SGI Varia, 100
SOME TEST METHODS AND INTERPRETATION OF TEST RESULTS
DAO VAN PHU, MAI VAN THANH Swedish Geotechnical Institute Linkoping, Sweden, January 1983
STATENS GEOTEKNISKA INSTITUT
ACKNOWLEDGEMENTS
This report is written during our stay at the Swedish geotechnical institute (SGI) from October 1982 to January 1983.
Great thanks to Dr Jan Hartlen, Director of SGI, for his great help during our studies in Sweden.
Great thanks to Dr Bo Berggren at SGI for his concern for our studying results.
Grateful thanks to Mr Goran Nilsson and Mr Rolf Larsson at SGI for their great help and for critical reading of the manuscript.
Gratitude is expressed to Mrs Eva Dyrenas for her expert typing of the manuscript.
We will also express our thanks to other memebers of SGI for their kindness and their help during our time at SGI.
Linkoping, January 1983
Dao van Phu Mai van Thanh
SGI nr 196 Klintland Grafiska, Linkoping
STATENS GEOTEKNISKA INSTITUT
SOME TEST METHODS AND INTERPRETATION OF TEST RESULTS
According to the SGI plan we have been working in the laboratory from November 1982 to January 1983.
During this period we have carried out laboratory tests such as
- fall-cone test
- oedometer test (incremental test apparatus) - consolidated drained direct shear test - clay-lime mixing test
A. TEST METHODS I. Fall-cone test
In Sweden the undrained shear strength is usually determined by fall-cone test.
Three different standard cones have been used: the 100 gm-
o O 0
30 cone; the 60 gm-60 cone and the 10 gm-60 cone.
The test is carried out as follows. A metal cone is placed vertically with the cone edge just in contact with the surface of the clay sample. The cone is then released to penetration into the clay and the penetration depth is measured.
1) Determination of Tk for undisturbed clay
The undrained shear strength Tk of clay can be expressed as:
= k g (kPa) ( 1 )
where: Tk is the unreduced value of the shear strength and usually i t is equal with Tfu' but if the liquid limit of the clay is higher than 80% we have to
reduce the Tk value to get Tfu· The reduction factor according to SGI you can see in Table 2; m is the cone weight; i is the penetration depth: g is the
I
gravity constant and k is a constant, whose magnitude
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2 STATENS GEOTEKNISKA INSTITUT
depends on the cone angle B. For "undisturbed"
clay, k also depends on degree of disturbance from the sampling.
When using the standard piston samplers SGI I and II (k ~ 1.0 for 30° cones and k ~ 0.25 for 60° cones
We can use Table 1 to evaluate.
Shear strength at different cone impressions Table 1.
/kPa, unreduced value/. According to SGF's Laboratory committee.
Ken ,_
tvo mm 0 l 2 3 4 5 6 7 8 9
5 !55 150 145 140 135 130 125 120 115 115
6 1;0 105 100 99 96 93 90 87 as 92
7 80 78 76 74 72 70 68 66 64 63
8 61 60 58 57 56 54 53 52 51 so
9 ~a 47 46 45 44 0 43 42 41 40 I
g 10 39 38 38 J7 36 36 35 34 34 33
ll 32 32 31 31 30 JO 29 29 23 28
"" 12 27 27 26 26 26 25 25 24 24 24
