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Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay 3 – Phase system Soil composition Volume Weight V AIR a Wa = 0 VV Vw WATER Ww = SOLIDS Vs Ws

V = VS+VW+Va W = WS + WW Partially Saturated Soil Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Volumetric ratios Volumetric ratios commonly used in soil mechanics are: -Void ratio e -Porosity n

-Degree of Saturation Sr

-Air content ac

-Air void ratio or Percentage air voids na

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Void ratio e is defined as the ratio of volume of voids to the volume of solids

e = Vv/Vs

 Volume of voids Vv refers to that portion of the volume of the soil not occupied by solid grains

 Since the relationship between Volume of air Va and Volume of water Vw usually changes with ground water conditions as well as imposed loads, it is convenient to designate all the volume not occupied by solid grains as void space Volume of voids Vv

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay e = 0 ⇒ absence of voids (solid material) e > 1 ⇒ Vv >> VS in the soil mass Soil type void ratio e Uniform sand, loose 0.85 Mixed-grain sand, dense 0.43 Soft glacial clay 1.20 Soft highly organic clay 3.00 Soft Bentonite 5.20

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay  In nature, even though the individual void spaces are larger in coarse-grained soils, the void ratios of fine-grained soils are generally higher than those of coarse-grained soils. The ratio of volume of voids Vv to total volume V is defined as Porosity n.

n = Vv/V

But V = Vv + Vs = (1+Vv/VS) VS = (1+e) VS ⇒ n = e/(1+e) The porosity provides a measure of the permeability of a soil

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay  Porosity n of soil cannot exceed 100 %

0 < n < 100

Porosity n of a natural deposit = f ( Shape of grains, uniformity of grain size, and the conditions of sedimentation.  n = 25 – 50 % (natural sands)  n = 30 – 60 % (soft natural clays)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Out of void ratio e and Porosity n, void ratio is used frequently in soil engineering because:

e = Vv/Vs n = Vv/V

Any change in V is a direct consequence of a similar changes in VV and while VS remains the same.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Water content

The water content w is given as WW/WS, expressed as a percentage.

- where WW = Weight of water

WS = Weight of solids (dry)

Natural water content of fine-grained soils > coarse-grained soils. [No upper limit to w]

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Degree of saturation Sr

For a fully saturated soil-water system, since all the voids will be completely filled with water:

VVγW = WW where γW = unit weight of water

For partial saturation: (VV - Va)γW = WW

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Hence the relationship for Sr the degree of saturation is given as:

Sr = [(VV - Va)γW] / VV γW = VW /VV

Sr is the ratio of the volume of water to the volume of void space (Generally expressed as a Percentage)

0 < Sr < 100

- For completely dry soil Sr = 0;

- For fully saturated soil Sr = 1 or (100 %) Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Soil can be partially saturated [ 0 < Sr < 100]

Sr ≈ 100

Thus, at a Sr of 100 % all the voids are filled with water.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Degree of Saturation of Sand in various states

Condition of Sand Sr [%] Dry 0 Humid 1-25 Damp 26-50 Moist 51-75 Wet 76-99 Saturated 100

- Valid only for sands

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Degree of saturation Sr

Fine or silty sands are moist, wet or saturated.

Clays are always completely or nearly saturated except in the layer of soil subjected to seasonal variation of temperature and moisture.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Air content ac = Va/VV

= [Va + VW – VW] /VV

= [VV-VW]/VV

= 1- Sr

ac = 0 for saturated soil

ac = 1 for dry soil

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Air void ratio or Percentage air voids na

= Va/V By writing as Va Vv/(V Vv)

= n ac Using n = Vv/V and ac = Va/Vv = n (1- Sr)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay  Unit weight γ = (W/V) is one of the most important physical properties of the soil.

-The unit weight must be expressed with due regard to the state of soil.

γ = f (unit weight of solid constituents, n, and Sr)

Earth Vertical stress ? pressure ?

