Example 5 – A 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 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 façade on all four faces Sway frequencies, na = nc = 0.2 Hertz. Hertz. Mode shapes are linear (k = 1.0). 3 Average building density: 160 Kg/m .
30 m
46 m
183 m
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.
Region Region al win d spee speed d
According to the Building Code of Australia (BCA), the structure should be treated as Level 3. Hence take average recurrence recurrence interval, interval, R, for loading and overall structural structural response equal to 1000 years. From Table 3.1 in 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 Win d dir dir ection multipli er
For Region B, Md = 0.95 for overturning forces and major structural system for all directions (Section (Section 3.3.2). 3.3.2). For cladding design, Md =1.0 . Terr Terr ain-height ain-height multiplier
z=h=183 m, For Terrain Category 3, Mz,cat = M183,cat3 = 1.23 (Table (Table 4.1 by 4.1 by interpolation) Shielding
There are no other buildings buildings of greater height in any direction. Take Ms, equal to 1.0 for all directions. Topography
Topographic Multiplier, Mt = Mh = 1.0 Site Sit e win d speed speed
Site wind speed for all directions for overall loads and main structural design, Vsit, = 60(0.95)(1.23)(1.0)(1.0) 60(0.95)(1.23)(1.0)(1.0) = 70.1 m/s ( Equation Equation 2.2) 2.2) For acceleration calculations (serviceability), Vsit, = 26(1.0)(1.23)(1.0)(1.0) = 32.0 m/s ( Equation Equation 2.2) 2.2) For cladding design,
Vsit, = 60(1.0)(1.23)(1.0)(1.0) 60(1.0)(1.23)(1.0)(1.0) = 73.8 m/s ( Equation Equation 2.2) 2.2)
D esign wi n d speeds speeds
For all wind directions, the design wind speeds, Vdes, = Vsit, = 70.1 m/s (for overall loads and main structure structure design) = 32.0 m/s (for accelerations) = 73.8 m/s (for cladding)
Aerodynamic shape factor factor
External pressure coefficients Windward walls : +0.8 for varying z (Table 5.2(A)) 5.2(A)) Leeward walls (normal to 46m wall): -0.5 (Table 5.2(B)) 5.2(B)) Leeward walls (normal to 30m walls): walls): -0.39 (Table 5.2(B) by 5.2(B) by interpolation) Side walls: -0.65 (Table (Table 5.2(C)) 5.2(C)) Roof : -1.3 (Table 5.3(A)) 5.3(A))
Area reduction factors (Section 5.4.2) 2
For elements greater than 100 m in area on roof or side walls, K a = 0.8 2 For small elements < 10m in area (e.g. glazed glazed curtain wall elements) K a = 1.0 Action combination factors (Section 5.4.3) K c,e e.g. lateral pressure on windward and leeward walls c,e= 0.9 for two effective surfaces – e.g. Case (g) in Table 5.5. 5.5. – Case Local pressure factors (Table 5.6) a = minimum of 0.2 30 m = 6 m, or 183 m. a = 6.0 m 2 2 limiting tributary areas for local pressure factors : 0.25a = 9 m 2 2 a = 36 m Internal pressures (Section 5.3) The building can be considered to be effectively sealed. In this case, C p,i = -0.2 or 0.0 (Table 5.1(A)) 5.1(A)) Action combination factor K c = 1.0 Dyn amic ami c response response f actor
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 C dyn
Turbulence intensity at z = h, Ih = 0.143 (Table 6.1 by 6.1 by interpolation) 1
