Weight and volume relationships

This chapter covers the following:

Weight and volume relationships
Soil compaction and classification parameters
Permeability and seepage

In soil mechanics, weight-volume relationships help you describe and predict soil behavior. They’re based on the three-phase system, which treats soil as a mixture of:

solids
water
air

A three-phase soil diagram illustrating the relationships between weights (or masses) and volumes of air, water, and solids. — Three-Phase diagram

Three-phase system

Definitions

Three-phase system: Soil is idealized as a system consisting of three components: solids, water, and air. Engineering relationships are derived by relating the weights and volumes of these components.

We’ll use the following symbols:

$V$ = total volume
$V_{s}$ = volume of solids
$V_{w}$ = volume of water
$V_{a}$ = volume of air
$W$ = total weight
$W_{s}$ = weight of solids
$W_{w}$ = weight of water

The total volume is the sum of the three phase volumes:

$V = V_{s} + V_{w} + V_{a}$

The volume of voids is the part of the soil volume not occupied by solids (so it includes both air and water):

$V_{v} = V_{w} + V_{a}$

Index properties

These properties describe the composition of a soil sample in terms of its solid, water, and air phases.

Water content

Water content is the ratio of water weight to solids weight:

$w = \frac{W _{w}}{W _{s}} \times 100%$

Total (bulk) unit weight

Total (bulk) unit weight is total weight per total volume:

$γ = \frac{W}{V}$

Dry unit weight

Dry unit weight uses only the weight of solids (but still divides by the total volume):

$γ_{d} = \frac{W _{s}}{V}$

Saturated unit weight

Saturated unit weight applies when the voids are completely filled with water (no air):

$γ_{s a t} = \frac{W _{s} + W _{w}}{V}$

Void ratio

Void ratio compares void volume to solids volume:

$e = \frac{V _{v}}{V _{s}}$

Porosity

Porosity is the fraction of the total volume that is void space:

$n = \frac{V _{v}}{V} \times 100%$

Degree of saturation

Degree of saturation is the fraction of the void space that is filled with water:

$S = \frac{V _{w}}{V _{v}} \times 100%$

Specific gravity of solids

Specific gravity of solids compares the density of soil solids to the density of water:

$G_{s} = \frac{ρ _{s}}{ρ _{w}} = \frac{W _{s} / V _{s}}{ρ _{w}}$

Quick conversion formula

These relationships are commonly used to convert between properties:

$γ = \frac{( 1 + w ) G _{s} γ _{w}}{1 + e}$

$γ_{d} = \frac{G _{s} γ _{w}}{1 + e}$

$S e = G_{s} w$

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

Unit systems: The unit weight of water is $γ_{w} = 9, 810 N/m^{3}$ in SI units, or $62.4 lb/ft^{3}$ in USCS units. Always check which system a problem uses and convert every value to one system before substituting into these equations - the FE exam regularly mixes SI and USCS figures in the same problem to test this. (The related distinction between total stress, pore water pressure, and effective stress is covered in the next chapter, Consolidation and stress.)

Example: Phase relationships

Given:

$V = 0.003 m^{3}$

$W_{s} = 40 N$

$w = 25%$

$G_{s} = 2.7$

Find: bulk unit weight, dry unit weight, volume of voids, and porosity (steps 3, 4, 6, and 7 below; steps 1, 2, and 5 find the intermediate values).

1. Weight of water:

Convert the water content to a decimal and multiply by the solids weight.

$W_{w} = w \cdot W_{s} = 0.25 \times 40 = 10 N$

2. Total weight:

Add the solids and water weights.

$W = W_{s} + W_{w} = 40 + 10 = 50 N$

3. Bulk unit weight:

Divide total weight by total volume.

$γ = \frac{W}{V} = \frac{50}{0.003} = 16, 667 N/m^{3}$

4. Dry unit weight:

Divide solids weight by total volume.

$γ_{d} = \frac{W _{s}}{V} = \frac{40}{0.003} = 13, 333 N/m^{3}$

5. Volume of solids:

Use the relationship between solids weight, specific gravity, and the unit weight of water. Carry at least 4-5 significant figures here rather than rounding $V_{s}$ early - answer choices in this kind of problem are often closely spaced, and an early rounding shifts the final porosity enough to pick the wrong option.

Assume $γ_{w} = 9, 810 N/m^{3}$ :

$V_{s} = \frac{W _{s}}{G _{s} \cdot γ _{w}} = \frac{40}{2.7 \times 9810} \approx 0.0015102 m^{3}$

6. Volume of voids:

Subtract the solids volume from the total volume.

$V_{v} = V - V_{s} = 0.003 - 0.0015102 \approx 0.0014898 m^{3}$

7. Porosity:

Compute void volume as a fraction of total volume.

$n = \frac{V _{v}}{V} \times 100% = \frac{0.0014898}{0.003} \times 100% \approx 49.7%$

Answer: $V_{v} \approx 0.00149 m^{3}$ , $n \approx 49.7%$ , $γ \approx 16, 667 N/m^{3}$ , $γ_{d} \approx 13, 333 N/m^{3}$

Soil compaction and classification parameters

Relative density

Relative density compares the in-place void ratio (or dry unit weight) of a granular soil to its loosest and densest possible states. Use the void-ratio form when $e_{max}$ and $e_{min}$ are known from lab testing; use the dry-unit-weight form when only field and lab unit-weight data are available - the two forms are equivalent.

$D_{r} = [\frac{( e _{max} - e )}{( e _{max} - e _{min} )}] \times 100$

$D_{r} = [\frac{( γ _{D field} - γ _{D min} )}{( γ _{D max} - γ _{D min} )}] [\frac{γ _{D max}}{γ _{D field}}] \times 100$

Relative compaction (%)

Relative compaction compares the field dry unit weight to the maximum dry unit weight from a compaction test.

$RC = (\frac{γ _{D field}}{γ _{D max}}) \times 100$

Plasticity index

Plasticity index is the range of water contents over which a fine-grained soil behaves plastically.

$P I = LL - P L$

$LL$ = liquid limit
$P L$ = plastic limit

Coefficient of uniformity

The coefficient of uniformity describes how spread out the particle sizes are.

$C_{U} = \frac{D _{60}}{D _{10}}$

Coefficient of concavity (or curvature)

The coefficient of concavity (curvature) describes the shape of the gradation curve.

$C_{C} = \frac{( D _{30} ) ^{2}}{D _{10} \times D _{60}}$

Under the USCS, a well-graded gravel needs $C_{U} \geq 4$ and $1 \leq C_{C} \leq 3$ ; a well-graded sand needs $C_{U} \geq 6$ and $1 \leq C_{C} \leq 3$ . A soil that fails either test is classified as poorly-graded.

Example: Gradation parameters

Given: a sand-size soil with $LL = 35%$ , $P L = 20%$ , $D_{10} = 0.1 mm$ , $D_{30} = 0.3 mm$ , $D_{60} = 0.6 mm$

Find: $P I$ , $C_{U}$ , $C_{C}$ , and whether the gradation is well-graded

$P I = LL - P L = 35 - 20 = 15%$

$C_{U} = \frac{D _{60}}{D _{10}} = \frac{0.6}{0.1} = 6$

$C_{C} = \frac{( D _{30} ) ^{2}}{D _{10} \times D _{60}} = \frac{( 0.3 ) ^{2}}{0.1 \times 0.6} = 1.5$

Since $C_{U} \geq 6$ and $1 \leq C_{C} \leq 3$ , this sand meets both well-graded criteria.

Answer: $P I = 15%$ , $C_{U} = 6$ , $C_{C} = 1.5$ , well-graded

Definitions

$D_{n}$: The particle diameter (in millimeters) at which $n %$ of the soil particles are finer by weight. The mean particle size ( $D_{50}$ ) is the diameter at which $50%$ of the particles are finer.

A grain-size distribution curve showing cumulative percent finer versus particle diameter for soil classification. — Particle size diameter example

Permeability and seepage

Hydraulic conductivity (coefficient of permeability)

Hydraulic conductivity $k$ measures how easily water flows through soil.

Constant head test:

$k = \frac{Q}{i A t _{e}}$

$i = \frac{d h}{d L}$

where, $Q = total quantity of water$

Falling head test:

$k = 2.303 (\frac{a L}{A t _{e}}) lo g_{10} (\frac{h _{1}}{h _{2}})$

Where:

$A$ = cross-sectional area of test specimen perpendicular to flow
$a$ = cross-sectional area of reservoir tube
$t_{e}$ = elapsed time
$h_{1}$ = head at time $t = 0$
$h_{2}$ = head at time $t = t_{e}$
$L$ = length of soil column

Discharge velocity and seepage velocity

Definitions

Discharge velocity: The velocity of flow calculated as if water moved through the entire cross-sectional area of the soil, including both solids and voids.
Seepage velocity: The actual velocity of water as it moves through the void spaces of the soil, found by dividing discharge velocity by porosity.

$v = ki$

$v_{s} = \frac{ki}{n}$

where, $v = discharge velocity$ and $v_{s} = seepage velocity$

Flow nets

Definitions

Flow net: A graphical representation of flow lines and equipotential lines used to analyze seepage through soils.

A flow net is a combination of flow lines and equipotential lines.

A flow line is a line along which a water particle travels.
An equipotential line connects points of equal total head.

Key properties used when constructing and interpreting a flow net:

There is no flow along equipotential lines, which are perpendicular to flow lines.
The total head along an equipotential line is equal at all points.
Flow lines cannot cross other flow lines and equipotential lines cannot cross other equipotential lines.
Equipotential lines intersect the flow lines at right angles.
Each element of a flow net must be a curvilinear square (sides may be curved, but a circle must be inscribed within it that touches all four sides).

The total flow rate through a flow net is found by:

$Q = k Δ h \frac{N _{f}}{N _{d}} L$

Where:

$Q$ = total flow rate
$N_{f}$ = number of flow channels in a flow net
$N_{d}$ = number of potential drops
$Δ h$ = head change from upstream to downstream
$k$ = coefficient of permeability
$L$ = length of structure (i.e., bank-to-bank)
Without $L$ , $q = k Δ h (\frac{N _{f}}{N _{d}})$ gives the flow rate per unit width of the structure; multiplying by $L$ (the out-of-plane width) scales this up to the total flow rate $Q$ .

Example: Flow net discharge

Given:

$N_{f} = 4$ flow channels

$N_{d} = 12$ potential drops

$Δ h = 6 m$

$k = 3 \times 1 0^{- 7} m/s$

$L = 50 m$ (out-of-plane width)

Find: total flow rate $Q$

Substitute the given values into the flow-net equation.

$Q = k Δ h \frac{N _{f}}{N _{d}} L = (3 \times 1 0^{- 7}) (6) (\frac{4}{12}) (50) = 3 \times 1 0^{- 5} m^{3} / s$

Answer: $Q = 3 \times 1 0^{- 5} m^{3} / s$

Factor of safety against seepage liquefaction

$F S_{s} = \frac{i _{c}}{i _{e}}$

$i_{c} = \frac{( γ _{sat} - γ _{w} )}{γ _{w}}$

where, $i_{e} = seepage exit gradient$

This chapter covers the following:

Weight and volume relationships
Soil compaction and classification parameters
Permeability and seepage