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5.9

Example 9A pitched-free roof

Design wind loads are required for a pitched-free roof over a minerals storage area nearHobart, Tasmania. The relevant information is as follows :

Location : Tasmania (Region A3)

Upwind terrain : Suburban-industrial terrain for 1 km in all directions,except for 300 metres of open water starting at 200 metres from the site

for the south to west quadrant.

Topography : ground slope less than 1 in 20 for greater than 5 kilometresin all directions.

Dimensions : average roof height : 15 metres

Horizontal dimensions: 20 metres 40 metres (rectangular planform).

Roof pitch : 10 degrees.

Orientation : the main axis of the structure is orientated NNW-SSE

Figure 5.18 Pitched-free roof over a storage area

Regional wind speed

According to the Building Code of Australia (BCA), the structure should be treated as

Level 2. Hence take average recurrence interval, R, for loading and overall structural

response equal to 500 years.

From Table 3.1in AS/NZS1170.2, V500= 45 m/s (Region A)

Wind dir ection multipli er

Wind direction multipliers for Hobart (Region A3) are given in (Table 3.2). Values range

from 0.80 (NE to S) to 1.0 (NW).

20 m

15 m

40 m

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Terr ain-height multiplier

z=h=15 m, Assume full Terrain Category 3 multipliers for north through east to south,

Mz,cat= M15,cat3= 0.89

For SW through NW directions, account should be taken of the open water terrain, usingSection 4.2.3. Since structure height is 15 metres, the terrain-height multipliers should beaveraged from a distance xito (1000+xi) upwind of the structure (seeFigure 4.1).

ByEquation (4.2), the lag distance, xi

metresmz

zz

r

r 2008.198

)2.0)(3.0(

152.0

)3.0(

25.125.1

0

0

The terrain for 1000m beyond 200m upwind for SW, W, NW, consists of 300m of Terrain

Category 2 followed by 700 metres of Terrain Category 3,

938.01000

)700)(89.0()300)(05.1(,

catzM (for south-west, west and north-west directions)

Shielding

There are no other buildings of greater height in any direction. Take Ms, equal to 1.0 for

all directions.

Topography

Topographic Multiplier, Mt= Mh = 1.0

Site wind speed

Site wind speed for North direction, Vsit,N = 45(0.85)(0.89)(1.0)(1.0) = 34.0 m/s

(Equation 2.2)

For all wind directions, site wind speeds are calculated in the following table.

Direction V500

(m/s)

Md Mz,cat Ms Mt Vsit,(m/s)

N 45 0.85 0.89 1.0 1.0 34.0

NE 45 0.80 0.89 1.0 1.0 32.0E 45 0.80 0.89 1.0 1.0 32.0

SE 45 0.80 0.89 1.0 1.0 32.0

S 45 0.80 0.89 1.0 1.0 32.0

SW 45 0.85 0.938 1.0 1.0 35.9

W 45 0.90 0.938 1.0 1.0 38.0

NW 45 1.00 0.938 1.0 1.0 42.2

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Design wind speeds

Vdes,can be obtained from Vsit,from the following graph, for the four wind directions

orthogonal to the main axes of the structure:

Vdes,ENE= 33.0 m/s (largest from NNE to ESE sector)Vdes,SSE= 34.0 m/s (largest from ESE to SSW sector)

Vdes,WSW= 40.1 m/s (largest from SSW to WNW sector)

Vdes,NNW= 42.2 m/s (largest from WNW to NNE sector)

Aerodynamic shape factor

Net pressure coefficients for pitched free roofs for wind normal to the ridge line areobtained from Table D5.

For empty under case :

Cp,w= -0.3, or +0.4 Cp,= -0.4, or 0.0

For blocked under case :

Cp,w= -1.2 Cp,= -0.9

For wind parallel to the ridge, pressure coefficients are obtained from Table D4(A)

For blocked under case :

Cp,n= -1.0 for first 20m from leading edge (windward half);

-0.8 for 20-40m from leading edge (leeward half)

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Area reduction factors (Section D1.2 and Table 5.4)

Tributary area of each roof half = 40 10 = 400 m2

Ka= 0.8 (Table 5.4)

Local net pressure factors (Section D1.3 and Table D1)

Cases 1 and 2 apply

a2= 4

2= 16 m

2; 0.25a

2= 4 m

2

For areas between 16m2or less within a distance of 4 m from a roof edge or the ridge,

K= 1.5

For areas of 4 m2or less within a distance of 2 m from a roof edge, or the ridge, K

= 2.0

Dynamic response factor

Cdyn= 1.0 (natural frequency greater than 1.0 Hertz) (Section 6.1)

Design wind pressure for major framing members and columns - ul timate limit states:

For WSW wind (normal to ridge line)

Load case 1(to give maximum drag empty under)

Net normal pressure on windward roof slope:Cfig = Cp,n Ka = +0.4 (0.8) = 0.32pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (40.1)

2(0.32)(1.0) = 309 Pa

Net pressure across roof surface = 0.31 kPa

Net normal pressure on leeward roof slope:Cfig = Cp,n Ka = -0.4 (0.8) = -0.32pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (40.1)

2(-0.32)(1.0) = -309 Pa

Net pressure across roof surface = -0.31 kPa

Load case 2(to give maximum downwards force empty under)

Net normal pressure on windward roof slope:Net pressure across roof surface = 0.31 kPa

Net normal pressure on leeward roof slope:

Cfig = Cp,n Ka = 0.0 (0.8) = 0.0Net pressure across roof surface = 0.0 kPa

Load case 3(to give maximum uplift blocked under)

Net normal pressure on windward roof slope:

Cfig = Cp,n Ka = -1.2 (0.8) = -0.96pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (40.1)

2(-0.96)(1.0) = -926 Pa


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Net normal pressure on leeward roof slope:


2CfigCdyn= (0.5)(1.2) (40.1)

2(-0.72)(1.0) = -695 Pa


For NNW wind (parallel to ridge line)

Negative pressure case(assume blocked under)

For first 20 m from NNW end of roof:


2CfigCdyn= (0.5)(1.2) (42.2)

2(-0.80)(1.0) = -855 Pa


For 20-40 m from NNW end of roof:Cfig = Cp,n Ka = -0.8 (0.8) = -0.64

pn= (0.5 air) Vdes,2

CfigCdyn= (0.5)(1.2) (42.2)2

(-0.64)(1.0) = -684 PaNet pressure across roof surface = -0.68 kPa

Positive pressure case(assume blocked under)

For first 30m from NNW end of roof:

Cfig = Cp,n Ka = +0.4 (0.8) = +0.32pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (42.2)

2(+0.32)(1.0) = +342 Pa

Net pressure across roof surface = +0.34 kPa

For 30-40m from NNW end of roof:Cfig = Cp,n Ka = +0.2 (0.8) = +0.16pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (42.2)

2(+0.16)(1.0) = +171 Pa

Net pressure across roof surface = +0.17 kPa

For 30-40m from NNW end of roof:Cfig = Cp,n Ka = -0.4 (0.8) = -0.32pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (42.2)

2(-0.32)(1.0) = -342 Pa


For SSE winds, factor above loads by (34.0/42.2)2= 0.65

This may be the dominant load case for some parts of the structure (e.g. columns at SSEend).

Fri ctional drag

Frictional drag should be calculated according to Clause D3.2,for wind directions (NNWand SSE) parallelto the ridge.

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For top surface roof sheeting , assume Cfig = Cf= 0.02

For bottom roof surface, with exposed members, take Cfig = Cf= 0.04 (Table D3)

Then frictional drag force per unit area, f = (0.5)(air)[Vdes,]2Cfig.Cdyn

For upper surface for NNW wind direction,f = (0.5)(1.2)[42.2]2(0.02) (1.0) = 21.4 Pa

Drag force on upper surface = 21.4 40 20/cos 10o= 17384 N = 17.4 kN

For lower surface,f = (0.5)(1.2)[42.2]

2(0.04) (1.0) = 42.7 Pa

Drag force on lower surface = 42.7 40 20/cos 10o= 34720 N = 34.7 kN

Total horizontal force due to frictional drag = 17.4 + 34.7 = 52.1 kN

Design wind pressure for cladding elements and purl ins - ultimate limit states:

For areas between 4 m2and 16 m

2within a distance of 4 m from WSW roof edge or the

ridge.

Cfig = Cp,n Ka K= +0.4 (1.0) (1.5) = 0.6 pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (40.1)

2(0.6)(1.0) = 579 Pa

Net downward pressure across cladding = 0.58 kPa

Cfig = Cp,n Ka K= -1.2 (1.0) (1.5) = -1.8 pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (40.1)

2(-1.8)(1.0) = -1737 Pa

Net upward pressure across cladding = -1.74 kPa

For areas of 4 m2or less within a distance of 2 m from WSW roof edge, or the ridge.

Cfig = Cp,n Ka K= +0.4 (1.0) (2.0) = 0.80 pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (40.1)

2(0.80)(1.0) = 772 Pa

Net downward pressure across cladding = 0.77 kPa

Cfig = Cp,n Ka K= -1.2 (1.0) (2.0) = -2.4 pn= (0.5 air) Vdes,

2CfigCdyn= (0.5)(1.2) (40.1)

2(-2.4)(1.0) = -2316 Pa

Net upward pressure across cladding = -2.32 kPa

chapter 5.9r

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