Soil Flashcards
1
Q
How to characterise soils? *
A
- particle die distribution curves
- atterberg limits
- moisture
- colour
- fabric
- strength
- compressibility
2
Q
What is a particle size distribution curves?
A
Percentage passing through sives vs. Particle size (mm). Silt -> Sand -> Gravel
3
Q
What is an Atterberg limits Casagrande chart?
A
Plasticity vs. Liquid limit. A line across with clays at the top left and silts bottom right.
4
Q
What is the problem if silt and sand partings in rock?
A
When you dig a hole you relieve stress from the blocks which means that they will move along the silt and sand partings where water is.
5
Q
Vv
A
Volume of liquid
6
Q
Vs
A
Volume of solids
7
Q
V
A
Volume
8
Q
Fabric
A
Spatial and geometric configuration of all the elements that make up a rock
9
Q
M
A
Overall mass of soil sample
10
Q
Vs= Ms/ (Gs*ρw)
A
Solid volume = solid mass / specific gravity * water density
11
Q
ν = V/ Vs
A
Specific volume = total volume/ solid volume
12
Q
e= Vv/ Vs
A
Void ratio = liquid volume/ solid volume
13
Q
n= Vv/ V
A
Porosity = liquid volume/ total volume
14
Q
ν = 1 + e
A
Specific volume = 1 + void ratio
15
Q
e= n/ (1-n)
A
void ratio = porosity/ (1-porosity)
16
Q
Saturated soil?
A
All voids completely fill with water
17
Q
Partly saturated soil (unsaturated soil)?
A
Some voids contain water
18
Q
γ = ρg
A
Unit weight (kN/m^3) = density * gravitational acceleration
19
Q
γw =
A
9.81 Kn/m^3
20
Q
σa= F/A
A
Axial stress = force / area
21
Q
E= σa / εa
A
Young’s modulus = stress/ strain
22
Q
εa = δL/ L
A
Strain = change in length / original length
23
Q
Principal stress?
A
Total force on a soil point
24
Q
Effective principal stress?
A
Total force - pore pressure
25
Pore pressure?
(u) stress caused by the water acting in every direction with equal intensity
26
Suction?
Negative pore water pressure
27
S= ua - uw
Suction = pore air pressure - pore water pressure
28
What is pore water pressure at the water table or phreatic surface?
Zero
29
Perched water table?
Water lies on a stratum of low permeability above the level of the normal water table
30
Artesian conditions?
Water is confined by an overlying layer of low permeability and fed from a distant source where the water table is at a higher elevation
31
Seepage?
Flow through a soil
32
During steady state seepage what happens to pore pressure?
Pore pressures remain constant and no soil deformations occur
33
What controls the flow in seepage?
The hydraulic driving head and in particular its gradient
34
h= u/γw + z + v^2/2*g
h= total head (potential head)
u= pore water pressure
γ= bulk unit weight
z= elevation above a chosen datum
v= flow/seepage velocity
g= gravitational acceleration
35
q= Aki or. v=q/A = ki (Darcy’s law)
q= volume of water flowing per unit time (m^3/s)
A= cross- sectional area of soil corresponding to flow q (m^2)
k= coeff of permeability (m/s)
i= hydraulic gradient
v= discharge velocity (m/s)
36
What is the relationship between coefficient of permeability and particle size of soil.
The larger the particle size the larger the coefficient of permeability
37
Hydraulic gradient (i)?
The rate of change of total head with distance in the direction of flow
38
i= -dh/ds
i= hydraulic gradient
dh= change in total head
ds= the distance between the flow line between the two points being compared
39
u=p γw (finding pore pressure using standpipe)
pore pressure = height of water in the standpipe * bulk unit weight of water
40
Quick condition?
The upward water pressure gradient and water flow reduce the effective stress
41
icrit - critical hydraulic gradient?
The hydraulic gradient when quick conditions occur
42
icrit= (γs/ γw) - 1
icrit = critical hydraulic gradient
γs = bulk unit weight of soil
γw= water unit weight
43
Assumptions for Darcy’s law (3)?
