Wind Design

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A Guide to AS/NZS 1170.2: 2002 Wind Actions

John Holmes Andrew King

Warreen Publishing

ii I Wind Actions

ISBN: 0-646-44705-X

Published in 2005 by:

Warreen Publishing P.O. Box 269 Mentone Victoria 3194 Australia Telephone: +61 3 9584 5885 FAX: t61 3 9585 3815

Printed by:

Design and Print Services University of Canterbury Christchurch New Zealand

Cover Design:

Justin Sharpe, Lucy Holmes and John Holmes

A Guide to ASINZS 1170.2: 2002

-1 Chapter 5 21 Chapter 5: Examples

5.1 Example 1 - Single-storey house in an urban area Design wind loads are required for a single-storey house in the Sydney metropolitan area. The relevant information is as follows :

Location: western suburbs of Sydney (Region A2). Terrain: Suburban terrain for all directions. Topography: ground slope less than 1 in 20 for greater than 5 kilometres in all directions. Dimensions: average roof height : 4.35 metres Horizontal dimensions, including entrance porch on south side (Figure 5.1): 24 metres x 12 metres Porch : 2.4 X 14.5 m Gable roof with 15 degrees pitch. Hip on west end. 0.6 m eaves. Building orientation: major axis is East-West. Timber frame construction. Wall studs are spaced at 0.4 m. Roof trusses are spaced at

The building walls are considered equally permeable. 1.2 m.

8

9

' Figure 5.1 Plan of single-storey house

This house is used as an example in the Guide to the Use of the Wind Load provisions of ASCE 7-02 [21]. Its use in the present Design Guide allows for a comparison of the two Standards for this building.

Regional wind speed According to the Building Code of Australia (BCA), the structure should be treated as Importance Level 2. Hence take average recurrence interval, R, equal to 500 years.

From Table 3.1 in ASl"ZS1170.2, V, = 45 mls (Region A)

Wind direction multipliers for Sydney (Region A2) are given in Table 3.2. Values range from 0.80 (N, NE, E) to 1.0 (W).

~ h 4 . 3 5 m, For Terrain Category 3, MZ,,, = M4,35,caU = 0.83 (Table 4.1(A))

Shielding Since the building is surrounded by suburban development, assume a Shielding Multiplier, M, of 0.85 for all directions.

Topography Topographic Multiplier, M, = M, = 1 .O

j Site wind speed

1 1 (Equation 2.2) i

\

Site wind speed for North direction, Vsit,N = 45(0.80)(0.83)(0.85)(1.0) = 25.4 m/s

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

E SE S

sw

I N I 45 I 0.80 I 0.83 I 0.85 I 1.0 I 25.4 I

45 0.80 0.83 0.85 1 .o 25.4 45 0.95 0.83 0.85 1 .o 30.2 45 0.90 0.83 0.85 1 .o 28.6 45 0.95 0.83 0.85 1 .o 30.2

I I I I I

NE 1 45 I 0.80 I 0.83 I 0.85 I 1 .o I 25.4

I I

W I 45 I 1.00 I 0.83 I 0.85 I 1 .o I 31.7 I NW I 45 I 0.95 I 0.83 I 0.85 I 1.0 I 30.2 I

I Design wind speeds i 1 1 i

j j

In this case, the building is orientated exactly N-S and E-W, and the design wind speeds, Vdes,B can be obtained by inspection from VsitB in the above table, for the four wind directions orthogonal to the building walls: i

Vdes,N = 30.2 m/s (largest from NW to NE sector)

Vdes,B = 30.2 m/s (largest from NE to SE sector)

Vdes,S = 30.2 m/s (largest from SE to SW sector)

Vdes,W = 3 1.7 m/s (largest from SW to NW sector)

j

j

j i 1 ~ Aerodynamic shape factor j I External pressures I j / Leeward walls : I ! (normal to long axis): -0.3 (Table 5.2(B)) / (normal to short walls): -0.3 j Side walls ~ (first 4.35 m from windward edge): -0.65 (Table 5.2(C)) I (4.35 m to 8.7 m): -0.5 I (8.7 m to 13.05 m): -0.3 j (beyond 13.05 m): -0.2

Roof upwind slope (normal to long axis: h/d = 4.35/12=0.36): -0.59, -0.13 (Table 5.3(B)) downwind slope and hip end (normal to long axis: h/d = 4.35/12=0.36): -0.5 (Table 5.3(C)) upwind slope (normal to short walls: h/d = 4.35/24=0.18): -0.5,O.O (Table 5.3(B)) downwind slope (normal to short walls: h/d = 4.3Y244.18): -0.5 (Table 5.3(C)) crosswind slope, R (normal to short walls: first 4.35 m from w.e): -0.9,0.4 (Table 5.3(A)) crosswind slope, R (normal to short walls: 4.35 m to 8.7m from w.e): -0.5,O.O (Table 5.3(A)) crosswind slope, R (normal to short walls: 8.7 m to 13.05 m from w.e): -0.3, M.1 (Table 5.3(A)) crosswind slope, R (normal to short walls: beyond 13.05 m from w.e): -0.2, M.2 (Table 5.3(A))

~

Windward walls : +0.7 (Table 5.2(A))

i

i

1 !

.. ..,...(......... . ..... j r ..... l..,...,..,,.....,...... '..l....j ....... ~ .,..,,... ./ 1. 1.1.1... ....... I I .,.....,......... l.... ...... * ,... ...,. ~.~ ,.,. I / / A Guide to ASINZS 11 70.2: 2002

-1 Chapter 5 23 Area reduction factors (Table 5.4) For wall studs (windward or leeward wall) Ka = 1.0 (Section 5.4.2) For roof trusses: tributary areas 1.2 X 9.75 = 11.7 m2 , Hence K, = 0.99 =U (by interpolation) For foundation loads : tributary area = 24 x 12 = 288 m2. Hence K, = 0.8

Local pressure factors (Table 5.6) a = minimum of 0.2 X 12 m = 2.4 m, or 4.35 m. a = limiting tributary areas for local pressure factors : 0.25a2 = 1.44 m2

a2 = 5.76 m2

Internal pressures (Section 5.3)

The building can be considered to have all walls equally permeable. In this case, Cp,i = -0.3 or 0.0 (Table 5.I(A))

Action combination factors

Case (c) in Table 5.5 is applicable for positive pressures on roof with negative internal pressures : K, = 0.8 Otherwise, K, = 1 .O

Dynamic response factor Cdyn = 1 .O (natural frequency greater than 1 .O Hertz) (Section 6. I )

Design wind pressure for major framing members - ultimate limit states:

North wall As windward wall (north wind direction): Cfig (external) = +0.7 (1 .O) (1 .O) = 0.7 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,i Cfig CdF = (0.5)(1.2) (30.2)2 (0.7)(1.0) = 383 Pa = 0.38 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pin, = (0.5 psi) Vdes,i Cfig CdY, = (0.5)( 1.2) (30.2)2 (-0.3)( 1 .O) = -164 Pa = -0.16 kPa Net pressure across windward wall surface = 0.38-(-0.16) = 0.54 kPa

As leeward wall (south wind direction): Cfig (external) = -0.3 (1 .O) (1 .O) = -0.3 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,i Cfig Cdyn = (0.5)(1.2) (30.2)2 (-0.3)(1.0) = -164 Pa= -0.16 kPa Cfig (internal) = 0.0 (1.0) = 0.0 pint = 0 kPa Net pressure across wall surface = -0.16 kPa (acting outwards)

As side wall -west wind direction: For length 0 to I h (0 to 4.35m)fi-om west edge: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,,, = (0.5 psi$ Vdes,i Cfig Cdyn = (OS) ( 1.2) (3 1.7)* (-0.65)( 1 .O) = -392 Pa = -0.39 kPa Cfig (internal) = 0.0 (1.0) = 0.0 pint = 0 kPa Net pressure across wall surface = -0.39 kPa (acting outwards)

For length I h to 2h (4.35 m to 8.7 m)&m west edge: Cfig (external) = -0.5 (1 .O) (1 .O) = -0.5 p,,, = (0.5 psi) Vdcs,i Cfig Cdyn = (0.5)(1.2) (31.7)2 (-0.5)(1.0) = -301 Pa = -0.30 kPa pht = 0 kPa Net pressure across wall surface = -0.30 kPa (acting outwards)

For length 2h to 3h (8.7 m to 13.05 m)from west edge: CfiP (external) = -0.3 (1 .O) (1 .O) = -0.3 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,i CfiP Cdyn = (0.5)(1.2) (31.7) (-0.3)(1.0) = -181 Pa = -0.18 kPa pin, = 0 kPa Net pressure across wall surface = -0.18 kPa (acting outwards)

As side wall -east wind direction: For length 0 to Ih (0 to 4.35m) from east edge: Cfig (external) = -0.65 (1.0) (1.0) = -0.65 p,,, = (0.5 psi$ Vdes,i CfiP Cdyn = ( O S ) ( 1.2) (30.2) (-0.65)( 1 .O) = -356 Pa = -0.36 kPa pi,, = 0 kPa Net pressure across wall surface = -0.36 kPa (acting outwards)

For length Ih to 2h (4.35 m to 8.7 m) from east edge: C,io (external) = -0.5 (1.0) (1.0) = -0.5 pe l = (0.5 paiJ Vdes,i Cfig Cdyn = (OS) ( 1.2) (30.2) (-0.5)(1 .O) = -274 Pa = -0.27 kPa pen, = 0 kPa Net pressure across wall surface = -0.27 kPa (acting outwards)

For length 2h to 3h (8.7 m to 13.05 m) from east edge: CfiP (external) = -0.3 (1 .O) (1 .O) = -0.3 p,,, = ( 0 . 5 ~ ~ ~ ~ ) Vdes,t CfiP Cdyn = (0.5)(1.2) (30.2) (-0,3)(1,0) = -164 Pa = -0.16 kPa

Net pressure across wall surface = -0.16 kPa (same as leeward wall case)

East wall

As windward wall (east wind direction): Cfig (external) = +0.7 (1 .O) (1 .O) = 0.7 p,,, =. (0.5 pa,,> Vdes,i Cfig Cdyn = (OS) ( 1.2) (30.2) (0.7)( 1 .O) = 383 Pa = 0.38 kPa Cfig (internal) = -0.3 (1.0) = -0.3 p,,, = (0.5pal,> Vdes,: Cfis Cdyn = (0.5)(1.2) (30.2) (-0.3)(1.0) = -164 Pa = -0.16 kPa Net pressure across wall surface = 0.38-(-0.16) = 0.54 kPa

As side wall - north wind direction: For length 0 to Ih (0 to 4.35m) from north edge: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,,, = (0.5pa,$ Vdes,i Cfig Cdyn = (0.5)(1.2) (30.2) (-0.65)(1.0) = -356 Pa = -0.36 kPa Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across wall surface = -0.36 = -0.36 kPa (acting outwards)

For length Ih to 2h (4.35 m to 8.7 m) from north edge: Cfig (external) = -0.5 (1 .O) (1 .O) = -0.5 p,,, = (0.5 pal,> Vdes,: Cfie Cdyn = ( O S ) ( 1.2) (30.2) ( -OS) ( 1 .O) = -274 Pa = -0.27 kPa

Net pressure across wall surface = -0.27 kPa (acting outwards)

As side wall - south wind direction: For length 0 to Ih (0 to 4.35m) from south edge: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,,, = (0.5 pa,,> Vdes,i Cfig Cdyn = ( O S ) ( 1.2) (30.2)* (-0.65)( 1.0) = -356 Pa = -0.36 kPa


As leeward wall - west wind direction: CfiP (external) = -0.3 (1.0) (1.0) = -0.3 p,,, = (OSp,,,) Vdes,t Cfig Cdy, = (0.5)(1.2) (31.7) (-0,3)(1.0) = -181 Pa = -0.18 kPa

Net pressure across wall surface = -0.18 kPa (ignore - side wall case governs)

P,,, = 0 kPa

PI,, = 0

P,,, = 0 kPa

P,,, = 0

PI,, = 0 kPa

West wall As windward wall (west wind direction): Cfig (external) = 44.7 (1 .O) (1 .O) = 0.7 peXt = (0.5 paiJ Vda,i CfiP Cdyn = (0.5)( 1.2) (3 1 .7)2 (0.7)( 1 .O) = 422 Pa = 0.42 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pint = (0.5pai$ V,,; C, Cdyn = (0.5)(1.2) (31.7)2(-0.3)(1.0) = -181 Pa= -0.18 kPa Net pressure across wah surface = 0.42-(-0.18) = 0.60 kPa

As side wall - northwind direction: For 0 to Ih (0 to 4.35 m) from north edge - same as east wall: -0.36 kPa

For Ih to 2h (4.35m to 8.7m) from north edge of east wall Cfig (external) = -0.5 (1 .O) (1 .O) = -0.5 p,,, = (0.5 pa,,> Vdes,i Cfig Cdyn = (0.5)( 1.2) (30.2)2 (-0.5)( 1 .O) = -274 Pa = -0.27 &a pint = 0 kPa Net pressure across wall surface = -0.27 kPa (acting outwards)

As side wall - south wind direction: For 0 to Ih (0 to 4.35 m) from south edge - same as east wall: -0.36 kPa.

South wall As windward wall (south wind direction): Cfig (external) = +0.7 (1 .O) (1 .O) = 0.7 p,,, = (0.5pa,,) Vde,: Cfig Cdyn = (0.5)(1.2) (30.2)2 (0.7)(1.0) = 383 Pa = 0.38 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pin, = (0.5pai,) Vdes,: Cfig Cdy,, = (0.5)(1.2) (30.2)2 (-0.3)(1.0) = -164 Pa = -0.16 kPa Net pressure across wall surface = 0.38-(-0.16) = 0.54 kPa

As leeward wall (north wind direction): Cfig (external) = -0.3 (1.0) (1.0) = -0.3 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,: Cfig Cdyn = (0.5)(1.2) (30.2)2 (-0.3)(1.0) = -164 Pa = -0.16 kPa pint = 0 kPa Net pressure across wall surface = -0.16 kPa (acting outwards) This case governs except for following side wall cases.

