Non-watertight longitudinal subdivision (longitudinal bulkheads or wash bulkheads that are not watertight) of tanks serves to:
- Reduce the free-surface effect: by dividing a wide tank into narrower compartments, the second moment of area of the free surface (i = L B^3/12) is greatly reduced, so the free-surface loss of GM is reduced.
- Reduce the sloshing of the liquid: the wash bulkheads damp the movement of the liquid in a partially full tank, reducing the dynamic loads on the tank structure and the effect on stability.
- Provide structural support: the longitudinal bulkheads act as girders that stiffen the tank and the hull, and support the deck and bottom.
- Reduce the free-surface effect during ballasting and cargo operations.
The subdivision is non-watertight so that the liquid can flow between the compartments (for filling/emptying and for equalising), while still reducing the free-surface and sloshing effects.
Ship 160 m long, 8,700 t displacement, floats at a waterline with half-ordinates 0, 2.4, 5.0, 7.3, 7.9, 8.0, 8.0, 7.7, 5.5, 2.8 and 0 m (at 16 m intervals, 10 intervals = 160 m). The ship develops a list of 10 deg due to instability. Calculate the negative metacentric height when the vessel is upright.
Waterplane area: A = 2 x (h/3)[y0 + y10 + 4(y1+y3+y5+y7+y9) + 2(y2+y4+y6+y8)]
= 2 x (16/3)[0 + 0 + 4(2.4+7.3+8.0+7.7+2.8) + 2(5.0+7.9+8.0+5.5)]
= 2 x 5.333[4 x 28.2 + 2 x 26.4] = 10.667[112.8 + 52.8] = 10.667 x 165.6 = 1766.4 m2.
Second moment of area about the longitudinal axis (transverse stability): I = 2 x (h/3) x sum of (weighted y^3).
y^3 values: 0, 13.824, 125, 389.017, 493.039, 512, 512, 456.533, 166.375, 21.952, 0.
weighted sum = 1x0 + 4x13.824 + 2x125 + 4x389.017 + 2x493.039 + 4x512 + 2x512 + 4x456.533 + 2x166.375 + 4x21.952 + 1x0
= 0 + 55.3 + 250 + 1556.1 + 986.1 + 2048 + 1024 + 1826.1 + 332.8 + 87.8 + 0 = 8166.2.
I = 2 x (16/3) x 8166.2 = 10.667 x 8166.2 = 87,105 m4.
Volume of displacement V = 8700/1.025 = 8487.8 m3.
BM = I/V = 87,105/8487.8 = 10.26 m.
At the angle of loll (10 deg) the righting lever is zero. For a wall-sided ship, tan(theta) = sqrt(-2 GM/BM):
tan(10 deg) = 0.1763, tan^2 = 0.0311.
-2 GM/BM = 0.0311, so GM = -0.0311 x 10.26/2 = -0.1596 m.
Answer: the negative metacentric height is about -0.16 m.