SOIL PHASE RELATIONSHIPS, INDEX PROPERTIES AND CLASSIFICATION
Case 4: When the soil is submerged
3.14 PLASTICIT Y CHAR T
Soil Phas e Relationships, Inde x Propertie s an d Soi l Classificatio n 59
Sensitivity, Sr = qu, undisturbed
q'u, remolde d (3.53)
where q'u is the unconfmed compressive strength of remolded cla y at the same water content as that of the undisturbed clay.
When q'u is very low as compared to qu the clay is highly sensitive. When qu = q'u the clay is said to be insensitive to remolding. On the basis of the values of St clays can be classified as in Table 3.14.
The clay s that have sensitivit y greater tha n 8 should be treated wit h care durin g constructio n operations because disturbance tends to transform them, at least temporarily, into viscous fluids. Such clays belong to the montmorillonite group and possess flocculent structure .
Thixotropy
If a remolde d cla y sampl e wit h sensitivity greate r tha n on e i s allowe d t o stan d withou t furthe r disturbance an d chang e i n wate r content , i t may regai n a t least par t o f its origina l strengt h an d stiffness. This increase in strength is due to the gradual reorientation o f the absorbed molecule s of water, and is known as thixotropy (fro m th e Greek thix, meaning 'touch' and tropein, meanin g 'to change'). Th e regainin g o f a par t o f th e strengt h afte r remoldin g ha s importan t application s i n connection wit h pile-driving operations , an d other type s of construction i n which disturbance of natural clay formations is inevitable.
corresponding liqui d limit wr The chart is divided into six regions, three above and three below lin e A, The equation t o the line A is
7p = 0.73 (W /- 20) (3.51 )
If a soil is known to be inorganic its group affiliatio n ca n be ascertained o n the basis of the values of/ an d wl alone. However, points representing organic clays are usually located withi n the same regio n a s those representing inorgani c silts of high compressibility, an d points representin g organic silt s in the region assigne d to inorganic silts of medium compressibility .
Casagrande (1932 ) studied the consistency limits of a number of soil and proposed the plasticity chart shown in Fig. 3.19. The distribution of soils accordin g to the regions are given below .
Region Above /4-line
21 3 Below /4-line
4 5 6
Liquid limi t wt
Less than 30 30 < W j < 5 0 w,>50 wl<30 30<w,< 50 w,>50
Type o f soi l
Inorganic clays of low plasticity and cohesionless soil s Inorganic clay s of medium plasticity
Inorganic clays of high plasticity Inorganic silts of low compressibility
Inorganic silts of medium compressibility and organic silts Inorganic silt s of high compressibility an d organic cla y The uppe r limi t of the relationship betwee n plasticit y index an d liquid limi t is provided b y another line called the [/-line whose equation is
I = 0.9(w-&) (3.52 )
Example 3. 9
Determine th e times (? ) required fo r particles o f diameters 0.2 , 0.02 , 0.0 1 an d 0.005 mm t o fall a depth of 1 0 cm from the surface in water.
Given: \JL = 8.15 x 10~3 poises, G = 2.65. (Note : 1 poise = 10~3 gm-sec/cm2) Solution
H = 8.15 x 10~3 x 10~3 = 8.15 x lO^6 gm-sec/cm2 . Use Eq. (3.24)
30// L 30X8.15X10"- x 6 1 0 1.48 2 x!0~ — - = - - -3 . m m (Gs-l) D 2 (2.65-1 ) D 2 D 2
The time s required fo r the various values of D are as given below.
D (mm) t 0.2 0.02 0.01 0.005
2.22 se c 3.71 mi n 14.82 mi n 59.28 min
Soil Phas e Relationships , Inde x Propertie s an d Soil Classification 6 1
Example 3.1 0
A sedimentatio n analysis by th e hydrometer metho d (15 2 H) was conducted wit h 50 g (= A/s)of oven dried soil. The volume of soil suspension is V = 103 cm3. The hydrometer reading Ra = 19.50 after a lapse of 60 minutes after the commencement of the test.
Given: Cm (meniscus) = 0.52, L (effective) = 14.0 cm, Co (zero correction) = +2.50, Gs = 2.70 and [i = 0.01 poise .
Calculate the smalles t particl e size , whic h woul d have settled a depth o f 14. 0 c m an d th e percentage finer than this size. Temperature of test = 25° C.
Solution
From Eq. (3.24) D(mm) =
where ^ = 0.01 x 10~3 (gm-sec)/cm2. Substituting
SOxO.OlxlO-3 1 4
£>- - XJ — = 0.0064 mm .
V (2.7-1 ) V6 0
From Eq. (3.31 )
From Table 3.6 for T= 25 °C, Cr= +1.3. Therefore , Rc =19.5-2.5 + 1.3=18.3
From Eqs (3.32) an d (3.31), w e have
CR 1.65G .
