14 - 1 CHAPTER 12 DYNAMIC ANALYSIS AND RESPONSE OF LINEAR SYSTEMS Modal equations for undamped systems We have derive the equations of motion for MDOF system as ( ) t + = mu ku p && These differential equations are coupled when one of the mass or stiffness matrices is not diagonal. Then, all equations must be solved simultaneously, which is difficult to carry out. The problem can be solved easier if we convert the equations in term of displacements () t u into equations in terms modal coordinates () t q . The displacement vector ( ) t u can be expressed as () () () 1 N r r r t q t t φ = = = ∑ u Φq Substitute this in the equation of motion () () () 1 1 N N r r r r r r q t q t t φ φ = = + = ∑ ∑ m k p && Pre-multiply by T n φ () () () 1 1 N N T T T n r r n r r n r r q t q t t φ φ φ φ φ = = + = ∑ ∑ m k p && Because of the orthogonality properties of modes, the only nonzero term in the summation is when r n = , so () ( ) ( ) T T T n n n n n n n q t q t t φ φ φ φ φ + = m k p && or ( ) ( ) ( ) n n n n n M q t Kq t P t + = &&
Transcript
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CHAPTER 12 DYNAMIC ANALYSIS AND
RESPONSE OF LINEAR SYSTEMS Modal equations for undamped systems We have derive the equations of motion for MDOF system as
( )t+ =mu ku p&&
These differential equations are coupled when one of the mass or stiffness matrices is not diagonal. Then, all equations must be solved simultaneously, which is difficult to carry out. The problem can be solved easier if we convert the equations in term of displacements ( )tu into equations in terms modal coordinates ( )tq . The displacement vector ( )tu can be expressed as
( ) ( ) ( )1
N
r rr
t q t tφ=
= =∑u Φq
Substitute this in the equation of motion
( ) ( ) ( )1 1
N N
r r r rr r
q t q t tφ φ= =
+ =∑ ∑m k p&&
Pre-multiply by Tnφ
( ) ( ) ( )1 1
N NT T Tn r r n r r n
r r
q t q t tφ φ φ φ φ= =
+ =∑ ∑m k p&&
Because of the orthogonality properties of modes, the only nonzero term in the summation is when r n= , so
( ) ( ) ( )T T Tn n n n n n nq t q t tφ φ φ φ φ+ =m k p&&
or ( ) ( ) ( )n n n n nM q t K q t P t+ =&&
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where T
n n nM φ φ= m Tn n nK φ φ= k ( ) ( )T
n nP t tφ= p
nM is called generalized mass for the nth natural mode
nK is called generalized stiffness for the nth mode and
( )nP t is called generalized force for the nth mode
Divide the equation by nM . The equation that governs the nth modal coordinate ( )nq t for the nth mode becomes
( )2 nn n n
n
P tq q
Mω+ =&&
( )nq t is the only unknown in this equation and the solution can be obtained as for the response of a SDOF system. The modal coordinate for all modes can be obtained from such equation ( 1,2,...,n N= ). The matrix form of all equations for 1,2,...,n N= is
( )t+ =Mq Kq P&&
where M and K are diagonal matrices consisting of nM and nK , respectively, on the main diagonal. Recall that
T=M Φ mΦ T=K Φ kΦ
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1 1 1 2 2 1
2 2 2 2 2
0sin
0 0om u k k k u p
tm u k k u
ω+ −⎡ ⎤⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫
+ =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥− ⎩ ⎭⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭
&&
&&
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Modal equations for damped systems
When damping is included, the equations of motion for MDOF system are
( )t+ + =mu cu ku p&& &
These are coupled equations. The set of equations could be uncoupled by transforming the equation in term of displacements ( )tu into equations in terms modal coordinates ( )tq . The displacement vector ( )tu can be expressed as
( ) ( ) ( )1
N
r rr
t q t tφ=
= =∑u Φq
Substitute this in the equation of motion
( ) ( ) ( ) ( )1 1 1
N N N
r r r r r rr r r
q t q t q t tφ φ φ= = =
+ + =∑ ∑ ∑m c k p&& &
Pre-multiply by Tnφ
( ) ( ) ( ) ( )1 1 1
N N NT T T Tn r r n r r n r r n
r r rq t q t q t tφ φ φ φ φ φ φ
= = =
+ + =∑ ∑ ∑m c k p&& &
Because of the orthogonality properties of modes, the only nonzero term in the 1st and 3rd summation is when r n= , so
( ) ( ) ( )1
N
n n nr r n n nr
M q t C q K q t P t=
+ + =∑&& &
where T
nr n rC φ φ= c
The nth equation may still involve rq& of other modes.
