Centre of pressure on a deep tank bulkhead.

The bulkhead is a vertical plane whose top edge is 12 m wide and lies at a depth equal to the head, 2 m below the waterline. The vertical heights of the bulkhead at seven equidistant points across the 12 m breadth are 0, 3, 5, 6, 5, 3 and 0 m. With six equal intervals over 12 m the spacing is h=2 m.

For a submerged vertical plane surface, the centre of pressure lies at a depth below the waterline given by

depth of centre of pressure = (second moment of area about the waterline)/(first moment of area about the waterline), i.e. d = M2/M1.

For each station across the breadth the bulkhead bottom edge lies at depth (2 + height) below the waterline, while the top edge is at depth 2 m throughout.

First-moment integrand, element about the waterline (per unit breadth), integrating from depth 2 to (2+H):

M1_element = integral of z dz from 2 to (2+H) = [(z^2)/2] = 0.5[(2+H)^2 - 4] = 2H + 0.5H^2.

Second-moment integrand:

M2_element = integral of z^2 dz from 2 to (2+H) = [(z^3)/3] = (1/3)[(2+H)^3 - 8] = 4H + 2H^2 + H^3/3.

Values of H: 0,3,5,6,5,3,0:

For M1 (2H + 0.5 H2): 0,10.5,22.5,30,22.5,10.5,0.

For M2 (4H+2H2+H3/3): 0, 39, 111.67, 168, 111.67, 39, 0.

Using Simpson's rule with spacing 2 m and 7 ordinates:

M1 = (2/3)[ (0+0) + 4(10.5+30+10.5) + 2(22.5+22.5) ]

= (2/3)[0 + 4x51 + 2x45] = (2/3)[204+90] = (2/3)(294) = 196.

M2 = (2/3)[0 + 4(39+168+39) + 2(111.67+111.67)]

= (2/3)[4x246 + 2x223.33] = (2/3)[984+446.67] = (2/3)(1430.67) = 953.8.

Depth of centre of pressure below the waterline = M2/M1 = 953.8/196 = 4.87 m.

Answer: the centre of pressure is approximately 4.87 m below the waterline. (Since the top edge is 2 m below the waterline, the centre of pressure is about 2.87 m below the top edge of the bulkhead.)