Consolidation
Oedometer Test
disc-shaped
loading piston “floats”
CLAY SAMPLE
water squeezed from sample drains freely
into chamber
Load
Schedule
time intervals (minutes) double:
oedometer and sample are
over
each
load
cycle
placed in a loading frame
loads
(N)
also
double:
(61.16,
first load applied is 61.16 N
122.32, 244.64, 489.28, 978.56, 1957.12, 3914.24, 7828.48)
from
time
load
deployed,
Then loads are halved every 24
strain
are taken
hours, dial
and areadings
24 hour strain
dial
for
24 hours
reading
recorded
(0.1, 0.25, 0.5, 1, 2, 4, 8, 16, 30, 60, 120, 240, 1440)
Why Double?
There are two types of plots
equal
spacing
on
a
produced from the data:
logarithmic
scale
is
acheived
e versus log σ and
by doubling the interval
dial reading versus log time
when plotting data it helps to
view the plot with equal
intervals of x
Other data:
ring dimensions:
diameter, (mm),
height, H0 (mm)
specimen area, A (mm2)
water content, initial: w0, final: w1
final dry mass, Ms (g) of sample
specific gravity of sample, Gs
Void Ratios
Δecontent,
 e 1ee  (7.1)
Method 1: use final water

ΔH  H  
ΔH 1  H
e 
w1 and Gs
1  e  Δe
Δe ΔH
If
Sr = 1 e
at0-eend
of
test,
e1 =the
w1G
Remember
Change
this
blast
and
from
you’ve
got:
past?
s
1 toΔe
0
0
10
00
Sincerearrange
Now
e0 = e1 + Δe
to find Δe
1
1  e1 
Δe  
ΔHH0  ΔH
H0
Δ H = Initial Dial Reading – Final Dial Reading
and H0 = Initial height of the specimen
Now with ΔH, H0 and 1+e1, find Δe
Δe
and e0 = e1 + Δe, find the constant ratio:
ΔH
Then multiply the Δ H for each load increment by this ratio
to find the corresponding Δe
The void ratio for each load increment is then e 0 - Δe
Void Ratios
The height of the specimen at the end of any load
increment is H1 = H0 –Initial Dial Reading +Dial Reading at
end of load increment
Method 2: use final dry weight, M
and G
The
void
the end
of any load
is:
At end
ofratio,
test, edry
of specimen
= Mincrement
1 atmass
s
s
H1 - Hs H1 Ms
Knowing A and eG
:
Hs   1 (7.2)
1 s
Hs
H
AG
s s ρw
s
Compressibility Characteristics
The
Compression
Index,
Index,
Ce C
isc Compressibility, mv is defined as
The Expansion
Coefficient
of
Volume
is
the
calculated
approximated
asstrain
the
slope
slope
the
volumetric
perofunit increase in effective stress.
between
the
expansion
any two
part
points
of the
one
2/MN (i.e., the inverse of pressure)
this
linear
portion
vs
logσ’
curve:
The
units
of
mv areofmthe
Virgin Compression Line:
e0  e1
mv can be expressed
in
terms of void ratio:
(7.5)
Cc 
 σ'1 

log
 σ'0 
1
mv 
1  e0
 e0  e1 


 σ'1 σ'0 
(7.3)
or in terms of specimen thickness:
1  H0  H1 


mv 
(7.4)
These
Their
arelogσ’
typical
reflect
plots
the
ofstress
void
history
e
versus
the
clay
effective
The
e shapes
vs
relationship
for aratio,
normally
consolidated
clay
H0of
σ'

σ'
 1 0 
is
nearlyσ’linear
shown clay
stress,
for aas
saturated
Preconsolidation Pressure
6. Draw
1.
2.
3.
4.
5.
Produce
Vertical
Bisect
Determine
tangent
line
the
straight-line
the
to
horizontal
tangent
through
curve
point,
and
atD
line
D.
of
portion of
maximum
through
horizontal
intersection
D.line
the
of
curve, BC
curvature
through
bisector
D.
and
on
the
production
of
Professor
recompression
CB
is the Arthur Casagrande taught
Soil
Mechanics
portion
preconsolidation
of the and Foundation
Engineering
at
Harvard
University
and
curve, ABσ’c.
pressure,
developed this empirical method to
determine the preconsolidation
pressure, σ’c using the e-logσ’ curve.
In-situ e-logσ’ Curve
For
slope
initial
two
The
σ’
the of the in-situ compression
0 isin-situ
overconsolid
line will
void
ratio,
be e
slightly
greater than that of
compression
present
0
ated
clays
the
at
lines
virgin
are
start
linethe
can
becompression line produced
effective
the
in-situ
from
of
expected
the
testing
lab
define
by to a disturbed field sample in
overburden
condition
is a
the
test
intersect
lab.
point
E at:at
pressure
estimated
by
approximates
void
( σ’c ,ratio
e0) of
the
G:
thatpoint
0.42e
of
0 the
in-situ
( σ’0 , e0void
)
ratio