CHAPTER 11
MOHR CIRCLE, FAILURE THEORIES, AND STRENGTH
TESTING OF SOIL AND ROCKS
11-1. Given an element with stresses as indicated in the figure, find: (a) The major and minor
principal stresses and the planes on which they act. (b) The stresses on a plane inclined at 30°
from the horizontal. (c) The max. shear stress and the inclination of the plane on which it acts.
SOLUTION:
113.15 6.85 20, 35 60, 53.15 49.69, 52.14 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 0 10 20 30 40 50 60 70 80 90 100 110 120 Normal stress (kPa)S h e a r s tr e s s (k P a
pole
solution continued on
next page
11-1 continued
x y xy x y 2 2 x y 2 2 xy 1Solve using Mohr's circle and the pole method (see plot on next page).
20 kPa,
100 kPa,
35 kPa
20 100
center
60 kPa
2
2
20 100
radius
(35)
53.15 kPa
2
2
(a)
center
radiu
3 y x 1 p1 1 o 1 y x y x x1 xys
60
53.15
113.15 kPa
center
radius
60
53.15
6.85 kPa
1
Pr incipal angle,
cos
2
2R
1
100
20
cos
20.59 measured cw from the x-axis at .
2
2 53.15
(b)
cos 2
sin 2
2
2
x1 y x x1y1 xy max x y 1 s xy100
20
100
20
cos(2 30) (35) sin(2 30)
49.69 kPa
2
2
100
20
sin 2
cos 2
sin(2 30) (35) cos(2 30)
52.14 kPa
2
2
(c)
radius
53.15 kPa
1
Orientation of max shear stress,
tan
2
2
1 o s1
20 100
tan
24.41 measured ccw from horizontal
2
2 35
11-2. Work Problem 11.1 with the element rotated 30° clockwise from the horizontal.
SOLUTION:
x y xy x y 2 2 x y 2 2 xy 1Solve using Mohr's circle and the pole method (see plot on next page).
20 kPa,
100 kPa,
35 kPa
20 100
center
60 kPa
2
2
20 100
radius
(35)
53.15 kPa
2
2
(a)
center
radiu
3 y x 1 p1 1 o 1s
60
53.15
113.15 kPa
center
radius
60
53.15
6.85 kPa
1
Pr incipal angle,
cos
2
2R
1
100
20
cos
30
50.6 measured cw from the x-axis at
.
2
2 53.15
(b) At the given orientation of 3
x1 x1y1 y1 x1y1 max x y 1 s xy 1 o s0 deg:
20 kPa
35 kPa
and
100 kPa
35 kPa
(c)
radius
53.15 kPa
1
Orientation of max shear stress,
tan
2
2
1
20 100
tan
30
5.6
2
2 35
11-2 continued
113.15 6.85 70, 52 60, 53.15 9.3, 16 pole -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 0 10 20 30 40 50 60 70 80 90 100 110 120Normal stress (kPa)
S hear st re ss ( k P a
11-3. With the element of Problem 11.1 rotated 40° clockwise from the horizontal, find the
magnitude and direction of the stresses on the vertical plane.
SOLUTION:
x y xy x y 2 2 x y 2 2 xy 1Solve using Mohr's circle and the pole method (see plot on next page).
20 kPa,
100 kPa,
35 kPa
20 100
center
60 kPa
2
2
20 100
radius
(35)
53.15 kPa
2
2
center
radius
6
0
53.15
113.15 kPa,
3center
radius
60
53.15
6.85 kPa
From the Mohr circle:
pole
(98,37)
stresses on a vertical plane
(98, 37)
113.15 6.85 100, 35 pole (98, 37) 20, -35 on vertical plane: (98, -37) -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 0 10 20 30 40 50 60 70 80 90 100 110 120 Normal stress (kPa)S hear s tr e ss (k P a
11-4. Work Example 11.3 with the element rotated 30° clockwise from the horizontal. In addition,
find the stresses (magnitude and direction) on the horizontal plane.
SOLUTION:
x y xy x y 2 2 x y 2 2 xy 1Solve using Mohr's circle and the pole method (see plot on next page).
