ConsolidationConsolidationOedometer TestOedometer Test

CLAY SAMPLECLAY SAMPLE

disc-disc-shapedshaped

loading piston “floats”

water squeezed from sample drains freely water squeezed from sample drains freely into chamberinto chamber

Load Load ScheduleSchedule

oedometer and sample are oedometer and sample are placed in a loading frameplaced in a loading frame

first load applied is 61.16 Nfirst load applied is 61.16 N

from time load deployed, from time load deployed, strain dial readings are strain dial readings are taken for 24 hourstaken for 24 hours

time intervals (minutes) time intervals (minutes) double: double: (0.1, 0.25, 0.5, 1, 2, 4, 8, 16, 30, 60, (0.1, 0.25, 0.5, 1, 2, 4, 8, 16, 30, 60,

120, 240, 1440)120, 240, 1440) over each load cycleover each load cycle

loads (N) also double:loads (N) also double: (61.16, 122.32, 244.64, 489.28, 978.56, 1957.12, (61.16, 122.32, 244.64, 489.28, 978.56, 1957.12, 3914.24, 7828.48)3914.24, 7828.48)

Then loads are halved every Then loads are halved every 24 hours, and a 24 hour strain 24 hours, and a 24 hour strain dial reading recordeddial reading recorded

Why Double?Why Double?equal spacing on a equal spacing on a

logarithmic scale is acheived logarithmic scale is acheived by doubling the intervalby doubling the interval

There are two types of plots There are two types of plots produced from the data:produced from the data:

e versus log e versus log σσ and and

dial reading versus log timedial reading versus log time

when plotting data it helps to when plotting data it helps to view the plot with equal view the plot with equal intervals of xintervals of x

Other data:Other data:ring dimensions: ring dimensions:

diameter,diameter, (mm), (mm),

height, height, HH00 (mm) (mm)

specimen area, specimen area, AA (mm (mm22))

water content, initial: water content, initial: ww00, final: , final: ww11

final dry mass, final dry mass, MMss (g) of sample (g) of sample

specific gravity of sample, specific gravity of sample, GGss

Void RatiosVoid RatiosMethod 1: use final water Method 1: use final water content, content, ww11 and G and Gss

If Sr = 1 at end of test, e1 = w1Gs

Remember this blast from the past?

0

100 e1

eeHΔH

Change e0-e1 toΔe and you’ve got: 0

0

He1

ΔHΔe

Δ H = Initial Dial Reading – Final Dial Reading

0

1

HΔee1

ΔHΔe

(7.1)

10

e1ΔHH

ΔHΔe

Now with ΔH, H0 and 1+e1, find Δe

Then multiply the Δ H for each load increment by this ratio to find the corresponding Δe

Since e0 = e1 + Δe

and e0 = e1 + Δe, find the constant ratio:

Now rearrange to find Δe

and H0 = Initial height of the specimen

ΔHΔe

The void ratio for each load increment is then e0 - Δe

Void RatiosVoid RatiosMethod 2: use final dry Method 2: use final dry weight, Mweight, Mss and G and Gss

At end of test, dry mass of specimen = Ms Knowing A and

Gs: ws

ss ρAG

MH (7.2)1

HH

HH-H

es

1

s

s11

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

The void ratio, e1 at the end of any load increment is:

Compressibility Compressibility CharacteristicsCharacteristics

These are typical plots of void ratio, e versus effective stress, σ’ for a saturated clayTheir shapes reflect the stress history of the clayThe e vs logσ’ relationship for a normally consolidated clay is nearly linear as shown

The Compression Index, Cc is calculated as the slope between any two points on this linear portion of the Virgin Compression Line:

0

1

10c

σ'σ'

log

eeC (7.5

)

The Expansion Index, Ce is the approximated slope of the expansion part of the e vs logσ’

curve:

The Coefficient of Volume Compressibility, mv is defined as the volumetric strain per unit increase in effective stress.The units of mv are m2/MN (i.e., the inverse of pressure)

mv can be expressed in terms of void ratio:

01

10

0v σ'σ'

eee1

1m (7.3)

or in terms of specimen thickness:

01

10

0v σ'σ'

HHH1

m (7.4)

Preconsolidation PressurePreconsolidation Pressure

Professor Arthur Casagrande taught Soil Mechanics and Foundation Engineering at Harvard University and developed this empirical method to determine the preconsolidation pressure, σ’c using the e-logσ’ curve.

1. Produce the straight-line portion of the curve, BC

2. Determine the point, D of maximum curvature on the recompression portion of the curve, AB

3. Draw tangent to curve at D.

4. Draw horizontal line through D.

5. Bisect tangent and horizontal line through D.

6. Vertical line through intersection of bisector and production of CB is the preconsolidation pressure, σ’c.

In-situ e-logIn-situ e-logσσ’ Curve’ CurveThe slope of the in-situ compression line will be slightly greater than that of the virgin compression line produced from testing a disturbed field sample in the lab.

The initial void ratio, e0 at the start of the lab test approximates that of the in-situ void ratio

The two compression lines are expected to intersect at a void ratio of 0.42e0

The in-situ compression line can be define by point E at: ( σσ’’cc , e0)

For overconsolidated clays the in-situ condition is estimated by the point G: ( σσ’’00 , e0)

σσ’’00 is the present effective overburden pressure