i
BORED PILE FOUNDATION DESIGN AT PROJECT OF
APARTEMEN TAMAN MELATI YOGYAKARTA
submitted to fulfill the requirement
to achieve Bachelor Degree of Civil Engineering Faculty of Engineering
by:
SHOLICHATUN NISA
D 100 164 008
CIVIL ENGINEERING DEPARTMENT
ENGINEERING FACULTY
UNIVERSITAS MUHAMMADIYAH SURAKARTA
2020
i
VALIDATION SHEET
BORED PILE FOUNDATION DESIGN AT PROJECT OF
APARTEMEN TAMAN MELATI YOGYAKARTA
SCIENTIFIC PUBLICATION
by:
SHOLICHATUN NISA
D 100 164 008
Has been checked and approved to be tested by:
Supervisor Lecturer
Anto Budi Listyawan, S.T., M.Sc.
NIDN. 0622036101
ii
VALIDATION SHEET
BORED PILE FOUNDATION DESIGN AT PROJECT OF
APARTEMEN TAMAN MELATI YOGYAKARTA
by:
SHOLICHATUN NISA
D 100 164 008
Has been defended at Final Examination Test before the Examiners
Engineering Faculty
Universitas Muhammadiyah Surakarta
On the day 2020
and declared qualified
Board of Examiners:
1. Anto Budi Listyawan, S.T., M.Sc. (NIDN. 0622036101) (…………………..)
(Chair of Examiner)
2. Agus Susanto,S.T.,M.T. (NIDN. 0611087101) (…………………..)
(1st Member of Examiner Board)
3. Ir. Renaningsih, M.T. (NIDN. 0624096301) (…………………..)
(2nd Member of Examiner Board)
Dean,
Ir. Sri Sunarjono, M.T., PhD., IPM
NIDN. 0630126302
iii
STATEMENT
I hereby declare that in this publication's text no work has ever been submitted to obtain a
degree at a tertiary institution and to the best of my knowledge there are no works or opinions
that have been written or published by others, unless written in reference to the text and
mentioned in References.
If later there is proven untruth in my statement above, then I will take full responsibility.
Surakarta, 2020
Writer
SHOLICHATUN NISA
D 100 164 008
1
BORED PILE FOUNDATION DESIGN AT PROJECT OF
APARTEMEN TAMAN MELATI YOGYAKARTA
Abstrak
Apartemen Taman Melati Yogyakarta berlokasi di Jalan Prof. Dr. Sardjito No. 66, Terban
Gondokusuman, Kota Yogyakarta. Proyek Apartemen Taman Melati Yogyakarta adalah sebuah
bangunan baru yang terdiri dari 5 lantai. Pekerjaan fondasi adalah salah satu pekerjaan penting
dalam konstruksi, karena fondasi memiliki fungsi mendukung dan menahan semua beban yang
bekerja dari struktur atas. Lapisan tanah di Proyek Apartemen Taman Melati Yogyakarta adalah
pasir dengan kedalaman dari ± 0 sampai 40 meter. Muka air tanah berada di kedalaman 14,50
meter. Fondasi yang digunakan dalam tugas akhir ini adalah pondasi tiang bor dengan diameter
0,4 m dan 0,7 m dengan kedalaman 12 m. Penelitian ini bertujuan untuk menganalisis beban
yang bekerja dari struktur atas, menghitung jumlah kapasitas daya dukung dari pondasi tiang
bor, dan menganalisis dimensi serta merencanakan tulangan dari pile cap dan pondasi tiang bor.
Metode penelitian yang digunakan adalah pengumpulan data dan studi pustaka. Berdasarkan
hasil analisis struktur atas menggunakan program SAP2000 memiliki beban aksial terbesar (P)
sebesar 3779,897 kN. Kapasitas daya dukung pondasi tiang bor menggunakan metode Reese and
Wright (1977) dengan diameter 0,4 m mendapatkan nilai 2108,134 kN dan diameter 0,7 m
mendapatkan nilai 4338,414 kN. Pondasi tiang bor untuk diameter 0,4 m membutuhkan 4 tiang
dimana kapasitas daya dukung kelompok adalah 3871,555 kN dan untuk diameter 0,7 m
membutuhkan 2 tiang dimana kapasitas daya dukung kelompok adalah 4289,926 kN. Pile cap
berdasarkan SNI 2847-2013 memiliki dimensi 2,20 x 2,20 x 1,30 m dan 3,80 x 1,70 x 1,30 m.
Tulangan pile cap arah x dan arah y didapatkan D25-110 untuk tulangan utama dan D19-110
untuk tulangan sengkang. Analisis pondasi tiang bor menggunakan program SP column untuk
diameter 0,4 m membutuhkan tulangan utama 6-D20 dengan area tulangan adalah 1884 mm2 dan
untuk diameter 0,7 m membutuhkan tulangan utama 16-D20 dengan area tulangan adalah 5024
mm2. Analisis tulangan sengkang berdasarkan SNI 2847-2013 untuk diameter 0,4 m
menggunakan Ø10-150 dan untuk diameter 0,7 m menggunakan Ø10-300.
Kata Kunci : Fondasi, Kapasitas Dukung, Pile Cap, Tiang Bor, Tulangan.
Abstract
Taman Melati Yogyakarta Apartment Project located on Prof. Dr. Sardjito No. 66 road, Terban
Gondokusuman, Yogyakarta City. Taman Melati Yogyakarta Apartment Project is a new
structure consisting of a 5-story building. Foundation work is one of the most important jobs in
a construction, because the foundation has the function of bearing and holding all the loads that
work on it. The soil layer in Taman Melati Yogyakarta Apartment Project is sand soil with a
depth of ± 0 to 40 meters. Groundwater level at that location is 14,50 meters. The foundation
used in this final project is the bored pile foundation with a diameter of 0,4 m and 0,7 m at a
depth of 12 m. The objective of the project are analyze the load of the upper structure that works,
calculating the amount of bearing capacity of the bored pile foundation, and analyze the
dimensions and reinforcement design of pile cap and bored pile foundation. Research method
that used are collecting data and study literature. Based on the results of the upper structure
analysis used SAP2000 program get value of largest axial load (P) 3779,897 kN. The bearing
capacity of a single bored pile used Reese and Wright method (1977) with diameter of 0,4 m
obtained value of 2108,134 kN and diameter of 0,7 m obtained value of 4338,414 kN. The pile
for diameter 0,4 m needed are 4 piles where the bearing capacity of group piles is 3871,555 kN
and for diameter 0,7 m needed are 2 piles where the bearing capacity of group piles is 4289,926
2
kN. The pile cap based on SNI 2847-2013 used dimension 2,20 x 2,20 x 1,30 m and 3,80 x 1,70
x 1,30 m. The x-direction and y-direction obtained D25-110 for the main reinforcement and
D19-110 for the stirrup reinforcement. The bored pile analysis used SP column program, for
diameter 0,4 m need main reinforcement 6-D20 with the reinforcement area is 1884 mm2 and for
diameter 0,7 m need main reinforcement 16-D20 with the reinforcement area is 5024 mm2. The
stirrup reinforcement analysis based on SNI 2847-2013 for diameter 0,4 m obtained Ø10-150
and for diameter 0,7 m obtained Ø10-300.
Keywords: Bearing Capacity, Bored Pile, Foundation, Pile Cap, Reinforcement.
1. INTRODUCTION
Taman Melati Yogyakarta Apartment Project located on Prof. Dr. Sardjito No. 66 road,
Terban Gondokusuman, Yogyakarta City. Taman Melati Yogyakarta Apartment Project is a new
structure consisting of a 5-story building. Foundation work is one of the most important jobs in
a construction, because the foundation has the function of bearing and holding all the loads that
work on it, namely the structure load on top, then the stresses that occur due to the structure load
will be channeled into the hard soil layer can carry the burden of construction.
In addition, to investigate the presence of hard soil layers, soil investigations such as the
Standard Penetration Test (SPT) are carried out. The soil layer in Taman Melati Yogyakarta
Apartment Project is sand soil with a depth of ± 0 to 40 meters. Groundwater level at that location
is 14,50 meters.
The foundation used in this final project is the bored pile foundation with a diameter of 0,4
meters and 0,7 meters at a depth of 12 meters. The reasons for planning the bottom structure
using a bored pile foundation are as follows (Halibu, 2015):
1. The building is in a densely populated area that is not possible to do the erection with
heavy equipment because it will cause noise and vibration that will affect the surrounding
structures,
2. The depth of the pile can be done using bored pile foundation,
3. Pile foundations that require a lot of heavy equipment may have to reconsider if
implemented on a highway in a very crowded city because it will cause congestion,
4. Can be installed through rocks (gravel).
The objective of the project are analyze the load of the upper structure that works,
calculating the amount of bearing capacity of the bored pile foundation, and analyze the
dimensions and reinforcement design of pile cap and bored pile foundation.
3
2. METHOD
The data obtained from the Taman Melati Yogyakarta Apartment Project is used to
achieve the objectives of this study. To design the bored pile foundation the first stage namely
the study of literature, study literature is intended to find information related to the same research
topic. The second stage namely data collection, like soil investigation data (N-SPT), building
structure data, and earthquake zone data. The third stage is calculate and analysis of upper
structure load using SAP2000 program. After get the result, determine the largest axial load that
occurs in the column. The fourth step is to analyze the bearing capacity of the single bored pile
foundation using the Reese and Wright Method. And then analyze how many piles needed, the
efficiency of the bored pile groups, the maximum load, and the bearing capacity of the bored
pile group. The fifth step is calculate the reinforcement on the pile cap and the bored pile
foundation based SNI 1726-2012. The last stage consists of conclusion and recommendation.
3. RESULT AND DISCUSSION
The planning of the bored pile foundation at the Taman Melati Yogyakarta Apartment
Project includes the calculation of the upper structure load, the calculation of soil the bearing
,and the reinforcement of the pile cap and bored pile.
