design - piles foundation rock
TRANSCRIPT
BH1 BH2 BH3GL 0 MPa GL 0 MPa GL 0 MPadepth level ucs depth level ucs depth level ucs11.5 -11.500 1.02 11.3 -11.300 2.16 11.7 -11.700 0.98 12.6 -12.600 1.96 12 -12.000 1.99 12.9 -12.900 2.61 13.3 -13.300 7.34 13 -13.000 4.2 13.8 -13.800 4.61 14.2 -14.200 4.21 14.5 -14.500 5.61 14.5 -14.500 1.35 15 -15.000 5.19 15.3 -15.300 1.08 15.9 -15.900 1.41
16.3 -16.300 2.42 16.6 -16.600 1.86 16.8 -16.800 6.31 17.1 -17.100 1.21 17.5 -17.500 0.75 17.3 -17.300 1.73 18 -18.000 1.08 18.2 -18.200 8.31 18 -18.000 2.16
19.5 -19.500 1.69 19.6 -19.600 1.31 19.5 -19.500 4.18
Rock level Rock level Rock level11 -11.000 11 -11.000 11 -11.000
Water level Water level Water level1.7 -1.700 1.8 -1.800 1.65 -1.650
1020 1990 980
UCS Shaft min 980 (for toe level -12 m)Toe min 1.96 (for toe level -12 m)
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PFP/040/09/04/A012 April 22, 2009
DESIGN CALCULATIONS OF 500 MM DIA PILES, 100 TONS (C) & 0 TONS (T)
1.0 GENERAL
2.0 PROPOSED PILE DETAILS
3.0 STRUCTURAL DESIGN CALCULATIONS OF THE PILE
3.1 MAXIMUM ALLAWABLE COMPRESSION LOAD
3.2 DESIGN REQUIREMENTS OF REINFORCEMENT
3.3 BENDING MOMENT CALCULATIONS IN PILE DUE TO H-FORCE &
CONSTRUCTIONAL ALLOWANCES
3.4 STRESS IN CONCRETE
3.5 CHECK FOR CLEAR SPACING
3.6 CHECK FOR BOND LENGTH
3.7 CHECK FOR STEEL IN TENSION
3.8 CHECK FOR SPACING OF STIRRUPS
4.0 GEOTECHNICAL DESIGN CALCULATIONS OF THE PILE
4.1 RESUME OF SOIL DATA
4.2 TOTAL PILE CAPACITY
4.2.1 CAPACITY OF PILE IN ROCK LAYER4.2.2 SHAFT FRICTIONAL RESISTANCE OF PILE IN SOIL ABOVE ROCK LAYER
4.3 CALCULATION OF SETTLEMENT OF PILE
4.3.1 SETTLEMENT OF PILE SHAFT4.3.2 SETTLEMENT OF PILE CAUSED BY THE LOAD AT END BEARING4.3.3 SETTLEMENT OF PILE CAUSED BY THE LOAD SUBMITTED ALONG THE PILE SHAFT4.3.4 TOTAL PILE SETTLEMENT
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PHEONIX FOUNDATIONPILING LLC
CONTENTS:
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Project : Villas - 63 CFA PILES
Location : Plot No. 98 – Sector Z 22 - MBZ CityABU DHABI
Owner : الحجري محمد سعيد سالم محمد السيد
PILINGContractor : العامة للمقاوالت برنت بلو
Consultants : الهندسية لالستشارات العالمي التصميم مكتب
References : BS 8110 : 1997 ; Structural use of ConcreteBS 8004 : 1986 ; BS Code of practice for FoundationsCharts for design of circular columns to BS 8110Principles of Foundation Engineering - Braja M DasPile Design and construction practice - M J Tomlinson, Third editionElements of Soil Mechanics for Civil and Mining Engineers
- By G N SmithPressure Meter and Foundation Engineering
- By F Baguelin, J F Jezequel, D H Shields
2.0 PROPOSED PILE DETAILS Ground Level ± 0.000 m G.Lvl
Pile Type : Bored Piles Using CFA Method 750 mmmax 1 m C.O.Lvl
Pile Diameter : 500 mm (Nominal) below platform
Design Capacity : 100 Tons (C) & 0 Tons (T)
Number of Piles : 63 Piles
Pile Toe Level : -8.60 m Pile length : 8.60 m
Cub Concrete Strength : 40.00 N/mm 2
-7.6 m Rock Lvl: 8.0m Steel Bars
Main Bars : 6 No: 16 mm & fy = 460 N/mm2 Links : 8 mm @ 150mm & fy = 250 N/mm2 -8.6 m Toe Lvl Cover : 75 mm
:
Nbr of Working test : 2 Piles (1.5%)
Nbr of Integrity test : 63 Piles (100.0%)
All Dimensions are in mm unless otherwise notedC- Compression, S- Shear, T - Tension
2
1.0 GENERAL
Reinforcement
Tests
Note:
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3.0 STRUCTURAL DESIGN CALCULATIONS OF THE PILE
