The document discusses different types of soil settlement including immediate, primary, and secondary consolidation settlements. It provides formulas to calculate settlement, defines concepts like void ratio, compression index, coefficient of consolidation, and overconsolidation ratio. It also includes sample calculations for estimating primary consolidation settlement of a clay layer under a surcharge load based on laboratory consolidation test results and given soil properties.

1. GEOTECHNICAL ENGINEERING 2

2. Compressibility of Soil
The increase in stress caused by foundation and other
loads compresses a soil layer. Possible causes are:
1. deformation of soil particles
2. relocations of soil particles
3. expulsion of air or water from the void spaces
Categories of soil settlement
1. Immediate settlement – due to elastic deformation of dry, moist, and
saturated soil without change in moisture content .
2. Primary consolidation settlement – due to volume change in saturated
cohesive soils due to expulsion of water that occupied the void spaces
3. Secondary consolidation settlement – due to plastic adjustment of soil
fabrics. It is an additional form of compression that occurs at constant
effective stress.

3. Solids Solids
ΔH
HS
eHS
Voids Voids
e’HS
H’=HS(1+e’)
H=HS(1+e)
Original state Compressed state
Basic settlement formula
ONE DIMENSIONAL PRIMARY CONSOLIDATION SETTLEMENT
)
1
( e
H
H S 

)
1
( e
H
HS


)
'
1
(
' e
H
H S 

)
'
1
(
)
1
(
' e
e
H
H 


'
H
H
H 


)
'
1
(
)
1
(
e
e
H
H
H 





4. )]
'
1
(
)
1
(
1
1
[ e
e
H
H 




]
)
1
(
)
'
1
(
)
1
(
[
e
e
e
H
H






)
1
(
)
'
(
e
e
e
H
H




Let e =e0 =void ratio before the vertical load is applied
)
1
(
)
'
(
0
0
e
e
e
H
H




H = thickness of stratum
e0 =void ratio before the vertical load is applied
e’ =void ratio after the vertical load is applied
stratum
the
of
settlement

H
Where:
'
e
e
e 


)
1
( e
e
H
H




)
1
( 0
e
e
H
H




ratio
in void
change

e

5. A. Primary Consolidation Settlement
1. Normally Consolidated Clays
Normally consolidated clays are those whose present effective overburden
pressure is the maximum pressure that the soil was subjected to in the past.
The maximum effective past pressure is called the preconsolidation
pressure.
ΔH= primary consolidation settlement
Cc = compression index
Cc = 0.009 (LL – 10)
eo = in situ void ratio
H = thickness of clay layer
Δp = average increase of effective stress on clay layer
po = average effective stress at the mid-height of clay layer
0
0 )
(
log
)
1
( p
p
p
e
H
C
H
o
C 





6. 2. Over Consolidated Clays
Over consolidated clays are those whose present effective overburden
pressure is less than that which the soil experienced in the past.
1. when Po + ΔP < Pc
2. when Po + ΔP > Pc
Where:
Pc = preconsolidation pressure
Cs = swell index (ranges from 1/5 to 1/10 of Cc)
0
0 )
(
log
)
1
( p
p
p
e
H
C
H
o
S 




c
o
C
c
o
S
p
p
p
e
H
C
p
p
e
H
C
H
)
(
log
)
1
(
)
(
log
)
1
(
0
0








7. Compression Index
Skempton
remolded clay )
7
(
007
.
0 
 LL
CC
Undisturbed clay )
10
(
009
.
0 
 LL
CC
38
.
2
2
.
1
)
1
(
141
.
0
G
e
G
C o
C


Rendon - Herreo
Nishida
all clays )
27
.
0
(
15
.
1 0 
 e
CC
Swell Index
Swell index is usually smaller than the compression index and in most cases
C
C
S C
C
C
10
1
to
5
1

Nagaray and Murty )
100
%
(
0463
.
0
LL
G
CS 

8. B. Secondary Consolidation Settlement
1
2 log
log t
t
C e
a



layer
clay
of
thickness
H
e
ion,
consolidat
primary
of
end
at the
ratio
void
e
required
is
settlement
where
,
settlement
primary
of
completion
after
time
t
settlement
primary
of
completion
for
time
t
ratio
in void
change
index
n
compressio
secondary
0
p
p
2
1










e
e
e
Ca
1
2
'
log
t
t
H
C
H a
S 

p
a
a
e
C
C


1
'

9. C. Immediate Settlement
Immediate or elastic settlement of foundation occur directly after
the application of the load without change in moisture content. This
depends on the flexibility of the foundation and the type of material
which it is resting.
Immediate settlement of foundation resting on elastic material of
finite thickness
f
i I
E
u
pB
H )
1
(
2



less)
(dimension
factor
influence
I
psf
or
kPa
in
soil
of
elasticity
of
modulus
E
ratio
s
poisson'
ft
or
m
in
foundation
of
diameter
or
width
B
psf
or
kPa
in
applied
pressure
net
p
:
where
f 






10. Table 1: Influence Factors for Foundation
Table 2: Values of Poisson’s Ratio
Table 3:Values of Modulus of Elasticity
Total settlement of foundations
The total settlement of foundation is the sum of the primary, secondary and
Immediate settlement thus:
i
s
T H
H
H
H 







11. Overconsolidation ratio
soil
ed
consolidat
normally
is
soil
1,
OCR
if
stress
effective
overburden
p
stress
dation
preconsoli
p
ratio
idation
overconsol
OCR
:
Where
o
c





o
c
p
p
OCR
Calculation of consolidation settlement under a foundation
layer
the
of
bottom
at the
pressure
in
increase
P
layer
the
of
middle
at the
pressure
in
increase
P
layer
the
of
top
at the
pressure
in
increase
P
:
Where
6
4
Rule)
s
impson'
pressure(S
in
increase
Average
b
m
t












 b
m
t
ave
P
P
P
P

12. Time rate of consolidation
The time required to achieve a certain degree of consolidation is evaluated
as a function of the shortest drainage path within the compressible zone ,
coeffecient of consolidation and a dimensionless time factor.
layer
the
of
bottom
and
top
at the
occur
drainage
if
layer
the
of
thickness
the
of
half
one
layer;
the
of
bottom
or
top
at the
occur
drainage
if
drainage
the
of
thickness
to
equal
zone
le
compressib
H
less)
(dimension
factor
time
T
ion
consolidat
of
t
coeffecien
ion
consolidat
of
degree
U
ion
consolidat
of
rate
time
t
:
where
dr
v





v
C
v
dr
v
C
H
T
t
2
)
(


13. Table 4 : Variation of Tv with U

14. %
U
of
ion
consolidat
a
reach
to
time
t
%
U
of
ion
consolidat
a
reach
to
time
t
where
t
t
)
H
(
)
(H
t
t
expirement
laboratory
from
result
field
settlement
between
Relation
%)
100
log(
933
.
0
781
.
1
T
%
60
U
100
%
4
60%
to
0
U
factor
time
of
value
e
approximat
2
2
1
1
2
2
2
1
2
1
2
drlab
2
drfield
lab
field
v
2
















U
U
U
U
Tv


15. Degree of consolidation at a distance z at any time
wo
wz
z
p
p
U 
1
Average Degree of consolidation for entire depth at any time
max
H
H
U t
z



ion
consolidat
primary
to
due
layer
the
of
settlement
ultimate
H
t
at time
layer
the
of
settlement
H
t
at time
pressure
pore
excess
initial
p
t
at time
pressure
pore
excess
p
where
max
t
wo
wz







16. Coeffecient of consolidation
ion
consolidat
50%
for
time
t
ion
consolidat
90%
for
time
t
:
where
)
(
197
.
0
C
Method
Time
Log
)
(
848
.
0
C
Method
Time
Root
50
90
50
2
v
90
2
v




t
H
t
H
dr
dr
Coeffecient of Volume compressibility
pressure
in
rise
p
ratio
void
final
e
ratio,
void
initial
e
ility
compressib
volume
of
t
coeffecien
m
:
where
C
k
:
range
loading
for the
ty
conductivi
Hydraulic
2
e
)
1
(
)
(
m
0
V
v
0
ave
0
v












w
v
ave
m
e
e
e
p
e
e


17. Problem
The soil profile shown in figure is to carry a surcharge of 60 kPa applied at the
ground surface. The result of laboratory consolidation test conducted on a
specimen taken from the middle of the clay layer is also shown. Calculate the
settlement in the field caused by primary consolidation due to surcharge.
Clay
60 kPa
6 m
1
.
1
e
,
2
.
18 0
3


m
kN
sat

Figure:

18. 1.12
1.10
1.08
1.06
1.04
1.02
1.00
10 20 40 60 80 100
Void
ratio
e’
Pressure, kPa
Laboratory consolidation Test results

19. Solution:
Given : e=1.1, Δp=60 kPa, H = 6m, γsat=18.2 kN/m3
Pressure at mid-height of clay layer
3m
h
where 
 h
p b
o 
kPa
17
.
25
3
)
81
.
9
2
.
18
( 


o
p
How to determine e’( void ratio after the surcharge load is applied)
a) compute Po+ΔP
b) Project vertically (Po + ΔP) on the laboratory result graph
c) Project horizontally e’ on the laboratory result graph
60
17
.
25 


 p
po
kPa
17
.
85


 p
po

20. 1.12
1.10
1.08
1.06
1.04
1.02
1.00
10 20 40 60 80 100
Void
ratio
e’
Pressure, kPa
Laboratory consolidation Test results
85.17 kPa
e’=1.047
Settlement due to primary consolidation
)
1
(
)
'
(
0
0
e
e
e
H
H




)
1
.
1
1
(
)
047
.
1
1
.
1
(
6



H m
1514
.
0

H

21. Δp=50 kPa
sand
γdry=16 kN/m3
γsat=18.5 kN/m3
sand
clay
sand
γsat=19 kN/m3
eo=0.95
LL =50
3m
6m
8m
Problem
A soil profile is shown below. A uniformly distributed load Δp=50 kPa is applied at
the ground surface. Assume Cs = 1/5Cc. Determine the settlement of clay caused
by primary consolidation if: a) the clay is normally consolidated b) the preconsolidation
pressure pc is 210 kPa c) the preconsolidation pressure pc = 150 kPa.
Figure:

22. Solution:
a) Normally consolidated Clay
)
log(
)
1
( 0
0 p
p
p
e
C
H
H o
c 



 95
.
0
e
8m,
H
kPa,
50 0 


p
Initial effective stress at mid-height of clay layer
clay
b
sand
b
sand
dry h
h
h
p 

 

 2
1
0
4
)
81
.
9
19
(
6
)
81
.
9
5
.
18
(
)
3
(
16
0 




p
kPa
9
.
136
0 
p
)
9
.
136
50
9
.
136
log(
)
95
.
0
1
(
36
.
0
8



H
)
10
(
009
.
0 
 LL
Cc 36
.
0
)
10
50
(
009
.
0 


c
C
m
2
.
0

H mm
200

H

23. b) 0ver consolidated Clay pc=210 kPa
po+Δp < pc
)
log(
)
1
( 0
0 p
p
p
e
C
H
H o
s 




c
s C
C
5
1
 072
.
0
36
.
0
5
1


s
C
)
9
.
136
50
9
.
136
log(
)
95
.
0
1
(
072
.
0
8



H m
04
.
0

H mm
40

H
c) 0ver consolidated Clay pc=150 kPa
po+Δp > pc
)
log(
)
1
(
)
log(
)
1
( 0
0
0 c
o
c
c
s
p
p
p
e
C
H
p
p
e
C
H
H







)
150
50
9
.
136
log(
)
95
.
0
1
(
36
.
0
8
)
9
.
136
150
log(
)
95
.
0
1
(
072
.
0
8





H
m
1528
.
0

H mm
8
.
152

H

24. Problem
A 2 m clay layer (G=2.72, e =0.92,Cc=1/3) is overlain with 3 m thick sand
(G=2.62, e=0.5,ω=0). The water table is 1.5 m below the ground(sand) surface.
If a 3m thick landfill (γ=17.3 kN/m3) is placed over the existing ground surface,
compute the consolidation settlement of the clay layer.
Landfill
sand
e=0.5
Clay
G=2.62
e=0.92
G=2.72
3 m
1.5 m
1.5 m
2 m
Figure:
γ=17.3 kN/m3

25. Solution
Unit weight of soil
Sand above water table
w
s
e
G



)
1
(
)
1
(


 81
.
9
)
5
.
0
1
(
)
0
1
(
62
.
2



s
 3
13
.
17
m
kN
s 

Sand below water table (submerged)
w
bs
e
G


)
1
(
)
1
(


 81
.
9
)
5
.
0
1
(
)
1
62
.
2
(



bs
 3
59
.
10
m
kN
bs 

Clay (submerged)
w
bc
e
G


)
1
(
)
1
(


 81
.
9
)
92
.
0
1
(
)
1
72
.
2
(



bc
 3
79
.
8
m
kN
bc 


26. )
log(
)
1
( 0
0 p
p
p
e
C
H
H o
c 



 92
.
0
e
2m,
H 0 

Initial effective stress at mid-height of clay layer
clay
b
sand
b
sand
dry h
h
h
p 

 

 2
1
0
kPa
37
.
50
)
1
(
79
.
8
)
5
.
1
(
59
.
10
)
5
.
1
(
13
.
17
0 



p
Average increase of effective stress on clay layer(due to landfill)
L
Lh
p 

 )
3
(
3
.
17

p kPa
9
.
51

p
)
37
.
50
9
.
51
37
.
50
log(
)
92
.
0
1
(
3
1
2



H
m
1608
.
0

H mm
8
.
160

H

27. Problem
A normally consolidated clay 3 m thick has the following properties:
Initial void ratio e0 = 0.8
Compression index Cc = 0.25
Average effective pressure po= 125 kPa
Expected pressure increase Δp=45 kPa
Secondary Compression index Ca = 0.02
Time for completion of primary settlement=1.5 years
What is the total settlement of the clay layer five years after the completion
of the primary settlement ?
Solution:
Primary consolidation settlement
)
log(
)
1
( 0
0 p
p
p
e
C
H
H o
c
P





kPa
45
p
kPa,
125
,
25
.
0
,
8
.
0
e
3m,
H 0
0 




 p
Cc

28. )
125
45
125
log(
)
8
.
0
1
(
25
.
0
3



 P
H m
0557
.
0

 P
H
mm
7
.
55

 P
H
Secondary consolidation settlement
1
2
'
log
t
t
H
C
H a
S 

p
a
a
e
C
C


1
'
e
e
ep 

 0 02
.
0
C
3m,
H a 

Solve for Δe using primary consolidation settlement equations
1
.
)
log(
)
1
( 0
0
EQ
p
p
p
e
C
H
H o
c
P 





2
.
)
1
( 0
EQ
e
e
H
H p 





29. 2
.
1
. EQ
EQ 
0
0
0 1
)
log(
)
1
( e
e
H
p
p
p
e
C
H o
c






)
log(
0
p
p
p
C
e o
c



 )
125
45
125
log(
25
.
0


e 0334
.
0

e
e
e
ep 

 0 0334
.
0
8
.
0 

p
e 7666
.
0

p
e
p
a
a
e
C
C


1
'
776
.
0
1
02
.
0
'


a
C 01132
.
0
'

a
C
1
2
'
log
t
t
H
C
H a
S 
 years
5
.
1
years
5 1
2 
 t
t
5
.
1
5
log
)
3
(
01132
.
0

 S
H m
01776
.
0

 S
H mm
76
.
17

 S
H
Total consolidation settlement s
P
T H
H
H 



 mm
46
.
73

 T
H

30. Problem
A rigid column footing 1.2 m in diameter is constructed on unsaturated clay layer.
The load on the footing is 170 kN. If E = 6900 kPa and μ=0.20, compute its
immediate settlement.
f
i I
E
u
pB
H )
1
(
2



Solution
1.2m
B
kN,
170
F
,
0.2
kPa,
6900
E 


 
A
F
p  2
)
6
.
0
(
170


p kPa
31
.
150

p
Determine If using Table 1
0.79

f
I
Immediate settlement
79
.
0
)
6900
2
.
0
1
)(
2
.
1
(
31
.
150
2


 i
H
m
0198
.
0

 i
H mm
8
.
19

 i
H

31. Problem
It is desired to calculate the consolidation settlement 0f a 3 m thick clay layer shown
in figure that would result from the load carried by the footing measuring 3 m by 1.5 m.
Assume the clay to be normally consolidated and the load of footings result to an
increase of pressure of 23.2 kPa, 12.22 kPa, and 6.1 kPa at the top, midheight and
bottom of the clay layer. Assume that the pressure increase varies parabolically
And use Simpson’s rule to calculate the average pressure increase.
Determine the following:
a) Overburden pressure at the base of the footing
b) Average pressure increase in the clay layer
c) Initial effective stress at the midheight of the clay layer
d) Primary consolidation settlement of the clay layer

32. 1.5 m
3.0 m
1.5 m
3.0 m
Footing size 3 m x 1.5 m
Dry Sand
γdry=14.8 kN/m3
Sand
γsat=18.5 kN/m3
clay
γsat=19 kN/m3
LL = 50
eo=1.0
1500 kN
Figure: Solution
a) Overburden Pressure at the footing base
the pressure at the base of the footing due
to height of soil above the base of the footing
m
1.5
D
)
(
0 
 D
q sand
dry

)
5
.
1
(
8
.
14
0
q
kPa
2
.
22
0
q
b) Average increase in pressure in the clay
layer
6
4 b
m
t
ave
P
P
P
P







kPa
1
.
6
kPa
22
.
12
kPa
2
.
23 




 b
m
t P
P
P

33. 6
1
.
6
)
22
.
12
(
4
2
.
23 


 ave
P kPa
03
.
13

 ave
P
c) initial effective pressure at midheight of clay layer
clay
b
sand
b
sand
dry h
h
h
p 

 

 2
1
0
5
.
1
)
81
.
9
19
(
5
.
1
)
81
.
9
5
.
18
(
)
5
.
4
(
8
.
14
0 




p
kPa
42
.
93
0 
p
d) Primary consolidation settlement of clay layer
)
log(
)
1
( 0
0 p
p
p
e
C
H
H o
c 



 kPa
03
.
13
0
.
1
50
LL 0 


 p
e
10)
-
0.009(LL
Cc  10)
-
0.009(50
Cc  0.36
Cc 
)
42
.
93
03
.
13
42
.
93
log(
)
1
1
(
36
.
0
3



H m
0306
.
0

H mm
6
.
30

H

34. 120 kPa
6 m
4.6 m
3.6 m
h
Sand
Clay
D
Problem
A surcharge of 120 kPa is applied on the ground surface on the soil profile
shown in figure.
a) How high will the water rise in the piezometer tube immediately after the
application of the load ?
b) What is the degree of consolidation at point D when h = 6m ?
c) Find h when the degree of consolidation at D is 80%

35. a) Height that water rise in the piezometer tube immediately after the
application of the load
w
p
h


81
.
9
120

h m
232
.
12

h
b. Degree of consolidation at point D when h = 6m
%
100
]
1
[
wo
wz
D
p
p
U 

m
232
.
12
h0 
 o
w
wo h
p  )
232
.
12
(
81
.
9

wo
p kPa
120

wo
p
m
6
hz 
 z
w
wz h
p  )
6
(
81
.
9

wz
p kPa
58.86

wz
p
%
100
]
120
86
.
58
1
[ 

D
U %
95
.
50

D
U

36. b. h when the degree of consolidation at point D is 80%
%
100
]
1
[
wo
wz
D
p
p
U 

%
100
]
120
1
[
%
80 wz
p


kPa
24

wz
p kPa
24

h
w
 24
81
.
9 
h m
446
.
2

h
Problem
Under a given surcharge, a 5m thick clay layer has a consolidation settlement of
Of 305 mm. Cv=0.003 cm2/sec.
a) What is the average degree of consolidation for the clay layer when the
settlement is 75 mm ?
b) How long will it take for 50% consolidation to occur if the layer is drained at
the top only ?
c) How long will it take for 50% consolidation to occur if the layer is drained at
the both ends ?

37. a) average degree of consolidation for the clay layer when the
settlement is 75 mm ?
Solution:
mm
75
H
mm,
305 t
max 


H
max
H
H
U t
D



305
75

D
U 2459
.
0

D
U %
59
.
24

D
U
b) Time for 50% consolidation to occur if the layer is drained at
the top only
v
dr
v
C
H
T
t
2
)
(

197
.
0

v
T
cm
m
Hdr 500
5
:
drainage
Single


. Cv=0.003 cm2/sec

38. 003
.
0
)
500
(
197
.
0
2

t
sec
16416667

t )
24
(
3600
16416667

t days
190

t
b) Time for 50% consolidation to occur if the layer is drained at both ends
cm
m
Hdr 250
5
2
1
:
drainage
way
Two


003
.
0
)
250
(
197
.
0
2

t
sec
4104167

t )
24
(
3600
4104167

t days
5
.
47

t
Problem:
A 3.2 m thick layer of saturated clay under a surcharge loading underwent 90%
Primary consolidation in 80 days with two way drainage.
a) Determine the coeffecient of consolidation for the pressure range
b) For a 10-cm thick specimen of the said clay, how long will it take to undergo
90% consolidation in the laboratory for the same pressure range ?

