m03 partially saturated soils[1]

CVNG 2002 M03 Compaction and Strength of Partially Saturated Soils of 37

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THE UNIVERSITY OF THE WEST INDIES

ST AUGUSTINE, TRINIDAD AND TOBAGO, WEST INDIES

FACULTY OF ENGINEERING

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING

CVNG 2002 Soil Mechanics

Module 3. Compaction and Strength of Partially Saturated Soils Lectures by Richard Dean, Semester 1, 2006/7

Proctor test equipment Moisture content

Dry unit

weight

Unconfined compression test machine Vertical strain, % 0 20

Axial total

stress

brittle

ductile



Objectives After completing this module and its labs and assignment, you will

1. understand the differences between dry, partially saturated, and fully saturated soils 2. understand how and why soils need to be compacted 3. be able to calculate densities and unit weights for soils 4. be able to specify and control the compaction of partially saturated soils 5. be able to use Mohr's Circle to calculate total stresses on any plane in a soil element 6. be able to measure the strength of partially saturated soils

Contents What are Compaction, Strength, and Partially Saturated Soils? ....................................................................3 Compaction equipment ...................................................................................................................................7 Lifts, materials and equipment ......................................................................................................................13 Controlling compaction (1) the Phase Diagram for Soils ..............................................................................14 Controlling compaction (2) Compaction curves ............................................................................................19 More about compaction curves.....................................................................................................................21 Controlling compaction (3) Checking compaction in the field .......................................................................24 Unconfined compression testing...................................................................................................................26 The Mohr's Circle Construction for Total Stress ...........................................................................................28 References....................................................................................................................................................32 Technical Standards .....................................................................................................................................32 Assignment for Module 3 ..............................................................................................................................33 Previous exam questions ..............................................................................................................................35



What are Compaction, Strength, and Partially Saturated Soils? A body of soil consists of a large number of soil particles, with air, water, and sometimes other

fluids in the void spaces. Partially saturated soil is soil that has both air and water in the void spaces. Partially saturated soils have special properties because of surface tension effects at air-water interfaces. Also, pathways for the flow of water and air through the soil skeleton are constrained in ways that do not apply for dry or fully saturated soils.

Some effects of having air as well as water in the void spaces

Partially saturated soils are relevant to civil engineering because the vadose zone between the ground surface and the water table is usually partially saturated, and because construction material used as fill or for earthworks such as embankments is usually partially saturated.

Compaction is the process of removing some of the air from partially saturated soils, using mechanical means. This increases the strength and stiffness of the soil. The bearing capacity of the ground is increased. Settlements of structures are reduced compared to structures on un-compacted material. Compaction also reduces the permeability of the soil, and stabilizes soil particles against erosion.

In sands, surface tension effects can hold particles together

In clays, the remaining water shrinks back against the particle surfaces, forming a soil/water skeleton

Sand grain Water

Air

Clay particle Water in semi-solid state on surface of clay particle

Air



Typical problems caused by poor or no compaction, from www.concrete-catalog.com/soil_compaction.html

Many methods can be used to compact a soil. They can be classed as pressure methods such as the use of rollers, kneading methods such as the use of a sheepsfoot rollers, impact methods such as dropping weights onto ground (dynamic compaction) or by use of explosives, and vibration methods.

Roller compactor

Dynamic compaction, from www.densification.com



A geotechnical engineer will normally specify the results needed for compaction, and will sometimes specify how those results are to be obtained. The engineer will use a compaction curve. This is a relationship between the dry unit weight of compacted material, and the water content at which it is compacted. Different materials have different compaction curves, and the curve for a given material depends on the nature of the compaction and the compactive effort used.

For a given compactive effort, the optimum water content is the water content which produces the greatest compaction, as measured by the maximum dry unit weight of the compacted material. The zero-air-voids (ZAV) curve is the theoretical maximum if all the air is removed from the soil. In the field, the engineer will check that a required unit weight has been achieved. This is done by field density tests.

Compaction increases the strength of partially saturated soil, but the main aim is in the future. The soil may be partially saturated during construction, but may become fully saturated later. For instance, this can occur if the water table rises, or if water seeps into the ground from rain or flood. For a fully saturated soil, the strength of the soil is larger for denser samples. Strength can be measured in many ways. In the unconfined compression test, a cylindrical sample is placed between platens and compressed vertically, without any support or "confinement" in the radial direction.

Results of Standard Proctor test for a silty clay. From page 66 of Das (2004)



The results of an unconfined compression test can be interpreted using the Mohr's Circle diagram, which is a plot of stresses on all of the planes in the soil. The stress on the failure plane, for example, can be found from the Mohr's Circle construction as shown above. The shear strength of the material is the radius of the circle, which is one half of the maximum stress achieved in the unconfined compression test.

