Stress and Strains in Soil and Rock Hsin-yu Shan Dept. of Civil Engineering National Chiao Tung University
Stress and Strain ε 1 1 2 ε 2 ε
Dimension 1 2 0 ε ε ε 0 1 2 ε 1 1 2 ε 2 ε
Plane Strain = 0 1 2 ε ε ε 1 2 ; x 0 = 0 ε 1 ; 1 If the material is linearly isotropic: = ν + ( ) 2 1 2 ε 2 ε
Axisymtric Conditions Axisymetic Triaxial Uniaxial ε 1 1 2 ε 2 ε
Axisymetric 1 0 ; 2 = θ sr ε1 ε 0 ; ε2 = r s r = redial displacement r = radial coordinate of the point ε 1 1 For linearly isotropic materials: = ν ( + ) + Eε 2 1 2 2 ε 2 ε
Triaxial Compression z > r = ; = = 1 z 2 ε = ε ; ε = ε = ε 1 z 2 Extension r r ε 1 1 z < r = = ; = 1 2 r ε = ε = ε ; ε = ε 1 2 r z z 2 ε 2 ε
Uniaxial Same as triaxial condition except r = 0 Compression z > r = ; = = 1 z 2 ε = ε ; ε = ε = ε 1 z 2 r r ε 1 1 ε 2 2 ε
Ko Stress State z > r 1 = z ; 2 = = r ε = ε ε = ε 0 1 z ; 2 = ε 1 1 z < r 1 = 2 = r ; = z ε = ε = 0 ε = ε 1 2 ; z 2 ε 2 ε r = ν ν z 1 If the material is linearly isotropic
2 ε 2 = 0 u 2 ε 2 = 0 r 1 = z 2 = = r ε2 = ε= ε r 1 = z 2 = = 0 ε2 = ε = ε r ε = ε = 1 = z 2 = = r 2 0
Rendulic Plot Axial Stress a' Radial Stress 2 r a' 2 r
q-p Plot Roscoe, Schofield, and Wroth (1958) q = 1 = 1 1 q p' = ( 1+ 2 + ) Lambe and Whiman (1969) 1 1 q = ( 1 ) = ( 1 ) 2 2 1 p' = ( 1+ ) 2 p or p
τ - Plot τ or
Ideal Shear Strength Test Independent control of stress and strain in each direction Direction of principal stress Rate of stress or strain application Uniform distribution of applied stress Control over pore (air or water) pressure
Parameters Obtained Shear strength parameters Pore pressure parameters Deformation parameters
Determination of Shear Strength from Test Results Define failure criteria 1. Peak stress difference ( 2. Maximum stress ratio. Maximum shear stress τ max 4. Specific strain level 1 ) max Determine principal stresses and shear stress 1 max Drained tests: 1. and 2. give same c, φ Undrained tests: 1. and 2. give different sets of c, φ
Selection of failure criteria depends on: Testing condition Field condition (Drained vs. Undrained) Method of analysis (Total stress vs. Effective stress)
Determining c and φ Mohr-Coulomb diagram, τ and Tangent line of Mohr circles τ φ c 1 2 or
q-p plot Linear regression of (q, p) sets 1 1 q = ( 1 ) = ( 1 ) 2 2 1 p' = ( 1+ ) 2
1 2 ( q 1 ) q = c cosφ + p' sinφ ψ tanψ = sinφ d = c cosφ d p 1 2 ( + 1 )
Curved Failure Envelope Dilatancy effect at lower stress level Crushing of particles at high stress level Rearrangement of particle orientation under higher stress (tend to be more parallel) Higher φ Lower φ
q p Compacted kaolinite Compacted Higgens clay =100 psi φ = 26 =100 psi φ = 25 = 2500 psi φ =12 =1200 psi φ =19
( 1 ) Stress path tangency Effective stress path for undrained test u increases Eff. stress path for undrained test u decreases Effive and total stress path of drained test Total stress path for undrained test or
( 1 ) max ( 1 ) Normally Consolidated and Lightly Overconsolidated Clay Loose Sand 1 max ε a
( 1 ) max ( ) 1 Overconsolidated Clay Dense Sand 1 max ε a
Deformation Properties Equivalent Elastic Parameters E i = Initial tangent modulus (initial linear section) E f = Secant modulus (0 to strain at failure) E 50 = Modulus from (0 to 50% strength)
Laboratory Shear Strength Tests Direct shear test Ring shear Simple shear, direct simple shear Unconfined compression test Triaxial test Compression Extension Plane strain test seldom Dynamic strength tests
Direct Shear ASTM D080 Normal Force Shear Force Soil Specimen
Stress Path τ or
Problems with Direct Shear Test Nonuniform stress Normal stress Shear stress Changing area of shear plane Contact between soil and shear box Squeezing of soil Contact between soil and shear box Changing area of shear plane
Direct Shear Advantages Easy to perform Low cost A lot of experience Drawback Small-size specimens Limited shear displacement Forced failure plane
