This document discusses wind loads on structures. It begins by explaining that wind is moving air with mass and velocity that exerts kinetic energy. The intensity of wind load depends on wind velocity squared and the dimensions of resisting members. Wind velocity varies by location, structure height, terrain, and surroundings. Wind pressure depends on velocity, building shape/surface, terrain protection, and air density. Structures respond to wind loads through static deflection and dynamic vibration. Both external and internal wind pressures must be considered to calculate net pressure. Internal pressure depends on opening distribution. Several examples are provided to demonstrate calculating wind loads on buildings.

1. Wind Loads on Structure
(Buildings)
Chapter 1
Structural Design (CEng 5123)
Mamush Teklie(MS.c), 2022

2. Wind Load on Structure
 Wind is moving air. Air has mass/density and moves in a particular
direction at a particular velocity (has Kinetic Energy)
 Are variable loads which act on the structure externally and internally
 The intensity of the wind load is related to:
 The square of wind velocity
 The dimension of the member that are resisting the wind
 Wind velocity is dependent on:
 Geographical location
 The height of structure
 Topography of the area
 Roughness of surrounding terain
 Wind pressure on building surface depends on:
 Velocity
 Shape and surface structure of the building
 The protection from wind offered by surrounding natural terrain or man
made structure
 Density of air

3. Wind loads on structure

4. Cont’d
 Response of the structure to wind load
 Background component
-static deflection of structure
 Resonant component
-dynamic vibration of the structure in response to
change in wind pressure
*Wind loads are particularly significant on tall building

5. Cont’d
 Wind pressure act on the structure both internal and
externally.
 External Pressure, We (EBCS 1, cls 3.5.2)
 Internal Pressure, Wi (EBCS 1, CLS 3.5.3)
 Net Pressure, Wn (EBCS 1, cls 3.5.4)
 The net wind pressure across a wall or an element is the
difference of the pressure on the surface taking due account
of their sign. Wn = We-Wi
Where: Cpe = external pressure coefficient
Cpi = internal Pressure coefficient
Ce(z) =exposure coefficient for the terrain (cls 3.8.5)
qref = reference mean wind velocity pressure

6. Cont’d
Pressure on surface

7. 1. Reference Wind pressure, qref (EBCS 1,cls 3.7.1)
Where  = air density
(kg/m3)
ref = reference wind
velocity (m/s)
Air density is affected by
altitude
Table 3.3. Values of air
density
Site altitude (m)
above sea level
 (kg/m3)
0 1.20
500 1.12
1000 1.06
1500 1.00
2000 0.94

8. Reference wind velocity
Is mean velocity 10m above farm land averaged over
a period of 10min with a return period of 50yr.
Vref = CDIRCTEMCALTVref, 0 EBCS 1, cls 3.7.2
Vref, 0 – basic reference wind velocity = 22m/s
CDIR –direction factor =1.0
CTEM –temp.(seasonal) factor = 1.0
CALT – altitude factor = 1.0

9. Reference Height, z (EBCS 1, Appendix A)

10. Cont’d

11. 2. Exposure Coefficient
 The exposure coefficient take account of the variation
from the reference wind velocity due to:
 The roughness around the structure
 The local topography and the height of the structure
 Exposure coefficient at height z meter, EBCS 1,cls 3.8.5
 Wind velocity tends to decrease at ground level
Where
Cr =roughness coefficient
Ct =topography coefficients
kT =terrain factor

12. Terrain Categories and Related Parameters (EBCS1 Table 3.2)
Category Terrain description kr
Z0
(m)
zmin
(m)
1
Lakeshore with  5 km fetch
upwind and smooth flat
country without obstacles
0.17 0.01 2
2
Farmland with boundary
hedges, occasional small farm
structures, houses or trees
0.19 0.05 4
3
Suburban or industrial areas
and permanent forests
0.22 0.3 8
4
Urban areas in which  15% of
the surface is covered with
buildings and their average
height exceeds 15m
0.24 1 16

