DYNAMIC ANALYSIS OF COMPLIANT BASED PSEUDO … · 297 Int. J. Mech. Eng. & Rob. Res. 2013 Amit...

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Int. J. Mech. Eng. & Rob. Res. 2013 Amit Gupta et al., 2013

DYNAMIC ANALYSIS OF COMPLIANT BASEDPSEUDO-RIGID-BODY CONSTANT FORCE SLIDERCRANK MECHANISM USING THE ENVIRONMENT

LIKE ANSYS

Amit Gupta1*, S S Rattan1 and Sudipto Mukherjee2

*Corresponding Author: Amit Gupta,[email protected]

A mathematical dynamic model is derived for the compliant constant-force mechanism, basedon the pseudo-rigid-body model simplification of the device. The compliant constant-forcemechanism is a slider mechanism incorporating large-deflection beams, which outputs near-constant-force across the range of its designed deflection. The equation of motion is successfullyvalidated with empirical data. The dynamic model is cast for one out of 28 possible configurationsof compliant constant-force slider crank mechanism, identified by type synthesis techniques(Howell, 2001). The author derived the dynamic model through Lagrange’s Equation for theselected configuration of compliant constant-force slider crank mechanism. An unexpecteddynamic trait of the constant-force mechanism is discovered: there exists a range of frequenciesfor which the output force of the mechanism accords nearer to constant-force than does theoutput force at static levels.

Keywords: Compliant mechanism, Pseudo rigid body slider crank mechanism, Constant forcemechanism

INTRODUCTIONCompliant Joint or Flexure Hinge

A flexure hinge is a thin member that providesthe relative rotation between two adjacent rigidmembers through flexing (bending). However,the clearance between mating parts of rigidconventional joints causes backlash in

ISSN 2278 – 0149 www.ijmerr.comVol. 2, No. 3, July 2013

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Int. J. Mech. Eng. & Rob. Res. 2013

1 Department of Mechanical Engineering, NIT, Kurukshetra 136119, India.2 Department of Mechanical Engineering, IIT, Delhi, India.

mechanical assemblies. Further, in all the

above joints there is relative motion causing

friction that leads to wear and increasedclearances. A kinematic chain of such joints

compounds the individual errors from backlashand wear, resulting in poor accuracy and

repeatability.

Research Paper

297


Compliant Mechanisms and ThePseudo-Rigid-Body Model

The emergence of the pseudo-rigid-bodymodel in recent years coincides with thebui lding realization of the aptness ofcompliance in engineering. Natural world alsoreveals the strength of compliance, such asthe flexible wings of a bee or hummingbird, orthe versatility of the human hand and wrist(Vogel, 1995). However, nature is difficult tomimic because of the complex nonlinearanalysis required to understand all but small-deflection compliance. The pseudo-rigid-bodymodel is a method of circumventing some ofthe obstacles that arise when engineeringcompliance into mechanical devices. Itprovides a welcome middle ground betweendesigning merely by trial and error, anddevising exact mathematical formulations.

Compliant Constant-ForceMechanisms

A constant-force mechanism yields a constantoutput force over a range of inputdisplacements. Compliant constant-forcemechanisms are essentially compliant slidermechanisms with flexible and rigid segmentdimensions optimized to minimize thevariation in the output force over a designedrange of displacement. Consider the compliantslider mechanism depicted in Figure. Givenan input displacement x, a compliantconstant-force mechanism will yield the sameforce F over the full range of its designeddeflection, plus or minus a small variation.

Figure 1: Compliant Joint

Figure 2: Conventional Joint

Figure 3: Compliant Slider Mechanism

There exist 28 possible configurations of thecompliant constant-force mechanism,identified by type synthesis techniques (Howell,2001). Of these, fifteen configurations areillustrated in Figure (Millar et al., 1996).Experimental validation of dynamic models ofone of these configurations is performed in thispaper.

