1. Lecture 4 Rock Testing part 2 1 1hr 20 mins. Fix errors on 2 slides

2. The strength of rock material Tensile Strength Uniaxial or Unconfined Compression Strength Triaxial (Compression) Test Durability 2

3. Triaxial Strength Most below 500 MPa, a few go up to 1000 MPa. May add pore pressure system and/or temperature system. Temperatures generally operate at below 500 °C 3

4. Hoek Cell specimen Rubber sleeve cell 11  2 =  3 www.ele.org.uk/Mtd/cat11/rm70.pdf 4

5. The most common test procedure consists of holding the confining pressure and temperature constant whilst varying the axial load to maintain a constant displacement rate. After Hoek & Franklin, 1968 5

6. Elliot and Brown cell The Elliot and Brown design includes a balancing piston that consists of a donut-shaped refuge chamber of equal cross- sectional area as the loading piston. After Elliot & Brown, 1985 6

7. Elliot and Brown cell The balancing piston works to: avoid changes in confining pressure due to piston movement; and counterbalance the piston load due to the confining pressure alone, so that the axial loading system need only apply the deviator stress plus the friction on the piston seals. After Elliot & Brown, 1985 7

8. Applying heat Use at elevated temperatures is important for projects such as high level radioactive waste disposal or oil recovery from deep reservoirs. For temperatures up to around 500 °C, two options are available: an external furnace or heating element can be used to heat the entire cell and its contents, or an internal furnace or heating element can be used to heat only the sample and the confining fluid immediately surrounding it. 8

9. Radial displacement After Elliot & Brown, 1988 9

10. Radial strains from volume Radial strains can be inferred from volumetric displacements, but are prone to inaccuracies. The radial strain is given by:  r =½ (   v -   a ) provided that : (i) the sample deforms as a right cylinder, (ii) a homogenous state of stress exists in the sample, (iii) volumetric strain,  v, is determined from measurements of volumetric displacement (measured how?), divided by the original sample volume; and (iv) the axial strain,  a, has been determined. 10

11. True Triaxial Tests σ1σ1 σ3σ3 σ2σ2 11

12. Smart et als, Cell Smart, Somerville & Crawford, 1999 12

13. Muddy SiltstoneLocharbriggs sandstone B.G.D. Smart, J.M. Somerville & B.R. Crawford, 1999. A rock test cell with true triaxial capability.Geotechnical and Geological Engineering 17: 157–176. 13

14. Slake Durability This apparatus provides a means of predicting the durability of rock to weakening and disintegration when subjected to the simulated effects of climatic slaking. The system incorporates a motor drive unit capable of revolving two or four specimen test drums at a speed of 20 revolutions per minute. www.ele.com/euro/pdfs/rockpdf.pdf 14

15. 10 rock lumps, each 40-60 g, total weight of 450-550 g. 10 mins. Rotation; dry; repeat. A= weight of the drum plus sample. C = weight of the drum plus retained portion after second cycle. D = weight of dry, clean drum. slake-durability index (2 nd cycle), 15

16. Slake durability I d %Classification 0 – 25Very low 25 – 50Low 50 – 75Medium 75 – 90High 90 – 95Very high 95 - 100Extremely high J. A. Franklin & R. Chandra, 1972. The Slake-Durability Test. Int. J. Rock Mech. Min. Sci, 9, pp. 325-341. 16

17. Several attempts have been made to link the slake durability index with other properties, including density, porosity, P-wave velocity and uniaxial compressive strength, e.g. Z.A. Moradian, A.H.Ghazvinian, M.Ahmadi & M.Behnia, 2010. Technical Note : Predicting Slake Durability Index of Soft Sandstone Using Indirect Tests. Int.J.Rock Mech. & Min. Sci. 47, 666– 671 17

18. Young’s modulus The stress level for which the tangent or secant modulus is given must be chosen to be appropriate for the particular field problem. However, where no particular reason exists to use a secant modulus to a defined stress level it is quite common practice to quote the tangent modulus at a stress level close to 50% of the UCS. 18

