#Jameel_Academy_Soil_Tests

1. University of Duhok
College of Engineering
Civil Department
Soil Mechanics – Practical
Soil Laboratory
A Report About :
Standard Compaction Test
Submitted By : Jameel Masoud
Lab. No. : 1
Group Name : ----
Testing Date : 28 / 1 / 2019
Submitting Date : 13 / 2 / 2019
Supervised By : Hussein
2018 – 2019

2. Introduction
Soil compaction consists of closely packing the thus increasing the soil dry unit
weight. compaction on water content and has no effect compaction process
should remove the air fraction completely. However, in practice, compaction
cannot completely eliminate the air fraction, but only reduces it to a minimum,
provided that appropriate techniques are used.
Purpose:
To determine the relation between water content and dry unit weight and to find
the maximum dry density and optimum moisture content. Also, improve the
engineering properties of soil mass. Its important because it will:
a) Increase in shear strength.
b) Increase in density.
c) Decrease in shrinkage.
d) Decrease in permeability.
e) Decrease in settlement (compressibility).
f) Decrease void ratio.
Supplies and materials:
1- Compaction device standard (Cylindrical metal mold, internal dimension 4
inch in diameter and 4.6-inch-high (volume 1/30 ft'^3 or 944 cm^3). Removable
mold collar 2.5" high and 4" diameter, Hammer 2" diameter face, 5.5 lb weight).
2- Balance. (Mechanical or Electrical).
3- Moisture cans.
4- Drying oven.
5- Large mixing pan.
6- Sieve No. 4.
7- Straight edge and knife.
8- Graduated cylinder.
9- Sample of soil of different water content (3 kg).

3. Sketch:
Compaction Device Moisture Can Sensitive Balance
Oven Dry
Large mixing Pan
Sieve No. 4 Soil Sample Graduated Cylinder

4. Test Procedure
1-The original bulk sample is air dried and weighed. The large particles are
removed by passing the sample through a NO. 4 sieve. The mass of
material required for the test is about 3 kg.
2- Clean and dry the mold, base and extension collar. Determine the weight
of compaction mold with its base (without the collar). Measure the
compaction mold to determine its volume (assume the volume is 1000
cm3
)
3-Compute the amount of initial water to add by the following method :
(a) Assume water content for the first test to be 8 percent
(b) Compute water ta add from the following equation:
Water to add (in ml) = (soil mass in grams × 8)/ 100
- Where "water to add" and the "soil mass" are in grams.
4-Measure out the water, add it to the soil, and then mix it thoroughly into
the soil using the trowel until the soil gets a uniform color.
5- Place the mold assembly on a solid base such as a concrete floor Add
loose soil to the mold so that it is about half full. Compact the soil by
applying 25 blows of the rammer dropped from controlled height of 300
mm. The drops should be applied at a uniform rate not exceeding around
15 seconds per drop, and the rammer should provide uniform coverage of
the specimen surface. Try to avoid rebound of the rammer from the top of
the guide sleeve. To avoid injury, the hand that holds the tube must be
kept dear from the falling hammer.
6- Place a second, app approximately equal layer of soil in the mold, and
compact it with 25 blows as before. Repeat with a third layer, which
should bring the compacted soil level in the extension collar to about 6
mm above the level of the mold body. If the compacted soil level in the
extension collar is much higher, the test becomes inaccurate because the
compacting energy per unit volume of soil is no longer constant (Note:
For the last layer, watch carefully, and add more sol after about 10 drops
if it appears that the soil will be compacted below the collar joint).

5. 7-Remove the extension collar careful y. Cut away the excess soil and level
off to the top of the mold. Any small cavity resulting from the removal of
stones should be filled with fine materials.
8-Weigh the compacted soil while it's in the mold and to the base, and record
the mass. Determine the wet mass of the soil by subtracting the weight of
the mold and base.
9-Remove the soil from the mold using a mechanical extruder and take soil
moisture content samples from the top and bottom of the specimen. Fill
the moisture cans with soil and determine the water content.
10- Place the soil specimen in the large tray and break up the soil until it
appears visually as if it will pass through the # 4 sieve, add 2 percent more
water based on the original sample mass, and re-mix as in step 4. Repeat
steps 5 through 9 until, based on wet mass, a peak value is reached
followed by two slightly lesser compacted soil masses (the procedures are
carried out to obtain five compaction points).
Calculations and Results :
To determine the Max. Dry Density, we need to determine the moisture content as
follow:
For Example: Wt of can = 22.57 g ,Wt of can + wet soil = 93.8 g
Wt of can + dry soil = 88.64 g
Moisture content (ω) =
Wt of water
Wt of dry soil
=
93.8 − 88.64
88.64 − 22.57
× 100 = 7.81%
- Moisture content for all sample can be calculated by using the above equation and
make average between them as shown in table below:

