25-Nav-A1 Fundamentals of Naval Architecture · May-98-Nav-A1 2016

Question 4 of 7: Trim Angle of a Box Barge After Flooding One Compartment

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

Question 4: Trim Angle of a Box Barge After Flooding One 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. One of ten identical, previously-empty (permeability 100%) watertight compartments floods completely and remains open to the sea — solved by the lost-buoyancy method (constant displacement, intact waterplane provides all the buoyancy).

Find. The trim angle after flooding.

Approach. Use the lost-buoyancy method: find the parallel rise from the volume lost up to the original draft over the intact waterplane area, locate the new centre of flotation, compute the intact waterplane's longitudinal inertia about it, then find the trimming moment produced by the lost buoyancy about the new $F$.

1. Waterplane areas. Each compartment is $60.00/5=12.0\text{ m}$ long by $10.50/2=5.25\text{ m}$ wide, area $=63.0\text{ m}^2$. Original waterplane $A_W=60.00(10.50)=630.0\text{ m}^2$, intact waterplane $A_W'=630.0-63.0=567.0\text{ m}^2$.
2. Parallel rise (constant-displacement/lost-buoyancy). Buoyancy lost up to the original draft: $v=63.0(3.00)=189.0\text{ m}^3$. $$\text{rise}=\frac{v}{A_W'}=\frac{189.0}{567.0}=0.3333\text{ m}\ \Rightarrow\ T'=3.00+0.333\approx3.333\text{ m}$$
3. New centre of flotation. Taking $x$ positive forward from midship, the lost area's centroid is at $x_c=+24.0\text{ m}$: $$x_F'=\frac{0(630.0)-24.0(63.0)}{567.0}\approx-2.667\text{ m (i.e. 2.667 m aft of midship)}$$
4. Intact waterplane's longitudinal inertia about the new $F$. Whole-box $I_L=\dfrac{BL^3}{12}=\dfrac{10.50(60.00)^3}{12}=189{,}000\text{ m}^4$ about midship. The lost compartment's own inertia is $\dfrac{5.25(12.0)^3}{12}=756.0\text{ m}^4$, so about midship it contributes $756.0+63.0(24.0)^2=37{,}044\text{ m}^4$. Intact $I_L$ about midship $=189{,}000-37{,}044=151{,}956\text{ m}^4$; shifting to the new centroid ($x_F'=-2.667\text{ m}$) by the parallel-axis theorem: $$I_L'=151{,}956-567.0(-2.667)^2\approx147{,}924\text{ m}^4$$
5. Trimming moment and trim angle. $MCT1cm=\dfrac{\rho\,I_L'}{100L}=\dfrac{1.025(147{,}924)}{100(60.00)}\approx25.27\text{ tonnef}\cdot\text{m/cm}$. Lever between the lost buoyancy's centroid and the new $F$: $24.0-(-2.667)=26.67\text{ m}$. $$\text{Trimming moment}=v\rho\,d=189.0(1.025)(26.67)\approx5166\text{ tonnef}\cdot\text{m}\ \Rightarrow\ COT=\frac{5166}{25.27}\approx204.4\text{ cm}=2.044\text{ m}$$ $$\boxed{\tan\theta_{trim}=\frac{2.044}{60.00}\approx0.03408\ \Rightarrow\ \theta_{trim}\approx1.95^\circ\text{ (by the head)}}$$