lecture35 seismic slope stability part2

8/12/2019 Lecture35 Seismic Slope Stability Part2

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Seismic Slope Stability Analysis

Part - II

Lecture-35

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Newmarks rigid Block Analysis Newmark (1965) developed a simple procedure to estimate permanent

slope displacement due to earthquake shaking by extending thepseudostatic approach to a sliding rigid block and considering theacceleration-time history of the sliding mass within the slope

The method assumes that permanent displacement at some well-defined slip surface begins when the inertial forces induced by theseismic excitation exceed the resisting forces that result from the fullmobilization of interfacial shear strength along the potential failuresurface

Since the servicability of the slope after the earthquake depends on thedeformations induced in the slope during earthquake, this method iscomparatively more useful than the pseudostatic analysis.

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Newmarks rigid block analysis

Fig: Analogy between (a) potential landslide and (b) rigid block sliding on a plane

Newmark considered the behaviour of sliding slope analogous to a rigidblock sliding on an inclined plane. This analogy is used to compute the permanent deformations in the

slope for any specific ground motion.

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Fig: Forces acting on a rigid block resting on an inclined plane under(a) Static conditions (b) Dynamic conditions

Under static conditions, equilibrium of the block requires that the available staticresisting force R s exceed the static driving force D s. Assuming that the blocksresistance is purely frictional (c=0),

Where f is the angle of friction between the block and the plane.

f

f

tantan

sintancos

W W

DR

FOSs

s

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Under dynamic conditions, considering the effect of inertial forces transmitted tothe block by the horizontal vibration of the inclined plane with acceleration a h(t) =kh(t)g (neglecting the effect of vertical accelerations), resolving the forcesperpendicular to the inclined plane gives the dynamic factor of safety at time t as

Where f is the angle of friction between the block and the plane.

f

cos)(sintan)sin)((cos

)()(

)(t k

t k t Dt R

t FOSh

h

d

d d

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Because of inertia, the block will tend to move in a direction opposite to theacceleration of the base. However, the block will begin to move relative to thebase only when the sum of the static and dynamic driving forces exceed theresisting forces. This equilibrium condition, when the block begins to slide, isexpressed in terms of a yield acceleration (a y)


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The dynamic factor of safety decreases as the horizontal seismic coefficient k h increases.

The value of k h for which the factor of safety becomes exactly 1.0 is termed asyield coefficient k y .

Yield acceleration a y (=ky g) is the minimum pseudostatic acceleration required toproduce instability of the block.

Substituting 1.0 for FOS in the above equation, we get

f

cos)(sin

tan)sin)((cos

)(

)()(

t k

t k

t D

t R t FOS

h

h

d

d d

)tan(tantan1

tantan f

f f

y k

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Fig: Variation of pseudostatic factor of safety with horizontal seismic coefficient

for a rigid block on a plane inclined at 20

Note: For f of 20 , factor of safety is 1.0 and the block is at the point of failureunder static conditions hence, k y is zero. For f of 30 and 40 , k y is 0.17 and0.36 respectively.

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Source: Kramer (1996)


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Computing permanent displacements

Fig: Computing permanent displacements in Newmarks rigid block analysis9


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Given the yield acceleration, the total accumulated displacement is

computed by double integration of the base acceleration record asshown in figure in the previous slide.

When the base acceleration exceeds the yield value, the block slidesand the first integration gives the velocity of the block. The velocity

reaches a peak after the base acceleration reverses direction andeventually the velocity declines to zero.

The second integration, of the velocity record, produces the netdisplacement during each pulse where the yield acceleration issurpassed. Following this procedure for the entire base motion recordyields the total, accumulated displacement.


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Influence of a y on permanent displacements

When soil slopes are considered, the yield acceleration is computed as a

function of the shear strength of the soil and the failure mechanism or slipsurface.

In a conventional Newmarks analysis, sliding is assumed to exhibit a rigid,perfectly-plastic response implying a yield acceleration that doesnot change with displacement.

Sliding block method predicts zero displacement if earthquake inducedaccelerations do not exceed the yield acceleration.

Since the permanent displacements are obtained by double integration ofexcess acceleration, the computed displacements for slope with low yieldacceleration will be greater than the displacements for slope with highyield acceleration.

The relation between the slope displacements and a y/a max is investigated

by several researchers. 11


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09.1

max

53.2

max1log90.0log a

a

a

aD

y y

where D is the predicted displacement (cm), a y is the yield acceleration(g) and a max is the peak horizontal acceleration of the designearthquake

Ambraseys and Menu (1988)

Ambraseys and Menu (1988) found a simple relation to find out thepermanent displacement of the slope D as

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Yegian et al. (1991)

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546.1642.6log460.1log y a aI DJibson (1993)

Jibson (1993) proposed a simpler approach where the ground motion

record is represented by a single, quantitative measure of thetotal shaking intensity. Using the Arias intensity ( Ia=[ p /2g]a 2dt ) torepresent shaking intensity, and analyzing several strong motionrecords, Jibson developed a relationship for displacementspredicted by a Newmark-type analysis.

where D is the predicted displacement (cm), Ia is the Arias intensity(m/s), and a y is the yield acceleration ( g). Thus, displacements for agiven yield acceleration can be estimated directly from Arias intensity,which can be estimated from other empirical attenuation equations.