13.  Roughness Coefficient, Cr (z)
 accounts for the variability of mean wind velocity due to the height
of the structure above ground level and the roughness of the terrain
Cr(z) = kr Ln(z/z0) for 200 z  zmin
Cr(z) = Cr(zmin) for z  zmin
Where: kT is the terrain factor
Zo is the roughness length depends on terrain category
Zmin is the minimum height

14.  Topography Coefficient, Ct [EBCS 1,cls 3.8.4]
 accounts for the increase in mean wind speed over isolated
hills and escarpments.
 Details for its calculation in such cases are given in EBCS1
(Figure 3.6 and 3.7). For all other situations, Ct may be
taken as unity.

15. Cont’d

18. Lu = Le = 500 m
H = 30 m
200 m
Wind
0.75 Le = 375 m
0.5 Le =
250 m
S = 1
S = 0
S = 0
S = 0 Boundary of topography
affected zone
Example

19. 3. External Pressure Coefficient, Cpe [EBCS 1,apedix A]
 Depends on:
 Dynamic response on different zone of the structure due to its
geometry, area and proximity to other structure.
The external pressure coefficient cpe for buildings and individual
parts of buildings depend on the size of the loaded area A.
Fig Variation of external pressure coefficient
cpe = cpe,1 A  1m2
cpe = cpe,1 + (cpe,10 - cpe,1) log10A 1m2 < A < 10m2
cpe = cpe,10 A  10m2

21. Cont’d
F
0.5e
d
0.8e
0.1e
0.2e
G
F
0.25e 0.25e
h
H
I
A
B
C
D
b
0.5e
d
0.1e
0.2e
G
F
0.25e 0.25e
h
F
H
I
A
B*
D
b
E
E

23. d=16m
b=20m
D
h=12m
E
b=20m and 2h=24m
e=20m
d=16m
e> d
A B*
Area of A=48m2
B*=144m2
d/h=16/12=1.33
0.2e=0.2*20
=4m 12m
A B* D E
CPe -1.0 -0.8
Interpolate b/n
0.8 and 0.6
-0.3

24. d=16m
b=20m
D
h=7m
E
b=20m and 2h=14m
e=14m
d=16m
e<d
A B
Area of A=48m2
B*=144m2
d/h=16/7=2.286
0.2e=0.2*14
=2.8m 11.2m
A B C D E
CPe -1.0 -0.8 -0.5
Interpolate b/n
0.8 and 0.6
-0.3
C
2m

25. Cont’d

26. Cont’d

27. Cont’d

28. Cont’d

29. Cont’d

30. Cont’d

31. Cont’d

32. Cont’d

33. Cont’d

34. Cont’d

35. Cont’d

36. Cont’d

37. Cont’d

38. Cont’d

39. Cont’d

40. 4.Internal Pressure Coefficient
 Internal pressure arises due to openings, such as windows,
doors and vents.
 if the windward pane has a greater proportion of opening than
the leeward panel, then the interior of the structure is subjected to
positive (outward) pressure as illustrated in Fig.(a).
 Conversely, if the leeward face has more openings, then the
interior is subjected to a negative (inward) pressure as illustrated
in Fig. 4.5(b).
 Like external pressure internal pressure is considered positive
when acting on to the surface of the structure.

41. Cont’d
Internal pressures in structures

42. Cont’d
 Internal pressure on a building or panel is given by:
wi =ce(zi)cpiqref
 Where zi is the reference height for internal pressure equal to
the mean height of the openings
 cpi depends on the distribution of openings around the
building.
 The values recommended by EBCS1 are given in Fig 4.6 for a
building without internal partitions.
 For buildings with internal partitions the extreme values, cpi =
0.8 and cpi = -0.5, may be used.

43. Cont’d

44. Wind force on structures
 The total wind force action on individual zones of clad structures is
proportional to the difference in pressure between the external and
internal faces. That is:
Fw = (we – wi) Aref
 When calculating the total force on (all zones of) a building, the
forces on each zone can be calculated using the above equation and
summed.
 Alternatively, the total force on an entire structure (or an exposed
individual member) can be expressed as:
Fw = ce(ze)cf qref Aref
 Where cf is a force coefficient. equal to the algebraic sum of the
external pressure coefficients on the windward and leeward faces.

45. Example
The structure illustrated in Fig is to be located in the centre
of Paris on a site surrounded by buildings of similar height.
It is an apartment building with internal partitions.
12
N
E
W
10
20

46. Cont’d
 Wind from the east and west is transmitted from clad faces
to the north and south masonry walls. Each external panel has
opening windows equal in area to one tenth of the total wall
area.
(a) Determine the total moment due to wind at the base of the
north and south masonry walls.
(b) Calculate the maximum pressure on the east masonry
wall.

47. Solution
 Reference pressure
 The basic reference wind velocity for Paris can be taken
from the map and is 26m/s . Assuming cDIR = cTEM = cALT =
1.0
 the reference wind velocity = 26 m/s.
 Hence the reference wind pressure is,
Exposure coefficient
h> b < 2b, the building is considered in two parts,
The reference heights for external pressure are thus:
ze = h = 20m
and
ze = b = 12m
Qref =
2
1
v2
ref
=
2
1
(1.25)(26)2
= 423 Nm2

48. Cont’d
As the building is located in an area of Roughness Category 4 (refer to Table 4.1), kT =
0.24, z0 = 1m and smin = 16m. Equation of roughness coefficient gives:
cr(20) = kTLn(20/z0)
= 0.24 Ln(20/1)
= 0.719
cr(12) = cr(zmin)
= 0.24 Ln(16/1)
= 0.666

49. Cont’d
Taking a topography coefficient of unity, the exposure coefficients become:
ce(20) = c 2
r (20) c 2
t (20) 






)
20
(
)
20
(
7
1
t
r
T
c
c
k
= (0.719)2
(1)2 








1
719
.
0
24
.
0
7
1
= 1.725
ce(12) = c 2
r (12) c 2
t (12) 






)
12
(
)
12
(
7
1
t
r
T
c
c
k
= (0.666)2
(1)2









1
666
.
0
24
.
0
7
1
= 1.563

50. Cont’d
External pressure
It can be seen from Fig. 4.3 that only zones D and E are of interest in this example. The
ratio d/h is 10/20 = 0.5. Hence from Table 4.3:
cpe(Zone D) = +0.8
cpe(Zone E) = -0.3
At the reference height of 20m, the external pressure on zone D is (equation (3.7)):
we = ce(20)cpeqref
= 1.7250.80.423
= 0.584kN/m2
The corresponding force on the upper part of zone D is 0.584(128) = 56kN.

51. Cont’d
At the reference height of 12m, the external pressure on zone D is
we = ce(12)cpeqref
= 1.5630.80.423
= 0.529 kN/m2
and the corresponding force is 0.529(1212) = 76kN. The corresponding force for zone
E are -21kNand -29kN for the upper and lower parts respectively. These forces are
illustrated in Fig. 4E-2.

52. Cont’d
Fig. 4E-2 Forces due to east wind
20
8
10
16
6
76
29
56
21
12
E
W
N
12

53. Cont’d
(a) Internal pressure within a structure is self equilibrating.
Thus, while it can cause significant pressures on individual wall
panels it results in no net force on the structure overall.
Accordingly, the overturning moment at the base of the north
and south walls due to wind is unaffected by internal pressure
and is given by:
Moment = (56+21)16 + (76+29)6
= 1862 kNm
Of this, half will apply at the base of each of the two walls.