39. Solution:
t
H
T
C dr
v
v
2
)
(

cm
m
Hdr 160
6
.
1
2
.
3
2
1
:
drainage
way
Two



t=80 days
t=80x24x3600 =6912000 sec
6912000
)
160
(
848
.
0
2

v
C
sec
00314
.
0
2
cm
Cv 
b) Time to undergo 90% consolidation in the laboratory
)
H
(
)
(H
t
t
2
drlab
2
drfield
lab
field

sec
6912000
tfield 
cm
160
Hdrfield 
cm
5
2
10
Hlab 

)
5
(
)
(160
t
6912000
2
2
lab

sec
6750

lab
t hrs
875
.
1

lab
t
0.848

v
T

40. Problem
Under normal loading condition, a 3.6 m thick clay (normally consolidated) has
po=190 kPa and e0=1.22 The hydraulic conductivity of the clay for the loading range
is 6.1 x10-5 m/day. A surcharge of 190 kPa reduces its void ratio to 0.98.
a) What is the coeffecient of volume compressibility of clay ?
b) What is the coeffecient of consolidation of the clay ?
c) How long will it take for this clay layer to reach 60% consolidation if it is
drained on one side only
a) Coeffecient of volume of compressibility of clay
Solution:
)
1
(
)
(
m 0
v
ave
e
p
e
e




2
e 0
ave
e
e 
 22
.
1
e0  98
.
0
e  kPa
190

p
2
22
.
1
98
.
0
eave

 1
.
1
eave 
)
1
.
1
1
(
190
98
.
0
22
.
1
mv



kN
m /
0006015
.
0
m 2
v 

41. b) coeffecient of consolidation of clay
w
v
m 
v
C
k  day
m/
6.1x10
k -5
 kN
m /
0.0006015
m 2
V 
)
81
.
9
(
0006015
.
0
C
)
6.1(10 v
-5

day
m
01034
.
0
C
2
v 
c) Time for 60% consolidation of clay
v
dr
v
C
H
T
t
2
)
(

0.286

v
T
m
Hdr 6
.
3
:
drainage
way
One

day
m
01034
.
0
C
2
v 
01034
.
0
)
6
.
3
(
286
.
0
2

t
days
358.5
t 

42. Problem
The following data were obtained from a laboratory consolidation test on a 30 mm
thick clay specimen drained on both sides.
Pressure (kPa) Void ratio
60 0. 87
150 0.72
time for 60% consolidation = 3.2 minutes
a) Determine the coeffecient of volume compressibility in m2/kN
b) Determine the hydraulic conductivity for the loading range
a) Coeffecient of volume of compressibility of clay
Solution:
)
1
(
)
(
m 0
v
ave
e
p
e
e




2
e 0
ave
e
e 
 87
.
0
e0  72
.
0
e 
60
-
50
1

p kPa
90

p
2
87
.
0
72
.
0
eave

 795
.
0
eave 
)
795
.
0
1
(
90
)
72
.
0
87
.
0
(
mv



kN
m
0009285
.
0
m
2
v 

43. b) hydraulic conductivity for the loading range
w
v
m 
v
C
k 
kN
m
0009285
.
0
m
2
v 
t
H
T
C dr
v
v
2
)
(
 min
2
.
3

t drainage)
(double
m
0.015
mm
15
2
30



dr
H
0.286

v
T
2
.
3
)
015
.
0
(
286
.
0
2

v
C
min
m
0000211
.
0
2

v
C
9.81
0.0009285)
0.0000211(
k 
min
m
1.83x10
k 7
-


44. Δp=60 kPa
sand
γdry=15kN/m3
γsat=18.0 kN/m3
sand
clay
sand
γsat=19 kN/m3
eo=0.95
LL =50
3.1m
6.2m
8.2m
PROBLEM SET #1
1. A soil profile is shown below. A uniformly distributed load Δp=60 kPa is applied at
the ground surface. Assume Cs = 1/5Cc. Determine the settlement of clay caused
by primary consolidation if: a) the clay is normally consolidated b) the preconsolidation
pressure pc is 220 kPa c) the preconsolidation pressure pc = 140 kPa.
Figure:

45. 2. A normally consolidated clay 3.6 m thick has the following properties:
Initial void ratio e0 = 0.85
Compression index Cc = 0.29
Average effective pressure po= 128 kPa
Expected pressure increase Δp=52 kPa
Secondary Compression index Ca = 0.02
Time for completion of primary settlement=2 years
What is the total settlement of the clay layer five years after the completion
of the primary settlement ?
3. It is desired to calculate the consolidation settlement 0f a 4 m thick clay layer shown
in figure that would result from the load carried by the footing measuring 3 m by 2.5 m.
Assume the clay to be normally consolidated and the load of footings result to an
increase of pressure of 25.2 kPa, 18.22 kPa, and 9.1 kPa at the top, midheight and
bottom of the clay layer. Assume that the pressure increase varies parabolically
And use Simpson’s rule to calculate the average pressure increase.

46. Determine the following:
a) Overburden pressure at the base of the footing
b) Average pressure increase in the clay layer
c) Initial effective stress at the midheight of the clay layer
d) Primary consolidation settlement of the clay layer
1.5 m
3.0 m
1.5 m
4.0 m
Footing size 3 m x 2.5 m
Dry Sand
γdry=14.8 kN/m3
Sand
γsat=16.5 kN/m3
clay
γsat=18 kN/m3
LL = 50
eo=1.0
1800 kN
Figure:

47. 4. A rigid column footing 1.5 m in diameter is constructed on unsaturated clay layer.
The load on the footing is 190 kN. If E = 7800 kPa and μ=0.30, compute its
immediate settlement.
5. The following data were obtained from a laboratory consolidation test on a 30 mm
thick clay specimen drained on both sides.
Pressure (kPa) Void ratio
70 1.0
180 0.80
time for 70% consolidation = 5 minutes
a) Determine the coeffecient of volume compressibility in m2/kN
b) Determine the hydraulic conductivity for the loading range
6. A 3.6 m thick layer of saturated clay under a surcharge loading underwent 80%
Primary consolidation in 90 days with two way drainage.
a) Determine the coeffecient of consolidation for the pressure range
b) For a 10-cm thick specimen of the said clay, how long will it take to undergo
90% consolidation in the laboratory for the same pressure range ?

48. 150 kPa
6.5 m
4.8 m
3.8m
h
Sand
Clay
D
7. A surcharge of 150 kPa is applied on the ground surface on the soil profile
shown in figure.
a) How high will the water rise in the piezometer tube immediately after the
application of the load ?
b) What is the degree of consolidation at point D when h = 7.6m ?
c) Find h when the degree of consolidation at D is 80%

49. Shear Strength of Soil
The shear strength of soil maybe attributed to three basic components.
1. Frictional resistance to sliding between soil particles
2. Cohesion and adhesion between particles.
3. Interlocking and bridging of solid particles to resist deformation.
A material fail because of the critical combination of normal and
shearing stress. Thus a failure plane can be expressed a function of normal
and shearing stress thus,
)
(
 f
f 
Mohr-Coulomb failure criteria


 tan

 c
f
stress
shearing
stress
normal
cohesion
c
friction
internal
of
angle
:
where
f 







50. Normal stress σ
Shearing stress τ
Y
X
Z

c Mohr- Coulomb
Failure criteria
Mohr Failure envelope
A
0,0
Note: if A passes through (0,0)
then c=0
Mohr Failure envelope and Mohr- Coulomb Failure Criteria
Failure Plane
1

3



Applied stress on soil


51. Normal stress σ
Shearing stress τ

2


c

1

0,0
MOHR CIRCLE
2
450 
 

)
2
45
tan(
2
)
2
45
(
tan 0
0
2
3
1



 


 c
3


52. Normal Force
Soil Specimen
Soil Specimen
Shear Force
Loading plate
Porous stone
Porous stone
Direct Shear Test Arrangement
Direct Shear test
Direct shear test is the simplest shear test. The test equipment consist of a metal
Sheet box in which the soil sample is placed. The size of the sample is usually
50 mm by 50 mm or 100 mm x 100 mm across and 25 mm high. The box is split
horizontally into halves. A normal force is applied from the top of the shear box.
shear force is applied by moving half of the box relative to each other to cause
failure in the soil sample.

