MDOF Dynamics

Structural Dynamics of LinearElastic MDOF Systems

u1

u2

u3

MDOF Dynamics 1Revised 3/09/06

Structural Dynamics of ElasticMDOF Systems

• Equations of Motion for MDOF Systems• Uncoupling of Equations through use of

Natural Mode Shapes• Solution of Uncoupled Equations• Recombination of Computed Response• Modal Response History Analysis• Modal Response Spectrum Analysis• Equivalent Lateral Force Method

MDOF Dynamics 2

Relevance to ASCE 7-05

ASCE 7-05 Provides guidance for three specific analysis procedures:

• Equivalent Lateral Force Analysis (ELF)• Modal Superposition Analysis (MSA)• Response History Analysis (RHA)

ELF Usually Allowed ELF Not Allowed

Cs

See ASCE 7-05Table 12.6-1

MDOF Dynamics 3

3.5Ts TTs

ux

uyrz

Majority of Massis in Floors

Typical Nodal DOF

Motion isPredominantlyLateral

Planar Frame with 36 Degrees of Freedom

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 16

MDOF Dynamics 4

Planar Frame with 36 Static Degrees of Freedombut with only THREE Dynamic DOF

u1

u2

u3

1

2

3

uU = u

u

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

MDOF Dynamics 5

Development of Flexibility Matrix

d1,1

d2,1

d3,1

f1=1 kip

d1,1

d2,1

d3,1

MDOF Dynamics 6


d1,2

d2,2

d3,2

f2=1 kipd1,1 d1,2

d2,1 d2,2

d3,1 d3,2

MDOF Dynamics 7


d1,3

d2,3

d3,3

f3=1 kip

d1,1

d2,1

d3,1

d1,2

d2,2

d3,2

d1,3

d2,3

d3,3

MDOF Dynamics 8

Concept of Linear Combination of Shapes (Flexibility)

1,1 1,2 1,3

2,1 1 2,2 2 2,3 3

3,1 3,2 3,3

d d dU = d f + d f + d f

d d d

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭

1,1 1,2 1,3 1

2,1 2,2 2,3 2

3,1 3,2 3,3 3

d d d fU d d d f

d d d f

⎡ ⎤⎧ ⎫⎢ ⎥⎪ ⎪= ⎨ ⎬⎢ ⎥

⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦-1K = DD F = U K U = F

MDOF Dynamics 9

Static Condensation

DOF with mass

Massless DOF , ,

, ,

K K FUK K 0U

m m m n mm

n m n n n

⎡ ⎤ ⎧ ⎫⎧ ⎫=⎨ ⎬ ⎨ ⎬⎢ ⎥

⎩ ⎭ ⎩ ⎭⎣ ⎦

, ,K U K U Fm m m m n n m+ =1

, ,K U K U 0n m m n n n+ =2

MDOF Dynamics 10

Static Condensation(continued)

2 1, ,U K K U−= −n n n n m mRearrange

1 1, , , ,K U K K K U F−− =m m m m n n n n m m m

Plug into

1, , , ,K K K K U F−⎡ ⎤− =⎣ ⎦m m m n n n n m m mSimplify

MDOF Dynamics 11

Condensed Stiffness Matrix

1, , , ,K K K K K−= −m m m n n n n m

Idealized Structural Property Matricesm1

m3

m2k1

k2

k3

f1(t), u1(t)

f2(t), u2(t)

f3(t), u3(t)

1 11 1 2 2

2 2 3

k -k 0K -k k + k -k

0 -k k + k

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

12

3

m 0 0M 0 m 0

0 0 m

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

1

2

3

( )( ) ( )

( )

f tF t f t

f t

⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

1

2

3

( )( ) ( )

( )

u tU t = u t

u t

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

MDOF Dynamics 12

Note: Damping to be shown later

Coupled Equations of Motion for Undamped Forced Vibration

MU(t) KU(t) F(t)+ =

1 1 1

2 2 2

3 3 3

1 0 0 u ( ) 1 1 0 u ( ) f ( )0 2 0 u ( ) 1 1 2 2 u ( ) f ( )0 0 3 u ( ) 0 3 2 3 u ( ) f ( )

m t k k t tm t k k k k t t

m t k k k t t

−⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥+ − + − =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎩ ⎭ ⎩ ⎭

1 1 2 1

2 1 2 2 3 2

3 2 3 3 3

1u ( ) k1u ( ) k1u ( ) f ( )2u ( ) 1u ( ) 1u ( ) 2u ( ) 2u ( ) f ( )3u ( ) 2u ( ) 2u ( ) 3u ( ) f ( )

m t t t tm t k t k t k t k t tm t k t k t k t t

+ − =− + + − =

− + + =

DOF 1

DOF 2

DOF 3

MDOF Dynamics 13

We need to develop a way to solve theequations of motion.

