access steel wind
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Document Ref: SX016a-EN-EU Sheet 1 of 8 Title
Example: Determination of loads on a building envelope
Eurocode Ref EN 1991-1-3, EN 1991-1-4 Made by Matthias Oppe Date June 2005
CALCULATION SHEET
Checked by Christian Müller Date June 2005
Example: Determination of loads on a building envelope
This worked example explains the procedure of determination of loads on a portal frame building. Two types of actions are considered: wind actions and snow actions.
30,00
5,98
8
α
[m]
7,20
7,30 72,00
Basic data
• Total length : b = 72,00 m
• Spacing: s = 7,20 m
• Bay width : d = 30,00 m
• Height (max): h = 7,30 m
• Roof slope: α = 5,0°
Height above ground:
h = 7,30 m
α = 5°
leads to:
h´ = 7,30 – 15 tan 5° = 5,988 m
Example: Determination of loads on a building envelopeC
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Document Ref: SX016a-EN-EU Sheet 2 of 8 Title
Example: Determination of loads on a building envelope
Eurocode Ref EN 1991-1-3, EN 1991-1-4 Made by Matthias Oppe Date June 2005
CALCULATION SHEET
Checked by Christian Müller Date June 2005
1 Wind loads
Basic values Determination of basic wind velocity: EN 1991-1-4 vb = cdir × cseason × vb,0 § 4.2
Where: vb basic wind velocity cdir directional factor cseason seasonal factor vb,0 fundamental value of the basic wind velocity
Fundamental value of the basic wind velocity (see European windmap):
vb,0 = 26 m/s (for Aachen - Germany)
EN 1991-1-4 Terrain category II ⇒ z0 = 0,05 m § 4.3.2 z > zmin
Table 4.1
⇒ vb = cdir × cseason × vb,0 = 26 m/s
For simplification the directional factor cdir and the seasonal factor cseason are in general equal to 1,0.
Basic velocity pressure
EN 1991-1-4 2bairb 2
1 vq ××= ρ § 4.5 eq. 4.10
25,1air =ρwhere: kg/m³ (air density)
5,4222625,121 2
b =××=q⇒ N/m²
Peak pressure
[ ] 2mvp (z)
21(z)71(z) vlq ×××+= ρ EN 1991-1-4
§ 4.5, eq. 4.8Calculation of vm(z)
vm(z) mean wind velocity
vm(z) = cr(z) × co(z) × vb
Example: Determination of loads on a building envelopeC
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Document Ref: SX016a-EN-EU Sheet 3 of 8 Title
Example: Determination of loads on a building envelope
Eurocode Ref EN 1991-1-3, EN 1991-1-4 Made by Matthias Oppe Date June 2005
CALCULATION SHEET
Checked by Christian Müller Date June 2005
Where: co(z) is the orography factor
cr(z) is the roughness factor
maxmin0
rr forln(z) zzzzzkc ≤≤⎟⎟
⎠
⎞⎜⎜⎝
⎛×=
minminrr for)((z) zzzcc ≤=
Where: z0 is the roughness length
kr is the terrain factor, depending on the roughness length z0 calculated using
07,0
II0,
0r 19,0 ⎟
⎟⎠
⎞⎜⎜⎝
⎛×=
zz
k
EN 1991-1-4 Where: z0,II = 0,05 (terrain category II) §4.3.2
zmin is the minimum height
Table 4.1 zmax is to be taken as 200 m
Calculation of Iv(z)
Iv(z) turbulence intensity
EN 1991-1-4
minminvv
maxmin0o
Iv
for)(
for)/ln()(
zzzII
zzzzzzc
kI
<=
≤≤×
= §4.4 eq. 4.7
Where: kI is the turbulence factor recommended value for kI is 1,0 z = 7,30 m so: zmin < z < zmax
( 20r
2b
0o
Ip )/ln(
21
)/ln()(71(z)
profilewindpressurebasicfactorgustsquared
zzkvzzzc
kq ×××××⎥⎦
⎤⎢⎣
⎡×
+= ρ )
Corrigendum
Example: Determination of loads on a building envelopeC
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Document Ref: SX016a-EN-EU Sheet 4 of 8 Title
Example: Determination of loads on a building envelope
Eurocode Ref EN 1991-1-3, EN 1991-1-4 Made by Matthias Oppe Date June 2005
CALCULATION SHEET
Checked by Christian Müller Date June 2005
2070
2p
)05,0/307(ln050050190
2625121
)050307(ln71)307(
⎟⎟⎠
⎞⎜⎜⎝
⎛×⎟
⎠⎞
⎜⎝⎛××
×××⎥⎦
⎤⎢⎣
⎡+=
,,,,
,,/,
,q
,
911,010947,05,422)050307(ln
71 32 =×××⎥⎦
⎤⎢⎣
⎡+= −
,/, kN/m²
Wind pressure on surfaces EN 1991-1-4 § 7.2(pressure coefficients for internal frame)
A positive wind load 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.
External pressure coefficients
The wind pressure acting on the external surfaces, we should be obtained from the following expression: EN 1991-1-4 we = qp(ze) × cpe §5.2 eq. 5.1
where: ze is the reference height for the external pressure
cpe is the pressure coefficient for the external pressure depending on the size of the loaded area A. = cpe,10 because the loaded area A for the structure is larger than 10 m²
a) vertical walls EN 1991-1-4 § 7.2
for 25,024,000,3030,7
≤==dh
Table 7.1
D: cpe = 0,7
E: cpe = - 0,3
Example: Determination of loads on a building envelopeC
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Document Ref: SX016a-EN-EU Sheet 5 of 8 Title
Example: Determination of loads on a building envelope
Eurocode Ref EN 1991-1-3, EN 1991-1-4 Made by Matthias Oppe Date June 2005
CALCULATION SHEET
Checked by Christian Müller Date June 2005
b) duopitch roofs EN 1991-1-4 § 7.2with α = 5,0°, Table 7.4a
θ = 0° (wind direction)
e = min (b; 2h) = min (72,00; 14,60) = 14,60 m
1) upwind face G: cpe = - 1,2 2) downwind face
H: cpe = - 0,6
I: cpe = - 0,6
J: cpe = 0,2 / - 0,6
⇒ cpe = - 0,6
(see Table 7.4a , Note 1)
External pressure coefficients cpe (for zone D, E, G, H, I and J):
I: = -0,6J: = -0,6H: = -0,6G: = -1,2
E: = -0,3D: = 0,7
cccc
cc
pe
pepepe
pepe
Internal pressure coefficient EN 1991-1-4 §5.2The wind pressure acting on the internal surfaces of a structure, wi should be
obtained from the following expression
eq.5.2
wi = qp(zi) × cpi
where: zi is the reference height for the internal pressure
cpi is the pressure coefficient for the internal pressure
Example: Determination of loads on a building envelopeC
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Document Ref: SX016a-EN-EU Sheet 6 of 8 Title
Example: Determination of loads on a building envelope
Eurocode Ref EN 1991-1-3, EN 1991-1-4 Made by Matthias Oppe Date June 2005
CALCULATION SHEET
Checked by Christian Müller Date June 2005
The internal pressure coefficient depends on the size and distribution of the openings in the building envelope.
Within this example it is not possible to estimate the permeability and opening ratio of the building. So cpi should be taken as the more onerous of + 0,2 and – 0,3. In this case cpi is unfavorable when cpi is taken to + 0,2.
EN 1991-1-4 § 7.2.9 (6) Note 2
Wind loads
The wind loadings per unit length w (in kN/m) for an internal frame are calculated using the influence width (spacing) s = 7,20 m:
w = (cpe + cpi) × qp × s Internal and external pressures are considered to act at the same time. The worst combination of external and internal pressures are to be considered for every combination of possible openings and other leakage paths.
EN 1991-1-4 § 7.2.9
Characteristic values for wind loading in [kN/m] for an internal frame:
- zones D, E, G, H, I and J
D: = 4,59 E: = 3,28
I: = 5,25J: = 5,25H: = 5,25G: = 9,18
w w
wwww
30,00e/10 = 1,46 1,46
7,30
[m]
Example: Determination of loads on a building envelopeC
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Document Ref: SX016a-EN-EU Sheet 7 of 8 Title
Example: Determination of loads on a building envelope
Eurocode Ref EN 1991-1-3, EN 1991-1-4 Made by Matthias Oppe Date June 2005
CALCULATION SHEET
Checked by Christian Müller Date June 2005
2 Snow loads
General Snow loads on the roof should be determined as follows: EN 1991-1-3 s = μi × ce × cz × sk §5.2.2 eq.5.1
where: μi is the roof shape coefficient
ce is the exposure coefficient, usually taken as 1,0
ct is the thermal coefficient, set to 1,0 for normal situations
sk is the characteristic value of ground snow load for the relevant altitude
Roof shape coefficient
Shape coefficients are needed for an adjustment of the ground snow load to a snow load on the roof taking into account effects caused by non-drifted and drifted snow load arrangements.
EN 1991-1-3 The roof shape coefficient depends on the roof angle. §5.3 Table 5.1 °≤≤° 300 α ⇒ μ1 = 0,8
Snow load on the ground
The characteristic value depends on the climatic region.
For a site in Aachen (Germany) the following expression is relevant:
EN 1991-1-3 ( )
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+×−×=
2
k 2561002,0264,0 Azs kN/ m² Annex C
Table C1
Where: z is the zone number (depending on the snow load on sea level), here: z = 2
A is the altitude above sea level, here A = 175 m
( ) 772,02561751002,02264,0
2
k =⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+×−×=s kN/m²
Example: Determination of loads on a building envelopeC
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Document Ref: SX016a-EN-EU Sheet 8 of 8 Title
Example: Determination of loads on a building envelope
Eurocode Ref EN 1991-1-3, EN 1991-1-4 Made by Matthias Oppe Date June 2005
CALCULATION SHEET
Checked by Christian Müller Date June 2005
Snow load on the roof
s = 0,8 × 1,0 × 1,0 × 0,772 = 0,618 kN/m²
spacing = 7,20 m
⇒ for an internal frame:
s = 0,618 × 7,20 = 4,45 kN/m
α
7,30
30,00
s = 4,45 kN/m
[m]
Example: Determination of loads on a building envelopeC
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