Jembatan Balok T

PERANCANGAN JEMBATAN BETON BERTULANG DENGAN TAMPANG BALOK T

Gambar 1. Penampang melintang jembatan

1. Kondisi Jembatan • • • • • •

Panjang bentang Lebar jembatan Lebar perkerasan Tipe jembatan Jumlah balok gelagar Panjang bersih gelagar

: 17,5 m :9m :7m : beton bertulang dengan gelagar balok T : 6 buah : 16,5 m

2. Spesifikasi Pembebanan a. Beban hidup : PPJJR No. 12/1970 (BM 100 %) • • •

Beban roda T Beban garis P Beban merata q

b. Beban kejut, k = 1 +

: 100% x 10 t = 10 t : 100% x 12 t/m = 12 t/m : 100% x 2,2 t/m2 = 2,2 t/m2 20 20 = 1+ = 1,2963 50 + L 50 + 17,5

3. Spesifikasi beton dan baja tulangan a. Beton • Kuat tekan, fc’ = 25 MPa • Kuat tekan ijin, fc’ = 10 MPa • Modulus elastis, Ec = 4700√25 = 23500 MPa b. Baja tulangan • •

Kuat leleh, fy = 400 MPa Modulus elastis, Es = 2x105 MPa

1

PERANCANGAN 1. Tiang sandaran momen lentur, Mu = 1,2×2×100×1,0 = 240 kg-m = 2400 N-m V = 1,2 × 2 × 100 = 240 kg = 2400 N

gaya geser,

Mn = φ bd2k d=130 mm

h=160 mm

Mu = Mn

b=160 mm

k=

Mu 2400 × 10 3 = = 1,1095 Mpa φ × b × d 2 0,8 × 160 × 130 2

ρ perlu = 0,85

ρ min =

f c' fy

⎛ ⎜ 1 − 1 − 2k ⎜ 0,85 f c' ⎝

⎞ ⎛ ⎞ ⎟ = 0,85 25 ⎜1 − 1 − 2 × 1,1095 ⎟ = 2,8502 × 10 −3 ⎜ ⎟ 400 ⎝ 0,85 × 25 ⎟⎠ ⎠

1,4 1,4 = = 3,5 × 10 −3 f y 400

As = ρ x b x d = 3,5×10-3 ×160×130 = 72,8 mm2 Dipakai tulangan 2∅10 (As = 157,0796 mm2) Kontrol kapasitas momen balok Dianggap baja tulangan telah luluh pada saat beton mulai retak (εc = 0,003) a= c=

As × f y 0,85 × a

β1

=

f c'

×b

=

157,0796 × 400 = 18,5 mm 0,85 × 25 × 160

18,5 = 21,7647 mm 0,85

⎛ 130 − 21,7647 ⎞ ⎛d −c⎞ f s = 600⎜ ⎟ = 2983,7847 MPa > fy ⎟ = 600⎜ ⎝ c ⎠ ⎝ 21,7647 ⎠

OK

a⎞ 18,5 ⎞ ⎛ ⎛ M n = As × f y ⎜ d − ⎟ = 157,0796 × 400⎜130 − ⎟ = 7586944,68 N-mm 2 2 ⎠ ⎝ ⎠ ⎝

=7586,9447 N-m > Mu (2400 N-m)

OK

Perencanaan tulangan geser Vu = 2400 N Vc =

1 6

f c' × b × d =

1 20 × 160 × 130 = 17333,3333 N 6

1 1 φVc = × 0,6 × 17333,3333 = 5199,9999 N > Vu 2 2

(secara teoritis tidak perlu sengkang) 2

walaupun secara teoritis tidak perlu sengkang, tetapi untuk kestabilan struktur dan peraturan mensyaratkan dipasang tulangan minimum smaksimum = ½ d = ½ x 130 = 65 mm luas tulangan geser minimum 1 Av min = 3

f c' × b × s fy

1 25 × 160 × 65 = 3 = 43,3333 mm2 400

dipakai tulangan ∅8 (As = 100,5310 mm2), maka jarak sengkang s=

Av × f y 100,5310 × 400 = = 150,7965 mm 1 1 ' fc × b 25 × 160 3 3

untuk penulangan geser dipakai sengkang ∅8-100

2. Perhitungan plat kantilever

Gambar 2. Pembebanan pada plat kantilever a. momen lentur (bending moment) Perhitungan momen lentur No. 1 2 3 4 5 6 7 P

