Problem 5. An earthfil dam with a volume of 3,500,000 yd3 is to be constructed.
ID: 1885354 • Letter: P
Question
Problem 5. An earthfil dam with a volume of 3,500,000 yd3 is to be constructed. The specifications for the dam require that the soil be compacted to a dry unit weight of 115 lb/ft, at a moisture content of 16%. The soil at the borrow site has a specific gravity for the solids of 2.68, a degree of saturation of 44%, and a void ratio of 0.7, and a 0.2 swell factor The soil is to be transported using trucks with a 10 yd3 volume capacity, and 14.5 ton weight capacity. Assuming the water content is constant from the borrow site to the construction site, determine the volume of borrow material that must be excavated Assuming all trucks are loaded to capacity, determine the number of truckloads required to move the soil from the borrow site to the construction site. If the degree of saturation at the borrow site increased from 44% to 51%, what volume of borrow material would need to be excavated from the borrow site? A. B. C.Explanation / Answer
SOLUTION:
We know ,
Gw= eS
and, y= (G+Gw)*62.43 / 1+e
dry unit weight= unit weight/ 1+w
115= unit weight/ 1+0.16
therefore, unit weight=115* 1.16
=133.4 lb/ft3
again, 133.4= ( 2.68 + 2.68*0.16) 62.43 / 1 +e
e= 0.45 , this is the void ratio of existing dam material
1+0.45 = 350000/ volume of solids
volume of solids= 240567.94 yd3
again this volume of solids won't change for the borrow material.
so 1+e= volume/ 240567.94
1.7 * 240567.94= total volume
=408965.504 yd3 volume of excavated material.
B. 1 truck= 10yd3
no. of truckloads= 408965.504 / 10
= 40896.55
=40897 truckloads
C. if the saturation changes then the void ratio changes,
now as water content remains the same,
Gw= eS
2.68 * 0.16 = e *0.51
e=0.841
volume of solids still remain same for construction material.
so volume of borrow material
1+ e = volume/ volume of solids
1+ 0.841 = volume / 240567.94
volume= 442833.69 yd3