PERENCANA
PERENCANAAN AN JEMBATAN PRATEGANGJEMBATAN PRATEGANG
PERENCANAAN JEMBATAN PRATEGANG PERENCANAAN JEMBATAN PRATEGANG Data Teknis P
Data Teknis Perencanaaerencanaan Jembatann Jembatan a. Jembatan
a. Jembatan Kelas
Kelas jalan jalan : : Kelas Kelas 11 Jumlah
Jumlah jalur jalur : : 2 2 jalurjalur Panjang
Panjang jembatan jembatan : : 40 40 metermeter Lebar
Lebar jembatan jembatan : : 9 9 metermeter Lebar
Lebar lantai lantai kendaraan kendaraan : : 7 7 metermeter Tipe
Tipe gelagar gelagar : : Balok Balok II Tebal
Tebal Perkerasan Perkerasan : : 5 5 cmcm
Gambar Bentang Jembatan
Gambar Bentang Jembatan
b. Trotoir b. Trotoir Jenis
Jenis konstruksi konstruksi : : Beton Beton bertulangbertulang Pipa
Pipa sandaran sandaran : : Circular Circular Hollow Hollow Sections Sections D D 60.5 60.5 mmmm Dimensi
Dimensi tiang tiang sandaran sandaran : : 20/15 20/15 cmcm Jarak
Jarak antar antar tiang tiang : : 2 2 mm Mutu beton, f‟c
Mutu beton, f‟c : 30 Mpa: 30 Mpa
Mutu
Mutu baja baja tulangan, tulangan, fy fy : : 240 240 Mpa Mpa (polos)(polos) Mutu
Mutu baja baja pipa pipa sandaran sandaran : : 1600 1600 MpaMpa Lebar
Lebar trotoir trotoir : : 100 100 cmcm Tebal
Tebal trotoir trotoir : : 25 25 cmcm Balok
Balok kerb kerb : : 20/25 20/25 cmcm Jenis
Jenis plat plat trotoir trotoir : : Beton Beton TumbukTumbuk c. Plat lantai kendaraan
c. Plat lantai kendaraan Tebal
Tebal plat plat : : 20 20 cmcm Mutu beton, f‟c
Mutu beton, f‟c : 30 Mpa: 30 Mpa
Mutu
Mutu baja baja tulangan, tulangan, fy fy : : 350 350 Mpa Mpa (ulir)(ulir) d. Gelagar
d. Gelagar Jenis
Jenis konstruksi konstruksi : : Beton Beton prategang prategang tipe tipe balok balok II Mutu beton, f‟c
Mutu beton, f‟c : 50 Mpa: 50 Mpa
Mutu
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Tipe
Tipe tendon tendon & & angkur angkur : : Angker Angker hidup hidup VSL VSL tipe tipe ScSc e. Abutment
e. Abutment Tinggi
Tinggi Abutment Abutment : : 6 6 metermeter Lebar
Lebar Abutment Abutment : : 11.6 11.6 metermeter Tipe
Tipe Abutment Abutment : : Type Type KantileverKantilever Mutu beton, f‟c
Mutu beton, f‟c : 30 Mpa: 30 Mpa
Mutu
Mutu baja baja tulangan, tulangan, fy fy : : 240 240 Mpa Mpa (polos)(polos) Mutu
Mutu baja baja tulangan, tulangan, fy fy : : 350 350 Mpa Mpa (ulir)(ulir)
Gambar Abutment
Gambar Abutment
Tegangan Yang Diijinkan (SNI 03
Tegangan Yang Diijinkan (SNI 03 – –28472847 – – 2002)2002) Tegangan Ijin Beton
Tegangan Ijin Beton PrategangPrategang
Mutu beton prategang (f‟c) 50 Mpa. Tegangan ijin sesuai dengan kondisi gaya pratekan dan
Mutu beton prategang (f‟c) 50 Mpa. Tegangan ijin sesuai dengan kondisi gaya pratekan dan
tegangan beton pada tahap beban kerja, tidak boleh melampaui nilai berikut: tegangan beton pada tahap beban kerja, tidak boleh melampaui nilai berikut: 1.
1. Keadaan awal, sesaat sesudah penyaluran gaya prategang (sebelum terjadinya kehilanganKeadaan awal, sesaat sesudah penyaluran gaya prategang (sebelum terjadinya kehilangan tegangan) (pasal 20.4.1)
tegangan) (pasal 20.4.1) 2.
2. Tegangan serat tekan terluarTegangan serat tekan terluar Untuk
Untuk Gelagar Gelagar ~Untuk ~Untuk PlatPlat f‟
f‟bb= 0.6 f‟= 0.6 f‟ccf‟f‟b‟b‟= 0.6 f‟c‟= 0.6 f‟c‟
= 0.6
= 0.6 x x 50 50 = = 0.60.6x x 3030 =
= 30 30 Mpa Mpa = = 18 18 MpaMpa ~Untuk
~Untuk Gelagar Gelagar ~Untuk ~Untuk PlatPlat 1. 1. fftt= ¼ = ¼ fft‟t‟= ¼= ¼ = ¼ = ¼x x = ¼ = ¼x x =
= 1.768 1.768 Mpa Mpa = = 1.369 1.369 MpaMpa 2.
