Questions Flashcards

1
Q

1 How does the concept of intact rock differ from rock mass?

A
  • Intact rock: Joint-free sample from core drilling used for testing
  • Rock mass: Consists of discontinuities, which include one or more of joints, faults, fissures, or bedding planes which occur in parallel
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2
Q

2 What are the CHILE and DIANE concepts?

A
  • CHILE: Continuous, homogeneous, isotropic, linearly elastic
  • DIANE: Discontinuous, inhomogeneous, anisotropic, not elastic
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3
Q

3 Explain the complete stress-strain curve in uniaxial compression.

A
  • Strain increases until the peak strength, where the material breaks, is reached.
  • Afterwards the material still retains a residual strength.
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4
Q

4 Explain what is meant by Class I and Class II post peak behavior.

A

• Class I behavior:
– “Stable” fracture propagation
– Work must be done on the specimen to effect further reduction in load-bearing ability.
– Some strength after the compressive strength has been exceeded

• Class II behavior:
– Unstable or self-sustaining behavior
– The elastic strain energy stored in the sample when the applied stress equals the compressive strength is sufficient to maintain fracture propagation until the specimen has lost virtually all strength.
– To control the fast failure process the surplus strain energy must be removed from the system.

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5
Q

Why are Brazilian test more often used to estimate the tensile strength than direct tensile test?

A

Direct tensile test difficult to conduct; indirect tensile strength by brazialian test are more comon

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6
Q

6 What is the difference between Mohr-Coulomb and Hoek-Brown?

A
  • Mohr-Coulomb is a classic constitutive model, while the Hoek-Brown is a failure criterion.
  • The Hoek-Brown model can not relate stress and strain in a general way as the Mohr-Coulomb model.(Hoek-Brown: intact rock: mb = mi; s=1; a=0,5)
  • Hoek-Brown assumes that the rock mass is characterized by an elastic-brittle-plastic behaviour while Mohr- Coulomb assumes that it is characterized by an elastic-perfectly-plastic behaviour.
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7
Q

7 Name and explain Mohr-Coulomb parameters with units and describe how these parameters can be defines in laboratory.

A
  • Coulomb postulated that, as the stresses increase, failure would first occur on the plane that satisfies the condition τ = c + σn tan φ for intact rock strength. These parameters can be defines in a triaxial test (shear test) plus stress-strain curve (UCS)
  • c = Cohesion [MPa]
  • σn = Normal stress [Mpa]
  • φ = Friction angle [°]
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8
Q

9 Describe the displacement related to shear strength in a shear test of rock

A

• Graph: shear stress t (y) to displacement (x) > first linear expansion to peak strength and then reduce to residual strength // compare to normal stress I ´n the mohr coulomb strength criterion: with cohesion its peak strength [Graph]

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9
Q

10 Explain JRC and JSC

A
  • JRC: joint roughness coefficient; on a scal 1 (for the smoothest) to 20 (for the roughest surface)
  • JSC: joint wall compressive strength
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10
Q

11 Explain three roughness components of a discontinuity

A
  • Asperity failure component
  • Geometrical component
  • Residual component
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11
Q

12 Difference between “constant normal load conditions” and constant normal stiffness conditions

A
  • Constant normal load conditions (CNL): normal load is constant, normal fracture stiffness equals 0
  • Constant normal stiffness conditions (CNS): normal load is not constant, normal fracture stiffness unequal 0
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12
Q

13 Show using a diagram the relations of shear stress and displacement in a shear test of a rock joint

A

Graph: shear stress t (y) to displacement (x) > first linear expansion to peak strength and then reduce to residual strength // compare to normal stress I ´n the mohr coulomb strength criterion: with cohesion its peak strength [Graph]

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13
Q

14 Why rock mass classification?

A
  • Identify critical parameters
  • Sectioning the rock mass in similar rock mechanics units
  • Numerical data for modelling
  • Joint language between specialists (e.g. geologist and engineers)
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14
Q

15 Describe the parameters used in the Q method.

A
•	Q = RQD∗Jr ∗Jw
Jn∗Ja∗SRF
•	RQD = Rock Quality Designation (drill core quality)
•	Jn = joint set number
•	Jr = joint roughness number
  • Ja = joint alteration number
  • Jw = joint water reduction factor
  • SRF = stress reduction factor
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15
Q

16 RQD, how is it defined? Where do we apply RQD?

A

• Rock Quality Designation
Length of core pieces > 10 cm length Total length of core run
• Used for evaluation of drill core qualities
• Is part of the Q method

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16
Q

17 SRF, how is it defined and where to use it?

A
  • Stress Reduction Factor
  • Weakness zones intersecting excavation, which may cause loosening of rock mass when tunnel is excavated
  • Is used in the Q method
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17
Q

18 Explain how Q-value can be used to evaluate the need of rock support.

A
  • With the help of Equivalent Dimension (De) and Excavation Support Ratio (ESR) the required rock support can be determined in a diagram.
  • It also allows to determine the recommended bolt spacing and length.
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18
Q

19 Describe the six parameters in RMR method.

A
  • UCS (Uniaxial compressive strength of intact rock)
  • Rock Quality Designation (RQD)
  • Spacing of discontinuities
  • Condition of discontinuities
  • Groundwater conditions
  • Orientation of discontinuities
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19
Q

20 How can UCS be estimated without laboratory testing?

A

• Field estimates with geological hammer, pocket knife or thumbnail (Grades: R0-R6)

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20
Q

21 Explain how the stand-up time and span are related.

A
  • They are used for the “Meaning of Rock Classes” in RMR
  • If a small span only stays open for a short amount of time, the cohesion of the rock mass is low and vice versa
  • If the stand-up time is long, there is less support required
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21
Q

22 Describe the two main factors the GSI system is focused on.

A
  • Rock structure

* Block surface conditions

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22
Q

23 Fractures and their properties are important in varying rock mass classifications systems. List out the main properties used in Q, RMR, GSI systems.

A
  • Joint sets/number/roughness/alteration/water reduction
  • Stress reduction
  • Spacing and condition of discontinuities
  • Groundwater conditions
  • Orientation of discontinuities
  • Uniaxial compressive strength of intact rock
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23
Q

24 Describe how rock mass classification is utilized to define the rock mass strength.

A
  • The strength and stiffness of the intact rock is not the same as that of the rock mass.
  • Scaling of the intact rock data needed, commonly by empirical relationships based on classification systems.
  • Strength and deformation is controlled by discontinuities
  • Presence of discontinuities is scale dependent
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24
Q

25 Explain the parameters in H&B criterion and describe how they can be determined.

A

• Parameters
– σ1r ; σ3r = Max. and min. effective principal stress at failure
– mb = H&B constant m for rock mass
– s and a = Constants for rock mass
– σci = UCS (for intact rock)
• Determination
– UCS = Uniaxial compressive strength
– mi = H&B constant for intact rock for mb
– GSI = Geological Strength Index for s, a

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25
Q

26 Applicability of the H&B criterion at different scales?

A

• Hoek-Brown is only based on individual samples, therefore it can not easily be scaled.

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26
Q

27 How is GSI utilized in H&B criterion? How can RMR and Q systems be utilized?

A

• Utilization of GSI
– GSI is used for the determination of s and a
– For RMR89r > 23 →− GSI = RMR89 − 5
– For RMR89r < 23; use Q’ classification value
• Utilization of RMR and Q systems
– For RMR76r > 18 →− GSI = RMR76r
– For RMR76r < 18; use Q’ classification value

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27
Q

28 Explain Young’s modulus, shear modulus and Poisson’s ratio.

A
  • Young’s modulus is a mechanical property that measures the stiffness of a solid material. It defines the relationship between stress (force per unit area) and strain (proportional deformation) in a material in the linear elasticity regime of a uniaxial deformation.
  • The shear modulus is defined as the ratio of shear stress to shear strain.

• Poisson’s ratio is a measure of the Poisson effect, the phenomenon in which a material tends to expand in directions perpendicular to the direction of compression.

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28
Q

29 Explain elasticity and plasticity.

A
  • Elasticity is defined as the property which enables a material to get back to (or recover) its original shape, after the removal of applied force.
  • Plasticity is defined as the property which enables a material to be deformed continuously and permanently without rupture during the application of force.
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29
Q

30 Explain what is “ In situ stress”” and how does it differ from “induced stress”?

A
  • In situ = pre-mining state of stress. The true state of stress on the earth´s crust is always a superposition of several stress components; essential to know for excavation and rock support design (magnitude and orientation); tectonic and gravitation are the two most important contributions to in situ stress
  • Difference to inducted stress is that is naturally exists and induced is made by mining for example
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30
Q

31 Kirsch equations: what can you calculate with them? Describe the parameters.

A

• With the kirsch equation you can calculate the stresses and displacements around a circular opening in elastic medium Sigma v = vertical stress, r and a are radius of boundries; k is stress ratio; phi is shear angle

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31
Q

32 For a future metro station you need to know approximate in situ state of rock mass at depth of 50 meter. How would you estimate the stress magnitude and orientation without doing stress measurements?

A
  • Calculation of the weight of the overlying strata: σv = γ(weight of overburden) ∗ z(50m)
  • Calculation of stresses of tectonic origin: σh = k ∗ σv = k ∗ γ ∗ z or Kirsch equation
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32
Q

33 Explain methods to evaluate and to measure the stress in the earth´s crust (both magnitude and direction)

A
  • Calculation of the weight of the overlying strata: σv = γ(weight of overburden) ∗ z(50m)
  • Calculation of stresses of tectonic origin: σh = k ∗ σv = k ∗ γ ∗ z
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33
Q

34 Explain what “In situ stress” is. What are the relevant components and how are they estimated and measured?

A

• In situ stress is the weight of the overlying strata combined with the stresses of tectonic origin (Pre-mining state of stress)
• Relevant components
– γ = Unit weight of overlying rock
– z = Depth below surface
– Eh = Average deformation modulus, based on stress measures
• Estimation and measurement
– Not direct stress measurements, only the effect can be measured
– Overcoring – Based on determination of strains in the wall of the borehole (stress relief)
– Hydraulic fracturing - Formation of a tensile crack in deep boreholes
– Borehole breakout – Analyses of stress failure of boreholes

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34
Q

35 Describe the rock conditions that increase the risk for structurally controlled failure. What are the conditions for stress induced fail- ure?

A

• Structurally controlled failure
– Low stress regime, common for low depth underground structures
– Relatively small loads, only key-blocks to be supported
• Stress induced failure
– Typical at great depth
– Induced stress/rock strength ratio important
– Time dependency of failure

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35
Q

36 Describe the types of failure that may occur for intact rock in compressional and tensional loading.

A
  • Axial splitting
  • Shearing
  • Crushing
  • Tensile failure
  • Axial splitting (Point load)
36
Q

37 What is spalling?

A
  • Flakes of a material breaking off a larger solid body
  • Spalling initiates at points of maximum compressive stress concentration which occur at right angles to the major principal stress direction
37
Q

38 Types of failure which occur in different rock masses under low and high in situ stress level.

A

• Low stress
– Massive rock, no fractures = Linear elastic response with little or no rock failure
– Massive rock, few fractures (discontinuities) = Blocks or wedges released by intersecting discontinuities fall or slide due to gravity loading

– Heavy jointed rock = The opening surface fails as a result of unravelling of small interlocking blocks and wedges. Failure can propagate a long way into the rock mass if it is not controlled
• High stress
– Massive rock, no fractures = Spalling, slabbing and crushing initiates at high stress concentration points on the boundary and propagates into the surrounding rock mass
– Massive rock, few fractures (discontinuities) = Failure occurs as a result of sliding on discontinuity surfaces and also by crushing and splitting of rock blocks
– Heavy jointed rock = The rock mass surrounding the opening fails by sliding on discontinuities and crushing of rock pieces. Floor heave and sidewall closure are typical results of this type of failure.

38
Q

39 How are safety and economics related to rock mechanics design?

A
  • The safety against rock failure has to be on right level: Too risky vs. too expensive constructions
  • In mining it is acceptable to allow some failures of the mine stopes: To get more money by decreasing support constructions and/or increasing stope dimensions
  • Often empirical safety factors: Excavation Support Ratio (ESR) used for calculation of Equivalent Dimension (De)
39
Q

40 What do you know about the level of accuracy for rock mechanical design based on phases of a construction/mining project?

A
  • Accuracy of RM design depends on the stage and size of the project
  • Appropriate level depends upon rock quality and size of the excavation work.
  • Conceptional valuation = Empirical design methods
  • Mine design stage = Analytical and empirical design methods
  • Excavation of rock = Numerical methods
  • Later years of mining = Fine tuning of support with analytical models and empirical methods
40
Q

41 Give examples of empirical, analytical and numerical methods for rock mechanics analyses.

A

• Empirical
– NATM
– RQD
– RMR

– Q
– GSI
• Analytical
– Based on calculations of strength and effective forces
– Pen and paper/computers
– Block failure and stress failure
• Numerical
– Can be based on empirical or analytical approach
– Solution is approximated by discretizating the problem in elements

41
Q

42 Describe the Room and Pillar method. Tell about the ore geometry, rock mechanic properties of the ore, roof and floor.

A
  • Room-and-Pillar methods are commonly applied to relatively flat orebodies and employ natural support (pillars).
  • The orebody is excavated as completely as possible leaving ore/waste as pillars to support the hanging wall (roof).
  • Dimensions of the rooms and pillars depend upon factors such as the stability of the roof, stability of the ore, thickness of the deposit and rock stresses.
  • Competent ore and host rock
  • Risk for subsidence and caving can be limited with backfill
42
Q

43 Describe the Sublevel Stoping method. Tell about the ore geometry, rock mechanic properties of the ore, hanging wall and the foot wall.

A
  • Used for steeply dipping, regular orebodies
  • Requires stable rock in both hanging wall and footwall
  • Sublevels are excavated to allow access for drilling equipment
  • Longholes are drilled along the sublevels and blasted in slices
  • Afterwards stopes can be left open or backfilled
43
Q

44 Elaborate on displacements and stored strain energy in different mining methods.

A
  • In unsupported methods the strain energy is stored in pillars and the surrounding host rock; usually low to no displacements
  • In artificially supported methods the strain energy is stored in in the blasted ore or in the used backfill; usually low to no displacements
  • In caving methods the strain energy causes the rock mass to collapse, the energy is afterwards stored in the broken rock; high displacements occur due to subsidence
44
Q

45 Describe the procedure to select a suitable mining method.

A

• Examination of technical questions like ore geometry, RM conditions (Technical feasibility)
– Qualitative methods
– Quantitative methods
• Economic evaluation
• Selection of two or three suitable methods and evaluatio

45
Q

46 What are the main geologic characteristics of a deposit that affect the mining method selection?

A

• Ore size, thickness, orientation, strength, grade distribution, depth

46
Q

47 Describe the parameters in Nicholas method

A
  • First classify the ore geometry and the grade distribution
  • Second the rock mechanics characteristics of the ore zone, haning wakkm and footwall
  • A numerical ranking by adding up the values of each mining method
  • The values of the tables represent the suitability of a given characteristic for a particular mining method
47
Q

49 Explain why it is important to apply the tunnel support close to the tunnel face.

A

• At the tunnel face; radial displacement is around 1/3 of its final value

48
Q

50 Two important mechanical features the support must fulfill to sufficiently support the tunnel?

A
  • Uniform internal support pressure should be higher than (hydrostatic stress) critifcal support pressure
  • Loosing of rock mass caused by e.g. water flow and dissolution of fracture fillings, slidings of blocks and stress relaxation needs
49
Q

51 What is the aim of rock reinforcement and rock support?

A

trengthen the rock mass to carry the load which it’s exposed to
• Attach loose rock blocks to the solid rock, or to keep (layered) rock slabs together
• Build up yielding structure: Broken rock and the reinforcement is stable, takes sufficient load and keeps rock deformations within accepted limits.

50
Q

52 What is the difference between rock reinforcement and support?

A

• Reinforcement
– Reinforcement induces stabilizing forces within the rock mass
– E.g. a bolted rock face
• Support
– Structural elements apply external stabilizing forces directly
– E.g. a ring of pre-cast concrete segments

51
Q

53 Why does it matter when and in what stage of the tunnel excava- tion the rock support is installed?

A
  • Rock support should be installed as soon as the load of the rock mass lowers to the maximum the support can handle
  • Shifting the load into the rock mass
  • Prevention of progressive failure
  • Installing the support too late does not have an effect, since the rock mass is already caved at that point
52
Q

54 Why is rock reinforcement important for the efficiency of the tun- neling process?

A
  • Excavation changes the virgin stress state.
  • Immediate reinforcement restricts rock movement and loosening of the material.
  • It helps to maintain the inherent strength of the rock mass.
  • Safety and continuity of production: Failures, rock dilution, ore wastes, i.e. improved selectivity of extraction
53
Q

55 Describe varying types of methods for design of rock support.

A
  • Empirical methods: Tables and diagrams
  • Analytical methods
  • Numerical methods (wedgege program)
  • Observational methods
54
Q

56 Explain the parameters in the Stability Graph.

A

• N r = Qr ∗ A ∗ B ∗ C
• N’= Stability number, reflects the ability of the rock mass to stand up under a given stress condition
• Q’= Modified Tunneling Quality Index Q (Mapping/Cores), except joint water reduction factor and the stress reduction factor
• A= Rock stress factor, reflects the stresses acting on the free surfaces of open stopes
• B= Joint orientation adjustment factor, influence of joints on the stability of the stope faces
• C= Gravity adjustment factor, adjustment for the effects of gravity (Failure of the roof by gravity induced
falls ad failure of the stope walls by slabbing or sliding

55
Q

57 Describe the major components defining the total inclination of the open pit wall

A
  • BENCH CONFIGURATION: face angle, the bench height, bench weight
  • INTER-RAMP SLOPE: number of benches
  • OVERALL SLOPE ANGLE (incl. haul roads, working levels etc.)
56
Q

58 Tell about factors that influence the slope stability.

A
  • Higher pit slope angles lead to decreased stability, lost production, dilution and clean up costs due to slope stability problems
  • Variability in geologic structures, variation in need for stability
57
Q

59 Describe (plane, wedge, toppling etc.) failure

A
  • Plane: in rock containing persistent joints dipping out of the slope face, and striking parallel to the face
  • Wedge: Occurs as sliding of a mass of rock on two intersecting planes, generally discontinuity planes.
  • Toppling: in strong rock containing discontinuities dipping steeply into the face
  • Circular: In rock fill, very weak rock or closely fractured rock with randomly oriented discontinuities
58
Q

60 Tell about geological data that is required to identify potential failure areas.

A
  • Joint orientation
  • In situ stress
  • Shear strength of discontinuities
  • Hydrology
59
Q

61 How to decrease the risk of instability of the slope caused by groundwater?

A

To decrease the risk of instability of the slope caused by groundwater drainage and pumping system are used.

60
Q

62 Rockfalls from high rock slopes and related risk horizontal roll of boulders. How can the risk be mitigated (reduced) by design.

