Question 2 of 5: Hydrostatic Loads

Question 2: Hydrostatic Loads (18 marks)

(a) Displacement and LCG from simplified Bonjean data [10]

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. $L_{bp}=60$ m, 4 stations (0=FP, 1, 2, 3=AP) spaced $\Delta x = L_{bp}/3 = 20$ m apart. Bonjean curves are supplied at the three half-stations only, and each is stated to apply to the whole 20 m block adjacent to it. Draft $T=4.5$ m, fresh water $\rho=1.0\ \text{t/m}^3$, level trim. Reading the supplied Bonjean chart at $z=4.5$ m gives the sectional areas below.

Find. (i) The longitudinal position of G (=B, since the vessel floats level with no trim) measured from AP; (ii) the displacement in tonnes at $T=4.5$ m.

Approach. Because each half-station's area is explicitly stated to represent its whole 20 m block (a 3-point block/prismatic idealisation, not Simpson's Rule), the displaced volume is simply $\Delta x\sum A_i$, and — since the vessel floats level with no trim — the longitudinal centre of buoyancy LCB equals the longitudinal centre of gravity LCG, found as the area-weighted mean of each block's own midpoint (which coincides with its half-station location).

1. Displaced volume. Each block is 20 m long with area equal to its half-station's Bonjean reading: $$V=\Delta x\,(A_{1/2}+A_{1\frac12}+A_{2\frac12}) = 20\,(21.0+35.5+29.0) = 20\times85.5 = \boxed{1710\ \text{m}^3}.$$
2. Displacement. In fresh water, $\rho=1.0\ \text{t/m}^3$, so mass displacement equals the volume numerically: $$\Delta = \rho V = 1.0\times1710 = \boxed{1710\ \text{t}}.$$ That is part (ii).
3. Centroid distance of each block from AP. The FP-end block (represented by station ½) spans $x=0$–20 m from FP, so its own midpoint is 10 m from FP $=50$ m from AP. The midship block (station 1½) spans 20–40 m from FP, midpoint 30 m from FP $=30$ m from AP. The AP-end block (station 2½) spans 40–60 m from FP, midpoint 50 m from FP $=10$ m from AP.
4. LCB (= LCG, level trim, no correction needed). Take the area-weighted mean of the three block centroids (the equal 20 m block length cancels top and bottom): $$\text{LCB}=\frac{21.0(50)+35.5(30)+29.0(10)}{21.0+35.5+29.0}=\frac{1050+1065+290}{85.5}=\frac{2405}{85.5}=\boxed{28.1\ \text{m fwd of AP}}.$$ That is part (i): G sits 28.1 m forward of AP (i.e. just aft of amidships, at $L_{bp}/2=30$ m, consistent with the fuller midship/stern sections pulling the centroid slightly aft of centre).

(b) Shear force and bending moment in a floating wood block [8]

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. Block: $L=1$ m long, $0.20$ m wide, $0.10$ m thick, uniform density, self-weight $W=100$ N; a $P=50$ N weight placed at the centre ($x=0.5$ m); block floats freely (no discrete supports) in fresh water.

Find. The shear force $V(x)$ and bending moment $M(x)$ distributions along the block's long axis.

Approach. A freely floating body has no discrete supports, so its "reaction" is the distributed buoyant pressure: because the added weight sits exactly at the centre (symmetric load, level trim), the block sinks bodily and the buoyant support is a uniform upward line load equal in total to the total weight. The self-weight is likewise uniform (uniform density), so the net loading is a uniform net-upward line load plus one downward point load at the centre — solved as a free body from each free end inward.

1. Uniform loads. Total downward weight $=W+P=150$ N over $L=1$ m of waterplane, so the buoyant support (matching it exactly, since the block floats in equilibrium) is $w_{buoy}=150$ N/m upward; self-weight is $w_{self}=W/L=100$ N/m downward. Net distributed load: $$w_{net}=w_{buoy}-w_{self}=150-100=\boxed{50\ \text{N/m upward}}.$$
2. Shear force. Cutting a free body from the left (free) end at $x$, $V(x)=w_{net}\,x$ minus the 50 N point load once $x>0.5$ m: $$V(x)=50x-50\,[x>0.5].$$ At $x=0.5^-$: $V=25$ N; at $x=0.5^+$: $V=25-50=-25$ N (a 50 N drop, matching the point load); at $x=1$: $V=50(1)-50=0$, confirming equilibrium at the free right end.
3. Bending moment. Integrating $V(x)$ with $M(0)=0$ (free end): $$M(x)=\tfrac12 w_{net}x^2\ (x\le0.5),\qquad M(x)=\tfrac12 w_{net}x^2-P(x-0.5)\ (x\ge0.5).$$ Both branches meet at the centre giving the maximum: $$M(0.5)=\tfrac12(50)(0.5)^2=\boxed{6.25\ \text{N}\cdot\text{m (sagging, at centre)}}.$$ $M(1)=25(1)^2-50(0.5)=0$, confirming the right free end is moment-free as required.