MDOF Dynamic Systems

ASEN 3112 - Structures

19MDOF

DynamicSystems

ASEN 3112 Lecture 1 – Slide 1

A Two-DOF Mass-Spring-Dashpot Dynamic SystemASEN 3112 - Structures

��

c

k

p (t) 1

2

k1

u = u (t)22

u = u (t)11

1

p (t) 2

c2

1Mass m

2Mass m

Static equilibriumposition

Static equilibriumposition

Consider the lumped-parameter, mass-spring-dashpot dynamic systemshown in the Figure. It has two pointmasses m and m , which are connectedby a spring-dashpot pair with constantsk and c , respectively. Mass m is linkedto ground by another spring-dashpot pairwith constants k and c , respectively.The system has two degrees of freedom (DOF). These are the displacements u (t) and u (t) of the two masses measuredfrom their static equilibrium positions.Known dynamic forces p (t) and p (t)act on the masses. Our goal is to formulatethe equations of motion (EOM) and to placethem in matrix form.

1

1

1

1

1

1

2

2

2

2 2


Dynamic Free-Body-Diagrams to Derive EOMASEN 3112 - Structures

F = m u..

F = k us1

s2

s2

.F = c ud1

p (t) 1

1 1

1 11

I1 1 1

F = m u..

Ip (t) 2 2 2 2

1

F = k (u −u ) . .F = c (u −u )d2 222

Fd2

2

F

x

The EOM are derived from the dynamicFree Body Diagrams (DFBD) shown inthe figure. Isolate the masses and applythe forces acting on them. Spring anddashpot forces are denoted by F andF for the ith spring and ith dashpot,respectively. The inertia force acting onthe ith mass is F .

Positive force conventions are thoseshown in the Figure. Summing forcesalong x we get the two force equilibriumequations

si

di

Ii

DFBD #1:∑

Fx at mass 1 : − FI 1 − Fs1 − Fd1 + Fs2 + Fd2 + p1 = 0

DFBD #2:∑

Fx at mass 2 : − FI 2 − Fs2 − Fd2 + p2 = 0


EOM in Scalar FormASEN 3112 - Structures

Replace now the expression of motion-dependent forces in terms of thedisplacement DOF, their velocities and accelerations:

Note that for the spring and dashpot that connect masses 1 and 2, relative displacements and velocities must be used. Replacing nowinto the force equilibrium equations furnished by the two DFBD gives

Finally, collect all terms that depend on the DOF and their time derivativesin the LHS, while moving everything else (here, the given applied forces)to the RHS. Sign convention: m u terms on the LHS must be positive.

Fs1 = k1 u1 Fs2 = k2 (u2− u1) Fd1 = c1 u1

Fd2 = c2 (u2−u1) FI 1 = m u1 FI 2 = m u21 2

m1 u1 + c1 u1 + c2 u1 − c2 u2 + k1 u1 + k2 u1 − k2 u2 = p1

m2 u2 − c2 u1 + c2 u2 − k2 u1 + k2 u2 = p2

−m1 u1 − k1 u1 − c1 u1 + k2 (u2 − u1) + c2 (u2 − u1) + p1 = 0

−m2 u2 − k2 (u2 − u1) − c2 (u2 − u1) + p2 = 0

..


EOM in Matrix FormASEN 3112 - Structures

Rewrite now the scalar EOM derived in the previous slide in matrix form:

Passing to compact matrix notation

Here M, C and K denote the mass, damping and stiffness matrices, respectively, p, u, u and u are the force, displacement, velocity and acceleration vectors, respectively. The latter four are functions of time: u = u(t), etc., but the time argument will be often omitted for brevity.

Note that matrices M, C and K are symmetric, and that M is diagonal. The latter property holds when masses are "point lumped" at the DOF locations, as in this model.

m1 00 m2

u1

u2+ c1 + c2 −c2

−c2 c2

u1

u2+ k1 + k2 −k2

−k2 k2

u1

u2= p1

p2

M u + C u + K u = p


Numerical ExampleASEN 3112 - Structures

m1 = 2 m2 = 1 c1 = 0.1 c2 = 0.3

k1 = 6 k2 = 3 p1 = 2 sin 3t p2 = 5 cos 2t

For the two-DOF example of Slide 2, assume the numerical values

Then the matrix EOM are

The known matrices and vectors (the givens for the problem) are

[2 00 1

] [u1

u2

]+

[0.4 −0.3

−0.3 0.3

] [u1

u2

]+

[9 −3

−3 3

] [u1

u2

]=

[2 sin 3t5 cos 2t

]

