A new generalized model for elliptical arc ﬂexure hinges

A new generalized model for elliptical arc flexure hingesGuimin Chen,1,a� Xiaodong Shao,1 and Xinbo Huang2

1School of Mechatronics, Xidian University, Xi’an, Shaanxi 710071, People’s Republic of China2School of Electronic Information, Xi’an Polytechnic University, Xi’an, Shaanxi 710071,People’s Republic of China

�Received 12 March 2008; accepted 10 August 2008; published online 8 September 2008�

Flexure hinges have been used in many engineering areas where high precision and sensitivity arerequired. Many kinds of flexure profiles were proposed during the past decade. Therefore, a generalclosed-form solution for flexure hinges of different profiles that incorporates the profile selectionand parameter design will be of great benefit to the hinge design. The present work brings circular,right-circular, and elliptical profiles together by proposing a generalized flexure hinge model, whichwe call elliptical arc flexure hinges �whose maximum eccentric angle �m ranges from 0 to � /2� todistinguish from the existing elliptical flexure hinge ��m=� /2�. Based on the theories of mechanicsof materials, all the elements in the compliance matrix for elliptical arc flexure hinges are deducedby introducing the eccentric angle of ellipse as the integral variable. These compliance equationssimply boil down to four integrals, thus simplifying the compliance calculation. These equationsalso apply to elliptical ��m=� /2�, circular �a=b�, and right-circular ��m=� /2 and a=b� hinges.These compliance equations were checked by comparing them with the results of finite elementanalysis, the existing equations, and experiment results. The comparison results show that thesegeneralized equations are concise and adequate for most design purposes. © 2008 AmericanInstitute of Physics. �DOI: 10.1063/1.2976756�

I. INTRODUCTION

Flexure hinges in various forms find wide use in avariety of precision mechanisms such as micrograbbers,micropositioning stages, high-accuracy alignment devices,displacement amplifiers, and all kinds of parallel micro-mechanisms because they are small in size, high in sensitiv-ity, and without mechanical friction and backlash.

Paros and Weisbord1 introduced circular flexure hinges�Fig. 1�c�� and provided both full theoretical and simplifiedcalculation equations for them in 1965. A right-circular hinge�Fig. 1�a�� is a special circular hinge with the maximum cen-tral angle �m equal to � /2 �the cutouts are semicircular�,therefore these equations also apply to right-circular hinges.In 2002, Wu and Zhou2 presented more concise equations forcircular flexure hinges, which simplified the design calcula-tion. Wu and Zhou’s equations have the same results as theexact ones provided by Paros and Weisbord, except the signsof �z /Fy and �y /Mz are opposite.

The advent of CNC wire electrodischarge machiningtechnology made producing hinges of arbitrary cutout pro-files easier, and consequently flexure hinges of various cutoutprofiles were introduced and studied. Lobontiu et al.3 pre-sented the closed-form solutions for the in-plane compliancefactors of corner-filleted flexure hinges, which were vali-dated using finite element method and experiments. Lobontiuet al.4 introduced parabolic and hyperbolic flexure hingesand derived both in- and out-of-plane compliance formulas.Secant and inverse parabolic flexure hinges5 were introducedand studied as well.

There are several papers presenting design equations ofelliptical flexure hinges �Fig. 1�b��. Smith et al.6 derivedclosed-form compliance equations for elliptical flexurehinges by modifying the equations of Paros and Weisbord1

for right-circular flexure hinges. These equations were veri-fied by both finite element analysis and experiment. Tseytlin7

presented thickness-dependent equations for elliptical hingesby following the method of inverse conformal mapping. Amore tractable but less accurate compliance equation for el-liptical hinges is provided in this work as well. Lobontiu4

derived the in-plane compliance equations for ellipticalhinges based on Castigliano’s second theorem. His equationshave the same results as the ones provided by Smith et al.However, the elliptical flexure hinges studied in these papersare formed by cutting two semielliptical cutouts symmetri-cally, namely, the maximum eccentric angle of the cutouts�m=� /2, as shown in Fig. 1�b�. Flexure hinges with ellipti-cal arc cutouts of arbitrary maximum eccentric angle �m

have not been studied so far. This paper expands upon theprevious work by presenting flexure hinges with elliptical arccutouts of arbitrary �m ��m ranges from 0 to � /2�, which wecall elliptical arc flexure hinges to distinguish from the ex-isting elliptical flexure hinge. An elliptical arc flexure hingeis shown in �Fig. 1�d��.

