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

Vs

c

e

Vv = (e - e)Vs

Vs

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)

Civil Engineering - Texas Tech University 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

Time (min 0.5)

Dis

plac

emen

t (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

)

Dis

plac

emen

t (in

)

t90 = 2.5 min0.5

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

Verti

cal D

ispl

acem

ent (

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

Time (min) - log Scale

Vert

ical

Dis

plac

emen

t (in

)

d100 = 0.42065

t 1

t 2= 4t 1

1

2

A

3

5

4 X 6

X7

d0 = 0.42305

8

d50=0.5(d0+d100)=0.42185

d50 9

d0

d100

t50 = 10.2 min

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

ss

GD

WH

2

4

sftv HHH )(

s

v

H

He


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

tcT v

v

2

50

H

tcT 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)