Dynamic of structures

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks

The HomogeneousProblem

Modal Analysis

Examples

Multi Degrees of Freedom Systems

Giacomo Boffi

Dipartimento di Ingegneria Civile e Ambientale, Politecnico di Milano

April 23, 2015

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis

Examples

OutlineIntroductory Remarks

An ExampleThe Equation of Motion, a System of LinearDifferential EquationsMatrices are Linear OperatorsProperties of Structural MatricesAn example

The Homogeneous ProblemThe Homogeneous Equation of MotionEigenvalues and EigenvectorsEigenvectors are Orthogonal

Modal AnalysisEigenvectors are a baseEoM in Modal CoordinatesInitial Conditions

Examples2 DOF System

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarksAn Example

The Equation of Motion

Matrices are LinearOperators

Properties of StructuralMatrices

An example


Modal Analysis

Examples

Introductory Remarks

Consider an undamped system with two masses and two degrees of freedom.

k1 k2 k3m1 m2

x1 x2

p1(t) p2(t)

We can separate the two masses, single out the spring forces and, using theDAlembert Principle, the inertial forces and, finally, write an equation ofdynamic equilibrium for each mass.

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

Introductory RemarksConsider an undamped system with two masses and two degrees of freedom.

k1 k2 k3m1 m2

x1 x2

p1(t) p2(t)

We can separate the two masses, single out the spring forces and, using theDAlembert Principle, the inertial forces and, finally, write an equation ofdynamic equilibrium for each mass.

p2

m2x2k3x2k2(x2 x1)

m2x2 k2x1 + (k2 + k3)x2 = p2(t)

m1x1k2(x1 x2)k1x1

p1

m1x1 + (k1 + k2)x1 k2x2 = p1(t)

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

The equation of motion of a 2DOF system

With some little rearrangement we have a system of twolinear differential equations in two variables, x1(t) and x2(t):{

m1x1 + (k1 + k2)x1 k2x2 = p1(t),m2x2 k2x1 + (k2 + k3)x2 = p2(t).

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

The equation of motion of a 2DOF system

Introducing the loading vector p, the vector of inertialforces f I and the vector of elastic forces fS ,

p ={p1(t)p2(t)

}, f I =

{fI ,1fI ,2

}, fS =

{fS,1fS,2

}we can write a vectorial equation of equilibrium:

fI + fS = p(t).

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

fS = Kx

It is possible to write the linear relationship between fS andthe vector of displacements x =

{x1x2

}Tin terms of a

matrix product.In our example it is

fS =[k1 + k2 k2k2 k2 + k3

]x = Kx

introducing the stiffness matrix K.The stiffness matrix K has a number of rows equal to thenumber of elastic forces, i.e., one force for each DOF and anumber of columns equal to the number of the DOF.The stiffness matrix K is hence a square matrix K

ndofndof

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

f I =Mx

Analogously, introducing the mass matrix M that, for ourexample, is

M =[m1 00 m2

]we can write

f I = Mx.

Also the mass matrix M is a square matrix, with number ofrows and columns equal to the number of DOFs.

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

Matrix Equation

Finally it is possible to write the equation of motion inmatrix format:

Mx + Kx = p(t).

Of course, we can consider the damping forces too, taking intoaccount the velocity vector x, introducing a damping matrix Cand writing

Mx + C x + Kx = p(t),

however it is now more productive to keep our attention onundamped systems.

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

Properties of K

I K is symmetrical, because the elastic force that acts onmass i due to an unit displacement of mass j , fS,i = kijis equal to the force on mass j due to unit diplacementof mass i , fS,j = kji in virtue of Bettis theorem.

I The strain energy V for a discrete system can bewritten

V =12xT fS =

12xTKx,

because the strain energy is positive it follows that K isa positive definite matrix.

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

Properties of M

Restricting our discussion to systems whose degrees offreedom are the displacements of a set of discrete masses,we have that the mass matrix is a diagonal matrix, with allits diagonal elements greater than zero. Such a matrix issymmetrical and definite positive, as well as the stiffnessmatrix is symmetrical and definite positive.

