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CivilSolve

Geotechnical

Soil Phase Relationships

Find void ratio, porosity, water content, saturation, and soil unit weights from a sufficient set of known values, with a worked phase table.

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Inputs

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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

  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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)

1

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

  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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.