dynamical analysis of soil and structures

1. Earthquake engineering: ground motion, seismic waves, near or
far, foundation design, dynamic seismic soil-structure interaction
(SSI)
1. Sources of Dynamic Loads and Engineering Relevance
5

6
3. Blast and shock engineering: refinery design, homeland
security, defense industry, air and ground shocks, explosion,
mining, ...
2. Wind engineering: high-rises, long bridges, transmission
towers, tall chimneys

7
4. Dynamic foundation design or foundation isolation of
machines/equipment: pump stations, compressors and turbines, high-
tech buildings, wind turbines, …
5. Pile driving & capacity: impact and stress waves, hammer cushions,
bearing capacity determination, NDT, ‘wave equation’ method

6. Impact loading transmitted to foundation and
structures
8

9
7. Offshore foundation engineering
8. Geophysical methods & in-situ sounding

9. Dynamic design for high-speed transportations, tunnels,
buried pipelines
10
10. Construction induced vibrations
….

2. Theory of Dynamics & Vibrations
Basic principles:
• Newton’s laws in physics
• Free-body diagram for dynamic problems
2 types of systems:
• Discrete systems (finite number of degree of freedoms)
• Continuous systems (infinite d.o.f.s)
11
. ., 2 :
nd
e g law m
 
F a m

 i
f a

Discrete systems (finite number of degree of freedoms)
• Lumped mass system:
(a) Single or 1-dof system
12
x(t)
f(t)

(b) Multi-dof system (c) Continuum system
13

14
Analysis of a Single d.o.f. system (SDOF):
(a) sign convention
(b) Free-body diagram (FBD) for
x(t)
f(t)
m
k
c
(c) Write the equation implied by FBD:
( ): ( )
LHS RHS f t k x c x m x
     
   
( )
m x c x k x f t
  
  … a simple 2nd-order differential equation
cx

kx
=
m x

f(t)
m
i
F a



15
1st step: Free Vibration Problem:
Common canonical form of the ‘Equation of Motion’ by
dividing by m,
.... a 2nd-order homogeneous differential equation
0
m x c x k x
  
 
2
2 0 ................ (*)
n n
x x x
 
  
 
.......... ( / )
..........
2
2 2 2 ......
n
n
c n
k
where circular rad s
m
c
m
k
c m m km
m
natural frequency
fraction ofcritical damping / damping ratio
Critical damping constant






 
  
 
 
 

Solution of
Take
• Special case 1: =0 (i.e., undamped case)
16
2
2 0 ......... (*)
n n
x x x
 
  
 
2 2
(*) ( 2 ) 0
t
n n
A e
   
   
2 2
2
1,2
For non-trivial solution, must have ( 2 ) 0
.............. a simple quadratic equation whose 2 roots are
1
n n
n n
   
   
  
   
1,2 1 , , = 1
n n
i i
  
     
( )= ' '
n n
i t i t
x t a e b e
 

 
cos( t) sin( t
s )
cos in n
i t
n
i
n
e i
e i
 
 
 

  
 
Recall
   
( )= ' cos( t)+ sin ( t) ' cos( t)- sin ( t)
( ' ')cos( t) ( ' ')sin ( t)
= cos( t) sin ( t) !
n n n n
n n
n n
x t a i b i
a b i a b
a b
   
 
 

   

( ) t
x t A e


17
( ) cos( ) sin( ) (*)
sin( ) sin cos cos sin .
n n
x t a t b t
trigonometricidentity A A A
 
     
 
  
General solution
:
( ) ( ) 2
2 / ... "natural period"
1/ ... "natural frequency"
( / sec)
2 ... " natural frequency"
( / sec)
/ ....( / sec)
n n n
n n
n n
n n
n
t T t
T
f T
Hz cycle
f circular
rad
with k m rad
  
 
 

  
 
 

 

2 2
(*)... , sin cos , (*)
( )
where " " , " " .
To determine the 2 constants ( , ) or ( , ) in the general solution,
n
n
Taking t a A and b A then is equivalent to
x t A t
a
A a b
b
a b A
   
 


  
 
  
 
-1
= sin( + )
amplitude = phase angle = tan
use initia
0
0
:
(i) initialdisplacement x(0)= x
(II) initialvelocity x(0) = x
l conditions
 

Case (ii) >1 (over-damped case)
18
both are pure exponential decaying terms (no oscillation at all)
2
1,2 1,2
are both -ve from 1
n n
    
   
1 2
( )=
t t
x t a e be
 
 
 
Case (iii) =1 (Critically-damped case)
1 2
= ( . . )
n i e a double root
  
 
   
1
( )= +b +b n t
t
x t a t e a t e

 

 

19
   
2 2
2 2
1 1
1 1
( )= '
' '
n n n n
n n
n n
i t i t
i t i t
t t
x t a e e
a e e b e e
     
   
   
     
  
 

 
Case (iv) 0 <  < 1 (Under-damped case)
- most relevant to geotechnical & structural problems
2 2
1,2 1 1
n n n n
      
       
i
0
0
As in previous cases, can find the 2 constants :
1) initialdisplacement x(0)= x
2) initialvelocity x(0) = x
by initial conditions
 
 
 
 
2 2
1 1
2
= ' '
= ' '
sin ( )
1 ... " "
n n
n
n d d
n
i t i t
t
t i t i t
t
d
d n
e a e b e
e a e b e
e A t
where damped natural frequency
   
 
   
 
 
  
  

 



 
 

20
0 0
Imposing theinitial conditions of
(1) initialdisplacement x(0)= x , (2) initialvelocity x(0) = x ,
one finds
 

Applications: Determination of damping ratio by Logarithmic Decrement method
(for small )
21
1
1
from solution,
n N
n N
t
N
t
N
x Ae
x Ae
Note :
free vibration
zw
zw 






1
1
( )
1
n N
n N N
n N
t
t t
N
t
N
x e
Then e
x e
zw
zw
zw






 
 
 
 
nt
Ae zw


22
ᇱ
ே
ேାଵ
௡ ேାଵ ே ௡ ௗ
ௗ ௗ ௗ ௡
ଶ
ଶ

(II) Forced vibration
23
( )
m x c x k x F t
  
 
2
2 ( ) / ................ (*)
n n
or x x x F t m
 
  
 
0
0
Consider a common and basic form of dynamic load:
................ (*)
.... force magnitude
.... excitation frequency
( ) sin( )
F
F t F t




homog particular
( ) ( ) ( )
x t x t x t
 
General Solution to (*):
( ) ( ) forced vibration
free vibration
or x t x t
 

24
:
sin( ) cos( )
Take A t B t
P f
P
To find x (t) or x (t)
x (t)  
 
 
 
2 2 2 2
0
2 2
0
2 2
2 sin( ) ( 2 )cos( ) ( / )sin( )
/
2
0
2
n n n n
n n
n n
gives
A B A t B A B t F m t
A F m
B
          
   
   
       
 
      
 
     
      
 
2
P n P n P 0
Equation of motion x + 2ςω x +ω x = F / m sin(ωt)
 
2 2
cos( ) sin( )
sin( ) cos( )
then A t B t
then A t B t
P
P
x (t)
x (t)
   
   
 
  


   
 
   
 
2 2
0 2 2
2 2
2 2
2 2
2 2
0
.
/ 2
det / 2
0
/
det / 2
2 0
n
n n
n
n
n n
n
F m
A
F m
B
 
   
 
 
   
 

 
  
 
 
 
 
 
  
 

 
2x2 equations :Solution is (e g. by Cramer's rule)
[ ]
[ ]

25
   
 
     
 
   
 
     
 
2 2
0 2 2
2 2
2 2
2 2 2 2
0
2 2
2 2
2 2
0
2 2
2 2
0
2 .
/ 2
det / 2
0
/ ( ) / 2
/
det / 2
2 0
2 / / 2
n
n n
n
n n n
n
n n
n
n n n
F m
A
F m
F m
B
F m
x2 equations : Solution is (e g. by Cramer's rule)
 
   
 
     
 
   
 
     

 
  
 
 
 
    
 
 
  
 
 
   
 
   
 
   
2
0
2 2
2
0
2 2
2
/ (1 )
1 2
/ 2
1 2
n
F k
A
F k
B
with

 

 





 


 
 
  
 
  where
 
 
 
 

 
   
     
   
     
 
   
 
 
0 n
P 2
2 2
2
n
n n
ω
2ς
F / k ω
x (t) = sin(ωt - ) tan =
ω
ω ω 1 -
1 - + 2ς ω
ω ω
sin( ) cos( )
in A t B t
 
 
P
x (t)

26
 
 
1
( , )
w h ere
S
x



   
   
   
   
 
   
 
 
 
   
   
   
   
 
   
 
 
 

 
0
P 2 2
2
n n
0
2 2
2
n
n n
n
F / k
x (t) = sin (ω t - )
ω ω
1 - + 2 ς
ω ω
D y n a m ic A m p lifica tio n F a cto r Q u a si - S ta tic D eflectio n
ω
A .F . ς F / k
ω
ω ω
1 - + 2 ς
ω ω
ω
2 ς
ω
ta n
f
=
D e in itio n s :

 
 
 
2
n
ω
1 -
ω
: this means the forced response amplitude is
• ( , )
P S
n
Note
x x AF





Half-power method for 
using steady-state vibration test data
28
/
1
max
max
and
ark a horizontal line with and find its 2 intesection points
2
with the AF curve and their corresponding frequencies and
1. Identify peak/natural frequency peak amplitude
2. M
peak natural
A
A
w
w
Steps :
2
2 1
.
-
3. Then
2 2
peak peak
w
w w w
V
w w
D

;

36
A simple example of application:
Base isolation design for foundation
0
( ) sin( )
ground motion y t y t


( ) absolute displacement w.r.t. a fixed frame
x t 
Free-body diagram:
=
( )
k x y
 ( )
c x y

 
m x


37
FBDgives the equation: ( ) ( )
k x y c x y mx
    
  
0 0 )
Keeping unknown terms on LHS and prescribed term on RHS,
one gets ( ) sin( ) cos( ) .................(*
mx cx kx k y c y t y k t c y t
  
     
  
   
2 2
0 0 sin( ) ( )
ky c y t F t
  
   
0
Recall the forced response amplitude is
• ( , ), /
P S S
x x AF x F k
 
 
       
   
2 2 2 2
0 0 0 0
2 2
0
max 2 2
2
For (**), .
1 /
Thus
1- 2
p
n n
F ky c y y k c
y c k
x
 

 

 
   


   
   
    
   
 
   
 
 
2
2 2
2
0
1 2
1- 2
2
n
p
Max
n n
crit
x
y
noting c k m



 

 
 
 
  
 
 
 
 
   
   
    
   
 
   
 
 

Ground motion transmissibility

Ground motion transmissibility & design implications:
38

Dynamic Analysis by Complex Notation
39
Recall cos( ) sin( )
i t
e t i t

 
 
 
0 0
2
0
= (cos( ) sin( ))
or 2 .... (*)
/
i t
i t
n n
t i t
x x x
m x c x k x F e F
F m e


 
 

  
  
Consider
 
 
Take ( )=X ei t
x t 

Solution : 2
( )=i X e , ( )= X e !
i t i t
x t x t
 
 
   
 
   
 
0
2 2
0
0
2 2
2
(*)
(
)
( 2
) 2
( 2
)
1
/
n
n
n
n
n
n
t i
hen gives
m
whose solutio
i
X
n is X
i
k
F
F
m
F

   
 

 
   
 
 
 

 
 
 
 

 
   
 
 
   
 
  

 



Note:
1
its form is
a ib


40
0
2
2
2 2
2 2
2 2
:
1/
1 2
1 2
1
1 2 1 2
n n
n n
n n n n
Note
k
X
i
i
k
F
 

 
 

 
   
 
   

 
 
   
 
 
 
   
 
 
   
 
 
 
 
 
 
 
 
 
 
 
 
 
 
   
 
 
   
   
    
       
   
   
   
    
       
   
   
        
   
   
   
 
0
F




41
0 0
2
2
0 0
1
( ) ( ) ( )*
(1 ) 2
( ) is called the function.
Inversely,
(1 ) 2 ( ) (
n n
n n
k
X X F
i
where
k i X or
F or
F F
  
 

 

 
  
 
 
 
 
 
 
 
 
 
 
 
   
 
 
     
 
   
 
 
 
 
 
   
 
 
 
   
 
 
   
 
 
