eds sanitary water pump 15-9-2015

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


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)


* 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


= 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


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



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)


* 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


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



Vs = Vn - Vc = -191.68 KN ;shear forces detained stirrup

Shear Requirements:

Vn >

170.57 > 217.35 → (don’t use sear reinforcement)

< Vn <




* 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


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 = = 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


= 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 <




* 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 = = 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

eds sanitary water pump 15-9-2015

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