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

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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

Figure 01 - Frame Dimensions

Basic velocity v b = 25.0 m/s
Basic velocity pressure q b = 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 trusses:
${\mathrm c}_\mathrm f\;=\;{\mathrm c}_{\mathrm f,0}\;\cdot\;{\mathrm\Psi}_\mathrm\lambda$

Determination of the basic force coefficient c f, 0 for trusses with infinite slenderness with the degree of solidity φ

Solidity ratio:
$\begin{array}{l}\mathrm\varphi\;=\;\frac{\mathrm A}{{\mathrm A}_\mathrm C}\;\\\mathrm{mit}\\\mathrm A\;=\;\mathrm{Summe}\;\mathrm{der}\;\mathrm{projizierten}\;\mathrm{Flächen}\;\mathrm{der}\;\mathrm{Stäbe}\\{\mathrm A}_\mathrm C\;=\;\mathrm l\;\cdot\;\mathrm b\;=\;\mathrm{umschlossene}\;\mathrm{Fläche}\;\mathrm{des}\;\mathrm{betrachteten}\;\mathrm{Bereichs}\end{array}$

Area ratio of the truss:
$\begin{array}{l}\mathrm A\;=\;2,828\;\mathrm m\;\cdot\;0,1\;\mathrm m\;\cdot\;5\;+\;2,0\;\mathrm m\;\cdot\;0,05\;\mathrm m\;\cdot\;4\;+\;2,0\mathrm m\;\cdot\;0,1\;\mathrm m\;\cdot\;2\;+\\+\;10\;\mathrm m\;\cdot\;0,2\;\mathrm m\;\cdot\;2\;=\;6,214\mathrm m²\\{\mathrm A}_\mathrm C\;=\;10\;\mathrm m\;\cdot\;2\;\mathrm m\;=\;20\;\mathrm m²\end{array}$

Figure 02 - Displaying Parameters for Determination of Solidity in RFEM/RSTAB

Solidity ratio:
$\mathrm\varphi\;=\;\frac{6,214\;\mathrm m²}{20\;\mathrm m²}\;=\;0,3107$

Once the degree of completeness is known, the basic force coefficient c f, 0 can be read as 1.6 , for example, in Figure 7.33 of the standard DIN EN 1991-1-4 [1] .

Figure 03 - Force Coefficient cf,0

Furthermore, the effective slenderness of the structural component must be determined to determine the reduction factor des λ .

Effective slenderness λ (Tab. 7.16 → DIN EN 1991-1-4 [2] )

$\mathrm\lambda\;=\;2\;\cdot\;\frac{10\;\mathrm m}{2\;\mathrm m}\;=\;10\;<\;70\;\rightarrow\;10\;\mathrm{ist}\;\mathrm{maßgebend}.$

With the previously calculated values, the reduction factor Ψ λ can be read as 0.95 in the diagram shown in Figure 7.36 of the standard.

Figure 04 - End-Effect Factor Ψλ

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

Option 1: static equivalent load F w
$\begin{array}{l}{\mathrm F}_\mathrm w\;=\;{\mathrm c}_\mathrm f\;\cdot\;{\mathrm q}_\mathrm p(\mathrm z)\;\cdot\;{\mathrm A}_\mathrm{ref}\\\mathrm{mit}\\\;{\mathrm A}_\mathrm{ref}\;=\;\mathrm{projizierte}\;\mathrm{Fläche}\\\;{\mathrm F}_\mathrm w\;=\;1,52\;\cdot\;0,596\;\mathrm{kN}/\mathrm m²\;\cdot\;6,214\;\mathrm m²\;=\;5,63\;\mathrm{kN}\end{array}$

Option 2: Load as Member Loads from Surface Load
${\mathrm F}_{\mathrm w1}\;=\;1,52\;\cdot\;0,596\;\mathrm{kN}/\mathrm m²\;=\;0,91\;\mathrm{kN}/\mathrm m²$

To ensure that this area load in RFEM/RSTAB is only distributed to the members, it is necessary to select the area of the load application to "Not filled, only on members". After entering the load and clicking [OK], the sum of the load to be applied is displayed once more in an info window.


[1]   Eurocode 1: Actions on structures - Part 1-4: General actions - Wind loads; 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 loads; DIN EN 1991-1-4/NA: 2010-12



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