Course
Year
Version
: S0484/Foundation Engineering
: 2007
: 1/0
Session 7 – 8
SETTLEMENT OF SHALLOW
FOUNDATION
SHALLOW FOUNDATION
Topic:
• General
• Immediate Settlement
• Consolidation Settlement
GENERAL
The settlement of shallow foundation may be divided into
three broad categories:
1. Immediate settlement, which is caused by the elastic
deformation of dry soil and of moist and saturated soils
without any change in the moisture content. Immediate
settlement are generally based on equations derived
from the elasticity theory
2. Primary consolidation settlement, which is the result
of a volume change in saturated cohesive soils
because of expulsion of the water that occupies the
void spaces.
3. Secondary consolidation settlement, which is
observed in saturated cohesive soils and is the result of
the plastic adjustment of soil particles.
This course will focus at immediate and primary consolidation settlement only.
IMMEDIATE SETTLEMENT
IMMEDIATE SETTLEMENT
General Equation (Harr, 1966)
• Flexibel Foundation
– At the corner of foundation Se 




L
B
; H=
B.qo

1   s2
Es
2
B.qo
1   s2 
Es
B.q o

Se 
1   s2  av
Es
– At the center of foundation Se 
– Average
 1  m 2  1 
1   1  m 2  m 



ln
 m. ln 
 1  m 2  1 
   1  m 2  m 



;
m
• Rigid Foundation
Se 


B.qo
1   s2 r
Es
Es = Modulus of elasticity of soil
B = Foundation width
L = Foundation length
IMMEDIATE SETTLEMENT
IMMEDIATE SETTLEMENT
If Df = 0 and H < , the elastic settlement of foundation
can be determined from the following formula:


(corner of rigid foundation)


(corner of flexible foundation)
2
2
B.qo
2 1   s F1  1   s  2  s F2
1  s 
Se 
Es
2
Se 
B.qo

1   s2 1   s2 F1  1   s  2 s2 F2
Es
The variations of F1 and F2 with H/B are given in the graphs of next slide
IMMEDIATE SETTLEMENT
IMMEDIATE SETTLEMENT
EXAMPLE
Problem:
A foundation is 1 m x 2 m in plan and carries
a net load per unit area, qo = 150 kN/m2.
Given, for the soil, Es = 10,000 kN/m2, s
0.3. Assuming the foundation to be flexible,
estimate the elastic settlement at the center
of the foundation for the following conditions:
a. Df = 0 and H = 
b. Df = 0 and H = 5 m
EXAMPLE
Solution:
Part a.
Se 


B.qo
1   s2 
Es
For L/B = 2/1 = 2    1.53, so
Se 


(1)(150)
1  0.32 (1.53)  0.0209m  20.9mm
10,000
Part b.
Se 

B'.qo

1   s2 1   s2 F1  1   s  2 s2 F2
Es

For L’/B’ = 2, and H/B’ = 10  F1  0.638 and F2  0.033, so
Se 






(0.5)(150)
1  0.32 1  0.32 (0.638)  1  0.3  2(0.3) 2 (0.033) x 4  0.0163m  16.3mm
10,000
IMMEDIATE SETTLEMENT
General Equation (Bowles, 1982)
1   s2
S e  q o .B'.
.F1
Es


L'
M
B'


1
1  M2  1 M2  N2
M  M2  1 1  N2 
F1  M . ln
 ln

2
2
2
2
 
M 1 M  N 1
M  M  N  1 


N
H
B'
Es = Modulus of elasticity of soil
H = effective layer thickness, ex. 2 - 4B below foundation
At the center of Foundation
At the corner of Foundation
L' 
L
2
L'  L
B
2
and F1 time by 4
B'  B
and F1 time by 1
B' 
IMMEDIATE SETTLEMENT
• For saturated clay soil
qo .B
S e  A1 .A 2
Es
IMMEDIATE SETTLEMENT
• For sandy soil


