ENV-2E1Y FLUVIAL GEOMORPHOLOGY
Examples of Previous Examination Questions and notes on how to solve
numeric questions for Slope Stability and Related Topics.
Exam Questions Pre-1999
ENV-2E1y -GEOMORPHOLOGY
Selected Examples of previous examinations questions.
The following are a selection of the questions which have appear over the years in Geomorphology.
You should note that the previous Course Number was ENV-2B07, before that ENV 264, and prior
to that it was ENV 210
Note: in several years similar types of question have been set. Accordingly the wording and values in any one
question will not necessarily be exactly as set, but will cover the full range of possibilities. Sometimes only parts
of former questions are given below, as actual full questions have been developed from aggregating different
combinations in various years.
YOU SHOULD ALSO NOTE THAT SEVERAL WORKED EXAMPLES
WERE GIVEN IN THE HANDOUTS, AND THESE SHOULD BE CONSULTED TO.
A. Numeric Questions - as actually set
1.
Describe how you would assess the consolidation characteristics of a soil.
A layer of homogeneous clay (unit weight 18 kNm-3) is 9 m thick and underlies a 7m layer of sand (unit weight 19
kNm-3). The water table is 3m below the surface, while beneath the clay is a permeable, incompressible layer of
sand.
If the water table falls to the interface between the upper sand and the clay, estimate the reduction in thickness of the
clay layer given the following data relating the coefficient of volume compressibility (mvc) to stress level:stress
(kPa)
mvc
(kPa-1)
100
120
140
160
200
250
0.00768
0.00597
0.00461
0.00358
0.00210
0.00170
The unit weight of water may be approximated to 10 kPa.
[Numeric Part 60%] - this question was set in 1992 and solution follows example given in section 3.9.2 of handouts.
2.
A 20m high slope is shown in Fig. 1. Estimate the factor of safety along the potential failure surface if the soil has
the following properties:f
c
g
= 20o
= 20 kPa
= 19.3 kNm-3
The actual figure was just a general profile of a slope
Estimate the depth to which tension cracks could develop at the slope crest, and the effect these would have on the
stability of the slope.
Clearly state the assumptions and limitations of the method of analysis you use, and indicate how you would
improve the estimates of the factor of safety.
[ numeric part 70%] This question was set in 1992 and a very similar one in 1989 In 1989, (40 minutes allowed)
the areas of the slices were given as shown in Fig. 1 as were the positions of the slices. In 1992, neither of these
were given as 60 minutes was now allowed for the question. The solution begins by following exactly the procedure
used in the Slope Stability Practical. However, one can estimate the depth of tension cracks from
d
2c

1
and if one is cunning, one of the slice boundaries would intersect the failure surface at exactly this depth below the
surface - see sections 5.7.7 and 5.8.6 of handout. Two effects of tension crack. (1) a reduction of the length of the
failure surface - i.e. neglect all slices to right of tension crack and recalculate; (2) water filling crack will cause a
lateral pressure tending to increase likelihood of failure.
Fig. 1 Question 2 Profile and Geometry of Slope
In 1989 the following data about the weight of the various slices was given. In 1992 you were expected to evaluate
these as 50% more time was allowed.
Slice Number
1
2
3
4
5
Area of Slice (m2)
18.2
43.2
47.2
27.8
2.4
2
3. A 4 m thick layer of fine sand of unit weight 19 kN m-3 overlies a 1.2 m thick layer of clay (16.67 kN m-3). The
water table is at the surface. The results from a one-dimensional consolidation test on a sample of the clay from 4.6
m below the surface are as follows:effective stress (kPa)
10
20
40
60
90
160
320
480
voids ratio
1.450
1.425
1.400
1.388
1.340
1.170
0.970
0.850
What is the overconsolidation ratio of the soil?
If an additional layer of sand 5 m thick (unit weight 16.4 kN m-3) were now deposited on the surface, how much
settlement could be expected in the clay layer?
(Take the unit weight of water to be 10 kN m-3).
see Question 5: 1980. First plot up graph on log paper. Where there is a kink in the curve, this indicates the
previous maximum stress in the soil. Now work out the present in situ stress (see section entitled Estimation of
effective vertical stress at depth in handout). The ratio of maximum to in situ stress gives the over consolidation
ratio. Now work out the new stress (after adding sand) and use graph to work out change in voids ratio. Finally
convert this change in voids ratio to a change in thickness
4. A 6 m layer of a sandy sediment is underlain by a thick layer of clay. The area is known to have been glaciated at
some point in the past. Results from a standard consolidation test on a sample of clay from 10 m below the present
ground surface are summarised in the table below.
Comment on the shape of the curve and estimate the over consolidation ratio of the sample in the field if the water
table is 7 m below ground surface. The unit weight of the clay is 15 kN m-3 and the degree of saturation of the sand
is 6.3%, and its voids ratio is 0.8. Estimate the overlying thickness of ice if, during glaciation, the water table was at
the sediment surface. The following assumptions may be made:(i) The sand sediment is incompressible for stresses up to 5000 kPa,
(ii) The unit weight of water is 10 kN m-3
kPa
60
120
240
480
960
1920
3000
300
30
voids ratio
1.402
1.380
1.357
1.279
1.173
1.068
1.000
1.075
1.150
see Question 1: 1981. This question is similar to question 4 above except that there is an added complication that
the sand is only partially saturated. The initial in situ stresses can be found using the various formulae in the basic
set of definitions. Once both the in situ and previous maximum stresses have been found, the difference can be
related to the overlying burden of ice (unit weight of ice is in data sheet). One point to watch though is that the sand
is fully saturated during the glaciation period, so allowance for this extra buoyancy must be made.
