Jembatan Balok T

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PERANCANGAN JEMBATAN BETON BERTULANG DENGAN

TAMPANG BALOK T

Gambar 1. Penampang melintang jembatan

1. Kondisi Jembatan

• Panjang bentang : 17,5 m • Lebar jembatan : 9 m • Lebar perkerasan : 7 m

• Tipe jembatan : beton bertulang dengan gelagar balok T • Jumlah balok gelagar : 6 buah

• Panjang bersih gelagar : 16,5 m 2. Spesifikasi Pembebanan

a. Beban hidup : PPJJR No. 12/1970 (BM 100 %) • Beban roda T : 100% x 10 t = 10 t • Beban garis P : 100% x 12 t/m = 12 t/m • Beban merata q : 100% x 2,2 t/m2_{= 2,2 t/m}2 b. Beban kejut, 1,2963 5 , 17 50 20 1 50 20 1 = + + = + + = L k

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

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PERANCANGAN 1. Tiang sandaran momen lentur, Mu = 1,2×2×100×1,0 = 240 kg-m = 2400 N-m gaya geser, V = 1,2 × 2 × 100 = 240 kg = 2400 N Mn = φ bd2k Mu = Mn 1095 , 1 130 160 8 , 0 10 2400 2 3 2 _× _× = × = × × = d b M k u φ Mpa 3 ' ' 10 8502 , 2 25 85 , 0 1095 , 1 2 1 1 400 25 85 , 0 85 , 0 2 1 1 85 , 0 _⎟⎟= × − ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = c y c perlu f k f f ρ 3 min 3,5 10 400 4 , 1 4 , 1 ₌ ₌ _× − = y f ρ 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)

5 , 18 160 25 85 , 0 400 0796 , 157 85 , 0 ' × × = × = × × × = b f f A a c y s _mm 7647 , 21 85 , 0 5 , 18 1 = = = β a c mm 7847 , 2983 7647 , 21 7647 , 21 130 600 600 ⎟= ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = c c d f_s MPa > fy O K 68 , 7586944 2 5 , 18 130 400 0796 , 157 2 ⎟⎠= ⎞ ⎜ ⎝ ⎛ ₋ × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × =A f d a M_n _s _y N-mm =7586,9447 N-m > Mu (2400 N-m) O K

Perencanaan tulangan geser Vu = 2400 N 3333 , 17333 130 160 20 6 1 6 1 '_× _× ₌ _× _× ₌ = f b d V_c _c N 9999 , 5199 3333 , 17333 6 , 0 2 1 2 1 ₌ _× _× ₌ c V

φ N > Vu (secara teoritis tidak perlu sengkang)

b=160 mm

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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

3333 , 43 400 65 160 25 3 1 3 1 ' min = × × = × × = y c v f s b f A mm2

dipakai tulangan ∅8 (As = 100,5310 mm2_{), maka jarak sengkang}

7965 , 150 160 25 3 1 400 5310 , 100 3 1 ' = × × = × × = b f f A s c y v _mm

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. Volume (m3₎ _γ (kg/m3₎ W (kg) Lengan (m) Momen (kg-m) 1 0,10 × 0,16 × 0,50 = 0,008 2400 19,2 1,8 34,5600 2 0,10×(0,70×0,110)/2 = 0,00385 2400 9,24 1,04 9,6096 3 0,10×0,05×0,50 = 0,0025 2400 6 1,025 6,1500 4 0,10 × (0,15 × 0,50)/2 = 0,00375 2400 9 0,95 8,5500 5 1,00 × 1,00 × 0,20 = 0,2 2400 480 0,5 240,0000 6 1,00 × (1,00 × 0,10)/2 = 0,05 2400 120 0,33 39,6000 7 1,00 × 0,90 × 0,07 = 0,063 2200 138,6 0,375 51,9750 P 2,0 × 100 kg/m 200 1,2 240,0000

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T 1,2963 × 10000 12963 0,5 6481,5000 Air hujan = 2 × 0,90 × 0,05 = 0,0625 1000 62,5 0,375 23,4375

Railing = 2 × 2m× 6 kg/m = 24 24 1,08 25,9200

Total momen, M 7161,3021

Total momen, M (N-m) 71613,0210

b. Gaya geser (shear force)

Berat tiang sandaran = 1 + 2 + 3 +4 + railing = 67,4400 Kg Slab kantilever dan perkerasan = 5 + 6 +7 = 738,6000 Kg

Beban roda = 12963,0000 Kg

Beban genangan air hujan = 62,5000 Kg

Toal gaya lintang = 13831,5400 Kg

= 138315,4000 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 d = 300-40 = 260 mm 5890 , 1 260 1000 8 , 0 10 85935,6252 2 3 2 _× _× = × = × × = d b M k u φ MPa 027094 , 0 200000 400 003 , 0 003 , 0 400 85 , 0 25 85 , 0 003 , 0 003 , 0 85 , 0 1 ' = + × × = + × × = s y y c b E f f f β ρ ρmaks = 0,75 ρb = 0,75 x 0,027094 = 0,0203205 3 ' ' 10 1333 , 4 25 85 , 0 5890 , 1 2 1 1 400 25 85 , 0 85 , 0 2 1 1 85 , 0 _⎟⎟= × − ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = c y c perlu f k f f ρ 3 min 3,5 10 400 4 , 1 4 , 1 ₌ ₌ _× − = y f ρ 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

4686 , 195 658 , 1074 1000 0619 , 210 × ₌ = perlu s mm dipakai tulangan ∅16-125 mm kontrol terhadap geser beton

7296 , 0 260 1000 165978,48 8 7 8 7× × = × × = = h b V c τ MPa < 0,45 fc = 11,25 MPa O K

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

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

P 20 cm 21 21 6 cm 15 cm 15 cm 50 cm 21 21 P

Gambar 4. Penyebaran beban hidup pada slab

lx = 1,4 m

ly = ∞

tx = 0,92 m

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Beban roda, T = 10000 kg

Bidang kontak = 0,92 m × 0,62 m

Penyebaran beban roda, 22726,1571 62 , 0 92 , 0 2963 , 1 10000 ₌ × × = T kg/m2

Dipakai tabel-Bittner (dari Dr. Ing Ernst Bittner)

Dengan lx = 1,4 , ly = ∞ (lantai tidak menumpu pada diafragma) 657 , 0 4 , 1 92 , 0 ₌ = x x l t fxm = 0,1233 443 , 0 4 , 1 62 , 0 ₌ = x y l t fym = 0,0661 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

Berat slab = 0,30 × 2400 = 720 kg/m2

Berat perkerasan = 0,06 × 2200 = 132 kg/m2

Berat air hujan = 0,05 × 1000 = 50 kg/m2

Total qDL = 902 kg/m2 7920 , 176 4 , 1 902 10 1 10 1 _× _× 2₌ _× _× 2₌ = _DL _x xm q l M kgm = 1767,920 Nm 9307 , 58 7920 , 176 3 1 3 1_× ₌ _× ₌ = _xm ym M M kgm = 589,307 Nm 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 d = 300-40 = 260 mm 3267 , 0 260 1000 8 , 0 10 17661,299 2 3 2 _× _× = × = × × = d b M k φ MPa 027094 , 0 200000 400 003 , 0 003 , 0 400 85 , 0 25 85 , 0 003 , 0 003 , 0 85 , 0 1 ' = + × × = + × × = s y y c b E f f f β ρ ρmaks = 0,75 ρb = 0,75 x 0,027094 = 0,0203205

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4 ' ' 10 2313 , 8 25 85 , 0 3267 , 0 2 1 1 400 25 85 , 0 85 , 0 2 1 1 85 , 0 _⎟⎟= × − ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = c y c perlu f k f f ρ 3 min 3,5 10 400 4 , 1 4 , 1 ₌ ₌ _× − = y f ρ As = ρ x b x d = 3,5 x10-3 x 1000 x 260 = 910 mm2

8373 , 230 910 1000 0619 , 210 × ₌ = perlu s mm dipakai tulangan ∅16-125 mm arah memanjang ly M = 9157,85 Nm h = 300 mm d = 300-40 = 260 mm 1693 , 0 260 1000 8 , 0 10 9157,85 2 3 2 _× _× = × = × × = d b M k φ MPa 027094 , 0 200000 400 003 , 0 003 , 0 400 85 , 0 25 85 , 0 003 , 0 003 , 0 85 , 0 1 ' = + × × = + × × = s y y c b E f f f β ρ ρmaks = 0,75 ρb = 0,75 x 0,027094 = 0,0203205 4 ' ' 10 2495 , 4 25 85 , 0 1693 , 0 2 1 1 400 25 85 , 0 85 , 0 2 1 1 85 , 0 _⎟⎟= × − ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = c y c perlu f k f f ρ 3 min 3,5 10 400 4 , 1 4 , 1 ₌ ₌ _× − = y f ρ As = ρ x b x d = 3,5 x10-3 x 1000 x 260 = 910 mm2