0 13 23 23 23 22 22 22 21 21 21 20
~ 14 20 :9, 5 19,5 19 19 !8,5 i8,5 !8 18 l 7 ,5
!..5 17,5 l 7 !7 l 7 ~6,5 16,5 16 16 15,5 15,5
16 15,S 15 15 15 14, 5 14, 5 14 14 14 13,5
17 13,5 13 ,5 13,5 13 13 13 12,5 12,S 12,5 12
18 12 12 '" t l, S 11,5 1:,.3 : 1 ,5 11 11 11
19 11 11 10,5 10,5 10,5 10,5 :o :o 10 9,9
5 39 38 36 35 34 32 31 30 29 28
6 27 26 26 25 24 23 23 22 21 21
7 20 19,5 19 18 ,5 18 17 ,5 17 16,5 16 15,5
8 15,5 15 14,S 14 14 13,5 13,5 13 12,5 12,5
I 9 12 12 11,5 11,5 11 ll 10,5 10,5 10 10
10 9,8 9,6 9,4 9,2 9, l 8,9 3,7 8,0 8,4 8,3
I g 11 8,1 a,o 7,8 7,7 7,3 7,4 7 ,3 7 ,2 7,0 6,9
12 6,8 6,7 6,6 6,5 6,4 6,3 6,2 6, l 6,0 5,9
"' !3 5,8 5,7 5,6 5, 5 5,5 5,4 5,3 5 ,2 5,2 5,1
0
14 5,0 4,9 4,9 4,8 4,7 4, 7 4,6 4,5 4,5 4,4
15 4,4 4,3 -i,2 4,2 4, l 4, l 4,0 4,0 3 ,9 3,9
16 3 I a 3, a J, 7 ;, 7 },6 3,6 3 ,6 3,5 3,5 3,4
17 3,4 3,4 3,3 3,3 3, 2 3, 2 3 ,2 3,1 3, 1 3,i
18 3,0 3,0 3,0 2,9 2,9 2,9 2,a 2,8 2,8 2. i'
19 2,7 2,7 2,7 2,6 2,6 2,6 2,6 2,5 2,5 2,5
2
5 5,9 5,7 5,4 5,2 5,0 4,9 ... , .. 4,5 4,4 4,2
6 4, I 4,0 3,8 3,7 3,6 3,5 3,4 3 ,3 3,2 3,1
7 3,0 2,9 2,8 .,c' ' 2,7 2,6 2,5 2,5 2 ,.; 2,4
8 2,3 2,2 2,2 2,1 2,1 2,0 2,0 I ,95 l ,9 1,85
I 9 1,8 1,8 : , 75 : '7 : ,65 I ,65 1,6 1,55 l, 55 l,S
10 I ,45 1,45 1,4 1,4 1,JS l,35 l, 3 1,3 t,25 l,~S
';:, l! I ,2 l ,2 I, 15 1,15 I, 15 l, l l , l l,05 1,05 l ,05
"' 12 LO I ,0 0,99 0,97 0,96 0,94 0,93 0,91 0,90 0,88
0 "' 13 0,87 0,86 0,85 0,83 0,82 0,81 0,80 J,78 0, 77 0,76
14 0,75 0,74 0,73 0,72 0,71 o, 70 0,69 0,68 0 ,07 0,66
15 0,65 0,64 0,64 0,63 0,62 0,61 0,60 0,00 0,59 0,58
16 0,57 0,57 0 ,56 0,55 0,55 0 ,$4 0,53 0,53 0,52 0,52
17 0,51 0,50 0,50 0,49 0,49 0,48 0,48 0,47 0,46 0,46
18 0,45 0,45 0,44 0,44 0,43 ,::) ,43 0,43 0 ,42 0,42 0,41
19 0,41 0,40 C,40 J,40 0,39 0,39 0,38 o, 38 0,38 0,37
"'
5 0,98 0,94 0,91 0,87 0,H4 o,a1 0,78 0,75 0,73 0, 70
6 0,68 0,66 0,64 0,62 0,60 0,58 0,56 0,55 0,53 0,52
7 0,50 0,49 0,47 0,46 ,; ,~ s ) ,4,; 0,42 0,41 0,40 0,39 8 0,38 0,37 0,36 0, }6 C I )S 0, 34 J,33 0,32 -J,32 0,31
9 0,30 0,30 0,29 0,28 0,28 0,27 o,:7 0,26 0,2E G,25
00 w 0,2.5 0,24 0,24 0,23 0,23 0,22 0,22 0,21 '.),21 0,21
·.o 1, 0,20 0,20 0, 195 0, 19 0, 19 0, 185 0, 18 0, 16 0, 175 0, 175
;j
12 13 0, l 7 0 ,145 a, a, 11 14s 0, 165 0, i.4 0, 16 0, 14 0, 16 0, 135 0, 155 0, 135 O, 155 0, 135 0,15 0, 13 0, 15 o, ;3 0, 145 o, 125 14 0, 125 9, ::!5 C, 12 0, 12 0, 12 0, ! 15 0, l 15 0, 115 o, 11 G,11I ! 15 0 ,11 0' ;,_ •1, tO'i 0, 105 0, 105 J, 1:::> 0, 10 0,099 0,098 0,097
:6 i 0,096 0,095 ,l ,')93 C,092 J,091 C,090 0,089 0,088 ?,087 0,086 0,085 0,084 C,J8} 0,082 0,081 0,000 0,079 0,076 'J,O77 0,077 l 71
lS 0,076 O,C75 0,074 0 ,073 J,072 J ,C72 0,071 0,070 O,Jb9 0,069
i
19 I 0,068 0,067 0 ,06 7 J,066 0,06S 0,J64 0,064 O ,C63 0,0CJ 0,062-·
SGI nr 196 Klintland Grafiska, Linkoping
STATENS GEOTEKNISKA INSTITUT 3
2) Determination of
TR
for remoulded clayThe remoulding of the clay should be continued until the depth of penetration becomes a maximum and you have two
on each other following cone-impressions with the same value.
Usually the 60 g-60° cone or the 10 g-30° cone is used. We can calculate the remoulded shear strength as in Eq (1) or look in Table 1.
We can calculate the sensitivity St = T
3) Determination of liquid limit (Wr,) with fall-cone apparatus
Definition of WL: the water content of a remoulded clay sample where the 60 g-60° cone is used makes an impression of 10 mm. The liquid limit of clay (WL) can then be cal
culated with the formula (2) at different impressions of 60 g-60° cone.
= N•W + M (2)
n
W = natural water content, % n
Table 2. Reduction factorµ at different values of the liquid limit (WL).
-1.2
I I
( I
' ! I J
I I
I
I I I l I'
I i I I I
i
I
1.0 ! i ! I i I !
- l I
i ' : ' i I I ' I I Ii I I
...
I 'I
II
l
I
' I
0 0.8 '
I i
i ' I0 I I I !
-
-ci a: "('0 Q) 0.6-
jI
' I i I-+-
I II
i iI
I II I ' ' '!
I !I
I I I I i I I 'I'
·-4, '! i : !
I
I II
i I i
I
I0.4, I ! I
i
!-
40 80 80 100 120 140 160' 180 200liquid limit, %
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4 STATENS GEOTEKNISKA INSTITUT
Table 3. Values of Mand N for different cone impressions.
Kon- Sjunk.
typ nir1c;
.
0.
1.
2 J.
4.
5.
6.
7.
8.
9mm
6o g 1 • 21 1.20 1.19 1. 18 1.17 1.16 1. 15 1.14 1.14 1.1J
600 7. -J-5 -J)~ -J.2 -J.O -2.9 -2.7 -2.6 -2.5 -2.J -2.2
1 • 12 1 • 11 l • 11
.
0 1 • 10 1.09 1.09 1.07 1.07 1.068. -2.1 - 1 .9 -1.8 -1.7 -1.6 -1 .4 -1.J -l.2 -1 • 1 -1.0
1 .05 1.05 1.04 1.04 1.0J 1.0J 1.02 1.01 1.01 1.00
9. -0.9
-o.8
-0.7 -<D. 6 -0.5 -0.4 ·-0.J -0.J -0.2 --0. 1 1.00 1.00 0.99 0.99 0.98 0.98 0.97 0.97 0:96 0.9610. 0
o.
1 0.2 0.20.J
o.4o.s
0.5 o.6 0.7 0.96-
0.95 0.95 o. <)4 0.94 0.94 0.93o.9J
0.93 0.9211 • 0.7 o.8 o_.9 0~9 1 • 0 1 • 1 1.1 1 .2 1 .J 1.J 0.92 0.92 0.91 0.91 0. 91 0.90 0.90 0.90 0.89 0.89
12. 1.4 1.4 1.5 1.5 1.6 1.7 1.7 1.8 1.8 1.9 0.89 o.88 o.88 0.88 0.88 0.87 0,87 0.87 0.87 0.86 lJ. 1.9 2.0 2.0.. 2. 1 2. 1 2.2 2.2 2.2 2.J 2.3
o.8'6 0.86 0.86 0.85 0.85 0.85 0.85 o.84 o.84
o.84
14. 2.4 2.4 2.5 2.5 2.5 2.6 2.6 2.7 2.7 2.7
4) Determination of natural water content (W0 )
w
n = ~ -wWd 100
( 3)
= weight of water
weight of dry sample.:
5) Determination of bulk density (p) p = w
A•h ( 4)
h = might of sample (cm) A = area of sample {cm2 ) W = weight of sample (g)
H N M N M N M N H N M N M N M
~
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III. Incremental loading test (oedometer test)
Oedometer tests on natural fine-grained soils are usually performed on "undisturbed" samples fitted into a con
fining ring. A test procedure with incremental loading, each increment being equal to the previous load and a new increment every 24 hours, was suggested by Terzaghi in 1925 and has been widely used since then. During the test the sample is drained from both ends (Fig. 1). The piston on top of the sample is loaded and moves into the ring which is fitted in position.
Fig. 1
The results from incremental oedometer tests are usually presented in a diagram where the strain at the end of each step is plotted versus the consolidation pressure.
In this plot the vertical effective pressure is in log-scale.
Fig 2. Typical results from an oedometer test on clay.
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From this diagram settlement may be calculated by reading off the relative compression between the vertical in situ pressure in the ground 0' and the calculated final pressure
0
0 1 and multiplying by the thickness of the soil layer.
1) The Casagrande method
We can use the Casagrande method for calculating the pre
consolidation pressure (0 1 ) and the coefficient of con-
e
solidation (Cv).
a) The_Casa~rande_method_to_evaluate_the_rreconsolidation
EE§§§::!E§
For determination of the preconsolidation 0 1 the Casagrande
C
method illustrated in Fig. 3 has been widely used.
Fig. 3
In this method a horizontal line and a tangent to the oedometer curve at the point with the smallest radius of curvature is drawn. The angle between the horizontal line and the tangent is bisected. The straight portion
of the oedometer curve is extended and the preconsolidation
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STATEN$ GEOTEKNISKA INSTITUT 7
pressure is evaluated as the pressure at the intersection of this line and the bisection.
b) The_Casagrande_method_to_evaluate_the_coefficient of consolidation
From incremental oedometer test the Casagrande method involves plotting the deformation versus the logarithm of time (see Fig. 4).
Fig. 4. Casagrande construction of Cv.
u = 0 is constructed by assuming a parabolic shape of the first part of the curve. u = 100% is constructed as the intersection between the tangent to the curve at its point of inflextion and the extention of the straight end part of the curve. Eso at
u
= 50% is then calculated, tso is constructed and C is calculated fromV 2
C Vso = Tso Hso
tso (5)
For odeometers with drainage from both ends Hso = Ho;sso where Ho is initial sample height and the time factor
Tso= 0.197.
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2) The Taylor method to evaluate Cv
We can use the Taylor method for calculating the coefficient of consolidation (Cv).
In the Taylor construction the deformation is plotted versus the square root of time, Fig. 5.
,o
Fig. 5 Taylor construction of Cv.
The straight portion of the curve is extended. Line A is then drawn at a distance of 15% "outside" that line (see Fig. 5).
u
= 90% is taken from the intersection between the curve and line A. s a t u=
50% and u=
100% can now be calculated and t 90 constructed.The coefficient preconsolidation Cv is evaluated from
2
C _ T90Hso (G)
V - t9 o
where the time factor T90 = 0.848.
The Cv-value is dependent on permeability which varies with temperature.
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STATEN$ GEOTEKNISKA INSTITUT 9
IIV. Direct shear test 1) Direct_shear_aEEaratus
Shear strength is often determined by direct shear tests.
The sample has a diameter of 50 mm and the sample height after consolidation should be between 10 mm in drained tests and 20 mm in undrained tests.
The sample is placed on a fixed pedestal ~50 mm which has a filter stone on top and an internal drainage channel.
If the compression during consolidation is expected to be small the sample is surrounded by a rubber membrane.
Thin metal rings are fitted outside the rubber membrane to keep the sample diameter constant during the test. The rubber membrane is sealed against the pedestal and the top part by clamps and drainage is provided by the filter stones and the drainage channels.
Consolidation stage Shearing stage
o) (<> o') (o o)---1,.-
Fig. 6 Direct shear apparatus
The test is performed by first applying the vertical load to the sample and allowing i t to consolidate.
After the consolidation the sample is sheared.
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The bolts fixing the top part are removed and the ball bearing which transmits the vertical load allows the top part to move horizontally with a tolerable amount of friction.
In shearing the top part is moved horizontally at a con
stant speed while the bottom pedestal is fixed. The sample thereby undergoes a fairly uniform angular distortion.
During the test the horizontal shear stress, the horizontal displacement of the top part and the height of the sample are measured by two indicators.
