Review of Relative Density
Principles
 Relative
Density principles apply to
compaction of relatively clean, coarsegrained soils.
 Relatively clean usually taken to be less 12
% or less finer than the #200 sieve.
 Important for compaction study of filters
Objectives
Explain basic principles
of compacting clean
sands and gravels
 Understand basic tests to
obtain reference
densities.
 Use 1 point compaction
test in design and quality
control

Summarize minimum
and maximum index
density tests
 Detail the importance
of water content in
compacting clean
sands and gravels

Review of Compaction Principles
 Compaction Tests
are not commonly
performed on soils with 12 % or fewer fines
 Small percentage of fines means soils
cannot easily hold water to examine range
of water and effect on dry density
Review of Compaction Principles
 Compaction
tests performed on clean sands
may have this appearance
w%
Compacting Clean Sands
 Clean
sands are compacted most easily at
either very dry or very wet water contents
 At intermediate water contents, capillary
stresses in voids resist compaction
 Bulking is term for this phenomenon
Compacting Clean Sands
 Vibration
most effective energy for sands
 Use smooth-wheeled vibratory roller
Relative Density
 Alternative
to traditional compaction test is
relative density tests
 Minimum Index Density
 Maximum Index Density
 Relative Density
Minimum Index Density

Minimum index
density of clean
sand is that
resulting from
very loosely
filling a steel
mold. ASTM
Method D4254
Sand
dropped no
more than 1”
Minimum Index Density

After filling the
mold, excess soil
is carefully screed
off. The volume
of this mold is 0.1
ft3. Knowing the
weight of soil in
the mold, the dry
density is easily
computed
Maximum Index Density
 Example
Minimum dry density = 96 pcf
 Maximum index density of clean sand
results from vibration at high amplitude on
vibratory table for 10 minutes.
ASTM D4253
 Example Maximum dry density = 117.5 pcf
Maximum Index Density
Weight on
sample inside
sleeve
Vibratory
table
Maximum Index Density
Weight on
sample inside
sleeve
Vibratory
table
Maximum Index Density
Sample
densified by
vibration
Measure D
height to
determine
new gd
Plate on which weight
sits during vibration
Void Ratio and Dry Density
The void Ratio is calculated for each state of
denseness of sample.
 Maximum void ratio occurs at minimum index
density - For Example Min.gd = 96.0 pcf
 Minimum void ratio occurs at maximum index
density For Example Maximum gd = 110.0 pcf

Gs  g water
e
1
g dry
Minimum and Maximum Void Ratios
 First
emax
Calculate void ratio at Minimum gd
Gs  g water
2.65  62.4

1 
 1  0.7225
g dry
96.0
 Next
emin
Calculate void ratio at Maximum gd
Gs  g water
2.65  62.4

1 
 1  0.5033
g dry
110.0
Relative Density Equation
emax  emeasured
Rd (%) 
x100
emax  emin
Diagram below illustrates a
relative density of about 40 %
emin
gdmax
emeasured
gd measured
increasing density
emax
gd min
Calculate Void Ratio of Compacted Sand
 Now,
assume that the density of this
sand was measured in a compacted fill
and it was 102.5 pcf. Calculate a value
for relative density of the fill. First,
calculate the void ratio of the fill:
Gs  g water
2.65  62.4
e
1 
 1  0.6133
g dry
102.5
Compute Relative Density
 Now,
use the values of void ratio in the
relative density equation:
em ax  emeasured
Rd (%) 
x100
em ax  em in
0.7225  0.6133
Rd (%) 
x100  49.9%
0.7223  0.5033
Compute Relative Density
 Relative
Density Equation
(rewritten in dry density terms)
 Solve for Example:
g d max g d  g d min 
Rd (%) 
100
g d g d max  g d min 
110.0102.5  96.0
Rd (%) 
 100  49.8%
102.5110.0  96.0

