Degree of saturation (%)

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 — Degree of saturation (%)

A soil has Gs = 2.72, void ratio e = 0.79 and water content w = 22.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.72$
$e = 0.79$
$w = 22.0%$
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 2 — schematic for Phase relations for a compacted fill — Degree of saturation (%)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.72(62.4)/(1+0.79) = 94.82 lb/ft³
Moist unit weight — γ = γd(1+w) = 94.82(1+0.220) = 115.7 lb/ft³
Porosity
$n = e/(1+e) = 0.79/1.79 = 0.441 = 44.1%$
Saturation
$S = wGs/e = 0.220(2.72)/0.79 = 0.757 = 75.7%$

Answer:

γd ≈ 94.8 lb/ft³, γ ≈ 115.7 lb/ft³, n ≈ 44.1%, S ≈ 75.7%

Why the other options are there

γd = 169.7 lb/ft³ (voids ignored)
S = 132.0% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Degree of saturation (%)

Example 3

Phase relations for a compacted fill — Degree of saturation (%) (2)

A soil has Gs = 2.66, void ratio e = 0.90 and water content w = 15.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.66$
$e = 0.90$
$w = 15.5%$
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 3 — schematic for Phase relations for a compacted fill — Degree of saturation (%) (2)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.66(62.4)/(1+0.90) = 87.36 lb/ft³
Moist unit weight — γ = γd(1+w) = 87.36(1+0.155) = 100.9 lb/ft³
Porosity
$n = e/(1+e) = 0.90/1.90 = 0.474 = 47.4%$
Saturation
$S = wGs/e = 0.155(2.66)/0.90 = 0.458 = 45.8%$

Answer:

γd ≈ 87.4 lb/ft³, γ ≈ 100.9 lb/ft³, n ≈ 47.4%, S ≈ 45.8%

Why the other options are there

γd = 166.0 lb/ft³ (voids ignored)
S = 218.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Degree of saturation (%)

Example 4

Phase relations for a compacted fill — Degree of saturation (%) (3)

A soil has Gs = 2.66, void ratio e = 0.91 and water content w = 9.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.66$
$e = 0.91$
$w = 9.0%$
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 4 — schematic for Phase relations for a compacted fill — Degree of saturation (%) (3)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.66(62.4)/(1+0.91) = 86.90 lb/ft³
Moist unit weight — γ = γd(1+w) = 86.90(1+0.090) = 94.72 lb/ft³
Porosity
$n = e/(1+e) = 0.91/1.91 = 0.476 = 47.6%$
Saturation
$S = wGs/e = 0.090(2.66)/0.91 = 0.263 = 26.3%$

Answer:

γd ≈ 86.9 lb/ft³, γ ≈ 94.7 lb/ft³, n ≈ 47.6%, S ≈ 26.3%

Why the other options are there

γd = 166.0 lb/ft³ (voids ignored)
S = 380.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Degree of saturation (%)

Example 5

Phase relations for a compacted fill — Degree of saturation (%) (4)

A soil has Gs = 2.71, void ratio e = 0.80 and water content w = 12.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.71$
$e = 0.80$
$w = 12.0%$
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 5 — schematic for Phase relations for a compacted fill — Degree of saturation (%) (4)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.71(62.4)/(1+0.80) = 93.95 lb/ft³
Moist unit weight — γ = γd(1+w) = 93.95(1+0.120) = 105.2 lb/ft³
Porosity
$n = e/(1+e) = 0.80/1.80 = 0.444 = 44.4%$
Saturation
$S = wGs/e = 0.120(2.71)/0.80 = 0.407 = 40.7%$

Answer:

γd ≈ 93.9 lb/ft³, γ ≈ 105.2 lb/ft³, n ≈ 44.4%, S ≈ 40.7%

Why the other options are there

γd = 169.1 lb/ft³ (voids ignored)
S = 246.0% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Degree of saturation (%)

Example 6

Phase relations for a compacted fill — Degree of saturation (%) (5)

A soil has Gs = 2.64, void ratio e = 0.59 and water content w = 11.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.64$
$e = 0.59$
$w = 11.0%$
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 6 — schematic for Phase relations for a compacted fill — Degree of saturation (%) (5)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.64(62.4)/(1+0.59) = 103.6 lb/ft³
Moist unit weight — γ = γd(1+w) = 103.6(1+0.110) = 115.0 lb/ft³
Porosity
$n = e/(1+e) = 0.59/1.59 = 0.371 = 37.1%$
Saturation
$S = wGs/e = 0.110(2.64)/0.59 = 0.492 = 49.2%$

