Active Aeroelasticity and Rotorcraft Lab.

Aeroelasticity

2020

Prof. SangJoon Shin

 Modeling

 Continuous and Discrete Systems

 Modal Methods

• Eigenmodes

• Rayleigh-Ritz

• Galerkin

 Discrete Point Methods

• Finite Difference

• Finite Element

 Solution of Dynamic Problems

• Mass Condensation – Guyan Reduction

• Component Mode Synthesis

Structural Dynamics Overview

Modeling Levels

 Real structural dynamics system (structures)

 Continuous representation of the structure

 Discrete representation of the structure

Real structures, in 3-D space,

comprised of different material, and subject to external excitation

Assumption : - material (linear elastic) - geometry

- loads

More assumptions

Modeling Levels

 Continuous representation of the structure

• Idealized model (infinite d.o.f)

1-D (continuous beam) representation of the blade

More assumptions, for example: low frequency behavior

1-D finite element

representation of the blade

 Discrete representation of the structure

• Idealized model (finite d.o.f.)

Structural System Representation

 Methods for describing structural systems

• Continuous system : infinite D.O.F.  exact solution only available for special cases

(e.g., vibration of uniform linear beams)

• Approximate solution : finite D.O.F.  two basic approaches 1) Modal methods

2) Discrete point methods

Discrete System

• Systems represented by finite number of degrees of freedom from the outset

• Properties described at certain locations can be obtained from (mass, stiffness) influence coefficient functions, or simply lumping techniques

• General mass-spring system represented by

M ሷ𝑢 + 𝐾 𝑢 = {𝐹 ሽ

Mass matrix Stiffness matrix Forcing vector

Discrete System

[Example] Lumped parameter formulation for a beam

[M]

Applied force Inertial force

Total force : 𝐹_{𝑖,𝑇𝑜𝑡} = 𝑓_𝑖 − 𝑚_𝑖𝑤ሷ_𝑖 (D’Alembert’s principle)

𝐹

_{𝑖 𝑡𝑜𝑡}

= 𝑓

_𝑖

−

𝑚

₁

𝑚

₂

𝑚

₃

𝑚

₄

ሷ

𝑤

₁

ሷ

𝑤

₂

ሷ

𝑤

₃

ሷ

𝑤

₄

Applied force 𝑓(𝑥, 𝑡) 𝐸𝐼(𝑥)

m(𝑥) : mass/unit length 𝑥

𝑥 𝑓₁(𝑡) 𝑓₂(𝑡) 𝑓₃(𝑡) 𝑓₄(𝑡)

𝑚₁𝑤₁ሷ 𝑚₂𝑤₂ሷ 𝑚₃𝑤₃ሷ 𝑚₄𝑤₄ሷ

Discrete System

Deflection 𝑤_𝑖 is

Repose

flexibility influence coefficient, Deflection @ i due to a unit load @ j

load

𝑤

_𝑖

= 𝑐

_𝑖𝑗

𝐹

_{𝑗 𝑡𝑜𝑡} ^𝑐𝑖𝑗

deflection

M 𝑤 + 𝐾 𝑤 = {𝑓 ሷ ሽ

This can also be extended to a full 2-D, 3-D structures

M

ሷ𝑢

⋮ ሷ𝑣

⋮

ሷ

𝑤

⋮

+ 𝐾

𝑢

⋮ 𝑣

⋮ 𝑤

⋮

=

𝐹

_𝑢

⋮ 𝐹

_𝑣

⋮ 𝐹

_𝑤

⋮

Note : Generally both [M] and [K] have coupled structures (off-

= 𝑐

_𝑖𝑗

𝑓

_𝑗

− 𝑀 𝑤 ሷ

_𝑖

Discrete System

Set of simultaneous, coupled DE subject to IC’s @ t=0

- First solve homogeneous equations for the lowest (few) eigenvalues (ω) and eigenvectors ([∅]: mode shape matrix)

Set

M ሷ 𝑤 + 𝐾 𝑤 = 𝐹

𝑤

_𝑖

= 𝑤

_𝑖^𝑜

ቋ

ሶ

𝑤

_𝑖

= 𝑤

_𝑖

ሶ

^𝑜

@ t = 0

M ሷ 𝑤 + 𝐾 𝑤 = 0

characteristic eqn.

w = ഥ 𝑤𝑒

^𝑖ω𝑡

൯

−ω M + K ෥ 𝑤𝑒

^𝑖ω𝑡

= 0 ⋯ (∗

eigenvector

Discrete System

4 eigenvalues , natural frequency

Eigenvectors are obtained by placing any root into (*)

𝜆

_𝑖

= ω

_𝑖²

𝑓

_𝑖

= ω

_𝑖

2𝜋

𝑘

₁₁

− 𝑚

₁₁

ω

₁²

𝑘

₁₂

− 𝑚

₁₂

ω

₂²

⋯

⋱ 𝜙

^𝑖

= 0

𝜙

^𝑖

Need to set at least one value of

A N-D.O.F system has N natural frequencies and N mode shapes associated to these natural frequencies.

𝜙

¹

𝜙

²

ω

₁

ω

₂

⋯

Discrete System

- Orthogonality Relations

ω

_𝑗

, 𝜙

_𝑖^𝑗 set of free vibration mode shapes

−ω

²

𝑀 𝜙 + 𝐾 𝜙 = 0

൯

−ω

_𝑟²

𝑀 𝜙

^𝑟

= 𝐾 𝜙

^𝑟

⋯ (1 ൯

−ω

_𝑠²

𝑀 𝜙

^𝑠

= 𝐾 𝜙

^𝑠

⋯ (2

Multiply (1) by and (2) by

𝜙

^{𝑠 𝑇}

𝜙

^{𝑟 𝑇}

ω

_𝑟²

𝜙

^{𝑠 𝑇}

[𝑀]𝜙

^𝑟

= 𝜙

^(𝑠)𝑇

𝐾 𝜙

^𝑟

ω

_𝑠²

𝜙

^{𝑟 𝑇}

[𝑀]𝜙

^𝑠

= 𝜙

^(𝑟)𝑇

𝐾 𝜙

^𝑠

⋯ (3) ω

_𝑟²

𝜙

^{𝑟 𝑇}

𝑀

^𝑇

𝜙

^𝑠

= 𝜙

^(𝑟)𝑇

𝐾

^𝑇

𝜙

^𝑠

ω

_𝑟²

𝜙

^{𝑟 𝑇}

[𝑀]𝜙

^𝑠

= 𝜙

^{𝑟 𝑇}

[𝐾]𝜙

^𝑠

⋯ (4)

Take transpose of both sides

[M], [K]

symmetric

Each satisfies

Discrete System

Subtract (4) from (3)

If r ≠ s 𝜙

^{𝑟 𝑇}

𝑀 𝜙

^𝑠

= 0 r = s 𝜙

^{𝑟 𝑇}

𝑀 𝜙

^𝑠

= 𝑀

_𝑟^∗

Also note that

(some value : modal stiffness)

𝜙

^{𝑟 𝑇}

[𝑀]𝜙

^𝑠

= 𝛿

_𝑟𝑠

𝑀

_𝑟^∗

Kronecker delta 𝛿_𝑟𝑠 = ቊ0 ∶ 𝑟 ≠ 𝑠 1 ∶ 𝑟 = 𝑠

𝜙

^{𝑟 𝑇}

[𝐾]𝜙

^𝑠

= ω

_𝑟²

𝑀

_𝑟^∗

𝛿

_𝑟𝑠 (modal stiffness)

(ω

_𝑠²

-ω

_𝑟²

) 𝜙

^{𝑟 𝑇}

𝑀

^𝑇

𝜙

^𝑠

=0

Discrete System

- Complete solution

𝜙

^𝑇

Pre-multiply by

Generalized coordinate

M ሷ 𝑤 + 𝐾 𝑤 = 𝐹 𝑤

_𝑖

𝑡 = ෍

𝑖=1 4

𝜙

_𝑖^𝑟

𝜂

_𝑖

(𝑡)

let

M 𝜙 ሷ𝜂 + 𝐾 𝜙𝜂 = 𝐹

𝜙

^𝑇

M 𝜙 ሷ𝜂 + 𝜙

^𝑇

𝐾 𝜙𝜂 = 𝜙

^𝑇

𝐹

Orthogonality Decoupled equations

𝑀

₁^∗

ሷ𝜂 + 𝑀

₁^∗

ω

₁²

𝜂

₁

= 𝑄

₁

, 𝑄

₁

= 𝜙

¹

𝐹 𝑀

_𝑛^∗

𝜂

_𝑛

ሷ + 𝑀

_𝑛^∗

ω

_𝑛²

𝜂

_𝑛

= 𝑄

_𝑛

⋮ ⋮

Generalized mass Generalized stiffness

Generalized or normalized coordinate

Generalized force

Discrete System

- Initial conditions

@ t=0, given

n x 1

𝑤 0 , ሶ 𝑤(0 )