The pseudo-rigid-body modelapproximates the deflection and forcecharacteristics of a complaint mechanism’sflexible members by assigning them torsionalspring and rigid-link counterparts. Applying themodel to the entire compliant mechanism

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using a series of heuristic rules results in arigid-link model. Herein lies the power of thepseudo-rigid-body model: its ability to converta difficult-to-analyze compliant mechanisminto a familiar rigid-body mechanism whichcan be analyzed using traditional kinematics.The model does not represent compliantmechanisms perfectly, but it has been shownto represent them very well, making it apowerful design tool. Though the pseudo-rigid-body model has been shown to be valid forthe static analysis of compliant mechanisms,very little research has been performed toexplore the usefulness of the pseudo-rigid-body model in dynamic analysis. If the modelcan be shown to approximate well the dynamicresponse of compliant mechanisms, then itsusefulness is extended even further.

Dynamic Model

The derivation of a closed-form dynamic modelfor constant-force mechanism configuration“Class 1A-d,” as designated in Figure.

Figure 4: Fifteen Configurations of theCompliant Constant-Force MechanismFlexible Segments are Depicted by aSingle Line, and Rigid Segments are

Depicted by Two Parallel Lines

Figure 5: Compliant Constant-ForceMechanism, Configuration Class 1A-d

First, the compliant mechanism is modeledas a rigid-body mechanism with lumpedcompliance using the pseudo-rigid-bodymodel. Converting the mechanism to its rigid-body counterpart greatly simplifies kinematicand dynamic analysis by allowing the use ofrigid-body modeling techniques. Lagrange’smethod is then used to obtain an equation ofmotion for the mechanism. The constant-forcemechanism has only one degree of freedom,so only one generalized coordinate and oneinstance of Lagrange’s equation are required.The choice of non-conservative forcesincluded in the generalized force is explained,as well as any assumptions made by themodel.

The mechanism is converted to its rigid-body counterpart by using the pseudo-rigid-body model rule for a cantilever beam with aforce at the free end as described below.

The flexible segment of length l is replacedby two rigid links, links 3 and 6, with lengths r

3

and r6. Link 3’s length is determined by the

relation for the pseudo-rigid-body link’scharacteristic radius, r

3 = l, where is the

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characteristic radius factor. The length of link6 is then r

6 = l – r

3. The compliance of the

flexible segment is represented by a torsionalspring at the new pin (“characteristic pivot”)joining links 3 and 6. The torsional springconstant k for a cantilever beam with a forceat the free end is given by K = K

0EI/l.

When the pseudo-rigid-body angle k is

zero, the torsional spring is undeflected andstores no strain energy. It is assumed that noplastic deformation occurs as the mechanismcycles and the flexible segment deflects.

Formulating the Lagrangian

Given an input slider displacement xb(t), the

dynamic model should indicate the reactionforce F

b(t) at the slider. The independent

coordinate xb(t) could be chosen as the

generalized coordinate for Lagrange’sequation and would result in a generalizedforce corresponding to F

b(t). Instead

2(t) will

be chosen as the generalized coordinate,corresponding to a generalized force Q2

(t).Choosing

2 as the generalized coordinate

simplifies much of the derivation. Note thatbecause

2 is an angle, the generalized force

Q2 has units of moment; a relation to transform

between Q2 and F

b will be given at the end.

The Lagrangian L is formed by taking thedifference of the scalar quantities of kineticenergy T and potential energy V of the system,L = T – V.

where K0 is the stiffness coefficient (a

nondimensionalized torsional spring constant),E is the modulus of elasticity of the flexiblesegment, and I is the moment of inertia of theflexible segment. The average values of andKo over a wide range of loading conditionsare used: = 0.85, K

0 = 2.65

For a more accurate k that changes withdeflection of the flexible segment, and K

0 can

be expressed as functions of end-load angle,but equation gives the average values mostcommonly used in pseudo-rigid-body modelcalculations.

