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Saturday, June 8, 2024

Ship Stability - I Trim and List

 

Trim

Learning Objectives
After completing this topic you will be able to:

Explain
  • LCG & LCB
  • Effect of Loading Discharging and shifting weights on LCG
  • Effect of change of volumetric displacement on LCB
  • MCTC

  • Why  is used instead of 
  • Effect of change of density on MCTC

  • Change of Trim = Change in draft Ford + Change in draft Aft
  • Use of Trim Tables
State
  • F is the Centroid of the water plane area
  • COF is the point about which the vessel changes her trim
  • Change in Draft Aft 
  • Change in Draft Forward
Calculate
  • Final Forward & Aft Drafts
  • Quantity of cargo loaded / discharged to produce a required trim
  • Quantity of cargo loaded / discharged to keep forward/ aft draft constant
  • Quantity of cargo loaded / discharged to reach a desired  forward/ aft draft
  • Final draft Forward and aft using M V Hindship


LCG, LCB


The center of gravity is the point through which the weight (Displacement) of the vessel acts vertically downward. The distance of this point from the aft perpendicular (or in some cases the center of flotation) is called the Longitudinal center of Gravity of LCG.

The center of buoyancy as we know is the center of the underwater volume. The longitudinal distance of this point from the Aft perpendicular is called LCB.

For a vessel that is upright the G and B lie on the centerline plane of the vessel. 

For a ship to be on even keel, the LCG = LCB


Effect of Loading/Discharging and shift of weight on LCG

The following media explains about the Trim By Shifting Of Weight



Effect of change in underwater volume on LCB

The following media explains about the Trim Effect due to Change in Underwater Volume.



Trimming Moment

Relationship between LCG and LCB
The vessel is in equilibrium in the longitudinal plane when the LCG and the LCB lie in a vertical line. If this equilibrium is disturbed by the shift of G to or a shift of B to  a couple develops which we call the trimming moment. If the LCG lies aft of LCB then the vessel will trim by the stern and vice versa.

(This is a different approach to trim that is used when we have access to ships stability data, where we can find the value of LCB for a series of Hydrostatic drafts, The LCG is calculated by considering the longitudinal trimming moments)

G can shift to  if there is a shift of weight or load/ discharge of weight in the longitudinal direction. 

W  d = Trimming moment

The shift of B with the G steady may be caused in a seaway due to wave/ swell action.

The difference between LCG and LCB multiplied by Final displacement gives us the trimming moment, which when divided by the MCTC gives the change in trim.




Moments required to change trim by 1 cm (MCTC)

Please refer to the media under the topic Trim= Trimming moment/MCTC.

MCTC=(∆×/100×LBP)

Please refer to the media under the topic Trim= Trimming moment/MCTC.

Why  is used instead of  for MCTC

 and  


The longitudinal metacenter or ML is the point where the vertical through the COB when the ship is in even keel condition, and the vertical through the COB when the ship is in trimmed condition, meet. The distance between  and the Center of Gravity is called the longitudinal Metacenter height  or . 

The vertical distance between the Center of Buoyancy and the longitudinal metacenter is called the  and is found from the formula:

 = 2nd Moment of Inertia about the axis through the center of flotation 

This is used for all shapes of vessels be they box shaped or ship shapes.

For box shaped vessels it is 

Since the distance BG is comparatively less than  or , and  is smaller than  we can substitute the relatively easily found  for  in the formulae of MCTC.


Effect of change in density on MCTC



MCTC of a vessel is given in the stability particulars and is given for the condition wherein the vessel is floating in SW.

When the density of the water changes, the underwater volume changes and with it the . Therefore the data needs to be modified for the water density in which the vessel is operating.



Trim= Trimming moment/MCTC

The following media explains about the trim



Change in trim = Change in draft forward + Change in draft aft

Please refer to the media under the topics "Trim = Trimming moment/MCTC" and ‘Change in Draft fwd   = [ * (LBP-LCF)]/LBP.

Use of trim table

Trim Tables
These are ready reckoner tables which provides the ships officer a rapid if not very accurate way of finding out what the effect of adding shifting or discharging a weight on board will have on the vessels trim.

They are to be found in the stability particulars or as is very often the case in the form of a separate booklet. The weight is usually 100 mt and the tables are for various drafts. In between conditions are to be interpolated. The trimming effect of filling or emptying various tanks are also provided. With loadicators in common use these tables have lost their utility to a large extent.


Centre of Floatation is centroid of water plane area

Please refer to the media under the topic "Trim = Trimming moment/MCTC".

LCF is the tipping centre or the pivoting point about which the vessels change her trim

Please refer to the media under the topic Trim= Trimming moment/MCTC.



Change in draft aft Ta = (Tc * LCF)/LBP

Please refer to the media under the topic ‘Change in Draft fwd  = [ * (LBP-LCF)]/LBP


Change in Draft fwd Tf = [Tc * (LBP-LCF)]/LBP

The following media explains about the Change of Trim


Calculation of quantity of cargo to be loaded/discharged/shifted to produce a requried trim

Quantity of cargo loaded / discharged to produce a required trim
A ship has to load 600 t of cargo. The drafts are   .7 m. Space is available 60 m F and 40 m aft of CF which is amidships. MCTC = 250 tm, TPC = 20 mt. Distribute the cargo to finish loading at a final trim of 0.5 m by the stern. Also calculate the Final drafts F & A.

Initial trim = 9.7 - 8.5 = 1.2 m by stern
Desired trim = 0.5 m by stern
Change in trim required = 1.2 - 0.5 = 0.7 m by the Head
Required TM = Trim  MCTC = 70  250 = 17500 tm by Head

 Let cargo loaded 60 m Fwd of Cf = X mt. TM caused = 60 X mt
Then cargo loaded 40 m aft of CF = (600  X) mt. TM caused = (600 - X)40 mt

Total TM caused = Desired TM
60X - 40(600 - X) = 17500
100X = 17500 + 24000
X = 415 mt

Cargo loaded fwd = 415 mt
Cargo loaded aft = 600 - 415) = 185 mt


Since CF is amidships
 


 

 

F

A

 

Draft

8.5

9.7

 

Sinkage

+ 0.30

+ 0.30

 

Change in Trim

+0.35

-0.35

Ans

Final Draft

9.15 m

9.65 m





Calculation of Final FWD/AFT Drafts

Loading/ Discharging/ Shifting of weights
A vessel length 170 m, LCF 82 m from APP, TPC 25 mt, MCTC 175 mt, Draft  m   m. The following operations are carried out.
  • Loads 200 mt 30 m fwd of m/ships
  • Disch 250 mt 20 m fwd of amidships
  • Shifts 100 mt from 20 m aft of amidships to 20 m fwd of midships
Find the final draft.

CF from amidships = 3 m aft of amidships

Weight

Distance from COF

TM

 

 

F

A

200 (L)

33

6600

 

250 (D)

23

 

5750

100 (S)

40

4000

 

 

                    Total

10600

5750

Trimming moment created = 10600 ~ 5750 = 4850 tm Fwd

Total weight Loaded / Disch = 50 discharged. Total Rise 

Total trim 





 

 

F

A

 

Draft

5.8

7.0

 

Rise

- 0.02

- 0.02

 

Change in Trim

+0.143

-0.134

               ANS:

Final Draft

5.923 m

6.846 m


Calculation of quantity of cargo to be loaded/discharged to keep forward/aft draught constant

Quantity of cargo loaded / discharged to keep forward/ aft draft constant
A ship 100 m long is floating at draft  m,  m. MCTC = 180 tm TPC = 20 mt, LCF 46 m from APP. Find a location where a weight of 60 mt can be loaded so as to keep the aft draft constant.

Sinkage 
Since this is bodily sinkage the draft F & A will increase by 0.03 m.

Let the weight be loaded at a distance of X mtrs fwd of CF
TM = 60X tm


However Aft draft is required to remain the same. Therefore the aft trim has to change (reduce) by 0.03 m

  = 0.03 m


  X = 0.196 m

ANS: The cargo of 60 mt is to be loaded at a point 0.196 m Fwd of CF


Calculation of quantity of cargo to be loaded/discharged to reach desired forward/aft draft

Quantity of cargo loaded / discharged to reach a desired  forward/ aft draft
A ship 160 m long arrives at the mouth of a river with drafts    m. MCTC 210 tm, TPC 20 mt. CF is 2m aft of amidships. The ship has to cross a bar where the maximum permitted draft is 6.2 m. To achieve No1 DBT, CG 60 m Fwd of COF, will have to be ballasted with SW to reduce the aft draft to 6.2 m. Find the quantity of water to be ballasted and the final drafts Fwd and Aft.

