9_chapter 9 shear strenght of discontinuities in rock

24
CHAPTER NINE Shear strength of discontinuities in rock Overview Most failure modes in rock slopes involve sliding and/or toppling. Sliding generally occurs along discontinuities and hence the study of shear strength of discontinuities is critical to any rock slope design. A simple way to estimate the shear strength of a single discontinuity is to take a sample of rock containing the discontinuity and then to incline the discontinuity from the horizontal until the upper half slides (Fig. 9.1). If the weight of the upper block is W and the gross area of contact is A, at an angle of inclination the effective normal stress 0 n and the shear stress down the line of maximum dip are respectively given by 0 n W cos A ÿ u and W sin A 9:1 where u is the water pressure within the discontinuity, generally zero in this particular test. If the angle of inclination at the point of sliding is f , and the associated effective normal and shear stresses are 0 n and f , the ratio of shear stress to effective normal stress at the point of slip is, applying equations (9.1), given by f 0 n sin f cos f tan f This simple result forms the basis of the Mohr±Coulomb failure criterion used in soil mechanics: f 0 n tan 9:2 The friction angle of a discontinuity can be de®ned using the following terms which refer to the displacements obtained during shear movements (Krahn and Morgenstern, 1979, 1980; Barton, 1980): . Peak friction angle p is evaluated on natural discontinuities, in correspondence with the maximum shear strength determined by roughness. 262

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Page 4: 9_Chapter 9 Shear Strenght of Discontinuities in Rock

Measured normal stress (vertical) = "n

Measured shearstress (horizontal) = #

dv measured verticaldisplacement

dh measured horizontaldisplacement

i

As shearing takes place, top block dilates with workbeing done against normal load through dv

To calculate stresses relative to actual plane of shearing

Then, i = arc Tandvdh

#i = (# cosi – " sin i) cosi"i = (" cosi + # sin i) cosi

#i = (# cosi + " sin i) cosi"i = (" cosi – # sin i) cosi

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JRC = 0 – 2

JRC = 2 – 4

JRC = 4 – 6

JRC = 6 – 8

JRC = 8 – 10

JRC = 10 – 12

JRC = 12 – 14

JRC = 14 – 16

JRC = 16 – 18

JRC = 18 – 20

0 5 10 cm

Scale

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Page 8: 9_Chapter 9 Shear Strenght of Discontinuities in Rock

Confined block

Unconfined block

Circular opening

Rock slope

(a)

(b)

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Transition curve

Effective normal stress "&n

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Page 15: 9_Chapter 9 Shear Strenght of Discontinuities in Rock

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1B2

Page 16: 9_Chapter 9 Shear Strenght of Discontinuities in Rock

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1BB

Page 17: 9_Chapter 9 Shear Strenght of Discontinuities in Rock

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1BC

Page 18: 9_Chapter 9 Shear Strenght of Discontinuities in Rock

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Jack for shear load

Specimen cast in concrete or plaster

Upper shear box

Lowershear box

Shear displacementgauge

Shear surface

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1B:

Page 19: 9_Chapter 9 Shear Strenght of Discontinuities in Rock

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Hanging weights

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Page 20: 9_Chapter 9 Shear Strenght of Discontinuities in Rock

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1CE

Page 21: 9_Chapter 9 Shear Strenght of Discontinuities in Rock

1500

1000

500

0

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1

2

3

5 10 15

She

ar s

tres

s: k

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" = 500 kPa

" = 750 kPa

" = 1000 kPa

" = 1500 kPa

Horizontal displacement: mm

Vert

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dis

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Effective normal strength "&n: kPa

c = 6·5 kPa

Equation (9.9)

$ = 42˚

"d = 30·2 MPa$b = 32˚JRC = 4·83

80 kPa

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