1. EAG 345 – GEOTECHNICAL ANALYSIS
(iii) Mohr-Coulomb Failure
Criterion
By: Dr Mohd Ashraf Mohamad Ismail

2. SIMPLE MODEL FOR THE SHEAR STRENGTH OF SOIL
Coulomb’s Model

3. Why we need a model to represent the shear strength of the soil ?

4. A lot of models in our life
Weight loss model
Face whitening cream
BEFORE
AFTER
BEFORE
AFTER

5. Why we need a model to represent the shear strength of the soil ?
“A mathematical representation of a process by means of a number of variables which are defined to represent the inputs, outputs, and internal states of the process, and a set of equations and inequalities describing the interaction of these variables”
We will use the outcomes to represent our soil for the purpose of soil investigation, engineering design (slope, retaining wall, foundation and etc) and for the geotechnical evaluation.

6. In the previous lecture we already learned the typical response of soils to shearing forces
So now, we would like to interpret these responses using a simple model where we will learn and discuss a simple model to gain insight into the soil behaviors that later will help us to interpret the shear strength of soils

7. A slip plane in a soil mass
Coulomb’s Model
Failure surface
Mobilized shear resistance
Retaining wall
A slip plane in a soil mass

8. A slip plane in a soil mass
Coulomb’s Model
A slip plane in a soil mass
Slip of a wooden block
A block of mass 2.1 kg rests on a horizontal table. The coefficient of friction is 0.2. what horizontal force H will start the block moving ?
Solution: Coulomb frictional law (physics course during high school)
H = horizontal force required to initiate the movement
W= weight of the block
µ =coefficient of static sliding friction between the block and the surface
Angle of friction, φ' = tan-1 * (coeff) = tan-1 * 0.2 = 11.31o w H T φ' N R φ' Fr
tan φ‘ = Ff / Fn Fn Ff

9. Coulomb’s Model σ‘n w τ H T N R R N
In term of stress
Slip of a wooden block
σ‘n τ φ' N R φ' w T H N R
((T/A) where T = shear force at impending slip and A is the area of the plane parallel to T) is the shear stress when slip is initiated
Normal effective stress on the plane on which slip is initiated
Friction angle

10. Coulomb’s Model
Coulomb’s law requires the existence or the development of a critical sliding plane also called slip plane.
In the case of the wooden block, the slip plane is horizontal plane at the interface between the wooden block and the table
Unlike the wooden block we do not know where the sliding plane is located for soils until it fail.
Failure surface
Mobilized shear resistance
Slip of a wooden block
Retaining wall φ' w T H N R

11. Effect of increasing the normal effective stress
Shear stress vs. normal stress
Void ratio vs. shear strain
If we were to plot the peak shear stress and the critical state shear stress for each constant normal effective stress for type 1 and 11 soils. We would get:

12. Friction angle Row 1 Row 2 Row 1 Row 2
From previous course soil mechanics we already learned that soils have different unit weights depending on the arrangement of the particles.
Let us simulate two extreme arrangements of soil particles for coarse-grained soils
One loose and the other dense
We will assume that the soil particles are spheres.
In two dimension the sphere will become disk.
The loose arrays is obtained by stacking the disk one on top of another while the dense packing is obtained by staggering the rows as illustrated here.
Row 2

13. Friction angle a a For simplicity, let us consider the first two rows
If we push row 2 relative to row 1 in the loose assembly, sliding would be initiated on the horizontal plane a-a,
Consistent with Coulomb’s frictional law
Once the motion is initiated the particles in the loose assembly would tend to move into void spaces.
The direction of movement would have a downward component that is compression.
In the dense packing the relative sliding of row 2 with respect to row 1 is restrained by the interlocking disks.
Sliding for these dense assembly would be initiated on an inclined plane rather than on horizontal plane.
For the dense assembly, the particle must ride up over each other of be pushed aside or both.
The direction of movement of the particles would have an upward component that is expansion.

14. Simple shear deformation of Type I soil
Loose sands, normally consolidated and lightly over consolidated clays (OCR ≤2)
Constant vertical effective stress
+ ve
Shearing under constant volume
Shear stress vs. shear strain
Volumetric strain vs. shear strain
Void ratio vs. shear strain

15. Simple shear deformation of Type I soil
Type II:
Dense sands and heavily over consolidated clays (OCR >2)
- ve
Constant vertical effective stress
Shearing under constant volume
Shear stress vs. shear strain
Volumetric strain vs. shear strain
Void ratio vs. shear strain

16. Simulation of failure in dense sand
(1)
(2)
(3)
Simulated shearing of a dense array of particles
Stresses on failure plane

17. Simulation of failure in dense sand
Solving for H and W we obtain
(4)
(5)
Dividing Eq. (4) and (5) and simplifying, we can obtain
By analogy with loose assembly, replace H by and W by resulting in

18. Simulation of failure in dense sand
Therefore:
Let investigate the sensitivity of to the shear strength
Assume and is constant. Then for we get
But if we get
That is an increase of 45% in shear strength for a 10% increase in

19. Friction angle
Shearing of a given volume of soil would cause impending slip of some particles to occur up the plane while others occur down the plane.
So the general equation will be:

20. Dilatancy
Dilation angle – measure of the change in volumetric strain with respect to the change in shear strain.
The lower normal effective stress, the greater the value alpha
Ability of the soil to expand depends on the magnitude of the normal effective stress.

21. Cohesion
For cemented soils, Coulomb’s frictional law can be written as:
C0 = cohesion

22. Cohesion

23. Cohesion

25. Revision: Coulomb’s frictional law (static or physics course)
Newton’s 3 Law of motions.
Scalar, vector and tensor

26. Mohr-Coulomb Failure Criterion (in terms of total stresses)  c 
failure envelope
Cohesion
Friction angle f
f is the maximum shear stress the soil can take without failure, under normal stress of .

27. Mohr-Coulomb Failure Criterion
Field conditions z
svc
shc
A representative soil sample

28. How to determine strength parameters c and f
Mohr-Coulomb Failure Criterion
Deviator stress, Dsd
Axial strain
(Dsd)fc
Confining stress = s3c
How to determine strength parameters c and f
s1 = s3 + (Dsd)f s3
(Dsd)fb
Confining stress = s3b
(Dsd)fa
Confining stress = s3a f
Mohr – Coulomb failure envelope
Shear stress, t
s or s’
s3c
s1c
s3b
s1b
(Dsd)fb
s3a
s1a
(Dsd)fa

29. Mohr-Coulomb Failure Criterion (in terms of effective stresses) ’ c’ ’
failure envelope
Effective cohesion
Effective
friction angle f
u = pore water pressure
f is the maximum shear stress the soil can take without failure, under normal effective stress of ’.

30. Mohr-Coulomb Failure Criterion
Shear strength consists of two components: cohesive and frictional.
’f f ’  ' c’
’f tan ’
frictional component c’
cohesive component

31. c and  are measures of shear strength.
Higher the values, higher the shear strength.

32. Mohr Circles & Failure Envelope
Soil elements at different locations
Failure surface Y
~ stable X
~ failure ’

33. Mohr Circles & Failure Envelope
The soil element does not fail if the Mohr circle is contained within the envelope
Initially, Mohr circle is a point  GL Y c
c+  c

34. Mohr Circles & Failure Envelope
As loading progresses, Mohr circle becomes larger…
.. and finally failure occurs when Mohr circle touches the envelope  GL c Y c

35. Mohr circles in terms of total & effective stressesX
total stresses t
s or s’ X
v’
h’ u + =
v’
h’
effective stresses u

36. Failure envelopes in terms of total & effective stressesh X
total stresses t
s or s’ X
v’
h’ u + = f’ c’
Failure envelope in terms of effective stresses c f
Failure envelope in terms of total stresses
If X is on failure
v’
h’
effective stresses u