Ge i-module2-rajesh sir

Geotechnical Engineering IGeotechnical Engineering - I
Soil water, Permeability,
Shear StrengthShear Strength
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKN
1

Module II
Occurrence adsorbed and capillary water types effective stress total
Module II
Soil water
Occurrence - adsorbed and capillary water types - effective stress - total
stress - pore pressure - pressure diagrams
Permeability
Definition - Darcy's law - factors affecting permeability - laboratory
determination stratified soils average permeability Seepagedetermination - stratified soils - average permeability. Seepage –
downward and upward flow -quick sand
Shear Strength
Definition - Mohr-Coulomb strength theory - Measurement of shear
strength – Types of Triaxial compression tests - measurement of pore
l d ff i U fi d C i
Shear Strength
pressure - total and effective stress – Unconfined Compression test - vane
shear tests –Direct shear test- strength parameters - choice of test
conditions for field problems.

Soil water
Modes of occurrence –
Effective stress - total stress - pore pressure –p p
Pressure diagrams under various conditions –
Flow of water through soil -

Modes of occurrence of soil water
Free water (Gravitational water)
• Water that is free to move through soil mass under gravity
Held water
• Structural water – chemically combined in the soil mineraly
• Adsorbed water (Hygroscopic, contact moisture or surface bound
moisture) – held by adhesion
• Capillary waterp y
Pore water
• Water that is free of strong soil attractive forces
• Capillary water and free water are pore waters

Stress conditions in soilSt ess co d t o s so
Effective and neutral pressures
• At any plane of soil mass, the total pressure is the total load per unit
area
• Pressure due to:
• Self-weight
• Over-burden
• Inter-granular pressure (effective pressure) and neutral pressure (pore
)pressure)

I t l ( ff ti ) ′Inter-granular pressure (effective pressure)
• Pressure transmitted from particle to particle through their contact
points, through the soil mass above the plane
′σ
points, through the soil mass above the plane
• Effective in decreasing the voids ratio and in mobilising the shear
strengthg
Neutral pressure (pore pressure) u w wu h= γ
• Pressure transmitted through pore fluid
• Equal to water load per unit area above the plane• Equal to water load per unit area above the plane
• No influence on voids ratio or shear strength
u′σ = σ +Hence, total pressure

Effective pressure under different conditions of soil-
water system
1. Submerged soil mass1. Submerged soil mass
u′σ = σ − WaterEffective pressure
z
1sat w w wz z h′σ = γ + γ − γAt A,
1z
wh
( )1 1sat w wz z z z= γ + γ − + γ B
z( )sat wz= γ − γ
z ′= γzγ
A
Here, effective pressure does not depend on 1z
Saturated soil
1 1 0w wu z z′σ = σ − = γ − γ =At B,
Saturated soil

1z1
wh
B
+ =
z
A
z ′γ
Eff ti P
( )1 wz z+ γ
T t l
1sat wz zγ + γ
Effective pressure
diagram
Pore pressure
diagram
Total pressure
diagram

2. Soil mass with surcharge
u′σ = σ −Effective pressure
1 sat wq z z z′σ = + γ + γ − γAt A,
Surcharge, q
z
1 sat wq γ γ γ
1q z z ′= + γ + γ
Moist
1z
B1 10q z q z′σ = + γ − = + γAt B,
Moist
soil
z whSaturated
soil
A

q
Moist
l
1z
q q
+
soil
1
B
=
1z γ
Saturated
soil
zwh
wzγ
P
A
Eff ti
( )1 sat wq z z+ γ + γ − γ
T t l
1 satq z z+ γ + γ
Pore pressure
diagram
Effective pressure
diagram
Total pressure
diagram

3. Saturated soil with capillary fringe
u′σ = σ −
( )1 sat wz z z′σ = + γ − γ 1 satz z′= γ + γAt A,
Effective pressure
( )1 sat wγ γ 1 satγ γ
Capillary
fringeAt x1,
z1x( ){ }1 1 1sat w
u
x z x
′σ = σ −
′σ = γ − − − γ
1z
Bx
1( ){ }
( )
1 1 1
1 1 1
sat w
sat wx z x
γ γ
′σ = γ + − γ
z whSaturated
soil
′ ′
Capillarity induces downward pressure
due to weight of water hanging below.
A
( )1sat wx x z′σ = γ − − γAt x,
1 1 wx z′ ′∴σ = γ + γ
1 satz′σ = γAt B,
( )1sat w
1 wx z′= γ + γ

x xσ = γ = γz γ
( )1 1 wz x− − γ
1 wz− γ
1z
1 sat satx xσ = γ = γ1 wz γ
x
+ ( )1 wx z− γ
S t t d
B
=
x z
′σ =
′γ + γz Saturated
soil
satxσ = γ
1 wx zγ + γ
Pore pressure
wzγ
A
T t l
( )1 satz z+ γ1 satz z′γ + γ
( )z z z′= + γ + γ diagram Total pressure
diagram
( )1 sat wz z z= + γ − γ
( )1 1 wz z z= + γ + γ
Effective pressure
diagram

