Soil Settlement
By
Kamal Tawfiq, Ph.D., P.E., F.ASCE
Fall 2010
Soil Settlement:
Total Soil Settlement = Elastic Settlement + Consolidation Settlement
Stotal = Se + Sc
Elastic Settlement or Immediate Settlement depends on
Elastic Settlement
{
{
Load Type (Rigid; Flexible)
Settlement Location (Center or Corner)
Theory of Elasticity
Time Depended Elastic Settlement (Schmertman & Hartman Method (1978)
Elastic settlement occurs in sandy, silty, and clayey soils.
By: Kamal Tawfiq, Ph.D., P.E.
By: Kamal Tawfiq, Ph.D., P.E.
Consolidation Settlement (Time Dependent Settlement)
* Consolidation settlement occurs in cohesive soils due to the expulsion of the water from the voids.
* Because of the soil permeability the rate of settlement may varied from soil to another.
* Also the variation in the rate of consolidation settlement depends on the boundary conditions.
SConsolidation = Sprimary + Ssecondary
Primary Consolidation
Secondary Consolidation
Volume change is due to reduction in pore water pressure
Volume change is due to the rearrangement of the soil particles
(No pore water pressure change, Δu = 0, occurs after the primary consolidation)
Water Table (W.T.)
Expulsion of
the water
Water
Voids
Solids
When the water in the voids
starts to flow out of the soil
matrix due to consolidation of
the clay layer. Consequently,
the excess pore water
pressure (Du) will reduce,
and the void ratio (e) of the
soil matrix will reduce too.
Elastic Settlement
Bqo
Se =
Es
Bqo
Se =
Where α =
Es
1
p
2
(1 - μs) α
(corner of the flexible foundation)
(1 - μs) α
(center of the flexible foundation)
2
2
[ ln ( √1 + m2 + m / √1 + m2 - m ) + m. ln ( √1 + m2 + 1 / √1 + m2 - 1 )
m = B/L
B = width of foundation
L = length of foundation
By: Kamal Tawfiq, Ph.D., P.E.
Se =
Bqo (1 - μs) α
Es
3.0
2.5
α
αav
αr
α, αav, αr
2.0
1.5
For circular foundation
α=1
αav = 0.85
αr = 0.88
1.0
3.0
1
2
3
4
5
6
7
8
9
10
L/B
Values of α, αav, and αr
By: Kamal Tawfiq, Ph.D., P.E.
Consolidation Settlement
Consolidation Settlement (Primary Consolidation) = Sc = (Cc/1+e o) Hc . log [(Po + P)/Po]
Qdesign = Column Load
Normally
Consolidated
Clay
Stressed Zone
Sand
B
Caly
Hc
Hc/2
Overburden
Pressure
Po
2
1
2
1
Stress
Distribution
Sand
By: Kamal Tawfiq, Ph.D., P.E.
Consolidation Settlement
Loading
Normally
Consolidated Soil
Unloading
p
Sand
Sand
H clay/2
Hsand
2
1
Dp
p
Hclay
Void Ratio
H clay/2
Sand
Sand
Void Ratio
Void Ratio
Hsand
2
1
Clay
Sand
Sand

Dpp
Hclay
Dp
P
Dp
P
eo
Cc
Po
Log P
Po
Po + Dp
P
Log P
Po
Dp
Po + 
P Log P
Cc H log(po +p)
Sultimate = H =
Po
Cc H 1 + eo Po + DP
DH =
log (
)
P0
1 + eO
Po = 
sand . Hsand + (
clay - 
water ) . Hclay/2
By: Kamal Tawfiq, Ph.D., P.E.
Re loading
with Heavy Load
p2
Hsand
2
1
H clay/2
1

p
Dp
2
The soil become
overconsolidated
soil
Dp

P

p2
Dp
2
H clay/2
Hclay
V oid Ratio
Hsand
2
Hclay
Dp
P22
V oid Ratio
Dp
eo
Cs
Cs
Po
Po + Dp
P
=
Pc
DH =
Po
Log P
Pc
Po + Dp
P2
2
Log P
CS H
P Cs CHC H
Cc H log Po + P2
2
log log (PPco + DP+
( H C= ) +
)
Sultimatelog
=
Po 1 +1e+ eO
1 + eO
1 + eo
Pc
Po PC
o
(
)
(
)
By: Kamal Tawfiq, Ph.D., P.E.
Re loading
with light Load
p2
Hsand
2
1
H clay/2
Dpp
The soil become
overconsolidated
soil
Void Ratio
Dp

