· Mul$ple’Degree’of’Freedom’Systems’ (MDoF) • More than one coordinate is needed to describe the motion of systems with more than one degree of freedom

ARO406: Advanced Dynamics &

Vibrations Part 3

Slides Borrowed From Prof. Don Edberg

May 8, 2014 © Donald Edberg 2010-present. All rights reserved.

1

Mul$ple Degree of Freedom Systems (MDoF)

•  More than one coordinate is needed to describe the motion of systems with more than one degree of freedom

•  This does not apply to continuous systems, which are “continuous” and require an infinite number of DoFs

•  Does apply to “lumped-parameter” systems (created by breaking continuous systems into “chunks” and FEMs (finite element models)

•  Animations may be found at: http://www.acs.psu.edu/drussell/Demos/multi-‐dof/multi-‐dof.html http://www.acs.psu.edu/drussell/Demos/multi-‐dof-‐springs/multi-‐dof-‐springs.html http://www.youtube.com/watch?v=kvG7OrjBirI

5/8/14 © Donald Edberg 2010-present. All rights reserved.

2

2 DoF System (1 of 2) •  Consider the two-mass

system shown to right •  Needs two coordinates to

completely describe motion •  Apply Newton’s law to

mass 1 (+ upwards):


3

F1 −m1z1 − k1z1 − k2 z1 − z2( ) = 0Force in Spring 1

Inertial reaction

Force upwards

Force in Spring 2

2 DoF System (2 of 2)


4

•  Both mass matrix [M] and stiffness matrix [K] are symmetric.

•  Is there an EASIER way to derive the equations of motion?

YES!

For mass 2: F2 −m2z2 − k2 z2 − z1( ) = 0

In matrix form: m1 00 m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1

z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+

k1 + k2 −k2

−k2 k2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1

z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

F1

F2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

Mass matrix Displacement vector

Force vector

Stiffness matrix

Lagrange’s Equa$ons (1 of 5) [see Thorby 1.5.3, Ex. 2.1, 6.1.3]

•  Based on energy methods •  Write down T, the kinetic energy of the system

Generally includes terms like ½ mivi2 and ½ Iiωi

2 •  Write down U, the potential energy of the system

Generally includes terms like mgxi, ½ kixi2 and ½ kθiθi

2 •  Compute Lagrange’s equation for all degrees of freedom:

•  Here qi represent “generalized displacements.” These can uniquely model all possible motion.

•  Qi are generalized external forcing functions corresponding to qi. They represent work done by external forces and actual displacements


5

•  For structures (non-rotating)

•  Damping terms added after transforming usually, but if viscous damping exists in terms of the generalized coordinates, can introduce a “dissipation function” D which produces approximate terms for a single damper c, and Lagrange equation is (c.f. Thorby 6.4.1)

•  If no damping, can write Lagrangian L = T – U, and:


6

Lagrange’s Equa$ons (2 of 5)

•  Apply Lagrange’s equations to two-mass system: (6.1.3) •  Choose coordinates that are independent •  Generalized forces must do same work as actual forces


7


Lagrange’s Equa$ons (4 of 5) •  Calculation of derivatives of T and U


8

• 

•  Same result as from free body diagrams, except don't have to worry about getting signs and directions right! It happens automatically


9


Example: Pendulum on a Cart (1 of 3) •  Can see two DoFs, so use x and θ as general coordinates. •  Position of cart is x. •  Position & velocity of pendulum

mass:


10

x = x + Lsinθ,Lcosθ( )v = x + L θ cosθ,−L θ sinθ( )The kinetic energy T = KE of cart + KE of pendulum mass

T = 12 Mx

2 + 12 m vx

2 + vy2( )

= 12 Mx

2 + 12 m x + L θ cosθ( )2

+ −L θ sinθ( )2⎡⎣⎢

⎤⎦⎥

= 12 M +m( ) x2 +mLx θ cosθ + 1

2 mL2 θ 2

•  Potential energy: from string and pendulum rise

•  If generalized forces = 0, Lagrange’s equations are

•  For x:


11

Example: Pendulum on a Cart (2 of 3)

U = 12 kx

2 +mgL(1− cosθ )L = T −U = 1

2 (M +m) x2 +mLx θ cosθ + 12mL

2 θ 2 − 12 kx

2 −mgL(1− cosθ )

•  For θ:

•  Coupling between x and θ motions


12

Example: Pendulum on a Cart (3 of 3)

The beam is massless & rigid; two concentrated masses m1 & m2; k & c are connected to ground. Force F applied to m2.


13

Ex. 2.1 Thorby

The problem has one DoF, z. Let q = z be the generalized coordinate. The external force Q = F since work done is Fz = Qq.

T = 12m1

&z2!

"#$

%&2

+12m2 &z

2 =&z2

2 m14+m2

!

"#

$

%&

U =12k z2!

"#$

%&2

=kz2

8

D =12c &z2!

"#$

%&2

=c&z2

8

Ex. 2.1 Thorby (cont…)


14

ddt

∂T∂q"

#$

%

&'=

ddt

∂T∂ &q!

"#

$

%&=

ddt

m14+m2

!

"#

$

%& &z

!

"#

$

%&=

m14+m2

'

()

*

+,&&z

∂U∂q

=∂U∂q

=kz4

∂D∂ &q

=∂D∂&z

=cz4

Q = F

Lagrange’s equation gives

m14+m2

!

"#

$

%&&&z +

c1&z4+kz4= F

Vibration Reduction (Thorby Ch. 11)

We saw the washing machine example where flexible suspension reduces vibrations & loads.

This is a constant problem in aerospace: •  Reciprocating engines •  Helicopter •  General vibration due to turbulence & shocks

(AC & SC)


15

Vibration Reduction Methods •  Dynamic absorber: connect an auxiliary mass with a

spring, minimum damping. Absorbs energy at one frequency only.

