This presentation is all about Shear Strength of Soil and it's importance in Civil Engineering, application of shear strength, direct shear test, mohr's circle, mohr's coulomb, shear strength, triaxial shear test, unconfined compression test, vane shear test

1. SHEAR
STRENGTH OF
SOIL
Prepared by:
Arbaz M. Kazi
Assistant Professor
VCET, Vasai (W)

2. Contents:
 What is Shear Force?
 What is Shear Stress?
 What is Shear Strength?
 What Causes Shear Stresses to be induced in Soil.
 Factors on which Shear Strength of Soil Depends.
 Importance of Shear Strength Knowledge for Civil Engineers
 Principal Plane
 Methods of Finding Normal & Shear Stress on any Plane
 Mohr’s-Coulomb Theory
 Mohr Coulomb failure criterion with Mohr circle of Stress.
 Factors Affecting Shear Strength of Soil.
 Measurement of Shear Test

3. What is Shear Force?
• Shear Force may be defined as “unaligned
forces, which when acts on a body causes
the body to move away in the opposite
direction”.
• Examples:
 Cutting of pages by Scissors.
 Tearing of pages by applying two
opposite forces.
 Failure of steel beams due to application
of two opposite forces.

4. What is Shear Stress?
• Shear Stress may be defined as a force that
causes layer or parts to slide upon each other in
opposite directions.
• The shear stress is developed in soil when it is
subjected to maximum compressions
What is Shear Strength?
• Shear Strength may be defined as the maximum resistance to shear
stress just before failure.

5. Factors on which Shear Strength of Soil Depends.
• The structural resistance to the displacement of soil because of interlocking of
particles.
• The frictional resistance between individual soil particles at their contact points.
• Cohesion or Adhesion between surface of soil particles.
Hence Shear strength is cohesionless soil results from
intergranular friction while in case of clay it results due to
cohesion between solid particles

6. Importance of Shear Strength Knowledge for Civil Engineers
The knowledge of shear strength is very important some of the uses are
provided below:
• In the design of foundations the evaluation of bearing capacity is
dependent on the shear strength.
• For the design of embankments for dams, roads, pavements,
excavations, levees etc. The analysis of the stability of the slope is done
using shear strength.
• In the design of earth retaining structures like retaining walls, sheet pile
coffer dams, bulks heads, and other underground structures etc.

7. Principal Plane
• A soil mass is subjected to a three dimensional stress system. But
stresses in third direction are not relevant, hence stress system is
simplified as two-dimensional.
• At every point in a stressed body, there are three planes on which shear
stress is zero, this planes are called as principal planes.
• The compressive stresses on the principal plane can be maximum or
minimum.
• The plane with maximum compressive stress (𝝈 𝟏) is called as major
principal plane.
• The plane with minimum compressive stress (𝝈 𝟑) is called as minor
principal plane.

8. Methods of Finding Normal & Shear Stress on any Plane
The two methods which is used for finding normal and shear stress on
any plane are:
1. Analytical Method
2. Graphical Method (Mohr’s Circle Method)
The Analytical method has became obsolete because it is tedious
and time consuming.

9. Analytical Method
Case I: When Soil is subjected to Major and Minor Principal Stress on Vertical
and Horizontal Plane
Soil element
s’1
s’1
s’3
s’3
q
s’
t
q
ssss
s
q
ss
t
2
22
2
2
'
3
'
1
'
3
'
1'
'
3
'
1
Cos
Sin






Resolving forces in s and t directions,
2'
3
'
1
2'
3
'
1'2
22 






 





 

ssss
st

10. Case II: When Soil is not subjected to Major and Minor Principal Stress on Vertical
and Horizontal Plane
Soil element
s’y
s’y
s’x
s’x
q
s’
t
qtq
ssss
s
qtq
ss
t
2*2
22
2*2
2
''''
'
''
SinCos
CosSin
xyxy
xy








Resolving forces in s and t directions,
2
2''2''
'2
22
t
ssss
st 







 








 

xyxy
txy = tyx

11. Graphical Method (Mohr’s Circle)

12. Steps to draw Mohr’s Circle:
a) Choose a convenient scale, for eg. 1cm = 20 kN/m2
b) Plot two axis representing normal stress and shear stress
respectively.
c) Plot σ1 and σ3 values using the scale on X-axis.
d) Take (
σ1+ σ3
2
) as centre and draw a circle.
e) Make a plane parallel to one which is represented in the fig. or
whose angle of inclination is known. The plane will cut Mohr’s
circle at certain point, the stresses corresponding to those points
will give σ & τ values for that plane.