 
C
C Dynamic Compliance
K )
( ) is called the / function
for the model.
X
where  Dynamic Impedance Stiffness
K
i t
F e 
 ( ) i t
x t X e 
 
( )

C

42
0 0
0 0
:
( ) ( ) F sin(ωt) which is the of .
then the corresponding response x(t) must be the Imaginary part of X .
( ) ( ) F cos(ωt), which is the of .
then th
i t
i t
i t
a If F t F e
e
b If F t F e





Results
imaginary part
real part
e corresponding response x(t) must be the Real part of X i t
e 
0
= (cos(ωt)+isin(ωt))as to the equation of moti
a forced excitation of (cos(ωt)+i sin(ωt)), with
i.e., now we got the solution
F
i t
X X
e
0
2
2
0
2 2
2 2
2 2
/
1 2
1 2
1 2 1 2
( ) i t i t
n n
n n
n n n n
k
X e e
i
i
k
F
x t
F
 
 

 
 

 
   
 
   
 
 
 
   
 
 
 
   
 
 
   
 
 

 
 
  
 
 
  
 
 
    
 

   
   
    
       
   
   
   
   
       
   
   
       
   
   
   

(cos(ωt)+i sin(ωt))






 
 


43
2
0
2 2
2 2
2 2
,
1 2
( ) ( * ) sin( ) cos( )
1 2 1 2
i.e
n n
i t
n n n n
x t X e t t
k
F

 

 
 
   
 
   
 
 
 
 
 
 
 
 
 
 
 
 
   
  
 
   
   
   
       
 
   
   
   
   
       
 
   
   
       
   
   
 
   
 
0
For F(t) = sin(ωt)
Im
F
., exactly the same result we had before!

Method of Fundamental Solution (Duhamel integral)
for arbitrary forcing for SDOF
44

f()
t

f(t)
Concept: consider the loading as a sequence of many small impulses.
Tool needed: a Unit-impulse response function g(t,) which gives the
displacement of a dynamic system due to an impulse of unit magnitude at
time  with zero initial conditions.
1
0
We can then write
( ) ( , ) ( )
In the limit of & 0,
( ) ( , ) ( )
N
i
t
x t g t f
N
x t g t f d
  

  

  
   
 


f()
t

45
Next, need g(t,):
: ( ) ( )
.
0 for t except 0
, ( ) 0
( ) 1
.
Solution m x c x k x t or m g c g k g t
with zero initial conditions
at t
In the above the function t at t
t dt
is called the
   




       
 
 
 
 
 
  
 
 
 

 
 

deltafunction
   
t
1/


46
(1) 0 ,
For t 
  0
m g c g k g
  
 
with zero initial conditions
(2) , 0... !
For t m g c g k g free vibration problem also

   
 
( , ) 0
g t 
 
.
o let's use the general free-vibration solution of
( , ) sin( ) .
nt
d
But likely non zero conditions at t after the impulse at time
S
g t A e t for t

 
   


 
  

47
 
+
+
(3) Theimpulse effect going from t= to :
: Consider ( )
Idea m g c g k g t dt


 
 


   
  
+
+
+ +
+
:
1st term: ( , )
2nd term ( , ) 0
3rd term ( ) 0
LHS
m g dt m g m g
m g dt m g m g
m g dt m g dt m g
 


 


 
 
 
 
 






 



 
  
   


 
  

+
:
( ) 1
RHS
t dt


 

 


48
LHS:
1
( , ) 1 ( , )
( , ) 0
m g or g
m
g
   
 
 


 

 
( )
:
0
( , ) 1
sin( ( )
n t
d
d
Summary
for t
g t
e t for t
m
 


  

 

 
 
  
 
 
 
( )
0
giving the solution under arbitrary forcing as
1
( ) sin( ( ) ( )
.
n
t
t
d
d
x t e t f d
m
the
 
   

 
  

Duhamel Integral
1
0
d
A
m

 


Practical Numerical Evaluation of
49
( )
0
1
( ) sin( ( ) ( ) for arbitrary ( )
n
t
t
d
d
x t e t f d f t
m
 
   

 
  

( )
0 0
( ) sin( ( ) sin( )cos( ) cos( )sin( )
( )
sin( ) cos( )
( ) cos( ) ( ) sin( ) ( )
n n n
n n
n n
d d d d d
t t
t t
t t
d d
d d
d d
Note
i t t t
ii e e e
e t e t
x t e f d e f d
m m
    
 
   
       
 
       
 
   
 
 
  

   
 
The two integrals can be numerically by trapezoidal rule or Simpson rule for arbitrary ( ).
f t
Remarks:
1. Good for arbitrary loading
2. Dumhamel integral (a convolution integral)
3. A time-domain method (vs frequency-domain methods)

Numerical Step-by-Step Time Integration Methods
 Direct time-domain procedures
 Allow for nonlinear problems (e.g., a change in k, c, or m as a function
of time and deformation)
 Arbitrary loading time function
 Variety of algorithms and propoals
 Commonly used in FEM, FDM, BEM
 Underlying numerical and discretization issues
50
x(t)
f(t)
m
k
c
f(t)
t
t

51
Idea:
f(t)
t
t
x(t)
t
t
t
x
x
x
 
 
 
 
 

 t+ t
x
x
x 
 
 
 
 
 


forcing
t+2 t
x
x
x 
 
 
 
 
 


forcing
( ) .
m x c x k x f t with some prescribed initial conditions
  
 
t t+t

52
Assumption: the acceleration varies linearly in the time interval of t.

Example : Linear acceleration method :
Writing ( ) , ( ) , ( ) ,
the assumption ( ) 0
t t t
t t t
t
x t x x t x x t x
x x
x t x for t
t
  

  

 
      
 

 
   
 
 
0
2
( ) ( )
2
t
t t t
t t
x t x x t d
x x
x x
t

  

 
 
    
 

    

 

  
 
 
0
2 3
( ) ( )
2 6
0 .
t
t t t
t t t
x t x x t d
x x
x x x
t
for t

  
 



 
   
 

     

 
  
 
 
 

t
x

t t
x 

t t
 
t

53
Taking = t, eq (1),(2), (3) become
 
( ) (1)
t t
x x t t
   
 
( ) (2)
2 2 2
t t t t t t t t t t t
t t t
x x x t x x x x x
  
  
       
       
2 2
(3)
3 6
t t t t t t
t t
x x x t x x

 
 
    
  
i.e., the primary unknown is t t
x 

To find it,
( ) ( ) !
t t t t t t t t
m x c x k x f t at t t m x c x k x f t
   
        
   

54
( )
t t t t t t t t
m x f t c x k x
   
  
 
2 2
( ) ( ) ( )
2 2 3 6
t t t t t t t t t t t
t t t t
f t c x x x k x x t x x
  
   
        
     
2 2
,
[ ] ( ) ( ) ( )
2 6 2 3
t t t t t t t t t
Collecting the terms
t t t t
m c k x f t c x x k x x t x
 
   
        
    
2 2
1
leading to
[ ] ( ) ( ) ( ) (*)
2 6 2 3
t t t t t t t t t
t t t t
x m c k f t c x x k x x t x

 
 
   
        
 
 
    


55
      
1 1
With and computable bymeans of (2) and (3) upon obtaining , one can
collectivelydetermine the displacement, velocityand acceleration at according to
w
[ ] [ ]
t t t t
t t t t t t
t
x x x
t t
 
  
 

 
x f B x
A A
 
   
here
upon defining .
[ ], t t
t
t
x
x
x
and 
 
 
 
 
 
x
B f
A


2
This is a step-by-step algorithm by which
.... !
t t t t t t n t
x x x x
x x x x
x x x x
    
       
       
  
       
       
       
   
   

56
Note:
1) Concept of stability, accuracy, efficiency
a) Conditionally and unconditionally algorithms: a critical time step tcrit
if conditionally stable.
b) Accuracy: Level of numerical amplitude decay/damping & phase shifts
due to accumulation of errors.
c) Explicit and Implicit schemes.
d) Short duration versus long-duration problems.
e) Applicable to MDOF systems (see posted reading)
a) Other methods e.g., Central difference, Newmark method in many
commercial FEM codes (see posted reading)

5
Response Spectrum (a spectral method)
The maximum response of a Single degree of freedom system with
general to a specific excitation (e.g., earthquake ground
motion) :
( , )
n
 
 
2
2 ( )
g
n n g
c x k x m x x
or
x x x x t
 
   
   
  
  
=
-kx-cx m (x+ x )
g
  
( )
g
x t

/
g
x g


6
( )
0
( )
2 0
Analytically, thedisplacement response is expressible in the Duhamel integralform as
1
( ) sin( ( ) ( )
1
sin( ( ) ( ) ..... (*)
1
n
n
t
t
d
d
t
t
d g
n
x t e t f d
m
e t x d
 
 
   

   
 
 
 
  
   


 
max v
pv
( ) ( , ) ....
( , ) ....
d
d
x t S
S
 
 
 Spectral velocity
Spectralpsuedo velocity


 
( )
0
1
( ) ( ) sin( ( ) ( )
n
t
t
d g
d
d
b x t e t x d
dt
 
   

 
   

 
 
( )
pv max
0
( , ) sin( ( ) ( ) is
n
t
t
n d g
S e t x d called
 
     
 
  
 SpectralPsuedo- Velocity

Leibniz rule :
 
0 0
0
( , )
( , ) ( ) ( , ) ( ) ( )
( , )
( )
t t
g g g
t
g
d dt g t
g t x d g t t x t x d
dt dt t
g t
x d
t

    

 

  




 

  

  a
max
(c)
( ) ( ) ( , ) ...
g d
x t x t S  
  Spectralacceleration
 
max
pv
( ) ( ) ( , ) ....
1
( , )
d d
n
n
a x t S
S where
 
 


Maximum response quantities :
Spectral Displacement


7
   
: eqn of motion g
gives m x x c x k x
   
Note   
   max
max
1
g
implies x x c x k x
m
   
  
  2
max
g n d n
k
x x S
m
 
    
d pv
S S
 
Summary of Approximate relations of
Spectral displacement, velocity and acceleration to
Spectral pseudo-velocity
d pv
n
V pv
a n pv
1
S = S
ω
S = S
S = ω S
2
n
n
with
T

 

8
2
2
n
n
T
T




d pv pv
n
V pv
a n pv pv
1
S = S S
ω
S = S
S = ω S S
log log log log(2 )
log log log log(2 )
n
n
T
T


   
   
d pv
a pv
S S
S S
y = m x +b
Line of constant Sd
Line of constant Sa

9
Tripartile plot of response spectrum

11
Frequency domain analysis
0 0 0
1 0 0
Recall for periodic functions of period T:
2 2
f(t)= sin( ) cos( ), ( )
n n
n n
a n t b n t or T
T
 
  

 
 
  
 
Fourier series
0
0
0
0
0
0
0
0
where
f(t) sin( )
f(t) cos( )
n
n
a n t dt
b n t dt




















0
0 0
0
/
0
/
A much more compact form is
f(t)= = f(t)
2
in t in t
n n
n
F e with F e dt
 
 
 






 
 
0
0 0
0
/
0
/
or
f(t)= = f(t)
2
in t in t
n n
n
F e with F e dt
 
 
 






 
 
t
f(t)

12
0

0
2
0
 0
3
2 2
n n
a b

Can determine the phase angle of magnitude of
each sinusoidal component of f(t)
ie f(t) can be looked upon as a sum of sin( ) !
n n n
A n t
 

Can determine the amplitude/magnitude of each
sinusoidal component of f(t) as 2 2
n n
a b


13
( )= f(t) i t
F e dt







(e.g. an earthquake ground motion or response)
Next : How to treating as non -perioidc function?
0 0 0 0
2
( ) , ( ) / , ( ) , ( ) ( )
n
a d b c n d F F
T

      
      
 
1
f(t)= ( )
2
i t
F e d

 



 
0
1 1
then (*) becomes f(t) ( ) (2)
2 2
in t i t
n
n
F e d F e d
 
  
 
 



 
 

The Fourier transform of f(t) Inverse Fourier transform of ( )
F 

0
0 0
0
/
0 /
Note that
1
f(t)= = f(t) (*)
2
in t in t
n n
n
F e with F e dt
 
 
 






 
 