Figure 6: (a) Compliant Constant-ForceMechanism, Configuration Class 1A-d,and (b) Its Pseudo-Rigid-Body Model

(a)

(b)Figure 7: Reaction Force Fb(t) at the Slider(Output Port); Independent GeneralizedCoordinate 2(t) and Generalized Force

Q2(t)

One way to formulate T is to separate themotion of the mechanism inertias into both

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translation and rotation. The center of mass ofeach link translates along a predefined pathas the mechanism moves, and each linkrotates about its center of mass. The mass oflink 6 can be lumped together with the massof the slider since both travel along a linearpath and neither rotates: ms = m6 + mslider.

The first three terms of the kinetic energyexpression represent the translational energyof the system, and the last two represent therotational energy:

21

2

33

2

22 2

1

2

1

2

1rmvmvmT scc

233

2

22 2

1

2

1 Jcjc

where mi is the mass, vci is the velocity of the

center of mass, Jci is the mass moment of

inertia, and ·i is the angular velocity of links 2and 3; r

1 is the velocity of the slider.

2

2

1kKV

The Lagrangian L= T – V must beexpressed in terms of the generalized

coordinate 2 and its time derivative

2 before

forming Lagrange’s equation. The followingequations recast the variables in T and V in

terms of 2 and

2

22

22 4

1rVc

32323232

322

22

23 cos

4

1 rrrrVc

23332222

1 sinsin rrr

2

3

2 sinsin r

rak

2

3

23 sinsin

r

ra

22

22

23

223

sin

cos

rr

r

It is also useful to give xb as a function of

2:

xb = r

1 + r

6

where,

22

22

3221 sincos rrrr

Lagrange’s Equation

Using Lagrange’s formulation, the equation ofmotion for the system is expressed as

2

22

Q

LL

dt

d

The left side of the equation is a statementof the principle of conservation of energy, and

Figure 8: Translational and RotationalMotion of the Mechanism Links

The mass moments of inertia of links 2 and3 are:

21 12

1iic rmJ

Assuming the mechanism lies in a planeperpendicular to gravity, the potential energyof the system is simply the torsional springenergy.

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the right side represents the non-conservativegeneralized force (Thomson and Dahleh,1998). When the derivatives of the Lagrangianare expanded out, the equation of motion forthe system becomes

where

222

22

3 sin rr

The generalized force 2

Q consists of amoment due directly to the force Fb acting onthe slider

bF , a torque due to Coulomb pinfriction

c, and a term to compensate for

unmodeled torque in the mechanism m:

mcFbQ

2

Though more elaborate expressions for theCoulomb friction term

c are possible, the

following simple relation gives sound results

22 sin Cc

Multiplying by 2 is a departure from

classical Coulomb friction formulation, but itgives better results, and is based on the ideathat the classical Coulomb friction coefficientis likely to be proportional to the angle

2.

The values of the Coulomb frictioncoefficient C and the unmodeled torque

um are

chosen using experimental data from statictests, as described in the next section.

Finally, to transform from the torque bF to

the mechanism’s output force Fb, use the

expression

2

b

Fb x

F b

where,

222

222

2

cossinsin

rr

xb

Experimental Setup

To test the validity of the dynamic model, aconstant-force test mechanism Class 1A-dconstant-force mechanisms mounted to thesame ground and sharing the same slider.Mounting two mechanisms opposite each otheris useful because each cancels the momentinduced by the other, and the issue of frictionbetween slider and ground is eliminated.

To apply the dynamic model to the pair ofconstant-force springs, simply imagine the

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device split down its line of symmetry. Theparameters of one of the halves (i.e., linklengths, masses, etc.), are used in the dynamicmodel equations to solve for the predictedforce F

b(t) of a single mechanism. The force

expected at the output port of the constant-force spring pair is then twice F

b(t). Note that

the value of mslider

is half the mass of the sharedslider, not its entire mass.

Each rigid link of the test mechanism ismade of 1.19 mm)-thick steel sheet metal, andeach flexible segment is made of two strips of0.64 mm-thick steel shim stock. Utility hingesserve as pins between ground and each rigidlink, and between each rigid link and flexiblesegment. Both ground pins mount to a steelground plate, and both flexible segmentsclamp to the shared slider, another steel plate.

designed to exhibit constant-force for amaximum of 40% deflection, or a deflection ofx

b = 0.40(r

2 + r

3), or 4.757 cm. The

mechanism’s total extended length is r2 + l , or

13.007 cm.

