in Fig. 26 1 206 1

U.

Rm,m Rm,A,b

i

u

u = N ⋅Im 1 i = ⋅V m,0 N

Planar coil

Im

F T F1

I m,1

L 1 V V= m,0 =m Y ⊥⋅⋅ F2T m,1 V m,11 N I m,1 = ⋅v1 n Y 1 ⊥ V m,0 L

R

V m,0 V m,1

R⊥ =· 10 11MA 4 MAs =1 Y4.4 Nm

N = 5 −11 m2 N−1 sH 33 = 3.8 · 10 ϕ

= 24 · 10−11 m2 N−1 rad Marschner et al.: 1.9 Equivalent MA

Z = =− i

t

Rm,2

=

V m,2

v1

i

⋅ 2 V m,m N

u

= 11 · 10−11 m2 N−1 F circuit models of two-layer Tbeams

V m,m = Y ⊥ ⋅ F T t

I m,1 =

1

Y⊥

⋅ v1

t

v1

Ω = ωϕ j

MnΛ =

1 V=m−=114 YR⊥⋅ ⋅NM Λ 2.3 10− m kA Y⊥ ΩN 1 t

u = N ⋅Im

1

1

⋅

8

1 I m,R = ⋅ ⋅V m m,A,b m,m N YR⊥ nR = with Fig. 31. Network parameters of the piezomagnetic unimorph =5 YR⊥ = −3 rad planarNconductor setup from Fig. 26 9.3 ⋅10 i=

R

R

Figure 30. Reluctance calculation based on the inductance. The

−320

Fig. 30. Reluctance calculation based on the magnetomechanical system is switched off inductance. by blockingThe the maguniFig. 29. Circuit model for uniform transversal electromagnetic counetomechanical system is switched off by blocking the unimormph. morph. pling applying an ideal planar conductor setup on a single piezomagnetic layer as shown in Fig. 26 840

−1 and R transformation gives MA = related reluctances Rm,mRand R20 areWb connected in m,2 paralm,1 = m,A,b −1 R ||R = 3 MA Wb . Assuming a homogeneous 820 lel.m,A,b The equivalent circuit of Fig. 24 V thus be extended by m,m L 1 can fieldmagnetic in the magnetic layer, its dimension-based reluctance = m,0reluctance ⋅ 2 the voltage divider including Rm,A,a of V N Im i m,1 m,1 is the air above and beside the conductors, as shown in Fig. 29. Contrary the coupling, the magnetic voltu =toNw ⋅ aI mideal solenoid −1 = 4 MA Wb , 1 V m,0 L (108) R = theV m,1 magnetic layer cannot beV m,m supchange uagem,m µ01µVr m h1 lacross = ⋅ V 2 be mea= setting ⋅V m,0 i = 0.m,0Therefore, nRm,2and Vnm,2 825 pressed iby cannot N 845 whereseperately wa =NN · (w dw ) is theand width of the fielddomain concentrat+ sured and rotational translational cantionbe underneath the this N turns width wt the and magnetic distance drew not decomposed way with but only when from each other. It follows, that luctances are taken into account (Marschner et al., 2014). Fig. 30. Reluctance The magRm,mcalculation · Rm,2 based on the inductance. d) Reluctances Rm,A,b = system is switched = 11.1 Wb−1 . the unimormph. (109) netomechanical offMA by blocking Rm,m − can Rm,2be determined on the basis of the to830 The reluctances

