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

2. 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

3. Shear failure mechanism 
So we can see that, for soils failure may be initiated at a point within a soil mass and then propagate through it this is known as progressive failure.
The stress state at a point in a soil mass due to applied boundary forces may be equal to the strength of the soil, thereby initiating failure.
Therefore as engineers we need to know the stress state at a point de to applied loads
And we will discuss stress state by using the mohr circle.
At failure, shear stress along the failure surface () reaches the shear strength (f).

4. Mohr Circle of stress tzx
100 kpA sZ sx
Soil element
tzx
40 kpA
50 kpA
Suppose a cubical sample of soil is subjected to the stresses shown in the above figure
We would like to know what the stresses at any point within the sample due to the applied stresses.

5. Mohr Circle of stress tzx
100 kpA
τ (kPa) sZ sx 50
tzx
40 kpA
σ (kPa)
100
50 kpA
-50
The first step is to plot a normal stress axis (abscissa) and shear stress axis (ordinate)

6. Mohr Circle of stress tzx
100 kpA
τ (kPa) sZ sx 50
tzx
40 kpA
σ (kPa)
100
50 kpA
-50
2. Assign a sign convention. We will use anti-clockwise shear and compression as +ve
A (100,40)
B (50,-40)

7. Mohr Circle of stress tzx
τ (kPa)
100 kpA sZ sx 50
A (100,40)
tzx
40 kpA
σ (kPa)
100
50 kpA
-50
B (50,-40)
3. Plot the stresses. Point A (+100,+40) represents the stresses on the horizontal plane while point B (+50,-40) represents the stresses on the vertical plane. Join AB

8. Mohr Circle of stress tzx
100 kpA
τ (kPa) sZ sx 50
A (100,40)
tzx
40 kpA
σ (kPa) O
100
50 kpA
-50
B (50,-40)
4. With point 0 as origin and OA or OB as radius draw a circle.

9. Mohr Circle of stress tzx
τ (kPa)
100 kpA sZ sx 50
A (100,40)
tzx
40 kpA 3 1
σ (kPa)
100
50 kpA
-50
B (50,-40)
5. The major principal stress is the value of the normal stress at point 1. The minor principal stress is the value of the normal stress at point 3.

10. Mohr Circle of stress tzx 100 kpA sZ 40 kpA sx 50 kpA3 1
σ (kPa)
100
50 kpA
-50
B (50,-40)
Alternatively: the principal stresses are related to the stress component σz σx τzx by:

11. Mohr Circle of stress tzx Alternatively: 100 kpA sZ 40 kpA sx 50 kpA3 1
σ (kPa)
100
50 kpA
-50
B (50,-40)
Alternatively:

12. Mohr Circle of stress tzx
τ (kPa)
100 kpA sZ sx 50
A (100,40)
tzx
40 kpA 3 1
σ (kPa)
100
50 kpA
-50
B (50,-40)
5. We will now determine the pole of Mohr’s circle. Draw a line through A to represent the horizontal plane and a line through B to represent the vertical plane. The intersection of these planes is the pole, denoted by P.

13. Mohr Circle of stress tzx
100 kpA
τ (kPa) sZ sx 50
A (100,40) P
tzx
40 kpA 3 1
σ (kPa)
100
50 kpA
-50
B (50,-40)
6. We will now determine the pole of Mohr’s circle. Draw a line through A to represent the horizontal plane and a line through B to represent the vertical plane. The intersection of these planes is the pole, denoted by P.

14. Pole method of finding stresses along a plane
Mohr Circle of stress
Pole method of finding stresses along a plane
τ (kPa) D C B Q F
σ (kPa) θ A E B A: θ
(Pole) P A B:
Point A on the Mohr’s circle represents the stresses on the plane AB. So the line AP is drawn parallel to AB. Point P become the Pole (P) in this case
If we need to find stresses on a plane EF, we draw a line from the pole parallel to EF, The point of intersection of this line with the Mohr’s circle is Q
The coordinates of Q give the stresses on the plane EF

15. Orientation of Failure Plane
Failure envelope q
(s’, tf)
(90 – q)
PD = Pole w.r.t. plane f’ s’
Therefore,
90 – q + f’ = q
q = 45 + f’/2

16. Mohr Circle of stress tzx
Horizontal plane
τ (kPa)
100 kpA sZ sx 50
A (100,40) P
tzx
40 kpA 3 1
σ (kPa)
100
50 kpA
Major principal plane
-50
B (50,-40)
7. Let’s find the direction of the major principal stress plane. A line is drawn from the pole, P to the point representing the principal stress, i.e…point 1. the angle 1PA is the inclination of the major principal plane to the horizontal and it is positive as shown (anti-clockwise direction to the horizontal plane)

17. Mohr Circle of Stress tzx 100 kPa sZ 40 kPa sx 50 kPa ψ τ (kPa) 50 P 3
-50
B (50,-40)
Horizontal plane ψ
Major principal plane

18. Mohr Circle of Stress tzx 100 kpA sZ 40 kpA sx 50 kpA3 1
σ (kPa)
100
50 kpA
-50
B (50,-40)
Alternatively: the angle between the major principal stress plane and the horizontal plane (ψ) is:

19. Mohr Circle of Stress tzx
τ (kPa)
100 kPa 50
A (100,40) sZ sx P -θ
tzx
40 kPa 3 1
σ (kPa)
100
50 kPa θ
-50
B (50,-40)
8. To determine the stresses on any plane oriented at θ from the horizontal plane (clockwise from the horizontal plane is negative), draw a line from the pole, P at an angle θ to the horizontal plane, AP here we show a plane oriented at -θ

