References [PDF]

References 1. Ansys 12.1 User Manual, Ansys Inc, 2009 2. Australian and New Zealand standards: Structural design actions Part 2: wind actions; AS/NZS 1170.2:2002, Standards Australia. 3. Australian standards: Structural design actions Part 2: wind actions; AS 1170.2:1989, Standards Australia 4. Becker, S., Lienhart, H., Durst, F., Flow around the three dimension obstacles in boundary layer, Journal of Wind Engineering and Industrial Aerodynamics, Vol. 90, pp. 265-279, 2002 5. Blocken, B., Carmeliet, J., Stathopoulos, T., CFD evaluation of the wind speed conditions in passages between buildings — effect of wall-function roughness modifications on the atmospheric boundary layer flow. Journal of Wind Engineering and Industrial Aerodynamics, 2007(a). 6. Blocken, B., Stathopoulos, T., Carmeliet, J., CFD simulation of the atmospheric

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Environment. 41 (2), 2007(b), pg 238–252 7. British Standard: Eurocode 1: Actions on Structures – Part1- 4: General actions - wind actions; BS EN 1991-1-4:2005 , British Standard Institution, London 8. British Standard: Loading for building- Part 2: Code of Practice for wind loads; BS 6399- 2:1997, British Standard Institution, London 9. Building Code of Australia (BCA), Australian Building Codes Board, 2007 10. Caracoglia, L., Jones, P. J., Analysis of full-scale wind and pressure measurements on a low-rise building, Journal of Wind Engineering and Industrial Aero Dynamics, 97, 2009, pp 157 –173 11. Caracoglia, L., Sangree, R. H., Jones, P. J., Schafer, B. W., Interpretation of full-scale strain data from wind pressures on a low-rise structure, Journal of Wind Engineering and Industrial Aero Dynamics, 96, 2008, pp 2363 –2382 12. Charney, F. A., Wind drift serviceability limit state design for multi storey building, Journal of Wind Engineering and Industrial Aero Dynamics, 36, 1990, pp 203–212

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13. Chen, X., Kareem, A., Validity of Wind Load Distribution based on High Frequency Force Balance Measurements, Journal of Structural Engineering, June 2005, pp 984 – 987 14. Clarke, A.G, Swane, R.A, Schneider, L.M, Shaw, P.J.R, Technical assistance to Sri Lanka on Cyclone Resistant Construction, Vol 1, Part 1 -4, 1979 15. Cook, N. J., Designer guide to wind loading of building structures, Butterworth, 1985 16. Cook, N. J., Wind loading, A practical guide to BS 6399-2 Wind loads for buildings, Thomas Telford, 1999 17. Cooney R.C., King, A.B. , Serviceability criteria of buildings, 1988 18. COST Action 732, Quality assurance and improvement of micro-scale meteorological models, 2007 19. Cowan, I.R., Castro, I.P., and Robins, A.G., Numerical considerations for simulations of flow and dispersion around buildings, Journal of Wind Engineering and Industrial Aerodynamics, Vol. 67 & 68, pp. 535-545, 1997 20. CP 3 Chapter V: 1972, Code of Basic data for the design of buildings chapter V. Loading, Part 2. Wind Loads, British Standard Institution, London 21. Cvitan, L., Determining wind gusts using mean hourly wind speed, GEOFIZIKA, Vol 20, 2003, pp 63 – 73 22. Database

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www.desinventar.lk, last accessed on 03/08/2010 23. Dyrbye, C., Hansen, S., O., Wind Loads on Structures, John Wiley & Sons, 1999 24. Franke, J., Recommendations of the cost action c14 on the use of CFD in predicting pedestrian wind environment. In: The 4th International Symposium on Computational Wind Engineering, Japan Association

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Engineering, Japan, 2006,pp. 529–532 25. Fu, J. Y., Li, Q. S., Wu, J. R., Xiao, Y. Q., Song, L. L, Field measurements of boundary layer wind characteristics and wind-induced responses of super-tall buildings, Journal of Wind Engineering and Industrial Aero Dynamics, 96, 2008, pp 1332 –1358 26. Goliger, A. M., South African wind loading specifications: The Euro way? , Journal of Wind Engineering and Industrial Aero Dynamics, 95, 2007, pp 1053 – 1064 120

27. Gu, M., Zhou, X. Y., An approximation method for resonant response with coupling modes of structures under wind action, Journal of Wind Engineering and Industrial Aero Dynamics, 97, 2009, pp 573 – 580 28. Holmes, J. D., Effective static wind load distribution in wind enginnering, Journal of Wind Engineering and Industrial Aero Dynamics, 90, 2002, pp 91 – 109 29. Holmes, J. D., Wind loading of structures (Second Edition), Taylor and Francis, 2007 30. Ilgin, H. E., Gunel, M. H., The Role of Aerodynamic Modifications in the Form of Tall Buildings Against Wind Excitation, METU Journal of Faculty of Architecture, vol 2, 2007, pp 17 – 25 31. Irwin, A.W. Human Response to Dynamics Motion of Structures, The Structural Engineer, London, 1978 32. Irwin, P. A., Wind engineering challenges of the new generation of super-tall buildings, Journal of Wind Engineering and Industrial Aero Dynamics, 97, 2009, pp 328 –334 33. Kappos, A. J., Dynamic loading and design of structures, Spon press, London. 34. Karunaratne, S.A., High rise buildings in Colombo (Structural Engineer’s View), Proceedings of the international conference on “Advances in Continuum Mechanics, Materials Science, Nano science and Nano technology: Dedicated to Professor Munidasa P. Ranaweera”, University of Peradeniya, Sri Lanka, 2008. pp 291- 302 35. Kasperski, M., Specification of the design wind load—A critical review of code concepts, Journal of Wind Engineering and Industrial Aero Dynamics, 97, 2009, pp 335 –357 36. Kijewski, T., Kareem. A., Dynamic wind effects, A comparative study of provision in codes and standards with wind tunnel data, 2001 37. Kolouske, V., Pirner, M., Fischer, O., Naprstek, J., Wind effects on civil engineering structures, Elsevier, 1984 38. Kwok, K. C. S., Hitchcock P. A., Burton, D. B, Perception of vibration and occupant comfort in wind-excited tall buildings, Journal of Wind Engineering and Industrial Aero Dynamics, 97, 2009, pp 368 –380 39. Kwok, K.C.S., Campbell, S., Hitchcock, P.A., “Dynamic characteristic and wind-induced response of two high-rise residential buildings during typhoon”, 121

