INDEX
JUDUL NO. DOKUMEN SHEET LITERATURE
GABLE FRAMEGudang baja BJ/PPBBI/Ys/01 GudangBajaJembatan Jalan BJ/PPBBI/Tbr/02 Jembatan1Kolam Renang BJ/PPBBI/Tbr/03 KolamrenangTandon BJ/PPBBI/Ys/04 tandon
TRUSSKontrol Tegangan Rangka Baja BJ/PPBBI/TS/06 tegtruss
PERHITUNGAN STRUKTUR BAJA GUDANG
I. DATA-DATA PERENCANAANI.1. DATA UMUM BANGUNAN
NAMA = LUAS LANTAI = 24 m X 41 mJARAK PORTAL = 6.00 mTINGGI KOLOM = 10.00 mSUDUT ATAP = 25.00 degree = 0.44 radian
II. DESIGN STRUKTUR SKUNDER :2.1. DESIGN GORDING
COBA = C 125x50x20x4,5Wx = 38.00 cm3Wy = 10.10 cm3JARAK GORDING (Jg) = 125.00 cm
a. BEBAN TETAP TEKANAN HUJAN = 20.00 kg/m2t ATAP = 0.40 mm = 3.14 m2BERAT ATAP = 3.93 kg/m'BERAT GORDING = 8.32 kg/m'BERAT HUJAN = 25.00 kg/m' +TOTAL BEBAN TETAP = 37.25 kg/m'
MtetapX =1/8 Q.cos25 L^2 = 151.90 kg.mMtetapY =1/8 Q.sin25 Ly^2 = 17.71 kg.m
b. BEBAN HIDUP = 100.00 kgMhidupX =1/4 P.cos25 Lx = 135.95 kg.mMhidupY =1/4 P.sin25 Ly = 31.70 kg.m
c. BEBAN ANGIN , SUDUT ATAP = 25 derajatTEKANAN ANGIN = 25.00 kg/m2KOF. TEKAN ANGIN ATAP = 0,02(SUDUT ATAP) - 0,4 = 0.10KOF. TEKAN HISAP ATAP = - 0,4 = -0.40KOF. TEKAN ANGIN DINDING = + 0,9 = 0.90KOF. TEKAN HISAP DINDING = - 0,4 = -0.40
UNTUK GORDING, AMBIL TEKAN ATAP :Q angin = [0,02(SUDUT ATAP) - 0,4 ] *P*Jg = 3.12 kg/mM angin= 1/8 * Q*L^2 = 14.06 kg.m
d. TOTAL TEGANGAN :Mx = MtetapX + MhidupX + Mangin = 301.91My = MtetapY + MhidupY = 49.40
TEGANGAN = Mx/Wx + My/Wy = 1283.65 < 1600 ? OK !
GORDING BISA DIPAKAI !
III. DESIGN STRUKTUR UTAMA :
III.1. PEMBEBANAN KUDA-KUDA a. PEMBEBANAN GRAVITASI
BERAT ATAP = 18.84 kg/mBERAT GORDING = 41.60 kg/m +
No. Doc : BJ/PPBBI/Ys/01Rev :Judul Dokument Gudang Baja
No. Doc : BJ/PPBBI/Ys/01Rev :Judul Dokument Gudang Baja
60.44 kg/mb. BEBAN HIDUP TERPUSAT = 100.00 kgc. BEBAN ANGIN , SUDUT ATAP = 25 derajat
TEKANAN ANGIN = 25.00 kg/m2KOF. TEKAN ANGIN ATAP = 0,02(SUDUT ATAP) - 0,4 = 0.10KOF. TEKAN HISAP ATAP = - 0,4 = -0.40KOF. TEKAN ANGIN DINDING = + 0,9 = 0.90KOF. TEKAN HISAP DINDING = - 0,4 = -0.40
BEBAN AKIBAT ANGIN :Q tekan atap = [0,02(SUDUT ATAP) - 0,4 ] *P*Jk = 15.00 kg/mQ hisap atap = - 0,4 *P*Jk = -60.00 kg/mQ tekan dinding = 0,9 *P*Jk = 135.00 kg/mQ hisap dinding = - 0,4 *P*Jk = -60.00 kg/mBEBAN DI RAK :Q = BETON + HIDUP = 2400 kg/m3 * 12cm*Jk + 400kg/m2 = 4128.00 kg/msehingga :Q di batang kuda2 = GRAV. + ANGIN = 75.44 kg/m + P 100 kg.Q di kolom kiri = ANGIN tekan = 135.00 kg/mQ di kolom kanan = ANGIN hisap = -60.00 kg/mQ di balok rak = 4128.00 kg/m