4 kPa,
6 kPa,
2 kPa
4
6
center
1kPa
2
2
4
6
radius
(2)
5.38 kPa
2
2
center
radius
1 5.38
6.38
3 1 3 p 1kPa,
center
radius
1 5.38
4.38 kPa
The following values can be determined directly from the Mohr circle
Pole = (0.6, 5.4)
(a) Stress on a horizontal plane = (0.6, 5.4)
(b)
6.38 kPa
and
4.38 kPa
(c)
o o p 3 max43 (rotated cw from x-axis)
and
47
(rotated ccw from x-axis)
(d)
5.38 kPa
Mohr circle shown on next page
11-4 continued
6, 2 6.38 -4.38 pole (0.6, 5.4) -4, -2 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11Normal stress (kPa)
S he a r st re ss ( k P a
11-6. The state of plane stress in a body is described by the following stresses:
1= 8500 kN/m
2compression,
3= 1500 kN/m
2tension. Determine by means of the Mohr circle the normal stress
and shear stress on a plane inclined at 20° to the plane on which the minor principal stress acts.
Check the results analytically. (After A. Casagrande.)
SOLUTION:
1 3 xy x y 2 2 x y 2 2 xy 3Solve using Mohr's circle and the pole method (see plot on next page).
8.5 MPa,
1.5 MPa,
0
8.5 1.5
center
3.5 MPa
2
2
8.5 1.5
radius
(2)
5 MPa
2
2
Assuming
acts in th
1 3 1 3 1 3e horizontal direction on a vertical plane. Pole = (-1.5, 0)
8.5 1.5
8.5 1.5
cos 2
cos(2 20)
7.33 MPa
2
2
2
2
8.5 1.5
sin 2
sin(2 20)
3.21 MPa
2
2
8.5
-1.5
pole (-1.5, 0)
(7.33, 3.21)
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
Normal stress (MPa)
S
hear
st
re
ss
(M
p
a
11-7. At a certain critical point in a steel beam, on a vertical plane the compressive stress is 115
MPa and the shearing stress is 31.5 MPa. There is no normal stress on the longitudinal
(horizontal) plane. Find the stresses acting on the principal planes and the orientation of principal
planes with the horizontal. (After Taylor, 1948.)
SOLUTION:
x y xy x y 2 2 x y 2 2 xy 1Solve using Mohr's circle and the pole method (see plot below).
115 MPa,
0,
31.5 MPa
115
0
center
57.5 MPa
2
2
115
0
radius
(31.5)
65.56 MPa
2
2
center
radius
57.5
65.
3 o o p 1 p 256
123.1 MPa
center
radius
57.5
65.56
8.06 MPa
76 (rotated ccw from x-axis)
and
14
(rotated cw from x-axis)
123 -8.1 pole (115, -31.5) (115, 31.5) -80 -60 -40 -20 0 20 40 60 80 -10 10 30 50 70 90 110 130 150Normal stress (MPa)
S h e a r s tre s s (M p a
11-8. A soil sample is under a biaxial state of stress. On plane 1, the stresses are (13, 4), while
on plane 2, the stresses are (5.8, -2). Find the major and minor principal stresses.
SOLUTION:
Solve using Mohr's circle and the pole method (see plot below).
Plot given points (13, 4) and (5.8, -2), and construct a perpendicular bisector.
The perpendicular bisector crosses the x-axis at the cent
1 3
er of the circle.
center = (9.8, 0)
Draw the circle and graphically determine the principal stresses.
15.1 and
5.3
15.1
5.3
(5.8, -2)
(13, 4)
9.8
-5
-3
-1
1
3
5
7
9
0
2
4
6
8
10
12
14
16
18
11-9. For the element shown in the figure: (a) Find the magnitude of the unknown stresses
hand
hon the horizontal plane. (b) Find the orientation of the principal stresses; clearly indicate
their orientation in a small sketch. (c) Show the orientation of the planes of maximum as well as
minimum shear.
SOLUTION:
Solve using Mohr's circle and the pole method using the given stresses.
Plot (2, -2) and (5, 3), and construct a perpendicular bisector.
The perpendicular bisector crosses the x-axis at the center of th
h h
1
e circle.
Center = (4.4, 0),
Radius = 3.1
Draw the circle and graphically determine the pole at (7.5, 0).
(a)
1.3
and
0
acting on a horizontal plane
(b) Principal stresses:
7.5 (on vertical plane), and
3 o max o min1.3 (on horizontal plane)
(c)
3.1 on a plane oriented 45 cw from the horizontal
3.1 on a plane oriented 45 ccw from the horizontal
(2, -2) (5, 3) Pole = (7.5,0) 1.3 4.4 (4.4, 3.1) (4.4, -3.1) -4 -3 -2 -1 0 1 2 3 4 0 1 2 3 4 5 6 7 8 Normal stress S hear s tr es s11-10. Given the element with stresses as shown in the figure: (a) Find the magnitude and
direction of
Hand
H. (b) Find the magnitude and direction of
1and
3.
SOLUTION:
Solve using Mohr's circle and the pole method using the given stresses.
Plot (4, 3) and (8, -2), and construct a perpendicular bisector.
The perpendicular bisector crosses the x-axis at the center of th
H H
1
e circle.
Center = (5.35, 0),
Radius = 3.3
Draw the circle and graphically determine the pole at (8.65, 0.3).
(a)
2.05
and
0.3
on the horizontal plane
(b) Principal stresses:
8.65 (on a plane oriented 8
oo 3
7.4 cw from the horizontal)
2.05 (on a plane oriented 2.6 ccw from the horizontal)
8, -2 8.65, 0.3 4, 3 2.05 5.35 Pole: 8.65, 0.3 2.05, 0.3 -5 -4 -3 -2 -1 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 11 Normal stress She ar s tr e s s11-11. Given the data of Example 11.5. (a) Find the magnitude of the stresses on the horizontal
plane. (b) Find the maximum shear stress, and determine the angle between the plane on which
it acts and the major principal plane.
SOLUTION:
x y x y xy 2 2 x y 2 2 xy h hSolve using Mohr's circle and the pole method (see plot below).
8
4
8 kPa,
4 kPa,
2 kPa,
center
6 kPa
2
2
8
4
radius
(2)
2.83 kPa
2
2
(a)
8 kPa,
2 kPa
(on the horizontal
o maxplane)
(b)
2.83 kPa
(on a plane rotated 45 ccw from the major principle plane )
8, 2 8.83 3.17 pole (4, 2) 8, -2 6, 2.83 -5 -4 -3 -2 -1 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 11
Normal stress (kPa)
S hear s tr e ss ( k P a
11-12. The state of stress on a small element is
v= 21 kPa,
h= 10 kPa, and the shear stress
on the horizontal plane is +3 kPa. (a) Find the magnitude and directions of the major and minor
principal stresses. (b) If the material is a loose sand, can you say whether the element is in a
state of failure? If it isn’t, how close is it? Why? State your assumptions clearly. (Assume
’ = 28
ofor the loose sand.)
SOLUTION:
x y xy x y 2 2 x y 2 2 xy 1Solve using Mohr's circle and the pole method (see plot below).
21kPa,
10 kPa,
3 kPa
21 10
center
15.5 kPa
2
2
21 10
radius
(3)
6.265 kPa
2
2
(a)
center
radius
15.5
6.2
3 o o p 1 p 365
21.76 kPa
center
radius
15.5
6.265
9.24 kPa
14.3 (rotated cw from horizontal)
and
104.3
(rotated cw from horizontal)
(b) The element is not in a state of failure. The shear stress on the fail
ure plane is about
1.2 kPa less than the shear stress on the failure plane at failure.
This could be expressed as a factor of safety:
shear stress on the failure plane
FS
shear stress on the failure pl
6.44
1.2
ane at failure
5.28
21.76 9.24 10, -3 21, 3 Pole: (10, 3) -12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12 0 2 4 6 8 10 12 14 16 18 20 22 24 26Normal stress (kPa)
S h ear st re ss ( k P a
11-13. Given the vertical and horizontal normal stresses of Problem 11.12. Find the maximum
values of shear stress on the horizontal and vertical planes to cause failure in a medium dense
sand. Assume the angle of internal friction for the sand is 32°.
SOLUTION:
As shown in the Mohr circle diagram, a value of = 6.1 kPa on the vert. and horz. planes
will result in a failure condition.
23.71 7.29 10, -6.1 21, 6.1 10, 6.1 -12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12 0 2 4 6 8 10 12 14 16 18 20 22 24 26Normal stress (kPa)
S h e a r s tre s s (k P a
11-14. The state plane stress in a mass of dense cohesionless sand is described by the following
stresses:
Normal stress on horizontal plane = 296 kPa
Normal stress on vertical plane = 160 kPa
Shear stress on horizontal and vertical planes = +/- 64 KPa
Determine by means of the Mohr circle the magnitude and direction of the principal stresses. Is
this state of stress safe against failure? (After A. Casagrande.)