The device’s nominal constant-force Fnom

doubled for the mechanism pair), as derivedin Howell (2001) is

Nr

kFnom 19.502

3

The average non-dimensionalized constant-force for a 40% deflection Class 1Aconstant-force mechanism, tabulated in Millaret al. (1996), is = 0.4773.

Figure 9 shows a photograph of theexperimental setup used to validate theconstant-force mechanism dynamic model.The setup was designed to allow testing of themechanism by sinusoidally cycling it (throughcompression and expansion) at differentfrequencies. Comparing the force dataobtained from these tests with the forcepredicted by the dynamic model wil lauthenticate or invalidate the model. Theexperimental setup is described here.

Figure 9: Constant-Force Mechanismfor Dynamic Testing Divided Along its Line

of Symmetry

The relevant dimensions, masses(incorporating the masses of the hinges andclamps), and properties of the test mechanismare listed in Table. Parameters used directlyin the dynamic model equations areemphasized. The variables b, h, and I are thewidth, thickness, and area moment of inertiaof the flexible segment’s cross section; E andm

fare the modulus of elasticity and mass of

the flexible segment. The test mechanism was

Figure 10: Photograph of ExperimentalSetup for Dynamic Testing of Constant-

Force Mechanisms

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The constant-force test mechanism detailedin the proceeding section bolts to a thickaluminum ground, mounted perpendicular toa large aluminum table. In this manner bothground pins of the test mechanism are fixedwith respect to the Table 1.

The test mechanism slider is actuated by asmall aluminum block free to move across alinear bearing. Driving the actuator block is avelocity-controlled 2-hp motor with a rotor andcrank-arm. The center of the motor shaft andthe point of attachment of the crank-arm to theactuator block share the same height y frombase table, as depicted in Figure. As such,assuming constant angular velocity of themotor and a small crank radius r, the actuatorblock drives the mechanism with anapproximately sinusoidal velocity.

The rotor is drilled with a series of tappedholes, each located at a different radius, forvariable positioning of the crank arm. This allowsfor larger or smaller total linear displacementsof the mechanism as the rotor cycles.

Bolted in-line between the actuator blockand test mechanism, a load cell measures

force exerted on the slider. Half of the mass ofthe load cel l is lumped with the testmechanism’s slider. The instrument used is astrain-gauge force transducer. Forcemeasurement errors due to unwanted torquefrom the load cell’s placement between theactuator block and slider were found to be lessthan 0.311 N, and so deemed to be negligible.

Parameter Value Parameter Value

r2 5.490 cm I = bh3/12 5.420 X 10–13 × m4

l 7.517 cm E 206.8 GPa

r3 = l 6.390 cm m2 13.8 g

r6 = l – r3 1.128 cm mf 10.7 g

b 2.540 cm m3 = mf 9.1 g

h 0.064 cm m6 = mf – m3 1.6 g

mslider 84.7 g K 3.359 N-m

ms = mslider + m6 86.3 g C 0.055 N-m

m –0.235 N-m

Table 1: Test Mechanism Dimensions,Material Properties, and Masses

(Parameters Used Directly in DynamicModel Equations)

Figure 11: Line Diagram of ExperimentalSetup

A linear potentiometer measuresmechanism deflection. The potentiometerhousing is mounted to the base table, with itspositioning rod attached to the actuator block.Since the actuator block, load cell, and sliderare bolted together, the potentiometermeasures the position x

b(t), a measurement

of the point where mechanism meets the slider,located with respect to where the mechanismattaches to ground. Both force and positiondata were acquired through a Measurementand Instrumentation lab of Rawal Institute ofEngineering and Technology Faridabad.

This setup allows for the testing of theconstant-force mechanism prototype over arange of frequencies, up to about 85 rad/s,above which there is danger of harming thesetup equipment and/or the mechanism.

