0132497468-Ch07_ISM.pdf

 Fluid Flow in Soils and Rock

Chapter 7

CHAPTER 7 FLUID FLOW IN SOILS AND ROCK -3

7-1.  A clean sand having a permeability of 4.5 x 10  cm/s and a void ratio of 0.45 is placed in a horizontal permeability apparatus, as shown in Fig. 7.2. Compute the discharge velocity and the seepage velocity as the head Δh goes from 0 to 80 cm. The cross-sectional area of the horizontal 2 pipe is 95 cm , and the soil sample is 0.65 m long. long . SOLUTION: Eq. 7. 7.5 : v = ki

⎛ 80cm ⎞ −3 cm s ⎟ = ( 4.5)(10) (1.231) = 0.0055 cm ⎝ 65 cm ⎠

disch arg e ve v elocity : v = ( 4.5)(10)−3 ⎜ v

= 0.55 m s

seepage velocity : vs vs

=

0.55 0.31

v

= ;

e

n=

n

1+ e

=

0.45 1 + 0.45

= 0.31

= 1.77 m s

7-2.  A sample of medium medium quartz sand is tested in a constant head head permeameter. The sample’s 3 diameter is 60 mm and its length is 130 mm. Under an app lied head of 60 cm, 119 cm  flows through the sample in 5 min. The M s of the sample is 410 g. Calculate (a) the Darcy Darcy coefficient of permeability, (b) the discharge velocity, and (c) the seepage velocity. (After A. Casagrande.) SOLUTION: (a) Eq. 7.9 : k =

QL hAt

=

(119 cm cm3 )(13.0 cm cm)( 4) (60 cm c m)π(6.0 cm c m) (5 mi m in)(60 2

s

min

)

cm s = 0.003 cm s = 3 × 10−3 cm

⎛ 60 cm ⎞ cm s ⎟ = 0.0138 cm ⎝ 13 cm ⎠

(b) discharge velocity: v = ki = (0.003 cm cm s) ⎜ (c ) seepage velocity: v s Vs Vt

=

ρs

=π

n= vs

Ms

=

410 g 2.70 g cm3

(6.0)2

v

= ; n

det er er mi m in e n, assume ρs

= 151.85 cm3

× 13 = 367.566 cm3 ;

4 Vv 215.716 = = 0.587 Vt 367.566

=

v n

=

0.0138 cm cm s 0.587

= 2.70 g cm3

Vv

= Vt − Vs = 367.566 − 151.85 = 215.716 cm3

= 0.024 cm s

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 Fluid Flow in Soils and Rock

Chapter 7

7-3.  A permeability test was run on a compacted compacted sample of dirty dirty sandy gravel. The sample was 175 mm long and the diameter of the mold 175 mm. In 90 s the discharge under a constant head 3 3 of 38 cm was 405 cm . The sample had a dry mass of 4950 g and its  ρ s was 2710 kg/m . Calculate (a) the coefficient of permeability, (b) the seepage velocity, and (c) the discharge velocity during the test. test. SOLUTION: (a) Eq. 7.9 : k =

QL

=

hAt

( 405 cm c m3 )(17.5 cm c m)( 4) (38 cm)π(17.5 cm)2 (90 s)

1 0−3 cm s = 0.00862 cm s = 8.62 × 10

⎛ 38 cm ⎞ cm s ⎟ = 0.0187 cm 17.5 cm ⎠ ⎝ 17.5

(c ) discharge velocity: v = ki = (0.00862 cm cm s) ⎜ (b) seepage seepage velocit velocity: y: v s

=

det er mine n, assume assume ρs Vs

=

Ms

ρs

=

4950 4950 g 3

2.71g cm

(17.5)2 Vt = π 4

n = 2.71 g cm3

= 1826.568 cm3

× 17.5 = 4209.243 cm3

cm3 = Vt − Vs = 4209.243 − 1826.568 = 2382.675 cm

Vv n= vs

v

Vv Vt

=

v n

=

2382.675 4209.243

=

0.0187 cm cm s

= 0.566

0.566

= 0.033 cm s

7-.4. During a falling-head permeability test, the head fell from 49 to 28 cm in 4.7 min. The specimen was 8 cm in diameter and had a length of 85 mm. The area of the standpipe was 0.45 2 cm . Compute the coefficient of permeability of the soil in cm/s, m/s. and ft/d. What was the probable classification of the soil tested? (After A. Casagrande.) SOLUTION: h aL Falling head test, use Eq. 7.10b: k = 2.3 log10 1  AΔt h2  A = π

(8)2

k = 2 .3

4

= 50.265 cm2 (0.45 cm2 )(8.5 cm) 2

(50. (50.26 265 5 cm )(4. )(4.7 7 min) min)(6 (60 0 s min) min)

k = 6.558 × 10

−7

log10

49 cm 28 cm

= 6.558 × 10−5 cm s

ms

⎛ 3.281 ftft ⎞ ⎛ 3600 s ⎞ ⎛ 24hr ⎞ ⎟ = 0.186 ft day ⎟⎜ ⎟⎜ ⎝ m ⎠ ⎝  hr   ⎠ ⎝ day ⎠

k = 6.558 × 10−7 m s ⎜

From Fig. 7.7, this is likely a fine sand and/or silt.

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 Fluid Flow in Soils and Rock

Chapter 7

7-5.  A falling-head permeability test is performed on a soil whose permeability is estimated to be -6 2.8 x 10  m/s. What diameter standpipe for head to drop from 31.2 cm to 19.4 cm in about 5 2 min? The cross section is 12 cm  and its length is 7.4 cm. (Taylor, 1948.) SOLUTION: h aL Falling head test, use Eq. 7.10b: k = 2.3 log10 1  AΔt h2  A

=π

(12)2 4

= 113.097 cm2

2.8(10)−4 cm s = 2.3

(a)(7.4 cm) 2

(113.097 cm )(5 min)(60 s min)

log10

31.2 cm 19.4 cm

= 6.558 × 10−5 cm s

a = 2.70 cm2

7-8. In Example 7.1, the void ratio is specified as 0.43. If the void ratio of the same soil were 0.35, evaluate its coefficient of permeability. First estimate in which direction k would go, higher or lower; then proceed. SOLUTION: If all other items remain the same, k will decrease if e decreases because the sample with the lower e is more dense.  Assume the test details are unchanged; i.e., 945 .7 g of water were collected in 1 min. The 2nd soil is more dense. Assume the diameter is fixed at 7.3 cm, and compute the new length. For the initial sample at e1 = 0.43, determine Vs .

= (41.9)(16.8) = 703.92 cm3 = Vv + Vs V 703.92 − Vs → Vs = 492.25 cm3 (assume Vs does not change) e1 = v = 0.43 = Vt

Vs

e2

= 0.35 =

the new Vt

Vs

Vv 492.25

= 172.288 + 492.25 = 664.538 cm3

the length, L = Eq.7.9: k

→ Vv = 172.288 cm3

=

Vt  A

QL hAt

= =

664.538 41.9

= 15.86 cm

(945.7 cm3 )(15.86 cm) 2

(75 cm)(41.9 cm )(60 s)

= 0.0795 cm s = 7.95 × 10−2 cm s

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 Fluid Flow in Soils and Rock

Chapter 7 2

7-9.  A falling-head permeability test on a specimen of fine sand 12.5 cm  in area and 10 cm long -4 gave a k of 6.2 x 10  cm/s. The dry mass of the sand specimen was 195 g and  ρ s was 2.71 3 Mg/m . The test temperature was 23°C. Compute the coefficient of permeability of the sand for a void ratio of 0.67 and the standard temperature of 20°C. (After A. Casagrande.) SOLUTION: Compute e1 Vs

=

Ms

ρs

195

=

2.71

= 71.956 cm3

= 12.5 cm2 Vt = (12.5)(10) = 125.0 cm3 = Vv + Vs → Vv = 125 − 71.956 = 53.044 cm3

 A

e1 = e2

Vv Vs

=

53.044 cm3 71.956 cm3

= 0.67 =

the new Vt

Vv − 2 Vs

= 0.737

→ Vv − 2 = (0.67)(71.956 cm3 ) = 48.21cm3

= 71.956 cm3 + 48.21cm3 = 120.17 cm3

the length, L =

Vt  A

=

120.17 cm3

= 9.61cm

12.5 cm2

h aL Falling head test, use Eq. 7.10b: k = 2.3 log10 1  AΔt h2 6.2(10)−4 cm s = 2.3 a

Δt

log10

h1 h2

(a)(10 cm) (12.5 cm )( Δt)

= 3.37(10)−4

2

log10

h1 h2

→ values on the left side do not change

for e2 , use L 2 to compute k at 23 deg C and at e2 k 23

= 3.37(10)−4 (2.3)(9.61)

k 20

= kT

1

= 0.67

= 5.958(10)−4 cm s at 23 deg C

12.5 viscosity of water at T

viscosity of water at 20 C

viscosity of water at 23 C = 0.00936 Poise viscosity of water at 20 C = 0.01005 Poise k 20

= ( 5.958(10)−4 )

0.00936 0.01005

= 5.55 × 10−4 cm s

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 Fluid Flow in Soils and Rock

Chapter 7

7-10.  A constant head permeability test is performed on a soil that is 2 cm x 2 cm square and 2.5 3 cm long. The head difference applied during the test is 18 cm, and 5 cm  is collected over a time of 100 sec. (a) Compute the permeability based on these test conditions and results. (b) A falling head test is to be done on the same soil specimen in the same time (t1 – t2 = 100 s) and the standpipe diameter is 0.8 cm. If the average head during the test should be 18 cm what are h1 and h2? SOLUTION: (a) Eq. 7.9 : k  A

=

QL

=

hAt

(405 cm3 )(17.5 cm)(4) (38 cm)π(17.5 cm) (90 s) 2

= 0.00862 cm s = 8.62 × 10−3 cm s

= 2 × 2 = 4 cm

2

k=

QL hAt

=

(5 cm3 )(2.5 cm) 2

(18 cm)(4 cm) (100 s)

= 0.001736 cm s = 1.74 × 10−3 cm s

aL h (b) Falling head test, use Eq. 7.10b: k = 2.3 log10 1  AΔt h2  A

= 4 cm2 ;

a=π

(0.8 cm)2

0.001736 cm s = 2.3 log10 h1 h2

h1 h2

= 0.503 cm2

4 (0.503)(2.5 cm) 2

(4cm )(100 s)

log10

h1 h2

= 0.240

= 1.2714

h1 = 1.2714h2 h1 + h2

= 18(2)

h1 = (1.2714)(36 − h1 ) h1 = 20.15 cm,

h2

= 15.85 cm

-4

7-11. The coefficient of permeability of a clean sand was 389 x 10  cm/s at a void ratio of 0.38. Estimate the permeability of this soil when the void ratio is 0.61. SOLUTION:  Assumptions : S = 100%, e1 = e2

Vv Vs

= 0.38 =

= 0.61 =

thus; Vt − 2

Vt

1000 − Vs Vs Vv 3

724.64 cm

= 1000 cm3 = Vv + Vs ,

= 389(10)−4 cm s

→ Vs = 724.64 cm3 (assume Vs does not change)

→ Vv = 442.03 cm3

= 442.03 + 724.64 = 1166.67 cm3

k is proportional to the volume of the sample, Vt k 0.61

k0.38

= 389(10)−4 ×

1166.67 1000

→ k0.61 = k0.38

Vt − 2 Vt −1

= 0.045 cm s

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 Fluid Flow in Soils and Rock

Chapter 7

7-12. Permeability tests on a soil supplied the following data: -4 e1 = 0.70, Temp1 = 25 C, k1 = (0.32)  cm/s -4 e2 = 1.10, Temp2 = 40 C, k1 = (1.80)  cm/s Estimate the coefficient of permeability at 20°C and a void ratio of 0.85. (After Taylor, 1948.) SOLUTION:  Assumptions : S = 100%, k 0.70

= kT

Vt −1 = 1000 cm3

= Vv + Vs ,

k0.70

= 0.32(10)−4 cm s

viscosity of water at T viscosity of water at 20 C

viscosity of water at 25 C = 0.00894 Poise viscosity of water at 20 C = 0.01005 Poise k 0.70

= ( 0.32(10)−4 )

e1 =

Vv

e2

Vs

= 0.70 =

= 0.85 =

thus; Vt − 2

0.00894

0.01005 1000 − Vs Vs Vv 3

588.235 cm

= 2.847 × 10−5 cm s → Vs = 588.235 cm3 (assume Vs does not change)

→ Vv = 500.0 cm3

= 588.235 + 500.0 = 1088.235 cm3

k is proportional to the volume of the sample, Vt k 0.85

= 2.847 × 10−5 ×

1088.235 1000

→ k0.85 = k0.70

Vt − 2 Vt −1

= 3.1× 10−5 cm s

7-13. For the initial case in Problem 6.33, compute the head of water required at the top o f the silty clay layer to cause a quick condition. SOLUTION:

σv = 195.71kPa, σ 'v = 146.66 kPa, u = 49.05 kPa  A quick condition occurs when σ 'v = 0 σ 'v = σv − u Thus, a quick condition occurs when: u = 195.71kPa 195.71 kPa = (9.81 m s2 )(1.0 M g m3 )hp hp = 19.95 m Initial conditions at 10 m:

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 Fluid Flow in Soils and Rock

Chapter 7

7-14. Sand is supported on a porous disc and screen in vertical cylinder, as shown in Fig. P7.14. These are equilibrium conditions. (a) For each of the five cases, plot the total, neutral, and effective stresses versus height. These plots should be approximately to scale. (b) Derive formulas for those three stresses in terms of the dimensions shown and e,  ρ s, and pw for each case at both the top and bottom of the sand layer. For case IV, assume the sand is 100% saturated to the upper surface by capillarity. For case V, assume the sand above level hc is completely dry and below is completely saturated. (After A. Casagrande.)

continu ed next page

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 Fluid Flow in Soils and Rock

Chapter 7

7-14 SOLUTION:

Case I

σtop = Dgρw

utop

= Dgρ w

σ 'top = 0

σbot = Lg ρsat + Dgρ w σ 'bot = Lgρ '

ubot

= (L + D)gρ w

= Dgρ w

σ 'top = 0

σ 'bot = ⎡⎣Lρ sat + Dρ w − Lρ w − Dρ w ⎤⎦ g = Lg(ρ sat − ρ w ) = Lgρ '

Case II

σtop = Dgρw

utop

σbot = Lgρsat + (L + D)gρ w σ 'bot = Lgρsat − hgρw

ubot

= (L + D + h)gρ w σ 'bot = Lgρ sat + ⎡⎣L + D − L − D − h⎤⎦ gρw = Lgρ sat − hgρw

Case III

σtop = Dgρ w σ 'top = 0 utop = Dgρ w ubot = (L + D − h)gρ w σbot = Lgρsat + (L + D)gρ w

σ 'bot = Lgρ sat + hgρ w

Case IV utop = 0 σtop = 0 σ 'top = 0 σbot = Lgρsat ubot = (L − h)gρ w

σ 'bot = Lgρsat − (L − h)gρ w

Case V utop = 0 σtop = 0 σ 'top = 0 ubot = (L − d − hc )gρ w σbot = (d)gρdry + (L − d)gρsat

σ 'bot = ⎡⎣dρdry + Lρsat − dρsat − Lρ w + dρ w + hcρ w ⎤⎦ g = ⎣⎡ d (ρ dry − ρ sat + ρ w ) + L (ρ sat − ρ w ) + hcρ w ⎦⎤ g

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 Fluid Flow in Soils and Rock

Chapter 7

7-15. Given the soil cylinder and test setup of Example 7.4, with actual dimensions as follows:  AB = 5 cm, BC = 10 cm, CD = 10 cm, and DE = 5 cm. Calculate the pressure, elevation, and total heads at points A through E in cm of water, and plot these values versus elevation. SOLUTION: Establish the datum at elevation corresponding to point E. Point

Elevation Head, z (cm)

Pressure Head, h p (cm)

Total Head, h (cm)

 A

30

0

30

B

25

5

30

C

15

0

15

D

5

-5

0

E

0

0

0

hydraulic gradient (B to D) = i = hC

Δh L

=

30 − 0 cm 20 cm

= 1.5

= hD + i × LD − C = 0 cm + (1.5)(10 cm) = 15.0 cm = hC − he − C = 15.0 cm − 15 cm = 0.0

hp − C

35

30

25    )   m   c    (   n   o    i    t   a   v   e    l    E

20

15

10 h-elev h-pressure

5

h-total 0 -10

-5

0

5

10

15

20

25

30

35

Pressure Head (cm)

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 Fluid Flow in Soils and Rock

Chapter 7

7-16. For each of the cases I, II, and III of Fig. P7.16, determine the pressure, elevation, and total head at the entering end, exit end, and point A of the sample. (After Taylor, 1948.) SOLUTION: For this solution, the datum is located at the tailwater elevation shown by the dashed line in the sketch. The exit end of the sample is the end closest to the smiley face symbol.

Elevation Head, z

Pressure Head, hp

Total Head, h

entering end

6

4

10

point A

3

-0.5

2.5

exit end

2

-2

0

entering end

-5

9

4

point A

-4

7

3

exit end

-1

1

0

entering end

-5

9

4

point A

-5

6

1

exit end

-5

5

0

CASE I

CASE II

CASE III

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 Fluid Flow in Soils and Rock

Chapter 7

7-17. For each of the cases shown in Fig. P7.16, determine the discharge velocity, the seepage velocity, and the seepage force per unit volume for (a) a permeability of 0.14 cm/s and a porosity of 46% and (b) a permeability of 0.0013 cm/s and a void ratio of 0.71. (After Taylor, 1948.) SOLUTION: (a) Case I Eq. (7.1): hydraulic gradient = i = i=

10 − 0 m 4m

= 2.5,

Δh L

,

Eq. (7.2): discharge velocity = v = ki

v = (0.14 cm s)(2.5) = 0.35 cm s

Eq. (7.8): seepage velocity = vs Eq. (7.17c): seepage force =

v

= ,

n j = iρ w g,

=

0.35 cm s

= 0.76 cm s 0.46 j = (2.5)(1.0 Mg m3 )(9.81m s2 ) = 24.52 kN m3

vs

(a) Case II i= vs

4−0m 4m

=

= 1.0,

0.14 cm s 0.46

v = (0.14 cm s)(1.0) = 0.14 cm s

= 0.30 cm s;

j = (1.0)(1.0 Mg m 3 )(9.81m s 2 ) = 9.81kN m3

(a) Case III i= vs

4−0m 4m

=

= 1.0,

0.14 cm s 0.46

v = (0.14 cm s)(1.0) = 0.14 cm s

= 0.30 cm s;

j = (1.0)(1.0 Mg m 3 )(9.81m s 2 ) = 9.81kN m3

(b) Case I i= n=

10 − 0 m 4m e

=

v = (0.0013 cm s)(2.5) = 3.25 × 10−3 cm s

= 2.5,

0.71

1 + e 1 + 0.71

=0.415,

vs

=

0.00325 cm s 0.415

= 7.83 × 10−3 cm s

 j = (2.5)(1.0 Mg m )(9.81m s ) = 24.52 kN m3 3

2

(b) Case II i= vs

4−0m 4m

=

= 1.0,

0.0013 cm s 0.415

v = (0.0013 cm s)(1.0) = 1.30 × 10−3 cm s

= 3.13 × 10−3 cm s;

j = (1.0)(1.0 Mg m3 )(9.81m s2 ) = 9.81kN m3

(b) Case III i= vs

4−0m 4m

=

= 1.0,

0.0013 cm s 0.415

v = (0.0013 cm s)(1.0) = 1.30 × 10−3 cm s

= 3.13 × 10−3 cm s;

j = (1.0)(1.0 Mg m3 )(9.81m s2 ) = 9.81kN m3

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 Fluid Flow in Soils and Rock

Chapter 7

7-18.  An inclined permeameter tube is filled with three layers of soil of different permeabilities as in Fig. P7.18. Express the head at points A, B, C, and D (with respect to the datum indicated) in terms of the different dimensions and permeabilities. (After A. Casagrande.)

SOLUTION: The problem cannot be readily solved without some additional assumptions. The following o solution is presented assuming the tube is inclined at an angle of 45  and the soil is centered in the inclined segment. For this solution, the following numeric values are implemented: H1 = 6 ft, -5 -5 -5 H2 = 2 ft, D = 2 ft, L = 0.5 ft, k 1 = 0.5 x 10  ft/s, k 2 = 1.5 x 10  ft/s, and k 3 = 0.75 x 10 ft/s. It is further assumed that the total length of the incl ine is 2.83 ft, measured along the top of the incline, from bend to bend. Determine how much Δh is lost in each soil section. Total head driving the system: h = 2 ft = Δh1 + Δh2 Velocity in each section is the same: v1 = v2 (0.5 × 10 −5 )

Δh1

= (1.5 × 10−5 )

0.5 Δh1 = 6Δh2 = 0.75Δh3

2 = Δh1 +

Δh2 0.25

+ Δh3

= v3 ;

= (0.75 × 10−5 )

vi

= ki

∑ Δh ; i

Li

k1

Δh1 L1

= k2

Δh2 L2

= k3

Δh3 L3

Δh3 1.0

1

Δh1 + 1.33Δh1 6 Thus, Δh1 = 0.8 ft, Δh2 = 0.13 ft,

Δh3 = 1.07 ft

Use the datum and trigonometry to determine h, hp ,and he . Point

h (ft)

h e (ft)

h p  (ft)

 A

6

0.38

5.62

B

5.2

0.73

4.47

C

5.07

0.91

4.16

D

4.0

1.62

2.38

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 Fluid Flow in Soils and Rock

Chapter 7 3

7-19.  Assume the soil of Fig. 7.10 has a saturated density of 1.89 Mg/m . If the head of water h above elevation B is 2.45 m, compute the effective stress at elevation A at the bottom of the soil sample during flow. What is the effective stress under these conditions at midheight in the soil column during steady-state flow? SOLUTION:

σ ' at Elev. A (bottom of soil sample) σ = ⎡⎣(5 m)(1.89 Mg m3 ) + (2 m)(1.0 Mg m3 ) ⎤⎦ (9.81m s2 ) = (11.45 Mg m2 )(9.81m s2 ) = 112.32 kPa u = (h + hw + L)gρw = (2.45 + 2 + 5 m)(9.81m s2 )(1.0 Mg m3 ) = (9.45 Mg m2 )(9.81m s2 ) = 92.70 kPa σ ' at Elev. A: σ ' = σ − u = 112.32 − 92.70 = 19.62 kPa σ ' at midheight of the soil sample Establish the datum at the bottom of the sample, Elev. A. Compute the total head at the bottom of the sample, h A Compute the total head at the top of the sample, htop The hydraulic gradient, i =

Δh L

=

9.45 − 2 m 5m

= 5 + 2 + 2.45 = 9.45 m

= 2m

= 1.49

Determine the water pressure at midheight of the sample, u M

= 9.45 m − (1.49)(2.5 m) = 5.725 m pressure head at midheight: hp −M = 5.725 m − 2.5 m = 3.225 m uM = gρw hp −M = (9.81)(1.0)(3.225) = 31.64 kPa total head at midheight: ht −M

σM = ⎡⎣(2.5 m)(1.89 Mg m3 ) + (2 m)(1.0 Mg m3 )⎤⎦ (9.81m s2 ) = (6.725 Mg m2 )(9.81m s2 ) = 65.97 kPa σ ' at midheight: σ ' = σ − u = 65.97 − 31.64 = 34.33 kPa

7-20. The foundation soil at the toe of a masonry dam has a porosity of 38% and a  ρ s of 2.73 3 Mg/m . To assure safety against piping, the specifications state that the upward gradient must not exceed 30% of the gradient at which a quick condition occurs. What is the maximum permissible upward gradient? (After Taylor, 1948.) SOLUTION: e=

n 1− n

=

0.38 1 − 0.38

= 0.613

ρs − ρ w 2.73 − 1.0 = = 1.072 Mg m3 1+ e 1 + 0.613 ρ − ρw 2.73 − 1.0 = = 1.072 Eq. 7.15: critical gradient = ic = s (1 + e)ρ w (1 + 0.613)(1.0)  Allowable exit gradient, iall = 0.30 × ic = (0.30)(1.072) = 0.32 Eq. 2.19:

ρ' =

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 Fluid Flow in Soils and Rock

Chapter 7

7-23.  A contractor plans to dig an excavation as show in Fig. P7.23. If the river is at level A, what is the factor of safety against quick conditions? Neglect any vertical shear. To what elevation can the water rise before a quick condition will develop? (After D. N. Humphrey.)

SOLUTION: Hsρg = ρw ghquick :

solve for hquick

(10 m )(2000 k g m3 )(9.81m s2 ) = (1000 k g m3 )(9.81m s2 )hquick hquick FS =

= 20 m hquick hactual

=

20 m 18 m

= 1.11

3

7-24. Given the excavation as shown in Example 7.13, with h = 18 m and  ρ  = 1915 kg/m . Calculate the minimum allowable Hs. SOLUTION: Hsρg = ρw ghquick :

solve for hquick

Hs (1915 k g m3 )(9.81m s2 ) = (1000 k g m3 )(9.81m s2 )(18 m) Hs

= 9.4 m for FS = 1.0

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 Fluid Flow in Soils and Rock

Chapter 7

7-25.  A sheet pile wall has been installed partially through a silty sand layer, similar to the one shown in Fig. 7.13(b). Assume a sheet pile 12 m long penetrates 6 m (halfway) into the silty sand layer of thickness 12 m. For this condition: (a) Draw a flow net using three (or four at most) flow channels. Note that the flow net is completely symmetrical about the bottom of the sheet pile. (This part is needed for the solution of Problem 7.35.) (b) If the water height on the upstream side is 5 m and on the downstream side 1 m, compute the amount of water flowing under the -4 sheet pile per meter of wall if the coefficient of permeability is 3.1 x 10  cm/s. (c) Compute the maximum hydraulic gradient at the downstream side of the sheet pile. SOLUTION: (a) The sketch below contains a ‘ 5-minute’ flow net that is symmetric about a vertical axis through the sheet pile.

(b) Eq. 7.20: q = khL

Nf  Nd

From the flow net: Nf

= 4, Nd = 4 × 2 = 8, hL = 5 − 1 = 4 m

q = (3.1× 10)−6 m s × (4 m) ×

(c) i =

ΔhL L

;

ΔhL =

hL Nd

=

4

3

= 6.20 × 10−6 m s per m of wall 8 4m 8

= 0.5 m

Scale the distance L from the flow net. L is the flow length in the first exit square immediately D/S of the sheet pile. L ≈ 2.375 m i=

ΔhL L

≈

0.5 m 2.375 m

≈ 0.21

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 Fluid Flow in Soils and Rock

Chapter 7

7-26. Using the data of Fig. 7.16, compute the total head, piezometric head, pressure head, and elevation head for points C and C’. Assume any convenient datum. SOLUTION:  Assume a datum at the elevation of the imp ervious boundary. Total head at point C: ht − C

= ht @U/S − Nd(U/ S− C) × ΔhL

where, ht @ U / S is the total head at the U/S reservoir, and Nd(U /S − C ) is the number of equipotential drops from the U/S reservoir to point C. Total head at point C: ht − C Total head at point C': ht − C'

= 42 − 5 ×

10.4 12

= 36.2 m above the datum (impervious boundary)

= 42 − 5 ×

Elevation head at point C: he − C

= 36.2 m above the datum (impervious boundary) 10.4 = 28 m (obtained directly from dam x-section)

Elevation head at point C': he − C ' pressure head, hp

12

= 0.0 m (obtained directly from dam x-section)

= ht − he

Pr essure head at point C: hp − C Pr essure head at point C': hp − C'

= 36.2 − 28 = 8.2 m = 36.2 − 0 = 36.2 m.

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 Fluid Flow in Soils and Rock

Chapter 7

7-27.  Assuming that you have completed the flow net of Problem 7.24, compute the total head, piezometric head, pressure head, and elevation head for a point halfway up the sheet pile from its base, on either side of the sheet pile. Assume the datum is at the bottom of the silty sand layer. Plot gradient versus depth of piling and extrapolate to find the exit gradient. SOLUTION:

Compute the heads at the midpoint of the embedded sheet pile section, on both sides. Point B in the sketch is on the right side and point C is on the left side of the sheet pile. PointB

= ht − A − Nd( A −B) × ΔhL is the total head at point A = 12 + 5 = 17 m, and

Total head at midpoint (B): ht −B where, ht − A

Nd( A−B) is the number of equipotential drops from the U/S reservoir to point B = 6.6 m.

= 17 − 6.6 ×

4

Elevation head at point B: he −B

= 13.7 m above the datum (bottom of silty sand) 8 = 9 m (obtained directly from x-section)

Pressure head at point B: hp −B

= ht −B − he −B = 13.7 − 9 = 4.7 m

Total head at point B: ht −B

Point C

= ht − A − Nd(A − C) × ΔhL is the total head at point A = 12 + 5 = 17 m, and

Total head at midpoint (C): ht − C where, ht − A

Nd( A−C) is the number of equipotential drops from the U/S reservoir to point C = 1.4 m.

= 17 − 1.4 ×

4

Elevation head at point C: he − C

= 16.3 m above the datum (bottom of silty sand) 8 = 9 m (obtained directly from x-section)

Pressure head at point C: hp − C

= ht − C − he− C = 16.3 − 9 = 7.3 m

Total head at point C: ht − C

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 Fluid Flow in Soils and Rock

Chapter 7

7-30. For the completed flow net of Fig. P7.30, compute the flow under the dam per meter of -4 dam if the coefficient of permeability is 4.2 x 10  cm/s.

SOLUTION:

Eq. 7.20: q = khL

Nf  Nd

From the flow net: Nf

= 3, Nd = 9.3, hL = 6.3 m

q = (4.2 × 10)−6 m s × (6.3 m) ×

3

3

= 8.53 × 10−6 m s per m of wall 9.3

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 Fluid Flow in Soils and Rock

Chapter 7

7-31. Given the data of Problem 7.25. Using the method of fragments, determine: (a) The amount of water flowing under the sheet pile per meter of wall. (b) The exit gradient. SOLUTION: Divide the flow regime into two fragments by drawing a vertical line at the sheet pile. Both fragments are type II. s = 6 m, T = 12 m kh

(a) q =

n

∑Φ

Φ1 = Φ 2 =

;

K

= f(m)

K'

m =1

For type II fragment: m

= sin

πs 2T

From Appendix C table, for m2 q=

kh

=

n

∑Φ

= sin

π(6 m) 2(12 m)

= 0.707,

= 0.5, Φ1 = Φ 2 =

(3.1× 10 −6 m s)(4 m) (1.0 + 1.0)

K K'

m2

= 0.5

= 1.000, and K = 1.854

= 6.20 × 10−6 m3 s per m of wall

m =1

(b) h2

Φ 2h

=

n

∑Φ

=

(1.000)(4 m) 2.000

= 2.0 m

m =1

iE

=

h2 π 2KTm

=

(2.0 m)π 2(1.854)(12 m)(0.707)

= 0.20

7-32. For the dam of Fig. 7.15, solve for q using the method of fragments. SOLUTION: Divide flow regime into 3 fragments by drawing vertical lines at U/S and D/S ends of the dam.  Assume all dimensions in Fig. 7.15 are in m eters. The dam embedment depth = 3 m. Distance from the impe rvious geo-boundary to the top of the foundation is 25 m, and the dam length, L = 58 m. fragment 1 = type II, fragment 2 = type I, fragment 3 = type II q=

kh n

∑Φ

;

Φ1 = Φ 2 =

K K'

= f(m)

m =1

Fragment 1 and 3 (Type II): s = 3 m, T = 25 m For type II fragment: m = sin

πs 2T

From Appendix C table, for m2

= sin

π(3 m) 2(25 m)

= 0.1874,

= 0.0351, Φ1 = Φ 3 =

K K'

m2

= 0.0351

= 0.514, and K = 1.585

Fragment 2 (Type I) :

Φ2 =

L a

=

58 m 25 − 3 m

= 2.636;

q=

kh n

∑Φ

=

k(26 − 7 m) 0.514 + 2.636 + 0.514

= (0.2729)hk = 5.18k

m =1

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 Fluid Flow in Soils and Rock

Chapter 7

7-37.  A protective three-layer filter is proposed between the foundation and rock drain located near the toe of a compacted earth-fill dam. Is this filter acceptable?

SOLUTION: Use Terzaghi's criteria for piping (Eq. 7.27) and permeability (Eq. 7.28) to evaluate the acceptability of the proposed filters. Use FS = 5.

< (4 to 5)D85 soil and Eq.7.28 (permeability): D15 filter > (4 to 5)D15 soil Foundation - Filter No. 1: D15 filter#1 = 0.3 mm (Eq.7.27): D15filter  < 5 × 0.1mm = 0.5 mm > 0.3 mm →∴ (OK) (Eq.7.28): D15filter  > 5 × 0.12 mm = 0.6 mm < 0.3 mm →∴ (No Good) Filter No. 1 - Filter No.  2 : D15 filter#2 = 2.0 mm (Eq.7.27): D15filter  < 5 × 1.0 mm = 5.0 mm > 2.0 mm →∴ (No Good) (Eq.7.28): D15filter  > 5 × 0.3 mm = 1.5 mm < 2.0 mm →∴ (OK) Filter No. 2 - Filter No. 3 : D15 filter#3 = 5.0 mm (Eq.7.27): D15filter  < 5 × 3.5 mm = 17.5 mm > 5.0 mm →∴ (OK) (Eq.7.28): D15filter  > 5 × 2.0 mm = 10.0 mm < 5.0 mm →∴ (No Good) Filter No. 3 - Rock Drain : D15 rockdrn = 15.0 mm (Eq.7.27): D15filter  < 5 × 10.0 mm = 50.0 mm > 15.0 mm →∴ (OK) (Eq.7.28): D15filter  > 5 × 5.0 mm = 25.0 mm < 15.0 mm → ∴ (No Good) Eq.7.27 (piping): D15 filter

CONCLUSION: The proposed filters do not meet Terzaghi's filter criteria for piping and permeability.

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 Fluid Flow in Soils and Rock

Chapter 7

7-38. In an attempt to reduce minor surface instability and maintenance problems on the backslopes of a rural highway, interceptor trench drains are to be installed at the top of the slope of to intercept surface and infiltrating groundwater from the hillsides above the road. The drains are 1 to 1.5 m deep, and the drainage trench lined with a geotextile filter. A perforated drain pipe is placed in the bottom of the trench, and the trench is backfilled with coarse drainage aggregate. Sieve analyses were performed on samples of soils typical of the problem areas along the highway alignment, and the following average data (percent passing) were obtained: Design the geotextile filter for the interceptor drains.

SOLUTION: Following are grain size distribution plots for the three soil samples. This information can be used to design the geotextile filter using the FHWA filter design procedure outlined in Section 7.10.4

Grain Size Distribu tion Plot 100

2"

1"

1/2"

#4

#10

#20

#40

#100

#200

Sample A

90

Sample B Sample C

80 70   g   n 60    i   s   s   a 50    P    %

40 30 20 10 0 100.00

2"

1"

1/2"

10.00

#4

#10

#20

#40

1.00

#100

#200

0.10

0.01

Grain Diameter (mm)

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