Allowing for eccentricity of loading due to deviations during construction, the ultimate axial load should not exceed the value of N given by:
40 , = ### (Net Area)
460 , = ### N = ### N Or 353.85 TONS Load Factor : 1.4 Safe N = 252.75 TONS Hence the pile is structurally safe to carry the load 100
BS 8004:1986 Art 7.4.2.5.4 refers OUT OF POSITION ALLOWED : 75 OUT OF PLUMB ALLOWED : 1:75 The reinforcement shall be designed for this Article requirement### : 50.00 kN Horizontal load from plumb on the pile : 13.33 kN (Including construction allowances) B.M produced by eccentricity : 75.00 kN.m
Since the piles are restrained by the pilecaps and Tie Beams the pile head shall be considered to be restrained from rotational moment. Hence the moment on the pile occurs mainly due to horizontal force at the pile head. Total horizontal force on the pile head (Ultimate) H = 83.33 kN
Reese and Matlock have established a series of curves for normally consolidated and cohesionless soils for which the elastic modulus of the soil Es is assumed to increase from zero at the pile head in direct proportion to the depth assuming that the pile behaves as an elastic beam on a soil which also behaves elastically. Refer pages 219 to 223 of " Pile Design And Construction Practice by M J Tomlinson Third Edition; 1987
Using the fixed Pile head equation (6.58) = =
= Bending Moment
= Bending Moment Coefficient from figure 6.39bT = Stiffness Factor
=E = Reference: BS 8110:1997 Part 2, Table 7.2
= 28000 MN/m² or N/mm²
I = ###
= Coefficient of subgrade modulus variation
= 24 (Refer figure 6.20)T = 1.3
= L/TL = Pile length
= 6.62 » 7
3
3.1 Maximum allawble compression load:
N = 0.4 fcu A c + 0.75 As fy (Eq 38 of BS 8110:1997)
fcu = N/mm2 A c mm2
fy = N/mm2 A s mm2
3.2 Design Requirements of Reinforcement :
3.3 Bending Moment calculations in Pile due to H-force and constructional allowances:
Elastic Analysis:
MF FmHT
MF
Fm
5ÖEI/nh
m4
nh
MN/m3
Zmax
Zmax
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Depth T Z=x/T x (m) (m) (kN.m)
0 1.30 0.00 -0.93 -100.75 1.3 1.30 1.00 -0.05 -5.42 2.6 1.30 2.00 0.25 27.08 3.9 1.30 3.00 0.17 18.42 5.2 1.30 4.00 0.05 5.42 6.5 1.30 5.00 0.00 0.00
75 + 100.75 = 175.75 kN.m
318 mm318
= 0.64 500
For the Ultimate State Design apply a load factor of 1.4 for working load: 1.4 N
= 5.60 ;M
= 1.41 h²
From Chart 13 for design of circular columns to BS 8110 , for this condition:
= 0.40%
= 0.61% Therefore safe
Main Bars, 6 Nos, Dia: 16 mm Area 0.61% of concrete Therefore O.K. Links 8 mm @ 150 mm
According to BS 8004:1986, Section 7.4.3.3.1 the working stress in compression should not exceed 25% of the characteristic concrete strength at 28 days;
Working stress =100 x 10000
= ###x 500 ²
< < 10 N/mm² Therefore safe
As an example of the foregoing calculations stresses, the Moment at 5.2 M = 5.42 kN.m
5.2 m = ±M
xI 2
= 1000 x ± 5.42 500 p / 4 x 250000 ### x 2
= 5.53 OR 4.65 5.09
As per BS 8004 Article 7.4.3.3.1:
10 N/mm² From the above results it can be inferred that stressses in the concrete are acceptable
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Fm MF =FmHT
Therefore maximum bending moment on pile = Mmax =
hs =
Then hs /h =
h3
Area of Steel required Areq (Asmin = 0.40%)
Area of Steel provided As
Proposed reinforcement:
3.4 Stress in Concrete:
N/mm2
p/4
0.25 x fcu
s at Nc f
Ac
103 x 106 x
N/mm², Which give sm =
these stresses are < 0.25 fcu or
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For 6 bars of 16 mm dia, c.c spacing = 166.50 Therefore clear spacing is 150.50 mm Art 3.12.11.2.4 of BS 8110 (Part 1) : 1997 max clear spacing is 300 mm Thus 6 bars provided are adequate.
Table 3.27 of BS 8110:1997 shows the following values of anchorages for various grades of concrete:
Grade Type 2 Deformed Bars30 40 f35 38 f40 35 f
40 N/mm² Anchorage length = 35 x 16 = 560 mm We propose 750 mm into the pile cap
As = ### mm² for 6 Bars, Dia 16 mm Ultimate Tensile Force = 1206.37 x 437 (460 X 0.95 = 437)
= ### N = 52.718 Tons Load Factor = 1.4 Allowable Tensile Force = 37.66 Tons > 0 Tons Therefore safe
40 N/mm²
400.00 mm 1.25
Area of Reinforcement Provided = 1206.37
=
= 0.575
= 0.575 0.673
Working Shear Stress in pile = v =
v = 83.33 = 0.424 N/mm² (Ultimate)500 ²
however we provide nominal 8 mm links @ 150 mm
For this condition use the following equation:
0.40 x 500 x 150 = 68.650 mm²0.95 x 460
< 100.53 mm² Therefore safe.5
3.5 Check for Clear Spacing:
3.6 Check for Bond Length:
For fcu =
3.7 Check for steel in tension:
3.8 Check for Spacing of Stirrups:
From Table 3.8 BS 8110 (Part 1:1997) , for fcu =
ds = gm =
mm2
Shear Stress = vc 0.79 x (100 x As/(b x d))1/3 x (400/ds)1/4/gm
N/mm2 for concrete grade 25
vc [40/25]1/3 = N/mm2 for concrete grade 40
Hw
p/4 xf ²
x 103
p/4 x
As v < vc the permissible shear stress is more than the actual shear stress, No Shear reinforcement is not required.
Minimum Shear reinforcement required = Asv ³ 0.4 bv sv / 0.95 fyv
Asv =
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4.1 Soil Report by : GULF LABORATORYPlatform Level : ± 0.000 m
Platform Level
Cut-off Level : max 1 m Cut-Off Level: below platform
Min UCS of shaft : 1770 kN/m²
Min UCS at Toe : 2175 kN/m²Zone of interest of reading for UCS Values at the Toe Level for End Bearing Capacity [25% from shaft (0.5D) & 75% from Toe (1.5D)] Rock LevelToe Level : -8.60 m Toe Level
Length of Piles:
4.2 Total Pile Capacity :4.2.1 Pile Capacity in Rock Layer :
Taking the worst case, Rock Level : -7.60 mUCS of rock at Toe of Pile : -8.60 m 2175 kN/m²UCS in Rock Socket : 1770 kN/m²Length of Rock Socket : 1 m
_a. 1st assumption:
= 196349.5 mm²
0.46 N/mm²
=
40 = 20 20 (see annex - page 4)30 ° (see annex - pages 6 & 7 to take an idea)
76.00 kN/m² ,
9.00 D = Diameter of Piles =500 mm
= ### mm²
= 0.500 (Pile Design and construction practice - M J Tomlinson,= 0.577 Third edition: Cast-in-place; concrete/sand)
25467 & ###
### x 25.467 = ### N 99.66%
4.0 GEOTECHNICAL DESIGN CALCULATIONS OF THE PILE:RESUME OF SOIL DATA:
It should be emphasized that the length of pile assumed in these calculations is based on Lower cut-off levels, Loadings and results of tests run on a few boreholes carried out by a third party. At any rate such information contained in the Boreholes results affecting the design may not be exactly representative of the whole area and particularly that the hard strata reported may differ from pile location to the other. The length of pile(s) shall be determined on site while boring by our supervisor on site as he is the best judge of the soil conditions because he watching every bore all the time.Of course the attention of the Engineer's Representative shall be drawn to each case as it arises and a decision taken regarding where to found the pile in each case.
Ultimate Bearing Capacity of Pile Qu
Reference: Braja M Das (Principles of Foundation Engineering - Pages 487 - 504) - Estimation of shaft & end bearing according to TERZAGHI which can be calculated as follows:
Qu = Qb + Qs = Ab qb +As qs = Ab (1.3 C Nc + Nq s'v0 + 0.3 g' D Ng) + 1/2 Ks s'v0 tand As
Ab = Pile base area
C = Cohesion at base = ½ quc of UCS of Rock (but less than 0.46) =
(see annex - page 6 for Mudstone - to be at the safe side since the soil report give a higher value, we adopt 0.46 N/mm2)
Nc , Ng & Nq Dimensionless Bearing Capacity Factors depneding of f (Tomlinson,3rd ed page 123)
Nc = Nq Ng = f = Angle of friction at the base pile level =
s'v0 = Effective overburden pressure at base pile level =
g' = Average Effective unit weight of shaft pile in rock = kN/m3 , (see annex - page 3 to take an idea)
As = Area of Socket friction in rock
Ks = Coefficient of the horizontal soil stress; d = Pile/Soil friction angle
Ks = K0 = (1-sinf)tand = tanf
qb = qs = kN/m2
Qb =
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### x ### = ### N 0.34%= ### N = 501.77 Tons
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Qs = Qu
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b. 2nd assumption:
For UCS of 2175 kN/m² ;
30 °, 3.00 ;
2 x 3.00 x 2175 = 13050.00 kN/m²= 13.05 N/mm²
Shaft Frictional Resistance of Pile in Rock according to Tomlinson (After Williams & Pells):
Where,= 0.28 ;= 0.65 ; (Dependent on Mass factor/RQD)
Average RQD < 50%
1.770 N/mm² (Average UCS Value)
0.28 x 0.650 x 1.770 = 0.322 N/mm²
Shaft Frictional Resistance of Pile in Rock according to according to Tomlinson (After Rosenberg & Journeaux) :
; a = 0.20 1.00 ;
0.2 x 1 x 1770 = 0.35 N/mm²
From the above comparative calculations it can be inferred that the minimum values are:
13.05 N/mm² ; 0.32 N/mm²
Therefore
### x 13.05 = ### N 83.51%
### x 0.32 = ### N 16.49% = ### N = 306.84 Tons
From the above comparative calculations it can be inferred that the minimum value for bearing in rock is:
306.84 Tons
4.2.2 Shaft Frictional Resistance of Pile in Soil above Rock Layer:Friction angle of shaft in the Soil Lay : 36.0Length of Pile shaft in Soil Layer - 1m for Cap Pile : 6.6 m
= As qs = ; Where
76.00 kN/m² ,
= ### mm²
; Where :
= 0.412 (Pile Design and construction practice - M J Tomlinson,= 0.727 Third edition: Cast-in-place; concrete/sand)
11.38 kN/m²
### x ### = ### N = ### Tons
Therefore, the total capacity of the pile is 318.64 Tons
7
(Reference: Tomlinson's "Pile Design and Construction" Pages 131 - 136 Unit end bearing according to Tomlinson can be calculated as follows:
qb = 2 Nf quc (4.26)Where Nf = tan² (45 + f/2) f = Nf =
qb =
qs = ab quc
a = Rock Socket Reduction factorb = Rock Socket correction factor
quc =
qs =
qs = ab quc ; b =
qs =
qb = qs =
Qb =
Qs =Qu
Qu =
° (see annex - page 1)
Qs (S) 1/2 Ks s'v0 tand As
s'v0 = Effective overburden pressure at the base of soil level =
Ks = Coefficient of the horizontal soil stress; d = Pile/Soil friction angle
As = Area of Socket friction
qs = 1/2 Ks s'v0 tand As
Ks = K0 = (1-sinf)tand = tanf
qs =
Qs (S) =
Qu =
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In compression :
F.S = 3 ; = 106.21 Tons> 100 Tons ; Therefore safe
In Tension :
51 Tons ; The weight of pile is not considered
F.S = 2.5 ; = 24.96 Tons
4.3 Calculation of Settlement of Pile :
Nomenclature:
4.3.1 Settlement of Pile shaft :
376 kN
624 kN
= 196349.54 mm²L = Length of the pile from cut-off level = 7600.00 mm
= 28 kN/mm²= 0.5
= 0.952 mm
4.3.2 Settlement of pile caused by the load at end bearing :
x
D = diameter of pile = 500.0 mm
= 1914.95 kN/m²
(see annex - pages 7, 9 & 10) 350000 kN/m²
= 0.30 (see annex - page 11)
= 0.88
2.19 mm
4.3.3 Settlement of pile caused by the load transmitted along the pile shaft :
x D x
p = perimeter of the pile = 1570.80 mm
= 1.00 m
3.36 mm
1.74 mm
4.3.4 Total pile settlement :
4.88 mmThe above figures, it must be appreciated, are calculated using data available from the soil report
Qallow
Qu = Qs =
Qallow
Reference: "Principles of Foundation Engineering", second edition by Braja M Das.
S = Total pile settlement = S1 + S2 + S3
S1 = Settlement of pile shaft
S2 = Settlement of pile caused by the load at end bearing
S3 = Settlement of pile caused by the load transmitted along the pile shaft
Settlement of pile shaft = S1 = (Qwp + x Qws )L
Ap Ep
Qwp = load carried by end bearing under working load condition =
Qws = load carried by shaft friction under working load condition =
Ap = area of pile cross section
Ep = Young's modulus of the pile materialSkin Friction resistance Distribution factor " x "x = 0.5 for uniform & parabolic ; x = 0.67 for triangular (Vesic, 1977)
Settlement of pile shaft = S1
S2 = qwp D (1 - ms² ) Iwp
Es
qwp = point load per unit area at pile point
Es = Young's modulus of rock at pile tip
µs = Poisson's ratio of soil
Iwp = influence factor
S2 = Settlement of pile caused by the load at end bearing =
S3 = Qws (1- µS2) Iws
pLe Es
Le = embedded length of pile
Iws = influence factor = 2 + 0.35 Ö Le / D (Vessic, 1977) =
S3 =
Total settlement anticipated = S = S1 + S2 + S3 =
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and the actual conditions may vary taking into consideration the practical aspects and limitationsof the piling operation.
We cannot predict residual settlements as these should better be left to the actual testing of piles.
8
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PILE DESIGN
JOB NUMBER: PFP/040/09/04/A012
PROJECT: Villas - 63 CFA PILES
LOCATION:Plot No. 98 – Sector Z 22 - MBZ City
ABU DHABI
OWNER: الحجري محمد سعيد سالم محمد السيد
CONSULTANT: الهندسية لالستشارات العالمي التصميم مكتب
CONTRACTOR: العامة للمقاوالت برنت بلو
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P.O. Box: 40287 - Abu Dhabi Mob: + 971 50 96 76 136
PHEONIX FOUNDATIONPILING LLC
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