a comprehensive elliptic integral solution to the large deflection

Aimei Zhang

Guimin Chen1

e-mail: [email protected]

Key Laboratory of Electronic Equipment

Structure Design of Ministry of Education,

School of Electro-Mechanical Engineering,

Xidian University,

Xi’an Shaanxi 710071, China

A Comprehensive EllipticIntegral Solution to the LargeDeflection Problems of ThinBeams in CompliantMechanismsThe elliptic integral solution is often considered to be the most accurate method for ana-lyzing large deflections of thin beams in compliant mechanisms. In this paper, a compre-hensive solution based on the elliptic integrals is proposed for solving large deflectionproblems. By explicitly incorporating the number of inflection points and the sign of theend-moment load in the derivation, the comprehensive solution is capable of solvinglarge deflections of thin beams with multiple inflection points and subject to any kinds ofload cases. The comprehensive solution also extends the elliptic integral solutions to besuitable for any beam end angle. Deflected configurations of complex modes solved bythe comprehensive solution are presented and discussed. The use of the comprehensivesolution in analyzing compliant mechanisms is also demonstrated by examples.[DOI: 10.1115/1.4023558]

1 Introduction

Compliant mechanisms, which achieve at least some of theirmobility from the deflection of flexible segments rather than fromarticulated joints only, offer many advantages such as energy stor-age, increased precision, and reduced wear, backlash and partnumber [1]. However, the nonlinearity associated with the largedeflection problem often complicates the design and analysis ofcompliant mechanisms. One of the major difficulties lies in accu-rately modeling the large deflections in compliant mechanisms.

Several methods are currently available for solving the largedeflection problems in compliant mechanisms, e.g., the circle-arcmethod [2], the finite element method, the chain algorithm[1,3–5], the Adomian decomposition method [6], the elliptic inte-gral solution [7], and various pseudo-rigid-body model (PRBM)methods [8–12]. Among these methods, the elliptic integral solu-tion is often considered to be the most accurate method for model-ing large deflections of beams that are so thin and flexible that theeffects of axial elongation and shear are negligible. Bisshopp andDrucker derived an elliptic integral solution for beams subject tovertical forces [13]. Howell and Lyon et al. [1,9,10,14] presentedthe elliptic integral solutions for a few load cases where no inflec-tion point is produced in a beam. An elliptic integral solution forthe large deflection with an inflection point was derived by Kim-ball and Tsai [15]. These solutions have the limitation that theslope h of the deflected beam is in the range of �pþ / � h < /(/ denotes the direction of the end force). Shoup presented thesolutions to the undulating and nodal elastica based on the ellipticintegrals [16,17]. Chen and Zhang [18] derived the elliptic inte-gral solution for stain energy in large-deflection beams and usedthe solution to evaluate the accuracy of PRBM. The elliptic inte-grals have also been used to solve the fixed-guided problem inwhich two inflection points might occur [19–21].

In this paper, a comprehensive elliptic integral solution is pro-posed for solving the large deflections of beams of any end angleand with multiple inflection points. By incorporating the numberof inflection points (m) and the sign of the end moment load (SM)in the derivation, the comprehensive solution is capable of locat-ing all the possible deflected configurations of the beam for agiven tip load or tip deflection. The comprehensive solution alsoextends the elliptic integral solutions to be suitable for any beamend angle. The relationship between the deflection angles atinflection points is revealed for the first time. The comprehensivesolution also encompasses the undulating and nodal elasticasolutions.

The rest of the paper is organized as follows. In Sec. 2, thebasic equations of the large deflection problem are reviewed andthe angles at inflection points are further discussed. Section 3presents the comprehensive elliptic integral solution and dis-cusses a few deflected configurations of complex modes. Twoexamples are presented to demonstrate the use of the compre-hensive solution in Sec. 4. Section 5 presents two case studies toshow the capabilities of the comprehensive solution in solvingproblems with the slope h out of the range of �pþ / � h < /and with multiple inflection points. The last section has conclud-ing remarks.

2 Large Deflection Beam Equations

Figure 1 shows an initially straight beam subject to an end forcegP and an end moment Mo. The end force can be divided into a ver-tical component P and a horizontal component nP and we have

g ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ n2

p(1)

Without loss of generality, we assume P is always positive, whileSM is introduced to denote the sign of Mo as

SM ¼1 Mo � 0

�1 M0 � 0

�(2)

1Corresponding author.Contributed by the Mechanisms and Robotics Committee of ASME for

publication in the JOURNAL OF MECHANISMS AND ROBOTICS. Manuscript received July22, 2012; final manuscript received November 15, 2012; published online March 26,2013. Assoc. Editor: Anupam Saxena.

Journal of Mechanisms and Robotics MAY 2013, Vol. 5 / 021006-1Copyright VC 2013 by ASME

Downloaded From: http://mechanismsrobotics.asmedigitalcollection.asme.org/ on 03/29/2013 Terms of Use: http://asme.org/terms

According to the Bernoulli-Euler beam theory, the bendingmoment at an arbitrary point A (x, y) on the beam is proportionalto the curvature at that point, that is

M ¼ EIdhds¼ Pða� xÞ þ nPðb� yÞ þMo (3)

where EI is the flexural rigidity of the beam, M is the moment anddh=ds the curvature.

