dynamic of structures
DESCRIPTION
Dynamic of structures. Different methods of estimation the vibrations of single degree of freedom.TRANSCRIPT
-
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
The HomogeneousProblem
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
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
IntroductoryRemarksAn Example
The Equation of Motion
Matrices are LinearOperators
Properties of StructuralMatrices
An example
The HomogeneousProblem
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
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
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblemThe HomogeneousEquation of Motion
Eigenvalues andEigenvectors
Eigenvectors areOrthogonal
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
The HomogeneousProblem
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
The HomogeneousProblem
Modal AnalysisEigenvectors are a base
EoM in Modal Coordinates
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
The HomogeneousProblem
Modal AnalysisEigenvectors are a base
EoM in Modal Coordinates
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
The HomogeneousProblem
Modal AnalysisEigenvectors are a base
EoM in Modal Coordinates
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
The HomogeneousProblem
Modal AnalysisEigenvectors are a base
EoM in Modal Coordinates
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
The HomogeneousProblem
Modal AnalysisEigenvectors are a base
EoM in Modal Coordinates
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
The HomogeneousProblem
Modal Analysis
Examples2 DOF System
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
Examples2 DOF System