COLD REGIONS TECHNICAL DIGEST NO. 96-1
28
ni3 / 2
(0.03)3/2
(24.5) = 1.59 ft
yt =
yice =
3/ 2
3/2
(0.22) + (0.03)
nb + ni
3/2
3/ 2
The shear stress on the underside of the ice is computed using eq 2:
fw = γwater yiceSf5B = (62.4)(15.9)(0.000053)(5)(900) = 237 lb/ft
The wind drag forces are then determined from eq 3, assuming a
critical wind of 25 mi/h (or 37 ft/s) in the direction of the river
flow. Using a drag coefficient Cd of 2.2 103 and a mass density
of air ρair of 2.5 103 slugs /ft3:
fa = CdρairU25B = (2.2103)(2.5103)(37)2(5)(900) = 34 lb/ft
Equation 4 gives the gravity forces exerted by the ice cover on a
unit width of boom:
fg = yice(1e)tiSf5B = (57.4)(10.5)(0.75)(0.000053)(5)(900) = 5 lb/ft
With a 0.75-ft-thick ice cover, the ice bottom will be roughly even
with the bottom of a typical boom unit. Water drag on the boom
unit is neglected in this design case.
For impact loading, it is assumed that the boom units displace
an average distance of 10 ft while stopping a 2-acre 1-ft-thick
ice floe moving with an initial velocity equal to the average cur-
rent velocity at 2.3 ft/s. The impact force fi is considered sepa-
rately, since a large floe would probably not impact the boom if
an ice cover is already in place. From eq 6b:
mice ∆V 2 2 (43, 560) (1) (0.92) (1.93) (2.3)2
fi =
=
= 40, 900 lb
2 (10)
2d
If the floe is roughly square in shape, the length of one side would
be approximately 300 ft. Distributing the impact load over this
length gives:
40, 900 lb
= 136 lb/ft along the contact area
300 ft
Neglecting vessel-induced forces, water drag on the boom unit,
and the ice impact forces, the total force acting on the boom fb is
the sum of these forces:
fb = fw + fa + fg = 237 + 34 + 5 = 276 lb/ft
(say 280 lb/ft)
Previous Page