13. NOTE: In case soil is subjected to stresses from all faces along
with shear stresses the procedure to draw Mohr’s circle is bit
different.
Steps a) to c) will be the same but step d) will be as
d) Plot τxy & τyx perpendicular to σx & σy values and join those
points so obtained by a straight line. The line of those points
will cut X-axis at certain point which will be the centre of
Mohr’s circle and remaining steps will be same as above.

14. Mohr-Coulomb Theory
• The Mohr – Coulomb criterion is the outcome of inspiration of two great
men, Otto Mohr born on 1835 and passed away on 1918 and Charles-
Augustin de Coulomb born on 1736 and passed away on 1806.
• The two men never coexisted but their brilliant minds contributed
significantly in the scientific knowledge. The combination of two
hypotheses gave us the Mohr – Coulomb failure surface.
• Chronologically, Coulomb was involved in military defense works (how
much knowledge have we gained due to war!) trying to built higher walls
for the French. In order to investigate why taller walls than usual were
failing and try to built them to stand, he wanted to understand the lateral
earth pressure against retaining walls and the shear strength of soils.

15. • He devised a shear strength test and
observed (at that time, with his tests) that
soil shear strength was composed of one
parameter that was stress – independent
named cohesion (c) and one that was stress
dependent, similar to friction of sliding solid
bodies named angle of internal friction (∅).
• Probably he executed shear strength tests
and found for different normal stresses (σ)
different shear stresses (τ). By plotting these
data on a (τ-σ) diagram he obtained the
straight line denoted by the equation τ = c +
σ*tan(∅) as can be seen in the next figure.
Mohr-Coulomb Theory

16. • Mohr (1900) presented a theory for
rupture in materials that held “a
material fails through a critical
combination of normal stress (σ)
and shear resistance (𝜏 𝑓), and not
through either maximum normal or
shear stress alone.
• The functional relationship on a
failure plane can be expressed in
the form.
𝜏 𝑓 = 𝑓(𝜎)
Mohr-Coulomb Theory

17. Mohr-Coulomb Theory
It is not known (Holtz et al, 1981)
who first combined both theories but
combining the Mohr failure criterion
with the Coulomb equation gave a
straight line tangent (to most of the
Mohr circles) and the Mohr –
Coulomb strength criterion was born.

18. Mohr Coulomb failure criterion with Mohr circle
of Stress
X
s’v = s’1
s’h = s’3
X is on failure s’1s’3
effective stresses
t
s’
f’ c’
Failure envelope in terms
of effective stresses
c’cotf’ (s’1 s’3)/2
(s’1  s’3)/2





 











 

2
'sin
2
'cot'
'
3
'
1
'
3
'
1 ss
f
ss
fc
Therefore,

19. Mohr Coulomb failure criterion with Mohr circle of
Stress





 











 

2
'sin
2
'cot'
'
3
'
1
'
3
'
1 ss
f
ss
fc
( ) ( ) 'cos'2'sin'
3
'
1
'
3
'
1 ffssss c
( ) ( ) 'cos'2'sin1'sin1 '
3
'
1 ffsfs c
( )
( ) ( )'sin1
'cos
'2
'sin1
'sin1'
3
'
1
f
f
f
f
ss




 c













2
'
45tan'2
2
'
45tan2'
3
'
1
ff
ss c

20.  The shearing strength of the soil is affected by:
• Soil composition: mineralogy, grain size and its distribution,
shape of particles, pore fluid, etc.
• Initial state: Whether loose, dense, or over consolidated.
• Structure: Arrangement of particles whether packed or
distributed.
• Loading conditions: Effective stress path, (drained, and
undrained) and type of loading.
Factors affecting Shear Strength of Soil

