The Houska method¶
The Houska method [1] results from the modification of the Terzaghi method [2] for a tunnel. Houska assumes that the ground does not always fully plastify, that the horizontal displacements are not large enough to mobilize the active earth pressure and that a full failure does not occur, so that the full shear resistance is not mobilized, as assumed by Terzaghi. As a result, lateral sliding wedges are formed which offer greater resistance to the vertical loading. The method is phenomenological and is briefly outlined in what follows. For the sake of simplicity only a tunnel with a circular crosssection is considered and it is assumed that no additional overburden is present at the surface. The approach is valif for depths above the roof larger than three times the tunnel radius.
The stresses to be calculated are the ones of the left half of the following figure and the geometry considered is the one of the right half.
The shear strength of the soil is reduced, i.e.
$c' = 2/3 \cdot c $
$\phi' = 2/3 \cdot \phi$
Following the silo theory, a reduced overburden pressure $ p_z$ acts over the tunnel roof:
$ p_{z} = \dfrac{\gamma \cdot B' - c'}{K_{\sigma} \tan(\phi')} \left( 1- \exp\left(-K_{\sigma}\cdot\tan(\phi') \dfrac{z}{B'}\right)\right)$
where $K_{\sigma}$ is as a rule assumed equal to unity following a recommendation by Terzaghi. Due to load redistribution over the springlines (an assumption made by Houska and based on observations), $p_{z}$ increases there over a width of $d'$ by $\sigma'$ :
$\sigma' = \left(\gamma \cdot h - p_z\right) \dfrac{R}{d'}$
where
$ d' = \begin{cases} B'- R_T + h_u\tan(\phi'),& h \cdot \tan(\phi') \geq B'-R \\ R\left(\tan(a')+\dfrac{1}{\cos(a')}-1\right), & h \cdot \tan(\phi') < B'-R \end{cases} $
with $a' = \dfrac{\pi}{4} - \dfrac{\phi'}{2}$. Outside of this area, the overburden determines the vertical load $\sigma_{\textrm{v}} = \gamma \cdot h$.
The horizontal load up to a depth of $h+z_1$, where $z_1 = R\cdot \sin(a')$ is given by
$p_x = \left( \gamma\cdot h +\sigma' + \gamma\cdot z'\right)\tan^2 (a') -2c'\tan(a')$
where $z'$ is assumed zero at the tunnel roof and increases downwards. Past that depth its value remains constant.
In the widget below the acting loads are shown for varying parameters of the problem.
