# Determining Force Coefficient of Resulting Member Loads for Plane Lattice Structures from Wind Load

### Technical Article

7 November 2017

001491

This article presents a simple example of a lattice structure to explain how to determine wind loading as a function of the lattice solidity.

#### Wind Perpendicular to Structure

Basic velocity vb = 25.0 m/s
Basic velocity pressure qb = 0.39 kN/m²
Peak velocity pressure ${\mathrm q}_\mathrm p(\mathrm z)\;=\;1.7\;\cdot\;{\mathrm q}_\mathrm b\;\cdot\;\frac{\mathrm z}{10}^{0.37}\;=\;1.7\;\cdot\;0.39\;\cdot\;\frac{7.5}{10}^{0.37}\;=\;0.596\;\mathrm{kN}/\mathrm m²$

Force coefficient cf for lattice structures:

Formula 2

$$cf = cf,0 · Ψλ$$

#### Determination of Force Coefficient cf,0 for Lattice Structures Without End-Effect Using Solidity Ratio φ

Solidity ratio:

Formula 3

$$φ = AAC$$

where
A = sum of projected areas of members
AC = l ⋅ b = enclosed area of considered face

Area ratio of the lattice:

Formula 4

$$A = 2.828 m · 0.1 m · 5 2.0 m · 0.05 m · 4 2.0 m · 0.1 m · 2 10 m · 0.2 m · 2 = 6.214 m²AC = 10 m · 2 m = 20 m²$$

Solidity ratio:

Formula 5

$$φ = 6.214 m²20 m² = 0.3107$$

After the solidity ratio is obtained, the force coefficient cf,0 of 1.6 can be read off the standard EN 1991‑1‑4, Figure 7.33 [1], for example.

It is also necessary to define the effective slenderness of the structural component in order to determine the end‑effect factor Ψλ.

#### Effective slenderness λ (Table 7.16 → BS EN 1991‑1‑4 [2])

Formula 6

$$λ = 2 · 10 m2 m = 10 < 70 → 10 is governing$$

Using the previously calculated values, the end-effect factor Ψλ of 0.95 can be read off the diagram in Figure 7.36 of the standard.

Using this factor, the following force coefficient is obtained:

Formula 7

$$cf = cf,0 · Ψλ = 1.6 · 0.95 = 1.52$$

#### Calculation of Resulting Wind Load of Lattice Structure

Option 1: Equivalent static load Fw

Formula 8

$$Fw = cf · qp(z) · Aref$$

where
Aref = projected area

Formula 9

$$Fw = 1.52 · 0.596 kN/m² · 6.214 m² = 5.63 kN$$

Formula 10

$$Fw1 = 1.52 · 0.596 kN/m² = 0.91 kN/m²$$

In order to distribute this area load in RFEM/RSTAB to the members only, it is necessary to select the ‘Empty, on members only’ option under Area of Load Application. After entering the load and clicking [OK], the sum of the load to be applied is displayed again in an information window.

#### Literature

 [1] Eurocode 1: Actions on structures - Part 1-4: General actions - Wind actions; EN 1991-1-4:2005 + A1:2010 + AC:2010 [2] National Annex - Nationally determined parameters - Eurocode 1: Actions on structures - Part 1-4: General actions - Wind actions; BS EN 1991‑1‑4:2005+A1:2010

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• Updated 15 December 2020

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