5. With the use of sketches to illustrate your answer, describe how a clay consolidates.
Partially processed data from a consolidation test undertaken on a soil sample taken from a depth of 16m below the
surface at the position show in the simplified bore hole log (Fig. 2) are show in the following data.
3
normal stress
(kPa)
10
25
50
100
200
400
800
1600
3200
sample thickness
(mm)
28.00
27.80
27.65
27.50
27.16
26.40
25.65
24.90
24.14
At the end of the test the voids ratio was 1.414.
Determine the over-consolidation ratio of the soil, and the voids ratio associated with the previous maximum
pressure.
If previous changes in the vertical pressure have only occurred through changes in the water level, determine the
lowest level the water table has reached.
Describe how the shear behaviour of a sample located at position A in Fig. 2 would differ from that tested.
see Question 3: 1990 - this question requires the manual solution of the consolidation data as done in the practical
write up by many of you - see practical sheet handout. The key point for the last past of the question is that the
overconsolidation ratio of the sample at A will be much higher than that at the base - you will need to compute the
values. Then noting the comments in section 4.9 of the handout you can make statements about the likely behaviour
of the respective samples under shear. Remember the critical value of OCR is 1.7
Fig. 2 [Question 5]
6. Indicate how the Atterberg Limits of soil may be determined. Indicate how a knowledge of these limits may be used
to predict the shear and consolidation behaviour of the soil.
A uniform bed of clayey soil having a specific gravity of 2.65 covers an area which has been glaciated in the past.
Apart from partial desiccation of the top 0.625 m to a degree of saturation of 61%, the soil stratum has remained
undisturbed since the retreat of the glacier. A sample is taken from 2.125m below the surface and found to have a
moisture content of 40.0%. Estimate the in situ unit weight of the sample of soil, and also the in situ stress on the
sample (you may assume that the water table is also 0.625m below the surface, and that the unit weight of water is
10 kN m-3).
The liquid limit is measured at 54.7% while the plastic limit is 33.2%. Plot these points on a suitable graph, and
hence, or otherwise, estimate the undrained shear stength of the sample of soil in situ. Would you expect a drained
test on the sample starting from the in-situ stress conditions to expand or contract?
In a laboratory consolidation test, the following results were obtained. If the water table was initially at the surface
during cover by the glacier, estimate the maximum thickness of ice.
4
TABLE 1
stress
(kPa)
20.0
40.0
60.0
80.0
100.0
120.0
140.0
160.0
180.0
125.0
98.0
void ratio
1.060
1.048
1.041
1.039
1.011
0.989
0.970
0.953
0.939
0.946
0.949 ------- end of test
At the final stress indicated in the consolidation test, the sample is sheared in a drained test. Estimate the change in
volume during the shearing.
Comment on the results.
see Question 3: 1993 First evaluate in situ unit weight of the soil using the general definitions formulae (the
soil is full saturated), and then work out the current in - situ stress. Plot the Liquid Limit and Plastic Limit
information as a Liquidity index plot against shear strength remembering the key values of 1.7 and 170 kPa for the
shear strengths at the two limits. From this get the in situ shear strength. Now plot the consolidation data, and plot
the point corresponding to the in situ voids ratio and stress. You will find this plots below the normal consolidation
line. Now plot the Critical Stress Line which is parallel to the normal consolidation line and displaced towards axis
by a factor of 1.7 (the critical OCR). In a drained test, the stress path will move either vertically upwards or
downwards until the critical state line is reached. Read of the change in voids ratio and convert this to a change in
thickness (i.e. proportion reduction in thickness
=
e
) and hence from initial thickness of clay layer, work out settlement.
1e
7) Describe how you would assess the consolidation characteristics of a soil.
[1995 paper]
A region of Holocene deposits between Acle and Yarmouth has a borehole section as shown in Fig. 3. Originally the
water level was at the surface, but as some time in the past drainage took place and the current mean water table is
located at the base of the upper sand. There is no evidence to suggest that further shrinkage is taking place in the upper
clay, while the upper sand may be considered as being incompressible.
The borehole was drilled with a piston sampler and several undisturbed samples were taken for unit weight
determination, the results of which are also shown on Fig. 3. A sample for consolidation was taken from mid depth in
the lower clay layers and a piezometer installed at the same depth. The equilibrium water level in the piezometer tube
was initially found to be 1.548m below present ground level and found to drop to 1.706m below ground level after 10
years. Results from the consolidation test are shown in Fig. 3.
As part of the environmental reconstruction of the area, a sample of the surface clay was reconstituted at high water
content in the laboratory and allowed to settle in a tall sedimentation tube. The mean bulk unit weight of this sample
was measured at 14 kN m-3.
Estimate how high above the present ground level was the original ground surface before drainage started, and also
estimate approximately how long ago this was. Have far into the future can it be expected that 90% of consolidation has
been completed, and how much will be this additional settlement.? Clearly state all assumptions you make.
The unit weight of water may be assumed to be 10 kN m-3
5
Fig. 3 Borehole log through sediment
TABLE 1. Consolidation characteristics of the soil.
Effective Stress
(kPa)
5
10
20
40
80
160
320
640
Voids Ratio
2.202
2.172
2.142
2.112
1.918
1.638
1.359
1.080
[Hint: divide the lower clay into three sections] - we have covered most of this question in lectures
[Description 20%; Assumptions 10%; Calculation 70%]
===============================================================
8. Describe the tests you would carry out to determine the consolidation characteristics of a soil.
[30%]
Fig. 4 shows the profile of a holocene deposit in an estuary. The mean tide level is at the ground surface. It is proposed
to enclose an area with a sea wall and to drain it so that the mean water table will be 3.4 metres bewlo the ground
surface. At the location of the proposed sea wall, which is to be constructed as an earth embankment, the overlying clay
deposit is stipped and fill of unit weight 20 kN m-3 added to a height of 5 metres.
Determine the settlement which is likely to take place beneath the embankment. The consolidation characteristics of the
middle and lower clay are similar; these results are shown in Table 2.
If the highest tisdes reach 2 m above the mean sea level, and if sea level is predictecto rise by 0.3 m in the next century
through global warming, detemine whether the sea wall adequately perform its task to prevent flooding of the area.
[70%]
This question was set in January 1996.
Solution
To estimate settlement we need to compute the values of mvc and plot these on a graph of mvc against mean
effective stress. It is possible to avoid plotting graph and use linear interpolation, but it is expected that most people
will prefer to plot graph as the computation can get tedious.
By definition:-
mvc  
1
e
.
1  eo 
hence values in column 6 can be evaluated.
6
Fig. 4 [Question 8] Profile of sediment
pressure
(kPa)
10.0
voids ratio
20.0
1.264
40.0
1.174
60.0
1.121
e

mean
pressure
mvc
0.090
10.0
15.0
0.003823
0.090
20.0
30.0
0.001968
0.053
20.0
50.0
0.001219
0.067
40.0
80.0
0.000790
0.052
50.0
125.0
0.00506
0.038
50.0
175.0
0.00380
1.354
100.0
1.054
150.0
1.002
200.0
0.964
0.061
120.0
260.0
320.0
0.903
[30 marks ex 70 to this point] with graph plotted if using graphical method, or [20 if not]
7
0.00259
Split clay layers up into 2 m thick sections and work out initial stresses. Thus we analyse behaviour 1m, 3m, and 5m
below top of middle clay, and same distances below top of lower clay. Call these layers 1 - 6 respectively (see figure
overleaf).
Thus 1m below top of middle clay, initial stress will be
2 * (17-10) + 1.5* (19 - 10) + 1*(17.5-10) = 35 kPa
Other two mid points of middle clay will be 35 + 2*(17.5-10) = 50 kPa and 65 kPa respectively.
Stress at base of lower sand will be 65 +1*(17.5-10)+2*(19.75-10) = 92 kPa
and stress 1 m below top of lower clay will be 92 + (18 - 10) = 100 kPa, while at 3 and 5m the initial stress will be
100 + 2*(18-10) = 116 kPa and 132 kPa respectively.
Layer
Initial Stress
(kPa)
Final Stress
(kPa)
Mean Stress
(kPa)
1
2
3
4
5
6
35.0
50.0
65.0
100.0
116.0
132.0
135.0
150.0
165.0
200.0
216.0
232.0
85.0
100.0
115.0
150.0
166.0
182.0
mvc from
graph or by
interpolation
0.000758
0.000664
0.000569
0.000443
0.000403
0.000370
thickness
(m)
settlement
(m)
2.0
2.0
2.0
2.0
2.0
2.0
Total
0.152
0.133
0.114
0.089
0.081
0.074
0.643
Stripping clay layer and lowering water table by 3.4 m will change initial stresses by:- 2* (17 - 10) + 1.4 * 10 = 0 kPa
, so stress increment after completion of wall =
0 +100 kPa = 100 kPa. This is added to original value and placed in table above.
We now read of values of mvc from the graph and work out settlement from:settlement =
mvc t 
where t is the thickness of the layer.
Summing all the settlements gives the total settlement of 0.768 m
Thus we can expect the final height of the sea wall after settlement to be 5 - 2 - 0.643m above
present mean sea level = 2.357 m.
This value is less than the 2.4 m predicted for high tide plus the sea level rise, and thus the proposed sea
wall will be insufficient to prevent flooding in the future.
8
Marks for this section:Working out initial stresses will generate 20 marks: the stress increment 5 marks.
the values of mvc correctly read will give 5 marks (if read from graph) or 15 marks if obtained by
interpolation). Ten marks will be generated from calculation of settlement giving a total of 70 for the numeric
part.
==========================================================
9.