The vertical stress remains constant. Consolidation is usually allowed for 24 hours and the rate of shear is such that an angular distortion of the sample of 0.15 radians is obtained in another 24 hours.
2) Results_from_direct_shear_tests
Consolidated drained direct shear tests gave results following the standard patterns described by Larsson (1977). Some of the stress-strain curves are plotted in Fig. 7. Sincethe vertical deformation is measured during the shearing we
have to correct the height of sample when we are calculating the horizontal deformation.
j
Fig 7. Shear stress versus horiz deformation
(corresponding with the different vertical load).
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STATEN$ GEOTEKNISKA INSTITUT 1 1
Shear failure is obtained only at very low vertical stress.
In Swedish practice failure is evaluated at the peak or at an angular distortion of 0.15 radians if no peak is ob
tained.
The "shear strengths" thus evaluated are plotted in Fig. 8.
I
Fig. 8 Shear strength versus vertical stress in con
solidated drained direct shear tests.
V. Clay lime mixing test
During the last years a new method has been developed for deep stabilization of soft normally consolidated or slightly overconsolidated clays. The soft clay is in situ mixed with unslaked lime in columns. The lime causes an increase in the shear strength and a reduction of the compressibility of soft clays. The lime columns can be up to 10 m long and have a diameter of 500 mm.
The laboratory tests are required in order to study the influence of different parameters such as nature of soil, percentage of lime, temperature and the shear strength of
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STATENS GEOTEKNISKA INSTITUT 12
the stabilized clay (to calculate the bearing capacity of the lime columns).
The shear strength of the stabilized clay has normally been determined by fall-cone, vane or unconfined com
pression tests.
The shear strength of lime stabilized soils can according to laboratory tests be derived from the following relation
'T
=
e + o'tan cp- <
'Tmax where 'T=
shear strenghte
=
cohesioncp
=
angle of internal friction o'=
effective vertical stress'T = shear strength of the stronger aggregates max in a lime stabilized soil
The lime stabilized clay is not homogeneous even when the mixing of the lime with the clay is carefully performed.
Stronger aggregates are formed due to the chemical reac
tions. Because of these aggregates the measured shear strength will vary with the test method used and with the size of the sample.
The shear strength determined by the Swedish fall-cone test or laboratory vane test will normally be the shear strength of the aggregates ('T ) •
max
B. INTERPRETATION OF TEST RESULTS
1) Fall cone test results
We have worked wiht the fall cone test on clay from 5 m and the 13 m depth. The test results were written down in Table 4 and 5J6,
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13 STATENS GEOTEKNISKA INSTITUT
2) Incremental loading test results a) The_Casagrande_method
The test results of the Casagrande method were written down in Table 9y10 Diagram 1 and Diagram 2, 3 7 4.
- corresponding with the load step from 80 to 160 kPa the coefficient of consolidation is calculated as:
=
T H50 2=
0.197x0.942 2=
10 -8 2 ;50 tso 1200 1 x m s
where Ho
=
19.44 mm; Tso=
0.197E5 o
=
0.60 mmHo-Eso 19.44-0.60
Hso
=
2=
2=
0.942 cmtso
=
1200 seconds- corresponding with the load step from 160 to 320 kPa the coefficient of consolidation is calculated as:
__ Tso Hso 2 _ 0.197 x 0.8422 __ -8 2
C 3 x 10 m /s
V tso - 420
where Tso
=
0.197; Ho=
17.68 mmEso
=
0.85 mmHo-Eso 17.68-0.85
Hso
=
2=
2 0.842 cmtso = 420 seconds
From Diagram 2 we can determine the value of preconsolidation pressure.
