1. Mohr’s Circles GLE/CEE 330 Lecture Notes Soil Mechanics
William J. Likos, Ph.D. Department of Civil and Environmental Engineering University of Wisconsin-Madison

2. Sign Convention normal stress (s) shear stress (t) (+s) (-s) (+s) (-s)
(USC)
(-t)
(+t)
(+t)
Element must have both pos. and neg. shear stress for equilibrium
(-t)

3. Principal Stresses & Principal Planes
Principal stress: normal stress on plane where shear stress is zero
s1 = maximum (major) principal stress
s3 = minimum (minor) principal stress
s2 = intermediate principal stress (3D)
If s1 = s2 = s3, then “isotropic stress” (thus no shear stress)
(s1 - s3 ) = “deviator stress” (results in shear stress)
Principal Plane: plane on which principal stress acts
Two principal planes are perpendicular s1 s1
t = 0
Not necessarily horizontal
And vertical! s3 s3

4. Derivation of Mohr’s Circle
Consider a 2D element where s1 ≠ s3
Sub-element: s1 t s3 s sa s3 ds dy ta a dx s1
Substituting...

5. sa s3
Divide by ds… ds dy ta a dx
Solve simultaneously… s1
(Note: if s1 = s3, ta = 0)
Rewrite (2)…
Square (1) and (3) and add together…

6. y
Is an equation for a circle in the form r x x0 t
½(s1–s3) s s3 s1
½(s1+s3)

7. t
Mohr’s Circle: graphical representation of state of stress on every plane in a 2D element
½(s1–s3) s s3 s1
½(s1+s3)
Typically only plot top half (+t) in geotechnical practice....
Typically only have stresses (+s) in compressive regime… t +t -s +t +s s3 s1 s -t -s -t +s

8. Mohr’s Circle for effective stress (s’ = s – u)
Circle for total stress (s) u
s’ or s
Shifts to left by pore pressure (u)
We will see later that this is a less stable state of stress (soil is closer to failure conditions)

9. Finding principal stresses if given stresses on perpendicular planes…sy
txy = -tyx
tyx
sx > sy
txy sx t
tmax
(sx, txy) sy s3 sx s1 s
(sy, tyx)
1/2(sx+sy)

10. Example: Find principal stresses and max shear stress (Analytical Solution)
sy = 3000 psf
txy = -300 psf
sx = 2100 psf

11. Graphical Solution:
draw to scale with x-scale = y-scale
plot known points and construct circle
“pick-off” principal stresses and max shear
sy = 3000 psf
txy = -300 psf
sx = 2100 psf

12. Finding stresses on any plane…
Graphical solution (Reference Plane Method)
1) Establish angle a from a reference plane (e.g,. horizontal).
2) Locate reference plane in Mohr’s circle
(center of circle to stress on ref plane)
3) Measure 2a from ref plane in same direction (CW or CCW)
4) This intersects circle at state of stress on angled plane sy
tyx
txy sa sx ta a t
Analytical solution:
Reference
Plane
(sx, txy) sy s3 sx s1 s 2a
(sa, ta)
*See also “Pole Point” or “Origin of Planes” Methods
(sy, tyx)

13. Example
Graphical solution
s1 = 2.5 kPa (C)
Reference Plane
(sa, ta) = (1.5, 1 kPa)
2a = 90 CCW sa
s3 = 0.5 kPa (C) ta
a = 45
Analytical solution:

14. Find orientation of principal planes: sy = 3000 psf
Let horizontal be reference plane
Find reference plane on circle
Measure angle to principal plane (2a = 33.7 CCW)
So major principal plane is 17 deg. CCW from horizontal
tyx = -300 psf
txy = 300 psf
sx = 2100 psf
s3 = 2009 psf
17 deg.
s1 = 3091 psf

15. Example: Find stresses on plane 30˚ as shown
40 psi
30˚
20 psi
Let major principal plane be reference plane
Find reference plane on circle
Measure angle (2a = 60˚ CCW)
40 psi
30˚
20 psi 35
8.7

16. Example: Find magnitude and orientation of principal stresses
20 psi
40 psi
10 psi
-10 psi
30˚
Plot known points and draw circle.
Find s1 = 44 psi, s3 = 16 psi
Let plane 30˚ from horizontal be reference plane, (s,t) = (20,10)
Find reference plane on circle
Measure angle to minor principal plane (2a = 45˚ CCW)
So minor principal plane is 22.5˚ CCW from reference plane, or 52.5˚ CCW from horizontal.
44 psi
52.5˚
16 psi

17. Graphical solution (Pole Method)
Draw line(s) on Mohr circle through known stress at known orientation. sy
tyx
Intersection of that line(s) with circle is the “Pole”
Pole
txy sa sx ta
Draw a line from Pole at any angle of interest. This line intersects the circle at the state of stress on the plane defined by that angle a
(sa, ta) a t
(sx, txy) s1 s
(sy, tyx)

18. So what happened to soil mechanics?
s’h t uw
s’v
Is there a plane within the soil mass where the induced shear stresses exceed the shear strength of the soil (c’, f’) ?
Will a “failure plane” develop?
How does shear strength depend on effective stress?

19. Mohr-Coulomb Failure Criterion
c’ = effective cohesion intercept
f’ = effective friction angle
s’ = effective stress on failure plane
This is a straight line in s’-t space t
failure
M-C Failure Envelope
no failure
s’1 t s’
s’3 f’
Mohr’s Circle c’
s’3
s’1 s’

20. Bearing Capacity Failure
Plane t
s’v t
s’h
Stable f’ c’ s’

22. Active Earth Pressure Failure
Retaining
Wall
Soil t
s’v t
s’h f’ c’
Stable s’