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Static Data Processing

Since a static test does not require the motorto be on, no filtering of electrical noise isnecessary. Also, because a static test consistsof only one cycle, no averaging is required.Position and force signals for static tests wereboth found to be clean, reliable, and free ofnoise, so the data was used as collected.

In every plot that displays measured forcedata (whether dynamic or static), spacingbetween plotted points is the time intervalbetween samples.

Given a position input xb(t), and all physical

parameters correctly defined, the dynamicmodel predicts the force expected at theconstant-force mechanism’s output port. Aside-by-side comparison of modeled vs.measured force from an actual mechanismshows that the model predicts the dynamicresponse of the test mechanism quitesatisfactorily.

Modeled vs. Measured Force

Figures 12 to 14 show position and force plotsof three dynamic tests of increasing frequency,. For each figure, the predicted force cycle iscalculated directly from the input sinusoid cyclex

b shown in the position plot, using the dynamic

model equations. The measured force cyclein each figure is the result of force dataprocessed as described in Section DynamicData Processing. The measured data isbanded by ±3 pooled sample standarddeviations, or ±3 , representing the 99.74%confidence interval of the measurement.

While the modeled force does not matchthe measured force point for point, it doespredict the average force, the peak-to-peakforce difference, and the general shape of the

Dynamic and Static Testing

Both dynamic and static tests can beperformed using the experimental setupdescribed above. Dynamic tests areperformed by setting the motor controller to runat a constant velocity, waiting for the system toreach steady-state, then collecting data at asampling rate well above the input sinusoidfrequency. The position of the mechanismmount is adjusted so that at maximumexpansion the mechanism has a preload.

Static tests are performed by removing thecrank arm from the setup and again preloadingthe mechanism, this time using a long barclamp. The clamp is slowly screwed tightcompressing the mechanism, and thenunscrewed allowing it to expand back to itsinitial position. Depending on whether the testis dynamic or static, the data is processeddifferently.

Dynamic Data Processing

Unlike a static test, a dynamic test must beperformed with the motor running, whichintroduces a sizable amount of electricalinterference. To eliminate electrical noise fromthe data, both position and force signals arelow-pass fi l tered through an 8th-orderButterworth filter with a cutoff frequency of 314rad/s.

As previously explained, for every dynamictest the input posi tion x

b(t) is always

constrained to be a sinusoid. Themechanism’s force output during one cycle,from its expanded state through compressionand back to expansion, at any frequency. Tocancel measurement errors from cycle to cyclein a given data set and to condense the datafor ease of viewing, 30 force cycles areaveraged together.

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force profile at any given frequency. These areimportant and useful elements to gather froma dynamic analysis of the system.

Dynamic Characterization of theModel

Although nonlinearities make it impossible toexpress an exact algebraic transfer functionfor the system, and no simple magnitude andphase plot can be shown, two useful plots canbe analyzed. The median force and peak-to-peak force magnitude difference of thedynamic model as functions of frequency areshown in Figure 15. Each frequency assumesa sinusoidal position input with amplitudeequal to the full 40% designed mechanismdeflection (with a slight “pre-displacement” togive a preload at full expansion). For good plotresolution, the dynamic model response iscalculated at 400 separate and equally spacedfrequencies.

The heavy solid line represents the forcepredicted by the dynamic model with allparameters as defined in Table. The next twolines are purely theoretical; what happens

Figure 12: Predicted and Measured Forcefor Sinusoidal Input xb of = 4 rad/s



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Figure 15: Frequency Plots Depicting theMedian Force and Peak-to-Peak

Magnitude Difference Exhibited by theConstant-Force Mechanism

when the mass of the slider is set to zero, orthe mechanism has no inertia at all (all massesset to zero). Setting all the inertias to zeroprovides a baseline useful for comparison ofthe other curves, and setting the end mass tozero shows the dynamic response of theconstant-force spring isolated as a separate“module”. The fourth and fifth lines in the figureshow the effect of multiplying the rigid link massby four (which represents a possibleimprovement to the test mechanism, thickeningthe rigid link to ensure it doesn’t flex), and theeffect of reducing the end mass by 75%. Lastly,the modeled force with Coulomb friction

C and

unmodeled torque um

set to zero is given.

