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12/09/2025
# Top 100 Rock Mechanics Problems for Mining Engineering
# # Section 1: Stress and Strain Analysis (Problems 1-20)
# # # Problem 1: Vertical Stress at Depth ✓
**Problem:** Calculate the vertical stress at a depth of 500m in a rock mass with average density of 2.7 g/cm³.
**Solution:**
- σᵥ = ρ × g × h
- σᵥ = 2700 kg/m³ × 9.81 m/s² × 500 m
- σᵥ = 13.24 MPa
# # # Problem 2: Horizontal Stress Ratio ✓
**Problem:** If the vertical stress is 15 MPa and the horizontal stress coefficient K₀ = 0.5, calculate the horizontal stress.
**Solution:**
- σₕ = K₀ × σᵥ
- σₕ = 0.5 × 15 MPa
- σₕ = 7.5 MPa
# # # Problem 3: Principal Stress Calculation ✓
**Problem:** Given σₓ = 20 MPa, σᵧ = 15 MPa, τₓᵧ = 5 MPa, find the principal stresses.
**Solution:**
- σ₁,₂ = (σₓ + σᵧ)/2 ± √[((σₓ - σᵧ)/2)² + τₓᵧ²]
- σ₁,₂ = (20 + 15)/2 ± √[((20 - 15)/2)² + 5²]
- σ₁,₂ = 17.5 ± √(6.25 + 25) = 17.5 ± 5.59
- σ₁ = 23.09 MPa, σ₂ = 11.91 MPa
# # # Problem 4: Elastic Modulus from Stress-Strain ✓
**Problem:** A rock sample experiences 0.002 strain under 40 MPa stress. Calculate Young's modulus.
**Solution:**
- E = σ/ε
- E = 40 MPa / 0.002
- E = 20,000 MPa = 20 GPa
# # # Problem 5: Poisson's Ratio Calculation ✓
**Problem:** Axial strain = 0.001, lateral strain = -0.0003. Find Poisson's ratio.
**Solution:**
- ν = -εₗₐₜₑᵣₐₗ/εₐₓᵢₐₗ
- ν = -(-0.0003)/0.001
- ν = 0.3
# # # Problem 6: Biaxial Stress State ✓
**Problem:** Calculate strain in x-direction when σₓ = 30 MPa, σᵧ = 20 MPa, E = 25 GPa, ν = 0.25.
**Solution:**
- εₓ = (1/E)[σₓ - ν(σᵧ + σᵤ)]
- εₓ = (1/25000)[30 - 0.25(20 + 0)]
- εₓ = (1/25000)(30 - 5) = 0.001
# # # Problem 7: Shear Modulus ✓
**Problem:** If E = 30 GPa and ν = 0.2, calculate the shear modulus G.
**Solution:**
- G = E/[2(1 + ν)]
- G = 30,000/[2(1 + 0.2)]
- G = 30,000/2.4 = 12.5 GPa
# # # Problem 8: Bulk Modulus ✓
**Problem:** Calculate bulk modulus K if E = 40 GPa and ν = 0.3.
**Solution:**
- K = E/[3(1 - 2ν)]
- K = 40,000/[3(1 - 2×0.3)]
- K = 40,000/[3(0.4)] = 33.33 GPa
# # # Problem 9: Stress Around Circular Opening ✓
**Problem:** Calculate tangential stress at θ = 0° around a circular tunnel with radius 3m, at distance r = 6m, under σₒ = 10 MPa.
**Solution:**
- σₜ = σₒ(1 + (a²/r²))
- σₜ = 10(1 + (3²/6²))
- σₜ = 10(1 + 0.25) = 12.5 MPa
# # # Problem 10: Stress Concentration Factor ✓
**Problem:** Maximum stress around a circular hole is 30 MPa, far-field stress is 10 MPa. Find stress concentration factor.
**Solution:**
- SCF = σₘₐₓ/σₒ
- SCF = 30/10 = 3
# # # Problem 11: Triaxial Stress State ✓
**Problem:** Calculate octahedral normal stress for σ₁ = 40 MPa, σ₂ = 25 MPa, σ₃ = 15 MPa.
**Solution:**
- σₒᶜᵗ = (σ₁ + σ₂ + σ₃)/3
- σₒᶜᵗ = (40 + 25 + 15)/3
- σₒᶜᵗ = 26.67 MPa
# # # Problem 12: Octahedral Shear Stress ✓
**Problem:** Find octahedral shear stress for the above principal stresses.
**Solution:**
- τₒᶜᵗ = (1/3)√[(σ₁-σ₂)² + (σ₂-σ₃)² + (σ₃-σ₁)²]
- τₒᶜᵗ = (1/3)√[(40-25)² + (25-15)² + (15-40)²]
