expo machines vibrating.pdf
Transcript of expo machines vibrating.pdf
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Vibrationsof
Machine Foundations
Richard P. Ray, Ph.D., P.E.Civil and Environmental Engineering
University of South Carolina
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ATST Telescope and FE Model
Fundamentals-Modeling-Properties-Performance
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Summary and Conclusions (Cho, 2005)1.
High fidelity FE models were created
2.
Relative mirror motions from zenith to horizon pointing: about 400 m intranslation and 60 rad in rotation.
3.
Natural frequency changes by 2 Hz as height changes by 10m.
4.
Wind buffeting effects caused by dynamic portion (fluctuation) of wind
5.
Modal responses sensitive to stiffness of bearings and drive disks
6. Soil characteristics were the dominant influencesin modal (dynamic) behavior of the telescopes.7.
Fundamental Frequency (for a lowest soil stiffness):
OSS=20.5hz; OSS+base=9.9hz; SS+base+Coude+soil=6.3hz8.
A seismic analysis was made with a sample PSD
9. ATST structure assembly is adequately designed:1.
Capable of supporting the OSS2.
Dynamically stiff enough to hold the optics stable
3.
Not significantly vulnerable to wind loadings
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Topics for Today
Fundamentals Modeling
Properties Performance
Alapok Modellezs
Tulajdonsgok Gyakorlati
Alkalmazs
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Foundation Movement
Alapok Mozgslehetsgei
X
Z
Y
Fundamentals-Modeling-Properties-Performance
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Design Questions (1/4) Tervezs
How Does It Fail? Static Settlement
Dynamic Motion TooLarge (0.02 mm)
Settlements Caused By
Dynamic Motion Liquefaction
What Are Maximum
Values of Failure?(Acceleration,Velocity,
Displacement)
Hogy rongldik/megytnkre?
Statikus sllyeds Dinamikus mozgs tl
nagy (0,02 mm)
Sllyeds dinamikusmozgs kvetkeztben
Megfolysods
Ronglds maximlisrtkei (gyorsuls,sebessg, eltolds)
Fundamentals-Modeling-Properties-Design-Performance
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Velocity Requirements Sebessg Kvetelmnyek
Massarch
(2004) "Mitigation of Traffic-Induced Ground Vibrations"
Fundamentals-Modeling-Properties-Performance
0,40
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300 800
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Design Questions (2/4) Tervezs
What Are RelationsBetween Loads AndFailure Quantities?
Loads -Harmonic,Periodic, Random
Load Structure
Foundation Soil Neighboring Structures
Model: Deterministic or
Probabilistic
Mi a kapcsolat terhelsis trsi mennyisgekkztt?
Terhelsek- Harmnikus,Peridikus, Vletlenszer
Terhels pletAlapozs Talaj Kzeli
pletek Model: Determinisztikus s
probabilisztikus
Fundamentals-Modeling-Properties-Performance
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Harmnikus
Peridikus
Vletlenszer
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Design Questions (3/4) Tervezs
Hogy hatrozzuk meg a tervezshez szksgesparamtereket? (How do we measure what isnecessary?)
Teljes mretarny teszt (Full scale Test) Prototpus teszt (Prototype Test) Kis mretteszt (Small Scale Tests (Centrifuge))
Laboratriumi teszt (Laboratory Tests (SpecificParameters)) Szmtgpes program (Computer Model)
Fundamentals-Modeling-Properties-Performance
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Design Questions (4/4) Milyen biztonsgi tnyezt hasznljunk? (What
Factor of Safety Do We Use? ) Van a biztonsgi tnyeznek rtelme? (Does FOS Have
Meaning) Mi trtnik trs utn (What Happens After There Is
Failure) letveszts (Loss of Life) Tulajdonvesztd (Loss of Property)
Gyrts kihagys (Loss of Production) Mi a munka clja, tervezett lettartalma, rtke (Purpose of
Project, Design Life, Value)
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r -2 r -2 r -0.5
r-1
r -1
r
Nyrhullm
Verticalcomponent
Horizontalcomponent
Shearwindow
Rayleigh wave
Relativeamplitude+
+
+
+
- -
+
+
Wave TypeHullm tpus
sszes energiaszzalka
Rayleigh 67
Shear 26Compression 7
Waves
Fundamentals-Modeling-Properties-Performance
Compresszi
hullmNyrablak
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Alapok Modellezse (Modeling Foundations)
Egyestett Tmb (m,c,k)Lumped Parameter (m,c,k) BlockSystem
Parameters Constant, Layers, Special Ellenllsi Fggvnyek Impedance Functions
Function of Frequency (), Layers Peremrtk Feladatok Boundary Elements (BEM)
Infinite Boundary, Interactions, Layers Vges Elemes (Finite Element/Hybrid (FEM, FEM-BEM))
Complex Geometry, Non-linear Soil
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Lumped Parameter (Egyestett tmb))sin( tPP o =
m
G
k
m
c
)sin(0 tPkzzczm =++ &&&
r
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Egy szabadsgfok (Single Degree ofFreedom)
)()(
)(
)()(
0
2
Norm
m
NForceSpringzk
Nor
sec
m
m
secNForceDampingzc
Nor
sec
mkgForceInertiazm
zkzczm
z
z
z
zzz
=
=
=
=++
&
&&
&&&
k
m
c
z
Tehetetlensgi er
Rug er
Csillapt Er
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Egy szabadsgfokSingle Degree of Freedom
condependssforsolution
sm
csthen
m
k
setandembydivide
ekcsmswhere
constantezformtakewillsolution
zkzczm
n
n
st
st
st
zzz
0
0)(
.......
0
22
2
2
=++
=
=++
==
=++
&&&
c=0Undamped
c=2mCriticallyDamped
c
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Single Degree of Freedom
)cos()0()sin(
)0(
)(
)0()0(
)(,)cos()sin()()sin()cos('
.).(,)(...0
2121
22
tzt
z
tz
AzandBz
conditioninitialfBAtBtAtz
titeidentitysEuler
condinitfwhereeetz
issm
c
s
nnn
n
n
nnti
titi
nn
n
nn
+
=
==
=+=+=
=+===++
&
&
undamped
)0(z&
z(0)
t
Kiindul felttel
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[ ] tn
t
nncrit
n
n
n
n
etztztz
condinitfwhereettz
m
c
sandmccthenif
m
c
m
c
sthen
presentdampingifsm
c
s
++=
=+======
=
=++
)0()1)(0()(
.).(,)()(220
22
0
2121
22
22
&
critical
Single Degree of Freedom
z(0)
)0(z&
t
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( )
( )
+
+=
+=+=+=
==
===
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Single Degree of Freedom
)sin(0 tPzkzczm Pzzz =++ &&&
k
m
c
)sin( tPP Po =( ) ( ){ }
( )( )
22
2222
0
1
2tan
sin
cos)sin(
==
+
+
+=
n
P
n
P
P
P
P
PP
DD
Dt
D
mk
c
t
cmk
P
tBtAez n
critc
c
D=
m
kn= 21 DnD = kmc
crit 2=
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SDOF tmenti s lland Transient
and Steady-State
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( ) ( )
222
0max
2222
0
21
1
sin)(
+
=
+=
n
P
n
P
P
PP
D
k
P
z
tcmk
P
tz
222
21
1
+
=
n
P
n
P
staticmax
D
zz
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Dynamic Magnification (Logarithmic)
0.1
1
10
100
0.1 1 10
Frequency Ratio (P/
n)
M