Background factor, B s 1
h Lh 85 10
0.25
0.26(h s ) 0.46b sh 2
183 85 10
2
( Equation Equation 6.2(2)) 6.2(2))
Lh
0.25
176m
For b = 46 m, s = 0 (for base bending moment), 1 B s 0.641 2 2 0.26(183) 0.46( 46) 1 176
For b = 30 m, s = 0 (for base bending moment), 1 B s 0.648 2 2 0.26(183) 0.46(30) 1 176
Hs = 1.0 g R = 1.2 + (2log e (600na ) = 1.2 + (2log e (600(0.20) ) = 3.28 Size reduction factor, S
1
3.5na h(1 g v I h ) 4na b0h (1 g v I h ) 1 1 V V des , des ,
For b = 46 m,
For b = 30 m,
Reduced frequency, N
E t
=
N
1 70.8N
2 5/6
na Lh 1 g v I h V des,
( Equation Equation 6.2(4)) 6.2(4))
π(0.746)
(1+70.8(0.746 )2 )5/6
= 0.107
(ratio of structural structural damping to critical) critical) : take as 0.03 (maximum value for reinforced concrete buildings – see see Notes to Clause 6.2.2) 6.2.2) 1 2 I h C dyn
2 2 H s g R SE t g v B s 1 2 g v I h
( Equation Equation 6.2(1)) 6.2(1))
For b = 46 m, 1 + 2(0.143) C dyn = =
2
(3.4) (0.641) +
(1.0)(3.28 )2 (0.151)(0.107) 0.03
(1 + 2(3.4)(0.143))
1 + 0.286 [7.410 + 5.794 ] 1.972
= 1.023
For b = 30 m, 1 + 2(0.143) C dyn = =
2
(3.4) (0.648) +
(1.0)(3.28 )2 (0.178)(0.107)
(1 + 2(3.4)(0.143))
1 + 0.286 [7.491 + 6.830 ] 1.972
= 1.056
0.03
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 = C p,e. K a. K c,e c,e. K = +0.8 (1.0) (0.9) (1.0) = 0.72 for windward wall Cfig = C p,e. K a. K c,e c,e. K = -0.5 (1.0) (0.9) (1.0) = -0.45 for leeward wall Height of sector (m)
Mz,cat 3
171.5 150 130 110 90 70 50 30 10
1.223 1.210 1.190 1.170 1.144 1.110 1.070 1.000 0.830
b.b.moment contribution
torsion contribution
(kN)
(MN.m)
(MN.m)
-1436 -1249 -1249 -1249 -1249 -1249 -1249 -1249 -1249
636 477 406 336 268 201 138 77 22
34.1 29.3 28.7 28.1 27.4 26.5 25.4 23.6 19.9
windward qz.Cfig
leeward qh.Cfig
windward qz.Cfig.Cdyn.A
leeward qh.Cfig.Cdyn.A
(kPa)
(kPa)
(kN)
2.099 2.055 1.988 1.921 1.837 1.729 1.607 1.404 0.967
-1.327 -1.327 -1.327 -1.327 -1.327 -1.327 -1.327 -1.327 -1.327
2272 1934 1871 180 8 1729 1628 1512 1321 910
2
Note - qz denotes : 0.6 Vdes,
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 appl ying an eccentricity of 9.2 m (i.e. 0.2b as required by Section 2.5.4) 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 = C p,e. K a. K c,e c,e. K = +0.8 (1.0) (0.9) (1.0) = 0.72 for windward wall Cfig = C p,e. K a. K c,e c,e. K = -0.39 (1.0) (0.9) (1.0) = -0.35 for leeward wall Height of sector (m)
Mz,cat 3
171.5 150 130 110 90 70 50 30 10
1.223 1.210 1.190 1.170 1.144 1.110 1.070 1.000 0.830
b.b.moment contribution
torsion contribution
(kN)
(MN.m)
(MN.m)
-752 -655 -655 -655 -655 -655 -655 -655 -655
391 294 249 206 164 123 84 46 13
13.7 11.7 11.5 11.2 10.9 10.5 10.0 9.3 7.6
windward qz.Cfig
leeward qh.Cfig
windward qz.Cfig.Cdyn.A
leeward qh.Cfig.Cdyn.A
(kPa)
(kPa)
(kN)
2.099 2.055 1.988 1.922 1.837 1.729 1.607 1.404 0.967
-1.032 -1.032 -1.032 -1.032 -1.032 -1.032 -1.032 -1.032 -1.032
1530 1302 1259 1217 1164 1096 1018 889 613
Total along-wind base bending moment obtained ob tained by summing contributions from last column = 1570 MN.m Contributions to the torsional moment are obtained by appl ying an eccentricity of 6.0 m (i.e. 0.2b as required by Section 2.5.4) 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 Win d normal t o 46 m face