If we take some soil and prepare samples at different water contents, then compact the samples using the same compactive effort for each, then test their strengths, we can plot two kinds of results. One is the relation between moisture content during compaction and shear strength of the partially saturated soil. This generally increases as the moisture content is reduced, up to a maximum which usually occurs at a water content drier than optimum. However, if we plot the strength of fully saturated soils, the maximum occurs at the optimum moisture content. This is why the optimum is so important.

ψ

2ψ

Normal stress

Shear stress

Water content during compaction

Dry density

Strength

Partially saturated, after compaction

Fully saturated



Compaction can occur naturally, without the use of machines. A track may be compacted a little more each time someone walks along it. The soil beneath a road may be compacted a little more each time a vehicle passes. Compaction can also occur due to cycles of wetting and drying, which effectively produce cycles of stress in the soil. The cycles produce a gradual compaction and strengthening of the material. As a result of these various processes, soil near the ground surface is sometimes stronger than the underlying soil. This is referred to as an overconsolidated dessication crust.

Wetting and drying cycles can also produce many problems. In expansive clays in tropical regions, drying causes the soil to shrink and crack, and the air temperature can be sufficient to suck moisture out of the soil. The active zone can be 10m to 20m deep. When the rains come, the soil sucks up water again. This causes expansion, which can cause much damage to foundations, structures, and earthworks.

Compaction equipment The most common ways of compacting soil is to place the un-compacted material is layers or "lifts"

of typically 1 foot thickness, and to pass machines over the surface to compact the soil. Before placing a lift, water may be sprayed onto the uncompacted material to achieve a desired water content, or this can be done after placement and before compaction. Compaction is achieved by:

• Pressure • Kneading • Vibration • Impact

strength

Depth below

ground surface



Smooth steel-wheeled rollers can be walk-behind, drive-on, or towed types. They achieve compaction primarily by the application of pressure to the soil, but also with some kneading as the roller passes over a place.

Pneumatic rollers compact by a combination of pressure and kneading as the wheel passes over a place. Wheels are typically offset front and back, so as to achieve maximum coverage of the ground.

From page 74 of Das (2004)

From www.constructionequipment.com/article/CA6359370.html



Sheepsfoot rollers are only suitable for clayey soils. As the roller passes over the soil, the projections or "feet" push into the soil, creating a severe kneading action which tends to force air out of the soil. The kneading action remoulds the soil, so its remoulded undrained shear strength is used for design in the short term.

Pictures from www.sbe.napier.ac.uk/projects/compaction/chapter2d.htm



Vibration systems can be fitted to smooth-wheeled, pneumatic, and sheepsfoot rollers to increase the effectiveness of the compaction – and so reduce the number of passes that a machine has to make on each lift. Vibration is usually achieved by attaching eccentric weights to the wheels.

Vibratory plates are hand-operated machines that apply low amplitude, high frequency vibrations to the ground. They are effective for granular fills, and also for asphalt, so they are commonly used in small road construction. Typical frequencies are 2000 to 6000 vibrations per minute. The engine and handle are generally partially isolated from the base plate to prevent damage to the machine or the operator, who walks behind the machine.


From

www.constructioncomplete.com/SingleDirectionPlateCompactorsGas/WackerWP1550APlateCompactor.html



Rammers deliver high impact force top the ground. They are good for cohesive and semi-cohesive soils. A typical frequency range is 500 to 750 blows per minute. Rammers have three effects – impact, vibration, and kneading.

Other types of compaction include dynamic compaction, where a heavy weight is dropped onto the ground. Dynamic compaction is good for granular and silty granular soils.

From www.macrental.net/plant_hire/compaction_and_rollers



In vibroflotation, granular material is compacted by a process involving liquefaction and re-solidification of denser material.

From p.82 of Das (2004), see also Brown (1977)



Granular material – sands and gravels – can also be compacted by blasting – explosive compaction. Holes are drilled into a soil body to a depth of about 2/3 of the depth to be compacted. Small explosive charges are fired at the bottoms of the holes. The shear wave from the explosion liquefies the soil, joggling the particles. The material then settles, water is expelled, and the particles come into contact again – the material solidifies at a greater density than previously.

Lifts, materials and equipment

The effects of rollers, vibration systems, and small rammers are generally felt close to the surface of the layer that is being compacted. For rollers, the effects depend on the wheel weights and areas, the amount of kneading action that is involved, the lift height, and the number of passes.

The maximum effect typically occurs a short distance below the ground surface. Above this, the equipment tends to disturb the soil. Below, the compactive effect is dissipated. Consequently, lifts are typically of the order of 6 inches to 1 foot.




Materials

Vibrating Sheepsfoot Rammer

Static Sheepsfoot Grid Roller Scraper

Vibrating Plate Compactor Vibrating Roller Vibrating Sheepsfoot

Scraper Rubber-tired Roller Loader Grid Roller

Lift Thickness Impact Pressure (with kneading) Vibration Kneading

(with pressure)

Gravel 12+ Poor No Good Very Good Sand 10+/- Poor No Excellent Good Silt 6+/- Good Good Poor Excellent Clay 6+/- Excellent Very Good No Good

From www.concrete-catalog.com/soil_compaction.html

Gravels tend to be excellent fill materials, with good drainage characteristics and suitable for foundation support and pavement subgrade. Sands are good. Silts are generally poor materials. Clays can give adequate foundation support but are not good for pavement subgrades, and can be difficult to compact. Organic soils are poor foundation materials and poor pavement supports.

Controlling compaction (1) the Phase Diagram for Soils To control compaction, the engineer needs to specify and measure moisture contents and densities of

uncompacted and compacted material. The phase diagram is a way of remembering how to calculate some of these quantities, and related quantities. In a real soil, particles, air, and water are mixed together. In the phase diagram we think of the particles as packed together in one region, without voids between them, and we think of all the voids together in another region, with air in one part and water in another.

AIR

WATER

SOIL

PARTICLES

Void spaces – the degree of saturation S of the soil is the volume of water divided by the volume of void spaces

Solids – the unit weight of the solids is Gs times the unit weight of water, where Gs is the specific gravity of the solids



There are three commonly used ways of measuring the relation between volumes of the fluid and solid phases of the material. The void ratio e is the ratio of the volume of air plus water to the volume of solids:

e = solids of Volume

waterair of Volume + =

nn−1

= V − 1

The porosity n is the ratio of the volume of air plus water to the total volume of the soil:

n = solidswaterair of Volume

water air of Volume++

+ =

ee+1

= V

V 1−

The specific volume V is the ratio of the volume of everything divided by the volume of solids:

V = solids of Volume

solids water air of Volume ++ = 1+e =

n−11

The void ratio is the most commonly used quantity for granular soils – sand and gravels. It is usually expressed as a fraction. Porosity is sometimes preferred in calculations for the flow of water through soils (seepage). Specific volume is sometimes used for clays instead of void ratio.

As noted in the sketch above, the degree of saturation is the ratio of the volume of water in the voids divided by the volume of the voids (air + water). It is usually expressed as a percentage. Another common measure is the gravimetric water content w, usually just called the water content by geotechnical engineers. This is the ratio of the weight of water to the weight of dry solids:

w = solidsdry of weight

waterof weight

Water content can be related to degree of saturation and void ratio by considering the relative volumes of water and solids, and their unit weights. The relative volume of the water in the soils is the degree of saturation S multiplied by the void ratio e. Hence the relative weight is Se times the unit weight of water. The relative volume of dry solids is 1, and their unit weight is Gs times the unit weight of water. Hence:

w = sGeS.

Engineers use several measures of density and weight. The word density is used to denote mass per unit volume, so that the density of water is typically around ρw = 1000 kg/m3. The phrase unit weight is used to denote weight divided by volume, so the unit weight of water is γw = 9.81 x 1000 N/m3 = 9.81 kN/m3 in earth's gravity of 9.81 m/s2. Unit weight depends on gravity, so is different on the moon, for instance.



The dry density ρdry of a soil is the mass of the dry soil divided by the volume of the soil + water + air. The dry unit weight γdry is this multiplied by the acceleration of gravity. Now the relative volume of solids to total volume is 1/(1+e), so:

ρdry = waterairsolids of volume

solidsdry of mass++

= e

G ws+ρ

1.

γdry = waterairsolids of volume

solidsdry of weight++

= e

G ws+

γ1

.

Dry unit weight is used as part of the method of controlling compaction – see later. The bulk density ρbulk of a soil is the mass of the soil and water divided by the volume of the soil + water + air. The bulk unit weight γdbulk is this multiplied by the acceleration of gravity. Considering the relative volumes of solids, water, and total volume, and considering their densities and unit weights, gives:

ρbulk = waterairsolids of volume

water solids of mass++

+ =

eSeG ws+

ρ+1

).(

γbulk = waterairsolids of volume

watersolids of weight++

+ =

eSeG ws+

γ+1

).(

We also often consider buoyancy effects. The buoyant density ρ′, also called the effective density or the submerged density, is the bulk density less the density of water. The buoyant unit weight γ′, also called the effective unit weight and the submerged unit weight, is the bulk unit weight less the unit weight of water:

ρ′ = ρbulk − ρw = ws

eeSG

ρ+

−+− .1

).1()1(

γ′ = γbulk − γw = ws

eeSG

γ+

−+− .1

).1()1(

The table gives the specific gravities of some minerals of which soil particles may be made (see page 17 of Das, 2005). If in doubt, a value of Gs = 2.65 is usually fairly accurate.



