Transcript Slide 1

Wave Travel and Attenuation
and
Machine Foundations
Richard P. Ray, Ph.D., P.E.
Civil and Environmental Engineering
University of South Carolina
USC
Topics for Today




Waves in Elastic Media
Waves in the Earth
Surface Excitations
Machine Foundations
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Waves
Compression, P
Primary (1-D)
Shear,S
Secondary
(1-D)
Rayleigh, R
Surface (2-D)
http://paws.kettering.edu/~drussell/demos.html
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Discrete Properties
Resonant Column - MOC - Wavelets
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Surface Block Mass






Δz1
Soil 1: G1,ρ1,μ1
..
..
Soil j: Gj,ρj,μj
..
..
Computational
Reaches Nodes
Vertical
Propagation
Soil m: Gm,ρm,μm
Horizontal
Polarization
τi,Vi
Δzi
A
P
B
Δzn-1
Rock Motion
Resonant Column - MOC - Wavelets
t=0......1.......2......3
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D z1
B.C. : τ P  0
C
A11
P
+
C
1
A2
C
B
2
D z2
A
C
B
3
D z3
3
C
BA4
D z4
4
CC+
CC+
P
2
P
3
CC+
P
4
CB55
P
BC.C
. : V P  f (t )
Dt
5
C+ characteristic:
dτ
dV
 ρνs
 0 and
dt
dt
dz
 νs
dt
 P -  A -  AP v AP ( V P - V A ) = 0
 = shearing stress;
V = particle velocity.
S = phase (shear wave) velocity;
 = mass density;
t
= time;
C- characteristic:
dτ
dV
dz
 ρνs
 0 and
 νs
dt
dt
dt
 P -  B +  BP v BP ( V P - V B ) = 0
Resonant Column - MOC - Wavelets
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Nonlinear Interpolation
τ 
τ
γ
1 α
Go 
C1 τ m

γγ i
ττ
and
2
2
Gtan 
i
R 1
dτ
dV
 ρνs
0
dt
dt




τ P - τ R - ρRP vRP ( V P - V R ) = 0
for γ and τ A3
A2
Rtime
Go

1  αR τ

C1τ max

R 1
and
dz
 νs
dt




Gtan
νs 
ρ
C3
Resonant Column - MOC - Wavelets
Rspace
C2
Dz
Dt
B3
A
C+
C
Sspace
Stime
B2
P
C-
B
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Stress Strain at Reach 30
3.00E+03
2.00E+03
30
Δz
31
τ ave  τ30  τ31  / 2
γave  δ30  δ31  / Dz
Stress
1.00E+03
0.00E+00
Reach 30
-1.00E+03
-2.00E+03
-3.00E+03
-2.50E-03
-1.50E-03
-5.00E-04
5.00E-04
1.50E-03
2.50E-03
Strain
Resonant Column - MOC - Wavelets
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Moc vs Shake Surf Velocity
2.00E+00
Moc Surf Velocity
Edushake Vel Outcrop
1.50E+00
Velocity (ft/sec)
1.00E+00
5.00E-01
0.00E+00
-5.00E-01
-1.00E+00
-1.50E+00
-2.00E+00
-2.50E+00
10
12
Resonant Column - MOC - Wavelets
14
16
Time (sec)
18
20
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 P -  A -  AP v AP ( V P - V A ) = 0
 P -  B +  BP vBP ( V P - V B ) = 0
dW 
t  Dt

t
dSH  
V

  D
 V j  V j 1 
1

dKE  Dz

2
2


 d dV

m
Dz j
j 1
4
SH  

Dt
Fdt 
Vbaset  Dt  Vbase t  base t  Dt   base t
4

2
( τ j  τ j 1  τ j old  τ j 1 old )
(δ
j 1
 δ j )  (δ j 1 old  δ j old )
Resonant Column - MOC - Wavelets

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Work-Energy
1.80E+03
1.20E+00
1.60E+03
1.00E+00
1.40E+03
Energy (ft-lb)
1.20E+03
8.00E-01
1.00E+03
Total Work
6.00E-01
8.00E+02
Hysteretic Energy
6.00E+02
Kinetic Energy
4.00E-01
Energy Ratio
4.00E+02
2.00E-01
2.00E+02
0.00E+00
0.00E+00
0
5
10
15
20
25
30
Time (sec)
Resonant Column - MOC - Wavelets
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Cumulative
Hysteretic Energy
Reach Number
Strain 400
Resonant Column - MOC - Wavelets
Time (sec)
Hyst 400
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Wavelets
A1
A2
A3
A4
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Resonant Column - MOC - Wavelets
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Profile View
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MEMS Accelerometer
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Data Acquisition
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Wavelets
Resonant Column - MOC - Wavelets
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Wavelets
Resonant Column - MOC - Wavelets
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 ( n '  n ) δt 
Wn ( s )   xn '  ψ * 