Equation (3) can be rewritten as (see Ref. [1] for detailedderivation)

dhds¼ 6

ffiffiffiffiffiffi2P

EI

r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi� sin hþ n cos hþ sin ho � n cos ho þ

M2o

2PEI

r(4)

where ho is the deflected angle of the beam end. The sign inEq. (4) is chosen as follows: positive for concave upward curva-ture and negative for convex downward curvature. Equation (4)can be rewritten as

dhds¼ 6

ffiffiffiffiffiffi2P

EI

r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi� sin hþ n cos hþ kp

(5)

where

k ¼ sin ho � n cos ho þ j (6)

and j is the load ratio [1] given as

j ¼ M2o

2PEI(7)

Integrating Eq. (5) results inffiffiffiffiffiP

EI

r ðL

0

ds ¼ 61ffiffiffi2pðho

0

dhffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi� sin hþ n cos hþ kp (8)

For the force index a defined in Ref. [1] given as

a ¼ffiffiffiffiffiffiffiffiPL2

EI

r

we have

a ¼ 61ffiffiffi2pðho

0

dhffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi� sin hþ n cos hþ kp (9)

Note that

dhds¼ dh

dx

dx

ds¼ dh

dxcos h

dhds¼ dh

dy

dy

ds¼ dh

dysin h

9>>>=>>>;

(10)

substituting Eq. (10) into Eq. (5), separating variables and inte-grating yields

a

L¼ 6

1

affiffiffi2pðho

0

cos hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi� sin hþ n cos hþ kp dh

b

L¼ 6

1

affiffiffi2pðho

0

sin hffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi� sin hþ n cos hþ kp dh

9>>>=>>>;

(11)

where a/L and b/L are the nondimensional coordinates of the tippoint along the x- and y-axes, respectively. The signs of Eqs. (9)and (11) are chosen in the same way as for Eq. (4). Equations (9)and (11) together serve as the general equations for the largedeflection problem.

The elastica theory [16] shows that a deflected beam may pos-sess an arbitrary number of inflection points. Let m ðm � 0Þdenote the number of inflection points. Knowing that M¼ 0 atinflection points, Eqs. (5) and (6) are reduced to the followingform at inflection points

k ¼ sin h� n cos h (12)

Equation (12) has an infinite set of solutions for h and each of thesolutions (denoted as h) represents the deflection angle at aninflection point, as shown in Fig. 2. The solution is given as

hj ¼ 2kpþ /6 cos�1ðk=gÞ k ¼ 0; 61; 62 … (13)

where j indicates the jth inflection point numbering from the fixedend of the deflected beam and / represents the angle of the forceapplied at the free end (as marked in Fig. 1)

/ ¼ p=2þ tan�1 n

Equation (13) also reveals the relationship between the deflectionangles at inflection points.

The deflection curves can be divided into two groups accordingto the sign of Mr (the resulting moment at the fixed end of thebeam, as marked in Fig. 1). Figure 3(a) presents the deflected con-figurations with positive curvature at the fixed ends, i.e., Mr > 0,and examples of m¼ 1, 2, 3. Because the curvature changes signat inflection points, we know Sr ¼ ð�1ÞmSM ¼ 1, where Sr

denotes the sign of Mr.

Fig. 1 Deflected configuration of a thin beam subject tocombined force and moment loads (with the positive directionsof P, nP and Mo shown)

Fig. 2 Solutions of Eq. (12) for n 5 0:1; j 5 0:1 and ho

5 0:1ðk 5 sin ho � n cos ho þ j 5 0:1003Þ. The values of h wherek 5 0:1003 represents the magnitude of the beam angle at theinflection points

021006-2 / Vol. 5, MAY 2013 Transactions of the ASME


Let’s consider m¼ 3 as an example to illustrate the change ofangle h along a deflected beam. The angle increases from 0 to h1

(the angle at the first inflection point D), then decreases from h1 toh2 (the angle at the second inflection point E), then increases fromh2 to h3 (point F) and finally, decreases from h3 to the tip angleho, as shown in Fig. 4. In this example, the deflection angles are inthe range of �2pþ / < h � /, thus we have

h1 ¼ h3 ¼ /� cos�1ðk=gÞ (14)

and

h2 ¼ �2pþ /þ cos�1ðk=gÞ (15)

Figure 3(b) plots a few deflected configurations having negativecurvature at the fixed ends, i.e., Mr < 0 and Sr ¼ ð�1ÞmSM ¼ �1.

Again, because the curvature changes sign at inflection points,Eqs. (9) and (11) can be rewritten as

a ¼ Srffiffiffi2pXm

j¼0

ð�1Þjð hjþ1

hj

dhffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi� sin hþ n cos hþ kp

a

L¼ Sr

affiffiffi2pXm

j¼0

ð�1Þjðhjþ1

hj

cos hdhffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi� sin hþ n cos hþ kp

b

L¼ Sr

affiffiffi2pXm

j¼0

ð�1Þjðhjþ1

hj

sin hdhffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi� sin hþ n cos hþ kp

9>>>>>>>>>>>=>>>>>>>>>>>;

(16)

Without loss of generality, we define h0 ¼ 0 and hmþ1 ¼ ho.Equation (16) will be solved using the elliptic integrals in Sec. 3.

