25-Nav-A1 Fundamentals of Naval Architecture · May-16-Nav-A1 2017
Question 4 of 5: Trim of a Box Barge After Flooding a Fore-End Compartment
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Tupper, Introduction to Naval Architecture; Lewis (ed.), Principles of Naval Architecture (PNA); IMO, International Code on Intact Stability (IS Code).
Question 4: Trim of a Box Barge After Flooding a Fore-End Compartment
Given. A box-shaped barge has a full-width, previously-empty fore-end compartment holed at sea; because the compartment stays open to the sea, this is solved by the lost-buoyancy method — displacement and $KG$ are unchanged, but the intact waterplane (and hence the centre of flotation and $MT1m$) shrink.
Given data
Quantity
Symbol
Value
Length
$L$
60 m
Beam
$B$
10 m
Original (even-keel) draft
$d_0$
3.0 m
$KG$
$KG$
2.5 m
Flooded compartment length (fore-end)
$l_c$
8.0 m
Permeability
$\mu$
90% (0.9)
Water density
$\rho$
1.025 t/m$^3$
Find. The new forward and aft drafts after flooding.
Profile schematic (not to scale): the fore-end compartment (length $l_c=8$ m, $\mu=0.9$) is open to the sea, shifting the intact centre of flotation $F_1$ aft and trimming the barge by the head.
Approach. Displacement and $KG$ stay fixed; find the intact waterplane area to get the new parallel (mean) draft, locate the new centre of flotation $F_1$ from the lost area, find $GM_L$ of the intact waterplane to get the new $MT1m$, then trim the barge about $F_1$ until its centre of buoyancy realigns under the unmoved centre of gravity.
Displacement and displaced volume (constant throughout).
$$V_0=L\,B\,d_0=60(10)(3.0)=1800\text{ m}^3,\qquad \Delta=V_0\,\rho=1800(1.025)=1845\text{ t}$$
Intact waterplane area and new parallel (mean) draft. The flooded portion removes $\mu\,l_c\,B=0.9(8)(10)=72\text{ m}^2$ of effective waterplane, leaving $A_1=L\,B-\mu\,l_c\,B=600-72=528\text{ m}^2$. Since the hull is wall-sided, the same displaced volume is regained over this smaller area at a deeper, uniform draft:
$$d_1=\frac{V_0}{A_1}=\frac{1800}{528}\approx3.409\text{ m}$$
New centre of flotation $F_1$. Taking moments of area about amidships ($x=0$, $+$ forward), the flooded strip's centroid is $x_c=L/2-l_c/2=26\text{ m}$ forward of amidships, with area $72\text{ m}^2$; the full plan (area 600 m$^2$) is already centred on $x=0$:
$$x_{F_1}=\frac{0-72(26)}{528}\approx-3.545\text{ m}\ \ (\text{i.e. }3.545\text{ m aft of amidships})$$
Longitudinal inertia of the intact waterplane and $MT1m$. Using $I=B\,l^3/12$ for a rectangle and scaling the lost strip's own inertia by $\mu$ (its "porosity"), then shifting by the parallel-axis theorem from amidships to $F_1$:
$$I_{full}=\frac{B\,L^3}{12}=180{,}000\text{ m}^4,\quad I_{lost,own}=\mu\frac{B\,l_c^3}{12}=384\text{ m}^4,\quad I_{L}\approx124{,}307\text{ m}^4\text{ about }F_1$$
With $KB_1=d_1/2\approx1.705\text{ m}$ (a wall-sided prism) and $BM_L=I_L/V_0\approx69.06\text{ m}$:
$$GM_L=KB_1+BM_L-KG\approx1.705+69.06-2.5\approx68.26\text{ m},\qquad MT1m=\frac{\Delta\,GM_L}{L}\approx\frac{1845(68.26)}{60}\approx2099\text{ t}\cdot\text{m/m}$$
Change of trim. Because the hull is a prism, the horizontal centre of buoyancy of the parallel-sunk volume coincides with the intact waterplane's own centroid, $x_{F_1}=-3.545\text{ m}$, while $G$ (unmoved) is still at $x=0$. The resulting moment trims the barge by the head until $B$ realigns under $G$:
$$\text{Trimming moment}=\Delta\,|x_{F_1}|=1845(3.545)\approx6541\text{ t}\cdot\text{m}\ \Rightarrow\ COT=\frac{6541}{MT1m}\approx3.116\text{ m}$$
Apportion the trim about $F_1$ and state the new drafts. $l_F=L/2-x_{F_1}\approx33.545\text{ m}$, $l_A=L/2+x_{F_1}\approx26.455\text{ m}$ (check: $l_F+l_A=60\text{ m}$):
$$\Delta d_F=COT\cdot\frac{l_F}{L}\approx1.742\text{ m (increase, bow down)},\qquad \Delta d_A=COT\cdot\frac{l_A}{L}\approx1.374\text{ m (decrease)}$$
$$\boxed{d_F=d_1+\Delta d_F\approx5.151\text{ m},\qquad d_A=d_1-\Delta d_A\approx2.035\text{ m}}$$
