2 An earthfill dam requires a compacted volume of 35 million

2. An earth-fill dam requires a compacted volume of 3.5 million yd of fill. The soil to be used is wet clay (see properties below). Determine amount of loose soil that must be hauled to construct the dam. TABLE 4.4 Representative properties of earth and rock Bank weight Loose weight Percent Swell lb/cy kg/m3 Ib/cy kg/m3swell factor 0.74 0.74 0.80 0.80 0.83 0.89 0.88 0.63 0.63 0.87 0.87 0.71 Material Clay, dry Clay, wet Earth, dry Earth, wet 2,700 1,600 2,000 1,185 3,000 ,780 2,200 1,305 2,800 1,660 2,240 1,325 3,200 1,8952,580 1,528 35 35 25 25 Earth and gravel 3,200 1,895 2,600 1,575 20 12 14 60 60 15 15 40 2,800 1,660 2,490 1,475 3,400 2,020 2,980 1,765 4,400 2,610 2,750 1,630 Rock, well blasted 4,200 2,490 2,640 1,565 2,600 1,542 2,260 1,340 2,700 1,600 2,360 1,400 3,500 2,075 2,480 1,470 Gravel, dry Gravel, wet Limestone and y olasted Sand, dry Sand, wet Shale The swell factor is equal to the loose weight divided by the bank weight per unit volume. The wet clay soil in Problem Number 2 is to be hauled in a truck which can handle a heaped load of 9 cy3. Assuming weight is not a problem, determine (a) the Shrinkage Factor for the soil, and (b) the amount of in-place soil the truck can carry per trip. 3.

Solution

Given : Shrinkage % = 15.3

Shrinkage factor = 1- 0.153 = 0.847

By definition : Compacted volume = Bank volume x Shrinkage factor

Given compacted vloume as 3.5 million unit

therefore, 3.5 million = Bank volume x 0.847

Bank volume = 4.132 million units

therefore Volume of haul (loose measure) = Bank volume x swell factor

for wet clay, volume of haul = 4.132 x 0.74 (from table)

vloume of haul = 3.057 units

2. a. Shrinkage factor = 0.847

b. in-place soil is bank soil