Ring Shear Advantages Cost is not very high Unlimited shear displacement Drawbacks Anisotropy Small specimen size Forced failure plane Limited experience
Unconfined Compression Test Soil Specimen
Triaxial Test Soil Specimen
No drainage 1 u
1-1 u a + u d = u
Pore Pressure Coefficients In an undrained test, the change of pore pressure can be related to the changes of stresses as: u = B[ + A( )] 1 A and B are the pore pressure coefficients u a = B
Change of total volume: V c = c c V ( = c V ua ) Change of volume of voids: V V v = c v nv u c c = compressibility of the soil skeleton c v = compressibility of the fluid in voids = c V v a c c c v c ε v = v = ε u
B = 1+ 1 n c c c v For saturated soils, if c c >> c v, B 1 (soil skeleton is much more compressible than the fluid) For very stiff clays whose c c is also very small, B may not be close to 1 even if the soil is almost saturated
A and B can only be determined from test results When ( 1 - ) remains 0, B can be determined from the relationship between the increasing and u For saturated specimens, B is close to 1 Get A by from the changes of u and ( 1 - ) A is not a constant, it changes through out the test ud 1 A = B 1
Use B to check the saturation of soil specimens Incremental increase of check the increase of u compute B If B is close to 1, the specimen is near saturation
Although B is an indication of the degree of saturation, there is no way of knowing the degree of saturation from values of B
u = B[ + A( )] 1 u = B A B( )] + 1 A = A B For B = 1 A = u 1 During shearing = 0 A = u 1
A depends on: the state of the specimen (soil type), the stress history (overconsolidation), the stress state (anisotropy), the stress level, loading conditions (loading or unloading), and other conditions A can vary from negative values to greater than 1 A at failure is defined as A f A f
Back Pressure Increase the degree of saturation Increase pore water pressure, u, in the specimen while maintaining the effective confining (consolidation) stress For example: 100 kpa 150 kpa 200 kpa 250 kpa u 50 kpa 100 kpa 150 kpa 200 kpa 50 kpa 50 kpa 50 kpa 50 kpa Check B at every stage
How Does Backpressure Works? Elevated pore water pressure can squeeze on the trapped air bubbles and make them decrease in size Elevated pore water pressure would make the trapped air more likely to dissolve in water Some would apply a small hydraulic gradient to help flush out the air during the saturation process
1 = 0 = 0 1 = p0 '1 = p0 = p0 ' = p0 1 = p0 = p0 '1 = ui ' = ui '1 = p0 ' = p0 1 = p1 = p1 1 = p0+p1 = p0+p1 '1 = ui ' = ui '1 = p0 ' = p0 1 = p1+p2 = p1 1 = p0 + p1+ p2 = p0+p1 1 = p0+p2 = p0 '1 = p2(1-af)+ ui ' = -Af p2 + ui '1 = p2(1-af)+p0 ' = -Af p2 + p0 '1 = p0+p2 ' = p0
Unconsolidated Undrained Test UU test or Q (quick) test 1 = 0 1 - = 0 u 0 u 1 + + u 2 1 = 0 + ( u1) = 0 = u0
Reason for the Confining Stress More similar to actual field condition where all around pressure exists Increase the degree of saturation Can make the fissures (formed during the extrusion of tube samples) in the specimens close up
( 1 ) d
Consolidated Drained Test CD test or S (slow) test = 0 c 1 - u 0 = 0 u c c + u=0 u c = back pressure
( 1 ) d ( )
Consolidated Undrained Test Without pore pressure measurement CU test or R (rapid) test With pore pressure measurement CU or test Isotropic consolidation CIU Anisotropic consolidation CAU R
= 0 c 1 - = 0 c u 0 u c + u 2 + u u c = back pressure
( 1 ) Effective stress failure envelope Effective stress path for undrained test u increases Total stress failure envelope Eff. stress path for undrained test, u decreases or
CD, S Tests ( 1 ) Effective stress failure envelope failure or
CU, R Tests CU, R ( 1 ) High sensitivity Low sensitivity ( 1 ) max ( 1 ) max Effective stress path for undrained test u increases or
UU, Q Tests (CU, R Tests) ( 1 ) ( 1 ) max Failure R envelope Q envelope
Energy Corrections Direct shear tests Triaxial tests
Direct Shear Tests v h v > 0 for expansion τ Total work done by applied shear stress = τ A d h = τ * A d h + A d v Energy used to expand (+) or contract (-) against normal stress Shear stress used to shear the soil to failure: d v τ* = τ d h
Triaxial Tests Work done by axial stress: d 1 ε 1 Work done to expand the specimen laterally against the confining stress: d d ε ε dw=work required to shear the specimen (subject to dε )
1 1 ε ε ε d d dw d = Dilate if dε >0 ) ( ) ( 1 1 1 ε ε ε ε d d d dw d + + = 2 1 strain Volumetric ε ε ε ε d d d d v + + = = Bishop assumed: 1 1 * ) ( ε d dw = v d d d ε ε ε 1 1 1 1 * ) ( ) ( =
1 1 1 ) ( )* ( ε ε d d v + = < 0 expand > 0 contract Say c = 0 1 1 2 ) ( ) ( sin φ + = 1 1 sin φ + = 1 1 2 )* ( )* ( * sin φ + =
φ Measured φ Particle orientation effect Dilatancy effect φ u φ cv : constant volume 46% Loose Initial porosity, n i φ u = mineral-to-mineral friction angle 4% Dense
1 ) (1 * sin 1 1 + = ε ε φ a v d d Bishop ) (1 ) (1 * sin 1 1 a v a v d d d d ε ε ε ε φ + = < 0 expand > 0 contract Rowe If soil tends to dilate, φ * Rowe < φ * Bishop (Rowe considers particle rearrangement effect)
Effect of Energy Corrections S test Uncorrected φ = 42 Bishop s correction φ = 8 more appropriate Rowe s correction φ = 24
φ Measured φ Dilatancy effect Particle orientation effect φ cv : constant volume φ u
Some Other Corrections or Considerations Measured stress Membrane correction Effect of friction between soil specimen and filter paper/porous disk/cap and pedestal B value Flexibility of drainage tubing Use metal tubing
Stress-Strain Properties At low to medium confining pressure 1 Dense sand Loose sand expansion ~4% ~20% ε a ε v contraction ε a
At high confining stress 1 Dense sand Loose sand expansion ~4% ~20% ε a ε v contraction ε a
At larger strains dense sand and loose sand tend to have the same void ratio They tend to have the same strength
Critical Void Ratio A. Casagrande Void ratio, e Loose sand Contraction Drained shear Where shearing starts Critical void ratio line Dense sand Dilate Where shearing starts Confining stress
Void ratio, e Change of pore water pressure Loose sand Where shearing starts Undrained shear Where shearing starts Confining stress Higer shear strength Dense sand Critical void ratio line
Dense Sand R test, V = 0 1 ( ) 1 increase ~4% ~20% ε a u decrease ε a
Loose Sand R test, V = 0 ) ( 1 1 increase ~4% ~20% ε a u decrease ε a
For normally consolidated clay, when ( 1 - ) reaches maximum, its shear strength has not been fully mobilized yet the shear strength is still increasing For overconsolidated clay, although the shear strength has been fully mobilized, ( 1 - ) keeps increasing due to the increase of effective confining stress the difference is not as significant as for N.C. clay
( 1 ) Stress path tangency ( 1 ) max Dense sand, OC clay Loose sand, NC clay
Steady-State Strength The steady-state deformation for any mass of particles is that the state in which the mass is continuously deforming at constant volume, constant normal effective stress, constant shear stress, and constant velocity
Grain Size Diameter of specimens should be at least 6 x (largest particle size) Can use specimens with the same shape of particle distribution curve for testing Also they have to have the same relative density φ increases as average effective stress increases and as max. particle size decreases
100 Percent passing 0 Particle size
Effect of State of Stress φ ~4 Plane strain Triaxial Dense Initial porosity, n i Loose
Direct Shear Test Direct shear is a plane strain test higher friction angle The stress on the failure plane is not uniform, not all parts reached peak stress at failure
ε f, tx is about x ε f, ps φ f, TC is about the same as φ f, TE ( 1 ) Plane strain Dense sand Triaxial ( 1 ) Plane strain Loose sand Triaxial ~1.5% ~4% ε a ε a
Effect of Overconsolidation ( 1 ) Overconsolidated Effective stress failure envelope Normally consolidated Total stress failure envelope Overconsolidated Normally consolidated Due to u < 0 or