Journal of Wind Engineering and Industrial Aero Dynamics, 93, 2005, pp 461–485 40. Lu, L. T., Chiang, W. L., Tang, J. P., Liu, M. Y., Cheng, C. W., “Active control for a benchmark building under wind excitations”, Journal of Wind Engineering and Industrial Aero Dynamics, 91, 2003, pp 469–493 41. Mendis, P., Ngo, T., Hariots, N., Hira, A., Samali, B., Cheung, J., Wind Loading on Tall buildings, EJSE special issue; Loading on structures, 2007. pp 41-54 42. Naval research laboratory, Web site URL:www.nrlmry.navy.mil, Last accessed on 28/07/2010 43. Ngo, T. T., Letchford C. W., Experimental study of topographic effects on gust wind speed, Journal of Wind Engineering and Industrial Aero Dynamics, 97, 2009, pp 426 – 438 44. Pagnini, L., Reliability analysis of wind-excited structures, Journal of Wind Engineering and Industrial Aero Dynamics, 98, 2010, pp 1 –9 45. Pantelides, C. P., “Active control of wind excited buildings”, Journal of Wind Engineering and Industrial Aero Dynamics, 36, 1990, pp 189–202 46. Pham, L., Actions on structures: Regulations and Standards, EJSE special issue: Loading on Structures, 2007, pp 4 - 8 47. Premachandra, W.R.N.R., “Study of new wind loading code to be adopting on Sri Lanka”, M.Sc Thesis, Graduate school, Kasetsart University, 2008 48. Reddy J.N., An introduction to finite element method, 3 rd edition, McGrawHill, 2005 49. Report on the Calibration of Euro code for wind loading (BS EN 1991 - 4) and its UK National Annex against the current UK wind code (BS 6399: Part 2:1997) 50. Richards, P. J., Hoxey, R. P., Appropriate boundary condition for computational wind engineering models using k – ε turbulence models, Journal of wind engineering and industrial aerodynamics, 46 & 47, 1993, pg 145 -153 51. Sachs, P., Wind forces in Engineering, 2nd edition, Pergamon Press, 1978 52. Simiu, E., Sadek, F., Riley, M. A., Discussion of “Definition of Wind Profiles in ASCE 7” by Yin Zhou and Ahsan Kareem, Journal of Structural Engineering, November 2003, pp 1564 – 1565 122

53. Simiu, E., Scanlan, R. H., Wind effects on Structures, John Willey & Sons, 1978. 54. Tominagaa,Y., Mochida, A., Yoshiec, R., Kataokad, H., Nozue, T., Yoshikawa, M., Shirasawa, T., AIJ guidelines for practical applications of CFD to pedestrian wind environment around buildings , Journal of wind engineering and industrial aerodynamics, 96, 2008, pg 1749 – 1761 55. Vickery, P.J., Masters, F. J., Powell, M. D., Wadhera, D., Hurricane hazard modeling: The past, present, and future, Journal of Wind Engineering and Industrial Aero Dynamics, 97, 2009, pp 392 –405 56. Walshaw, D., Anderson, C. W., A model for extreme wind gusts, Journal of Applied Statistics, 49, Part 4, 2000, pp 499 – 508 57. Weggel, J., R., Maximum Daily Wind Gust Related to Mean Daily Wind Speed, Journal of Structural Engineering, April 1999, pp 465 – 468 58. Wijeratne, M.D., Jayasinghe, M.T.R., “Wind loads for high-rise buildings constructed in Sri Lanka”, Transactions Part 2- Institution of Engineers, Sri Lanka, 1998, pp 58-69. 59. Yang, W., Quan, Y., Jin, X., Tamura, Y., Gu, M., Influences of equilibrium atmosphere boundary layer and turbulence parameter on wind loads of lowrise buildings, Journal of wind engineering and industrial aerodynamics, 96, 2008, pg 2080 – 2092 60. Zhang, L. L., Li, J., Peng, Y., Dynamic response and reliability analysis of tall buildings subject to wind loading, Journal of Wind Engineering and Industrial Aero Dynamics, 96, 2008, pp 25–40 61. Zhou, Y., Kareem, A., Closure to “Definition of Wind Profiles in ASCE 7” by Yin Zhou and Ahsan Kareem, Journal of Structural Engineering, November 2003, pp 1565 – 1566

123

APPENDIX A

A.1.0 Design of 48 m Height Building A.1.1 Determination of storey height

It is assumed that beam slab construction is used. Head room = 3.0m Slab thickness = 200mm Beam depth = 600mm Beam width = 400mm or with of the column whichever is less Space for services and ceiling = 300mm Below the beam soffit level. Total floor height = 3.0+0.6m+0.3m=3.9m Use a floor height = 3.1 + 0.6 + 0.3 = 4.0m Use a floor height of 4.0m The building layout in plan

A.1.2 Determination of number person per floor Consider 25% area allocated for the services. Therefore the rentable area is about 75% of the total area of the building. The area allocated for each person is 10 m2. The population handled in 5 minutes is 14% Average interval between two lifts is 35 seconds The car speed is equal 4ms-1. The population per floor = 60  30 

0.75  135 persons per floor 10

It is decided to use hard zoning arrangement where the lifts starting at ground floor and continue to the upper most floor.

125   

A.1.3 Determination of numbers of lifts Total population in 1st floor to 12th floor = 135  12  1620 persons The lift system is designed for a five minute peak minute capacity. The peak capacity is assumed as 10% of the total population in these floors. Population handled in five minute period = 0.10  1620  162 persons Capacity of the lift cars is selected as 20 passengers with 2ms-1 speed, which will give an average interval of about 30 seconds. The round trip time for 3.6m floor to floor height) = 115 S. Time required to travel additional distance

= (2 x 0.4 x 12) / 2 = 4.8S

Total round trip time

= 115 + 4.8S=119.8 S

Thus, the number of lift required is 119.8/30 = 4. four lifts cars can be used to serve the 12 floors. 2550mm 150mm

2400mm

150mm

 24 Passenger Lift 

Figure A.1: Approximate size of the lift shafts  

A 1.4 Stair case

The stair case selected has a width of 1.5m, a rise of 0.15m, a thread of 0.30m. Thus, the number of steps required is 22, giving a flight length of 3.3m and a landing width of 1.5 m. Total internal space required for staircase is 3.0 x 4.8m. 126   

A 1.5 Lateral load resisting system Lateral load stiffness is provided by using shear walls in the service core of the building. the length of the shear wall parallel to the 60 m wall is 48 m and parallel to the 30 m wall is 6.0m. The main feature of the wall arrangement selected that a considerable attention has been paid to ensure the structure will be proportionate non-twisting as far as possible. A.1.6 Selection of section dimensions 

The section dimensions for slabs and beams are selected so that deflection criterion could be satisfied.