21. Other laboratory tests include,
Direct simple shear test, torsional ring shear test, plane strain Tri-axial test, Laboratory vane
shear test, laboratory fall cone test
Determination of shear strength parameters of soils
(c, f or c’, f’)
Laboratory tests on specimens
taken from representative
undisturbed samples
Field tests
Most common laboratory tests to
determine the shear strength
parameters are,
1.Direct shear test
2.Triaxial shear test
1. Vane shear test
2. Torvane
3. Pocket penetrometer
4. Fall cone
5. Pressuremeter
6. Static cone penetrometer
7. Standard penetration test

22. Laboratory Tests
Field conditions
z
svc
svc
shcshc
Before construction
A representative
soil sample
z
svc + Ds
shcshc
After and during
construction
svc + Ds

23. Laboratory Tests
Simulating field conditions in
the laboratory
Step 1
Set the specimen in
the apparatus and
apply the initial stress
condition
svc
svc
shcshc
Representative
soil sample
taken from the
site
0
00
0
Step 2
Apply the
corresponding field
stress conditions
svc + Ds
shcshc
svc + Ds
svc
svc
t
t

24. Direct Shear Test
Schematic diagram of the direct shear apparatus

25. Direct Shear Test
Preparation of a sand specimen
Components of the shear box Preparation of a sand specimen
Porous
plates
Direct shear test is most suitable for consolidated drained tests specially on
granular soils (e.g.: sand) or stiff clays

26. Direct Shear Test
Leveling the top surface of
specimen
Preparation of a sand specimen
Specimen preparation
completed
Pressure plate

27. Direct Shear Test
Test procedure
Porous
plates
Pressure plate
Steel ball
Step 1: Apply a vertical load to the specimen and wait for consolidation
P
Proving ring
to measure
shear force
S
Step 2: Lower box is subjected to a horizontal displacement at a constant rate

28. Direct Shear Test
Shear box
Loading frame to
apply vertical load
Dial gauge to
measure vertical
displacement
Dial gauge to measure
horizontal displacement
Proving ring to
measure shear
force

29. Direct Shear Test
Analysis of test results
sampletheofsectioncrossofArea
(P)forceNormal
stressNormal s
sampletheofsectioncrossofArea
(S)surfaceslidingat thedevelopedresistanceShear
stressShear t
Note: Cross-sectional area of the sample changes with the horizontal displacement

30. Direct Shear Tests on Sands
Shearstress,t
Shear displacement
Dense sand/ OC
clay
tf
Loose sand/ NC
claytf
Dense sand/OC Clay
Loose sand/NC Clay
Changeinheightof
thesample
ExpansionCompression
Shear displacement
Stress-Strain Relationship

31. tf1
Normal stress = s1
Direct Shear Tests on Sands
How to determine strength parameters c and f
Shearstress,t
Shear displacement
tf2
Normal stress = s2
tf3
Normal stress = s3
Shearstressatfailure,tf
Normal stress, s
f
Mohr – Coulomb failure Envelope

32. Direct Shear Tests on Sands
Some important facts on strength parameters c and f of sand
Sand is
cohesionless
hence c = 0
Direct shear tests are
drained and pore water
pressures are dissipated,
hence u = 0
Therefore,
f’ = f and c’
= c = 0

33. Failure envelopes for clay from drained direct shear tests
Shearstressatfailure,tf
Normal force, s
f’
Normally consolidated clay (c’ = 0)
In case of clay, horizontal displacement should be applied at a very slow rate to
allow dissipation of pore water pressure (therefore, one test would take several
days to finish)
Over consolidated clay (c’ ≠ 0)
Direct Shear Tests on Clays

34. Interface Tests on Direct Shear Apparatus
In many foundation design problems and retaining wall problems, it is required to
determine the angle of internal friction between soil and the structural material
(concrete, steel or wood)
fst tan' cf
Where,
c = cohesion,
f = angle of internal friction
Foundation material
Soil
P
S
Foundation material
Soil
P
S

