Page 59 - Mechanics of Microelectromechanical Systems

46 Chapter 1
The bending moment can be found by following the path detailed for the
homogeneous cross-section, and it becomes simple to show that the bending
moment is given by.

where and are the moments of inertia of the two segments, each taken
with respect to its own principal axis (passing through its center of
symmetry). By comparing Eq. (1.179) to Eq. (1.173), it can be concluded
that the bending rigidity of the composite beam is:

where the subscript denotes equivalent and is given in Eq. (1.177).
When more than two layers are included into the composite cross-section, the
equivalent bending rigidity can be expressed as:

where is calculated by means of Eq. (1.178). This formulation is quite
convenient as any of the bending-related stiffnesses/compliances that can be
formulated for various geometries in the case of homogeneous materials are
able to be utilized by simply using the bending rigidity provided by Eq.
(1.181).
Bending about the z-axis, which for classical MEMS devices where w > t
is not a very frequent occurrence, can be treated in a similar fashion, and the
generic Eqs. (1.178) and (1.181) just need to use the subscript z instead of the
subscript y.

6.1.2 Axial Loading and Torsion

For a homogeneous cross-section bar that is loaded axially by a force
this force can be expressed as:

Consider now that the composite cross-section of Fig. 1.25 (a) belongs to a
fixed-free bar subjected to an axial force This force will divide into two
components, and each of them acting on one of the two distinct
portions. By assuming that the two segments are of equal length, they will
also deform axially by the same quantity, and therefore their strains will be
identical. In this case, the total axial force is: