structural-dynamics.pptx

U.S. Army Corps of Engineers Structural Dynamics 2
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

Linear Single Degree of
Freedom Systems
PART I

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 t
I ( )
f t
D ( )
0 5
. ( )
f t
S
0 5
. ( )
f t
S
mu t c u t k u t F t
( ) ( ) ( ) ( )
  
)
t
(
F
)
t
(
f
)
t
(
f
)
t
(
f S
D
I =
+
+
EQUATION OF DYNAMIC
EQUILIBRIUM

MASS
• Includes all dead weight of structure
• May include some live load
• Has units of FORCE/ACCELERATION
INERTIAL
FORCE
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
DAMPING
FORCE
VELOCITY
1.0
C
PROPERTIES OF
STRUCTURAL DAMPING

• Includes all structural members
• May include some “seismically nonstructural” members
• Has units of FORCE/DISPLACEMENT
SPRING
FORCE
DISPLACEMENT
1.0
K
STIFFNESS
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
SPRING
FORCE
DISPLACEMENT
STIFFNESS
AREA =
ENERGY
DISSIPATED
PROPERTIES OF
STRUCTURAL STIFFNESS
(2)

)
t
cos(
u
)
t
sin(
u
)
t
(
u ω
+
ω
ω
= 0
0

mu t k u t
( ) ( )
  0
Equation of Motion:
0
0u
u 
Initial Conditions:
Solution:
m
k
=
ω
UNDAMPED FREE
VIBRATION

m
k
=
ω
π
2
ω
=
f
ω
π
2
=
1
=
f
T
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
Displacement,
inches
T = 0.5 sec
u0

u0
1.0
UNDAMPED FREE
VIBRATION (2)

)]
t
sin(
u
u
)
t
cos(
u
[
e
)
t
(
u D
D
D
t
ω
ω
ξω
+
+
ω
= 0
0
0
ξω

mu t c u t k u t
( ) ( ) ( )
   0
Equation of Motion:
0
0u
u 
Initial Conditions:
Solution:
c
c
c
m
c
=
ω
2
=
ξ
2
ξ
-
1
ω
=
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, x  1.0 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
Displacement,
inches
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.
-4.00
-2.00
0.00
2.00
4.00
-20.00 -10.00 0.00 10.00 20.00
Velocity, In/sec
Damping
Force,
Kips
Velocity, in/sec.

Welded Steel Frame x = 0.010
Bolted Steel Frame x = 0.020
Uncracked Prestressed Concrete x = 0.015
Uncracked Reinforced Concrete x = 0.020
Cracked Reinforced Concrete x = 0.035
Glued Plywood Shear wall x = 0.100
Nailed Plywood Shear wall x = 0.150
Damaged Steel Structure x = 0.050
Damaged Concrete Structure x = 0.075
Structure with Added Damping x = 0.250
(2)

Natural Damping
Supplemental Damping
ξ is a structural (material) property,
independent of mass and stiffness
critical
0
7
to
5
0
=
ξ %
.
.
NATURAL
ξ is a structural property, dependent on
mass and stiffness, and
damping constant C of device
critical
30
to
10
=
ξ %
AL
SUPPLEMENT
C
(3)

)
t
sin(
p
)
t
(
u
k
)
t
(
u
m ω
=
+ 0


Equation 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
Force,
Kips
po=100 kips
= 0.25 sec
= Frequency of the Forcing Function
ω
π
2
=
T
ω
T
UNDAMPED HARMONIC LOADING

Solution:
)]
t
sin(
)
t
[sin(
)
/
(
k
p
)
t
(
u ω
ω
ω
-
ω
ω
ω
-
1
1
= 2
0
)
t
sin(
p
)
t
(
u
k
)
t
(
u
m ω
=
+ 0


Equation of Motion:
Assume system is initially at rest:
UNDAMPED HARMONIC
LOADING (2)

