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Last update: 2010 Ahmed Elgamal
Multi-Degree-Of-Freedom (MDOF) Systems and Modal Analysis
Ahmed Elgamal
1
Ahmed Elgamal
SDOF Shear Building (rigid roof)
m = lumped mass = mroof + 2 (1/2 mcol)
3
c
3
ccol
h
24EI
h
12EI22kk
gumuckuum
2
2
Ahmed Elgamal
2-Story Shear Building (2-DOF system)
m1
m2
k1
k2
u1
u2
col22 m
2
12roofmm
col11 m
2
14floormm
col22 2kk col11 2kk
g212222 um)u(ukum
g
2
1
2
1
22
221
2
1
2
1u
m
m
u
u
kk
kkk
u
u
m0
0m
g11121211 umuk)u(ukum
gu1mukum or,
where 1 or 1 is the Identity Matrix
gu m1kuum 3
Ahmed Elgamal
m1
m3
k1
k2
u1
u2
u3
k3
m2
g323333 um)u(ukum
g212232322 um)u(uk)u(ukum
g11121211 umuk)u(ukum
g
3
2
1
3
2
1
33
3322
221
3
2
1
3
2
1
u
m
m
m
u
u
u
kk0
kkkk
0kkk
u
u
u
m00
0m0
00m
3-Story Shear Building (3-DOF system)
4
3
Ahmed ElgamalFor a N-DOF system,
n
m
2
1
m
m
m
m
nn
n
1m1mmm
3322
221
kk000
k|
0|
0k-kkk-00
|
|
0
0
00kkkk
00k-kk
m
k
5
Ahmed ElgamalEquation of motion for a N-DOF system,
n
m
2
1
n
k
2
1
u
|
|
u
|
|
|
|
u
u
m
m
m
m
n
m
2
1
nn
n
1m1mmm
3322
221
u
|
|
u
|
|
|
|
u
u
kk000
k|
0|
0k-kkk-00
|
|
0
0
00kkkk
00k-kk
g
n
m
2
1
u
m
|
|
m
|
|
|
|
m
m
with initial conditions:
gu m1kuucum
uu
uu
(t=0)
(t=0)
or (with damping included):
6
4
Natural Frequencies of a N-DOF system
Similar to the SDOF system, MDOF systems have natural frequencies. A 2-DOF
has 2 natural frequencies w1 and w2, and a n-DOF system has natural frequencies
w1 , w2 , …, wn
Similar to the SDOF, free vibration involves the system response in its
natural frequencies. The corresponding Free Vibration Equation is (with
no damping):
0kuum
In free vibration, the system will oscillate in a steady-state harmonic
fashion, such that:
uu2ω
tcosbtsinau ωω e.g. gives uu 2-ω7
Ahmed Elgamal
substituting for u , we get:
0 u km-2ω
or
0u m-k2ω Equation 1
The above equation represents a classic problem in
Math/Physics, known as the Eigen-value problem.
The trivial solution of this problem is u = 0 (i.e., nothing is
happening, and the system is at rest). 8
5
Ahmed Elgamal
For a non-trivial solution (which will allow for computing the
natural frequencies during free vibration), the determinant of
0 m-k 2ω 0 m-k λor 2λ ωwhere
For a 2-DOF system for instance (see next page), the
above determinant calculation will result in a quadratic
equation in the unknown term l . If this quadratic equation
is solved (by hand), two roots are found (l1 and l2), which
define w1 and w2 (the natural resonant frequencies of this
2-DOF system).
m-k2ω must be equal to zero such that:
9
Ahmed Elgamal
2-Story Shear Building (2-DOF system)
m1
m2
k1
k2
u1
u2 g
2
1
2
1
22
221
2
1
2
1u
m
m
u
u
kk
kkk
u
u
m0
0m
0
0
u
u
mkk
kmkk
2
1
222
2121
l
l
0u m-k l
0
mkk
kmkk
222
2121
l
l
0)k)(k()mk)(mkk 2222121( ll
0)kk())kk(mkm(()mm( 2121221
2
21 ll
02 cba lla
acbb
2
4
2
21l
Solve for the l1 and l2 using the standard approach
ZY
XW
Determinant = WZ-XY
Note: For
Set Determinant = 0:
or,
10
6
Ahmed Elgamal
For a general N-DOF system:
Matlab or similar computer program can be used to solve the
determinant equation (of order equal to the NDOF system), defining
NDOF roots or NDOF natural frequencies
w1 , w2 ,…, wNDOF
Note:
These resonant (natural) frequencies w1 , w2 , … are conventionally
ordered lowest to highest (e.g., w1 = 8 radians, w2 = 14 radians, and
so forth).