c. ANALISA STRUKTUR Dengan bantuan program komputer SAP90 didapat :Elemen kuda2 = Moment = 2539.00 kg.m
Axial = 1634.00 kgElemen balok = Moment = 6985.00 kg.m
Axial = 458.00 kgPondasi utama = Fx = 2136.00 kg
Fy = 11690.00 kgMz = 3016.00 kg.m
Pondasi rak = Fx = 1574.00 kgFy = 8094.00 kgMz = 1257.00 kg.m
d. DESIGN ELEMENElemen kuda2 = WF 300x150x9x6.5Elemen kolom = WF 300x150x9x6.5Elemen balok rak = WF 300x150x9x6.5Elemen kokok & cantilever = WF 150x75
III.2. Design Struktur1. Standart and Reference :
- Peraturan Pembebanan Indonesia utk. Gedung 1983- AISC
2. Material - Steel fy = 2400.00 kg/cm2- Concrete fc' = 240.00 kg/cm2
3. Design metode- Elastic design : for steel structure - Ultimate design : for concrete structure
No. Doc : BJ/PPBBI/Ys/01Rev :Judul Dokument Gudang Baja
4. Structure analysis- Manual : for structure - SAP 90 : for structure- COSMIC : for roof
III.2.1. Design balok rak :Try :Beam = WF 300x150x6.5x9fy = 240.00 MpaWeight = 36.70 kg/mIx = 7210.00 cm4flens width (bf) = 150.00 mmflens thickness (tf) = 9.00 mmweb deepth (d) = 300.00 mmweb thickness (tw) = 6.50 mmradius of giration ('r) = 12.40 cmSect. Area (A) = 46.78 cm2Span (L) = 600.00 cma. Loading :Roof panel 0.3x0.001x30 m3Roof weight / panel = 141.30 kgnumber of panels = 20.00 panelsRoof length = 30.00 mTotal roof weight = 2826.00 kgTotal Beam weight = 220.20 kgLive load = 25.00 kg/m2
Total Weight = 647.20 kg/mMax. moment = 1/8.Q.L^2
= 6985.00 kg.cm
b. Lateral support check :L < 76. Bf / fy^0.5 (in.)
76.08 (in.)> 193.25 cm
Use Fb = [ 2/3 - ( fy(L/r)^2 )/(1530e3.Cb) ] fyFb = 0.61 fy
= 145.62 Mpac. Compact check :- Flens bf / 2.tf < 65 / fy^0,5
8.33 < 11.02 OK- Web d / tw < 640 / fy^0,5
46.15 < 108.49 OK
d. Stress checkfb = Mu c / I = 14.53 kg/cm2
No. Doc : BJ/PPBBI/Ys/01Rev :Judul Dokument Gudang Baja
fb / Fb = 0.01 OK
III.2.2. Design of Column :Try :Column = WF 300x150x6.5x9fy = 240.00 MpaWeight = 36.70 kg/mI = 7210.00 cm4flens width (bf) = 150.00 mmflens thickness (tf) = 9.00 mmweb deepth (d) = 300.00 mmweb thickness (tw) = 6.50 mmradius of giration ('r) = 12.40 cmSect. Area (A) = 46.78 cm2Length (L) = 900.00 cma. Loading :Axial (Pu) = 2009.00 kgMoment (Mu) = 3864.00 kg.m
=b. Beam - column check :b.1. Lateral support check :
L < 76. Bf / fy^0.5 (in.)76.08 (in.)