agnification
D=0.02D=0.05
D=0.10
D=0.20
D=0.50
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Lumped Parameter System
Kx
Z
KzCz
Cx
K
C
/2 C
/2
X
)sin(0 tPzkzczm Pzzz =++ &&&
mI
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Rendszer paramterek Lumped Parameter Values
Mode Verticalz
Horizontalx
Rocking
Torsion
Stiffness
k
Mass Ratio
mDampingRatio, D
1
4Gr
2
8Gr
)1(3
8 3
Gr
3
16 3Gr
5r
I
3
8
)2(
r
m
34
)1(
r
m
2/1425.0
m2/1
288.0
m2/1)1(
15.0
mm+ m21
50.0
+
m
D=c/ccr
G=Shear Modulus =Poisson's Ratio r=Radius
=Mass Density I
,I
=Mass Moment of Inertia
5
8
)1(3
r
I
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Design Example 1 (Plda)VERTICAL COMPRESSORUnbalanced Forces
(kiegyenslyozatlan erk)
Vertical = 45 kN
Horizontal Primary = 0,5 kNOperating Speed= 450 rpm
Wt Machine + Motor = 5 000 kg
Soil Properties
Shear Wave Velocity Vs
= 250 m/sec
Density,
= 1600 kg/m3
Shear Modulus, G = 1,0e8 Pa
Poisson's Ratio,
= 0,33
DESIGN CRITERION:Smooth Operation At Speed
Velocity
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rGr
Qmm
k
QZ
z
static =
===
800
100,14
1000)00045(667,0
4
)1(05,0
mr 5.105.0
075.0==
( )
mmZZ
DM
mD
m
rm
staticdynamic
z
05,0
2
10,153,0
425,0
65,055,116004
0002367,0
4
)1(
33
==
==
=
=
=
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
Jump to Figure
Fundamentals-Modeling-Properties-Performance
667,0
100,14
)1(
4 8 rGrk
==
34
)1(
r
mm
=
tmeg
tmbalaptest
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Horizontal Translation Only
mmk
QXMag
mD
r
mmm
lwrtEquivanlen
x
staticx
37
02/1
3
103,199,3108,6
33,02
8
18002,141,0
288,0
49,0
8
299,3
510
=
====
===
==
Rocking About Point "O"
0,25019,0
29,3)29,31(
15,0
)1(
15,0
29,3)39,3(1700
100,1
8
)67,0(3
8
)1(3
/4,32100,1
10054,1
/10054,1)33,01(3 39,31080,68)1(38
/5,3332039,33
510
3
5
7
5
7
10
10
373
4
3
4
3
==
+
=
+
=
=
==
=
==
= ==
===
==
Mag
mm
D
r
Im
secradI
k
radNGrk
secradrpmmlw
rEquivalent
n
Fundamentals-Modeling-Properties-Performance
Ax = 40x10-3
mm
csak vzszintes mozgs
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mmResonanceAt
mmhXMotionHorizontal
rad
k
MDeflectionAngularStatic
mNMBaseAboutMomentStatic
s
os
33
37
710
0
100,85)104,3(0,25
104,341054.8
1054.8
10054.1
9000
900051800
==
====
=
===
===
Fundamentals-Modeling-Properties-Performance
X
X = 40x10-3
mm
Dynamic Magnification (Linear)
0
5
10
15
20
25
30
0,0 0,5 1,0 1,5 2,0Frequency Ratio ( P/ n)
Magnification
0,02
0,05
0,1
0,2
0,5
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Impedance Methods
Based on Elasto-Dynamic Solutions
Compute Frequency-Dependent ImpedanceValues (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
Ellenllsi fggvnyek
Frekvencitl fggellenllsi rtkek
Peremrtk integrl mdszer
Egyenletes, egy rteg, specilistalajrtk eloszls szksges
Tbbfajta alap tpusra megoldott
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Impedance Functions
( ))sin()cos( titPePP oti
o +==
( )
++=+== SOILSTATIC
z
zz D
KCikKCiK
A
RS
2)(
Radiation DampingSoil Damping
Jump Wave
Sz
Fundamentals-Modeling-Properties-Performance
Energia csillaptsTalaj csillapts
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Impedance Functions
Luco
and Westmann
(1970)
sVr
Gra ==0
Fundamentals-Modeling-Properties-Performance
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Impedance Functions
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Boundary Element
Stehmeyer
and Rizos, 2006
Fundamentals-Modeling-Properties-Performance
Peremrtk feladatok
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B-Spline Impulse Response Approach