54. Cont’d
(b) To determine the total pressure on the east wall, it is necessary to calculate the
internal as well as the external pressure. As there are internal partitions, the worst value
for cpi is assumed, that is , cpi = -0.5. The maximum pressure will occur in the upper part
of zone D. In this part of the building the mean height of the windows will be assumed
to equal the mean height of the part. Hence:
Zi = 16m
The exposure coefficient at this height is calculated as before and is 0.666. Thus:
wi = ce(zi)cpiqref
= 0.666(-0.5)(0.423)
= -0.141
The net pressure on the upper part of zone D is the difference between the external and
the internal pressures, that is:
we – wi = 0.584 –(-0.141) = 0.725 kN/m2

55. Example 2
 The building shown in below is to be built in a sloped
terrain in Ambo. The details of the terrain and the position
of the building are shown in the figure. The building is
meant for a lathe shop inside of which has no partition
walls. Six windows of 2.5m * 2.75m size are provided in
each of the longer sides. Two windows of 2.0m * 2.75m
size and one door of 2.5m * 3m are provided in each of the
shorter sides of the building. Elevation of the existing
ground level at the building site is 1720m.
 Calculate the wind load acting on the middle truss in the
roof truss assembly. The trusses are spanning in the short
direction of the building at 3m centers. The shape of the
roof truss is also shown in the figure.

56. Cont’d
12 m
30 m
Plan
12 m
Elevation
4.5 m
1.6 m
30 m
12 m
1.6 m
Side elevation
Truss
500 m
30 m
200 m
Wind
Terrain detail with position of the building

57. Cont’d
(1) Basic value of wind velocity for Ethiopia vref,o = 22 m/sec
(2) Reference wind velocity vref = CDIR CTEM CALT vref,o = 1 * 1 * 1 * 22 = 22 m/sec
CDIR = CTEM = CALT = 1 → shall be taken as per EBCS 1
(3) Reference wind pressure qref
To calculate the air density, the altitude is considered at the mid height of the
building.
The altitude = 1720 m + (4.5+1.6)/2 = 1723.05m
For an altitude of 1500 m → Air density ρ = 1.0 kg/m3
For an altitude of 2000 m → Air density ρ = 0.94 kg/m3
For an altitude of 1723.05 m → Air density ρ = 0.9732 kg/m3
(interpolated
value)
Reference wind pressure qref = ½ ρ vref
2
= ½ * 0.9732 * 222
= 235.5 N/m2

58. Cont’d
(4) External wind pressure We = qref Ce(ze) Cpe
Exposure coefficient Ce(ze):
Ce(ze) = 






)
(
*
)
(
7
1
)
(
*
)
(
2
2
z
C
z
C
K
z
C
z
C
t
r
T
t
r
KT = Terrain Factor = 0.22 (for terrain category III)
)
(z
Cr = Roughness Coefficient = 





o
T
Z
Z
K ln for Z > Zmin
= )
( min
z
Cr for Z < Zmin
In our problem, Ambo can be considered as a sub urban area and
therefore the terrain category falls as category III
For terrain category III, Zmin = 8 m
Z = Reference height - can be taken as the overall height of the building
= 6.1 m
Therefore, Z < Zmin ; )
(z
Cr = )
( min
z
Cr
From table 3.3 in page no. 58 of EBCS 1, )
(z
Cr = 0.72

59. Cont’d
)
(z
Ct = Topography Coefficient:
Φ = Upwind slope H/Lu in the wind direction = 30/500 = 0.06
Le = Effective length of upwind slope = Lu, if 0.05 < Φ < 0.3
The topography affected zone is marked in the diagram shown below.
s = A factor to be obtained by interpolation from the value of s = 1.0 at
the crest of escarpment and the value s = 0 at the boundary of the topography
affected zone.

60. Cont’d
Lu = Le = 500 m
H = 30 m
200 m
Wind
0.75 Le = 375 m
0.5 Le =
250 m
S = 1
S = 0
S = 0
S = 0 Boundary of topography
affected zone

61. Cont’d
Here s = 0.4667 (Horizontally interpolated value)
Therefore, Ct = 1 + 2s Φ = 1 + 2(0.4667)(0.06) = 1.056
Ce(ze) = 






)
(
*
)
(
7
1
)
(
*
)
(
2
2
z
C
z
C
K
z
C
z
C
t
r
T
t
r
= 






056
.
1
*
72
.
0
22
.
0
*
7
1
056
.
1
*
72
.
0 2
2
= 1.749
Alternatively Ce(ze) can be taken from table 3.5 (page 62) for calculated values
of )
(z
Ct = 1.056 and ze = h = 6.1 m.

