1. FREE VIBRATION OF LINEAR MDOF
SYSTEMS
𝑚𝑚−1 =
1/𝑚𝑚 0 0
0 1/𝑚𝑚 0
0 0 1/𝑚𝑚
; 𝑚𝑚−1𝜅𝜅 =
2 −1 0
−1 2 −1
0 −1 2
= 𝐸𝐸
To obtain the eigenvalues, we solve: 𝐸𝐸 − 𝜔𝜔2Ι = 0 which leads to finding the roots
of a cubic polynomial in 𝜔𝜔2. The roots are:
𝜔𝜔1
2
= 0.5858 𝜔𝜔1 = 0.7654 rad/sec
𝜔𝜔2
2
= 2.0000 𝜔𝜔2 = 1.4142 rad/sec
𝜔𝜔3
2
= 3.4142 𝜔𝜔3 = 1.8478 rad/sec
And for each 𝜔𝜔𝑖𝑖 we can obtain eigenvectors by setting the first component in the
vector 𝑍𝑍1 = 1 and solving the remaining linear equations. Following this procedure
gives the eigenvectors:
𝑍𝑍(1) =
1.0
2
1.0
𝑍𝑍(2) =
1.0
0.0
−1.0
𝑍𝑍(3) =
1.0
− 2
1.0
𝜔𝜔1 𝜔𝜔2 𝜔𝜔3
So the 1st element is arbitrarily normalised to 1.

2. FREE VIBRATION OF LINEAR MDOF
SYSTEMS
The eigenvectors show the relative
amplitudes when the structure is vibrating
only at that corresponding natural frequency.
We can show these eigenvectors graphically in
the form of mode shapes, which give relative
vibrations for each of the modes.

3. FREE VIBRATION OF LINEAR MDOF
SYSTEMS
The mode shapes:
1
.
0
1
.
0
Mode 1
1
.
0
1
.
0
Mode 2
at 𝜔𝜔1
at 𝜔𝜔2
at 𝜔𝜔3
1
.
0
Mode 3
1
.
0

4. FREE VIBRATION OF LINEAR MDOF
SYSTEMS
Normalisation of Eigenvectors
It is often convenient to rescale the eigenvectors to different lengths. Sometimes
this might be (as in FE code) to make the largest amplitude 1 (as we have already
done). Another procedure is to make the eigenvectors have length 1 (as in Matlab)
i.e. to scale the eigenvectors as follows:
𝑢𝑢(1) =
𝑍𝑍(1)
𝑍𝑍 1
, 𝑢𝑢(2) =
𝑍𝑍(2)
𝑍𝑍(2)
, 𝑢𝑢(3) =
𝑍𝑍(3)
𝑍𝑍(3)
, … , 𝑢𝑢(𝑁𝑁) =
𝑍𝑍(𝑁𝑁)
𝑍𝑍(𝑁𝑁)
For the previous 3 x 3 system example, the unit-length normalised eigenvectors
are:
𝑢𝑢(1) =
𝑍𝑍(1)
𝑍𝑍 1
=
1
12+ 2
2
+12
�
1
2
1.0
2
1.0
=
1
2
1.0
2
1.0
=
⁄
1.0
2
�
1
2
⁄
1
2
𝑖𝑖. 𝑒𝑒. 𝑢𝑢(1) = 1

5. FREE VIBRATION OF LINEAR MDOF
SYSTEMS
And
𝑢𝑢(2) =
𝑍𝑍(2)
𝑍𝑍 2
=
1
12 + 02 + 12 �
1
2
1.0
0
−1.0
=
1
2
1.0
0
−1.0
=
�
1.0
2
0
�
−1
2
𝑢𝑢(2) = 1
And
𝑢𝑢(3) =
𝑍𝑍(3)
𝑍𝑍 3
=
1
12 + 2
2
+ 12
�
1
2
1.0
− 2
1.0
=
1
2
1.0
− 2
−1.0
=
�
1.0
2
�
−1.0
2
�
1
2
𝑢𝑢(3) = 1

