TRIMTRIM may be considered as the longitudinal equivalent of LIST. TRIM is also known as LONGITUDINAL STABILITY. TRIM PERBEDAAN DRAFT DEPAN DAN DRAFT BELAKANG Draft depan Draft belakang Trimming by the bow (trim depan) Draft depan Draft belakang Trimming by the stern (trim belakang)

Draft depan = Draft belakang Even keel (trim nol / kapal rata)

Agung Hernowo

TRIMMING MOMENTTrimming moment = W x GG1 = w x d BML = IL ---V L2 ----12d

BML = IL V = =

the longitudinal second moment of the water-plane about the centre of flotation Volume of displacement

BL3 IL = ----12 L B = the length of the water-plane = the breadth of the waterplane BL3 Untuk bidang air (waterplane) berbentuk persegi panjang BML = -------12 VAgung Hernowo

MCTCThe moment to change trim one centimetre (MCT 1 cm or MCTC) adalah moment yang diperlukan untuk merubah trim sebesar 1 cm

MCTC =

x GML -------------100 L

GML L

= displacement kapal dalam ton = the longitudinal GM in metres = panjang kapal dalam meter

Agung Hernowo

CHANGE OF DRAFT DUE TO CHANGE OF TRIM

Menghitung perubahan draft depan dan belakang sebagai akibat adanya perubahan trim (change of trim): l ---- x Change of trim in cm L

Change of aft draft in cm =

l L

= jarak centre of flotation dari buritan dalam meter = panjang kapal dalam meter

Change of drat F (foreward) = Change of trim Change of draft A

Agung Hernowo

DAMPAK SHIFTING MUATAN:SOAL NO.1: Sebuah kapal memiliki panjang = 126 m, F = 5.5 m, A = 6.5 m, C.O.F berada 3 m di belakang amidships. MCTC = 240 ton.m. displacement kapal = 6,000 ton. Hitung draft baru setelah 120 ton muatan di kapal dipindahkan ke depan sejauh 45 m. PENYELESAIAN: Trimming moment

= w x d = 120 ton x 45 m

= 5,400 ton.m 5,400 = -------240

Change of trim in cm

Trimming moment = -----------------------MCTC = 22.5 cm l = ---- x Change of trim LAgung Hernowo

Change of draft aft

DAMPAK SHIFTING MUATAN:

PENYELESAIAN SOAL NO.1 (LANJUTAN):(126: 2) - 3 = ----------------- x 22.5 = 10.7 cm 126

Change of draft aft

(126: 2) + 3 Change of draft F = ----------------- x 22.5 = 11.8 cm 126 A draft F draft Initial draft : 6.500m 5.500 m Change of draft due to trim : - 0.107m + 0.118 m New draft : 6.393m 5.618 mAgung Hernowo

DAMPAK SHIFTING MUATAN:

SOAL NO.2:

Sebuah kapal berbentuk kotak berukuran 90 m x 10 m x 6 m, berada di air laut dengan draft 3 m even keel. Hitunglah draft baru jika 64 ton muatan di kapal dipindahkan sejauh 40 m ke belakang.PENYELESAIAN: BML = L2 : (12 x d)

Agung Hernowo

DAMPAK LOADING & DISCHARGING

Loading

Bodily sinkage =

w ------TPC w --------TPC

Discharging

Bodily rise

=

Change of trim

=

wxd --------MCTC

Agung Hernowo

DAMPAK LOADING & DISCHARGING SOAL NO.1: Sebuah kapal dengan panjang 90 m, F draft = 4.5 m dan A draft = 5.0 m. C.O.F berada 1.5 m di belakang amidships. TPC = 10 ton, MCTC = 120 ton.m. Hitung draft baru jika kapal memuat 450 ton di posisi 14 m di depan amidships. PENYELESAIAN:

Agung Hernowo

DAMPAK LOADING & DISCHARGING SOAL NO.2: Sebuah kapal berbentuk kotak berukuran 40 m x 6 m x 3 m berada di air laut dengan draft 2 m (even keel). MCTC = 8.4 ton.m; Hitung draft baru jika kapal membongkar 35 ton di posisi 6 m dari haluan.

PENYELESAIAN:

Agung Hernowo

DAMPAK LOADING & DISCHARGING SOAL NO.3: Sebuah kapal memiliki panjang = 100 m. Draft saat tiba di pelabuhan adalah F = 3 m, A = 4.3 m. TPC = 10 ton, MCTC = 120 ton.m, C.O.F berada 3 m di belakang amidships. Kapal memuat 80 ton cargo di posisi 24 m depan amidships, membongkar 40 ton cargo di posisi 12 m di belakang amidships. Hitung draft baru kapal setelah melakukan kegiatan tsb!. PENYELESAIAN:

Agung Hernowo

DAMPAK LOADING & DISCHARGINGSOAL NO.4: Sebuah kapal dengan displacement = 6,000 ton pada draft F = 7 m, A = 8 m. MCTC = 100 ton.m, TPC = 20 ton. C.O.F berada di tengah-tengah panjang kapal. Dari empat palkah masing-masing dibongkar 500 ton muatan tiap palkah dengan keterangan sbb: Palkah no.1, centre of gravity terletak 40 m di depan amidships Palkah no.2, centre of gravity terletak 25 m di depan amidships Palkah no.3, centre of gravity terletak 20 m di belakang amidships Palkah no.4, centre of gravity terletak 50 m di belakang amidships Kapal juga isi bahan bakar sbb: 150 ton berada di posisi 12 m di depan amidships 50 ton berada di posisi 15 m di belakang amidships Hitung draft baru kapal setelah melakukan kegiatan tsb!.Agung Hernowo

DAMPAK LOADING & DISCHARGINGSOAL NO.5 A ship arrives in port trimmed 25 cm by the stern. The centre of flotation is amidships. MCT 1cm = 100 tonnes.m. A total of 3,800 tonnes cargo is to be discharged from 4 holds, and 360 tonnes of bunkers loaded in No.4 double bottom tank; 1,200 tonnes of cargo is to be discharged from No.2 hold and 600 tonnes from No.3 hold. Find the amount to be discharged from Nos.1 and 4 holds if the ship is to complete on an even keel. Centre of gravity of No.1 hold is 50 m forward of the centre of flotation Centre of gravity of No.2 hold is 30 m forward of the centre of flotation Centre of gravity of No.3 hold is 20 m abaft of the centre of flotation Centre of gravity of No.4 hold is 45 m abaft of the centre of flotation Centre of gravity of No.4 DB tank is 5 m abaft of the centre of flotation Total cargo to be discharged from 4 holds 3,800 tonnes Total cargo to be discharged from Nos.2 and 3 1,800 tonnes Total cargo to be discharged from Nos.1 and 4 2,000 tonnes Let x tonnes of cargo be discharged from No.1 hold. Let (2000 - x) tonnes of cargo be discharged from No.4 hold.

Agung Hernowo

USING TRIM TO FIND THE POSITION OF C.O.FSOAL NO.1: Draft kapal saat tiba di pelabuhan adalah F = 3.80 m, A = 4.50 m. Kapal akan melakukan pemuatan dengan rincian sbb: 100 ton pada posisi 24 m di belakang amidships 30 ton pada posisi 30 m di depan amidships 60 ton pada posisi 15 m di depan amidships Draft kapal pada akhir pemuatan adalah F = 4.40 m, A = 5.10 m. Jika posisi C.O.F di belakang amidships, hitunglah jarak C.O.F terhadap amidship!. PENYELESAIAN SOAL NO.1: Misalkan jarak C.O.F dari amidships = x meter Trim saat kapal tiba = 4.50 m 3.80 m = 0.70 m = 70 cm. Trim selesai pemuatan = 5.10 m 4.40 m = 0.70 m = 70 cm. Change of trim = 0, berarti: Moment change of trim by head = Moment change of trim stern {30 x (30 + x)} + {60 x (15 + x)} = 100 x (24 x) 900 + 30x + 900 + 60x = 2,400 100x 190x = 2,400 1,800 = 600 x = 3.156 m C.O.F terletak pada 3.158 meter di belakang amidships.

Agung Hernowo

LOADING A WEIGHT TO KEEP THE AFTER DRAFT CONSTANTl w --- x Change of trim = ------L TPC w L Change of trim = ------- x ----- (1) TPC l w x d Change of trim = ------------MCTC w x d -----------MCTC (2)

=

w L -------- x ----TPC l

L x MCTC d = ----------------l x TPC

dL l

= The distance forward of the C.O.F to load a weight to keep the draft aft constant = The ships length, LBP = The distance of the C.O.F to stern

Agung Hernowo

LOADING A WEIGHT TO KEEP THE AFTER DRAFT CONSTANTSOAL NO.1: Sebuah Box-shaped vsl berukuran 60 m x 10 m x 6 m berada dilaut dengan draft F = 4 m dan A = 4.4 m. Kapal akan memuat 30 ton cargo. Jika draft belakang tetap 4.4 m, dimana muatan tersebut harus ditempatkan? PENYELEYAIAN SOAL NO.1: TPCSW = WPA : 97.56 = (60 x 10) ; 97.56 = 6.15 ton = 60 x 10 x {(4 + 4.4) : 2} x 1.025 = 2583 ton BML = L2 :12d = 602 : (12 x 4.2) = 71.42 m Karena GML = BML, maka MCTC = ( x BML) : (100 x L) = (2583 x 71.42) : (100 x 60) = 30.75 ton.m Karena bidang air berbentuk persegi panjang, maka C.O.F ditengah d = (L x MCTC) : (l x TPC) = (60 x 30.75) : (30 x 6.15) = 10 m dari C.O.FAgung Hernowo

