Soil Model of Foundation Overlap

Technical Article

001433

05/03/2017

previous article presented different variants of surface elastic foundations in addition to the traditional subgrade reaction modulus method. The following article describes another method for surface foundation. This method considers the adjacent ground areas by means of a foundation overlap. In this case, foundation parameters refer to the continuing works by Pasternak and Barwaschow.

Equation According to Pasternak

$$\begin{array}{l}{\mathrm c}_1\;=\;\frac{\displaystyle{\mathrm E}_0}{\mathrm H\;⋅\;(1\;-\;2\;⋅\;\mathrm\mu²)}\\{\mathrm c}_2\;=\;{\mathrm E}_0\;⋅\;\frac{\displaystyle\mathrm H}{6\;⋅\;(1\;+\;\mathrm\mu)}\end{array}$$
where
$${\mathrm E}_0\;=\;\mathrm{elastic}\;\mathrm{modulus}\;=\;{\mathrm E}_\mathrm s\;⋅\;\frac{\displaystyle1\;-\;\mathrm\mu\;-\;2\;⋅\;\mathrm\mu²}{1\;-\;\mathrm\mu}$$
H = foundation thickness
μ = Poisson's ratio

Equation According to Barwaschow

$$\begin{array}{l}{\mathrm c}_1=\;\frac{\displaystyle{\mathrm E}_0}{\mathrm H\;⋅\;(1\;-\;\mathrm\mu²)}\\{\mathrm c}_2\;=\;{\mathrm E}_0\;⋅\;\frac{\displaystyle\mathrm H}{20\;⋅\;(1\;-\;\mathrm\mu²)}\end{array}$$
where
$${\mathrm E}_0\;=\;{\mathrm E}_\mathrm s\;⋅\;\frac{\displaystyle1\;-\;\mathrm\mu\;-\;2\;⋅\;\mathrm\mu²}{1\;-\;\mathrm\mu}$$
H = foundation thickness
μ = Poisson's ratio

The foundation overlaps applied to this method should ideally reach far enough until the settlement on the edge of the foundation overlap is close to zero. Moreover, the additional area should not have any additional governing stiffness, which is why the foundation overlap thickness should be kept very low.

In addition to a short calculation time, a further advantage of this variant is its consideration of the shear resistance. Furthermore, this method allows you to graphically display the settlement behavior outside of the foundation edge. In this way, it is also possible to represent the interaction between several separate buildings which have an influence on each other via the subsidence basin.

Example

E0 = 10,000 kN/m2
μ = 0.2
H = 3 m

$$\begin{array}{l}{\mathrm c}_{1,\mathrm z}\;=\;\frac{\displaystyle{\mathrm E}_0}{\mathrm H\;⋅\;(1\;-\;2\;⋅\;\mathrm\mu²)}\;=\;\frac{\displaystyle10,000}{3\;⋅\;(1\;-\;2\;⋅\;0.2²)}\;=\;3,623.19\;\mathrm{kN}/\mathrm m²\\{\mathrm c}_{2,\mathrm v}\;=\;{\mathrm E}_0\;⋅\;\frac{\displaystyle\mathrm H}{6\;⋅\;(1\;+\;\mathrm\mu)}\;=\;1,000\;⋅\;\frac{\displaystyle3}{6\;⋅\;(1\;+\;0.2)}\;=\;4,166.67\;\mathrm{kN}/\mathrm m²\end{array}$$

Reference

 [1] Barth, C. & Rustler, W. (2013). Finite Elemente in der Baustatik-Praxis (2nd ed.). Berlin: Beuth. [2] Kolar, V.; Nemec, I.: Modelling of Soil-Structure Interaction. Amsterdam: Elsevier Science Publishers, 1989