How to model a column, a foundation pile, a wall or a foundation pit as a spring?

Contents

The goal of this article is to translate the stiffness of a vertical elements (column, wall, surface support) to a spring constant $k$ which can be applied in the Y-direction in Diamonds.

For a point support (column or foundation pile)

With Hooke’s Law $F=k \cdot \Delta x$ and the formula for stress $\sigma=\frac{F}{A}$ , a formula for the spring constant can be deducted. This formula assumes that the element is supported at the buttom.

$k=\frac{F}{\Delta x} = \frac{\sigma A}{\Delta x} = \frac{E \varepsilon A}{\Delta x} = \frac{E (\Delta x /l) A}{\Delta x} = \frac{E A}{l}$

with

$E$ the modulus of elasticity of the column
$A$ the cross-section area of the column
$l$ the length of the column

Worked example for a column

Find the spring constant for a concrete column C25/30 of 300x300mm and 3m high.

$\begin{align*} k &=\frac{E A}{l} = \frac{30 000N/mm^2 \cdot 300mm \cdot 300mm}{3000mm}\\ &=900 000N/mm=900 000kN/m\\ \end{align*}$

This value must be entered in

Worked example for a foundation pile

Find the spring constant for a circular foundation pile C25/30 with a diameter of 500mm, 8m deep.

In case of a foundation pile, there are two springs in series: the one for the pile and one for the soil

$\frac{1}{k}= \frac{1}{k_p} + \frac{1}{k_g}$

If there is no data available on the soil, but you still want to take the effect into account, you could assume that $k_g=k_p$

$k= \frac{k_p}{2} = 0.5 \cdot \frac{E A}{l}$

The spring constant than becomes:

$\begin{align*} k &=0.5 \cdot \frac{E A}{l} = 0.5 \cdot \frac{30 000N/mm^2 \cdot \pi \cdot (250mm)^2}{8000mm}\\ &=468 750N/mm=468 750kN/m\\ \end{align*}$

This value must be entered in .

For a line support (wall)

Again Hook’s law and the formula for stress are used. But since the length the wall will be modelled as a line support, this dimension is no longer in the formula:

$k= \frac{E b}{l}$

with

$E$ the modulus of elasticity of the wall
$b$ the thickness of the wall
$l$ the height of the wall

Worked example for a wall

Find the spring constant for a masonry wall (6170N/mm^2) 14cm thick and 2,8m high.

$\begin{align*} k &=\frac{E b}{l} = \frac{6170N/mm^2 \cdot 140mm}{2800mm}\\ &=308500kN/m/m = 308500kN/m^2\\ \end{align*}$

This value must be entered in

For a surface support (foundation pit)

Again, same principle, but both dimensions of the surface support will be modelled. Which gives this as the formula for the spring constant:

$k= \frac{E}{l}$

with

$E$ the modulus of elasticity of the surface support
$l$ the height of the pit

Remark

Replacing surface supports (like foundation pits) with springs, implies that you to model the circumference of the foundation pit. Therefor it is advised to apply this type of spring constant only to smaller models. For example: in a model that only contains the foundation slab and not all floors above it.

Worked example for a foundation pit

Find the spring constant for a circular concrete foundation pit C12/15 (E=27000N/mm²), diameter 1.8m and 5m deep.

To take the rigidity of the soil below the foundation pit into account, we multiply the formule with 0,5 (see worked example of foundation pile above).

$k= 0.5 \cdot \frac{E}{l} = 0.5 \cdot \frac{27 000N/mm^2}{5000mm}=2.7N/mm^3=2.7 \cdot 10^6kN/m^3$

This value must be entered in

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For a point support (column or foundation pile)

For a line support (wall)

For a surface support (foundation pit)

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