8/2/2019 Chapter 13-Shear Strength
1/37
CHAPTER 13
SHEAR STRENGTH OFSOIL
8/2/2019 Chapter 13-Shear Strength
2/37
Shear Strength
The shear strength of the soil is the internal resistance perunit area that the soil mass can offer to resistfailing/failure and sliding along any plane inside it.
Mohr (1900) presented a theory saying that rupture (orfailure) in material is because of a critical combination of
normal stress and shearing stress and not from eithermax. normal or shear stress alone.
For most soil mechanics problems, it is sufficient toapproximate the shear stress on the failure plane as linearfunction of the normal stress (Coulomb, 1776). This linearfunction is :
)('tan''
tan
stresseffectiveoftermsincor
c
f
f
8/2/2019 Chapter 13-Shear Strength
3/37
8/2/2019 Chapter 13-Shear Strength
4/37
Factors That Affect The ShearStrength of Soil
Cohesionless soil
1. Soil type
2. Soil density
3. Grain size distribution
4. Mineral type, angularity and particle size
5. Deposit variability-because of variations insoil types, particle arrangements etc., theeffective friction angle is rarely uniform withdepth.
6. Etc.
8/2/2019 Chapter 13-Shear Strength
5/37
Factors That Affect The ShearStrength of Soil
Cohesive soil
1. Soil type the more plastic the soil (higherPI), the lower the values of c and .
2. Particle Bonding
3. Stress history
4. Peak and ultimate shear strength
5. Etc.-such as sample disturbance, strainrate, anisotropy etc.
8/2/2019 Chapter 13-Shear Strength
6/37
Shear Strength
8/2/2019 Chapter 13-Shear Strength
7/37
Inclination of The Plane of FailureCaused by Shear
As stated by the Mohr-Coulomb failure criterion, failure fromshear will occur when the shear stress on a plane reaches a valuegiven by:
'tan'' cf
8/2/2019 Chapter 13-Shear Strength
8/37
Inclination of The Plane of FailureCaused by Shear
From Figures 11.2 and 11.3, angle bad = 2 = 90+
or
and
2
'45
2
'
45tan'22
'
45tan''
2
'45tan
'sin1
'sin1
2
31
2
cGeneralformula
8/2/2019 Chapter 13-Shear Strength
9/37
Drained and Undrained Conditions
The drained condition exists when the rate ofloading is slow compared to the rate ofdrainage. Thus, water is able to easily flow into
or out of the voids and essentially no excesspore water pressures develop in the soil.
The undrained condition exists when the rateof loading is rapid compared to the rate ofdrainage. Thus, water is not able to flow into orout of the voids quickly enough, and excesspore water pressures develop in the soil.
8/2/2019 Chapter 13-Shear Strength
10/37
Laboratory Tests For Detemination ofShear Strength Parameters
We do tests to get shear strength parameterssuch as c, , c, , etc. There are several testsconducted in the laboratory.
1. Direct shear test
2. Triaxial test
3. Unconfined compression test
4. Vane shear test
5. etc.
8/2/2019 Chapter 13-Shear Strength
11/37
Direct Shear Test
Size of specimen is 100mm X 100mm
Can be either stress or strain controlled
Normal stress of specimen max. is approximately 1050 kN/m2
Resisting shear force of the soil corresponding to any sheardisplacement can be measured by a horizontal proving ring or load
cell.
8/2/2019 Chapter 13-Shear Strength
12/37
Direct Shear Test
In the direct shear test arrangement, the shearbox that contains the soil specimen is generallykept inside a container that can be filled with
water to saturate the specimen. A drained test is made on a saturated soil
specimen by keeping the rate of loading slowenough so that the excess pore water pressuregenerated in the soils is dissipated completelyby drainage.
Pore water from the specimen is drainedthrough two porous stone.
8/2/2019 Chapter 13-Shear Strength
13/37
Direct Shear Test
8/2/2019 Chapter 13-Shear Strength
14/37
Direct Shear Test
8/2/2019 Chapter 13-Shear Strength
15/37
Direct Shear Test
8/2/2019 Chapter 13-Shear Strength
16/37
Direct Shear Test
8/2/2019 Chapter 13-Shear Strength
17/37
Triaxial Test
In triaxial test, specimen issubjected to a confining pressure(3) by compression of the fluid inchamber.
To cause shear failure, axial stressis applied through a verticalloading ram. This axial stress issometimes called deviator stress.
