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.