Beton Jembatan

A. PERHITUNGAN JEMBATAN PELATDIKETAHUI- MUTU BETON (fc) = 37 Mpa - LEBAR KERB = 95 cm- MUTU BAJA (fy) = 370 Mpa = 0.95 m- MUATAN JEMBATAN KELAS II = 7 Ton- TEBAL LANTAI JEMBATAN = 30 cm = 0.3 m- PANJANG JEMBATAN (lp) = 3.55 m- LEBAR JEMBATAN (lb) = 7.5 m - Diafragma = 20/40 cm- BERAT JENIS BETON ( ɣc) = 2400 Kg/m³ - Gelagar = 35/75 cm- BERAT JENIS BAJA TULANGAN ( ɣy) = 7800 Kg/m³- BERAT JENIS ASPAL ( ɣp) = 2200 Kg/m³- BERAT JENIS AIR ( ɣw) = 1000 Kg/m³- TEBAL GENANGAN AIR = 0.04 m- TEBAL LAPISAN ASPAL = 0.1 m

1) MENGHITUNG PEMBEBANAN LANTAI JEMBATANa. Beban Mati (wd)- Berat Sendiri pelat = 0.3 x 2400 = 720 Kg/m²- Berat Lapisan Perkerasan = 0.1 x 2200 = 220 Kg/m²- Berat Genangan Air = 0.04 x 1000 = 40 + Kg/m²

wd = 980 Kg/m²

MOMEN AKIBAT BEBAN MATI (MD)- Daerah Lapangan ( MD )⁺ = 1 x wd x L²

8= 1 x 980 x 12.6025

8= 1543.806 = 1.54381 tm

- Daerah Lapangan ( VD ) = 1 x wd x l2

= 1 x 980 x l2

= 1739.5 = 1.7395 tm

- DaerahTumpuan ( MD ˉ) = 1 x MD ⁺3

= 1 x 1.543813

= 0.514602 tm

- momen arah lebar lebar jembatan MD Lb⁺ x = 0.2 x MD ⁺ = 0.30876

MD Lbˉ x = 0.2 x MD ˉ = 0.10292

b. Beban Hidup (WL)- Beban Kendaraan Akibat Muatan Jembatan Kelas II

adalah T = 7 tona = 35 cm = 0.35 mb = 14 cm = 0.14 m

- Luas Bidang Kontak roda Kendaraan Terhadap Permukaan Lantai Jembatan

Tebal Lapisan Aspal t2 = 0.1 mTebal Lapisan Pelat t1 = 0.3 m

- Luas Bidang Kontak = Fx = Ax x BxAx = a + 2 ( 0.5 t1 + t2 )

= 0.35 + 2 ( 0.5 x 0.3 + 0.1 )= 0.35 + 0.5= 0.85 m

Bx = b + 2 ( 0.5 t1 + t2 )= 0.14 + 2 ( 0.5 x 0.3 + 0.1 )= 0.14 + 0.5= 0.64 m

Fx = Ax x Bx= 0.85 x 0.64= 0.544 m²

- Beban Hidup MerataWl = T + Pn Pn = t x l x Wn

Fx 1.75= 7 + 0.28571 = 2 x 5 x 50

0.544 1.75= 13.3929 t/m² = 500

1.75= 285.714 kg = 0.28571

ML⁺ = 0.125 x Wl x L² MLˉ = 0.33333 x ML⁺= 0.125 x 13.39286 x 12.6025 = 7.03265= 21.0979 tm

VL = 0.5 x WL x L= 0.5 x 21 x 3.55= 37.4488 tm

2) Menghitung Pelat dengan Beban terpusat> Menghitung Beban Hidup

Lebar Kerja Beban Terpusat saDimana Nilai r = 1

Lb < 3 x r x Lp

Lb < 3 x 1 x 3.557.5 < 10.65

> Beban ditengah-tengah BentanganU/Lb < 3 x r x Lp

sa = Ax + r x Lp x Lb = 0.85 + 1 x 3.55 x 7.5Lb + r x Lp 7.5 + 1 x 3.55

= 4.4 x 7.511.05

= 2.98643 m

> Menghitung Momen Beban Hidup Dalam awal panjang jembatan Lp

= = 7.06461sa

> Momen Tumpuan'= 0.33333 x= 2.35487

> Momen Positif arah Lebar Jembatan LbU/Lb > 2 x Lp MLb = = Mlp⁺

7.5 > 2 x 3.55 1 + 4 Ax7.5 > 7.1 3 Lp

= 7.06461165787341 + 4 0.85

3 3.55= 7.064611658 = 5.35503 t/m

1.319248826

> Momen Negatif arah Lebar Jembatan Lb= 0.1 x MLb= 0.1 x 5.355026= 0.5355 t/m

3) Perhitungan Penulangan> Penulangan arah x Lp- Tulangan Lapangan

= 1.2 MDLp + 1.6 MLLp= 1.2 x 1.543806 + 1.6 x 7.06461= 13.1559 tm

- Tulangan Tumpuan= 1.2 + 1.6 MLLp⁻= 1.2 x 0.535503 + 1.6 x 2.35487= 4.4104

> Penulangan arah y (Lb)- Tulangan Lapangan

= 1.2 MDLb + 1.6 MLLb= 1.2 x 0.308761 + 1.6 x 5.35503= 8.93855 tm

- Tulangan Tumpuan

Mlp⁺ Ml⁺

Mlp⁻ Mlp⁺

MLb⁻

Mlp⁺

Mlp⁻ MDLp⁻

Mlb⁺

= 1.2 + 1.6 MLLb⁻= 1.2 x 0.10292 + 1.6 x 0.5355= 0.98031

Mlp⁻ MDLb⁻

> Menghitung Penulangan Pelat Lantai Kendaraanb = 100 cm = 1000 mm fc = 37h = 30 cm = 300 mm fy = 370d = 0.9 x h B1 = 0.85 - 0.008 ( fc' - 30 )

= 0.9 x 300 = 0.85 - 0.008 ( 7 )= 270 mm = 0.85 - 0.056

B1 = 0.794

- Menghitung Persentase Penulangan PbPb = B1 x [ 0.85 x fc ] [ 600 ]

fy 600 + fy= 0.794 x [ 0.85 x 37 ] [ 600 ]

