Final KG

Chapter 5

FINAL KG

Ketika suatu kapal telah selesai dibangun, maka beberapa informasi penting tentang stabilitas kapal tersebut juga harus disampaikan kepada pemilik kapal (owner). Beberapa informasi tersebut adalah berat kapal kosong (lightweight), VCG dan LCG serta informasi lain seperti posisi titik berat muatan dan bunker space. Hal ini sangat diperlukan untuk mengetahui kondisi awal dimana displacement dan KG untuk berbagai kondisi dapat diketahui. Final KG dapat diketahui dari nilai momen sejumlah muatan yang dimuat ataupun yang akan dikeluarkan, terhadap lunas kapal (keel).

Final Moment = Final KG x Final Displacement

Contoh 1

Sebuah kapal dengan displacement 6000 ton mempunyai KG = 6 m dan KM = 7.33 m. Berikut adalah beberapa muatan yang diangkut (loading) dan yang akan diturunkan (discharge)

loaded cargo :                       1000 ton,       KG 2.5 m

                                                500 ton,          KG 3.5 m

                                                750 ton,          KG 9.0 m

discharge cargo :                 450 ton,          KG 0.6 m

                                                800 ton           KG 3.0 m

Tentukan besarnya nilai GM

Jawab

Berat muatan	KG	Momen thd keel
+6000 +1000 +500 +750 -450 -800	6.0 2.5 3.5 9.0 0.6 3.0	+36000 +2500 +1750 +6750 -270 -2400
+7000		+44330

Final KG        = Final moment/final displacement

                        = 44330/7000 = 6,33 m

GM      = KM – KG

            = 7.33 – 6.33

            = 1 m

Contoh 2

Kapal dengan displacement 5000 ton mempunyai KG 4.5 m dan KM 5.3 m. Beberapa muatan berikut dinaikkan ke atas kapal

            2000 ton, KG 3.7 m                          1000 ton, KG 7.5 m

Tentukan berapa besarnya muatan (KG 9 m) yang harus dinaikkan ke atas kapal, sehingga kapal dapat berlayar dengan kondisi GM minimum 0.3 m

Asumsi muatan yang akan ditambahkan adalah ’x’ ton.

Berat muatan	KG	Momen thd keel
5000 2000 1000 x	4.5 3.7 7.5 9.0	22500 7400 7500 9x
8000 + x		37400 + 9x

diketahui       KM      = 5.3 m

                        GM      = 0.3 m

                        KG      = KM – GM

                                    = 5.3 – 0.3

                                    = 5 m

Final KG        =37400+9x/8000+x

            5          = 37400+9x/8000+x

40000 + 5x    = 37400 + 9x

            2600   = 4x

            x          = 650 ton

Jadi besarnya muatan yang harus ditambahkan adalah 650 ton

MOMENT OF STATIC STABILITY

Moment of static stability may be defined as the moment to return the ship to the initial position when inclined by an external force (such as wind and wave action).

we can see a ship which has been inclined by an external force. The centre of Buoyancy has moved from B to B₁ parallel to gg₁, and the force of buoyancy (w) acts vertically upward through B₁.

The weight of the ship (W) acts vertically downwards through the centre of gravity (G). The perpendicular distance between the lines of action of forces (GZ) is called the righting lever.

Finally we can find the Moment of static stability by multiplying the righting lever and Displacement.

Moment of Static Stability = W x GZ   (eq.01)

The Moment of Static Stability at a Small Angle of Heel

Remember:

Moment of Static Stability = W x GZ

Triangle GZM: GZ = GMsinq^o

So the Moment of Static Stability:

= W x GM x sinq^o                      (eq.02)

Example 01.

A ship of 4000 tonnes displacement has KG 5.5 m and KM 6.0 m. Calculate the moment of statical stability when heeled 5 degrees.

Answer:

GM = KM – KG = 6.0-5.5 = 0.5 m

Moment of statical stability = W x GM x sinq

                                                                      = 4000 x 0.5 x sin 5^o

                                                                      = 174.4 tonnes m

Example 02.

When a ship of 12000 tonnes displacement is heeled 6 ½ degrees the moment of statical stablity is 600 tonnes m. Calculate the initial metacentric height.

Answer:

Moment of statical stability = W x GM x sinq

GM = (Moment of statical stability)/(W x sinq)

   = 600/(12000 x sin 6 ½ ^o)

   = 0.44 m

The Moment of Static Stability at a Large Angle of Heel

At large angle of heel (More than 15 degrees), the force of  buoyancy can no longer be considered to act vertically upwards through the initial metacentre (M).  As shown in figure 03, the centre of buoyancy has moved further out to the low side, and the vertical through B₁ no longer passes through M. The righting lever (GZ) is once again the perpendicular distance between the vertical through G and the vertical through B₁, and the moment of statical stability is equal to W x GZ. But GZ is no longer equal to GMsinq. Up to the angel at which the deck edge is immersed, it may be found by using a formula known as the Wall-sided formula.

GZ = (GM + ½ BM tan²q)sinq         (eq.03)

Example 03.

A ship of 6000 tonnes displacement has KB 3 m, KM 6 m, and KG 5.5 m. Find the moment of statical stability at 25 degrees heel.

Answer:

     GZ = (GM + ½ BM tan²q)sinq

        = (0.5 + ½ x 3 x tan²25^o)sin25^o

        = 0.8262 sin25^o

        = 0.35 m

     Moment of statical stability = W x GZ

        = 6000 x 0.35

        = 2100 tonnes m

Example 04.

A box-shaped vessel 65 m x 12 m x 8 m has KG 4 m, and is floating in salt water upright on even keel at 4 m draft F and A. Calculate the moment of statical stability at (a) 5 degrees and (b) 25 degrees heel.

Answer:

     W = L x B x draft x 1.025    KB = ½ draft

       = 65 x 12 x 4 x 1.025      KB = 2 m

       = 3198 tonnes

     BM = B²/12d

        = (12 x 12)/12 x 4

        = 3 m

     KB = 2 m

     BM = 3 m

     KM = KB + BM = 2 + 3 = 5 m

     KG = 4 m

     GM = KM – KG

        = 5 – 4

        = 1 m

at 5^o heel          GZ = GMsinq

                 = 1 x sin 5^o

                 = 0.0872 m

               Moment of statical stability

                 = W x GZ

                 = 3198 x 0.0872

                 = 278.9 tonnes m

at 25^o heel    GZ = (GM + ½ BM tan²q)sinq

                 = (1 + ½ x 3 x tan²25^o)sin25^o

                 = (1 + 0.3262)sin25^o

                 = 1.3262 sin25^o

                 = 0.56 m

              Moment of statical stability

                 = W x GZ

                 = 3198 x 0.56

                 = 1790.9 tonnes m