AS1170.2 Examples

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

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

Building orientation : major axis is East-West

Reinforced concrete construction. Curtain wall façade on all four faces

Sway frequencies, na = nc = 0.2 Hertz. Mode shapes are linear (k = 1.0).

Average building density: 160 Kg/m3.

Figure 5.13 183-m tall building

This building (known as the CAARC building), or variations of it, have been used as

benchmarks for wind-tunnel testing.

183 m

30 m 46 m

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/NZS1170.2, V1000 = 60 m/s (Region B)

For calculation of accelerations, use a 1-year return period, then V1 = 26 m/s

Wind direction multiplier

For Region B, Md = 0.95 for overturning forces and major structural system for all

directions (Section 3.3.2).

For cladding design, Md =1.0 .

Terrain-height multiplier

z=h=183 m, For Terrain Category 3, Mz,cat = M183,cat3 = 1.23 (Table 4.1 by interpolation)

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 all directions for overall loads and main structural design,

Vsit, = 60(0.95)(1.23)(1.0)(1.0) = 70.1 m/s (Equation 2.2)

For acceleration calculations (serviceability),

Vsit, = 26(1.0)(1.23)(1.0)(1.0) = 32.0 m/s (Equation 2.2)

For cladding design, Vsit, = 60(1.0)(1.23)(1.0)(1.0) = 73.8 m/s (Equation 2.2)

Design wind speeds

For all wind directions, the design wind speeds,

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

= 32.0 m/s (for accelerations)

= 73.8 m/s (for cladding)

Aerodynamic shape factor

External pressure coefficients

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 m2 in area on roof or side walls, Ka = 0.8

For small elements < 10m2

in area (e.g. glazed curtain wall elements) Ka = 1.0

Action combination factors (Section 5.4.3)

Kc,e= 0.9 for two effective surfaces – e.g. lateral pressure on windward and leeward walls

– Case (g) in Table 5.5.

Local pressure factors (Table 5.6)

a = minimum of 0.2 30 m = 6 m, or 183 m. a = 6.0 m

limiting tributary areas for local pressure factors : 0.25a2 = 9 m

2

a2 = 36 m

2

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

Kc = 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, Ih = 0.143 (Table 6.1 by interpolation)

Background factor,

h

sh

s

L

bshB

22 46.0)(26.01

1

(Equation 6.2(2))

mh

Lh 17610

18385

1085

25.025.0

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

641.0

176

)46(46.0)183(26.01

122

sB

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

648.0

176

)30(46.0)183(26.01

122

sB

Hs = 1.0

3.28=)(600(0.20)(2log+1.2=)(600n(2log+1.2=g eaeR

Size reduction factor,

,

0

,

)1(41

)1(5.31

1

des

hvha

des

hva

V

Igbn

V

IghnS

For b = 46 m,

For b = 30 m,

Reduced frequency,

,

1

des

hvha

V

IgLnN

6/528.701 N

NEt

(Equation 6.2(4))

( )

0.107=)70.8(0.746+1

π(0.746)5/62

=

(ratio of structural damping to critical) : take as 0.03

(maximum value for reinforced concrete buildings –see Notes to Clause 6.2.2)

hv

tRssvh

dynIg

SEgHBgI

C21

21

22

(Equation 6.2(1))

For b = 46 m,

( )

[ ]1.023=

1.972

5.794+7.4100.286+1

43)2(3.4)(0.1+1

0.03

107)(0.151)(0.)(1.0)(3.28+(0.641)(3.4)2(0.143)+1

C

2

2

dyn

=

=

For b = 30 m,

( )

[ ]1.056=

1.972

6.830+7.4910.286+1=

43)2(3.4)(0.1+1

0.03

107)(0.178)(0.)(1.0)(3.28+(0.648)(3.4)2(0.143)+1

=C

2

2

dyn

Calculation of base moment

Calculations were carried out by spreadsheet. Summaries of the results are given in the

following tables.

Wind normal to 46 m wall:

Cfig = Cp,e. Ka. Kc,e. K = +0.8 (1.0) (0.9) (1.0) = 0.72 for windward wall

Cfig = Cp,e. Ka. Kc,e. K = -0.5 (1.0) (0.9) (1.0) = -0.45 for leeward wall

Height

of sector (m)

Mz,cat 3 windward qz.Cfig

(kPa)

leeward qh.Cfig

(kPa)

windward qz.Cfig.Cdyn.A

(kN)

leeward qh.Cfig.Cdyn.A

(kN)

b.b.moment contribution

(MN.m)

torsion contribution

(MN.m)

171.5 1.223 2.099 -1.327 2272 -1436 636 34.1

150 1.210 2.055 -1.327 1934 -1249 477 29.3

130 1.190 1.988 -1.327 1871 -1249 406 28.7

110 1.170 1.921 -1.327 1808 -1249 336 28.1

90 1.144 1.837 -1.327 1729 -1249 268 27.4

70 1.110 1.729 -1.327 1628 -1249 201 26.5

50 1.070 1.607 -1.327 1512 -1249 138 25.4

30 1.000 1.404 -1.327 1321 -1249 77 23.6

10 0.830 0.967 -1.327 910 -1249 22 19.9

Note - qz denotes : 0.6 Vdes,2

Total along-wind base bending moment obtained by summing contributions from second

last column

= 2560 MN.m

Contributions to the torsional moment are obtained by applying an eccentricity of 9.2 m

(i.e. 0.2b as required by Section 2.5.4) to the sectional forces.

Total base torsion obtained by summing contributions from last column = 243 MN.m

Wind normal to 30 m wall:

Cfig = Cp,e. Ka. Kc,e. K = +0.8 (1.0) (0.9) (1.0) = 0.72 for windward wall

Cfig = Cp,e. Ka. Kc,e. K = -0.39 (1.0) (0.9) (1.0) = -0.35 for leeward wall

Height

of sector (m)

Mz,cat 3 windward qz.Cfig

(kPa)

leeward qh.Cfig

(kPa)

windward qz.Cfig.Cdyn.A

(kN)

leeward qh.Cfig.Cdyn.A

(kN)

b.b.moment contribution

(MN.m)

torsion contribution

(MN.m)

171.5 1.223 2.099 -1.032 1530 -752 391 13.7

150 1.210 2.055 -1.032 1302 -655 294 11.7

130 1.190 1.988 -1.032 1259 -655 249 11.5

110 1.170 1.922 -1.032 1217 -655 206 11.2

90 1.144 1.837 -1.032 1164 -655 164 10.9

70 1.110 1.729 -1.032 1096 -655 123 10.5

50 1.070 1.607 -1.032 1018 -655 84 10.0

30 1.000 1.404 -1.032 889 -655 46 9.3

10 0.830 0.967 -1.032 613 -655 13 7.6

Total along-wind base bending moment obtained by summing contributions from last

column

= 1570 MN.m

Contributions to the torsional moment are obtained by applying an eccentricity of 6.0 m

(i.e. 0.2b as required by Section 2.5.4) to the sectional forces.

Total base torsion obtained by summing contributions from last column

= 96 MN.m

Cross-wind response

Turbulence intensity at z= 2h/3 (122 m) = 0.159

Wind normal to 46 m face

For wind normal to 46 m face (b=46 m), reduced velocity,

( ) 5.13=(3.4)0.143+1(0.2).46.