3 Comprehensive Elliptic Integral Solution

In this section, the elliptic integrals are used to solve Eq. (16).The incomplete elliptic integrals of the first and second kinds aredefined as [22]

Fðc; tÞ ¼ðc

0

ddffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� t2 sin dp (17)

and

Eðc; tÞ ¼ðc

0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� t2 sin dp

dd (18)

respectively, where c is called the amplitude and t (�1 � t � 1)the modulus. When c ¼ p=2, they become the complete ellipticintegrals of the first and second kinds and are denoted as F(t) andE(t), respectively.

Due to the limitation of the range of elliptic integrals, the ellip-tic integral solutions to Eq. (16) are divided into two parts: jkj > gand jkj � g.

3.1 Case I: k>g. When jkj > g, the deformed curve changesmonotonously [15] and there is therefore no inflection point(m¼ 0). Equation (16) can be written as

a¼ SM

ffiffiffi2pffiffiffiffiffiffiffiffiffiffi

kþgp f

a

L¼ SM

ffiffiffi2p

g2affiffiffiffiffiffiffiffiffiffikþgp ½�nkf þnðkþgÞeþ

ffiffiffiffiffiffiffiffiffiffikþg

pc�

b

L¼ SM

ffiffiffi2p

g2affiffiffiffiffiffiffiffiffiffikþgp ½kf �ðkþgÞeþn

ffiffiffiffiffiffiffiffiffiffikþg

pc�

9>>>>>>>>>>=>>>>>>>>>>;

for jkj> gðm¼ 0Þ

(19)

where

f ¼ Fðc2; tÞ � Fðc1; tÞe ¼ Eðc2; tÞ � Eðc1; tÞc ¼

ffiffiffiffiffiffiffiffiffiffiffikþ np

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffik� sin ho þ n cos ho

pt ¼

ffiffiffiffiffiffiffiffiffiffiffi2g

kþ g

s

c1 ¼ sin�1

ffiffiffiffiffiffiffiffiffiffiffig� n

2g

r

c2 ¼ � �0:5kb cpþ ð�1Þk sin�1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffigþ sin ho � n cos ho

2g

s

for ðk � 1Þpþ / < ho � kpþ / ðk ¼ 0;61;62 � � �Þ

and �0:5kb c gives the largest integer less than or equal to –0.5k.

Fig. 3 The deflected curves with multiple inflection points

Fig. 4 The angle of a deflected curve for m 5 3 and Mo0<0

Journal of Mechanisms and Robotics MAY 2013, Vol. 5 / 021006-3


3.2 Case II: jkj£ g. For jkj � g, the deformed curve mayhave an arbitrary number of inflection points (m � 0). Equation(16) can be expressed using the elliptic integrals as

a ¼ Srffiffiffigp f

a

L¼ Sr

ag5=2½�ngf þ 2ngeþ

ffiffiffiffiffi2g

pc�

b

L¼ Sr

ag5=2½gf � 2geþ n

ffiffiffiffiffi2g

pc�

9>>>>>>>=>>>>>>>;

for jkj � g and m � 0

(20)

where

f ¼ð�1ÞmFðc2; tÞ � Fðc1; tÞ � 2

Xm

j¼1

ð�1ÞjFðchj; tÞ m � 1

Fðc2; tÞ � Fðc1; tÞ m ¼ 0

8><>:

e ¼ð�1ÞmEðc2; tÞ � Eðc1; tÞ � 2

Xm

j¼1

ð�1ÞjEðchj; tÞ m � 1

Eðc2; tÞ � Eðc1; tÞ m ¼ 0

8><>:

c ¼ffiffiffiffiffiffiffiffiffiffiffikþ np

� ð�1Þmffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffik� sin ho þ n cos ho

pt ¼

ffiffiffiffiffiffiffiffiffiffiffikþ g

2g

s

Sr ¼ ð�1ÞmSM

c1 ¼ sin�1


kþ g

r

c2 ¼ � �0:5kb cpþ ð�1Þk sin�1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffigþ sin ho � n cos ho

kþ g

s

for ðk � 1Þpþ / < ho � kpþ /ðk ¼ 0;61;62 � � �Þ

chj¼ � �0:5kb cpþ ð�1Þk p

2

for ðk � 1Þpþ / < hj � kpþ /ðk ¼ 0;61;62 � � �Þ

where �0:5kb c is the largest integer less than or equal to –0.5k.Equations (19) and (20) include seven unknown parameters:

three load parameters (a, j, and n), three deflection parameters (a,b, and ho) and shape parameter (m). During the calculation, theshape parameter m should be given in advance. When there is noinflection point (m¼ 0), the deflection can be solved by Eq. (19)for jkj > g and Eq. (20) for jkj � g. When m � 1, the deflection issolved exclusively by Eq. (20). Among the other six parameters,given any three of them, the other three can be solved usingEqs. (19) or (20). Nevertheless, an ideal way is to directly solveEqs. (19) or (20) for a, a/L and b/L for given n, j, and ho. If threeparameters other than n, j and ho are given, a numerical iterationprocess is required to solve the deflection. It should be noted thatboth SM ¼ 1 and SM ¼ �1 should be considered when solvingdeflection problems where Mo is not specified.

The above derivation is based on the assumption of P > 0. Inthe case that P < 0, the signs of n, j, ho, SM and b (but not a) arereversed in Eqs. (19) and (20).