LOADING A WEIGHT TO PRODUCE A REQUIRED DRAFTSOAL NO.1: Sebuah kapal dengan panjang 150 m, draft saat tiba di muara : F = 5.5 m & A = 6.3 m. MCTC = 200 ton.m, TPC = 15 ton. C.O.F terletak 1.5 m di belakang amidships. Kapal akan masuk sungai dimana maksimum draft adalah 6.2 m. Jika kapal mengisi ballast di FPT, dimana titik G berada 60 m di depan C.O.F.;Hitunglah jumlah ballas yang diisi,berapa draft depan setelah pengisian ballas? PENYELESAIAN SOAL NO.1: Bodily sinkage = w : TPC = (w : 15) cm New A draft = 6.3 m + (w : 15) ..(1) Required draft = 6.2 m (2) Pengurangan draft belakang = 0.1 m + (w :15) = 10 cm + (w : 15) cm .(3) Change of trim = (w x d) : MCTC = 60 w : 200 = 0.3 w cm by the head Change of draft A due to C.O.T = (l : L) x C.O.T = [{(150 : 2) 1.5} : 150] x 0.3 w = 0.147 w cm .(4)

Pengurangan draft belakang = Change of draft A due to C.O.T 10 + (w : 15) = 0.147 w 150 + w = 2.205 w 1.205 w = 150 w = 124.481 = 124.5 tonAgung Hernowo

LOADING A WEIGHT TO PRODUCE A REQUIRED DRAFT

PENYELESAIAN SOAL NO.1 (LANJUTAN): Menghitung draft depan Bodily sinkage = w : TPC = 124.5 : 15 = 8.3 cm C.O.T = (w xd) : MCTC = (124.5 x 60) : 200 = 37.35 cm C.O.D = (76.5 : 150) x 37.35 cm = 19.05 cm. (F draft)

Initial draft Bodily sinkage C.O.D due to C.O.T New draft

A 6.3 + 0.08 6.38 - 0.18 6.20 mAgung Hernowo

F 5.5 + 0.08 5.58 + 0.19 5.77 m

USING C.O.T TO FIND GML

GML : GG1 = L : TRIM SOAL NO.1: Pada saat kapal memindahkan muatan ke belakang, G bergeser 0.2 m mendatar dan C.O.T = 0.15 m. Jika panjang kapal = 120 m, hitunglah GML kapal tsb.

Agung Hernowo

KOMBINASI LIST & TRIMTrim awal = 15 cm trim depan Required Trim = 30 cm trim belakang Change of Trim = 45 cm trim belakang Trimming moment = C.O.T x MTCT = 45 x 120 = 5400 ton.m by stern Misalkan minyak yang dipindah ke tanki no.5 = w ton Trimming Moment = w x d = w x (23.5 + 21.5) = 45 w 5400 = 45 w w = 5400 : 45 = 120 ton KM KG GM = 7.00 m = 6.40 m = 0.60 m

Misalkan minyak yang dipindahkan ke tanki 5 kiri = Y ton Listing moment = Y x d = Y x (6 +6) = 12 Y GG1 = GM x Tan 5o = 0.05Agung Hernowo

KOMBINASI LIST & TRIMTrim awal = 15 cm trim depan Required Trim = 30 cm trim belakang Change of Trim = 45 cm trim belakang Trimming moment = C.O.T x MTCT = 45 x 120 = 5400 ton.m by stern Misalkan minyak yang dipindah ke tanki no.5 = w ton Trimming Moment = w x d = w x (23.5 + 21.5) = 45 w 5400 = 45 w w = 5400 : 45 = 120 ton KM KG GM = 7.00 m = 6.40 m = 0.60 m

Misalkan minyak yang dipindahkan ke tanki 5 kiri = Y ton Listing moment =Y xd = Y x (6 +6) = 12 Y o GG1 = GM x Tan 5 = 0.0525

Initial listing moment Syarat tegak : Moment ke kanan 315

= x GG1 = 6000 x 0.05 = 314.959 = 315

= Moment ke kiri = 12 Y Y = 315 : 12 = 26.25 ton

Jadi pendistribusian bahan bakar adalah sbb : tanki no.2 kiri = 200 ton, Tanki no.2 kanan = 200 120 = 80 ton Tanki no.5 kiri = 26.25 ton, Tanki no.5 kanan = 120 26.25 = 93.75 ton

Agung Hernowo

TRIM

MAU KISI-KISI?

Agung Hernowo