0 1 = 106 kPa
C
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STATEN$ GEOTEKNISKA INSTITUT 14
b) The_Taylor_method
The test results of the Taylor method were written down in Table 41 ; 2 Diagram S, 6 and 7, B, 9 :J 10
- Corresponding with the load step from 80 to 160 kPa the coefficient of consolidation is calculated as:
=
T90xH5o2
=
0.848 X 0.9185 2 = 10-8 2/C V 900 8X m S
t90
where Ho = 20 mm 1.02 mm= 18.98 mm
€so= 0,61 mm
t 9 0 = 900
18.98-0.61
Hso= = 0.9185 cm
2
- Corresponding with the load step from 160 to 320 kPa the coefficient of consolidation is calculated as:
C = TgoXHso 2_ 0.848 X 0~822 =
V tgo - 900
where Ho = 20 - 2.90 = 17.10 mm sso = 0.667 mm
17.10-0.667
Hs o = = 0.822 cm
2
From diagram we can determine the value of preconsolidation pressure
0 1 = 109 kPa
C
3) The shear test results
The shear test results were written down in Table and !1-,1SJ6 Diagram M and f2.,
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STATEN$ GEOTEKNISKA INSTITUT 15
4) The clay lime test results
We have worked with the clay lime test as:
- Mixing and remoulding the soil of grey clay with thin s i l t layers between 3 and 8 m depth.
- Taken out the small sample to determine water content (Wn = 47%)
- From Diagram 12 we find out the value of unslaked lime (0.004 gcao/g wet soil)
- The weight of wet soil is 2913 g - 0,044 x 2913 = 128 g lime
= weight of wet soil= 2913
Weight of dry soil = 1968.2
1+ water content 1+0.47
= weight of lime 128
Lime content 0 0 6.5%
weight of dry soil '6
=
1968.2'6=
- To compact the mixture in plastic tubes with a special compaction equipment (five clay lime samples).
The compacted lime samples were put in the storage in a temperature of 70 c.
To determine the shear strength by unconfined compression test.
The test results were written down in Table 17, 18, 19, 21 and Diagram 10
1 I 6, 17 and HL
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1_6 - - - -.. ---- - - --
TABLE~
FALL
COI\JE TESTRESULTS
RUTIN TEST ( AT 5 m DEPTH)
2. Bulk density p
Height of sample
.A:f... .
cm2 p= ~
Area of sample=
... . :J.,, f
cm wWeight of sample= ..'S 4 0. g p
3. Natural water content Wn (soil . . . . . . . ... .. . . .. .. . .. ... .. . . .. ... . . . . . ... )
cup no ... .1... ....2.....
wet sample+ cup ..:?-.f.·..f.2 g ..2fi~.6~
dry sample + cup .4-S ..o.6 g •••• .••••• g
..46
-5°:1 g . . . g weight of water..9. . • . o .6 .
g .Jf0 •.1.4. gweight of cup .. 2...2.S g ..2. ..10. g
weight of dry sample wd =
.1.2 " · R1
g.14:
A-.-f gWW
natural water content Wn = Wa · 100
. .-+~ . . .
'f.. % .1.Q•.3.6 i4. Liquid limit WL (soil •.. . .•..
? . "- ft . .. . c1.c '1 .. ... . ... .. . . . . . . .. - ... )
fall cone t est
cup no
... . 1 .... .
....2. ....impression
60 g 60 ° wet sample+ cup ..i..-4.~
. u.
g .2.6...6.S"gdry sample+ cup .4 5;..01, g ...... .... g
.16
,..fi:1 . g . . . g we i ght of water Ww = ••9
o.O.6. g .1.0•.1!t. gweight of cup .2 ..35". g .2. to .. g
weight of dry sample Wd = .1. 2. .. .$J .1 g .14-~.+1. g
water content W = -Wd Ww • 100
.. 1-.o• . r.. ' . r:O. ..
f".3 iwL = M • w + N = .. a,.9.6_ II ••• .t.0.,.("3+.. fl•.C. .. WL = • •
6.8 , .3
%5 . Unconfined compression test Height of sample .•.. . . .. . ...... cm
1 fu ...... ... . kPa
l.Li··· ··· · ··· · · ·>
type of undisturbed sample remoulded sample
.. . 2. . 2: ..
kPacone
60g60° 100g30° 400g30° !Og60° 60g60° 100g30 o . .1 .2. '-. kPa
Sensitivity 1fu/1R .. :f8....
impression ... ..~~ .If: ..... .... ....
..1 . 1...
. .. ......mm .. .....6.~.'f. .....
.1 . o. .6.
..... ....1-,
0average 6.
t-
10. gRecommended reduction factor . . .. . .... .
Remarks ••••••• • • •• • •• •• •• •• •• • ••• . • • • . . . .• . . . • . . . . • . •.•. .• • •. • . . . • • • . . . • . . . . . . •.•. .. •..•• •• ... • •• ••. . ••..•..• . . .. .. . _ . .. ... .