3.1 Load Analysis
Making a building model using SAP 2000 version 15 program shown in Figure 2 then entering
all existing loads such as dead loads, live loads, and earthquake loads then run analysis so that
the largest axial load located on frame 161 the value can be seen as follow:
P (axial load ) = 3779,897 kN
Vx (shear force) = -428,747 kN , Vy (shear force) = -113,259 kN
Mx (moment) = -88,8694 kNm , My (moment) = -417,0733 kNm
4
Figure 1. Modeling 3D structure use SAP 2000 program
Figure 2. Location of the maximum axial load on the 1st floor
3.2 Soil Bearing Capacity Analysis
1. Calculation of Bearing Capacity based N-SPT Data
a) Ultimate Friction Resistance (Qs)
According to Reese and Wright method (1977), ultimate end resistance (Qs) for the
bored pile can be calculated as follow:
Medium sand at depth ± 0 m until 4,50 m
Diameter (d) = 0,4 m
As = π.d.t (1)
= π.0,4.(4,50-0)
= 5,655 m2
Naverage = 45+35
2
5
= 38,5
N60 =𝐸𝑚 𝑥 𝐶𝐵 𝑥 𝐶𝑠 𝑥 𝐶𝑅
0,6 x Naverage (2)
= 0,8 𝑥 1,15 𝑥 1 𝑥 1
0,6 x 38,50
= 59,033
fs = 𝑁60−53
450+ 1,6 (𝑡𝑠𝑓) (3)
= 59,033−53
450+ 1,6 (105,6)
= 168,973 kN/m2
Control of friction resistance maksimum (fs ≤ 107 kN/m2)
fs = 168,973 kN/m2 ≥ 107 kN/m2 (not ok)
So, fs used 107 kN/m2
Qs = As.fs (4)
= 5,655.107
= 605,071 kN
Fine sand at depth 4,50 m until 7,50 m
Diameter (d) = 0,4 m
As = π.d.t
= π.0,4.(7,5-4,5)
= 3,770 m2
Naverage = 39
N60 =𝐸𝑚 𝑥 𝐶𝐵 𝑥 𝐶𝑠 𝑥 𝐶𝑅
0,6 x Naverage
= 0,8 𝑥 1,15 𝑥 1 𝑥 1
0,6 x 39
= 59,8
fs = 𝑁60−53
450+ 1,6 (𝑡𝑠𝑓)
= 59,8−53
450+ 1,6 (105,6)
= 168,975 kN/m2
Control of friction resistance maksimum (fs ≤ 107 kN/m2)
fs = 168,975 kN/m2 ≥ 107 kN/m2 (not ok)
So, fs used 107 kN/m2
Qs = As.fs
= 3,770. 107
= 403,380 kN
6
Dense sand at depth 7,50 m until 10 m
Diameter (d) = 0,4 m
As = π.d.t
= π.0,4.(10-7,5)
= 3,142 m2
Naverage = 43
N60 =𝐸𝑚 𝑥 𝐶𝐵 𝑥 𝐶𝑠 𝑥 𝐶𝑅
0,6 x Naverage
= 0,8 𝑥 1,15 𝑥 1 𝑥 1
0,6 x 43
= 65,933
fs = 𝑁60−53
450+ 1,6 (𝑡𝑠𝑓)
= 65,933−53
450+ 1,6 (105,6)
= 168,989 kN/m2
Control of friction resistance maksimum (fs ≤ 107 kN/m2)
fs = 168,989 kN/m2 ≥ 107 kN/m2 (not ok)
So, fs used 107 kN/m2
Qs = As.fs
= 3,142. 107
= 336,150 kN
Fine sand at depth 10 m until 12 m
Diameter (d) = 0,4 m
As = π.d.t
= π.0,4.(12-10)
= 2,513 m2
Naverage = 60
N60 =𝐸𝑚 𝑥 𝐶𝐵 𝑥 𝐶𝑠 𝑥 𝐶𝑅
0,6 x Naverage
= 0,8 𝑥 1,15 𝑥 1 𝑥 1
0,6 x 60
= 92
fs = 𝑁60−53
450+ 1,6 (𝑡𝑠𝑓)
= 92−53
450+ 1,6 (105,6)
= 169,047 kN/m2
Control of friction resistance maksimum (fs ≤ 107 kN/m2)
7
fs = 169,047 kN/m2 ≥ 107 kN/m2 (not ok)
So, fs used 107 kN/m2
Qs = As.fs
= 2,513.107
= 268,920 kN
The total of Qs = 605,071 + 403,380 + 336,150 + 268,920
= 1613,522 kN
b) Ultimate End Resistance (Qb)
According to Reese and Wright method (1977), ultimate end resistance (Qb) for the
bored pile can be calculated as follow:
Diameter (d) = 0,4 m
Ab = ¼.π.d2 (5)
= ¼.π.(0,4)2
= 0,126 m2
8D = 8.0,4
= 3,2 m
4D = 4.0,4
= 1,6 m
N-SPT = (1,2𝑥43)+(2𝑥60)+(1,6𝑥55,5)
4,8
= 54,25
N60 = 𝐸𝑚 𝑥 𝐶𝐵 𝑥 𝐶𝑠 𝑥 𝐶𝑅
0,6 x N-SPT
= 0,8 𝑥 1,15 𝑥 1 𝑥 1
0,6 x 54,25
= 83,183
fb = 40.(tsf) (6)
= 40.105,6
= 4224 kN/m2
Control of friction resistance maksimum (fb ≤ 10700 kN/m2)
fb = 4224 kN/m2 ≤ 10700 kN/m2 (ok)
Qb = Ab.fb (7)
= 0,126. 4224
= 530,803 kN
8
c) Ultimate Bearing Capacity (Qu)
Diameter (d) = 0,4 m
Weight of bored pile foundation :
Wp = ¼.π.d2.L.γconcrete (8)
= ¼.π.0,42.12.24
= 36,191 kN
Qu = Qb + Qs – Wp (9)
= 530,803 + 1613,522 – 36,191
= 2108,134 kN
d) Allowable Bearing Capacity (Qa)
Qa =𝑄𝑢
𝑆𝐹 (10)
= 2108,134
2
= 1054,067 kN
e) Number of Piles Needed
The number of piles needed is calculated by dividing the axial forces that occur with
the bearing capacity of the piles, the formula as follow (Harianti and Pamungkas,
2013).
