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MINI PILE FOUNDATION ANALYSIS
TOWER 86 m 5 Ton BEKASI
1) ALLOWABLE SOIL PRESSURE at 1.5 m
qc S1
qc1 27.5kgf
cm2
:=
qc S2
qc2 12.5kgf
cm2
:=
qc S3
qc3 22.5kgf
2:=
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2) LOAD (from Structural Design Calculation )
Tension Force (Uplift)T 1085.031kN:= Fzmax T:=
Compression Force C 1247.115kN:= Fzmax1 C:=
Fx 96.919kN:= Fx 9882.988 kgf ⋅= Horizontal Force
Fy 96.951kN:= Fy 9886.251 kgf ⋅=
Fa max Fx Fy,( ):= Fa 96951N= Fa 9886.251 kg⋅=Maximum Horizontal Force
3) DIMENSION FOUNDATION
Concrete
Height of the top of pedestal to bottom pad h 1.80m:=
Height of pedestal above ground level ht 0.30m:=
Pad
Thickness of pad at the face of pedestal hm 0.5m:=
Width of Pad B 2.1m:=
Length of Pad L 3m:=
Pedestal
Width of pedestal bb 0.60m:= bt bb:=Height of pedestal at the bottom
hts h hm−:= hts 1.3m=
Tie Beam
btb 0.3m:=Width of Tie Beam
Height of Tie Beam htb 0.50m:=
hg 0.15m:=Depth of Tie Beam under ground level
Length of Tie Beam Wbase 10.00m:= overlaps
hs 0m( ) h ht hg+ htb+ hm+( )− 0
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4) MATERIAL SPECIFICATION
Specified Yield Strength of Steel
f y1 240N
mm2
:= for diameter < φ 12 mm (undeformed rebar)
f y2 390N
mm2
:= for diameter > φ 13 mm (deformed rebar)
Specified Compressive Stregth of Concrete
K 22.5N
mm2
:= f c 0.83K:= f c 18.675N
mm2
⋅=
Unit weight of Water
γw 9810 N
m3
:=
Unit weight of Concrete
γc 24000N
m3
:=
Unit weight of Steel
γs 78500N
m3
:=
Unit weight of Soil
γsoil 15000N
m3
:=
β1 0.85:=Concrete strength of factor
Minimum Reinforcement ratio
ρmin1.38
f y2mm
2
N⋅
:= ρmin 0.00354=
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5) CHECK STABILITY
WEIGHT OF CONCRETE
Weight of Foot Plate
Wfp B L⋅ hm⋅ γc⋅:= Wfp 7709.055 kgf ⋅=
Weight of Pedestal
Wped bb bt⋅ hts⋅ γc⋅:= Wped 1145 kgf ⋅=
Weight of tie Beam
Wtb btb htb⋅ Wbase( )⋅ γc⋅:= Wtb 3671 kgf ⋅=
Weight of Concrete
Wcon Wfp Wped+ Wtb+:= Wcon 12525 kgf ⋅=
Volume of Concrete
Vcon
Wcon
γc
:= Vcon 5.118 m3
⋅=
VOLUME of CONCRETE
Vpadd
Wfp
γc
:= Vpadd 3.15 m3
⋅=
Vpedestd
Wped
γc
:= Vpedestd 0.468 m3
⋅=
Vtiebmd
Wtb
γc:= Vtiebmd 1.5 m
3
⋅=
VOLUME of CONCRETE Actual
Vconcrtd Vpadd Vpedestd+ Vtiebmd+:= Vconcrtd 5.118 m3
⋅=
VOLUME of CONCRETE Ground Water Table Effect
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VOLUME of SOIL
VOLUME of SOIL Ground Water Table Effect
Vsoild hdepth B⋅ B⋅( ):= Vsoild 6.615 m3
⋅=
Vbfd Vsoild Vpadd−:= Vbfd 3.465 m3
⋅=
VOLUME of SOIL ActualVsoilw 0.000m
3:=
WEIHGT of SOIL
Wsoil γsoil Vsoild⋅ γsoil γw−( ) Vsoilw⋅+:= Wsoil 10118.134 kgf ⋅=Wbf γsoil Vbfd⋅:= Wbf 5299.975 kgf ⋅=
DATA REQUIRED for CALCULATION BORE PILES CAPACITY
DATA BORED PILES
Quantity of Pile npile 6:= Piles
Maximum distance of pile from center of mass
