Multj Degrees of Freedom Systems MDOF Giacomo Boffj - - PowerPoint PPT Presentation

Multj Degrees of Freedom Systems

MDOF Giacomo Boffj

htup://intranet.dica.polimi.it/people/boffj‐giacomo Dipartjmento di Ingegneria Civile Ambientale e Territoriale Politecnico di Milano

March 27, 2020

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

Outline

Introductory Remarks An Example The Equatjon of Motjon is a System of Linear Difgerentjal Equatjons Matrices are Linear Operators Propertjes of Structural Matrices An example The Homogeneous Problem The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal Modal Analysis Eigenvectors are a base EoM in Modal Coordinates Initjal Conditjons Examples 2 DOF System

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Sectjon 1 Introductory Remarks

Introductory Remarks An Example The Equatjon of Motjon is a System of Linear Difgerentjal Equatjons Matrices are Linear Operators Propertjes of Structural Matrices An example The Homogeneous Problem Modal Analysis Examples

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Introductory Remarks

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

𝑙1 𝑙2 𝑙3 𝑛1 𝑛2 𝑦1 𝑦2 𝑞1(𝑢) 𝑞2(𝑢)

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Introductory Remarks

We can separate the two masses, single out the spring forces and, using the D’Alembert Principle, the inertjal forces and, fjnally. write an equatjon of dynamic equilibrium for each mass.

𝑞2 𝑛2 ̈ 𝑦2 𝑙3𝑦2 𝑙2(𝑦2 − 𝑦1)

𝑛2 ̈ 𝑦2 − 𝑙2𝑦1 + (𝑙2 + 𝑙3)𝑦2 = 𝑞2(𝑢)

𝑛1 ̈ 𝑦1 𝑙2(𝑦1 − 𝑦2) 𝑙1𝑦1 𝑞1

𝑛1 ̈ 𝑦1 + (𝑙1 + 𝑙2)𝑦1 − 𝑙2𝑦2 = 𝑞1(𝑢)

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The equatjon of motjon of a 2DOF system

With some litule rearrangement we have a system of two linear difgerentjal equatjons in two variables, 𝑦1(𝑢) and 𝑦2(𝑢): 𝑛1 ̈ 𝑦1 + (𝑙1 + 𝑙2)𝑦1 − 𝑙2𝑦2 = 𝑞1(𝑢), 𝑛2 ̈ 𝑦2 − 𝑙2𝑦1 + (𝑙2 + 𝑙3)𝑦2 = 𝑞2(𝑢).

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The equatjon of motjon of a 2DOF system

Introducing the loading vector 𝐪, the vector of inertjal forces 𝐠𝐽 and the vector of elastjc forces 𝐠𝑇, 𝐪 = 𝑞1(𝑢) 𝑞2(𝑢) , 𝐠𝐽 = 𝑔

𝐽,1

𝑔

𝐽,2 ,

𝐠𝑇 = 𝑔

𝑇,1

𝑔

𝑇,2

we can write a vectorial equatjon of equilibrium: 𝐠𝐉 + 𝐠𝐓 = 𝐪(𝑢).

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

𝐠𝑇 = 𝐋 𝐲

It is possible to write the linear relatjonship between 𝐠𝑇 and the vector of displacements 𝐲 = 𝑦1𝑦2

𝑈 in terms of a matrix product, introducing the so called

stjfgness matrix 𝐋. In our example it is 𝐠𝑇 = 𝑙1 + 𝑙2 −𝑙2 −𝑙2 𝑙2 + 𝑙3 𝐲 = 𝐋 𝐲 The stjfgness matrix 𝐋 has a number of rows equal to the number of elastjc forces, i.e.,

• ne force for each DOF and a number of columns equal to the number of the DOF.

The stjfgness matrix 𝐋 is hence a square matrix 𝐋

ndof×ndof

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

𝐠𝑇 = 𝐋 𝐲

It is possible to write the linear relatjonship between 𝐠𝑇 and the vector of displacements 𝐲 = 𝑦1𝑦2

𝑈 in terms of a matrix product, introducing the so called

stjfgness matrix 𝐋. In our example it is 𝐠𝑇 = 𝑙1 + 𝑙2 −𝑙2 −𝑙2 𝑙2 + 𝑙3 𝐲 = 𝐋 𝐲 The stjfgness matrix 𝐋 has a number of rows equal to the number of elastjc forces, i.e.,

• ne force for each DOF and a number of columns equal to the number of the DOF.

The stjfgness matrix 𝐋 is hence a square matrix 𝐋

ndof×ndof

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

𝐠𝑇 = 𝐋 𝐲

It is possible to write the linear relatjonship between 𝐠𝑇 and the vector of displacements 𝐲 = 𝑦1𝑦2

𝑈 in terms of a matrix product, introducing the so called

stjfgness matrix 𝐋. In our example it is 𝐠𝑇 = 𝑙1 + 𝑙2 −𝑙2 −𝑙2 𝑙2 + 𝑙3 𝐲 = 𝐋 𝐲 The stjfgness matrix 𝐋 has a number of rows equal to the number of elastjc forces, i.e.,

• ne force for each DOF and a number of columns equal to the number of the DOF.

The stjfgness matrix 𝐋 is hence a square matrix 𝐋

ndof×ndof

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

𝐠𝐽 = 𝐍 ̈ 𝐲

Analogously, introducing the mass matrix 𝐍 that, for our example, is 𝐍 = 𝑛1 𝑛2 we can write 𝐠𝐽 = 𝐍 ̈ 𝐲. Also the mass matrix 𝐍 is a square matrix, with number of rows and columns equal to the number of DOF’s.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Matrix Equatjon

Finally it is possible to write the equatjon of motjon in matrix format: 𝐍 ̈ 𝐲 + 𝐋 𝐲 = 𝐪(𝑢).

Of course it is possible to take into consideratjon also the damping forces, taking into account the velocity vector ̇ 𝐲 and introducing a damping matrix 𝐃 too, so that we can eventually write 𝐍 ̈ 𝐲 + 𝐃 ̇ 𝐲 + 𝐋 𝐲 = 𝐪(𝑢). But today we are focused on undamped systems...

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Matrix Equatjon

Finally it is possible to write the equatjon of motjon in matrix format: 𝐍 ̈ 𝐲 + 𝐋 𝐲 = 𝐪(𝑢).

Of course it is possible to take into consideratjon also the damping forces, taking into account the velocity vector ̇ 𝐲 and introducing a damping matrix 𝐃 too, so that we can eventually write 𝐍 ̈ 𝐲 + 𝐃 ̇ 𝐲 + 𝐋 𝐲 = 𝐪(𝑢). But today we are focused on undamped systems...

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Matrix Equatjon

Finally it is possible to write the equatjon of motjon in matrix format: 𝐍 ̈ 𝐲 + 𝐋 𝐲 = 𝐪(𝑢).

Of course it is possible to take into consideratjon also the damping forces, taking into account the velocity vector ̇ 𝐲 and introducing a damping matrix 𝐃 too, so that we can eventually write 𝐍 ̈ 𝐲 + 𝐃 ̇ 𝐲 + 𝐋 𝐲 = 𝐪(𝑢). But today we are focused on undamped systems...

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Propertjes of 𝐋

𝐋 is symmetrical. 𝐋 is a positjve defjnite matrix.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Propertjes of 𝐋

𝐋 is symmetrical. 𝐋 is a positjve defjnite matrix.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Propertjes of 𝐋: symmetry

The elastjc force exerted on mass 𝑗 due to an unit displacement of mass 𝑘, 𝑔

𝑇,𝑗 = 𝑙𝑗𝑘 is

equal to the force 𝑙𝑘𝑗 exerted on mass 𝑘 due to an unit diplacement of mass 𝑗, in virtue of Bettj’s theorem (also known as Maxwell‐Bettj reciprocal work theorem).

The strain energy associated with an imposed displacement vector 𝐲 is 𝑊 = 1/

2 𝐲𝑈 ⋅ 𝐠 where 𝑔

is the vector of the elastjc forces that cause the displacement, so we can write 𝑊 = 1/

2 𝐲𝑈𝐋𝐲.

Now consider two sets of displacements, 𝐲𝑏 and 𝐲𝑐 and write ① the strain energy associated with fjrst applying 𝐲𝑏 and later 𝐲𝑐: 𝑊

𝑏𝑐 = 1/ 2 𝐲𝑈 𝑏𝐋𝐲𝑏 + 1/ 2 𝐲𝑈 𝑐𝐋𝐲𝑐 + 𝐲𝑈 𝑐𝐋𝐲𝑏

and ② the strain energy associated with fjrst applying 𝐲𝑐 and later 𝐲𝑏: 𝑊

𝑐𝑏 = 1/ 2 𝐲𝑈 𝑐𝐋𝐲𝑐 + 1/ 2 𝐲𝑈 𝑏𝐋𝐲𝑏 + 𝐲𝑈 𝑏𝐋𝐲𝑐.

Because 𝑊

𝑏𝑐 = 𝑊 𝑐𝑏 (the fjnal deformed confjguratjon is the same) it is 𝐲𝑈 𝑐𝐋𝐲𝑏 = 𝐲𝑈 𝑏𝐋𝐲𝑐 and,

because 𝐲𝑈

𝑐𝐋𝐲𝑏 = 𝐲𝑈 𝑏𝐋𝑈𝐲𝑐 we can conclude that 𝐋𝑈 = 𝐋, i.e., 𝐋 is a symmetrical matrix.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Propertjes of 𝐋: symmetry

The elastjc force exerted on mass 𝑗 due to an unit displacement of mass 𝑘, 𝑔

𝑇,𝑗 = 𝑙𝑗𝑘 is

equal to the force 𝑙𝑘𝑗 exerted on mass 𝑘 due to an unit diplacement of mass 𝑗, in virtue of Bettj’s theorem (also known as Maxwell‐Bettj reciprocal work theorem).

The strain energy associated with an imposed displacement vector 𝐲 is 𝑊 = 1/

2 𝐲𝑈 ⋅ 𝐠 where 𝑔

is the vector of the elastjc forces that cause the displacement, so we can write 𝑊 = 1/

2 𝐲𝑈𝐋𝐲.

Now consider two sets of displacements, 𝐲𝑏 and 𝐲𝑐 and write ① the strain energy associated with fjrst applying 𝐲𝑏 and later 𝐲𝑐: 𝑊

𝑏𝑐 = 1/ 2 𝐲𝑈 𝑏𝐋𝐲𝑏 + 1/ 2 𝐲𝑈 𝑐𝐋𝐲𝑐 + 𝐲𝑈 𝑐𝐋𝐲𝑏

and ② the strain energy associated with fjrst applying 𝐲𝑐 and later 𝐲𝑏: 𝑊

𝑐𝑏 = 1/ 2 𝐲𝑈 𝑐𝐋𝐲𝑐 + 1/ 2 𝐲𝑈 𝑏𝐋𝐲𝑏 + 𝐲𝑈 𝑏𝐋𝐲𝑐.

Because 𝑊

𝑏𝑐 = 𝑊 𝑐𝑏 (the fjnal deformed confjguratjon is the same) it is 𝐲𝑈 𝑐𝐋𝐲𝑏 = 𝐲𝑈 𝑏𝐋𝐲𝑐 and,

because 𝐲𝑈

𝑐𝐋𝐲𝑏 = 𝐲𝑈 𝑏𝐋𝑈𝐲𝑐 we can conclude that 𝐋𝑈 = 𝐋, i.e., 𝐋 is a symmetrical matrix.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Propertjes of 𝐋: defjnite positjvity

The strain energy 𝑊 for a discrete system is 𝑊 = 1 2𝐲𝑈𝐠𝑇, and expressing 𝐠𝑇 in terms of 𝐋 and 𝐲 we have 𝑊 = 1 2𝐲𝑈𝐋 𝐲, and because the strain energy is positjve for 𝐲 ≠ 𝟏 it follows that 𝐋 is defjnite positjve.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Propertjes of 𝐍

Restrictjng our discussion to systems whose degrees of freedom are the displacements of a set of discrete masses, we have that the mass matrix is a diagonal matrix, with all its diagonal elements greater than zero. Such a matrix is symmetrical and defjnite positjve. Both the mass and the stjfgness matrix are symmetrical and defjnite positjve. Note that the kinetjc energy for a discrete system can be writuen 𝑈 = 1 2 ̇ 𝐲𝑈𝐍 ̇ 𝐲.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Propertjes of 𝐍

Restrictjng our discussion to systems whose degrees of freedom are the displacements of a set of discrete masses, we have that the mass matrix is a diagonal matrix, with all its diagonal elements greater than zero. Such a matrix is symmetrical and defjnite positjve. Both the mass and the stjfgness matrix are symmetrical and defjnite positjve. Note that the kinetjc energy for a discrete system can be writuen 𝑈 = 1 2 ̇ 𝐲𝑈𝐍 ̇ 𝐲.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Generalisatjon of previous results

The fjndings in the previous two slides can be generalised to the structural matrices of generic structural systems, with two main exceptjons.

1 For a general structural system, in which not all DOFs are related to a mass, 𝐍

could be semi‐defjnite positjve, that is for some partjcular displacement vector the kinetjc energy is zero.

2 For a general structural system subjected to axial loads, due to the presence of

geometrical stjfgness it is possible that, for some partjcular confjguratjon of the axial loads, a displacement vector exists, for which the strain energy is zero and consequently the matrix 𝐋 is semi‐defjnite positjve.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Generalisatjon of previous results

The fjndings in the previous two slides can be generalised to the structural matrices of generic structural systems, with two main exceptjons.

1 For a general structural system, in which not all DOFs are related to a mass, 𝐍

could be semi‐defjnite positjve, that is for some partjcular displacement vector the kinetjc energy is zero.

2 For a general structural system subjected to axial loads, due to the presence of

geometrical stjfgness it is possible that, for some partjcular confjguratjon of the axial loads, a displacement vector exists, for which the strain energy is zero and consequently the matrix 𝐋 is semi‐defjnite positjve.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Generalisatjon of previous results

The fjndings in the previous two slides can be generalised to the structural matrices of generic structural systems, with two main exceptjons.

1 For a general structural system, in which not all DOFs are related to a mass, 𝐍

could be semi‐defjnite positjve, that is for some partjcular displacement vector the kinetjc energy is zero.