• This will be done by a transformation of coordinatesfrom Normal Coordinates (displacements at the nodes)To Modal Coordinates (amplitudes of the naturalMode shapes).

• Because of the Orthogonality Property of the natural modeshapes, the equations of motion become uncoupled,allowing them to be solved as SDOF equations.

• After solving, we can transform back to the normalcoordinates.

MDOF Dynamics 14

Solutions for System in Undamped Free Vibration(Natural Mode Shapes and Frequencies)

MU(t) KU(t) 0+ =

Assume U( ) sint tφ ω=

2K M 0φ ω φ− =Then has three (n) solutions:

2U( ) ω sint tφ ω= −

1,1

1 2,1 1

3,1

,φ

φ φ ωφ

⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

1,2

2 2,2 2

3,2

,φ

φ φ ωφ

⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

1,3

3 2,3 3

3,3

,φ

φ φ ωφ

⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

Natural Mode ShapeNatural Frequency

MDOF Dynamics 15

Solutions for System in Undamped Free Vibration

For a SINGLE Mode2K ω Mφ φ=

2K MΦ= ΦΩ For ALL Modes

[ ]Φ = φ φ φ1 2 3Where:

21

2 22

23

ω

ω

ω

⎡ ⎤⎢ ⎥

Ω = ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

MDOF Dynamics 16

TM IΦ Φ=

1,i 1.0φ =orNote: Mode shape has arbitrary scale; usually

MODE 1 MODE 3MODE 2

1,1φ

2,1φ

3,1φ 3,2φ

2,2φ

1,2φ

3,3φ

2,3φ

1,3φ

Mode Shapes for Idealized 3-Story Frame

Node

Node

Node

MDOF Dynamics 17

Concept of Linear Combination of Mode Shapes(Transformation of Coordinates)

U Y=Φ

1,1 1,2 1,3

2,1 1 2,2 2 2,3 3

3,1 3,2 3,3

U y y yφ φ φφ φ φφ φ φ

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭

1,1 1,2 1,3 1

2,1 2,2 2,3 2

3,1 3,2 3,3 3

yU y

y

φ φ φφ φ φφ φ φ

⎡ ⎤⎧ ⎫⎪ ⎪⎢ ⎥= ⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥⎩ ⎭⎣ ⎦

MDOF Dynamics 18

Mode Shape

Modal Coordinate = amplitude of mode shape

Orthogonality Conditions

[ ]Φ = φ φ φ1 2 3

*1

T *2

*3

m

M m

m

⎡ ⎤⎢ ⎥

Φ Φ= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Generalized Mass*1

T *2

*3

k

K k

k

⎡ ⎤⎢ ⎥

Φ Φ= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Generalized Stiffness

*1*2*3

f ( )F( ) f ( )

f ( )

⎧ ⎫⎪ ⎪Φ =⎨ ⎬⎪ ⎪⎩ ⎭

T

tt t

t

Generalized Force*1

T *2

*3

c

C c

c

⎡ ⎤⎢ ⎥

Φ Φ= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Generalized Damping

MDOF Dynamics 19

Development of Uncoupled Equations of Motion

MU+CU+ KU F( )t=MDOF Equation of Motion:

U Y= ΦTransformation of Coordinates:

MFY+ CFY+ KFY F( )t=Substitution:T T T TM Y C Y K Y F(t)Φ Φ +Φ Φ +Φ Φ =ΦPremultiply by TΦ :

* * * *1 1 1 11 1 1

* * * *2 2 2 2 2 2 2

* * * *3 3 33 3 3 3

m c k f ( )y y ym y + c y + k y f ( )

y y ym c k f ( )

t

t

t

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎧ ⎫⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

⎩ ⎭ ⎩ ⎭ ⎩ ⎭⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭

Using Orthogonality Conditions: Uncoupled Equations of Motion are:

MDOF Dynamics 20

Development of Uncoupled Equations of Motion(Explicit Form)

* * * *1 1 1 1 1 1 1m y +c y + k y = f ( )tMODE 1

* * * *2 2 2 2 2 2 2m y +c y + k y = f ( )tMODE 2

* * * *3 3 3 3 3 3 3m y +c y + k y = f ( )tMODE 3

MDOF Dynamics 21

Simplify by Dividing Through by m* and defining*i

i *i i

c2m

ξω

=

Development of Uncoupled Equations of Motion(Explicit Form)

2 * *1 1 1 1 1 1 1 1y 2 y y f (t) / mξ ω ω+ + =MODE 1

2 * *2 2 2 2 2 2 2 2y 2 y y f (t) / mξ ω ω+ + =MODE 2

2 * *3 3 3 3 3 3 3 3y 2 y y f (t) / mξ ω ω+ + =MODE 3

MDOF Dynamics 22

ug r,1U

,1

,2

,3

,1

,2

,3

( ) u ( )F ( ) M ( ) u ( )

( ) u ( )

1.0 u ( )M 1.0 ( ) M u ( )

1.0 u ( )