> 193.25 cmUse Fb = [ 2/3 - ( fy(L/r)^2 )/(1530e3.Cb) ] fy
Fb = 0.54 fy= 129.64 Mpa
b.2. Compact check :- Flens bf / 2.tf < 65 / fy^0,5
8.33 < 11.02 OK- Web d / tw < 640 / fy^0,5
46.15 < 108.49 OK
c. Stress checkfb = Mu c / I = 803.88 kg/cm2
fb / Fb = 0.62 OK
fa = Pu / A = 42.95 kg/cm2k factor = 0.80k l /r = 58.06From tab. AISC , Fa = 17.62 (ksi)
1214.90 kg/cm2fa / Fa = 0.04 < 0.15
OKTotal stress ratio :
No. Doc : BJ/PPBBI/Ys/01Rev :Judul Dokument Gudang Baja
fa/Fa + fb/Fb = 0.04 + 0.62= 0.66 OK
III.2.3. Design kuda-kuda :Try :Column = WF 300x150x6.5x9fy = 240.00 MpaWeight = 36.70 kg/mI = 7210.00 cm4flens width (bf) = 150.00 mmflens thickness (tf) = 9.00 mmweb deepth (d) = 300.00 mmweb thickness (tw) = 6.50 mmradius of giration ('r) = 12.40 cmSect. Area (A) = 46.78 cm2Length (L) = 900.00 cma. Loading :Axial (Pu) = 1634.00 kgMoment (Mu) = 2539.00 kg.m
=
b. Beam - column check :b.1. Lateral support check :
L < 76. Bf / fy^0.5 (in.)76.08 (in.)
> 193.25 cmUse Fb = [ 2/3 - ( fy(L/r)^2 )/(1530e3.Cb) ] fy
Fb = 0.54 fy= 129.64 Mpa
b.2. Compact check :- Flens bf / 2.tf < 65 / fy^0,5
8.33 < 11.02 OK- Web d / tw < 640 / fy^0,5
46.15 < 108.49 OK
c. Stress checkfb = Mu c / I = 528.22 kg/cm2
fb / Fb = 0.41 OK
fa = Pu / A = 34.93 kg/cm2k factor = 0.80k l /r = 58.06From tab. AISC , Fa = 17.62 (ksi)
1214.90 kg/cm2fa / Fa = 0.03 < 0.15
OK
No. Doc : BJ/PPBBI/Ys/01Rev :Judul Dokument Gudang Baja
Total stress ratio :fa/Fa + fb/Fb = 0.03 + 0.41
= 0.44 OK
III.2.4. Design of Bolt Connection : Beam to beam connect. :a. Loading :Geser vertical (Pv) = 1634.00 kgMoment (Mu) = 2539.00 kgmForce due to Mu/l = 126.95 kgResultante geser = 1638.92 kg
b. shear check :Resultan of shear = 1638.92 kgAllow. Shear = 0,6 (fy/1.5) = 950.40 kg/cm2
1074.33 kgn = 1.53 pcs.
Allow. Shear Load 1f1,2 =
8 / 25
A PERENCANAAN BALOK GELAGAR JEMBATAN
Potongan :
Denah
Pembebanan :Beban Mati :
- Berat sendiri pelat 20cm = 0.2*2400*2 = 960 kg/m'- Berat aspal = 0.07*2300*2 = 322 kg/m'- Berat air hujan = 0,01 * 1000 *2 = 20 kg/m'
1302 kg/m'
Beban Hidup :- Beban Hidup = 400 *2 = 800 kg/m'
Kombinasi pembebanan = 1.2 B.Mati + 1.6 B.HidupQ = 1.2 1302 + 1.6 800 = 2842.4 kg/m'
Berat sendiri balok = 0.4 * (0.8 - 0.2)*2400 = 576 kg/m'
Beban Merata pada balok Gelagar : 3418.4 kg/m'
10,5
15,00
10,5
2,00
2,00
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
Beban Kendaraan :
- Berat kendaraan Truck = 10000 / 2 = 5,000 kg- Beban kejut roda belakang 20% = 0,2 * 5000 = 1,000 kg
6,000 kg
Momen pada Balok Gelagar :P = 6,000 kg
w = 3418.4 kg/m'
10.5 m
M max =
= 1/8 3418.4 10.5 1/4 6,000 10.5= 47,109.83 + 15,750 = 62,859.83 kg.m= 62,859.83 E4 N.mm
Data balok :- lebar b = 400 mm- tinggi h = 800 mm- beton decking = 25 mm- tebal efektif d = 775 mm- Mutu beton fc' = K-300 = 24.61 Mpa- Mutu baja fy = U-39 = 390 MPa