The axial load applied by the
loading ram corresponding to agiven axial deformation ismeasured by a proving ring orload cell attached to the ram.
8/2/2019 Chapter 13-Shear Strength
18/37
Triaxial Test
There are three standard types ofTriaxial Test:
1. Consolidated-drained test or drained test(CD Test)
2. Consolidated-undrained test or
undrained test (CU Test)3. Unconsolidated Undrained test (UU Test)
C lid t d D i d T t
8/2/2019 Chapter 13-Shear Strength
19/37
Consolidated-Drained Test(CD)
Because the pore water pressure developedduring the test is completely dissipated,therefore u=0.
From 3(Total) = u + 3
when u=0 3 = 0 + 3
3 = 3
Total and effective axial stress at failure
= 3 + ( d)f= 1 = 1
f= tan
Consolidated-drained triaxial test on clayeysoil may take several days to complete. Thistime period is required because deviatorstress must be applied very slowly to ensure
full drainage. For this reason the CD type isuncommon.
8/2/2019 Chapter 13-Shear Strength
20/37
Consolidated-Drained Test (CD)
Loose sand andnormallyconsolidated clay
Dense sand andover consolidatedclay
Loose sand andnormallyconsolidated clay
Dense sand andover consolidatedclay
8/2/2019 Chapter 13-Shear Strength
21/37
Consolidated-Undrained Test(CU)
Major principal stress at failure (total): 3(t)+(d)f=1(t)
Major principal stress at failure (effective): 1(t)-(ud)f= 1
Minor principal stress at failure (total): 3(t)
Minor principal stress at failure (effective): 3(t) -(ud)f=3
Where (ud)f= pore water pressure at failure
1(t)- 3(t) = 1- 3
f= tan(cu)
Consolidated-undrained is the most common type oftriaxial test. Because drainage is not allowed in these testsduring the application of deviator stress, they can beperformed quickly.
(a)Specimen underchamber confiningpressure
(b) Deviator stressapplication
8/2/2019 Chapter 13-Shear Strength
22/37
Consolidated Undrained Test (CU)Loose sand and normallyconsolidated clay
Dense sand andoverconsolidated clay
Loose sand andnormally consolidatedclay
Dense sand andoverconsolidated clay
8/2/2019 Chapter 13-Shear Strength
23/37
Unconsolidated Undrained Test(UU)
8/2/2019 Chapter 13-Shear Strength
24/37
Unconsolidated Undrained Test(UU)
8/2/2019 Chapter 13-Shear Strength
25/37
Discussion on Triaxial Test
Laboratory test generally is divided into 2 categories.
1. Shear strength tests based on total stress:
Also known as undrained shear strength test
To get undrained shear strength, Su, c, - total stress
Test type vane shear test, unconfined compression,
unconsolidated undrained (UU), consolidated undrained (CU) Performed exclusively on plastic (cohesive) soil
Normally for evaluation of foundation and embankment supportedby cohesive soil
Analysis for rapid loading or unloading conditions
Conditions applicable to field situations where change in shear
stress occurs quickly enough that soft cohesive soil does not havetime to consolidate
Known as short term analysis
Will give a bigger value of strength carried out with the same cellpressure in triaxial test.
8/2/2019 Chapter 13-Shear Strength
26/37
Discussion on Triaxial Test
2. Shear strength test based on effective stress.
To get effective strength c, ,etc.
Referred to as drained shear strength test.
Test include direct shear test, CD, CU, etc.
Performed on plastic (cohesive) soil and nonplastic (cohesionless)soil
Effective shear parameters used for long-term analysis wherecondition are relatively constant.
Eg. includes long term stability of slopes, embankments, earthsupporting structure, foundation etc.
Effective shear stress analysis fundamentally models the shearstrength of soil
8/2/2019 Chapter 13-Shear Strength
27/37
Discussion on Triaxial Test
Strength tests conducted on samples of a stiffoverconsolidated clay gave lower strengths for CDtests than CU tests because since anoverconsolidated specimen tends to expand duringshear (in undrained condition), the pore waterpressure decreases or even goes negative and thus
the effective stress is increased.
Because 3(f) = 3(f)-(-uf)or 1(f) = 1(f)-(-uf)the effective stresses are greater than the totalstress.Thus the undrained strength is greater than thedrained strength, which is opposite to the behavior
of normally consolidated clay.
Therefore, in normally consolidated clayShear strength of CD>CUIn CD, soil becomes stiffer due to the reduction ofvolume because (1-3)fcd>(1-3)fcu
