CE 240 Soil Mechanics & Foundations Lecture 11

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Jul 18, 2012 (5 years and 5 months ago)

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CE 240
Soil Mechanics & Foundations
Lecture 11.1
Shear Strength of Soil I
(Das, Ch. 11)

Shear strength in soils

Introduction

Definitions

Mohr-Coulomb criterion

Introduction

Lab tests for getting the shear strength

Direct shear test

Introduction

Procedure & calculation

Critical void ratio
Class Outlines
Shear Strength

The strength of a material is the
greatest stress it can sustain;

So that the unit of strength is the same
as stress (Pa in SI unit system);
Significance of Shear Strength

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.

Understanding shear strength is the
basis to analyze soil stability problems
like:

lateral pressure on earth retaining
structures (Chs. 12, 13),

slope stability (Ch. 14), and

bearing capacity (Ch. 15).
Shear Failure in Soils
Slope Failure in Soils
Failure due to inadequate
strength at shear interface
Static: Transcosna
Grain Elevator
Canada (Oct. 18, 1913)
West side of foundation sank 24-ft
Bearing Capacity Failure
Dynamic: Foundation failure by liquefaction
after the 1964 Niigata Earthquake. (USGS)
Dynamic: Lateral Spreading caused by the 1906 San
Francisco Earthquake at Moss Landing, CA
(USGS Professional Paper 993)
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 cl
ay
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.
Mohr-Coulomb Failure
Criterion
Shear
Strength,S
φ
= φ′
C′
Normal Stress,
σn
=
σ′
= γ
h
''
tan
(
11.2)
tan
'
(
11.3
)
fn
n
fn
n
cc
cc
τ
σφ
µ
σ
τσ
φ
µ
σ
=+
=+
′′

=+
=+
f
where

shea
r str
ength
c =
cohesion; c
=effective coh
esi
on
φ = angl
e of internal f
riction; φ
= effecti
ve an
g
le of in
ternal friction
= coefficient
of fri
ction;
'
= effectiv
e co
effici
ent of frictio
τ
µµ
=


n.
µ=tanφ’
0.51-0.58
0.58-0.70
0.70-0.78
0.58-0.70
0.70-0.84
0.84-1.00
0.67-1.11
0.49-0.70
Mohr-Coulomb shear failure criterion
σ1
σ1
σ3
σ3
σ
n
τ
f
σ1
σ3
σ
τ

τ
f
=c

+
µ’
σn’
c’
φ
φ
Failed
Zone
(σ,
τ)
σff
From trigonometric equalities we have
Way 1: Increase the normal
stress in one direction
σ1
σ1
major principle stress
σ
n
τ
f
σ3
σ3
Minor principle stress
Confining stress
Way 2: directly apply the shear stress
Consider the following situation:
-
A normal stress is applied
vertically and held constant
-
A shear stress is then applied
until failure
Shear
stress τ
Normal stress σn
Normal stress σn
Determination of Shear Strength Parameters
The shear strength parameters of a soil are
determined in the lab primarily with two types
of tests: 1) Direct Shear Test; and 2) Triaxial
Shear Test.
(1)
(2)
Soil
Normal stress σn
Shear stress σ3
σ3
σ1
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

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 Data
Shear stress
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.
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
1.
Determine the dry
unit weight, γd
2.
Calculate the void
ratio, e
3.
Calculate the normal
stress & shear
stress
1

=
d
w
Gs
e
γ
γ
A
V
A
N
=
=
τ
σ
;
Figures
Peak Stress
s3
s2
s1
Shear stress, s
N3
= 30 kg
N2
= 20 kg
N1
= 10 kg
Horizontal displacement, ∆H
Figures (cont)
ShearStress, s (psf)
C′
φ
(σ1,s1)

3,s3)
(σ2,s2)
Normal Stress
σ, psf
Figures (cont)
Verticaldisplacement
Horizontal displacement
Reading Assignment:
Das, Ch. 11
HW: Problem 11.1