preface - mengenal ilmu teknik sipil · preface to design high load building foundation in less...
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Preface To design high load building foundation in less soil bearing capacity, we need to compute the most
effective foundation type before it’s determined. It’s possible to use shallow foundation, but it
wouldn’t be efficient to spend a lot of money in designing a huge dimension of the foundation to
bear the load in the less soil bearing capacity. So in this condition deep foundation is preferred,
because the dimension of pile is smaller than shallow foundation, even the installation process is
need more effort and skill.
In this second foundation engineering assignment, N-SPT data bore hole 2 given, using 20 x 20 cm
driven piles and 30 cm for bore piles diameter, and it need to compute the most efficient pile depth,
piles configuration, lateral load, moment and settlement, to hold the load from the building. Before
computing the specification of pile, interpreted soil strength data is the base information to design.
In case of soil parameter is not that easy to be determined, the geotechnical engineer must have a
knowledge and skill to make interpretation of soil parameter. Because soil parameter is one of the
important value in design.
Interpreted Standard Penetration Data From the drilling log SPT data shown that the depth of driving pile is 20 meter in case the soil
strength isn’t good enough to hand the load from the building.
Interpreting SPT data
It’s assumed that the soil properties are divided by some layers which each of them have any
similarities of N value.More layer are divided, the chance to get an accurate N even value is higher.
The SPT data should be corrected by:
a. Overburden correction
It’s assumed that the soil properties along the depth are normally consolidated loose sand,
thus the equation of overburden correction should be:
, which
CN = overburden correction
Po’ = effective overburden pressure (KN/m2)
Pr = effective reference strain = 100 KN/m2
Thus, the N value
N = CN . N’
Where N’ = N
b. Correction for loose sand and saturated silty sand
This correction is just given to the depth of soil below the water table. So for the depth
above water table is needn’t to be corrected. It means that for the soil below the water
table is need to be corrected twice, which is overburden correction and the correction for
this condition.
If the N value is bigger than 15, so it should be reducted/corrected by:
N’ = 15 + 0,5 . (N’ – 15)
After it has corrected by the equation above, it should be corrected by overburden pressure.
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
From the given SPT data, it can be shown that the water table is at water surface, so it’ll use
ϒsat along the depth of soil.
Because the layer is dominated by silt and clay, so it can be assumed that the data given is needn’t
to be corrected by two corrections above and the layer is divided by it’s depth where the N value
shown at the SPT log data. Hence, the data could be shown as table below:
depth N Cu (t/m2) φ' ϒsat q' σv' Nc Nq
1 5 2 0 1.8 0.8 0.8 5.7 1
2 5 0 21 2 1.8 1.8 5.7 9.5
3 5 2 0 1.9 2.7 2.7 5.7 1
4 1 1 0 1.9 3.6 3.6 5.7 1
5 1 1 0 1.9 4.5 3.6 5.7 1
6 2 1.3 0 1.9 5.4 3.6 5.7 1
7 2 1.3 0 1.9 6.3 3.6 5.7 1
8 2 1.3 0 1.9 7.2 3.6 5.7 1
9 2 1.3 0 1.9 8.1 3.6 5.7 1
10 3 0 20 2 9.1 3.6 12 9
11 2 1.3 0 1.7 9.8 3.6 5.7 1
12 2 1.3 0 1.7 10.5 3.6 5.7 1
13 2 1.3 0 1.7 11.2 3.6 5.7 1
14 2 1.3 0 1.7 11.9 3.6 5.7 1
15 2 1.3 0 1.7 12.6 3.6 5.7 1
16 2 1.3 0 1.7 13.3 3.6 5.7 1
17 2 1.3 0 1.7 14 3.6 5.7 1
18 17 6 0 1.9 14.9 3.6 5.7 1
19 17 6 0 1.9 15.8 3.6 5.7 1
20 18 6.5 0 1.9 16.7 3.6 5.7 1
Table 1. soil parameter
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Figure 1. vertical strenght
The value of density type of soil and friction angle is given from figure 2.13 relationship between SPT
N-values and angle of shearing resistance (Tomlinson, 1977)
Figure 1. SPT N on Angle of internal friction
Nc and Nq can be determined by meyerhoft graph below:
0
5
10
15
20
25
0 5 10 15 20
q'
q'
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Figure 1. bearing capacity factor Nc, Nq, Nϒ (after Terzaghi and Meyerhoft)
The driven pile will be held above the pile cap, so the embedment of piles would be 20 meter length
at maximum. To know the most efficient piles length, the bearing capacity of pile must be calculated
as it depth, so it will give the information to determined the length.
Calculating the shaft resistance The calculation of shaft resistance for sand is based on Brom method which it’s based on the
approximated value of Kd and δ. The Brom’s method is chosen because it gives more representative
result when used in site. The shaft resistance is diverse by the base resistance and friction resistance.
for this case, it’s assumed that silt can be chaged as clay soil, so λ and α method is took to calculate
the friction resistance and for base resistance, brom’s method is prefered.
Friction resistance To calculate the value of overburden pressure σv=ϒ.h, ϒ value is need to be determined. SPT data
has already shown the soil properties of every layer, so ϒ value can be determined by the table of
relationship between soil properties and ϒ value.
Equation of pile friction resistance for sand:
Kd value is given from Tomlison 1977 that based on soil density or 1-sinφ, and the δ value for the
concrete pile is δ = 0,75 . φ (Mayerhof).
For clay:
Where fs can be calculated by method below :
LAMBDA METHOD:
Fs ave =
Which is :
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
= constant
’ = average of vertical effective strength
= average of un drained shear strength
Fsave = average of shaft friction
ALPHA METHOD
Fs = α . cu
Which is:
Fs = shaft friction
α = adhesive factor
cu = undrained cohession
Base Resistance
Brom’s base resistance equation for sand:
Qb = Ab . q' . Nq
Which,
Qb = base resistance (KN)
q' = effective vertical loads at the base of pile
Nq = bearing capacity factor (given by table 2.7, Tomlinson)
Ab = base pile area
For clay :
Qb = Ab . (Cu . Nc)
Cu = cohession of soil
Nc = bearing capacity factor
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
So the calculation of friction resistance and base resistance can be shown by table below:
Driving piles : square concrete piles
Dimension 20 x 20 cm
As = 0,2 . 4 = 0,08 m2
σv' ave Cu ave K (low Dr) δ As
sand clay Qs layer
Depth fs=K . σv . Tanδ
metode α metode λ Qs = As . Fs
α fs (t/m2) λ fs = λ . (σv+2Cu) sand α method λ method
1 0.40 2.00 1.00 0.00 0.80 0.00 0.75 1.50 0.47 2.07 0.00 1.20 1.65
2 1.33 1.00 0.64 15.75 0.80 0.33 1.25 0.00 0.45 1.51 0.26 0.00 1.21
3 1.93 1.33 1.00 0.00 0.80 0.00 0.75 1.50 0.44 2.02 0.00 1.20 1.62
4 2.35 1.25 1.00 0.00 0.80 0.00 1.15 1.15 0.42 2.06 0.00 0.92 1.65
5 2.43 1.20 1.00 0.00 0.80 0.00 1.15 1.15 0.41 1.97 0.00 0.92 1.58
6 3.00 1.22 1.00 0.00 0.80 0.00 1.05 1.37 0.39 2.14 0.00 1.09 1.71
7 3.47 1.23 1.00 0.00 0.80 0.00 1.05 1.37 0.38 2.24 0.00 1.09 1.79
8 3.88 1.24 1.00 0.00 0.80 0.00 1.05 1.37 0.36 2.31 0.00 1.09 1.85
9 4.25 1.24 1.00 0.00 0.80 0.00 1.05 1.37 0.35 2.34 0.00 1.09 1.87
10 4.59 1.12 0.66 15.00 0.80 1.60 1.25 0.00 0.33 2.27 1.28 0.00 1.82
11 4.91 1.14 1.00 0.00 0.80 0.00 1.05 1.37 0.32 2.28 0.00 1.09 1.82
12 5.21 1.15 1.00 0.00 0.80 0.00 1.05 1.37 0.30 2.27 0.00 1.09 1.82
13 5.49 1.16 1.00 0.00 0.80 0.00 1.05 1.37 0.29 2.24 0.00 1.09 1.79
14 5.76 1.17 1.00 0.00 0.80 0.00 1.05 1.37 0.27 2.20 0.00 1.09 1.76
15 6.01 1.18 1.00 0.00 0.80 0.00 1.05 1.37 0.26 2.14 0.00 1.09 1.72
16 6.25 1.19 1.00 0.00 0.80 0.00 1.05 1.37 0.24 2.08 0.00 1.09 1.66
17 6.49 1.19 1.00 0.00 0.80 0.00 1.05 1.37 0.23 2.00 0.00 1.09 1.60
18 6.72 1.46 1.00 0.00 0.80 0.00 0.50 3.00 0.21 2.03 0.00 2.40 1.62
19 6.94 1.70 1.00 0.00 0.80 0.00 0.50 3.00 0.20 2.02 0.00 2.40 1.62
20 7.17 1.94 1.00 0.00 0.80 0.00 0.50 3.25 0.18 1.99 0.00 2.60 1.59
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Ap=0,2^2=0,04 m2
Depth sand clay Cummulative value as depth
Wc (ton)
Qallow alpha
Q allow lambda
Qp = Ap.q'.Nq
Qp=Ap.cu.Nc Qs cummulative
alpha + sand Qs cum lambda
Qp cummulative
1 0.78 0.46 1.20 1.65 0.46 0.10 0.85 1.15
2 7.45 0.00 1.46 1.96 0.46 0.20 0.93 1.26
3 0.78 0.46 2.66 3.58 0.91 0.30 1.78 2.39
4 0.78 0.23 3.58 5.22 1.14 0.40 2.37 3.46
5 0.78 0.23 4.50 6.80 1.37 0.50 2.96 4.49
6 0.78 0.30 5.59 8.51 1.66 0.60 3.68 5.63
7 0.78 0.30 6.68 10.31 1.96 0.70 4.41 6.82
8 0.78 0.30 7.78 12.15 2.26 0.80 5.14 8.05
9 0.78 0.30 8.87 14.03 2.55 0.90 5.86 9.30
10 7.06 0.00 10.15 15.54 5.83 1.00 7.71 11.30
11 0.78 0.30 11.24 17.36 6.38 1.10 8.52 12.60
12 0.78 0.30 12.34 19.18 6.93 1.20 9.33 13.89
13 0.78 0.30 13.43 20.97 7.47 1.30 10.14 15.17
14 0.78 0.30 14.52 22.73 8.02 1.40 10.95 16.43
15 0.78 0.30 15.61 24.44 8.57 1.50 11.77 17.65
16 0.78 0.30 16.70 26.11 9.12 1.60 12.58 18.84
17 0.78 0.30 17.80 27.71 9.67 1.70 13.39 20.00
18 0.78 1.37 20.20 29.33 11.36 1.80 15.45 21.54
19 0.78 1.37 22.60 30.95 13.05 1.90 17.51 23.08
20 0.78 1.48 25.20 32.54 14.86 2.00 19.75 24.65
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Bored piles : circle concrete piles
Diameter = 0,3 m As = phi . 0,3 = 0,94 The different between driving pile and bore pile is on the alpha coefficient because the soil will be remolded after boring.
σv' ave
Cu ave K (low Dr) δ As
sand clay Qs layer
fs=K . σv . Tanδ
metode α metode λ Qs = As . Fs
Depth α fs (t/m2) λ fs = λ . (σv+2Cu) sand α method λ method
1 0.40 2.00 1.00 0.00 0.94 0.00 1.00 2.00 0.47 2.07 0.00 1.88 1.94
2 1.33 1.00 0.64 15.75 0.94 0.33 1.00 0.00 0.45 1.51 0.31 0.00 1.42
3 1.93 1.33 1.00 0.00 0.94 0.00 1.00 2.00 0.44 2.02 0.00 1.88 1.90
4 2.35 1.25 1.00 0.00 0.94 0.00 1.00 1.00 0.42 2.06 0.00 0.94 1.93
5 2.43 1.20 1.00 0.00 0.94 0.00 1.00 1.00 0.41 1.97 0.00 0.94 1.85
6 3.00 1.22 1.00 0.00 0.94 0.00 1.00 1.30 0.39 2.14 0.00 1.22 2.01
7 3.47 1.23 1.00 0.00 0.94 0.00 1.00 1.30 0.38 2.24 0.00 1.22 2.11
8 3.88 1.24 1.00 0.00 0.94 0.00 1.00 1.30 0.36 2.31 0.00 1.22 2.17
9 4.25 1.24 1.00 0.00 0.94 0.00 1.00 1.30 0.35 2.34 0.00 1.22 2.20
10 4.59 1.12 0.66 15.00 0.94 1.60 1.00 0.00 0.33 2.27 1.51 0.00 2.13
11 4.91 1.14 1.00 0.00 0.94 0.00 1.00 1.30 0.32 2.28 0.00 1.22 2.14
12 5.21 1.15 1.00 0.00 0.94 0.00 1.00 1.30 0.30 2.27 0.00 1.22 2.13
13 5.49 1.16 1.00 0.00 0.94 0.00 1.00 1.30 0.29 2.24 0.00 1.22 2.11
14 5.76 1.17 1.00 0.00 0.94 0.00 1.00 1.30 0.27 2.20 0.00 1.22 2.07
15 6.01 1.18 1.00 0.00 0.94 0.00 1.00 1.30 0.26 2.14 0.00 1.22 2.02
16 6.25 1.19 1.00 0.00 0.94 0.00 1.00 1.30 0.24 2.08 0.00 1.22 1.95
17 6.49 1.19 1.00 0.00 0.94 0.00 1.00 1.30 0.23 2.00 0.00 1.22 1.88
18 6.72 1.46 1.00 0.00 0.94 0.00 0.75 4.50 0.21 2.03 0.00 4.23 1.91
19 6.94 1.70 1.00 0.00 0.94 0.00 0.75 4.50 0.20 2.02 0.00 4.23 1.90
20 7.17 1.94 1.00 0.00 0.94 0.00 0.70 4.55 0.18 1.99 0.00 4.28 1.87
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Depth sand clay Cummulative value as depth
Wc (ton)
Qallow alpha
Q allow lambda
Qp = Ap.q'.Nq
Qp=Ap.cu.Nc Qs cum
alpha + sand Qs cum lambda
Qp cummulative
1 0.78 0.46 1.88 1.94 0.46 0.18 1.23 1.27
2 7.45 0.00 2.19 2.25 0.46 0.35 1.26 1.30
3 0.78 0.46 4.07 4.15 0.91 0.53 2.48 2.54
4 0.78 0.23 5.01 6.08 1.14 0.71 3.01 3.73
5 0.78 0.23 5.95 7.94 1.37 0.88 3.54 4.87
6 0.78 0.30 7.17 9.95 1.66 1.06 4.27 6.13
7 0.78 0.30 8.39 12.06 1.96 1.24 5.01 7.45
8 0.78 0.30 9.61 14.22 2.26 1.41 5.75 8.82
9 0.78 0.30 10.83 16.43 2.55 1.59 6.48 10.21
10 7.06 0.00 12.34 17.94 5.83 1.77 8.41 12.14
11 0.78 0.30 13.56 20.08 6.38 1.94 9.23 13.57
12 0.78 0.30 14.79 22.21 6.93 2.12 10.05 15.00
13 0.78 0.30 16.01 24.32 7.47 2.30 10.87 16.41
14 0.78 0.30 17.23 26.39 8.02 2.47 11.69 17.79
15 0.78 0.30 18.45 28.40 8.57 2.65 12.51 19.14
16 0.78 0.30 19.67 30.36 9.12 2.83 13.33 20.45
17 0.78 0.30 20.90 32.24 9.67 3.00 14.15 21.71
18 0.78 1.37 25.13 34.15 11.36 3.18 17.36 23.37
19 0.78 1.37 29.36 36.05 13.05 3.36 20.57 25.03
20 0.78 1.48 33.63 37.92 14.86 3.53 23.84 26.70
We used the smallest value of Q allowable between alpha method and lambda method. In calculation table above shown that alpha method gives Q
allowable smaller than lambda method. So, Q allowable in alpha method is taken.
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
To determine the length of the pile, it’s need to be compared with the stiffness of the pile and the
load that need to hold. Stiffness of pile will give the information about the characteristic of the pile.
Is it long flexible or short and stiff. So it will be used to calculate lateral force and moment.
The table above is calculated the value of Qs and Qp for both driving piles and bored piles that
depend by its depth, so the depth of pile can be determined as efficient as needed.
Here is the example of calculation for 7 m depth driving piles:
Permitted bearing capacity over the shaft resistance is use equation below:
, which
Qa = bearing capacity over shaft resistance
Wpile = weight of driven pile
SF = safety factor (SF1 = 3, SF2 = 1,5)
The bigger value of SF1 than SF2is used because the maximum value of pile friction resistance can be
rise if the settlement of the pile about 2 to 7 mm, in case of base resistance is need bigger
settlement to make the resistance of base can work properly.
In this case, the weight of concrete is 2,5 ton/m3, and lenght of pile is 20 – 2 = 18 m, where 2 = the
depth of pile cap.
Wpile = 2,5 . 0.22 . 7 = 0,7 ton
Thus, shaft resistance for alpha method
Bored pile has the same procedure of calculation as driving piles, because it’s cast with the same
material, concrete. bored pile designed without a bell, so Ds = Db = 0,3 m. It’s makethe procedure is
the same as designing driving piles.
Pile configuration Quantity of piles that used and the depth of embedment is depend on the load that stressed the
piles vertically. Also, from the table of Qall for driving piles and bored piles above has known the
value of allowable bearing capacity (Qallowable) each depth. Thus, an efficient quantity and deph of
piles can be determined.
Here is the piles configuration design for column k-6 and k-5.
Loads at column K-5 = 30 ton, and K-6 = 25 ton
Driving piles
Pile that needed at K-5 is : Qcolumn / Qa = 30 / 4,41 = 6,8 = 7 piles 7 m depth
Pile that needed at K-6 is : Qcolumn / Qa = 25 / 4,41 = 5,66 = 6 piles pile at 7 m depth
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Pile spacing that used is 3d = 3 . 0,2 = 0,6 m between each pile. Hence, the pile configuration of K-6
can be shown as the figure below:
Pullout Resistance of piles Under certain construction condition
Permitted bearing capacity over strain (Ta) -driving piles Coyle and Castello’s (1981) method can be used to calculate strain capacity of pile :
3d
3d
7 m
3d
0,5 m
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Pile with lateral force Determine criteria of short and long pile.
In the calculation of pile foundation that takes lateral force, beside the condition of pile head, it’s
needed to be evaluated as the characteristic of the short and long pile.
Because the soil characteristic is normally consolidated soft clay, soil subgrade value is raise linearly
as it depth. Hence, it should use stiffness factor as below:
, which is
Ep = pile elasticity modulus = 4700 x where fc assumed 50 mpa.
Ip = inertia of pile = 1/12 . b . h ˆ3 for square, and
Nh = 500 KN/m3 = 50 Ton/m3
Hence,
m for driving piles and with same procedure T for bore
pile is 2.540157 m
L > 4 T is long pile (elastic), thus for bore pile 4T = 10.16 m , and 4T = 6.18 m for driving pile.
Pile length chosen is more than 4T for both driving pile and bore pile and use pile cap in the surface
of pile, so the head is not free (fixed head).
Horizontal force is 10% from column load. The column that will be calculated is column K-5 = 30 Ton,
and K-6 = 25 ton, so the Horizontal force each K-5 = 3 ton, and K-6 = 25 ton.
Brom’s method It’s determined that pile cap height is 1 meter above the ground surface and the pile get into the pile
cap.
Calculation for K-5 = 30 Ton
Hu = 3 ton
= 2,8 m
, where e = ½ . pile cap height = ½ . 1 = 0,5 meter
So, Mmax = 3 (0,5 + 0,67 . 2,8) =7.128 Ton m
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Reese and Matlock’s method To calculate deflection, moment and soil reaction in Reese and Matlock’s method is need the value of Fy, Fm, and Fx, that can be determined from the
graphic given at manual pondasi tiang book.
Z value can be determined by calculate the depth (x) over stiffness (T). the calculation can be shown as table below.
Driven pile
x 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0
z 0.6 1.3 1.9 2.6 3.2 3.9 4.5 5.2 5.8 6.5 7.1 7.8 8.4 9.0 9.7 10.3 11.0 11.6 12.3 12.9
For K-5 = 30 Ton, the length of pile has determined in 7 meter depth and 6 piles (3 x 2) with Qallowable 6,82 ton each, as consideration of load that hold
and the characteristic of long pile elastic. so it takes z = from 0.6 to 4,5. For K-6 = 25 Ton, the length has determined in 7 meter depth with 4 piles (2 x 2).
From the deflection (Fy), moment (Fm), and soil reaction (Fp) graphic, it can be determined the value of Fy, Fm, and Fp coefficient to calculate deflection
(yx), moment (Mx), and soil reaction (Px). The value of slope (Sx) and shear force (Vx) is 0 in case of piles head is fixed.
driven pile Hu = 3 ton Hu = 2.5 ton
z fy fm fp Yf Mf Pf Yf Mf Pf
0.646383 0.8 -0.4 0.5 0.020055 -1.85649 0.969574 0.016712 -1.54707 0.807978
1.292765 0.5 0.1 0.7 0.012534 0.464121 1.357404 0.010445 0.386768 1.13117
1.939148 0.2 0.25 0.5 0.005014 1.160303 0.969574 0.004178 0.96692 0.807978
2.585531 0.4 0.22 0 0.010027 1.021067 0 0.008356 0.850889 0
3.231913 0 1.5 -0.5 0 6.961821 -0.96957 0 5.801517 -0.80798
3.878296 -0.2 0.5 -0.1 -0.00501 2.320607 -0.19391 -0.00418 1.933839 -0.1616
4.524678 -0.1 0 -0.08 -0.00251 0 -0.15513 -0.00209 0 -0.12928
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Bore Pile
x 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0
z 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.1 3.5 3.9 4.3 4.7 5.1 5.5 5.9 6.3 6.7 7.1 7.5 7.9
Bore pile at K-5 = 30, the length is 9 meter depth and 3 piles with Q allowable 10,21 Ton each. K-6 = 25 ton used this pile configuration too. So the z takes
from 0,4 to 3,5 m.
bore pile 3 ton 2.5 ton
z fy fm fp Yf Mf Pf Yf Mf Pf
0.39 0.9 -0.7 0.2 0.008369 -5.33433 0.236206 0.018801 -4.44527 0.196838
0.79 0.8 -0.2 0.6 0.007439 -1.52409 0.708618 0.016712 -1.27008 0.590515
1.18 0.6 -0.05 0.62 0.005579 -0.38102 0.732238 0.012534 -0.31752 0.610199
1.57 0.45 0.18 0.6 0.004184 1.371685 0.708618 0.009401 1.143071 0.590515
1.97 0.2 0.25 0.4 0.00186 1.905118 0.472412 0.004178 1.587598 0.393677
2.36 0.15 0.24 0.3 0.001395 1.828913 0.354309 0.003134 1.524094 0.295257
2.76 0.05 0.2 0.1 0.000465 1.524094 0.118103 0.001045 1.270078 0.098419
3.15 0 0.15 0 0 1.143071 0 0 0.952559 0
3.54 -0.2 0.1 -0.4 -0.00186 0.762047 -0.47241 -0.00418 0.635039 -0.39368
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
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Gian Wilda Satria / 2008410166 Foundation Engginering 2nd Assignment
Conclusion
In case of the pile diameter or side is small and it’s not so deep, it makes settlement isn’t have
significant effect in the design. So it can be ignored in design.
Shaft resistance is depend on friction resistance and base resistance of pile. Base resistance is
depend on dimension of base pile, bigger dimension, bigger resistance can get. Friction resistance is
also depend on dimension of the pile and rough of shaft. Soil properties also give a big deal of
bearing resistance that can hold load of building above.
The load of column that transferred to foundation can be hold by piles that configure efficiently to
give optimum strength that can give by pile.
From calculation between driven pile and bore pile above can be shown that bore pile with higher
dimension give more strength to bear the load from building, even it base doesn’t modify with bell.
It means that the bigger dimension applied, more strength can give.