Prediction of settlement trough induced by tunneling in cohesive ground

Abstract

Surface settlements of soil due to tunneling are caused by stress relief and subsidence due to movement of support by excavation. There are significant discrepancies between empirical solutions to predict surface settlement trough because of different interpretations and database collection by different authors. In this paper, the shape of settlement trough caused by tunneling in cohesive ground is investigated by different approaches, namely analytical solutions, empirical solutions, and numerical solutions by the finite element method. The width of settlement trough was obtained by the finite element method through establishing the change in the slope of the computed settlement profile. The finite element elastic-plastic analysis gives better predictions than the linear elastic model with satisfactory estimate for the displacement magnitude and slightly overestimated width of the surface settlement trough. The finite element method overpredicted the settlement trough width i compared with the results of Peck for soft and stiff clay, but there is an excellent agreement with Rankin’s estimation. The results show that there is a good agreement between the complex variable analysis for Z/D = 1.5, while using Z/D = 2 and 3, the curve diverges in the region faraway from the center of the tunnel.

Appendix

1.1 Mindlin’s problem

In the ξ-plane, the function F is to be considered along the boundary $\xi = \alpha \sigma = \alpha \exp (i\theta )$

$$ F = i\int\limits_{0}^{s} {(t_{x} } + it_{z} )\,{\text{d}}s $$

(15)

where the integration path should be such that the material (inside the cavity) lies to the left when traveling along the integration path. This means that ds = −rdβ. Because on the boundary of the cavity z = −h−r cos β, it follows that the boundary function F for this part of the solution, which will be denoted by F ₁, is:

$$ \frac{{F_{1} }}{\gamma \cdot r} = - \int\limits_{0}^{\beta } {\left[ {h\cos \beta + r\cos^{2} \beta } \right]\,{\text{d}}\beta - ik_{0} } \int\limits_{0}^{\beta } {\left[ {h\sin \beta = r\sin \beta \cos \beta } \right]} \,{\text{d}}\beta. $$

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Elaboration of the integrals gives

$$ \frac{{F_{1} }}{\gamma\cdot r} = - \frac{1}{2}r\beta - \frac{1}{4}r\sin 2\beta - h\sin \beta - ik_{0} \left[ {h(1 - \cos \beta ) + \frac{1}{4}r(1 - \cos 2\beta )} \right] $$

(17)

The expression (17) gives the integrated surface traction F ₁ as a function of β, the angular coordinate along the circular boundary in the z plane. This quantity is needed, however, as a function of θ, the angle along the inner circular boundary in the ξ-plane, for that reason the relation between these two angles:

$$ \theta = w(\xi ) = - ih\frac{{1 - \alpha^{2} }}{{1 + \alpha^{2} }}\frac{1 + \xi }{{1 - \xi^{2} }} $$

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where α is the geometric parameter r/h, and the ratio of the radius of the circular cavity to its depth is as follows:

$$ \frac{r}{h} = \frac{2\alpha }{{1 + \alpha^{2} }} $$

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Along the inner circular boundary in ξ-plane ξ = ασ = α.exp (iθ), with equation (18) this gives, after some mathematical manipulations:

$$ \frac{x}{r} = \frac{{(1 - \alpha^{2} )\sin \theta }}{{(1 + \theta^{2} ) - 2\alpha \cos \theta }} $$

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$$ \frac{z + h}{r} = - \frac{{(1 + \alpha^{2} )\cos \theta - 2\alpha }}{{(1 + \alpha^{2} ) - 2\alpha \cos \theta }} $$

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These equations represent a circle of radius r around a point at a depth h, because:

$$ x^{2} + \left( {z + h} \right)^{2} = r. $$

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The angle β can be expressed as:

$$ \tan \beta = \frac{x}{z + h} = \frac{{(1 - \alpha^{2} \sin \theta )}}{{(1 + \alpha^{2} )\cos \theta - 2\alpha }} $$

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The value of β is defined such that it varies continuously from β = 0 to β = 2π in the interval from θ = 0 to θ = 2π. This can be accomplished by modifying the standard function l = arc tan (z/x) in such a way that for values of x and z in four quadrants, the value of the function varies continuously from 0 to 2π when θ = 0 to θ = 2π.

The function F ₁ can be calculated from the formula (17), which can be written as:

$$ \frac{{F_{1} }}{{\gamma \pi r^{2} }} = \frac{\beta }{2\pi } - \frac{\sin 2\pi }{4\pi } - \frac{1}{\pi }\frac{h}{r}\sin \beta - \frac{{iK_{0} }}{\pi }\left[ {\frac{h}{r}(1 - \cos \beta ) + \frac{1}{4}(1 - \cos 2\beta )} \right] $$

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