Basic Soil Mechanics Chapter 2

Physical Soil State

Author: Eze-Odikwa Tochukwu Jed

Note: All articles posted here are accurate, up-to-date and drafted from real university curriculums. Proper references will be added at the bottom of this article upon its completion. 

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College Reg Number: MOUAU/CME/14/18475

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2.1 Phase Relationship:

Soil is comprised of solids, liquids and gases. The solid phase may be minerals, organic matter or both.  The spaces between the solid particles are called voids. Water is often the predominant liquid and air the predominant gas. Therefore, the terms “Water” and “Air” will be used instead of liquid and gases.

The soil water is called Pore water and plays a very important role in the behavior of soils under load. If all the voids are filled with air, the soil is said to be a dry soil. If all the voids are filled with water, the soil is said to be saturated soil, otherwise it is unsaturated.

The notations are defined as follows: V_{V =} Volume of Solids, V_V = Volume of void, V_a = volume of air, While V_V = Volume of water.

The physical properties of soil are influenced by relative proportions of each three phases.

The total volume of the soil V from the diagram (Fig.2.1) is the sum of the volume of air V_a, volume of water V_w, and volume of solid V_s. this is given by the formula

V = V_s+V_w+V_a                                  (2.1)

Recall that V_V = V_a+V_w                    (2.2)

Therefore V = V_V+V_a                        (2.3)

Weight of soil, is the sum of the weight of solid (W_a), and weight of water (W_w) because weight of air, (W_a) is negligible.

W = W_a+W_w+W_s                            (2.4)

Since W_a is negligible, i.e. W_a = 0

W = W_w+W_a                                  

W = \frac{Ww}{Ws}\times100\% (2.5)

2.2 Water or Moisture Content:

Moisture content is the ratio of weight of water to the weight of solids often expressed in percentage. It is represented with w.

Moisture content,                            (2.6)

The particle size distribution of a soil affects its moisture content, for instance fine grained soils such as silts and clays will generally have more moisture content than that of the coarse grained soils, such as gravels and sands. The water content of a soil is a very important parameter in both agricultural and civil engineering as it is used in the determination of the plasticity, liquidity, shrinkage limit and other parameters of the soil, such as field capacity and permanent writing point which aids irrigation system design.

The water content of a soil is found by weighing a sample of the soil and placing it in an oven at 110 + 5°C until the weight of the sample remains constant (in about 24 hours). The soil is removed from the oven, cooled and re-weighed.

The detailed procedure to determine the moisture content can be obtained in American Society for Testing of Materials (ASTM) D2216. It is a common mistake to use the total weight as the denominator. Remember, it is weight of solids as the denominator; but in geology and other disciplines moisture content is defined as the ratio of the mass of water to the total mass.

Procedure for determining moisture content of soil.

• Weigh a soil sample in a can
• Oven dry the sample
• Weigh oven dry sample for 24hours

Note:

Weight of can = W1

Weight of wet soil = W2

Weight of dry soil = W3

Ww = W2 – W3 (weight of water)

Ws = W3 – W1 (weight of solids)

W = \frac{W2-W3}{W3-W1}\times100\% (2.7)

2.3 Volumetric Relationships

The following are the volumetric relationships parameters of soil widely used in soil engineering.

Viod ratio (e)

This is the ratio of the volume of void space to the volume of solids. Void ratio is usually expressed as e.

e = \frac{Vv}{Vs} (2.8)

The viod ratio of soil is also affected by the particle size distribution of the soil (coarse grained soils will most likely have greater void ratios than that of fine grained soils).

Porosity (n)

Porosity is the ratio of the volume of voids to the total volume. It is usually expressed as n in percentage. The porosity of soil is also known as percentage of void and it cannot exceed 100% because it will mean that the volume of the void is more than the total solids which is not possible. Like the particle size distribution, the denseness or looseness of a soil also affects its porosity. The more the dense a soil the less porous and vice versa.

n = \frac{Vv}{V} (2.9a)

Prosody and void ratio are related by the expression

n = \frac{e}{1+e}

And this is shown below

Recall that n = \frac{Vv}{V}=\frac{Vv}{Vs+Vv}

Dividing the numerator and denominator by the volume of solids, V_s

n = \frac{\frac{Vv}{Vs}}{\frac{Vs}{Vs}+\frac{Vv}{Vs}} (2.9b)

Recall that e = \frac{Vv}{Vs}

Therefore we replace \frac{Vv}{Vs} as e in the equation (2.9b) above.

Therefore n = \frac{e}{1+e}

Also if we make e the subject of the formula from eqn. (2.10) we have

n(1+e) = e

n + ne = e

e – ne = n

e(1-n) = n

e = \frac{n}{1-n} (2.11)

Degree of saturation (S)

It is the ratio of the volume of water to the volume of voids often expressed in percentage.

S = \frac{vw}{Vv}\times\frac{100}{1} (2.12)

It is also given as

S = \frac{WGs}{e}\times\frac{100}{1}=\frac{Water Content\times Specific Graviy }{Void ratio}

This is proved below

S = \frac{Vw}{Vv}, since, e = \frac{Vv}{Vs}

V_v = eV_s (a)

Recall that density p = \frac{M}{V}

While unite weight y = \frac{W}{V}

If unit weight of water Yw = \frac{Ww}{Vw}

Therefore the volume of the water will be Vw = \frac{Ww}{Yw} (b)

Recall that moisture content, W = \frac{Ww}{Ws}

Ww = wWs (c)

Putting equation (c) into (b)

Ww = \frac{wWs}{Yw} (d)

Recall that S = \frac{Vw}{Vv}

S = \frac{\frac{wWs}{Yw}}{eVs}

S = \frac{{wWs}}{Yw}\times\frac{1}{eVs}, or, S = \frac{{wWs}}{eVsYw}

Recall that Gs = \frac{{Ws}}{VsYw}

S = \frac{{WGs}}{e} (2.13)

If S = 1 or 100%, the soil is saturated. If S = 0, the soil is completely dry, it is practically impossible to obtain a soil with S = 0.

Swell Factor (SF)

Swell or free swell factor is the ratio of the volume of excavated material to the volume of in situ material. Volume of container filled with the soil is the volume of excavated material. Volume of the space left behind after excavation is the volume of in-situ material.

If SF decreases, it shows that the soil shrinks when excavated. If SF increases, it shows that the soil expands when excavated.

2.4 Volume-Weight Relationships

Unit weight: is the weight of soil per unit volume

Y=\frac{W}{V}=\frac{TotalWeight}{Total Volume} (2.15)

The unit weight of a soil is given by the formula Y=\frac{Gs+Se}{1+e}

Where the parameters given in the equation is the same as earlier stated while Y = ordinary unit weight or bulk or moist unit weight. The relationship is proved below:

Y=\frac{W}{V}=\frac{Ws+Ww}{Vs+Vv}=\frac{Ws+wWs}{Vs+Vv} (a)

Recall that:

W_w = wW_s and e=\frac{Vv}{Vs}

S=\frac{wGs}{e},Vv=eVs

w=\frac{Se}{Gs}

Substituting equation (b) and equation (c) into equation (a)

Y=\frac{Ws+wWs}{Vs+Vv}

Y=\frac{Ws+\frac{Se}{Gs}Ws}{Vs+eVs}

Factorizing we have:

Y=\frac{Ws(1+\frac{Se}{Gs})}{Vs(1+e)} (d)

Recall that

G_{s} = \frac{Ws}{VsYw}

G_{s}Y_{w} = \frac{Ws}{Vs} (e)

Substituting equation (e) into equation (d) we have

S=\frac{\frac{wWs}{Yw}}{eVs}

S = \frac{wWs}{Yw}\times\frac{1}{eVs}, or, S = \frac{wWs}{eVsYw}

Y = \frac{(Gs+Se)Yw}{1+e} (2.16)

Note: in a situation where the soil is saturated the formular in equation (2.16) will be reduced to

Y_{sat} = \left(\begin{array}{c} \frac{Gs+e}{1+e}\end{array}\right)Y_{w}(S=1)

Dry Unit Weight of Soil (yd)

The ratio of weight of dry soil per unit volume is known as dry unit weight

Y_{d} = \frac{Ws}{V} (2.17)

The value of the dry unit weight can also be given by the equation

Y_{d} = \frac{Y}{1+W}

Where Y = unit weight or bulk or moist unit weight. W = water content, this is proven below.

Y_{d} = \frac{Y}{1+W}

Y_{d} = \frac{Ws}{V}=\frac{W-wWw}{V}=\frac{W-wWs}{V}

Y_{d} = \frac{W}{V}-\frac{wWs}{V}

Y_{d} = Y-WY_{d}

Y = Y_{d}+WY_{d}

Y = Y_{d}(1+W)

Y_{d} = \frac{Y}{1+W} (2.18)

Saturated Unit weight of Soil (Y_sat)

Saturated unit weight is the unit weight of the soil when it is fully submerged in water.

Y_{sat} = \frac{Wsat}{V} (2.19)

2.5 Volume-Mass Relationship

Bulk Mass Density: The bulk mass density (which is also known as wet mass density, bulk density or simply density) is defined as:

P = \frac{M}{V} (2.20)

When M = Total Mass and V = Unit Total Volume its unit of measurement are:

\frac{Kg}{m^{3}},\frac{g}{m^{1}},\frac{KN}{m^{3}}

It is very important to understand how to convert from one unit to another. Conversion unit can be seen in the appendix.

Dry Density: This is defined as the mass of solids per unit total volume.

p_{d} = \frac{Ms}{V} (2.21)

Note: The mass of the soil measured before the drying of the soil and a high value of the dry mass density or dry density shows that the soil is compact.

Relative Density: The term “relative density” is commonly used to indicate the in-situ (on site) denseness or looseness (how compacted or loose soil material is). Relative density is defined as:

D_{r} = \frac{e_{max}-e}{e_{max}-e_{min}} (2.22)

Where D_r = relative density

e = in-situ void ratio of the soil (natural state)

e_max = void ratio of soil in the loosest condition

e_min = void ratio of soil in the most dense condition

The values of D, may vary from a minimum of 0 (zero) for a very loose soil to a maximum of 1 (one) for a very dense soil.

