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ROCK STRENGTH (Text Ch. 3) Lecture 9 – Introduction to Rock Strength David Hart University of Wisconsin ecow.engr.wisc.edu/cgi- bin/get/gle/474/hartdavid/notes/lectu re9strengthintro.ppt - ROCK STRENGTH • Shear fracture is the dominant mode of failure for rocks under all but the lowest confining stress. Extension Compression Paterson, Experimental Rock Deformation – The Brittle Field Example from Goodman, Intro to Rock Mechanics ROCK STRENGTH • The peak stress is the strength of the rock. – It may fail catastrophically if the load frame is “soft”. Example below is for a “stiff” frame. • The compressive strength of rock is a function of the confining pressure. • As the confining pressure increases so does the strength. Goodman, Intro to Rock Mechanics ROCK STRENGTH • The variation of peak stress σ1, peak (at which failure occurs) with the confining pressure (for which σ2 = σ3 ) is referred to as the rock CRITERION OF FAILURE. • The simplest and the best known failure criterion of failure is the MOHR-COULOMB (M-C) criterion: the linear approximation of the variation of peak stress σ1, peak with the confining pressure. MOHR-COULOMB Criterion of Failure • It has been established that rock fails in σ1,p compression by shearing along a ‘failure’ plane oriented at an angle θ with respect to σ1 that is specific for a particular rock. Rock • The M-C linear strength criterion cylinder σ =σ implies that θ stays the same regardless 2 3 of the confining pressure applied. τ 2θ 2θ 2θ σ 0 σ2=σ3 1,p σ MOHR-COULOMB Criterion τ = Si + σ tanφ τ cr φ A D 2θ Sι φ 90o C 0 σ2=σ3 Co B σ1,p σ MOHR-COULOMB Criterion in terms of shear and normal stress on the plane of failure τ cr = Si + σ tanφ where Iτlcr is the shear strength, Si (cohesion) is the intercept with the τ axis of the linear envelope, and φ ('angle of friction') is the slope angle of the linear envelope of failure. The relationship between the angle θ (the angle between the normal to the plane of failure -point A- and the direction of the max. principal stress) and φ is π φ =2θ + φ or θ =o 45 + 2 2 MOHR-COULOMB Criterion can also be given The Mohr-Coulomb criterion do this we oin terms of the principal stresses. T recall the following relationships: σ −σ σ +σ σ −σ τ = 1 2sin 2θ σ =1 2+ 1cos 2 2θ 2 2 2 Plug these relationships into the Mohr-Coulomb criterion: τ crS= i +σ tanφ and you get: σ1, p−σ 2 1⎛σ, p+σ 2σ, 1 p−σ 2 ⎞ sin2θ=i S +⎜ + cosθ2 ⎟ tan φ 2 ⎝ 2 2 ⎠ MOHR-COULOMB Criterion in Terms of Principal Stresses By separating all the terms containing σ1 from the rest of the expression, and recalling the above relationship between θ and φ (and also performing some trigonometric manipulations), we finally obtain the Mohr- Coulomb criterion in terms of the principal stresses only: ⎛ φ ⎞ 2⎛ φ ⎞ σ,1 p2=S i tan⎜ ° 45 + ⎟ +σ 2 tan ⎜45° + ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ or: 2⎛ φ ⎞ 1σ,= pC 0 +σ 2 tan⎜45° + ⎟ ⎝ 2 ⎠ MOHR-COULOMB Criterion in Terms of Principal Stresses ⎛ φ ⎞ 2⎛ φ ⎞ σ,1 p2=S i tan⎜ ° 45 + ⎟ +σ 2 tan ⎜45° + ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ or: 2⎛ φ ⎞ σ 1σ,= pC 0 +σ 2 tan⎜45° + ⎟ 1 ⎝ 2 2⎛ o φ⎞ ⎠ tan⎜ 45+ ⎟ ⎝ 2⎠ Remember: the angle 1 45+φ/2 is equal to θ, where θ is the angle ⎛ o φ⎞ C 2o = Si tan⎜ + ⎟ 45 between the direction ⎝ 2⎠ 0 of σ1 and the normal σ2= σ 3 to the plane of failure. Example The criterion of failure of a limestone is given by (in MPa): τ = 10 + 0.8σ cr Represent the same criterion in terms of principal stresses. Solution From the given equation we deduce that: Si = 10 MPa, and tan φ = 0.8, or φ = 38.7 °. The criterion of failure in terms of principal stresses is: ⎛ φ ⎞ 2⎛ φ ⎞ σ,1 p2=S i tan⎜ ° 45 + ⎟ +σ 2 tan ⎜45° + ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ By plugging the given values in the above equation we obtain (in MPa): ⎛ 38. ° 7⎞ 2⎛ 38. ° 7⎞ σ=,12 p × 10tan⎜ 45 ° + ⎟ +σ 2 tan ⎜45° + ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ EXAMPLES OF MOHR- τ COULOMB CRITERION: 1. Hydrostatic state of stress: O P σ σ1 = σ2 = σ3 = P When does rock fail under this condition? 