Types of piles and their structural characteristics

JABATAN KEJURUTERAAN INFRASTRUKTUR DAN GEOMATIK
FAKULTI KEJURUTERAAN AWAM DAN ALAM SEKITAR

3.0

PILE FOUNDATION

3.1

Types of piles and their structural characteristics

1. Steel piles, Figure 3.1
 Consist of pipe piles or rolled steel H-section piles
 The allowable structural capacity of steel piles :
Qall  As f s

Where :

As – cross-sectional area of steel
fs – allowable stress of steel

 Use of additional thickness and epoxy coating are used to
avoid corrosion, and typical condition of splicing (sambat) when
needed is shown in Figure 3.1.

Figure 3.1 Steel Piles
2. Concrete piles
 Two categories of concrete piles are (a) precast and (b) cast-insitu
 Precast piles, Figure 3.2:
-

prepared with ordinary reinforcement
1


-

in the shape of square or octagonal

Figure 3.2 Precast piles with ordinary reinforcement
 Cast-in-situ or cast-in-place, Figure 3.3 :
made by driving a steel casing with mandrel into the ground
-

-

upon reaching the desired depth, mandrel is pulled out and the
casing remain
with or without pedestal
uncased piles :
- casing is driven to the desired depth, and filled with
fresh concrete later gradually withdrawn
- with or without pedestal
allowable loads :
cased pile : Qall  As f s  Ac f c
uncased pile : Qall  Ac f c
where :
As – cross sectional area of steel
Ac - cross sectional area of concrete
fs – allowable stress of steel
fc - allowable stress of concrete

Figure 3.3 Cast in place concrete piles
2


3. Timber piles
 Three classifications are :
o
o
o

Class A : to carry heavy loads; min butt dia. = 14in
(356mm)
Class B : to carry medium loads; min butt dia. = 12-13in
(305-330mm)
Class C : used as temporary works but permanently for
submerged structure; min butt dia. = 12in (305mm)

 Splicing can be done by means of pipe sleeves or metal straps
or bolts, Figure 3.4
 The allowable load-carrying capacity :
Qall  Ap f w

Where :

Ap – average cross-sectional area of the pile
fw – allowable stress for the timber

Figure 3.4 Splicing of timber piles (a) use of pipe sleeves (b) use of
metal straps and bolts
4. Composite piles
 Upper and lower portions of composite piles are made of
different material
 They may in the form of : steel-cast-in-place concrete or
timber-concrete piles
3


5. Pile in term of their function support capacity, Figure 3.5:
(a) Bearing pile, (b) friction pile, (c) piles under uplift,
(d) piles under lateral loads, (e) batter piles under lateral loads



Figure 3.5
 Requirements and conditions for pile foundations, Figure 3.6 :


Figure 3.6 Conditions for use of pile foundations
4


- transmit load to the stronger underlying bedrock, 3.6(a)
- gradually transmitting the load to the surrounding soil by
means of frictional resistance at the soil-pile interface, 3.6(b)
- subjected to horizontal load while supporting the vertical load
transmitted by superstructure, 3.6(c)
- built extended into hard stratum under collapsible soil (loess) to
avoid the zone of moisture change that lead to swell and
shrink, 3.6(d)
- to resist uplifting forces for basement mats under water table,
3.6(e)
- to resist scouring at the bridge abutments and piers that can
lead to possible loss of bearing capacity of soil underneath,
3.6(f)
3.2

Estimating Pile Length, Figure 3.7

Figure 3.7 (a) and (b) Point Bearing Piles; and (c) Friction Piles
 Length of pile estimation depending upon the mode of load
transfer to the soil ; namely :
o

Point Bearing Piles

- the ultimate capacity of the piles depends entirely on the
bearing capacity of the hard stratum
- hence the length, L of the pile is fairly well established
5


- the ultimate pile load is then; Qu  Q p  Qs (Figure 3.7a)
where :
Qp – load carried at the pile point
Qs – load carried by skin friction developed at the side of the
pile
- piles can be extended into hard stratum with Qu  Q p (Figure
3.7b)
o

Friction Piles

- if no hard stratum presence, piles are driven through softer soil
to specified depths
- resistance to vertical loading, is provided mainly by the skin
friction; (in clayey soil is called adhesion)
- the ultimate load is given by : Qu  Qs
o

Compaction Piles

- piles are driven in granular soil to achieve proper compaction of
soil close to ground surface
- the length depends on :relative density before and after
compaction as well as required depth of compaction

6


3.3

Installation of Piles, Figure 3.8

Figure 3.8 Pile driving equipment
 Four method used in piles driving are ; drop hammer, single
acting air or steam hammer, double-acting and differential air
or steam hammer, and diesel hammer
- drop hammer, Figure 3.8a
o raised by a winch, and allowed to drop at a certain height
H
o slow rate of hammer blows
- single acting air or steam hammer, Figure 3.8b
o ram is raised by air or steam pressure and then drops by
gravity
- double-acting and differential air or steam hammer, Figure 3.8c
7


o ram is raised and pushed downward by air or steam
pressure
- diesel hammer, Figure 3.8d
o consist of ram, an anvil block and a fuel-injection system
o ram is raised, fuel is injected near the anvil, ram is
released, drops and compresses air-fuel mixture and
ignites it
o this causes; pile to be pushed downward and ram raised
 Vibratory pile driver, Figure 3.8e; consists of counter-rotating
weights that produces centrifugal force that cancel each other
but sinusoidal dynamic vertical force produced pushes the pile
downward
3.4

Pile Load Transfer Mechanism

 Frictional resistance, f(z) with depth is given by :
f z  

Q z 

pz 

Where :
Q z  - increase in pile load

Δz – increase in depth

P – perimeter of pile
 Nature of variation of pile load is as given by Figure 3.9 and
Woo and Juang(1970) has obtained actual variation of load
transfer by a bored concrete pile in Taiwan as in Figure 3.10

8


Figure 3.9 Load transfer
mechanism for piles

3.5

Figure 3.10 Load transfer
curves for a concrete
bored pile, Woo and
Juang (1975)

Equations for Estimating Pile Capacity

 Ultimate load-carrying capacity of pile, Qu is :
Qu  Q p  Qs

Where :
Qp – load-carrying capacity of the pile point
Qs – frictional resistance
 Point bearing capacity, Qp is :
*
*
Q p  Ap q p  Ap cN c  q' N q 

Where :
Ap – cross sectional area of pile tip
c – cohesion of the soil supporting the pile tip
qp - unit point cohesion
9


q’ =γ’L – effective vertical stress at the level of the pile tip
L- pile length
*
*
N c , N q - the bearing capacity factors
 Frictional resistance, Qs is :
Qs   pLf

Where :
p – perimeter of the pile section
ΔL – incremental pile length where, p and f is constant
f – unit friction resistance at any depth z
 There are many other methods for estimating Qp and Qs
3.6

Meyerhof’s Method – Estimation of Qp

 The value of unit point resistance qp remains constant beyond
the critical embedment ratio, (Lb/D)cr, Figure 3.11

Figure 3.11 Nature of variation of unit point resistance in a
homogeneous sand
 Figure 3.12 is the relationship of (Lb/D)cr and Ø(degree) where
at Ø = 45°, (Lb/D)cr = 25

10


 For piles in sand, c=0; but Qp should not exceed Apql,
*
*
Q p  Ap q p  Ap q' N q and Q p  Ap q' N q  Ap ql

 The limiting point resistance is :
SI unit : ql kN / m   50 N tan  ; or English
2

*
q

*
ql lb / ft 2   1000 N q tan 
*
ql kip / ft 2   N q tan 

Where : Ø – soil friction angle in the bearing stratum

Figure 3.12 Nature of variation of unit
point resistance in sand

Figure 3.13 Variation of the
maximum values of

*
N q with Ø

 Using SPT method (Meyerhof, 1976):
q p kN / m 2   40N L / D  400 N

where N - average SPT number at 10D above and 4D below
the pile point.
 For piles in clay, with saturated and undrained conditions
(Ø=0)
*
Q p  N c cu Ap  9cu Ap

Where : cu – undrained cohesion (undrained shear strength) of
the soil below the pile tip

11


3.7

Vesic’s Method – Estimation of Qp

 Vesic (1977) proposed value of Qp as :
*
'
*
Q p  Ap q p  Ap cN c   o N 

Where :
 1  2K o 
'
 o - mean normal ground effective stress = 
q '
 3 

Ko – earth pressure coefficient = 1 – sin Ø
*
N , N  - bearing capacity factors (see Table D.6 of Das textbook)
*
c

3.8

Janbu’s Method – Estimation of Qp

NOT to be covered

 Janbu (1976) proposed value of Qp as :
*
*
Q p  Ap cN c  q' N q 

Where :
*
*
N c , N q - bearing capacity factors, Figure 9.14

Figure 3.14 (a)Meyerhof’s and (b) Janbu’s bearing capacity factors

12


3.9

Coyle and Castello’s Method (Estimation of Qp in Sand) NOT TO
BE COVERED

 Coyle and Castello (1981) proposed value of Qp as :
*
Q p  q' N q Ap

Where :
q’ – effective vertical stress at the pile tip
*
N q - bearing capacity factor, Figure 3.15

13

*
Figure 3.15 Variation of N q with L/D, unit frictional resistance and K
value for piles in sand (Coyle and Castello, 1981)

3.10 Frictional Resistance, Qs in Sand
 Frictional resistance is, Qs   pLf
 Factors to be kept in mind while estimating unit frictional, f
-

the nature of pile installation
unit skin friction increases with depth
at similar depth, bored or jetted piles has a lower unit
skin friction compared to driven piles

 Approximation of f : (Figure 3.15)
For z = 0 to L’ :
For z = L’ to L :

f  K v' tan 

f  f z L'

Where :
K – effective earth coefficient
 v' - effective vertical stress at specified depth
 - soil-pile friction angle
L’ = 15d
 Read text for values of K, fav and Qs between 1976 and 1982
3.11 Frictional Resistance, Qs in Clay
 Three method of estimating Qs in Clay :
1.  Method :
- proposed by Vijayvergia and Focht (1972)
- assumption : displacement of soil caused by pile driving results
in a passive lateral pressure at any depth
- average unit skin resistance as :

14




'

f av    v  2cu



Where :
 v' - mean effective vertical stress for entire embedment length,


A1  A2  A3  ......
L

cu – mean undrained shear strength (Ø=0)
 - refer to Figure 3.16b
- total frictional resistance is : Qs  pLf av

Figure 3.16a Critical embedment ratio and bearing capacity factors
for various soil friction angles, (Meyerhof, 1976).