Examination made I 19 . . . by
RUTIN TEST 7A-BLE S
1. Soil classification . . . .G:r.-e.'(j, ...$.0-£-l...
(>.f .... ;;,:1-t. .... C.f~... ; ... .
2. Bulk density p
Height of sample . ./.I{-:.... cm
P = _w_
Area of sample= .J.!J.~
r-:.
cm 2 A• HWeight of sample -~~.ff. .. g 0
3. Natural water content Wn (soil ....•.•.•.•.••...•... )
cup no .. .I. ... .
wet sample+ cup
1.i,.:rr;..
g . l/f.,3:l. gdry sample+ cup '/o.,.{C ... g / •• g iC.dJ.£. g .l:.,2 {.... g -- /' !1 t,
weight of water Ww i:;, .. • .7;. • • g ..f.'·,?;?;. . g
weight of cup ,ft,,_./t/. .. g
.~.-.1:...(.
gweight of dry sample .!.e
.'-f.5:.
gJf..,.
$. ?. . gnatural water content Wn
.~1...
%..-1.~ ... }?Jf:....
4. Liquid limit WL (soil . . • . • . . . • . • . • . . . ) fall cone test
cup no
... J. ...
...Limpression
60 g 60° wet
dry sample sample + + cup
l.o...
,S.Q
1.6. ..
..
g g
g
{
l.1.-.r3.1.(). ,.t>.i..
g • . . I. • . . • . i 2 ( gcup g
...
weight of water Ww = . ! .. .G'
9··"t: .. g. f.:. ~.3.
g... weight of cup ~
.:.:r.:-f. .
g . . . g 2 2(weight of dry sample
J. .'· . 't
$". g ..'if:..ff2. gWw 1
wutcr content W = Wd • 100 .i'l.:L ...
_g _,,
wL = M • w + N
.a/1.¥:.. . . J.r.:....
+a.,.J.o. ..
WL - •• .;:>•••••••5. Unconfined compression test Height of sample . . . cm
1fu . . . kPa
6. Fall cone test (soil .•.•.•••...••.••..•••....•.•.••••..••..•...•.••••••.. )
type of undisturbed sample remoulded sample
.l.r.1.S:-..
kPacone
60g60° 100g30° 400g30° 10g60° 60g6o 0 100g30° .(•. 15'"".. kPa Sensitivity 'fulTR f.
-P.-...2....
impression
... ..r.,.Q ..
.... ..../{.t!. ...mm ... . ]-::.lf:.. .... ...
.I.e.
:_J. ...8,o i2., O
average
?- 4t
//,3Recommended reduction factor . . .
Remarks .•..•....•..•....••..••..•.•.•...••..••...•••...•...••.•...•...
Examination made I 19 ••••. by
18
FA.II
cone:, lest
1ieSulf:aee-t-1 m ie;✓ bi,
RUTIN TEST
1. Soil classification ••••••..••..••.•••••..
~7:4
.t~.{. • • •i •_,(J_if.-t: •
.<:.(7· · ·2. Bulk density p
Height of sample
.. .tf.:...
cm wI.Cl
2 0 = -
Area of sample =
.1.y., r,.
cm ' A• HWeight of sample .5/fo... g 0 ./,. { . / . t/m 3
3. Natural water content Wn (soil ...•••...•.•.••.••.•..•...••... )
cup no ... :I...
wet sample+ cup
"<
~i.l.~ g 2.t.L(IJ.. gdry sample+ cup f.5:dJ.6 .. g
.i...
2.S-.. g1.6..
S:ff. g ...:t. .:. :1.~-. g weight of water Ww =.f/.Q. 6...
g1 tL.C!1..
g~ , t:-
weight of cup 2, •l .-:-.•.J. • g 4.,.:fP .. g
weight of dry sample
!~~8.1. ..
g I_ft... 9:."!:
gnatural water content Wn =
l:t ...
%_(::;J. ...
%..b ....
ice, '
4. Liquid limit WL (soil
····'ff'·~7J··
···~···)fall cone test
cup no
.... I. ....
impression
60 g 60° wet sample + cup .i!-1...!-?<.. g
4f..
f.o .. gdry sample + cup 1.!f'..,.t>.?... g
?:1.?-..S-..
g f{J., ..!f.r.. g ..~.-.!.~ .. g...