Define
ω
ω
=
β
( )
)
t
sin(
)
t
sin(
k
p
)
t
(
u ω
β
-
ω
β
-
1
1
= 2
0
Static Displacement,
The Steady State
Response is always at
the structure’s loading frequency
Transient Response
(at structure’s frequency)
LOADING FREQUENCY
Structure’s NATURAL FREQUENCY
Dynamic Magnifier,
UNDAMPED HARMONIC
LOADING
S
u
D
R

-10
-5
0
5
10
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Displacement,
in.
-10
-5
0
5
10
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Displacement,
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
Displacement,
in.
-200
-100
0
100
200
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Force,
Kips
rad/sec
π
2
=
ω
rad/sec
π
4
=
ω uS  50
. .
in
5
0
=
β .
LOADING,kips
STEADY
STATE
RESPONSE, in.
TRANSIENT
RESPONSE, in.
TOTAL
RESPONSE, in.

STEADY
STATE
RESPONSE, in.
TRANSIENT
RESPONSE, in.
TOTAL
RESPONSE, in.
LOADING,
kips
rad/sec
π
4
=
ω
rad/sec
π
4
≈
ω .
in
0
.
5

S
u
99
0
=
β .
-500
-250
0
250
500
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Displacement,
in. -150
-100
-50
0
50
100
150
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Force,
Kips
-500
-250
0
250
500
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Displacement,
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
Displacement,
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
Displacement,
in.
S
u
π
2
Linear Envelope
UNDAMPED RESONANT
RESPONSE CURVE

STEADY
STATE
RESPONSE, in.
TRANSIENT
RESPONSE, in.
TOTAL
RESPONSE, in.
LOADING, kips
rad/sec
π
4
=
ω
rad/sec
π
4
≈
ω uS  50
. .
in
01
1
=
β .
-150
-100
-50
0
50
100
150
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Force,
Kips
-500
-250
0
250
500
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Displacement,
in.
-500
-250
0
250
500
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Displacement,
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
Displacement,
in.

STEADY
STATE
RESPONSE, in.
TRANSIENT
RESPONSE, in.
TOTAL
RESPONSE, in.
LOADING, kips
rad/sec
π
8
=
ω
rad/sec
π
4
=
ω uS  50
. .
in
0
2
=
β .
-150
-100
-50
0
50
100
150
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Force,
Kips
-6
-3
0
3
6
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Displacement,
in.
-6
-3
0
3
6
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
Displacement,
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
Displacement,
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 
Magnification
Factor
1/(1-

2
)
Resonance
Slowly
Loaded
1.00
Rapidly
Loaded
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
Displacement
Amplitude,
Inches
0% Damping %5 Damping
HARMONIC LOADING AT RESONANCE
EFFECTS 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, 
Dynamic
Response
Amplifier
0.0% Damping
5.0 % Damping
10.0% Damping
25.0 % Damping
2
2
2
ξβ
2
+
β
-
1
1
=
)
(
)
(
RD
Resonance
Slowly
Loaded
Rapidly
Loaded

SUMMARY REGARDING
VISCOUS DAMPING
IN HARMONICALLY LOADED
SYSTEMS
• 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 DAMPING
IN 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
ENERGY
ABSORBED
ENERGY
DISSIPATED
ENERGY
RECOVERED
TOTAL
ENERGY
DISSIPATED
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
GROUND
ACC,
g
DEVELOPMENT 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.

ut
ug ur
EARTHQUAKE FORCE
ug = Ground displacement
ur = Relative displacement
ut = Total displacement
= ug + ur
ut = Total acceleration
= ug + ur
:
:
:

m u t u t cu t k u t
g r r r
[ ( )  ( )]  ( ) ( )
    0
mu t cu t k u t mu t
r r r g
 ( )  ( ) ( )  ( )
   
EARTHQUAKE 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)
Ground
Acceleration
(g's)
-30
-20
-10
0
10
20
30
40
0 10 20 30 40 50 60
Time (sec)
Ground
Velocity
(cm/sec)
-15
-10
-5
0
5
10
15
0 10 20 30 40 50 60
Time (sec)
Ground
Displacement
(cm)
EARTHQUAKE GROUND MOTION -
1940 EL CENTRO

)
(
)
(
)
(
)
( t
u
m
t
u
k
t
u
c
t
u
m g
r
r
r




 



)
(
)
(
)
(
)
( t
u
t
u
m
k
t
u
m
c
t
u g
r
r
r




 



ξω
2
=
m
c 2
ω
=
m
k
Divide through by m:
Make substitutions:
)
t
(
u
)
t
(
u
)
t
(
u
)
t
(
u g
r
r
r




 -
=
ω
+
ξω
2
+ 2
Simplified form:
“SIMPLIFIED” FORM OF EQUATION OF
MOTION:

“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 g
r
r
r




 -
=
ω
+
ξω
2
+ 2
Ground motion acceleration history
Structural frequency
Damping ratio

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)
Structural
Displacement
(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)
Ground
Acceleration
(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
DISPLACEMENT,
inches
5% Damped Response
Spectrum for Structure
Responding 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.
PEAK
DISPLACEMENT,
inches

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
Displacement,
Inches
COMPLETE 5% DAMPED ELASTIC DISPLACEMENT
RESPONSE SPECTRUM FOR EL CENTRO GROUND MOTION

DEVELOPMENT OF PSEUDOVELOCITY
RESPONSE 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
Pseudoacceleration,
in/sec
2
D
)
T
(
PSA 2
ω
≡
5% Damping
DEVELOPMENT OF
PSEUDOACCELERATION
RESPONSE SPECTRUM

NOTE ABOUT THE RESPONSE
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

0.00
50.00
100.00
150.00
200.00
250.00
300.00
350.00
0.1 1 10
Period (sec)
Acceleration
(in/sec
2
)
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
Pseudoacceleration,
g
0%
5%
10%
20%
Damping
PSEUDOACCELERATION RESPONSE
SPECTRA
FOR 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
Pseudoacceleration,
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
PSEUDOVELOCITY,
in/sec
FOUR-WAY LOG PLOT OF RESPONSE
SPECTRUM

0.1
1
10
100
0.01 0.1 1 10
Period, Seconds
Pseudo
Velocity,
In/Sec
0% Damping
5% Damping
10% Damping
20* Damping
For a given earthquake,
small variations in structural
frequency (period) can produce
significantly 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
Pseuso
Velocity,
in/sec
Period, seconds
El Centro
Loma Prieta
North Ridge
San Fernando
Average
Different earthquakes
will 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
TS
T0
Spectral
Acceleration,
g
SD1
SDS
TS = SD1 / SDS
T0 = 0.2TS

Linear Multiple Degree
of Freedom Systems
PART II

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 COORDINATES
FROM 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 NORMAL
COORDINATES.

EQUATIONS OF MOTION
MULTI-DEGREE-OF-FREEDOM SYSTEM
- MASS CONCENTRATED AT FLOOR LEVELS WHICH ARE SUBJECT TO LATERAL
DISPLACEMENTS ONLY
          
p(t)
u
k
u
c
u
m r
r
r 

 



EQUATIONS OF MOTION
• FREE VIBRATION
[C] = [0], {P(T)} = {0}
       
0

 r
r u
k
u
m 


EQUATIONS OF MOTION
MOTION OF A SYSTEM IN FREE VIBRATION IS SIMPLE HARMONIC
{ } { }
{ } { }
{ } { } t
sin
A
u
t
cos
A
u
t
sin
A
u
r
r
r
ω
ω
-
=
ω
ω
=
ω
=
2




EQUATIONS OF MOTION
[ ]{ } [ ]{ } { }
[ ]{ } { }
0
=
ω
-
0
=
+
ω
2
2
A
m
k
A
k
A
m
-

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.
 











