Ch4-1_to4-2

© D.J. Inman 1/58 Mechanical Engineering at Virginia Tech

Chapter 4 Multiple Degree of Freedom Systems

Extending the first 3 chapters to more then one degree of freedom

The Millennium bridge required many degrees of freedom to model and design with.


The first step in analyzing multiple degrees of freedom (DOF) is to look at 2 DOF •  DOF: Minimum number of coordinates to specify the

position of a system •  Many systems have more than 1 DOF •  Examples of 2 DOF systems

–  car with sprung and unsprung mass (both heave) –  elastic pendulum (radial and angular) –  motions of a ship (roll and pitch)

Fig 4.1


4.1 Two-Degree-of-Freedom Model (Undamped)

A 2 degree of freedom system used to base much of the analysis and conceptual development of MDOF systems on.


Free-Body Diagram of each mass

x1 x2

m1 m2 k1 x1

k2(x2 -x1)

Figure 4.2


Summing forces yields the equations of motion:

m1!!x1(t) = −k1x1(t) + k2 x2 (t) − x1(t)( )m2!!x2 (t) = −k2 x2 (t) − x1(t)( ) (4.1)

Rearranging terms:m1!!x1(t) + (k1 + k2 )x1(t) − k2x2 (t) = 0m2!!x2 (t) − k2x1(t) + k2x2 (t) = 0

(4.2)


Note that it is always the case that

•  A 2 Degree-of-Freedom system has – Two equations of motion! – Two natural frequencies (as we shall see)!


The dynamics of a 2 DOF system consists of 2 homogeneous and coupled equations

•  Free vibrations, so homogeneous eqs. •  Equations are coupled:

– Both have x1 and x2. –  If only one mass moves, the other follows – Example: pitch and heave of a car model

•  In this case the coupling is due to k2. – Mathematically and Physically –  If k2 = 0, no coupling occurs and can be solved

as two independent SDOF systems


Initial Conditions •  Two coupled, second -order, ordinary

differential equations with constant coefficients

•  Needs 4 constants of integration to solve •  Thus 4 initial conditions on positions and

velocities

x1(0) = x10 , !x1(0) = !x10 , x2 (0) = x20 , !x2 (0) = !x20


Solution by Matrix Methods

x(t) =x1(t)x2 (t)⎡

⎣⎢

⎤

⎦⎥ , !x(t) =

!x1(t)!x2 (t)

⎡

⎣⎢

⎤

⎦⎥ , !!x(t) =

!!x1(t)!!x2 (t)⎡

⎣⎢

⎤

⎦⎥

M =m1 00 m2

⎡

⎣⎢

⎤

⎦⎥ , K =

k1 + k2 −k2

−k2 k2

⎡

⎣⎢

⎤

⎦⎥

M!!x + Kx = 0

�

m1˙ ̇ x 1(t) + (k1 + k2)x1(t) − k2x2(t) = 0m2˙ ̇ x 2(t) − k2x1(t) + k2x2(t) = 0

The two equations can be written in the form of a

single matrix equation (see pages 272-275 if matrices and

vectors are a struggle for you) :

(4.4), (4.5)

(4.6), (4.9)


Initial Conditions

x(0) =

x10

x20

⎡

⎣⎢

⎤

⎦⎥, and !x(0) =

!x10

!x20

⎡

⎣⎢

⎤

⎦⎥

IC’s can also be written in vector form


The approach to a Solution:

Let x(t) = ue jω t

j = −1, u /= 0, ω , u unknown

⇒ -ω 2M + K( )ue jω t = 0 ⇒ -ω 2M + K( )u = 0

For 1DOF we assumed the scalar solution aeλt Similarly, now we assume the vector form:

(4.15)

(4.16)

(4.17)


This changes the differential equation of motion into algebraic vector equation:

-ω 2M + K( )u = 0 (4.17)

This is two algebraic equation in 3 uknowns ( 1 vector of two elements and 1 scalar):

u=u1

u2

⎡

⎣⎢

⎤

⎦⎥, and ω


The condition for solution of this matrix equation requires that the the matrix inverse does not exist:

If the inv -ω 2M + K( ) exists ⇒ u = 0: which is the

static equilibrium position. For motion to occur

u /= 0 ⇒ -ω 2M + K( )−1 does not exist

or det -ω 2M + K( ) = 0 (4.19)

