PT POSO ENERGY SATU PAMONAHead Office:Cyber 2 Tower Lt.32, Jl. Rasuna Said Blok X5 No. 13, Kuningan Timur
Setiabudi, Jakarta Selatan, Indonesia Phone: +6221-82498438
Branch Office: Jl. Pulau Seram No. 464, Poso, Sulawesi Tengah, Indonesia.
Phone: (62 452) 23548
PROJECT NAME POSO HYDROPOWER PLANT (2 x 35 MW)
LOCATION SULEWANA, POSO-SULAWESI TENGAH
DOC. NO. REVISION NO. 0
AREA MESS & OFFICE
TITLE EQUIPMENT DATASHEET - SANITARY WATER
REV. DATE PREPARED CHECKED APPROVED DESCRIPTION
Approved Ali M. Toha Sign Date
Checked Iswar Rauf Sign Date
Prepared Fachrul Hidayat Sign Date
EQUIPMENT DATASHEET SPECIFICATIONSANITARY WATER PUMP
No Description Requirement Remark
1 GENERAL
Manufacturer VENDOR DATA
Quantity 2 Unit
Capacity
Head 33 m
Type Submersible Sewage Pump
Liquid Characteristic
- Density of water 997 kg/m3
- Viscousity 8.94E-07
Working Pressure - bar VENDOR DATA
Outlet Diameter - mm DN 100
Drive transmission Direct coupling
Noise level at 1 m distance 70 dB
Weight - kg VENDOR DATA
Total Pump Length VENDOR DATA
2 MOTOR
Rated Power - kW 15 kW Refers to SAMJIN
Speed rpm VENDOR DATA Industial Catalogue
Lubricating System - Water/Oil VENDOR DATA
Protection Level / Insulation Class IP55 / F
Voltage/Frequency/Phase VENDOR DATA
Weight - Kg VENDOR DATA
Dimension (L x W x H) - mm VENDOR DATA
3 OTHER
Minimum Warranty 1 year
Prepared By : Fachrul HidayatSign: Date:
Reviewed By : Iswar RaufSign: Date:
60 m3/h
Liquid Temperature (oC) 20 ~ 30 oC
Ambient Temperature (oC) 28 ~ 32 oC
DESAIGN BEAM (B1 1100x800)
Data Plan
f'c = 25
fy = 400
fy = 245 MPa ( N/mm2)
Dimention Beam And Reinforcement Estimated
h = 1100 mm ; high beam = 1100 1100
b = 800 mm ; width beam = 800 mm ~ 800
dc = 50 mm ; beton decking
= 25 mm ; diameter reinforcement estimated
= 19 mm ; diameter Stirrup
d = 1018.5 mm
Moment And Shear CalculationMoment And Shear value can be used by output SAP 2000
There Are :M3 (+) = 998 KNmM3 (-) = 1123 KNmV2 max = 1049 KN (Suppor-V2 max = 584 KN (mid) +
Calculation Of Longitudinal Reinforcement
= = 0.027094
ρ max = = 0.02032 0.840798
ρ min = = 0.0035
MPa ( N/mm2)
MPa ( N/mm2)
Æ tul.
Æ sengk.
; d = h - dc - 1/2 Æ tul. - Æ seng. ; effective high beam
rb
0,75 rb
mm ~
0.399001
Negative Reinforcement (Support)
Mu (-) = 1123 kNm 0.840798
Mn = (Mu/0.8) = 1403.75 kNm
Rn = Mn/(bd^2) = 1.692 N/mm2
m = fy/(0.85*f'c) = 18.82353
r = 1/m[1-(1-(2*Rn*m)/fy)] = 0.004412 0.0035
0.02032
r = Pakai = 0.0044 0.159
As perlu = = 3594.909 mm2 0
n (jlm tul) = = 7.327203 8 Æ 25
As' = n* = 3925 → As' > → OKE
Check / Beam Analisis
a = ( As' . fy ) / ( 0,85 f'c . b ) ; Hight Strenght Concrete Beams Equivalence
= 92.35 mm
Mn ada = As' .fy( d - 1/2 a ) *10^-6 ; Nominal Moment
= 1,526.55 Knm
Mu = 1,123.00 KNm
Mu < ( OK )
Positive Reinforcement (Mid Span)
Mu (+) = 998 kNm
Mn = (Mu/0.8) = 1247.5 kNm
Rn = Mn/(bd^2) = 1.503241 N/mm2
m = fy/(0.85*f'c) = 18.82353
r = = 0.003901 0.0035
0.02032
r = Pakai = 0.0039
As perlu = = 3178.823 mm2
