Shear strength
5.1 General
giant landslide in Tibet 2000
Shear strength
5.1 General
granary foundation failure on clay
Shear strength
5.1 General
Foundation soil liquefaction caused by earthquake
Shear strength
5.1 General
Embankment
Retaining wall
Foundation
Key problem: shear strength
Shear strength
5.1 General
Collapse
Rotational
slip
Translation
slip
flow slide
Shear strength
5.2 Coulomb law
τf = c + σ’ tan φ
τf = shear strength
c = cohesion
φ = angle of internal friction
f

粘土
c
σ1

major principle stress
σ3
σ3
Minor principle stress
Confining stress
σ1
Shear strength
Consider the following situation:
A normal stress is applied vertically and held constant
A shear stress is then applied until failure
3
3

1


3

1
dlcos
dlsin
1
Shear strength
• For any given normal stress, there will be one value of shear stress
• If the normal stress is increased, the shear stress will typically increase in sands
and stay the same in clays

O
A(, )
3

2
1/2(1 +3 )
2

1
1


1

2















1
3 
3 

2 1
2




2
Shear strength
5.3 Mohr–Coulomb failure criterion

A
 c
3
 f 2 f
1
cctg 1/2(1 +3 )

1
 1  3 
2
sin  
1
c cot    1  3 
2


 o 
  2c tan  45  
2
2



 o 
2
o
 3   1 tan  45    2c tan  45  
2
2




 1   3 tan 2  45o  
2

For sandy soil: c=0

2
o
 3   1 tan  45  
2

 1   3 tan 2  45o 
Shear strength

A
 c
3
max
 f 2 f
cctg 1/2(1 +3 )
1

1

 f  90     45 
2
2
  45
max
In the case represented by the figures in this chapter, in which it
is assumed that the vertical direction is the direction of the major
principal stress, the planes on which the stresses are most critical
make an angle π/4 −  /2 with the vertical direction. Thus it can be
expected that sliding failure will occur in planes that are somewhat
steeper than 45. If for instance  = 30, which is a normal value for
sands, failure will occur by sliding along planes that make an angle
of 30 with the vertical direction.
Shear strength
5.4 Shear test
(1) Direct shear test
Normal stress σn
Shear stress σ3
Soil
Shear strength
Direct shear test is Quick and
Inexpensive
Shortcoming is that it fails the
soil on a designated plane which
may not be the weakest one
Shear strength
Shear strength
Shear strength
• The discussion thus far have referenced failure of the soil.
• Failure is indicated by excessive strain with little to no increase (even
decrease) in stress.
• After failure, the soil strength does not go to 0
• The soil retains residual strength
Peak Strength
Shear
stress
Residual Strength
Shear displacement
Shear strength
Shear
stress
c
Overconsolidated
OCR >1
normallyconsolidated
OCR=1
φ
normal stress
Typical plot for clays - drained condition
Shear strength
(2) Triaxial shear test
•The test is designed to as closely as possible
mimic actual field or “in situ” conditions of
the soil.
△
3
3
3
3
3
3
△
Shear strength
Shear strength
Triaxial tests are run by:
saturating the soil
applying the confining
stress (called σ3)
Then applying the vertical
stress (sometimes called
the deviator stress) until
failure
3 main types of triaxial tests:
Consolidated – Drained
Consolidated –
Undrained
Unconsolidated Undrained
Shear strength
The specimen is saturated
Confining stress (σ3) is applied
This squeezes the sample causing volume decrease
Drain lines kept open and must wait for full consolidation (u = 0) to continue wit
test
Once full consolidation is achieved, normal stress applied to failure with drain lines
still open
Normal stress applied very slowly allowing full drainage and full consolidation o
sample during test (u = 0)

f=f
=’
Shear strength
(3) Unconfined compression test
qu
qu
• The specimen is not placed in the
cell
• Specimen is open to air with a σ3 of
0
• Test is similar to concrete
compression test, except with soil
(cohesive – why?)
• Applicable in most practical
situations – foundations for
example.
• Drawing Mohrs circle with σ3 at 0
and the failure (normal) stress σ3
defining the 2nd point of the circle –
often called qu in this special case
• c becomes ½ of the failure stress
Shear strength
Shear strength
Shear strength
unconfined compression apparatus
量表
量力环
qu
加压
框架
升降
螺杆
qu
无侧限压缩仪
Shear strength
unconfined compression apparatus
Shear strength
unconfined compression apparatus

u=0
qu
 f  cu 
2
cu
qu 
Shear strength
sensitivity
the effects of disturbance of
soil constitutive property on
soil strength
qu
St 
qu '
Low sensitivity
1<St≤2
Middle sensitivity 2< St≤4
High sensitivity
St>4
Shear strength
(4) Vane shear test
This test is used for the in-situ
determination of the undrained strength of
intact, fully saturated clays; the test is not
suitable for other types of soil. In
particular, this test is very suitable for soft
clays, the shear strength of which may be
significantly altered by the sampling
process and subsequent handling.
Generally, this test is only used in clays
having undrained strengths less than 100
kN/m2. This test may not give reliable
results if the clay contains sand or silt
laminations.
f 
2 M max
D

D 2  H 

3 

Shear strength
Example
Shear strength
Shear strength
Shear strength
Remark
• Triaxial tests rarely run
• The unconfined test is very common
• In most cases, clays considered φ = 0 and c is used as the
strength
• Sands are considered c = 0 and φ is the strength parameter
• Direct shear test gives us good enough data for sand / clay mixes
(soils with both c and φ)
• Tables showing N value vs strength very commonly used (page
567 for clays for example).