RocPlane Planar sliding stability analysis for rock slopes
Theory Manual
2001
Rocscience Inc.
PLANAR FAILURE
Geometry – No Tension Crack
Known Parameters: H = Slope Height
β = α = ψ =
Slope Dip
Unknown Parameters: B = intersection point, slope & bench C = intersection point, failure plane & bench
Failure Plane Dip Upper Bench Dip
N = slope length, origin ! B M = bench length, B! C
O = Origin (0,0) γ = Rock Unit Weight
L = failure plane length, origin ! C A = wedge area W = wedge weight
N =
H sin β
B = { N cos β , H } = { H cot β , H }
[1] [2]
To solve for distances L&M, use vector addition:
OB + BC = OC & H cot β # & M cosψ # & L cos α # % "+% "=% " H M sin ψ $ ! $ ! $ L sin α ! This gives two equations:
H cot β + M cosψ = L cos α H + M sin ψ = L sin α
[3] [4]
From equations [4]:
M =
L sin α − H sinψ
[5]
Substituting [5] into [3]:
H cot β + ( L sin α − H ) cotψ = L cosα H (cot β − cotψ ) = L (cos α − sin α cotψ ) H (1 − cot β tanψ ) L = sin α − cosα tanψ
[6]
1
PLANAR FAILURE From equation [3]:
M =
L cos α − H cot β
[7]
cosψ
To calculate L and M, use equations [6] & [7]. Do not use equation [5] because ψ = 0 is common & M is irresolvable using [5].
C = { L cosα , L sin α }
[8]
Area Calculation:
A = A =
1 2 1
B × C
B x C y − B y C x 2 W = A ⋅ γ
[9]
[10]
Flow Chart: 1. 2. 3. 4. 5. 6. 7.
Solve Solve Solve Solve Solve Solve Solve
for for for for for for for
N B L M C A W
(eq. (eq. (eq. (eq. (eq. (eq. (eq.
[1]) [2]) [6]) [7]) [8]) [9]) [10])
2
PLANAR FAILURE
Water Forces – No Tension Crack Case 1:
Maximum Pressure Mid Height
0 ! Zw ! L sin" Lw = wetted length =
Zw sin α
P = Maximum water pressure =
U = water force =
1 2
P ⋅ Lw =
1 2
Z wγ w
), Z ) * Z w ⋅ γ w '* w ' 2 + 2 (+ sin α (
1 , 1
2
U =
Case 2:
Z w ⋅ γ w
[11]
4 sin α
Maximum Pressure at Toe
Lw =
Z w
sin α P = γ Z w
1 1 , Z ) U = P ⋅ Lw = (γ ⋅ Z w )* w ' 2 2 + sin α ( 2
U =
Z w ⋅ γ w 2 sin α
[12]
3
PLANAR FAILURE
Geometry – Tension Crack
Known Parameters: H = Slope Height # = Slope Dip " = Failure Plane Dip
$ = T = % = O = & =
Upper Bench Dip Tension Crack Distance Tension Crack Dip Origin (0,0) Rock Unit Weight
Unknown Parameters: B = Slope/Bench intersection point C = Tension Crack/Bench intersection point D = Failure Plane/Tension Crack intersection point N = Slope Length, O ! B M = Bench Length, B ! C L = Failure Plane Length, O ! D Q = Tension Crack Length, D ! C A = Wedge Area W = Wedge Weight
As in the no tension crack case:
N =
H sin β
B = { H cot β , H } Now,
C = B + {T ,T tanψ } T M = cosψ
[13]
[14]
D = C − {Q cosθ , Q sin θ }
[15]
D = { L cosα , L sin α }
[16]
Let’s solve for D,Q,L:
Equate equations [15]&[16]:
C x , C y − {Q cosθ , Q sin θ } = { L cos α , L sin α } or
L =
C x − Q cosθ cos α
[17]
4
PLANAR FAILURE
L =
and
C y − Q sin θ
sin α
[18]
Equate equations [17]&[18] and solve for Q:
Q=
C y cot α − C x
sin θ cot α − cosθ
[19]
Area Calculation:
A = A =
1 2 1 2
B × D +
1 2
( D − B) × (C − B )
B x D y − B y D x +
1 2
( D x − B x )(C y − B y ) − ( D y − B y )(C x − B x )
[20]
W = A ⋅ γ Flow Chart: 1. 2. 3. 4. 5. 6. 7. 8. 9.