3.3 Discussion of Comprehensive Solution. Generallyspeaking, a deflected configuration with more inflection pointscorresponds to a higher strain-energy level in the beam. Unlessotherwise constrained, a deflected beam will have the least num-ber of inflection points possible that still satisfies the governingequations and the boundary conditions. Therefore, when solvinglarge deflection problems in compliant mechanisms, an iterativeprocess can be employed to incrementally increase m from 0 to anumber that yields a feasible solution (in this work, the build-in

function “fsolve” in MATLAB was used with “TolFun” was set to10�8). As a starting point, it is reasonable to assume

� 2pþ / < h � / (21)

thus only the following two solutions of hj in Eq. (13) need to beconsidered

hj ¼ �2pþ /þ cos�1ðk=gÞ or hj ¼ /� cos�1ðk=gÞ (22)

It is interesting to note that, for the range of Eq. (21), all theangles at odd inflection points are equal, and the angles at eveninflection points are also equal. Substituting Eq. (22) into Eq. (20)yields

a ¼ Srffiffiffigp f

a

L¼ Sr

ag5=2½�ngf þ 2ngeþ

ffiffiffiffiffi2g

pc�

b

L¼ Sr

ag5=2½gf � 2geþ n

ffiffiffiffiffi2g

pc�

9>>>>>>>=>>>>>>>;

for � 2pþ / < h � / and jkj � g (23)

where

f ¼ ð�1ÞmFðc2; tÞ � Fðc1; tÞ þ 2mSrFðtÞe ¼ ð�1ÞmEðc2; tÞ � Eðc1; tÞ þ 2mSrEðtÞc ¼

ffiffiffiffiffiffiffiffiffiffiffikþ np

� ð�1Þmffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffik� sin ho þ n cos ho

pt ¼

ffiffiffiffiffiffiffiffiffiffiffikþ g

2g

s

c1 ¼ sin�1


kþ g

r

c2 ¼sin�1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffigþ sin ho � n cos ho

kþ g

r�pþ / < ho � /

�sin�1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffigþ sin ho � n cos ho

kþ g

r�2pþ / < ho � �pþ /

8>>><>>>:

There exists a special case worth mentioning in Eq. (23). Whenho ¼ 0 and m is even, f reduces to 2mSrFðtÞ, e to 2mSrEðtÞ and cto 0, thus Eq. (23) is rewritten as

a

L¼ 1

gFðtÞ ½�nFðtÞ þ 2nEðtÞ�

b

L¼ 1

gFðtÞ ½FðtÞ � 2EðtÞ�

9>>=>>; for ho ¼ 0 and m is even

(24)

This indicates that a/L and b/L are independent of SM and m, thatis to say, the deflected beam has multiple possible configurationsfor the same tip deflection, as illustrated in Fig. 5. Even for agiven m that is even, there still exist two deflection configurations,with one corresponding to SM ¼ 1 and the other to SM ¼ �1. Thisspecial case corresponds to the fixed-guided condition, which isan important case that can be found in many compliant mecha-nisms, e.g., fully compliant bistable mechanism [20,21] and paral-lel guided mechanisms.

Now consider a comprehensive solution for a deflection whereh is beyond the scope outlined in Eq. (21) and is therefore alsobeyond the range of solutions presented in Refs. [1] and [15],which are valid for �pþ / � h < /. Figure 6 illustrates thedeflected curves for SM ¼ 1, n¼ 0, and ho ¼ 3p obtained byEq. (19). It is shown that the deflected curve tends to a circlewith increasing j, which concurs with the results in the nodal



elastica solution [17]. The graphs in Fig. 7 plot the deflectionscalculated using Eq. (20) for h ¼ 2pþ /þ cos�1ðk=gÞ andh ¼ 4pþ /þ cos�1ðk=gÞ.

The comprehensive solution is also useful for pure-force andpure-moment load cases. When the end load is a pure force(Mo ¼ 0), SM can be both þ1 and –1, and Eq. (20) is used tosolve for the force index a and the coordinates of the tip pointfor both cases. When the end load is a pure moment, i.e., P¼ 0, thecomprehensive solution can be used to approximate thedeflection by using a very large value for j instead of1, e.g., 105.Figure 8 compares the tip point locus approximated by thecomprehensive solution and that of [1] in the range of h0 2 ½0; 2p�.The two loci agree well and the maximum root mean square

error is ðerror=LÞmax ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi½a=L� ða=LÞH�

2 þ ½b=L� ðb=LÞH�2

q¼ 3:52� 10�4 (subscript “H” indicates that the value is obtainedusing the solution on Page 45 of Ref. [1]).

4 Verification

In this section, two examples taken from the literature are pre-sented to verify the effectiveness of the comprehensive solution.