SOLUTION:
x y xy x y 2 2 x y 2 2 xy 1Solve using Mohr's circle and the pole method (see plot below).
160 kPa,
296 kPa,
64 kPa
160
296
center
228.0 kPa
2
2
160
296
radius
(64)
93.38 kPa
2
2
(a)
center
radius
3 o o p 1 p 3 o228
93.38
321.38 kPa
center
radius
228
93.38
134.62 kPa
21.63 (rotated cw from horizontal)
and
111.63
(rotated cw from horizontal)
(b) For dense sand, assume '
38 .
The element is not in a state
of failure. The shear stress on the failure plane is about
61 kPa less than the shear stress on the failure plane at failure.
This could be expressed as a factor of safety:
shear stress on the fa
FS
ilure plane
125
2.0
shear stress on the failure plane at failure
64
321.38 134.62 160, -64 296, 64 160, 64 -180 -140 -100 -60 -20 20 60 100 140 180 0 40 80 120 160 200 240 280 320 360
Normal stress (kPa)
S hear st re ss ( k P a
11-15.
At a given point within a sand deposit the major, intermediate, and minor principal
stresses are 10, 6, and 4 Mn/m
2, respectively. Construct the Mohr diagram, and from it scale the
normal and shearing stresses and the obliquity angles on planes at 35°, 50°, 65°, and 80° from
the major principal plane. (After Taylor, 1948.)
SOLUTION:
Angle (deg)
Normal stress
(MN/m
2)
Shear Stress
(MN/m
2)
35 8.5 2.60
50 7.75 2.89
65 7.30 2.97
80 7.05 3.00
35 deg
50 deg
65 deg
80 deg
Pole: (4, 0)
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
0
1
2
3
4
5
6
7
8
9
10
11
12
Normal stress (MN/m^2)
S
hear
st
re
ss
(M
n/
m
^
2
11-16.
A 1-m cube within a mass of stressed soil has a stress of 200 kPa on its top and bottom
faces, 100 kPa on one pair of vertical faces, and 60 kPa on the other pair of vertical faces. There
is no shear stress on any face. Fill in the following table. (After Taylor, 1948.)
SOLUTION:
1 2 3 x y 2 2 x y 2 2 xy max o ff ff200 kPa,
100 kPa,
60 kPa
200
60
Center
130.0 kPa
2
2
200
60
Radius
(0)
70.0 kPa
2
2
R
sin '
'
32.58
C
x
cos(90
')
;
C
x
R
x
cos(90
32.58)
x
37.693,
92.3 kPa
70
sin(9
ff ff0
')
;
70 sin(90
32.58)
58.98 kPa
R
(kPa) (kPa) (deg)Major principal plane 200 0 0
Intermediate principal plane 100 0 90
Minor principal plane 60 0 90
Plane of maximum shearing stress 130.0 70.0 45
Plane of maximum obliquity 92.3 58.98 32.58
200, 0 Pole:(60, 0) 130, 70 92.3, 58.98 -100 -80 -60 -40 -20 0 20 40 60 80 100 0 20 40 60 80 100 120 140 160 180 200
Normal stress (kPa)
S hear s tr e ss ( k Pa
11-17. In Problem 11.16 what is
, assuming c = 0?
SOLUTION:
1 2 3 x y 2 2 x y 2 2 xy max o200 kPa,
100 kPa,
60 kPa
200
60
Center
130.0 kPa
2
2
200
60
Radius
(0)
70.0 kPa
2
2
R
sin '
'
32.58
C
200, 0 Pole:(60, 0) 130, 70 92.3, 58.98 -100 -80 -60 -40 -20 0 20 40 60 80 100 0 20 40 60 80 100 120 140 160 180 200Normal stress (kPa)
S hear s tr e ss ( k Pa
11-19.
(a) Draw the Mohr circle for this point, showing the pole location. (b) What are the
stresses acting on a horizontal plane passing through this point? (c) The cohesion intercept for
this soil is and the friction angle is If the major principal stress remains the same, what would the
minor principal stress have to be to cause failure?
SOLUTION:
Plot (40, 10) and (20, 10), and construct a perpendicular bisector.
