Rock Mechanics Fundamentals Rock Mechanics – Understanding the mechanical behavior of the rock mass Ground Control – Controlling the ground around the mining/civil excavation Rock Mechanics / Ground Control basically deals with two fundamental topics:

Material Properties: Strength, Stiffness/Modulus, Poisson’s Ratio, Weight/Density, etc.

Stress: Stress Field, Stress Concentrations, etc. Specific Gravity (sg) – The ratio of the weight of a substance to the weight of an equal volume of a standard substance (typically water).

Note: Water = 62.4 lb/ft3, 8.345 lb/gal, 7.48 gal/ft3, 1 gm/cm3, 1000 kg/m3 Porosity (n) – The ratio of the volume of voids in a material to the total volume of the material. Void Ratio (e) – The ratio of the volume of voids in a material to the volume of solids in the material. (Problem - Specific Gravity) A sedimentary rock weighs 165 lb/ft3 (pounds per cubic foot). What is its specific gravity? Stress (σ) – The force applied per unit area that produces deformation in a body.

waterofweight

substance ofweight sggravity Specific

V

V

volumetotal

voidsof volumePorosity vn

s

v

V

V

solids of volume

voidsof volume Ratio Void e

2.64

)(lb/ft 62.4

)(lb/ft 165

waterofweight

substance ofweight sggravity Specific

3

3

area

force)( Stress

(Problem – Overburden Stress) The rock mass over a mine has an average specific gravity of 2.55. What is the stress in psi (pounds per square inch) that is exerted per foot of depth? (Problem – Overburden Stress) (Cont’d) Note: 145 psi/MPa, 3.28 ft/m Strain (ε) – The deformation resulting from stress, represented as a dimensionless number (in/in). (Problem – Strain) A 75 ft long building is subjected to an increase in length of 0.3 ft due to subsidence. What is the corresponding strain?

)lb/ft( 159

)(lb/ft 62.4 *2.55

waterofweight *gravity Specific substance ofweight

waterofweight

substance ofweight gravity Specific

3

3

depth ofpsi/ft 1.1

)/ft(in 144

)(lb/ft 159

area acting

overburden ofweight )( stress overburden

i

22

3

i

i

depth of MPa/m 025.0m/ft .3048

MPa/psi 0.006895 *depth ofpsi/ft 1.1

depth ofpsi/ft 1.1)( stress overburden

i

i

i

length initial

lengthin change)(Strain

strains-micro4000or 0.004ft 75

ft 0.3

length initial

lengthin change)(Strain

ε

ε

Rock Properties: Elastic Properties:

Elastic Modulus Shear Modulus Poisson’s Ratio

Strength: Compressive, Tensile, Shear Mohr-Coulomb: Friction Angle, Cohesion Hoek-Brown: m, s Uniaxial Compression Testing: Unconfined/Uni-Axial Compressive Test (UCS): Most common rock property Relatively easy to obtain

Figure. Uniaxial-Compressive Test specimen. 1. Standards by the International Society for Rock Mechanics (ISRM). 2. A right circular cylinder a. Diameter > 54 mm b. Height/Diameter > 2.0 c. Flat Ends (< 0.02 mm) d. Perpendicular Ends (< 0.05/50 mm) 3. Stored so as to maintain water content no greater than 30 days. 4. Constant strain rate 0.5 – 1.0 MPa/sec.

Elastic Modulus (E) (or Young’s Modulus) – The ratio of the axial stress to the axial strain produced in the same direction. Note: Elastic Modulus of Steel = 30,000,000 psi or 200 GPa Elastic Modulus of Rock = 1,000,000 - 15,000,000 psi (typical) Shear Modulus (G) (or Modulus of Rigidity) – The ratio of the shear stress to the shear strain Poisson’s Ratio (v) – The ratio of the complementary strain induced in an orthogonal direction to the primary strain generated by the applied force (uni-axial stress, plane stress). Note: Poisson’s Ratio of Water = 0.5 (hydrostatic) Poisson’s Ratio of Rock = 0.1 - 0.4 (typical)

a

a

ε

σ

strain axial

stress axial E

E where a

a

r a

ν12

E

γ

τG

xy

xy

(0.001050, 0.0)

= 131 MPa

- Secant Modulus

Axial Strain ( ) %

Axi

al S

tres

s (

)

(-0.000625, 65.5)/2c

tE

sE

rRadial Strain ( ) %

40

-0.2 -0.1 0.10

a

c

120

80 (0.001775, 65.5)

- Tangent Modulus

a

0.30.2

(0.002500, 131.0)

Figure Stress-strain curves for a typical uniaxial compressive test. Tangent Elastic Modulus (Et) – The slope of the axial stress-strain curve at a specific point, generally at 50% of the peak strength. Secant Elastic Modulus (Es) – The slope of the axial stress-strain curve from zero stress to a specific stress, generally at the peak strength or 50% of the peak strength. (Problem – Modulus of Core) -Determine the Tangent Elastic Modulus at 50% of failure strength from the graph?

a

a

Δε

Δσ

changestrain

change stress Et

0

0s changestrain

change stress E

εε

σσ

a

a

GPa .390.00105 - 0.00250

MPa 0131

Et

a

a

Δε

Δσ

(Problem – Poisson’s Ratio) - Determine the Secant Poisson’s Ratio at 50% failure strength from the graph?

(Problem – Elastic Modulus)A surveyor’s 100 ft long steel tape has a cross section of 0.250 in by 0.03 in. What is its elastic modulus if its elongation is 0.064 in when held by a force of 12 lb? Note: The elastic modulus of steel stays the same irregardless of strength!

35.0001775.0

000625.0

a

r

psi 1600

in 0.03*in 0.250

lb 12area

forceStress

053300.0

in/ft 12 *ft 100

in 064.0

length initial

lengthin changeStrain

psi 10 X 30

000533.0

psi 1600strain

stress E

6

Hooke’s Law: In a tri-axial stress field, the strain in any given direction can be determined by:

Horizontal Stress - In a gravity loading situation without any tectonic or residual stresses, the “natural” horizontal stress is determined by the overburden’s Poisson’s Ratio

Note: In general, this is not the case and the tectonic/residual stresses determine the magnitude of the horizontal stress. (Problem – Horizontal Stress) - Determine the ratio of horizontal to vertical stress for a gravity loaded material with a Poisson’s Ratio of 0.25?

vh *)1(

v

33.025.01

25.0

)1(

*)1(

v

h

vh

v

v

Stress Concentrations:

r

r

u

a

u

r

rr

yy

Figure. Schematic for Kirsch's equations for stresses around a circular opening.

Stress around a circular opening according to Kirsch: Stress at the edge (r = a) of a circular opening according to Kirsch: (Problem – Stress Concentration) - For a unit vertical stress, what is the tangential stress at the side of a circular tunnel?

θ r

a

r

aσ σ

θ r

aσ

r

aσ σ

θ r

a

r

aσ

r

aσ σ

yyrθ

yyyyθθ

yyyyrr

2sin32

12

2cos3

12

12

2cos34

12

12

4

4

2

2

4

4

2

2

4

4

2

2

2

2

0

2cos2

0

σ

θσσ σ

σ

rθ

yyyyθθ

rr

3

1121

2cos2

))((

θσσ σ yyyyθθ

(Problem – Stress Concentration) - For a unit vertical stress, what is the tangential stress at the top of a circular tunnel? (Problem – Stress Concentration) - For a unit vertical stress, what is the vertical stress 2 radii from the edge of the circle (r = 3a)? Biaxial Stress: For a bi-axial stress field, the uni-axial solutions can be combined? For instance, a uni-axial vertical stress field of 3 would give a stress concentration of 9 at the side of the hole. A uni-axial horizontal stress of 1, would give a stress concentration of -1 at the side of the hole. The combination of a 3 vertical stress and 1 horizontal stress would result in a stress concentration of 8 at the side of the hole (and 0 at the top of the hole).

1

1121

2cos2

-

))((

θσσ σ yyyyθθ

0741

180cos81

31

2

1

91

2

1

2cos3

12

12

4

4

2

2

4

4

2

2

. σ

)( a

a

a

a σ

θ r

aσ

r

aσ σ

θθ

θθ

yyyyθθ

Figure The radial (r) and tangential(t) stress at the top and side of a circular opening.

Figure.Vertical stress around a hole in a 200 psi vertical stress field

1 2 3 4 50

1

2

3

-10+1

1

2

3

4

5

z/r

s /szz

s /szz

x/r

s r

s t

s r

s t

r

Y X

Z

SZZ(Psi)

-50-75-100-125-150-175-200-225-250-275-300-325-350-375-400-425-450-475-500-525-550-575-600

Stress Concentrations:

For the previous stress concentration charts, a safety factor of 2 (short term) and 4 (long term) is recommended for compression and a safety factor of 4 (short term) to 8 (long term) is recommended for tension

Figure.The vertical stress concentrations at the top and side of a rectangular opening.

