the pore water presure acting on the base (A.w.[z+hcrit]) begins to exceed the weight of the
block of soil (A.z):
A.w.[z+hcrit] = A.z
z( - w) = w.hcrit
or
icrit = hcrit/z = ( - w)/w
(main text Equation 3.33)
In the present case, we will assume = 2w giving
q = A.k.i k = q/Ai, where A = 1122 mm2. (With A in mm2 and the flowrate q in mm3/sec, the
permeability k is calculated in mm/s)
Hydraulic gradient, i 1
2
5
10
-3
-3
-3
kA, mm/s (from test 1) 6.3 10 6.3 10 6.3 10
-4
kB, mm/s (from test 2) 2.55 10 2.63 10-4
Q3.4 Solution
You will need to derive the formulae (main text Equations 3.21 and 3.22) for the equivalent
bulk horizontal and vertical permeabilities of an alternating layer system, as in the main text
Section 3.6.
For horizontal flow (ie flow parallel to
Q3.3 In the constant head permeameter test described in Example E3.2, the sample was found
to fluidize in upward flow at a hydraulic gradient of 0.84. Estimate the unit weight of the soil
in its loosest state.
[University of London 2nd year BEng (Civil En
Q3.2 Solution
Diagram of falling head permeameter: see main text Figure 3.10
The derivation of the equation follows the main text Section 3.4.2.
At the start of the test (time t=0), the water level in the upper (small-bore) tube is at a height
h0 above th
QUESTIONS AND SOLUTIONS: CHAPTER 3
Laboratory measurement of permeability; fluidization; layered soils
Q3.1 Describe by means of an annotated diagram the principal features of a constant head
permeameter. Give three reasons why this laboratory test might
the given value of 'apparent. Remember that the rotation on the Mohr circle must be divided by
2 to give the actual rotation in the physical plane.
apparent
= 18
P
T
S
4
apparent
O
t
1
21
90 + apparent
90 - apparent
18
s
C
R
Q
Figure Q2.8a: Mohr circle o
The failure envelopes sketched in Figure Q2.7c show that
'crit is constant (= 32.5, closer to 'peak max = 32 than the estimate of 35 based
on 'peak 0.8 max) because there is no dilation at the critical state
'peak reduces as the normal effective stress
/2 %
V
0.167
%
0.083
max = 14
H
Figure Q2.7b: Mohr circle of strain increment leading up to peak
From Figure Q2.7b, the angle of dilation at peak is given by
max = y/x = 2.5/10 max = 14
We might expect 'crit ~ 'peak 0.8 max (main text Equation 2.14), giv
kN/m2
60
40
20
46
90 - 46
0
100
kN/m2
Figure Q2.7a: Mohr circle of stress
The first plane of maximum stress ratio is horizontal (this is an assumption that has to be
made to draw the Mohr circle of stress). From Figure Q2.7a, the second plane of maximum
0.5 'v.tan
which increases linearly from zero at the top of the pile to
0.5 [(18 kN/m3 20 m) - (9.81 kN/m3 20 m)] tan20 = 29.81 kPa at the base
Hence
T = [(0 + 29.81 kPa) 2] [( 0.5 m) 20 m] = 468 kN
Stress analysis and interpretation of shearbox test data
At the soil surface
At the water table
At the base of the pile
z, m
0
5
25
v, kPa
0
95
475
u, kPa
0
0
196.2
'v, kPa
'h, kPa
, kPa
0
95
278.8
0
47.5
139.4
0
23.17
67.99
The surface area of the upper 5m of the pile is ( 0.3)m 5m = 4.71m2, and the average
sh
Use of strength data to calculate friction pile load capacity
Q2.5 A friction pile, 300 mm in diameter, is driven to a depth of 25 m in dense sand of unit
weight 19 kN/m3. The ratio of horizontal to vertical effective stresses is 0.5. The angle of
frictio
(b) At a depth of 3m below the top of a dry embankment made of this sand, the vertical
effective stress is 3m18kN/m3 = 54kPa. This corresponds to a hanger load of 20kg, at which
the peak shear stress is approximately 36.4 kPa
(c) In the 1:10 scale model,
Determination of peak strengths
Q2.4 The following results were obtained from a shearbox test on a 60 mm 60 mm sample
of dry sand of unit weight 18 kN/m3.
Zero force
Peak shear force for a hanger load of 3kg
Peak shear force for a hanger load of 10kg
Peak
Development of a critical state model
Q2.3 Mining operations frequently generate large quantities of fine, particulate waste known
as tailings. Tailings are generally transported as slurries, and stored in reservoirs impounded
by embankments or dams made
Q2.2 Solution
Typical graphs of (a) shear stress against shear strain ; (b) volumetric strain vol against
shear strain ; and (c) specific volume v against shear strain are as shown in main text
Figure 2.21.
In the test carried out on the initially dense s
QUESTIONS AND SOLUTIONS: CHAPTER 2
The shearbox test
Q2.1 Describe with the aid of a diagram the essential features of the conventional shearbox
apparatus. Stating clearly the assumptions you need to make, show how the quantities
measured during the test
hence
v = 2.65 (1.14) (1000/2018) = 1.497
(The void ratio e = v-1 = 0.497)
The saturation ratio Sr is calculated using main text Equation 1.10,
Sr = w.Gs/e = w.Gs/(v-1)
(main text Equation 1.10)
Sr = 0.14 2.65 0.497 = 0.746 or 74.6%
14
Hence
Mass of tin empty, g
Mass of tin + sample wet, g
Mass of tin + sample dry, g
w, %
Density, kg/m3
Dry density, kg/m3
14
88
81
10.45
1730
1566
14
68
62
12.5
1950
1733
14
98
87
15.07
2020
1755
14
94
82
17.65
1930
1640
14
93
80
19.70
1860
1554
Figure Q1
Compaction
Q1.9 The following results were obtained from a standard (2.5 kg) Proctor compaction test:
Mass of tin empty, g
Mass of tin + sample wet, g
Mass of tin + sample dry, g
Density, kg/m3
14
88
81
1730
14
68
62
1950
14
98
87
2020
14
94
82
1930
14
93
Mass of tin
empty, g (mt)
Mass of tin +
sample wet, g
(ms+mw+mt)
Mass of tin +
sample dry, g
(ms+mt)
ms/mw, %
Cone
penetration d,
mm
ln(d)
18.2
19.1
17.7
18.6
51.5
45.5
50.7
43.4
37.8
35.6
39.7
36.3
69.9
25.0
60.0
14.2
50.0
8.5
40.1
5.1
3.219
2.653
2.14
1
Soil PSD
100
Filter PSD
90
Percentage passing
80
70
E
60
50
40
30
20
B
A
10
C
0
0.001
0.01
D
0.1
1
10
100
Particle size (m m)
Fine
Medium
Coarse
Fine
Medium
Coarse
Fine
Medium
Coarse
CLAY
SILT
SAND
GRAVEL
Figure Q1.7: Particle size distribution curves for
size ranges. To convert this to a percentage of the total sample, we must multiply by 121g and
divide by 399g. Hence
Size, m
% of pan sample
<2
33
2
Size, m
% of pan sample 33
smaller than this size
% of total sample 10.0
smaller than this size
2-10
24
10
Reading off from the curve,
D10 0.0035 mm (3.5m)
D60 0.52 mm
Hence the uniformity coefficient U = D60/D10 150 (148.6)
The soil is approximately 40% silt, 50% sand and 10% fine gravel: this makes it a sandy SILT
according to the system given in Table 1.5.
The vertical total stress v = 19.99kN/m3 0.4245m = 8.486kPa. Hence the vertical
effective stress 'v = v - u = 8.486kPa - 4.164kPa 'v = 4.322kPa
(c) In the dense, saturated state, the weights of water and soil grains above the base do not
change. Hence the
(a) Beneath the office block, the vertical total stress is increased by the surcharge of 90kPa,
giving
v = 132kPa + 90 kPa v = 222 kPa
The pore water pressure u is unchanged, u = 29.4kPa
The vertical effective stress 'v = v - u = (222kPa - 29.4kPa)
'v = 1