The document discusses wind loads and earthquake loads on structures according to Eurocode standards. It provides classifications of loads and an overview of wind loads, defining wind as moving air with mass and kinetic energy. Wind speeds are typically measured 10m above the ground. It describes modelling wind actions through peak velocity pressure, force coefficients, and a structural factor. Terrain categories, roughness factors, and orography factors are defined for calculating mean wind speeds at different heights. Pressure coefficients are used to determine wind pressures on external and internal surfaces. Structural response is assessed through wind forces calculated from surface pressures and force coefficients.

1. UNIT THREE
Wind Loads &Earth Quake
Loads
As per
ES EN 1991-1-4:2015 & ES EN 1998:2015

2. Introduction
 Classification of loads
Area of application: Concentrated, Distributed (UDL)
Direction: Vertical (Gravity), Horizontal (Lateral)
Response: Static, Dynamic
Variation with time: Permanent (Dead), Variable (Live)
 Classification of loads in Building Codes
Permanent (Dead)
Variable (Live)/ Transient
Environmental Loads
• Wind
• Earthquake (Seismic)
• Snow
• Rain
• Earth pressure
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2

3.  Wind Load :
Wind is moving air
Air has mass/density and moves in a particular direction at a
particular velocity (has Kinetic Energy)
Obstacles that destructs the flow will transfer partially in to
potential energy of wind pressure /Wind load on Structures/
Winds are measured regularly at a large number of locations
The standard measurement is at 10m above the surrounding
terrain, which provides a fixed reference with regard to the
drag effects of the ground surface 3

4.  Are variable loads which act directly on the internal and
external surface
 The response of a structure to the variable action can be
components,
1. Background component
 It involves static deflection of the structure under the
wind pressure
2. Resonant component
 It involves dynamic vibration of the structure in response
to changes in pressure 4

5.  In most structures the resonant component is relatively small
and structural response to wind forces is treated using static
methods of analysis alone
Wind action:
• Wind loads act normal to the surface of the structure
• Negative/suction/_ when it acts away from the structure
• Positive/pressure – when it acts in to the structure
 Even though the wind loads are dynamic and highly variable,
the design approach is based on a maximum static load (i.e.,
pressure) equivalent
 It shall be calculated for each of the loaded areas under may be:
• The whole structure
• Parts of the structure, i.e. components, cladding units
and their fixings 5

6.  Wind effects induce: 6
 forces
 vibrations, and
 in some cases instabilities in the overall structure as
well as its non-structural components

7. Modelling of wind actions
 The wind actions calculated using ES EN 1991-1-4:2015 are
characteristic values.
 The basic values are characteristic values having annual
probabilities of exceedance of 0.02, which is equivalent to a mean
return period of 50 years.
 The effect of the wind on the structure depends on
• Size
• shape
• dynamic properties of the structure
 The response of structures should be calculated from
• Peak velocity pressure, qp
• Force and pressure coefficients and
• Structural factor, cscd
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8. Wind Velocity and Velocity Pressure
 Peak velocity pressure qp depends on the wind climate, the terrain
roughness and orography, and the reference height.