- τₒᶜᵗ = (1/3)√[225 + 100 + 625] = (1/3)√950 = 10.27 MPa
# # # Problem 13: Elastic Settlement ✓
**Problem:** A foundation load of 1000 kN acts over area 4 m². Rock has E = 20 GPa, ν = 0.25. Calculate settlement.
**Solution:**
- σ = P/A = 1000/4 = 250 kPa
- Settlement ≈ σ×B×(1-ν²)/E (simplified)
- s ≈ 0.25×2×(1-0.25²)/20,000,000 = 0.00002 m = 0.02 mm
# # # Problem 14: Thermal Stress ✓
**Problem:** Rock heated by 50°C, α = 10×10⁻⁶/°C, E = 30 GPa, constrained. Find thermal stress.
**Solution:**
- σₜₕₑᵣₘₐₗ = -α×ΔT×E
- σₜₕₑᵣₘₐₗ = -10×10⁻⁶×50×30×10⁹
- σₜₕₑᵣₘₐₗ = -15 MPa (compression)
# # # Problem 15: Strain Energy Density ✓
**Problem:** Calculate strain energy density for σ = 20 MPa, E = 25 GPa.
**Solution:**
- U = σ²/(2E)
- U = (20×10⁶)²/(2×25×10⁹)
- U = 8×10³ J/m³ = 8 kJ/m³
# # # Problem 16: Mohr Circle Construction ✓
**Problem:** σₓ = 25 MPa, σᵧ = 15 MPa, τₓᵧ = 8 MPa. Find center and radius of Mohr circle.
**Solution:**
- Center: (σₓ + σᵧ)/2 = (25 + 15)/2 = 20 MPa
- Radius: √[((σₓ - σᵧ)/2)² + τₓᵧ²] = √[(5)² + 8²] = √89 = 9.43 MPa
# # # Problem 17: Maximum Shear Stress ✓
**Problem:** From Problem 16, find maximum shear stress.
**Solution:**
- τₜₘₐₓ = Radius of Mohr circle = 9.43 MPa
# # # Problem 18: Plane Stress Transformation ✓
**Problem:** Transform stress to plane at 30° using data from Problem 16.
**Solution:**
- σₙ = (σₓ + σᵧ)/2 + (σₓ - σᵧ)cos(2θ)/2 + τₓᵧsin(2θ)
- σₙ = 20 + 5×cos(60°) + 8×sin(60°)
- σₙ = 20 + 2.5 + 6.93 = 29.43 MPa
# # # Problem 19: Effective Stress ✓
**Problem:** Total stress = 25 MPa, pore pressure = 5 MPa. Calculate effective stress.
**Solution:**
- σ' = σ - u
- σ' = 25 - 5 = 20 MPa
# # # Problem 20: Hydrostatic Stress ✓
**Problem:** Find hydrostatic stress for σ₁ = 30 MPa, σ₂ = 20 MPa, σ₃ = 10 MPa.
**Solution:**
- σₘ = (σ₁ + σ₂ + σ₃)/3
- σₘ = (30 + 20 + 10)/3 = 20 MPa
# # Section 2: Rock Strength and Failure Criteria (Problems 21-40)
# # # Problem 21: Mohr-Coulomb Failure ✓
**Problem:** Rock has c = 5 MPa, φ = 30°. Find shear strength at σₙ = 20 MPa.
**Solution:**
- τ = c + σₙ tan φ
- τ = 5 + 20 × tan(30°)
- τ = 5 + 20 × 0.577 = 16.54 MPa
# # # Problem 22: Uniaxial Compressive Strength ✓
**Problem:** Cylindrical sample (D=50mm, L=100mm) fails at 125 kN. Find UCS.
**Solution:**
- UCS = P/A = 125,000 N / (π × (0.025)² m²)
- UCS = 125,000 / 0.00196 = 63.69 MPa
# # # Problem 23: Tensile Strength from Brazilian Test ✓
**Problem:** Disk (D=50mm, t=25mm) fails at 15 kN in Brazilian test. Find tensile strength.
**Solution:**
- σₜ = 2P/(π×D×t)
- σₜ = 2×15,000/(π×0.05×0.025)
- σₜ = 30,000/0.00393 = 7.64 MPa
# # # Problem 24: Hoek-Brown Criterion ✓
**Problem:** σcᵢ = 50 MPa, mᵢ = 15, s = 1. Find σ₁ when σ₃ = 10 MPa.
**Solution:**
- σ₁ = σ₃ + σcᵢ√(mᵢσ₃/σcᵢ + s)
- σ₁ = 10 + 50√(15×10/50 + 1)
- σ₁ = 10 + 50√(3 + 1) = 10 + 100 = 110 MPa
# # # Problem 25: Griffith Criterion ✓
**Problem:** Tensile strength = 3 MPa. Find compressive strength using Griffith criterion.
**Solution:**
- σc = 8σₜ (for Griffith criterion)