M =[

2 00 1

]C =

[0.4 −0.3

−0.3 0.3

]K =

[9 −3

−3 3

]p =

[2 sin 3t5 cos 2t

]


Vibration Eigenproblem: Assumptionson Damping and Forces


Suppose that the example two-DOF system is undamped: c = c = 0, whence C = 0, the null 2 x 2 matrix. It is also unforced: p = p = 0, whence p = 0, the null 2 x 1 vector. The matrix EOM reduces to

This is the MDOF generalization of the free, undamped SDOF oscillator covered in Lecture 17. Next, assume that this unforced and undamped dynamic system is undergoing in-phase harmonic motions of circular frequency ω.

M u + K u = 0..

1

1

2

2


Vibration Eigenproblem: Assumed Harmonic Motion


The assumption of harmonic motion can be mathematically stated using eithertrigonometric functions, or complex exponentials:

in which U is a nonzero 2-vector that collects amplitudes of the motions of the point masses as entries, and (in the first form) α is a phase shift angle. Both expressions lead to identical results. For the ensuing derivations we select the first form. The corresponding velocities and accelerations are

2

u(t) = U cos (ω t − α) or u(t) = U e iω t

u(t) = − ω U sin (ω t − α).

u(t) = − ω U cos (ω t − α)..


Vibration Eigenproblem: Characteristic EquationASEN 3112 - Structures

Substitute the accelerations into the matrix EOM, and extract U cos(ω t − α) as common post-multiplier factor:

If this vector expression is to be identically zero for any time t, the product ofthe first two terms (matrix times vector) must vanish. Thus

This D is called the dynamic stiffness matrix.

The foregoing equation states the free vibrations eigenproblem for an undamped MDOF system. For nontrivial solutions U �= 0 the determinantof the dynamic stiffness matrix must be zero, whence

This is called the characteristic equation of the dynamic system.

M u + K u = [K − ω2M

]U cos ωt − α = 0

[K − ω2M

]U = D ω U = 0

C ω2 = det D ω = det[K − ω2M

] = 0


Vibration Eigenproblem: Natural FrequenciesASEN 3112 - Structures

For a two-DOF system C(ω) is a quadratic polynomial in ω , which will yield two roots: ω and ω . We will later see that under certainconditions on K and M, which are satisfied here, those roots are realand nonnegative. Consequently the positive square roots

are also real and nonnegative. Those are called the undamped natural circular frequencies. (Qualifiers "undamped" and "circular" are often omitted unless a distinction needs to be made.) As usual they aremeasured in radians per second, or rad/sec. They can be converted tofrequencies in cycles per second, or Hertz (Hz) by scaling through 1/(2π),and to natural periods in seconds by taking the reciprocals:

2 2

2

1 2

ω1 = + ω21 ω2 = + ω2

2

fi = ω i

2π, Ti = 1

fi= 2π

ω i, i = 1, 2


Eigenvectors As Vibration ModesASEN 3112 - Structures

How about the U vector? Insert ω = ω in the eigenproblem to get

and solve this homogeneous system for a nonzero U (This operationis always possible because the coefficient matrix is singular by definition.)Repeat with ω = ω , and call the solution vector U . Vectors U and U are mathematically known as eigenvectors. Since they can be scaledby arbitrary nonzero factors, we must normalize them to make themunique. Normalization criteria often used in structural dynamics arediscussed in a later slide.

For this application (i.e., structural dynamics), these eigenvectors are called undamped free-vibrations natural modes, a mouthful often abbreviated to vibration modes or simply modes, on account of the physicalinterpretation discussed later. The visualization of an eigenvectoras a motion pattern is called a vibration shape or mode shape.

2 2

2

11

1

1

1

1 22222

(K - ω M) U = 0


Generalization to Arbitrary Number of DOF


2

2i

i

Thanks to matrix notation, the generalization of the foregoingderivations to an undamped, unforced, Multiple-Degree-Of-Freedom(MDOF) dynamical system with n degrees of freedom is immediate.

In this general case, matrices K and M are both n x n. Thecharacteristic polynomial is of order n in ω. This polynomial providesn roots labeled ω , i = 1, 2, ... n, arranged in ascending order.Their square roots give the natural frequencies of the system.