2 For a general structural system subjected to axial loads, due to the presence of

geometrical stjfgness it is possible that, for some partjcular confjguratjon of the axial loads, a displacement vector exists, for which the strain energy is zero and consequently the matrix 𝐋 is semi‐defjnite positjve.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The problem

Steady–state solutjon: graphical statement of the problem

𝑞(𝑢) = 𝑞0 sin 𝜕𝑢. 𝑙1 = 2𝑙, 𝑙2 = 𝑙; 𝑛1 = 2𝑛, 𝑛2 = 1𝑛; 𝑙1 𝑦1 𝑦2 𝑛2 𝑙2 𝑛1 𝑞(𝑢)

The equatjons of motjon

𝑛1 ̈ 𝑦1 + 𝑙1𝑦1 + 𝑙2 (𝑦1 − 𝑦2) = 𝑞0 sin 𝜕𝑢, 𝑛2 ̈ 𝑦2 + 𝑙2 (𝑦2 − 𝑦1) = 0. ... but we prefer the matrix notatjon ...

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The problem

Steady–state solutjon: graphical statement of the problem

𝑞(𝑢) = 𝑞0 sin 𝜕𝑢. 𝑙1 = 2𝑙, 𝑙2 = 𝑙; 𝑛1 = 2𝑛, 𝑛2 = 1𝑛; 𝑙1 𝑦1 𝑦2 𝑛2 𝑙2 𝑛1 𝑞(𝑢)

The equatjons of motjon

𝑛1 ̈ 𝑦1 + 𝑙1𝑦1 + 𝑙2 (𝑦1 − 𝑦2) = 𝑞0 sin 𝜕𝑢, 𝑛2 ̈ 𝑦2 + 𝑙2 (𝑦2 − 𝑦1) = 0. ... but we prefer the matrix notatjon ...

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The problem

Steady–state solutjon: graphical statement of the problem

𝑞(𝑢) = 𝑞0 sin 𝜕𝑢. 𝑙1 = 2𝑙, 𝑙2 = 𝑙; 𝑛1 = 2𝑛, 𝑛2 = 1𝑛; 𝑙1 𝑦1 𝑦2 𝑛2 𝑙2 𝑛1 𝑞(𝑢)

The equatjons of motjon

𝑛1 ̈ 𝑦1 + 𝑙1𝑦1 + 𝑙2 (𝑦1 − 𝑦2) = 𝑞0 sin 𝜕𝑢, 𝑛2 ̈ 𝑦2 + 𝑙2 (𝑦2 − 𝑦1) = 0. ... but we prefer the matrix notatjon ...

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The steady state solutjon

We prefer the matrix notatjon because we can fjnd the steady‐state response of a SDOF system exactly as we found the s‐s solutjon for a SDOF system. Substjtutjng 𝐲(𝑢) = 𝝄 sin 𝜕𝑢 in the equatjon of motjon and simplifying sin 𝜕𝑢, 𝑙 3 −1 −1 1 𝝄 − 𝑛𝜕2 2 1 𝝄 = 𝑞0 1 dividing by 𝑙, with 𝜕2

0 = 𝑙/𝑛, 𝛾2 = 𝜕2/𝜕2 0 and Δst = 𝑞0/𝑙 the above equatjon can

be writuen 3 −1 −1 1 − 𝛾2 2 1 𝝄 = 3 − 2𝛾2 −1 −1 1 − 𝛾2 𝝄 = Δst 1 0 .

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The steady state solutjon

We prefer the matrix notatjon because we can fjnd the steady‐state response of a SDOF system exactly as we found the s‐s solutjon for a SDOF system. Substjtutjng 𝐲(𝑢) = 𝝄 sin 𝜕𝑢 in the equatjon of motjon and simplifying sin 𝜕𝑢, 𝑙 3 −1 −1 1 𝝄 − 𝑛𝜕2 2 1 𝝄 = 𝑞0 1 dividing by 𝑙, with 𝜕2

0 = 𝑙/𝑛, 𝛾2 = 𝜕2/𝜕2 0 and Δst = 𝑞0/𝑙 the above equatjon can

be writuen 3 −1 −1 1 − 𝛾2 2 1 𝝄 = 3 − 2𝛾2 −1 −1 1 − 𝛾2 𝝄 = Δst 1 0 .

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The steady state solutjon

We prefer the matrix notatjon because we can fjnd the steady‐state response of a SDOF system exactly as we found the s‐s solutjon for a SDOF system. Substjtutjng 𝐲(𝑢) = 𝝄 sin 𝜕𝑢 in the equatjon of motjon and simplifying sin 𝜕𝑢, 𝑙 3 −1 −1 1 𝝄 − 𝑛𝜕2 2 1 𝝄 = 𝑞0 1 dividing by 𝑙, with 𝜕2

0 = 𝑙/𝑛, 𝛾2 = 𝜕2/𝜕2 0 and Δst = 𝑞0/𝑙 the above equatjon can

be writuen 3 −1 −1 1 − 𝛾2 2 1 𝝄 = 3 − 2𝛾2 −1 −1 1 − 𝛾2 𝝄 = Δst 1 0 .

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The steady state solutjon

The determinant of the matrix of coeffjcients is Det 3 − 2𝛾2 −1 −1 1 − 𝛾2 = 2𝛾4 − 5𝛾2 + 2 but we’ll fjnd convenient to write the polynomial in 𝛾 in terms of its roots Det = 2 × (𝛾2 − 1/2) × (𝛾2 − 2). Solving for 𝝄/Δst in terms of the inverse of the coeffjcient matrix gives 𝝄 Δst = 1 2(𝛾2 −

1 2)(𝛾2 − 2)

1 − 𝛾2 1 1 3 − 2𝛾2 1 0 = 1 2(𝛾2 −

1 2)(𝛾2 − 2)

1 − 𝛾2 1 → 𝐲s‐s = Δst 2(𝛾2 −

1 2)(𝛾2 − 2)

1 − 𝛾2 1 sin 𝜕𝑢.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The steady state solutjon

The determinant of the matrix of coeffjcients is Det 3 − 2𝛾2 −1 −1 1 − 𝛾2 = 2𝛾4 − 5𝛾2 + 2 but we’ll fjnd convenient to write the polynomial in 𝛾 in terms of its roots Det = 2 × (𝛾2 − 1/2) × (𝛾2 − 2). Solving for 𝝄/Δst in terms of the inverse of the coeffjcient matrix gives 𝝄 Δst = 1 2(𝛾2 −

1 2)(𝛾2 − 2)

1 − 𝛾2 1 1 3 − 2𝛾2 1 0 = 1 2(𝛾2 −

1 2)(𝛾2 − 2)

1 − 𝛾2 1 → 𝐲s‐s = Δst 2(𝛾2 −

1 2)(𝛾2 − 2)

1 − 𝛾2 1 sin 𝜕𝑢.

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

The solutjon, graphically

1 0.5 1 2 5 Normalized displacement β2=ω2/ω2

• steady-state response for a 2 dof system, harmonic load

ξ1/Δst ξ2/Δst

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Comment to the Steady State Solutjon

The steady state solutjon is 𝐲s‐s = Δst 1 2(𝛾2 −

1 2)(𝛾2 − 2)

1 − 𝛾2 1 sin 𝜕𝑢. As it’s apparent in the previous slide, we have two difgerent values of the excitatjon frequency for which the dynamic amplifjcatjon factor goes to infjnity. For an undamped SDOF system, we had a single frequency of excitatjon that excites a resonant response, now for a two degrees of freedom system we have two difgerent excitatjon frequencies that excite a resonant response. We know how to compute a partjcular integral for a MDOF system (at least for a harmonic loading), what do we miss to be able to determine the integral of motjon?

