*5.9 CAPILLARITY
5.10 ILLUSTRATIVE EXAMPLES
Example 5.1. Determine the neutral and effective stress at a depth of 16 m below the ground level for the following conditions: Water table is 3 m below ground level ; G = 2.68; e = 0.72;
average water content of the soil above water table is 8%.
(S.V.U.—B. Tech., (Part-time)—April, 1982) The conditions are shown in Fig. 5.21:
GWT w = 8%
G = 2.68
e = 0.72 13 m13 m 3 m 3 m
Fig. 5.21 Soil profile (example 5.1) G = 2.68
e = 0.72
w = 8% for soil above water table.
γ = G w
e w
( )
(1 ) . 1
+
+ γ
= 2.68 × 108 172
.
. × 9.81 kN/m3 = 16.51 kN/m3.
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γsat = G e e w +
F
+HG I
KJ
1 . γ
= ( . . ) . 2 68 0 72
172
+ × 9.81 kN/m3 = 19.39 kN/m3
Total pressure at a depth of 16 m : σ = (3 × 16.51 + 13 × 19.39) = 301.6 kN/m2. Neutral pressure at this depth : u = 13 × 9.81 = 127.5 kN/m2
∴ Effective stress at 16 m below the ground level : σ = (σ – u) = (301.6 – 127.5) = 174.1 kN/m2.
49.53 3 m
16 m
264.16
264.16 127.5 174.1
37.44
(a) Total stress (b) Neutral pressure (c) Effective stress All pressures
are in kN/m2
49.53
Fig. 5.22 Pressure diagrams (Example 5.1)
Example 5.2. A saturated sand layer over a clay stratum is 5 m in depth. The water is 1.5 m below ground level. If the bulk density of saturated sand is 17.66 kN/m3, calculate the effective and neutral pressure on the top of the clay layer. (S.V.U.—B.E., (R.R.)—Nov., 1969)
The conditions given are shown in Fig. 5.23.
1.5 m Sand-dry (assumed) WT
3.5 m Sand saturated
Clay
1.5 m 1.5 m
3.5 m 3.5 m
63.10 63.10
17.64
18.93 18.93
46.40 34.34
All pressures are in kN/m2
(a) Total pressure (b) Neutral pressure (c) Effective pressure Fig. 5.23 Soil profile (Example 5.2) Fig. 5.24 Pressure diagrams
Let us, in the absence of data, assume that the sand above the water table is dry. Bulk density of saturated sand,
γsat = 17.66 kN/m3. γsat = G e
e w +
F
+HG I
KJ
1 γ
SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 149 Let us assume: G = 2.65
or 17.66 = ( . )
( ) ( . ) 2 65
1 + 9 81 +
e e
whence e = 1.06
γd = G e γw
( )
. ( . ) 1
2 65 1 106
+ =
+ × 9.81 kN/m3 = 12.62 kN/m3 Total stress at the top of clay layer:
σ = 1.5 × 12.62 + 3.5 × 17.66 = 80.74 kN/m2 Neutral stress at the top of clay layer:
u = 3.5 × 9.81 = 34.34 kN/m2 Effective stress at the top of clay layer:
σ = (σ – u) = 80.74 – 34.34 = 46.40 kN/m2.
Example 5.3. Compute the total, effective and pore pressure at a depth of 15 m below the bottom of a lake 6 m deep. The bottom of the lake consists of soft clay with a thickness of more than 15 m. The average water content of the clay is 40% and the specific gravity of soils may be
assumed to be 2.65. (S.V.U.—B.E., (R.R.)—April, 1966)
The conditions are shown in Fig. 5.25:
58.86 58.86
All pressures are in kN/m2
206 117.9 206 117.9
(a) Total pressure (b) Neutral pressure (c) Effective pressure Water
Lake bed
Saturated clay 15 m15 m 6 m 6 m
Fig. 5.25 Clay layer below lake bed (Example 5.3) Fig. 5.26 Pressure Diagrams (Example 5.3) Water content wsat = 40%
Specific gravity of solids, G = 2.65
Void ratio, e = wsat . G
= 0.4 × 2.65 = 1.06 γsat = G e
e w +
F
+HG I
KJ
1 γ
= ( . . ) ( . ) 2 65 106
1 106 +
+ × 9.81 kN/m3 = 17.67 kN/m3
Total stress at 15 m below the bottom of the lake:
σ = 6 × 9.81 + 15 × 17.67 = 323.9 kN/m2
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Neutral stress at 15 m below the bottom of the lake:
u = 21 × 9.81 kN/m3 = 206.0 kN/m2 Effective stress at 15 m below the bottom of the lake:
σ′ = 323.9 – 206.0 = 117.9 kN/m2 Also, σ = 15 × γ′ = 15 × (γsat – γw)
= 15(17.67 – 9.81) = 117.9 kN/m2 The pressure diagrams are shown in Fig. 5.26.
