Weight-Volume Relations Soil can be considered as a 3-phased material Air, Water, Solids.

Weight-Volume Relations Soil can be considered as

a 3-phased material

Air, Water, Solids

Soil Structure

Soil Structure

3-Phase Idealization

3-Phase Soil Block

Weightlb g kg kN

SoilPhase

Volumeft3 cc m3

WA = 0 Air VA

WT WW

Water VW

VV

VT

WS Solids VS

Weight Relations Water content, w

w = [WW/WS] x 100%

may be > 100% for clays Total (Moist,Wet) Unit Weight

= T = WET = WT / VT

Dry Unit Weightd = WS / VT

Table 2.2

Volumetric Relations Void Ratio, e

e = VV / VS

may be > 1, especially for clays Porosity, n

n = [VV / VT] x 100%

0% < n < 100% Degree of Saturation, S

S = [VW / VV] x 100%

0% < S < 100%

Inter-relationships Wet -> Dry Unit Weight

d = WET / (1+w/100)

WS = WT / (1+w/100)

Dry Unit Weight @ Saturation (Zero Air Voids)zav = W / (w/100+1/Gs)

Soil Block Analysis Use given soil data to completely fill

out weight and volume slots Convert between weight and volume

using specific gravity formula

Known Weight: V = W / Gs w

Known Volume: W = V Gs w

w=1g/cc=9.81kN/m3=1000kg/m3=62.4lb/ft3

Example Soil Block Analysis

kg m3

0 A

4.0W

0.002

S

Given: WT=4kg, VT=0.002m3, w=20%, Gs=2.68

Example Soil Block AnalysisGiven: WT=4kg, VT=0.002m3, w=20%, Gs=2.68

WS = WT / (1+w/100)

WS = 4kg / (1+20/100) = 3.333 kg

WW = WT – WS

WW = 4kg – 3.333 kg = 0.667 kg

Checkw = WW/WS x 100%

w = 100% x 0.667 / 3.333 = 20.01% OK

Example Soil Block

kg m3

0 A

4.0

0.667 W

0.002

3.333 S

Example Soil Block AnalysisVS = WS / GSw

VS = 3.333kg / (2.68 x 1000 kg/m3) = 0.00124 m3

VW = WW/GSw

VW = 0.667kg / (1 x 1000kg/m3) = 0.00067 m3

VA = VT – VS - VW

VA = 0.00200–0.00124–0.00067 = 0.00009m3

VV = VA + VW

VV = 0.00067+0.00009 = 0.00076m3

Example Soil Block

kg m3

0 A 0.00009

4.0

0.667 W

0.00067 0.00076

0.002

3.333 S

0.00124

Example Soil Block AnalysisT = 4.0kg/0.002m3 = 2000 kg/m3=19.62 kN/m3=124.8lb/ft3

D = 3.333kg/0.002m3=1666.5kg/m3=16.35 kN/m3

D = 19.62kN/m3 / 1.20 =16.35 kN/m3

e = 0.00076/0.00124 = 0.613n = 100x0.00076/0.002 = 38.0%S = 100x0.00067/0.00076 = 88.2%

Modified Soil Block Analysis

kg m3

0 A

120 20W

100S

Given: WT=4kg, VT=0.002m3, w=20%, Gs=2.68

Modified Soil Block AnalysisGiven: WT = 4 kg, VT = 0.002 m3

WT / VT ratio must remain unchanged

4 kg / 0.002 m3 = 120 kg / X

X = 0.06 m3 = VT

Modified Soil Block AnalysisVS = WS / GSw

VS = 100kg / (2.68 x 1000 kg/m3) = 0.0373 m3

VW = WW/GSw

VW = 20kg / (1 x 1000kg/m3) = 0.0200 m3

VA = VT – VS - VW

VA = 0.0600–0.0373–0.0200 = 0.0027m3

VV = VA + VW

VV = 0.0027+0.0200 = 0.0227m3

Modified Soil Block

kg m3

0 A 0.0027

120

20 W

0.0200 0.0227

0.06

100 S

0.0373

Modified Soil Block AnalysisT = 120kg/0.06m3 = 2000 kg/m3=19.62 kN/m3

D = 100kg/0.06m3=1666.7kg/m3=16.35 kN/m3

e = 0.0227/0.0373 = 0.609 (0.613)n = 100x0.0227/0.06 = 37.8% (38.0%)S = 100x0.02/0.0227 = 88.1% (88.2%)

Saturation Assumption

If a soil is partially saturated, we can get to full saturation by direct replacement of air with water.

It is further assumed that there will be

no increase in total volume.

3-Phase Idealization

Solids

Water

Air

Modified Soil Block

kg m3

2.7

AW 0.0027

122.7

20 W

0.0200 0.0227

0.06

100 S

0.0373

In Situ Comparators

Relative Density, Dr

Dr=100% x [emax – ein situ] / [emax – emin]

O% < Dr < 100%

Relative Compaction, R%R% = [d-in situ / d-max,lab] x 100%

R% may be > 100%

Consistency of Soil

Atterberg Limits

Liquid Limit, LL Plastic Limit, PL Shrinkage Limit, SL

Atterberg Limits

Liquid Limit

Liquid Limit Plot

Shear strength of soil @ LL is approx. 2.5 kN/m2 (0.36 psi)

Liquid Limit Europe & Asia

Fall Cone Test

BS1377

Plastic Limit

3mm DiameterThread

Shrinkage Limit

Consistency of Soil Plasticity Index, PI

PI = LL - PL Activity, A

A = PI / % Clay Liquidity Index, LI

LI = [w – PL] / [LL – PL]

Activity (Skempton, 1953)

A = PI / % Clay

Clays

Liquidity Index

LI = [w-PL] / [LL-PL]