⎧ ⎫+⎪ ⎪= + =⎨ ⎬⎪ ⎪+⎩ ⎭

⎧ ⎫⎧ ⎫⎪ ⎪ ⎪ ⎪+⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

g r

I g r

g r

r

g r

r

u t tt u t t

u t t

tu t t

t

Move to RHS as EFFF ( ) = - M R ( )gt u t

Earthquake “Loading” for MDOF System

MDOF Dynamics 23

MDOF Dynamics 24

* TgF (t) MR u (t)= −Φ

1R= 1

1

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

m1=2

m2=3

m3=1u1

u2

u3 2 1M 3

1

+⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

1R= 1

0

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

u1

u2

u3

m1=2

m2=3

m3=1

2M 3

1

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Modal Earthquake Loading

( )u tg

( )u tg

Definition of Modal Participation Factor

For earthquakes: * Ti i gf (t) MRu (t)φ= −

Typical Modal Equation:

Modal Participation Factor pi

i

* T2 i i

i i i i i i g* *i

f (t) MRy 2 y y u (t)m m

φξω ω+ + = = −

MDOF Dynamics 25

Caution Regarding Modal Participation Factor

Ti iMφ φ

Ti

i *i

MRpm

φ=

It’s value is dependent on the (arbitrary) method used to scale the mode shapes.

MDOF Dynamics 26

Variation of First Mode Participation Factorwith First Mode Shape

p1=1.0 p1=1.4 p1=1.6

MDOF Dynamics 27

1.0 1.0 1.0

Concept of Effective Modal Mass

2 *i i im p m=For each Mode i

• The sum of the effective modal mass for all modes isequal to the total structural mass.

• The value of effective modal mass isindependent of mode shape scaling.

• Use enough modes in the analysis to providea total effective mass not less than 90% of thetotal structural mass

MDOF Dynamics 28

Variation of First Mode Effective Masswith First Mode Shape

1m / M1m / M1m / M

MDOF Dynamics 29

1.0

=0.71.0

=1.01.0

=0.9

Derivation of Effective Modal Mass (1)For each mode:

2i i i i i i i gy 2 y y p uξ ω ω+ + = −

SDOF system:

2i i i i i i gq 2 q q uξω ω+ + = −

Modal response history, qi(t) is obtained by first solving the SDOF system.

MDOF Dynamics 30

Derivation of Effective Modal Mass (2)

y ( ) p q ( )i i it t=From Previous Slide

u ( ) y ( )i i it tφ=Recall

u ( ) p q ( )i i i it tφ=Substitute

MDOF Dynamics 31


Applied “static” forces required to produce ui(t):

( ) ( ) ( )i i i i iV t Ku t PK q tφ= =

2i i iK Mφ ω φ=Recall:

Substitute:

2( ) ( )i i i i iV t M P q tφ ω=

MDOF Dynamics 32

Derivation of Effective Modal Mass (4)T

i iV V R=Total shear in mode:

2 2( ) ( )T Ti i i i i iV M RP q t MRPφ ω φ= = ( )iq tω

“Acceleration” in modeDefine effective modal mass: T

i i iM MRPφ=

and2 ( )i i iV M q tω=

MDOF Dynamics 33


TT Ti

i i i i i iTi i i

MRM MRP M PM

φφ φ φφ φ⎡ ⎤

= = ⎢ ⎥⎣ ⎦

*2iii mPM =

MDOF Dynamics 34

Development of a Modal Damping Matrix

In previous development, we have assumed:

Φ ΦTCc

cc

=

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1

2

3

*

*

*

• Rayleigh “Proportional Damping”• Wilson “Discrete Modal Damping”

Two Methods Described Herein:

MDOF Dynamics 35

Rayleigh Proportional Damping

MASSPROPORTIONALDAMPER

STIFFNESSPROPORTIONALDAMPER

C M K= +α β

MDOF Dynamics 36


C M K= +α β

MDOF Dynamics 37

For Modal Equations to Be Uncoupled:

2ω ξ φ φn n nT

nC=

Using Orthogonality Conditions:

2 2ω ξ α βωn n n= +

ξω

α ω βnn

n= +1

2 2

Φ ΦT M I=Assumes


Select Damping value in two modes, ξm and ξn

Compute Coefficients α and β:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−−

=⎭⎬⎫

⎩⎨⎧

n

m

mn

mn

mn

nm

ξξ

ωωωω

ωωωω

βα

/1/12 22

C M K= +α βForm Damping Matrix

MDOF Dynamics 38

Rayleigh Proportional Damping (Example)

5% Critical in Modes 1 and 3

MDOF Dynamics 39

0.00

0.05

0.10

0.15

0 20 40 60Frequency, Radians/Sec.

