Unit weight of solids

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 — Unit weight of solids

A soil has Gs = 2.70, void ratio e = 0.46 and water content w = 11.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.70$
$e = 0.46$
$w = 11.5%$
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 — Unit weight of solids

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+0.115) = 128.7 lb/ft³
Porosity
$n = e/(1+e) = 0.46/1.46 = 0.315 = 31.5%$
Saturation
$S = wGs/e = 0.115(2.70)/0.46 = 0.675 = 67.5%$

Answer:

γd ≈ 115.4 lb/ft³, γ ≈ 128.7 lb/ft³, n ≈ 31.5%, S ≈ 67.5%

Why the other options are there

γd = 168.5 lb/ft³ (voids ignored)
S = 148.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Unit weight of solids

Example 3

Phase relations for a compacted fill — Unit weight of solids (2)

A soil has Gs = 2.72, void ratio e = 0.63 and water content w = 17.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.72$
$e = 0.63$
$w = 17.5%$
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 3 — schematic for Phase relations for a compacted fill — Unit weight of solids (2)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.72(62.4)/(1+0.63) = 104.1 lb/ft³
Moist unit weight — γ = γd(1+w) = 104.1(1+0.175) = 122.3 lb/ft³
Porosity
$n = e/(1+e) = 0.63/1.63 = 0.387 = 38.7%$
Saturation
$S = wGs/e = 0.175(2.72)/0.63 = 0.756 = 75.6%$

Answer:

γd ≈ 104.1 lb/ft³, γ ≈ 122.3 lb/ft³, n ≈ 38.7%, S ≈ 75.6%

Why the other options are there

γd = 169.7 lb/ft³ (voids ignored)
S = 132.4% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Unit weight of solids

Example 4

Phase relations for a compacted fill — Unit weight of solids (3)

A soil has Gs = 2.72, void ratio e = 0.91 and water content w = 11.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.72$
$e = 0.91$
$w = 11.0%$
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 4 — schematic for Phase relations for a compacted fill — Unit weight of solids (3)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.72(62.4)/(1+0.91) = 88.86 lb/ft³
Moist unit weight — γ = γd(1+w) = 88.86(1+0.110) = 98.64 lb/ft³
Porosity
$n = e/(1+e) = 0.91/1.91 = 0.476 = 47.6%$
Saturation
$S = wGs/e = 0.110(2.72)/0.91 = 0.329 = 32.9%$

Answer:

γd ≈ 88.9 lb/ft³, γ ≈ 98.6 lb/ft³, n ≈ 47.6%, S ≈ 32.9%

Why the other options are there

γd = 169.7 lb/ft³ (voids ignored)
S = 304.1% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Unit weight of solids

Example 5

Phase relations for a compacted fill — Unit weight of solids (4)

A soil has Gs = 2.69, void ratio e = 0.72 and water content w = 16.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.69$
$e = 0.72$
$w = 16.5%$
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 — Unit weight of solids (4)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.69(62.4)/(1+0.72) = 97.59 lb/ft³
Moist unit weight — γ = γd(1+w) = 97.59(1+0.165) = 113.7 lb/ft³
Porosity
$n = e/(1+e) = 0.72/1.72 = 0.419 = 41.9%$
Saturation
$S = wGs/e = 0.165(2.69)/0.72 = 0.616 = 61.6%$

Answer:

γd ≈ 97.6 lb/ft³, γ ≈ 113.7 lb/ft³, n ≈ 41.9%, S ≈ 61.6%

Why the other options are there

γd = 167.9 lb/ft³ (voids ignored)
S = 162.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Unit weight of solids

Example 6

Phase relations for a compacted fill — Unit weight of solids (5)

A soil has Gs = 2.75, void ratio e = 0.69 and water content w = 21.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.75$
$e = 0.69$
$w = 21.0%$
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 6 — schematic for Phase relations for a compacted fill — Unit weight of solids (5)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.75(62.4)/(1+0.69) = 101.5 lb/ft³
Moist unit weight — γ = γd(1+w) = 101.5(1+0.210) = 122.9 lb/ft³
Porosity
$n = e/(1+e) = 0.69/1.69 = 0.408 = 40.8%$
Saturation
$S = wGs/e = 0.210(2.75)/0.69 = 0.837 = 83.7%$

