联系我们： 手动添加方式: 微信>添加朋友>企业微信联系人>13262280223 或者 QQ: 1483266981

1
CENV2006 SOIL MECHANICS. SEMESTER 1 EXAMINATION AY 2022-2023: SOLUTIONS
QUESTION 1
A new building, which is square in plan view, is being constructed. The ground conditions are
summarised in Figure Q1 and comprise a layer of compressible clay, sandwiched by sand and
gravel layers that can be assumed to be incompressible and highly permeable relative to the clay.
Table Q1 shows results from three compression stages of an oedometer test on a sample from the
mid-depth of the clay layer. The building settlement requires assessment. Assume that the
properties of the clay sample represent the full clay layer.
Figure Q1: Ground conditions and planned building
Table Q1: Oedometer test data for three loading stages
Stress
increment
Time, t
minutes 0 0.5 1 2 4 8 16 32 64
60-120 kPa
Settle_xfffe_ment, ρ
mm
0 0.104 0.151 0.210 0.260 0.279 0.281 0.281 0.281
120-180 kPa 0 0.178 0.252 0.361 0.521 0.696 0.805 0.829 0.830
180-240 kPa 0 0.153 0.216 0.313 0.441 0.555 0.604 0.609 0.609
At the start of first stage: sample height, h = 25 mm.
At end of final stage: moisture content, w = 49%
(a) For each oedometer test stage, plot the settlement, ρ, against the square root of time, t.
Find the one-dimensional modulus, 0
′
, and consolidation coefficient, cv, for each stage.
State the units you have used.
[8 marks]
Sand and gravel
Water table
New building
Clay
Sand and gravel
10 m
6 m M C
20 m
γ = 20 kN/m3
γ = 17 kN/m3
2
2 marks for plot.
Calculation of stiffness: 0
′ =
Δ
′
Δ
= Δ
′
/

Calculation of consolidation coefficient: = 3
4
2

Answers need not be exact – variation include whether to update sample height h between stages
in calculation of strain increment, and also the approximate fit of tx to the data.
(b) Using the sample height, h, at the start and end of the stages, develop a figure showing the
specific volume of the sample, v, at different effective vertical stresses,
′
, and estimate the
one-dimensional virgin compression stiffness, 0, the unload-reload stiffness, 0, and the
intercept of the one-dimensional normal compression line, 0, for the clay.
[5 marks]
Use phase relations to convert from measured moisture content and sample height at end of stage
3, through to values of specific volume at other stages and effective stress levels.
Calculation of voids ratio from moisture content: = ,
Calculation of specific volume from voids ratio: = 1 + ,
Calculation of specific volume at all values of sample height:

=

2 marks for plot
1 mark for each of the three parameters: 0 = 0.2, 0 = 0.04, 0 = 3.4
E0 3 marks ρult tx (min)
Stage 1 5346 16.9 32.1 0.281 3.6 1.9
Stage 2 1807 5.7 10.8 0.830 10.6 3.2
Stage 3 2463 7.8 14.8 0.609 7.6 2.7
tx^0.5 (min^0.5) cv 3 marks
3
It is valid to state that 0 could be lower (more stiff) because there is no intermediate point between
the start and end of stage 1, and the path may shift onto the NCL at a lower stress than 120 kPa.
(c) Calculate the in situ vertical total and effective stresses, and
′ , at the mid-depth of
the clay layer (i.e. points M and C), before construction of the building.
[2 marks]
= 20 × 10 + 17 × 3 = 251
= 10 × 13 = 130

