qub : Unconfined compressive strength of the socket zone = qu (lab) 5 Ab : Area of pile base.
Qs = α βqus As Where
α : Reduction factor relating to qus. β : Correction factor related to the discontinuity spacing in the rock mass. qus : Unconfined compressive strength along the shaft area. As : Area of pile shaft.
Page No.
Friction Capacity of Pile Depth Range No.
From
To m
Length RQD qucs m
%
MPa
α
β
As
Qs
m2
kN
1
-12.77
-18.05
5.28
50
0.96
0.50
0.65
16.59
5164
2
-18.05
-21.75
3.70
50
1.51
0.40
0.65
11.62
4537
3
-21.75
-23.35
1.60
50
1.78
0.35
0.65
5.03
2050
4
-23.35
-26.15
2.80
50
3.45
0.22
0.65
8.80
4276
5
-26.15
-32.25
6.10
50
4.11
0.21
0.65
19.16
10811
6
-32.25
-32.50
0.25
50
3.85
0.21
0.65
0.79
418
3
*
Qs
Ultimate Skin Friction Capacity
=
27255 kN
Page No.
End Bearing Capacity of Pile
φ
=
27
Nφ
=
2.66
qub
=
0.77
MPa
Ab
=
0.79
m2
* Ultimate End Bearing Capacity
Qb
=
3217
kN
* Ultimate Total Capacity
Qu
=
30472
kN
=
3.05
* Actual Factor of Safety
o
Thus, the pile can safely carry the applied load
4
`
Page No.
5
3.0 Elastic Analysis In granular soil-rock environment where the soil modulus is assumed to increase linearly with depth, stiffness factor (T) is given by: -
EI nh Where
T =5
E : Modulus of elasticity of concrete E = K o + 0.2 f c u , where Ko = 20 4 I : Moment of inertia of pile = πD 64 D : diameter of pile. nh : Modulus of subgrade reaction
* Stiffness Factor * Net Pile Length
E
=
32.00
GPa
I
=
0.05
m4
nh
=
45
T
=
2.04
m
=
29.50
m
H1
=
500.00 kN
H2
=
133.33 kN
H
=
633.33 kN
MN/m3
Shear Forces
* Due to 5% of Axial Load * Due to Out-of-Verticality * Total Shear Force
=
N 75
* The following evaluation is based on the elastic analysis of laterally-loaded fixed pile head (pile cap) and linearly increasing soil modulus (Reese & Matlock in Tomlinson 1994).
* Bending Moment M f = Fm HT Where
Fm : Moment coefficient
Page No.
Z
Moment
(x)
Fm
m
x T
0.00
0.00
-0.92
-1185.77
1.00
0.49
-0.46
-590.44
2.00
0.98
-0.08
-107.11
3.00
1.47
0.17
215.34
4.00
1.97
0.25
328.00
5.00
2.46
0.24
311.55
6.00
2.95
0.18
228.44
7.00
3.44
0.11
139.77
8.00
3.93
0.05
62.22
9.00
4.42
0.01
8.00
10.00
10.00
4.91
-0.01
-10.67
12.00
11.00
5.41
0.00
0.00
Mf kNm
-1500 0.00
-1000
-500
0
2.00
Depth (m)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Depth
6
4.00 6.00 8.00
Moment (kNm)
N.B. (x) is the depth from the cut-off level
500
17 18 19 20 21 22 23 24 25 26 27 28 Page No.
7
4.0 Longitudinal Reinforcement
* * * *
M1
=
1185.77 kNm
[From the elastic analysis]
Out-of-Position Moment
M2
=
750.00 kNm
= 0.075 N
Ultimate Moment
Mult
=
2903.66 kNm
[Factored to 1.50]
Ultimate Axial Load
Nult
=
15000
kN
Pile Diameter
h
=
1000
mm
d'
=
fcu
=
60
N/mm2
fy
=
460
N/mm2
1] From PROKON
=
1.45
%
------ (1)
2] 0.30% Ac
=
0.30
%
------ (2)
[BD 74/00, C9.2]
3] Minimum of 6 bars
=
0.61
%
------ (3)
[BD 74/00, C9.2]
4] 0.15N/fy
=
0.62
%
------ (4)
[BD 74/00, C9.2]
=
1.45
%
[Factored to 1.50]
101.00 mm
* Required Steel Percentage
Top 12.00 m
* Maximum of (1), (2), (3) & (4)
ρreq.
* * * *
Provided Steel Diameter
=
32
Required No. of Steel Bars
=
15
Provided No. of Steel Bars
=
15
Spacing Between Bars c/c
=
167
mm
mm
[Maximum 300 mm]
OK Rest of Pile
* * * *
ρreq.
Maximum of (2), (3) & (4)
=
0.62
Provided Steel Diameter
=
32
Required No. of Steel Bars
=
7
Provided No. of Steel Bars
=
7
% mm
OK
Page No.
8
5.0 Stress in Concrete
* * * *
Pile Diameter
D
=
1000
mm
Compressive Strength of Concrete
fcu
=
60
Working Axial Load
N
=
10000
Area of Concrete
Ac
=
785398 mm2
N/mm2 kN
* Permissible service stress should not exceed 25 % of the specified cube strength as per BS8004: 7.4.4.3 * Actual Stress in Concrete, σ = N
A
σ1
=
15.00
N/mm2
σ2
=
12.73
N/mm2
OK
6.0 Ultimate Vertical Load
* Proposed Vertical Reinforcement
:
15 Nos. Φ 32 mm
As per BS8110, Part 1 - 1997, The ultimate axial load should not exceed the value of "N" given by:
N =0.3 5f c uAc 0. 7 As c f y * * * *
Yield strength of steel.
fy
=
460
Area of steel
Asc
=
12064
Compressive strength of concrete.
fcu
=
60
Area of concrete
Ac
=
773334 mm2
N
=
20125
kN
Nall
=
13416
kN
Napp
=
10000
kN
* Allowable Axial Load * Applied Axial Load
N/mm2 mm2 N/mm2
[Divided by 1.50]
OK
Page No.
9
7.0 Settlement Calculations The settlement of piles under a vertical working load is calculated as follows: -
s = s1 + s2 + s3 s : Total settlement. s1 : Settlement of pile shaft due to elastic shortening. s2 : Settlement of pile caused by pile point load. s3 : Settlement of pile caused by pile shaft-transmitted load.
s1 =
(Qwp + ξQws ) L
where
Ap E p Qwp
: Load carried at the pile point under working load condition.
ξ
: Unit skin resistance distribution along pile shaft.
Qws
: Load carried by frictional (skin) resistance under working load condition.
L
: Length of pile.
Ap
: Area of pile cross-section.
Ep
* * * *
: Modulus of elasticity of pile material.
Ultimate End Bearing Capacity
Qb
=
3217
kN
Ultimate Skin Friction Capacity
Qs
=
27255
kN
Factor of Safety of End Bearing
=
3.05
Factor of Safety of Skin Friction
=
3.05
Qwp
=
1056
kN
Qws
=
8944
kN
ξ
=
0.67
L
=
29.50
m
Ap
=
0.79
m2
Ep
=
32.00
GPa
s1
=
8.27
mm
Page No.
s2 =
qwp D Es
Where
(1 − µ s2 ) I wp
Qwp
qwp
: Point load per unit area at the pile point =
D
: Diameter of pile.
Es
: Modulus of elasticity of soil at or below pile point.
µs
: Poisson's ratio of soil.
Iwp
: Influence factor = αr for circular foundations. qwp
=
1344
kN/m2
D
=
1000
mm
Es
=
1.00
GPa
µs
=
0.30
Ap
10
Q D s3 = ( ws ) (1 − µ 2 ) I ws pL Es where
Iwp
=
0.85
s2
=
1.04
p
: Perimeter of pile.
Iws
: Influence factor =
Total Settlement
2 + 0.35
mm
L D
p
=
3142
mm
Iws
=
3.90
s3
=
0.34
mm
s
=
9.66
mm
Page No.
8.0 Shear Calculations on Piles Assumptions and Considerations
* * * * * * *
Concrete Compressive Strength
fcu
=
60
N/mm2
Yield Strength of Stirrups
fyv
=
460
N/mm2
Applied Shear Force
=
633.33 kN
Applied Normal Force
=
10000
kN
Factored Shear Force
V
=
950
kN
[Factored to 1.50]
Factored Normal Force
N
=
15000
kN
[Factored to 1.50]
Pile Diameter
dp
=
1000
mm
11
Ac * Cross-sectional Area of Pile * Main Reinforcement : No. of Bars Diameter of Bars
* Area of Tension Reinforcement * Concrete Cover
As
=
785398 mm2
=
15
=
32
mm
=
6032
mm2
=
75
mm
[Half the Total Reinforcement].
Shear Stress Calculations
* * * * *
Width of Pile
b
=
1000
mm
Centroid of Tension Zone
c
=
239
mm
Effective Depth
d
=
739
mm
Shear Stress
v
=
1.29
N/mm2
Maximum Shear Stress
vmax
=
4.75
N/mm2
[BD 74/00, C5.1]
Shear Stress is OK
Page No.
* Calculating the factor ξ s vc
ξs
* Shear Stress in Concrete
(1 +
=
0.91
(Table 9, BS 5400)
γm
=
1.25
(Table 8, BS 5400)
vc
=
0.69
N/mm2
ξ s vc
=
0.63
N/mm2
0.05N ) Ac
=
1.95
12
ξ s vc * Shear Reinforcement Criterion
=
1.22
N/mm2
(BD 74/00)
* The criterion of shear reinforcement is based on the following cases: Case (1) v <