R

R

tal inductance of the setup as described by Marschner et al.840850 (e) Completed equivalent (2010). The reluctances arecircuit separated by a disabled magnerelated reluctances Rm,m and R m,A,b are connected in paral0 and tomechanical interaction, which is achieved Ω = trans820 lel.The Themechanical equivalent parameters circuit of Fig. can thus beby extended by and24magnetomechanical υ = 0 or I = 0 and I = 0, respectively. With a m,R theduction magnetic voltage divider including reluctance of coefficients werem,1 determined with Table R 4 m,A,a andpure the 835 FE-analysis these ideal boundary are themagnetic air above and beside conductors, as in Fig. material parameters in the Table 3 as well asshown E2 conditions = 71 GPa 29. and −1 matched. Once the inductance L of the arrangement is deterContrary to = the0 m ideal magnetic voltd33 = d31 A solenoid for the coupling, aluminumthe carrier. Figure 31 mined byall FE-simulations, the totalparameters. reluctance can be displays determined It can besupdeacross thenetwork magnetic layer cannot berelated age change Vm to the by individual by applying thecannot voltduced that the airireluctances path the coil a magnetic distinct 825 pressed setting = 0. above Therefore, nR causes and n be reducmea845 tionseperately of the magnetic field strength in the magnetic layer sured and rotational and translational domain can-to % of the MMF force). higher valnot12.9 be decomposed this(magnetomotive way but only when theFor magnetic reues of µare or a larger thickness of the magnetic layer the ratio luctances taken into account (Marschner et al., 2014). r of the magnetic voltages is approx. 1/µr . Solenoid condid) Reluctances tions, except the transverse coupling, can be achieved either 830 Theclosing reluctances can be path determined basis of the toby the magnetic around on the the turns or eliminating talRinductance of the setup as described by Marschner et al. 850 m,1 . (2010). The reluctances are separated by a disabled magnetomechanical interaction, which is achieved by Ω = 0 and (f) Electromechanical transfer function υ = 0 or I m,R = 0 and I m,1 = 0, respectively. With a pure The included interaction or feedback of the magnetomechan835 magnetic FE-analysis these ideal boundary conditions are ical transduction causes a further magnetic load. transmatched. Once the inductance L of the arrangementThe is deterfer function Z = ϕ/i can easily be deduced from the elecmined by FE-simulations, the total reluctance can be related trical impedance. For this reason, the circuit description to the individual reluctances by applying the magnetic volt-is simplified in a stepwise manner as shown in Fig. 33. The reluctances Rm,1 = Rm,A,a and Rm,2 were transformed in the first step to the electrical side. Then the chained transducers can be combined to one transducer with transduction coefficient X. Finally, the bending compliance is transformed to the electrical side, too. If no external moment or deflection is applied, then the transducer can be eliminated. The inductance network represents the magnetic voltage divider and

Ω

t

1 Wb 1 Wb 11MA 4 MA 1

Piezomagnetic unimorph

J. Sens. Sens. Syst., 3, 187–211, 2014

MA

rad Nm −1

sH 66

1 1 Wb Rm,1 = 20 MA

t

2 −1−3 m9.3 N⋅10

−11 −320 sH m2 N 13 = −1.65 · 10

−11 sH m2 N−1 12 = −1.1 · 10

sH 44

−11

H 11

Nm MA

Nm

age divider ϕ ruleradas shown in Fig. 30. The total reluctance FT = −1.9 Rm Z==Ri m,1 +R m,2 and total inductance L are related to MA V m,m = Y ⊥ ⋅ F T each other by v1 1 I m,1 = ⋅v1 N2 Y⊥ = L. (107) Rm n= t

t

m

1 =electromagnetic N 2.3⋅10 −8 A total inductance of 1.1 µH of the system −114 ⋅ N kA Y is calculated with FE-analysis from⊥ the static magnetic field energy. The magnetic voltages across Rm,1 and Rm,2 reach Figure 31. Network parameters of the piezomagnetic unimorph 0.871 ANetwork and 0.129 A, as noted in Fig. 27. Backunimorph transformaFig. parameters of the with with31. planar conductor setup from Fig.piezomagnetic 26. tion gives Rm,1setup = 20from MA/Wb and Rm,2 = Rm,A,b ||Rm,m = planar conductor Fig. 26 3 MA/Wb. Assuming a homogeneous field in the magnetic layer its dimension-based reluctance the bending compliance. By applying is the current divider rule the analytical transfer function follows age divider rule as shown in Fig. 30. directly: The total reluctance MA wa RRmm,m =ϕR to 4 total inductance L are related (108) =m,1 + Rm,2= and nWb L1 µ0 µrM h13l R each . (110) Z = other = nby · = · R i i X L1 + L2 + nR /X2 where wa = N ·(wt +dw ) is the width of the field concentra2 N tion= underneath the N inturns dw The value Fig.with 31. width wt and distance L . is displayed (107) R m each other. It follows, that from t

t

Model validity A(g)total inductance ofm,2 1.1 µH of MA the electromagnetic system Rm,m · R R = = 11.1 . static magnetic(109) m,A,b isThe calculated with FE-analysis from field Rm,m − Rmodel Wbthe low-frequency is valid up to a frequency bem,2 energy. The magnetic voltages across R and R reach m,1 m,2 low the first natural frequency of the beam, which follows 0.871 0.129 A, in Fig. 27. differential Back transformafrom A theand solutions of as thenoted Euler–Bernoulli equation gives R = 20 MA/Wb and R = R m,1 m,2 m,A,b ||Rm,m = tion with 3 MA/Wb. s Assuming a homogeneous field in the magnetic layer2its dimension-based reluctance is E·I , (111) (βl) = ω MA w(%a · A) Rm,m = =4 (108) µr h1 l modulus, Wb I the moment of inertia, % the where E µis0 Young’s density and A the cross sectional area. Considering a fixed– where wa = N ·(wt +dw ) is the2 width of the field concentrafree beam as an example, (β l) = 3.52 applies, giving a nattion underneath the N turns1 with width wt and distance dw ural frequency of from each other. It follows, that s 3.52 Rm,m · REm,2· I MA f1m,A,b = = (112) R = 11.1= 372. Hz . (109) 2π Rm,m (h1 %−1 R +m,2 h2 %2 )wl 4 Wb

This natural frequency is also found when the high-frequency model of the beam from Fig. 13 is used to simulate the velocity υˆ II of the free beam end. Figure 32 shows the circuit description using the program LTSPICE when the beam is discretized into seven pieces. For the ideal rods, ideal controlled sources were applied. If two 1x/2-transducers are connected in series then they can be combined into one transducer with www.j-sens-sens-syst.net/3/187/2014/

Z=

= nR ·

i

i

=

·

X

. L 1 + L 2 + nR X 2

(110)

applies, giving a natural frequency of

The value is displayed in Fig. 31.

885

3.52 f1 = g) Model validity 2π U. Marschner et al.: Equivalent circuit models of two-layer beams

L2

L3

{Del_n}

{Del_n}

I1

L7

L6

{Del_n}

{Del_n}

{Del_n}

{Del_n}

{Del_n}

F3

F4

F5

F6

F7

E1 {Del_x}

E2 {Del_x}

E3 {Del_x}

E4 {Del_x}

E5 {Del_x}

E6 {Del_x}

E7 {Del_x}

}x 2E C2 }x 3E C3 _l l_e e D D { {Del_m} { {Del_m}

}x 4E C4 }x 5E C5 _l l_e e D D { {Del_m} { {Del_m}

.PARAM Del_x=0.014

.PARAM Del_m=7m

Figureet32. LTSpicemodels model of two-layerbeams beam. U. Marschner al.:High-frequency Equivalent circuit ofthe two-layer Fig. 32. High frequency LTSpice model of the two-layer beam

cal trans4 and the GPa and g. 31 disdeduced uction of 12.9 % of ckness of ges is aperse couh around

L1

(110)

AC 1

+

n

1

L2 = N2

Rm,A,a

L2

i

u = N ⋅Im 1 i = ⋅V m N

X = YR⊥ N i

L1

u

= X ⋅ MT =

1 X

⋅Ω

Vm

V m = YR⊥ ⋅ M Λ I m,R =

⋅Ω

YR⊥

nR

MΛ M nR

Ω

i

u

MΛ X

L2

100°

40dB

60°

0dB

20°

-40dB

-20°

-80dB

-60°

Ω

Ω = ωϕ j

895

1x = 1.4 cm for the beam of Fig. 26. The free end is stimu-900 lated a moment source with infinitely mechanical Thebylow-frequency model is an valid up to a large frequency below source impedance. In an electric network, this corresponds to the first natural frequency of the beam, which follows from a high-impedance current source. The result of the frequency the solutions of the Euler-Bernoulli differential equation with domain simulation is depicted in Fig. 34. Only six instead of s seven natural frequencies are described by this model since905 E ·I 2 the left-side of both, the rotational and the (βl) = ω short-circuiting . (111) (% · A) is transformed to a short circuit, which translational domain, is blocking mass C1. This eliminates C1. 2 Despite the number of finite network in Considering as small an example a fixed-free beam, (βelements 1 l) = 3.52 the LTSpice model of Fig. 32 and the applies, giving a natural frequency of partial influence of the910 boundary conditions on the adjacent node, the mechanical s network turns out to be a good approximation for the descrip3.52 EI the372 flexure ftion = Hz . beam including (112) 1 =of the dynamic behavior of 2π (h % + h2 %2 )wl4To improve the results, more 1 1 the lower natural frequencies. network elements (typically nine or more per wavelength)915 should be used.

{Del_n}

{Del_n}

{Del_n}

F4

F5

F6

F7

J. Sens. Sens. Syst., 3, 187–211, 2014 open

940

This frequency is also found when the high frequency 11 natural Summary model of the beam from Fig. 13 is used to simulate the velocity free beam aend. Fig. 32 shows II of the In thisυˆpaper a generalized generalized theory the for LTSpice the mecircuit description when the beam is strips discretized into seven chanical behavior of laminated plate or bimorph elepieces. For the rods, ideal controlled sources were ap- 950 ments under theideal influence of strain-inducing physical quanplied. If presented. two ∆x/2-transducers are connected in series then tities is These conditions were formulated so that they candescription be combined to one transducer withuniaxial ∆x = 1.4 cm for the of the bimorph only the stress for the beam oftoFig. 26. The free is stimulated by a conditions need be considered. On end this basis, it was possimoment source with anofinfinitely large mechanical ble to determine the set differential equations of thesource twoimpedance. an electric corresponds to a highlayer beam In element and network, to derive this a circuit interpretation of 955 impedance current source. The result of the frequency dothe linear case. main is depicted in an Fig.internally 34. Onlyinduced six instead of Thesimualation neutral layers caused by bending moment by an externally applied are since difseven natural and frequencies are described by moment this model ferent. The midpoints of theoflocation of rotational the neutraland layer the left-side short-circuiting both, the the at the bimorph ends is occurring for an moment are 960 translational domain, transformed to external a short circuit, which as points for applying an external ischosen blocking mass C1. This eliminates C1.longitudinal force. AtDespite this location the bimorph length changes when it is ac-in the small number of finite network elements tuated. For this reason the equivalent circuit includes three the LTSpice model of Fig. 32 and the partial influence of mechanical domains: translational deflection direction, lonthe boundary conditions on the adjacent node, the mechangitudinal direction The circuit representation of 965 ical network turns and out bending. to be a good approximation for the the bimorph element assumes small bending angles as well description of the dynamic behavior of the flexure beam inas smallthe longitudinal forces and neglects quadratic cluding lower natural frequencies. To improve thesolution results, terms for the longitudinal direction. It is also assumed that more network elements (typically 9 or more per wavelength) should be used.

L6 {Del_n}

10KHz

945

Figure 33. Simplification steps for the setup in Fig. 29 to obtain the Fig. 33. Simplification steps for the ϕ/i setup in/i. Fig. 29 to obtain the electromechanical transfer functions and electromechanical transfer functions ϕ/i and Ω/i, respectively

{Del_n}

-100° 1KHz

X

X2

SINE(0 1 1)

935

Figure 34. Amplitude and phase spectrum of υˆ II of the left-side Fig. Spectrum of υˆII of the left-side fixed bimorph fixed34. bimorph.

⋅

L1 n R

vII

80dB

vˆ II

Ω

}x 7E C7 l_e D { {Del_m}

.PARAM Del_n=0.66m

-120dB 100Hz Frequency f

I1 www.j-sens-sens-syst.net/3/187/2014/ L5 L7 L4

L3

1

MΛ M


open


t

N 2 Rm,2

u

u

gnetomeoad. The from the scription uctances first step s can be ficient X. e electriapplied, ance nete bending e analyti- 890

1

Rm,2 Rm,A,b Rm,m + Y 2⊥

i

=

207

L5

F2

.ac dec 10000 100 100000

(112)

L4

F1


1

EI = 372 Hz . (h1 %1 + h2 %2 )wl4

SINE(0 1 1)

AC 1 L1

s

11 Summary

970

ends oc for appl the bimo son the e translati bending assumes forces a tion wh nal com which fu Full b response Quasi-st that are where c mass eff If the early an source a source a fine a r pairs of other. Fo beams w ducer m are deve with ele metric tr circuit r tion by a sis, sim behavio fast time are used mechan units, e. versibili physical and elim mined e Netw combine

208


Table 4. Piezoelectric (PE) and piezomagnetic (PM) two-layer beam network parameters. PE Transversal (symmetric)

PE Longitudinal

N

h1

C1 C2

E 3 D3 E 3 D3 w

P P

pu te S

l

,

e

,

=2

h2

3

w

2

2 ⎞

V

=

ωI

S ,T

1

α

=

h2 ⋅ w ⎛ T d332 ⎜μ − H l ⎜⎝ 33 s33

R m,1 1

Rm,2

l

Yt

1 E s

⎞ ⎟ ⎟ ⎠2

j

V

ωI R V = YR ⋅ M Λ jωI R =

C2

j

ωI

1

1

YR

⋅

jω I 1 =

1

Y

⋅

MΛ

v

h2 ⋅ l ⎛ T d132 ⎜μ − w ⎜⎝ 33 s11H

⎞ ⎟ ⎟ ⎠2

w

V m xΛ H 3 ⋅

d31

2

2 2

w E1E2h1h2 (h1 + h2 ) (α − α ) 1 2 2x Λ E1h1 + E2h2

M nR

1 H s

n

11

1

F1,v1

Ω

M, Ω c = cS

F1

FT

t

=

33

Ω

V = Y ⋅ FT

Rm,2

⎞ ⎟ ⎟ ⎠1

1 H s

t

1)

