study guide for an exam (geotechnical)
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5Stressesinsoil_Shearstress.pptxGeotechnical Engineering:
Stresses in soil – shear stress
Dr Martin Pritchard
Contents
Stresses within soils – total stress, pore pressure & effective stress
Shear strength
Failure criterion for soils
Mohr-Coulomb – failure criterion for soils
Shear strength tests
Stress path – failure envelope
Tutorial questions
Learning outcome:
Assess, from first principles, the importance of the relevance of shear strength parameters.
Specific text:
http:// environment.uwe.ac.uk/geocal/SoilMech/stresses/stresses.htm
http://environment.uwe.ac.uk/geocal/glossary/GLOSS_A.HTM
Stresses in soils:
Total stress (σ): is the stress acting on a plane within a soil mass (force per unit area), as if the soil was a solid material.
NB in a dry soil forces act between soil particles at contact points. Contact area is very small and more-or-less impossible to measure therefore the traditional notation of stress is used.
In fully saturated soils ‘total stress’ is defined as (see effective stress below of derivation):
Pore Pressure (uw): is the pressure in the water, which surrounds the soil particles.
where
ρw = density of water (fresh water 1 Mg/m3)
g = gravity (9.81 m/s2)
hw = depth below water table
γw = unit weight of water (ρw.g)
The total force, F = F’ + uwAw or F = F’ + uwA - uwAc
Ac is very small compared to A, thus Ac/A tends to 0 and can be ignored, thus the formula can be re-written as:
The area of particle contact Ac
The area of water Aw = A – Ac
The pore water pressure = uw
The amount of the vertical force carried by the inter-particle contact = F’
Effective stress (σ’): is the particle-to-particle stress that is transmitted through the soil structure (for fully saturated soils).
Represents the average inter-particle stress and is called effective stress (s’)
The formula can be re-written in terms of effective stress:
In a saturated soil the load is carried:
Partly by the soil particles themselves (as above)
Partly by the water in between the soil particle (pore water pressure ‘uw’)
Inter-particle stresses act in one direction – normal to the plane of contact between the particles
Stresses from uw is are hydrostatic pressures and acts equally in all directions
Karl Terzaghi (1882–1963)
“All the measurable effects of a change of stress, such as compression, distortion and a change in the shearing resistance are exclusively due to changes in effective stress.” (Terzaghi, 1936)
This is the fundamental principle of effective stress (after Terzaghi in 1936), which states that ‘the effective stress at any depth is given by the total vertical stress minus the pore water pressure at that point’. The value of effective stress is important as it controls certain aspects of soil behaviour, notably deformation and strength.
9
Expressions can be derived for various stress conditions at depth z in a soil mass and water table at depth zw
Total stress σ = gsath + gb(z – h)
Pore water pressure uw = γwh
Effective stress σ’ = σ - uw
= h(gsat – gw) + γb(z-h)
= g’h + gb(z - h)
Revision: Density (r) and Unit Weight (g)
Unit weight = density x gravity
9.81m/s2
Density = Mass / Volume
Mass = no gravity
Weight = when gravity is taken into account
Bulk density (ρ) Mg/m3 x 9.81 m/s2 = Bulk unit weight (g) kN/m3
Dry density (ρd) Mg/m3 x 9.81 m/s2 = Dry unit weight (gd) kN/m3
Saturated density (ρ sat) Mg/m3 x 9.81 m/s2 = Saturated unit weight (γsat) kN/m3
Submerged unit weight (g’) kN/m3 = gsat - gw ,
Density of water (1000 kg/m3 or) 1 Mg/m3 x 9.81m/s2 = Unit weight of water (gw) = 9.81kN/m3 (NB density of sea water 1025 kg/m3)
Bulk unit weight gb (kN/m3)
Is the natural in situ unit weight of the soil
gb = total weight / total volume
Saturated unit weight gsat (kN/m3)
Is when the weight of the water is included in the weight of the soil
gsat = saturated weight / total volume
Buoyant (submerged) unit weight g’ (kN/m3)
Is when the weight of the water is not included in the weight of the soil
g’ = saturated unit weight – unit weight of water
Unit weight of water γw = 9.81kN/m3
Revision on unit weights
Example 1 (total and effective stress): A site is underlain by a substantial thickness of sand, which possesses a unit weight, g , of 19 kN/m3, above and below the water table. Calculate the total and effective stresses at a depth of 5 m below the ground surface for the following cases: a) When the water table is coincident with the ground surface. b) When the water table is situated at a depth of 1.5 m. c) When the water table is 2.5 m above ground level.