19. Young’s modulus Can either be determine from triaxial testing of rock samples or from UCS test. Strain gauges are cemented to the sample and strain recorded as the sample is loaded and unloaded incrementally. 19

20. Typical values for are: Fine-grained limestone: 0.25 Apilite: 0.2 Oolitic limestone: 0.18 Granite: 0.11 Calcareous shale: 0.02 Biotite schist: 0.01 20

21. Deere-Miller plot Note – both scales are logarithmic After Pells, 1993 Med E/UCS 500:1 200:1 Med E/UCS 21

22. Design of Rock Pillars using UCS. Pillars are usually designed on the basis of the average stress in the pillar, given by the tributary area theory: A t is the area supported by one pillar, A p is the area of the pillar, and  v is the vertical stress at the level of the roof opening 22

23. Example for square & long pillars W per unit length of pillar 23

24. Factor of safety To determine the allowable dimensions of a pillar, or to evaluate its factor of safety, the average pillar stress must be compared with the pillar strength,  p. It should be noted that the pillar strength is not the same as the ucs of the rock,  c, because of the shape and size effects discussed earlier. 24

25. Hustrulid’s equation Hustrulid’s equation offers an estimate of strength in compression for rectangular pillars of square cross- section:  p is the strength of the pillar (assumed to have a height greater than h crit) W and H are the width and height of the pillar respectively  c is the ucs of the rock performed on cylinders with a height (h): diameter ratio of 2 h crit is the minimum height of a cubical specimen of rock such that an increase in the specimen dimension will produce no further reduction in strength. Experiments have shown that h crit  1 m. 25

26. Demonstration Mining plans show a room and pillar style excavation, such that pillars of 1 m x 1 m and height 2.5 m, support an excavation of span 4 m in each direction. Laboratory tests on standard specimens of height 150 mm and diameter 75 mm give a uniaxial compressive strength of 50 MPa. The overburden rock is 20 m deep and has a unit weight of 27 kN m -3. Determine the factor of safety against uniaxial failure of the pillars. 26

27. Solution Determine the vertical stress due to the overburden: MPa Determine the average pillar stress: MPa 5 5 1 4 27

28. Therefore the factor of safety is given by: Determine the pillar strength using Hustrulid’s equation: MPa 28

29. Demonstration In the design of a new airport complex for a major city the designers propose, for environmental reasons, to build an underground car park within the bedrock at a depth of 50m. 100m long rock pillars left in situ during and after construction will support the excavation. The rock mass is homogeneous, with few discontinuities and has a unit weight of 27 kN m -3. The substantial body of data regarding the design of pillars in this rock mass show that their compressive strength may be calculated from the expression below, where A is the plan area of the pillar in square metres. The pillars are required to have a factor of safety of 2.5 against compressive failure. You may assume that there is no surface loading. 29

30. (a) Determine the uniaxial compressive strength of the rock as measured in the laboratory (b) Determine the uniaxial compressive strength of the rock pillars. (c) Using the tributary area theory show that the opening width, W o, is given by (d) Hence plot the relationship between W o and W p and determine the extraction ratio. 30

31. Solution (a) When A -> 0,  c -> 60 MPa (b) When A -> very large,  c -> 15 MPa (c) For a 100 m long pillar of width, W p using the tributary area theory, and gives, 31

32. hence, MPa (d) 32

33. W p m W o m The extraction ratio = W o /W p = 3.44 33

34. Rock Strength Criteria 34

35. Coulomb's shear strength criterion In 1776 Coulomb first proposed that the shear strengths of rock and of soil are made up of two parts - a constant cohesion and a normal stress- dependent frictional component: 35 C = cohesion  = angle of internal friction

36. and, From Mohr’s circle of stress: 36

37. Substitute into the Coulomb eqn. to give the limiting stress condition on any plane defined by , 37 For the critical plane, sin 2  = cos , cos 2  = -sin , and

38. Hence: and, 38 and,

39. The uniaxial tensile strength of rock is generally lower than those predicted from a plot of  1 vs  3. For this reason, a tensile cut-off T 0 is usually applied. 39

40. Problems with Coulomb's criterion It implies that a major shear fracture exists at peak strength. Experimental observations show that this is not always the case. It implies a direction of shear failure that does not always agree with experimental observations. Experimental peak strength envelopes are generally non-linear. They can be considered linear only over limited ranges of  n or  3. 40

41. Griffith crack theory In 1921 Griffith modeled the fracture of brittle materials, such as steel and glass. He proposed that fracture is initiated at tensile stress concentrations at the tips of minute, thin cracks (now referred to as Griffith cracks) distributed throughout an otherwise isotropic, elastic material. Griffith obtained the following criterion for crack extension in plane compression: Where T 0 is the uniaxial tensile strength of the uncracked material (a positive number). 41

42. This criterion can also be expressed in terms of the shear stress, , and the normal stress,  n, acting on the plane containing the major axis of the crack: 42

43. Empirical criteria Bieniawski Hoek & Brown 43

44. Bieniawski criteria or where, and Bieniawski found that, for the range of rock types tested, k  0.75 and c  0.90. The corresponding values of A and B are given in the table. Note that both A and B take relatively narrow ranges for the rock types tested. 44

45. Brady’s use of Bieniawski’s equation This criterion was based on studies of rock fracture development around a bored raise in a pillar in mineralised shale in an Australian mine. Using a boundary element analysis to calculate the elastic stresses induced around the raise as the pillar was progressively mined, Brady found that fracture of the rock could be accurately modelled using Bieniawski’s equation with A = 3.0, k = 0.75 and  c = 90 MPa. This value for  c is approximately half the mean value of 170 MPa measured in laboratory tests. 45

46. Hoek and Brown criterion Where m varies with rock type. 46 For intact rock:

47. 47

48. Demonstration Using the triaxial data given below, develop a peak strength criterion for use in underground excavation design. Experience has shown that the in situ uniaxial compressive strength is 50% of that obtained in the laboratory. 48

49. Solution using Bieniawski’s criterion 49

50. (  1 /  c )-1 (  3 /  c ) 0.75 gradient is 5.67 50

51. Solution using Hoek and Brown’s criterion 51

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53. Demonstration Laboratory tests on specimens of limestone have produced unconfined compressive and tensile strengths of 80 MPa and 10 MPa, respectively. Using the Hoek- Brown and plane Griffith Criteria, estimate the maximum principal stress at failure for two biaxial tests in which  2 = 20 MPa and  2 = 40 MPa. Which of these two criteria would best predict peak strength under these conditions? From question  c = 80 MPa and  t = -10 MPa 53

54. Solution using Hoek and Brown’s criterion substitute  3 =  t and  1 = 0, As the tests are biaxial, we can presume  3 = 0. However the Hoek- Brown Criterion ignores the effect of the intermediate principal stress. For  2 =  3 = 20 MPa,  1 = 20 + (7.88 x 20 x 80 + 1 x80 2 ) 0.5 = 157.8 MPa For  2 =  3 = 40 MPa,  1 = 40 + (7.88 x 40 x 80 + 1 x80 2 ) 0.5 = 217.8 MPa However, it is likely these will both be over-estimates! 54

55. Solution using Griffith criterion Again the Griffith Criterion ignores the effect of the intermediate principal stress, and so neither of these criteria is valid for biaxial compression. So again we assume that we can represent the biaxial condition using the triaxial case. Rearranging the first equation gives: For  2 =  3 = 20 MPa,  1 = 20 + 4 x 10  4(10 x 20 + 10 2 ) 0.5 = 129.3 MPa For  2 =  3 = 40 MPa,  1 = 40 + 4 x 10  4(10 x 40 + 10 2 ) 0.5 = 169.4 MPa 55

56. Comment Determining which of these criteria is the best predictor of peak strength is difficult. Strictly neither of the criteria are valid for biaxial conditions, but the Hoek-Brown criterion is for strength, whereas the Griffith Criterion relates to the onset of fracturing. Since the onset of fracturing occurs before peak strength and usually well before complete failure, perhaps the Hoek-Brown Criterion is the better of the two. However it is more likely that the peak strength will lie somewhere between the two sets of results. 56