Soil Comp Action Test report by yasas89

March 9, 2018 | Author: yasas89 | Category: Density, Soil, Physical Quantities, Physics, Physics & Mathematics
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SOIL COMPACTION TEST Introduction In general, the soil bearing capacity will be increased simultaneously with the increment of density or unit weight of the soil. The increment of soil density can be obtained via compaction process i.e. the process of mechanically reducing the air void. Objective Determine the maximum dry density at the optimum moisture content under laboratory condition. Theory The test consists of compacting the soil or aggregate to be tested into a standard mould using a standardized compactive energy at several different levels of moisture content. The maximum dry density and optimum moisture content is determined from the results of the test. Soil in place is tested for in-place dry bulk density, and the result is divided by the maximum dry density to obtain a relative compaction for the soil in place. In the other hand, soil compaction test is carried out in the laboratory in determining the ideal volume of water to be poured while compaction the soil on site so that the required compaction degree can be obtained. The important characteristics of soil compacted with an ideal compaction degree are: a) High shear strength b) Low permeability coefficient and capacity c) Reduce settlement when additional load is applied The moisture content recorded when the maximum dry unit weight is achieved is known as the optimum moisture content. There are two types of compaction i.e.: 1. Standard Proctor 2. Modified Proctor Standard Proctor will be used in undertaking the experiment where the standard data are recorded as the following:

Standard Proctor

Volume of mould

Mass of Hammer

Drop Distance

No. of Blows Per Layer

No. of Layer

944 cm3

2.5 kg

305 mm

25

3

Equipment a) Sieve 5.0 mm b) Weighing machine c) Empty mould with inner diameter of 101.6 mm, inner height of 16.43 mm and volume of 944 cm together with the base plate d) Hammer with diameter of 50 mm and mass of 25 kg e) Other equipment in determining the soil moisture content Procedures 1 Small soil sample from jobsite collected. 2 5kg of dry soil passing through 4.75mm sieve opening prepared. 3 Empty mould, collar and base plate weighed. The empty containers also weighed. 4 The sample mixed thoroughly with approximately 9% water of the total soil volume. 5 The soil sample divided into three sections. 6 The any first section placed in the mould and compacted. Distribute 25 blows uniformly over the surface and ensure that rammer always falls freely and is not obstructed by soil in the guide tube. 7 The second section place into the mould and being compacted followed by the last section of soil sample. 8 The attached collar removed when the compaction completes. 9 The compacted soil trimmed using the straightedge until it is even with the top of the mould. 10 Small amount of soil from the upper mould taken and placed into a container. 11 Then the container with the soil sample being weighed. 12 Small amount of soil from the bottom mould taken and placed into a container. 13 Then the container with the soil sample being weighed. 14 Both containers placed into an oven to determine its moisture contain. 15 The compacted soil’s sample unit weight determined by dividing the weight of the compacted soil in the mould with the soil sample volume (volume of the mould). 16 The experiment repeated with three varying water content (12%, 15% and 18%).

17 The dry density computed by using the compacted soil’s wet (bulk) density and the moisture content known. 18 The soil’s dry density versus moisture content graph plotted. Data Table 1 No. of Test

1 (9.5%)

2 (12.5%)

Mass of empty mould (kg)

5.305

4.175

Mass of mould + wet soil(kg)

7.015

5.935

Mass of wet soil, M (kg)

1.710

1.760

Volume of mould, V (m3)

9.433x10-4

9.433x10-4

Bulk density (ρ) = M/V (kg/m3)

1812.785

1865.790

Dry density, (ρ) = ρ b / (1+m) (kg/m3)

1226.512

1404.962

3 (15.5%)

4 (18.5%)

6.385

5.200

8.270

7.150

1.885

1.950

9.433x10-4

9.433x10-4

1998.304

2067.211

569.480

421.622

Table 2 No. of Container

1(9.5%)

2(12.5%)

3(15.5%)

4(18.5%)

Upp er

Botto Upp m er

Botto Upp m er

Botto Upp m er

Botto m

Mass of empty container (g)

16.5 4

16.00

15.97

15.90

16.40

16.48

15.89

16.68

Mass of empty container + wet soil (g)

46.5 0

45.99

56.06

63.56

38.79

39.43

69.30

57.77

Mass of container + dry soil (g)

46.3 7

45.84

55.94

63.39

38.26

38.85

67.42

56.13

Mass of water, Mw (g)

0.13

0.15

0.12

0.17

0.53

0.58

1.88

1.64

Mass of dry soil, Ms (g)

29.8 3

29.84

39.97

47.49

21.86

22.37

51.53

39.45

Moisture content, m =Mw/ Ms (%)

0.43 6

0.520

0.300

0.356

2.425

2.593

3.648

4.157

Average Moisture content (%)

0.478

0.328

2.509

3.903

Gs = 2.70 γw = 1000kg/m3

Table 3 A= 0%

A= 5%

M%

10

15

20

25

10

15

20

25

Ρd(kg/m 3 )

2125.9 84

1921.7 08

1753.2 47

1611.94 0

2019.68 5

1825.62 3

1665.58 4

1531.3 43

A= 10% M%

10

15

20

25

ρd(kg/m3)

1913.386

1729.537

1577.922

1450.746

Calculation Example for test no 1 From table 1: Mass of wet soil (M)

= 1.710 kg

Volume of mould (V)

= 9.433x10-4 m 3

Bulk density, ρb = (M/ V)

= 1.710 /9.433x10-4

Dry density, ρd = ρpb / (1+m)

= 1812.785 /(1+0.478)

= 1812.785 kg/ m 3

= 1226.512 kg/m3

From table 2: Moisture content,m

= 0.478%

Mass of water, Mw (g)

= Mass of empty container + wet soil (g) - Mass of container + dry soil (g) = 46.50-46.37 = 0.13 g

Mass of dry soil, Ms (g) empty container (g)

= Mass of container + dry soil (g) - Mass of

= 46.37 - 16.54 = 29.83 g From table 3: When A =0% Dry density, ρd

= pw (1-A)/ (1/Gs +m) = 1000(1-0)/ (1/2.7+0.1) = 2125.984

Result Maximum Density (ρdry) = Optimum moisture content (V) =

Discussion

Conclusion Daripada ujikaji yang dijalankan, nilai ketumpatan maksimum dan nilai kelembapan optimum yang diperolehi ialah masing masing. Oleh itu, objektif ujikaji ini iaitu untuk menentukan ketumpatan kering yang maksimum pada kandungan kelembapan optimum dibawah keadaan makmal tercapai.

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