np =𝑃
𝑄𝑎 (11)
= 3779,897
1054,067
= 3,568 rounded into 4 pile
f) Pile Group Efficiency (Eg)
The data needed is as follows:
Diameter (d) = 0,4 m
s = 3.d (distance of each pile in groups/spacing used 3.d (Wulandari, 2019))
= 3.0,4
= 1,2 m
m = 2
n = 2
θ = arctg 0,4/1,2
= 18,434o
The formula for estimate the pile group efficiency calculate used formula as follow
(Listyawan, A.B., et all. 2017):
9
Eg = 1 − 𝜃(𝑛−1)𝑚+(𝑚−1)𝑛
90𝑚𝑛 (12)
= 1 − 18,434(2−1)2+(2−1)2
90.2.2
= 0,795
2. Bearing Capacity of Pile Groups
Qg = Eg x np x Qa (13)
= 0,795 x 4 x 1217,164
= 3871,555 kN
Qg = 3871,555 kN > Pu = 3779,897 (Ok)
3. Maximum Load on Piles Group
Figure of the location pile and force that work on the pile can be seen on the Figure 4 as
follow:
Figure 3. Load that Work
Pu = Axial Load + Pile Cap Load
= 3779,897 + (1,3.2,2.2,2.24)
= 3779,897 + 151,008
= 3930,905 kN
Mx = -88,8694 kNm
My = -417,0733 kNm
Xmaks = 0,6 m
Ymaks = 0,6 m
∑X2= 2x2x0,62 = 1,44 m2
∑Y2= 2x2x0,62 = 1,44 m2
nx = 2 pile
10
ny = 2 pile
n = 4 pile
To find the maximum and minimum load on these pile groups can be seen through the
following equation (Harianti and Pamungkas, 2013):
Pmin = 𝑃𝑢
𝑛+
𝑀𝑦.𝑋 𝑚𝑎𝑥
𝑛𝑦.∑X2 +𝑀𝑥.𝑌 𝑚𝑎𝑥
𝑛𝑥.∑Y2 (14)
= 3930.905
4+
-417,0733 .0,6
2.1,44+
-88,8694 .0,6
2.1,44
= 877,322 kN/pile > 0 (Safe)
Pmax = 𝑃𝑢
𝑛−
𝑀𝑦.𝑋 𝑚𝑎𝑥
𝑛𝑦.∑X2 −𝑀𝑥.𝑌 𝑚𝑎𝑥
𝑛𝑥.∑Y2 (15)
= 3930.905
4-
-417,0733 .0,6
2.1,44-
-88,8694 .0,6
2.1,44
= 1088.131 kN/pile < Qa = 1217,164 kN/pile (Safe)
For the bored pile foundation diameter 0,7 m used same formula like above. The result
of calculation soil bearing capacity can be seen Table 1.
Table 1. Recapitulation of the Soil Bearing Capacity
Diameter 0,4 m 0,7 m
Qs 1613,522 kN 2863,663 kN
Qb 530,803 kN 1625,586 kN
Qu 2108,134 kN 4338,414 kN
Qa 1054,067 kN 2169,207 kN
np 4 piles 2 piles
Eg 0,795 0,898
Qg 3871,555 kN 4289,926 kN
Pmax 1088,131 kN/pile 2189,331 kN/pile
3.3 Pile Cap Design
1. Review of the Shear
a) Overview of 1-way Shear Stress
The location of 1-way shear stress from pile cap design can be seen at the Figure 5 as
follow:
11
Figure 4. 1-way Shear Stress
1-way shear stress occurs on one side only, then it is calculated on the bearing
capacity of the largest bored pile on one side.
Data Analysis :
Pu = 3779,897 kN
bcolumn = 300 mm = 0,30 m
hcolumn = 500 mm = 0,50 m
B = 2200 mm = 2,20 m
L = 2200 mm = 2,20 m
hpile cap = 1300 mm = 1,30 m
Dreinforcement = 22 mm
σ = P/A = 3779,897 / (2,20 . 2,20) = 780,970 kN/m2
Sb = 75 mm (Based SNI 03-2847-2002 page 41 clause 9.7(1))
ds = 75 + 22/2 = 86 mm
dpile cap = h – ds = 1300 – 86 = 1214 mm = 1,214 m
G’ = L-(L/2+bk/2+d) = 2,20 - (2,2/2 + 0,3/2 + 1,214) = -0,264 m
Shear stress, Vu :
kN 453,585- 0,264-2,20. . 780,970 'G.L.u
V (18)
Shear force, Vc :
Vc = 0,17. dBcf ..' (19)
= 0,17. 05,29 . 2200.1214
= 2447165,6 N = 2447,166 kN
Control Vu = -453,585 kN < .Vc = 0,75. 2447,166 = 1853,374 kN (Safe)
So the pile cap construction for the bored pile foundation is safe against 1-way shear
stresses.
