Kuliah-8 TBK Stabilitas Memanjang

7/28/2019 Kuliah-8 TBK Stabilitas Memanjang

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© Adi Wirawan Husodo2012

Kuliah-8 Teori Bangunan Kapal

Stabilitas Memanjang (TRIM)

Dosen : Adi Wirawan Husodo, ST, MT

1




• Trim is also known as ‘longitudinal stability’

• Trim is measured as the difference between the draft forward and

aft.

– If difference is zero ship is on even keel

– If forward draft is greater than aft draft trimming by bow

– If aft draft is greater than forward draft trimming by stern




• Consider a ship to be floating at rest still water and on an even keel(fig 15.1)

• Now let a weight w, already on board, be shifted aft through adistance d. The center of gravity of the ship will shift from G to G1,parallel to the shift of center of gravity of the weight shifted.

GG1= (w x d)/WW x GG1=w x d…….trimming moment




• When trimmed (fig 15.2), the wedgeof buoyancy LFL1 emerges andwedge WFW1 is immersed

• The volume of immersed wedge =the volume of emerged wedges

• F , the point which the ship trim, is

the center of gravity of water planearea (center of flotation or tippingcentre)

• ML (longitudinal metacentre) is thepoint of intersection betweenverticals through the longitudinal

positions of centre of buoyancy. Thevertical distance between the center of gravity and the longitudinalmetacentre is called longitudinal metacenrric height (GML)




Moment to change Trim one cm (MCT 1 cm)

• Moment required to change trim

by 1 cm

• MCT 1 cm = (W x GML)/100L

W : vessel’s displacement, ton

GML

: long. Metacentric height, m

L : Vessel’s length, m

• Fig 15.3 (a) ship in equilibrium

• Fig 15.3 (b) weight w is shifted

through d distance. The center of

gravity will shift from G to G1

• Causing the trimming moment of

W x GG1





• See fig 15.3 (c)

• The ship will trim to bring thecentre of gravity and buoyancy int

othe same vertical line the ship isagain in equilibrium

• The tipping centre (F) is l metrefrom aftGG1 = (w x d)/W = GML tan θ

tan θ =(w x d)/(W x GML)

tan θ = t/L…….fig 15.4.(b)





• Let the change of trim due to shifting is 1 cm

• Then w x d is the moment to change trim 1 cm

• tan θ= 1/100 L

But

• tan θ=(w x d)/(W x GML)

• tan θ= MCT 1 cm/(W x GML)

• MCT 1 cm/(W x GML) = 1/100 L

MCT 1 cm = (W x GML)/100 L




To find the change of draft due to change of

trim

• Trim will cause a change in the draft forward and draft aft. One will be

increased and one will decreased.

• A is a new draft aft, F is a new draft forward. Trim = A-F

• x is the change of draft aft, y is the change draft forward

• See the triangles WW1F1 and WW1C




To find the change of draft due to change of

trim

• Change of draft aft in cm

= (l /L) x Change of trim in cm – l = distance F from aft, m

– L = ship’s length, m

– x + y = t

• Change of draft F in cm = Change of trim – Changeof draft A

mL

cmx tmcmx

mLcmt

mcmx

l

l




The effect of shifting weight already on board

A ship 126 m long is floating at draft of 5.5 m F and 6.5 m A. The centre of flotation is 3 m aft of amidships. MCT 1 cm = 240 ton. m, displacement 6000 tons. Find the new drafts if a weight 120 ton already on board isshifted forward a distance of 45 m.

Trimming moment = w x d= 120 x 45

= 5400 ton m by the head

Change of trim = trimming moment/MCT 1 cm

= 5400/240

= 22.5 cm by the headChange of draft aft = (l /L) x Change of trim

= (60/126) x 22.5

= 10.7 cm




The effect of shifting weight already on board

Change of draft Forward = Change of trim – Change of draft Aft

= 22.5 – 10.7

= 11.8 cm

Original draft 6.500 m A 5.500 m F

Change due trim -0.107 m +0.118 m

New draft 6.393 m 5.618 m




The effect of loading and/or discharging weight

• When a weight is loaded at the centre of flotation it will produce no trimming

moment, but the ship’s draft will increase uniformly so that the ship displaces

an extra weight of water equal to the weight loaded. If the weight is now

shifted forward or aft away from the center of flotation, it will cause a change

of trim. cause both the bodily sinkage and a change of trim

• When a weight is being discharged, if the weight is first shifted to the centre of flotation it will produce a change of trim, and if it is then discharged from the

centre of flotation, the ship will rise bodily.

TPC = the mass which must be

loaded or discharged to change

the ship’s mean draft by 1 cm

(ton)

WPA = water plane area, m2

ton97.56

WPATPC

cmTPC

w riseor sinkageBodily





A ship 90 m long is floating at drafts 4.5 m F and 5.0 m A. The centre of flotation is 1.5 m aft of amidships. TPC 10 tons. MCT 1 cm is 120 ton m.Find the new drafts if a total weight of 450 tons is loaded in a position14 m forward of amidships

Bodily sinkage = w/TPC Change of trim = trim moment/MCT 1

cm= 450/10 = (450 x 15.5)/120

= 45 cm = 58.12 cm by the head

Change of draft aft = (l /L) x Change of trim

= (43.5/90) x 58.12

= 28.09 cm

Change of draft Forward = Change of trim – Change of draft Aft

= 58.12 – 28.09

= 30.03 cm





Original draft 5.000 m A 4.500 m F

Bodily sinkage + 0.450 m + 0.450 m

5.450 m 4.950 m

Change due trim - 0.281 m + 0.300 m

NEW DRAFT 5.169 m A 5.250 m F




TERIMA KASIH

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Kuliah-8 TBK Stabilitas Memanjang