It is assumed that the flexural and shear resistance required will be provided by using sufficient amount of reinforcement.

A.1.6.1 Selection of slab thickness Slab thickness has been selected as 0.175m. This gives an effective depth of about 0.144m with 25mm cover and 12mm diameter reinforcement.

Span 6000mm   40.2 effectivedepth 144mm The deflection criterion could be easily satisfied for continuous slabs for the above ratio with high yield steel reinforcement (20x1.68=43.68)

A 1.6.2 Selection of beam dimensions The depth selected is 600mm. this gives an effective depth of about 550mm. Span /effective depth for 6m long beam =

6000mm  10.91 which is reasonable value. 550mm

The width of the beam is 400mm.

A 1.6.3 Selection of column dimensions In this building lateral loads are carried by shear walls, the frame primarily carry the vertical loads. In the lower stories, the effect of moment transferred from the beams will have little influence on the amount of reinforcement since the column sizes required to 127   

resist heavy axial loads increases significantly. Thus, the column sizes are selected by considering them as axially loaded members with 1%, 1.5% and 2% reinforcement. The grade of concrete will have a considerable effect on the size of the column. Therefore, two grades, grade 40 and 50 concrete have been used for the calculation. A typical internal column carries a load from 6m x 5m area. The loads are evaluated below. Self weight of slab

= 6  5  0.175  24  151kN

Weight of finishes at 0.5kN/m2  6  6  0.5  18kN Weight of partition at 1.0kN/ m2  6  6  1.0  36 kN Weight of services at 1.0 kN/ m2  6  6  1.0  36 kN Weight of beam (0.6m x 0.375m)  [6  6]  0.6  0.375  24.0  64.8kN Imposed load on the slab at 2.5kN/m2  6  6  2.5  90 kN

Design dead and imposed load on a column form one floor considered = 1.4151  18  36  36  64.8  1.6  90  514.52kN Trial column size from ground floor to 12th floor is 600mm x 600mm Total column load at 12th floor = 514.52  12  0.6  0.6  4  12  24  1.4  6754 .85kN

Total column is axially loaded and the reinforcement ratio is 1.5% N  0.35 Ac f cu  0.67 Ac f y N  0.35 Ac f cu  0.670.015 Ac  f y

For Grade 40:

N  0.35 Ac  40  0.670.015 Ac 460 N = 18.62 Ac with 1.5% r/f N = 17.08 Ac with 1% r/f N = 20.16 Ac with 2% r/f Load at Ground floor with 600mm x 600m columns with 2% reinforcement between ground floor and 12th = 7257.6kN 128   

Therefore, from ground floor to 12th floor, 600 mm x 600 mm columns are used. Shear wall thickness is 300 mm for the entire height of the building.

129   

Figure A.2: Plan view of the 48 m height building 130   

A.2.0 Design of 183 m Height Building A.2.1 Determination of storey height

It is assumed that beam slab construction is used. Head room = 2.7m Slab thickness = 200mm Beam depth = 600mm Beam width = 400mm or with of the column whichever is less Space for services and ceiling = 300mm Below the beam soffit level. Total floor height = 2.7+0.6m+0.3m=3.6m Use a floor height = 2.7 + 0.6 + 0.3 = 3.6m Use a floor height of 3.6

A 2.2 Determination of number person per floor Consider 25% area allocated for the services. Therefore the rentable area is about 75% of the total area of the building. The area allocated for each person is 10 m2. The population handled in 5 minutes is 14% Average interval between two lifts is 35 seconds The car speed is equal 4ms-1. The population per floor = 30  46 

0.75  104 persons per floor 10

It is decided to use hard zoning arrangement where the lifts starting at ground floor are curtailed at a suitable level. The arrangement selected as follows 1st to 20th floor 20th to 35th floor 35th to 50th floor

131   

A 2.3 Determination of numbers of lifts A 2.3.1 For the floors between 1st floor and 20th Total population in 1st floor to 20th floor = 104  20  2080 persons The lift system is designed for a five minute peak minute capacity. The peak capacity is assumed as 12% of the total population in these floors. Population handled in five minute period = 0.12  2080  250 persons Capacity of the lift cars is selected as 28 passengers, which will give an average interval of about 26 seconds. The round trip time is about 175 seconds. Thus, the number of lift required is 175/26 = 7. Seven lifts cars can be used to serve the 20 floors.

A 2.3.2 For the floors between 20th and 35th The number of floor served = 15 Since lifts cars serving 1st to 20th floors have been curtailed, the useful area in a floor has now been increased to 78.5%. Therefore, the population per floor is given by 30  46 

0.785  108 persons 10

Total population = 108  15  1620 persons With 12% of population handled within 5 minutes. 5 minute capacity = 0.12  1620  194 persons Select 24 capacity lift cars, which give an average interval of 30 seconds. The round trip time is 145 seconds. The lift has to travel express 4 ms-1 from ground floor to 20th floor and come back. The total distance travelled is 20  3.6  2  144 m . The time taken in 144m/4 = 36 seconds The total round trip time = 145  36  181sec ond The number of lifts required is 181/30 = 6, thus six lifts should be provided.

A 2.3.3 For the floors between 36th to 50th A lift of 4ms-1 will be used. No of floors served including access to roof = 16

132   

Since lift cars serving 21st to 35th floors have been curtailed. The useful area in a floor has now been increased to 81.6%. Therefore, the population per floor is given by 30  46 

0.816  113 persons 10

Total population = 113  15  1695 persons With 12% of population handled within 5 minutes. 5 minute capacity = 0.12  1695  203 persons Select 24 capacity lift cars, which give an average interval of 26 seconds. The round trip time is 145 seconds The time taken for express travel is 35  3.6  2 6  42 sec onds The total round trip time = 145  42  187 sec ond The number of lifts required is 187/30 = 6, thus seven lifts should be provided Arrangement of lifts is as follows Ground to 20th floor = 6, 28 passenger lifts + 1 service lift 21st to 35th floor

= 6, 24 passenger lifts + 1 service lift

36th to 50th floor

= 6, 24 passenger lifts + 1 service lift

2550mm

2850mm 150m

150m

150m

2400mm

2400 mm

150m

 

 24 Passenger Lift 

        28  Passenger  f

Figure A.3: Approximate size of the lift shafts

133   

A.2.4 Stair case The stair case selected has a width of 1.5m, a rise of 0.15m, a thread of 0.30m. Thus, the number of steps required is 22, giving a flight length of 3.3m and a landing width of 1.5m. Total internal space required for staircase is 3.0 x 4.8m.

