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

### Technical Article

001491

11/08/2017

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

Force coefficient cf for lattice structures:
$${\mathrm c}_\mathrm f\;=\;{\mathrm c}_{\mathrm f,0}\;\cdot\;{\mathrm\Psi}_\mathrm\lambda$$

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

Solidity ratio:

$$\begin{array}{l}\mathrm\varphi\;=\;\frac{\mathrm A}{{\mathrm A}_\mathrm C}\;\end{array}$$

where

 A is the sum of the projected area of the members Ac is the enclosed area of the examined face calculated as:
$${\mathrm A}_\mathrm C\;=\;\mathrm l\;\cdot\;\mathrm b$$

Area ratio of the lattice:

$$\begin{array}{l}\mathrm A\;=\;2.828\;\mathrm m\;\cdot\;0.1\;\mathrm m\;\cdot\;5\;+\;2.0\;\mathrm m\;\cdot\;0.05\;\mathrm m\;\cdot\;4\;+\;2.0\;\mathrm m\;\cdot\;0.1\;\mathrm m\;\cdot\;2\;+\\+\;10\;\mathrm m\;\cdot\;0.2\;\mathrm m\;\cdot\;2\;=\;6.214\;\mathrm m^2\end{array}$$ $$\begin{array}{l}{\mathrm A}_\mathrm C\;=\;10\;\mathrm m\;\cdot\;2\;\mathrm m\;=\;20\;\mathrm m^2\end{array}$$

Solidity ratio:

$$\mathrm\varphi\;=\;\frac{6.214\;\mathrm m²}{20\;\mathrm m²}\;=\;0.3107$$

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])

$$\mathrm\lambda\;=\;2\;\cdot\;\frac{10\;\mathrm m}{2\;\mathrm m}\;=\;10\;<\;70\;\rightarrow\;10\;\mathrm{is}\;\mathrm{governing}$$

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:

$${\mathrm c}_\mathrm f\;=\;{\mathrm c}_{\mathrm f,0}\;\cdot\;{\mathrm\Psi}_\mathrm\lambda\;=\;1.6\;\cdot\;0.95\;=\;1.52$$

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

##### Variant 1: Equivalent static load Fw
$$\begin{array}{l}{\mathrm F}_\mathrm w\;=\;{\mathrm c}_\mathrm f\;\cdot\;{\mathrm q}_\mathrm p(\mathrm z)\;\cdot\;{\mathrm A}_\mathrm{ref}\end{array}$$

where

 Aref is the projected area
$${\mathrm F}_\mathrm w\;=\;1.52\;\cdot\;0.596\;\mathrm{kNm}²\;\cdot\;6.214\;\mathrm m²\;=\;5.63\;\mathrm{kN}$$
$${\mathrm F}_{\mathrm w1}\;=\;1.52\;\cdot\;0.596\;\mathrm{kN}/\mathrm m²\;=\;0.91\;\mathrm{kN}/\mathrm m²$$

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.

#### Reference

 [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