n (jlm tul) = = 6.47913 7 Æ 25
As' = n* = 3434.375 → As' > → OKE
Check / Beam Analisis
a = ( As' . fy ) / ( 0,85 f'c . b ) ; Hight Strenght Concrete Beams Equivalence
= 80.81 mm
rmin < r < rmaxrmin
rmax
r*b*d
dipakai →
f Mn
1/m-[1-(Ö1-(2*Rn*m)/fy)] rmin < r < rmaxrmin
rmax
r*b*d
Taken →
Mn ada = As' .fy( d - 1/2 a ) *10^-6 ; Nominal Moment
= 1,343.66 Knm
Mu = 998.00 KNm
Mu < ( OK )
Shear Reinforcement CalculationV2 max = 1049 KN (Support)V2 max = 584 KN (Mid)
Shear Support Reinforcement
Vu = 1049 kN
Vc = = 679.00 KN ;Borne Shear Capacity Of Concrete
Vn = = 1748.33 KN ;Stronger Shear Capacity (f=0,6) SNI 1991 Section 3.2.3
Vs = Vn - Vc = 1069.33 KN ;shear forces detained stirrup
Shear Requirements:
Vn >
1748.33 > 407.4 → (Need Shear Reinforcement )
< Vn <
203.70 < 1748.33 < 407.4 → (Don’t Use Shear Minimum)
Cek apakah penampang mampu memikul gaya geser rencana :
* Vs ≤1069.33 ≤ 2716.00 → bear a strong cross-section shear force plan, do not need to be enlarged
* Vs ≤
f Mn
1/6*Öfc’*b*d
Vu/f
f Vc
0,5 f Vc f Vc
2/3.Öfc'.b.d
1/3.Öfc'.b.d
1069.33 ≤ 3395 → between minimum value 600 mm or ½ d
* Vs >
1069.33 > 3395 → beetween minimum value of 300 mm or ¼ d
Shear Reinforcement Determine ;(s=1000) :
Av,u = Vs.S/fy.d = 4285 mm
Av,u = = 1020 mm taken of the = 4285.347 mm
Av,u = b.s/3.fy = 1088 mm
Stirrups Space (s) = = 132 mm
Control :
s ≤ d/2 dan s ≤ 600132 ≤ 509.25 132 ≤ 600 125 Æ 19
Mid Span Shear Reinforcement
Vu = 584 kN
Vc = = 679 KN ;Borne Shear Capacity Of Concrete
Vn = = 973.33 KN ;Shear Nominal Stronger (f=0,6) SNI 1991 Section 3.2.3
Vs = Vn - Vc = 294.33 KN ; Shear Reinforcemenet detained Stirrup
Persyaratan Geser :
Vn >
973.33 > 407.4 → (Need Shears Capacity )
< Vn <
203.70 < 973.33 < 407.4 → (Don’t Use Minumum Shear Capacitty)
Cek apakah penampang mampu memikul gaya geser rencana :
* Vs ≤294.33 ≤ 2716.00 → bear a strong cross-section shear force plan, do not need to be enlarged
* Vs ≤
294.33 ≤ 3395 → between minimum value 600 mm or ½ d
* Vs >
294.33 > 3395 → beetween minimum value of 300 mm or ¼ d
Shear Reinforcement Determiner;(s=1000) :
Av,u = Vs.S/fy.d = 1180 mm
Av,u = = 1020 mm Taken Of Th= 1179.539 mm
Av,u = b.s/3.fy = 1088 mm
1/3.Öfc'.b.d
1/6*Öfc’*b*d
Vu/f
f Vc
0,5 f Vc f Vc
2/3.Öfc'.b.d
1/3.Öfc'.b.d
1/3.Öfc'.b.d
Stirrup Space (s) = = 481 mm
Control :
s ≤ d/2 dan s ≤ 600481 ≤ 509.25 481 ≤ 600 OK 250 Æ 19
DESIGN BEAM (B1 800x600)
Data Plan
f'c = 25
fy = 400
fy = 245 MPa ( N/mm2)
Beam Dimention And Expacted Reinforcement
h = 800 mm ; high of beam = 1100 800
b = 600 mm ; Width of bea = 800 mm ~ 600
dc = 50 mm ; beton decking
= 25 mm ; diameter rainforcement
= 13 mm ; diameter stirrup
d = 724.5 mm
Reinforcement And Shear Calculation Moment and Shear Ultimate Reinforcement can be use by output SAP2000,
So, can be used : M3 (+) = 173 KNmM3 (-) = 291 KNmV2 max = 114 KN (Suppor-V2 max = 93 KN (Mid) +
= = 0.0270938
ρ max = = 0.0203203
ρ min = = 0.0035
MPa ( N/mm2)
MPa ( N/mm2)
Æ tul.
Æ sengk.
; d = h - dc - 1/2 Æ rein. - Æ stirr. ; effective height of beam
rb
0,75 rb
mm ~
Negative Reinforcement (Support)
Mu (-) = 291 kNm
Mn = (Mu/0.8) = 363.750 kNm
Rn = Mn/(bd^2) = 1.155 N/mm2
m = fy/(0.85*f'c) = 18.823529
r = 1/m[1-(1-(2*Rn*m)/fy)] = 0.0029705 0.0035
0.02032
r = Used = 0.0030
As perlu = = 1291.2771 mm2
n (jlm tul) = = 2.6319023 Taken 3 Æ 25
As' = n* = 1471.875 → As' > As → OKE
Check / Beam Analisis
a = ( As' . fy ) / ( 0,85 f'c . b ; Hight Strenght Concrete Beams Equivalence
= 46.18 mm
Mn ada = As' .fy( d - 1/2 a ) *10^- ; Nominal Moment
= 412.96 Knm
Mu = 291.00 KNm
Mu < ( OK )
Positive Reinforcement (Mid Span)
Mu (+) = 173 kNm
Mn = (Mu/0.8) = 216.25 kNm
Rn = Mn/(bd^2) = 0.6866384 N/mm2
m = fy/(0.85*f'c) = 18.823529
r = = 0.0017453 0.0035
0.02032
r = Pakai = 0.0035
As perlu = = 1521.45 mm2
n (jlm tul) = = 3.1010446 4 Æ 25
As' = n* = 1962.5 → As' > As → OKE
rmin < r < rmax rmin
rmax
r*b*d
f Mn
1/m-[1-(Ö1-(2*Rn*m)/fy)] rmin < r < rmax rmin
rmax
r*b*d
dipakai →
Check / Beam Analysis
a = ( As' . fy ) / ( 0,85 f'c . b ; Hight Strenght Concrete Beams Equivalence
= 61.57 mm
Mn ada = As' .fy( d - 1/2 a ) *10^- ; Nominal Moment
= 544.57 Knm
Mu = 173.00 KNm
Mu < ( OK )
CalculationOF Shear ReinforcementV2 max = 114 KN (Support)V2 max = 93 KN (Mid Span)
Support Shear Reinforcement
Vu = 114 kN
Vc = = 362.25 KN ;Borne Shear Capacity Of Concrete
Vn = = 190.00 KN ;Stronger Shear Capacity (f=0,6) SNI 1991 Section 3.2.3
Vs = Vn - Vc = -172.25 KN ;shear forces detained stirrup
Shear Requirements:
Vn >
190.00 > 217.35 → (don’t use shear reinforcement)
< Vn <
108.67 < 190.00 < 217.3 → (don’t use minimum shear)
Cek apakah penampang mampu memikul gaya geser rencana :
* Vs ≤-172.25 ≤ 1449.00 → bear a strong cross-section shear force plan, do not need to be enlarged
f Mn
1/6*Öfc’*b*d
Vu/f
f Vc
0,5 f Vc f Vc
2/3.Öfc'.b.d
* Vs ≤
-172.25 ≤ 1811.25 → between minimum value 600 mm or ½ d
* Vs >
-172.25 > 1811.25 → beetween minimum value of 300 mm or ¼ d
Shear Reinforcement Determine ;(s=1000) :
Av,u = Vs.S/fy.d = -970 mm
Av,u = = 765.3 mm taken of the most' = 816.327 mm
Av,u = b.s/3.fy = 816.3 mm
Stirrups Space (s) = = 325 mm
Control :
s ≤ d/2 dan s ≤ 600325 ≤ 362.25 325 ≤ 600 OK 250 Æ 13
Mid Span Shear Reinforcement
Vu = 102.344 kN
Vc = = 362.25 KN ;Borne Shear Capacity Of Concrete
Vn = = 170.57 KN ;Stronger Shear Capacity (f=0,6) SNI 1991 Section 3.2.3
Vs = Vn - Vc = -191.68 KN ;shear forces detained stirrup
Shear Requirements:
Vn >
170.57 > 217.35 → (don’t use sear reinforcement)
< Vn <
108.67 < 170.57 < 217.3 → (don’t use minimum shear)
Cek apakah penampang mampu memikul gaya geser rencana :
* Vs ≤-191.68 ≤ 1449.00 → bear a strong cross-section shear force plan, do not need to be enlarged
* Vs ≤
-191.68 ≤ 1811.25 → between minimum value 600 mm or ½ d
* Vs >
