Examples With Solutions

CVEN3OO1O SYSTEMS

MODELLING AND DESIGN

GEOTECIINICAL COMPONENT

WORI(ED EXAMPLES

Lecturer Professor Ian Johnston Email: ianwilØunimelb.edu.au

PROBLEMS

l.

For the soil profile given in the figure below, plot the variation in vertical and horizontal total and effective stresses as well as the pore-water pressure between the ground surface and the upper surface of the rock. Assume the sand above the water table is dry. Assume that the unit weight of water is 9.81kN/m2. Ground surface SANDT¿,y: 16 kN/m' y""¡ 18 kNim'

y*:9.81kN/m' Ko:0.5

ROCK 2.

A soil profile

consists of 5m of medium dense sand, overlying a 7m thick layer of dense gravel, over a very deep formation of stiff clay. The ground water table is at 3m below the ground surface. For the soil properties shown in the figure, what is the distribution of total and effective, vertical and horizontal stresses and pore-water pressures to a depth of 20m below ground level. Take the unit weight of water as 9.81 tlm3.

T¡"= 17 kN/ml

y.,: Ko

19

:0

kN/m'

45

Yoo:20 kN/m'

T*:

22 kN/m'

Ko = 0'3

Tq: l8 kN/mr yo:20 kN/rnr Ko = 0.5

J.

A constant head permeability test was conducted on a specimen of sand. The specimen has a diameter of 50mm, a length of 100mm. The head across the test specimen was maintained at 200mm and I litre of water was collected in 7 minutes and 40 seconds. Estimate the coefficient of permeability of the sand specimen.

4.

A sandy clay specimen, 50mm in diameter and 100mm long, was placed in a falling head permeameter which had a standpipe with an internal diameter of 10mm. The initial head in the standpipe was set at 500mm and, in 24 hours, the head fell to 320mm. Estimate the permeability of the clay.

5.

The in-situ permeability of a very deep silty sand layer is to be determined at the base of a cased investigation borehole of 100mm internal diameter using a single packer. The borehole is 6.5m deep and the test section, at the base of the borehole, is 500mm long. The packer pipes are set at a level of lm above ground level and the pressure gauge is set at this level. The results obtained are provided below.

readins lkPal

Flow of water (litres/min)

0

2.56

20 40 60 80

3.r2

Pressure gauge

100

r20

4.03 4.76

s.48 7.76 10.98

For the ground water table located at a depth of 5m below ground level, estimate the permeability of the silty sand. Can you make any comment about what might also be occurring at the latter stages of the test? 6.

A layer of soil is 14m deep and is underlain by an impervious rock. A well is constructed to the rock and water is pumped out of the well at a rate of 5000 litres/day. Observation wells have been placed at rcdial distances of 5m and 10m to the pump well. If the ground water level was originally at the ground surface, and the water levels in the two observation wells were 4m and 2m respectively below the ground surface, estimate the permeability of the soil.

7.

A wide excavation in an isotropic sand is supported by a sheet pile wall as shown in the figure below. For a lm length of wall, estimate the amount of water which will seep into the excavation for the sand having a permeability of 1xl0-s m/sec. lnrperuioN shect pilc wall

Imperuious ræk

8.

If the excavation considered in Question 7 was only 9m wide and supported on both sides by a sheet pile wall as shown below, for a lm length of wall, what water seepage would enter the excavation for the sand having the same permeability of lx10-5 m/sec.

Lnpewiou shæt pile wall

ImperyioB rock

9.

An impavious concrete gravity dam of length 100m and width 30m is to be constructed on a uniform soil deposit of depth 30m which overlies a very thick impervious layer. The dam is to be provided with a 15m deep cut-off wall at its mid-point. If the water depth at the upstream face is 18m and the soil has a permeability of 2 x 10-6 m/sec, sketch the flow net and estimate a.

b.

The seepage quantity, and The uplift pressure distribution on the underside of the dam.

