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Design of wind actions on a portal frame building
according to EN 1991 -1-4(2005)
Basic in format ion
1 Total length 2 bay width 3 spacing 4 height(max)
5 roof slope H. aboveground
b=70 m d=25 m s= 7 m h= 7.5 m slope α=5° h’=7.5 -12.5
tan5°=6.41m
Fig1: one storey building frame
Fig 2 : transversal
6.4m
5°
25 m
7.5m
70.0 m
25.0 m
7.5 m
6.4 m
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Basic values
Determinat ion of b asic wind ve lo c i ty
Fig 3 : Actions of wind on surfaces
,0b dir season bv c c v
EN 1991-1-4(2005) 4.2
bv is the basic wind velocity
dir c is the directional factor
seasonc is the seasonal factor
,0bv
is the fundamental value of the basic wind velocity
(obtained from meteo center national annex)
take ,0bv =30m /s
terrain category III → z 0 =0.3 m and z min=5 m EN 1991-1-4(2005) 4.3.2 (table 4.1)
z=7.5 m we have z≥ z min
,0 30 /b dir season bv c c v m s
we take dir c = seasonc =1.0 for the unfavorable case because c dir and c season are used generallyas reduction factors
Basic v eloci ty pr essure
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21
2 p air b
q v EN 1991-1-4(2005) 4.5 eq 4.10
Where recommended valueair =1.25kg/m
3 (air density)
2 21 *1.25*30 562.5 /2
pq N m
Peak pressure
21
1 72
p v mq z I z v z EN 1991-1-4(2005) 4.5 eq 4.8
Calculation of mv z
0m r bv z c z c z v
0c z is the orography factor
r c z is the roughness factor
0
logr r
z c z k
z
case where min max z z z → 5 7.5 200m m m
r k Is the terrain factor depending on the roughness length z 0 calculated using
0.07
0
0,
0.19r II
z k
z
Calculation of the turbulence intensity v I z
00
log
I v
k I
z c z
z
for z min ≤ z≤z max EN 1991-1-4(2005) 4.4 eq 4.7
minv v I I z for z≤ z min
where I k is the turbulence factor (recommended value is I k =1.0)
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Fig 4: description of (+) and(-) wind
1/ nigh t si tuat ion → all the building is closed
External pressu re coeff ic ients
Wind 1 θ=0°
The wind pressure acting on the external surface,w e
Should be obtained from the following expression
.e p e pew q z c EN 1991-1-4(2005) 5.2 eq 5.1
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e z is the reference height for the external pressure
pec is the pressure coefficient for the external pressure depending on the size of the loaded
area A
Fig 5: euro curve for c pe
= ,10 pec EN 1991-1-4(2005) 7.2.1 2 21 10if m A m then ,1 ,1 ,10 10log pe pe pe pec c c c A
a)Vert ical walls
For7.5
0.3 0.2525
h
d then a linear interpolation is needed
1st point A(0.25;0.7) and 2 nd point B(1;0.8)
f x mx p we have0.8 0.7 2
1 0.25 15
B A
B A
y ym
x x
2 2
0.25 *0.25 0.715 3
f p p
2 2
15 3 f x x then
2 20.3 0.3 0.71
15 3 f
This value will be neglected to the table value EN 1991-1-4(2005) table 7.4a
D c pe=0.7 and E c pe=-0.3
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fig 6: euro fig top vew + zone actions
b)Duo pitch roofs
α=5.0° ; θ=0°(wind direction) ; e=min(b;2h) =min(75;15) =15m
Fig 7: euro fig transverversal vew
zon e F G H I J
c pe -1.7 -1.2 -0.6 -0.6 -0.6
J: two cases are considered c pe=+ 0.2 /-0.6
but c pe=-0.6 EN 1991-1-4(2005) table 7.4a note1
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Fig 8 : top vew and zone details
Internal pressure coeff ic ients
The wind pressure acting on the internal surfaces of the structure w i is obtained from the
expression EN 1991-1-4(2005) (5.2) eq 5.2
i p i piw q z c
i z is the reference height for thr internal pressure