A.2.5 Lateral load resisting system A.2.5.1 Ground floor to 20th floor

46 m side: The lift shaft walls with a length of (2.55 x 3) = 7.65 and the wall behind the gents toilet and staircase can be used to resist the lateral loads in X direction.

30 m side: There are four shear walls of length 12.21m and 5.51 m length side wall of gent toilet resist the wind load acting perpendicular to 30 m side. These shear walls can be allowed to resist the lateral loads independently or those can be coupled.

A.2.5.2. From 21st floor to 35th floor 46 m side: The shear walls are as for 0-20th floor. 30 m side: The numbers of shear walls have been reduced. But the central shear wall has been kept to provide sufficient stiffness to the lateral wind load.

A.2.5.3. From 36th to 50th floor 46 m side: The shear walls are as for 0-20th floor 30 m side: The numbers of shear walls have been reduced. But the central shear wall has been kept to provide sufficient stiffness to the lateral wind load

A.2.6 Selection of section dimensions 

The section dimensions for slabs and beams are selected so that deflection criterion could be satisfied.



It is assumed that the flexural and shear resistance required will be provided by using sufficient amount of reinforcement.

134   

A.2.6.1 Selection of slab thickness Slab thickness has been selected as 0.175m. This gives an effective depth of about 0.144m with 25mm cover and 12mm diameter reinforcement.

Span 6000mm   40.2 effectivedepth 144mm The deflection criterion could be easily satisfied for continuous slabs for the above ratio with high yield steel reinforcement (20x1.68=43.68)

A.2.6.2 Selection of beam dimensions The depth selected is 600mm. this gives an effective depth of about 550mm. Span /effective depth for 6m long beam =

6000mm  10.91 which is reasonable value. 550mm

The width of the beam is 400mm.

A.2.6.3 Selection of column dimensions In this building lateral loads are carried by shear walls, the frame primarily carry the vertical loads. In the lower stories, the effect of moment transferred from the beams will have little influence on the amount of reinforcement since the column sizes required to resist heavy axial loads increases significantly. Thus, the column sizes are selected by considering them as axially loaded members with 1%, 1.5% and 2% reinforcement. The grade of concrete will have a considerable effect on the size of the column. Therefore, two grades, grade 40 and 50 concrete have been used for the calculation. A typical internal column carries a load from 6m x 5m area. The loads are evaluated below. Self weight of slab

= 6  5  0.175  24  126 kN

Weight of finishes at 0.5kN/m2  6  5  0.5  15kN Weight of partition at 1.0kN/ m2  6  5  1.0  30 kN Weight of services at 1.0 kN/ m2  6  5  1.0  30 kN Weight of beam (0.6m x 0.4m)  [6  5]  0.6  0.4  24.0  63.36kN Imposed load on the slab at 2.5kN/m2  6  5  2.5  75kN 135   

Design dead and imposed load on a column form one floor with only 60% imposed load considered = 1.4126  15  30  30  63.3  1.6  75  0.6  442.02kN Trial column size from 40th to 49th floor is 600mm x 600mm Total column load at 40th floor = 442 .02  10  0.6  0.6  3.6  10  24  1.4  4855 .66 kN

Total column is axially loaded and the reinforcement ratio is 1.5% N  0.35 Ac f cu  0.67 Ac f y N  0.35 Ac f cu  0.670.015 Ac  f y

For Grade 40:

N  0.35 Ac  40  0.670.015 Ac 460 N = 18.62 Ac with 1.5% r/f N = 17.08 Ac with 1% r/f N = 20.16 Ac with 2% r/f

For Grade 50: N = 22.12 Ac with 1.5% r/f N = 20.58 Ac with 1% r/f N = 23.66 Ac with 2% r/f Load at 30th floor with 0.7m x 0.7m columns between 30th and 39th = 9868.56kN Load at 20th floor with 0.8m x 0.8m columns between 20th and 29th = 14862.90kN Load at 10th floor with 0.9m x 0.9m columns between 10th and 19th = 20262.88kN Load at Ground floor with 1.0m x 1.0m columns between ground floor and 9th = 25892.68kN

136   

Table A.1: column sizes at different height levels Floor

Concrete Grade 40

50

1%

1.5%

2%

1%

1.5%

2%

40th floor

533 x533

511 x511

491 x491

486 x486

469 x469

453 x 453

30th floor

760 x760

728 x728

700 x700

692 x692

668 x668

646 x646

20th floor

933 x933

893 x893

859 x859

850 x850

820 x820

793 x793

10th floor

1089x1089

1043x1043 1003x1003

992 x992

957 x957

925 x925

Ground floor

1231x1231

1179x1179 1133x1133 1122x1122

1082x1082

1046x1046

The following sizes have been selected for the columns on the basis above calculation with grade 50 concrete.

Location

Column sizes

Ground floor – 10th floor

1050mm x 1050mm

10th floor to 20th floor

950mm x 950mm

20th floor to 29th floor

800mm x 800mm

30th floor to 39th floor

700mm x 700mm

40th floor to 49th floor

500mm x 500mm

The shear wall thicknesses selected are as follows Location

Shear wall thickness

Ground floor – 20th floor

300mm

21st floor to 35th floor

250mm

36th floor to 50th floor

200mm

            137   

     

A.5: Plan view of the ground floor of 183 m height building 138 

 

   

Figure A.5: Plan view of the 25th floor of 183 m height building

 

139   

Figure A.6: Plan view of the 46th floor of 183 m height   140   

APPENDIX B B.1. Wind Load Calculation for 48m height building 60 m  30m 

48 m 

Figure B.1: dimension of the 48 m height building B.1.1. Wind load calculation of 48m building according to CP3 Chapter V-Part 2: 1972 B.1.1.1. Regional wind speed According to the report on “Technical Assistance to Sri Lanka on Cyclone Resistant construction” From Table 3.1 Post disaster wind speed on Zone 3 is 120 mph (38m/s)