Example 5.4. A uniform soil deposit has a void ratio 0.6 and specific gravity of 2.65. The natural ground water is at 2.5 m below natural ground level. Due to capillary moisture, the average degree of saturation above ground water table is 50%. Determine the neutral pres-sure, total pressure and effective pressure at a depth of 6 m. Draw a neat sketch.
(S.V.U.—B. Tech., (Part-time)—April, 1982) The conditions are shown in Fig. 5.27:
45.22
115 115
Uncertain
Uncertain Uncertain 24.53 24.53
–
+ +
45.22
80.66 34.34
All pressures in kN/m2
(a) Total pressure (b) Neutral pressure (c) Effective pressure S = 50% due to
capillary moisture
Saturated soil 2.5 m 2.5 m
3.5 m 3.5 m
Fig. 5.27 Soil profile (Example 5.4) Fig. 5.28 Pressure diagrams (Example 5.4) Void ratio, e = 0.6
Specific gravity G = 2.65
γsat = G e e w +
F
+HG I
KJ
= ++1
2 65 0 60 1 0 60 . γ ( . . )
( . ) × 9.81 kN/m3 = 19.93 kN/m3
γ at 50% saturation
= G Se
e w
+
F
+HG I
KJ
= ×+ +1
2 65 0 5 0 60 1 0 60
. ( . . . )
( . )
γ × 9.81 kN/m3 = 18.09 kN/m3. Total pressure, σ at 6 m depth = 2.5 × 18.09 + 3.5 × 19.93
= 115 kN/m2
Neutral pressure, u at 6 m depth = 3.5 × 9.81 = 34.34 kN/m2 Effective pressure, σ at 6 m depth = (σ – u)
= 115.00 – 34.34 = 80.66 kN/m2 The pressure diagrams are shown in Fig. 5.28.
SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 151 It may be pointed out that the pore pressure in the zone of partial capillary saturation is difficult to predict and hence the effective pressure in this zone is also uncertain. It may be a little more than what is given here.
Example 5.5. Estimate the coefficient of permeability for a uniform sand where a sieve analy-sis indicates that the D10 size is 0.12 mm.
D10 = 0.12 mm = 0.012 cm.
According to Allen Hazen’s relationship, k = 100 D102
where k is permeability in cm/s and D10 is effective size in cm.
∴ k = 100 × (0.012)2 = 100 × 0.000144 = 0.0144 cm/s
∴ Permeability coefficient = 1.44 × 10–1 mm/s.
Example 5.6. Determine the coefficient of permeability from the following data:
Length of sand sample = 25 cm
Area of cross section of the sample = 30 cm2
Head of water = 40 cm
Discharge = 200 ml in 110 s. (S.V.U.—B. Tech., (Part-time)—June, 1981) L = 25 cm
A = 30 cm2
h = 40 cm (assumed constant) Q = 200 ml. t = 110 s
q = Q/t = 200/110 ml/s = 20/11 = 1.82 cm3/s i = h/L = 40/25 = 8/5 = 1.60
q = k . i . A k = q/iA = 20
11 16 30× . × cm/s = 0.03788 cm/s
= 3.788 × 10–1 mm/s.
Example 5.7. The discharge of water collected from a constant head permeameter in a period of 15 minutes is 500 ml. The internal diameter of the permeameter is 5 cm and the measured difference in head between two gauging points 15 cm vertically apart is 40 cm. Calculate the coefficient of permeability.
If the dry weight of the 15 cm long sample is 4.86 N and the specific gravity of the solids is 2.65, calculate the seepage velocity. (S.V.U.—B.E., (N.R.)—May, 1969)
Q = 500 ml ; t = 15 × 60 = 900 s.
A = (π/4) × 52 = 6.25π cm2 ; L = 15 cm ; h = 40 cm;
k = QL
At h = ×
× × ×
500 15
6 25. π 900 40cm/s = 0.106 mm/s Superficial velocity v = Q/At = 500
900 6 25× . πcm/s = 0.0283 cm/s
= 0.283 mm/s
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Dry weight of sample = 4.86 N
Volume of sample = A . L = 6.25 × π × 15 cm3 = 294.52 cm3 Dry density, γd = 4 86
294 52 .
. N/cm3 = 16.5 kN/m3 γd = G
e γw
(1 + ) (1 + e) = 2 65 10
16 5 .
.
× = 1.606, since γw ≈ 10 kN/m3 e = 0.606
n = e e
(1 + ) = 0.3773 = 37.73%
∴ Seepage velocity, vs = v/n = 0 283 0 3773
.
. = 0.750 mm/s.
Example 5.8. A glass cylinder 5 cm internal diameter and with a screen at the bottom was used as a falling head permeameter. The thickness of the sample was 10 cm. With the water level in the tube at the start of the test as 50 cm above the tail water, it dropped by 10 cm in one minute, the tail water level remaining unchanged. Calculate the value of k for the sample of the soil. Comment on the nature of the soil. (S.V.U.—B.E., (R.R.)—May, 1969)
Falling head permeability test:
h1 = 50 cm; h2 = 40 cm
t1 = 0; t2 = 60s ... t = t2 – t1 = 60s A = (π/4) × 52 = 6.25π cm2 ; L = 10 cm Since a is not given, let us assume a = A.
k = 2 303. aL. log ( / )10 1 2
At h h
= 2.303 × (10/60) log10 (50/40) cm/s
= 0.0372 cm/s
= 3.72 × 10–1 mm/s The soil may be coarse sand or fine graved.
Example 5.9. In a falling head permeability test, head causing flow was initially 50 cm and it drops 2 cm in 5 minutes. How much time required for the head to fall to 25 cm ?
(S.V.U.—B.E., (R.R.)—Feb., 1976) Falling head permeability test:
We know: k = 2 303. aL. log ( / )10 1 2
At h h
Designating 2 303. aL
At as a constant C k = C
.1t
. log10 (h1/h2) When h1 = 50 ; h2 = 48, t = 300 s
∴ k
C= 1
300log ( / )10 50 48
SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 153 When h1 = 50; h2 = 25; substituting:
1
300 log10 (50/48) = (1/t) log10 (50/25)
∴ t = 300 2
25 24
10 10
log
log ( / ) = 5093.55s = 84.9 min.
Example 5.10. A sample in a variable head permeameter is 8 cm in diameter and 10 cm high.
The permeability of the sample is estimated to be 10 × 10–4 cm/s. If it is desired that the head in the stand pipe should fall from 24 cm to 12 cm in 3 min., determine the size of the standpipe
which should be used. (S.V.U.—B.E., (R.R.)—Dec., 1970)
Variable head permeameter:
Soil sample diameter = 8 cm height (length) = 10 cm
Permeability (approx.) = 10 × 10–4 cm/s
h1 = 24 cm, h2 = 12 cm, t = 180 s Substituting in the equation
k = 2 303. aLlog ( / )10 1 2 At h h , 10–3 = 2 303 10
16 180 10 24 12 . × × log ( / )
× ×
a π
∴ a = π × ×
×
16 180
2 303 10. 4(log102)cm2 = 1.305 cm2 If the diameter of the standpipe is d cm
a = (π/4) d2
∴ d = 4 1 305× .
π cm = 1.29 cm
∴ The standpipe should be 13 mm in diameter.
Example 5.11. A horizontal stratified soil deposit consists of three layers each uniform in itself. The permeabilities of these layers are 8 × 10–4 cm/s, 52 × 10–4 cm/s, and 6 × 10–4 cm/s, and their thicknesses are 7, 3 and 10 m respectively. Find the effective average permeability of the deposit in the horizontal and vertical directions.
(S.V.U.—B. Tech., (Part-time)—April, 1982) The deposit is shown in Fig. 5.29:
First layer
Second layer
Third layer 10 m10 m 3 m 7 m 7 m
Fig. 5.29 Soil profile (Example 5.11)
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k1 = 8 × 10–4 cm/s h1 = 7 m k2 = 52 × 10–4 cm/s h2 = 3 m k3 = 6 × 10–4 cm/s h3 = 10 m kh (or kx) = ( )
( )
k h k h k h h h h
1 1 2 2 3 3
1 2 3
+ +
+ +
= (8 7 52 3 6 10) 20
× + × + ×
× 10–4 = 13.6 × 10–4 cm/s
∴ Effective average permeability in the horizontal direction = 13.6 × 10–3 mm/s
kv(or kz) = h h k
h k
h k
1 1
2 2
3 3
+ +
F HG I
KJ
= 1 20
10−4 [ /7 8 3 52 10 6+ / + / ] = 7.7 × 10–4 cm/s
∴ Effective average permeability in the vertical direction = 7.7 × 10–3 mm/s.