1 1

33

j

C1

1

1 E s

ωI

1

h1 ⋅ l ⎛ T d132 ⎜μ − w ⎜⎝ 33 s11H

d33

−

11

Rm,1

=

w (α E h + α E h )

xΛ

E

3

Im

0

wh1α1E1

1

=

d33

1 YR

1

w 2

h1 ⋅ w ⎛⎜ μ T − d332 ⎞⎟ l ⎜⎝ 33 s33H ⎟⎠1

⋅

d31

,

1

=

1

S ,T

B3 H 3

h2

3

i

j

i

2

2 ⎞ ⎛ C2 = h2 ⋅ w ⎜⎜ ε 33T − dE33 ⎟⎟ l Ne ⎝ s33 ⎠2 l Ne u xΛ E 3

h1

l

h1

w

/

1)

l

h1

2

/

xΛ

PM Transversal

,

h2

2 ⎞ ⎛ C1 = h1 ⋅ w ⎜⎜ ε 33T − dE33 ⎟⎟ l Ne ⎝ s33 ⎠1

C1 = C2 = l ⋅ w ⎜⎜ ε 33T − d31E ⎟⎟ h1 ⎝ s11 ⎠ ⎛

h2

1)

B3 H 3

i

S ,T

1

3

h1

l

S ,T

,

PM Longitudinal

P E 3 D3

v1

Ω = Ω −Ω I

I

1 = 2

F1,v1

2

E h − E2 h E1h1 + E2h2 2 1 1

II

Ω M II

2 2

E1h1 + E2h2 nR = l = 12 2 4 w EI E1 h1 + E1E2h1h2 ( 4 h12 + 6 h1h2 + 4 h22 ) + h24 E22 l n= l = EA w ( E1h1 + E2h2 )

The single wire with current i symbolizes the section of a coil

longitudinal compression forces cause only functions of coordinates which fulfill the demand for biuniqueness. Full bimorph solutions are derived for both quasi-static responses and dynamic responses without pressure load. Quasi-static responses are dynamic responses at frequencies that are well below the first natural frequency of the bimorph, where compliance effects dominate dynamics responses and mass effects are negligible.

J. Sens. Sens. Syst., 3, 187–211, 2014

If the active layer(s) of the flexure beam behave linearly and reversibly regarding thermodynamics, then an ideal source acts in the interacting domain, too. This additional source and the moment or rotational velocity source define a reversible transducer, which relates the coordinate pairs of interacting domain and mechanical domain to each other. For piezoelectric and piezomagnetic two-layer flexure beams with typical arrangements, reversible multi-port transducer



models are developed. Piezomagnetic unimorph models are coupled with electromagnetic coil transducers. Except for an asymmetric transversally coupled piezoelectric bimorph, a unified circuit representation is summarized. The structural description by an equivalent circuit is the key to an efficient analysis, simulation and understanding of the dynamic bimorph behavior. Sophisticated circuit simulators like SPICE offer fast time and frequency domain simulations. When SI units are used, simulation results can be directly read in the related mechanical, magnetic or other domain with their respective units, e.g., Ampère → Newton. Because of linearity and reversibility, network elements can be transformed into other physical domains. Subsequent


209

transducers can be combined and eliminated. This way transfer coefficients can be determined easily. Network methods and finite-element methods can be combined on the one hand in order to determine network parameters. On the other hand completed network models concentrate properties of system parts. They can be integrated in FE models which describe the continua of other system parts. The equivalent circuit of the planar-coil-driven piezomagnetic unimorph, for example, can be integrated into the acoustic FE model of a piezomagnetic loudspeaker to determine electroacoustic function parameters. Such a combined simulation saves time and computational resources.

