Hitung Str Gudang Jembatan Kolam Tandon

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


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


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


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 :


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


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

No. Doc : BJ/PPBBI/Tbr/02Rev :Judul Dokument Jembatan Jalan

9 / 25


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

10 / 25


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

11 / 25


= 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

12 / 25


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

13 / 25


- 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

14 / 25


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

15 / 25


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

16 / 25


= 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 =

17 / 25


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

18 / 25


1500

D16 - 125D22 - 125 600

19 / 25


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

21 / 25


= 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

22 / 25


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


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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