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Strength criteria for rock

Intermediate Rock Mechanics Reviewed 24 Oct 2026 2 min read

A strength criterion states the combination of stresses at which rock fails. The simplest, Mohr-Coulomb, uses a cohesion and a friction angle, and in principal-stress form gives the major principal stress at failure as the uniaxial strength plus the minor principal stress times a passive-type coefficient. It is linear, easy to use and built into most software, but real rock shows a curved strength envelope: confinement helps a lot at low stress and less at high stress.

The Hoek-Brown criterion captures that curvature. For intact rock it uses the uniaxial compressive strength and a material constant m_i, which ranges from single figures for fine-grained, weak rocks to above 30 for coarse granites. For rock masses, the generalised form reduces the constants according to the Geological Strength Index and a disturbance factor for blast damage or stress relief, so that a heavily jointed mass is much weaker than its intact blocks.

Diagram: Indiacon original, not to scale.

Discontinuities have their own criteria. A smooth planar joint behaves roughly frictionally. Rough joints dilate as they slide, so their strength is higher at low normal stress; Barton's relation expresses this through a joint roughness coefficient and the joint wall strength. Infilled joints may be governed by the infill clay.

Which criterion governs depends on the problem: intact rock for pillars and small blocks, the rock mass for large slopes and tunnels in jointed ground, and a single discontinuity for wedge and planar failures.

Tip

Fit Mohr-Coulomb parameters to Hoek-Brown over the actual stress range of your problem, not over all stresses.

Key formulas
  • Mohr-Coulomb: σ₁ = σ_c + σ₃ tan²(45° + φ / 2)
  • Hoek-Brown (intact): σ₁ = σ₃ + σ_ci √(m_i σ₃ / σ_ci + 1)
  • Joint shear strength (frictional): τ = σ_n tan φ_j
Common mistake

Using intact rock parameters for a large excavation in jointed rock.

Solved example

Problem. An intact granite has σ_ci = 100 MPa and m_i = 32. Find its strength at a confining stress of 5 MPa by Hoek-Brown.

  1. m_i σ₃ / σ_ci = 32 × 5 / 100 = 1.6.
  2. √(1.6 + 1) = √2.6 = 1.612.
  3. σ₁ = 5 + 100 × 1.612 = 166.2 MPa.
  4. Confinement of only 5 MPa raises the strength by about 66 percent over the uniaxial value.

Answer. About 166 MPa

Quick quiz

1Why is Mohr-Coulomb only approximate for rock?

Real rock has a curved strength envelope.

2What does m_i reflect?

Rock type and texture; coarse crystalline rocks have high values.

3Which strength governs a planar slide along bedding?

The shear strength of that discontinuity.

Code note
  • BS 5930 (UK): Code of practice for ground investigations.
  • EN 1997 (Eurocode 7) (Europe): Geotechnical design.

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • R. E. Goodman, Introduction to Rock Mechanics, Wiley, 2nd edition.
  • E. Hoek, E. T. Brown, Underground Excavations in Rock, Institution of Mining and Metallurgy, 1980.

Free and official sources

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