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H
δδδδ c
∆∆∆∆σσσσz
′
σσσσz0′
σσσσz0′
}σσσσz f′
� Soil Mechanics Review of Consolidation
VoidsVoids
Solids
Vv = eVs
Vs
∆∆∆∆ e
Vv = (e - ∆∆∆∆ e)Vs
Vs Solids
∆∆∆∆σσσσz
′
σσσσz0′
σσσσz0′
}σσσσz f′
Before After
σσσσ0
Before Loading
××××u0
•
Point, P
σσσσ0 + ∆σ∆σ∆σ∆σ
Immediately After Loading
××××u0+∆∆∆∆u
•
Point, P
σσσσ0 + ∆σ∆σ∆σ∆σ
σσσσ0 + ∆σ∆σ∆σ∆σ
Shortly after LoadingNo settlement
Long after LoadingSettlement Complete
××××u0+∆∆∆∆u
××××u0
σσσσ0 + ∆σ∆σ∆σ∆σ
Settlement
� Distortion Settlement (Immediate)
� Consolidation (Time Dependent)
� Secondary Compression time
Settlement
Laboratory Consolidation Test
Test Results
Test ResultsConsolidation Plot-Idealized
Normally and Over-Consolidated Soils
czo σσ ′=′ ….. Normally consolidated
czo σσ ′<′ ….. Over consolidated
czo σσ ′>′ ….. Under consolidated
When soil is loaded to a stress level greater than it ever ‘experienced’ in the past, the soil structure is no longer able to sustain the increased load, and start to breakdown.
� Preconsolidation Pressure - Pc:� Maximum pressure experienced by soil in the past
� Normal Consolidation: OCR = 1� when the preconsolidation pressure is equal to the existing effective vertical overburden pressure Pc = p’o
� present effective overburden pressure is the maximum pressure that soil has been subjected in the past
p’o
Pc
Effective Consolidation Stress p’o
Void ratio e
Pc
� Over Consolidation: OCR > 1� when the preconsolidation pressure is greater than the existing effective vertical overburden pressure Pc > p’o
� present effective overburden pressure is less than that which soil has been subjected in the past
� It also said soil is in preconsolidated condition
� OCR (over consolidation ratio) =
� Under Consolidation: OCR < 1� when the preconsolidation pressure is less than the existing effective vertical overburden pressure Pc < p’o,
� e.g : recently deposited soil geologically or manually.
Pc = p’o
'
c
o
P
p
Pc > p’o
p’o
Pc
p’o
Pc < p’o
Pc
Over-Consolidation Margin & Over-consolidation Ratio
zccm σσσ ′−′=′ ….. Over-consolidation Margin
zo
cOCRσ
σ
′
′= ….. Over consolidation ratio
Settlement PredictionsN.C. Clays
′
′
+=∑
00
log1 z
zfcc H
e
C
σ
σδ
Settlement Calculation:Normally consolidated clay
L H s eorε
∆ ∆ ∆= = =
soil + water
voids
solids
Hf
∆∆∆∆H = s
voids
solids
∆∆∆∆e
ef
1
Ho
eo
1
H =
2.4
2.6
2.8
1 2 1 2
22 1
1
2
1
1
1
'log ' log ' log 'log
'
'log
1 '
v
o o o o
o v o
o
c
o
oc c
o
L H s eor
L H H e
es H H
e
e e e eeC
pp p p
p
H pS C
e p
ε
ε
∆ ∆ ∆= = =
+
∆= =
+
− −−∆= = =
∆ −
=+
1
1.2
1.4
1.6
1.8
2
2.2
2.4
1 10 100
Pressure, p (log scale)
Vo
id r
ati
o,
e
Cc
1
Effective Consolidation Stress p’o
Pc
� For normally consolidated clay
� For layered normally consolidated clay:
'H p p +∆=∑
Settlement Calculation (cont’d):
'log
1 '
'log
'
o oc c
o o
oc ce o
o
H p pS C or
e p
p pS C H
p
+ ∆=
+
+ ∆=
p’1 = p’o, and p’2 include the additional stress ∆p applied by the structure
when computing settlement using percentage vertical strain vs log effective pressure
� In overconsolidated clay
� Cr is the slope of rebound curve (swell index); Cr ≈ 20% to 10% Cc
'log
1 '
o oc r
o o
H p pS C
e p
+∆=
+
'log
1 '
o oc c
o o
H p pS C
e p
+∆=
+ ∑
'log log
1 ' 1 '
o c o oc r c
o o o o
H P H p pS C C
e p e p
+ ∆= +
+ +
p’1 = p’o, and p’2 =po+∆p < Pc
p’1 = p’o, and p’2 =po+∆p > Pc
Time Rate of Consolidation
2%
0 to 60%,4 100
60%, 1.781 0.933log(100 %)
tt
tt v
t v
SU
S
Ufor U T
for U T U
π
=
= =
> = − −
2
2 or
t v
v v drv v v
dr v
t T HT c t
H c= =
� Ut = average degree of consolidation (%)
� St = settlement of the layer at time t
� S = ultimate settlement of the layer due to primary consolidation
� Tv = time factor
� Hdr= average longest drainage path during consolidation
� cv = coefficient of consolidation
� tv = time for consolidation
Example
� Calculate the settlement due to primary con-solidation for 5m clay layer due to a surcharge of 50kPa
2m
Sand
Sand50% saturation
Ground water table
Surcharge = 50kPa
surcharge of 50kPa applied at the ground level. The clay is normally consolidated.
� Calculate the time rate of settlement when cv is given as 0.85m2/yr
5m
5m
Rock
SandGs=2.65, e=0.7
ClayCc=0.45, eo=0.9γsat=15kPa
� Submerged unit weight of clay
� So
Solution
� Calculation of Average effective Overburden Pressure (po)� The moist unit weight of sand
above the ground water table( )2.65 0.5 0.7 9.81
1 1 0.7
s w wsand
G Sr e
e
γ γγ
+ ⋅ ⋅⋅ + ⋅ ⋅= =
+ +
( )
o sand sand clay
' '
15 9.81 5.19 kPa
5p' 2 +3 ' + '
2
clay sat clay wγ γ γ
γ γ γ
= −
= − =
= ⋅ ⋅ ⋅
� Calculation of Settlement
� Submerged unit weight of sand below the ground water table
( )
( )
( )' '
1
1 1
2.65 1 9.819.516kPa
1 0.7
sand sat sand w
s ws w ww
GG e
e e
γ γ γ
γγ γγ
= −
−⋅ + ⋅= − =
+ +
− ⋅= =
+
1 1 0.7
22.21kPa
e+ +
=
o sand sand clay2
52 22.21+3 9.516+ 5.19 85.94 kPa
2= ⋅ ⋅ ⋅ =
'log
1 '
2.5 85.94 500.45 log
1 0.9 85.94
0.592m 0.199m 0.792m 0.8m
o oc c
o o
H p pS C
e p
+ ∆=
+
+=
+
= + = �
Time rate of settlement
0.520.240.07130
0.230.160.03120
0.060.080.00810
t (yr)St(m)
TvUavg
0
0.1
0 1 2 3 4 5 6 7 8 9
Time (yr)
8.550.761.16395
∞0.80∞100
6.240.720.84890
4.170.640.56780
2.960.560.40370
2.110.480.28760
1.450.400.19750
0.930.320.12640
0.520.240.07130 0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Sett
ele
men
t (m
)