Course
Year
: S0825/Foundation Engineering
: 2009
SHALLOW FOUNDATION
Session 5 – 10
SHALLOW FOUNDATION
Topic:
• BEARING CAPACITY
–
–
–
–
–
–
General
Terzaghi Model
Meyerhoff Model
Influence of ground water elevation
Influence of multi layer soil
Shallow Foundation Bearing by N-SPT value
• SETTLEMENT
– General
– Immediate Settlement
– Consolidation Settlement
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SHALLOW FOUNDATION
SESSION 5 – 6
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TYPES OF SHALLOW FOUNDATION
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TYPES OF SHALLOW FOUNDATION
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TERZAGHI MODEL
Assumptions:
• Subsoil below foundation structure is homogenous
• Shallow foundation Df < B
• Continuous, or strip, footing : 2D case
• Rough base
• Equivalent surcharge
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TERZAGHI MODEL
FAILURE ZONES:
1. ACD : TRIANGULAR ZONES
2. ADF & CDE : RADIAL SHEAR ZONES
3. AFH & CEG : RANKINE PASSIVE ZONES
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TERZAGHI MODEL
(GENERAL FAILURE)
• STRIP FOUNDATION
qult = c.Nc + q.Nq + 0.5..B.N
• SQUARE FOUNDATION
qult = 1.3.c.Nc + q.Nq + 0.4..B.N
• CIRCULAR FOUNDATION
qult = 1.3.c.Nc + q.Nq + 0.3..B.N
Where:
c = cohesion of soil
q =  . Df ; Df = the thickness of foundation
embedded on subsoil
 = unit weight of soil
B = foundation width
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Nc, Nq, N = bearing capacity factors


 e 23 / 4 / 2 tan

Nc  cot  
 1




 2. cos 2
  


 4 2  
e 23 / 4 / 2 tan
Nq 
  
2. cos 2   
 4 2

1  K py
 tan 
N  

1
2
2  cos  
BEARING CAPACITY FACTORS
GENERAL
FAILURE
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BEARING CAPACITY FACTORS
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GENERAL
FAILURE
TERZAGHI MODEL
(LOCAL FAILURE)
•
•
•
STRIP FOUNDATION
qult = 2/3.c.Nc’ + q.Nq’ + 0.5..B.N’
SQUARE FOUNDATION
qult = 0.867.c.Nc’ + q.Nq’ + 0.4..B.N’
CIRCULAR FOUNDATION
qult = 0.867.c.Nc’ + q.Nq’ + 0.3..B.N’
Where:
c = cohesion of soil
q =  . Df ; Df = the thickness of foundation
embedded on subsoil
 = unit weight of soil
B = foundation width
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Nc, Nq, N = bearing capacity factors


 e 23 / 4 '/ 2 tan '

Nc  cot  ' 
 1


'


 2. cos 2
  


 4 2  
e 23 / 4 '/ 2 tan '
Nq 
  ' 
2. cos 2   
4 2

1  K py
 tan  '
N  

1
2
2  cos  ' 
’ = tan-1 (2/3. tan)
BEARING CAPACITY FACTORS
LOCAL FAILURE
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BEARING CAPACITY FACTORS
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GROUND WATER INFLUENCE
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GROUND WATER INFLUENCE
• CASE 1
0  D1 < Df  q = D1.dry + D2 . ’
• CASE 2
0  d  B  q = dry.Df
the value of  in third part of equation is replaced with
 = ’ + (d/B).(dry - ’)
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FACTOR OF SAFETY
qall
qu

FS
qall( net ) 
qnet ( u )  qu  q
qnet ( u )
q   .D f
FS
Where:
qu = gross ultimate bearing capacity of shallow foundation
qall = gross allowable bearing capacity of shallow foundation
qnet(u) = net ultimate bearing capacity of shallow foundation
qall = net allowable bearing capacity of shallow foundation
FS = Factor of Safety (FS  3)
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NET ALLOWABLE BEARING CAPACITY
PROCEDURE:
1.
Find the developed cohesion and the angle of friction
cd 
2.
c
FS shear
 tan 
 FS shear
d  tan 1 