Example 5.12. An unconfined aquifer is known to be 32 m thick below the water table. A constant discharge of 2 cubic metres per minute is pumped out of the aquifer through a tubewell till the water level in the tubewell becomes steady. Two observation wells at distances of 15 m and 70 m from the tubewell show falls of 3 m and 0.7 m respectively from their static water levels. Find the permeability of the aquifer. (S.V.U.—B.Tech., (Part-time)—April, 1982)
The conditions given are shown in Fig. 5.30 :
q = 2 m /min3
Original WT
Drawdown curve 32 m
32 m
29 m 29 m 3 m 3 m
31.3 m 31.3 m
0.7 m
70 m 70 m 15 m
Fig. 5.30 Unconfined aquifer (Example 5.12)
SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 155 We know,
k = q r r z z log ( / )e
( )
2 1 22
12
π −
= 2 303 2 70 15 100
60 313 29
10 2 2
. log ( / )
( . )
× ×
×π − cm/s
= 1.18 × 10–1 mm/s.
Example 5.13. Determine the order of magnitude of the composite shape factor in the Poiseulle’s equation adapted for flow of water through uniform sands that have spherical grains and a void ratio of 0.9, basing this determination on Hazen’s approximate expression for permeabil-ity.
Poiseulle’s equation adapted for the flow of water through soil is:
k = D e
e C
s2 3
. . 1
( ). γ
µ +
with the usual notation, C being the composite shape factor.
By Allen Hazen’s relationship, k = 100 D102
D10 is the same as the diameter of grains, Ds, for uniform sands.
∴ k = 100 Ds2. Here Ds is in cm while k is in cm/s.
Substituting –
100 Ds2 = D e e C
s2 3
. . 1
( ). γ
µ +
C = 100 1 . .µ ( 3 )
γ +e
e ; here µ is in N-Sec/cm2 and γ is in N/cm3. C = 100 10 10 19
9 81 10 0 9
7 6
3 3
× × ×
× ×
− .
. ( . )
since µ = 10–3 N-sec/cm2 (at 20°C) and γ = 9.81 kN/m3 = 0.002657.
Example 5.14. A cohesionless soil has a permeability of 0.036 cm per second at a void ratio of 0.36. Make predictions of the permeability of this soil when at a void ratio of 0.45 according to the two functions of void ratio that are proposed.
k1 : k2 = e e
e e
13 1
23
1 1 2
( ):
( )
+ +
0.036 : k2 = ( . )
. :( . ) . 0 36
136
0 45 145
3 3
= 0.546 : 1
∴ k2 = 1
0 546. × 0.36 mm/s = 6.60 × 10–1 mm/s Also, k1 : k2 = e12 : e22
0.036 : k2 = (0.36)2 : (0.45)2 = 0.1296 : 0.2025
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∴ k2 = 0 2025
0 1296 .
. × 0.36 = 5.625 × 10–1 mm/s.
Example 5.15. Permeability tests on a soil sample gave the following data:
Test No. Void ratio Temperature °C Permeability in 10–3
mm/s
1 0.63 20 0.36
2 1.08 36 1.80
Estimate the coefficient of permeability at 27°C for a void ratio of 0.90.
Viscosity at 20 C = 1.009 10 N.sec/mm Viscosity at 36 C = 0 10 N.sec/mm Viscosity at 27 C = 0 10 N.sec/mm
Unit weight = 9.81 10 N/mm
2 2 2
6 3
° ×
° ×
° ×
U V|
W|
×−
−
−
− 9
9 9
706 855 . .
γ
According to Poisuelle’s equation adapted to flow through soil,
k = D e
e C
s2 3
. . 1
( ). γ
µ +
From test 1,
0 36 10 9 81 10 1 009 10
0 63 163
3 2
6 9
. 3
.
.
. .( . )
( . )
× = ×
×
− −
Ds C − = 1491.46
∴ Ds2 . C = 2.414 × 10–7 From test 2,
1 80 10 9 81 10 0 706 10
1 08 2 08
3 2
6 9
. 3
.
.
. .( . )
.
× = ×
×
− −
Ds C − = 8415.35
∴ Ds2 . C = 2.139 × 10–7 Average value of Ds2 . C = 2.2765 × 10–7
∴ At 27°C and a void ratio of 0.90, k = 2.2765 × 10–7 × 9 81 10
0 855 10
0 90 19
6 9
. 3
. .( . )
.
×
×
−
− = 1.002 × 10–3 mm/s.
Example 5.16. To what height would water rise in a glass capillary tube of 0.01 mm diam-eter ? What is the water pressure just under the meniscus in the capillary tube ?
Capillary rise, hc = 4T ds
γw c, where Ts is surface tension of water and γw is its unit weight and dc is the diameter of the capillary tube.
∴ hc = 4 73 10
9 81 10 0 01
6 6
× ×
× ×
−
. − . mm, assuming Ts = 73 × 10–6 N/mm and = 30/0.01 mm γw = 9.81 × 10–6 N/mm3.
= 3000 mm or 3 m.
Water pressure just under the meniscus in the tube
= 3000 × 9.81 × 10–6 N/mm2
SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 157
= 3 × 9.81 kN/m2
= 29.43 kN/m2.
Example 5.17. The D10 size of a soil is 0.01 mm. Assuming (1/5) D10 as the pore size, estimate the height of capillary rise assuming surface tension of water as 75 dynes/cm.
(S.V.U.—B.Tech., (Part-time)—April, 1982) The effective size D10 of the soil = 0.01 mm
Pore size, dc = (1/5)D10 = (1/5) × 0.01 mm = 0.002 mm.
Surface tension of water = 75 dynes/cm = 75 × 10–-6 N/mm.
Height of Capillary rise in the soil:
hc = 4 T d
s w c
γ
= 4 75 10 9 81 10 0 002
6 6
× ×
× ×
−
. − . mm, since γw = 9.81 × 10–6 N/mm3
= 4 75
9 81 0 002 1000
×
× ×
. . m
= 15.3 m.
Example 5.18. What is the height of capillary rise in a soil with an effective size of 0.06 mm and void ratio of 0.63 ?
Effective size = 0.06 mm Solid volume ∝ (0.06)3
∴ Void volume per unit of solid volume ∝ 0.63(0.06)3
Average void size, dc = (0.63)1/3 × 0.06 mm = 0.857 × 0.06 = 0.0514 mm Capillary rise, hc = 4T
d
s
γw c
= 4 73 10 9 81 10 0 0514
6 6
× ×
× ×
−
. − . mm
≈ 0.58 m.
Example 5.19. The effective sizes of two soils are 0.05 mm and 0.10 mm, the void ratio being the same for both. If the capillary rise in the first soil is 72 cm, what would be the capillary rise in the second soil ?
Effective size of first soil = 0.05 mm
∴ Solid volume ∝ (0.05)3
∴ Void volume ∝ e(0.05)3
Average pore size, dc = e1/3 × 0.05 mm Capillary rise of hc = 4T
ds γw c
∴ hc ∝ 1/dc
Since the void ratio is the same for the soils, average pore size for the second soil
= e1/3 × 0.10 mm.
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Substituting, hc = 4T ds
γw c = 36 cm,
since dc for the second soil is double that of the first soil and since hc ∝ 1 dc.
Example 5.20. The figure (Fig. 5.31) shows a tube of different diameter at different sections.
What is the height to which water will rise in this tube? If this tube is dipped in water and inverted, what is the height to which water will stand? What are the water pressures in the tube at points X, Y and Z ?
Let us denote the top level of each section above the water level as h and the height of capillary rise based on the size of the tube in that section, as hc.
Using hc (mm) = 30 dc(mm),
we can obtain the following as if that section is independently immersed in water :
dc (mm) hc(mm) h(mm) Remarks
2 15 10 Water enters the next section
1 30 20 Water enters the next section
0.75 40 35 Water enters the next section
0.50 60 70 Water does not enter the next section
15 mm 15 mm2 mm
f x y 1 mmf 10 mm
10 mm 40 mm
40 mm
60 mm 60 mm
15 mm 0.75
mmf 35 mm
5 mm 0.2 mmf Free water level 0.5
mmf z
Fig. 5.31 Tube with varying section (Example 5.20) Therefore water enters and stands at 60 mm above free water level.
If the tube is dipped in water and inverted,
hc = 30/dc = 30/0.2 = 150 mm.
Since this is greater than the height of the tube, it will be completely filled.
SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 159
Pressure at X = + hcγw = 15 9 81 100 . × .
= 0.147 kN/m2 = 147 N/m2 Pressure at Y = – hcγw = − ×1 9 81
100
. = – 0.098 kN/m2 = – 98 N/m2
Pressure at Z = – hcγw = − ×4 9 81 100
. = – 0.392 kN/m2 – 392 N/m2
Example 5.21. Sketch the variation in total stress, effective stress, and pore water pres-sure up to a depth of 6 m below ground level, given the following data. The water table is 2 m below ground level. The dry density of the soil is 17.66 kN/m3, water content is 12% ; specific gravity is 2.65. What would be the change in these stresses, if water-table drops by 1.0 m ?
(S.V.U.—B.Tech., (Part-time)—May, 1983) γd = 17.66 kN/m3
G = 2.65
w = 12% (assumed that it is at saturation) γd = G
e γw
(1 + ) 1.8 = 2 65
1 .