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Comment to the Steady State Solutjon

The steady state solutjon is 𝐲s‐s = Δst 1 2(𝛾2 −

1 2)(𝛾2 − 2)

1 − 𝛾2 1 sin 𝜕𝑢. As it’s apparent in the previous slide, we have two difgerent values of the excitatjon frequency for which the dynamic amplifjcatjon factor goes to infjnity. For an undamped SDOF system, we had a single frequency of excitatjon that excites a resonant response, now for a two degrees of freedom system we have two difgerent excitatjon frequencies that excite a resonant response. We know how to compute a partjcular integral for a MDOF system (at least for a harmonic loading), what do we miss to be able to determine the integral of motjon?

Multj DoF Systems Giacomo Boffj Introductjon

An Example The Equatjon of Motjon Matrices are Linear Operators Propertjes of Structural Matrices An example

The Homogeneous Problem Modal Analysis Examples

Comment to the Steady State Solutjon

The steady state solutjon is 𝐲s‐s = Δst 1 2(𝛾2 −

1 2)(𝛾2 − 2)

1 − 𝛾2 1 sin 𝜕𝑢. As it’s apparent in the previous slide, we have two difgerent values of the excitatjon frequency for which the dynamic amplifjcatjon factor goes to infjnity. For an undamped SDOF system, we had a single frequency of excitatjon that excites a resonant response, now for a two degrees of freedom system we have two difgerent excitatjon frequencies that excite a resonant response. We know how to compute a partjcular integral for a MDOF system (at least for a harmonic loading), what do we miss to be able to determine the integral of motjon?

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Sectjon 2 The Homogeneous Problem

Introductory Remarks The Homogeneous Problem The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal Modal Analysis Examples

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Homogeneous equatjon of motjon

To understand the behaviour of a MDOF system, we have to study the homogeneous solutjon. Let’s start writjng the homogeneous equatjon of motjon, 𝐍 ̈ 𝐲 + 𝐋 𝐲 = 𝟏. The solutjon, in analogy with the SDOF case, can be writuen in terms of a harmonic functjon of unknown frequency and, using the concept of separatjon of variables, of a constant vector, the so called shape vector 𝝎: 𝐲(𝑢) = 𝝎(𝐵 sin 𝜕𝑢 + 𝐶 cos 𝜕𝑢) ⇒ ̈ 𝐲(𝑢) = −𝜕2 𝐲(𝑢). Substjtutjng in the equatjon of motjon, we have 𝐋 − 𝜕2𝐍 𝝎(𝐵 sin 𝜕𝑢 + 𝐶 cos 𝜕𝑢) = 𝟏

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Homogeneous equatjon of motjon

To understand the behaviour of a MDOF system, we have to study the homogeneous solutjon. Let’s start writjng the homogeneous equatjon of motjon, 𝐍 ̈ 𝐲 + 𝐋 𝐲 = 𝟏. The solutjon, in analogy with the SDOF case, can be writuen in terms of a harmonic functjon of unknown frequency and, using the concept of separatjon of variables, of a constant vector, the so called shape vector 𝝎: 𝐲(𝑢) = 𝝎(𝐵 sin 𝜕𝑢 + 𝐶 cos 𝜕𝑢) ⇒ ̈ 𝐲(𝑢) = −𝜕2 𝐲(𝑢). Substjtutjng in the equatjon of motjon, we have 𝐋 − 𝜕2𝐍 𝝎(𝐵 sin 𝜕𝑢 + 𝐶 cos 𝜕𝑢) = 𝟏

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Homogeneous equatjon of motjon

To understand the behaviour of a MDOF system, we have to study the homogeneous solutjon. Let’s start writjng the homogeneous equatjon of motjon, 𝐍 ̈ 𝐲 + 𝐋 𝐲 = 𝟏. The solutjon, in analogy with the SDOF case, can be writuen in terms of a harmonic functjon of unknown frequency and, using the concept of separatjon of variables, of a constant vector, the so called shape vector 𝝎: 𝐲(𝑢) = 𝝎(𝐵 sin 𝜕𝑢 + 𝐶 cos 𝜕𝑢) ⇒ ̈ 𝐲(𝑢) = −𝜕2 𝐲(𝑢). Substjtutjng in the equatjon of motjon, we have 𝐋 − 𝜕2𝐍 𝝎(𝐵 sin 𝜕𝑢 + 𝐶 cos 𝜕𝑢) = 𝟏

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Homogeneous equatjon of motjon

To understand the behaviour of a MDOF system, we have to study the homogeneous solutjon. Let’s start writjng the homogeneous equatjon of motjon, 𝐍 ̈ 𝐲 + 𝐋 𝐲 = 𝟏. The solutjon, in analogy with the SDOF case, can be writuen in terms of a harmonic functjon of unknown frequency and, using the concept of separatjon of variables, of a constant vector, the so called shape vector 𝝎: 𝐲(𝑢) = 𝝎(𝐵 sin 𝜕𝑢 + 𝐶 cos 𝜕𝑢) ⇒ ̈ 𝐲(𝑢) = −𝜕2 𝐲(𝑢). Substjtutjng in the equatjon of motjon, we have 𝐋 − 𝜕2𝐍 𝝎(𝐵 sin 𝜕𝑢 + 𝐶 cos 𝜕𝑢) = 𝟏

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Eigenvalues

The previous equatjon must hold for every value of 𝑢, so it can be simplifjed removing the tjme dependency: (𝐋 − 𝜕2𝐍) 𝝎 = 𝟏. The above equatjon, the EQUATION OF FREE VIBRATIONS, is a set of homogeneous linear equatjons, with unknowns 𝜔𝑗 and whose coeffjcients depend on the parameter 𝜕2. Speaking of homogeneous systems, we know that there is always a trivial solutjon, 𝝎 = 𝟏, and non‐trivial solutjons are possible if the determinant of the matrix of coeffjcients is equal to zero, det (𝐋 − 𝜕2𝐍) = 0. The EIGENVALUES of the MDOF system are the values of 𝜕2 for which the above equatjon (the EQUATION OF FREQUENCIES) is verifjed or, in other words, the frequencies of vibratjon associated with the shapes for which we have equilibrium: 𝐋𝝎 = 𝜕2𝐍𝝎 ⇔ 𝐠𝑇 = 𝐠𝐽.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Eigenvalues

The previous equatjon must hold for every value of 𝑢, so it can be simplifjed removing the tjme dependency: (𝐋 − 𝜕2𝐍) 𝝎 = 𝟏. The above equatjon, the EQUATION OF FREE VIBRATIONS, is a set of homogeneous linear equatjons, with unknowns 𝜔𝑗 and whose coeffjcients depend on the parameter 𝜕2. Speaking of homogeneous systems, we know that there is always a trivial solutjon, 𝝎 = 𝟏, and non‐trivial solutjons are possible if the determinant of the matrix of coeffjcients is equal to zero, det (𝐋 − 𝜕2𝐍) = 0. The EIGENVALUES of the MDOF system are the values of 𝜕2 for which the above equatjon (the EQUATION OF FREQUENCIES) is verifjed or, in other words, the frequencies of vibratjon associated with the shapes for which we have equilibrium: 𝐋𝝎 = 𝜕2𝐍𝝎 ⇔ 𝐠𝑇 = 𝐠𝐽.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Eigenvalues

The previous equatjon must hold for every value of 𝑢, so it can be simplifjed removing the tjme dependency: (𝐋 − 𝜕2𝐍) 𝝎 = 𝟏. The above equatjon, the EQUATION OF FREE VIBRATIONS, is a set of homogeneous linear equatjons, with unknowns 𝜔𝑗 and whose coeffjcients depend on the parameter 𝜕2. Speaking of homogeneous systems, we know that there is always a trivial solutjon, 𝝎 = 𝟏, and non‐trivial solutjons are possible if the determinant of the matrix of coeffjcients is equal to zero, det (𝐋 − 𝜕2𝐍) = 0. The EIGENVALUES of the MDOF system are the values of 𝜕2 for which the above equatjon (the EQUATION OF FREQUENCIES) is verifjed or, in other words, the frequencies of vibratjon associated with the shapes for which we have equilibrium: 𝐋𝝎 = 𝜕2𝐍𝝎 ⇔ 𝐠𝑇 = 𝐠𝐽.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Eigenvalues, cont.