B.1.1.2. Design Wind Speed

V  VB S1 S 2 S 3 Topography factor (S1) = 1.0 Ground roughness, building size and height above ground factor (S2) Ground roughness 3 – Country with many windbreaks average roof height is about 10 m . Because of greatest horizontal dimension (60 m) exceeds 50m, building class is Class C

From Table 3 S2 = 1.02 Statistical factor for permanent building S3 = 1.00 141   

Design wind speed = 38 1.0 1.02 1.00  38.76ms 1 B.1.1.3. Dynamic pressure of the wind

q  kV 2 q  0.60  38.76 2  901.40 N / m 2

The load F acting in a direction normal to the individual structural member or cladding F  (C pe  C pi ) qA

External pressure coefficient Cpe h =48m w = 30m

h 48   1.6 w 30

3 h  6 2 w

l 60  2 w 30

3 l  4 2 w

C pe  0.7 for windward direction and Cpe = -0.4 for Leeward direction Assume four faces are equally permeable According to Appendix E in CP 3: Chapter V C pi  0.3

B.1.1.4. Wind normal to 60m long wall Wind force acting on windward wall of building F  [(0.7  0.2 )901.40  60] / 1000  27.04kN / mheight

Wind force on leeward wall F  [(0.4  0.2 )901.40  60] / 1000  32.45kN / mheight Total force = 27.04kN – (-32.45kN) = 59.49 kN/m height

B.1.1.5. Wind normal to 30 m long wall Wind force acting on windward wall of building F  [(0.8  0.2)901 .40  30] / 1000  16.22 kN / mheight

Wind force on leeward wall 142   

F  [( 0.1  0.2)901 .40  30] / 1000  8.11kN / mheight

Total force = 16.22kN – (-8.11)kN = 24.33 kN/m height

B.1.2. Wind load calculation of 48m building (BS 6399-2:1997) B.1.2.1. Dynamic classification of the building

Building height above the base = 48 m Building type factor Kb = 1 Dynamic augmentation factor Cr = 0.05< 0.25 So BS 6399-2:1997 can be used B.1.2.2. Design wind speed and dynamic pressure

The site is categorized as in town terrain with an average level of roof tops at least H0=6.0m Reference roof height Hr= 48 m If X < 2H0 The effective height He is greater of He=Hr -0.8H0 or He=0.4Hr He= 48 – 0.8 x 6 = 43.2 or He= 0.4 x 48 = 19.2; He=43.2 m B.1.2.3.Site wind speed

Vs=Vb x Sa x Sd x Ss x Sp B.1.2.3.1. Altitude factor Sa

When topography is not considered significant

S a  1  0.001 s

 s is the site altitude (MSL) = 3m S a  1  0.001  3  1.003 B.1.2.3.2. Directional factor Sd

The orientation of the building is ignored. Sd= 1.0 B.1.2.3. 3. Seasonal factor Ss 143   

For the permanent building SS=1.0 B.1.2.3.4. Probability factor

The standard value of risk Sp=1.0 B.1.2.3.5. Basic wind speed

Vb= 21 ms-1 Vs= 21 x 1.003 x 1 x 1 x 1 = 21.06 ms-1 60 m 

30 m 

48m 48 m  30m

Normal to 60 m wall

Normal to 30 m wall

Figure B.2: Building divided according to division-by-parts rule

B.1.2.4. Effective wind speed

Ve=Vs x Sb Sb is the terrain and building factor appropriate to the wind direction being considered Sb = ScTc{1 + (gt x St x Tt) + Sh} Sc is the fetch factor Tc is the fetch adjustment factor Tt is turbulence adjustment factor gt is gust peak factor Sh is topographic increment 144   

From Table 22 and Table 23 by assuming site is within 1 km away from the sea B.1.2.4.Diagonal dimension

Normal to 30m wall Hr = 30m; a = 42.43 m; gt = 3.44(Directional wind speed used with standard pressure coefficient and size effect factor ) Hr = 48m; a = 34.99 m; gt = 3.44 Normal to 60m wall a = 76.84 m; gt=3.44 Normal to 30m wall

Hr = 30m ;Sb = 1.35 x 0.965{1+ (3.44x 0.132 x 1.06)} = 1.93 Hr = 48m ;Sb = 1.449 x0.982 {1+ (3.44 x 0.119 x 1.006)} = 2.01

Normal to 60m wall Sb = 1.449 x0.982 {1+ (3.44 x 0.119 x 1.006)} = 2.01 B.1.2.5.Effective wind speed Ve=Vs x Sb

Normal to 30m wall

Hr = 30m; Ve = 21.06 x 1.93 = 40.65ms-1 Hr = 48m; Ve = 21.06 x 2.01 = 42.33ms-1

Normal to 60m wall Ve = 21.06 x 2.01 = 42.33 ms-1

B.1.2.6.Dynamic wind pressure

qs = 0.6Ve2 Normal to 30m wall

Hr= 30m; qs = 0.6 x 40.652 =991.45 Pa Hr= 48m; qs = 0.6 x 42.332 =1075.13 Pa

Normal to 60m wall qs = 0.6 x 42.332 =1075.13Pa B.1.2.7.External pressures

pe= qsCpeCa B.1.2.7.1.External pressure coefficient Cpe

Wind normal to 30m Cpe,windward = +0.78 , Cpe,Leeward = -0.28

145   

Wind normal to 60m Cpe,windward = +0.8 , Cpe,Leeward = -0.3 B.1.2.7.2.Size effect factor Ca

Wind normal to 30m Ca=0.85 Wind normal to 60m Ca=0.83 B.1.3. Wind load calculation for 48m building according to BS EN 1991-1-4:2005(E) (Compliance with National Annex for UK)

Building locates in suburban area in zone 3. The terrain is flat and site is 1km away from the shore line. B.1.3.1. Basic wind speed

From Clause 4.2

Vb = Cdir. Cseason. Vb,0

From N.A.2.4 Vb,0 = Vb,map. Calt Vb,map – value of the fundamental basic wind velocity 22 ms-1 Calt - Altitude factor ; Calt = 1 + 0.001A (10/Z)0.2 for Z > 10m Assuming site elevation is 3m above the MSL. Calt = 1 + 0.001(3) (10/48)0.2 = 1.002 Vb,0 = 22 x 1.002 = 22.04 ms-1 Taking Cdir = 1 and Cseason =1 Vb,0 = 22.04 ms-1 B.1.3.2. Mean wind velocity

From Clause 4.3.1

Vm = Cr(z).C0(z).Vb

From Figure NA.3

Cr(z) = 1.4 Vm = 1.4 x 1 x 22.04 = 30.86 ms-1

B.1.3.3. Effect of neighbouring buildings

From Annex A ; A-4 Assuming the building is more than twice as high as the average height has of the neighbouring structures. 146   

Then peak velocity pressure at height Zn (Ze = Zn) Height of the building hhigh < 2 davg Then r = hhigh = 48m

1 1 x  r : Z n  r   48  24m 2 2 B.1.3.4. Wind turbulence Clause 4.4 The turbulence intensity Iv(z)

Because of distance upwind to shoreline is 1km Z- hdis = 48 – 0 = 48m (disturbance height for suburban terrain is 0m) l(48) = 0.122

From Figure NA 5

B.1.3.5.Peak velocity pressure Clause 4.5. The peak velocity pressure qp(z) at height z, which includes mean and short – term velocity

From NA.2.17

qp(z) = Ce(z)Ce,Tqb

From Figure NA. 7

Ce(z) = 3.68 Ce,T = 1.0 qp(z) = 3.68 x 0.5 x 1.2 x 1.0 x 22.042 = 1072.56 pa

B.1.3.6.Wind actions B.1.3.6.1.Wind pressures on surfaces Clause 5.2 the wind pressure acting on the external surface is We

We =qp(ze).Cpe The wind pressure acting on the internal surfaces of a structure Wi=qp(ze).Cpi

147   

B.1.3.6.2.Pressure coefficient

From Clause 7.2.2 The reference height ze, for windward walls of rectangular plan buildings depend on the aspect ratio h/b

(i)

Wind normal to 60m wall 60  qp(z) = qp(ze) 

48 

h< b then ze = h=48m qp (z) = qp(ze)

Figure B.3: Pressure distribution when wind flow normal to 60m wall

(ii)

Wind normal to 30m wall ze= h = 48m

qp(z) = qp(h)

ze= b = 30m

qp(z) = qp(b) 

Figure B.4: Pressure distribution when wind flow normal to 30m wall B.1.3.6.3.External pressure coefficient Cpe

From Table 7.1 When wind flow normal to 60m wall 148   

h < d = 48/30 = 1.6 For windward wall

Cpe = +0.8

For Leeward wall

Cpe = -0.53

When wind flow normal to 30m wall h zmin = 5 m

Wind turbulence Turbulence length scale 

z L z   Lt   for z =48 m > zmin = 5 m  zt  Where zt = 200 m Lt=300 m and   0.67  0.05 ln 0.3  0.61

 48  L z   300   200 

0.61

 125.62m

Non dimensional power spectrum density function SL(z,n) S L z, n  

n.S v  z , n 



Where f L 

2 v

=

6.8 f L  z , n  1  10.2 f L ( z, n) 5 / 3

n.L z  vm z 

Natural frequency of vibration of the building

n

46 46   0.958 h 48

fL 

0.958.125.62  3.99 30.17

Then S L 

6.83.99   0.054 1  10.2(3.99) 5 / 3

150   

From Annex B; B.2. (2) The background factor B2 For the lack of correlation of the of the pressure on the structure surface

B2 

B2 

1 bh 1  0.9   L z s  

0.63

1  60  48  1  0 .9   91.99 

0.63

 0.50

Iv(28.8) = 0.142

Cs 

1  7.(0.142). 0.50  0.853 1  7.0.142

Cd  1.00 B.1.3.8. External force

For windward direction We=1098.02(0.8) = 878.42 Pa For leeward direction We=1098.02(-0.53) = -581.95 Pa Total pressure (by considering non simultaneous action) We,Total =(0.873)[878.42-(-581.95)]=1274.90 Pa Reference area = 60 x 4 = 240 m2 Fw,e= (0.853)(1)(1274.90)240 = 261 kN Internal force Wi = 1098.02(-0.3) = 329.41 Pa Fw,i = -329.41 x 240 = -79.06 kN 151   

B.1.3.9. Wind flow normal to 30m wall

buildings is divided in to two parts as shown in figure B.5. ze= h = 48m

ze= b = 30m

Figure B.5: Building divided according to division-by –parts rule. B.1.3.9.1. Basic wind speeds

At 30m height = Calt = 1+0.001.(3).(10/30)0.2 = 1.0024 Vb = 22 x 1.0024 = 22.04

B.1.3.9.2. Mean wind velocities

Mean wind velocity at 30m height = 1 x 1.3 x 22.04 = 28.65 ms-1 Mean wind velocity at 48m height = 30.86 ms-1 B.1.3.9.3.Wind turbulence

From Figure NA 5 in N.A. l(48) = 0.122 and l(30) = 0.142 B.1.3.9.4. Peak velocity pressure

At 48m height qp(48) = 1072.56 Pa At 30m height qp(30) = 3.42 x 0.5 x 1.2 x 22.042 = 996.78 Pa

B.1.3.9.5. External pressure coefficient

h/d = 48/60 = 0.8 152   

windward wall

Cpe = +0.77

Leeward wall

Cpe = -0.45

B.1.3.9.6.Wind forces

Consider one storey, which has 4.0m storey height From Clause 5.3 External forces Fw,e  C s .C d .

 w .A e

ref

surface

Internal forces Fw,i 

 w .A i

ref

surface

Friction forces F fr  C fr .q p  z e . Aref The background factor B2 B2 

Cs 

1  30  48  1  0.9  91.99 

0.63

 0.55

1  7.(0.142). 0.55  0.87 1  7.0.142

Cd  1 B.1.3.9.7. External force For windward direction At 30m height

We=996.78(0.77) = 767.52 Pa

At 48m height

We=1072.56(0.77) = 825.87 Pa

For leeward direction At 30m height

We=996.78(-0.45) = -448.55 Pa

At 48m height

We=1072.56(-0.45) = -482.65 Pa

Total pressure (by considering non simultaneous action) At 30m height

We,Total =(0.872)[767.52-(-448.55)]=1060.41 Pa 153 

 

At 48m height

We,Total =(0.872)[825.87-(-482.65)]=1141.03 Pa

Reference area = 30 x 4 = 120 m2

Total force At 30m height

Fw,e= (0.87)(1.00)(1060.41)120 = 110.71 kN

At 48m height

Fw,e= (0.87)(1.00)(1141.03)120 = 119.12 kN

Internal force Wi = 996.78(-0.3) = -299.03 Pa Fw,i = -299.03 x 120 = -35.88 kN

B.1.4. Wind load calculation for 48m building according to the AS/NZS 1170.2:1989 Height of the building = 48m Length of the building = 60m Width of the building = 30 m

Height 48   1.53  5 and width 30 A first mode of frequency n 

46 46   0.953  1Hz h 48

Therefore no need a dynamic analysis.

