A sample calculation done for determination of wind load according to BS6399:1997 to a rectangular building. This document determines the wind load values to be exerted to the structure
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Digests - Special Civil Actions
Design of wind actions on a portal frame building according to EN 1991 -1-4(2005) B a s i c i n fo fo r m a t i o n
1 Total length
b=70 m
2 bay width
3 spacing
4 height (max)
5 roof slope
H. above ground
d=25 m
s= 7 m
h= 7.5 m
slope α=5°
h’=7.5 -12.5 -12.5 tan5°=6.41m
6.4 m
7.5 m
70.0 70.0 m
25.0 25.0 m
Fig1: one storey building frame
5°
6.4m
7.5m
25 m
Fig 2 : transversal transversal
Basic values D e te r m in a t io n o f b a s i c w i n d v e l o c i ty
Fig 3 : Actions of wind o n surfaces
vb cdir cseasonvb,0 vb
EN 1991-1-4(2005) 4.2
is the basic wind velocity
cdir is the directional factor c season is the seasonal factor
vb ,0 is the fundamental value of the basic wind velocity (obtained from meteo center national annex) take
vb ,0
=30m /s
terrain category III → z 0 =0.3 m and
z=7.5 m
we have
z min=5 m
EN 1991-1-4(2005) 4.3.2 (table 4.1)
z≥ z min
vb cdir cseasonvb,0 30m / s we take
cdir = c season =1.0 for the unfavorable case because
as reduction factors
B a s i c v el o c i t y p r es s u r e
c dir and c season are used generally
q p
1
2
2
air vb
EN 1991-1-4(2005) 4.5 eq 4.10 3
Where recommended value air =1.25kg/m (air density)
q p
1 2
*1.25*30 2 562.5 N / m2
Peak pressure
q p z 1 7 I v z
1 2
vm
z
2
EN 1991-1-4(2005) 4.5 eq 4.8
Calculation of vm z
vm z cr z c0 z vb c0 z is the orography factor cr z is the roughness factor
z case z 0
cr z k r log
where zmin
z z max → 5m 7.5m 200m
k r Is the terrain factor depending on the roughness length z 0 calculated using z 0 k r 0.19 z 0, II
0.07
Calculation of the turbulence intensity I v z I v
k I
z c0 z log z 0
I v I v z min
for z min ≤ z≤z max
EN 1991-1-4(2005) 4.4 eq 4.7
for z≤ z min
where k I is the turbulence factor (recommended value is k I =1.0)
Z=7.5 m then
2 7 k 1 z 2 r * vb q p z 1 * k r log 2 z 0 z c0 z log basic pressure wind profile z 0
A positive wind loads stands for pressure whereas a negative wind load indicates suction on the surface.this definition applies for the external wind action as well as for the internal wind action.
Fig 4: description of (+) and(-) wind
1 / n i g h t s i t u a t i o n → all the building is closed E x t e r n a l p r e s s u r e c o e ff i c i e n t s
Wind 1 θ=0° The wind pressure acting on the external surface,w e Should be obtained from the following expression
we q p ze .c pe
EN 1991-1-4(2005) 5.2 eq 5.1
z e is the reference height for the external pressure c pe is the pressure coefficient for the external pressure depending on the si ze of the loaded
area A
Fig 5: euro curve for c pe
c pe,10
EN 1991-1-4(2005) 7.2.1
=
if 1m2 A 10m2 then c pe c pe,1 c pe,1 c pe,10 log10 A a)Vertical walls
For
h d
7.5 25
0.3 0.25 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 have m
f 0.25
f x
2 15
2 15
y B yA x B xA
*0.25 p 0.7 p
x
2 3
then f 0.3
2 15
0.8 0.7 1 0.25
2 15
2 3
0.3
2 3
0.71
This value will be neglected to the table value D
c pe=0.7
and
E
EN 1991-1-4(2005) table 7.4a
c pe=-0.3
fig 6: euro fig top ve w + zone actions b ) D u o p it c h r o o f s
Table 2 wind 1( θ=0°) horizontal and vertical forces to Aref Zone
Horizontal
vertical
H (KN)
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
X H 14.33
Y H
35.0 Z H 3.29
R z
990.7
X
V
11.79 Y V 35.0
Z R z
11.79 m
12.5m
W
6.88 m
R x B
A 25.0 m 14.33 m
Fig 12 force positions
f o r ce s a c t io n s
3.29 m
Z V 6.88
O v er t u r n in g m o m e n t
M ov =R x *3.29+R z *(25-11.79) =315.36*3.29+990.7*13.21 =14125 KNm s t ab i li z in g m o m e n t
take 0.6 KN/m 2 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 Stability