pic is the pressure coefficient for the internal pressure
The internal pressure coefficient depends on the size and distribution of the opening in the
structure envelope
We calculate it from the EN 1991-1-4(2005) (7.2.9 (6))
h/d=7.5/25=0.3 →assimilation to h/d=0.25 (see section above)
when we haven’t information about opening we take the more unfavorable
of (+0.2) and (-0.3) EN 1991-1-4(2005) (7.2.9 (6)note 2)
In this case 0.2 pic
Wind loads
pe pi pw c c q s S= 7 m (spacing)
7 * 0.668 4.676 pe pi pe piw c c c c applied to a portal frame (KN/m)
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*0.668 pe piw c c applied to surfaces(KN/m2 )
Table 1 θ=0°
zone A B C D E F G H I J
c p e -1.2 -0.8 -0.5 0.7 -0.3 -1.7 -1.2 -0.6 -0.6 -0.6
c p i +0.2 +0.2 +0.2 +0.2 +0.2 +0.2 +0.2 +0.2 +0.2 +0.2
c p e -c p i -1.4 -1.0 -0.7 +0.5 -0.5 -1.9 -1.4 -0.8 -0.8 -0.8
W(KN/m)
Port.frame
-6.55 -4.68 -3.28 +2.34 -2.34 -8.88 -6.55 -3.74 -3.74 -3.74
W(KN/m ) 0.935
0.668
0.468 0.334
0.334
-1.269 -0.935 0.534 0.534 0.534
Surfaces
(m 2 )
75*6.4
=48075*6.4
=480
15152
4 10
=11.25
62.5*1.5
=93.75
70*11=
770
70*11=
770
70*1.5
=105
Value
KN
160.32
160.32
14.28 87.66 411.18 411.18 56.07
The A,B,or C zone is not considered because of the symetric actions
Fig 9: wind distribution actions on duopitch roofs
1.5
w1
-6.55 (G)
-8.88(F)
11111.5
-2.34 (E )
-3.74 (I )
-3.74 (H )
+2.34(D)
-3.74 ( J )
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d=25m
EN 1991(2005)7.2.2 fig 7.4
→+0.7
→wind D b=70m E →-0.3
A B C
A=e/5 B=4e/5 C=d-e
=3m =12m =10m EN 1991-1-4:2005 7.2.2 (2)
Fig 10 wind distributions on vertical walls
Determination of wind act ions forces
. .w s d f p e ref éléments
F c c c q z A EN 1991-1-4(2005) 5.2.2( eq5.3)
1.0 s d
c c EN 1991-1-4(2005) (6.2 note 1.c)
The decomposition is V =zone*cosα=zone*0.996 and H=zone*sinα=zone*0.087
Point appl ication of forces
i i
H
i
xT X
T
160.32* 0 160.32* 25 2*1.25 7.66 *0.75 35.77* 7 35.77*19.5 4.88*13.25315.36
14.33m
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35.0i i
H
i
m yT
Y T
i i
H
i
T z Z
T
2*160.32*3.2 2*1.25 7.66 *6.06 35.77*7.02 35.77* 6.89 4.88*7.44315.36
3.29m
i i
V
i
xT X
T
14.23*0.75* 2 87.31*0.75 409.54*7 409.54*19.5 55.85*13.25
990.7
11.79m
i i
V
i
yT Y
T
35.0m
i i
V
i
T z Z
T
14.23*6.06* 2 87.31*6.06 409.54*7.02 409.54*6.89 55.85* 7.44990.7
6.88m
Fig 11 application forces
x
yz
ED
I
J
H
F2
F1
G
w1
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Table 2 wind 1( θ=0°) horizontal and vertical forces to Aref
Zone Horizontal
H (KN)
vertical
V (KN)
Coordinates (m)
X Y Z
D 160.32 0 0 35 3.20
E 160.32 0 25.0 35 3.20
F 1 14.28sin5 1.25 14.28cos5 14.23 0.75 1.875 6.06
F 2 14.28sin5 1.25 14.28cos5 14.23 0.75 68.125 6.06
G 87.66*0.087 7.66 87.66*0.996 87.31 0.75 35 6.06
H 411.18*0.087 35.77 411.18*0.996 409.54 7.0 35 7.02
I 411.18*0.087 35.77 411.18*0.996 409.54 19.5 35 6.89
J 56.07*0.087 4.88 56.07* 0.996 55.85 13.25 35 7.44
R x 315.36 14.33
H X 35.0
H Y 3.29 H Z
R z
990.7 11.79V X 35.0V Y 6.88V Z
12.5m
W
Fig 12 force positions
fo rces act ion s
14.33 m
6.88 mR x
3.29 m
R z
Z
25.0 m
BA
11.79 m
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Overturn in g mom ent
M ov =R x *3.29+R z *(25-11.79)
=315.36*3.29+990.7*13.21
=14125 KNm
stab i liz ing mom ent
take 0.6 KN/m2 as an approximative value of self-weight then(but not final)
W=0.6*70*25=1050 KN
M w =1050*12.5=13125 KNm
Stabil i ty : 14125-13125=1000 KNm
The structure is not stable then the difference will be taken by anchors