53. Problem
Direct shear test were performed on a dry, sandy soil. The specimen is 50 mm in
diameter and 25 mm in height. Test results are as follows
Test No. Normal force, N Shear force,N
1 243 124
2 268 137
3 352 179
4 412 210
specimen
of
area
sectional
Cross
Force
Normal
Stress
Normal 

specimen
of
area
sectional
Cross
Force
shear
Resisting
Stress
Shear 

Determine the cohesion and angle of internal friction.
Solution: Graph the test results using Mohr-Coulomb Failure criteria
cross sectional area of specimen
4
(0.05)2


A 2
m
001963
.
0

A

54. Test Normal stress,kPa Shear stress, kPa
specimen
of
area
sectional
Cross
Force
Normal
Stress
Normal 

specimen
of
area
sectional
Cross
Force
shear
Resisting
Stress
Shear 

1 Pa
123790
0.001963
243

1000
123790
8
.
123 Pa
63168
0.001963
124

1000
63168
1
.
63
2 Pa
136525
0.001963
268

1000
136525
5
.
136 69791Pa
0.001963
137

1000
69791
8
.
69
3 Pa
179317
0.001963
352

1000
179317
3
.
179 91187Pa
0.001963
179

1000
91187
2
.
91
4 Pa
209882
0.001963
412

1000
209882
8
.
209 106980Pa
0.001963
210
 9
.
106
1000
106980

55. 120
100
80
60
40
20
0 50 100 150 200 250
Normal stress,kPa
Shear
stress,kPa
123.8
63.1
Test 1
136.5
69.8
Test 2
179.3
91.2
Test 3
209.8
106.9
Test 4

Since the graph passes through (o,o)
123.8
63.1
8
.
123
1
.
63
tan 
 0
27


0
c 
Angle of internal friction:
Cohesion

56. Problem
A direct shear test is performed in a specimen of dry sand. The shear box is
circular in cross section with a diameter of 50 mm. Then normal force imposed
on the specimen is 250 N when the shear force is 150N. Determine the angle
of internal friction of this sand.
Solution:
cross sectional area of specimen
4
(0.05)2


A 2
m
001963
.
0

A
specimen
of
area
sectional
Cross
Force
Normal
Stress
Normal 

0.001963
250

 Pa
127324


specimen
of
area
sectional
Cross
Force
shear
Resisting
Stress
Shear 

0.001963
150

 Pa
76394



57. For dry sand c=0 (cohesionless)


 tan

 c 
tan
127324
0
76394 
 0
96
.
30


Problem
A 6 m thick soil has a water table 4 m below the ground surface. The soil
above the water table has degree of saturation of 45%, void ratio of the
soil is 0.4 and the solids have a specific gravity of 2.7. Test shows that
The soil have angle of internal friction of 320 and cohesion of 14.6 kPa.
What is the potential shear a strength on a horizontal plane at a depth of
2 m below the ground surface ?
eo=0.4 c=14.6 kPa
S =45% ø=320
Gs=2.7
4 m
2 m
Figure

58. Solution:
Normal stress is equal to effective stress 2 m from the ground surface
eo=0.4 c=14.6 kPa
S =45% ø=320
Gs=2.7
4 m
2 m
2 m
e
Se
G w
s
m



1
)
( 

Moist unit weight of soil 4 m below
the ground surface
 
4
.
0
1
81
.
9
]
40
.
0
45
.
0
7
.
2
[



m

3
18
.
20
m
kN
m 
 h
m

  )
2
(
18
.
20

 kPa
36
.
40


Potential shear strength


 tan

 c 0
32
tan
36
.
40
6
.
14 


kPa
82
.
39



59. (b)
a) Schematic Diagram of Triaxial Test Apparatus
b) State of stress on an incremental element in the soil sample

60. 
2
R


Deviator stress
O
C

 max


3

1

TRIAXIAL SHEAR TEST ( SINGLE TEST)
SINGLE TEST ON COHESIONLESS SOIL
Plane of Failure
3

1

3
 3

1

1

Mohr’s Strength
envelope



 2
tan RSin



 
max
3
1
2
1






R
 
3
1
3
1








Sin
2
450 
 


61. 
2
R


Deviator stress
O
C

 max


3

1

TRIAXIAL SHEAR TEST ( SINGLE TEST)
SINGLE TEST ON COHESIVE SOIL
Plane of Failure
3

1

3
 3

1

1

Mohr’s Strength
envelope



 2
tan RSin
c 


 c plane
principal
h the
wit
makes
plane
failure
that the
angle
friction
internal
of
angle
soil
of
cohesion
c
stress
shear
failure
at
stress
principal
Minor
failure
at
stress
principal
Major
:
where
3
1












62. Problem
In a triaxial test for a soil sample , when the principal stress are 270 kPa and 40
kPa, the soil fails along a plane making an angle of 680 with the horizontal.
What is the cohesion of the soil in kPa.
Solution:
c
∅
0
136
2 

0
68


R


270
40
0
0
0
44
136
180 

0
44
Deviator stress: 270-40 = 230
230
R=1/2(230)=115
=115
115
A
B
C
D
E
0
90
ABC
in 
0
0
0
46
44
90 



AC
R
Sin 

AC
Sin
115
460

87
.
159

AC
AED
in 
AE=AC-40-115
AE=159.87-40-115 =4.869
AE
DE


tan
869
.
4
46
tan 0 c

c=5.04 kPa

63. Problem
In a triaxial test, a specimen of saturated sand was consolidated under a
chamber confining pressure of 80 kPa. The axial stress on the specimen was
then increased by allowing the drainage of the specimen. The specimen fails at
120 kPa. What is the consolidated undrained friction angle ?
Deviator stress
∅
R


80
120
Solution: 120-80=40
40
R= ½(40) =20
=20
20
80+20=100
0
90
undrained friction angle
100
20
sin 

0
54
.
11



64. Problem
A cohensionless soil sample is subjected to a triaxial test. The critical state
friction angle of the soil is 280 and the normal effective stress at failure is 200 kPa.
Determine the following :
a) Critical state shear stress
b) Deviator stress
Solution:
∅ = 280
AO
R
Tan 
0
28
R A

1

3



200
A
280
200
A

O C
O
A
B
Critical state shear stress
ABO
in 
200
tan A

 
200
28
tan 0 A

 kPa
34
.
106

A

0
90
CAO
in 
AO
200
280

Cos 51
.
226
AO 
51
.
226
280 R
Tan 
R=120.44 kPa
2
2
)
(AO
R
OC 

kPa
54
.
256

OC
kPa
98
.
376
1 

 R
OC

kPa
1
.
136
3 

 R
OC

Deviator stress
3
1 

 

