Saturated unit weight

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

Effective stress below a water table

A profile has 3 m of moist sand (γ = 18.0 kN/m³) over saturated sand (γ_sat = 20.5 kN/m³) with the water table at 3 m. Find the effective stress at 8 m depth.

Given

0–3 m:
$\gamma = 18.0 kN/m^{3}$
3–8 m:
$\gamma_sat = 20.5 kN/m^{3}$
Water table at 3 m

Find

σ′ at 8 m

Start with the thinking

Total stress accumulates layer by layer.
Pore pressure is measured from the water table, not the ground surface.

Step-by-step solution

Total stress
$\sigma = 18.0(3) + 20.5(5) = 54.0 + 102.5 = 156.5 kPa$
Pore pressure — u = γw(8 − 3) = 9.81(5) = 49.1 kPa
Effective stress
$\sigma' = \sigma - u = 156.5 - 49.1$
Result
$\sigma' = 107.4 kPa$

Answer:

\sigma' = 107 kPa

Why the other options are there

156.5 kPa (pore pressure ignored)
78.0 kPa (u measured from the surface)

Reference: FE Reference Handbook — Geotechnical — Effective stress

Example 3

Phase relations for a compacted fill — Saturated unit weight

A soil has Gs = 2.73, void ratio e = 0.79 and water content w = 11.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.73$
$e = 0.79$
$w = 11.0%$
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 3 — schematic for Phase relations for a compacted fill — Saturated unit weight

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.73(62.4)/(1+0.79) = 95.17 lb/ft³
Moist unit weight — γ = γd(1+w) = 95.17(1+0.110) = 105.6 lb/ft³
Porosity
$n = e/(1+e) = 0.79/1.79 = 0.441 = 44.1%$
Saturation
$S = wGs/e = 0.110(2.73)/0.79 = 0.380 = 38.0%$

Answer:

γd ≈ 95.2 lb/ft³, γ ≈ 105.6 lb/ft³, n ≈ 44.1%, S ≈ 38.0%

Why the other options are there

γd = 170.4 lb/ft³ (voids ignored)
S = 263.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Saturated unit weight

Example 4

Phase relations for a compacted fill — Saturated unit weight (2)

A soil has Gs = 2.65, void ratio e = 0.58 and water content w = 19.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.65$
$e = 0.58$
$w = 19.5%$
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 4 — schematic for Phase relations for a compacted fill — Saturated unit weight (2)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.65(62.4)/(1+0.58) = 104.7 lb/ft³
Moist unit weight — γ = γd(1+w) = 104.7(1+0.195) = 125.1 lb/ft³
Porosity
$n = e/(1+e) = 0.58/1.58 = 0.367 = 36.7%$
Saturation
$S = wGs/e = 0.195(2.65)/0.58 = 0.891 = 89.1%$

Answer:

γd ≈ 104.7 lb/ft³, γ ≈ 125.1 lb/ft³, n ≈ 36.7%, S ≈ 89.1%

Why the other options are there

γd = 165.4 lb/ft³ (voids ignored)
S = 112.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Saturated unit weight

Example 5

Phase relations for a compacted fill — Saturated unit weight (3)

A soil has Gs = 2.68, void ratio e = 0.77 and water content w = 22.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.68$
$e = 0.77$
$w = 22.5%$
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 — Saturated unit weight (3)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.68(62.4)/(1+0.77) = 94.48 lb/ft³
Moist unit weight — γ = γd(1+w) = 94.48(1+0.225) = 115.7 lb/ft³
Porosity
$n = e/(1+e) = 0.77/1.77 = 0.435 = 43.5%$
Saturation
$S = wGs/e = 0.225(2.68)/0.77 = 0.783 = 78.3%$

Answer:

γd ≈ 94.5 lb/ft³, γ ≈ 115.7 lb/ft³, n ≈ 43.5%, S ≈ 78.3%

Why the other options are there

γd = 167.2 lb/ft³ (voids ignored)
S = 127.7% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Saturated unit weight

Example 6

Phase relations for a compacted fill — Saturated unit weight (4)

A soil has Gs = 2.71, void ratio e = 0.61 and water content w = 13.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.71$
$e = 0.61$
$w = 13.0%$
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 — Saturated unit weight (4)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.71(62.4)/(1+0.61) = 105.0 lb/ft³
Moist unit weight — γ = γd(1+w) = 105.0(1+0.130) = 118.7 lb/ft³
Porosity
$n = e/(1+e) = 0.61/1.61 = 0.379 = 37.9%$
Saturation
$S = wGs/e = 0.130(2.71)/0.61 = 0.578 = 57.8%$

Answer:

γd ≈ 105.0 lb/ft³, γ ≈ 118.7 lb/ft³, n ≈ 37.9%, S ≈ 57.8%

Why the other options are there

γd = 169.1 lb/ft³ (voids ignored)
S = 173.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Saturated unit weight

Example 7

Phase relations for a compacted fill — Saturated unit weight (5)

A soil has Gs = 2.72, void ratio e = 0.60 and water content w = 23.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.72$
$e = 0.60$
$w = 23.0%$
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 — Saturated unit weight (5)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.72(62.4)/(1+0.60) = 106.1 lb/ft³
Moist unit weight — γ = γd(1+w) = 106.1(1+0.230) = 130.5 lb/ft³
Porosity
$n = e/(1+e) = 0.60/1.60 = 0.375 = 37.5%$
Saturation
$S = wGs/e = 0.230(2.72)/0.60 = 1.043 = 104.3%$

Answer:

γd ≈ 106.1 lb/ft³, γ ≈ 130.5 lb/ft³, n ≈ 37.5%, S ≈ 104.3%

Why the other options are there

γd = 169.7 lb/ft³ (voids ignored)
S = 95.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Saturated unit weight

Example 8

Phase relations for a compacted fill — Saturated unit weight (6)

A soil has Gs = 2.70, void ratio e = 0.93 and water content w = 14.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.70$
$e = 0.93$
$w = 14.0%$
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 — Saturated unit weight (6)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.70(62.4)/(1+0.93) = 87.30 lb/ft³
Moist unit weight — γ = γd(1+w) = 87.30(1+0.140) = 99.52 lb/ft³
Porosity
$n = e/(1+e) = 0.93/1.93 = 0.482 = 48.2%$
Saturation
$S = wGs/e = 0.140(2.70)/0.93 = 0.406 = 40.6%$

Answer:

γd ≈ 87.3 lb/ft³, γ ≈ 99.5 lb/ft³, n ≈ 48.2%, S ≈ 40.6%

Why the other options are there

γd = 168.5 lb/ft³ (voids ignored)
S = 246.0% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Saturated unit weight

Example 9

Phase relations for a compacted fill — Saturated unit weight (7)

A soil has Gs = 2.66, void ratio e = 0.94 and water content w = 21.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.66$
$e = 0.94$
$w = 21.0%$
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 — Saturated unit weight (7)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.66(62.4)/(1+0.94) = 85.56 lb/ft³
Moist unit weight — γ = γd(1+w) = 85.56(1+0.210) = 103.5 lb/ft³
Porosity
$n = e/(1+e) = 0.94/1.94 = 0.485 = 48.5%$
Saturation
$S = wGs/e = 0.210(2.66)/0.94 = 0.594 = 59.4%$

Answer:

γd ≈ 85.6 lb/ft³, γ ≈ 103.5 lb/ft³, n ≈ 48.5%, S ≈ 59.4%

Why the other options are there

γd = 166.0 lb/ft³ (voids ignored)
S = 168.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Saturated unit weight

Example 10

Phase relations for a compacted fill — Saturated unit weight (8)

A soil has Gs = 2.74, void ratio e = 0.45 and water content w = 15.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.74$
$e = 0.45$
$w = 15.0%$
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 10 — schematic for Phase relations for a compacted fill — Saturated unit weight (8)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.45) = 117.9 lb/ft³
Moist unit weight — γ = γd(1+w) = 117.9(1+0.150) = 135.6 lb/ft³
Porosity
$n = e/(1+e) = 0.45/1.45 = 0.310 = 31.0%$
Saturation
$S = wGs/e = 0.150(2.74)/0.45 = 0.913 = 91.3%$

Answer:

γd ≈ 117.9 lb/ft³, γ ≈ 135.6 lb/ft³, n ≈ 31.0%, S ≈ 91.3%

Why the other options are there

γd = 171.0 lb/ft³ (voids ignored)
S = 109.5% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Saturated unit weight

Core formulas for this FE topic

Worked exam-style examples