Ycef presentaion derivation of wind loading upload

PRESENTED ON YOUNG CIVIL ENGINEERS FORUM
(YCEF)
BY
OMOTORIOGUN VICTOR. F (B.ENG., GMICE.)
DERIVATION OF WIND ACTIONS
ON STRUCTURES

LEARNING OUTCOMES
▪ At the end of this session, you should be able to :
▪ Apply Eurocode 1: 1991-1-4:Wind Actions and the U.K National Annex to evaluate wind
actions on roofs.
▪ Apply Eurocode 1: 1991-1-4:Wind Actions and the U.K National Annex to Assess building
frames for wind actions.
2

SYNOPSIS
▪ Introduction
▪ BS EN 1992-1-1-4:Wind Actions on Structures
▪ Steps for Assessing Wind Load on Buildings
▪ Worked Example - Ware-house Building
▪ Worked Example - High- rise Building
3

INTRODUCTION
▪ Wind can be defined as a flowing mass of air. Buildings and other
structures are obstacles that deflect or obstruct the wind.
▪ Wind load result from forces exerted by a moving mass of air, in the form
of kinetic energy which is consequently converted to potential energy or
pressure.
▪ Wind effect on buildings an be classified as either
▪ Static: use of equivalent static forces for ordinary buildings with conventional structural
system.
▪ Dynamic: dynamic response of super-tall and slender structures.
▪ Wind Tunnelling
4

EUROCODE 1 PART 1-4: WIND ACTIONS
Eurocode 1 on wind loads, published in 2005 (British Standards Institution,
2005), is a European Standard (EN) intended for use in most European
countries.The version in each Member State includes the 'National
Annex,' which refers only to that specific country. EN 1991-1-4 represents
many years of research by members from several European Union
countries and two independent committees, and is the closest document
to a genuinely multi-national wind-loading standard that exists today.
This code refers to buildings and other structures up to 200 m in height
and to bridges with spans less than 200 m.
5

EUROCODE 1 PART 1-4: WIND ACTIONS
▪ To apply wind loads on
buildings using the Eurocode,
you have to become familiar
with this two document.
▪ BS EN 1991-1-4:2005
▪ Eurocode 1: Actions on Structures, Part 1-
4: General Actions:Wind Actions
▪ N.A to BS EN 1991-1-4: 2010
▪ U.K National Annex
6

STARTING POINT
vb,o = Fundamental value of basic wind velocity
▪ Is the 10-minute wind velocity
---- at 10m above the ground
----of a flat terrain
----with probability of exceedance of 1/10 over a period of 50
years.
▪ It value is given in the National Annex as: Where:
• Vb,map = map values of
fundamental wind velocity
• calt = Altitude factor
𝑣𝑏,𝑜 = 𝑣𝑏,𝑚𝑎𝑝 × 𝑐𝑎𝑙𝑡
8

MAP VALUES OF WIND VELOCITY-NIGERIA
9

CALT= ALTITUDE FACTOR
▪ Depends on
✓ Altitude of the site above mean sea level, A (m)
✓ The reference height of the structure zs (m). Its value is defined in BS EN 1991-1-4: 2005.
✓ The first expression i.e. for zs=10m is conservative and can be used for all cases:
𝑐𝑎𝑙𝑡=ቐ
1 + 0.001 ∙ 𝐴 ; 𝑧𝑠 ≤ 10𝑚
1 + 0.001 ∙ 𝐴 ∙
10
𝑧𝑠
0.2
; 𝑧𝑠 > 10𝑚
10

ZS= REFERENCE HEIGHT
(BS EN 1991-1-4:2005, FIGURE 6.1)
▪ Zmin depends on the terrain category
as defined in Table 4.1 of BS EN 1991-1-14
For buildings zs =0.6h
Where: h=height of the building. 11

VB= BASIC WIND VELOCITY
▪ It is possible to reduce the fundamental value of the basic wind
velocity to obtain the base wind velocity, by taking into account:
▪ Wind Direction, c,dir : Accounts for the wind direction. It values is defined in Table
NA.1 in the National Annex Eurocode 1-1-4. It value may be conservatively taken
as 1.0
▪ Season c,season: Used for only temporary structures. It values are defined in table
NA.2 in Clause NA. 2.7 of the UK National Annex. For permanent
structures, it value is taken as 1.0
𝑣𝑏=𝑣𝑏,𝑜 ∙ 𝑐𝑑𝑖𝑟 ∙ 𝑐𝑎𝑙𝑡
13