Mod

al D

ampi

ng R

atio

MASSSTIFFNESSTOTAL

TYPEα = 0.41487β = 0.00324

Mode ω1 4.942 14.63 25.94 39.25 52.8

Structural Frequencies

Rayleigh Proportional Damping (Example)5% Damping in Modes 1 & 2, 1 & 3, 1 & 4, or 1 & 5

MDOF Dynamics 40

Modes α β1 & 2 .36892 0.005131 & 3 .41487 0.003241 & 4 .43871 0.002271 & 5 .45174 0.00173

0.00

0.05

0.10

0.15

0 20 40 60Fequency, Radians/sec

Mod

al D

ampi

ng R

atio

1,21,31,41,5

MODES

Proportionality Factors (5% each indicated mode)

Wilson Damping

Directly Specify Modal Damping Values ξi*

Φ ΦTCc

cc

mm

m=

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1

2

3

1 1 1

2 2 2

3 3 3

22

2

*

*

*

* *

* *

* *

ωξω ξ

ω ξ

MDOF Dynamics 41

FORMATION OF EXPLICIT DAMPING MATRIXFROM “WILSON” MODAL DAMPING

(NOT Usually Required)

cC

nn

nn

T =

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

•••=ΦΦ

−−

ωξωξ

ωξωξ

22

22

11

22

11

( )Φ ΦT c C− − =

1 1

MDOF Dynamics 42

C M Mi i iT

ii

n

=⎡

⎣⎢

⎤

⎦⎥

=∑2

1

ξω φ φ

5 5 5

0

10

0

2

4

6

8

10

12

4.94 14.57 25.9 39.2 52.8

Frequency, Radians per second

Mod

al D

ampi

ng R

atio

Wilson Damping (Example)5% Damping in Modes 1 and 2, 310% in Mode 5, Zero in Mode 4

MDOF Dynamics 43

Wilson Damping (Example)5% Damping in all Modes

MDOF Dynamics 44

5 5 5 5 5

0

2

4

6

8

10

12

4.94 14.57 25.9 39.2 52.8

Frequency, Radians per second

Mod

al D

ampi

ng R

atio

Solution of MDOF Equations of Motion

• Explicit (Step by Step) Integration of Coupled Equations

• Explicit Integration of FULL SET of Uncoupled Equations

• Explicit Integration of PARTIAL SET of Uncoupled

Equations (Approximate)

• Modal Response Spectrum Analysis (Approximate)

MDOF Dynamics 45

Computed Response for Piecewise Linear Loading

Time, t

Forc

e, V

(t)

t1 t3t2t0

Vt2

Vt1 1δ

Time, τ

δ = −−

V Vt tt t2 2

2 1

MDOF Dynamics 46

EXAMPLE of MDOF Response of Structure Responding to 1940 El Centro Earthquake

MDOF Dynamics 47

k3=180 k/in

u1(t)

u2(t)

u3(t)

k1=60 k/in

k2=120 k/in

m1=1.0 k-s2/in

m2=1.5 k-s2/in

m3=2.0 k-s2/in

Assume WilsonDamping with 5%critical in each mode.

N-S Component of 1940 El Centro earthquakeMaximum acceleration = 0.35 g

Example # 1

10 ft

10 ft

10 ft

Example # 1 (Continued)

Form Property Matrices:

k3=180 k/in

u1(t)

u2(t)

u3(t)

k1=60 k/in

k2=120 k/in

m1=1.0 k-s2/in

m2=1.5 k-s2/in

m3=2.0 k-s2/in

M =⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−10

152 0

..

.kip s / in2

K =−

− −−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

60 60 060 180 1200 120 300

kip / in

MDOF Dynamics 48


K MΦ ΦΩ= 2

Solve Eigenvalue Problem:

k3=180 k/in

u1(t)

u2(t)

u3(t)

k1=60 k/in

k2=120 k/in

m1=1.0 k-s2/in

m2=1.5 k-s2/in

m3=2.0 k-s2/in

Ω2 2

210966

212 4=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−

..

.sec

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=Φ

47.2676.0300.057.2601.0644.0

000.1000.1000.1

MDOF Dynamics 49

Normalization of Modes using Φ ΦT M I=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=Φ

514.0431.0223.0534.0384.0478.0

208.0638.0749.0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

47.2676.0300.057.2601.0644.0

000.1000.1000.1vs

MDOF Dynamics 50

MODE 1 MODE 2 MODE 3

T=1.37 sec T=0.639 sec T=0.431 sec

Example # 1 (Continued)Mode Shapes and Periods of Vibration

ω=4.58 rad/sec ω=9.83 rad/sec ω=14.57 rad/sec

MDOF Dynamics 51


k3=180 k/in

u1(t)

u2(t)

u3(t)

k1=60 k/in

k2=120 k/in

m1=1.0 k-s2/in

m2=1.5 k-s2/in

m3=2.0 k-s2/in

M MT*

..

.sec /= =

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−Φ Φ1801

2 4552310

2kip in

ωn =⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

458983

1457

.

..