4. When flow takes place through soil
u′σ σ
a. Flow from top to bottom
Effective pressure uσ = σ −
( )1 wu z z h= + − γAt A,
Effective pressure
Pore pressure
z zσ = γ + γ 1z h
w
Total pressure
1sat wz zσ = γ + γ
B
S t t d
Flow
h
Effective pressure
( )1 1sat w wz z z z h= γ + γ − + − γ
z Saturated
soil
wh
A
( )sat w wz h= γ − γ + γ
h′
wz h′= γ + γ Increase in effective pressure (downward seepage pressure)

1 wu z= γAt B, Pore pressure
1 wzσ = γ
1 wγp
Total pressure
1 1 0w wz z′σ = γ − γ =Effective pressure h
i
z
=
w
x
x h
z
′ ′σ = γ + γAt x from B,
wx ix′= γ + γ Hydraulic gradient
z
1z h
w
x
h
z
γ
1
B
Flow
h
x
x ′γ
z Saturated
soil
wh
Dept. of CE, GCE Kannur Dr.RajeshKNPressure diagram
z ′γwhγ A

b. Flow from bottom to top
u′σ σ
p
Effective pressure uσ = σ −
( )1 wu z z h= + + γAt A,
Effective pressure
Pore pressure h
z zσ = γ + γ
Total pressure
1z
h1sat wz zσ = γ + γ
Effective pressure
B
S t t d
Flow
wh
( )1 1sat w wz z z z h= γ + γ − + + γ
z Saturated
soil
( )sat w wz h= γ − γ + γ
h′
A
wz h′= γ − γ Decrease in effective pressure (upward seepage pressure)

1 wu z= γAt B, Pore pressure
1 wzσ = γ
1 wγp
Total pressure
1 1 0w wz z′σ = γ − γ =Effective pressure
w
x
x h
z
′ ′σ = γ − γAt x from B,
wx ix′= γ − γ
h
1z
w
whx
′
z
B
Saturated
Flow
x
w
x
x h
z
′γ − γ
x z
A
Satu ated
soilw
x
h
z
γ
h′h
A
Pressure diagram
wz h′γ − γwhγ

x ′γWhen submerged soil pressure balances seepage pressure ,ixγγWhen submerged soil pressure balances seepage pressure ,wixγ
0′σ =
′γ
wx ix′γ = γ c
w
i i
γ
⇒ = =
γ
Critical
hydraulic gradient
When , cohesionless soils loose their shear strength, and a visible
it ti (b ili lik h ) h > i k d
ci i=
1G
agitation (boiling-like phenomenon) happens => quick sand.
( )1Gγ 1
1
c
G
i
e
−
∴ =
+
( )1
1
w G
e
γ −
′γ =
+
We have,
• To prevent quick sand during excavations below water table, pumping
can be done before excavation, to lower water table.
• Quick sand doesn’t occur in clays, since cohesive forces prevent boiling.

• Problem 1: Water table is located at a
d h f 3 b l d f idepth of 3m below ground surface in a
deposit of sand of 11m thickness. A
capillary fringe is present above water
t bl t d l l S t t d it
Water table
3 m
Sa
table, upto ground level. Saturated unit
weight of sand is 20kN/m3. Calculate
the effective pressure at depths 0, 3, 7 and
11m from ground surface
4 m
anddepo
11m from ground surface.
4 m
osit
Depth Total pressure Porewater
pressure
Effective pressure
0 0 –3x9.81= – 29.43 29.430 0 3x9.81 29.43 29.43
3 3x20=60 0 60
7 7x20=140 4x9.81=39.24 140 – 39.24=100.76
11 11x20=220 8x9.81=78.48 220 – 78.48=141.52

29.43−
1z
3m
29.43
29.43
z Saturated
60
3m
7m
– = 60
soil
11m
140
220
39.24 100.76
11m
Total pressure
di
220
Pore pressure
diagram
78.48
Effective pressure
141.52
diagram diagram diagram