- σc = 8 × 3 = 24 MPa
# # # Problem 26: Factor of Safety ✓
**Problem:** Applied stress = 15 MPa, rock strength = 45 MPa. Calculate factor of safety.
**Solution:**
- FOS = Strength/Applied Stress
- FOS = 45/15 = 3
# # # Problem 27: Cohesion from Triaxial Tests ✓
**Problem:** Two tests: (σ₁=40, σ₃=5) and (σ₁=60, σ₃=15). Find cohesion and friction angle.
**Solution:**
- From Mohr circles: slope m = tan φ
- (σ₁-σ₃)₁ = 35, (σ₁+σ₃)₁ = 45
- (σ₁-σ₃)₂ = 45, (σ₁+σ₃)₂ = 75
- tan φ = Δ(σ₁-σ₃)/Δ(σ₁+σ₃) = 10/30 = 0.333
- φ = 18.4°, c ≈ 8.7 MPa
# # # Problem 28: Point Load Index ✓
**Problem:** Point load test on 50mm diameter core fails at 2.5 kN. Find point load index.
**Solution:**
- Isₒ = P/De² (where De = equivalent diameter)
- Isₒ = 2500/(50²) = 2500/2500 = 1.0 MPa
# # # Problem 29: Estimate UCS from Point Load ✓
**Problem:** Using result from Problem 28, estimate UCS.
**Solution:**
- UCS ≈ 24 × Isₒ (empirical correlation)
- UCS ≈ 24 × 1.0 = 24 MPa
# # # Problem 30: Schmidt Hammer Rebound ✓
**Problem:** Schmidt hammer gives rebound value of 45. Estimate UCS using correlation.
**Solution:**
- UCS = 2.51e^(0.0635×R) (typical correlation)
- UCS = 2.51 × e^(0.0635×45)
- UCS = 2.51 × e^2.86 = 43.8 MPa
# # # Problem 31: Rock Quality Designation ✓
**Problem:** Core run of 1.5m yields pieces: 0.3m, 0.15m, 0.4m, 0.25m, 0.4m. Calculate RQD.
**Solution:**
- RQD = (Sum of pieces ≥ 0.1m)/Total length × 100%
- Valid pieces: 0.3, 0.15, 0.4, 0.25, 0.4 = 1.5m
- RQD = 1.5/1.5 × 100% = 100%
# # # Problem 32: Slake Durability Index ✓
**Problem:** Initial dry mass = 500g, final dry mass after two cycles = 425g. Find Id₂.
**Solution:**
- Id₂ = (Final mass/Initial mass) × 100%
- Id₂ = (425/500) × 100% = 85%
# # # Problem 33: Triaxial Test Analysis ✓
**Problem:** σ₃ = 8 MPa, failure at σ₁ = 48 MPa, φ = 25°. Find cohesion.
**Solution:**
- (σ₁ - σ₃)/2 = c×cos φ + (σ₁ + σ₃)×sin φ/2
- 20 = c×cos(25°) + 28×sin(25°)
- 20 = 0.906c + 11.83
- c = 9.02 MPa
# # # Problem 34: Direct Shear Test ✓
**Problem:** Normal stress = 12 MPa, shear stress at failure = 9 MPa, c = 2 MPa. Find φ.
**Solution:**
- τ = c + σₙ tan φ
- 9 = 2 + 12 tan φ
- tan φ = 7/12 = 0.583
- φ = 30.3°
# # # Problem 35: Crack Initiation Stress ✓
**Problem:** In UCS test, first crack appears at 35% of peak strength. Peak strength = 80 MPa. Find crack initiation stress.
**Solution:**
- σcᵢ = 0.35 × σc
- σcᵢ = 0.35 × 80 = 28 MPa
# # # Problem 36: Crack Damage Stress ✓
**Problem:** Crack damage occurs at 70% of peak strength. Peak = 60 MPa. Find crack damage stress.
**Solution:**
- σcd = 0.70 × σc
- σcd = 0.70 × 60 = 42 MPa
# # # Problem 37: Brittleness Index ✓
**Problem:** σc = 80 MPa, σₜ = 6 MPa. Calculate brittleness index B₁.
**Solution:**
- B₁ = (σc - σₜ)/(σc + σₜ)
- B₁ = (80 - 6)/(80 + 6) = 74/86 = 0.86
# # # Problem 38: Strength Anisotropy ✓
**Problem:** Strength parallel to foliation = 70 MPa, perpendicular = 45 MPa. Find anisotropy ratio.
**Solution:**
- Anisotropy ratio = σc‖/σc⊥
- Ratio = 70/45 = 1.56
# # # Problem 39: Confinement Effect ✓
**Problem:** σc at σ₃=0 is 50 MPa, at σ₃=10 MPa is 85 MPa. Find strength increase per MPa confinement.
**Solution:**
- Strength increase = (85-50)/(10-0) = 3.5 MPa per MPa confinement
# # # Problem 40: Failure Envelope ✓
**Problem:** Plot failure envelope for c = 8 MPa, φ = 32°. Find shear strength at σₙ = 25 MPa.
**Solution:**
- τ = c + σₙ tan φ
- τ = 8 + 25 × tan(32°)
