This is the most complete discussion of Soil Mechanics
CHAPTER 1. INTRODUCTION
In a general technical sense, soil is defined as a material consisting of aggregates (granules) of solid minerals that are not cemented (chemically bound) to each other from weathered organic matter (which has solid particles) accompanied by liquid and gas fills the empty spaces between the solid particles. Soil is useful as a building material in various types of civil engineering work, besides that soil also functions as a support for the foundation of the building.
The term geotechnical engineering is defined as the science and practice of civil engineering that deals with natural materials found on (and close to) the earth's surface. In a general sense, geotechnical engineering also includes the application of basic soil mechanics and rock mechanics applications to foundation design problems.
1.2 PROBLEM FORMULATION
1.2.1 What is the rock cycle and soil origin?
1.2.2 What is the composition of the soil?
1.2.3 How is land classified?
1.2.4 What is the permeability and seepage of the soil?
1.2.5 What is the concept of effective stress?
1.2.6 What are the stresses in a soil mass?
1.2.7 What is the compressibility of the soil?
1.2.8 How is soil compacted?
The purpose of making this paper is:
1.3.1 To know the rock cycle and the origin of the soil
1.3.2 To determine the composition of the soil
1.3.3 To know the classification of land
1.3.4 To determine the permeability and seepage of soil
1.3.5 To know the concept of effective stress
1.3.6 To find out the stresses in a soil mass
1.3.7 To determine the compressibility of the soil
1.3.8 To determine soil compaction
The benefits of making this paper are:
1.4.1 We can know the rock cycle and the origin of the soil
1.4.2 We can know the composition of the soil
1.4.3 We can know the classification of soil
1.4.4 We can know the permeability and seepage of the soil
1.4.5 We can know the concept of effective stress
1.4.6 We can find the stresses in a soil mass
1.4.7 We can know the compressibility of the soil
1.4.8 We can know the compaction of the soil
CHAPTER 2. DISCUSSION
2.1 Soil And Rock
2.1.1 The Rock Cycle and Soil Origins
Soil comes from the weathering of rocks with the help of organisms, forming a unique body that covers the rocks. The process of soil formation is known as pedogenesis. This unique process forms soil as a natural body consisting of layers or is referred to as a soil horizon. Based on its origin, rocks can be divided into three basic types: igneous rocks, sedimentary rocks and metamorphic rocks. Igneous rock This rock is formed from cooling magma. Rocky magma melts deep within the earth. Magma in the earth's crust is called lava. Sedimentary rock is formed as pushed together or cemented by the weight of water and the overlying layers of sediment. The process of settling into the lower layers took thousands of years. Metamorphic rocks are rocks that originate from pre-existing rocks, such as igneous or sedimentary rocks, then undergo physical and chemical changes so that their properties differ from those of their parent rock. Physical changes include the destruction of rock grains, the increase in the size of the mineral grains that make up the rock, the flattening of the mineral grains that make up the rock, and so on. Chemical changes are related to the emergence of new minerals as a result of recrystallization or due to the addition/reduction of certain chemical compounds. The causative factor of the metamorphosis process is a change in conditions of high pressure, high temperature or due to fluid circulation. Pressure can come from the force of the load or the weight of the rock that is pressing or from the tectonic movements of the earth's crust when mountains form. An increase in temperature can occur due to the intrusion of magma, liquid or magma gas that infiltrates the earth's crust through local heating cracks due to friction of the earth's crust or a temperature increase associated with a geothermal gradient (increase in temperature as a result of its deeper location).
2.1.2 Soil Particles
The size of the soil particles is very diverse with considerable variation. Soil can generally be referred to as gravel, sand, silt, clay, depending on the size of the predominant particle in the soil. To explain soil based on particle sizes, several organizations have developed soil size limits that have been developed by MIT (Massachussetts Institute of Technology), USDA (US Department of agriculture), AASHTO (America Association of State Highway and Transportation Officials) and by the US Army Corps of Engineers and the US Bureau of Reclamation which then produced what is known as the USCS (Unified Soil Classification System)
Gravel is pieces of rock that sometimes also contain particles of the minerals quartz, feldspar, and other minerals.
The sand is mostly composed of the minerals quartz and feldspar. Granules from other minerals may also exist in this group.
Silt is mostly a microscopic fraction of soil consisting of very fine quartz grains, and a number of particles in the form of flat plates which are fragments of mica minerals.
Clay mostly consists of microscopic and submicroscopic particles in the form of flat plates and are particles of mica, clay minerals, and other very fine minerals.
2.1.3 Specific Weight
The value of the specific weight of the soil particles (solid part) is often needed in various computational purposes in soil mechanics. These values can be accurately determined in the laboratory. Most of these minerals have a specific weight ranging from 2.6 to 2.9. The specific weight of the denser part of light-colored sand, generally consisting mostly of quartz, can be estimated as 2.65, for loamy or silty soils, it ranges from 2.6 to 2.9.
2.1.4 Mechanical Analysis of Soil
Mechanical analysis of soil is the determination of the variation in the sizes of the particles present in the soil. These variations are expressed as a percentage of the total dry weight. There are two methods commonly used to obtain the size distribution of soil particles, namely: sieve analysis (for particle sizes larger than 0.075mm in diameter), and hydrometer analysis (for particle sizes smaller than 0.075mm in diameter). Mechanical analysis (sieve analysis and hydrometer) is generally plotted on a semilogarithmic paper known as a grain size distribution curve.The particle diameter is plotted on a logarithmic scale, and the percentage of grains that pass the sieve is plotted on an ordinary arithmetic scale.
2.2 Soil Composition
2.2.1 Volume-Weight Relationship
In order to establish the volume-weight relationship of soil aggregates, three phases (ie: granular solids, water, and air) are separated. So, the investigated soil sample can be expressed as:
V = Vs + Vv = Vs + Vw + Va
Vs = volume of solid particles
Vv = pore volume
Vw = volume of water in the pores
Va = volume of air in the pore
If the air is considered to have no weight, then the total from the ground can be expressed as:
W = Ws + Ww
Ws = weight of solid grains
Ww = weight of water
Commonly used volume relationships for soil elements are void ratio, porosity, and degree of saturation. The void ratio is defined as the ratio between the pore volume and the volume of solid particles. So:
℮ = void ratio
Porosity is defined as the ratio between the pore volume and the total soil volume, or
n = porosity
The degree of saturation is defined as the ratio between the volume of water and the pore volume or
S = degree of saturation. Generally, the degree of saturation is expressed in percent.
The relationship between pores and porosity can be derived from the equation above, as follows:
Commonly used terms related to weight are water content and unit weight. The definitions of these terms are as follows:
The water content (w) which is also referred to as water content is defined as the ratio between the specific gravity and the weight of the solid particles of the volume of soil investigated.
The unit weight is the weight of the soil per unit volume. So,
Volume weight can also be expressed in terms of solid granular weight, moisture content, and total volume.
sometimes it is necessary to know the dry weight per unit volume of soil. This ratio is called the dry unit weight. So,
ᵧ ͩ = to ᵧ ͩ =
Volume weight is expressed in English units as: pounds per cubic foot (1b/ft3). In SI, the units used are newtons per cubic meter (N/m3). We can write the equations for density as follows:
ρ = and ρ ͩ =
ρ = soil density (kg/m3)
ρ ͩ = dry soil density (kg/m3)
m = total soil mass tested (kg)
ms = mass of solid particles of soil tested (kg)
The unit of total volume, V, is m3.
The unit weight of soil in units of N/m3 can be obtained from the density which has units of kg/m3 as follows:
ᵧ = ρ . g = 9.81 ρ and ᵧ ͩ = ρ ͩ .g = 9.81 m/s2
2.2.2 Relationship Between Volume Weight, Pore Number, Moisture Content, and Specific Weight
To obtain the relationship between unit weight, void ratio, and moisture content, consider a soil element in which the volume of solids is one. Since the volume of the solid particles is 1, the pore volume is equal to the void ratio. The weight of solid particles and water can be expressed as:
Ws = Gs ᵧʷ and Ww = wWs = w Gs ᵧʷ
Gs = specific weight of granular solids
w = water content
ᵧʷ = the unit weight of water
Based on the English system, the unit weight of water is 62.4 1b/ft3; in SI, the unit weight of water is 98.1 kN/m3.
Using the definitions of unit weight and dry unit weight, we can write:
ᵧ ͩ =
Since the weight of water in the element under consideration is wGsᵧʷ, the volume occupied by water is:
Therefore, the saturation weight is:
Or Se = wGs
If the soil sample is water saturated, that is, the pore space is completely filled with water, the unit weight of the saturated soil can be determined in the same way as above, namely:
ᵧsat = unit weight of unsaturated soil
2.2.3. Relationship Between Volume Weight, Porosity and Moisture Content
The relationship between unit weight, porosity and moisture content can be developed in the same way as before.
If V is equal to 1, then Vv is equal to n. So, Vs = 1- n . Solid grain weight (Ws) and water weight (Ww) can be expressed as follows:
Ws = Gsᵧʷ (1 – n)
Ww = wWs = wGsᵧʷ (1 – n)
So, the dry unit weight is equal to:
The unit weight of the soil is equal to:
The water content of a water-saturated soil can be expressed as:
2.2.4 Relative Density
The term relative density is generally used to indicate the degree of density of granular soils in the field. Relative density is defined as:
Dr = relative density
℮ = void ratio of soil in the field
℮max = void ratio of soil in the loosest state
℮min = void ratio of the soil in the densest state
Relative density values vary from the lowest price = 0 for very loose soils, to the highest price = 1 for very dense soils.
2.2.5 Soil Consistency
If the fine-grained soil contains clay minerals, then the soil can be kneaded without causing cracks. This cohesive property is due to the presence of water adsorbed around the surface of the clay particles. When the water content is very high, the mixture of soil and water will become very soft like a liquid. Therefore, on the basis of the water contained in the soil, soil can be separated into four basic states, namely: solid, semi-solid, plastic and liquid.
The water content is expressed in percent, where the transition from solid to semi-solid state occurs is defined as the shrinkage limit. The water content at which the transition from a semi-solid state to a plastic state occurs is called the plastic limit, and from a plastic state to a liquid state is called the liquid limit. These boundaries are also known as Atterberg limits.
Since the plastic properties of a soil are caused by water adsorbed around the surface of the clay particles, it can be expected that the type and amount of clay minerals contained in a soil will affect a plastic limit and a liquid limit of the soil in question. The relationship between PI and the clay-sized fraction for each soil has a different line. This situation is due to the different types of clay minerals contained in each soil. On the basis of the results of these studies, Skempton defines a quantity called activity which is the slope of the line which represents the relationship between PI and the percent of grains that pass the 2µ sieve, or it can also be written as follows:
A = activity
Activity is used as an index to identify the swelling capacity of a clay soil. The activity values for the various clay minerals are given in the table below.
Mineral Activity Table
Seed, Woodward, and Lundgren studied the plastic properties of several soils prepared by mixing different percentages of sand and clay. They concluded that although the relationship between the plastic index and the percentage of grains smaller than 2µ is a straight line, as studied by Skempton, the lines do not always pass through the center of the axis. Therefore, activity can be defined as follows:
Where C' is the constant of the studied soil.
For the results of the experiments carried out, C' = 9
Follow-up studies by Seed, Woodward, and Lundgren show that the relationship between the plasticity index and the percentage of clay-sized fraction in the soil can be represented by two straight lines. For soils containing a clay-sized fraction greater than 40%, the straight line will pass through the center of the axis when reprojected.
2.2.7 Soil Structure
Soil structure is defined as the geometric arrangement of the soil grains. Among the factors that influence soil structure are the shape, size, and mineral composition of soil particles and the nature and composition of groundwater. In general, soils can be classified into two groups: non-cohesive soils and cohesive soils. The soil structure for each group will be explained below.
