SECTION 21 -DRY-DOCKING

INTRODUCTION
It is a requirement that all ships be dry-docked for inspection and maintenance below the waterline.
When a ship is being dry-docked additional forces acting at the keel take effect, being the reaction
or upthrust afforded by the blocks onto which the ship is being landed. These forces can create
undue loads on the stern structure and cause loss of stability of the ship. This section investigates
these effects.

Learning Objectives
On completion of this section the learner will achieve the following:
1. Understand the sequence of events that takes place whilst a ship is being dry-docked.2.
Calculate the upthrust at the blocks (P force) at any stage during dry-docking of the ship.3.
Understand the loss of stability during dry-docking and calculate the loss of stability as
either a rise of the ship's centre of gravity (increase in KG) or as a fall of the metacentre
(reduction in KM).4.
Conduct dry-docking calculations.5.
Understand the practical considerations during the dry-docking of a ship.

CLASS 2/1 STABILITY -SECTION 21 -Dry-docking

253
 21.1 SEQUENCE OF EVENTS DURING DRY-DOCKING
Figures 21.1 to 21.3 illustrate what happens as the ship enters the dry dock and the water is
pumped out of the dock.

The ship enters the dry dock with a small trim by the stern and is floated into position.

The gates are closed and water is pumped out of the dock until the ship touches the blocks
aft. Immediately the ship touches the blocks aft this denotes the start of the critical period (it
is now that the ship will start to experience a loss of stability, hence the term).

1.2. Ship enters dock with a small trim by the stern.

Water is pumped out of the dock until the ship touches
the blocks aft.

Fig. 21.1

As more water is pumped out of the dock the true mean draught will start to reduce as the
ship experiences more and more support at the stern. The upthrust afforded by the blocks
at the stern is termed the 'P force', this continues to increase as the buoyancy force
reduces. Throughout the docking process the ship will displace a progressively lessening
volume of water as the true mean draught reduces and the P force increases to provide
more support for the ship (in effect, the P force takes over supporting the ship and the role
of the buoyancy force in supporting the ship reduces). At this stage the aft draught will be
reducing at a greater rate than what the forward draught is increasing, the ship will be
trimming by the head as the overall true mean draught reduces. For reasons discussed
later, the loss of stability will also be increasing as the P force increases.

Eventually the ship will come to rest on the blocks along it's entire length, this critical instant
denotes the end of the critical period, since for a flat bottomed ship the problem of stability
loss is no longer of concern.

CLASS 2/1 STABILITY -SECTION 21 -Dry-docking 254

2.
3.
~
4.
 3.

4.

Fig. 21.2

6.

5.

r 6.

Fig.21.3

CLASS 2/1 STABILITY -SECTION 21 -Dry-docking

255

5.
21.2 CALCULATING THE P FORCE

21.2.1 Calculation of P force at any stage during dry-docking

Throughout the dry-docking procedure the true mean draught reduces as it would if the ship were
rising out of the water due to weights being discharged.

Consider the formula: I Rise (cms) = ~ I

I TPC I

The P force may be considered to have the same effect on true mean draught as if a weight had
actually been discharged, therefore: Reduction in TMD (cms) = P force (1)
TPC
or: P force (t) = Reduction in TMD (cms) x TPC

This formula may be used to calculate the upthrust at the blocks at any stage in the docking
process since the true mean draught is always reducing as water is taken out of the dock.

21.2.2 Calculation of P force during the critical period when dry-docking

In the period between the ship touching the blocks aft (start of the critical period) and touching the
blocks forward and aft (the critical instant) the ship undergoes a change of trim.

The change of trim at any stage during the critical period may be considered to be the same as the
change of trim that would have occurred had a weight 'w' been discharged from a position at the
aft perpendicular equivalent to the upthrust P in tonnes (if it is assumed that the ship is on the
blocks aft at the aft perpendicular).

Consider the formula: COT (cms) = ~

MCTC

If the force P is considered to have the same effect as a weight discharged at the aft perpendicular,
then: COT (cms) = ~
MCTC

or: P = COT (cms) x MCTC

Dist LCF foap

This can be used to calculate the P force during the critical period only.

Use of both of these formulae will be seen in subsequent dry docking calculation examples.

