Formwork for Concrete Structures by R.L.Peurifoy and G.D- By EasyEngineering.net

Downloaded From : www.EasyEngineering.net
Failures of Formwork 189
Example 7-1
Determine the resulting force when a bucket containing 3,000 lb of
concrete is emptied in 5 sec with a maximum height of fall of 4 ft.
Applying Eq. (7-1), we get:
F = ⎡ ⎤ ⎣ W 2 gh ⎦
/32.2T
= ⎡ 2 ⎤
⎣3, 000 lb 2(32.2 ft per sec )(4 ft) ⎦ /(32.2 ft per sec 2 )(5 sec)
= 300 lb
If this force acts on an area of 2 sq ft of deck, the resulting increase
in pressure, equivalent to an increase in weight, will average 150 lb
per sq ft. Assume that the decking is 1-in. nominal thickness lumber
supported by 2 × 8 S4S joists spaced 24 in. apart. The effect of this
force on the decking lumber will be a temporary increase in the
deflection and in the bending moment in the decking. Assume that
the entire force is applied to a strip of decking 12 in. wide and 24 in.
long, all of which is located between two joists. Consider this as a
simple beam 12 in. wide, ¾ in. deep, and 24 in. long, subjected to a
uniform load of 150 lb per lin ft.
The bending moment for a simple span beam with a uniform load
will be:
ww.EasyEngineering.n
From Eq. (5-5), M = wl 2 /96 in.-lb
= 150 lb per ft (24 in.) 2 /96
= 900 in.-lb (a)
Rearranging the terms in Eq. (5-10), the resisting moment
will be:
Equating (a) and (b),
M = f b
bd 2 /6
= f b
(12 in.)(3/4 in.) 2 /6
= 1.125f b
(b)
1.125f b
= 900
f b
= 800 lb per sq in.
Thus, for the stated conditions the decking will be subjected to a
temporary additional fiber stress in bending equal to 800 lb per sq in.
This is a substantial increase in stress for lumber. However, at the
time this impact force acts, the assumed live load, in excess of the
Downloaded From : www.EasyEngineering.net