## LESSON 16. Field Compaction

16.1. Introduction

The laboratory compaction test gives an ideal that what compacted dry unit weight to be achieved in field. Generally in the field, 90 % - 95% of laboratory obtained dry unit weight is attained. The amount of compaction achieved in the field depends on the thickness of the layer, type of roller used, pressure intensity applied on the soil, number of passes of the roller. Depending upon the type of soils, compaction can be done by vibration, rolling or ramming. Roller can be smooth wheel roller, sheepsfoot rollers and pneumatic rubber rollers.

16.2. Compaction of Cohesion less Soil

In case of cohesionless soils, vibration is the most effective method of compaction. Best results can be obtained when the frequency of vibration is near to the natural frequency of the soil to be compacted. The vibrating equipments can be hydraulic type or dropping weight type.

16.3. Compaction of Cohesive Soil

In case of moderately cohesive soils, compaction can be done in layers to get best possible results. The compaction is done by roller. For silts of low plasticity, pneumatic rollers are preferred. In case of soils with moderate plasticity, sheepsfoot rollers are preferred.

The types of equipment used for compaction of various soils are reported in Table 16.1 (Ranjan and Rao, 2000):

Table 16.1: Type of equipment used for compaction of different soils

 Soil type Equipment Use Sands Vibratory rollers Embankments for oil storage tanks Sand, silts, clay Pneumatic rubber rollers Base, Sub-base, embankments for highway, airfield Clay Sheepsfoot rollers Core of the earth dam Crushed rock, gravel, sand Smooth wheeled rollers Road construction All soils Rammer Fills behind the retaining walls, trench fills

The field compaction can be expressed by relative compaction which is the ratio of dry unit weight [ $\gamma$ d(field)] of soil in filed and maximum dry unit weight [ $\gamma$dmax] of soil in laboratory. Thus,

${{\rm Re}\nolimits} lative\;compaction = {{{\gamma _{d(field)}}} \over {{\gamma _{d\max }}}}$                     (16.1)

Problem 1

The in situ void ratio of a granular soil is 0.45. The maximum (emax) and minimum (emin) void ration of the soil is 0.7 and 0.4, respectively. The specific gravity (Gs) of the soil is 2.7. Determine the relative compaction.

Solution:

${\gamma _{d(\max )}}={{{G_s}} \over {1 + {e_{\min }}}}{\gamma _w}={{2.7} \over {1 + 0.4}} \times 10=19.29\,kN/{m^3}$

${\gamma _{d(\min )}}={{{G_s}} \over {1 + {e_{\max }}}}{\gamma _w}={{2.7} \over {1 + 0.7}} \times 10=15.88\,kN/{m^3}$

${\gamma _{d(in\,\,situ)}}={{{G_s}} \over {1 + {e_{in\;situ}}\;}}{\gamma _w}={{2.7} \over {1 + 0.45}} \times 10=18.62\,kN/{m^3}$

Thus, the relative compaction is:

${{{\gamma _{d(in\;situ)}}} \over {{\gamma _{d\max }}}}={{18.62} \over {19.29}} \times 100 = 96.5\%$

Problem 2

An embankment of volume 50,000 m3 is constructed (compacted) with maximum dry density of 18 kN/m3 and optimum moisture content of 10%. Determine how much weight of soil (wet weight) will be required to construct the embankment.

Solution:

The unit weight of wet soil can be written as:

${\gamma _{wet}}=(1 + w){\gamma _d}$

where w is the water content, $\gamma$d is the dry unit weight of the soil and $\gamma$w is the wet unit weight of the soil. Thus, wet unit weight can be written as:

${\gamma _{wet}}=(1 + 0.1) \times 18 = 19.8\;kN/{m^3}$

Thus, the weight of the dry and wet soil can be written as:

Weight of dry soil required = 50,000 × 18 = 9 × 105 kN.

Weight of wet soil required = 50,000 × 19.8 = 9.9 × 105 kN.

References

Ranjan, G. and Rao, A.S.R. (2000). Basic and Applied Soil Mechanics. New Age International Publisher, New Delhi, India.