( )t - + = mu ku p xx

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 ^u

( )

^t into equations in terms modal coordinates

( )

^t

q . The displacement vector u

( )

^t can be expressed as

( ) ( ) ( )

1 N

r r r

t φ q t t

=

=

∑

=

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 φ_n^T

( ) ( ) ( )

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 nq tn n nq tn n t

φ mφ && +φ φk =φ p

or M q tn&&n

( )

+ K q tn n

( )

= P tn

( )

where

T

n n n

M =φ mφ K_n =φ φ_n^Tk _n P tn

( )

=φn^Tp

( )

t Mn is called generalized mass for the nth natural mode Kn is called generalized stiffness for the nth mode and

n

( )

P t is called generalized force for the nth mode

Divide the equation by M_n. The equation that governs the nth modal coordinate q tn

( )

for the nth mode becomes

( )

2 n

n n n

n

q q P t ω M

+ =

&&

n

( )

q 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 (n =1, 2,...,N). The matrix form of all equations for n=1, 2,...,N is

( )

^t

+ =

Mq Kq P&&

where M and K are diagonal matrices consisting of M_n and K_n, respectively, on the main diagonal. Recall that

= T

M Φ mΦ K =Φ^TkΦ

14 - 3

1 1 1 2 2 1

2 2 2 2 2

0 sin

0 0

m u k k k u po

m u ⁺k ⁻k u ωt

⎡ ⎤ ⎧ ⎫ ⎡⎨ ⎬+ ⎤ ⎧ ⎫ ⎧ ⎫⎨ ⎬ ⎨ ⎬=

⎢ ⎥ ⎢ − ⎥ ⎩ ⎭

⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭

&&

&&

14 - 5

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 ^u

( )

^{t into}

equations in terms modal coordinates ^q

( )

^t . The displacement vector

( )

^t

u can be expressed as

( ) ( ) ( )

1 N

r r r

t φ q t t

=

=

∑

=

u Φq

Substitute this in the equation of motion

( ) ( ) ( ) ( )

1 1 1

N N N

r r r r r r

r r r

q t q t q t t

φ φ φ

= = =

+ + =

∑

^{m &&}

∑

^{c &}

∑

^{k p}

Pre-multiply by φ_n^T

( ) ( ) ( ) ( )

1 1 1

N N N

T T T T

n r r n r r n r r n

r r r

q t q t q t t

φ φ φ φ φ φ φ

= = =

+ + =

∑

^{m &&}

∑

^{c &}

∑

^{k p}

Because of the orthogonality properties of modes, the only nonzero term in the 1^st and 3^rd summation is when r =n, so

( ) ( ) ( )

1 N

n n nr r n n n

r

M q t C q K q t P t

=

+

∑

+ =

&& &

where

T

nr n r

C =φ φc

The nth equation may still involve q&_r 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&& &

C will be diagonal when the system has classical damping.

Cnr=0 when n≠ r. Then, we will have an uncoupled equation.

( ) ( ) ( )

n n n n n n n

M q t&& +C q& +K q t = P t

Divide the equation by M_n. The equation that governs the nth modal coordinate q tn

( )

for the nth mode becomes

( )

2 2 ⁿ

n n n n n n

n

q q q P t

ζ ω ω M

+ + =

&& &

ζn is the damping ratio for the nth mode. The solution of this equation is the response of a damped SDF system.

14 - 7

Displacement response

Once the modal coordinate ^{q t}_n

( )

have been determined by solving the modal equations. The contribution of the nth mode to

displacement of MDOF system is

( ) ( )

n t =φnq tn

u

By superposition the response contribution from all modes, the total displacement is then

( ) ( ) ( ) ( )

1 1

N N

n n n

n n

t t φ q t t

= =

=

∑

=

∑

=

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

n n

r t r t

=

=

∑

where r tn

( )

is the contribution of nth mode to that response quantities. It is the response due to an equivalent static force.

( ) ( )

²

( )

n t = n t =ω φn nq tn

f ku m

Note that

( ) ( ) ( ) ( )

1 1

N N

n n

n n

t t t t

= =

= =

∑

=

∑

f ku ku f

14 - 9

14 - 11

14 - 13

Modal contribution of excitation vector ^p

( )

^{t =sp t}

( )

Suppose the applied force vector ^p

( )

^t has fixed spatial distribution s (pattern vector which is independent of time), but the magnitude of

( )

^t

p 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 r

r r

φ

= =

=

∑

=

∑

Γ

s s m

If we pre-multiply by φ_n^T and use mode orthogonality, we obtain

T T

n n n n nMn

φ s = Γ φ mφ = Γ or

T n n

Mn

Γ = φ s

The contribution of nth mode to the excitation vector s is

n = Γn φn

s m

The generalized force is

( ) ( ) ( ) ( ) ( )

1 N

T T

n n r n r n n

r

P t φ p t φ φ p t M p t

=

= ^s =

∑

Γ ^m = Γ

Example

14 - 15

Consider two sets of forces

( ) ( )

0 0 0 0 1

t p t

⎧ ⎫⎪ ⎪

⎪ ⎪⎪ ⎪

= ⎨ ⎬

⎪ ⎪⎪ ⎪

⎪ ⎪⎩ ⎭

p and

( ) ( )

0 0 0

1 2

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

Modal analysis for ^p

( )

^{t =sp t}

( )

From the modal equation

( ) ( )

2 2 ⁿ

n n n n n n n

n

q q q P t p t

ζ ω ω M

+ + = = Γ

&& &

The factor Γ_n 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 q tn

( )

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).

