Course
Year
Version
: S0484/Foundation Engineering
: 2007
: 1/0
Session 5 – 6
BEARING CAPACITY OF SHALLOW
FOUNDATION
SHALLOW FOUNDATION
Topic:
• General
• Terzaghi Model
• Meyerhoff Model
• Brinch Hansen Model
• Influence of multi layer soil
• Influence of ground water elevation
• Shallow Foundation Bearing by N-SPT
value
TYPES OF SHALLOW FOUNDATION
TYPES OF SHALLOW FOUNDATION
TERZAGHI MODEL
Assumptions:
• Subsoil below foundation structure is
homogenous
• Shallow foundation Df < B
• Continuous, or strip, footing : 2D case
• Rough base
• Equivalent surcharge
TERZAGHI MODEL
FAILURE ZONES:
1. ACD : TRIANGULAR ZONES
2. ADF & CDE : RADIAL SHEAR ZONES
3. AFH & CEG : RANKINE PASSIVE ZONES
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
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

N  
 1 tan 
2
2  cos  
BEARING CAPACITY FACTORS
GENERAL
FAILURE
BEARING CAPACITY FACTORS
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
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
BEARING CAPACITY FACTORS
GROUND WATER INFLUENCE
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 - ’)
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)
NET ALLOWABLE BEARING CAPACITY
PROCEDURE:
1. Find the developed cohesion and the angle of friction
cd 
2.
c
FS shear
 tan  

 FSshear 
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.
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
EXAMPLE – SOLUTION
Foundation Type: Square Foundation
EXAMPLE – SOLUTION
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
BEARING CAPACITY FACTOR


Nq  tan 2  45  e . tan
2

Nc   Nq  1 cot 
N  2( Nq  1) tan 
SHAPE, DEPTH AND INCLINATION FACTOR
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.
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)
ECCENTRICALLY LOADED FOUNDATIONS
ECCENTRICALLY LOADED FOUNDATIONS
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
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
EXAMPLE – SOLUTION
With c = 0, the bearing capacity equation becomes
TWO-WAY ECCENTRICITY
TWO-WAY ECCENTRICITY – CASE 1
TWO-WAY ECCENTRICITY – CASE 2
TWO-WAY ECCENTRICITY – CASE 3
TWO-WAY ECCENTRICITY – CASE 4
BEARING CAPACITY OF LAYERED SOILS
STRONGER SOIL
UNDERLAIN BY
WEAKER SOIL
BEARING CAPACITY OF LAYERED SOILS
 2D f
 2c H 
qu  qb   a    1H 2 1 
H
 B 

 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
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
 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
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
SOLUTION
Correction of standard penetration number
(Liao and Whitman relationship)
SOLUTION