Volume (m3) 0,10 × 0,16 × 0,50 = 0,008 0,10×(0,70×0,110)/2 = 0,00385 0,10×0,05×0,50 = 0,0025 0,10 × (0,15 × 0,50)/2 = 0,00375 1,00 × 1,00 × 0,20 = 0,2 1,00 × (1,00 × 0,10)/2 = 0,05 1,00 × 0,90 × 0,07 = 0,063 2,0 × 100 kg/m

γ (kg/m3)

W (kg)

Lengan (m)

2400 2400 2400 2400 2400 2400 2200

19,2 9,24 6 9 480 120 138,6 200

1,8 1,04 1,025 0,95 0,5 0,33 0,375 1,2

Momen (kg-m) 34,5600 9,6096 6,1500 8,5500 240,0000 39,6000 51,9750 240,0000

3

1,2963 × 10000 Air hujan = 2 × 0,90 × 0,05 = 0,0625 Railing = 2 × 2m× 6 kg/m = 24 Total momen, M Total momen, M (N-m) T

1000

12963 62,5 24

0,5 0,375 1,08

6481,5000 23,4375 25,9200 7161,3021 71613,0210

b. Gaya geser (shear force) Berat tiang sandaran = 1 + 2 + 3 +4 + railing Slab kantilever dan perkerasan = 5 + 6 +7 Beban roda Beban genangan air hujan Toal gaya lintang

= = = = = =

67,4400 738,6000 12963,0000 62,5000 13831,5400 138315,4000

Kg Kg Kg Kg Kg N

c. perhitungan baja tulangan Mu = 1,2×71613,021 =85935,6252 N-m Vu = 1,2×138315,400 = 165978,48 N h = 300 mm k=

d = 300-40 = 260 mm

Mu 85935,6252 × 10 3 = 1,5890 MPa = φ × b × d 2 0,8 × 1000 × 260 2

ρ b = 0,85

f c' × β1 × fy

0,003 0,003 +

fy

= 0,85

Es

25 × 0,85 0,003 × = 0,027094 400 400 0,003 + 200000

ρmaks = 0,75 ρb = 0,75 x 0,027094 = 0,0203205 ρ perlu = 0,85

f c' fy

⎛ ⎜ 1 − 1 − 2k ⎜ 0,85 f c' ⎝

⎞ ⎛ ⎞ ⎟ = 0,85 25 ⎜1 − 1 − 2 × 1,5890 ⎟ = 4,1333 × 10 −3 ⎜ ⎟ 400 ⎝ 0,85 × 25 ⎟⎠ ⎠

1,4 1,4 = = 3,5 × 10 −3 f y 400

ρ min =

As = ρ x b x d = 4,1333x10-3 x 1000 x 260 = 1074,658 mm2 Dipakai tulangan ∅16 (As = 210,0619 mm2), dengan jarak antar tulangan s perlu =

210,0619 × 1000 = 195,4686 mm 1074,658

dipakai tulangan ∅16-125 mm kontrol terhadap geser beton τc =

7 8

V 165978,48 = 7 = 0,7296 MPa × b × h 8 × 1000 × 260

<

0,45 fc = 11,25 MPa

OK

4

3. Perhitungan plat bagian dalam (inner slab) a. Momen lentur akibat beban hidup

Gambar 3. posisi roda Penyebaran beban hidup (roda) pada slab

P

P 6 cm 15 cm 15 cm

21

20 cm

21

21

50 cm

21

tx = 0,92 m ly = ∞ ty =0,62 m

lx = 1,4 m

Gambar 4. Penyebaran beban hidup pada slab

5

Beban roda, T = 10000 kg Bidang kontak = 0,92 m × 0,62 m Penyebaran beban roda, T =