 The basic wind velocity shall be calculated
Vb = Cdir ⋅ Cseason ⋅ V b0
where:
vb is the basic wind velocity, defined as a function of wind
direction and time of year at 10 m above ground of terrain
category II
vb0 is the fundamental value of the basic wind velocity
cdir is the directional factor, recommended value is 1.0
cseason is the season factor , recommended value is 1.0
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9.  If 10 minutes mean wind velocity having the probability p for
annual exceedance vb is multiplied by the probability factor, Cprob
𝐶𝑝𝑟𝑜𝑏 =
1−𝐾.ln − ln 1−𝑝
(1−𝐾.ln − ln 0.98
𝑛
where:
K is the shape parameter, recommended value is 0.2
n is the exponent, recommended value is 0.5
 mean wind velocity vm(z) at a height z above the terrain depends on
• terrain roughness
• orography and
• basic wind velocity, vb
vm(z) = Cr (z) ⋅Co(z) ⋅Vb
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10. where:
Cr(z) is the roughness factor,
Co(z) is the orography factor, taken as 1.0 otherwise specified in 4.3.3
Cr(z)=𝑘𝑟. ln
𝑍
𝑍0
… … … … … … … 𝑧𝑚𝑖𝑛 ≤ 𝑧 ≤zmax
Cr(z)=𝐶𝑟. ln 𝑧𝑚𝑖𝑛 … … … … … … … 𝑧 ≤ 𝑧𝑚𝑖𝑛
where:
z0 is the roughness length
kr terrain factor calculated using
kr=0.19
𝑧0
𝑧0
,
𝐼𝐼
0.07
where:
z0,II = 0.05 m (terrain category II, Table 4.1)
zmin is the minimum height defined in Table 4.1
zmax is to be taken as 200 m, unless specified in the National Annex
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11. Terrain category Z0 m Zmin m
0. Sea or coastal area exposed to the open sea 0.003 1
I. I Lakes or flat and horizontal area with
negligible vegetation and without obstacles
0.01 1
II. Area with low vegetation such as grass and
isolated obstacles (trees, buildings) with
separations of at least 20 obstacle heights
0.05 2
III. Area with regular cover of vegetation or
buildings or with isolated obstacles with
separations of maximum 20 obstacle heights
(such as villages, suburbain terrain, permanent
Forest)
0.3 5
IV. Area in which at least 15 % of the surface is
covered with buildings and their average height
exceeds 15 m
1.0 10
Table 4.1 — Terrain categories and terrain parameters
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12. Orography factor (Co(z) )
 It accounts for the increase of mean wind speed over isolated
hills and escarpments
Figure A.1 — Illustration of increase of wind velocities over orography
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13.  The effects of orography should be taken into account in the
following situations
a) For sites on upwind slopes of hills and ridges:
- where 0.05 < Φ ≤ 0.3 and 𝑥 ≤ Lu / 2
b) For sites on downwind slopes of hills and ridges:
- where Φ < 0.3 and x < Ld/2
-where Φ ≥ 0.3 and x < 1.6H
c) For sites on upwind slopes of cliffs and escarpments:
- where 0.05 < Φ ≤ 0.3 and 𝑥 ≤ Lu / 2
d) For sites on downwind slopes of cliffs and escarpments:
- where Φ < 0.3 and x < 1.5Le
- where Φ ≥ 0.3 and x < 5H
 It is defined by:
Co =1 for ∅ <0.05
Co=1+2S.∅ for 0.05 < ∅ <0.3
Co=1+0.6S for ∅ >0.3
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14. Where:
S the orographic location factor,
∅ the upwind slope H/Lu in the wind direction
Le the effective length of the upwind slope,
Lu the actual length of the upwind slope in the wind direction
Ld the actual length of the downwind slope in the wind direction
H the effective height of the feature
X the horizontal distance of the site from the top of the crest
z the vertical distance from the ground level of the site
 Values of the effective length Le.
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15. 6/4/2023
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16. 6/4/2023
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17.  Wind Turbulence Iv(z): is the standard deviation of the turbulence
divided by the mean wind velocity
σv = kr ⋅ Vb ⋅ kl
Iv(z)=(
σv
𝑉𝑚
(𝑧)
) =
𝑘𝑙
𝐶𝑜
𝑍 .ln(z/𝑧0
)
… … … … … … 𝑓𝑜𝑟 𝑧𝑚𝑖𝑛 ≤ 𝑧 ≤ 𝑧𝑚𝑎𝑥
Iv(z)=Iv(zmin)...............................................for z< 𝑧𝑚𝑖𝑛
where:
Iv(z) is turbulence intensity at height z
σv is standard deviation
kl is the turbulence factor, recommended value is k1 = 1.0.
co is the orography factor
z0 is the roughness length
 Peak velocity pressure qp(z) at height z, which includes mean and
short-term velocity fluctuations
qp z = 1 + 7. Iv z .
1
2
𝜌. Vm
2(z)=Ce(z).qp
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18.  𝑤ℎ𝑒𝑟𝑒:
𝝆 is the air density, which depends on the altitude, temperature
and barometric pressure to be expected in the region during
wind storms
 The value of the air density ρ may be given in the National
Annex. The recommended value is 1.25 kg/m3.
ce(z) is the exposure factor
ce(z)=
𝑞𝑝
(𝑧)
𝑞𝑏
qb is the basic velocity pressure
qb=
1
2
. 𝜌. 𝑣𝑏
2
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19. For flat terrain where co(z) = 1.0, kl=1.0, exposure factor Ce(z) is:
Figure 4.2 Illustrations of the exposure factor ce(z) for co=1.0, kl=1.0
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20. Wind Pressure On Surfaces
 Wind pressure acting on external surfaces, we , should be obtained
We = qp (ze ) ⋅ cpe
where:
qp(ze) is the peak velocity pressure
ze is the reference height for external pressure
cpe is the pressure coefficient for external pressure
 Wind pressure acting on internal surfaces, wi , should be obtained
Wi = qp (zi ) ⋅ cpi
where:
zi is the reference height for internal pressure
cpi is the pressure coefficient for internal pressure
Net pressure is the difference between the pressures on the opposite
surfaces taking due account of their signs
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21.  Pressure is directed towards the surface is taken as positive
 Suction is directed away from the surface as negative
Figure 5.1 — Pressure on surfaces
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22. Wind Forces
 The wind forces for the whole structure or a structural component
should be determined
1. by calculating forces using force coefficients
Fw = cscd . cf . qp (ze ) . Aref or Fw = cscd . 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 cf .qp (ze ) . Aref
2. summation of the forces Fwe, Fwi and Ffr by calculating forces
from surface pressures
External force: Fwe = cscd. 𝐬𝐮𝐫𝐟𝐚𝐜𝐞𝐬 We . Aref
Internal force: Fwi = 𝐬𝐮𝐫𝐟𝐚𝐜𝐞𝐬 Wi . Aref
friction forces: Ffr = cfr ⋅ qp (ze ) ⋅ Afr
where:
cscd is the structural factor
cf is the force coefficient for the structure/structural element
we is the external pressure on the individual surface at height ze
wi is the internal pressure on the individual surface at height zi
Aref is the reference area of the individual surface
cfr is the friction coefficient
Afr is the area of external surface parallel to the wind
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23. Structural Factor cscd
 Structural factor cscd should take into account the effect on wind
actions from the non-simultaneous occurrence of peak wind
pressures on the surface together with the effect of the vibrations
of the structure due to turbulence
 Take cscd 1.0 for:
• buildings height less than 15 m
• facade and roof elements natural frequency greater than 5 Hz
• framed buildings which have structural walls less than 100 m
high and whose height is less than 4 times the in-wind depth,
• chimneys with circular cross-sections whose height is less
than 60 m and 6.5 times the diameter
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24.  Otherwise
cscd =
1+2𝑘𝑝.𝑙𝑣 𝑧𝑒
. 𝐵2+𝑅2
1+7.𝑙𝑣(𝑧𝑒)
where:
• ze is the reference height.
• kp is the peak factor
• Iv is the turbulence intensity defined
• B2 is the background factor see in ES EN 1991:2015 Annex B-2
• R2 is the resonance response factor see in ES EN 1991:2015Annex B-2
 Size Factor (cs) : account reduction effect due to the non-
simultaneity of peak wind pressures
 Dynamic Factor (cd): account effect from vibrations due to
turbulence in resonance
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25. The Reference Height
Figure 6.1 — General shapes of structures covered by the design procedure.
The structural dimensions and the reference height used are also shown.
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26. cscd values for different types of structures
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27. 6/4/2023
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28. Friction coefficients
 Wind friction disregarded when the total area of all surfaces parallel
with (small angle to) the wind is equal to or less than 4 times the
total area of all external surfaces perpendicular to the wind.
 Friction forces should be applied on the part of the external surfaces
parallel to the wind, located beyond a distance from the upwind
eaves or corners, equal to the smallest value of 2b or 4h.
Surface Friction coefficient cfr
Smooth
(i.e. steel, smooth concrete)
0.01
Rough
(i.e. rough concrete, tar-boards)
0.02
very rough
(i.e. ripples, ribs, folds)
0.04
Table 7.10 — Frictional coefficients cfr for walls, parapets and roof surfaces
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29. Figure 7.22 — Reference area for friction
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30. Pressure coefficients
 Depend on the size of the loaded area
Cpe = Cpe,1 ……………………………….for A≤1m2
Cpe = Cpe,1 - ( Cpe,1 - Cpe,10)log10A…….…for 1m2<A<10m2
Cpe = Cpe,10……………………………….for A≥10m2
 Cpe,10 and Cpe,1 should be used for the orthogonal wind directions
0°, 90°, 180°
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31. Vertical walls of rectangular plan buildings
 Values of external pressure coefficients for different cases are given
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32. NOTE: The velocity pressure should be assumed to be uniform over each horizontal strip
considered.