In the limit of T  
leading to

15
Remarks:
 is generally a complex-valued function of
frequency, i.e., with a real and an imaginary part.
 is the Fourier amplitude spectrum (real value)
 Exposes the frequency content of the loading or
response
 It is a full characterization of f(t) in the frequency
domin
 i.e.,
 FFT algorithms (Fast Fourier Tranform)
( ) gives ( ) and ( ) gives back ( ) !
f t F F f t completely
 
 
( )
F 

( )
F 


13
( )= f(t) i t
F e dt







(e.g. an earthquake ground motion or response)
Next : How to treating as non -perioidc function?
0 0 0 0
2
( ) , ( ) / , ( ) , ( ) ( )
n
a d b c n d F F
T

      
      
 
1
f(t)= ( )
2
i t
F e d

 



 
0
1 1
then (*) becomes f(t) ( ) (2)
2 2
in t i t
n
n
F e d F e d
 
  
 
 



 
 

The Fourier transform of f(t) Inverse Fourier transform of ( )
F 

0
0 0
0
/
0 /
Note that
1
f(t)= = f(t) (*)
2
in t in t
n n
n
F e with F e dt
 
 
 






 
 
In the limit of T  
leading to
A Fourier transform pair

15
Remarks:
 is generally a complex-valued function of
frequency, i.e., with both real and imaginary parts.
 is the Fourier amplitude spectrum (real value)
 Exposes the frequency content of the loading or
response
 It is a full characterization of f(t) in the frequency
domin
 i.e.,
 FFT algorithms (Fast Fourier Tranform)
( ) gives ( ) and ( ) gives back ( ) !
f t F F f t completely
 
 
( )
F 

( )
F 


16
Basic operational properties:
Fourier transformof derivatives:
[ ( )] ( ) ( )= ( ) w
w
¥
-
-¥
Á = = ò
  i t
recall f t f t F f t e dt
( )
( )
2
( ) [ ( )]= ( ).
[ ( )]
( )
w
w
w w
¥
-
-¥
Á
= Á
=
ò
 


i t
d
b f t f t e dt
dt
i f t
i F
F
( )
( )
( ) [ ( )]= ( )!
w w
Á 
n
n
c f t i F
( )
In integration by part, ( ) ( )
( ) ( )
( ) ( ) 0@
w
w
w w
w
w w
-
¥
-
-¥
¥
¥
- -
-¥ -¥
= -
= +
=  ¥
ò
ò

i t
i t
i t i t
de
f t e f t dt
dt
f t e i f t e dt
i F if f t
( )
) [ ( )]= ( ) ( )= ( ) w w
w
¥ ¥
- -
-¥ -¥
Á = =
ò ò

  
 i t i t
df t
a Consider f t f t F f t e dt e dt
dt

17
:
( )
Solution of SDOF equation of motion under arbitrary forcing
m x c x k x f t
  
Applications :
 
 
To apply Fourier tranform method, consider
( ) i t
m x c x k x f t e dt

  
 
2
( ) ( )
( )
m i c k X F
 
 
 
    
 
 
2
( ) ( ) ( ) ( )
( )
( ) ( )
( ) ( ) .
i t
with
i m X i c X k X F
X x t e dt as the Fourier tranform of x t etc

   

 



   

   

 
( ) i t
m x c x k x f t e dt



  
  

18
2
( ) ( )
1
X F
m i c k
 
 

 
  
  
 
 
or
2
2 ( ) / ................ (*)
n n
x x x f t m
 
  
 
 
2 2
( ) ( )
(*) ( ) 2 /
n n
then gives i X F m
 
   
   
 
 
2 2
2
( )
( )
( )
( ) 2
1
(1 ) 2
n n
n n
m
X
i
k
i
F
F



   
 

 
 
 
 
 
 
 
 
 
 
 
 
  
 
 
   
 
 
     
 
   
 
 
 




19
2
: :
. : 2 ( )
n n g
Example Earthquake ground motion problem
Eq of motion x x x x t
 
   
  
 
 
2 2
/
( ) ( )
( ) ( )
1
( ) 2
n n
x x
X x
i
x
 
 
   
 
 
  
 
 
 
 H 
 



Concept of a Transfer function
Excitation Response
Transfer function
H()

19
2
: Earthquake ground motion problem:
. : 2 ( )
n n g
Recall Eq of motion x x x x t
Example
-
 
  
  
 
 
2 2
/
( ) ( )
( ) ( )
1
( ) 2
g
n n
x x g
X x
i
x
H
 
 
   
 
 
  
 
 
 
 
 



Concept of a Transfer function
Excitation Response
Transfer function
H() of dynamical
system

20

( )
g
x w



( )
X w


( )
w
H

21
Remarks:
 Theoretical modeling characterization of dynamic and
vibratory systems in frequency domain
 Other transfer functions of engineering
interest
/ v/ /
( ), ( ), ( )
. ., g g g
d a a a a
e g H H H
  
 Experimental determination of transfer functions
of physical systems

22

( )
w
H
In more complex systems:

23
2
: (2 )
g n n
a x x x x
Note  
    
  
2
(2 )
n n
a x x
 
   

  
:
: !
g g
g
a x x a x x
Note a x x
Note     
 
 
    
 
  
 
2 2
( ) )
1
( ( )
( ) 2
n n
With X
i
x
 
   
 
 
 
 
 
 
 

 
2
2 2
( )
1
( ) 2
n n
a x
i


   
 

 
   
 
 
 

 
 
 
2
2 2
( )
2
( ) 2
n n
n n
i
a x
i

  
   
 
 

 

 
 
 

 
Re
Im
Z1+ Z2
Z2
Z1

Method of Energy Balance
29
 
2 2
1 1
( )
2 2
Kinetic energy m x Potential elastic Energy k x
For a 1DOF :
 
&
x(t)
f(t) A
B
0
( ) sin( ) ( ) sin (
F t F t x t A t for
+ ) undamped free vibration
  
  
 
 
2
2
2
2
Note that KE+PE=
At time A, KE+PE=0 ,
1
2
1
At time B, KE+PE=
2
0
2
1 1
2
n
n n
k A
m A
k
k A m A
m
Conservation of energy
w
w w


 


constant for undamped case

Method of Energy Balance
30
2 / 2 / 2 /
0
Idea: Compare Input nergy & energy dissipated per cycle
( ) sin( ), ( ) sin( ), ( ) cos( )
( ) ( ) sin( )
T T T
input o o o
E
F t F t x t A t x t A t
dx
W F t dx F t dt F t A
dt
p w p w p w
w w f w w f
w w
D
  

    
   
  
Damped 1DOF under Forced Steady State Vibration :
&
 
2 /
0
2 /
2
0 0
cos( )
sin( ) cos( )cos( ) sin( )sin( ))
sin( ) sin ( ) sin( )
T
o
T
o
t dt
F A t t t dt
F A t dt F A
p w
p w
w f
w w w f w f
f w p f



  
 


x(t
) f(t)
 
2 / 2 /
2
2 /
2
( ) ( )
cos( ))
.....
T T
dissip damping
o o
T
o
dx dx
W F t dx t c dt
dt dt
c A t dt
cA
p w p w
p w
w w f
wp
D
 

 
  
 
 
 


 


31
0
2
0
os( )
sin( )
................ (*)
: input dissip
input
dissip
W W
W F Ac
For steady state must ha
W c A
F
A
c
ve
p f
pw
f
w
D D
D
D



 
0
2
max 2
2
/
C a (*) is ident
: ical t
-
o
l
2
im
1
p
n n
F k
x
w w
V
w w

   
   
    
   
 
   
 
 




Hysteretic cyclic stress-strain behavior
in many real materials
32
Area  energy dissipation



Linear viscous model vs physical cyclic stress-
strain behavior :
33
Hysteretic loop’s area’s strong frequency dependency in
linear viscous damper not seen in real materials
x
F

Equivalent Viscous Damping
34
exp
2
exp
2
( ' )
................ (*)
erim
dissip
erim
W c A W hyster loop s area
W
and define
ak
A
T e
equiv
Idea
c
:
pw
pw
D D
D
 

2 2
exp
2
2
(2 )
(2 )
2
equiv equiv
erim crit
crit crit
equiv n
n
equiv
n
c c
W
c
Ta A c A km
c
A km
A
ke
k
Minor - arrangement :
pw pw
w
z p w
w
w
p z
w
D  

 
  
 
exp
2
!
1
4
2
erim
equiv
n
W
kA
z
w
p
w
D
 
   
   
 
 
2
exp
1
4
2
erim equiv
n
W kA
w
pz
w
D
   
    
 
 

• If experimental measurement is done at
35
exp exp
2
!
1 4
4
2
erim erim
equiv
W W
Triangular Area
kA
z
p
p
D D
 
 
 
 
n
 



A

Shear Modulus and Damping:
Stress‐strain response is hysteretic and strain level dependent
Basic engineering parameters:
‐secant modulus G
‐equivalent viscous damping 
both of which are dependent on level of confinement/stress state
2
1
4
2
loop
equiv
G
Area
G



 


 
 
 

• Soil is a particulate/porous
medium
• Under a water table, void
space in the ‘soil skeleton, is
filled by water with pore
water pressure u
SOLID GRAINS
WATER

Importance of effective stresses:
They control:
(a) Shear strength/failure condition in soil
(b) Stiffness/deformability/settlement of foundations
v v
vertical soil effective stress
horizontal soil effective stress h h
u
u
 
 
  
  
v
h



Concept of Effective Stress in Soil
u

In 3D:
Note: Sign convention in soil mechanics:
Compressive stress is positive

Field Measurement Methods:
(I) For low  level (<10‐6 to 10‐4 ) ….. Gmax
(II) Geophysical/wave‐based methods (downhole, cross hole seismic
measurements)
(III) Impulse or low‐level vibration methods
(IV) Shear wave speed: Compressional P‐wave speed:
2 1
1 2
p
G
V

 

 
  

 
s
G
V



Typical range of Vs:
200’/s to 1500’/s for unconsolidated soil.
2000’/s for stiff soils
3000’/s for sandstone
>6000’/s for rocks
E.g., =100 lb/s, Vs=300/s:
Gmax =(100/32.2)(300)^2 = 0.28x106 lb/sqft(psf)
Gmax =(100/32.2)(1500)^2 = 7x106 lb/sqft(psf)
1psf=47.85Pa
1 psi=6.891 kPa

DIFFERENT LABORATORY TESTS FOR DIFFERENT STRAIN LEVELS

Laboratory methods:
Resonant column tests:
‐excited torsionally (or axially) at the top or base
into resonant modes
‐transducers used to measure motion’s time
histories
‐sinusoidal or random excitation
‐use of solid or hollow samples
‐ Equiv viscous damping ratio from hysteretic loop
(r)

Lab tests for higher strain level:
(a) Cyclic simple shear test
Can be stress‐control or strain‐controlled
Drained or undrained (pore pressure development)
Simple shear (stack rings) vs direct shear test
cyclic
h0

cyclic

v0
(v0 ,‐cyclic)
(h0 ,cyclic)

(b) Cyclic triaxial test:
Repetitive axial compression/extension ‐ different stress path
stress control or strain controlled
Drained or undrained (pore pressure development)
0

0+d‐cyclic

d‐cyclic
(0 ,0)
, ,
2(1 )
E
E G G
   

  

~ 0.15 to 0.35 (sand)
0.5 (saturated clay)



(c) Bender element test
Piezoelectric sensors (transmitter and receiver)
Measure average wave speed between 2 points in specimens
s measured
L G
V
t 

 

+ ‐

(d) Shake Table Test
others….

