6. Compaction (Das, chapter 6)

1

What is Compaction?

 Compaction- Densification of soil by reducing air voids by application of mechanical energy.  Compaction is used in construction of highway embankments, earth and many other engineering structures, loose must be compacted to improve their strength by increasing their unit weight  The degree of compaction is measured in terms of its dry unit weight.

Air Air Compaction Water Water

Solid Solid

2 Purpose of Compaction

 Increases of soils  Increases the of foundations  Decreases the undesirable settlement of structures  Reduction in  Increasing the stability of slopes on embankments.

3

Role of Water in Compaction

 Water is added to the soil during compaction which acts as a lubricating agent on the soil particles. The soil particles slip over each other and move into a densely packed position.

 In soils, compaction is a function of o Consider 0% moisture -Only compact so much o Add a little water - compacts better o A little more water -a little better compaction o Even more water – ???

4 Role of Water in Compaction o For a given soil, the dry unit weight increases as water is added to the soil. This continues up to certain moisture content. o Beyond this moisture content, more water added will fill the void space with water so further compaction is not possible. o The moisture content at which the maximum dry unit weight is attained is generally referred to as the optimum moisture content.

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Laboratory Compaction tests

 Purpose: Find the maximum dry unit weight and corresponding optimum moisture content.

 The standard was originally developed to simulate field compaction in the laboratory

 There are two types of standard tests (ASTM – 698) : 1. Standard Proctor test 2. Modified Proctor test

6 Standard Proctor Test

Layer 3

Layer 2

Layer 1

No. of blows = 25

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Modified Proctor Test

Layer 5 Layer 4 Layer 3 Layer 2 Drop = 457.2 mm (18 in) Layer 1

Drop = 304.8 mm (12 in)

No. of blows = 25

hammer hammer = 2.5 kg (5.5 lb) = 4.9 kg (10 lb)

8 Summary of Laboratory Compaction Tests

Test Hammer Height of #Blows per # Layers Volume of weight drop layer Mold

Standard 2.5 kg 304.8 mm 25 3 944 cm3 proctor Test (5.5 lb) (12 in) (1/30 ft3)

Modified 4.9 kg 457.2 mm 25 5 944 cm3 Proctor test (10 lb) (18 in) (1/30 ft3)

Standard test compaction effort = 600 kN-m/m3 (12,400 lb-ft/ft3) Modified test compaction effort = 2700 kN-m/m3 (56,000 lb-ft/ft3)

9

Compaction Test- Procedure W1

Oven drying @ 105oC Get water content (w %)10 W2 Presentation of Results For each test, the moist unit weight of compacted soil is W W   2 1 Vmold Then the dry unit weight is calculated as   d   w%  1   100 

Repeat the previous procedure for several water content and calculate corresponding d.

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Presentation of Results

Dry of optimum wet of optimum

12 Presentation of Results To better understand and check the compaction curve, it is helpful to plot the zero air void curve.

The zero air void curve is a plot of the dry unit weight against water content at 100% degree of saturation.

This variation of dry as function of water content and degree of saturation can be calculated using: G    s w d G w 1 s S Form definition of zero air voids, S =1

Gs  w  d zav  1 Gs w 13

Presentation of Results To plot the zero air void curves, assume values for w ( say 5%, 10%. 15%, etc) and calculate d zav

Gs  w  d zav  1 Gs w

The actual compaction curve will be always below the zero air void curve.

14 Factors Effecting Compaction Compaction effort

. Compaction effort is the energy per unit volume

. As the compaction effort increases:  The maximum dry unit weight of compaction increase.  The optimum moisture content decreases.

15

Factors Effecting Compaction

• Grain-size distribution, shape of the soil grains, specific gravity of soil solids, and amount and type of minerals present

• Fine grain soil needs more water to reach optimum; while coarse grain soil needs less water to reach optimum.

16 Example

 Example 6.1  Das, Chapter 6.

17

Field Compaction Equipment

1. Smooth wheel rollers 2. Pneumatic rubber-tired rollers Suitable for proof rolling o o Better than the smooth-wheel subgrades and for finishing rollers. operation of fills with sandy and o can be used for sandy and clayey clayey soils. soil compaction. They are not suitable for o o Compaction is achieved by a producing high unit weights of combination of pressure and compaction when used on thicker kneading action. layers. 18 Field Compaction Equipment

3. Sheepsfoot roller o most effective in compacting clayey soils.

6. Vibratory rollers 19

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Specification of Field Compaction

 During construction of soil structures (dams, ), there is usually a requirement to achieve a specified dry unit weight.

 In most specifications, it is required to achieve a dry density of 90 to 98% of maximum dry density obtained from standard or modified Proctor tests.

 This is expressed as relative compaction as follows:

 R  d field 100  d max- lab

d field = dry unit weight in the field

d max-lab = max. dry unit weight from laboratory test Specification of Field Compaction

d max - lab

Target Accept area 95% d max - lab Reject

w1 w2 23

Example 1 Find (a) the maximum dry unit weight of crushed limestone fill to be used as base material, (b) its OMC, and (c) the moisture range for 95% of Standard Proctor.

24 Example 2 Find (a) The dry unit weight and water content at 95%Std. Proctor; (b) The degree of saturation at maximum dry density, and (c) plot the zero air voids line.

25

Example 3

26 27

Compaction Field Control

Calculate Relative compaction, R

If R> specification  ok d field If R< specification remove lift and recompact

28 Compaction Field Control

The standard procedures for determining the field unit weight of compaction include: 1. cone method 2. Rubber balloon method 3. Nuclear method

29

Determination of Field unit weight

Sand Cone Test Procedure

W1 =Wt. sand cone before test

1 2

3 4 30 Determination of Field unit weight

Sand Cone Test Procedure

5 6 7 Oven dry @ 105oC  w%

W2=moist wt. of soil from hole

W4 =Wt. sand cone after test

W2 Weight of sand to fill the cone and hole, Dry weight of soil  W3  (1 w/100) W5 = W1-W4 W3 W5 Wc  d field  Volume of soil  V  V  d (sand ) Now you can calculate: Wc = weight of sand to fill the cone  R  d field 100 dsand = unit weight of sand  31 Both obtained from laboratory calibration of sand cone d max- lab

Example

 Example 6.3  Das, Chapter 6.

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