′ = = 121
1 mark each for and
′ .
(d) Estimate the over-consolidation ratio of the clay, before construction of the building.
[2 marks]
OCR = 1 (1 mark) because the oedometer response (part (b)) shows a change (fall) in stiffness at
120 kPa, which is approximately the in situ vertical effective stress (1 mark).
(e) The building will apply a vertical stress of 100 kPa to the ground surface. Estimate the
increment of vertical effective stress, Δ
′ that this will cause at the mid-depth of the clay
layer at points M and C, which are under the middle and the corner of the building
respectively. State any assumptions involved in your approach.
[4 marks]
Using Fadum’s chart or solution, the increment of stress. As a fraction of the building stress, can be
calculated. The building is square, with length and breadth = = 20 and the depth of interest
is = 13 .
For point C, consider a single foundation of = = 20 , so
=
= 20
13
= 1.54, for which = 0.218.
So, the increment of vertical effective stress, Δ
′ = 0.218 × 100 = 21.8 .
For point M, consider superposition of four foundations of = = 10 , so
=
= 10
13
= 0.77, for
which = 0.141. So, the increment of vertical effective stress, Δ
′ = 0.141 × 100 = 14.1 .
1 mark for recognising to use Fadum (or Newmark, if used instead). 1 mark for each stress.
1 mark for giving an assumption: e.g. elastic behaviour, no influence of stiffness heterogeneity.
(f) Estimate the long term settlement of the building above points M and C due to compression
of the clay layer.
[4 marks]
Assume that the clay follows the NCL. Initially, at
′ = 121 , = 0 ln(
′ ) = 2.441. Using
the same calculation, for the increased stress of
′ + Δ
′
the final after consolidation is found.
The vertical strain in the clay layer is given by = / and the settlement is =
where = 6 . This leads to = 81.4 and = 54.2 .
Alternative approaches are acceptable if explained, e.g. use stiffness ′0 from the second stage of
the oedometer test; or calculate ′0 directly from differentiation of the NCL (i.e. 0
′ =
′
/ ). These
will give no more than 2-3% difference to the values calculated here.
3 marks for a generally correct approach; 1 mark for numerical values correct.
4
Q2 A specimen of saturated Kimmeridge Clay having an initial overconsolidation ratio (based on
average effective stress p′) of 2 is subjected to an undrained shear test in a triaxial cell from an
effective cell pressure of 100 kPa.
(a) Explain briefly how the Cam clay model can be used to determine the state paths followed during
shear. [6 marks]
(b) Using the Cam clay model with Γ = 2.50, Μ = 1.2, λ = 0.16 and κ = 0.05, calculate and plot these
state paths in the q vs p and p′ and v vs ln p′ planes. Your axes should extend from 0 to 200 kPa for p,
p′ and q on the q vs p, p′ plot, and the scales for the q and p, p′ axes should be the same.
[8 marks]
(c) A second, identical specimen is subjected to a drained (rather than an undrained) test from an
effective cell pressure of 100 kPa. Calculate the values of q and p’ and the specimen volume change
at failure, if the volume of dry solids in the specimen was 875 ml (millilitres). [4 marks]
(d) A third specimen having the same previous stress history as the first two was subjected to an
undrained shear test from an effective cell pressure of 50 kPa. Use the Cam clay model to calculate
the volume of water taken into the specimen on reducing the cell pressure from 100 kPa to 50 kPa,
and the theoretical values of q and p′ at yield. Sketch the stress path (q vs p′) on your graph from
part (b). [5 marks]
(e) Why is this calculated value of q unlikely to be achieved in reality [2 marks]
Solution
(a) Explain briefly how the Cam clay model can be used to determine the state paths followed during
shear. [6 marks]
With the drainage taps closed, the specimen is forced to follow a path at constant specific volume v
from its state at the start of the shear test to failure on the critical state line.
The specific volume v may be calculated from the stress history of the specimen, which has been
consolidated isotropically to a cell pressure of 200 kPa before being allowed to swell back to its
current cell pressure along an unloading (κ) line of 100 kPa.
= ( + ) .
′ + .

′
′ (Eq 1)
where initially p′o = 200 kPa and p′ = 100 kPa
p′ remains constant from the start of the test until the initial yield surface is reached, given by

′ +
′

′ = 0 (Eq 2)
that is, yield occurs when
= ′

′
′ (Eq 3)
Following yield, the stress state is on an expanding yield surface characterised by an increasing value
of p′o. At any stage of the test, the specimen can be considered to have reached its current stress
state on the v, lnp′ plot by isotropic normal compression along the isotropic normal compression line
5
from a specific volume v = (Γ + λ – κ) at p′ = 1 kPa to an apparent maximum isotropic compression
pressure p′o, followed by swelling back along a κ-line to the current value of p′.
= (Γ + λ κ) .
′ + .