BASIC WIND PRESSURE
▪ The dynamic pressure is given as the kinetic energy per unit volume of the
flowing air. It value is defined as:
- Where ρ= density of air. It recommended value is 1.226kg/m3 in the National Annex
This is just a “nominal” value of the wind pressure. It assumes a uniform
laminar flow of air. It doesn’t necessarily account for the effect of wind
turbulence or variation of pressure with height.To account for this, we must
determine the Peak Wind Pressure
𝑞𝑏 =
1
2
𝜌 ∙ 𝑣𝑏
2
14

PEAK WIND PRESSURE
▪ To account for the effect of wind turbulence and variation of
pressure with height.The base pressure can be transformed into
the “peak” value qp(z) by applying factors that account for
▪ The site’s proximity to the coast
▪ How far the site is located from the boundary of a town (if applicable)
▪ Influence of orography (if applicable)
▪ Before the peak wind velocity can be determine, the terrain
category must be known.This is classified in the National Annex
into:
▪ Sea
▪ Open Country
▪ Town
16

PEAK WIND PRESSURE- CONT’D
• The figure shown, shows the shape of the
velocity profile of buildings
• The shape of the velocity profile changes
with width/ height as a result the peak
wind pressure also changes
• Clause 7.2.2 of BS EN 1991-1-4 requires
that:
• A building whose height h, is less than
b should be considered to be one
part as shown in the figure
• A building whose height is greater
than b but less than 2b should be
considered in two parts as shown in
the figure
• A building whose height is greater
than 2b should be considered in
multiple parts as shown in the figure 17

PEAK WIND PRESSURE CONT’D
▪ Where influence of orography is insignificant.The peak wind
pressure is calculated by applying two coefficients:
▪ “exposure coeffient” ce(z): Account for site’s proximity to shoreline
▪ “Exposure modifier” ce,(T): Account for sites within town
▪ For sites located in a country terrain ce,(T) =1
▪ Both factors depend on the average height of the surrounding structures.
This is usually determine through the displacement height hdis
𝑞𝑝(𝑧) = 𝑐𝑒(𝑧)𝑐𝑒(𝑇)𝑞𝑏
18

HDIS = DISPLACEMENT HEIGHT
▪ This quantity is required to account for the reduction in the wind velocity
due to the presence of closely spaced buildings or presence of other
obstructions. In the National Annex it is defined as:
▪ Where data about the surrounding buildings is not known, it is appropriate to
assume a value of 3m in a Town terrain. If the building is in the country side, it value
may be taken as 0
൞
𝑚𝑖𝑛 0.6ℎ, 0.8ℎ𝑎𝑣𝑒 , 0 ≤ 𝑥 ≤ 2ℎ𝑎𝑣𝑒
𝑚𝑖𝑛 0.6ℎ, 1.2ℎ𝑎𝑣𝑒 − 0.2𝑥 , 2ℎ𝑎𝑣𝑒
0, 𝑥 ≥ ℎ𝑎𝑣𝑒
19

CE(Z)-EXPOSURE COEFFICIENT
The values of
ce(z) are read by
interpolating
between the
contour of the
chart given as
Figure N.A 7 in
the National
Annex
N:B
z= the height at
which the peak
wind pressure is
being sought
20

CE(T)- EXPOSURE MODIFIER
The values of
ce(T) are read by
interpolating
between the
contour of the
chart given as
Figure N.A 7 in
the National
Annex
N:B
z= the height at
which the peak
wind pressure is
being sought
21

IS OROGRAPHY SIGNIFICANT ?
The orography factor
account for increase in the
peak wind pressure and
needs to be considered only
if the structure falls within
one of the shadowed areas
on the figure.
The value of co is determined using the detailed procedure given in
A3 of BS-EN-1991-1-1-4
𝑞𝑝 𝑧 =𝑐𝑒 𝑧 𝑐𝑒 𝑇 𝑐𝑜(𝑧 + 0.6 /1.6)2
∙ 𝑞𝑏
22