4.1 Partial Compliant Four-Bar Mechanism. Figure 9shows a partially compliant four-bar mechanism taken from Ref.[11]. In the mechanism, Link DQ is compliant while links AB and

Fig. 5 The deflected configurations corresponding to m 5 2and m 5 4 when ho 5 0

Fig. 6 The deflected curves for SM 5 1, n 5 0, and ho 5 3p

Fig. 7 The curves of m 5 1;SM 5 � 1;j 5 0:01;n 5 10, andho 5 p

Fig. 8 Comparison of the tip locus of the comprehensive solu-tion with that of Ref. [1] for a beam subject to a pure moment

Fig. 9 A partially compliant four-bar mechanism containing aflexible beam DQ



BQ are rigid. The lengths of link AB, BC, CQ and DQ are denotedas LAB, LBC, LCQ, and LDQ, respectively. Link AB has revolutejoints at each end and serves as the input link subject to a puremoment input Tin. The loop closure equations are

Qx � LCQ sin ho � LBC sinðh2 � hoÞ ¼ LAB cos h1

Qy þ LCQ cos ho � LBC cosðh2 � hoÞ ¼ LAB sin h1 þ LDA

(25)

Figure 10 shows the free-body diagrams for links DQ, AB, andBQ. Applying the static equilibrium for link BQ yields

Mo þ P½LAC sinðh2 � hoÞ þ LCQ sin ho�þ nP½LAC cosðh2 � hoÞ � LCQ cos ho� ¼ 0 (26)

The input moment Tin can be obtained by applying moment equi-librium for link AB

Tin ¼ PLABðcos h1 þ n sin h1Þ (27)

The parameters in Ref. [11] are used LDQ ¼ LBC ¼ L;LAB

¼ ð1� 1=ffiffiffi2pÞL; LCQ ¼ L=20; h2 ¼ 135 deg; and h10 ¼ 0 deg.

The motion of the flexible beam and the coupler link is deter-mined by simultaneously solving the loop closure equation(Eq. (25)), the static equilibrium equation (Eq. (26)) and Eq. (19)or Eq. (20). Link DQ carries an inflection point at positionsh1 ¼ 45 deg; 135 deg; and 315 deg, and the inflection point disap-pears at h1 ¼ 245 deg. These positions are shown in Fig. 11. To

further obverse the change of the inflection point, the moments atboth ends of DQ (Mo and MD) and the forces applied at Q are plot-ted as functions of h1 in Fig. 12. The deflection of Link DQ startswith one inflection point (m¼ 1). The inflection point disappearsat h1 ¼ 233 deg where M0 ¼ 0, and returns at h1 ¼ 265 deg whereMD ¼ 0. The kinetostatic behavior of the mechanism achieved byEq. (27) is plotted in Fig. 13. Tin equals zero at four positions: A(h1 ¼ 0 or 360 deg) and C are stable equilibrium positions and Band D are unstable equilibrium positions. The results obtainedusing a finite element analysis (FEA) model built with the ANSYS

software (Link DQ was meshed into 200 elements with Beam188) are also shown in Fig. 13 for comparison. It shows that theresult of the comprehensive solution agrees well with FEA. On apersonal computer with a processor of 2.4 GHz, it takes 25.9 s forthe comprehensive solution implemented in MATLAB to obtain theresults, while the FEA model consumes 387.8 s on average tocomplete the same calculation.

4.2 Fixed-Guided Compliant Mechanism. The comprehen-sive elliptic integral solution was also used to analyze the kineto-static behavior of a fixed-guided compliant mechanism such asused in many full compliant bistable mechanisms [20,21]. Figure14 illustrates an initially-straight fixed-guided beam, of which thefixed end has an angle b with the horizontal and the guided endmoves vertically. A few authors have contributed to this problem,for example, Zhao et al. [19] presented a numerical method, Holstet al. [21] considered axial deflection in their elliptic integralsolution, and Kim [23] proposed a curve decomposition method tosimplify the derivation and calculation.

When a vertical force Fv is applied at the guided end, the beamis deflected to arrive at a static equilibrium state. During bending,the end slope is constrained to be constant, thus we have ho � 0.The coordinates of the guided end are given as

a ¼ L� d sin b

b ¼ �d cos b(28)

Fig. 10 The free-body diagram of each link in the mechanism

Fig. 11 The deflected shape calculated by the comprehensivesolution

Fig. 12 Plots of the moments and end forces of the flexiblebeam versus crank angle h1

Fig. 13 Plots of input moment Tin versus crank angle h1



where d is the displacement of the guided end. Fv can be solved as

Fv ¼ nP sin b� P cos b (29)

Given the displacement d, the coordinates of the guided end,a and b, can be solved by Eq. (28). Substituting a and b intoEq. (23), j and n can be obtained by solving Eq. (24) for bothSM ¼ 1 and SM ¼ �1.

The parameters of the beam are the same as those used inRef. [19]: E ¼ 160GPa; b ¼ 1mm; h ¼ 0:1mm;L ¼ 30mm; andb ¼ 30 deg. Figure 15 shows two possible deflection paths of thebeam. In general, both deflection paths start with configurationswith two inflection points (m¼ 2), and end with one of the inflec-tion points having disappeared (m¼ 1). For the deflection processshown in Fig. 15(a), SM ¼ 1 (Sr ¼ 1 for m¼ 2 and Sr ¼ �1 form¼ 1) and the first inflection point moves toward the fixed end ofthe beam until it disappears. For the deflection process shown inFig. 15(b), Sr ¼ �1 (SM ¼ �1 for m¼ 2 and SM ¼ 1 for m¼ 1),the second inflection point moves toward the guided end until itdisappears. As listed in Table 1, the load-deflection relationshipsof the two deflection paths are identical except the end momentshave opposite signs where two inflection points exist. Reference

[21] shows a photograph of a device consisting of two parallelfixed-guided beams, in which one beam is deflected with SM ¼ 1,the other deflected with SM ¼ �1, and both carry two inflectionpoints (m¼ 2). For the purpose of comparison, we overlay theresults of the comprehensive elliptic integral solution on thephotograph of Ref. [21] (photo courtesy of Brian Jensen), asshown in Fig. 16, with good agreement.