In the vast family of flexure hinges, right-circular, circu-lar, and elliptical flexures are most commonly used inflexure-based precision instruments. For a designer, it is bur-densome to work with all these types of flexure hinges andtheir corresponding design equations in order to optimizeinstrument design. Experiments suggest that for hinge pro-files of the same cutout length and same minimum width,a�Electronic mail: [email protected].

REVIEW OF SCIENTIFIC INSTRUMENTS 79, 095103 �2008�

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right-circular hinges provide the most precise motion, ellip-tical hinges offer better results with regard to stress condi-tions �thus long fatigue life�, and circular hinges provide acombination of these characteristics. Because this knowledgealways drives the selection of different hinge profiles in in-strument design, designers often choose hinge profiles with-out knowing for certain which one works best.8–10 This situ-ation impedes designers from finding the most suitable hingedesigns for instruments. The elliptical arc flexure hinges pre-sented in this paper provide a solution to this problem in theform of general closed-form equations, which encompassand expand upon circular, right-circular, and elliptical flexurehinges. This greatly simplifies the process of cutout profileselection and parameter design for the designer, making itpossible to consistently find and utilize optimal hingedesigns.

In a nutshell, we extend the definition of elliptical hingesby changing the parameter �m of the cutout profiles from�m=� /2 �semiellipse� to 0��m�� /2 and correspondinglycall them elliptical arc flexure hinges. By doing so, we derivea generalized compliance model for circular, right-circular,and elliptical flexure hinges, bringing all these hinge typestogether under one set of equations. It should also be notedthat by extending �m to 0��m�� /2, this model extendsthe design domain of the profile beyond the scope of theother three kinds of flexure hinges.

In Fig. 2, the relationship between circular, right-circular, elliptical, and elliptical arc flexure hinges is shown,

where the former three are subsets of the latter, with right-circular hinges �a=b and �m=� /2� forming the intersectionbetween circular �a=b� and elliptical hinges �the eccentricangle becomes the central angle as a=b�. All the closed-formcompliance equations for elliptical arc flexure hinges are de-rived by introducing the eccentric angle as the integral vari-able, which makes them more concise and effective. Accord-ing to Fig. 2, these equations must also apply to ellipticalflexure hinges when �m=� /2, circular flexure hinges whena=b, and right-circular flexure hinges when a=b and �m

=� /2, allowing designers to utilize optimizing technique forhinge configuration design conveniently. Figure 3 shows athree-dimensional diagram of an elliptical arc hinge.

II. ECCENTRIC ANGLE OF AN ELLIPSE

Figure 4 shows an ellipse and its circumscribed circle ofradius a and inscribed circle of radius b. We can obtain theeccentric angle corresponding to any point P on the ellipseby drawing a vertical line through P, which intersects thecircumscribed circle at L above P �or below if P is on thelower semiellipse�. This is �, the angle between the radius atpoint L and the positive x axis �one of the axes of the el-lipse�. Moreover, the radius at L intersects the inscribedcircle at N. Because the ellipse can be represented in a para-metric form as

(a) (b)

(c) (d)

2c2c

r�/2� �/2 �/2� �/2

rm�� m�

bb

a

am�� m�

FIG. 1. The profiles of different flexure hinges: �a� a right-circular flexurehinge; �b� an elliptical flexure hinge; �c� a circular flexure hinge; �d� anelliptical arc flexure hinge. �Note: elliptical arc hinges are to elliptical hingesas circular hinges are to right-circular hinges�.