11
Ahmed Elgamal
Mode ShapesSteady State vibration at any of the resonant frequencies wn
takes place in the form of a special oscillatory shape, know
as the corresponding mode shape fn
To define these mode shapes (one for each identified wn), go
ahead and substitute the value of wn for w in Eq. 1
and solve for the vector u which will define the
corresponding mode shape fn :
0 m-k nn f2ω
0u m-k2ω
12
7
Ahmed Elgamal2-DOF system ( 2 mode shapes f1 and f2)
u1
u2m2
m1
11φ
21φ
f12
f22
u1
u2m2
m1
11φ
21φ
f1 f2
Note: Any mode shape fn only defines relative amplitudes of motion of the
different degrees of freedom in the MDOF system. For instance, if you are solving
a 2-DOF system, you might end up with something like (when solving for the first
mode):
f11 - 2f21 = 0, only defining a ratio between amplitudes of f11 and f21
(for instance, if f11 = 1, then f21 = 0.5, or if you choose f11 = 2, then f21 = 1, and so
forth).
Generally, go ahead and make fmn= 1 (where m is top floor Dof and n is mode
shape number) and solve for the other degrees of freedom in the vector fn
21
11
1f
ff
22
12
2f
ff
13
Ahmed Elgamal
Note: When you substitute any of the wn values into Eq. 1, the determinant of the matrix (k-
wn2m) automatically becomes = 0, since this wn is a root of the determinant equation (i.e.,
the matrix becomes singular).
The determinant being zero is a necessary condition for obtaining a vector u (the mode shape
fn) that is not equal to zero (i.e., a solution other than the trivial solution of u = 0.
2-Story shear building
(one node in mode 2)4-story shear building (4-DOF system)
Note one node in mode 2, two in mode 3, and 3 in mode 4
3-Story Shear Building
node
nodenode
node
Sample Mode shape Configurations
14
8
Ahmed Elgamal
nn
T
n Mφφ m
n
2
nnn
T
n MK ωφφ k
1.0n
T
n φφ m2
nn
T
n ωφφ k
Properties of fn
b) For any mode fn , modal mass Mn is defined by:
d) If then
c) For any mode fn , modal stiffness Kn is defined by:
To do that, multiply each component of mode fn bynM
1
(not )0r
T
nr
T
n φφφφ mk 0r
T
n φφ
a) Mode shapes are orthogonal such that (for any nr)
15
Ahmed Elgamal
Solution of by Mode superposition
Example of a 2-DOF system ( 2 mode shapes and )1φ 2φ
u1
u2m2
m1
11φ
21φ f22
f12
16
9
To benefit from the mode orthogonality property, multiply by FT to get:
Substituting
Results in
Ahmed Elgamal
Φqu
2
1
2221
1211
2
22
12
1
21
11
2
1
q
qqq
u
u
φφ
φφ
φ
φ
φ
φ Φqq 21 φφ
in the equation of motiongu m1kuum
gu m1kΦqmΦ q
gu m1ΦkΦΦqmΦΦTTT q
gT
2
T
1
2
1
2
T
21
T
2
2
T
11
T
1
2
1
2
T
21
T
2
2
T
11
T
1 uq
q
q
q
m1
m1
kk
kk
mm
mm
φ
φ
φφφφ
φφφφ
φφφφ
φφφφor
Modal Analysis (Solution of MDOF equation of motion by Mode Superposition)
The solution u will be represented by a summation of the mode shapes fn, each
multiplied by a scaling factor qn (known as the generalized coordinate) . For instance,
for the 2-DOF system:
In the above, F is known as the modal matrix. As such, changes in the displaced shape
of the structure u with time will be captured by the time histories of the vector q
u =
17
Ahmed Elgamal
Due to the orthogonality property of mode shapes (see previous slide), the
matrix equation becomes un-coupled and we get:
g22
2
2222
g11
2
1111
uLqMqM
uLqMqM
ω
ω
or,
g
2
22
2
22
g
1
11
2
11
uM
Lqq
uM
Lqq
ω
ω
The terms L1/M1 and L2/M2 are known as modal participation factors. These terms
control the influence of on the modal response. You may notice that (if both modes
are normalized to 1.0 at roof level for example) L1/M1 > L2/M2 since f11 and f21 are
of the same sign while f12 and f22 are of opposite signs. Therefore, the first mode is
likely to play a more prominent role in the overall response (frequency content of the
input ground motion also affects this issue).