, C =
Ms sg 2.65(G-1 )
= 1.65X2. 7 p .%
sg 2.65(2.7-1 ) 5 0
Example 3.1 1
A 500 g sample o f dry soil was used fo r a combined siev e and hydrometer analysi s (15 2 H type hydrometer). Th e soi l mas s passin g throug h the 7 5 fi siev e = 12 0 g. Hydromete r analysi s wa s carried out on a mass of 40 g that passed through the 75 (Ji sieve. The average temperature recorde d during the test was 30°C.
Given: Gs = 2.55, Cm (meniscus) = 0.50, Co = +2.5, n = 8.15 x 10~3 poises.
The actual hydrometer reading Ra = 15.00 after a lapse of 12 0 min after the start of the test.
Determine the particle siz e D and percent fine r P'% and P%.
Solution
From Eq. (3.29 )
L =16.3-0.16417?
where, R = Ra + Cm = 15.0 + 0.5 = 15. 5 L = 16.3 - 0.1641 x 15.5 = 13.757 From Eq . (3.24 )
30x8.15xlO-6 13.75x.| =0.0047 ^ 03 mm 2.55-1
From Eq . (3.32 )
Percent finer , P'% = S8 cC R x 100 M
From Table 3.7 , C = 1.02 for Gs =2.55 From Table 3.6, Cr = +3.8 for T= 30 °C
Now, R c = Ra- Co + CT = 15 - 2.5 + 3.8 = 16. 3 Now, /"=L 0 2 x l 6 3x 100 = 41.57%
P% = 41.57 x
40
—500 Example 3.1 2
500 g of dry soil was used for a sieve analysis. The masses of soil retained on each sieve are given below:
US standard siev e 2.00 mm 1 .40 mm 1.00mm
Mass i n g 10 18 60
US standard sieve 500 fj.
250 j U 125/1 75 fji
Mass i n 135 145
56 45
g
Plot a grain size distribution curve and compute the following:
(a) Percentages o f gravel, coars e sand , medium sand, fine sand an d silt , as per th e Unified Soil Classification System, (b) uniformity coefficien t (c) coefficient of curvature.
Comment o n the type of soil.
Solution
Computation of percen t fine r
US stand- Diameter , D ard siev e o f grain s i n m m
2.00 m m 1 .40 mm 1.00mm 500/1 250 fj,
125/1 75 p.
2.00 1.40 1.00 0.500 0.25 0.125 0.075
Mass retained i n g
10 18 60 135 145 56 45
% retained
2.0 3.6 12.0 27.0 29.0 11.2 9.0
Cumulative
% retaine d 2.0 5.6 17.6 44.6 73.6 84.8 93.8
% finer P
98.0 94.4 82.4 55.4 26.4 15.2 6.2
Soil Phas e Relationships , Inde x Propertie s and Soil Classification 63
100 90 80 70 g 6 0
<5
<| 5 0
<u
£ 4 0
OH
30 20 10 n
Gravel Sand
Coarse to medium
«
\ \
60%
^\ V \
A*r
1
\
r\
309
£>3o
Fine
>,
N, A
10%
\
D ~** ^(
e
Silt + clay
108 6 4 2 1 .8 .6 . 4 0. 2 0.1.08.06.0 4 .0 2 Grain diameter, D in mm
Figure Ex. 3.12
(a) Percentage coars e t o medium sand = 98 - 4 8 = 50 percent Percentage fin e san d = 48 - 6. 2 = 41.8 percen t
Percentage sil t and clay = 6.2 percent.
(b) Uniformity coefficient C = ZXn
DIQ 0.09 8 (c) Coefficient of curvature C =
= 5.92 (0.28)2
i yxD6 0 0.098x0.5 8 The soil is just on the border line of well graded sand .
= 1.38
Example 3.1 3
Liquid limit tests on a given sample of clay were carried out. The data obtained are as given below.
Test No . 1
Water content, % 7 0 Number of blows, N 5
64 47
30
44 45
Draw the flow curve on semi-log pape r and determine the liquid limit and flow index of the soil.
Solution
Figure Ex. 3.1 3 give s the flow curv e for the given sample of clay soil . As per the curve, Liquid limit , \v{ = 50%
Flow index , /, = 29
70
60
S3 5 0 I
40
30
\
\
No\
2 4 6 81 0 202 5 4 0 608010 0 Number o f blow s
Figure Ex. 3.13
Example 3.14
The laborator y test s on a sample of soil gav e the following results:
wn - 24% , w, = 62%, wp = 28%, percentage of particles les s than 2 JJL - 23 % Determine: (a ) The liquidit y index , (b) activity (c ) consistency an d nature of soil . Solution
(a) Plasticit y index, Ip = wl- wp = 62 - 2 8 = 34%
wn -w 24-2 8
Liquidity index , 7 , = — p - = — = -0.12.
(b) Activity , A -
p 34
*P = 3 4
of particles < 2/u 2 3= 1.48.
(c) Comments :
(i) Since I : is negative, the consistency of the soil is very stiff to extremely stiff (semisolid state).
(ii) Since I is greater than 17% the soil is highly plastic.