• Problem 2: For the given soil profile, compute the effective pressure at a
depth of 12 mdepth of 12 m.
To find the total pressure:
L 1 0 t 3
n 0 4
Layer 1: 0 to 3 m.
1
n
e
n
=
−
0.4
0.667
1 0.4
= =
−
Gγ
1
w
d
G
e
γ
γ =
+
3
32.65 9.81kN m
15.595kN m
1 0.667
×
= =
+

( )G e+ γ ( ) 3
2 65 0 667 9 81kN m+ ×( )
1
w
sat
G e
e
+ γ
γ =
+
( ) 32.65 0.667 9.81kN m
19.52kN m
1 0.667
+ ×
= =
+
( )d r sat dSγ = γ + γ − γ ( ) 3
15.595 0.3 19.52 15.595 16.772kN m= + × − =
L r 2 3m to 4mLayer 2: 3m to 4m.
w
d
Gγ
γ =
3
32.68 9.8kN m
16.432kN m
×
= =
( ) 3
32.68 0.6 9.81kN m
20 111kN
+ ×
1
d
e
γ
+ 1 0.6+
( ) 3
20.111kN m
1 0.6
satγ = =
+
( )S ( ) 3
16 432 0 8 20 111 16 432 19 375kN( )d r sat dSγ = γ + γ − γ ( ) 3
16.432 0.8 20.111 16.432 19.375kN m= + × − =
Layer 3: 4m to 6m
Layer 3: 4m to 6m. 3
20.111kN msatγ =

Layer 4: 6m to 9m.
( ) 3
32.1 3 9.8kN m
12.508kN m
1 3
sat
+ ×
γ = =
+1 3+
Layer 5: 9m to 12m.
sate w G= 0.35 2.7 0.945= × =
( ) 3
32.7 0.945 9.81kN m
18.384kN m
1 0.945
sat
+ ×
γ = =
+
Total pressure at a depth of 12 m 3 16.772 1 19.375 2 20.111= × + × + ×
2
3 12.508 3 18.384 202.59kN m+ × + × =
Effective pressure at a depth of 12 m 2
202 59 8 9 8 124 11kN m= × =
Effective pressure at a depth of 12 m 202.59 8 9.8 124.11kN m= − × =

• Problem 3:
Bottom heave, when effective
stress at top of gravel layer isp g y
zero

( ) wG e+ γ ( )2.7 1.2 9.81+ × 3
( )
1
w
sat
e
γ
γ =
+
( )
1 1.2
=
+
3
17.39kN m=For the clay layer,
At the top of gravel strata, 12 17.39 15 9.81′σ = × − × 2
61.53kN m=
Let the bottom of excavation be h m above the top of gravel layer, so
that the effective stress at top of gravel layer is zero.
0u′σ = σ − =
p g y
17.39 15 9.81u h∴σ = ⇒ = ×
8 5 mh∴ = 8.5 mh∴ =
12 8.5 3.5md∴ = − =

Permeability

Definition
Property of porous material that permits passage of water
through interconnecting voids
• Material having continuous voids - permeable
• Gravels - highly permeable; stiff clays - impermeable
• Mostly in soils flow is laminar• Mostly in soils, flow is laminar

Head, gradient and potential, g p
When water flows through soil, the total head consists of:
1. Piezometric head or pressure head
2. Velocity head
3. Position head Negligible, for flow through soil
Rise of water in piezometric tube
3. Position head g g , g
With respect to any datum

A
( )h
A
H
wh
( )w a
h
B DATUM
H
h
az
z
w
( )w b
h
a
bz
b
b
Point Piezo
head
Position
head
Total
head
b
head head head
a (hw)a za H
b (hw)b zb 0
H: Initial hydraulic head
h: Hydraulic head (potential)
at any point h h z= ±
( w)b b
c hw z hw – z i=h/l: hydraulic gradient
at any point wh h z= ±

Darcy's law
Quantity of water q flowing through a cross sectional area of soil
mass under a hydraulic gradient can be expressed as:
Darcy s law
q kiA= (1)
Hydraulic gradient
k
i
Coefficient of permeability Darcy, 1856
Total area of cross section
Hydraulic gradienti
A
• Greater k, greater is the flow through soil
• Average discharge velocity
q
v ki= = (2)v ki
A
= =
Coefficient of permeability is the average discharge velocity under
(2)
p y g g y
unit hydraulic gradient

Discharge and Seepage VelocityDischarge and Seepage Velocity
• In eqn. (1), is the Total area of cross sectionA
VOIDS SOLIDSA A A= +VOIDS SOLIDS
VOIDSA A<
Actual velocity s
VOIDS
q q
v
A A
= >
Seepage velocity

VOIDSq vA v A= = s VOIDSq vA v A
1
s
VOIDS VOIDS VOIDS
A v A V
v v
A n A V n
⎧ ⎫
= = = =⎨ ⎬
⎩ ⎭
∵
VOIDS VOIDS VOIDS⎩ ⎭
k i (3)
s pv k i= (3)Also,
Coefficient of percolation
1p ps
p
k i kv k
k
v ki k n n
= = = ∴ =
v ki k n n

Coefficient of permeability influences:
• Water retaining capacity and stability of earth dams
• Capacity of pumping installations for the lowering of ground waterp y p p g g g
table during excavations
• Rate of settlement of buildingsg
• etc.

Factors affecting permeability
1. Effect of size and shape of particles:
( )
2
Permeability Grain size∝
k
2
10.k C D=
A constant
k
C
Coefficient of permeability
in cm/s
Effective size, in cm
(= 100, if D10 is in cm)
C
10D
Allen Hazen 1892Allen Hazen, 1892

2. Effect of property of pore fluid
Coefficient of permeability is directly proportional to unit weight
f fl id d i l ti l t it i it
w
k
γ
∝
of fluid and inversely proportional to its viscosity
η
( )
2.
0 1wC
k e
γ
= C is a constant( )0.1k e= −
η
C
3. Effect of voids ratio
3
e
1
e
k
e
∝
+
Taylor, 1948

4. Effect of structural arrangement of particles & stratificationg p
• Horizontal flow:• Horizontal flow:
is same for all stratai
1 1 2 2 3 3hQ k iZ k iz k iz k iz= = + + +
( )1 1 2 2 3 3
1
hk k iz k iz k iz
Z
∴ = + + +
Z

• Vertical flow:
From principle of continuity of flow
For the same area,
1 1 2 2Av A v= =
,
1 2v v= =
i.e., downward velocity is same for all stratav
1 1 2 2 3 3
vk h
v k z k z k z
Z
= = = = =
Total loss of head, 1 2 3
1 1 2 2 3 3
h h h h
z i z i z i
= + + +
= + + +
1 1 2 2 3 3z i z i z i+ + +

Z Z Z
( )v
Z Z
k v
hh
v
= =
1 2 3
v
Z
k
z z z
k k k
=
⎛ ⎞
+ + +⎜ ⎟
⎝ ⎠
That is,
1 2 3k k k⎝ ⎠
5. Effect of Degree of saturation: More entrapped air causesg pp
less permeability
6. Effect of Adsorbed water: Reduces permeabilityp y

Thus, coefficient of permeability can be expressed as:
3
2
. . .
1
w
s
e
k D C
e
γ
=
η +
(4)
D i th di t f h i l i hi h h th ti f
sD is the diameter of spherical grain which has the same ratio of
volume to surface area collectively for the grains in a soil
is a shape constantC

Determination of permeability
Constant head test
Falling head test
Laboratory methods
Falling head test
Pumping testsField methods p g
Bore hole tests
Indirect method
C t ti f i i di t ib tiComputation from grain size distribution

Constant head test:
Suitable for coarse grained soils
Suitable for coarse grained soils

C t t h d t i t f t b t t i il lConstant head permeameter consists of a tube to contain soil sample
Tube can be of any convenient dimension
The head h is kept a constant during the test
S il i t t d b f i th t tSoil is saturated before commencing the test
Test is performed by allowing water to flow through the soil sample
Measure the quantity of discharge Q in time t
Q
k iA
1Q L
k
Q
q k iA
t
= =
Q
k
t h A
∴ =

Falling head test Suitable for fine grained soils
Soil sample is kept in a vertical
cylindercylinder
A transparent stand pipe of cross-
sectional area a is attached to thesectional area a is attached to the
cylinder
Soil is saturated before commencingSoil is saturated before commencing
the test
Test is performed by allowing waterp y g
to flow through the soil sample
Measure the elapsed time t2 – t1 for theMeasure the elapsed time t2 t1 for the
head to fall from initial h1 to final h2

• Let -dh be the change in head during a small time interval dt.
.dQ kiA dt=From Darcy’s Law,
kh
kiA dt dh Adt dh
.dQ a dh= −Also,
h
i =∵. . .kiA dt a dh Adt a dh
L
= − ⇒ = −
Integrating both sides,
i
L
=∵
2 2t hAk dh
dt
L h
= −∫ ∫
g g ,
1 1t haL h∫ ∫
12.3
li
aL h
k1
l
aL h
k
( )
1
10
2 1 2
lo. gi.e , k
A t t h
=
−( )
1
2 1 2
logek
A t t h
=
−

Constant head test was used to find coefficient of permeability of a
sand sample Diameter of sample = 10 cm length of sample = 20 cm
Problem 1:
sand sample. Diameter of sample = 10 cm, length of sample = 20 cm,
head of water = 35 cm, 110 cm3 of water was collected in 1 min 20 s,
determine coefficient of permeability.
1Q L
k
t h A
=Coefficient of permeability
3
110Discharge, cmQ = 1min,20 80Time, s st = =g
20Length of sample, cmL = 35Constant water head, cmh =
1 110 20 1Q L
k 0 01009 8 648 d2
80 35 10
4
Q
k
t h A
∴ = = × × =
⎛ ⎞π ×
⎜ ⎟
⎝ ⎠
0.01009 8.648cm s m day= =

Falling head test was used to find coefficient of permeability.
Diameter of sample = 6 cm diameter of stand pipe = 2 cm initial
Problem 2:
Diameter of sample = 6 cm, diameter of stand pipe = 2 cm, initial
head = 45 cm, final head after 2 min = 30 cm. Determine coefficient
of permeability. 15Length of sample, cmL =
( )
1
10
2 1 2
2.3
log
aL h
k
A t t h
=
−
2
22
4
Area of stand pipe, cma
π ×
=
( )
4
2
26
4
Area of sample, cmA
π ×
=
4
2 1 2 mint t− =
1 245 , 30cm cmh h= =
12.3
l
aL h
k i
( )
1
10
2 1 2
logk
A t t h
∴ = =
−
cm min 5.5 ?m day=

Problem 3:
4 cm has k3=7×10-4 cm/s. Assume that the flow is taking place perpendicular to the layers.

Z
k =
20
=
1 2 3
1 2 3
vk
z z z
k k k
=
⎛ ⎞
+ + +⎜ ⎟
⎝ ⎠
4 4 4
8 8 4
2 10 5 10 7 10− − −
=
⎛ ⎞
+ + +⎜ ⎟
× × ×⎝ ⎠
4
3.24 10 cm s−
= ×
( )
1
10
2.3
log
aL h
k
A t t h
=
−
( )2 1 2A t t h
( ) 1
2 1 10
2.3
log
aL h
t t
kA h
− =( ) 0
2
2
104 2
2.3 2cm 20cm 25
log
3 24 10 cm s 24cm 12
kA h
−
× ×
=
× ×3.24 10 cm s 24cm 12× ×
3775.37s=
=1hr, 2min, 55.37s

• Assignment!
Explain field tests for determination of permeability (pumpingExplain field tests for determination of permeability (pumping
tests).

Shear Strength

Definition
• Ability to resist sliding along an internal surface within a soil mass:
A very important property that determines the strength of a soile y po ta t p ope ty t at dete es t e st e gt o a so
mass
• Stability of foundations, slopes, embankments
etc. depend on shear strength
Shear strength consists of:g
– Frictional resistance
– Cohesion
50

Mohr-Coulomb failure theoryy
• Material fails essentially by shear
• Shearing stress on the failure plane is a unique function of the normal
stress acting on that plane
( )F
i i
( )s F σ=
51Friction
tanF N= φ
tan
F N
A A
= φ
F N
F N
∝
= μtans = σ φ (1)

tans = σ φ for purely granular soils(1)
• If cohesion also present,
for purely granular soils(1)
tans c= +σ φ (2) Coulomb Equation
experimentally determined
i i l t
c ⎫
⎬
E (2) th t i i d d t fc σ
empirical parametersφ
⎬
⎭
At zero σ s c=
Eqn. (2) assumes that is independent ofc σ
At zero ,σ s c
That is, cohesion is the shear resistance at zero
l
52
normal pressure

Plot between s and σ at failure (Coulomb envelope)
53

s s
φ
c
σ
φ
σ
Coulomb envelope for ideal (pure)
friction material
Coulomb envelope for purely
cohesive material
54

Mohr’s stress circle
55

σy
τxyτxy
τyxσn
σx
σx
τyx
τn
α
σ
τxy
yx
y xσ σ>
σy
cos2 sin2
2 2
y x y x
n xy
σ σ σ σ
σ α τ α
− +⎛ ⎞
= + +⎜ ⎟
⎝ ⎠⎝ ⎠
sin2 cos2
2
y x
n xy
σ σ
τ α τ α
−
= −
56
2
n xy

2
0 sin2 cos2 0
2
y x
n xy
σ σ
τ α τ α
−
= ⇒ − =
2
tan2 xy
y x
τ
α
σ σ
=
−
2
2
1 3
2 2
y x y x
xy
σ σ σ σ
σ τ
+ −⎛ ⎞
= ± +⎜ ⎟
⎝ ⎠
Principal
stresses1,3
2 2
xy⎜ ⎟
⎝ ⎠
stresses
2 2
2 2y x y xσ σ σ σ
σ τ τ
+ −⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟2 2
y y
n n xyσ τ τ− + = +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
,0
2
y xσ σ+⎛ ⎞
⎜ ⎟
⎝ ⎠
2
2
2
y x
xy
σ σ
τ
−⎛ ⎞
+⎜ ⎟
⎝ ⎠
57
⎝ ⎠ 2⎝ ⎠

( ),σ τ
2
2y xσ σ−⎛ ⎞
( ),y xyσ τ
τxyτ
2
2
y x
xyτ
⎛ ⎞
+⎜ ⎟
⎝ ⎠
σy
σx σσ σσ σ+
θ
τ
σσ3 σ1
2
y xσ σ+
τxy
( )( ),x xyσ τ−
1 2
tan 2xy
y x
τ
θ α
σ σ
−
= =
−

( )
σ σ
( ),σ τ
τ
1 3
2
σ σ−
1 3
sin2
2
σ σ
α
−
σσ3 σ1
2α
2
σσ3 σ1
1 3
2
σ σ+ 1 3
cos2
2
σ σ
α
−
σ1
σ3
σ33
σ
59
σ1

1 3 1 3σ σ σ σ+ −1 3 1 3
1 3
cos2
2 2
sin2
σ σ σ σ
σ α
σ σ
τ α
+
= +
−
=
Stresses on any plane in terms of
principal stresses
01 3−σ σ
sin2
2
τ α=
01 3
max , 45
2
when= =
σ σ
τ α
1 3
2
on this pla e., n
+
=
σ σ
σ
60

τ
Coulomb envelope
s
φ
failureτ
φ
c
σ 1σ3σ
c
61

φ′ ′ ′ Shearing strength based on
effective stress
( )
tan
tan
s c
s c u
σ φ
σ φ
′ ′ ′= +
′ ′= + −
(3)
tanu us c σ φ= +
Shearing strength based on total
stress
(4)
uc Apparent cohesion based on total stress
Apparent angle of shearing resistance
based on total stress
uφ
u Apparent cohesion based on total stress
62

′ ′ ′ ′1 3 1 3
cos2
2 2
σ σ σ σ
σ α
′ ′ ′ ′+ −
′ = +
′ ′
Stresses on any plane in terms
of principal stresses
(5)
1 3
sin2
2
σ σ
τ α
′ ′−
=
p p
tans c σ φ′ ′ ′+ (3)
(6)
s is the shear strength
tans c σ φ= + (3)
τ is the shear stress on a planeg
s τ− is a minimum on the plane of failure
p
( ) 0
d s
d
−
=
τ
αdα
1 3 1 3 1 3
cos2 tan sin2
2 2 2
s c
⎛ ⎞ ⎛ ⎞′ ′ ′ ′ ′ ′+ − −
′ ′− = + + −⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
σ σ σ σ σ σ
τ α φ α
63
2 2 2⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
φ

( )d s ⎛ ⎞ ⎛ ⎞′ ′ ′ ′− − −
⎜ ⎟ ⎜ ⎟
τ σ σ σ σ( ) 1 3 1 3
0 tan 2sin2 2cos2 0
2 2
d s
d
⎛ ⎞ ⎛ ⎞
′= ⇒ − − =⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
τ σ σ σ σ
φ α α
α
( ) ( )( ) ( )1 3 1 3tan sin2 cos2 0′ ′ ′ ′′⇒ − − − − =σ σ φ α σ σ α
tan sin2 cos2′⇒ − =φ α α
t t 2 2 9 450′ ′+
′
′+
φ
φφtan cot 2 2 9 450
2
f
′ ′⇒ − = ⇒ = + ⇒ = ′=+
φ
αα α φ αφ

Failure stress and maximum stress
•Failure plane doesn’t carry maximum shear stress
•Plane that has maximum shear stress is not the failure plane
Coulomb envelope
failureτ
maxτ
s
φ
c
σ 1σ3σ
65

Measurement of shear strengthMeasurement of shear strength
• Direct shear test
• Triaxial compression test
• Unconfined compression test
• Vane shear test• Vane shear test
Based on drainage conditions,
• Undrained test – no drainage allowed – no dissipation of pore
pressure during the test
• Consolidated undrained test – drainage allowed only with initial
normal stress
• Drained test - drainage allowed throughout the testDrained test - drainage allowed throughout the test
66

Disadvantages of direct shear test
• Shear and normal stress distribution along failure surface is not
if ti t th f il i t bili d if luniform entire strength of soil is not mobilised uniformly
• Plane of shear failure is pre-determined it may not be the weakest
plane
• Effect of lateral restraint by the side walls of the shear boxy
• No control over drainage
67

Direct shear test
N
F
68

• Strain controlled test
• At failure, shear force F corresponding to the normal force N is
noted
• Repeated for a number of identical samples
• From the set of values of σ and τ at failure, envelope is drawnFrom the set of values of σ and τ at failure, envelope is drawn
• Slope of the envelope represents angle of shearing resistance
τ
φφ
c
69
σ

Triaxial compression test
Casagrande & Terzaghi – 1936
70

σ1- σ3 Deviator stress
σ1
σ3
σ3σ3
σ3 σ3 Cell pressure
σ3
σ1
σ1- σ3
71

• The cylindrical soil specimen is subjected to an all round pressure
(cell pressure,σ3) initially and then to a vertical pressure, σ1(cell pressure,σ3) initially and then to a vertical pressure, σ1.
• (σ1- σ3) is known as deviatoric stress
• During the test the deviatoric stress and vertical deformation of the
sample are measured till failure
• Deviatoric stress and pore pressure corresponding to failure are noted
• Failure: at max value of stress or 20% axial strainFailure: at max value of stress or 20% axial strain
• A number of Mohr’s circles can be drawn from different sets of
observations (σ1 and σ3)from which the failure envelope can beobservations (σ1 and σ3)from which the failure envelope can be
determined
72

Coulomb envelope
ss
φ
c
σ
73

Stress state in soil specimen during triaxial compression
tanc′ ′ ′= +τ σ φ
τ
tanf c= +τ σ φ
FF
c’
′φ 45
2
f
φ
α
′
′ = +
C A sin
FC
KC
φ′ =
σ′ 1
′σ
3
′σK O
KC
74

But FC = Radius of Mohr’s circle
M i h t ( )1 ′ ′= Maximum shear stress ( )1 3
1
2
σ σ′ ′= −
( )1
φ ′ ′′ ′
( )1 3
1
FC σ σ′ ′−
Also, KC=KO+OC ( )1 3
1
cot
2
c φ σ σ′ ′′ ′= + +
( )
( )
1 3
1 3
2sin
1
cot
2
FC
KC c
σ σ
φ
φ σ σ
′∴ = =
′ ′′ ′+ +
2
1 3 tan 45 2 tan 45
2 2
c
′ ′⎛ ⎞ ⎛ ⎞′ ′ ′= + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
φ φ
σ σSimplifying,
2
1 3 tan 2 tanf fcσ σ α α′ ′ ′ ′ ′= +
2 2⎝ ⎠ ⎝ ⎠
1 3 2N c Nφ φσ σ′ ′ ′ ′ ′= + 2 2 0
tan tan 45where, fNφ
φ
α
′⎛ ⎞′ ′= = +⎜ ⎟
⎝ ⎠
1 3 φ φ
2
, fφ ⎜ ⎟
⎝ ⎠

Stress state during triaxial compression …g p
I t f t t lI t f ff ti In terms of total
stresses
In terms of effective
stresses
2
1 3 tan 2 tanf fcσ σ α α′ ′ ′ ′ ′= +
2
1 3 tan 2 tanf u fcσ σ α α= +
2N N
1 3 2N c Nφ φσ σ′ ′ ′ ′ ′= +
2 2 0
tan tan 45
2
fNφ
φ
α
′⎛ ⎞′ ′= = +⎜ ⎟
⎝ ⎠
1 3 2 uN c Nφ φσ σ= +
2 2 0
tan tan 45
2
u
fNφ
φ
α
⎛ ⎞
= = +⎜ ⎟
⎝ ⎠
(7)
2
fφ ⎜ ⎟
⎝ ⎠
(8)
2
fφ ⎜ ⎟
⎝ ⎠
76

Advantages of triaxial test
Sh h ll h 3 d i di i b d i h• Shear strengths at all the 3 drainage conditions can be done with
complete control
• Stress distribution on the failure plane is uniform
• State of stress at any stage of test is determinabley g
77

Unconfined compression test
3 0σ =• A special case of triaxial test in which
11 3 2 tan 45
2
2 u
uN c N cφ φ σσ
φ
σ
⎛ ⎞
=′ ′ ′ ′ ′= + ⇒ +⎜ ⎟
⎝ ⎠
Hence,
Only one value of σ1 . Therefore, only one Mohr’s circle
So, this test can be applied only for saturated clays ( )0uφ =
, for saturated clays10 2u ucφ σ= ⇒ =
1
2 2
u
u
q
c
σ
∴ = = qu Unconfined compressive strength at failure
78

s
Coulomb envelope
failureτ
cu
1
2 2
failu e
uq σ
= =
σ 1σ3 0σ =
2 2
79

Problem 1. An unconfined cylindrical specimen of clay fails under an axial
f 240 kN/ 2 Th f il l i li d l f 550 hstress of 240 kN/m2. The failure plane was inclined at an angle of 550 to the
horizontal. Find the shear strength parameters c and ϕ for the soil.
2
tan 2 tancσ σ α α= +1 3 tan 2 tanf u fcσ σ α α= +
3 0σ = 1 2 tanu fcσ α∴ =
0
45
2
f
φ
α = + ( )0
2 45fφ α∴ = −
0
20=
2 ( )f
1
1 2 tanu f uc c
σ
σ α= ⇒ =
240
=
2
84 kN m=1
2tan
u f u
fα 2tan55

Problem 2. Two identical specimens of dry sand are tested in the triaxial
apparatus with confining pressure of 150 N/mm2 and 250 N/mm2 respectivelyapparatus with confining pressure of 150 N/mm2 and 250 N/mm2 respectively.
If the angle of friction for sand is 350 what are the values of axial stresses at
failure of the specimens?
2
1 3 tan 2 tanf u fcσ σ α α= +
For cohesionless dry sand, 0uc =
2 2 0
1 3 3tan tan 45
2
f
φ
σ σ α σ
⎛ ⎞
∴ = = +⎜ ⎟
⎝ ⎠
0
2 0
1
35
150tan 45
2
σ
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
(1)
2⎝ ⎠
2
553.53 N mm=

0
2 0 35⎛ ⎞2 0
1
35
250tan 45
2
σ
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
(2)
2
922.543 N mm=

Problem 3. Two identical specimens of a soil were tested in the triaxial
Fi i f il d d i f 770 kN/ 2 h h llapparatus. First specimen failed at a deviator stress of 770 kN/m2 when the cell
pressure was 200 kN/m2. Second specimen failed at a deviator stress of 1370
kN/m2 when the cell pressure was 400 kN/m2. Find c and ϕ for the soil.
2
1 3 tan 45 2 tan 45
2 2
c
φ φ
σ σ ⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
We have,
First specimen
2
1 3 770 kN mσ σ− = 2
3 200 kN mσ =
2
1 970 kN mσ∴ =
2
970 200tan 45 2 tan 45
2 2
c
φ φ⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
(1)
2 2⎝ ⎠ ⎝ ⎠
Second specimen
2
1370 kN 2
400 kN
2
1770 kN mσ∴ =2
1 3 1370 kN mσ σ− = 2
3 400 kN mσ = 1 1770 kN mσ∴ =
2
1770 400tan 45 2 tan 45c
φ φ⎛ ⎞ ⎛ ⎞
+ + +⎜ ⎟ ⎜ ⎟ (2)
1770 400tan 45 2 tan 45
2 2
c= + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
(2)

From (1) and (2)From (1) and (2),
2
42.5 kN mc = 0
36.87φ =
Problem 4. The same soil as in the previous problem if tested in a direct shear
apparatus estimate the shear stress at which sample will fail under a normalapparatus, estimate the shear stress at which sample will fail under a normal
stress of 600 kN/m2.
tanf cτ σ φ= +
42 5 600tan36 87τ = +
2
492 5 kN m=42.5 600tan36.87fτ = + 492.5 kN m

Assignment: Vane shear testAssignment: Vane shear test
85

SummarySummary
Soil water
Occurrence - adsorbed and capillary water types - effective stress - total
stress - pore pressure - pressure diagrams
Soil water
Permeability
Definition - Darcy's law - factors affecting permeability - laboratoryDefinition Darcy s law factors affecting permeability laboratory
determination - stratified soils - average permeability. Seepage –
downward and upward flow -quick sand
Definition - Mohr-Coulomb strength theory - Measurement of shear
strength Types of Triaxial compression tests measurement of pore
Shear Strength
strength – Types of Triaxial compression tests - measurement of pore
pressure - total and effective stress – Unconfined Compression test - vane
shear tests –Direct shear test- strength parameters - choice of test
conditions for field problems
86
conditions for field problems.

Ge i-module2-rajesh sir

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