En passant, take note that the kinetic energy for a discretesystem is

T =12xTMx.

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

Properties of M

Restricting our discussion to systems whose degrees offreedom are the displacements of a set of discrete masses,we have that the mass matrix is a diagonal matrix, with allits diagonal elements greater than zero. Such a matrix issymmetrical and definite positive, as well as the stiffnessmatrix is symmetrical and definite positive.En passant, take note that the kinetic energy for a discretesystem is

T =12xTMx.

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

Generalisation of previous results

The findings in the previous two slides can be generalised tothe structural matrices of generic structural systems, withone exception.

For a general structural system, M could be semi-definitepositive, that is for some particular displacement vector thekinetic energy could be zero.

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

Generalisation of previous results

The findings in the previous two slides can be generalised tothe structural matrices of generic structural systems, withone exception.For a general structural system, M could be semi-definitepositive, that is for some particular displacement vector thekinetic energy could be zero.

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

The problem

Graphical statement of the problem

k1 = 2k , k2 = k ; m1 = 2m, m2 = m;p(t) = p0 sint.

k1

x1 x2

m2k2

m1

p(t)

The equations of motion

m1x1 + k1x1 + k2 (x1 x2) = p0 sint,m2x2 + k2 (x2 x1) = 0.

... but we prefer the matrix notation ...

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

The steady state solution

because using the matrix notation we can follow the same stepswe used to find the steady-state response of a SDOF system.First, the equation of motion

m[2 00 1

]x + k

[3 11 1

]x = p0

{10

}sint

substituting x(t) = sint and simplifying sint, dividing by k ,with 20 = k/m,

2 = 2/20 and st = p0/k the above equationcan be written([

3 11 1

] 2

[2 00 1

]) =

[3 22 11 1 2

] = st

{10

}solving for /st gives

st=

[1 2 1

1 3 22]{

10

}24 52 + 2 =

{1 2

1

}(22 1)(2 2) .

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples



m[2 00 1

]x + k

[3 11 1

]x = p0

{10

}sint


2 = 2/20 and st = p0/k the above equationcan be written

([3 11 1

] 2

[2 00 1

]) =

[3 22 11 1 2

] = st

{10


st=

[1 2 1

1 3 22]{

10

}24 52 + 2 =

{1 2

1

}(22 1)(2 2) .

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples



m[2 00 1

]x + k

[3 11 1

]x = p0

{10

}sint



3 11 1

] 2

[2 00 1

]) =

[3 22 11 1 2

] = st

{10

}

solving for /st gives

st=

[1 2 1

1 3 22]{

10

}24 52 + 2 =

{1 2

1

}(22 1)(2 2) .

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples



m[2 00 1

]x + k

[3 11 1

]x = p0

{10

}sint



3 11 1

] 2

[2 00 1

]) =

[3 22 11 1 2

] = st

{10


st=

[1 2 1

1 3 22]{

10

}24 52 + 2 =

{1 2

1

}(22 1)(2 2) .

MultiDOFSystems

Giacomo Boffi





An example


Modal Analysis

Examples

The solution, graphically

01

0 0.5 1 2 5

Normalized

Displacem

ent

2 = 2/20

Steady-State Response for a 2 DOF System, Harmonic Load

1/st2/st

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks

The HomogeneousProblemThe HomogeneousEquation of Motion

Eigenvalues andEigenvectors

Eigenvectors areOrthogonal

Modal Analysis

Examples

Homogeneous equation of motion

To understand the behaviour of a MDOF system, we startwriting the homogeneous equation of motion,

Mx + Kx = 0.

The solution, in analogy with the SDOF case, can bewritten in terms of a harmonic function of unknownfrequency and, using the concept of separation of variables,of a constant vector, the so called shape vector :

x(t) = (A sint + B cost).

