466 proje2
TRANSCRIPT
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DESIGN OF PILE GROUP
First of all we chose our piles diameter as 1 meter. Than we took the length of the
pile as 48 meter. We decided cross-section of pile as circle and factor of safety as 2,5. Due to
circle cross-section pile must be displacement (driven) pile. We used C30 concrete forproduction of piles. Frictional resistance of a pile in a clay deposit can be calculated by three
different methods. These are:
methodF methodP method
Our soil layers are clay and sand. So we must used these method to calculate axial
load capacity of pile, we choosed to evaluate the load capacity of pile with using method.
Work To Do:
yCapacity of a single pileySettlement analysisyR.C. design of single pileyR.C. design of pile cap
PileProperties
Type: Driven Pile (concrete) 1000 mm
Load from tower and rose = 125 kN
Diameter: 1000 mm
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a=10 m
b=10 m
L=48 m
h=1 m
s=4 m
s=4 m
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SoilProfile
Air 1 m
Water = 9,81 kN/m3 15 m
Clay s = 17 kN/m3 Cu=20 =0,5 7 m
Sand s = 20 kN/m3 = 30 =0,35 25 m
yVolume of Foundation ( V ) = a*b*h = 10*10*1 = 100 m 3yWeight of Foundation = V * = 100* 25 = 2500 kNyLenght of pile in water and soil ( saturated zone) = 47 myVolume of pile in saturated zone = Base area of pile * lenght in saturated zone
= (*D2)/4 * L = (*12)/4 * 47 = 36,9 m3
yVolume of pile in air = Base area of pile * lenght in air= (*D
2)/4 * L = (*1
2)/4 * 1 = 0,785 m
3
yWeight of a pile = = 36,9*(25-9,81) + 0,785 * 25 =580 kN
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Bg =8 m
n =3 (row)
m= 3 ( number of piles in a row)
for clay and sand min spacing is 2D
N = 9 (number of piles)
Qtotal = 2500 + 580 + 125 = 3205 kN
Mx = My = 61000 kNm = = 6 * 42 = 96 m2
Qn=
For Mx
Q1 = = 2897 kN = 289,7 ton (in compression)
For My
Q1 = = -2185,5 kN = 218,55 ton (in tension)
4 m 4 m 1 m1 m
1 m
4 m
4 m
1 m
My
Mx
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CALCULATION OF A SINGLE PILE LOAD CARRYING CAPACITY
1 m
15 m
0
1 7 m
7*(17-9,81)=50,33 kN/m2
12,5 m
2
177,705 kN/m2 (average) 25 m
3
25*(20-9,81)+50,33 = 305,8 kN/m2
0,3 < = 0,35 < 0,4 so dense sand
Circumference of pile = * D = 3,14 m
A1 = (50,33*7) /2 = 176,155 kN/m2
A2 = (50,33*177,705)/2*12,5 =1425,22kN/m2
A3 = 177,705 *12,5 =2221,3 kN/m2
Area of Pv Diagram = A1 + A2 + A3
= 176,155 + 1425,22 + 2221,3
= 3822,68 kN/m2
Area of Pv Diagram in Sand Region = 1425,22 + 2221,3 = 3646,45 kN/m2
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Nq = 22,5 for = 30o
Pv = 177,705 kN/m2
from Pv diagram
Qend = Pv*Nq *Ab
Qend = 177,705*22,5* (*12)/4
Qend = 3140 kN
y Assume K = 1 (0,6 < K < 1,25 )y tan = 0,45 ( for concrete pile )
Qfriction (in sand) = Circumference of pile * Area of Pv Diagram in Sand Region* K * tan
Qfriction (in sand) = 3,14 * 3646,45 * 1 * 0,45
Qfriction (in sand) = 5152,5 kN
Qfriction in clay with methodCu = 20 kN/m2 < 50 kN/m2 so = 1
Qfriction (in clay) = Cu * * As
Qfriction (in clay) = 20* 1 * *1*7
Qfriction (in clay) = 439,8 kN
Qult (in compression ) = Qend + Qfriction (in sand) + Qfriction (in clay) Qult (in tension ) = Qfriction (in sand) + Qfriction (in clay)
Qult (in compression ) = 3140 + 5152,5 + 439,8 ult (in tension ) = 5152,5 + 439,8
Qult (in compression ) = 8732,3 kN ult (in tension ) = 5592,3 kN
Qall (in compression) = Qult (in compression ) / FS Qall ( in tension) = Qult ( in tension) / FS
Qall ( in compression) = 8732,3 / 2,5 all ( in compression) = 5592,3 / 2,5
Qall ( in compression) = 3500 kN = 350 ton all ( in compression) = 2237 kN = 223,7 ton
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CHECKS
Qall for a single pile in compression = 350 ton Qall for a single pile in tension = 223,7 ton
Total load for a pile in compression = 289,7 ton Total load for a pile in tension = 218,5 ton
350 > 289,7 safe
223,7 > 218,5 safe
CALCULATION OF SETTELEMENT
S = S1 + S2 + S3
S1 =
S1 =
S1 = 4,89 mm
S2 =
S2 =
S2 = 9,49 mm
= Q = coefficient= Q = Length of pile
= Piles base area
= Modulus of elasticty of pile
S1 = Elastic settlement of pile
S2 = Settelement caused by load in tip
S3 = Settelement caused by friction load
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S3 = * * *
= 2 + 0,35 = 2 + 0,35 = 2,93 (clay)
= 2 + 0,35 = 2 + 0,35 = 3,75 (sand)
For Clay
S3 (clay) = * * * 2,93
S3 (clay)
= 3,5 mm
For Sand
S3(sand) = * * * 3,75
S3 (sand) = 4,32 mm
S3 =
S3 (clay) +
S3 (sand)
S3 = 3,5 + 4,32
S3 = 7,82 mm
Total Settelement
S = S1 + S2 + S3 = 4,89 + 9,49 + 7,82 S = 22,2 mm
Group Pile SettelementSg = * S = * 22,2 Sg =62,79 mm
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PULLOUT RESISTANCE OF PILES
= + W
W L
D
Pile in Clay :=L * P * * 80 =0,9 0,0025 * (= 20 )
> 80 =0,4
=0,9 0,0025 * 20 = 0,85
L = 7 m
P = * D = 3,14 m
=7 * 3,14 * 0,85 * 20= 373,66 kN
Pile in Sand:yDense sand has 60 80 Dr we choose 70 Dry From table = 14,5 = 14,5y From figure
= 0,97 = 0,97 * 30 = 29
y From figure Ku = 1,3y < L
= Net uplift capacity =Gross uplift capacityW= Effective weight of pile
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y =0,5* P** () * Ku* tan + P * * Ku * tan *
y =0,5* *1*20* () * 1,3 * tan29 + *1* 20* 1,3 * tan29 *
= 11653,1 kN
= + =373,66 + 11653,1
= 12026,76 kN
W = Weight of Pile = 580 kN
= + W = 12026,76 + 580
= kN
= =
= 5042,7 kN
One Pile Capacity at the Tension Zone = 2185,5 kN
We have 3 piles at tension zone = 3 * 2185,5
= 6556,5 kN
6556,5 > 5042,7 SAFE
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COMMENTSAt the beginning of the project we started to make some assumptions. First of all we
chose our piles cross-section as circle. Then we took the length of pile 48 meter. We used
factor of safety 2,5. Because our soil layers are clay and sand. These assumptions are for
economical and safety design.
Firstly we made assumption of the number of piles. Then we started to calculate load
carrying capacity of single pile and we found the load carrying capacity of single pile. Firstly
we took number of piles as 9 then we calculated Qdesing and we saw that our Qdesign value is
bigger than Qapplied. This means that piles can carry applied load and we dont need to
increased the number of pile. At this condition our caps dimensions are 10 m x 10 m.
After that we started to calculate settlement of pile groups. Firstly we calculated the
settlement. Sum of the settlement were 6,279 cm. for the offshore wind platform. This value is
normal. To get less settlement we must increase the dimensions of the pile cap. If we cannot
increase the dimensions of pile cap, we must increase length of piles or cross-section of piles.