Non-cohesive soil structures can generally be divided into two main categories: single grain structures and honeycomb structures. In a single grain structure, the soil grains are in a stable position and each grain is in contact with one another. The shape and size distribution of soil grains and their position affect the density of the soil. For an arrangement in a very loose state, the void ratio is 0.91. However, the void ratio decreases to 0.35 when the spherical grains of the same size are arranged in such a way that the bed becomes very dense. The condition of the original soil is different from the model above because the original soil grains do not have the same shape and size. In natural soils, the smallest sized grains occupy the voids between the larger grains. This situation shows a tendency towards a reduction in soil pore size. However, the unevenness of the grain shape generally causes a tendency towards increasing the void ratio of the soil. As a result of the two factors mentioned above, the original soil void ratio is approximately in the same range as the void ratio obtained from the soil model where the grain shape and size are the same.
In the honeycomb structure, fine sand and silt form small arcs to form a chain of grains. Soils that have a honeycomb structure have a large void ratio and are usually able to carry relatively low static loads. However, when the structure is subjected to heavy loads or when subjected to vibration loads, the soil structure will be damaged and cause large settlements.
2.3 Soil Classification
The soil classification system is a system for organizing several types of soil that are different but have similar characteristics into groups and sub-groups based on their uses. Most of the soil classification systems that have been developed for engineering purposes are based on simple soil index properties such as size distribution and plasticity.
2.3.1 Classification Based on Texture
In a general sense, what is meant by soil texture is the condition of the soil surface in question. Soil texture is influenced by the size of each grain in the soil. In general, native soil is a mixture of grains that have different sizes. In a soil classification system based on texture, soils are named on the basis of the main components they contain, for example sandy loam, silty clay and so on.
2.3.2 Classification Based on Usage
Classification based on texture is relatively simple because it is based only on the size distribution of the soil. In fact, the amount and type of clay minerals contained in the soil greatly affect the physical properties of the soil in question. Therefore, it is necessary to take into account the plasticity of the soil due to the presence of clay minerals, in order to interpret the characteristics of a soil. Because the textural classification system does not take into account soil plasticity and overall does not reflect any important soil properties, it is considered inadequate for most engineering purposes. At present there are two soil classification systems that are always used by civil engineering experts. These systems are:
The AASHTO Classification System was developed in 1929 as the Plublic Road Administration Classification System. This system has undergone several improvements. This classification is based on the following criteria:
1) Grain size :
Gravel: that part of the soil that passes through a 75 mm diameter sieve and is retained through the No.20 (2mm) sieve.
Sand: that part of the soil that passes the No. 10 (2mm) sieve and is retained on the No. 200 (0.075mm).
Silt and clay: the part of the soil that passes the No. sieve. 200.
The name silt is used when the fine parts of the soil have a plasticity index of 10 or less. The name loamy is used when the fine parts of the soil have a plastic index of 11 or more.
3) If rocks (size larger than 75mm) are found in the soil sample for which the soil classification will be determined, then these rocks must be removed first. However, the percentage of rock removed should be recorded.
The Unified Classification System was introduced by Casagrande in 1942 for use in the airfield loading works carried out by The Army Corps of Engineering during World War II. In cooperation with the United States Bureau of Reclamation in 1952, this system was perfected. This system classifies land into two major groups, namely:
1) Coarse-grained soil, namely: gravel and sand where less than 50% of the total weight of the soil sample passes sieve No.200. The symbols for this group start with the initial letter G or S. G stands for gravel or gravelly soil and S stands for sand or sandy soil.
2) Fine-grained soil (fine-granied-soil), namely soil where more than 50% of the total weight of the soil sample passes sieve No.200. The symbols for this group start with the initial letter M for inorganic silt, C for inorganic clay, and O for organic silt and organic clay.
Other symbols used for USCS classification:
W : Well Graded (soil with good gradation)
Q: Poorly Graded (soil with bad grades)
L : Low Plasticity (low plasticity) (LL<50)
H : High Plasticity (LL>50)
2.3.3 Comparison between the AASHTO System and the Unified System
The two classification systems, AASHTO and Unified, are based on soil texture and plasticity. Also both systems divide the soil into two main categories, namely: coarse-grained and fine-grained, which are separated by a No. sieve. 200. According to the AASHTO system, a soil is considered a fine-grained soil when more than 35% passes the No. sieve. 200. According to the Unified system, a soil is considered fine-grained soil if more than 50% passes through the No. sieve. 200. A coarse-grained soil containing about 35% fines will behave like a fine-grained material.
2.4 Permeability and Seepage
Soil is an arrangement of solid grains and pores that are interconnected with each other so that water can flow from one point that has a higher energy to a point that has a lower energy. The study of the flow of water through the pores of the soil is needed in mechanics, this is very useful in analyzing the stability of an earthen dam and retaining wall construction that is exposed to seepage forces.
2.4.1 Hydraulic Gradient
According to Bernaoulli's equation, the total energy head at a point in flowing water can be expressed as the sum of the pressure head, velocity head, and elevation height, or
high high high
elevation velocity pressure
h = total energy height
p = pressure
v = speed
g = acceleration due to gravity
= volumetric weight of water
If the Bernaulli equation above is used for water flowing through the pores of the soil, the part of the equation that contains the velocity head can be neglected. This is because the speed of seepage of water in the soil is very small. Therefore, the total energy head at a point can be expressed as follows:
The loss of energy between two points can be written by the following equation:
The energy loss ∆h can be expressed in the form of a dimensionless equation as shown below:
i = hydraulic gradient
L= the distance between points A and B, i.e. the length of the water flow over which the pressure loss occurs.
2.4.2 Darcy's law
In 1856, Darcy introduced a simple equation that is used to calculate the velocity of water flowing in saturated soil, expressed as follows:
in = which
v = flow rate
k = seepage coefficient
The seepage coefficient has the same unit as velocity. The term seepage coefficient is mostly used by soil engineering experts, the experts refer to it as hydraulic conductivity. Where English units are used, the seepage coefficient is expressed in ft/minute or ft/day, and the total volume in ft3. In SI units, the seepage coefficient is expressed in cm/sec, and the total volume is in cm3.