Let the quantity of ballast be X mt

Bodily sinkage 

Aft draft will increase by .

With an existing draft of 6.3 m, the aft draft will have to be reduced by  m to attain 6.2 m
Desired 

Trimming moment should create the above 

TM caused by Ballast = (60  X) tm

  by Head 





X = 112 mt

ANS: Amount of ballast to be taken into No1 DBT is 112 mt

Sinkage 






 

 

F

A

 

Draft

5.70

6.30

 

Rise

+ 0.056

+ 0.056

 

Change in Trim

+0.164

-0.156

               ANS:

Final Draft

5.92 m

6.20 m


List

Learning Objectives
After completing this topic you will be able to:

Explain
  • Cross curves of Stability & KN curves
  • How to determine GZ from cross/ KN curves
  • Effect on GZ values due to shift of weights (Vertical and Horizontal)
  • Range of Stability
Describe
  • Effect of length, Breadth and Freeboard on the curve of Statical Stability
Calculate
  • Angle of List resulting from Transverse and  Vertical movement of weight using GZ curve
  • Area under GZ curve using GZ  curves


Cross curves of stability and KN curves

The value of GZ is a major factor which effects the vessels Statical stability. It has a bearing on all aspects of the vessels safety in a seaway and in port while carrying out cargo operations. While GZ can be calculated using various formulae eg Wall sided, Attwood etc, there is a simpler form of arriving at a vessels GZ for a particular angle of heel and displacement. 

These are Cross curves of Stability/ KN curves which are provided to the vessel as a part of her stability data.


How to determine GZ from cross/KN curves

Cross curves of Stability
GZ = GM(F)  Sin (For angles of heel < )

Wall-sided formula --- GZ = {GM(F) + (0.5  BM  )}  Sin (For angles of heel < )

The above two formulae assumes that the values of GM and BM remain constant as the ship heels. However, due to change in the breadth of the water-plane during heeling, the value of KM increases, and when the deck edge submerges, it starts decreasing. Hence the two formulae will not give accurate results for GZ for higher angles of heel.

Attwood’s formula is valid for all angles of heel but it cannot be used on board ship for calculating GZ.

KG is a very important factor in getting GM, Therefore shipyards supply what are known as Cross curves of Stability for an assumed value of KG.


With reference to the above diagram, if the actual operational KG is more than the assumed value of KG used for compiling the data then a correction GX has to be applied to the GZ obtained from the Cross curves of Stability.

 sin 

Where  is the difference between the tabulated value of KG and the actual value and  is angle of heel.

GX is the correction, and it is +ive if the actual value is less than the assumed value and vice versa.

In the diagram.

Actual GZ for angle of heel , is GZ - GG1 sin


KN Curves 
The disadvantage of using the cross curves is that the correction that has to be applied is either -ive or +ive. There is scope for error in the application of the correction.

The naval architect in shipyard calculates the value of KN (as shown in the sketch) by applying the Attwood’s formula, assuming that KG = 0. They calculate KN values for different displacements and angles of heel and tabulate the results in the form of table and graph, which are included in the ship’s stability particulars. Knowing the value of KG, the ship’s officer calculates the correction to be applied to KN values to obtain GZ as follows ---

KG + FSC = KG(F) ; Correction = KG(F)  Sin  ; GZ = KN - KG  Sin 
 
CROSS CURVES OF STABILITY PARTICULARS
NOTE:-
  • The format of this table is universally adopted on all ships.
  • It is prepared for various displacements such that there should not be any need for interpolation.
  • As per standard practice, the angles of heel are --- , , , , , , , ,  and . The KN values and hence the GZ values are obtained for these angles only, without any interpolation for other angles in-between.
  • One more column is provided for the angle of heel at which the deck edge will submerge.
  • Using the GZ values for the above angles, a graph is plotted. Various information are calculated by this graph, which helps to decide if the ship’s stability complies with the prescribed criteria.
POOP and FCSLE INCLUDED by M.V.HindShip Stability Particulars


  • D.G. Shipping specially prepared this book for examination purposes, by abridging the actual stability book of a ship built in Vishakhapatnam. To reduce the size of the KN tables, the data was selected for every 500 tons of displacement. Hence the student has to interpolate for the actual displacement to obtain the correct value of KN.
  • Furthermore the table gives KN values for the ship without superstructure (column A) and with superstructure (column B), for angles of heel from  upwards. This practice has been discontinued on present ships. Hence the student should only use column B for calculating the GZ values.
  • The student should carry this book to the examination hall for calculating all examples based on this ship.

Effect on GZ values due to shift of weights

Effect on GZ values due to shift of weights (Vertical and Horizontal)
Vertical Shift


Any vertical shift of weight or discharge of weight from below G or the addition of weight above G will cause G to rise. ie the KG will increase.

In the above diagram G will rise to ,  and the GZ will be 

Therefore it can be positively stated that any change in weights on board ship which causes G to rise will result in a reduction in GZ's and a depressed GZ curve i.e. the maximum GZ value and the range of stability will reduce.


In case of an extreme shift of weight causing a negative GM the effect on the GZ curve will be as follows


Horizontal Shift
If a weight is loaded or discharged off the centreline of the ship, or if a weight is shifted transversely, then G will shift transversely to a new position on one side of the ship, which will cause the ship to list on that side. Now, assuming that the ship is listed to starboard side, and if the ship experiences a beam-wave from port side, it will cause the ship to heel further to starboard side.



As seen from the sketch, the final angle of inclination  is a combination of list and heel. The righting lever GZ will not be true because G has shifted to . The correct righting lever is , which can be calculated by formula --- 


w --- weight loaded, discharged or shifted.
d --- Transverse distance of w loaded or discharged from centerline, or total distance of w shifted transversely.
W --- Final displacement of the ship.

Values of  are calculated for  =  and , which will be  and  respectively. After drawing the GZ graph in the normal way, values of GG1 and  are plotted at  and  respectively, and a straight slopping line is drawn to cut the GZ graph.



The intersection of the two graphs is the angle of list of the ship. The graph of  is called the Heeling arm graph.

The area under the heeling arm graph is considered to be loss of stability. The area above the heeling arm graph is the residual stability of the ship.


Pure loss of stability

The range in which the GZ is a righting lever. Beyond that the GZ turns into an upsetting lever and will capsize a vessel. This is independent of the condition of the vessels intact stability.

The following media explains about the Curves of Stability




Effect of increased length/breath, freeboard on the curve of stability curves

Effect of length, Breadth and Freeboard on the curve of Statical Stability
There are a number of variables that affect ships stability, and we can also be tempted to say that changing a certain variable will improve vessels stability. However the most important factor which affects the vessels stability is the position of G. The lower the KG, the better the stability, owing to the affect of GM on KG.

Having said that we can individually consider various parameters of a vessel like its length, breadth and freeboard and what affect they have on G. In doing so we have to bear in mind that for all ship shapes

, and I is the 2nd Moment of Area about the centerline.

Freeboard
An increase in freeboard will result in reduced draft a resultant reduction in volumetric displacement (V). This results in an increased BMT, resulting in an increase in KM, with its beneficial effect on GZ. A vessel with higher freeboard will also have a higher range of stability since she will remain wall sided for a larger range of heel than a comparable vessel with smaller freeboard.


Breadth
An increase in breadth will result in a increase in the 2nd Moment of Area (I), which as in freeboard will result in an increased KM and higher values of GZ. There is however a downside to increasing the breadth, it results in a decrease of the range of stability


Length
There is not much of an effect of increase in length, since it does not affect the GM. It however does increase the volumetric displacement which in turn will lead to a small but not significant reduction in GZ.

Calculation of Angle of list resulting from transverse and vertical movement of weight using GZ curves

Angle of List resulting from Transverse and Vertical movement of weight using GZ curve

Example
A vessel in SW at displacement 13700 t KG is 7.0 m, FSM 1400 tm. 300 t cargo is loaded at a KG of 12 m , 6m to port of the center line. Find the list caused.

Given:

0

5

10

20

KN (m)

0.0

0.793

1.581 

3.130

Final KM = 8.073 m

In this problem the cargo is loaded at a position above the vessels G, and to port of the centerline. This results in a vertical and atwartship shift in the position of the vessels G.

Using Formula for list





KG                    =  7.00 m Fluid KG = 7 + 0.107 + 1.0 = 7.207 m
FSC                  = 0.100 m
Vertical      = 0.107 m
KM                   = 8.073 m
Initial GM Fluid   = 0.866 m
Tan q = GG1÷GM(F) = 0.129 ÷ 0.866 ; q = 8.473°

 = 8.44º Comparison purposes

Using GZ curve 

A listing arm curve is superimposed on the ships GZ curve, the point of intersection providing the list of the vessel.

   (⁰)

0

   5

10

20

KN (m)

0.0

0.793

1.581

3.130

KG Sin  (-)   

0

0.628

1.251

2.465 

Listed GZ

0

0.165

0.330

0.665

Listing lever at 

Listing lever at 
From the graph, list = 
Student should note that the list obtained from the GZ graph will always be less than that obtained by formula.





Calculation of Area under GZ curve using Simpson's rule

Area under GZ curve using GZ curves

1. A ship of 4000 mt displacement, whose GZ values are as follows:

Angle of heel in⁰

10

20

30

40

GZ (mtrs)

0.15

0.27

0.38 

0.40

Calculate the dynamical stability up to 40⁰ angle of heel.


In the above example the  cannot be used directly, hence we convert degrees  to radians

Hence  radians

GZ

SM

Product of Area

0

1

0

0.15

4

0.60

0.27

2

0.54

0.38

4

1.52

0.40

1

0.40

 

 

 Area = 3.06



ANS: Dynamical stability = Area under curve  W = 0.178  4000 = 712 tm


    

Dynamical Stability

Introduction:
Dynamical Stability
We have so far examined Transverse Statical stability, which was defined as the ability of the vessel to return to the upright position after having been inclined from it. An important aspect was that the forces were static i.e. the environment (sea and wind),  were not considered, only buoyancy and weight were taken into account.

In Dynamic stability however the forces that the waves and wind have on the vessels stability are considered. It is defined as the work done to incline a ship. It is the product of the area under the GZ curve in meter radians multiplied by the vessels displacement in tonnes. The Dynamical Stability is expressed in  tonnes meters.

The Dynamical Stability requirements are given in the IMO publication "International Code on Intact Stability 2008" Part A consists of the Mandatory requirements and are being reproduced here from the Code. For further information the student will do well to study the Code itself. 

Oil tanker requirements may be found in MARPOL Annex 1 and are not dealt with in this module, and requirements for ships carrying grain will be dealt with in a later section.

Learning Objectives
After completing this topic you will be able to:

State 
  • Dynamical Stability requirements as per SOLAS
  • Dynamical Stability at stated angle of heel represents the potential energy of the ship
  • Potential energy is used in overcoming resistance to rolling and producing rotational energy
  • Dynamical Stability =   Area under GZ curve


Statical stability requirments as per SOLAS



Dynamical Stability at stated angle of heel represents potential energy of the ship



The area under the curve represents the potential energy of the vessel, and to incline her work has to be done to vertically separate the G and B and incline her to a particular angle of heel. That is the Dynamical Stability for that angle of heel.

Potential energy is used in overcoming resistance to rolling and in producing rotational energy

This potential energy is used partly in overcoming resistance to rolling and partly in producing rotational energy as the ship returns to the upright. It is this rotational energy when the ship is upright causes it to continue rolling to the other side, and if there are no other disturbing forces, the ship will roll to an angle where the sum of the energy used in overcoming resistance to rolling and the dynamical stability are equal to the rotational energy when upright.


The vertical separation of  from G may be stated to be , and if the position of M (Metacenter) is assumed to be fixed then  will oscillate about M in an arc. This is the rotational energy into which the vessels potential energy is converted into once she is displaced from the vertical.


You must take the quiz only after you study the eBook contents in this chapter, including watching the videos. The self-assessment quizzes are drawn from the complete chapter. Please take this as often as you wish. Please make sure that you return here after you have finished reviewing this topic.

Dynamical stability = W × Area under GZ Curve

Example 1:
A vessel floating in SW at a displacement of 15400 mt, KM 8.034 m, KG 6.1 m, FSM 3050 tm. Given the following data

 

KN

0.796

1.575

2.355

3.112

3.887

4.641

6.023

6.546

7.615

7.923


Draw the GZ curve and Calculate the dynamical stability at 40⁰ angle of Heel. Also state if the vessel complies with the general criteria of the Intact stability code 2008.

FSC = FSM  W = 3050  15400 = 0.198m. ; KG (F) = 6.1 + 0.198 = 6.298m.

KN

KG(F)  Sin q

GZ

Mult. 1

Prod. 1

Mult. 2

Prod. 2

0

0

0

0

1

0

1

0

0.796

0.549

0.247

4

0.988

 

 

1.575

1.094

0.481

2

0.962

4

1.924

2.355

1.630

0.725

4

2.900

 

 

3.112

2.154

0.958

2

1.916

2

1.916

3.887

2.662

1.225

4

4.900

 

 

4.641

3.149

1.492

1

1.492

4

5.968

6.023

4.048

1.975

 

 

1

1.975

6.546

4.453

2.093

 

 

 

 

7.615

5.454

2.161

 

 

 

 

7.923

6.083

1.840

 

 

 

 


 

 

 

 

13.158

 

11.783


Area ( to ) = 5  3  13.158  57.3 = 0.383 mR> 0.055 mR
Area ( to ) = 10  3  11.783  57.3 = 0.685 mR> 0.09 mR
Area ( to ) = 0.685 – 0.383 = 0.302 mR> 0.03 mR
GM(F) = KM – KG(F) = 8.034 – 6.298 = 1.736m. > 0.15m.
Max. GZ = 2.161m. > 0.2m.
Max. GZ occurs at  =  > 

Hence all the stability criteria are satisfied.
Dynamical stability at  = 15400  0.685 = 10549 tm.


Intact stability requirements for carriage of grain


Intact stability requirements for the carrige of grains

Intact Stability Requirements for the carriage of Grain
The International Code for the Safe Carriage of Grain in Bulk (International Grain Code), adopted by resolution MSC.23 (59), has been mandatory under SOLAS chapter VI since 1 January 1994.

Selected sections of the International Grain Code are excerpted here; however the student should study the Code in full for a complete understanding of the regulations governing the carriage of grain.

Grain Stability Calculations
Stability on a Grain carrying vessel.
Students interested in studying how the stability calculations are done practically on board a grain carrying vessel may go through the Annex to this module.



  1. The angle of heel due to the shift of grain shall not be greater than  or in the case of ships constructed on or after 1 January 1994 the angle at which the deck edge is immersed. whichever is the lesser; 
  2. In the statical stability diagram, the net or residual area between the heeling arm curve and the righting arm curve up to the angle of heel of maximum difference between the ordinates of the two curves, or  or the angle of flooding , whichever is the least, shall in all conditions of loading be not less than 0.075 metre-radians; and 
  3. The initial metacentric height, after correction for the free surface effects of liquids in tanks. shall be not less than 0.30 m. 




Volumetric  heeling moments (VHM) caused due to shift of grain in partially filled/full compartments

Cargo holds are classified according to how the grain is loaded in them. They can be either Filled or Partly filled. Then again a filled compartment may be trimmed or untrimmed. These categories are required for the consideration of the effect on G during the carriage of grain.

During the carriage of grain, grain settles and creates “voids”.  Then during the passage of the vessel the grain shifts into these voids. This shift of grain creates a heeling moment which is superimposed on the GZ curve.

Obviously the partly filled/ untrimmed holds will cause a larger shift of grain than filled/ trimmed compartments.

Certain assumptions are made with regard to the shift of grain.

In a filled compartment, the transverse shift of grain is supposed to take place at an angle to the horizontal of 25⁰ and in filled compartments it is assumed to be 15⁰ to the horizontal.


These assumptions result in the availability of Volumetric Heeling Moment (VHM) curves from which Volumetric heeling moment is obtained against the ullage / depth of grain. These tables are available for all compartments and also provide the KG for that particular cargo level.

After the Volumetric Heeling Moments VHM) are obtained, they are adjusted for vertical shift by multiplying the VHM by 1.06 in the case of FILLED & TRIMMED condition. For FILLED/UNTRIMMED there is no multiplication factor.

In PARTLY FILLED compartments the VHM is multiplied by 1.12.

The VHM is now converted into Heeling Moments (HM)by dividing by the Stowage Factor (SF)

The Heeling Arm is obtained by dividing the HM by Displacement (W). This is in upright position and equivalent to the vertical shift  that we have studied in previous sections. We will name it 
Since the Grain Code mandates the calculation of residual stability up to a 40⁰ angle of heel, the Heeling arm in upright position  is multiplied by where  i.e. 0.76604 which the Code rounds off to 0.8 for convenience.



A summary of the sequence of calculations:

VHM as obtained from the tables 

VHM adjusted for vertical shift by multiplying by 1.06/1.12/1.0

Total VHM for all holds obtained.

VHM converted to 

HM converted to 


The ships GZ curve is drawn as in standard cargo vessels by obtaining GZ values from the KN curves and correcting them

 values are superimposed on the GZ curve (See HM Diagram) and various data obtained from the curve and compared to the intact stability requirements.



Use of Maximum permissible VHM curves

Allowable Heeling Moments
Tables and curves are available which show the allowable Heeling Moments that take into account the 3 Mandatory Intact Stability Criteria. It is less time consuming and affords a quick and accurate method of fulfilling grain stability requirements as required by the Code.

In the following example we will use the Allowable heeling moments as given in the table.

1. Vessel Displacement 29250 mt, KG 8.35 m, KM 11.0 m, FSM 1950tm. Find the allowable heeling moments.
KM = 11.0 m
KG   =  8.35 m
FSM = 1950 tm
KG(Fluid) = 8.417 m

From the table

ANS: Allowable Heeling moments = 21183.08 tm

* ** ALLOWABLE HEELING MOMENT ** *

KGo (m)

Dispt. (t)

