Hi,
This is an example of a problem that deals with the application of the Sand Cone Test to determine the density of the compacted soil. The solution is given in details.
Problem:
A sand cone density test was conducted on soil compaction at a land development project. It has the following data:
the volume of the standard Proctor mould = 1/30 cubic feet
Mass of the sand drained to fill the mould = 3.30 lbs
Mass of the sand to fill the cone = 0.53 lbs
Mass of the soil dug from the hole = 8.11 lbs (and have water content w = 11.11%)
Mass of jug + Sand + Cone before used to fill the hole = 10.20 lbs
Mass of jug+ Sand cone after used = 3.20 lbs
Determine: (A) The density of the Ottawa Sand(pcf)
(B) The volume of the cone (cubic feet)
(C) The volume of the hole dug + Cone (Cubic ft)
(D) The volume of the hole dug
(E) The dry density of the compacted soil
Solution (see image):
Thanks!
This is an example of a problem that deals with the application of the Sand Cone Test to determine the density of the compacted soil. The solution is given in details.
Problem:
A sand cone density test was conducted on soil compaction at a land development project. It has the following data:
the volume of the standard Proctor mould = 1/30 cubic feet
Mass of the sand drained to fill the mould = 3.30 lbs
Mass of the sand to fill the cone = 0.53 lbs
Mass of the soil dug from the hole = 8.11 lbs (and have water content w = 11.11%)
Mass of jug + Sand + Cone before used to fill the hole = 10.20 lbs
Mass of jug+ Sand cone after used = 3.20 lbs
Determine: (A) The density of the Ottawa Sand(pcf)
(B) The volume of the cone (cubic feet)
(C) The volume of the hole dug + Cone (Cubic ft)
(D) The volume of the hole dug
(E) The dry density of the compacted soil
Solution (see image):
(A)
The volume of Standard Proctor mould = 1/30 ft3. Mass in mould = 3.30 lbs hence, density = Mass/volume = 3.30 *30/1 = 99 pcf
(B)Volume of Cone = Mass of Sand Filled in Cone/ Density of Sand hence, density = 0.53/99 = 5.35 * 10-3 ft3.
(C)Mass of sand poured into the hole = 10/20 – 3.20 – 0.53 = 6.47 lbs so, volume of holw = 6.47/99 = 0.065 ft3.
(D)Volume of hole = 0.065 ft3.
(E)Dry Density = Bulk Density/(1+w) = 8.11/[0.065(1+0.111)] = 112.293 pcf
Thanks!
1. The sand cone apparatus is filled with dry uniform sand up to 5 litre mark; corresponding mass of sand is determined to be 7.72 kg. The specific gravity of sand is 2.65.
Compute: Ms,Vs,Vv,e,n,Ms,Vs,Vv,e,n, and ρdρd.