•  Damped absorber: like dynamic absorber but damped. •  Centrifugal dynamic absorber: uses a self-tuning

pendulum coupled with rotation speed. •  Rotational inertial w/damping: “Lanchester” damper, or

“Houdaille” damper. •  Active control: use an actuator to apply equal/opposite

excitation (noise-cancelling headphones, µgravity isolation system on ISS) or higher harmonic control (on helis) or active flutter control /aeroelastic wing.


16

Isolation Used in Two Ways 1.  Protect sensitive items (µg experiments or SC on LV)

from high environments. 2.  Reduce disturbances applied to a structure, i.e. imbalance

from a rotating machine. Isolate from base environment. Recall:

T = displacement transmissibility

Works for disp., vel., accel.


17

zx= 1+ (2ζΩ)2

(1−Ω2 )2 + (2ζΩ)2= T

T = zx=zx=zx

Isolation (2 of 3)

Notes on isolation: 1. Always attenuation if Ω > √2 2. Always amplification if Ω < √2

BAD (large amplification) near Ω = 1 So the trade is: get high-frequency attenuation without too much low-frequency response.

3. Smaller ζ = higher high frequency attenuation, low frequency response.


18

Thorby Fig. 11.2

Isolation (3 of 3) •  This is dealt with by assuming sinusoidal inputs or

random input. –  If random, expressed as a power spectral density (PSD)

plot

•  The response PSD = (input PSD)×(T 2) graph •  We will talk about random excitation later


19

Dynamic Absorber (Thorby 11.2)


20

Assume an object, such as a sensitive optical table, mass m1 is excited by a force f(t) = F sin ωt.

The table’s motion is excessive, so add “absorber” ma and ka to reduce. How should the absorber be designed, i.e. what ma and ka ?

Courtesy: Cha

Dynamic Absorber (2 of 7)


21

m1 00 m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1z2

⎫⎬⎪

⎭⎪

⎧⎨⎪

⎩⎪+

k1 + k2 −k2−k2 k2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1z2

⎫⎬⎪

⎭⎪

⎧⎨⎪

⎩⎪=

F0 sinωt0

⎫⎬⎪

⎭⎪

⎧⎨⎪

⎩⎪

Assume steady response z1

z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

z1

z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪sinωt,

z1

z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=ω 2 z1

z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

Note: different notation from previous slide.



22

−ω 2 m1 00 m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥+

k1 + k2 −k2−k2 k2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎫⎬⎪

⎭⎪

z1z2

⎫⎬⎪

⎭⎪

⎧⎨⎪

⎩⎪=

F 0

0

⎫⎬⎪

⎭⎪

⎧⎨⎪

⎩⎪

⎧⎨⎪

⎩⎪EoM becomes

Define ω12 = k1

m1

, ω22 = k2

m2

Non-dimensional: Ω1 =ωω1

, Ω2 =ωω2

Mass ratio µ = m2

m1

zs =F0

k1

= Static displacement of m1 with F0 applied

Then rearrange to get z1, the response of m1:



23

z1k1F0

= z1zs

= 1−Ω22

1−Ω12( ) 1−Ω2

2( )− µΩ12

Note z1 = 0, if Ω2 = 1 or ω = ω2: if excitation freq. = natural freq. of absorber m2, the displacement of m1 is ZERO! What happens to little mass m2?

Use z2k1

F0

= z1

zs= 1

1−Ω12( ) 1−Ω2

2( )− µΩ12

If ω =ω2, z2k1

F0

= −1µΩ1

2 = −Mω12

mω 2 = −Mω12

mω22 = − k1

k2

So z2 = –F0/k2, potentially much larger displacement


Non-dimensional response of the main system when the dynamic absorber is tuned to the same frequency as the main system. Mass ratio = 0.2. Thorby Fig. 11.6


24

Dynamic Absorber (6 of 7) The force on m exactly balances the force

on m1 at Ω2 =1.0. What happens off of Ω2 = 1.0? •  Larger displacements outside of Ω1 = 1.0. •  These are where the natural frequencies

are. In this case, 0.8 and 1.25 ω2.


25



26

•  Usually want m as small as possible—natural frequencies. move closer to absorber freq. and attenuation band is reduced too.

•  Absorber does not have to be tuned to system frequency

•  Best attenuation if absorber natural frequency = excitation frequency.

•  These are useful for machine operating at a single frequency — most helicopters.

•  Animations at: http://www.acs.psu.edu/drussell/Demos/absorber/DynamicAbsorber.html

Matrix Methods (Thorby 6.2) Larger Multi-DoF (MDoF) systems are analyzed by: 1)  Converting individual elements from their individual

mass + stiffness into global (system) coordinates—done by FEM program.

2)  Transform from global to modal coordinates, or “physical” to “modal” coordinates.

Look at #1 for mass data. Assume there is a relationship between individual masses and global displacement.


27

r}{ = Xm[ ] z}{ (6.11)

vector of displacement of individual masses

Global coordinate displacements

Matrix Methods (2 of 3) Note The system KE is the sum of individual KEs: Where is a diagonal matrix of masses {ṙ} is a column vector of the individual mass velocity If we substitute 6.12 into 6.14 and note (6.12)ᵀ


28

&r}{ = Xm[ ] &z}{ (6.12)

T = 12m1&r1

2 +12m2 &r2

2 +........+ 12mn &rn

2 =12

&r{ }T m!" #

$ &r{ } (6.14)

[ ]m

→ r{ }T = z{ }T Xm[ ]T

Then T = 12z{ }T Xm[ ]T m⎡⎣ ⎤⎦ Xm[ ] z{ } where all these are vectors

Matrix Methods (3 of 3)


29

where T = 12z{ }T M[ ] z{ } M[ ] = Xm[ ]T m⎡⎣ ⎤⎦ Xm[ ]

is the mass matrix in global coordinates

Similar development for stiffness, let si compression or tension. (6.19) Assume compression extensions {s} = [Xs]{z} (6.20) Then where

U = 12 k1s1

2 + 12 k2s2

2 + .... + 12 kn

2 = 12 s{ }T k⎡⎣ ⎤⎦ s{ }

U = 12 z{ }T K[ ] z{ } K[ ] = Xs{ }TT k⎡⎣ ⎤⎦ Xs{ }

stiffness matrix of system in global coordinates.

Thorby Example 6.1 (1 of 6)


30

The equipment box is isolated by two springs, k1 & k2; empty box has mass m & CG at point G, and inertia I. Three equipment boxes, treated as point masses m1, m2, and m3 are attached to the box. The box motion is defined by vertical translation z and rotation θ about point C, the box’s CG. The point G is the CG of the entire system. Derive equations of motion for the assembled box in terms of z & θ, assume zero damping.

We want solutions of , or eqn. of

M[ ] X{ }+ K[ ] X}{ = 0

X}{ = zθ

⎫⎬⎪

⎭⎪

⎧⎨⎪

⎩⎪Where

[M] and [K] are 2 × 2

Thorby Example 6.1 (2 of 6) We start with a mass matrix [m] that is 5 x 5 (four masses & 1

inertia), and have to create a global mass matrix, [M], for the two degrees of freedom, z & θ, that is assembled from [m] and a transformation matrix [Xm], which is 5 x 2.

[M], (2x2), comes from equation 6.17:


31

M[ ] = Xm[ ]T m[ ] Xm[ ]

where m[ ] =

m 0 0 0 00 I 0 0 00 0 m1 0 00 0 0 m2 00 0 0 0 m3

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

Thorby Example 6.1 (3 of 6)


32

•  There are no non-diagonal terms in [m], since m and I are about G (CG).

•  {Xm} is defined by (6.11) and transforms the five displacements of the mass elements – rm, rI, rm1, rm2, rm3 – into the two global coordinates, z & θ. Hence it must be 5 × 2.

Thorby Example 6.1 (4 of 6) Transformation matrix [Xm]: •  Its 1st column,

{rm}1, is obtained from the displacements obtained when z moves one unit upwards.

•  The 2nd column {rmi}2 is obtained when θ rotates 1 unit CW. May 8, 2014 © Donald Edberg 2010-present. All

rights reserved. 33

rmrIrm1rm2rm3

⎧

⎨

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎫

⎬

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪1

=

10111

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

}{

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

=

cba

f

rm

1

2

Xm}{ = rm{ }1 rm{ }2{ } =1 − f0 11 a1 b1 −c

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

Disp. has no rotation effect

Example 6.1 (5 of 6)


34

[ ]

[ ] ⎥⎦

⎤⎢⎣

⎡

++++−−+

−−++++=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

32

22

122

321

321321

3

2

1

111

101

00000000000000000000

111

101

mcmbmamfIfmcmbmamfmcmbmammmmm

M

cba

f

mm

mI

m

cba

f

M

T

Units

⎥⎦

⎤⎢⎣

⎡2mLmL

mLm

The global mass matrix is M[ ] = Xm{ }T m⎡⎣ ⎤⎦ Xm{ }

Example 6.1 (6 of 6) Stiffness matrix: where and {Xs} = matrix of spring deformations. Look at spring extension s1 & s2: Find from setting z = 1, then θ = 1. Now Equations of motion are


35

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−=

⎭⎬⎫

⎩⎨⎧

θ

zed

ss

11

2

1

[ ] ⎥⎦

⎤⎢⎣

⎡

+−

−+=⎥

⎦

⎤⎢⎣

⎡

−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−=

22

12

21

2121

2

1

11

00

11

kekdekdkekdkkk

ed

kk

ed

KT

02

21

221

2121

32

22

122

321

321321 =⎭⎬⎫

⎩⎨⎧

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡

+−

−++

⎪⎭

⎪⎬⎫

⎥⎦

⎤⎢⎣

⎡

++++−−+

−−++++••

••

θθ

zkekdekdk

ekdkkkzmcmbmamfIfmcmbmam

fmcmbmammmmm

k[ ] = Xs{ }T k!" #$ Xs{ } [ ] ⎥

⎦

⎤⎢⎣

⎡=

2

1

00k

kk

Modal Coordinates (Thorby 6.22) Can represent motion of a system by physical coordinates by

“modal” coordinates. A mode is a set of displacements for the entire structure. “Normal modes” are special modes that make the mass & stiffness

matrices DIAGONAL—that means motion of each mode can be treated independently of other modes.

Each mode can then be treated as a single DoF system. This is easier mathematically.

Start with a set of equations in global coordinates: We want a form


36

M[ ] q{ }+ K[ ] q}{ = Q}{

M[ ] z}{ + K[ ] z}{ = F}{

Modal Coordinates (2 of 8)


37

First, we transform global coordinates z to model coordinates q: •  If the number of qs is the same as the number of zs, [X] is

square and no information is lost, true for smaller systems. •  Usually number of qs is << number of zs, and we assume that

high frequency information is lost.

z{ } = X[ ] q{ } and z}{ T = q}{ T X[ ]T

z{ } = X[ ] q{ }z{ } = X[ ] q{ }

Modal Coordinates (3 of 8)


38

Mass matrix transformation Where is the modal mass matrix: Stiffness matrix is done the same way: External forces: we transform F to Q using virtual work. •  The work W done by the external forces F during virtual

displacement Ẑ is W = ẐT F. •  If Q is the set of external modal forces providing the same

loading as F, and are the virtual modal displacements.

T = 12z{ }T M[ ] z{ } = 1

2q{ }T X[ ]T M[ ] X[ ] q{ } = 1

2q{ }T M[ ] q{ }

[ ]M M[ ] = X[ ]T M[ ] X[ ]K[ ] = X[ ]T K[ ] X[ ]

q

Modal Coordinates (4 of 8) Then Therefore, So Q = XT F Can use these relations to translate from physical to modal

coordinates. We said before, a special set of coordinates will diagonalize

[M] & [K]. This happens if the columns of the transformation matrix X are the “eigenvectors” or natural mode shapes of the system.

May 8, 2014 © Donald Edberg 2010-present. All

rights reserved. 39

zTF = qTQ

Since z = Xq, z = Xq and zT = qTXT

QqFXqW TTT ˆˆ ==

Modal Coordinates (5 of 8) Eigenvectors means the characteristic motion of the system,

and eigenvalues are the characteristic frequencies of vibration.

Eigenvectors and eigenvalues can be real or complex: •  If no damping, they are both real. •  If proportional damping, eigenvalues are complex but

vectors are real. •  If damping is small, they’re “nearly” real (Imaginary parts

<< Real parts) and work well for many structures. •  If gyroscopic coupling (i.e., turntable), get complex modes.


40

Modal Coordinates (6 of 8) How to find them? Start with any physical system. We assume a solution,


41

Mz + Kz = 0

M =

m11 m12 . . .m21 m22 . . .. . . . .. . . . .. . . . .

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

,K =

k11 k12 . . .k21 k22 . . .. . . . .. . . . .. . . . .

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

z = zeiωt → &&z = −ω 2z

So −ω 2 M[ ] z{ }+ K[ ] z{ } = 0

Assuming z ≠ 0, K[ ]−ω 2 M[ ] = 0.

Modal Coordinates (7 of 8) If n is the number of DoFs, there are n values of ω2 that solve

this problem. The normal form is [K] – λ[M] = 0 and numerical solvers

exist to do this problem. One way is to solve the determinant |K – λM| = 0 for values of λn (n-th order polynomial solver). λn are known as “eigenvalues” or characteristic values.

The eigenvectors or mode shapes are found by substituting the eigenvalues back into the characteristic equation, one at a time. Since they are homogenous, they can be multiplied by any constant and still work.


42

Modal Coordinates (8 of 8) There are two ways to “normalize” or scale eigenvectors: 1)  Set largest component in each column = 1.0, useful for

looking at relative deflections. 2)  Scale elements so that XTMX is the unit diagonal matrix.

These are called “orthonormal modes.” In these cases, the modal stiffness is also diagonal, and the

physical system is transformed into a set of completely independent 1-DoF equations.


43

Thorby Example 6.3 m1 = 1 kg, m2 = 2 kg k1 = 10 N/m, k2 = 10 N/m Find eigenvalues and vectors. Scale so that the largest element = 1.0.


44

m1 00 m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+

k1 + k2 −k2−k2 k2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

F1F2

⎫⎬⎪

⎭⎪

⎧⎨⎪

⎩⎪

1 00 2

⎡

⎣⎢

⎤

⎦⎥z1z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+ 20 −10

−10 10⎡

⎣⎢

⎤

⎦⎥

z1z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 0

0⎫⎬⎭

⎧⎨⎪

⎩⎪

20 −10−10 10

⎡

⎣⎢

⎤

⎦⎥ − λ

1 00 2

⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

z1z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 0

0⎧⎨⎩

⎫⎬⎭

Recall

In the form of K – λM = 0,



45

20 − λ −10−10 10 − 2λ

⎡

⎣⎢

⎤

⎦⎥

z1z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 0{ } (E)

20 − λ −10−10 10 − 2λ

= 20 − λ( ) 20 − λ( )−100 = 0

λ 2 − 25λ + 50 = 0 ⇒ λ1,2 = 2.1922, 22.807

ω1 = λ1 =1.480 rad/s

ω2 = λ2 = 4.775 rad/s

So determinant = 0:

Now, find eigenvector by substituting back into (E), one at the time. Note, we can find the ratio . Can use either row of (E).

2

1

zz



46

20 − λ( )z1 −10z2 = 0 or −10z1 + 10 − 2λ( )z2 = 0z1z2

= 1020 − λ

or 10 − 2λ10

For λ1 = 2.1922,z1z2

⎛⎝⎜

⎞⎠⎟1

= 0.5615. Set z2 =1, z1 = 0.5615

For λ2 = 22.807,z1z2

⎛⎝⎜

⎞⎠⎟ 2

= −3.56. Set z1 =1, z2 = −0.2807

So ϕ{ }1 =0.56151.0

⎧⎨⎩

⎫⎬⎭

and ϕ{ }2 =1.0

−0.2807⎧⎨⎩

⎫⎬⎭

{ϕ}1 and {ϕ}2 are the eigenvectors, or “normal modes.”



47

Transform Equations of Motion: The transformation matrix [X] is formed using the eigenvectors from the original equations as columns.

X[ ] = ϕ{ }1 ϕ{ }2⎡⎣⎢

⎤⎦⎥ =

0.5615 1.01.0 −0.2807

⎡

⎣⎢

⎤

⎦⎥

The modal mass matrix is M[ ] = X[ ]T M[ ] X[ ]

M[ ] = 0.5615 1.01.0 −0.2807

⎡

⎣⎢

⎤

⎦⎥

T1 00 2

⎡

⎣⎢

⎤

⎦⎥

0.5615 1.01.0 −0.2807

⎡

⎣⎢

⎤

⎦⎥ =

2.315 00 1.157

⎡

⎣⎢

⎤

⎦⎥

The modal stiffness matrix is K[ ] = X[ ]T K[ ] X[ ]

K[ ] = 0.5615 1.01.0 −0.2807

⎡

⎣⎢

⎤

⎦⎥

T20 −10−10 20

⎡

⎣⎢

⎤

⎦⎥

0.5615 1.01.0 −0.2807

⎡

⎣⎢

⎤

⎦⎥ =

5.075 00 26.40

⎡

⎣⎢

⎤

⎦⎥



48

M[ ] q{ }+ K[ ] q{ } = Q{ }2.315 00 1.157

⎡

⎣⎢

⎤

⎦⎥q1q2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+ 5.075 0

0 26.40⎡

⎣⎢

⎤

⎦⎥

q1q2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

Q1Q2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

The complete equations in normal coordinates

Both mass & stiffness matrices are DIAGONAL—no cross coupling. m11q1 + k11q1 =Q1 & m22q2 + k22q2 =Q2

Note : k11m11

=ω12 & k22

m22

=ω22

Note that magnitudes of generalized of modal mass & stiffness are arbitrary since they depend on how eigenvectors are SCALED.



49

Sometimes eigenvectors are scaled so that [M] is equal to the identity matrix.

M[ ] = m11 00 m22

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1 0

0 1⎡

⎣⎢

⎤

⎦⎥ = I[ ]

K[ ] = k11 00 k22

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

λ1 00 λ2

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

ω12 0

0 ω22

⎡

⎣⎢⎢

⎤

⎦⎥⎥

In this case, eigenvectors are derived as “weighted normal” or “orthonormal.” Stiffness becomes:



50

Ways to do scaling: If already have mass and stiffness modal matrices, rescale by multiplying each eigenvector {φ}i by a factor αi where

iii m

1=α

In this example,α1 =1

2.315, α2 =

11.157

ϕ{ }1

new=

12.315

0.56151.0

!"#

$%&= 0.3690

0.6572

!"#

$%&

, ϕ{ }2

new=

11.157

= 1.0−0.2807

!"#

$%&= −0.9294

0.2609

!"#

$%&

Then X[ ]new = ϕ{ }1ϕ{ }2

()*

+,-=

0.3690 −0.92940.6572 0.2609

(

)*

+

,-

Can show M[ ] = X[ ]T M[ ] X[ ] = 1 00 1

(

)*

+

,-

and K[ ] = X[ ]T K[ ] X[ ] = 2.192 00 22.807

(

)*

+

,-=

ω12 0

0 ω22

(

)

**

+

,

--



51

Another way: The generalized mass for mode i, mii, in the modal equations is mii = {Φ}i

T[M]{Φ}i where{Φ}i can be eigenvector in any arbitrary form and [M] is the original mass matrix.

If α i =1mii

, thenα i =1

ϕ{ }iTM[ ] ϕ{ }i

For this example: α1 = 0.6572, α2 = 0.2609

ϕ{ }1= 0.5615

1.0⎧⎨⎩

⎫⎬⎭, ϕ{ }2

= −3.5611.0

⎧⎨⎩

⎫⎬⎭

and the orthonormal eigenvectors are:

ϕ{ }1 =α1 ϕ{ }1= 0.6572 0.5615

1.0⎧⎨⎩

⎫⎬⎭= 0.369

0.6572⎧⎨⎩

⎫⎬⎭

ϕ{ }2 =α2 ϕ{ }2= 0.2609 −3.561

1.0⎧⎨⎩

⎫⎬⎭= −0.9294

0.2609⎧⎨⎩

⎫⎬⎭

Same as before.

Scaling Methods


52

These methods may be extended to undamped systems of any size. The model system always has diagonal [M] and [K] matrices and the system becomes a series of independent 1-DoF equations:

m11&&q1 + k11q1 =Q1m22&&q2 + k22q2 =Q2

M

mnn&&qn + knnqn =Qn

Then, calculating the response of a large multi-DoF system relatively simple, where we find the response of each SDoF and use superposition to sum the responses. This is known as “normal mode summation” method.

Summary—Using Normal Modes 1.  Equations are vastly simplified, mass and stiffness

matrices are diagonal. 2.  Response = Σ normal mode responses if damping small

(often). 3.  Normal modes have known natural frequencies, and we

can omit those with natural frequencies above the excitation frequencies è reduces # of equations.

4.  Can validate a structural model by comparing calculated modes with measured modes.


53

Orthogonality Properties of Eigenvectors •  The EoM can be written {-λ[M]+[K]}{z}= 0 where λ = ω2

and are the roots of the characteristic equation. For one eigenvalue λi and corresponding eigenvector , it becomes


54

{ }iz

K[ ] z{ }i = λi M[ ] z{ }i 6.49( )and for another = j, K[ ] z{ } j = λ j M[ ] z{ } j 6.50( )

premult. (6.49) by z{ } jT : z{ } j

T K[ ] z{ }i = λi z{ } jT M[ ] z{ }i

postmult. (6.50) by z{ }i : z{ }iT K[ ] z{ } j = λ j z{ }i

T M[ ] z{ } j

•  So any two eigenvector are orthogonal with respect to mass matrix.

•  Can show that they are also orthogonal with respect to stiffness matrix. The procedure is called a “cross-orthogonality” check or “XOR.”


55

Orthogonality Properties of Eigenvectors (Cont.)

Subtract to get 0 = λi − λ j( ) z{ } jT M[ ] z{ }i

If λi ≠ λ j, then z{ } jT M[ ] z{ }i = 0

Damping in Multi-DoF System (Thorby 6.4) Use any method to get dif. eqs. i.e.

Lagrange


56

D =12c1&z1

2 +12c2 ( &z2 − &z1)

2

In matrix form

m1 00 m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1z1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+

c1 + c2 −c2−c2 c2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1z1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+

k1 + k2 −k2−k2 k2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1z2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

F1F2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪General form is:

m11 m21

m21 m22

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

z1

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪+

c11 c21 c21 c22

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

z1

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪+

k11 k21 k21 k22

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

z1

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪=

F1

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

Damping in Multi-DoF System (2 of 4)


57

M[ ] z{ }+ c[ ] z{ }+ k[ ] z{ } = F{ }M[ ], c[ ]& k[ ] are symmetric & square

Can solve one of two ways: •  Normal mode summation—preferred because it’s completely analytic, but can have issues with damping matrix •  Direct solution of global eqs. We’ve seen that in normal coordinates with no damping, mass & stiffness matrices are diagonal. If there is damping, the damping matrix [C] does not always become diagonal, so modes are coupled.

Damping in Multi-DoF System (3 of 4) Can use complex eigenvalues & vectors, which diagonalize all

those. But this is avoided when possible because it’s more “complex.”

Cannot avoid complex eigenvalues when dealing with •  Rotating systems where centrifugal & Coriolis forces couple

the modes •  Flutter analysis where damping terms arise from

aerodynamics •  Vehicle suspensions that are heavily damped


58

Damping in Multi-DoF System (4 of 4)

Generally we use real eigenvalues if we can, justified by: 1. Damping is generally very small, and undamped modes

area good approximation to actual modes. 2. Damping is often “proportional” and will diagonalize, but

eigenvalues are complex. Using measured damping

Usually we do normal modes ignoring damping, then add a diagonal damping matrix based on measurements of structure (if available) or similar structures (if not).

Want to underestimate damping, since lower damping increases response (more conservative)


59

Thorby Example 6.4 CONVAIR damping schedule? A structure has analytical & tested modes at f1 = 10.1 Hz, f2 =

17.8 Hz, f3 = 23.5 Hz Measured damping ratios were ζ1 = 0.018, ζ2 = 0.027, ζ3 =

0.035 Write the EOMs, assuming modes were scaled to be

orthonormal (mass matrix is unit matrix) Recall: May 8, 2014 © Donald Edberg 2010-present. All

rights reserved. 60

miiq + ci q + kiiqi =Qi for i =1, 2, 3mii =1

kii = miiω i2 = mii (2π fi )

2

cii = 2ζ iimiiω i = 4πζ i fimii



61

mode # fi ζ i kii cii1 10.1 0.018 4027 2.282 17.8 0.027 12508 6.043 25.3 0.035 25269 11.13

EoMs:

1 0 00 1 00 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

q1

q2

q2

⎧

⎨⎪

⎩⎪⎪

⎫

⎬⎪

⎭⎪⎪

+2.28 0 0

0 6.04 00 0 11.13

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

q1

q2

q2

⎧

⎨⎪

⎩⎪⎪

⎫

⎬⎪

⎭⎪⎪

+4027 0 0

0 12508 00 0 25269

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

q1

q2

q2

⎧

⎨⎪

⎩⎪⎪

⎫

⎬⎪

⎭⎪⎪

=Q1

Q2

Q3

⎧

⎨⎪

⎩⎪⎪

⎫

⎬⎪

⎭⎪⎪

Make a table

Response of Multi-DoF systems by Normal Node Summation (6.5)

We have a series of equations: Use the SDOF response method to solve each equation. And

then sum the results: this process is called ‘normal mode summation’

In larger systems, the usual practice is to omit modes that have frequencies well above any excitation frequency


62

miiq i+cii qi + kiiqi =Qi (i =1,..., 4)

Response of M-DoF systems by Normal Node Summation 6.5

Procedure: 1.  Convert applied forces F (individual time histories

applied to grid points of system) in modal forces 2.  Calculate modal responses from the system modal eqs.

Apply to each SDoF eqn. 3. Convert modal response back to actual response zi:


63

iQ

Q{ }= X[ ]T F{ }Modal matrix

z{ }= x[ ] q{ }

Example 6.5 (Thorby p. 148) Massless beam with two concentrated masses m1 & m2. L = 4 m, EI = 2E6 Nm2, m1 = 10 kg, m2 = 8 kg Use ζ = 0.02 for both normal modes. Find system normal modes, check orthogonality, write EoMs

in normal coordinates, calculate time history where F1 = 1000 H(t) , F2 = 0. Plot z(t).


64

Example 6.5 (2 of 13) Flexibility matrix is in appendix, for a uniform beam. (note flexibility matrix = inverse stiffness) System EoMs: Undamped system EoMs:

May 8, 2014 © Donald Edberg 2010-present. All

rights reserved. 65

K[ ]−1

= L3

EI

124

548

548

13

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

[M ] z{ }+ [K ]{z} = {0}

[M ] z{ }+ C[ ] z{ }+ [K ]{z} = {0}

Example 6.5 (3 of 13) With Since we know K-1, premultiply:


66

K −ω 2M( )zi = 0K −1 K −ω 2M( )zi = 0I −ω 2K −1M( )zi = 0(K −1M − 1

ω 2 I )zi = 0

zi = zieiωt,



67

M =m1 00 m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 10 0

0 8⎡

⎣⎢

⎤

⎦⎥, K −1 = L3

48EI2 55 16

⎡

⎣⎢

⎤

⎦⎥

Subs in to get Rewrite as Λ is defined as Characteristic equation is determinant of above equation = 0: |[…]-Λ […]| = 0

L3

48EI2 55 16

⎡

⎣⎢

⎤

⎦⎥10 00 8

⎡

⎣⎢

⎤

⎦⎥ −

1ω 2

1 00 1

⎡

⎣⎢

⎤

⎦⎥

⎛

⎝⎜⎞

⎠⎟Zi = 0

20 4050 128

⎡

⎣⎢

⎤

⎦⎥ − Λ

1 00 1

⎡

⎣⎢

⎤

⎦⎥

⎛

⎝⎜⎞

⎠⎟z{ } = 0

Λ = 48EIL3ω 2 = 1.5E6 N m

ω 2

Using definition of Λ, ω1 = 102.02, ω2 = 621.315 To get amplitude ratios, subtract back into characteristic equation. Often use superscript in parenthesis to show the mode number.



68

20 − Λ 4050 128− Λ

= (20 − Λ)(128− Λ)− 2000 = 0

Λ2 −148Λ + 560 = 0

Λ12=148± 1482 − 4(560)

2= 74 ± 70.114 = 144.11

3.885

{ }czz

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡

Λ−

Λ−

2

1

1

1

128504020



69

From Line 1, 20 − Λ1( )z11( ) + 40z2

1( ) = 0

→ z2

z1

⎛⎝⎜

⎞⎠⎟

(1)

= 20 − Λ1

−40= 20 −144.11

−40= 3.103

From Line 2, 50z1(1) + (128− Λ1)z2

(1) = 0

z2

z1

⎛⎝⎜

⎞⎠⎟

(1)

= −50128− Λ1

= −50128−144.11

= 3.103, so z1

z2

⎛⎝⎜

⎞⎠⎟

(1)

= 0.3222

Use either line to get second ratio: z1

z2

⎛⎝⎜

⎞⎠⎟

(2)

= −2.482

Then, the two modes are: ϕ1 =0.3222

1.0⎛⎝⎜

⎞⎠⎟

, ϕ2 =−2.482

1.0⎧⎨⎩

⎫⎬⎭

•  These are mode shapes. •  Note:

mode 1 has 1 – 1 = 0 nodes; mode 2 has 2 – 1 = 1 nodes, etc.

•  Scale for orthonormality: multiply by αi derived earlier:



70

α i = ϕ{ }iT M[ ] ϕ{ }i⎡⎣ ⎤⎦

−12NODE

Example 6.5 (8 of 13) For mode 1: For mode 2: Orthonormal

eigenvectors: Modal matrix May 8, 2014 © Donald Edberg 2010-present. All

rights reserved. 71

α1 =z1z2

⎧⎨⎩

⎫⎬⎭1

T m1 00 m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

z1z2

⎧⎨⎩

⎫⎬⎭1

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−12

=0.32221

⎛⎝⎜

⎞⎠⎟

T10 00 8

⎡

⎣⎢

⎤

⎦⎥0.3221

⎧⎨⎩

⎫⎬⎭

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−12

= 0.3326

α2 =−2.4821

⎛⎝⎜

⎞⎠⎟

T10 00 8

⎡

⎣⎢

⎤

⎦⎥−2.4821

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−12

= 0.1198

ϕ1( ) =α1 ϕ 1( ) = 0.3326 0.32221

⎛⎝⎜

⎞⎠⎟=0.10710.3326

⎛⎝⎜

⎞⎠⎟

ϕ2( ) =α2 ϕ 2( ) = 0.1198 −2.48221

⎛⎝⎜

⎞⎠⎟=

−0.29750.1198

⎛⎝⎜

⎞⎠⎟

x[ ] = ϕ1( ) ϕ2( )⎡⎣ ⎤⎦ =0.1071 −0.29750.3326 0.1198

⎡

⎣⎢

⎤

⎦⎥

Example 6.5 (9 of 13) Note: normalized mass matrix = {X}T [M] {X} = Undamped normal equations are Damping is assumed to be such that there is no coupling. So

damping matrix is diagonal, terms are: May 8, 2014 © Donald Edberg 2010-present. All

rights reserved. 72

= 0.1071 0.3326−0.2975 0.1198

⎡

⎣⎢

⎤

⎦⎥10 00 8

⎡

⎣⎢

⎤

⎦⎥

0.1071 −0.29750.3326 0.1198

⎡

⎣⎢

⎤

⎦⎥ =

1 00 1

⎡

⎣⎢

⎤

⎦⎥ = I[ ]

1 00 1

⎡

⎣⎢

⎤

⎦⎥q1q2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪+

ω12 0

o ω22

⎡

⎣⎢⎢

⎤

⎦⎥⎥+

q1q2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

Q1Q2

⎛⎝⎜

⎞⎠⎟

cii = 2ζ iω imii (note that mii = 1 in this case)



73

C[ ] = 2ζ1ω1 00 2ζ 2ω2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 4.08 0

0 24.85⎡

⎣⎢

⎤

⎦⎥

ω12 =10408, ω2

2 = 386020

Now, we need to convert the modal displacements (q) to global displacements (z) using the transformation matrix [X]: And external load F to model forces Q

z( ) = X[ ] q( ) or z1z2

⎛⎝⎜

⎞⎠⎟= 0.1071 −0.2975

0.3326 0.1198⎡

⎣⎢

⎤

⎦⎥q1q2

⎛⎝⎜

⎞⎠⎟

Q( ) = X[ ]T F( )Q1Q2

⎛⎝⎜

⎞⎠⎟= 0.1071 0.326

−0.2975 0.1198⎡

⎣⎢

⎤

⎦⎥F1F2

⎛⎝⎜

⎞⎠⎟⇒

Q1 = 0.1071F1 + 0.3326F2Q2 = −0.2975F1 + 0.1198F 2



74

q1 + 2ζ1ω1 q1 +ω12q1 =Q1

q2 + 2ζ 2ω2 q2 +ω22q2 =Q2

is a 1 kN step function. So Response of 1-DoF system to a step input of magnitude a is

1F F1 =1000 H (t)F2 = 0

Q1 = 0.1071×1000 H (t), Q2 = −0.2975×1000 H (t)

z = amωn

2 1− e−ζωnt cosω dt +

ζ1−ζ 2

sinω dt⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

We now have two uncoupled equations:



75

Recall the definition of damped natural frequency: ω d =ωn 1−ζ 2

So: q1 =107.1ω1

2 1− e−ζ1ω1t (cosω d1 t +ζ1

1−ζ12

sinω2dt⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

q2 =−297.5ω2

2 1− e−ζ2ω2t (cosω d2 t +ζ 2

1−ζ 22

sinω2dt⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

The physical coordinate z1 is: z1 = 0.1071q1 − 0.2975q2

Now assemble a spreadsheet with time, modal coords, physical coords:

time0

0.050.10

q1

0

q2

0

z1

0

z2

0

Example 6.5 (13 of 13) The physical response z1 contains the sum of a low-frequency mode

#1 and a higher-frequency mode #2. Note that the high-frequency second mode damps out rapidly,

leaving the first mode to slowly decay. Can also solve by direct integration: read Thorby 6.6


76

High-freq 2nd mode Low-freq 1st mode

Eigenvalues For Larger (n > 2) Systems (Thorby Ch. 7)

In general we’re trying to solve M & K are n × n symmetric Methods for solution: Solving characteristic equation/ Gaussian elimination

(used for previous example), see 7.2.1 Matrix iteration, (7.2.2) Jacobi diagonalization, (7.2.3) Others: Choleski factorization

QR method Lanczos method—NASTRAN uses, or is option.

Since this isn’t a math class, we won’t cover these. But they will be a homework problem to use numerical methods to solve.


77

K[ ]−ω 2 M[ ]( ) z( ) = 0

Systems Admitting Rigid-Body Motions Systems with positive-definite mass and stiffness can have

harmonic oscillations in all modes. If stiffness matrix is positive-semidefinite, behavior is different.

This implies that the system is supported in such a manner that rigid-body motion is possible. If PE is due to elastic effects alone and body undergoes pure rigid-body motion (no elastic deflections), the PE is zero without all coordinates = 0.


78

The kinetic energy is simply: T = 1

2m1 x1

2 +m2 x22 = 1

2x{ }T M[ ] x{ }

And M[ ] = m1 00 m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Systems Admitting Rigid-Body Motions (2 of 4)


79

The potential energy is: U = 1

2k(x2 − x1)2 = 1

2x{ }T K[ ] x{ } where K[ ] = k −k

−k k⎡

⎣⎢

⎤

⎦⎥

Let xi (t) = Xi f t( ) for i =1, 2

Obtain K[ ] X{ } =ω 2 M[ ] X{ }, or K[ ]−ω 2 M[ ]( ) X{ } = 0{ }Note that det(…) = 0 is singular. PE > 0, so the system is positive semi-definite. Therefore, system admits rigid-body mode of form:

X =X0 = X0

11

⎡

⎣⎢

⎤

⎦⎥ = X01

KX0 = X0

k −k−k k

⎡

⎣⎢

⎤

⎦⎥11

⎡

⎣⎢

⎤

⎦⎥ =

00

⎧⎨⎩

⎫⎬⎭

for all k!



80

Indicates a rigid body mode with ω = 0. If it is a valid solution to eigenvalue problem, so other eigenvector must be orthogonal, ie.

X0{ }T M[ ] X{ } = X0 (m1x1 +m2x2 ) = 0Since X0 ≠ 0, m1x1 +m2x2 = 0 and also, m1 x1 +m2 x2 = 0

The system’s linear momentum = 0 for the elastic mode. Its general motion is a combination of rigid-body and elastic mode, true for unrestricted system (i.e. vehicles). [K] is singular, so to solve and transform problem into elastic one only, use the relationship x2 = − m1

m2

x1



81

Form a transformation matrix [C] such that original vector :

Substitute [C] into the problem to get transformed system.

X{ }T = { }Tand arbitrary vector: ′X{ } = x1{ }T

Then X{ } = C[ ] ′X{ } where C[ ] =1

− m1

m2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Torsional Example (1 of 4)


82

M[ ] =I 0 00 I 00 0 I

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

K[ ] =k −k 0−k 2k −k0 −k k

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥, k = GJ

L

Use I1 = I2 = I; J1 = J2 = J; L1 = L2 = L

Fundamentals of Vibration, Meirovitch, §7.8, Example 7.6.



83

θ0{ } =θ0

111

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪=θ01 is the rigid-body (rotation) mode.

In general, the displacement vector θ =

θ1

θ2

θ3

⎧

⎨⎪

⎩⎪⎪

⎫

⎬⎪

⎭⎪⎪

.

Orthogonality gives {θ0}T [M]{θ} = (Iθ1 + Iθ2 + Iθ3) = 0 I ≠ 0, so θ1 + θ2 + θ3 = 0 for all elastic modes. From this, we can write θ3 = – (θ1 + θ2).

Thus θ3 is known as soon as θ1 & θ2 are known: not independent.



84

If we define a reduced displacement vector ′θ{ } = θ1

θ2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪,

We can make a transformation θ{ } = C[ ] ′θ{ }.

Noting that θ1 =1θ1 + 0θ2

θ2 = 0θ1 +1θ2

θ3 = −1θ1 −1θ2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

, we see that the matrix C[ ] =1 00 1−1 −1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥.

The reduced mass matrix is calculated from

M[ ]′ = C[ ]T M[ ] C[ ] = 2I II 2I

⎡

⎣⎢

⎤

⎦⎥.

The reduced stiffness matrix is calculated from

K[ ]′ = C[ ]T K[ ] C[ ] = 2k kk 5k

⎡

⎣⎢

⎤

⎦⎥.



85

Solution found by: ′K[ ] ′θ{ } =ω 2 M[ ] ′θ{ } having solutions

ω1 =kI

, ′θ{ }1 =10

⎧⎨⎩

⎫⎬⎭

and ω2 =3kI

, ′θ{ }2 =0.5−1

⎧⎨⎩

⎫⎬⎭

The original eigenvectors are θ1{ } = C[ ] ′θ1{ } =1 00 1−1 −1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

10

⎧⎨⎩

⎫⎬⎭=

10−1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

θ2{ } = C[ ] ′θ2{ } =1 00 1−1 −1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

0.5−1

⎧⎨⎩

⎫⎬⎭=

0.5−10.5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

and rigid mode θ0{ } =111

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪



86

ω3 =3kI

ω2 =kI

ω1 = 0, Rigid-Body Mode

= Antisymmetric Mode (θ2 = 0, one node)

= Symmetric Mode (two nodes)

Documents

· Mul$ple’Degree’of’Freedom’Systems’ (MDoF) • More than one coordinate is needed to describe the motion of systems with more than one degree of freedom