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CIVL841-Analytical and numerical methods in Geomechanics
Name: Ioannis Vazaios-St.Number: 10123567
Assignments 3 and 4
Question 1: Footing bearing capacity for a cohesive, frictionless soil
In order to examine the accuracy of the numerical solution for the bearing capacity of a footing
using FEA code Phase2 from Rocscience, the model illustrated in figure 1 was used. The main
geometrical features of the model and its boundary conditions are illustrated in this particular
figure as well.
Figure 1 FEA model of a footing on a cohesive, frictionless soil
In the figure above it can be observed that the mesh has been refined in the area around the
footing in order to increase the accuracy of the result. Although 3-node triangular elements are
not suitable for the numerical analysis of purely cohesive soils, with refinement of the mesh
around the area of interest additional degrees of freedom will be added in order to overcome
problems regarding the desired response of the analysis. However, a very refined mesh can result
is various problems due to the low stiffness of the very small elements and thus refinement of the
mesh must be done with caution. Additional geometrical features of the model include the width
of the footing which is B=5.0m and its height is H=1.0m. The properties of the materials assigned
to the different components of the model are listed in table 1.
Properties Soil Concrete
Young’s Modulus E (MPa) 30 30,000
Poisson’s ratio v 0.25 0.2
Cohesion (kPa) 10 Not applicable
Friction angle φ (0) 0 Not applicable
Dilation angle ψ (0) 0 Not applicable
Table 1 Material Properties
The soil is assumed to have an elasto-plastic behaviour obeying the Mohr-Coulomb failure
criterion for a purely cohesive material, while the footing which is made of concrete is assumed
50m
100m
to have an elastic response. The footing can be considered to be rigid as Econcrete/Esoil=1,000 which
indicates that the relative stiffness between the footing and the soil is expected to be relatively
high. In the analysis conducted an initial, geostatic stress field was not applied; hence the only
stresses developing in the soil are the result of the loading of the footing.
Continuing with the analysis, it was divided in 31 stages in order to examine thoroughly the
response of the stresses applied on the ground due to the loading sequence applied on the footing.
However, in order to prevent numerical instabilities and possible convergence problems by
exceeding the bearing capacity of the soil, the loading sequence applied was in the form of
prescribed displacements applied on the footing rather than in the form of a distributed load. The
final magnitude of the prescribed displacements was 0.75m starting from 0 and was increased at
each stage by 5%.
At each single step the pressure applied on the ground surface was examined along the entire
width of the footing. The distribution of the pressure under the footing at the final stage of the
analysis is illustrated in figure 2.
Figure 2 Stress distribution below the footing
From the figure above it can be observed that the stress is approximately maintained constant at
the central points of the footing but at the edges of it there are rather significant stress
concentrations. These stress concentrations are the result of the cohesive nature of the soil in this
case. The rigid footing is moving downwards due to the prescribed displacements, therefore the
displacement is uniform. The points close to the centre of the footing are moving downwards due
the displacement of it. However, the points of the ground close to the edges of the footing do not
have the same response in order to follow the displacement of it, thus stress concentrations occur
in order to make these points follow the same displacement of the rest of the ground surface
below the footing. Continuing the assessment of the data, these stress concentrations will not be
taken into account and the average stress developing below the footing will be estimated for each
stage as illustrated in table 2. Additionally, from the average stress acting below the footing, the
total force is calculated by multiplying the stress with the width of it.
v/vtotal Deflection v per step
(m) p(MPa) F(MN/m)
0 0 0 0
0.05 0.025 0.007027 0.035137
0.1 0.05 0.014644 0.073221
0.15 0.075 0.022553 0.112767
0.2 0.1 0.029957 0.149783
0.25 0.125 0.033911 0.169557
0.3 0.15 0.036201 0.181005
0.35 0.175 0.038012 0.19006
0.4 0.2 0.039498 0.197488
0.45 0.225 0.040815 0.204077
0.5 0.25 0.041979 0.209893
0.55 0.275 0.043058 0.215288
0.6 0.3 0.044016 0.220079
0.65 0.325 0.044877 0.224386
0.7 0.35 0.045693 0.228467
0.75 0.375 0.046436 0.232178
0.8 0.4 0.047107 0.235533
0.85 0.425 0.047718 0.23859
0.9 0.45 0.048259 0.241293
0.95 0.475 0.048724 0.243619
1 0.5 0.049115 0.245577
1.05 0.525 0.049466 0.24733
1.1 0.55 0.049757 0.248786
1.15 0.575 0.049999 0.249995
1.2 0.6 0.050194 0.250971
1.25 0.625 0.050373 0.251866
1.3 0.65 0.050561 0.252806
1.35 0.675 0.050727 0.253633
1.4 0.7 0.050899 0.254497
1.45 0.725 0.051063 0.255313
1.5 0.75 0.051213 0.256064 Table 2 Average stress and force for each deflection step
From the table above the following figure is produced.