Notice that each curve in the peak-to-peakforce plot first curves down, then sus-tains alinear range before it starts to increase (allexcept for the third curve and last curve). Thisdip in magnitude difference is demonstratednicely by Figures 12 to 14. Clearly, the force

profile of Figure 13 at = 42 rad/s has a lowerpeak-to-peak force difference than theproceeding and following figures at = 4 rad/s and = 94 rad/s.

This very interesting and unexpecteddiscovery of the peak-to-peak force plot is thatthere exists a range of frequencies over whicha constant-force mechanism exhibits betterconstant-force behavior than at static levels.This range of frequencies coincides with theinitial magnitude difference drop and most ofthe linear portion for each of the cases plottedin Figure 15. This unexpected findingsignificantly improves the likelihood that thecompliant constant-force mechanism could beviable in industry.

For instance, if a designer were to use thetest mechanism in an application and wantedto output as close to constant-force aspossible, he or she would run the mechanismat a frequency of 30 rad/s (see the heavy solidline of Figure 15). This would result in aconstant-force mechanism with a median forceof 40 N and a force variance of ±3.5 N, muchbetter than the ±6 N force variance the devicedemonstrates statically. Or if a designerwanted to maximize the range of frequenciesover which the mechanism exhibits “better-than-static” constant-force, he or she couldminimize the slider mass as much as possible.Figure 15 shows that the smaller the inertia ofthe slider, the higher the frequency before theforce magnitude difference starts to rise.

Depending on what attributes are mostdesirable—a wide frequency band withmoderately low peak-to-peak force, a singlefrequency with very low peak-to-peak force, orsome other similar effect—the constant-forcemechanism parameters can be optimized to

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achieve the desired results. It was thought thatthis better-than-static constant-forcephenomenon was caused in part by inclusionof ôC and ôum in the dynamic model, and thelast curve of Figure 15 indicates as much. Thepeak-to-peak force difference of the dynamicmodel with

C and ôum set to zero does not

manifest a dip over the initial range offrequencies, as do the other curves. Instead, itcurves up sharply. However, the phenomenonis not strictly due to

C and

um only, as

evidenced by the third curve of Figure 15. Thiscurve, which represents the dynamic modelwith all of the inertias set to zero, does include

C and

um, and it exhibits no dip in peak-to-

peak force. So, the phenomenon is likely dueto some combination of inertial effects and theeffects modeled by

C and

um. A linearization

of the dynamic model about several operatingpoints may hint at the physical reasons for thisbetter-than-static constant-force effect, andmay be a fruitful area for further research.

The peak-to-peak force and median forceplots end at 150 rad/s for two reasons: (1) mosteverything of interest in the two plots occursbelow this frequency, and (2) there is an upperlimit (not necessarily 150 rad/s) above whichthe constant-force mechanism starts to yielda negative force (i.e., will start to pull insteadof push). This occurs when the force cycleexceeds a frequency where the peak-to-peakforce equals twice the median force. For thetest device (heavy solid line), this occurs atabout 99 rad/s.

In few applications will it be useful to give aconstant-force mechanism a displacementinput by attaching an actuator or surface directlyto the slider; the two will usually be touching,but not rigidly connected. When the force

becomes negative, this represents a situationwhere the slider breaks contact with theactuator or surface, possibly to cause animpact later. Of course these frequencieswould be undesirable in most applications andshould be avoided.

Evaluation of the Dynamic Model

How well does the dynamic model representthe constant-force mechanism? And, if themodel anticipates the physical system well,can it be simplified, perhaps by omittinginsignificant terms?

