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