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Dynamics of multiple degree of freedom linear systems, eigenvalue problem, natural frequency and mode shapes, two DOF systems
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© Eng. Vib, 3rd Ed.1/58 @ProfAdhikari, #EG260
Chapter 4 Multiple Degree of Freedom Systems
Extending the first 3 chapters to more then one degree of freedom
The Millennium bridge requiredmany degrees of freedom to modeland design with.
© Eng. Vib, 3rd Ed.2/58 College of Engineering
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
© Eng. Vib, 3rd Ed.3/58 College of Engineering
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.
© Eng. Vib, 3rd Ed.4/58 College of Engineering
Free-Body Diagram of each mass
x1 x2
m1m2k1 x1
k2(x2 -x1)
Figure 4.2
© Eng. Vib, 3rd Ed.5/58 College of Engineering
Summing forces yields the equations of motion:
© Eng. Vib, 3rd Ed.6/58 College of Engineering
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)!
© Eng. Vib, 3rd Ed.7/58 College of Engineering
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
© Eng. Vib, 3rd Ed.8/58 College of Engineering
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
© Eng. Vib, 3rd Ed.9/58 College of Engineering
Solution by Matrix MethodsThe 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)
© Eng. Vib, 3rd Ed.10/58 College of Engineering
Initial Conditions
IC’s can also be written in vector form
© Eng. Vib, 3rd Ed.11/58 College of Engineering
The approach to a Solution:
2
2
Let ( )
1, , , unknown
-
-
j t
j t
t e
j
M K e
M K
x u
u 0 u
u 0
u 0
For 1DOF we assumed the scalar solution aeλt
Similarly, now we assume the vector form:
(4.15)
(4.16)
(4.17)
© Eng. Vib, 3rd Ed.12/58 College of Engineering
This changes the differential equation of motion into algebraic vector equation:
2
1
2
- (4.17)
This is two algebraic equation in 3 uknowns
( 1 vector of two elements and 1 scalar):
= , and
M K
u
u
u 0
u
© Eng. Vib, 3rd Ed.13/58 College of Engineering
The condition for solution of this matrix equation requires that the the matrix inverse does not exist:
2
12
2
If the inv - exists : which is the
static equilibrium position. For motion to occur
- does not exist
or det - (4.19)
M K
M K
M K
u 0
u 0
0
The determinant results in 1 equation in one unknown ω (called the characteristic equation)
© Eng. Vib, 3rd Ed.14/58 College of Engineering
Back to our specific system: the characteristic equation is defined as
2
21 1 2 2
22 2 2
4 21 2 1 2 2 1 2 2 1 2
det - 0
det 0
( ) 0
M K
m k k k
k m k
m m m k m k m k k k
Eq. (4.21) is quadratic in so four solutions result:
(4.20)
(4.21)
2 21 2 1 2 and and
© Eng. Vib, 3rd Ed.15/58 College of Engineering
Once ω is known, use equation (4.17) again to
calculate the corresponding vectors u1 and u2
21 1
22 2
( ) (4.22)
and
( ) (4.23)
M K
M K
u 0
u 0
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.
© Eng. Vib, 3rd Ed.16/58 College of Engineering
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 becomesω4-6ω2+8=(ω2-2)(ω2-4)=0 ω2 = 2 and ω2 =4 or
1,3 2 rad/s, 2,4 2 rad/s
Each value of ω2 yields an expression for u:
© Eng. Vib, 3rd Ed.17/58 College of Engineering
Computing the vectors u
For 12 =2, denote u1
u11
u12
then we have
(-12M K )u1 0
27 9(2) 3
3 3 (2)
u11
u12
0
0
9u11 3u12 0 and 3u11 u12 0
2 equations, 2 unknowns but DEPENDENT!(the 2nd equation is -3 times the first)
© Eng. Vib, 3rd Ed.18/58 College of Engineering
u11
u12
1
3 u11
1
3u12 results from both equations:
only the direction, not the magnitude can be determined!
This is because: det( 12M K )0.
The magnitude of the vector is arbitrary. To see this suppose
that u1 satisfies
( 12M K )u1 0, so does au1, a arbitrary. So
( 12M K )au1 0 ( 1
2M K )u1 0
Only the direction of vectors u can be determined, not the magnitude as it remains arbitrary
© Eng. Vib, 3rd Ed.19/58 College of Engineering
Likewise for the second value of ω2
For 22 = 4, let u2
u21
u22
then we have
(-12M K )u 0
27 9(4) 3
3 3 (4)
u21
u22
0
0
9u21 3u22 0 or u21 1
3u22
Note that the other equation is the same
© Eng. Vib, 3rd Ed.20/58 College of Engineering
What to do about the magnitude!
13
12 1
13
22 2
1 1
1 1
u
u
u
u
Several possibilities, here we just fix one element:
Choose:
Choose:
© Eng. Vib, 3rd Ed.21/58 College of Engineering
Thus the solution to the algebraic matrix equation is:
13
1,3 1
13
2,4 2
2, has mode shape 1
2, has mode shape 1
u
u
Here we have introduce the name mode shape to describe the vectorsu1 and u2. The origin of this name comes later
© Eng. Vib, 3rd Ed.22/58 College of Engineering
Return now to the time response:
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 2
1 1 1 1 2 2 2 2
1 2 1 2
( ) , , ,
( )
( )
sin( ) sin( )
where , , , and are const
j t j t j t j t
j t j t j t j t
j t j t j t j t
t e e e e
t a e b e c e d e
t ae be ce de
A t A t
A A
x u u u u
x u u u u
x u u
u u
ants 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 exponentialform to to sine requires Euler’s formula for trig functions and uses up the +/- sign on omega
© Eng. Vib, 3rd Ed.23/58 College of Engineering
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
© Eng. Vib, 3rd Ed.24/58 College of Engineering
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 1 with magnitudes proportional to u1 the 1st mode shape
x(t)x1(t)
x2 (t)
A1
u11
u12
sin1t A1u1 sin1t
© Eng. Vib, 3rd Ed.25/58 College of Engineering
A graphic look at mode shapes:
Mode 1:
k1m1
x1
m2
x2k2
Mode 2:
k1
m1
x1
m2
x2k2
x2=A
x2=Ax1=A/3
x1=-A/3
u1 1
3
1
u2 1
3
1
If IC’s correspond to mode 1 or 2, then the response is purely in mode 1 or mode 2.
© Eng. Vib, 3rd Ed.26/58 College of Engineering
Example 4.1.7 given the initial conditions compute the time response
© Eng. Vib, 3rd Ed.27/58 College of Engineering
1 21 2
1 1 2 2
1 21 2
1 1 2 2
sin sin1 mm3 3
0 sin sin
2 cos 2 cos03 3
02 cos 2 cos
A A
A A
A A
A A
At t = 0 we have
© Eng. Vib, 3rd Ed.28/58 College of Engineering
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 2 1 2
3 sin sin
0 sin sin
0 2 cos 2cos
0 2 cos 2cos
1.5 mm, 1.5 mm, rad2
A A
A A
A A
A A
A A
4 equations in 4 unknowns:
Yields:
© Eng. Vib, 3rd Ed.29/58 College of Engineering
The final solution is:
1
2
( ) 0.5cos 2 0.5cos2
( ) 1.5cos 2 1.5cos2
x t t t
x t t tThese initial conditions gives a response that is a combination of modes. Both harmonic, but their summation is not.
Figure 4.3
(4.34)
© Eng. Vib, 3rd Ed.30/58 College of Engineering
Solution as a sum of modes
1 1 1 2 2 2( ) cos cost a t a t x u u
Determines how the first frequency contributes to theresponse
Determines how the second frequency contributes to theresponse
© Eng. Vib, 3rd Ed.31/58 College of Engineering
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
1 2 k1
m1
1.63,2 2k2
m2
1.732
© Eng. Vib, 3rd Ed.32/58 College of Engineering
Some matrix and vector reminders
Then M is said to be positive definite
© Eng. Vib, 3rd Ed.33/58 College of Engineering
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
© Eng. Vib, 3rd Ed.34/58 College of Engineering
Some matrix results to help us use available computational tools:A matrix M is defined to be symmetric if
M M T
A symmetric matrix M is positive definite if
xT Mx 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
© Eng. Vib, 3rd Ed.35/58 College of Engineering
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 M
If M is diagonal, then the matrix square root is just the root
of the diagonal elements:
L M 1/2 m1 0
0 m2
(4.35)
© Eng. Vib, 3rd Ed.36/58 College of Engineering
A change of coordinates is introduced to capitalize on existing mathematicsFor a diagonal, positive definite matrix M:
© Eng. Vib, 3rd Ed.37/58 College of Engineering
How the vibration problem relates to the real symmetric eigenvalue problem
© Eng. Vib, 3rd Ed.38/58 College of Engineering
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
© Eng. Vib, 3rd Ed.39/58 College of Engineering
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)
© Eng. Vib, 3rd Ed.40/58 College of Engineering
Normal and orthogonal vectors
x x1
M
xn
, y
y1
M
yn
, inner product is xT y xi yi
i1
n
x orthogonal to y if xT y 0
x is normal if xT x 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 xT x (4.43)
© Eng. Vib, 3rd Ed.41/58 College of Engineering
Normalizing any vector can be done by dividing it by its norm:
x
xT x has norm of 1
To see this compute
(4.44)
x
xT x
xT
xT x
x
xT x
xT x
xT x1
© Eng. Vib, 3rd Ed.42/58 College of Engineering
Examples 4.2.2 through 4.2.4
© Eng. Vib, 3rd Ed.43/58 College of Engineering
The first normalized eigenvector
© Eng. Vib, 3rd Ed.44/58 College of Engineering
v2 1
2
1
1
, v1
T v2 1
2(1 1)0
v1T v1
1
2(11)1
v2T v2
1
2(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
© Eng. Vib, 3rd Ed.45/58 College of Engineering
Modes u and Eigenvectors v are different but related:
u1 v1 and u2 v2
x M 1/2q u M 1/2v
Note
M 1/2u1 3 0
0 1
13
1
1
1
v1
(4.37)
© Eng. Vib, 3rd Ed.46/58 College of Engineering
This orthonormal set of vectors is used to form an Orthogonal Matrix
P is called an orthogonal matrix
P is also called a modal matrix
called a matrix of eigenvectors (normalized)
(4.47)
© Eng. Vib, 3rd Ed.47/58 College of Engineering
Example 4.2.3 compute P and show that it is an orthogonal matrix
From the previous example:
1 1
1 11
1 12
1 1 1 11 1
1 1 1 12 2
1 1 1 1 2 01 1
2 1 1 1 1 2 0 2
T
P
P P
I
v v
© Eng. Vib, 3rd Ed.48/58 College of Engineering
Example 4.2.4 Compute the square of the frequencies by matrix manipulation
1 2 rad/s and 2 2 rad/s
In general:
© Eng. Vib, 3rd Ed.49/58 College of Engineering
Example 4.2.5
Figure 4.4
The equations of motion:
In matrix form these become:
© Eng. Vib, 3rd Ed.50/58 College of Engineering
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
© Eng. Vib, 3rd Ed.51/58 College of Engineering
Next compute the eigenvectorsFor 1 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)2 v11
2
v11 0.1091, and v21 0.9940
v1 0.1091
0.9940
, likewise v2
0.9940
0.1091
© Eng. Vib, 3rd Ed.52/58 College of Engineering
Next check the value of P to see if it behaves as its suppose to:
Yes!
© Eng. Vib, 3rd Ed.53/58 College of Engineering
A note on eigenvectorsIn the previous section, we could have chosed v2 to be
v2 0.9940
0.1091
instead of v2
-0.9940
0.1091
because one can always multiple an eigenvector by a constant
and if the constant is -1 the result is still a normalized vector.
Does this make any difference?
No! Try it in the previous example
© Eng. Vib, 3rd Ed.54/58 College of Engineering
All of the previous examples can and should be solved by “hand” to learn the methodsHowever, they can also be solved on calculators with matrix functions and with the codes listed in the last sectionIn 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
© Eng. Vib, 3rd Ed.55/58 College of Engineering
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=)
© Eng. Vib, 3rd Ed.56/58 College of Engineering
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))
© Eng. Vib, 3rd Ed.57/58 College of Engineering
An alternate approach to normalizing mode shapes
From equation (4.17) M 2 K u 0, u 0
Now scale the mode shapes by computing such that
iui T M iui 1 i 1
uiT ui
wi iui is called mass normalized and it satisfies:
i2Mwi Kwi 0 i
2 wiT Kwi , i 1,2
(4.53)
© Eng. Vib, 3rd Ed.58/58 College of Engineering
There are 3 approaches to computing mode shapes and frequencies
(i) 2 Mu Ku (ii) 2u M 1Ku (iii) 2v M
12 KM
12 v
(i) Is the Generalized Symmetric Eigenvalue Problemeasy for hand computations, inefficient for computers
(ii) Is the Asymmetric Eigenvalue Problemvery expensive computationally
(iii) Is the Symmetric Eigenvalue Problemthe cheapest computationally