Above is equation for nth modal coordinate. If we put equations for all n together in matrix, we get
( )t+ + =Mq Cq Kq P&& &
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C will be diagonal when the system has classical damping. nrC =0 when n r≠ . Then, we will have an uncoupled equation.
( ) ( ) ( )n n n n n n nM q t C q K q t P t+ + =&& &
Divide the equation by nM . The equation that governs the nth modal coordinate ( )nq t for the nth mode becomes
( )22 nn n n n n n
n
P tq q q
Mζ ω ω+ + =&& &
nζ is the damping ratio for the nth mode. The solution of this equation is the response of a damped SDF system.
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Displacement response Once the modal coordinate ( )nq t have been determined by solving the modal equations. The contribution of the nth mode to displacement of MDOF system is
( ) ( )n n nt q tφ=u By superposition the response contribution from all modes, the total displacement is then
( ) ( ) ( ) ( )1 1
N N
n n nn n
t t q t tφ= =
= = =∑ ∑u u Φq
This method to determine the response of MDF system to excitation is known as the classical modal analysis or classical mode superposition method, or modal analysis. Modal analysis can be used to solve linear system with classical damping only. Damping must be classical to obtain modal equations that are uncoupled. Element forces To determine other response quantities ( )r t , the concept of superposition is used
( ) ( )1
N
nn
r t r t=
=∑
where ( )nr t is the contribution of nth mode to that response quantities. It is the response due to an equivalent static force.
( ) ( ) ( )2n n n n nt t q tω φ= =f ku m
Note that ( ) ( ) ( ) ( )1 1
N N
n nn n
t t t t= =
= = =∑ ∑f ku ku f
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Modal contribution of excitation vector ( ) ( )t p t=p s Suppose the applied force vector ( )tp has fixed spatial distribution s (pattern vector which is independent of time), but the magnitude of ( )tp varies with time as a scalar value ( )p t .
( ) ( )t p t=p s
We can expand the vector s using combination of modes
1 1
N N
r r rr r
φ= =
= = Γ∑ ∑s s m
If we pre-multiply by Tnφ and use mode orthogonality, we obtain
T Tn n n n n nMφ φ φ= Γ = Γs m or
Tn
nnM
φΓ =
s
The contribution of nth mode to the excitation vector s is
n n nφ= Γs m
The generalized force is
( ) ( ) ( ) ( ) ( )1
NT T
n n r n r n nr
P t p t p t M p tφ φ φ=
= = Γ = Γ∑s m
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Example
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Consider two sets of forces
( ) ( )
00001
t p t
⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
p and ( ) ( )
000
12
t p t
⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪−⎪ ⎪⎪ ⎪⎩ ⎭
p
The first one [0 0 0 0 1]T can be expanded as
The second one [0 0 0 –1 2]T can be expanded as
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Modal analysis for ( ) ( )t p t=p s
From the modal equation
( ) ( )22 nn n n n n n n
n
P tq q q p t
Mζ ω ω+ + = = Γ&& &
The factor nΓ is called modal participation factor. It is a measure of the degree to which the nth mode participates in the response.
This modal participation factor is not a useful definition because it depends on how the modes are normalized.
We can write the modal coordinate ( )nq t in term of response of SDF system with a unit mass, and vibration properties of the nth mode (natural frequency and damping ratio of the nth mode).