Example 1. The water content of a fully saturated quartz sand is 25%. Calculate its: (a) Void ratio (b) Porosity (c) Dry density (d) Bulk density (e) Submerged unit weight

Answer: 1: (a) Since the sand is quartz, we shall assume Gs=2.65. Since it is fully saturated, S=100%. Hence e = w.Gs/S = 0.25 x 2.65 / 1 = 0.6625

(b) The porosity is the volume of voids divided by the total volume, so is e/(1+e) = 0.6625/1.6625 = 39.8%

(c) The dry density is the dry mass divided by the volume. The relative volumes are 1 (solids) to 1+e (total). Hence ρdry = 1 x 2.65x 1000 / (1.6625) = 1594 kg/m3.

(d) The bulk density is the total mass divided by the volume. The relative volumes are 1 (solids) to 1+e (total). Hence ρbulk = (1 x 2.65 + 0.6625x1)x 1000 / (1.6625) = 1992 kg/m3.

(e) The submerged unit weight is the bulk unit weight less the unit weight of water. Hence γ′ = (1992 – 1000) x 9.81 N/m3 = 9.73 kN/m3.

Example 2. A material has a bulk unit weight of 18.5 kN/m3 and a water content of 23%. What is its dry density?

Answer 2: The bulk unit weight includes water. Considering the definition of water content, the dry unit weight is therefore the bulk unit weight divided by 1 plus the moisture content.. Hence γdry = 18.5/1.23 = 15.04 kN/m3. The dry density is therefore this divided by 9.81, which comes to 15040 / 9,.81 = 1533 kg/m3.

Example 3. A site investigation has been carried out at a site and the soil has been found to consist of

2m of uniform quartz sand with a void ratio of 0.75, overlying a uniform kaolin clay. The water table is 0.5 m below the ground surface, and the sand may be assumed to be fully dry above the water table and fully saturated below it. The clay is fully saturated and has a moisture content of 47%. For a depth of 6m below the ground surface, calculate:



(a) The total vertical stress at that depth (b) The water pressure at that depth (c) The effective vertical stress at that depth, equal to the total stress less the water

pressure Answer 3: To calculate the stresses in the soil, we need to first calculate the unit weights of the

materials. And before that it is useful to draw a diagram.

For the dry quartz sand above the water table, the specific gravity is 2.65 and the void ratio is 0.75m, so the dry unit weight is 2.65 x 9.81 / 1.75 = 14.86 kN/m3.

For the fully saturated sand below the water table, the bulk unit weight is (2.65+0.75) x 9.81 / 1.75 = 19.06 kN/m3.

For the kaolin clay, the specific gravity is 2.6 (from the table of specific gravities). The moisture content is 47%, so the void ratio is 0.47 x 2.6 / 1 = 1.222. Hence the bulk unit weight is (2.6+1.222) x 9.81 / (1+1.222) = 16.87 kN/m3

The total vertical stress at 6m below the ground surface is the sum of

1) stress due to 0.5m of dry sand, i,. 0.5x14.86 = 7.43 kN/m2 2) stress due to 1.5m of saturated sand, ie. 1.5 x 19.06 = 28.59 kN/m2 3) stress due to 4m of clay,. ie. 4 x 16.87 = 67.48 kN/m2

Hence the total stress at 6m depth is 7.43 + 28.59 + 67.48 = 103.5 kN/m2. This depth is 5.5m below the water table, so the pore water pressure is 5.5 x 9.81 = 53.95 kN/m2. Hence the vertical effective stress is 103.5 – 53.95 = 49.55 kN/m2

Dry sand, Gs=2.65, e=0.7

Fully saturated sand, Gs=2.65, e=0.7

Fully saturated clay, Gs=2.6, w=0.47

0.5m

1.5m

4m



Controlling compaction (2) Compaction curves

If a soil is compacted under a certain amount of compactive effort, then the density of the soil that is achieved due to the compaction depends on the moisture content of the soil. The relation between moisture content and dry density is called a compaction curve.

In the standard Proctor test, the apparatus shown below is used, together with moisture content tins and an oven. The apparatus consists of a mold of volume 943.3 cm3, into which soil is placed in layers for compaction, and a hammer or rammer of a standard weight and standard drop.

From p.64 of Das (2005)

A few kilograms of the soil are first sieved to remove oversize particles. Depending on the size of the particles remaining, the remaining material is then compacted in several steps, using either of three methods called A to C. In each method, the material is compacted in a mold in three layers. Each layer is placed at the same moisture content, and subjected to a number of blows of a standard Proctor hammer. The hammer drops 304.8mm and its weight is 24.4N (its mass is 5.5 lb). The compactive energy for each drop is 24.4 x 0.3048 = 7.437 Joules. The compactive effort for the three methods is:



Method No.of blows per layer

Volume of mold

Compactive effort

A 25 944.3 cm3 7.437 x 25 x 3 / (944.3 x 10−6) ≈ 593 kN.m / m3

B 25 944.3 cm3 7.437 x 25 x 3 / (944.3 x 10−6) ≈ 593 kN.m / m3

C 56 2124 cm3 7.437 x 56 x 3 / (2124 x 10−6) ≈ 589 kN.m / m3

The moisture content and dry density of the compacted material is measured. The material is then removed from the mold, water is added, and the process is repeated. After five or six specimens have been compacted and measured in this way, a curve is drawn of dry unit weight versus moisture content.

