Mechanics of Microelectromechanical Systems - N.Lobontiu and E.Garcia Part 11 potx

Figure 5.21 Lever-based displacement amplification with flexure hinge: a flexure parallelto lever; b flexure perpendicular to lever As detailed in Chapter 1, the rotation angle slope and

Figure 5.21 Lever-based displacement amplification with flexure hinge: (a) flexure parallel

to lever; (b) flexure perpendicular to lever

As detailed in Chapter 1, the rotation angle (slope) and horizontaldisplacement at point 3 (the tip of the flexure hinge) of the design in Fig.5.21 (a) can be found when the compliances of the flexure are known in theform:

The rigid lever is tangent to the deformed flexure hinge at the junction point

3, and therefore the position of the lever’s free tip can be calculated as:

Similarly, the displacement at point 2 about the force direction can becalculated as:

such that the displacement amplification becomes:

For the amplification scheme of Fig 5.21 (b), point 3 will move due tobending and axial deformations as follows:

The horizontal displacements and as well as the displacementamplification a are given by Eqs (5.67), (5.68) and (5.69), respectively Thevertical displacement of point 1 in Fig 5.21 (b) is:

Example 5.12

Assess the errors that appear in the amplification produced by themicrodevice of Fig 5.21 (b) when neglecting the axial deformation of theflexure hinge Consider that the flexure has a constant rectangular cross-section and that and

Solution:

When the axial deformation is accounted for, in addition to the produced deformation, the horizontal displacement at point 3 can becalculated by means of Castigliano’s displacement theorem as:

bending-and the slope at the flexure’s tip is:

For small displacements, the horizontal motions at points 1 and 2 can beapproximated as:

and the corresponding displacement amplification becomes:

By following a similar approach, the amplification for the case where theaxial deformations are neglected is:

The relative error between the amplifications given in Eq (5.76) versus Eq.(5.75) can be expressed as:

and Fig 5.22 is a plot showing this error as a function of the flexure’s sectional width

cross-Figure 5.22 Relative errors in the amplification of the microdevice of Fig 5.21 (b) when axial and bending deformation are considered versus the case when only bending deformation

is accounted for

Example 5.13

Compare the final positions of the two amplification devices shown inFigs 5.21 (a) and (b) when the flexure hinge has constant square cross-

Also compare the amplifications produced by the two devices

Solution:

The needed compliances are:

The notation of the compliances in Eqs (5.78) corresponds to the local frame

of the flexure, where the y-axis is perpendicular to the flexure’s longitudinalaxis By using Eqs (5.78), in combination to Eqs (5.66) through (5.71), thedisplacements of point 1 in the two configurations of Figs 5.21 (a) and (b)can be determined If the following substitution is used:

the ratio of the displacements is plotted in Fig 5.23

Figure 5.23 Ratio of horizontal displacements for point 1: Fig 5.21 (a) versus Fig 5.21 (b)

The ratio is larger than 1 for smaller values of and larger values ofwhich means when the force F acts closer to the flexure hinge and when thelength of the flexure hinge is larger

Figure 5.24 Amplification ratio: Fig 5.21 (a) versus in Fig 5.21 (b)

The amplifications of the two mechanisms are calculated by means of thecorresponding equations, and a comparison between the two amplifications isperformed through the two device’s amplification ratio, as shown in Fig 5.24.The amplification of the mechanism in Fig 5.21 (a) is larger than the oneproduced by the mechanism in Fig 5.21 (b) for the set of numerical values ofthis problem, as shown in Fig 5.24.

Example 5.14

The microdevice of Fig 5.25 is actuated linearly by a thermal actuator oflength Determine the horizontal displacement at the free tip 1 of the rigidvertical link for a temperature increase Known are all geometricalamounts as well the elastic and thermal properties

Figure 5.25 Thermally-actuated displacement-amplification microdevice

Solution:

The horizontal displacement at point 2 is given in Eq (5.68) as afunction of the horizontal displacement at point 3 and the slope at the samepoint As shown previously for the mechanism of Fig 5.21 (a), thedisplacement and rotation can be calculated as:

where is the force generated by thermal actuation at the interface betweenthe horizontal actuator and the vertical rigid link The value of is given in

Eq (4.7), Chapter 4 and is rewritten here for convenience:

where is the bloc force – Eq (4.5) and is the free displacement – Eq.(4.6) By substituting Eqs (5.81) and (5.80) into Eq (5.68), an algebraicequation is formed, which can be solved for as:

By combining Eq (5.82) with the first Eq (5.67), the horizontaldisplacement at point 1 becomes:

5.2 Sagittal Displacement-Amplification Microdevices

Another type of amplification microdevice is the sagittal design, whichwas presented in Chapter 3 as a suspension component Figures 5.26, 5.27and 5.28 are sketches of three sagittal amplifiers

Figure 5.26 Sagittal displacement-amplifying microdevice with four straight flexure hinges

Figure 5.27 Sagittal displacement-amplifying microdevice with four curved flexure hinges

The operation principle of all these devices is rather simple: application of aninput force F will generate deformation of the flexure-based mechanisms,and the input displacement corresponding to the force F will be amplified at

the output port, about a direction perpendicular to the input one, due to theinclination in the compliant legs, as shown in Fig 5.26.