B.1.4.1. Design hourly mean wind speed From Clause 4.2.2. 





Vz  V M ( z ,cat ) M s M t M i 

Vz = The design hourly mean wind speed at height z, in meters per second (z = 48m)

V = The basic wind speed = 47 ms-1 

M z = An hourly mean wind speed multiplier for a terrain category at height z 154   

Building locates in sub urban area, therefore terrain category is 3. 

M z =1.064

Ms =

Shielding multiplier Assume Shielding buildings are in more than 12m distance.

M s =1.0 

M t = Topographic multiplier Building locates in flat land. 

Therefore M t =1.0

M i = structure importance multiplier Building has post disaster functions

M i = 1.0 

V z  38  1.064  1  1  1.0  40.43ms 1

B. 1.4.2. Dynamic Wind Pressure From Clause 4.3. 



q z  0.6Vz  10 3 

q z = The free stream hourly mean dynamic wind pressure at height z, in kilopascals 

Vz = The design hourly mean wind speed at height z, in meters per second 

q z  0.6  40.432  10 3  0.98kpa

B.1.4.3. Horizontal force acting on a building

From Clause 4.4.2 



Fz  C p ,e q z Az 155   



Fz = The hourly mean net horizontal force acting on a building or structure at height z C p ,e = The pressure coefficients for both windward and leeward surfaces 



q z = The free stream hourly mean dynamic wind pressure resulting from Vz , in

kilopascals Az = The area of a structure or a part of a structure, at height, in square meters

B.1.4.3.1. For wind normal to 30m wall

d 60  2 b 30 C p ,e = 0.8 + 0.3 = 1.10 

Fz  1.10  0.98  30  32.34kN / mheight

B.1.4.3.2.For wind normal to 46m wall

d 30   0.50 b 60 C p ,e = 0.8 + 0.5 = 1.3 

Fz  1.3  0.98  60  76.44kN / mheight

B.1.5. Wind load calculation for 48m building according to AS1170.2:2002 B.1.5.1. Regional wind speed According to the report on “Technical Assistance to Sri Lanka on Cyclone Resistant Construction” From Table 3.1 Post disaster wind speed on Zone2 is 85 mph (38m/s)

156   

B.1.5.1.1. Wind direction multiplier According to clause 3.3 Wind direction multiplier for region A M d  1.0 for overturning moment and major structural system for all directions B.1.5.1.2. Terrain- height multiplier According to Clause 4.2.1 Terrain category is category 3 From Table 4.1(B) Z=h=48m, for terrain category 3, M z ,cat  M 48.cat 3  1.06 B.1.5.1.3. Shielding According to clause 4.3.1 There are no other buildings of greater height in any direction. Therefore M s  1.0 for all directions. B.1.5.1.4. Topography According to clause 4.4.1 Topography multiplier M t  M h  1.0 B.1.5.2. Site wind speed Site wind speed for all directions for overall loads and main structural design

Vsit ,  38(1.0)(1.06)(1.0)(1.0)  40.28ms-1

B.1.5.3. Design wind speed For all wind directions, the design wind speeds

157   

` Vdes ,  Vsit ,  = 40ms-1 (for overall loads and main structure design) B.1.5.4. Aerodynamic shape factor B.1.5.4.1. External pressures From Table 5.2(A) External pressure coefficient for windward wall C pe  0.8 (wind speed vary with height) From Table 5.2(B) d/b =0.50 10m Assuming site elevation is 3m above the MSL. Calt = 1 + 0.001(3) (10/183)0.2 = 1.002 Vb,0 = 22 x 1.002 = 22.04 ms-1 Taking Cdir = 1 and Cseason =1 Vb,0 = 22.04 ms-1

C.3.2 Mean velocity From Clause 4.3.1

Vm = Cr(z).C0(z).Vb

From Figure NA.3

Cr(z) = 1.71 Vm = 1.71 x 1 x 22.04 = 37.69 ms-1

166   

C.3.3 Effect of neighbouring buildings From Annex A ; A-4 Assuming the building is more than twice as high as the average height has of the neighbouring structures. Then peak velocity pressure at height Zn (Ze = Zn) Height of the building hhigh > 2 davg Then r = 2.dlarge = 2 x 46 m = 92 m

1 1 x  r : Z n  r   92  46m 2 2 C. 3.4 Wind turbulence From Clause 4.4 The turbulence intensity Iv(z) Because of distance upwind to shoreline is 1km and assuming have = 10 m for terrain category 3 From A.5 where 2 .have < x < 6.have hdis is lesser of 1.2 have – 0.2 X or 0.6h 1.2 x 10 – 0.2 x Z- hdis = 183 – 0 = 183 m (disturbance height for suburban terrain is 0m) From Figure NA 5

l(183) = 0.13

C.3.5 Peak velocity pressure From Clause 4.5 The peak velocity pressure qp(z) at height z, which includes mean and short – term velocity From NA.2.17

qp(z) = Ce(z)Ce,Tqb From Figure NA. 7 Ce(z) = 4.21 (At 183 m height)

167   

Ce,T = 1.0 qp(z) = 4.21 x 0.5 x 1.2 x 1.0 x 22.042 = 1227.03 Pa C.3.6 Wind actions C.3.6.1 Wind pressures on surfaces From Clause 5.2 the wind pressure acting on the external surface is We We =qp(ze).Cpe The wind pressure acting on the internal surfaces of a structure Wi=qp(ze).Cpi C.3.6.2 Pressure coefficient From Clause 7.2.2 The reference height ze, for windward walls of rectangular plan buildings depend on the aspect ratio h/b (i)

Wind normal to 60m wall b =46m qp(z)= qp(h)  b=46m

qp(z)=  hstrip=18.2m qp(z)= qp(b)  b = 46m

168   

Figure C.3: division-by-parts rule for 183 m building wind flows normal to 46 m wall h< b then ze = h=48m qp (z) = qp(ze) (ii)

Wind normal to 30m wall

b =30m qp(z)= qp(h)  b=30m

qp(z)= qp(zstrip)  hstrip=17.6m

qp(z)= qp(b)  b = 30m

Figure C.4: division-by-parts rule for 183 m building wind flows normal to 30 m wall C.3.6.2.1 External pressure coefficient Cpe From Table 7.1 When wind flow normal to 46m side h / d = 183/30 = 6.1 For windward wall