( )22n n n n n nD D D p tζ ω ω+ + =&& &
where
( ) ( )n n nq t D t= Γ
This form is convenient because we can construct a response spectrum for excitation ( )p t and the response of SDF system ( )nD t could be directly read from the spectrum.
The contribution to displacement of the nth mode is
( ) ( ) ( )n n n n n nt q t D tφ φ= = Γu
The equivalent static force on the nth mode is
( ) ( ) ( ) ( ) ( ) ( )2 2 2n n n n n n n n n n n n n nt t q t D t D t A tω φ ω φ ω⎡ ⎤= = = Γ = =⎣ ⎦f ku m m s s
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The contribution of nth mode to response ( )r t is
( ) ( )2stn n n nr t r D tω⎡ ⎤= ⎣ ⎦
where stnr is the static response due to external forces ns
( )nr t is a product of results from two analyses:
(1) Static analysis of the structure subjected to external forces ns
(2) Dynamic analysis of the nth mode SDF system subjected to excitation ( )p t
The combined response due to contribution from all modes is
( ) ( ) ( )2
1 1
N Nst
n n n nn n
r t r t r D tω= =
⎡ ⎤= = ⎣ ⎦∑ ∑
Modal contribution to response
The contribution to response from the nth mode is
( ) ( )2stn n n nr t r r D tω⎡ ⎤= ⎣ ⎦
where str is the static value of r due to external forces s and st
nn st
rrr
=
is the nth modal contribution factor. nr is dimensionless and independent of how the modes are normalized. And
11
N
nn
r=
=∑
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CHAPTER 13
EARTHQUAKE ANALYSIS OF LINEAR SYSTEMS
RESPONSE HISTORY ANALYSIS
Equation of motion
( )eff t+ + =mu cu ku p&& &
where ( ) ( )eff gt u t= −p mι&&
The spatial distribution of ( )eff tp is =s mι
Modal expansion of displacement and forces
( ) ( ) ( )1
N
r rr
t q t tφ=
= =∑u Φq
1
N
n nn
φ=
= Γ∑mι m
where nn
n
LM
Γ = Tn nL φ= mι T
n n nM φ φ= m
The contribution of the nth mode to excitation vector mι is
n n nφ= Γs m
which is independent of how modes are normalized.
Modal equations
( )22n n n n n n n gq q q u tζ ω ω+ + = −Γ&& & &&
Rewrite ( )nq t as ( ) ( )n n nq t D t= Γ
Then,
( )22n n n n n n gD D D u tζ ω ω+ + = −&& & &&
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Modal responses
Displacement due to the nth mode is
( ) ( ) ( )n n n n n nt q t D tφ φ= = Γu
The equivalent static force in the nth mode is
( ) ( )n n nt A t=f s
where ( ) ( )2n n nA t D tω=
The response contribution from the nth mode is
( ) ( )stn n nr t r A t=
( ) ( )2n n
n nn
t A tφωΓ
=u
Total responses
Total displacement is
( ) ( ) ( )1 1
N N
n n n nn n
t t D tφ= =
= = Γ∑ ∑u u
Total response is
( ) ( ) ( )1 1
N Nst
n n nn n
r t r t r A t= =
= =∑ ∑
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Multistory buildings with symmetric plan
( )gu t+ + = −mu cu ku m1&& & &&
1 1
N N
n n nn n
φ= =
= = Γ∑ ∑m1 s m
where hn
nn
LM
Γ = 1
Nhn j jn
j
L m φ=
=∑ 2
1
NT
n n n j jnj
M mφ φ φ=
= =∑m
jm is the mass of the jth story.
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Effective modal mass and heights
( ) ( ) ( )*stbn bn n n nV t V A t M A t= =
where ( ) 2
*hnst h
n bn n nn
L hM V L
M= = Γ = is called the effective modal mass.
