Chapter (12)

Instructor : Dr. Jehad Hamad

2017-2018

Shear strength in soils Direct shear test Unconfined Compression Test Tri-axial Test

Chapter Outlines

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 (φ), 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

φσ ′′+′= tancsfriction internal of angle

cohesioncstrengthshear s

=′

=′=

φ

General State of Stress

σ1

σ1 major principle stress

σ3 σ3. Minor principle stress Confining stress

Shear Strength,S

Normal Stress, σn = σ′ = γ h

C′

φ = φ′

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

σ3

σ1

Direct Shear TestDirect 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

ASTM D 3080

Direct Shear Test (cont.) 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

Direct Shear Test

Direct Shear Test DataSh

ear s

tres

s

Residual Strength

Peak Strength

Direct Shear Test Data Volume change

∆H

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.

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 rate9. 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.

Figures

Shea

r st

ress

, sPeak Stress

N1 = 10 kg

N2 = 20 kg

N3 = 30 kg

Horizontal displacement, ∆H

s3

s2

s1

Figures (cont)

Shea

rSt

ress

, s (p

sf)

C′

φ

(σ1,s1)

(σ3,s3)(σ2,s2)

Normal Stress σ, psf

Figures (cont)

Verti

cald

ispl

acem

ent

Horizontal displacement

27 Triaxial Shear Test

28 Triaxial Shear Test• The test is designed to as

closely as possible mimic 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

Triaxial Shear TestSpecimen preparation (undisturbed sample)

Sampling tubes

Sample extruder

Triaxial Shear TestSpecimen preparation (undisturbed sample)

Edges of the sample are carefully trimmed

Setting up the sample in the triaxial cell

Triaxial Shear Test

Sample is covered with arubber membrane andsealed

Cell is completelyfilled with water

Specimen preparation (undisturbed sample)

Triaxial Shear TestSpecimen preparation (undisturbed sample)

Proving ring tomeasure thedeviator load

Dial gauge tomeasure verticaldisplacement

In some tests

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 pressureBack pressure Pore pressure or

volume change

Water

Soil sample

Types of Triaxial Tests

Is the drainage valve open?

yes no

Consolidatedsample

Unconsolidatedsample

Is the drainage valve open?

yes no

Drained loading

Undrainedloading

Under all-around cell pressure σc

σcσc

σc

σcStep 1

deviatoric stress (∆σ = q)

Shearing (loading)

Step 2

σc σc

σc+ q

Types of Triaxial Tests

Is the drainage valve open?

yes no

Consolidatedsample

Unconsolidatedsample

Under all-around cell pressure σc

Step 1

Is the drainage valve open?

yes no

Drained loading

Undrainedloading

Shearing (loading)

Step 2

CD test

CU test

UU test

36Consolidated – 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 simluatingoverconsolidated soil.

37Consolidated – 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)

38Consolidated – 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

39Unconsolidated – 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

40 Shear Strength of Soil

c

Shearstress

normal stress

Typical UU plot for clays

41Unconfined 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

42

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)

Consolidated- drained test (CD Test)

Step 1: At the end of consolidationσVC

σhC

Total, σ = Neutral, u Effective, σ’+

0

Step 2: During axial stress increase

σ’VC = σVC

σ’hC = σhC

σVC + ∆σ

σhC 0

σ’V = σVC + ∆σ = σ’1

σ’h = σhC = σ’3

Drainage

Drainage

Step 3: At failureσVC + ∆σf

σhC 0

σ’Vf = σVC + ∆σf = σ’1f

σ’hf = σhC = σ’3fDrainage

Deviator stress (q or ∆σd) = σ1 – σ3

Consolidated- drained test (CD Test)

σ1 = σVC + ∆σ

σ3 = σhC

Volu

me

chan

ge o

f the

sa

mpl

e

Expa

nsio

nC

ompr

essi

on

Time

Volume change of sample during consolidationConsolidated- drained test (CD Test)

CD tests How to determine strength parameters c and φD

evia

tor s

tress

,∆σ d

Axial strain

Shea

r stre

ss,τ

σ or σ’

φMohr – Coulomb failure envelope

(∆σd)fa

Confining stress = σ3a(∆σd)fb

Confining stress = σ3b

(∆σd)fc

Confining stress = σ3c

σ3c σ1cσ3a σ1a

(∆σd)fa

σ3b σ1b

(∆σd)fb

σ1 = σ3 + (∆σd)f

σ3

CD tests Failure envelopesSh

ear s

tress

,τ

σ or σ’

φdMohr – Coulomb failure envelope

σ3a σ1a

(∆σd)fa

For sand and NC Clay, cd = 0

Therefore, one CD test would be sufficient to determine φd of sand or NC clay