Cek Ukuran Balok
0.003
Ccxb
d
Tb=As.fy
b
Letak garis netral pada regangan berimbang :
l =
1/8 w l2 + 1/4 P l
2 +
a=b1.xb
es=ey
P = 10000 kg
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
xb=d
xb = .d
=0.003
7750,003 + 390200000
= 469.70 mm
= 0,75 . Xb= 0,75 469.70 = 352.2727
a = b= 0,85 352.2727= 299.43 mm
Cc = 0,85.fc.b.a= 0.85 24.61 400 299.43 = 2,505,466 N
Mn = Cc . (d - (a/2))= 2,505,466 ( 775 - 299 / 2)= 1,566,627,902 Nmm > 628,598,250 Nmm
Momen kapasitas balok > Momen yang terjadi …… OK
Pembesian Lentur Balok :
m = fy / (0,85 .fc')
=390
0.85 24.61= 18.64
Rn
= 628,598,250.00
0.85 400 775 2= 3.0782
= (1/m) * (1 - ( 1 - ((2 Rn m)/fy)) )
=1
( 1 - 1 -2 3.0782 18.64
18.64 390
= 0.00857876
= 600
fy 600 + fy
ecuecu + (fy/Es)
ecuecu + (fy/Es)
xmax
xmax
= M / (f b d2)
r
r b0,85*f'c'*b1
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
= 0.85 24.61 0.85 600
390 600 + 390 = 0.02763127
= 0,75= 0,75 0.02763127= 0.02072346 0.00857876
= 1,4 / fy = 0.00359 0.00857876
= 0.00858 400 775 = 2,659.41 mm2
Dipasang tulangan :
A = 7 D 25 3,436.12 mm2)
A' = 4 D 25 1,963.50 mm2)
Gaya Geser pada Balok Gelagar :kondisi Geser max terjadi saat :
6,000 kg 1,500 kg
w = 3418.4 kg/m'
10.5 m
V max = 1/2 w l + P1 + P2= 1/2 3418.4 10.5 + 6,000 + 1,500 = 17,946.60 + 6,000 + 1,500 = 25,446.60 kg= 254,466.00 N
Pembesian Geser Balok :Kekuatan Geser Beton :
=
= 0.6 .1/6. pb 24.61 400 775= 153,786.25 N
r max r b
> r =
r min
< r =
Aperlu = r .b .d
( Aact =
( Aact =
P1 = P2 =
l =
f Vc f .1/6. pbfc'. bw .d
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
Gaya Geser yang harus diterima oleh Tulangan Geser :== 254,466.00 - 153,786.25 = 100,679.75 N
Dipakai Tulangan Geser :f = 12
Av ada = 2 12 2= 226.19 mm2
= Av .fy .d / s
s =
=0.6 226.19 240 775
100,679.75
= 25,243,325.29 100,679.75
= 250.73 mm
dipasang tulangan geser dia.12 mm - jarak 150 mm
B PERHITUNGAN PONDASI TIANG PANCANG
Berat Jembatan :- Berat pelat beton = 0,2*15*10,5*2400 = 75,600.00 kg- Berat Aspal = 0,2*15*10,5*2300 = 72,450.00 kg- Berat Balok 40x80 = 0,4*0,6*8*10,5*2400 = 48,384.00 kg- Berat Balok 30x50 = 0,3*0,3*7*15*2400 = 22,680.00 kg- Berat air hujan = 0,01*15*10,5*1000 = 1,575.00 kg- Berat railing = 1,500.00 kg- Berat abutment (2bh) = 2,2*16*0,75*2*2400 = 126,720.00 kg- Beban Hidup = 400*15*10,5 = 63,000.00 kg- Berat Kendaraan = 10000*2 = 20,000.00 kg
= 431,909.00 kg
Rencana Pondasi :
Dipakai pondasi TIANG PANCANG dia.40 - 19m'- A = 0.25*3.14*40*40 = 1256 cm2- K = 3.14*40 = 125.6 cm
f Vs Vu - f Vc
p/4
f Vs
f Av .fy .df Vs
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
- Nilai Conus dan JHP pada kedalaman 19 m' sebesar :- Cn = 100 kg/cm2- JHP = 1000 kg/cm
- Daya dukung 1 tiang pondasi :
P =A. Cn