15


Figure 3.16b Variation of  with pile embedment length and its
application, (McCleland – 1974).
2.  Method :
- unit skin resistance in clayey soil is : f  cu
 - empirical adhesion factor, Figure 3.17

α

16


Figure 3.17 Variation of  with undrained cohesion of clay
- total frictional resistance is : Qs   fpL  cu pL
3.  Method :
- assumption : excess pore water pressure in normally
consolidated clay for driven pile shall dissipates gradually
- thus unit frictional resistance for the pile is :
f   v'

Where :
 v' - vertical effective stress = γ’z
  K tan  R

ØR – drained friction angle of remolded clay
K – earth pressure coefficient
Where : K  1 sin  R for normally consolidated clays
K  1 sin  R  OCR for overly consolidated clays
- total frictional resistance is : Qs   fpL
3.12 Point Bearing Capacity of Piles Resting on Rock
 Goodman (1980) has approximate the ultimate unit point
resistance in rock as :
q p  qu N  1

Where :
N  tan 2 45   / 2

qu – unconfined compression strength of rock
 - drained angle of friction

17


 After taking care of scale effect, qu ( design) 

qu (lab)
5

 Table 3.1 is the typical value of qu(lab) for rocks and Table 3.2
the value of angle of friction respectively
Table 3.1 Typical unconfined compressive strength of rocks
qu
lb/in2

MN/m2

10,000 – 20,000
15,000 – 30,000
5,000 – 10,000
20,000 – 30,000
8,500 – 10,000

Rock type
Sandstone
Limestone
Shale
Granite
Marble

70 – 140
105 – 210
35 – 70
140 – 210
60 – 70

Table 3.2 Typical Values of angle of friction, Ø, of rocks
Rock type
Sandtone
Limestone
Shale
Granite
Marble

Angle of friction, Ø
27 – 45
30 – 40
10 – 20
40 – 50
25 - 30

 Hence, with FS = 3, the allowable point bearing capacity, Qp is :
Q p ( all) 

q

u ( design)

N





 1 Ap

FS

Table 3.3 Typical pre-stressed concrete pile in use

18


Table 3.4 : Bearing capacity factors for deep foundations, N*c and N*σ,
Vesic’s, 1977.

19


Table 3.5 Janbu’s bearing capacity factors

20


Example 3.1
Given : A square 305 mm x 305 mm concrete pile and 12 m long.
Fully embedded in homogeneus sand layer, γd = 16.8 kN/m3 ,
c=0 and Øavg=35°. The average SPT value near pile tip is 16.
Find : a. Qp using Meyerhof’s, Vesic’s, Janbu’s and SPT method.
b. Qs using Qs   pLf and

f  K v' tan  ..( for ..z  0  L' )
f  f z  L ' ...( for ...z  L'  L)

if K=1.3

and   0.8 .
c. Estimate the load-carrying capacity of pile, Qall if FS=4.
d. Qall using Coyle and Costello’s method
Solution :
a. Meyerhof’s :
Because it is a homogeneous soil, Lb=L. For Ø=35°,
(Lb/D)cr =(L/D)cr ≈ 10 (Figure 3-16a). So for this pile, Lb/D = 39.34 >
*
(Lb/D)cr. Hence, from the same figure N q  120
*
Q p  Ap q p  Ap q' N q  0.0929201.6120  2247.4kN

*
ql kN / m 2   50 N q tan   50120tan 35  4201.25kN / m 2

*
Q p  Ap ql  0.09294201  390.3kN  Ap q' N q

Qp = 390 kN
*
Vesic’s : use I rr  90 ; with Ø=35°; N  79.5 so :

1  21  sin  
'
*
*
Q p  A p o N   A p 
 q' N 
3


1  21  sin 35
 0.0929
 201.679.5  923kN
3


*
Janbu’s : with c=0; use  '  90;..and ..  35;..N q  41.3 by int erpolation 
*
Q p  Ap q' N q  0.0929m 2 201.6kN / m 2 41.3  773.5kN

SPT method :

q p kN / m 2   40N L / D  400 N

Q p  Ap q p  0.0929m 2 401639.34  2339kN

21


Limiting value = Q p  Ap 400 N  0.0929m 2 40016  595kN
For design purpose : Q p 

595kN  773.5kN  390kN
 586kN
3

b. from sub-topic 3.10 from the note :
L'  15D  150.305m  4.58m

For z = 0 :  v'  0; f  K v' tan   0
For z = L’ to L :

 v'  L'  16.8kN / m 3 4.58m  76.94kN / m 2

f  K v' tan   1.376.94tan0.8  35  53.2kN / m 2

Thus
 f  f z  4.58m 
Qs   z  0
 pL' f z  20 ft pL  L'
2


2
 0  53.2kN / m 
4  0.305m 4.58m   53.2kN / m 2 4  0.305m 12  4.58m 
: 


2







 149  482  631kN

c. thus load carrying capacity of pile, Qu = Qp(avg) + Qs
Q p ( avg )  586kN and Qs  631kN;.....Qall 

Qult 586  631

 304.25kN
FS
4

d. Coyle and Castello’s
*
Qult  Q p  Qs  q' N q Ap  K v' tan0.8  pL;....and ...

L
12

 39.3
D 0.305

*
For Ø=35° and L/D=39.3; N q  40 K≈1.0
Thus :

*
Qult  Q p  Qs  q' N q A p  K v' tan0.8  pL



 



 201.6kN / m 2 40  0.0929m 2
 1.016.8  12  tan0.8  354  0.30512 
 749  1569  2318kN

And Qall 

Qult 2318

 579.6kN
FS
4

22


Example 3.2
Given : A driven pile in clay
as in Figure E9.2. The pipe
pile has outside diameter of
406mm and wall thickness
of 6.35mm.

Find :

a. Net point bearing capacity.
b. Skin resistance using α, λ and β method if ØR =30°; the
top 10m is normally consolidated clay and the bottom
clay layer has OCR=2.
c. Net allowable pile capacity, Qall if FS=4.

Solution :
a. Cross section of pile, Ap 



D2 



0.4062  0.1295m 2

4
4
*
Q p  Ap q p  Ap N c cu ( 2)  0.12959100  116.55kN

b. Skin resistance, Qs :
(α method) :

Qs   fpL  cu pL

From Figure α vs cu : cu(1)=30kN/m2 α=1.0; cu(2)=100 α=0.5
Thus :
Qs   fpL   cu pL   1cu (1)  0.40610   2 cu ( 2)  0.40620
 130 0.40610  0.5100 0.40620  1658.2kN

(λ method) : where f av    v'  2cu ( av) 
cu ( avg ) 

cu (1) 10  cu ( 2) 20
30



3010  10020
 76.7kN / m 2
30

Use the plotted Figure E9.2b, for σ’v vs depth;
'

v 

A1  A2  A3 225  552.38  4577

 178.48kN / m 2
L
30

23


From Figure λ vs L; λ=0.14 for L=30m; so

f av    v'  2cu ( av)   0.14178.48  276.7  46.46kN / m 2

Hence; Qs  pLf av   0.4063046.46  1777.8kN
(β method) : where ØR =30°; f   v' ;   K tan  R ; K  1 sin  R
K  1 sin  R  OCR

For z=0-5m :
 0  90 
2
f av(1)  1  sin  R  tan  R v' ( av)  1  sin 30tan 30
  13.0kN / m
2 


For z=5-10m :
 90  130.95 
2
f av( 2)  1  sin  R  tan  R v' ( av)  1  sin 30tan 30
  31.9kN / m
2



For z=10m-30m , OCR=2:
 130.95  326.75 
2
f av(3)  1  sin  R  tan  R OCR v' ( av)  (1  sin 30)tan 30 2 
  93.43kN / m
2



so





Qs  p f av(1) 5  f av( 2) 5  f av(3) 20   0.406135  31.95  93.4320  2669.7kN

c. So use α and λ method which produced almost similar results,
Qs 

1658.1  1777.8
 1718kN
2

Qult  Q p  Qs  116.46  1718  1834.46kN;....hence...Qall 

Qult 1834.46

 458.6kN
FS
4

Example 3.3
Given : An H-pile (size HP 310 x 1.226), length of embedment =
26m, driven through soft clay and rest on sandstone, qu(lab) for
sandstone = 76 MN/m2, Ø=28°, FS=5.
Find : The allowable point bearing capacity, Qp(all)
Solution : Since q p  qu N  1 ; N  tan 2 45   / 2 and qu ( design) 

qu (lab)
5



  
 qu ( lab)  
2

 tan  45    1  A p
5 
2
q p Ap
qu N   1A p





Q p ( all ) 

 
FS
FS
FS
 76  10 3 kN / m 2  

28 
2
3
2
  1  15.9  10 m

  tan  45 
5
2 





 
 182kN
5





24


EXAMPLE OF FINAL EXAMINATION QUESTION
Q4 The most common function of piles is to transfer a load that cannot be adequately
supported at shallow depths to a depth where adequate support becomes available.
Hence, the piles can also be categorized based on its function/ support capacity.
(a)

Briefly describe with relevant sketches the five (5) functions / support
capacity of piles.
(5 marks)

(b)

Reinforced concrete piles 18 m long, of square section and width 400 mm
are driven through 8 m of loose fill with unit weight of 13 kN/m3 to
penetrate 10 m into an underlying firm to stiff saturated clay. The
groundwater table is found at a depth of 2 m below ground surface.

(i)

Determine the ultimate bearing capacity, Qult, of pile by the given formula,
if the undrained shear strength of the clay increases linearly with depth
from 65 kN/m2 at the top of the clay to 100 kN/m2 at a depth of 10 m
below the surface of the clay.
Assuming that the unit weight of stiff saturated clay is 17 kN/m3
throughout the layer and the frictional capacity of the loose fill is
negligible.
(10 marks)

(ii)

Assuming that it is necessary to provide a number of such piles to carry
the total foundation load, explain the bearing capacity of the pile group is
estimated? Discuss your answer with the help of relevant sketches.
(5 marks)

25

ANSWER
Q4

The most common function of piles is to transfer a load that cannot be adequately
supported at shallow depths to a depth where adequate support becomes available.
Hence, the piles can also be categorized based on its function/ support capacity.
(a)

Briefly describe with a relevant sketch what are the five (5) function/
support capacity of piles.
(5 marks)

(a) Bearing pile, (b) friction pile, (c) piles under uplift,
(d) piles under lateral loads, (e) batter piles under lateral loads
(b)

A reinforced concrete piles 18 m long, of square section and width 400
mm is driven through 8 m of loose fill with unit weight of 13 kN/m3 to
penetrate 10 m into the underlying firm to stiff saturated clay. The
groundwater table is found at a depth of 2 m below ground surface.
(i)

Determine the ultimate bearing capacity, Qult of pile by the given
formula, if the undrained shear strength of the clay increases

26

linearly with depth from 65 kN/m2 at the top of the clay to 100
kN/m2 at a depth of 10 m below the surface of the clay.
Assuming that the unit weight of firm to stiff saturated clay is 17
kN/m3 throughout the layer and the frictional capacity of the loose
fill is negligible.
Given that:-





qtip = cu Nc (Based on Meyerhof’s equation); f s ( avg )    v '  2cu
(10 marks)
Answer:To determine Qp:qtip = cu Nc = 100 kN/m2 x 9 = 900 kN/m2

[1M]

Ap = 0.4 x 0.4 = 0.16 m2

[1M]

Qp = Apqtip = 0.16 x 900 = 144 kN

[0.5M]

To determine Qs:-

(45.14  117.04)(10)
2
v' 
 81.09kN / m 2
10
Elevation (m)
0
2
8
18

[1M]

Effective Vertical Pressure (kN/m2)
0
26
45.14
117.04
[1M]

(65  100)(10)
2
cu 
 82.5kN / m 2
10
Based on Figure 1,  = 0.185



[1M]
[1M]



f s ( avg )    v '  2cu
= (0.185)[81.09+2(82.5)]
= 45.53kN/m2
As = 4 x 0.4 x 10 = 16 m2

[1M]
[0.5 M]

27

Qs = As. fs = 16 x 45.53 = 728.48 kN

[1M]

Qult = Qs + Qp = 728.48 + 144
= 872.48 kN

[1M]

(ii)

Assuming that it is necessary to provide a number of such piles to
carry the total foundation load, how could the bearing capacity of
the pile group be estimated? Discuss your answer with a relevant
sketch.
(5 marks)

Answer:For most practical purposes, the ultimate load of pile group, (QvG)ult, can be estimated
based on the smaller value of the following two values:(a) Group Action – block failure (Figure A) of pile group by
breaking into the ground along an imaginary perimeter and bearing
at the base. The ultimate capacity for the group failure can be
estimated from the following relationship:(QvG)ult =  x n x (Qv)ult
[2M]
(b) Individual Action (Figure B) – if there is no group action
(when the center to center spacing, s, is large enough, >1), in that
case, the piles will behave as individual piles. The total load of the
group can be taken as n times the load of the single pile, in which
(QvG)ult = n x (Qv)ult =  (Qv)ult
[2M]

28


Figure : (A) Individual action, (B) Group action
[2 x 0.5M = 1M

3.13 Pile Load Test
 Pile load test arrangement by means of hydraulic jack is shown
in Figure 3.18a
 Step loads are applied to the pile, so that a small amount of
settlement is allowed to occur
 Settlement from field test is recorded as in Figure 3.18b
 Net settlement calculation for any load Q :
- When Q = Q1 : Net settlement, snet(1)  st (1)  se(1)
29


- When Q = Q2 : Net settlement, snet( 2)  st ( 2)  se( 2)
Where :
snet – net settlement
se – elastic settlement of the pile itself
st – total settlement
 The values of Q then plotted against se produces diagram in
Figure 3.18c

Figure 3.18 (a) Test arrangement (b) load vs total settlement
(c) load vs net settlement
3.14 Failure criteria of a pile
 The ultimate failure load for a pile is defined as the load when
the pile plunges or the settlements occur rapidly under
sustained load and the amount of settlement exceed the
acceptable soil-pile system
 Or
 Besides it, many engineers define the failure load at the point
of intersection of the initial tangent to the load-settlement
curve and the tangent to or the extension of the final portion of
the curve.

30


 Arbitrary settlement limits that the pile is considered to have
failed when the pile head has moved 10 percent of the pile end
diameter or the gross settlement of 1.5 in. (38 mm) and net
settlement of 0.75 in. (19 mm) occurs under two times the
design load. (JKR standard)

 However, all of these definitions for defining failure are
judgemental.
3.15 Pile Driving Formulas
 Due to varying soil profiles layers a point bearing pile cannot
always satisfied the capability of penetrating the dense soil to a
predetermined depth; therefore several equations have been
developed by many to calculate the ultimate capacity of pile
during driving.
 According to Engineering News Record (ENR), Qu is :
Qu 

WR h
S C

Where :
WR – weight of the ram
h – height of fall of the ram
S – penetration of pile per hammer blow (from last few
driving blows)
C – a constant
(for drop hammers : C = 1 in. ; S and h are in inches)
(for steam hammers : C = 0.1 in. ; S and h are in inches)
FS = 6
For single and double-acting hammers WRh is replaced by EHE
Thus :
Qu 

EH E
S C

Example 3.4
A precast concrete pile 12 in. x 12 in. in cross section is driven by a
hammer. Given :
Maximum rated hammer energy = 30 kip-ft
Hammer efficiency = 0.8
31


Weight of ram = 7.5 kip
Pile length = 80 ft
Coefficient of restitution = 0.4
Weight of pile cap = 550 lb
Ep = 3 x 106 kip/in2
Number of blows for last 1 in. of penetration = 8
Estimate the allowable pile capacity by the
a. Modified ENR formula (use FS=6)
b. Danish formula (use FS = 4)
c. Gates formula (use FS = 3)
Solution :
a.
Weight of pile + cap = 12  12  80150lb / ft 3   550  12.55kip
12
12
and WR h  30kip  ft
2
EW R h WR  n W p 0.830  12kip  in  7.5  0.42 12.55
Qu 

 607kip
1
S  C WR  W p
7.5  12.55
8  0.1

Qall 

b.

Qu 607

 101kip
FS
6
EH E
Qu 
EH E L
S
2 Ap E p

Use Ep = 3 x 106 lb/in2
And

Qu 
Qall

c.

EH E L

2 Ap E p

0.830  1280  12
 3  10 6

212  12
kip / in 2 
 1000




 0.566in.

0.830  12  417kip

 0.566
417

 104kip
4
1
8

Qu  a EH E b  log S   27 0.8301  log1   252kip
8
252
Qall 
 84kip
3

3.16 Hiley’s Formula for estimating single RC pile capacity.
32


The Hiley’s formula gives the simplest method of calculating the final
setting or the ultimate load of a pile while driving depending upon
the given parameter.
It is usually written as :
BW H h WH  Pe 2
C
s 

2 FS  WL
WH  P

And
C  Cc  C p  C q

where :
s
C
WH
h
P
P1
P2
WL
FS
e
Cc

- Set value /1 blow (mm/blow)
- Temporary compression of pile & soil (mm)
- Weight of hammer (kN)
- Drop of hammer (mm)
- Total load (P1 + P2) (kN)
- Weight of pile (kN)
- Weight of driving assembly (kN)
- Pile working load (kN)
- Factor of safety
- Coefficient of restitution
- Temporary compression coefficient due to pile
head and cap (mm), Table 3.3
Cp, - Temporary compression coefficient due to pile
length (mm), Table 3.3
Cq, - Temporary compression coefficient due to ground
or quake (mm), Table 3.3

Note :
(a)

(b)
(c)
(d)

This formula was developed by Hiley (1925). The formula
assumes the energy of the falling hammer during pile
driving is proportional resisted by the pile. This method is
widely considered to be one of the better formulas that
intended to be applied to cohesionless, well-drained soils
or rock.
Weight of the hammer shall be about 0.5 to 2.0 times of
the total pile weight.
The term mass and weight are interchangeably
The term Cp and Cq are shown in Figure 3.19 after a pile
set measurement of pile are made.