Figure 5.28 Sagittal displacement-amplifying microdevice with eight straight flexure hinges

Figure 5.29 Quarter model of sagittal displacement amplifier with rotation joints

Figure 5.29 depicts the quarter model of an amplification mechanism of thetype sketched in Fig 5.26 under the assumption that the flexure hinges arepure rotational joints

An input displacement translates into the amplified outputdisplacement as indicated in Fig 5.29, where the initial position isindicated by thicker lines For and being the lengths of the three rigidlinks 1-2, 2-3 and 3-4 respectively, it can be shown that the followinggeometric relationships do apply:

By solving Eqs (5.84) for it can be shown that the displacementamplification of this device is:

As Eq (5.85) indicates, the displacement amplification is not a linearfunction of the input displacement Figure 5.30 is a plot of theamplification in terms of the length of the mid link and its original inclinationangle (for an input displacement whereas Fig 5.31 shows thevariation of the same amount as a function of the input displacement (when

Figure 5.31 Displacement amplification as a function of the input magnitude

Figure 5.32 Physical quarter-model of sagittal displacement amplifier: (a) spring-based

model; (b) displacement diagram

The three mechanisms shown in Figs 5.26 and 5.27 will be studied next

in terms of their displacement amplification capacity, but also they will be

characterized in terms of two other important qualifiers: the input stiffness and the output stiffness Figure 5.32 (a) is a simplified physical model of the

real microdevices of Figs 5.26, 5.27 and 5.28 describing the spring features

by means of two sets of matching wedges that can relatively slip without

friction along their mating surfaces For convenience, the subscript in has been used to denote the input (horizontal) direction of Fig 5.32 (a), whereas

out signifies the output (vertical) direction in the same figure.

The actuating (input) force F encounters elastic resistance which can bemodeled by the horizontal spring of stiffness At the same time, due tothe relative inclination of the two rigid links, elastic resistance is also setabout the perpendicular direction, and this is modeled by the spring ofstiffness The same inclination amplifies the input displacement to avalue about the direction perpendicular to the input one, as pictured inFig 5.32 (b)

When the work introduced in the system by the action force F entirelybalances the work of the resistance force and the potential energy stored inthe two elastic springs, the following equation applies:

where division by the factor of 2 in the work terms has been applied becausethe respective forces are applied quasi-statically At the same time, thefollowing relationship holds true:

from Fig 5.32 (b) By substituting now Eq (5.87) into Eq (5.86), thefollowing equation is obtained:

In the particular case where both the active force and the resistance force arezero (the mechanism deforms through application of the input displacement

Eq (5.88) simplifies to:

which shows that the displacement amplification a, the input stiffness andthe output stiffness are related This condition is accurate for a devicewith pure rotation joints, but is only an approximation for devices utilizingmicrohinges, as shown in the following

Figure 5.33 Quarter-model of displacement-amplification microdevice with one straight

flexure hinge

The micromechanism of Fig 5.26 will further be studied by formulatingthe three corresponding qualifiers mentioned above, namely the displacementamplification, input stiffness and output stiffness A quarter-model will beagain employed, as sketched in Fig 5.33 Essentially, the design of Fig 5.33

is statically-equivalent to the simplified model of Fig 5.34, where the tworigid links have been eliminated

Figure 5.34 Reduced quarter-model of displacement-amplification microdevice with one

straight flexure hinge for input stiffness calculation

The approach followed here is the one based on the stiffness approach ofChapter 1 for a straight flexible member As shown in Fig 5.34, two sets ofreference frames are utilized here: one is the global frame XY, and the other

is the local reference frame with its x-axis aligned with the straight flexurehinge The actuation force F decomposes locally into the and andtherefore, the following matrix equation can be written, according toCastigliano’s first theorem:

The supplemental part of the subscript which has been used in Chapter 1 todenote the extremity of the flexible member which is assumed free (point 3here) was eliminated from the notation, because the 2-3 flexure is symmetric

As Eq (5.90) indicates, axial and bending effects are both taken into account

In order to determine the input stiffness, a relationship between the force

F and the corresponding displacement (taken about the global directionX) is needed This displacement results from adding up the two localdeformations, and namely:

The local deformations and can be expressed from the first two rows

of the matrix Eq (5.90) as:

where it has been taken into consideration that and are the projections

of F onto the local x-and y-axes The rotation (slope) at point 2 is zero,because the flexure hinge is rigidly attached to the link 1-2 at that particularpoint By combining now Eqs (5.91) and (5.92), results in the followingequation giving the input stiffness:

It should be mentioned that Eq (5.93) is generic in the sense that it canaccommodate any shape of a flexure hinge whose required stiffnesses areknown

A similar approach is followed in order to determine the other importantstiffness, the output stiffness Figure 5.35 pictures the reduced quarter-modelthat corresponds to this case.

Figure 5.35 Reduced quarter-model of displacement-amplification microdevice with one

straight flexure hinge for output stiffness calculation

The force is originally applied at point 4 of Fig 5.33, and is transferred atpoint 3 on the reduced quarter-model of Fig 5.35 A relationship similar tothe matrix Eq (5.90) can be written, connecting the load components

and to the deformations, according to the following equations:

The displacement at 4 about the Y-direction (the direction of the appliedforce can be expressed as:

The output stiffness is found by combining Eqs (5.94) and (5.95), namely:

The displacement amplification is determined by following a path similar

to the one used in finding the output stiffness, the only differences here beingthat there is no force and the actuation force F (which is applied at point 1,

as shown in Fig 5.34) will generate a horizontal reaction at point 4, whichcan be transferred at point 3, as it was done previously In this case, theequations connecting local forces to local displacements (deformations) are:

By combining Eqs (5.97) and Eq (5.95) gives the output displacementwhich is produced by application of the force F at the input port 1, namely:

The displacement amplification is found by means of Eqs (5.91), (5.92) and(5.98) as:

Example 5.15

Check whether the input stiffness output stiffness anddisplacement amplification a of the quarter-model of a microdevice with fourstraight hinges, as the one pictured in Fig 5.26, satisfy the condition of Eq.(5.89), which is valid for a similar device with pure rotation joints

The full microdevice comprises four flexures and its spring-based model

is shown in Fig 5.36 (a) There are two branches between the nodal points Aand B, the upper one and the lower one, each formed of two identical springsconnected in series The equivalent spring stiffness for each branch isaccording to the series connection rule, presented in this chapter As aconsequence, the equivalent spring shown in Fig 36 (b) results from

combining two springs in parallel, each having the stiffness and itsstiffness is the sum of these two intermediate stiffnesses, namely:

Figure 5.36 Spring-based models: (a) Full four-flexure sagittal amplification microdevice

model; (b) Equivalent model

As a consequence, the input stiffness of the whole micromechanism is equal

to the stiffness of one quarter of it Similar reasoning can be applied todemonstrate that the total output stiffness is:

by following the left and right branches between points C and D in Fig 5.36(a)

The following reasoning can be developed in order to determine thedisplacement amplification of the full microdevice A simplified quarter-model amplification device of the type discussed here is sketched in Fig 5.37(a), where the input and output displacements are shown (the thicker lineindicates the flexure in its final position) The corresponding full microdevicemodel is shown in Fig 5.37 (b)

When the full mechanism is analyzed, the input displacement is actuallyapplied from both sides, as indicated schematically in Fig 5.37 (b) and theoutput can also be collected at two ports about the direction perpendicular tothe input direction As a consequence, the amplification of the wholemicrodevice can be calculated as:

As the case was with the input and output stiffnesses, Eq (5.104) indicatesthat the displacement amplification of the entire sagittal micromechanism isequal to the amplification produced by one quarter model

Figure 5.37 Input and output displacements for a sagittal displacement amplification

microdevice: (a) quarter model; (b) full model

Example 5.17

Express the input and output stiffnesses, as well as the displacementamplification of the microdevice with four curved flexure hinges sketched inFig 5.27 The quarter-model of this sagittal microdevice, together with thedefining geometry, are shown in Fig 5.38

Figure 5.38 Quarter-model of displacement-amplification microdevice with four curved

flexure hinges

Solution:

The curvature radius of the flexure is R and the corresponding centerangle (not indicated in Fig 5.38) is The lengths of the two adjoining rigidlinks are and respectively The input stiffness, output stiffness anddisplacement amplification are calculated in a manner similar to the one usedfor the straight-flexure quarter-model analyzed previously

The reaction moment which is set when the force F acts at point 1,can be found by considering that the rotation at point 2 is zero (the circularflexure remains rigidly connected to the link 1-2 at that point) It isconsidered that the local frame indicated in Fig 1.20 of Chapter 1, it is also