The determinant results in 1 equation in one unknown ω (called the characteristic equation)


Back to our specific system: the characteristic equation is defined as

�

det -ω2M + K( ) = 0⇒

det−ω2m1 + k1 + k2 −k2

−k2 −ω2m2 + k2

⎡

⎣ ⎢

⎤

⎦ ⎥ = 0⇒

m1m2ω4 − (m1k2 +m2k1 + m2k2 )ω2 + k1k2 = 0

Eq. (4.21) is quadratic in ω2 so four solutions result:

(4.20)

(4.21)

ω12 and ω2

2 ⇒ ±ω1 and ±ω2


Once ω is known, use equation (4.17) again to calculate the corresponding vectors u1 and u2

(−ω12M + K )u1 = 0 (4.22)

and (−ω2

2M + K )u2 = 0 (4.23)

This yields vector equation for each squared frequency:"

Each of these matrix equations represents 2 equations in the 2 unknowns components of the vector, but the coefficient matrix is singular so each matrix equation results in only 1 independent equation. The following examples clarify this."


Examples 4.1.5 & 4.1.6:calculating u and ω •  m1=9 kg,m2=1kg, k1=24 N/m and k2=3 N/m •  The characteristic equation becomes

ω4-6ω2+8=(ω2-2)(ω2-4)=0 ω2 = 2 and ω2 =4 or ω1,3 = ± 2 rad/s, ω2,4 = ± 2 rad/s

Each value of ω2 yields an expression for u:"


Computing the vectors u For ω1

2 =2, denote u1 =u11

u12

⎡

⎣⎢

⎤

⎦⎥ then we have

(-ω12M + K )u1 = 0⇒

27 − 9(2) −3−3 3− (2)

⎡

⎣⎢

⎤

⎦⎥u11

u12

⎡

⎣⎢

⎤

⎦⎥ =

00⎡

⎣⎢

⎤

⎦⎥ ⇒

9u11 − 3u12 = 0 and − 3u11 + u12 = 0

2 equations, 2 unknowns but DEPENDENT!"(the 2nd equation is -3 times the first) "


u11

u12

=13⇒ u11 =

13u12 results from both equations:

only the direction, not the magnitude can be determined!This is because: det(−ω1

2M + K ) = 0.The magnitude of the vector is arbitrary. To see this suppose that u1 satisfies(−ω1

2M + K )u1 = 0, so does au1, a arbitrary. So (−ω1

2M + K )au1 = 0⇔ (−ω12M + K )u1 = 0

Only the direction of vectors u can be determined, not the magnitude as it remains arbitrary


Likewise for the second value of ω2

For ω22 = 4, let u2 =

u21

u22

⎡

⎣⎢

⎤

⎦⎥ then we have

(-ω12M + K )u = 0⇒

27 − 9(4) −3−3 3− (4)

⎡

⎣⎢

⎤

⎦⎥u21

u22

⎡

⎣⎢

⎤

⎦⎥ =

00⎡

⎣⎢

⎤

⎦⎥ ⇒

−9u21 − 3u22 = 0 or u21 = −13u22

Note that the other equation is the same"


What to do about the magnitude!

u12 = 1 ⇒ u1 =1

3

1⎡

⎣⎢

⎤

⎦⎥

u22 = 1 ⇒ u2 =−1

3

1⎡

⎣⎢

⎤

⎦⎥

Several possibilities, here we just fix one element: Choose:"

Choose:"


Thus the solution to the algebraic matrix equation is:

ω1,3 = ± 2, has mode shape u1 =1

3

1⎡

⎣⎢

⎤

⎦⎥

ω2,4 = ± 2, has mode shape u2 =−1

3

1⎡

⎣⎢

⎤

⎦⎥

Here we have introduce the name mode shape to describe the vectors u1 and u2. The origin of this name comes later


Return now to the time response:

x(t) = u1e− jω1t , u1e

jω1t , u2e− jω2 t ,u2e

jω2 t ⇒

x(t) = au1e− jω1t + bu1e

jω1t + cu2e− jω2 t + du2e

jω2 t

⇒ x(t) = ae− jω1t + be jω1t( )u1 + ce− jω2 t + de jω2 t( )u2

= A1 sin(ω1t + φ1)u1 + A2 sin(ω2t + φ2 )u2

where A1,A2 ,φ1, and φ2 are constants of integration

We have computed four solutions:"

Since linear, we can combine as:"

determined by initial conditions. "

(4.24)

(4.26)

Note that to go from the exponential form to to sine requires Euler’s formula for trig functions and uses up the +/- sign on omega


Physical interpretation of all that math! •  Each of the TWO masses is oscillating at TWO

natural frequencies ω1and ω2 •  The relative magnitude of each sine term, and

hence of the magnitude of oscillation of m1 and m2 is determined by the value of A1u1 and A2u2

•  The vectors u1 and u2 are called mode shapes because the describe the relative magnitude of oscillation between the two masses


What is a mode shape? •  First note that A1, A2, Φ1 and Φ2 are determined by

the initial conditions •  Choose them so that A2 = Φ1 = Φ2 =0 •  Then:

•  Thus each mass oscillates at (one) frequency w1 with magnitudes proportional to u1 the 1st mode shape

x(t) =x1(t)x2 (t)⎡

⎣⎢

⎤

⎦⎥ = A1

u11u12

⎡

⎣⎢

⎤

⎦⎥sinω1t = A1u1 sinω1t


A graphic look at mode shapes:

Mode 1: k1

m1

x1

m2

x2 k2

Mode 2: k1

m1

x1

m2

x2 k2

x2=A

x2=A x1=A/3

x1=-A/3

u1 =13

1⎡

⎣⎢

⎤

⎦⎥

u2 =−13

1⎡

⎣⎢

⎤

⎦⎥

If IC’s correspond to mode 1 or 2, then the response is purely in mode 1 or mode 2.


Example 4.1.7 given the initial conditions compute the time response

( ) ( )( ) ( )

( ) ( )( ) ( )⎥

⎥⎦

⎤

⎢⎢⎣

⎡

+++

+−+=⎥⎦⎤

⎢⎣⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+++

+−+=⎥⎦⎤

⎢⎣⎡

⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡

2211

22

11

2

1

2211

22

11

2

1

2cos22cos2

2cos23

2cos23)(

)(

2sin2sin

2sin3

2sin3)(

)(

00

)0( mm, 01

=(0)consider

φφ

φφ

φφ

φφ

tAtA

tAtA

txtx

tAtA

tAtA

txtx

!!

!xx


�

1 mm0

⎡

⎣ ⎢

⎤

⎦ ⎥ =

A1

3sin φ1( ) − A2

3sin φ2( )

A1 sin φ1( ) + A2 sin φ2( )

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

00⎡

⎣ ⎢ ⎤

⎦ ⎥ =

A1

32 cos φ1( ) − 2 A2

3cos φ2( )

A1 2 cos φ1( ) + 2A2 cos φ2( )

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

At t = 0 we have"


3 = A1 sin φ1( ) − A2 sin φ2( )0 = A1 sin φ1( ) + A2 sin φ2( )0 = A1 2 cos φ1( ) − A2 2cos φ2( )0 = A1 2 cos φ1( ) + A2 2cos φ2( )

A1 = 1.5 mm,A2 = −1.5 mm,φ1 = φ2 =π2

rad

4 equations in 4 unknowns: "

Yields:"


The final solution is: x1(t) = 0.5cos 2t + 0.5cos2t

x2 (t) = 1.5cos 2t −1.5cos2tThese initial conditions gives a response that is a combination of modes. Both harmonic, but their summation is not.

Figure 4.3

(4.34)


Solution as a sum of modes

�

x(t) = a1u1 cosω1t + a2u2 cosω2t

Determines how the first "frequency contributes to the"response"

Determines how the second "frequency contributes to the"response"


Things to note •  Two degrees of freedom implies two natural

frequencies •  Each mass oscillates at with these two frequencies

present in the response and beats could result •  Frequencies are not those of two component

systems

•  The above is not the most efficient way to calculate frequencies as the following describes

ω1 = 2 ≠ k1m1

= 1.63,ω2 = 2 ≠k2m2

=1.732


Some matrix and vector reminders

A =a bc c⎡

⎣⎢

⎤

⎦⎥ ⇒ A−1 =

1ad − cb

d −b−c a⎡

⎣⎢

⎤

⎦⎥

xTx = x12 + x2

2

M =m1 00 m2

⎡

⎣⎢

⎤

⎦⎥ ⇒ xTMx = m1x1

2 + m2x22

M > 0 ⇒ xTMx > 0 for every value of x except 0

Then M is said to be positive definite


4.2 Eigenvalues and Natural Frequencies •  Can connect the vibration problem with the

algebraic eigenvalue problem developed in math

•  This will give us some powerful computational skills

•  And some powerful theory •  All the codes have eigen-solvers so these

painful calculations can be automated


Some matrix results to help us use available computational tools: A matrix M is defined to be symmetric if

M = MT

A symmetric matrix M is positive definite if xTMx > 0 for all nonzero vectors x

A symmetric positive definite matrix M can be factored M = LLT

Here L is upper triangular, called a Cholesky matrix


If the matrix L is diagonal, it defines the matrix square root

The matrix square root is the matrix M 1/2 such that M 1/2M 1/2 = MIf M is diagonal, then the matrix square root is just the rootof the diagonal elements:

L = M 1/2 =m1 0

0 m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

(4.35)


A change of coordinates is introduced to capitalize on existing mathematics

M =m1 00 m2

⎡

⎣⎢

⎤

⎦⎥, M −1 =

1m1

00 1

m2

⎡

⎣⎢

⎤

⎦⎥, M −1/2 =

1m1

0

0 1m2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Let x(t) = M −1/2q(t) and multiply by M −1/2 : M −1/2MM −1/2

I identity! "## $## %%q(t) + M −1/2KM −1/2

&K symmetric! "## $## q(t) = 0 (4.38)

or %%q(t) + &Kq(t) = 0 where &K = M −1/2KM −1/2

&K is called the mass normalized stiffness and is similar to the scalar km

used extensively in single degree of freedom analysis. The key here is that&K is a SYMMETRIC matrix allowing the use of many nice properties and

computational tools

For a diagonal, positive definite matrix M:"


How the vibration problem relates to the real symmetric eigenvalue problem

Assume q(t) = ve jω t in !!q(t) + "Kq(t) = 0−ω 2ve jω t + "Kve jω t = 0, v ≠ 0 or"Kv =ω 2v

vibration problem

(4.40)

# $% &% ⇔ "Kv = λv real symmetric eigenvalue problem (4.41)

#$% &% v ≠ 0

Note that the martrix !K contains the same type of informationas does ωn

2 in the single degree of freedom case.


Important Properties of the n x n Real Symmetric Eigenvalue Problem •  There are n eigenvalues and they are all real

valued •  There are n eigenvectors and they are all real

valued •  The set of eigenvectors are orthogonal •  The set of eigenvectors are linearly

independent •  The matrix is similar to a diagonal matrix

Window 4.1 page 285


Square Matrix Review •  Let aik be the ikth element of A then A is symmetric

if aik = aki denoted AT=A •  A is positive definite if xTAx > 0 for all nonzero x

(also implies each λi > 0) •  The stiffness matrix is usually symmetric and

positive semi definite (could have a zero eigenvalue)

•  The mass matrix is positive definite and symmetric (and so far, its diagonal)


Normal and orthogonal vectors

x =x1

!xn

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥, y =

y1

!yn

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥, inner product is xTy = xiyi

i=1

n

∑

x orthogonal to y if xTy = 0 x is normal if xTx = 1

⎫⎬⎪

⎭⎪if a the set of vectores is is both orthogonal and normal it is called an orthonormal set

The norm of x is x = xTx (4.43)


Normalizing any vector can be done by dividing it by its norm:

xxTx

has norm of 1

To see this compute

(4.44)

xxTx

=xT

xTxxxTx

=xTxxTx

= 1


Examples 4.2.2 through 4.2.4

!K = M −1/2KM −1/2 =1

3 00 1

⎡

⎣⎢

⎤

⎦⎥

27 −3−3 3⎡

⎣⎢

⎤

⎦⎥

13 0

0 1⎡

⎣⎢

⎤

⎦⎥

so !K =3 −1−1 3⎡

⎣⎢

⎤

⎦⎥ which is symmetric.

det( !K − λI ) = det3-λ -1-1 3-λ

⎡

⎣⎢

⎤

⎦⎥ = λ2 − 6λ + 8 = 0

which has roots: λ1 = 2 =ω12 and λ2 = 4 =ω2

2


( !K − λ1I )v1 = 0⇒3− 2 −1−1 3− 2

⎡

⎣⎢

⎤

⎦⎥v11

v12

⎡

⎣⎢

⎤

⎦⎥ =

00⎡

⎣⎢

⎤

⎦⎥ ⇒

v11 − v12 = 0 ⇒ v1 = α11⎡

⎣⎢⎤

⎦⎥

v1 = α 2 (1+1) = 1⇒ α = 12

v1 =12

11⎡

⎣⎢⎤

⎦⎥ The first normalized eigenvector"


v2 =12

1−1⎡

⎣⎢

⎤

⎦⎥, v1

Tv2 =12

(1−1) = 0

v1Tv1 =

12

(1+1) = 1

v2Tv2 =

12

(1+ (−1)(−1)) = 1

⇒ v i are orthonormal

Likewise the second normalized eigenvector is computed and shown to be orthogonal to the first, so that the set is orthonormal


Modes u and Eigenvectors v are different but related: u1 ≠ v1 and u2 ≠ v2

x = M −1/2q⇒ u = M −1/2vNote

M 1/2u1 =3 00 1⎡

⎣⎢

⎤

⎦⎥

13

1

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

11⎡

⎣⎢⎤

⎦⎥ = v1

(4.37)


This orthonormal set of vectors is used to form an Orthogonal Matrix

P = v1 v2[ ]PTP =

v1Tv1 v1

Tv2v2Tv1 v2

Tv2

⎡

⎣⎢

⎤

⎦⎥ =

1 00 1⎡

⎣⎢

⎤

⎦⎥ = I

PT !KP = PT !Kv1 !Kv2⎡⎣ ⎤⎦ = PT λ1v1 λ2v2[ ]

=λ1v1

Tv1 λ2v1Tv2

λ1v2Tv1 λ2v2

Tv2

⎡

⎣⎢

⎤

⎦⎥ =

λ1 00 λ2

⎡

⎣⎢

⎤

⎦⎥ = diag(ω1

2 ,ω22 ) = Λ

P is called an orthogonal matrix"

P is also called a modal matrix"

called a matrix of eigenvectors (normalized)"

(4.47)


Example 4.2.3 compute P and show that it is an orthogonal matrix From the previous example:

P = v1 v1[ ] = 12

1 11 −1⎡

⎣⎢

⎤

⎦⎥ ⇒

PTP =12

12

1 11 −1⎡

⎣⎢

⎤

⎦⎥

1 11 −1⎡

⎣⎢

⎤

⎦⎥

= 12

1+1 1−11−1 1+1⎡

⎣⎢

⎤

⎦⎥ =

12

2 00 2⎡

⎣⎢

⎤

⎦⎥ = I


Example 4.2.4 Compute the square of the frequencies by matrix manipulation

PT !KP =12

1 11 −1⎡

⎣⎢

⎤

⎦⎥

3 −1−1 3⎡

⎣⎢

⎤

⎦⎥

12

1 11 −1⎡

⎣⎢

⎤

⎦⎥

= 12

1 11 −1⎡

⎣⎢

⎤

⎦⎥

2 42 −4⎡

⎣⎢

⎤

⎦⎥

= 12

4 00 8⎡

⎣⎢

⎤

⎦⎥ =

2 00 4⎡

⎣⎢

⎤

⎦⎥ = Λ =

ω12 0

0 ω22

⎡

⎣⎢

⎤

⎦⎥

⇒ω1 = 2 rad/s and ω2 = 2 rad/sIn general:

Λ = PT !KP = diag λi( ) = diag(ω i2 ) (4.48)


Example 4.2.5

Figure 4.4 The equations of motion:

m1!!x1 + (k1 + k2 )x1 − k2x2 = 0m2!!x2 − k2x1 + (k2 + k3)x2 = 0

(4.49)

In matrix form these become:

m1 00 m2

⎡

⎣⎢

⎤

⎦⎥ !!x +

k1 + k2 −k2

−k2 k2 + k3

⎡

⎣⎢

⎤

⎦⎥x = 0 (4.50)


Next substitute numerical values and compute P and Λ

m1 = 1 kg, m2 = 4 kg, k1 = k3 = 10 N/m and k2 =2 N/m

⇒ M =1 00 4⎡

⎣⎢

⎤

⎦⎥, K =

12 −2−2 12⎡

⎣⎢

⎤

⎦⎥

⇒ !K = M −1/2KM −1/2 =12 −1−1 12⎡

⎣⎢

⎤

⎦⎥

⇒ det !K − λI( ) = det12 − λ −1−1 12 − λ

⎡

⎣⎢

⎤

⎦⎥ = λ2 −15λ + 35 = 0

⇒λ1 = 2.8902 and λ2 = 12.1098⇒ω1 = 1.7 rad/s and ω2 = 12.1098 rad/s


Next compute the eigenvectors For λ1 equation (4.41 ) becomes:

12 - 2.8902 −1

−1 3 - 2.8902⎡

⎣⎢

⎤

⎦⎥v11

v21

⎡

⎣⎢

⎤

⎦⎥ = 0

⇒ 9.1089v11 = v21

Normalizing v1 yields

1 = v1 = v112 + v21

2 = v112 + (9.1089)2v11

2

⇒ v11 = 0.1091, and v21 = 0.9940

v1 =0.10910.9940⎡

⎣⎢

⎤

⎦⎥, likewise v2 =

−0.99400.1091

⎡

⎣⎢

⎤

⎦⎥


Next check the value of P to see if it behaves as its suppose to:

P = v1 v2[ ] = 0.1091 −0.99400.9940 0.1091⎡

⎣⎢

⎤

⎦⎥

PT !KP =0.1091 0.9940−0.9940 0.1091⎡

⎣⎢

⎤

⎦⎥12 −1−1 3⎡

⎣⎢

⎤

⎦⎥0.1091 −0.99400.9940 0.1091⎡

⎣⎢

⎤

⎦⎥ =

2.8402 00 12.1098

⎡

⎣⎢

⎤

⎦⎥

PTP =0.1091 0.9940−0.9940 0.1091⎡

⎣⎢

⎤

⎦⎥0.1091 −0.99400.9940 0.1091⎡

⎣⎢

⎤

⎦⎥ =

1 00 1⎡

⎣⎢

⎤

⎦⎥

Yes!


A note on eigenvectors In the previous section, we could have chosed v2 to be

v2 =0.9940−0.1091⎡

⎣⎢

⎤

⎦⎥ instead of v2 =

-0.99400.1091

⎡

⎣⎢

⎤

⎦⎥

because one can always multiple an eigenvector by a constantand if the constant is -1 the result is still a normalized vector.

Does this make any difference?

No! Try it in the previous example


All of the previous examples can and should be solved by “hand” to learn the methods However, they can also be solved on calculators with matrix functions and with the codes listed in the last section In fact, for more then two DOF one must use a code to solve for the natural frequencies and mode shapes.

Next we examine 3 other formulations for solving for modal data


Matlab commands •  To compute the inverse of the square matrix

A: inv(A) or use A\eye(n) where n is the size of the matrix

• [P,D]=eig(A) computes the eigenvalues and normalized eigenvectors (watch the order). Stores them in the eigenvector matrix P and the diagonal matrix D (D=L)


More commands •  To compute the matrix square root use sqrtm(A)

•  To compute the Cholesky factor: L= chol(M)!•  To compute the norm: norm(x)!•  To compute the determinant det(A)!•  To enter a matrix:

K=[27 -3;-3 3]; M=[9 0;0 1];!•  To multiply: K*inv(chol(M))!


An alternate approach to normalizing mode shapes

From equation (4.17) −Mω 2 + K( )u = 0, u ≠ 0

Now scale the mode shapes by computing α such that

α iui( )T M α iui( ) = 1⇒α i =1uiTui

wi = α iui is called mass normalized and it satisfies:−ω i

2Mwi + Kwi = 0 ⇒ω i2 = wi

T Kwi , i = 1,2

(4.53)


There are 3 approaches to computing mode shapes and frequencies (i) ω

2 Mu = Ku (ii) ω 2u = M −1Ku (iii) ω 2v = M−1

2 KM−1

2 v

(i)  Is the Generalized Symmetric Eigenvalue Problem easy for hand computations, inefficient for computers

(ii) Is the Asymmetric Eigenvalue Problem

very expensive computationally (iii) Is the Symmetric Eigenvalue Problem

the cheapest computationally

Ch4-1_to4-2

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Transcript of Ch4-1_to4-2