D
kPa
88
.
240

D


67. R
0
3 

1

0


1

1

UNCONFINED COMPRESSION TEST (UNIAXIAL)
u
c
2
1


 R
cu
u
u C
q 2

strength
n
compressio
unconfined
qu 

68. Problem
A consolidated undrained compression test was conducted on a saturated clay soil
by isotropically consolidating the soil using a cell pressure of 150 kPa and then
incrementally applying the load on the plunger while keeping the cell pressure
constant. Failure was observed when the stress exerted by the plunger is 160 kPa
and the pore water pressure was recorded to be 54 kPa. Determine the following:
a) Undrained friction angle
b) Undrained shear strength of clay
c) Drained friction angle
Solution:
a) Undrained friction angle
kPa
150
3 
 kPa
310
160
150
1 



∅u


150
310
160
R=80
0
90
Deviator stress = 310-150=160
R= 160/2 =80
150
R
R


U
Sin
150
80
80


U
Sin
0
35
.
20

U

b) Undrained shear strength of clay R
u 
 kPa
80

u


69. c) Drained Friction angle
'
-
pressure
water
Pore 3
3 


'
-
150
54 3

 kPa
96
'
3 

'
-
pressure
water
Pore 1
1 


'
-
310
54 1

 kPa
256
'
1 



∅𝐷
Undrained condition
310
150
256
96
Drained condition
0
90 R’
Deviator stress = 256-96=160
R’= 160/2 =80
96
R
R
'
'


D
Sin
96
80
80


D
Sin
0
03
.
27

D


70. Problem
A sample of moist sand was subjected to a series of triaxial test. The soil fail
under the following stresses:
Cell Pressure, Plunger stress,
Sample 1 16 kPa 36 kPa
Sample 2 27 kPa 58 kPa
What is the angle of internal friction of the soil ?
3
 1

Solution


16
36
27
58
R1
R2
10
2
16
36
2
3
1
1 






R
5
.
15
2
27
58
2
'
' 3
1
2 






R
26
2
16
36
2
3
1
1 






C
5
.
42
2
58
27
2
'
' 3
1
2 






C
1
2
1
2
C
C
R
R
Sin




26
5
.
42
10
5
.
15




Sin
0
47
.
19



71. Problem Set # 2
1. Direct shear test were performed on a dry, sandy soil. The specimen is 50 mm in
diameter and 50 mm in height. Test results are as follows
Test No. Normal force, N Shear force,N
1 263 145
2 288 158
3 371 201
4 432 230
Determine the cohesion and angle of internal friction.
2. A consolidated undrained compression test was conducted on a saturated clay soil
by isotropically consolidating the soil using a cell pressure of 170 kPa and then
incrementally applying the load on the plunger while keeping the cell pressure
constant. Failure was observed when the stress exerted by the plunger is 190 kPa
and the pore water pressure was recorded to be 7`4kPa. Determine the following:
a) Undrained friction angle
b) Undrained shear strength of clay
c) Drained friction angle

72. 3. A cohensionless soil sample is subjected to a triaxial test. The critical state
friction angle of the soil is 320 and the normal effective stress at failure is 240 kPa.
Determine the following :
a) Critical state shear stress
b) Deviator stress
4. A 5 m thick soil has a water table 3.5m below the ground surface. The soil
above the water table has degree of saturation of 61%, void ratio of the
soil is 0.4 and the solids have a specific gravity of 2.68. Test shows that
The soil have angle of internal friction of 280 and cohesion of 13.2 kPa.
What is the potential shear a strength on a horizontal plane at a depth of
2 m below the ground surface ?
eo=0.4 c=13.2 kPa
S =61% ø=280
Gs=2.68
3.5m
1.5 m
Figure

73. 5. A direct shear test is performed in a specimen of dry sand. The shear box is
circular in cross section with a diameter of 100 mm. Then normal force imposed
on the specimen is 350 N when the shear force is 210N. Determine the angle
of internal friction of this sand.
6. In a triaxial test for a soil sample , when the principal stress are 300 kPa
and 60 kPa, the soil fails along a plane making an angle of 560 with the
horizontal. What is the cohesion of the soil in kPa.

74. LATERAL EARTH PRESSURE
DEFINITION OF TERMS
Coefficient of earth pressure at rest, ko – the ratio of the horizontal to the vertical stress
on a soil mass when an earth retaining structure does not move.
Coefficient of active earth pressure, ka – the ratio of the lateral to the vertical principal
effective stresses when an earth retaining structure moves away from the retained soil.
Coefficient of passive earth pressure, kp – the ratio of the lateral to the vertical principal
effective stresses when an earth retaining structure is forced against a soil mass.
FORMULAS

sin
1

o
k
Coefficient of at earth pressure at rest
angle
friction
drained


For dense sand backfill
5
.
5
1
sin
1
min












d
d
o
k


 where
wall
the
behind
sand
of
t
unit weigh
dry
compacted
actual

d

state
loosest
in the
sand
of
t
unit weigh
dry
min 
d


75. For fine grained normally consolidated soils








100
%
42
.
0
44
.
0
PI
ko
For overconsolidated clays
0CR
)
(
)
( ed
consolidat
normally
o
idated
overconsol
o k
k 
pressure
overburden
effective
present
pressure
dation
preconsoli
OCR 
H
k
p o
h 

2
o H
k
2
1
F 

ph
v

h

F
H
H/3
Unit weight

76. Rankine passive earth pressure coefficient for horizontal backfill











2
45
tan
sin
1
sin
1 2 


p
k











2
45
tan
sin
1
sin
1 2 


a
k
Rankine active earth pressure coefficient for horizontal backfill
Fa
Fp
mass
soil
of
friction
internal
of
angle


where

77. Rankine active earth pressure coefficient for inclined backfill







cos
cos
cos
cos
cos
cos
cos
2
2
2
2






a
k
Rankine passive earth pressure coefficient for inclined backfill







cos
cos
cos
cos
cos
cos
cos
2
2
2
2






p
k
Fa


Fp


backfill
inclined
the
of
slope


mass
soil
of
friction
internal
of
angle


where

78. RETAINING WALLS
Retaining walls are structures which hold back masses of earth or other loose
materials where conditions make it impossible to let these masses assume their
natural slopes. Such condition occur when for example the width of an
excavation, cut or enbankment is restricted by condition of ownership, use of the
structure or economy. For instance
in railway or highway construction the width of the right of way is
fixed so that cut or fill must be located within this width. Basement
walls of buildings must be located within the property and must retain
the soil surrounding the building.
TYPES OF RETAINING WALLS
a) Gravity wall – this retains the earth entirely due to its own weight.
b) Concrete cantilever wall – consists of a vertical arm that retains
the earth and is held in position by a footing or base slab
c) Counterfort – a cantilever retaining wall provided by counterforts
spaced approximately to half of its height whose purpose is to
reduced the effect of bending moment in the vertical portion of
the wall.

79. ▓
Heel
Toe Base slab
Arm
Tile drain
▓
Crushed stones
Cantilever Retaining wall
▓
A A
Weep
Holes
4” pipe
counterfort
key
Section A - A
Counterfort Retaining Wall
▒
▒
▒
▒
▒
▒
▒
▒
Continuous
Back drain,
Crushed stones
Tile drain
Gravity wall


80. Soil Unit weight Angle of internal coeffecient
lb/ft3 friction (degrees) of friction
w/ concrete


1. Sand or gravel without
fine particles, highly
permeable
2. Sand or gravel with silt
mixture, low permeability
3. Silty sand, sand and
gravel with high clay
content
4. Medium or stiff clay
5. Soft clay, Silt
110-120
120-130
110-120
100-120
90-110
33-40
25-35
23-30
25-35
20-25
0.5 - 0.6
0.4 - 0.5
0.3 - 0.4
0.2 - 0.4
0.2 - 0.3

81. h
Ca
2
2
1
h
C
P a

EARTH PRESSURE FOR COMMON LOADING CONDITION
h



P
2
2
1
h
C
P a

3
h
y 
a) Horizontal surface of fill at top
of the wall
2
2
1
h
C
P a

3
h
y 
h
y y
b) inclined surface of fill sloping up
and back from the top of the wall
h
Ca

82. )
( '
h
h
Ca 

)
2
(
2
1 '
h
h
h
C
P a 
 
h
y
P
s 
s
h 
'
c) Horizontal surface of fill carrying a uniformly distributed additional
load s ( surcharge) due to traffic or goods from storage yard
)
2
(
3
3
'
'
2
h
h
hh
h
y




83. Problem
A gravity type retaining wall is subject to passive earth pressure on
level fill on one side (1.5 m) with sandy soil whose unit weight is 17.6 kN/m3
With an angle of internal friction of 290 as shown in the figure. Determine
the total earth pressure per meter length of wall when :
a) The other side (7.5m) is subject to horizontal surface of fill (medium stiff clay
unit weight of 18.2 kN/m3, angle of internal friction of 330) at top of the wall.
b) The other side (7.5m) is subject to an inclined surface of fill (medium stiff
clay unit weight of 18.2 kN/m3, angle of internal friction of 330) sloping up
and back 250 from the top of the wall.
c. The other side (7.5m) is subject to horizontal surface of fill (medium stiff clay
unit weight of 18.2 kN/m3, angle of internal friction of 330) at top of the wall
with uniform surcharge of 10.92 kN/m2.
7.5 m
1.5 m

84. 2
1 )
5
.
7
)(
2
.
18
(
295
.
0
2
1

P
0
29


3
6
.
17
m
kN


h
Ca
1
P
7.5 m
y
1.5 m
0
33


3
2
.
18
m
kN


h
Cp
P2


Sin
Sin
Ca



1
1
295
.
0
33
1
33
1
0
0




Sin
Sin
Ca
2
1
2
1
h
C
P a

kN
P 151
1 
2
1 )
5
.
1
)(
6
.
17
)(
88
.
2
(
2
1

P


Sin
Sin
Cp



1
1
88
.
2
29
1
29
1
0
0




Sin
Sin
Cp
2
2
2
1
h
C
P p

kN
P 57
2 
2
1 P
P
H 

57
151

H
kN
H 97

Coeffecient of active soil pressure horizontal backfill
Coeffecient of passive soil pressure horizontal backfill
the total earth pressure per
meter length of wall

85. 0
25
P1 7.5 m
y
0
25
0
25
1.5 m
h
Cp
P2=57kN
0
33


3
2
.
18
m
kN


0
29


3
6
.
17
m
kN


b)
)
(
)
(
2
2
2
2







Cos
Cos
Cos
Cos
Cos
Cos
Cos
Ca





2
1
2
1
h
C
P a

2
1 )
5
.
7
)(
2
.
18
(
408
.
0
2
1

P
408
.
0
)
33
25
25
(
)
33
25
25
(
25
0
2
0
2
0
0
2
0
2
0
0






Cos
Cos
Cos
Cos
Cos
Cos
Cos
Ca
kN
P 9
.
208
1 
kN
H 9
.
151

2
1 P
P
H 


86. )
( '
h
h
Ca 

y
P

s
h 
'
7.5 m
h
Cp
P2=57 kN
1.5 m
0
29


3
6
.
17
m
kN


0
33


3
2
.
18
m
kN


3
92
.
10
m
kN
s 
c)
)
2
(
2
1 '
h
h
h
C
P a 
 

s
h 
'
m
h 6
.
0
2
.
18
92
.
10
'


})
6
.
0
{
2
5
.
7
(
5
.
7
)
2
.
18
(
295
.
0
2
1
1 

P
295
.
0
33
1
33
1
0
0




Sin
Sin
Ca
kN
P 16
.
175
1 
2
1 P
P
H 

57
16
.
175 

H
kN
H 16
.
118


87. EXTERNAL STABILITY OF RETAINING WALLS
For a retaining wall to be externally stable( the wall must not undergo
bodily displacement without breaking up internally) it must be
a) safe against sliding
b) safe against overturning
c) Permissible soil bearing pressure at the base must not be exceeded
1
P
2
P
Ws1
Wc1
Wc2
Ws2
HEEL
TOE
Forces acting per m width
1. Weight of soil
A
W s
s 

2. Weight of Wall
A
W c
c 

3. Passive and active earth
Pressure
4. Surcharge load
A1
A2
A3
A4
B = width of base

88. 1
P
2
P
Ws1
Wc1
Wc2
Ws2
HEEL
TOE
Resultant Force on the Wall:
Horizontal Component : RH = ΣFH
Vertical Component : RV = ΣFV
Total Resultant Force :
Location of Resultant :
= location of the resultant from the toe
2
V
2
H )
R
(
)
R
(
R 

 

 M
0
RM
x
RV
x
RV
RH
x
R

89. Factor of Safety against sliding : 5
.
1
R
R
FS
H
V
S 


Where: μ= coeffecient of friction at the base
Factor of Safety against overturning :
Where : ΣRM = sum of all righting moment (counterclockwise about the toe)
ΣOM = sum of all overturning moments (clockwise about the toe
0
.
2
OM
RM
FSO 



Distribution of pressure at the base :
where : B = width of the base
e = B/2 -
Note : use the positive sign at the point ( heel or toe ) nearer the resultant.
)
B
e
6
1
(
B
R
f V


x

90. Problem
Analyze the external stablity of the retaining wall with soil properties
as shown in the figure. The wall is also subject to a surcharge load of
5.5 kPa. Allowable soil bearing capacity at the base is 350 kPa.
Concrete weighs 24 kN/m3 and coeffecient of friction at the base is
0.3.
0.6 m
1.0 m
3.0 m
1.5 m
0.4 m
4.4 m
0
30


3
5
.
18
m
kN


0
25


3
2
.
17
m
kN


2
55
.
5
m
kN
s 

91. 1
P
Ws1
Wc1
Wc2
Ws2
TOE HEEL
A1
A2
A3
A4
3.0 m

s
h 
'
2
55
.
5
m
kN
s 
0.4 m
0.6 m
1.3 m
1.4 m
4.4 m
y1
2.3 m
1.5 m
0.5 m
2
P
y2
1
1 A
W s
s 

2
1 A
W c
c 

kN
Ws 36
.
195
)
4
.
4
)(
4
.
1
(
5
.
18
1 

kN
Wc 36
.
63
)
4
.
4
)(
6
.
0
(
24
1 

kN
Ws 92
.
18
)
1
.
1
)(
1
(
2
.
17
2 

4
2 A
W c
c 

m
s
h 3
.
0
5
.
18
5
.
5
'




m
h 8
.
4
4
.
0
4
.
4 


33
.
0
30
1
30
1
0
0




Sin
Sin
Ca
kN
Wc 8
.
28
)
4
.
0
)(
3
(
24
2 

)
2
(
2
1 '
1 h
h
h
C
P a 
 
kN
P 12
.
79
})
3
.
0
{
2
8
.
4
(
8
.
4
)
5
.
18
(
33
.
0
2
1
1 



92. 1
c
W
1
s
W
2
s
W
m
h
h
hh
h
y 69
.
1
})
3
.
0
{
2
8
.
4
(
3
)
3
.
0
)(
8
.
4
(
3
)
8
.
4
(
)
2
(
3
3 2
'
'
2
1 






46
.
2
25
1
25
1
0
0




Sin
Sin
Cp
kN
P 6
.
47
)
5
.
1
(
2
.
17
)
46
.
2
(
2
1 2
2 

195.36 2.3 449.32
63.36 1.3 82.34
18.92 0.5 9.46
2
c
W 28.8 1.5 43.2
1
P 79.12 1.69 133.7
2
P 47.6 0.5 23.8
Force (kN) Dist. From toe Righting Moment Overturning Moment
(m) kN.m kN.m
RV =Ws1+Wc1+Ws2+Wc2 = 306.44
RH = P1 – P2 =31.52 kN ∑RM = 608.12 kN.m ∑OM = 133.7 kN.m

93.  

 M
0
RM
x
RV 7
.
133
12
.
608
44
.
306 

x 55
.
1

x
Factor of Safety against sliding
H
V
S
R
R
FS

 91
.
2
52
.
31
)
44
.
306
(
3
.
0


S
FS



OM
RM
FSO 54
.
4
7
.
133
12
.
608


O
FS
Factor of Safety against overturning
Distribution of pressure at the base
)
B
e
6
1
(
B
R
f V


05
.
0
2
3
55
.
1
2





B
x
e
)
(
36
.
112
)
3
)
05
.
0
(
6
1
(
3
44
.
306
n
Compressio
kPa
fH 


)
(
93
.
91
)
3
)
05
.
0
(
6
1
(
3
44
.
306
n
Compressio
kPa
fT 


at the heel
at the toe

94. Problem Set # 3
One side of a retaining wall 8 m high supports a soil having a unit weight of 16
kN/m3, angle of shearing internal friction of 30 . The surface of this side of
wall is horizontal and level with the top of the wall, the other side has soil
having a unit weight of 15kN/m3 and angle of internal friction of 25 level to a
height of 2m. Neglect wall friction and use Rankine’s formula for active and
passive pressure .
Analyze the external stablity of
the retaining wall with soil
properties as shown in the figure.
The wall is also subject to a
surcharge load of 6 kPa.
Allowable soil bearing capacity at
the base is 300 kPa. Concrete
weighs 24 kN/m3 and coeffecient
of friction at the base is 0.4
0.6 m
1.0 m
3.6 m
1.8 m
0.5 m
5.5 m
0
30


3
5
.
18
m
kN


0
25


3
2
.
17
m
kN


2
6
m
kN
s 

95. 1. A 7 ft clay layer is buried beneath a 10 ft stratum of very compact granular soil.
Compact sand underlies the clay. The layer of granular soil is composed of
material having a unit weight of 130 pcf( lb/ft3). The clay unit weight is 105 pcf.
A laboratory compression test indicates a compression index of 0.40. A planed
building loading will cause a 550 psf (lb/ft2) increase at the middle of the clay
layer. If e0=1.3, Determine the following:
a) primary consolidation settlement in inches of the clay layer for the indicated
condition
b) primary consolidation settlement in inches of the clay layer if the ground
water table is at the ground surface. Water weighs 62.4 pcf and all other
data remain unchanged.
2. At a planned construction site, a 2 m thick buried clay layer lies beneath a
superficial stratum of free- draining granular soil. Free draining granular soil
also underlies the clay layer. Double drainage from the clay layer can therefore
occur when construction loads cause consolidation. The coeffecient of
consolidation for the clay is 0.001 m2/day. Settlement calculations indicate
that the clay layer will eventually compress 4 cm(primary consolidation) due
to the effect of building loads.
a) How long will it take to attain 90% consolidation ?
b) How much settlement will occur in the first 12 months ?
c) What period of time is required to attain 2 cm settlement ?
Prelim Exam

96. Landfill
sand
e=0.5
Clay
G=2.60
e=0.90
G=2.70
4 m
1.5 m
1.5 m
3 m
Figure:
γ=17.3 kN/m3
3. A 3m clay layer (G=2.70, e =0.90,Cc=1/3) is overlain with 3 m thick sand
(G=2.60, e=0.5,ω=0). The water table is 1.5 m below the ground(sand) surface.
If a 4m thick landfill (γ=17.3 kN/m3) is placed over the existing ground surface,
compute the consolidation settlement of the clay layer

97. 4. A direct shear test is performed in a specimen of dry sand. The shear box is
circular in cross section with a diameter of 50 mm. Then normal force imposed
on the specimen is 400 N when the shear force is 200N. Determine the angle
of internal friction of this sand.
5. The following data were obtained from a laboratory consolidation test on a 50 mm
thick clay specimen drained on both sides.
Pressure (kPa) Void ratio
80 1.0
210 0.80
time for 60% consolidation = 5 minutes
a) Determine the coeffecient of volume compressibility in m2/kN
b) Determine the hydraulic conductivity for the loading range

98. 1. A 7 m thick soil has a water table 3 m below the ground surface. The soil
above the water table has a degree of saturation of 45%, void ratio of the soil
is 0.4 and the solids have specific gravity of 2.7. Test show that the soil have
an angle of internal friction of 320 and cohesion of 14.6 kPa. What is the potential
shear strength on a horizontal plane at a depth of 2 m below the ground surface?
Prelim Exam
2. A triaxial shear test was performed on a well drained sand sample. The normal
and shear sresses on the failure plane are 6300 psf and 4200 psf respectively.
a) What is the angle of internal friction of the sand ?
b) What is the angle of failure of the plane ?
c) What is the maximum principal stress ?
3. A cohensionless soil sample is subjected to a triaxial test. The critical state
friction angle of the soil is 280 and the normal effective stress at failure is 200 kPa.
Determine the following :
a) Critical state shear stress
b) Deviator stress

99. 4. A 7 ft clay layer is buried beneath a 10 ft stratum of very compact granular soil.
Compact sand underlies the clay. The layer of granular soil is composed of
material having a unit weight of 130 pcf( lb/ft3). The clay unit weight is 105 pcf.
A laboratory compression test indicates a compression index of 0.40. A planed
building loading will cause a 550 psf (lb/ft2) increase at the middle of the clay
layer. If e0=1.3, Determine the following:
a) primary consolidation settlement in inches of the clay layer for the indicated
condition
b) primary consolidation settlement in inches of the clay layer if the ground
water table is at the ground surface. Water weighs 62.4 pcf and all other
data remain unchanged.
5. At a planned construction site, a 2 m thick buried clay layer lies beneath a
superficial stratum of free- draining granular soil. Free draining granular soil
also underlies the clay layer. Double drainage from the clay layer can therefore
occur when construction loads cause consolidation. The coeffecient of
consolidation for the clay is 0.001 m2/day. Settlement calculations indicate
that the clay layer will eventually compress 4 cm(primary consolidation) due
to the effect of building loads.
a) How long will it take to attain 90% consolidation ?
b) How much settlement will occur in the first 12 months ?
c) What period of time is required to attain 2 cm settlement ?