gu
NDOF
1j
φ2
jiji mM
NDOF
1j
φ jiji mLFor a diagonal mass matrix:
u1
u2m2
m1
11φ
21φ
f12
f22
u1
u2m2
m1
11φ
21φ
f1 f2
18
10
Ahmed Elgamal
Note that the original coupled matrix Eq. of motion, has now become
a set of un-coupled equations. You can solve each one separately (as
a SDOF system), and compute histories of q1 and q2 and their time
derivatives. To compute the system response, plug the q vector back
into Equation 2 and get the u vector
(the same for the time derivatives to get relative velocity and
acceleration).
The beauty here is that there is no matrix operations involved, since
the matrix equation of motion has become a set of un-coupled
equation, each including only one generalized coordinate qn.
Φqu
19
Ahmed Elgamal
Now, you can add any modal damping you wish (which is
another big plus, since you control the damping in each mode
individually). If you choose = 0.02 or 0.05, the equations
become:
Damping in a Modal Solution
i
g
i
ii
2
iiiii uM
Lqq2q ωω , i = 1, 2, … NDOF
OK, go ahead now and solve for qi(t) in the above uncoupled equations
(using a SDOF-type program), and the final solution is obtained from:
Φqu
qΦu
qΦu
g
t u 1uu 20
11
Ahmed Elgamal
Φqqu
21
2
1
32
22
31
21
1211
2
32
22
12
1
31
21
11
3
2
1
q
qqq
u
u
u
φφ
φ
φ
φ
φφφ
φ
φ
φ
φ
φ
φ
Modal Analysis (3-DOF system)
The solution u will be represented by a summation of the mode shapes fn, each
multiplied by a scaling factor qn (known as the generalized coordinate) . For instance,
for the 3-DOF system:
In the above, F is known as the modal matrix. As such, changes in the displaced shape
of the structure u with time will be captured by the time histories of the vector q
Note: If a two mode solution is sought, the system above becomes:
Φqqu
321
3
2
1
333231
232221
131211
3
33
23
13
2
32
22
12
1
31
21
11
3
2
1
q
q
q
qqq
u
u
u
φφφ
φφφ
φφφ
φφφ
φ
φ
φ
φ
φ
φ
φ
φ
φ
111
31
21
11
3
2
1
u
u
u
φ
φ
φ
φ
u
Note: If a single (1st or fundamental) mode solution is sought, the system above
becomes:
21
Ahmed Elgamal
Multi-Degree-Of-Freedom (MDOF) Response Spectrum Procedure
1. Once you have generalized coordinates and uncoupled equations, use response
spectrum to get maximum values of response (ri)max for each mode separately.
Calculate expected max response ( ) usingr 2
maximax rr root sum
square formula
where i = 1, 2, … N degrees of freedom of interest (maybe first 4 modes at most) and
r is any quantity of interest such as |umax| or SD
(note that summing the maxima from each mode directly is typically too
conservative and is therefore not popular; because the maxima occur at different
time instants during the earthquake excitation phase)
See A. Chopra “Dynamics of Structures” for improved formulae to estimate .maxr22
12
Ahmed Elgamal
Response Spectrum Modal Responses
Max relative displacement |un| or |ujn| (jth floor, nth mode)
jnnd
n
njn S
M
Lu φ (Sdn is Sd evaluated at frequency or period Tn)nω
Estimate of maximum floor displacement
M
1n
2
jnj uu (M = number of modes of interest)
Maximum Equivalent static force fn or fjn (jth floor, nth mode)
jnjna
n
njn mS
M
Lf φ
23
Ahmed Elgamal
Therefore, modal base shear V0n and moment M0n
N
1j
jn0n fV
# of floors
base
N
1j
jjn0n dfM
where dj = Distance from floor j to base
Estimate of maximum base shear and moment:
M
1n
2
0n0 VV
M
1n
2
0n0 MM
24
13
Ahmed Elgamal
Damping Matrix for MDOF Systems
Mass-proportional damping
c = aom
Stiffness-proportional damping
c = a1k
Classical damping (Rayleigh damping)
Stiffness proportional damping appeals to intuition
because it generates damping based on story
deformations. However, mass proportional damping may
be needed as will be shown below.
