bray ucanterbury seismic displ
TRANSCRIPT
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Simplified Seismic SlopeDisplacement Procedures
Jonathan D. Bray, Ph.D., P.E.
Univ. of California at Berkeley
Thanks to Dr. T. Travasarou, Prof. E. Rathje, & others,
with support from PEER & the Packard Foundation
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Simplified Seismic Slope Displacement Procedures
OUTLINE
I. Introduction
II. Seismic Slope Displacement Analysis
a. Earthquake Ground Motion
b. Dynamic Resistance
c. Dynamic Response
d. Seismic Displacement Calculation
III. Simplified Slope Displacement Procedures
IV. Conclusions
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I. Seismic Slope Stability
4th Ave. Slide, 1964 Alaska EQ 1999 Chi-Chi, Taiwan EQ
EERC Slide Collecti on
Landslide Stabilization
CSASolid-Waste Landfi ll
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Two Critical Design Issues
1. Are there materials that will lose significantstrength as a result of cyclic loading?
Flow Slide
2. If not, will the system undergo significant
deformations that may jeopardize system
performance?Seismic Displacements
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Seismic Slope Displacement
Slip along a distinct surface
Distributed shear deformation
Add volumetric-induced deformation, when appropriate
Newmark-type
seismic displacement
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FEA of Shaking Table Test of Clay Slope
0 2 4 6 8 10-2.5
-2
-1.5
-1
-0.5
0
0.5
Time (sec)
Displacement(inches)
Horizontal displacement (D2)
PLAXIS
RECORD
0 2 4 6 8 10-3
-2
-1
0
1
Displacement(inches)
Horizontal displacement (D4)
0 2 4 6 8 10-1.5
-1
-0.5
0
0.5
1
Time (sec)
Horizontal displacement (D5)
0 2 4 6 8 10-1.5
-1
-0.5
0
0.5Vertical displacement (D1)
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II. Seismic Displacement Analysis:
Critical Components
a. Earthquake Ground Motion
b. Dynamic Resistance
c. Dynamic Response
d. Seismic Displacement Calculation
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a. Earthquake Shaking:
Acceleration Time History
acceleratio
n(g)
-0.50
-0.25
0.00
0.25
0.50
time (s)
0 5 10 15 20 25 30
Izmit (180 Comp) 1999 Kocaeli EQ (Mw
=7.4) scaled to MHA = 0.30 g
PGA = 0.3 g Tm = 0.63 s & D5-95 = 15 s
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Acceleration Response Spectrum(provides response of SDOF of different periods at 5% damping,
i.e., indicates intensity and frequency content of ground motion)
0 1 2 3 4Period s
0.0
0.5
1.0
1.5
Spe
ctralAcceleration(g)
5% Damping
Sa(0.5)
Sa(1.0)
PGA
Sa(0.2)
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b. Dynamic Resistance:Simplified Estimates of Yield Coefficient (ky)
(seismic coefficient that results in FS=1.0 in pseudostatic stabili ty analysis)
H
c = cohesion
= friction angle
kc
Hy= +
+ tan( )
cos ( tan tan )
2 1
1S2
1 H
L
S1
kFS S H
H S S Ly
static=
+ +
( ) cos sin
( )
1 2
2
1 1 1
1 2
( )FS
S H L S H
S Hstatic
= + +
tan cos
cos sin
1
2
1 2
1 1 1
2 2
2
with 11
11=tan ( )S
Shallow SlidingDeep Sliding
Shewbridge 1996Bray et al. 1998
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Factors Affecting
Dynamic Strength of Clays
Rate of loading: Crate > 1
Number of significant cycles: Ccyc < 1
Progressive failure: Cprog < 1
Sdynamic = Sstatic (Crate) (Ccyc ) (Cprog)
Also, post-peak strength reduction if EQ-induced strain exceeds strain at peak stress
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Vane Shear and Direct Shear tests
Small vane 2.5 cm diameter
Large vane 5.7 cm diameter
Large direct shear box 30 cm
by 30 cm square
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-10
0
10
20
30
40
50
60
70
displacement (in)
shearstress(p
sf)
original vane shear data
derived from vane shear
direct shaer data
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Seed and Martin 1966
c. Dynamic Response of Sliding Mass
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Dynamic Response: Equivalent Acceleration Concept
accounts for cumulative effect of incoherent motion in deformable slid ing mass
Horz. Equiv. Accel.: HEA = (h/v) g MHEA = max. HEA value
kmax = MHEA / g
H vh
Seed and Mart in 1966
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0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
Ts-waste/Tm-e
0.00.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
M
HEA/(MH
A,rock*NR
F) Rock site median, and 16th and 84th
0.1 1.350.2 1.200.3 1.09
0.4 1.000.5 0.920.6 0.870.7 0.820.8 0.78
MHA,rock (g) NRF
probability of exceedance lines
(Bray & Rathje 1998)
FACTORS AFFECTING MAXIMUM SEISMIC LOADING
kmax
/(MHA
roc
k
*NRF/g)
Ts-sliding mass / Tm-EQ
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kmax depends on stiffness and geometry of the
sliding mass (i.e., its fundamental period)
Ts,1-D = 4 H / Vs
Ts , 1-D = Initial Fundamental Period of Sliding Mass
H = Height of Sliding Mass
Vs = Average Shear Wave Velocity of Sliding Mass
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d. Seismic Displacement CalculationNewmark (1965) Rigid Sliding Block Analysis
Assumes:
Rigid sliding block
Well-defined slip surface develops
Slip surface is rigid-perfectly plastic
Acceleration-time history defines EQ loading
Key Parameters:
ky - Yield Coefficient (max. dynamic resistance)
kmax - Seismic Coefficient (max. seismic loading)
ky /
kmax (if > 1, D = 0; but if < 1, D > 0)
C
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Rigid
Sliding
Block:
use
accel.-
time hist.
