Electrostatic Beam Model

Jan Lienemann and Jan G. Korvink

IMTEK – Institute for Microsystem Technology, Albert Ludwig UniversityGeorges Kohler Allee 103, D-79 110 Freiburg, Germany

Tel. +49 761 203 7390, Fax. +49 761 203 7382, lieneman@imtek.de

March 30, 2003

Abstract

In this short report we detail how to derive the nonlinear dynamic system equations for a beamsubject to electrostatic actuation. We start from the FEM discretization of a linear beam and a pointcharge approach for the electrostatics. Lagrange methods are used to couple the energy domains.

Contents1 The System 1

2 FEM Discretization of the Beam 22.1 FEM setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Stress, strain and displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.3 Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Electrostatic Actuation 6

4 Nonconservative Work 7

5 Equations of Motion 7

6 Implementation 86.1 Node numbering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86.2 Building of matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86.3 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

7 Summary 11

1 The System

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Figure 1: System setup

1

The system to be simulated consists of two beams (see fig. 1). Both beams are ideal conductors. The beamat the bottom is fixed and connected to ground, the beam at the top is pin supported on both sides andconnected to a voltage source Vin. We assume that the upper and the lower beam have n nodes each.

2 FEM Discretization of the Beam

The FEM discretization is calculated according to [1], p. 511.

2.1 FEM setup

We start with the Lagrangian formulation of Newton’s law

ddt

∂L

∂ xi− ∂L

∂xi= Fi (1)

L = T ∗−V, (2)

with t the time, xi the complete and independant generalized coordinates of the system, L the Lagrangian,Fi the generalized forces, T ∗ the kinetic coenergy and V the potential energy.

When gravity can be neglected as common in micromechanics, the only contribution to the potential energyis the elastic energy stored in the deformation of the beam.

2.2 Stress, strain and displacement

The deformation is determined by the stress-strain relationships

σ = Eε , (3)

where σ = (σx,σy,σz,τxy,τyz,τzx) is the vector of stresses (see fig 2), and ε = (εx,εy,εz,γxy,γyz,γzx) is thevector of strains,

E =E

(1−ν)(1−2ν)

1−ν ν ν 0 0 0ν 1−ν ν 0 0 0ν ν 1−ν 0 0 00 0 0 1−2ν

2 0 00 0 0 0 1−2ν

2 00 0 0 0 0 1−2ν

2

, (4)

E is the modulus of elasticity and ν is Poisson’s ratio.

τ zx

τ zy

σz

xyτ

τ yz

τ yx

σy

τ xz

σx

τ zx

σz

τ zy

σxxyτ

τ xz

τ yx

σy

τ yz

z

x

y

Figure 2: Stresses acting on a volume element

2

The strain can be related to the geometric displacement by means of the strain-displacement relationships

ε(t) = d u(t). (5)

The operator d is determined by the beam geometry and will be calculated in the next section on page 4.

2.3 Discretization

For discretizing the problem, i.e. transforming the infinite dimensional partial differential equation into anordinary differential equation, the beam is split into finite length elements. Each beam element has twovertices with degrees of freedom qi on each side. Between these vertices, the displacement is interpolatedby shape functions,

u(x,q, t) = f(x)q(t). (6)

Two adjacent elements share the nodes on their sides. The strain-displacement and stress-displacementrelationships now read

ε(t) = d f q(t) = B q(t) (7)σ(t) = E B q(t) (8)

The potential energy can then be written as

V =12

∫

VεT σ dV =

12

qT∫

VBT E B dV q =

12

qT Kq. (9)

The kinetic coenergy of the distributed mass can be expressed as

T ∗ =12

∫

Vρ u2 dV =

12

qT∫

Vρ fT f dV q =

12

qT Mq. (10)

The electrostatic force will come in later in section 3 due to the change in electrostatic energy.

The many possible deflections of a beam can be classified by

Torsional displacements: A rotation about the beam’s longitudinal axis.

Axial displacements: Compressing or expanding the beam along its longitudinal axis.

Flexural displacements: Deflecting the beam out of its plane undeformed axis.

Since the flexural displacement is the most important operation mode for actuation, we will focus on that inthe further calculation. Each element vertex has two degrees of freedom q: The deflection y perpendicularto the beam and the rotation θ in the deformation plane, i.e. for small displacement the slope of the beam,yielding

qe =(y1,θ1,y2,θ2

). (11)

The hermite shape functions for the one-dimensional linear element with length L are (see fig. 3)

f =

1L3

(2x3 −3Lx2 +L3)

1L2

(x3 −2Lx2 +L2x

)

1L3

(−2x3 +3Lx2)

1L2

(x3 −Lx2)

. (12)

The derivation of the strain-displacement relationships is shown in fig. 4. With v the displacement in the ydirection, we can write for the displacement u in x direction:

u = −yθz ≈−ydvdx

(13)

3

Figure 3: Hermite shape functions for one-dimensional finite element (from [1])

and thus for the strain

ε =dudx

= −yd2

dx2 v = dv. (14)

Including d in (7), we see that

B = df = −yd2

dx2 f = − yL3

[12x−6L 6Lx−4L2 −12x+6L 6Lx−2L2] . (15)

The beam is not stressed in the y nor in the z direction, i.e. σx = σy = 0; the strain-stress relationship

εx =1E

(σx −νσy −νσz) (16)

therefore reduces toEε = σx (17)

and thusE = E. (18)

4

Figure 4: The strain-displacement relationships for a flexural displacement (from [1])

Inluding this in (9), we get

K =

∫

VBT EB dV

=

∫ L

0

∫

A

Ey2

L6

12x−6L6Lx−4L2

−12x+6L6Lx−2L2

[12x−6L 6Lx−4L2 −12x+6L 6Lx−2L2]dAdx (19)

with A the cross-sectional area. For a constant cross section and modulus of elasticity, we get

K =2EIL3

6 3L −6 3L3L 2L2 −3L L2

−6 −3L 6 −3L3L L2 −3L 2L2

where I =∫

Ay2dA. (20)

For the kinetic energy of a extended body, two contributions must be considered: Rotational and transla-tional inertia.

Translational inertia

From (10), we get

Mt =∫

VρfT fdV =

∫ L

0ρAfT fdx, (21)

which evaluates to

Mt =ρAL420

156 22L 54 −13L22L 4L2 13L −3L2

54 13L 156 −22L−13L −3L2 −22L 4L2

(22)

Rotational inertia

The x translation of a point in the cross section due to rotation about the neutral axis is

u =−yθz = −yddx

v = −yddx

fq. (23)

5

The speed of that point is

u = −yddx

fq. (24)

Inserting this into (10), we get

Mr =∫

Vρy2

(dfdx

)T ( dfdx

)

dV =∫ L

0ρI(

dfdx

)T ( dfdx

)

dx. (25)

This finally yields

Mr =ρI

30L

36 3L −36 3L3L 4L2 −3L −L2

−36 −3L 36 −3L3L −L2 −3L 4L2

(26)

The generalized inertial mass is now found by

M = Mt +Mr. (27)

3 Electrostatic Actuation

Due to the energy stored in the electric field, the Lagrangian (2) is augmented to

L = T ∗−V −We. (28)

A full treatment of the electrostatics would involve evaluating integrals over the beam for every time step.Although these integrals could be integrated into the system by means of Gauß integration, this would resultin complicated and unhandy expressions. We therefore decided to approximate the charge distribution byassuming that the charges are concentrated at the interfaces between the finite elements [2].

The electric potential for two point charges can be calculated by integrating Coulomb’s law, taking a testcharge from infinity to near the charge under consideration. In 3D, this is

Ue,i j = −∫ ri j

∞

Qi4πε0r2 dr =

Qi4πε0ri j

, (29)

where Qi is the charge, ε0 is the permittivity of free space and ri j is the distance.

Another contribution to the energy comes from the self capacity of the point charge. Since the charge isin reality distributed over the beam element, we can calculate the voltage at the center by folding pointsources over the rectangular element area Ai = wh, where w and h are the dimensions of the rectangle:

Ue,ii =Qi

4πε0Ai

∫

Ai

1r− ri

dA′i

=Qi

2πε0wh

(

h lnw+

√w2 +h2

h+w ln

h+√

w2 +h2

w

)

(30)

The integral can also be evaluated by switching to polar coordinates. The accuracy of the lumping increasesby making the elements smaller for a given beam geometry.