′
′ (Eq 1 bis)
or
′ = (Γ+λ κ) ( ).
′
(Eq 4)
The test ends on the critical state line, at a value of p′ (p′c) that can be calculated from the (constant)
specific volume v and the equation of the critical state line,
= Γ .
′
(Eq 5)
at which the apparent preconsolidation pressure, p′o,c, is given by
= (Γ + λ κ) .
′
, + .
, ′

′ (Eq 6)
Equating these,

, ′

′ = 1 (Eq 7)
The deviator stress on the critical state line, qc, can be calculated from the equation of the critical
state line,
= Μ ′
(Eq 8)
Selecting a series of values of p′o between the initial value of 200 kPa and the final value p′o,c, the
corresponding values of p′ at the various stages of the test can be calculated from Eq 1 and the
values of q from Eq 3. [6 marks]
(b) Using the Cam clay model with Γ = 2.50, Μ = 1.2, λ = 0.16 and κ = 0.05, calculate and plot these
state paths in the q vs p and p′ and v vs ln p′ planes. Your axes should extend from 0 to 200 kPa for p,
p′ and q on the q vs p, p′ plot, and the scales for the q and p, p′ axes should be the same.
[8 marks]
The calculations are carried out in a spreadsheet using the equations and process outlined above. The
results are given below and the plots are at the end of the solution.
The specific volume v is calculated using Eq 1 with p′o = 200 kPa, p′ = 100 kPa and the Cam clay
parameters as given.
v = 1.797
p′ = constant = 100 kPa until yield, so the value of q at yield can be calculated from Eq 3 as
qy = 83.2 kPa
The value of p′ on the critical state line, p′c, is calculated from Eq 5, as
p′c = 81 kPa
Corresponding values of qc and p′o,c are qc = 97.2 kPa (from Eq 8) and p′o,c = 220.1 kPa from Eq 7.
6
The state path between yield and failure is calculated by selecting a number of values of p′o between
the initial value (200 kPa) and the value at the critical state (220.1 kPa); and calculating the
corresponding value of p′ using Eq 4, and the corresponding value of q using Eq 3 for the current
values of p′o and p′ (Table 1). [8 marks]
p’o 200 200 204.0256 208.0512 212.0768 216.1023 220.1279
p’ 100 100 95.71055 91.68359 87.89845 84.33642 80.98054
q 0 83.17766 86.93392 90.15511 92.90157 95.22644 97.17665
Start Yield crit state
Table 1: Values of q, p’o and p’ calculated using the Cam clay model for Test 1
The total stress path follows the line
′ = 100 +
(derivation not required) and the pore water pressure takes up the difference between p and p
3