For a system with 𝑂 degrees of freedom the expansion of det 𝐋 − 𝜕2𝐍 is an algebraic polynomial of degree 𝑂 in 𝜕2 that has 𝑂 roots, these roots either real or complex conjugate. In Dynamics of Structures those roots 𝜕2

𝑗 , 𝑗 = 1, … , 𝑂 are all real because the

structural matrices are symmetric matrices. Moreover, if both 𝐋 and 𝐍 are positjve defjnite matrices (a conditjon that you can always enforce for a stable structural system) then all the roots, all the eigenvalues, are strictly positjve: 𝜕2

𝑗 ≥ 0,

for 𝑗 = 1, … , 𝑂.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Eigenvectors

Substjtutjng one of the 𝑂 roots 𝜕2

𝑗 in the characteristjc equatjon,

𝐋 − 𝜕2

𝑗 𝐍 𝝎𝑗 = 𝟏

the resultjng system of 𝑂 − 1 linearly independent equatjons can be solved (except for a scale factor) for 𝝎𝑗, the eigenvector corresponding to the eigenvalue 𝜕2

𝑗 .

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Eigenvectors

The scale factor being arbitrary, you have to choose (arbitrarily) the value of one of the components and compute the values of all the other 𝑂 − 1 components using the 𝑂 − 1 linearly indipendent equatjons. It is common to impose to each eigenvector a normalisatjon with respect to the mass matrix, so that 𝝎𝑈

𝑗 𝐍 𝝎𝑗 = 𝑛

where 𝑛 represents the unit mass.

Please understand clearly that, substjtutjng difgerent eigenvalues in the equatjon of free vibratjons, you have difgerent linear systems, leading to difgerent eigenvectors.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Eigenvectors

The scale factor being arbitrary, you have to choose (arbitrarily) the value of one of the components and compute the values of all the other 𝑂 − 1 components using the 𝑂 − 1 linearly indipendent equatjons. It is common to impose to each eigenvector a normalisatjon with respect to the mass matrix, so that 𝝎𝑈

𝑗 𝐍 𝝎𝑗 = 𝑛

where 𝑛 represents the unit mass.

Please understand clearly that, substjtutjng difgerent eigenvalues in the equatjon of free vibratjons, you have difgerent linear systems, leading to difgerent eigenvectors.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Eigenvectors

The scale factor being arbitrary, you have to choose (arbitrarily) the value of one of the components and compute the values of all the other 𝑂 − 1 components using the 𝑂 − 1 linearly indipendent equatjons. It is common to impose to each eigenvector a normalisatjon with respect to the mass matrix, so that 𝝎𝑈

𝑗 𝐍 𝝎𝑗 = 𝑛

where 𝑛 represents the unit mass.

Please understand clearly that, substjtutjng difgerent eigenvalues in the equatjon of free vibratjons, you have difgerent linear systems, leading to difgerent eigenvectors.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Initjal Conditjons

The most general expression (the general integral) for the displacement of a homogeneous system is 𝐲(𝑢) =

𝑂

• 𝑗=1

𝝎𝑗(𝐵𝑗 sin 𝜕𝑗𝑢 + 𝐶𝑗 cos 𝜕𝑗𝑢). In the general integral there are 2𝑂 unknown constants of integratjon, that must be determined in terms of the initjal conditjons.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Initjal Conditjons

Usually the initjal conditjons are expressed in terms of initjal displacements and initjal velocitjes 𝐲0 and ̇ 𝐲0, so we start deriving the expression of displacement with respect to tjme to obtain ̇ 𝐲(𝑢) =

𝑂

• 𝑗=1

𝝎𝑗𝜕𝑗(𝐵𝑗 cos 𝜕𝑗𝑢 − 𝐶𝑗 sin 𝜕𝑗𝑢) and evaluatjng the displacement and velocity for 𝑢 = 0 it is 𝐲(0) =

𝑂

• 𝑗=1

𝝎𝑗𝐶𝑗 = 𝐲0, ̇ 𝐲(0) =

𝑂

• 𝑗=1

𝝎𝑗𝜕𝑗𝐵𝑗 = ̇ 𝐲0. The above equatjons are vector equatjons, each one corresponding to a system of 𝑂 equatjons, so we can compute the 2𝑂 constants of integratjon solving the 2𝑂 equatjons 𝑦0,𝑘 =

𝑂

• 𝑗=1

𝜔𝑘𝑗𝐶𝑗, ̇ 𝑦0,𝑘 =

𝑂

• 𝑗=1

𝜔𝑘𝑗𝜕𝑗𝐵𝑗 =, 𝑘 = 1, … , 𝑂.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Initjal Conditjons

Usually the initjal conditjons are expressed in terms of initjal displacements and initjal velocitjes 𝐲0 and ̇ 𝐲0, so we start deriving the expression of displacement with respect to tjme to obtain ̇ 𝐲(𝑢) =

𝑂

• 𝑗=1

𝝎𝑗𝜕𝑗(𝐵𝑗 cos 𝜕𝑗𝑢 − 𝐶𝑗 sin 𝜕𝑗𝑢) and evaluatjng the displacement and velocity for 𝑢 = 0 it is 𝐲(0) =

𝑂

• 𝑗=1

𝝎𝑗𝐶𝑗 = 𝐲0, ̇ 𝐲(0) =

𝑂

• 𝑗=1

𝝎𝑗𝜕𝑗𝐵𝑗 = ̇ 𝐲0. The above equatjons are vector equatjons, each one corresponding to a system of 𝑂 equatjons, so we can compute the 2𝑂 constants of integratjon solving the 2𝑂 equatjons 𝑦0,𝑘 =

𝑂

• 𝑗=1

𝜔𝑘𝑗𝐶𝑗, ̇ 𝑦0,𝑘 =

𝑂

• 𝑗=1

𝜔𝑘𝑗𝜕𝑗𝐵𝑗 =, 𝑘 = 1, … , 𝑂.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Orthogonality ‐ 1

