Geotechnical

Soil Phase Relationships

Complete soil phase (weight–volume) table — void ratio e, porosity n, water content w, degree of saturation S, moist/dry/saturated unit weights and buoyant unit weight γ′ — from any sufficient subset of those quantities plus specific gravity Gs. Use for: 'find the void ratio / unit weight / saturation', converting between phase quantities, borrow-pit and compaction phase problems. Not for: compaction curves or relative density specs.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

specific gravity (optional)Specific gravity of solids Gs (dimensionless, e.g. 2.65 for quartz sand, 2.70 for clay)water content (optional)Gravimetric water content w as a DECIMAL (0.15 for 15%): weight of water / weight of solidsvoid ratio (optional)Void ratio e (dimensionless): volume of voids / volume of solidsporosity (optional)Porosity n as a decimal (0.35 for 35%): volume of voids / total volumesaturation (optional)Degree of saturation S as a DECIMAL (1.0 = fully saturated, 0.6 for 60%)moist unit weight (optional)Moist (total, bulk) unit weight γ of the soil [dimension: unitweight]dry unit weight (optional)Dry unit weight γ_d of the soil [dimension: unitweight]saturated unit weight (optional)Saturated unit weight γ_sat of the soil [dimension: unitweight]gamma water (optional)Unit weight of water γ_w; defaults to 9.81 kN/m^3 — pass 62.4 pcf to match US textbook tables [dimension: unitweight]total mass (optional)Total (wet) mass of the sample, from the lab — give with dryMass and/or totalVolume [dimension: mass]dry mass (optional)Oven-dry mass of the sample (with totalMass gives the water content) [dimension: mass]total volume (optional)Total sample volume (with the masses gives the unit weights) [dimension: volume]

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SI saturated clay: Gs=2.70, w=20%, S=100%

Given

specific gravity: 2.7
water content: 0.2
saturation: 1

Assumptions

Unit weight of water taken as 9.81 kN/m³ (62.4 pcf differs by under 0.1%).
Phase identities: S·e = w·Gs; γ_d = Gs·γ_w/(1+e); γ = γ_d(1+w); γ_sat = (Gs+e)γ_w/(1+e); n = e/(1+e); γ′ = γ_sat − γ_w.

Solution steps

Given phase quantities
Everything else follows from these via the closed-form phase identities.
Gs = 2.7; w = 0.2; S = 1; γ_w = 9.81 kN/m^3
Void ratio e from S·e = w·Gs
The fundamental link between the water and void phases: S·e = w·Gs.
e = w·Gs / S
e = 0.2 × 2.7 / 1 = 0.54
= 0.54
Dry unit weight from Gs and e
Solids weight spread over the total (solids + voids) volume.
γ_d = Gs·γ_w / (1 + e)
γ_d = 2.7 × 9.81 kN/m^3 / 1.54 = 17.2 kN/m^3
= 17.2 kN/m^3
Moist unit weight from γ_d and w
Adding the pore water's weight to the dry soil.
γ = γ_d·(1 + w)
γ = 17.2 kN/m^3 × 1.2 = 20.64 kN/m^3
= 20.64 kN/m^3
Saturated unit weight from Gs and e
Unit weight with every void filled with water.
γ_sat = (Gs + e)·γ_w / (1 + e)
γ_sat = 3.24 × 9.81 kN/m^3 / 1.54 = 20.64 kN/m^3
= 20.64 kN/m^3
Porosity n from γ_sat and γ_d
Saturating the soil adds exactly one pore volume of water, so the unit-weight gain is n·γ_w.
n = (γ_sat − γ_d) / γ_w
n = 3.44 kN/m^3 / 9.81 kN/m^3 = 0.351
= 0.351
Buoyant (submerged) unit weight
Below the water table, buoyancy carries one water-unit-weight of every unit volume.
γ′ = γ_sat − γ_w
γ′ = 20.64 kN/m^3 − 9.81 kN/m^3 = 10.83 kN/m^3
= 10.83 kN/m^3