The associated vibration modes, which are called U before normalization, are the eigenvectors of the vibration eigenproblem.Normalization schemes are studied later.


Linkage to Linear Algebra Material


To facilitate linkage to the subject of algebraic eigenproblems covered inLinear Algebra sophomore courses, as well as to capabilities of matrixoriented codes such as Matlab, we effect some notational changes.First, rewrite the vibration eigenproblem as

where i is a natural frequency index that ranges from 1 through n. Second, rename the symbols as follows: K and M become A and B, respectively, ω becomes λ , and U becomes x . These changes allow us to rewrite the vibration eigenproblem as

This is the canonical form of the generalized algebraic eigenproblem thatappears in Linear Algebra textbooks. If both A and B are real symmetric, this is called the generalized symmetric algebraic eigenproblem.

K Ui = ω2

2

i M Ui

A xi

i

= λi

i

B xi

i i

i = 1, 2, . . . n

i = 1, 2, . . . n


Properties of the Generalized Symmetric Algebraic Eigenproblem


If, in addition, B is positive definite (PD) it is shown in Lineaar Algebra textbooks that

(i) All eigenvalues λ are real(ii) All eigenvectors x have real entries

Property (i) can be further strengthened if A enjoys additional properties:

(iii) If A is nonnegative definite (NND) and B is PD all eigenvalues are nonnegative real(iv) If both A and B are PD, all eigenvalues are positive real

The nonnegativity property is obviously important in vibration analysis sincefrequencies are obtained by taking the positive square root of the eigenvalues.If a squared frequency is negative, its square roots are imaginary. Thepossibility of negative squared frequencies arises in the study of dynamicstability and active control systems, but those topics are beyond the scopeof the course.

ii


The Standard Algebraic EigenproblemASEN 3112 - Structures

If matrix B is the identity matrix, the generalized algebraic eigenproblem reduces to the standard algebraic eigenproblem

If A is real symmetric this is called the standard symmetric algebraiceigenproblem, or symmetric eigenproblem for short. Since the identityis obviously symmetric and positive definite, the aforementioned properties(i) through (iv) also hold for the standard version.

Some higher level programming languages such as Matlab, Mathematicaand Maple, can solve eigenproblems directly through their built-in libraries.Of these 3 commercial products, Matlab has the more extensive capabilities,being able to process generalized eigenproblem forms directly. On the other hand, Mathematica only accepts the standard eigenproblem as built-infunction. Similar constraints may affect hand calculators with matrixprocessing abilities. To get around those limitations, there are transformationmethods that can convert the generalized eigenproblem into the standard one.

A xi = λi xi i = 1, 2, . . . n


Reduction to the Standard Eigenproblem (cont'd)


˜ ˜

Two very simple schemes to reduce the generalized eigenproblem to standard form are

(I) If B is nonsingular, premultiply both sides by its inverse.

(II) If A is nonsingular, premultiply both sides by its inverse.

For further details, see Remark 19.1 in Lecture 19.

The matrices produced by these schemes will not generally be symmetric,even if A and B are. There are fancier reduction methods that preservesymmetry, but those noted above are sufficient for this course.


Using the Characteristic PolynomialTo Get Natural Frequencies


For a small number of DOF, say 4 or less, it is often expedient to solve forthe eigenvalues of the vibration eigenproblem directly by findingthe roots of the characteristic polynomial

in which λ = ω and D = K − ω M = K − λ M is the dynamic stiffness matrix. This procedure will be followed in subsequent Lectures for two-DOF examples. It is particularly useful with hand calculators thatcan compute polynomial roots but cannot handle eigenproblems directly. Eigenvectors may then be obtained as illustrated in the next Lecture.

C(λ) = det(D) = 0

2 2


Eigenvector UniquenessASEN 3112 - Structures

We go back to the mass-stiffness notation for the vibration eigenproblem,which is reproduced below for convenience:

For specificity we will assume that

(I) The mass matrix is positive definite (PD) while the stiffness matrix is nonnegative definite (NND). As a consequence, all eigenvalues λ = ω are nonnegative real, and so are their positive square roots ω .

(ΙΙ) All frequencies are distinct:

Under these assumptions, the theory behind the algebraic eigenproblemsays that for each ω there is one and only one eigenvector U , whichis unique except for an arbitrary nonzero scale factor.