70.1=

)Ig+b(1n

V=V

hvc

θdes,

n

Building dimensions are 6 : 1.5 : 1

For 6 : 1 : 1 and I2h/3 = 0.12 from Equation 6.3(7),

-2.510=2.76-0.603V-0.201V+0.0165V-0.000406V=Clog n

2

n

3

n

4

nfs10


-2.075=2.36-0.384V-0.141V+0.0125V-0.000334V=Clog n

2

n

3

n

4

nfs10

By interpolation for 6 : 1 : 1 and I2h/3 = 0.159, log10 Cfs = -2.298


-2.996=0.559

1.675-=

0.000123V+0.02V-1

0.000394V-0.0683V+-3.2=Cog 4

n

2

n

4

n

2

n

fs10l


-2.821=0.560

-1.580=

0.000124V+0.02V-1

0.00037V-0.0637V+-3.0=Clog 4

n

2

n

4

n

2

n

fs10


By interpolation for 6 : 1.5 : 1 and I2h/3 = 0.159, log10 Cfs = -2.605 hence, Cfs = 0.00248

fs

k

hv

mRdynfig

C

h

z

Ig

K

d

bgCC

21

5.1 (Equation 6.3(2))

( ) 0.03

π(0.00248)

183

z

3)(3.4)(0.14+1

1

30

461.5(3.28)

2= = 0.00951 z

).(.5.0)(2

, dynfigdesaireq CCdVzw (Equation 6.3(1))

( ) 51z).30.(0.00970.10.6= 2 = 841 z N/m = 0.841 z (kN/m)

Note : this is an inertial force distribution, proportional to the mode shape

Equivalent cross-wind load per unit height is evaluated in the following table.

Height of sector, z

(m)

Force/unit height (kN/m)

Sector height

(m)

Sector force (kN)

moment contribution

(MN.m) 171.5 144.2 23 3317 568.9

150 126.2 20 2523 378.5

130 109.3 20 2187 284.3

110 92.5 20 1850 203.5

90 75.7 20 1514 136.2

70 58.9 20 1177 82.4

50 42.1 20 841 42.1

30 25.2 20 505 15.1

10 8.4 20 168 1.7

Total cross-wind base bending moment obtained by summing contributions from last

column

= 1710 MN.m

Alternatively, using Equation 6.3(3)

fs

m

hv

desair

Rc

CK

kh

Ig

VbgM

2

3

1

5.05.0 2

2

2

,

( )( )

0.03

π(0.00248)(1)1183

3.4(0.143)+1

0.6(70.1)46)0.5(3.28)(= 2

2

2

N.m

= 1.719 109 N.m = 1720 MN.m



( ) 7.86(3.4)0.143+1(0.2).30.

70.1=

)Ig+b(1n

V=V

hvc

θdes,

n =

Building dimensions are 6 : 1 : 1.5


-2.155=4.093-0.396V+0.0226V-0.000457V=Cog n

2

n

3

nfs10l


-1.999=3.82-0.363V+0.0197V-0.00038V=Clog n

2

n

3

nfs10



-1.544=2.76-0.603V-0.201V+0.0165V-0.000406V=Clog n

2

n

3

n

4

nfs10


-1.462=2.36-0.384V-0.141V+0.0125V-0.000334V=Clog n

2

n

3

n

4

nfs10


By interpolation for 6 : 1 : 1.5 and I2h/3 = 0.159, log10 Cfs = -1.792 hence, Cfs = 0.0162

fs

k

hv

mRdynfig

C

h

z

Ig

K

d

bgCC

21

5.1 (Equation 6.3(2))

( ) 0.03

π(0.0162)

183

z

3)(3.4)(0.14+1

1

46

301.5(3.28)= 2 = 0.01034 z

).(.5.0)(2

, dynfigdesaireq CCdVzw (Equation 6.3(1))

( ) (0.01034z)×46×70.10.62

= = 1402 z N/m = 1.40 z (kN/m)

Equivalent cross-wind load per unit height is evaluated in the following table

Height of sector , z

(m)

Force/unit height (kN/m)

Sector height

(m)

Sector force (kN)

Moment contribution

(MN.m) 171.5 240.1 23 5522 947.1

150 210.0 20 4200 630.0

130 182.0 20 3640 473.2

110 154.0 20 3080 338.8

90 126.0 20 2520 226.8

70 98.0 20 1960 137.2

50 70.0 20 1400 70.0

30 42.0 20 840 25.2

10 14.0 20 280 2.8

Total cross-wind base bending moment is obtained by summing contributions from last

column

= 2850 MN.m

Alternatively, using Equation 6.3(3)

fs

m

hv

desair

Rc

CK

kh

Ig

VbgM

2

3

1

5.05.0 2

2

2

,

( )( )

0.03

π(0.0162)(1)1183

3.4(0.143)+1

0.6(70.1)30)0.5(3.28)( 2

2

2

= N.m

= 2.86 109 N.m = 2860 MN.m

Calculation of accelerations

To calculate accelerations for assessment of serviceability of the building, Appendix G is

used.