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[ ]{ } [ ]{ } { } tie puKuM =+&&
Fundamentals-Modeling-Properties-Performance
{ } { }
[ ] [ ]{ }{ } { }pUMK
Uu
=
=
2
thene ti
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G1
,1
,1
u1
u2
u7
u8
[ ] ),,(1111
GfnK =
8
7
2
1
8,87,82,81,8
8,77,72,71,7
8,27,22,21,2
8,17,12,11,1
u
u
u
u
kkkk
kkkk
kkkk
kkkk
[ ] )( 11 fnm =
linear
linear
dcxyybxau iii
==
+++=
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[ ]{ } [ ]{ } { } tie puKuM =+&&
tie
p
p
p
pp
u
u
u
uu
kkk
kkkk
kkkkk
kkkkkkk
u
u
u
uu
m
m
m
mm
=
+
5
4
3
2
1
5
4
3
2
1
5,54,53,5
5,44,43,42,4
5,34,33,32,31,3
4,23,22,21,2
3,12,11,1
5
4
3
2
1
5
4
3
2
1
&&
&&
&&
&&&&
{ } { } { } { }[ ] [ ]{ }{ } { } { }
[ ]{ } valuedcomplexare
forsolvegiven
andethenezif titi
===
ZK
ZpZMKZzZ
,
,22
&&
22 1221* DiDDGG +=
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C h l T tiOscilloscope
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Crosshole TestingOscilloscope
PVC-casedBorehole
PVC-casedBorehole
Downhole
Hammer
(Source) VelocityTransducer
(GeophoneReceiver)
t
x
Shear Wave Velocity:Vs
= x/t
Test
Depth
ASTM D 4428
Pump
packer
Note: Verticality of casingmust be established byslope inclinometers to correct
distances x with depth.
Slope
Inclinometer
Slope
Inclinometer
R C l T
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Resonant Column Test
G, D for Different
Fundamentals-Modeling-Properties-Performance
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Torsional
Shear Test
Schematic Stress-Strain
Fundamentals-Modeling-Properties-Performance
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Hollow Cylinder RC-TOSS
Fundamentals-Modeling-Properties-Performance
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TOSS Test Results
Fundamentals-Modeling-Properties-Performance
Steam Turbine-Generator
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Steam Turbine-Generator (Moreschi
and Farzam, 2003)
Fundamentals-Modeling-Properties-Performance
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Machine Foundation Design Criteria
Deflection criteria: maintain turbine-generatoralignment during machine operating conditions
Dynamic criteria: ensure that no resonancecondition is encountered during machineoperating conditions
Strength criteria: reinforced concrete design
Fundamentals-Modeling-Properties-Performance
Jump to Resonance
El/kihajlsi kritrium: a turbina-genertor szintben maradjon mkdse alatt
Dinamikus felttel: nincs rezonancia a gp mkdse alatt
Erssgi felttel: elfesztett beton tervezs
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STG Pedestal Structure
Fundamentals-Modeling-Properties-Performance
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Vibration Properties Evaluation
Identification of the foundation natural
frequencies for the dominant modes Frequency exclusion zones for the natural
frequencies of the foundation system andindividual structural members (20%)
Eigenvalue analysis: natural frequencies,mode shapes, and mass participationfactors
Fundamentals-Modeling-Properties-Performance
Az alap sajt frekvenciinak meghatrozsa a dominns lengsekre/mdokra
Kihagysi frekvencia znk a termszetes frekvencikra
Eigenrtk
elemzs:termszetes frekvencik,
lengsmdnl alakok, tmeg egyttdolgozsa
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XY
Z
XY
Z
Finite Element Model
Structure and Base
Fundamentals-Modeling-Properties-Performance
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Low Frequency Modes
1st mode6.5 Hz95 % m.p.f.