( +e) ∴ (1 + e) = 2 65 180
.
. = 1.472
∴ e = 0.472
γsat = 2 65 0 472 1472
. .
.
+ × 9.81 kN/m3 = 20.81 kN/m3 Total stress at 2 m below GL = 2 × 17.66 kN/m2 = 35.32 kN/m2
Total stress at 6 m below GL = (2 × 17.66 + 4 × 20.81) = 118.56 kN/m2 Neutral stress at 2 m below GL = zero.
Neutral stress at 6 m below GL = 4 × γw = 4 × 9.81 kN/m2 = 39.24 kN/m2 Effective stress at 2 m below GL, σ = σ – u = 35.32 – 0 = 35.32 kN/m2 Effective stress at 6 m below GL, σ = σ – u = 118.56 – 39.24 = 79.32 kN/m2
The variation in the total, neutral, and effective stresses with depth is shown in Fig. 5.32.
G-L
GWL 2 m 2 m
4 m 4 m
35.32 35.32
39.24 118.56
118.56 79.3279.32
All pressures are in kN/m2
(a) Initial conditions (b) Total stress (c) Neutral stress (d) Effective stress Fig. 5.32 Soil profile and pressure diagrams (Example 5.21)
Immediately after the water table is lowered by 1 m, the conditions are shown in Fig. 5.33.
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2 m 2 m
3 m
3 m Submerged Capillary conditions (assume 100% saturation)
Dry (assumed) Original WT
1 m 35.32
56.13
118.56 118.56
29.43 89.13
+ +
56.13 45.13
?9.81
– + ?
(a) Conditions after lowering the water table by 1 m( (b) Total stress (c) Neutral stress (d) Effective stress
Fig. 5.33 Conditions and pressure diagrams (Example 5.21) The top 2 m is assumed to be dry.
The next 1 m is under capillary conditions.
With suspended water is may be assumed to be 100% saturated.
The next 3 m is submerged Total stress:
σ at 2 m below GL = 2 × 17.66 kN/m2 = 35.32 kN/m2 σ at 3 m below GL = (2 × 17.66 + 1 × 20.81) = 56.13 kN/m2 σ at 6 m below GL = (2 × 17.66 + 4 × 20.81) = 118.56 kN/m2 Variation is linear.
Neutral stress:
u up to 2 m below GL is uncertain.
u at 2 m below GL is due to capillary meniscus.
It is given by – 1 × 9.81 kN/m2. u at 3 m below GL is zero.
u at 6 m below GL = + 3 × 9.81 kN/m3 = 29.43 kN/m2 Effective stress :
σ at 2 m below GL = 35.32 – (– 9.81) = 45.13 kN/m2 σ upto 2 m below GL is uncertain.
σ at 3 m below GL = 56.13 kN/m2.
σ at 6 m below GL = 118.56 – 29.43 = 89.13 kN/m2
The variation of total, neutral, and effective stresses is shown in Fig. 5.33.
The variation of the letter two from the surface up to 2 m depth is uncertain because the capillary conditions in this zone cannot easily be assessed.
SUMMARY OF MAIN POINTS
1. Soil moisture or water in soil occurs in several forms, the free water being the most important.
2. The total stress applied to a saturated soil mass will be shared by the pore water and the solid grains ; that which is borne by pore water is called the ‘neutral stress’ (computed as γw . h), and
SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 161 that which is borne through grain-to-grain contact is called the ‘effective stress’ ; the effective stress, obtained indirectly by subtracting the neutral stress from the total stress, is significant in the mobilisation of shear strength.
3. Permeability is the property of a porous medium such as soil, by virtue of which water or any other fluid can flow through the medium.
4. Darcy’s law (Q = k . i . A) is valid for the flow of water through most soils except in the case of very coarse gravelly ones. The macroscopic velocity obtained by dividing that total discharge by the total area of cross-section is called ‘superficial velocity’. In contrast to this, the microscopic veloc-ity obtained by considering the actual pore space available for flow is referred to as the ‘seepage velocity’.
5. Energy may be expressed in the form of three distinct energy heads, i.e., the pressure head, the elevation head, and the velocity head. The direct of flow is determined by the difference in total head between two points.
6. The constant head permeameter and the variable head permeameter are used in the laboratory for the determination of the coefficient of permeability of a soil.
Pumping tests are used in the field for the same purpose, using the principles of well hydraulics.
7. Permeant properties, such as viscosity and unit weight, and soil properties, such as grain-size, void ratio, degree of saturation, and presence of entrapped air, affect permeability.