- τ = 8 + 25 × 0.625 = 23.6 MPa
# # Section 3: Rock Mass Classification (Problems 41-60)
# # # Problem 41: RMR Basic Calculation ✓
**Problem:** UCS=65MPa, RQD=75%, Spacing=0.8m, Condition=slightly rough, GW=damp, Orientation=fair. Calculate basic RMR.
**Solution:**
- UCS (65 MPa): 7 points
- RQD (75%): 13 points
- Spacing (0.8m): 10 points
- Condition: 25 points
- Groundwater (damp): 10 points
- Basic RMR = 7+13+10+25+10 = 65
# # # Problem 42: RMR Orientation Adjustment ✓
**Problem:** Basic RMR = 65 (from Problem 41), unfavorable orientation for tunnels. Find adjusted RMR.
**Solution:**
- Orientation adjustment for tunnels (unfavorable): -10
- Adjusted RMR = 65 - 10 = 55
# # # Problem 43: Q-System Basic ✓
**Problem:** RQD=80%, Jn=9, Jr=3, Ja=2, Jw=1.0, SRF=1.0. Calculate Q value.
**Solution:**
- Q = (RQD/Jn) × (Jr/Ja) × (Jw/SRF)
- Q = (80/9) × (3/2) × (1.0/1.0)
- Q = 8.89 × 1.5 × 1.0 = 13.33
# # # Problem 44: Rock Mass Rating to GSI ✓
**Problem:** RMR₈₉ = 58. Convert to GSI.
**Solution:**
- GSI ≈ RMR₈₉ - 5 (for good conditions)
- GSI ≈ 58 - 5 = 53
# # # Problem 45: Hoek-Brown for Rock Mass ✓
**Problem:** σcᵢ=60MPa, mᵢ=12, GSI=45, D=0. Find mb, s, and a.
**Solution:**
- mb = mᵢ × exp((GSI-100)/28)
- mb = 12 × exp((45-100)/28) = 12 × 0.176 = 2.11
- s = exp((GSI-100)/9) = exp(-6.11) = 0.002
- a = 0.5 + (1/6)(e^(-GSI/15) - e^(-20/3)) = 0.516
# # # Problem 46: Rock Mass Strength ✓
**Problem:** Using data from Problem 45, find rock mass UCS.
**Solution:**
- σc = σcᵢ × s^a
- σc = 60 × (0.002)^0.516
- σc = 60 × 0.032 = 1.92 MPa
# # # Problem 47: Deformation Modulus ✓
**Problem:** σcᵢ=50MPa, GSI=40, D=0, Ei=20GPa. Find rock mass deformation modulus.
**Solution:**
- Em = Ei × (0.02 + (1-D/2)/(1+e^((60+15D-GSI)/11)))
- Em = 20 × (0.02 + 0.5/(1+e^((60-40)/11)))
- Em = 20 × (0.02 + 0.5/7.39) = 20 × 0.088 = 1.76 GPa
# # # Problem 48: Tunnel Support Pressure ✓
**Problem:** Q=4.5, tunnel span=8m, ESR=1.6. Estimate support pressure using Q-system.
**Solution:**
- Proof = 2×Jn×Q^(-1/3)/Jr (approximately)
- For Q=4.5, typical support pressure ≈ 0.1-0.4 MPa
- Estimated support pressure ≈ 0.2 MPa
# # # Problem 49: Stand-up Time ✓
**Problem:** Tunnel span = 6m, RMR = 45. Estimate stand-up time.
**Solution:**
- Stand-up time = RMR₈₉/(span^k) where k≈0.4
- Stand-up time ≈ 45/6^0.4 ≈ 45/2.3 ≈ 20 hours
# # # Problem 50: Rock Load Height ✓
**Problem:** Q = 2.5, tunnel width = 10m. Calculate rock load height.
**Solution:**
- Ht = B/Q^0.33 (approximately)
- Ht = 10/2.5^0.33 = 10/1.36 = 7.35 m
# # # Problem 51: Joint Condition Rating ✓
**Problem:** Joint separation = 1mm, slightly weathered, smooth surfaces, hard infill 40 MPa (failure occurs)
- EDZ extent ≈ 1.5×R = 1.5×3 = 4.5 m from center
- EDZ thickness ≈ 1.5 m beyond tunnel wall
# # # Problem 71: Squeezing Ground ✓
**Problem:** σ₀ = 25 MPa, σcₘ = 8 MPa, tunnel radius = 4 m. Check for squeezing.
**Solution:**
- Squeezing parameter: σ₀/σcₘ = 25/8 = 3.125
- Since ratio > 2.5, severe squeezing expected
- Strain ≈ (σ₀/σcₘ - 2)/2 = (3.125-2)/2 = 0.56 = 56%
# # # Problem 72: Ground Support Interaction ✓
**Problem:** Support stiffness = 5 GN/m/m, ground stiffness = 2 GN/m/m. Find load sharing ratio.
**Solution:**
- Total stiffness = Ks + Kg = 5 + 2 = 7 GN/m/m
- Support takes: Ks/(Ks+Kg) = 5/7 = 71.4% of load
- Ground takes: 2/7 = 28.6% of load
# # # Problem 73: Time-Dependent Convergence ✓
**Problem:** Initial convergence = 15 mm, time constant = 30 days, time = 60 days. Find total convergence.
**Solution:**
- u(t) = u₀(1 + t/τ) for linear model
- u(60) = 15(1 + 60/30) = 15×3 = 45 mm
- Additional convergence = 45 - 15 = 30 mm
# # # Problem 74: Ventilation Pressure Drop ✓
**Problem:** Airway: 12 m² area, 500 m length, airflow = 50 m³/s, friction factor = 0.01. Find pressure drop.
**Solution:**
- ΔP = f×L×ρ×v²/(2×Dh)
- v = Q/A = 50/12 = 4.17 m/s
- Dh = 4A/P ≈ 4×12/14 = 3.43 m (assuming rectangular)