'
s
n 0


N 1
Localized Time Index
Wavelet Scale
Fourier Transform
Wavelet via Fourier
Transform
By varying the wavelet scale s and translating along the
localized time index n, one can construct a picture
showing both the amplitude of any features versus the
scale and how this amplitude varies with time.
Resonant Column - MOC - Wavelets
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Resonant Column - MOC - Wavelets
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Wavelets
Resonant Column - MOC - Wavelets
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r -2
r -2 r -0.5
+
-
+
Rayleigh wave
Vertical
Horizontal
component component
+
Shear
wave
+
Relative
amplitude
-
r -1
+
+
Shear
window
r
r -1
Waves
Fundamentals-Modeling-Properties-Performance
Wave Type
Percentage of
Total Energy
Rayleigh
67
Shear
26
Compression
7
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Free-Field Analytical Solutions
ur
uz
 r 
 L V 
u z (r , ,0)  i  03  RV (a0 ) H 02  
 2  
 CR 
  M 0V 
2  r 
 

ur (r , ,0)  i 
R
(
a
)
H
V
0
1 
3

 2  
 CR 
Fundamentals-Modeling-Properties-Performance
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Free-Field Analytical Solutions
ur
uz
 r 
 L V 
u z (r , ,0)  i  03  RV (a0 ) H 02  
 2  
 CR 
  M 0V 
2  r 
 

ur (r , ,0)  i 
R
(
a
)
H
V
0
1 
3

 2  
 CR 
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Trench
Isolation
Karlstrom and Bostrom 2007
Fundamentals-Modeling-Properties-Performance
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Chehab and Nagger 2003
Fundamentals-Modeling-Properties-Performance
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Celibi et al (in press)
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ATST Telescope and FE Model
Fundamentals-Modeling-Properties-Performance
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Summary and Conclusions (Cho, 2005)
1.
2.
3.
4.
5.
High fidelity FE models were created
Relative mirror motions from zenith to horizon pointing: about 400 mm in
translation and 60 mrad in rotation.
Natural frequency changes by 2 Hz as height changes by 10m.
Wind buffeting effects caused by dynamic portion (fluctuation) of wind
Modal responses sensitive to stiffness of bearings and drive disks
6. Soil characteristics were the dominant influences
in modal (dynamic) behavior of the telescopes.
7.
8.
9.
Fundamental Frequency (for a lowest soil stiffness):
OSS=20.5hz; OSS+base=9.9hz; SS+base+Coude+soil=6.3hz
A seismic analysis was made with a sample PSD
ATST structure assembly is adequately designed:
1.
Capable of supporting the OSS
2.
Dynamically stiff enough to hold the optics stable
3.
Not significantly vulnerable to wind loadings
Fundamentals-Modeling-Properties-Performance
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Foundation Movement
Z
Y
θ
φ
X
ψ
Fundamentals-Modeling-Properties-Performance
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Design Questions (1/4)
How Does It Fail?
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

Static Settlement
Dynamic Motion Too
Large (0.02 mm)
Settlements Caused By
Dynamic Motion
Liquefaction
What Are Maximum
Values of Failure?
(Acceleration,
Velocity,
Displacement)
Fundamentals-Modeling-Properties-Design-Performance
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Velocity Requirements
0,40
Massarch (2004) "Mitigation of Traffic-Induced Ground Vibrations"
Fundamentals-Modeling-Properties-Performance
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300
800
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Design Questions (2/4)

What Are Relations
Between Loads And
Failure Quantities?
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Loads -Harmonic,
Periodic, Random
Load→ Structure →
Foundation → Soil →
Neighboring Structures
Model: Deterministic or
Probabilistic
Fundamentals-Modeling-Properties-Performance
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Design Questions (3/4)

How Do We Measure What Is Necessary?





Full Scale Tests
Prototype Tests
Small Scale Tests (Centrifuge)
Laboratory Tests (Specific Parameters)
Computer Model
Fundamentals-Modeling-Properties-Performance
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Design Questions (4/4)

What Factor of Safety Do We Use?