Sand
g = 19 kN/m3
5 m
σ & σ’
water table
a) Total stress (s) at 5 m
5 m x 19 kN/m3 = 95 kN/m2
σ = g.z
σ’ = σ - uw
σ’ = 5 m x 19 kN/m3 = 95 kN/m2
uw = 5 m x 9.8 1kN/m3 = 49 kN/m2
σ’ = 95 kN/m2 - 49 kN/m2 = 46 kN/m2
or σ’ = 5 m x (19 kN/m3 - 9.81 kN/m3 ) = 46 kN/m2
13
b) When the water table is situated at a depth of 1.5 m
Sand
g = 19 kN/m3
5 m
s & s’
water table
1.5 m
3.5 m
Total stress σ at 5 m
Effective stress σ’ at 5 m
5 m x 19 kN/m3 = 95 kN/m2
σ = 5 m x 19 kN/m3 = 95 kN/m2
u = 3.5 m x 9.81 kN/m3 = 34.3 kN/m2
σ’ = 95 kN/m2 - 34.3 kN/m2 = 60.7 kN/m2
σ = g.z
σ’ = σ - u
c) When the water table is 2.5 m above ground level
Sand
g = 19 kN/m3
5 m
σ & σ’
water table
2.5 m
Total stress s at 5 m
Effective stress s at 5 m
(5 m x 19 kN/m3) + (9.81 kN/m3x 2.5 m) = 119.5 kN/m2
σ = 119.5 kN/m2
uw = 7.5 m x 9.81 kN/m3 = 73.6 kN/m2
σ’ = 119.5 kN/m2 - 73.6 kN/m2 = 45.9 kN/m2
σ = g.z
σ’ = σ - u
Clay
Sand
water table
3 m
5 m
9 m
(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 2 (plotting total and effective stress profiles):
A layer of saturated clay 4 m thick is overlain by sand 5 m deep, the water table being 3m below the surface. The saturated unit weights of the clay and sand are 19 kN/m3 and 20 kN/m3 respectively: above the water table the unit weight of sand is 17 kN/m3.
Plot values of total vertical stress and effective vertical stress against depth
ground level
σ’ = σ - uw
σ (kN/m2) = depth (m) x unit weight of soil (kN/m3)
Shear strength of soils
Defined 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
The pore water has no shear strength
Shear surface
Relative displacement of soil mass
F
F
N
N
There are two components of shear strength; these are friction (f) and cohesion (c). NB pore water has no shear strength.
The values of c and f are known as the shear strength parameters.
In 1773 Coulomb developed an expression for these parameters, which defines a straight-line failure envelope.
Coulomb’s equation:
where:
c = apparent cohesion
f = angle of shearing resistance
sn = normal stress on failure plane
The first scientific study of soil mechanics was undertaken by French physicist Charles-Augustin de Coulomb, who published a theory of earth pressure in 1773. Coulomb’s work and a theory of earth masses published by Scottish engineer William Rankine in 1857 are still primary tools used to quantify earth stresses. These theories have been amended in the 20th century to take into account the influence of cohesion, a more recently discovered property of soils that causes them to behave somewhat differently under stress than Rankine and Coulomb predicted.
He formulated the Coulomb’s law, which deals with the electrostatic interaction between electrically charged particles. The coulomb, SI unit of electric charge, was named after him.
Born in Angoulême, France to a wealthy family, Charles-Augustin de Coulomb was the son of Henri Coulomb, an inspector of the Royal Fields in Montpellier. The family soon moved to Paris, where Coulomb studied mathematics at the famous Collège des Quatre-Nations. A few years later in 1759, he was enrolled at the military school of Mézières. He graduated from Ecole du Génie at Mézières in 1761.
Coulomb worked in the West Indies as a military engineer for almost nine years. When he came back to France, he was quite ill. During the French Revolution, Coulomb lived in his estate at Blois, where he mostly carried out scientific research. He was made an inspector of public instruction in 1802.
18
tf = c + σn tan f
Shear stress (tf) kN/m2
Normal stress (σn) kN/m2
f
c
Coulomb’s equation (1773)
sn tan f
Cohesionless soils
Shear stress (tf) kN/m2
Normal stress (sn) kN/m2
f
tf = tan f
e.g. sand & gravels
(only friction)
Shear stress (tf) kN/m2
Normal stress (sn) kN/m2
Frictionless 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
c
f = 0 undrained strength envelope
tf = c
e.g. undrained saturated clays & silts
(only cohesion)
tf = c + sn tan f
Shear stress (tf) kN/m2
Normal stress (sn) kN/m2
f
c
Partially saturated mixtures of cohesive & frictional material
e.g. clayey sands, silty sands
(both friction & cohesion)
Mohr-Coulomb failure criterion for soils
The force acting within a soil at a given point can be resolved into three principal stresses. These stresses act at right angles to each other and are termed:
Major principal stress (σ1)
Intermediate principal stress (σ2)
Minor principal stress (σ3)
Definition:
Principal stress is the normal stress acting on a principal plane
A Principal plane is the plane on which no shear stress will occur
If a soil sample is subject to an all-round pressure (sh or in a triaxial test s3) and the vertical pressure (sv or in a triaxial test s1 = s3+ DS) then the shear and normal stresses on any plane within the soil can be represented by points on the circumference of a circle, i.e. using Mohr circle construction
3.1 Stresses on the plane of failure:
Coulomb’s failure envelope:
Test conditions
Foundations built on a saturated silty clay
bearing capacity required
the soil must be able to take the load before the uw has had time to dissipate
soil will become stronger with time as uw dissipate & effective stress increases
settlement (dissipation of uw) 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 uw
Type of test depends on information required e.g.
Undrained test
Drained test or undrained with measurement of uw
Effective stress
Shear stress (tf) kN/m2
Normal stress (sn) kN/m2
Shear strength of a soil can be expressed in terms of effective stress
tf’ = c’ + sntan f’
f
f’
c
c’
Measurement of shear strength parameters:
Shear box test:
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.
Shear box test (BS 1377; Part 7; 1990: 4 & 5)
Triaxial Compression Test (BS 1377; Part 7 & 8: 1990)
Shear Vane Test (In-situ test)
Shear box test:
Typical results from a shear box test
Shear Stress (t) kN/m2
Horizontal Displacement (mm)
sn = 20 kN/m2
sn= 40 kN/m2
sn= 80 kN/m2
Loose
Dense
tf (20kN/m2)
tf (40kN/m2)
tf (80kN/m2)
tf (20kN/m2)
tf (40kN/m2)
tf (80kN/m2)
Normal Stress (sn) kN/m2
Shear Stress (t) kN/m2
20 40 80
Dense
Loose
Medium
Medium
Ultimate or residual shear strength
f’
Volumetric Displacement
Horizontal Displacement (mm)
Vertical Displacement (mm)
Compression – volume decrease
Dilation – volume increase
sn=20 kN/m2
sn=40 kN/m2
sn=80 kN/m2
At the maximum shearing resistance (tf) = the rate of maximum volume change (dilation)
A thin rupture zone of the soil at critical density is produced
Typical results
Example 4 (shear box)
The following results were recorded during a shear box test on a cohesive soil:
If the specimen size was 60 mm x 60 mm, plot the failure envelope and determine the apparent cohesion and angle of shearing resistance.
Area of shear box = 60 x 60 = 3600 mm2
or 3600/(1000x1000) = 3.6x10-3 m2
= 20.3 kN/m2
= 30.3 kN/m2
cohesive soil therefore obtain a ‘c’ value
C = 25 kN/m2
ϕ = 14o
Example 5 (shear box):
Two specimens of sand were tested in a shear box at the same constant normal stress of 200 kN/m2. Specimen A was prepared in a loose state, and specimen B was compacted into a dense state.
Plot the stress/strain and vertical displacement curves for the two tests.
Determine the angle of shearing resistance corresponding to the loose and dense states.
Loose sand: τ rises slowly to reach a value of 88 kN/m2.
Dense sand: τ rises steeply to reach a peak value of 125 kN/m2, before falling slowly to level off at the same approx. ultimate value of 88 kN/m2.
88 kN/m2
125 kN/m2
The horiz. disp. vs vertical disp. plot shows that:
As the loose specimen is sheared, a reduction in volume takes place.
As the dense specimen is sheared, after an initial amount of small compression, volume expansion (dilation) takes place.
Horiz. disp (mm)
Shear stress (kN/m2)
Loose sand, ϕ = 24o
Dense sand, ϕ= 32o
Normal stress (kN/m2)
Shear stress (kN/m2)
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 Equipment
Triaxial Test Cell
Test Sample
Stresses on specimen in triaxial cell
Cell Pressure
Deviator Stress =P/A
s1=s3+P/A
s1 = Major principal stress
s3 = Minor principal stress
P/A = (s1-s3) = Deviator Stress (DS)
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.
where P = vertical load, which is measured by a proving ring (kN)
A = Area calculated using the following method
Brittle failure (shear)
Plastic failure (barrelling)
failure at 20% strain
Strain (e) %
Deviator Stress (kN/m2)
Typical stress/strain graphs
The triaxial compression tests are commonly undertaken on undrained specimens, where a rubber membrane seals the specimen within the cell. These results 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.
Undrained
Consolidated undrained with uw measurement
Consolidated drained
The various triaxial compression tests area described in detail in the following parts of BS 1377, 1990;
24hrs
1). Undrained test*:
Drainage is prevented throughout the test, so that no dissipation of pore pressure is possible.
Parameters obtained: cu and fu
Typical site problem: immediate bearing capacity of foundations in saturated clay.
2). Consolidated — undrained 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 f’ (i.e. referred to effective stress) and ccu and fcu (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.
3). 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 f’d
Typical site problem: long-term slope stability
Shear stress (tf) kN/m2
normal stress (sn) kN/m2
Undrained shear strength of saturated clays
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
f = 0 undrained strength envelope
t = cu
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 6 (triaxial test) 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 (e) %
Deviator Stress (kN/m2)
672
573
425
425
573
672
425+100 = 525
573+200 = 773
672+300 = 972
A
B
C
Mohr circle diagram
425
573
672
425+100 = 525
573+200 = 773
672+300 = 972
Shear Stress (τ) kN/m2
Cell Pressure kN/m2
400
300
200
100
0
0 200 400 600 800 1000
s1=525
s3=100
Sample A
Sample B
s1=773
s3=200
s1=972
s3=300
Sample C
σ1-σ3
Diameter of Circle
Radius of Circle =
{
cu = 130 kN/m2
f = 16o
Shear vane test (in-situ test)
The vane test is a simple in-situ test suitable for use on saturated clay at the bottom of a trial pit 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
where:
T = Torque
c = Cohesion
d = Diameter
h = height of vane
t
f
= c + s
n
tan f
Shear stress
(t
f
) kN/m
2
Normal stress (s
n
) kN/m
2
f
c
s
n
tan f
Author/ Editor Craig, RE. Title Craig’s Soil Mechanics Edition 7th Publication Year 2004 Publisher Taylor & Francis ISBN 9780415561266 Recommended reading ✓
Author/ Editor Whitlow, R. Title Basic Soil Mechanics Edition 4th Publication Year 2000 Publisher Prentice Hall ISBN 9780582381094 Recommended reading ✓
Author/ Editor Vickers, B. Title Laboratory Works in Soil Mechanics
(Scanned copy of Chapter 4: ‘Shear-strength Tests’ is available to students on X-stream)
Edition 2nd Publication Year 1983 Publisher Granada Recommended reading ü
Author/ Editor Vickers, B.
Title Laboratory Works in Soil Mechanics
(Scanned copy of Chapter 4: ‘Shear-strength Tests’ is
available to students on X-stream)
Edition 2
nd
Publication Year 1983
Publisher Granada
Recommended reading
ü
|
Author/ Editor |
Vickers, B. |
|
Title |
Laboratory Works in Soil Mechanics (Scanned copy of Chapter 4: ‘Shear-strength Tests’ is available to students on X-stream) |
|
Edition |
2nd |
|
Publication Year |
1983 |
|
Publisher |
Granada |
|
Recommended reading |
|
|
Author/ Editor |
Craig, RE. |
|
Title |
Craig’s Soil Mechanics |
|
Edition |
7th |
|
Publication Year |
2004 |
|
Publisher |
Taylor & Francis |
|
ISBN |
9780415561266 |
|
Recommended reading |
|
Author/ Editor |
Whitlow, R. |
|
Title |
Basic Soil Mechanics |
|
Edition |
4th |
|
Publication Year |
2000 |
|
Publisher |
Prentice Hall |
|
ISBN |
9780582381094 |
|
Recommended reading |
✓ |
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s=s'+u
w
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Water table level
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Saturated density g
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Bulk density b
Water table level
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Embankment construction
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Failure
Pore pressure (uw)
Effective stress
Total stress
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where:
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𝛼 = 45! + 𝜙 2
Values of , & used in design will normally have been obtained from laboratory or in-situ tests. It is essential that the conditions of the tests are reported:
1. Undrained = total stresses (u, u & u)
2. Drained or Undrained with pore pressure measured = effective stresses (’, ’ & ’)
where uf = pore pressure at failure
A subscript ‘d’ can also be used to further denote that the parameters have been obtained under fully drained conditions (as in the ‘triaxial consolidated drained test’), e.g.:
Material
f
' peak
f
' ult
Dense well graded SAND or angular GRAVEL5535
Medium desne uniform SAND4032
Dense slightly clayey SILT4732
Sandy silty CLAY3530
Shaley CLAY3535
Silty CLAY (London Clay)2115
Sheet1
| Material | f' peak | f' 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 |
Sheet2
Sheet3
Normal load (N)73191309427545
Shear load at failure (N)109139170197227
(
)
Load
stress
Area
s
=
(
)
3
3
7310
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t
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Normal stress (kN/m
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)20.3
Shear stress at failure (kN/m
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)30.3
Sheet1
| Normal load (N) | 73 | 191 | 309 | 427 | 545 |
| Shear load at failure (N) | 109 | 139 | 170 | 197 | 227 |
| Normal stress (kN/m2) | 20.3 | ||||
| Shear stress at failure (kN/m2) | 30.3 |
Sheet2
Sheet3
Sheet1
| Normal load (N) | 73 | 191 | 309 | 427 | 545 |
| Shear load at failure (N) | 109 | 139 | 170 | 197 | 227 |
Sheet2
Sheet3
0
10
20
30
40
50
60
70
050100150200
Normal stress (kN/m
2
)
Shear stress at failure (kN/m
2
)
Normal stress (kN/m
2
)20.353.185.8118.6151.4
Shear stress at failure (kN/m
2
)30.338.647.254.763.1
Chart1
| 20.3 |
| 53.1 |
| 85.8 |
| 118.6 |
| 151.4 |
Sheet1
| Normal load (N) | 73 | 191 | 309 | 427 | 545 |
| Shear load at failure (N) | 109 | 139 | 170 | 197 | 227 |
| 20.3 | 53.1 | 85.8 | 118.6 | 151.4 | |
| 30.3 | 38.6 | 47.2 | 54.7 | 63.1 |
Sheet1
Sheet2
Sheet3
Sheet1
| Normal load (N) | 73 | 191 | 309 | 427 | 545 |
| Shear load at failure (N) | 109 | 139 | 170 | 197 | 227 |
| 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 |
Sheet2
Sheet3
Loose state Horiz. Disp (10-‐2 mm) -‐ L 0 50 100 150 200 250 300 350 400 450 Vert. Disp (10-‐2 mm) -‐ L 0 -‐6 -‐9 -‐11 -‐13 -‐14 -‐14 -‐14 -‐14 -‐14 Shear stress (kN/m2) -‐ L 0 35 57 72 79 84 84 88 89 90
Dense state Horiz. Disp (10-‐2 mm) -‐ D 0 50 100 150 200 250 300 350 400 450 Vert. Disp (10-‐2 mm) -‐ D 0 -‐3 3 11 18 23 26 27 28 28 Shear stress (kN/m2) -‐ D 0 67 111 124 120 103 94 88 88 87
Loose state
Horiz. Disp (10
-2
mm) - L050100150200250300350400450
Vert. Disp (10
-2
mm) - L0-6-9-11-13-14-14-14-14-14
Shear stress (kN/m
2
) - L0355772798484888990
Dense state
Horiz. Disp (10
-2
mm) - D050100150200250300350400450
Vert. Disp (10
-2
mm) - D0-3311182326272828
Shear stress (kN/m
2
) - D06711112412010394888887
(
)
100%
o
X
Strain
L
e
=´
(
)
13
P
Deviatorstress
A
ss
æö
=-
ç÷
èø
(
)
())
oooo
VolumeVALALALX
===-
(
)
(
)
1
oo
o
VA
orAorA
LX
e
==
--
Test TypePart No Para.
1Undrained triaxial compression test78 & 9
2Consolidated undrained triaxail compression test with pore water pressure measurement87
3Consolidated drained triaxail compression test 88
Sheet1
| Test Type | Part No | Para. | |
| 1 | Undrained triaxial compression test | 7 | 8 & 9 |
| 2 | Consolidated undrained triaxail compression test with pore water pressure measurement | 8 | 7 |
| 3 | Consolidated drained triaxail compression test | 8 | 8 |
Sheet2
Sheet3
Sample Cell Pressure (
s
3)
A100
B200
C300
A100
B200
C300
Cell Pressure (
s
3
)
Deviator Stress (
s
1
-
s
3
)=P/A
Sample
Major principal Stress
s
1
13
P
A
ss
=+
Sheet1
| Sample | Cell Pressure (s3) | |
| A | 100 | |
| B | 200 | |
| C | 300 |
Sheet2
Sheet3
Sheet1
| Sample | Deviator Stress (s1-s3)=P/A | Cell Pressure (s3) | Major principal Stress s1 | |
| A | 100 | |||
| B | 200 | |||
| C | 300 |
Sheet2
Sheet3
A100
B200
C300
Cell Pressure (
s
3
)
Deviator Stress (
s
1
-
s
3
)=P/A
Sample
Major principal Stress
s
1
13
2
ss
-
æö
ç÷
èø
Sheet1
| Sample | Deviator Stress (s1-s3)=P/A | Cell Pressure (s3) | Major principal Stress s1 | |
| A | 100 | |||
| B | 200 | |||
| C | 300 |
Sheet2
Sheet3
Example 7:
The results given below were obtained from a consolidated undrained triaxial compression test on clay soil:
Plot the failure envelopes with respect to both total stress and effective stress and determine the values of ccu, ϕcu, c’ and ϕ’.
Cell pressure (kN/m2) 100 200 400 DS at failure (kN/m2) 248 294 382 Uw at failure (kN/m
2) 13 96 264
Cell pressure (kN/m
2
)100200400
DS at failure (kN/m
2
)248294382
U
w
at failure (kN/m
2
) 1396264
(
)
(
)
2
/3
2
d
TorqueTchd
p
=+
6. STRESS PATH
Determining the best common tangent on the Mohr circle diagram for three or more circles can introduce discrepancies, from person to person. Therefore an alternative method of plotting the topmost point of each circle can be employed:
6. STRESS PATH
Determining the best common tangent on the Mohr circle diagram for three or more circles
can introduce discrepancies, from person to person. Therefore an alternative method of
plotting the topmost point of each circle can be employed:
6. Stress path
Determining the best common tangent on the Mohr circle diagram for three or more circles can introduce discrepancies, from person to person. Therefore an alternative method of plotting the topmost point of each circle can be employed:
τ
σ
t
s
θ
Example 8 (Mohr circles & stress path)
Three specimens of the same soil were subjected to the pressures given below:
Specimen 1 2 3
σ3 (kN/m 2) 100 200 300
σ1 (kN/m 2) 400 600 800
Obtain c and φ as well as the stresses on the three slip surfaces from:
1. Mohr circle construction method 2. Stress path method
Example 8 (Mohr circles & stress path)
Three specimens of the same soil were subjected to the pressures given below:
Specimen 1 2 3
s
3
(kN/m
2
)
100 200 300
s
1
(kN/m
2
) 400 600 800
Obtain c and f as well as the stresses on the three slip surfaces from:
1. Mohr circle construction method
2. Stress path method
Example 8 (Mohr circles & stress path)
Three specimens of the same soil were subjected to the pressures given below:
|
Specimen |
1 |
2 |
3 |
|
3 (kN/m2) |
100 |
200 |
300 |
|
1 (kN/m2) |
400 |
600 |
800 |
Obtain c and as well as the stresses on the three slip surfaces from:
1. Mohr circle construction method