Figure 7.4 — Reference height, ze, depending on h and b, and corresponding velocity
pressure profile
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33. The external pressure coefficients cpe,10 and cpe,1 for zone A, B, C, D and E
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34. Figure 7.5 — Key for vertical walls
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35. Table 7.1 — Recommended values of external pressure coefficients for vertical walls
of rectangular plan buildings
 For buildings with h/d ≥ 5,the resulting force is multiplied by 1.
 For buildings with h/d ≤ 1,the resulting force is multiplied by 0.85.
 For intermediate values of h/d, linear interpolation may be applied.
NOTE:
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36. Flat roofs
 Flat roofs are defined as having a slope (𝛼) of –5°< 𝛼 < 5°
Figure 7.6 — Key for flat roofs
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37. Table 7.2 — External pressure coefficients for flat roofs
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38. Monopitch roofs
Figure 7.7 — Key for monopitch roofs
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39. Table 7.3a — External pressure coefficients for monopitch roofs
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40. Table 7.3b — External pressure coefficients for monopitch roofs
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41. Duo pitch roofs
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42. Table 7.4a — External pressure coefficients for duopitch roofs
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43. Table 7.4b — External pressure coefficients for duopitch roofs
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44. Hipped roofs
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45.  Multispan Roofs
 Vaulted roofs and domes
Reading assignment
Table 7.5 — External pressure coefficients for hipped roofs of buildings
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46. Internal Pressure
 Internal and external pressures shall be considered to act at the same
time
 The worst combination of external and internal pressures shall be
considered for every combination of possible openings and other
leakage paths.
 The internal pressure coefficient, cpi, depends on the size and
distribution of the openings in the building envelope.
 When in at least two sides of the buildings (facades or roof) the
total area of openings in each side is more than 30 % of the area of
that side, the face is dominant.
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47. Cont…..
 Area of the openings at the dominant face is twice the area of the
openings in the remaining faces
Cpi = 0.75 ⋅ Cpe
 Area of the openings at the dominant face is at least 3 times the area
of the openings in the remaining faces
Cpi = 0.90 ⋅ Cpe
 Area of the openings at the dominant face is between 2 and 3 times
use interpolation
 For buildings without a dominant face, the opening ratio μ for each
wind direction θ
𝝁 =
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑝𝑒𝑛𝑖𝑛𝑔𝑠 where cpe is negative or − 0.0
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎𝑙𝑙 𝑜𝑝𝑒𝑛𝑖𝑛𝑔
NOTE: This applies to façades and roof of buildings with and without internal partitions
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48.  If not possible to estimate 𝜇 for a particular case then Cpi should be
taken as the more onerous of +0.2 and -0.3.
 The internal pressure coefficient of open silos and chimneys
Cpi=-0.6
 The internal pressure coefficient of vented tanks with small openings
Cpi=-0.4
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49. Local Effects of Wind Pressure
• Wind around a corner
49
Images from FEMA Multi Hazard Seminar
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50. • Uplift on roof
50
Images from FEMA Multi Hazard Seminar
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51. Wind Load Example
 The building shown in below is to be built in a sloped terrain in
Debre Markos town. 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.
 Provide
• Six windows of 2.25m * 2.5m size on each longer sides
• Two windows of 1.75m * 2.5m size on each shorter sides
• One door of 2.5m * 3m on each shorter sides.
 Elevation of the existing ground level at the building site is 1820m
 Trusses spacing 3m centers
 Calculate the wind load acting on the roof trusses

52. 52
Side elevation
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53. 6/4/2023 53

54. Solution
 basic wind velocity:
vb = Cdir × Cseason × vb,0
 Fundamental value of basic wind velocity (see Ethiopian wind map)
vb,0 = 22 m/s (for Debre markos – Ethiopia)
vb = Cdir × Cseason × vb,0=1*1*22=22m/s
 Basic velocity pressure, qp
The value of the air density ρ in the National Annex is 1.25 kg/m3
Basic velocity pressure, qp= ½ ρ vb
2 = ½ * 1.25* 222
=302.5N/m2
 Peak pressure qp(z)
qp z = 1 + 7. Iv z .