P

p2
Dp
2
H clay/2
Hclay
2
Hsand
2
1
Hclay
Po + Dp
P2 2
Void Ratio
Dp

P22
eo
Cs
Po
Pc
DH =
Po
Log P
CS H
1 + eO
Pc
Cs H log
S
=

H
=
ultimate
P + DP2
1 + eo
log ( o
)
Po
Log P
(
Pc
Po
)
By: Kamal Tawfiq, Ph.D., P.E.
Determining The Preconsolidation Pressure (Pc)
Cassagrande Graphical Method
Void Ratio
5
6
3
1
4
2
7
Po
OCR = Pc/Po
OCR = 1
OCR > 1
OCR > 4
Pc
Log P
Normally Consolidated
Over Consolidated
Heavily Over Consolidated
By: Kamal Tawfiq, Ph.D., P.E.
Dr. Kamal Tawfiq - 2010
Figure 1
Example:
1.
2.
3.
4.
gsand = 96 pcf
Soil sample was obtained from the clay layer
Conduct consolidation test [9 load increments ]
Plot e vs. log (p) (Figure 2)
Determine Compression Index (Cc ) & Swelling Index (Cs)
0.9
Sand
Void Ratio (e)
0.7
0.6
Soil Sample
eo = 0.795
Cc = 0.72
Cs = 0.1
In the lab and before
the consolidation test
the stresses on the
sample = 0.
During testing, the
geostatic stress is
gradually recovered
0.4
0.3
100
Determine
Po
16 ft
wc = 0.3
In the ground, the
sample was
subjected to
geostatic stresses.
0.5
5.
Clay
gclay = 110 pcf
4 ft
Dp1 Dp
2
0.8
In the lab the stresses are
added to the soil sample
3 ft
W.T.
1
In the lab and after removing
the soil sample from the ground,
the stresses on the soil sample = 0
Dp1
Dp
Stress
Dp7
Increments Dp
Dp9
G.S.
Po = 3.(96) + 4.(96-62.4) + 8.(110-62.4) = 803.2 lb/ft2
Cc
Cs
Dp9
Po 1000
Log (p)
10000
Figure 2
Dr. Kamal Tawfiq - 2010
Tangent to point 1
Example:
6.
G.S.
gsand = 96 pcf
Using Casagrande’s Method to determine Pc
3 ft
W.T.
Sand
gclay = 110 pcf
Pc = 800 lb/ft2
4 ft
Clay
Po
16 ft
wc = 0.3
Overconsolidation Ratio
Pc
Po
1
=1
The soil is
Normally Consolidated
N.C. soil
0.9
Dp1
2
6
0.8
Void Ratio (e)
OCR =
Point of
maximum curvature
X
X
1
4
0.7
5
3
0.6
0.5
0.4
7
0.3
100
Po = Pc1000
Log (p)
10000
Dr. Kamal Tawfiq - 2010
Casagrande’s Method to Determine Preconsolidation Pressure (Pc)
1
Normally Consolidated Soil
Point of
maximum curvature
1
0.9
2 Horizontal line
Void Ratio (e)
0.8
X
X
1
0.7
3
0.6
0.5
0.4
7
0.3
Po = Pc1000
100
Log (p)
Overconsolidation Ratio OCR =
Pc
Po
=1
The soil is Normally Consolidated (N.C.) soil
10000
Dr. Kamal Tawfiq - 2010
2
Casagrande’s Method to Determine Pc
Overconsolidated Soil
Point of
maximum curvature
2 Horizontal line
X
X
Void Ratio (e)
1
3
7
Po
Overconsolidation Ratio OCR =
Pc
Po
The soil is oversonsolidated (O.C.) soil
>1
Pc
Log (p)
Dr. Kamal Tawfiq - 2010
Building
qdesign
Example:
gsand = 96 pcf
A 150’ x 100’ building will be constructed at the site.
The vertical stress due to the addition of the building
qdesign =1000 lb/ft2
G.S.
W.T.
Sand
Clay
3 ft
4 ft
DP1
Po
Po
16 ft
The weight of the building Qdesign will be transferred