20. Mohr Circle of Stress tzx 100 kPa sZ 40 kPa sx 50 kPa θ-θ
tzx
40 kPa 3 1
σ (kPa)
100
50 kPa θ
-50
B (50,-40)
Alternatively: the angle between the major principal stress plane and the horizontal plane (ψ) is:

21. Mohr Circle of Stress tzx 100 kPa sZ 40 kPa sx 50 kPa θ-θ
tzx
40 kPa 3 1
σ (kPa)
100
50 kPa θ
-50
B (50,-40)
Alternatively: the angle between the major principal stress plane and the horizontal plane (ψ) is:

22. Example 1
A sample of soil (0.1 m x 0.1 m) is subjected to the forces shown in Fig. 1. Determine
σ1 ,σ3 and ψ
The maximum shear stress
The stresses on a plane oriented at 30o counterclockwise from the major principal stress plane
5 kN sZ sx
tzx
1 kN
3 kN
Fig. 1

23. Example 1
A sample of soil (0.1 m x 0.1 m) is subjected to the forces shown in Fig. 1. Determine
σ1 ,σ3 and ψ
The maximum shear stress
The stresses on a plane oriented at 30o counterclockwise from the major principal stress plane
5 kN sZ sx
tzx
1 kN
3 kN
Fig. 1

24. Example 1 tzx Procedure:
There are two approaches to solve this problem. You can either use Mohr’s circle or the appropriate equation. Both approaches will be used here.
(1) Find the area: Area:
(2) Calculate the stress:
500 kPa sZ sx
tzx
100 kPa
300 kPa
Fig. 1

25. Example 1 (3) Draw Mohr’s circle and extract σ1 ,σ3 and τmax τ σ 300
200
A (500,100)
100 σ
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

26. Example 1 (3) Draw Mohr’s circle and extract σ1 ,σ3 and τmax τ σ 300
200
A (500,100)
100 σ
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

27. Example 1
(4) Draw the pole on Mohr’s circle. The pole of Mohr’s circle is shown by point P in the Fig. below τ
300
200 P
A (500,100)
100 σ
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

28. Example 1
A sample of soil (0.1 m x 0.1 m) is subjected to the forces shown in Fig. 1. Determine
σ1 ,σ3 and ψ
The maximum shear stress
The stresses on a plane oriented at 30o counterclockwise from the major principal stress plane
5 kN sZ sx
tzx
1 kN
3 kN
Fig. 1

29. Major principal stress plane
Example 1
(5) Determine ψ. Draw line from P to σ1 and measure the angle between the horizontal plane and this line. τ
300
Horizontal Plane
200
A (500,100)
100 ψ
Major principal stress plane σ
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

30. Example 1
(5) Alternatively the angle between AOC can be used to determine the ψ τ
300
200
A (500,100)
100 ψ θ σ O C
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

31. Example 1
(6) Determine the stresses on a plane inclined at 30o from the major principal stress plane. Draw a line M’ and N’ through P with an inclination of 30o from the major principal stress plane, angle CPN. The coordinate at point N is (470, 120) τ
300
Coordinate at point N is (470,120)
200 N’ P
A (500,100)
100 M’
30o
Plane inclined at 300 from the major principal stress plane θ σ C
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

32. Example 1 (Alternative solution)
(3) Draw Mohr’s circle and extract σ1 ,σ3 and τmax τ
300
200
A (500,100)
100 σ
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

33. Example 1 (Alternative solution)
Alternatively, we can use analytical solution from the described equation explained before: τ
300
200
A (500,100)
100 σ
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

34. Example 1 (Alternative solution)
Analytical solution:

35. Example 1 (Alternative solution)
Analytical solution:
Graphical solution:

36. Example 1 (Alternative solution)
Check Equilibrium:
500 kPa
100 kPa 3
(+) Y 2
22.5o
300 kPa
(+) X
100 kPa
541.4 kPa 1
Length of 2-3 = 0.1 m
Length of 3-1 = 0.1 x (tan 22.5o) = m
Length of 1-2 = 0.1 / (cos 22.5o) = m

37. Example 1 (Alternative solution)
Check Equilibrium:
500 kPa
100 kPa 3
(+) Y 2
22.5o
300 kPa
(+) X
100 kPa
541.4 kPa 1

38. Major principal stress plane
Example 1 (Alternative solution)
Alternatively, we can use analytical solution from the described equation explained before: τ
300
Horizontal Plane
200
A (500,100)
100 ψ
Major principal stress plane σ
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

39. Example 1 (Alternative solution)
Alternatively, we can use analytical solution from the described equation explained before: τ
300
Coordinate at point N is (470,120)
200 N’ P
A (500,100)
100 M’
30o
Plane inclined at 300 from the major principal stress plane θ σ C
100
200
300
400
500
600
-300
B (300,-100)
-200
-100

40. Example 1 (Alternative solution)
Alternatively, we can use analytical solution from the described equation explained before:

41. Submit on Wednesday (28/9/2011) during class
Homework
1035 kN/m2
414 kN/m2
621 kN/m2
414 kN/m2
For the stressed soil element shown in Fig. above determine:
Major principal stress
Minor principal stress
Normal and shear stresses on the plane AE
By using graphical and analytical solution
Refer to: Principles of Geotechnical Engineering (Braja, M. Das) Page 259
Submit on Wednesday (28/9/2011) during class