- ΔP = 0.01×500×1.2×(4.17)²/(2×3.43) = 152 Pa
# # # Problem 75: Crown Settlement ✓
**Problem:** Tunnel: width = 10m, height = 8m, Em = 1.5 GPa, σ₀ = 12 MPa. Estimate crown settlement.
**Solution:**
- Settlement ≈ σ₀×B×(1-ν²)/Em (approximation)
- Assuming ν = 0.3: Settlement ≈ 12×10×(1-0.09)/(1.5×10⁹)
- Settlement ≈ 109.2/(1.5×10⁹) = 7.28×10⁻⁸ m = 0.073 mm
# # # Problem 76: Fault Zone Stability ✓
**Problem:** Fault dip = 60°, cohesion = 0, φ = 25°, σₙ on fault = 10 MPa. Check stability.
**Solution:**
- Required τ for sliding = σₙ tan φ = 10×tan(25°) = 4.66 MPa
- Available shear stress from gravity ≈ σₙ×tan(60°) = 17.3 MPa
- Since 17.3 > 4.66, fault is unstable - sliding expected
# # # Problem 77: Wedge Stability ✓
**Problem:** Rock wedge: weight = 50 kN, sliding plane dip = 35°, φ = 30°, cohesion = 5 kPa. Find safety factor.
**Solution:**
- Driving force = W sin(35°) = 50×0.574 = 28.7 kN
- Normal force = W cos(35°) = 50×0.819 = 40.95 kN
- Resisting force = c×A + N tan φ
- Assuming unit area: Resist = 5 + 40.95×tan(30°) = 5 + 23.6 = 28.6 kN
- FS = 28.6/28.7 = 0.996 ≈ 1.0 (marginally stable)
# # # Problem 78: Tunnel Invert Heave ✓
**Problem:** Swelling clay, swell pressure = 2 MPa, invert slab thickness = 0.5 m, concrete strength = 30 MPa. Check adequacy.
**Solution:**
- Slab stress = P×L²/(8×t) for simply supported
- For unit width: σ = 2×10⁶×(width)²/(8×0.5)
- Assuming 10m width: σ = 2×100×10⁶/4 = 50 MPa
- Since 50 > 30 MPa, slab inadequate - increase thickness
# # # Problem
12/09/2025
# Rock Mechanics Problems for Mining Engineering
# # Section 1: Stress and Strain Analysis (Problems 1-20)
# # # Problem 1: Vertical Stress at Depth ✓
**Problem:** Calculate the vertical stress at a depth of 500m in a rock mass with average density of 2.7 g/cm³.
**Solution:**
- σᵥ = ρ × g × h
- σᵥ = 2700 kg/m³ × 9.81 m/s² × 500 m
- σᵥ = 13.24 MPa
# # # Problem 2: Horizontal Stress Ratio ✓
**Problem:** If the vertical stress is 15 MPa and the horizontal stress coefficient K₀ = 0.5, calculate the horizontal stress.
**Solution:**
- σₕ = K₀ × σᵥ
- σₕ = 0.5 × 15 MPa
- σₕ = 7.5 MPa
# # # Problem 3: Principal Stress Calculation ✓
**Problem:** Given σₓ = 20 MPa, σᵧ = 15 MPa, τₓᵧ = 5 MPa, find the principal stresses.
**Solution:**
- σ₁,₂ = (σₓ + σᵧ)/2 ± √[((σₓ - σᵧ)/2)² + τₓᵧ²]
- σ₁,₂ = (20 + 15)/2 ± √[((20 - 15)/2)² + 5²]
- σ₁,₂ = 17.5 ± √(6.25 + 25) = 17.5 ± 5.59
- σ₁ = 23.09 MPa, σ₂ = 11.91 MPa
# # # Problem 4: Elastic Modulus from Stress-Strain ✓
**Problem:** A rock sample experiences 0.002 strain under 40 MPa stress. Calculate Young's modulus.
**Solution:**
- E = σ/ε
- E = 40 MPa / 0.002
- E = 20,000 MPa = 20 GPa
# # # Problem 5: Poisson's Ratio Calculation ✓
**Problem:** Axial strain = 0.001, lateral strain = -0.0003. Find Poisson's ratio.
**Solution:**
- ν = -εₗₐₜₑᵣₐₗ/εₐₓᵢₐₗ
- ν = -(-0.0003)/0.001
- ν = 0.3
# # # Problem 6: Biaxial Stress State ✓
**Problem:** Calculate strain in x-direction when σₓ = 30 MPa, σᵧ = 20 MPa, E = 25 GPa, ν = 0.25.
**Solution:**
- εₓ = (1/E)[σₓ - ν(σᵧ + σᵤ)]
- εₓ = (1/25000)[30 - 0.25(20 + 0)]
- εₓ = (1/25000)(30 - 5) = 0.001
# # # Problem 7: Shear Modulus ✓
**Problem:** If E = 30 GPa and ν = 0.2, calculate the shear modulus G.
**Solution:**
- G = E/[2(1 + ν)]
- G = 30,000/[2(1 + 0.2)]
- G = 30,000/2.4 = 12.5 GPa
# # # Problem 8: Bulk Modulus ✓
**Problem:** Calculate bulk modulus K if E = 40 GPa and ν = 0.3.
**Solution:**
- K = E/[3(1 - 2ν)]
- K = 40,000/[3(1 - 2×0.3)]
- K = 40,000/[3(0.4)] = 33.33 GPa
# # # Problem 9: Stress Around Circular Opening ✓
**Problem:** Calculate tangential stress at θ = 0° around a circular tunnel with radius 3m, at distance r = 6m, under σₒ = 10 MPa.