6. FREE VIBRATION OF LINEAR MDOF
SYSTEMS
The Modal Matrix
It is common to arrange the Normalised eigenvectors into a special matrix as
follows:
U = 𝑢𝑢(1), 𝑢𝑢 2 , … , 𝑢𝑢(𝑁𝑁)
For the 3 x 3 example, the modal matrix [U] is:
U =
⁄
1
2 �
1
2
⁄
1
2
�
1
2
0 �
−1.0
2
⁄
1
2 �
−1
2
⁄
1
2
The modal matrix for the previous example.
The modal matrix [U] is an important matrix which allows a system of
equations, under certain conditions, to be diagonalised. We will examine this
feature in this lecture.

7. Lecture 9 Forced Response of MDOF
Linear Systems
ORTHOGONALITY OF NORMAL MODES
Here the properties of Normal modes are derived using a simpler proof
that assumes distinct Natural frequencies i.e. no repeated eigenvalues.
Consider the standard eigenvalue problem for the case when 𝑚𝑚 and 𝑘𝑘
are symmetric matrices. An important property of Normal modes can be
established which will be very important for the analysis of systems using
Normal Coordinates. Now if:
𝑘𝑘 − 𝜔𝜔2 𝑚𝑚 ̂
𝑍𝑍 = 0
For any given eigenvalue 𝜔𝜔𝑖𝑖
2
and corresponding eigenvector 𝑍𝑍 𝑖𝑖 then:
𝜔𝜔𝑖𝑖
2
𝑚𝑚 𝑍𝑍 𝑖𝑖 = 𝑘𝑘 𝑍𝑍 𝑖𝑖 (5.4)
and for some other eigenvalue 𝜔𝜔𝑗𝑗
2
𝜔𝜔𝑗𝑗
2
𝑚𝑚 𝑍𝑍 𝑗𝑗 = 𝑘𝑘 𝑍𝑍 𝑗𝑗 (5.5)

8. ORTHOGONALITY OF NORMAL MODES
Now if we pre-multiply 𝜔𝜔𝑖𝑖
2
𝑚𝑚 𝑍𝑍 𝑖𝑖 = 𝑘𝑘 𝑍𝑍 𝑖𝑖 by 𝑍𝑍(𝑗𝑗)
𝑇𝑇
, and 𝜔𝜔𝑗𝑗
2
𝑚𝑚 𝑍𝑍 𝑗𝑗 = 𝑘𝑘 𝑍𝑍 𝑗𝑗 by 𝑍𝑍(𝑖𝑖)
𝑇𝑇
, we
get:
𝜔𝜔𝑖𝑖
2
𝑍𝑍(𝑗𝑗)
𝑇𝑇
𝑚𝑚 𝑍𝑍 𝑖𝑖 = 𝑍𝑍(𝑗𝑗)
𝑇𝑇
𝑘𝑘 𝑍𝑍(𝑖𝑖) ≡ 𝑍𝑍 𝑖𝑖
𝑇𝑇
𝑘𝑘 𝑍𝑍 𝑗𝑗 (5.6)
and
𝜔𝜔𝑗𝑗
2
𝑍𝑍(𝑖𝑖)
𝑇𝑇
𝑚𝑚 𝑍𝑍 𝑗𝑗 = 𝜔𝜔𝑗𝑗
2
𝑍𝑍(𝑗𝑗)
𝑇𝑇
𝑚𝑚 𝑍𝑍(𝑖𝑖) ≡ 𝑍𝑍 𝑖𝑖
𝑇𝑇
𝑘𝑘 𝑍𝑍 𝑗𝑗 (5.7)
where the last term in (5.6) and the 2nd term in (5.7) have been transposed (also
using (5.5) and (5.4)), and where the rhs is typically some scalar 𝛼𝛼, which, for 𝑚𝑚
and 𝑘𝑘 being symmetric give (XT
AY)T
= 𝑌𝑌𝑇𝑇
𝐴𝐴𝐴𝐴 = 𝛼𝛼 = 𝛼𝛼𝑇𝑇
. Now if 𝜔𝜔𝑖𝑖 ≠ 𝜔𝜔𝑗𝑗 when
𝑖𝑖 ≠ 𝑗𝑗 i.e. no repeated eigenvalues then subtracting (5.7) from (5.6), gives:
𝜔𝜔𝑖𝑖
2
− 𝜔𝜔𝑗𝑗
2
𝑍𝑍(𝑗𝑗)
𝑇𝑇
𝑚𝑚 𝑍𝑍(𝑖𝑖) = 0