Briefly review the methods available for analysing the stability of slopes indicating the advantages and
disadvantages of the various methods.
[30%]
You are asked by a National Park Board to comment on plans to construct a footpath 10m from the creat of 1 60 o slope
which is 20m high (Fig. 5.). Test of the soil reveal that it has a cohesion of 30 kPa, and angle of fraiction of 19 o and
a unit weight of 16 kN m-3. Review the stability of the slope, by considering ptoetntial striaghtline failure surfaces,
and comment on the proposed location of the footpath. You may assume that the footpath imposes negligible
surcharge on the slope.
This question was set in January 1996.
[70%]
Solution
Empirically, it is expected that the most critical straight line failure will occur when the failure surface is
approximately (  +  ) /2, i.e. about 40o. Therefore initially examine potential failures for case with no tension
crack at 35o, 40o, and 45o, and then insert a fourth value depending on initial results. The most critical failure
surface found will then be analysed with a tension crack present, and also one filled with water. There is no need to
repeat trials for these other two situations as a good approximation of the most critical surface will already have been
found.
The solution is best solved in tabular form,
and may be done graphically, by
measurement from the figure, or
trigonometrically.
We need the length of the failure surface
(L), and also the weight of the sliding mass.
failure
surface
angle
[1]
35
40
45
38




failure
surface
length
[2]
34.9
31.1
28.3
32.5
area of
wedge
weight of
wedge
cohesion
xL
Normal
Force
Shear
Force
Factor of
Safety
[3]
170.2
122.9
84.5
140.5
[4]
2722
1966
1352
2248
[5]
1046
933
849
975
[6]
768
519
329
610
[7]
1562
1264
956
1384
[8]
1.162
1.149
1.232
1.145
the weight (W) of the wedge is area x unit weight (16 kN m-3)
the Normal Force is W cos  tan 
the Shear Force is W sin 
the factor of safety is then ([5] + [6]) / [7]
9
or from the Data Sheets
Fs 
c . L  W . cos  . tan 
W . sin 
[45 marks out of 70] up to this point
After first three trials, critical value must
line between 35o and 40o, so try another
value at 38o, this gives a factor of safety of
1.145 so critical failure angle is about 38o
(see also graph below).
Note, strictly, one would expect to refine
the calculation further by additional trials,
but the increased accuracy hardly warrants
extra time spent unless one has access to a
computer (In fact the critical Fs is still
1.145 to three significant figures (although
the most critical angle is actually 38.5o.
Tension crack depth =
2 . c 20 x 30

 3. 75 m

16
[5
marks out of 70}
Now repeat above, but this time taking
failure surface only as far as the tension
crack. The failure surface length and the
area (and hence weight are this reduced).
failure
surface
angle
[1]
38
failure
surface
length
[2]
26.4
area of
wedge
[3]
131.5
weight of
wedge
[4]
2104
cohesion
xL
[5]
792
Normal
Force
[6]
571
Shear
Force
Factor of
Safety
[7]
1296
[8]
1.052
The slope is still stable, but is now becoming somewhat critical as Fs is only just over 1.000
[10 marks out of 70]
We finally repeat the calculation with the crack filled with water. This will exert a hydrostatic pressure so that the
total horizontal water force (U) will be:-
U  0. 5  w hc2
where hc is the depth of the tension crack. There is now a modification to the factor of safety relationship.
The normal force is now ( W cos  - U sin ) tan 
and the Shear Force is ( W sin  + U cos )
failure
surface
angle
[1]
38
[10 marks out of 70]
failure
surface
length
[2]
26.4
area of
wedge
[3]
131.5
weight of
wedge
[4]
2104
cohesion
xL
Normal
Force
[5]
792
[6]
556
Shear
Force
Factor of
Safety
[7]
1351
[8]
0.997
The factor of safety is less than unity and failure is likely. Since the critical failure surface intersects the top of the slope
beyond the footpath, the path is likely to be lost in any failure.
Two preventative measures are called for.
move the footpath further from the crest
minimise the risk of tension crack formation by encouraging vegetation growth and keeping surface moist to prevent
shrinkage and hence creation of tension cracks.
10
B. Modified versions of Numeric Questions set
1. Fig. 6 shows a cross-section through a slope in a frictionless soil. Estimate the factor of safety against failure along
the circular arc shown if the unit weight of the material is 16 kN m3 and the cohesion (c) = 20 kPa. A cardboard
template of the outline of the slope is provided.
Versions of this question have been set twice in the last 15 years. Section 5.5 of the handout indicates how to
solve this question. [Use the template to find centre of gravity, and use squares on graph paper to find area of
sliding section].
Fig. 6 [ Question B1]
2. The top of a 4.5m thick layer of clay of unit weight 16 kN m-3 is 9 m below the surface. Below and above the clay
is silt of unit weight 18 kN m-3. The water table is 2 m below the ground surface. By sub-dividing the clay layer
into three layers, estimate the settlement which will occur when 3 m of sand of unit weight 17 kN m-3 is deposited
on the surface.
