When a tank is part-full the free liquid surface is able to move as the ship heels, transferring liquid to the low side. This makes the centre of gravity of the tank's contents shift towards the low side, creating an additional heeling (virtually raising the centre of gravity) and thereby reducing the effective metacentric height. The loss of GM is:

reduction in GM = rho_liquid x i / Delta

where i is the second moment of area of the free liquid surface about its longitudinal centroidal axis (i = L B^3/12 for a rectangular surface) and Delta the ship displacement. The free-surface effect:

• is independent of the quantity of liquid, depending only on the shape (area and breadth) of the free surface, so even a small amount can be damaging;
• is proportional to the cube of the tank breadth, so wide tanks have the largest effect;
• is eliminated when the tank is completely full (pressed up) or completely empty;
• is a virtual loss of GM and does not affect the true KG, appearing as a reduction of effective GM and of righting levers (GZ).

It must therefore be taken into account in the metacentric and cross-curves, usually by reducing GM by the free-surface correction. In practice tanks are divided with longitudinal bulkheads (wash/centreline bulkheads) to reduce the breadth i, and press up or strip out tanks when the free surface is significant.

Box barge 45 m long, 15 m wide, floats at a level keel draught of 2 m in sea water, load uniformly distributed over the full length. Two 30 tonne masses are loaded at 10 m from each end, and 50 tonne distributed evenly between them.

Buoyancy per metre, uniform over the full length: the box displaces LB.d.rho = 45x15x2 = 1350 m3 = 1350 x 1.025 = 1383.75 t. Buoyancy per metre b = 1383.75/45 = 30.75 t/m.

Weights:

Uniform (still floating hull and level-keel load) weight per metre w0 = 30.75 t/m (to give even keel before adding the point masses). The three added loads are then point loads.

The two 30 t masses at 10 m from ends and the 50 t at the middle (evenly distributed across the central 25 m length at 50/25 = 2 t/m between x=10m and x=35m). Shear force is the algebraic sum of the net load (weight - buoyancy) to one side.

For x from 0 to 10 m: only the uniform base load acts; since w0 balances b with no net load, the shear is essentially from the base alone - shear = 0 over the ends and the net jumps occur at the point loads.

Treated as a simply supported beam on uniform buoyancy with load diagram:

Net point loads cause shear jumps: at x=10 m a downward 30 t (jump in SF = -30 t); distributed 2 t/m over 10-35 m adds slope; at 35 m end of distributed load the accumulated shear returns and the symmetric load at 45-10=35m balances.

By symmetry, the maximum shear force occurs adjacent to one of the 30 t end masses. Consider the left side: total buoyancy of the left 10 m strip = 10 x 30.75 = 307.5 t; total weight of left 10 m (uniform base) = 10 x 30.75 = 307.5 t, so they exactly balance and the shear due to the base over the outer 10 m is zero. The net load diagram is therefore a series of point/partial loads sitting on an exactly balanced uniform basis.

Loaded mass summary: base uniform load = buoyancy (balanced). Added: 30 t end loads and a 50 t uniform load over the middle 25 m at 2 t/m.

For the left half (symmetry axis at 22.5 m): total added weight left of mid = 30 (at x=10) + 2 x 12.5 (50 t spread to mid) = 30+25 = 55 t. The reaction portion of buoyancy on the left half = 1383.75/2 = 691.9 t, of which the base weight matches. The point 30 t at 10 m therefore creates a shear to its right of SF = -30 t, and the 2 t/m distributed load adds -2 per metre, so just left of midspan the shear is -30 - 2x12.5 = -55 t.

By symmetry the maximum shear force occurs at the ends of the distributed 50 t section, and equals 30 + 50/2 = 55 t. Hence maximum shear force = 55 tonnes (55 t). Sketch: the shear-force diagram shows a step of -30 t at x=10 m, then a linearly falling line over x=10 to 35 m at -2 t/m reaching -55 t at midspan, and a mirror image rising on the after half -50-?with symmetric positive values, maximum 55 t.