If all the modes are retained in solution, that is,

𝜙 ሶ𝜂 0 = ሶ 𝑤(0 ) 𝜙𝜂 0 =

𝑤

₁

0 𝑤

₂

0 𝑤

₃

0 𝑤

₄

(0 )

and

w = ෍

𝑖=1 𝑛

൯ 𝜙

^𝑖

𝜂

_𝑖

(𝑡 𝜂 0 = 𝜙

⁻¹

𝑤(0 )

m x n n x 1

Discrete System

- Truncation

where m<<n

mx1

Problem can be truncated by using only a few selected number of modes w(x, t) = ෍

𝑖=1 𝑚

൯ 𝜙 ^𝑖 (𝑥)𝜂_𝑖(𝑡

But now calculation of initial condition on

𝜂

is not straightforward.

𝜂 0 = 𝜙

⁻¹

𝑤(0 )

𝜙𝜂 0 = 𝑤(0 )

mxn nx1

nxm mx1 nx1

not invertible!

Discrete System

[Note] The normal equations of motion are uncoupled on the left-hand side due to the modal matrix composed of eigenvectors.

→ Solve for subject to and

𝜂 𝑡 𝜂 0 ሶ𝜂 0

w(x, t) = ෍

𝑖=1 𝑚

൯ 𝜙 ^𝑖 (𝑥)𝜂_𝑖(𝑡

𝜂

_𝑖

0 = 1

𝑀

_𝑖^∗

𝜙

₁^𝑖

⋯ 𝜙

_𝑛^𝑖

[𝑀]

𝑤

₁

0 𝑤

₂

0

⋮

𝑤

_𝑛

(0 ) 𝑀

^∗

𝜂 0 = 𝜙

^𝑇

𝑀 𝑤(0 )

𝜙

^𝑇

𝑀 𝜙𝜂 0 = 𝜙

^𝑇

𝑀 𝑤(0 )

mxn nxn nxm mx1 mxn nxm nx1

𝑀_𝑚×𝑚^∗ ∶ 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙

Premultiply by ,

𝜙

^𝑇

𝑀

and find w from

Discrete System

- Motion Dependent Forces

Forces 𝐹_𝑖 may be dependent on position, velocity, acceleration after structure @ its nodes i, as well as time

Consider a general case

Consider an N degree of freedom system

𝐹

_𝑖

= 𝐹

_𝑖

(𝑤

₁

, 𝑤

₂

, ⋯ ሶ 𝑤

₁

, ሶ 𝑤

₂

, ⋯ ሷ 𝑤

₁

, ሷ 𝑤

₂

⋯ , 𝑡 )

𝐹

_𝑖

= ෍

𝑘=1 𝑁

𝑎

_𝑖𝑘

𝑤

_𝑘

+ 𝑐

_𝑖𝑘

𝑤

_𝑘

ሶ + 𝑒

_𝑖𝑘

𝑤

_𝑘

ሷ + 𝐹

_𝑖

𝑡

M 𝑤 + 𝐾 𝑤 = 𝑎 𝑤 + 𝑐 ሷ 𝑤 + 𝑒 ሶ 𝑤 + {𝐹 ሷ

_𝑖

𝑡 ሽ

Discrete System

Let

Can also write it as

𝑤

_𝑖

= ෍

𝑗 𝑛=3

𝜙

_𝑖^𝑗

𝜂

_𝑗

𝑡

𝑀

^∗

ሷ𝜂 + [𝑤

²

𝑀

^∗

]𝜂 = 𝜙

^𝑇

𝑎 𝜙𝜂 + 𝜙

^𝑇

𝑐 𝜙 ሶ𝜂 + 𝜙

^𝑇

𝑒 𝜙 ሷ𝜂 + 𝑄

The terms on the summation on the right-hand side couple (in general) the equations of motion. This is typical in aeroelastic problem.

