Wind Force Due to Friction
Technical Article
The wind, which blows parallel to the surfaces of a structure, can generate friction forces on these surfaces. This effect is mainly important for very large structures.
The standard distinguishes between the free-standing wall, the free-standing roof, and the long closed building for friction effects resulting from the wind [1].
The percentage of the friction force due to wind of the total wind force is determined according to the following formula:
${\mathrm F}_{\mathrm{fr},\mathrm j}\;=\;{\mathrm c}_{\mathrm{fr},\mathrm j}\;\cdot\;{\mathrm q}_{\mathrm p(\mathrm{ze})\mathrm j}\;\cdot\;{\mathrm A}_{\mathrm{fr},\mathrm j}$ [1] (5.7)
where
cfr = friction coefficient
q p(ze) = is the peak velocity pressure at reference height ze
Afr = area of external surface parallel to the wind
The proportion due to friction is to be superimposed by vectorial summation with the other wind forces Fw,e (external wind pressure) and Fw,i (internal wind pressure). The resulting friction forces only act in the direction of the wind components parallel to external surfaces.
The effects of wind friction on the surface can be disregarded when the total area of all surfaces parallel with (or at a small angle to) the wind is equal to or less than 4 times the total area of all external surfaces perpendicular to the wind (windward and leeward) [1] 5.3 (4).
- For smooth surfaces such as steel or smooth concrete, the friction coefficient cfr is 0.01.
- On rough surfaces such as rough concrete or tar-boards, the friction coefficient cfr is 0.02.
- For surfaces that are very rough (ripples, ribs, folds), the friction coefficient cfr is 0.04.
The reference height ze is the height h of the top edge of the wall and the roof height for free-standing roofs.
Example of a Wall with Rippled Surface
Wind zone 2
Terrain category 2
cfr = 0.04
Length d = 20 m
Reference height ze = 2.5 m
$\begin{array}{l}{\mathrm A}_{\mathrm{fr}}\;=\;2\;\cdot\;2.5\;\mathrm m\;\cdot\;20\;\mathrm m\;=\;100\;\mathrm m²\\{\mathrm q}_{\mathrm p(\mathrm{ze})}\;=\;1.7\;\cdot\;\mathrm{qb}\;=\;1.7\;\cdot\;0.39\;\mathrm{kN}/\mathrm m²\;=\;0.663\;\mathrm{kN}/\mathrm m²\\{\mathrm F}_{\mathrm{fr}}\;=\;0.04\;\cdot\;0.663\;\mathrm{kN}/\mathrm m²\;\cdot\;100\;\mathrm m²\;=\;2.65\;\mathrm{kN}\end{array}$
Example of a Free-Standing Roof with Ribbed Surface
Wind zone 2
Terrain category 2
cfr = 0.04
Length d = 7 m
Width b = 4 m
Reference height ze = 3 m
$\begin{array}{l}{\mathrm A}_{\mathrm{fr}}\;=\;2\;\cdot\;4\;\mathrm m\;\cdot\;7\;\mathrm m\;=\;56\;\mathrm m²\\{\mathrm q}_{\mathrm p(\mathrm{ze})}\;=\;1.7\;\cdot\;\mathrm{qb}\;=\;1.7\;\cdot\;0.39\;\mathrm{kN}/\mathrm m²\;=\;0.663\;\mathrm{kN}/\mathrm m²\\{\mathrm F}_{\mathrm{fr}}\;=\;0.04\;\cdot\;0.663\;\mathrm{kN}/\mathrm m²\;\cdot\;56\;\mathrm m²\;=\;1.49\;\mathrm{kN}\end{array}$
Example of a Hall with Ribbed Surface
Wind zone 2
Terrain category 2
cfr = 0.04
Length d = 30 m
Width b = 10 m
Reference height ze = 5.5 m
Area of all surfaces parallel to the wind Atotal = 2 ⋅ 30 m ⋅ 4 m + 2 ⋅ 30 m ⋅ 5.22 m = 553.2 m²
The smaller value of 2 ⋅ b or 4 ⋅ h is to be used for the distance y from the windward leading edge.
$\begin{array}{l}\mathrm y\;=\;2\;\cdot\;10\;=\;20\;\mathrm m\\{\mathrm A}_{\mathrm{fr}}\;=\;2\;\cdot\;(30\;\mathrm m\;–\;20\;\mathrm m)\;\cdot\;4\;\mathrm m\;+\;2\;\cdot\;(30\;\mathrm m\;–\;20\;\mathrm m)\;\cdot\;5.22\;\mathrm m\;=\;184.4\;\mathrm m²\\{\mathrm q}_{\mathrm p(\mathrm{ze})}\;=\;2.1\;\cdot\;{\mathrm q}_{\mathrm b}\;\cdot\;\left(\frac{\mathrm z}{10}\right)^{0.24}\;=\;2.1\;\cdot\;0.39\;\mathrm{kN}/\mathrm m²\;\cdot\;\left(\frac{5.5\;\mathrm m}{10}\right)^{0.24}\;=\;0.711\;\mathrm{kN}/\mathrm m²\\{\mathrm F}_{\mathrm{fr}}\;=\;0.04\;\cdot\;0.711\mathrm{kN}/\mathrm m²\;\cdot\;184.4\;\mathrm m²\;=\;5.245\;\mathrm{kN}\end{array}$
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