11
DESIGN OF ICE BOOMS
cDhs ρw V 2
fd =
(5)
2
where cD = drag coefficient for the submerged portion of the
boom unit (1.52.0 for rectangular shapes)
hs = submerged depth of boom unit
ρw = density of water (1.93 slugs/ft3, 1000 kg/m3)
V = average water velocity.
Impact load from
In general, the impact loads on an ice boom from a broken
broken cover
cover are low. An exception occurs when an intact sheet upstream
of the boom is released in the form of floes that can be as large as
several acres. In most cases, large floes submerge the boom units
and override the ice boom. Relatively large forces can act on the
boom before it submerges, however. The impact of a large ice floe
typically does not occur when any of the other loads are present,
since large floes will be free to hit the boom only when the river is
relatively free of other ice. Impact loads may be distributed along
the length of the boom or analyzed as point loads. Distributed
over the boom width, the impact force can be expressed as:
mice ∆V
fi =
(6a)
L∆t
where L = length of the boom, perpendicular to the direction of
flow.
mice = mass of the ice floe
∆V = change in floe velocity, usually considered to end up
at rest
∆t = estimated time for the ice floe to come to rest.
Estimating the time required for the floe to come to rest may be
difficult. If the designer has an idea of the boom's compliance, an
energy transfer approach can be used to estimate impact force per
unit width of river:
mice ∆V 2
E = fi ∆x =
2L
where E is the energy transferred to the boom in the process of
stopping the floe, and ∆x is the distance required to stop the floe.
If we rearrange the equation, the impact force can be expressed
as:
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