25-Nav-A2 Hydrodynamics of Ships (I)_ Resistance and Propulsion · December 2017
Question 5 of 5: Correcting and Reading the Statical Stability (GZ) Curve
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.
Question 5: Correcting and Reading the Statical Stability (GZ) Curve (20 marks)
Given. A cross-curve GZ table computed for an assumed $KG=6.5$ m, and the ship's actual departure loading (masses and KGs above).
Find. The corrected GZ curve, GM, the angle of vanishing stability, the angle of maximum stability, and the maximum righting moment.
As-supplied vs. KG-corrected GZ curves, with the initial tangent used to read GM.
Approach. Find the actual departure $KG$ from the weight table, apply the standard wall-sided correction $\Delta GZ=(KG_{assumed}-KG_{actual})\sin\theta$ to every tabulated point, then read GM from the corrected curve's initial slope, the vanishing angle from where it recrosses zero, and the maximum stability angle/GZ from a local quadratic fit near the peak.
Actual departure KG. $$\Delta=4200+9100+1500+200=15{,}000\ \text{t}$$ $$KG_{actual}=\dfrac{4200(6.0)+9100(7.0)+1500(1.1)+200(7.5)}{15{,}000}=\dfrac{92{,}050}{15{,}000}=6.137\ \text{m}$$
Correction to apply. Since the actual $KG$ (6.137 m) is lower than the assumed 6.5 m used for the tabulated curve, the ship is more stable than shown, and GZ must be corrected upward: $$\Delta KG=KG_{assumed}-KG_{actual}=6.5-6.137=0.363\ \text{m}$$ $$GZ_{corr}(\theta)=GZ_{given}(\theta)+0.363\sin\theta$$
Corrected GZ table.
θ
0°
15°
30°
45°
60°
75°
90°
GZgiven (m)
−0.055
0.11
0.36
0.58
0.12
−0.05
−0.63
+0.363 sinθ
0
0.094
0.182
0.257
0.315
0.351
0.363
GZcorr (m)
−0.055
0.204
0.542
0.837
0.435
0.301
−0.267
GM from the initial slope. For small angles $GZ\approx GM\sin\theta$, so the chord between the two smallest tabulated angles gives $$\boxed{GM=\dfrac{GZ_{corr}(15^\circ)-GZ_{corr}(0^\circ)}{\sin15^\circ-\sin0^\circ}=\dfrac{0.204-(-0.055)}{0.2588}=1.00\ \text{m}}$$ (this is the ordinate reached by the tangent to the curve at the origin when extended to $57.3^\circ$ — the graphical GM construction shown dashed in the figure).
Angle of vanishing stability. $GZ_{corr}$ is still positive at $75^\circ$ (0.301 m) and negative at $90^\circ$ ($-0.267$ m); linear interpolation between them gives $$\boxed{\theta_{vanish}=75+15\times\dfrac{0.301}{0.301+0.267}=83.0^\circ}$$
Angle and value of maximum stability. The curve peaks near $45^\circ$; fitting a parabola through the three points at $30^\circ,45^\circ,60^\circ$ (0.542, 0.837, 0.435 m) locates the vertex at $$\theta_{max}=45+\dfrac{15}{2}\times\dfrac{(0.542-0.435)}{(0.542-2(0.837)+0.435)}=43.8^\circ,\qquad GZ_{max}=0.839\ \text{m}$$
Maximum righting moment. $$\boxed{RM_{max}=\Delta\times GZ_{max}=15{,}000\times0.839=12{,}585\ \text{t}\cdot\text{m}}$$
