Transcript Natural Soil Deposits

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
Weight
lb g kg kN
WA = 0
WT
Soil
Phase
Air
Volume
ft3 cc m3
VA
VT
WW
WS
Water
Solids
VW
VS
VV
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 Weight
(d = 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
Given: WT=4kg, VT=0.002m3, w=20%, Gs=2.68
kg
0
4.0
m3
A
W
S
0.002
Example Soil Block Analysis
Given: 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
Check
w = WW/WS x 100%
w = 100% x 0.667 / 3.333 = 20.01% bOK
Example Soil Block
kg
0
m3
A
4.0
0.002
0.667
3.333
W
S
Example Soil Block Analysis
VS = WS / GS(w
VS = 3.333kg / (2.68 x 1000 kg/m3) = 0.00124 m3
VW = WW/GS(w
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
0
m3
A
0.00009
4.0
0.00076
0.667
3.333
W
S
0.00067
0.00124
0.002
Example Soil Block Analysis
(T = 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 = 0.613
n = 100x0.00076/0.002 = 38.0%
S = 100x0.00067/0.00076 = 88.2%
Modified Soil Block Analysis
Given: WT=4kg, VT=0.002m3, w=20%, Gs=2.68
kg
0
120
20
100
m3
A
W
S
Modified Soil Block Analysis
Given: 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 Analysis
VS = WS / GS(w
VS = 100kg / (2.68 x 1000 kg/m3) = 0.0373 m3
VW = WW/GS(w
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
0
m3
A
0.0027
120
0.0227
20
100
W
S
0.0200
0.0373
0.06
Modified Soil Block Analysis
(T = 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
Air
Water
Solids
Modified Soil Block
kg
2.7
m3
A
W
0.0027
122.7
0.0227
20
100
W
S
0.0200
0.0373
0.06
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 Diameter
Thread
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]