Lesson 5 Structural Dynamics

U.S. Army Corps of Engineers Structural Dynamics 1

Lesson 5

Structural Dynamics


Lesson Objectives

Upon conclusion, participants should have:

1. A clear understanding of seismic structural response in terms of structural dynamics

2. An appreciation that code-based seismic design provisions are based on the principles of structural dynamics


Part I

Linear Single Degree of Freedom Systems


Structural Dynamics of SDOF Systems: Topic Outline

Equations of Motion for SDOF Systems

Structural Frequency and Period of Vibration

Behavior under Dynamic Load

Dynamic Amplification

Effect of Damping on Behavior

Linear Elastic Response Spectra


Mass

Stiffness

Damping

F t u t( ), ( )

t

F(t)

t

u(t)

Idealized Single Degree of Freedom System


F t( )f tI ( )

f tD( )0 5. ( )f tS0 5. ( )f tS

mu t c u t k u t F t( ) ( ) ( ) ( )

)t(F)t(f)t(f)t(f SDI =++

Equation of Dynamic Equilibrium


MASS

• Includes all dead weight of structure• May include some live load• Has units of FORCE/ACCELERATION

INE

RT

IAL

FO

RC

E

ACCELERATION

1.0

M

Properties of Structural MASS


DAMPING

• In absence of dampers, is called Natural Damping• Usually represented by linear viscous dashpot• Has units of FORCE/VELOCITY

DA

MP

ING

FO

RC

E

VELOCITY

1.0

C

Properties of Structural DAMPING


• Includes all structural members• May include some “seismically nonstructural” members• Has units of FORCE/DISPLACEMENT

SP

RIN

G F

OR

CE

DISPLACEMENT

1.0

K

ST

IFF

NE

SS

Properties of Structural STIFFNESS


• Is almost always nonlinear in real seismic response• Nonlinearity is implicitly handled by codes• Explicit modelling of nonlinear effects is possible

SP

RIN

G F

OR

CE

DISPLACEMENT

ST

IFF

NE

SS

AREA =ENERGYDISSIPATED

Properties of Structural STIFFNESS (2)


)tcos(u)tsin(u

)t(u ω+ωω

= 00

mu t k u t( ) ( ) 0Equation of Motion:

00uu Initial Conditions:

Solution:

m

k=ω

Undamped Free Vibration


m

k=ω

π2

ω=f

ω

π2=

1=

fT

Period of Vibration(sec/cycle)

Cyclic Frequency(cycles/sec, Hertz)

Circular Frequency (radians/sec)

-3

-2

-1

0

1

2

3

0.0 0.5 1.0 1.5 2.0

Time, seconds

Dis

pla

ce

me

nt,

in

ch

es

T = 0.5 sec

u0

u01.0

Undamped Free Vibration (2)


)]tsin(uu

)tcos(u[e)t(u DD

Dt ω

ω

ξω++ω= 00

0ξω

mu t c u t k u t( ) ( ) ( ) 0Equation of Motion:

00uu Initial Conditions:

Solution:

cc

c

m

c=

ω2=ξ 2ξ-1ω=ωD

Damped Free Vibration


Time, seconds

Displacement, inches

x = Damping ratio

When x = 1.0, the system is called critically damped.

Response of Critically Damped System, = 1.0x or 100% critical

Damping in Structures


-3

-2

-1

0

1

2

3

0.0 0.5 1.0 1.5 2.0

Time, seconds

Dis

pla

ce

me

nt,

in

ch

es

0% Damping

10% Damping

20% Damping

Damped Free Vibration


True damping in structures is NOT viscous. However, for low damping values, viscous damping allows for linear equations and vastly simplifies the solution.

Damping in Structures

-4.00

-2.00

0.00

2.00

4.00

-20.00 -10.00 0.00 10.00 20.00

Velocity, In/sec

Da

mp

ing

Fo

rce

, K

ips

Velocity, in/sec.