8. The overall permeability of a layered deposit depends not only on the strata thicknesses and their permeabilities, but also on the direction of flow that is being considered. It can be shown that the permeability of such a deposit in the horizontal direction is always greater than that in the vertical direction.
9. The phenomenon of ‘capillary rise’ of moisture in soil has certain important effects such as satu-ration of soil even above the ground water table, desiccation of clay soils and increase in the effective stress in the capillary zone.
REFERENCES
1. Alam Singh and B.C. Punmia: Soil Mechanics and Foundations, Standard Book House, Delhi-6.
2. A.W. Bishop: The Measurement of Pore pressure in the Triaxial Test, Pore pressure and Suction in soils, Butterworths, London, 1961.
3. A.W. Bishop, I. Alpan, E.E. Blight and I.B. Donald: Factors controlling the strength of Partly Saturated Cohesive Soils, Proc. ASCE Research conference on shear strength of cohesive soils, Boulder, Colorado, USA, 1960.
4. H. Darcy: Les fontaines pulaliques de la ville de Dijon, Paris : Dijon, 1856.
5. J. Dupuit: Etudes théoretiques et pratiques sur la mouvement des eaux dans les canaux découvert et a travers les terrains perméables, 2nd edition, Paris, Dunod, 1863.
6. A Hazen: Some Physical Properties of Sand and Gravels with Special Reference to Their Use in Filtration, Massachusetts State Board of Health, 24th Annual Report, 1892.
7. A Hazen: Discussion of ‘Dams on Sand Foundations’, by A.C. Koenig, Transactions, ASCE, 1911.
8. IS : 2720 (Part XVII)—1986 : Methods of test for soils – Laboratory Determination of Permeability.
9. IS : 2720 (Part XXXVI)—1987 : Methods of test for soils—Laboratory Determination of Perme-ability of Granular Soils (constant head).
10. A.R. Jumikis: Soil Mechanics, D. Van Nostrand Co., Princeton, NJ, USA, 1962.
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11. J.S. Kozeny: Über Kapillare Leitung des wassers in Boden, Berlin Wein Akademie, 1927.
12. T.W. Lambe: The Measurement of Pore Water Pressures in Cohesionless Soils, Proc 2nd Internal Conference SMFE, Rotterdam, 1948.
13. T.W. Lambe: Soil Testing for Engineers, John Wiley and Sons, Inc., NY, USA, 1951.
14. T.W. Lambe and R.V. Whitman: Soil Mechanics, John Wiley and Sons, Inc., NY, USA, 1969.
15. A.G. Loudon: The Computation of Permeability from Simple Soil Tests, Geotechnique, 1952.
16. D.F. McCarthy: Essentials of Soil Mechanics and Foundations, Reston Publishing Co., Reston, VA, USA, 1977.
17. A.S. Michaels and C.S. Lin: The Permeability of Kaolinite—Industrial and Engineering Chemis-try, 1952.
18. M. Muskat: The Flow of Homogeneous Fluids through Porous Media, McGraw-Hill Book Co., New York, USA, 1937.
19. M. Muskat: The Flow of Homogeneous Fluids Through Porous Media, J.W. Edwards, 1946.
20. A.E. Scheidegger: The Physics of Flow Through Porous Media, The MacMillan Co., New York, USA, 1957.
21. S.B. Sehgal: A Testbook of Soil Mechanics, Metropolitan Book Co. Pvt. Ltd., Delhi-6, 1967.
22. G.N. Smith: Elements of Soil Mechanics for Civil and Mining Engineers, 3rd edition, Metric, Crosby Lockwood Staples, London, 1974.
23. M.G. Spangler: Soil Engineering, International Test Book Company, Scranton, USA, 1951.
24. D.W. Taylor: Fundamentals of Soil Mechanics, John Wiley and Sons, Inc., New York, USA, 1948.
25. K. Terzaghi and R.B. Peck: Soil Mechanics in Engineering Practice, John Wiley and Sons, Inc., 1948.
26. A. Thiem: Über die Ergiebig Keit artesicher Bohrlocher, Schachtbrunnen und Filtergalerien, Jour-nal für Gasbeleuchtung und Wasseracersorgung, 1870.
27. R.V. Whitman, A.M. Richardson, and K.A. Healy: Time-lags in Pore pressure Measurements, 5th International Conference SMFE, Paris, 1961.
