COLD REGIONS TECHNICAL DIGEST NO. 96-1
30
The horizontal load on the midstream anchors Tanchor will be
the load acting on a tributary width of half of the wire rope span
on either side of the anchor:
Tanchor = fb(tributary width) = 280(168)(0.5 + 0.5) = 47,040 lb
Use a 1-1/2-in. 6 19 GIPS-FC wire rope with a breaking capac-
ity of 82.8 tons. With a depth of 25 ft and a scope of 7 horizontal
to 1 vertical, the anchor cable length would be 175 ft.
Floats must be designed to support the half of the weight of the
anchor cable and adjacent boom cables, as well as the weight of
the junction plates and connections. Figures 1 and 9 show typical
float arrangements. Table 2 gives a weight of 3.18 lb per lineal
foot for 1-3/8 in. 9 16 GIPS-FC wire rope. The boom cable
weight is then:
205 ft 3.78 lb/ft = 775 lb
and the anchor cable weighs:
175 ft 3.78 lb/ft = 662 lb
Assume the junction plate with wire rope connections weighs
100 lb and the float weighs 150 lb. The float at each junction must
supply a buoyant force (Fb) adequate to support its own weight,
the submerged weight of the junction plate and half the sub-
merged weight of the anchor cable and adjacent boom cables:
γ -- γ a tr
γ
γ
f 150 + steel wwtaeer1100+ (6662) + (7775) + (7775 = 1200bb
11
11
11
00 + ( 62) + ( 75) + ( 5) 120 l l
)=
fbb== 150 + steel
γ γtstelel
22
22
22
s ee
where γsteel = 490 lb/ft3 and γwater = 62.4 lb/ft3.
The volume of water (Vw) displaced to provide this buoyant force is
Ww 1200
Vw =
=
= 19.2 ft3
γ w 62.4
Assuming that 40% of the float's volume is submerged un-
der static conditions, the total volume of the float must be 19.2
ft3/40%, or 48 ft3. A 3-ft-diam. 6-ft-long cylindrical float, with
a volume of 42.4 ft3 would work in this case.
Conventional ice boom units are generally 2030 ft long, with
a spacing between units of 36 times the ice thickness. For each
205-ft-long main span cable, eight 20-ft-long boom units with
5.0- ft gaps between units would work well. A smaller gap width
could be achieved by using nine 20-ft booms with 2.5-ft gaps.