The seepage coefficient of soil depends on several factors, namely: liquid viscosity, pore size distribution, grain size distribution, void ratio, surface roughness of soil particles, and degree of soil saturation. In clay soils, the soil structure, ion concentration and thickness of the water layer adhering to the clay particles determine the seepage coefficient.
The price of the seepage coefficient for each soil is different. Some prices of seepage coefficients are given in the table below:
Type of soil
Less than 0.000001
Less than 0.000002
The seepage coefficient of unsaturated soil is low, this value will increase rapidly with increasing degree of saturation of the soil in question.
The seepage coefficient can also be related to the properties of the liquid flowing through the soil in question by the following equation:
= volumetric weight of water
= water viscosity
= absolute seepage
Absolute seepage, has units of L2 (ie cm2, ft2, etc.)
2.4.4 Determination of Seepage Coefficient in the Laboratory
There are two standard laboratory tests used to determine the seepage coefficient of a soil, namely: the constant height test and the falling height test. The drop height test is most suitable for fine-grained soils with small seepage coefficients.
2.4.5 Effect of Water Temperature on Prices k
The seepage coefficient is a function of the volume weight and thickness of the water, which means it is also a function of the temperature during the experiment, so it can be written:
kT1 , kT2 = seepage coefficient at temperatures T1 and T2
ηT1 , ηT2 = viscosity of water at temperatures T1 and T2
(T1) , (T2) = weight of water volume at temperature T1 and T2
2.4.6 Empirical Relationship for Seepage Coefficient
Several empirical equations for estimating the soil seepage coefficient have been introduced in the past.
For sandy soils with uniform grain size, hazen introduces an empirical relationship for the seepage coefficient in the following form:
k (cm/second) = cD210
c = a constant that varies from 1.0 to 1.5
D10 = effective size, in millimeters.
The above equation is based on the results of investigations carried out by Hazen on clean loose sand soils.
2.4.7 Equivalent Seepage in Layered Soils
The seepage coefficient of a soil may vary according to the direction of flow depending on the behavior of the soil in the field. For layered soils where the seepage coefficient of flow in a certain direction will change from layer to layer, it is necessary to determine the equivalent seepage value to simplify the calculation. So the following equation is obtained:
2.4.8 Seepage Test in the Field by Pumping from Wells
In the field, the average seepage coefficient in the direction of flow from a layer of soil can be determined by carrying out a pumping test from the well. The seepage coefficient in the direction of flow can be written as follows:
2.4.9 Seepage Coefficient of Auger Bore
The seepage coefficient in the field can also be estimated by making auger holes. This type of test is commonly referred to as a slug test. Holes are made in the field to a depth of L below the groundwater level. First of all the water is weighed out of the hole. This situation will cause groundwater to flow into the hole through the perimeter and bottom of the hole. The addition of water level in the auger hole and the time is recorded. The seepage coefficient can be determined from the data.
r = radius of the auger hole
y = average value of the distance between the water level in the auger hole and the groundwater level during the time interval ∆t (minutes).
Determination of the seepage coefficient of the auger hole is usually not able to provide accurate results, but it can provide a value of the power of k.
2.4.10 Continuity Equation
In actual circumstances, water flows in the soil not only in one direction and also not uniformly for the entire area perpendicular to the direction of flow. For such problems, groundwater flow calculations are generally made using graphs called flow networks. The flow network concept is based on the Laplace Continuity equation which describes the steady flow state for a point in the soil mass. The continuity equation for flow in the two dimensions above can be simplified to:
2.4.11 Water Network
The combination of several flow lines and equipotential lines is called a flow network. Flow networks are made to calculate groundwater flow, in making flow networks. The flow lines and equipotentials are drawn in such a way that:
1) The equipotential lines cut perpendicular to the flow lines
2) Flow elements are made to approximate a square shape.
2.4.12 Gradient at Exit and Safety Factor Against Boiling
If seepage under waterworks is not properly controlled, it will result in a large hydraulic gradient at the exit near the construction. The high gradient at the exit means that the seepage force is large, causing it to swell upward or cause the soil to lose strength. This situation will affect the stability of the water structure concerned.
2.5 Concept of Effective Stress
In a soil with a certain volume, the pore grains are related to each other so that they form a channel such as the compressibility of the soil, the bearing capacity of the foundation, the stability of the embankment, and the horizontal earth pressure in the construction of retaining walls.
2.5.1 Stress in Saturated Soil without Seepage
The total stress at point A can be calculated from the unit weight of the water-saturated soil and the unit weight of the water above it.
= H γw + (HA – H) γsat
= total stress at point A.
γw = volume weight of water.
γsat = unit weight of water-saturated soil.
H = height of the water level measured from the ground surface in the tube.
HA = distance between point A and the water surface.
2.5.2 In Water Saturated Soils with Seepage
The effective stress at a point in the soil mass will change due to the seepage of water through it. This effective stress will increase or decrease depending on the direction of seepage.
1) Water seepage upward.
Figure 5.3a shows a layer of granular soil inside a cylinder where there is upward seepage of water caused by the addition of water through channels at the bottom of the cylinder. The speed of adding water is fixed. The pressure loss due to upward seepage between points A and B is h. It should be remembered that the total stress at any point in the soil mass is due to the weight of the water and soil above that point.
At point A
Total voltage: A = H1 γw
Pore water pressure: uA = H1 γw
Effective stress: A' = A - uA = 0
At point B
Total voltage: B = H1 γw + H2γsat
Pore water pressure: uB= (H1 + H2 + h )γw
Effective stress: B' = H2γ' - h γw
In the same way, the effective stress at point C which lies at a depth z below the soil surface can be calculated as follows:
At point C.
Total voltage: C = H1 γw + zγsat
Pore water pressure: uC = γw
Effective stress: C' = zγ' - z
2) Seepage of Water Down.