Both questions can be answered byanalyzing how well the modeled force fits themeasured force at each frequency tested. Themodeled force is first compared to themeasured force with the model untouched.Then various model parameters are set to zeroto test their relative importance to the model.As a measure of goodness of fit, the errorformula

measured

measuredmolded

F

FFE

where Fmeasured

is the average measured force,is applied to each data point in a force cycleand the average error E over the full force cycleis calculated. For better presentation,goodness of fit G

fitis cast as a percentage,

Gfit = 100 (1 – E) Figure 16 shows how well

the modeled force fits the measured force, foreach test frequency. The first plot, representedby small circles, shows the goodness of fit ofthe dynamic model with all terms intact. Thesucceeding plots each eliminate one or twomodel parameters. Even with the completemodel, the fit worsens as frequency increases.This is probably due to viscous damping, whichwas not modeled.

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Observe that the dynamic model representsthe constant-force mechanism very well. Overthe range of frequencies tested, the modeledforce is within about 3% relative error of themeasured force.

With the first question answered, the secondquestion remains; can the omission of certainminor terms simplify the model? Not surprisingis that the worst of these is the model with theend mass set to zero. The end masspossesses most of the system mass uponwhich inertial forces act. What is surprising isthe result of omitting the unmodeled torque term

um. This is interesting compared to the effect

of ignoring the Coulomb friction term C. The

contribution of um

in comparison to C is more

predominate than anticipated.

Setting the link masses, m2 and m

3, to zero

represents a fair reduction in the equation ofmotion for the mechanism, equation. Doingthis only sacrifices accuracy at higher

frequencies (see Figure 16), so the dynamicmodel could be simplified by omitting the linkinertias, but removing these terms does notpoint to a simpler model derivation.

Figure 16 seems to be more useful inillustrating the relative importance of thedynamic model parameters, rather than as atool to simplify the model.

CONCLUSIONThe constant-force mechanism dynamicmodel, based on the pseudo-rigid-body modelof the mechanism, proves to be a useful,viable abstraction of the physical system. Thedynamic model of approximates distributedcompliance as a point compliance, whileconverting the device into a rigid-bodymechanism; yet, a dynamic model of thesimplified mechanism yields very satisfactoryresults. This further validates the usefulness ofthe pseudo-rigid-body model as a dynamicsmodeling tool, in conjunction with the researchof Lyon et al. (1997).

Not only does the dynamic model effectivelypredict the output force of an actual constant-force mechanism, but it illuminates a veryuseful dynamic property of the mechanism:over certain frequencies it exhibits betterconstant-force behavior dynamically thanstatically. The knowledge of such a propertymakes the constant-force mechanism muchmore attractive for application in dynamicsystems.

REFERENCES1. Ananthasuresh G K and Kota S (1995),

“Designing Compliant Mechanisms”,Mechanical Engineering, Vol. 117,No. 11, pp. 93-96.

Figure 16: Goodness of Fit of ModeledForce to Measured Force; Goodness of Fitof Modeled Force, Minus Various Terms,

to Measured Force

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2. Atanackovic T M and Cveticanin L J(1996), “Dynamics of Plane Motion of anElastic Rod”, ASME Journal of AppliedMechanics, Vol. 63, pp. 392-398.

3. Bahgat B M, Osman M O M and SankarT S (1981), “An Approach for DynamicAnalysis of Mechanical Systems withMultiple Clearances Using LagrangianMechanics”, Contributed by the DesignEngineering Division of the ASME forPresentation at the Design EngineeringTechnical Conference, pp. 1-8.

4. Bisshopp K E and Drucker D C (1945),“Large Deflection of Cantilever Beams”,Quarterly of Applied Mathematics,Vol. 3, No. 3, pp. 272-275.

5. Burns R H and Crossley F R E (1968),“Kinetostatic Synthesis of Flexible LinkMechanisms”, ASME Paper No. 68-Mech-36.

6. Darcovich J, Angeles J and Misra A K(1992), “Dynamics of Single-LoopMechanisms with Flexible Links”, FlexibleMechanisms, Dynamics, and Analysis,DE, Vol. 47, pp. 453-460.

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