# Effective Mass

In suspended MEMS structures moving at high speeds there is a velocity distribution in the structure from the anchor, which is at rest, to the moving structure that has a velocity v. Assuming sinusoidal motion, the bending displacement of the beam from its equilibrium position is

$y \left ( x \text{,} t \right ) = y \left ( x \right ) \sin {\omega t}$.

The velocity distribution can also be viewed as a distribution of kinetic energy. Rayleigh's method equates the maximum kinetic energy the beam has as it moves through the equilibrium position with the maximum potential energy stored in the flexure when it has its maximum displacement.

Illustration of the how the energy in a moving cantilever sloshes between potential energy at the maximum displacement of the beam and kinetic energy as the cantilever moves through the equilibrium position
Illustration of the velocity distribution along a cantilever beam moving at frequency ω

The effect of the kinetic energy distribution is that structures with low velocity appear to have reduced mass compared to those moving at higher velocities. This apparent mass of a part of the structure is called its effective mass. When calculating the resonant frequency of MEMS devices, the effective mass of the structure must be used.

$\omega_\text{o} = \sqrt{\frac{k}{m_\text{eff}}}$

## Effective Mass Calculation

Effective masses are calculated using Rayleigh's method. Rayleigh assumed that for a suspended structure the potential energy stored in the flexure at maximum displacement is equal to the kinetic energy of the entire structure when it is moving at its maximum velocity.

KEmax = PEmax

For a flexure of of length l, width w, thickness t, density ρ spring constant ky the maximum potential energy is

$\text{PE}_\text{max} = \int\limits_{0}^{y \left ( l \right )_\text{max} } k_\text{y} y\, dy = \frac{1}{2} k_\text{y}y \left ( l \right )_\text{max}^2$.

The maximum kinetic energy is found by integrating the kinetic energy of differential mass elements along the length of the beam.

$\text{KE}_\text{max} = \int \limits_{0}^{l} \frac{1}{2} \rho w t v \left ( x \right )^2 \, dx = \int \limits_{0}^{l} \frac{1}{2} \rho w t \omega^2 y \left ( x \right )^2 \, dx$.

The shape function $y \left ( x \right )$ of the flexure depends on the way it is connected to the other structures in the device and the boundary conditions it is subject to. In general,

$\text{KE}_\text{max} = \frac{1}{2} \rho w t \omega^2 \int \limits_{0}^{l} y \left ( x \right )^2 \, dx$.

Applying Rayleigh's method,

$\frac{1}{2} k_\text{y}y \left ( l \right )_\text{max}^2 = \frac{1}{2} \rho w t \omega^2 \int \limits_{0}^{l} y \left ( x \right )^2 \, dx$,

which can be rearranged to form an expression for the resonant frequency

$\omega_\text{o}^2 = \frac{k_\text{y} y \left (l \right )_\text{max}^2}{\rho w t \int \limits_{0}^{l} y \left ( x \right )^2 \, dx} = \frac{k_y}{m_\text{eff}}$

and defines the effective mass as

$m_\text{eff} = \rho w t \int \limits_{0}^{l} \left [ \frac{y \left ( x \right )}{y \left ( l \right )_\text{max}} \right ]^2 \, dx$.

### Cantilever Effective Mass

The shape function for a cantilever with a force F applied at its free end is

$y \left( x \right) = \frac{F}{EI} \left( -\frac{x^3}{6} + \frac{lx^2}{2} \right)$,

where E is the Young's modulus of the cantilever material and $I = \frac{wt^3}{12}$ is the bending moment of inertia of the beam.

The displacement at the free end of the cantilever is

$y \left( l \right) = \frac{F l^3}{3 E I}$.

So, $\frac{y \left ( x \right )}{y \left ( l \right )_\text{max}} = -\frac{1}{2} \left( \frac{x}{l} \right)^3 + \frac{3}{2} \left( \frac{x}{l} \right)^2$

and $\left( \frac{y \left ( x \right )}{y \left ( l \right )_\text{max}} \right)^2 = \frac{1}{4} \left( \frac{x}{l} \right)^6 - \frac{3}{2} \left( \frac{x}{l} \right)^5 + \frac{9}{4} \left( \frac{x}{l} \right)^4$.

Subbing the expression above into the equation for effective mass gives

$m_\text{eff} = \rho w t \int \limits_{0}^{l} \left [ \frac{1}{4} \left( \frac{x}{l} \right)^6 - \frac{3}{2} \left( \frac{x}{l} \right)^5 + \frac{9}{4} \left( \frac{x}{l} \right)^4 \right] \, dx$.

After integrating and factoring out an l the expression becomes

$m_\text{eff} = \rho w t l \left[ \frac{1}{28} \left( \frac{x}{l} \right)^7 - \frac{1}{4} \left( \frac{x}{l} \right)^6 + \frac{9}{20} \left( \frac{x}{l} \right)^5 \right]_0^l$,

which simplifies and evaluates to

$m_\text{eff} = m \left[ \frac{1}{28} - \frac{1}{4} + \frac{9}{20} \right] = \frac{33}{140} m$.

### Guided-end Effective Mass

The shape function for a guided-end beam with a force F applied at its free end is

$y \left( x \right) = \frac{F}{EI} \left( -\frac{x^3}{6} + \frac{lx^2}{4} \right)$.

The displacement at the free end of the cantilever is

$y \left( l \right) = \frac{F l^3}{12 E I}$.

So, $\frac{y \left ( x \right )}{y \left ( l \right )_\text{max}} = -2 \left( \frac{x}{l} \right)^3 + 3 \left( \frac{x}{l} \right)^2$

and $\left( \frac{y \left ( x \right )}{y \left ( l \right )_\text{max}} \right)^2 = 4 \left( \frac{x}{l} \right)^6 - 12 \left( \frac{x}{l} \right)^5 + 9 \left( \frac{x}{l} \right)^4$.

Subbing the expression above into the equation for effective mass gives

$m_\text{eff} = \rho w t \int \limits_{0}^{l} \left [ 4 \left( \frac{x}{l} \right)^6 - 12 \left( \frac{x}{l} \right)^5 + 9 \left( \frac{x}{l} \right)^4 \right] \, dx$.

After integrating and factoring out an l the expression becomes

$m_\text{eff} = \rho w t l \left[ \frac{4}{7} \left( \frac{x}{l} \right)^7 - 2 \left( \frac{x}{l} \right)^6 + \frac{9}{5} \left( \frac{x}{l} \right)^5 \right]_0^l$,

which simplifies and evaluates to

$m_\text{eff} = m \left[ \frac{4}{7} - 2 + \frac{9}{5} \right] = \frac{39}{105} m$.

### Resonator Effective Mass

The folded flexure combdrive resonator consists of two sets of folded flexure beams that suspend a plate mass with the rotors of two combdrives attached to it, called the shuttle. The effective mass of the moving structure is determined using Rayleigh's method by considering the kinetic energy of each part of the structure.

The plate mass and the rotors of the combdrives move through a maximum displacement of ymax as a rigid body. Under a sinusoidal variation in displacement at frequency ω the velocity of the plate mass and the rotors as they move through their equilibrium position is ωymax.

The folded flexure truss by comparison has a maximum displacement of $\frac{1}{2} y_\text{max}$ so the velocity is also half that of the plate mass and combdrive rotors. The flexure beams attached to the plate mass move with a rigid body motion and bending motion to a maximum of the plate displacement, while the flexure beams attached to the anchor have a maximum displacement of half the plate mass.

For a structure of arbitrary composition and geometry the effective mass is found using

$m_\text{eff} = \int \rho \left( x \right) A \left( x \right) \left [ \frac{y \left ( x \right )}{y_\text{max}} \right ]^2 \, dx$,

where $A \left( x \right)$ is the cross-sectional area of the differential element being integrated.

To determine effective mass of the resonator it is broken down into the shuttle, four guided-end flexure beams connected to the anchor, six flexure trusses ( see the folded flexure article for the definition of truss geometry) and four guided-end flexure beams connected to the plate.

$m_\text{eff} = 4 \rho w_\text{fb} t \int \limits_{0}^{l} \left[ \frac{y_\text{fba} \left ( x \right )}{y_\text{max}} \right ]^2 \, dx + 4 \rho w_\text{fb} t \int \limits_{0}^{l} \left[ \frac{y_\text{fbp} \left ( x \right )}{y_\text{max}} \right ]^2 \, dx + 6 m_\text{t} \left[ \frac{y_\text{t}}{y_\text{max}} \right]^2 + m_\text{s} \left[ \frac{y_\text{s}}{y_\text{max}} \right]^2$

By substituting in the values of the displacements of the truss and the shuttle, the equation above becomes

$m_\text{eff} = 4 \rho w_\text{fb} t \int \limits_{0}^{l} \left[ \frac{y_\text{fba} \left ( x \right )}{y_\text{max}} \right ]^2 \, dx + 4 \rho w_\text{fb} t \int \limits_{0}^{l} \left[ \frac{y_\text{fbp} \left ( x \right )}{y_\text{max}} \right ]^2 \, dx + 6 m_\text{t} \left[ \frac{1}{4} \right] + m_\text{s} \left[ 1 \right]$.

For the guided-end flexure connected to the anchor

$y_\text{fba} \left( x \right) = \frac{F/4}{EI} \left( -\frac{x^3}{6} + \frac{lx^2}{4} \right)$

and for the guided-end flexure connected to the plate

$y_\text{fbp} \left( x \right) = \frac{F/4}{EI} \left( -\frac{x^3}{6} + \frac{lx^2}{4} \right) + \frac{1}{2} y_\text{max}$.

The force on each beam in the flexure is 1/4 the force applied to the shuttle.

Because the trusses are rigid, the maximum displacement of the plate is

$y_\text{max} = \frac{F l^3}{24 E I}$.

When these are substituted into the fractional terms in the integrals above they become

$\frac{y_\text{fbp} \left ( x \right )}{y_\text{max}} = -\left(\frac{x}{l} \right)^3 + \frac{3}{2} \left( \frac{x}{l} \right)^2$ and

$\frac{y_\text{fba} \left ( x \right )}{y_\text{max}} = -\left(\frac{x}{l} \right)^3 + \frac{3}{2} \left( \frac{x}{l} \right)^2 + \frac{1}{2}$.

After integration and some tedious algebra

$m_\text{eff} = 4 m_\text{fb} \left[ \frac{13}{140} \right] + 4 m_\text{fb} \left[ \frac{83}{140} \right] + 6 m_\text{t} \left[ \frac{1}{4} \right] + m_\text{s} \left[ 1 \right]$,

which is further simplified to

$m_\text{eff} = \frac{96}{35} m_\text{fb} + \frac{3}{2} m_\text{t} + m_\text{s}$

## Contributors

The content of this article was contributed by the following Serendi-CDI members:

1. Peter Gilgunn, MEMS Lab, Carnegie Mellon University, 02-01-11

2. add your name, affiliation and date of contribution as the next entry and increment the value of this line by one.