12
b) Overview of 2-way Shear Stress
The location of 2-way shear stress from pile cap design can be seen at the Figure 6 as
follow:
Figure 5. 2-way Shear Stress
The calculation steps are carried out as follows:
P = 3779,897 kN
bcolumn = 300 mm = 0,30 m
hcolumn = 500 mm = 0,50 m
B = 2200 mm = 2,20 m
L = 2200 mm = 2,20 m
hpile cap = 1300 mm = 1,30 m
Dreinforcement = 22 mm
σ = P/A = 3779,897 / (2,20 . 2,20) = 780,970 kN/m2
Sb = 75 mm (Based SNI 03-2847-2002 page 41 clause 9.7(1))
ds = 75 + 22/2 = 86 mm
dpile cap = h – ds = 1300 – 86 = 1214 mm = 1,214 m
B’ = bk+2.(1/2).d = 300+2.(1/2). 1214 = 1514 mm = 1,514 m
L’ = hk+2.(1/2).d = 500+2.(1/2). 1214 = 1714 mm = 1,714 m
βc = Ratio from the long side to the short side (m)
= bcolumn / hcolumn
= 300/500 = 0,60
bo = )dh()db(.2 kk
= 2.{(300+1214)+(500+1214)}=6456 mm
13
αs = constants whose value depends on the column in the building
= 40 for the foundation with the location of the column in the building
= 30 for foundations with columns on the edge of buildings
= 20 for foundations with columns in the corner of the building
Stress that occurs in the Vu (all reactions that occur in the x direction and y direction).
kN 1753,282 1,714) . 1,514 - 2,20 . ,20780,970.(2)B'.L'-B.L.(uV (20)
Calculate the smallest shear stress that can be held by Vc, namely:
Vc = 0,17. .β
21
c
cf ' .bo.d (21)
= 0,17. 1214. 6456.05,29.6,0
21
= 31119048 N = 31119,048 kN
Vc = .12
1.
0
.2
b
dscf ' .bo.d (22)
= 1214.6456.05,29.12
1.
6456
1214.402
= 33518753 N =33518,753 kN
Vc = 1/3. cf ' bo.d (23)
= 1/3. 05,29 .6456. 1214 = 14081016,98 N = 14081,016 kN
The smallest used ,Vc = 14081,016 kN
Control Vu = 1753,282 kN < .Vc = 0,75. 14081,016 kN = 10560,763 kN (Safe)
So the pile cap construction for the bored pile foundation is safe against 2-way shear
stress.
2. Reinforcement of Pile Cap
The calculation of pile cap reinforcement based on the formula as follow (Wulandari,
2019):
a) Calculation of reinforcement x-direction
Data analysis :
Pmax = 1088,131 kN
bcolumn = 300 mm = 0,30 m
hcolumn = 500 mm = 0,50 m
B = 2200 mm = 2,20 m
14
L = 2200 mm = 2,20 m
hpoer = 1300 mm = 1,30 m
Dmain reinforcement = 25 mm
Dstirrup reinforcement = 19 mm
ds = 75 + 25/2 = 87,5 mm
dpoer = hpoer – ds = 1300 – 87,5 = 1212,5 mm = 1,2125 m
Dpile = 400 mm = 0,4 m
fc’ = 29,05 MPa
fy = 390 MPa
ϒconcrete = 24 kN/m3
Xmaks = 0,6 m
∑X2 = 2.2.0,62 = 1,44 m2
β1 = 0,85-0,05.(𝑓′𝑐−28
7) = 0,85-0,05.(
29,05−28
7)
= 0,843
Pile Cap Weight (qu) = 2,20 m . 2,20 m . 24 kN/m3
= 116,16 kN/m
The moment that works on the Pile Cap
Mu = (nx.Pmax.Xmax)-(1/2 . qu . ∑X2) (24)
= (2 . 1088,131 . 0,6) – (1/2 . 116,16 . 1,44)
= 1222,122 kN.m
Mu every 1 meter = 𝑀𝑢
𝐿 =
1222,122 𝑘𝑁.𝑚
2,2 𝑚 = 555,510 kN.m/m
Calculate the bearing factor K (with b = 1000 mm) :
MPa 0,4722
212,50,9.1000.1
6.10 555,510
2φ.b.d
uMK (25)
Kmaks = 382,5.𝛽1.(600+𝑓𝑦−225.𝛽1).𝑓′𝑐
(600+𝑓𝑦)2 = 382,5.0,843.(600+390−225.0,843).29,05
(600+390)2
= 7,645 Mpa
So, K < Kmaks (qualify)
ω = 0,85 - √0,72 − 0,17 𝐾
𝑓′𝑐 (26)
= 0,85 - √0,72 − 0,17 0,472
29,05
= 0,018
15
ρ = ω - 𝑓′𝑐
𝑓𝑦 (27)
= 0,018 - 29,05
390
= 0,00133
ρb = 0,85.𝑓′𝑐
𝑓𝑦 . β1 . (
600
600+𝑓𝑦) (28)
= 0,85.29,05
390 . 0,843 . (
600
600+390)
= 0,03232
ρmax = 0,75 . ρb (29)
= 0,75 . 0,03232
= 0,02425
ρmin = 1,4
𝑓𝑦 (30)
= 1,4
390
= 0,00359
Control of strain reinforcement ratio :
ρ = 0,00133 < ρmin = 0,00359 < ρmax = 0,02425
So, strain reinforcement ratio that used is ρmin = 0,00359
As,u = ρ.b.d (31)
= 0,00359 . 1000 . 1212,5
= 4352,564 mm2
Spacing main reinforcement (used D25) :
mm 112,778 4352,564
.10002
.251/4.
us,A
.S2
.D1/4.
s (31)
mm 3900 130033 hpoers (32)
mm 450s (33)
Choose smallest value, s = 112,778 mm = 110 mm
Reinforcement area = (Ok)uAs2
mm490,4462110
.10002
.251/4.
s
.S2
.D1/4.
(34)
So, main reinforcement used D25-110 = 4462,490 mm2
Control moment capacity of moment ultimite
Calculate the height of the compressed concrete stress block:
a = 𝐴𝑠 .𝑓𝑦
0,85 . 𝑓′𝑐 . 𝑏 (35)
= 4462,490 . 390
0,85 . 29,05 . 1000 = 70,482 mm
16
Mn = As . fy . (d - 𝑎
2) (36)
= 4462,490 . 10-6 . 390 . (1212,5 - 70,482
2)
= 2048,868 kN.m
ØMn = 0,85. 2048,868 kN.m
= 1741,537 kN.m > Mu = 1222,122 kN.m (Ok)
Calculation of stirrup reinforcement (top):
Asb = 20%.As,u (37)
= 20%.4352,564
= 870,513 mm2
Asb,min = {0,002-(fy-350)/350000}.b.h (38)
= {0,002-(390-350)/350000}.1000.1300
= 2451,429 mm2
Choose biggest value, Asb,u =2451,429 mm2
Spacing stirrup reinforcement (used D19) :
mm 115,659 2451,429
.10002
.191/4.
us,A
.S2
.D1/4.
s
mm 6500 1300.5.5 hpoers (39)
mm 450s
Choose smallest value, s = 115,659 mm = 110 mm
Reinforcement area = (Ok)usb,
A2
mm534,2577110
.10002
.191/4.
s
.S2
.D1/4.
So, stirrup reinforcement used D19-110= 2577,534 mm2
b) Calculation of reinforcement y-direction
Data analysis :
Pmax = 1088,131 kN
bcolumn = 300 mm = 0,30 m
hcolumn = 500 mm = 0,50 m
B = 2200 mm = 2,20 m
L = 2200 mm = 2,20 m
hpoer = 1300 mm = 1,30 m
Dmain reinforcement = 25 mm
Dstirrup reinforcement = 19 mm
ds = 75 + 25/2 = 87,5 mm
17
dpoer = hpoer – ds = 1300 – 87,5 = 1212,5 mm = 1,2125 m
Dpile = 400 mm = 0,4 m
fc’ = 29,05 MPa
fy = 390 MPa
ϒconcrete = 24 kN/m3
Ymaks = 0,6 m
∑Y2 = 2.2.0,62 = 1,44 m2
β1 = 0,85-0,05.(𝑓′𝑐−28
7) = 0,85-0,05.(
29,05−28
7)
= 0,843
Pile Cap Weight (qu) = 2,20 m . 2,20 m . 24 kN/m3
= 116,16 kN/m
The moment that works on the Pile Cap
Mu = (ny.Pmax.Ymax)-(1/2 . qu . ∑Y2)
= (2 . 1088,131 . 0,6) – (1/2 . 116,16 . 1,44)
= 1222,122 kN.m
Mu every 1 meter = 𝑀𝑢
𝐵 =
1222,122 𝑘𝑁.𝑚
2,2 𝑚 = 555,510 kN.m/m
Calculate the bearing factor K (with b = 1000 mm) :
MPa 0,4722
212,50,9.1000.1
6.10 555,510
2φ.b.d
uMK
Kmaks = 382,5.𝛽1.(600+𝑓𝑦−225.𝛽1).𝑓′𝑐
(600+𝑓𝑦)2 = 382,5.0,843.(600+390−225.0,843).29,05
(600+390)2
= 7,645 Mpa
So, K < Kmaks (qualify)
ω = 0,85 - √0,72 − 0,17 𝐾
𝑓′𝑐
= 0,85 - √0,72 − 0,17 0,472
29,05
= 0,018
ρ = ω - 𝑓′𝑐
𝑓𝑦
= 0,018 - 29,05
390
= 0,00133
ρb = 0,85.𝑓′𝑐
𝑓𝑦 . β1 . (
600
600+𝑓𝑦)
18
= 0,85.29,05
390 . 0,843 . (
600
600+390)
= 0,03232
ρmax = 0,75 . ρb
= 0,75 . 0,03232
= 0,02425
ρmin = 1,4
𝑓𝑦
= 1,4
390
= 0,00359
Control of strain reinforcement ratio :
ρ = 0,00133 < ρmin = 0,00359 < ρmax = 0,02425
So, strain reinforcement ratio that used is ρmin = 0,00359
As,u = ρ.b.d
= 0,00359 . 1000 . 1212,5
= 4352,564 mm2
Spacing main reinforcement (used D25) :
mm 112,778 4352,564
.10002
.251/4.
us,A
.S2
.D1/4.
s
mm 3900 130033 hpoers
mm 450s
Choose smallest value, s = 112,778 mm = 110 mm
Reinforcement area = (Ok)uAs2
mm490,4462110
.10002
.251/4.
s
.S2
.D1/4.
So, main reinforcement used D25-110 = 4462,490 mm2
Control moment capacity of moment ultimit
Calculate the height of the compressed concrete stress block:
a = 𝐴𝑠 .𝑓𝑦
0,85 . 𝑓′𝑐 . 𝑏
= 4462,490 . 390
0,85 . 29,05 . 1000
= 70,482 mm
Mn = As . fy . (d - 𝑎
2)
= 4462,490 . 10-6 . 390 . (1212,5 - 70,482
2)
= 2048,868 kN.m
19
ØMn = 0,85. 2048,868 kN.m
= 1741,537 kN.m > Mu = 1222,122 kN.m (Ok)
Calculation of stirrup reinforcement (top):
Asb = 20%.As,u
= 20%.4352,564
= 870,513 mm2
Asb,min = {0,002-(fy-350)/350000}.b.h
= {0,002-(390-350)/350000}.1000.1300
= 2451,429 mm2
Choose biggest value, Asb,u =2451,429 mm2
Spacing stirrup reinforcement (used D19) :
mm 115,659 2451,429
.10002
.191/4.
us,A
.S2
.D1/4.
s
mm 6500 1300.5.5 hpoers
mm 450s
Choose smallest value, s = 115,659 mm = 110 mm
Reinforcement area = (Ok)usb,
A2
mm534,2577110
.10002
.191/4.
s
.S2
.D1/4.
