Soil aggregate, plasticity, and classification
1.17 EMPIRICAL RELATIONSHIPS FOR PROCTOR COMPACTION TESTS
1.17 EMPIRICAL RELATIONSHIPS FOR PROCTOR
wopt(%) [ .= 1 95 0 38− . (log )](E PL) (1.82)
γd 3
kN/m opt
(max)
. (%)
( )=22 68. e0 0183w (1.83)
where
PL is the plastic limit, %
E is the compaction energy, kN-m/m3
For modified Proctor test, E ≈ 2700 kN/m3. Hence, wopt(%)≈0 65. (PL)
γd(max)(kN/m3)≈22 68. e−0 012. (PL)
Osman et al. (2008) analyzed a number of laboratory compaction test results on fine-grained (cohesive) soil, including those provided by Gurtug and Sridharan (2004). Based on this study, the following cor- relations were developed:
wopt(%)≈( .1 99 0 165− . ln )( )E PI (1.84)
γd 3
kN/m opt
(max)( )≈ −L Mw (%) (1.85)
where
L=14 34 1 195ln. + . E (1.86)
M= −0.19+0 0. 73lnE (1.87)
wopt is the optimum moisture content, % PI is the plasticity index, %
γd(max) is the maximum dry unit weight, kN/m3 E is the compaction energy, kN-m/m3
DiMatteo et al. (2009) analyzed the results of 71 fine-grained soils and pro- vided the following correlations for optimum moisture content wopt and max- imum dry unit weight γd(max) for modified Proctor tests (E = 2700 kN-m/m3)
wopt LL) 3.04 LL
(%)= − . ( + .
+
0 86 2 2
Gs
(1.88)
γd(max)kN m opt. PI .
( / 3)=40 316.
(
w−0 295)
( 0 32)−2 4. (1.89)where
LL is the liquid limit, % PI is the plasticity index, %
Gs is the specific gravity of soil solids
Mujtaba et al. (2013) conducted laboratory compaction tests on 110 sandy soil samples (SM, SP-SM, SP, SW-SM, and SW). Based on the test results the following correlations were provided for γd(max) and wopt (opti- mum moisture content):
γd 3
(kN/m ) 4.49 log( ) 151 log( ) 10.2u
(max) = C + E + (1.90)
logwopt=1 67 0 193. − . log( )Cu −0 153. log( )E (1.91) where
Cu = uniformity coefficient
E = compaction energy (kN-m/m3)
Example 1.7
For a sand with 4% finer than No. 200 sieve, estimate the maximum relative density of compaction that may be obtained from a modified Proctor test. Given D50 = 1.4 mm.
Solution
For the modified Proctor test, E = 2696 kN-m/m3. From Equation 1.80
A=0.216lnE−0.850=( .0216) ln 2696)( −0.850=0.856 From Equation 1.81
B= −0 0. 3lnE+0 0.3 6= −( .0 03 ln 2696)( )+0 0.3 6=0 0. 69 From Equation 1.79
Dr=AD50−B=( .0 856 1 4)( . )−0 069. =0 836. =83 6. %
Example 1.8
For a silty clay soil given LL = 43 and PL = 18. Estimate the maximum dry unit weight of compaction that can be achieved by conducting modified Proctor test. Use Equation 1.85.
Solution
For the modified Proctor test, E = 2696 kN-m/m3. From Equations 1.86 and 1.87
L=14 34 1 195ln. + . E=14 34 1 195ln 2696. + . ( )=23 78. M= −0.19+0 0. 73lnE= −0.19+0 0. 73ln 2696( )=0.387 From Equation 1.84
wopt(%) ( . . ln )( )E PI
[ . . ln( )]( )
= −
= − −
=
1 99 0 165
1 99 0 165 2696 43 18 17..16%
From Equation 1.85
γd(max)= −L Mwopt=23 78. −( .0 387 17 16)( . )=17 14. kN/m3
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