K Ui = ω2

2

i M Ui

ωi

i i i

i i

�= ω j if i �= j



What Happens If There Are Multiple Frequencies?

Life gets more complicated. See Remark 19.3 in Lecture Notes.


Normal Modes and Their Orthogonality Properties


An eigenvector U scaled as per some normalization condition (for example,unit Euclidean length) is called a normal mode. We will write

Here φ is the notation for the ith normal mode. This is obtained from U ondividing through some scaling factor c , chosen as per some normalization criterion. (Those are discussed in a later slide).

Normal modes enjoy the following orthogonality properties with respect tothe mass and stiffness matrices:

Ui

i

= ci φi

i

i = 1, 2, . . . n

φTi M φ j = 0 φT

i K φ j = 0 i �= j


Generalized Mass and Stiffness


If i = j, the foregoing quadratic forms provide two important quantitiescalled the generalized mass M , and the generalized stiffness K :

These are also called the modal mass and the modal stiffness, respectively,in the structural dynamics literature.

The squared natural frequencies appear as the ratio of generalized stiffnessto the corresponding generalized masses:

This formula represents the generalization of the SDOF expression forthe natural frequency: ω = k/m, to MDOF.

Mi = φTi M φi > 0 Ki

ii

= φTi K φi ≥ 0 i = 1, 2, . . . n

ω2

2

i

n

= Ki

Mii = 1, 2, . . . n


Mass-Orthonormalized EigenvectorsASEN 3112 - Structures

The foregoing definition of generalized mass provides one criterion foreigenvector normalization, which is of particular importance in modalresponse analysis. If the scale factors are chosen so that

the normalized eigenvectors (a.k.a. normal modes) are said to bemass-orthogonal. If this is done, the generalized stiffnesses becomethe squared frequencies:

Mi = φTi M φi = 1 i = 1, 2, . . . n

ω2i = Ki = φi K φi i = 1, 2, . . . n


Eigenvector Normalization CriteriaASEN 3112 - Structures

Three eigenvector normalization criteria are in common use in structural dynamics. They are summarized here for convenience. The ith unnormalized eigenvector is generically denoted by U , whereas a normalized one is called φ . We will assume that eigenvectors are real, and that each entry has the same physical unit. Furthermore, for (III) we assume that M is positive definite (PD).

(I) Unit Largest Entry. Search for the largest entry of U in absolute value, and divide by it. (This value must be nonzero, else U would be null.)

(II) Unit Length. Compute the Euclidean length of the eigenvector, and divide by it. The normalized eigenvector satisfies U U = 1.

(III) Unit Generalized Mass. Get the generalized mass M = U M U , which must be positive under the assumption that M is PD, and divide U by the positive square root of M . The normalized eigenvector satisfies φ M φ = 1. The resulting eigenvectors are said to be mass-orthogonal.

If the mass matrix is the identity matrix, the last two methods coalesce. Forjust showing mode shapes pictorially, the simplest normalization method (I) works fine. For the modal analysis presented in following lectures, (III) offerscomputational and organizational advantages, so it is preferred for that task.

i

i

i

i i

i

i

i

i

i i

i

i

T

T

T


Frequency Computation ExampleASEN 3112 - Structures

Consider the two-DOF system introduced in an earlier slide. As numeric values takem = 2, m = 1, c = c = 0, k = 6, k = 3, p = p = 0. The mass and stiffness matricesare

whereas the damping matrix C and dynamic force vector p vanish. The free vibrations eigenproblem is

The characteristic polynomial equation is

The roots of this equation give the two undamped natural squared frequencies

The undamped natural frequencies (to 4 places) are ω = 1.225 and ω = 2.449.If the physical sets of units is consistent, these are expressed in radians per second.To convert to cycles per second, divide by 2π: f = 0.1949 Hz and f = 0.3898 Hz.The calculation of eigenvectors is carried out in the next Lecture.

121 21 1 22

M =[

2 00 1

]K =

[9 −3

−3 3

]

[9 −3

−3 3

] [U1U2

]= ω2

[2 00 1

] [U1U2

]or

[9 − 2ω2 −3

−3 3 − ω2

] [U1U2

]=

[00

]

det[

9 − 2ω2 −3−3 3 − ω2

]= 18 − 15ω2 + 2ω4 = (3 − 2ω2)(.6 − ω2) = 0

ω21

1

1

= 32

= 1.5 ω22

2

2

= 6


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