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 # 17
27-Mar-2018

2. 2
CONSOLIDATION OF SOIL
Load/stress application on soil
→ causes soil compression
Reasons for soil compression
 Compression/expulsion of air in soil
voids
 Soil compaction (already discussed)
 Distortion/crushing of soil grains
 Negligible under normal structural loads
 Expulsion/compression of water from
the voids
 Soil consolidation

3. 3
CONSOLIDATION OF SOIL
Which soils have high water holding ability?
Phenomenon associated with saturated fine
grained soils only.
Consolidation → compression/volume
reduction of soil mass due to expulsion of
water when subjected to external load/stress.

4. 4
Before Consolidation
Solids
Water
After Consolidation
Soil volume reduction due to expulsion of water upon
application of external load/stress.
fully saturated soil, so all voids filled with water only (no air)
Solids
Water
CONSOLIDATION OF SOIL
Saturated Fine-grained Soil

5. 5
Soil volume reduction due to expulsion of water upon
application of external load/stress.
→ Settlement of structures
→ Cracks in walls, foundations, etc.
Consolidation Damages

6. 6
MECHANISM OF CONSOLIDATION
Spring-Cylinder Model / Hydro-mechanical Analogue
Spring
Cylinder
Cross-sectional area = A
Water
Piston
(Frictionless, water-tight)
Pressure
Gauge

7. 7
Consolidation Model
(Spring-Cylinder Model / Hydro-mechanical Analogue)
Load (P) applied on the piston.
Load, P
P = PS + PW
PS = Load carried by the spring
PW = Load carried by water
With the valve closed
PS = 0, &
PW = P
Piston

8. 8
When the valve is opened → water flow outward
Decrease in excess hydrostatic pressure
Increase in compression of spring
Load, P
With the valve opened
PS > 0, &
PW < P
P = PS + PW
PS = Load carried by the spring
PW = Load carried by water
Consolidation Model
(Spring-Cylinder Model / Hydro-mechanical Analogue)

9. 9
After some time → equilibrium is reached
Load, P
With the valve opened; after some
time span.
Excess hydrostatic pressure, Δu = 0
PW = 0, &
PS = P
P = PS + PW
PS = Load carried by the spring
PW = Load carried by water
Consolidation Model
(Spring-Cylinder Model / Hydro-mechanical Analogue)

10. 10
Valve
Closed
Spring-Cylinder Model – Summary
Time dependent response of saturated fine-grained soils.
With the valve closed
PS = 0, &
PW = P
With the valve opened
PS > 0, &
PW < P
After t >>> 0
PW = 0, &
PS = P
Valve
open
In case of soil
Stress acting on soil mass → Total Stress = σ
Stress carried by water → Pore water pressure = u
Stress carried by soil particles → Effective stress = σ’
σ = σ’+ u OR σ' = σ - u
Spring-cylinder assembly
Total load acting on the system = P
Load carried by water = PW
Load carried by Spring = PS
P = PS + PW OR PS = P - PW

11. 11
 Similar phenomenon occurs when load is applied on a saturated
clay deposit (very low permeability).
 Load is first taken by water only.
 Pore water pressure slowly dissipates,
 Soil particles start taking load gradually
 After some time excess water pressure is completely
dissipated through voids, and the load is carried only by soil
particles.
Spring-Cylinder Model → Application to Soil

12. 13
Consolidation Model
(Hydro-mechanical Analog)
 = ’ + u
Valve
opened
Initially
u = 
’ = 0
Finally
u = 0
’ = 
Time
,u,’
Valve
closed

13. 14
Consolidation vs Compaction
Compaction Consolidation
Applicable to unsaturated soils. Applicable to saturated soils.
Decrease in air voids (not water voids) Decrease in water voids (air voids do
not exist).
Applicable for both fine-grained and
coarse-grained soils
Only applicable for fine-grained soils
Instantaneous process Time-dependent process
Can occur over 100s of year.
May be accomplished by rolling,
tamping, or vibration.
In general, caused by static loading.