10000 × 1,2963 = 22726,1571 kg/m2 0,92 × 0,62

Dipakai tabel-Bittner (dari Dr. Ing Ernst Bittner) Dengan lx = 1,4 , ly = ∞ (lantai tidak menumpu pada diafragma) t x 0,92 = = 0,657 1,4 lx

fxm = 0,1233

ty

fym = 0,0661

lx

=

0,62 = 0,443 1,4

Mxm = 0,1233 × 22726,1571 × 0,92 × 0,62 = 1598,3379 kgm = 15983,379 Nm Mym = 0,0661 × 22726,1571 × 0,92 × 0,62 = 856,8543 kgm = 8568,543 Nm

b. momen lentur akibat beban mati = 0,30 × 2400 = 0,06 × 2200 = 0,05 × 1000

Berat slab Berat perkerasan Berat air hujan Total qDL

= = = =

720 132 50 902

kg/m2 kg/m2 kg/m2 kg/m2

M xm =

1 1 × q DL × l x2 = × 902 × 1,4 2 = 176,7920 kgm = 1767,920 Nm 10 10

M ym =

1 1 × M xm = × 176,7920 = 58,9307 kgm = 589,307 Nm 3 3

c. momen total Mx = 15983,379 + 1767,920 = 17661,299 Nm My =8568,543 + 589,307 = 9157,85 Nm d. perhitungan baja tulangan arah melintang lx M = 17661,299 Nm h = 300 mm k=

d = 300-40 = 260 mm

M 17661,299 × 10 3 = = 0,3267 MPa φ × b × d 2 0,8 × 1000 × 260 2

ρ b = 0,85

f c' × β1 × fy

0,003 0,003 +

fy Es

= 0,85

25 × 0,85 × 400

0,003 = 0,027094 400 0,003 + 200000

ρmaks = 0,75 ρb = 0,75 x 0,027094 = 0,0203205 6

ρ perlu = 0,85 ρ min =

f c' fy

⎛ ⎜ 1 − 1 − 2k ⎜ 0,85 f c' ⎝

⎞ ⎛ ⎞ ⎟ = 0,85 25 ⎜1 − 1 − 2 × 0,3267 ⎟ = 8,2313 × 10 −4 ⎜ ⎟ 400 ⎝ 0,85 × 25 ⎟⎠ ⎠

1,4 1,4 = = 3,5 × 10 −3 f y 400

As = ρ x b x d = 3,5 x10-3 x 1000 x 260 = 910 mm2 Dipakai tulangan ∅16 (As = 210,0619 mm2), dengan jarak antar tulangan s perlu =

210,0619 × 1000 = 230,8373 mm 910

dipakai tulangan ∅16-125 mm arah memanjang ly M = 9157,85 Nm h = 300 mm k=

d = 300-40 = 260 mm

M 9157,85 × 10 3 = = 0,1693 MPa φ × b × d 2 0,8 × 1000 × 260 2

ρ b = 0,85

f c' × β1 × fy

0,003 0,003 +

fy

= 0,85

25 × 0,85 × 400

Es

0,003 = 0,027094 400 0,003 + 200000

ρmaks = 0,75 ρb = 0,75 x 0,027094 = 0,0203205 ρ perlu = 0,85

ρ min =

f c' fy

⎛ ⎜ 1 − 1 − 2k ⎜ 0,85 f c' ⎝

⎞ ⎛ ⎞ ⎟ = 0,85 25 ⎜1 − 1 − 2 × 0,1693 ⎟ = 4,2495 × 10 −4 ⎜ ⎟ 400 ⎝ 0,85 × 25 ⎟⎠ ⎠

1,4 1,4 = = 3,5 × 10 −3 f y 400

As = ρ x b x d = 3,5 x10-3 x 1000 x 260 = 910 mm2 Dipakai tulangan ∅16 (As = 210,0619 mm2), dengan jarak antar tulangan s perlu =

210,0619 × 1000 = 230,8373 mm 910

dipakai tulangan ∅16-125 mm

4. Perhitungan Gelagar a. beban mati (dead load) Hand rail Railing Perkerasan Air hujan Pelat lantai

= {(0,10 × 0,16 × 1,00 × 2400)/2} × 1,1871 = 2 × 1,00 × 6 × 1,1871 = 0,06 × 2200 × 4,5716 = 0,05 × 1000 × 4,5716 = 0,30 × 2400 × 4,5716

= = = = =

22,7923 14,2452 603,4512 228,5800 3291,5520

kg/m kg/m kg/m kg/m kg/m

7

= 1,00 × 0,50 × 2400 × 1,00

= 1200,0000 kg/m = 5360,6207 kg/m Balok melintang (diafragma), Tb = 0,30 × 0,60 × 2400 × 0,9 = 388,8 kg Gelagar Total

Gambar 5. Garis pengaruh momen

Gambar 6. Potongan memanjang balok pada perhitungan momen lentur b. momen lentur akibat beban mati M qDL → M x =

⎧x⎛ x ⎞⎫ 1 q DL × L2 ⎨ ⎜1 − ⎟ ⎬ 2 ⎩ L ⎝ L ⎠⎭

Momen pada potongan 1, x = 2,0 m (M1 DL) ⎧ 2 ⎛ 1 2 ⎞⎫ M qDL = × 5360,6207 × 16,52 ⎨ ⎜1 − ⎟⎬ 2 16 , 5 16 ,5 ⎠ ⎭ ⎝ ⎩ MTb= ½ × 388,8 × 2 M1 DL

=

77729,0002 kgm

= =

388,8000 kgm 78117,8002 kgm 781178,0020 Nm

8

Momen pada potongan 2, x = 4,0 m (M2 DL) ⎧ 4 ⎛ 1 4 ⎞⎫ M qDL = × 5360,6207 × 16,52 ⎨ ⎜1 − ⎟⎬ 2 ⎩16,5 ⎝ 16,5 ⎠ ⎭ MTb= ½ × 388,8 × 4 M2 DL

Momen pada potongan 3, x = 6,0 m (M3 DL) ⎧ 6 ⎛ 1 6 ⎞⎫ M qDL = × 5360,6207 × 16,52 ⎨ ⎜1 − ⎟⎬ 2 ⎩16,5 ⎝ 16,5 ⎠ ⎭ MTb= ½ × 388,8 × 6 M3 DL

Momen pada potongan 4, x = 8,25 m (M4 DL) ⎧ 8,25 ⎛ 8,25 ⎞ ⎫ 1 M qDL = × 5360,6207 × 16,52 ⎨ ⎜1 − ⎟⎬ 2 ⎩ 16,5 ⎝ 16,5 ⎠ ⎭ MTb= ½ × 388,8 × 8,25 M4 DL

=

134015,5175 kgm

= =

777,6000 kgm 134793,1175 kgm 1347931,1750 Nm

=

168859,5521 kgm

=

1166,4000 kgm 170025,9521 kgm 1700259,5210 Nm

=

182428,6232 kgm

=

1603,8000 kgm 184032,4232 kgm 1840324,2320 Nm

c. Beban hidup (live load) koefisien kejut = 1,2963 beban garis, P = 1,2963 ×

12000 × 4,5716 = 25859,6294 kg 2,75

beban terbagi merata, q =

2200 × 4,5716 = 3657,28 kg/m 2,75

d. Momen lentur akibat beban hidup ⎧x⎛ x ⎞⎫ M x (P ) = P × L ⎨ ⎜ 1 − ⎟ ⎬ L L ⎠⎭ ⎩ ⎝ M x (q ) =

⎧x⎛ x ⎞⎫ 1 q × L2 ⎨ ⎜1 − ⎟ ⎬ 2 ⎩ L ⎝ L ⎠⎭

Momen pada potongan 1, x = 2,0 m (M1 LL) ⎧ 2 ⎛ 2 ⎞⎫ M x (P ) = 25859,6294 × 16,5⎨ ⎜1 − ⎟⎬ ⎩16,5 ⎝ 16,5 ⎠ ⎭ M x (q ) =

M1 LL

⎧ 2 ⎛ 1 2 ⎞⎫ × 3657,28 × 16,52 ⎨ ⎜1 − ⎟⎬ 2 ⎩16,5 ⎝ 16,5 ⎠ ⎭

=

45450,2577 kgm

=

53030,5600 kgm

=

98480,8177 kgm 984808,1770 Nm

9

Momen pada potongan 2, x = 4,0 m (M2 LL) ⎧ 4 ⎛ 4 ⎞⎫ M x (P ) = 25859,6294 × 16,5⎨ ⎜1 − ⎟⎬ ⎩16,5 ⎝ 16,5 ⎠ ⎭

M x (q ) =

⎧ 4 ⎛ 1 4 ⎞⎫ × 3657,28 × 16,52 ⎨ ⎜1 − ⎟⎬ 2 ⎩16,5 ⎝ 16,5 ⎠ ⎭

M2 LL

Momen pada potongan 3, x = 6,0 m (M3 LL) ⎧ 6 ⎛ 6 ⎞⎫ M x (P ) = 25859,6294 × 16,5⎨ ⎜1 − ⎟⎬ ⎩16,5 ⎝ 16,5 ⎠ ⎭

M x (q ) =

⎧ 6 ⎛ 1 6 ⎞⎫ × 3657,28 × 16,52 ⎨ ⎜1 − ⎟⎬ 2 16 , 5 16 ,5 ⎠ ⎭ ⎝ ⎩

=

78362,5133 kgm

=

91432,0000 kgm

=

169794,5133 kgm 1697945,1330 Nm

=

98736,7668 kgm

=

115204,3200 kgm

M3 LL

213941,0868 kgm 2139410,8680 Nm

Momen pada potongan 4, x = 8,25 m (M4 LL) ⎧ 8,25 ⎛ 8,25 ⎞ ⎫ M x (P ) = 25859,6294 × 16,5⎨ ⎜1 − ⎟⎬ ⎩ 16,5 ⎝ 16,5 ⎠ ⎭

M x (q ) =

⎧ 8,25 ⎛ 8,25 ⎞ ⎫ 1 × 3657,28 × 16,52 ⎨ ⎜1 − ⎟⎬ 2 ⎩ 16,5 ⎝ 16,5 ⎠ ⎭

M4 LL

=

106670,9713 kgm

=

124461,8100 kgm 231132,7813 kgm 2311327,8130 Nm

Tabel. Momen lentur total Pembebanan Beban mati, DL Beban hidup, LL Total, Mu (1,2MD+1,6ML)

M.1 781178,0020 984808,1770

M.2 1347931,1750 1697945,1330

M.3 M.4 1700259,5210 1840324,2320 2139410,8680 2311327,8130

2513106,6856

4334229,6228

5463368,8140 5906513,5792

e. Gaya geser (shearing force) Beban mati terbagi merata Balok melintang Beban hidup garis P Beban hidup terbagi merata q Total V

= = = =

0,5 × 5360,6207 × 16,5 1,4 × 388,8 0,5 × 25859,6294 0,5 × 3657,28 × 16,5

44225,1208 544,3200 12928,8147 30172,5600 87870,8155 878708,1550

kg kg kg kg kg N

10

f. Perhitungan baja tulangan Pada tumpuan V

=

878708,1550 N

h

=

1300 mm

b

=

500 mm

d

=

1300 - 60 = 1240 mm

Perencanaan tulangan geser Vu = 878708,1550 N Vc =

1 6

f c' × b × d =

1 25 × 500 × 1217,5 = 507291,6667 N 6

1 1 φVc = × 0,6 × 507291,6667 = 152187,5 N < Vu 2 2 878708,1550

698350.141

(perlu sengkang)

517992.127 517447.807

CL 337089.793 111642.2755

Gambar 7. Diagram gaya geser (SFD) Hasil perhitungan dapat dilihat pada tabel berikut No. 1 2 3

Penampang kritis Vu (N) Vc (N) ½ φ Vc (N)

4 5 6 7

Vs s s mak Dipakai

(N) (mm) (mm)

titik 1 0-2m 878708.1550 507291.6667 152187.5 Perlu sengkang 957221.925 79.91645314 608.75 D10 - 75



titik 4 6 - 8,25 m 337089.793 507291.6667 152187.5 Perlu sengkang 54524.655 1402.994317 608.75 D10 - 500

Potongan I-I (8,25 m dari tumpuan) bE = 1400 mm hf = 300 mm

lebar efektif, diambil nilai terkecil dari : bE = 14 L = 14 × 17,5 = 4,375 m h = 1300 mm

bE = bw + 16h f = 500 + (16 × 300) = 5300 mm

bE = jarak gelagar = 1400 mm bw = 500 mm

11

Mu = 5906513,5792 N-m cb = db

0,003 0,003 +

fy

0,003

a b = 0,85

0,003 +

Es

fy

db

Es

ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm dalam keadaan setimbang (ΣH = 0) Ab × f y = 0,85 × f c' × {a b × bw + (b f − bw )× t }

Ab =

0,85 × f c' × {a b × bw + (b f − bw )× t } 0,85 × 25 × {756 × 500 + (1400 − 500) × 300} = = 34425 mm2 fy 400

kemampuan sayap mendukung momen t⎫ 300 ⎫ ⎧ ⎧ M = b f × t × 0,85 × f c' × ⎨d − ⎬ = 1400 × 300 × 0,85 × 25 × ⎨1260 − ⎬ = 9906750000 Nmm 2 2 ⎭ ⎩ ⎭ ⎩

M = 9906750 Nm > 5906513,5792 Nm → blok beton a ada di dalam sayap Letak garis netral, c a⎫ ⎧ M = b f × a × 0,85 × f c' × ⎨d − ⎬ 2⎭ ⎩ a⎫ ⎧ 5906513579,2 = 1400 × a × 0,85 × 25 × ⎨1260 − ⎬ 2⎭ ⎩

a2 – 2520a + 397076,5431 = 0 a = 168,8889 mm, c = 168,8889/0,85 = 198,6928 mm luas tulangan yang diperlukan A=

0,85 × f c' × bw × a 0,85 × 25 × 500 × 168,8889 = = 4486,1114 mm2 < 0,75×Ab = 0,75×34425 400 fy

Dipakai tulangan ∅30 (As = 706,8583 mm2), jumlah tulangan yang dibutuhkan n=

4486,1114 = 6,3 706,8583

dipakai 8∅30 (As = 5654,8664 mm2)

Potongan II-II (6 m dari tumpuan) Mu = 5463368,8140 N-m cb = db

0,003 0,003 +

fy Es

a b = 0,85

0,003 0,003 +

fy

db

Es

ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm

12

dalam keadaan setimbang (ΣH = 0) Ab × f y = 0,85 × f c' × {a b × bw + (b f − bw )× t } 0,85 × f c' × {a b × bw + (b f − bw )× t } 0,85 × 25 × {756 × 500 + (1400 − 500 ) × 300} = = 34425 mm2 Ab = fy 400

kemampuan sayap mendukung momen t⎫ 300 ⎫ ⎧ ⎧ M = b f × t × 0,85 × f c' × ⎨d − ⎬ = 1400 × 300 × 0,85 × 25 × ⎨1260 − ⎬ = 9906750000 Nmm 2⎭ 2 ⎭ ⎩ ⎩

M = 9906750 Nm > 5463368,8140 Nm → blok beton a ada di dalam sayap Letak garis netral, c a⎫ ⎧ M = b f × a × 0,85 × f c' × ⎨d − ⎬ 2⎭ ⎩ a⎫ ⎧ 5463368814,0 = 1400 × a × 0,85 × 25 × ⎨1260 − ⎬ 2⎭ ⎩


0,85 × f c' × bw × a 0,85 × 25 × 500 × 155,3214 = = 4125,7247 mm2 < 0,75×Ab = 0,75×34425 fy 400


4125,7247 = 5,8 706,8583

dipakai 6∅30 (As = 4241,1501 mm2)

Potongan III-III (4 m dari tumpuan) Mu = 4334229,6228 N-m cb = db

0,003 0,003 +

fy Es

a b = 0,85

0,003 0,003 +

fy

db

Es

ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm dalam keadaan setimbang (ΣH = 0) Ab × f y = 0,85 × f c' × {a b × bw + (b f − bw )× t } Ab =

0,85 × f c' × {a b × bw + (b f − bw )× t } 0,85 × 25 × {756 × 500 + (1400 − 500 ) × 300} = = 34425 mm2 fy 400

13

kemampuan sayap mendukung momen 300 ⎫ t⎫ ⎧ ⎧ M = b f × t × 0,85 × f c' × ⎨d − ⎬ = 1400 × 300 × 0,85 × 25 × ⎨1260 − ⎬ = 9906750000 Nmm 2 ⎭ 2⎭ ⎩ ⎩

M = 9906750 Nm > 4334229,6228 Nm → blok beton a ada di dalam sayap Letak garis netral, c a⎫ ⎧ M = b f × a × 0,85 × f c' × ⎨d − ⎬ 2⎭ ⎩

a⎫ ⎧ 4334229622,8 = 1400 × a × 0,85 × 25 × ⎨1260 − ⎬ 2⎭ ⎩


0,85 × f c' × bw × a 0,85 × 25 × 500 × 121,4820 = = 3226,8656 mm2 < 0,75×Ab = 0,75×34425 fy 400


3226,8656 = 4,6 706,8583

dipakai 6∅30 (As = 4241,1501 mm2)

Potongan IV- IV (2 m dari tumpuan) Mu = 2513106,6856 N-m cb = db

0,003 0,003 +

fy Es

a b = 0,85

0,003 0,003 +

fy

db

Es

ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm dalam keadaan setimbang (ΣH = 0) Ab × f y = 0,85 × f c' × {a b × bw + (b f − bw )× t } Ab =

0,85 × f c' × {a b × bw + (b f − bw )× t } 0,85 × 25 × {756 × 500 + (1400 − 500 ) × 300} = = 34425 mm2 fy 400

kemampuan sayap mendukung momen 300 ⎫ t⎫ ⎧ ⎧ M = b f × t × 0,85 × f c' × ⎨d − ⎬ = 1400 × 300 × 0,85 × 25 × ⎨1260 − ⎬ = 9906750000 Nmm 2⎭ 2 ⎭ ⎩ ⎩

M = 9906750 Nm > 2513106,6856 Nm → blok beton a ada di dalam sayap

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Letak garis netral, c a⎫ ⎧ M = b f × a × 0,85 × f c' × ⎨d − ⎬ 2⎭ ⎩ a⎫ ⎧ 2513106685,6 = 1400 × a × 0,85 × 25 × ⎨1260 − ⎬ 2⎭ ⎩


0,85 × f c' × bw × a 0,85 × 25 × 500 × 68,9284 = = 1830,9106 mm2 < 0,75×Ab = 0,75×34425 fy 400


1830,9106 = 2,6 706,8583

dipakai 3∅30 (As = 2120,5750 mm2)

Tabel Penulangan balok Pembebanan Beban mati, DL Beban hidup, LL Total, Mu (1,2MD+1,6ML) tulangan

M.1 781178,0020 984808,1770

M.2 1347931,1750 1697945,1330

M.3 M.4 1700259,5210 1840324,2320 2139410,8680 2311327,8130

2513106,6856

4334229,6228

5463368,8140 5906513,5792

3∅30

6∅30

6∅30

8∅30

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DAFTAR PUSTAKA

Agus Iqbal Manu, Ir.,Dipl. Heng., 1995, Dasar-Dasar Perencanaan Jembatan Beton Bertulang, Cetakan I,P.T. Mediatana Saptakarya, Jakarta Bambang Supriyadi, DR.,Ir., CES.,DEA., 2000, Jembatan, Edisi pertama, Beta Offset, Jogjakarta Departemen Pekerjaan Umum, Standar Bangunan Atas Jembatan Gelagar Beton Bertulang Tipe T, 1993, Departemen Pekerjaan Umum Ditjen Bina Marga Dit. Bina Program Jalan Subdit. Perencanaan Teknik Jembatan

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Jembatan Balok T

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