2. Uniaxial compression: τ σ1 ≠ 0 ; σ2 = σ3 = 0 φ D Failure occurs when σ1 A attains its peak value C . o S Failure plane (point A) i φ normal is at θ=(45 + φ/2) to =C σ1 direction. O B σ1,p o σ MOHR-COULOMB Criterion- Tensile Stresses Question: What if the normal stress on the plane of failure is tensile? This implies that there is no contact between the two created surfaces and hence there is no frictional resistance. Thus, the Mohr-Coulomb (M-C) criterion loses its validity. Note, however, that σ2 can be tensile as long as the normal stress σ remains compressive: Mohr- Coulomb criterion will still be valid. Remember that σ2=σ3 can never be less than the tensile strength of the rock T, since T implies tensile failure. MOHR-COULOMB Criterion- Tensile Stresses Case I: σ2 and σ are both tensile, so M-C criterion does not apply. Specimen fails in tension when σ2 = T. Case II: σ2 is tensile, but σ is zero or larger, so M-C criterion DOES apply! Case III: All stresses are compressive so M-C does apply. NOTE: Typically T = 0.1Co to 0.05Co MOHR-COULOMB CRITERION IN TERMS OF THE PRINCIPAL STRESS RATIO K Engineers/practitioners like to to work with principal stress ratio K = σ2 / σ1 (typically σ2 represents the horizontal stress σh, and σ1 represents the vertical stress σv ). It is thus more convenient on occasion to express the Mohr-C criterion in terms of K. MOHR-COULOMB CRITERION IN TERMS OF THE PRINCIPAL STRESS RATIO K Given: 2 φ σ1,p = Co + σ2 tan (45° + ) 2 We divide both sides by σ1,p: 2 φ 1 = Co + σ2 tan (45° + ) σ1,p σ1,p 2 Using K and rearranging terms, we obtain: Co 2 φ Co = 1 - K tan (45° + ) σ1,p = σ1,p 2 φ 1 - K tan2 (45° + ) 2 MOHR-COULOMB CRITERION IN TERMS OF THE PRINCIPAL STRESS RATIO K With the M-C criterion written this way we can determine a limit for the ratio K at failure. It is 2 φ K < cot (45° + ) 2 This condition makes the criterion feasible. Otherwise the failure stress is infinite or negative, which is not physically possible. MOHR-COULOMB CRITERION IN TERMS OF THE PRINCIPAL STRESS RATIO K This condition for K is a useful tool in determining stability. For example if the angle φ = 40° then K < 0.22 Or σ2 < 0.22 σ1 . CLASS: What if the angle φ = 0°? Determining the Critical State of Stress at a Point in a Rock Formation When the M-C Criterion of Failure and One of the Principal Stresses Are Known In this example the φ an strength material τ σt i + S properties of the rock = τ cr Si and φ are given, Given and we also know the σ at failure. σ 1,p σ What is then the 1,p complete state of stress at the point in question? φ Let’s imagine τ first that we have solved the A problem, i.e. we R have found the C B R σ σ2 σ1,p CRITICAL Mohr 90 + φ circle denoting the state of 180 − (90 + φ) = 45 − φ / 2 stress at failure. 2 We notice that in the triangle ABC two of the sides are equal to the radius R. Also, recall φ the value of the angle τ ABC. Hence the angle A ACB is 45° - φ/2. R Thus, in order to C B R σ σ2 σ1,p construct the critical 90 + φ Mohr circle we fist plot 180 − (90 + φ) the M-C criterion and = 45 − φ / 2 2 the point representing σ1,p. Then, we construct the angle ACB and φ draw the line CA, τ where A intersects A the M-C criterion. Now we construct IIa II I D C σ an identical angle B σ2 σ1,p CAB and draw AB. (90 +φ) Point B on the σ 180 − (90 + φ) = 45 − φ / 2 axis is the center 2 of the critical Mohr circle. An alternate way is to construct at point A a 90° angle and draw line AD. Point D has to be equal to σ2. OTHER EMPIRICAL MOHR ENVELOPES Goodman, Intro to Rock Mechanics A straight line doesn’t always fit the data well Mohr-Coulomb is an approximation OTHER EMPIRICAL MOHR ENVELOPES Several examples of other expressions defining the Mohr envelope in rocks for which the Coulomb linear relationship does not appear to be representative, can be found in the literature. These are all EMPIRICAL relationships. The Hoek and Brown criterion is particularly popular now: σ σ σ 1,p = 2 + m 2 + s C o Co Co where m and s are constants of the material and are determined empirically. SIZE EFFECT ON COMPRESSIVE STRENGTH It has been found through experimentation that rock strength is affected by the SIZE of the specimen tested. This can be explained simply as being the result of the many flaws found in rock in the form of fissures, pores, cleavage, bedding planes, foliation, etc. The larger the size of the specimen tested the higher the probability that a weak flaw will be available, capable of initiating failure.
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