14. 15
Inferences from Spring-Cylinder Model
Magnitude of consolidation settlement
dependent on compressibility of soil (i.e. the stiffness of the spring)
 expressed in term of compression index (Cc)
Rate of consolidation/settlement
dependent on
i. permeability, &
ii. compressibility of soil.
 expressed in term of co-efficient of consolidation (Cv)

15. 16
Inferences from Spring-Cylinder Model





 

VC
HT
t
2
%60;
1004
2






 ufor
u
T

%60
);100(log933.0781.1 10


ufor
uT
%90;848.0
%50;197.0
90
50


uforT
uforT
Magnitude of consolidation → compression index (Cc)
Rate of consolidation → co-efficient of consolidation (Cv)
 Time required for consolidation can be determined?
Derivation
– SELF STUDY –
(next two slides)
where,
t = time required for any degree of
consolidation
CV = coefficient of consolidation
H = length of the drainage path
(H = t → for one-way drainage
H = t/2 → for two-way drainage)
t = thickness of consolidating soil layer
T = constant known as ‘Time Factor’
u = degree of consolidation

16. 17
Inferences from Spring-Cylinder Model
Magnitude of consolidation → compression index (Cc)
Rate of consolidation → co-efficient of consolidation (Cv)
 Time required for consolidation can be determined?
Vt  v
t 1
kiv Darcy’s equation →
Hhi  wh 
H
k
v
w 




)1(
)()(
1






 Hk
t
w

Permeability / Velocity of
flow through soil
Volume of water required
to be squeezed out
)2(Hmt v 
t = time required for any degree of consolidation
Δσ = change in stress
mV = coefficient of volume compressibility
H = length of the drainage path
(H = t → for one-way drainage
H = t/2 → for two-way drainage)
t = thickness of consolidating soil layer
Next Chapter
(Permeability & Seepage)

17. 18
Inferences from Spring-Cylinder Model
Combining (1) and (2).
)3(
2





 
k
Hm
t wv 









wv
V
m
k
C







VC
H
t
2






 

VC
HT
t
2
)1(
)()(
1






 Hk
t
w

)2(Hmt v 
Replacing CV in (3);
Magnitude of consolidation → compression index (Cc)
Rate of consolidation → co-efficient of consolidation (Cv)
 Time required for consolidation can be determined?

18. 19
Consolidation Time (t)
Time required for consolidation (consolidation time) is
independent of the magnitude of stress change (Δσ).





 

VC
HT
t
2








wv
v
m
k
C






 

k
mHT
t wv 2
&
where,
t = time required for any degree of
consolidation
CV = coefficient of consolidation
H = length of the drainage path
T = constant known as ‘Time Factor’
u = degree of consolidation

19. 20
Consolidation Settlement in the Field
External stress (Δσ) applied on a soil stratum in the field.
 SAND→ Quick drainage of water → Immediate settlement
 CLAY → Slow drainage → Consolidation settlement (time dependent)
H
depth
SandG.W.T
Clay
Sand


20. 21
One-Dimensional Consolidation
Drainage and deformations occur in vertical direction only.
(none laterally)
A reasonable simplification for solving consolidation problems

21. 22
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.

22. 23
1-D Lab Consolidation
NSL
metal ring
SOIL
Porous Stones
Field
Undisturbed soil
specimen
Lab
Consolidometer / Oedometer
Stopwatch

23. 24
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.

24. 25
Deformation ~ Time Plot
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.

25. 26
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

26. 27
Consolidation Settlement in the Field
Immediately after load application (t = 0)
All the applied stress carried by pore water only, Δu = Δσ
Effective stress, Δσ’ = 0
Remember
Δσ = Δu + Δσ’

27. 28
Consolidation Settlement in the Field
Some time after load application (0 < t < ∞)
Pore water pressure starts dissipating, Δu < Δσ
Additional stress start getting transferred to soil particles,
Δσ’ > 0
Remember
Δσ = Δu + Δσ’

28. 29
Consolidation Settlement in the Field
Long time after load application (t = ∞)
Pore water pressure dissipated completely, Δu = 0
All the applied stress being taken by soil particles, Δσ’ = Δσ
Remember
Δσ = Δu + Δσ’