Take into consideratjon two distjnct eigenvalues, 𝜕2

𝑠 and 𝜕2 𝑡 , and write the

characteristjc equatjon for each eigenvalue: 𝐋 𝝎𝑠 = 𝜕2

𝑠𝐍 𝝎𝑠

𝐋 𝝎𝑡 = 𝜕2

𝑡 𝐍 𝝎𝑡

premultjply each equatjon member by the transpose of the other eigenvector 𝝎𝑈

𝑡 𝐋 𝝎𝑠 = 𝜕2 𝑠𝝎𝑈 𝑡 𝐍 𝝎𝑠

𝝎𝑈

𝑠𝐋 𝝎𝑡 = 𝜕2 𝑡 𝝎𝑈 𝑠𝐍 𝝎𝑡

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Orthogonality ‐ 1

Take into consideratjon two distjnct eigenvalues, 𝜕2

𝑠 and 𝜕2 𝑡 , and write the

characteristjc equatjon for each eigenvalue: 𝐋 𝝎𝑠 = 𝜕2

𝑠𝐍 𝝎𝑠

𝐋 𝝎𝑡 = 𝜕2

𝑡 𝐍 𝝎𝑡

premultjply each equatjon member by the transpose of the other eigenvector 𝝎𝑈

𝑡 𝐋 𝝎𝑠 = 𝜕2 𝑠𝝎𝑈 𝑡 𝐍 𝝎𝑠

𝝎𝑈

𝑠𝐋 𝝎𝑡 = 𝜕2 𝑡 𝝎𝑈 𝑠𝐍 𝝎𝑡

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Orthogonality ‐ 2

The term 𝝎𝑈

𝑡 𝐋 𝝎𝑠 is a scalar, hence

𝝎𝑈

𝑡 𝐋 𝝎𝑠 = 𝝎𝑈 𝑡 𝐋 𝝎𝑠 𝑈 = 𝝎𝑈 𝑠𝐋𝑈 𝝎𝑡

but 𝐋 is symmetrical, 𝐋𝑈 = 𝐋 and we have 𝝎𝑈

𝑡 𝐋 𝝎𝑠 = 𝝎𝑈 𝑠𝐋 𝝎𝑡.

By a similar derivatjon 𝝎𝑈

𝑡 𝐍 𝝎𝑠 = 𝝎𝑈 𝑠𝐍 𝝎𝑡.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Orthogonality ‐ 3

Substjtutjng our last identjtjes in the previous equatjons, we have 𝝎𝑈

𝑠𝐋 𝝎𝑡 = 𝜕2 𝑠𝝎𝑈 𝑠𝐍 𝝎𝑡

𝝎𝑈

𝑠𝐋 𝝎𝑡 = 𝜕2 𝑡 𝝎𝑈 𝑠𝐍 𝝎𝑡

subtractjng member by member we fjnd that (𝜕2

𝑠 − 𝜕2 𝑡 ) 𝝎𝑈 𝑠𝐍 𝝎𝑡 = 0

We started with the hypothesis that 𝜕2

𝑠 ≠ 𝜕2 𝑡 , so for every 𝑠 ≠ 𝑡 we have that the

corresponding eigenvectors are orthogonal with respect to the mass matrix 𝝎𝑈

𝑠𝐍 𝝎𝑡 = 0,

for 𝑠 ≠ 𝑡.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Orthogonality ‐ 3

Substjtutjng our last identjtjes in the previous equatjons, we have 𝝎𝑈

𝑠𝐋 𝝎𝑡 = 𝜕2 𝑠𝝎𝑈 𝑠𝐍 𝝎𝑡

𝝎𝑈

𝑠𝐋 𝝎𝑡 = 𝜕2 𝑡 𝝎𝑈 𝑠𝐍 𝝎𝑡

subtractjng member by member we fjnd that (𝜕2

𝑠 − 𝜕2 𝑡 ) 𝝎𝑈 𝑠𝐍 𝝎𝑡 = 0

We started with the hypothesis that 𝜕2

𝑠 ≠ 𝜕2 𝑡 , so for every 𝑠 ≠ 𝑡 we have that the

corresponding eigenvectors are orthogonal with respect to the mass matrix 𝝎𝑈

𝑠𝐍 𝝎𝑡 = 0,

for 𝑠 ≠ 𝑡.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Orthogonality ‐ 4

The eigenvectors are orthogonal also with respect to the stjfgness matrix: 𝝎𝑈

𝑡 𝐋 𝝎𝑠 = 𝜕2 𝑠𝝎𝑈 𝑡 𝐍 𝝎𝑠 = 0,

for 𝑠 ≠ 𝑡. By defjnitjon 𝑁𝑗 = 𝝎𝑈

𝑗 𝐍 𝝎𝑗

and consequently 𝝎𝑈

𝑗 𝐋 𝝎𝑗 = 𝜕2 𝑗 𝑁𝑗.

𝑁𝑗 is the modal mass associated with mode no. 𝑗 while 𝐿𝑗 ≡ 𝜕2

𝑗 𝑁𝑗 is the respectjve

modal stjfgness.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Orthogonality ‐ 4

The eigenvectors are orthogonal also with respect to the stjfgness matrix: 𝝎𝑈

𝑡 𝐋 𝝎𝑠 = 𝜕2 𝑠𝝎𝑈 𝑡 𝐍 𝝎𝑠 = 0,

for 𝑠 ≠ 𝑡. By defjnitjon 𝑁𝑗 = 𝝎𝑈

𝑗 𝐍 𝝎𝑗

and consequently 𝝎𝑈

𝑗 𝐋 𝝎𝑗 = 𝜕2 𝑗 𝑁𝑗.

𝑁𝑗 is the modal mass associated with mode no. 𝑗 while 𝐿𝑗 ≡ 𝜕2

𝑗 𝑁𝑗 is the respectjve

modal stjfgness.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem

The Homogeneous Equatjon of Motjon Eigenvalues and Eigenvectors Eigenvectors are Orthogonal

Modal Analysis Examples

Orthogonality ‐ 4

The eigenvectors are orthogonal also with respect to the stjfgness matrix: 𝝎𝑈

𝑡 𝐋 𝝎𝑠 = 𝜕2 𝑠𝝎𝑈 𝑡 𝐍 𝝎𝑠 = 0,

for 𝑠 ≠ 𝑡. By defjnitjon 𝑁𝑗 = 𝝎𝑈

𝑗 𝐍 𝝎𝑗

and consequently 𝝎𝑈

𝑗 𝐋 𝝎𝑗 = 𝜕2 𝑗 𝑁𝑗.

𝑁𝑗 is the modal mass associated with mode no. 𝑗 while 𝐿𝑗 ≡ 𝜕2

𝑗 𝑁𝑗 is the respectjve

modal stjfgness.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis

Eigenvectors are a base EoM in Modal Coordinates Initjal Conditjons

Examples

Sectjon 3 Modal Analysis

Introductory Remarks The Homogeneous Problem Modal Analysis Eigenvectors are a base EoM in Modal Coordinates Initjal Conditjons Examples

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis

Eigenvectors are a base EoM in Modal Coordinates Initjal Conditjons

Examples

Eigenvectors are a base

The eigenvectors are reciprocally orthogonal, so they are linearly independent and for every vector 𝐲 we can write 𝐲 =

𝑂

• 𝑘=1

𝝎𝑘𝑟𝑘. The coeffjcients are readily given by premultjplicatjon of 𝐲 by 𝝎𝑈

𝑙𝐍, because

𝝎𝑈

𝑗 𝐍 𝐲 = 𝑂

• 𝑘=1

𝝎𝑈

𝑗 𝐍 𝝎𝑘𝑟𝑘 = 𝝎𝑈 𝑗 𝐍 𝝎𝑗𝑟𝑗 = 𝑁𝑗𝑟𝑗

in virtue of the ortogonality of the eigenvectors with respect to the mass matrix, and the above relatjonship gives 𝑟𝑗 = 𝝎𝑈

𝑗 𝐍 𝐲

𝑁𝑗 .

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis

Eigenvectors are a base EoM in Modal Coordinates Initjal Conditjons

Examples

Eigenvectors are a base

Generalising our results for the displacement vector to the acceleratjon vector and explicitjng the tjme dependency, it is 𝐲(𝑢) =

𝑂

• 𝑘=1

𝝎𝑘𝑟𝑘(𝑢), ̈ 𝐲(𝑢) =

𝑂

• 𝑘=1

𝝎𝑘 ̈ 𝑟𝑘(𝑢). Introducing 𝐫(𝑢), the vector of modal coordinates and 𝛀, the eigenvector matrix, whose columns are the eigenvectors, we can write 𝑦𝑗(𝑢) =

𝑂

• 𝑘=1

Ψ𝑗𝑘𝑟𝑘(𝑢), ̈ 𝑦𝑗(𝑢) =

𝑂

• 𝑘=1

Ψ𝑗𝑘 ̈ 𝑟𝑘(𝑢),

• r, in matrix notatjon

𝐲(𝑢) = 𝛀 𝐫(𝑢), ̈ 𝐲(𝑢) = 𝛀 ̈ 𝐫(𝑢).

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis

Eigenvectors are a base EoM in Modal Coordinates Initjal Conditjons

Examples

EoM in Modal Coordinates...

Substjtutjng the last two equatjons in the equatjon of motjon, 𝐍 𝛀 ̈ 𝐫 + 𝐋 𝛀 𝐫 = 𝐪(𝑢) premultjplying by 𝛀𝑈 𝛀𝑈𝐍 𝛀 ̈ 𝐫 + 𝛀𝑈𝐋 𝛀 𝐫 = 𝛀𝑈𝐪(𝑢) introducing the so called starred matrices, with 𝐪⋆(𝑢) = 𝛀𝑈𝐪(𝑢), we can fjnally write 𝐍⋆ ̈ 𝐫 + 𝐋⋆ 𝐫 = 𝐪⋆(𝑢). The vector equatjon above corresponds to the set of scalar equatjons 𝑞⋆

𝑗 = 𝑛⋆ 𝑗𝑘 ̈

𝑟𝑘 + 𝑙⋆

𝑗𝑘𝑟𝑘,

𝑗 = 1, … , 𝑂.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis

Eigenvectors are a base EoM in Modal Coordinates Initjal Conditjons

Examples

... are 𝑂 independent equatjons!

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

𝑗𝑘 = 𝝎𝑈 𝑗 𝐍 𝝎𝑘

= 𝜀𝑗𝑘𝑁𝑗, 𝑙⋆

𝑗𝑘 = 𝝎𝑈 𝑗 𝐋 𝝎𝑘

= 𝜕2

𝑗 𝜀𝑗𝑘𝑁𝑗.

where 𝜀𝑗𝑘 is the Kroneker symbol, 𝜀𝑗𝑘 = 1

𝑗 = 𝑘 𝑗 ≠ 𝑘

Substjtutjng in the equatjon of motjon, with 𝑞⋆

𝑗 = 𝝎𝑈 𝑗 𝐪(𝑢) we have

a set of uncoupled equatjons 𝑁𝑗 ̈ 𝑟𝑗 + 𝜕2

𝑗 𝑁𝑗𝑟𝑗 = 𝑞⋆ 𝑗 (𝑢),

𝑗 = 1, … , 𝑂

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis

Eigenvectors are a base EoM in Modal Coordinates Initjal Conditjons

Examples

... are 𝑂 independent equatjons!

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

𝑗𝑘 = 𝝎𝑈 𝑗 𝐍 𝝎𝑘

= 𝜀𝑗𝑘𝑁𝑗, 𝑙⋆

𝑗𝑘 = 𝝎𝑈 𝑗 𝐋 𝝎𝑘

= 𝜕2

𝑗 𝜀𝑗𝑘𝑁𝑗.

where 𝜀𝑗𝑘 is the Kroneker symbol, 𝜀𝑗𝑘 = 1

𝑗 = 𝑘 𝑗 ≠ 𝑘

Substjtutjng in the equatjon of motjon, with 𝑞⋆

𝑗 = 𝝎𝑈 𝑗 𝐪(𝑢) we have

a set of uncoupled equatjons 𝑁𝑗 ̈ 𝑟𝑗 + 𝜕2

𝑗 𝑁𝑗𝑟𝑗 = 𝑞⋆ 𝑗 (𝑢),

𝑗 = 1, … , 𝑂

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis

Eigenvectors are a base EoM in Modal Coordinates Initjal Conditjons

Examples

Initjal Conditjons Revisited

The initjal displacements can be writuen in modal coordinates, 𝐲0 = 𝛀 𝐫0 and premultjplying both members by 𝛀𝑈𝐍 we have the following relatjonship: 𝛀𝑈𝐍 𝐲0 = 𝛀𝑈𝐍 𝛀 𝐫0 = 𝐍⋆𝐫0. Premultjplying by the inverse of 𝐍⋆ and taking into account that 𝐍⋆ is diagonal, 𝐫0 = (𝐍⋆)−1 𝛀𝑈𝐍 𝐲0 ⇒ 𝑟𝑗0 = 𝝎𝑈

𝑗 𝐍 𝐲0

𝑁𝑗 and, analogously, ̇ 𝑟𝑗0 = 𝝎𝑗

𝑈𝐍 ̇

𝐲0 𝑁𝑗 .

Note that 𝑟𝑗0 and ̇ 𝑟𝑗0 depend only on the single eigenvector 𝝎𝑗.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Sectjon 4 Examples

Introductory Remarks The Homogeneous Problem Modal Analysis Examples 2 DOF System

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

2 DOF System

𝑙1 = 2𝑙, 𝑙2 = 3𝑙; 𝑛1 = 2𝑛, 𝑛2 = 4𝑛; 𝑞(𝑢) = 𝑞0 sin 𝜕𝑢. 𝑙1 𝑦1 𝑦2 𝑛2 𝑙2 𝑛1 𝑞(𝑢)

𝐲 = 𝑦1 𝑦2 , 𝐪(𝑢) = 0 𝑞0 sin 𝜕𝑢, 𝐍 = 𝑛 2 4 , 𝐋 = 𝑙 5 −3 −3 3 .