xmax 0.45m:= ymax 0.45m:=
xmax 0.45m= ymax 0.45m=
Number of Piles in the direction
npilex 3:= npiley 3:=
Total Quadratic distance for number of piles in the direction :
ΣX npile xmax( )2
⋅:= ΣX 1.215 m2
=
ΣY npile ymax( )2
⋅:= ΣY 1.215 m2
=
Center distance of pile to pile: sp 0.90m:=
Diameter of pile ∆pile 0.6m:=
Cross section Area of pile:
Apile1
4π ∆pile⋅ ∆pile⋅:= Apile 0.283m
2=
Circumference of pile:
C ∆ C 1 885m
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A) ALLOWABLE BEARING CAPACITY PER PILE - COMPRESSIVE
Total Compression Force :
Ptotcomp Wconcrt npileWpile+ C+ Vpadd γsoil⋅−:= Ptotcomp 186774.792 kgf ⋅=
End Bearing Capacity Based on DCPT at -16.00 m:
qce1 165kgf
cm2
:= qce2 180kgf
cm2
:= qce3 125kgf
cm2
:=
Average Allowable Soil Pressure at -16.00 m
qend
qce1 qce2+ qce3+
2
20
⎛
⎜⎝
⎞
⎟ ⎠
∆pile
1m⋅ 1
Ldepth
∆pile
⎛
⎝
⎞
⎠+
⎡
⎣
⎤
⎦⋅
⎡
⎢
⎣
⎤
⎥
⎦3
:= qend 57.183kgf
cm2
⋅=
End Bearing Capacity
Qend qend Apile⋅:= Qend 161682.066 kgf ⋅=
Skin Friction Capacity Based on CPT for -16.00 m (Meyerhof) :
qstot
444 568+ 560+( )
3
kgf
cm
5:=
Skin Friction Capacity per pile
Qskin qstot( ) Cpile⋅:= Qskin 19754.335 kgf ⋅=
Skin Friction Capacity All pile
Qskint npile Qskin⋅:= Qskint 118526.008 kgf ⋅=
Maximum Allowable Bearing Capacity per pile :
Qut Qend Qskin+:= Qut 181436.401 kgf ⋅=
Maximum Allowable Bearing Group Capacity per pile :
Qug Qut npile⋅( ):= Qug 1.089 106
× kgf ⋅=
Q 1 089 106
× kgf⋅= Must be > OK!!!!Pt t 1 868 105
× kgf⋅=
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B) ALLOWABLE BEARING CAPACITY PER PILE - TENSION
Total Tension Force :
PtotTens T:= PtotTens 1.106 105
× kgf ⋅=
Skin Friction Capacity
Qskint 1.185 105
× kgf ⋅= Must be > OK!!!!PtotTens 1.106 105
× kgf ⋅=
SFtens
Qskint Wconcrt npileWpile+ Wbf +( )+PtotTens
:=
SFtens 1.701= Must be > 1.5 OK!!!!
C) F ACTOR SAFETY for SLIDING
Total Lateral Force :
Fxy Fx2
Fy2
+:= Fxy 13978.963 kgf ⋅=
Friction Coeffecient : ϕfr 0.45:=
SFsliding ϕfr
Wcon C+( )Fxy
⋅:=
SFsliding 4.497= Must be > 1.5 OK!!!!
D) FACTOR SAFETY for PUNCHING SHEAR
Magnified for structute under 213 m μf 1.30:=
Factored shear force at section
Fzcmax μf C Wcon+( )⋅ Wfp−:= Fzcmax 1.739 105
× kgf ⋅=
Strenght reduction factor for shear ϕsh
0.85:=
Ratio of long side to short side
βcB
B:= βc 1=
Normal diameter of Bar dbpad 19mm:=
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Shear strenght of concrete
ϕsh * Fzcpsh = Fzcpsh1 but not greater than Fzcpsh2
For two direction punch shear
Fzcpsh1 ϕsh 12
βc
+⎛
⎝
⎞
⎠
⋅
f cN
mm2
⋅
6
⋅ bo⋅ dpad⋅
⎡
⎢
⎣
⎤
⎥
⎦
:= Fzcpsh1 3.653 105
× kgf ⋅=
But not greater than
Fzcpsh2 ϕsh
f cN
mm2
⋅
3⋅ bo⋅ dpad⋅
⎛
⎜
⎝
⎞
⎟
⎠:= Fzcpsh2 2.435 10
5× kgf ⋅=
Shear strenght of concrete
Fzcpsh min Fzcpsh1 Fzcpsh2,( ):=Fzcpsh 2.435 10
5× kgf ⋅=
SFpshear
ϕsh Fzcpsh⋅
Fzcmax:=
SFpshear 2.381= Must be > 1.5 OK!!!!
E) DETERMINE the REINFORCEMENT in EACH DIRECTION
Top and Bottom Reinforcement
Factor Uniform Load
qbot
μ
f
W
con
Fzcmax+
( )⋅
B2
:= qbot 0.539 Nmm
2⋅=
Critical section for moment is at the face of pedestal
B b( )⎡ ⎤2⎡ ⎤
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ρmin
1.38
f y2mm2
N⋅
:= ρmin
0.00354=
Balanced Reinforcement ratio
ρbal
0.85 f c⋅ β1⋅( )f y2
600N
mm2
600N
mm2
f y2+
⋅:= ρbal 0.021=
Maxmum Reinforcement ratio
ρmax3
4ρbal⋅:= ρmax 0.016=
Nominal Moment strenght
Mnbot
Mbot
ϕflex
:= Mnbot 353662.504 N m⋅⋅=
Rnbot
Mnbotmm
2
N⋅
B dpad( )2
⋅:= Rnbot 0.796=
mf y2
0.85 f c⋅:= m 24.569=
Required reinforcement ratio
ρreqbot1
m1 1
2 m⋅ Rnbot⋅
f y2mm
2
N⋅
−⎛ ⎜
⎝
⎞⎟
⎠
−⎡⎢
⎣
⎤⎥
⎦
⋅:= ρreqbot 0.002095=
Use reinforcement ratio
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Abar
1
4π⋅ d
bpad
2⋅:= A
bar 2.835 cm
2⋅=
Number of Bar
n
barbot
round Areqbot
Abar
1+⎛
⎝
⎞
⎠0,
⎡
⎣
⎤
⎦:= n
barbot
13=
Spacing of Bar
SbarbotB
nbarbot 1−
⎛
⎝
⎞
⎠:= Sbarbot 175 mm⋅=
Used Spacing
S1B
nbarbot 1−:= S1 175 mm⋅=
Use nbarbot 13= with diameter dbpad 19 mm⋅= Or with spacing S1 175 mm⋅=
for Foot p late used D19 - 150 mm
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F) REINFORCEMENT for PEDESTAL
Magnified for structure under 213 m μf 1.30:=
Tension
Factored maximum tension forces
FZTmax μf T⋅:= FZTmax 1410540.3N=
Factored moment forces
Mytm μf Fy hts⋅( )⋅:= Mytm 163847.19 m N⋅=
CompressionFactored maximum compression forces
FZC1max μf C⋅:= FZC1max 1621249.5 N=
Factored moment forces
Mxtm μf Fx hts⋅( )⋅:= Mxtm 163793.11 m N⋅=
From the Interaction Diagram, Use Reinforcement as follow :
REINFORCEMENT for Pedestal used 22 D19
Stirrups for Pedestal used 10 - 150 mm
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x
y
600 x 600 mm
f'c = 23 MPa
fy = 390 MPa
Confinement: Tied
clr cover = 51 mm
spacing = 61 mm
22 N-19 at 1.73%
As = 6226 mm^2
Ix = 1.080e+010 mm^4
Iy = 1.080e+010 mm^4
Xo = 0 mm
Yo = 0 mm
PCACOL V2.30
Pn k
N
Mn (0°) (kN-m)
6436
5149
-2185
620
1
2
1993 PCAc
Licensed To: Licensee name not yet specified.
File name: G:\DOCYUD~1\DATA\MASTER\PCACOL~1\TWR86 COL 08/21/13 PCACOL(tm)V2.30 Proprietary Software of PORTLAND CEMENT ASSN. Page 109:41:54 Licensed to: Licensee name not yet specified.
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y p
OOOOOOO OOOOO OOOOO OOOOO OOOOO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OO OOOOOOO OO OO OO OO OOOOOOO OO OO OO OO OO OO OO OO OO
OO OO OO OO OO OO OO OO OO OOOO OOOOO OO OO OOOOO OOOOO OOOOO (TM)
======================================================================== Computer program for the Strength Design of Reinforced Concrete Sections ========================================================================
Licensee stated above acknowledges that Portland Cement Association (PCA) is not and cannot be responsible for either the accuracy or adequacy of the material supplied as input for processing by the PCACOL(tm) computer program. Furthermore, PCA neither makes any warranty expressed nor implied with respect to the correctness of the output prepared by the PCACOL(tm) program. Although PCA has endeavored to produce PCACOL(tm) error free, the program is not and can't be certified infallible. The final and only responsibility for analysis, design and engineering documents is the licensees. Accordingly, PCA disclaims all responsibility in contract, negligence or other tort for any analysis, design or engineering documents prepared in connection with the use of the PCACOL(tm) program.
08/21/13 PCACOL(tm)V2.30 Proprietary Software of PORTLAND CEMENT ASSN. Page 209:41:54 Licensed to: Licensee name not yet specified.
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General Information: ==================== File Name: G:\DOCYUD~1\DATA\MASTER\PCACOL~1\TWR86.COL Project: ZYDHI Code: ACI 318-89 Column: TOWER 86 m Units: SI Metric Engineer: Yudhy Date: 08/21/13 Time: 09:25:03
Run Option: Investigation Short (nonslender) column Run Axis: Biaxial Column Type: Structural
Material Properties: ==================== f'c = 22.5 MPa fy = 390 MPa Ec = 23981.5 MPa Es = 199955 MPa fc = 19.125 MPa erup = 0 mm/mm eu = 0.003 mm/mm
Stress Profile: Block Beta1 = 0.85
Geometry: ========= Rectangular: Width = 600 mm Depth = 600 mm
Gross section area, Ag = 360000 mm^2 Ix = 1.08e+010 mm^4 Xo = 0 mm Iy = 1.08e+010 mm^4 Yo = 0 mm
Reinforcement:
============== Rebar Database: User-defined Size Diam Area Size Diam Area Size Diam Area ------------------------------------------------------------------- 10 11 100 15 16 200 19 19 283 20 20 300 25 25 500 30 30 700 35 36 1000 45 44 1500 55 56 2500
Confinement: Tied; phi(c) = 0.7, phi(b) = 0.9, a = 0.8 N-10 ties with N-30 bars, N-10 with larger bars.
Layout: Rectangular Pattern: Equal Bar Spacing [Cover to transverse reinforcement (ties)]
Total steel area, As = 6226 mm^2 at 1.73%
22N-19 Cover = 40 mm
08/21/13 PCACOL(tm)V2.30 Proprietary Software of PORTLAND CEMENT ASSN. Page 309:41:54 Licensed to: Licensee name not yet specified.
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Applied Loads Computed Strength Computed/ P Mx My P Mx My Applied Pt. (kN) (kN-m) (kN-m) (kN) (kN-m) (kN-m) Ray length ---------------------------------------------------------------- 1 1621 176 0 4116 459 -0 2.539 2 -1411 176 0 -1478 186 -0 1.048
Program completed as requested!
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G) REINFORCEMENT for TIE BEAM
Total Compression Force
Ctot
2 C⋅
1.40
⎛
⎝
⎞
⎠:= Ctot 1.782 106
× N⋅=
Length of Tie Beam Ltie Wbase:= Ltie 10 m=
Uniform Load for Tie Beam
qtb
Ctot
Ltie
:= qtb 178159.2861
mN⋅=
qtb
Ltie
M A Mtbmax M A
Ltie 2x1−x1 x1
Maximum Moment forces
Mtbmax
qtb Ltie2
⋅
24:= Mtbmax 742330.357m N⋅=
Moment forces at fix side
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Concrete Coverppad 40mm:=
Effective depth of Tie Beam
dtie htb ppad−:= dtie 46 cm⋅=
Rn
Mntie
ϕm btb⋅ dtie2
⋅
⎛
⎝
⎞
⎠
mm2
N
⎛
⎝
⎞
⎠⋅:=
Rn 12.993=
m1
f y2
0.85 f c⋅:= m1 24.569=
Required Reinforcement ratio
ρreq1
m1
1 12 m1⋅ Rn⋅
f y2mm
2
N⋅
⎛
⎜
⎝
⎞
⎟
⎠
−−⎡
⎢
⎣
⎤
⎥
⎦
⋅:=
ρreq 0.028=
Used Reinforcement ratio
ρused ρmin ρmin ρreq≥if
ρreq ρreq ρmin≥if
:=
ρused 0.028=
Required Area for Reinforcement
Atiereq ρused btb⋅ dtie⋅:= Atiereq 3864 mm2
⋅=
Ause Atiereq:=
Normal diameter of Bar dbpad 16mm:=
Area of Bar
Abar 1
4π⋅ dbpad
2⋅:= Abar 201.062 mm
2⋅=
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Shear Reinforcement
Nominal diameter for stirrups dstr 10mm:=
Quantity of section nstr 2:=
For Shear Reinforcement, use minimum steel shear reinforcement
Avtie1 nstr 1
4⋅ π⋅ dstr
2⋅:=
Avtie1 157.08 mm2
⋅=
Required Stirrup Spacing
Stie
3 Avtie1⋅ f y2⋅mm
2
N⋅
btb
:= Stie 612.611 mm⋅=
Maximum Stirrup Spacing
Stie2
htb ppad−
2:= Stie2 230 mm⋅=
So, Using dstr 10 mm⋅= with Stie2 230 mm⋅=
Tie Beam Reinforcement :
for Top and bottom used 4 D16
for Middle used 2 10
Stirrup used 10 - 200 mm
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