Excel Perhitungan Struktur Jembatan [PDF]

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5.4.1.1 Kondisi Perancangan 1. Kondisi Jembatan (lihat gambar) > Panjang jembatan total > Jumlah bentang > Panjang bentang > Lebar jembatan > Lebar perkerasan > Tipe jembatan > Jumlah balok/gelagar > Panjang bersih gelagar



= = = = = = = =



105 m 6 buah 17.5 m 4.32 m 3.5 m Beton bertulang dengan gelagar balok T 2 buah 16.5 m



2. Spesifikasi Pembebanan a. Beban hidup : PPJJR No.12/1970 (BM 70%) > Beban roda T = 70% x 10 t > Beban garis P = 70% x 12 t/m > Beban merata q = 70% x 2.2 t/m2 1 + 20 : 50 L 𝑘=1+20/(50+𝐿) b. Beban kejut, 20 = 1 + 50 + 17.5 3. Spesifikasi Beton dan Baja Tulangan a. Beton > Kuat tekan, fc' > Kuat tekan ijin, fc' > Modulus Elastisitas, Ec b. Baja Tulangan > Kuat leleh, fy > Modulus Elastisitas, Es



= = = =



7t 8.4 t/m 1.54 t/m2 1.0011396



=



1.30



= = =



20 Mpa 10 Mpa 4700 x √20



=



21019.039 Mpa



= =



200 Mpa 2 x 10⁵



=



200000 Mpa



= =



2 2



5.4.1.2 Perencanaan 1.



Tiang sandaran (lihat gambar 5.4) Momen lentur, M Gaya geser, V b d h Ø



= = = =



100 100



x



1



= =



160 mm 130 mm 160 mm 0.8



𝑘=𝑀𝑢/(∅𝑏𝑑^2 = )



ρperlu



x x



2000 0.8 x



200 kg.m 200 kg



= =



𝑀𝑛=∅.𝑏.𝑑^2.𝑘 Mu =Mn



x 160



10^3 x 16900



=(0,85𝑓𝑐^′)/𝑓𝑦 (1−√(1−2𝑘/0,85𝑓𝑐 ′))



=



0.925



Mpa



0.5 =



0.85



=



x 200



20



1



-



0.007



=



1.4 fy



=



1.4 200



=



ρperlu




Kuat tekan ijin, fc' > Modulus Elastisitas, Ec Baja Tulangan > Kuat leleh, fy > Modulus Elastisitas, Es



Momen lentur, M Gaya geser, V k=Mu/0,8bd^2 Pperlu Pmin As



NT = ND a = As.fy/0,85.fc'.b c = a/b1



= = = = =



= = = = = Pperlu < Pmin , P = Pmin = 0,007 =



= =



fs = 600(d-c/c) Mn = As.fy(d-a/2)



Vu = 2000 N Vc = 1/3 akar fc' .b.d 1/2DVc



= = Mn/Mu = 1,625



= = (teoritis tidak perlu sengkang)



tetapi untuk kestabilan struktur dan peraturan mensyaratkan dipasang tulangan minimum (spasi maksimum) Smaksimum = 1/2 d = atau Smaksimum = 600 mm Digunakan spasi = 65 mm, dengan luas tulangan minimum : Av = 1/3 akar fc'.b.s/fy



=



akai tulangan D 8 mm (Av = 100,531 mm2), maka jarak sengkang : S = Av.fy/1/3 akar fc'.b i dipakai tulangan D8-80 mm untuk geser, dan 2D10 untuk lentur.



=



Momen lentur (Bending moment)



84 x 54



7000 kg cm 20046.2962962963 kg/cm2 0.200462962962963 Mpa 1.65 ; Iy = oc



0,84/1,65 0.509090909090909 0,54/1,65 0.327272727272727



15510 N.m 9120 N.m



15510 N.m 200 mm 160 mm 15510 x 10^3/0,8x1000x160^2 0.75732421875 Mpa 0.025371342100882 1820 mm2



110.473626373626 mm



200-40-16



9120 Nm 200 mm =



Mu/D.bd^2 0,85 fc'/fy (1-akar(1-2k/0,85 fc'))



= =



p.b.d



=



201,062/1008x1000



=



Hand rail



= = = = = =



Railing Perkerasan Air hujan Pelat lantai



Gelagar Total qDL



=



Balok melintang (diafragma), Tb = 0,3x0,6x2400x0,5



Mx



=



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



1/2x2464,5x16,5^2{2/16,5(1-2/16,5)} 1/2x356,4x2



M1.DL



men pada potongan 2, x = 4 m (M2.DL) = =



1/2x2464,5x16,5^2{4/16,5(1-4/16,5)} 1/2x356,4x4



M2.DL



men pada potongan 3, x = 6 m (M3.DL) = =



1/2x2464,5x16,5^2{6/16,5(1-6/16,5)} 1/2x356,4x6



M3.DL



en pada potongan 4, x = 8,25 m (M4.DL) = = M2.DL



1/2x2464,5x16,5^2{8,25/16,5(1-8,25/16,5)} 1/2x356,4x8,25



= =



= =



1.299 1,299x8400/2,75x2,031



P.L{x/L(1-x/L)} 1/2 q.L^2{x/L(1-x/L)}



men pada potongan 1, x = 2 m (M1.LL) = =



8059x16,5{2/16,5(1-2/16,5)} 1/2x1137,36x16,5^2{2/16,5(1-2/16,5)}



M1.LL



men pada potongan 2, x = 4 m (M2.LL) = =



8059x16,5{4/16,5(1-4/16,5)} 1/2x1137,36x16,5^2{4/16,5(1-4/16,5)}



M2.LL



men pada potongan 3, x = 6 m (M3.LL) = =



8059x16,5{6/16,5(1-6/16,5)} 1/2x1137,36x16,5^2{6/16,5(1-6/16,5)}



M3.LL



men pada potongan 4, x = 8,25 m (M4.LL) = = M4.LL



8059x16,5{6/16,5(1-8,25/16,5)} 1/2x1137,36x16,5^2{8,25/16,5(1-8,25/16,5)}



Tabel 5.2 Momen lentur total (Nm) M1 360920 355180 716100



ing force)



tulangan



= = = = = =



562820 374380 450 1150 1150-60 Mu/D.b.d^2



0,85 fc'/fy(1-akar 1-2k/0,85.fc') 0,85x20/200(1-akar 1-2x1,3158/0,85x20) 0.668858729478804 1,4/fy Pperlu Vu



144 mm 9120x10^3/0,8x1000x144^2 0,85x20/200(1-akar(1-2x0,5498/0,85x20)



0.5497685185185 Mpa = 0.0028



0,007x1000x144



=



1008 mm2



199.46626984127 mm



0,1x0,16x1x2400/2 x 1,580 2x1x61,580 0,07x2200x2,031 0,05x1000x2,031 0,20x2400x2,031



30.336 kg/m = = = =



123.36 kg/m 312.77 kg/m 101.55 kg/m 974.88 kg/m



0,95x0,54x2400x1



= =



1231.2 kg/m 2774.1 kg/m 216 kg



1/2qDL.L^2(x/L(1-x/L))