35. Advantages of Direct Shear Apparatus
 Due to the smaller thickness of the sample, rapid drainage can be achieved
 Can be used to determine interface strength parameters
 Clay samples can be oriented along the plane of weakness or an identified
failure plane
Disadvantages of Direct Shear Apparatus
 Failure occurs along a predetermined failure plane
 Area of the sliding surface changes as the test progresses
 Non-uniform distribution of shear stress along the failure surface

36. Tri-axial Shear Test
Soil sample at
failure
Failure plane
Porous
stone
impervious
membrane
Piston (to apply deviatoric stress)
O-ring
pedestal
Perspex
cell
Cell pressure
Back pressure Pore pressure or
volume change
Water
Soil
sample

37. Tri-axial Shear Test
Specimen preparation (undisturbed sample)
Sampling tubes
Sample extruder

38. Tri-axial Shear Test
Specimen preparation (undisturbed sample)
Edges of the sample are
carefully trimmed
Setting up the sample in the
triaxial cell

39. Tri-axial Shear Test
Sample is covered with a
rubber membrane and sealed
Cell is completely
filled with water
Specimen preparation (undisturbed sample)

40. Tri-axial Shear Test
Specimen preparation (undisturbed sample)
Proving ring to
measure the
deviator load
Dial gauge to
measure vertical
displacement

41. Types of Tri-axial Tests
Is the drainage valve open?
yes No
Consolidated
sample
Unconsolidated
sample
Is the drainage valve open?
yes No
Drained
loading
Undrained
loading
Under all-around cell pressure sc
scsc
sc
scStep 1
Deviator stress
(Ds = q)
Shearing (loading)
Step 2
sc sc
sc+ q

42. Types of Tri-axial Tests
Is the drainage valve open?
yes No
Consolidated
sample
Unconsolidated
sample
Under all-around cell pressure sc
Step 1
Is the drainage valve open?
yes No
Drained
loading
Undrained
loading
Shearing (loading)
Step 2
CD test
CU test
UU test

43. Deviator stress (q or Dsd) = s1 – s3
Consolidated- Drained test (CD Test)
s1 = sVC + Ds
s3 = shC

44. Volumechangeofthe
sample
ExpansionCompression
Time
Volume change of sample during consolidation
Consolidated- drained test (CD Test)

45. Deviatorstress,σd
Axial strain
Dense sand or
OC clay
(Dsd)f
Dense sand or
OC clay
Loose sand or
NC clay
Volumechangeof
thesample
ExpansionCompression
Axial strain
Stress-strain relationship during shearing
Consolidated- Drained test (CD Test)
Loose sand or
NC Clay
(Dsd)f

46. CD Tests How to determine strength parameters c and f
Deviatorstress,σd
Axial strain
Shearstress,t
σ or σ’
f
Mohr – Coulomb
failure envelope
(σd)fa
Confining stress = σ3a(σd)fb
Confining stress = σ3b
(Dsd)fc
Confining stress = σ3c
σ
3c
σ1cσ3a σ1a
(∆σd)fa
σ3b σ1b
(∆σd)fb
σ1 = σ3 + (σd)f
σ3

47. CD Tests
Strength parameters c and f obtained from CD tests
Since u = 0
in CD tests,
σ = σ’
Therefore,
c = c’ and
f = f’
cd and fd
are used
to denote
them

48. CD Tests Failure envelopesShearstress,t
σ or σ’
fd
Mohr – Coulomb
failure envelope
σ3a σ1a
(σd)fa
For sand and NC Clay, cd = 0
Therefore, one CD test would be sufficient to determine fd of sand or
NC clay

49. CD Tests Failure envelopes
For OC Clay, cd ≠ 0
t
σ or σ’
f
σ3 σ1
(σd)f
c
σc
OC NC

50. • Unconfined Compression Test is a special type of Unconsolidated – Undrained
(UU) test that is commonly used for clay specimens.
• It is special case of a tri-axial compression test. In this test the confining pressure
(𝜎3) is 0.
• In this, cylindrical soil specimen (with height to diameter ratio of 2 to 2.5) is
loaded axially by a compressive force until failure takes place. No rubber
membrane is necessary to encase the specimen. The vertical compressive stress
is the major principal stress (𝜎1) and the other two principal stresses are zero.
• This test may be conducted on undisturbed or remoulded cohesive soils. It
cannot be conducted on coarse-grained soils such as sands and gravels as these
cannot stand without lateral support. Also the test is essentially a quick or
Undrained one because it is assumed that there is no loss of moisture during the
test, which is performed fairly fast.
Unconfined Compression Test

51. Procedure:-
 Push the sampling tube into the sample, remove the sampling tube
along with the Saturated the soil sample in sampling tube.
 Coat the inside of the split mould with a thin layer of grease/oil to
prevent adhesion of the soil.
 Extrude the specimen from the sampling tube to the split mould with
the help of sample extractor and knife.
 Trim the two ends of the mould.
 Weight the soil sample and mould.
 Remove the sample from the mould by splitting it in two parts.
 Measure the length and dia. of the specimen.

52. Procedure:-
 Place the specimen on the bottom plate of the compression machine.
 Raise the bottom plate of the machine to make contact of the specimen
with the upper plate.
 Adjust the strain dial gauge and proving ring dial gauge to read zero.
 Apply the compression load by raising the bottom plate of the machine
to produce axial strain at a rate of ½ to 2% per minute.
 Record the strain and proving ring dial gauges readings every 30
seconds.
 Compress the specimen till it fails or 20% vertical deformation is
reached, whichever is earlier.
 Note the least count of strain gauge and load dial gauge

53. The Mohr circle can be draw for stress conditions at failure. As the minor principal stress is
zero, the Mohr circle passes through the origin .The failure envelop is horizontal. The
cohesion intercept is equal to the radius of the circle.
t
σ or σ’
σ3 σ1
(σ1 = qu )
c
c = 𝒒 𝒖/2
c = t

54. Due to this test we can check the various parameters of the soil, like
 Unconfined compressive strength
 Sensitivity of soil
 Shear parameters of the soil, etc.
 The Unconfined Compressive Strength (qu) is defined as the ratio of failure load to
the cross sectional area of the soil sample, if it is not subjected to any lateral
pressure.
 Where:-
 𝑞𝑢 = Unconfined Compressive Strength
 P = Failure Load
 Ac = Corrected Area at failure.
 𝑞 𝑢 = 𝑃/𝐴𝑐
 Ac = Ao/1−e Ao = Initial Area, e = Strain
 e = ∆ 𝐿/𝐿 𝑜 ∆ 𝐿= Change in length, 𝐿 𝑜 = Initial Length of the sample

55. Vane Shear Test
• If suitable undisturbed for remoulded
samples cannot be got for conducting
tri-axial or unconfined compression
tests, the shear strength is determined by
a device called the Shear Vane.
• The vane shear test may also be
conducted in the laboratory. The
laboratory shear vane will be usually
smaller in size as compared to the field
vane.
• The shear vane usually consists of four
steel plates welded orthogonally to a
steel rod, as shown in figure.
• The applied torque is measured by a
calibrated torsion spring, the angle of
twist being read on a special gauge.

56. • A uniform rotation of about 1° per minute is used. The vane is forced
into the soil specimen or into the undisturbed soil at the bottom of a
bore-hole in a gentle manner and torque is applied.
• The torque is computed by multiplying the angle of twist by the spring
constant.
• The shear strength S of the clay is given by:
𝑠 =
𝑇
𝜋𝐷2( Τ𝐻 2 + Τ𝐷 6)
• If both the top and bottom of the vane partake in shearing the soil.
Here, T = torque
D = diameter of the vane
H = height of the vane
If only one end of the vane partakes in shearing the soil, then
𝑠 =
𝑇
𝜋𝐷2( Τ𝐻 2 + Τ𝐷 12)
• The shearing resistance is mobilised at failure along a cylindrical
surface of diameter D, the diameter of the vane, as also at the two
circular faces at top and bottom.