Figure 17 shows the change of force Fv versus d for differentvalues of b (b ¼ 5 deg; 10 deg; 15 deg; 20 deg; 25 deg; 30 deg, and35 deg). Equilibrium positions occur where Fv ¼ 0, with the twostable equilibrium positions of the bistable mechanism occurringat d ¼ 0 and at the largest value of d for Fv ¼ 0.

5 Case Studies

Two mechanisms are employed as case studies to demonstratethe unique capabilities of the comprehensive solution to solveflexible beam problems that are outside the range of othermethods.

Fig. 14 Fixed-guided compliant mechanism

Fig. 15 Two possible deflection paths of the fixed-guidedbeam (b 5 30 deg) achieved by the comprehensive solution. (a)SM 5 1 when m 5 2 and (b) SM 5 � 1 when m 5 2.

Table 1 The end forces and moments for the two deflectionpaths

Case (a) Case (b)

d (m) P (N) n Mo (Nm) P n Mo (Nm)

0.004 0.592 8.083 0.011 0.592 8.083 �0.0110.008 1.271 3.753 0.013 1.271 3.753 �0.0130.012 1.988 2.309 0.014 1.988 2.309 �0.0140.016 2.673 1.588 0.011 2.673 1.588 �0.0110.020 3.106 1.184 7.97� 10�4 3.106 1.184 7.97� 10�4

0.024 0.320 5.180 0.014 0.320 5.180 0.014

Fig. 16 Comparison of the two deflected configurations(SM 5 1 and SM 5 � 1) obtained by the comprehensive solutionwith the experimental results in Ref. [21]

Fig. 17 Force Fv versus d for different b



5.1 Circular-Guided Compliant Mechanism. A circular-guided compliant mechanism shown in Fig. 18 has motion that isoutside the range of other methods. Link OA is initially straight,flexible, and has one end attached to ground and the other rigidlyconnected to crank AB at point A. Crank AB is rigid and is pinnedto ground at point B, thus guiding point A to follow a circularpath. The lengths of link OA and AB are denoted as LOA and LAB,respectively. The x-axis is oriented along link OA and the angleof link AB with respect to the x-axis is denoted as b. The loopclosure equations are given as

a ¼ LOA � LAB cos b0 þ LAB cos b

b ¼ �LAB sin b0 þ LAB sin b

)(30)

where b0 is the initial angle of link AB. The tip angle of beam OAis equal to b� b0. The kinetostatics of the mechanism can beobtained by simultaneously solving Eqs. (20) and (30).

For the parameters of the mechanism given in Table 2, Fig. 19shows the deflected shapes of link OA at different values of b (bis increased from b0 to 190 deg). OA starts to be deflected withtwo inflection points, where the first inflection point movestowards the root of beam OA and disappears when b exceeds164 deg, while the second one moves slightly around s¼ 2 L/3.Figures 20(b) and 20(c) show photographs of the deflected beamat b ¼ 90 deg and b ¼ 170 deg, respectively. They agree wellwith the results of the comprehensive solution shown in Fig. 19.

The kinetostatics of the mechanism are plotted in Fig. 21.Figure 21 shows that beam OA buckles at the beginning when nPreaches 47.7 N, which is slightly smaller than the critical bucklingforce predicted by Euler’s formula for long column with fixedends (Pcr ¼ 4p2EI=L2 ¼ 48:23N). This difference is due to theboundary conditions applied at the tip of OA.

5.2 Coupler-Curve Guided Compliant Mechanism. Thissubsection presents a case study of a coupler-curve guidedcompliant mechanism to show the capability of the comprehen-sive solution in solving large deflections of flexible beams withmore than two inflection points. The mechanism is comprised of afour-bar crank-rocker mechanism and a cantilever beam (OE)whose free end is rigidly connected to the coupler of the crank-rocker mechanism via rigid link BE, as shown in Fig. 22.

The x-axis is oriented along link OE. As link AB rotates clock-wise, the tip deflections and the tip angle of beam OE are given as

a ¼ r2ðcos h2 � cos h20Þ þ r5½cosðh50 þ h3 � h30Þ � cos h50� þ L

b ¼ r2ðsinh2 � sin h20Þ þ r5½sinðh50 þ h3 � h30Þ � sin h50�ho ¼ h3 � h30

9>=>;

(31)

Fig. 18 Circular-guided compliant mechanism containing aflexible beam OA

Table 2 Parameters of three-link mechanism

LOA (m) LAB (m) b0ðdegÞ E (Pa) I (m4)

0.03 0.02 30 1:4� 109 7:8540� 10�13

Fig. 19 The deflected configurations of beam OA

Fig. 20 Photographs of the circular-guided compliant mecha-nism. (a) b 5 b0, and OA is straight, (b) b 5 90 deg and thedeflected curve of OA has two inflection points, (c) b 5 170 deg,and the deflected curve of OA has one inflection point

