NOTES ON MOHR’S CIRCLES AND SHEAR STRENGTH
  
  

A
A
 cos    n 
 sin 
cos
cos 
A
A
 FH  0   n  cos  sin    n  cos  cos
F
v
 0   1  A   n 
From (2):
 n A tan    n A
From (1):
1  A
1
=  n A   n A tan 2 
  n 1  tan 2  
 cos 2   sin 2  
1
n 

cos 2 


2
 1 cos    n
σn
=
σ1
1+cos2θ 
2
τn
=
σ1cos 2θtanθ
τn
=
σ1sinθcosθ
n

1
2
 sin 2
1
 2
Plot of σn vs τn for various values of θ.
From the geometry of the circle,
n 
n 
1 1
2
1
2

2
 cos 2
 sin 2
When the soil sample has confining pressure, σ3 , as well as axial pressure, σ1 ; the normal
and shear stresses are:
   3    1   3  cos 2
n  1
2
2
   
 n  1 3 sin 2
2
The normal and shear stress curves move apart
Plot of σn vs τn
The equations for normal and shear stress can be found from the geometry of the circle
   3    1   3  cos 2
n  1
2
2
   
 n  1 3 sin 2
2
Shear strength comes from two sources
1. CohesionCohesive shear strength, c
2. Friction Angle of internal friction, φ
or Angle of shearing resistance
For Unconfined Test,

 n   1 1 2  1 2 cos 2
 n   1 2 sin 2
2 f  90   '

For Confined Test,
   3 1   3
n  1

cos 2
2
2
 
 n  1 3 sin 2
2
Strength of Clay
Effective Stress:
 '= -u
 1'= 1 -u
 3'= 3 -u
if  3' >  3 , then,
 3'- 3  u is positive
u = negative
x
c
sin  '
 3'  x  c

c
c
sin  '
 1

 1
 sin  ' 
 3'  c 
3  0
1  sin  ' 

 sin  ' 
 3'   3  u  c 
Suppose,  ' = 30
c  5 psi
1  0.5 
u   5 psi 
 0.5 
u   5 psi
In fact, the cohesion in clay (all soils, actually) is due to the negative porewater pressure
in that soil.
Strength of Sand
In the absence of pore water, the strength of sand is due almost entirely to friction.
SAND STRENGTH
Strength Tests for Sand
1. Direct Shear (Poor test)
2. Triaxial Test (Much better test)
-Uniform stress: fields in the center of the sample
-Uniform strain: fields in the center of the sample
Most soils are a mixture of fine and coarse grained sizes so they usually have both c and
φ properties. How much c depends on whether the sample is loaded slowly enough to
allow the pore water pressure to dissipate. Tests are classified by the rate with which
-Confining pressure
-Deviator stress
are applied.
Rate of Application
Confining
Symbol for
Deviator Stress
Type of Test
Pressure
Test
(drainage permitted
(consolidation
during test)
prior to test)
Q
(UU)
Rapidly
Rapidly
Unconsolidated – Undrained c, φ
R
(CU)
Slowly
Rapidly
Consolidated- Undrained (may
make measurements of pore water
pressures)
S
(CD)
Slowly
Slowly
Consolidated- Undrained c', φ'
Total Stress properties are measured in the UU test
Effective pressure properties are measured in the CD test
Effective pressure properties may be determined in the CU test if pore water pressures are
measured during the undrained portion of the test.
There are 3 pairs of quantities: σ1, σ3 ;
σf, τf (failure plane);
c,φ
The radiusof the circle is r = (σ1-σ3)/2
The projection of the failure line is x=c/tanφ
Normally, if you are given any three of the quantities you can find the other three.
1. Suppose you know σ1, σ3 and c
Find expressions for σf, τf and φ or tanφ
2. Suppose you know σ1, σf and τf
Find expressions for σ3, c and φ
3. Suppose you know σ1, σf and c
Find expressions for σ3, τf and φ
Problem 01
Given :  1  15 kPa
 3  5 kPa
c  2 kPa
Find :  f  8.45 kPa
 f  4.76 kPa
  18
Solution:
r
 1   3   15  5  5 kPa
2
 f  5cos 
2
 f  10  5sin 
x
2
tan 

2 
R sin   sin  (10  x)  10 
 sin   5
tan  

10sin   2 cos   5
2
5
cos  
10
10
  20 :
0.529  0.5
sin  
  18 :
0.499  0.5
 f  5cos   4.76 kPa
 f  10  5sin   8.45 kPa
Problem 02
Given :  1  20 kPa
 f  9 kPa
 f = 7 kPa
c  2.80 kPa
Find :
  25
 3  4.54 kPa
Solution:
c  7  9 tan 
20   3
2
c
7
x

9
tan  tan 
20  r  r sin   9
r
20  r  r 1  sin    9
 20   3
20   3
c 
R sin   

sin   r 

tan  
2
 2
20   3
20   3
 sin   c cos  
2
2
20   3
20 
(1  sin  )  9
1
2
20   3
20   3
sin    7  9 tan   cos  
2
2
From 1 :
20   3
(1  sin  )
2
2(20  9)

 20   3
(1  sin  )
 20-9  
From  2  :
3
2
3
2
sin  
3
2
 10sin   7 cos   9sin   10
1  sin    10  10sin   9sin   7 cos 
2
3
2
1  sin    10 1  sin    9sin   7 cos 
 1  sin  
 sin  
 cos  
  18 
  14 

