Solution Naval Numerical 8

Numerical 8: A box shaped vessel, 50 metres long × 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A centre line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres.
Solution:
Given L = 50m
B = 10m
d = 4m
l = 15m
μ = 30 %.
KG = 3m.

to calculate list:

we know that if a compartment is bilged there is an increase in draught.

which is given by: $= \frac{Vol Of Lost Buoyancy}{Area Of Intact Water plane}$

Given that the compartment is having only half the breadth of the ship thus,
Volume of lost Buoyancy can be calculated as:
= $\mu l(B/2)d$
= 0.3 x 15 x (10/2) x 4
= 90 cubic-meter
and, The area of intact water plane can be find as:
$(L-l)B +l(B/2) +l(B/2)(1-\mu )$
= (50-15)10 + 15(10/2) + 15x(10/2)(1-0.3)
=350+75+52.2
=477.5 sq-m

Thus, increased in draught = 90/477.5=0.18848m
New draught = 4 + 0.18848 =4.18848m

we know $BB_1$ = BMtanθ

Since the BM and GM are assumed to be having a very small difference the above formula can be written as $BB_1$ = GMtanθ

Now we need to find GM and $BB_1$ .

Calculation for $BB_1$ :-

$BB_1$ is the distance to which the centre of buoyancy has shifted.

$BB_1$ can be found as:

$BB1 =\frac{a x y}{A}$

but since permiability is given it will be:

$BB1 =\frac{\mu xa x y}{A}$

where 'μ' = 0.3 (given)

'a' is area of transverse plane of bilged compartment =l xB/2 = 15 x 5 =75 m-sq

'A' is the area of intact water plane i.e 477.5 m-sq (calculated above)

and from the figure we see y = B/4 = 2.5m (it is the distance of centroid of bilged chamber from centre line.

Thus, $BB1 =\frac{0.3 x 15 x 5 x 2.5}{477.5}=0.12m$

Calculation for GM:-

We know KM= KG+GM

thus, GM= KM-KG

KG = 3m (given)

we still do not know KM, Which we need to calculate,

lets use KM = KB+BM

here KB= draught/2 =4.18848/2 = 2.094m ( New draught is calculated above).

And now BM = I/⛛ ---(i)

Here I is the moment of inertia about the new axis, which can be calculated by using parallel axis theorem.

$I=I_{00} - Ax H^{2}$ .---(ii)

'H' is the distance of new axis from original axis and 'A' is the area of intact water plane.

$H=\frac{B}{2}+BB_1$

Thus H=0.12 m.---(iii)

A=477.5 m-sq---(iv)

need to calculate $I_{00}$

$I_{00}=\frac{LB^{3}}{12}-\mu \frac{l.\left ( \frac{B}{2} \right )^{3}}{3}$

Note in the above expression it is confusing that denominator is different for both the moment of inertia but it is simply because of the following facts:

The moment of inertia of a rectangle with respect to an axis passing through its centroid, is given by the following expression: $\frac{LB^{3}}{12}$