7 Shear Strength

PBW N302

Credit Hours

CEM/WEE/STE

Dr. Asmaa ModdatherSoil Mechanics and Foundations

Faculty of Engineering Cairo University

Spring 2014

Dr. Asmaa Moddather PBW N302 Spring 2014

Soil fail in shear.

Shear strength of soil is the internal resistance that the

soil mass offer to resist failure/sliding along any plane

inside it.

Shear failure will occur at points where shear stresses

() exceeds soils shear strength (S).

Introduction


Shear stresses are generated into the soil mass due to

adding external loads and/or excavations.

The engineer needs to know the nature of shearing

resistance in the soil mass to analyze problems such as:

o bearing capacity of foundations.

o stability of slopes.

o lateral pressure on retaining walls.

Introduction


Introduction

Bearing capacity of Foundation

Failure surfaceStable mass


Introduction

Stability of slopes

Failure surface

Stable mass


Introduction

lateral pressure on retaining walls

Failure surface

Stable mass

Failure surface

Direction of movement


1. Local failure takes place at points where shear stresses() > shear strength (S).

2. When local failure occurs at sufficiently large numberof points within the soil mass, a general failure takesplace.

3. Failure takes the form of sliding of a soil block over aFailure/Sliding/Slip surface within the soil mass.

To study shear failure at a point, we need to calculate:

1. Stresses (, ) on any plane through this point

2. Shear resistance (S) at this point

Failure Mechanism


Considering a certain point insidethe soil mass and knowing thenormal () and shear () stressesacting on two planes at this point:

o What is the maximum andminimum normal stresses(magnitude and direction)?

o What is the maximum shearstresses (magnitude anddirection)?

o What is the normal () and shear() stresses acting on any plane?

xx

y

y

xy

xy

xy

xy

Normal and Shear Stresses in a Soil Mass


It is a graphical method to presentthe state of stress along any planepassing through any point within thesoil mass.

Need to define and signconventions:

o For :

Compression +ve sign

Tension -ve sign

o For :

Rotation clockwise -ve sign

Rotation anticlockwise +ve sign

xx

y

y

xy

xy

xy

xy

Mohr Circle Presentation


Stresses on plane a: (x, xy)

Stresses on plane b: (y, -xy) xx

y

y

xy

xy

xy

xy

Plane b

Plane a

a

b

Mohrs circle

ox

yxy

xy

Mohr Circle


Every point on the circle represents the state of stress acting on a

plane passing through the soil element.

There are infinite number of planes passing through the element.

a

b

Mohrs circle

ox

yxy

xy

xx

y

y

xy

xy

xy

xy

Plane b

Plane a

Mohr Circle


Maximum and minimum normal stresses:

o Major principal plane (1,0)

o Minor principal plane (3,0)

Maximum and minimum shear stresses:

o Plane c ((1+3)/2, max)

o Plane d ((1+3)/2, min)

a

b

Minor principal plane (3, 0)

o

max

min

Major principal plane (1, 0)

c

d

xx

y

y

xy

xy

xy

xy

Plane b

Plane a

Mohr Circle


Define: Pole

A unique point on Mohrs circle from which if we draw a lineparallel to any plane, the line will intersect the circle at a pointwhose co-ordinates are and acting on this plane.

a

b

xx

y

y

xy

xy

xy

xy

Plane b

Plane a

(,)

Pole

Mohr Circle


How to find the Pole?

Need to know the following about a single plane:

o State of stress ( and )o Direction of plane

a

b

xx

y

y

xy

xy

xy

xy

Plane b

Plane a

Pole

Vertical plane

Horizontal plane

Mohr Circle


Knowing the location of the pole, determine the direction of:

o Major principal plane (1, with horizontal, clockwise)o Minor principal plane (2, with horizontal, clockwise)o Plane with maximum shear stress (3, with horizontal, clockwise)o Plane with minimum shear stress (4, with horizontal, anticlockwise)

a

b


max

min


c

d

xx

y

y

xy

xy

xy

xy

Plane b

Plane a

Pole

1

234

Mohr Circle


The state of stress on 2 perpendicular

planes within a soil element is shown in

the opposite figure. Draw Mohrs circle,

and determine:

i. The principal stresses and the planes on

which they act.

ii. Maximum shear stress and the

inclination of the plane on which it acts.

iii. The state of stress on a plane inclined

20o anticlockwise with horizontal.

Soil Element

2 t/m2

2 t/m2

5 t/m25 t/m2

1.5 t/m2

Example


Stresses on plane a: (+5, -1.5) t/m2

Stresses on plane b: (+2, +1.5) t/m2

a

b

o

5 t/m2

2 t/m2

2 t/m2

5 t/m21.5 t/m2

Plane b

Plane a

Pole

Note: horizontal and vertical axes MUST be drawn

with the same scale

(t/m2)

(t/m2)

Example


i.

Principle stresses ( = 0):

1: Major principle stress = 5.62 t/m2, 1 = 68o clockwise, with horizontal

3: Minor principle stress =1.37 t/m2, 2 = 158o clockwise, with horizontal

ii.Maximum shear stress: max = 2.12 t/m

2, 3 = 23o anticlockwise, with horizontal

a

b

o 1

3

max

1

3

2

(t/m2)

(t/m2)

Pole

Example


iii.State of stress at plane inclined 20o anticlockwise with horizontal: = 1.39 t/m2

= 0.18 t/m2

(t/m2)

a

b20

(1.39, 0.18) (t/m2)

(t/m2)

Pole

Example


Shear resistance in soils is due to:

o Friction and interlocking between soil particles Friction component

o Inter-particle attraction forces (due to electro-chemical effects) Cohesion component

The shear strength (S) of soil at a point is expressed as a linear

function of the effective normal stress () acting on plane of

failure:

where, c and are the shear strength parametersc: effective cohesion

: effective angle of shearing resistance

''' tancS +=

Shear Failure Criterion in Soil


c

Mohr-Coulomb shear strength failure envelope

'''

f tancS +==



Shear strength failure envelope

Plastic equilibrium

Elastic equilibrium

(f, f)

c



o Relative density:

v. loose versus v. dense

o Gradation:

poorly graded versus well graded

o Particle shape:

rounded versus angular

o Particle surface roughness (as roughness increases, increases)

small large

small large

small large

For coarse grained soils, S depends on friction and interlocking

between particles ():

Factors affecting Soil Shear Strength (S)


For coarse grained soil:

o ranges from 27o to 45o

o dry ~ wet



o Stress history (overconsolidation ratio):

as OCR increases, S increases

o Soil fabric (floculated, dispersed):

floculated has higher S

o Soil disturbance (affects soil fabric):

as disturbance increases, S decreases

o Soil permeability (water drainage):

Shear strength of soil loaded under drained conditions (slowly) is different from that loaded under undrained conditions (quickly)

Drained shear strength versus undrained shear strength

c

For fine grained soils, S depends on friction and interlocking between particles () as well as cohesion (c):



Shear strength parameters for a particular soil are

determined by means of laboratory tests on specimens

sampled from in-situ soil.

Great care is required in sampling, storage, and handling

of samples prior to testing, especially in case of

undisturbed samples where it is necessary to preserve the

in-situ structure and water content of soil.

Shear Strength Tests


Shear box apparatus.

Weights to apply normal load

Proving ringShear box

Vertical displacement dial gage

Motor, applies horizontal displacement at constant rate

Direct Shear Test


Direct Shear Test


Is a device used to measure force. It consists

of an elastic ring of known diameter with a

measuring device located in the center of the

ring that measures its deflection.

Each proving ring has its calibration constant,

by which we convert the readings of the

measuring device (deflection) into force.

Force = proving ring reading x calibration

constant.

Proving Ring


Schematic of shear box apparatus:

Sample may be circular or square

Area of sample is ~ 3 to 4 in2, 1 inch high

Box is split into 2 halves

Loading plate

Direct Shear Test


1. Assemble the apparatus (sample, porous stones, filter paper (if needed), shear box parts) and fill shear box with water if needed.

2. Apply normal force on top of shear box (N1).

3. Shear force is applied to specimen by moving one half of the box relative to the other to cause failure in the soil specimen. Shearing force is measured by proving ring or load cell.

4. Vertical (v) and horizontal (h) displacements are monitored during the test.

5. Repeat steps 2, 3, and 4 for different normal forces (N2, N3).

Picture of sample after failure

Direct Shear Test Procedure


1. For each test, calculate:

Normal stress () = normal force/area of sample

Shear stress () = shear force/area of sample

2. Plot:

Loose sand and normally consolidated clay result in similar (-h and v-hrelationships)

Dense sand and overconsolidated clay result in similar (-h and v-hrelationships)

Direct Shear Test AnalysisS

h

e

a

r

s

t

r

e

s

s

,

Shear displacement, h

Loose sand

f

Dense sand

Peak shear

strength

f

= = constant

Dense sand

Loose sandC

h

a

n

g

e

i

n

h

e

i

g

h

t

o

f

s

p

e

c

i

m

e

n

,

v

E

x

p

a

n

s

i

o

n

C

o

m

p

r

e

s

s

i

o

n Shear displacement, h


3. For test 1, get 1 and max1 = f1.

For test 2, get 2 and max2 = f2.

For test 3, get 3 and max3 = f3.

4. Plot:

c

Test (1)

Test (2)

Test (3)

(1, f1)

(2,f2)

(3, f3)

Note: horizontal and vertical axes MUST be drawn with the same scale

Direct Shear Test Analysis


1. Generally, direct shear tests are conducted on dry sand, saturated sand,

and saturated clay.

2. For sands, soil has high permeability, water drains instantaneously

during the test, u = 0 during shearing, therefore, = throughout

the test.

3. For clays, to maintain = throughout the test (similar to sands), the

test is conducted slowly so excess pore water pressure (u) can drain

during shearing.

4. Measured shear strength parameters: c, called drained/effectiveshear strength parameters.

Remarks on Direct Shear Test


c

(,f)

For each test, we know and f on failure plane, which is horizontal.

Pole



Mohr Circle for Direct Shear Test


A shear-box test carried out on a silty clay soil, gave the

following results:

Example

If the shear box area is 36 cm2 and the proving ring

constant is 15 kg/mm, determine the cohesion intercept

and the angle of shearing resistance.

235.6176.6117.758.9Vertical load (N)

8.97.15.33.2Proving ring dial gauge reading (mm)


Example

235.6176.6117.758.9Vertical load (kg)

8.97.15.33.2Proving ring dial gauge reading

(mm)

132.75105.7578.75=3.2 x15

=47.25Horizontal force, H (kg)

6.544.913.27=58.9/36

=1.64 (kg/cm2)

3.692.942.19=47.25/36

=1.31 (kg/cm2)


Example

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 6.00 6.50 7.00

S

h

e

a

r

S

t

r

e

s

s

,

(

k

g

/

c

m

2

)

Normal stress, (kg/cm2)

c

c = 0.60 kg/cm2

= 25o


Advantages:

1. Simple

2. Not expensive

Disadvantages:

1. Failure occurs at a predetermined plane, not necessarily theweakest plane (might overestimate strength).

2. Cant measure u during the test, therefore, test conditionsare adjusted such that u = 0.

3. Shear stress distribution over the shear surface of thespecimen is not uniform (stress concentration at corners).

Advantages and Disadvantages of Direct Shear Test


Test specimen

Load frame

Proving ring

Dial gage for vertical displacement

Triaxial cell

Triaxial cell

Triaxial Shear Test


O-rings

Porous stones

Top/bottom platens

Filter paper

Rubber membrane

Triaxial Shear Test


O-rings

Transparent cylinder

Vertical load

Confining pressure

Triaxial Shear Test


Triaxial Shear Test


Top drainage

Open valve

All around Pressure supply

Open valve

Air is released

Porous stone

Membrane

Porous stone

O ring Loading cap

Loading ram

Transparent cylinder

Air release valve

Specimen

Bottom drainage

Sample Placement

Test Procedure


Sample Placement:

1. Trim sample to required dimensions (h/d ~ 2).

2. Place sample on bottom filter paper and porous stone.

3. Place top filter paper, porous stone, and loading cap.

4. Place thin rubber membrane around sample and o-rings.

5. Fill cell with water.

6. Lower loading bar just to rest on loading cap.

Test Procedure


Open valve


Loading Stages:

o Stage I

Test Procedure

Applying confining pressure (c)

Open/closed valve


Open valve


Loading Stages:

o Stage II

Test Procedure

Applying confining pressure (c)

Open/closed valve

Applying axial load (N)Deviator stress (d)


Loading Stages:

o Stage 1:

Apply confining pressure (c) = cell pressure.

o Stage 2:

Increase axial load (N) up to failure. Axial load/area of specimen is called deviator stress (d).

c

c

cc

Stage (1)

c+df

c+df

cc

Stage (2)

Test Procedure


Drainage conditions during stages 1 and 2 are set according torequired type of test.

c

c

cc

Stage (1)

Valve (open/closed)

cc

c + df

c + dfStage (2)

Valve (open/closed)

Valve openedDrainage allowedConsolidated, C

Valve closedDrainage not allowedUnconsolidated, U

Valve openedDrainage allowedDrained, D

Valve closedDrainage not allowed

Undrained, U

Types of Triaxial Tests


Stage (1)

c

c

cc

Valve (open/closed)

Stage (2)

cc

c + df

c + df

Valve (open/closed)

Consolidated, C Unconsolidated, U Drained, D Undrained, U

Possible test types:

Stage 1 Stage 2 Test Type

C D Consolidated-Drained (CD)

C U Consolidated-Undrained (CU)

U U Unconsolidated-Undrained (UU)

U D Unconsolidated-Drained (UD)Never done



Pole

Minor principal plane Major principal plane

c

c

c

c

Stage (1)

c+df

c+df

cc

Stage (2)

(c, 0) (c+df, 0)

Note:c = 3c+df = 1

Stage 1

Stage 2

Mohr Circle for Triaxial


Stage (1)

c

c

cc

Valve (open)

Stage (2)

cc

c + df

c + df

Valve (open)

Consolidated, C Drained, D

Consolidated-Drained Test (CD)

3 = c cc c + df1 =

u = 0

3 = 31 = 1

3 =

1 =

u = 0

3 = 31 = 1


D

e

v

i

a

t

o

r

S

t

r

e

s

s

,

d

Axial Strain, v

NCC

OCC

df

OCC

NCC

C

h

a

n

g

e

i

n

v

o

l

u

m

e

o

f

s

p

e

c

i

m

e

n

,

V

E

x

p

a

n

s

i

o

n

C

o

m

p

r

e

s

s

i

o

n Axial Strain, vdf

Dense sand and overconsolidated clay result in similar (d-v and V-vrelationships)

Loose sand and normally consolidated clay result in similar (d-v and V-vrelationships)

Stress-strain Relationships:



d

1fdf

Mohr circle at failure

3

cStage I

Stage II

1



1f(2)

3(2)

1f(1)3(1)c=0

Normally Consolidated Clay (NCC) and Sands




Overconsolidated Clay (OCC)

OCC NCC

pc

c

2

1


Stage (1)

c

c

cc

Valve (open)

Stage (2)

cc

c + df

c + df

Valve (closed)

Consolidated, C Undrained, U

Consolidated-Undrained Test (CU)

3 = c cc c + df1 =

u = 0

3 = 31 = 1

3 =

1 =

u = value measured during test

3 = 3 - u1 = 1 - u


D

e

v

i

a

t

o

r

S

t

r

e

s

s

,

d

Axial Strain, v

NCC

OCC

df

OCC

NCC

C

h

a

n

g

e

i

n

p

o

r

e

w

a

t

e

r

p

r

e

s

s

u

r

e

,

u

-

v

e

+

v

e

Axial Strain, vdf

Dense sand and overconsolidated clay result in similar (d-v and u-vrelationships)

Loose sand and normally consolidated clay result in similar (d-v and u-vrelationships)

Stress-strain Relationships:



1f

Mohr circle at failure(Total stresses, measured)

3

'1f3

u u

Normally Consolidated Clay (NCC)

u = +ve3 = 3 u decreases

1 = 1 u decreases

Same diameter

c=0

Mohr circle at failure(Effective stresses, calculated)




o If cpc, soil behaves as NCC, u = +ve

3 = 3 u decreases

1 = 1 u decreases



OCC NCC

pc

c

2

1


Mohr circle at failure(Effective stresses, calculated)


u1 u2

u3 u4u = -ve

u = +ve



Drainage conditions during stages 1 and 2 are set according to

required type of test.c

c

cc

Stage (1)

Valve (open/closed)

cc

c + df

c + dfStage (2)

Valve (open/closed)

Valve openedDrainage allowedConsolidated, C

Valve closedDrainage not allowedUnconsolidated, U

Valve openedDrainage allowedDrained, D

Valve closedDrainage not allowed

Undrained, U



Stage (1)

c

c

cc

Valve (closed)

Stage (2)

cc

c + df

c + df

Valve (closed)

Unconsolidated, U Undrained, U

Unconsolidated-Undrained Test (UU)

3 = c cc c + df1 =

3 =

1 =

ud = increase in p.w.p due to duc = increase in p.w.p due to c

3 = 3 - u 1 = 1 - u

Measured u during shearing = uc + ud +ve (NCC), -ve (OCC)


1

One effective stress circle to all total stress circles (calculated)

'3

u= 0

1(1)3(1)

u(1) (-ve)u(2) (+ve)


Same diameter

cu

3(2) 1(2)

cu = undrained shear strength

Note: increasing c doesnt result in any increase in 3

Since all 3 prior to shearing is the same in all samples, the strength of the

samples when sheared will be the same.