The hydraulic gradient caused by downward seepage of water is equal to h/H2. The total stress, pore water pressure and effective stress at point C are:
C = H1 γw + zγsat
uC = (H1 + z – iz )γw
C' = (H1 γw + zγsat ) – (H1 + z – iz )γw
= zγ' - from γw
2.5.3 Seepage Force
In the previous sub-chapter, it has been explained that seepage can result in an increase or decrease in the effective stress at a point in the soil. It is shown that the effective stress at a point located at a depth z from the soil surface placed in the cylinder , where there is no seepage of water, is equal to zγ'. So the effective force on an area A is
P1' = zγ' A
If there is upward seepage of water through the soil layer in Figure 5.3, the effective force on area A at depth z can be written as follows:
P2' = ( zγ' - iz γw)A
Therefore, the reduction in net force as a result of seepage is:
P1' - P2' = iz γwA
The volume of soil on which the effective force acts is equal to zA. So the effective force per unit volume of soil is
= = i γw
The force per unit volume, iγw, for this situation acts upward, that is, in the direction of flow. Likewise for water seepage downwards, the seepage force per unit volume of soil is iγw.
2.5.4 Bulging in the Soil Caused by Seepage in the Pile Piling
The seepage force per unit volume of soil can be calculated to examine the possibility of failure of a sheet pile where seepage in the soil may cause heave downstream. After conducting many experimental models, Terzaghi (1922) concluded that swelling generally occurs in areas as far as D/2 from the sheet pile (where D is the depth of sheet piling). Therefore, we need to investigate the stability of the soil in the area D).
2.5.5 Effective Stress in Partially Saturated Soil
In a partially saturated soil, water does not fill all of the pore space in the soil. So, in this case there are 3 phase systems, namely solid grains, pore water and pore air. Therefore, the total stress at each point in the soil consists of the inter-grain pressure, pore water pressure, and pore air pressure. From the results of experiments in the laboratory , Bishop, Alpan, Blight, and Donald (1960) provide an effective stress equation for a partially saturated soil.
σ' = σ - ua + χ (ua – uw)
σ' = effective stress
σ = total stress
ua = pore air pressure
uw = pore water pressure
In the equation above, χ is the part of the cross-sectional area occupied by water. For dry soils χ = 0 and for water-saturated soils, χ = 1.
Bishop, Alpan, Blight, and Donald have shown that the mean value of χ is dependent on the degree of saturation (S) of the soil. But the price is also influenced by other factors such as soil structure.
The pore spaces in the soil that are connected to each other can behave as collections of capillary tubes with varying cross-sectional areas. The height of the water rise in the capillary tube can be written by the following formula:
Τ = surface tension
α = contact angle between the water surface and the capillary wall
d = diameter of the capillary tube
= volumetric weight of water
from the equation above it can be seen that the prices of Τ α and γw are fixed, then:
Although the concept of capillary rise as demonstrated by an ideal capillary tube can be used in soil, it should be noted that the capillary tube formed in the soil has a varying cross-sectional area. The result of non-uniformity of capillary water rise can be seen when a dry sandy soil inside the cylinder is placed in contact with water.
Hazen (1930) provides a formula for determining the height of the capillary water rise by approximation, namely:
= effective size (in mm)
e = void ratio
C = a constant that varies from 10 mm2 to 50 mm2
Effective stress in the capillary water rise zone
The general relationship between the total stress, effective stress, and pore water pressure is given in the following equation:
= ' + u
The pore water pressure u at a point in a layer of soil that is 100% saturated with capillary water is equal to -γwh (h=height of a point in terms of the groundwater table) with atmospheric pressure taken as a datum. If there is a partially saturated layer of water caused by capillary action, then the pore water pressure can be written as follows:
u = -
S = degree of saturation, in percent.
2.6 Stresses in a Soil Mass
In soils that must support foundations of various shapes, stress increases generally occur. The increase in stress on the soil depends on the load per unit area where the foundation is located, the depth of the soil under the foundation where the stress is considered, and other factors.
2.6.1 Normal Stress and Shear Stress in a Plane
Normal stresses and shear stresses acting in any plane can be determined by drawing a Mohr circle. The sign convention used in Mohr's circle here is: the normal compressive stress is considered positive, the shear stress is considered positive if the shear stress acting on opposite sides of the square stress element rotates in a counterclockwise direction.
There is still another important way to determine the stresses in a plane using the Mohr circle, namely the method of poles, or the method of the center of the plane.
2.6.2 Stresses Due to Concentrated Loads
Boussinesq has solved the problem associated with determining the stresses at any point in a homogeneous, elastic, and isotropic medium where the medium is of infinite area and a concentrated load acts on its surface. The Boussinesq formula for the normal stress at point A due to concentrated load P is:
It should be remembered that the equations, which are normal stresses in the horizontal direction, are dependent on the medium's Poisson number. In contrast, the vertical stress, ∆pz as above is independent of the Poisson number.
2.6.3 Vertical Stress Due to Line Loads
The increase in vertical stress, ∆p, in the soil mass can be calculated using the basics of elastic theory as follows:
The above equation can be written in the following form:
The above equation is a dimensionless form of the equation. With this equation, the variation of ∆p /(q/z) with respect to x/z can be calculated. The value of ∆p calculated from the above equation is the additional stress on the ground caused by the line load.
2.6.4 Vertical Stress Due to Lane Loads
The basic equation for the vertical stress increase at a point in a soil mass due to line loading can also be used to determine the vertical stress at a point due to bending strip loading.
2.6.5 Vertical Stress Under the Circular Even Load Center Point
By using the Boussinesq solution for the vertical stress ∆pz resulting from a concentrated load, we can also determine the magnitude of the vertical stress below the center of the bending circle which is subjected to a uniformly distributed load.
2.6.6 Vertical Stress Caused by Rectangular Loads
The Boussnesq formula can also be used to calculate the added vertical stress under a rectangular bending load
2.6.7 Influence Diagram for Voltage
The procedure used to obtain the vertical stress at any point under a load area is as follows:
1) Determine the depth of point z below the area that gets the load evenly distributed where the vertical stress increase at that point is to be determined.
2) Describe the area of the load with the length of a graph (AB).
