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

288 Chapter 5 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 horizontal displacement 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 the form: 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 be calculated as: such that the displacement amplification becomes: 5. Static response of MEMS 289 For the amplification scheme of Fig. 5.21 (b), point 3 will move due to bending and axial deformations as follows: The horizontal displacements and as well as the displacement amplification a are given by Eqs. (5.67), (5.68) and (5.69), respectively. The vertical displacement of point 1 in Fig. 5.21 (b) is: Example 5.12 Assess the errors that appear in the amplification produced by the microdevice of Fig. 5.21 (b) when neglecting the axial deformation of the flexure 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 bending- produced deformation, the horizontal displacement at point 3 can be calculated by means of Castigliano’s displacement theorem as: and the slope at the flexure’s tip is: For small displacements, the horizontal motions at points 1 and 2 can be approximated as: and the corresponding displacement amplification becomes: 290 Chapter 5 By following a similar approach, the amplification for the case where the axial 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 cross- sectional width. 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 in Figs. 5.21 (a) and (b) when the flexure hinge has constant square cross- section. Consider that and E = 160 GPa. Also compare the amplifications produced by the two devices. Solution: The needed compliances are: 5. Static response of MEMS 291 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 longitudinal axis. By using Eqs. (5.78), in combination to Eqs. (5.66) through (5.71), the displacements 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 of which means when the force F acts closer to the flexure hinge and when the length of the flexure hinge is larger. Figure 5.24 Amplification ratio: Fig 5.21 (a) versus in Fig. 5.21 (b) 292 Chapter 5 The amplifications of the two mechanisms are calculated by means of the corresponding equations, and a comparison between the two amplifications is performed 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 one produced by the mechanism in Fig. 5.21 (b) for the set of numerical values of this problem, as shown in Fig. 5.24. Example 5.14 The microdevice of Fig. 5.25 is actuated linearly by a thermal actuator of length Determine the horizontal displacement at the free tip 1 of the rigid vertical link for a temperature increase Known are all geometrical amounts 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 a function of the horizontal displacement at point 3 and the slope at the same point. As shown previously for the mechanism of Fig. 5.21 (a), the displacement and rotation can be calculated as: where is the force generated by thermal actuation at the interface between the horizontal actuator and the vertical rigid link. The value of is given in Eq. (4.7), Chapter 4 and is rewritten here for convenience: 5. Static response of MEMS 293 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 algebraic equation is formed, which can be solved for as: By combining Eq. (5.82) with the first Eq. (5.67), the horizontal displacement at point 1 becomes: 5.2 Sagittal Displacement-Amplification Microdevices Another type of amplification microdevice is the sagittal design, which was presented in Chapter 3 as a suspension component. Figures 5.26, 5.27 and 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 an input force F will generate deformation of the flexure-based mechanisms, and the input displacement corresponding to the force F will be amplified at 294 Chapter 5 the output port, about a direction perpendicular to the input one, due to the inclination 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 the type sketched in Fig. 5.26 under the assumption that the flexure hinges are pure rotational joints. An input displacement translates into the amplified output displacement as indicated in Fig. 5.29, where the initial position is indicated by thicker lines. For and being the lengths of the three rigid links 1-2, 2-3 and 3-4 respectively, it can be shown that the following geometric relationships do apply: By solving Eqs. (5.84) for it can be shown that the displacement amplification of this device is: 5. Static response of MEMS 295 As Eq. (5.85) indicates, the displacement amplification is not a linear function of the input displacement Figure 5.30 is a plot of the amplification in terms of the length of the mid link and its original inclination angle (for an input displacement whereas Fig. 5.31 shows the variation of the same amount as a function of the input displacement (when and Figure 5.30 Displacement amplification as a function of the main link’s length and inclination angle As expected, the amplification increases with the length increasing and with the inclination angle decreasing, as shown in Fig. 5.30. Also, an interesting feature of this device consists in the fact that larger input displacements produce smaller amplifications, as indicated by the plot of Fig. 5.31. Figure 5.31 Displacement amplification as a function of the input magnitude 296 Chapter 5 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 be modeled by the horizontal spring of stiffness At the same time, due to the relative inclination of the two rigid links, elastic resistance is also set about the perpendicular direction, and this is modeled by the spring of stiffness The same inclination amplifies the input displacement to a value about the direction perpendicular to the input one, as pictured in Fig. 5.32 (b). When the work introduced in the system by the action force F entirely balances the work