9. ORTHOGONALITY OF NORMAL MODES
Therefore from 𝜔𝜔𝑖𝑖
2
− 𝜔𝜔𝑗𝑗
2
𝑍𝑍(𝑗𝑗)
𝑇𝑇
𝑚𝑚 𝑍𝑍(𝑖𝑖) = 0 we have:
𝑍𝑍 𝑗𝑗
𝑇𝑇
𝑚𝑚 𝑍𝑍 𝑖𝑖 = 0 𝑖𝑖 ≠ 𝑗𝑗 (5.9)
And from (5.6):
𝑍𝑍 𝑗𝑗
𝑇𝑇
𝑘𝑘 𝑍𝑍 𝑖𝑖 = 0 𝑖𝑖 ≠ 𝑗𝑗 (5.10)
Equation (5.9) is a statement that the normal modes 𝑍𝑍(𝑖𝑖) are orthogonal with respect
to the mass matrix for all i. Equation (5.10) states that 𝑍𝑍(𝑖𝑖) are orthogonal with
respect to the stiffness Matrix 𝒌𝒌 for all i.
When i=j, (and if eigenvectors 𝑍𝑍(𝑖𝑖) are normalised to magnitude = 1) then in general:
𝑍𝑍(𝑖𝑖)
𝑇𝑇
𝑚𝑚 𝑍𝑍(𝑖𝑖) = 𝑀𝑀𝑖𝑖𝑖𝑖 𝑖𝑖 = 1, … , 𝑁𝑁 known as the Generalised Mass
and
𝑍𝑍(𝑖𝑖)
𝑇𝑇
𝑘𝑘 𝑍𝑍(𝑖𝑖) = 𝐾𝐾𝑖𝑖𝑖𝑖 𝑖𝑖 = 1, … , 𝑁𝑁 known as the Generalised Stiffness
Now all of the above equations can be written in the form:

10. ORTHOGONALITY OF NORMAL MODES
. . .
. 𝑴𝑴 .
. . .
=
𝑀𝑀11 . 0
. 𝑀𝑀22 .
0 . .
= U 𝑇𝑇
𝑚𝑚 U
and
. . .
. 𝑲𝑲 .
. . .
=
𝐾𝐾11 . 0
. 𝐾𝐾22 .
0 . .
= 𝑈𝑈 𝑇𝑇 𝑘𝑘 𝑈𝑈
where : 𝑍𝑍(𝑖𝑖)
𝑇𝑇
𝑚𝑚 𝑍𝑍(𝑖𝑖) = 𝑀𝑀𝑖𝑖𝑖𝑖 and 𝑍𝑍(𝑖𝑖)
𝑇𝑇
𝑘𝑘 𝑍𝑍(𝑖𝑖) = 𝐾𝐾𝑖𝑖𝑖𝑖, and 𝑈𝑈 is the modal matrix i.e. U =
𝑍𝑍 1 , 𝑍𝑍 2 , 𝑍𝑍 3 , … , 𝑍𝑍𝑛𝑛 ; note 𝑴𝑴 and 𝑲𝑲 are diagonal matrices.
These highly important relations only apply when 𝑚𝑚 and 𝑘𝑘 are symmetric, and the
eigenvalues are unique. When the eigenvalues are repeated (i.e. two are the same)
they do not apply (unless the dynamic matrix 𝐸𝐸 or (𝐸𝐸−1
= 𝐷𝐷) is symmetric (i.e.
lumped-mass model) in which case [U]T
= [U]−1
i.e. modes are orthogonal to each
other.