370 600 + 370= 0.794 x 0.085 x 0.61856= 0.04175

- Menghitung Persentase Penulangan Max PmaxPmax = 0.75 x Pb

= 0.75 x 0.041746= 0.03131

- Menghitung Persentase Penulangan Min PminPmin = 1.4 = 0.003784

370

- Menghitung Koefisien Tahanan K

K = MLp x = 13.1559 x 100000000.8 x b x 0.8 x 1000 x 72900

= 2.2558206708843 Mpa

m = fy = 370 = 11.76470.85 x fc 0.85 x 37

10⁷d²

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.7647 x 2.2558211.7647 370

= 0.085 [ 1 - √ 1 - 0.14345 ]= 0.085 [ 1 - 0.925497 ]= 0.00633 > 0.00378 dipakai P perlu = Tulangan Tunggal

a. Tulangan Lapangan Arah x LpDiambil As = P x b x d

= 0.00633 x 1000 x 270= 1709.83

16 - 100 = 2010.6

Mn = As x fy [ d - a/2 ] a = As x fy= 2010.6 x 370 [ 270 - 11.8271 ] 0.85 x fc x b= 743922 [ 258.1729 ] = 2010.6 x 370= 19.20605307 KnM 0.85 x 37 x 1000

= 743922Mn > MLp 31450

Q = 23.654119.2061 > 13.1559

0.819.2061 > 16.4449

b. Tulangan Tumpuan Arah x Lp


K = MLp x = 4.4104 x 100000000.8 x b x 0.8 x 1000 x 72900

= 0.7562407397859 Mpa

m = fy = 370 = 11.76470.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.7647 x 0.7562411.7647 370

= 0.085 [ 1 - √ 1 - 0.04809 ]= 0.085 [ 1 - 0.975658 ]= 0.00207 < 0.00378 dipakai P min = Tulangan Tunggal

Dari Tabel Diperoleh Ø

10⁷d²

Karena P < Pmin maka dipakai Pmin

As = P x b x d= 0.00378 x 1000 x 270= 1021.62

16 - 150 = 1340.4


= 495948Mn > MLp 31450

Q = 15.769412.9996 > 4.4104

0.812.9996 > 5.51299

C. Tulangan Lapangan Arah y Lb


K = MLb x = 8.93855 x 100000000.8 x b x 0.8 x 1000 x 72900

= 1.5326740345958 Mpa

m = fy = 370 = 11.76470.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.7647 x 1.5326711.7647 370


Diambil As = P x b x d

= 0 x 1000 x 270= 1021.62

16 - 150 = 1340.4

Mn = As x fy [ d - a/2 ] a = As x fy= 1340.4 x 370 [ 270 - 7.88 ] 0.85 x fc x b= 495948 [ 262 ] = 1340.4 x 370= 12.9996 KnM 0.85 x 37 x 1000

= 495948Mn > MLb 31450

Q = 15.769412.9996 > 8.93855


10⁷d²


0.812.9996 > 11.1732

d. Tulangan Tumpuan Arah y Lb


K = x = 0.98031 x 100000000.8 x b x 0.8 x 1000 x 72900

= 0.16809 Mpa

m = fy = 370 = 11.76470.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.7647 x 0.1680911.7647 370


Karena P < Pmin maka dipakai Pmin

As = P x b x d= 0.00378 x 1000 x 270= 1021.62

16 - 150 = 1340.4


= 495948Mn > 31450

Q = 15.769412.9996 > 0.98031

0.812.9996 > 1.22539 KNm

MLb⁻ 10⁷d²


MLb⁻

B. TIANG SANDARANa. Beban Mati- Pipa sandaran : tebal = 3 - 5- Diameter Pipa yang digunakan: 2 - 3- Tebal : 4 mm = 0.4 cm- Diameter Pipa D : 3 = 3 x 2.54 = 7.62 cm- Berat Volume ɣp : 7800 kg/m³- diamter Pipa (d) dalam d : 7.62 - 2 x 0.4 = 6.82 cm- Jarak antara Tiang : 2 m- Dimensi Tiang : b = 10 cm = 0.1 m- : h = 20 cm = 0.2 m- : t = 100 cm = 1 m- Berat jenis beton ( ɣc) : 2400 Kg/m³- Panjang Jembatan : 13.55 m

A. Beban MatiLuas Penampang Pipa Ap Berat Pipa PermeterAp = 0.25 ∏ ( D²-d² ) WP = Ap x ɣp

= 0.25 x 3.14 x 58.064 - 46.512 = 9.0683 x 10⁻⁴x 7800= 0.25 x 3.14 x 11.552 = 7.0733 Kg/m= 9.0683 x 10⁻⁴ m²

Berat Seluruh Pipa sepanjang Jarak Tiang Berat Pipa yang dipikul tiang sandaranQp = wp x L x n Vp = 0.5 Qp

= 7.0733 x 2 x 2 = 14.147= 28.293

Berat sendiri tiang sandaran Berat aksial tiang sandaranVt = b x h x ɣc Vu = 1.2 ( 2 x Vp + Vp )

= 0.1 x 0.2 x 2400 = 1.2 ( 2 x 14.147 + 48 )= 48 kg = 91.552 kg

B. Beban HidupWh = 100 - 200 Kg/mDipakai= 200 Kg/m

Q = √ Wp²+ Wh²= √ 50.031 + 40000= 200.13 kg/m

Jarak tiang SandaranMmax = 0.5 x Q x L²

= 0.5 x 200.13 x 4= 400.25 kg/m = 4003 Kg/cm

Wpipa = ∏/32 [ D⁴ - d⁴ ] = 3.14 [ 3371 - 2163 ]D 32 7.62

= 0.0981 [ 158.54 ]= 15.557

c. Perhitungan Penulangan tiang SandaranMu = 320 + 0.3333 Mmax Pn Perlu= Pu

= 320 + 133.42 0.65= 453.42 kg = 4.5342 Kn = 0.9155

0.65= 1.4085

Momen Dan gaya aksial rencana

Mn = Mu = 4.5342 = 5.6677 Knm P = Pu = 1.4085 = 1.7606 KnmQ 0.8 Q 0.8

jarak eksentritas e

e = Mn = 5.6677 = 4.024 mPn 1.4085

Luas Tulangand = 0.9 x h = 0.9 x 200 = 180d' = 0.1 x h = 0.1 x 200 = 20P = 1.7 % = 0.017

As = As' = 0.5 P x b x d= 0.5 x 0.017 x 100 x 180= 153 mm²

Dari tabel didapat = 2 Ø 12 = 226.2 mm² dipasang di masing-masing sisi kolom

eksentritas yang diperlukan

m = fy = 370 = 11.7650.85 fc 31.45

e' = [ e + ( d - h/2 ) ]= [ 4.024 + ( 180 - 100 ) ]= [ 4.024 + ( 80 )

1000= 4.104 mm= 4104 m

( 1 - e'/d ) = ( 1 - 4104 ) ( 1 - d'/d) = ( 1 - 20 )180 180

= -21.8 = 0.8889

P= P' = As = 226.2 = 226.2 = 0.0126

b x d 100 x 180 18000

Pn = 0.85 fc b d [ ( 1-e'/d) + √ ( 1-e'/d)² + 2 mp ( 1-d'/d)= 6E+05 [ -21.8 + √ 475.23 + 2 x 11.765 x 0.0126 x 0.8889 ]= 6E+05 [ -21.8 + √ 475.23 + 0.2628= 6E+05 [ -21.8 + √ 475.49= 6E+05 [ -21.8 + 21.806 ]= 3.4122 KnM

