SETTLEMENT
Of
SHALLOW FOUNDATION
2 types of settlement
Elastic ( Immediate ) settlement
Time independent
Causes:
Elastic deformation of dry soil particles
Consolidation settlement
Time dependant
Causes:
Water expulsion
CONSOLIDATION SETTLEMENT
To calculate the consolidation settlement we have to perform
ONE DIMENSIONAL
CONSOLIDATION
TEST
Consolidometer
Audiometer
time
P1
P2
P3
P4
deformation
P4 << P3 << P2 << P1
I
II
III
I – Initial compression
II – Primary Consolidation
III – Secondary Settlement
I – Initial compression S1
Causes: imperfect contact surface, loading, particles relocation
II – Primary Consolidation S2
Causes: water expulsion
III – Secondary Settlement S3
Causes: Particles relocation, Particles deformation, Particles destruction
Total settlement S = S1 + S2 + S3
e
eo
e1
e2
While P2 is greater than P1
e2 is smaller than e1
P1 P2
Log P
e
Cc
Compression Index
We can use any
curve because the
slope is the same
The slope from the curve
Cc = 0.009 ( LL – 10 ) for undisturbed samples
Cc = 0.007 ( LL – 10 ) for remolded samples
Cc
Log P
Soil History
We can find Soil History by finding what is called
PRECONSOLIDATION PRESSURE
Or
Max Past Effective Overburden Pressure
Pc
e
Pc
Log P
Based on values of Pc
Clay soils may be
NCC – Normally Consolidated Clay
OCC – Over Consolidated Clay
NCC – Normally Consolidated Clay.
Where Po is greater than Pc
P
The slope of
the curve in
this area is Cc
Pc = Max Past Effective
Overburden Pressure
Po = Max Present Effective
Overburden Pressure
Cc
Pc
Po
OCC –
Over Consolidated Clay.
Where Po is smaller than Pc
P
The slope of
the curve in
this area is Cs
Cs
Po
Pc
Pc = Max Past Effective
Overburden Pressure
Po = Max Present Effective
Overburden Pressure
Cs
Cc
OCC
Pc
NCC
Total settlement S = Si + Sp + Ss
•INITIAL COMPRESSION ( Si )
•PRIMARY CONSOLIDATION SETTLEMENT
•SECONDARY SETTLEMENT ( Ss )
ep
( Sp )
Cα=
Ss = C’α H log(
/log(
)
)
C’α =
ep
Cα
t1
t2
INITIAL COMPRESSION ( Si )
Si = q B
Iρ
Where:
q- net applied pressure
B- width of footing
-Poisson’s ratio (tab. 6.6 p 167)
-E- modulus of elasticity (tab. 6.5 p 167)
Iρ- influence factor (tab. 6.4 p 167)
TIME RATE OF SETTLEMENT
summary
Immediate settlement
Initial compression
Si = q B
Iρ
Si = q B
NCC
Primary consolidation
S=
P0 + ΔP
Pc
OCC
Consolidation
settlement
P0 + ΔP < Pc
Secondary settlement
Ss = C’α H log(
)
Iρ
Settlement due to surcharge
S = f ( P, soil, Z, X, load application type )
Load application types:
point
line
strip
circle
square
rectangle
irregular
The settlement of a foundation can be divided
into two major categories:
A- Elastic or immediate settlement
B- Consolidation settlement
Settleme
nt
Immediate
or Elastic
Settlement
Consolidatio
n Settlement
Time
Secondary
Settlement
For the calculation of foundation settlement, it is
required to determine the vertical stress increase in
the soil mass due to the load applied on the
foundation.
This chapter is divided into the following three
parts:
1. Procedure for calculation of vertical stress
increase
2. Elastic settlement calculation
3. Consolidation settlement calculation
dr. isam jardaneh / foundation engineering 61303 /
2010
dr. isam jardaneh / foundation engineering 61303 /
2010
As
suggested
by
Griffiths 1984, to find
average
pressure
increase between z =
H1 and z = H2 below
the
corner
of
a
uniformly
loaded
rectangular area
dr. isam jardaneh / foundation engineering 61303 /
2010
dr. isam jardaneh / foundation engineering 61303 /
2010
Continuous
Footing
Need Interpolation ???
Example
Find
Elastic Settlement
¯ = 160 KN/m²
q
1.5m
33x3m
x3m
1.0m
After 5 years
Using
Influence Factor
Method
γ = 17.8 KN/mᶟ
3.0m
E KN/m²
8000
10000
16000
0.1
1m
0.5m
2.5m
2m
0.5
Depth
ΔZ
Es
m
0.0 – 1.0
1.0 – 1.5
1.5 – 4.0
4.0 – 6.0
m
KN/m²
8000
10000
10000
16000
1.0
0.5
2.5
2.0
Average
Iz
0.233
0.433
0.361
0.111
Iz
----- ΔZ
Es
mᶟ/KN
0.291x10¯⁴
0.217x10¯⁴
0.903x10¯⁴
0.139x10¯⁴
∑ = 1.55x10¯⁴
q
C1 = 1-0.5 (_
q----q
)
17.8x1.5
= 1-0.5
160 – (17.8x1.5)
[-------------] = 0.9
C2 = 1 + 0.2 log (5/0.1) = 1.34
‒
Se = C1 C2 ( q – q ) ∑ (Iz/E)Δz
Se = 24.9 mm
Range of Materials Parameters for Computing Settlement
dr. isam jardaneh / foundation engineering 61303 /
2010
Example 1
Example 2
Problem # 1
Problem # 2
Problem # 3
Problem # 4