Class 7 Consolidation Test ( Geotechnical Engineering )

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)

 Consolidation
 Definitions & Introduction
 Significance
 Consolidation vs Compaction
 Type of Consolidations
 One-Dimensional Consolidation Test
 Definition
 Procedure
 Graphs and results
Class Outlines

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

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

Consolidation - Case Study
Palacio de las Bellas, Artes, Mexico City
Total settlement = 9ft
The Leaning Tower of Pisa

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.

Consolidation vs. Compaction

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.

Voids
Voids
Solids
H
Vv = eVs
V
s
 c
 e
Vv = (e -  e)Vs
V
s
Solids
sz
′
sz
′
sz0
′
sz0
′
sz0
′
sz0
′
}sz f
′
}sz f
′
Before After
Consolidation (cont.)

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

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.

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.

Test Results
B
sp
Cr
Cc
Recompression
Index
Compression
Index
Pre-Consolidation
Stress

Consolidation Test
 Two types of consolidometers (oedometers)
commonly used:
 Floating-ring
 Fixed ring
 This lab uses the fixed-ring consolidometer
 ASTM D 2435

Laboratory Consolidation Test

Consolidation Test

Laboratory Consolidation Test
sv
sv
Solids
Voids
Solids
Voids
Vs
Vv
Vv
Vs
1 Place sample in ring
2 Apply load
3 Measure height change
4 Repeat for new load
 V
Confining
stress
Before After

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)

http://www.uic.edu/classes/cemm/cemmlab/Experiment%2011-Consolidation.pdf#search='consolidation%20test'

Calculations and Graphs - v 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
Tim e (m in 0.5
)
Displacement(in)

v 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
)
Displacement(in)
t90 = 2.5 m in0.5
1
2 3
4
5
t90
d0
B DC
CD = 1.15 BC
A

Calculation and Graph – v 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
VerticalDisplacement(in)

v vs log(time) Graph – Find t50
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
Tim e (m in) - log Scale
VerticalDisplacement(in)
d100 = 0.42065
t 1
t 2 = 4t 1
1
2
A
3
5
4 X 6
X
7
d0 = 0.42305
8
d50=0.5(d0+d100)=0.42185
d50 9
d0
d100
t50 = 10.2 m in
10

Calculation
 Determine the height of
solids (Hs) of the
specimen in the mold
 Determine the change
in height (H)
 Determine the final
specimen height, Ht(f)
 Determine the height of
voids (Hv)
 Determine the final void
ratio
ws
s
s
GD
W
H









2
4
sftv HHH  )(
s
v
H
H
e 

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
2
90
H
tc
T v
v 
2
50
H
tc
T v
v 

Calculation Sample (Ex. pp.121)
Eq 17.2
1(in) - Hs
Hv = Hi - Hs
Hi
e = Hv / Hs
(1.0 + 0.9917) / 2 (0.848 x 0.99592)/(4 x 302)
t90

Plot e vs log p
sc
R min
Cc
Cr

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)

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