gu m1kuucum
kmc 10 aa
25
In any modal equation, we have
where,
Therefore, ao can be specified to obtain any desired zn for
a given mode n such that Cn = a0 Mn
or
(e.g. at w1 = 2 radians/s, z1 = .05) find a0
Ahmed Elgamal
0qKqCqM nnnnnn
nnnn M2ζC ω
n0nnn MaM2ζ ω nn0 2ζa ω
Mass-proportional damping: c = ao m
Defining a0 to obtain a desired modal damping zn in mode n
26
14
Ahmed Elgamal
With a0 defined by a0 = 2 zn wn, this form of mass proportional
damping will change with frequency according to z = a0 / 2w as
shown in the figure below.
o
2
az
ω
z
nω
mc oa
1ω 2ω 4ω3ω
o
2
a
nω
mc oa
1ω 2ω 4ω3ω
27
In any modal equation, we have
where, and
Therefore, ao can be specified to obtain any desired n for
a given mode n such that Cn = a1Kn , or:
or
(e.g. at w1 = 2 radians/s, z1 = .05) find a1
Ahmed Elgamal
0qKqCqM nnnnnn
n
2
nn MK ωnnnn M2ζC ω
n1n MaMζ 2
nn ωω2 n nn1 2ζa /ω
Stiffness-proportional damping: c = a1 k
Defining a1 to obtain a desired modal damping zn
28
15
Ahmed Elgamal
With a1 defined by a1 = 2 zn / wn , this form of stiffness proportional
damping will change with frequency according to z = a1w / 2 as shown
in the figure below (damping increases linearly with frequency.
kc 1a
2
az 1ω
z
ω1ω 2ω
3ω29
Physically, we often observe (in first approximation) a nearly equal
value of damping for the first few modes of structural response
(e.g., first 1- 4 modes or so), and we want to model that. Therefore,
we use (Classical or Rayleigh damping):
Now we choose damping ratios zi and zj for two modes (natural
frequencies wi and wj) and solve for the coefficients a0 and a1 (two
equations in two unknowns).
Ahmed Elgamal
kmc 10 aa
n1n0n MaMaMζ 2
nn ωω2 n
n
2
n ωω 2/)aa(ζ 10 n
)2/a()2/a(ζ 10 nn ωω n
30
16
Ahmed Elgamal
z
ω
Frequency range
of interest
iω jω
w1 for example 2nd or , 3rd resonance for example
Combined
Stiffness-proportional damping
Mass-proportional damping
nearly uniform damping
kc 1a
mc 0a
kmc 10 aa
Variation of Classical (Rayleigh) Damping with Frequency
Damping defined by z = (a0/2w)+(a1w/2) results in the variation shown by the
combined curve below, which has the desirable feature of being somewhat uniform
within a frequency range of interest (say 1 Hz to 7 Hz or 2 to 14 in radians/s).
31
Notes
1) For a choice of zi = zj = zsame damping ratio in the
two modes, we get
,
2) Classical damping and is attractive because of
combination of mass and stiffness, allowing the no-
damping free-vibration mode shapes to un-couple the
matrix equation of motion.
Ahmed Elgamal
ji
ji
0
2ζa
ωω
ωω
ji
1
2ζa
ωω
32
17
Ahmed Elgamal
Caughey damping
The above procedure was generalized by Caughey to allow for more
control over damping in the specified modes of interest (i.e. to be
able to specify z for more than 2 modes i and j)
In this generalization, you can stay within the scope of classical
damping by using
1N
0i
i 1
ia kmmc
to find coefficients to match zi modal damping ratios, see for
instance “Dynamics of Structures” by A. Chopra.ia
33
Ahmed Elgamal
Disadvantages:
1. c can become a full matrix instead of being a banded
matrix (if m and k are banded) as with c = a0m + a1k
2. You must check to ensure that you don’t end up with a
negative zi in some mode where you have not specifically
specified damping (because damping variation with
frequency might display sharp oscillations).
In summary, c = a0m + a1k is the usual choice at present
despite the limitations discussed above.
34