Phase Relationships

Example 1

Soil phase relationships from field data

A sample has moist unit weight 19.2 kN/m³, water content 16% and Gs = 2.68. Find the dry unit weight, void ratio and degree of saturation.

Given

$\gamma = 19.2 kN/m^{3}$
$w = 0.16$
$Gs = 2.68$
γw = 9.81 kN/m³

Find

γ_d, e, S

Start with the thinking

Dry unit weight first — everything else follows.
Se = wGs closes the saturation calculation.

Figure 1 — schematic for Soil phase relationships from field data

Step-by-step solution

Dry unit weight
$\gamma_d = \gamma/(1 + w) = 19.2/1.16 = 16.55 kN/m^{3}$
Void ratio — e = Gsγw/γ_d − 1 = 2.68(9.81)/16.55 − 1
Evaluate
$e = 26.29/16.55 - 1 = 1.589 - 1 = 0.589$
Saturation
$S = wGs/e = 0.16(2.68)/0.589$
Result
$\gamma_d = 16.6 kN/m^{3}, e = 0.589, S = 0.728 (72.8%)$

Answer:

\gamma_d = 16.6 kN/m^{3}, e = 0.589, S = 72.8%

Why the other options are there

γ_d = 22.3 kN/m³ (multiplied by 1 + w)
S = 100% (assumed saturated)

Reference: FE Reference Handbook — Geotechnical — Phase relationships

Example 2

Phase relations for a compacted fill — Phase Relationships

A soil has Gs = 2.66, void ratio e = 0.55 and water content w = 13.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.66$
$e = 0.55$
$w = 13.5%$
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure 2 — schematic for Phase relations for a compacted fill — Phase Relationships

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.66(62.4)/(1+0.55) = 107.1 lb/ft³
Moist unit weight — γ = γd(1+w) = 107.1(1+0.135) = 121.5 lb/ft³
Porosity
$n = e/(1+e) = 0.55/1.55 = 0.355 = 35.5%$
Saturation
$S = wGs/e = 0.135(2.66)/0.55 = 0.653 = 65.3%$

Answer:

γd ≈ 107.1 lb/ft³, γ ≈ 121.5 lb/ft³, n ≈ 35.5%, S ≈ 65.3%

Why the other options are there

γd = 166.0 lb/ft³ (voids ignored)
S = 153.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Phase Relationships

Example 3

Phase relations for a compacted fill — Phase Relationships (2)

A soil has Gs = 2.68, void ratio e = 0.95 and water content w = 23.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.68$
$e = 0.95$
$w = 23.0%$
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure 3 — schematic for Phase relations for a compacted fill — Phase Relationships (2)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.68(62.4)/(1+0.95) = 85.76 lb/ft³
Moist unit weight — γ = γd(1+w) = 85.76(1+0.230) = 105.5 lb/ft³
Porosity
$n = e/(1+e) = 0.95/1.95 = 0.487 = 48.7%$
Saturation
$S = wGs/e = 0.230(2.68)/0.95 = 0.649 = 64.9%$

Answer:

γd ≈ 85.8 lb/ft³, γ ≈ 105.5 lb/ft³, n ≈ 48.7%, S ≈ 64.9%

Why the other options are there

γd = 167.2 lb/ft³ (voids ignored)
S = 154.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Phase Relationships

Example 4

Phase relations for a compacted fill — Phase Relationships (3)

A soil has Gs = 2.75, void ratio e = 0.48 and water content w = 20.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.75$
$e = 0.48$
$w = 20.0%$
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure 4 — schematic for Phase relations for a compacted fill — Phase Relationships (3)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.75(62.4)/(1+0.48) = 115.9 lb/ft³
Moist unit weight — γ = γd(1+w) = 115.9(1+0.200) = 139.1 lb/ft³
Porosity
$n = e/(1+e) = 0.48/1.48 = 0.324 = 32.4%$
Saturation
$S = wGs/e = 0.200(2.75)/0.48 = 1.146 = 114.6%$

Answer:

γd ≈ 115.9 lb/ft³, γ ≈ 139.1 lb/ft³, n ≈ 32.4%, S ≈ 114.6%

Why the other options are there

γd = 171.6 lb/ft³ (voids ignored)
S = 87.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Phase Relationships

Example 5

Phase relations for a compacted fill — Phase Relationships (4)

A soil has Gs = 2.63, void ratio e = 0.88 and water content w = 23.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.63$
$e = 0.88$
$w = 23.5%$
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure 5 — schematic for Phase relations for a compacted fill — Phase Relationships (4)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.63(62.4)/(1+0.88) = 87.29 lb/ft³
Moist unit weight — γ = γd(1+w) = 87.29(1+0.235) = 107.8 lb/ft³
Porosity
$n = e/(1+e) = 0.88/1.88 = 0.468 = 46.8%$
Saturation
$S = wGs/e = 0.235(2.63)/0.88 = 0.702 = 70.2%$

Answer:

γd ≈ 87.3 lb/ft³, γ ≈ 107.8 lb/ft³, n ≈ 46.8%, S ≈ 70.2%

Why the other options are there

γd = 164.1 lb/ft³ (voids ignored)
S = 142.4% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Phase Relationships

Example 6

Phase relations for a compacted fill — Phase Relationships (5)

A soil has Gs = 2.72, void ratio e = 0.53 and water content w = 23.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.72$
$e = 0.53$
$w = 23.5%$
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure 6 — schematic for Phase relations for a compacted fill — Phase Relationships (5)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.72(62.4)/(1+0.53) = 110.9 lb/ft³
Moist unit weight — γ = γd(1+w) = 110.9(1+0.235) = 137.0 lb/ft³
Porosity
$n = e/(1+e) = 0.53/1.53 = 0.346 = 34.6%$
Saturation
$S = wGs/e = 0.235(2.72)/0.53 = 1.206 = 120.6%$

Answer:

γd ≈ 110.9 lb/ft³, γ ≈ 137.0 lb/ft³, n ≈ 34.6%, S ≈ 120.6%

Why the other options are there

γd = 169.7 lb/ft³ (voids ignored)
S = 82.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Phase Relationships

Example 7

Phase relations for a compacted fill — Phase Relationships (6)

A soil has Gs = 2.71, void ratio e = 0.89 and water content w = 23.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.71$
$e = 0.89$
$w = 23.0%$
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure 7 — schematic for Phase relations for a compacted fill — Phase Relationships (6)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.71(62.4)/(1+0.89) = 89.47 lb/ft³
Moist unit weight — γ = γd(1+w) = 89.47(1+0.230) = 110.1 lb/ft³
Porosity
$n = e/(1+e) = 0.89/1.89 = 0.471 = 47.1%$
Saturation
$S = wGs/e = 0.230(2.71)/0.89 = 0.700 = 70.0%$

Answer:

γd ≈ 89.5 lb/ft³, γ ≈ 110.1 lb/ft³, n ≈ 47.1%, S ≈ 70.0%

Why the other options are there

γd = 169.1 lb/ft³ (voids ignored)
S = 142.8% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Phase Relationships

Example 8

Phase relations for a compacted fill — Phase Relationships (7)

A soil has Gs = 2.63, void ratio e = 0.72 and water content w = 10.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.63$
$e = 0.72$
$w = 10.0%$
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure 8 — schematic for Phase relations for a compacted fill — Phase Relationships (7)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.63(62.4)/(1+0.72) = 95.41 lb/ft³
Moist unit weight — γ = γd(1+w) = 95.41(1+0.100) = 105.0 lb/ft³
Porosity
$n = e/(1+e) = 0.72/1.72 = 0.419 = 41.9%$
Saturation
$S = wGs/e = 0.100(2.63)/0.72 = 0.365 = 36.5%$

Answer:

γd ≈ 95.4 lb/ft³, γ ≈ 105.0 lb/ft³, n ≈ 41.9%, S ≈ 36.5%

Why the other options are there

γd = 164.1 lb/ft³ (voids ignored)
S = 273.8% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Phase Relationships

Example 9

Phase relations for a compacted fill — Phase Relationships (8)

A soil has Gs = 2.74, void ratio e = 0.53 and water content w = 9.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.74$
$e = 0.53$
$w = 9.0%$
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure 9 — schematic for Phase relations for a compacted fill — Phase Relationships (8)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.53) = 111.7 lb/ft³
Moist unit weight — γ = γd(1+w) = 111.7(1+0.090) = 121.8 lb/ft³
Porosity
$n = e/(1+e) = 0.53/1.53 = 0.346 = 34.6%$
Saturation
$S = wGs/e = 0.090(2.74)/0.53 = 0.465 = 46.5%$

Answer:

γd ≈ 111.7 lb/ft³, γ ≈ 121.8 lb/ft³, n ≈ 34.6%, S ≈ 46.5%

Why the other options are there

γd = 171.0 lb/ft³ (voids ignored)
S = 214.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Phase Relationships

Example 10

Phase relations for a compacted fill — Phase Relationships (9)

A soil has Gs = 2.62, void ratio e = 0.65 and water content w = 9.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.62$
$e = 0.65$
$w = 9.5%$
γw = 62.4 lb/ft³

Find

γd, γ, n and S

Start with the thinking

Every phase relation follows from the unit-volume-of-solids diagram.
Compute γd first — the moist unit weight is just γd(1 + w).

Figure 10 — schematic for Phase relations for a compacted fill — Phase Relationships (9)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.62(62.4)/(1+0.65) = 99.08 lb/ft³
Moist unit weight — γ = γd(1+w) = 99.08(1+0.095) = 108.5 lb/ft³
Porosity
$n = e/(1+e) = 0.65/1.65 = 0.394 = 39.4%$
Saturation
$S = wGs/e = 0.095(2.62)/0.65 = 0.383 = 38.3%$

Answer:

γd ≈ 99.1 lb/ft³, γ ≈ 108.5 lb/ft³, n ≈ 39.4%, S ≈ 38.3%

Why the other options are there

γd = 163.5 lb/ft³ (voids ignored)
S = 261.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Phase Relationships

Core formulas for this FE topic

Worked exam-style examples