Id, OJweight of water Ww .;/...P.( .. g ••••.••••. g
··· weight of cup Z.,.t.ft. g ..fL.lo. .. g
weight of dry sample Wd J~.,/i:f .. g
(f,.frg
wu.ter content w = Ww Wa • 100 ff... .
C.5... .
wL = M • w + N =
o.,.:}.6... '1P...
+.o.l .... .
WL = ••f.? ...
5. Unconfined compression test
Height of sample . . . cm
Tfu . . . kPa
6. Fall cone test (soil •••
.A~C. i::f..4if t . r:J.~.... ...)
type of undisturbed sample remoulded sample 'fu ~~-·-· kPa
cone
60g60° 100g30° 400g30° 10g60° 60g60° 100g30° 'R /.:~£. kPa
Sensitivity TfulTR . ✓..f.: ....
impression ...
/i:.'f...
.... ... ... ...mm ...
.~:.r.. ..
··· ....1.1...
....:;-,
0 /o.6
average {.';:;-
lo
gRecommended reduction factor . . . .
Remarks
Examination made I 19 ••••• by
TABLE 7-
FALL CONE TEST
f<.E5ULT5RUTIN TEsT ( AT 11 m D E P "fH )
1. Soi l classification •••.. . .
· M · ..
/2.cr.i,l.•. )' •../2'.lo-:J. t • • - ~ · · · · · · · ·· · ··· · · ··· ·· ·· · · · ··· · · · ·· · · ·· · · ·· · ···
2. Bulk density p
Height of sample
...1 . 1-...
cm ! ,4-2.p w 1. 6 2..
Area of sampl e=
.A . fL . f
cm 2 A• H -HJ. ::,. ff
Weight of sample ..5:'11-.2.. CJ p .1.,.6.2. . t /m3
3. Natural water content Wn (soil . .. . . . . . • . . . . . . . ... )
cup no
.... 1.... .
... 2...wet sample+ cup ..4fj.,.G.B g .2. 0 ..4.-. g
dry sample+ cup ..t2.Jt.'f g ..•...••.. g .1. 2 ..'i.6. g . . . . • . . . g weight of water
..1:..
2.1. g ..1:, .5'. 3. . gweight of cup .. 2•.12.. CJ ..2 ..1.=l-. g
weight of dry sample Wa = .1.0 ..3S: g f0... .6,9. g
WW
natural water content Wn = wd • 100 .6.9.h.6 , .l-.0.~Lf:3. % .ffl-'·0. ~ %
4. Liquid limit WL (soil • . • . . . • . . . . ... . . . . • . . . . . . ) fall cone test
cup no ....1...
.... Z ....
i mpress ion
60 g 60° wet sample+ cup ..1.9.6. lJ. g .:2.o..~ . CJ
dry sample+ cup .1.2.. !l:1 g • •• • ••••.• g .f /4..K.6. g . . . g
weight of water
.. T. . .
2..1. CJ.- *~
~3. gweight of cup
. 2.. . .
12. g .2 ..11-; . gweight of dry sample wd f.Q •.3.5". . g 1.o...6j.. g
11 Ww
water content W = -Wd • 100
.b. 9 .. . 66 . , "'f.a ..
4-J. i. .o. .9.6.. • ..
lo. ....
+ . . a..f.o.
WL =. 6 . r. .!j ..
%5. Unconfined compression test Height of sample . . . . ..•...• . .. • cm
Tfu . . . kPa
6 . Fall cone test (soil . . . . . . ..
5 . o . -f.-t. ..
~.. ... ... ... )
type of undisturbed sample remoulded sample ' fu
.. . t5"... .
kPacone
60g60° 100g30° 400g30° 10g60° 60CJ60° 100g30° ' R .1... 2.. ... . kPa
Sensitivity ' ful 'R ..2..9. •.6.
impressi on
...
.. .0. ...... ... .1.0,.8. ....mm
.. ... .5 ,b a ..
.... ..1.1... ....'5.:2. -11 . 2
average 5.,~o
11
Recommended reduction factor .. . . .
Remarks ••••..•.....•... .•.... ........ . . • . . . ..... . . .• . . . ... . . • • • • • • • • •. • • • .. • • • • • • • • • • • • • • • • • • • • • •
Examination made I 19 •• .•. by