Module 5 · Lesson 5.2

The Se = wGs identity and unit weights

Derive the master identity and the four unit weights, then compute a real soil state.

Why this matters

Two results do most of the work in phase problems: the identity Se = wGs, which links how wet a soil is to how much void space it has, and the family of unit weights that turn a soil's state into the load it applies. Both come straight out of the phase diagram.

From first principles

The master identity: Se = wGs

We want to show: Link degree of saturation, void ratio, water content and specific gravity in one identity.

Water content is a mass ratio; saturation and void ratio are volume ratios. Specific gravity is exactly the bridge between mass and volume for the solids, so combining them collapses to a single clean identity.

Unit weights convert the soil's state into weight per unit volume. Taking $V_{s} = 1$ (so $V = 1 + e$) and weighing each phase, the total weight is the solids weight $G_{s} γ_{w}$ plus the water weight $S e γ_{w}$ . Dividing by the volume $1 + e$ gives the general moist unit weight, and the special cases follow:

Bulk (moist) unit weight

γ = \frac{G _{s} + S e}{1 + e} γ_{w}

What it calculates: the weight per unit volume at any degree of saturation

$γ$: bulk unit weight (kN/m³)
$γ_{w}$: unit weight of water ≈ 9.81 (kN/m³)
S: degree of saturation (fraction)

In plain terms: Using Se = wGs, this is equivalent to γ = Gs(1+w)/(1+e)·γw — the form used when the water content is known.

Dry, saturated and submerged unit weights

γ_{d} = \frac{G _{s}}{1 + e} γ_{w}, γ_{s a t} = \frac{G _{s} + e}{1 + e} γ_{w}, γ^{'} = \frac{G _{s} - 1}{1 + e} γ_{w}

What it calculates: the three limiting/derived unit weights

$γ_{d}$: dry unit weight (S = 0) (kN/m³)
$γ_{s a t}$: saturated unit weight (S = 1) (kN/m³)
$γ^{'}$: submerged (buoyant) unit weight = γsat − γw (kN/m³)

This assumes

γsat sets S = 1; γd sets S = 0; γ′ applies below the water table where buoyancy acts.

In plain terms: The submerged unit weight is the one to use for effective stress below the water table — it already subtracts the buoyant effect of the porewater.

Try it

Three-phase explorer

Solids, water and air. Move the sliders and watch the state and unit weights follow.

Void ratio e0.70

Degree of saturation S80 %

Specific gravity G_s2.70

What stays fixed

Solids volume is taken as 1 unit, so the total volume is 1 + e.
Water content follows from Se = wGs — you cannot make S exceed 1.

Porosity n: 0.412
Water content w: 20.7 %
Bulk γ: 18.8 kN/m³
Dry γ_d: 15.6 kN/m³
Saturated γ_sat: 19.6 kN/m³
Submerged γ': 9.8 kN/m³

Partly saturated (S = 80%): the bulk unit weight sits between γ_d and γ_sat.

Worked example

State and unit weights of a saturated clay

Given

Saturated clay (S = 1)
Water content w = 45%
Specific gravity Gs = 2.70
γw = 9.81 kN/m³

Find

Void ratio, porosity, and the saturated and submerged unit weights.

Assumptions

The clay is fully saturated, so all voids are water-filled (S = 1).

Practice

A moist sand has void ratio e = 0.65, water content w = 12% and Gs = 2.65. Taking γw = 9.81 kN/m³, what is its bulk unit weight? (Answer in kN/m³.)

Practice

For the same sand (e = 0.65, w = 12%, Gs = 2.65), what is the degree of saturation, as a percentage?

Summary

Se = wGs links saturation, void ratio, water content and specific gravity; for saturated soil e = wGs.
Bulk γ = (Gs + Se)/(1+e)·γw = Gs(1+w)/(1+e)·γw.
Dry γd, saturated γsat and submerged γ′ = γsat − γw follow from the same diagram.
Use γ′ below the water table for effective stress.

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint