Assignment 2 -Solution

1

ASSIGNMENT No. 2 Due: 26th of October, 2012

1. The maximum dry unit weight determined in the laboratory for a quartz sand is 16.0

3m

kN. In the field, if the relative compaction is 90%, determine the hydraulic

conductivity of the sand in the field compaction condition (given that k for the sand at the maximum dry unit weight condition is 0.03 cm/sec and 7.2=sG ).

3)(max)( 4.14169.09.0 mkN

labdfieldd ===

sec1067.5

655.0655.1

839.1839.003.01

1

839.014.14

81.97.21

655.0116

81.97.21

23

3

31

1

2

32

12

2

1

cm

e

e

e

ekk

Ge

Ge

d

ws

d

ws

=

=

+

+=

=

=

=

=

=

=

Concordia University Department of Building, Civil & Environmental Engineering

CIVI 432: Soil Mechanics

Instructor: Adel Hanna Fall 2012

2

2. A constant head permeameter has an internal diameter of 62.5mm and is fitted in the side with three mamometer tapping at point A, B, and C. During tests on a specimen of sand the following data were recorded:

Test No. Quantity of water collected in 5min Manometer level (mm) above datum A B C 1 136.2ml 62 90 117 2 184.5 ml 84 122 164 3 309.4 ml 112 175 244

Length between tapping points: A-B= 120mm B-C= 125mm Determine the average of permeability of the soil (average of six values).

Test No.

Time (sec)

hAB (cm)

hBC (cm)

Q (cm3)

q=Q/T (cm3/sec)

D (cm)

Area (cm2)

LAB (cm)

LBC (cm) kAB=qLAB/AhAB kBC=qLBC/AhBC

1 300 2.8 2.7 136.2 0.454 6.25 30.664 12 12.5 0.063 0.069 2 300 3.8 4.2 184.5 0.615 6.25 30.664 12 12.5 0.063 0.060 3 300 6.3 6.9 309.4 1.031 6.25 30.664 12 12.5 0.064 0.061

Average: 0.064 0.063 k (cm/sec) 0.063335979

3

3. Draw the flow net pattern for the concrete dam shown in Fig. 1, in each of the following cases:

Case a: No sheet piles. Case b: One sheet pile 3.0 m deep measured from the bottom floor of the dam. The sheet pile is located 0.5 m from the upstream face of the dam. Case c: One sheet pile 3.0 m deep measured from the bottom floor of the dam. The sheet pile is located 0.5 m from the downstream face of the dam. Case d: Two sheet piles 3.0 m deep each, measured from the bottom floor of the dam. The sheet piles are located 0.5 m from the upstream face and 0.5 m from the downstream face of the dam.

a) Calculate the volume of seepage beneath the dam in each of the above cases. The coefficient of permeability of the soil below the dam is k = 1.5x10-5 m/sec.

b) Comment and describe the effect of the existence of the sheet pile and its location on the calculated seepage volume in each of the above cases.

c) How high would be the rise of water in piezometer tubes placed at points A, B, and C in each case of the above cases. Provide the solution in a tabulated form.

d) Calculate the uplift pressure on the base of the dam in each of the above cases. Tabulate the results and comment on the outcome in terms of the effect of the sheet pile on the uplift pressure.

Fig. 1

A

C

6.0

B

7.5

5.0

1.0

10.0

4.0

2.0

2.0

0.5

Impervious Stratum

U.S.W.L.

D.S.W.L.

Concrete Dam

Dims. in m0.5

Slope 1 : 3

1.0

Figure 1

4

Case a:

( )

mNHh

sm

NN

kHq

NN

d

d

f

d

f

778.097

106.391.35.05.7105.1

91.3

55

===

===

=

=

Point Height of water in piezometer tube (m) A 7.7778.015.8 = B 4.10778.045.13 = C 7.4778.04.75.10 = or 7.4778.06.15.3 =+

5

Point Dissipated head (m) Pressure Head (m) Uplift Force (U)

(kN/m2) 1 0 7.5 6.7381.95.7 = 2 778.01 = h 7.7778.015.8 = 5.7581.97.7 = 3 5.35.4 = h 5778.05.45.8 = 05.4981.95 = 4 224.68 = h 28.2778.085.8 = 5.7581.97.7 = 5 79 = h 0.5 9.481.95.0 =

6

Case b:

( )

mNHh

sm

NN

kHq

NN

d

d

f

d

f

538.0137

108.213

5.35.05.7105.1

135.3

55

===

===

=

=

Point Height of water in piezometer tube (m)

A 96.7538.015.8 = B 52.9538.04.75.13 = C 26.4538.06.115.10 =

7


(kN/m2) 1 0 7.5 6.7381.95.7 = 2 538.01 = h 962.7538.015.8 = 1.7881.9962.7 = 3 842.49 = h 658.3538.095.8 = 9.3581.9658.3 = 4 456.612 = h 044.2538.0125.8 = 1.2081.9044.2 = 5 713 = h 0.5 9.481.95.0 =

8

Case c:

( )

mNHh

sm

NN

kHq

NN

d

d

f

d

f

538.0137

108.213

5.35.05.7105.1

135.3

55

===

===

=

=


A 96.7538.015.8 = B 13.11538.04.45.13 = C 53.4538.01.115.10 =

9


(kN/m2) 1 0 7.5 6.7381.95.7 = 2 538.01 = h 962.7538.015.8 = 1.7881.9962.7 = 3 152.24 = h 348.6538.045.8 = 3.6281.9348.6 = 4 456.612 = h 044.2538.0125.8 = 1.2081.9044.2 = 5 713 = h 0.5 9.481.95.0 =

10

Case d:

( )

mNHh

sm

NN

kHq

NN

d

d

f

d

f

438.0167

103.216

5.35.05.7105.1

165.3

55

===

===

=

=


A 06.8438.015.8 = B 30.10438.03.75.13 = C 32.4438.01.145.10 =

11


(kN/m2) 1 0 7.5 6.7381.95.7 = 2 438.01 = h 062.8438.015.8 = 1.7981.9062.8 = 3 504.38 = h 996.4438.085.8 = 01.4981.9996.4 = 4 57.615 = h 93.1438.0155.8 = 93.1881.993.1 = 5 716 = h 0.5 9.481.95.0 =

12

4. The section through a long cofferdam is shown in Fig. 2. The coefficient of permeability of the soil is 4x10-7 m/sec. Draw the flow net of this system and determine the quantity of seepage entering zone of the cofferdam. Check the safety of the structure against piping. The saturated unit weight of the soil is 19 kN/m3.

3.0

5.0

5.0

2.0

Impervious Stratum

Water Surface

Dims. in m

Sheet Piles

9.0

G.L.

Figure 2

13

a)

daymm

mmmH

NN

kq

NN

mH

mk

d

f

d

f

3347

7

148.0sec

1071.115)7/6(sec

107.4

76

5sec

104

====

=

=

=

=

b) F.S against piping:

w

cr

cr

i

iiSF

'

.

max

=

=

The head loss per equipotential is:

mNm

d

71.07/55 ==

We are interested in the exit hydraulic gradient near the sheet pile (shortest path) The length of travel is approximately:

18.243.0

936.0.

936.08.9

8.919

43.0371.0

667.13/5

max

===

=

=

=

==

==

iiSF

i

i

mL

cr

w

wsatcr

exit

14

5. a) Calculate the values and pot the variation of total stress, effective stress, and pore

water pressure with depth for the soil profile shown in Figure. Consider stresses at points A, B, C, and D.

dry=14.25kN/m3sat=18.74kN/m3



A

B

C

D

Sand

Silty sand

Silty sand

Figure 1

3m

4m

5m

Rock

GWT

Figure 3

b) An isolated square foundation has been designed for carrying the 150kPa. The dimension of foundation is 2mx2m. Determine the vertical stress increase ( ) at points A, B, C, and D by implementing area method (2:1)

a) Stress at A:

0' === AAA u

Stress at B:

275.42325.14' mkN

zsanddryBB ==== Stress at C:

2

2

2

59.80'

24.394

8.119)27.19(474.4243

m

kNu

m

kNu

m

kN

CCC

wC

siltysanddrysanddryC

==

==

=+=+=

Stress at D:

15

2

2

22

94.130'

29.889

23.219543

m

kNu

m

kNu

m

kN

DDD

wD

siltysandsatsiltysandsatsanddryD

==

==

=++=

Plots:

16

b) 2 to 1 Method: Load: 150kPa

))((' ZLZBLBq++

=

B=Z (square foundation) Point Z B+Z q

A 0 2 150 150 B 3 5 150 24 C 7 9 150 7.41 D 13 15 150 2.66

17

6. Figure 4 shows the profile of a soil subjected to upward seepage flow. The reading of a piezometer tube inserted to point D is 14.5 m. The coefficient of permeability in the vertical direction of the sand, silty sand, and sandy silt layers are 1x10-4, 2x10-4, and 3x10-4 cm/sec. Determine the values of the effective stress at point A, B, C, and D, and draw the pattern of effective stress variation with depth.

d ry=1 4 .2 5k N/ m 3s a t=18 .7 4k N/ m 3

d ry=1 4 .9 6k N/ m 3s a t=19 .2 7k N/ m 3

d ry=1 5 .8 7k N/ m 3s a t=19 .8 8k N/ m 3

A

B

C

D

Sand

Silty sand

Silty sand

3m

4m

5m

Water level1m

Figure 4

18

Stress at A:

0'

81.9

81.9

3

31

=

=

==

A

A

wA

m

kNu

m

kNH

Stress at B:

2

22

1

21

70.21'

92.4281.9)312

5.131()(

62.6481.5481.9)3(

m

kNu

m

kNz

Hh

zHu

m

kNmH

BB

wB

sandsatwB

==

=++=++=

=+=+=

Stress at C:

21 7.14127.19462.6443 mkNH siltysandsatsandsatwC =+=++=

2

22

1

6.421.987.141'

1.9881.9)12

85.181()(

m

kNm

kNz

Hh

zHu

C

wC

==

=

++=++=

Stress at D:

2

22

1

2

26.10025.1425.241'

25.14281.9)12

125.1121()(

5.24188.19572.19474.18381.9

5431

m

kNm

kNz

Hh

zHu

m

kN

D

wD

sandysiltsatsiltysandsatsandsatwD

==

=

++=++=

=+++=

+++=

Plot of effective stress: Note that zero depth in the graph is the water surface.

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7. A 8 m-thick layer of stiff saturated clay is underlain by a layer of sand. The sand is under artesian pressure. Calculate the maximum depth of cut, H, that can be made in the clay.

Figure 5

Consider the stability of point A: ( )

wA

claysatA

u

H

=

=

6.38 )(

Heave is occurred when 0=A ( ) ( )

mHH

HHHu

A

wclaysatAAA

04.60187.1080

187.10881.96.31886.38 )(

=

==

====

21

8. Refer to Figure 6. The circular flexible area is uniformly loaded. Given: 2320m

kNq = .

Determine the vertical stress increase ( ) at point A.

Figure 6

R: Radius of the circle r: Distance from center of the circle z: Depth of the point

22

24

4

4

04.103320322.0

322.01

44

144

4,4,4

mkNqI

I

Rr

Rz

mrmRmzqI

===

=

==

==

===

=

23

9. Refer to Figure 7. The flexible area is uniformly loaded. Given: 2300m

kNq = .

Determine the vertical stress increase ( ) at point A.

Figure 7

24

22

32

1

2

23

21

321

67.42

135.1

112

300

3,5.1,300

1

1121

mkN

mzmRm

kNq

z

Rq

=

+

=

===

+

=

++=

2232

2

232

4.41138.0300

138.067.2

38

5.035.1

8,5.1

mkNIq

I

z

Ln

z

Bm

mLmBIq

====

=

===

===

==

==

2321 47.1254.414.4167.42 mkN

=++=++=

Assignment 2 -Solution

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