Experimental Synthesis and empirical relationships:
 Gives trends with respect to basic factors
 Different functional forms proposed
 Useful for preliminary assessments

Primary factors affecting shear modulus & damping ratio
Strain amplitude
Mean effective stress
Void ratio
Relative density Dr (sand)
Secondary factors
# of cycles, OCR,  & c, grain characteristics, soil fabric &
structure, frequency, degree of saturation, cementation

Empirical relationships on shear modulus:
 Useful for preliminary assessments
 Gives trends with respect to basic factors
 
 
1/2
2
max 1/2
2
. ., (1972)
(2.973 )
1230 ... angular grains
1 1
( ) ( ~ )
(2.17 )
2630 ... round grains
1 1
0 0
20 .18
40 .30
60 .41
k mean
k mean
e g
e
OCR
e psi
G psi
e
OCR
e psi
PI k
where
-4
Hardin & Drnevich
10



 

 

 
 

 
 
  

 
 

 
 
  
 







 
 
 
 


units of G and p = kg/cm2
Iwasaki et al. (1978)

1/ 2
2
4
2
0..44
2
2 5
max 2
0..4
2
6
2
(1978)
(2.17 )
700 , @10
1 1 /
(2.17 )
( / ) 850 , @10
1 1 /
(2.17 )
900 , @10
1 1 /
mean
mean
mean
e
e kg cm
e
G kg cm for sand
e kg cm
e
e kg cm









 

 

 
 

 
 
 

 

 
  
 
  
 
 

 

 
 
 
  
 
Iwasaki et al

Seed and Idriss (1970)
Separate sand and clay
     
 
1/ 2
max 2 2
max max
2
max
( ) 1000 , 15 0.6 ,
1
0.6 , 2 ,....
( / ) ( , )
mean
u
G psf K K Dr F
psf
F for silt F for gravel
G
kg cm f OCR PI
S
Sand :
Clay :
 
 
  
 
 
 

𝐾

Strain dependence of cyclic shear modulus
• Basic observations: (a) γ
τ
γ
m
m c
g
g
t
= +
max
( ) ,
i As g t t
 ¥ 
m c
g
t
g
 =
+
max
1
m
t
 =

max
0
( ) 0,
d
i As G
d g
t
g
g =
 
( )
2
1
d m
d m c m c
t g
g g g
æ ö
æ ö ÷
ç
÷
ç ÷
ç
÷
 = - ÷
ç ç
÷ ÷
ç ÷
ç ç
+ ÷
è ø ç +
è ø
max
0
1
d
G
d c
g
t
g =
 = =
1 1
1
( ) ( )
max max
max max max
G G
G G G
g
g
g t g t
= +  = +
Summary :
( ) 1 1
1
1
max
ref
max
max max
ref
max
G
where
G
G G
t
g
g
g g
g
t
 = = =
+
+



(v ,ult)
(K0v ,‐ult)
cyclic
 c’
max/ult
v
K0v


Basic observation: (b)
1
(i) reverse slope Gmax
(ii) half area of loop constant% * triangle a_b_c’s area
triangle abc’s area
k
»
»
= 
max
1
triangle abc’s area= (ad)(cd-cb)
2
where ad =2
2 /
2 /
cd G
bd G
t
t
t
⋅
=
=
2
1 1
max max
1 1 1 1
area of hyeteretic loop= (2 )2 = 4
k k
G G G G
t t t
æ ö æ ö
÷ ÷
ç ç
÷ ÷
⋅ - -
ç ç
÷ ÷
ç ç
÷ ÷
ç ç
è ø è ø
2
1
1
max
2
max
max
max
1 1
4
2 ( )
= 1
1
4 (
2
( )
1
k
k
G G G
Equivalent damping ratio
G
G
G
G
t
g
t p
p
g
z z
æ ö
÷
ç ÷
-
ç ÷
ç æ ö
÷
ç
è ø ÷
ç ÷
= -
ç ÷
ç ÷
ç
è ø
æ ö
÷
ç ÷
 = -
ç ÷
ç ÷
ç
è ø

g
1
( )
0
max
G
G
g
1
0
( )
max
z g
z

Concept of Strain‐compatible shear modulus and damping
To recognize the significant strain‐dependence of the shear modulus and damping ratio,
one generally needs to use a pair of G and  in the dynamic analysis of the model so that
their values are consistent with the strain/deformation level of the response.
Approach:
1. Assume an initial estimate of strain, say, 0 & obtain G(0)
and  (0).
2. Perform dynamic analysis to find the predicted strain level,
say, 1.
3. Check if 1 ‐ 0 is close enough.
4. If not, replace 0 by 1 and repeat step 1 to 4
until difference of the different n+1 and n is
acceptable
5. Then G( n+1) and  (n+1) are called the
Strain‐compatible shear modulus and damping ratio.
* The iterative approach is built in the program SHAKE 91,
2000 for 1‐D equivalent linear ground response analysis
1
( )
0
max
G
G
g
1
0
( )
max
z g
z
0

Multi‐Degree of Freedom Systems
 Soil layering and profile variation
 Improve accuracy and lower degree of
approximation
 Higher‐order wave motion at high
frequencies

Multi-D-O-F system  Continuum system
2

m1
f1(t)
x2(t)
f2(t)
H/2
H/2
, G
, 2G
FBD for Equations of motion for undamped case:
x1(t)
k1
k2
1 1
m x

f1 = 1 1 1 1 2 1
( )
m x k x f
x
 + - =

1 1 2
( )
k x x
-
2 2 1 1 2 2 2 2
2 2 1 1 1 2 2 2
( )
( )
m x k x x k x
or m x k x k x
f
f
k
 + =
+
- -
- + =


f2
2 2
k x 2 2
m x

=

FBD for Equations of motion for undamped case:
1 1 1 1 2 1 (1)
( )
m x k f
x x
+ - =

2 2 1 1 1 2 2 2
( ) (2)
m x k x k k x f
- =
+
+

{ } { } { }
In matrix form  [M] x + f
=
[K] x

{ } { }
1 1
2 2
For the 2-DOF model of the soil layer,
0
....
0
....
3
, ....
m
Mass matrix
m
k k
Stiffness matrix
k k
x f
forcing vector
x f
é ù
ê ú
=
ê ú
ë û
é ù
-
ê ú
=
ê ú
-
ë û
ì ü ì ü
ï ï ï ï
ï ï ï ï
= =
í ý í ý
ï ï ï ï
ï ï ï ï
î þ î þ
[M]
[K]
x f
Note:
M and K are symmetric

{ } { } { } { }
For a damped system, the equation of motion is
+
[M] x [C] x +[K] x f
=
 
{ } { } { }
Case 1: Free vibration of undamped MDOF:
+
[M] x [ 0
=
K] x

{ } { }
{ } { }
) (0)
) (0)
subject to
a
b
=
=
Initial Conditions :
x x
x x
 
Note: Same matrix equation of motion from finite element
methods

{ } { } { } { }
For a damped system, the equation of motion is
+
[M] x [C] x +[K] x f
=
 
{ } { } { }
Case 1: Free vibration of undamped MDOF:
+
[M] x [ 0
=
K] x

{ } { }
{ } { }
) (0)
) (0)
subject to
a
b
=
=
Initial Conditions :
x x
x x
 
Note: Same matrix equation of motion from finite element
methods
{ } { }
As in SDOF:
!
i t
e w
I
=
dea :
x η

{ } { } { } [ ]{ } { }
[ ]
{ }
[ ]
2
2
where the matrix
For non-trivial solution of , must have
det 0.
or
one
w
w
 - + =
é ù
= - +
ê ú
ë û
=
)
=
[M] η [K] η 0 H(ω η 0
H(ω) [M] [K]
η
H(ω)
[ ]
[ ]
2
2
2
2 2 2
0
For the 2-dof example,
0 3 3
with the determinant det ( )( 3 )
m k k m k k
m k k k m k
m k m k k
w
w
w
w w
é ù
é ù é ù
- - + -
ê ú
ê ú ê ú
=- + =
ê ú
ê ú ê ú
- - - +
ë û ë û ë û
= - + - + -
H(ω)
H(ω)

[ ]
( )
( )
2
4 2 2
2
1
2
2
Settingdet 0 4 2 0, a quadratic equation in .
: (2 2) 0.585
(2 2) 3.414
Theyare the of t
k k
m m
k k
Roots
m m
k k
m m
w w w
w
w
æ ö æ ö
÷ ÷
ç ç
=  - + =
÷ ÷
ç ç
÷ ÷
ç ç
è ø è ø
æ ö æ ö
÷ ÷
ç ç
= - =
÷ ÷
ç ç
÷ ÷
ç ç
è ø è ø
æ ö æ ö
÷ ÷
ç ç
= + =
÷ ÷
ç ç
÷ ÷
ç ç
è ø è ø
H(ω)
naturalfrequencies
1
he MDOFsystem.
Put in ascending order, the lowest frequency, i.e., is called the Fundamental natural frequency.
w

Modal shapes associated with each natural frequency
{ } { } { }
( )
( )
{ } ( )
2 1
1 1
1
2
: ... (*) will allow non-unique solutions
2 , (*)
1 2 1 1 2
0
normalize it in
0
1 1 2 1
For a
For the dof system can be reduced to
or
w w
h
h
é ù
- +
ê ú
ë û
é ù ì ü
- + - ï ï
ì ü ì ü +
ê úï ï ï ï ï ï
ï ï ï ï ï ï
ê ú =  =
í ý í ý í ý
ï ï ï ï ï ï
ê ú ï ï
- + î þ
ï ï ï ï
î þ ï ï
ê ú î þ
ë û
1
1
1
[M] [K] η = 0 η
η
{ } { } { }
( )
( )
{ } ( )
2 2
2 2
1
2
different ways
: will allow non-unique solution
1 2 1 1 2
0
0
1 1 2 1
For a
w w
h
h
é ù
- +
ê ú
ë û
é ù ì ü
- + - ï ï
ì ü ì ü -
ê úï ï ï ï ï ï
ï ï ï ï ï ï
ê ú
 =  =
í ý í ý í ý
ï ï ï ï ï ï
ê ú ï ï
- + î þ
ï ï ï ï
î þ ï ï
ê ú î þ
ë û
1
2
1
[M] [K] =
η 0 η
η

{ } { } { }
{ } { } { }
{ } { } { } { } { } { }
{ } { } { } { } { } { }
2
2
2
2
..... (1)
..... (2)
0 ..... (1)
..... (2)
But since the matirx
i i
i
j j
j
T T T
j i j i j
i
T T T
i j i j i
j
and
is also true
w
w
w
w
- + =
- + =
- + = =
- + =
Orthogonality of Mode Shapes
Note : [M] η [K] η 0
[M] η [K] η 0
Observation :
η [M] η η [K] η η 0
η [M] η η [K] η η 0
[K]
{ } { } { } { } { } { }
( ){ } { }
{ } { }
{ } { }
2 2
!
( , too) is symmetic, it is always true that
.
,
(1) (2)
0
0
, !
0
T
j i
i j
T
j
j i
i j
T i
j i
f
and
Ta
i
k
f
i g
i
n w w
w w
w w
-  =
= ¹
=
-
 ¹

T T
T T
[M]
a [K] b = b [K] a a [M] b = b [M]
η
η [M] η
η [K]
a
[M] η
η
{ } { }
We say that the mode shape
with respect the the mass matrix and the stiffness matrix .
i j
is orthogon
]
al
η η
[M
t
[K]
o

Usefulness:
If we put the mode shape vectors in a matrix form
{ } { } { }
{ } { }
{ } { }
1 2
... N
T
T
i i
ii
T
T
i i
ii
then must be
with the diagonal element
must be
with the diagonal element
é ù
= ê ú
ë û
 =
L =
=
Q =
[T] η η η
[T] [K][T] [Λ] a diagonal matrix
η [K] η
[T] [M][T] [Θ] a diagonal matrix
η [M] η
. . diagonalizes both the !
i e and matrices
[T] [K] [M]

0
&
0 3
1 2 1
1 2 1 2
,
1 1 1 2 1
m k k
m k k
é ù é ù
-
ê ú ê ú
= =
ê ú ê ú
-
ë û ë û
é ù
é ù +
+ - ê ú
ê ú
= = ê ú
ê ú
-
ê ú
ë û ë û
T
E.g. For the 2- DOF example of the 2- layered soil,
[M] [K]
[T] [T]
( )
( )
1 2 1 1 0 1 2 1 2
0 1 1 1
1 2 1
T
m
é ù
é ù ê ú
é ù
é ù
+ + -
ê ú ê ú
ê ú
ê ú
= = =
ê ú ê ú
ê ú
ê ú
-
ê ú ë û ê ú
ë û
ë û ê ú
ë û
2
2
1+ 2 +1 0
[T] [M][T] m [Θ]
0 1- 2 +1
1 2 1 1 1 1 2 1 2
[ [ ]
1 3 1 1
1 2 1
T
k
é ù é ù
é ù é ù
+ - + -
ê ú ê ú
ê ú ê ú
= = =
ê ú ê ú
ê ú ê ú
-
-
ê ú ë û ë û
ë û
ë û
4 0
T] [K][T] k Λ
0 4

Modal Analysis of MDOF:
Observation and Ideas:
(i) If the mode shapes are orthogonal to each other. If so, they must be
independent of each other!
(ii) That means any x can be decomposed into a linear combination of
the mode shapes , 1, 2 ...,
i
i N
=
η
{ } { } { } { }
{ }
1 2
1 2
1
2
1
. . ( ) ( ) ( ) ... ( )
( )
..
N
N
N
i
i
i
N
i e t q t q t q t
q
q
q t
q
=
= + + +
ì ü
ï ï
ï ï
ï ï
ï ï
ï ï
ï ï
é ù
= = í ý
ê ú
ë ûï ï
ï ï
ï ï
ï ï
ï ï
ï ï
î þ
å 1 2 N
x η η η
η η η ... η
{ } { }
( ) ( )
t t
=
i.e. let'swrite x [T] q
The s are called the 'Principal' or 'Normal' coordinates of the dynamical system
i
q

With , the equation of motion
{ } { }
( ) ( )
t t
=
x [T] q
{ } { } { } for free vibration of undamped MDOF becomes
+
[M] x [K] x = 0

( )
t
 +
[M][T]q [K] =
[T]q 0

( ) ... (*)
t +
T T
[T] [M][T]q [T] [K] =
[T]q 0

( )
( ) ( )
( )
( )
( ) ( )
( )
1 1 1 1
2 2 2 2
4
4 0 0
4
4 0 0
k
m q k q q q
m
k
m q k q q q
m
+ =  + =
- + =  + =
-
2
2
2
2
1+ 2 +1
1+ 2 +1
1 2 +1
1 2 +1
 
 
1 1
1 1
0.5858 0
3.414 0
k
q q
m
k
q q
m
æ ö
÷
ç
 + =
÷
ç ÷
ç
è ø
æ ö
÷
ç
 + =
÷
ç ÷
ç
è ø


( ) 0!
t + =
 [Θ] q [Λ]q

[ ]
T
Now multiply it by T
( )
( )
& [ [ ]!
T T
é ù
ê ú é ù
ê ú ê ú
= = = =
ê ú ê ú
ë û
ê ú
ê ú
ë û
2
2
1+ 2 +1 0 4 0
Recall [T] [M][T] m [Θ] T] [K][T] k Λ
0 4
0 1- 2 +1
2
1 1 1 1
2
2 2 2 2
0, . . ( )
0, . . ( )
. ., ... (*)
q q i e a SDOF for
i
q t
q q i e a SDOF for q
e gives
t
w
w
+ =
+ =



2
1 1 1
2
2 2 2
( ) ( ) 0
( ) ( ) 0
q t q t
q t q t
of
w
w
+ =
+ =
Solution


i i
A and B
To find the constants , need Initial Conditions :
{ } { } { } { }
{ } { }
( ) ( ) (0) (0)
(0) (0)
t t
=  =
=
Note x [T] q x [T] q
x [T] q


{ } { }
{ } { }
1
1
(0) (0)
(0) (0) !
-
-
=
=
q [T] x
q [T] x
 

1 1 1 1 1
2 2 2 2 2
( ) cos( ) sin( )
( ) cos( ) sin( )
q t A t B t
q t A t B t
w w
w w
= +
= +


Forced Vibration of Undamped MDOF
{ } { } { }
: (t)
Equation of motion [M] x + [K] x f
=

{ } { }
( ) ( )
Take t t
=
Idea : x [T] q
( )
t n t
he +
[M][T]q [ [ =
K T]q f
]

( )
t
T T T
then [T] [M][T]q + [T] [K][T]q [T
= f
]

( )
t
T
then [Θ]q + [Λ]q [T]
= f

1 1
( )
t
- - T
then q + [Θ] [Λ]q [Θ] [T]
= f

1 2
1
is diagonal with diagonal eleme t
(
n s !
is called th
)
e
i
where
t
w
-
-

=
= T
q + [Ω]q [P]
[Ω] [Θ] [Λ]
[P] [Θ] [T] Participation factor matrix
= f

and each is called a
ij
P Participation factor

Meaning of Pij :
Pij gives an indication of how the forcing at the j th mass or node
contribute to the vibration of the ith mode.
1
th
2
e.g., For the i mode's coefficient ,
( ) ( ( )
)
N
i
i i i i j i
j
j
q
q t q P f t F t
w
=
+ =
= å

1 1 2 2 3 3
( ( ( . (
) ) ) )
..
i i i iN N
P f t P f t P f t P f t
= + + + +
{ } { } { } { }
{ }
{ } { }
1 2
1 2
1
1
2
2
With ( ) ( ) ( ) ... ( )
one should note the entire product ( ) represents the 1st mode' contribution,
( ) represents the 2nd mode' contribution, etc to ( )
because the
N
N
t q t q t q t
that q t
q t t
= + + +
ATTENTION :
x η η η
η
η x
{ }
size of depends the chosen normalization of .
i
i
q η

Damped MDOF system under Forced Vibration
{ } { } { } { }
The equation of motion is
+
[M] x [C] x +[K] x f
=
 
:
Solution by modal analysis
{ }
{ } { }
1 2 2
1. Find ... from solving the eigenvalue poblem of [ ]
2. Take ( ) ( )
3. Get [ ... (*)
[
n
t t
w
é ù
= - +
ê ú
ë û
=
+ +
 + +
T T T T
T T
[T] η η η M K η 0
x [T] q
[T] [M][T]q T] [C][T]q [T] [K][T]q [T] f
[Θ]q T] [C][T]q [Λ]q [T] f
=
=
=


1 1 1
[
- - -
 + +
T T
q [Θ] T] [C][T]q [Θ] [Λ] f
=
q [Θ] [T]


[ ] [ ]
[ ] [ ] [ ]
1
1
0
( ) ..... (*)
( ) ( ) ..... (**)
N
j
j
a
b
a b
-
-
=
= +
= å
Particulardamping forms that work :
[C] M K Rayleigh Damping
[C] M M K Caughey damping
1 1
1
4.
2
Will be diagonal generally? If Yes, can write it as 2
and treat the MDOF as N number of SDOF systems, i.e.,
[ i i
etc
z w
z w
-
é ù
ê ú
ê ú
ê ú
ê ú
ë û
T
T] [C][T]
Key Question :
[Θ]
But the Answer is NO generally.
2
1
2 , 1,2,...,
N
i i i i i i ij j
j
q q q P f i N
z w w
=
+ + = =
å
 

[ ] [ ]
, a b
= +
For Rayleigh Damping [C] M K
[ ]
[ ]
1 1
[ [ [
[ [
then
a
a a
a a
- -
- =
 = =
= =
T T
T
Case 1: Mass proportionaldamping [C] M
T] [C][T] T] M [T] Θ]
[Θ] T] [C][T] [Θ] Θ] [I]
[ ]
[ ]
1 1
[ [ [
[ [ ....(**)
then
b
b b
b b
- -
- =
 = =
= =
T T
T
Case 2 : Stiffness proportionaldamping[C] K
T] [C][T] T] K [T] Λ]
[Θ] T] [C][T] [Θ] Λ] [Ω]
Equating to the diagonal matrix of 2 ,
2
2
i i
i i i
i
a z w
a
a z w z
w
 =  =
[I]
2
Equating to the diagonal matrix of 2 ,
one gets 2
2
i i
i
i i i i
b z w
bw
b w z w z
=  =
[Ω]

[ ]
[ ]
2 2
i
i
i
a b
bw
a
z
w
= +
= +
Rayleigh damping Case with [C] M K gives
Modal damping ratio
Note the difference in frequency dependency

[ ]
[ ]
[ ]
[ ] 1
1 2 ...
)
is the
(
N
i i i ij j
j
N
o
wher
r
q q q P
e
f
=
ààà
+ + =
=
å
-1 1
T T
i i i i
T T
i i i i
T - T
[P] = [Θ] [T] [Θ] [ ] Partcipation Factor matrix
η C η η K η
η M η η M η
η η η
 
[ ]
[ ]
[ ]
[ ] [ ]
( )
)
1,2,.., (
i i i
q q q i N
e à
+ àà
+ = =
T T T
i i i i i
T T T
i i i i i i
η C η η K η η f
η M η
N
η M η η M
ot that can also be written as
η
 
1 1 1
[ ( )
- - -
+ + à
T T
q [Θ] T] [C][T]q [Θ] [Λ]q [Θ] [T]
= f
 
Modal form of governing equations for MDOF:

23
Importance of Modal Decomposition in dynamic analysis
• One can choose/select only modes that are important to analyze (e.g., the lower modes in earthquake
problems)
• Can use larger time steps in using direct step-by-step integration methods (for stability and accuracy)
• Without the modal reduction analytically, small time steps will be needed as all higher modes esp. for
a complicated numerical model, are automatically included in the integration but high accuracy is
not quaranteed .
• Artificial numerical damping is thus often employed to damped out high frequency componentwsies

24
Particular Example: MDOF structures subject to Earthquake base motion

{ } { } { } { }
Recall the general equation of motion for MDOF is:
+
[M] x [C] x +[K] x f
=
 
{ } { }
( )
a =
Adaptation to earthquake ground motionproblem?
f 0
{ } { } { } { }
1
1
( )
1
1
ground
b x where
ì ü
ï ï
ï ï
ï ï
ï ï
ï ï
=í ý
ï ï
ï ï
ï ï
ï ï
ï ï
î þ
abs
x = x + 1 1
  
{ } { } { } { } { }
MDOF equation of motion for ground motion is
( ) !
ground
x t effectively
+ =
-
[M] x [C] x +[K] x [M] 1 f
=
  

{ } { } { } { }
With the Equation of motion for MDOF due to base ground motion
ground
x
-
+
[M] x [C] x +[K] x [
= M] 1
  
[ ]{ }
[ ]
2
2 ( )
i i i i i i grou
i i
nd
o q q q x t w ere
r h
V w w a a
æ ö
÷
ç ÷
- ç ÷
ç ÷
÷
=
ç
è
+
ø
T
i
T
i i
η M 1
+ =
η M η
  
{ }
2 1
!
then by the modal decomposition via , one gets
2 ( )
i i i ground
x t
V w w -
+ + = =-
T
x = Tq
q [ ]q [ ]q [Θ] [T] f [P]f [P] ]
= [M 1
  
[ ]
can be diagonalized by the modal shape/modal matrix ,
and assuming [C] T
[ ]
[ ]
[ ]
[ ]
[ ]{ }
[ ]
1 1 2
... , then one can w
,
r
1,2,. ,
ite
( ) ..
o
i i i
N
gr und
M
q q q i N
and
x t
-
-
é ù
= =
æ ö
÷
ç ÷
+ + = ç =
û
÷
ê ú
ë
ç ÷
÷
ç
è ø
T T T
i i
T
i i i
T T T
i i i i i
T
T
i
Recall the Participation facto
η
r matrix i
η C η K η η 1
η M η η M η η M
s
[P] [Θ] [T] [T] η η η
η
  

as the “Seismic
participation factor”

[ ]{ }
[ ]
:
i
i
i i w
a
a
æ ö
÷
ç ÷
ç ÷
ç ÷
÷
ç
è ø
=
T
i
T
i i
Variation of
η M
as a function of or
1
η M η
typically becomes smaller and smaller
at larger and larger natural frequencies .
i
i
numerically
a
w
General observation :
Reason why?

Response spectrum method for MDOF:
2
from solving
2 ( ) ( )
is g
by the response s
.
pectrum, the max
r
ive
response of
a
n by (
( )
due to particu
, )
la
n n ground ground
n
x x
t
x x t x t
x
V w w
w V
-
+
d
R
+
the spectral d
a
isplace
l
ment S
ec l
=
   
[ ]{ }
[ ]
2
Now, in a modal decomposition, the modal coefficient of each mode is gove
(
rned by
2 )
i
i i i i i i gr d
i i
oun
q
q q q x t where
V w a a
w
æ ö
÷
ç ÷
- ç ÷
ç ÷
÷
è
+ =
ç ø
T
i
T
i i
η M 1
=
η M η
+
  
max
the maximum c h
an be given b ( , ) for eac mode !
y = i
i i i
i
q q w V
a
 d
S
How about the maximum value of xi ?