11. Uncoupling Equations with Damping
using Normal Coordinates
The real benefit of the orthogonality properties of normal modes is in
analysing systems with proportional damping. Consider the system:
𝑚𝑚 ̈
𝑧𝑧 + 𝑐𝑐 ̇
𝑧𝑧 + 𝑘𝑘 𝑧𝑧 = 𝑝𝑝(𝑡𝑡)
Now suppose that the damping matrix is proportional to both the mass and
stiffness matrix i.e.
𝑐𝑐 = 𝛼𝛼 𝑚𝑚 + 𝛽𝛽[𝑘𝑘]
where α and β are constants. Then substitution of this damping model into
the equation:
𝑚𝑚 ̈
𝑧𝑧 + 𝛼𝛼 𝑚𝑚 + 𝛽𝛽 𝑘𝑘 ̇
𝑧𝑧 + 𝑘𝑘 𝑧𝑧 = 𝑝𝑝(𝑡𝑡)
Now we use the modal matrix [U] to do a normal coordinate transformation
i.e. by transforming the physical coordinates 𝑧𝑧 to modal coordinates 𝑞𝑞 i.e.
𝑧𝑧 → 𝑞𝑞 by setting:

12. Uncoupling Equations with Damping
using Normal Coordinates
𝑧𝑧 = [U]𝑞𝑞
̇
𝑧𝑧 = [U] ̇
𝑞𝑞
̈
𝑧𝑧 = [U] ̈
𝑞𝑞
Substituting 𝑧𝑧, ̇
𝑧𝑧, and ̈
𝑧𝑧 above gives:
𝑚𝑚 U ̈
𝑞𝑞 + 𝛼𝛼 𝑚𝑚 + 𝛽𝛽 𝑘𝑘 U ̇
𝑞𝑞 + 𝑘𝑘 U 𝑞𝑞 = 𝑝𝑝(𝑡𝑡)
Now pre-multiply through by [U]𝑇𝑇
i.e.
[U]𝑇𝑇
[𝑚𝑚][U] ̈
𝑞𝑞 + 𝛼𝛼[U]𝑇𝑇
𝑚𝑚 + 𝛽𝛽[U]𝑇𝑇
𝑘𝑘 U ̇
𝑞𝑞 + [U]𝑇𝑇
𝑘𝑘 U 𝑞𝑞 = [U]𝑇𝑇
𝑝𝑝(𝑡𝑡) (5.11)
But we know that the modal matrix is orthogonal with respect to [m] and [k] i.e.
[U]T
𝑚𝑚 U and [U]T
𝑘𝑘 U are both diagonal matrices. Therefore, equation (5.11)
now reduces to a system of uncoupled SDOF equations i.e.
⋱ 0 0
0 𝑀𝑀 0
0 0 ⋱
̈
𝑞𝑞 + 𝛼𝛼
⋱ 0 0
0 𝑀𝑀 0
0 0 ⋱
+ 𝛽𝛽
⋱ 0 0
0 𝐾𝐾 0
0 0 ⋱
̇
𝑞𝑞 +
⋱ 0 0
0 𝐾𝐾 0
0 0 ⋱
𝑞𝑞 = [U]𝑇𝑇
𝑝𝑝 𝑡𝑡