The perpendicular bisector crosses the x-axis at the center of the circle.
Center = (30, 0),
Radius = 14.14 psi
(a) Draw the circle (see
1 3
H H
3
circle 11-19a) and graphically determine the pole at (26.2, 13.5).
44.1psi,
15.9 psi
(b)
34.1psi
and
13.5 psi
on the horizontal plane
(c)
2.05 psi
(see Mohr circle 11-19b)
30, 0 44.14, 0 15.86, 0 20, 10 Pole (26.2, 13.5) 40, 10 34.1, 13.5 -10 0 10 20 0 10 20 30 40 50 44.14, 0 11.00, 0 -10 0 10 20 0 10 20 30 40 50Circle 11-19a
values in psi
Circle 11-19b
values in psi
11-20.
The figure shows an element of soil at the interface between two dry sand layers on a 28°
slope. The interface is 10 ft below the ground surface, and for both sand layers the friction angle
is 34° and K
o= 0.44. Assume that the shear stress is zero on both the vertical and horizontal
planes. (a) Draw the Mohr circle for this point, and determine the pole location. (b) Determine
the normal and shear stresses on the soil interface (i.e., on the 28° plane). (c) What is the shear
stress on the failure plane (
f) and what is the shear stress on the failure plane at failure (
ff)?
Use these values to determine the factor of safety.
SOLUTION:
v h o v
'
(112 pcf )(10 ft)
1120 psf,
'
K
'
(0.44)(1120)
492.8 psf
Plot (1120, 0) and (492.8, 0), and draw the Mohr circle (see below).
Center = (806.4, 0),
Radius = 313.6 psf
(a) Draw the circle and graphically
1 3 o 28 28 o o 3determine the pole at (492.8, 0).
44.1 psf,
15.9 psf
(b)
1008 psf
and
241 psf
at 28 from the horizontal plane
'
34
(c) 45
45
62
draw the failure plane from
at 62 from the horizontal
2
2
from the M
f ff ff fohr circle,
259 psf
and
352 psf
352
FS
1.36
259
1120.0, 0.0 492.8, 0.0 Pole (492.8, 0) 1008, 241 631, 259 631, 352 -400 -200 0 200 400 600 0 200 400 600 800 1000 1200 Normal stress (psf) S h e a r s tre s s (p s f11-26.
In a direct shear test on a specimen of cohesionless sand, the vertical normal stress on
the specimen is 240 kN/m
2and the horizontal shear stress at failure is 160 kN/m
2. (a) Assuming
uniform stress distribution within the failure zone and a straight line failure envelope which goes
through the origin, determine by means of the Mohr circle the magnitude and direction of the
principal stresses at failure.
SOLUTION
o
Plot (240, 160) and (0, 0). This defines the failure envelope at 33.69 .
A normal to the failure envelope crosses the x-axis at the center of the circle.
Center = (346.66, 0),
Radius = 192.29 kPa
(a) Dra
o 1
o 3
w a horizontal line from (240, 160) to locate the pole at (453.32, 160).
538.95 kPa
at 61.8 cw from the horizontal
154.37 kPa
at 28.2 ccw from the horizontal
346.66
538.95
154.37
240, 160
Pole (453.32,
160)
-150
-100
-50
0
50
100
150
200
250
300
0
50
100 150 200 250 300 350 400 450 500 550 600
Normal stress (kPa)
S
hear
st
re
ss
(
k
P
a
11-27. A specimen of sand is tested in direct simple shear. The stress conditions are shown.
Initial conditions:
v= 3.12 kg/cm
2, K
o= 0.5 At failure:
v= 3.12 kg/cm
2,
hv= 1.80 kg/cm
2(a) Draw the Mohr circles for both initial and final stress conditions. (b) Show clearly the locations
of the poles of these circles. (c) Determine the magnitude and orientation of the principal
stresses at failure. (d) What is the orientation of the failure plane? (e) If the shear strain at failure
is 10° as shown in the figure, what are the stresses and on the sides of the specimen at failure?
SOLUTION:
2 2 2 kg h o v cm kg cm kg cmK
(0.5)(3.12)
1.56
(a) See Mohr circle plot below.
(b) Initial Circle: Center = (2.34, 0),
Radius = 0.78
,
Pole
(1.56, 0)
Final Circle: Center = (2.34, 0),
Radius = 1.96
,
Pole
(1.56,
-
2 2 2 2 o kg 1 cm 1 o kg 3 cm 3 o kg kg s cm s cm1.8)
(c)
4.30
,
33.3 ccw from horizontal
1.96
,
56.7 cw from horizontal
(d) refer to the Mohr circle diagram
(e) As shown in the diagram,
2.22
,
1.96
at 10
3.12, -1.8
4.30
1.56, -1.8
Pole: 1.56
1.56, 1.8
0.38
Pole: (1.56
-1.8)
2.22, 1.96
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
Normal stress (kPa)
S
h
e
a
r s
tre
s
s
(k
P
a
Initial condition
Final condition
11-28.
Two conventional CD triaxial compression tests were conducted on a dense angular dry
sand at the same void ratio. Test A had a confining pressure of 150 kPa, while in test B the
confining pressure was 600 kPa; these stresses were held constant throughout the test. At
failure, tests A and B had maximum principal stress differences of 600 and 2550 kPa,
respectively. (a) Plot the Mohr circles for both tests at initial conditions and at failure. (b)
Assuming c = 0, determine
. (c) What is the shear stress on the failure plane at failure for both
tests? (d) Determine the theoretical orientation of the failure plane in each specimen. (e) What is
the orientation of the plane of maximum obliquity?
SOLUTION:
1 3 xy
1 3 xy
x y
(a) Solve using Mohr's circle and the pole method (see plot).
Test A :
750,
150,
0,
center
435,
radius
285,
pole
(150, 0)
Test B :
3150,
600,
0,
center
1875,
radius
1275,
pole
(600, 0)
center
,
r
2
2 x y 2 xy o 1 3 o o o o maxadius
2
(b) From the plot, '
43
(c and d)
sin 2
2
750 150
Test A :
66.5 ,
sin(2 66.5)
219.4
2
3150
600
Test B :
66.5 ,
sin(2 66.5)
932.5
2
(e)
66.5 ,
'
43
3150 750 600 150 -1600 -1200 -800 -400 0 400 800 1200 1600 0 400 800 1200 1600 2000 2400 2800 3200Normal stress (kPa)
S h e a r s tre s s ( k P a
11-29. Two consolidated–drained triaxial tests were performed on specimens of the same clay,
with the following results at failure:
Test 1:
’
1= 73.4 psi,
’
3= 26.6 psi
Test 2:
’
1= 48.0 psi,
’
3= 12.0 psi
Determine the effective Mohr–Coulomb failure envelope (
’ and c) based on these test results.
SOLUTION:
o
From the M-C plot shown below:
17 and c
10 psi.
73.4
12
26.6
48
-50
-40
-30
-20
-10
0
10
20
30
40
50
0
10
20
30
40
50
60
70
80
90
100
Normal stress (kPa)
S
h
e
a
r s
tre
s
s
(k
P
a
Mohr-Coulomb
failure envelope
11-30.
A triaxial specimen of loose sand is first consolidated nonhydrostatically, with
1= 15 kPa
and
3= 10 kPa. The sample is then failed by holding the vertical stress constant and decreasing
the horizontal stress (this is a lateral extension test). The angle of internal friction is 30° (c = 0).
(a) Draw the Mohr circles for both initial and “at failure” conditions. (b) What will be the major and
minor principal stresses at failure?
SOLUTION:
1 3
(a) See plot below.
(b) At failure:
15 kPa,
5 kPa
15
5
10
-10
-5
0
5
10
0
5
10
15
20
Normal stress (kPa)
S
hear
st
re
ss (
k
P
a
Initial condition
Failure condition
11-31.
Another sample of the same sand tested in Problem 11.30 (consolidated
nonhydrostatically, with
1= 15 kPa and
3= 10 kPa) is tested by holding the vertical stress
constant and increasing the horizontal stress (this is a lateral compression test). The angle of
internal friction is 30° (c = 0). (a) Draw the Mohr circles for both initial and “at failure” conditions.
(b) What will be the major and minor principal stresses at failure?
SOLUTION:
1 3
(a) See plot below.
(b) At failure:
45 kPa,
15 kPa
15 10 45 -30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30 0 5 10 15 20 25 30 35 40 45 50 55 60 65
Normal stress (kPa)
S h e a r s tre s s (k P a