[𝐴] [𝐶] [𝐸]

fully populated (in general)

𝑀

_𝑟^∗

ሷ𝜂

_𝑟

+ 𝑤

_𝑟²

𝑀

_𝑟^∗

𝜂

_𝑟

= ෍

𝑠=1 𝑚

𝐴

_𝑟𝑠

𝜂

_𝑠

+ 𝐶

_𝑟𝑠

𝜂

_𝑠

ሶ + 𝐸

_𝑟𝑠

𝜂

_𝑠

ሷ + 𝑄

_𝑟

not necessarily positive definite

Discrete System

- For proportional damping,

Then, due to orthogonality on [K] and [M]

C = α K + β[M ]

⋯ damping matrix is proportional to a linear combination of the mass and stiffness matrices

any value, constants

𝐶

_𝑟𝑠

= 0 when r ≠ 𝑠

𝑆𝑒𝑡 𝐶

_𝑟𝑟

= 2𝜍

_𝑟

𝜔

_𝑟

𝑀

_𝑟^∗

No coupling

Critical damping ratio: obtained from experiments or guess

ቊ𝑀

^𝑟∗

ሷ𝜂

_𝑟

+ 2𝜍

_𝑟

𝜔

_𝑟

𝜂

_𝑟

+ 𝜔

_𝑟²

𝜂

_𝑟

= 𝑄

_𝑟

𝑡

⋮

m set of uncoupled

equations

Continuous System

• At this point, a distinction between two main classes of approaches for approximating the solution of structural systems needs to be made.

• The two basic approaches are

1) modal methods: represent displacements by overall motion of the structure

2) discrete point methods: represent displacement by motion at

many discrete points distributed along the structures

Continuous System

• Consider a basic high-aspect ratio wing modeled as a cantilever beam for symmetric response

Partial differential equation for continuous beam

m ሷ 𝑤 − 𝑇𝑤

^{′ ′}

+ 𝐸𝐼𝑤

^{′′ ′′}

= 𝑓

_𝑧

𝑚 𝑥 ∶𝑚𝑎𝑠𝑠 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ (Τ 𝑘𝑔 𝑚)Τ 𝑤 𝑥, 𝑡 : 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑚 𝑇 ∶ 𝑎𝑥𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑁

𝐸𝐼 𝑥 : 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 (𝑁 ∙ 𝑚²) 𝑓_𝑧 ∶ 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑁 𝑚Τ 𝑓_𝑥 ∶ ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 ( Τ𝑁 𝑚)

• Pinned end

• Fixed end

𝑤 = 0 𝑀 = 𝐸𝐼𝑤^′′ = 0

• Free end

𝑤 = 0 𝑤^′ = 0

• Vertical spring

𝑀 = 𝐸𝐼𝑤^′′ = 0 𝑆 = 𝐸𝐼𝑤^{′′ ′} = 0

𝑀 = 𝐸𝐼𝑤^′′ = 0 𝑆 = 𝐸𝐼𝑤^{′′ ′} = 𝑘_𝑣𝑤 𝑥

𝑤, 𝑧

w(𝑥, 𝑡)

Response of a Uniform Cantilevered Beam

Same solution procedure as before

i) find solution to homogeneous equation

ii) then determine complete solution as expansion of homogeneous solution

𝑚 = 𝐸𝐼 = 𝑐𝑜𝑛𝑠𝑡.

𝑇 = 0

B. C. @𝑥 = 0 ቊ 𝑤 = 0 𝑤

^′

= 0

I. C. @𝑡 = 0 ቊ 𝑤 = 0

ሶ

𝑤 = 0

@𝑥 = 𝑙 ቊ𝑀 = 𝐸𝐼𝑤′′ = 0 𝑆 = 𝐸𝐼𝑤

^′′′

= 0

geometric B.C.

natural B.C.

(Rest I.C.’s) 𝑥

𝑤, 𝑧

𝑓(𝑥, 𝑡)