Unconsolidated-Undrained Test (UU)


Special case of UU test

o c = 0

o Very simple and quick

Proving ring (measures axial load)

Dial gage (measures axial displacement)

Sample

Loading frame

Unconfined Compression Test


Sample Placement:

1. Trim sample to required dimensions (h/d ~ 2).

2. Place sample on the loading device. Begin test immediately

as drying will alter samples characteristics considerably.

3. Lower loading piston until it contacts specimen.

4. Begin the test, continue until load values decrease or until

20% strain is reached.



Specimen before test

Ho

Ao

Specimen after test

Bulging

H

H

A

)(1A

)HH(1

AH)(H

HAA

HAH)A(HHAAHVV

ConstantV

o

o

o

o

oo

ooo

oo

testbeforeafter test

=

=

=

=

=

=

=

)100(%)(1

AA o

=



Readings: Load (F), vertical displacement (H)

Data Reduction:

o Axial strain: a = (H/Ho) x 100

o where: Ho = initial sample height

o Stress: = F/Ac

where: Ac corrected area = Ao/(1-a)

Ao = initial x-sectional area of the sample

Data Plotting:

o Plot stress () versus axial strain (a)



Strain, (%)

S

t

r

e

s

s

,

qu

qu = unconfined strength

cu = undrained shear strength = qu/2

a,f



qu = unconfined strength

cu = undrained shear strength = qu/2

qu

qu

31

== qu

0

c = 0 c = 0

0.0

qu

u = 0

cu



1

One effective stress circle to all total stress circles (calculated)

'3

u= 0

1(1)3(1)

u(UU1)u(UU2)


Same diameter

cu

3(2) 1(2)

cu = undrained shear strength

c = 3 = 0

u(UC)

Unconsolidated-Undrained (UU) Test versus Unconfined Compression Test


Drained shear strength parameters:

o c', OCC

o : NCC, sand

These parameters are obtained from:

o CD Test

o CU Test (with pore pressure measurements)

Applications

o Used to design structures for long term condition Drained.

Final Remarks


Undrained shear strength parameters:

o cu, u=0 OCC, NCC

These parameters are obtained from:

o UU Test

o Unconfined Compression Test

Applications

o Used to design structures for short term condition

Undrained.

Final Remarks


Example (1)

The following results were obtained at failure in a series of

CD triaxial tests on specimens of fully saturated clay.

i. Determine the shear strength parameters of this soil.

ii. If a specimen is subjected to a confining pressure of 150 kPa

and a deviator stress of 200 kPa, would it fail? Comment.

Confining pressure

(kN/m2)

Deviator stress

(kN/m2)Test

1003501

2104102


Example (1)

1 = c +df3 = c

c

(kN/m2)df

(kN/m2)Test

4501001003501

6202102104102

450100 210

620

'

c '

i. Shear strength parameters:

c' = 119 kN/m2

' = 12o

Note: since c>0, this clay is OCC,

and pc > 210 kN/m2


Example (1)

350150

c

ii. If a specimen is subjected to a confining pressure of 150 kPa and a

deviator stress of 200 kPa, would it fail? Comment.

Elastic equilibriumThe specimen will not fail


Example (2)

A clay specimen failed at unconfined strength value of 400 kN/m2

when placed in an unconfined test apparatus, what would be the

undrained shear strength for the tested specimen?

0.0

400

u = 0

cuUndrained Shear strength:

cu = qu/2 = 400/2 = 200 kN/m2


1112171613940Load (kg)

0.3500.3000.1110.0800.0530.0340.0130.0H (cm)

3.3652.8851.0670.7690.5100.327= 0.013x100/10.4

= 0.125

0.0 = H/Ho (%)

14.3414.2714.0013.9613.9213.90

=13.87

13.85A (cm2)

0.770.841.211.150.930.65= 0.29

0Stress

(kg/cm2)

In an unconfined compression test, the following data were collected

(sample height = 10.4 cm, sample diameter = 4.2 cm), Find the unconfined

strength and the undrained shear strength for the tested sample. (Ao = 13.85

cm2, Ho = 10.4 cm).

Example (3)

)100

0.125(113.85

=

87.134.0

=

)100(%)(1

Ao

=


Example (3)

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00Axial Strain, a (%)

A

x

i

a

l

S

t

r

e

s

s

,

(

k

g

/

c

m

2

)

qu

Unconfined strength: qu = 1.21 kg/cm2

Undrained Shear strength: cu = qu/2 = 0.61 kg/cm2

7 Shear Strength

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