Answer:

γd ≈ 103.6 lb/ft³, γ ≈ 115.0 lb/ft³, n ≈ 37.1%, S ≈ 49.2%

Why the other options are there

γd = 164.7 lb/ft³ (voids ignored)
S = 203.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Degree of saturation (%)

Example 7

Phase relations for a compacted fill — Degree of saturation (%) (6)

A soil has Gs = 2.71, void ratio e = 0.80 and water content w = 22.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.71$
$e = 0.80$
$w = 22.0%$
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 — Degree of saturation (%) (6)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.71(62.4)/(1+0.80) = 93.95 lb/ft³
Moist unit weight — γ = γd(1+w) = 93.95(1+0.220) = 114.6 lb/ft³
Porosity
$n = e/(1+e) = 0.80/1.80 = 0.444 = 44.4%$
Saturation
$S = wGs/e = 0.220(2.71)/0.80 = 0.745 = 74.5%$

Answer:

γd ≈ 93.9 lb/ft³, γ ≈ 114.6 lb/ft³, n ≈ 44.4%, S ≈ 74.5%

Why the other options are there

γd = 169.1 lb/ft³ (voids ignored)
S = 134.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Degree of saturation (%)

Example 8

Phase relations for a compacted fill — Degree of saturation (%) (7)

A soil has Gs = 2.65, void ratio e = 0.73 and water content w = 22.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.65$
$e = 0.73$
$w = 22.0%$
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 — Degree of saturation (%) (7)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.65(62.4)/(1+0.73) = 95.58 lb/ft³
Moist unit weight — γ = γd(1+w) = 95.58(1+0.220) = 116.6 lb/ft³
Porosity
$n = e/(1+e) = 0.73/1.73 = 0.422 = 42.2%$
Saturation
$S = wGs/e = 0.220(2.65)/0.73 = 0.799 = 79.9%$

Answer:

γd ≈ 95.6 lb/ft³, γ ≈ 116.6 lb/ft³, n ≈ 42.2%, S ≈ 79.9%

Why the other options are there

γd = 165.4 lb/ft³ (voids ignored)
S = 125.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Degree of saturation (%)

Example 9

Phase relations for a compacted fill — Degree of saturation (%) (8)

A soil has Gs = 2.62, void ratio e = 0.57 and water content w = 13.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.62$
$e = 0.57$
$w = 13.5%$
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 9 — schematic for Phase relations for a compacted fill — Degree of saturation (%) (8)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.62(62.4)/(1+0.57) = 104.1 lb/ft³
Moist unit weight — γ = γd(1+w) = 104.1(1+0.135) = 118.2 lb/ft³
Porosity
$n = e/(1+e) = 0.57/1.57 = 0.363 = 36.3%$
Saturation
$S = wGs/e = 0.135(2.62)/0.57 = 0.621 = 62.1%$

Answer:

γd ≈ 104.1 lb/ft³, γ ≈ 118.2 lb/ft³, n ≈ 36.3%, S ≈ 62.1%

Why the other options are there

γd = 163.5 lb/ft³ (voids ignored)
S = 161.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Degree of saturation (%)

Example 10

Phase relations for a compacted fill — Degree of saturation (%) (9)

A soil has Gs = 2.71, void ratio e = 0.67 and water content w = 19.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.71$
$e = 0.67$
$w = 19.5%$
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 — Degree of saturation (%) (9)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.71(62.4)/(1+0.67) = 101.3 lb/ft³
Moist unit weight — γ = γd(1+w) = 101.3(1+0.195) = 121.0 lb/ft³
Porosity
$n = e/(1+e) = 0.67/1.67 = 0.401 = 40.1%$
Saturation
$S = wGs/e = 0.195(2.71)/0.67 = 0.789 = 78.9%$

Answer:

γd ≈ 101.3 lb/ft³, γ ≈ 121.0 lb/ft³, n ≈ 40.1%, S ≈ 78.9%

Why the other options are there

γd = 169.1 lb/ft³ (voids ignored)
S = 126.8% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Degree of saturation (%)

Core formulas for this FE topic

Worked exam-style examples