**Solution:**
- σₜ = σₒ(1 + (a²/r²))
- σₜ = 10(1 + (3²/6²))
- σₜ = 10(1 + 0.25) = 12.5 MPa
# # # Problem 10: Stress Concentration Factor ✓
**Problem:** Maximum stress around a circular hole is 30 MPa, far-field stress is 10 MPa. Find stress concentration factor.
**Solution:**
- SCF = σₘₐₓ/σₒ
- SCF = 30/10 = 3
# # # Problem 11: Triaxial Stress State ✓
**Problem:** Calculate octahedral normal stress for σ₁ = 40 MPa, σ₂ = 25 MPa, σ₃ = 15 MPa.
**Solution:**
- σₒᶜᵗ = (σ₁ + σ₂ + σ₃)/3
- σₒᶜᵗ = (40 + 25 + 15)/3
- σₒᶜᵗ = 26.67 MPa
# # # Problem 12: Octahedral Shear Stress ✓
**Problem:** Find octahedral shear stress for the above principal stresses.
**Solution:**
- τₒᶜᵗ = (1/3)√[(σ₁-σ₂)² + (σ₂-σ₃)² + (σ₃-σ₁)²]
- τₒᶜᵗ = (1/3)√[(40-25)² + (25-15)² + (15-40)²]
- τₒᶜᵗ = (1/3)√[225 + 100 + 625] = (1/3)√950 = 10.27 MPa
# # # Problem 13: Elastic Settlement ✓
**Problem:** A foundation load of 1000 kN acts over area 4 m². Rock has E = 20 GPa, ν = 0.25. Calculate settlement.
**Solution:**
- σ = P/A = 1000/4 = 250 kPa
- Settlement ≈ σ×B×(1-ν²)/E (simplified)
- s ≈ 0.25×2×(1-0.25²)/20,000,000 = 0.00002 m = 0.02 mm
# # # Problem 14: Thermal Stress ✓
**Problem:** Rock heated by 50°C, α = 10×10⁻⁶/°C, E = 30 GPa, constrained. Find thermal stress.
**Solution:**
- σₜₕₑᵣₘₐₗ = -α×ΔT×E
- σₜₕₑᵣₘₐₗ = -10×10⁻⁶×50×30×10⁹
- σₜₕₑᵣₘₐₗ = -15 MPa (compression)
# # # Problem 15: Strain Energy Density ✓
**Problem:** Calculate strain energy density for σ = 20 MPa, E = 25 GPa.
**Solution:**
- U = σ²/(2E)
- U = (20×10⁶)²/(2×25×10⁹)
- U = 8×10³ J/m³ = 8 kJ/m³
# # # Problem 16: Mohr Circle Construction ✓
**Problem:** σₓ = 25 MPa, σᵧ = 15 MPa, τₓᵧ = 8 MPa. Find center and radius of Mohr circle.
**Solution:**
- Center: (σₓ + σᵧ)/2 = (25 + 15)/2 = 20 MPa
- Radius: √[((σₓ - σᵧ)/2)² + τₓᵧ²] = √[(5)² + 8²] = √89 = 9.43 MPa
# # # Problem 17: Maximum Shear Stress ✓
**Problem:** From Problem 16, find maximum shear stress.
**Solution:**
- τₜₘₐₓ = Radius of Mohr circle = 9.43 MPa
# # # Problem 18: Plane Stress Transformation ✓
**Problem:** Transform stress to plane at 30° using data from Problem 16.
**Solution:**
- σₙ = (σₓ + σᵧ)/2 + (σₓ - σᵧ)cos(2θ)/2 + τₓᵧsin(2θ)
- σₙ = 20 + 5×cos(60°) + 8×sin(60°)
- σₙ = 20 + 2.5 + 6.93 = 29.43 MPa
# # # Problem 19: Effective Stress ✓
**Problem:** Total stress = 25 MPa, pore pressure = 5 MPa. Calculate effective stress.
**Solution:**
- σ' = σ - u
- σ' = 25 - 5 = 20 MPa
# # # Problem 20: Hydrostatic Stress ✓
**Problem:** Find hydrostatic stress for σ₁ = 30 MPa, σ₂ = 20 MPa, σ₃ = 10 MPa.
**Solution:**
- σₘ = (σ₁ + σ₂ + σ₃)/3
- σₘ = (30 + 20 + 10)/3 = 20 MPa
# # Section 2: Rock Strength and Failure Criteria (Problems 21-40)
# # # Problem 21: Mohr-Coulomb Failure ✓
**Problem:** Rock has c = 5 MPa, φ = 30°. Find shear strength at σₙ = 20 MPa.
**Solution:**
- τ = c + σₙ tan φ
- τ = 5 + 20 × tan(30°)
