1 kh training april 2005. 2 gudiance on the selection of kh soil soil to rock rock soil soil to rock...
TRANSCRIPT
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Kh Training
April 2005
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Gudiance on the selection of Kh
• Soil
• Soil to Rock
• Rock
• Soil
• Soil to Rock
• Rock
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SOIL
• The following slides relate Relative density and Consistency of soils to the Material Strength in Chapter 52 – Tables 52-2 and 52-3.
• The following slides relate Relative density and Consistency of soils to the Material Strength in Chapter 52 – Tables 52-2 and 52-3.
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Soil
• Massive Jointed Cohesive
• Cohesive
• Low Plastic Cohesionless
• Non-Plastic Cohesionless
• Massive Jointed Cohesive
• Cohesive
• Low Plastic Cohesionless
• Non-Plastic Cohesionless
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Cohesive
• Field Tests
• Lab Tests
• The PI of the Soils is greater than 10
• Field Tests
• Lab Tests
• The PI of the Soils is greater than 10
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Lab. Data
• Natural moisture content %
• Specific Gravity Gs
• Liquid Limit LL
• Plasticity Index PI
• Clay Fraction (%<0.002mm)
• Unconfined Compression Strength (psf)
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Characterizing Clays
Plasticity Index - The numerical difference between the Liquid Limit water content and the Plastic Limit water content
PI = LL - PL
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Consistency of Clays
• To evaluate, clays must be saturated
• Saturation establishes base line for comparisons -even soft clays will become firm when dried
• Saturated consistency strong indicator of engineering behavior
• Water has zero shear strength
• To evaluate, clays must be saturated
• Saturation establishes base line for comparisons -even soft clays will become firm when dried
• Saturated consistency strong indicator of engineering behavior
• Water has zero shear strength
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Steps in Evaluating Consistency of Saturated Clay
• Obtain data - LL, PI, either dry or wsat(%)
• Calculate PL (not usually reported) PL = LL - PI
Liquid LimitPlastic Limit
Establish location of wsat % on diagram
v/softsoftmediumstiff
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Evaluating Consistency of Saturated Clay
• Steps to Evaluate– For saturated deposit, obtain sample and
measure oven dry w(%)– For unsaturated deposit, measure dry density,
assume value for Gs and calculate theoretical saturated w(%)
1001
(%)
sd
wsat Gw
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1001
(%)
sd
wsat Gw
Example
What is the saturated water content of the following clay compared to its LL and PL?
Given: LL = 59, PI = 36, dry = 1.28 g/cm3Assume Gs = 2.70
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Consistency and Atterberg Limits
Begin Constructing a consistency diagram by locating the LL value of 59
Liquid Limit 59
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Consistency and Atterberg Limits
Next, calculate the Plastic Limit, which is equal to the LL minus the PI and plot that
PL = 59 - 36 = 23
Liquid LimitLiquid Limit 59 59
Plastic Limit 23
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Consistency and Atterberg Limits
Now, divide the range between the LL and PL into thirds (PI 3) - (36 3 = 12)
Liquid Limit 59
Plastic Limit 23
v/soft
35 47
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Consistency and Atterberg Limits
• Now label the ranges as shown
Liquid Limit 59
Plastic Limit 23
v/softsoftmediumstiffv/stiff
35 47
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Consistency and Atterberg Limits
Next, Calculate the saturated water content, given that dry = 1.28 g/cm3 and Gs = 2.70
%41.11002.70
1
1.28
1.0(%)wsat
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Consistency and Atterberg Limits
• Plot the saturated water content on the diagram to identify its consistency
Liquid Limit 59
Plastic Limit 23
wsat % = 41 %, medium consistency
v/softsoftmediumstiffv/stiff35 47
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Liquidity Index
• Another method for quickly assessing the saturated consistency of a clay uses a term Liquidity Index
• Liquidity Index is defined from a simple equation that expresses numerically where the saturated water content is in relation to the Atterberg Limits
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Liquidity Index Equation
PI
PILLw
PI
PLwLI satsat
If wsat = Liquid Limit, then LI = 1.0
If wsat = Plastic Limit, then LI = 0
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Summary of Clay Problem Soils
• Problem Clays are very soft clays with liquidity index 1, or
• Stiff to very stiff clays with liquidity index 0
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Liquidity Index and Saturated Consistency
Liquidity Index < 0 0 to 0.330.33 to 0.670.67 to 1.00 > 1.00
Liquidity Index < 0 0 to 0.330.33 to 0.670.67 to 1.00 > 1.00
SaturatedConsistencyvery stiffstiffmediumsoftvery soft
SaturatedConsistencyvery stiffstiffmediumsoftvery soft
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What is the liquidity index of the Example Soil?