The following shows a typical example. The weights of compacted material were measured by tipping the material from the mold onto a tray, weighing the tray, and subtracting the weight of the tray. The moisture contents were measured by taking moisture content samples, weighing them, drying them in an oven for 24 hours at 105C, then weighing again and accounting for the weights of the tins. Based on the results, it is possible to calculate the remaining columns.

Sample Weight in

mold, N Moisture

content, % γbulk,

kN/m3 γdry,

kN/m3 Void ratio (Gs=2.66)

γdry if S=100%, kN/m3

1 17.23 12.0

2 18.13 13.7

3 18.78 15.6

4 19.07 18.0

5 18.89 21.5

6 18.45 23.8 The void ratio in the fifth column is the void ratio that would correspond to the listed moisture content if the soil was saturated. Thus e = w/Gs. The maximum unit weight in the last column is then calculated as Gs.γw/(1+e). The results are plotted on the following page. The optimum moisture content is the value of moisture content at the maximum dry density.



The degree of saturation at the optimum can also be read from the graph. First, plot the optimum point. Next, plot the point at the same dry density but on the zero air voids line, i.e the line corresponding to full saturation. Next read the moisture content off that point on zero air void line. This is the maximum moisture content for the given dry density. The degree of saturation at optimum is the optimum moisture content divided by this maximum moisture content.

More about compaction curves Typical compaction curves for a variety of materials are shown on the next page. Most materials

have the characteristic bell-shaped curve. Other types of curve are also possible. Type A curves are for silty and clayey materials where the fines have liquid limits between about 30% and 70%. Types B and C occur for materials with low liquid limits, lower than about 30%. Soils with liquid limits greater than about 70% may have compaction curves like C or D – but such curves are not very common.

Moisture content of compacted sample, %

12 14 16 18 20 22 24 15

16

17

18

Dry unit weight, kN/m3

19



Typical curves for various soil types – note that dry density is being plotted vertically instead of dry unit

weight. The result for optimum moisture content is unchanged. From page 79 of Sarsby (2000)

Various types of compaction curve. From page 67 of Das (2004)



The compaction curve for a soil varies with the compactive effort. For higher efforts, the curve is displaced leftwards (towards smaller water contents) and upwards (towards higher densities).

Effect of compactive effort, from page 76 of Sarsby (2000)

The effect of compactive effort has two effects. First, heavier compaction equipment developed over the last 40 years or so, and the original Proctor test has become less useful. In the modified Proctor test, the soil in the mold is compacted in five layers instead of three, and each layer is compacted with a 44.5N hammer instead of a 24.4N one. The compactive effort in the modified Proctor test is about 2696 kN.m/m3, which is about 4.5 times the effort for the standard Proctor test.

The Proctor tests do not necessarily match the compaction curves that can be achieved using practical equipment. There are two approaches to specifying field compaction. In a method specification, the engineer specifies the moisture content for the compacted material, and the method by which the material is to be compacted (including the equipment, the lift thickness, and the number of passes, which is typically around 10 to 12). Problems with method specification are (a) it requires the consultant to have a high degree of contracting knowledge, and (b) it is not necessarily the most best or most economical way of using the contractor's equipment and skills. In end-product specification, the engineer specifies the range of final densities to be achieved, and leaves the contractor to work out how to achieve densities in that range.



Controlling compaction (3) Checking compaction in the field If soil is trucked to a site, placed, and compacted, after adding water if necessary, the result is termed

an engineered fill. The main objective of the compaction is to achieve a certain range of dry densities, which will give a certain range of strength for the fill. A secondary objective may be to achieve a certain range of moisture contents too. To verify compaction, the engineer will carry out moisture content and density tests on samples taken from the compacted fill.

Moisture content tests are straightforward, The engineer arranges that a small mobile laboratory be available on site, including a weighing balance and an oven. Measuring moisture content involves taking samples, measuring their wet weights, drying them for 24 hours at 105°C, measuring the dry weights, and taking account of the weights of the tins.

Density tests are more time consuming. There are four methods to choose from. In the sand replace-ment method, a small hole is dug into the compacted material. The weight of material W1 removed from the hole is measured. A standard cone is then used to fill the hole with a standard dry sand. The purpose of the cone is to ensure that the method of pouring is repeatable. The weight Ws of sand needed to achieve this is measured. The volume of the hole is deduced by dividing Ws by the density or unit weight of the sand – which is measured in a separate calibration exercise. The bulk unit weight of the compacted material is then W1 divided by the volume of the hole. The dry unit weight is this divided by 1+w.