62. External pressure coefficient Cpe:
e/4 = 3.05m
23.9 m
e/4 = 3.05m
4.78m 4.78m
1.22m
30 m
J
G H I
F
F
e/10 = 1.22 m
e = b or 2h which
ever is smaller.
b = 30 m;
2h = 2*6.1 = 12.2m
Hence e = 12.2 m
α = Tan-1
(1.6/6)
= 15°
Area F = 3.721 m2
Area G = 29.158 m2
Area H = 143.4 m2
Area I = 143.4 m2
Area J = 36.6 m2

63. Cont’d
Cpe for Area F = Cpe1 + (Cpe10 - Cpe1) log10A; For area F, Cpe1 = -2.0 & Cpe10 = -0.9
(Table A4)
Cpe for Area F = -2.0+ (-0.9– (-2.0)) log103.721 = -0.5136
All the other areas are more than 10 m2
, therefore, Cpe for all other areas = Cpe10
The corresponding values are taken from the table A4.

64. Internal pressure coefficient, Cpi:
Internal pressure coefficient Cpi for buildings without internal partitions is based on the
value
ΣArea of openings at the leeward and wind parallel side
= (6*2.5*2.75) + (4*2.0*2.75) + (2*2.5*3.0) = 78.25m2
ΣArea of openings at the windward, leeward and wind parallel side
= 78.25 + (6*2.5*2.75) =119.5m2
5
.
119
25
.
78

 = 0.6548; For μ = 0.6548, the value of Cpi from figure A11, = -0.125

65. Cont’d
5
.
119
25
.
78

 = 0.6548; For μ = 0.6548, the value of Cpi from figure A11, = -0.125
The summary of the Cpe and Cpi values are shown below in a table.
Zone F G H I J
Area (m2) 3.721 29.158 143.4 143.4 36.6
for =00
Cpe (-ve) -0.5136 -0.8 -0.3 -0.4 -1.0
Cpe (+ve) 0.2 0.2 0.2 ------ -----
Cpi -0.125 -0.125 -0.125 -0.125 -0.125
(Cpe - Cpi) +ve 0.325 0.325 0.325 0.125 0.125
(Cpe - Cpi) -ve -0.3886 -0.675 -0.175 -0.275 -0.875
for =900
Cpe (-ve)  -0.49 -1.3 -0.6 -0.5 -
Cpi  -0.375  -0.375  -0.375  -0.375

66. Cont’d
 In the problem it is asked to calculate the wind load on the
roof truss which is at the middle of the roof truss assembly.
The middle truss is spanned across the zones G,H,I and J.
Out of the zones, coefficient for G is critical; hence the load
on roof truss is considered based on the critical value.
 Wnet = We – Wi = qref Ce(z) [Cpe – Cpi] = 0.2355 * 1.749 * [-1.3 –
(-0.375)] = -0.381 kN/m2 (Negative)
 Wnet = We – Wi = qref Ce(z) [Cpe – Cpi] = 0.2355 * 1.749 *
[0.2 – (-0.125)] = 0.134 kN/m2 (Positive)
 This Wnet acts on the roof covering, which is supported by
purlins. Purlins are supported by the truss. Consider purlins
to be supported at each and every joint of the principal rafter
of the truss. The figure below shows the load transfer path.

67. Cont’d
Purlins
Roof truss
Roof covering
6.21 m
Center to center distance between
the purlins = 6.21/4 = 1.5524 m

68. Cont’d
 Center to center distance between the truss = 3m
 Load transferred to the purlin from the roof covering = - 0.381 *
1.5524 = -0.5915kN/m
 Load transferred over the intermediate joints of principal rafter of truss
= -0.7514*3 = -1.774 kN
 Load transferred over the end joints of principal rafter of truss = -
2.254/2 = -0.887 kN
 These loads are perpendicular to the principal rafter of the truss.

69. Cont’d

71. Thank you!