Intact rock versus rock mass
A hand specimen of granite may be stronger than the best concrete. A cliff of the same granite can still collapse. The difference is between intact rock, the solid material between fractures, and the rock mass, the intact blocks together with the joints, bedding planes, faults and weathered zones that cut through them. Engineering works deal with the rock mass.
Intact rock strength is measured in the laboratory: uniaxial compressive strength on cylindrical cores, and the quicker point load index, which crushes a core or lump between two cones and correlates roughly with compressive strength. Intact rock is also characterised by its density, porosity, modulus and durability, for example its resistance to slaking in water.
The discontinuities control the mass. Their orientation relative to a slope or tunnel decides whether blocks can slide or fall; their spacing decides the block size; their roughness, aperture and infilling decide their shear strength; and water in them adds pressure and can soften infilling clays. A rock mass is therefore weaker, more deformable and more permeable than its intact pieces, and its properties depend on scale: a small sample may contain no joints, a large excavation many.
Engineers bridge the gap with rock mass classification systems and with strength criteria that downgrade intact properties according to the quality of the mass. Field mapping of joints, core logging, and in-situ tests such as plate loading and pressuremeter tests give the information needed.
Log discontinuity orientation, spacing and condition as carefully as intact strength.
Point load index: I_s = P / D_e² (D_e = equivalent core diameter)Size-corrected index I_s(50) for a 50 mm coreRough correlation: UCS ≈ (20 to 25) × I_s(50)
Designing a slope or tunnel from intact core strength alone.
Problem. A 50 mm diameter core fails in a diametral point load test at 9.0 kN. Find the point load index and a rough estimate of uniaxial compressive strength using a factor of 22.
- I_s = P / D² = 9 000 / 50² = 9 000 / 2 500 = 3.6 MPa; for a 50 mm core no size correction is needed, so I_s(50) = 3.6 MPa.
- UCS ≈ 22 × 3.6 = 79 MPa.
- The factor varies with rock type, so it should be calibrated against a few uniaxial tests.
Answer. I_s(50) = 3.6 MPa; UCS roughly 80 MPa
Quick quiz
1Why is a rock mass weaker than intact rock?
Discontinuities such as joints and faults cut it into blocks with weak surfaces.
2What does the point load test measure?
A strength index from crushing a core or lump between two cones.
3Why does scale matter in rock?
Larger volumes contain more discontinuities.
- BS 5930 (UK): Code of practice for ground investigations.
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
Found an error? Write to indiacongroupofengineers [at] gmail [dot] com. For corrections and feedback. We never ask for payment or personal details.