On a Class of Contact Problems in Rock Mechanics Exadaktylos George, Technical University of Crete

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On a Class of Contact Problems in Rock Mechanics Exadaktylos George, Technical University of Crete

Acknowledgements We would like to thank the financial support from the EU 5 th Framework Project “Integrated tool for in situ characterization of effectiveness and durability of conservation techniques in historical structures” (DIAS) with Contract Number: DIAS-EVK4-CT

A class of plane contact problems in Rock Mechanics includes: 1.The interaction of a thin liner with a circular opening in an elastic, isotropic, homogeneous rock (design of tunnel support) 2.The indentation of rocks (design of cutting tools (contact stresses), characterization of elasticity of rocks) 3.The cutting of rocks (design of cutting tools, characterization of strength of rocks) Mixed Boundary Value Problems

Why considering these 3 problems simultaneously ? Because in underground construction the rock excavation precedes every other work (design of proper cutting tools, i.e. picks, discs, drag bits etc. & operational parameters of machine). Rock cutting gives precious information for the strength of the intact rock whereas indentation for its elasticity at mesoscale (~ 1mm -100 mm). Support must be manufactured by considering rock deformability and strength.

Problem #1: Elastic interaction of a thin shell in perfect contact with a circular opening Add BC’s Equilibrium eqn’s for the shell (Kirchhoff-Love) Constitutive relations

Method of solution Kolosov-Muskhelishvili complex variable method 1) 2) 3)

Numerical implementation Ivanov (1976) System of (8n+4) eqns with (8n+4) unknowns n=20

Comparison with classical analytic solution by Savin (1961) for ‘welded’ elastic ring Discre-pancy ? Other References: Einstein and Schwartz (1979), Bobet (2001)

Why choosing the Complex Variable technique ? Stress Intensity factors at crack tips: K I, K II

Interaction of 2 straight cracks with supported hole System of (16n+8) eqns with (16n+8) unknowns

Problem #2: Rock indentation by DIAS portable indentor where k is the ‘penetration stiffness’ with dimensions

Elasticity of mtl from back-analysis of indentation test data (analytical solution by Lur’e, 1964) Surface waves Indentation Ø=2.5 mm

Recurrent loading-unloading cycles Ø=2.5 mm More complex σ-ε paths ?

Problem #3: Rock cutting by drilling 2 nd generation of DFMS with tripod 3 rd generation of DFMS [light instrument with jackleg (like jackhammer)] Ø=5 mm

WOB-Torque measurements Normal & tangential forces during drilling are linearly constrained Each point is a test with different cutting depth δ

Numerical modeling of rock cutting by drilling (Stavropoulou, 2005) DIAS EU R&D Project ( comminution Elasto-visco-plastic cutting model (FLAC 2D ) vxvx

Comparison of numerical simulations with experimental drilling data (Stavropoulou, 2005) Remark #1: Initiation of strain localization Remark #2: c,φ for numerical modeling estimated from triaxial compression tests in lab (ψ=0 o )

An approach to design structures in brittle rock masses: 2.Elasticity & strength of intact rock (L =.001 – 0.1 m) - Fast drilling/indentation/ acoustic measurements 3.Rock transected by cracks (L=.1 – 100 m) - LEFM (fast algorithms) for stress analysis, K I,K II,K III estimations & check of micromech – damage models ! -Stiffness and strength of joint walls (another contact problem) 4.Support -DIAS measurements -Modeling Hoek, Kaiser & Bawden (1995) 1. Excavation

END

System of complex integro-differential eqns Note from the 1 st eqn that for the limit of zero relative rigidity or thickness of the shell the radial and tangential stresses vanish Boundary element method

Limit for relative rigidity of the thin shell tending to infinity

Frictional contact of a gently dipping rigid slider with an elastic half-plane Boundary conditions 1) 2) 3)

Analytical solution (Muskhelishvili, 1963) Normal force varies proportionally with indentation depth

Graphical illustration of the solution Remark #1 Remark #2 tanφ=0, tanφ=0.5 tanφ=1 tanφ=0, tanφ=0.5 tanφ=1 Remark #3