FSshear = 1.4 – 1.6
Calculate the gross allowable bearing capacity (qall) according to terzaghi
equation with cd and d as the shear strength parameters of the soil
Ex.: qall = cd.Nc + q.Nq + ½ .B.N
Where Nc, Nq, N = bearing capacity factor for the friction angle, d
3.
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Find the net allowable bearing capacity (qall(net))
qall(net) = qall - q
EXAMPLE – PROBLEM
A square foundation is 5 ft x 5 ft in plan. The soil supporting the
foundation has a friction angle of  = 20o and c = 320 lb/ft2. The unit
weight of soil, , is 115 lb/ft3. Assume that the depth of the foundation (Df)
is 3 ft and the general shear failure occurs in the soil.
Determine:
- the allowable gross load on the foundation with a factor of safety (FS)
of 4.
- the net allowable load for the foundation with FSshear = 1.5
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EXAMPLE – SOLUTION
Foundation Type: Square Foundation
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EXAMPLE – SOLUTION
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SHALLOW FOUNDATION
SESSION 7 – 8
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GENERAL BEARING CAPACITY EQUATION
Meyerhof’s Theory
Df
qu  c.Nc.Fcs .Fcd .Fci  q.Nq.Fqs .Fqd .Fqi  (0.5). .B.N .Fs .Fd .Fi
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BEARING CAPACITY FACTOR


Nq  tan 2  45  e . tan
2

Nc   Nq  1 cot 
N  2( Nq  1) tan 
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SHAPE, DEPTH AND INCLINATION FACTOR
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EXAMPLE 2
Determine the size (diameter) circle foundation of tank structure as
shown in the following picture
P = 73 ton
dry = 13 kN/m3
sat = 18 kN/m3
c = 1 kg/cm2
 = 20o
Tank
2m
Foundation
GWL
With P is the load of tank, neglected the weight of foundation and use
factor of safety, FS = 3.5.
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EXAMPLE 3
SQUARE FOUNDATION
B = 4m
dry = 13 kN/m3
DETERMINE THE FACTOR OF SAFETY FOR:
-CASE 1 : GWL LOCATED AT 0.3m (MEASURED FROM THE SURFACE OF SOIL)
-CASE 2 : GWL LOCATED AT 1.5m (MEASURED FROM THE SURFACE OF SOIL)
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ECCENTRICALLY LOADED FOUNDATIONS
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ECCENTRICALLY LOADED FOUNDATIONS
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ONE WAY ECCENTRICITY
Meyerhof’s step by step procedure:
• Determine the effective dimensions of the foundation as :
B’ = effective width = B – 2e
L’ = effective length = L
Note:
–
–
•
If the eccentricity were in the direction of the length of the foundation, the value of L’ would be equal to L-2e
and the value of B’ would be B.
The smaller of the two dimensions (L’ and B’) is the effective width of the foundation
Determine the ultimate bearing capacity
qu '  c.Nc.Fcs.Fcd.Fci  q.Nq.Fqs.Fqd.Fqi  0,5..B'.N.Fs.Fd.Fi
•
•
•
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to determine Fcs, Fqs, Fs use effective length and effective width
to determine Fcd, Fqd, Fd use B
The total ultimate load that the foundation can sustain is
Qult = qu’.B’.L’ ; where B’xL’ = A’ (effective area)
The factor of safety against bearing capacity failure is
FS = Qult/Q
Check the factor of safety against qmax, or,
FS = qu’/qmax
EXAMPLE – PROBLEM
A Square foundation is shown in the following figure. Assume
that the one- way load eccentricity e = 0.15m. Determine the
ultimate load, Qult
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EXAMPLE – SOLUTION
With c = 0, the bearing capacity equation becomes
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TWO-WAY ECCENTRICITY
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TWO-WAY ECCENTRICITY – CASE 1
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TWO-WAY ECCENTRICITY – CASE 2
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TWO-WAY ECCENTRICITY – CASE 3
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TWO-WAY ECCENTRICITY – CASE 4
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BEARING CAPACITY OF LAYERED SOILS
STRONGER SOIL UNDERLAIN BY
WEAKER SOIL
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BEARING CAPACITY OF LAYERED SOILS
2D f
 2ca H 
2

qu  qb  
   1H 1 
B
H



 K s tan 1 

   1H
B


1
qb  c2 N c ( 2 ) Fcs ( 2 )   1 D f  H N q ( 2) Fqs( 2)   2 BN  ( 2) Fs ( 2 )
2
1
q1  c1 N c (1)   1 BN  (1)
2
1
q2  c2 N c ( 2 )   2 BN  ( 2 )
2
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BEARING CAPACITY OF LAYERED SOILS
 2D f
 2c H 
qu  qb   a    1H 2 1 
H
 B 