Figure 3 Numerical versus Analytical solution of the bearing capacity of the footing
In figure 3 the closed solution derived by Prandtl for the bearing capacity of a footing is
compared to the staged, numerical solution. Prandtl’s expression is the following:
BcP *)2(* in which, c, is the cohesion of the soil and, B, is the width of the footing
For this specific case the bearing capacity of the footing is qu=10*5.14=51.4kPa, thus the force
respectively is P=257kN/m=0.257MN/m. From figure 3 it can be inferred that the numerical
solution converges to the analytical one at the limit state of the footing.
Question 2: UCS test for a cohesive, frictionless soil specimen
In the following example the simulation of an unconfined compressive strength (UCS) test will be
discussed. The numerical simulation of this particular example was conducted using FEA
software Phase2 from Rocscience. The configuration of the model is illustrated in figure 4. The
model consists of the soil specimen, which has the same properties as the soil in the previous
question and they are listed in table 1, and the steel plates of the UCS test apparatus, which are
made of steel with Young’s modulus E=210GPa and Poisson’s ratio v=0.3. The steel plates are
2cm thick while the diameter of the specimen is 5cm and its height 10cm. The steel plates are
assumed to have an elastic response during the analysis, while the soil specimen is assumed to
have an elasto-plastic response obeying the Mohr-Coulomb criterion. Again, as in the previous
question, the loading sequence is applied indirectly via prescribed displacements assigned to the
upper plate while the lower plate maintains its initial position. The loading sequence is divided
into 64 stages in order to obtain the full response of the specimen during loading until it reaches
its yield stress, unloading and reloading again until it reaches its yield stress. Analytically the
yield stress of a frictionless, cohesive material is obtained by the following expression:
kPacUCS 2010*2*2
In order to make the specimen reach its peak strength the final, compressive displacement value
assigned is u≈7.5e-05m. Again, this displacement value was staged starting from 0 and increased
by 5% at each step. The results of the analysis are illustrated in figure 5.
Figure 4 FEA model for a UCS test
Figure 5 Axial stress versus axial strain for one loading cycle
From figure 5 it can be observed that the response of the specimen is elastic until it reaches
approximately its analytically defined peak strength of 20 kPa. Then its behaviour is perfectly
plastic. The specimen is then unloaded by reducing the prescribed displacement to zero and then
it is reloaded again it reaches its peak strength. In the post peak region the material behaves as
perfectly plastic and the stress applied on the specimen is equal to the peak strength of the soil.
Additionally, it can be observed that the linear branches during loading, unloading and reloading
are approximately parallel, indicating that the elasticity modulus remains constant through the
whole analysis and it is not affected by the loading condition. Finally, it has to be noted that the
2D simulation of the UCS test should be better performed by applying an axisymmetric analysis
rather than a plane strain analysis, as the simulation of this kind of problems involves three
dimensional phenomena.
Question 3: Derive the expressions of Mohr-Coulomb criterion in (s,t) and
(p,q) space
In the first part of this section the expressions of Mohr-Coulomb criterion will be derived for the
(s,t) space in which:
2
31 s
2
31 t
In figure 6 the above mentioned are illustrated.
Figure 6 Schematic of the Mohr-Coulomb failure criterion
The expression of the Mohr circle can also be written in the following form:
022
Therefore, the centre of the Mohr circle can be written in the following form as well:
31
21
,0
0,2
2,
2
S
BAS
And the radius of the circle respectively:
3131
2
3
2
131
31
22
42
2
2
4
t
BAt
φ
σ
τ
Therefore, the Mohr circle can be written in the form below by substituting the quantities above
and because the circle of failure is tangential to failure locus the following system of equations
can be produced:
0tan
tan
03131
223131
22
c
c
Thus from this equation the following 2nd
order polynomial is derived:
0tan2tan1 3131
22 c (αx2+βx+γ=0)
But this polynomial although is a polynomial of 2nd
order it has only one solution as the failure
locus and the Mohr circle have only one, unique point in common as the failure locus is the
tangent of the Mohr circle. Therefore:
2
31
22
31
tan1
1
2tan1
tan
)tan1(2
)(tan2
2
ccsolution
And by substituting the solution to the expression of the failure locus the following expression is
obtained:
2
31
2
2
2
31
tan1
tan
2tan1
tan
tantan12
tan2tan
cc
ccc solutionsolution
Additionally, from figure 6 it can be observed that:
coscos2
31 tsolution
Thus, the following expression can be derived:
2
31
2
2
tan1
tan
2tan1
tancos
cct but
cos
sintan and
2
31 s
So
sincos sct
But the expression above can also be written in the following form:
AsBt tan in which
sintansintan
cos
1
A
cB
In the second part of this section the expressions of Mohr-Coulomb criterion will be derived for
the (p,q) space in which:
3
321 p
31 q
Assuming that σ2=σ3 (compression) the expressions above can be written in the following form:
3
2322323
231111
13
31 qpqpqp
q
p
Thus 3
3
3
32313
qpqqpq
Therefore,
3
6
3
32331
qpqpqp
From the derivation of the expressions for the Moh-Coulomb criterion in the (s,t) space the
following equation was used:
sin3
cos6
sin3
sin6sin
3
6cos2
sin6
6cos
2sin
2cos
2
3131
cpq
qpcq
qpc
qc
Therefore, the slope of the Mohr-Coulomb failure locus in (p,q) space for compression is:
sin3
sin6
c
Assuming that σ2=σ1 (extension) the expressions above can be written in the following form:
3
6
3
23
3
3
3
23
3
33
3
3
3
33332
32
31
3
1111
13
31
qpqpqp
qpqqpq
qp
qpqpqp
q
p
Thus,
pc
q
qpcqqp
cq
qpc
qc
sin3
sin6
sin3
cos6
sinsin6cos63sin3
6cos2
sin6
6cos
2sin
2cos
2
3131
Therefore, the slope of the Mohr-Coulomb failure locus in (p,q) space for extension is:
sin3
sin6
e
Question 4: Deriving the plastic matrix for two loading cases for a purely
cohesive material (υ=00, c≠0)
Table 3 Calculus for the plastic matrix for a purely cohesive soil
Properties
E(kPa) 30000
v 0.25
c(kPa) 10
υ(0) 0
ψ(0) 0
Stress Field Case1 Case2
σxx(kPa) 0 20
σyy(kPa) 20 20
τxy(kPa) 0 10
α(0) 90 0
k1 1 1
k2 1 1
Case1 Case2
R1C1 0.25 0.00
R1C2 -0.25 0.00
R1C3 0.00 0.00
R1C4 0.00 0.00
R2C1 -0.25 0.00
R2C2 0.25 0.00
R2C3 0.00 0.00
R2C4 0.00 0.00
R3C1 0.00 0.00
R3C2 0.00 0.00
R3C3 0.00 0.25
R3C4 0.00 0.00
R4C1 0.00 0.00
R4C2 0.00 0.00
R4C3 0.00 0.00
R4C4 0.00 0.00
Plastic
matrix Case1 Case2
B 48000
C1 0.5 0
C2 -0.5 0
C3 0 0.5
C4 0 0
R1 0.5 0
R2 -0.5 0
R3 0 0.5
R4 0 0
Table 4 Elastic Constitutive matrix
De 1 2 3 4
1 36000 12000 0 12000
2 12000 36000 0 12000
3 0 0 12000 0
4 12000 12000 0 36000
Table 5 Plastic Constitutive matrix for unconfined compression
Case1 Dp 1 2 3 4
1 12000 -12000 0 0
2 -12000 12000 0 0
3 0 0 0 0
4 0 0 0 0
Table 6 Plastic constitutive matrix for the shear loading case
Case2 Dp 1 2 3 4
1 0 0 0 0
2 0 0 0 0
3 0 0 12000 0
4 0 0 0 0
Table 7 Total constitutive matrix for unconfined compression
Total matrix Case1 1 2 3 4
1 24000 24000 0 12000
2 24000 24000 0 12000
3 0 0 12000 0
4 12000 12000 0 36000
Table 8 Total constitutive matrix for the shear loading case
Total matrix Case2 1 2 3 4
1 36000 12000 0 12000
2 12000 36000 0 12000
3 0 0 0 0
4 12000 12000 0 36000
References
Griffiths D.V., Willson S.M., 1986, An explicit form of the plastic matrix for Mohr-
Coulomb material. Communications is Applied Numerical Methods. Vol. 2, pp.
523-529.
Smith I.M., Griffiths D.V., 2004, Programming the Finite Element Method, Willey,
4th
ed., West Sussex, England.
Moore, I.D., 2013, CIVL 841: Analytical and Numerical methods in Geomechanics,
course material, Department of Civil Engineering, Queen’s University, Kingston
ON, Canada.
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