Results

Void ratio e

0.54

Porosity n

0.351

Water content w (decimal)

0.2

Degree of saturation S (decimal)

Moist (total) unit weight γ

20.64kN/m^3

Dry unit weight γ_d

17.2kN/m^3

Saturated unit weight γ_sat

20.64kN/m^3

Buoyant (submerged) unit weight γ′

10.83kN/m^3

Where this answer was checked

source: Closed-form phase identities (NCEES FE Reference Handbook, soil phase relationships) — e = w·Gs/S, then unit weights from (Gs, e, w) with γw = 9.81 kN/m³
verified by: hand-recomputed
derivation: e = 0.20·2.70/1.0 = 0.54; n = 0.54/1.54 = 0.35065; γd = 2.70·9.81/1.54 = 26.487/1.54 = 17.1994 kN/m³; γ = 17.1994·1.20 = 20.6392 kN/m³; γsat = (2.70+0.54)·9.81/1.54 = 31.7844/1.54 = 20.6392 (= γ, as it must at S=1); γ′ = 20.6392 − 9.81 = 10.8292 kN/m³.

Example 2

SI partially saturated sand: Gs=2.65, e=0.60, S=50%

Given

specific gravity: 2.65
void ratio: 0.6
saturation: 0.5

Assumptions

Unit weight of water taken as 9.81 kN/m³ (62.4 pcf differs by under 0.1%).
Phase identities: S·e = w·Gs; γ_d = Gs·γ_w/(1+e); γ = γ_d(1+w); γ_sat = (Gs+e)γ_w/(1+e); n = e/(1+e); γ′ = γ_sat − γ_w.

Solution steps

Given phase quantities
Everything else follows from these via the closed-form phase identities.
Gs = 2.65; e = 0.6; S = 0.5; γ_w = 9.81 kN/m^3
Porosity n from void ratio
Porosity is the void fraction of the TOTAL volume, so the denominator is 1 + e.
n = e / (1 + e)
n = 0.6 / 1.6 = 0.375
= 0.375
Water content w from S·e = w·Gs
The fundamental link between the water and void phases: S·e = w·Gs.
w = S·e / Gs
w = 0.5 × 0.6 / 2.65 = 0.113
= 0.113
Dry unit weight from Gs and e
Solids weight spread over the total (solids + voids) volume.
γ_d = Gs·γ_w / (1 + e)
γ_d = 2.65 × 9.81 kN/m^3 / 1.6 = 16.25 kN/m^3
= 16.25 kN/m^3
Moist unit weight from γ_d and w
Adding the pore water's weight to the dry soil.
γ = γ_d·(1 + w)
γ = 16.25 kN/m^3 × 1.11 = 18.09 kN/m^3
= 18.09 kN/m^3
Saturated unit weight from Gs and e
Unit weight with every void filled with water.
γ_sat = (Gs + e)·γ_w / (1 + e)
γ_sat = 3.25 × 9.81 kN/m^3 / 1.6 = 19.93 kN/m^3
= 19.93 kN/m^3
Buoyant (submerged) unit weight
Below the water table, buoyancy carries one water-unit-weight of every unit volume.
γ′ = γ_sat − γ_w
γ′ = 19.93 kN/m^3 − 9.81 kN/m^3 = 10.12 kN/m^3
= 10.12 kN/m^3

Results

Void ratio e

0.6

Porosity n

0.375

Water content w (decimal)

0.113

Degree of saturation S (decimal)

0.5

Moist (total) unit weight γ

18.09kN/m^3

Dry unit weight γ_d

16.25kN/m^3

Saturated unit weight γ_sat

19.93kN/m^3

Buoyant (submerged) unit weight γ′

10.12kN/m^3

Where this answer was checked

source: Closed-form phase identities (NCEES FE Reference Handbook, soil phase relationships) — w = S·e/Gs, then unit weights from (Gs, e, S) with γw = 9.81 kN/m³
verified by: hand-recomputed
derivation: w = 0.5·0.6/2.65 = 0.113208; n = 0.6/1.6 = 0.375; γd = 2.65·9.81/1.6 = 25.9965/1.6 = 16.2478 kN/m³; γ = 9.81(2.65+0.5·0.6)/1.6 = 9.81·2.95/1.6 = 28.9395/1.6 = 18.0872 kN/m³ (cross-check: 16.2478·1.113208 = 18.0872 ✓); γsat = (2.65+0.6)·9.81/1.6 = 31.8825/1.6 = 19.9266 kN/m³; γ′ = 10.1166 kN/m³.