2. Keadaan akhir, setelah kehilangan gaya prategang (pasal 20.4.2)Keadaan akhir, setelah kehilangan gaya prategang (pasal 20.4.2) 1.
1.Tegangan serat tekan terluarTegangan serat tekan terluar ~Untuk
~Untuk Gelagar Gelagar ~Untuk ~Untuk PlatPlat f‟
f‟bb= 0.45 f‟= 0.45 f‟cc f‟f‟b‟b‟= 0.45 f‟= 0.45 f‟c‟c‟
= 0.45
= 0.45 x x 50 50 = = 0.450.45x x 3030 =
= 22.5 22.5 Mpa Mpa = = 13.5 13.5 MpaMpa 2.
2. Tegangan serat tarik terluarTegangan serat tarik terluar ~Untuk
~Untuk Gelagar Gelagar ~Untuk ~Untuk PlatPlat f
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=
= 3.536 3.536 Mpa Mpa = = 2.739 2.739 MpaMpa 3.
3. Mutu beton pada saat peneganganMutu beton pada saat penegangan f‟ f‟cici= 0.8 f‟= 0.8 f‟cc = 0.8 = 0.8 x x 5050 = 40 Mpa = 40 Mpa
Modulus elastisitas beton Modulus elastisitas beton 1.
1. Beton prategang f‟Beton prategang f‟cc= 50 Mpa= 50 Mpa
E Ecc= 4700= 4700 = 4700 = 4700x x = 33234.02 Mpa = 33234.02 Mpa 2.
2. Beton konvensional f‟Beton konvensional f‟c‟c‟= 30 Mpa= 30 Mpa
E Ec‟c‟= 4700= 4700 = 4700 = 4700x x = 25742.96 Mpa = 25742.96 Mpa Dimana: E
Dimana: Ecc= modulus elastisitas beton prategang (Mpa)= modulus elastisitas beton prategang (Mpa)
E
Ec‟c‟= modulus elastisitas beton konvensional (Mpa)= modulus elastisitas beton konvensional (Mpa)
f‟
f‟cc= mutu beton prategang (Mpa)= mutu beton prategang (Mpa)
f‟
f‟c‟c‟= mutu beton konvensional (Mpa)= mutu beton konvensional (Mpa)
1.
1. Tegangan Ijin Tendon Tegangan Ijin Tendon PrategangPrategang
Digunakan tendon VSL dengan sifat-sifat: Digunakan tendon VSL dengan sifat-sifat:
Diameter Diameter nominal nominal = = 12.5 12.5 mmmm
Luas Luas tampang tampang nominal nominal = = 98.7 98.7 mmmm22
Beban Beban putus putus minimum minimum = = 18.75 18.75 tonton = 18750 kg = 18750 kg = (18750 = (18750x x 9.81) N9.81) N = 183937.5 N = 183937.5 N
Beban Beban leleh leleh (20%) (20%) = = 1875018750x x 0.80.8 = 15000 kg = 15000 kg = (15000 = (15000x x 9.81) N9.81) N = 147150 N = 147150 N
Tegangan putus minimum (f
Tegangan putus minimum (fpupu) ) ==
= 1863.6 Mpa = 1863.6 Mpa Tegangan leleh (f Tegangan leleh (fpypy) ) == = 1490.88 Mpa = 1490.88 Mpa Modulus elastisitas (E
Modulus elastisitas (Ess) ) = = 200000 200000 MpaMpa
Tegangan tarik pada tendon prategang tidak boleh melampaui: Tegangan tarik pada tendon prategang tidak boleh melampaui: 1. Akibat gaya pengangkuran tendon
1. Akibat gaya pengangkuran tendon f
fpp = 0.94 f= 0.94 fpypy
= 0.94
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2. Sesaat setelah penyaluran gaya prategang 2. Sesaat setelah penyaluran gaya prategang f fpp = 0.82 f= 0.82 fpypy = 0.82 = 0.82 x x 1490.881490.88 = 1222.52 Mpa = 1222.52 Mpa
Tetapi tidak lebih dari Tetapi tidak lebih dari f fpp = 0.74 f= 0.74 fpupu = 0.74 = 0.74 x x 1863.61863.6 = 1379.06 Mpa = 1379.06 Mpa
3. Tendon pasca tarik, pada daerah angkur dan sambungan, segera setelah penyaluran gaya 3. Tendon pasca tarik, pada daerah angkur dan sambungan, segera setelah penyaluran gaya f fpp = 0.70 f= 0.70 fpupu = 0.70 = 0.70 x x 1863.61863.6 = 1304.52 Mpa = 1304.52 Mpa
Perencanaan Trotoir dan Plat Lantai Perencanaan Trotoir dan Plat Lantai Perencanaan Trotoir
Perencanaan Trotoir
Gambar Rencana Trotoir
Gambar Rencana Trotoir
Pendimensian Sandaran Pendimensian Sandaran
Sandaran direncanakan menumpu pada tiang sandaran dengan bentang 2 m, yang di Sandaran direncanakan menumpu pada tiang sandaran dengan bentang 2 m, yang di rencanakan menahan beban merata vertikal sebesar 0.75 kN/m. Direncanakan Sandaran rencanakan menahan beban merata vertikal sebesar 0.75 kN/m. Direncanakan Sandaran dengan penampang pipa bulat, data sebagai berikut:
dengan penampang pipa bulat, data sebagai berikut: D D (diameter) (diameter) = = 60.5 60.5 mmmm t t (tebal) (tebal) = = 3.2 3.2 mmmm G G (berat) (berat) = = 4.52 4.52 kg/mkg/m
W W (momen (momen tahanan) tahanan) = = 7.84 7.84 cmcm33
σ
σ
(tegangan (tegangan ijin) ijin) = = 1600 1600 kg/cmkg/cm22 Pembebanan: Pembebanan: ~ beban mati (qd) = 4.52 kg/m ~ beban mati (qd) = 4.52 kg/m beban ultimate qd beban ultimate qduu= 4.52= 4.52x x 1.1 1.1 = = 5 5 kg/mkg/m ~ beban hidup (ql) = 0.75 kN/m = 75 kg/m ~ beban hidup (ql) = 0.75 kN/m = 75 kg/m beban ultimate ql beban ultimate qluu= 75= 75 x x 2 2 = = 150 150 kg/mkg/m ~~ beban beban ultimate ultimate (qu) (qu) = = qdqduu + ql+ qluu = 5 + 150
= 5 + 150 Qu = 155 kg/m Qu = 155 kg/m
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Dari hasi analisa statika dengan mengunakan program STAAD PRO, diperoleh momen Dari hasi analisa statika dengan mengunakan program STAAD PRO, diperoleh momen maksimum , yaitu sebesar 0.642 kNm.
maksimum , yaitu sebesar 0.642 kNm. MMmaxmax= 0.642 kNm= 0.642 kNm = 6420 kgcm = 6420 kgcm
σ
σ
== = = = 818.878 kg/cm = 818.878 kg/cm22 <<σ =
σ =
1600 kg/cm1600 kg/cm22Jadi, dipakai pipa baja diameter 60.5 mm sebagai sandaran. Jadi, dipakai pipa baja diameter 60.5 mm sebagai sandaran. Perencanaan Tiang Sandaran
Perencanaan Tiang Sandaran
Tiang sandaran direncanakan menerima beban terpusat dari sandaran sebesar w
Tiang sandaran direncanakan menerima beban terpusat dari sandaran sebesar wx x L, yangL, yang bekerja horisontal pada ketinggian 0.9 m dari permukaan trotoir. Direncanakan dimensi tiang bekerja horisontal pada ketinggian 0.9 m dari permukaan trotoir. Direncanakan dimensi tiang sandaran dengan lebar 15 cm, dan tinggi 20 cm, dengan asumsi tiang sandaran sebagai balok sandaran dengan lebar 15 cm, dan tinggi 20 cm, dengan asumsi tiang sandaran sebagai balok kantilever.
kantilever.
Gaya Yang Bekerja Pada Tiang Sandaran
Gaya Yang Bekerja Pada Tiang Sandaran
Pembebanan Pembebanan ~ beban mati (pd) ~ beban mati (pd)
berat sendiri tiang (atas/pdberat sendiri tiang (atas/pd11) = 0.15) = 0.15x x 0.20.2x x 0.650.65x x 24 24 = = 0.468 0.468 kNkN beban
beban ultimate ultimate pdpd11uu = 46.8= 46.8 x x 1.3 1.3 = = 0.6084 0.6084 kNkN
berat sendiri tiang (bawah/pdberat sendiri tiang (bawah/pd22) = 0.15) = 0.15x x 0.20.2 x x 0.380.38x x 24 = 0.274 kN24 = 0.274 kN beban
beban ultimate ultimate pdpd22uu = 27.4= 27.4 x x 1.3 1.3 = = 0.3562 0.3562 kNkN
berat 1 pipa sandaran (pdberat 1 pipa sandaran (pd33) = 0.0452) = 0.0452x x 2 = 0.0904 kN2 = 0.0904 kN beban
beban ultimate ultimate pdpd33uu = 0.0904= 0.0904x x 1.1 1.1 = = 0.0995 0.0995 kNkN
~
~ beban beban hidup hidup (pl) (pl) = = 0.75 0.75 kNkN beban ultimate pl
beban ultimate pluu= 0.75= 0.75 x x 2 = 1.5 kN2 = 1.5 kN Momen yang terjadi
Momen yang terjadi MMmaxmax= pd= pd11uu x x XX22 – – pdpd22uu x x XX11+ pd+ pd33uu x x XX22+ pl+ pluu x x 90 + pl90 + pluu x x 4545 = 0.6084 = 0.6084 x x 55 0.3562 0.3562 3.63.6
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Start Free Trial Cancel Anytime. = 2 = 2 x x 1.5 kN = 3000 N1.5 kN = 3000 N Perhitungan penulangan Perhitungan penulangan Data perencanaan: Data perencanaan: b b = = 150 150 mmmm h h = = 200 200 mmmm f‟c f‟c = 30 Mpa= 30 Mpa fy fy = = 240 240 MpaMpa
Direncanakan tulangan pokok Ø 10, sengkang Ø 6 Direncanakan tulangan pokok Ø 10, sengkang Ø 6 d
d = h= h – – selimut betonselimut beton – –ØØ sengkang
sengkang – –(½(½x x Ø Tul. Tarik)Ø Tul. Tarik) = 200 = 200 – –2020 – –66 – –(½(½x x 10)10) = 169 mm = 169 mm A. Penulangan lentur A. Penulangan lentur Mu Mu = = 205.255 205.255 kNcm kNcm = = 205.255205.255x x 101044NmmNmm Mn Mn = = = = 256.569256.569 x x 101044 NmmNmm Rn Rn = = = = 0.59888 0.59888 MpaMpa m m = = = = 9.4129.412
Rasio penulangan keseimbangan (ρb);
Rasio penulangan keseimbangan (ρb);
ρρbb == = = = 0.0645 = 0.0645 ρρmaxmax= 0.75= 0.75x x ρρbb = 0.75 = 0.75 x x 0.0645 = 0.0483750.0645 = 0.048375 ρρminmin= = = = = = 0.0058340.005834 Rasio penulangan perlu Rasio penulangan perlu
ρρ == = = = 0.002525 = 0.002525 ρ < ρ
ρ < ρminmin 0.002525 < 0.005834 (digunakan ρ0.002525 < 0.005834 (digunakan ρminmin))
AsAsperluperlu= ρ= ρminmin x
x bbx x dd = 0.005834
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bbminmin= 2= 2 x x selimut beton + 2selimut beton + 2 x x ØØsengkang + nsengkang + nx x D Tul. Tarik + (nD Tul. Tarik + (n – –1)1) x x 2525 = 2
= 2 x x 40 + 240 + 2x x 6 + 26 + 2x 10 + ( 2x 10 + ( 2 – –1 )1 )x x 2525 = 137 mm < b
= 137 mm < b = 150 mm ………….( O.K )= 150 mm ………….( O.K )
As‟ As‟tekantekan= 20 %= 20 %x x AsAsperluperlu = 0.2
= 0.2 x x 131.265 = 26.253 mm131.265 = 26.253 mm22 Dipakai tulangan 2 Ø 10 mm Dipakai tulangan 2 Ø 10 mm
As‟ As‟adaada= 2= 2 x x ( ¼( ¼x x ππx x ØØ22 )) = 2
= 2 x x ( ¼( ¼x x ππx x 101022 )) = 157.08 mm
= 157.08 mm22 > As‟> As‟tekantekan= 26.253 mm= 26.253 mm22………….( O.K )………….( O.K )
B. Penulangan geser B. Penulangan geser Vc Vc = = 1/61/6x x x x bbx x dd = 1/6 = 1/6 x x x x 150150 x x 149149 = 20402.67 N = 20402.67 N ½½
ø
ø
Vc Vc = = ½½x x 0.60.6x x 20402.6720402.67= 6120.8 N > Vu = 1500 N (tidak diperlukan tulangan geser) = 6120.8 N > Vu = 1500 N (tidak diperlukan tulangan geser) Cukup dipasang sengkang praktis. Digunakan Ø 6
Cukup dipasang sengkang praktis. Digunakan Ø 6 – –150 mm yang dipasang disepanjang tiang.150 mm yang dipasang disepanjang tiang.
Gambar Penulangan Tiang Sandaran Gambar Penulangan Tiang Sandaran Perencanaan Kerb
Perencanaan Kerb
Kerb direncanakan untuk menahan beban tumbukan arah menyilang sebesar 100 kN, yang Kerb direncanakan untuk menahan beban tumbukan arah menyilang sebesar 100 kN, yang bekerja sebagai beban titik. Direncanakan kerb terbuat dari beton bertulang, dengan dimensi bekerja sebagai beban titik. Direncanakan kerb terbuat dari beton bertulang, dengan dimensi lebar 20 cm dan
lebar 20 cm dan tinggi 25 cm, menggunakan beton dengan mutu f‟c 30 Mpa, tulangan baja mututinggi 25 cm, menggunakan beton dengan mutu f‟c 30 Mpa, tulangan baja mutu
fy 240 Mpa, yang dipasang 2 Ø 10 pada masing-masing sisinya, dan sengkang Ø 6
fy 240 Mpa, yang dipasang 2 Ø 10 pada masing-masing sisinya, dan sengkang Ø 6 – –200 mm200 mm sepanjang kerb.
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Beban merata Beban merata ~
~ berat berat plat plat lantai lantai = = 0.200.20x 1x 1 x x 24 24 = = 4.8 4.8 kN/mkN/m beban
beban ultimate ultimate = = 4.84.8 x x 1.3 1.3 = = 6.24 6.24 kN/mkN/m ~
~ berat berat plat plat lantai lantai trotoir trotoir = = 0.250.25x x 11 x x 23 = 5.75 kN/m23 = 5.75 kN/m beban
beban ultimate ultimate = = 5.755.75 x x 1.3 1.3 = = 7.475 7.475 kN/mkN/m ~
~ berat berat air air hujan hujan = = 0.050.05x 1x 1 x x 10 10 = = 0.5 0.5 kN/mkN/m Beban
Beban ultimate ultimate = = 0.50.5 x x 1.2 1.2 = = 0.6 0.6 kN/m kN/m ++ qd qd11uu = = 14.315 14.315 kN/mkN/m Beban terpusat Beban terpusat pd pduu= pd= pd11uu+ pd+ pd22uu + 2.pd+ 2.pd33uu = 0.6084 + 0.3562 = 0.6084 + 0.3562 + (2 + (2 x x 0.0995)0.0995) = 1.1636 kN = 1.1636 kN 1.
1. Beban pada plat lantai kendaraanBeban pada plat lantai kendaraan ~
~ berat berat plat plat lantai lantai = = 0.200.20x 1x 1 x x 24 24 = = 4.8 4.8 kN/mkN/m beban
beban ultimate ultimate = = 4.84.8 x x 1.3 1.3 = = 6.24 6.24 kN/mkN/m ~
~ berat berat aspal aspal = = 0.050.05x x 11 x x 22 = 1.1 kN/m22 = 1.1 kN/m beban
beban ultimate ultimate = = 1.11.1 x x 1.2 1.2 = = 1.32 1.32 kN/mkN/m ~
~ berat berat air air hujan hujan = = 0.10.1x 1x 1 x x 10 10 = = 1 1 kN/mkN/m beban
beban ultimate ultimate = = 11 x x 1.2 1.2 = = 1 1 kN/m kN/m ++ qd
qd22uu = = 8.56 8.56 kN/mkN/m
1.
1. Beban mati tambahanBeban mati tambahan
Beban mati tambahan berupa pelapisan ulang lapisan aspal dengan tebal 50 mm Beban mati tambahan berupa pelapisan ulang lapisan aspal dengan tebal 50 mm ~
~ berat berat aspal aspal = = 0.050.05x x 11 x x 22 = 1.1 kN/m22 = 1.1 kN/m beban ultimate qd
beban ultimate qd33uu = 1.1= 1.1x x 2 2 = = 2.2 2.2 kN/mkN/m
Beban hidupBeban hidup
Beban pada plat trotoirBeban pada plat trotoir Beban merata
Beban merata ~
~ beban beban pejalan pejalan kaki kaki = = 5 5 kPakPax x 1 m = 5 kN/m1 m = 5 kN/m beban ultimate ql beban ultimate ql11uu = 5= 5x x 2 2 = = 10 10 kN/mkN/m Beban terpusat Beban terpusat pl pluu = 1.5 kN= 1.5 kN
Beban pada plat lantai kendaraanBeban pada plat lantai kendaraan #
# Faktor Faktor beban beban dinamis dinamis (DLA)(DLA) K = 1 + DLA ,
K = 1 + DLA ,
Faktor beban dinamis untuk truk adalah 0.3 (BMS ‟92, hal
Faktor beban dinamis untuk truk adalah 0.3 (BMS ‟92, hal2-20)2-20) maka K = 1 + 0.3 = 1.3
maka K = 1 + 0.3 = 1.3 #
# Beban truk “T”Beban truk “T”
Beban truk “T” sebesar 200 kN, maka
Beban truk “T” sebesar 200 kN, maka tekanan untuk satu roda:tekanan untuk satu roda:
P Puu ==
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Gambar Skema Pembebanan Kondisi II
Gambar Skema Pembebanan Kondisi II
Kondisi IIIKondisi III
Gambar Skema Pembebanan Kondisi III
Gambar Skema Pembebanan Kondisi III
Kondisi IVKondisi IV
Gambar Skema Pembebanan
Gambar Skema Pembebanan
Kondisi IV
Kondisi IV
Kondisi VKondisi V
Gambar Skema Pembebanan Kondisi V
Gambar Skema Pembebanan Kondisi V
Kondisi VIKondisi VI
Gambar Skema Pembebanan
Gambar Skema Pembebanan
Kondisi VI
Kondisi VI
Penulangan Plat Lantai
Penulangan Plat Lantai KendaraanKendaraan
Dari hasi analisa statika dengan mengunakan program STAAD PRO, diperoleh momen Dari hasi analisa statika dengan mengunakan program STAAD PRO, diperoleh momen maksimum pada kondisi II, yaitu:
maksimum pada kondisi II, yaitu:
MMmaxmaxtumpuan tumpuan = = 77.976 77.976 kNmkNm
MMmaxmaxlapangan lapangan = = 71.471 71.471 kNmkNm Data perencanaan: Data perencanaan: f‟c f‟c = 30 Mpa= 30 Mpa fy fy = = 350 350 MpaMpa Tebal
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m m = = = = 13.725513.7255 Rasio penulangan keseimbangan (ρb);
Rasio penulangan keseimbangan (ρb);
ρρbb == = = = 0.0391128 = 0.0391128 ρρmaxmax= 0.75= 0.75x x ρρbb = 0.75 = 0.75 x x 0.0391128 = 0.029334590.0391128 = 0.02933459 ρρminmin= = = = = = 0.0040.004 Rasio penulangan perlu Rasio penulangan perlu
ρρ == = = = 0.010115 = 0.010115 ρ > ρ
ρ > ρminmin 0.010115 > 0.004 (digunakan ρ)0.010115 > 0.004 (digunakan ρ)
AsAsperluperlu= ρ= ρx x bb x x dd = 0.010115 = 0.010115x x 10001000 x x 172172 = 1739.78 mm = 1739.78 mm22
Digunakan tulangan pokok D 16 mm Digunakan tulangan pokok D 16 mm Perhitungan jarak (S) dan As
Perhitungan jarak (S) dan Asadaada
As As = = ¼¼x x ππx x DD22 = ¼ = ¼x x ππx x 161622 = 201.06 mm = 201.06 mm22 S S == = 115.5 mm ≈ 100 mm= 115.5 mm ≈ 100 mm AsAsadaada= = = = 2010.6 2010.6 mmmm22 Diperoleh As
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Diperoleh As
Diperoleh Asadaada> As> Asperluperlu, maka dipakai tulangan bagi Ø 10, maka dipakai tulangan bagi Ø 10 – –200200
Gambar Penulangan Plat Lantai Kendaraan
Gambar Penulangan Plat Lantai Kendaraan
Perencanaan Struktur Gelagar Perencanaan Struktur Gelagar
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Gambar Penampang Balok Prategang
Gambar Penampang Balok Prategang
Perhitungan
Perhitungan Section Properties Section Properties Penampang Balok Tengah
Penampang Balok Tengah
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Start Free Trial Cancel Anytime. = = = = 2891.94 2891.94 cmcm22 = = = = 35.05 35.05 cmcm = = = = 35.05 35.05 cmcm
Setelah kompositSetelah komposit
Jarak efektif antar gelagar sebesar 175 cm. Karena mutu beton plat dan balok berbeda, maka Jarak efektif antar gelagar sebesar 175 cm. Karena mutu beton plat dan balok berbeda, maka lebar efektif plat komposit dengan balok prategang adalah:
lebar efektif plat komposit dengan balok prategang adalah: b
beffeff
x
x n (n adalah rasio perbandingan antara mutu beton, n = 0.77)n (n adalah rasio perbandingan antara mutu beton, n = 0.77) 175
175 x x 0.77 = 134.75 cm0.77 = 134.75 cm
Tabel
Tabel PerhitunganPerhitunganSection Properties Section Properties Balok Tengah Setelah KompositBalok Tengah Setelah Komposit
Bag. Bag. AA (cm (cm22)) y y (cm) (cm) A A x xyy (cm (cm33)) Momen
Momen Inersia ‘I’Inersia ‘I’ (cm
(cm44))
3
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= = = = 45.91 45.91 cmcm Penampang Balok Ujung
Penampang Balok Ujung 1.
1. Sebelum kompositSebelum komposit
AApp = b= bx x h h = 80= 80x x 165 165 = = 13200 13200 cmcm22 IIpp = 1/12= 1/12x x bb x x hh33 = 1/12= 1/12x x 8080 x x 16516533= 29947500 cm= 29947500 cm44 = = = = 82.5 82.5 cmcm = 165= 165 – –82.5 82.5 = = 82.5 82.5 cmcm 1.
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Start Free Trial Cancel Anytime. Bj Bj beton beton = = 24 24 kN/mkN/m33 Bj Bj aspal aspal = = 22 22 kN/mkN/m33 Bj Bj air air = = 10 10 kN/mkN/m33
Jarak efektif antar gelagar = 175 cm = 1.75 m Jarak efektif antar gelagar = 175 cm = 1.75 m Tebal plat = 20 cm = 0.2 m Tebal plat = 20 cm = 0.2 m Tebal aspal = 5 cm = 0.05 m Tebal aspal = 5 cm = 0.05 m Tebal air = 10 cm = 0.1 m Tebal air = 10 cm = 0.1 m Luas penampang plat (A
Luas penampang plat (A11) = 1.75) = 1.75x x 0.2 = 0.35 m0.2 = 0.35 m22
Luas penampang aspal (A
Luas penampang aspal (A22) = 1.75) = 1.75x x 0.05 = 0.0875 m0.05 = 0.0875 m22
Luas penampang air (A
Luas penampang air (A33) = 1.75) = 1.75 x x 0.1 = 0.175 m0.1 = 0.175 m22
qd
qd22= Bj beton= Bj beton x x AA33+ Bj aspal+ Bj aspal x x AA22+ Bj air+ Bj airx x AA33
= 24
= 24 x x 0.35 + 220.35 + 22x x 0.0875 + 100.0875 + 10x x 0.1750.175 = 12.075 kN/m
= 12.075 kN/m
Akibat diafragmaAkibat diafragma Bj
Bj beton beton = = 25 25 kN/mkN/m33
Tebal diafragma (t) = 15 cm = 0.15 m Tebal diafragma (t) = 15 cm = 0.15 m
Gambar Penampang Diafragma Gambar Penampang Diafragma Luas
Luas penampang penampang (A) (A) = = (135(135x x 105)105) – –(2(2 x x (A(AIVIV+ A+ AVV))))
= 13975 cm
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Gambar Beban Yang Bekerja Pada Arah Melintang Jembatan
Gambar Beban Yang Bekerja Pada Arah Melintang Jembatan
a.
a. Besarnya beban Besarnya beban terbagi rterbagi rata (ata (UDL) UDL) tergantung pada tergantung pada panjang total panjang total yang dibebani yang dibebani (L).(L). L = 40 m > 30 m, maka: L = 40 m > 30 m, maka: q q == = = = 7 kPa = 7 kPa
Jarak efektif antar gelagar = 175 cm = 1.75 m, maka beban merata yang bekerja di sepanjang Jarak efektif antar gelagar = 175 cm = 1.75 m, maka beban merata yang bekerja di sepanjang gelagar adalah: gelagar adalah: ql ql11 = 1.75= 1.75x x qq = 1.75 = 1.75 x x 77 = 12.25 kNm = 12.25 kNm b.
b. Beban terpusat Beban terpusat P yang P yang ditempatkan tegak ditempatkan tegak lurus lurus arah lalu arah lalu lintas pada lintas pada jembatan adalahjembatan adalah sebesarnya 44.0 kN/m.
sebesarnya 44.0 kN/m.
Faktor Beban Dinamik untuk “KEL” lajur
Faktor Beban Dinamik untuk “KEL” lajur “D”“D”, untuk bentang (L, untuk bentang (LEE) = 40 m, nilai DLA = 0.4.) = 40 m, nilai DLA = 0.4.
Maka:
Maka: K K = = 1 1 + + DLADLA K = 1 + 0.4 = 1.4 K = 1 + 0.4 = 1.4
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Dimana:
Dimana: Vw Vw = = kecepatan kecepatan angin angin rencana rencana = = 30 30 m/detm/det Cw
Cw = = koefisien koefisien Seret Seret = = 1.21.2 T TEWEW = 0.0012= 0.0012x x 1.21.2 x x 303022 = 1.296 kN/m = 1.296 kN/m Analisa Statika Analisa Statika Beban Tetap Beban Tetap
Gambar Diagram Momen dan Gaya Lintang Akibat Berat Sendiri Gambar Diagram Momen dan Gaya Lintang Akibat Berat Sendiri 1.
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VA
VA =241,5 =241,5 kN kN VB VB = = 241,5 241,5 kNkN Gambar Diagram Momen dan Gaya Lintang Akibat Beban Mati
Gambar Diagram Momen dan Gaya Lintang Akibat Beban Mati
Reaksi tumpuan: Reaksi tumpuan: R RAA= R= RBB= ½= ½x x qq x x LL = ½ = ½x x 12.07512.075x x 4040 = 241.5 kN = 241.5 kN
Momen & Gaya Lintang pada setiap titik: Momen & Gaya Lintang pada setiap titik:
Momen pada titik X dengan jarak setiap 2.0 m; Momen pada titik X dengan jarak setiap 2.0 m; M
Mxx= (R= (RAA
x
x X)X) – –(½(½x x qqx x XX22))
Gaya Lintang pada titik X dengan jarak setiap 2.0 m; Gaya Lintang pada titik X dengan jarak setiap 2.0 m;
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Gambar Diagram Momen dan Gaya Lintang Akibat Diafragma Gambar Diagram Momen dan Gaya Lintang Akibat Diafragma 1.
1. Akibat diafragmaAkibat diafragma Reaksi tumpuan: Reaksi tumpuan: R RAA= R= RBB= ½= ½x x ∑ P∑ P = ½ = ½x x 5.245.24x x 1111 = 28.823 kN = 28.823 kN
Momen & Gaya Lintang pada setiap titik: Momen & Gaya Lintang pada setiap titik:
Momen pada titik X dengan jarak setiap 2.0 m; Momen pada titik X dengan jarak setiap 2.0 m;
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Start Free Trial Cancel Anytime. Titik 7, X = 14 m Titik 7, X = 14 m M M77= (28. 823= (28. 823x x 14)14) – –(5.24(5.24x x 14)14) – –(5.24(5.24x x 10)10) – –(5.24(5.24x 6)x 6) – –(5.24(5.24x x 2)2) = = 235.828 235.828 kNmkNm V V77 = V= V66= = 13.102 13.102 kNkN Titik 8, X = 16 m Titik 8, X = 16 m M M88= (28. 823= (28. 823x x 16)16) – –(5.24(5.24x x 16)16) – –(5.24(5.24x x 12)12) – –(5.24(5.24x x – –(5.24(5.24 x x 4)4) = = 251.55 251.55 kNmkNm V V88 = 13.102= 13.102 – –5.245.24 = = 7.861 7.861 kNkN Titik 9, X = 18 m Titik 9, X = 18 m M M99= (28. 823= (28. 823x x 18)18) – –(5.24(5.24x x 18)18) – –(5.24(5.24x x 14)14) – –(5.24(5.24x x 10)10) – –(5.24(5.24 x x 6)6) – –(5.21(5.21x x 2)2) = = 256.791 256.791 kNmkNm V V99 = V= V88= = 7.861 7.861 kNkN Titik 10, X = 20 m Titik 10, X = 20 m M M1010= (28. 823= (28. 823 x x 20)20) – –(5.24(5.24 x x 20)20) – –(5.24(5.24 x x 16)16) – –(5.24(5.24 x x 12)12) – –(5.24(5.24 x x – –(5.21(5.21 x x 4)4) = = 262.031 262.031 kNmkNm V V1010= 7.861= 7.861 – –5.245.24 = = 2.62 2.62 kNkN
Beban Lalu Lintas Beban Lalu Lintas
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Start Free Trial Cancel Anytime. Titik Titik 3, 3, X X = = 6 6 m m YY33 = = = = 5.1 5.1 mm A A33 = ½= ½x x 5.15.1x 40 x 40 = = 102 102 mm22 Titik Titik 4, 4, X X = = 8 8 m m YY44 = = = = 6.4 6.4 mm A A44 = ½= ½x x 6.46.4x 40 x 40 = = 128 128 mm22 Titik Titik 5, 5, X X = = 10 10 m m YY55= = = = 7.5 7.5 mm A A55 = ½= ½x x 7.57.5x 40 x 40 = = 150 150 mm22 Titik Titik 6, 6, X X = = 12 12 m m YY66= = = = 8.4 8.4 mm A A66 = ½= ½x x 8.48.4x 40 x 40 = = 168 168 mm22 Titik Titik 7, 7, X X = = 14 14 m m YY77= = = = 9.1 9.1 mm A A77 = ½= ½x x 9.19.1x 40 x 40 = = 182 182 mm22 Titik Titik 8, 8, X X = = 16 16 m m YY = = = = 9.6 9.6 mm
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Start Free Trial Cancel Anytime. V V99 = = 132.3 132.3 kNkN Titik Titik 10, 10, X X = = 20 20 m m MM1010= = 3528 3528 kNmkNm V V1010= = 107.8 107.8 kNkN
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Start Free Trial Cancel Anytime. V VAA= = 25.92 25.92 kNkN Titik Titik 1, 1, X X = = 2 2 m m MM11 = = 49.248 49.248 kNmkNm V V11 = = 23.328 23.328 kNkN Titik Titik 2, 2, X X = = 4 4 m m MM22 = = 93.312 93.312 kNmkNm V V22 = = 20.736 20.736 kNkN Titik Titik 3, 3, X X = = 6 6 m m MM33 = = 132.192 132.192 kNmkNm V V33 = = 18.144 18.144 kNkN Titik Titik 4, 4, X X = = 8 8 m m MM44 = = 165.888 165.888 kNmkNm V V44 = = 15.552 15.552 kNkN Titik Titik 5, 5, X X = = 10 10 m m MM55 = = 194.4 194.4 kNmkNm V V55 = = 12.96 12.96 kNkN
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Start Free Trial Cancel Anytime. 1 1 2 2 3 3 4 4 5 5 6 6 77 Mo Mo MMGG MMTT 8 9 10 8 9 10 (2+3+ (2+3+ 4) 4) (5+6+7 (5+6+7 +9) +9) (kNm) (kNm) (kNm) (kNm) (kNm) (kNm) (kNm) (kNm) (kNm)(kNm) (kNm(kNm ) ) (kNm) (kNm) (kNm) (kNm) (kNm)(kNm) M MAA 0 0 0 0 0 0 0 0 653.857 653.857 0 0 0 0 00 653.85653.85 7 7
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Tebal karet dalam = 18 mmTebal karet dalam = 18 mm
Tinggi keseluruhan = 92 mmTinggi keseluruhan = 92 mm
Beban ternilai pada perputaran nol, pada geser maksimum = 7353 kNBeban ternilai pada perputaran nol, pada geser maksimum = 7353 kN
Beban ternilai pada perputaran maksimum, pada geser maksimum = 3377 kNBeban ternilai pada perputaran maksimum, pada geser maksimum = 3377 kN Gaya lintang maksimum yang terjadi pada satu gelagar
Gaya lintang maksimum yang terjadi pada satu gelagar V
VUU= 1718.824 kN < V= 1718.824 kN < Vperletakanperletakan= 3377 kN ………(O.K)= 3377 kN ………(O.K)
Perencanaan Abutment Perencanaan Abutment
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Panjang
Panjang bentang bentang jembatan jembatan = = 40 40 mm Bj
Bj beton beton prategang prategang = = 25 25 kN/mkN/m33 A
App = 9200 cm= 9200 cm22 = 0.92 m= 0.92 m22
~
~ berat berat gelagar gelagar = = 55x x (40(40x x 0.920.92 x x 25) Pd25) Pd44= = 4600 4600 kNkN
3.
3. Beban diafragmaBeban diafragma Panjang
Panjang bentang bentang jembatan jembatan = = 40 40 mm Jarak
Jarak antar antar diafragma diafragma = = 4 4 mm Bj
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