 1  sin  
 1  sin  
 1  sin  
 3  20 

22 

 1  sin  
 3  20  
20 1  sin    22 20 1  sin    18sin   14 cos 

1  sin  
1  sin  
20  20sin   22  20  20sin   18sin   14 cos 
22sin   22  14 cos 
 f
2 f
1  sin  14



cos 
22 2  1   f    1   f
1  sin  7
  0.636
cos 
11
1  sin   0.636 cos 
  30 :
1  1.051
  29 :
1  1.041
  25 :
1  0.999
c  7  9 tan 25
c  2.80 kPa
 3  20 
22
1  sin 25
 3  4.54 kPa



Problem 03
Given :  1  20 kPa
 f  8 kPa
Find :
c = 3 kPa
 3  3.65 kPa
 f  7.24 kPa
  27.9
Solution:
1
 2
 f  3  8 tan 
20  r  r sin   8
20   3
2
3
x
tan 
 3
r
 4
 20   3
20   3
3 
R sin   

sin  

tan  
2
 2
20   3
f 
cos 
 6
2
From  2  :
20  r (1  sin  )  8
 20   3 
20  
 (1  sin  )  8
 2 
 20   3 

 (1  sin  )  20  8  12
 2 
20(1  sin  )   3 (1  sin  )  24
From  5  :
 20   3 
 20   3 

 sin   3cos   

 2 
 2 
10sin  
3
2
sin  
3
2
 3cos   10
 5
3
2
1  sin    3cos   10  10sin   10 1  sin  
 3 1  sin    20 1  sin    6 cos 
 3 1  sin    20 1  sin    24
 B
From  A 
20 1  sin    6 cos   20 1  sin    24
20  20sin   20  20sin   6 cos   24
40sin   6 cos   24
20sin   3cos   12
sin   0.15cos   0.6
Trial and Error:
  30 :
0.630  0.6
  27 :
0.588  0.6
  28 :
0.602  0.6
  27.9 : 0.6005  0.6
 f  8 tan   3
 f  7.24 kPa
 3  20 
24
1  sin 
 3  3.65 kPa
Problem 01
Draw the sketch of the Mohrs Circle, marking all that is known and all that is to be found.
Solution:
 1   3 
r
2
r cos   f
 1   3   r sin   
2
x
f
c
tan 
  1   3 
 1   3 
c 


 sin   r 
2
tan  
2

 1   3  sin   c cos    1   3 
2
2


2c
sin   
 cos  
  1   3  
f 
f 
 1   3 
2
 1   3 
2
 1   3 
 1   3 
cos 

 1   3  sin 
2
Find  by trial and error
Problem 02
Given :  1 ,  f ,  f
Draw Sketch
Find : c,  ,  3
Solution:
c   f   f tan 
r
1   3
2
c
x
tan 
 1  r 1  sin     1  r  r 1  sin     f   1 
1   3
2
1  sin  
 1   3
1   3
c 
 2  tan   sin   2


1   f 

 f
1   3
2
(1  sin  )
 
1 3
(1  sin  )
1   3
 3
sin   c cos   1
2
2
2
1
2
1
2
1
2
1
sin  
1
2
 c cos   
 sin   1  c cos   
3
2
3
2
sin  
3
2
(1  sin  )
 
 sin   1   f   f tan   cos    1 

Then  3   1 
And
c=
2
 1   f
1  sin 
1   3
2


 3
1
 1
tan 
cos 
2
2

1
 1   f   (1  sin  )

1  sin 

Find  by trial and error
 f cos    f sin   
1
2
1  sin   
1
2
 sin   1   1   f 
1
 1  sin   sin  1    1   f

2 
=
=
1
2
 2 sin    1   f
 f cos    f sin    f   1 1  sin  
 f cos    f 1  sin     1 1  sin  

cos  

 1  sin   
f 
  1   f
cos 
1  sin  


1
 f


f
Problem 03
Given :  1 ,  f , c
 3, f , 
Find :
Solution:
r
1   3
1
2
c
x
tan 
 2
   3
 3
c 
R sin    1

sin   1

tan  
2
 2
c   f   1 tan 
 4
f

1   3
2
cos 
 5
    3   1   3 
 1

 sin 
 2   2 
From  6  :
f
1 1
sin  
3

6
3
sin    f
2
2
2
 1 1  sin     3 1  sin    2 f
2

 3
A
(1  sin  )
(1  sin  )
From (3):
c cos  
 1   3  sin    1   3 
c cos  
1
2
1
2
sin  
1  3
2

sin  
3
3
2
0
1  sin    0
2
2
 3 1  sin     1 1  sin    2c cos 
c cos  
1  sin   
2
2
 B
From (A):
 3 1  sin     1 1  sin    2 f
From (B):
 3 1  sin     1 1  sin    2c cos 
 1 1  sin    2 f   1 1  sin    2c cos 
2 1 1  sin    2 f  2c cos 
  f  c cos  

1


1  sin    
Find  by trial and error
Then,  f  c   f tan 
And,  3 
2 f   1 1  sin  
1  sin  
Check Derivation
From(6):
 1   3   1   3 

 sin 
 2   2 
     
 f   1  1  3   1 3  sin 
2
2  2 
 f  1  
 1   3   1   3 

 sin 
 2   2 
f 
f 
1 1
2

2
sin  
3
2

3
2
sin 
2 f   1 1  sin     3 1  sin  
Alternatively,
 1   3 
 1  sin  
 2 
 f  1  
2  1   f    1   3 1  sin  
2  1   f    1 1  sin     3 1  sin  
 3 1  sin     1 1  sin    2  1   f 
 3 1  sin     1 1  sin    2c cos 
 1 1  sin    2  1   f    1 1  sin    2c cos 
 1   1 sin    1   1 sin   2  1   f   2c cos 
2 1 sin   2  1   f   2c cos 
sin  
sin  

1
 f
1
c cos 
1
  c cos 
1


1
 f
1