Response of a Uniform Cantilevered Beam

EIw

^′′′′

+ m ሷ 𝑤 = 0 ⋯ 1 let w x, t = ഥ 𝑤(𝑥)𝑒

^𝑖ω𝑡

൫EI ഥ 𝑤

^′′′′

− mω

²

𝑤)𝑒 ഥ

^𝑖ω𝑡

= 0 ⋯ 2 ഥ

𝑤

^′′′′

− mω

²

𝐸𝐼 𝑤 = 0 ഥ ⋯ 3

To solve, let ഥ 𝑤 = 𝑒

^𝑝𝑥

( 𝑠𝑖𝑛, 𝑐𝑜𝑠, 𝑠𝑖𝑛ℎ, 𝑐𝑜𝑠ℎ ) 𝑝

⁴

𝑒

^𝑝𝑥

− 𝑚ω

²

𝐸𝐼 𝑒

^𝑝𝑥

= 0 𝑛𝑜𝑛𝑡𝑟𝑖𝑣𝑖𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

4 𝑟𝑜𝑜𝑡𝑠

𝑝

⁴

= 𝑚ω

²

𝐸𝐼

p = λ, −λ, iλ, −iλ where λ

²

= 𝜔 𝑚

𝐸𝐼

Response of a Uniform Cantilevered Beam

Determine A, B, C, D from B.C.’s in matrix form

ഥ

𝑤 𝑥 = 𝐶

₁

𝑒

^λx

+ 𝐶

₂

𝑒

^−λx

+ 𝐶

₃

𝑒

^iλx

+ 𝐶

₄

𝑒

^−iλx

ഥ

𝑤 𝑥 = 𝐴𝑠𝑖𝑛ℎλx + Bcoshλx + Csinλx + Dcosλx

or

For a nontrivial solution,

Transcendental equation

Δ = 0

Δ = 2𝑐𝑜𝑠ℎλlcosλl + 𝑠𝑖𝑛

²

λl + 𝑐𝑜𝑠

²

λl + 𝑐𝑜𝑠ℎ

²

λl − 𝑠𝑖𝑛ℎ

²

λl = 0

cosλl = −1 𝑐𝑜𝑠ℎλl

= 1 = 1

Response of a Uniform Cantilevered Beam

many solutions possible

Ref. : Blevins “Formulas for Natural Frequency and Mode Shapes”

λl = 0.597π, 1.49π, 5

2 π, 7 2 π 3

2 π 0.5π

~ ~

𝜔

_𝑟

= λl

²

𝐸𝐼 𝑚𝑙

⁴

For eigenvectors (mode shapes), place

λl

into first three equations

ቇ ഥ

𝑤

_𝑟

𝑥 = 𝑐𝑜𝑠ℎλ

_𝑟

𝑥 − 𝑐𝑜𝑠λ

_𝑟

𝑥 − ( 𝑐𝑜𝑠ℎλ

_𝑟

𝑙 + 𝑐𝑜𝑠λ

_𝑟

𝑙

𝑠𝑖𝑛ℎλ

_𝑟

𝑙 + 𝑠𝑖𝑛λ

_𝑟

𝑙 )(𝑠𝑖𝑛ℎλ

_𝑟

𝑥 − 𝑠𝑖𝑛λ

_𝑟

𝑥

cosλl

−1 coshλl

Response of a Uniform Cantilevered Beam

w x, t = ෍

𝑟=1

∞

𝜙

_𝑟

(𝑥)𝑒

^{𝑖𝜔𝑟𝑡}

𝜔

₂

= 22 𝐸𝐼 𝑚𝑙

⁴

𝜔

₃

= 61.7 𝐸𝐼 𝑚𝑙

⁴

𝜔

₁

= 3.52 𝐸𝐼

𝑚𝑙

⁴

𝑟𝑎𝑑 𝑠 Τ

Orthogonality

Since each solution satisfies

w x, t = 𝜙

_𝑟

(𝑥)𝑒

^{𝑖𝜔𝑟𝑡}

−mω

_𝑟²

𝜙

_𝑟

+ 𝐸𝐼𝜙

_𝑟^{′′ ′′}

= 0 ⋯ (1 )

−mω

_𝑠²

𝜙

_𝑠

+ 𝐸𝐼𝜙

_𝑠^{′′ ′′}

= 0 ⋯ (2 )

Multiply (1) by

𝜙

_𝑠 and integrate

ω

_𝑟²

න

0 𝑙

𝜙

_𝑠

𝑚𝜙

_𝑟

𝑑𝑥 = න

0 𝑙

𝜙

_𝑠

𝐸𝐼𝜙

_𝑟^{′′ ′′}

𝑑𝑥 ⋯ 3

ω

_𝑠²

න

0 𝑙

𝜙

_𝑟

𝑚𝜙

_𝑠

𝑑𝑥 = න

0 𝑙

𝜙

_𝑟

(𝐸𝐼𝜙

_𝑠^′′

)′′𝑑𝑥 ⋯ 4

and (2) by

𝜙

_𝑟 and integrate

m ሷ 𝑤 + 𝐸𝐼𝑤

^{′′ ′′}

= 0

Orthogonality

Subtract (4) from (3), and integrate by parts

(ω

_𝑟²

− ω

_𝑠²

) න

0 𝑙

𝜙

_𝑟

𝑚𝜙

_𝑠

𝑑𝑥 = 𝜙

_𝑠

𝐸𝐼𝜙

_𝑟^{′′ ′}

|

₀^𝑙

− 𝜙

_𝑠

′𝐸𝐼𝜙

_𝑟^′′

|

₀^𝑙

+ න

0 𝑙

𝜙

_𝑠

𝐸𝐼𝜙

_𝑟^′′

𝑑𝑥

−𝜙

_𝑟

𝐸𝐼𝜙

_𝑠^{′′ ′}

|

₀^𝑙

+ 𝜙

_𝑟

′𝐸𝐼𝜙

_𝑠^′′

|

₀^𝑙

− න

0 𝑙

𝜙

_𝑟

′′𝐸𝐼𝜙

_𝑠^′′

𝑑𝑥

defection shear slope moment

Note that all the constant terms on RHS=0 because of BC’s for example :

• pinned → w = 0 ⇒ ϕ = 0

w′′ = 0 ⇒ ϕ′′ = 0

• fixed → w = 0 ⇒ ϕ = 0 w′ = 0 ⇒ ϕ′ = 0

• free → ϕ

^′′

= 0 and EIϕ

^{′′ ′}

= 0

Orthogonality

For

𝑟 ≠ 𝑠

, we have

න

0 𝑙

𝜙

_𝑟

𝑥 𝑚 𝑥 𝜙

_𝑠

𝑥 𝑑𝑥 = 0 න

0 𝑙

𝜙

_𝑟

𝑥 𝑚 𝑥 𝜙

_𝑠

𝑥 𝑑𝑥 = 𝛿

_𝑟𝑠

𝑀

_𝑟^∗

Also,

න

0 𝑙

𝜙

_𝑠

(𝐸𝐼𝜙

_𝑟^′′

)′′𝑑𝑥 = 𝛿

_𝑟𝑠

𝑀

_𝑟^∗

𝑤

_𝑟²

⟹ can transform to normal coordinates

Complete solution

Place (6) into (5) and integrate after multiplying with

𝜙

_𝑠

let

w x, t = σ

_𝑟=1^∞

𝜙

_𝑟

(𝑥)η

_𝑟

(𝑡) ⋯ (6) m ሷ 𝑤 + 𝐸𝐼𝑤

^{′′ ′′}

= 𝑓 𝑥, 𝑡 ⋯ (5 )

෍

𝑟=1

∞

ሷ𝜂 න

0 𝑙

𝑚𝜙

_𝑠

𝜙

_𝑟

𝑑𝑥 + ෍

𝑟=1

∞

𝜂

_𝑟

න

0 𝑙

𝜙

_𝑠

(𝐸𝐼𝜙

_𝑟^′′

)′′𝑑𝑥 = න

0 𝑙

𝜙

_𝑠

𝑓 𝑥, 𝑡 𝑑𝑥

because of orthogonality

𝑀

_𝑟

𝜂

_𝑟

ሷ + 𝑀

_𝑟

ω

_𝑟²

𝜂

_𝑟

= 𝑄

_𝑟

⋮ 𝑀

_𝑟

= න

0 𝑙

𝜙

_𝑟²

𝑥 𝑚 𝑥 𝑑𝑥 𝑄

_𝑟

= න

0 𝑙

𝜙

_𝑟

𝑥 𝑓(𝑥, 𝑡)𝑑𝑥

Complete solution

To find I.C.’s on

𝜂

_𝑟,

and

ሶ

𝑤 x, 0 = ෍

𝑟=1

∞

𝜙

_𝑟

𝑥 𝜂

_𝑟

ሶ (0 = ) 𝑤

₀

ሶ 𝑥

@ t = 0, w x, 0 = ෍

𝑟=1

∞

𝜙

_𝑟

𝑥 𝜂

_𝑟

(0 = 𝑤 )

₀

𝑥

Multiply by

𝑚𝜙

_𝑠

(𝑥)

and integrate

න

0 𝑙

𝑚 𝜙

_𝑠

𝑤

₀

𝑑𝑥 = ෍

𝑟=1

∞

𝜂

_𝑟

(0 න )

