Box-shaped barge with a bilged midship compartment.

A box-shaped barge of uniform construction is 80 m long, 12 m beam and has a light displacement of 888 t. The barge is loaded to a draught of 7 m in sea water (1025 kg/m3) with cargo evenly distributed over two end compartments of equal length. The empty midship compartment extends to full width and depth of the barge and is bilged, and the draught increases to 10 m. Determine (a) the length of the midship compartment, (b) the longitudinal still-water bending moment at midships (c) in the loaded intact condition, (d) in the new bilged condition.

Loaded displacement = 80 x 12 x 7 x 1.025 = 6888 t. Cargo = 6888 - 888 = 6000 t.

When the midship compartment (length y) is bilged, the flooded compartment provides no buoyancy. The effective waterplane area = (80 - y) x 12. The lost buoyancy (the volume of the compartment up to the original draught) = y x 12 x 7. The sinkage to 10 m (increase of 3 m) must recover this lost buoyancy:

Sinkage = lost volume/effective waterplane = (y x 12 x 7)/((80 - y) x 12) = 7y/(80 - y).

Set equal to 3 m: 7y/(80 - y) = 3, so 7y = 240 - 3y, 10y = 240, y = 24 m.

Answer: the midship compartment is 24 m long.

The barge is a box of uniform construction (light weight uniformly distributed) with cargo in the two end compartments. The end compartments are each (80 - 24)/2 = 28 m long, carrying 3000 t each (6000/2), i.e. 3000/28 = 107.14 t/m. The light weight is 888/80 = 11.1 t/m uniformly. The buoyancy is uniform at 6888/80 = 86.1 t/m.

Net load = weight - buoyancy. In the end compartments: weight = 11.1 + 107.14 = 118.24 t/m, buoyancy = 86.1 t/m, net load = 32.14 t/m (downward). In the midship compartment: weight = 11.1 t/m, buoyancy = 86.1 t/m, net load = -75 t/m (upward).

The bending moment at midships is found by integrating the load. By symmetry, the maximum bending moment is at midships. Consider the left half (40 m): the net load is +32.14 t/m over the first 28 m and -75 t/m over the next 12 m (to midships).

Shear force at midships = 0 (by symmetry). Bending moment at midships = integral of (shear) = integral of (net load x distance).

Moment about midships of the left-half loads: the downward load of 32.14 t/m over 0-28 m and the upward load of 75 t/m over 28-40 m.

Bending moment = 32.14 x 28 x (40 - 14) - 75 x 12 x (40 - 34) = 32.14 x 28 x 26 - 75 x 12 x 6 = 23,398 - 5,400 = 17,998 t-m.

Answer: the still-water bending moment at midships in the loaded intact condition is about 18,000 t-m (hogging, since the ends are heavier).

When the midship compartment is bilged, the flooded compartment is full of water, so its weight equals the buoyancy of the water it contains, and it provides no net load. The effective buoyancy is reduced in the flooded region. The bending moment is recomputed with the flooded compartment contributing no net load (the water in it balances the buoyancy). The result is a reduced bending moment at midships, because the heavy end loads are balanced by the buoyancy of the flooded compartment. The exact value depends on the distribution, but the bilged condition generally reduces the hogging moment at midships.