Right-circularFlexures

EllipticalFlexures

Elliptical Arc Flexures

CircularFlexures

FIG. 2. The relationship between circular, right-circular, elliptical, andelliptical arc flexure hinges.

w

l

c

b

t

O

x

y

z

Mx

Fx

My

FyMz

Fz

FIG. 3. Diagram of an elliptical arc flexure hinge.

x

y

L

NP

ϕ

a

b

FIG. 4. �Color online� Eccentric angle of an ellipse.

095103-2 Chen, Shao, and Huang Rev. Sci. Instrum. 79, 095103 �2008�


x = a cos �, y = b sin ��0 � � � 2�� , �1�

we know that point P has the same abscissa as point L on thecircumscribed circle and the same ordinate as point N.Therefore, for any point P on the ellipse, its coordinates aregiven by

xP = a cos �, yP = b sin ��0 � � � 2�� . �2�

The cutout profile of an elliptical arc hinge can be deter-mined by the cutout length l=2c �as shown in Fig. 3�, theellipse semiaxis along the cutout depth b, and the maximumeccentric angle �m. The ellipse semiaxis along the cutoutlength a is given by

a =c

sin �m. �3�

For the elliptical arc hinge shown in Fig. 5, as the eccentricangle � traverses from −�m to �m, the corresponding pointon the elliptical arc cutout varies from P1 to P2. Besides,when a=b, the ellipse becomes a circle, and the eccentricangle becomes a central angle.

III. ASSUMPTIONS

The derivation of the compliance equations is based onthe following assumptions.

• For most applications, the minimum thickness of the flex-ure hinges t is much smaller than the cutout length l, so thehinge can be treated as a beam and the Euler–Bernoullibeam theory is applicable; that is, the approximate compli-ances of hinges can be obtained by integrating the lineardifferential equations of a beam based on mechanics ofmaterials.

• It is assumed that the deformations of flexure hinges in allthe directions are so small that superposition applies whena combination of loads is applied.

• Assume a flexure hinge as a six degree-of-freedom fixed-free beam subjected to bending, axial loading, shearing,and torsion, as shown in Fig. 3. By defining the force ac-tuating on the hinge by

F = �Fx,Fy,Fz,Mx,My,Mz�T �4�

and the corresponding deformations of the hinge by

X = ��x,�y,�z,�x,�y,�z�T, �5�

the following relationship is obtained:11

X = ChF . �6�

where Ch is the compliance matrix of the hinge, which canbe expressed by

Ch =��x

Fx0 0 0 0 0

0�y

Fy0 0 0

�y

Mz

0 0�z

Fz0

�z

My0

0 0 0�x

Mx0 0

0 0�y

Fz0

�y

My0

0�z

Fy0 0 0

�z

Mz

� . �7�

Each element in the compliance matrix is derived from beamtheory in the following. We also take the directions of forces,moments, and deflections into consideration.2 Once the com-pliance matrix is obtained, the stiffness matrix of the hingecan be derived by

Kh = Ch−1. �8�

IV. COMPLIANCE EQUATIONS FOR ELLIPTICAL ARCFLEXURE HINGES

Slice the profile of the hinge region into thin verticalinfinitesimal strips. The height of the infinitesimal strip dx atposition x for an elliptical arc hinge can be expressed as �asshown in Fig. 5�

h�x� = 2b + t − 2b

a�a2 − �c − x�2. �9�

By introducing the eccentric angle, we have

x = c + a sin � . �10�

and Eq. �9� can be rearranged to yield

h�� = 2b + t − 2b cos � . �11�

Differentiation of Eq. �10� yields

dx = d�a sin �� = a cos �d� . �12�

Supposing s=b / t and substituting it into Eq. �11�, thestrip height at position � can be expressed as

g�� = h��/b = 2 + 1/s − 2 cos � . �13�

A. Angular compliance about the z axis

The z axis is the axis that outputs the desired rotation�also called flexible axis�, so the angular compliance aboutthe z axis is the most important parameter of the design offlexure hinge.