• Given that wsat = 41%, LL = 59, and PI = 36, entering these terms in the LI equation:
PI
PILLw
PI
PLwLI satsat
5036
365941.
PI
PLwLI sat
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Conclusion - Medium Consistency
Liquidity Index < 0< 0 0 to 0.330 to 0.330.33 to 0.670.67 to 1.000.67 to 1.00 > 1.00> 1.00
SaturatedConsistencyvery stiffvery stiffstiffstiffmediumsoftsoftvery softvery soft
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Field Estimate of Consistency
Consistency
– Very Soft
– SoftSoft
– MediumMedium
– StiffStiff
– V/Stiff to HardV/Stiff to Hard
Rule of Thumb
Thumb will penetrate soil more than 1-inch. Extrudes between fingers when squeezed in fist
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Field Estimate of Consistency
Consistency
Very SoftVery Soft
Soft
MediumMedium
StiffStiff
V/Stiff to HardV/Stiff to Hard
Rule of Thumb
Thumb will penetrate soil about 1-inch. Easily molded in fingers
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Field Estimate of Consistency
Consistency
Very SoftVery Soft
SoftSoft
Medium
StiffStiff
V/Stiff to HardV/Stiff to Hard
Rule of Thumb
Thumb will not penetrate soil, but will indent about 1/4 inch. Molded by finger pressure
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Field Estimate of Consistency
Consistency
Very SoftVery Soft
SoftSoft
MediumMedium
Stiff
V/Stiff to HardV/Stiff to Hard
Rule of Thumb
Thumb will not indent soil, but soil can be indented with thumbnail.
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Field Estimate of Consistency
Consistency
Very SoftVery Soft
SoftSoft
MediumMedium
StiffStiff
V/Stiff to Hard
Rule of Thumb
Can only be marked with knife - not indented with thumbnail.
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Soft Clays - Field EstimatesSPT,
BLOWS/FOOT CONSISTENCY C VALUE, PSF
< 2 Very Soft – extrudedbetween fingers
< 250
2 – 4 Soft – Molded by lightfinger pressure
250-500
4 – 8 Medium – Molded bystrong finger pressure
500 – 1000
8 – 15 Stiff – Indented by thumbbut not penetrated
1000 – 2000
15 – 30 Very Stiff – Indented byThumbnail
2000 – 4000
> 30 Hard – Indented withKnife
> 4000
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Estimating Effective angle for claysKinney's Correlation
20
25
30
35
0 10 20 30 40 50 60 70 80 90
Plasticity Index (PI)
Effe
ctiv
e a
ngle
, de
gree
s
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SOILS WITH PI LESS THAN 10SOILS WITH PI LESS THAN 10
• Methods to estimate RELATIVE DENSITY for nonplastic soils
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Relative Density
• Studies for the Alaska pipeline established empirical estimates of relative density for various soil types, based on the measured in place density of the soils
• Average values of minimum and maximum index density for the project were used
• May be useful for preliminary estimates
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80
90
100
110
120
130
140
0 10 20 30 40 50 60 70 80 90 100
Relative Density, %
Dry
Den
sity
, p
cf
sand and silty sand
Gravelly sand
Reference - Donovan, N.C. and Sukhmander Singh, "Liquefaction Criteria for the Trans-Alaska Pipeline." Liquefaction Problems in Geotechnical Engineering, ASCE Specialty Session, Philadelphia, PA, 1976.
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Relative Density
• Because saturated water content is related to dry density, a chart can be derived relating in place saturated water content to relative density, as shown on the following slide
1001
(%)
sd
wsat Gw
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5
10
15
20
25
30
35
40
45
0 10 20 30 40 50 60 70 80 90 100
Relative Density, %
Sat
ura
ted
Wat
er C
on
ten
t, %
Reference Donovan, N.C. and Sukhmander Singh, "Liquefaction Criteria for ભthe Trans-Alaska Pipeline." Liquefaction Problems in Geotechnical Engineering, ASCE Specialty Session, Philadelphia, PA, 1976.
