Question

Earth, whose in situ weight is 105 lb/cf and whose compacted weight is 122 lb/cf, is placed in a fill at the rate of 260 cy/hr, measured as compacted earth. The thickness of the compacted layers is 7 in. A compactor with a roller drum width of 4.16 ft is used to compact the material at a speed of 2 mph. Assume a 55-min hour operating efficiency. Determine the number of rollers.

Answer 1

Answer:

2

Explanation:

To find the solution we must first make the transformations to international units,

In this way:

$260cy / hr * 27ft ^ 3 / cy = 7020 ft ^ 3 / hr$

Inches of length (L) to feet =

$7in * ft / 12in = 0.588ft$

So we can identify the area by hour:

$A / h = 7020 / 0.588 = 12034.28 ft ^ 3 / hr$

Once the area is identified, we now identify the number of passes required,

Passes required = Area x number of passes

$P_{r} = 12034.28 * 4$

$P_{r} = 48136ft ^ 2 / hr$

We transform speed to international units

$V = 2mph * 5280ft / mi$

$V = 10560ft / hr$

So we can identify the covered area

$covered_ {area} = v * m * p$

where

m = drum widht

p = efficiency

$covered_{area} = 10560 * 4.16 * (55/60) = 40269ft ^ 2$

In this way we obtain the number of rollers given by:

No. of Rollers = Passes required / covered area

No of rollers = 48136/40269

No of rollers = 1.19, that is, 2.

Number of rollers required are 2.