Fort Worth Relative Density Study
 NRCS
lab in Fort Worth studied 28 filter
sands and used some published data
 Minimum and Maximum Index Densities
were performed on each sample
 A 1 point dry Standard Proctor energy mold
was also prepared for each sample.
 Values of 50% and 70% relative density were
plotted against the 1 point Proctor value
70 % Relative Density vs. 1 Point Proctor
130
125
70 % Relative Density
120
115
110
Best fit correlation
105
100
70 %RD = 1 Point line
95
90
90
95
100
105
110
115
120
Field 1 Point Proctor Test Dry Density, pcf
125
130
70 % Relative Density vs. 1 Point Proctor
Conclusion
is that the field 1 point
Proctor dry test is about equal to 70
% relative density
50 % Relative Density vs. 1 Point Proctor
125
120
best fit line
50 % Rd
115
110
105
100
95 % of 1
point
95
90
90
95
100
105
110
115
Field 1 pointdry density
120
125
130
50 % Relative Density vs. 1 Point Proctor
Conclusion
is that the 95 % of the
field 1 point Proctor dry test is
about equal to 50 % relative
density
Relative Density Estimates from FW
SML Study
gD
70=
1.075 x gd 1pt -9.61,
for RD70 and gd 1pt in lb/ft3
gD
50
= 1.07 x gd 1pt - 12.5,
for RD50 and gd 1pt in lb/ft3
Relative Density Estimates from FW
SML Study
Example
Relative Density Estimates
– Given: 1 Point Proctor Test
gd = 105.5 pcf
– Estimate 70 % and 50% Relative Density
– Given that measured gd is 98.7, evaluate
state of compaction of sand.
Review of Relative Density
 Class
Problem - Relative Density
– A soil’s minimum index density is 96.5
pcf and its maximum index density is
111.5 pcf. The Gs value is 2.65
– Calculate the emin and emax
– Compute the void ratio and dry density
corresponding to a relative density value
of 70 %
Class Problem Solution
 Given:
Minimum index density is 96.5 pcf
 Maximum index density is 111.5 pcf.
emax
Gs  g water
2.65  62.4

1 
 1  0.7136
g min_dry
96.5
Gs  g water
2.65  62.4
emin 
1 
 1  0.4831
g max_dry
111.5
Class Problem Solution
 Now,
substitue a value for RD of 70(%) in
the relative density equation
emax  emeasured
Rd (%) 
x100
emax  emin
0.7136  emeasured
70 
x100
0.7136  0.4831
Class Problem Solution
 Solving
and Rearranging the equation:
0.7136  emeasured
70 
x100
0.7136  0.4831
0.7136  emeasured
70

100
.2305
0.16135  0.7136  emeasured
emeasured  0.7136  0.16135  0.5225
Class Problem Solution
 Now,
calculate a value for dry density at this void
ratio:
Gs  g water
e
1
g dry
2.65  62.4
1.55225 
g dry
 Summary
2.65  62.4
0.55225 
1
g dry
g dry
165.36

 106.5 lb
1.55225
ft 3
- The dry density corresponding to
70(%) relative density for this sample is 106.5 pcf
140
Other information on Relative Density
130
Dry Density, pcf
Gravelly sand
120
sand and silty
sand
110
100
90
Reference - Donovan, N.C. and Sukhmander Singh, "Liquefaction Criteria
for the Trans-Alaska Pipeline." Liquefaction Problems in Geotechnical
Engineering, ASCE Specialty Session, Philadelphia, PA, 1976.
80
0
10
20
30
40
50
60
Relative Density, %
70
80
90
100
Other information on Relative Density
45
Chart is for silty
sands (SM)
Saturated Water Content, %
40
35
30
Average
25
20
15
Reference ભ Donovan, N.C. and Sukhmander Singh,
"Liquefaction Criteria for the Trans-Alaska Pipeline." Liquefaction
Problems in Geotechnical Engineering, ASCE Specialty Session,
Philadelphia, PA, 1976.
10
5
0
10
20
30
40
50
60
Relative Density, %
70
80
90
100
Class Problem
 Given
that the water content of a silty
sand that was obtained from a saturated
zone of a channel bank measured 24.5
percent
 What is the estimated relative density
of the sand?
Class Problem Solution
Reading
from the chart, the
estimated Rd value is about 42
percent.