+K. JHP
3 5
=125600
+125600
3 5
= 41,866.67 + 25,120 = 66,986.67 kg
- 1 tiang menerima beban :
= 431,909.00
= 21,595 kg < P= 66,987 kg ….. (OK) 20
D PERENCANAAN ABUTMENT JEMBATAN
Denah
Tampak Samping
350 350 350 350 5050
120 200
200
60
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
Tampak Depan
PembebananPembebanan pada 1 Balok Gelagar (l=10.5m)kondisi Beban max terjadi saat :
6,000 kg 1,500 kg
w = 0 kg/m'
10.5 m
P max = 1/2 w l + P1 + P2= 1/2 3418.4 10.5 + 6,000 + 1,500 = 17,946.60 + 6,000 + 1,500 = 25,446.60 kg
W = berat sendiri balok = 0.60 2 2400= 2,880 kg/m'
P = 25,446.60 kgw = 2,880 kg/m'
P1 = P2 =
l =
350 350 350 350 5050
60
200 200 200 200 200200 200
BALOK GELAGAR 40x80
Tiang Pancang d.40-19m
ABUTMENT
350
60
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
untuk balok menerus tumpuan sendi :
M max =
= 1/12 2,880 3.5 1/8 25,447 3.5= 2,940.00 + 11,133 = 14,072.89 kg.m= 14,072.89 E4 N.mm
Data Abutment :- lebar b = 1200 mm- tinggi h = 600 mm- beton decking = 25 mm- tebal efektif d = 575 mm- Mutu beton fc' = K-300 = 24.61 Mpa- Mutu baja fy = U-32 = 320 MPa
Cek Ukuran Abutment :
0.003
Ccxb
d
Tb=As.fy
b
Letak garis netral pada regangan berimbang :
xb=d
xb = .d
=0.003
5750,003 + 320200000
= 375.00 mm
= 0,75 . Xb= 0,75 375.00 = 281.25
a = b= 0,85 281.25= 239.06 mm
Cc = 0,85.fc.b.a= 0.85 24.61 1200 239.06 = 6,000,995 N
Mn = Cc . (d - (a/2))
1/12 w l2 + 1/8 P l
2 +
a=b1.xb
es=ey
ecuecu + (fy/Es)
ecuecu + (fy/Es)
xmax
xmax
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
= 6,000,995 ( 575 - 239 / 2)= 2,733,265,549 Nmm > 140,728,875 Nmm
Momen kapasitas balok > Momen yang terjadi …… OK
Pembesian Lentur Arah Memanjang :
m = fy / (0,85 .fc')
=320
0.85 24.61= 15.30
Rn
= 140,728,875.00
0.85 1200 575 2= 0.4173
= (1/m) * (1 - ( 1 - ((2 Rn m)/fy)) )
=1
( 1 - 1 -2 0.4173 15.30
15.30 320
= 0.00131733
= 600
fy 600 + fy
= 0.85 24.61 0.85 600
320 600 + 320 = 0.03623789
= 0,75= 0,75 0.03623789= 0.02717842 0.00131733
= 1,4 / fy = 0.00437 0.00131733
= 0.00437 1200 575 = 3,018.75 mm2
Dipasang tulangan : 380.133
A = D 22 - 125 3,041.06 mm2)
= M / (f b d2)
r
r b
0,85*f'c'*b1
r max r b
> r =
r min
> r =
Aperlu = r min .b .d
( Aact =
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
Pembesian Lentur Arah Melintang :
P = 25,446.60 kgw = 2,880 kg/m'
M max =
= 1/8 2,880 1.2 1/4 25,447 1.2= 518.40 + 7,634 = 8,152.38 kg.m= 8,152.38 E4 N.mm
m = fy / (0,85 .fc')
=320
0.85 24.61= 15.30
Rn
= 81,523,800.00
0.85 1000 575 2= 0.2901
= (1/m) * (1 - ( 1 - ((2 Rn m)/fy)) )
=1
( 1 - 1 -2 0.2901 15.30
15.30 320
= 0.00091290
= 0.00091 1000 575 = 524.92 mm2
Dipasang tulangan :
A = D 16 - 125 1,608.50 mm2)
1/8 w l2 + 1/4 P l
2 +
= M / (f b d2)
r
Aperlu = r .b .d
( Aact =
120
60
D16 - 125
D16 - 125
D22- 125
D22 - 125 600