So, stirrup reinforcement used D19-110= 2577,534 mm2
c) Calculation of the Length of the Distribution of Reinforcement Stress (ld)
The length of the distribution of reinforcement stress (ld), calculated using the following
formula (Asroni, 2014):
ld = bd
bd
trKb
c
set
cf
yf.
...
'.1,1
and ld should ≥ 300 mm
With :
Ψt =1,0 (fresh concrete under reinforcement is only 75 mm <300 mm)
Ψe = 1,0 (reinforcement not epoxy coated)
Ψs = 1,0 (used reinforcement D22).
db = 25 mm
λ = 1,0 (used normal concrete).
cb = 75 mm (cover of concrete, Sb = 75 mm).
Ktr = 0 (for simplification: Arcticle 12.2.4 SNI 03-2847-2013)
20
(cb+Ktr)/db = (75+0)/25 = 3,00 > 2,5 → used (cb+Ktr)/db = 2,5.
ld = 25.5,2
1.1.1.
05,29.1.1,1
390 = 657,808 mm > 300 mm. (40)
Used ld = 657,808 mm = 0,658 m.
Available length,
lt = L/2-bk/2–75 = 2200/2–300/2–75 = 875 mm = 0,875 m. (41)
Because lt = 0,875 m > ld = 0, 658 m, then the width of the pile cap is enough.
For the bored pile foundation diameter 0,7 m used same formula like above. The result
of calculation pile cap reinforcement can be seen Table 2.
Table 2. Recapitulation of the Pile Cap Reinforcement
Diameter 0,4 m 0,7 m
Main Reinforcement
of x-direction
D25 – 110 D25 – 110
Main Reinforcement
of y-direction
D25 – 110 D25 – 110
Stirrup Reinforcement
of x-direction
D19 – 110 D19 – 110
Stirrup Reinforcement
of y-direction
D19 – 110 D19 – 110
Based on analysis result of pile cap reinforcement, so we can draw the reinforcement like
Figure 7, Figure 8, Figure 9, and Figure 10 as follow:
Figure 6. Detail Reinforcement for Pile Cap 2,20x2,20 m
21
Figure 7. Cross Section I-I for Pile Cap 2,20 x 2,20 m
Figure 8. Detail Reinforcement for Pile Cap 3,80 x 1,70 m
22
Figure 9. Cross Section I-I for Pile Cap 3,80x1,70 m
3.4 Bored Pile Design
1. Calculation of Main Reinforcement
Calculation of bored pile reinforcement calculated like a column using SP Column
program. The data that used as follow:
P (axial load ) = 1224,358 kN
V (shear force) = 72,083 kN
Mx (moment) = 3,166 kNm
My (moment) = -3,025 kNm
The values of the above forces are entered into the SP Column program and the
reinforcement ratio value is 1,50%. This ratio is sufficient because the column
reinforcement ratio is between 1% to 4%. After getting the ratio, the reinforcement of the
bored pile is obtained: 6 – D20 (As = 1884 mm2).
Figure 10. Diagram Interaction for Bored Pile Diameter 0,4 m
23
2. Calculation of Stirrup Reinforcement
Data analysis :
Vu = 72,083 kN = 72083 N
Pu = 1088,131 kN = 1088131 N
f’c = 29,05 Mpa
fy = 390 Mpa
DPile = 400 mm
Dmain reinforcement = 20 mm
Dstirrup reinforcement = 10 mm
Number of stirrup =2
h = 400 mm
ds = 75 + ½. Dmain reinforcement + Dstirrup reinforcement (42)
= 75 + ½.20 + 10 = 95 mm
d = h – ds (43)
= 400 – 95 = 305 mm
Ag = ¼.π.Dpile2 = ¼.π.4002 = 125663,706 mm2 (44)
bw = Ag / (0,8.h) (45)
= 125663,706 / (0,8.400) = 392,699 mm
Av = ¼.π.Dstirrup reinforcement2.Number of stirrup (46)
= 1/4.π.102.2
= 157,080 mm2
Vn need = 𝑉𝑢
Ø =
72083
0,75 = 96110,667 N (47)
Calculate the shear force that is able to be held by concrete (Vc) :
Vc = 0,17. (1 +𝑃𝑢
14.𝐴𝑔) . 𝜆. √𝑓′𝑐. 𝑏𝑤. 𝑑 (48)
= 0,17. (1 +1088131
14.125663,706) . 1. √29,05. 392,699.305
= 177621,602 N
Ø.Vc = 0,75. 177621,602
= 133216,202 N
Vs need = Vn need – Vc (49)
= 96110,667 – 177621,602
= -81510,936 N
24
Because the value of Ø.Vc > Vu and the value of Vs minus, so no need shear reinforcement
and used minimum shear reinforcement.
Calculate the spacing of stirrup (s) :
s ≤ d/2 = 305/2 = 152,5 mm (50)
s ≤ 16.Dmain reinforcement = 16.20 = 320 mm (51)
s ≤ 48.Dstirrup reinforcement = 48.10 = 480 mm (52)
s ≤ b min = 392,699 mm (53)
Choose the smallest value of s and rounded down, s = 150 mm < 152,5 mm
So, used stirrup Ø10 – 150
Area stirrup used :
Av used = 𝑛.
