Supported Axisymmetric Tunnels Within Linear Viscoelastic Burgers Rocks

Abstract

An exact closed form solution is derived for the mechanical behaviour of a linear viscoelastic Burgers rock around an axisymmetric tunnel, supported by a linear elastic ring. Analytical formulae are provided for the displacement of the rock/lining interface and for the pressure exerted by the rock on the lining, taking into account the stiffness and its installation time. Results calculated from these formulae do validate the corresponding numerical results of a 2D finite differences code. Further, comparison to previous existing solutions for the same viscoelastic model indicates similarities and differences. A parametric study is performed to investigate the effect of the viscoelastic constants, the stiffness and installation time of the support. The derived closed form solution is used to construct the time-dependent Supported Ground Reaction Curves of the viscoelastic rock, i.e. the time contour plots on the convergence confinement diagram. The importance of the effect of the support on the restrained rock creep and the exerted pressure on the lining, during the design life of a structure, is examined.

Appendix

The constitutive equations of a linear viscoelastic Burgers rock in cylindrical coordinates are given by Eqs. 6 through 8. For plane strain conditions:

$$ \varepsilon_{zz} (t) = 0 $$

(34)

Inserting (34), (9) and (10) in (8) and solving for σ _zz (t) yield

$$ \sigma_{zz} (t) = \frac{3KP - Q}{6KP + Q}\left[ {\sigma_{rr} (t) + \sigma_{\theta \theta } (t)} \right] $$

(35)

From (6) and (7) using (34) and (9):

$$ \sigma_{\theta \theta } (t) = \left( {K + \frac{2}{3}\frac{Q}{P}} \right)\varepsilon_{\theta \theta } (t) + \left( {K - \frac{1}{3}\frac{Q}{P}} \right)\varepsilon_{rr} (t) $$

(36)

$$ \sigma_{rr} (t) = \left( {K + \frac{2}{3}\frac{Q}{P}} \right)\varepsilon_{rr} (t) + \left( {K - \frac{1}{3}\frac{Q}{P}} \right)\varepsilon_{\theta \theta } (t) $$

(37)

(35), (36) and (37) are the constitutive equations for plane strain. Writing these equations with strain as a function of stress yields:

$$ \frac{Q}{P}\varepsilon_{\theta \theta } (t) = \frac{2Q + 3KP}{Q + 6KP}\sigma_{\theta \theta } (t) + \frac{Q - 3KP}{Q + 6KP}\sigma_{rr} (t) $$

(38)

$$ \frac{Q}{P}\,\varepsilon_{rr} (t) = \frac{2Q + 3KP}{Q + 6KP}\sigma_{rr} (t) + \frac{Q - 3KP}{Q + 6KP}\sigma_{\theta \theta } (t)\, $$

(39)

$$ \varepsilon_{zz} (t) = 0 $$

(40)

The equilibrium equation for axisymmetry in cylindrical coordinates is

$$ \frac{{{\text{d}}\sigma_{rr} (t)}}{{{\text{d}}r}} + \frac{{\sigma_{rr} (t) - \sigma_{\theta \theta } (t)}}{r} = 0,\quad \sigma_{\theta \theta } (t) = \frac{{{\text{d}}^{2} \Upphi (t)}}{{{\text{d}}r^{2} }},\quad \sigma_{rr} (t) = \frac{1}{r}\frac{{{\text{d}}\Upphi (t)}}{{{\text{d}}r}} $$

(41)

where Φ(t) is the Airy stress function for the viscoelastic problem. The compatibility condition is given by

$$ \frac{{{\text{d}}\varepsilon_{\theta \theta } (t)}}{{{\text{d}}r}} + \frac{{\varepsilon_{\theta \theta } (t) - \varepsilon_{rr} (t)}}{r} = 0 $$

(42)

From (38), (39):

$$ \frac{{{\text{d}}\varepsilon_{\theta \theta } (t)}}{{{\text{d}}r}} = \frac{P}{Q}\left( {\frac{2Q + 3KP}{Q + 6KP}\frac{{{\text{d}}\sigma_{\theta \theta } (t)}}{{{\text{d}}r}} + \frac{Q - 3KP}{Q + 6KP}\frac{{{\text{d}}\sigma_{rr} (t)}}{{{\text{d}}r}}} \right) $$

(43)

$$ \frac{{\varepsilon_{\theta \theta } (t) - \varepsilon_{rr} (t)}}{r} = \frac{P}{Q}\left( {\frac{{\sigma_{\theta \theta } (t) - \sigma_{rr} (t)}}{r}} \right) $$

(44)

Using the equation of equilibrium in (44):

$$ \frac{{\varepsilon_{\theta \theta } (t) - \varepsilon_{rr} (t)}}{r} = \frac{P}{Q}\left( {\frac{{{\text{d}}\sigma_{rr} (t)}}{{{\text{d}}r}}} \right) $$

(45)

Replacing the strains with stresses from the constitutive equations into the equation of compatibility, we derive:

$$ \frac{\text{d}}{{{\text{d}}r}}[\sigma_{\theta \theta } (t) + \sigma_{rr} (t)] = 0 $$

(46)

or, after some mathematical manipulation:

$$ \left( {\frac{{{\text{d}}^{2} }}{{{\text{d}}r^{2} }} + \frac{1}{r}\frac{\text{d}}{{{\text{d}}r}}} \right)[\sigma_{\theta \theta } (t) + \sigma_{rr} (t)] = 0 $$

(47)

Replacing σ _θθ (t) and σ _rr (t) from (41) into (47):

$$ \left( {\frac{{{\text{d}}^{2} }}{{{\text{d}}r^{2} }} + \frac{1}{r}\frac{\text{d}}{{{\text{d}}r}}} \right)\left( {\frac{{{\text{d}}^{2} }}{{{\text{d}}r^{2} }} + \frac{1}{r}\frac{\text{d}}{{{\text{d}}r}}} \right)\Upphi (t) = 0 \to \nabla^{2} (\nabla^{2} \Upphi ) = 0 $$

(48)

From (48) it is concluded that the stress distribution in the rock is the same whether the rock material is linear elastic or linear viscoelastic.