bray ucanterbury seismic displ

7/27/2019 Bray UCanterbury Seismic Displ

1/46

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


2/46

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


3/46

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


4/46

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


5/46

Seismic Slope Displacement

Slip along a distinct surface

Distributed shear deformation

Add volumetric-induced deformation, when appropriate

Newmark-type

seismic displacement


6/46

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)


0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

Time (sec)


0 2 4 6 8 10-1.5

-1

-0.5

0

0.5Vertical displacement (D1)


7/46

II. Seismic Displacement Analysis:

Critical Components

a. Earthquake Ground Motion

b. Dynamic Resistance

c. Dynamic Response

d. Seismic Displacement Calculation


8/46

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


9/46

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)


10/46

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


11/46

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


12/46

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


13/46

Seed and Martin 1966

c. Dynamic Response of Sliding Mass


14/46

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


15/46

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


16/46

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


17/46

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


18/46

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


19/46

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


20/46

Calculated Seismic Displacement

Expected ky


21/46

SAFE

UNSAFE

?

Think About It as a Cliff

Calculated Seismic Displacement is an Index of Performance


22/46

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.


23/46

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


24/46

3. Estimate max. crest acceleration

MHA at Top vs. Base Rock MHA for Some Solid-Waste Landf ills(Bray & Rathje 1998)

MHA,top = ?


25/46

4. Estimate kmax for sliding mass using plot of

max. accel. ratio vs. (y/h)

Kmax varies with Ts


26/46

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


27/46

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)


28/46

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


29/46

EFFICIENCY

of Ground Motion

Intensity Measures

STIFF WEAK SLOPE


30/46

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


31/46

(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


32/46

(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


33/46

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


34/46

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


35/46

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


36/46

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


37/46

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:


38/46

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


39/46

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


40/46

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?


41/46

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


42/46

Calculate k as function ofDa, Sa, Ts, Mw, &

Seismic Coefficient from Allowable DisplacementBray &Travasarou (2009)

+

Seismic coefficient, k:

Seismic Coefficient depends on:


43/46

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


44/46

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


45/46

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


46/46

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

bray ucanterbury seismic displ

Documents