Fig. 21 Plots of input moment Tin versus crank angle b



where h20; h30, and h50 are the initial angle of link AB,BC, and BE, respectively. The tip loads can be obtained bysimultaneously solving Eqs. (20) and (31). Figure 23 shows

the free-body diagram for link AB, BC, BE, and OE.Applying the static equilibrium for link AB, BC, and BEyields

Tin ¼ FBxr2 sin h2 � FByr2 cos h2

FBx ¼ �nP� Pr5½cosðh50 þ h3 � h30Þ þ n sinðh50 þ h3 � h30Þ� þMo

r3ðtan h4 cos h30 � sin h3ÞFBy ¼ P� fPr5½cosðh50 þ h3 � h30Þ þ n sinðh50 þ h3 � h30Þ� þMog tan h4

r3ðtan h4 cos h30 � sin h3Þ

9>>>>=>>>>;

(32)

For the parameters of the mechanism given in Table 3, point Etraces a trajectory shown in Fig. 24. OE starts to be deflected withtwo inflection points. The third inflection point appears from theroot of the beam when the moment at the root (Mr) changes sign

from negative to positive. The results also show that the angles atthe first and the third inflection points are equal when the beamcontains three inflection points (e.g., inflection points A and C

Fig. 22 The coupler-curve guided mechanism in its initial posi-tion (solid lines) and a deflected position (dashed lines)

Fig. 23 The free-body diagram of each link in the coupler-curve guided mechanism

Table 3 Parameters of planar linkage mechanism

r1ðmÞ r2ðmÞ r3ðmÞ r4ðmÞ r5ðmÞ L (m) E (Pa) I (m4) h20ðdegÞ h30ðdegÞ h40ðdegÞ h50ðdegÞ

0.1 0.052 0.13 0.117 0.115 0.11 1:4� 109 4:9� 10�14 0 63.8 85.4 90

Fig. 25 Photographs of the coupler-curve guided compliantmechanism. (a) OE has two inflection points (m 5 2) and (b) OEhas three inflection points (m 5 3)

Fig. 24 Locus of Point E (the coupler curve) and the deflectedconfigurations of beam OE



Nono

高亮

shown in Fig. 24). Figures 25(a) and 25(b) show photographs ofthe deflected beam with two inflection points and three inflectionpoints, respectively. They agree well with the results of the com-prehensive solution shown in Fig. 24. From the input torquecurve plotted in Fig. 26, we can see that beam OE buckles atthe beginning when nP reaches 0.2216 N, which is approxi-mately equal to the critical buckling force predicted by Euler’sformula for long column with fixed ends (Pcr ¼ 4p2EI=L2

¼ 0:2242N).

6 Conclusions

A comprehensive solution based on the elliptic integrals wasproposed for solving large deflection problems. By explicitlyincorporating the number of inflection points (m) and the sign ofthe end-moment load (SM) in the derivation, the comprehensivesolution is capable of solving large deflections of thin beams withmultiple inflection points and subject to any kinds of end loads.The comprehensive solution also extends the elliptic integralsolutions to be suitable for any beam end angle. A few deflectedconfigurations of complex modes solved by the comprehensivesolution were presented and discussed. Two examples taken fromthe literature were presented to verify the effectiveness of thecomprehensive solution. Lastly, two mechanisms were employedas case studies to demonstrate the unique capabilities of the com-prehensive solution to solve large deflection problems of flexiblebeams that are outside the range of other methods. The compre-hensive solution also shows promise for use in compliant mecha-nism synthesis, which will be an important topic for our futurework.

Acknowledgment

The authors gratefully acknowledge the financial supportfrom the National Natural Science Foundation of China underGrant No. 51175396, the program for new century excellent tal-ents in university under Grant No. NCET-11-0689, and the Fun-damental Research Funds for the Central Universities underGrant No. K5051204021. The permission to use the photographin Ref. [21] by Professor Brian D. Jensen is also gratefullyacknowledged.

References[1] Howell, L. L., 2001, Compliant Mechanisms, Wiley-Interscience, New York,

NY.[2] Campanile, L. F., and Hasse, A., 2008, “A Simple and Effective Solution of the

Elastica Problem,” J. Mech. Eng. Sci., 222(12), pp. 2513–2516.[3] Hill, T. C., and Midha, A., 1990, “A Graphical, User-Driven Newton-Raphson

Technique for Use in the Analysis and Design of Compliant Mechanisms,”J. Mech. Des., 11(1), pp. 123–130.

[4] Coulter, B. A., and Miller, R. E., 1988, “Numerical Analysis of a GeneralizedPlane Elastica With Non-Linear Material Behavior,” Int. J. Numer. MethodsEng., 26, pp. 617–630.

[5] Chase, R. P. Jr., Todd, R. H., Howell, L. L., and Magleby, S. P., 2011, “A 3-DChain Algorithm With Pseudo-Rigid-Body Model Elements,” Mech. BasedDes. Struct. Mach., 39(1), pp. 142–156.

[6] Banerjee, A., Bhattacharya, B., and Mallik, A. K., 2008, “Large Deflection ofCantilever Beams With Geometric Non-Linearity: Analytical and NumericalApproaches,” Int. J. Non-Linear Mech., 43(5), pp. 366–376.