-191.68 > 1811.25 → beetween minimum value of 300 mm or ¼ d
Shear Reinforcement Determine ;(s=1000) :
1/3.Öfc'.b.d
1/3.Öfc'.b.d
1/6*Öfc’*b*d
Vu/f
f Vc
0,5 f Vc f Vc
2/3.Öfc'.b.d
1/3.Öfc'.b.d
1/3.Öfc'.b.d
Av,u = Vs.S/fy.d = -1080 mm
Av,u = = 765.3 mm taken of the most = 765.306 mm
Av,u = b.s/3.fy = 816.3 mm
Stirrup Space (s) = = 346.7 mm
Control:
s ≤ d/2 dan s ≤ 600347 ≤ 362.25 347 ≤ 600 OK 250 Æ 13
200 D 13
600
800
250 D 13
600
800200
D 13
200 D 13
DESAIGN COLUMN K1(1500X1000)
Data Plan
f'c = 25
fy = 400
fy = 245 MPa ( N/mm2)
Dimensi Column
high column = 6 m
h = 1500 mm
b = 1000 mm
Ag = 1500000 mm ; Section Area
dc = 50 mm ; beton decking
= 32 mm ; reinsforcement diameter
= 16 mm ; stirrups diameter
d = 918 mm
Assuming the Column Reinforcement
Calculation of Longitudinal Reinforcement
= 15000 mm2
n = = 18.6604 mm2 19 Æ 32
Ast = n* = 15273 mm2 → As' > → OKE
Check Column
Section 12.3 of SNI, Maximum compressive strength of the field can not exceed:== 20,806.95 KN
0,1xf'cxAg = 3750 KN
> 0,1xf'cxAg
20,806.95 > 3,750.00 → (OKE )
> Pu
20,806.95 > 5210.8 → (OKE )
5210800
MPa ( N/mm2)
MPa ( N/mm2)
Æ tul.
Æ sengk.
; d = h - dc - 1/2 D tul. - D seng.
• The area of reinforcement obtained from SAP output table 'Concrete Design 1 - Summary Column Data - ACI 318-99'
• Luas Tulangan didapatkan dari tabel output SAP 'Concrete Design 1 - Column Summary Data - ACI 318-99'
• Default on SAP2000 design as earthquake resistant structure (Sway Special)
• SAP2000 has calculated that the minimum requirement of the amount of reinforcement required in flexura;
• While the shear reinforcement, SAP2000 not take into account the needs of minimum shear reinforcement,it be calculated manually
As (output SAP2000)
taken →
f Pn (max) 0,8.f.[0,85.f'c(Ag-Ast)+Ast.fy] ; f =0.65 column sliding
f Pn (max)
f Pn (max) ; (Pu, maximum value by SAP2000)
Calculation Of Shear ReinforcementV3 max 1 = 566 KN
Vu = 566 kN
Vc = = 1898.221 KN ;concrete shear capacity
Vn = = 943.33 KN
Vs = Vn - Vc = -954.89 KN ;shear force detained of stirrup
Shear Requirement
Vn >
943.33 > 1138.9326 → (don’t need shear reinforcement )
< Vn <
569.47 < 943.33 < 1138.93 → (don’t use minimum shear)
Cek apakah penampang mampu memikul gaya geser rencana :
* Vs ≤-954.89 ≤ 3060.00 → bear a strong cross-section shear force plan, do not need to be enlarged
* Vs ≤
-954.89 ≤ 3825 → minimum value of 600 mm or ½ d
* Vs >
(the greatest output SAP V3 )
(1+Nu/14Ag)(Öfc’/6)*b*d
Vu/f ;Nominal shear Strength (f=0,6) SNI 1991 Section 3.2.3
f Vc
0,5 f Vc f Vc
2/3.Öfc'.b.d
1/3.Öfc'.b.d
1/3.Öfc'.b.d
-954.89 > 3825 → minimum value of 300 mm or ¼ d
Determine Shear Reinforcement ;(s=1000) :
Av,u = Vs.S/fy.d = -4246 mm
Av,u = = 1275.51 mm been the most = 1360.54 mm
Av,u = b.s/3.fy = 1360.54 mm
space of strrup (s) = = 295 mm
Control :
s ≤ d/2 dan s ≤ 600295 ≤ 459 295 ≤ 600 Take 250 Æ 16