IS OROGRAPHY SIGNIFICANT ? & Z > 50M
▪ Where orography is significant and the building height exceed 50m.The
effect of mean speed wind velocity and turbulence must be considered.
▪ The peak wind pressure is given as:
▪ Where vm = mean speed wind velocity defined as:
𝑞𝑝 𝑧 = (1 + 3𝐼𝑣 𝑧 ,𝑓𝑙𝑎𝑡 ∙
𝑘𝐼,𝑇
𝑐𝑜 𝑧
2
0.613 ∙ 𝑣𝑚
2
For sites in Town terrain
𝑞𝑝 𝑧 = (1 + 3𝐼𝑣 𝑧 ,𝑓𝑙𝑎𝑡 ∙
1
𝑐𝑜 𝑧
2
0.613 ∙ 𝑣𝑚
2
For sites in Country
terrain
𝑉
𝑚(z)=𝑐𝑟 𝑧 ∙ 𝑐𝑟,𝑇 ∙ 𝑐𝑜 (𝑧)𝑣𝑏 For sites in Town terrain
𝑉
𝑚(z)=𝑐𝑟(𝑧) ∙ 𝑐𝑜 (𝑧) ∙ 𝑣𝑏 For sites in Country terrain 23

IS OROGRAPHY SIGNIFICANT ? & Z > 50M CONT’D
▪ Up to four charts (Figs NA.3 to NA.6) are required to calculate qp(z)
23
Cr(z
)
Cr,T(z
)
IV,flat(z
)
kI,T(
z)

Flow Charts For Wind Calculation
Sites in country Sites in Town
25

COEFFICIENTS OF WIND ACTION
▪ Once the peak wind pressure has been determined, the wind pressure
from it can be determine.
▪ The wind action varies depending on the part of the structure been
assessed for the wind loading.The following coefficients are used to
determine the magnitude of wind loads on sections of a building.They
are reliant on what part is been assessed for the wind loads.
▪ External Pressure Coefficients
▪ Internal Pressure Coefficients
▪ Lack of Correlation Factor
▪ Structural Factor
▪ Net Pressure Coefficient
27

EXTERNAL PRESSURE COEFFICIENTS
▪ The external pressure coefficients are of two forms cpe,1 and cpe,10
▪ Cpe,1 applies to discrete portion of the structure. (Areas < 1m2)
▪ Cpe,10 applies to larger portion of the structure. (Areas > 10m2)
▪ For intermediate situations i.e. (1<Area<10m2) logarithmic interpolation is
suggested.
▪ Table NA.4 in the UK Annex provides the values of these pressure
coefficients.These are based on zones within vertical walls of a building
that are defined in Figure 7.5 of Eurocode 1-1-4.
▪ NA.6 –NA.8 provides external coefficients for roofs in line with figures 7.6
-7.9 of BS EN 1991-1-4.
𝐶𝑃𝑒 = 𝐶𝑃𝑒,1 − 𝐶𝑃𝑒,1 −𝐶𝑃𝑒,10 ∙ log 𝐴
28

EXTERNAL PRESSURE COEFFICIENTS/ EXAMPLE
Side Walls
windward leeward
The zones for a typical rectangular building with
flat roof is shown: External pressure coefficient
may be obtained by interpolating from table using
h/d ratios.
𝑒 = 𝑚𝑖𝑛 ቊ
𝑏
2ℎ
29

INTERNAL PRESSURE COEFFICIENTS
▪ The internal pressure coefficient Cpi depends on the openings within the
walls of the structure assessed for wind loads. Clause 7.2.9 of Eurocode
1-1-4 gives guidance on how to determine it value.
▪ If there are openings within the envelope within the opposite sides of the building
that are greater than 30% of each surface area, the building must be treated like a
canopy in accordance with clauses 7.3 & 7.4 of BS EN 1991-1-4
▪ For structures with a dominant openings.The value of the internal pressure is taken
as a proportion of the external pressure.The term ‘dominant’ is defined as one face
of a building having a single large opening
▪ Cpi= 0.75Cpe (if the dominant face are twice that of all other opening in the structure)
▪ Cpi = 0.9Cpe (if the dominant face are three times that of all other opening s in the structure)
▪ Where dominant face are not present in the structure figure 7.13 is combined with
NA.9 to determine it value
▪ If there is a lack of certainty regarding the openings within the structure, it value may
be taken as either +0.2 or -0.3.Whichever gives the onerous result 30