Impervious concrete dam

Impermeable layer

10. One side of a wide cofferdam is formed by a sheet pile

wall driven 10m into alayer of sand which is 20m deep and overlies an impervious rock. The sand has a coefficient of permeability of 1 x l0-3 m/sec If the level of water on one side of the cofferdam is 5m above the upper surface of the sand and on the other side it is at ground level, estimate

a. the flow rate of water into the inside of the cofferdam, b. the pore-water pressure at the point indicated which is 8m below the ground surface

c.

at the upstream side of the cofferdam, and the pore-water pressures that are acting horizontally on the sheet pile wall.

lm¡rrvlal 11.

reck

A confined artesian

sandy gravel aquifer is located at a site which consists ofa surface sand layer of 4m thickness, overlying a clay of 8m thickness which in turn overlies the very deep aquifer. If the ground water level in the surface sand is 2m below the ground surface and the piezometric level of the water in the artesian aquifer is lm above ground level, estimate a. The pore-water pressure at the base, the middle and the top of the clay layer, and b. The rate of seepage from the aquifer to the upper sand layer. +The permeability of the clay is 3 x 10-7 m/sec.

12.For the geotechnical conditions described in Question 11, it is proposed to construct a multistorey building with a 2level basement extending down into the clay layer. To achieve this, it is intended that sheet piles will be driven into the clay layer (with ground anchors for lateral stability) and the soil inside the sheet pile wall excavated to a depth of 7m below the ground surface to accommodate the double basement. The resulting excavation will be dewatered by means of pumping from a sump. What will be the rate of seepage into the excavation? Also, what safety concerns would you have about the stability of the excavation? Assume a unit weight for all soits of 20kN/m3. 13.

An earth dam is to be 100m wide and will have a cross-section as shown in the figure below. The foundation is effectively impervious and the water level on the upstream face is 35m. What would be the seepage loss through the dam if the permeability of the dam material were6xl0-7m/sec?

14.

For the dam described in Question 13, wh_at would be the seepage loss if it was only the horizontal permeability which was 6 x 10-7 m/sec with the vertical permeability 2.5 x 10t m/sec.

15. For the dam described in_Question 13, what would be the seepage loss all directions was 6 x 10-7 m./sec, but the toe drain was removed.

if the permeability in

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Question 10 The flow net for this problem is provided below. It can be used to provide all answers to the question.

InrpcrvÍous rock

a.

Water flow rate

For the flow net, there are 22 potentialdrops and 10 stream tubes. For

'224xl0-3x5

O=

k:

10-3 m/sec,

=2.27x10-3 m3/sec/m

For a 10 m length of wall, the total flow in an hour is given by

Q:

b.

2.27

x l0-3 x

10

x 60 x 60

:

82 m3.

The pore-water Eessure at the point at depth 8m

The water pressure at any point in a flow net is made up of a combination of static pressure and the changes brought about by seepage. For the point at a depth of 8m below the ground surface in the above figure, the head under static conditions would be 13m. However, seepage under a head of 5m is occurring and some of this head is lost by the time the water reaches the point under consideration. However, by its position 8m under the ground surface, the head would have increased by 8m. So the water pressure at the point is made up of the pressure head of 8m plus the original 5m head, less the loss in pressure due to seepage. Based on the flow net, there are 22 steps between the upstream and downstream seepage surfaces over which the seepage head of 5m is dissipated. The point 'üe are interested in is located on an equipotential line that is 5 drops from the upstream seepage surface. It follows that at this point, the 5 m head of water causing seepage will have dropped 5 of the 22 equal steps. Therefore, the piezometric head is 5 - (5122) x 5 *8: 5-1.14+8 1 1.86m of water

:

And the pore-water pressure is

(11.86)x 10:1l8.6kPa The same calculation can be carried out in terms of total head, elevation head and pressure head using the expression

H1:

Hp

* ]Iu

The elevation head is based on a selected datum which is usually taken

as the

exit point for the seepage.

The total head at the surface where the seepage starts (and He : 0) is 5m of water but by the time the water has seeped to a depth of 8m, it has passed through 5 of the 22 eqtipotentials. Therefore the total head is given by

Hr:

5-5(5122)

:

3.86m of

water:

38.6kPa

The elevation head is given by

H¡ Therefore,

as Hp

:

Hr

-

:

water:

-80kPa

HB,

Hp:

c.

- 8m of

38.6

-

C80)

kPa:

1l8.6kPa

Pore-water pressures on the wall

For the sheet pile wall, a close-up of the wall itself with the flow net as drawn is shown in the frgure below. The equipotential lines have been numbered 0 to 22 which corresponds to the number of equal steps that the 5m total seepage head has dissipated in its journey from the upstream to the downstream seepage surface. Each step represents a seepage pressure drop of (5122) metres of water or (5122) x 10kPa. The points where these equipotential lines meet the sheet pile wall have been given the descriptors a to w as shown in the figure.

The spreadsheet provided below, shows the calculations for each of the points a to w. Note that the depth of each of the points below the upstream (and downstream) seepage surface have been scaled from the figure

and entered into the spreadsheet. This has been done directly from the flow net and involves simply scaling the depth off the drawing. It follows that the flow net is critical in getting this dimension. It is not calculated, it is drawn by trial and error.

Point on

Depth (m)

Potential drops

Total (H1)

Elevation (H¿)

Pressure (Ho)

wall Head (m)

(kPa)

Head (m)

(kPa)

Head (m)

50.0 47.7 45.5 43.2 40.9 38.ó

0.0 -1.4 -3.1

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5.00

-t4

6.t7

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=5-0x(5122)): 5.00 =5-(Ix(5122\\: 4.77 :5-(2x(5/22)): 4.55 :5-ßx(5122)\= 4.32 =5-(4x(5122)): 4.09 =5-(5x(5122)): 3.86 =5-(6x(5122)): 3.64 =5-(7x(5122)): 3.41 :5-(8x(5122)): 3.18 :5-(9x(5122)): 2.9s :5-(l0x(5122\\: 2.73 =S-llx(5122)): 2.50 :5-( 2x(5122ù: 2.27 :5-( 3x(5122)): 2.05 =5-( 4x(5122\\: 1.82 =5-( 5x(5122)):1.59 =5-( 6x(5122\\: 1.36 :5-( 7x(5122\\: l.l4 =5-( 8x(5122)): 9.91 =5-( 9x(5122\\: 0.68 :5-(20x(5122)): 0.45

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It might

be worth noting that while the flow net drawn above is a reasonable flow net satisfying most conditions quite well, there does seem to be a minor error in the size of the squares close to the wall, which only really becomes more obvious when looking closely at the net. For example, the squares formed from a to b and c to d are smaller than the square formed from b to c. It would have been thought that they would be of the same size at this location, only becoming smaller as the toe of the wall is approached. As noted, avery minor error. The distribution of horizontal pressure on both sides of the sheet piles wall is shown in the figure below by the lines with the crosses (x). The long dashed lines are the pressure distributions if there were no seepage (i.e. if there was a vertical impervious barrier between the base of the sheet pile and the impervious rock at 20m depth).

It is of interest to note that the seepage case pressure distribution on the upstream side of the wall is less than the equivalent "no seepage" pressure distribution. This is because the 5m seepage head dissipates as flow moves to the base of the sheet pile. On the downstream side of the wall, the seepage case pressure distribution is greater than the "no seepage" pressure distribution. This is because with no seepage, the 5m seepage head is not present, but once seepage starts, it adds to the static head on the downstream side ofthe wall.

It should be noted that the pressure distributions on either side of the wall are not equal. This is something that has to be chocked when sheet pile walls .are designed so that there is an adequate factor of safety against the sheet pile failing laterally.

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