0 𝑙

𝑚𝜙

_𝑠

𝜙

_𝑟

𝑑𝑥 = 𝜂

_𝑠

(0)𝑀

_𝑠^∗

𝜂

_𝑟

0 = 1

𝑀

_𝑟^∗

න

0 𝑙

𝑚𝜙

_𝑟

𝑤

₀

𝑥 𝑑𝑥

ሶ

𝜂

_𝑟

0 = 1 𝑀

_𝑟^∗

න

0 𝑙

𝑚𝜙

_𝑟

𝑤 ሶ

₀

𝑥 𝑑𝑥

Rayleigh-Ritz Method

 Energy-based method

• Form of the solutions is assumed to be as :

w(x, t) ≈ ෍

𝑟=1 𝑁

𝛾 𝑡 𝑞

_𝑟

𝑡

assumed modes need to satisfy at least geometrical boundary conditions

assume w x, t = ෍

𝑖=1 𝑁

𝛾

_𝑖

𝑡 𝑞

_𝑖

𝑡

satisfy

𝑤 = 0 ቅ

𝑤

^′

= 0 @𝑥 = 0

𝑥 𝑓(𝑥, 𝑡)

𝛾

₁

𝛾

₂

𝑥

²

𝑙

²

𝑥

³

𝑙

³

Rayleigh-Ritz Method

T = 1 2 න

0 𝑙

𝑚 𝑥 ෍

𝑖=1 𝑀

𝛾

_𝑖

𝑞

_𝑖

ሶ ෍

𝑗=1 𝑀

𝛾

_𝑗

𝑞

_𝑗

ሶ 𝑑𝑥 = 1 2 ෍

𝑖=1 𝑀

෍

𝑗=1 𝑀

න

0 𝑙

𝑚𝛾

_𝑖

𝑥 𝛾

_𝑗

𝑥 𝑑𝑥 ሶ 𝑞

_𝑖

𝑞

_𝑗

ሶ

V = 1 2 න

0 𝑙

𝐸𝐼 𝑤

^{′′ 2}

𝑑𝑥 = 1 2 ෍

𝑖=1 𝑀

෍

𝑗=1 𝑀

න

0 𝑙

𝐸𝐼 𝑥 𝛾

_𝑖

′′ 𝑥 𝛾

_𝑗

′′ 𝑥 𝑑𝑥𝑞

_𝑖

𝑞

_𝑗

δW = න

0 𝑙

𝑓𝛿𝑤𝑑𝑥 = ෍

𝑖=1 𝑀

න

0 𝑙

𝑓(𝑥)𝛾

_𝑖

𝑑𝑥𝛿𝑞

_𝑖

𝑚

_𝑖𝑗^∗

𝑘

_𝑖𝑗^∗

Plug into Lagrange’s equations,

𝑑 𝑑𝑡

𝜕𝑇

𝜕𝑞

_𝑖

− 𝜕𝑇

𝜕𝑞

_𝑖

+ 𝜕𝑉

𝜕𝑞

_𝑖

= 𝑄

_𝑖

which gives

෍

𝑖=1 𝑀

𝑚

_𝑖𝑗^∗

𝑞

_𝑗

ሷ + ෍

𝑖=1 𝑀

𝑘

_𝑖𝑗^∗

𝑞

_𝑗

ሷ = 𝑄

_𝑖 coupled set of equations!

Rayleigh-Ritz Method

For a quick and “dirty” way to find the first natural frequency, assume only one mode shape,

𝑚

₁₁^∗

𝑞

₁

ሷ + 𝑘

₁₁^∗

𝑞

₁

= 𝑄

₁

Rayleigh quotient with

q = ത 𝑞𝑒

^𝑖ω𝑡

Clearly we can obtain higher modes by assuming more than one mode

ω

²

= ׬

₀^𝑙

𝐸𝐼 𝛾

₁^{′′ 2}

𝑑𝑥

׬

₀^𝑙

𝑚𝛾

₁²

𝑑𝑥

⋯ upper bound for the actual frequency

ω

_𝑟²

= 𝛾

_𝑟^𝑇

𝐾 𝛾

_𝑟

𝛾

_𝑟^𝑇

𝑀 𝛾

_𝑟

Galerkin’s Method

• Galerkin’s method applies to P.D.E. directly – residual method

Look at general beams

Domain

• Assumed modes must satisfy all the boundary conditions (geometric and natural ones)