8373 , 230 910 1000 0619 , 210 × ₌ = perlu s mm dipakai tulangan ∅16-125 mm 4. Perhitungan Gelagar a. beban mati (dead load)

Hand rail = {(0,10 × 0,16 × 1,00 × 2400)/2} × 1,1871 = 22,7923 kg/m Railing = 2 × 1,00 × 6 × 1,1871 = 14,2452 kg/m Perkerasan = 0,06 × 2200 × 4,5716 = 603,4512 kg/m

Air hujan = 0,05 × 1000 × 4,5716 = 228,5800 kg/m

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Gelagar = 1,00 × 0,50 × 2400 × 1,00 = 1200,0000 kg/m

Total = 5360,6207 kg/m

Balok melintang (diafragma), Tb = 0,30 × 0,60 × 2400 × 0,9 = 388,8 kg

Gambar 5. Garis pengaruh momen

Gambar 6. Potongan memanjang balok pada perhitungan momen lentur b. momen lentur akibat beban mati

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × = → L x L x L q M M_qDL _x _DL 1 2 1 2

Momen pada potongan 1, x = 2,0 m (M1 DL)

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × × = 5 , 16 2 1 5 , 16 2 5 , 16 6207 , 5360 2 1 2 qDL M _{= 77729,0002}_kgm MTb= ½ × 388,8 × 2 = 388,8000 kgm M1 DL = 78117,8002 kgm 781178,0020 Nm

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Momen pada potongan 2, x = 4,0 m (M2 DL) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × × = 5 , 16 4 1 5 , 16 4 5 , 16 6207 , 5360 2 1 2 qDL M _{= 134015,5175}_kgm MTb= ½ × 388,8 × 4 = 777,6000 kgm M2 DL = 134793,1175 kgm 1347931,1750 Nm Momen pada potongan 3, x = 6,0 m (M3 DL)

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × × = 5 , 16 6 1 5 , 16 6 5 , 16 6207 , 5360 2 1 2 qDL M _{= 168859,5521}_kgm MTb= ½ × 388,8 × 6 = 1166,4000 kgm M3 DL 170025,9521 kgm 1700259,5210 Nm Momen pada potongan 4, x = 8,25 m (M4 DL)

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × × = 5 , 16 25 , 8 1 5 , 16 25 , 8 5 , 16 6207 , 5360 2 1 2 qDL M = 182428,6232 kgm MTb= ½ × 388,8 × 8,25 = 1603,8000 kgm M4 DL 184032,4232 kgm 1840324,2320 Nm

c. Beban hidup (live load) koefisien kejut = 1,2963 beban garis, 4,5716 25859,6294 75 , 2 12000 2963 , 1 × × = = P kg

beban terbagi merata, 4,5716 3657,28 75 , 2 2200_× ₌ = q kg/m

d. Momen lentur akibat beban hidup

( )

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × = L x L x L P P M_x 1

( )

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

Momen pada potongan 1, x = 2,0 m (M1 LL)

( )

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × = 5 , 16 2 1 5 , 16 2 5 , 16 6294 , 25859 P Mx = 45450,2577 kgm

( )

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × × = 5 , 16 2 1 5 , 16 2 5 , 16 28 , 3657 2 1 2 q M_x _{= 53030,5600}_kgm M1 LL = 98480,8177 kgm 984808,1770 Nm

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Momen pada potongan 2, x = 4,0 m (M2 LL)

( )

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × = 5 , 16 4 1 5 , 16 4 5 , 16 6294 , 25859 P M_x _{= 78362,5133}_kgm

( )

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × × = 5 , 16 4 1 5 , 16 4 5 , 16 28 , 3657 2 1 2 q M_x _{= 91432,0000}_kgm M2 LL = 169794,5133 kgm 1697945,1330 Nm Momen pada potongan 3, x = 6,0 m (M3 LL)

( )

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × = 5 , 16 6 1 5 , 16 6 5 , 16 6294 , 25859 P Mx = 98736,7668 kgm

( )

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × × = 5 , 16 6 1 5 , 16 6 5 , 16 28 , 3657 2 1 2 q M_x _{= 115204,3200}_kgm M3 LL 213941,0868 kgm 2139410,8680 Nm Momen pada potongan 4, x = 8,25 m (M4 LL)

( )

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × = 5 , 16 25 , 8 1 5 , 16 25 , 8 5 , 16 6294 , 25859 P M_x _{= 106670,9713}_kgm

( )

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × × = 5 , 16 25 , 8 1 5 , 16 25 , 8 5 , 16 28 , 3657 2 1 2 q Mx = 124461,8100 kgm M4 LL 231132,7813 kgm 2311327,8130 Nm

Tabel. Momen lentur total

Pembebanan M.1 M.2 M.3 M.4

Beban mati, DL

Beban hidup, LL 781178,0020 984808,1770 1347931,1750 1697945,1330 1700259,5210 2139410,8680 1840324,2320 2311327,8130 Total, Mu

(1,2MD+1,6ML) 2513106,6856 4334229,6228 5463368,8140 5906513,5792

e. Gaya geser (shearing force)

Beban mati terbagi merata = 0,5 × 5360,6207 × 16,5 44225,1208 kg

Balok melintang = 1,4 × 388,8 544,3200 kg

Beban hidup garis P = 0,5 × 25859,6294 12928,8147 kg

Beban hidup terbagi merata q = 0,5 × 3657,28× 16,5 30172,5600 kg

Total V 87870,8155 kg

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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 = 878708,1550N 6667 , 507291 1217,5 500 25 6 1 6 1 '_× _× ₌ _× _× ₌ = f b d V_c _c N 5 , 152187 6667 , 507291 6 , 0 2 1 2 1 ₌ _× _× ₌ c V φ N < Vu (perlu sengkang)

Gambar 7. Diagram gaya geser (SFD) Hasil perhitungan dapat dilihat pada tabel berikut

No. Penampang titik 1 titik 2 titik 3 titik 4

kritis 0 - 2 m 2 - 4 m 4 - 6 m 6 - 8,25 m

1 Vu (N) 878708.1550 698350.141 517992.127 337089.793

2 Vc (N) 507291.6667 507291.6667 507291.6667 507291.6667

3 ½ φ Vc (N) 152187.5 152187.5 152187.5 152187.5

Perlu sengkang Perlu sengkang Perlu sengkang Perlu sengkang

4 Vs (N) 957221.925 656625.235 356028.545 54524.655

5 s (mm) 79.91645314 116.5014334 214.8641793 1402.994317

6 s mak (mm) 608.75 608.75 608.75 608.75

7 Dipakai D10 - 75 D10 - 110 D10 - 200 D10 - 500

Potongan I-I (8,25 m dari tumpuan)

lebar efektif, diambil nilai terkecil dari : 375 , 4 5 , 17 4 1 4 1 = × = = L b_E m

(

16 300

)

5300 500 16 = + × = + = w f E b h b mm 1400 = = jarak gelagar b_E mm C L 878708,1550 698350.141 _517992.127 517447.807 337089.793 111642.2755 bw = 500 mm bE = 1400 mm hf = 300 mm h = 1300 mm

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Mu = 5906513,5792 N-m s y b b E f d c + = 003 , 0 003 , 0 b s y b d E f a + = 003 , 0 003 , 0 85 , 0 ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm

dalam keadaan setimbang (ΣH = 0)

(

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{

a b b b t

}

f f A_b× _y =0,85× _c'× _b× _w+ _f − _w ×

(

)

{

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{

(

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34425 400 300 500 1400 500 756 25 85 , 0 85 , 0 ' = × − + × × × = × − + × × × = y w f w b c b f t b b b a f A mm2

kemampuan sayap mendukung momen

9906750000 2 300 1260 25 85 , 0 300 1400 2 85 , 0 ' = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ₋ × × × × = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − × × × × =b t f d t M _f _c Nmm

M = 9906750 Nm > 5906513,5792 Nm → blok beton a ada di dalam sayap

Letak garis netral, c

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − × × × × = 2 85 , 0 f' d a a b M _f _c ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ₋ × × × × = 2 1260 25 85 , 0 1400 ,2 5906513579 a a a2_{– 2520a + 397076,5431 = 0} a = 168,8889 mm, c = 168,8889/0,85 = 198,6928 mm luas tulangan yang diperlukan

1114 , 4486 400 8889 , 168 500 25 85 , 0 85 , 0 × '× × ₌ × × × ₌ = y w c f a b f A mm2 < 0,75×Ab = 0,75×34425

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

3 , 6 8583 , 706 1114 , 4486 ₌ = n dipakai 8∅30 (As = 5654,8664 mm2)

Potongan II-II (6 m dari tumpuan) Mu = 5463368,8140 N-m s y b b E f d c + = 003 , 0 003 , 0 b s y b d E f a + = 003 , 0 003 , 0 85 , 0 ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm

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(

)

{

a b b b t

}

f f A_b× _y =0,85× _c'× _b× _w+ _f − _w ×

(

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{

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{

(

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9906750000 2 300 1260 25 85 , 0 300 1400 2 85 , 0 ' = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ₋ × × × × = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − × × × × =b t f d t M _f _c Nmm

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − × × × × = 2 85 , 0 f' d a a b M f c ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ₋ × × × × = 2 1260 25 85 , 0 1400 ,0 5463368814 a a a2_{– 2520a + 367285,2984 = 0} a = 155,3214 mm, c = 155,3214/0,85 = 182,7311 mm luas tulangan yang diperlukan

7247 , 4125 400 155,3214 500 25 85 , 0 85 , 0 × '× × ₌ × × × ₌ = y w c f a b f A mm2 < 0,75×Ab = 0,75×34425

8 , 5 8583 , 706 7247 , 4125 ₌ = n dipakai 6∅30 (As = 4241,1501 mm2)

Potongan III-III (4 m dari tumpuan) Mu = 4334229,6228 N-m s y b b E f d c + = 003 , 0 003 , 0 b s y b d E f a + = 003 , 0 003 , 0 85 , 0 ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm

(

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{

a b b b t

}

f f A_b× _y =0,85× _c'× _b× _w+ _f − _w ×

(

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{

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

⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − × × × × = 2 85 , 0 f' d a a b M _f _c ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ₋ × × × × = 2 1260 25 85 , 0 1400 ,8 4334229622 a a a2_{– 2520a + 291376,7813 = 0} a = 121,4820 mm, c = 121,4820/0,85 = 142,92 mm luas tulangan yang diperlukan

8656 , 3226 400 121,4820 500 25 85 , 0 85 , 0 × '× × ₌ × × × ₌ = y w c f a b f A mm2 < 0,75×Ab = 0,75×34425

6 , 4 8583 , 706 8656 , 3226 ₌ = n dipakai 6∅30 (As = 4241,1501 mm2)

Potongan IV- IV (2 m dari tumpuan) Mu = 2513106,6856 N-m s y b b E f d c + = 003 , 0 003 , 0 b s y b d E f a + = 003 , 0 003 , 0 85 , 0 ab = 0,6 db = 0,6 (1300-40) = 756 mm > 300 mm

(

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{

a b b b t

}

f f A_b× _y =0,85× _c'× _b× _w+ _f − _w ×

(

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9906750000 2 300 1260 25 85 , 0 300 1400 2 85 , 0 ' = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ₋ × × × × = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − × × × × =b t f d t M _f _c Nmm

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Letak garis netral, c ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − × × × × = 2 85 , 0 f' d a a b M f c ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ₋ × × × × = 2 1260 25 85 , 0 1400 ,6 2513106685 a a a2_{– 2520a + 168948,3486 = 0} a = 68,9284 mm, c = 68,9284/0,85 = 81,0922 mm luas tulangan yang diperlukan

9106 , 1830 400 68,9284 500 25 85 , 0 85 , 0 × '× × ₌ × × × ₌ = y w c f a b f A mm2 < 0,75×Ab = 0,75×34425

6 , 2 8583 , 706 9106 , 1830 ₌ = n dipakai 3∅30 (As = 2120,5750 mm2)

Tabel Penulangan balok

Pembebanan M.1 M.2 M.3 M.4 Beban mati, DL Beban hidup, LL 781178,0020 984808,1770 1347931,1750 1697945,1330 1700259,5210 2139410,8680 1840324,2320 2311327,8130 Total, Mu (1,2MD+1,6ML) 2513106,6856 4334229,6228 5463368,8140 5906513,5792 tulangan 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