Skip to content

Water content (%)

Geotechnical · FE Reference Handbook section

Geotechnical
1 formulas
10 exam-style examples
~47 min
All Geotechnical lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

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

  • γ=19.2kN/m3\gamma = 19.2 kN/m^{3}
  • w=0.16w = 0.16
  • Gs=2.68Gs = 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.
Moist sand, γ = 19.2 kN/m³Saturated sandWT

Figure 1 — schematic for Soil phase relationships from field data

Step-by-step solution

  1. Dry unit weight

    γd=γ/(1+w)=19.2/1.16=16.55kN/m3\gamma_d = \gamma/(1 + w) = 19.2/1.16 = 16.55 kN/m^{3}

  2. Void ratio — e = Gsγw/γ_d − 1 = 2.68(9.81)/16.55 − 1

  3. Evaluate

    e=26.29/16.55−1=1.589−1=0.589e = 26.29/16.55 - 1 = 1.589 - 1 = 0.589

  4. Saturation

    S=wGs/e=0.16(2.68)/0.589S = wGs/e = 0.16(2.68)/0.589

  5. Result

    γd=16.6kN/m3,e=0.589,S=0.728(72.8\gamma_d = 16.6 kN/m^{3}, e = 0.589, S = 0.728 (72.8%)
Answer:
γd=16.6kN/m3,e=0.589,S=72.8\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 — Water content (%)

A soil has Gs = 2.66, void ratio e = 0.62 and water content w = 21.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

  • Gs=2.66Gs = 2.66
  • e=0.62e = 0.62
  • w=21.5w = 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).
Compacted fill (γ = 124.5 pcf)

Figure 2 — schematic for Phase relations for a compacted fill — Water content (%)

Step-by-step solution

  1. Dry unit weight — γd = Gsγw/(1+e) = 2.66(62.4)/(1+0.62) = 102.5 lb/ft³

  2. Moist unit weight — γ = γd(1+w) = 102.5(1+0.215) = 124.5 lb/ft³

  3. Porosity

    n=e/(1+e)=0.62/1.62=0.383=38.3n = e/(1+e) = 0.62/1.62 = 0.383 = 38.3%

  4. Saturation

    S=wGs/e=0.215(2.66)/0.62=0.922=92.2S = wGs/e = 0.215(2.66)/0.62 = 0.922 = 92.2%
Answer:

γd ≈ 102.5 lb/ft³, γ ≈ 124.5 lb/ft³, n ≈ 38.3%, S ≈ 92.2%

Why the other options are there

  • γd = 166.0 lb/ft³ (voids ignored)
  • S = 108.4% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Water content (%)

Example 3
Phase relations for a compacted fill — Water content (%) (2)

A soil has Gs = 2.74, void ratio e = 0.87 and water content w = 21.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

  • Gs=2.74Gs = 2.74
  • e=0.87e = 0.87
  • w=21.5w = 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).
Compacted fill (γ = 111.1 pcf)

Figure 3 — schematic for Phase relations for a compacted fill — Water content (%) (2)

Step-by-step solution

  1. Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.87) = 91.43 lb/ft³

  2. Moist unit weight — γ = γd(1+w) = 91.43(1+0.215) = 111.1 lb/ft³

  3. Porosity

    n=e/(1+e)=0.87/1.87=0.465=46.5n = e/(1+e) = 0.87/1.87 = 0.465 = 46.5%

  4. Saturation

    S=wGs/e=0.215(2.74)/0.87=0.677=67.7S = wGs/e = 0.215(2.74)/0.87 = 0.677 = 67.7%
Answer:

γd ≈ 91.4 lb/ft³, γ ≈ 111.1 lb/ft³, n ≈ 46.5%, S ≈ 67.7%

Why the other options are there

  • γd = 171.0 lb/ft³ (voids ignored)
  • S = 147.7% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Water content (%)

Example 4
Phase relations for a compacted fill — Water content (%) (3)

A soil has Gs = 2.68, void ratio e = 0.46 and water content w = 9.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

  • Gs=2.68Gs = 2.68
  • e=0.46e = 0.46
  • w=9.5w = 9.5%

  • 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).
Compacted fill (γ = 125.4 pcf)

Figure 4 — schematic for Phase relations for a compacted fill — Water content (%) (3)