For wind normal to 46 m face (b=46 m), reduced velocity, V des,θ 70.1 V n = = = 5.13 ncb(1 + g v I h ) (0.2).46.(1 + (3.4)0.143) Building dimensions are 6 : 1.5 : 1 For 6 : 1 : 1 and I2h/3 = 0.12 from Equation from Equation 6.3(7), 6.3(7), log log 10C fs = 0.000406V n 4 - 0.0165V n 3 + 0.201V n 2 - 0.603V n - 2.76 = -2.510 For 6 : 1 : 1 and I2h/3 = 0.20 from Equation from Equation 6.3(8), 6.3(8), 4 3 log log 10C fs = 0.000334V n - 0.0125V n + 0.141V n 2 - 0.384V n - 2.36 = -2.075 By interpolation for 6 : 1 : 1 and I2h/3 = 0.159, log10 Cfs = -2.298
For 6 : 2 : 1 and I2h/3 = 0.12 from Equation from Equation 6.3(9), 6.3(9), 2 4 -3.2 + 0.0683V n - 0.000394V n - 1.675 log 10C fs = = = -2.996 1 - 0.02V n 2 + 0.000123V n 4 0.559 For 6 : 2 : 1 and I2h/3 = 0.20 from Equation from Equation 6.3(10), 6.3(10), 2 4 -3.0 + 0.0637V n - 0.00037V n -1.580 log 10C fs = = = -2.821 1 - 0.02V n 2 + 0.000124V n 4 0.560 By interpolation for 6 : 2 : 1 and I2h/3 = 0.159, log10 Cfs = -2.911
By interpolation for 6 : 1.5 : 1 and I2h/3 = 0.159, log10 Cfs = -2.605
C fig C dyn
b K m z 1.5 g R 2 d 1 g v I h h
= 1.5(3.28)
46
1
30
(1 + (3.4)(0.143))
2
k
C fs
( Equation Equation 6.3(2)) 6.3(2))
z
π(0.00248)
183
0.03
weq ( z ) 0.5 air V des, .d .(C fig C dyn ) 2
hence, Cfs = 0.00248
= 0.00951 z
( Equation Equation 6.3(1)) 6.3(1))
= 0.6 (70.1) 2 .30.(0.009 51z) = 841 z N/m = 0.841 z (kN/m)
Note : this is an inertial force distribution, distribution, proportional to the mode shape Equivalent cross-wind load per unit height is evaluated in the following table. Height of sector, z (m) 171.5 150 130 110 90 70 50 30 10
Force/unit height (kN/m) 144.2 126.2 109.3 92.5 75.7 58.9 42.1 25.2 8.4
Sector height (m) 23 20 20 20 20 20 20 20 20
Sector force (kN) 3317 2523 2187 1850 1514 1177 841 505 168
moment contribution (MN.m) 568.9 378.5 284.3 203.5 136.2 82.4 42.1 15.1 1.7
Total cross-wind base bending moment obtained b y summing contributions from last column = 1710 MN.m
Alternatively, using Equation using Equation 6.3(3)
0.5 air V des, 2 2 3 C fs M c 0.5 g R b h K m 2 1 g v I h k 2 = 0.5(3.28)( 46)
0.6(70.1)2
1832 (1)(1)
(1 + 3.4(0.143))2
π(0.00248) 0.03
N.m
9
= 1.719 10 N.m = 1720 MN.m Win d normal t o 30 m face
For wind normal to 30 m face (b=30 m), reduced velocity, V des,θ 70.1 V n = = = 7.86 ncb(1 + g v I h ) (0.2).30.(1 + (3.4)0.143) Building dimensions are 6 : 1 : 1.5 For 6 : 1 : 2 and I2h/3 = 0.12 from Equation 6.3(11), log 10C fs = 0.000457V n 3 - 0.0226V n 2 + 0.396V n - 4.093 = -2.155 For 6 : 1 : 2 and I2h/3 = 0.20 from Equation 6.3(12), log log 10C fs = 0.00038V n 3 - 0.0197V n 2 + 0.363V n - 3.82 = -1.999 By interpolation for 6 : 1 : 2 and I2h/3 = 0.159, log10 Cfs = -2.079
For 6 : 1 : 1 and I2h/3 = 0.12 from Equation 6.3(7), log log 10C fs = 0.000406V n 4 - 0.0165V n 3 + 0.201V n 2 - 0.603V n - 2.76 = -1.544 For 6 : 1 : 1 and I2h/3 = 0.20 from Equation 6.3(8), log log 10C fs = 0.000334V n 4 - 0.0125V n 3 + 0.141V n 2 - 0.384V n - 2.36 = -1.462 By interpolation for 6 : 1 : 1 and I and I 2h/3 2h/3 = 0.159, log10 Cfs = -1.504 By interpolation for 6 : 1 : 1.5 and I and I 2h/3 2h/3 = 0.159, log10 C fs = -1.792
C fig C dyn
b K m z 1.5 g R 2 d 1 g v I h h
= 1.5(3.28)
30
1
46
(1 + (3.4)(0.143))
2
k
C fs
( Equation Equation 6.3(2)) 6.3(2))
z
π(0.0162)
183
0.03
= 0.01034 z
hence, C fs = 0.0162
weq ( z ) 0.5 air V des, .d .(C fig C dyn ) 2
( Equation Equation 6.3(1)) 6.3(1))
2
= 0.6 (70.1) × 46 ×(0.01034z) = 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) 171.5 150 130 110 90 70 50 30 10
Force/unit height (kN/m) 240.1 210.0 182.0 154.0 126.0 98.0 70.0 42.0 14.0
Sector height (m) 23 20 20 20 20 20 20 20 20
Sector force (kN) 5522 4200 3640 3080 2520 1960 1400 840 280
Moment contribution (MN.m) 947.1 630.0 473.2 338.8 226.8 137.2 70.0 25.2 2.8
Total cross-wind base bending moment is obtained by summing contributions from last column = 2850 MN.m Alternatively, using Equation using Equation 6.3(3)
0.5 air V des, 2 2 3 C fs M c 0.5 g R b h K m 2 1 g v I h k 2 = 0.5(3.28)( 30)
0.6(70.1)2
(1 + 3.4(0.143))
2
1832 (1)(1)
π(0.0162) 0.03
N.m
9
= 2.86 10 N.m = 2860 MN.m
Calculation of accelerations
To calculate accelerations for assessment of serviceability of the building, Appendix building, Appendix G is G is used. 5
Mass per unit height, mo = 160 46 30 Kg/m = 2.21 2.2110 Kg/m 1.3
1.3
5
h /mo = 183 /(2.21 (2.2110 ) = 0.00395. Since this exceeds 0.0016, cross-wind accelerations may be excessive ( Equation Equation G1) G1) and should be checked.
Al ong-win ong-win d acce acceleration leration calculati calculati on
Size reduction factor, S
1
3.5na h(1 g v I h ) 4na b0h (1 g v I h ) 1 1 V V des , des ,
For b = 46 m,
For b = 30 m,
Reduced frequency, N
E t
=
N
1 70.8N
2 5/6
na Lh 1 g v I h V des,
( Equation Equation 6.2(4)) 6.2(4))
π(1.635)
(1 +70.8(1.635 )2 )5/6
= 0.0647
(ratio of structural structural damping to critical) critical) : take as 0.01 (maximum value for resonant response for reinforced concrete buildings – see see Notes to Section 6.2.2) 6.2.2) Calculate dynamic response factor - for resonant response only:
C dyn,res
H s g R 2 SE t 2 I h 1 2 g v I h
(derived from Equation from Equation 6.2(1)) 6.2(1))
For b = 46 m, 2(0.143) C dyn,res = =
(1.0)(3.28 )2 (0.0531)(0.0647) 0.01
(1 + 2(3.4)(0.143))
0.286 [3.698 ] 1.972
= 0.279
For b = 30 m, 2(0.143) C dyn,res = =
0.286 [4.736 ] 1.972
(1.0)(3.28 )2 (0.0680)(0.0647) 0.01
(1 + 2(3.4)(0.143)) = 0.316
Calculation of resonant base moments: These are most easily obtained from previous calculations of p eak base bending moments, by adjusting for changes in V des,θ dyn des,θ and C dyn For b = 46 m, Resonant base moment = For b = 30 m, Resonant base moment =
= 145 MN.m = 98 MN.m
Peak along-wind accelerations: For b = 46 m, 3 3 2 6 xmax resonant peak base moment = 5 2 × 145× 10 m/s max 2 2.21× 10 × 183 mo h 2
= 0.0588 m/s = 0.0588 ×
1000 9.8
mg = 6.0 mg
For b = 30 m, 3 3 2 6 xmax resonant peak base moment = 5 2 × 98 × 10 m/s max 2 2.21× 10 × 183 mo h 2 = 0.0397 m/s = 0.0397 ×
1000
mg = 4.1 mg 9.8 These accelerations are well within accepted comfort criteria, e.g. for office buildings 2 ISO 10137:2007 allows 0.13 m/s for office buildings at the frequency of 0.2 Hertz.