Example 3

US borrow soil: γ=120 pcf, w=15%, Gs=2.70 (γw = 62.4 pcf)

Given

moist unit weight: 120 pcf
water content: 0.15
specific gravity: 2.7
gamma water: 62.4 pcf

Assumptions

Unit weight of water as given in the input.
Phase identities: S·e = w·Gs; γ_d = Gs·γ_w/(1+e); γ = γ_d(1+w); γ_sat = (Gs+e)γ_w/(1+e); n = e/(1+e); γ′ = γ_sat − γ_w.

Solution steps

Given phase quantities
Everything else follows from these via the closed-form phase identities.
Gs = 2.7; w = 0.15; γ = 120 pcf; γ_w = 62.4 pcf
Dry unit weight from γ and w
Stripping the pore water's weight from the moist soil.
γ_d = γ / (1 + w)
γ_d = 120 pcf / 1.15 = 104.3 pcf
= 104.3 pcf
Void ratio e from the dry unit weight
Inverting γ_d = Gs·γ_w/(1+e) for the void ratio.
e = Gs·γ_w / γ_d − 1
e = 2.7 × 62.4 pcf / 104.3 pcf − 1 = 0.615
= 0.615
Saturated unit weight from Gs and e
Unit weight with every void filled with water.
γ_sat = (Gs + e)·γ_w / (1 + e)
γ_sat = 3.31 × 62.4 pcf / 1.61 = 128.1 pcf
= 128.1 pcf
Porosity n from γ_sat and γ_d
Saturating the soil adds exactly one pore volume of water, so the unit-weight gain is n·γ_w.
n = (γ_sat − γ_d) / γ_w
n = 23.75 pcf / 62.4 pcf = 0.381
= 0.381
Degree of saturation S from γ, Gs and e
Solving the moist unit weight identity γ = γ_w(Gs + S·e)/(1 + e) for S.
S = (γ·(1 + e)/γ_w − Gs) / e
S = (120 pcf × 1.61 / 62.4 pcf − 2.7) / 0.615 = 0.659
= 0.659
Buoyant (submerged) unit weight
Below the water table, buoyancy carries one water-unit-weight of every unit volume.
γ′ = γ_sat − γ_w
γ′ = 128.1 pcf − 62.4 pcf = 65.7 pcf
= 65.7 pcf

Results

Void ratio e

0.615

Porosity n

0.381

Water content w (decimal)

0.15

Degree of saturation S (decimal)

0.659

Moist (total) unit weight γ

120pcf

Dry unit weight γ_d

104.3pcf

Saturated unit weight γ_sat

128.1pcf

Buoyant (submerged) unit weight γ′

65.7pcf

Where this answer was checked

source: Closed-form phase identities (NCEES FE Reference Handbook, soil phase relationships) — γd = γ/(1+w), e = Gs·γw/γd − 1, S = w·Gs/e, γsat = (Gs+e)γw/(1+e)
verified by: hand-recomputed
derivation: γd = 120/1.15 = 104.3478 pcf; e = 2.70·62.4/104.3478 − 1 = 168.48/104.3478 − 1 = 1.61460 − 1 = 0.61460; n = 0.61460/1.61460 = 0.38065; S = 0.15·2.70/0.61460 = 0.405/0.61460 = 0.65896; γsat = (2.70+0.61460)·62.4/1.61460 = 206.831/1.61460 = 128.100 pcf; γ′ = 128.100 − 62.4 = 65.700 pcf.