- τ = 5 + 20 × 0.577 = 16.54 MPa
# # # Problem 22: Uniaxial Compressive Strength ✓
**Problem:** Cylindrical sample (D=50mm, L=100mm) fails at 125 kN. Find UCS.
**Solution:**
- UCS = P/A = 125,000 N / (π × (0.025)² m²)
- UCS = 125,000 / 0.00196 = 63.69 MPa
# # # Problem 23: Tensile Strength from Brazilian Test ✓
**Problem:** Disk (D=50mm, t=25mm) fails at 15 kN in Brazilian test. Find tensile strength.
**Solution:**
- σₜ = 2P/(π×D×t)
- σₜ = 2×15,000/(π×0.05×0.025)
- σₜ = 30,000/0.00393 = 7.64 MPa
# # # Problem 24: Hoek-Brown Criterion ✓
**Problem:** σcᵢ = 50 MPa, mᵢ = 15, s = 1. Find σ₁ when σ₃ = 10 MPa.
**Solution:**
- σ₁ = σ₃ + σcᵢ√(mᵢσ₃/σcᵢ + s)
- σ₁ = 10 + 50√(15×10/50 + 1)
- σ₁ = 10 + 50√(3 + 1) = 10 + 100 = 110 MPa
# # # Problem 25: Griffith Criterion ✓
**Problem:** Tensile strength = 3 MPa. Find compressive strength using Griffith criterion.
**Solution:**
- σc = 8σₜ (for Griffith criterion)
- σc = 8 × 3 = 24 MPa
# # # Problem 26: Factor of Safety ✓
**Problem:** Applied stress = 15 MPa, rock strength = 45 MPa. Calculate factor of safety.
**Solution:**
- FOS = Strength/Applied Stress
- FOS = 45/15 = 3
# # # Problem 27: Cohesion from Triaxial Tests ✓
**Problem:** Two tests: (σ₁=40, σ₃=5) and (σ₁=60, σ₃=15). Find cohesion and friction angle.
**Solution:**
- From Mohr circles: slope m = tan φ
- (σ₁-σ₃)₁ = 35, (σ₁+σ₃)₁ = 45
- (σ₁-σ₃)₂ = 45, (σ₁+σ₃)₂ = 75
- tan φ = Δ(σ₁-σ₃)/Δ(σ₁+σ₃) = 10/30 = 0.333
- φ = 18.4°, c ≈ 8.7 MPa
# # # Problem 28: Point Load Index ✓
**Problem:** Point load test on 50mm diameter core fails at 2.5 kN. Find point load index.
**Solution:**
- Isₒ = P/De² (where De = equivalent diameter)
- Isₒ = 2500/(50²) = 2500/2500 = 1.0 MPa
# # # Problem 29: Estimate UCS from Point Load ✓
**Problem:** Using result from Problem 28, estimate UCS.
**Solution:**
- UCS ≈ 24 × Isₒ (empirical correlation)
- UCS ≈ 24 × 1.0 = 24 MPa
# # # Problem 30: Schmidt Hammer Rebound ✓
**Problem:** Schmidt hammer gives rebound value of 45. Estimate UCS using correlation.
**Solution:**
- UCS = 2.51e^(0.0635×R) (typical correlation)
- UCS = 2.51 × e^(0.0635×45)
- UCS = 2.51 × e^2.86 = 43.8 MPa
# # # Problem 31: Rock Quality Designation ✓
**Problem:** Core run of 1.5m yields pieces: 0.3m, 0.15m, 0.4m, 0.25m, 0.4m. Calculate RQD.
**Solution:**
- RQD = (Sum of pieces ≥ 0.1m)/Total length × 100%
- Valid pieces: 0.3, 0.15, 0.4, 0.25, 0.4 = 1.5m
- RQD = 1.5/1.5 × 100% = 100%
# # # Problem 32: Slake Durability Index ✓
**Problem:** Initial dry mass = 500g, final dry mass after two cycles = 425g. Find Id₂.
**Solution:**
- Id₂ = (Final mass/Initial mass) × 100%
- Id₂ = (425/500) × 100% = 85%
# # # Problem 33: Triaxial Test Analysis ✓
**Problem:** σ₃ = 8 MPa, failure at σ₁ = 48 MPa, φ = 25°. Find cohesion.
**Solution:**
- (σ₁ - σ₃)/2 = c×cos φ + (σ₁ + σ₃)×sin φ/2
- 20 = c×cos(25°) + 28×sin(25°)
- 20 = 0.906c + 11.83
- c = 9.02 MPa
# # # Problem 34: Direct Shear Test ✓
**Problem:** Normal stress = 12 MPa, shear stress at failure = 9 MPa, c = 2 MPa. Find φ.
**Solution:**
- τ = c + σₙ tan φ
- 9 = 2 + 12 tan φ
- tan φ = 7/12 = 0.583
- φ = 30.3°
# # # Problem 35: Crack Initiation Stress ✓
**Problem:** In UCS test, first crack appears at 35% of peak strength. Peak strength = 80 MPa. Find crack initiation stress.
