ce 240 soil mechanics & foundations lecture...
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CE 240Soil 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: CohesionCohesion (C), is a measure of the forces that cement particles of soils
Dry sand with no cementationDry sand with some cementationSoft clayStiff 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.
Mohr-Coulomb Failure Criterion
Shear Strength,S
φ = φ′
C′
Normal Stress, σn = σ′ = γ h
' '
tan (11.2)
tan ' (11.3)f n n
f n n
c c
c c
τ σ φ µσ
τ σ φ µ σ
= + = +
′ ′ ′= + = +
f
where shear strength
c = cohesion; c =effective cohesionφ = angle of internal friction; φ = effective angle of internal friction
= coefficient of friction; ' = effective coefficient of frictio
τ
µ µ
=′
′
n.
µ=tanφ’
0.51-0.580.58-0.700.70-0.78
0.58-0.700.70-0.840.84-1.000.67-1.110.49-0.70
Mohr-Coulomb shear failure criterion
σ1
σ1
σ3σ3σ n
τ f
σ1σ3 σ
τ
2θ
τf = c’ + µ’σn’
c’φ
φFailed 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 fσ3 σ3
Minor principle stressConfining 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 ParametersThe shear strength parameters of a soil are
determined in the lab primarily with two types of tests: 1) Direct Shear Test; and 2) TriaxialShear 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 DataSh
ear s
tres
s
Residual Strength
Peak Strength
Direct Shear Test Data:Volume change
∆H
Direct Shear Test: Procedure1.Measure inner side or diameter of shear box and find the area2.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 sand6.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: Procedure8. 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.
Calculations1. Determine the dry
unit weight, γd
2. Calculate the void ratio, e
3. Calculate the normal stress & shear stress
1−=d
wGseγγ
AV
AN
== τσ ;
Figures
Peak Stress
s3
s2
s1
Shea
r st
ress
, s
N3 = 30 kg
N2 = 20 kg
N1 = 10 kg
Horizontal displacement, ∆H
Figures (cont)Sh
ear
Stre
ss, s
(psf
)
C′
φ
(σ1,s1)
(σ3,s3)(σ2,s2)
Normal Stress σ, psf
Figures (cont)
Ver
tical
disp
lace
men
t
Horizontal displacement
Reading Assignment:
Das, Ch. 11
HW: Problem 11.1
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