Cross-win Cross-win d acce accelerati on calcul ation Win d normal t o 46 m face
For wind normal to 46 m face (b=46 m), reduced velocity, Vdes, 32.0 Vn = = = 2.34 ncb( 1 + gvIh ) ( 0.2 ) . 46.(1 +( 3.4 )0.143) Building dimensions are 6 : 1.5 : 1 For 6 : 2 : 1 and I2h/3 = 0.12 from Equation from Equation 6.3(9), 6.3(9), 2 -3.2 + 0.0683Vn - 0.000394Vn 4 -2.838 log10 Cfs = = = -3.174 1- 0.02Vn2 + 0.000123Vn4 0.894 For 6 : 2 : 1 and I2h/3 = 0.20 from Equation from Equation 6.3(10), 6.3(10), 2 3.0 + 0.0637Vn 0.00037Vn4 -2.662 log10 Cfs = = = -2.978 1- 0.02Vn2 + 0.000124Vn4 0.894 By interpolation for 6 : 2 : 1 and I2h/3 = 0.159, log10 Cfs = -3.078 For 6 : 1 : 1 and I2h/3 = 0.12 from Equation from Equation 6.3(7), 6.3(7), 4 3 log10 Cfs = 0.000406Vn - 0.0165Vn + 0.201Vn2 - 0.603Vn - 2.76 = -3.270 For 6 : 1 : 1 and I2h/3 = 0.20 from Equation from Equation 6.3(8), 6.3(8), log10 Cfs = 0.000334Vn4 - 0.0125Vn3 + 0.141Vn2 - 0.384Vn - 2.36 = -2.637 By interpolation for 6 : 1 : 1 and I2h/3 = 0.159, log10 Cfs = -2.961 By interpolation for 6 : 1.5 : 1 and I2h/3 = 0.159, log10 Cfs = -3.02. Hence, Cfs = 0.000956
max y
2 C fs 1.5bg R 0.5 air V des , K 2 m mo 1 g v I h
1.5( 46 ) ( 3.28 ) = 2.21× 105
( Equation Equation G3(1)) G3(1))
0.6( 32.0 )2 ( 0.000956) 2 m/s 2 ( 1) (1 +( 3.4 ) ( 0.143 )) 0.01
2 = 0.156 m/s = 0.156.(
1000 ) mg = 15.9 mg 9.8
Win d normal t o 30 m face
For wind normal to 30 m face (b=30 m), reduced velocity, Vdes, 32.0 Vn = = = 3.59 ncb( 1 + gvIh ) ( 0.2 )30.(1 +( 3.4 )0.143) Building dimensions are 6 :1: 1.5
For 6 : 1 : 2 and I2h/3 = 0.12 from Equation 6.3(11), log10 Cfs = 0.000457Vn3 - 0.0226Vn2 + 0.396Vn - 4.093 = -2.941 For 6 : 1 : 2 and I2h/3 = 0.20 from Equation 6.3(12), log10 Cfs = 0.00038Vn3 - 0.0197Vn2 + 0.363Vn - 3.82 = -2.753 By interpolation for 6 : 1 : 2 and I2h/3 = 0.159, log10 Cfs = -2.849 For 6 : 1 : 1 and I2h/3 = 0.12 from Equation 6.3(7), log10 Cfs = 0.000406Vn4 - 0.0165Vn3 + 0.201Vn2 - 0.603Vn - 2.76 = -3.030 For 6 : 1 : 1 and I2h/3 = 0.20 from Equation 6.3(8), log10 Cfs = 0.000334Vn4 - 0.0125Vn3 + 0.141Vn2 - 0.384Vn - 2.36 = -2.444 By interpolation for 6 : 1 : 1 and I2h/3 = 0.159, log10 Cfs = -2.744 By interpolation for 6 : 1.5 : 1 and I2h/3 = 0.159, log10 Cfs = -2.797
max y
