Download - CEG 801 - Consolidation Test
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Civil Engineering - Texas Tech University
CE 3121: Geotechnical Engineering Laboratory
Class 7
Consolidation Test on Cohesive Soil
Sources:
Soil Mechanics – Laboratory Manual, B.M. DAS (Chapter 17)
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Consolidation Definitions & Introduction Significance
Consolidation vs Compaction Type of Consolidations One-Dimensional Consolidation Test
Definition Procedure Graphs and results
Class Outlines
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Consolidation - Definition
Consolidation refers to the compression or settlement that soils undergo as a response of placing loads onto the ground
These loads produce corresponding increases in the vertical effective stress, sv’
Consolidation is a time-dependent process, in some soils it may take long time (100 years ?) to achieve complete settlement
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Significance
The amount of soil volume change that will occur is often one of the governing design criteria of a project
If the settlement is not kept to tolerable limit, the desire use of the structure may be impaired and the design life of the structure may be reduced
It is therefore important to have a mean of predicting the amount of soil compression or consolidation
It is also important to know the rate of consolidation as well as the total consolidation to be expected
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Consolidation - Case Study
Palacio de las Bellas, Artes, Mexico City
Total settlement = 9ftThe Leaning Tower of Pisa
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Consolidation - Introduction
The compression is caused by: Deformation of soil particles Relocations of soil particles Expulsion of water or air from void spaces
Most of the settlement of a structure on clay is mainly due to volumetric changes and rarely due to shear strain.
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Consolidation vs. Compaction
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Consolidation (cont.)
During consolidation, pore water or the water in the voids of saturated clay gets squeezed out – reducing the volume of the clay – hence causing settlement called as consolidation settlement
The spring analogy to consolidation.
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VoidsVoids
Solids
H
Vv = eVs
Vs
c
e
Vv = (e - e)Vs
Vs Solids
z′
z′
z0′
z0′
z0′
z0′
}z f′
}z f′
Before After
Consolidation (cont.)
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Types of Consolidation
There are three types of consolidation: Immediate consolidation; caused by elastic
deformation of dry soil or moist and saturated soil without change in moisture content
Primary consolidation; caused as a result of volume change in saturated cohesive soils due to exclusion of water occupied the void spaces
Secondary consolidation; occurs in saturated cohesive soils as a result of the plastic adjustment of soil fabrics
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Types of Consolidation (cont.)
Clayey soils undergo consolidation settlement not only under the action of “external” loads (surcharge loads) but also under its own weight or weight of soils that exist above the clay (geostatic loads).
Clayey soils also undergo settlement when dewatered (e.g., ground water pumping) – because the effective stress on the clay increases
Coarse-grained soils DO NOT undergo consolidation settlement due to relatively high hydraulic conductivity compared to clayey soils. Instead, coarse-grained soils undergo IMMEDIATE settlement.
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1- D Consolidation Test
The main purpose of consolidation tests is to obtain soil data which is used in predicting the rate and amount of settlement of structures founded on clay.
The four most important soil properties determined by a consolidation test are: The pre-consolidation stress, sp’, This is the maximum
stress that the soil has “felt” in the past. The compression index, Cc , which indicates the
compressibility of a normally-consolidated soil. The recompression index, Cr , which indicates the
compressibility of an over-consolidated soil. The coefficient of consolidation, cv , which indicates the
rate of compression under a load increment.
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Test Results
B
sp
Cr
Cc
Recompression Index
CompressionIndex
Pre-Consolidation Stress
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Consolidation Test
Two types of consolidometers (oedometers) commonly used: Floating-ring Fixed ring
This lab uses the fixed-ring consolidometer ASTM D 2435
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Laboratory Consolidation Test
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Consolidation Test
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Laboratory Consolidation Test
v
v
Solids
Voids
Solids
Voids
Vs
Vv Vv
Vs
1 Place sample in ring2 Apply load3 Measure height change4 Repeat for new load
V
Confiningstress
Before After
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Procedure
Measure the inner diameter and height of the consolidation cutter/ring and record its mass
Prepare a soil specimen for the test by trimming and placing the soil in the ring
Determine the mass of ring + soil Collect some excess soil for moisture content Assume Gs = 2.7 Saturate the lower (larger) porous stone on the base of
the consolidometer Place the specimen and ring and place the upper
stone/disk Follow the rest in your lab manual Place 1.5 kg (1st day), 3kg (2nd day), 6kg (3rd day), 12kg
(4th day)
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Civil Engineering - Texas Tech Universityhttp://www.uic.edu/classes/cemm/cemmlab/Experiment%2011-Consolidation.pdf#search='consolidation%20test'
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Civil Engineering - Texas Tech University
Calculations and Graphs - dv vs w(time)
Plot of Vertical Displacement vs. Time(P = 1000 psf)
0.42
0.4205
0.421
0.4215
0.422
0.4225
0.423
0.4235
0.00 5.00 10.00 15.00 20.00 25.00
Time (min 0.5)
Dis
pla
cem
en
t (i
n)
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dv vs wtime Graph – Find t90
Plot of Vertical Displacement vs. Time(P = 1000 psf)
0.42
0.4205
0.421
0.4215
0.422
0.4225
0.423
0.4235
0.00 5.00 10.00 15.00 20.00 25.00
Time (min 0.5)
Dis
pla
ce
me
nt
(in
)
t90 = 2.5 min0.5
1
2 3
4
5
t90
d0
B DC
CD = 1.15 BC
A
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Calculation and Graph – dv vs log(time)
Logarithm of time curve fitting
0.42
0.4205
0.421
0.4215
0.422
0.4225
0.423
0.4235
0.1 1 10 100 1000 10000
Time (min) - log Scale
Verti
cal
Dis
pla
cem
en
t (i
n)
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dv vs log(time) Graph – Find t50Logarithm of time curve fitting
0.42
0.4205
0.421
0.4215
0.422
0.4225
0.423
0.4235
0.1 1 10 100 1000 10000
Time (min) - log Scale
Vert
ical D
isp
lacem
en
t (i
n)
d100 = 0.42065
t 1
t 2= 4t 1
1
2
A3
5
4 X 6
X7
d0 = 0.423058
d50=0.5(d0+d100)=0.42185
d50 9
d0
d100
t50 = 10.2 min
10
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Calculation
Determine the height of solids (Hs) of the specimen in the mold
Determine the change in height (DH)
Determine the final specimen height, Ht(f)
Determine the height of voids (Hv)
Determine the final void ratio
ws
ss
GD
WH
2
4
sftv HHH )(
s
v
H
He
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Calculation (cont.)
Calculate the coefficient of consolidation (cv) from t90
Calculate the coefficient of consolidation (cv) from t50
Plot e-log p curve and find: sc, Cc, Cr
Plot cv – log p curves
290
H
tcT vv
250
H
tcT vv
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Calculation Sample (Ex. pp.121)
Eq 17.2
1(in) - Hs
Hv = Hi - HsHie = Hv / Hs
(1.0 + 0.9917) / 2 (0.848 x 0.99592(/)4 302x)
t90
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Plot e vs log p
sc
R min
Cc
Cr
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In Your Report
Plot all curves find t90 and t50 (10 plots) Show your calculations in a table and find
e, cv (t90), cv (t50)
Plot e vs. log (p) and determine: Pc
Cc
Cr
Plot cv vs. log (p) (2 plots)