PRACTICAL APPLICATION.
427
Fig. 404 is the skeleton diagram of a swing bridge over a passage 100 feet
wide. P is the position of the pivot upon which the bridge turns, and
A, B, and C are the blocks which support the bridge in the closed position.
Their respective distances apart are shown in the figure.
It is, first of all, necessary to assume a value for the anticipated dead and
live loads. Let us take the former at 30 cwts. and the latter at 70 cwts. per
foot run.
50
- - 71
-4
*- — Span b-----4 -------Span a - -
Fig. 404.
1. To find the amount of hallast required.—Suppose the ballast box to
occupy a length of 16 feet at the tail-end of the bridge. Then the centre of
gravity of the counterpoise will be 42 feet from the pivot, and, by taking
moments about P,
42 B + 50 x 1| xy= 125 x 11 x -Q0
.•. B = 234 tons,
where B is the quantity in tons of ballast required. To afford a margin of
stability it will be as well to say 250 tons.
2. To find the pivot reaction —■
Bridge structure, 175 x 1|, . . . = 262^ tons.
Ballast,...........................................=250 „
Rp= 51g „
3. To find the reactions qf the bearing blocks.—It will be convenient to
consider the dead and live loads in combinations, adding the ballast later.
This admits of the taking of four cases to cover the principal dispositions of
the load :—
Case I.—Dead load throughout.
Case II. —Dead load on spån a ; dead plus live load on span b.
Case III. —Dead plus live load on span a; dead load on span b.
Case IV.—Live load throughout.
For the purpose in view it is only necessary to deal with one of these
cases. Accordingly, we will select Case II. as typical of the group.
From formula (106)—
, /3 a b\ b3
^a(a + b)= w1 a-+ 2) ~ w2fi
x 104 71\ c713
RA (104 x 175) = 1| x 1042(--^ --+ 2/~ 5“8'
RA = 54-1 tons.