Question 2 of 6: Hydrostatic Loads

Question 2: Hydrostatic Loads (20 marks)

(a) Bonjean curves for three hull cross-sections [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. Three idealized midship cross-sections, all 15 m deep: (a) a trapezoid with a 20 m wide base narrowing toward the deck; (b) a wall-sided (vertical-side) form with a rounded bilge, constant 20 m beam; (c) a V-shaped wedge, pointed at the keel, opening to 20 m beam at the deck.

Find. The shape of the Bonjean curve (sectional area $A(z)$ vs. draft $z$, measured up from the keel) for each of the three forms.

Approach. A Bonjean curve is simply the running integral of the local half-breadth: $A(z)=\int_0^z 2\,b(z^{\prime})\,dz^{\prime}$, so its slope at any draft equals the waterplane half-breadth there, and its curvature (concave up/down) is set entirely by how that half-breadth varies with draft for each hull form.

1. (a) Flared trapezoid — half-breadth decreases with draft. With bottom half-breadth $b_0=10\text{ m}$ and (assumed) deck half-breadth $b_1=5\text{ m}$ varying linearly over $D=15\text{ m}$: $$b(z)=b_0-\frac{b_0-b_1}{D}z,\qquad A(z)=2\!\left[b_0z-\frac{b_0-b_1}{2D}z^2\right]=20z-\frac{z^2}{3}$$ Since $dA/dz=2b(z)$ decreases as $z$ increases (the section narrows going up), the curve is concave down — steep near the keel, flattening toward the deck.
2. (b) Wall-sided, rounded bilge — constant half-breadth. Above the small bilge radius the sides are vertical, so $b(z)=b_0=10\text{ m}$ is constant and $$A(z)=2b_0z=20z$$ a straight line through the origin with slope = the beam (20 m²/m); only a short nonlinear "toe" appears right at the keel where the bilge radius fills in.
3. (c) V-wedge — half-breadth increases with draft. Pointed at the keel ($b=0$ at $z=0$) and full beam at the deck ($b_1=10\text{ m}$ at $z=D=15\text{ m}$), varying linearly: $$b(z)=\frac{b_1}{D}z,\qquad A(z)=\frac{b_1}{D}z^2=\frac{2}{3}z^2$$ Since $dA/dz$ increases linearly with $z$, the curve is concave up — flat near the keel, steepening toward the deck (a pure parabola).

(b) Still-water shear force and bending moment diagrams [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. $L=100\text{ m}$, $\Delta=10{,}000\text{ t}$, 10 stations of $\Delta x=10\text{ m}$ each, bow (station 1) to stern (station 10):

Both rows sum to exactly 10,000 t, confirming the vessel is in vertical equilibrium ($\Delta=$ total buoyancy) — a necessary check before the load curve can be integrated.

Find. The still-water shear force diagram (SFD) and bending moment diagram (BMD) along the hull.

Approach. Take the net load at each station as $n_i=B_i-W_i$ (buoyancy minus weight, positive up); the running sum of $n_i$ gives the shear force at each station boundary, and the trapezoidal running integral of the shear gives the bending moment.

1. Net load per station. $n_i=B_i-W_i$ (t): $$n=[\,50,\,-400,\,-200,\,50,\,350,\,350,\,300,\,0,\,-250,\,-250\,]$$ summing to zero, consistent with the equal 10,000 t totals.
2. Shear force — cumulative sum. $F_j=\sum_{i=1}^{j}n_i$, starting from $F_0=0$ at the bow: $$F=[0,\,50,\,-350,\,-550,\,-500,\,-150,\,200,\,500,\,500,\,250,\,0]\ \text{t}$$ The shear correctly returns to $F_{10}=0$ at the free stern end, confirming overall equilibrium. Maximum shear magnitude is $\boxed{F_3=-550\text{ t}=-5395\text{ kN}}$ (using $g=9.81\,\text{m/s}^2$), just aft of the bow.
3. Bending moment — trapezoidal integration of shear. $M_j=M_{j-1}+\tfrac12(F_{j-1}+F_j)\Delta x$, giving the raw curve $$M_{raw}=[0,250,-1250,-5750,-11000,-14250,-14000,-10500,-5500,-1750,-500]\ \text{t}\!\cdot\!\text{m}$$ The small residual $-500\text{ t}\!\cdot\!\text{m}$ at the free stern end is the expected discretization error of the coarse 10-station trapezoidal rule (true equilibrium demands $M=0$ there); a standard linear "closing" correction $M_j^{c}=M_{j,raw}-(\text{residual})\,j/10$ removes it: $$M^{c}=[0,300,-1150,-5600,-10800,-14000,-13700,-10150,-5100,-1300,0]\ \text{t}\!\cdot\!\text{m}$$
4. Extreme (design) moment. The corrected curve peaks at station 5 (midships, $x=50\text{ m}$): $$\boxed{M_{max}=-14{,}000\text{ t}\!\cdot\!\text{m} = -137{,}340\text{ kN}\!\cdot\!\text{m}}$$ Physically, buoyancy exceeds weight amidships (stations 5–7) while weight exceeds buoyancy near the bow (stations 2–3) and stern (9–10) — net support concentrated amidships and net weight concentrated at the ends is exactly the classic hogging condition (as defined in Q1b.ii): the ends tend to droop relative to a well-supported midbody, putting the deck in compression and the bottom hull in tension.