As side wall - west wind direction: For length 0 to Ih (0 to 4.35) m from west edge: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,,, = (0.5 pa,> Vdcs,i C, CdP = (0.5)( 1.2) (3 1 .7)2 (-0.65)( 1 .O) = -392 Pa = -0.39 kPa cfig (internal) = 0.0 (1 .o) = 0.0 pint = 0 kPa Net pressure across wall surface = -0.39 kPa (acting outwards)

For length I h to 2h (4.35 m to 8.7 m)jPom west edge: Cfig (external) = -0.5 (1 .O) (1 .O) = -0.5 p,,, = (OSp,,,) Vdcs,t Cfip Cdyn = (0.5)(1.2) (31.7)2 (-OS)(l .O) = -301 Pa = -0.30 kPa pint = 0 kPa Net pressure across wall surface = -0.30 kPa (acting outwards)

For length 2h to 3h (8.7 m to 13.05 m ) h m west edge: Cfig (external) = -0.3 (1.0) (1.0) = -0.3 p,,, = (0.5 pa,,> Vde,i Cfig Cdyn = (0.5)( 1.2) (3 1 .7)2 (-0.3)( 1 .O) = - 18 1 Pa = -0.18 kPa pint = 0 kPa Net pressure across wall surface = -0.18 kPa (acting outwards)

As side wall - east wind direction For 0 to 4.35 m+om east edge (i.e. gable end): Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,,, = (0.5 paiJ Vda,: Cfig Cdyn = (0.5)( 1.2) (30.2)2 (-0.65)( 1 .O) = -356 Pa = -0.36 kPa

1 p,,, = 0 kPa

1 1 l

~

1


For 4.35 m to 8.7 m from east edge: Cfig (external) = -0.5 (1.0) (1.0) = -0.5 p,,, = (0.5 pa,$ Vdes,; Cfig Cdyn = (0.5)(1.2) (30.2)2 (-OS)(l.O) = -274 Pa = -0.27 kPa


I

P,,, = 0

1 Roof ~ I I north wind direction ' north roof slope (surface 8).

Cfig (external) = -0.59 (1 .O) (1 .O) = -0.59 p,,, = (0.5 pa,$ Vdes,; Cfig Cdyn = (0.5)( 1.2) (30.2)2 (-0.59)( 1 .O) = -323 Pa = -0.32 kPa Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across roof surface = -0.32-(0.0) = -0.32 kPa South roof slope (surface 9) and hip end (surface 7). Cfig (external) = -0.5 (1.0) (1.0) = -0.5 p,,, = (0.5pa,,> Vdes,; Cfig Cdyn = (0.5)(1.2) (30.2)2 ( -OS) ( l .O) = -274 Pa = -0.27 kPa

Net pressure across roof surface = -0.27-(0.0) = -0.27 kPa

east wind direction east roof slope (surface 10). Cfig (external) = -0.59 (1 .O) (1 .O) = -0.59 p,,, = (0.5 pa,,> Vdes,l Cfig Cdyn = (0.5)(1.2) (30.2)2 (-0.59)( 1 .O) = -323 Pa = -0.32 kPa Cfig (internal) = 0.0 (1 .O) = 0.0 p,,, = 0 kPa Net pressure across roof surface = -0.32-(0.0) = -0.32 kPa west roof slope (surface 11). CfiU (external) = -0.5 (1.0) (1.0) = -0.5

P,,, = 0 kPa

P,,, = 0 kPa

P,,, = (0.5 pa,,> Vdes,; Cfig Cdyn = ( O S ) ( 1.2) (30.2)2 (-0.5)( 1 .O) = -274 Pa = -0.27 kPa pi,, = 0 kPa Net pressure across roof surface = -0.27-(0.0) = -0.27 kPa

south wind direction south roof slope (surface 9). Cfig (external) = -0.59 (1 .O) (1 .O) = -0.59 p,,, = (0.5 pa,,> Vdes,; Cfig Cdyn = (0.5)(1.2) (30.2)2 (-0.59)(1.0) = -323 Pa = -0.32 kPa Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across roof surface = -0.32-(0.0) = -0.32 kPa north roof slope (surface 8) and hip end (surface 7). Cfig (external) = -0.5 (1 .O) (1 .O) = -0.5 p,, = (0.5 pa,$ Vde,,; Cfi8 Cdyn = (0.5)( 1.2) (30.2)2 (-0.5)(1 .O) = -274 Pa = -0.27 kPa p,,, = 0 kPa Net pressure across roof surface = -0.27-(0.0) = -0.27 kPa crosswind roof slopes (surfaces I O and 11). For length 0 to l h (0 to 4.35 m) from windward edge: Cfig (external) = -0.9 (1 .O) (1 .O) = -0.9 p,,, = (0.5 pa,$ Vdes,; Cfig Cdyn = (0.5)( 1.2) (30.2)2 (-0.9)( 1 .O) = -493 Pa = -0.49 kPa

Net pressure across roof surface = -0.49-(0.0) = -0.49 kPa For length l h to 2h (4.35m to 8.7m)from windward edge: Cfiv (external) = -0.5 (1.0) (1.0) = -0.5

P,,, = 0 kPa

P,,, = 0 kPa

p,,;= (0.5 pa,) Vdes,; Cfig Cdyn = (0.5)( 1.2) (30.2)2 (-0.5)( 1 .O) = -274 Pa = -0.27 kPa pi,, = 0 kPa

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27 Chapter 5

Net pressure across roof surface = -0.27-(0.0) = -0.27 kPa For length 2h to 3h (8.7m to 13.05m)fiom windward edge: C6- (external) = -0.3 (1 .O) (1 .O) = -0.3 (negative case)

--c

p,, = ( 0 . 5 ~ ~ ~ ) Vdes,: Cfig Cdyn = (0.5)(1.2) (30.2)2 (-0.3)(1.0) = -164 Pa = -0.16 kPa pin, = 0 kPa Net pressure across roof surface = -0.16-(0.0) = -0.16 kPa Cfig (external) = +O. 1 (1 .O) (1 .O) = +O. 1 (positive case) p,, = ( 0 . 5 ~ ~ ~ ) Vdes,l Cfig C,,,, = (0.5)( 1.2) (30.2)2 (+O. 1)(0.8) = +44 Pa = +0.04 Wa C (internal) = -0.3 (1.0) = -0.3 p,, = (0.5pa,,> Vdes,: Cfig Cdyn = (0.5)(1.2) (30.2)2 (-0.3)(0.8) = -131 Pa = -0.13 kPa Net pressure across roof surface = +0.04-(-0.13) = t-0.17 kPa

fig

(note use of K, equal to 0.8 in this case)

west wind direction west roof slopes (suifaces 7 and 11). Cfig (external) = -0.59 (1 .O) (1 .O) = -0.59 p,,, = (0.5 pa,,) Vdes,: Cfig Cdyn = (0.5)( 1.2) (3 1 .7)2 (-0.59)(1 .O) = -356 Pa = -0.36 kPa

Net pressure across roof surface = -0.36-(0.0) = -0.36 kPa crosswind roof slopes (surfaces 8 and 9). For length 0 to 4.35 m from windward edge: C (external) = -0.9 (1 .O) (1 .O) = -0.9 p,,, = (0.5 pa,,) Vdes,: Cfig Cdyn = (0.5)( 1.2) (3 1 .7)2 (-0.9)( 1.0) = -543 Pa = -0.54 kPa

Net pressure across roof surface = -0.54-(0.0) = -0.54 kPa For length 4.35 m to 8.7m from windward edge: Cfiv (external) = -0.5 (1 .O) (1 .O) = -0.5

P,,, = 0 kPa

fig

PI,, = 0 kPa

..D

p,,, = (0.5 pa,,) Vdes,: Cfig Cdyn = (0.5)(1.2) (3 1 .7)2 (-0.5)(1 .O) = -301 Pa = -0.30 Wa pin, = 0 kPa Net pressure across roof surface = -0.30-(0.0) = -0.30 kPa For length 8.7 m to 13.05 m from windward edge: Cfig (external) = -0.3 (1 .O) (1 .O) = -0.3 (negative case) p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,t Cfig Cdyn = (0.5)(1.2) (31.7)2 (-0.3)(1.0) = -181 Pa = -0.18 kPa

Net pressure across roof surface = -0.18-(0.0) = -0.18 kPa Cfig (external) = +O. 1 (1 .O) (1 .O) = +O. 1 (positive case) p,,, = (0.5 pa,$ Vdes,: C,. Cdyn = (0.5)(1.2) (3 1.7)2 (+0.1)(0.8) = +48 Pa = +0.05 kPa C (internal) = -0.3 (1 .O) = -0.3 p,,, = (0.5pa,,) Vdes,; C,. Cdyn = (0.5)(1.2) (31.7)2 (-0.3)(0.8) = -145 Pa = -0.14 kPa Net pressure across roof surface = +0.05-(-0.14) = i-0.19 kPa For length 13.05 m to 24 m fiom windward edge: Cfig (external) = -0.2 (1 .O) (1 .O) = -0.2 (negative case) p,,, = (0.5 pa,,) Vdes,: Cfig C,, = (0.5)( 1.2) (3 1 .7)2 (-0.2)( 1 .O) = -12 1 Pa = -0.12 kPa

Net pressure across roof surface = -0.12-(0.0) = -0.12 kPa Cfig (external) = +0.2 (1 -0) (1 ,O) = +0.2 (positive case) p,,, = (0.5 p,,,) Vdes,: Cfig Cdyn = (0.5)( 1.2) (3 1 .7)2 (+0.2)(0.8) = +96 Pa = +O. 10 kPa C (internal) = -0.3 (1.0) = -0.3 p,, = (0.5 pal) Vdes,; Cfig Cdyn = (0.5)( 1.2) (3 1 .7)2 (-0.3)(1 .O) = -145 Pa = -0.14 kPa Net pressure across roof surface = +0.10-(-0.14) = M.24 kPa east roof slopes (surface IO). Cfig (external) = -0.5 (1 .O) (1 .O) = -0.5 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,t Cfig Cdm= (0.5)(1.2) (31.7)2 (-0.5)(1.0) = -301 Pa = -0.30 kPa


PI,, = 0 kPa

fig

PI,, = 0 kPa

fig

PI,, = 0 @a

Note: under-eaves pressures are same as adjacent walls.

i

Design loadings for foundations - ultimate limit states Use roof pressures as above multiplied by 0.8 (Ka)

Design loadings for windows - ultimate limit states (note: for permissible stress design - divide loads by 1.5)

For windows less than 1.44 m2 in area on north, east and south walls: Cfig (external) = t0.7 (1.0) (1.25) = +0.875 (positive case) p,,, = (0.5paiJ Vdes,i Cfig Cdyn = (0.5)(1.2) (30.2)2 (+0.875)(1.0) = +479 Pa = +0.48 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pin, = (0.5 pair) V,,,,, Cfig Cdyn = (0.5)( 1.2) (30.2)2 (-0.3)( 1 .O) = -164 Pa = -0.16 kPa Net pressure across window = +0.48-(-0.16) = +0.64 kPa

For windows less than 1.44 m2 in area on west wall: Cfig (external) = -1-0.7 (1.0) (1.25) = +0.875 (positive case) p,,, = (0.5p,J Vdes,: CfiP Cdy, = (0.5)(1.2) (31.7)2 (+0.875)(1.0) = +528 Pa = +0.53 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pint = (0.5paiJ Vdes,i Cfig Cdyn = (0.5)(1.2) (31.7)2 (-0,3)(1.0) = -181 Pa = -0.18 kPa Net pressure across window = +0.53-(-0.18) = +0.7 1 kPa

For windows less than 1.44 m2 in area on east and west walls within a distance of 1.2 m (0. Id) from south and north corners, or on north or south wall within 1.2 m from east corners: Cfig (external) = -0.65 (1 .O) (2.0) = -1.3 (negative case) p,,, = (0.5paiJ Vdes,, Cfig Cdy, = (0.5)(1.2) (30.2)2 (-1.3)(1.0) = -711 Pa = -0.71 kPa

Net pressure across window = -0.71-(0.0) = -0.71 kPa Pi,,, = 0

For windows less than 1.44 m2 in area on north and south walls within a distance of 1.2 m from west corners: Cfig (external) = -0.65 (1.0) (2.0) = -1.3 (negative case)

pi,,t = 0 kPa Net pressure across window = -0.78-(0.0) = -0.78 kPa

p,,, - - (0.5 p,,) Vdcs,i Cfig Cdyn = (OS)( 1.2) (3 1 .7)2 (-1.3)(1 .O) = -784 Pa = -0.78 kPa

For windows (or doors) more than 1.44 m2 in area on north, east and south walls: Cfig (external) = +0.7 (1 .O) (1 .O) = +0.7 (positive case) pext = (0.5 pair) Vdes,: Cfig Cdyn = (0.5)(1.2) (30.2)2 (+0.7)(1 .O) = +383 Pa = +0.38 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pi,, = ( 0 . 5 ~ ~ ~ ) Vdes,: Cfig Cdyn = (0.5)(1.2) (30.2)2 (-0.3)(1.0) = -164 Pa= -0.16 kPa Net pressure across window = +0.38-(-0.16) = +0.54 kPa

For windows (or doors) more than 1.44 m2 in area on west wall: Cfig (external) = +0.7 (1 .O) (1 .O) = +0.7 (positive case) p,,, = (0.5 pa,) Vdes,: Cfig Cdyn = (OS)( 1.2) (3 1.7)2 (+0.7)( 1 .O) = +422 Pa = +0.42 kPa Cfig (internal) = -0.3 (1.0) = -0.3 pi,, = ( 0 . 5 ~ ~ ~ ) Vdcs,: Cfig Cdyn = (0.5)(1.2) (31.7)2 (-0.3)(1.0) = -181 P a = -0.18 kPa Net pressure across window = +0.42-(-0.18) = +0.60 kPa

For windows or doors up to 5.76 m2 in area on east and west walls within a distance of 2.4 m from south and north corners, or on north or south wall within 2.4 m from east corners: Cfiu (external) = -0.65 (1 .O) (1.5) = -0.975 (negative case) p,,p= ( 0 . 5 ~ ~ ~ ) Vdes,i Cfig Cdyn = (0.5)(1.2) (30.2)2 (-0.975)(1.0) = -534 Pa = -0.53 kPa pin, = 0 kPa Net pressure across window = -0.53-(0.0) = -0.53 kPa

~,.,. .. A Guide to ASINZS 11 70.2: 2002

Wind Actions Chapter 5

For windows or doors up to 5.76 m2 in area on north and south walls within a distance of 2.4 m from west corners: Cfig (external) = -0.65 (1 .O) (1.5) = -0.975 (negative case) p,,, = (0.5 psi) Vdes,: CfiP Cdyn = (0.5)( I .2) (3 1 .7)2 (-0.975)(1 .O) = -588 Pa = -0.59 H a pint = 0 kPa Net pressure across window = -0.59-(0.0) = -0.59 kPa

For any other windows or doors, loads for major framing members apply.

Design loadings for roof cladding and battens - ultimate limit states

(note: for permissible stress design - divide loads by 1 S )

For roof cladding elements less than 1.44 m2 in area, on roof surfaces 8 and 9, up to 1.2 m from east edge of roof, and on roof surfaces 10 and 11, up to 1.2 m from south edge of roof: CfiP (external) = -0.9 (1 .O) (2.0) = -1.8 p,,, = (0.5 pa,,> Vdes,i Cfig Cdyn = ( O S ) ( 1.2) (30.2)* (-1 .8)( 1 .O) = -985 Pa = -0.99 kPa pin, = 0 kPa Net pressure across element = -0.99-(0.0) = -0.99 kPa

For roof cladding elements less than 1.44 m2 in area, on roof surfaces 8 and 9, up to 1.2 m from west edge of roof : Cfi. (external) = -0.9 (1 .O) (2.0) = -1.8

-.c

p,,,=(0.5pai,> Vde,,t Cfig Cdyn = (0.5)(1.2) (31.7)2(-1.8)(1.0) = -1085 Pa= -1.09 kPa pi,, = 0 kPa Net pressure across element = -1.09-(0.0) = -1.09 kFa

For roof cladding elements less than 5.76 m2 in area, on roof surfaces 8 and 9, up to 2.4 m from east edge of roof, and on roof surfaces 10 and 11, up to 2.4 m from south edge of roof: Cfig (external) = -0.9 (1.0) (1.5) = -1.35 p,, = (OSp,,,) Vdes,; Cfig Cdyn = (0.5)(1.2) (30.2)(-1,35)(1.0) = -739 Pa= -0.74 kPa pi,, = 0 kPa Net pressure across element = -0.74-(0.0) = -0.74 kPa

For roof cladding elements less than 5.76 m2 in area, on roof surfaces 8 and 9, up to 2.4 m from west edge of roof : C (external) = -0.9 (1.0) (1.5) = -1.35 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,: CfiP Cdyn = (0.5)(1.2) (31.7)2 (-1.35)(1.0) = -814 Pa = -0.81 kPa pint = 0 kPa Net pressure across element = -0.81-(0.0) = -0.81 kPa

fig

For roof cladding elements less than 1.44 m2 in area, on roof surfaces 8, 9,lO and 11, up to 1.2 m from windward edge of roof : Cfig (external) = -0.59 (1 .O) (2.0) = -1.18 (negative case) p,,, = (0.5 paiJ Vdes,t Cfig Cdy,, = (0.5)( 1.2) (30.2)2 (- 1. IS)( 1 .O) = -646 Pa = -0.65 kPa pint = 0 kPa Net pressure across element = -0.65-(0.0) = -0.65 kPa

For roof cladding elements less than 1.44 m2 in area, on roof surface 7, up to 1.2 m from west edge of roof: Cfig (external) = -0.59 (1 .O) (2.0) = - 1.18 (negative case) p,,,=(0.5p,i,)Vdes,~CfisCdyn= (0.5)(1.2) (31.7)2(-1.18)(1.0)=-711 Pa=-0.71 kPa pin, = 0 kPa Net pressure across element = -0.71-(0.0) = -0.71 kPa

For roof cladding elements less than 5.76 m2 in area, on roof surfaces 8, 9,lO and 11, up to 2.4 m from windward edge of roof: Cfig (external) = -0.59 (1 .O) (1.5) = -0.89 (negative case) p,,, = (0.5 paiJ Vd,,i Cfig Cdyn = (0.5)(1.2) (30.2)2 (-0.89)(1 .O) = -487 Pa = -0.49 kPa pint = 0 H a Net pressure across element = -0.49-(0.0) = -0.49 kPa

For roof cladding elements less than 5.76 m2 in area, on roof surface 7, up to 2.4 m from west edge of roof : Cfig (external) = -0.59 (1.0) (1.5) = -0.89 p,,, = (O.5paiJ Vdes,t Cfig CdYn = (0.5)(1.2) (31.7)2 (-0.89)(1.0) = -537 Pa = -0.54 kPa pint = 0 kPa Net pressure across element = -0.54-(0.0) = -0.54 kPa

Loading on battens should be applied as moving 'patch' loads, with a tributary area equal to the batten spacing times the truss rafter spacing, with the appropriate local pressure factor for that area. For example, in this case, if the batten spacing is 0.6 m, the tributary area would be 1.2 X 0.6 = 0.72 mz. Since this area is less than 0.25a2, the local pressure factor would be 1 .O, 1.5 or 2.0, depending on the distance of the centre of the 'patch' from the roof edge. Over the remainder of the batten, K, should be taken as 1 .O; i.e. the loading should be the same as used for major structural members.

~

5.2 Example 2 - Single-storey house on a cliff top I

I overlooking the sea I

The single-storey house in Example 1 is now assumed to be located on a cliff top at a coastal location.

Location : coast of Victoria (Region Al) Terrain : Open sea to south, Open country for all other directions. Topography : building is located 10m from the edge of a cliff facing south Dimensions : as for Example 1. Building orientation : major axis is east-west Construction: as for Example 1.

Regional wind speed From Table 3. I in ASINZS 1170.2, V,, = 45 d s (Region A) Wind direction multipliers for Region A1 are given in Table 3.2. Maximum value is 1.0 (W). z=h=4.35 m, For Terrain Category 2, Mz,ca, = M4,35,ca: = 0.91 (Table 4.1(A))

Shielding Since there are no surrounding buildings, Ms = 1 .O

Topography Prohle through topography for the south, south-east and south-west directions are shown in Figure 5.2 (a), (b) and (c) respectively.

.......................... I ...I. I .,.,... I I...... ~.-., .... I.... .._I... I...lj ..... ' ..I.. "...X..."..".I.~~ ..-.,,..--. ~ .......... .. ....,..... A Guide to ASINZS 11 70.2: 2002

Chapter Actions 5 I 31

N-s/ ~ 6 0 m wm ~ /*m section section - 36 m

NW-SE section

Figure 5.2. Sections through site for three wind directions

Hill-shape multiplier Mh for S winds ;

H=60m L u = 3 6 m x = lOm -=0.83 H

2-4, The house is within the separation zone (Figure 4.4) .

L, = greater of 0.36Lh or 0.4H = max{ 13.0 m, 24 m} = 24 m L,= 10 x 2 4 m = 2 4 0 m

(Equation 4.4(3))

Hill-shape multiplier M,, for SW winds : H

H=60m L u = 4 5 x = 1 6 m . - = 0.67 244

The house is within the separation zone (Figure 4.4) . L, = greater of 0.36Lh or 0.4H = max{ 16.2 m, 24 m} = 24 m L,= 10 X 2 4 m = 2 4 0 m

(Equation 4.4(3))

Hill-shape multiplier Mh for SE winds H H=60m L u = 8 5 m x = 1 5 m -=0.35

L, = greater of 0.36Lh or 0.4H = max(30.6 m, 24 m} = 30.6 m L, = 10 X 30.6 m = 306 m

2-44

60 (Equation 4.4(2)) 3.5 (4.3 5 + 30.6)

Site wind speed Site wind speed for North direction, Vsit,N = 45(0.90)(0.91)(1.0)(1 .O) = 36.9 m/s (Equation 2.2)

Wind Actionq 32 /Chnpter5

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

Design wind speeds Again, the building is orientated exactly N-S and E-W, and the design wind speeds, Vdes,@ can be obtained by inspection from VsitB in the above table, for the four wind directions orthogonal to the building walls:

Vdes,* = 38.9 m/s (largest from NW to NE sector)

Vdes,E = 48.2 m/s (largest from NE to SE sector)

VdesSS = 64.6 m/s (largest from SE to SW sector)

Vd,s,w = 64.6 m/s (largest from SW to NW sector)

(note the much higher design wind speeds compared with those in Example 1 for this very exposed site)

Aerodynamic shape factor As for Example 1

Dynamic response factor Cdyn = 1 .O (natural frequency greater than 1 .O Hertz) (Section 6.1)

Design wind pressure (major framing members) - ultimate limit states:

North wall

As windward wall (north wind direction): Cfig (external) = +0.7 (1 .O) (1 .O) = 0.7 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,: Cfig Cdyn = (0.5)(1.2) (38.9)2 (0.7)(1.0) = 636 Pa = 0.64 kPa C (internal) = -0.3 (1 .O) = -0.3 pi,, = (0.5 pG) Vdes,: CfiP Cdy, = (0.5)(1.2) (38.9)2 (-0.3)(1.0) = -272 Pa = -0.27 kPa Net pressure across wall surface = 0.64-(-0.27) = 0.91 kPa

fig

As leeward wall (south wind direction): Cog (external) = -0.3 (1.0) (1.0) = -0.3 p,,, = (0.5 p,J Vdes,: Cfig Cdyn = ( O S ) ( 1.2) (64.6)* (-0.3)( 1 .O) = -75 1 Pa = -0.75 kPa Cfig (internal) = 0.0 (1.0) = 0.0 pint = 0 kPa Net pressure across wall surface = -0.75 kPa (acting outwards)

~.-'.. A Guide to ASlNZS 11 70.2: 2002

Wind Chapter Actions 5 I 33 As side wall -west wind direction: Forjrs t 4.35 mfrom west edge: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,, = (0.5 paiJ Vdes,i Cfig Cdyn = (0.5)(1.2) (64.6)2 (-0.65)(1.0) = -1628 Pa = -1.63 kPa Cfig (internal) = 0.0 (1.0) = 0.0 pint = 0 kPa Net pressure across wall surface = -1.63 kPa (acting outwards)

For 4.35 m to 8.7 m from west edge: Cfig (external) = -0.5 (1.0) (1.0) = -0.5 p,, = (0.5 paJ Vdes,i Cfig Cdyn= (0.5)(1.2) (64.6) (-0.5)(1.0) = -1252 Pa = -1.25 kPa pint = 0 kPa Net pressure across wall surface = -1.25 kPa (acting outwards)

For 8.7 m to 13.05 mfrom west edge: Cfig (external) = -0.3 (1.0) (1.0) = -0.3 p,,, = (0.5 pa,$ Vdes,: Cfig Cdyn = (0.5)(1.2) (64.6) (-0.3)(1.0) = -751 Pa = -0.75 kPa pint = 0 kPa Net pressure across wall surface = -0.75 kPa (same as leeward wall case)

As side wall - east wind direction: Forfirst 4.35 mfrom east edge: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,,, = (0.5 pai$ Vdes,i Cfig Cdyn = (0.5)(1.2) (48.2) (-0.65)( 1 .O) = -906 Pa = -0.91 kPa pint = 0 kPa Net pressure across wall surface = -0.91 kPa (acting outwards)

For 4.35 m to 8.7 mfrom east edge: Cfie (external) = -0.5 (1 .O) (1 .O) = -0.5

-.- p,,, = (0.5 pa,> Vdes,i Cfig Cdyn = (0.5)(1.2) (48.2) (-0.5)(1.0) = -697 Pa = -0.70 kPa pi,, = 0 kPa Net pressure across wall surface = -0.70 kPa (ignore - leeward wall case governs)

For 8.7 m to (24-13.05=)10.95 mfrom east edge: Cfig (external) = -0.3 (1.0) (1.0) = -0.3

pint = 0 kPa Net pressure across wall surface = -0.42 kPa (ignore - leeward wall case governs)

p,,, - - (0.5 paiJ Vdes,i Cfig CdY, = (0.5)( 1.2) (48.2) (-0.3)( 1 .O) = -418 Pa = -0.42 kPa

East wall As windward wall (east wind direction): Cfig (external) = +0.7 (1.0) (1.0) = 0.7 p,, = (0.5 pai$ Vdes,i Cfig Cdyn = (0.5)(1.2) (48.2) (0.7)(1.0) = 976 Pa = 0.98 kPa Cfig (internal) = -0.3 (1.0) = -0.3 pint = ( 0 . 5 ~ ~ ~ ) Vdes,i Cfig Cdyn = (0.5)(1.2) (48.2) (-0.3)(1.0) = -418 Pa = -0.42 kPa Net pressure across wall surface = 0.98-(-0.42) = 1.40 kPa

As side wall - north wind direction: Forfirft 4.35 mfiorn north edge: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,,, = (0.5pai,) Vdes,: Cfig Cdyn = (0.5)(1.2) (38.9) (-0.65)(1.0) = -590 Pa = -0.59 kPa Cfig (internal) = 0.0 (1 .O) = 0.0 pint = 0 kPa Net pressure across wall surface = -0.59 kPa (ignore - leeward wall case governs)

j

j

For 4.35 m to (9.75-4.35=)5.4 mfiom north edge of west wall: Cfig (external) = -0.5 (1.0) (1.0) = -0.5 p,,, = (0.5 pair) Vdes,: C,. CdP = (0.5)(1.2) (38.9)2 (-0.5)( 1 .O) = -454 Pa = -0.45 kPa pint = 0 kPa Net pressure across wall surface = -0.45 kPa (ignore - leeward wall case governs)

As side wall - & wind direction: Forjrs t 4.35 mfiom south edges: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,,, = (0.5 pair) Vdes,: Cfig Cdyn = (0.5)( 1.2) (64.6)2 (-0.65)( 1 .O) = -1628 Pa = -1.63 kPa pint = 0 kPa Net pressure across wall surface = -1.63 kPa (acting outwards) As leeward wall - west wind direction: Cfig (external) = -0.3 (1.0) (1.0) = -0.3 p,,, = (0.5 pa,,> Vdes,: Cfig Cdyn = (0.5)( 1.2) (64.6)2 (-0.3)( 1 .O) = -751 Pa = -0.75 kPa pint = 0 kPa Net pressure across wall surface = -0.75 kPa (acting outwards)

West wall

As windward wall (west wind direction): Cfig (external) = +0.7 (1 .O) (1 .O) = 0.7 p,,, = (0.5 pair) Vdes,: Cfig Cdyn = (0.5)( 1.2) (64.6)2 (0.7)( 1 .O) = 1753 Pa = 1.75 kPa C (internal) = -0.3 (1 .O) = -0.3 pi,, = (0.5 pair) Vdes,: Cfig Cdyn = (0.5)( 1.2) (64.6)2 (-0.3)( 1 .O) = -75 1 Pa = -0.75 Net pressure across wall surface = 1.75-(-0.75) = 2.50 kPa

fig

As side wall - Forjrs t 4.35 m from north edge - same as east wall: -0.59 kPa.

wind direction:

For 4.35 m to (12.0-4.35=)7.65 mfiom north edge of east wall: C,. (external) = -0.5 (1 .O) (1 .O) = -0.5 p,,, = (0.5 pa,,> Vdes,t Cfig Cdyn = (0.5)(1.2) (38.9)2 (-0.5)(1,0) = -454 Pa = -0.45 kPa pi,, = 0 kPa Net pressure across wall surface = -0.45 kPa (acting outwards)

As side wall - & wind direction: Forjrs t 4.35 mfiom south edge - same as east wall: -1.63 H a .

As leeward wall - east wind direction: Cfio (external) = -0.3 (1 .O) (1 .O) = -0.3 p,:= (0.5pai$ Vdes,: Cfig Cdyn = (0.5)(1.2) (48.2)* (-0.3)(1.0) = -418 Pa= -0.42 kPa pin, = 0 kPa Net pressure across wall surface = -0.42 kPa (ignore - side wall case governs)

South wall

As windward wall (south wind direction): C,. (external) = +0.7 (1 .O) (1 .O) = 0.7 p,,, = (0.5 pair) Vdes,: Cfig Cdy, = (0.5)( 1.2) (64.6)2 (0.7)(1 .O) = 1753 Pa = 1.75 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pin, = (0.5 pair) Vdes,; Cfig Cdyn = (0.5)( 1.2) (64.6)2 (-0.3)( 1 .O) = -75 1 Pa = -0.75 kPa Net pressure across wall surface = 1.75-(-0.75) = 1.50 kPa

..~.... A Guide to ASlNZS 11 70.2: 2002


As leeward wall (north wind direction): Cfig (external) = -0.3 (1 .O) (1.0) = -0.3 p,,, = (0.5 psi) VdS,: C,& Cdy, = (0.5)(1.2) (38.9)2 (-0.3)(1.0) = -272 Pa = -0.27 kPa pin, = 0 kPa Net pressure across wall surface = -0.27 kPa (ignore - side wall case governs)

As side wall - WeSt wind direction: Forfirst 4.35 mfrom west edge: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,: Cfig Cdyn = (0.5)(1.2) (64.6)2 (-0.65)(1.0) = -1628 Pa = -1.63 kPa pi,, = 0 kPa Net pressure across wall surface = -1.63 kPa (acting outwards) For 4.35 m to 8.7 mJi-Om west edge: Cfi8 (external) = -0.5 (1 .O) (1 .O) = -0.5 p,,, = ( 0 . 5 ~ ~ ~ ) V,,,,: C,& Cdyn = (0.5)(1.2) (64.6)2 (-0.5)(1.0) = -1252 Pa = -1.25 kPa pint = 0 kPa Net pressure across wall surface = -1.25 kPa (acting outwards)

For 8.7 m to 13.05 mfrom west edge: Cfig (external) = -0.3 (1 .O) (1 .O) = -0.3 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,: Cfig Cdy,, = (0.5)(1.2) (64.6) (-0.3)(1.0) = -751 Pa = -0.75 kPa pin, = 0 kPa Net pressure across wall surface = -0.75 kPa (acting outwards)

For 13.05 rn to (24-8.7=)15.3 m from west edge: Cfig (external) = -0.2 (1 .O) (1 .O) = -0.2 p,,, = (0.5 paiJ Vdes,i Cfig Cdyn = (0.5)( 1.2) (64.6) (-0.2)( 1 .O) = -501 Pa = -0.50 kPa pint = 0 kPa Net pressure across wall surface = -0.50 kPa (acting outwards)

As side wall - Forjirst 4.35 m from east edge (i.e. gable end): Cfig (external) = -0.65 (1.0) (1.0) = -0.65 p,,, = (0.5 pa,> Vdes,: Cfig Cdyn = (0.5)( 1.2) (48.2) (-0.65)(1 .O) = -906 Pa = -0.91 kPa pint = 0 kPa Net pressure across wall surface = -0.91 kPa (acting outwards)

wind direction

For 4.35 m to 8.7 m from east edge: CfiP (external) = -0.5 (1.0) (1.0) = -0.5 p,,, = (0.5 pa,> Vdes,: Cfig Cdyn = (0.5)(1.2) (48.2) (-0.5)(1.0) = -697 Pa = -0.70 kPa pint = 0 kPa Net pressure across wall surface = -0.70 kPa (acting outwards)

Roof north wind direction north roof slope (surface 8). C,. (external) = -0.59 (1 .O) (1 .O) = -0.59 p,, = ( 0 . 5 ~ ~ ~ ) Vdes,: Cfig Cdyn = (0.5)(1.2) (38.9) (-0.59)(1.0) = -536 Pa = -0.54 kPa Cfig (internal) = 0.0 (1 .O) = 0.0 pint = 0 kPa Net pressure across roof surface = -0.54-(0.0) = -0.54 kPa

1 /

1 j p,,, = 0 kPa 1 ;

South roofslope (surface 9) and hip end (surface 7). C,. (external) = -0.5 (1.0) (1.0) = -0.5

j , p,,, = (OSp,,,) Vdes,i Cfig Cdyn = (0.5)(1.2) (38.9)' (-OS)(l.O) = -454 Pa = -0.45 kPa


east wind direction east roof slope (surface IO). CfiP (external) = -0.59 (1 .O) (1 .O) = -0.59 p,,, = (0.5 pa,$ Vdes,i CfiP Cdyn = (0.5)(1.2) (48.2)' (-0.59)(1.0) = -822 Pa = -0.82 @a C,. (internal) = 0.0 (1 .O) = 0.0

Net pressure across roof surface = -0.82-(0.0) = -0.82 kPa PI,, = 0 kPa

I I

west roof slope (surface 11). Cfin (external) = -0.5 (1 .O) (1 .O) = -0.5

. .D 1 j p;", = O kPa

p,,, = (0.5 pa,,) Vdes,i CfiP Cdyn = (0.5)( 1.2) (48.2)' ( - O S ) ( 1 .O) = -697 Pa = -0.70 kPa

I Net pressure across roof surface = -0.70-(0.0) = -0.70 kPa south wind direction south roofslope (surface 9). C,. (external) = -0.59 (1 .O) (1 .O) = -0.59 p,, = (0.5 paid Vdes,i C,. Cdyn = (0.5)( 1.2) (64.6)2 (-0.59)( 1 .O) = -1477 Pa = -1.48 kPa CfiP (internal) = 0.0 (1 .O) = 0.0 pint = 0 @a Net pressure across roof surface = -1.48-(0.0) = -1.48 kPa

north roof slope (surface 8) and hip end (surface 7). Cfig (external) = -0.5 (1 .O) (1 .O) = -0.5 pCxr = ( 0 . 5 ~ ~ ~ ) Vdcs,: CfiP Cdyn = (0.5)(1.2) (64.6)' (-OS)(l.O) = -1252 Pa = -1.25 kPa pin, = 0 kPa Net pressure across roof surface = -0.27-(0.0) = -1.25 kPa

1 ] Forjrs t 4.35 m ~

crosswind roof slopes (surfaces 10 and 11).

CfiE (external) = -0.9 (1 .O) (1 .O) = -0.9 I j p,,, = 0 kPa

pexy = (0.5 pa,,> Vdes,; Cfig Cdyn = (0.5)( 1.2) (64.6)' (-0.9)( 1 .O) = -2254 Pa = -2.25 H a

1 j

1 j I I pint = 0 H a j I


For 4.35 m to 8.7mfiom windward edge C,. (external) = -0.5 (1 .O) (1 .O) = -0.5 p,,, = (0.5pai,> Vdes,i CfiP CdyD = (0.5)(1.2) (64.6)' (-0.5)(1.0) = -1252 Pa = -1.25 kPa


For 8.7 m to 12 mfiom windward edge CfiP (external) = -0.3 (1 .O) (1 .O) = -0.3 (negative case) p,,, = (0.5 pa,,> Vdes,i Cfig Cdyn = (0.5)( 1.2) (64.6)' (-0.3)( 1 .O) = -75 1 Pa = -0.75 kPa pint = 0 kPa Net pressure across roof surface = -0.16-(0.0) = -0.75 kPa C,. (external) = +O. 1 (1 .O) (1 .O) = +O. 1 (positive case) p,,, = (0.5 pa,,> Vdes,i (2,. Cdyn = (0.5)( 1.2) (64.6)2 (+O. 1)(0.8) = +200 Pa = +0.20 kPa C (internal) = -0.3 (1 .O) = -0.3 pi,, = (0.5 pai) Vdes,: CfiP Cdyn = (0.5)( 1.2) (64.6)' (-0.3)(0.8) = -601 Pa = -0.60 kPa

i

~

fig

Actions I 37 Chapter 5 Net pressure across roof surface = M.20-(-0.60) = +OB0 kPa

(note use of K, equal to 0.8 in this case)

west wind direction west roof slopes (sufaces 7 and 11). C,. (external) = -0.59 (1 -0) (1 .O) = -0.59 p,,, = (0.5 psi$ Vdes,: C,. Cdyn = (0.5)(1.2) (64.6)2 (-0.59)(1.0) = -1477 Pa = -1.48 kPa pint = 0 kPa Net pressure across roof surface = -1.48-(0.0) = -1.48 kPa

crosswind roof slopes (surfaces 8 and 9). Forfirst 4.35 m Cfig (external) = -0.9 (1 .O) (1 .O) = -0.9 p,,, = (O.5pa,,) Vdes,l Cfig Cdy, = (0.5)(1.2) (64.6)2 (-0.9)(1.0) = -2253 Pa = -2.25 kPa pint = 0 kPa Net pressure across roof surface = -2.25-(0.0) = -2.25 kPa

For 4.35 m to 8.7m from windward edge C,. (external) = -0.5 (1 .O) (1 .O) = -0.5 p,,, = (0.5 p, , , ) Vdea,l Cfig Cdy, = (0.5)(1.2) (64.6)2 (-OS)(l.O) = -1252 Pa = -1.25 kPa

Net pressure across roof surface = -1.25-(0.0) = -1.25 kPa P,,,, = 0 kPa

For 8.7 m to 13.05 m from windward edge C,. (external) = -0.3 (1.0) (1.0) = -0.3 (negative case) p,,, = (0.5 p,,,. Vdes,: Cfig Cdyn = (0.5)( 1.2) (64.6) (-0.3)( 1 .O) = -75 1 Pa = -0.75 kPa

Net pressure across roof surface = -0.75-(0.0) = -0.75 kPa Cfig (external) = +O. 1 (1 .O) (1 .O) = +O. 1 @ositive case) p,,, =, (0.5 pa,,) Vdes,: Cfis Cdyn = (0.5)(1.2) (64.6)2 (+0.1)(0.8) = +200 Pa = +0.20 kPa Cfig (internal) = -0.3 (1.0) = -0.3 pint = (0.5 pa,,> Vdes,: Cfig Cdyn = (0.5)(1.2) (64.6) (-0.3)(0.8) = -601 Pa = -0.60 kPa Net pressure across roof surface = +0.20-(-0.60) = +0.80 kPa

P,,,, = 0 kPa

For 13.05 m to 24 m from windward edge C,. (external) = -0.2 (1 .O) (1 .O) = -0.2 (negative case) p,,, = (0.5 p,J Vdes,: C,. Cdyll = (0.5)(1.2) (64.6) (-0.2)(1 .O) = -501 Pa = -0.50 kPa pint = 0 kPa Net pressure across roof surface = -0.50-(0.0) = -0.50 kPa C,. (external) = +0.2 (1 .O) (1 .O) = +0.2 (positive case) p,,, = (0.5 psi$ Vdes,t C,. Cdyn = (0.5)(1.2) (64.6) (+0.2)(0.8) = +401 Pa = +0.40 kPa C,. (internal) = -0.3 (1 .O) = -0.3 pi,, = (0.5 pa,> Vdes,: C,. Cdy, = (0.5)(1.2) (64.6)2 (-0.3)( 1 .O) = -75 1 Pa = -0.75 kPa Net pressure across roof surface = +0.40-(-0.75) = +1.15 kPa

east roof slopes (surface IO) . Cfig (external) = -0.5 (1.0) (1.0) = -0.5 p,,, = (O.5palJ Vdes,: Cfig Cdyn = (0.5)(1.2) (64.6) (-0.5)(1.0) = -1252 Pa = -1.25 kPa pint = 0 kPa Net pressure across roof surface = -1.25-(0.0) = -1.25 kPa

Note: under-eaves pressures are same as adjacent walls.

Design loadings for foundations - ultimate limit states Use roof pressures as above multiplied by 0.8 (K,)

I

Design loadings for windows - ultimate limit states (note: for permissible stress design - divide loads by 1 S)

For windows less than 1.44 m2 in area on north wall: Cfig (external) = +0.7 (1.0) (1.25) = +0.875 (positive case) p,,, = (0.5 pair) Vdes,i Cfig Cdyn = (0.5)(1.2) (38.9)2 (+0.875)(1.0) = +794 Pa = +0.79 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pint = (OSp,,) Vdes,i Cfig Cdyn = (0.5)(1.2) (38.9)' (-0.3)(1.0) = -272 Pa = -0.27 kPa Net pressure across window = +0.79-(-0.27) = +1.06 kPa

For windows less than 1.44 mz in area on east wall: Cfig (external) = 10.7 (1 .O) (1.25) = +0.875 (positive case) p,,, = (0.5 paiJ Vdes,; Cfig Cdyn = (0.5)(1.2) (48.2)' (+0.875)(1.0) = +I220 Pa = +1.22 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pint = (0.5 paiJ Vdes,i Cfig Cdyn = ( O S ) ( 1.2) (48.2)2 (-0.3)( 1 .O) = -418 Pa = -0.42 kPa Net pressure across window = +1.22-(-0.42) = +1.64 kPa

For windows less than 1.44 m2 in area on south and west walls: Cfig (external) = +0.7 (1 .O) (1.25) = +OX75 (positive case) p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,t Cfig Cdyn = (0.5)(1.2) (64.6)' (+0.875)(1.0) = +2191 Pa = +2.19 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pint = (O.5paiJ Vdes,82 Cfig Cdyn = (0.5)(1.2) (64.6)2 (-0.3)(1.0) = -751 Pa = -0.75 kPa Net pressure across window = +2.19-(-0.75) = +2.94 kPa

For windows less than 1.44 m2 in area on east and west walls within a distance of 1.2 m from north corners: Cfig (external) = -0.65 (1 .O) (2.0) = -1.3 (negative case) pex,=(0.5paiJVdes,~CfigCdyn = (0.5)(1.2) (38.9)2(-1.3)(1.0)=-1180Pa=-1.18kPa pint = 0 kPa Net pressure across window = -1.18-(0.0) = -1.18 kPa

For windows less than 1.44 m2 in area on north and south walls within a distance of 1.2 m from east comers: Cfig (external) = -0.65 (1 .O) (2.0) = -1.3 (negative case) p,,, = (0.5pJ Vdcs,t Cfig Cdyn = (0.5)(1.2) (48.2)2 (-1.3)(1.0) = -1812 Pa= -1.81 kPa pint = 0 kPa Net pressure across window = -1.81-(0.0) = -1.81 kPa

For windows less than 1.44 m2 in area on east and west walls within a distance of 1.2 m from south corners, and on north and south walls within a distance of 1.2 m from west comers: Cfig (external) = -0.65 (1.0) (2.0) = -1.3 (negative case) p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,i Cfig Cdyn = (0.5)(1.2) (64.6)2 (-1.3)(1.0) = -3255 Pa = -3.26 kPa pint = 0 kPa Net pressure across window = -3.26-(0.0) = -3.26 kPa

For windows (or doors) more than 1.44 m2 in area on north wall: Cfig (external) = +0.7 (1 .O) (1 .O) = +0.7 (positive case) p,,, = (0.5 psi) Vdes,i Cfig Cdyn = (0.5)(1.2) (38.9)* (+0.7)(1.0) = +636 Pa = +0.64 kPa Cfig (internal) = -0.3 (1.0) = -0.3 pi,, = (0.5 psi) Vdes,i C.fig Cdyn = ( O S ) ( 1.2) (38.9)2 (-0.3)( 1 .O) = -272 Pa = -0.27 kPa Net pressure across window = +0.64-(-0.27) = +0.91 kPa

i i i

I ..... . ... 11, I/. .... ...,.,... ..... ... ......... " ...... "..l.l.*..."...........~...l.* -_,. __......_.r .,..,. ...l.."l* ,......... ~ , * ...,...... A Guide to ASINZS 11 70.2: 2002

Chapter 5 Wind Actions I 39 For windows (or doors) more than 1.44 m2 in area on east wall: C,. (external) = +0.7 (1 .O) (1 .O) = +0.7 (positive case) p,,, = (0.5 paiJ Vdes,: Cfig Cdyn = (0.5)(1.2) (48.2)2 (+0.7)( 1 .O) = +976 Pa = M.98 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pi,, = (0.5 pa,,) Vdes,: Cfig Cdyn = (0.5)(1.2) (48.2)2 (-0.3)(1.0) = -418 Pa = -0.42 kPa Net pressure across window = +0.98-(-0.42) = +1.40 kPa

For windows (or doors) more than 1.44 m2 in area on south and west walls: C,. (external) = +0.7 (1 .O) (1 .O) = +0.7 (positive case) p,,, = (0.5 paiJ Vdes,i Cfig Cdyn = (0.5)(1.2) (64.6)2 (+0.7)(1.0) = +1753 Pa = +1.75 kPa C,. (internal) = -0.3 (1 .O) = -0.3 pint = ((0.5 pa,> Vdes,i Cfig Cdyn = (0.5)( 1.2) (64.6)2 (-0.3)( 1 .O) = -75 1 Pa = -0.75 kPa Net pressure across window = +1.75-(-0.75) = +2.50 kPa

For windows up to 5.76 m2 in area on east and west walls within a distance of 2.4 m from north corners: Cfig (external) = -0.65 (1 .O) (1.5) = -0.975 (negative case) p,,, = (0.5 pa,,> Vdes,i Cfig Cdyn = (0.5)(1.2) (38.9)2 (-0.975)(1.0) = -885 Pa = -0.89 kPa pint = 0 kPa Net pressure across window = -0.89-(0.0) = -0.89 kPa

For windows up to 5.76 m2 in area on north and south walls within a distance of 2.4 m from east corners: Cfig (external) = -0.65 (1 .O) (1.5) = -0.975 (negative case) p,,, = (0.5 pa,$ Vdes,i C,. Cdy, = (0.5)( 1.2) (48.2)2 (-0.975)( 1 .O) = -1359 Pa = -1.36 kPa pin, = 0 kPa Net pressure across window = -1.36-(0.0) = -1.36 kPa

For windows up to 5.76 m2 in area on east and west walls within a distance of 2.4 m from south comers, and on north and south walls within a distance of 2.4 m from west corners: Cfig (external) = -0.65 (1.0) (1.5) = -0.975 (negative case) p,,, = (0.5 paJ Vdes,: Cfig Cdy, = (0.5)( 1.2) (64.6)2 (-0.975)( 1 .O) = -2441 Pa = -2.44 kPa pi,, = 0 kPa Net pressure across window = -2.44-(0.0) = -2.44 kPa

For any other windows or doors, loads for major framing members apply.

Design loadings for roof cladding and battens - ultimate limit states

(note: for permissible stress design - divide loads by 1.5)

For roof cladding elements less than 1.44 m2 in area, on roof surfaces 8 and 9, up to 1.2 m from east edge of roof: Cfig (external) = -0.9 (1 .O) (2.0) = -1.8 p,,, = (0.5 pa,,> Vdes,: Cfig Cdyn = (0.5)(1.2) (48.2)2 (-1.8)( 1 .O) = -2509 Pa = -2.5 1 kPa pint = 0 @a Net pressure across element = -2.5 1 -(0.0) = -2.5 1 kPa

For roof cladding elements less than 1.44 m2 in area, on roof surfaces 8 and 9, up to 1.2 m from west edge of roof, and on roof surfaces 10 and 1 1, up to 1.2 m from south edge of roof : Cfig (external) = -0.9 (1 .O) (2.0) = -1.8 p,,, = (0.5 psi,) Vdes,: C,. Cdyn = (0.5)( 1.2) (64.6)2 (- 1.8)( 1 .O) = -4507 Pa = -4.5 1 kPa pi,, = 0 kPa Net pressure across element = -4.5 1-(0.0) = -4.5 1 kPa

For roof cladding elements less than 5.76 m2 in area, on roof surfaces 8 and 9, up to 2.4 m from east edge of roof: Cfig (external) = -0.9 (1.0) (1.5) = -1.35 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,t Cfig Cdyn = (0.5)(1.2) (48.2)2 (-1.35)(1.0) = -1882 Pa = -1.88 kPa pint = 0 kPa Net pressure across element = -1.88-(0.0) = -1.88 kPa

For roof cladding elements less than 5.76 m2 in area, on roof surfaces 8 and 9, up to 2.4 m from west edge of roof, and on roof surfaces 10 and 11, up to 2.4 m from south edge of roof: Cfis (external) = -0.9 (1.0) (1.5) = -1.35 -.- p,,, = (O.5pai) Vdes,; CfiP Cdyn = (0.5)(1.2) (64.6)2 (-1.35)(1.0) = -3380 Pa = -3.38 kPa pTn, = 0 kPa Net pressure across element = -0.81-(0.0) = -3.38 kPa For roof cladding elements less than 1.44 m2 in area, on roof surface 8, up to 1.2 m from north edge of roof : Cfig (external) = -0.59 (1.0) (2.0) = -1.18 (negative case) p,,,= ( 0 . 5 ~ ~ ~ ) Vdes,t Cfig Cdyn = (0.5)(1.2) (38.9)2 (-1.18)(1.0) = -1071 P a = -1.07 kPa pi,, = 0 kPa Net pressure across element = -1.07-(0.0) = -1.07 kPa For roof cladding elements less than 1.44 m2 in area, on roof surface 7, up to 1.2 m from west edge of roof, and on roof surface 9, up to 1.2 m from south edge of roof : Cfig (external) = -0.59 (1.0) (2.0) = -1.18 (negative case) p,,, = (O.5pai) Vdes,; Cfig Cdyn = (0.5)(1.2) (64.6)2 (-1.18)(1.0) = -2955 Pa= -2.95 kPa pint = 0 kPa Net pressure across element = -2.95-(0.0) = -2.95 kPa

For roof cladding elements less than 5.76 m2 in area, on roof surface 8, up to 2.4 m from north edge of roof : Cfig (external) = -0.59 (1 .O) (1.5) = -0.89 (negative case) p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,t C,,, Cdy, = (0.5)(1.2) (38.9)2 (-0.89)(1.0) = -808 Pa = -0.81 kPa pint = 0 kPa Net pressure across element = -0.81-(0.0) = -0.81 kPa

For roof cladding elements less than 5.76 m2 in area, on roof surface 7, up to 2.4 m from west edge of roof, and on roof surface 9, up to 2.4 m from south edge of roof: Cfig (external) = -0.59 (1 .O) (1 S ) = -0.89 p,,, = (0.5 pa,> Vdes,: Cfig Cdyn = (0.5)(1.2) (64.6)2 (-0.89)(1 .O) = -2228 Pa = -2.23 E a pint = 0 kPa Net pressure across element = -2.23-(0.0) = -2.23 kPa

Loading on battens should be applied as moving 'patch' loads with a tributary area equal to the batten spacing times truss rafter spacing, with the appropriate local pressure factor for that area. For example, in this case, if the batten spacing is 0.6 m, the tributary area would be 1.2 X 0.6 = 0.72 m2. Since this area is less than 0.25a2, the local pressure factor could be 1 .O, 1.5 or 2.0, depending on the distance of the centre of the 'patch' from the roof edge. Over the remainder of the batten, K, should be taken as 1 .O; i.e. the loading should be the same as used for major structural members.

A Guide to ASINZS 11 70.2: 2002


5.3 Example 3 - A steel-framed warehouse on the edge of an urban area

41

Design wind loads are required for a steel portal-framed warehouse in Manukau at the southern edge of the Auckland metropolitan area. The relevant information is as follows:

Location: southern edge of Auckland (Region A6) Terrain: Suburban terrain for westerly through northerly to south-easterly directions.

Topography: ground slope less than 1 in 20 for greater than 5 kilometres in all directions. Dimensions: eaves height : 5.85 m Horizontal dimensions, (Figure 5.3) : 25 metres X 15 metres. Gable roof with 5 degrees pitch. average roof height = 5.85 + OS(7.5 tan 5") = 6.2 metres Building orientation: major axis is North-west - South-east Steel portal frame construction. Frames are spaced at 5 m. The building has a large (4m wide by 3 m high) roller door on the south-west wall. There are windows (1 m') near the south-east end, and a small (2.5 m') door near the

Open water to south and south-west.

north-west end, of the north-east wall.

Figure 5.3 Framing of industrial warehouse

Regional wind speed According to AS/NZSll70.0 Table 3.1, [8], the structure should be treated as Importance Level 2. Hence take average recurrence interval, R, equal to 500 years (Table 3.2 in [4]).

From Table 3.1 in AS/NZS 1170.2, V, = 45 m l s (Region A)

Wind direction multipliers for Auckland (Region A6) are given in (Table 3.2). Values range from 0.85 (N, S) to 1 .O (E, W).

Terrain-height z=h=6.2 m, for Terrain Category 3, MZ,,t = M6,2,aat3 = 0.83 (Table 4.1(A))

for Terrain Category 2, Mz,cat = M6,Z,aa = 0.93 (Table 4.1 (A), by interpolation)

Shielding The building has other industrial development to the north and west, assume a Shielding Multiplier, Ms, of 0.85 for N, W, NW directions. Take Ms =1.0 for other directions.

~~~1 ~~~1 m/s) m/s)

N 45 0.85 0.83 0.85 1 .o 27.0 NE 45 0.95 0.83 1 .o 1 .o 35.5 E 45 1 .oo 0.83 1 .o 1 .o 37.4 SE 45 0.95 0.83 1 .o 1 .o 35.5 S 45 0.85 0.93 1 .o 1 .o 35.6

SW 45 0.95 0.93 1 .o 1 .o 39.8 W 45 1 .oo 0.83 0.85 1 .o 31.7

NW 45 0.95 0.83 0.85 1 .o 30.2


Design wind speeds In this case, the building is orientated NE-SW and NW-SE, and the design wind speeds, Vdes,B can be obtained by inspection from VsitS in the above table, for the four wind directions orthogonal to the building walls:

Vdes,NE = 37.4 m/s (largest from N to E sector)

Vdes,SE = 37.4 m/s (largest from E to S sector)

Vdes,SW = 39.8 m/s (largest from S to W sector)

Vdes.NW = 3 1.7 m/s (largest from W to N sector)

Aerodynamic shape factor

External pressures

Windward walls : +0.7 taking z = h =6 m (Table 5.2(A))

Leeward walls (normal to long axis): -0.5 (Table 5.2(B)) Leeward walls (normal to short walls): -0.37 (Table 5.2(B) by interpolation for d b =1.67)

Side walls (0 to 6.2 m from windward edge): -0.65 (Table 5.2(C)) (6.2 m to 12.4 m): -0.5 (12.4 m to 18.6 m): -0.3 (beyond 18.6 m): -0.2

A Guide to ASINZS 1170.2: 2002


Roof (0 to 6.2 m from windward edge): -0.9, -0.4 (Table 5.3(A)) (6.2 m to 12.4 m): -0.5 (12.4 m to 18.6 m): -0.3, +0.1 (beyond 18.6 m): -0.2, +0.2

Area reduction factors (Table 5.4) For wall loads on end frame columns (as side wall): tributary areas 5.85 X 2.5 = 14.6 m2

For wall loads on central frame columns (as side wall): tributary areas 5.85 X 5 = 29.25 m2

For roof loads on end frame rafters: tributary areas 15 x 2.5 = 37.5 m2.

For roof loads on central frame rafters: tributary areas 15 X 5 = 75 m2.

Ka = 0.97 (Table 5.4 by interpolation)

K, = 0.89

K, = 0.88

K, = 0.83

Local pressure factors (Table 5.6)

a = minimum of 0.2 X 15 m, 6.2 m = Min(3.0 m, 6.2 m} = & limiting tributary areas for local pressure factors : 0.25a2= 2.25 m2

a2 = h2

Internal pressures (Section 5.3)

The roller door will be treated as a potential dominant opening (12 m2) for SW wind only. The background permeability of the other walls is estimated as 0.3% of the wall area, giving a total area of 5.85 x (25 + (2)(15)) X 0.003= 0.965 m2. Then ratio of dominant opening area to total open area due to permeability = 12/0.965=12.4

In this case, for SW winds, Cp,i = +0.7 (Table 5. I@))

For other wind directions, the roller door will be assumed closed with the building considered to have all walls equally permeable.

Cp,i = -0.3 or 0.0 (Table 5.1(A))

Action combination factors

Case (b) in Table 5.5 is applicable for NE and SW winds: K, = 0.8

Case (c) is applicable, for all wind directions, to positive pressures on downwind end of roof with negative internal pressures: Kc = 0.8

Case (d) in Table 5.5 is applicable to external negative roof and leeward wall pressures and positive internal pressures, for SW winds: K, = 0.95. However, Kc= 0.8 from Case (b) governs for external roof and leeward pressures.

Otherwise, Kc = 1 .O

Dynamic response factor CdY, = 1 .O (natural frequencies greater than 1 .O Hertz) (Section 6.1)

Design wind pressure for portal frames - ultimate limit states North-east wind direction (roller door closed Cp,i = 0.0)

All Frames

windward wall (north east wall): Cfig (external) = Cp,, Ka Kc = +0.7 (1.0) (0.8) = 0.56 p,,, = (0.5 pa,,> Vdes,: Cfig Cdyn = ( O S ) ( 1.2) (37.4)2 (+0.56)( 1 .O) = 470 Pa = +0.47 kPa Cfig (internal) = 0.0 (1.0) = 0.0

Net pressure across wall surface = +0.47-(0.0) = +0.47 kPa Pi,, = (0'5 pa,,> 'des,: 'fig 'dyn = O'O

leeward wall (south west wall): Cfig (external) = CP,, K, K, = -0.5 (1 .O) (0.8) = -0.40 p,,, = (0.5 pad Vdes,: Cfig Cdyn = (0.5)(1.2) (37.4)2 (-0.40)(1.0) = -336 Pa = -0.34 kPa Cfig (internal) = 0.0 (1.0) = 0.0

Net pressure across wall surface = -0.34-(0.0) = -0.34 kPa Pint = (0'5 Pair) 'des,: 'fig 'dynn = O'O

I

j j Endframes j

Roof(0 to 6.2 m from north east wall): Cfig (external) = Cp,, Ka Kc = -0.9 (0.88) (0.8/0.88) = -0.72

p,,, = (O.5paJ Vdes,: Cfig Cdynn = (0.5)(1.2) (37.4)2 (-0.72)(1.0) = -604 Pa =-0.60 kPa Cfig (internal) = 0.0 (1 .O) = 0.0


I

(note that K, cannot be less than 0.8/Ka - Section 5.4.3)

Pin, = (0'5 pa,,> 'des,: 'fig 'dyn = O'O kPa

Roof (6.2 m to 12.4 m from north east wall): Cfig (external) = Cp,c Ka Kc = -0.5 (0.88) (ON0.88) = -0.40 p,,, = (0.5 pa,,> Vdes,: Cfig Cdyn = (0.5)( 1.2) (37.4)2 (-0.40)( 1 .O) = -336 Pa =-0.34 kPa Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across roof surface = -0.34-(0.0) = -0.34 kPa Pint = (0'5 pa,> 'des,: 'fig 'dy, = o*o kPa

Roof (12.4 m to 15 m p o m north east wall): Cfig (external) = Cp,, Ka Kc = -0.3 (0.88) (0.8/0.88) = -0.24 p,,, = (0.5 pa,,> Vdes,: Cfig Cdy, = (OS)( 1.2) (37.4)2 (-0.24)( 1 .O) = -201 Pa =-0.20 kPa Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across roof surface = -0.20-(0.0) = -0.20 kPa Pint = (0'5 pa,> 'des,: 'fig 'dYn = o*o

Central frames

Same as end frames (combined Ka K, for roof pressures is equal to 0.8)

-0.34 kPa -0.60 kPa -0.20 kPa

@ -0.34 kPa +0.47 kPa

Figure 5.4 Load case I - NE wind

A Guide to ASlNZS 11 70.2: 2002

Chapter Actions 5 I 45 South-west wind direction (dominant windward opening on SW wall, Cp,i=+O, 7)

All Frames windward wall (south west wall): Cfig (external) = Cp,, K, K, = +0.7 (1.0) (0.8) = 0.56 p,,, = (0.5pai,) Vdcs,i Cfig Cdyn = (0.5)(1.2) (39.8)2 (+0.56)(1.0) = 532 Pa = +0.53 kPa Cfig (internal) = 0.7 ( 1 .O) = 0.7 pint = (0.5 paiJ Vdes,i Cfig Cdyn = (0.5)(1.2) (39.8)2 (+0.7)(1.0) = 665 Pa = +0.67 kPa Net pressure across wall surface = +0.53-(0.67) = -0.14 kPa

leeward wall (north east wall): Cfig (external) = Cp,, Ka K, = -0.5 (1 .O) (0.8) = -0.40 p,,, = (0.5 pa,,.> Vdcs,i C,,, Cdyll = (0.5)(1.2) (39.8) (-0.40)(1 .O) = -380 Pa = -0.38 kPa Cfig (internal) = 0.7 (0.95) = 0.67 pin = (0.5 pa,$ Vdes,i Cfig Cdyn = +0.64 @a Net pressure across wall surface = -0.38-(0.64) = -1.02 kPa

End frames Roof first 6.2 m from south west wall): Cfig (external) = Cp,, K, Kc = -0.9 (0.88) (0.8/0.88) = -0.72 p,,, = (O.5pa,,.) Vdes,i Cfig Cdyn = (0.5)(1.2) (39.8) (-0.72)(1.0) = -684 Pa =-0.68 kPa Cfig (internal) = Cp,, Kc = 0.7 (0.95) = 0.67 pint = (0.5 pa,,) Vdes,i Cfig Cdyll = (0.5)( 1.2) (39.8)* (0.67)( 1 .O) = +637 Pa = +0.64 kPa Net pressure across roof surface = -0.68-(0.64) = -1.32 kPa

Roof (6.2 m to 12.4 m from south west wall): C,,, (external) = Cp,, Ka Kc = -0.5 (0.88) (0.8/0.88) = -0.40 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,i Cfig Cdyn = (0.5)(1.2) (39@ (-0.40)(1.0) = -380 Pa =-0.38 kPa Cfig (internal) = 0.7 (0.95) = 0.67 pint = (0.5 pair) Vdes,i Cfig dyn = +0.64 @a Net pressure across roof surface = -0.38-(0.64) = -1.02 kPa

Roof (12.4 m to 15 mfFom south west wall): Cfig (external) = Cp,, Ka Kc = -0.3 (0.88) (0.8/0.88) = -0.24 p,,, = (0.5 pa,$ Vdes,i Cfig Cdyn = (0.5)( 1.2) (39.8) (-0.24)( 1 .O) = -228 Pa =-0.23 kPa Cfig (internal) = 0.7 (0.95) = 0.67

Net pressure across roof surface = -0.23-(0.64) = -0.87 kPa Pint = (05 Pair) V d e s , i fig dYn =

Central frames

Same as end frames (combined KaKc for roofpressures is equal to 0.8)

-1.32 kPa -1.02 kPa

a 4 -0.87 kPa

-0.14 kPa -1.02 kPa

Figure 5.5 Load case 2 - SW wind

North-west wind direction (roller door closed, Cp,i = 0.0)

End frame at NWend side walls (south west and north east walls): Cfig (external) = Cp,, Ka K, = -0.65 (0.97) (1 .O) = -0.63

(note K, = 1 .O in this case) p,,, = (0.5 paiJ Vdes,; Cfig Cdyn = ( O S ) ( 1.2) (3 1 .7)2 (-0.63)( 1 .O) = -380 Pa = -0.38 kPa Cfig (internal) = 0.0 (1.0) = 0.0

Net pressure across wall surface = -0.38-(0.0) = -0.38 kPa roof Cfig (external) = Cp,, Ka Kc = -0.9 (0.88) (1 .O) = -0.79 p,,, = (0.5 pair) Vdes,; Cfig Cdyn = ( O S ) ( 1.2) (3 1.7)2 (-0.79)(1 .O) = -476 Pa = -0.48 E a Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across wall surface = -0.48-(0.0) = -0.48 kPa

Pint = (0.5 Pair) Vdes,; cfig Cdyn = 0.0 kPa

Pi,, = (0.5 pa& Vdes,; cfig Cdyn = 0.0 @a

-0.48 kPa -0.48 kPa b 4

-0.38 kPa -0.38 kPa

Figure 5.6 Load case 3 - NW wind - end frame

First inboard frame from SE end

(downward roof load case, roller door closed, Cp,i = -0.3) The tributary areas for this frame are between 17.5 m and 22.5m from the windward end. i.e. between 2.8h and 3.6h from the windward end. Fraction of area in 2h to 3h zone (1.1/5) = 0.22 Fraction of area in >3h zone (3.96) = 0.78

side walls (south west and north east walls): Cfig (external) = Cp,, K, Kc = (-0.222)(0.89)( 1 .O) = -0.20 p,,, = (0.5pair) Vdes,; CfiP Cdyn = (0.5)(1.2) (31.7)2 (-0.20)(1.0) = -121 Pa= -0.12 kPa Cfig (internal) = -0.3 (1 .O) = -0.30 pint = (0.5 paiJ Vdes,t C, Cdyn = (OS)( 1.2) (3 1 .7)2 (-0.30)( 1 .O) = -1 81 Pa = -0.18 kPa Net pressure across wafl surface = -0.12-(-0.18) = +0.06 kPa

roo$ Cfig (external) = CP,, Ka Kc = (+0.178)(0.83)(0.8/0.83) = +0.14 p,,, = (0.5 pair) Vdes,; Cfig Cdyn = ( O S ) ( 1.2) (3 1 .7)2 (+O. 14)( 1 .O) = +84 Pa = +0.08 kPa Cfig (internal) = -0.3 (0.8) = -0.24 pint = (0.5paiJ Vdes,; C, Cdyn = (0.5)(1.2) (31.7)2 (-0.24)(1.0) = -145 Pa = -0.14 kPa Net pressure across wafl surface = +0.08-(-0.14) = +0.22 kPa

+0.22 kPa +0.22 kPa

Figure 5.7 Load case 4 - NW wind - inboard frame

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Wind Act Chapter 5

South-east wind direction (roller door closed, Cp,i = 0.0)

End frame at SE end side walls (south west and north east walls): Cfig (external) = CP,, K, K, = -0.65 (0.97) (1 .O) = -0.63

(note: K, = 1 .O in this case) p,,, = (0.5 pa,,> Vdes,i Cfig C,,, = (0.5)(1.2) (37.4)2 (-0.63)(1.0) = -529 Pa = -0.53 kPa Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across wall surface = -0.53-(0.0) = -0.53 kPa PI,, = (Oe5 pal,> 'des,: 'fig ',-jyn = O'O kPa

roof Cfig (external) = Cp,, K, Kc = -0.9 (0.88) (1 .O) = -0.79 p,,, = (0.5 pad Vdes,: Cfig Cdyn = (0.5)( 1.2) (37.4)' (-0.79)( 1 .O) = -663 Pa = -0.66 kPa Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across wall surface = -0.66-(0.0) = -0.66 kPa Pint = (0'5 pa,,) 'des,: 'fig 'dy, = O'O kPa

-0.66 kPa -0.66 kPa 4 4

-0.53 kPa -0.53 kPa

Figure 5.8 Load case 5 - SE wind - end frame

First inboard frame from NWend (downward roof load case, roller door closed, Cp,i = -0.3)

The tributary areas for this frame are between 17.5 m and 22.5m from the windward end. i.e. between 2.81 and 3.6h from the windward (SE) end. Fraction of area in 2h to 3h zone (1.115) = 0.22 Fraction of area in >3h zone (3.915) = 0.78

side walls (south west and north east walls): Cfig (external) = Cp,, K, K, = (-0.222)(0.89)( 1 .O) = -0.20 p,,, = (O.5paJ Vdes,i Cfig C,,, = (0.5)(1.2) (37.4)' (-0.20)(1.0) = -168 Pa = -0.17 kPa C (internal) = -0.3 (1 .O) = -0.30 pin, = (0.5 pa,> Vdes,i Cfig Cdy, = (0.5)(1.2) (37.4)' (-0.30)(1.0) = -252 Pa = -0.25 kPa Net pressure across wall surface = -0.17-(-0.25) = +0.08 kPa

fig

roof Cfig (external) = Cp,, K, K, = (+0.178)(0.83)(0.8/0.83) = +0.14 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,i Cfig CdY, = (0.5)(1.2) (37.4)' (+0.14)(1.0) = +117 Pa = +0.12 kPa Cfig (internal) = -0.3 (0.8) = -0.24 pint = (0.5 paiJ Vdes,: Cfig Cdyn = (0.5)( 1.2) (37.4)2 (-0.24)(1 .O) = -201 Pa = -0.20 kPa Net pressure across wall surface = +0.12-(-0.20) = +0.32 kPa

ind Actions 48 /6,,ter5

j +0.32 kPa +0.32 kPa

+0.08 kPa

~

~ +0.08kPa

Figure 5.9 Load case 6 - SE wind - inboard frame

Design wind pressures for doors and windows - ultimate limit states


For roller door (12 mz) on the south west wall:

positive case - SW wind 39.8 mhec Cfig (external) = Cp,, (K,) (K,) = +0.7 (1 .O) (1 .O) = +0.7

(note: K,= 1 .O, since A> a') p,,, = (0.5 paiJ Vdes,z Cfig Cdyn = ( O S ) ( 1.2) (39.8)2 (+0.7)( 1 .O) = +665 Pa = +0.67 kPa C (internal) = -0.3 (1 .O) = -0.3 pint = (0.5pai$ Vdes,; Cfig Cdyn = (0.5)(1.2) (39.8)2 (-0.3)(1.0) = -285 Pa = -0.29 kPa Net pressure across window = +0.67-(-0.29) = t-0.96 kPa

fig

negative case - SE wind 3 7.4 Lsec Cfig (external) = Cp,, (Ka) (K,) = -0.5 (1 .O) (1 .O) = -0.5

p,,, = (0.Spai,> Vdes,: Cfig Cdyn = (0.5)(1.2) (37.4)' (-0.5)(1 .O) = -420 Pa = -0.42 kPa Cfig (internal) = 0.0 (1.0) = 0.0

Net pressure across window = -0.42-(0.0) = -0.42 kPa

(note: K,= 1 .O, since A> 0.25a2)

Pint = (0'5 pa,> 'des,: 'fig 'dyn = O'O kPa

For small door (2.5 m2 in area) on north-east wall: Positive case - NE wind 3 7.4 m/sec Cfip (external) = +0.7 (1 .O) (1.0) = +0.7 p,,, = (0.5 pa,$ Vdes,t Cfig Cdyn = (OS)( 1.2) (37.4)2 (+0.7)( 1 .O) = +587 Pa = +0.59 kPa Cfig (internal) = -0.3 (1 .O) = -0.3 pint = (0.5 paiJ Vdes,i Cfig Cdyn = ( O S ) ( 1.2) (37.4)2 (-0.3)( 1 .O) = -252 Pa = -0.25 kPa Net pressure across window = +0.59-(-0.25) = +0.84 kPa

Negative case - NW wind 31.7 mhec Cfig (external) = Cp,, (Ka) (K,) = -0.65 (1 .O) (1 S) = -0.975 p,,, = (0.5 pad Vdes,l Cfig Cdyn = (0.5)(1.2) (3 1.7)2 (-0.975)(1.0) = -588 Pa = -0.59 kPa Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across door = -0.59-(0.0) = -0.59 kPa Pint = ( O e 5 p a k ) Vdes,82 'fig 'dyn = kPa

Negative case - SW wind 39.8 mhec (with roller door open) Cfig (external) = Cp,, (Ka) (K,) = -0.50 (1 .O) (1 .O) = -0.50 p,, = (0.5 paiJ Vdes,i Cfig Cdyn = ( O S ) ( 1.2) (39.8)2 (-0.50)( 1 .O) = -475 Pa = -0.48 kPa Cfig (internal) = 0.7 (1.0) = +0.70 pint = (0.5 pa,> Vdes,l Cfig Cdyn = (0.5)(1.2) (39.8)' (+0.70)(1.0) = +665 Pa = +0.67 kPa Net pressure across door = -0.48-(0.67) = - 1.15 kPa

(SW wind case with positive internal pressure is clearly dominant)

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Wind Act Chapter 5

For windows (I mz in area) on north-east wall: Positive case - NE wind 37.4 mhec Cfig (external) = +0.7 (1 .O) (1.25) = +0.875 p,,, = (0.5 p,) Vds,i Cfig CdY, = (0.5)(1.2) (37.4)* (+0.875)(1.0) = +734 Pa = +0.73 kPa C (internal) = -0.3 (1.0) = -0.3 pint = (0.5 psi$ Vdes,t Cog CdY, = (0.5)(1.2) (37.4)2 (-0.3)(1.0) = -252 Pa = -0.25 kPa Net pressure across window = +0.73-(-0.25) = W.98 kPa

fig

Negative case - SE wind 3 7.4 mhec Cfig (external) = -0.65 (1.0) (2.0) = -1.3 (negative case - SE wind) p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,: Cfig Cdyn = (0.5)(1.2) (37.4)2 (-1.3)(1.0) = -1091 Pa = -1.09 kPa Cfig (internal) = 0.0 (1 .O) = 0.0

Net pressure across window = -1.09-(0.0) = -1.09 kPa Pint = (05 pa;,> des,: pp dyn=

Negative case - SW wind 39.8 mhec (with roller door open) Cfig (external) = Cp,, (K,) (K,) = -0.50 (1 .O) (1 .O) = -0.50 p,, = (0.5 pJ Vdes,i Cfig Cdyn = (0.5)(1.2) (39.8)2 (-0.50)(1.0) = -475 Pa = -0.48 kPa Cfig (internal) = 0.7 (1.0) = +0.70 pi,, = (0.5 p i , ) Vdes,l Cfig Cdyn = (0.5)(1.2) (39.8)2 (+0.70)(1.0) = +665 Pa = +0.67 kPa Net pressure across window = -0.48-(0.67) = -1.15 kPa

(SW wind case with positive internal pressure is dominant)

Design wind pressures for wall and roof claddings and supporting structure - ultimate limit states


Wall cladding For cladding on north west or south east walls Area< 2.25 m2, between 0 and 1.5 m from the north east comers: (NE Wind) Cfig (external) = -0.65 (1.0) (2.0) = -1.3 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,t Cog Cdyn = (0.5)(1.2) (37.4)* (-1.3)(1.0) = -1091 Pa = -1.09 kPa pin, = 0 kPa Net pressure across element = -1.09-(0.0) = -1.09 kPa

For cladding on north west or south east walls Area< 2.25 m2, between 0 and 1.5 m from the north east corners: (SW Wind with roller door open) Cfig (external) = -0.30 (1 .O) (1 .O) = -0.3 p,,, = (0.5 paiJ Vdes,t Cfig Cdyn = (0.5)(1.2) (39.8)* (-0.3)(1.0) = -285 Pa = -0.29 kPa pint = ( O S ) ( 1.2) (39.8)2 (+0.70)(1 .O) = +665 Pa = +0.67 kPa Net pressure across element = -0.29-(+0.67) = -0.96 kPa (NE wind case is dominant)

For cladding on north west or south east walls Area< 2.25 m2, between 0 and 1.5 m from the south west comers: (SW Wind with roller door open) Cog (external) = -0.65 (1 .O) (2.0) = -1.3 peXt = (0.5paiJ Vdes,i Cog Cdy, = (0.5)(1.2) (39.8)2 (-1.30)(1.0) = -1236Pa = -1.24 kPa pint = ( O S ) ( 1.2) (39.8)2 (+O.70)(1 .O) = +665 Pa = W.67 kPa Net pressure across element = -1.24-(W.67) = -1.91 kPa

For cladding on north west or south east walls Area< 9 m2, between 0 and 3.0 m from the north east comers: (NE Wind) Cog (external) = -0.65 (1.0) (1.5) = -0.975 p,,, = (O.5pai) Vdes,i Cfig Cdyn = (0.5)(1.2) (37.4)2 (-0.975)(1.0) = -818 Pa = -0.82 kPa

pint = 0 kPa Net pressure across element = -0.82-(0.0) = -0.82 kPa

For cladding on north west or south east walls Area< 9 m', between 0 and 3.0 m from the north east comers: (SW Wind with roller door open) (same as cladding with Area Vdes,: Cfig Cdyn = (0.5)(1.2) (37.4)' (-1.3)(1.0) = -1091 Pa= -1.09 kPa pint = 0 kPa Net pressure across element = -1.09-(0.0) = -1.09 kPa

For cladding on south west walls Area

Chapter Actions 5 1 51 For cladding on south west walls Area< 9 m2 between 0 and 3.0 m from the north west corners: (NW Wind) Cfig (external) = -0.65 (1 .O) (1.5) = -0.975 p,,, = (0.5 psi$ Vdes,: Cfig Cdyn = (0.5)(1.2) (31.7)2 (-0.975)(1.0) = -588 Pa = -0.59 kPa pint = 0 kPa Net pressure across element = -0.59-(0.0) = -0.59 kPa

Roof cladding For roof cladding elements area< 2.25 m2, on roof between 0 and 1.5 m from north east, or south east, edges of roof: (NE or SE Wind) Cfig (external) = -0.9 (1.0) (2.0) = -1.8 peXt = ( 0 . 5 ~ ~ 3 Vdes,: Cfip Cdyn= (0.5)(1.2) (37.4)2 (-1.8)(1.0) = -1511 Pa= -1.51 kPa pint = 0 kPa Net pressure across element = -1.51-(0.0) = -1.51 kPa

For roof cladding elements between 2.25 and 9 m2 in area, on roof between 0 and 3.0 m from north east, or south east, edges of roof: Cfig (external) = -0.9 (1.0) (1.5) = -1.35 p,,,=(0.5pai,)Vdes,~ CfipCdyn= (0.5)(1.2) (37.4)2 (-1.35)(1.0)=-1133 Pa=-1.13 kPa pi,, = 0 kPa Net pressure across element = - 1.13-(0.0) = - 1.13 kPa

For roof cladding elements area< 2.25 m2, on roof between 0 and 1.5 m from south west edge of roof: (SW Wind with roller door open) Cfig (external) = -0.9 (1.0) (2.0) = -1.8 pext=(0.5pai,)Vdes,~CfigCdyn= (0.5)(1.2) (39.8)2 (-1.8)(1.0)= -1711 Pa=-1.71 Wa pint = (0.5)(1.2) (39.8)2 (+0.70)(1.0) = +665 Pa = +0.67 kPa Net pressure across element = -1.71-(0.67) = -2.38 Wa

For roof cladding elements between 2.25 and 9 m2, on roof between 0 and 3.0 m from south west edge of roof: (SW Wind with roller door open) Cfig (external) = -0.9 (1.0) (1.5) = -1.35 p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,: Cfig Cdyn = (0.5)(1.2) (39.8)2 (-1.35)(1.0) = -1283 Pa = -1.28 kPa pin, = (0.5)( 1.2) (39.8)2 (+0.70)( 1.0) = +665 Pa = +0.67 kPa Net pressure across element = -1.28-(0.67) = -1.95 kPa

For roof cladding elements all areas, on roof between 3.0 and 6.2 m from south west edge of roof: (SW Wind with roller door open) Cfig (external) = -0.9 (1 .O) (1 .O) = -0.9 pext = (0.5 pair) Vdes,: Cfig Cdyn = (0.5)(1.2) (39.8)2 (-0.9)(1.0) = -855 Pa = -0.86 kPa pint = (0.5)(1.2) (39.8)2 (+0.70)(1.0) = +665 Pa = +0.67 kPa Net pressure across element = -0.86-(0.67) = -1.53 kPa

For roof cladding elements all areas, on roof between 6.2 and 12.4 m from south west edge of roof: (SW Wind with roller door open) Cfig (external) = -0.5 (1.0) (1.0) = -0.5 p,,, = (0.5 pa;$ Vdes,: Cfig Cdyn = (0.5)(1.2) (39.8)2 (-0.5)(1.0) = -475 Pa = -0.475 kPa pint = (0.5)(1.2) (39.8)2 (+0.70)(1.0) = +665 Pa = +0.665 kPa Net pressure across element = -0.475-(0.665) = -1.14 kPa

For roof cladding elements all areas, on roof between 12.4 and 15 m from south west edge of roof: (SW Wind with roller door open) Cfig (external) = -0.5 (1 .O) (1 .O) = -0.3 p,,, = (0.5 pa,$ Vdes,: Cfig Cdp = (0.5)(1.2) (39.8)2 (-0.3)(1.0) = -285 Pa = -0.29 kPa p,,, = (0.5)(1.2) (39.8)2 (+0.70)(1.0) = +665 Pa = +0.67 kPa Net pressure across element = -0.29-(0.67) = -0.96 kPa (cases above govern for this zone)

For roof cladding elements area< 2.25 m2, on roof between 0 and 1.5 m from north west edge of roof: (NW Wind) Cfig (external) = -0.9 (1 .O) (2.0) = -1.8 p,,, = (OSp,,,) Vdes,i Cfig Cdyn = (0.5)(1.2) (31.7)2 (-1.8)(1.0) = -1085 Pa= -1.09 kPa pint = 0 kPa Net pressure across element = -1.09-(0.0) = -1.09 kPa (cases above govern for this zone)

For roof cladding elements between 2.25 and 9 m2, on roof between 0 and 3.0 m from north west edge of roof: Cfig (external) = -0.9 (1.0) (1.5) = -1.35 p,,, = (O.5paJ Vdes,: CfiP Cdyn= (0.5)(1.2) (31.7)2 (-1.35)(1,0) = -814 Pa= -0.81 kPa pint = 0 kPa Net pressure across element = -0.81-(0.0) = -0.81 kPa (cases above govern for this zone)

Governing design pressures for roof cladding elements, less than 2.25m2 in area, are shown in Figure 5.10

1.5 m

-1.1 , ;I.

I

I I

South west wall

Figure 5. I O Design pressures for roof cladding (kPa)

For other wall and roof cladding elements, use loads for portal frames, with positive loads multiplied by 1.25 (K,) for elements less than 2.25 m2 in area.

Loading on roof purlins should be applied as moving 'patch' loads with the appropriate local pressure factor for that area. For example, in this case, if the purlin spacing is 0.6 m, the tributary area for a patch 2.5 m long (equal to one half the frame spacing) would be 2.5 X 0.6 = 1.5 mz. Since this area is less than 0.25a2, the local pressure factor could be 1.0, 1.5 or 2.0, depending on the distance of the centre of the 'patch' from the roof edge. Over the remainder of the purlin, Kp should be taken as 1.0 for that load case; i.e. the loading should be the same as that used for major structural members.

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5.4 Example 4 - A medium-height office building in a tropical city

Design wind loads are required for a proposed office building in the Darwin C.B.D. The relevant information is as follows :

Location : CBD Darwin (Region C) Terrain : Suburban terrain for all directions. Topography : ground slope less than 1 in 20 for greater than 5 kilometres in all directions. Dimensions : average roof height : 48 metres

Reinforced concrete frame construction. Mullions for glazing panels are spaced at 1.5 m. Roof is flat with a 0.9m high parapet. Sway frequencies are greater than 1 Hertz.

Horizontal dimensions: 30 metres X 60 metres (rectangular cross-section).

/

/ Figure 5.11 48-metre high office building

This building was used as an example in the Guide to ASCE 7-02 [21]. Its use in the present Design Guide allows for a comparison of the two Standards for this building.

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. I in AS/NZS1170.2, V,,, = F,. 66 d s (Region C) = (1.05)(66) = 69.3 m / s

Wind direction multiplier For Region B, M, = 0.95 for overturning forces and major structural system for all directions (Section 3.3.2).

For cladding design, M, =1 .O .

Terrain- heig h t multiplier z=h=48 m, For Terrain Category 3, Mz,cat = M48,cat3 = 1.28 (Table 4.I(B) by interpolation)

Shielding There are no other buildings of greater height in any direction. Take Ms, equal to 1 .O for all directions.


Site wind speed Site wind speed for all directions for overall loads and structural design of main supporting members,

VsitS = 69.3(0.95)( 1.28)( 1 .O)( 1.0) = 84.3 d s (Equation 2.2)

For cladding design,

VsitB = 69.3( 1 .O)( 1.28)(1 .O)( 1 .O) = 88.7 d s (Equation 2.2)

Design wind speeds For all wind directions, the design wind speeds,

Vdes,B = VsitB = 84.3 d s (for overall loads and main structure design)

= 88.7 d s (for cladding)


External pressures

Windward walls : +0.8 for varying z (Table 5.2(A))

Leeward walls (normal to 60m walls): -0.5 (Table 5.2(B)) Leeward walls (normal to 30m walls): -0.3 (Table 5.2(B))

Side walls: -0.65 (Table 5.2(C))

Roof: -1.3 for first 24m from windward edge -0.7 for 24 to 48 m from windward edge -0.9 beyond 48 m from windward edge (Table 5.3(A))

Area reduction factors (Section 5.4.,2)

For elements greater than 100 m2 in area on roof or side walls, Ka = 0.8 For small elements < 10m2 in area, Ka = 1 .O

Local pressure factors (Table 5.6) a = minimum of 0.2 X 30 m = 6 m, or 48 m. a = 6.0 m limiting tributary areas for local pressure factors : 0.25a2 = 9 m2

a2 = 36 mz

Reduction factor due to parapet (Table 5.7) Height of parapet, h,, is 0.9m = 0.03 w, where w is the shortest horizontal dimension (30 m). K, = 0.8 for roof cladding areas RA1 and RA2. For all other areas, Kr = 1 .O.

Internal pressures (Section 5.3) The building is air-conditioned and can be considered to be effectively sealed. In this case, Cp,i = -0.2 or 0.0 (Table 5. I (A))

AGuide to ASINZS 1170.2: 2002

-1 Chapter 5 55

Height of

sector (m) 44 35 25 15 5

Action combination factor

Kc= 1.0

Dynamic response factor CdP = 1.0 (Section 6.1)

Calculation of base moment Calculations were carried out by spreadsheet. Summaries of the results are given in the following tables.

Wind normal to 60 m wall :

M,,,t3 windward leeward windward leeward moment qz.C, qh,Cfig q,.C,g.Cdyn.A qZ.Cfig.Cdyn.A contribution (kPap (kPa) (kN) (kN) (MN.m)

1.266 3.337 -2.132 1601.7 - 1023.3 115.5 1.20 2.998 -2.132 1798.8 -1279.2 107.7 1.10 2.519 -2.132 1511.5 -1279.2 69.8 0.95 1.879 -2.132 1127.4 - 1279.2 36.1 0.80 1.332 -2.132 799.5 - 1279.2 10.4

Height of

sector (m) 44 35 25 15 5

Note - q, denotes : 0.6 X Vdes,i

Total along-wind base bending moment obtained by summing contributions from last column of table = 340 MN.m

Mz,& 3 windward leeward windward leeward moment qz.c, qh.CRg qZ.C,g.Cdy,.A qZ.Cfig.Cdyn.A contribution (kPa! (kPa) (kN) (kN) (MN.m)

1.266 3.337 - 1.279 800.9 -307.0 48.7 1.20 2.998 -1.279 899.4 -383.8 44.9 1.10 2.519 -1.279 755.8 -383.8 28.5 0.95 1.879 - 1.279 563.7 -383.8 14.2 0.80 1.332 -1.279 399.7 -383.8 3.9

Wind normal to 30 m wall

Roof structure loads

For,first 24 m fiom leading edges (any orthogonal wind direction) - tributary areas greater than 100m2: Cfig (external) = -1.3 (0.8) (1 .O) = -1.04 peXt = (0.5 pair) Vdes,t Cfig CdW = (O.q(l.2) (84.3)* (-1.04)( 1 .O) = -4434 Pa = -4.43 kPa CfiP (internal) = 0.0 (1.0) = 0.0 pint = 0 kPa Net pressure across roof surface = -4.43-(0.0) = -4.43 kPa

For 24 m to 48 m from leading edges (any orthogonal wind direction) - tributary areas greater than 1 OOm2: Cfig (external) = -0.7 (0.8) (1.0) = -0.56 p,,, = (0.5paiJ Vdes,: Cfig Cdyn = (0.5)(1.2) (84.3)2 (-0.56)(1.0) = -2388 Pa = -2.39 kPa Cfig (internal) = 0.0 (1.0) = 0.0 pint = 0 kPa Net pressure across roof surface = -2.39-(0.0) = -2.39 kPa

For 48 m to 60 m from leading edges (wind directions normal to 30m walls) - tributaly areas greater than 1 OOm2: Cfig (external) = -0.5 (0.8) (1 .O) = -0.40 p,,, = (0.5pair) Vdes,: Cfig Cdyn = (0.5)(1.2) (84.3)2 (-0.40)(1.0) = -1706 Pa = -1.71 kPa Cfig (internal) = 0.0 (1 .O) = 0.0 pi,, = 0 kPa Net pressure across roof surface = -1.7 1 -(O.O) = - 1.7 1 kPa

Cladding pressures (note: for permissible stress design - divide loads by 1 S )

Wall elements For elements (WA1) less than 9 m2 in area on walls at height z: Cfip (external) = +0.8 (1 .O) (1.25) = +1 .O (positive case) p,,, = (0.5 pa,> Vdes,: Cfig Cdy, = (0.5)(1.2) (88.7)2(MZ,ca,3/1.28)2 ( + l W . O ) = +4721(MZ,cao/l .28)2 Pa = +4.72(,Z,cat3/1 .28)2 kPa Cfig (internal) = -0.2 (1 .O) = -0.2 pint = (0.5pair) Vdes,: Cfig Cdyn = (0.5)(1.2) (88.7)2 (-0.2)(1.0) = -944 Pa = -0.94 kPa Net pressure across element = 4.72(Mz,cat3/1 .28)2 + 0.94 kPa

For elements (SA5) less than 9 m2 in area on walls within a distance of 3 m from corners: Cfig (external) = -0.65 (1 .O) (3.0) = -1.95 (negative case) p,,, = (0.5pa,,> Vdes,: Cfig Cdyn = (0.5)(1.2) (88.7)2 (-1.95)(1.0) = -9205 Pa= -9.21 kPa pint = 0 kPa Net pressure across element = -9.21-(0.0) = -9.21 kPa

For elements (SA4) less than 36 m2 in area on walls within a distance of 6 m from comers: Cfig (external) = -0.65 (1 .O) (2.0) = -1.3 (negative case) p,,, = (OSp,,,) Vdes,: Cfig Cdyn = (0.5)(1.2) (88.7)2 (-1.3)(1.0) = -6137 Pa = -6.14 kPa pi,, = 0 kPa Net pressure across element = -6.14-(0.0) = -6.14 kPa

For elements (SA3) less than 9 m2 in area on walls greater than a distance of 6 m ikom my comer: Cfig (external) = -0.65 (1 .O) (1 S) = -0.975 (negative case) p,,, = (0.5 pair) Vdes,: Cfig Cdyn = (0.5)(1.2) (88.7)2 (-0.975)(1.0) = -4603 Pa = -4.60 kPa pint = 0 kPa Net pressure across element = -4.60-(0.0) = -4.60 kPa

For elements greater than 9 m2 in area on walls greater than a distance of 6 m from any corner: Cfig (external) = -0.65 (1 .O) (1 .O) = -0.65 (negative case) p,,, = (0.5 pair) Vdes,; Cfig Cdyn = (0.5)(1.2) (88.7)2 (-0.65)(1.0) = -3068 Pa = -3.07 kPa pint = 0 kPa Net pressure across element = -3.07-(0.0) = -3.07 H a

Roof elements For elements (RA2) less than 9 m2 in area on roof, less than a distance of 3 m from edges: Cfig (external) = -1.3 (1.0) (2.0)(0.8) = -1.6 (negative case; K, of 0.8 included; note limit of -2.0 on K,.Cp,)

A Guide to ASINZS 11 70.2: 2002


p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,t Cfig Cdyn = (0.5)(1.2) (88.7)2 (-1.6)(1.0) = -7553 Pa= -7.55 kPa pint = 0 kPa Net pressure across element = -7,55-(0.0) = -7.55 kPa

For elements (RA1) less than 36 mz in area on roof, less than a distance of 6 m from edges: Cfig (external) = -1.3 (1.0) (1.5)(0.8) = -1.56 (Ka 10 m) p,,, = (0.5 psi$ Vdes,; Cfig Cdy, = (0.5)(1.2) (88.7)2 (-1.56)(1.0) = -7364 Pa = -7.36 kPa pi,, = 0 kPa Net pressure across element = -7.36-(0.0) = -7.36 kPa

Other roof elements, less than a distance of 24 m from edges: Cfig (external) = -1.3 (1.0) (l.O)(l.O) = -1.3 (Ka 4 . 0 can be used for A> 10 m2; no reduction due to parapet in this case) p,,, = ( 0 . 5 ~ ~ ~ ) Vdes,i Cfig CdY, = (0.5)(1.2) (88.7) (-1.3)(1.0) = -6137 Pa = -6.14 kPa pint = 0 kPa Net pressure across element = -6.14-(0.0) = -6.14 kPa

5.5 Example 5 - A tall building 183 m high in a city centre

Design wind loads are required for a tall office building in the Brisbane metropolitan area. The relevant information is as follows :

Location : CBD Brisbane (Region B) Terrain : Suburban terrain for all directions. Topography : ground slope less than 1 in 20 for greater than 5 kilometres in all directions. Dimensions : average roof height : 183 metres

Building orientation : major axis is East-West Reinforced concrete construction. Curtain wall faqade on all four faces Sway frequencies, na = nc = 0.2 Hertz. Mode shapes are linear (k = 1 .O). Average building density: 160 Kg/m3.

Horizontal dimensions: 46 metres X 30 metres (rectangular cross-section).

183 m

3P57& 46 m This building (known as the CAARC building) was used as an example in the Commentary to AS1170.2-1989 [4]. Its use in the present Design Guide allows for a comparison of the two Standards for this building.

Regional wind speed According to the Building Code of Australia (BCA), the structure should be treated as Level 3. Hence take average recurrence interval, R, for loading and overall structural response equal to 1000 years.

From Table 3.1 in AS/NZS 1 170.2, V,, = 60 d s (Region B)

For calculation of accelerations, use a 5-year return period, then V, = 28 d s

Figure 5.12 183-m tall building

Wind direction multiplier For Region B, M, = 0.95 for overturning forces and major structural system for all directions (Section 3.3.2). For cladding design, M, =1.0 .

Terrain-height multiplier z=h=183 m, For Terrain Category 3, Mz,cat = M,83,ca0 = 1.23 (Table 4.I(A) by interpolation)

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


Site wind speed Site wind speed for all directions for overall loads and main structural design,

Vsitp = 60(0.95)( 1.23)( 1 .O)( 1 .O) = 70.1 m/s (Equation 2.2)

For acceleration calculations (serviceability),

VsitB = 28( 1 .O)( 1.23)( 1 .O)( 1 .O) = 34.4 m/s (Equation 2.2)

For cladding design, VsitB = 60( 1 .O)( 1.23)(1.0)( 1 .O) = 73.8 m/s (Equation 2.2)

Design wind speeds For all wind directions, the design wind speeds,

- Vdes,B - Vsitp = 70.1 m/s (for overall loads and main structure design)

= 34.4 d s (for accelerations)

= 73.8 m/s (for cladding)


External pressures

Windward walls : +0.8 for varying z (Table 5.2(A))

Leeward walls (normal to 46m wall): -0.5 (Table 5.2(B)) Leeward walls (normal to 30m walls): -0.39 (Table 5.2(B) by interpolation)

Side walls: -0.65 (Table 5.2(C))

Roof: -1.3 (Table 5.3(A))

Area reduction factors (Section 5.4.2) For elements greater than 100 mz in area on roof or side walls, K, = 0.8 For small elements < 10m2 in area (e.g. glazed curtain wall elements) K, = 1 .O

Local pressure factors (Table 5.6)

a=minimumof0.2 x 3 0 m = 6 m , o r 1 8 3 m . a=- limiting tributary areas for local pressure factors : 0.25a2 = bz

a2 = am2

AGuide to ASINZS 1170.2: 2002

wind Actions Chapter 5

Internal pressures (Section 5.3) The building can be considered to be effectively sealed. In this case, Cp,i = -0.2 or 0.0 (Table 5.1(A))

Action combination factor

K,= 1.0

Dynamic response factor Cdyn to be obtained from Section 6.2.2 for along-wind response Cfig. Cdyn to be obtained as a product from Section 6.3.2 for cross-wind response

Calculation of along-wind Cdyn Turbulence intensity at z = h, I,, = 0.143 (Table 6.1 by interpolation) Background factor, I (Equation 6.2(2))

B, = J0.26(h -s)' +0.46b,,,2 1 + _ _ _ _ . ~ _ ~ - ~

L h

Forb = 46 m, s = 0 (for base bending moment),

= 0.641 B, = \j[O.26(183iZ + 0.46(46)2]

I76

1

I +

Forb = 30 m, s = 0 (for base bending moment),

= 0.648 B, = \j[ 0.26(1 83)2 + 0 ~ 6 ( 3 O ) ~ ] 1

I +

Hs= 1.0 176

gR = ~ ~ = ~ 2 1 ~ g , ( 6 0 0 ( 0 . 2 0 ) ) =3.OY

I Size reduction factor, S =

3..5nah(I 4- g v I h ) ] [ I + 4nabOh(1+gvihl] [ I + vdes,8 vdes ,8

Forb = 46 m, I

3. j(0.2) (1 83)( 1 + (3.7)( 0.143)) I + 4(0.2)( 46)( I + (3.7)(0.143)) 1 " [ I , 70.1 li 70.1 =0.146 1

[3.794][ 1.8031 - -

Wind Actions 6o 1 Chapter 5

Forb = 30 m, 1

1 3.5(0.2)(18

Wind Design

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