Substituting in the equation of free vibrations, we have(K 2M)(A sint + B cost) = 0

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Eigenvalues

The previous equation must hold for every value of t, so itcan be simplified removing the time dependency:(

K 2M) = 0.

This is a homogeneous linear equation, with unknowns iand the coefficients that depends on the parameter 2.

Speaking of homogeneous systems, we know that there isalways a trivial solution, = 0, and that different non-zerosolutions are available when the determinant of the matrixof coefficients is equal to zero,

det(K 2M) = 0

The eigenvalues of the MDOF system are the values of 2

for which the above equation (the equation of frequencies)is verified.

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Eigenvalues

The previous equation must hold for every value of t, so itcan be simplified removing the time dependency:(

K 2M) = 0.This is a homogeneous linear equation, with unknowns iand the coefficients that depends on the parameter 2.

Speaking of homogeneous systems, we know that there isalways a trivial solution, = 0, and that different non-zerosolutions are available when the determinant of the matrixof coefficients is equal to zero,

det(K 2M) = 0

The eigenvalues of the MDOF system are the values of 2

for which the above equation (the equation of frequencies)is verified.

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Eigenvalues, cont.

For a system with N degrees of freedom the expansion ofdet(K 2M) is an algebraic polynomial of degree N in

2.A polynomial of degree N has in general N real or complexconjugate roots.When the matrices involved are symmetric, as it is alwaysthe case for structural matrices, the roots, 2i , i = 1, . . . ,Nare all real.If both K and M are positive definite matrices, conditionthat is always satisfied by stable structural systems, theroots (the eigenvalues) are also positive,

2i > 0 for i = 1, . . . ,N.

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Eigenvectors

Substituting one of the N roots 2i in the characteristicequation, (

K 2i M)i = 0

the resulting system of N 1 linearly independent equationscan be solved (except for a scale factor) for i , theeigenvector corresponding to the eigenvalue 2i .

A common choice for the normalisation of the eigenvectorsis normalisation with respect to the mass matrix,Ti Mi = 1

Please consider that substituting different valuesfor 2i in the characteristic equation you obtaindifferent linear systems, with different solutions.

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Initial Conditions

The most general expression (the general integral) for thedisplacement of a homogeneous system is

x(t) =Ni=1

i(Ai sini t + Bi cosi t).

In the general integral there are 2N unknown constants ofintegration, that must be determined in terms of the initialconditions.

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Initial Conditions

Usually the initial conditions are expressed in terms of initialdisplacements and initial velocities x0 and x0, so we start derivingthe expression of displacement with respect to time to obtain

x(t) =Ni=1

ii (Ai cosi t Bi sini t)

and evaluating the displacement and velocity for t = 0 it is

x(0) =Ni=1

iBi = x0, x(0) =Ni=1

iiAi = x0.

The above equations are vector equations, each onecorresponding to a system of N equations, so we can computethe 2N constants of integration solving the 2N equations

Ni=1

jiBi = x0,j ,Ni=1

jiiAi = x0,j , j = 1, . . . ,N.

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Orthogonality - 1

Take into consideration two distinct eigenvalues, 2r and 2s ,

and write the characteristic equation for each eigenvalue:

Kr = 2rMrKs = 2sMs

premultiply each equation member by the transpose of theother eigenvector

Ts Kr = 2r

Ts Mr

Tr Ks = 2s

Tr Ms

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Orthogonality - 2

The term Ts Kr is a scalar, hence

Ts Kr =(Ts Kr

)T= Tr K

T s

but K is symmetrical, KT = K and we have

Ts Kr = Tr Ks .

By a similar derivation

Ts Mr = Tr Ms .

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Orthogonality - 3

Substituting our last identities in the previous equations, wehave

Tr Ks = 2r

Tr Ms

Tr Ks = 2s

Tr Ms

subtracting member by member we find that

(2r 2s ) Tr Ms = 0

We started with the hypothesis that 2r 6= 2s , so for everyr 6= s we have that the corresponding eigenvectors areorthogonal with respect to the mass matrix

Tr Ms = 0, for r 6= s.