z2
Iz
S e  C1 .C 2 q  q 
z
0 Es
where:
– Iz = factor of strain influence
– C1 = correction factor to thickness of embedment
foundation = 1 – 0.5x[q/(q-q)]
– C2 = correction factor due to soil creep
= 1+0,2.log(t/0,1)
– t = time in years
– q = stress caused by external load
– q =  . Df
IMMEDIATE SETTLEMENT
Young Modulus
Circle Foundation or L/B =1
z=0
 Iz = 0.1
z = z1 = 0,5 B  Iz = 0.5
z = z2 = 2B
 Iz = 0.0
Foundation with L/B ≥ 10
z=0
 Iz = 0.2
z = z1 = B  Iz = 0.5
z = z2 = 4B  Iz = 0.0
EXAMPLE
A shallow foundation 3 m x 3 m (as shown in the following drawing). The
subgrade is sandy soil with Young modulus varies based on N-SPT value
(use the following correlation: Es = 766N)
Determine the settlement
occur in 5 years (use strain
influence method)
EXAMPLE
EXAMPLE
Depth
(m)
z
(m)
Es
(kN/m2)
Iz
(average)
(m3/kN)
0.0 – 1.0
1.0
8000
0.233
0.291 x 10-4
1.0 – 1.5
0.5
10000
0.433
0.217 x 10-4
1.5 – 4.0
2.5
10000
0.361
0.903 x 10-4
4.0 – 6.0
2.0
16000
0.111
0.139 x 10-4
Iz
z
Es

 q 
 17.8 x1.5

  1  0.5
  0.9
C1  1  0.5


160

17
.
8
x
1
.
5
q

q




1.55 x 10-4
 t 
 5 
C2  1  0.2. log 
  1  0.2. log 
  1.34
0
.
1
0
.
1






2B
S e  C1.C2 . q  q 
0
Iz
.z
Es
S e  (0.9)(1.34)(160  17.8 x1.5)(1.55 x10  4 )
S e  24.8mm
CONSOLIDATION SETTLEMENT
CONSOLIDATION SETTLEMENT
• Normal Consolidation
pc  po
or
pc
1
po
po  p
Cc
Sc 
.Hc . log
1  eo
po
• Over consolidation
pc
p c  p o or
1
po
po + p < pc
po < pc < po+p
Sc 
p  p
Cs
.Hc . log o
1  eo
po
Sc 
p
p  p
Cs
Cc
.H c . log c 
.H c . log o
1  eo
po 1  eo
pc
CONSOLIDATION SETTLEMENT
where:
–
–
–
–
–
eo = initial void ratio
Cc = compression index
Cs = swelling index
pc = preconsolidation pressure
po = average effective pressure on the clay layer before the construction of
the foundation
=  ’.z
– p = average increase of pressure on the clay layer caused by the
foundation construction and other external load, which can be determine
using method of 2:1, Boussinesq, Westergaard or Newmark.
Alternatively, the average increase of pressure (p) may be approximated
by:
1
p  pt  4pm  pb 
6
pt = the pressure increase at the top of the clay layer
pm = the pressure increase at the middle of the clay layer
pb = the pressure increase at the bottom of the clay layer
CONSOLIDATION SETTLEMENT
EXAMPLE
A foundation 1m x 2m in plan is shown in the following figure.
Estimate the consolidation settlement of the foundation.
Assume the clay is normally consolidated.
EXAMPLE
po  p
Cc
Sc 
.Hc . log
1  eo
po
po = (2.5)(16.5) + (0.50)(17.5-10) +(1.25)(16-10) = 52.5 kN/m2
qo .B.L
2:1 method
B  z L  z 
150.1.2
p 
 13.45 kN / m 2
1  3.252  3.25
p 
Sc 
0.32
52.5  13.45
2.5 x log
 44 mm
1  0.8
52.5
ALLOWABLE SETTLEMENT