undated (16-Nav-A1) - Fundamentals of Naval Architecture

Question 1: Maximum Deck Load and Hydrostatics of a Triangular Wall-Sided Barge

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A wall-sided (vertical-sided) barge whose waterplane, in plan, is a triangle: a point apex at the bow and the full 16 m beam at the stern, 80 m long. Because the sides are vertical, this triangular waterplane shape — and hence its area, LCF and moments of inertia — is the same at every draft up to the 6 m depth.

Find. (i) the maximum deck load and the highest permissible $KG$ of that load; (ii) at $d=4$ m, the tons-per-metre immersion and the longitudinal centre of flotation.

Approach. Get the (draft-independent) waterplane area and its transverse/longitudinal moments of inertia from the triangle formulas on the paper's own reference sheet; use Archimedes' principle at the freeboard-limited maximum draft and at the light-ship draft to get the maximum deck load, then use the metacentre at that maximum draft as the ship's absolute ceiling on overall $KG$ to back out the highest permissible $KG$ of the load itself.

1. Waterplane area (constant for all drafts $\le D$). A triangle of base $B_{max}=16$ m and height $L=80$ m: $$A_W=\tfrac12 B_{max}L=\tfrac12(16)(80)=640\text{ m}^2$$
2. Maximum permissible draft and displacements. Summer draft is depth less minimum freeboard: $$d_{max}=D-fb=6-1.2=4.8\text{ m}$$ $$\Delta_{light}=\rho A_W d_{light}=1.025(640)(2.1)=1377.6\text{ t},\qquad \Delta_{max}=\rho A_W d_{max}=1.025(640)(4.8)=3148.8\text{ t}$$ $$\boxed{W_{load}=\Delta_{max}-\Delta_{light}=3148.8-1377.6\approx1771.2\text{ t}}$$
3. Transverse $BM$ and $KM$ at the maximum draft. Using the reference triangle with base $b=16$ m, apex offset $a=b/2=8$ m (isosceles), height $h=L=80$ m, the centroidal transverse moment of inertia (about the ship's centreline) is $\bar I_y=\tfrac{bh}{36}(a^2+b^2-ab)$: $$I_T=\frac{16(80)}{36}\!\left(8^2+16^2-8(16)\right)=6826.7\text{ m}^4$$ At $d_{max}=4.8$ m, $V_{max}=A_W d_{max}=3072\text{ m}^3$, and since the hull is wall-sided (rectangular cross-section at every station) $KB=d/2$: $$KB_{max}=2.4\text{ m},\quad BM_{T,max}=\frac{I_T}{V_{max}}=\frac{6826.7}{3072}\approx2.222\text{ m},\quad KM_{T,max}=KB_{max}+BM_{T,max}\approx4.622\text{ m}$$
4. Highest permissible $KG$ of the load. The absolute ceiling on the loaded ship's $KG$ is $KM_{T,max}$ (the $GM=0$ limit); balance moments of weight about the keel between the light ship and the deck load: $$\Delta_{max}\,KM_{T,max}=\Delta_{light}\,KG_{light}+W_{load}\,KG_{load}$$ $$\boxed{KG_{load}=\frac{\Delta_{max}KM_{T,max}-\Delta_{light}KG_{light}}{W_{load}}=\frac{3148.8(4.622)-1377.6(3.10)}{1771.2}\approx5.81\text{ m above the keel}}$$
5. Part (a): tons per metre immersion at $d=4$ m. Because $A_W$ is the same at every draft, $TPM$ is too — the classic wall-sided-hull property that TPM does not depend on draft: $$\boxed{TPM=\rho A_W=1.025(640)=656\text{ t/m}}$$
6. Part (b): centre of flotation at $d=4$ m. The centroid of a triangle lies $h/3$ from its base; here the "base" (widest end) is the stern, so: $$\boxed{LCF=\frac{L}{3}=\frac{80}{3}\approx26.67\text{ m forward of the stern}\ (\approx53.33\text{ m aft of the bow})}$$

Question 2: Trim Angle from Shifting a Weight Toward the Bow

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The same triangular wall-sided barge, at the light-ship draft, floating at even keel. An 80 t weight already aboard, at amidships, is moved 25 m toward the bow.

Find. The resulting trim angle.

Approach. Get $MT1m$ from the longitudinal $BM$ and $KG$ at the light-ship draft, find the total change of trim from the shift's moment, then convert to an angle over the barge's length.