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1500
D16 - 125D22 - 125 600
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No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan
Gaya Geser pada Balok Abutment :kondisi Geser max terjadi saat :
2,880 kg 720 kg
w = 0 kg/m'
10.5 m
V max = 1/2 w l + P1 + P2= 1/2 3418.4 10.5 + 2,880 + 720 = 17,946.60 + 2,880 + 720 = 21,546.60 kg= 215,466.00 N
Kekuatan Geser Beton :
=
= 0.6 .1/6. pb 24.61 1200 575= 342,298.42 N
Gaya Geser yang harus diterima oleh Tulangan Geser :== 215,466.00 - 342,298.42 = (126,832.42) N
----> Balok tanpa tulangan geser mampu untuk menahan geser
Dipakai Tulangan Geser praktis :
dipasang tulangan geser dia.16 mm - jarak 125 mm
600
P1 = P2 =
l =
f Vcf .1/6. pbfc'. bw .d
f Vs Vu - f Vc
2000
D22 - 125
D22 - 125
D16 - 125
D16 - 125
20 / 25
PERENCANAAN STRUKTUR KOLAM RENANG
DATA - DATA :- panjang kolam = 7 m- lebar kolam = 3 m- tinggi kolam = 1.5 m- tebal dinding = 15 cm- tebal pelat dasar = 15 cm
= 1.6 t/m3
= 1 t/m3
A PERENCANAAN DINDING KOLAM
A.1 Gaya yang bekerja pada dinding
A.2 Pada dasar kolam bekerja tekanan sebesar :
- P akibat tanah =
- g tanah
- g tanah
1/2 g tanah h2 ka
tekanan tanah tekanan air tanah
+
+
7.00
3.00
1.50
No. Doc : BJ/PPBBI/Tbr/03Rev :Judul Dokument Kolam Renang
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No. Doc : BJ/PPBBI/Tbr/03Rev :Judul Dokument Kolam Renang
= 1/2 (1.6) . 1.5^2 . 0,333= 0.5994 t/m2
- P akibat air = = 1/2 (1) . 1.5^2= 1.125 t/m2
A.3 Momen yang terjadi pada dinding bagian bawah
= (0.5994 + 1.125) . 1/3 . 1.5 = 0.8622 t.m
A.4 Pembesian dinding kolam :- tebal dinding = 150 mm- beton decking = 20 mm- tebal efektif d = 130 mm- Mutu beton fc' = K-225 = 17.89 Mpa- Mutu baja fy = U-24 = 240 MPa
m = fc / (0,85 .fy) = 240 / (0.85 . 17.89) = 15.78
= ( 0.8622.10^7) / (0,85 1000 . 130^2) = 0.600209
= (1/15.78) * (1 - 1 - ((2 0.60 . 15.78)/240))
= 0.00255
331.80 mm2
Menurut PBI-71 psl 9.1(2) , tulangan minimum untuk pelat adalah 0,25% dari luas beton yang ada, atau :
Amin = 0,25% . 1000 . 150 = 375 mm2Apakai = 375 mm2
Dipasang tulangan 393 mm2 (ok)Tulangan pembagi = 20% A = 20% 393 = 78.6 mm2Dipasang tulangan 251 mm2 (ok)
1/2 g air h2
M = (Ptanah + Pair) . 1/3 h
Rn = M / (f b d2)
r = (1/m) * (1 - ( 1 - ((2 Rn m)/fy)) )
Aperlu = r .b .d =
f10 - 200
f8 - 200
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No. Doc : BJ/PPBBI/Tbr/03Rev :Judul Dokument Kolam Renang
B PERENCANAAN PELAT DASAR KOLAM
B.1 Gaya yang bekerja pada pelat, ketika kolam dalam keadaan kosong
B.2 Pada pelat dasar bekerja tekanan sebesar :
- P akibat air = = 1 . 1.5= 1.5 t/m2
B.3 Momen yang terjadi pada dinding bagian bawah
ly/lx = 3,5/3,0 = 1,167
Berdasarkan tabel 13.3.2 PBI-71 , diperoleh momen :
lx 3.00 mly 3.50 mly/lx 1.17Jepit penuh/elastis {1/2] 2
44.6737.67
Mlx = -Mtx = 0.0447 . Q .lx2 = 0.0447 . 1500 . 3^2 603 kg.mMly = -Mty = 0.0000 . Q .lx2 = 0.0377 . 1500 . 3^2 508.50 kg.m
g air h
wlx (tabel setelah interpolasi)wly (tabel setelah interpolasi)
ly = 3,5
lx = 3,0
7.00
3.00
balo
k sl
oof
15x2
5
23 / 25
No. Doc : BJ/PPBBI/Tbr/03Rev :Judul Dokument Kolam Renang
B.4 Pembesian pelat dasar kolam :- tebal pelat = 150 mm- beton decking = 20 mm- tebal efektif d = 130 mm- Mutu beton fc' = K-225 = 17.89 Mpa- Mutu baja fy = U-24 = 240 MPa
m = fc / (0,85 .fy) = / (0.85 . 240) = 15.78
= ( 603.10^4) / (0,85 1000 . 130^2) = 0.42
= (1/15.78) * (1 - 1 - ((2 0.42 . 15.78)/240))
= 0.00177
230.60 mm2
Menurut PBI-71 psl 9.1(2) , tulangan minimum untuk pelat adalah 0,25% dari luas beton yang ada, atau :
Amin = 0,25% . 1000 . 150 = 375 mm2Apakai = 375 mm2
Dipasang tulangan 393 mm2 (ok)
C GAMBAR PENULANGAN
f10 - 200
f8 - 200 f10 - 200
SLOFF 15x25
Rn = M / (f b d2)
r = (1/m) * (1 - ( 1 - ((2 Rn m)/fy)) )
Aperlu = r .b .d =
f10 - 200
3 f 12
f 8 - 150
3 f 12
PERHITUNGAN TANDON AIR BAWAH
1. Plat dinding tandon :Pa (tekanan tanah aktif) = ca .w .h^2/2ca (coefisien tekanan aktif) = 0.3w (berat jenis tanah) = 1800 kg/m3h (kedalaman bangunan) = 2 m
Pa = 1080 kg/mMu = 1,2 x Pa x 1/3h 864 kg.m 8.4672 kN.m
2. Plat lantai tandon : Tebal plat = 0.15 mPanjang arah x (lx) = 2 mPanjang arah y (ly) = 2 mly / lx = 1.00
Berat volume beton = 2400Tebal selimut beton = 20 mm d = t*1000 - 20 - 5 = 125 mm b = 1 m
Beban-beban yang bekerja :1. Beban mati
Plat = 0.15 * 2400 = 360
Finishing lantai (keramik) = 0
Plafond + instalasi 0
Total 369
2. Beban hidup
Beban hidup yang bekerja = 2000
Wu = 1.2 * qd + 1.6 * ql = 3642.8lihat tabel PBI 71 :X untuk Mlx = 44X untuk Mly = 44X untuk Mtx = 0X untuk Mty = 0
641.133 kgm = 6.28 kNm
641.133 kgm = 6.28 kNm
0 kgm = 0.00 kNm
0 kgm = 0.00 kNm
Mutu beton = 20 MPaMutu baja = 240 MPa f = 0.8
Rn perlu = Mn perlu / ( b . d^2 ) 0.050595240.03794643
kg/m3
kg/m2
kg/m2
kg/m2
kg/m2
kg/m2
kg/m2
Mlx = 0.001 * Wu * lx2 * X =
Mly = 0.001 * Wu * lx2 * X =
Mtx = - 0.001 * Wu * lx2 * X =
Mty = - 0.001 * Wu * lx2 * X =
ρ balance = 0.85 . fc'/fy x 600/(600+fy) = ρ max = 0.75 . ρ balance =
No. Doc : BJ/PPBBI/Ys/04Rev :Judul Dokument Tandon
No. Doc : BJ/PPBBI/Ys/04Rev :Judul Dokument Tandon
0.0025
m = fy/0.85 . fc' = 14.11765
Arah Mu Mn As perlu tul.pakai As adakNm kNm N/mm2 > ρmin mm2 Ø (mm)
1. Plat vertikal x ( lap ) 8.47 10.584 0.847 0.00353 0.0035 441 5032. Plat lantai y ( lap ) 6.28 7.8538768 0.503 0.00209 0.0025 313 335
r min u/ plat dg.Tul.fy240Mpa (altf SKSNI dan CUR) =
ρ min < ρ perlu < ρ max
As perlu = ρ perlu . b . d
Rn=Mn/bd2 r perlu cek r
f8 - 100f8 - 150
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