2.2 Application of the Mohr Circle
Method to Soil Stress Calculations
2.2 - 1
Stresses acting on a Soil Element in the Ground
Consider a typical soil element A in the ground as shown
below.
The stresses acting on the soil are denoted as σx and σy .
ground surface
σv
A
2.2 - 2
σx
Stresses acting on a Soil Element in the Ground
At the given point A in the soil mass, the general state of stresses in
two-dimensions with respect to the x,y coordinate axes are:
•
•
•
•
σy is the normal stress on a plane having a unit normal in the y-direction.
σx is the normal stress on a plane having a unit normal in the x-direction.
τxy is the y-directed shear stress on a plane having a unit normal in the
x-direction.
τyx is the x-directed shear stress on a plane having a unit normal in the
y-direction.
While the signs of τxy and τyx will be different, their magnitudes will
always be the same.
σy
τyx
x
y
τxy
σx
A
τxy
τyx
σy
2.2 - 3
σx
Stresses acting on a Soil Element in the Ground
If we cut a plane P-P’ through element A which is inclined at an angle of θ to
the vertical, as shown:
σy
P
τyx
x
τxy
y
σx
σx
θ
τxy
τyx
σy
P’
What is the stress σθ and shear stress τθ acting on the plane P-P’?
σθ
σx
τθ
τxy
τyx
σy
2.2 - 4
Stresses acting on a Soil Element in the Ground
Note that σθ and τθ will not have the same values as σx (or σy) and τxy.
σθ
σx
τθ
τxy
τyx
σy
In fact, as the angle θ of the plane changes, the value of σθ and τθ will also
change continuously.
For any soil element, it is possible to find an angle θ such that the shear
stress τθ acting on the plane is zero.
The plane on which the shear stress τθ is zero is called the principal plane.
Hence, on a principal plane, there is only a normal stress σθ acting, and there
is no shear stress.
The normal stress σθ acting on a principal plane is called a principal stress.
2.2 - 5
Stresses acting on a Soil Element in the Ground
Questions:
How many principal planes (and principal stresses) are
there for a typical soil element shown below?
What can you say about magnitude of the principal stresses,
compared to the normal stresses on other non-principal
planes?
σy
P
τyx
τxy
σx
θ
τxy
τyx
σy
2.2 - 6
P’
Stresses acting on a Soil Element in the Ground
Let’s consider again the soil element A from slide 2.1-2.
•
For a soil element located in a level or horizontal ground,
there are no shear stresses on the horizontal and vertical
planes.
•
Hence, σv and σh are the principal stresses
•
•
In general cases, σv > σh under insitu conditions
Hence, σv = σ1 (σ1 denotes major principal stress)
and σh = σ3 (σ3 denotes minor principal stress)
σ1 = σv
σ3 = σh
θ
Next, how do we find the stresses acting on any arbitrary plane?
2.2 - 7
General Stress State on Any Plane
•
Consider the major principal plane, acted upon by the
major principal stress σ1 , and the minor principal plane,
acted upon by the minor principal stress σ3 .
σ1
σ3
τθ
θ
σn
•
By considering the equilibrium of the element shown, it can
be shown that on any plane, inclined at an angle of θ to the
direction of the major principal plane, there is a shear stress
τ and a normal stress σn . The magnitudes of these stresses
are :
σ n = σ3 + (σ1 − σ3 )cos 2 θ
2.2 - 8
σ1 − σ3
τ=
sin 2θ
2
(1a)
(1b)
Mohr Circle (in terms of Principal Stresses)
Re-writing and simplifying Eqs (1a) and (1b), we obtain:
σ nθ =
σ1 + σ3 σ1 − σ3
cos 2θ
+
2
2
σ1 − σ3
sin 2θ
τθ =
2
(2a)
(2b)
which relates σnθ and τθ to σ1 and σ3 following the figure below:
σ1
σ3
τθ
θ
σnθ
SIGN CONVENTION
• Stresses will be considered positive
compressive
when ___________.
• τ is positive when counterclockwise
_______________.
• θ is measured counterclockwise
_______________
from the direction of σ1 .
For a given σ1 and σ2, Eqs. 2a and 2b describe a circle when
plotted in the σn (x-axis) and τ (y-axis) space for 0<θ<180°.
This graphical representation of the state of stress is known as
the Mohr circle.
2.2 - 9
Mohr Circle (in terms of General Stress Components)
If the given stress state is not in terms of principal stresses,
then the normal stress σnθ and τθ can be expressed in terms of
the general normal stress components σx , σy and τxy as:
σ nθ =
τθ =
σy + σx
2
σy − σx
2
+
σy − σx
2
cos 2θ + τ xy sin 2θ
(3a)
sin 2θ − τ xy cos 2θ
Visually, this is represented as
(3b)
σy
θ
σx
τxy
τyx
τθ
σnθ
For a given σy, σx and τxy, Eqs. 3a and 3b describe a circle when
plotted in the σn (x-axis) and τ (y-axis) space for 0<θ<180°.
2.2 - 10
Mohr Circle – General Plane
Hence, for a soil element shown below with the stresses acting on the
two orthogonal (principal) planes and one arbitrary inclined plane:
σ1
τθ
θ
σ3
σnθ
By plotting the normal stress σn on the x-axis and τ on the y-axis for all
possible planes associated with any arbitrary θ, the equivalent Mohr
circle shown below is below:
τ
Note that
D
τθ
O
σ3
2.2 - 11
A
σnθ
σ1
2θ
C
B
σn
1. The point A corresponds to the
stress state (σ3 , 0) acting on the
vertical plane
2. The point B corresponds to the
stress state (σ1 , 0) acting on the
horizontal plane
3. The point D corresponds to the
stress state (σnθ , τθ ) acting on
the inclined plane
Mohr Circle – Failure Envelope
How does the Mohr circle play a part in providing information
about soil failure?
Recall that soil strength can be expressed in terms of
(i) effective friction angle φ’ and effective cohesion c’, and
(ii) the undrained shear strength su.
If using φ’ and c’, then we
can draw the following
failure envelope on the τ-σn
space
failure
envelope
τ
O
2.2 - 12
τ
failure
envelope
su
φ’
c’
If using su, then we can
draw the following failure
envelope on the τ-σn
space
σn
O
σn
Mohr Circle – Failure Stress State
The failure envelope represents the line or boundary which the
Mohr circle representing the soil stress state cannot cross.
The Mohr circle representing the soil stress state can, in the
limit, touch the failure envelope tangentially, but it cannot
cross or intersect the envelope.
Hence, the following Mohr circles represent possible soil
states when the soil element fails:
τ
D
τf
c’
A
σ1f
σnf
E
τf
φ
σ3f
2.2 - 13
τ
2θf
C
B
σn
c’
90°
σ3f
A
σ1f
C
B
σn
Mohr Circle – Failure Plane
For a soil with friction angle φ, when a given soil element fails, only
point D on the Mohr circle touches the failure envelope.
τ
D
τf
O
φ
σ3
A
σnf
2θf
C
B σn σ3
σ1
The stress state at point D occurs on the failure plane that is at an
angle of θf counterclockwise to the plane at which the major
σ1
principal stress acts.
σ3
τf
θf
σnf
2.2 - 14
Strength Envelopes
For a given friction angle φ, how do different soil stress states
plot on the Mohr circle space, from a ‘stable’ state to a ‘just
yielded’ state?
Mohr strength envelope
Y τ
O
2.2 - 15
φ
φ
A
B
C
σn
X
Circle B :
tangential to the strength envelope
condition of incipient failure
Circle A :
completely within the strength envelope
quite stable
Circle C :
beyond the strength envelope
cannot exist
Relationship between φ and θf
It is also helpful to obtain a relationship between the soil
strength parameter φ and the orientation of the failure plane θf in
the soil element.
τ
D
τf
O
φ
σ3
A
σnf
σn
2θf
C
B
σ1
Angle DCO = 180° – 2θf
Angle DOC = φ
Angle ODC = 90°
Sum of angles in ∆ODC
φ + 90° + 180° – 2θf = 180°
2.2 - 16
Hence θf = φ/2 + 45°
Example 1
On a failure plane in a purely frictional mass of dry sand the total
stresses at failure were:
shear = 3.5 kN/m2 ; normal = 10.0 kN/m2
Determine the resultant stress on the plane of failure, the angle of
shearing resistance of the soil, and the angle of inclination of the
failure plane to the major principal plane.
Y τ
D
3.5
O
φ
Resultant stress = OD =
σn
2θf
A
10
C
B
X
3.5 2 + 10 2 = 10.6 kN/m2
Angle of shearing resistance φ : tan φ = 3.5/10 = 0.35 ⇒ φ = 19.3°
2.2 - 17
Angle of inclination of failure plane θ = φ/2 + 45° = 54.6°
Example 2
40 kN/m2
Given
σ1
B
30°
20 kN/m2
20 kN/m2
σ3
τθ
θ
σnθ
B
40 kN/m2
Calculate the stresses on the plane B-B.
