Page 61 - Mechanics of Microelectromechanical Systems

48 Chapter 1
Solution:
The position of the neutral axis is given by Eq. (1.177). It can be shown
that this equation reduces to:

The equivalent rigidity in bending is calculated by means of Eq. (1.180), and
an equivalent homogeneous cross-section beam can be found whose bending
rigidity is:

By equating Eqs. (1.180) and (1.189), the equivalent thickness is:

The rigidity that corresponds to axial equivalence is determined by using Eq.
(1.184) and by considering that the same axial rigidity should be produced by
an equivalent homogeneous bar, namely:

and the axial-related thickness for this problem is:

which results from equating Eqs. (1.184) and (1.191).

Figure 1.26 Thickness ratio in terms of thickness factor and elastic factor