Download presentation

Presentation is loading. Please wait.

Published byAlexandra Fitzpatrick Modified over 2 years ago

1
ENV-2E1Y: Fluvial Geomorphology: Slope Stability and Geotechnics Landslide Hazards River Bank Stability Section 4 - Shear Strength of Soils N.K. Tovey Н.К.Тови М.А., д-р технических наук Landslide on Main Highway at km 365 west of Sao Paulo: August 2002

2
Introduction Seepage and Water Flow through Soils Consolidation of Soils ENV-2E1Y: Fluvial Geomorphology: Shear Strength ~ 1 lecture Slope Stability ~ 4 lectures River Bank Stability ~ 2 lectures Special Topics –Decompaction of consolidated Quaternary deposits –Landslide Warning Systems –Slope Classification –Microfabric of Sediments

3
Section 4 - Shear Strength of Soils Definitions: a normal load or force is one which acts parallel to the normal (i.e. at right angles) to the surface of an object a shear load or force is one which acts along the plane of the surface of an object the stress acting on a body (either normal or shear) is the appropriate load or force divided by the area over which it acts. Stress and Force must NOT be confused

4
EQUILIBRIUM There are three conditions: –the net effect of all forces parallel to one direction must be zero –the net effect of all forces orthogonal (at right angles) to the above direction must be zero –the sum of the moments of the forces must be zero the first two conditions can be checked by resolving forces (e.g. see Fig. 4.1) Section 4 - Shear Strength of Soils

5
Resolution of Forces Section 4 - Shear Strength of Soils P1P1 P3P3 P2P2 2 3 At Equilibrium: Resolve forces parallel to P 1 :- P 1 = P 2 cos 2 + P 3 cos Similarly at right angles to P 1 P 2 sin 2 = P 3 sin

6
Section 4 - Shear Strength of Soils Coulomb: a French Military Engineer Problem: Why do Military Fortifications Fail?

7
Section 4 - Shear Strength of Soils Coulomb: a French Military Engineer Problem: Why do Military Fortifications Fail? N F F = N tan is the angle of internal friction F N Is there a relationship between F and N?

8
Section 4 - Shear Strength of Soils Suppose there is some glue between block and surface Initially - block will not fail until bond is broken N F F = C + N tan C is the cohesion F N C Block will fail Block is stable

9
Three types of material –granular (frictional) materials - i.e. c = 0 (sands) = tan –cohesive materials - i.e. = 0 (wet clays) = c –materials with both cohesion and friction = c + tan Section 4 - Shear Strength of Soils F = C + N tan above equation is specified in forces In terms of stress: = c + tan

10
Stress Point at B - stable Stress Point at A - stable only if cohesion is present if failure line changes, then failure may occur. Section 4 - Shear Strength of Soils F N F - F G - G B A

11
Section 4 - Shear Strength of Soils F N F - F N NNNNNN N Displacement dense loose Peak in dense test is reached at around 1 - 3% strain

12
Section 4 - Shear Strength of Soils displacement Increasing normal stress Displacement / dense loose Normalising curves to normal stress leads to a unique set of curves for each soil.

13
Types of Shear Test –Stress controlled test –Strain controlled test (as done in practical) Section 4 - Shear Strength of Soils Failure in stress controlled test Displacement F Readings cannot be taken after peak in a stress controlled test N N NNNN BANG!

14
Section 4 - Shear Strength of Soils displacement displacement V V Dense Test Loose Test Medium Dense

15
All tests eventually come to same Void Ratio Section 4 - Shear Strength of Soils Plot volume changes as Void Ratio Void Ratio displacement medium Critical void ratio loose dense

16
= c + tan Does not allow for water pressure. Principal of Effective Stress From Consolidation Total Stress = effective stress + pore water pressure or = - u In terms of stresses involved water cannot take shear so = c + ( - u ) tan or = c + tan Mohr - Coulomb failure criterion if pore water pressure = 0 then original equation applies Section 4 - Shear Strength of Soils Effects of Water Pressure

17
Distance stress point is from failure line is a measure of stability. Greater distance > greater stability Section 4 - Shear Strength of Soils Mohr - Coulomb A -ve pwp moves stress point to right Moves point further from failure line greater stability Moves point closer to failure line less stability +ve pwp Slopes near Hadleigh Essex are only stable because of -ve pwp

18
Problems with Standard Shear Box Shear zone is complex Difficult to get undisturbed samples which are square Difficult to do undrained or partially drained tests –sands - always will be drained –clays - may be partially drained - depends of strain rate. Section 4 - Shear Strength of Soils The Triaxial Test

19
Section 4 - Shear Strength of Soils The Triaxial Test Load Cell Pressure Sample in rubber membrane Porous stone

20
Cell pressure can be varied to match that in ground cylindrical samples can be obtained sample can be sealed to prevent drainage or to allow partial drainage can perform both undrained and drained tests Section 4 - Shear Strength of Soils The Triaxial Test

21
Drained Test –allow complete dissipation of the pore water pressure. –speed of the test must allow for the permeability of the material. –for clays time is usually at least a week. –measure the volume of water extruded from or sucked into the sample in such tests. Undrained Test –no drainage is allowed. –measure the pore water pressures during the test. Section 4 - Shear Strength of Soils

22
Drained Test –response to load and volume change is similar to standard shear box. Undrained Test –burette is replace by a pore water pressure measuring device. –Since drainage is not required, test can be rapid. –Shear stress will be lower than in drained test if positive pore water pressures develop Section 4 - Shear Strength of Soils

23
In undrained dense tests pwp goes negative In drained dense tests volume increases displacement water pressure -ve +ve displacement water pressure +ve -ve Dense Loose

24
Section 4 - Shear Strength of Soils 4.8 Failure modes in the Triaxial Test. Loading – its length will shorten as the strain increases – some bulging towards the end. Over consolidated samples (and dense sands), –usually a very definite failure plane as peak strength is reached. Normally consolidated clays and loose sands, –failure zone is not visible –usually numerous micro failure zones criss-crossing the bulging region. Undrained test –orientation of the failure zone is at 45 o to the horizontal, Drained test –orientation will be at (45 + /2), - often not as well defined.

25
Diagram gives an insight into why some slopes appear to fail soon after they have formed, while in other cases they are initially stable, but fail much later. Section 4 - Shear Strength of Soils e log -ve pwp +ve pwp Water squeezed out Critical State Line Water sucked in

26
4.9 Unifying remarks on the behaviour of soils under shear. Drained –Some soils expand –Some soils contract –Depends on initial compaction. Undrained –Some samples +ve pwp develop –Some samples -ve pwp develop All samples move towards Critical State Line (CSL) What happens if sample has OCR consistent with CSL? –sample shears with no volume change in dense test –or no pore water change in undrained test. Section 4 - Shear Strength of Soils

27

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google