Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
An in-depth analysis of Multi-Degree-Of-Freedom (MDOF) systems, covering equations, modal responses, and damping. It includes information on natural frequencies, mode shapes, and response spectrum modal responses. The document also discusses various types of damping, including mass-proportional, stiffness-proportional, and Rayleigh damping.
Typology: Assignments
1 / 17
This page cannot be seen from the preview
Don't miss anything!
Last update: 2010 Ahmed Elgamal
1
Ahmed Elgamal
3
c 3
c col
g mu kucumu
2
m 1
m 2
k 1
k 2
u 1
u 2
2 ^2 mcol 2
1 m mroof 2
1 ^1 mcol 2
m mfloor 4
k 1 2kcol1 k^2 2kcol
m 2 u^2 k 2 (u 2 u 1 )m 2 ug
g 2
1
2
1
2 2
1 2 2
2
1
2
1 u m
m
u
u
k k
k k k
u
u
0 m
m 0
m 1 u^1 k 2 (u 1 u 2 )k 1 u 1 m 1 ug
or,
where 1 or 1 is the Identity Matrix
3
Ahmed Elgamal
m 1
m 3
k 1
k 2
u 1
u 2
u 3
k 3
m 2
m 3 u^3 k 3 (u 3 u 2 )m 3 ug
m 2 u^2 k 3 (u 2 u 3 )k 2 (u 2 u 1 )m 2 ug
m 1 u^1 k 2 (u 1 u 2 )k 1 u 1 m 1 ug
3
2
1
3
2
1
3 3
2 2 3 3
1 2 2
3
2
1
3
2
1 u
m
m
m
u
u
u
0 k k
k k k k
k k k 0
u
u
u
0 0 m
0 m 0
m 0 0
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 w 1 and w 2 , and a n - DOF system has natural frequencies
w 1 , w 2 , …, w n
mu ku 0
u u
2 ω
e.g. u asinω t bcos ωt gives u u
2
7
Ahmed Elgamal
substituting for (^) u , we get:
- m k u 0
2 ω
or
k - m u 0
2 ω
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
For a non-trivial solution (which will allow for computing the
natural frequencies during free vibration), the determinant of
k - m 0
2 ω or k^ - λ m ^0 2 where λ ω
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 (l 1 and l 2 ), which
define w 1 and w 2 (the natural resonant frequencies of this
2 - DOF system).
k - m
2 ω^ must be equal to zero such that:
9
Ahmed Elgamal
2 - Story Shear Building (2-DOF system)
m 1
m 2
k 1
k 2
u 1
g 2
1
2
1
2 2
1 2 2
2
1
2
1 u m
m
u
u
k k
k k k
u
u
0 m
m 0
0
0
u
u
k k m
k k m k
2
1
2 2 2
1 2 1 2 l
l
0 k k m
k k m k
2 2 2
1 2 1 2
( k 1 k 2 lm 1 )(k 2 lm 2 )(k 2 )(k 2 ) 0
2
2
a
b b ac
2
2
2
l 1
Y Z
W X
Determinant = WZ-XY
Note: For
Set Determinant = 0:
or,
10
u 1
m 2 u 2
m 1
φ 11
φ 21
f 12
f 22
u 1
m 2 u 2
m 1
φ 11
φ 21
f
f
Note: Any mode shape f n 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):
f 11 - 2 f 21 = 0 , only defining a ratio between amplitudes of f 11 and f 21
(for instance, if f 11 = 1 , then f 21 = 0. 5 , or if you choose f 11 = 2 , then f 21 = 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 f n
21
11 1
22
12 2 f
f
(^) 13
Ahmed Elgamal
Note: When you substitute any of the wn values into Eq. 1 , the determinant of the matrix (k - wn^2 m) 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 f n ) 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
node node
node
Sample Mode shape Configurations
14
n n
T
n
2 n n n
T
T
2 n n
T
Properties of f
b) For any mode f n , modal mass Mn is defined by:
d) If then
c) For any mode f n , modal stiffness Kn is defined by:
To do that, multiply each component of mode f n by
0 (not^ ) r
T r n
T
T
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 ) 1 φ 2 φ
m 1
16
n
u Φq
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
i i
2 i i i i i u M
L q 2 ωq ωq ^ , i = 1, 2, … NDOF
OK, go ahead now and solve for qi(t) in the above uncoupled equations
u Φq
u Φq
g
t
20
u (^) q Φq
2
1
32
22
31
21
11 12
2
32
22
12
1
31
21
11
3
2
1
q
q q q
u
u
u
φ φ
φ
φ
φ
φ
φ φ
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:
u q Φq
3
2
1
31 32 33
21 22 23
11 12 13
3
33
23
13
2
32
22
12
1
31
21
11
3
2
1
q
q
q
q q q
u
u
u
φ φ φ
φ φ φ
φ φ φ
φ φ φ
φ
φ
φ
φ
φ
φ
φ
φ
φ
1 ^1 ^1
31
21
11
3
2
1 q q
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
Calculate expected max response ( (^) r) using (^)
2 rmax rimax 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)
22
Damping Matrix for MDOF Systems
Mass-proportional damping
c = a o m
Stiffness-proportional damping
c = a 1 k
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.
mu ^ cu ku m1 u
c a 0 m a 1 k
25
In any modal equation, we have
where,
Therefore, a o can be specified to obtain any desired z n for
a given mode n such that Cn = a 0 Mn
or
(e.g. at w 1 = 2 radians/s, z 1 = .05) find a 0
Ahmed Elgamal
M (^) n qn CnqnKnqn 0
n n n n C 2 ζωM
2 ζ (^) n ωnMna 0 Mn 0 n n
o
26
o
z
ω n
ω 1 ω 2 ω 3 ω 4
o
ω n
ω 1 ω 2 ω 3 ω 4
27
In any modal equation, we have
where, and
Therefore, a o can be specified to obtain any desired n for
a given mode n such that Cn = a 1 Kn , or:
or
(e.g. at w 1 = 2 radians/s, z 1 = .05) find a 1
Ahmed Elgamal
M (^) n qn CnqnKnqn 0
n
2 Cn 2 ζnωnMn K (^) n ωnM
ζ Mn a 1 Mn
2 (^2) n ωn ωn a 1 2 ζn/ωn
1
28
Frequency range of interest
ω
nearly uniform damping
Variation of Classical (Rayleigh) Damping with Frequency
Damping defined by z = (a 0 / 2 w)+(a 1 w/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
= z j
= z same damping ratio in the
two modes, we get
,
combination of mass and stiffness, allowing the no-
damping free-vibration mode shapes to un-couple the
matrix equation of motion.
Ahmed Elgamal
i j
i j 0
2 a ζ ω ω
ωω
i j
1
2 a ζ ω ω
32
Caughey damping
^
N 1
i 0
1 i
ai
33
Ahmed Elgamal
Disadvantages:
matrix (if m and k are banded) as with c = a 0 m + a 1 k
negative zi in some mode where you have not specifically
specified damping (because damping variation with
frequency might display sharp oscillations).
In summary, c = a 0 m + a 1 k is the usual choice at present
despite the limitations discussed above.
34