( )

2 2

n n n n n n

D&& + ζ ω D& +ω D = p t

where

( ) ( )

n n n

q t = Γ D t

This form is convenient because we can construct a response spectrum for excitation ^{p t}

( )

and the response of SDF system D tn

( )

could be directly read from the spectrum.

The contribution to displacement of the nth mode is

( ) ( ) ( )

n t =φnq tn = Γn nφ D tn

u

The equivalent static force on the n^th mode is

( ) ( )

²

( )

²

( )

²

( ) ( )

n t = n t =ω φn nq tn =ω φn nΓnD tn = n ⎡⎣ωnD tn ⎤⎦ = nA tn

f ku m m s s

14 - 17

The contribution of nth mode to response ^{r t}

( )

^is

( )

^{st 2}

( )

n n n n

r t = r ⎡⎣ω D t ⎤⎦

where r_n^st is the static response due to external forces s_n

( )

r tn is a product of results from two analyses:

(1) Static analysis of the structure subjected to external forces s_n (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 N

st

n n n n

n n

r t r t r ω D t

= =

⎡ ⎤

=

∑

=

∑

⎣ ⎦

Modal contribution to response

The contribution to response from the nth mode is

( )

^{st 2}

( )

n n n n

r t = r r ^⎡⎣ω D t ^⎤⎦

where r^st is the static value of r due to external forces s and

st n

n st

r r

= r

is the nth modal contribution factor. r_n is dimensionless and independent of how the modes are normalized. And

1

1

N n n

r

=

∑

=

CHAPTER 13

EARTHQUAKE ANALYSIS OF LINEAR SYSTEMS

RESPONSE HISTORY ANALYSIS Equation of motion

eff

( )

t

+ + =

mu cu ku p_{&& &}

where peff

( )

t = −mιu_&&g

( )

t The spatial distribution of peff

( )

t is s mι= Modal expansion of displacement and forces

( ) ( ) ( )

1 N

r r r

t φ q t t

=

=

∑

=

u Φq

1 N

n n

n

φ

=

=

∑

Γ

mι m

where _n ⁿ

n

L

Γ = M L_n =φ_n^Tmι M_n =φ_n^Tmφ_n The contribution of the nth mode to excitation vector mι is

n = Γn φn

s m

which is independent of how modes are normalized.

Modal equations

( )

2 2

n n n n n n n g

q&& + ζ ω q& +ω q = −Γ u&& t

Rewrite q tn

( )

as q tn

( )

= ΓnD tn

( )

Then,

( )

2 2

n n n n n n g

D&& + ζ ω D& +ω D = −u&& t

14 - 19

Modal responses

Displacement due to the nth mode is

( ) ( ) ( )

n t =φnq tn = Γn nφ D tn

u

The equivalent static force in the nth mode is

( ) ( )

n t = nA tn

f s

where A tn

( )

=ωn²D tn

( )

The response contribution from the nth mode is

( )

^st

( )

n n n

r t =r A t

( )

^{n n}₂

( )

n n

n

t φ A t ω

= Γ u

Total responses Total displacement is

( ) ( ) ( )

1 1

N N

n n n n

n n

t t φ D t

= =

=

∑

=

∑

Γ

u u

Total response is

( ) ( ) ( )

1 1

N N

st

n n n

n n

r t r t r A t

= =

=

∑

=

∑

14 - 21

14 - 23

Multistory buildings with symmetric plan

g

( )

u t + + = −

mu cu ku_{&& &} m1_&&

1 1

N N

n n n

n n

φ

= =

=

∑

=

∑

Γ

m1 s m

where

h n n

n

L

Γ = M

1 N h

n j jn

j

L m φ

=

=

∑

²

1 N T

n n n j jn

j

M φ φ m φ

=

= ^m =

∑

mj is the mass of the jth story.

14 - 25

Effective modal mass and heights

( )

^st

( )

^*

( )

bn bn n n n

V t =V A t = M A t

where *

( )

²

h

st h n

n bn n n

n

M V L L h

= = Γ = M is called the effective modal mass.