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Modal Analysis

Examples

Orthogonality - 4

The eigenvectors are orthogonal also with respect to thestiffness matrix:

Ts Kr = 2r

Ts Mr = 0, for r 6= s.

By definitionMi = Ti Mi

andTi Ki =

2i Mi .

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal AnalysisEigenvectors are a base

EoM in Modal Coordinates

Initial Conditions

Examples

Eigenvectors are a base

The eigenvectors are linearly independent, so for everyvector x we can write

x =Nj=1

jqj .

The coefficients are readily given by premultiplication of xby Ti M, because

Ti Mx =Nj=1

Ti Mjqj = Ti Miqi = Miqi

in virtue of the ortogonality of the eigenvectors with respectto the mass matrix, and the above relationship gives

qj =Tj MxMj

.

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Initial Conditions

Examples

Eigenvectors are a base

Generalising our results for the displacement vector to theacceleration vector, we can write

x(t) =Nj=1

jqj(t), x(t) =Nj=1

j qj(t),

xi(t) =Nj=1

ijqj(t), xi(t) =Nj=1

ij qj(t).

Introducing q(t), the vector of modal coordinates and ,the eigenvector matrix, whose columns are the eigenvectors,

x(t) = q(t), x(t) = q(t).

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Initial Conditions

Examples

EoM in Modal Coordinates...

Substituting the last two equations in the equation ofmotion,

M q + Kq = p(t)

premultiplying by T

TM q +TKq = Tp(t)

introducing the so called starred matrices we can finallywrite

M? q + K? q = p?(t)

where p?(t) = Tp(t), and the scalar equation are

p?i =

m?ij qj +

k?ijqj .

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Initial Conditions

Examples

... are N independent equations!

We must examine the structure of the starred symbols.The generic element, with indexes i and j , of the starredmatrices can be expressed in terms of single eigenvectors,

m?ij = Ti Mj = ijMi ,

k?ij = Ti Kj =

2i ijMi .

where ij is the Kroneker symbol,

ij =

{1 i = j

0 i 6= j

Substituting in the equation of motion, with p?i = Ti p(t)

we have a set of uncoupled equations

Mi qi + 2i Miqi = p?i (t), i = 1, . . . ,N

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks




Initial Conditions

Examples

Initial Conditions Revisited

The initial displacements can be written in modalcoordinates,

x0 = q0

and premultiplying both members by TM we have thefollowing relationship:

TMx0 = TM q0 = M?q0.

Premultiplying by the inverse of M? and taking intoaccount that M? is diagonal,

q0 = (M?)1TMx0 qi0 =

Ti Mx0Mi

and, analogously,

qi0 =i

TMx0Mi

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


2 DOF System

k1 = k , k2 = 2k ; m1 = 2m, m2 = m;p(t) = p0 sint.

k1

x1 x2

m2k2

m1p(t)

x ={x1x2

}, p(t) =

{0p0

}sint,

M = m[2 00 1

], K = k

[3 22 2

].

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


Equation of frequencies

The equation of frequencies isK 2M = 3k 22m 2k2k 2k 2m = 0.

Developing the determinant

(2m2)4 (7mk)2 + (2k2)0 = 0, (1)with 20 = k/m and

2 = 20,

22 7 + 2 = 0 = 733

4. (2)

Solving the algebraic equation in 2

21 =km

7334

22 =km

7 +33

4

21 = 0.31386km

22 = 3.18614km

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


Eigenvectors

Substituting 21 for 2 in the first of the characteristic

equations gives the ratio between the components of thefirst eigenvector,

k (3 2 0.31386)11 2k21 = 0

while substituting 22 gives

k (3 2 3.18614)12 2k22 = 0.

Solving with the arbitrary assignment 21 = 22 = 1 gives

the unnormalised eigenvectors,

1 =

{+0.84307+1.00000

}, 2 =

{0.59307+1.00000

}.