Kontrol

Pn > Pu/Q3.4122 > 0.9155

0.653.4122 > 1.4085 Knm = aman

dipasang di masing-masing sisi kolom

LANTAI TROTOAR

1) Beban Mati- Beban tiang Sandaran Vt

M1 = Vu x L= 91.5518 x 0.95= 86.9742 kgm

V1 = Vt = 86.9742 kg- berat sendiri lantai trotoar

qt = h1 x ɣc - M2 = 0.5 x qt x= 0.3 x 2400 = 0.5 x 720 x 0.9025

720 kg/m = 324.9 kgm- Berat pasangan kerb

qkerb = h2 x ɣkerb V2 = qt x L = 720 x 0.95= 0.2 x 2000 = 684 Kg.m= 400

M3 = 0.5 x qkerb x v3 = qkerb x L= 180.5 kg/m = 400 x 0.95

= 380

2) Beban Hidup- Beban hidup berguna qh = 100 - 200 kg/m

dipakai = 200 kg/m- M4 = 0.5 x qh x V4 = qh x L

= 0.5 x 200 x 0.95² = 200 x 0.95= 90.25 kg.m = 190 kg.m

- Beban Horisontal- Pada Tiang sandaran

M5 = Wh x L= 200 x 1= 200 kg.m

- Pada kerb ditetapkan 500 kg/mM6 = 500 ( h1+h2 )

= 500 ( 0.3 + 0.2 )= 250 kg.m

- Momen Trotoar yang diperhitungkan- Beban Mati

MD = M1 + M2 + M3 VD = V1 + V2 + V3= 87 + 324.9 + 180.5 = 86.9742 + 684 + 380= 592.3742 kg.m = 1150.97 kg.m

L²

kg/m²kg/m²

L²

L²

ML = M4 + M5 + M6 VL = V4= 90.25 + 200 + 250 = 190 kg.m= 540.25 kg.m

- Momen Rancang untuk Penulangan lantai trotoarMu = 1.2 MD + 1.6 ML Vu = 1.2 VD + 1.6 VL

= 710.849 + 864.4 = 1381.17 + 304= 1575.249 = 15.7525 KnM = 1685.17 = 16.8517 KnM

3) Penulangan TrotoarDik :

fc = 37 Mpa Pmax = 0.03131fy = 370 Mpa Pmin = 0.00378d = 180 mm m = 11.7647Pb = 0.017

K = Mu x = 15.7525 x 10000000.85 x b x 0.85 x 1000 x 32400

= 0.5719858535553 Mpa

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.7647 x 0.5719911.7647 370

= 0.085 [ 1 - √ 1 - 0.03637 ]= 0.085 [ 1 - 0.98164 ]= 0.00156 > 0.00378 dipakai P min = Tulangan Tunggal

a. Tulangan Lapangan Arah x LpDiambil As = P x b x d

= 0.00378 x 1000 x 180= 681.081

12 - 100 = 1131

Mn = As x fy [ d - a/2 ] a = As x fy= 1131 x 370 [ 180 - 6.65294 ] 0.85 x fc x b= 418470 [ 173.3471 ] = 1131 x 370= 72.54054371 KnM 0.85 x 37 x 1000

= 418470Mn > Mu 31450

Q = 13.3059

10⁶d²

Dari Tabel Diperoleh Ø mm²

72.5405 > 15.75250.8

72.5405 > 19.6906 KnM = Aman

C. PERHITUNGAN LANTAI JEMBATAN

1) Beban Mati- WD = 980- Pelat Tersebut dianggap menumpu secara menerus pada balok gelagar yang sejajar- Pelat Tersebut dapat dihitung menurut PBI 1971-NI-2 pada tabel 13.3.1 keadaan IVB dengan

Ly/Lx > 2.5Dimana Ly = Panjang Pelat/ gelagar = 13.55

Lx = Jarak gelagar 1 - 2 m = Lb/n-1 = 1.25- Jarak gelagar dengan membagi rata lebar jembatan

lebar jembatan = 7.5 m maka = Ly/Lx > 2.5jarak gelagar = 1.25 m 10.84 > 2.5

- Momen yang terjadiArah Lx = MLx = 0.001 wd x

= 0.001 x 980 x 1.6 x 42 koef x = 42= 64.3125 kg/m

MLx = 0.001 wd x= 0.001 x 980 x 1.6 x 83 koef x = 83= 127.094 kg/m

Arah Ly = Mly = 0.001 wd x= 0.001 x 980 x 1.6 x 8 koef x = 8= 12.25 kg/m

2) Beban Hidupa. Luas bidang Kontak I = 0.544b. Beban Muatan jembatan Kelas II( tiga) = 7 tonc. Reaksi tumpuan Akibat Muatan Angin = 150 ( pedoman perancangan

pembebanan jembatan jalan raya 13.3.28 tahun 1987, yang bekerja setinggi 2 mdiatas lantai kendaraan (sesuai tinggi truck). Panjang Truck 5 m denganjarak roda = 1.75 m

R = h L 1501.75

= 2 x 5 x 1501.75

= 857.143 Kg = 0.85714 ton

d. Beban Gandar

T = 5 + 0.85714= 5.85714 ton

e. Beban Hidup yang Tersebar Merata

Kg/m²

Lx²

Lx²

Lx²

m²

kg/m²

WL = T = 5.85714 = 10.7668 t/m = 10767 kg/mI 0.544

- Ada 3 kemungkinan Beban Hidup T bekerja pada lantai jembatan1) Beban T bekerja ditengah-tengah jarak gelagar

Dik =Lx = 1.25 m tx = 0.85 = 0.68 koef. Bitnertx = 0.85 m lx 1.25 Fx = 0.11671ty = 0.64 m

ty = 0.64 = 0.512 koef. BitnerT lx 1.25 Fy = 0.05184

interpolasi dari tabel bitnerLx = 1.3 dari tabel

1.25 ty tx:lxtx lx 0.7 0.68 0.6

ty 0.64 0.6 0.1142 0.125 III0.512 0.11671 II

0.85 0.5 0.117 0.1283 I

Didapat : I = 0.117 + 0.7 - 0.68 [ 0.1283 - 0.117 ]0.7 - 0.6

= 0.117 + 0.2 [ 0.0113 ]= 0.11926

Didapat : III = 0.1142 + 0.7 - 0.68 [ 0.125 - 0.1142 ]0.7 - 0.6

= 0.1142 + 0.2 [ 0.0108 ]= 0.11636

Didapat : II = 0.11926 - 0.6 - 0.512 [ 0.11926 - 0.11636 ]0.6 - 0.5

= 0.11926 - 0.88 [ 0.0029 ]Fx = 0.11671

interpolasi dari tabel bitnerdari tabel

ty tx:lxlx 0.7 0.68 0.6

0.6 0.0501 0.0533 III0.512 0.05184 II

0.5 0.0585 0.0657 I

Didapat : I = 0.0585 + 0.7 - 0.68 [ 0.0657 - 0.0585 ]