J. Sens. Sens. Syst., 3, 187–211, 2014

210


Appendix A: Nomenclature Symbol

Unit

A B cA

Vs m−2 m

cS

m

Ca , Cb C1 , C2 d33 d31 ∗ d31

F m V−1 , m A−1 m V−1 , m A−1 m V−1 , m A−1

D3

As m−2

E = 1/s EA EI E F1 , F 3 FI , FII

Pa

F∗

N

h1 , h2 H i Im I

m A m−1 A V

V m−1 N N

k K l L M M3 M I , M II

m H Nm Nm Nm

n n0 nF1 nR N

m N−1 m N−1 rad Nm−1 rad Nm−1

Ne p Q Rm s sijE

Pa As A Wb−1 m2 N−1 m2 N−1

Description Influence coefficients ratio Magnetic flux density Distance between neutral layer and material interface for actuation Distance between the neutral layer and material interface for sensing Deflection function constants Capacitances of layers 1 and 2 Piezoelectric or piezomagnetic constant for longitudinal coupling Piezoelectric or piezomagnetic constant for transversal coupling Piezoelectric or piezomagnetic constant for transversal coupling and large beam width Displacement current in active electrical field direction Young’s modulus Averaged extensional stiffness Averaged bending stiffness Electric field strength Forces in directions 1 and 3 Forces at left and right side of a beam in direction 3 Force at the beam cross section for large angles Thicknesses of layers 1 and 2 Magnetic field strength Electrical current Magnetic flux rate Generalized time integral of the flow Coupling factor Generalized second field quantity in the interacting domain Beam length Inductance Moment Source moment Moment at left and right side of a beam Compliance Reference compliance Rotational compliance related to F1 Rotational compliance Number of turns; number of discrete elements Number of interdigital electrodes minus one Pressure Electric charge Magnetic reluctance Elasticity constant Elasticity constant for E = 0

J. Sens. Sens. Syst., 3, 187–211, 2014

Symbol

Unit

sijE∗

m2 N−1

sijH S S0 t T TB u υ 1, υ 3 υ I , υ II Vm V w x1 , x 2 , x 3 X Xt⊥

XR⊥

Yt YR Z α1 , α 2 β1 , β2 ε S εmn T εmn T∗ εmn ζ ϑ 3 µr µTmn µTmn∗ ν ϕ 8 ξ1 , ξ3 ξ1∗ , ξ3∗ χ 9 ω I , II

Description

Elasticity constant for E = 0 and large beam width m2 N−1 Elasticity constant for H = 0 Strain Strain at cS and x3 = 0 s Time N m−2 Stress N m−2 Mean-free part of the stress distribution V Voltage m s−1 Velocity in directions 1 and 3 m s−1 Velocity at left and right side of a beam in direction 3 A Magnetomotive force (MMF) or magnetic voltage Generalized influence field strength m Beam width m Mechanical coordinate axes Interacting domain Translational electromechanical transduction coefficient for transverse coupling Rotational electromechanical transduction coefficient for transverse coupling V N−1 , Translational transduction A N−1 coefficient V (Nm)−1 , Rotational transduction coefficient A (Nm)−1 rad A−1 Electromechanical transfer function Generalized influence coefficient of layers 1 and 2 Generalized material coefficient of layers 1 and 2 F m−1 Permittivity F m−1 Permittivity for S = 0 F m−1 Permittivity for T = 0 F m−1 Permittivity for T = 0 and large beam width Layer thickness ratio K Temperature Generalized influence quantity Relative permeability Vs (Am)−1 Permeability for T = 0 Vs (Am)−1 Permeability for T = 0 and large beam width Poisson’s ratio rad Angle Wb Magnetic flux m Deflections in directions 1 and 3 m Transformed deflections ξ1 , ξ3 Young’s modulus ratio % Humidity rad s−1 Rotational velocity rad s−1 Circular frequency rad s−1 Rotational velocity at left and right side of a beam


U. Marschner et al.: Equivalent circuit models of two-layer beams Acknowledgements. This Work was supported in part by the

Deutsche Forschungsgemeinschaft (German Research Foundation) within the Collaborative Research Center (SFB) 639 “Textilereinforced composite components for function-integrating multimaterial design in complex lightweight applications”, subproject D3. The authors thank Günther Pfeifer from the Technische Universität Dresden for his advice concerning Combined Simulation. We also thank Alison B. Flatau from the University of Maryland and Matthias Plötner from the Technische Universität Dresden, who gave insightful comments and suggestions. Axel Renner from the Technische Universität Dresden and Jin-Hyeong Yoo from the University of Maryland helped solving ANSYS-problems, which is gratefully acknowledged, too. Edited by: A. Schütze Reviewed by: three anonymous referees

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J. Sens. Sens. Syst., 3, 187–211, 2014