Example # 01 A Proctor compaction test was conducted on a soil sample, and the following observations were made: Water content % 8 11.5 14.5 17.5 19.5 21.5 Mass of wet soil (kg) 1.70 1.90 2.00 1.98 1.95 1.92 If the volume of the mold used was 950 c.c., and the specific gravity of the soil was 2.65, draw the dry density versus moisture content curve, (i) Also plot the 100% and 80% saturation lines, (ii) Find the maximum dry density and the optimum moisture content, (iii) Also find the degree of saturation at OMC.
Transcript
Example # 01 A Proctor compaction test was conducted on a soil sample,
and the following observations were made:
Water content % 8 11.5 14.5 17.5 19.5 21.5
Mass of wet soil (kg) 1.70 1.90 2.00 1.98 1.95 1.92
If the volume of the mold used was 950 c.c., and the specific gravity of the soil was 2.65, draw the dry density versus moisture content curve, (i) Also plot the 100% and 80% saturation lines, (ii) Find the maximum dry density and the optimum moisture content, (iii) Also find the degree of saturation at OMC.
Solution: Table is extend to compute to d (dry density)
for each value of w, by using the relation.
cc/gmw.
MwccmoldofVolume
kginsoilwetofMasswV
Mw
d
d
11
950
11
9501000
11
1
ρ
ρρ
These experimental values of d for each value of w are computed in column (4). A graph is then plotted between the values of column (1) and column (4), so as to obtain the requisite compaction curve.
Now, for the given values of w, d values are computed by using the above equation, a graph is then plotted between column (1) and column (5), so as to obtain 100% saturation line, or Zero -Air Void line,
Using the equation, various values of d are worked out in column (06) of table above for different values of w. A graph is then plotted between values of column (1) and column (5), so as to obtain 80% saturation line.
(iii) Maximum dry density form compaction curve
= 1.84 gm/cc. Ans
33
3131652
8006521
1652 c/gmw.
.
.
.w
cm/gm.d
ρ
This maximum dry density corresponds to w = 14.5%, and hence, O.M.C is 14.5% Ans.(iii) at O.M.C w = 14.5% = 0.145, G = 2.65
d = 1.84 gm/cc
%.C.M.Oatsaturationofreedegthe,Hence
...Sor
..S.or
.
.S
..or
S..cc/gm.cc/gm.
havewe,
SGw
.G
r
rr
r
r
wd
486
8640440380
440144138084165265214501
65214501
1652841
1
ρρ
Example # 02 The following results were obtained from
a standard compaction test on a sample of soil. Water Content (%) 0.12 0.14 0.16 0.18 0.20 0.22Mass of wet soil (kg) 1.68 1.85 1.91 1.87 1.87 1.85
The volume of the mold used was 950 ml. Make necessary calculations and plot the compaction curve and obtain the maximum dry density and the optimum water content. Also calculate the void ratio, the degree of saturation and the theoretical maximum dry density (G = 2.70).
Degree of saturation 0.46 0.65 0.77 0.78 0.83 0.86
Theoretical maximum dry density
2.04 1.96 1.89 1.82 1.75 1.69
9500.M
1d
wGeρρ
ewGSr
Gw
Gmaxtheo wd
1
ρρ
The compaction curve2.00'
1.90'
1.80'
1.70'
1.60'
1.50'
1.40'
1.30'
1.20'
1.10'
0 0.10' 0.20' 0.30'
Water content From the Plot. (Pd) max = 1.74 gm/ml, O.W.C. = 15.2%
Example # 03 Laboratory compaction test results for a clayey silt are given
Moisture Content
(%)
Dry unit weight
(kN/m3)
6 14.808 17.459 18.52
11 18.912 18.514 16.9
Following are the result of a field unit weight determination test performed on the same soil by means of the sand-cone method: Calibrated dry density of Ottawa sand
= 1570 kg/m3
Calibrated mass of Ottawa sand to fill the cone = 0.545 kg Mass of jar + cone sand (before use) = 7.59 kg Mass of jar + cone sand (after use) = 4.78 kg Mass of moist soil from hole = 3.007 kg Moisture content of moist soil = 10.2%
Determine a. Dry unit weight of compaction in the field b. Relative compaction in the field
Solutiona. In the field,
Mass of sand used to fill the hole and cone = 7.59 kg – 4.78 kg =2.81 kg
Mass of sand used to fill the hole = 2.81 kg – 0.545 kg = 2.265 kg
Continue
= 20.45 kN/m3
35618
1002101
4520
1001
m/kN..
.%w
wetd
γγ
d (k
N/m
3 )20
18
16
14
0 4 8 12 16 20
19kN/m3
W (%)Plot of laboratory compaction test resultThe results of the laboratory compaction test are plotted.From the plot, we see the d(max) = 19 kN/m3.
%.
..
maxfieldR
d
d 7971000195618
γγ
Tutorial # 01
Problem # 01For the data given below (Gs 2.64 Mg/m3
a) Plot the compaction curves.b) Establish the maximum dry density and
optimum water content for each test.c) Compute the degree of saturation at the
optimum point for data in column Ad) Plot the 100% saturation (zero air voids)
curve. Also plot the 70, 80, 90% saturation curves. Plot the line of optimums.
Problem # 02 The results of standard proctor test are given in the table. Determine the d(max) and O.M.C. Also determine the moisture content required to achieve 95% of d(max)
Problem # 03 Laboratory compaction test results on a clayey silt are listed in the table.
W% Dry unit weight (kN/m3)6 14.808 17.459 18.52
11 18.912 18.514 16.9
Following are the result of field unit weight determination test on the same soil with the sand cone method: Dry density of Ottawa sand = 1570
kg/m3
Mass of Ottawa sand to fill the cone = 0.545 kg Mass of Jar+Cone+Sand(before use) = 7.59 kg Mass of Jar+Cone+Sand(after use) = 4.78 kgMass of moist soil from hole = 3.007 kg Moisture content of moist soil = 10.2%
Determine:a. Dry unit weight of compaction in the field.b. Relative compaction in the field.