0.7 - 0.6= 0.0585 + 0.2 [ 0.0072 ]= 0.05994

Didapat : III = 0.0501 + 0.7 - 0.68 [ 0.0533 - 0.0501 ]0.7 - 0.6

= 0.0501 + 0.2 [ 0.0032 ]= 0.05074

Didapat : II = 0.05994 - 0.6 - 0.512 [ 0.05994 - 0.05074 ]0.6 - 0.5

= 0.05994 - 0.88 [ 0.0092 ]Fy = 0.05184

- Momen Arah Lx dan Ly

Mxm = Fx WL I= 0.11671 x 10767 x 0.544= 683.575

Mym = Fy WL I= 0.05184 x 10767 x 0.544= 303.658

- 2) Terdapat 2 beban T yang bekerja dengan Jarak 1.75

Tx cx1.75 ty A ty B

- Cx = 1.75 - 2 x 0.5 x tx= 1.75 - 0.85

1.25 = 0.9Tx

- Tx = 2 tx + cxty ty = 2 x 0.85 + 0.9

= 2.6 > Lx = 1.25tx tx Dipakai = 1.25

Keadaan A

Tx = 1.25 = 1Lx 1.25

ty = 0.85 = 0.68Lx 1.25


ty tx:lxlx 1 0.68 0.6

0.7 0.0841 0.0533 III0.68 0.08553 II0.6 0.0862 0.0657 I

Didapat : I = [ 0.0862 - 0.0841 ] x 0.68 + 0.0841= 0.0021 x 0.68 + 0.0841= 0.00143 + 0.0841

Fx = 0.08553


ty tx:lxlx 1 0 0.6

0.7 0.033 0.0533 III0.68 0.03694 II0.6 0.0388 0.0657 I

Didapat : I = [ 0.0388 - 0.033 ] x 0.68 + 0.033 ]= 0.0058 x 0.68 + 0.033= 0.00394 + 0.033

Fy = 0.03694

- Momen Arah Lx dan Ly

Mxm = Fx WL tx.ty= 0.08553 x 10767 x 0.544= 500.95

Mym = Fy WL tx.ty= 0.03694 x 10767 x 0.544= 216.386

Keadaan B

Cx = 0.9 = 0.72 ty tx:lxLx 1.25 lx 0.8 0.72 0.7

0.6 0.1043 0.1142ty = 0.64 = 0.512 0.512 0.11255Lx 1.25 0.5 0.1067 0.117

Didapat : I = 0.1067 + 0.8 - 0.72 [ 0.117 - 0.1067 ]0.8 - 0.7

= 0.1067 + 0.8 [ 0.0103 ]= 0.11494

Didapat : III = 0.1043 + 0.8 - 0.72 [ 0.1142 - 0.1043 ]0.8 - 0.7

= 0.1043 + 0.8 [ 0.0099 ]= 0.11222

Didapat : II = 0.11494 - 0.6 - 0.512 [ 0.11494 - 0.11222 ]0.6 - 0.5

= 0.11494 - 0.88 [ 0.00272 ]Fx = 0.11255

Cx = 0.9 = 0.72 ty tx:lxLx 1.25 lx 0.8 0.72 0.7

0.6 0.0434 0.0501ty = 0.64 = 0.512 0.512 0.04982Lx 1.25 0.5 0.0541 0.0585

Didapat : I = 0.0541 + 0.8 - 0.72 [ 0.0585 - 0.0541 ]0.8 - 0.7

= 0.0541 + 0.8 [ 0.0044 ]= 0.05762

Didapat : III = 0.0434 + 0.8 - 0.72 [ 0.0501 - 0.0434 ]0.8 - 0.7

= 0.0434 + 0.8 [ 0.0067 ]= 0.04876

Didapat : II = 0.05762 - 0.6 - 0.512 [ 0.05762 - 0.04876 ]0.6 - 0.5

= 0.05762 - 0.88 [ 0.00886 ]Fy = 0.04982

Mxm = Fx WL tx.ty= 0.11255 x 10767 x 0.544= 659.2

Mym = Fy WL tx.ty= 0.04982 x 10767 x 0.544= 291.822

Momen Arah Lx dan Ly

Mx = Mxm A - Mxm B Mx = Mxm A - Mxm B= 500.95 - 659.2 = 216.386 - 291.822= 158.251 kg.m = 75.4353 kg.m

- 3) Terdapat 2 beban T yang bekerja dengan Jarak 1.00

Tx cx1 ty A ty B

- Cx = 1 - 2 x 0.5 x tx= 1 - 0.85

1.25 = 0.15Tx

- Tx = 2 tx + cxty ty = 2 x 0.85 + 0.15

= 1.85 > Lx = 1.25tx tx Dipakai = 1.25

Keadaan A

Cx = 1.25 = 1 ty tx:lxLx 1.25 lx 1 0 0.6

0.7 0.0841 0.0533 IIIty = 0.64 = 0.512 0.512 0.08518 IILx 1.25 0.6 0.0862 0.0657 I

Didapat : I = [ 0.0862 - 0.0841 ] x 0.512 + 0.0841 ]= 0.0021 x 0.512 + 0.0841

= 0.00108 + 0.0841Fx = 0.08518

`

Cx = 1.25 = 1 ty tx:lxLx 1.25 lx 1 0 0.6

0.7 0.033 0.0533 IIIty = 0.64 = 0.512 0.512 0.03597 IILx 1.25 0.6 0.0388 0.0657 I

Didapat : I = [ 0.0388 - 0.033 ] x 0.512 + 0.033 ]= 0.0058 x 0.512 + 0.033= 0.00297 + 0.033

Fy = 0.03597

Mxm = Fx WL tx.ty= 0.08518 x 10767 x 0.544= 498.883

Mym = Fy WL tx.ty= 0.03597 x 10767 x 0.544= 210.679

Keadaan B

Cx = 0.15 = 0.12 ty tx:lxLx 1.25 lx 0.2 0.12 0.1

0.6 0.1796 0.1975ty = 0.64 = 0.512 0.512 0.19502Lx 1.25 0.5 0.1882 0.2068

Didapat : I = 0.1882 + 0.2 - 0.12 [ 0.2068 - 0.1882 ]0.2 - 0.1

= 0.1882 + 0.8 [ 0.0186 ]= 0.20308

Didapat : III = 0.1796 + 0.2 - 0.12 [ 0.1975 - 0.1796 ]0.2 - 0.1

= 0.1796 + 0.8 [ 0.0179 ]= 0.19392

Didapat : II = 0.20308 - 0.6 - 0.512 [ 0.20308 - 0.19392 ]