Solve Solve Solve Solve Solve Solve Solve Solve Solve
for for for for for for for for for
N B C M Q L D A W
(eq. (eq. (eq. (eq. (eq. (eq. (eq. (eq. (eq.
[1]) [2]) [13]) [14]) [19]) [17]) [16]) [20]) [10])
5
PLANAR FAILURE
Water Forces – Tension Crack Case 1:
Maximum Pressure Mid Height
Zt = Zw – Zf Zf = Dy = L s i n " U = Water force on failure plane V = Water force on tension crack
Type A.
If Zw ! Zf 2
Z w ⋅ γ w
U =
Type B.
4 sin α
;
V = 0
[20]
If Zw > Zf and Zw/ 2 < Zf
L1 =
Zw 2 sin α
L2 = L − L1 1 P 1 = Z w ⋅ γ w 2 P 2 = γ w ⋅ Z t
[21]
1 1 U = P 1 ⋅ L1 + ( P 1 + P 2 ) L2 2 2 2
V =
Type C.
Z t ⋅ γ w 2 sin θ
If Zw > Zf and Zw/2 ' Zf
6
PLANAR FAILURE
P 3 = γ ⋅ Z f P 4 =
L3 = L4 =
1
γ ⋅ Z w 2 , Z w ) − Z f ' * + 2 ( sin θ Z w
[22]
2 sin θ 1 U = L ⋅ P 3 2 1 1 V = ( P 3 + P 4 ) L3 + P 4 L4 2 2
Case 2:
Maximum Pressure at Toe
P 5 = γ ⋅ Z t P 6 = γ ⋅ Z w L5 =
Z t
sin θ 1 U = ( P 5 + P 6 ) L 2 1 V = P 5 ⋅ L5 2
[23]
7
PLANAR FAILURE Case 3:
Maximum Pressure at Base of Tension Crack
Zt = Zw - Zf
P 7 = γ ⋅ Z t L7 =
Z t
sin θ 1 U = P 7 ⋅ L 2 1 V = P 7 ⋅ L7 2
[24]
8
PLANAR FAILURE
External Force
E x = E ⋅ cos δ E y = E ⋅ sin δ
Seismic Force
S = W y ⋅ α s
α s =
Seismic Coefficient
W = Weight of Wedge Wy = Directional Weight Component
W y = −W S x = S ⋅ cos Ω S y = S ⋅ sin Ω Active Bolt Force
J = Active Bolt Force
J x = J ⋅ cos ∆ a J y = − J ⋅ sin ∆ a
Passive Bolt Force
K x = K ⋅ cos ∆ p K y = − K ⋅ sin ∆ p Active Water Force
V x = −V ⋅ sin θ V y = V ⋅ cosθ
K = Passive Bolt Force
(Tension Crack)
V = Tension Crack Water Force
9
PLANAR FAILURE
Normal/Shear Force on Failure Plane
W = Wedge Weight Wy = -W
Active Forces Only:
- F + ↑ y
F y = W y + E y + S y + J y + V y F y = − A ⋅ γ − E ⋅ sin δ − S ⋅ sin Ω − J ⋅ sin ∆ a + V ⋅ cosθ
→ - F x
+
[24]
F x = E x + S x + J x + V x ∆ a − V ⋅ sin θ F x = E ⋅ cos δ − S ⋅ cos Ω + J ⋅ cos
[25]
N = − F y + K y cos α + ( F x + K x ) sin α − U
[26]
S = − F y ⋅ sin α − F x ⋅ cos α
[27]
Shear Strength on Failure Plane
Strength Criterion = Mohr Coulomb C = Cohesion N = Normal Force
φ =
Friction Angle
L = Length of Failure Surface
τ = c ⋅ L + N ⋅ tan φ + K x ⋅ cos α + K y ⋅ sin α
[28]
$!! ! #!! ! "
Passive Bolt
Factor of Safety
F =
Re sisting Forces
F =
Shear Strength
Driving Forces Shear Force
=
τ S
[29]
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