foundation design for the united states penitentiary

Chapter (12)

Instructor : Dr. Jehad Hamad

2017-2016

Chapter Outlines

Shear strength in soils

Direct shear test

Unconfined Compression Test

Tri-axial Test

Shear Strength

The strength of a material is the greatest stress it

can sustain

The safety of any geotechnical structure is

dependent on the strength of the soil

If the soil fails, the structure founded on it can

collapse

Significance of Shear Strength

Engineers must understand the nature of shearing resistance in order to analyze soil stability problems such as;

Bearing capacity

Slope stability

Lateral pressure on -retaining structures

Pavement

Shear Strength in Soils The shear strength of a soil is its resistance to

shearing stresses.

It is a measure of the soil resistance to deformation by continuous displacement of its individual soil particles

Shear strength in soils depends primarily on interactions between particles

Shear failure occurs when the stresses between the particles are such that they slide or roll past each other

Shear Strength in Soils (cont.)

Soil derives its shear strength from two

sources:

Cohesion between particles (stress

independent component)

Cementation between sand grains

Electrostatic attraction between clay particles

Frictional resistance between particles

(stress dependent component)

Shear Strength of Soils; Cohesion

Cohesion (C), is a measure of the forces that cement particles of soils

Dry sand with no cementation

Dry sand with some cementation

Soft clay

Stiff clay

Shear Strength of Soils; Internal Friction

Internal Friction angle (f), is the measure of the shear strength of soils due to friction

Mohr-Coulomb Failure Criteria

This theory states that a material fails because of

a critical combination of normal stress and shear

stress, and not from their either maximum

normal or shear stress alone.

The relationship between normal stress and

shear is given as

f tancsfriction internal of angle

cohesionc

strengthshear s

f

General State of Stress

σ1

σ1 major principle stress

σ3

σ3 Minor principle

stress, Confining

stress

Mohr–Coulomb Failure Criterion

Shear

Strength,S

Normal Stress, n = = g h

C

f = f

Mohr-Coulomb Failure Criterion

State of Stresses in Soils

σ1

Shear stress σ3 σ3

Normal stress σn

Consider the following situation:

- A normal stress is applied vertically

and held constant

- A shear stress is then applied until

failure

Determination of Shear Strength Parameters

The shear strength parameters of a soil are determined

in the lab primarily with two types of tests;

Direct Shear Test

Triaxial Shear Test

Soil

Normal stress σn

Shear stress σ3

3

1

Direct Shear Test

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

Used to determine the shear strength of both

cohesive as well as non-cohesive soils

Direct Shear Test The test equipment consists of a

metal box in which the soil specimen is placed

The box is split horizontally into two halves

Vertical force (normal stress) is applied through a metal platen

Shear force is applied by moving one half of the box relative to the other to cause failure in the soil specimen

Soil

Normal stress σn

Shear stress σ3

Direct Shear Test

Direct Shear Test Data S

hear

str

ess

Residual Strength

Peak Strength

Direct Shear Test Data Volume change

DH

Direct Shear Test (Procedure) 1.Measure inner side or diameter of shear box and find the area

2.Make sure top and bottom halves of shear box are in contact and fixed together.

3.Weigh out 150 g of sand.

4.Place the soil in three layers in the mold using the funnel. Compact the soil with 20 blows per layer.

5.Place cover on top of sand

6.Place shear box in machine.

7.Apply normal force. The weights to use for the three runs are 2 kg, 4 kg, and 6 kg if the load is applied through a lever arm, or 10 kg, 20 kg, and 30 kg, if the load is applied directly.

Note: Lever arm loading ratio 1:10 (2kg weight = 20 kg)

Direct Shear Test (Procedure) 8. Start the motor with selected speed (0.1 in/min) so that the rate of shearing is at a

selected constant rate

9. Take the horizontal displacement gauge, vertical displacement gage and shear load gage

readings. Record the readings on the data sheet.

10. Continue taking readings until the horizontal shear load peaks and then falls, or the

horizontal displacement reaches 15% of the diameter.