1
2
𝜌. Vm
2(z)
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55.  Mean wind velocity vm(z)
vm(z) = Cr (z) ⋅Co(z) ⋅Vb
 Debre Markos considered as a sub urban area and therefore the
terrain category falls as category III
 Z0,II=0.05m , Z0 =0.3m , Zmin = 5m, Z=6.1m and Zmax=200m
 So, z=6.1m > zmin =5m
 Kr=0.19
𝑧0
𝑧0
,
𝐼𝐼
0.07
= 0.19
0.3
0.05
0.07
=0.215
Cr(z)=𝑘𝑟. ln
𝑍
𝑍0
… … … … … … … 𝑧𝑚𝑖𝑛 ≤ 𝑧 ≤zmax
=0.215*ln(
6.1
0.3
)=0.648
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56. Orography factor, Co(z)
 The sites are downwind slopes of cliffs and escarpments:
- where Φ < 0.3 and x < 1.5Le
- where Φ ≥ 0.3 and x < 5H
 0.06 < 0.3 and 200m < 1.5*500=750 so, account effects of
orography
Co(z ) =1 for ∅ <0.05
Co(z ) =1+2S.∅ for 0.05 < ∅ <0.3
Co(z ) =1+0.6S for ∅ >0.3
∅ = downwind slope H/Lu in the wind direction = 30/500 = 0.06
Le = Effective length of downwind slope = Ld , if 0.05 < Φ < 0.3
X=200m
𝑋
𝐿𝑒
=
200
500
= 0.4 and Z=6.1m,
𝑍
𝐿𝑒
=
6.1
500
= 0.0122
S =0.6 read from the graph Figure A.2 — Factor s for cliffs and
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57. Co(z ) =1+2S.∅=1+2*0.6*0.06=1.072
∴vm(z) = Cr (z) ⋅Co(z) ⋅Vb =0.648*1.072*22= 15.28m/s
Wind Turbulence:
Iv(z)=
𝑘𝐼
𝐶𝑜
𝑍 .ln(z/𝑧0
)
𝑧𝑚𝑖𝑛 ≤ 𝑧 ≤ 𝑧𝑚𝑎𝑥
=
1
1.072∗𝑙𝑛(6.1/0.3 )
= 0.301
 Peak velocity pressure qp(z)
 𝐪𝐩 𝒛 = 1 + 7. Iv z .
1
2
𝜌. Vm
2(z)
=[1+7*0.301]*
1
2
∗ 1.25*15.282
= 453.39N/m2
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58. External pressure coefficient Cpe:
I. Wind direction θ=00
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59. 6/4/2023 59

60.  The Values of Cpe are given in Table 7.4a for different values of
pitch angle α,in our case α=150
 For zone F
Cpe= Cpe1-(cpe1-cpe10)log10A for 1m2<A<10m2
= -2.0-(-2.0-(-0.9))log103.721= -1.372
zone A(m2) cpe1(-ve) cpe1(+ve) cpe10(-ve) cpe10(+ve) cpe(-ve) cpe(+ve)
F 3.721 -2 0.2 -0.9 0.2 -1.372 0.200
G 29.158 -1.5 0.2 -0.8 0.2 -0.800 0.200
H 143.4 -0.3 0.2 -0.3 0.2 -0.300 0.200
I 143.4 -0.4 0 -0.4 0 -0.400 0.000
J 36.6 -1.5 0 -1 0 -1.000 0.000
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61. Face area(m
2
) Area of openings(m2) Percentage(%)
Long pan 00 135 33.75 25
Long pan 1800 135 33.75 25
Pinion -900 63.6 16.25 25.55
Pinion +900 63.6 16.25 25.55
00 slope 180 0.00 0.00
1800 slope 180 0.00 0.00
 Determination of the percentage of the opening areas of the openings
When examining the areas of the openings, no face has more than
30% open area. So we are in the presence of a building with out a
dominant face.
