This article presents a worked example on wind load on a building of 10 storeys to EN 1991-1-4-2005 + A1: 2010, and the UK National Annex to the standard. The site on which the medium-rise building is to be constructed is situated within Birmingham at a distance of 168km from Weston-Super-Mare which is the nearest sea, and also 20Km away from the town’s boundary. The site is of low elevation and orography is not significant. We shall determine
- Wind load on the building walls
- Overall wind load on the building
The building plan is similar throughout the 1o floors, and each storey height is 3m. The plan of the building is produced below

The steps required to determine the wind load shall be broadly categorized into three as explained in “Wind Actions on Structures”. We shall first determine the wind velocity, then wind pressure, and then the wind load on the building. Now let start with the first step!
Calculation of Wind Velocity
Fundamental basic wind velocity (vb,o)
Fundamental basic wind velocity (vb,o) = vb,map × calt
vb,map for Birmingham = 22m/s (From figure NA.1 of UK NA)
Altitude factor (calt) = 1 + 0.001. A.(10/Zs )0.2 (for Zs ≥ 10m)
Altitude of the site above mean sea level (A) = 66m
Reference height as defined in figure 6.1 of the standard (Zs) = 0.6h
(Zs) = 0.6h = 0.6 x 30 = 18m
calt = 1 + 0.001 x 66 x (10/18 )0.2 = 1.05
vb,o = 22 × 1.05 = 23.17m/s
Basic Wind Velocity (vb)
Basic wind velocity (vb) = vb,o . Cdir . Cseason
Take Cdir and Cseason as 1
vb = vb,o . Cdir . Cseason
vb = 23.17 x 1 x 1
vb = 23.17m/s
We are done with the first category which is determining the wind velocity, then we move to the second category which is determining the wind pressure.
Calculation of Wind Pressure
Basic Velocity Pressure (qb)
qb = 1/2 ρ vb2 (from Clause NA.2.18 of the UK National Annex)
ρ = 1.226kg/m3
qb = (1/2 x 1.226 x 23.172 )/1000
qb = 0.329KN/m²
Peak Velocity Pressure
Since the site is located in town terrain and orography is not significant, the peak velocity pressure shall be determined using equation NA.3a in UK National Annex:
qpz = Cez •ce,(T) • qb (for town terrain) equation (NA.3a)
The building windward wall shall be considered in two parts because b < h ≤ 2b (h = 30m, b= 18m) in accordance with figure 7.4 of the standard.
The lower part = b = 18m
The remaining upper part = h – b = 30 -18 = 12m
The information about surrounding is unknown, so we shall be conservative and take displacement height (hdis ) as 0, hence (z-hdis) = z when using both figure NA.7 and NA.8 of UK National Annex for both exposure factor and exposure correction factor.
Exposure factor
The nearest sea from the site is taken as Weston-super-Mare which is 168Km away.
From figure NA.7 of the UK National Annex which is reproduced below:
Exposure factor ( Cez) for lower part (at Z = 18) = 2.7
Exposure factor (Cez ) for upper part (at Z = 30) = 3.05

Exposure correction factor
The site is 20KM from town boundary.
From figure NA.8 of the UK National Annex which is reproduced below:
Exposure correction (CeT ) factor for lower part (at Z = 18) = 0.82
Exposure correction factor (CeT ) (for upper part (at Z = 30) = 0.88

qp(30) = Cez •ce,(T) • qb = 0.329 x 2.7 x 0.82 = 0.73KN/m²
qp(18) = Cez •ce,(T) • qb = 0.329 x 3.05 x 0.88 = 0.88KN/m²
We now move to third category of determining the wind load on the building.
Calculating Wind Load on Building
The objective of this article is to determine both the wind load on individual walls, and the overall wind load on the building. We shall start off with determining the wind load on walls.
Wind Load on Building Walls
Division of building walls into Zones
We shall first define the building walls into zones according to figure 7.5 of the standard. So, we first determine the parameter “e”.
e = min(b,2h) = min (18, 2x 30=60) = 18m
since e ≥ d, according to figure 7.5 of the standard, the building walls are divided into zones A, B, D, and E. (figure 7.5 is reproduced below for better conceptualization).

External pressure coefficients
The area of each zone is greater than 1m2 then cpe,10 values are used.
We look up the external pressure coefficients for each zone in table 7.1 of the standard depending on h/d
h/d = 30/18 = 1.67
Zone A
Since h/d = 1.67 which is between 5 and 1, we adopt linear interpolation to find the value of cpe,10 for 1.67
$\frac{5-1}{1.67-1}=\frac{-1.2-(-1.2)}{x-(-1.2)}$
4x = – 4.8
x = – 1.2
cpe,10 (zone A) = -1.2
Zone B
Using linear interpolation
$\frac{5-1}{1.67-1}=\frac{-0.8-(-0.8)}{x-(-0.8)}$
4x = – 3.2
x = – 0.8
cpe,10 (zone B) = -0.8
Zone D
Using linear interpolation
$\frac{5-1}{1.67-1}=\frac{0.8-0.8}{x-0.8}$
4x = 3.2
x = 0.8
cpe,10 (zone D) = 0.8
Zone E
Using linear interpolation
$\frac{5-1}{1.67-1}=\frac{-0.7-(-0.5)}{x-(-0.5)}$
4x + 2 = -0.134
x = – 0.53
cpe,10 (zone E) = -0.53
Internal Pressure Coefficients
Although there is sparse information about openings in the building, it is certain there is no dominant opening on any of the faces, so Cpi is taken as the more onerous of -0.3 or +0.2.
The most onerous of internal pressure coefficient for every instance of wind load will always be that with alternate sign with the external pressure coefficient.
Net pressure coefficients (CPnet = CPe – CPi)
cpnet,10 (zone A) = (cpe,10 – cpi,10) = -1.2 – 0.2 = -1.4
cpnet,10 (zone B) = (cpe,10 – cpi,10) = -0.8 – 0.2 = -1.0
cpnet,10 (zone D) = (cpe,10 – cpi,10) = 0.8 – (-0.3) = 1.1
cpnet,10 (zone E) = (cpe,10 – cpi,10) = -0.53 – 0.2 = -0.73
Structural factor
For framed buildings less than 100m high and whose height is less than 4 times the in-wind depth, structural factor can be taken as 1.
Calculating Wind Load on each zone of Building walls
wk = cscd . qp(Z). cnet
Wall A = 1 x 0.88 x (-1.4) = -1.1 KN/m²
Wall B = 1 x 0.88 x (-1.0) = -0.88 KN/m²
Wall D (at z = 30) = 1 x 0.88 x (1.1) = 0.97 KN/m²
(at z = 18) = 1 x 0.73 x (1.1) = 0.80 KN/m²
Wall E = 1 x 0.88 x (-0.73) = -0.64 KN/m²
Next is to calculate the overall wind load on the building
Overall Wind Load on Building
The overall wind load on the building shall be calculated using net pressure coefficients in NA.4 of UK National Annex.
Net Pressure Coefficient
Using linear interpolation
$\frac{5-1}{1.67-1}=\frac{1.3-1.1}{x-1.1}$
4x – 4.4 = 0.134
x = 4.534/4
x = 1.13
cpe,10 (overall) = 1.13
Overall Wind Load
wk = cscd . qp(Z). cnet
Wall D (at z = 30) = 1 x 0.88 x (1.13) =1.0 KN/ m²
(at z = 18) = 1 x 0.73 x (1.13) = 0.82 KN/m²