Calculations

Determine the dry unit

weight, gd

Calculate the void ratio, e

Calculate the normal stress

& shear stress

1d

wGse

g

g

A

V

A

N ;

Shear Stress vs. Horizontal Displacement

Sh

ear

stre

ss, s

Peak Stress

N1 = 10 kg

N2 = 20 kg

N3 = 30 kg

Horizontal displacement, DH

s3

s2

s1

Results of a Direct shear Test

Sh

ear

Str

ess,

s (

psf)

C

f

(1,s1)

(3,s3) (2,s2)

Normal Stress , psf

Figures (cont)

Ver

tica

l d

ispla

cem

ent

Horizontal displacement

The drained angle of Friction, f' , of normally Consolidated Clays Generally

Decreases with the plasticity of Soil.

39

Triaxial Shear Test

40

Triaxial Shear Test • The test is designed to as closely as

possible to the actual field or “in situ”

conditions of the soil.

• 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

41

Consolidated – Drained Triaxial Test • 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 with 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 of sample during test (u = 0)

• Test can be run with varying values of σ3 to create a Mohrs

circle and to obtain a plot showing c and φ

• Test can also be run such that σ3 is applied allowing full

consolidation, then decreased (likely allowing some

swelling) then the normal stress applied to failure simluating

overconsolidated soil.

42

Consolidated – Drained Triaxial Test

• In the CD test, the total and effective stress is the

same since u is maintained at 0 by allowing drainage

•This means you are testing the soil in effective stress

conditions

•Applicable in conditions where the soil will fail under

a long term constant load where the soil is allowed to

drain (long term slope stability)

43

Consolidated – Undrained Triaxial Test

• The specimen is saturated

• Confining stress (σ3) is applied

− This squeezes the sample causing volume decrease

− Again, must wait for full consolidation (u = 0)

• Once full consolidation is achieved, drain lines are closed (no

drainage for the rest of the test), and normal stress applied to

failure

− Normal stress can be applied faster since no drainage is

necessary (u not equal to 0)

• Test can be run with varying values of σ3 to create a Mohrs

circle and to obtain a plot showing c and φ

• Applicable in situations where failure may occur suddenly such

as a rapid drawdown in a dam or levee

44

Unconsolidated – Undrained Test

• The specimen is saturated

• Confining stress (σ3) is applied without drainage or consolidation

(drains closed the entire time)

• Normal stress then increased to failure without allowing drainage

or consolidation

• This test can be run quicker than the other 2 tests since no

consolidation or drainage is needed. Test can be run with varying

values of σ3 to create a Mohrs circle and to obtain a plot showing

c and φ

• Applicable in most practical situations – foundations for example.

• This test commonly shows a φ = 0 condition

Triaxial Shear Test

Soil sample at

failure

Failure plane

Porous

stone

impervious

membrane

Piston (to apply deviatoric stress)

O-ring

pedestal

Cell pressure

Back pressure

Pore pressure or

volume change

Water

Soil

sample

46

The Real World

• 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)

Triaxial Shear Test

Specimen preparation (undisturbed sample)

Sampling tubes

Sample extruder

Triaxial Shear Test


Edges of the sample are

carefully trimmed Setting up the sample in

the triaxial cell

Triaxial Shear Test

Sample is covered with a rubber

membrane and sealed

Cell is completely

filled with water


Triaxial Shear Test

Specimen preparation (undisturbed sample

Proving ring to measure

the deviator load

Dial gauge to

measure vertical

displacement

In some tests

Triaxial 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

Wate

r

Soil

sample

Types of Triaxial Tests

Is the drainage valve open?

yes no

Consolidated

sample Unconsolidated

sample


yes no

Drained

loading

Undrained

loading

Under all-around cell pressure c

c c

c

c Step 1

deviatoric stress

(D = q)

Shearing (loading)

Step 2

c c

c+ q

Types of Triaxial Tests


yes no

Consolidated

sample Unconsolidated

sample

Under all-around cell pressure c

Step 1


yes no

Drained

loading

Undrained

loading

Shearing (loading)

Step 2

CD test

CU test

UU test

Consolidated- drained test (CD Test)