Equipment and procedure for the sand replacement method, from pages 78-79 of das (2005)



Example 4. An engineer scrapes flat the surface of a small area of a compacted engineered fill, then digs a hole through a template approximately 4 inches in diameter and 4 inches deep. The mass of material removed from the hole is 1.434 kg, and its moisture content is later found to be 13%. The sand replacement method is used to refill the hole with sand, and 1.323 kg of sand are needed. The poured density of the sand is 1.730 grams/cc. What dry density has been achieved for the compacted fill?

Answer 4: The volume of the hole was 1323 / 1.73 = 764.7 cm3. hence the bulk density of the clay

was 1434 / 764.7 gm/cm3 = 1.875 gm/cm3, equivalent to 1875 kg/m3. Hence the dry density of the compacted fill was 1875 / (1+0.13) = 1659 kg/m3.

Other methods for measuring the compacted density include (1) the rubber balloon method, in which the volume of the hole is measured using a rubber balloon filled with water, (2) the Shelby tube method in which a tube is pushed into the fill and pulled out again - the volume of material extracted is measured directly or indirectly, and (3) the nuclear method, in which a nuclear density meter is used to measure density. A summary of the various methods is given below.

Field Density Testing Method

Sand Cone Balloon Dens meter Shelby Tube Nuclear Gauge

Advantages * Large sample * Accurate

* Large sample * Direct reading obtained * Open graded material

* Fast * Deep sample * Under pipe haunches

* Fast * Easy to redo * More tests (statistical reliability)

Disadvantages * Many steps * Large area required * Slow * Halt Equipment * Tempting to accept flukes

* Slow * Balloon breakage * Awkward

* Small Sample * No gravel * Sample not always retained

* No sample * Radiation * Moisture suspect * Encourages amateurs

Errors * Void under plate * Sand bulking * Sand compacted * Soil pumping

* Surface not level * Soil pumping * Void under plate

* Overdrive * Rocks in path * Plastic soil

* Miscalibrated * Rocks in path * Surface prep required * Backscatter

Cost * Low * Moderate * Low * High

From www.concrete-catalog.com/soil_compaction.html



Unconfined compression testing In an unconfined compression test, a

cylindrical sample of soil is placed between platens and subjected to compression without any lateral support or confinement. The test gives a quick way of measuring the unconfined compressive strength qu of the soil.

The load F applied to the soil is measured by a proving ring. This is a string ring which is compressed in a diametrical direction. The ring deforms and the deformation is measured by a dial gauge placed at the center of the ring. A calibration factor or curve is then used to calculated the load on the sample from the dial gaage reading. The nominal stress on the sample is then equal to F divided by the initial cross-sectional area Ao of the sample.

A second dial gauge is used to measure the relative movement of the platens. If the reduction in height is ∆h at some time during the test, then the axial compressive strain is ∆h/Ho, where Ho is the initial height of the specimen.

As long as a soil has sufficient cohesion to stay together, it can be tested in an unconfined compression machine. If the soil is a fully saturated clay, the volume of the specimen will stay sensibly constant during the test, so the area increases from the initial value of Ao to a value of A = Ao/(1−∆h/Ho) when the sample height has reduced by h. In some applications, the stress on the sample is taken to be corrected ("true") stress F/A rather than the nominal stress F/Ao.

Typical stress-strain curves are shown in the following sketches. The response is ductile if it involves large strain with the compressive stress continually increasing. For a ductile material, the unconfined compressive strength is taken as the stress at 20% strain. The response is brittle if the material is initially quite stiff (large increase in stress for only a small vertical strain), but subsequently the stress peaks and reduces at larger strains. For a brittle material, the strength is taken to be the peak stress.



It is generally possible to obtain an idea of the elastic properties of the soil from the initial part of the stress-strain curve. Subsequently, the curve becomes significant non-linear. Plasticity develops, so that an unload-reload cycle will not return along the same stress-strain curve as the initial loading cycle.

For clays, the following descriptions are used for clays of various consistencies:

Range of unconfined compressive strengths, kN/m2

Consistency (Das, 2004) British system

Very soft 0 – 25 0-40

Soft 25 – 50 40 – 80

Medium (firm) 50 – 100 80 – 150

Stiff 100 – 200 150 – 300

Very stiff 200 – 400 300 – 600

Hard > 400 > 600

Vertical strain

Vertical total

stress

0% 20%

ductile

brittle



As the maximum stress is approached, the sample may start to bulge or "barrel" if it is ductile, or may shear along an inclined rupture plane if brittle, or may fail in a combination of these ways. Barreling is due to friction on the platens which prevents the ends of the sample from expanding outwards like the middle. The angle of the rupture plane to the vertical is an interesting parameter, and is discussed below.

The Mohr's Circle Construction for Total Stress In general, stresses act on a variety of different planes in a soil body. We are interested in all of

those planes. The Mohr's Circle Construction allows us to plot the stresses on different planes, and to quickly calculate the stresses on one plane once the principal stresses have been determined.

Suppose that principal total stresses σ1 and σ3 act on an element of a soil body, as shown. Note that the lateral total stress is zero in the unconfined compression test, but it is useful here to develop a general analysis that can be used for other applications as well as for the unconfined compression test (which is obtained by putting σ3=0).

stresses on a soil element, view in the x, z plane, and view of an inclined plane at angle θ to the x,y plane

Suppose we want to calculate the stresses acting on some plane AB in the material at an angle θ to the horizontal plane. To calculate the stresses on plane AB, we construct a triangular wedge of soil as shown in the next sketch.

σ1

σ1

σ3 σ3 x

z

θ

A

B



Let us suppose that a stress σ acts normally on the inclined plane, and a shear stress τ acts tangentially on that plane. We consider a unit depth of wedge into the paper. Let L be the length of the inclined plane from A to B. Then the inclined plane has area L.1. The vertical plane on the left has area Lsinθ.1 The horizontal plane on the base has area Lcosθ.1. The force on any plane is the stress on that plane multiplied by the area of the plane. Hence we can resolve in the directions normal and tangential to the inclined plane AB:

Resolving forces normal to AB : σ.L.1 = σ3.(Lsin.θ).sinθ + σ1.(Lcos.θ).cosθ

Resolving tangential to AB: τ.L.1 = σ3.(Lsin.θ).cosθ − σ1.(Lcos.θ).sinθ

Dividing the equations by L, and using the trigonometric identities 1+cos2θ=2cos2θ, 1−cos2θ=2sin2θ, and sin2θ = 2sinθ.cosθ, gives:

σ = 2

31 σ+σ +

231 σ−σ

.cos2θ

τ = 2

31 σ−σ.sin2θ

This means that, if we plot the normal stress σ on the horizontal axis of a graph, and the shear stress τ on the vertical axis, the result is:

σ1

σ3

A

B σ

τ

θ

L.cosθ

L.sinθ

L



If we now consider all possible planes in the material that are inclined at some angle to the horizontal and contain a tangent in the direction of the y-axis, we will get a circle on the graph:

The normal stresses at the intersection of the Mohr's Circle with the normal stress axis are the principal total stresses. The major principal total stress is the one that is the largest. The minor principal total stress is the one that is smallest (including consideration of out-of-plane stresses). In 3D there are three principal stresses. If the value of one of them is between the other two, it is the intermediate principal total stress.

Normal stress σ

Shear stress τ

Normal stress σ

Shear stress τ

231 σ+σ

231 σ−σ

.cos2θ

231 σ−σ

.sin2θ 231 σ−σ

σ3 σ1

2θ



In the unconfined compression test, the lateral total stresses are zero, so the Mohr's Circle passes through the origin of the graph. If we have measured the angle of the failure plane to the horizontal, we can calculate the stresses on the rupture plane directly from the Mohr's Circle.

The radius of the circle is half of the unconfined compressive strength. It equals the shear stress on a plane inclined at 45° to the horizontal. For the failure plane, the normal and shear stresses can be calculated as:

σ = 2uq

.(1−cos2ψ) = (qu.sinψ).sinψ

τ = 2uq

.sin2ψ = (qu.sinψ).cosψ

For partially saturated soils, the unconfined compressive strength generally increases for decreasing water content, up to a maximum. The increase is due to the fact that, at relatively large water contents, the meniscuses in the material get tighter as the water content reduces, so that the effects of the surface tension forces increase. At lower water contents, the size of the wetted areas at inter-particle contacts starts to reduce, so that there is less strengthening effect.

ψ

2ψ

Normal stress

Shear stress

qu



References • Craig, R.F. (2005). Craig's Soil Mechanics. 7th edition. E&F Spon • Craig, R.F. (2004). Craig's Soil Mechanics. 7th edition, Solutions Manual. E&F Spon • Das, B.M. (2004). Principles of Foundation Engineering, Thomson (Brooks Cole) • Fredlund, D.G. and Rahardjo, H. (1993). Soil Mechanics for Unsaturated Soils. Wiley • Lambe, P.W., and Whitman, R.V. (1979). Soil Mechanics – SI version. Wiley • Sarsby, R. (2000). Evironmental Geotechnics. Thomas Telford • Terzaghi, K., Peck, R., and Mesri, G. (1996). Soil Mechanics in Engineering Practice. Wiley • Hilf, J.W. (1991). Compacted fill. Chapter 8 of Foundation Engineering Handbook, 2nd edition, ed H-Y