A

• Ditch and/or berm on the edge of the bench (berms, rockshed, ditch, fill, fence), rock trap, free hanging mesh suspended from above, warning signs

61
Q

63 How can slope instability be detected by monitoring? Tell about monitoring methods.

A
  • Identification of potential failure areas using geological data
  • Monitoring of potentially unstable areas using tension crack mapping and shear movement
62
Q

64 Assess and compare the stability characteristics of these rock slopes

A
  • A) discontinuos with a displacement faced a potential risk -> step path
  • B) in rock containing discontinuities dipping steeply into the face -> might occur toppling failure (flexural)
  • C) containing persistent joints dipping out of the slope face and striking parallel to the face; sliding rock -> plane failure
  • D) widley spaced orthogonal joints -> block toppling of columns of the rock from the west
63
Q

65 Explain the potential reinforcement methods for a traffic tunnel with large span and weak rock mass.

A
  • Concrete (shotcrete, shotcrete with fibres, invertlining, shotcrete or cast concrete)
  • Rock bolting
  • Invert struts
  • Steel sets
  • Fiberglass bolts
  • Forepoling (prebolting)
64
Q

66 How are rock bolts grouped by their function?

A
•	Bolts can be classified based on
–	Type of anchoring
–	Active or passive bolts
–	Short or long term support
–	Prestressed (pretensioned) or unstressed
65
Q

67 How are rock bolts anchored?

A
  • Grouted bolts
  • Point anchored bolts
  • Friction bolts
66
Q

68 Explain how mechanically anchored bolts are installed, their pros and cons.

A

• Mmechanically anchored bolts are worked best in semi hard rock and hard rock (pro)
• Sensitive to a detonation wave -> loosening of anchor (con)
• Usually not classified as stable reinforcement(con)
• If also concrete grouted -> stable reinforcement (pro)
– Worked: with the nut the bearing plate get stucked to the rock and hold the face- with the tension rod the cone opens the the shell and held it to the rock in the hole

67
Q

69 Tell about the behaviour of different types of rock bolts in rock bolt loading tests.

A
  • Resin grouted fibreglass rods show the lowest deformatio on high load
  • Followed by cement grouted steel rebar and resin grouted steel rebar
  • Up to 150 mm deformation on EXL Swellex dowel and Type SS 39 Split Set stabilizer
68
Q

70 Describe components of the shotcrete (sprayed concrete). Also tell about the additives and their function.

A

• Components
– Binding agent (cement, silica, fly ash)
– Aggregate (sieved natural rock material)
– Water

• Additives
– Accelerators
∗ Fasten the hardening of concrete
∗ Makes it possible to increase the thickness of “in one step sprayed” concrete layer
∗ Are added in the nozzle
– Plasticizers
∗ Improve the strength
∗ The amount of water can be reduced significantly
∗ Improved workability
– Air entrainments - pore-forming agents
∗ To reduce damage from freeze-thaw
∗ Distribute tiny air bubbles through the concrete which provide space for frozen water to expand
∗ In fresh concrete the air bubbles act as filler reducing the need of water and improving the workability and durability

69
Q

71 What is the effect of the accelerator on the strength of concrete?

A
  • Increased strength in the first 10 hours
  • Lower final strength
  • Are added in the nozzle
70
Q

72 What is the effect of steel fibers and wire mesh on concrete’s de- formation properties? Explain the load–deformation curve.

A

• Steel fibers
– To improve toughness and deformation properties of the concrete structure
– Replaced meshing in many applications (safety/cost reduction)
• Wire mesh
– Mesh helps concrete stick to rock where shotcrete adheres poorly

– Mesh prevents falling of small rocks between bolts
– Instead of shotcreting, when shotcrete is impractical
– Support for heavy boulders or collapsing rock mass
– Shotcrete + meshing especially in difficult rock conditions (e.g. spalling)
• Load-Deformation curve
– Shows which resistance the support can take in relation to the displacement of the underlying rock

71
Q

73 Explain shotcrete failure modes. What is the support mechanism of shotcrete mainly based on?

A
•	Failure modes
–	Adhesive failure
–	Direct shear failure
–	Punching shear failure
–	Flexural failure
•	Support mechanisms
–	Adhesion between rock and shotcrete
–	Shear strength of shotcrete
–	Bending strength of shotcrete
72
Q