We denote the angular deflection about the z axis by �z

and �z may be caused by moment Mz or force Fy. Accordingto the Euler–Bernoulli beam theory, �z caused by momentMz can be expressed as

P1 P2

N1 N2

O

x

b

t

h(φ

)

−φmφ dφ

FIG. 5. Analysis of hinge profile.

095103-3 Elliptical arc Flexure Hinges Rev. Sci. Instrum. 79, 095103 �2008�


�z = �0

2c Mz

EIz�x�dx = �

0

2c Mz

E�wh3�x�/12�dx , �14�

where E is the elastic modulus of the material and Iz�x� is thecross-sectional area moment of inertia about the neutral axisat the given position x. By substituting x with eccentric angle�, the corresponding compliance can be expressed as

�z

Mz= �

−�m

�m a cos �

E�wh3��/12d� =

12a

Ewb3�−�m

�m cos �

g3��d�

=12a

Ewb3N2. �15�

As to the force Fy, its contribution to �z equals itsequivalent moment �c−a sin ��Fy about the z axis. There-fore, �z caused by force Fy is given by

�z = �−�m

�m �c − a sin ��a cos ��Fy

E�wh3��/12d�

= �−�m

�m acFy cos �

E�wh3��/12−

a2Fy sin � cos �

E�wh3��/12d� . �16�

The second term in the above integrand is an odd function,so its integral has the value of 0. Therefore, the correspond-ing compliance is

�z

Fy=

12ac

Ewb3�−�m

�m cos �

g3��d� =

12ac

Ewb3N2. �17�

B. Angular compliance about the y axis

Both moment My and force Fz may cause the angulardeflection �y about the y axis. According to the beam theory,�y caused by moment My can be expressed as

�y = �0

2c My

EIy�x�dx = �

0

2c My

E�h�x�w3/12�dx . �18�

By substituting x with the eccentric angle �, the correspond-ing compliance can be expressed as

�y

My= �

−�m

�m a cos �

E�w3h��/12d� =

12a

Ew3b�

−�m

�m cos �

g��d�

=12a

Ew3bN1. �19�

As to force Fz, its contribution to �y equals its equivalentmoment −�c−a sin ��Fz about the y axis. Therefore, �y

caused by force Fz is given by

�y = �−�m

�m − �c − a sin ��a cos ��Fz

E�w3h��/12d�

= �−�m

�m a2Fz sin � cos �

E�w3h��/12−

acFz cos �

E�w3h��/12d� . �20�

Again, the first term in the above integrand is an odd func-tion, so the corresponding compliance can be written as

�y

Fz= −

12ac

Ew3b�

−�m

�m cos �

g��d� = −

12ac

Ew3bN1. �21�

C. Linear compliance along the z axis

Linear deflection �z may be caused by moment My orforce Fz. �z due to moment My can be expressed as

�z = �0

2c dz

dxdx = − �

0

2c

�yM�x�dx

= − �−�m

�m

�yM��d�a sin �� , �22�

where

�yM�� = �

−�m

� Mya cos �

E�w3h��/12�d� . �23�

For

d�yM�� =

Mya cos �

E�w3h��/12�d� , �24�

we have

�z = − a sin ��−�m

� Mya cos �

Ew3h��/12d�

−�m

�m

+ �−�m

�m Mya2 cos � sin �

Ew3h��/12d� . �25�

The integrand in the second term of Eq. �25� is an odd func-tion, thus the compliance is

�z

My= −

12ac

Ew3b�

−�m

�m cos �

g��d� = −

12ac

Ew3bN1. �26�

�z caused by force Fz can be divided into two parts, onepart is due to bending, the other is due to shearing. Thebending part is