(Hi) Since A is greater than 1.40, the soil is active and is subject to significant volume change (shrinkage and swelling).
Example 3.1 5
Two soil sample s teste d i n a soil mechanics laboratory gav e the followin g results:
Sample no . 1 Sampl e no. 2 Liquid limi t 50 %
Plastic limi t 30 % Flow indices , /, 2 7
40%
20%17
Soil Phas e Relationships, Inde x Propertie s and Soil Classificatio n 6 5 (a) Determine the toughness indices and
(b) comment on the types of soils.
Solution
w, - w
S~\ 7 _ ' P
Sample ,, ,, = Z . = = 0.74 ; Sampl e 2, /, = . = = 1.18
F ' 2 7 2 7 1 7 1 7
(b)(i) Bot h the soils are clay soils as their toughness indices li e between 0 and 3.
(ii) Soi l one is friable at the plastic limit since its It value is less than one.
(iii) Soi l two is stiffer than soil one at the plastic limit since the It value of the latter is higher.
Example 3.1 6
The natural moisture content of an excavated soil is 32%. Its liquid limit is 60% and plastic limit is 27%. Determine th e plasticity index of the soil and comment about the nature of the soil.
Solution
Plasticity index, I = \vt - wp = 60 - 2 7 = 33%
The nature of the soil can be judged by determining its liquidity index, /; from Eq . (3.45)
W»-W 32 "27
IP 3 3
since the value of It i s very close to 0, the nature of the soil as per Table 3.10 is very stiff . Example 3.17
A soil with a liquidity index of-0.20 has a liquid limit of 56% and a plasticity index of 20%. What is its natural water content? What is the nature of this soil?
Solution As per Eq. (3.45 )
Liquidity index,
'p
Wp = w{ -1 = 56 - 2 0 = 36,
wn = ltlp + wp=-0.20 x 20 + 36 = 32.
Since /, is negative, the soil is in a semisolid o r solid stat e as per Table 3.10.
Example 3.1 8
Four differen t type s o f soil s wer e encountere d i n a large project . Th e liqui d limit s (wz), plasti c limits ( w ) , and the natural moisture content s (wn) o f the soils ar e given below
I
Soil typ e 1 2 3 4
w,%
120 80 60 65
wp% 40 35 30 32
wn% 150
70 30 25
Determine: (a) the liquidity indices lt of the soils, (b) the consistency o f the natural soils and (c) the possible behavio r o f the soils under vibrating loads.
Solution
(a) / , = /
By substituting the appropriate values in this equation, we have Type
1 2 3 4
I, 1.375 0.778 0 -0.21
(b) Fro m Table 3.10 , Type 1 is in a liquid state, Type 2 in a very soft state , Type 3 in very stiff state , and Type 4 in a semisolid state .
(c) Soi l types 3 and 4 are not much affected by vibrating loads. Type 1 is very sensitive even for small disturbance and as such is not suitable for any foundation. Type 2 is also very soft , with greater settlement of the foundation or failure of the foundation due to development of pore pressure under saturated condition taking place due to any sudden application of loads.
Example 3.1 9
A shrinkag e limi t tes t o n a cla y soi l gav e th e followin g data . Comput e th e shrinkag e limit . Assuming tha t the total volume o f dry soil cake is equal t o its total volume a t the shrinkage limit , what is the degree o f shrinkage? Comment o n the nature of soil
Mass of shrinkage dish and saturated soil M , = 38.7 8 g Mass o f shrinkage dish and oven dry soil M 2 = 30.4 6 g
Mass o f shrinkage dish M 3 = 10.6 5 g
Volume of shrinkage dish V o - 16.2 9 cm3 Total volum e of oven dry soil cak e V d - 10.0 0 cm3 Solution
Refer to Fig. 3.1 5
The equation fo r shrinkage limi t ws = ——M
where Mw = mass of water in the voids at the shrinkage limit.
Mo = mass of sample at the plastic state = Ml -M3 = 38.78- 10.65 = 28.13 g
Soil Phas e Relationships , Inde x Propertie s and Soil Classificatio n 6 7 Volume of water lost from the plastic state to the shrinkage limit AV = (Vo - V d)
or AV = 16.29 - 10.0 0 = 6.29 cm3
Mass of dry soil = Ms = M2-M2 = 30.46 - 10.6 5 = 19.81 g
Now, Mw = Mo - Ms -(Vo-Vd)pw = 28.13 -19.81- (6.29) (1) = 2.03 g (M - M )-( V -V,)p M
From Eq. (3.41) , v v = — -2 - s-— ^ - ^^ = — ^ = -- = 0.102 = 10.2%
4 ' M ^ M s 19.8 1
As per Eq. (3.48a), the degree of shrinkage, Sr is
Sf = V^L x ,„„ = (16.29- 10.0) x 100 =
V0 16.2 9
From Table 3.11 the soil is of very poor quality.
3.15 GENERA L CONSIDERATION S FO R CLASSIFICATION O F SOILS