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay (for a partially saturated soil) Bulk unit weight γb = Total weight of soil mass / Total Volume

= (WW + WS)/ (VW + VS + Va)

For a saturated soil γb = γsat ⇒ Va = 0 = (WW + WS)/ (VW + VS)

Where γsat = Saturated unit weight of the soil

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Dry unit weight γd ⇒ Vw = 0 For a dry soil γd = (WS)/ (Va + VS)

γd = Ws/V

= (W – WW)/V

= [W/V – wWS/V]

= γbulk – w γd

γd = γb/(1+w)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Typical values of Unit Weight for Soils

3 3 Soil type γsat (kN/m ) γd (kN/m )

Gravel 20 - 22 15 - 17

Sand 18 - 20 13 - 16

Silt 18 - 20 14 - 18

Clay 16 - 22 14 - 21

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Specific Gravity Specific Gravity is the ratio of the unit weight of a substance to the unit weight of water γw at 4˚C. In soil mechanics, specific gravity generally refers to the specific gravity of solid particles Gs, and is defined as the unit weight of solid particles to the unit weight of water.

Gs = γS / γW Unit weight of solid

constituents γS = WS/VS = WS/VSγW

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay The value of the specific gravity can be determined from laboratory tests.

W2 W3 W1 W4

Where W1 = Wt. of empty sp. gravity bottle

W2 = W1 + dry soil

W3 = W2+ water (without any entrapped air)

W4 = W1 + water

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Specific Gravity

Gs = Weight of soil solids / Weight of water volume equivalent to that of water

= (W2 – W1) / [(W4-W1) – (W3-W2) ]

For most of soils Gs ranges from 2.5 – 2.9 Mineral G Gs = 2.65 for sands s Kaolinite 2.62-2.66 G = 2.2 – 2.4 for Pond ash s Illite 2.60 – 2.86

Gs = 4.4 – 5.2 for Iron ore Montmorillonite 2.75-2.78

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay For a partially saturated soil:

Mass Specific Gravity Gm = γb / γw

Gm (dry) = γd / γw – for dry soil

Gm (saturated) = γsat / γw – for saturated soil

Gm (dry) = Mass Specific gravity (dry state)

Gm (sat.) = Mass Specific gravity (saturated state)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Submerged (Buoyant unit weight) γ'

= Weight of soil inside the water / Total volume Treating whole soil mass as one unit = [ (WS + WW) - VγW] /V

= (WS + WW)/V - γW

= γsat - γW

VW WW γ′ = γsat - γW VS WS

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Phase relations Two approaches:

I. Specific Volume approach (VS = 1) II. Unit Volume approach (V = 1)

Using specific volume approach, VS is put as unit volume.

Specific Volume ν = 1 + e = V/VS (which is nothing but Total volume per unit Volume of solids)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay For dry soil – 2 phase system

VV = Air e Wa = 0 Va

V W Solids VS 1 WS = GSVSγW

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Soil is dry:

From the definition of void ratio e = VV/VS

VV = e

VS = 1

Gs = γS / γW = WS/VSγW

WS = GSγW

γd = WS/V = GSγW/(1+e) γd = GSγW/ (1 + e)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Basic relationships

VV = Water e WW = eγW Vw

V W Solids VS 1 WS = GSγW

For Saturated soil – 2 phase system

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Soil is fully saturated: From the definition of water content

W = WW/WS

= eγW / GSγW

e = W GS  For fully saturated case

γsat = W/V = (GSγW + eγW)/(1+e)

γsat = (GS + e)γW/(1 + e)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Basic relationships

Va AIR Wa = 0 VV = e Vw = wGS WATER Ww = wGSγW

SOLIDS 1 Ws = GSγW

For Partially Sat. soil – 3 phase system

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay From the definition of degree of saturation Sr = VW/VV

= wGS/e e = wGS/Sr

For Sr = 1, e = wGS γd = WS/V

= GSγW/(1+wGS/Sr)

-Valid for a partially saturated soil

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Relationship between γd, Gs, w and na

Using VS = WS/GSγW V = VS + VW + Va

VW = WW/γW 1 = VS/V + VW/V + na & by writing Ww = w Ws 1 – na = VS/V + VW/V

= γd/γW (w + 1/GS)

γd = (1-na) GSγW / (1 + wGS)

When soil becomes completely saturated na = 0

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Unit Volume approach (V = 1)

wG (1-n) S AIR Wa = 0 n

WATER wGS(1-n)γW V = 1 SOLIDS 1-n GS(1-n)γW

For Partially Sat. soil – 3 phase system

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay For Partially Sat. soil – 3 phase system

n = Vv/V

With V = 1 n = Vv

e = VV/Vs = n / 1 – n

γd = Ws/V = (1-n)Gs γw

γbulk = W/V = (1-n)Gs γw + wGs (1-n)γw

Percentage air voids na = Va/V

= (VV-Vw)V = [n-wGs(1-n)] Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay For a completely Saturated soil