{ } { } { } { }
{ }
1 2
1 2
th
1 2
we defined ( ) ( ) ( ) ... ( ) .
1. the max response of the i mode depends also on the normalization of the
2.the maximums ( ), ( ), ... typically occur at different time
Max
N
N
i
i
t q t q t q t
of q t q t
x q
= + + +
£
R x η η η
η
ecall
Note :
max
1 2
1 max 2 max ... . .
N
i i N i
q q i e RHS is generally an over estimation
h h h
+ + + -
A more optimal and commonly used maximum's estimate is by the " rule":
SRSS
max
1 2 2 2 2
1 max 2 max
max ( ) ( ) ... ( )
N
i i i N i
x q q q
h h h
+ + +


Frequency Domain Analysis of MDOFs
2
)
By modal decomposition, it has been shown that each m
2
ode is g
(
overned by
i i i i i i i
q q q x t
V w w a
+ -
+ =
  
 
2 2
2 2
( )
( )
( )
( ) ( ) ,
( ) )
)
2 (
(
2
i i i ground
ground
i i i
i t
i i i
i i
i
i
q q t e dt as the Fourier tran
q
x
form of q t
then i q
i
x




   

  




 





    
 

  
 
 




 


Define the Fourier transform
 
2
/ / 2
( ) ( ( (
) )
2
)
ground ground
i i i
i q r
a g ound
i i
q a
i
o
e
r
x er
i
wh
q H H


  

   

 

 

 
 
 
 

2
( ) ( )
& & ( )
( )
i i i i
q q
i
q i q
 
 
 
 
   

[ ] { } { } { } { }
1 2
1 2
1
( ) ( ) ( ) ( ) ... ( ) ( )
N
N i
N i
i
t t q t q t q t q w
=
= = + + + = å
Recall x T q η η η η
[ ] { } { } { }
1 2
1 2
( ) ( ) ( )( ) ( ) ... ( ) N
N
q t q q
w w w w w
 = = + + +
x T q η η η

   
{ } { } { }
1 1 2 2
/ / /
( ) ( ) ... ( ) ( )
ground ground ground
N N
q a q a q a x
w w w w
é ù
= + + + ⋅
ê ú
ë û
H η H η H η 

{ }
/
1
( ) ( )
ground
N
i i
q a
i
x
w w
=
é ù
ê ú
= ⋅
ê ú
ë û
å H η 

 
/ 2 2
)
(
2
ground
q a
i
i
i i
i
where
i
H
  




 

 

 
 
 

[ ] { } { } { }
1 2
1 2
( ) ( ) ( )( ) ( ) ... ( ) N
N
q t q q
w w w w w
 = = + + +
x T q η η η

   
{ } { } { }
1 1 2 2
/ / /
( ) ( ) ... ( ) ( )
ground ground ground
N N
q a q a q a x
w w w w
é ù
= + + + ⋅
ê ú
ë û
H η H η H η 

{ }
/
1
,
( ) ( ) ( )
ground
N
i i
q a
i
Concisely
x
w w w
=
é ù
ê ú
=
ê ú
ë û
å
x H η 



Recall g g
x in MDOF x
= =  = =
absolute absolute absolute absolute
Absolute Accelerationsin MDOF :
a x x +1 a x x +1
  
      
1
. .
N
j
j
j
Good to expand in terms of
i e a
=
= å
i
1 η
1 η
Idea? Pre-multiply bothside by ?
i T
η M
1
. .
N
i T j
j
j
i e a
=
= å
T
η M1 η Mη

!
i
j i i
same as the seismic participation factor
a =
T
T
η M1
η M η

( )
1 1 1
( ) ( )
g
N N N
i i i
i i g i i g
i i i
x
q x q x
w w
a a
= = =
+ +
= =
æ ö
÷
ç
= =
÷
ç ÷
ç ÷
è ø
å å å
absolute absolute
a x x +1
η η η
  
   
   
   
2
Recall 2
i i i i i i ground
i
q q q x
V w w a
-
+ + =
  
2 2
2 (2 ) !
i g i i i i i i i i i
i
q x i q q i q
V
a w w w V w w w
 + =- - =- +
 
    
2
2
i g i i i i i
i
q x q q
V w
a w
 + =- -
  
  /
2 2
( ) ( ) ( ) ( )!
Note that solution for all are similar to the1-dof system's solution, i.e.,
1
( ) 2 g
i
g x x g
n n
q
x x x
i
H
   
   
 
 
 
  
 
 
 
 


 
  

 
2
/ / 2
( )
Recallwe had ) ) )
( ( (
2
ground ground
ground
i i i
i q
i
a a
i i
q
q i
i
x w th
H H


 
  

 

 

 

 
 
 

 
( )
2
2
/
2
2 2
2
(2 )( ( ( )
(2 )
)
)
(
)
2
ground
i ground i i i i i
i i i ground
i i i
ground
i
i
i
i
q a
i
i
q x i q q
i x
i
x
i
V w w w
V w w w
V w w w
w w V
w w
w
a
a
w
w
 + =- -
=- +
+ ö
æ ÷
ç ÷
ç ÷
ç ÷
ç ÷
- + ÷
ç ø
=
è
H
 
   




( )
( )
2
2
1
2
1
( )
(2 )
)
(
2
i i i
ground
i
N N
i i
i
i g
i i
i i i
q x
i
i
x
w
V w w w
w w w
w
a
V
a w
= =
+
æ ö
æ ö ÷
ç ÷
ç ÷
÷
ç ç ÷
= = ÷
ç ç ÷
÷
ç ç ÷
÷
- + ÷
ç ÷
ç è ø ø
+
è
å å
absolute
a η η 

 
  

( )
( )
( )
{ }
1
2
2 2
2
/ 2 2
1 1
( )
(2 )
(
2
(2 )
where
2
is the vector-va
)
ground
N N
i i
o
i i i
ground
i i
i
i i g
i i
N
i
i
i
md f i i i
a a
i i i
i
q x
i
i
i
x
i
w
V w w w
w w V w
V w w
a
w
w w V
a
w
a
w
w
w
= =
=
+
æ ö
æ ö ÷
ç ÷
ç ÷
÷
ç ç ÷
= = ÷
ç ç ÷
÷
ç ç ÷
÷
- + ÷
ç ÷
ç è ø
è ø
æ ö
÷
ç ÷
ç ÷
ç ÷
ç ÷
- + ÷
ç
è ø
+
+
=
å å
å
absolute
H
a η η
η
 


  
lued transfer function for a MDOF.
( )
{ }
( )
{ }
( )
{ }
2
2
/ 2 2
2 2
1 1 1 2 2 2
2 2 2 2
1
2
1
1
1 2
1 1 2 2 2
(
2
2 )
2
(2 ) (2 )
2
,
2
ground
mdof i i
i
i
i
i
a
i i
i
a
For a DOF as example
i
i
i
i
i
i
a
w
w
V w w
w w V w
V w w w V w w w
w w V w w w V w
a a
w w
=
æ ö
+
÷
ç ÷
ç ÷
ç ÷
ç ÷
- + ÷
ç
è ø
æ ö æ ö
÷ ÷
ç ç
÷ ÷
ç ç
= +
÷ ÷
ç ç
÷ ÷
ç ç
÷ ÷
- + - +
÷ ÷
ç ç
è ø è
+
=
ø
+
å η
η η
H

( )
Typical profiles of transfer functions H
for MDOFsystems in the frequency domain
w

Dynamics of Continuous System
 Lumped‐mass models (SDOF, MDOF) are inherently an
approximation to many physical systems.
 In geotechnical engineering, the soil region supporting a
building or bridge is inherently a continuous (3D) domain.
 In structural and mechanical engineering:
plate/shell/dome structures,…
 In foundation engineering: earth dams, embankments ….

Multi-D-O-F system  Continuum systems
2

3
Consider a soil or solid stratum/layering
in pure horizontal or simple shear motion y(x,t):
=
x
2
2
y
dx in free vibration
t
r
¶
¶
d
t t
+
t
G, ρ
dx H
( )
2
2
( )
y
a Equation of Motion d dx
t
t t t r
¶
 + - =
¶
2
2
...... (1)
y
Dividing by dx
x t
t
r
¶ ¶
 =
¶ ¶
( ) :
...... (2)
b Constitutive relation for shear deformation
G
t g
=

( ) :
................ (3)
c Strain displacement relation for shear deformation
y
x
g
-
¶
=
¶
2
2
...... (1) ...... (2)
y
Together with and G
x t
t
r t g
¶ ¶
= =
¶ ¶
one gets
2
2
( ) ......... (*)
y y
G
x x t
r
¶ ¶ ¶
=
¶ ¶ ¶
For a homogeneous and medium in free vibration,
G r
2 2
2
2 2
........ 1D classical
........
s
s
y y
C
x t
G
with C
r
¶ ¶
=
¶ ¶
=
Wave Equation
shear - wave speed
2 2
2 2
, ( )
y G y y
Generally G
x x x t
r
¶ ¶ ¶ ¶
 + =
¶ ¶ ¶ ¶

2 2
2 2
( )
g
y y
G y t
x t
r
æ ö
¶ ¶ ÷
ç ÷
 = +
ç ÷
ç ÷
ç
¶ ¶
è ø

( ) ?
g
y t
Case of Horizontal Earthquake GroundMotion Acceleration 
2 2
2
2 2
:
0
( )
s
g
Summary
y y
C
y t
x t
the equation of motion for free vibration or forced vibration
of a soil layer inhorizontal shear motion
ì
ï
¶ ¶ ï
- = í
ï
¶ ¶ ï
î


0: (0, ) 0 .
: ( , ) 0
At x y t fixed base condition
At x h H t traction free top surface condition
t
= =
= = -
a) Boundary conditions :
2 2
2
2 2
( ) ( : find the basic dynamic characteristics)
Equation of motion: 0
subject to the above boundary conditions
s
I purpose
y y
C
x t
¶ ¶
- =
¶ ¶
Free - Vibration Case
0 0
0: ( ,0) ( ) ( ,0) ( )
At t y x y x and y x y x
= = =
b) Initial conditions :
 
Assume steady-state motion
Approach :

2 2 2
2 2 2
2 2 2
2
2
2
Assume steady-state motion, i.e. take ( , ) ( )
( )
0 ( ) 0
0.
i t
s s
s
y x t X x e
y y d X x
C becomes C X x
x t dx
d X
X
dx C
w
w
w
=
¶ ¶
 - = + =
¶ ¶
æ ö
÷
ç ÷
 + =
ç ÷
ç ÷
ç
è ø
Approach :
( ) sin cos for some constants A and B.
s s
X x A x B x
C C
w w
æ ö æ ö
÷ ÷
ç ç
÷ ÷
= +
ç ç
÷ ÷
ç ç
÷ ÷
ç ç
è ø è ø
Solution :
Apply Boundary conditions :
( ) @ 0: (0, ) 0 0 sin 0 cos 0 ( ) sin
s s s
i x y t A B B X x A x
C C C
w w w
æ ö æ ö æ ö
÷ ÷ ÷
ç ç ç
÷ ÷ ÷
= =  = + =  =
ç ç ç
÷ ÷ ÷
ç ç ç
÷ ÷ ÷
ç ç ç
è ø è ø è ø
( )@ : ( , ) 0 (0) cos cos 0
s s
X
ii x H H t G G A H H
x C C
w w
t
æ ö
æ ö æ ö
¶ ÷
ç ÷ ÷
ç ç
÷
÷ ÷
= =  = ç  =
ç ç
÷
÷ ÷
ç ç ç
÷
÷ ÷
ç ç
÷
ç
¶ è ø è ø
è ø

cos 0 ...... (*)
Characteristic/Eigen equation .
s
H
C
is the on the frequency
w
w
æ ö
÷
ç ÷=
ç ÷
ç ÷
ç
è ø
3 5 (2 1)
, , ,.. , 1,2,....
2 2 2 2
s
H i
i giving
C
w p p p p
-
= = =
Implication :
(2 1), 1, 2, ....
2
as the naturalfrequencies of the homogeneous soil layer in shear motion
s
i
C
i i
H
p
w = - =
: ( ) sin sin (2 1) ,
2
i
i
s
x x
Mode Shapes X x i
C H
w p
æ ö æ ö
÷
ç ÷
ç
÷
= = -
ç ÷
ç
÷ ÷
ç ç
÷
ç è ø
è ø

1 1
2 1 2
3 1 2
(2 1),
2
as the frequency & ( ) sin ,
2 2
3
3 as the 2 frequency & ( ) sin ,
2
5
5 as the 3 frequency & ( ) sin
2
s
i
s
C
With i it means
H
C x
fundamental natural X x
H H
x
nd natural X x
H
x
rd natural X x
H
p
w
p p
w
p
w w
p
w w
= -
æ ö
÷
ç
= = ÷
ç ÷
ç
è ø
æ ö
÷
ç
= = ÷
ç ÷
ç
è ø
æ ö
ç
= = ç
ç
è
...... for a stratum.
÷
÷
÷
ø
homogeneous

