Numerical example worked by hand¶
Step-by-step reconstruction of a reference case to document the traceability of the calculation in RockPlane NX. Values are obtained analytically with a scientific calculator, independent of the software, and compared to the software output.
Serves as the audit reference for compliance reviews and for anyone who wants to manually check the consistency between the theoretical manual formulas (Hoek-Bray / RocPlane) and the implementation.
1. Base case input¶
| Parameter | Value |
|---|---|
| H · slope height | 30 m |
| β · slope face dip | 60° |
| α · failure plane dip | 30° |
| ψ · upper face dip | 0° |
| γ · unit weight | 26 kN/m³ |
| c · cohesion | 50 kPa |
| φ · friction angle | 30° |
| Approach | Characteristic (γ = 1.0) |
| Water / Seismic / Reinforcement | None |
2. Geometry¶
N = H / sin β = 30 / 0.86603 = 34.641 m
L = H / sin α = 30 / 0.50000 = 60.000 m
M = (L·cos α − H·cot β) / cos ψ = 34.641 m
A = ½·H²·(cot α − cot β) = ½·900·(1.732 − 0.577) = 519.62 m²
W = γ · A = 26 · 519.62 = 13 510.2 kN/m
3. Forces and equilibrium¶
F_x = 0 (no E, no seismic, no reinforcement)
F_y = W_y = −γ_G · W = −13 510.2 kN/m
N = −F_y·cos α + F_x·sin α − U
= 13 510.2 · 0.866 + 0 − 0 = 11 700.6 kN/m
S = −F_y·sin α − F_x·cos α
= 13 510.2 · 0.500 − 0 = 6 755.1 kN/m
4. Mohr-Coulomb resistance and FS¶
τ_res = c·L/γ_c + N·tan φ/γ_φ
= 50·60 + 11 700.6 · 0.57735
= 3 000 + 6 754.8 = 9 754.8 kN/m
FS = τ_res / |S| = 9 754.8 / 6 755.1 = 1.4441
5. Manual vs software comparison¶
| Quantity | Hand | Software | Δ |
|---|---|---|---|
| N (normal) | 11 700.6 kN/m | 11 700.6 kN/m | < 0.001 % |
| S (shear) | 6 755.1 kN/m | 6 755.1 kN/m | < 0.001 % |
| τ_res | 9 754.8 kN/m | 9 754.8 kN/m | < 0.001 % |
| FS | 1.4441 | 1.4441 | 0.0000 |
Exact match to 4 decimals. The software formulation faithfully reproduces the classical Hoek-Bray formula:
6. Extension with seismic kh = 0.15¶
S_mod = k_h · W = 0.15 · 13 510.2 = 2 026.5 kN/m
F_x = −2 026.5 kN/m F_y = −13 510.2 kN/m
N = 13 510.2·0.866 − 2 026.5·0.500 = 10 687.3 kN/m
S = 13 510.2·0.500 + 2 026.5·0.866 = 8 510.1 kN/m
τ_res = 3 000 + 10 687.3 · 0.57735 = 9 170.0 kN/m
FS = 9 170.0 / 8 510.1 = 1.078
Software comparison: FS = 1.0776, match to 4 decimals.
7. Extension with water in the discontinuity (uplift U)¶
Base case + Zw = 10 m, triangular distribution max at toe.
u_max = γ_w · Z_w = 9.81 · 10 = 98.1 kPa
s_wet = Z_w / sin α = 10 / 0.5 = 20.0 m
U = ½ · u_max · s_wet = ½ · 98.1 · 20 = 981.0 kN/m
N_eff = 11 700.6 − 981.0 = 10 719.6 kN/m
S = 6 755.1 kN/m (unchanged)
τ_res = 3 000 + 10 719.6 · 0.57735 = 9 188.4 kN/m
FS = 9 188.4 / 6 755.1 = 1.360
Software comparison: FS = 1.3603 and U = 981.0 kN/m. Exact match.
8. Extension with passive nail (Clouterre N-V)¶
Base case + passive nail F = 500 kN/m, Δ = 15° below horizontal.
K_x = F · cos Δ = 500 · 0.96593 = 482.96 kN/m
K_y = −F · sin Δ = −500 · 0.25882 = −129.41 kN/m
N = 13 639.6 · 0.866 + 482.96 · 0.500 = 12 054.2 kN/m
S = 6 755.1 kN/m (unchanged — passive K does not enter driving S)
contrib_K = K_x · cos α + K_y · sin α = 353.7 kN/m
τ_res = 3 000 + 12 054.2 · 0.57735 + 353.7 = 10 314.5 kN/m
FS = 10 314.5 / 6 755.1 = 1.527
Clouterre check — Φ = α + Δ = 45°:
| Mode | Contribution |
|---|---|
| Axial | T_max·(cos Φ + sin Φ · tan φ) = 557.6 |
| Shear | V_max·(sin Φ − cos Φ · tan φ) = 74.8 |
Axial mode wins → applied force corresponds to K components computed above. Software comparison: FS = 1.5267.
9. Extension with active anchor¶
Base case + active anchor pre-load F = 500 kN/m, Δ = 20°.
The anchor enters as an ACTION (J) in the resultants F_x, F_y, not in the resistance.
J_x = F · cos Δ = 500 · 0.93969 = 469.85 kN/m
J_y = −F · sin Δ = −500 · 0.34202 = −171.01 kN/m
F_x = 469.85 kN/m F_y = −13 681.2 kN/m
N = 13 681.2 · 0.866 + 469.85 · 0.500 = 12 082.8 kN/m
S = 13 681.2 · 0.500 − 469.85 · 0.866 = 6 433.6 kN/m (reduced)
τ_res = 3 000 + 12 082.8 · 0.57735 = 9 977.4 kN/m
FS = 9 977.4 / 6 433.6 = 1.551
Software comparison: FS = 1.5506, match to 3 decimals.
Active anchor vs passive nail (same F = 500 kN/m, Δ ≈ 15-20°)
- Active anchor (F=1.551): reduces driving shear S directly (pre-applied force).
- Passive nail (F=1.527): acts through resisting τ (mobilised during sliding). The anchor is slightly more effective for the same F.
10. Summary¶
| Case | Addition to base | FS | Internal test |
|---|---|---|---|
| 1–5 | base case dry | 1.4441 | T03 |
| 6 | seismic k_h = 0.15 | 1.078 | T04 |
| 7 | water Z_w = 10 m, max at toe | 1.360 | T05 |
| 8 | passive nail F = 500, Δ = 15° | 1.527 | T07 |
| 9 | active anchor F = 500, Δ = 20° | 1.551 | T06 |
All cases match to 3-4 decimal places with the software output, confirming the traceability of the calculation for each modelled mechanism (weight W, seismic S, water U, reinforcement J/K).
For the complete suite of 97 automated regression tests, see the repository README or contact the GeoStru team.