Answer:

γd ≈ 101.5 lb/ft³, γ ≈ 122.9 lb/ft³, n ≈ 40.8%, S ≈ 83.7%

Why the other options are there

γd = 171.6 lb/ft³ (voids ignored)
S = 119.5% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Unit weight of solids

Example 7

Phase relations for a compacted fill — Unit weight of solids (6)

A soil has Gs = 2.65, void ratio e = 0.59 and water content w = 18.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.65$
$e = 0.59$
$w = 18.0%$
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 — Unit weight of solids (6)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.65(62.4)/(1+0.59) = 104.0 lb/ft³
Moist unit weight — γ = γd(1+w) = 104.0(1+0.180) = 122.7 lb/ft³
Porosity
$n = e/(1+e) = 0.59/1.59 = 0.371 = 37.1%$
Saturation
$S = wGs/e = 0.180(2.65)/0.59 = 0.808 = 80.8%$

Answer:

γd ≈ 104.0 lb/ft³, γ ≈ 122.7 lb/ft³, n ≈ 37.1%, S ≈ 80.8%

Why the other options are there

γd = 165.4 lb/ft³ (voids ignored)
S = 123.7% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Unit weight of solids

Example 8

Phase relations for a compacted fill — Unit weight of solids (7)

A soil has Gs = 2.74, void ratio e = 0.46 and water content w = 15.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.74$
$e = 0.46$
$w = 15.0%$
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 8 — schematic for Phase relations for a compacted fill — Unit weight of solids (7)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.46) = 117.1 lb/ft³
Moist unit weight — γ = γd(1+w) = 117.1(1+0.150) = 134.7 lb/ft³
Porosity
$n = e/(1+e) = 0.46/1.46 = 0.315 = 31.5%$
Saturation
$S = wGs/e = 0.150(2.74)/0.46 = 0.893 = 89.3%$

Answer:

γd ≈ 117.1 lb/ft³, γ ≈ 134.7 lb/ft³, n ≈ 31.5%, S ≈ 89.3%

Why the other options are there

γd = 171.0 lb/ft³ (voids ignored)
S = 111.9% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Unit weight of solids

Example 9

Phase relations for a compacted fill — Unit weight of solids (8)

A soil has Gs = 2.70, void ratio e = 0.64 and water content w = 24.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.70$
$e = 0.64$
$w = 24.0%$
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 9 — schematic for Phase relations for a compacted fill — Unit weight of solids (8)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.70(62.4)/(1+0.64) = 102.7 lb/ft³
Moist unit weight — γ = γd(1+w) = 102.7(1+0.240) = 127.4 lb/ft³
Porosity
$n = e/(1+e) = 0.64/1.64 = 0.390 = 39.0%$
Saturation
$S = wGs/e = 0.240(2.70)/0.64 = 1.013 = 101.3%$

Answer:

γd ≈ 102.7 lb/ft³, γ ≈ 127.4 lb/ft³, n ≈ 39.0%, S ≈ 101.3%

Why the other options are there

γd = 168.5 lb/ft³ (voids ignored)
S = 98.8% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Unit weight of solids

Example 10

Phase relations for a compacted fill — Unit weight of solids (9)

A soil has Gs = 2.68, void ratio e = 0.64 and water content w = 12.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

$Gs = 2.68$
$e = 0.64$
$w = 12.5%$
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 10 — schematic for Phase relations for a compacted fill — Unit weight of solids (9)

Step-by-step solution

Dry unit weight — γd = Gsγw/(1+e) = 2.68(62.4)/(1+0.64) = 102.0 lb/ft³
Moist unit weight — γ = γd(1+w) = 102.0(1+0.125) = 114.7 lb/ft³
Porosity
$n = e/(1+e) = 0.64/1.64 = 0.390 = 39.0%$
Saturation
$S = wGs/e = 0.125(2.68)/0.64 = 0.523 = 52.3%$

Answer:

γd ≈ 102.0 lb/ft³, γ ≈ 114.7 lb/ft³, n ≈ 39.0%, S ≈ 52.3%

Why the other options are there

γd = 167.2 lb/ft³ (voids ignored)
S = 191.0% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Unit weight of solids

Core formulas for this FE topic

Worked exam-style examples