Welded Steel Frame x = 0.010Bolted Steel Frame x = 0.020

Uncracked Prestressed Concrete x = 0.015Uncracked Reinforced Concrete x = 0.020Cracked Reinforced Concrete x = 0.035

Glued Plywood Shear wall x = 0.100Nailed Plywood Shear wall x = 0.150

Damaged Steel Structure x = 0.050Damaged Concrete Structure x = 0.075

Structure with Added Damping x = 0.250

Damping in Structures (2)


Natural Damping

Supplemental Damping

ξ is a structural (material) property,independent of mass and stiffness

critical07to50=ξ %..NATURAL

ξ is a structural property, dependent onmass and stiffness, anddamping constant C of device

critical30to10=ξ %ALSUPPLEMENT

C

Damping in Structures (3)


)tsin(p)t(uk)t(um ω=+ 0Equation of Motion:

-150

-100

-50

0

50

100

150

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Time, Seconds

Fo

rce

, Kip

s

po=100 kips = 0.25 sec

= Frequency of the Forcing Function

ω

π2=T

ω

T

Undamped Harmonic Loading


Solution:

)]tsin()t[sin()/(k

p)t(u ω

ω

ω-ω

ωω-1

1= 2

0

)tsin(p)t(uk)t(um ω=+ 0Equation of Motion:

Assume system is initially at rest:

Undamped Harmonic Loading (2)


Define ω

ω=β

( ))tsin()tsin(k

p)t(u ωβ-ω

β-1

1= 2

0

Static Displacement, The Steady State

Response is always atthe structure’s loading frequency

Transient Response(at structure’s frequency)

LOADING FREQUENCY

Structure’s NATURAL FREQUENCY

Dynamic Magnifier,

Undamped Harmonic Loading

Su

DR


-10

-5

0

5

10

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00Dis

pla

ce

me

nt,

in.

-10

-5

0

5

10

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00Dis

pla

ce

me

nt,

in.

-10

-5

0

5

10

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Time, seconds

Dis

pla

ce

me

nt,

in.

-200-100

0

100200

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00F

orc

e, K

ips

rad/secπ2=ωrad/secπ4=ω uS 50. .in50=β .

LOADING,kips

STEADYSTATERESPONSE, in.

TRANSIENTRESPONSE, in.

TOTALRESPONSE, in.




TOTALRESPONSE, in.

LOADING, kips

rad/secπ4=ωrad/secπ4≈ω .in 0.5Su990=β .

-500

-250

0

250

500

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Dis

pla

ce

me

nt,

in.

-150

-100

-50

0

50

100

150

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00F

orc

e, K

ips

-500

-250

0

250

500

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Dis

pla

ce

me

nt,

in.

-80

-40

0

40

80

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Time, seconds

Dis

pla

ce

me

nt,

in.


-80

-40

0

40

80

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Time, seconds

Dis

pla

ce

me

nt,

in.

Suπ2

Linear Envelope

Undamped Resonant Response Curve




TOTALRESPONSE, in.

LOADING, kips

rad/secπ4=ωrad/secπ4≈ω uS 50. .in011=β .

-150

-100

-50

0

50

100

150

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00F

orc

e, K

ips

-500

-250

0

250

500

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Dis

pla

ce

me

nt,

in

.

-500

-250

0

250

500

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Dis

pla

ce

me

nt,

in

.

-80

-40

0

40

80

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Time, seconds

Dis

pla

ce

me

nt,

in

.




TOTALRESPONSE, in.

LOADING, kips

rad/secπ8=ωrad/secπ4=ω uS 50. .in02=β .

-150

-100

-50

0

50

100

150

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00F

orc

e, K

ips

-6

-3

0

3

6

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Dis

pla

ce

me

nt,

in

.

-6

-3

0

3

6

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Dis

pla

ce

me

nt,

in

.

-6

-3

0

3

6

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

Time, seconds

Dis

pla

ce

me

nt,

in

.


0.00

2.00

4.00

6.00

8.00

10.00

12.00

0.00 0.50 1.00 1.50 2.00 2.50 3.00

Frequency Ratio

Mag

nifi

catio

n F

acto

r 1/

(1-2

)

Resonance

SlowlyLoaded

1.00

RapidlyLoaded

Response Ratio: Steady State to Static(Absolute Values)


-200

-150

-100

-50

0

50

100

150

200

0.00 1.00 2.00 3.00 4.00 5.00

Time, Seconds

Dis

pla

cem

en

t A

mp

litu

de

, In

che

s

0% Damping %5 Damping

Harmonic Loading at ResonanceEffects of Damping


0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0.00 0.50 1.00 1.50 2.00 2.50 3.00

Frequency Ratio,

Dy

na

mic

Re

sp

on

se

Am

plif

ier

0.0% Damping

5.0 % Damping

10.0% Damping

25.0 % Damping

222 ξβ2+β-1

1=

)()(RD

Resonance

SlowlyLoaded Rapidly

Loaded


For system loaded at a frequency less than √2 or 1.414 times its natural frequency, the dynamic response exceeds the static response. This is referred to as Dynamic Amplification.

An undamped system, loaded at resonance, will have an unbounded increase in displacement over time.

Summary Regarding Viscous Dampingin Harmonically Loaded Systems


Summary Regarding Viscous Dampingin Harmonically Loaded Systems

Damping is an effective means for dissipating energy in the system. Unlike strain energy, which is recoverable, dissipated energy is not recoverable.

A damped system, loaded at resonance, will have a limited displacement over time, with the limit being (1/2x) times the static displacement.

Damping is most effective for systems loaded at or near resonance.


LOADING YIELDING

UNLOADING UNLOADED

F F

F

u

F

u

u u

ENERGYABSORBED

ENERGYDISSIPATED

ENERGYRECOVERED

TOTALENERGYDISSIPATED

Concept of Energy Absorbed and Dissipated


-0.40

-0.20

0.00

0.20

0.40

0.00 1.00 2.00 3.00 4.00 5.00 6.00

TIME, SECONDS

GR

OU

ND

AC

C,

gDevelopment of Effective

Earthquake Force Unlike wind loading,

earthquakes do not apply any direct forces on a structure

Earthquake ground motion causes the base to move, while masses at the floor levels try to stay in their places due to inertia.

This creates stresses in the resisting elements.


utug ur

Development of Effective Earthquake Force

ug = Ground displacement

ur = Relative displacement

ut = Total displacement

= ug + ur

ut = Total acceleration

= ug + ur

:

: :


m u t u t c u t k u tg r r r[ ( ) ( )] ( ) ( ) 0

mu t c u t k u t mu tr r r g ( ) ( ) ( ) ( )

Development of EffectiveEarthquake Force

Inertia force depends on the total acceleration of the masses.

Resisting forces from stiffness and damping depend on the relative displacement and velocity.

Thus, the equation of motion can be written as:


Many ground motions now available via the Internet

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0 10 20 30 40 50 60

Time (sec)

Gro

un

d A

cce

lera

tio

n (

g's

)

-30

-20

-10

0

10

20

30

40

0 10 20 30 40 50 60

Time (sec)

Gro

un

d V

elo

city

(cm

/se

c)

-15

-10

-5

0

5

10

15

0 10 20 30 40 50 60

Time (sec)

Gro

un

d D

isp

lace

me

nt

(cm

)

Earthquake Ground Motion - 1940 El Centro


)()()()( tumtuktuctum grrr

)()()()( tutum

ktu

m

ctu grrr

ξω2=m

c 2ω=m

k

Divide through by m:

Make substitutions:

)t(u)t(u)t(u)t(u grrr -=ω+ξω2+ 2

Simplified form:

“Simplified” form of Equation of Motion:


For a given ground motion, the response history ur(t) is a function of the structure’s frequency w and

damping ratio x

)t(u)t(u)t(u)t(u grrr -=ω+ξω2+ 2

Ground motion acceleration history

Structural frequency

Damping ratio

“Simplified” form of Equation of Motion:


Change in ground motion

or structural parameters x

and w requires re-calculation of structural response

-6

-4

-2

0

2

4

6

0 10 20 30 40 50 60

Time (sec)

Str

uct

ura

l D

isp

lace

me

nt

(in

)

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0 10 20 30 40 50 60

Time (sec)

Gro

un

d A

cce

lera

tio

n (

g's

) Excitation applied to

structure with given x and w

Peak Displacement

Computed Response

SOLVER

Response to Ground Motion (1940 El Centro)


0

4

8

12

16

0 2 4 6 8 10

PERIOD, Seconds

DIS

PL

AC

EM

EN

T, in

ches

5% Damped Response Spectrum for StructureResponding to 1940 El Centro Ground Motion

The Elastic Response Spectrum

An Elastic Response Spectrum is a plot of the peak computed relative displacement, ur, for an elastic structure with

a constant damping x and a varying fundamental frequency w (or period T=2p/w), responding to a given ground motion.

PE

AK

DIS

PL

AC

EM

EN

T, i

nc

he

s


Computation of Deformation (or Displacement) Response Spectrum


0.00

2.00

4.00

6.00

8.00

10.00

12.00

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Period, Seconds

Dis

pla

ce

me

nt,

Inc

he

s

Complete 5% Damped Elastic Displacement Response Spectrum for El Centro Ground Motion


Development of PseudovelocityResponse Spectrum

D)T(PSV ω≡

5% Damping


0.0

50.0

100.0

150.0

200.0

250.0

300.0

350.0

400.0

0.0 1.0 2.0 3.0 4.0

Period, Seconds

Ps

eu

do

ac

ce

lera

tio

n, i

n/s

ec

2

D)T(PSA 2ω≡

5% Damping

Development of PseudoaccelerationResponse Spectrum


The Pseudoacceleration Response Spectrum represents the TOTAL ACCELERATION of the system, not the relative acceleration. It is nearly identical to the true total acceleration response spectrum for lightly damped structures.

5% Damping

Note about the Response Spectrum


0.00

50.00

100.00

150.00

200.00

250.00

300.00

350.00

0.1 1 10Period (sec)

Ac

ce

lera

tio

n (

in/s

ec2 )

Total Acceleration Pseudo-Acceleration

Difference Between Pseudo-Acceleration and Total Acceleration

System with 5% Damping


0.00

1.00

2.00

3.00

4.00

0.0 1.0 2.0 3.0 4.0 5.0

Period, Seconds

Ps

eu

do

ac

ce

lera

tio

n,

g

0%

5%

10%

20%

Damping

Pseudoacceleration Response Spectrafor Different Damping Values


Example Structure

K = 500 kips/in.

W = 2,000 kips

M = 2000/386.4 = 5.18 kip-sec2/in.

w = (K/M)0.5 =9.82 rad/sec

T=2p/ w = 0.64 sec

5% Critical Damping

@T=0.64 sec, Pseudoacceleration = 301 in./sec2

0.0

50.0

100.0

150.0

200.0

250.0

300.0

350.0

400.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Period, Seconds

Pse

ud

oac

cele

rati

on

, in

/sec

2

Base Shear = M x PSA = 5.18(301) = 1559 kips

Use of an Elastic Response Spectrum


0.10

1.00

10.00

100.00

0.01 0.10 1.00 10.00

PERIOD, Seconds

PS

EU

DO

VE

LOC

ITY,

in/s

ec

1.0

10.0

0.1

0.01

Acceleration, g

0.00

1

10.0

0.10

1.0

0.01

0.001

Displa

cem

ent,

in.

Four-Way Log Plot of Response Spectrum


0.1

1

10

100

0.01 0.1 1 10Period, Seconds

Pse

ud

o V

elo

city

, In

/Sec

0% Damping

5% Damping

10% Damping

20* Damping

1.0

10.0

0.1

0.01

Acceleration, g

0.00

1

10.0

0.10

1.0

0.01

0.001

Displa

cem

ent,

in.

For a given earthquake,small variations in structural frequency (period) can producesignificantly different results.

1940 El Centro, 0.35 g, N-S


0.1

1.0

10.0

100.0

0.01 0.10 1.00 10.00

Pse

uso

Ve

loci

ty,

in/s

ec

Period, seconds

El Centro

Loma Prieta

North Ridge

San Fernando

Average

Different earthquakeswill have different spectra.

5% Damped Spectra for Four California Earthquakes Scaled to 0.40 g (PGA)


Relative Displacement

Total Acceleration

Ground Displacement

Zero

VERY FLEXIBLE STRUCTURE(T > 10 sec)


Total Acceleration

Zero

Ground Acceleration

Relative Displacement

VERY STIFF STRUCTURE(T < 0.01 sec)


ASCE 7-10 Uses a Smoothed Design Acceleration Spectrum

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7

Period, seconds

0.4SDS

Sa = SD1 / T

Sa = SDS(0.4 + 0.6 T/T0)

Sa = SD1 TL / T2

TST0

Sp

ectr

al A

ccel

era

tion,

g

SD1

SDS

TS = SD1 / SDS

T0 = 0.2TS


Part II

Linear Multiple Degree of Freedom Systems


Structural Dynamics of MDOF Systems

Uncoupling of Equations through use of Natural Mode Shapes

Solution of Uncoupled Equations

Recombination of Computed Response

Modal Response Spectrum Analysis

Equivalent Lateral Force Procedure


Solving Equations of Motion

We need to develop a way to solve the equations of motion.

