No Slide TitlePa or N/m2 bar 1 x 10-5 Pa or N/m2 atmosphere 9.869 x 10-6 To convert Into Multiply...

EEL5225: Principles of MEMS Transducers (Fall 2003)1

EEL5225: Principles of MEMS Transducers (Fall 2003)

Elasticity

Agenda:Elasticity

– Force– Stress– Strain– Young’s modulus– Shear modulus– Bulk modulus

Reading: Senturia, Chapter 8, pp. 184-200.

Lecture 10 by H.K. Xie 9/17/2003


Microfabrication: Discussion

Process design issues– Device geometry– Backside processing– System partitioning and packaging– Process partitioning and contamination constraints– Thermal constraints– Material property control– Process accuracy– Alignment features– Wafer architecture– Die separation– Packaging

Reading: Senturia p. 85-91


Forces

Simple Physics:

“Two-types of forces may act on a solid body”

A body force is distributed over the volume of a body, example: gravity forces.A surface force is distributed over the surface of a body and can be further decomposed into...

– Forces that act normalto a surface, example: hydrostatic pressure.

– Forces that act tangentialto a surface, example:

shear stress

/; N kg m sF Ma = ⋅ =r r

Ref. Cook & Young, “Advanced Mechanics

of Materials”, p.2


Stress

Stress: Force per unit area acting on the surface of a differential surface element.

One atmosphere ≈ 14 psi ≈ 100 kPa

Normal Stress:

Shear Stress:

2/Pa N m =

lim0

F dFn nA dAA

σ =∆

=∆∆ →

Ref. Senturia, p. 184.

lim0

F dFt tA dAA

τ =∆

=∆∆ →

Sign Convention: or , where " " is the normal plane on which the stress acts

and "j" is the direction of the stress. Typically, , since the normal plane and the direction are the same.

iij ij

ii i

τ σ

σ σ=


Pressure unit conversion

7.5028 x 10-3torr or mm HgPa or N/m2

1.4508 x 10-4psi or lb/in2Pa or N/m2

7.5028mtorr or micron HgPa or N/m2

0.01mbarPa or N/m2

4.018 x 10-3in. waterPa or N/m2

2.954 x 10-4in. HgPa or N/m2

1.020 x 10-5kg/cm2Pa or N/m2

10dynes/cm2Pa or N/m2

1 x 10-5barPa or N/m2

9.869 x 10-6atmospherePa or N/m2

Multiply byIntoTo convert


Deformation

Decomposition of deformation: Rigid body translation: “center of mass translation”Rigid body rotation: “center of mass rotation”Axial deformation:Shear deformation:

Deformation notation: or is the x-direction deformation or is the y-direction deformation

or is the z-direction deformation

u uxu vyu wz



Strain

Strain: “differential change in length per unit length”, microstrain = relative change in length of one part per million.

Uniaxial Normal Strain:

Pure Shear Strain:

( ) ( )lim

0

x x xx x xx

u u uxxx

ε+∆ −

= =∂

∂→ ∆∆

lim0,

y yx xxy

u uu uy xy xx y

γ = + = +∂∂

∂ ∂→

∆ ∆ ∆ ∆ ∆ ∆

Strain Notation: for axial strain and for shear straini ijε γ= =



Strain

yx zx y z

uu ux y z

ε ε ε= = =∂∂ ∂

∂ ∂ ∂

Strain-deformation relationships:

note that if there is a linear relationship between stress and strain, symmetry exists

and and

y yx x z zxy xz yz

xy yx xz zx yz zy

u uu u u uy x z x z y

γ γ γ

γ γ γ γ γ γ

= + = + = +

= = =

∂ ∂∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

Young’s Modulus: E (Pa, or N/m2)

xx Eεσ =


Stress-Strain Behavior

Ref. Cook & Young, “Advanced Mechanics of Materials”, p. 10.


Poisson’s Ratio

Poisson Contraction:When a deformable body is subjected to an axial stress, not only will it elongate, but it will also contract laterally. In general,

, where is Poisson's ratio. Poisson 's ratio is a material

property that for elastic materials

lateral

axial

ευ

ε=−

(0≤υ≤0.5)

υ

Uniaxial Stress Example:


y xε υε= −


Compressibility

Volume Change: Bulk modulus or modulus of volume expansion

( ) ( ) ( ) ( )So as there is no volume change regardless of the axial loading.

Therefore, indicates an incompressible material.

Another way to express this is to def

1 1 1 1 2

0.5

0.5

x x x xV x y z x y zε υε υε υ ε

υ

υ

∆ = ∆ + ∆ − ∆ − ≈ ∆ ∆ ∆ −

→

=

( )

ine the bulk modulus of volume expansion,

, which relates the volume strain to an applied hydrostic pressure.3 1 2E V

VK

υ∆

−=

Compressibility: 1/K1/K 0 when υ 0.5: Incompressible


Hooke’s Law

Isotropic materials:Material properties are invariant with position.

Generalized Hooke’s Law: Assume a linear relationship between stress and strain.

( ) ( ) ( ) 1 1 1

x y z y x z z x yy zx E E Eε σ υ σ σ ε σ υ σ σ ε σ υ σ σ = − + = − + = − +

= Modulus of Elasticity or Young's ModulusE

1 1 1

xy xy yz yz xz xzG G Gγ τ γ τ γ τ= = =

( )E

2 1+ = Shear Modulus. Note: =G G

υ


Plane Stress

Plane Stress Assumption:Assume zero shear stress contribution. This is valid for thin films, far away from edge attachment.

( )

( )

1

1

x y

y xy

x E

E

σ υσ

σ υσ

ε

ε

−

−

=

=

If and

then, "biaxial plane stress" 1

x y x y

E

σ σ σ ε ε ε

σ ευ

= = = =

=−

Ref. Senturia, pg 190.


Stress Concentration

Edge Conditions:In general the edge behavior is very complex and usually requires FEM modeling

The edge region of a tensile film attached to a substrate.Ref. Senturia, pg 191.


Anisotropic Materials

Additional elastic constants are needed for anisotropic material.The six independent components of stress and strain are related in the following matrix notation.

i ij jjCσ ε= ∑ where Cij are the stiffness coefficients of the material.

Inverted form is ε i ij jjS σ= ∑ where Sij are the compliance coefficients.


Anisotropic Material – Si

For materials with cubic symmetry such as single-crystal siliconThere are three non-zero coefficients and a high degree of symmetry in the matrix.

where the experimentally determined values for single-crystal silicon are:

C11=165 GPaC12=64 GPaC44=80 GPa

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