Step 1: At the end of consolidation

VC

hC

Total, = Neutral, u Effective, ’ +

0

Step 2: During axial stress increase

’VC = VC

’hC = hC

VC + D

hC 0

’V = VC + D = ’1

’h = hC = ’3

Drainage

Drainage

Step 3: At failure

VC + Df

hC 0

’Vf = VC + Df = ’1f

’hf = hC = ’3f Drainage

Deviator stress (q or Dd) = 1 – 3


1 = VC + D

3 = hC

Vo

lum

e c

ha

ng

e o

f th

e s

am

ple

Expansio

n

Co

mp

ressio

n

Time

Volume change of sample during consolidation


Devia

tor

str

ess,

D

d

Axial strain

Dense sand or

OC clay

(Dd)f

Dense sand or

OC clay

Loose sand or

NC clay

Vo

lum

e c

ha

ng

e

of th

e s

am

ple

Exp

an

sio

n

Com

pre

ssio

n

Axial strain

Stress-strain relationship during shearing


Loose sand or

NC Clay (Dd)f

CD tests How to determine strength parameters c and f D

evia

tor

str

ess,

D

d

Axial strain

Sh

ear

str

ess,

or ’

f Mohr – Coulomb

failure envelope

(Dd)fa

Confining stress = 3a (Dd)fb

Confining stress = 3b

(Dd)fc

Confining stress = 3c

3c 1c 3a 1a

(Dd)fa

3b 1b

(Dd)fb

1 = 3 + (Dd)f

3

CD tests Failure envelopes S

hear

str

ess,

or ’

fd

Mohr – Coulomb

failure envelope

3a 1a

(Dd)fa

For sand and NC Clay, cd = 0

Therefore, one CD test would be sufficient to determine fd of

sand or NC clay

(Dd)fa

CU tests How to determine strength parameters c and f

Shear

str

ess,

or ’ 3b 1b 3a 1a

(Dd)fa

fcu

Mohr – Coulomb

failure envelope in

terms of total stresses

ccu ’3b ’1b

’3a ’1a

Mohr – Coulomb failure

envelope in terms of

effective stresses

f’

C’ ufa

ufb

’1 = 3 + (Dd)f - uf

’3 = 3 - uf

Effective stresses at failure

uf

CU tests Failure envelopes

For sand and NC Clay, ccu and c’ = 0

Therefore, one CU test would be sufficient to determine fcu

and f’(= fd) of sand or NC clay

Shear

str

ess,

or ’

fcu Mohr – Coulomb

failure envelope in

terms of total stresses

3a 1a

(Dd)fa

3a 1a

f’

Mohr – Coulomb failure

envelope in terms of

effective stresses

Unconsolidated- Undrained test (UU Test)

Step 1: Immediately after sampling

0

0

Total, = Neutral, u Effective, ’ +

-ur

Step 2: After application of hydrostatic cell pressure

’V0 = ur

’h0 = ur

C

C

-ur Duc = -ur c

(Sr = 100% ; B = 1)

Step 3: During application of axial load

C + D

C

No

drainage

No

drainage -ur c ± Du

’VC = C + ur - C = ur

’h = ur

Step 3: At failure

’V = C + D + ur - c Du

’h = C + ur - c Du

’hf = C + ur - c Duf =

’3f

’Vf = C + Df + ur - c Duf = ’1f

-ur c ± Duf C

C + Df No

drainage


Total, = Neutral, u Effective, ’ + Step 3: At failure


’3f

’Vf = C + Df + ur - c Duf = ’1f

-ur c ± Duf C

C + Df No

drainage

Mohr circle in terms of effective stresses do not depend on the cell pressure.

Therefore, we get only one Mohr circle in terms of effective stress for different

cell pressures

’ ’3 ’1 Df

3b 1b 3a 1a Df ’3 ’1


Total, = Neutral, u Effective, ’ + Step 3: At failure


’3f

’Vf = C + Df + ur - c Duf = ’1f

-ur c ± Duf C

C + Df No

drainage

or ’

Mohr circles in terms of total stresses

ua ub

Failure envelope, fu = 0

cu

3b 1b


Effect of degree of saturation on failure envelope

3a 1a 3c 1c

or ’

S < 100% S > 100%

66

Unconfined Compression Test • 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

foundation design for the united states penitentiary

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