kmax = PGA/g
Cover
Base
Rock
Deformable
Sliding
Block:
use equiv.
accel.-time
hist.
kmax = MHEA/g
Mid-Ht
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Decoupled & Coupled
Deformable Sliding Analysis
Earth Fill
Potential Slide Plane
Decoupled
Analysis
Coupled
Analysis
Flexible System
Dynamic Response
Rigid Block
Sliding Response
Flexible SystemFlexible System
Dynamic Response and
Sliding Response
Max Force at
Base = ky W
Calculate HEA-
time history
assuming nosliding along base
Double integrate
HEA-time
history given kyto calculate U
0.1 1 10 100 1000
U (cm)
-40
-20
0
20
40
60
80
100
DisplacementDifference(cm):
U
-U
decoupled
decoupled
co
upled
k = 0.05
k = 0.1
k = 0.2
y
y
y
(b)
Decoup
ledConserva
tive
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Calculated Seismic Displacement
Expected ky
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SAFE
UNSAFE
?
Think About It as a Cliff
Calculated Seismic Displacement is an Index of Performance
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Evaluate Seismic Performance
Given seismic displacement estimates:
Small (e.g., D 15 cm)
Moderate
Large (e.g., D 1 m)
Evaluate the ability of earth structure and systems founded on it to
accommodate this level of deformation.
Consider:
Consequences of failure and conservatism of hazard assessmentand stabili ty analyses, e.g., the consequences of shallow slid ingare usually less severe than deep sliding & more easily repaired.
Defensive measures that provide "ductility" or " shield" criticalcomponents from damage, e.g., slip layer above critical linersystem.
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III. Simplif ied Seismic Slope Procedures:
Makdisi & Seed (1978)
1. Evaluate materials strength loss potential
If significant, do not use this method
If slight, use slightly reduced strength (in many
cases, use ~ 10% to 20% strength reduction)
2. Calculate ky
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3. Estimate max. crest acceleration
MHA at Top vs. Base Rock MHA for Some Solid-Waste Landf ills(Bray & Rathje 1998)
MHA,top = ?
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4. Estimate kmax for sliding mass using plot of
max. accel. ratio vs. (y/h)
Kmax varies with Ts
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5. Estimate seismic displacement (U) basedon ky/kmax & EQ Magnitude
WARNING: Do not use
figure in original
Makdisi & Seed (1978)
paper; it is off by anorder of magnitude
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Limitations It must be noted that the design curves are derived from alimited number of cases. These curves should be updated and
refined as analyt ical results for more embankments are obtained.Makdisi & Seed (1978)
Very limited number of earthquake ground motions used.
Estimating MHA at the crest to estimate kmax is problematic.
Relatively simple analysis employed, e.g., shear slice method.
Decoupled analysis to calculate seismic displacement.
Upper and lower bounds are not t rue upper and lower bounds.
No estimate of uncertainty.
B & T (2007)
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INTENSITY MEASURES, e.g.:
amplitude: PGA, PGV, PGD, Sa, Svfrequency content: Tp, Tm
duration: D5_95,combination: Arias Intensity
Housner Spectral Intensity
Bray & Travasarou (2007)
1. SLOPE MODEL
nonlinear soil response
fully coupled deformable stick-slip
stiffness (Ts) & strength (ky)
8 Ts values & 10 ky values Tsky
D2. EARTHQUAKE DATABASE41 EQ - 688 records
Over 55,000 analyses
EFFICIENCY
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EFFICIENCY
of Ground Motion
Intensity Measures
STIFF WEAK SLOPE
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PGASA
SA(1.3Ts)
SA(1.5Ts)
SA(2.0Ts)
arms
AriasIc
PGVEPVSI
PGDTm
D5_
95
Intensity Measure
0
1
2
S
tandardDeviatio
n
0
1
2
StandardD
eviation
ky = 0.10 Ts = 0.5 s
ky = 0.20 Ts = 0.5 s
0
1
2
PGASA
SA(1.3Ts)
SA(1.5Ts)
SA(2.0Ts)
arms
Arias Ic
PGV
EPVSI
PGDTm
D5_
95
Intensity Measure
0
1
2 ky = 0.20 Ts = 1.0 s
ky = 0.10 Ts = 1.0 s
ST
R
O
NG
E
R
MORE FLEXIBLE
EFFICIENCY RESULTS
Sa (1.5Ts) Arias Intensity Housner Spectral Intensity
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(1) Model for D = 0 P (D = 0 ) = f(Sa(1.5Ts), ky, Ts)
Note: D = 0 is negligible displacements, i.e. D < 1 cm
MIXED RANDOM VARIABLE APPROACH
0.01 0.1 1
Yield Coefficient
0.00
0.25
0.50
0.75
1.00
P(D
="0")
0 1 2 3
Fundamental Period (s)
0.00
0.25
0.50
0.75
1.00
P(D="0")
0.001 0.01 0.1 1 10
SA(1.5Ts) (g)
0.00
0.25
0.50
0.75
1.00
P(D="0")
(a) (b) (c)
MIXED RANDOM VARIABLE APPROACH
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(2) Model for D > 0 Ln (D) = f(Sa(1.5Ts), ky, Ts, M) + [ = 0.66]MIXED RANDOM VARIABLE APPROACH
B & T (2007) S i i Di l t
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1) Zero Displacement Estimate
( ) ++= ))5.1(ln()ln(566.0)ln(333.0)ln(83.210.1)ln( 2 sayyy TSkkkD
))5.1(ln(52.3)ln(484.0)ln(22.376.11)"0"(sasyy TSTkkDP +==
2) NonZero Displacement Estimate
( ) ++ )7(278.050.1))5.1(ln(244.0))5.1(ln(04.3 2 MTTSTSssasa
is the standard normal cumulative distribution function (NORMSDIST in Excel)
Bray & Travasarou (2007) Seismic Displacement
is a normally-distributed random variable with zero mean and standard deviation = 0.66
*Replace first term (i.e., -1.10) with -0.22 for cases where Ts < 0.05 s
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MODEL TRENDS
0 0.5 1 1.5 2
Ts (s)
0
0.5
1
SA(1.5Ts)(g) 5% damping
Scenario Event:
M 7 at 10 km
Soil SS fault
0 0.5 1 1.5 2
Ts (s)
0
0.2
0.4
0.6
0.8
1
P(D="0")
ky= 0.3
ky= 0.2
ky= 0.1
ky= 0.05
0 0.5 1 1.5 2
Ts (s)
1
10
100
Displac
ement(cm) ky= 0.05
ky= 0.1
ky= 0.2
ky= 0.3
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PROBABILITY OF EXCEEDANCE
0 0.5 1 1.5 2
Ts (s)
0
0.2
0.4
0.6
0.8
1
P(D
="0")
ky= 0.3
ky= 0.2
ky= 0.1
ky= 0.05
0 0.5 1 1.5 2
Ts (s)
1
10
100
Displace
ment(cm) ky= 0.05
ky= 0.1
ky= 0.2
ky= 0.3
0 0.5 1 1.5 2
Ts (s)
0
0.2
0.4
0.6
0.8
1
P(D
>30cm)
ky= 0.3ky= 0.2
ky= 0.1
ky= 0.05
Scenario Event:
M 7 at 10 km
Soil SS fault
Validation of Bray & Travasarou Simplif ied Procedure
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System EQ
Obs.
Dmax
(cm)
Bray & Travasarou 2007
P (D = 0 ) Est. Disp (cm)
Makdisi &
Seed 1978
D (cm)
Bray &
Rathje 1998
D (cm)
BuenaVista LF LP None 0.75 0.8 - 3 0 0
Guadalupe LF LP Minor 0.95 0.3 - 1 0 0 4
Pacheco Pass LF LP None 1.0 0 - 0.1 0 0
Marina LF LP None 0.9 0.5 - 2 0 0
Austrian Dam LP 50 0.0 20 - 70 1 30 20 100
Lexington Dam LP 15 0.0 15 - 65 0 10 30 110
Chiquita Canyon C LF NR 24 0.0 10 - 30 1 40 3 20
Chiquita Canyon D LF NR 30 0.0 6 - 22 0 10 2 15
Sunshine Canyon LF NR 30 0.0 20- 70 0 0
OII Section HH LF NR 15 0.1 4 - 15 3 30 2 25
La Villita Dam S3 1 0.95 0.3 - 1 0 0
La Villita Dam S4 1.4 0.5 1 - 5 0 0
La Villita Dam S5 4 0.25 3 - 10 0 1 0
Validation of Bray & Travasarou Simplif ied Procedure
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Hypothetical Example: 57 m-High Earth Dam
Located 1.1 km from Hayward Fault
s
sm
m
V
HT
s
s 33.0
/450
576.26.2
=
=
No liquefaction or soils that will undergo significant strength loss
Undrained Strength: c = 14 kPa and = 21o so ky = 0.14Triangular Sliding Block with avg. Vs ~ 450 m/s & H = 57 m
Deterministic Analysis:
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Deterministic Analysis:
Mean Max. Mw ~ 6.9 at R ~ 1.1 km; Ts = 0.33 s; & ky = 0.14
Using 1.5 Ts = 1.5 (0.33 s) = 0.5 s & Abrahamson & Silva (97) &Sadigh et al. (97) rock attenuation relations for strike-slip fault:avg. Sa(0.5 s) = 1.07 g
Probability of Zero Displacement (i.e., D < 1 cm):
Nonzero Seismic Displacement Estimate:
Design Seismic Displacement Estimate (16% and 84 %):
D ~ 0.5 to 2 times median D = 20 cm to 80 cm
( ) 0)07.1ln(52.3)33.0)(14.0ln(484.0)14.0ln(22.376.11)"0"( =+==DP
( ) ++= )07.1ln()14.0ln(566.0)14.0ln(333.0)14.0ln(83.210.1)ln( 2D
( ) ++ )79.6(278.0)33.0(50.1)07.1ln(244.0)07.1ln(04.3 2
cmDD 40)77.3exp())exp(ln( ==
For fully probabilist ic implementation see Travasarou et al. 2004
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Pseudostatic Slope Stability Analysis
1. k = seismic coefficient; represents earthquake loading
2. S = dynamic material strengths
3. FS = factor of safety
Selection of acceptable combination ofk, S, & FS
requires calibration through case histories or
consistency with more advanced analyses
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A Prevalent Pseudostatic Method
Seed (1979)
appropriate dynamic strengths
k = 0.15
FS > 1.15
FS > 1.15 does not mean the system is safe!
BUT this method was calibrated for earth dams where1 m of displacement was judged to be acceptable
WHAT about other systems?
WHAT about other levels of acceptable displacement?
IS k = 0.15 reasonable for all sites?
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Seismic Coefficient
k should depend on level of shaking, i.e.,distance to fault, earthquake magnitude,
dynamic response of earth structure
k should also depend on crit icality of structure
and acceptable level of seismic performance, i.e.,
amount of allowable seismic displacement
How does one then select k?
S i i C ffi i f All bl Di l
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Calculate k as function ofDa, Sa, Ts, Mw, &
Seismic Coefficient from Allowable DisplacementBray &Travasarou (2009)
+
Seismic coefficient, k:
Seismic Coefficient depends on:
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Ts= 0.2s
0.00
0.10
0.20
0.30
0.40
0 0.5 1 1.5 2
Sa(@1.5Ts) (g)
S
eismicCoefficient
Seismic Coefficient depends on:
Da, EQ (Sa & Mw), & Ts
5 cm
15 cm
30 cm
M=7.5 M=6
Pseudostatic Slope Stability Procedure
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1. Are there materials that could lose significant
strength? If so, use post-shaking reduced strengths.Otherwise, use strain-compatible dynamic strengths
2. Select allowable displacement: Da and select
exceedance probability (e.g., use = 0 for 50%)3. Estimate initial period of sliding mass: Ts
4. Characterize seismic demand: Sa(1.5T
s) & Mw
5. Calculate seismic coefficient: k = f(Da, Sa, Ts, Mw & )& perform pseudostatic analysis using k. If FS 1,
then Dcalc < Da at specified level of exceedance
Pseudostatic Slope Stability Procedure
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Simplified Seismic Coefficient Estimates(used in code applications, e.g., B.C., Canada)
Stiff Slopes: Ts 0.1 s & Using: Da = 15 cm
For M 7.5:
k15 cm 0.25 Sa 0.25 (2.5 PGA) 0.65 PGA
Conclusions
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Conclusions
First question: will materials lose strength?
If not, evaluate seismic slope stabil ity interms of seismic displacements
Bray & Travasarou (2007) approach withdeformable sliding mass captures:
a. Earthquake shaking - Sa(1.5 Ts) & Mw
b. Dynamic resistance of slope - ky
c. Dynamic response of sliding mass - Tsd. Coupled seismic displacement - D
Earthquake and material characterization
are most important