LACK OF CORRELATION COEFFICIENT
▪ Eurocode 1-1-4 clause 7.2.2(3) Accounts for the lack of correlation
between the wind pressures of walls exposed to windward and leeward
winds
▪ It depends on the slenderness of the buildings and only relatively stocky
buildings receive the full benefit, most slender buildings do not benefit
at all.
▪ This factor only applies to walls that are within the windward (zone D)
and leeward (E) wind areas and not those within the prevailing wind
areas
• Linear interpolation may be
carried out, provided the value of
h/d falls within the range
31

STRUCTURAL FACTOR
▪ The final coefficients applied is the structural factor.This can be
determined using clause 6.2 (1) in Eurocode 1-1-4 or clause NA 2.20 in
the National Annex.
▪ In the National Annex, the structural factor is divided into size cs and
dynamic coefficients cd.
▪ The structural factor accounts for the effects of:
▪ Non simultaneous occurrence of peak wind pressures on the surface Cs
▪ The vibration of the structure due to wind turbulence
▪ Table NA.3 and Figure NA.9 are used to derive the value of cs and cd
▪ Where the height of the structure is less than 15m.The value of cscd may
be conservatively taken equal to 1.0
32

STRUCTURAL FACTOR –SIZE FACTOR
• The size factors are
determined on the basis of
the zone in which the
structure falls when
estimating the exposure
coefficient in figures NA.7
& NA.8
• Linear interpolation within
the table is permitted.
33

STRUCTURAL FACTOR –DYNAMIC FACTOR
• The dynamic factors are determined
from figures NA.9.
• It value depends on the structural
damping δs, of the structure under
consideration.
• Values of the structural damping for
typical classes of structures are given in
Annex F3 of BS EN 1991-1-14:2005.
Typical values of δs for buildings
34

NET PRESSURE COEFFICIENT
• Clause NA 2.27 allows for the
generation of overall wind
loads by applying the net
pressure coefficients instead
of summing the pressures on
the windward and leeward
zones.
• The net pressure coefficient
replaces the external and
internal pressure coefficients.
35

WIND FORCES
▪ Haven determined the coefficients, the wind pressure is calculated by applying
the factors to the peak velocity pressure qp(z)
▪ The wind pressure acting on any segment of the building is given as:
▪ The overall wind load on the building is given as:
▪ The lack of correlation coefficient may be applied to zone D and E or when determining
the overall wind load on the building if the conditions for it application are satisfied.
▪ Partial factors must be applied to the wind loads.Wind loads are treated as imposed
loading within the Eurocode.
𝑤𝑘 = 𝑞𝑝(𝑧) ∙ 𝐶𝑝𝑒 −𝐶𝑝𝑖 ∙ 𝑐𝑠∙ 𝑐𝑑
𝑤𝑘 = 𝑞𝑝(𝑧) ∙ 𝐶𝑛𝑒𝑡 ∙ 𝑐𝑠∙ 𝑐𝑑
36

STRUCTURAL ANALYSIS
▪ The structural analysis aspect depends on the chosen lateral stability
system for resisting the wind loads.
• Steel bracing and shear walls are
alternative methods that can be
adopted to ensure lateral stability
of a building structure under
design wind loads
• The wind load can be converted
from pressure to point loads by
multiplying by a reference area
• This point loads can then be
apportioned to levels as shown in
the figure.
• For roofs, the wind loads are
treated in similar fashion to
imposed loading.
𝐹𝑘 = 𝑤𝑘 ∙ 𝐴𝑟𝑒𝑓
Aref = is the reference area
i.e. ( Area of the zone as
defined earlier)
37

A 25m x 50 x 10m, 14.5m at
heaves warehouse is to be
constructed 20km outside the city
of Lagos and 40 km from the
shoreline. Determine the wind
actions on all faces including roof
of this building assuming the
influence of orography is
insignificant.
Take site’s Altitude = 105 m above
MSL
According to the National Annex, The terrain of this site can be
classified as ________?
39

▪ Step 1- Fundamental VelocityVb,o
= 40 × 1.105 =
44.2𝑚/𝑠
40m/
s
𝑣𝑏,𝑚𝑎𝑝 = 40𝑚/𝑠
𝑐𝑎𝑙𝑡 = 1 + 0.001𝐴
=1+0.001(105)=1.105
𝒗𝒃,𝒐 = 𝟒𝟒. 𝟐𝒎/𝒔
40