74 Describe three main purposes of backfilling.

A
  • To prevent caving, surface subsidence, rock bursting etc.
  • Acts as a structural component in pillar recovery (eg. sublevel stoping)
  • Environmental aspects to avoid waste areas on the surface.
73
Q

75 Tell about how the backfill works as a support medium.

A

• By imposing a constraint on the key blocks in a stope boundary, backfill prevents spatially progressive disintegration of the near-field rock mass (low stress settings).
– Kinematic constraint on surface blocks in destressed rock
• Body displacements of stope wall rock mobilise the passive resistance of the ftll.
– Support forces mobilized locally in fractured and joined rock
• As a global support element in the mine structure, ie. mining-induced displacements at the fill-rock interface
– Global support due to compression of the fill mass by wall closure
→− reflected as reductions in the stress state throughout the mine near-field domain.

74
Q

76 Tell about the differences between the simple wedge failure model and the tension wedge failure model.

A

• Simple wedge failure model
– assumes failure to initiate from the toe of a fill block
– Wedge block may fail along the plane of lowest shearing resistance
• Tension wedge failure model
– Tensile cracks often seen at a distance equivalent to one third the height of failure
– assumes a vertical tension crack to intersect the shear failure plane

75
Q

77 Three types of backfill: Tell about classification of backfill based on water content.

A
  • Hydraulic fill: Water-rich, in most cases fine-grained mixtures, gravity flow, pumping
  • Paste backfill: Mixtures with less water (optimum water content for hydration), sufficiently fine grained material, suitable for pumping
  • Rock fill: Mixtures with low water content and less fine-grained material, transportation with trucks and LHD
76
Q

78 Explain the main purposes of numerical modelling in rock mechan- ics?

A
  • To visualize the problem and the input data
  • To analyze input data (sensitivity analyses)
  • To automatize routine calculations
  • To solve a complicated, laborious problem
  • To predict rock mass behaviour
  • To explain phenomena
77
Q

79 Prediction of rock mass behaviour is an important task where nu- merical models are used. Tell about typical applications in mining and rock engineering.

A
  • Pillar, roof, excavation stability analyses (With and in without rock support)
  • Mechanical stress-strain-analyses (deformation of the tunnel, ground heave etc.)
78
Q

80 What is THM modelling?

A
  • THM = Thermo-hydro-mechanical modelling

* Couples rock fracturing, fluid flow and temperature change

79
Q

81 What is the difference between continuum or discontinuum models? Give examples.

A

• Continuum approach
– No fractures or large number of fractures
→− Homogenization assumption can be made.
– Examples
∗ Finite Difference Method
∗ Finite Element Method
∗ Boundary Element Method
• Discontinuum approach
– The fracture response/fracture propagation to loading is of interest (e.g. DEM, DDM)
– Large-scale displacements of individual blocks are of interest (e.g. DEM)
– Flow/transport in fracture systems is of interest (DFN)
– Examples
∗ Discrete Element Method
∗ Discrete Fracture Network
∗ Displacement Discontinuity Method

80
Q

82 What are the reasons why monitoring is needed?

A
  • Monitoring the natural state before the project
  • Ensures safety during construction and operation
  • To check the validitity of assumptions or models
  • To control the implementation of ground treatment or support actions
  • Key to observational method
81
Q

83 What kind of monitoring techniques exist?

A
  • Distance measurement
  • GPS measurement
  • Long-range radar
  • Temperature evolution
  • Visual observation
  • Microseismics
82
Q

84 Can every phenomenon be monitored directly?

A

• No, e.g. stress can’t be measured directly

83
Q

85 In which kind of projects could seismic monitoring be used?

A
  • In mining projects to detect potential of rockburst

* To locate areas which cause seismic events in mines

84
Q

86 How would you monitor an unstable rock slope?

A
  • Long-range radar
  • Microseismics
  • Tension crack mapping
  • Shear movement
85
Q

87 Can D&B excavation be monitored? How about TBM and how?

A

• D&B
– Yes, e.g. microseismics or seismic network
• TBM
– Yes, e.g. extensometers