EM 1110-2-1100 (Part II)
30 Apr 02
EXAMPLE PROBLEM II-7-8
FIND:
The wind drag for a Beaufort 5 fresh breeze of 20 knots (10.31 m/s) at an angle of 30 deg to the wind.
Assume gustiness can be accounted for by applying a factor of 1.20 to the wind speed. Beaufort 5 is usually the
limiting wind for port operations of loading and unloading.
GIVEN:
A C9 containership is moored at a pier with an asymmetrical container distribution of 5 high in the bow half
of the vessel and none in the aft portion. The useable length of the ship for cargo handling is 246 m. The height
above the water is 10.9 m. The containers are 2.7 m tall.
SOLUTION:
The forward and aft areas exposed to the wind are
246
10.9 ' 1,340 m 2
Aaft '
2
Aforward ' 123 10.9 % 5(2.7) ' 3,000 m 2
The net effective area is then
Aeff ' 1,340 % 3,000 sin30E ' 2,170 m 2
The wind speed, including a gustiness factor, is
Vw ' 1.20 (10.31) ' 12.4 m/s
Substituting these values into Equation II-7-31, we get
1
(1.2)(1.0)(2,170)(12.4)2 ' 200 kN
Fw '
2
where we have assumed ρa = 1.2 kg/m3 at 20 EC and CD = 1.0 as a reasonable value since the mean drag coefficient
for bow and transverse winds varies between 0.82 to 1.11.
(h) Longitudinal current loads on ships are taken from procedures by NAVFAC Design Manual DM26.5.
The total longitudinal current load Fc,tot is composed of form drag Fc,form, skin friction Fc,fric, and propeller
Fc,prop drag components. All three components have similar equations.
Fc , tot ' Fc , form % Fc , fric % Fc , prop
(II-7-33)
(i) Form drag is due to the flow of water past the vessel's cross-sectional area and is defined as
1
2
Fc , form ' &
(II-7-34)
ρw Cc , form B T Vc cos θc
2
Harbor Hydrodynamics
II-7-71
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