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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.