Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show. Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.

1. 1
Geotechnical Engineering–I [CE-221]
BSc Civil Engineering – 4th Semester
by
Dr. Muhammad Irfan
Assistant Professor
Civil Engg. Dept. – UET Lahore
Email: mirfan1@msn.com
Lecture Handouts: https://groups.google.com/d/forum/2016session-geotech-i
Lecture # 18
30-Mar-2018

2. 2
One-Dimensional Consolidation
Drainage and deformations occur in vertical direction only.
(none laterally)
A reasonable simplification for solving consolidation problems

3. 3
1-D Consolidation Theory
(Terzaghi, 1936)
Assumptions of one-dimensional consolidation theory
1. Soil is homogenous.
2. Soil is fully saturated.
3. Coefficient of consolidation (CV) remains constant throughout the soil
mass and also remains constant with time.
4. Coefficient of permeability (k) is constant throughout.
5. Darcy’s law for flow of water through the soil mass is valid,
i.e., v = k.i
6. Consolidation is a one-dimensional problem i.e., water flows in only
one direction and the resulting settlement also occur in one direction
only.
7. Soil particles are assumed to be incompressible i.e., all the settlement
is due to the expulsion of water.

4. 4
1-D Lab Consolidation
NSL
metal ring
SOIL
Porous Stones
Field
Undisturbed soil
specimen
Lab
Consolidometer / Oedometer
Stopwatch

5. 5
1-D Lab Consolidation
 Devised by Carl Terzaghi.
 The apparatus is called Consolidometer /
Oedometer
 Soil specimen placed inside a metal ring
 Two porous stones, one at the top and
other at the bottom of specimen
SOIL
Porous
Stones
 Diameter of specimen = 50-75 mm (2”-3”)
 Diameter/Height: between 2.5 & 5
 Specimen kept submerged in water throughout the test
 Load is applied through a lever arm
 Each load is usually applied for 24hrs (or till deformations become
negligible)
 Each loading increment is usually double the previous load.
 After complete loading, unloading is done in steps.

6. 6
Magnitude of settlement → compression index (Cc)
Rate of consolidation → co-efficient of consolidation (Cv)
1. Time ~ Deformation curve
i. Cv (Coefficient of consolidation)
2. Pressure ~ Deformation curve
i. Cc (Compression index)
ii. Cr (Recompression index)
iii. aV (Coefficient of compressibility)
iv. mV (Coefficient of volume change)
SOIL
Porous
Stones
CONSOLIDATION TEST
- Interpretation of Test Results -

7. 7
Stage–I: Initial compression →
mainly due to preloading.
Stage–II: Primary Consolidation
→ due to dissipation of pore
water pressure (expulsion of
water)
Stage–III: Secondary
Consolidation → due to plastic
readjustment of soil fabric.
CONSOLIDATION TEST
Deformation ~ Time Plot

8. 8
Methods of Determining Cv
1. Casagrande’s log-time
method (1938)
2. Taylor’s Square root of
time method (1948)
CONSOLIDATION TEST
Deformation ~ Time Plot
Used to determine Cv (coefficient
of consolidation)
 Rate of consolidation
 Consolidation Time

9. 9
DEFORMATION ~ TIME PLOT
Casagrande’s log-time method





 

t
HT
CV
2
Used to determine Cv (coefficient
of consolidation)
 Rate of consolidation
 Consolidation Time
U = 100%R100
t1 t2=4t1
x
x
R0
U = 0%
t50
U = 50% R50 = (R0+R100)/2
t50: obtained from deformation ~ time plot
H = t/2 (for two-way drainage)
t = height of specimen
T = Time factor = 0.197 (for u=50%)
R2
R1





 

50
50
2
t
HT
CV

10. 10
 Plot specimen deformation against log-of-time
 R0 and R100 represent dial readings for zero percent
(U0) and 100% (U100) degree of consolidation
respectively.
 For determining U100
 Draw two tangents to the bottom part of curve.
Intersection of these tangents represent U100.
 The corresponding abscissa represent t100.
 For determining U0
 Choose any two times t1 and t2, such that t2 = 4t1
 The corresponding dial readings are R1 and R2
 Mark off vertical distance between R1 & R2 (say ‘x’)
 R0 lies ‘x’ distance above R1.
 R50 represents 50% degree of consolidation (U50)
 R50 lies mid-way between R0 and R100.
 The time (t50) is used for determination of Cv.
U = 100%R100
t1 t2=4t1
x
x
R0
U = 0%
t50
U = 50% R50 = (R0+R100)/2
DEFORMATION ~ TIME PLOT
Casagrande’s log-time method
R2
R1

11. 11
DEFORMATION ~ TIME PLOT
Taylor’s Square-root of Time Method
Uses t90 for computing Cv
Plot deformation ~ square root of time.
 Project the straight line portion of
the initial part of curve backward to
zero time to define R0 or U0.
 Draw a second line from R0 with
abscissa 1.15 times as large as the
corresponding values on the first
line.
 The intersection of the second line
with the laboratory curve defines R90
or U90.
Deformation
time
x
1.15x
U90
U0R0
√t90





 

90
2
90
t
HT
CV
T90 = Time factor for 90% degree of
consolidation (U90).
T90 = 0.848

12. 12
DEFORMATION ~ TIME PLOT
Summary
Used for determining Cv
Cv is used for determining rate of consolidation, and
consolidation time.
Casagrande’s log-time fitting method
Determines Cv corresponding to U=50%.
Deformation ~ log-time plot
Taylor’s square root of time method
Determines Cv corresponding to U=90%.
Deformation ~ square root of time plot
Cv determined from √(time) method is often slightly
greater than the log-time method.





 

50
2
50
t
HT
CV





 

90
2
90
t
HT
CV

13. 13
CONCLUDED
REFERENCE MATERIAL
Principles of Geotechnical Engineering – (7th Edition)
Braja M. Das
Chapter #11
An Introduction to Geotechnical Engineering (2nd Edition)
By R. D. Holtz, W. D. Kovacs and T. C. Sheahan
Chapter #8 & 9