0.6 - 0.5= 0.20308 - 0.88 [ 0.00916 ]

Fx = 0.19502

Cx = 0.15 = 0.12 ty tx:lxLx 1.25 lx 0.2 0.12 0.1

0.6 0.0639 0.0651ty = 0.64 = 0.512 0.512 0.06637Lx 1.25 0.5 0.0761 0.0778

Didapat : I = 0.0761 + 0.2 - 0.12 [ 0.0778 - 0.0761 ]0.2 - 0.1

= 0.0761 + 0.8 [ 0.0017 ]= 0.07746

Didapat : III = 0.0639 + 0.2 - 0.12 [ 0.0651 - 0.0639 ]0.2 - 0.1

= 0.0639 + 0.8 [ 0.0012 ]= 0.06486

Didapat : II = 0.07746 - 0.6 - 0.512 [ 0.07746 - 0.06486 ]0.6 - 0.5

= 0.07746 - 0.88 [ 0.0126 ]Fy = 0.06637

Mxm = Fx WL tx.ty= 0.19502 x 10767 x 0.544= 1142.26

Mym = Fy WL tx.ty= 0.06637 x 10767 x 0.544= 388.75

Momen Arah Lx dan Ly

Mx = Mxm A - Mxm B My = Mym A - Mym B= 498.883 - 1142.26 = 210.679 - 388.75= 643.372 kg.m = 178.071 kg.m

3) Momen total untuk menghitung tulanganMomen yang digunakan adalah momen terbesarArah Lx = Mu = 1.2 MDx + 1.6 MLx

= 77.175 + 1093.72= 1170.9 kg.m = 11.709 KnM

= 1.2 MDx + 1.6 MLx= 152.513 + 364.574= 517.086 kg.m = 5.17086 KnM

Arah Ly = Mu = 1.2 MDx + 1.6 MLx= 14.7 + 284.914= 299.614 kg.m = 2.99614 KnM

= 0.33333 Mu= 0.99871

4) Menghitung Kebutuhan Tulangan

- Penulangan arah Lx fc = 37b = 1000 mm fy = 370h = 20 cm = 200 mm Pmax = 0.03131 KnMd = 0.9 h = 180 Pmin = 0.00378 KnM

a. Tulangan Lapangan arah Lx


Mu⁻

Mu⁻

K = Mu x = 11.709 x 10000000.8 x b x 0.8 x 1000 x 32400

= 0.451734446649 Mpa

m = fy = 370 = 11.76470.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.7647 x 0.45173444711.7647 370


- Menghitung luas Penampang Tulangan

As = P x b x d= 0.00378 x 1000 x 180= 681.081

12 - 100 = 1131

Mn = As x fy [ d - a/2 ] a = As x fy= 1131 x 370 [ 180 - 6.65294 ] 0.85 x fc x b= 418470 [ 173.347 ] = 1131 x 370= 72.5405 KnM > 11.709 KnM 0.85 x 37 x 1000

= 41847031450

= 13.3059

b. Tulangan Tumpuan arah Lx

10⁶d²



K = x = 5.17086 x 10000000.8 x b x 0.8 x 1000 x 32400

= 0.1994930794386 Mpa

m = fy = 370 = 11.76470.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.7647 x 0.19949307911.7647 370



As = P x b x d= 0.00378 x 1000 x 180= 681.081

12 - 100 = 1131


= 41847031450

= 13.3059

c. Tulangan Lapangan arah Ly

Mu⁻ 10⁶d²



K = Mu x = 2.99614 x 10000000.8 x b x 0.8 x 1000 x 32400

= 0.1155917901235 Mpa

m = fy = 370 = 11.76470.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.7647 x 0.1155917911.7647 370



As = P x b x d= 0.00378 x 1000 x 180= 681.081

12 - 100 = 1131


= 41847031450

= 13.3059

10⁶d²


b. Tulangan Tumpuan arah Ly


K = x = 0.99871 x 10000000.8 x b x 0.8 x 1000 x 32400

= 0.0385305967078 Mpa

m = fy = 370 = 11.76470.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.7647 x 0.03853059711.7647 370



As = P x b x d= 0 x 1000 x 180= 681.081

12 - 100 = 1131


= 41847031450

= 13.3059

Mu⁻ 10⁶d²


PERHITUNGAN BALOK DIAFRAGMA

1. Balok Diafragma Ujung

wd wd wd

Lantai jembatan

Balok Diafragma

Balok Gelagar

1.25 1.25 1.25A. Beban Mati

Dimensi Balok Diafragma20/40

Dik =b = 20 : 0.2 m d = 0.9 x hh = 40 : 0.4 m = 36 cm = 360 mm

- Berat Sendiri Balok Diafragmaqb = b x h x ɣc

= 0.2 x 0.4 x 2400= 192 Kg/m

Beban Pelat Lantai Kendaraan WD dimana Penyebaran Pelat ke balok diafragma sebagai beban segitiga. Untuk memudahkan Perhitungan maka beban segitiga tersebut dimodifikasimenjadi beban merata

Qeqp = 0.33333 x Wd x Lx= 0.33333 x 980 x 1.25= 408.333 Kg.m

- Beban Mati qd = qb + Qeqp

= 192 + 408.33= 600.33 Kg/m

b. Beban HidupBeban Hidup qL = 100 - 250 Kg/m dipakai = 250 Kg/m

Beban balok diafragma ujung

Qu = 1.2 qd + 1.6 ql= 1.2 x 600.33 + 1.6 x 250= 1120.4 kg.m

- Menghitung Momen

Mu = 0.125 Qu Lx²= 0.125 x 1120.4 x 1.5625= 218.83 kg/m = 2.1883 KnM

- Mengihtung Gaya Geser/ lintang

Vu = 0.5 x Qu x Lx= 0.5 x 1120.4 x 1.25= 700.25 kgm = 7.0025

Karena Balok diafragama menerus pada Perletakannya di balok gelagar, maka momen2nya yang diperhitungkan adalah PBI 71.200

Momen Lapangan Mu⁺ = 0.75 x Mu

= 0.75 x 2.1883= 1.6412 KnM

Momen TumpuanMu⁻ = 0.6667 x Mu

= 0.6667 x 2.1883= 1.4589 KnM

C. Perhitungan Penulangan

a. Tulangan Lapangan


K = Mu x 10⁶ = 1.6412 x 10000000.8 x b x d² 0.8 x 200 x 1E+05

= 0.079147904 Mpa

m = fy = 370 = 11.7650.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.765 x 0.079111.765 370



As = P x b x d= 0.0038 x 200 x 360= 272.43

Dari Tabel Diperoleh Ø 2 Ø 14 = 308 mm²

Mn = As x fy [ d - a/2 ] a = As x fy= 308 x 370 [ 360 - 9.0588 ] 0.85 x fc x b= 113960 [ 351 ] = 308 x 370= 39.993 KnM > 1.6412 KnM 0.85 x 37 x 200