′ at
(Eq 9)
each stage (not asked for).
The graph of v vs lnp′ progresses at v = constant (= 1.797) from the start of the test at p′ = 100 kPa to
the end of the test on the critical state line at p′ = 81 kPa as already calculated.
The plots are shown at the end of the solution.
[4 marks for the plot of q vs p’; and 2 each for the plots of q vs p’ and v vs lnp’: 8 marks in total for
part (b)]
(c) A second, identical specimen is subjected to a drained (rather than an undrained) test from an
effective cell pressure of 100 kPa. Calculate the values of q and p’ and the specimen volume change
at failure, if the volume of dry solids in the specimen was 875 ml (millilitres). [4 marks]
There is no generation in pre pressure in a drained test; the effective stress follows the total stress
path given by Eq 9. Failure is reached when this intersects the critical state line (Eq 8). Equating these
and solving gives
qc = 200 kPa, p’c = 167 kPa [2 marks]
The critical sate line is the specific volume on the critical state line at p’c = 167 kPa is calculated using
Eq 5 as
vc = 1.6781 [1 mark]
Now, the total volume Vt = Vs + Vv = Vs.v
where Vs is the volume of soil solids and Vv is the volume of voids.
As Vs is constant, the change in total volume Vt = Vs × the change in specific volume v
That is, Vt = 875 ml × (1.797 – 1.681) = 101 ml [1 mark]
(d) A third specimen having the same previous stress history as the first two was subjected to an
undrained shear test from an effective cell pressure of 50 kPa. Use the Cam clay model to calculate
the volume of water taken into the specimen on reducing the cell pressure from 100 kPa to 50 kPa,
7
and the theoretical values of q and p′ at yield. Sketch the stress path (q vs p′) on your graph from
part (b). [5 marks]
The volume of water taken in by the specimen on reducing the cell pressure from 100 kPa to 50 kPa is
calculated from the increase in specific volume as the state of the specimen moves up the unloading
(κ) line,
= .
2
′ 1
′ (Eq 10)
with p′2 = 50 kPa and p′1 = 100 kPa, giving v = 0.035. Multiplying this by the volume of solids Vs =
875 ml gives a volume of water taken in of 30.3 ml. [1 mark]
According to Cam clay, yield will occur at p′ = 50 kPa, and q at yield can be calculated from the
equation of the yield locus (Eq 2) with p′ = 50 kPa and p′o = 200 kPa (unchanged from the original
value as the specimen has only swollen), giving
p′y = 50 kPa, qy = 83.2 kPa [2 marks]
The stress path up until yield is shown on the diagram at the end of the solution. [2 marks]
(e) Why is this calculated value of q unlikely to be achieved in reality [2 marks]
The calculated stress state is well in excess of the critical state stress ratio. the specimen is now quite
heavily overconsolidated, and is likely to fail by rupture along a slip plane as stresses and strains
localise, rather than the continuum failure (stresses and strains uniform and continuous throughout
the specimen) assumed by Cam clay. [2 marks]
Fig. 1: (a) q vs p′ and q vs p plots; (b) v vs lnp′ plots Total: 25 marks
0
50
100
150
200
250
0 50 100 150 200 p’ and p, kPa
q vs p’, Test 3
q vs p’,
Test 1
q vs p,
Test 2
CSL
1.5
1.6
1.7
1.8
1.9
2
2.1
3.5 4.5 5.5
lnp’ (p’ in kPa)
Test 1
kappa
line
q vs p, Test 1
ISO NCL
CSL
q, kPa
v
8
Q3. Figure Q3 shows a cross section through one half of a canal embankment. The embankment is
underlain by permeable chalk, with a discontinuous layer of clay between the embankment and the
chalk as indicated. The canal is lined with a low permeability material such that there is a 0.5 m head
drop across it.
Figure Q3: Cross section through one half of a canal embankment (NOT TO SCALE)
(a) Construct a flownet for seepage from water from the canal into the underlying chalk. Assume
that the embankment material remains saturated so there is no need to find the phreatic surface
(top flowline) while drawing the flownet. Label the head value of each equipotential. Calculate the
rate of leakage from the canal, in litres per hour per metre length of the embankment. [15 marks]
(b) Explaining your reasoning, determine the position of the line of zero gauge pore water pressure
within the embankment. Calculate the maximum negative pore water pressure (suction) within the
embankment, according to your flownet. Using a quantitative argument, discuss whether the
assumption that the soil above this line remains saturated is reasonable. Take the surface tension of
water T = 7 × 10-5 kN/m. [6 marks]
(c) Comment on the assumption that the flow regime is symmetrical about the centreline of the
embankment, with reference to the likely continuity of the clay layer. What would be the effect on
the flow if there were no discontinuity in the clay layer on the other side of the embankment, and
what might be the consequences [4 marks]
Solution
(a) Construct a flownet for seepage from water from the canal into the underlying chalk. Assume
that the embankment material remains saturated so there is no need to find the phreatic surface
(top flowline) while drawing the flownet. Label the head value of each equipotential. Calculate the
rate of leakage from the canal, in litres per hour per metre length of the embankment. [15 marks]
Embankment fill, permeability k = 10-6 m/s
Impermeable clay layer
Permeable fractured chalk
Retaining wall
(impermeable)
Canal 2 m 1 m
1 m
6.5 m
10 m
3.5 m
1.5 m
2 m
GWL in
chalk
1.5 m
9
The flownet, drawn according to the usual rules, is below. As stated in the question, the usual
“phreatic surface” requirement is ignored, which means that the upper surface of the embankment is
assumed to be the bounding flowline.
Fig. 2: Flownet for Q3 (cross section must be drawn to scale)
[10 marks for the quality of the flownet, plus 2 for correct labelling of the equipotentials]
The flowrate is calculated from the formula
= . .

where the symbols have their usual meaning. In this case, k = 10-6 m/s; the head drop H = 4.5 m; NF =
2 × 2 for symmetry = 4, NH = 6, giving
= (10 6 / ) × (4.5 ) ×
4
6
= 3 × 10 6 3

Multiply by 3600 to convert from per second to per hour, and by 1000 to convert from m3 to litres,
giving
q = 10.8 litres per hour, per metre run
[3 marks for correct identification of parameters and calculation]
(b) Explaining your reasoning, determine the position of the line of zero gauge pore water pressure
within the embankment. Calculate the maximum negative pore water pressure (suction) within the
embankment, according to your flownet. Using a quantitative argument, discuss whether the
10
assumption that the soil above this line remains saturated is reasonable. Take the surface tension of
water T = 7 × 10-5 kN/m. [6 marks]
In general, the point of zero gauge pore water pressure on the x m equipotential is at a height of x m
above the datum for measurement of potential. (This is because the manometer height above the
point of interest, which indicates the pressure head, is then zero). These points are indicated on the
flownet and joined up to denote the line of zero gauge pore water pressure. [3 marks]
The maximum suction corresponds approximately to the maximum height above the zero pressure
elevation at which an equipotential intersects the soil surface. From the flownet, this increases as the
edge of the embankment is approached. Taking the 0.75 m equipotential as the last calculable
location gives a height (scaling off the flownet) of about 1.5 m, corresponding to a maximum suction
of about 15 kPa. [1 mark]
The simple air entry analysis for a contact angle α = 0 during drying gives a suction at air entry
= 4

while Hazen’s formula can be used to relate the pore size d (in mm) to the permeability k (in m/s):

= 0.01 × 10
2 ( 2)
In the present case, k = 10-6 m/s implying d10 = 10-2 mm and ue = 28 kPa. Hence the assumption that
the embankment remains saturated is just about reasonable in this case. [2 marks]
(c) Comment on the assumption that the flow regime is symmetrical about the centreline of the
embankment, with reference to the likely continuity of the clay layer. What would be the effect on
the flow if there were no discontinuity in the clay layer on the other side of the embankment, and
what might be the consequences [4 marks]
If the clay layer is natural, it will likely continue at about the same angle of dip below the other side
of the embankment. This would be enough to make the flownet unsymmetrical. It is in addition likely
that the clay missing from the toe of slope on the right hand side (indicated in the diagram) was
removed during construction of the embankment. Hence it is likely that the clay layer on the other
side of the embankment is intact. If the natural ground surface slopes, it is unclear what the depth of
the embankment would be on the left hand side, or what the toe drainage arrangements might be.
Thus the assumption that the flownet is symmetrical is highly questionable. [2 marks]
Without a “gap” in the clay layer, the bottom flowline would continue along the upper surface of the
clay layer. Depending on the topography and any drainage provision, the flowline could exit the
embankment at the toe (assuming the embankment continues to meet the clay layer). Other
flowlines might also exit the embankment at the soil surface, which could then become the line of
zero gauge pore water pressure. The associated increased pore pressures could cause instability of
the embankment slope, and the emergence of uncontrolled seepage could cause erosion of the
embankment surface. [2 marks]
Total: 25 marks
11
QUESTION 4
Q4 Figure Q4 shows a cross section through the foundation of a bridge abutment, located
on stiff clay overlying bedrock on the side of a valley. A potential failure mechanism,
consisting of two sliding blocks, B and C, is shown. In this question you will develop an
upper bound plasticity solution for this mechanism.
The bridge abutment, A, is subjected to a vertical load per unit length (into the page), ,
from the bridge. The bridge structure constrains the abutment to move straight
downwards if the foundation fails.
Figure Q4: Foundation of bridge abutment
a Calculate the angles and and construct a hodograph (velocity diagram) showing the
movement of the abutment, A, and blocks B and C on failure, based on the mechanism
shown. Scale the diagram such that the vertical movement of the abutment is one unit.
[5 marks]
Distance RS – from trigonometry (also 5-12-13 right angle triangle), RS = 13w/6 [1 mark]
Angles: = tan 1
3
4
= 36.87°, = tan 1 5
12 = 14.25° [1 mark]
[3 marks for hodograph]
b Derive an expression for the energy dissipated on the failure planes for a unit movement
of the abutment as a function of and . Assume that the clay-abutment (PQ) and
clay-bedrock (RS) interfaces mobilise the full shear strength of the clay, . It is
recommended to begin by listing the failure planes and tabulating their parameters.
[4 marks]
Stiff
clay
Bedrock
Abutment, A
B
C
V
Distances
PQ = w
PR = QR = 5w/6
QS = 2w
P Q
R
S
α α
β α
w
Origin,
bedrock, o
a b
α
c
β
α α
1
0.75
0.85
1.3
0.32
12
Failure plane Length Sliding displacement Work component
A vs. B → PQ 0.75 3
4 = 0.75
O vs. B → RP 5
6
1.25 25
24 = 1.04
B vs. C → RQ 5
6
0.85 17
24 = 0.71
O vs. C → RS 13
6
1.3 169
60 = 2.82
Summing components: work done per unit movement of abutment =
319
60
= 5.32
[2 marks for correct approach, additional 2 marks for correct numerical result]
c Calculate the vertical load, (kN/m), that will cause the failure mechanism to occur, for
= 50 and = 6 . Assume that the soil is weightless.
[3 marks]
Writing the work equation, i.e. input = dissipation, but cancelling the unit displacement
on each side:
= 5.32 = 5.32 × 6 × 50 = 1595 /
[2 marks for correct approach, additional 1 mark for correct numerical result]
d Now consider that the influence of the soil weight. Calculate the potential energy
released by the movement of the clay blocks, B and C, for a unit movement of the
abutment as a function of soil unit weight γ.
[4 marks]
Clay block Area Vertical movement PE component
B 1
2
× ×
2
3
= 2
3
1 2
3
C 1
2 × 2 ×
5
6
= 5 2
6
8
25
4 2
15
Total release of potential energy =
3 2
5
[1 mark for areas, 2 marks for vertical movements, 1 mark for correct final result]
e Recalculate the vertical load, V, that will cause the failure mechanism to occur, allowing
for the influence of the soil weight. Assume that = 50 , = 6 and =
20 / 3.
[3 marks]
Writing the work equation, i.e. inputs (from V and PE) = dissipation, but cancelling the
unit displacement on each side:
+
3
5
2
= 5.32
= 5.32 3
5
2
= 5.32 × 6 × 50 (3 × 20 × 62)
5 = 1163 /
[2 marks for attempting work equation with all terms, 1 mark for correct result]
13
f Unexpectedly, the bridge abutment collapses under a vertical load of = 600 / .
Subsequent investigations suggest that this is caused by a weakening of the clay at the
clay-bedrock interface (RS), so that the available shear strength on this plane is , =
, where 0 < < 1. To assess the credibility of this explanation, calculate the apparent reduction factor on shear strength, , by extending the solution developed in part (e). [3 marks] The component of the work dissipation from RS is modified to 169 60 = 2.82 so that the updated work dissipation is given by: 5 2 + 169 60 The updated work equation becomes: + 3 5 2 = 5 2 + 169 60 Substituting in for specified values = 20 / 3, = 6 , = 50 and = 600 / the interface strength reduction factor is found as: = 60 169 + 3 5 2 5 2 = 0.33 [2 marks for correctly modifying work equation, 1 mark for correct result] g The upper bound failure mechanism developed in this question may not be optimal, and so could over-estimate the foundation capacity. Sketch two simple alternative failure mechanisms that could be considered, indicating their geometry and kinematics. [3 marks] Three examples are shown below. Any credible and admissible options are acceptable. [2 marks for one correct suggestion, 1 extra mark for a second suggestion] END OF PAPER Stiff clay Bedrock Stiff clay Bedrock Shear fan (not rigid) Rigid block Stiff clay Bedrock Rigid block Rigid block