QUESTIONS AND PROBLEMS
5.1 Define ‘neutral’ and ‘effective’ pressure in soils. (S.V.U.—B.E., (R.R.)—Nov., 1969) 5.2 Write short notes on ‘neutral’ and ‘effective’ pressure. What is the role of effective stress in soil
mechanics ? (S.V.U.—B.E., (R.R.)—Dec., 1970)
(S.V.U.—B.E., (R.R.)—Nov., 1975) 5.3 A uniform homogeneous sand deposit of specific gravity 2.60 and void ratio 0.65 extends to a large depth. The ground water table is 2 m from G.L. Determine the effective, neutral, and total stress at depths of 2 m and 6 m. Assume that the soil from 1 m to 2 m has capillary moisture leading to degree of saturation of 60%. (S.V.U.—B.Tech., (Part-time)—Sep., 1962) 5.4 Describe clearly with a neat sketch how you will determine the coefficient of permeability of a clay sample in the laboratory and derive the expression used to compute the permeability coeffi-cient. Mention the various precautions, you suggest, to improve the reliability of the test results.
(S.V.U.—B.Tech., (Part-time)—May, 1983) 5.5 Define ‘permeability’ and explain how would you determine it in the field.
(S.V.U.—B.Tech., (Part-time)—May, 1983)
SOIL MOISTURE–PERMEABILITY AND CAPILLARITY 163 5.6 What are the various parameters that effect the permeability of soil in the field ? Critically
discuss. (S.V.U.—B.Tech., (Part-time)—April, 1982)
5.7 What are the various factors that affect the permeability of a soil stratum ? If k1, k2, k3 are the permeabilities of layers h1, h2, h3 thick, what is its equivalent permeability in the horizontal and vertical directions ? Derive the formulae used. (S.V.U.—Four-year B.Tech.,—June, 1982) 5.8 Differentiate between ‘Constant Head type’ and ‘Variable Head type’ permeameters. Are both of them required in the laboratory ? If so, why ? Derive the expression for the coefficient of perme-ability as obtained from the variable head permeameter.
(S.V.U.—B.Tech., (Part-Time)—June, 1981) 5.9 Estimate the coefficient of permeability for a uniform sand with D10 = 0.18 mm.
5.10 Explain the significance of permeability of soils. What is Darcy’s law ? Explain how the perme-ability of a soil is affected by various factors. (S.V.U.—B.R., (R.R.)—Feb., 1976) 5.11 Distinguish between superficial velocity and seepage velocity. Describe briefly how they are determined for sand and clay in the laboratory. (S.V.U.—B.E., (R.R.)—Nov., 1974) 5.12 List the factors that affect the permeability of a given soil. State the precautions that should be
taken so that satisfactory permeability data may be obtained.
Explaining the test details of a falling head permeameter, derive the formula used in the compu-tation. Also evaluate the types of soil material for which falling head and variable head permeameters are used in the laboratory. Compare the relative merits and demerits of labora-tory and field methods of determining the coefficient of permeability.
(S.V.U.—B.E., (R.R.)—Nov., 1973) 5.13 Calculate the coefficient of permeability of a soil sample, 6 cm in height and 50 cm2 in cross-sectional area, if a quantity of water equal to 430 ml passed down in 10 minutes, under an effective constant head of 40 cm. (S.V.U.—B.E., (R.R.)—Dec., 1971) 5.14 Define coefficient of permeability and list four factors on which the permeability depends.
A falling head permeability test is to be performed on a soil sample whose permeability is esti-mated to be about 3 × 10–5 cm/s. What diameter of the standpipe should be used if the head is to drop from 27.5 cm to 20.0 cm in 5 minutes and if the cross-sectional area and length of the sample are respectively 15 cm2 and 8.5 cm ? Will it take the same time for the head to drop from
37.7 cm to 30.0 cm ? (S.V.U.—B.E., (R.R.)—Nov., 1973)
5.15 (a) What are the conditions necessary for Darcy’s law to be applicable for flow of water through soil ?
(b) Why is the permeability of a clay soil with flocculated structure greater than that for it in the remoulded state ?
5.16 (a) State the principle of Darcy’s law for laminar flow of water through saturated soil.
(b) Demonstrate that the coefficient of permeability has the dimension of velocity.
(c) The discharge of water collected from a constant head permeameter in a period of 15 minutes is 400 ml. The internal diameter of the permeameter is 6 cm and the measured difference in heads between the two gauging points 15 cm apart is 40 cm. Calculate the coefficient of permeability and comment on the type of soil. (S.V.U.—B.E., (N.R.)—Sept., 1967) 5.17 A glass cylinder 5 cm internal diameter with a screen at the bottom is used as a falling head permeameter. The thickness of the sample is 10 cm. The water level in the tube at the start of the test was 40 cm above tail water level and it dropped by 10 cm in one minute while the level of tail water remained unchanged. Determine the value of the coefficient of permeability.
(S.V.U.—B.E., (N.R.)—Sep., 1968)