Fang. Chapman & Hall, Chapter 3, pp.249−316

Technical Standards • Annual Book of ASTM Standards, Volume 04.08 Soil and Rock (I): D 420 - D 5611 and Volume 04.09

Soil and Rock (II): D 5714 - latest, American Society for Testing and Materials

D698-00 Methods for Laboratory Compaction Characteristics of Soil Using Standard Effort

D1556-00 Method for Density and Unit Weight of Soil in Place by the Sand-Cone Method

D1557-00 Method for Laboratory Compaction Characteristics of Soil Using Modified Effort

D2167-94 Density and Unit Weight of Soil in Place by the Rubber Balloon Method

D2168-90(1996) Methods for Calibration of Laboratory Mechanical-Rammer Soil Compactors

D2937-00 Method for Density of Soil in Place by the Drive-Cylinder Method

D4564-93 Method for Density of Soil in Place by the Sleeve Method

D4718-87 Standard Practice for Correction of Unit Weight and Water Content for Soils Containing Oversize Particles

D4914-99 Methods for Density of Soil and Rock in Place by the Sand Replacement Method in a Test Pit

D4959-00 Method for Determination of Water (Moisture) Content of Soil By Direct Heating

D5080-00 Standard Test Method for Rapid Determination of Percent Compaction

• BS 1377. Methods of test of soils for civil engineering purposes. (7 vols). British Standards Institution • BS 6031. Code of practice for earthworks. British Standards Institution • BS 8004. British Standard Code of Practice for Foundations (formerly CP2004). 1986EN1997

Eurocode 7: Geotechnical design, www.eurocodes.co.uk • ISO 17892. Geotechnical investigation and testing. International Standards Organization • PCTTS 599. Guide to the design and construction of small buildings, TT Bureau of Standards • US Navy. Design Manual – Soil Mechanics, Foundations, and Earth Structures. NAVDOCKS DM-7



Assignment for Module 3 Please staple your answer sheets firmly together at one corner, and write your student id at the top right on the first sheet. There is no need to bind the sheets in a folder, or to use a plastic spine. Hand your assignment in by giving it to Dr Dean, or by putting it in his letter box or through/under his door. 1. What is compaction? How does it differ from compression? Give four reasons to compact soil.

Give five examples of problems that occur if soil is not compacted.

2. Describe four types of compaction equipment. Under what circumstances would you use which type?

3. The water content of a saturated quartz sand is 25%, and its degree of saturation is 75%. Calculate its:

(a) Void ratio (b) Porosity (c) Dry density (d) Bulk density (e) Submerged unit weight

4. (a) A material has a dry unit weight of 16.5 kN/m3 and a water content of 17%. What is its bulk density?

(b) A material has a dry unit weight of 15.8 kN/m3 and a moisture content of 24% when fully saturated. Calculated the average specific gravity of the soil particles.

5. A site investigation has been carried out at a site and the soil has been found to consist of 3m of uniform quartz sand with a void ratio of 0.85, overlying a uniform illite clay. The water table is 0.8m below the ground surface. The sand may be assumed to be fully dry above the water table and fully saturated below it. The clay is fully saturated and has a moisture content of 63%. For a depth of 7m below the ground surface, calculate:

(a) The total vertical stress at that depth (b) The water pressure at that depth (c) The effective vertical stress at that depth, equal to the total stress less the water pressure

6. The following results were obtained from a compaction test using a mold of 944.3 cm3 volume. Plot the compaction curve and the zero air voids line, assuming the material has a specific gravity



of 2.68. What is the optimum water conetnt and the maxmum dry density for this compactive effort? What is the degree of saturaton at the optimum point?

Sample Weight in mold, N Moisture content, %

1 17.7 13.5

2 18.4 15.3

3 18.8 17.1

4 19 19.5

5 18.7 22

6 18.3 25.5

7. What laboratory equipment is needed in the field to carry out quality control tests on compaction? Describe:

(a) a procedure for measuring moisture content

(b) a procedure for mesuring dry density of compacted fill

8. An engineer scrapes flat the surface of a small area of a compacted engineered fill, then digs a hole through a template. The mass of material removed from the hole is 1.42 kg, and its moisture content is later found to be 17%. The sand replacement method is used to refill the hole with sand, and it is found that 1.35 kg of sand are needed. The poured density of the sand is 1.69 grams/cc. What dry density has been achieved for the compacted fill?

9. (a) How is the unconfined compressive strength of a soil specimen measured?

(b) What is the unconfined compressive strength of dry, uncemented sand?

(c) The unconfined compressive strength of a sandy clay is measured as 175 kN/m2. The rupture surface was at 61° to the horizontal. Calculate the total normal stress and the shear stress on the rupture surface when the soil reached its compressive strength.



Previous exam questions

2006

2001



2000 (July)

2000 (May)



1999