Solution θ = 30° ⇒ 2θ = 60°
Y τ
τθ
10
O
20
60°
σnθ
σn
40
X
10cos 60°
2.2 - 18
Normal stress on B-B σn = 30 + 10 cos 60° = 35 kN/m2
Shear stress on B-B τ = 10 sin 60° = 8.7 kN/m2
Example 3
20 kN/m2
B
τθ
30°
40 kN/m2
40 kN/m2
σ1
θ
Given
σnθ
- 60°
σ3
B
20 kN/m2
Calculate the stresses on the plane B-B.
Solution
θ = -60° ⇒ 2θ = -120°
Y τ
10cos 60°
O
τθ
2.2 - 19
σn
σnθ
20
120°
40
X
10
Normal stress on B-B σn = 30 - 10 cos 60° = 25 kN/m2
Shear stress on B-B τ = -10 sin 60° = -8.7 kN/m2
The Method of Poles
for Mohr Circle Solutions
2.2 - 20
Example 2 (using Method of Poles)
40 kN/m2
Given
B
30°
20 kN/m2
20 kN/m2
B
40 kN/m2
Calculate the stresses on the plane B-B (using the origin of planes or
the method of pole).
Solution
a) Draw the Mohr circle
b) Draw line A’-A’ through point (40,0) and
parallel to plane on which (40,0) acts.
c) Intersection of A’-A’ with Mohr circle at
point (20,0) is the origin of planes, OP.
d) Draw line B’-B’ through OP parallel to
B-B.
e) Read coordinates of point X where B’-B’
intersects Mohr circle.
2.2 - 21
Normal stress on B-B σn = 35 kN/m2
(35, 8.7)
X
B’
OP
30°
A’
60°
20
B’
A’
40
10cos 60°
Shear stress τ = 8.7 kN/m2
Example 3 (using Method of Poles)
20 kN/m2
Given
B
30°
40 kN/m2
40 kN/m2
B
20 kN/m2
Calculate the stresses on the plane B-B (using the origin of planes or
the method of pole).
Solution
a) Draw the Mohr circle
b) Draw line A’-A’ through point (20,0) and
parallel to plane on which (20,0) acts.
c) Intersection of A’-A’ with Mohr circle at
point (40,0) is the origin of planes, OP.
d) Draw line B’-B’ through OP parallel to
B-B.
e) Read coordinates of point X where B’-B’
intersects Mohr circle.
2.2 - 22
Normal stress on B-B σn = 25 kN/m2
OP B’
10cos 60°
A’
A’
20
B’
60°
30°
40
X
(25, -8.7)
Shear stress τ = -8.7 kN/m2
Example 4
40 kN/m2
Given
20 kN/m2
60°
B
B
20 kN/m2
40 kN/m2
Calculate the stresses on the horizontal plane B-B (using the origin
of planes or the method of pole).
Where is the pole (origin of planes)?
2.2 - 23
Example 4
40 kN/m2
Given
20 kN/m2
60°
B
B
20 kN/m2
40 kN/m2
Calculate the stresses on the horizontal plane B-B (using the origin
of planes or the method of pole).
Solution
O
a) Draw the Mohr circle
b) Draw line A’-A’ through point (40,0) and
parallel to plane on which (40,0) acts.
c) Intersection of A’-A’ with Mohr circle is
the origin of planes, OP.
d) Draw line B’-B’ through OP parallel to
B-B.
e) Read coordinates of point X where B’-B’
intersects Mohr circle.
2.2 - 24
P
A’
B’
Normal stress on B-B σn = 35 kN/m2
20
(35, 8.7)
X
40
B’
A’
Shear stress τ = 8.7 kN/m2
Example 5
Given
40 kN/m2
B
30°
10 kN/m2
B
10 kN/m2
20 kN/m2
Calculate the magnitude and direction of the principal stresses.
Solution
OP
B’
A’
a) Draw the Mohr circle
(20, 10)
b) Draw line A’-A’ through point (20,10) and
parallel to plane on which (20,10) acts.
A’
37.5°
c) Intersection of A’-A’ with Mohr circle
44.14
15.86
gives the origin of planes, OP.
B’
d) Draw line B’-B’ through OP so that it
intersects the major principal stress
(40, -10)
at (44.1, 0).
e) Measure the angle the line B’-B’ makes with the horizontal = 37.5°
f) The direction of the principal stress = 52.5° to the horizontal.
2.2 - 25