to the mid height of the clay layer
eo = wc . Gs = 0.3 x 2.65 = 0.795
1
Qdesign = 15,000,000 lb
The added stress at 15’ from
the ground surface is
0.9
Dp1
Dp =
15,000,000 lb
(150+15) x (100+15)
DP = 790.51 lb/ft2
Void Ratio (e)
0.8
0.7
0.6
0.5
DP + Po =
790.51 + 803 = 1593.51 lb/ft2
0.4
0.3
100
Po1000
Log (p)
10000
Dr. Kamal Tawfiq - 2010
Building
Example:
qdesign
DP + Po = 790.51 + 803 = 1593.51 lb/ft2
gsand = 96 pcf
Consolidation Settlement
DH =
DH =
Cc H
1 + eO
log (
G.S.
W.T.
Sand
Po + DP
)
Po
0.72 x 16
1593.51
log (
)
1 + 0.795
803
DH = 1.9 ft
Clay
3 ft
4 ft
DP1
Po
Po
16 ft
eo = wc . Gs = 0.3 x 2.65 = 0.795
1
0.9
Dp1
Void Ratio (e)
0.8
DP1
0.7
0.6
0.5
0.4
0.3
100
Po1000
Po + DP
Log (p)
10000
Demolished
When the building was removed, the soil has become
an overconsolidated clay.
qdesign
gsand = 96 pcf
The rebound has taken place through
swelling from pint 1 to point 2
Dr. Kamal Tawfiq - 2010
G.S.
W.T.
Sand
Clay
3 ft
4 ft
Po
Po
16 ft
eo = wc . Gs = 0.3 x 2.65 = 0.795
1
0.9
Dp1
Void Ratio (e)
0.8
DP1
0.7
2
0.6
1
0.5
0.4
0.3
100
Po1000
Po + DP
Log (p)
10000
Dr. Kamal Tawfiq - 2010
Constructing a new building
Scenario #1
The soil now is overconsolidated Soil:
qdesign
gsand = 96 pcf
DH =
W.T.
Sand
The new building is heavier in weight
CS H
1 + eO
G.S.
Clay
3 ft
4 ft
DP2
Po
Po
16 ft
P
C CH
P + DP
log ( C ) +
log ( o
)
Po
PC
1 + eO
eo = wc . Gs = 0.3 x 2.65 = 0.795
1
eo = 0.61
0.9
DH =
0.1 x 16
1 + 0.61
log (
1593.51
803
+
0.72 x 16
log ( 2100
1 + 0.61
1593.51
)
DP2
0.8
Void Ratio (e)
Assume Po + Dp2 = 2100 psf
Dp1
DP1
0.7
0.6
CS
0.5
CC
)
0.4
=
0.3
100
Po1000
Pc
Log (p)
Po + DP
New Building
10000
Dr. Kamal Tawfiq - 2010
Constructing a new building
Scenario # 2
The soil now is overconsolidated Soil:
qdesign
gsand = 96 pcf
G.S.
W.T.
Sand
Clay
The new building is lighter in weight
3 ft
4 ft
DP2
Po
Po
16 ft
eo = wc . Gs = 0.3 x 2.65 = 0.795
CS H
P + DP
log ( o
)
P0
1 + eO
DH =
1
0.9
eo = 0.61
=
log ( 1600
1593.51
)
Void Ratio (e)
0.1 x 16
1 + 0.61
DP2
0.8
Assume Po + Dp2 = 1600 psf
DH =
Dp1
DP1
0.7
0.6
CS
0.5
0.4
0.3
100
Po1000Po
+ DP
Log (p)
10000
New Building
Dr. Kamal Tawfiq - 2010
Example of Semi-log Scale
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.1
1
10
100
1000
Rate of Consolidation
Settlement at any time = Stime
Stime = Sultimate * U%
Sultimate= (Cc/1+eo) Hc . log [(Po + P)/P o]
U% = f (Tv) ....
Tv = f (cv) ......
Tv =
cv . t
(Hdr)2
Qdesign = Column Load
u =Excess Pore Water Pressure
Sand
Caly
Overburden
Pressure
Po
Hdr = Hc /2
P
P
Stress Distribution
2: 1 method
Uo
Hc = Layer Thickness
Sand
u =Excess Pore Water Pressure
By: Kamal Tawfiq, Ph.D., P.E.