*
1 1
N N
n jn n
M m= =
=∑ ∑
( ) ( ) ( )*stbn bn n n bnM t M A t h V t= =
* nn h
n
LhL
θ
= is called the effective modal height where 1
N
n j j jnj
L h mθ φ=
=∑
* *
1 1
N N
n n j jn n
h M h m= =
=∑ ∑
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Example
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( ) ( )n n n nt D tφ= Γu
( ) ( )rn n rn nu t D tφ= Γ
( ) ( ) ( )*stbn bn n n nV t V A t M A t= =
( ) ( ) ( )*stbn bn n n bnM t M A t h V t= =
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Response Spectrum Analysis (RSA)
To determine the peak value of response due to earthquake
excitation, the modal response history analysis requires computation
of response at all time instants during vibration, i.e, ( )nD t and ( )nA t
by solving the equations of motion for modal SDF systems,
( )22n n n n n nD D D p tζ ω ω+ + =&& &
( ) ( ) 2n n nA t D t ω=
and ( )nr t and ( )r t , from the equations .
Modal response ( ) ( )stn n nr t r A t=
Total response ( ) ( ) ( )1 1
N Nst
n n nn n
r t r t r A t= =
= =∑ ∑
Finally, the peak response ( )maxo tr r t
∀=
To avoid solving the equation of motions for modal SDF
systems, the peak value of ( )nD t and ( )nA t can be conveniently read
from response spectrum, so the peak modal response can be
calculated as
Peak modal response stno n nr r A=
where ( )maxn ntA A t
∀=
The peak response could be estimated by combining the peak
modal response using modal combination rule, such as square-root-of-
sum-of-squares (SRSS) or complete quadratic combination (CQC)
rule
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SRSS 2
1
N
o non
r r=
≈ ∑
CQC 1 1
N N
o in io noi n
r r rρ= =
≈ ∑∑
where ( )( ) ( ) ( )
3/ 2
22 2 2 2 2
8
1 4 1 4i n i in n in
in
in i n in in i n in
ζ ζ ζ β ζ βρ
β ζ ζ β β ζ ζ β
+=
− + + + +
Response history analysis (RHA): peak base shear = 73.728 kips.
Response spectrum analysis (RSA): peak base shear is
The difference is because the peaks of different modes do not
occur at the sane time, so the peak of total response can not be
computed exactly.
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Rayleigh Damping Matrix (section 11.4.1)
Damping matrix should not be computed from dimension
structural member. Instead, it should be constructed from modal
damping ratios.
To construct a classical damping matrix, which give a diagonal
matrix T=C Φ c Φ , Rayleigh damping matrix 0 1a a= +c m k uses
combination of mass-proportional damping matrix and stiffness-
proportional damping matrix. Each term retains the orthogonal
properties of mode shapes.
For mass-proportional damping 0T T
oa a= = =C Φ c Φ Φ m Φ M
having diagonal entries 0 2n n n n nC a M Mζ ω= =
0 2 n na ω ζ= or 0
2nn
aζ
ω=
For stiffness-proportional damping 1 1T Ta a= = =C Φ c Φ Φ k Φ K
having diagonal entries 21 1n n n nC a K a Mω= =
21 2n n naω ω ζ= or 1
2n
naω
ζ =
Thus for 0 1a a= +c m k 0 1
2 2n
nn
a aωζ
ω= +
The coefficient 0a and 1a depends on the modal damping ratios selected for two modes. Suppose we want the damping ratios of mode i and j to be iζ and jζ , and the modal frequencies are iω and
jω . The coefficients can be determined by solving
0
1
1/11/2
i i i
j j j
aa
ω ω ζω ω ζ
⎡ ⎤ ⎧ ⎫⎧ ⎫ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦ ⎩ ⎭
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If modal damping ratios are to be specified for all modes and the classical damping matrix is to be obtained, the generalized damping matrix is
1 1 1 1
2 2 2 2
22
.. ..2n n n n
C MC M
C M
ζ ωζ ω
ζ ω
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
C
And we know that
T=C Φ cΦ Therefore, we multiply ( ) 1T −
Φ from the left and 1−Φ from the right; we get
( ) 1 1T − −=c Φ CΦ This is the classical damping matrix which results in modal damping ratios that we specified.
