Question 1
a. Taking into consideration various ground and loading conditions, discuss four
methods which may be used to reduce the order of magnitude of the total and
differential settlement beneath large flexible rafts. 8 marks
b. The ground conditions at the proposed location of the construction of an 8m by
4m flexible foundation are given in Table Q1. During the site investigation it was
also established that the water table varies between ground level and 1m below
ground level.
The base of the foundation will be located at a depth of 1m below ground level
and the foundation will carry a maximum gross loading intensity of 120kN/m2
.
Stating appropriate assumptions and using the information given in Figures Q1A
and Q1B, for a design life of 30 years calculate the total settlement beneath the
centre and corner of the foundation. 13 marks
c. Comment on the values of total and differential settlement established in part (b).
In addition, if considered appropriate, make a recommendation to reduce these
settlement values in accordance with one of the methods discussed in part (a).
4 marks
(Total 25 marks)
Depth below ground level (m) Soil Type Soil Properties
0.0 – 5.0 Silty CLAY E = 40 MN/m2
. mv = 1.00 x 10-4 m2
/kN
= 0.5 b = sat = 20kN/m3
.
??
′ = 0.0008
5.0 – 11.0 CLAY E = 50 MN/m2
. mv = 0.85 x 10-4 m2
/kN
= 0.5. b = sat = 21.0kN/m3
.
??
′ = 0.0006
11.0+ Well jointed
ROCK
INCOMPRESSIBLE STRATUM
Table Q1
Page 2 of 12
Question 2
a. In layered soils, show how the following expressions can be used to
determine the average coefficients of permeability parallel and
perpendicular to the bedding planes.
??
̅̅̅ =
?1?1 + ?2?2 + ?3?3 … … … . ????
?
4 marks
??
̅̅̅ =
?
?1
?1
+
?2
?2
+
?3
?3
- … … …
??
??
5 marks
b. Discuss a 10 Point Plan for a Systematic Approach to planning a monitoring
program using Geotechnical Instrumentation.
16 marks
Page 3 of 12
Question 3
3a Review the methods available for circular slip slope stability analysis in terms of both
total and effective stress and explain the circumstances in which each method would
be used. 10 Marks
b Figure Q3 illustrates a trial slip surface in clay (not to scale). Sufficient data is
included within the table on this figure (Table Q3) to complete the table and analyse
the trial slip surface using Bishop’s Simplified analysis. Only two rounds of the
iteration process need be completed. Comment on the results obtained. 15 marks
(Total 25 marks)
Return Figure Q3 with your answer script.
Soil Properties
c' = 20kN/m2
ɸ' = 33
b = sat = 20kN/m3
See Equation Sheet
Question 4
a. Determine ϕ’, the effective angle of friction along the discontinuity shown in
Figure Q4, assuming the slope is on the point of failure when the tension
crack is full of water, c’=27kN/m2 and γ(rock)=24kN/m3
.
12 marks
b. Demonstrate the importance of controlling the supply of water to the tension
crack and discontinuity.
5 marks
c. Suggest methods of stabilising the slope and explain in detail how the slope
could be drained.
8 marks
(Total 25 marks)
Page 4 of 12
Question 5
a. Sketch and explain the effective stress paths developed for any three of
the four common types of stress/strain tests.
6 marks
b. Table Q5 shows the results of consolidated undrained tests on clay.
i. Using the graph paper provided plot these results on a t/s(s’) graph.
10 marks
ii. Plot the Kf line and the modified total failure envelope and determine
the effective shear parameters.
3 marks
iii. Plot the total and effective stress paths and plot the drained stress
path for the specimen tested at a confining pressure of 300kN/m2
.
3 marks
iv. Plot the Ko stress path assuming Ko=0.6.
3 marks
Confining pressure
(σ3) kN/m2
Deviator stress at
failure (σ1-σ3) kN/m2
Pore water pressure
at failure (u) kN/m2
100 80 59.9
200 145 105.1
300 214 144.8
Table Q5
Put your name on and return the stress path diagram plotted on graph paper with your
answer script
(Total 25 marks)
Page 5 of 12
Question 6
a. With the aid of a diagram show the assumed distribution of the strain
influence factor with depth for a square and strip foundation in granular soil
according to Schmertsmann.
6 marks
b. A 4m square foundation is to be founded at 2m below ground level in a
deep deposit of granular soil. The soil has a unit weight of 19kN/m3 and the
water table is at a depth of 10m. The results from a Dutch cone
penetrometer test are presented in Table Q6 and the foundation will impose
a net foundation pressure of 190kN/m2 at embedment depth.
Determine the settlement under the foundation
i. Immediately
10 marks
ii. After a design life of 25years
6 marks
iii. Comment on your results
3 marks
Depth (m) 1.4-2.8 2.8-4.0 4.0-5.8 5.8-7.7 7.7-11.5
Cone Resistance (MN/m2
) 2.6 4.9 7.0 7.5 7.3
Table Q6
(Total 25 marks)
Put your name on and return the stain factor diagram plotted on graph paper with your
answer script.
Page 6 of 12
UNIVERSITY OF SOUTH WALES
Prifysgol De Cymru
Faculty of Computing, Engineering
and Science
NG3S109
Geotechnics – Analysis and Design
Scale: N/A
Figure Q1A
Page 7 of 12
UNIVERSITY OF SOUTH WALES
Prifysgol De Cymru
Faculty of Computing, Engineering
and Science
NG3S109
Geotechnics – Analysis and Design
Scale: N/A
Figure Q1B
Page 8 of 12
Return with answer script
Student Number ………………………………
Slice
No.
b
(m)
W
(kN/m
run)
u.b
(kN/m
run)
c'b
(kN/m
run)
c’b+
tan'(W-ub)
(kN/m run)
Wsin
(kN/m run)
1 .
(cosα+(sinαtanφ/F))
Page 9 of 12
UNIVERSITY OF SOUTH WALES
Prifysgol De Cymru
Faculty of Computing, Engineering
and Science
NG3S109
Geotechnical & Environmental
Engineering
Scale: N/A
Figure Q4
Page 10 of 12
Equation Sheets
Consolidation
??,?+1 = ??.? + ?(??−1,? + ??+1,? − 2??,?)
??
??,?+1 = ??.? + ?(2??−1,? − 2??,?)
? = ??
∆?
∆?
2
(1 − ?) = (1 − ??)(1 − ??)
?ℎ?? ?? ≤ 60%:
?? =
?
4
(
??%
100 )
2
?ℎ?? ?? ≥ 60%:
?? = 1.781 − 0.933???10(100 − ??%)
?? =
??
?
?
2
?? =
?ℎ?
4?2
? = 0.525? ?? ? = 0.564?
???? = ?? ? ?? ? ∆?′
?? =
??−??
1+??
?
1
∆?′
Steinbrenner
?? = ?
?
?
[(1 − ?
2
)?1 + (1 − ? − 2?
2
)?2
]
Schmertmann
??? = 0.5 + 0.1 (
??
?
′
?
)
0.5
?? = ?1?2∆?∑(
??
??
) ∆?
2?
0
?1 = 1 − 0.5 (
?
′
?
??
)
?2 = 1 + 0.2???10 (
?
0.1
)
Burland and Burbridge
? = ????????
[(?′? −
2
3
?′?) ??
0.7???
]
? = ????????
[(?′?
)?
0.7???
]
?? = (
1.25?
?
?
?
? - 0.25)
2
?? =
?
??
(2 −
?
??
)
Page 11 of 12
?? = 2.5?? ??? ?
?
⁄ = 1
?? = 3.5?? ??? ?
?
⁄ = 10
?? = 1 + ?3 + (??????10
?
3
)
?3 = 0.3 ?? 0.7 & ?? = 0.2 ?? 0.8
Secondary Settlement
???? =
??
1 + ??
? ?????10 (
?2
?1
)
???? = ?′?? ?????10 (
?2
?1
) ?ℎ??? ?′? =
??
1 + ??
Permeability
?
′ = √????
? = ?′?
??
??
Parabolic Equations
? =
?
2 − 4??
2
4??
?ℎ??? ?? = √?
2+?
2 − ? ??? ∆? = 0.35(? + ∆?)
Rock Slope Stability
? =
?? + (? cos ?? − ? − ? sin ??)????
? sin ?? + ? cos ??
p A (H z) cos ec
w w V z
2
2
1 w w p U z (H z) cos ec 2
1
[(1 ( ) ) cot cot ] 2
1 2 2
H p f W H z
Page 12 of 12
Slope Stability
Fellenius Method
? =
?′ ∑ Δ? + ????′ ∑(Δ????? − ?Δ?)
∑ Δ????? +
???
?
Bishop’s simplified equation
? =
∑[?
′? + (Δ? − ??)????′]
∑ Δ????? (???? +
???? ????′
?
)
??????? ?????, ?? =
2?
?√??
Δ? =
2???
360
Janbu Wedge analysis
? =
?
′? + (? − ??)????′
?? − ??
Translational Slip
? = [(1 − ?)? + ?????]????2?
? = [(1 − ?)? + ?????]?????????
? = ???????2?
Stress Paths
????????
′
????????? ∶ ∆? = ?[∆?3 + ?(∆?1 − ∆?3)]
? =
∆?
∆?3
? =
∆?
?(∆?1)
?
′ =
?′1 + ?′3
2
??? ?
′ =
?′1 − ?′3
2
?
′ = ? − ?
?′3 = ????′1
Sample Solution