CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 1 Lecture 05 Soil Compression
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 1
Lecture 05Soil Compression
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 2
Soil CompressionSoil compression: what is it?
Total volume decreases what is reduced?
Soil compression: what causes it? Stress increase and/or self internal reaggrangement of soil particles
Types of soil compressionElastic settlement (also termed immediate settlement)
Sands
Consolidation Clays
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 3
Elastic settlementElastic settlement
Caused by the stress incrementCauses no change in the pore pressure pore water flows out!Occurs in dry soils and saturated soils with high permeabilityCan solve with elastic stress – strain relationshipCan solve for the settlement if stress increments are known (e.g. provide in previous lecture)Integrate over the depth of influence
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 4
Elastic settlementA special case: 1D compression
Radial displacement not allowed Displacement only allowed in vertical direction
Determine vertical stress – strain relationshipDetermine vertical stress – radial stress relationship
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 5
ConsolidationConsolidation: used to describe time dependent settlement that occurs in claysRemember in sands, compression occurs due to air / pore water flowing out of the soil structureIn clays, due to very low permeability, it takes a lot of time for the pore water to flow outThink of the case belowIf in sand, the pore water is
not allowed to drain,would settlement occur?
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ConsolidationWhen valve is closed
When valve is open
Note:
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 7
Time dependent variation of stresses and pressure
isochrone
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 8
1D Laboratory consolidation test - OedometerTubular steel mold size: 64 mm in diameter and 25 mm thickTwo porous stones at the boundariesApply load in increments (usually loaded for 24 hrs)
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 9
1D Laboratory consolidation test - OedometerDeveloping void ratio – vertical stress plot
Step 1: calculate Hs
Step 2: calculate Hv
Calculate initial void ratio e0
Calculate subsequent changes in void ratios and effective vertical stresses
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Normally vs. Overconsolidated clays
Effect of Stress history
Overconsolidation ratio
More common to use σ’p
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 11
Test Results
B
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Methods to determine ’p1. Casagrande
1. Choose the point of maximum curvature (point A)2. Draw horizontal line from point A3. Draw a line tangent to point A4. Bisect the angle made by steps 2 and 35. Extend the straight line portion of the virgin curveThe point of intersection of these two lines is the ’p
A
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Field curve Schmertmann (1953)
1. Select ’v02. Calculate e0. Draw a horizontal line from e0. Select Point A.3. From the point 0.42e0, draw a horizontal line. The pointthat intersects the lab virgin curve is point B.4. Connect points A and B. The line is the field virgin compression line.
Due to disturbance, the lab compression curve has a slope less than that of the field virgin curve.
NC Clay
’v0
e0
0.42e0
A
B
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Field curve Schmertmann (1953)
1. Select ’p using Casgrande’s method2. Calculate e0. Draw a horizontal line from e0 up to ’v0. Then drawA line parallel to the lab rebounding curve.Select Point A. This the field recompression curve.3. From the point 0.42e0, draw a horizontal line. The pointthat intersects the lab virgin curve is point B.4. Connect points A and B. The line is the field virgin compression line.
Due to disturbance, the lab compression curve has a slope less than that of the field virgin curve.
OC Clay
’p
e0
0.42e0
A
B
Lab rebounding curve
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 15
Settlement PredictionsN.C. Clays
S Cc
1 eo H log
zf
z0
Cc: compression indexav: coefficient of compressibilitymv: coefficient of volume change
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 16
Settlement PredictionsN.C. Clays
V=1+eo
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Settlement PredictionsO.C. Clays…… Case I
r zf
0 z0
CS H log1 e
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Settlement PredictionsO.C. Clays…… Case II
r c c zf
0 z0 0 c
C CS H log H log1 e 1 e
Final
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Cc and Cr
Undisturbed ClayCc = 0.009(LL-10)
Disturbed ClayCc = 0.007(LL-10)
Cr = 1/5 – 1/10 Cc
Cr
Cc
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 20
Secondary compression
0
0
/ log1
c rs
p
C C C tS He t
settlement
u=0
0
0
log1s
p
C tS He t
logeC
t
C/Cc = constantC/Cc = 0.01 – 0.07
average: 0.04
0
0
/ log1
c cs
p
C C C tS He t
w0
C
Mesri, 1973Inorganic clays and silts: 0.04 0.01Organic clays: 0.05 0.01Peat : 0.06 0.01
CIE_3008 토질역학 및 실험 Lec.5 Soil Compression 21
Meaning of Consolidation Curve
e1
2
v
EOP curve (End of Primary Curve)
12
settlement
u=0
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Uniqueness of EOP curve
e
v
EOP curve (End of Primary Curve)
tp for laboratory test is very shorttp in the field is very longAre the curves from laboratory identical to the curve from field?
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Secondary compressionIf tp is small, then secondary compression can be large.Suppose the clay is NC.
How can we reduce the secondary compression?
Surcharge
QuizHow can we plot the secondary compression in an EOP curve?
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Preconsolidation Pressure’p Largest pressure that the soil has experienced? Is it true?
Pressure at which major structural changes including the break-down of Interparticle bonds and interparticle displacement begin to occur.
-Boundary between stiff and soft deformation response of a soil to loading-Expressed as the overconsolidation ratio (OCR)-Mostly 1.2 – 3-Mechanism contributing to ’p
-Geological loading and unloading-Freezing and thawing-Secondary compression-Weight of ice
Sketch the preconsolidation pressure of a soil that has undergone secondary compression