/ secrad Tn =⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

13706390431

...

sec

Compute Generalized Mass:

MDOF Dynamics 52


k3=180 k/in

u1(t)

u2(t)

u3(t)

k1=60 k/in

k2=120 k/in

m1=1.0 k-s2/in

m2=1.5 k-s2/in

m3=2.0 k-s2/in

Compute Generalized Loading:

)()(* tvMRtVg

TΦ−=

*

2.5661.254 ( )2.080

n gV v t⎧ ⎫⎪ ⎪=− −⎨ ⎬⎪ ⎪⎩ ⎭

MDOF Dynamics 53


Write Uncoupled (Modal) Equations of Motion:

( ) /* *y y y V t m1 1 1 1 12

1 1 12+ + =ξω ω

k3=180 k/in

u1(t)

u2(t)

u3(t)

k1=60 k/in

k2=120 k/in

m1=1.0 k-s2/in

m2=1.5 k-s2/in

m3=2.0 k-s2/in

( ) /* *y y y V t m2 2 2 2 22

2 2 22+ + =ξ ω ω

( ) /* *y y y V t m3 3 3 3 32

3 3 32+ + =ξ ω ω

)(425.10.21458.0 111 tvyyy g−=++

)(511.06.96983.0 222 tvyyy g=++

)(090.04.212457.1 333 tvyyy g−=++

MDOF Dynamics 54

Modal Participation Factors

1.425

0.511

0.090

−

1.911

0.799

0.435

−ModeModeMode

123

φ φiT

iM =10.φi, .1 10=Modal Scaling

MDOF Dynamics 55

Modal Participation Factors

1.000 0.7441.425 0.644 1.911 0.480

0.300 0.223

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

MDOF Dynamics 56

usingφ11 1, = usingφ φ1 1 1T M =

Effective Modal Mass

MDOF Dynamics 57

2 *n n nM P m=

Mode 1Mode 2Mode 3

3 66 81 810 64 14 950 20 5 100%

.

.

.4.50 100%

%nM Accum%

Example # 1 (Continued)Solving Modal Equation via NONLIN:

For MODE 1:( ) /* *y y y V t m1 1 1 1 1

21 1 12+ + =ξω ω

)(425.10.21458.000.1 111 tvyyy g−=++

M = 1.00 kip-sec2/in

C = 0.458 kip-sec/in

K1 = 21.0 kips/inch

Scale Ground Acceleration by Factor 1.425

MDOF Dynamics 58

MDOF Dynamics 59

MODE 1


MODE 2

MODE 3

-2.00

-1.00

0.00

1.00

2.00

0 1 2 3 4 5 6 7 8 9 10 11 12

MODAL Displacement Response Histories (From NONLIN)

-6.00

-3.00

0.00

3.00

6.00

0 1 2 3 4 5 6 7 8 9 10 11 12

-0.20

-0.10

0.00

0.10

0.20

0 1 2 3 4 5 6 7 8 9 10 11 12

Time, Seconds

T1=1.37 sec

T2=0.64

T3=0.43

Maxima


-5

-4

-3

-2

-1

0

1

2

3

4

5

0 2 4 6 8 10 12

Time, Seconds

Mod

al D

ispl

acem

ent,

Inch

es

MODE 1MODE 2MODE 3

MODAL Response Histories:

MDOF Dynamics 60

MDOF Dynamics 61

-6-4-20246

0 1 2 3 4 5 6 7 8 9 10 11 12

-6-4-20246

0 1 2 3 4 5 6 7 8 9 10 11 12

Time, Seconds

u1(t)

u2(t)

u3(t)

=0.300 x MODE 1 - 0.676 x MODE 2 + 2.47 x MODE 3

Example #1 (Continued)Compute Story Displacement Response-Histories:

-6-4-20246

0 1 2 3 4 5 6 7 8 9 10 11 12

u t y t( ) ( )= Φ

MDOF Dynamics 62

u1(t)

u2(t)

u3(t)

Example #1 (Continued)Compute Story Shear Response-Histories:

-400-200

0200400

0 1 2 3 4 5 6 7 8 9 10 11 12

-400-200

0200400

0 1 2 3 4 5 6 7 8 9 10 11 12

-400-200

0200400

0 1 2 3 4 5 6 7 8 9 10 11 12

Time, Seconds

=k2[u2(t)-u3(t)]

Example #1 (Continued)

Displacements and Forces at time of Maximum Displacements (t = 6.04 seconds)

134.8

196.9

220.4

1348

3317

5521

0

Story Shear (k) Story OTM (ft-k)

134.8 k

62.1 k

23.5 k

5.11”

2.86”

1.22”

MDOF Dynamics 63

Example # 1(Continued)Displacements and Forces at time of Maximum Shear

(t = 3.18 seconds)

38.2 k

182.7 k

124.7 k

3.91”

3.28”

1.92”

38.2

162.9

345.6

382

2111

5567

0

Story Shear (k) Story OTM (ft-k)

MDOF Dynamics 64

Modal Response Response Spectrum Method• Instead of solving the time history problem for each

mode, use a response spectrum to compute theMAXIMUM response in each mode.

• These maxima are generally NONCONCURRENT

• Combine the maximum modal responses using somestatistical technique, such as Square Root of the Sum ofthe Squares (SRSS), or Complete Quadratic Combination(CQC)

• The technique is APPROXIMATE

• It is the basis for the Equivalent Lateral Force (ELF) Method

MDOF Dynamics 65

Example #1 (Response Spectrum Method)

MDOF Dynamics 66

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00

Period, Seconds

Spec

tral

Dis

plac

emen

t, In

ches

1.20”

3.04”

3.47”

MODE 3T=0.431 Sec. MODE 2

T=0.639 Sec.

MODE 1T=1.37 Sec.

Displacement Response Spectrum1940 El Centro, 0.35g, 5% Damping

MODAL RESPONSE

)(425.10.21458.0 111 tvyyy g−=++

)(511.06.96983.0 222 tvyyy g=++

)(090.04.212457.1 333 tvyyy g−=++

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00

Period, Seconds

Spec

tral

Dis

plac

emen

t, In

ches

1.20”

3.04”

3.47”

y1 1425 347 4 94= =. * . . "y2 0511 304 155= =. * . . "y3 0090 120 0108= =. * . . "

Modal Equations of Motion Modal Maxima

Example #1 (Continued)

MDOF Dynamics 67

MDOF Dynamics 68

MODE 1

MODE 2

MODE 3

-2.00

-1.00

0.00

1.00

2.00

0 1 2 3 4 5 6 7 8 9 10 11 12

-6.00

-3.00

0.00

3.00

6.00

0 1 2 3 4 5 6 7 8 9 10 11 12

-0.20

-0.10

0.00

0.10

0.20

0 1 2 3 4 5 6 7 8 9 10 11 12

Time, Seconds

T=1.37

T=0.64

T=0.43

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00

Period, Seconds

Spec

tral

Dis

plac

emen

t, In

ches

1.20 x 0.090”

3.04 x 0.511”

3.47x 1.425”The Scaled ResponseSpectrum values givethe Same Modal Maximaas the Previous ResponseHistories

Example #1(Continued)

Example #1 (Continued)Computing NONCONCURRENT Story Displacements

10000 6440 300

4 9404 94031811482

.

.

..

.

..

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

=⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

MODE 1

10000 6010 676

155015500 9311048

...

....

−−

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

= −−

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

MODE 2

10002 570

2 4700108

01080 278

0 267

..

..

..

.−⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

= −⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

MODE 3

MDOF Dynamics 69

Example #1 (Continued)Modal Combination Techniques (For Displacement)

MDOF Dynamics 70

At time of Max. Displacement

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

++++++

80.239.460.6

267.0048.1482.1278.0931.0181.3108.0550.1940.4

Sum of Absolute Values

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

22.186.215.5

“Exact”

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

++++++

84.133.318.5

267.0048.1482.1278.0931.0181.3108.0550.1940.4

222

222

222

Square Root of the Sum of the Squares Envelope of story displacement

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

93.118.315.5

Response Spectrum Results Response History Results

Example #1 (Continued)Computing Interstory Drifts

4 940 31813181 1482

1482 0

175916991482

. .

. ..

.

.

.

−−−

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪=⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

MODE 1

1550 09310931 1048

1048 0

248101171048

. ( . ). ( . )

.

.

..

− −− − −

− −

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪=−

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

MODE 2

0108 02780278 02670267 0

03860545

0267

. ( . ). ..

..

.

− −− −

−

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪= −⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

MODE 3

MDOF Dynamics 71

Example #1 (Continued)Computing Interstory Shears (Using Drift)

1759 601699 1201482 180

105520392668

. ( ). ( ). ( )

.

.

.

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪=⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

MODE 1

2481 600117 1201048 180

14891401886

. ( ). ( ). ( )

..

.−

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪=

−

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

MODE 2

0386 600545 120

0267 180

232654

481

. ( ). ( )

. ( )

...

−⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪= −⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

MODE 3

MDOF Dynamics 72

106 149 232204 14 65 4267 189 481

220215331

2 2 2

2 2 2

2 2 2

+ ++ ++ +

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪=⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

...

Example #1 (Continued)Computing Interstory Shears: SRSS Combination

207203346

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

“Exact”135197220

⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

“Exact”38 2163346

.⎧

⎨⎪

⎩⎪

⎫

⎬⎪

⎭⎪

“Exact”

At time ofMax. Shear

At time of Max.Displacement

Envelope = Maximumper Story

MDOF Dynamics 73

Caution:

Do NOT compute story shears from the storydrifts derived from the SRSS of the storydisplacements.

Calculate the story Shears in each mode(using modal drifts) and then SRSSthe results.

MDOF Dynamics 74

Using Less than Full (Possible)Number of Natural Modes

-5

-4

-3

-2

-1

0

1

2

3

4

5

0 2 4 6 8 10 12

Time, Seconds

Mod

al D

ispl

acem

ent,

Inch

es

MODE 1MODE 2MODE 3

MODAL Response Histories:

MDOF Dynamics 75

Using Less than Full Number of Natural Modes

Time-History for MODE 1

y ty t y t y t y t y t y t y t y t y tny t y t y t y t y t y t y t y t y tny t y t y t y t y t y t

( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .... ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .... ( )( ) ( ) ( ) ( ) ( ) (

=1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2

3 3 3 3 3 3

1 2 3 4 5 6 7 81 2 3 4 5 6 7 81 2 3 4 5 6 7 83 3 3) ( ) ( ) .... ( )y t y t y tn

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

[ ]u t y t( ) ( )= φ φ φ1 2 3

3 x nt 3 x 3 3 x nt

3 x nt

Transformation:

Time-History for DOF 1

u tu t u t u t u t u t u t u t u t u tnu t u t u t u t u t u t u t u t u tnu t u t u t u t u t u t

( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .... ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .... ( )( ) ( ) ( ) ( ) ( ) (

=1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2

3 3 3 3 3 3

1 2 3 4 5 6 7 81 2 3 4 5 6 7 81 2 3 4 5 6 7 83 3 3) ( ) ( ) .... ( )u t u t u tn

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

MDOF Dynamics 76

Using Less than Full Number of Natural Modes

Time-History for MODE 1

y ty t y t y t y t y t y t y t y t y tny t y t y t y t y t y t y t y t y tn

( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .... ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .... ( )

=⎡

⎣⎢

⎤

⎦⎥

1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2

1 2 3 4 5 6 7 81 2 3 4 5 6 7 8

NOTE: Mode 3 NOT Analyzed

MDOF Dynamics 77

u tu t u t u t u t u t u t u t u t u tnu t u t u t u t u t u t u t u t u tnu t u t u t u t u t u t

( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .... ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .... ( )( ) ( ) ( ) ( ) ( ) (

=1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2

3 3 3 3 3 3

1 2 3 4 5 6 7 81 2 3 4 5 6 7 81 2 3 4 5 6 7 83 3 3) ( ) ( ) .... ( )u t u t u tn

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

3 x nt 3 x 2 2 x nt

3 x nt

[ ]u t y t( ) ( )= φ φ1 2

Transformation:

Time-History for DOF 1

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

++++++

82.131.318.5

84.133.318.5

267.0048.1482.1278.0931.0181.3108.0550.1940.4

222

222

222

Square Root of the Sum of the Squares

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

++++++

53.2112.449.6

80.239.460.6

267.0048.1482.1278.0931.0181.3108.0550.1940.4

Sum of Absolute Values

Envelope of StoryDisplacement

Using Less than Full Number of Natural Modes(Modal Response Spectrum Technique)

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

93.118.315.5

MDOF Dynamics 78

2 modes3 modes

ASCE 7 allow an ApproximateModal Analysis Technique Called the

“EQUIVALENT LATERAL FORCE PROCEDURE”

• Empirical Period of Vibration• Smoothed Response Spectrum• Compute Total Base Shear V as if SDOF• Distribute V Along Height assuming

“Regular” Geometry• Compute Displacements and Member

Forces using Standard Procedures

MDOF Dynamics 79

EQUIVALENT LATERAL FORCE PROCEDURE

• Method is based on FIRST MODE response

• Higher modes can be included empirically• Has been calibrated to provide a

reasonable estimate of the envelopeof story shear, NOT to provide accurate estimates of story force

• May result in overestimate of overturningmoment.

MDOF Dynamics 80

EQUIVALENT LATERAL FORCE PROCEDURE• Assume first mode effective mass = Total Mass = M = W/g

• Use Response Spectrum to obtain Total Acceleration at T1

Acceleration, g

Sa1

Period, Sec.T1

WSg

WgSMgSV aaaB 111 )()( ===

MDOF Dynamics 81

MDOF Dynamics 82

dr

EQUIVALENT LATERAL FORCE PROCEDURE

Assume Linear First Mode Response

gWtd

hhtf x

rx

x21)()( ω=

hx

Wxfx

i

nstories

ii

rnstories

iiB Wh

hgtdtftV ∑∑

==

==1

21

1

)()()( ωh

Portion of base shear applied to story i

∑=

= nstories

iii

xx

B

x

Wh

WhtVtf

1

)()(

ELF EXAMPLE

k3=180 k/in

k1=60 k/in

k2=120 k/in

m1=1.0 k-s2/in

m2=1.5

m3=2.0

MDOF Dynamics 83

3h

RecallT1=1.37 seconds

ELF EXAMPLETotal Weight = M x g = (1.0 + 1.5 + 2.0) 386.4 = 1738 kips

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00

Period, Seconds

Spec

tral

Dis

plac

emen

t, In

ches

3.47”

