25-Nav-A2 Hydrodynamics of Ships (I)_ Resistance and Propulsion · December 2017
Question 2 of 5: Trim from a Longitudinal Weight Shift
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 2: Trim from a Longitudinal Weight Shift (20 marks)
Given. Same barge as Q1 ($L=70$ m, $B=17$ m), at the light-ship draft $d_0=2.2$ m, initially on an even keel. A weight $w=20$ t is shifted $\ell=27$ m towards the bow.
Find. The resulting trim angle.
Approach. A weight shift produces a trimming moment $w\ell$; dividing by the moment-to-change-trim-one-centimetre at the light draft gives the change of trim, which (with $LCF$ at midships for this box form) is split evenly between the ends and converted to an angle through $\tan\theta=\text{trim}/L$.
MCT1cm at the light draft. $\Delta_0=\rho A_W d_0=1.025\times1190\times2.2=2683.45\ \text{t}$ (as in Q1). $BM_{L,0}=L^2/(12d_0)=4900/(12\times2.2)=4900/26.4=185.61\ \text{m}$. $$MCT1cm_0=\dfrac{\Delta_0\,BM_{L,0}}{100L}=\dfrac{2683.45\times185.61}{7000}=71.15\ \text{t}\cdot\text{m/cm}$$ (Interestingly this equals Q1's design-draft value exactly — for a box barge $\Delta\,BM_L=\rho A_W d\times L^2/(12d)=\rho A_W L^2/12$ is independent of draft, so MCT1cm is the same at every waterline.)
Change of trim. $$t=\dfrac{w\ell}{MCT1cm_0}=\dfrac{20\times27}{71.15}=7.59\ \text{cm}=0.0759\ \text{m}$$
Trim angle. With the centre of flotation at midships, this change of trim is the full difference between the forward and aft drafts over the length $L$: $$\boxed{\theta=\tan^{-1}\!\left(\dfrac{t}{L}\right)=\tan^{-1}\!\left(\dfrac{0.0759}{70}\right)=0.062^\circ\ \text{(by the head)}}$$ Equivalently, the forward draft rises by $t/2=3.79\ \text{cm}$ and the aft draft falls by the same amount, since $F$ is at $L/2$.