of the resistance force and the potential energy stored in the two elastic springs, the following equation applies: where division by the factor of 2 in the work terms has been applied because the respective forces are applied quasi-statically. At the same time, the following relationship holds true: 5. Static response of MEMS 297 from Fig. 5.32 (b). By substituting now Eq. (5.87) into Eq. (5.86), the following equation is obtained: In the particular case where both the active force and the resistance force are zero (the mechanism deforms through application of the input displacement Eq. (5.88) simplifies to: which shows that the displacement amplification a, the input stiffness and the output stiffness are related. This condition is accurate for a device with pure rotation joints, but is only an approximation for devices utilizing microhinges, 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 formulating the three corresponding qualifiers mentioned above, namely the displacement amplification, input stiffness and output stiffness. A quarter-model will be again 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 two rigid links have been eliminated. Figure 5.34 Reduced quarter-model of displacement-amplification microdevice with one straight flexure hinge for input stiffness calculation [...]... displacement of a fixed-free microbar as a function of the extension load according to: (a) large-deformation theory; (b) small-deformation theory The two plots of Fig 5.43 show the tip displacements according to the large-deformation theory – Fig 5.43 (a) – and the small-deformation theory – Fig 5.43 (b) –, both in terms of the applied force It can be noticed that the predictions of the small-displacement... Schmidt and Sidebottom [3] An example will be analyzed next in order to better contrast the differences between the small- and large-displacement theories Example 5.18 Consider a fixed-free bar of constant cross-section that is acted upon by an axial force at its free end Compare the maximum displacements corresponding to small- and large-displacement theories Given are the force the cross-sectional... local deformations, and namely: The local deformations and of the matrix Eq (5.90) as: can be expressed from the first two rows 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 particular point By combining now Eqs (5.91) and (5.92), results... response of MEMS 313 Equation (5.137) is solved numerically in order to determine the tip deflections of the microcantilever by means of the large-deformation model, and Fig 5.46 (a) plots the values of versus the force F The non-linear relationship between load and deformation can be seen again The tip deflection is calculated by the small-displacement assumptions by means of the known relationship: and. .. situations that are stability-related 316 Figure 5.49 Figure 5.50 Figure 5.51 Chapter 5 Stability-related conditions: (a) Stable; (b) Neutral equilibrium; (c) Unstable Finite element model showing out -of- the-plane buckling of a thin column Buckling of a thin ring: (a) Undeformed; (b) Finite element model of the outof-the-plane deformed shape A perturbation will displace the ball, as shown in Fig 5.49 (a), but... end pinned and a spring and a load at the other end While the separation between small and large displacements is rather flexible, the mathematical description and solutions of the two theories are quite different The small-displacement theory considers that the loading and resulting deformations/displacements of a microcomponent are independent and can be superimposed linearly The large-displacement... that the curvature at the free end is zero, becomes: where: 5 Static response of MEMS 311 The length of the beam can be expressed by integrating the differential length ds of Eq (5.133), namely: The right-hand side of Equation (5.135) can be expressed in terms of an integral of the form: which is known as an elliptic integral of the first kind where c is a constant Equation (5.135) enables to find the... nonlinear, and effects of different loads do combine and affect each other and, together, they affect the deformed state of a MEMS component In many instances, an originally non-linear mathematical model can be linearized, especially when the deformations are small, as shown in the example of Fig 5.42, where the axial force produces an angular rotation of the pinned bar of length l, because of an initial... previously In the present case, the stiffness at the output port of the side device is composed of the output stiffness of this device, coupled in parallel to the input stiffness of the middle microdevice, and therefore, the following equation applies (the minus sign of the original Eq (5.89) is ignored): By combining Eqs (5 .117 ), (5 .118 ) and (5 .119 ), yields the absolute output displacement, namely: In order... shows the angle when calculated by means of the small-deflection hypotheses, according to which: Figure 5.45 Tip rotation angle according to the: (a) large-displacement theory; (b) smalldisplacement theory It can be noticed that the predictions of the small-displacement theory are always higher than the ones of the large-displacement model, and for large values of the force F the difference in results . the loading and resulting deformations/displacements of a microcomponent are independent and can be superimposed linearly. The large-displacement theory is non- linear, and effects of different. upper one and the lower one, each formed of two identical springs connected in series. The equivalent spring stiffness for each branch is according to the series connection rule, presented in. combine and affect each other and, together, they affect the deformed state of a MEMS component. In many instances, an originally non-linear mathematical model can be linearized, especially when