1. Longitudinal moment of inertia (constant, as in Q1). Using $\bar I_x=\tfrac{bh^3}{36}$ with $b=B_{max}=16$ m, $h=L=80$ m (independent of the apex offset $a$): $$I_L=\frac{16(80)^3}{36}\approx227{,}556\text{ m}^4$$
2. $BM_L$, $KM_L$, $GM_L$ at the light-ship draft. $V_{light}=A_W d_{light}=640(2.1)=1344\text{ m}^3$, $KB_{light}=d_{light}/2=1.05$ m: $$BM_{L,light}=\frac{227{,}556}{1344}\approx169.31\text{ m},\qquad GM_{L,light}=KB_{light}+BM_{L,light}-KG_{light}\approx1.05+169.31-3.10\approx167.26\text{ m}$$
3. Moment to trim 1 m and change of trim. $$MT1m=\frac{\Delta_{light}\,GM_{L,light}}{L}=\frac{1377.6(167.26)}{80}\approx2880.3\text{ t}\cdot\text{m/m}$$ $$\text{Trimming moment}=w\,s=80(25)=2000\text{ t}\cdot\text{m}\ \Rightarrow\ COT=\frac{2000}{2880.3}\approx0.694\text{ m}$$
4. Trim angle. $$\boxed{\theta=\arctan\!\left(\frac{COT}{L}\right)=\arctan\!\left(\frac{0.694}{80}\right)\approx0.497^\circ\text{ (by the head)}}$$

Question 3: KG from an Inclining Experiment (Least-Squares Fit)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Five (moment, list-angle) pairs from moving a known weight across the deck, plus the ship's $KM$ and its displacement as tested (with the 35 LT inclining gear aboard, at 39 ft above the keel).

Find. $GM$ from the plotted data, then the ship's $KG$ corrected for removal of the inclining gear after the test.

Approach. Correct every reading for the small permanent list observed at zero applied moment, fit a straight line of $\tan\theta$ against moment through the corrected origin, take $GM$ from its slope, recover $KG$ from $KM-GM$, then strip out the temporary inclining gear's contribution to get the ship's true $KG$.

1. Correct for the initial list. At zero applied moment the ship already lists $0.2^\circ$ to port; subtract this offset (add $0.2^\circ$, taking starboard positive) from every reading: $$\begin{aligned} (910,\,2.7)&\to(910,\,2.9^\circ)\qquad(575,\,1.4)\to(575,\,1.6^\circ)\qquad(0,\,-0.2)\to(0,\,0^\circ)\\ (-541,\,-1.8)&\to(-541,\,-1.6^\circ)\qquad(-882,\,-2.5)&\to(-882,\,-2.3^\circ) \end{aligned}$$
2. Least-squares slope through the origin. For small angles $\tan\theta=M/(\Delta\,GM)$, so plotting $\tan\theta$ against $M$ gives a line of slope $1/(\Delta\,GM)$ through the origin: $$\text{slope}=\frac{\sum M_i\tan\theta_i}{\sum M_i^2}\approx5.055\times10^{-5}\ \text{ft}^{-1}\cdot\text{LT}^{-1}$$
3. Experimental $GM$ and $KG$ as inclined. $$GM=\frac{1}{\Delta\cdot\text{slope}}=\frac{1}{3900(5.055\times10^{-5})}\approx5.07\text{ ft},\qquad KG_{inclined}=KM-GM=24.5-5.07\approx19.43\text{ ft}$$
4. Correct for removal of the inclining gear. The 35 LT of temporary gear (part of the 3900 LT tested) is removed after the test; take moments about the keel to strip it out: $$KG_{ship}=\frac{\Delta\,KG_{inclined}-w_g\,kg_g}{\Delta-w_g}=\frac{3900(19.43)-35(39)}{3865}\approx19.25\text{ ft}$$ $$\boxed{GM\approx5.07\text{ ft},\qquad KG_{ship}\approx19.25\text{ ft (gear removed)}}$$

Question 4: Trim of the Triangular Barge After Flooding a Stern-End Compartment

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The same triangular barge as Q1, at draft 4 m, with an empty, full-width, 5 m compartment at the extreme stern opened to the sea. Because the flooded compartment stays open to the sea, displacement and $KG$ are unchanged — this is a lost-buoyancy problem — but because the hull tapers, the flooded strip is a trapezoid (beam 16 m at the transom, 15 m at 5 m forward of it), not a rectangle.

Find. The new forward and aft drafts after flooding, by the lost-buoyancy method.

Approach. Displacement and $KG$ stay fixed. Using bow-referenced station $\xi$ ($0$–$80$ m, beam $B(\xi)=0.2\xi$), find the flooded strip's area and centroid by integration, subtract $\mu$ times the strip from the full triangle to get the intact waterplane, find the new parallel draft, the new $LCF_1$, and the intact $I_L$ about $F_1$, then trim about $F_1$ until the (unmoved) $G$ realigns under the shifted centre of buoyancy.

