25-Nav-A2 Hydrodynamics of Ships (I)_ Resistance and Propulsion · December 2017
Question 1 of 5: Barge Freeboard, Maximum Deck Load & Design-Draft Hydrostatics
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, Dec 2017 — 98-Mar-A2 Fundamentals of Naval Architecture, 3 hours, open book (any non-communicating calculator permitted). Five questions constitute a complete exam paper and each is of equal value.
Reference texts: Tupper, Introduction to Naval Architecture; Lewis (ed.), Principles of Naval Architecture (PNA); IMO, International Code on Intact Stability (IS Code).
Check: this paper, although listed under Hydrodynamics of Ships (I): Resistance and Propulsion, is headed "98-Mar-A2, Fundamentals of Naval Architecture" and its five questions are entirely barge/ship stability and hydrostatics (freeboard, trim, inclining experiment, flooding, GZ curve) with no resistance or propulsion content whatsoever. Solved here as the exam that was actually printed.
Given. A wall-sided, box-shaped barge with length $L=70$ m, beam $B=17$ m, depth $D=7.5$ m, minimum summer freeboard $f=1.8$ m, light-ship $KG_0=3.75$ m at light draft $d_0=2.2$ m. Sea-water density $\rho=1.025\ \text{t/m}^3$.
Given data
Quantity
Symbol
Value
Length
L
70 m
Beam
B
17 m
Depth
D
7.5 m
Min. summer freeboard
f
1.8 m
Light-ship KG
KG₀
3.75 m
Light draft
d₀
2.2 m
Check: water density is not stated for Q1; taken as sea water $\rho=1.025\ \text{t/m}^3$, the value the same barge is explicitly given in Q4.
Find. (i) the maximum summer deck load and the height above keel of its centre of gravity; (ii) at the design draft of 4 m: TPM, MCT1cm, $KB$, $BM_T$, $BM_L$, and the centre of flotation.
Barge midship section: light draft, summer maximum draft, and light-ship KG.
Approach. Treat the barge as a rectangular box: displacement $\Delta=\rho L B d$, $KB=d/2$, $BM_T=B^2/(12d)$; find the freeboard-limited summer draft, subtract the light-ship displacement to get the deck load, then set the fully-loaded overall $KG$ equal to $KM$ (the $GM=0$ limit) to back out the load's own vertical centre of gravity; repeat the coefficient formulas at the stated 4 m design draft.
Waterplane area and summer load draft. The rectangular waterplane area is $A_W=L\times B=70\times17=1190\ \text{m}^2$. The maximum permissible draft is set by the minimum freeboard: $d_{max}=D-f=7.5-1.8=5.7\ \text{m}$.
Light-ship and maximum displacement. $\Delta_0=\rho A_W d_0=1.025\times1190\times2.2=2683.45\ \text{t}$. At the summer draft, $\Delta_{max}=\rho A_W d_{max}=1.025\times1190\times5.7=6952.575\ \text{t}$. The maximum deck load is the difference: $$\boxed{W_{load}=\Delta_{max}-\Delta_0=6952.575-2683.45=4269.13\ \text{t}}$$
Metacentric height at the summer draft. $KB_{max}=d_{max}/2=5.7/2=2.85\ \text{m}$. $BM_{T,max}=B^2/(12d_{max})=17^2/(12\times5.7)=289/68.4=4.225\ \text{m}$. So $KM_{max}=KB_{max}+BM_{T,max}=2.85+4.225=7.075\ \text{m}$.
Height of the load's centre of gravity. The heaviest deck load a box barge may safely carry is the one that just uses up all the available $GM$ — i.e. the loaded ship's overall $KG$ equals $KM_{max}$. Taking moments of weight about the keel, $\Delta_{max}\,KM_{max}=\Delta_0 KG_0+W_{load}\,KG_{load}$, so $$\boxed{KG_{load}=\dfrac{\Delta_{max}KM_{max}-\Delta_0 KG_0}{W_{load}}=\dfrac{6952.575\times7.075-2683.45\times3.75}{4269.13}=9.17\ \text{m}}$$ i.e. about $9.17-7.5=1.67$ m above deck level — a physically reasonable height for a stack of deck cargo.
(a) TPM and MCT1cm at the 4 m design draft. Tons-per-metre immersion is simply the waterplane area times density (the mass that raises the draft by a full metre): $TPM=\rho A_W=1.025\times1190=1219.75\ \text{t/m}$. At $d=4$ m, $\Delta=\rho A_W d=1.025\times1190\times4=4879.0\ \text{t}$ and $BM_L=L^2/(12d)=70^2/(12\times4)=4900/48=102.08\ \text{m}$. Using $GM_L\approx BM_L$ (the $KG$-$KB$ term is negligible next to $BM_L$ for a shallow box form), the moment to change trim 1 cm is $$\boxed{MCT1cm=\dfrac{\Delta\, BM_L}{100L}=\dfrac{4879.0\times102.08}{100\times70}=71.15\ \text{t}\cdot\text{m/cm}}$$
(b) KB, BMT, BML at the 4 m design draft. $KB=d/2=4/2=2.00\ \text{m}$. $BM_T=B^2/(12d)=289/48=6.02\ \text{m}$, so $KM_T=KB+BM_T=8.02\ \text{m}$. $BM_L=102.08\ \text{m}$ (Step 5), so $$\boxed{KM_L=KB+BM_L=2.00+102.08=104.08\ \text{m}}$$
(c) Centre of flotation. Because beam and freeboard are constant over the full length, the waterplane is a plain rectangle, symmetric fore-and-aft and port-to-starboard. Its centroid — the centre of flotation, F — therefore sits exactly at midships: $$\boxed{LCF=L/2=35\ \text{m from either perpendicular, on the centreline}}$$
