NATIONAL UNIVERSITY OF SINGAPORE

DEPARTMENT OF CIVIL ENGINEERING

CE2112 SOLUTIONS FOR CONSOLIDATION II

1. An oil tank, 10 m high and 50 m diameter, is to be constructed on a site where 1 m of gravel

overlies 4 m of soft clay. Below the clay is dense sand. In the initial filling and proof-loading

of the tank, water is to be used. Filling will progress in weekly increments, and the

increments must be of such a magnitude to ensure that the mean excess pore water pressures

in the soft clay does not exceed 60 kPa. Superposition may be assumed valid. Given that cv =

52 m2/year, calculate the sequence of weekly water levels for the test loading. What factors

could limit the validity of the analysis?

Solution:

Soft clay layer has 2-way drainage, drainage path H = 2m.

Cv = 52m2/year = 1m2/week

Assuming unit weight of water w = 10kN/m3

Oil tank

Permeable Gravel

Soft Clay

Permeable dense sand

Cv = 52m2/yr 4m

1m

Since allowable excess pore pressure = 60kPa, initial allowable loading = 60kPa => 6m of water.

After the 1st week, T = 2H

tcv = 22

1*1= 0.25 => T = 0.5

Initial excess pore pressure profile is rectangular. Hence use central curve,

U = 0.56 or 56% => 1 - U = 0.44 or 44%

i.e. 44% of the initial excess pore pressure still remains, on average.

=> uave = 0.44 * 60 = 26.4kPa

Thus, on average, we can add another 60 26.4 = 33.6kPa 3.36m of water.

Just after the 2nd loading of 33.6kPa, we expect the resultant excess pore pressure profile to look

something like this:

33.6kPa

Initial

excess pore

pressure

Excess pore

pressure

just before

2nd

loading

Excess pore

pressure

just before

2nd

loading

Excess pore

pressure

due to 2nd

loading

60kPa

A B

After the 2nd week: There is no solution for the initial excess pore pressure shown in B above.

Hence, we need to split the excess pore pressure up into 2 portion, one arising from the initial

loading of 60kPa (i.e. the area under the curved portion) and the other arising from the 2nd

loading (after 1st week) (i.e. the area under the rectangular portion).

Excess pore pressure from the 1st loading: T = 2H

tcv = 22

2*1= 0.5 => T = 0.7071

=> U = 0.76 or 76% => 1 - U = 0.24 or 24%

i.e. 24% of the excess pore pressure from the first loading still remains.

uaveA = 0.24 * 60kPa = 14.4kPa

Excess pore pressure from the 2nd loading: T = 2H

tcv = 22

1*1= 0.25 => T = 0.5

U = 0.56 or 56% => 1 - U = 0.44 or 44%

=> uaveB = 0.44 * 33.6 = 14.8kPa

Total average excess pore pressure still remaining = 14.4 + 14.8kPa = 29.2kPa

Hence, we can add another 60 29.2 = 30.8kPa or 3.08m of water => up to 10m.

2. The diagram below shows a 6 m deep excavation which is to be made in a 20 m thick layer of

residual soil overlying sand which in turn overlies bedrock. The permeabilities are (i) residual

soil: 10-8 m/s and (ii) sand: 10-4 m/s. The compression index Cc for the residual soil is 0.05

and its unit weight is 19 kN/m3. The excavation is to be made over a relatively short duration

of about 20 days but it is to be left open for another 120 days. The excavation was effectively

dewatered. Taking the elevation datum to be at the bottom of the excavation, the hydraulic

head at the base of the excavation is 0m. During the excavation, groundwater recharge was

undertaken via shallow sub-soil drains near the ground surface which has the effect of

maintaining the phreatic surface virtually at the ground surface, thus maintaining a hydraulic

head of 6m at the retained soil surface. Seepage analysis shows that, in the long-term, after

steady-state seepage has established itself, the streamlines in the vicinity of the retaining wall

(both in the retained soil and in the excavation) are approximately vertical straight lines (i.e.

one-dimensional seepage flow) and that the hydraulic head within the sand layer is 2.5m.

20

m

6m

6m

Residual soil:

= 19kN/m3

k = 10-8

m/sResidual

soil

Sand Layer (highly permeable)

Hydraulic head h = 0m

Hydraulic head h = 6m

Hydraulic head h = 2.5m

(a) Discuss qualitatively the transient ground response arising from the construction work,

with appropriate sketches of the initial, short-term, long-term and transient (i.e. interim)

profiles of total and effective vertical stresses as well as pore pressures.

(b) By making appropriate simplifying assumptions, estimate

(i) the final settlement at the ground surface if the excavation is left open for a long time,

(ii) the ground settlement after 120 days.

Solution:

(a) Initial state, before excavation: Groundwater is in hydrostatic equilibrium.

s u s 0 0 0 gwt

380kPa 200kPa 180kPa

At ground surface: = 0 & u = 0 => = 0

At the base of the residual soil layer:

= 20 * 19 kPa = 380kPa

u = 20 * 10 kPa = 200kPa

= 380 200 = 180kPa

Short-term i.e. just after excavation:

Inside the excavation, at the excavation level:

= 0, = 20

180*6= 54kPa (Effective stress cannot changed in the short-term; hence

we can just use the initial effective stress at 6m depth)

u = 0 54 = -54kPa

At the base of the residual soil layer:

= 14 * 19 kPa = 266kPa

and = 180kPa (i.e. unchanged)

u = = 266 180 = 86kPa

Inside the retained soil, , and therefore u remain unchanged.

Hydraulic head h = 2.5m in the sand layer and the base of the residual soil layer

Elevation z = -14m

In the long term, in the sand layer and base of the residual soil,

u = w(h z) = 10*(2.5 (-14)) = 165kPa

Inside the excavation, at the base of the residual soil,

= 266kPa (same as short-term)

= 266 165 = 101kPa

s u s

380 164.7 200 180 215.3

s u s

266 86 64.7 101.3 180

165 165

In the retained soil, at the ground surface, , u and remain unchanged.

At the base of the residual soil layer,

= 380kPa (no change in total stress since there is no loading)

u = 165kPa

= 380 165 = 215kPa

Notice that there is

(i) a general increase in effective stress in the retained soil in the long-term; this will lead to

settlement.

(ii) a general decrease in effective stress in the excavation area in the long-term, leading to heave

(or swelling) of the soil.

(c) Since the coefficient of volume change is not known, we need to use the compression index:

e = Cc

'

'lg

1

0

s

s

At the mid-depth of the retained soil,

0 = 90kPa (i.e. short-term)

1 = 107.5kPa (i.e. long-term)

e = 0.05

5.107

90lg = -3.858x10-3

Since no other information is given to allow us to find the in-situ value of void ratio, we need to use

the bulk weight to give an estimate

=

e

eGsw

1

Assuming Gs = 2.65,

1.9 = e

e

1

65.2 => in-situ e = 0.833

mv = '1 s

e

e=

905.107*833.110858.3 3

x= 1.203x10-4 /kPa or 120.3strains/kPa

cv = vwm

k

=

4

8

10203.1*10

10

xm2/s = 8.313x10-6m2/s

or 0.718m2/day

Caution! This is not an accurate way of determining cv. Wherever possible, cv should be measured.

Directly measured values are often more representative and reliable! We only do this because no

other information is available.

(i) = D

v udzm = mv*(area under excess pwp profile)

The initial excess pore pressure u can be obtained by considering the difference in the short-

and long-term pore pressure profiles, i.e.

u u

165 200 (200-165)

= 35kPa

Hence, excess pwp profile is triangular with maximum of 35kPa at the base of the soil layer.

Hence,

= 1.203x10-4* *35 * 20kPa = 0.042m or 42mm.

(ii) This is a double-drainage consolidation problem, so we should use the centre curve.

Drainage path D = 10m

t = 120days.

T = 210

120*718.0= 0.862 => T = 0.928

U = 0.91 from chart.

Alternatively, use the approximate formula 3T) - exp( -1 U41

32

= 0.862)*3 - exp( -1 41

32 = 0.936

120 = 0.936*42mm = 39.3mm

3. A seabed soil profile consists 5 m of soft, normally consolidated marine clay underlain by dense

sand. The seabed is at a depth of 4 m below mean sea level (msl). It is required to reclaim this site

to a height of 2m above msl with sand (also known as hydraulic fill). The properties of the soil

are as follows:

Marine clay: compression index = 0.7

swelling index = 0.2

coefficient of consolidation = 2 m2/year

in-situ water content at 4 m below seabed level = 70%

bulk unit weight = 16 kN/m3

Gs = 2.68

Hydraulic fill: bulk unit weight = 18 kN/m3 (below waterline)

15 kN/m3 (above waterline)

(a) Estimate the final settlement of the reclaimed land arising from consolidation of the soft clay.

(b) How long will it take for 90% of this settlement to occur?

(c) It is desired to enforce this 90% settlement within a period of 2 years by placing an additional

surcharge consisting of hydraulic fill on top of the final reclamation level. Estimate what

height of surcharge will be required to achieve this?