CLASS 2/1 STABILITY -SECTION 21 -Dry-docking

256
 21.3 LOSS OF STABILITY WHEN DRY-DOCKING

Loss of stability commences as soon as the ship touches the blocks aft and continues to worsen as
the value of the P force increases. The maximum loss of GM of concern occurs the instant
immediately prior to the ship settling on the blocks forward and aft -this time being termed the
critical instant. Once the ship is flat on the blocks it will be in a safe condition as the risk of heeling
over as a result of becoming unstable will have passed (most ship's having a substantial area of
flat bottom). For ships that have a relatively small percentage of flat bottom area additional
measures must also be taken such as using side shores to support the ship in the upright condition
when in the dry dock.

Either of two methods of calculation of the loss of GM may be used.

21.3.1 Loss of GM as a result of a rise in G (increase in KG)

Consider the upward movement of G that would occur if a weight
'w' is discharged from a position at the keel (Kg = 0 m). When
discharging a weight the centre of gravity of the ship, G, will move
directly away from the centre of gravity of the discharged weight to
Gv as shown in figure 21.4.

GGvwill be equalto the loss of GM where: I GGv = ~ I

I W-w I

Fig. 21.4
'd' is the distance between the centre of gravity of the ship (G) and the centre of gravity of the
discharged weight which was at the keel 'K', Therefore distance 'd' is the initial KG of the ship.

(It should be noted also that KM changes as a result of a reduction in the ship's draught.)

If the P force is considered to have the same effect as discharging an equivalent weight from the
keel then:
I LossofGM=~ I
IW-~I
The effect on the ship's stability is made r '--~ , t ~_. -
1'f_+ ilW~Pj
clearer if the available righting moment at a !R"idM BI)

L particular angle of heel is considered. Figure

21.5 shows a ship during the critical period
where it has taken the blocks at the aft end
only. During docking the ship becomes heeled. t~mQ" WI.,
0 to a small angle of inclination by an external
force such as the wind.
A!!~I- WI.,

0 The forces acting are as follows:

Wf is the total weight force acting downwards

rI through the centre of gravity at G; twf

Fig. 21.5
(W -P) is the remaining, or residual, buoyancy force acting upwards through the geometric centre
of the underwater volume at 81;

P is the upthrust of the blocks exerted at the keel aft.

(W -P) x GZ represents a righting moment;

PxGZ1 represents a capsizing moment.

CLASS 2/1 STABILITY -SECTION 21 -Dry-docking 257

n
 Therefore the available righting moment is given by:
Available righting moment = (W -P) ( x GZ J -P( X GZ1 J

It is essential that the righting moment afforded by the upward acting (remaining) buoyancy force
remains greater than the capsizing moment afforded by the upthrust of the P force acting at the
keel at all times prior to the ship touching the blocks forward and aft. If the ship should become
unstable during the critical period it will lurch off the blocks to one side resulting in structural
damage to the ship, movement of the blocks and great embarrassment!

It is for this reason that the loss of GM is calculated for the critical instant (when the ship touches
the blocks forward and aft) to ensure that adequate stability is maintained prior to the ship taking
the blocks overall.

21.3.2 Loss of GM as a result of a fall in M (decrease in KM)

Consider figure 21.6 that illustrates the ship heeled

by an external force such as the wind during the
critical period where the ship has taken the blocks at
R"~~

.
"~n!i"vtQ-
,,
04b<Itj\

P~4! :, t lW~"t
tA~a.fI
the aft end only. j ,

The total weight force of the ship acts downwards

through G. Counteracting this are the two upward
forces; the P force acting upwards at the keel and I~ _11 WI.
the residual buoyancy force (W -P) acting upwards MIIaBI _Wl

through the centre of buoyancy (B,). The resultant

of the two upward acting forces acts through the
new metacentre (~) such that:

p x x = (W -P) xy !
(1) 'WI
Fig. 21.6
MM1 represents the resulting fall of the transverse metacentre (or loss of GM).
Consider the two similar triangles:
Sine e = y therefore: y = Sine e x MM1 (2)
MM1
Also: Sine e = ~ therefore: x = Sine e x KM1 (3)
KM1

Combining formulae 1, 2 and 3 above gives: (W -P) x Sine e x MM1 = P x Sine e x KM1

Divide both sides by Sine 6: (W -P) X MM1 = P X KM1

Expanding this gives: (W x MM1) -(P x MM1) = P X KM1

:. (W x MM1) = (P x KM1) + (P x MM1)
:. (W x MM1) = P X (KM1+ MM1)
:. (W x MM1) = P x Initial KM
:. MM1 = P x Initial KM
W

~ In this formula the KM value is that which corresponds to the true mean draught for the
instant that the loss of GM is being calculated and not that for the initial true mean draught that the
ship has prior to docking. It is found by entering the hydrostatic data with a displacement value that
corresponds to that given by (W -P).
W in this formula is the ship's initial displacement.
CLASS 2/1 STABILITY -SECTION 21 -Dry-docking
. 258

i.e.
 21.4 TYPICAL DRY-DOCKING PROBLEMS

In the following example both methods of calculating the loss of GM will be used. An explanation is
included to prove that both methods are equally valid.

Example 1
Prior to entering dry dock M. V. Almarhas draughts F4.86 mA 5.24 m and an effective KG of 9. 16m.
Calculate:
(a) the GM when the ship takes the blocks forward and aft (at the critical instant);
(b) the draughts at the same instant;

Solution (a)
(1) Calculate initial TMD
AMD = 5.05 m
Enter data with AMD and obtain LCF position.

LCF = 86.32 + (-0.02 xQJ!§) = 86.31 m foap

0.1
Use this to calculate the TMD.

TMD = dA -((dA -dF) x LCF fcap

TMD = 5.24 -LLBP[

Q..3.§ x 86.31
167.87
TMD = 5.045 m

(2) Enter data with TMD 5.045 m

Displacement = 15120 + (320 xW§J = 15264 t
0.1
MCTC = 312 t-m LCF = 86.31 m foap

(3) Calculate the P force at the critical instant

The initial trim of the ship is 38 cms by the stern. When the ship touches the blocks forward
and aft the effective trim will be zero i.e. the ship will be on even keel. The ship will have
therefore experienced a change of trim of 38 cms by the head -as if a weight equal to the
P force had been discharged from the aft perpendicular.

p = COT (cms) x MCTC p = 38 x 312 = 137.4 tannes (138 tannes)

Dist LCF foap 86.31

(4) Enter data with the effective displacement (W -P) at the critical instant to obtain KM

Effective displacement = W -P = 15264 -138 = 15126 tannes

KM= 10.70+ (-0.10 x§) = 10.698m

320

(5) Calculate the loss of GM (both methods used)

Method 1
Loss ofGM = P xKG Loss ofGM = 138 x9.16 = 0.084 m
W-P (15264-138)

Method 2
Loss of GM = P x KM Loss of GM = 138 x 10.698 = 0.097 m
W 15264

CLASS 2/1 STABILITY -SECTION 21 -Dry-docking

259

[J
 (6) Calculate the GM at the critical instant
Method 1 Method 2
at critical instant I 10.698 at critical instant I 10.698
9.160 9.160
GM 1.538
ILoss initial GM I 1.538
of GM I 0.084 i LOSS of GM 0.097
at critical instant 1.454 i I~crit.ical instant 1.441 !

Both answers are different but are both valid since a true measure of a ship's stability is it's
righting moment value at any given angle of heel.
Within small angles of heel the righting moment is given by:

Righting moment (t-m) = Displacement x GMx Sine9

By method 1 I
At the critical instant the effective displacement = W -P = 15126 t since the P force acts as
a weight being discharged from the keel.
RM = Displacement x GM x Sine e
RM = 15126 x 1.454 x Sine e = 21993Sine e t-m

By method 2
By considering the loss of GM as a result of the fall of the metacentre:
RM = Displacement x GM x Sine e
RM = 15264 x 1.441 x Sine e = 21995Sine e t-m
(The slight difference arises due to rounding up of values in the calculation.)

Solution (b)
At the critical instant the ship will be on an even keel. The draught at the same instant may
be calculated by one of two methods.

Method 1
The initial TMD has already been calculated as being 5.045 m.
Entering the data with this obtain the TPC value. TPC = 31.96 + (0.04 x Q,Q4.§J= 31.98 t
0.1

Reduction in TMD = 11§. = 4.3 cms

31.98
.'. Draught at critical instant = 5.045 -0.043 = 5.002 m (Ans)

Method 2
If the effective displacement at the critical instant is (W -P)
Effective displacement = W- P = 15264 -138 = 15126 tonnes
Enter the data with this displacement value to obtain the TMD at the critical instant.
TMD = 5.00 + (0.1x §. ) = 5.002 m
320
Therefore the draught at the critical instant = 5.002 m (Ans)
(Clearly method 2 is much easier!)

During the docking operation it is essential that the 'critical instant' draught is determined as both
draughts forward and aft will be constantly being read. As the ship's draught approaches that as
calculated for the critical instant, also evidenced by the fact that the ship will be in a near even keel
condition at that time, the rate at which the water is pumped out of the dock will be slowed down to
allow final adjustment of the ship's fore and aft alignment prior to the ship taking the blocks overall.
Once on the blocks the rate of pumping will be increased again.

CLASS 2/1 STABILITY -SECTION 21 -Dry-docking

260

IGM
KM
Initial
21.5 PRACTICAL CONSIDERATIONS DURING DRY-DOCKING
The major considerations that should be borne in mind are:

(1) that the P force is kept to an acceptable level, and;

(2) that the ship maintains an acceptable positive GM during the critical period.

21.5.1 The requirement to limit the P force

During the critical period prior to taking the blocks fully forward and aft the P force will be acting at
a single point on the stern frame of the ship. The stern frame is specially strengthened to accept
the forces exerted on it during dry- docking but there will be a maximum limit that must not be
exceeded. If the P force becomes too great structural damage will occur. It is usual to have
acceptable near-light conditions of loading for dry-docking specified in the ship's stability data
book. If an actual P-force value is not quoted then it may be approximated from the recommended
condition(s) given by rearranging the dry-docking formulae and calculating it. Under normal
circumstances the ship's classification society will investigate any proposed dry-docking condition
and verify that it is appropriate. Under exceptional circumstances a ship may be dry-docked in a
part-loaded condition but this will only ever be cbne after taking classification society advice. It
would often be more prudent to discharge any cargo on board prior to entering dry dock.

An obvious method to limit the P force during the critical period is to keep the initial trim by the
stern small, consider the formula for calculating the P force during the critical period:

It is clear from the above that P force is directly proportional to the change of trim that the ship will
undergo. Limiting the trim will therefore limit the maximum loads that will be experienced by the
stern frame. The greater the displacement of a given ship, the more important will be the need to
limit the docking trim.

21.5.2 Limiting the loss of GM

Consideration of the formulae will indicate that the greater the trim of the ship when docking, the
greater will be the loss of GM.

Loss of GM =.Px KG
w-p
Clearly, the greater the trim, the greater the P force; the greater the P force, the greater the loss of
GM!
Alternatively, the ship should dry-dock with a greater effective GM that will ensure that stability is
maintained. Improving the ship's initial GM will be achieved by:

(1) Lowering the effective KG by lowering weights within the vessel, discharging weights from
high up or taking on an acceptable amount of ballast in double bottom tanks, or;

(2) Minimising free surface effects by topping up slack tanks wherever possible.

ExamfJle 2
M. V. Almar about to dry cbck requires a minimum GM of 0.3 m at the time the ship takes the
blocks folWard and aft. Current draughts are F 6.89 m and A 8.47 m. KG is 8.86 m.

Calculate the maximum permissible trim by the stem on entering the dry dock.
.,
CLASS 2/1 STABILITY -SECTION 21 -Dry-docking
261
 Solution
(1) initial TMD

AMD = 7.68 m
Enter data with AMD and obtain LCF position.

LCF = 85.01 + (-0.09 xQQ§) = 84.94 m foap

0.1
Use this to calculate the TMD.

(2) Enter data with TMD 7.671 m

Displacement = 23580 + (340 x QQZV = 23821 t
0.1

-
MCTC = 347 + (2 x 0.071) = 348 t-m
0.1
LCF = 85.01 + (-0.09 x Q.QZ1)= 84.95 m foap
0.1
KM = 9.37 + (-0.01 x QQZ1) = 9.363 m
0.1

We must assume that KM rem ains constant in this case.

(3) Calculate the allowed loss of GM
9.363
8.860
Initial GM 0.503
Minimum critical GM 0.300
GM at critical instant 0.203

(4) Calculate the maximum allowed P force and hence the maximum intial trim

Method 1
Loss of GM = .e x KG 0.203 = P x 8.860
w-p (23821 -P)

0.203(23821 -P) = 8. 860P

4835.663 -0.203P = 8.860P
4835.663 = 8. 860P + 0.203P
4835.663 = 9.063P P = 534 tannes

p = COT x MCTC 534 = COT x 348

d 84.95
COT = 534 x 84.95 = 130 cms (Ans)
348

CLASS 2/1 STABILITY -SECTION 21 -Dry-docking 262

Calculate
 To ensure that a GM of 0.3 m is maintained at the critical instant the trim of the ship must
not exceed 1.30 m by the stern.

Method 2

Loss of GM = P x KM 0.203 = P x 9.363

W 23821

p = 0.203 x 23821 P = 516 tannes

9.363

p = COT x MCTC 516 = COT x 348

d 84.95

COT = 516 x 84.95 = 126 cms (Ans)

348

CLASS 2/1 STABILITY -SECTION 21 -Dry-docking

263