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Tools · Soil mechanics

Three-phase diagram

Enter any three known soil properties. You get every other one with its formula and working, a phase diagram that redraws as you type, and where your soil sits against the saturation lines.

Known values

Fill in any three (leave the rest blank). Fill more and the tool checks they agree.

Worked examples

From lab readings

 

Fills in w = (M − Md) / Md and γ = (M / V)·γw. Add G (or one more value) to finish.

Settings

Phase diagram

GivenDerived so farStill to findTop of the ratioBottom of the ratio
Volume
Weight
Va = e(1 − S)0.270
Air
Wa = 00
Vw = S·e0.322
Water
Ww = S·e·γw0.322 ×γw
Vs = 11.000
Solids
Ws = G·γw2.680 ×γw
V = 1 + e = 1.592
Vv = e = 0.592
W = (G + S·e)·γw = 3.002 ×γw

e = 0.592 · n = 37.17% · S = 54.4% · w = 12.00%

Solve it step by step · start

What is given, what is to be found

Given: G, w, γ. To find: e, n, S, ac, na, γd, γsat, γ'.

Each step uses only values you already know (orange = given, green = found earlier) to find one new value. Tap any chip above to see what it means on the diagram.

All parameters

ParameterFormulaValue
G Specific gravity of solidsgiven
G = WsVs·γw
2.680
e Void ratio
e = VvVs = n1 − n = wGS
0.5917
n Porosity
n = VvV = e1 + e
= 0.5921 + 0.592 = 0.3717
37.17 %
w Water contentgiven
w = WwWs = S·eG
12.00 %
S Degree of saturation
S = VwVv = wGe
= 0.12 × 2.680.592 = 0.5436
54.36 %
ac Air content
ac = VaVv = 1 − S
= 1 − 0.5436 = 0.4564
45.64 %
na Percentage air voids
na = VaV = n·ac = e(1 − S)1 + e
= 0.592 × (1 − 0.5436)1 + 0.592 = 0.1697
16.97 %
γ Bulk unit weightgiven
γ = (G + S·e)·γw1 + e = γd (1 + w)
18.500 kN/m³
γd Dry unit weight
γd = G·γw1 + e = γ1 + w
= 2.68 × 9.811 + 0.592 = 16.518
16.518 kN/m³
γsat Saturated unit weight
γsat = (G + e)·γw1 + e
= (2.68 + 0.592) × 9.811 + 0.592 = 20.164
20.164 kN/m³
γ' Submerged unit weight
γ' = γsat − γw = (G − 1)·γw1 + e
= (2.68 − 1) × 9.811 + 0.592 = 10.354
10.354 kN/m³

How it is solved: every parameter depends on just three, G, e and S, so any three independent knowns fix the soil. The tool finds G = 2.680, e = 0.5917 and S = 54.36%, then works out the rest from the formulas above. Check: S·e = 0.3216 and w·G = 0.3216, so Se = wG holds.

Dry unit weight and water content

051015202530354091113151719212325Water content w (%)γd
S = 100% (zero air voids)80%60%Your soil: w = 12.00%, γd = 16.518 kN/m³

Each line is γd = Gγw / (1 + wG/S) for G = 2.680. No soil can plot above the zero air voids line; compaction curves always sit below it.

Key relationships

Se = wG
The basic link between saturation, void ratio and water content
n = e1 + e
e = n1 − n
γ = (G + Se)·γw1 + e
Bulk unit weight
γd = G·γw1 + e = γ1 + w
Dry unit weight
γsat = (G + e)·γw1 + e
S = 1
γ' = γsat − γw = (G − 1)·γw1 + e
Submerged (buoyant)
γd = G·γw1 + wGS
Use with S = 1 for the zero air voids line
γd = (1 − na)·G·γw1 + wG
In terms of air voids
na = n·ac
ac = 1 − S
Gm = γγw
Mass specific gravity