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Induced Deformations

For a rigid block, the accelerations are equal at all times to the groundacceleration until the slip phase of motion; therefore, the acceleration-time history of an earthquake motion in excess of the yield accelerationmay be directly integrated twice to obtain the relative displacement

Difficulty is selecting an appropriate accelerogram that simulates motionsin the slide mass and, more importantly, accounting for the deformablenature of the soil embankment and waste fill materials

Materials that comprise most slopes are compliant rather than rigid,

especially waste fill materials

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The applicability of Newmarks sliding block analysis to liquefaction

induced lateral spreading is inhibited by several shortcomings:

1. An obvious problem is defining an appropriate shear strengthfor a liquefiable soil. In an approximate manner, this problemcan be addressed with undrained shear strengths, although thisapproach requires special soil testing and is not applicable to allsoil conditions.

2. The lack of a well defined slip surface in a lateral spreadconfounds the definition of a simple yield acceleration. Inreality, the "yield acceleration" in a lateral spread will changedramatically with the occurrence of liquefaction andsubsequent deformations.

3. Lateral spreading can continue after earthquake motions stopand a Newmark-type analysis is incapable of modeling these

deformations

Liquefaction Induced Deformations

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Decoupled Procedures

The decoupled procedure was originally proposed by Makdisi & Seed in 1978, andis typically used for more critical civil structures such as earth dams.

Decoupled methods can be thought of as a modified rigid-block analysis thatincorporate dynamic response in a two -step procedure (shown in figure)

where the dynamic response of the slope is analyzed separately from the slidingresponse.

Unlike, rigid-block methods, here the slide mass is assumed to be responsive todynamic loading.

Step 1:

Dynamic-response analysis

Step 2:

Sliding-response analysis17


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Coupled ProceduresAs opposed to the original Newmark (1965) rigid sliding block model, which ignores thedynamic response of a deformable sliding mass, Makdisi and Seed (1978) introduced the

concept of an equivalent acceleration to represent the seismic loading of a potentialsliding mass.

The decoupled approximation results from the separate dynamic analysis that isperformed assuming that no relative displacement occurs along the failure plane and therigid sliding block calculation that is performed using the equivalent acceleration-time

history from the dynamic response analysis to calculate seismic displacement.

Dynamic response and the sliding response are analyzed as a single coupled behavior.

Elastic

Rigid Block

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Coupled Procedures

Coupled procedures are at the opposite end of the spectrum from rigid-block methods. These methods have evolved in parallel with advancednumerical modeling techniques such as finite-elements.

These methods are theoretically more accurate from the modelingstandpoint in that the dynamic response of the slope and the sliding

response of the failure mass are modeled as a single, coupled behavior.However, despite being a better representation of the true behavior,these methods can be burdensome from a computational and modelingstandpoint.

Use of coupled procedures in practice is uncommon.

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Example Problem

The slope shown in figure is intersected by two 15 cm thick seams of clay material.The intact slope materials can be characterized by the parameters, c= 120 kPa. f = 0,g = 20.4 kN/m 3. The clayey seams exhibit c = 36 kPa, f =0 and g = 18.8 kN/m 3. (i)Compute the minimum static factor of safety for the slope. Compute the yieldacceleration.

8 m 8 m8 m8 m

8 m

8 m

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Solution

Yield Acceleration:

Yield acceleration is the pseudostatic acceleration that corresponds to a factor ofsafety of 1.0

Weight of the sliding mass W = *16*16*18.8 kN/m = 2406.4 kN/m

Angle of slope = tan -1(16/32) = 26.6

Driving Force F D = W sin + kh W cos

Resisting Force F R = c l = 36* (16 2+322) = 1288 kN/m

Pseudostatic factor of safety against sliding = F R/ FD = 1.0

FD = FR

W sin + kh W cos = 1288 kN/m

On solving, k h = 0.0978

Yield acceleration a y = 0.0978 g

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Exercise Problem

The slope shown in figure is characterized by the parameters, c= 20 kPa. f = 20 , g =20 kN/m 3. Compute the yield acceleration.

10 m

15 m

40

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Kramer, S.L. (1996) Geotechnical Earthquake Engineering, Prentice Hall.

Day, R.W. (2001) Geotechnical Earthquake Engineering Handbook, McGraw-Hill.

Slope stability calculator: http://www.wise-uranium.org/cssth.html (Accessed

on 12 April 2012)

Newmark, N. M. (1965) Effects of earthquakes on dams and embankments.

Geotechnique, 15 (2) 139-160.

Leshchinsky, D., San, D. K.C. (1993) Pseudo-static slope stability analysis: design

charts ASCE Journal of Geotechnical Eng, 120 (9), 1514 1531.

References
http://www.wise-uranium.org/cssth.htmlhttp://www.wise-uranium.org/cssth.htmlhttp://www.wise-uranium.org/cssth.htmlhttp://www.wise-uranium.org/cssth.html

lecture35 seismic slope stability part2

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