Seismic Design of Slopes - IIT · PDF fileSeismic Design of Slopes Earth Dams and Embankments...
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1 Seismic Design of Slopes Earth Dams and Embankments II Amit Prashant Indian Institute of Technology Gandhinagar Short Course on Geotechnical Aspects of Earthquake Engineering 04 – 08 March, 2013 Deformation Analysis Basic steps: Determine the magnitude and source of the earthquake or earthquakes that should be considered Determine the time-history or time histories of the ground motion associated with the earthquake or earthquakes Determine the yield strength of the embankment and foundation materials Determine the dynamic response of embankment and foundation materials Predict the extent of structural deformations resulting from earthquake shaking If predicted deformations are not tolerable, explore design alternatives that would provide a tolerable response 2
Transcript of Seismic Design of Slopes - IIT · PDF fileSeismic Design of Slopes Earth Dams and Embankments...
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Seismic Design of SlopesEarth Dams and Embankments
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Amit Prashant
Indian Institute of Technology Gandhinagar
Short Course on
Geotechnical Aspects of Earthquake Engineering
04 – 08 March, 2013
Deformation Analysis
Basic steps:
Determine the magnitude and source of the earthquake or earthquakes that should be considered
Determine the time-history or time histories of the ground motion associated with the earthquake or earthquakes
Determine the yield strength of the embankment and foundation materials
Determine the dynamic response of embankment and foundation materials
Predict the extent of structural deformations resulting from earthquake shaking
If predicted deformations are not tolerable, explore design alternatives that would provide a tolerable response
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Equilibrium of Slope Wedge as a Block on inclined plane
As long as shear force FS is lower than shear resistance S, the block will remain stable. (FOS >1)
At FOS = 1, when FS = S, the block will reach impending motion condition. (no movement yet)
For FOS < 1, when FS > S, the block will have unbalanced force of (FS – S). Hence, the block will move with acceleration a/g = (FS – S)/W.
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N
S
W
hk W
sin cosS hF W k
cos sinhN W k
N
friction coefficient
S N
Newmark’s Sliding Block Method
First Method to assess stability in terms of deformations than factor of safety
Assumptions: Pre-existing slip surface: corresponding to kh for FOS = 1
Material is rigid-plastic
Permanent displacement occurs due to dynamic sheasrstress exceeding shear resistance.
The acceleration time history acting on the embankment is known
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Merits of the Newmark’s Method
Advantages Estimates deformation
Easy to use
Disadvantages Potential failure mass and embankment are assumed to be
rigid.
Lateral displacements may be out of phase with the inertial forces at different points in the moving wedge
Can significantly overpredit deformations.
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Acceleration-Time history for Analysis
Acceleration-Time history varies along the slip surface.
What history do we consider as representative for the whole moving block?
Can we use average acceleration-time history aavg(t) from our earlier discussion?........... Yes!!
Any other alternatives of averaging?
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a1(t)
Bed Rock
a2(t)
a3(t)
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(contd…)
Acceleration-Time history for Analysis
For a given bed rock motion at the site, consider acceleration time histories at three or more locations along the slip surface Using wave propagation model, e.g.,
SHAKE
Independent analysis for each location
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m3
m2
m1
1
2
3
a1(t)
Bed Rock
a2(t)
a3(t)
(contd…)
Acceleration-Time history for Analysis
We can now compute weighted average acceleration-time history for the wedge.
Weights can be assumed as mass of soil columns of fixed width above the slip surface at different locations
Hence……
Horizontal Equivalent Acceleration
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1
1
( )
( )
n
i i
i
n
i
i
m a t
HEA t
m
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Example: Find HEA(t) for the given three accelerograms
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t
1a
+1
-1
t
2a
+1
t
3a
+1
-1
-1
10im
20im
20im
1sec
Solution
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t
HEA
+1
-1
at 0 :
10( 1) 20(0) 20(1)0.2
50
t
HEA
at 1:
10(0) 20( 1) 20(0)0.4
50
t
HEA
t
1a
+1
-1
t
2a
+1
t
3a
+1
-1
-1
1sec
at 2 :
10( 1) 20(0) 20( 1)0.2
50
t
HEA
at 3:
10(0) 20( 1) 20(0)0.4
50
t
HEA
at 4 :
10( 1) 20(0) 20( 1)0.2
50
t
HEA
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Yield AccelerationNewmark’s Sliding Block Analysis
Yield acceleration ky is defined as Seismic coefficient corresponding to FOS = 1.
How to obtain? Perform Pseudo-static analysis at
several values of kh.
Plot the curve between kh and FOS
The corresponding slip surface is now used to compute HEA.
Example ky = 0.2
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W
hk W
hk
FOS
0 0.1 0.2 0.3 0.4
0.8
1.0
1.2
1.4
Newmark’sSliding Block Analysis
Integrate acceleration time history for movement of block starting from when acceleration exceeds ky velocity. Continue integrating until velocity becomes zero.
Integrate velocity now to compute displacement.
Note the shaded region.
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Permanent displacement calculation
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t
HEA
+0.4
-0.4
t
v
0.2yk
t
u
0.1
0.133
0.1
0.133
Some empirical relations
Ambraseys and Menu (1988) Equation
Yegian et al (1991)
Neq = equivalent number of cycles
T = predominant period of input motion
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])()1log[(90.0log 09.1
max
53.2
max
a
a
a
au
yy
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max
2
maxmax
2
max
)(48.11)(38.1612.1022.0)log(a
a
a
a
a
a
TNa
u yyy
eq
g
Permanent Seismic Deformation (Hynes and Franklin, 1984)
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Makdisi-Seed Simplified Procedure for Estimating Embankment Earthquake-Induced Deformations (1978)
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Determine yield acceleration Standard slip circle analysis.
Determine earthquake induced average acceleration
Determine permanent seismic vertical deformation
Makdisi-Seed Simplified Procedure for Estimating Embankment Earthquake-Induced Deformations (1978)
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From Harder, et al. 1990, as modified by Cameron (WES), 1996
Peak base acceleration known for the site
Obtain peak crest acceleration.
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Makdisi-Seed Simplified Procedure for Estimating Embankment Earthquake-Induced Deformations (1978)
Depth of sliding obtained from stability analysis.
Obtain Peak Average Acceleration
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Makdisi-Seed Simplified Procedure for Estimating Embankment Earthquake-Induced Deformations (1978)
Read permanent displacement: Not a unique number, but a range.
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Example: 1
A 10m high embankment is located in seismic zone IV. Take importance factor I as 1.0. The foundation conditions include a non-liquefiable crust of clayey silt extending to 1.5 m below ground surface underlain by a 1.5-m thick liquefiable layer and a non-liquefiable soils. The embankment is constructed with well compacted granular soils not expected to liquefy during the design earthquake. Evaluate the slope stability. Assume the magnitude of earthquake as 6.5 if required.
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Stability Analysis for Yield Acceleration
For seismic zone IV, Z = 0.24
Importance factor, I = 1.0 (given)
Horizontal peak ground acceleration = Z × I = 0.24
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Displacement calculation
From the relationship of Hynes-Griffin and Franklin, 1984, Upper bound permanent displacement = 0.8 m
Mean plus sigma displacement = 0.22 m
Mean displacement = 0.13 m
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Example: 2
PGA for Safety Evaluation = 0.60g
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Assuming No Liquefaction in Sub-Soil
Static load (only self weight): FOS=1.99
Pseudo-static: Yield coeff (FOS=1.0) is 0.275g
FOS Approach by: Terzaghi (1950): Yield coeff. should be >0.20g
Mercuson (1981): Yield coeff. > 0.2g-0.3g
Hynes and Franklin (1984): Yield coeff of 0.1g will give permanent deformation less than 1m.
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Assuming No Liquefaction in Sub-Soil
Permanent Deformation by Newmark’s Sliding Block Concept Makdisi and Seed (1978) approach: 5 -15 mm
Ambraseys and Menu (1988): 39 mm
Yegian et al. (1991): 30mm
Permanent deformation of about 40mm quite acceptable.
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Liquefaction of Sub-Soil
Liquefied soil layers may not transmit significant amount of shear waves.
Will the embankment be stable under its own weight?
Liquefied soil layers will loose considerable amount of strength.
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Liquefaction of Sub-Soil
Residual strength of liquefiable soil strata considered as per Seed and Harder (1990)
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Liquefaction of Sub-Soil
FOS against self weight: 1.39 for North Embankment
1.18 for South Rail Embankment
1.31 for South Road Embankment
Embankment will be stable due to its own weight after foundation soils have liquefied
Conservative Assumptions: Liquefaction occurs early during shaking
Base of embankment still sustains PGA of 0.60g
Deformations computed for 0.60g but with residual strength of liquefiable soils
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Post-Liquefaction Evaluation Using Residual Strength of Liquefiable Layers
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Post-Liquefaction Deformation
Deformation of 500mm: acceptable.
Deformation of 1300mm: on the higher side; but can be handled as an emergency measure in a relatively short time
These deformations are for maximum embankment height and with conservative assumptions
Remedial measures not recommended.
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More Reading Materials Recently, IITK-GSDMA Guidelines for Seismic
Design of Earth Dams and Embankments
Available at www.nicee.org
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Thank You
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Estimation of Crest SettlementBy: J. R. Swaisgood, Embankment dam deformations caused by earthquakes, 2003 Pacific conference on EQ Engg.
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Estimation of Crest Settlement (contd…)By: J. R. Swaisgood, Embankment dam deformations caused by earthquakes, 2003 Pacific conference on EQ Engg.
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Permanent slope displacement from Earthquake Ground Motion
Horizontal equivalent acceleration, HEA
Horizontal yield acceleration: ay (corresponding to FOS=1)
Calculation of velocity (considering acceleration above yield acceleration)
Calculation of permanent displacement over time
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1
t
t
V adt
2
1
t
t
S Vdt