Values of mvc are given in the table below.
stress
kPa
10
20
40
80
120
160
200
320
mvc
kPa -1
0.00405
0.00355
0.00300
0.00233
0.00200
0.00179
0.00169
0.00153
3. Fig. 7 shows (a) the profile and (b) the cross-section of the debris (unit weight 16 kN m-3 from a landslide
which flowed across a 20 m wide gully creating a dam. A lake formed upatream of the obstruction. Estimate the
discharge through the central 10 m of the Sand if the permeability is 10 m s 1-. Plot the variations in total stress and
effective stress along the line A-B.
11
The first part requires drawing a suitable flow net, counting the number of squares and pressure drops and working
out discharge as show in section 2.13. The is a slight catch here as the top flow line must be first drawn in. This
will be of parabolic form similar to that shown in Fig. 5.18 of handout. Once the flownet has been drawn, the
excess water pressure at select points along A - B can be estimated, and hence the total and effective stresses. For
this it is necessary to use slices to divide up the section above A - B.
Note, the provision of the section (b) is largely a red-herring since one is asked only to work out the discharge in the
central 10m which is of constant depth only, so the sloping sides can be neglected.
4.
Outline the Infinite Slope method for analysis of stability of a slope. Clearly indicate the assumptions involved and
the various situations to which this method may be applied. A slope of 45 0 is cut into a soil having a cohesion of 30
kPa, a unit weight of 15 kN m-3 and an angle of friction of tan-1(0,5). Estimate the maximum height for which the
slope will be stable.
Essentially, solution to this question is covered in Section 5.7 of the handouts
12
5. In the determination of the liquid limit of a soil by the fall cone method, the following readings were take:moisture content (%)
57.5
63.0
69.5
74.2
78.0
cone penetration (mm)
13.8
16.5
19.7
22.1
24.0
The plastic limit was determined as 30%
The soil is saturated and normally consolidated to 50 kPa, and has a unit weight of 17.10 kN m -3. If the soil is
consolidated to 500 kPa, estimate the shear strength of the soil.
Assume the Specific Gravity of the soil is 2.65, and the unit weight of water may be taken as 10 kN m-3.
Clearly state and assumption made in the analysis. Briefly describe other applications of the Atterberg Limits.
First plot up the data in the table to work out the liquid limit (i.e. at a penetration of 20mm). Hence get the slope of
the consolidation line Cc = 1.325 (LL - PL). From the initial stress of 50 kPa, find the initial voids ration from
the General Definition formulae and the data give about unit weight etc. (i.e. work the general definition formulae
backwards). Now plot the point on an e - log s plot and draw a line with the computed gradient of Cc. Where this
crosses the 500 kPa line, read of the voids ratio and convert this into a moisture content and finally into a liquidity
index. Finally plot the Liquidity Index vs Shear Strength diagram (section 1.6.5 of handout) and read off the shear
strength corresponding to the computed value of Liquidity Index.
Several Examples of calculations were given in the Lecture Notes and you should be familiar with these.
e.g. (1) calculation of effective vertical stresses
(2) calculations of quicksand formation
(3) U - ÖTv calculations for time of primary consolidation
(4) calculation of reduced thickness
(5) settlement calculations - see section 3.9.2
(6) calculations of settlement times in field - section 3.11 and also practical
(7) environmental reconstruction 3.13
(8) examples involving time - dependant pore-water dissipation section 3.13 - 3.14 and 3.15
(9) comparisons between Atterberg Limits and Cc - see section 3.16
Questions could be framed involving combinations of the above - see for example section 3.15 involves a
computation of effective vertical stress (1) before proceeding with main calculation.
NOTE: the following Question (i.e. number 2 in 1990) is beyond the scope of the topics covered in this years
course. but has been included in this list for reference purposes.
*.
Describe the possible mechanisms of failure in a composite river bank.
Fig. 8 shows the profile of a composite river bank.
Erosion of the non-cohesive bank is only effective at or near bankfull conditions when a mean erosion rate of 5mm
per hour has been measured.
A series of observations on the stability of the bank begins immediately after the stage has fallen to 1.3m below the
bank top where is remains until the onset of heavy rain 10 days later. The rain continues for 24 hours, and the
stage then rapidly rises to bankfull conditions where it remains for 20 hours before falling in a subsequent period
of prolonged dry weather.
Fig. 9 shows a simplified model of the change in unit weight through desiccation and wetting, and also the rate of
development of desiccation cracks upwards from the base of the overhang.
By the use of the Stability Charts, or otherwise, investigate the stability of the upper bank, and estimate when, if at
all during the study period, the bank fails.
13
Fig. 8 Profile of River Bank
Fig. 9 Drying, Wetting and Crack Propagation Curves
C. Descriptive Questions as set
Note: in some years, the questions were 45 minutes long (pre - 1987), some were 40 minutes long, and currently they
are 60 minutes long. Some times similar questions in different years were set but the allowable time was different.
Where longer times were allowed more detailed answers were expected. As previously, the wording has been modified
in some questions to cover similar questions in different years.
1.
What is meant by the terms:Effective stress
Total stress
14
Critically review the methods available to measure or estimate the shear strength of soils.
2. With particular reference to the Mam Tor Landslide in Derbyshire, discuss the possible causes of landslides.
Summarise the methods you would use to investigate the stability of a slope.