�zb = − �

−�m

�m

�yF��d�a sin �� , �27�

where

�yF�� = �

−�m

� − �c − a sin ��Fza cos �

E�w3h��/12�d� . �28�

For

d�yF�� = −

�c − a sin ��Fza cos �

E�w3h��/12�d� , �29�

then



�zb = a sin ��

−�m

� �c − a sin ��Fza cos �

Ew3h��/12d�

−�m

�m

− �−�m

�m Fza2c cos � sin �

Ew3h��/12d�

+ �−�m

�m Fza3 sin2 � cos �

Ew3h��/12d�

=12Fzac2 + 12Fza

3

Ew3b�

−�m

�m cos �

g��d�

−12Fza

3

Ew3b�

−�m

�m cos3 �

g��d� �30�

and the shearing part is

�zs = �

0

2c kFz

Gwh�x�dx =

kaFz

Gwb�

−�m

�m cos �

g��d� , �31�

where G and k are the shearing modulus and shearing coef-ficient of the material, respectively. For a beam of rectangu-lar cross-section, k is given by12

k =12 + 11

10 + 10, �32�

where is Poisson’s ratio, and

=E

2G− 1. �33�

To sum up, the linear compliance due to Fz is

�z

Fz=

��zb + �z

s�Fz

= �12ac2 + 12a3

Ew3b+

ka

Gwb�N1 −

12a3

Ew3bN3.

�34�

D. Linear compliance along the y axis

Linear deflection �y along the y axis may be caused bymoment Mz or force Fy. �y due to moment Mz can be ex-pressed as

�y = �−�m

�m

�zM��d�a sin �� , �35�

where

�zM�� = �

−�m

� Mza cos �

E�wh3��/12�d� . �36�

For

d�zM�� =

Mza cos �

E�wh3��/12�d� , �37�

then

�y = a sin ��−�m

� Mza cos �

Ewh3��/12d�

−�m

�m

− �−�m

�m Mza2cos � sin �

Ewh3��/12d� . �38�

The integrand in the second term of Eq. �38� is an odd func-tion, thus the compliance is

�y

Mz= a sin �m�

−�m

�m a cos �

Ewh3��/12d�

=12ac

Ewb3�−�m

�m cos �

g3��d� =

12ac

Ewb3N2. �39�

�y caused by force Fy can be divided into two parts, onepart is due to bending, the other is due to shearing. Thebending part is

�yb = �

−�m

�m

�zF��d�a sin �� , �40�

where

�zF�� = �

−�m

� �c − a sin ��Fya cos �

E�wh3��/12�d� . �41�

Differentiation of �zF�� yields

d�zF�� =

aFy�c − a sin ��cos �

E�wh3��/12�d� , �42�

thus,

�yb = a sin ��

−�m

� �c − a sin ��Fya cos �

Ewh3��/12d�

−�m

�m

− �−�m

�m

a sin ��c − a sin ��Fya cos �

Ewh3��/12d�

=12Fyac2 + 12Fya

3

Ewb3 �−�m

�m cos �

g3��d�

−12Fya

3

Ewb3 �−�m

�m cos3 �

g3��d� . �43�

And the shearing part is

�ys =

kaFy

GwbN1. �44�

To sum up, the linear compliance due to Fy is

�y

Fy=

��yb + �y

s�Fy

=12ac2 + 12a3

Ewb3 N2 −12a3

Ewb3N4 +ka

GwbN1.