6.  As shown in table we have five sample and each sample have two moisture
content therefore we make average between them to get the correct one.
Now, we can calculate the dry unit weight:
For Example : 𝜔 = 7.81 % , Wt of mold + Soil = 3757.2 g
Weight of mold = 1933 g ,Volume of mold =1000 𝑐𝑚3
, Gs = 2.65
𝑊t of Soil in Mold = 3757.2 – 1933 = 1824.2 g
Wet unit weight (γ𝑡) =
weight of soil in mold
volume of mold
=
1824.2
1000
× 9.81 = 17.895
kN
m3
Dry unit weight (γ 𝑑) =
γ 𝑡
1+𝜔
=
17.895
1+0.0781
= 16.592
kN
m3
Now, the line of zero void will be sketched by using this formula:
If we use 𝜔 = 7.81 % as shown the Zero void Unit weight will be :
Moisture content determination
Sample no. 1 2 3 4 5
Can no. A1 A2 A3 A4 A5 A6 A7 A8 A9 A10
Weight of
can (g)
22.57 22.02 22.00 22.32 17.85 17.80 17.32 17.46 17.21 17.25
Weight of
can + wet
soil (g)
93.80 98.24 93.96 101.76 97.62 88.47 95.22 102.53 100.65 97.07
Weight of
can + dry
soil (g)
88.64 92.66 87.37 94.46 89.00 80.94 85.47 91.83 88.83 85.68
Water
content (%)
7.81 7.90 10.08 10.12 12.12 11.93 14.31 14.39 16.50 16.64
Average
water
content (%)
7.85 10.10 12.02 14.35 16.57

7. 𝛾𝑧𝑎𝑣 =
𝛾 𝑤 × 𝐺𝑠
1 + 𝐺𝑠 × 𝜔
=
9.81 × 2.65
1 + 2.65 × 0.0781
= 21.518
kN
m3
There is a tabulate of our sample:
- The important one is the Compaction Curve and zero void Line:
- The Max. Dry Unit Weight and Optimum moisture
content will be (19.1 kN/m^3), (12.2 %) Respectively.
Unit Weight Determination
Water content (%) 7.855 10.100 12.021 14.347 16.574
weight of soil + mold (g) 3757.200 3921.400 4109.000 4082.500 4015.200
weight of mold (g) 1933.000 1933.000 1933.000 1933.000 1933.000
weight of soil in mold (g) 1824.200 1988.400 2176.000 2149.500 2082.200
wet unit weight (kN/m^3) 17.895 19.506 21.347 21.087 20.426
dry unit weight (kN/m^3) 16.592 17.717 19.056 18.441 17.522
dry unit weight (Zero void)
(kN/m^3)
21.518 20.508 19.716 18.835 18.063
15
16
17
18
19
20
21
22
6 8 10 12 14 16 18
DryUnitWeight(kN/m^3)
Moisture Content %
19.1
12.2

8. Discussion and Conclusion :
In this test we determine the Dry unit weight and moisture content of 5
sample which gives us the Maximum Dry Density and Optimum
moisture content, finally we get a result after doing all procedure
accurate which is (𝛾 𝑚𝑎𝑥 = 19.1
𝑘𝑁
𝑚3 and OMC = 12.2 %) , according to
standard and the field compaction we can compare our sample’s max.
density with Field density to know is the field compaction reach the
acceptable level of compaction or not. this result shows us with increase
Water content to a limit, the dry density will be increased but if passed
from limited (OMC) it will decrease the density because the solid soil
volume will replace with water volume (density of water less than
density of soil) therefore it decreases the density. The errors that maybe
occur during the test maybe in the compacting of sample during 25 drop
of un equal drop or not distributed over the sample area or insufficient
of layer thickness when filling the mold to 3 or 5 layer, or maybe in the
mold connector sometimes if not properly connected it will have
damaged. we know that if we do the test on many sample and then get
the average, the result will be very accurately.
In conclusion the purpose of this test is to determine the Max. dry density
and the corresponding moisture content which is OMC and compare it
with the field compaction, now we can’t compare the field compaction
because the field compaction is not tested therefore the relative density
not exist until we get the field compaction from another test. finally, we
know how to determine the lab compaction and how to check the field
compaction by using the relative density.

9. Q) what's the difference between compaction and consolidation?
A) As shown in table :
Q) What is the influence of compaction effort on compaction curve ?
A) the compaction effort depend on (number of layer, weight of hammer
, falling of hammer and number of drop) with increase of each of the following,
it will increase compaction effort it means that the curve of compaction will
gave us larger dry density with less moisture content.
Consolidation Compaction
Consolidation is a process where steady and static
pressure causes compression of saturated soil.
Compaction is a process where a
mechanical pressure is used to
compress the soil mass for the purpose
of soil improvement.
Static load will applied Dynamic load will applied
it reduce the volume by removing pore water
It reduce the volume by removing air
void
Used in clayey soil Used in sandy soil
It’s a natural process Its mechanical process