- homogeneous, porous, saturated, constant volume medium
- fluid homogeneous and incompressible
- flow is steady and continuous (steady state)
44
(dh^2dx^2) + (dh^2/dy^2) = 0
Laplace equation
dh= change in total head
dx= change in x direction
dy= change in y direction
45
How can you show flow lines and total head lines?
You can draw flow lines as orthogonal trajectories against equal total head lines.
46
EP lines?
Equipotential lines. These represent lines of equal total head
47
FL lines?
Flow lines enclose flow ‘channels’ within which flow is constant
48
H= Nd Δh (orthogonal trajectory flow lines)?
Δh= head drop between adjacent EP lines
H= total head drop across system
Nd= number of potential drops (6?)
49
Stages in flow net construction?*
- identify boundary condition relating to flow lines and equipotentials e.g. impermeable boundary usually coincides with a flow line; open water surface constitutes an equipotential
- sketch trial flow lines remembering that flow ‘channels’ are like tubes within which flow occurs. Flow lines do not cross one another, nor do they converge completely
- sketch trial equipotential lines such they are perpendicular to the flow lines and so that the enclosed areas between flow lines and equipotential lines form ‘curvilinear squares’
- it is convenient if EPs drawn with equal head losses between adjacent EPs
50
Δq = kH / Nd (flow through incremental channel)
Δq = change in vol of water flowing per unit time
K= coeff of permeability
H= total head across system
Nd= number of potential drops
51
q = kH Nf / Nd (flow through system)
q= vol of water flowing per unit time
k= coeff of permeability
H= total head drop across system
Nf= number of flow channels
Nd= number of potential drops
52
What is the impact on the change in total head over distance as EPs are closer of that given distance?
The closed the EPs are over a given distance, the greater the change in total head over that distance and hence the greater the hydraulic gradient
53
(Gs + S*e) * γw = γ
—————
(1+e)
Gs= specific gravity
S= saturation
e= void ratio
γw = unit weight of water
γ= unit weight of soil
54
e* S = w* Gs
e= void ratio
S= saturation
w= water content
Gs= specific gravity
55
γdry = (Gs) * γw
——
(1+e)
γdry = unit weight of dry soil
Gs= specific gravity
e= void ratio
γw = unit weight of water
56
w= Mw/ Ms
Water content = water mass / solid mass
57
Sr = Vw/ Vv
Saturated = water volume / liquid volume
58
Gs= ρs/ ρw
Specific gravity = density solid / water density
59
When soil is not isotopic with respect to permeability, can FL and EP lines be orthogonal?
No
60
Kx?
Coefficient of permeability in the x direction
61
What happens when you change principal stress in granular soil?
Δ σ=Δσ´ drained condition
62
What happens when you change principle stress for fine grained soil?
Δ σ = Δu undrained condition
63
Consolidation?
When effective stress increases in fine-grained soil, the pore water pressure will increase. The increase in pore water pressure Δu is termed an excess pore pressure which induces a pressure gradient, resulting in a transient flow towards free- drowning boundaries of the soil layer. Flow continues until the excess pore water pressure have dissipated. In the long term when t=tinf and the excess pore water pressure have dissipated, the increment of total stress is carried entirely by the soil skeleton Δ σ = Δσ´ and is once again drained conditions. As drainage takes place the particles re orientated themselves and the soil compresses as inter-particle forces increase and there is a reduction in volume
64
Oedometer test?
Measured compression and time on consolidation of settlement
65
ΔH/ H0 = Δe/ (1+e)
Change in height/ original height = change in void ratio/ 1+ void ratio
66
What do you plot from oedometer data to understand consolidation?
e vs σv’
67
How does compression change in over consolidated and normally consolidated clays?
The settlement/ compression of over- consolidated clays is much less than normally consolidated clays. Therefore beware of exceeding pre consolidation pressure in over consolidated clays
68
Over consolidated?
Clays that have been loaded up in the past to effective stresses greater than those acting at present
69
3 causes of unloading?
- melting of ice sheets
- erosion of upper layers
- rise in ground water table (affecting u rather than σv)
70
mv = 1/(1+e0) (Δe/ Δ σ´)
Mv = coeff of volume compressibility
e0= original void ratio
e1= new void ratio
σ1´= new effective stress
σ2´= old effective stress
71
mv= 1/H0 * (ΔH/Δσ´)
Μv= coeff of volume compressibility
H0= original height
H1= new height
σ´1= new effective stress
σ0´= original effective stress
72
Assumptions of theory of one-dimensional consolidation?
- soil is homogeneous
- soil is fully saturated
- solid particles & water are incompressible
- compression and flow are one- dimensional
- strains are small
- Darcy’s law valid at all hydraulic gradients
- the coefficient of permeability, k, and the coefficient of volume compressibility, mv, remain constant throughout the process
- there is a unique relationship, independent of time, between void ratio and effective stress
73
Cv= d2u/dx2 = du/dt
Cv= coeff of consolidation
du= change in pore water pressure
dz= change in distance along z
dt= change in time
74
uz = Σ (2ui/M) * (sin(Mz/H))*e^(-M^2*Tv)
limits= m=inf and m=0
uz= value of the excess pore water pressure at depth z at time t
ui = initial excess pore water pressure, uniform over depth z
M= 0.5*π*(2m+1) where m= positive integer varying from 0 to infinity
Tv= dimensionless time factor
H= length of drainage path
75
Tv= Cv * t / H^2
Tv= dimensionless time factor
Cv= coeff of consolidation
H= length of drainage path
t= time
76
What is the length drainage path for 2 way drainage - permeable of both sides?
H/2
77
What is the length drainage path of a 1 way drainage system - flow is one direction with one permeable side.
H
78
Where in a shape of soil does consolidation immediately occur?
Top and bottom of shape
79
U= (ui-ue)/ui = 1-ue/ui
U= average degree of consolidation
ui= initial excess pore water pressure, uniform over depth z
ue= value of the excess pore water pressure when σ´ is acting within the soil
80
Cc= Δe / log(σ1) -log(σ0)
Compressive index = change in void ratio/ over change in logs of effective stress
81
ρinf = mv* Δ σv’ * Η
Long term settlement = coefficient of volume compressibility * change in effective stress * drainage height
82
cv = kz/ γw*mv
Cv= coeff of consolidation
Kz= permeability in the z direction
γw = unit weight of water
mv = coeff of volume compressibility
83
U = ρt / pinf
U= average degree of consolidation
ρt= compression at time t
ρinf = total long term settlement
84
Degree of consolidation?
Considered in terms of the amount of excess pore water pressure that has dissipated.
Used to assess amount of settlement that has occurred
85
U = 2H Δui - Area(3) / 2HΔui
U= average degree of consolidation
H= drainage path height
ui= initial excess pore water pressure
Area(3) = pore water pressure excess circle
86
U = (e0-e)/(e0-e1)
U= average degree of consolidation
e0= initial void ratio
e= current void ratio at t
e1 = final void ratio after full consolidation
87
U = (σ-σ0)/(σ1-σ0)
U= average degree of consolidation
σ = current effective stress at time t
σ0 = initial effective stress
σ1 = effective stress after consolidation has finished
88
σ1=σ0 + ui = σ + ue
σ1 = effective stress at the end of consolidation
σ0 = initial effective stress
ui = change in pore water pressure over consolidation
σ = current effective stress
ue = the value of the excess pore water pressure when σ´ is acting within the soil
89
U = ui-ue / ui = 1- ue/ui
U= average degree of consolidation
ui = initial pore water pressure uniform over depth z
ue = value of the excess pore water pressure when σ´ is acting within the soil
90
U = 1 - Σ 2/M2 * e^(-M2 Tv)
U = average degree of consolidation
M = goes from inf to 0
e= void ratio
Tv= dimensionless time factor
91
How do you find the cv from an oedemeter test?
Plot sample compression and square root of time
92
What’s the first stage of consolidation?
- initial compression (bedding in)
- small but rapid settlement from compression of air, rearrangement of larger particles
- not from any loss of water
93
Second stage of consolidation?
- primary consolidation
- due solely to expulsion of water
94
Third stage of consolidation?
- secondary compression
- volume changes that are essentially independent of excess pwps
- occurs after U= 100%
95
U = 2sqrt( Tv/ π)
When works?
U = average degree of consolidation
Tv = dimensionless time factor
Works when U < 50%
96
cv = π* d^2 / 16*t50
cv = coeff of consolidation
d = half mean thickness of sample for the stress increment being considered
t50 = time when U is at 50%
97
cv = 0.848 d^2 / t90
cv = coeff of consolidation
d = half mean thickness of sample for the stress increment being considered
t90 = time when U is at 90%
98
Tf = c’ + σn’ tanψ ´
Only for drained conditions
Tf = shear strength at failure
Tf = τmax (peak shear stress) and c is 0 for sand
c’ = cohesion intercept or apparent cohesion
σn’ = a linear function of yhe effective normal stress
Ψ= angle of internal shearing resistance
99
Tf = cu = Su
Tf = the shear strength at failure
Cu= untrained shear strength
100
Two types of tests to establish shear strength of soil?
- direct shear test (square shear box)
- triaxial test (cylindrical sample)
101
Difference between shear box test and triaxial apparatus?
Boundary conditions are not as easy to control in the shear box test as the triaxial apparatus
102
Testing samples in undrained soil (short term)?
Sample is sealed so that no water leaves or enters, no overall volume change takes place during shearing
103
Testing drained soil samples?
Water can move freely in and out of sample during all stages of the test. If shearing is performed slowly, pore pressure u = 0 and so σ´ = σ
104
How does excess pwp happen over time? For foundations where loading is applied
Excess pwp increases from hydrostatic conditions during the construction period. Short term the undrained conditions apply. Then during the consolidation period the excess pwp decreases time. In the long term the pore water pressure returns to hydrostatic
105
What happens to the factor of safety ? For foundations when loading is applied
It decreases when pore water pressure is high. Conditions become more stable as dissipation takes place (but FOS does not return to original value because of applied loads)
106
F= Σ shearing resistance / Σ disturbing forces?
F = factor of safety
107
What happens to pwp for unloading (slopes)?
Pwp decreases during excavation period then returns almost to normal over selling period. It doesn’t return exactly to normal as permanent drop in u due to change in boundary drainage conditions Tf
108
What’s happens to F?
Decreases as excavation
109
σn’ = Fv / L^2
Normal stress = normal force / length ^2
110
τ = FH / L^2
Shear stress = shearing force / length ^2
111
Direct shear box test procedure?
A number of identical specimens are tested at different values of Fv
Then σn’ is plotted against τmax
112
How does strength change with density?
Strength increases with density.
Angle of interface shearing resistance increases
113
What is the relationship between τ αnd ε?
ΔL / L = ε
Shear stress peaks for dense sand with brittle attributes and then decreases to similar values of loose sand of T after large displacements. Loose sand increases like log to this point.
114
How do different densities of sand change for ΔH/ H and ΔL/L
Dense sand increases and then steadies. Maximum slope occurs at same strain as peak strength
115
What is the difference between densities of sand for e vs ΔL/L?
Loose sand starts higher but approximately same void ratio at ultimate conditions
116
What happens in a direct shear box apparatus?
Normal force is applied and a shearing force. The vertical ΔH and lateral ΔL displacements are measured
117
Disadvantages of the direct shear box test?
- drainage conditions are not controlled
- effective stress only known if pore pressure is zero
- non uniform shear stresses on plane
- area under shear and vertical loads does not remain constant
118
Advantages of direct shear box apparatus?
- simplicity
- good for setting up sands
- can perform interface tests
119
When is critical state reached?
When τ is constant and there is no further volume change. It applied to both drained and undrained conditions
120
How does consolidated - undrained shear box test happen?
Stage 1
- sample consolidated under some normal stress σn
- after dissipation of excess pore water pressure u-> 0 as t -> inf
Stage 2
- sample sheared with no drainage
- no drainage therefore no change in e
121
Critical strength?
The point at which there are no further changes in shear stress of volumetric strain
122
Residual strength?
With active clays strength can reduce even further after very large strains. Such strains are often generated by slope movements or from geological processes that cause large strains or distortions in a clay stratum.
In these situations the clay platelets re-orientate themselves so that become aligned in a direction parallel to the shearing plane, producing highly polished surfaces.
Such surfaces have a much lower strength than that at critical state, and this strength is referred to as the residual strength
The strength parameters are given an r subscript to denote it.
In many cases clays that have been sheared in the past contain polished shearing surfaces which can only mobilise themselves residual strength and many cut slopes that contain such features have failed because of this.
123
What are the two stages of shear stress tests?
1. A consolidation stage
2. A shearing stage which may be drained or undrained
124
Steps in triaxial tests?
1. Sample trimmed, 2:1 H:D ratio
2. Place on lower pedestal with porous stones top and bottom
3. Enclose sample with rubber latex membrane
4. Seal menbrane with o-rings
5. Fit cell chamber around sample with loading ram
6. Fill cell with water
7. Increase cell pressure without loading ran in contact Δ σ = Δu no drainage and therefore no volume change for fully saturated sample
8. Bring loading ram into contact with sample and start applying axial load Fa
9. Swell or consolidate sample to correct in-situ stress
10. Close drainage and start loading sample axially, measuring pore pressure, at a rate slow enough not to generate any differential pwp within sample
125
A = A0 / 1-εa
A = area after deformation
A0 = area of sample
εa = axial strain = ΔH/ H0
ΔH = change in height
H0 = initial height of sample
126
A = A0 (1-εv / 1-εa)
A = area after deformation
A0 = area of sample
εa = axial strain = ΔH/ H0
ΔH = change in height
H0 = initial height of sample
εv = volumetric strain = ΔV/V0
ΔV = change in volume
V0 = initial volume of sample
127
When does shearing continue until?
The untrained shear strength of the soil Su is reached
128
Su = σa - σf / 2
Shear stress = (major principal stress - minor principal stress) /2
129
What happens in triaxial test for consolidated - untrained test?
- drainage of sample takes place under specified CP until consolidation complete; deviatior stress than applied with drainage closed
- apply steps 1-7, increase cell pressure (CP) until a positive pwp is measured
130
131
132
How is a sample during triaxial tests swelled or consolidated sample to correct in situ stress?
A) if pore fluid linked through to atmospheric pressure -> restrict to varying CP -> (u goes to 0 to change σ´ to CP
B) Apply a back pressure to pore water at required value of u (and σ= CP) to obtain in situ σ´ so σ´ = CP-u
Important to measure the sample volume change either using a) a butter or b) a volume gauge
The advantage of using a back pressure is that it is easier to maintain full saturation in the sample
133
Which type of triaxial test do you use for untrained and drained soils?
U-U tests find Su for untrained soils
C-U tests find ψ´ and c’ for drained conditions
134
Moth’s circle?
τ vs σ
Plot two points (σz , τzx) & (σx, -τxz)
Create a circle with centre on the τ=0 / σ axis
Pole is where the two points are at intercept to create a right angle on the circle.
Applying a line at θ from P to circle to get N (σθ, τθ) - stressed acting on plant inclined at angle θ to the horizontal
135
136
Principal planes and principal stresses?
There are two points where a Mohr’s circle crosses the σ axis. These points represent the planes on which the shear stress is zero and the normal stress is either a maximum or a minimum. These planes are known as principal planes and the corresponding normal stresses as principal stresses.