2nd
mode
7.2 Hz76 % m.p.f
Fundamentals-Modeling-Properties-Performance
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High Frequency Modes
28th mode46.3 Hz0.3% m.p.f
42nd mode64.6 Hz0.03% m.p.f
Excitation frequency: 50-60 HzFundamentals-Modeling-Properties-Performance
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Local Vibration Modes
Identification of naturalfrequencies for individual
structural members
Quantification of changeson vibration properties due
to foundation modifications
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Assumptions in FE analyses
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Optics Lab mass/Instrument weight = 228 tons Wind mean force = 75 N, RMS = 89 N
Ground base excitation PSD = 0.004 g2
/hz Concrete Pier
High Strength Concrete (E=3.11010
N/m2
,=0.15)
Soil Stiffness, k Four different values using Arya & ONeils
formula based on the site test data (Shear
modulus:30~75ksi, Poissons ratio:0.35~0.45)
Assumptions in FE analyses
Fundamentals-Modeling-Properties-Performance
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Soil property range: Shear modulus (30~75ksi), Poissons ratio (0.35~0.45)
Pier Footing: Diameter (23.3m)
min
for shear modulus of 30 ksi; max
for 75 ksi
Frequency vs
Soil Stiffness
Stiffness m i n m in +33.3% m in +66.6% m ax
Kx 1.19E+10 1.83E+10 2.48E+10 3.12E+10
Ky 1.19E+10 1.83E+10 2.48E+10 3.12E+10
Kz 1.48E+10 2.45E+10 3.41E+10 4.38E+10Krx 1.34E+12 2.21E+12 3.09E+12 3.96E+12
Kry 1.34E+12 2.21E+12 3.09E+12 3.96E+12
Krz 1.74E+12 2.61E+12 3.49E+12 4.36E+12
6.3 7.0 7.4 7.5
6.4 7.1 7.5 7.7
9.4 9.7 9.9 10
9.4 10.3 11.1 11.8
10.4 11.9 12.6 13.3
11.2 13.0 13.6 13.7
4
5
6
MODE
1
2
3
Stiffness units = SI, frequency mode (hz)
Fundamentals-Modeling-Properties-Performance
d l i ( h )
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Summary and Conclusions (Cho, 2005)
1.
High fidelity FE models were created
2.
Relative mirror motions from zenith to horizon pointing: about 400 m intranslation and 60 rad in rotation.
3. Natural frequency changes by 2 hz as height changes by 10m.4. Wind buffeting effects caused by dynamic portion (fluctuation) of wind
5.
Modal responses sensitive to stiffness of bearings and drive disks
6. Soil characteristics were the dominant influences in modal
behavior of the telescopes.7.
Fundamental Frequency (for a lowest soil stiffness):
OSS=20.5hz; OSS+base=9.9hz; SS+base+Coude+soil=6.3hz8.
A seismic analysis was made with a sample PSD
9.
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
F Fi ld A l ti l S l ti
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Free-Field Analytical Solutions
=
R
VzCrHaRVLiru
2003
0 )(2
)0,,(
=
RVr C
rHaR
VMiru
2
103
0 )(2
)0,,(
uruz
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USCUSCUSCFundamentals-Modeling-Properties-Performance
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USCUSCUSCFundamentals-Modeling-Properties-Performance
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Karlstrom
and Bostrom
2007
Trench
Isolation
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Chehab
and Nagger 2003
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Celibi
et al (in press)
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Thank-you Questions?
2 2 0 5
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r -2 r -2 r -0.5
r -1
r -1
r
Shearwave
Verticalcomponent
Horizontalcomponent
Shearwindow
Rayleigh wave
Relativeamplitude+
+
+
+
- -
+
+
Wave Type Percentage ofTotal Energy
Rayleigh 67
Shear 26
Compression 7
Waves
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Waves
Rayleigh, R
Surface
Shear,S Secondary
Compression, PPrimary
Machine Performance Chart
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Machine Performance Chart
Performance ZonesA=No Faults, New
B=Minor Faults,
Good Condition
C = Faulty, CorrectIn 10 Days To Save
$$D = Failure Is Near,Correct In 2 Days
E = Stop Now
0.002