Mass per unit height, mo = 160 46 30 Kg/m = 2.21105 Kg/m

h1.3/mo = 183

1.3/(2.21105) = 0.00395. Since this exceeds 0.0016, cross-wind

accelerations may be excessive (Equation G1) and should be checked.

Along-wind acceleration calculation

Size reduction factor,

,

0

,

)1(41

)1(5.31

1

des

hvha

des

hva

V

Igbn

V

IghnS

For b = 46 m,

For b = 30 m,

Reduced frequency,

,

1

des

hvha

V

IgLnN

6/528.701 N

NEt

(Equation 6.2(4))

( )

0647.0=5/62)70.8(1.635+1

π(1.635)=

(ratio of structural damping to critical) : take as 0.01

(maximum value for resonant response for reinforced concrete buildings –see Notes to

Section 6.2.2)

Calculate dynamic response factor - for resonant response only:

hv

tRsh

resdynIg

SEgHI

C21

2

2

,

(derived from Equation 6.2(1))

For b = 46 m,

( )

[ ]279.0=

1.972

3.6980.286=

43)2(3.4)(0.1+1

0.01

.0647)(0.0531)(0)(1.0)(3.282(0.143)

=C

2

resdyn,

For b = 30 m,

( )

[ ]316.0=

1.972

4.7360.286=

43)2(3.4)(0.1+1

0.01

.0647)(0.0680)(0)(1.0)(3.282(0.143)

=C

2

resdyn,

Calculation of resonant base moments:

These are most easily obtained from previous calculations of peak base bending

moments, by adjusting for changes in Vdes,θ and Cdyn

For b = 46 m,

Resonant base moment =

= 145 MN.m

For b = 30 m,

Resonant base moment =

= 98 MN.m

Peak along-wind accelerations:

For b = 46 m,

2max

3

hmx

o

resonant peak base moment = 6

25 10×145×183×10×2.21

3 m/s

2

= 0.0588 m/s2 =

9.8

1000×0.0588 mg = 6.0 mg

For b = 30 m,

2max

3

hmx

o

resonant peak base moment = 6

25 10×98×183×10×2.21

3 m/s

2

= 0.0397 m/s2 =

9.8

1000×0.0397 mg = 4.1 mg

These accelerations are well within accepted comfort criteria, e.g. for office buildings

ISO 10137:2007 allows 0.13 m/s2 for office buildings at the frequency of 0.2 Hertz.

Cross-wind acceleration calculation



( )34.2

143.0)4.3(1.46).2.0(

0.32

)Ig1(bn

VV

hvc

,des

n =+

=+

=

Building dimensions are 6 : 1.5 : 1


174.3894.0

8382

V000123.0V02.01

V000394.0V0683.02.3Clog 4

n

2

n

4

n

2

n

fs10 -=.-

=+-

-+-=


978.2894.0

6622

V000124.0V02.01

V00037.0V0637.00.3Clog 4

n

2

n

4

n

2

n

fs10 -=.-

=+-

+=



270.376.2V603.0V201.0V0165.0V000406.0Clog n

2

n

3

n

4

nfs10 -=--+-=


637.236.2V384.0V141.0V0125.0V000334.0Clog n

2

n

3

n

4

nfs10 -=--+-=


By interpolation for 6 : 1.5 : 1 and I2h/3 = 0.159, log10 Cfs = -3.02.