13. Uncoupling Equations with Damping
using Normal Coordinates
i.e. 𝑀𝑀𝑖𝑖𝑖𝑖 ̈
𝑞𝑞𝑖𝑖 + 𝛼𝛼𝑀𝑀𝑖𝑖𝑖𝑖 + 𝛽𝛽𝐾𝐾𝑖𝑖𝑖𝑖 ̇
𝑞𝑞𝑖𝑖 + 𝐾𝐾𝑖𝑖𝑖𝑖𝑞𝑞𝑖𝑖 = U 𝑇𝑇𝑝𝑝(𝑡𝑡)
𝑖𝑖
𝑖𝑖 = 1, … , 𝑁𝑁
And on division by 𝑀𝑀𝑖𝑖𝑖𝑖 gives:
̈
𝑞𝑞𝑖𝑖 + 𝛼𝛼 + 𝛽𝛽
𝐾𝐾𝑖𝑖𝑖𝑖
𝑀𝑀𝑖𝑖𝑖𝑖
̇
𝑞𝑞𝑖𝑖 +
𝐾𝐾𝑖𝑖𝑖𝑖
𝑀𝑀𝑖𝑖𝑖𝑖
𝑞𝑞𝑖𝑖 =
U 𝑇𝑇𝑝𝑝(𝑡𝑡) 𝑖𝑖
𝑀𝑀𝑖𝑖𝑖𝑖
If we let:
𝐾𝐾𝑖𝑖𝑖𝑖
𝑀𝑀𝑖𝑖𝑖𝑖
= 𝜔𝜔𝑛𝑛𝑛𝑛
2
then we have:
̈
𝑞𝑞𝑖𝑖 + 2𝜉𝜉𝑖𝑖𝜔𝜔𝑛𝑛𝑛𝑛 ̇
𝑞𝑞𝑖𝑖 + 𝜔𝜔𝑛𝑛𝑛𝑛
2
𝑞𝑞𝑖𝑖 =
U 𝑇𝑇𝑝𝑝(𝑡𝑡) 𝑖𝑖
𝑀𝑀𝑖𝑖𝑖𝑖
(6.4)
where 𝜔𝜔𝑛𝑛𝑛𝑛 is the undamped modal natural frequency of the ith mode, and 𝜉𝜉𝑖𝑖 is
the ith modal damping factor where:
𝜉𝜉𝑖𝑖 =
𝛼𝛼 + 𝛽𝛽𝜔𝜔𝑛𝑛𝑛𝑛
2
2𝜔𝜔𝑛𝑛𝑛𝑛
=
𝛼𝛼
2𝜔𝜔𝑛𝑛𝑛𝑛
+
𝛽𝛽
2
𝜔𝜔𝑛𝑛𝑛𝑛

14. Uncoupling Equations with Damping
using Normal Coordinates
Equation (6.4) can be solved repeatedly for each of the terms q1(t), q2(t), ..., qN(t) using
the techniques discussed in the 1st three lectures as separate SDOF equations to obtain
qT =q1(t), q2(t),…,qN(t)). The physical displacement vector can then be obtained i.e.:
z(t)=[U]q
which now superposes the solution of the individual modes.
A note on the damping model
It is often not possible to construct the damping matrix theoretically. Indeed modal
damping factors are usually measured by exciting a real structure and measuring the
response. However we can see from the form of proportional damping, that as a
model, it describes modal damping which for damping proportional to stiffness,
increases with mode frequency and therefore is not a good model for high frequency
vibration.

15. Uncoupling Equations with Damping
using Normal Coordinates
EXAMPLE
Consider a 2DOF system with the following dynamic model:
𝑚𝑚 ̈
𝑧𝑧 + 𝑐𝑐 ̇
𝑧𝑧 + 𝑘𝑘 𝑧𝑧 = 𝑝𝑝(𝑡𝑡)
m1 = 8000 kg and m2 = 4000 kg
k1 = k2 = k3 = 1000 kN/m

16. Uncoupling Equations with Damping
using Normal Coordinates
EXAMPLE
Construct free body diagrams and using Newtonian mechanics and equations of
motion, get:
[𝑚𝑚] =
8000 0
0 4000
kg ; [𝑘𝑘] =
3000000 −1000000
−1000000 1000000
N/m
Proportional Damping is assumed, i.e. [c] is proportional to mass:
[c]=10[m]
Therefore in the Proportional Damping model 𝑐𝑐 = 𝛼𝛼 𝑚𝑚 + 𝛽𝛽[𝑘𝑘] here,
𝛼𝛼=10 and 𝛽𝛽=0. The task is to uncouple the model into SDOF equations.

