Transcript Foundation design soil

Foundation design
 Present by Mr. Sieng PEOU
 Master science of geotechnical
engineering
 Tel-011 874 974
 email: [email protected]
Type of foundation
 Shallow foundation
1-Spread footing : support the load from
building by column
2-Strip footing : support the load from
building by walls
3-Mat foundation: combined all footing
Type of foundation
 Deep foundation
1- End bearing pile : pile stand on
rocks or very dense soils, so we
have only end bearing capacity
2- Combined bearing pile : pile stand
on normal soils, so we have end
bearing capacity and skin friction
3- Floating pile : pile stand on very
loose or very soft soil, so we have
only skin friction
Spread footing
Q
B
Strip footing
q
B
Mat foundation
B
End bearing pile
Soft soil layer
Rock layer
Pile
Combined bearing pile
Soft soil layer
Stiff soil layer
Pile
Floating pile
Soft soil layer
Pile
Bearing capacity for Shallow
foundation
 Type of failure
1-General shear failure for dense soil,we
can use C & f for design soils bearing
capacity
2-Local shear failure for loose soil, we can
use C’=2/3 C & f’=arctg(2/3tgf) for
design soils bearing capacity
3-Punching shear failure for very loose
soil,not recommended
General shear failure
Q
D
Shear line
Local shear failure
Q
D
Shear line
Punching shear failure
Q
D
Failure mechanisms and derivation of
equations
Ultimate bearing capacity
qu
S
S
Settlement
qu
s
Failure mechanisms and derivation of
equations
 A relatively undeformed wedge of soil below the foundation




forms an active Rankine zone with angles (45º + f'/2).
The wedge pushes soil outwards, causing passive Rankine
zones to form with angles (45º - f'/2).
The transition zones take the form of log spiral fans.
For purely cohesive soils (f = 0) the transition zones become
circular for which Prandtl had shown in 1920 that the solution
is qf = (2 + p) Cu = 5.14 Cu
This equation is based on a weightless soil. Therefore if the
soil is non-cohesive (c=0) the bearing capacity depends on
the surcharge qo. For a footing founded at depth D below the
surface, the surcharge qo = gD. Normally for a shallow
foundation (D<B), the shear strength of the soil between the
surface and the founding depth D is neglected.
Semi-circular slip mechanism
 Moment causing rotation
= load x lever arm
= [(q - qo) x B] x [½B]
 Moment resisting rotation
= shear strength x length of arc x lever arm
= [Cu] x [p.B] x [B]
 At failure these are equal:
(q - qo ) x B x ½B = Cu x p.B x B
 Net pressure (q - qo ) at failure
= 2 p x Cu
 This is an upper-bound solution.
Circular arc slip mechanism
 Moment causing rotation
= load x lever arm
= [ (q - qo) x B ] x [B/2]
 Moment resisting rotation
= shear strength x length of arc x lever arm
= [Cu] x [2a R] x [R]
 At failure these are equal:
(q - qo) x B x B/2 = Cu x 2 a R x R
 Since R = B / sin a :
(q - qo ) = Cu x 4a /(sin a)²
 The worst case is when
 tana=2a at a = 1.1656 rad = 66.8 deg
 The net pressure (q - qo) at failure
 = 5.52 x Cu
Bearing capacity for strip footing
general equation
After Terzaghi (1943)
qd = CNc + gs DNq +0.5 gBNg
f
N q = tan (45  )ep tanf Re issner 1924
2
2
Nc = (Nq – 1 ) . Cotgf Prandtl 1921
Ng = 2(Nq + 1)tgf Caquot and Kerisel 1953 Vessic 1973
Bearing capacity for footing
From TSA equation
After Vessic (1973)
qd = 5.14 Cu(1+0.2B/L) + gs D
Cu:Undrained cohesion
B: Width of footing
L: Length of footing
Bearing capacity for footing
From TSA equation
After Skemton (1951)
qd = 5 Cu(1+0.2B/L)(1+0.2D/B) + gs D
D/B<2.5
Cu:Undrained cohesion
B: Width of footing
L: Length of footing
D: Depth of footing
Bearing capacity for footing
From TSA equation
After Meyerhof (1951 to 1963)
qd = 5.14 Cu(1+0.2B/L)(1+0.2D/B) + gs D
D/B<2.5
Cu:Undrained cohesion
B: Width of footing
L: Length of footing
Bearing capacity for footing
From ESA equation
After Vessic (1973)
qd = gs D Nq(1+B/L.tgf)+0.5gBNg(1-0.4B/L)
f: Internal friction angle
B: Width of footing
L: Length of footing
Bearing capacity for footing
From ESA equation
After Meyerhof (1951 to 1963)
qd = gs D Nq.Sq.dq+0.5gBNgSgdg
Sq=Sg=10.1KPB/L ; Kp= tg2(45+f/2)
dq=dg=1+0.1Kp0.5D/B
Nq the same Nq Terzaghi ; Ng=(Nq-1)tg(1.4f)
f: Internal friction angle
B: Width of footing
L: Length of footing
Bearing capacity for footing
From general equation
After Meyerhof (1963)
qd = C.Nc.Fcs.Fcd.Fci+gs D Nq. Fqs.Fqd.Fqi +0.5gBNg Fgs.Fgd.Fgi
qnet =C.Nc.Fcs.Fcd.Fci+gs D (Nq-1). Fqs.Fqd.Fqi +0.5gBNg Fgs.Fgd.Fgi
Nq by Reissner1924 ; Nc by Prandtl1921 ; Ng by Caquot and
Kerisel 1953 and by Vessic 1973
f: Internal friction angle
B: Width of footing
L: Length of footing
Bearing factor
 Shape factor by De Beer 1970




Fcs=1+B/L.Nq/Nc
Fqs=1+B/L.tgf
Fgs=1-0.4.B/L
Depth factor by Hansen 1970
Condition D/B<1
Fcd=1+0.4D/B
Fqd=1+2.tgf(1-sinf)2D/B
Fgd=1
Bearing factor
 Depth factor by Hansen 1970
Condition D/B>1
Fcd=1+0.4.arctg(D/B)
Fqd=1+2.tgf(1-sinf)2.arctg(D/B)
Fgd=1
 Inclined factor by Meyerhof 1963 Meyerhof and
Hanna 1981
Fci=Fqi=(1-a/90)2
Fgi=(1-a/f)2
Bearing capacity of mat foundation
 The gross ultimate bearing capacity of a mat
foundation can be determined by the same
equation used for shallow foundation.
 A suitable factor of safety should be used to
calculate the net allowable bearing capacity.For
rafts on clay, the factor of safety should not be
less than 3 under dead load and maximum live
load.However, under the most extreme
conditions,the factor of safety should be at least
1.75 to 2. For rafts constructed over sand,a
factor of safety of 3 should normally be used.
Ultimate bearing capacity equation
for mat foundation on saturated clay
qnet (u )
Df
0.195B
= 5.14Cu (1 
)(1  0.4 )
L
B
Structural
Design of
Mat
Foundation
Conventional Rigid Method
Step1: Calculate the total column load
Q = Qi
Step2: Determine the pressure on the soil (q)
below the mat at point A, B, C…by using the
equation
Q M y X M xY
q= 

A
Iy
Ix
 Where
A=BL
Ix=(1/12)BL3 : moment of inertia about the X
axis
IY=(1/12)LB3 : moment of inertia about the Y
axis
Mx : moment of the column load about the X
axis = Q.eY
MY : moment of the column load about the Y
axis = Q.ex
Step 3: Compare the values of the soil pressures
determine in step 2 with the net allowable soil
pressure to check if q<qall(net)
Step 4: Divide the mat into several strips in X and
Y direction. Let the width of any strip be B1.
Step 5: Draw the shear and moment diagrams for
each individual strip in X and Y direction. For
example, take bottom strip in the X direction its
average soil pressure can be given as:
qav=1/2(ql+qF)
Where ql and qF soil pressures at point I and F
The total soil reaction is equal to qavBB1 because the
shear between the adjacent strips has not been taken
into account. for this reason, the soil reaction and the
column load need to be adjusted
Qav =
qav BB1   Qi
2
qav (mod ified)
Qav
= qav
qav BB1
eX and eY are the load eccentricities in the direction
of the X and Y
Q X'

X '=
QY'

Y'=
B
e X = X '
2
L
e y = Y '
2
i
Q
i
i
Q
i
Also, the column load modification factor is
Qav
F=
 Qi
So the modified column load are FQi
Now the shear and moment diagram for this strip can
be drawn. this procedure can be repeated for all
strips in the X and Y direction.
Step 6: determine depth of the mat d. This can be done
by checking for diagonal tension shear near various
column. According to ACI Code 318-95(section
11.122.1c). For critical section

U = b0 d f (0.34) f 'c

Where:
-U : Factored column load (MN)=F
-f : Reduction factor =0.85
-f’c : Compressive strength of concrete 28 days
(MN/M2)
The unit of b0 and d in the preceding equation
are in meters. The expression of b0 in term of
d, which depends on the location of the
column with respect to the plan of the mat, can
be obtained from Figure 4.8c.
Step 7: from the moment diagrams of all strips in
a given direction (that is X or Y), obtain the
maximum positive and negative moments per
unit width M’=M/B1
Step8: Determine the areas of steel per unit width for
positive and negative reinforcement in X and Y
directions from the following equations.
a
M U = M ' (load . factor ) = fAS f y (d  )
2
a=
AS f y
0.85 f 'c b
Where:
-As: area of steel per unit width
-fY :Yield stress of reinforcement in tension
-Mu :Factored moment
Verify the stable of footing
Q
Qs
Qf
B&L
Qtotal=Q+Qf+Qs
Q- load apply by column
Qf –load of footing
Qs –load of soil above footing
Allowable bearing capacity
 Net ultimate bearing capacity
qnet=qd-gs.D
 Net allowable bearing capacity
qnetall=qnet/FS
FS-Safety factor =3
 Gross allowable bearing capacity
qall=qnetall+gs.D
Verify stable of footing
q
all
net
Q
=
BL
And verify the stable of
footing from equation
qall
Qtotal

BL
We find value of B
When effect water table
D
d
D1
Water level case I
D2
B
Water level case II
When effect water table
1-In case I if the water table is located so that
0<D1<D, so we will change the factor
gs.D
g.D1+D2(gsat-gw)
Also value g in the last term of the equation has
to be replaced by g’= (gsat-gw)
2-In case II for a water table located so 0<d<B
value g in the last term of the equation has to be
replaced by gcal= g’+d/B.(gg’)
Stable of footing when effect
inclined load
qall>V/(BL)
Q
H
Tall>H
a
D
V=Q.Cos a
H=Q.Sin a
T=V.tg(2/3f)+2/3.C.B.L.
Tall=T/1.5
T
B
Q
V
When effect 0ne way bending
moment
We change B to B’ for calculate bearing capacity
Q
B’=B-2eB
MB
eB=MB/Q
B
Verify stable of footing when
effect one way bending moment
When eB<B/6
qall  qmax
Q
MB
qmin
Q
6e B
=
(1 
)
BL
B
qmax
Q
6e B
=
(1 
)
BL
B
Verify stable of footing when
effect one way bending moment
When eB>B/6
qall  qmax
Q
MB
4Q
qmax =
3L( B  2eB )
Not recommended
Foundation with two way
Eccentricity
For calculate bearing capacity we have to change:
Q
B to B’=B-2eB
ML
L to L’=L-2eL
A’=B’*L’
MB
L
eB=MB/Q
eL=ML/Q
B
Verify stable of footing when
effect two way bending moment
Qult= qu’.A’
Case eL/L>1/6
eB/B>1/6
B1=B(1.5-3eB/B)
L1=L(1.5-3eL/L)
B’=A’/L
Verify stable of footing when effect
two way bending moment
Qult= qu’.A’
Case 1/6<eL/L<0.5
0<eB/B<1/6
A’=0.5(L1+L2)B
B’=A’/L1
Qult= qu’.A’
Case eL/L< 1/6
1/6<eB/B< 0.5
A’=0.5(B1+B2)L
B’=A’/L
Qult= qu’.A’
Case eL/L< 1/6
eB/B< 1/6
A’= L2B+0.5(B+B2)(L-L2)
B’=A’/L
Footing on two layer
gs
D
c1 g1 f1
c2 g2 f2
B
d1
Bearing capacity of footing
on two layer
1- Determine influenced thickness
H=0.5Btg(45+f1/2)
If H<d1 : our footing not effect on second layer,
so we calculate the soils bearing capacity by
using values C1,g1,f1
If H>d1 : our footing effect on second layer, so
we calculate the soils bearing capacity by using
condition as follows:
Bearing capacity of footing on
two layer
From TSA condition
1- Design CR=CU2/CU1
If CR<1 :
Nc =
1,5d1
 5,14CR
B
3d1
Nc =
 6,05CR
B
< 5.14 for strip footing
< 6.05 for spread footing
so we calculate the soils bearing capacity by using
equation
qnet=C u1NC
If CR>0.7 the value of NC is decrease 10%
Bearing capacity of footing
on two layer
If CR>1 :
for strip footing
for spread footing
N1  N 2
Nc =
2
N1  N 2
0,5B
 4,14
d1
1,1B
N2 =
 4,14
d1
0,33B
N1 =
 5,05
d1
0,66B
N2 =
 5,05
d1
N1 =
so we calculate the soils bearing capacity by
using equation
qnet=C u1NC
Bearing capacity of footing
on two layer
From general equation 1
1- Determine the average values of soils
parameter
d1f1  ( H  d1 )f 2
f'=
H
d1c1  ( H  d1 )c 2
c' =
H
2- Determine the soils bearing capacity by
using values
C’ and f’
Bearing capacity of footing
on two layer
From general equation 2
1- Determine the bearing capacity for first layer
qnet1 = C1.Nc.Fcs.Fcd.Fci+gs D (Nq-1). Fqs.Fqd.Fqi +0.5g1BNg Fgs.Fgd.Fgi
2- Determine the soils bearing capacity for
second layer
qnet2= C2.Nc.Fcs.Fcd.Fci+(gs D+g1d1) (Nq-1). Fqs.Fqd.Fqi +0.5.g2.BNg Fgs.Fgd.Fgi
Bearing capacity of footing
on two layer
From general equation 2
3- Determine the bearing capacity
P  PV K s tanf1 Pd1C1 < q
qnet = qnet 2 

net1
Af
P = 2(B+L)
Pv = 0.5 g1 d12+ gs D d1
Ks =1-sinf1
Af =BL
Af
Bearing capacity from in situ
test
 From static cone penetration test
1- for B<1.22m
qc
qallowable =
15
2- for B>1.22m
qallowable =
qc 3,28B  1

25
3,28B
 From dynamic cone penetration test
Rd
qallowable =
20
)2
Bearing capacity from in situ test
 From standard penetration test SPT by Meyerhof(1965)
1- for B<1.22m
N 60 
D  Se (m m) 
q net ( KN / m ) =
1  0.33 
0.05 
B  25 
all
2
2- for B>1.22m
N60  B  0.3 
q net ( KN / m ) =


0.08  B 
all
2
2
D  Se (mm) 

1  0.33 
B  25 

Combined footing
Rectangular combined footing
Q1
L1
Q1+Q2
X
L3
Q2
L2
Section
q
Plan
B
L
Design dimension of rectangular
combined footing
 Determine the area of the footing
Q1  Q2
A=
qall ( net )
 Determine the location of the resultant of the
column loads
Q2 .L3
X =
Q1  Q2
 For uniform distribution of soil pressure under
the foundation, the resultant of the column
loads should pass through the centroid of the
foundation.Thus,
L = 2( L1  X )
Design dimension of rectangular
combined footing
 Once the length L is determined,obtain the
value of L1
L1 = L  L2  L3
 Note that the magnitude of L2 will be known and
depends on the location of the property line
 The width of the foundation then is
A
B=
L
Combined footing
Trapezoidal combined footing
Q1
Section
L2
Q1+Q2
Q2
X
L3
L1
Plan
B1
L
B2
Design dimension of trapezoidal
combined footing
 Determine the area of the footing
 And we have relation
Q1  Q2
A=
qall ( net )
B1  B2
A=
L
2
 Determine the location of the resultant of the
column loads
Q2 .L3
X =
Q1  Q2
Design dimension of trapezoidal
combined footing
 From the property of a trapezoid, we have
 B1  2 B2  L

X  L2 = 
 B1  B2  3
 With Known values of A,L,X and L2 we can find
values of B1 and B2, Note that for a trapezoid,
L
L
 X  L2 
3
2
Combined footing
Cantilever footing
Q1
Q2
S
Section
e R
1
L1
Plan
R2
B2
Design dimension of
Cantilever footing
 Design arm moment for soils reaction
strength R1
S’=S-e (value of e is proposed by designer)
 Design soils reaction strength
S
R1 = Q1
S'
Q1.e
R2 = Q2 
S'
R2 = Q1  Q2  R1
Design dimension of
Cantilever footing
 Design the dimension of first footing
R1
A1 =
all
qnet
C

L1 = 2 e  
2

A1
B1 =
L1
 C is length of column
 Design the dimension of second footing
R2
A2 = all
qnet
A2
B2 =
L2
Rock quality
 Rock quality designation(RQD) is an index or
measure of the quality of a rock mass(Stagg and
Zienkiewicz 1968) used by many engineers.RQD
is computed from recovered core samples as
RQD =
 lengthof intactpieces of core  100mm
Lengthof coreadvance
 A core advance of 1500mm produced a sample
length of 1310mm consisting of dust,gravel,and
intact pieces of rock.The sum of length of pieces
100mm or larger in length is 890mm.The recovery
ratio Lr=1310/1500=0.87 and RQD=890/1500=0.59
Allowable Bearing capacity of
rock
 The allowable bearing capacity is
depending on geology,rock type,and
quality(as RQD).
 If RQD>0.8 would not require as high an
FS as for RQD=0.4.
 We take FS from 6 to 10 for RQD less
than about 0.75
Bearing capacity for sound rock
f

quc = 2c' tan 45  
2

N q = t an6 ( 45 
f
2
N c = 5 t an4 ( 45 
)
f
2
)
Ng = N q  1
For calculate bearing capacity
we use equation Terzaghi
F=45 degree for most rock except
limestone or shale where values
between 38 to 45 degree.
Similarly we could in most cases
estimate Cu=5MPa as a
conservative value.
And finally we may reduce the
ultimate bearing capacity base on
RQD as:
qult(cal)=qult(RQD)2
Table 4.2 Range of the Unconfined Compression Strength
of Various Types of Rocks
quc
Rock type
Granite
Limestone
Sandstone
Shale
MN/m2
65–250
30–150
25–130
5–40
Phi
kip/m2
9.5–36
4.0–22
3.5–19
0.75–6
(deg)
45–55
35–45
30–45
15–30
Rang of properties for selected rock
groups;data from several sources
Type of
rock
Unit wt.(KN/m3) E(MPa.103)
m
qu((Mpa)
Basalt
28
17-103
0.27-0.32
170-415
Granite
26.4
14-83
0.26-0.30
70-276
Schist
26
7-83
0.18-0.22
35-105
Limestone
26
21-103
0.24-0.45
35-170
Porous
limestone
-
3-83
0.35-0.45
7-35
Sandstone
22.8-23.6
3-42
0.20-0.45
28-138
Shale
15.7-2.2
3-21
0.25-0.45
7-40
concrete
15.7-23.6
variable
0.15
15-40
Settlement of shallow
foundation
 There are two types of settlement
1-Immediate settlement or elastic settlement Se
for sandy soils
2-Consolidation settlement Sc for fine grained
soils
2-1-Primary consolidation settlement for
inorganic soils
2-2-Secondary consolidation settlement for
organics soils
Immediate settlement on sandy soils
 Foundation could be considered fully flexible or
fully rigid
1-A uniformly loaded, perfectly flexible
foundation resting on an elastic material such
as saturated clay will have a sagging profile as
shown in figure 1,because of elastic
settlement.
2-If the foundation is rigid and is resting on an
elastic material such as clay,it will undergo
uniform settlement and the contact pressure will
be redistributed as shown in figure 2.
Type of foundation settlement
Settlement profile
Figure 1
Settlement profile
Figure 2
Calculate immediate
settlement
Q
D
q0
mPoisson’s ratio
E-Modulus of elasticity
Soil
Rock
H
Calculate immediate
settlement
 At corner of the flexible foundation S
 At center of the flexible foundation
 1  m 2  1 
1   1  m 2  m 

a = ln
 m ln
 1  m 2  1 
p   1  m 2  m 



e
Bq 0
2 a
=
(1  m )
E
2
Bq 0
Se =
(1  m 2 )a
E
L
m=
B
 Average settlement for flexible foundation
 Settlement for rigid foundation
Bq 0
Se =
(1  m 2 )a av
E
Bq 0
Se =
(1  m 2 )a r
E
Value of a
Shape of
foundation
Flexible foundation
Center
Corner
Average
Rigid
foundation
Circular
1
0.64
0.85
0.79
Square
1.12
0.56
0.95
0.82
Rectangular
L/B=1.5
1.36
0.68
1.15
1.06
L/B=5.0
2.1
1.05
1.83
1.7
L/B=10
2.54
1.27
2.25
2.1
Immediate settlement of
foundation on saturated clay
 Janbu et al.(1956)proposed an equation
for evaluating the average settlement of
flexible foundations on saturated clay soils
(Poisson’s ratio m=0.5)
q0 B
S e = A1. A2
E
Variation of A1 With H/B by Christian and
Carrier(1978)
H/B
A1
Circle
1
2
4
6
8
10
20
30
0.36
0.47
0.58
0.61
0.62
0.63
0.64
0.66
1
0.36
0.53
0.63
0.67
0.68
0.70
0.71
0.73
2
0.36
0.63
0.82
0.88
0.90
0.92
0.93
0.95
L/B
3
0.36
0.64
0.94
1.08
1.13
1.18
1.26
1.29
4
0.36
0.64
0.94
1.14
1.22
1.30
1.47
1.54
5
0.36
0.64
0.94
1.16
1.26
1.42
1.74
1.84
Variation of A2With D/B by Christian
and Carrier(1978)
D/B
0
2
4
6
8
10
12
14
16
18
20
A2
1
0.9
0.88
0.875
0.87
0.865
0.863
0.86
0.856
0.854
0.85
Consolidation settlement
 For normally consolidated clay s’0 ≥ s’p
s o  s
Cc  H
S=
. log.
1  eo
so
WL (%)
Cc = 0.2343
.s

 100 
1
s = (s t  4s m  s b )
6
s 'P = 22I P
0.48
s 'P = 7.04Cu
Cu
0.83
N
s 'P = 0.193s 0 '. 
s 0 ' 
By Mayne & Mitchell
By Mitchell(1988)
0.689
By Mayne & Kemper(1988)
Consolidation settlement
 For over consolidated clay s’0<s’p
1- s'O + s  s'P
s o  s
Cs . H
S =
log
1  eo
so
W (%)
Cs = 0,0463 L
.s

 100 
2- s'O + s > s'P
 s ' o  s
s ' P Cc.H
Cs.H
S=
. log

. log
1  eo
s o 1  eo
 s 'P



Tolerable Settlement of
building
 Settlement analysis is an important part of
the design and construction of foundation
 Large settlement of various component of
structure may lead to considerable
damage or may interfere with the proper
functioning of the structure.
Settlement of foundation
di-total displacement at
point i
dij-different settlement
between point i and j
 relative deflection
d ij
  angular
hij=
lij
distortion
/L=deflection ratio
Limiting angular distortion as recommended
by Bjerrum(Compiled from Wahls,1981)
damage Category of potential
Danger to machinery sensitive to settlement
h
1/750
Danger to frames with diagonals
Safe limit for no cracking of building
First cracking of panel walls
1/600
1/500
1/300
Difficulties with overhead cranes
Tilting of high rigid building becomes visible
Considerable cracking of panel and brick walls
1/300
1/250
1/150
Danger of structure damage to general building
Safe limit for flexible brick walls L/H>4
Safe limit include a factor of safety
1/150
1/150
Allowable settlement criteria:1955 U.S.S.R
Building code(compiled from walhls,1981)
Type of structure
Sand and hard clay Plastic clay
h
Civil and industrial building column foundation
For steel and reinforced concrete structure
0.002
0.002
For end rows of columns with brick cladding
0.007
0.001
For structure where auxiliary strain does not arise during
Nonuniform settlement of foundation
0.005
0.005
Tilt of smokestacks,tower,silos,and so on
0.004
0.004
Crane ways
0.003
0.003
/L
Plain brick walls
For multistory dwelling and civil building
At L/H<3
0.0003
0.0004
At L/H>5
0.0005
0.0007
For one-story mills
0.0010
0.0010
Allowable average settlement for different building
type(compiled from Wahls,1981)
Type of building
Allowable average
settlement(mm)
Building with plain brick walls
L/H>2.5
80
L/H<1.5
100
Building with brick walls,reinforced with
reinforced concrete or reinforced brick
150
Framed building
100
Solid reinforced concrete foundation of
smokestacks,silos,towers,and so on
300
Deep foundation
 Need for pile foundation
1-When the upper soils layers are highly compressible
and too weak to support the load transmitted by the
superstructure, piles are used to transmit the load to
underlying bedrock or stronger soil layer.
2-When subjected to horizontal force, pile foundations
resist by bending while still supporting the vertical load
transmitted by superstructure.This situation is
generally encountered in the design and construction
of earth-retaining structures and foundations of tall
structures that are subjected to strong wind and/or
earthquake forces.
Deep foundation
3-The expansive and collapsible soils may extend to a
great depth below the ground surface.These soils
swell and shrink as the water content increase and
decrease.If shallow foundations are used, the
structure may suffer considerable damage.The pile
have to extend into stable soil layer beyond the zone
of possible moisture change.
4-The foundation of some structures, such as
transmission towers,offshore platforms, and basement
mats below the water table, are subjected to uplifting
forces.Pile are sometime used for these foundations to
resist the uplifting force.
Deep foundation
5-Bridge abutments and piers are usually constructed
over pile foundations to avoid the possible loss of
bearing capacity that a shallow foundations might
suffer because of soil erosion at the ground surface.
Although numerous investigations, both theoretical
and experimental, have been conducted to predict the
behavior and the load-bearing capacity of piles in
granular and cohesive soils,the mechanisms are not
yet entirely understood and never be clear.The design
of pile foundations may be considered somewhat of
an”art”as a result of the uncertainties involved in
working with some subsoil condition.
Types of piles
 Different types of piles are used in construction
work,depending on the type of load to be
carried, the subsoil conditions,and the water
table.Pile can be divided into these categories:
-Steel piles
-Concrete piles
-Wooden(timber)piles
-Composite piles
Comparisons of piles made of different materials
Pile type
Usual
length of
pile(m)
Maximum
length of
pile(m)
Usual load Approximate
maximum
(KN)
load(KN)
Steel
15-60
Practically
unlimited
300-1200
-
Advantages: a-Easy to handle with respect to cutoff and extension to the
desired length
b-Can stand high driving stresses
c-Can penetrate hard layer such as dense gravel,soft rock
d-High load-carrying capacity
disadvantages: a-Relatively costly material
b-High level of noise during pile driving
c-Subject to corrosion
d-H-piles may be damaged or deflected from the vertical
during driving through hard layers or past major obstructions
Comparisons of piles made of different materials
Pile type
Usual length
of pile(m)
Maximum
length of
pile(m)
Precast precast::10-15 precast::30
concrete Prestressed: Prestressed:
60
10-35
Usual
load
(KN)
Approximate
maximum
load(KN)
3003000
precast::800900
Prestressed:
7500-8500
Advantages: a-Can be subjected to hard driving
b-Corrosion resistant
c-Can be easy combined with concrete superstructure
disadvantages: a-Difficult to achieve proper cutoff
b-Difficult to transport
Comparisons of piles made of different materials
Pile type
Usual
length of
pile(m)
Maximum
length of
pile(m)
Cased castin place
concrete
5-15
15-40
Usual load Approximate
maximum
(KN)
load(KN)
200-500
Advantages: a-Relatively cheap
b-Possibility of inspection before pouring concrete
c-Easy to extend
disadvantages: a-Difficult to splice after concreting
b-Think casings may be damages during driving
800
Comparisons of piles made of different materials
Pile type
Usual
length of
pile(m)
Maximum
length of
pile(m)
Usual load Approximate
maximum
(KN)
load(KN)
uncased
5-15
30-40
cast-in place
concrete
Advantages: a-Initially economical
300-500
700
b-Can be finished at any elevation
disadvantages: a-Voids may be created if concrete is placed rapidly
b-In soft soils,the sides of the hole may cave in thus
Squeezing the concrete
Comparisons of piles made of different materials
Pile type
Usual
length of
pile(m)
Maximum
length of
pile(m)
Wood
10-15
30
Usual load Approximate
maximum
(KN)
load(KN)
100-200
270
Advantages: a-Economical
b-Permanently submerged piles are fairly resistant to decay
c-Easy to handle
disadvantages: a- Decay above water table
b-Can be damaged in hard driving
c-Low load-bearing capacity
d-Low resistance to tensile load when splices
Typical concrete pile
Practical list of typical air and steam hammers
Maker of
hammer*
V
V
V
MKT
V
V
R
MKT
R
V
R
MKT
V
V
MKT
MKT
MKT
V
Model
no.
3100
540
060
OS-60
040
400C
8/0
S-20
5/0
200-C
150-C
S-14
140C
08
S-8
11B3
C-5
30-C
Type of
hammer
Single acting
Single acting
Single acting
Single acting
Single acting
Differential
Single acting
Single acting
Single acting
Differential
Differential
Single acting
Differential
Single acting
Single acting
Double acting
Double acting
Double acting
Rated energy
(kN-m)
407
271
244
244
163
154
110
82
77
68
66
51
49
35
35
26
22
10
Blows per
minute
58
48
62
55
60
100
35
60
44
98
95-105
60
103
50
55
95
110
133
Ram weight
(kN)
449
182
267
267
178
178
111
89
78
89
67
62
62
36
36
22
22
13
Practical list of typical diesel hammers
Maker of
hammer*
K
M
K
K
M
K
MKT
K
V
L
M
V
L
MKT
MKT
L
Model
no.
K150
MB70
K-60
K-45
M-43
K-35
DE70B
K-25
N-46
520
M-14S
N-33
440
DE20
DE-10
180
Rated energy
(kN-m)
379.7
191.2-86
143.2
123.5
113.9-51.3
96
85.4-57
68.8
44.1
35.7
35.3-16.1
33.4
24.7
24.4-16.3
11.9
11.0
Blows per minute
45-60
38-60
42-60
39-60
40-60
39-60
40-50
39-60
50-60
80-84
42-60
50-60
86-90
40-50
40-50
90-95
Piston weight
(kN)
147
71
59
44
42
34
31
25
18
23
13
13
18
9
5
8
Pile driven formulas
 To develop the desired load-carrying capacity,a point bearing
pile must penetrate the dense soil layer sufficiently or have
sufficient contact with a layer of rock.This requirement cannot
always be satisfied by driving a pile to a predetermined depth
because soil profile vary.For that reason, several equations
have been developed to calculate the ultimate capacity of a pile
during driving.These dynamic equations are widely used in the
field to determine whether the pile has reached a satisfactory
bearing value at the predetermined depth.One of the earliest of
these dynamic equations-commonly referred to as the
Engineering News Record (ENR) formula-is derived from the
work-energy theory;that is : Energy imparted by the hammer
per blow =(pile resistance)(penetration per hammer blow)
ENR equations
WR h
Qu =
S C
 Where WR-Weight of the ram
h-height of fall of ram(Cm)
S-penetration of the pile per
hammer blow(Cm)
C-a constant
C=2.54 Cm for drop hammer
C=0.254Cm for steam hammer
Factor of safety FS=6
ENR equations for single and double acting
hammer
E .H E
Qu =
S C
 Where E-hammer efficiency
HE-rated energy of the hammer
S-penetration of the pile per hammer
blow(Cm)
C-a constant
C=0.254 Cm
Factor of safety FS=4 to 6
Modified ENR equations
EWR h WR  n2WP
Qu = (
)(
)
S  C WR  WP
 Where E-hammer efficiency
h-height of fall of the ram(Cm)
S-penetration of the pile per hammer
blow(Cm)
WP-weight of the pile
n-coefficient of restitution between
the ram and the pile cap
C=0.254 Cm
Factor of safety FS=4 to 6
Michigan state highway commission equations
 After testing on 88 pile(1965)
 Where WR-weight of the ram
1,25EHE WR  n2WP
Qu =
S  C WR  WP
WP-weight of the pile
HE-rated energy of the hammer
S-penetration of the pile per hammer
blow(M)
C-a constant
C=2.54.10–3M
Factor of safety FS= 6
Danish equations
EHE
Qu =
EHE L
S
2 AP EP
 Where E-hammer efficiency
EP-modulus of elasticity of the pile
HE-rated energy of the hammer
S-penetration of the pile per hammer
blow(M)
L-length of the pile
AP-area of the pile cross section
Factor of safety FS= 6
Pacific Coast Uniform Building Code equations
After International Conference of building
 W  nW 
officials,1982

( EH )
R
P
 W W 
R
P 

Qu L
S
AP E P
E
Qu =
 Where E-hammer efficiency
HE-rated energy of the hammer
S-penetration of the pile per hammer
blow(M)
L-length of the pile
EP-modulus of elasticity of pile
n=0.25 for steel piles and n=0.1 for another
piles
Factor of safety FS= 4 to 5
Value of E & n
Hammer type
Efficiency,E
Single and double acting hammers
0.7-0.85
Diesel hammers
0.8-0.9
Drop hammers
0.7-0.9
Pile material
Cast iron hammer and concrete pile
without cap
Coefficient of restitution
n
0.4-0.5
Wood cushion on steel pile
0.3-0.4
Wooden pile
0.25-0.3
Equation for estimation of pile
capacity
 QU=QP+Qs
Where QU is ultimate load carrying capacity
of pile
QP is load carrying capacity of the pile
point
QS is frictional resistance
Pile foundation
Qu= Qp
Qu= Qp+Qs
Qs
Qs
Weak
soil
L
Qu= Qs
Weak
soil
L
Weak
soil
L
Lb
Qp
Qp
Rock
Qp
Strong soil layer
Strong soil layer
Minimum pile embedment depth
into founding soil strata
 From civil engineering association forum the
minimum pile embedment depth into bearing
stratum is 3 times diameter of pile.
 Replace the pile with one having a different helix
configuration. The replacement pile must not
exceed any applicable maximum embedment
length and either (A) meet the minimum effective
torsion resistance criterion and all applicable
embedment criteria shown in Table for the
design load type (s), or (B) pass proof testing.
Replacement pile embedment
criteria
Design Load type
Replacement Pile Embedment Criterion
Tension
The last helix must be embedded at
least three times its own diameter
beyond the position of the first helix
of the replaced pile.
Compression
The last helix must be embedded
beyond the position of the first helix
of the replaced pile.
Shear/Overturning
Embedment must satisfy the specified
minimum.
Load-carrying Capacity of the pile point,QP
from Terzaghi’s equation
 QP=AP.qP=AP(CN *c+q’N *q)
Where AP-area of pile tip
C-cohesion of the soil supporting the pile tip
qP-unit point resistance
q’-effective vertical stress at the level of the pile
tip
N*C,N*q-bearing capacity factor after Caquot &
Kerisel
Nq = e
*
7 tgf
N = ( N 1) cotf
*
C
*
q
Load-carrying Capacity of the pile point,QP
from Eric Gervreau in Euro code 2000
 QP=AP.qP=AP(1.3CN *c+50N *q)
Where AP-area of pile tip
C-cohesion of the soil supporting the pile
tip
qP-unit point resistance
N*C,N*q-bearing capacity factor after
Caquot & Kerisel
Nq = e
*
7 tgf
N = ( N 1) cotf
*
C
*
q
Critical depth
 In the case of calculation of q’, the normal
practice is to assume that q’ increases
linearly with depth from zero at ground
level to a maximum value q’(max) at the tip
of pile.
 However, extensive research carried out
by Vessic(1967) has indicated that q’
varies linearly from the ground surface up
to a limited depth only beyond which q’,
remains constant irrespective of the depth
of embedment of pile.
Critical depth
 This phenomenon was attributed to arching of
SAND.
 This depth within which q’ varies linearly with
depth may be called as the critical depth Dc.
 From the curves given by Poulos (1980), we
may write
 For 28<f<36.5 we have Dc/B=5+0.24(f-28)
 For 36.5<f<42 we have Dc/B=7+2.35(f-36.5)
Critical depth
 From Caquot & Kerisel Dc=B/4.N*q(2/3)
 In Bearing Capacity Technical Guidance by Career
Development and Resources for Geotechnical
Engineers
-Dc = 10B, for loose silts and sands
-Dc = 15B, for medium dense silts and sands
-Dc = 20B, for dense silts and sands
-loose when
N<10 or f<30
-medium dense when 10<N<30 or 30<f<36
-dense when
30<N or 36<f
Critical depth
 This critical concept implies that fs for cohesionless
soil for a driven pile varies linearly with depth up to
depth Dc only and beyond this depth fs remains
constant.
 Note that the application concept Dc in case the soil is
homogeneous for the whole depth of embedment D.
 Since no information is available on the layered
system of soil, this approach has to be used with
caution. Tomlinson(1986) Bowles(1988) has not use of
this concept .This indicates that this method has
not yet found favor with the designer.
Load-carrying Capacity of the pile point in sand
from ESA condition after Meyerhof (1976)
QP=AP.qP=APq’N *q
Where AP-area of pile tip
qP-unit point resistance
q’-effective vertical stress at the level of
the pile tip
N*q-bearing capacity factor
QP=Apq’N*q<Apqi
qi=50N*qtgf(KN/M2)
As per Tomlinson, the maximum base resistance
qp is normally limited 11000KPa.
Load-carrying Capacity of the pile point in
sand from ESA condition after Meyerhof
(1976)
 The angle f to be use for determination
Nq* are
 For driven pile f = f1
 For bored pile f = f1-3
Where f1 is angle of internal friction prior to
installation of pile.
Load-carrying Capacity of the pile point in
saturated clay from TSA condition
 QP=AP.qP=ApCU N *c= 9CUAP
Where AP-area of pile tip
qP-unit point resistance
N*c-bearing capacity factor for f=0 N*C=9
Carrying capacity of piles in layered soil
from meyerhof equation
Carrying capacity of piles in layered soil
 If the pile toe terminates in a layer of dense sand or
stiff clay overlying a layer of soft clay or loose sand
there is a danger of it punching through to the weaker
layer.
 To account for this, Meyerhof's equation is used.
 The base resistance at the pile toe is
qp = q2 + (q1 - q2)H / (10B) but < q1
 where
-B is the diameter of the pile
-H is the thickness between the base of the pile and
the top of the weaker layer
-q2 is the ultimate base resistance in the weak layer
-q1 is the ultimate base resistance in the strong layer.
Relation between ultimate point resistance of pile and
depth in sand stratum beneath weak soil layer from
Terzaghi lectures, 1974-1982
Relation between ultimate point resistance of pile and
depth in sand stratum beneath weak soil layer from
Terzaghi lectures, 1974-1982
Skin friction from b Method
From Meyehof 1976
f<28
we have b=0.44
28<f<35 we have0.44< b<0.75
35<f<37 we have 0.75<b<1.20
f=bs’0
s’0-effective vertical stress at center of layer
As Tomlinson, the maximum frictional resistance is
limited 110KPa
Skin friction
Skin friction from b Method
Skin friction from b Method
 The angle f to be use for determination b are
 For driven pile f = 0.75f1+10
 For bored pile f = f1-3
Where f1 is angle of internal friction prior to
installation of pile.
Skin friction from a Method
Skin friction for clayey soil for driven pile
f=axCu a=1
for Cu=<25KPa
a=0.5 for Cu=>70KPa
a=1-(Cu-25)/90 for
25KPa<Cu<70KPa
a=1
for Cu<=35KPa
a=0.5 for Cu=>80KPa
a=1-(Cu-35)/90 for
35KPa<Cu<80KPa
Skin friction for clayey soil for Bored pile or drilled shafts
f=axCu a=0.45 for London clay
a=0.7 time value for driven diplacement pile
a=0
for Z<1.5
API(1984)
Semple and Rigden(1984)
Skempton(1959)
Flaming et al(1985)
Reese and Oneill(1985)
for driven Pile
Tomlinson a method
 Case 1:pile driven through sands or sandy
gravels into stiff clay strata.
 Case 2:pile driven through soft clay into
stiff clay strata.
 Case 3:pile driven into a firm to stiff clay
without any overlying strata.
 The value of a vary with Cu and L/B ratio
Tomlinson a method
Negative skin friction
 Negative skin friction is a downward drag force exerted
on the pile by the soil surrounding it.This action can
occur under conditions such as the following:
1-if a fill of clay soil is placed over a granular soil layer
into witch a pile is driven, the fill will gradually
consolidate. This consolidation process will exert a
downward drag force on the pile during the period of
consolidation.
2-if a fill of granular soil is placed over a layer of soft
clay,it will induce the process of consolidation in the clay
layer and thus exert a downward drag on the pile
3-lowering of the water table will increase the vertical
effective stress on the soil at any depth,which will induce
consolidation settlement in clay.If a pile is located in the
clay layer,it will be subjected to a downward drag force.
Clay fill over granular soil
Granular soil fill over clay
Clay
Hf
Hf
fill
Sand
fill
L
Sand
L
L1
Neutral
plane
Clay
Clay fill over granular soil
f n = K 's 0 ' tand
 Where:
K’=earth pressure coefficient =Ko=1-sinf
s’o=vertical effective stress at any depth Z
= g’f.Z.
g’f =effective unit weight of fill Clay
d=soil-pile friction angle = 0.5f to 0.7f
H
Qn =  ( PK ' g ' f tand ) Zd z =
0
PK ' g ' f H 2 tand
2
Granular soil fill over clay
 In this case, the evidence indicates that the
negative skin stress on the pile may exist from
Z=0 to Z=L1,which is referred to as the neutral
depth.The neutral depth may be given as
(Bowles 1982)
L  H f  L  H f g 'f H f

L1 =

L1  2
g'
 2g ' f H f
 
g'

 Hence,the total drag force is
L1 PK ' g ' tand
Qn =  PK ' (g ' f H f  g ' Z ) tandd Z = PK ' L1g ' f H f tand 
2
0
L1
2
Determine End bearing capacity of
pile foundation from SPT test
Driven Method
qp=CN(Mpa)
C
0.45
qp=CN(Mpa)
0.4
qp=CN(Mpa)
0.04 Ls/D
qp=CN(Mpa)
0.35
Glacial Coarse to fine siltqp=CN(Mpa)
0.25
Residual sandy silt
qp=CN(Mpa)
0.25
Decourt(1982)
Residual Clayey silt
qp=CN(Mpa)
0.2
Decourt(1982)
Clay
qp=CN(Mpa)
0.2
Matin et al.(1987)
Clay
qp=CN(Mpa)
0.12
Decourt(1982)
All soil
qp=CN(Mpa)
0.3
Sand
Silt, sandy silts
N=average SPT value in By Martin et al(1987)
local failure zone
By Decourt(1982)
Ls=Length of pile in sand Mayerhof(1976)
D=width of pile C<=0.4
N=average SPT value in Matin et al.(1987)
local failure zone
ForL/D>=5
If L/D<5,C=0.1+0.04L/D
for closed end pile
and C=0.06L/D
for open end pile
Thorburn and Mac Vicar(1987)
Shioi and Fukui(1982)
Determine End bearing capacity of
pile foundation from SPT test
Cast in place method
qp=CN(Mpa)
0.15
qp<3.0MPa
Shioi and Fukui(1982)
qp=CN(Mpa)
0.15
qp<7.5MPa
Yamashita et al(1987)
Fine grained soil
qp=CN(Mpa)
0.15
qp=0.09(1+0.16Lt) Yamashita et al(1987)
Lt=pile length
Bored pile
Sand
qp=CN(Mpa)
0.1
Shioi and Fukui(1982)
Clay
qp=CN(Mpa)
0.15
Shioi and Fukui(1982)
Coarse grained soil
Determine skin friction from SPT
test
Driven Methode
Coarse grained soil
qf=A+BN(Kpa)
A
0
B
2
Coarse grained &fine soilqf=A+BN(Kpa)
10
3.3
qf=A+BN(Kpa)
Fine grained soil
Cast in place methode
qf=A+BN(Kpa)
Coarse grained soil
0
10
30
2
qf=A+BN(Kpa)
0
5
qf=A+BN(Kpa)
0
5
qf=A+BN(Kpa)
0
10
Shioi and Fukui(1982)
qf=A+BN(Kpa)
0
1
Findlay(1984)&Shioi & Fukui(1982)
qf=A+BN(Kpa)
0
3.3
qf=A+BN(Kpa)
10
3.3
Fine grained soil
Bored pile
Coarse grained soil
Fine graned soil
N=average SPT
along Shaft
3<N<50
Mayerhof(1956)
Shioi and Fukui(1982)
Decourt(1982)
Shioi and Fukui(1982)
qf<200Kpa
Yamashita et al(1987)
Shioi and Fukui(1982)
qf<150Kpa
Yamashita et al(1987)
Wright &Reese(1979)
qf<170Kpa
Decourt(1982)
Load-Carrying capacity of pile point resting on
rock
 The ultimate unit point resistance in
rock(Goodman,1980) is approximately
qp=qu-R(Nf+1)
Where Nf=tg2(45+f/2)
qu-R=unconfined compression strength of rock
f=drained angle of friction of rock
The allowable load-carrying capacity of the pile
point.thus
Q p ( all )

q
=
uR

( Nf  1) AP
FS
FS=3
Typical unconfined compressive strength of rock
Rock type
qu-R(Mpa)
Sandstone
70-140
Limestone
105-210
Shale
35-70
Granite
140-210
Marble
60-70
qu  R ( design ) =
qu  R (lab)
5
Drilled Shafts Extending into
Rock
 Based on the procedure developed by Reese and
O’Neill(1988-1989),we can estimate the bearing load
capacity of drilled shafts extending into Rock as
follows:
1-Calculate the ultimate unit side resistance as:
f=6.564qu0.5≤0.15qu
Where qu=unconfined compression strength or Rock
core
2-Calculate the ultimate capacity based on side
resistance only:
Qu=πDsLf
 Calculate the settlement Se of the shaft at the top of the Rock
socked:
Se=Se(s)+Se(b)
Where Se(s)=elastic compression of the drilled shaft within the
socket, assuming on side resistance
Se(b)=settlement of the base
However
And
QU L
Se(s)=
AC EC
Se(b)=
QU I f
DS Emass
Where Qu=Ultimate friction load
Ac=Cross-section area of the drilled shaft
in the socket
Ds=Diameter of the drilled shaft
Ec=Young’s modulus of the concrete
Emass=Young’s modulus of the rock mass
If=Elastic influence coefficient (read on
chart)
L=Depth of embedment in rock
If Se is less than 10mm, then the ultimate loadcarrying capacity from this way is correct.
If Se≥ 10mm, there may be rapid, progressive side
shear failure in the rock socket ,resulting in a
complete loss of side resistance. In that case the
ultimate capacity is equal to the point resistance :

CS

3
DS

Qu = 3qU Ac 

101  300 d
 
CS



0.5 
 


 
Where
Cs=Spacing of discontinuities
δ=Thickness of individual discontinuity
qu=unconfined compression strength of
the rock beneath the base of the socket or
drilled shaft concrete, whichever is smaller.
Note that applies for horizontally stratified
discontinuities with Cs>305 mm and δ<5mm
Typical values of angle of friction of rocks
Rock type
Sandstone
Angle of friction
f(deg)
27-45
Limestone
30-40
Shale
10-20
Granite
40-50
Marble
25-30
Group pile
Pile cap
d
d
d
L
Bg
d
Lg
d
d
Group pile efficiency
 Determination of the load bearing capacity of group
piles is extremely complicated and has not yet been
fully resolved.When the piles are placed close to each
other,a reasonable assumption is that the stress
transmitted by the piles to the soil will overlap,thus
reducing the load bearing capacity of the
pile.Ideally,the piles in a group should be spaced so
that the load bearing capacity of the group should be
no less than the sum of the bearing capacity of the
individual piles.In practice,the minimum center to
center pile spacing d is 2.5D and in ordinary situations
is actually about 3D to 3.5D.
Efficiency factor
 Many structural engineers used a simplified
analysis to obtained the group efficiency for
friction piles (ratio between Qs & Qu is over
80%),particularly in sand.The piles may act in one
of two way:
1-as a block with dimension Lg*Bg*L
2-as individual piles
If the piles act as the block, the frictional capacity is
Qg(u)=favPgL note Pg=2(n1+ n2-2)d+4D
For each pile acting individually
Q(u)=favLP
Efficiency factor
h=
Qg (u )
 Q(u )
 Where h=group efficiency
Qg(u)=ultimate load bearing capacity of
group pile
Q(u)=ultimate load bearing capacity of
each pile
2(n1  n2  2)d  4 D
h=
Pn1n2
Converse Labarre equation
 (n1  1)n2  (n2  1)n1 
h = 1 

90n1n2


 (deg) = arctg( D / d )
Pile in sand
 Model test results on group piles in sand have shown
that group efficiency can be greater than 1 because
soil compaction zones are created around the piles
during driving.Based on the experimental observations
of the behavior of group piles in sand to date,two
general conclusions may be drawn:
1-for driven group piles in sand with d>3D, Qg(u)=SQ(u)
2-for bored group piles in sand at conventional
spacing
d=3D,Qg(u) may be taken 2/3 to 3/4 time SQ(u)
Pile in clay
 The ultimate load bearing capacity of group piles in clay
may be estimated with the following procedure:
1-Determine SQu=n1n2(QP+Qs) ;
SQu=n1n2[9CuAp+SaPCuL]
2-determine the ultimate capacity by assuming that the
piles in the group act as a block with dimension
Lg*Bg*L.The skin resistance of the block is:
Qs(g)=S2aCuLg+Bg)L
Calculate the point bearing capacity from
QP(g)=N*cCuLgBg , N*C=5.14(1+0.2Bg/Lg)(1+0.2L/Bg)<9
SQ(u)=Qs(g)+QP(g)
3-Compare the 2 results,The lower of the two value is
Qg(u)
Piles in rock
 For point bearing piles resting on
rock,most building codes specify that
Qg(u)=SQ(u),provided that the minimum
center to center spacing of pile is
D+300mm.For H-piles and piles with
square cross sections,the magnitude of D
is equal to the diagonal dimension of the
pile cross section
Settlement of piles and groups in
sands and Gravels
 The present Knowledge is not sufficient to
evaluate of pile and pile groups. For most
engineering structures, the loads to be applied
to a pile group will be governed by consideration
of consolidation settlement rather than by
bearing capacity of the groups divided by an
arbitrary factor of safety of 2 or 3. It has been
found from field observation that the settlement
of a pile groups is many times the settlement of
a single pile at the corresponding working load.
Settlement of piles and groups in
sands and Gravels
 The settlement of a group is affected by the
shape and size of group, length of pile, method
of installation of pile and possibly many other
factors.
 There are no equations that would
satisfactorily predict the settlement of pile in
SAND. It is better to rely on load tests for
piles in SAND.
 In this chapter we try to show some equations
for estimation the settlement of pile in SAND.
Settlement of pile shaft
Se1 =
(Q pall  Q all
f )L
Ap E p
 Where : L-pile length
EP-elastic modulus of pile
material,for concrete pile EP=21000MPa
z=0.5
AP-area of pile tip
Settlement of pile cause by load at
the pile point
Se2 = 0.85
q all
p B
E
(1  m )
2
 Where : B-Width of pile
E-elastic modulus of soil
m-Poisson ratio
Settlement of pile cause by the load
transmitted along the pile shaft
Q all
B
f
Se3 =
(1  m 2 ) I f
PL E
L
I f = 2  0.35
B
 Where : B-Width of pile
E-elastic modulus of soil
m-Poisson ratio
L-pile length
P-perimeter of the pile section
For group piles in sand and gravel, for elastic
settlement, Meyerhof (1976) suggested the empirical
relation
S g ( e ) (mm ) =
0.96 q Bg I
N 60
q = Qg/ (Lg.Bg) (in kN/m2)
I = influence factor = 1 – L/8.Bg > 0.5
Consolidation settlement of group piles
 The settlement of pile group in clay can be estimated
by assuming that the total load is carried by an
equivalent raft located at depth of 2L/3 where L is the
length of the piles.It may be assumed,that the load is
spread from the perimeter of the group at a slope of 1
horizontal to 4 vertical to allow for that part of the
load transferred to the soil by skin friction.The vertical
stress increment at any depth below the equivalent
raft may be estimated by assuming in turn that the
total load is spread to the underlying soil at slope of 1
horizontal to 2 vertical.The consolidation settlement
is than calculated as the shallow foundation.
Equivalent raft concept
Q
Q
q=
B' L'
1:4
d
2L/3
L
q
1:2
d
d
d
Bg
Lg
d
B’=D+d+L/3
B’&L’
L’=D+2d+L/3
Thank you for your attention
 Mr. Sieng
PEOU
 Master
science of
geotechnical
engineering