Solution:

(a) Additional overburden stress from landfill = 2*15 + 4*(18-10) kPa = 62kPa

Coefficient of volume change is not given, so we need to use the compression index.

4m

5m

2m

Fill surface after reclamation

Sea level

Sand fill

Soft clay

Dense sand (highly permeable)

Consider mid-depth of soft clay layer:

Initially, 0 = 2.5 * (16 10) = 15kPa

Finally, after a long-term (after consolidation),

= 15 + 62 = 77kPa

Hence, change in void ratio e = Cc log

15

77 = 0.7 * 0.71 = 0.497

We do not know the initial void ratio, so we need to deduce it from indirectly,

= 16 = e

e

1

)68.2(*10 => e ~1.8

We can also deduce e from water content assuming fully saturation,

e = Gsw = 2.68 * 0.7 = 1.876

Hence, the two estimates agree reasonably well and we can adopt e = 1.876 (usually more reliable

to estimate e from water content than bulk unit weight).

Equivalent mv = '1 s

e

e=

62*876.2

497.0/kPa = 2.787x10-3/kPa

Initial excess pore pressure profile is rectangular, hence

= udzmv = 2.787x10-3* 62 * 5m = 0.86m or 86cm.

(b) U = 0.9 => T = 0.92 => T = 0.846

2H

tcv = 0.846 => t = 2

5.2*846.0 2years = 2.65years

(c) Absolute settlement corresponding to 90% consolidation = 0.9 * 86cm = 77.4cm

For t = 2years, T = 2H

tcv = 25.2

2*2 = 0.64 = T = 0.8 => U = 0.83 or 83%.

If 83% consolidation under surcharge corresponds to 77.486cm,

then full settlement under surcharge = 83.0

8677.4/0.83 = 0.931.036m or 9301036mm

Let be the required effective overburden stress of the surcharge.

Then 2.787x10-3 * 5 = 0.931.036 => = 66.7kPa74.3kPa ~ 75kPa

Hence, additional surcharge on top of the landfill is 66.775 62 = 413kPa

or 413/15m = 0.2687m of sand i.e. 2687cm of sand.

4. The figure below shows a site underlain by two layers of soft clay. In order to enforce early

settlement of the ground, a 5 m thick layer of loose sand (unit weight = 14 kN/m3 ) is placed on

top of the ground. Estimate how long it will take for the ground surface to settle 50 mm. The

groundwater table is at the ground surface.

Properties:

upper soft soil layer: cv = 0.5 m2/year

k = 10-9 m/s

= 15 kN/m3

sand seam: k = 10-3 m/s

= 17 kN/m3

lower soft soil layer: cv = 1.5 m2/year

k = 10-9 m/s

= 17 kN/m3

Solution:

mv = vwc

k

=

5.0*10

24*365*3600*10 9m2/kN = 6.31x10-3m2/kN

Surcharge = 14 * 5 = 70kPa

Final compression = 70 * 8 * 6.31x10-3m = 3.534m

Lower soil layer,

mv = vwc

k

=

5.1*10

24*365*3600*10 9m2/kN = 2.10x10-3m2/kN

Final compression = 70 * 8 * 2.1x10-3m = 1.176

Estimate settlement after 1 week: t = 7days

Ground surface

Upper soft soil layer 8 m

Sand seam 1 m

Lower soft soil layer 8 m

Bedrock (impermeable)

Upper soil layer: T = 2H

tcv = 16*365

7*5.0= 5.993x10-4

T = 0.02448 which is too small to read from the graph.

Using the approximate formula instead:

4T

U =

-45.993x10*4 = 0.0276

Compression after 7 days = 0.0276*3.534m = 97.6mm > 50mm

Hence, 50mm is likely to represent very small degree of consolidation. Hence we can use the

approximate relation to solve for the real time.

For the upper soil layer:

4T U =

16*365*

0.5t*4

= 0.0104t =

u

tu

Where the subscript u refers to the upper soil layer.

Hence, tu = 0.0104ut

Where u = 3.534m.

tu = 0.0369t

For the lower soil layer:

4T U =

64*365*

1.5t*4

= 9.253x10-3 t =

l

tl

Where the subscript l refers to the lower soil layer.

Hence, tl = 9.253x10-3lt

Where l = 1.176m.

Hence, tl = 0.01088t

Hence, total settlement at time t = t = tu + tl = (0.0369+0.01088) t

=0.0478t

For t = 0.05m, => t ~1.1days i.e. it takes ~1.1days to reach 50mm settlement.