▪ Step 2
▪ BasicWind VelocityVb
▪ BasicWind pressure qb
𝑣𝑏 = 𝑣𝑏,0 ∙ 𝑐𝑑𝑖𝑟∙ 𝑐𝑠𝑒𝑎𝑠𝑜𝑛
= 44.2 × 1 = 44.2𝑚/𝑠
𝒗𝒃 = 𝟒𝟒. 𝟐𝒎/𝒔
• The directional factor has been
taken equal to 1.0.
• The seasonal factor as also been
ignored since the structure is
deemed to be a permanent
structure
𝑞𝑏 =
1
2
𝜌 ∙ 𝑣𝑏
2
𝑊ℎ𝑒𝑟𝑒 𝜌 = 1.226
𝑞𝑝 =
1
2
× 1.226 × 44.22
= 1.198𝑘𝑁/𝑚2
𝒒𝒃 = 𝟏. 𝟏𝟗𝟖𝒌𝑵/𝒎𝟐
41

▪ Step 3
▪ PeakWind Pressure qb
• The building is considered to be
one part because h<b (14.5<50m)
• Since the effect of orography is not
significant, the simplified approach
can be used to determine the peak
velocity pressure.
• Site is located in a Open Country
Terrain Therefore ce,(T) =1 & hdis = 0
14.5
40km
ce,(Z) = 2.6
𝑞𝑝(𝑧) = 2.6 × 1.0 × 1.198 = 3.115𝑘𝑁/𝑚2
𝒒𝒃 𝒛 = 𝟑. 𝟏𝟏𝟓𝒌𝑵/𝒎𝟐
42

▪ Step 4: Coefficients of Wind Pressure
▪ External and Internal Pressure Coefficients
𝑏
2ℎ
• The building is rectangular
building with duo-pitch roof.The
wind zones are shown.
• By observation the zones are
greater than 1m2 therefore Cpe,10
are applicable
• There's no information regarding
the openings within this
structure therefore (assuming no
dominant opening) Cpi is taken
as the more onerous of -0.3 or
+0.2.
50
2(14.5) = 29
𝑒 = 29𝑚
43

▪ Step 4:
▪ External & Internal Pressure Coefficients Cpe,10
Roof pitch 𝛼 = 𝑡𝑎𝑛−1 4.5
12.5
= +19. 8°
ℎ
𝑑
=
15
30
=0.5
• Cpe, 10 values are obtained from table NA.4
and NA.7a
• NA.4 applies to the zones within the vertical
walls
• NA.7a applies to the zones in duo pitch roof
when the wind is acting perpendicular to
the length of the building.
• Zone F= -0.92 or + 0.39
• Zone G = -0.7 or +0.3
• Zone H= -0.35 or +0.24
• Zone I = -0.5 or -0.5
• Zone J = -1.18 or -1.18
Vertical
Walls
Cpe,10 values
• Zone A= -1.2
• Zone B =-0.8
• Zone D=+0.73
• Zone E=-0.37
Roof
• Cpi values = −0.3 𝑜𝑟 0.2
• Since building is less than 15m structural
factor CsCd =1.0
• Also as h/d <1 lack of correlation factor can
be applied to zones D and E. 44