3 6
1 1
1
: 100', 100 / , 0.25 10
283.7 / sec
2
4.457 / sec 1.41sec
s
Example Homogeneous soil layer H g lb t G psi
G
C ft
rad T
r
r
p
w
w
= = = ´
 = =
 =  = =

:
Forced Vibration due to ground motion
1
2
2
1
2
2
1
:
( , ) ( ) ( )
( ) ( )
( ) ( )
i i
i
i i
i
i i
i
Let y x t q t X x
y
q t X x
x
y
q t X x
t
=
=
=
¥
¥
¥
=
¶
¢¢
 =
¶
¶
 =
¶
å
å
å
Modal expansion approach

2 2
0
for all mode shapes
i i
i i i i
s s
X X X X
C C
w w
æ ö æ ö
÷ ÷
ç ç
¢¢ ¢¢
÷ ÷
+ =  =-
ç ç
÷ ÷
ç ç
÷ ÷
ç ç
è ø è ø

2
1
( ) ( )
i
i i
i s
q t X x
C
w
=
¥ æ ö
÷
ç ÷
=- ç ÷
ç ÷
ç
è ø
å

2 2
2
2 2
( )
s g
y y
C y t
x t
¶ ¶
- = 
¶ ¶

2
1 1
2
1
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
i
i i i i g
i i
s
i
i i i g
i s
q t X x q t X x y t
C
q t q t X x y t
C
w
w
= =
=
¥ ¥
¥
æ ö
÷
ç ÷
- - =
ç ÷
ç ÷
ç
è ø
æ ö
æ ö ÷
ç ÷
ç ÷
ç ÷
 + =-
÷
ç
ç ÷ ÷
ç
ç ÷
ç ÷
è ø ÷
ç
è ø
å å
å
 
 

i j
Orthogonality of X and X :
2
How? From 0
i
i i
s
X X
C
w
æ ö
÷
ç
¢¢ ÷
+ =
ç ÷
ç ÷
ç
è ø
2
0
0
0 0
0 .... (1)
H
i
i i j
s
H H
H
i j i j i j
Consider X X X dx
C
and X X dx X X X X dx
w
æ ö
æ ö ÷
ç ÷
ç ÷
ç ¢¢ ÷
+ =
÷
ç
ç ÷ ÷
ç
ç ÷
ç ÷
è ø ÷
ç
è ø
¢¢ ¢ ¢ ¢
= -
ò
ò ò
2
0
0
0 0
0 ...... (2)
H
i
j j i
s
H H
H
j i j i j i
Likewise X X X dx
C
and X X dx X X X X dx
w
æ ö
æ ö ÷
ç ÷
ç ÷
ç ¢¢ ÷
+ =
÷
ç
ç ÷ ÷
ç
ç ÷
ç ÷
è ø ÷
ç
è ø
¢ ¢ ¢ ¢ ¢
= -
ò
ò ò
0
( ) ( ) 0
H
i j
X x X x dx if i j
= ¹
ò
0
& 0
H
i j
X X
¢ =

2
1
0
2
0 0
( ) ( ) ( ) ( )
( ) !
( ) ( ) ( ) ( ) ( ) ( )
j
j j j g
i s
H
i
H H
i
i i i i g i
s
Then take q t q t X x y t
C
and multipleby X x dx and
q t q t X x X x dx y t X x dx
C
w
w
=
¥ æ ö
æ ö ÷
ç ÷
ç ÷
ç ÷
+ =-
÷
ç
ç ÷ ÷
ç
ç ÷
ç ÷
è ø ÷
ç
è ø
æ ö
æ ö ÷
ç æ ö
÷
ç ÷
ç ÷
ç
÷
 + =-
÷
ç
ç ÷
÷ ç ÷
÷
ç è ø
ç ÷
ç ÷
è ø ÷
ç
è ø
å
ò
ò ò
 
 
2
0
2
0
( )
( ) ( ) ( )
( )
4
( )
(2 1)
H
i
i
i i g H
s
i
i g i
X x dx
q t q t y t
C X x dx
y t where
i
w
a a
p
æ ö
÷
ç ÷
 + =-
ç ÷
ç æ ö
÷
ç
è ø ÷
ç ÷
ç ÷
è ø
=- =
-
ò
ò
 


Damping:
:
: , ...
G or E c
t g V g s e e
= + = +
Theory of Viscoelasticity
Kelvin - Voigt 

k
c
Maxwell :
G
t t
g
V
= +



2
2
( ) Equation ofMotion ...... (1)
y
a
x t
t
r
¶ ¶
=
¶ ¶
( ) Constitutive relation for viscoelastic model in shear deformation:
...... (2)
b
G
t g V g
= + 
( ) :
................ (3)
c Strain displacement relation for shear deformation
y
x
g
-
¶
=
¶
2 3 2
2 2 2
(1),(2) (3),
0
......... (*)
( )
g
Combining and the equation of motion with viscous damping is
y y y
G
y t
x x t t
V r
r
ì ü
ï ï
¶ ¶ ¶ ï ï
+ - = í ý
ï ï
¶ ¶ ¶ ¶ ï ï
î þ

: ( , ) ( ) ( )
... modal decomposition
or method of separation of variables in p.d.e.s
Take y x t q t X x
idea of
=
Solution

2 3 2
2 2 2
2
2
(*)
0
( )
0
......... (**)
( )
( ) ( ),
0
( )
g
S
g
S
g
Then leads to
G y y y
y t
x x t t
C q X q X q X
y t
Dividing by T t X x
X q X q
C
y t
X q X q
V
r r
V
r
V
r
ì ü
ï ï
¶ ¶ ¶ ï ï
+ - = í ý
ï ï
¶ ¶ ¶ ¶ ï ï
î þ
ì ü
ï ï
ï ï
¢¢ ¢¢
⋅ + ⋅ - ⋅ = í ý
ï ï
ï ï
î þ
ì ü
ï ï
¢¢ ¢¢ ï ï
+ ⋅ - =í ý
ï ï
ï ï
î þ

 

 


2
2
2
2
2
0 .... ( ) ( :1st term is a function of x only.
constant, say= ....( ),
( ) 0
with the boundary conditions of (0) 0 ( ) 0.
S
S
S
X q X q
C Note
X q X q
X
Taking
X C
then X X
C
X and X H
V
r
w
w
¢¢ ¢¢
+ ⋅ - = 
¢¢
= - 
¢¢
  + =
¢
= =
For free vibration :
 
Recall we found
(2 1) ( ) sin sin (2 1) 1, 2, ....
2 2
for a homogeneous soil layer in shear motion
s i
i i
s
C x x
i and X x i for i
H C H
p w p
w
æ ö æ ö
÷
ç ÷
ç
÷
= - = = - =
ç ÷
ç
÷ ÷
ç ç
÷
ç è ø
è ø

2
2 2
2
2 mode ,
2 2
2 0
.... mode free vibration !
i i i
i i i
S S i
i i i i i
th
Setting which increases with number
C C G
q q q
a damped SDOF equation for i in
w V w V w
V
V w V
r r
V w w
æ ö
÷
ç ÷
=  = =
ç ÷
ç ÷
ç
è ø
 + + =
 
2 2
2 2
2 2
( ),
( ) 0 0 ..... (***)
S S
With
q q
q q q
q q C C
V w V w
w w
r r

æ ö æ ö
÷ ÷
ç ç
÷ ÷
  + ⋅ + =  + + =
ç ç
÷ ÷
ç ç
÷ ÷
ç ç
è ø è ø
 
 

2
,
the governing equation
2 ( ) with the extra damping term
4
with for the homgeneous soil case.
(2 1)
i i i i i j g
i
becomes
q q q y t
i
V w w a
a
p
+ + =-
=
-
For the earthquake ground motion problem
  
(2 1), 1, 2, .... naturalfrequencies
2
s
i
C
i i
H
p
w = - =
( ) sin sin (2 1) , ...
2
i
i
s
x x
X x i mode shapes
C H
w p
æ ö æ ö
÷
ç ÷
ç
÷
= = -
ç ÷
ç
÷ ÷
ç ç
÷
ç è ø
è ø

v
1
max
Recall for a homogeneous layer,
( , ) 4
with (2 1)
(2 1)
i i
i j i i
i
S
q and i
i
w z
a a w w
w p
= = = -
-
:
Example : Modal contribution to response quantities by R
Applications and insights from Ex
esponse Spectrum :
act Solutions
v v
2
max max
1 1
v
2 2
( , ) ( , )
4 4
( , ) ( ) (1)
(2 1) (2 1) (2 1)
4 ( , ) 1 1 1 1
~ 1, , , ...
(2 1) (2 1) 9 25
i i i i
i i i
i i
S S
y H t q X H
i i i
S
where
i i
w z w z
p w pw
w z
pw
æ ö æ ö
æ ö ÷ ÷
÷ç ç
ç ÷ ÷
 = = ÷ =
ç ç
ç ÷ ÷
÷ç ç
ç ÷
ç ÷ ÷
ç ç
- - -
è øè ø è ø
æ öæ ö
÷ ÷
ç ç
÷ ÷ =
ç ç
÷ ÷
ç ç ÷
÷ç
ç - -
è ø
è ø
1
max a v
Similarly,
4
( , ) ( , ) ( , )
i j i i i i
a H t S S
w
a w z w z
p
= = 1

Note: there are now 2 regimes of dynamic behavior:
depending on whether  is > or < than cutcoff

( )
1 1
( , ) ( ) ( ) ( )
2 2
x ct
x ct
y x t f x c t f x c t g d
c
h h
+
-
= - + + + ò

Basic Wave Propagation Theory
in 3D for Soil Dynamics & Seismology
R. Y. S. Pak

2
Homogeneous & isotropic with 2 elastic constants ( )
1 2
Based on (i) Newton'slaw, (ii) stress-strain relationship & (iii) small strain theory
(Navier's equation):
( 2 )
G
and G
G

 


Equationof motion
- =
-
-
-
+
( )
2
1 2 3
, ,
2
2 2
2
( ) ( )
( ) ( )
In rectangular coordinates with , , ,Navier's equation becomes
( ) ,
( ) , (repeated index
k ki i kk i
i
k i
k i k k
i 1,2,3
i 1,2,3
G
or G G
u u u
G u Gu u
u
u u
G G
x x x x t

 
 
 
u u u
u u u
u
=
=
 ⋅ - ´ ´ =
+  ⋅ +  =
=
+ + =
¶
¶ ¶
 + + =
¶ ¶ ¶ ¶ ¶



3
1
2
2 2
3 3
2
1 1
,
)
( )
k
i
k i
k k
k i k k
i 1,2,3
u
u u
G G
x x x x t
 
=
= =
=

¶
¶ ¶
 + + =
¶ ¶ ¶ ¶ ¶
å
å å
Formulation for 2D/3D Soil Dynamics and SSI Problems
12
13 11

 

1 1 2 3
2
. ., sin ( ) , 0
d
d
e g u A x C t u u
L

 
   
 
 
• a sinusoidal plane compressional/dilatational or P-wave train of wave length Ld
• a solution of the Navier’s eqn of motion
2 1 1 3
2
. ., sin ( ) , 0
s
S
e g u A x C t u u
L

 
   
 
 
• a sinusoidal plane shear or S-wave train of wave length Ls
• a solution of Navier’s eqn of motion
Examples: Plane waves solutions:
x2
x1
x3
with ... shear wave speed
s
G
C

=
2
with ... compressional/dilatational wave speed
d
G
C


+
=

3 2
1
2 3
1 2 3
3
1
2 3, 2
1 2 3 1 2 3 3 1
1 2 3 2 1
3
1 1
2
x x
where
x x x x x x x x
x x
 


  

    
 
 
 
 
 
 
 


     
        
 
       
 
 
 

 
 
 
1
e
e e e
e e e ψ e
e
subject to boundary, initial and loading conditions
General Solution by Method of Potentials:
Take 
   
u ψ
2 2
2 2
2 2 2 2
2 2 2
2
2 2 2
1 2 3
Then Navier's equation becomes
1 1
, ( ave equations in 3D)
with
d s
w
C t C t
x x x


 
   
 
  
   
  
ψ
ψ

Plane waves propagating in arbitrary 3D directions:
In terms of the 2 potential functions:
( ) ( , ) ( ) ( , ) ( )
. ., is the direction of the unit vector which denotes the direction of wave propagation
with compressional/dilatational wave speed ... i.e. a P-wav
d d
d
a x t f C t x t f C t
ie
C
d d d
d
k x u k k x
u k
  
       
e.
( ) ( , ) ( ) ( , ) ( )
. ., is orthogonal to the unit vector which denotes the direction of wave propagation
with shear wave speed ..... i.e. a S-wave.
The sh
s s
s
b x t C t x t C t
ie
C
s s s
s
ψ g k x u ψ k g k x
u k
Note:

        
ear wave motion can be decomposed into
a Horizontally-polarized (SH) shear wave and a Vertically-polarized (SV) shear wave.

x
Wave propagation direction
SH wave motion (out of plane)
SV wave motion (in‐vertical plane)
z
y

O
x1 or x, u1 or u
x2 or y
u2 or v
x3 or z, u3 or w
2
2
3 2
1
1 2 3
2 1
3
3 1 2
0 ( ):
( )
( )
2
With u and plane strain
x
u u
x x x x z
u w
x x x z x
depend on only scalar potential functions and
 
  
 
  
 



 
  
     
    
 
  
     
    
Simplification for 2D Wave Propagation Problems (Plane Strain):
2 2
2 2
2 2
1 1
,
d s
C t C t
 
 
 
   
 
3 2
1
2 3
1 2 3
3
1
2
1 2 3 3 1
1 2 3 2 1
3
1 1
2
x x
x x x x x
x x
 


    
 
 
 
 
 
 
 


  
    
 
    
 
 
 

 
 
 
e
e e e
ψ e
e

Example: Time harmonic Incident Plane P- and SV-waves
in a Semi-Infinite Soil in 2D (Plane-strain)
n
l

sin 2(1 )
where = , the ratio of compressional to shear wave speed
sin (1 2 )
s d d
d s s
C C
k k
C C

 
  

   

Physical phenomena of wave
reflection, mode conversion,
diffraction, refraction,...

1 2
Other important wave types and solutions of boundary value problems of soil dynamics and
seismological interest:
- Rayleigh waves (a surface wave with motion decaying quickly with depth) with wave speed CR (< CS< Cd)
- Theoretical basis of site/soil characterization by wave-based methods and forced vibration testing
(eg dynamic load on finite area)
- Fundamental to seismic/dynamic soil-structure
- Extension and applications to multi-layer soil configurations
Dynamic foundation response from applied loads
(SSI problems)
Incoming SH wave (simpler than SV wave)
with 1 = 2

Dynamic Soil-Structure Interaction
(551) Problems
ExamDle: A Comoressor Foundation Dynamic Problem
Rigid body dynamics: F=ma, M=Icx
Soil
Soil
e at (0 rad/sec

Needed material information:
Establish soil/site/foundation/structure conditions
•Dynamic loading, nature, magnitudes and directions
•Soil profile: sand, clay, rock, thickness, layering,
•Density, void ratio, P1, OCR,
•Dynamic shear modulus G and damping with their strain
dependence
•Machine’s mass, moment of inertia, stiffness
‘Foundation type, shape, weight/bearing pressure
‘Design criteria
Example: Characterization of dynamic loads
Newton’s law: F= m a, M= ‘M
Mmachine
Soil

Resultant
dynamic loads:
F=me rw2 = m7v2/r
5 to 10% rotating mass
= Fsin(wt) or
F = Fsin(at)
F = FcosQot) F, = Fcos(cot)
Fundamental example problem:
A Rigid footing under dynamic loading
• A Mechanics Contact problem
• 3D elastodynamic theory and wave radiation phenomena
• Interface & boundary conditions
• Concept of radiation damping
• Analytical methods
• Finite element and boundary element methods
• Frequency and time domain
I
i-i I foundation
interface

Foundation Interfacial Impedance Matrix
V K(w) 0
H 0 0
M 0 0
T 0 KTT(co) ®
• From solution of 3D continuum boundary value problems
• In frequency domain / time-harmonic factor ebo)t
• Each function is generally complex-valued for nonzero (0
i.e., K1 (w) = Re[K (w)] + i Im[K (w)]
• ‘Dynamic foundation impedance functions’
• Contact stress distributions
• Horizontal-rocking motion coupling
Note:
1. Notion of ‘Frequency
(*)
dependent’ springs and
dashpots due to 3D wave
propagation phenomena
2. Radiation/geometric
damping
i.e. a mechanical analog of Impedance Functions (suggestion by Lysmer 1965)
0
Khh(o)
Kmh (0)
0
0
Khm (t0)
Kn?m (t0)
0
Recalifrom SDOF with spring-rnass-dasphot:
F(t)=kx(t)+c 1(t) F(o)=kx+kocx =‘F(o)=(k+iwc)x=K(w)x ....(*)
Comparing it with
K(co)=Re[K(w)]+ i Im[K(w)]
one can set
k =Re[K1(o)] =k(w)
c= Jm[K(o)]/w=c(o)
= 2 = Ccriticai

Foundation Interfacial Compliance Matrix
C (w) = Re[C (w)] + i Im[C1 (w)]
- //,‘
*I
‘
C(w) 0 0 0 V
Ah — 0 Chh(w) Chm(W) 0 H
Or — 0 CmhfrO) Cmm(CO) 0 M
°T 0 0 0 CTT(a)) T
• jCJ=[K]’
• Symmetric matrices due to reciprocity property in elastodynamics theory
• Depend on foundation shape, type, embedment, soil profile,
• Examples for circular and square surface foundations:
•
Classical Solution for Foundation Compliance Functions by Homogenous Half-space model:
Notation: a=radius, a=Poisson ratio, i=shear modulus (Luco & Westmann (1971)
16
/I
7
.6 -
__zr— ‘

Square foundation impedance functions by Homogenous Half-space model:
(Pak & Guzina 1999)
Rk SlfCsf of Sqs Sfc, F d.ss
Hp El. H,If-S
k =ic =-
‘ Gb’
hh
Gb
K,
k Kmm
mhGb2 mm
Gb3
r
,‘97Ib!fI3
.i
FIg 0-21 V,siI ,ibSi ( dI f,,Sim. f 4
Fry (1963), Richart, Hall & Woods (1970)
Results from centrifuge model tests
*
Pak et al. (2010) “Physical Characteristics of Dynamic Vertical-Horizontal-Rocking
Response of Surface Foundations on Cohesionless Soils,” Geotechnique
G=shear modulus, v=Poisson ratio, b=half-width
V,s1,,j SifY,p cf a Sql,aa Sda,a F,,,,d.,,,
HopSaCC ElsO, HaJf-0p.a
IC =
CpIi,.g Cf• Sq F,.,,d.U,,
H.
LS,aJ OCtl,,aa,, Sqaa 0IISCS PCCIiII
HCgaC E. Hi1-S
soc. 00.6
Results from experimental field tests
sEC. 0.7 R1XKf3JVIHR4rIoS5 30 FOIJ’4DATIONS 371
0
FI0C,a 0-06. CsopIad sskICga,d .1___i__Ji_..._. so1
,IICjC ,ibIoI,,,a ,f WES Ba,a . 8, 337
I. Taos 36. (a) CI5O5Cs,ao ‘t’ B,’C CI
C 0.33 2C
A,,.,,,

Reasons:
1: In-situ depthwise variation ofshear modulus ofreal soils
(i.e., generally not homogeneous)
e.g., sand, gravel, or NC clay
2: Theoretical vs physical contact stress distributions
small-strain theory
(a)
failure limit
(b)
actual service state
(c)

3: Effects of foundation weight and loading
If
Analytic Resolution by New Elastodynamic 551 Theory
(Pak and Ashlock, EESD 2011)
1.5
0 Depth ratio
E hf/b
0 Modulus ratio
,=
—
0.5
Standard elements
AG elements
2
10
y/a xla

R. Y. S. PAK AND J. C ASHLOCK
—1
0 1 2
cøa / (G0/p)°5
(d)
Vertical, Horizontal, Rocking Foundation Responses
—
— .254.0 14. c1.6
-
y45. c3.2 —j4.5. c1.5
‘ 2, 2.4 -:=5. ‘1.3
.
1=3.5)=19 =1 05
0 500 1000 1500
Frequency [Hzl
500 1000 1500 2000
Frequency 1Hz]
3
0
0
E
E
2’:
2
E
Cu
0
0
rn
1 2
wa I (G0/p)°5
Figure I I. Impedance functions K11() by 2-zone theory with h= I. (a) K,,:
(b) Ku,,: (C) K,,,,,: and (dl K,,,,, and Kh,,,.
/ -H
.-
2000
—
— C=.25.n=4.0 — 0=4. ã=1.6
n=2.5. 2=3.2 : C=4.5. 2=1.5
-0=3. G=2.4 —-0=5. 2=1.3
O 0 .)35. =19 .4=0,2=1.0S
- -.
—b

Mechanical Analogs of Impedance Functions for simplified analysis
F(t)=kx(t)+c±(t) F(do)=kx+iacx
=F(o)=(k+iac)x=K(oi)x ....(*)
Comparing it with
K11(o)=Re[K1(w)]+i [m[K1(eo)]
one can set in (*)
k = Re[K. (w)] = k(to) Note:
U 1. Notion of ‘Frequency-dependent’ springs
c = Im[K1 (w)] / tü = c(a) and dampers (3D wave propagation
phenomena
— C — C 2. Concept of Radiation/geometric
1i damping
4J !W1t critical

Use: Convert differential equations of motion
to simpler algebraic equations:
-1
2
( ):
F[ ( )] ( ) ( )
:
1
F [ ( )] ( )
2
:
F[ ( )] ( ) ( ), F[ ( )] ( ) ( ) , .
i t
i t
Fourier transform of a function f t
f t f t e dt
Inverse Fourier transform
e d
Key properties
f t i f t i etc



  

   





 

 


 
F
F F
F F
Frequency‐Domain Approach:

(1) Vertical vibration:
Time harmonic response,
i.e.
( ) , ( ) ,....
i t i t
F t e t e
 
  
F Δ
Frequency‐domain Dynamic Analysis of Foundations
0 vv 0 0 G
V =K (ω)Δ & Δ =Δ
......(1)
v e o
F V m
   

G

2
v-e o G
F -V =-ω mΔ
=
v
m

v

Fv‐e
V0
2
v-e vv G G
F -K (ω)Δ = -ω mΔ
...(2)
v-e
G 2
vv
F
Δ =
-ω m +K (ω)
଴

(a) Equations of Motion:
x x
F m
 
 
(2) Horiz‐rocking coupled vibration:
 
F ma, M I α
G G
M I

 
G
I 

o
H
= G
m

e
F e
M
G
O
1
2
d
d
2b
o
M
o
H
...............(1)
e o G
F H m
   
 2
e o G
F -H =-ω mΔ
1 2 .....(2)
e e o o G
M F d H d M I
    
 2
e e 1 o 2 o G
M +F d +H d -M =-ω I θ

...(3)
     
 
 
     
 
 
   
     
G o e
2
G o e e 1
2
Δ H F
1 0
m 0
-ω + =
θ M M +F d
-d 1
0 I
(b) Foundation‐soil interfacial impedance relations:
.............(4)
    
   
 
    
o hh hm o
o mh mm o
H K K Δ
=
M K K θ
.............(5)
      
 
 
     
 
 
 
   
      
G hh hm o e
2
G mh mm o e e 1
2
Δ K K Δ F
1 0
m 0
-ω + =
θ K K θ M +F d
-d 1
0 I

(c) Relation between motions at different points :
0
G


G
o


o

o

............(6)
   
 
   
 
 
   
o G G 2
o G
o G
2
o G
Δ =Δ -θ *d
θ =θ
Δ Δ
1 -d
=
θ θ
0 1
2
d

(d) Final equation assembly:
Substituting (6) to (5), one gets
2
( ) .............(7)


 
  
 
M K F
x

0
,
0
m
I
 
  
 
M
2
2
1 0 1
1 0 1
hh hm
mh mm
K K d
K K
d

 
   
  
   
  
   
K

1
, ,
G e
G e e
F
M Fd


   
 
   

   
x F
where
The matrix equation (7) for coupled lateral response motion can be readily
solved for any given loading F .

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dynamical analysis of soil and structures

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