By using the definition of dry unit weight, we can also express relative density in term of maximum or minimum dry unit weight.

Thus, D_{r}= \frac{{\begin{bmatrix}\frac{{1}}{Yd_{min}} \end{bmatrix}-\begin{bmatrix}\frac{{1}}{Yd} \end{bmatrix}}}{\begin{bmatrix}\frac{{1}}{Yd_{min}} \end{bmatrix}-\begin{bmatrix}\frac{{1}}{Yd_{max}} \end{bmatrix}} (2.23)

D_r is inversely proportional to Y_d

The equation (2.22) can be simplified further to obtain:

D_{r} = \frac{\frac{\frac{Yd-Yd_{min}}{YdYd_{min}}}{Yd_{max}-Yd_{min}}}{{Yd_{max}Yd_{min}}}

D_{r}=\frac{{Yd-Yd_{min}}}{YdYd_{min}}\times\frac{{Yd_{max}Yd_{min}}}{Yd_{max}-Yd_{min}}

D_{r}=\frac{{Yd-Yd_{min}}}{Yd(Yd_{max}-Yd_{min})} (2.24)

Where Y_{d min} = dry unit in the loosest condition

Y_d = in – situ dry unit weight

y_{d max} = dry unit weight

Saturated Density: is the mass density of the soil when its fully saturated.

P_{sat}=\frac{M^{sat}}{V} (2.25)

2.6 Specific Gravity (G_S): This is the ratio of the weight of soil solids to the weight of equal volume of water

G_{s}=\frac{W^{S}}{V^{W}} (2.26)

Since V_w = V_sY_w

therefore G_{s}=\frac{W^{S}}{V^{s}Y^{w}} (2.27)

Y_w = unit weight of water Y_w = 9.81KN/m³

The specific gravity of most soils ranges from 2.6 to 2.8. The procedure to determine the specific gravity of the soil is given in ASTM D854.

Two types of containers are used to determine the specific gravity:

1. Pycnometer (which is used for coarse grained soiled)
2. 50ml density bottle (which is used for fine soils)

It is obtained as follows:

W₁ = weight of empty density bottle or weight of empty pycnometer.

W₂ = weight of empty density bottle of pycnometer + sample of soil

W₃ = weight of density bottle/pycnometer + sample of soil + distilled water

Reason for using distilled water is to avoid elements or compounds that can or could react with the soil.

W₄ = weight of density bottle/pycnometer + distilled water

G_{s}=\frac{W_{2}-W_{1}}{(W_{4}-W_{1})-(W_{3}-W_{2})} (2.28)

Example 2.1

In a natural state, a soil has a volume of 0.0075m³ and weighs 155N. After oven drying, the soil weighs 136.4N. Calculate the following:

1. Moisture Content
2. Moist Unit Weight
3. Dry Unit Weight
4. Void ratio
5. Porosity
6. Degree of saturation

(Take G_s = 2.68, Y_w = 9.81KN/m³)

Solution:

1.

Moisture Content, W=\frac{W_{w}}{W_{s}}\times100\%

W = 155N

W_s = 136.4N

W_s = weight after oven drying

W=\frac{W-W_{s}}{W_{s}}\times100\%

W=\frac{155-136.4}{136.44}\times100\%

W = 13.6%

2.

Y =\frac{W}{V}=\frac{155N}{0.0075m^{3}}

Y = 20666.7KN/m^{3} = 20.7KN/m^{3}

3.

Y_{d} =\frac{W_{s}}{V}=\frac{136.4}{0.0075}

= 18,186.7N/m^{3};Y_{d} =18.2KN/m^{3}

4.

e =\frac{V_{v}}{V_{s}}

Recall that G_{s} =\frac{W_{s}}{V_{s}Y_{w}}

V_{s} =\frac{W_{s}}{G_{s}Y_{w}}

Y_{w} =9.81KN/m^{3}=9.81\times10^{3}N/m^{3}

V_{s} =\frac{136.4N}{2.69\times9.81\times10^{3}/m^{3}}

V_{s} = 0.0052m^{3}

Recall that V = V_v + V_v; Therefore V_v = V-V_s

V_{v} = 0.0075m^{3}-0.0052m^{3}

V_{v} = 0.0023m^{3}

e =\frac{V_{v}}{V_{s}}=\frac{0.0023m^{3}}{0.0052m^{3}}

e=0.44

5. Porosity

n =\frac{e}{1+e}

n =\frac{0.44}{1+0.44}

n =0.31

6. Degree of saturation

S =\frac{V_{w}}{V_{V}}\times100

Y_{w} =\frac{W_{w}}{V_{w}}

Y_{w} =\frac{W_{w}}{V_{w}}=\frac{155-136.4}{9.81\times10^{3}}

V_{w} = 0.0019m^{3}

S =\frac{V_{w}}{V_{v}}\times100\%

S =\frac{0.0019}{0.0023}\times100\%

S =82.6\%