[7] Frisch-Fay, R., 1962, Flexible Bars, Butterworth, Washington, D.C.[8] Saxena, A., and Kramer, S. N., 1998, “A Simple and Accurate Method for

Determining Large Deflections in Compliant Mechanisms Subjected to EndForces and Moments,” ASME J. Mech. Des., 120(3), pp. 392–400.

[9] Lyon, S. M., Howell, L. L., and Roach, G. M., 2000, “Modeling Flexible Seg-ments With Force and Moment End Loads Via the Pseudo-Rigid-Body Model,”Proceedings of the ASME International Mechanical Engineering Congress andExposition, Orlando, FL, Nov. 5–10, DSC-Vol. 69–72, pp. 883–890.

[10] Lyon, S. M., and Howell, L. L., 2002, “A Simplified Pseudo-Rigid-Body Modelfor Fixed-Fixed Flexible Segments,” Proceedings of the ASME DesignEngineering Technical Conferences, Montreal, Canada, Sept. 29–Oct. 3,DETC2002/MECH-34203, pp. 23–33

[11] Su, H., 2009, “A Pseudorigid-Body 3R Model for Determining Large Deflection ofCantilever Beams Subject to Tip Loads,” ASME J. Mech. Rob., 1(2), p. 021008.

[12] Chen, G., Xiong, B., and Huang, X., 2011, “Finding the Optimal CharacteristicParameters for 3R Pseudo-Rigid-Body Model Using an Improved ParticleSwarm Optimizer,” Precis. Eng., 35(3), pp. 505–511.

[13] Bisshopp, K. E., and Drucker, D. C., 1945, “Large Deflection of CantileverBeams,”, 33(3), pp. 272–275. Available at http://users.cybercity.dk/~bcc25154/Webpage/largedeflection.htm

[14] Howell, L. L., and Leonard, J. N., 1997, “Optimal Loading Conditions for Non-Linear Deflections,” Int. J. Non-Linear Mech., 32(3), pp. 505–514.

[15] Kimball, C., and Tsai, L. W., 2002, “Modeling of Flexural Beams Subjected toArbitrary End Loads,” ASME J. Mech. Des., 124, pp. 223–235.

[16] Shoup, T. E., and McLarnan, C. W., 1971, “On the Use of the Undulating Elas-tica for the Analysis of Flexible Link Mechanisms,” ASME J. Eng. Ind., 93, pp.263–267.

[17] Shoup, T. E., 1972, “On the Use of the Nodal Elastica for the Analysis of Flexi-ble Link Devices,” ASME J. Eng. Ind., 94(3), pp. 871–875.

[18] Chen, G., and Zhang, A., 2011, “Accuracy Evaluation of PRBM for PredictingKinetostatic Behavior of Flexible Segments in Compliant Mechanisms,”Proceedings of the ASME 2011 International Design Engineering TechnicalConferences and Computers and Information in Engineering Conference, Aug.28–31, Washington, DC, DETC2011-47117.

[19] Zhao, J., Jia, J., He, X., and Wang, H., 2008, “Post-Buckling and Snap-ThroughBehavior of Inclined Slender Beams,” IEEE Sens. J., 75(4), p. 041020.

[20] Zhao, J., Yang, Y., Fan, K., Hu, P., and Wang, H., 2010, “A Bistable ThresholdAccelerometer With Fully Compliant Clamped-Clamped Mechanism,” IEEESens. J., 10(5), pp. 1019–1024.

[21] Holst, G. L., Teichert, G. H., and Jensen, B. D., 2011, “Modeling and Experi-ments of Buckling Modes and Deflection of Fixed-Guided Beams in CompliantMechanisms,” ASME J. Mech. Des., 133, p. 051002.

[22] Byrd, P. F., and Friedman, M. D., 1954, Handbook of Ellipitic Integrals forEngineers and Physicists, Springer-Verlag, Berlin.

[23] Kim, C., 2011, “Curve Decomposition Analysis for Fixed-Guided Beams WithApplication to Statically Balanced Compliant Mechanism,” Proceedings ofthe 2011 ASME International Design Engineering Technical Conferences,DETC2011-47829.

Fig. 26 Plots of input moment Tin versus crank angle h2



http://dx.doi.org/10.1243/09544062JMES1244

http://dx.doi.org/10.1115/1.2912569

http://dx.doi.org/10.1002/nme.1620260307

http://dx.doi.org/10.1002/nme.1620260307

http://dx.doi.org/10.1080/15397734.2011.541783

http://dx.doi.org/10.1080/15397734.2011.541783

http://dx.doi.org/10.1016/j.ijnonlinmec.2007.12.020

http://dx.doi.org/10.1115/1.2829164

http://dx.doi.org/10.1115/1.3046148

http://users.cybercity.dk/~bcc25154/Webpage/largedeflection.htm

http://users.cybercity.dk/~bcc25154/Webpage/largedeflection.htm

http://dx.doi.org/10.1016/S0020-7462(96)00069-8

http://dx.doi.org/10.1115/1.1455031

http://dx.doi.org/10.1115/1.3427884

http://dx.doi.org/10.1115/1.3428264

http://dx.doi.org/10.1115/1.2870953

http://dx.doi.org/10.1109/JSEN.2010.2042712

http://dx.doi.org/10.1109/JSEN.2010.2042712

http://dx.doi.org/10.1115/1.4003922

a comprehensive elliptic integral solution to the large deflection

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