1
4.𝑑𝑝2.𝑆
𝑠 =
2.1
4.102.1000
150 = 333,333 mm2 > Av (Ok) (54)
For the bored pile foundation diameter 0,7 m used same formula like above. The result
of calculation bored pile renforcement can be seen Table 3.
Table 3. Recapitulation of the Bored Pile Reinforcement
Diameter 0,4 m 0,7 m
Main Reinforcement 6 – D20 16 – D20
Stirrup Reinforcement Ø10 – 150 Ø10 – 300
To be able to function properly the building, it must be planned as possible both in terms
of cost and strength. Foundation work is one of the most important jobs in a construction, because
the foundation has the function of bearing and holding all the loads that work on it, namely the
structure load on top, then the stresses that occur due to the structure load will be channeled into
the hard soil layer can carry the burden of construction.
Calculation of the upper structure load using SAP 2000 version 15 program by entering
dead load, live load, and dynamic earthquake load according to SNI 2847-2013 obtained the
largest axial load value on the frame 161. The value of largest axial load (P) is 3779,897 kN. The
bearing capacity of a single bored pile used Reese and Wright method has result for diameter of
0,4 m obtained value of 2108,134 kN and for diameter 0,7 m obtained value of 4338,414 kN.
The pile with diameter 0,4 m needed to be able to withstand the largest axial load of
columns are 4 piles where the bearing capacity of group piles is 3871,555 kN, greater than the
value of the largest axial load columns of 3779,897 kN. The pile with diameter 0,7 m needed to
be able to withstand the largest axial load of columns are 2 piles where the bearing capacity of
group piles is 4289,926 kN, greater than the value of the largest axial load columns of 3779,897
kN.
25
The bored pile for diameter 0,4 m used pile cap 2,20 x 2,20 x 1,30 m and the bored pile
for diameter 0,7 m used pile cap 3,80 x 1,70 x 1,30 m. Based on SNI 1726-1012 reinforcement
that used are x-direction reinforcement and y-direction reinforcement. The x-direction obtained
for the main reinforcement is D25-110 and for the stirrup reinforcement is D19-110. While the
y-direction obtained for the main reinforcement is D25-110 and for the stirrup reinforcement is
D19-110.
The bored pile reinforcement analysis used SP column software, with diameter 0,4 m
need main reinforcement 6-D20 with the reinforcement area is 1884 mm2 and reinforcement ratio
of 1,50%. While the pile with diameter 0,7 m need main reinforcement 16-D20 with the
reinforcement area is 5024 mm2 and reinforcement ratio of 1,31%. Stirrup reinforcement analysis
bored pile foundation based SNI 1726-2012 get value for diameter 0,4 m obtained Ø10-150 and
for diameter 0,7 m obtained Ø10-300.
4. CLOSING
Based on the results of the analysis bored pile foundation, it has some conclusions as follows:
1) Calculation of the upper structure load obtained the largest axial load value on the frame
161. The value of largest axial load (P) is 3779,897 kN.
2) The bearing capacity of a single bored pile used Reese and Wright method has result for
diameter of 0,4 m obtained value of 2108,134 kN and for diameter 0,7 m obtained value of
4338,414 kN.
3) The pile with diameter 0,4 m needed 4 piles where the bearing capacity of group piles is
3871,555 kN and pile with diameter 0,7 m needed 2 piles where the bearing capacity of
group piles is 4289,926 kN.
4) The pile for diameter 0,4 m used pile cap 2,20 x 2,20 x 1,30 m and the pile for diameter 0,7
m used pile cap 3,80 x 1,70 x 1,30 m. The reinforcement that used are x-direction
reinforcement and y-direction reinforcement. The x-direction obtained for the main
reinforcement is D25-110 and for the stirrup reinforcement is D19-110. While the y-
direction obtained for the main reinforcement is D25-110 and for the stirrup reinforcement
is D19-110.
5) The pile with diameter 0,4 m need main reinforcement 6-D20 with the reinforcement area
is 1884 mm2 and reinforcement ratio of 1,50%. While the pile with diameter 0,7 m need
main reinforcement 16-D20 with the reinforcement area is 5024 mm2 and reinforcement
ratio of 1,31%. And the result of analysis stirrup reinforcement of bored pile foundation for
diameter 0,4 m obtained Ø10-150 and for diameter 0,7 m obtained Ø10-300.
26
REFERENCE
Asroni, A.(2014). Kolom Fondasi & Balok T Beton Bertulang Berdasarkan SNI 2847- 2013.
Surakarta: Teknik Sipil Universitas Muhammadiyah Surakarta.
Badan Standardisasi Nasional.(2013). Persyaratan Beton Struktural Bangunan Gedung SNI
2847-2013. Jakarta: Badan Standardisasi Nasional.
Harianti, E. & Anugrah P. ( 2013). Desain Fondasi Tahan Gempa. Yogyakarta: ANDI.
Halibu, E.Z.(2015). Pelaksanaan Pondasi Bored Pile dan Metode Pelaksanaan Pada Proyek
Pembangunan Gedung RSJ Prof Dr. V.L. Ratumbuysang Manado. Manado: Politeknik
Negeri Manado.
Listyawan, A.B., et all. (2017). Mekanika Tanah dan Rekayasa Pondasi. Surakarta:
Muhammadiyah University Press.
Wulandari, D. P. (2019). Perencanaan Pondasi Tiang Bor (Bored Pile) Pada Proyek
Pembangunan Gedung Rawat Inap RSUD dr. M. Yunus Kota Bengkulu. Malang:Universitas
Muhammadiyah Malang.
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