Figure.The vertical stress around a (homogeneous, elastic) rectangular opening in a 200 psi vertical stress field.

1 2 3 4 5

2

3

45

10

1

2

3

s v/szz

s v/szz+10-1

z/r

x/r

s v

s v

r/2r

Y X

Z

SZZ(Psi)

0-25-50-75-100-125-150-175-200-225-250-275-300-325-350-375-400-425-450-475-500-525-550

(Problem – Stress Concentration) - A tunnel is being constructed 500 ft underground. The properties of the material are: specific gravity = 2.59, Poisson’s Ratio = .25, compressive strength = 13150 psi and tensile strength = 1225 psi. Calculate the vertical and horizontal stresses at the construction site. Calculate the stress concentration factor for the center top and corner for a rectangular opening with a width-to-height ratio of four. Calculate the safety factors. Vertical stress: Horizontal stress: Stress ratio (M): The stress concentration factor for M=1/3 and w/h=4, using the figure 5 above: Top center = -0.5 Corner = 5.5 Safety Factors: Note: Good for short term stability in tension and long term stability in compression.

psi 561

500*)/ft(in 144

)(lb/ft 4.62*59.2

area acting

overburden ofweight )( stress overburden

i

22

3

i

i

psi

psi*.

.

*σv)(

ν σ vh

187

5612501

250

1

3

12501

250

561

187

1

.

.

v)(

ν

σ

σM

v

h

4.37

561 * .5

1225

*SCF

strength tensile SFtension

vσ

4.26

561 * 5.5

13150

*SCF

strength ecompressiv SFcomp

vσ

Mohr’s Circle of Stress:

The normal stress on a plane oriented at angle θ to the principal axes:

θ τθ σσσσ

θθ τθ σθ σ σ

xzxzzx

xzzxθ

2sin2cos22

cossin2cossin 22

systemy -in x plane of angle

plane on stressshear

plane on stress normal

θ

τ

σ

θ

θ

face z-on x stressshear

directiony in stress normal

directionin x stress normal

xz

z

x

τ

σ

σ

The shear stress on a plane oriented at angle θ to the principal axes: The principal stress in the stress field: The principal stress direction in the stress field. (Problem - Mohr Circle) From the previous tunnel problem, the concentrated vertical stress at the mid-height of the side of the opening is 1,402 psi (2.5 * 561). If we have a 45 joint with no cohesion and a friction angle of 35 intersecting the tunnel side, will this joint slip? First, we calculate the normal stress on the joint:

θ τθ

σσ

θ θ τθθ σσ τ

xzxz

xzxzθ

2cos2sin2

cossincossin 22

systemy -in x plane of angle

plane on stressshear

plane on stress normal

θ

θτ

θ σ

θ

θ

face z-on x stressshear

directiony in stress normal

directionin x stress normal

xz

z

x

τ

σ

σ

face z-on x stressshear

directiony in stress normal

directionin x stress normal

xz

z

x

τ

σ

σ

xz

xzp σσ

τ θ

2arctan

2

1

face z-on x stressshear

directiony in stress normal

directionin x stress normal

xz

z

x

τ

σ

σ

psi 701

45)*(2sin 0 45)*(2 cos 2

01402

2

14020

2θsin τ2θ cos 2

σσ

2

σσ σ xz

xzzxθ

τ

σσσσ ,σσ xz

xzxz 22

31 22

stress principal minimum

stress principal maximum

3

1

σ

σ

Next, we calculate the shear stress on the joint: Finally, we calculate the shear resistance to sliding: The shear force is 701 psi and the resistance is 491, so the joint will slide. The Hoek-Brown failure criteria:

2331 cc sσσmσσ σ

psi 701

45)*(2 cos 045)*(2sin 2

01402

2θ cos τ2θsin 2

σσ τ xz

xzθ

psi 491

tan(35)* 7010

tanσcτ n

constants empiricalsm,

rockintact ofstrength ecompressiv Uniaxial

stress principalMinor

stress principalMajor

3

1

cσ

σ

σ

Hoek-Brown “m” and “s” values: