Dry 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 — Dry unit weight

A soil has Gs = 2.72, void ratio e = 0.70 and water content w = 10.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.72$
$e = 0.70$
$w = 10.5%$
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 2 — schematic for Phase relations for a compacted fill — Dry unit weight

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.72(62.4)/(1+0.70) = 99.84 lb/ft³
Moist unit weight — γ = γd(1+w) = 99.84(1+0.105) = 110.3 lb/ft³
Porosity
$n = e/(1+e) = 0.70/1.70 = 0.412 = 41.2%$
Saturation
$S = wGs/e = 0.105(2.72)/0.70 = 0.408 = 40.8%$

Answer:

γd ≈ 99.8 lb/ft³, γ ≈ 110.3 lb/ft³, n ≈ 41.2%, S ≈ 40.8%

Why the other options are there

γd = 169.7 lb/ft³ (voids ignored)
S = 245.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Dry unit weight

Example 3

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

A soil has Gs = 2.74, void ratio e = 0.66 and water content w = 10.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.74$
$e = 0.66$
$w = 10.0%$
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 3 — schematic for Phase relations for a compacted fill — Dry unit weight (2)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.66) = 103.0 lb/ft³
Moist unit weight — γ = γd(1+w) = 103.0(1+0.100) = 113.3 lb/ft³
Porosity
$n = e/(1+e) = 0.66/1.66 = 0.398 = 39.8%$
Saturation
$S = wGs/e = 0.100(2.74)/0.66 = 0.415 = 41.5%$

Answer:

γd ≈ 103.0 lb/ft³, γ ≈ 113.3 lb/ft³, n ≈ 39.8%, S ≈ 41.5%

Why the other options are there

γd = 171.0 lb/ft³ (voids ignored)
S = 240.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Dry unit weight

Example 4

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

A soil has Gs = 2.65, void ratio e = 0.62 and water content w = 13.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.65$
$e = 0.62$
$w = 13.0%$
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 4 — schematic for Phase relations for a compacted fill — Dry unit weight (3)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.65(62.4)/(1+0.62) = 102.1 lb/ft³
Moist unit weight — γ = γd(1+w) = 102.1(1+0.130) = 115.3 lb/ft³
Porosity
$n = e/(1+e) = 0.62/1.62 = 0.383 = 38.3%$
Saturation
$S = wGs/e = 0.130(2.65)/0.62 = 0.556 = 55.6%$

Answer:

γd ≈ 102.1 lb/ft³, γ ≈ 115.3 lb/ft³, n ≈ 38.3%, S ≈ 55.6%

Why the other options are there

γd = 165.4 lb/ft³ (voids ignored)
S = 180.0% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Dry unit weight

Example 5

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

A soil has Gs = 2.69, void ratio e = 0.80 and water content w = 21.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.69$
$e = 0.80$
$w = 21.5%$
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 — Dry unit weight (4)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.69(62.4)/(1+0.80) = 93.25 lb/ft³
Moist unit weight — γ = γd(1+w) = 93.25(1+0.215) = 113.3 lb/ft³
Porosity
$n = e/(1+e) = 0.80/1.80 = 0.444 = 44.4%$
Saturation
$S = wGs/e = 0.215(2.69)/0.80 = 0.723 = 72.3%$

Answer:

γd ≈ 93.3 lb/ft³, γ ≈ 113.3 lb/ft³, n ≈ 44.4%, S ≈ 72.3%

Why the other options are there

γd = 167.9 lb/ft³ (voids ignored)
S = 138.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Dry unit weight

Example 6

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

A soil has Gs = 2.73, void ratio e = 0.56 and water content w = 17.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.73$
$e = 0.56$
$w = 17.5%$
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 6 — schematic for Phase relations for a compacted fill — Dry unit weight (5)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.73(62.4)/(1+0.56) = 109.2 lb/ft³
Moist unit weight — γ = γd(1+w) = 109.2(1+0.175) = 128.3 lb/ft³
Porosity
$n = e/(1+e) = 0.56/1.56 = 0.359 = 35.9%$
Saturation
$S = wGs/e = 0.175(2.73)/0.56 = 0.853 = 85.3%$

Answer:

γd ≈ 109.2 lb/ft³, γ ≈ 128.3 lb/ft³, n ≈ 35.9%, S ≈ 85.3%

Why the other options are there

γd = 170.4 lb/ft³ (voids ignored)
S = 117.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Dry unit weight

Example 7

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

A soil has Gs = 2.67, void ratio e = 0.60 and water content w = 19.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.67$
$e = 0.60$
$w = 19.0%$
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 — Dry unit weight (6)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.67(62.4)/(1+0.60) = 104.1 lb/ft³
Moist unit weight — γ = γd(1+w) = 104.1(1+0.190) = 123.9 lb/ft³
Porosity
$n = e/(1+e) = 0.60/1.60 = 0.375 = 37.5%$
Saturation
$S = wGs/e = 0.190(2.67)/0.60 = 0.846 = 84.6%$

Answer:

γd ≈ 104.1 lb/ft³, γ ≈ 123.9 lb/ft³, n ≈ 37.5%, S ≈ 84.6%

Why the other options are there

γd = 166.6 lb/ft³ (voids ignored)
S = 118.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Dry unit weight

Example 8

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

A soil has Gs = 2.72, void ratio e = 0.88 and water content w = 24.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.72$
$e = 0.88$
$w = 24.0%$
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 — Dry unit weight (7)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.72(62.4)/(1+0.88) = 90.28 lb/ft³
Moist unit weight — γ = γd(1+w) = 90.28(1+0.240) = 111.9 lb/ft³
Porosity
$n = e/(1+e) = 0.88/1.88 = 0.468 = 46.8%$
Saturation
$S = wGs/e = 0.240(2.72)/0.88 = 0.742 = 74.2%$

Answer:

γd ≈ 90.3 lb/ft³, γ ≈ 111.9 lb/ft³, n ≈ 46.8%, S ≈ 74.2%

Why the other options are there

γd = 169.7 lb/ft³ (voids ignored)
S = 134.8% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Dry unit weight

Example 9

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

A soil has Gs = 2.70, void ratio e = 0.46 and water content w = 13.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.70$
$e = 0.46$
$w = 13.5%$
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 — Dry unit weight (8)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.70(62.4)/(1+0.46) = 115.4 lb/ft³
Moist unit weight — γ = γd(1+w) = 115.4(1+0.135) = 131.0 lb/ft³
Porosity
$n = e/(1+e) = 0.46/1.46 = 0.315 = 31.5%$
Saturation
$S = wGs/e = 0.135(2.70)/0.46 = 0.792 = 79.2%$

Answer:

γd ≈ 115.4 lb/ft³, γ ≈ 131.0 lb/ft³, n ≈ 31.5%, S ≈ 79.2%

Why the other options are there

γd = 168.5 lb/ft³ (voids ignored)
S = 126.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Dry unit weight

Example 10

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

A soil has Gs = 2.71, void ratio e = 0.91 and water content w = 17.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.71$
$e = 0.91$
$w = 17.5%$
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 10 — schematic for Phase relations for a compacted fill — Dry unit weight (9)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.71(62.4)/(1+0.91) = 88.54 lb/ft³
Moist unit weight — γ = γd(1+w) = 88.54(1+0.175) = 104.0 lb/ft³
Porosity
$n = e/(1+e) = 0.91/1.91 = 0.476 = 47.6%$
Saturation
$S = wGs/e = 0.175(2.71)/0.91 = 0.521 = 52.1%$

Answer:

γd ≈ 88.5 lb/ft³, γ ≈ 104.0 lb/ft³, n ≈ 47.6%, S ≈ 52.1%

Why the other options are there

γd = 169.1 lb/ft³ (voids ignored)
S = 191.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Dry unit weight

Core formulas for this FE topic

Worked exam-style examples

Dry unit weight

Core formulas for this FE topic

γD = VSNeeds technical review

Worked exam-style examples