1. Displacement (constant). $V_0=A_W d_0=640(4)=2560\text{ m}^3$, $\Delta=\rho V_0=1.035(2560)\approx2649.6\text{ t}$.
2. Flooded strip (stern-end, $\xi=75$ to $80$). $B(\xi)=0.2\xi$, so: $$a_{strip}=\int_{75}^{80}0.2\xi\,d\xi=0.1(80^2-75^2)=77.5\text{ m}^2,\qquad \bar\xi_{strip}=\frac{\int_{75}^{80}0.2\xi^2\,d\xi}{a_{strip}}=\frac{0.2(80^3-75^3)/3}{77.5}\approx77.53\text{ m from the bow}$$
3. Intact waterplane area and new parallel draft. Effective area lost is $\mu\,a_{strip}=0.85(77.5)=65.9\text{ m}^2$: $$A_1=A_W-\mu\,a_{strip}=640-65.9\approx574.1\text{ m}^2,\qquad d_1=\frac{V_0}{A_1}=\frac{2560}{574.1}\approx4.459\text{ m}$$
4. New centre of flotation $F_1$. The full triangle's own centroid is $\xi_{F0}=\tfrac{2}{3}L\approx53.33$ m from the bow; removing $\mu\,a_{strip}$ at $\bar\xi_{strip}$: $$\xi_{F_1}=\frac{A_W\,\xi_{F0}-\mu\,a_{strip}\,\bar\xi_{strip}}{A_1}=\frac{640(53.33)-0.85(77.5)(77.53)}{574.1}\approx50.56\text{ m from the bow}\ (\approx29.44\text{ m fwd of the stern})$$
5. Longitudinal inertia of the intact waterplane about $F_1$, and $MT1m$. Working all second moments about the bow ($I=\int\xi^2 B(\xi)\,d\xi$) then shifting once to $F_1$: $$I_{full,bow}=I_L+A_W\,\xi_{F0}^2\approx2{,}048{,}000\text{ m}^4,\quad I_{strip,bow}=\int_{75}^{80}0.2\xi^3\,d\xi\approx465{,}969\text{ m}^4$$ $$I_{L,1,about\ bow}=I_{full,bow}-\mu\,I_{strip,bow}\approx1{,}651{,}927\text{ m}^4,\qquad I_{L,1,about\ F_1}=I_{L,1,about\ bow}-A_1\,\xi_{F_1}^2\approx184{,}436\text{ m}^4$$ With $KB_1=d_1/2\approx2.229$ m and $BM_{L,1}=I_{L,1,about\ F_1}/V_0\approx72.05$ m: $$GM_{L,1}=KB_1+BM_{L,1}-KG\approx2.229+72.05-2.5\approx71.77\text{ m},\qquad MT1m=\frac{\Delta\,GM_{L,1}}{L}\approx\frac{2649.6(71.77)}{80}\approx2377\text{ t}\cdot\text{m/m}$$
6. Change of trim and new drafts. $G$ is unmoved at $\xi_{F0}=53.33$ m, aft of the new $F_1$ (50.56 m); the resulting moment trims the barge by the stern: $$\text{Trimming moment}=\Delta\,|\xi_{F0}-\xi_{F_1}|=2649.6(2.78)\approx7355\text{ t}\cdot\text{m}\ \Rightarrow\ COT=\frac{7355}{2377}\approx3.094\text{ m}$$ Apportioning about $F_1$ ($l_F\approx50.56$ m to the bow, $l_A\approx29.44$ m to the stern): $$\boxed{d_F=d_1-COT\cdot\frac{l_F}{L}\approx4.459-1.955\approx2.50\text{ m},\qquad d_A=d_1+COT\cdot\frac{l_A}{L}\approx4.459+1.139\approx5.60\text{ m}}$$

Question 5: Correcting the GZ Curve for the Actual Loading Condition

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A static-stability ($GZ$) curve computed for an assumed $KG=6.5\text{ m}$, and the ship's actual weight distribution before sailing, from which the true $KG$ differs from the assumed value.

Find. The corrected $GZ$ curve, $GM$ read from it, the angle of vanishing stability, the angle of maximum stability, and the maximum righting moment.

Approach. Find the actual $KG$ by taking moments of the loading condition about the keel, apply the standard small-angle $KG$-correction to every point of the given $GZ$ curve, then read $GM$, the vanishing-stability angle, and the maximum-$GZ$ angle off the corrected curve.

1. Actual $KG$ for the loading condition. $$KG_{actual}=\frac{4200(6.0)+9100(7.0)+1500(1.1)+200(7.5)}{15{,}000}=\frac{92{,}050}{15{,}000}\approx6.137\text{ m}$$
2. Correct the GZ curve for the KG difference. $KG_{actual}
3. $GM$ from the corrected curve. Since the origin must theoretically be $(0^\circ,0)$, take the slope from the first genuine reading: $$GM\approx\frac{GZ(15^\circ)}{\sin15^\circ}=\frac{0.204}{0.2588}\approx0.788\text{ m}$$
4. Angle of vanishing stability. $GZ_{actual}$ is still positive at $75^\circ$ ($0.301\text{ m}$) but negative at $90^\circ$ ($-0.267\text{ m}$); interpolating linearly for the zero crossing: $$\theta_{vanish}=75+15\left(\frac{0.301}{0.301+0.267}\right)\approx83.0^\circ$$
5. Angle of maximum stability and maximum righting moment. The corrected curve peaks at the tabulated $45^\circ$ point ($GZ_{max}\approx0.837\text{ m}$): $$\boxed{GM\approx0.79\text{ m},\quad \theta_{vanish}\approx83.0^\circ,\quad \theta_{max}\approx45^\circ,\quad M_{max}=\Delta\cdot GZ_{max}\approx15{,}000(0.837)\approx12{,}554\text{ t}\cdot\text{m}}$$