Calculation of the width e
e = min(b;2h) = min(30 m;2*6.1m)=12.2m
Conclusion: When e ≥ d (depth of the building is d = 12 m), there is
no C zone
Calculation of Internal pressure coefficient Cpi:
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62.  Determining of zones that cpe values are negative or -0.0(the
exception of zone D,all cpe values are negative)
𝝁 =
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑝𝑒𝑛𝑖𝑛𝑔𝑠 where cpe is negative or − 0.0
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎𝑙𝑙 𝑜𝑝𝑒𝑛𝑖𝑛𝑔
𝝁 =
6∗2.25∗2.5+4∗1.75∗2.5+2∗2.5∗3.0
6∗2.25∗2.5+4∗1.75∗2.5+2∗2.5∗3.0+6∗2.25∗2.5
=
66.25𝑚2
100𝑚2 = 0.6625
 Reading the Coefficient cpi
0.25 <
ℎ
𝑑 = 0.5083 < 1
 Interpolation for Determining cpi ,when 𝝁 = 𝟎. 𝟔𝟔𝟐𝟓
ℎ
𝑑 cpi
0.25 -0.03
0.5083 ? Interpolate cpi = -0.0541
1.0 -0.10
Calculation of
ℎ
𝑑
=
6.1
12
= 0.5083
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63. The summary of the Cpe and Cpi values are shown below in a table
zone A(m2) cpe(-ve) cpe(+ve) cpi (cpe -cpi)-ve (cpe -cpi)+ve
F 3.721 -1.372 0.200 -0.0541 -1.318 0.254
G 29.158 -0.800 0.200 -0.0541 -0.746 0.254
H 143.4 -0.300 0.200 -0.0541 -0.246 0.254
I 143.4 -0.400 0.000 -0.0541 -0.346 0.054
J 36.6 -1.000 0.000 -0.0541 -0.946 0.054
 Net wind pressure
Wnet = We +Wi = 𝑞𝑝 𝑧𝑒 [cpe + cpi]
cscd =1, since z=6.1m<15m
Disregarded Wind friction……. Check area of surfaces
 Calculate the wind load on the roof truss
• Truss spanned across the zones FHIJ and
• Truss spanned across the zones GHIJ
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64.  Truss spanned across the zones FHIJ: Out of the zones,
coefficient for F is critical:
Wnet = We – Wi = qp ze [cpe – cpi]
= 0.45339 * [-1.372 – (-0.0541)]
= - 0.598 kN/m2
Truss spanned across the zones GHIJ: Out of the zones,
coefficient for J is critical:
Wnet = We – Wi = qp ze [cpe – cpi]
= 0.45339 * [-1.0 – (-0.0541)]
= - 0.429 kN/m2
 This Wnet acts on the roof covering, which is supported by purlins.
Purlins are supported by the truss.
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65.  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.
Purlins
Roof truss
Roof covering
6.40 m
Center to center distance between
the purlins = 6.40/4 = 1.6m
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66.  Center to center distance between the truss = 3m
Loads on purlin and truss Truss spanned across 1st middle truss, spanned
across FHIJ and GHIJ
FHIJ GHIJ
End purlin load (kN/m) -0.598 *0.5* 1.6
= -0.4784
-0.429 *0.5*1.6
=-0.3432
-0.5*(0.4784+0.3432)
=-0.4108
Middle purlin load
(kN/m)
-0.598 * 1.6
= -0.9568
-0.429 * 1.6
= -0.6864
-0.5*(0.9568+0.6864)
=-0.8216
end joints load on rafter
truss (kN)
0.5*(-0.4784)*3
=-0.7176
0.5*(-0.3432)*3
=-0.5148
0.5*(-0.4108)*3
=-0.6162
middle joints load on
rafter truss (kN)
0.5*(-0.9568)*3
=-1.4352
0.5*(-0.6864)*3
=-1.0296
0.5*(-0.8216)*3
=-1.2324
1.4352
0.7176
0.7176 Figure: end truss analysis
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67. 67
1.2324
0.6162
0.6162 Figure: 1st middle truss analysis
1.0296
0.5148
0.5148 Figure: middle truss analysis
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68. From our project
Loads on purlin and truss Truss spanned across
End purlin load (kN/m) -2.134*0.5*1.25=-1.33
End purlin load (kN/m) -2.134*1.25=-2.67
end joints load on rafter truss (kN) 0.5*-1.33*5=-3.325
middle joints load on rafter truss (kN) -1.33*5=-6.65
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69. Load on the wall
• Maximum positive governing load =0.6606KN/m2
• Maximum negative governing load =-1.39KN/m2
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