17. Uncoupling Equations with Damping
using Normal Coordinates
EXAMPLE
So if we remind ourselves of Equation (5.11) i.e.:
[U]𝑇𝑇
[𝑚𝑚][U] ̈
𝑞𝑞 + 𝛼𝛼[U]𝑇𝑇
𝑚𝑚 + 𝛽𝛽[U]𝑇𝑇
𝑘𝑘 U ̇
𝑞𝑞 + [U]𝑇𝑇
𝑘𝑘 U 𝑞𝑞 = [U]𝑇𝑇
𝑝𝑝(𝑡𝑡)
where we know that [U]𝑇𝑇 𝑚𝑚 U and [U]𝑇𝑇 𝑘𝑘 U are both diagonal matrices.
For this system, the Modal Matrix (i.e. the normalised eigenvectors of [m]-1[k])
is:
[U] =
0.7071 0.4472
−0.7071 0.8944
(This is actually the eigenvector matrix for [m]-1[k] that Matlab returned)

18. Uncoupling Equations with Damping
using Normal Coordinates
EXAMPLE
The numerical magnitudes of U T
𝑚𝑚 U and U T
𝑘𝑘 U are:
U T
𝑚𝑚 U =
6.0 0
0 4.8
x 103
=
𝑀𝑀11 0
0 𝑀𝑀22
;
and
U T 𝑘𝑘 U =
3.0 0
0 0.6
x 106=
𝐾𝐾11 0
0 𝐾𝐾22
And since in this example:
̈
𝑞𝑞𝑖𝑖 + 𝛼𝛼 ̇
𝑞𝑞𝑖𝑖 +
𝐾𝐾𝑖𝑖𝑖𝑖
𝑀𝑀𝑖𝑖𝑖𝑖
𝑞𝑞𝑖𝑖 =
𝑈𝑈 𝑇𝑇𝑝𝑝(𝑡𝑡) 𝑖𝑖
𝑀𝑀𝑖𝑖𝑖𝑖
i =1,2
or
̈
𝑞𝑞𝑖𝑖 + 2𝜉𝜉𝑖𝑖𝜔𝜔𝑛𝑛𝑛𝑛 ̇
𝑞𝑞𝑖𝑖 + 𝜔𝜔𝑛𝑛𝑛𝑛
2
𝑞𝑞𝑖𝑖 =
U 𝑇𝑇𝑝𝑝(𝑡𝑡) 𝑖𝑖
𝑀𝑀𝑖𝑖𝑖𝑖
where:
𝜔𝜔𝑛𝑛𝑛𝑛
2
=
𝐾𝐾𝑖𝑖𝑖𝑖
𝑀𝑀𝑖𝑖𝑖𝑖
𝛼𝛼 = 2𝜉𝜉𝑖𝑖𝜔𝜔𝑛𝑛𝑛𝑛 or 𝜉𝜉𝑖𝑖 =
𝛼𝛼
2𝜔𝜔𝑛𝑛𝑛𝑛

19. Uncoupling Equations with Damping
using Normal Coordinates
EXAMPLE
Evaluating the natural frequencies and damping factors (starting with the lowest) we get:
𝜔𝜔1 =11.1803 rad/s 𝜔𝜔2 =22.3607 rad/s
and correspondingly:
𝜉𝜉1= 0.4472 or 44.72% critical and 𝜉𝜉2= 0.2236 or 22.36% critical.
So we now have two uncoupled 2nd order equations of the form given by equations (6.4)
which can be solved as SDOF systems for 𝑞𝑞1(t) and 𝑞𝑞2(𝑡𝑡) from which 𝑧𝑧 = [U]𝑞𝑞 allows us to
obtain the response in terms of the original physical variable vector 𝑧𝑧.
Note: Matlab sometimes gives the highest eigenvector as the first column using the eig
function. It is wholly correct but it may result in the order of the modal parameters not
starting with the first mode.

20. Matlab code for example
% Matlab code for example in Lecture 9
clear all
M = [8 0;0 4]*1e3;
k = [3 -1;-1 1]*1e6;
alpha = 10;
c = alpha*M;
E = inv(M)*k; % E = [m]-1*[k]
[U,eigvalues] = eig(E); % calculate U from (E-w^2I)=0
Mii = diag(U'*M*U) % take the diagonals of the matrix
Kii = diag(U'*k*U)
w = sqrt(Kii./Mii)
psi = alpha./(2*w)

21. Notes made in lecture

22. Notes made in lecture