1/2x2464,5x16,5^2{2/16,5(1-2/16,5)} 1/2x356,4x2



1/2x2464,5x16,5^2{4/16,5(1-4/16,5)} 1/2x356,4x4



1/2x2464,5x16,5^2{6/16,5(1-6/16,5)} 1/2x356,4x6



1/2x2464,5x16,5^2{8,25/16,5(1-8,25/16,5)} 1/2x356,4x8,25



= =



35735.25 kgm 356.4 kgm



= =



36091.65 kgm 360916.5 Nm



= =



61612.5 kgm 712.8 kgm



= =



62325.3 kgm 623253 Nm



= =



77631.75 kgm 1069.2 kgm



= =



78700.95 kgm 787009.5 Nm



= =



83870.01563 kgm 1470.15 kgm



= =



85340.16563 kgm 853401.6563 Nm



1,299x8400/2,75x2,031 =



1540/2,75x2,031



= =



8058.712582 kg 1137.36 kg/m



= =



14164.30303 kgm 16491.72 kgm



= =



30656.02303 kgm 306560.2303 Nm



= =



24421.21212 kgm 28434 kgm



= =



52855.21212 kgm 528552.1212 Nm



= =



30770.72727 kgm 35826.84 kgm



= =



66597.56727 kgm 665975.6727 Nm



= =



33243.375 kgm 38705.7825 kgm



= =



71949.1575 kgm 719491.575 Nm



P.L{x/L(1-x/L)} 1/2 q.L^2{x/L(1-x/L)}



8059x16,5{2/16,5(1-2/16,5)} 1/2x1137,36x16,5^2{2/16,5(1-2/16,5)}



8059x16,5{4/16,5(1-4/16,5)} 1/2x1137,36x16,5^2{4/16,5(1-4/16,5)}



8059x16,5{6/16,5(1-6/16,5)} 1/2x1137,36x16,5^2{6/16,5(1-6/16,5)}



8059x16,5{6/16,5(1-8,25/16,5)} 1/2x1137,36x16,5^2{8,25/16,5(1-8,25/16,5)}



M2 623260 613460



M3 787010 772690 2796420



= = = = =



M4 9E+05 8E+05 2E+06



1/2x2464,5x16,5 2,5x356,4 1/2x8059 1/2x1137,36x16,5 34635.845



= = = = =



### 891 4030 9383 ### N



Nm N mm mm Mu/D.b.d^2



1090 mm =



5628280x10^3/0,8x450x1090^2 =



c'/fy(1-akar 1-2k/0,85.fc') 0/200(1-akar 1-2x1,3158/0,85x20)



=



1,4/200



0,007x450x1090



85.fc'.b 92,4/0,85 >fy =



=



=



0.01



3433.5



mm2



= 3534,292x200/0,85x20x450= = 108.71 mm OK (3534,292x200)x(1090-92,4/2)



92.4



Nmm Nm



= =



(1/3akar20)x450x1090 = ### N 1040507.628 N > Vu(Teoritis tidak perlu sengkang)



struktur dan simum), 225 mm mm



Tabel 5.1 Perhitungan Momen Lentur



Leng Mom en an No Volumey(m3) (kg/m3) W (kg) (kg. (m) m)



l 2 3 4 5 6 7p T



0,10 2400 19,2 0,85 16,4 x 2400 0 5 20 0,16 2400 9,24 0,86 7,96 x 2400 6,00 2 0 0,50 2400 9,00 0,80 4,80 = 2400 0 0 0 0,00 2200 396, 0,72 6,53 8 00 5 0 0, 10 99,0 0,41 163, x (0, 0 3 550 70 x 1000 96,2 0,27 27,2 0, 5 5 3 110) 200, 0,31 30,1 /2 00 3 3 =0,0 9093 1,20 2400 04 ,00 0 ,000 0, l O 62,5 0,27 2500 30365,40 x 0 5 ,560 Nm 0,05 24,0 0,31 19,5 x 0 3 60 b. Gaya geser (Shear Force) 0,50 0,82 19,8 = Berat tiang = 67,4405 kg 00 0,00 Slab kantil = 591,250 kg 3036 3 ,540 Beban0,roda 10 = 9093,000 kg (0, Bebanxgenan = 62,500 kg 15 x Total gaya lintang, = 9814,190 V kg 0,50) /2 = = 98141,900 N 0,00 4 1,00 c. Perhitungan baja tulangan x 0,82 Mu = 30370,000 Nm 5 x Vu= 98150,000 0,20 N = 0, hf = 300 mm → d= 300-40 = 260 mm 165 1,00 x 2 k = Mu/ᴓbd = 30370 x 103/ 0,8 x 1000 x 2602= 0,5616 Mpa (0,82 5 x ρperlu= 0,85fc'/ fy(√1-2k/0,85 fc') = 0,85 x 20 / 200 (√1-2 x 0,5616/0,85 x 20) 0, = 10)/0 2= ρperlu0,04 < ρmin → ρ = ρmin = 0,007 1 As = ρbd 1,00= 0,007 x 1000 x 260 = 1830 mm2 x 0,62 5 x 0,07 =



Dipakai tulangan φ15 (As = 201,062 mm2), dengan jarak antar tulangan sperlu= 201,062/1820 x 1000 = 110,474 mm Dipakai tulangan φ16-100 mm, Kontrol terhadap geser beton : τc = V/ 7/8 bh = 98150 / OK 7/8 x 1000 x 260 = 0,431 Mpa < 0,45 fc'



B. Perhitungan pelat bagian dalam (inner slab) (lihat gambar 5.5) a. Momen lentur (pada PBI 1970) diperoleh : fxm= 0,1500 fym= 0,0933 Mxm= fxm x T x tx x ty = 0,1500 x 20046,3 x 0,84 x 0,54 = 1364 kg.m Mym= fym x T x tx x ty = 0,0933 x 20046,3 x 0,84 x 0,54 = 849 kg.m



b. Momen Lentur akibat beban mati Berat Slab = 0,20 x 2400 = 480 kg/m2 Berat perke = 154 kg/m2 Berat air hu = 0,05 x 1000 = 50 kg/m2 Total qDL = 684 kg/m2 Mxm = 1/10 x qDL x lx2= 1/10 x 684 x 1,652= 187 kgm= 1870 Nm Mym = 1/3 x Mxm = 1/3 x 187 = 63 kgm = 630 Nm



Perencanaan tulangan geser smaksimum= 600 mm digunakan spasi = 225 mm, dengan luas tulangan minimum Av min = 1/3(√fc')b.s / fy = (1/3√20) x 450 x 225 / 200 = 754,673 mm2 Dipakai tulangan φ12 mm (Av=226,195 mm2), maka jarak sengkang : s= Av.fy/ 1/3(√fc') b = 2 x 226,195 x 200 / (1/3√20) x 450 = 134,876 mm Jadi dipakai tulangan φ12-150 mm untuk geser, dan 5φ30 untuk lentur. Pada potongan 1 M1 = 716100 Nm



Lebar efektif balok (b) dipilih yang terkecil diantara : b = 1/4 x L = 1/4 x 16500 = 4125 mm b = bw + 16 hf = 450 + (16 x 200) = 3650 mm b = jarak p.k.p = 200 mm Kontrol penampang balok T Dianggap seluruh flens menerima desakan sepenuhnya Mnf = 0,85 fc' . Bhf (d- hf/2) = 0,85 x 20 x 2000 x 200 (1060-200/2) = 6528 x 103 Nm Mnf > Ml, maka balok berperilaku sebagai balok -T persegi, k = Mul / φ bd2= 716100 x 103 / 0,8 x 450 x 10602= 1,770 Mpa ρ perlu = 0,85 fc'/fy (1-√1-2k/0,85 fc') = 0,85 x 20/200 (1-√1-2 x 1,7704/0,85 x 20) = 0,00937 ρmin = 1,4/fy = 1,4/200 = 0,007



ρ perlu > ρ min → ρ = 0,00937 As = ρbd = 0,00937 x 450 x 1060 = 4468,530 mm2 Diapakai baja tulangan 6φ30 (As=4241,150 mm2) NT= ND a = As.fy / 0,85fc'.b = 4241,1501 x 200 / 0,85 x 20 x 450 = c = a/βl= 110,8797/0,85 = 130,447 fs = 600 (d-c/c) = 600 (1060-130,447/130,447) = 4275,555 Mpa > fy OK = 852,098 x 106 Nmm = 852,098 x 103 Nm Mn/Mu = 1,19



OK



Cek Daktailitas tulangan : As max = 0,0319hf {b+bw(0,510d/hf -1)} = 0,0319 x 200 {2000+450(0,510 x 1060/200 - 1)} = 17649,313 mm2 As min = ρ min bd = 0,007 x 450 x 1060 = 3339,000 mm2 Pada potongan 4 M4 = 1688470 Nm < Mnf = 6528 x 103 Nm Perilaku balok sebagai balok -T persegi k = Mu/φbd2= 1688470 x 103/0,8 x 450 x 10602= 4,174 ρ perlu = 0,85 fc'/fy(1-√1-2k/0,85 fc') = 0,85 x 20/200 (1-√1-2 x 41742/0,85 x 20) = 0,0244 ρ min = 1,4/fy = 1,4/200 = 0,007 ρ min = 0,75 x 0,85fc'βˌ/fy x 600/600+fy = 0,75 x 0,85 x 20 x 0,85/200 (600/600 + 200) = 0,0542 ρ min < ρ perlu < ρ max → ρ = 0,0244 As = ρbd = 0,0244 x 450 x 1060 = 11638,800 mm2 Diapakai baja tulangan 16 φ 30 (As = 11309,734 mm2) NT = ND a = As.Fy/0,85 fc'b = 11309,734 x 200 / 0,85 x 20 x 450 = 295,679 mm



c = a/βˌ = 295,679/0,85 = 347,858 mm fs = 600(d-c/c)= 600(1060 - 347,858/347,858) = 1228,332 Mpa > fy OK Mn = As fy (d-a/2) = (11309,734 x 200) x (1060-295,6793/2) = 2063,258 x 106 Nmm = 2063,258 x 10 3 Nm Mn/Mu = 1,22



OK



Tulangan disusun 4 lapis d aktual = 1150-40 - 25 x 2 = 1060 mm



OK