▪ Step 4:Wind Loads
▪ Zone A : 𝑤𝑘 = 3.115 ∙ −1.2 − 0.2 × 1.0 = 4.36𝑘𝑁/𝑚2
▪ Zone B : 𝑤𝑘 = 3.115 ∙ −0.8 − 0.2 × 1.0 = 3.12𝑘𝑁/𝑚2
▪ Zone D: 𝑤𝑘 = 3.115 ∙ +0.73 − −0.3 × 1.0 × 0.85 = 2.73𝑘𝑁/𝑚2
▪ Zone E: 𝑤𝑘 = 3.115 ∙ −0.37 − 0.20 × 1.0 × 0.85 = 1.51𝑘𝑁/𝑚2
▪ Zone F: 𝑤𝑘 = 3.115 ∙ −0.92 − 0.2 × 1.0 = 3.49𝑘𝑁/𝑚2
▪ Zone G: 𝑤𝑘 = 3.115 ∙ −0.7 − 0.2 × 1.0 = 2.80𝑘𝑁/𝑚2
▪ Zone H: 𝑤𝑘 = 3.115 ∙ −0.35 − 0.2 × 1.0 = 1.71𝑘𝑁/𝑚2
▪ Zone I: 𝑤𝑘 = 3.115 ∙ −0.5 − 0.2 × 1.0 = 2.18𝑘𝑁/𝑚2d
▪ Zone J: 𝑤𝑘 = 3.115 ∙ −1.18 − 0.2 × 1.0 = 4.30𝑘𝑁/𝑚2
▪ OverallWind Load:
𝑤𝑘 = 𝑞𝑝(𝑧) ∙ 𝐶𝑛𝑒𝑡 ∙ 𝑐𝑠∙ 𝑐𝑑 = 3.115 × 0.95 × 1 × 0.85 = 𝟐. 𝟓𝟐𝒌𝑵/𝒎𝟐
• Net Pressure Coefficient : from chart in
NA.2.2.2 𝒄𝒏𝒆𝒕 = 𝟎. 𝟗𝟓
45

▪ Step 4: Structural Analysis
▪ Assuming the warehouse building is to be constructed from portal frames, the wind
load, is converted to uniformly distributed load by multiplying by spacing.
▪ Take spacing between frames = 3.75m
• Partial factors should be
applied
• Use any method to analyse
the frames
• Elastic Analysis
• Plastic Analysis
Typical Internal Frame
46

A 20 storey building is to be
constructed in the city centre of
Lagos.The site is located 20km from
the boundary of the city and 18km
from the shore. Determine the wind
load on the walls of this building
assuming the influence of orography
is insignificant.
Take site’s Altitude = 15m above MSL
According to the National Annex, The terrain of this site can
be classified as ________? 48

▪ Step 1- Fundamental VelocityVb,o
= 42 × 1.011 =
42.46𝑚/𝑠
40m/
s
𝑣𝑏,𝑚𝑎𝑝 = 42𝑚/𝑠
𝑐𝑎𝑙𝑡 = 1 + 0.001 ∙ 𝐴 ∙
10
𝑧𝑠
0.2
𝑍𝑠 = 0.6ℎ = 0.6 × 75 = 45𝑚
𝑐𝑎𝑙𝑡 = 1 + 0.001 ∙ 15 ∙
10
45
0.2
=1.011
𝒗𝒃,𝒐 = 𝟒𝟐. 𝟒𝟔𝒎/𝒔
49

▪ Step 2
▪ BasicWind VelocityVb
▪ BasicWind pressure qb
𝑣𝑏 = 𝑣𝑏,0 ∙ 𝑐𝑑𝑖𝑟∙ 𝑐𝑠𝑒𝑎𝑠𝑜𝑛
= 42.46 × 1 = 42.46𝑚/𝑠
𝒗𝒃 = 𝟒𝟐. 𝟒𝟔𝒎/𝒔
• The directional factor has been
taken equal to 1.0.
• The seasonal factor is also
ignored because the structure is
not temporary.
𝑞𝑏 =
1
2
𝜌 ∙ 𝑣𝑏
2
𝑊ℎ𝑒𝑟𝑒 𝜌 = 1.226
𝑞𝑝 =
1
2
× 1.226 × 42.462
= 1.11𝑘𝑁/𝑚2
𝒒𝒃 = 𝟏. 𝟏𝟏𝒌𝑵/𝒎𝟐
50

▪ Step 3
▪ Peak Wind Pressure qb
• The building must be considered in two parts since h>b
(75m>50m).
• A lower part extending from the ground up to a height equal to b=50m
and a upper part consisting of the remainder
• Since the effect of orography is also insignificant, the simplified
approach can be used to determine the peak velocity pressure.
• Site is located in a Town Terrain
• Information about the height of existing buildings is unknown,
therefore the default value of hdis can be used ℎ𝑑𝑖𝑠=3m
51