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


Normalisation

We compute first M1 and M2,

M1 = T1 M1

={0.84307, 1

}[2m 00 m

]{0.84307

1

}={1.68614m, m

}{0.843071

}= 2.42153m

M2 = 1.70346m

the adimensional normalisation factors are

1 =2.42153, 2 =

1.70346.

Applying the normalisation factors to the respective unnormalisedeigenvectors and collecting them in a matrix, we have the matrix ofnormalised eigenvectors

=

[+0.54177 0.45440+0.64262 +0.76618

]

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


Modal Loadings

The modal loading is

p?(t) = T p(t)

= p0

[+0.54177 +0.642620.45440 +0.76618

] {01

}sint

= p0

{+0.64262+0.76618

}sint

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


Modal EoM

Substituting its modal expansion for x into the equation ofmotion and premultiplying by T we have the uncoupledmodal equation of motion{

mq1 + 0.31386k q1 = +0.64262 p0 sint

mq2 + 3.18614k q2 = +0.76618 p0 sint

Note that all the terms are dimensionally correct. Dividingby m both equations, we have

q1 + 21q1 = +0.64262p0m

sint

q2 + 22q2 = +0.76618p0m

sint

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


Particular Integral

We set1 = C1 sint, = 2C1 sint

and substitute in the first modal EoM:

C1(21 2

)sint =

p?1m

sint

solving for C1

C1 =p?1m

121 2

with 21 = K1/m m = K1/21:

C1 =p?1K1

2121 2

= (1)st

11 21

with (1)st =p?1K1

= 2.047p0k

and 1 =

1

of course

C2 = (2)st

11 22

with (2)st =p?2K2

= 0.2404p0k

and 2 =

2

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


Integrals

The integrals, for our loading, are thusq1(t) = A1 sin1t + B1 cos1t +

(1)st

sint1 21

q2(t) = A2 sin2t + B2 cos2t + (2)st

sint1 22

for a system initially at restq1(t) =

(1)st

11 21

(sint 1 sin1t)

q2(t) = (2)st

11 22

(sint 2 sin2t)

we are interested in structural degrees of freedom, too... disregardingtransientx1(t) =

(11

(1)st

1 21+ 12

(2)st

1 22

)sint =

(1.109261 21

0.1092711 22

)p0k

sint

x2(t) =

(21

(1)st

1 21+ 22

(2)st

1 22

)sint =

(1.315751 21

+0.1842451 22

)p0k

sint

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


The response in modal coordinates

To have a feeling of the response in modal coordinates, lets saythat the frequency of the load is = 20.This implies that 1 = 1 =

2.00.31386

= 6.37226 and

2 =2

= 2.03.18614

= 0.62771.

-2-1.5

-1-0.5

0 0.5

1 1.5

2 2.5

0 5 10 15 20 25 30

q i/ st

= o t

q1()/st q2()/st

In the graph above, the responses are plotted against anadimensional time coordinate with = 0t, while theordinates are adimensionalised with respect to st =

p0k

MultiDOFSystems

Giacomo Boffi

IntroductoryRemarks


Modal Analysis


The response in structural coordinates

Using the same normalisation factors, here are the responsefunctions in terms of x1 = 11q1 + 12q2 andx2 = 21q1 + 22q2:

-2-1.5

-1-0.5

0 0.5

1 1.5

2 2.5

0 5 10 15 20 25 30

x i/ st

= o t

x1()/st x2()/st

Introductory RemarksAn ExampleThe Equation of Motion, a System of Linear Differential EquationsMatrices are Linear OperatorsProperties of Structural MatricesAn example

The Homogeneous ProblemThe Homogeneous Equation of MotionEigenvalues and EigenvectorsEigenvectors are Orthogonal

Modal AnalysisEigenvectors are a baseEoM in Modal CoordinatesInitial Conditions


Documents

Dynamic of structures