Page 193 - Mechanical design of microresonators _ modeling and applications

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Microbridges: Lumped-Parameter Modeling and Design

192 Chapter Four
component, namely, the second (left) segment, if the reference frame
origin is chosen at the right fixed end, as shown in Fig. 4.15. A similar
approach was taken, as already mentioned, in analyzing two-segment
microhinges in Chap. 3, where only the properties of one basic segment
of the two were utilized. The subscripts (2) and (1) have been employed
to denote the left and the right segments, respectively. To simplify
notation, no subscript is used here, but all compliances, for instance,
are those of the second mirrored segment of Fig. 4.15, taken with
respect to point 2 (which is assumed free) when point 3 is fixed. In the
end, the generic formulation produces the bending and torsional
resonant frequencies that are associated with the midpoint (point 2) of
the two-segment microbridge of Fig. 4.15.
Bending resonant frequency. The lumped-parameter bending-related stiff-
ness and effective mass are determined here for the entire flexible
structure of Fig. 4.15. To express the lumped-parameter stiffness at
point 2, which is the ratio of a force applied at that point about the z
direction to the resulting out-of-the-plane deflection, the following two
steps have to be undertaken:

● Calculation of the force and moment reactions at fixed point 1 when
a force is applied about the z axis at midpoint 2 in terms of that force
by using compliances
● Calculation of the deflection produced at midpoint 2 under the action
of the force applied at that point
Both steps need to utilize the compliance transforms presented in
Chap. 3 which enable us to express compliances in terms of switched
and arbitrarily translated reference frames.
The lumped-parameter bending stiffness that is associated with
midpoint 2 can be expressed as

1
k b,e = (4.88)
1+2(c í 1)c C + (c í c l /2) 2C + (2c í c l)C
1 1 l 2 1 c 2 1 r
The constants c and c which enter Eq. (4.88) are
2
1
a b + a b a b + a b
12 1
12 2
22 1
11 2
c = c = (4.89)
1
2
2
a a í a 12 a a í a 2
12
11 22
11 22
ʾ ʿ ʾ ʿ ʾ ʿ
where a = C + C a = C + C a = C + C (4.90)
11 r r 12 c c 22 l l
l ʿ íl ʿ
and b = í C b = + C
1 ʿ c 2 ʿ l (4.91)
2C r 2C c
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