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

188 Chapter Seven
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g = acceleration of gravity = 32.2 ft per sec 2
V 2
= velocity of concrete at point 2
V 3
= velocity of concrete at point 3
V 2
= 2gh
V 3
= 0
The average velocity between points 2 and 3 is (V 2
+ V 3
)/2 =
V 2
/2.
t = y/[(V 2
)/2]
= 2y/V 2
a = (V 2
– V 3
)/t
ww.EasyEngineering.n
= (V 2
– 0)/t
= V 2
/t
( ) /t
= 2gh
Only that mass of concrete between points 2 and 3 will be undergoing
a deceleration at a given time. Only this mass need be considered
in determining the force resulting from deceleration. The weight
of this concrete will be:
w′ = wt
m = wt/g
= Wt/32.2T
When a given mass of concrete undergoes a change in velocity
between points 2 and 3, designated as an acceleration, the resulting
force F is equal to:
F = ma
Substituting for m and a, we get: 2 × 8
F = (Wt/32.2T) • ( 2gh)/
t
= ⎡ ⎤ ⎣ W 2 gh /32.2T lb (7-1)
⎦
Equation (7-1) shows that the force F depends on the original
weight of the concrete in the bucket, the vertical distance from the top
surface of the concrete in the bucket to the deck, and the time required
to empty the bucket. The vertical distance between points 2 and 3
does not affect this force. It should be noted that the force is directly
related to the rate of emptying a bucket, whereas the force is related
to the square root of the height of fall. Thus, reducing the rate of flow
is more effective than reducing the height of fall if the force is to be
reduced.
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