Shear Strength

29
Geotechnical Engineering A 5 Shear Strength

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Transcript of Shear Strength

Page 1: Shear Strength

Geotechnical Engineering A

5

Shear Strength

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Contents1. Shear Strength of Soils 2. Test Conditions3. Measurement of Shear Strength

Parameters• Shear Box Test• Triaxial Test

4. Shear Vane Test (Insitu Test)5. Vertical Stress

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1. Shear Strength of Soils

Shear strength• Define as the maximum shear

stress that can be applied to that soil in any direction

• When this maximum has been reached the soils yields & is regarded to have failed

• Shear strength is derived from the frictional resistance F generated from inter-particle forces , N

• The pore water has no shear strength

Shear surface

Relative displacement of soil mass

F F

N

N

Shear strength f at failure is some function of the total stress normal (n) to that plane f = c + ntan

where c & = experimentally determined constants

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Shear Strength of Dry Sand

shear stress (fkN/m

2

normal stress (nkN/m2

f= ntan

frictional soil no cohesion

Experimentally determined constants c &

f = c + ntan

ntan c = 0

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shear stress (fkN/m

2

normal stress (nkN/m2

Experimentally determined constants c &

Undrained Shear Strength of Saturate SoilsIn terms of total stress, any saturated soil, which is not allowed to drain, will exhibit no, or very little frictional resistance e.g. clays

c

= 0 undrained strength envelope

f= c

cohesive soil no friction

f = c + ntan

c

ntan = 0

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Experiment has shown that this general expression is correct for a wide range of soils for a limited range of stresses

f = c + n tan

shear stress (fkN/m

2

normal stress (nkN/m2

c

General Case Sand + Clay

Coulomb’s equation (1773)

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Effective Stress

shear stress (fkN/m

2

normal stress (nkN/m2

’ = u

Shear strength of a soil can be expressed in terms of effective stress

f’ = c’ + ntan ’

’c c’

Forcestress

Area

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2. Test Conditions

• Values of used in design will normally have been obtained from laboratory or insitu tests

• It is essential that the conditions of the tests are reported

– Undrained = total stresses (u uu

– Drained or Undrained with PWP measured = effective stresses (’ ’’

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• Foundations built on a saturated silty clay– bearing capacity required– the soil must be able to take the

load before the PWP has had time to dissipate

– soil will become stronger with time as PWP dissipate & effective stress increases

– settlement (dissipation of PWP) will occur which may itself cause problems

– critical stage from a strength point of view is immediately after the load is applied

• Long term stability of a cut slope or retaining wall– effective stress parameters

– obtained from a drained or undrained test with the measurement of PWP

Type of test depends on information required e.g.

Undrained test Drained test or Undrained with measurement of PWP

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3. Measurement of shear Strength parameters

• This is the simplest form of laboratory shear strength test and is often referred to as the direct shear test as it relates the shear stress at failure directly to the normal stress, thus the failure envelope may be plotted directly from the results.

• In the direct shear box the sample is caused to shear along the plane dividing the upper and lower pieces by applying a horizontal load to the upper piece while the lower piece is held in position. The load is generally applied via a proving ring, hence the load causing the sample to shear can be read directly and the shear stress, is the load causing shear divided by the plan area of the box.

• The test is repeated several times on different specimens of the same sample using different normal loads. The results are then be plotted, to give the shear strength envelope, form which a value of may be obtained.

• The apparatus, comprises a square box construction in two separate pieces, an upper piece and a lower piece. The vertical normal load is applied directly through the upper pressure plate and it is divided by the plan area of the box to give the normal stress .

• The volumetric behavior of the soil can also determine during the test by measuring the amount of horizontal displacement and vertical displacement using dial gauges.

• By using either porous or impervious platens above and below the sample, both drained and undrained tests may be performed. It is not possible to measure the pore water pressures during the test. Consequently, the shear box provides a simple method of measuring total stress in clay samples and the total and effective stress of free draining granular deposits.

• The shear strength of samples with a pre-formed shear plane can also be established. This parameter, which is known as residual shear strength, is of use when assessing the stability of failed slopes.

1) Shear Box Test (BS 1377; Part 7; 1990: 4 &5)

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Shear box

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Typical Results from a Shear box Test

Shear Stress ( kN/m2

Horizontal Displacement (mm)

n=20kN/m2

n=40kN/m2

n=80kN/m2

Loose

Dense

f (20kN/m2)

f (40kN/m2)

f (80kN/m2)

f (20kN/m2)

f (40kN/m2)

f (80kN/m2)

Normal Stress n

Shear Stress ( kN/m2

20 40 80

Dense

Loose

Medium

Medium

Ultimate or residual shear strength

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Volumetric Displacement

Horizontal Displacement (mm)

Vertical Displacement

(mm)

Compression – volume decrease

Dilation – volume increase

n=20kN/m2

n=40kN/m2

n=80kN/m2

At the maximum shearing resistance (f) = the rate of maximum volume change (dilation)

A thin rupture zone of the soil at critical density is produced

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Typical Results

Material ' peak ' ult

Dense well graded SAND or angular GRAVEL 55 35

Medium desne uniform SAND 40 32

Dense slightly clayey SILT 47 32

Sandy silty CLAY 35 30

Shaley CLAY 35 35

Silty CLAY (London Clay) 21 15

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Example 1 the following results were recorded during a shear box test on a cohesive soil:

Normal load (N) 73 191 309 427 545Shear load at failure (N) 109 139 170 197 227

If the specimen size was 60mm x 60mm, plot the failure envelope and determine the apparent cohesion and angle of shearing resistance.

Area of shear box = 60 x 60 = 3600mm2 or 3600/(1000x1000) = 3.6x10-3m2

Loadstress

Area

3

3

73 10'

3.6 10nnormal stress

= 20.3 kN/m2

3

3

109 10

3.6 10fShear stress

= 30.3 kN/m2

Normal stress (kN/m2) 20.3 53.1 85.8 118.6 151.4

Shear stress at failure (kN/m2) 30.3 38.6 47.2 54.7 63.1

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0

10

20

30

40

50

60

70

0 50 100 150 200

Normal stress (kN/m2)

Sh

ea

r s

tre

ss

at

failu

re (

kN

/m2 )

Normal stress (kN/m2) 20.3 53.1 85.8 118.6 151.4

Shear stress at failure (kN/m2) 30.3 38.6 47.2 54.7 63.1

cohesive soil therefore obtain a ‘c’ value

C = 25 kN/m2

= 14o

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2) Triaxial Compression Test (BS 1377; Part 7; 1990)

The apparatus consists of a cell, which is filled with water under pressure; the specimen is loaded vertically, via a proving ring to measure load.

The vertical load on the specimen is increased until failure occurs, the vertical strain being recorded at the same time using a dial gauge. The test is repeated on different specimens from the same soil, using different values of cell pressure.

Triaxial Test EquipmentTriaxial Test Equipment Triaxial Test CellTriaxial Test Cell Test SampleTest Sample

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Stresses on Specimen in Triaxial CellStresses on Specimen in Triaxial Cell

Cell Pressure Deviator Stress =P/A 1=3+P/A

1 = major principal stress

3 = minor principal stress

Therefore, P/A = (1-3) =Deviator stress

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The deviator stress is the load on the specimen, P, divided by the cross sectional area of the specimen. However, as the sample is compressed during the test, the cross sectional area will increase. Therefore, in calculating the deviator stress an allowance for the change in area must be considered.

For the calculation of deviator stress, it is assumed that the volume of the specimen remains constant and that the sample will deform as a cylinder,

100%o

XStrain

L 1 3

PDeviator stress

A

where P = vertical load, which is measured by a proving ring (kN)A = Area calculated using the following method

( ) )o o o oVolume V A L AL A L X 1o o

o

V Aor A or A

L X

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Brittle failure (shear)

Plastic failure (barrelling)

failure at 20% strain

Strain () %

Deviator Stress (kN/m2)

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Having tested three specimens from the same sample at three different cell pressures, the shear strength parameters may be assessed using a construction known as a Mohr circle diagram. This diagram may be explained by way of the following example.

Example 2; The stress/strain graphs for three specimens taken from a single sample of silty clay are shown below. Calculate the shear strength parameters for the clay.

Strain () %

Deviator Stress (kN/m2)

672

573

425

Sample Cell Pressure (3)A 100B 200C 300

A 100B 200C 300

Cell Pressure (3)Deviator Stress

(1-3)=P/ASample

Major principal Stress 1

425573

672

1 3P

A

425+100 = 525525573+200 = 773773672+300 = 972972

Plot Mohr Circle Diagram

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Mohr Circle Diagram

A 100B 200C 300

Cell Pressure (3)Deviator Stress

(1-3)=P/ASample

Major principal Stress 1

425573

672

425+100 = 525525573+200 = 773773672+300 = 972972

Shear Stress ( kN/m2

Cell Pressure kN/m2

400

300

200

100

0

0 200 400 600 800 100011=525=52533=100=100

Sample A

Sample B

11=773=77333=200=200 11=972=97233=300=300

Sample C

1-3

Diameter of CircleRadius of Circle =

1 3

2

{ccuu = 130 = 130

kN/mkN/m22

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Shear stress (fkN/m2

normal stress (nkN/m2

Undrained Shear Strength of Saturate Soils

In terms of total stress, any saturated soil, which is not allowed to drain, will exhibit no, or very little frictional resistance e.g. clays

cu

= 0 undrained strength envelope

= cu

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The triaxial compression test are commonly undertaken on undrained specimens, where a rubber membrane seals the specimen within the cell. These result are in terms of total stress. However, it is possible to measure the pore water pressures during the shearing stage, allowing the effective stress parameters to be recorded. Another method of assessing the effective stress of a sample is to apply the load at a very slow rate and allow the sample to drain.

UndrainedUndrained Consolidated undrained with pore Consolidated undrained with pore water pressure measurementwater pressure measurement

Consolidated drained Consolidated drained

Test Type Part No Para.1 Undrained triaxial compression test 7 8 & 92 Consolidated undrained triaxail compression test with pore water pressure measurement8 73 Consolidated drained triaxail compression test 8 8

The various triaxial compression tests area described in detail in the following parts of BS 1377, 1990;

24hrs

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• Undrained test:Undrained test:– Drainage is prevented throughout the test, so that no dissipation of pore pressure is

possible.– Parameters obtained: cu and u

– Typical site problem: immediate bearing capacity of foundations in saturated clay.

• Consolidated Consolidated — — undrained testundrained test: – Free drainage is allowed for (usually) 24 hours under cell pressure only to allow the

specimen to consolidate or to become saturated. Drainage is then prevented and pore-pressure readings taken during the application of axial load (i.e. shearing stage).

– Parameters obtained: c’ and ’ (i.e. referred to effective stress) and ccu and cu (i.e. referred to total stress)

– Typical site problem: sudden change in load, after an initial stable period, e.g. rapid drawdown of water behind a dam; or where effective stress analysis is required, e.g. slope stability.

• Consolidated - Drained test:Consolidated - Drained test:– Free drainage is allowed during a consolidation stage and drainage maintained during

the axial loading (which is carried out at a slow rate) so that no increase in pore pressure occurs.

– Parameters obtained: c’d and ’d

– Typical site problem: long-term slope stability

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4. Shear Vane Test (Insitu Test)• The vane test is a simple insitu test suitable for use on saturated clay at the

bottom of a trialpit or borehole on site: The vane is driven or pushed into the soil and a measured torque (T) applied to it until it rotates. The failure surface is the curved surface plus the flat ends of the ‘cylinder’ of soils whose diameter and height are that of the vane. The torque required to cause failure in a saturated clay is:

c 2

/ 32

dTorque T c h d

where:

T = Torque

c = Cohesion

d = Diameter

h = height of vane

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5. Vertical Stress

Total vertical stress = Depth x unit weight of soil

Effective vertical stress = Total stress – porewater pressure

kN/m2 = m x kN/m3

v = D x

’v = v - u

kN/m2 = kN/m2 - kN/m2

2

GL

depth = d

= unit weight of soil u = unit weight of water x h

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Example 3: Determine Total Stress, Pore Water Pressure and Effective Stress at the centre of the clay layer

Sand

Clay

Unit weight = 18kN/m3

Unit weight = 19.5kN/m3

Unit weight = 21.5kN/m3

2.5m3m

7m

water table

(2.5x18)+(19.5x0.5)+(2x21.5)

PWPPWP

(2.5x9.81)

or Effective stress

(2.5x18)+((19.5-9.81)x0.5)+((21.5-9.81)x2 ) = 73.2kN/m2

= 97.75kN/m2

= 24.5kN/m2

97.75 - 24.5 = 73.2kN/m2

’v = v - u

v (kN/m2) = depth (m) x unit weight of soil (kN/m3)

Effective stressEffective stress

Total stressTotal stress

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3m

5m

9m

(3 x 17) + (2 x 20) = 91 kN/m2

total stress

(3x17) + (2x20) + (4x19) = 167 kN/m2

total stress

3 x 17 = 51 kN/m2 2 x 9.81 = 19.6 kN/m2

6 x 9.81 = 58.8 kN/m2

91 -19.6 = 71.4 kN/m2

effective stress

167 -58.8 = 108.2 kN/m2

effective stress

Example 4: A layer of saturated clay 4m thick is overlain by sand 5m deep, the water table being 3m below the surface. The saturated unit weights of the clay and sand are 19kN/m3 and 20kN/m3 respectively: above the water table the unit weight of sand is 17kN/m3.

Plot values of total vertical stress and effective vertical stress against depth

Clay

Sand

water table

ground level

’v = v - uv (kN/m2) = depth (m) x unit weight of soil (kN/m3)