= 1139606290

= 18.118

- Tulangan Tumpuan


K = Mu⁻ x 10⁶ = 1.4589 x 1E+060.8 x b x d² 0.8 x 200 x 1E+05

= 0.0704 Mpa

m = fy = 370 = 11.7647060.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.765 x 0.070411.7647 370



As = P x b x d= 0.00378 x 200 x 360= 272.432



= 1139606290

= 18.118

b Tulangan geser

Tulangan geser yang digunakan ada 2 Ø 10perhitungan kekuatan geser oleh beton

vc = 0.1667 x √ 34 x b x d= 0.1667 x 5.83095 x 200 x 360= 69.971 KN

SK SNI T-15-1991-03 pasal 3.4.5 ayat 5, menetapkan perhitungan perlu tidaknya dipasang begel denganpemeriksaan terhadap nilai VU

Apabila Nilai Vu> 0.5 Q Vc Perlu tulangan geser> 0.5 x 0.6 x 69.971

7.0025 < 20.9914 Perlu tulangan geser

2) Balok diafragma tengah

Balok Diafragma Balok Gelagar

Lantai Jembatan

Pelat jembatan tidak menumpu pada balok diafragma dimensi balok yang digunakan b = 200 mmh = 400 mmd = 0.9 x h

= 360 mm

- Berat sendiri Balok Qb = b x h x ( ɣc)

= 200 x 400 x 2400= 192000000= 192 Kg/m

- Beban HidupBeban Hidup qL = 100 - 250 Kg/m dipakai = 250 Kg/m

Beban balok diafragma ujung

Qu = 1.2 qb + 1.6 ql= 1.2 x 192 + 1.6 x 250= 630.4 kg.m

- Menghitung Momen

Mu = 0.125 Qu Lx²= 0.125 x 630.4 x 1.5625= 123.13 kg/m = 1.2313 KnM

- Mengihtung Gaya Geser/ lintang

Vu = 0.5 x Qu x Lx= 0.5 x 630.4 x 1.25= 394 kgm = 3.94

Karena Balok diafragama menerus pada Perletakannya di balok gelagar, maka momen2nya yang diperhitungkan adalah PBI 71.200

Momen Lapangan Mu⁺ = 0.75 x Mu

= 0.75 x 1.2313= 0.9234 KnM

Momen TumpuanMu⁻ = 0.6667 x Mu

= 0.6667 x 1.2313= 0.8208 KnM

C. Perhitungan Penulangan

a. Tulangan Lapangan


K = Mu x 10⁶ = 0.9234 x 10000000.8 x b x d² 0.8 x 200 x 1E+05

= 0.044533058 Mpa

m = fy = 370 = 11.7650.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.765 x 0.044511.765 370



As = P x b x d= 0.0038 x 200 x 360= 272.43


Mn = As x fy [ d - a/2 ] a = As x fy= 308 x 370 [ 360 - 9.0588 ] 0.85 x fc x b= 113960 [ 351 ] = 308 x 370= 39.993 KnM > 0.9234 KnM 0.85 x 37 x 200

= 1139606290

= 18.118

- Tulangan Tumpuan


K = Mu⁻ x 10⁶ = 0.75 x 1E+060.8 x b x d² 0.8 x 200 x 1E+05

= 0.0362 Mpa

m = fy = 370 = 11.7647061 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.765 x 0.036211.7647 370

= 0.085 [ 1 - √ 1 - 0.0023 ]= 0.085 [ 1 - 0.9988 ]= 9.8E-05 < 0.0038 dipakai P min = Tulangan Tunggal


As = P x b x d= 0.00378 x 200 x 360= 272.432



= 1E+056290

= 18.118

b Tulangan geser

Tulangan geser yang digunakan ada 2 Ø 10perhitungan kekuatan geser oleh beton

vc = 0.1667 x √ 34 x b x d= 0.1667 x 5.83095 x 200 x 360= 69.971 KN


Apabila Nilai Vu> 0.5 Q Vc Perlu tulangan geser> 0.5 x 0.6 x 69.971


PERHITUNGAN BALOK GELAGAR

1) Beban Matia. Berat Sendiri Balok Gelagar

- Direncanakan jenis Penampang Balok

b = 0.35 = 350 d = 0.9 x hh = 0.75 = 750 = 0.9 x 0.75

= 0.675 = 675

- Berat Sendiri Balok Gelagarqg = b x h x ɣc

= 0.35 x 0.75 x 2400= 630 kg/m

- Perhitungan MomenM1 = 0.125 x qg x

= 0.125 x 630 x 183.6025= 14458.7

- Perhitungan gaya geser / lintangV1 = 0.5 x qg x L

= 0.5 x 630 x 13.55= 4268.25 Kg

b. Akibat Beban Lantai Jembatan- Gaya Lintang (Vp) menjadi beban Merata (gp)

VuPelat = 0.5 x qd x Lx= 0.5 x 980 x 1.25= 612.5 kg/m

- Untuk 2 arah qpqp = 2 Vu

= 2 x 612.5= 1225 kg/m

- Perhitungan momenM2 = 0.125 x qp x

= 0.125 x 1225 x 183.6025= 28114.1 kg/m

- Perhitungan gaya geser/LintangV2 = 0.5 x qp x L

= 0.5 x 1225 x 13.55= 8299.38

L²

L²

c. Akibat Beban Balok DiafragmaGaya Lintang (Vu) masing-masing balok Diafragma, menjadi beban titik pada tumpuannyapada tumpuannya di balok gelagar

Pu Pt Pt Pu

2.7875 2.7875 2.7875

13.55Lf = 2.23 Lx

= 2.23 x 1.25= 2.7875

Pu = 2 x Vu = 2 x 700.25 = 1400.5 Kg.mPt = 2 x Vt = 2 x 394 = 788 Kg.m

∑P = 2 Pu + 2 Pt= 2 ( 1400.5 ) + 2 ( 788 )= 2801 + 1576= 4377 Kg

R = 0.5 ∑P= 0.5 x 4377= 2188.5 Kg

- Perhitungan Momen

M3 = R 0.5 L - Pv 0.5 L - Pt 0.5 Lf= 2188.5 ( 6.775 ) - 1400.5 ( 6.775 ) - 788 ( 1.39375 )= 14827.09 - 9488.39 - 1098.28= 4240.425 kg.m

V3 = R = 2188.5 kg

Total Momen Akibat beban MatiMD = M1 + M2 + M3

= 14458.7 + 28114.1 + 4240.43= 46813.254688 kg.m

VD = V1 + V2 + V3= 4268.25 + 8299.38 + 2188.5= 14756.1 kg

2) Bebab Hidup

a. Perhitungan Koefisien Kejut ( KK )

KK = 1 + 20 = 1 + 20 = 2050 + L 50 + 13.55 63.55

= 1 + 0.314713= 1.31471

b. Beban Hidup Merata (Qu)

Beban Merata yang diperhitungkan ( L < 30 m ) dengan Qh= 2.2 t/mUntuk Kelas II = 50% x qh

= 50% x 2.2= 1.1 t/m

Qu = qh x Lx x KK2.75

= 1.1 x 1.25 x 1.31471 2.752.75

= 1.1 x 0.45455 x 1.31471= 0.657356 t/m = 657.356 kg/m

Dalam Penggunaan beban jalur kendaraan ditetapkan (PPJJR) 1987 bahwa bila lebar jembatan lebihBesar dari 5.5, maka besar selebihnya hanya dibebani sebesar 50% saja dari beban hidup yang diperhitungkan.