Hence, Cfs = 0.000956

fs

m

hv

desair

o

RC

KIg

V

m

bgy

2

2

,

max1

5.05.1 (Equation G3(1))

( ) 01.0

)000956.0()1(

)143.0)(4.3(1

)0.32(6.0

1021.2

)28.3)(46(5.12

2

5

+×= m/s

2

= 0.156 m/s2 = )

8.9

1000.(156.0 mg = 15.9 mg



( )59.3

143.0)4.3(1.30)2.0(

0.32

)Ig1(bn

VV

hvc

,des

n =+

=+

=

Building dimensions are 6 :1: 1.5


941.2093.4V396.0V0226.0V000457.0Clog n

2

n

3

nfs10 -=-+-=


753.282.3V363.0V0197.0V00038.0Clog n

2

n

3

nfs10 -=-+-=



030376.2V603.0V201.0V0165.0V000406.0Clog n

2

n

3

n

4

nfs10 .-=--+-=


444.236.2V384.0V141.0V0125.0V000334.0Clog n

2

n

3

n

4

nfs10 -=--+-=


By interpolation for 6 : 1.5 : 1 and I2h/3 = 0.159, log10 Cfs = -2.797 hence, Cfs = 0.00160

fs

m

hv

desair

o

RC

KIg

V

m

bgy

2

2

,

max1

5.05.1 (Equation G3(1))

( ) 01.0

)00160.0()1(

)143.0)(4.3(1

)0.32(6.0

1021.2

)28.3)(30(5.12

2

5

+×= m/s

2

= 0.132 m/s2 = )

8.9

1000.(132.0 mg = 13.4 mg

The peak cross-wind accelerations are on the limit of acceptability for office buildings

according to ISO 10137, at the natural frequency of 0.2 Hertz, some remedial action may

be required (wind-tunnel tests may reduce the predicted accelerations).

Cladding pressures

(note: these are ultimate limit states design loads)

For elements less than 9 m2

in area on walls at height z:

Cfig (external) = Cp,e. Ka. Kc,e. K = +0.8 (1.0) (0.9) (1.5) = +1.08 (positive case WA1)

(Note: Kc,e = 0.9 for two effective surfaces)

pext, z = (0.5 air) Vdes,2 Cfig Cdyn = (0.5)(1.2) (73.8)

2(Mz,cat3/1.23)

2 (+1.08)(1.0)

= +3529(Mz,cat3/1.23)2 Pa = +3.53(Mz,cat3/1.23)

2 kPa

Cfig (internal) = -0.2 (1.0) = -0.2

pint = (0.5 air) Vdes,2 Cfig Cdyn = (0.5)(1.2) (73.8)

2 (-0.2)(1.0) = -654 Pa = -0.65 kPa

Net pressure across element = 3.53(Mz,cat3/1.23)2

+ 0.65 kPa


in area on walls within a distance of 3 m from corners:

Cfig (external) = Cp,e. Ka. Kc,e. K = -0.65 (1.0) (1.0) (3.0) = -1.95 (negative case)

pext = (0.5 air) Vdes,2 Cfig Cdyn = (0.5)(1.2) (73.8)

2 (-1.95)(1.0) = -6372 Pa = -6.37 kPa

pint = 0 kPa

Net pressure across element = -6.37 - (0.0) = -6.37 kPa


in area on walls within a distance of 6 m from corners:



2 (-1.3)(1.0) = -4248 Pa = -4.25 kPa

pint = 0 kPa



in area on walls greater than a distance of 6 m from corners:

Cfig (external) = Cp,e. Ka. Kc,e. K = -0.65 (1.0)(1.0) (1.5) = -0.975 (negative case)


2 (-0.975)(1.0) = -3186 Pa = -3.19 kPa

pint = 0 kPa


For elements greater than 9 m2

in area on walls greater than a distance of 6 m from

corners:



2 (-0.65)(1.0) = -2124 Pa = -2.12 kPa

pint = 0 kPa


Documents

AS1170.2 Examples