Step 3 Cont’d
PeakWind Pressure qb
(75-3)m
18km
(50-3)m
(75-3)m
(50-3)m
>20km
• Exposure
coefficient Ce(z) at
50m and 75 height
for the building is
3.68 & 3.85
respectively.
• Exposure modifier
Ce(T) at 50m and 75
height for the
building is 0.91 &
0.95 respectively.
𝑞𝑝(50) = 3.68 × 0.91 × 1.11 = 𝟑. 𝟕𝟐𝒌𝑵/𝒎𝟐
𝑞𝑝(75) = 3.85 × 0.95 × 1.11 = 𝟒. 𝟎𝟔𝒌𝑵/𝒎𝟐
52

▪ Step 4: Coefficients of Wind Pressure
▪ External & Internal Coefficients
ℎ
𝑑
=
75
50
= 1.5
Vertical
Walls
Cpe,10 values
• Zone A= -1.2
• Zone B =-0.8
• Zone D=+0.8
• Zone E=-0.53
• The building is rectangular
building with flat roof.Therefore
the wind zones defined in slide
28 can be used.
• By observation the zones are
greater than 1m2 therefore Cpe,10
are applicable
• There's no information regarding
the openings within this
structure therefore (assuming no
dominant opening) Cpi is taken
as the more onerous of -0.3 or
+0.2.
𝑏
2ℎ
50
2(75) = 150
𝑒 = 50𝑚
𝑒 = 𝑑
• Net Pressure Coefficient : from chart in NA.2.2.2
𝒄𝒏𝒆𝒕 = 𝟏. 𝟏𝟓
• Lack of Correlation factor: for h/d @1.5, =0.86
53

▪ Step 4: Coefficients of Wind Pressure Cont’d
▪ Structural Factor CsCd
From table NA.3 we can determine the size
factor, cs using :
𝑏 + ℎ = 75 + 50 = 125𝑚
𝑧 − ℎ𝑑𝑖𝑠 = (50 − 3)/(75 − 3)
Zone C applies
𝑐𝑠 = 0.85
From figure NA.9c we can determine the
dynamic factor, cd using :
δs= 0.1 (assuming reinforced concrete
building)
ℎ
𝑏
=
75
50
= 1.5
𝑐𝑑 = 1.05
54

▪ Step 4:Wind Loads
Zone A : 𝑤𝑘(50𝑚) = 3.72 ∙ −1.2 − 0.2 × 0.85 × 1.05 = 4.65𝑘𝑁/𝑚2
𝑤𝑘(75𝑚)= 4.06 ∙ −1.2 − 0.2 × 0.85 × 1.05 = 5.08𝑘𝑁/𝑚2
Zone B : 𝑤𝑘(50𝑚) = 3.72 ∙ −0.8 − 0.2 × 0.85 × 1.05 = 3.32𝑘𝑁/𝑚2
𝑤𝑘(75𝑚)= 4.06 ∙ −0.8 − 0.2 × 0.85 × 1.05 = 3.62𝑘𝑁/𝑚2
Zone D : 𝑤𝑘(50𝑚) = 3.72 ∙ 0.8 − −0.3 × 0.85 × 1.05 × 0.86 = 3.14𝑘𝑁/𝑚2
𝑤𝑘(75𝑚)= 4.06 ∙ 0.8 − −03 × 0.85 × 1.05 × 0.86 = 3.43𝑘𝑁/𝑚2
Zone E : 𝑤𝑘(50𝑚) = 3.72 ∙ −0.53 − 0.2 × 0.85 × 1.05 × 0.86 = 2.08𝑘𝑁/𝑚2
𝑤𝑘(75𝑚)= 4.06 ∙ −0.53 − 0.2 × 0.85 × 1.05 × 0.86 = 2.27𝑘𝑁/𝑚2
OverallWind Loads:
𝑤𝑘(50𝑚) = 3.72 ∙ 1.15 × 0.85 × 1.05 × 0.86 = 𝟑. 𝟐𝟖𝒌𝑵/𝒎𝟐
𝑤𝑘(75𝑚)= 4.06 ∙ 1.15 × 0.85 × 1.05 × 0.86 = 𝟑. 𝟓𝟖𝒌𝑵/𝒎𝟐
𝑤𝑘 = 𝑞𝑝(𝑧) ∙ 𝐶𝑛𝑒𝑡 ∙ 𝑐𝑠∙ 𝑐𝑑
55

Step 4: Structural Analysis
1. n
• The load is shared
according to the
stiffnesses of the lateral
stabilizing elements
e.g. shear walls or steel
bracing
56

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