1
0
0
0
1
0
0
0
1
m

DETERMINE STIFFNESS MATRIX
k = 12EI/hs
3 = 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
 















1
1
0
1
2
1
0
1
2
250
k
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] - w2[M]
w1 = 7.036 RADIANS/SEC.
The period is equal
to 2p/w:
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:
   










w



w



w


w

2
2
2
2
250
250
0
250
500
250
0
250
500
m
k

FIND MODE SHAPES
First Mode:









































0
0
0
)
036
.
7
(
250
250
0
250
)
036
.
7
(
500
250
0
250
)
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
(
250
250
0
250
)
685
.
19
(
500
250
0
250
)
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
(
250
250
0
250
)
491
.
28
(
500
250
0
250
)
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
33
32
31
23
22
21
13
12
11
3
2
1
X
X
X
u
u
u
r
r
r
3
33
2
32
1
31
3
3
23
2
22
1
21
2
3
13
2
12
1
11
1
X
X
X
u
X
X
X
u
X
X
X
u
r
r
r


















or
or
     
     
X
X
ur 




 3
2
1

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 11.4.5
GENERAL 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 11.4.7
SITE-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
im
i
m g
w
M




n
i
im
i
m g
w
L
1

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
w
L i
i
i




3
1
1
1
= 1.0 kip-sec2/in. (11
2 + 21
2 + 31
2)
= 1.0 (0.4452 + 0.8022 + 1.02)
= 1.841 kip-sec2/in.
g
w
M i
i
i




3
1
2
1
1

MODE
= 1.0 KIP-SEC2/IN. (F12 + F22 + F32)
= 1.0 (-1.22 - 0.55 + 1.0)
= -0.77 KIP-SEC2/IN.
g
w
L i
i
i




3
1
2
2
= 1.0 kip-sec2/in. (12
2 + 22
2 + 32
2)
= 1.0 (1.222 + 0.552 + 1.02)
g
w
M i
i
i




3
1
2
2
2

MODE = 1.0 KIP-SEC2/IN. (F13 + F23 + F33)
= 1.0 (1.802 - 2.25 + 1.0)
= 0.552 KIP-SEC2/IN.
g
w
L i
i
i




3
1
3
3
= 1.0 kip-sec2/in. (13
2 + 23
2 + 33
2)
= 1.0 (1.8022 + 2.252 + 1.02)
g
w
M i
i
i




3
1
2
3
3

Response Spectrum Analysis
For each mode m, determine (cont’d):
Effective Weight
Participating Mass
g
M
L
W
m
m
m
2

W
W
PM m





n
i
i
w
W
1
Where weight at floor level i

Determine Effective Weight and
Participation Mass for Each Mode
kips
in
in
kip
g
M
L
W 72
.
1059
sec
.
.
sec
4
.
386
841
.
1
247
.
2
2
2
2
1
2
1
1 





≈ 3 x w = 1159.2 kips
kips
g
M
L
W 65
.
12
4
.
386
310
.
9
)
552
.
0
( 2
3
2
3
3 



kips
g
M
L
W 08
.
82
4
.
386
791
.
2
)
77
.
0
( 2
2
2
2
2 




  kips
Wi 45
.
1154

Determine Effective Weight and
Participation Mass for Each Mode
 












996
.
0
011
.
0
4
.
386
3
65
.
12
071
.
0
4
.
386
3
08
.
82
914
.
0
4
.
386
3
72
.
1059
3
3
2
2
1
1
PM
W
W
PM
W
W
PM
W
W
PM

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
m
im
i
im
i
im V
w
w
F
 



DISTRIBUTE BASE SHEAR FOR EACH MODE
OVER HEIGHT OF STRUCTURE
Level, i Weight, wi i1 wi i1 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

EXAMPLE
39.6
31.7 17.6
17.6
25.2
14.1
14.1
31.2
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 - APPROXIMATE MODAL
ANALYSIS
• 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

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 ENVELOPE
OF STORY SHEAR, NOT TO PROVIDE ACCURATE ESTIMATES
OF STORY FORCE
• MAY RESULT IN OVERESTIMATE OF OVERTURNING MOMENT.
ASCE 7 COMPENSATES.

PROCEDURE
• 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
W
S
g
W
g
S
M
g
S
V a
a
a
B 1
1
1 )
(
)
( 



1st Mode 2nd Mode Combined
+ =
PROCEDURE
HIGHER MODE EFFECTS

V
C
F vx
x 


 n
i
k
i
i
k
x
x
vx
h
w
h
w
C
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?

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