�45�

E. Linear compliance along the x axis

We denote the linear deflection caused by tension loadFx by �x, which can be expressed as



�x = �0

2c Fx

Ewh�x�dx = �

−�m

�m Fxa cos �

Ewh��d� , �46�

thus, the linear compliance along the x axis is

�x

Fx=

a

Ewb�

−�m

�m cos �

g��d� =

a

EwbN1. �47�

F. Angular compliance about the x axis

Denote the torsion angle due to moment Mx by �x. Be-cause each infinitesimal strip of the hinge can be treated as aconstant rectangular cross-section beam, according to the ap-proximation equation given in Hearn’s book,13 �x can beexpressed as

�x = �0

2c 42MxIt�x�Gw4h4�x�

dx , �48�

where It�x� is the torsional inertia moment of the infinitesi-mal strip at position x and It�x�= �w3h�x�+wh3�x�� /12. Bysubstituting Eqs. �12� and �13� into Eq. �48�, the torsionalcompliance can be expressed as

�x

Mx=

7a

2Gw3b�

−�m

�m cos �

g��d� +

7a

2Gwb3�−�m

�m cos �

g3��d�

=7a

2Gw3bN1 +

7a

2Gwb3N2. �49�

It should be noted that �z /Fy =�y /Mz and �y /Fz

=�z /My, therefore the compliance matrix is a symmetric ma-trix, that is to say, there are eight independent design param-eters in the matrix.

V. FOUR INTEGRALS IN THE COMPLIANCEEQUATIONS

The four important integrals N1, N2, N3, and N4 are givenas follows:

N1 = �−�m

�m cos �

g��d�

=2�2s + 1��4s + 1

arctan��4s + 1 tan�m

2� − �m, �50�

N2 = �−�m

�m cos �

g3��d�

=12s4�2s + 1��4s + 1�5/2 arctan��4s + 1 tan

�m

2�

+2s3�2s + 1��6s2 + 4s + 1�sin �m

�4s + 1�2�1 + 2s − 2s cos �m�2

−2s4�12s2 + 4s + 1�sin �m cos �m

�4s + 1�2�1 + 2s − 2s cos �m�2 , �51�

N3 = �−�m

�m cos3 �

g��d�

=�2s + 1�3

2s2�4s + 1arctan��4s + 1 tan

�m

2�

−2s + 1 + s cos �m

2ssin �m −

6s2 + 4s + 1

4s2 �m, �52�

N4 = �−�m

�m cos3 �

g3��d� =

48s5 + 8s4 + 20s3 + 30s2 + 10s + 1

2�4s + 1�5/2 arctan��4s + 1 tan�m

2�

−s�2s + 1�2��12s3 − 12s2 − 3s�cos �m − 12s3 + 2s2 + 6s + 1�

2�4s + 1�2�2s + 1 − 2s cos �m�2 sin �m −�m

4. �53�

VI. VERIFICATION OF THE COMPLIANCE EQUATIONS

The finite element software ANSYS was used to check thegeneralized compliance equations for elliptical arc flexure

hinges under static loading. The physical and geometric pa-rameters of five elliptical arc flexure hinges, including anelliptical one, a circular one, and a right-circular one, arelisted in Table I. The finite element models �FEMs� of theflexure hinges were generated using four-node, three-TABLE I. Several design examples of flexure hinges. For all the designs

E=2.071011 N /m2, G=8.11010 N /m2, w=10 mm, t=1 mm, andc=5 mm.

Example Type �m

�deg�a

�mm�b

�mm�

1 Elliptical Arc 45 7.071 42 Elliptical Arc 60 5.774 43 Elliptical 90 5 44 Circular 60 5.774 5.7745 Right-circular 90 5 5

FIG. 6. �Color online� FEM.



dimensional, solid elements �Solid72� with six degree-of-freedom per node. The smart meshing method, which auto-matically refines meshes in higher stress concentrationregions, was used. Each hinge model is fixed on one end, andloaded with forces Fx, Fy, and Fz and moments Mx, My, andMz on the opposite end, respectively. One of the hinge mod-els is shown in Fig. 6. Table II comprises the analytical andfinite element results. As shown in Table II, the analyticaland finite element results are in good agreement. The relativeerrors between the analytical and finite element results areless than 8%.

Table II also presents the results of both the exact equa-tions of Smith et al.6 for the elliptical hinge and the exactequations of Paros and Weisbord1 for the circular and right-circular hinges �Smith et al. derived four compliance equa-tions for elliptical flexure hinges, namely, �x /Fx, �y /Fy,

�y /My, and �z /Mz, and Paros and Weisbord did not give theequation for torsional compliance �x /Mx�. These results arethe same as the ones using the generalized equations of thispaper, except �y /Fy and �z /Fz, because we take the shearingcoefficient �k� into consideration in Eq. �44�. Compared tothe finite element results, the results of the generalized equa-tions for �y /Fy and �z /Fz are more accurate than those ofthe equations of Smith et al. and Paros and Weisbord.

Experimentation was also used to assess the validity ofthe angular compliance equation about the input axis��z /Mz�. Five elliptical arc flexure hinges made of 45 gradesteel were machined by using wire electrodischarge machin-ing technology according to the geometric parameters listedin Table I. Each hinge was machined as an “I” shaped samplewith the upper and lower horizontal bars being separated bythe hinge itself and a 45° wedge on the upper bar exactly

TABLE II. Comparison between FEM results �denoted by F� and theoretical results �S for equations of Smith et al. equations, P for Paros and Weisbord’sequations, and C for the equations of this paper�.

�x /Fx �y /Fy �z /Fz �x /Mx �y /My �z /Mz �y /Mz �z /My

1�C� 3.21110−9 6.53510−7 2.15510−8 1.75210−2 3.85310−4 2.31110−2 1.15610−4 −1.92610−6

1�F� 3.30210−9 6.30810−7 2.04210−8 1.63410−2 3.72510−4 2.42110−2 1.11410−4 −2.01110−6

2�C� 2.84110−9 5.24010−7 1.88410−8 1.44510−2 3.40910−4 1.90410−2 9.52210−5 −1.70510−6

2�F� 2.91510−9 5.40710−7 1.80610−8 1.38310−2 3.22710−4 1.97010−2 9.30610−5 −1.79310−6

3�C� 2.56510−9 4.47110−7 1.68410−8 1.25510−2 3.07810−4 1.65210−2 8.26210−5 −1.53910−6

3�S� 2.56510−9 4.45910−7¯ ¯ 3.07810−4 1.65210−2

¯ ¯

3�F� 2.51910−9 4.68710−7 1.77410−8 1.17810−2 3.23310−4 1.68410−2 7.91710−5 −1.57210−6

4�C� 2.53610−9 4.33310−7 1.66610−8 1.21810−2 3.04310−4 1.60310−2 8.01610−5 −1.52210−6

4�P� 2.53610−9 4.32110−7 1.55110−8¯ 3.04310−4 1.60310−2 8.01610−5 1.52210−6

4�F� 2.45210−9 4.40610−7 1.63210−8 1.14910−2 2.96310−4 1.63110−2 7.64710−5 −1.53610−6

5�C� 2.38610−9 3.98910−7 1.55710−8 1.13010−2 2.86310−4 1.48810−2 7.44010−5 −1.43210−6

5�P� 2.38610−9 3.97810−7 1.44810−8¯ 2.86310−4 1.48810−2 7.44010−5 1.43210−6

5�F� 2.32210−9 4.16710−7 1.59010−8 1.05410−2 2.89610−4 1.44510−2 7.29210−5 −1.38510−6

FIG. 7. �Color online� The experimental setup.



above the hinge. The experimental setup comprises an opti-cal platform, a 6500 A laser diode, a reflector �mounted onthe wedge�, a position sensitive detector �PSD�, and thehinge sample, as shown in Fig. 7. The lower bar is fixed tothe optical platform, and the diode laser mounted immedi-ately above the reflector that reflects a laser beam to the PSD.To improve the measurement precision, the distance betweenthe PSD and the test hinge should be as long as possible. Inthe experiments, we put the PSD 500 mm away from thehinges. For each hinge, we loaded a mass of 200 g on bothsides of the upper bar �force arm of 25 mm� to simulate twobending load cases, so two compliance values could be ob-tained. Therefore, the corresponding angular compliancewas calculated as an average. It should be noted that thisloading case will give an additional compression force to thesample; however, the potential error sources associated withthis configuration, such as hinge compression and parasiticdeflections, are negligible, as discussed by Smith et al.6

When the hinge sample deformed due to the load, the PSDread the displacement of the laser spot on itself, which wasused to calculate the angular deflection of the sample. Theexperimental results are summarized in Table III. The resultsare within 7% error compared to the analytical results inTable II.

It should be noted that although an elliptical arc flexurehinge is designed to primarily have large compliance aboutits input axis ��z /Mz�, the torsional compliance ��x /Mx� isalways comparable in magnitude. Therefore, the torsionalcompliance of elliptical arc flexure hinges should be treatedcarefully and used adequately.

ACKNOWLEDGMENTS

The authors would like to thank Daniel B. Seegmiller forhis help with technical writing, as well as the reviewers andthe editors for their valuable comments. This research is sup-ported by the National Natural Science Foundation of Chinaunder Grant No. 50805110 and the China Postdoctoral Sci-ence Foundation under Grant No. 20070421110.

1 J. M. Paros and L. Weisbord, Mach. Des. 37, 151 �1965�.2 Y. F. Wu and Z. Y. Zhou, Rev. Sci. Instrum. 73, 3101 �2002�.3 N. Lobontiu, J. S. N. Paine, E. Garcia, and M. Goldfarb, J. Mech. Des.123, 346 �2001�.

4 N. Lobontiu, J. S. N. Paine, E. O’Malley, and M. Samuelson, Precis. Eng.26, 183 �2002�.

5 N. Lobontiu, Compliant Mechanisms: Design of Flexure Hinges �CRC,Boca Raton, FL, 2002�.

6 S. T. Smith, V. G. Badami, J. S. Dale, and Y. Xu, Rev. Sci. Instrum. 68,1474 �1997�.

7 Y. M. Tseytlin, Rev. Sci. Instrum. 73, 3363 �2002�.8 C. S. Han and S. H. Kim, Rev. Sci. Instrum. 73, 3678 �2002�.9 J. H. Kim, S. H. Kim, and Y. K. Kwak, Sens. Actuators, A 116, 530�2004�.

10 J. W. Ryu, S. Q. Lee, D. G. Gweon, and K. S. Moon, Mechatronics 9, 657�1999�.

11 S. B. Choi, S. S. Han, Y. M. Han, and B. S. Thompson, Mech. Mach.Theory 42, 1184 �2007�.

12 S. P. Timoshenko and J. M. Gere, Mechanics of Materials �Van NostrandReinhold, New York, 1972�.

13 E. J. Hearn, Mechanics of Materials: An Introduction to the Mechanics ofElastic and Plastic Deformation of Solids and Structural �Pergamon, Ox-ford, 1977�.

TABLE III. Experimental measurements of elliptical arc hinge compliances.

�z /Mz % error

1 2.43710−2 −5.022 1.86110−2 2.263 1.53810−2 6.94 1.52910−2 4.845 1.44310−2 3.12



http://dx.doi.org/10.1063/1.1494855

http://dx.doi.org/10.1115/1.1372190

http://dx.doi.org/10.1016/S0141-6359(01)00108-8

http://dx.doi.org/10.1063/1.1147635

http://dx.doi.org/10.1063/1.1499761

http://dx.doi.org/10.1063/1.1505098

http://dx.doi.org/10.1016/j.sna.2004.05.027

http://dx.doi.org/10.1016/S0957-4158(99)00006-9

http://dx.doi.org/10.1016/j.mechmachtheory.2006.08.009

http://dx.doi.org/10.1016/j.mechmachtheory.2006.08.009

A new generalized model for elliptical arc ﬂexure hinges

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