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HAJIR SANATA

TUGAS MEKANIKA FLUIDA. HAJIR SANATA. 20080110027.

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HAJIR SANATA

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  1. TUGAS MEKANIKA FLUIDA HAJIR SANATA 20080110027

  2. 1. Balokkayudenganpanjang 1,0 m, lebar 0,4 mdantinggi 0,3 mmengapungdi air dengansisitingginyavertikal. RapatrelatifkayuS = 0,7. Hitung volume air yang dipindahkandanletakpusatapung.Penyelesaian : Volume balok, V = 1,0x0,4x0,3 = 0,12 m3Beratbalok, W = ρbalok g V = S ρair g V = 0,7x1000x9,81x0,12 = 824,04 NVolume air yang dipindahkanbenda,V = V = = = = 0,084 m3

  3. Kedalamanbagianbalok yang terendam,d = = = 0,21 mLetakpusatapung,OB = = = 0,105 mJadiletakpusatapungadalah 0,105 mdaridasarbalok

  4. 2. SuatubalokpontondenganlebarB=6,0 m, panjangL=12 mdansaratd=1,5 mmengapungdidalam air tawar (ρ= 1000kg/m3). Hitung : a. beratbalokponton b. saratapabilaberadadi air laut (ρ = 1025kg/m3) c. beban yang dapatdidukungolehpontondi air tawarapabilasaratmaksimum yang diijinkanadalah 2,0 m.Penyelesaian : a. dalamkeadaanterapung, beratbendasamadenganberat air yang dipindahkanbenda (FB)FG = FB = ρ1 .g .B .L .d= 1000x9,81x6,0x12,0x1,5= 1.059,480 N = 1059,48 kN

  5. JadiberatbendaadalahFG= 1059,48 kNb. Mencarisarat (draft) di air laut.padakondisimengapung, beratbendasamadengangayaapung FG = FB = ρ2 .g .B .L .dd = = = 1,463 m FG d FB

  6. c. Untuksaratmaksimumdmak = 2,0 m , gayaapung total :FBmak= ρ .g .B .L .dmak= 1000x9,81x6,0x12,0x2,0 = 1.412.640 N= 1412,64 kNbeban yang dapatdidukungadalah :Bmak = gayaapungmaksimum – beratponton= 1412,64 – 1059,48 = 353,16 kN

  7. 3. Balokterbutdaribahandenganrapatrelatif 0,8 mempunyaipanjangL=1,0 mdantampangliantangbujursangkardengansisi 0,8 mdiapungkandidalam air dengansumbupanjanngnyavertikal. Hitungtinggimetasentrumdanselidikistabilitasbenda. B=0,8 H=0,8 G L=1,0 d B O

  8. Penyelesaian :S = = 0,8γbenda= 0,8x1000 = 800 kgf/m3Luastampanglintangbalok, A = B H = 0,8x0,8 = 0,64 m2Beratbenda, FG = γair L A = 800x1,0x0,64 = 512 kgfBerat air yang dipindahkan,FB = γair L d = 1000x0,64xd = 640 d kgfDalamkeadaanmengapung, FG = FB512 = 640 dd= 0,8 m

  9. Jarakpusatapungterhadapdasarbalok,OB = = = 0,4 mJarakpusatberatterhadapdasarbalok,OG = = = 0,5 mBG = OG – OB = 0,5 – 0,4 = 0,1 mMomeninersiatampangbujursangakarI0 = B H3 = x0,8x0,83 = 0,03413 m4

  10. Volume air yang dipindahkan, V = A d = 0,64 x 0,8 = 0,512 m3BM = = = 0,06667 mTinggimetasentrum,GM = BM – BG = 0,06667 – 0,1 = – 0,03333 mKarenatinggimetasentrumGMbertandanegatif, makabendadalamkondisitidakstabil.

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