:
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 t wo cases are similar e90 =e180 =min(b;2h) =min(25;15) =15m
EN 1991-1-4:2005 (7.2.5 Figure 7.8) x
z E
I
I
H
H G
G F
F
D
y w2
Fig 13 application forces
F
e/4
H
I
G faite ou noue
G
H
b=25m
I
F
e/4
e/10
e/2
Fig 12 euro zone decompositions for duopitch roofs( θ =90°) E x t e r n a l p r e s s u r e c o e f fi c i e n t s
EN 1991-1-4(2005) table 7.1
a)Vertical w alls zo n e
A
B
C
D
E
C p e
-1.2
-0.8
-0.5
+0.7
-0.3
A is limited to e/5=15*0.2=3m B is limited to 4e/5
=12m
C is limited to d-e=70-15 =55 m
b=25 m
b=25 m and d=70 m(in accordance to θ=90°)
e/5(A)
e4/5(B)
d-e (C)
Fig 13 zone distributions b ) D u o p it c h r o o f s
α=5.0°
;
See fig above
θ=90°=180°(wind direction) EN 1991-1-4(2005) table 7.4b
zone
F
G
H
I
c pe
-1.6
-1.3
-0.7
-0.6
Internal pressure coefficients
The wind pressure acting on the internal surfaces of the structure w i is obtained from the expression
wi q p zi c pi
z i
is the reference height for thr internal pressure
c pi 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) In this case
EN 1991-1-4(2005) (7.2.9 (6)note 2)
c pi 0.2
Wind loads
w c pe c pi * q p = c pe c pi *0.668
EN 1991-1-4(2005) (7.2.9 (6)note 2)
Table 3 details in zone zo n e
A
B
C
D
E
F
G
H
I
c p e
-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/
-0.935
-0.668
+0.334
-0.334
-1.202
-1.0
-0.601
-0.534
2
m )
-0.468
1.202 (F )
0.534 (I)
0.534 (I)
0.601 (H)
0.601 (H)
1.0 (G)
C 0.668 (B) 0.47
0.935
1.0 (G)
1.202 (F
)
C 0.668 (B) 0.47
wind
A
0.935 (A) 1.5
22
Fig 14
1.5
distribution actions
Determination of fric 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°
F fr cFr q p ze Aref c Fr is the friction coefficient
q p z e
Aref
see above
Aref =
EN 1991-1-4(2005) (7.5 table 7.10)
q p z
is the area of external surface parallel to the wind
C fr =0.04
EN 1991-1-4(2005) (5.3 eq 5.7 )
EN 1991-1-4(2005) 7.5
(ripples,ribs,folds..)
width*lenght
Lenght L=min(2b ;4h) =min(2*25,4*7.5) =30m
EN 1991-1-4(2005) 7.5(3)
Width a) v e r t i c al w a l ls 6504*2=13 m b ) d u o p i t ch r o o f s 2*( 12.5/cos(5°))=25.1 m
take only a lineair distribution forces to all faces(apply only for q p (7.5))
F fr 0.04* 0.668* 30*13 10.42KN
Vertical w alls
D u o p i t c h r o o f s F fr
KN 0.04*0.668*30* 25.1 20.12
F Fr roofs F Fr walls//wind
d=70m
h=7.5m 6.4m
b=25m wind
Fig 15 friction forces
Determination o f
forc es
See procedure for θ=0°
wind 2
(θ=90° or θ=180° )
table 4: global forces by zone Zone
Surface
Vertical
2
(m )
R Z
R x ( K N )
1.1
173.75 2
0
0.334*173.75 58.033
1.1
0
0.334*173.75 58.033
1.202*11.25 13.522
0
25* 6.4
E
25* 6.4
2*
(KN)
D
F
Horizontal
173.75 2
15 15 * 11.25 10 4
G
H
15 15 25 26.25 2 10
26.25*1.0 26.25
0
15 15 150 2 10
150* 0.601 90.15
0
1562.5*0.534 834.375
0
25
I
25 70
15
1562.5 2
10.42 20.12 30.54
F fr R z =964.297
Z
R x=+146.606
R z
12.5 m
12.5 m
7.5 m
R x B
A
3.2 m
25.0 m
Fig16 force point f o r ce s a c t io n s O v er t u r n in g m o m e n t
M ov =R x *3.2+R z *35 =146.606*3.2+964.297*35 =43219.5 KNm s t ab i li z in g m o m e n t
take 0.6 KN/m 2 as an approximative value of self-weight then(but not final)
W=0.6*70*25=1050 KN M w= 1050*35=36750 KNm Stability
:
43219.5-36750=6469.5 KNm
Olso in this case the structure is not stable
Z
Rz 35.0m
35.0m
Rx 3.2m X
W
Fig 17 long span
2/ Day situatio n → the building is o p e n e d In this case rhe internal coeffi cient must be calculated from En 19 91-1-4 (2005)7.2.9 Take a 5*4.5 m 2 3*(2*1.5)m2
one door in front ( θ =90°) three doors in long spans and hasard disribution( θ =0°)
Dominant face :22.5m2 Other faces: 3m2
22.5 9
2.5 then
c p ,i 0.75c p, e En
1991 -1-4 (2005)7.2.9( eq7.1 )
So we re turn to the d e p a r t c a s e and recalculate the a l l w i n d a c t i o n s http://www.arab-eng.org/vb/users/355867.html