න𝛾

_𝑗

𝑃. 𝐷. 𝐸. 𝑑𝑥 = 0

w x, t = ෍

𝑖=1 𝑁

𝛾

_𝑖

𝑡 𝑞

_𝑖

𝑡

for a pinned-pinned beam,

m ሷ 𝑤 + 𝐸𝐼𝑤

^{′′ ′′}

− (𝑇𝑤

^′

)′ = 𝑓 𝑥, 𝑡

൰ 𝛾

_𝑗

= sin( 𝑗𝜋𝑥

𝐿

If

𝛾

_𝑗 is on exact mode shape, P.D.E. would be satisfied exactly But if not → error

𝑓𝑜𝑟 𝑗 = 1,2, … , 𝑁

Galerkin’s Method

Now set

: Average error in PDE with respect to some weighting function ℎ_𝑖 𝑥 that minimize the error in the interval, usually take ℎ_𝑖 𝑥 = 𝛾_𝑖 𝑥

approx

E = m ሷ 𝑤 + 𝐸𝐼𝑤

_{𝑎𝑝𝑝𝑟𝑜𝑥}^{′′ ′′}

− 𝑇𝑤

_{𝑎𝑝𝑝𝑟𝑜𝑥}^{′ ′}

− f

න

0 𝑙

ℎ

_𝑖

𝑥 𝐸 𝑥 𝑑𝑥 = 0

෍

𝑗=1 𝑀

቉

ሷ

𝑞

_𝑗

[න

0 𝑙

𝛾

_𝑖

𝑥 𝑚(𝑥)𝛾

_𝑗

𝑥 𝑑𝑥 + ෍

𝑗=1 𝑀

න

0 𝑙

𝛾

_𝑖

𝐸𝐼𝛾

_𝑗^{′′ ′′}

𝑑𝑥 − න

0 𝑙

𝛾

_𝑖

𝑇𝛾

_𝑗^{′ ′}

𝑑𝑥

= න

0 𝑙

𝛾

_𝑖

𝑓 𝑥, 𝑡 𝑑𝑥

Different from Rayleigh-Ritz

Galerkin’s Method

⋯ coupled set of DE’s

(except when 𝛾_𝑗is natural mode shape)

𝑚

_𝑖𝑗

𝑞

_𝑗

ሷ + 𝑘

_𝑖𝑗

𝑞

_𝑗

= 𝑄

_𝑗

Used standard technique, let

q = ത 𝑞𝑒

^𝑖ω𝑡

𝐼𝜔

²

− 𝑚

⁻¹

[𝑘 ത ] 𝑞 = 0

Eigenvalues →

approximate natural frequencies

Eigenvectors →

approximate natural mode shapes

For M different weighting function

𝛾

₁

, 𝛾

₂

, …𝛾

_𝑀

,

we have M equations to find M unknowns 𝑞₁, 𝑞₂, …𝑞_𝑀 To find M unknowns 𝑞₁, 𝑞₂, …𝑞_𝑀 in matrix form

Galerkin’s Method

Note :

iv) The closer

𝛾

_𝑗

(𝑥)

is to

𝜙 𝑥 ,

the less the coupling i) more assumed modes

→

better approximation

𝜙

₁

𝑥 = 𝐴𝑐𝑜𝑠𝜆

₁

𝑥 + 𝐵𝑠𝑖𝑛𝜆

₁

𝑥 + 𝐶𝑐𝑜𝑠ℎ𝜆

₁

𝑥 + 𝐷𝑠𝑖𝑛ℎ𝜆

₁

𝑥

= 𝑎

₀

+ 𝑎

₁

𝑥 + 𝑎

₁

𝑥

²

+ 𝑎

₃

𝑥

³

+ ⋯

ii) more accurate assumed shapes

→

better approximation iii) If 𝛾_𝑗(𝑥) is natural mode shapes, system will be uncoupled

Galerkin : very powerful, turn PDE’s into ODE’s

very general, can also be used in nonlinear problem !!

v) If Rayleigh-Ritz assumed mode shapes satisfy both geometric and natural B.C.’s, two methods are identical

(can be shown by integration by parts)

𝑚 ሷ 𝑤 + 𝐸𝐼𝑤

^{′′ ′′}

+ 𝐹 𝑤

^𝑛

= 𝑓