Oleh Mas Edy

Marine Higher Education Institute
In progress Updated 14 Feb, 2011

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Soal 1

A ship of 8500 tonnes displacement has TPC 10 tonnes, MCT 1cm 100 tonnes m and the centre of flotation is amidships. She is completing loading under coal tips. Nos. 2 and 3 holds are full, but space is available in No. 1 hold (centre of gravity 50 m forward of amidships), and in No. 4 hold (centre of gravity 45 m aft of amidships). The present drafts are 6.5 m F and 7 m A, and the load draft is 7.1 m. Find how much cargo is to be loaded in each of the end holds so as to put the ship down to the load draft and complete loading on an even keel.

Jawab:

 
Displ = 8500 ts
TPC = 10 ts / cm
MCT = 100 ts/cm
                       A        O           F
Initial Draft = 7.0     6.75     6.5      Initial trim = 50 cm by S
Final draft =   7.1     7.1       7.1      Final Trim = 0

AT =                       0.35 m              ch.of trim = 50 cm by H

Jml muatan yang di muat = AT x TPC
                                        = 35 x 10 t/cm
                                        = 350 ts

Jika di misalkan palka 1    = W ts

Maka jml. Muatan yang di muat di palka 4 = ( 350 – w ) ts

 

Ch of trim      =    T.mo
                           MCTC
T.mo              = Ch of trim x MCTC
                     = 50 x 100

                     = 5000 t.m … (2)

            ……1) = … (2)
95W – 15750 = 5000
95w               = 15750 + 5000
W                  = 20750
                           95

W                  = 218,4 ts

W di palka 1 = 218.4 ts
W di palka 4 = 350 – w
                     = 350 – 218.4
                     = 131.6 ts

 

 

Soal 2
A ship is floating at drafts of 6.1 metres F and 6.7 metres A. The following cargo is then loaded:
20 tonnes in a position whose centre of gravity is 30 metres forward of amidships.
45 tonnes in a position whose centre of gravity is 25 metres forward of amidships.
60 tonnes in a position whose centre of gravity is 15 metres aft of amidships.
30 tonnes in a position whose centre of gravity is 3 metres aft of amidships.
The centre of flotation is amidships, MCT 1cm 200 tonnes m and TPC 35 tonnes.

Find the new drafts forward and aft.

Jawab:

 

 

 

Bodily singkage = W = 155 ts         = 4.4 cm      = 0.044 m

                            Tpc    35 ts / cm
Change of trim = Trim moment
                               MCT 1 cm
                        = 735 ts
                           200 ts/cm
                        = 3.67 cm
Change of draft = l x change of trim
                             L
Change of draft = 1 x change of trim
                             2
                          = 1 x 3.67
                             2

                          = 1.83 cm

                                  A                       F
Initial draft            = 6.7                    6.1
Bodily singkage     = 0.044                0.044
Final Cond             = 6.744                6.144
Change due trim = – 0.0183            +0.0183
New draft            =   6.726 m            6.162 m

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