When simulated in 2D, we assume that the point charge is equivalent to a line charge in 3D. (29) then reads

Ue,i j =−∫ r j

ri

Qi4πε0r

dr =Qi

4πε0ln

rir j

. (31)

6

We used the 3D equations for the current implementation. The equations can now be combined to form thematrix expression

V = PQ with Pi j =

14πε0ri j

i 6= j

12πε0wh

(

h ln w+√

w2+h2

h +w ln h+√

w2+h2

w

)

i = j. (32)

The energy is then

We =12

VQ =12

QT PQ, (33)

and the complete Lagrangian is specified by

L =12

qT Mq− 12

qT Kq− 12

QT PQ. (34)

4 Nonconservative Work

Energy is introduced into the system by the voltage source, and dissipated by the damping of the structure.The variation of nonconservative work therefore reads

δW nc = δq(−Eq)︸ ︷︷ ︸

Fq

+ Vin︸︷︷︸

FQ

δQ. (35)

Fq and FQ are the generalized forces for the mechanical and electrical degrees of freedom. The vector Vinhas an entry Vin for all charge nodes on the upper beam, and an entry 0 for all charges on the lower beam.

The damping matrix E is usually calculated by a linear combination of the stiffness and mass matrix

E = ckK+ cmM (36)

using the mode-preserving Rayleigh damping formulation.

5 Equations of Motion

With (1), we can calculate the equations of motion. As shown before, all matrices are symmetric. For thedisplacements yi and the rotations θi, we get

ddt

∂L

∂ yi= ∑

jMyi j y j

∂L

∂yi= −∑

jKyi jy j −

12 ∑

j∑k

Q j

∂Pjk

∂yiQk

=⇒ ∑j

Myi j y j +Kyi jy j +12 ∑

kQ j

∂Pjk

∂yiQk

︸ ︷︷ ︸

nonlinear!

= ∑j−Eyi j y j (37)

ddt

∂L

∂ θi= ∑

jMθi jθ j

∂L

∂θi= −∑

jKθi jθ j

=⇒ ∑j

(

Mθi jθ j +Kθi jθ j

)

= ∑j−Eθi jθ j. (38)

7

For the charges, we get

ddt

∂L

∂ Qi= 0

∂L

∂Qi= −∑

jPi jQ j

=⇒ ∑j

Pi jQ j = Vin. (39)

Since Pi j depends on yi, (39) is also nonlinear.

6 Implementation

The implementation takes the necessary material and geometry parameters and calculates both the matricesand the nonlinear parts of the equations.

6.1 Node numbering

The state vector x is build as follows:

• The first 2n entries are the displacement and rotation parameters of the upper beam (see fig. 1), withyi and θi arranged successively; y is pointing upwards (away from the lower beam);

• the next n entries are the charges of the upper beam;

• the remaining n entries are the charges of the lower beam,

x =[

y1,θ1,y2,θ2, . . . ,yn,θn,qu,1, . . . ,qu,n,ql,1, . . . ,ql,n

]

. (40)

6.2 Building of matrices

The K, M and E matrices are built according to (9), (10) and (36). The capacity matrix P is built accordingto (32) and ri j =

√

∆x2 +∆y2. ∆y is left as function of the state vector and thus introduces a nonlinearity.For two nodes i and j on the upper beam, ∆yi j = |yi − y j|. If node j is on the lower beam, ∆yi j = |yi + s|,where s is the distance between the two beams in the non-deflected state. If both nodes are on the lowerbeam, ∆y = 0. ∆x evaluates to |i− j|L if both nodes are on the upper or lower beam, and |i− n− j|L ifNode j is on the upper beam and Node i is on the lower beam. Due to the boundary conditions treated inthe next section, several spatial degrees are freedom yi are set to 0. We denote the set of indices for whichthe yi are fixed by

D = {i|yi ≡ 0∨ i ≥ n+1} (41)

This further simplifies some capacitance relations.

From section 6.1, we know that the indices in the state vector x for the charges on the upper beam are inthe range [2n+1,3n], while the indices for the lower beam are in the range [3n+1,4n]. In conclusion, the

8

global capacity matrix (using indices for x) can be written as

P2n+i,2n+ j =1

4πε0·

2wh

(

h ln w+√

w2+h2

h +w ln h+√

w2+h2

w

)

(1∗) i = j

1/√

(i− j)2L2 +(yi − y j)2 (2) i 6= j, i, j ≤ n, i, j 6∈ D

1/√

(i− j)2L2 + y2i (3) i 6= j, i, j ≤ n, i 6∈ D, j ∈ D

1/√

(i− j)2L2 + y2j (4) i 6= j, i, j ≤ n, i ∈ D, j 6∈ D

1/|i− j|L (5∗) i 6= j, i, j ≤ n, i, j ∈ D or i, j > n

1/√

(i− j)2L2 + s2 (6∗) D 3 i ≤ n, j > n, i 6= j−n orD 3 j ≤ n, i > n, i 6= j +n

1/s (7∗) D 3 i ≤ n, j = i+n, orD 3 j ≤ n, i = j +n

1/√

(i− j)2L2 +(yi + s)2 (8) D 63 i ≤ n, j > n, i 6= j−n

1/√

(i− j)2L2 +(y j + s)2 (9) D 63 j ≤ n, i > n, i 6= j +n

1/|yi + s| (10) D 63 i ≤ n, j = i+n1/|y j + s| (11) D 63 j ≤ n, i = j +n.