Step-by-step solution

  1. Dry unit weight — γd = Gsγw/(1+e) = 2.68(62.4)/(1+0.46) = 114.5 lb/ft³

  2. Moist unit weight — γ = γd(1+w) = 114.5(1+0.095) = 125.4 lb/ft³

  3. Porosity

    n=e/(1+e)=0.46/1.46=0.315=31.5n = e/(1+e) = 0.46/1.46 = 0.315 = 31.5%

  4. Saturation

    S=wGs/e=0.095(2.68)/0.46=0.553=55.3S = wGs/e = 0.095(2.68)/0.46 = 0.553 = 55.3%
Answer:

γd ≈ 114.5 lb/ft³, γ ≈ 125.4 lb/ft³, n ≈ 31.5%, S ≈ 55.3%

Why the other options are there

  • γd = 167.2 lb/ft³ (voids ignored)
  • S = 180.7% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Water content (%)

Example 5
Phase relations for a compacted fill — Water content (%) (4)

A soil has Gs = 2.62, void ratio e = 0.61 and water content w = 15.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

  • Gs=2.62Gs = 2.62
  • e=0.61e = 0.61
  • w=15.0w = 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).
Compacted fill (γ = 116.8 pcf)

Figure 5 — schematic for Phase relations for a compacted fill — Water content (%) (4)

Step-by-step solution

  1. Dry unit weight — γd = Gsγw/(1+e) = 2.62(62.4)/(1+0.61) = 101.5 lb/ft³

  2. Moist unit weight — γ = γd(1+w) = 101.5(1+0.150) = 116.8 lb/ft³

  3. Porosity

    n=e/(1+e)=0.61/1.61=0.379=37.9n = e/(1+e) = 0.61/1.61 = 0.379 = 37.9%

  4. Saturation

    S=wGs/e=0.150(2.62)/0.61=0.644=64.4S = wGs/e = 0.150(2.62)/0.61 = 0.644 = 64.4%
Answer:

γd ≈ 101.5 lb/ft³, γ ≈ 116.8 lb/ft³, n ≈ 37.9%, S ≈ 64.4%

Why the other options are there

  • γd = 163.5 lb/ft³ (voids ignored)
  • S = 155.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Water content (%)

Example 6
Phase relations for a compacted fill — Water content (%) (5)

A soil has Gs = 2.68, void ratio e = 0.75 and water content w = 11.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

  • Gs=2.68Gs = 2.68
  • e=0.75e = 0.75
  • w=11.0w = 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).
Compacted fill (γ = 106.1 pcf)

Figure 6 — schematic for Phase relations for a compacted fill — Water content (%) (5)

Step-by-step solution

  1. Dry unit weight — γd = Gsγw/(1+e) = 2.68(62.4)/(1+0.75) = 95.56 lb/ft³

  2. Moist unit weight — γ = γd(1+w) = 95.56(1+0.110) = 106.1 lb/ft³

  3. Porosity

    n=e/(1+e)=0.75/1.75=0.429=42.9n = e/(1+e) = 0.75/1.75 = 0.429 = 42.9%

  4. Saturation

    S=wGs/e=0.110(2.68)/0.75=0.393=39.3S = wGs/e = 0.110(2.68)/0.75 = 0.393 = 39.3%
Answer:

γd ≈ 95.6 lb/ft³, γ ≈ 106.1 lb/ft³, n ≈ 42.9%, S ≈ 39.3%

Why the other options are there

  • γd = 167.2 lb/ft³ (voids ignored)
  • S = 254.4% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Water content (%)

Example 7
Phase relations for a compacted fill — Water content (%) (6)

A soil has Gs = 2.73, void ratio e = 0.46 and water content w = 9.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

  • Gs=2.73Gs = 2.73
  • e=0.46e = 0.46
  • w=9.0w = 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).
Compacted fill (γ = 127.2 pcf)

Figure 7 — schematic for Phase relations for a compacted fill — Water content (%) (6)

Step-by-step solution

  1. Dry unit weight — γd = Gsγw/(1+e) = 2.73(62.4)/(1+0.46) = 116.7 lb/ft³

  2. Moist unit weight — γ = γd(1+w) = 116.7(1+0.090) = 127.2 lb/ft³

  3. Porosity

    n=e/(1+e)=0.46/1.46=0.315=31.5n = e/(1+e) = 0.46/1.46 = 0.315 = 31.5%

  4. Saturation

    S=wGs/e=0.090(2.73)/0.46=0.534=53.4S = wGs/e = 0.090(2.73)/0.46 = 0.534 = 53.4%
Answer:

γd ≈ 116.7 lb/ft³, γ ≈ 127.2 lb/ft³, n ≈ 31.5%, S ≈ 53.4%

Why the other options are there

  • γd = 170.4 lb/ft³ (voids ignored)
  • S = 187.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Water content (%)

Example 8
Phase relations for a compacted fill — Water content (%) (7)

A soil has Gs = 2.74, void ratio e = 0.81 and water content w = 20.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

  • Gs=2.74Gs = 2.74
  • e=0.81e = 0.81
  • w=20.5w = 20.5%

  • 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).
Compacted fill (γ = 113.8 pcf)

Figure 8 — schematic for Phase relations for a compacted fill — Water content (%) (7)

Step-by-step solution

  1. Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.81) = 94.46 lb/ft³

  2. Moist unit weight — γ = γd(1+w) = 94.46(1+0.205) = 113.8 lb/ft³

  3. Porosity

    n=e/(1+e)=0.81/1.81=0.448=44.8n = e/(1+e) = 0.81/1.81 = 0.448 = 44.8%

  4. Saturation

    S=wGs/e=0.205(2.74)/0.81=0.693=69.3S = wGs/e = 0.205(2.74)/0.81 = 0.693 = 69.3%
Answer:

γd ≈ 94.5 lb/ft³, γ ≈ 113.8 lb/ft³, n ≈ 44.8%, S ≈ 69.3%

Why the other options are there

  • γd = 171.0 lb/ft³ (voids ignored)
  • S = 144.2% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Water content (%)

Example 9
Phase relations for a compacted fill — Water content (%) (8)

A soil has Gs = 2.65, void ratio e = 0.60 and water content w = 9.5%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

  • Gs=2.65Gs = 2.65
  • e=0.60e = 0.60
  • w=9.5w = 9.5%

  • 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).
Compacted fill (γ = 113.2 pcf)

Figure 9 — schematic for Phase relations for a compacted fill — Water content (%) (8)

Step-by-step solution

  1. Dry unit weight — γd = Gsγw/(1+e) = 2.65(62.4)/(1+0.60) = 103.3 lb/ft³

  2. Moist unit weight — γ = γd(1+w) = 103.3(1+0.095) = 113.2 lb/ft³

  3. Porosity

    n=e/(1+e)=0.60/1.60=0.375=37.5n = e/(1+e) = 0.60/1.60 = 0.375 = 37.5%

  4. Saturation

    S=wGs/e=0.095(2.65)/0.60=0.420=42.0S = wGs/e = 0.095(2.65)/0.60 = 0.420 = 42.0%
Answer:

γd ≈ 103.3 lb/ft³, γ ≈ 113.2 lb/ft³, n ≈ 37.5%, S ≈ 42.0%

Why the other options are there

  • γd = 165.4 lb/ft³ (voids ignored)
  • S = 238.3% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Water content (%)

Example 10
Phase relations for a compacted fill — Water content (%) (9)

A soil has Gs = 2.74, void ratio e = 0.50 and water content w = 9.0%. Find the dry unit weight, moist unit weight, porosity and degree of saturation.

Given

  • Gs=2.74Gs = 2.74
  • e=0.50e = 0.50
  • w=9.0w = 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).
Compacted fill (γ = 124.2 pcf)

Figure 10 — schematic for Phase relations for a compacted fill — Water content (%) (9)

Step-by-step solution

  1. Dry unit weight — γd = Gsγw/(1+e) = 2.74(62.4)/(1+0.50) = 114.0 lb/ft³

  2. Moist unit weight — γ = γd(1+w) = 114.0(1+0.090) = 124.2 lb/ft³

  3. Porosity

    n=e/(1+e)=0.50/1.50=0.333=33.3n = e/(1+e) = 0.50/1.50 = 0.333 = 33.3%

  4. Saturation

    S=wGs/e=0.090(2.74)/0.50=0.493=49.3S = wGs/e = 0.090(2.74)/0.50 = 0.493 = 49.3%
Answer:

γd ≈ 114.0 lb/ft³, γ ≈ 124.2 lb/ft³, n ≈ 33.3%, S ≈ 49.3%

Why the other options are there

  • γd = 171.0 lb/ft³ (voids ignored)
  • S = 202.8% (ratio inverted)

Reference: FE Reference Handbook — Geotechnical → Water content (%)

© 2026 Dr. Steve Efe. Civil Engineering Capstone Studio. All rights reserved.