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

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

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


Equations of Motion

Multi-degree-of-freedom system

- mass concentrated at floor levels which are subject to lateral displacements only

p(t)ukucum rrr


Equations of Motion

Free vibration

[C] = [0], {p(t)} = {0}

0 rr ukum


Equations of Motion

Motion of a system in free vibration is simple harmonic

{ } { }

{ } { }

{ } { } tsinAu

tcosA u

tsinAu

r

r

r

ωω-=

ωω=

ω=

2


Equations of Motion

[ ]{ } [ ]{ } { }

[ ]{ } { }0=ω-

0=+ω2

2

Amk

AkAm-


Example

Given

hs = 10 ft

w = 386.4 kips/floor

E = 4000 ksi

Icol = 4500 in.4 each column (0.7Ig )


Determine Mass Matrix

m = w/g = 386.4 / 386.4

= 1.0 kip-sec.2/in.

100

010

001

m


Determine Stiffness Matrix

k = 12EI/hs3 = 12 x 4000 x 9000 / (12x10)3 = 250 kips/in.

kij = force corresponding to displacement of coordinate i resulting from a unit displacement of coordinate j

110

121

012

250k

250

250

250

K13 = 0

K23 = -250

K33 = 250

K12 = -250

K22 = 500

K32 = -250

K11 = 500

K21 = -250

K31 = 0

1

1

1


Find Determinant for Matrix [k] - 2[m]

The period is equal to 2/:

T1 = 0.893 sec.

T2 = 0.319 sec.

T3 = 0.221 sec.

Setting the determinant of the above matrix equal to zero yields the following frequencies:

2

2

2

2

2502500

250500250

0250500

mk

1 = 7.036 radians/sec.




Find Mode Shapes

First Mode:

0

0

0

)036.7(2502500

250)036.7(500250

0250)036.7(500

11

21

31

2

2

2

31 = 1.0, 21 = 0.802, 11 = 0.445


Second Mode:

32 = 1.0, 22 = -0.55, 12 = -1.22

Find Mode Shapes

0

0

0

)685.19(2502500

250)685.19(500250

0250)685.19(500

12

22

32

2

2

2


Third Mode:

33 = 1.0, 23 = -2.25, 13 = 1.802

Find Mode Shapes

0

0

0

)491.28(2502500

250)491.28(500250

0250)491.28(500

13

23

33

2

2

2


Orthogonality Properties of Natural Mode Shapes

The natural modes of vibration of any multi degree of freedom system are orthogonal with respect to the mass and stiffness matrices. The same type of orthogonality can be assumed to apply to the damping matrix as well:

{m}T[m] {n} = 0

{m}T[c] {n} = 0 for m ≠ n

{m}T[k] {n} = 0


Mode Superposition Analysis of Earthquake Response

3

2

1

333231

232221

131211

3

2

1

X

X

X

u

u

u

r

r

r

3332321313

3232221212

3132121111

XXXu

XXXu

XXXu

r

r

r

or

or

XXur 321


Mode Superposition Analysis of Earthquake Response


Modal Response Spectrum Analysis

An elastic dynamic analysis of structure utilizing the peak dynamic response of all modes having a significant contribution to total structural response. Peak modal responses are calculated using the ordinates of the appropriate response spectrum curve which correspond to the modal periods. Maximum modal contributions are combined in a statistical manner to obtain an approximate total structural response.


ASCE 7-10 Section 11.4.5General Procedure Design Spectrum

A five percent damped elastic design response

spectrum constructed in accordance with ASCE 7-10

Figure 11.4-1, using the values of SDS and SD1

consistent with the specific site.


ASCE 7-10 Section 11.4.7Site-Specific Ground Motion Procedures

The site-specific ground motion procedures set forth in Chapter 21 are permitted to be used to determine ground motions for any structure.


Response Spectrum Analysis

For each mode m, determine:

Earthquake participation factor

Modal Mass

1

2

i

imim g

wM

n

i

imim g

wL1


Determine Modal Mass and Participation Factors for Each Mode

= 1.0 kip-sec2/in. (f11 + f21 + f31)

= 1.0 (0.445 + 0.802 + 1.0)

= 2.247 kip-sec2/in.

g

wL i

ii

3

11

1

= 1.0 kip-sec2/in. (f112 + f21

2 + f31

2)

= 1.0 (0.4452 + 0.8022 + 1.02)


g

wM i

ii

3

1

21

1



= 1.0 kip-sec2/in. (f12 + f22 + f32)

= 1.0 (-1.22 - 0.55 + 1.0)

= -0.77 kip-sec2/in.

g

wL i

ii

3

12

2

= 1.0 kip-sec2/in. (f122 + f22

2 + f32

2)

= 1.0 (1.222 + 0.552 + 1.02)


g

wM i

ii

3

1

22

2



= 1.0 kip-sec2/in. (f13 + f23 + f33)

= 1.0 (1.802 - 2.25 + 1.0)


g

wL i

ii

3

13

3

= 1.0 kip-sec2/in. (f132 + f23

2 + f33

2)

= 1.0 (1.8022 + 2.252 + 1.02)


g

wM i

ii

3

1

23

3



For each mode m, determine (cont’d):

Effective Weight

Participating Mass

gM

LW

m

mm

2

W

WPM m

n

iiwW

1

Where weight at floor level i


Determine Effective Weight and Participation Mass for Each Mode

kipsin

in

kipg

M

LW 72.1059

sec

.

.

sec4.386

841.1

247.22

22

1

21

1

≈ 3 x w = 1159.2 kips

kipsgM

LW 65.124.386

310.9

)552.0( 2

3

23

3

kipsgM

LW 08.824.386

791.2

)77.0( 2

2

22

2

kipsWi 45.1154


Determine Effective Weight and Participation Mass for Each Mode

996.0

011.04.3863

65.12

071.04.3863

08.82

914.04.3863

72.1059

33

22

11

PM

W

WPM

W

WPM

W

WPM



Determine number of modes to be considered... to

represent at least 90% of participating mass of

structure (ASCE 7-05 Sec. 12.9.1)

PM = (Wm/W) 0.90



Determine spectral acceleration for each mode from design response spectra

Mode 1: T1 = 0.893 sec → Sa1 = 0.084g

Mode 2: T2 = 0.319 sec → Sa2 = 0.24g

Mode 3: T3 = 0.221 sec → Sa3 = 0.34g



Determine base shear for each mode

Mode 1: V1 = 0.0840 x 1059.72 = 89.0 kips

Mode 2: V2 = 0.24 x 82.08 = 19.7 kips

Mode 3: V3 = 0.34 x 12.65 = 4.3 kips



Distribute base shear for each mode over height of structure

where Fim = lateral force at level i for mode m

Vm = base shear for mode m

mimi

imiim V

w

wF


Distribute Base Shear for Each Mode over Height of Structure

Level, i Weight, wi fi1 wi fi1 Fi1

3 386.4 1 386.4 39.6

2 386.4 0.802 309.9 31.7

1 386.4 0.445 171.9 17.6

= 868.2 88.9

Mode 1 V1 = 89.0 kips




3 386.4 1 386.4 -25.2

2 386.4 -0.55 -212.5 14.1

1 386.4 -1.22 -471.4 31.2

= -297.5 20.1





3 386.4 1 386.4 7.8

2 386.4 -2.25 -869.4 -17.6

1 386.4 1.802 696.3 14.1

= 213.3 4.3



Response Spectrum Analysis Example

39.6

31.7 17.6

17.6

25.2

14.1

14.131.2

7.8

Mode 1 Mode 2 Mode 3



Perform lateral analysis for each mode... to determine member forces for each mode of vibration being considered



Combine dynamic analysis results (moments, shears,

axial forces, and displacements) for all considered

modes using root mean square combination (SRSS)...

to approximate total structural response or resultant

design values


ASCE 7 Allows an Approximate Modal 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

ASCE 7 - Approximate Modal Analysis



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 overturning moment. ASCE 7 compensates.


Assume first mode effective mass = Total Mass = M = W/g

Use Response Spectrum to obtain Total Acceleration @ T1

T1

Sa1/g

Period, sec

Acceleration, g

WSg

WgSMgSV aaaB 111 )()(



1st

Mode 2nd

Mode Combined

+ =

Equivalent Lateral Force ProcedureHigher Mode Effects


VCF vxx

n

i

kii

kxx

vx

hw

hwC

1

Distribution of Forces along Height


k=1 k=2

0.5 2.5

2.0

1.0

Period, Sec

k

k = 0.5T + 0.75(sloped portion only)

k accounts for Higher Mode Effects


Thank You!!

Any Remaining Questions?

Documents

Lesson 5 Structural Dynamics