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Equatjon of frequencies

The equatjon of frequencies is 𝐋 − 𝜕2𝐍 = 5𝑙 − 2𝜕2𝑛 −3𝑙 −3𝑙 3𝑙 − 4𝜕2𝑛 = 0. Developing the determinant (8𝑛2) 𝜕4 − (26𝑛𝑙) 𝜕2 + (6𝑙2) 𝜕0 = 0 Solving the algebraic equatjon in 𝜕2 𝜕2

1 = 𝑙

𝑛 13 − √121 8 , 𝜕2

2 = 𝑙

𝑛 13 + √121 8 ; 𝜕2

1 = 1

4 𝑙 𝑛, 𝜕2

2 = 3 𝑙

𝑛.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Equatjon of frequencies

The equatjon of frequencies is 𝐋 − 𝜕2𝐍 = 5𝑙 − 2𝜕2𝑛 −3𝑙 −3𝑙 3𝑙 − 4𝜕2𝑛 = 0. Developing the determinant (8𝑛2) 𝜕4 − (26𝑛𝑙) 𝜕2 + (6𝑙2) 𝜕0 = 0 Solving the algebraic equatjon in 𝜕2 𝜕2

1 = 𝑙

𝑛 13 − √121 8 , 𝜕2

2 = 𝑙

𝑛 13 + √121 8 ; 𝜕2

1 = 1

4 𝑙 𝑛, 𝜕2

2 = 3 𝑙

𝑛.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Equatjon of frequencies

The equatjon of frequencies is 𝐋 − 𝜕2𝐍 = 5𝑙 − 2𝜕2𝑛 −3𝑙 −3𝑙 3𝑙 − 4𝜕2𝑛 = 0. Developing the determinant (8𝑛2) 𝜕4 − (26𝑛𝑙) 𝜕2 + (6𝑙2) 𝜕0 = 0 Solving the algebraic equatjon in 𝜕2 𝜕2

1 = 𝑙

𝑛 13 − √121 8 , 𝜕2

2 = 𝑙

𝑛 13 + √121 8 ; 𝜕2

1 = 1

4 𝑙 𝑛, 𝜕2

2 = 3 𝑙

𝑛.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Eigenvectors

Substjtutjng 𝜕2

1 for 𝜕2 in the fjrst of the characteristjc equatjons gives the ratjo

between the components of the fjrst eigenvector, 𝑙 (5 − 2 ⋅ 1 4)𝜔11 = 3𝑙𝜔21 while substjtutjng 𝜕2

2 gives

𝑙 (3 − 4 ⋅ 3)𝜔12 = 3𝑙𝜔22. Solving with the arbitrary assignment 𝜔11 = 𝜔22 = 1 gives the unnormalized eigenvectors, 𝝎1 = +1 +

3 2

, 𝝎2 = −3 +1 .

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Eigenvectors

Substjtutjng 𝜕2

1 for 𝜕2 in the fjrst of the characteristjc equatjons gives the ratjo

between the components of the fjrst eigenvector, 𝑙 (5 − 2 ⋅ 1 4)𝜔11 = 3𝑙𝜔21 while substjtutjng 𝜕2

2 gives

𝑙 (3 − 4 ⋅ 3)𝜔12 = 3𝑙𝜔22. Solving with the arbitrary assignment 𝜔11 = 𝜔22 = 1 gives the unnormalized eigenvectors, 𝝎1 = +1 +

3 2

, 𝝎2 = −3 +1 .

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Normalizatjon

We compute fjrst 𝑁1 and 𝑁2, 𝑁1 = 𝝎𝑈

1𝐍 𝝎1

= 𝑛 1,

3 2 2

4 1

3 2

• = 𝑛 2,

6 1

3 2

= 11 𝑛 and, in a similar way, we have 𝑁2 = 22 𝑛; the adimensional normalisatjon factors are 𝛽1 = √11 = 3.317, 𝛽2 = √22 = 4.690. Applying the normalisatjon factors to the respectjve unnormalised eigenvectors and collectjng them in a matrix, we have the matrix of normalized eigenvectors 𝛀 = +0.30151 −0.63960 +0.45227 +0.21320

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Modal Loadings

The modal loading is 𝐪⋆(𝑢) = 𝛀𝑈 𝐪(𝑢) = 𝑞0 1

3/2

−3 1 0 1 sin 𝜕𝑢 = 𝑞0

3/2

1 sin 𝜕𝑢

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Modal EoM

Substjtutjng its modal expansion for 𝐲 into the equatjon of motjon and premultjplying by 𝛀𝑈 we have the uncoupled modal equatjon of motjon

• 11 𝑛 ̈

𝑟1 + 1 4 11 𝑛 𝑙 𝑛 𝑟1 = 3 2𝑞0 sin 𝜕𝑢 22 𝑛 ̈ 𝑟2 + 3 22 𝑛 𝑙 𝑛 𝑟2 = 𝑞0 sin 𝜕𝑢 Note that all the terms are dimensionally correct. Dividing by 𝑁𝑗 both equatjons, we have

• ̈

𝑟1 + 1 4 𝜕2

0𝑟1 = 3/2 𝑞0

11𝑛 sin 𝜕𝑢 ̈ 𝑟2 + 3 𝜕2

0𝑟2 =

𝑞0 22𝑛 sin 𝜕𝑢

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Partjcular Integral

We set 𝜊1 = 𝐷1 sin 𝜕𝑢, ̈ 𝜊 = −𝜕2𝐷1 sin 𝜕𝑢 and substjtute in the fjrst modal EoM: 𝐷1 (𝜕2

1 − 𝜕2) sin 𝜕𝑢 = 3

22 𝑞0 𝑙 𝑙 𝑛 sin 𝜕𝑢 solving for 𝐷1 𝐷1 = 3 22Δ 𝜕2 𝜕2

1 − 𝜕2

and, analogously, 𝐷2 = 1 22Δ 𝜕2 𝜕2

2 − 𝜕2

with Δ = 𝑞0/𝑙.

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

Integrals

The integrals, for our loading, are thus

• 𝑟1(𝑢) = 𝐵1 sin 𝜕1𝑢 + 𝐶1 cos 𝜕1𝑢 + 𝐷1 sin 𝜕𝑢,

𝑟2(𝑢) = 𝐵2 sin 𝜕2𝑢 + 𝐶2 cos 𝜕2𝑢 + 𝐷2 sin 𝜕𝑢, and, for a system initjally at rest, it is

• 𝑟1(𝑢) = 𝐷1 (sin 𝜕𝑢 − 𝛾1 sin 𝜕1𝑢) ,

𝑟2(𝑢) = 𝐷2 (sin 𝜕𝑢 − 𝛾2 sin 𝜕2𝑢) , where 𝛾𝑗 = 𝜕/𝜕𝑗 We are interested in structural degrees of freedom, too...

• 𝑦1(𝑢) = (𝜔11 𝑟1(𝑢) + 𝜔12 𝑟2(𝑢))

𝑦2(𝑢) = (𝜔21 𝑟1(𝑢) + 𝜔22 𝑟2(𝑢))

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

The response in modal coordinates

To have a feeling of the response in modal coordinates, let’s say that the frequency of the load is 𝜕 = 2𝜕0, hence 𝛾1 =

2.0 1/4 = 4 and 𝛾2 = 2.0 √3 = 1.15470.

10 20 30 40 50

0t

0.1 0.0 0.1

qi/

st

Modal Response

q1 q2

In the graph above, the responses are plotued against an adimensional tjme coordinate 𝛽 with 𝛽 = 𝜕0𝑢, while the ordinates are adimensionalised with respect to Δst =

𝑞0 𝑙

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

The response in structural coordinates

Using the same normalisatjon factors, here are the response functjons in terms of 𝑦1 = 𝜔11𝑟1 + 𝜔12𝑟2 and 𝑦2 = 𝜔21𝑟1 + 𝜔22𝑟2:

10 20 30 40 50

0t

0.4 0.2 0.0 0.2

Xi/

st

Structural Response

x1 x2

Multj DoF Systems Giacomo Boffj Introductjon The Homogeneous Problem Modal Analysis Examples

2 DOF System

The response in structural coordinates

And the displacement of the centre of mass plotued along with the difgerence in displacements.

10 20 30 40 50

0t

0.4 0.2 0.0 0.2 0.4

Xi/

st

Structural Response Tweaked

(4x2 + 2x1)/6 (x2 x1)