33


Figure 3.19 : Example graph of pile set
Table 3.6 : Values of Cc, Cp and Cq
Form of
compression

Pile head
and cap, Cc

Pile length,
Cp

Quake, Cq

Material

Head of timber
pile
Short dolly in
helmet or
driving cap
3 in/76.2mm
packing under
helmet or
driving cap
1 in/25.4mm
pad only on
head of
reinforced
concrete pile
Timber pile
(E=1,500,000
lb/in2) or
(E=10,342,500
kPa)
Pre-cast pile
(E=2,000,000
lb/in2) or
(E=13,790,000
kPa)
Steel pile for
cast in place
(E=30,000,000
lb/in2) or
(E=206,850,000
kPa)
Ground
surrounding pile
and under pile

Easy
driving
(inch)

Medium
driving
(inch)

Hard
driving
(inch)

Very hard
driving
(inch)

0.05

0.10

0.15

0.20

0.05

0.10

0.15

0.20

0.07

0.15

0.22

0.30

0.03

0.05

0.07

0.10

0.004L

0.008L

0.012L

0.016L

0.003L

0.006L

0.009L

0.012

0.003L

0.006L

0.009L

0.012

0.05

0.10 –
0.20

0.15 –
0.25

0.05 –
0.15

34

point
Note :
Length, L measure in feet
1 feet = 0.3048 m
1 inch = 25.4 mm
Table 3.7 : Coefficient of restitution, e.
Description

Coefficient of restitution, e

Piles driven with double acting hammer
-

Steel piles without driving cap
Reinforced concrete pile without helmet but with
packing on top of pile
Reinforced concrete piles with short dolly in helmet
and packing
Timber pile

0.5
0.5
0.4
0.4

Piles driven with single acting and drop hammer
-

Reinforced concrete piles without helmet but with
packing on top of piles
Steel piles or steel tube of cast in place piles fitted
with driving cap and short dolly covered by steel plate
Reinforced concrete piles with helmet and packing,
dolly in good condition
Timber pile in good condition
Timber pile in poor condition

0.4
0.32

0.25
0.25
0.00

Example 3.5
Using Hiley’s formula calculate the final set of a 200mm X 200mm RC pile.
The pile driven with single acting and drop hammer with medium driving. The
type of pile is the reinforced concrete pile with helmet and packing, dolly in
good condition.
Other data and parameters are :
Pile working load,
Mass of hammer,
Factor of safety, FS
Pile length, L
Mass driving assembly,
Drop of hammer,
Hammer efficiency,

= 275 kN
= 25 kN
= 2.0
= 18 m
= 2.0 kN
= 400 mm
= 85%
35

Density of concrete,

= 24 kN/m3

Solution :
Mass of pile, P1

= Concrete density X Area X Length of pile
= 24 X (0.2 X 0.2) X 18
= 17.28 kN

Total load, P

= P1 + P2
= 17.28 + 2.0
= 19.28 kN

Value of e

= 0.25 (Table 3.7)

BW H h
0.8525400  15.454mm

FS  WL
2.0  275
WH  Pe 2 25  19.28  0.252

 0.592
WH  P
25  19.28

Value of C
Cc
Cp
Cq

:
= 0.15in X 25.4 = 3.81 mm
= 0.006(59ft) = 0.354in X 25.4 = 8.99 mm
= 0.10in X 25.4 = 2.54 mm

C = Cc + Cp + Cq = 3.81 + 8.99 + 2.54 = 15.34 mm
BW H h WH  Pe 2
C
s 

2 FS  WL
WH  P
15.34
 15.454  0.592
2
 s  1.48mm / blow
 or S  14.8mm / 10blow ( Final Set )

Using  s 

Example 3.6
Given :

36


A 200mm x 200mm RC square pile. The pile driven with single-acting
and drop hammer with hard driving. The type of pile is reinforced
concrete pile with helmet and packing, dolly in good condition.
Mass of hammer, Wn
Factor of safety, FS
Pile length, l
Mass Driving assembly,P2
Drop hammer, h
Hammer efficiency, B
Set value, S

=25kN
=2.0
=24m
=3.0 kN
=500mm
=85%
=19mm/10 blow (Figure 3.20)

Figure 3.20
Required : Ultimate load of pile
Solution :
Mass of pile, P1 = Concrete densityxAreaxlength
= 24x(0.2x0.2)x24=23.04kN
Total load, P2
Value of e
Cp + Cq
Cc
Temporary
compression, C
Set value, s

=
=
=
=

P1 + P2 = 23.04 + 3=26.04kN
0.25 (Table 3.7)
20mm (Figure 3.20)
0.22inx25.4=5.59mm

= 5.59 + 20 = 25.59mm
= 19mm/10 blow =1.9mm/blow

37


BW H h



0.8525500

1
 1



FS  s  C c  C p  C q  2.01.9  25.59
2
 2



2
2
WH  Pe
25  26.040.25

 0.5217
WH  P
25  26.04

 361.52kN

By using Hiley’s equation :
WL 

WH  Pe 2
 361.52  0.5217  188.6kN
 1
 WH  P
FS  s  C p  C q  C c 
 2

BW H h



Therefore, the pile working load must be less than 188.6kN
3.17 Settlement of Piles, Vesics (1969)
 Settlement of a pile under vertical working load, Qw is :
s  s1  s 2  s3

Where :
s – total pile settlement
s1 – elastic settlement of pile
s2 – settlement caused by the load at the pile tip
s3 – settlement caused by the load transmitted along pile shaft
 Formulae :
- elastic settlement, s1 :
s1 

Q

wp

 Qws L
Ap E p

Where :
Qwp – load carried at the pile point under working condition
Qws – load carried by frictional resistance under work load
Ap – area of pile cross section
L – length of pile
Ep – modulus of elasticity of the pile material
 - nature of unit skin friction (=0.5 or 0.67), Figure 3.21

38


Figure 3.21 Various types of unit friction resistance along pile shaft
- load at pile point, s2 :
s2 

q wp D
Es

1   I
2
s

wp

Where :
D – width or diameter of pile
qwp – point load per unit area = Qwp/Ap
Es – modulus of elasticity of soil at or below the pile point
μs – Poisson’s ratio of soil
Iwp – influence factor = 0.85
Or
s2 

Qwp C p
Dq p

Where :
qp – ultimate point resistance of the pile
Cp – an empirical coefficient, Table 3.8
Table 3.8 Typical Values of Cp
Soil type
Sand (dense to loose)
Clay (stiff to soft)
Silt (dense to loose)

Driven Pile
0.02-0.04
0.02-0.03
0.03-0.05

Bored Pile
0.09-0.18
0.03-0.06
0.09-0.12

- load carried by pile shaft, s3 :
Q  D
2
s3   ws 
 pL  E 1   s I ws

 s





Where :
39


p – perimeter of the pile
L – embedded length of pile
Iws – influence factor = 2  0.35
Or
s3 

L
D

and Cs (a constant) = 0.93  0.16 L / D C p

Qws C s
Lq p

Cp from Table 3.8
Example 3.7
Given : A pre-stressed concrete pile 21m long, being driven into
sand. Working load, Qw = 502 kN. The pile is octagonal in shape with
D = 356 mm, see Figure E9.4. Skin resistance, Qs carries 350 kN, and
Qp carries the rest. Use Ep = 21 x 106 kN/m2, Es = 25 x 103 kN/m2, μs
= 0.35 and ξ = 0.62.
Find : The settlement of the pile.
Solution :
From Table D3; for D=356mm, Ap=1045cm2, p=1.168mm and
Qws=350 kN; so Qwp=502-350=152 kN
Due to material :
s1 

Q

 Qws L

wp

Ap E p



152  0.6235021  0.00353m  3.35mm



0.1045m 2 21  10 6



q wp 

Qwp
Ap

Due to point load :
s2 

q wp D
Es

1   I
2
s

wp





 152  0.356 

1  0.35 2 0.85  0.0155m  15.5mm

3 
 0.1045  25  10 

Due to skin :
With I ws  2  0.35

L
21
 2  0.35
 4.69
D
0.356

40

Q  D
 350  0.356 
2
2
s3   ws 
 pL  E 1   s I ws  1.16821 25  10 3  1  0.35 4.69
And




 s
 0.00084mm  0.84mm









Therefore the total settlement is :

s  s1  s2  s3  3.35  15.5  0.84  19.69mm
3.18 Pullout Resistance of Piles
 The gross ultimate resistance of a pile subjected to uplifting
force, Figure 3.22 is :
Tug  Tun  W

Where :
Tug – gross uplift capacity
Tun – net uplift capacity
W – effective weight of pile

Figure 3.22 Uplift capacity of piles
a.

In Clay


Das and Seeley (1982), estimated Tun as :
41


Tun  Lp ' cu

Where :
L – length of the pile
p – perimeter of pile section
 ’- adhesion coefficient at soil-pile interface
cu – undrained cohesion of clay
Values of  ' :


-

for cast-in-situ: (for bore pile)
2
 '  0.9  0.00625cu for cu ≤ 80 kN/m
2
 ' 0.4 for cu > 80kN/m

-

for pipe piles :

 '  0.715  0.0191cu for cu ≤ 27 kN/m
 ' 0.2 for cu > 27 kN/m

2

2

b. In Sand
Das and Seeley (1975), estimated Tun as :


Tun  

L

0

 f u p dz with fu varies by

f u  K u v' tan  for (z≤Lcr) such

as in Figure 3.23a

-

Steps in finding Tun in dry soil;

find relative density and use Fig 3.23c to find Lcr
if L ≤ Lcr then :
Tun 

1
pL2 K u tan  with values Ku and  from Figure
2

3.23b&c
-

if L > Lcr then :
42

Tun  1 pL2 K u tan   pLcr K u tan  L  Lcr  with values Ku and
cr
2

 from Figure 3.23b&c

Where :
Ku – uplift coefficient
 v' - effective vertical stress at a depth z
 - soil-pile friction


Thus with FS=2 to 3, allowable uplift capacity Tu(all)
is :

Tu ( all) 

Tug
FS

Figure 3.23 (a) Variation of fu (b) Ku (c) Variation of  /Ø, (L/D)cr with
relative density of sand Dr
Example 3.8

43


Given : A 50 ft long concrete pile embedded in a saturated clay
cu=850 lb/ft2. 12 in x 12 in. in cross section. Use FS=4.
Find : Allowable pullout capacity, Tun(all)
Solution : with cu =850 lb/ft2 ≈ 40.73 kN/m2
 '  0.9  0.00625cu   0.9  0.0062540.73  0.645
504  10.645850
 109.7kip
1000
109.7 109.7


 27.4kip
FS
4

Tun  Lp ' cu 

And Tun( all)

Example 3.9
Given : A precast concrete pile, with cross section = 350mm x
350mm. Length of pile as 15m. Assume : γsand=15.8 kN/m3,
Øsand=35°, Dr=70%.
Find : Pullout capacity if FS=4.
Solution : From Figure 3.23; for Ø=35° and Dr=70%
L
   14.5;..Lcr  14.50.35m   5.08m
 D  cr


 1;..  135  35;.......K u  2


Hence : for L (15m) > Lcr (5.08m)
Tun  1 pL2 K u tan   pLcr K u tan  L  Lcr 
cr
2


1
2

0.35  415.85.082 2 tan 35  0.35  415.85.08215  5.08 tan 35

 1961kN

Tu ( all) 

Tug
FS



1961
 490kN
4

44


3.19 Group piles efficiency
 Converse –Labarre method of estimating pile-group efficiency
developed by Jumikis, 1971 using the following equation :
Eg  1  

n  1m  m  1n
90mn

Where :
Eg – pile-group efficiency
θ – tan-1(d/s), (deg)
n – number of piles in row
m – number of rows of piles
d – diameter of piles
s – spacing of piles, center to center, same unit as pile
diameter.
Example 3.10
Given :
A pile group consists of 12 friction piles in cohesive soil, Figure 3.24.
Each pies diameter is 300mm and center-to-center spacing is 1 m. By
means of a load test, the ultimate load of a single pile was found to
be 450 kN. Take SF as 2.0.
Required :
Design capacity of the pile group, using the Converse-Labarre
equation.
1
 3

  tan 1    18.4 ; E g  1  18.4

4  13  3  14  0.710
9034

Allowable bearing capacity of a single pile=450kN/2=225kN
Design capacity of the pile group = 0.710(12)(225kN)=1917kN.

45


Figure 3.24
 Another method of estimating efficiency of pile group as quoted
by Das (2007) as follows :
 A pile cap is normally constructed over group piles; either in
contact or well above the ground, Figure 3.25 a&b.
 In practice, minimum center-to-center pile spacing, d = 2.5D,
or 3-3.5D as in ordinary situations; where D - diameter of piles
 Thus, the efficiency of a group pile, η is :


Qg (u )

Q

u

Where :

Qg(u) – ultimate load-bearing capacity of the group pile
Qu – ultimate load-bearing capacity of each pile without group
effect
 If group as a block thus :


2n1  n2  2d  4d
pn1 n2

46


Figure 3.25 Pile groups

3.26

timate Capacity of Group
Piles in Saturated Clay

 For most practical purposes, the ultimate load of pile group,
(QvG)ult, can be estimated based on the smaller value of the
following two values, Figure 3.27 (a) and (b):(a) Group Action – block failure (Figure A) of pile group by
breaking into the ground along an imaginary perimeter and
bearing at the base. The ultimate capacity for the group failure
can be estimated from the following relationship:(QvG)ult =  x n x (Qv)ult
(b) Individual Action (Figure B) – if there is no group action
(when the center to center spacing, s, is large enough, >1), in
that case, the piles will behave as individual piles. The total load of
the group can be taken as n times the load of the single pile, in
which
(QvG)ult = n x (Qv)ult =  (Qv)ult

47


Figure 3.27 : (A) Individual action, (B) Group action
 Feld’s Method : in estimating group capacity of friction piles,
Qg(u)

Figure 3.28 Feld’s Method
Table 3.9 Arrangement of Feld’s Method
Pile
type
A
B
C

No. of Piles
1
4
4

No. of adjacent
piles
8
5
3

Reduction factor
for each pile
1-8/16 #
1-5/16
1-3/16

Note: 16 # no. of arrow

Ultimate capacity
Col.2 x Col.4
0.5Qu
2.75Qu
3.25Qu
Σ6.5Qu=Qg(u)

48


 Therefore, efficiency,  

Qg (u )

Q



u

6.5Qu
 72%
9Qu

3.20 Ultimate Capacity of Group Piles in Saturated Clay
 Figure 3.29 shows a group of pile in saturated clay, steps to
find the ultimate load-bearing capacity Qg(u) are :
Find Qu in pile group :
 As individual
From :
 Qu  n1n2 Q p  Qs ; Q p  Ap 9cu ( p)  and Qs  pcu L
So

Q

:

u



 n1n2 9 Ap cu ( p )  pcu L



(1)

 As pile group (dimensions of LgxBgxL):

p

c L   2Lg  Bg cu L

g u

 Point bearing capacity as : Ap q p  Ap cu ( p ) N c*  Lg Bg cu ( p ) N c*
 With N c* from Figure 3.29, thus :

Q

u

*
 Lg Bg cu ( p ) N c   2Lg  Bg cu L

(2)

Where :
D
D
Lg  n1  1d  2  and Bg  n2  1d  2 
2
2

 The lower value from (1) and (2) is Qg(u)

49


Figure 3.29 Ultimate group piles in clay

Figure 3.30 Variation of N c* with Lg/Bg and L/Bg

50


Example 3.11
Given : The section of 3 x 4 group pile in a layered saturated clay
is shown in Figure 3.31. The piles are square in cross section
(350mm x 350mm). The center-to-center spacing d, of the piles is
1220mm.
Required : The allowable load-bearing capacity of the pile group.
Use FS=4.

5m

40 kN/m2

10 m

70 kN/m2

1.22 m

Figure 3.31 Group pile in clay soil
If pile act as single pile:

Q

u



 n1 n2 9 Ap cu ( p )   pcu L





 n1 n2 9 Ap cu ( p )   1 pcu (1) L1   2 pcu ( 2) L2

2



2

With cu(1)=40 kN/m ;α1=0.86 and cu(2)=70 kN/m ;α2=0.63 thus;
Ap=0.350x0.350=0.093m2, p=4x0.350=1.22m

Q

u



n1 n2 9 A p cu ( p )   1 pcu (1) L1   2 pcu ( 2) L2



 3  490.09370  0.861.22405  0.631.227010
 3  458.6  209.8  538.02  9677.04kN

If pile as a group :
D
Lg  n1  1d  2   4  11.22  0.305  3.965m
2
D
Bg  n2  1d  2   3  11.22  0.305  2.745m
2

51

Lg
Bg



3.965
L
15
 1.44;.....

 5.46
2.745
Bg 2.745

From Figure 3.29: N c*  8.6
(assuming : that at the end of curve at right hand stays horizontal)

Thus :

Q

u

*
 Lg Bg cu ( p ) N c   2Lg  Bg cu L

 3.9652.745708.6  23.965  2.745405  7010
 6552  13.42200  700  18630kN

Hence, ΣQu=9677 kN,

Q

all



9677 9677

 2419kN
FS
4

3.21 Consolidation Settlement of group pile in clay by mean of 2:1
distribution method.
o
o
o
o
o
o

L=depth of pile embedment
Qg – total load of superstructure (–) weight of soil
excavated
Assume load Qg transmitted at depth of 2L/3
from top of pile.
The load Qg spread out at 2 : 1 horizontal line
from this depth
Line a-a’ and bb’ are two 2:1 lines
Stress increased at the middle of each soil layer :

pi 

B

Qg

g

 z i Lg  z i 

o
o
o
o

Lg and Bg – the length and width of pile group
zi – distance from z=0 to the middle of clay layer
For layer 2 : zi=L1/2; For layer 3 : zi=L1+L2/2
For layer 4 : zi=L1+L2+L3/2

o

Consolidation Settlement,

o

Where :

e  C c log

 e(i ) 
si  
Hi
1  e(i ) 



p0  p
;
p0

Layer 2 : Hi=L1; Layer 3 : Hi=L2; Layer 4 : Hi=L3

o

Total consolidation settlement,

s g   si

Figure 3.32
52


Example 3.12
A group of pile in clay is shown in Figure 3.33. Determine the
consolidation settlement of the pile groups. All clays are normally
consolidated.

Figure 3.33 Pile group in clay soil
Solution :
p(1) 
p( 2) 

p(3) 

B

Qg

B

g

g



2000
 14.52kN / m 2
2.2  93.3  9

Qg

B

2000
 51.6kN / m 2
2.2  3.53.3  3.5

Qg

g





2000
 9.2kN / m 2
2.2  123.3  12

 z i Lg  z i 
 z i Lg  z i 

 z i Lg  z i 

With s1 

Cc (1) H 1
1  e0(1)

 p0(1)  p(1) 
log 
 and ;
p0(1)





53


p0(1)  216.2  12.518  9.81  134.8kN / m 2
 p0(1)  p(1)  0.37   134.8  51.6 
log 
log
  0.1624m  162.4mm

1  e0(1)
p0(1)

 1  0.82  134.8 


 216.2  1618  9.81  218.9  9.81  181.62kN / m 2

s1 

p0 ( 2 )

s 2 

Cc (1) H 1

Cc ( 2) H 2
1  e0 ( 2 )

 p0( 2)  p( 2)  0.24  181.62  14.52 
log 
log
  0.0157m  15.7mm

p0( 2)
181.62



 1  0.7



p0(3)  181.62  218.9  9.81  119  9.81  208.99kN / m 2
s3 

C c ( 3) H 2
1  e0 ( 3 )

 p0(3)  p(3)  0.252  208.99  9.2 
log 
log
  0.0054m  5.4mm

p 0 ( 3)
 208.99 

 1  0.75



Therefore the total settlement :
Δsg = 162.4 + 15.7 + 5.4 = 183.5mm
3.22 Elastic settlement of pile group.
 Vesic (1969) developed the simplest relation of :
Elastic settlement of group pile, s g ( e) 

Bg
D

s

Bg – width of pile group
D – width or diameter of each pile in the group
s = s1 + s2 + s3 – total elastic settlement at working load
 Meyerhof (1976) developed elastic settlement of pile group in
sand and gravel.
Elastic settlement of group pile, s g ( e ) in  

2q B g I
N corr.

 Where :
q=Qg/(LgBg) in ton/ft2
Lg and Bg – length and width of pile group section (ft)
Ncor – average of SPT no. at Bg below pile tip (within seat of settlement)
Influence factor, I=1-L/8Bg ≥ 0.5
L – length of pile embedment

54


Example 3.13 (Cumulative)
A reinforced concrete piles 18m long, of square section (diameter)
and width 300 mm is driven through 6 m of loose fill with unit weight
of 15 kN/m3 to penetrate 12 m into the underlying firm to stiff
saturated clay. The groundwater table is found at a depth of 3 m
below ground surface.
(i)

Determine the ultimate bearing capacity, Qult of pile by
the given formula, if the undrained shear strength of the
clay increases linearly with depth from 80 kN/m2 at the
top of the clay to 120 kN/m2 at a depth of 12 m below
the surface of the clay.
Assuming that the unit weight of firm to stiff saturated
clay is 18 kN/m3 throughout the layer and the frictional
capacity of the loose fill is negligible.
Given that:qtip = cu



Nc

f s ( avg )    v '  2cu



(Based

on

Meyerhof’s

equation);

(ii)

Evaluate Qa if using total FS=2.5

(iii)

Evaluate Qa if using FS = 2 for skin and FS = 3 for tip.
3m
3m
12m

55


To determine Qp:qtip = cu Nc = 120 kN/m2 x 9 = 1080 kN/m2
Ap = 0.3 x 0.3 = 0.09 m2
Qp = Apqtip = 0.09 x 1080 = 97.2 kN
To determine Qs:Depth(m) Effective Vertical Pressure (kN/m2)
0
0
3
3x15=45
6
45 + 3(15-9.81) = 60.57
18
60.57 + 12(18-9.81) = 158.85
(60.57  158.85)(12)
2
v' 
 109.71kN / m 2
12
(80  120)(12)
2
cu 
 100kN / m 2
12

Based on Figure 1,  = 0.185 for L=18m



f s ( avg )    v '  2cu



= (0.185)[109.71+2(100)]
= 57.3 kN/m2
As = 4 x 0.3 x 12 = 14.4 m2
Qs = As. fs = 14.4 x 57.3 = 825.12 kN
Qult = Qs + Qp = 825.12 + 97.2 = 922.32 kN
(ii)
(iii)

Qa = 922.32/2.5 = 368.9kN
Qa = 825.12/2 + 97.2/3 = 444.96kN

56


3.23 Calculation of single, group pile capacity and settlement from
Prakash & Sharma - for sandy soil.
 Table used for values of Nq and Ø, Table 3.10.

Ø
Nq(driven)
Nq(drilled)

20
8
4

25
12
5

28
20
8

Table
30 32
25 35
12 17

3.10
34 36
45 60
22 30

38
80
40

40 42 45
120 160 230
60 80 115

 Table used for values of Ks for various pile types in sand, Table
3.11
Table 3.11
Pile type
Ks
Bored pile
0.5
Driven H pile
0.5 – 1.0
Driven displacement pile
1.0 – 2.0
 For most design purpose δ=2/3Ø (Meyerhof, 1976)
Example 3.14
A closed-ended 12-in (300mm) diameter steel pipe is driven into
sand to a 30ft (9m) depth. The water is at ground surface and
sand has Ǿ=36° and unit weight (γsat) is 125 lb/ft3 (19.8kN/m3).
Estimate the pipe pile’s allowable load.
Solution :
For circular pile : Ap 

 1 ft 2
4

 0.785 ft 2 , p   1  3.14 ft

Nq=60, Table 3.10; Ks=1.0, Table 3.11;     36  24
Using the formula of the ultimate capacity :

Qv ult  Q p  Q f

2
3

2
3

L L

'
 Ap v' N q  pK s tan    vl L
L 0

Where :
57

LL


L 0

'
vl

L   sub 20 B / 220 B    sat 20 B L  20 B 

This is with the assumption of : σ’vl increases with depth up to
20B. Below this depth, σ’vl remains constant.
With γsub or γ’ = 125 – 62.5 =62.5 lb/ft3, B=1ft, L=30ft.
LL


L 0

'
vl

L   sub 20 B / 220 B    sub 20 B L  20 B 

Then :  62.5  10  120  1  62.5  20  130  20  1lb
 12,500  12,500  25kips111.25kN 

Thus :


LL

'
L 0

 0.785 sub 20 B 60  3.141tan 2425  58.88  34.95  93.83kips

Therefore with FS=3:
(Qv)all=(Qv)ult/FS=93.83/3=31kips (137.95kN)
Example 3.15
For the pile described in example 3.14, estimate the pile
settlement. The pile has ¾ in. wall thickness and is closed at the
bottom.
Solution :
B=12 in. (outside diameter);
L=30x12=360 in.
(Qv)all=31,000 lb (from Example 3.14)


122  113in 2

Area of base



Pipe inside diameter
Area of steel section

  12 2  10.52 / 4  144  0.184 ft 2  26.496in 2

4
 12  (2  3 / 4)  10.5in.





1. Semiempirical method :

58


From the relation of : (from Example 3.14)


LL

'
L 0

 0.785 sub 20 B 60  3.141tan 2425  58.88  34.95  93.83kips

And (Qv)all=(Qv)ult/FS=93.83/3=31kips
Assuming allowable loads are actual loads; then
Q pa  Q p all  58.83 / 3  19.6kips;....
Q fa  Q f



all

 31  19.6  11.4......or....(34.95 / 3)kips...with..some..roundoff ..error..

Due to material :
Ss 

Q

pa

  s Q fa L

19.6  0.5  11.41000  360 

26.496  30  10 6

Ap E p

25.3  36  10 4
 0.011in
26.496  3  10 7

Vesic (1977) recommends αs = 0.5 for uniform or parabolic
skin friction distribution along pile shaft.
Ep = 30x106 psi for steel
Ep = 21 x 106 kN/m2 for concrete
Due to point :
Sp 

C p Q pa
Bq p



0.03  19.6  113
 0.094in
12  58.88

Cp=0.03 (Table 9.3); qp=Qp/Ap=58.88/113
Cs  0.93  0.16
Due to skin :
S ps 

C s Q fa
Df qp



Df
B

.C p  0.93  0.16

360
0.03  0.054
12

0.05411.4113  0.0033in
36058.88

Using St=Ss+Sp+Sps=0.011+0.094+0.0033=0.108in(2.7mm)
2. Empirical method :
Using :

St 

Q L
B
12
31  360  1000
 va 

 0.12  0.014
100 Ap E p 100 26.496  30  10 6

 0.134in.(3.35mm)

59


Example 3.16
Using data of example 3.14, find the allowable bearing capacity
based on standard penetration data as given in Figure 3.34.

Figure 3.34

Solution :
(b) Average N value near pile tip, Navg(tip)=(10+12+14)/3=12
(c) Point bearing, Qp
 v'  125  62.5lb / ft 3 30 ft   1875lb / ft 2  0.938ton / ft 2 (tsf )
1 ton = 2000 lb
Correction for depth of N values,
C N  0.77 log10 20 / 0.938  1.02
Therefore ; N  C N N  1.02  12  12
And 0.4 N D f Ap / B  0.4  12  30  0.785 / 1  113tons

4 N Ap  4  12  0.785  37.7tons

For driven piles :

The lower of these values is Qp=37.7 tons

Q p  0.4 N / B D f Ap  4 N Ap

(d) Shaft friction, Qf
Average N value along pile shaft,
Navg(shaft)= (4+6+6+8+10)/5=6.8
Use σ’v for average depth of L/2=30/2=15ft so
σ’v= 0.938/2=0.469tsf
C N  0.77 log10 20 / 0.469  1.25 Therefore ;





(Meyerhof,1976)

Q f   f s  p D f ;.. f s*  N / 50  1tsf

(Meyerhof,1976)

60


N  C N N  1.25  6.8  8.5 ; f s  N / 50  8.5 / 50  0.17tsf ( 1tsf )
So Q f  f s  p  L  0.17    1 30  16tons
(e) Allowable bearing capacity, Qall :
Qv ult  Q p  Q f  37.7  16  53.7tons

Qv all  Qv ult / FS  53.7 / 3  17.9tons  35.8say...36kips..(156kN)

 Pile group sample calculations
 Settlement of pile group and check on design :
1.

Vesic’s Method (1977) : S G  S t b / B 

2.

Meyerhof’s Method (1976) (if SPT N values available) :
S

2 pI b
N

where :
p

QG all

 Df 
  0.5
.....and ......I  1 

bb
8b 



61


Example 3.16
Using data from Example 3.14, calculate the pile group bearing
capacity if the piles are placed 4ft center to center and joined at
the top by a square pile cap supported by nine piles. Estimate pile
group settlement.

Figure 3.35

Solution :
(a) bearing capacity
B=1ft; s=4ft; b  4  4  1  9 ft , ; b=10ft; n=9
Qv ult  93.83kips for a single pile (from empirical method Ex 3.15)

QvG ult  nQv ult  9  93.83kips  844.47kips

QvG all  9  93.83  281kips(1250kN ),...with...FS  3.0
3

(b) settlement
B=1ft; b  4  4  1  9 ft , (square arrangement); n=9 piles;
(Qg)all=281kips; zone of influence, b =9ft below the group base;
Navg=(12+14+14)/3≈13; for single pile st=0.134in.(EX.3.14)
1. Vesic’s (1977): S G  S t b / B   0.134 9 / 1  0.40in
2. Meyerhof’s (1976): (N values)

62


p

QG all

bb
 Df
I  1 

8b




281
 3.47kips / ft 2  1.74tons / ft 2
99

 
30 
  1 

  8  9   0.58  0.5

 

where Df is pile length = 30 ft

So :
2 pI b 23.47 0.58 9

 0.93in
13
N
2q Bg I 23.47 
2q Bg I
sg ( e ) in  

sg ( e ) in  
N corr.
N corr.
S

Example 3.17
Given :
A 236-kip(1050kN) of vessel (water tank) is to be
supported on a pile foundation in an area where soil investigations
indicated soil profile Fig 3.36.
Required : Design a pile foundation so that the maximum allowable
settlement for the group does not exceed allowable settlement,
Sa=0.6in (15mm).

Figure 3.36 Soil profile and soil properties used : N-SPT value;

63


 v'  effective ...vertical..stress;...  36.. for ..sand ;.. clay  110lb / ft 3 ;..
 sand  125lb / ft 3 ;.... ' sand  125  62.5  62.5lb / ft 3

Solution :
1.
2.

-

Soil profile as in Figure 3.36
Pile dimensions and allowable bearing capacity
top 4 ft consist of top soil and soft clay – this layer has no
contribution to the side frictional resistance.
Increasing in N values except at 24ft – due to gravel –
neglected
Try 34ft(10.3m) long with 30ft(9.1m) penetration into sand and
12-in(305mm) diameter steel-driven frictional pile
This pile has 0.75in thickness and is closed at the bottom
Static analysis by utilizing soil strength :
LL

'
Qv ult  Ap v' N q  pK s tan   vl L and

Ap   / 41  0.785 ft 2
2

L 0

Nq=60 for Ǿ=36° from Table 9.5; perimeter, p=πB=3.14ft
Ks=1.0 from Table 9.6; δ=2/3Ǿ=2/3(36°)=24°
Thus :



lb 
 0.785 ft 2 1690 2 60 

ft 







 440  1690lb / ft 2

3.14 ft 1tan 24
 20 ft  1690  10
2


 79.6  43.7  123.3kip
Q 
Qv all  v ult  123.3  41.1kips( say41kips)or (182.5kN )
FS
3

- Empirical analysis by utilizing standard penetration test (SPT) :
Point bearing, Qp:
Navg near pile tip = (8+12+14+14)/4=12
σ’v near pile tip = 440+(125-62.5)30=2315lb/ft2=1.15t/ft2
Correction for depth of N values, C N  0.77 log10 20 / 1.15  1.0
Therefore ; N  C N N  1.0  12  12
64


And Q p  0.4 N / B D f Ap  4 N Ap
Q p  0.4 N D f Ap / B  0.4  12  30  0.785 / 1  113tons > than
4 N Ap  4  12  0.785  37.7tons

Therefore use Qp=38 tons = 76kips
Shaft friction, Qf:
Navg along shaft = (4+6+6+8+12)/5=7.2 say 7
And f s  N / 50  1tsf
f s  N / 50  7 / 50  0.14tsf  1tsf ; Q f  f s pL  0.14  3.14  30  13.2ton
Therefore :

Qv ult  Q p  Q f  (38  13.2)tons  102.4kips
Q 
Qv all  v ult  102.4  34kips..(151.3kN )
FS

3

Q 
Q 

p all

f

all

 2kip 
 38ton
 / 3  25.3kips
 1ton 
 13.22 / 3  8.8kips

will ..be..used ..in.. predicting ..settlement

3.

Number of piles and their arrangements
The number of piles required to support
236kip vessel load :
n

Qva
236

 6.9
Qv all 34

Try a group of 9 piles (Figure 3.37);
Piles at 4ft center-to-center
A 10ft x 10ft pile cap is required
Assume pile cap = 3ft thick
Pile cap width, b = 10ft
Outer periphery, b  b  1  10  1  9 ft
(see Figure 9.34)
Figure 3.37

65


Pile cap weight
Total weight
Load per pile
Pile group capa.
4.

=
=
=
=

(3 x 10 x 10)ft3 x 0.15kip/ft3 = 45 kips
236 + 45 = 281 kips
281/9 = 31kips<34kips
OK
34 x 9 = 306kips>281kips
OK

Settlement of single pile

Semiempirical Method
St=Ss+Sp+Sps
Where :
Ss 

Q

pa

  s Q fa L

A E 
p

Ss 

Q

pa

A E 

Sp 

C p Q pa

B  q 

p

Q 
Q 

p actual
f

p

  s Q fa L
p

and

actual

 25.331 / 34  23kips  Q pa
 8.831 / 34  8kips  Q fa

Ep=30 x 106psi; αs=  =0.5



23  0.5  830  12  1000  0.012in ; A =26.5 in2
p
  2
2
6
  12  10.5 30  10
4








and Cp=0.03; Qpa=23kips; B=12in; Ap=113.09 in2

p

qp =Qp/Ap=76/113.09=0.672kip/in2;
Sp 

C p Q pa



0.03  23kips

Ap   / 412  113.09in 2
2

B  q  12in  0.672kip / in   0.086in
2

p

S ps 

C s Q fa
Df qp

and Qfa=8kips; Df=30x12in; qp=0.672kips/in2


Df 

30  12 
C s  0.93  0.16
 C p   0.93  0.16
0.03  0.054
B 
12 




C s Q fa
0.054  8kips
2
S ps 

 0.0018in ; Ap=113.09 in
2
D f q p 30  12in  0.672kip / in





Therefore :
St=Ss+Sp+Sps=0.012in+0.086in+0.0018in=0.0998in
Say 0.1in (2.5mm)

66


Empirical Method
St 

Q L
B
12in 31kips  360in  1000lb / kip
 va 

100 A p E p  100
26.5in 2  30  10 6 lb / in 2





 0.12  0.014  0.134in..(3.35mm)

From the two results consider the larger : settlement for a
single pile St=0.134 in.
5.

Settlement of pile groups in cohesionless soils
With B=1ft; b  9 ft ; n=9 piles; within zone of influence of 9 ft;
Navg=(12+14+14)/3≈13; group load, Qg=281kips;
Total settlement of single pile; St=0.134 in;
By Vesic’s : S G  S t b / B  0.134in 9 ft / 1 ft   0.402..say..0.4in(10mm)
By Meyerhof’s (SPT) method :

I

QG

281
 3.47kips / ft 2  1.74tons / ft 2
9  9
bb
30 

   D /  b   1 
1
8 
 0.58
8 9



Where : p 







f

I 
 0.58 
SG  2 p b    2  1.74  9 
  0.5in (13mm)
N
 13 

The larger is SG=0.5in(13mm) < allowable settlement, Sa=0.6in
Therefore OK..
3.24 Distribution of load in pile groups
The load on any particular pile within a group may be computed by
using the elastic equation :
Qm 

Myx
Mxy
Q


2
n  x
 y2

 

 

Where :
Qm – axial load on any pile m
Q
– total vertical load acting at the centroid of the pile
group
n
- number of piles
Mx, My - moment with respect to x and y axis respectively
67


x, y

- distance from pile to y and x axes respectively

Example 3.18
Given : A pile cap consists of 9 pile as in Figure 3.38. A column load
of 2250 kN acts vertically on point A.
Required : Load on pile 1,6 and 8.

Figure 3.38

Solution :
Myx
Mxy
Q


2
n  x
 y2

Qm 

 

 

Q=2250kN; n=9

 x   61m
 y   61m
2

2

 6m 2

2

2

 6m 2

M x  2250kN 0.4  900kN.m
M y  2250kN 0.25  562.5kN.m

Load on pile no. 1:
Q1 

2250 562.5kN.m 1m 900kN.m 1m


 306.25kN
9
6m 2
6m 2

68


Load on pile no. 6:
Q6 

2250 562.5kN.m 1m 900kN.m0


 343.75kN
9
6m 2
6m 2

Load on pile no. 8:
Q6 

2250 562.5kN.m0 900kN.m 1m


 100kN
9
6m 2
6m 2

Figure 3.39

69


Example 3.19
Given : A pile cap with five piles. The pile cap is subjected to a 900
kN vertical load and a moment with respect to the y axis of 190
kN.m, Figure 3.39.
Required : Shear and bending moment on section a-a due to the pile
reacting under the pile cap.
Solution :
Q=1000kN; n=5; M y  190kN.m ; M x  0kN.m ;

 x   41m

2

2

Q2  Q4 

 4m 2

1000 190kN.m1m 0kN.m y


 247.5kN
5
4m 2
 y2

Shear at a-a
Moment at a-a

 

: (247.5kN)(2) = 495kN
: (2)(247.5kN)(1m-0.3m) = 173 kN.m

(Draw free body diagram of the pile cap and take summation of
shear and moment at section a-a)
Example 3.20
Given :
A pile group consists of four friction piles in cohesive soil, Figure
3.40. Each pile’s diameter is 300 mm and center-to-center spacing is
0.75m.
Required :
(a)
(b)
(c)

Block capacity of the pile group. Use safety factor of 3.
Allowable group capacity based on individual pile failure.
Use a factor of safety of 2, along with the ConverseLabarre equation for the pile-group efficiency.
Design capacity of the pile group.

70


Solution :
(a)

Figure 3.40
Block capacity: Since c-to-c spacing = 0.75 and < 0.90m;
Coyle and Sulaiman, 1970 suggested :

Qg  2DW  L f  1.3  c  N c  W  L

D=10.5m
W=0.75+0.15+0.15=1.05m
L=0.75+0.15+0.15=1.05m
f=αc
qu=200 kN/m2; c=200/2=100kN/m2; α=0.56 (Figure 3.17)
f=0.56x100=56kN/m2
71


Nc=5.14 (from Table 2.3 for shallow rectangular footing for
Ø=0˚- Vesic, 1973)
Qg  2 DW  L  f  1.3  c  N c  W  L





 210.51.05  1.0556  1.3 100kN / m 2 5.141.051.05
 2469.6  736.7  3206kN

Allowable block capacity 
(b)

3206
 1069kN
3

based on individual pile

Qult  Qs  Qtip

Qs  f Asurface   56    0.3m  10.5m   56  9.9  554kN
 0.32 
Qtip  cN c Atip   100kN / m 9
  64kN
4 

618
Thus Qult  554  64  618kN; Qall 
 309kN
2
*

2

With :
n=2, m=2, θ=tan-1(1/2.5)=21.8˚
Eg  1  

n  1m  m  1n  1  21.8 2  12  2  12  0.758
90mn
9022

Qall for group (based on individual pile) :
Qg ( all)  309kN40.758  937kN

(c)

Design capacity of group is the smaller of two = 937kN
(even using FS=2)

72


3.25 Conventional rigid method
Example 3.21
The allowable bearing capacity of vertical pile ( length 12 m and 30 cm in diameter )
against vertical load = 120 kN, against horizontal load = 30 kN dan 65 kN against pull
out load, Figure 3.41.
That pile group will retain vertical load V = 1500 kN, horizontal load H = 300 kN and
momen = 150 kNm at the centroid of the pile group. Design the proper pile lay out to
retain those of external load. For stability control, use this formula (conventional rigid
method):

Sn 

V [ M  Ve x ]e x [ M  Ve y ]e y


2
2
n
 ex
 ey

Answer
Number of piles = 1500 / 120 ~ 12 ; 300 / 30 ~ 10 ; 150 / 65 ~ 3
Efficiency take 0.7, so number of pile = 12/0.7 = 16 piles
2 d = 2 x 0.3 = 0.6 m ( minimum length for pile to edge of pile cap )
take 0.6 m
3 d = 3 x 0.3 = 0.9 m ( minimum length for centre to centre of pile )
take 1.0 m
Answer
4.2 m
Try this lay out :

a

yy
c

b

d

1
ey
xx

4.2 m

2
3
4

a

b
ex

c

d

Figure 3.41

73

Check stability to obtain how much the external load imposed to each piles, then each
piles should be compared to allowable bearing capacity.
ex1a = ex2a =ex3a =ex4a = ex1d = ex2d = ex3d = ex4d =1.5 m
ex1a2 = 2.25
ex1b = ex2b =ex3b =ex4b =ex1c =ex2c =ex3c =ex4c = 0.5 m
ex1b2 = 0.25
 ex2 = 8x2.25 + 8 x 0.25 = 18 + 2 = 20
ey1a = ey1b =ey1c =ey1d = ey4a = ey4b = ey4c = ey4d =1.5 m
ey1a2 = 2.25
ey2a = ey2b =ey2c =ey2d =ey3a =ey3b =ey3c =ey3d = 0.5 m
ey2a2 = 0.25



ey2 = 20

Mx only and V positioned at the centroid, formula is simplified to
Sn 

V [ M ]e x

n  ex 2

Q1a = 1500 / 16  150 x 1.5 / 20 = 93.75 – 11.25 = 82.5 kN < 120 OK
Q1d = 93.75 + 11.25 = 105 kN < 120 OK
Check all the piles !

74


3.26 DESIGN AND ANALYSIS OF PILE UNDER LATERAL
STATIC LOADS (CASE FROM PRAKASH AND SHARMA)
BRINCH HANSEN’S METHOD
The ultimate soil reaction at any depth is given by equation (6.3),
Pxu   vx K q  cK c

For cohesionless soil, equation becomes:
Pxu   vx K q

Where;
 vx is the effective vertical overburden pressure at depth x and
coefficient K q and Kc is determined from Figure 3.42.
The procedure for calculating ultimate lateral resistance consists of
the following steps:
1. Divide the soil profile into a number of layers.
2. Determine σvx and Kq and Kc for each layer and then calculate Pxu for
each layer and plot it with depth.
3. Assume a point of rotation at depth xr below ground and take the
moment about the point of application of lateral load Qu (Figure 6.2).
4. If this moment is small or near zero, then xr is the right value. If
not, repeat steps (1) through (3) until the moment is near zero.
5. Once xr (the depth of the point of rotation) is known, take moment
about the point (center) of rotation and calculate Qu.
This method is illustrated in Example 3.22.

75


Figure 3.42 Coefficients Kq and Kc (Brinch Hansen, 1961)

EXAMPLE 3.22
A 20 - ft (6.0m) long 20 - in. (500mm) - diameter concrete pile is
instated into sand that has Ø’ = 30' and γ = 120 lb/ft3 (I920kg/m3).
The modulus of elasticity of concrete is 5 x 105 kips/ft2 (24 x 106
kN/m2). The pile is 15 ft (4.5 m) into the ground and 5 ft (1.5 m)
above ground. The water table is near ground surface. Calculate the
ultimate and the allowable lateral resistance by Brinch Hansen’s
method.

76


0.6

SOLUTION
(a) Divide the soil profile in five equal layers, 3 ft long each (Figure
6.8).
(b) Determine σvx

σvx = γ’x = (120 – 62.5) x = 0.0575 x kips/ft2
1000

Where x is measured downwards from the ground level.
For each of the five soil layers, calculations for σvx and pxu are carried
out as

77


shown in Table 6.1. pxu is plotted with depth in Figure 6.8. The values
for pxu at the middle of each layer are shown by a solid dot.
(c) Assume the point of rotation at 9.0 ft below ground level and
take moment about the point of application of lateral load,
Qu. Each layer is 3 ft thick, which

Gives:
∑ M = 0.6 x 3 x 6.5 + 2 x 3 x 9.5 + 3.8 x 3 x 12.5 – 5.9 x 3 x 15.5 8 x 3 x 18.5
= 11.7 + 57 + 142.50 - 274.35 - 444 = 211.2 - 71 8.35
= - 507.2 kip-ft/ft width
Where :
(0.6 - from center point) x (3 – thickness of each layer) x
(6.5 – distance from center to Qu)
(d) This is not near zero; therefore, carry out a second trial by
assuming a point of rotation at 12ft below ground. Then, using the
above numbers,

78


∑ M = 11.7 + 57 + 142.50 + 274.35 - 444 = 41.6 kip ft/ft
The remainder is now a small number and is closer to zero.
Therefore, the point of rotation xr can be taken at 12 ft below
ground.
(e) Take the moment about the center of rotation to determine Qu:

Qu(5 + 12) = 0.6 x 3 x 10.5 + 2 x 3 x 7.5 + 3.8 x 3 x 4.5 + 5.9 x 3 x
1.5 – 8 x 3 x 1.5
= 18.9 + 45 + 51.3 + 26.55-36 = 105.8
Qult = 105.8/17 = 6.2 kips/ft width
Qult = 6.2 x B = 6.2 x 1.67 =10.4 kips
(where B = 20 in. = 1.67 ft)
Qall = 10.4/2.5 = 4.2 kips using a factor of safety 2.5

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Types of piles and their structural characteristics

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