 K s tan 1 

   1H  qt
B



1
qb  c2 N c ( 2 ) Fcs ( 2 )   1 D f  H N q ( 2 ) Fqs ( 2 )   2 BN  ( 2 ) Fs ( 2 )
2
1
qt  c1 N c (1) Fcs (1)   1 D f N q (1) Fqs (1)   1 BN  (1) Fs (1)
2
Rectangular Foundation
 B  2c H 
 B  2 D f
qu  qb  1   a    1H 2 1  1 
H
 L  B 
 L 
1
q1  c1 N c (1)   1 BN  (1)
2
1
q2  c2 N c ( 2 )   2 BN  ( 2 )
2
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 K s tan 1 

   1H  qt
B


BEARING CAPACITY OF LAYERED SOILS
SPECIAL CASES
– TOP LAYER IS STRONG SAND AND BOTTOM LAYER IS
SATURATED SOFT CLAY (2 = 0)
– TOP LAYER IS STRONGER SAND AND BOTTOM LAYER IS WEAKER
SAND (c1 = 0 , c2 = 0)
– TOP LAYER IS STRONGER SATURATED CLAY (1 = 0) AND
BOTTOM LAYER IS WEAKER SATURATED CLAY (2 = 0)
Find the formula for the above
special cases
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BEARING CAPACITY FROM N-SPT VALUE
A square foundation BxB has to be
constructed as shown in the following figure.
Assume that  = 105 lb/ft3, sat = 118 lb/ft3, Df
= 4 ft and D1 = 2 ft. The gross allowable
load, Qall, with FS = 3 is 150,000 lb. The
field standard penetration resistance, NF
values are as follow:
Determine the size of the
foundation
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SOLUTION
Correction of standard penetration number
(Liao and Whitman relationship)
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SOLUTION
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SHALLOW FOUNDATION
SESSION 9 – 10
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GENERAL
The settlement of shallow foundation may be divided into 3 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.
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IMMEDIATE SETTLEMENT
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IMMEDIATE SETTLEMENT
General Equation (Harr, 1966)
•
Flexibel Foundation
– At the center of foundation



 1  m 2  1 
1   1  m 2  m 

  ln
 m. ln 
2
2




   1 m  m 
 1  m  1  



B.q o
1   s2  av
Es
L
m

;
; H=
B
Se 
– Average
•

B.q o

1   s2
Es
2
B.q o
Se 
1   s2 
Es
Se 
– At the corner of foundation
Rigid Foundation
Se 


B.qo
1   s2 r
Es
Es = Modulus of elasticity of soil
B = Foundation width
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L = Foundation length
IMMEDIATE SETTLEMENT
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IMMEDIATE SETTLEMENT
If Df = 0 and H < , the elastic settlement of foundation can be determined
from the following formula:

 (corner of rigid foundation)



B.qo
1   s2 F1  1   s  2  s2 F2
2
1   s 
Se 
Es
2
Se 
B.qo

1   s2 1   s2 F1  1   s  2  s2 F2
Es
(corner of flexible foundation)
The variations of F1 and F2 with H/B are given in the graphs of next slide
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IMMEDIATE SETTLEMENT
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IMMEDIATE SETTLEMENT
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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
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EXAMPLE
Solution:
Part a.


B.q o
1   s2 
Es
Se 
For L/B = 2/1 = 2    1.53, so
Part b.
Se 
Se 


(1)(150)
1  0.32 (1.53)  0.0209m  20.9mm
10,000

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 
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





(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
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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
S e  A1 .A 2
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q o .B
Es
IMMEDIATE SETTLEMENT
• For sandy soil

S e  C1 .C 2 q  q
I
 E
z2
z
0
z
s
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
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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
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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)
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EXAMPLE
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EXAMPLE
Iz
z
Es
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

 t 
 5 
C2  1  0.2. log 
  1  0.2. log 
  1.34
 0.1 
 0.1 
 q 


17.8 x1.5
C1  1  0.5
 160  17.8 x1.5 
  0.9
qq
  1  0.5






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 )
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S e  24.8mm
1.55 x 10-4
CONSOLIDATION SETTLEMENT
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CONSOLIDATION SETTLEMENT
• Normal Consolidation
pc  po
or
pc
1
po
po  p
Cc
Sc 
.Hc . log
1  eo
po
• Over consolidation
pc  po
or
po + p < pc
po < pc < po+p
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pc
1
po
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:
p 
1
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
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pb = the pressure increase at the bottom of the clay layer
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.
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EXAMPLE
Sc 
p  p
Cc
.H c . log o
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 
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0.32
52.5  13.45
2.5 x log
 44 mm
1  0.8
52.5
ALLOWABLE SETTLEMENT
Bina Nusantara