Average
Chart is for silty sands (SM)
Other information on Relative Density
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Field Estimates for Relative Density
• Relative Density
– Very Loose
– LooseLoose
– MediumMedium
– DenseDense
– Very DenseVery Dense
Description
1/2” Reinforcing rod
can be pushed easily
by hand into soil
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Field Estimates for Relative Density
• Relative Density
– Very LooseVery Loose
– Loose
– MediumMedium
– DenseDense
– Very DenseVery Dense
Description
Can be excavated
with a spade. A
wooden peg 2”x
2”can easily be drive
to depth of
6 ”.
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Field Estimates for Relative Density
• Relative Density
– Very LooseVery Loose
– LooseLoose
– Medium
– DenseDense
– Very DenseVery Dense
Description
Easily penetrated
with a 1/2”
reinforcing rod
driven with a 5
pound hammer
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Field Estimates for Relative Density
• Relative Density
– Very LooseVery Loose
– LooseLoose
– MediumMedium
– Dense
– Very DenseVery Dense
Description
Requires a pick for
excavation. Wooden
peg 2”x 2”is hard to
drive beyond 6 ”.
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Field Estimates for Relative Density
• Relative Density
– Very LooseVery Loose
– LooseLoose
– MediumMedium
– DenseDense
– Very Dense
Description
Penetrated only a
few centimeters with
a 1/2” reinforcing
rod driven with a 5
pound hammer
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Fort Worth Relative Density Study
• NRCS lab in Fort Worth studied 28 filter sands and used some published data
• Minimum and Maximum Index Densities were performed on each sample
• A 1 point dry Standard Proctor energy mold was also prepared for each sample.
• Values of 50% and 70% relative density were plotted against the 1 point Proctor value
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70 % Relative Density vs. 1 Point Proctor
90
95
100
105
110
115
120
125
130
90 100 110 120 130
Field 1 Point Proctor Test Dry Density, pcf
70 %
Rel
ativ
e D
ensi
ty
70 %RD = 1 Point line
Best fit correlation
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• Conclusion is that 100 % of the Field
1 point Proctor dry test is about
equal to 70 % relative density
70 % Relative Density Vs. 1 Point Proctor
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50 % Relative Density Vs. 1 Point Proctor
90
95
100
105
110
115
120
125
90 95 100 105 110 115 120 125 130
Field 1 pointdry density
50 %
Rd
95 % of 1 point
best fit line
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• Conclusion is that 95 % of the field 1
point Proctor dry test is about
equal to 50 % relative density
50 % Relative Density Vs. 1 Point Proctor
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D70 = 1.075 x d 1pt -9.61,
for RD70 and d 1pt in lb/ft3
D50 = 1.07 x d 1pt - 12.5,
for RD50 and d 1pt in lb/ft3
Relative Density Estimates from FW SML Study
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• Example Relative Density Estimates
–Given: 1 Point Proctor Testd = 105.5 pcf
–Estimate 70 % and 50% Relative Density from the Fort Worth equations
–Given that measured d is 98.7, evaluate state of compaction of sand.
Relative Density Estimates from FW SML Study
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D70 = 1.075 x d 1pt -9.61, = 1.075 x 105.5 -9.61 = 103.8 pcf
D50 = 1.07 x d 1pt - 12.5, = 1.07 x 105.5 - 12.5
= 100.4
Solution to Example Problem
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Solution to Example Problem
• Conclusion is that measured dry
density of 98.7 pcf is less than the
estimated 50% Relative Density dry
density value of 100.4 pcf
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Solution to Example Problem
• Measured Dry Density is 98.7 pcf and Field 1 Point Dry Density is 105.5 pcf. Measured Dry Density then is 98.7 105.5 = 93.6 percent of the 1 Point Proctor -
• Again, by rule of thumb < 50% RD
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Empirical Estimates for Sands and Gravels
• Empirical Estimates of Shear Strength Estimating effective friction angle
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NRCS Correlations
25
30
35
40
45
50
15 25 35 45 55 65 75 85
Relative Density (%)
Eff
ecti
ve
an
gle
, deg
rees
Well-graded Sands and Gravels
Poorly Graded Sands
Silty Sands
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The End