3) Place the plan on the influence diagram in such a way that the projection of the point where the voltage increase will be sought coincides with the center point of the influence diagram.
4) Calculate the total number of area elements from the diagram included in the load area plan.
The value of the increase in voltage at the point under review can be found by the formula:
∆p = (AP)qM
AP = influence number
q = uniformly distributed load on the area under consideration (unit load/unit area)
2.7. Soil Compression
The addition of a load above a soil surface can cause the subsoil to experience compression. This compression is caused by the deformation of soil particles, the relocation of particles, the release of water or air from the pores, and other causes. In general, settlement in the soil caused by loading can be divided into two major groups, namely:
1) Consolidation settlement, which is the result of changes in the volume of water-saturated soil as a result of the release of water that occupies the pores of the soil.
2) Immediate settlement, which is the result of elastic deformation of dry, wet and water-saturated soils without any change in water content.
2.7.1 Basics of Consolidation
When a water-saturated soil layer is given an additional load, the pore water pressure will increase suddenly. In highly permeable sandy soils, water can flow quickly. The discharge of water from the pores is always accompanied by a decrease in soil volume, the reduced volume of the soil can cause subsidence of the soil layer. Because the pore water in sandy soils can flow out quickly, immediate settlement and consolidation settlement occur together.
When a layer of compressible water-saturated clay is given an added stress, settlement will occur immediately. The clay seepage coefficient is very small compared to the sand seepage coefficient so that the increase in pore water pressure caused by loading will decrease gradually over a very long time. So for soft clay soils, volume changes caused by the release of water from the pores (ie consolidation) will occur after immediate settlement. The consolidation settlement is usually much larger and slower and longer than the immediate settlement.
Deformation as a function of time of water-saturated clay can be understood easily when a simple rheological model is used. The rheological model consists of a linear elastic spring connected in parallel with a dashpot. The stress-stress relationship of the spring and dashpot can be given as follows:
Spring : σ =
Dashpot : σ = n
= a toy
= spring constant
η = dashpot constant
t = time
2.7.2 Graph of Pore Number
The following is a step by step sequence of execution:
1) Calculate the height of the solid grains Hs
= dry weight of soil sample
A = cross-sectional area of the soil sample
= specific weight of the soil sample
= volumetric weight of water
2) Calculate the initial height of the pore space Hv
Hv = H – Hs
Where: H = initial height of the soil sample
3) Calculate the initial pore number:
4) For the addition of the first load p1 (total load/sectional area of the soil sample), which causes a decrease in ΔH1, calculate the change in void ratio, Δe1 :
ΔH1 is obtained from the initial and final readings on the measuring scale for a load of p1.
5) Calculate the new void ratio, e1 after consolidation caused by the increased pressure p1 :
2.7.3 Normally Consolidated or Overly Consolidated Clays
A soil in the field at a certain depth has experienced the "maximum effective pressure due to the weight of the soil above it" in its geological history. This maximum effective overburden pressure may be equal to or less than the overburden pressure that existed at the time of sampling. The reduced pressure in the field may be caused by natural geological processes or processes caused by living things. During this time, the soil will expand as a result. When the soil sample is subjected to a consolidation test, a small amount of compression will occur if the total load applied during the experiment is less than the maximum effective overburden pressure previously experienced by the soil in question. If,
This situation can be proven in the laboratory by loading the soil sample beyond its maximum overburden pressure, then lifting the load and applying it again.
This situation leads us to two basic definitions based on stress history:
1) Normally consolidated, where the effective overburden pressure at this time is the maximum pressure ever experienced by the soil.
2) Overly consolidated, where the effective overburden pressure at this time is less than the pressure that the soil has experienced before. The maximum effective overburden pressure that has been previously experienced is called the preconsolidation pressure.
2.7.4 The Effect of Damage to Soil Structure on the Relationship Between Pore Number and Pressure
A soil sample is said to be "reformed" if the structure of the soil is disturbed. This situation will affect the shape of the graph showing the void ratio and pressure of the soil in question. For a normally consolidated clay with a low to moderate degree of sensitivity and void ratio eo and effective overburden pressure po, the void ratio changes as a result of added stress. rough field.
For highly consolidated clays with a low to moderate degree of sensitivity and have experienced preconsolidation pressure pc and void ratio eo and effective overburden pressure po.
With the knowledge gained from the analysis of the consolidation test results, we can now calculate the possible settlement caused by primary consolidation in the field by assuming that the consolidation is one dimensional.
Now let's consider a layer of saturated clay with thickness H and cross-sectional area A and an average effective overburden pressure of po. Caused by an increase in pressure of Δp, assume that the decrease in primary consolidation that occurs is S. So the change in volume can be given as follows:
ΔV = Vo – V1 = H . A – (H – S) . A = S . A
Where: Vo and V1 are the initial volume and final volume of the pore, ΔVv, so:
ΔV = S . A = Vv0 – Vv1 = ΔVv
Where: V v0 and V v1 are the initial volume and final volume of the pore, respectively.
2.7.5 Compression Index
The compression index used to calculate the amount of settlement that occurs in the field as a result of consolidation can be determined from the curve showing the relationship between void ratio and pressure obtained from consolidation tests in the laboratory.
1) Expansion index
The index of expansion is less than the index of compression and can usually be determined in the laboratory, in general. Liquid limit, plastic limit, compression index, and expansion index for soils that are not yet degraded
2) Decrease caused by secondary consolidation.
At the end of primary consolidation, settlement still occurs as a result of plastic adjustment of the soil grains. This consolidation stage is called secondary consolidation. During secondary consolidation, the relationship curve between deformation and log time is a straight line. Variation of void ratio and time for an additional load will be the same. The secondary compression index can be defined as.
= secondary compression index
= change in void ratio
t1 . t2 = time
2.7.6 Speed of Consolidation Time
The total settlement due to primary consolidation caused by the addition of stress above the ground surface can be calculated using the equations.
The mathematical derivation of the equation is based on the following assumptions:
1) The soil (clay-water system) is homogeneous.
2) The soil is completely saturated.
3) The compressibility of water is neglected.