Wind 2 a/ θ=90°
b/ θ=180°
we consider that the two cases are similar
e90 =e180 =min(b;2h) =min(25;15) =15m EN 1991-1-4:2005 (7.2.5 Figure 7.8)
Fig 13 application forces
G
H
y
xz
E
D
II
H
F F
G
w2
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zone F G H I
c pe -1.6 -1.3 -0.7 -0.6
Internal pressure coeff ic ients
The wind pressure acting on the internal surfaces of the structure w i is obtained from the
expression
i p i piw q z c
i z is the reference height for thr internal pressure
pic is the pressure coefficient for the internal pressure
The internal pressure coefficient depends on the size and distribution of the opening in the
structure envelope
h/d=7.5/70=0.1≤0.25
when we haven’t information about opening we take the more unfavorable
of (+0.2) and (-0.3) EN 1991-1-4(2005) (7.2.9 (6)note 2)
In this case 0.2 pic
Wind loads
* pe pi pw c c q = *0.668 pe pic c EN 1991-1-4(2005) (7.2.9 (6)note 2)
Table 3 details in zone
zon e A B C D E F G H I
c pe -1.2 -0.8 -0.5 +0.7 -0.3 -1.6 -1.3 -0.7 -0.6
c p i +0.2 +0.2 +0.2 +0.2 +0.2 +0.2 +0.2 +0.2 +0.2
Cpe-cpi -1.4 -1.0 -0.7 +0.5 -0.5 -1.8 -1.5 -0.9 -0.8
W(KN/
m 2 )
-0.935 -0.668 -0.468 +0.334 -0.334 -1.202 -1.0 -0.601 -0.534
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Fig 14 distribution actions
Determinat ion of fr ic tion forces EN 1991-1-4(2005) (5.3 note 4)
The effects of wind friction on the surface can be disregarded when the total area of all surfaces parallel
With the wind is equal or less than 4 times the total area of all external surfaces perpendicular to the wind
Considered only wind θ=90°
fr Fr p e ref F c q z A EN 1991-1-4(2005) (5.3 eq 5.7 )
Fr c is the friction coefficient EN 1991-1-4(2005) (7.5 table 7.10)
p eq z see above pq z
ref A
is the area of external surface parallel to the wind EN 1991-1-4(2005) 7.5
C fr =0.04 (ripples,ribs,folds..)
ref A
= width*lenght
Lenght L=min(2b ;4h) =min(2*25,4*7.5) =30m EN 1991-1-4(2005) 7.5(3)
Width
a) vert ical w al ls 6504*2=13 m
b) duop i tch roo fs 2*( 12.5/cos(5°))=25.1 m
take only a lineair distribution forces to all faces(apply only for q p(7.5))
1.0 (G)
0.601 (H)
0.534 (I)
1.202 (F )
1.5
0.935 A
0.668 (B)
0.47 C
1.202 (F )
0.668 (B)
0.47 C
1.0 (G)
0.601 (H)
0.534 (I)
221.5
0.935 (A)
wind
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G 15 1525 26.25
2 10
26.25*1.0 26.25 0
H 15 1525 150
2 10
150* 0.601 90.15 0
I 1525 70 1562.5
2
1562.5*0.534 834.375 0
F fr 10.42 20.12 30.54
R z =964.297 R x =+146.606
Fig16 force point
fo rces act ion s
Overturn in g moment
M ov =R x *3.2+R z *35
=146.606*3.2+964.297*35
=43219.5 KNm
stab i liz ing mom ent
take 0.6 KN/m2 as an approximative value of self-weight then(but not final)
7.5 m R x3.2 m
R zZ
25.0 m
BA
12.5 m
12.5 m
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W=0.6*70*25=1050 KN
M w =1050*35=36750 KNm
Stabil i ty : 43219.5-36750=6469.5 KNm
Olso in this case the structure is not stable
Fig 17 long span
2/ Day situat io n → the building is opened
In this case rhe internal coefficient must be calculated from En 1991-1-4 (2005)7.2.9
Take a 5*4.5 m2 one door in front ( θ =90°)
3*(2*1.5)m2 three doors in long spans and hasard disribution( θ =0°)
Dominant face :22.5m2
Other faces: 3m2 22.5
2.59
then , ,0.75 p i p ec c En 1991 -1-4 (2005)7.2.9( eq7.1 )
So we return to the depart case and recalculate the al l w ind act ions
http://www.arab-eng.org/vb/users/355867.html Ibnmessaoud10
W
Z
X3.2m
35.0m
Rx
Rz
35.0m
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