1.37sec

Spectral Acceleration = ω2SD = (2π/1.37)2 x 3.47 = 72.7 in/sec2

= 0.188g

Base Shear = SaW = 0.188 x 1738 = 327 kips

MDOF Dynamics 84

ELF EXAMPLE (Story Forces)

3386(3 ) 0.381 0.375(327) 125

386(3 ) 579(2 ) 722( ) Bhf V kips

h h h= = = =

+ +

125 k

MDOF Dynamics 85

W=386 k

3h

W=579 k

W=722 k

125 k

77 k

327 k

125 k(220 k)

250 k(215 k)

Story Shear (k)

327 k(331 k)

ELF EXAMPLE (Story Displacements)Units=inches

5.983.891.82

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

ELFTime History(Envelope)

Modal ResponseSpectrum

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

84.133.318.55.15

3.181.93

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

MDOF Dynamics 86

ELF EXAMPLE (Summary)

• ELF procedure gives GOOD CORRELATIONwith base shear (327 kips ELF vs 331 kps ModalResponse Spectrum)

• ELF story force distribution is NOT as GOOD.ELF underestimates shears in upper stories.

• ELF gives reasonable correlation with displacements

MDOF Dynamics 87

ASCE 7-05 ELF APPROACH

• Uses Empirical Period of Vibration• Uses Smoothed Response Spectrum• Has Correction for Higher Modes• Has Correction for Overturning Moment• Has Limitations on Use

MDOF Dynamics 88

Approximate Periods of Vibrationxnta hCT =

Ct = 0.028, x = 0.8 for Steel Moment FramesCt = 0.016, x = 0.9 for Concrete Moment FramesCt = 0.030, x = 0.75 for Eccentrically Braced FramesCt = 0.020, x = 0.75 for All Other SystemsNote: Buildings ONLY!

NTa

1.0=For Moment Frames < 12 stories in height, minimumstory height of 10 feet. N= Number of Stories.

MDOF Dynamics 89

Adjustment Factor on Approximate Period

computedua TCTT ≤=

SD1 Cu> 0.40g 1.4

0.30g 1.40.20g 1.50.15g 1.6< 0.1g 1.7

Applicable ONLY if Tcomputed comes from a “properlysubstantiated analysis”

MDOF Dynamics 90

ASCE 7 Smoothed Design Acceleration Spectrum(for use with ELF)

“Long Period” Acceleration2

1

“Short Period” Acceleration

2

1

3 3

WCV S= ⎟⎠⎞

⎜⎝⎛

=

IR

SC DSS

⎟⎠⎞

⎜⎝⎛

=

IRT

SC DS

1

Spe

ctra

l Res

pons

e A

ccel

erat

ion,

Sa

SDS

SD1

Varies

T = 1.0T=0.2 Period, T

MDOF Dynamics 91

R is the Response Modification Factor which is afunction of system inelastic behavior. This is coveredin the topic on inelastic behavior. For now, we willUse R=1 which implies linear elastic behavior.

I is the Importance Factor. This depends on the Seismic Use Group. I=1.5 for Essential Facilities,1.25 for important high occupancy structures,and 1.0 for normal structures. This is coveredin the topic on the ASCE 7 Provisions. For now, we will use I=1.

MDOF Dynamics 92

EQUIVALENT LATERAL FORCE PROCEDUREHigher Mode Effects

+ =

Combined1st Mode 2nd Mode

MDOF Dynamics 93

Distribution of Forces along Height

VCF vxx =

∑=

= n

i

kii

kxx

vx

hw

hwC

1

MDOF Dynamics 94

k accounts for Higher Mode Effects

0.5 2.5

2.0

1.0

Period, Sec

k

k = 0.5T + 0.75(sloped portion only)

k=2k=1

MDOF Dynamics 95

ELF EXAMPLE (Story Forces)V=327 kips T=1.37 seconds k=0.5(1.37)+0.75=1.435

146 k

120 k

61 k

327 k

W=386 k (125 k)

(125 k)

(77 k)

MDOF Dynamics 96

146 k

266 k

Story Shear (k)

327 k

W=579 k

3hW=722 k

ASCE 7 ELF Limitations

• Applicable ONLY to “Regular” structures with Tless than 3.5 Ts Note:(Ts=SD1/SDS)

Adjacent story stiffness does not vary more than 30%

Adjacent story strength does not vary more than 20%

Adjacent story masses does not vary more than 50%

If violated, must use more advanced analysis (typicallyModal response spectrum analysis)

MDOF Dynamics 97

ASCE 7 ELF Other considerations that effect loading

• Orthogonal Loading Effects• Redundancy • Accidential Torsion• Torsional Amplification• P-Delta Effects

• Importance Factor• Ductility and Overstrength

MDOF Dynamics 98

MDOF Dynamics

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