3.
Outline the erosion processes and mechanisms of failure associated with river banks which have a composite
structure. How would you assess the stability of such banks?
4.
Why do the undrained and drained shear-strengths of soils differ?
Outline the methods available for determining the drained and undrained strengths of soils in the field and in the
laboratory. For each method you should indicate its advantages and disadvantages.
5. What are the Atterberg Limits of a soil? Describe how they are measured and indicate how they are used to classify
soils. How may they be used to predict the consolidation and shear strength behaviour of a soil?
6. Critically review the different techniques for measuring the mechanical properties of soils. Briefly explain the value
of the parameters derived from such tests.
7. Discuss the mechanism by which river banks may fail. Explain why traditional slope stability analysis techniques are
often inappropriate in these situations.
8. The process of consolidation is frequently explained in terms of spring and dashpot models. To what extent do these
represent the true situation. Critically review the assumptions made in classical consolidation theory.
How may a knowledge of the consolidation characteristics of a soil be used to understand the past behaviour and
predict the future behaviour of a soil?
9. What field and laboratory investigations would you undertake to examine whether or not a slope was stable? How
would your approach differ if you were to examine a slope which had actually failed? Illustrate your answer with
reference to one or more specific landslides.
10. Discuss, giving examples, why it is essential to understand how water flows through a soil when analysing the
response of the soil to mechanical loading. What analytical techniques would you employ in such studies of ground
water flow?
11. What is meant by the following geotechnical terms
a) total stress, and
b) effective stress.
Describe how you would carry out an investigation to determine the cause of a landslide. Indicate in what
circumstances you would use a total stress analysis, and in which cases you would choose an effective stress
analysis.
12. Describe the mechanisms by which soil consolidates, indicating any deficiencies in the 'classical' theories of
consolidation. Describe the test you would carry out to determine the consolidation characteristics of a soil,
indicating how the characteristic properties may be used for other purposes.
13. Discuss the full cycle of events which cause river banks to fail. Explain why traditional slope stability analysis
methods are usually inappropriate in these situations. [1995 exam]
14
A knowledge of how water flows through a soil is essential in understanding the behaviour of the soil and river
engineering structures.
Discuss why it is essential to understand how water flows through a soil when analysing the response of the soil to
mechanical loading or to the performance of river engineering structures. What analytical techniques would you
employ in such studies of ground water flow? [1995 exam]
15. Describe the field and laboratory tests you would carry out to determine the shear characteristics of soils. Include a
review of any subsequent analyses which may be needed to predict the future behaviour of the soil.
15
ENV-2E1Y Fluvial Geomorphology - Previous Exam Questions. 1999 – onwards
In 1999, the Course Code was ENV-2B07 and the unit was 20 credits with a 2 hour exam in which 2 questions were
to be answered. There were three questions from the Slopes Section and 3 from the Rivers Sections. The number
of numeric questions in the two sections used to alternate form year to year (i.e. in 1997 there were two numeric
questions in the rivers Section, in 1999 there were two numeric questions in the Slopes Section.
1999 Paper
1) Discuss the similarities and differences in potential modes of failure between river banks and normal slopes. Why
do methods developed for the latter not always apply to the former?
Describe the field measurements, laboratory tests, and analyses you would make to assess the stability of of
composite river banks.
2). Describe how you would determine the consolidation behaviour of a sediment, and critically review any
assumptions made in classical consolidation theory.
Fig. 1 shows a simplified borehole taken in 1990 from within the site reclaimed for the new Hong Kong
International Airport. The sequences identified as M1 and M2 represent the marine deposits in the Holocene and
last inter-glacial periods, while T1 is the part of the M2 affected by desiccation and pedogenesis during the fall in
sea level during the last glaciation. The area itself was not glaciated, and it may be assumed that during the glacial
period, the local ground water level was at the base of the T1 unit, and that the unit weight and degree of saturation
of the T1 unit during that period were 16.50 kN m-3 and 0.4 respectively.
Estimate:
a) the unit weight of the desciccated layer after the sea level rises,
b) the effective overall stress increment when depositiion of the holocene sediments is complete
c) the consolidation of the pre-holocene layers as a result of the recent deposition.
The consolidation in the M2 and T1 units during the Holocene if the consoldation behaviour of the two units are as
shown in Table 1, while it may be assumed that the specific gravity of the sediment in all layers is 2.65..
normal stress (kPa)
20
40
80
160
320
640
1280
2560
5120
M2 unit
voids ratio
1.600
1.450
1.300
1.150
1.000
0.850
0.700
0.550
0.400
Table 1. Consolidation behaviour of M2 and T1 units
16
T1 unit
voids ratio
0.805
0.780
0.754
0.729
0.703
0.678
0.652
0.550
0.400
Fig. 1 Simplified borehole of sediments at Chek Lap Kok Airport site.
Part (a) - need to calculate initial voids ratio
( G s  S r e ).  w

(1  e )
or e (


 Sr )  ( G s 
)
w
w
if S r  0 . 4 , G s  2 . 65 , and   16 . 5
or e 
( 2 . 65  1. 65 )
 0. 8
(1. 65  0 . 4 )
Now calculate the unit weight of the saturated layer after it becomes saturated. Here we assume that resaturation takes
place with no change in voids ratio (valid unless smectitic).
( G s  S r e ).  w

(1  e )
or
( 2 . 65  0 . 4 ). 10
 19 . 17
(1  0 . 4 )
Part (b)
depth
top
mid
top
mid
top
mid
top
mid
T1
T1
M2a
M2a
M2b
M2b
M2c
M2c
0
1.25
2.5
3.75
5
6.25
7.5
8.75
initial unit
wt
(kNm-3)
0
16.5
16.5
16.6
16.6
17.05
17.05
17.48
total
water effective new unit
stress pressure stress
weight
(kPa)
(kPa)
(KpA)
(kNm-3)
0.00
0.00
0.00
20.63
20.63
19.17
41.25
0.00
41.25
19.17
62.00
12.50
49.50
16.60
82.75
25.00
57.75
16.60
104.06
37.50
66.56
17.05
125.38
50.00
75.38
17.05
147.23
62.50
84.73
17.48
Part (c) - using mid points of each layer as being representative.
17
Not needed
0.00
11.46
22.92
31.17
39.42
48.23
57.04
66.39
Stress
net stress
decrement increment
0.00
9.17
18.33
18.33
18.33
18.33
18.33
18.33
55.00
45.83
36.67
36.67
36.67
36.67
36.67
36.67
The key issue here is to estimate mvcand there are several methods to estimate this. One involves working out the
change in voids ratio with stress and correcting by the factor (1 + e o) followed by plotting a graph of mvc so that
values of mvc can be read off appropriately. Another method involves an incemremtal approach where the change in e
and change in stress are computed and explicit values of mvc are computed for each increment. All methods will give
approximately the same answer for mvc. All methods use the formula
m vc  
1
e
.
1  e o 
Using the latter method
Layer
1
2
3
4
T1
M2a
M2b
M2c
1.25
3.75
6.25
8.75
and total setllement =
20.63
49.50
66.56
84.73
 . m
66.46
86.17
103.23
121.39
vc
Stress
initial voids
increment
ratio
45.83
0.804
36.67
1.414
36.67
1.350
36.67
1.291
Final Void Change in
ratio
e
0.76
0.041
1.29
0.126
1.26
0.094
1.22
0.069
mvc
.mvc
0.000501
0.001423
0.001090
0.000818
Summation
0.022979
0.052162
0.03997
0.030007
0.145117
. z
All layers are 2.5m thick so total settlement will be
2.5 * 0.145117 = 0.363m
==========================================
3). How are "Factors of Safety" used to assess the stabilioty of slopes. Discuss wht some slopes with low factors of
safety are stable, while others with high factors fail.
[20%]
Fig. 2 shows a simplified cross section of the winter dyke on the south side of the Rhine opposite Wageningen in the
Netherlands. The landward side is an extensive area including many hundred houses. During floods in February 1995,
the river rose to within 0.5 m of the crest of the dyke.
Investigate the stability of the dyke based on the slip circle and slices shown. The meaen values of the unit weight,
cohesion, and angle of friction are 17.8 kN m-3, 20 kPa, and 11o respectively.
[70%]
In the circumstances prevailing at the peak river flow, what recommendations would you have given, both in the short
term and long term regarding the stability of the dyke?
[10%]
Fig 2 Section through winter dyke on south bank of Rhine near Wageningen, The Netherlands
18
=====================================================================================
ENV-2E1Y Fluvial Geomorphology Exam 2001
1.
Briefly discuss the validity of the assumptions made in Terzaghi’s Theory of consolidation.
[30%]
A study is to be made on the behaviour of a layer of marine clay 1.999m thick buried between two sandy-silty
layers. Initially the layer is in equilibrium with no excess pore pressure. During two storm events exactly one year
apart cause additional sedimentation leading to a stress increment of 10 kPa on both occasions. Estimate and plot
the excess pore water distribution with depth immediately preceding the second increment, and also after two years.
Data from a laboratory oedometer test done on a sample 19.5 mm thick is shown in table 1.
[70%]
Table 1
Time (minutes)
0.0
1.0
3.9
8.9
15.8
25.0
35.9
50.0
70.9
106.0
141.3
188.8
247.5
405.0
2.
Settlement (mm)
0.000
0.120
0.240
0.360
0.480
0.604
0.720
0.837
0.960
1.080
1.140
1.200
1.236
1.296
In the analysis of the stability of slopes, some assumptions lead to unsafe solutions while others lead to safe
solutions. Explain why these differences occur, and the approach you would take for an initial appraisal of the
stability of a slope.
[30%]
A 60o slope is 20m high. By considering potential planar failure surfaces, determine whether the slope is stable.
Data from standard shear box tests done on the material of the slope are shown in Table 2. The unit weight of the
material is 17 kN m-3.
You may assume that the water table is below any potential failure surface and you may ignore the effects of
tension cracks.
[70%]
Normal Force (N)
100
200
300
400
500
600
Shear Force (N)
134.8
161.6
188.4
215.2
242.0
268.8
The shear box is 6cm x 6cm in size
19
3.
You have been asked to investigate the causes of a major landslide in a slope approximately 100 m high and 150
wide. Describe the field and laboratory studies you would undertake to establish the causes of the failure.
In your answer you should include a discussion of the resources of manpower, time, and equipment you are likely
to require to complete the survey.
4.
Explain why the mechanisms of failure of river banks are often very different from those of normal slopes.
[30%]
What field and/or laboratory measurements would you do to ascertain the stability of such river banks?
[70%].
================================================================
1. Model Answer
Evaluate (time) and plot settlement against (time)
Time
(mins)
Time
(mins)1/2
Settlement
(mm)
0.0
1.0
3.9
8.9
15.8
25.0
35.9
50.0
70.9
106.0
141.3
188.8
247.5
405.0
0.000
1.000
1.969
2.979
3.969
5.000
5.990
7.071
8.419
10.296
11.885
13.739
15.732
20.125
0.000
0.120
0.240
0.360
0.480
0.604
0.720
0.837
0.960
1.080
1.140
1.200
1.236
1.296
% Settlement
(worked out after
construction of graph)
0.0
10.0
20.0
30.0
40.0
50.4
60.0
69.8
80.0
90.0
95.0
100.0
1/2
Time - (mins )
0
5
10
15
0.0
Experimental Data
0.2
Approximate Line
Construction Line
Settlement (mm)
0.4
0.6
0.8
1.0
A
B
1.2
1.4
20
First part of curve is linear. (see approximate line. Plot line with gradient 1.155 times that of original line.
Intersection at A defines 90% consolidation point, hence scale to find theoretical 100% consolidation point in absence
of secondary consolidation at point B corresponding with a settlement of 1.2 mm.
[suggest that 30 ex 70 marks are given for getting this graph and values]
So evaluate the proportions of consolidation for other values and enter them in table.
[35 marks]
Also since the material in field and lab are the same the following relationship is valid
Tlab
2
d lab
 Tfield 2
d field
So the equivalent of one year in field in terms of lab =
365 * 86400 * 0.01952 / 1.9992 = 3000 seconds = 50 minutes
This corresponds to one of the lab times and represents the 69.8% (~ 70%) consolidation..
[10 marks]
Representing a time factor of 0.4. So plot up values from graph corresponding to T v=0.4. These must be multiplied
by 10 as the annual increment is 10 kPa.
Further after 2 years, the corresponding value of T v would be 0.8 and this is the relevant line for the residual pressure
for the first increment after 2 years. Since there is a second increment after 12 months, at the end of 2 year from the
start, this increment will have dissipated to Tv = 0.4, so the final pore pressure distribution after 2 years would be the
equivalent of the curve from Tv=0.8 added to the curve for T v=0.4. This does assumes linearity and that increments
can be added which is a reasonable first approximation. A more detailed analysis would require a finite difference
approach.
Excess Pore Pressure (kPa)
0
1
2
3
4
5
6
7
0
Depth (m)
0.5
1 year
1
2 year
1.5
2
[10 marks for distribution at end of 1st year and 5 at end of 2nd year]
21
2.
Determine shear and normal stresses
Normal Force
(N)
100
200
300
400
500
600
Shear Force
(N)
134.8
161.6
188.4
215.2
242.0
268.8
Normal Stress
(kPa)
27.8
55.6
83.3
111.1
138.9
166.7
Shear Stress
(kPa)
37.4
44.9
52.3
59.8
67.2
74.7
Now plot Shear Stress against Normal Stress
80
70
Shear Stress (kPa)
60
50
40
30
20
10
0
0
50
100
150
200
Normal Stress (kPa)
This gives and intercept of 30 kPa for cohesion and 15o for angle of friction
Initially, one does not know the critical failure surface but it is likely to be around about 40 o so try angles of 35, 40
and 45o and then revise when results from these are known. Analysis is best done in Tabular form.
22
failure
surface
angle
35
40
45
38
37
37.5
failure
surface
length
34.9
31.1
28.3
32.5
33.2
32.9
area of
wedge
weight of
wedge
cohesion
xL
Normal
force
Shear
Force
Factor of
Safety
170.2
122.9
84.5
140.5
149.9
145.2
2892.7
2089.0
1437.0
2388.8
2549.0
2468.0
1046.1
933.4
848.5
974.6
997.0
985.6
634.9
428.8
272.3
504.4
545.5
524.6
1659.2
1342.8
1016.1
1470.7
1534.0
1502.4
1.0131
1.0145
1.1030
1.0056
1.0055
1.0052
1.12
Factor of Safety
1.1
1.08
1.06
1.04
1.02
1
0.98
0.96
35
37
39
41
43
45
Failure Angle
Critical failure surface lies between 35 and 40o so refine analysis leading to critical angle at 37.5o.
At this angle factor of safety is 1.005 and so only just stable.
Note depending on precision of analysis, some students my find slope is marginally unstable - this will get full marks
provided final value is within range of 0.98 to 1.02 and comment on stability is consistent. 100% of marks available
if gets precision to 0.5o, 95% for precision to 1o and 90% marks for this part if precision is to nearest 5 o. 80%
maximum marks if only one slip surface done.
23