Tugas Merancang Kapal 1 [PDF]

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OWNER REQUIREMENT Data Desain Keterangan Jenis Kapal Ro-Ro Jenis Muatan Penumpang dan Kendaraan Payload Penumpang 750 Pax Kendaraan 75 unit Unit Kecepatan Dinas 14 knots Radius Pelayaran 474 nm Rute



Surabaya Banjarmasin



314.07



Banjarmasin Surabaya



314.07



Daerah Pelayaran Bunkering



Muatan



Indonesia Surabaya



PERHITUNGAN DWT Jumlah



Penumpang Gol I Sepeda Gol II Sepeda Motor ( =500 cc) Gol IV A Mobil (L 100 2. CB ; CB > 0.68 Fb2 = F_b∙(C_B+0.68)/1.36 = = 3. Depth (D) L/15 = R



=



0.9413744 941.37438 mm 5.8446667 182.64583



untuk L < 120m ; R = L/0.48 untuk L > 120m ; R = 250 jika, D < L/15 ; tidak ada koreksi jika, D > L/15 ; Fb3 = Fb2 + (R(H-(L/15))) Fb3



=



941.37438 mm



L Standart Height [ m ] [m] Raised quarter - deck All other superstructure 30 or less 0.9 1.8 75 1.2 1.8 125 or more 1.8 2.3 Koreksi Bangunan Atas 1. Forecastle 2. Poop ⇨ h (m) st 75 ⇨ 1.8 125 ⇨ 2.3 interpolasi 87.67 ⇨ 1.9267



L1 (m)



tFC



=



⇨



L1 (m)



m 2.6 m



75 ⇨ 125 ⇨ interpolasi 87.67 ⇨ tPO



hst (m) 1.8 2.3 1.9267



=



2.6 m



karena tFC > hst maka



karena tPO > hst maka



EFC



EPO



= = =



SFC 10.5204 m 0.12 · L



= =



m



SPO 7.0136 m



Total Panjang Efektif E = EFC + EPO = 17.534 m =



0.2 · L



3. Pengurangan Akibat Bangunan Atas ⇨ hst (m) L1 (m) 24 ⇨ 350 ; regulation 37 ⇨ 85 860 Interpolasi 87.67 ⇨ 882.3229508 Pengurangan = 21% · hst ; regulation 37 table 37.1 = 185.2878197 mm Total Lambung Timbul Fb' = Fb3 - Pengurangan = 756.0865616 mm = 0.756 m Ketinggian Bow Minimum (BWM) CB min



=



0.69



CB



=



0.5678



BWM



= 56∙ L1∙(1−L1/500)∙(1.36/(CB+0.68))



= =



4019.13232 mm 4.019 m



Tabel Pengurangan Freeboard Total panjang efektif Superstructure [ E ] Pengurangan [ mm ] 24 350 85 860 122 1070



Bila panjang berada diantaranya maka harga pengurangan diperoleh dengan interpolasi linier.Jika E < 1.0 L maka harga pengurangan diperoleh dari prosentase tabel di bawah ini : Untuk kapal tipe “ A “ :



Batasan 1. Lambung Timbul Sebenarnya Fba = H-T = 1.48 m Lambung Timbul Sebenarnya harus lebih besar dari Lambung Timbul Total Kondisi



=



Accepted



2. Ketinggian Bow Bow Height



Fba + SFC + TFC = = 6.082 m Ketinggian Bow harus lebih besar dari Ketinggian Bow Minimum



Kondisi



=



Accepted



Perhitungan Tonase Input Data H T VPO



= =



5.28 m 3.80 m



=



271.006 m3



VFC



=



239.339 m3



VDH



= = =



15812.183 m3 4005.876 ton 28 orang



=



2 orang



=



26 orang



∆ Zc N1 N2



; asumsi jumlah penumpang dalam kabin



Gross Tonnage VU



= =



∆∙((1.25∙H/T)−0.115)



6500.809529988 m3 = VPO + VFC + VDH = 16322.528 m3 = VU + VH



VH V



; Volume dibawah geladak cuaca ; Volume ruang tertutup diatas geladak cuaca



= 22823.337 m3 = 0.2+0.02∙log_10V = 0.287167583 = V · K1 = 6554.1225934558 GT



K1 GT



Net Tonnage Vr'



=



10896.6829416984 = 0.2+0.02∙log_10 〖 V_C 〗



K2 K3 a



; Total Volume ruang muat



= 0.2807458863 〖 1.25=〗 ^((GT+10000)/10000) = 1.4468583786 〖 = K 〗 _2∙V_r′∙((4∙T)/(3∙H))^2 =



jadi, Kondisi



NT = a+K_3∙(N_1∙N_1/10)



2813.8203629025



a ≥ 0.25 ∙ GT = Diterima



=



jadi, Kondisi



2817.00345



NT ≥ 0.30 ∙ GT = Diterima



Regresi Kurva Faktor h dan Faktor CI Regresi Kurva Faktor h X = Cpv Y = faktor h f =0 X Y 0.500 0.333 0.522 0.340 0.581 0.360



X 0.500 0.516 0.561



Y 0.335 0.340 0.360



X 0.5000 0.5106 0.5447



Y 0.3351 0.34 0.36



0.643 0.706 0.768 0.830 0.891 0.954 1.000



0.613 0.668 0.725 0.789 0.856 0.933 1.000



0.380 0.400 0.420 0.440 0.460 0.480 0.500



0.5838 0.6302 0.6804 0.7393 0.8109 0.9000 1.0000



0.38 0.40 0.42 0.44 0.46 0.48 0.50



f = 0.5



0.380 0.400 0.420 0.440 0.460 0.480 0.500



f = 1.0



f=0 Y



f(x) = 1.240907059x^4 - 3.4551355082x^3 + 3.5356220095x^2 - 1.2507295648x + 0.4288283645 R² = 0.9999002048 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10



X f=0 Polynomial (f = 0) 0.60



f=0.5



Y



f(x)0.40 = 1.0971592524x^4 - 3.0684652869x^3 + 2.9550318898x^2 - 0.7889130919x + 0.3050260031 R² = 0.9999358549



0.20 0.00 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10



X f=0.5



f=1 0.60



Y



f(x) = 0.821549649x^4 - 1.8734526126x^3 + 0.9771628659x^2 + 0.602905867x - 0.0282375791 0.40 R² = 0.9999345118 0.20 0.00 0.40



0.50



0.60



0.70



0.80



0.90



X f=1



Polynomial (f = 1)



1.00



1.10



Hasil Regresi Kurva Faktor h 1. f=0 Y a b c d e



3. f = 1.0 =ax +bx +cx +dx+e = 1.2409 = -3.4551 = 3.5356 = -1.2507 = 0.4288 4



3



2



Y a b c d e



2. f = 0.5 4 3 2 Y =ax +bx +cx +dx+e a b c d e



= = = = =



1.0972 -3.0685 2.9550 -0.7889 0.3050



Regresi Kurva Faktor CI X = Cw Y = CI Garis 1



X = Cw" Y = CI' Garis 2



X 0.5000 0.5686



Y 0.0236 0.0300



X 0.5000 0.5775



Y 0.02 0.03



0.6683 0.7539 0.8332 0.9027 0.9581



0.0400 0.0500 0.0600 0.0700 0.0800



0.6555 0.7309 0.8072 0.8840 0.9536



0.04 0.05 0.06 0.07 0.08



garis 1



Y



0.09 0.08 f(x) = 0.07 0.5538742789x^4 - 1.4709182917x^3 + 1.5132558461x^2 - 0.6042558461x + 0.0967185953 R² = 0.9999619583 0.06 0.05 0.04 0.03 0.02 0.40 0.50 0.60 0.70 0.80 0.90 1.00



X garis 1



Polynomial (garis 1)



4 3 2 =ax +bx +cx +dx+e = 0.8215 = -1.8735 = 0.9772 = 0.6029 = -0.0282



Garis 2 0.10



Y



0.08 f(x) = 0.3449226898x^4 - 0.9626392315x^3 + 0.998694463x^2 - 0.3257811281x + 0.0320130257 0.06 R² = 0.9999838841 0.04 0.02 0.00 0.40



0.50



0.60



0.70



0.8 0



0.90



X Garis 2



Polynomial (Garis 2)



Hasil Regresi Kurva Faktor CI 1. Garis 1 Y = a x4 + b x 3 + c x 2 + d x + e a = 0.5539 b = -1.4709 c = 1.5133 d = -0.6043 e = 0.0967 2. Garis 2 4 3 2 Y =ax +bx +cx +dx+e a b c d e



= = = = =



0.3449 -0.9626 0.9987 -0.3258 0.032



1.00



Perhitungan Stabilitas Satuan Panjang



⇨ 1 feet



=



Berat



⇨ 1 long ton



=



Input Data LPP



=



287.6312336 f



B Bw



=



57.41469816 f



=



57.41469816 f



T HM



=



12.45652887 f



=



17.31774934 f



SF



=



0f



; Sheer fore



SA



=



0f



; Sheer aft



∆0



=



13142.63725 long ton



ℓST



=



57.52624672 f



; panjang bangunan atas



hST



=



8.530183727 f



; tinggi bangunan atas



CB



=



0.645118093



CWP



=



0.18



CX



= CM =



0.3048 m 1.016 ton



; maximum waterline breadth



0.980835038



Perhitungan Awal CPV = C_B/C_WP



; vertical prismatic coefficient



A0



= 3.583989 = LPP ∙ BW ∙ CWP ; luas bidang garis air



AM



= 2972.567 ft2 = BW ∙ CX ∙ T ; luas area midship



S



A2 D



F A1



= 701.4813 ft2 = (ℓ_ST∙h_ST )+(0.5∙L_PP∙S_F/3)+(0.5·L_PP·S_A/3)



; sheer rata-rata = 490.7095 ft = (0.98 ∙ LPP ∙ HM) + S ; area of vertical centerline plane to depth = 5372.213 ft2 = S/L_PP +H_M ; tinggi kapal rata-rata = = = = =



19.02379 ft D-T ; lambung timbul rata-rata 6.567257 ft 1.01 ∙ A0 ; area of waterline plane at depth D 3002.293 ft2 maybe estimate from A0 and nature



of stations above waterline



Perhitungan GZ DT



δ



Perhitungan h0 〖 = ∆ 〗 _0+((A_0+A_1)/2)∙(F/35) = 13703.186378968 = D_T/2−∆_0 =



C W'



= A_2/(L_PP∙D) =



CX'



8.3972995617



= (35∙D_T)/(A_2∙B) =



CW''



0.9874510309



= (35∙D_T)/(A_1∙D) =



CPV''



0.9817935828



= (A_M+(B∙F))/(B∙D) =



CPV'



-6291.0440645671



1.5549389272



= C_W^′−((140∙δ)∙(1−C_PV^′′))/(L_PP∙D∙B) =



-0.5739566307



f0



= (T∙(A_0/A_1 −1))/(2∙F∙(1−C_PV)) = -0.003670222



f1



= (D∙(1−A_0/A_1 ))/(2∙F∙(1−C_PV^′)) =



f2



KG



-0.001938601



= jika CX' ≥ 0.89, maka f2= 9.1 ∙ (CX' - 0.89) jika CX' ≤ 0.89, maka f2= 0 = =



0.886804381 23.30944143 ft



Perhitungan h1 Referensi : Regresi h1 untuk f= 0 = h1 untuk f= 0.5 = h1 untuk f= 1 = h1 interpolasi =



Kurva Faktor h 4363.505662 3840.715795 3049.337412 4365.532623 -43285.39095 ft



KG'



=



GG'



= KG’ – KG



=



-43308.70039 ft



Perhitungan h0 Referensi : Regresi Kurva Faktor h h0 untuk f= 0 = 87.041541 h0 untuk f= 0.5 = 75.203087 h0 untuk f= 1 = 63.977916 h0 interpolasi = 87.1284405 KB0 (1 - h0) ∙ T = = = =



G'B0



-1072.86141 KG’ – KB0 -42212.5295



Perhitungan h2 Referensi : Regresi Kurva Faktor h h2 untuk f= 0 = 1.29700051 h2 untuk f= 0.5 = 1.10087254 h2 untuk f= 1 = 1.03082554 h2 interpolasi = 0.94914622 G'B90



= (D_T∙h_2∙B)/(4∙D_0 )−(d^2/D_0 ∙17.5/((A_2−((70∙d)/8∙(1−C_PV^′ ′))) ))



=



CI BM0 CI' BM90



GM0 G'M0 G'M90 b1



b2



b3



-0.61895329 a x 4 + b x 3 + c x2 + d x + e



= = = =



0.02896009 (C_I∙L_PP∙B_W^3)/(35∙D_0 )



3.42732029 ft a x 4 + b x 3 + c x2 + d x + e



= =



0.76742704 ( 〖 C′ 〗 _I∙L_PP∙D^3)/(35∙D_0 )+(L_d∙d∙D^2)/(140∙D_0 )



= =



3.40031553 ft



KB0 + BM0 – KG



= = = =



-1092.74353 ft KB0 + BM0 – KG’ = = = =



42215.9569 ft BM90 – G’B90 4.01926882



(9∙( 〖 G`B 〗 _90− 〖 G`B 〗 _0))/8−( 〖 G`M 〗 _0− 〖 G`M 〗 _90)/32



47522.6733



(9∙(𝐺_^′)/��



= =



( 〖 G`M 〗 _0+ 〖 G`M 〗 _90)/8



=



3∙( 〖 G`M 〗 _0− 〖 G`M 〗 _90)/32−3∙(( 〖 G`B 〗 _90− 〖 G`B 〗 _0))/8



5277.49702



=



-11872.0973



Perhitungan Lengan Stabilitas Φ GG' ∙ sin (1 ∙ b1 ∙ sin (2 ∙ Φ) b2 ∙ sin (4 ∙ Φ) b3 ∙ sin (6 ∙ Φ) GZ (ft) GZ (m)



0°



5° 10° 15° 20° 0 -3772.69324 -7516.7031007 -11203.564 -14805.2459 0 8248.084615 16245.808745 23750.411 30534.0992 0 1804.132666 3390.8786411 4569.045 5196.66999 0 -5933.3193 -10278.385072 -11872.094 -10287.8348 0 346.2047394 1841.5992129 5243.7984 10637.6885 0 1135.842321 6041.9921684 17204.063 34900.5529



Φ 25° 30° 35° 40° GG' ∙ sin (1 ∙ -18294.4 -21644.3936 -24829.862414 -27826.552 b1 ∙ sin (2 ∙ Φ) 36391 41143.22197 44646.629826 46794.844 b2 ∙ sin (4 ∙ Φ) 5198.129 4573.24565 3397.3150251 1812.0292 b3 ∙ sin (6 ∙ Φ) -5949.69 -18.9081304 5916.934342 10268.909 GZ (ft) 17345 24053.16585 29131.016779 31049.23 GZ (m) 56906.29 78914.58611 95574.202031 101867.55 Φ GG' ∙ sin (1 ∙ b1 ∙ sin (2 ∙ Φ) b2 ∙ sin (4 ∙ Φ) b3 ∙ sin (6 ∙ Φ)



45° -30611.68 47522.6582 8.40522102 11872.0634 28791.4469 94460.1275



45° 50° 55° 60° 65° -30611.7 -33164.0703 -35464.317607 -37494.933 -39240.4794



GZ (ft)



47522.66 46807.98037 44672.503769 41181.048 8.405221 -1796.23154 -3384.4336559 -4564.8327 11872.06 10297.25842 5966.0440159 37.816213 28791.45 22144.93699 11789.796522 -840.90185



GZ (m)



94460.13 72653.99274 38680.434784 -2758.8643 -45592.5833



Φ GG' ∙ sin (1 ∙ b1 ∙ sin (2 ∙ Φ) b2 ∙ sin (4 ∙ Φ) b3 ∙ sin (6 ∙ Φ)



70° 75° 80° 85° 90° -40687.7 -41825.5446 -42645.409894 -43141.047 -43308.6867



GZ (ft) GZ (m)



30592.06 23815.93818 16316.915352 8322.6126 -5199.57 -4577.43473 -3403.7427917 -1819.9212 -10259.4 -11872.0032 -10306.655925 -5982.3837 -25554.6 -34459.0443 -40038.893259 -42620.739 -83840.6 -113054.607 -131361.19836 -139831.82



(GG′∙sin 〖 (Φ∙π) 〗 )/180 GG' ∙ sin (1 ∙= Φ)



b1 ∙ sin (2 ∙ Φ) = (b_1∙sin 〖 (Φ∙2∙π) 〗 )/180 b2 ∙ sin (4 ∙ Φ)



=(b_2∙sin 〖 (Φ∙4∙π) 〗 )/180



b3 ∙ sin (6 ∙ Φ) = (b_3∙sin 〖 (Φ∙6∙π) 〗 )/180



36439.5923 -5195.19799 -5900.53438 -13896.6194



75.6871242 -16.8104207 -56.7241994 -43306.5342 -142081.805



Perhitungan Lengan Dinamis (LD) ; h = 0.09 rad



L_Dn= 1/3∙h∙(GZ_(n−10)+4∙GZ_(n−5)+GZ_n)



Sudut [ o ] 10 20 30 40 LD Total



LD [ft.rad] 93.8527002 973.1514202 3027.302731 4992.411441 9086.718293



Hasil Perkalian a b c θmax



LD [m.rad] 28.60630302 296.6165529 922.7218725 1521.687007 2769.631736



Matrik = -332100.9962 = 21809.83491 = -274.0155296 =



Sudut [ o ] 0°



h = 50*3,14/1800



39.79671° ; sudut maximum Lengan Statis (m) 0



5° 10° 15° 20°



1135.842321 6041.992168 17204.06313 34900.55295



25° 30° 35°



56906.28896 78914.58611



40°



95574.20203 101867.553



45° 50° 55° 60° 65° 70° 75° 80° 85° 90°



94460.12747 72653.99274 38680.43478 -2758.864327 -45592.58335 -83840.57904 -113054.6074 -131361.1984 -139831.8207 -142081.805



25 20 15 10 5 0



0



10



20



30



40



50



60



70



80



90



100



-5 -10 -15 -20



Sudut Maksimum GZ max = 1E+05 m ; nilai maksimum GZ dari semua sudut (00 s.d. 900) Kolom Ke = 9 ; nilai terbesar tersebut pada kolom ke berapa Heel at GZ m = 40° ; pada sudut heel berapa GZ maksimum Titik X1



=



35



X2 X3



= =



40 45



1 1



35 40



1225 1600



36 -1.7



Y1 Y2 Y3



= 95574 = 1E+05 = 94460



1



45



2025



0.02 -0.04 0.02



Matriks



Invers Matrik -63 3.2



28 -1.5



Batasan Stabilitas Menurut IMO Resolution A. 749 (18) INPUT DATA e [ m . rad ] o



30 =



922.722



40o =



1521.687



30 -40 =



598.965



o



o



GZ 30o = GM0 =



Kriteria IMO Regulation A. 749 (18) e0.30o ≥ 0.055 = Accepted e0.40o ≥ 0.09 =



Accepted



e



≥ 0.03 =



Accepted



h30o ≥ 0.2 =



Accepted



ɸmax ≥ 25o =



Accepted



o 30,40



GM0 ≥ 0.15 = Accepted Status = OK Roll Period [ s ] B= 17.50 m G'Mo = 42215.96 Period=0.79*B / G'Mo^1/2=



0.1



78914.586



Ɵmax [X ] =



39.8



[ feet ] =



-1092.74



[m]=



0.30



o



Referensi Biaya Operasional Kapal Dharma Kencana Sby - Banjar Kapasitas Kapal (Ukuran 600 GT) Jumlah Penumpang Jumlah Kendaraan Roda 2 Jumlah Kendaraan Mobil Jumlah Kendaraan Mobil Barang Jumlah Kendaraan Bus, Truck (Kecil) Jumlah Kendaraan Bus, Truck (Besar)



= = = = = =



750 12 24 4 22 13



SUP = standarisasi harga penumpang Penumpang Sepeda Sepeda Motor ( =500 cc) Mobil (L