**Solution:**
- σcᵢ = 0.35 × σc
- σcᵢ = 0.35 × 80 = 28 MPa
# # # Problem 36: Crack Damage Stress ✓
**Problem:** Crack damage occurs at 70% of peak strength. Peak = 60 MPa. Find crack damage stress.
**Solution:**
- σcd = 0.70 × σc
- σcd = 0.70 × 60 = 42 MPa
# # # Problem 37: Brittleness Index ✓
**Problem:** σc = 80 MPa, σₜ = 6 MPa. Calculate brittleness index B₁.
**Solution:**
- B₁ = (σc - σₜ)/(σc + σₜ)
- B₁ = (80 - 6)/(80 + 6) = 74/86 = 0.86
# # # Problem 38: Strength Anisotropy ✓
**Problem:** Strength parallel to foliation = 70 MPa, perpendicular = 45 MPa. Find anisotropy ratio.
**Solution:**
- Anisotropy ratio = σc‖/σc⊥
- Ratio = 70/45 = 1.56
# # # Problem 39: Confinement Effect ✓
**Problem:** σc at σ₃=0 is 50 MPa, at σ₃=10 MPa is 85 MPa. Find strength increase per MPa confinement.
**Solution:**
- Strength increase = (85-50)/(10-0) = 3.5 MPa per MPa confinement
# # # Problem 40: Failure Envelope ✓
**Problem:** Plot failure envelope for c = 8 MPa, φ = 32°. Find shear strength at σₙ = 25 MPa.
**Solution:**
- τ = c + σₙ tan φ
- τ = 8 + 25 × tan(32°)
- τ = 8 + 25 × 0.625 = 23.6 MPa
# # Section 3: Rock Mass Classification (Problems 41-60)
# # # Problem 41: RMR Basic Calculation ✓
**Problem:** UCS=65MPa, RQD=75%, Spacing=0.8m, Condition=slightly rough, GW=damp, Orientation=fair. Calculate basic RMR.
**Solution:**
- UCS (65 MPa): 7 points
- RQD (75%): 13 points
- Spacing (0.8m): 10 points
- Condition: 25 points
- Groundwater (damp): 10 points
- Basic RMR = 7+13+10+25+10 = 65
# # # Problem 42: RMR Orientation Adjustment ✓
**Problem:** Basic RMR = 65 (from Problem 41), unfavorable orientation for tunnels. Find adjusted RMR.
**Solution:**
- Orientation adjustment for tunnels (unfavorable): -10
- Adjusted RMR = 65 - 10 = 55
# # # Problem 43: Q-System Basic ✓
**Problem:** RQD=80%, Jn=9, Jr=3, Ja=2, Jw=1.0, SRF=1.0. Calculate Q value.
**Solution:**
- Q = (RQD/Jn) × (Jr/Ja) × (Jw/SRF)
- Q = (80/9) × (3/2) × (1.0/1.0)
- Q = 8.89 × 1.5 × 1.0 = 13.33
# # # Problem 44: Rock Mass Rating to GSI ✓
**Problem:** RMR₈₉ = 58. Convert to GSI.
**Solution:**
- GSI ≈ RMR₈₉ - 5 (for good conditions)
- GSI ≈ 58 - 5 = 53
# # # Problem 45: Hoek-Brown for Rock Mass ✓
**Problem:** σcᵢ=60MPa, mᵢ=12, GSI=45, D=0. Find mb, s, and a.
**Solution:**
- mb = mᵢ × exp((GSI-100)/28)
- mb = 12 × exp((45-100)/28) = 12 × 0.176 = 2.11
- s = exp((GSI-100)/9) = exp(-6.11) = 0.002
- a = 0.5 + (1/6)(e^(-GSI/15) - e^(-20/3)) = 0.516
# # # Problem 46: Rock Mass Strength ✓
**Problem:** Using data from Problem 45, find rock mass UCS.
**Solution:**
- σc = σcᵢ × s^a
- σc = 60 × (0.002)^0.516
- σc = 60 × 0.032 = 1.92 MPa
# # # Problem 47: Deformation Modulus ✓
**Problem:** σcᵢ=50MPa, GSI=40, D=0, Ei=20GPa. Find rock mass deformation modulus.
**Solution:**
- Em = Ei × (0.02 + (1-D/2)/(1+e^((60+15D-GSI)/11)))
- Em = 20 × (0.02 + 0.5/(1+e^((60-40)/11)))
- Em = 20 × (0.02 + 0.5/7.39) = 20 × 0.088 = 1.76 GPa
# # # Problem 48: Tunnel Support Pressure ✓
**Problem:** Q=4.5, tunnel span=8m, ESR=1.6. Estimate support pressure using Q-system.
**Solution:**
- Proof = 2×Jn×Q^(-1/3)/Jr (approximately)
- For Q=4.5, typical support pressure ≈ 0.1-0.4 MPa
- Estimated support pressure ≈ 0.2 MPa
# # # Problem 49: Stand-up Time ✓
**Problem:** Tunnel span = 6m, RMR = 45. Estimate stand-up time.
**Solution:**
- Stand-up time = RMR₈₉/(span^k) where k≈0.4
- Stand-up time ≈ 45/6^0.4 ≈ 45/2.3 ≈ 20 hours
# # # Problem 50: Rock Load Height ✓
**Problem:** Q = 2.5, tunnel width = 10m. Calculate rock load height.
**Solution:**
- Ht = B/Q^0.33 (approximately)
- Ht = 10/2.5^0.33 = 10/1.36 = 7.35 m
# # # Problem 51: Joint Condition Rating ✓
**Problem:** Joint separation = 1mm, slightly weathered, smooth surfaces, hard infill
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