Does FOS Have Meaning
What Happens After There Is Failure
Loss of Life
 Loss of Property
 Loss of Production


Purpose of Project, Design Life, Value
Fundamentals-Modeling-Properties-Performance
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r -2
r -2 r -0.5
+
-
+
Rayleigh wave
Vertical
Horizontal
component component
+
Shear
wave
+
Relative
amplitude
-
r -1
+
+
Shear
window
r
r -1
Waves
Fundamentals-Modeling-Properties-Performance
Wave Type
Percentage of
Total Energy
Rayleigh
67
Shear
26
Compression
7
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r -2
r -2 r -0.5
+
-
+
Rayleigh wave
Vertical
Horizontal
component component
+
Shear
wave
+
Relative
amplitude
-
r -1
+
+
Shear
window
r
r -1
Waves
Fundamentals-Modeling-Properties-Performance
Wave Type
Percentage of
Total Energy
Rayleigh
67
Shear
26
Compression
7
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Modeling Foundations

Lumped Parameter (m,c,k) Block System


Impedance Functions


Function of Frequency (ω), Layers
Boundary Elements (BEM)


Parameters Constant, Layers, Special
Infinite Boundary, Interactions, Layers
Finite Element/Hybrid (FEM, FEM-BEM)

Complex Geometry, Non-linear Soil
Fundamentals-Modeling-Properties-Performance
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Lumped Parameter
P  Po sin( t )
r
m
m
c
Gνρ
k
mz  cz  kz  P0 sin(t )
Fundamentals-Modeling-Properties-Performance
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Single Degree of Freedom
mz z  c z z  k z z  0
 m 
mz z  Inertia Force (kg )
or ( N ) c
2 
 sec 
 N  sec  m 
c z z  Dam pingForce 

 or ( N )
 m  sec 
z
m
k
N
k z z  Spring Force  (m) or ( N )
m
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Single Degree of Freedom
mz z  cz z  k z z  0
solution will take form z   e .......  constant
st
where (m s2  cs  k ) e st  0
k
st
2
divideby m e and set n 
m
c
2
2
then s  s  n  0
m
solution for s dependson c
c=0…Undamped
c=2mω…Critically
Damped
c<2mω…Underdamped
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Single Degree of Freedom
c
2
s  s  n  0 undampe d
...s  in
m
in t
 i n t
z (t )  1e   2 e
where 1 ,  2  f (init. cond.)
2
Euler' s identity eint  cos(nt )  i sin(nt )
z (t )  A sin(nt )  B cos( t ) A, B  f (initial condition)
z (0)  B and z(0)  An
 z (0) 
 sin(nt )  z (0) cos(nt )
z (t )  
z(0)
 n 
z(0)
t
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Single Degree of Freedom
c
2
s  s  n  0 if dam pingpresent
m
2
2
c
 c 
2
then s  
 



n
2m
 2m 
critical
c
if
 0 then c  ccrit  2mn and s  
 n
2m
 n t
z (t )  (1   2t )e
where 1 ,  2  f (init. cond.)
z(0)
z(0)
z (t )  z (0)(1  nt )  z (0)t e nt
t
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Single Degree of Freedom
if c  2mnthen
suppose D 
c
ccrit
then s   Dn 
 0 u n de rdam pe
d
 dam pingratio and  D  n 1  D 2
Dn 
2
 n
2
  Dn  i D

z (t )  1e  Dnt i Dt   2 e  Dnt i Dt  e  Dnt 1e i Dt   2 e i Dt
z (t )  e
 D n t
z (t )  e
 D n t
( A sin( D t )  B cos( D t )
 z (0)  z (0) Dn

sin( D t )  z (0) cos( D t )

D


See Chart
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
Single Degree of Freedom
P  Po sin(P t )
mz z  cz z  kz z  P0 sin(P t )
ze

 n Dt 
A sin( Dt )  B cos Dt 
P0
k  m   c 
2 2
P
c P
tan 
2
k  m P
k
n 
m
2
2
m
sin  P t    c
k
P



P
2 D


n


2



P
1 


n

 D  n 1  D
2
D
c
ccrit
ccrit  2 km
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SDOF Transient and Steady-State
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z (t ) 
P0
k  m   c 
2 2
P
z max
P0

k
z max  z static
2
2
sin  P t   
P
1
  
1   P 
  n 
2
2
   2
  2 D P 
n 
 
1
  
1   P 
  n 
2
2
   2
  2 D P 
n 
 
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Dynamic Magnification (Logarithmic)
100
D=0.02
D=0.05
D=0.10
D=0.20
D=0.50
Magnification
10
1
0.1
0.1
1
Frequency Ratio (P/n)
Fundamentals-Modeling-Properties-Performance
10
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Lumped Parameter System
Z
Kz
Cz
Iψ
mz z  cz z  kz z  P0 sin(P t )
Kx
m
ψ
X
Cx
Kψ
Cψ/2
Cψ/2
Fundamentals-Modeling-Properties-Performance
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Lumped Parameter Values
Mode
Vertical
z
Horizontal
x
Rocking
ψ
Torsion
θ
Stiffness k
4Gr
1 
8Gr
2 
8Gr 3
3(1   )
16Gr 3
3
Mass Ratio
m
ˆ
m
m(1   )
4 r 3
m(2   )
8r 3
3I (1   )
8r 5
I
r 5
Damping
Ratio, D
0.425
mˆ 1 / 2
0.288
mˆ 1/ 2
0.15
(1  mˆ )mˆ 1/ 2
0.50
1  2 mˆ
D=c/ccr G=Shear Modulus ν=Poisson's Ratio r=Radius
ρ=Mass Density Iψ,Iθ=Mass Moment of Inertia
Fundamentals-Modeling-Properties-Performance
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Design Example 1
VERTICAL COMPRESSOR
Unbalanced Forces
•Vertical
= 45 kN
•Horzontal Primary
= 0,5 kN
•Operating Speed
= 450 rpm
•Wt Machine + Motor = 5 000 kg
DESIGN CRITERION:
Smooth Operation At Speed
Velocity <0,10 in/sec
Displacement < 0,002 in <0,05mm
Soil Properties
Shear Wave Velocity Vs = 250 m/sec
Density,
ρ = 1600 kg/m3
Shear Modulus,
G = 1,0e8 Pa
Poisson's Ratio,
ν = 0,33
Jump to Chart
Fundamentals-Modeling-Properties-Performance
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4Gr
4 1,0 108  r
k

(1  )
0,667
Q0
(1  )Q0 0,667(45 000)1000
 0,05m m 

kz
4Gr
4 1,0 108  r
0.075
r
 1.5m
0.05
Try a 3 x 2,5 x 1 foundation block, r = 1,55 m
Mass = 18 000 kg Total Mass = 18 000 + 5 000 = 23 000 kg
Z static 
m(1  )
mˆ 
4 r 3
mˆ 
(1  ) m 0,67  23 000

 0,65
3
3
4r  4 16001,55
0,425
 1 
 0,53 M z  1,0 

mˆ
 2D 
Z dynamic  Z static  0,05m m
D
Fundamentals-Modeling-Properties-Performance
Jump to Figure
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Design Example - Table Top
5m
Q0=1800 N
m=250 000 kg
Iψ=1,0 x 107 N-m-sec2
10m
X
5m
4m
ψ
DESIGN CRITERION
5.0 mm/sec Horizontal Motion at
Machine Centerline
X = 0,04 mm from combined
rocking and sliding
Speed = 160 rpm
Slower speeds, X can be larger
Soil Properties
Shear Wave Velocity Vs = 200 m/sec
Shear Modulus,
G = 6,80x107 Pa
Density,
γ = 1700 kg/m3
Poisson's Ratio,
ν = 0,33
Fundamentals-Modeling-Properties-Performance
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Horizontal Translation Only
Equivanlent r 
D
lw


10 5

0,288
 0,41  Magx  1,2
1/ 2
mˆ
2  m
 0,49
3
8 r
Q 1800 2  0,33
3
 0 

1
,
3

10
mm
7
kx
8 6,8 10  3,99
 3,99m mˆ 
X static
Rocking About Point "O"
Ax = 40x10-3 mm
lw3 4 10 53
Equivalent r 

 3,39m
  320 rpm  33,5 rad / sec
3
3
8Gr 3
8  6,80107  3,393
k 

 1,0541010 N / rad
3(1  )
3(1  0,33)
4
n 
k
I
1,0541010

 32,4 rad / sec
1,0 107
3(1  ) I
3(0,67) 1,0 107
mˆ  

 3,29
8
r 5
8 1700(3,39)5
0,15
0,15
D 

 0,019  Mag  25,0
(1  mˆ  ) mˆ  (1  3,29) 3,29
Fundamentals-Modeling-Properties-Performance
USC
Static Mom ent About Base  M 0  1800 5  9000 N  m
Static AngularDeflection  s 
Mo
9000
7


8
.
54

10
rad
10
k 1.05410
HorizontalMotion  X   s  h  8.54107  4  3,4 103 m m
At Resonance 25,0(3,4 103 )  85,0 103 m m
X = 40x10-3 mm
Dynamic Magnification (Linear)
X
30
0,02
ψ
Magnification
25
0,05
20
15
0,1
10
0,2
5
0,5
0
0,0
0,5
1,0
1,5
Frequency Ratio ( P / n)
Fundamentals-Modeling-Properties-Performance
2,0
USC
Impedance Methods





Based on Elasto-Dynamic Solutions
Compute Frequency-Dependent Impedance
Values (Complex-Valued)
Solved By Boundary Integral Methods
Require Uniform, Single Layer or Special Soil
Property Distribution
Solved For Many Foundation Types
Fundamentals-Modeling-Properties-Performance
USC
Impedance Functions
P  Poeit  Po cos( t )  i sin(t ) 
Sz


Rz
2K
Sz 
 K  iC  K STATIC  k ( )  i C 
DSOIL 
Az



Soil Damping
Radiation Damping
Jump Wave
Fundamentals-Modeling-Properties-Performance
USC