Total 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

Phase relations for a compacted fill — Total unit weight

A soil has Gs = 2.74, void ratio e = 0.85 and water content w = 9.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.74$
$e = 0.85$
$w = 9.0%$
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 2 — schematic for Phase relations for a compacted fill — Total unit weight

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.85) = 92.42 lb/ft³
Moist unit weight — γ = γd(1+w) = 92.42(1+0.090) = 100.7 lb/ft³
Porosity
$n = e/(1+e) = 0.85/1.85 = 0.459 = 45.9%$
Saturation
$S = wGs/e = 0.090(2.74)/0.85 = 0.290 = 29.0%$

Answer:

γd ≈ 92.4 lb/ft³, γ ≈ 100.7 lb/ft³, n ≈ 45.9%, S ≈ 29.0%

Why the other options are there

γd = 171.0 lb/ft³ (voids ignored)
S = 344.7% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Total unit weight

Example 3

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

A soil has Gs = 2.65, void ratio e = 0.88 and water content w = 8.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.65$
$e = 0.88$
$w = 8.5%$
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 — Total unit weight (2)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.65(62.4)/(1+0.88) = 87.96 lb/ft³
Moist unit weight — γ = γd(1+w) = 87.96(1+0.085) = 95.43 lb/ft³
Porosity
$n = e/(1+e) = 0.88/1.88 = 0.468 = 46.8%$
Saturation
$S = wGs/e = 0.085(2.65)/0.88 = 0.256 = 25.6%$

Answer:

γd ≈ 88.0 lb/ft³, γ ≈ 95.4 lb/ft³, n ≈ 46.8%, S ≈ 25.6%

Why the other options are there

γd = 165.4 lb/ft³ (voids ignored)
S = 390.7% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Total unit weight

Example 4

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

A soil has Gs = 2.63, void ratio e = 0.84 and water content w = 17.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.63$
$e = 0.84$
$w = 17.0%$
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 — Total unit weight (3)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.63(62.4)/(1+0.84) = 89.19 lb/ft³
Moist unit weight — γ = γd(1+w) = 89.19(1+0.170) = 104.4 lb/ft³
Porosity
$n = e/(1+e) = 0.84/1.84 = 0.457 = 45.7%$
Saturation
$S = wGs/e = 0.170(2.63)/0.84 = 0.532 = 53.2%$

Answer:

γd ≈ 89.2 lb/ft³, γ ≈ 104.4 lb/ft³, n ≈ 45.7%, S ≈ 53.2%

Why the other options are there

γd = 164.1 lb/ft³ (voids ignored)
S = 187.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Total unit weight

Example 5

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

A soil has Gs = 2.74, void ratio e = 0.64 and water content w = 17.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.74$
$e = 0.64$
$w = 17.5%$
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 — Total unit weight (4)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.64) = 104.3 lb/ft³
Moist unit weight — γ = γd(1+w) = 104.3(1+0.175) = 122.5 lb/ft³
Porosity
$n = e/(1+e) = 0.64/1.64 = 0.390 = 39.0%$
Saturation
$S = wGs/e = 0.175(2.74)/0.64 = 0.749 = 74.9%$

Answer:

γd ≈ 104.3 lb/ft³, γ ≈ 122.5 lb/ft³, n ≈ 39.0%, S ≈ 74.9%

Why the other options are there

γd = 171.0 lb/ft³ (voids ignored)
S = 133.5% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Total unit weight

Example 6

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

A soil has Gs = 2.62, void ratio e = 0.60 and water content w = 12.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.62$
$e = 0.60$
$w = 12.5%$
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 — Total unit weight (5)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.62(62.4)/(1+0.60) = 102.2 lb/ft³
Moist unit weight — γ = γd(1+w) = 102.2(1+0.125) = 115.0 lb/ft³
Porosity
$n = e/(1+e) = 0.60/1.60 = 0.375 = 37.5%$
Saturation
$S = wGs/e = 0.125(2.62)/0.60 = 0.546 = 54.6%$

Answer:

γd ≈ 102.2 lb/ft³, γ ≈ 115.0 lb/ft³, n ≈ 37.5%, S ≈ 54.6%

Why the other options are there

γd = 163.5 lb/ft³ (voids ignored)
S = 183.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Total unit weight

Example 7

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

A soil has Gs = 2.63, void ratio e = 0.52 and water content w = 23.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.63$
$e = 0.52$
$w = 23.5%$
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 7 — schematic for Phase relations for a compacted fill — Total unit weight (6)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.63(62.4)/(1+0.52) = 108.0 lb/ft³
Moist unit weight — γ = γd(1+w) = 108.0(1+0.235) = 133.3 lb/ft³
Porosity
$n = e/(1+e) = 0.52/1.52 = 0.342 = 34.2%$
Saturation
$S = wGs/e = 0.235(2.63)/0.52 = 1.189 = 118.9%$

Answer:

γd ≈ 108.0 lb/ft³, γ ≈ 133.3 lb/ft³, n ≈ 34.2%, S ≈ 118.9%

Why the other options are there

γd = 164.1 lb/ft³ (voids ignored)
S = 84.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Total unit weight

Example 8

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

A soil has Gs = 2.67, void ratio e = 0.71 and water content w = 10.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.67$
$e = 0.71$
$w = 10.5%$
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 8 — schematic for Phase relations for a compacted fill — Total unit weight (7)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.67(62.4)/(1+0.71) = 97.43 lb/ft³
Moist unit weight — γ = γd(1+w) = 97.43(1+0.105) = 107.7 lb/ft³
Porosity
$n = e/(1+e) = 0.71/1.71 = 0.415 = 41.5%$
Saturation
$S = wGs/e = 0.105(2.67)/0.71 = 0.395 = 39.5%$

Answer:

γd ≈ 97.4 lb/ft³, γ ≈ 107.7 lb/ft³, n ≈ 41.5%, S ≈ 39.5%

Why the other options are there

γd = 166.6 lb/ft³ (voids ignored)
S = 253.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Total unit weight

Example 9

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

A soil has Gs = 2.73, void ratio e = 0.55 and water content w = 19.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.73$
$e = 0.55$
$w = 19.5%$
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 9 — schematic for Phase relations for a compacted fill — Total unit weight (8)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.73(62.4)/(1+0.55) = 109.9 lb/ft³
Moist unit weight — γ = γd(1+w) = 109.9(1+0.195) = 131.3 lb/ft³
Porosity
$n = e/(1+e) = 0.55/1.55 = 0.355 = 35.5%$
Saturation
$S = wGs/e = 0.195(2.73)/0.55 = 0.968 = 96.8%$

Answer:

γd ≈ 109.9 lb/ft³, γ ≈ 131.3 lb/ft³, n ≈ 35.5%, S ≈ 96.8%

Why the other options are there

γd = 170.4 lb/ft³ (voids ignored)
S = 103.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Total unit weight

Example 10

Phase relations for a compacted fill — Total unit weight (9)

A soil has Gs = 2.68, void ratio e = 0.54 and water content w = 19.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.68$
$e = 0.54$
$w = 19.5%$
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 — Total unit weight (9)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.68(62.4)/(1+0.54) = 108.6 lb/ft³
Moist unit weight — γ = γd(1+w) = 108.6(1+0.195) = 129.8 lb/ft³
Porosity
$n = e/(1+e) = 0.54/1.54 = 0.351 = 35.1%$
Saturation
$S = wGs/e = 0.195(2.68)/0.54 = 0.968 = 96.8%$

Answer:

γd ≈ 108.6 lb/ft³, γ ≈ 129.8 lb/ft³, n ≈ 35.1%, S ≈ 96.8%

Why the other options are there

γd = 167.2 lb/ft³ (voids ignored)
S = 103.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Total unit weight

Core formulas for this FE topic

Worked exam-style examples