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 column

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

Input

Young’s modulus E [N/mm²]

Width b [mm]

Length h [mm]

Level height H [mm]

k = E × b × h H = 30 000 × 300 × 300 3 000 = 900 000kN/m

Note: E is in N/mm², b, h and H are in mm; the result is converted to kN/m.

This value must be entered in :

For a foundation pile

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

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

If there is no data available on the soil, but you still want to take the effect into account, you could assume that $k_g=k_p$ . The formula for the spring contant becomes:

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

The "2" is refered to as the stiffenss ratio $s$ in the calculator below.

Input

Young’s modulus E [N/mm²]

Width b [mm]

Length h [mm]

Level height H [mm]

Stiffness ratio s [-]

k = E · b · h s · H = 27 000 · 300 · 300 2 · 3 000 = 405 000kN/m

Note: E is in N/mm² and b, h and H are in mm, so (E·b·h)/(s·H) is in N/mm; 1 N/mm = 1 kN/m.

This value must be entered in :

For a 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

Input

Young’s modulus E [N/mm²]

Width b [mm]

Level height H [mm]

k = E × b H = 6 170 × 140 3 000 = 287 933kN/m²

Note: E is in N/mm², b and H are in mm; the result is converted to kN/m².

This value must be entered in :

For a foundation pit (supported area)

Again, there are two springs in series: the one for the pit $k_p$ and one for the soil $k_g$ .

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

If there is no data available on the soil, but you still want to take the effect into account, you could assume that $k_g=k_p$ . Both dimensions of the surface support will be modelled. Which gives this as the formula for the spring constant $k$ :

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

with

$E$ the modulus of elasticity of the surface support
$l$ the height of the pit
The "2" is refered to as the stiffenss ratio $s$ in the calculator below.

Input

Young’s modulus E [N/mm²]

Level height H [mm]

Stiffness ratio s [-]

k = E s · H = 27 000 2 · 3 000 = 4 500 000kN/m³

Note: E is in N/mm² and H is in mm; the result is converted to kN/m³.

This value must be entered in :

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

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For a column

For a foundation pile

For a wall

For a foundation pit (supported area)

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