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8/2/2019 Chapter 13-Shear Strength
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CHAPTER 13
SHEAR STRENGTH OFSOIL
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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
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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.
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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.
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Shear Strength
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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
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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
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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.
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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.
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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.
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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.
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Direct Shear Test
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Direct Shear Test
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Direct Shear Test
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Direct Shear Test
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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.
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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
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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.
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Consolidated-Drained Test (CD)
Loose sand andnormallyconsolidated clay
Dense sand andover consolidatedclay
Loose sand andnormallyconsolidated clay
Dense sand andover consolidatedclay
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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
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Consolidated Undrained Test (CU)Loose sand and normallyconsolidated clay
Dense sand andoverconsolidated clay
Loose sand andnormally consolidatedclay
Dense sand andoverconsolidated clay
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Unconsolidated Undrained Test(UU)
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Unconsolidated Undrained Test(UU)
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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.
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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
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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
But in overconsolidated clay shear strength CU>CD In CD, soil dilate and pore pressure reduces
CUeffective
CU totalCD effective
CU totalCU effective
CDeffectivef
n
f
n
Overconsolidated clay
Normally consolidated clay
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Use of CD Test in EngineeringPractice
CD conditions are the mostcritical for the long-term steadyseepage case for embankmentdams and the long-term stabilityof excavations or slopes in bothsoft and stiff clays.
Note: we should be aware that,practically speaking, it is noteasy to actually conduct a CDtest on a clay in the laboratorybecause of time factor. Therefore
CD test is not very popular.
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Use of CU Test in EngineeringPractice
CU strengths are used for stabilityproblems where the soils have firstbecome fully consolidated and are atequilibrium with the existing stresssystem. Then, for some reason,
additional stresses are applied quickly,with no drainage occuring. Practicalexamples include rapid drawdown ofembankment dams and the slopes ofreservoirs and canals.Since it is possible to measure theinduced pore pressures in a CU test
and thereby calculate the effectivestresses in the specimen, CU tests aremore practical for obtaining theeffective stress strength parameters.
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Use of UU Test in EngineeringPractice
Applicable in situations where the engineeringloading is assumed to take place so rapidly thatthere is no time for the induced excess pore waterpressure to dissipate or for consolidation to occurduring the loading period. We also assume thatthe change in total stress during construction
does not affect the in situ undrained shearstrength. Examples include end of construction ofembankment dams and foundations forembankment, piles, and footings on normallyconsolidated claysFor these cases, often the most critical designcondition is immediately after the application ofthe load (at the end of construction) where theinduced pore pressure is the greatest but beforeconsolidation has had the time to take place. Onceconsolidation begins, the void ratio and the watercontent naturally decrease and the strengthincreases. So the embankment or foundationbecomes increasing safer with time.
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Unconfined Compression Test
The unconfined compressiontest is a special type ofunconsolidated-undrainedtest that is commonly usedfor clay specimens.
In this test, the confiningpressure 3 is 0.
At failure, the total minorprincipal stress is zero and
the total major principalstress is 1 (Figure 11.33).
ufc
q
22
11
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Unconfined Compression Test
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Unconfined Compression Test
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Vane Shear Test
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Vane Shear Test
Fairly reliable results for theundrained shear strength, cu (=0concept), of very soft to mediumcohesive soils may be obtaineddirectly from vane shear tests.
If T is the maximum torque appliedat the head of the torque rod tocause failure, it should be equal tothe sum of the resisting moment ofthe shear force along the side
surface of the soil cylinder (Ms) andthe resisting moment of the shearforce at each end (Me) (Figure11.43).
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Vane Shear Test
T = Ms + Me + Me
The resisting moment can begiven as:
Ms = (dh)cu (d/2)
Two ends
Surfacearea
Momentarm
42
32dhd
cTu
42
32
dhd
Tcu
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Example