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

190 Chapter Seven
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weight of the concrete slab, is not acting on the slab, which provides
some factor of safety, or relief.
Consider the effect this force may have in increasing the unit
stress in horizontal shear in a joist. Assume that the full force acts on
the deck at one end of a single joist. This will produce a temporary
increase of approximately 300 lb in the vertical shear in that end of
the joist.
V = 300 lb
(a)
Rearranging the terms in Eq. (5-13), the shear force will be:
V = 2f v
bd/3
(b)
Equating (a) and (b), we get:
2f v
bd/3 = 300
f v
= 3 × 300/2bd
= 3(300 lb)/[2 (1½ in.) × (7¼ in.)]
= 41.4 lb per sq in.
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Motor-Driven Concrete Buggies
The current method of placing concrete is with buckets from a crane
or pumping from a concrete pump truck. However, in the past,
motor-driven concrete buggies were frequently used for placing concrete.
Although concrete buggies are not commonly used today, the
material contained in this section is presented for those isolated situations
where buggies may be used.
The weight of a loaded buggy can be as much as 3,000 lb. Some
states limited the speed of buggies to a maximum 12 mi/hr. If the
buggies are stopped quickly, substantial horizontal forces can be produced.
It is possible that the thrusts resulting from these horizontal
forces can exceed the resisting strength of horizontal and diagonal
bracing systems.
Consider the horizontal force that may be caused by a loaded
motor-driven buggy that is stopped quickly, based on the following
assumptions and conditions:
Weight of loaded buggy = 3,000 lb
Maximum speed, 10 mi/hr = 14.7 ft per sec
Assume buggy is stopped in 5 sec
The resulting force is given by the following equation:
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