Untuk Lb > 5.50 mQu Lb = Qu + 50% (Lb-2.75)QuLb = 657.3564 + 50% ( 657.356 ) ( 7.5 - 2.75 )

= 657.3564 + 328.678 x 4.75= 657.3564 + 1561.22= 2218.578 Kg/m' = 2.21858 tm

Perhitungan Momen

Mqu = 0.125 x Qu x= 0.125 x 2218.58 x 183.603= 50917.06 kg.m = 50.9171 tm

- Perhitungan Gaya Lintang/Geser

Vqu = 0.5 x Qu x L= 0.5 x 2218.58 x 13.55= 15030.87 kgm = 15.0309 ton.m

L²

c. Beban terhadap garisBeban Hidup garis ditetapkan sebesar 12 ton yang bekerja sebagai sebagai beban titik (Pd) ditengahtengah bentang jembatan

Pd = 12 x Lx KK - Jembatan Kelas II = 50% x 12 ton2.75 = 6 ton

= 6 1.25 x 1.314712.75

= 3.58558 ton= 3585.58 kg

Untuk Lb > 5.50 mPd Lb = Pd + 50% (Lb-5.50)PdLb = 3585.58 + 50% ( 3585.58 ) ( 7.5 - 2.75 )

= 3585.58 + 1792.79 x 4.75= 3585.58 + 8515.75= 12101.33 Kg = 12.1013 ton

Perhitungan MomenMpd = 0.25 x Pd x L

= 0.25 x 12101.3 x 13.55= 40993.268767 kgm = 40.9933 tm

Perhitungan gaya geser/lintang

Vpd = 0.5 x Pd= 0.5 x 12101.3= 6050.667= 6.050667 ton

Momen akibat beban hidupML = Mqu + Mpd

= 50917.06 + 40993.3= 91910.32 kgm= 91.91032 ton.m

Gaya Lintang/Geser akibat beban hidupVL = Vqu + Vpd

= 15030.9 + 6050.67= 21081.5 Kg= 21.0815 ton

Menghitung Momen Rencana

Mu = 1.2 Md + 1.6 Ml= 1.2 ( 46813.25 ) + 1.6 ( 91910.3 )= 56175.9 + 147057= 203232.42511 Kg.m

Mu = 0.75 x Mu= 0.75 x 203232= 152424.3= 152.4243 tm

= 0.333333 x Mu= 0.333333 x 203232= 67744.14= 67.74414 tm

Rencana Tulangan Lapangan

K = Mu x 10⁶ = 152.42 x 100000000.8 x b x d² 0.8 x 350 x 5E+05

= 11.9478204064 Mpa

m = fy = 370 = 11.7650.85 x fc 0.85 x 37

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 11.765 x 11.94811.7647 370

= 0.085 [ 1 - √ 1 - 0.7598 ]= 0.085 [ 1 - 0.4901 ]= 0.04334 < 0.0313 dipakai P Max = Tulangan Tunggal

Mu⁻

Perhitungan Tulangan Rangkap

Dik : b : 350 mm Mu+ = 152.4243 tm = 1524.24 KNmh : 750 mmFc : 37 Mpa Mu+ = 1524.243 tm = 1905.3 KNmFy : 370 Mpa Q 0.8

Perhitungan Selimut Beton (d') dan tinggi balok (d)

d' = 0.1 x h d = h - d'= 0.1 x 750 = 750 - 75= 75 = 675 mm

Perhitungan Konstanta keamananPb = B1 x [ 0.85 x fc ] [ 600 ]

fy 600 + fy= 0.794 x [ 0.85 x 37 ] [ 600 ]

370 600 + 370= 0.794 x 0.085 x 0.61856= 0.04175

Pmax = 75% Pb= 75% x 0.04175= 0.03131

Pasangan kopel gaya beton tekan dengan tulangan tari mempunyai rasio penulangan 90% dari Pmax

P = 90% x Pmax= 90% x 0.03131= 0.02818

m = 11.7647

K = P x fy ( 1 - 0.5 Pmax x m )= 0.02818 x 370 ( 1 - 0.5 x 0.03131 x 11.7647 )= 10.4262 x ( 1 - 0.18418 )= 10.4262 x 0.81582= 8.50592 < 11.9478

Mn = k x b x Syarat tulangan Rangkap= 8.50592 x 350 x 455625= 1356428491.90875 Mn < Mu/Q= 1356.4284919 1356.43 > 190.53

d²

PERHITUNGAN BALOK GELAGAR

1) Beban Matia. Berat Sendiri Balok Gelagar

- Direncanakan jenis Penampang Balok

b = 0.3 = 300 d = 0.9 x hh = 0.65 = 650 = 0.9 x 0.65

= 0.585 = 585

- Berat Sendiri Balok Gelagarqg = b x h x ɣc

= 0.3 x 0.65 x 2400= 468 kg/m

- Perhitungan MomenM1 = 0.125 x qg x L²

= 0.125 x 468 x 162.563= 9509.9063

- Perhitungan gaya geser / lintangV1 = 0.5 x qg x L

= 0.5 x 468 x 12.75= 2983.5 Kg

b. Akibat Beban Lantai Jembatan- Gaya Lintang (Vp) menjadi beban Merata (gp)

VuPelat = 0.5 x qd x Lx= 0.5 x 980 x 1.25= 612.5 kg/m

- Untuk 2 arah qpqp = 2 Vu

= 2 x 612.5= 1225 kg/m

- Perhitungan momenM2 = 0.125 x qp x L²

= 0.125 x 1225 x 162.563= 24892.383 kg/m

- Perhitungan gaya geser/LintangV2 = 0.5 x qp x L

= 0.5 x 1225 x 12.75= 7809.375

c. Akibat Beban Balok DiafragmaGaya Lintang (Vu) masing-masing balok Diafragma, menjadi beban titik pada tumpuannyapada tumpuannya di balok gelagar

Pu Pt Pt Pu

Lf = 2.23 Lx= 2.23 x 1.25= 2.7875

Pu = 2 x Vu = 2 x 548.8 = 1097.6 Kg.mPt = 2 x Vt = 2 x 340.1 = 680.2 Kg.m

∑P = 2 Pu + 2 Pt= 2 ( 1097.6 ) + 2 ( 680.2 )= 2195.2 + 1360.4= 3555.6 Kg

R = 0.5 ∑P= 0.5 x 3555.6= 1777.8 Kg

- Perhitungan Momen

M3 = R 0.5 L - Pv 0.5 L - Pt 0.5 Lf= 1777.8 ( 6.375 ) - 1097.6 ( 6.375 ) - 680.2 ( 1.39375 )= 11333.5 - 6997.2 - 948.029= 3388.25 kg.m

V3 = R = 1777.8 kg

Total Momen Akibat beban MatiMD = M1 + M2 + M3

= 9509.91 + 24892.383 + 3388.25= 37790.53531 kg.m

VD = V1 + V2 + V3= 2983.5 + 7809.375 + 1777.8

= 12570.7 kg

2) Bebab Hidup

a. Perhitungan Koefisien Kejut ( KK )

KK = 1 + 20 = 1 + 20 = 2050 + L 50 + 12.75 62.75

= 1 + 0.31873= 1.31873

b. Beban Hidup Merata (Qu)

Beban Merata yang diperhitungkan ( L < 30 m ) dengan Qh= 2.2 t/mUntuk Kelas II = 50% x qh

= 50% x 2.2= 1.1 t/m

Qu = qh x Lx x KK2.75

= 1.1 x 1.25 x 1.318732.75

= 1.1 x 0.4545455 x 1.31873= 0.65936 t/m = 659.363 kg/m

Dalam Penggunaan beban jalur kendaraan ditetapkan (PPJJR) 1987 bahwa bila lebar jembatan lebihBesar dari 5.5, maka besar selebihnya hanya dibebani sebesar 50% saja dari beban hidup yang diperhitungkan.

Untuk Lb > 5.50 mQu Lb = Qu + 50% (Lb-5.50)QuLb = 659.363 + 50% ( 659.363 ) ( 7.5 - 5.5 )

= 659.363 + 329.681 x 2= 659.363 + 659.363= 1318.73 Kg/m' = 1.31873 tm

Perhitungan Momen

Mqu = 0.125 x Qu x L²= 0.125 x 1318.73 x 162.563= 26796.9 kg.m = 26.7969 tm

- Perhitungan Gaya Lintang/Geser

Vqu = 0.5 x Qu x L= 0.5 x 1318.73 x 12.75= 8406.87 kgm = 8.40687 ton.m

c. Beban terhadap garisBeban Hidup garis ditetapkan sebesar 12 ton yang bekerja sebagai sebagai beban titik (Pd) ditengahtengah bentang jembatan

Pd = 12 x Lx KK - Jembatan Kelas II = 50% x 12 ton2.75 = 6 ton

= 6 1.25 x 1.318732.75

= 3.59652 ton= 3596.52 kg

Untuk Lb > 5.50 mPd Lb = Pd + 50% (Lb-5.50)PdLb = 3596.52 + 50% ( 3596.52 ) ( 7.5 - 5.5 )

= 3596.52 + 1798.26 x 2= 3596.52 + 3596.52= 7193.05 Kg = 7.19305 ton

Perhitungan MomenMpd = 0.25 x Pd x L

= 0.25 x 7193.05 x 12.75= 22927.8341181 kgm= 22.9278 tm

Perhitungan gaya geser/lintang

Vpd = 0.5 x Pd= 0.5 x 7193.05= 3596.52= 3.59652 ton

Momen akibat beban hidupML = Mqu + Mpd

= 26796.9 + 22927.8= 49724.7 kgm= 49.7247 ton.m

Gaya Lintang/Geser akibat beban hidupVL = Vqu + Vpd

= 8406.87 + 3596.523= 12003.4 Kg= 12.0034 ton

Menghitung Momen Rencana

Mu = 1.2 Md + 1.6 Ml= 1.2 ( 37790.535 ) + 1.6 ( 49724.7 )= 45348.6 + 79559.584= 124908.2268 Kg.m

Mu = 0.75 x Mu= 0.75 x 124908= 93681.2= 93.6812 tm

Mu⁻ = 0.33333 x Mu= 0.33333 x 124908= 41636.1= 41.6361 tm

Rencana Tulangan Lapangan

K = Mu x 10⁶ = 93.6812 x 10000000.8 x b x d² 0.8 x 300 x 342225

= 1.1405894035787 Mpa

m = fy = 300 = 14.11760.85 x fc 0.85 x 25

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 14.1176 x 1.1405914.1176 300

= 0.07083 [ 1 - √ 1 - 0.10735 ]= 0.07083 [ 1 - 0.9448 ]= 0.00391 < 0.0301 dipakai P Max = Tulangan Tunggal


As = P x b x d= 0.0039099 x 300 x 585= 686.18284

Dari Tabel Diperoleh Ø 7 Ø 40 = 8796.6 mm²

Mn = As x fy [ d - a/2 ] a = As x fy= 8796.6 x 300 [ 585 - 206.979 ] 0.85 x fc x b= 2638980 [ 378.02118 ] = 8796.6 x 300= 997.59 KnM > 93.6812 KnM 0.85 x 25 x 300

= 26389806375

= 413.958

- Tulangan Tumpuan


K = Mu⁻ x 10⁶ = 41.6361 x 10000000.8 x b x d² 0.8 x 300 x 342225

= 0.50693 Mpa

m = fy = 300 = 14.117647060.85 x fc 0.85 x 25

Pperlu = 1 [ 1 - √ 1 - 2 m k ]m fy

= 1 [ 1 - √ 1 - 2 x 14.1176 x 0.5069314.117647 300

= 0.0708333 [ 1 - √ 1 - 0.04771 ]= 0.0708333 [ 1 - 0.97585 ]= 0.0017104 < 0.00378 dipakai P min


As = P x b x d= 0.0037838 x 300 x 585= 664.05405

Dari Tabel Diperoleh Ø 2 Ø 25 = 931.8 mm²

Mn = As x fy [ d - a/2 ] a = As x fy= 931.8 x 300 [ 585 - 21.9247 ] 0.85 x fc x b= 279540 [ 563.07529 ] = 931.8 x 300= 157.402 KnM > 41.6361 KnM 0.85 x 25 x 300

= 2795406375

= 43.8494

b Tulangan geser

Tulangan geser yang digunakan adalah 2 Ø 10perhitungan kekuatan geser oleh beton

vc = 0.16667 x √ 34 x b x d= 0.16667 x 5.8309519 x 300 x 585= 170.555 KN


Apabila Nilai Vu > 0.5 Q Vc Perlu tulangan geser> 0.5 x 0.6 x 170.555


Beton Jembatan

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