4) The compressibility of the soil particles is neglected.
5) Water flow is only one way.
6) Darcy's law applies.
2.7.7 Consolidation Coefficient
The consolidation coefficient usually decreases as the liquid limit of the soil increases. The range of variations in cv values for a particular soil liquid limit is rather wide.
For an additional load applied to a soil sample there are two graphical methods commonly used to determine cv values from a laboratory one-dimensional consolidation test. One of the two methods is called the logarithmic time method introduced by Casagrande and Fadum, while the other method is called the time root method introduced by Taylor.
The addition of vertical stresses in the soil caused by loads with a limited area will decrease with increasing depth z as measured from the soil surface down. The calculation of the addition of Δp to these equations should be the addition of the average pressure , namely:
2.7.8 Calculation of Immediate Decrease Based on Elastic Theory
Immediate settlement for foundations that are on elastic material can be calculated from the equations derived using the basic principles of elastic theory. The form of the equation is as follows:
= elastic drop
= net applied pressure
B = width of foundation (= diameter of circular foundation)
= Poisson's number
= soil elasticity modulus (young's modulus)
= influence factors that do not have dimensions
2.7.9 Decrease in Total Foundation
The total settlement of a foundation can be given as follows:
ST = S + Ss + ρi
ST = total decrease
S = decrease due to primary consolidation
Ss = decrease due to secondary consolidation
ρi = immediate decline
example of a decline in the field
At present there are many available literature examples of incidents where the basic principle of soil compressibility is used to enrich the amount of settlement that occurs in a layer of soil in a field that is given an additional load. In some cases, the magnitude of the decline that occurred in the field was one or nearly the same as the estimated magnitude of the decline. In another incident, the estimated decline was far deviated from the actual decline that occurred on the ground. The discrepancy between the estimated decline and the actual decline in the field may be caused by several reasons, including:
1) the evaluation of soil properties that was carried out turned out to be incorrect.
2) the soil layer was not homogeneous and irregular.
3) error in evaluating the increase in net stress with depth, which in fact greatly affects the magnitude of settlement.
2.8 Soil Compaction
In compaction of highway embankments, earthen dams, and many other engineering structures, loose soil must be compacted to increase its unit weight. The compaction serves to increase the strength of the soil, thereby increasing the bearing capacity of the foundation above it. Compaction can also reduce the amount of unwanted soil settlement and increase the occupancy of embankment slopes.
2.8.1 Compaction and General Principles
The level of soil compaction is measured from the dry unit weight of the compacted soil. When water is added to a soil that is being compacted, it acts as a wetting agent on the soil particles. For the same compaction effort, the dry unit weight of the soil will increase as the water content in the soil increases. Please note that when the moisture content w = 0, the wet unit weight of the soil is the same as the dry unit weight.
If the water content is increased gradually with the same compaction effort, then the weight of the amount of solid matter in the soil per unit volume also increases gradually. The dry unit weight of the soil at water content can be expressed:
After reaching a certain water content w = w2, the addition of water content actually tends to reduce the dry unit weight of the soil. This is because the water then occupies pore spaces in the soil which can actually be occupied by solid particles from the soil. The water content at which the maximum dry unit weight of the soil is achieved is the optimal water content.
Laboratory experiments that are commonly carried out to obtain maximum dry unit weight and optimum moisture content are proctor compaction tests.
2.8.2 Factors Affecting Density
Water content has a large influence on the level of density that can be achieved by a soil. Besides water content, other factors that also affect compaction are soil type and compaction effort.
Lee and Sedkamp (1972) studied compaction curves of 35 soil types. They concluded that the compaction curves of these soils could be distinguished into only four general types.
The energy required for compaction in the Standard Proctor test can be written as follows:
From the compaction curves for the four soil types (ASTM D-698) it can be seen that:
1) When the compaction energy increases, the maximum dry unit weight of the compacted soil also increases, and
2) If the compaction energy increases, the optimum water content decreases.
2.8.3 Modified protector test
With the development of heavy rollers used for field compaction, the standard proctor test had to be modified to better represent site conditions. This modified proctor test is called the Modified proctor test. The compaction energy carried out in the modified test can be calculated as follows:
= 56.250 ft-1b/ft3 (≈2693.3 kJ/m3)
Because the compaction energy is greater, the modified proctor test also produces a higher maximum dry unit weight value. The increase in the maximum dry unit weight is accompanied by a decrease in the optimum water content.
2.8.4 ASTM and AASHTO Specifications for Compaction Tests
The specifications given for the Proctor test according to ASTM and AASHTO with a mold volume of 1/30 ft3 and 25 blows per layer are generally used for fine-grained soils that pass American No. sieve. 4. Actually, at each mold size there are still four other methods suggested, which vary according to the mold size, the number of impacts per layer, and the maximum soil particle size in the compacted soil aggregate.
2.8.5 Structures of Compacted Cohesive Soil
Lambe has investigated the effect of compaction on clay structure. At a certain moisture content, a higher compaction effort tends to produce more clay particles with a parallel orientation, so that more of the soil structure is dispersed. Soil particles are closer to each other and have a higher unit weight by themselves. The investigation by Seed and Chand also yielded similar results for compacted kaolin clay.
2.8.6 Effects of Compaction on the Properties of Cohesive Soils
Compaction causes changes to the cohesive soil structure. These changes include changes in the permeability, compressibility, and strength of the soil.
The one-dimensional compressibility properties of clay soil that is compacted on the dry side and the wet side of the optimum rate is at low pressure, a soil that is compacted on the wet side of the optimum rate will be easier to compress than the soil that is compacted on the dry side of the optimum water rate. The strength of compacted clay soils generally decreases with increasing water content. Please note that around the optimum water level, there is a large decrease in soil strength.
2.8.7 Compaction in the Field
Almost all compaction in the field is done with a roller. The types of rollers commonly used are:
1) Iron roller with a smooth surface
2) Rubber-tire roller (air)
3) Goat leg roller, and
4) Vibrating roller.
The smooth-faced iron roller is suitable for leveling the surface of subgrades and for finishing work on sand or clay fills.