*

1 1

N N

n j

n n

M m

= =

∑

=

∑

( )

^st

( )

^*

( )

bn bn n n bn

M t = M A t = h V t

* n

n h

n

h L L

= θ is called the effective modal height where

1 N

n j j jn

j

L^θ h m φ

=

=

∑

* *

1 1

N N

n n j j

n n

h M h m

= =

∑

=

∑

Example

14 - 27

( ) ( )

n t = Γn nφ D tn

u

( ) ( )

rn n rn n

u t = Γ φ D t

( )

^st

( )

^*

( )

bn bn n n n

V t =V A t = M A t

( )

^st

( )

^*

( )

bn bn n n bn

M t = M A t = h V t

14 - 29

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, D tn( ) and A tn( ) by solving the equations of motion for modal SDF systems,

( )

2 2

n n n n n n

D&& + ζ ω D& +ω D = p t

( ) ( ) ²

n n n

A t =D t ω

and r tn( ) and ^{r t}( ), from the equations . Modal response r tn

( )

= r A tn^st n

( )

Total response

( ) ( ) ( )

1 1

N N

st

n n n

n n

r t r t r A t

= =

=

∑

=

∑

Finally, the peak response o ^max

( )

r t r t

= ∀

To avoid solving the equation of motions for modal SDF systems, the peak value of D tn( ) and A tn( ) can be conveniently read from response spectrum, so the peak modal response can be calculated as

Peak modal response r_no =r A_n^st _n

where n ^max n

( )

t

A 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

14 - 31

SRSS ²

1 N

o no

n

r r

=

≈

∑

CQC

1 1

N N

o in io no

i n

r ρ r r

= =

≈

∑∑

where ^{( )}

( ) ( ) ( )

3/ 2

2 2 2 2 2 2

8

1 4 1 4

i n i in n in

in

in i n in in i n in

ζ ζ ζ β ζ β

ρ 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

SRSS ^{60.4692 +24.5332 +9.8672 +2.9432 +0.5952 =66.066} kips 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.

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 ^{C=ΦTc Φ}, Rayleigh damping matrix c= a0m+a1k 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 0

T T

a ao

= = =

C Φ c Φ Φ mΦ M having diagonal entries C_n =a M0 _n =2ζ ω_{n n}M_n

0 2 _{n n}

a = ω ζ or ⁿ ₂ ⁰

n

ζ a

= ω

For stiffness-proportional damping 1 1

T T

a a

= = =

C Φ cΦ Φ k Φ K having diagonal entries Cn =a K1 n =a1ωn²Mn

2

1 _n 2 _{n n}

aω = ω ζ or ¹

2

n n

aω ζ =

Thus for c= a0m+a1k Æ ^{0 1}

2 2

n n

n

a aω ζ ⁼ ω ⁺

The coefficient a0 and a1 depends on the modal damping ratios selected for two modes. Suppose we want the damping ratios of mode i and j to be ζi and ζj , and the modal frequencies are ωi and

ωj. The coefficients can be determined by solving

0 1

1 1/

2 1/

i i i

j j j

a a

ω ω ζ

ω ω ζ

⎡ ⎤⎧ ⎫ ⎪ ⎪⎨ ⎬ ⎨ ⎬=⎧ ⎫

⎢ ⎥

⎩ ⎭ ⎪ ⎪

⎣ ⎦ ⎩ ⎭

14 - 33

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

2

2

.. ..

n

2

n n n

C M

C M

C M

ζ ω

ζ ω

ζ ω

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥

= =

⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

C

And we know that

=

T

C Φ cΦ

Therefore, we multiply

( )

^{ΦT −1} from the left and ^Φ−1from the right; we get

( )

^{T −1 −1}

=

c Φ CΦ

This is the classical damping matrix which results in modal damping ratios that we specified.

14 - 35

Example

1 1

1

⎡ ⎤

⎢ ⎥

= ⎢ ⎥

⎢ ⎥

⎣ ⎦

m

2 1

1 2 1

1 1

⎡ − ⎤

⎢ ⎥

= − ⎢ − ⎥

⎢ − ⎥

⎣ ⎦

k

0.445, 1.247, 1.802 ω

n

=

0.328 0.737 0.591 0.591 0.328 0.737 0.737 0.591 0.328

⎡ ⎤

⎢ ⎥

= ⎢ − ⎥

⎢ − ⎥

⎣ ⎦

Φ

1 1

1

T

⎡ ⎤

⎢ ⎥

= = ⎢ ⎥

⎢ ⎥

⎣ ⎦

M Φ mΦ

1 1 1

2 2 2

3 3 3

2 0.0445

2 0.1247

2 0.1802

M

M

M ζ ω

ζ ω

ζ ω

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥

= ⎢ ⎥ ⎢= ⎥

⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

C

( )

^{1 1 0.08710.0483 0.12680.0483 0.03970.0086}

0.0086 0.0397 0.1355

T − −

− −

⎡ ⎤

⎢ ⎥

= = −⎢ − ⎥

⎢− − ⎥

⎣ ⎦

c Φ CΦ