2 C fs 1.5bg R 0.5 air V des , K 2 m mo 1 g v I h
1.5( 30 ) ( 3.28 ) = 2.21× 105
hence, Cfs = 0.00160
( Equation Equation G3(1)) G3(1))
0.6( 32.0 )2 ( 0.00160) 2 m/s 2 ( 1) (1 +( 3.4 ) ( 0.143 )) 0.01
2 = 0.132 m/s = 0.132.(
1000 ) mg = 13.4 mg 9.8
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) 2
For elements less than 9 m in area on walls at height z: z: Cfig (external) = C p,e. K a. K c,e c,e. K = +0.8 (1.0) (0.9) (1.5) = +1.08 (positive case WA1) (Note: K c,e c,e = 0.9 for two effective surfaces) 2 2 2 pext, z = (0.5 air ) Vdes, Cfig Cdyn = (0.5)(1.2) (73.8) (Mz,cat3/1.23) (+1.08)(1.0) 2 2 = +3529(Mz,cat3/1.23) Pa = +3.53(Mz,cat3/1.23) kPa Cfig (internal) = -0.2 (1.0) = -0.2 2 2 pint = (0.5 air ) Vdes, Cfig Cdyn = (0.5)(1.2) (73.8) (-0.2)(1.0) = -654 Pa = -0.65 kPa 2 Net pressure across element = 3.53(Mz,cat3/1.23) + 0.65 kPa 2
For elements less than 9 m in area on walls within a distance of 3 m from corners: Cfig (external) = C p,e. K a. K c,e (1.0) (3.0) = -1.95 (negative case) c,e. K = -0.65 (1.0) (1.0) 2 2 pext = (0.5 air ) Vdes, Cfig Cdyn = (0.5)(1.2) (73.8) (-1.95)(1.0) = -6372 Pa = -6.37 kPa pint = 0 kPa Net pressure across element = -6.37 - (0.0) = -6.37 kPa 2
For elements less than 36 m in area on walls within a distance of 6 m from corners: Cfig (external) = C p,e. K a. K c,e (1.0) (2.0) = -1.3 (negative case) c,e. K = -0.65 (1.0) (1.0) 2 2 pext = (0.5 air ) Vdes, Cfig Cdyn = (0.5)(1.2) (73.8) (-1.3)(1.0) = -4248 Pa = -4.25 kPa pint = 0 kPa Net pressure across element = -4.25 - (0.0) = -4.25 kPa 2
For elements less than 9 m in area on walls greater than a distance of 6 m from corners: Cfig (external) = C p,e. K a. K c,e -0.65 (1.0)(1.0) (1.5) = -0.975 (negative case) case) c,e. K = -0.65 2 2 pext = (0.5 air ) Vdes, Cfig Cdyn = (0.5)(1.2) (73.8) (-0.975)(1.0) = -3186 Pa = -3.19 kPa pint = 0 kPa Net pressure across element = -3.19 - (0.0) = -3.19 kPa 2
For elements greater than 9 m in area on walls greater than a distance of 6 m from corners: Cfig (external) = C p,e. K a. K c,e (1.0) (1.0) = -0.65 (negative case) c,e. K = -0.65 (1.0) (1.0) 2 2 pext = (0.5 air ) Vdes, Cfig Cdyn = (0.5)(1.2) (73.8) (-0.65)(1.0) = -2124 Pa = -2.12 kPa pint = 0 kPa Net pressure across element = -2.12 - (0.0) = -2.12 kPa
