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Moist Unit Weight

γ = \frac{G _{s} γ _{w} ( 1 + w )}{1 + e}

Bulk (moist) unit weight from phase relations. Dry unit weight γ_d = Gs γ_w/(1+e) = γ/(1+w). Saturated: set S = 1 so e = w Gs and γ_sat = (Gs+e)γ_w/(1+e). γ_w = 9.81 kN/m³ (SI) or 62.4 lb/ft³ (US).

γ

(kN/m³)

Moist unit weight

G_{s}

Specific gravity of solids

w

Water content

e

Void ratio

γ_{w}

(kN/m³)

Unit weight of water

Step-by-step solved example

Gs = 2.70, w = 0.12, e = 0.50, γ_w = 9.81 kN/m³. Find γ and γ_d.

γ = W/V from the three-phase soil diagram.

1. Moist
γ = 2.70×9.81×(1.12)/(1.50) = 26.487×1.12/1.50 = 19.78 kN/m³.
$γ = 19.78 kN/m^{3}$
2. Dry
γ_d = γ/(1+w) = 19.78/1.12 = 17.66 kN/m³, or Gs γ_w/(1+e) = 26.487/1.50 = 17.66 kN/m³.
$γ_{d} = 17.66 kN/m^{3}$

Answer: γ = 19.8 kN/m³, γ_d = 17.7 kN/m³

10 practice questions

0/10 correct

1.γ_d for Gs = 2.70, e = 0.50, γ_w = 9.81 kN/m³ is nearest

2.Moist and dry unit weights are related by

3.γ_sat = (Gs + e) γ_w / (1+e). For Gs = 2.65, e = 0.60, γ_sat is nearest

4.Buoyant unit weight γ' equals

5.US customary γ_w is

6.If w doubles and e is unchanged, moist γ

7.A soil with γ = 20.0 kN/m³ and w = 0.10 has γ_d =

8.Increasing e at fixed w and Gs causes γ to

9.Dry unit weight does not depend on

10.For Gs = 2.70, w = 0, e = 0.50, γ equals γ_d =