Tire-rubber rollers are in many ways better than smooth-faced metal rollers. A rubber-tyre roller is basically a heavily loaded, rubber-wheeled cart arranged in several closely spaced rows.
The goat leg roller is in the form of a cylinder which has many legs protruding from the drum. This tool is very effective for compacting clay soil.
Vibrating rollers are especially useful for compaction of granular soils (sand, gravel, etc.) Any vibrating tool is attached to a flat iron roller, a rubber-tire roller, or to a goat's foot roller to generate vibration in the soil.
2.8.8 Specifications for Field Compaction
In most specifications for earthworks, the contractor is required to achieve a field density of 90 to 95% of the maximum dry unit weight of the soil.
In compaction of granular soils, the compaction specification is sometimes given in terms of the relative density term Dr. Please do not confuse relative density with relative density. The definition of Dr is:
Based on observations of 47 soil samples, Lee and Singh provide a correlation between the R and Dr of granular soils:
R = 80 + 0,2Dr
2.8.9 Organic Soil Compaction
The presence of organic matter in a soil tends to reduce the strength of the soil. In general, soil with a high organic matter content is not used as fill soil. However, due to certain economic reasons, sometimes soil with a low organic matter content has to be used for compaction. The organic content (OC) of a soil is defined as follows:
In the investigation conducted by Franklin, Orozco, and Semrau in the laboratory to investigate the effect of organic content on soil composition properties, it can be concluded that soil with an organic content higher than 10% is not good for compaction work.
2.8.10 Determination of Volume Weight Due to Compaction in the Field
When the compaction work is underway, it is of course necessary to know whether the volume weight specified in the specifications can be achieved or not. The standard procedure for determining the unit weight due to compaction is:
1) Sand cone method
2) The rubber balloon method
3) Use of nuclear density measuring devices
The sand cone consists of a plastic or glass bottle with a metal cone attached to the top. These plastic bottles and cones are filled with poorly graded dry Ottawa sand. In the field, a small hole is dug in the compacted soil surface. If the weight of the wet soil excavated from the hole can be determined and the water content of the excavated soil is also known. After the hole has been dug, a cone with the bottle filled with sand is placed over the hole. The sand is allowed to flow out of the bottle filling the entire hole and cone. After that, the weight of the tube, cone, and remaining sand in the bottle is weighed. So,
W5 = W1 – W4
Ws = weight of sand filling the hole and the volume cone of the hole excavated can be determined as follows:
Wc = weight of sand that only fills the cone
= dry unit weight of Ottawa sand
The values of Wc and ᵧd(sand) are determined by calibration in the laboratory. So the dry unit weight resulting from field compaction can now be determined as follows:
The procedure for carrying out the rubber balloon method is the same as for the sand cone method, namely a test hole is dug and the original soil is taken from the hole and weighed. But the volume of the hole is determined by placing a rubber balloon filled with water in the hole. This water comes from a vessel that has been calibrated, so that the volume of water that fills the hole (same as the volume of the hole) can be read directly. The dry unit weight of compacted soil can be determined by the above equation.
The nuclear compaction meter is now used in some cases to determine the dry unit weight of compacted soil. This tool can be operated in a dug hole or soil surface. This tool can measure the weight of wet soil per unit volume and also the weight of water present in a unit volume of soil. The dry unit weight of soil can be determined by subtracting the wet weight of the soil by subtracting the wet weight. soil to water west per unit volume of soil.
2.8.11 Special Compaction Techniques
Several types of special compaction techniques have recently been developed, and these special types have been implemented in the field for large scale compaction works. Among the well-known methods are vibratory compaction, dynamic compaction, blast compaction, loading and pumping of water from the soil.
CHAPTER 3. CLOSING
1) Soil is defined as a material consisting of aggregates (granules) of solid minerals which are not cemented (chemically bound) to each other from decayed organic matter (which has solid particles) accompanied by liquid and gas filling the spaces. empty space between the solid particles. Soil comes from the weathering of rocks with the help of organisms, forming a unique body that covers the rock. The process of soil formation is known as pedogenesis. Two methods are commonly used to obtain the size distribution of soil particles, namely: sieve analysis and hydrometer analysis.
2) If the fine-grained soil contains clay minerals, then the soil can be kneaded without causing cracks. This cohesive property is due to the presence of water adsorbed around the surface of the clay particles. The plastic index of a soil increases along a line according to the increase in the percentage of the clay-sized fraction contained in the soil.
3) Soil classification system based on texture is considered inadequate for most of the engineering purposes. At present there are two soil classification systems, namely the AASHTO classification system and the Unified classification system.
4) The coefficient of seepage of soil depends on several factors, namely the viscosity of the liquid, the distribution of pore sizes, the distribution of grain sizes, the void ratio, the surface roughness of the soil particles, and the degree of soil saturation. The seepage coefficient is a function of the volume weight and viscosity of the water, which means it is also a function of the temperature during the experiment.
5) The effective stress at a point in the soil mass will change due to the seepage of water through it. This effective stress will increase or decrease depending on the direction of seepage
6) The increase in stress on the soil depends on the load per unit area where the foundation is located, the depth of the soil under the foundation where the stress is considered, and other factors.
7) For soft clay soils, volume changes caused by the release of water from the pores (i.e. consolidation) will occur immediately after settlement. These consolidation downturns are usually much larger and slower and longer than the immediate downturn. The compression index used to calculate the amount of settlement that occurs in the field as a result of consolidation can be determined from the curve showing the relationship between void ratio and pressure obtained from consolidation tests in the laboratory.
8) For the same compaction effort, the dry unit weight of the soil will increase if the water content in the soil increases. Water content has a large influence on the level of density that can be achieved by a soil. Compaction causes changes to the cohesive soil structure. These changes include changes in the permeability, compressibility, and strength of the soil.
M. Das Braja, Braja M. Das, Endah Noor, B. Mochtar 1985. Soil mechanics (Geotechnical Engineering Principles) Volume I. Surabaya: University Institute of technology 10 November.