γsat = Gs(1-n)γw + nγw

w = nγw/[Gs(1-n)γw

= e/Gs n WATER nγ e = wGs W

(for Sr = 1) SOLIDS 1-n GS(1-n)γW

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay For a dry soil

γd = Ws/V =

= (1-n)Gsγw

n AIR 0 V =1

SOLIDS 1-n GS(1-n)γW

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Some additional phase properties and relations  Porosity can be defined with respect to each of the phase of a soil

Soil particle porosity ns = Vs/V (expressed in %)

Water porosity nw = Vw/V (also referred as volumetric water Air porosity n = V /V a a content θw)  The water and air porosities represent their volumetric percentages in the soil; The soil particle porosity can be visualized as the percentage of total volume comprised of soil particles.

ns + n = ns + na + nw = 100%

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Some additional phase properties and relations

Volumetric water content θw = Vw/V

Using Sr = Vw/Vv ; θw = SrVV/V = nSr = eSr/1+e n = Vv/V; n = e/1+e The relationship between the gravimetric water contents, w and θw can be established by substituting the basic weight-volume relationship: For partially saturated soils:

θw = wGsSr / (Sr+wGs ) Using e = wGs / Sr

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem 1

A 0.9 m3 soil specimen weighs 17 kN and has a moisture content of 9%. The specific gravity of the soil solids is 2.7. Using the phase relations calculate:

(i) γ, (ii) γd, (iii) e, (iv) n, (v) Vw, and (vi) Sr.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution to example problem 1

3 Given: V = 0.9 m , W = 17 kN, w = 9%, and Gs = 2.7 From the definition of unit weight, γ = W/V γ = 17/0.9 = 18.89 kN/m3

Using γd = γ/(1+w) = 18.89/(1+0.09) = 17.33 kN/m3

Ws = γdV = 17.33 x 0.9 = 15.59 kN

Ww = W – Ws = 17 – 15.59 = 1.41 kN

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Using Gs = γs/γw = (Ws/Vs)/γw 3 Vs = Ws/(Gs γw) = 15.6 /(2.7 x 9.81) = 0.5889 m 3 V = Vv+Vs  Vv = 0.9 – 0.5889 = 0.311 m

Void ratio e = Vv/Vs = 0.311/0.5889 = 0.528

Porosity n = Vv/V = 0.311/0.9 = 0.346 (expressed as n = 34.6%) 3 Volume of water Vw = Ww/γw = 1.4/9.81 = 0.143 m

Degree of saturation Sr = Vw/Vv = 0.143/0.311 = 0.459

(expressed as Sr = 45.9%)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Soil composition Volume (m3) Weight (kN) V = a W = 0 0.168 AIR a = 0.311 V V Vw = WATER Ww = = 0.143 1.41

W = V = 0.589 SOLIDS s s 15.59

3 V = VS+VW+Va = 0.9 m W = WS + WW = 17 kN Partially Saturated Soil

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem 2

A soil has an “in-situ” in-place void ratio eo = 1.87; Natural moisture content wN = 60% and Gs = 2.75. Find γbulk and Degree of saturation Sr? m3 kN V = 0.22 a AIR Wa = 0 VV = e = 1.87 Vw = wGS WATER Ww = (0.6)*2.75*9.81 = 0.6*2.75

1 SOLIDS Ws = 2.75*9.81 (W = W + W = (V = 1+1.65+0.22) s w 43.16 kN)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution for example problem 2 3 ∴ γbulk = W/V = 43.17/2.87 = 15.04 kN/m

∴ Sr = Vw/Vv = 1.65/1.87 = 0.882 (88.2 %) Other relations…

3 ∴ γd = Ws/V = (W-Ww)/V = (43.16-16.19)/2.87 = 9.39 kN/m

∴Air content ac = Va/Vv = 0.22/1.87 = 0.1176 (11.76%)

(also using ac = 1-Sr = 1-0.882 = 11.8%)

∴ Percentage air voids (or Air voids) na = Va/V = 0.22/2.87 = 0.077

(also using na =nac = (Vv/V)*ac = (1.87/2.87)*0.1176 = 0.0766)

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Particle sizes, shapes and arrangement

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay