Course
Year
Version
: S0484/Foundation Engineering
: 2007
: 1/0
Session 19 – 20
PILE FOUNDATIONS
PILE FOUNDATIONS
Topic:
• Settlement of Piles
• Laterally Loaded Piles
• Pull Out Resistance of Piles
• Pile Driving Formula
• Negative Skin Friction
SETTLEMENT OF PILES
S = S1 + S2 + S3
Where:
S = total pile settlement
S1 = elastic settlement of pile
S2 = settlement of pile caused by the load at
the pile tip
S3 = settlement of pile caused by the load
transmitted along the pile shaft
SETTLEMENT OF PILES
S1

Q

wp
  .Qws L
A p .E p
Where:
Qwp = load carried at the pile point under working load condition
Qws = load carried by frictional (skin) resistance under working load condition
Ap = area of pile cross section
Ep = modulus of elasticity of the pile material
L = length of pile
 = the magnitude which depend on the nature of unit friction (skin) resistance
distribution along the pile shaft.
SETTLEMENT OF PILES
S2 
qwp .D
Es
1   .I
2
s
wp
Where:
qwp = point load per unit area at the pile point = Qwp/Ap
D = width or diameter of pile
Es = modulus of elasticity of soil at or below the pile point
s = poisson’s ratio of soil
Iwp = influence factor
= r
SETTLEMENT OF PILES
 Qws  D
2

S3  
1   s .I ws
 pL  Es

Where:
Qws = friction resistance of pile
L = embedment length of pile
p = perimeter of the pile
Iws = influence factor
I ws
L
 2  0.35
D

EXAMPLE
The allowable working load on a prestressed
concrete pile 21 m long that has been driven into
sand is 502 kN. The pile data are as follow:
- Diameter (D) = 356 mm
- The area of cross section (Ap) = 1045 cm2
- Perimeter (p) = 1.168 m
Skin resistance carries 350 kN of the allowable load,
and point bearing carries the rest. Use Ep = 21 x 106
kN/m2, Es = 25,000 kN/m2, s = 0.35 and  = 0.62)
Determine the settlement of the pile.
EXAMPLE
S1

Q

wp
S2 
qwp .D
Es
  .Qws L
A p .E p
1   .I
2
s

152  0.6235021  0.00353m  3.35mm
0.104521x106 
 152  0.356 

 1  0.352 0.85  0.0155m  15.5mm

 0.1045  25,000 

wp
I ws  2  0.35

L
21
 2  0.35
 4.69
D
0.356
Q  D
 350  0.356 
S3   ws 
1   s2 .I ws  

 1  0.352 4.69  0.00084m  0.84mm

 1.16821 25,000 
 pL  Es




S = S1 + S2 + S3 = 3.35 + 15.5 + 0.84 = 19.69 mm
LATERALLY LOADED PILE
LATERALLY LOADED PILE
ELASTIC SOLUTION – EMBEDDED IN GRANULAR SOIL
LATERALLY LOADED PILE
LATERALLY LOADED PILE
For L/T  5
LATERALLY LOADED PILE
LATERALLY LOADED PILE
LATERALLY LOADED PILE
ELASTIC SOLUTION – EMBEDDED IN COHESIVE SOIL
LATERALLY LOADED PILE
LATERALLY LOADED PILE
ULTIMATE LOAD ANALYSIS – MEYERHOF – PILES IN SAND
ULTIMATE LOAD RESISTANCE (Qu(g))
LATERALLY LOADED PILE
MAXIMUM MOMENT, Mmax DUE TO THE LATERAL LOAD Qu(g)
For long (flexible) piles
in sand
MAXIMUM MOMENT, Mmax DUE
TO THE LATERAL LOAD Qg
LATERALLY LOADED PILE
ULTIMATE LOAD ANALYSIS – MEYERHOF – PILES IN CLAY
ULTIMATE LOAD RESISTANCE (Qu(g))
LATERALLY LOADED PILE
MAXIMUM MOMENT, Mmax DUE TO THE LATERAL LOAD Qu(g)
For long (flexible) piles
MAXIMUM MOMENT, Mmax DUE TO
THE LATERAL LOAD Qg
PULL OUT RESISTANCE OF PILES
PULL OUT RESISTANCE OF PILES
PULL OUT RESISTANCE OF PILES
EXAMPLE:
A concrete pile 50 long is embedded in a saturated clay with
cu = 850 lb/ft2. The pile is 12 in. x 12 in. in cross section. Use
FS = 4 and determine the allowable pullout capacity of the
pile
Solution
Given cu = 850 lb/ft2  40.73 kN/m2
’ = 0.9 – 0.00625cu = 0.9 – (0.00625)(40.73) = 0.645
(50)( 4 x1)(0.645)(850)
Tun  L. p. '.cu 
 109.7 kip
1000
Tun 109.7
Tun( all) 

 27.4 kip
FS
4
PULL OUT RESISTANCE OF PILES
PULL OUT RESISTANCE OF PILES
For dry soils, the equation simplifies to
1
Tun  . p. .L2cr .K u . tan   p. .Lcr .K u .L  Lcr . tan 
2
Determine the value of Ku and  from figure 9.36b
and 9.36c.
Tun( all)
Tun

FS
Where Tun(all) = allowable uplift capacity and FS is Factor of Safety (a
value of 2 – 3 is recommended)
PULL OUT RESISTANCE OF PILES
EXAMPLE:
a precast concrete pile with a cross section 350 mm x 350 mm
is embedded in sand. The length of pile is 15 m. Assume that
sand = 15.8 kN/m3, sand = 35o, and the relative density of sand =
70%. Estimate the allowable pullout capacity of the pile (FS = 4)
Solution
From figure 9.36, for  = 35o and relative density = 70%
L
   14.5 ; Lcr  (14.5)(0.35m)  5.08m
 D  cr

 1 ;   135  35o

Ku  2
1
Tun  . p. .L2cr .K u . tan   p. .Lcr .K u .L  Lcr . tan 
2
Tun  1961 kN
Tun( all) 
Tun 1961

 490 kN
FS
4
PILE DRIVING FORMULA
NEGATIVE SKIN FRICTION
Can occur under condition such as:
- If a fill of clay soil is placed over a granular soil layer into which a
pile is driven, the fill will gradually consolidate. This consolidation
process will exert a downward drag force on the pile during a
period of consolidation
- 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
NEGATIVE SKIN FRICTION
CLAY FILL OVER GRANULAR SOIL
NEGATIVE SKIN FRICTION
GRANULAR SOIL FILL OVER CLAY
THE UNIT NEGATIVE SKIN FRICTION
AT ANY DEPTH FROM z = 0 TO z = L1
NEGATIVE SKIN FRICTION