# Soil Model of Foundation Overlap

### Technical Article

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
E0 = modulus of elasticity = ${\ mathrm E} _ \ mathrm s \; ⋅ \; \ frac {\ displaystyle1 \; - \; \ mathrm \ mu \; - \; 2 \; ⋅ \; \ mathrm \ mu²} {1 \; - \; \ mathrm \ mu}$
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 overlap to be applied for this method should ideally extend until the settlements at the edge of the foundation overlap approach zero. In addition, the additional area should not have any governing additional stiffness, which is why the thickness of the foundation collar should be kept very small.

In addition to a short calculation time, another advantage of this option is the consideration of the shear resistance. Furthermore, you can use this method to graphically display the settlement behavior outside the foundation edge. In this way, it is also possible to display the interactions of several independent structures that interact with each other via the settlement depression.

#### Example:

E0 = 10,000 kN/m 2
μ = 0.2
H = 3m

$$\begin{array}{l}{\mathrm c}_{1,\mathrm z}\;=\;\frac{\displaystyle{\mathrm E}_0}{\mathrm H\;⋅\;(1\;-\;2\;⋅\;\mathrm\mu²)}\;=\;\frac{\displaystyle10.000}{3\;⋅\;(1\;-\;2\;⋅\;0,2²)}\;=\;3.623,19\;\mathrm{kN}/\mathrm m²\\{\mathrm c}_{2,\mathrm v}\;=\;{\mathrm E}_0\;⋅\;\frac{\displaystyle\mathrm H}{6\;⋅\;(1\;+\;\mathrm\mu)}\;=\;1.000\;⋅\;\frac{\displaystyle3}{6\;⋅\;(1\;+\;0,2)}\;=\;4.166,67\;\mathrm{kN}/\mathrm m²\end{array}$$