25-Nav-B3 Finite Element Analysis for Ship Structures · May 2016

Question 2 of 7: Stability

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Stability (20 marks: a–8, b–8, c–4)

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.

Find. (a) $GM$, $KG$ from the inclining test, and whether the ship stays upright after the cargo move; (b) the heel angle taken up after the move; (c) the largest weight that can be shifted 14 m without instability.

Approach. Use the small-angle inclining-test formula to get $GM$ (and hence $KG$) in the arrival condition; use the parallel-axis weight-shift rule $\Delta KG=\sum w_id_i/\Delta$ to update $KG$ (and $GM$) after the cargo move; correct the given GZ curve for the resulting rise in $G$ ($GZ'=GZ-\Delta KG\sin\theta$) to read off the new equilibrium heel angle; and set $GM_{new}=0$ to solve for the limiting weight in part (c).

1. GM from the inclining test. For small angles, $GM=\dfrac{w\,d}{\Delta\tan\theta}=\dfrac{50\times20}{12{,}000\times\tan(6.35^\circ)}=\dfrac{1000}{12{,}000\times0.11128}=\boxed{0.749\ \text{m}}$.
2. KG in the arrival condition. $KG=KM-GM=10-0.749=\boxed{9.251\ \text{m}}$.
3. Rise in KG from the cargo move. Moving $w_1=3000$ t up by $d_1=14$ m (mass and hence $\Delta$, $KM$ unchanged) raises the overall centre of gravity by $\Delta KG=\dfrac{w_1d_1}{\Delta}=\dfrac{3000\times14}{12{,}000}=3.5\ \text{m}$, giving $KG_{new}=9.251+3.5=12.751\ \text{m}$.
4. New GM — stability check. $GM_{new}=KM-KG_{new}=10-12.751=\boxed{-2.751\ \text{m}}$. A negative $GM$ means $G$ has risen above $M$: the ship is initially unstable and will not remain upright — it heels (or lolls) to whatever angle restores a positive righting arm, assuming small-angle wall-sided theory still applies and no free-surface or trim effects complicate the picture (see callout).
5. Part (b) — correcting the GZ curve for the raised G. Raising $G$ by $\Delta KG=3.5$ m reduces the righting arm at every heel angle by $\Delta KG\sin\theta$: $GZ'(\theta)=GZ(\theta)-3.5\sin\theta$. Tabulating:

   $\theta$	GZ (orig.)	$3.5\sin\theta$	GZ′ (corrected)
   5°	0.08	0.305	−0.225
   10°	0.25	0.608	−0.358
   20°	0.34	1.197	−0.857
   30°	0.36	1.750	−1.390
   40°	0.32	2.250	−1.930

   $GZ'$ is negative and becomes steadily more negative from 5° through 40° — there is no angle in the surveyed range where a positive righting arm reappears.

6. Part (c) — maximum weight for zero margin. Setting $GM_{new}=0$ (the boundary of initial stability) with the same 14 m lift: $0=GM-\dfrac{w_{max}d_1}{\Delta}\ \Rightarrow\ w_{max}=\dfrac{GM\,\Delta}{d_1}=\dfrac{0.749\times12{,}000}{14}=\boxed{642\ \text{tonnes}}$.