(42)with i, j ∈ [1,2n]. The numbers in the second column correspong to the numbers in the flowchart (fig. 5).Terms with (∗) are constant, i.e. they can be included in the K matrix. If we constrain the yi to yi ≥−s asproposed in the next section, the | · | can be omitted.

Start i=j

1*yes

i<=nno

j<=n

yes

yes no

yes yes yesno no noi in D

j in D j in D

5* 4 3 2

yes no yes no

j<=nno

5*

no

j in D

i−j=n i−j=n

7* 6* 11 9

yes no yes

yes

i in D

j−i=n j−i=n

7* 6* 10 8

yes no yes no

Figure 5: Flowchart for (42). The numbers correspond to the second column of (42).

9

The coupling Jacobian with | · | omitted then can be written as

∂P2n+i,2n+ j

∂yk=

14πε0

·

− yi−y j√

(i− j)2L2+(yi−y j)2

3 (1) i 6= j, i, j ≤ n, i, j 6∈ D, k = i

yi−y j√

(i− j)2L2+(yi−y j)2

3 (2) i 6= j, i, j ≤ n, i, j 6∈ D, k = j

− yi√(i− j)2L2+y2

i3 (3) i 6= j, i, j ≤ n, i 6∈ D, j ∈ D, k = i

− y j√

(i− j)2L2+y2j3 (4) i 6= j, i, j ≤ n, i ∈ D, j 6∈ D, k = j

− yi+s√(i− j)2L2+(yi+s)23 (5) D 63 i ≤ n, j > n, i 6= j−n, k = i

− y j+s√

(i− j)2L2+(y j+s)23 (6) D 63 j ≤ n, i > n, i 6= j +n, k = j

− sign(yi+s)(yi+s)2 (7) D 63 i ≤ n, j = i+n, k = i

− sign(y j+s)

(y j+s)2 (8) D 63 j ≤ n, i = j +n, k = j

0 (9) otherwise.

(43)

Figure 6 shows the occupation of the matrices. The lines in the lower right part of the K matrix show theconstant terms in Pi j due to fixed yi on the upper beam.

M x E x K x F gB u

Figure 6: Schematic overview of the occupation of the system matrices.

6.3 Boundary conditions

Dirichlet boundary conditions are introduced by removing variables from the state vector. Since all fixeddegrees of freedom q f have the value 0, removing the respective rows and columns from the matricesand setting all occurences of q f to zero in the nonlinear part is sufficient. Two mechanical constraints areimplemented:

• Pinned support on both sides of the upper beam, as in fig. 1. This removes the displacements y1 andyn from the system.

• Fixed support on the left side of the upper beam. This removes the displacement y1 and the rotationθ1 from the system.

The implementation is flexible enough to easily allow for other boundary conditions.

The voltages should be specified with care. If the voltage is too large, the upper beam will penetrate thelower beam, which is unphysical. Furthermore, due to the rapid increase of the electrostatic potential nearthe point charge, the correct motion and acceleration of the beam become hard to track accurately.

This can be avoided by introducing additional constraints yi ≥−s, yielding a differential-algebraic equationsystem. However, then it will be necessary to consider impact. A general rule is to limit the deflection toyi ≥−s/3, which is the range where the snap-through of the upper beam is avoided.

10

7 Summary

We have implemented a system matrix generator for a dynamical coupled microelectromechanical beamsystem. The charge distribution is approximated by point charges. The system includes electrostatic non-linearities and rotational effects.

References

[1] William Weaver, Jr., Stephen P. Timoshenko, and Donovan H. Young, “Vibration problems in engi-neering”, 5th ed., Wiley (1990)

[2] L. Silverberg, and L. Weaver, Jr. “Dynamics and Control of Electrostatic Structures”, Journal ofApplied Mechanics, Vol. 63, p. 383–391 (June 1996)

11