8.00

8.10

8.20

8.30

8.40

8.50

8.60

8.70

8.80

8.90

7000

26536

26381

26226

26071

25917

25762

25607

25452

25297

25142

7500

26795

26629

26463

26298

26132

25966

25800

25634

25468

25302

8000

26919

26742

26565

26389

26212

26035

25858

25681

25504

25327

8500

26936

26748

26560

26372

26184

25996

25808

25620

25432

25244

9000

26871

26672

26473

26274

26075

25876

25677

25478

25278

25079

9500

26743

26533

26323

26113

25903

25693

25483

25273

25063

24852

10000

26571

26350

26129

25908

25687

25466

25244

25023

24802

24581

10500

26372

26139

25907

25675

25443

25211

24978

24746

24514

24282

11000

26162

25919

25676

25433

25189

24946

24703

24459

24216

23973

11500

25950

25696

25441

25181

24933

24678

24424

24170

23915

23661

12000

25715

25450

25184

24919

24654

24388

24123

23857

23592

23327

12500

25474

25197

24921

24645

24368

24092

23815

23539

23262

22986

13000

25241

24954

24666

24379

24091

23804

23516

23229

22941

22654

13500

25003

24704

24406

24107

23809

23510

23212

22913

22614

22316

14000

24761

24452

24142

23833

23523

23213

22904

22594

22284

21975

14500

24530

24210

23889

23568

23247

22927

22606

22285

21965

21644

15000

24305

23973

23641

23310

22978

22646

22314

21983

21651

21319

15500

24085

23742

23399

23056

22713

22371

22028

21685

21342

20999

16000

23875

23521

23167

22813

22459

22105

21751

21398

21044

20690

16500

23677

23312

22947

22582

22217

21852

21488

21123

20758

20393

17000

23493

23117

22742

22366

21990

21614

21238

20862

20486

20110

17500

23321

22934

22547

22160

21773

21386

20999

20612

20225

19838

18000

23162

22764

22366

21968

21569

21171

20773

20375

19977

19579

18500

23022

22613

22203

21794

21385

20976

20567

20158

19748

19339

19000

22895

22475

22054

21634

21214

20794

20374

19953

19533

19113

19500

22779

22348

21917

21485

21054

20623

20191

19760

19329

18898

20000

22684

22242

21799

21357

20915

20472

20030

19588

19145

18703

20500

22606

22152

21699

21246

20792

20339

19885

19432

18979

18525

21000

22537

22072

21608

21143

20679

20215

19750

19286

18821

18357

21500

22488

22012

21537

21061

20586

20110

19635

19159

18684

18208

22000

22459

21972

21486

20999

20513

20026

19539

19053

18566

18080

22500

22438

21941

21443

20945

20448

19950

19452

18955

18457

17960

23000

22433

21925

21416

20907

20399

19890

19381

18873

18364

17855

23500

22449

21929

21409

20890

20370

19850

19331

18811

18291

17771

24000

22476

21945

21415

20884

20353

19822

19291

18761

18230

17699

24500

22517

21975

21433

20891

20350

19808

19266

18724

18182

17640

25000

22578

22025

21472

20919

20366

19813

19260

18707

18154

17602

25500

22651

22087

21523

20959

20395

19831

19267

18703

18139

17575

26000

22736

22161

21586

21011

20435

19860

19285

18710

18135

17560

26500

22838

22252

21666

21079

20493

19907

19321

18735

18149

17563

27000

22955

22358

21760

21163

20566

19969

19372

18775

18178

17580

27500

23084

22476

21868

21260

20651

20043

19435

18827

18218

17610

28000

23230

22611

21992

21373

20753

20134

19515

18896

18276

17657

28500

23393

22762

22132

21502

20872

20241

19611

18981

18350

17720

29000

23566

22925

22283

21642

21001

20359

19718

19076

18435

17794

29500

23755

23102

22450

21797

21145

20492

19840

19188

18535

17883

30000

23959

23296

22632

21969

21305

20642

19978

19315

18651

17988

30500

24174

23500

22825

22151

21476

20802

20127

19452

18778

18103

31000

24402

23716

23031

22345

21660

20974

20288

19603

18917

18231

31500

24645

23948

23252

22555

21858

21162

20465

19768

19072

18375



You must take the quiz only after you study the eBook contents in this chapter, including watching the videos. The self-assessment quizzes are drawn from the complete chapter. Please take this as often as you wish. Please make sure that you return here after you have finished reviewing this topic.

Calculation of Heeling  arm =VHM/(SF) and  = 0.8 x

A vessel loaded to her marks with grain SF 1.5. The displacement is 19943 mt, KM 8.704 M, KG 7.679 M, FSM 1284 TM. Total VHM is 5414 . Verify whether the vessel complies with the International Grain Code.



KG Solid = 7.679 m
KG Fluid = 7.743 m
KM = 8.704 m
GM Fluid = 0.961 m………………………………………………………………………………………….Complies since  0.3 m




Given  KN values, GZ is calculated
GZ = KN - KG Sin

KN(m)

KGSin  (-ive)

GZ (m)

0

0

0

0

5

0.811

7.743 Sin5

0.136

12

1.925

7.743Sin12

0.315

20

3.12

7.743Sin20

0.742

30

4.475

7.743Sin30

0.604

40

5.608

7.743Sin40

0.631

60

7.268

7.743Sin60

0.562

75

7.727

7.743Sin75

0.248


From the curve Heel =  ………………………………………………………………………………….Complies since 
It is suggested that the list is worked out using

 , then the curve is adjusted to make this list.

Next the area between the heeling arm curve and the GZ curve is calculated.

The curvilinear section is found using Simpsons Rule and the triangular section by plain geometry

In Triangular section CFB

DC = FE = 0.18 m (From curve)
BF = EF – 0.145  = 0.18 – 0.145 = 0.035 m





(For simplicity we work in m degree and finally convert to m radians

For the curvilinear part,



Station()

Ordinate (GZ in m)

SM

POA

6.5

0

1

0

12.08

0.135

4

0.54

17.67

0.255

2

0.51

23.25

0.345

4

1.38

28.83

0.41

2

0.82

34.42

0.445

4

1.78

40

0.451

1

0.451

Sum POA=5.481


Total area under curve = 10.20014 + 0.58625 = 10.78639 m degrees =  = 0.18824

Complies with 3rd criteria  0.075 mr




Drawing - Heeling arm curve on Righting arm curve for given ship's condition and determining angle of heel

Please refer the topic ‘Calculation of Heeling arm  = VHM/(SF  ) and    '

Comparison of the results with criteria set in Reg 4 of Grain Code

Please refer the topic ‘Calculation of Heeling arm  = VHM/(SF  ) and    '
at

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