Cpe = +0.80

For Leeward wall

Cpe = -0.70

When wind flow normal to 30m wall h/d = 183/46 = 3.98 For windward wall

Cpe = +0.80 169 

 

For Leeward wall

Cpe = -0.65

C.3.6.2.2 Internal pressure coefficient Cpi From Figure 7.13, Note 2 Assuming that building has all four walls are permeable and impermeable roof.] Estimated  values are 0.69 for 46m wall and 0.81 for 30m wall Wind flows normal to 46 m wall h/d = 6.1 Cpi = -0.18 Wind flows normal to 30 m wall h/d = 0.8 Cpi = -0. 30 C.3.7 Wind forces Consider one storey, which has 3.6 m storey height From Clause 5.3 External forces Fw,e  C s .Cd .

 w .A e

ref

surface

Internal forces Fw,i 

 w .A i

ref

surface

Friction forces F fr  C fr .q p  z e . Aref C.3.7.1 Structural factors CsCd The structural factor CsCd should take into account the effect on wind actions from the non simultaneous occurrence of peak wind pressures on the surfaces (Cs) together with the effect of the vibrations of the structure due to turbulence (Cd)

C.3.7.1.1 The size factor Cs

170   

1  7.lv ( z s ). B 2 Cs  1  7.lv  z s  C.3.7.1.2 The dynamic factor Cd Cd 

1  2.k p .lv  z s . B 2  R 2 1  7.lv  z s  B 2

Where, zs – reference height zs = 0.6 (183) = 109.8 m > zmin = 5 m Wind turbulence

C.3.7.1.3 Turbulence length scale 

z L z   Lt   for z =183m > zmin = 5 m  zt  Where zt = 200m Lt=300m and   0.67  0.05 ln 0.3  0.61  183  L z   300   200 

0.61

 284.18m

C.3.7.1.4 Non dimensional power spectrum density function SL(z,n) S L z, n  

n.S v z , n 

Where f L 



2 v

=

6.8 f L  z , n  1  10.2 f L ( z, n) 5 / 3

n.L z  vm z 

C.3.7.1.5 Natural frequency of vibration of the building

n

46 46   0.251 h 183 171 

 

fL 

0.251.284.18  1.893 37.69

Then S L 

6.81.893  0.085 1  10.2(1.893) 5 / 3

From Annex B; B.2.

C.3.7.1.6 The background factor B2 For the lack of correlation of the of the pressure on the structure surface

B2 

B2 

1 bh 1  0.9   L z s  

0.63

1  46  183  1  0 .9   208 . 09 

0 . 63

 0 . 51

C.3.7.1.7 The peak factor (kp) The ratio of the maximum value of the fluctuating part of the response to its standard deviation

k p  2 lnv.T  

0.6 or kp= 3 whichever is lager 2 ln v.T 

Where v is up-crossing frequency

v  n1, x

R2 But v  0.08 Hz B2  R2

C.3.7.1.8 The resonance response factor (R2)

For allowing turbulence in resonance with the considered vibration made if the structure

172   

R2 

2 .S L z s , n1, x .Rh  h .Rb  b  2.

Where,

 - Total logarithmic decrement of damping. For reinforced concrete building  = 0.10 SL =0.085

h 

Rh 

b 

Rb 

R2 

4.6h 4.61831.893 . f L z s , h   7.66 L z  208.09 1

h



1

1  e   7.166   2 h

2 h

2





1 1  e 27.66   0.122 2 27.66

4.6b 4.6461.893 . f L z s , h    1.925 L z  208.09 1

b



1 2 b

2

2.0.10 

v  0.958

1 1  e   1.925   2b

2





1 1  e 21.925   0.387 2 21.925

.0.085 . 0.122 . 0.387   0.198

0.20  0.508  0.08 0.51  0.20

k p  2 ln 0.508 . 600 

0.6  3.560  3 2 ln 0.508 . 600

Iv(109.8) = 0.091

Cs 

1  7.(0.091). 0.51  0.889 1  7.0.091

173   

Cd 

1  2.3.560 . 0.091. 0.51  0.20  1.063 1  7.0.091 0.51

C.3.8 External force At 183 m height

C.3.8.1 For windward direction

We=1988.88(0.8) = 1591.10 Pa C.3.8.2 For leeward direction

We=1988.88(-0.7) = -1392.22 Pa C.3.8.3 Total pressure (by considering non simultaneous action)

We,Total =(1)[1591.10-(-1392.22)]=2983.32 Pa Reference area = 46 x 3.6 = 165.6 m2 Fw,e= (0.889)(1.102)(2983.32)165.6 = 484 kN C.3.8.4 Internal force

Wi = 1988.88(-0.18) = 358 Pa Fw,i = -358 x 165.6 = -59.28 kN C.3.8.5 Friction force

Total area parallel to the wind direction = 2 x 30 x183 = 10980 m2 < 0.4 ATotal Hence no need to calculate friction forces. C.3.9 Wind flow normal to 30m wall

Basic wind speed = 30.06 ms-1 174   

C.3.9.1 Mean wind velocities Height (m)

Cr(z)

Velocity (ms-1)

183

1.71

37.69

153

1.64

36.15

135.4

1.62

35.07

117.7

1.57

34.60

100.1

1.55

34.16

82.5

1.52

33.50

64.9

1.46

32.18

47.6

1.38

30.42

30

1.30

28.65

Height (m)

Ce(z)

Pressure (Pa)

183

4.21

1227.03

153

4.13

1203.72

135.4

4.08

1189.14

117.7

4.04

1177.49

100.1

3.97

1157.08

82.5

3.91

1139.60

64.9

3.81

1110.45

47.6

3.65

1063.82

30

3.42

996.78

C.3.9.1 Peak velocity pressure

C.3.9.2 External pressure coefficient

h/d = 183/60 = 0.8 windward wall

Cpe = +0.80

Leeward wall

Cpe = -0.65 175 

 

C.3.9.3 Wind Forces

Considering one storey, which has 4.0m storey height From Clause 5.3 External forces Fw,e  C s .Cd .

 w .A e

ref

surface

Internal forces Fw,i 

 w .A i

ref

surface

Friction forces F fr  C fr .q p  z e . Aref C.3.9.3.1 The background factor B2

B2 

1  30  183  1  0.9  208.09 

0.63

 0.52

C.3.9.3.2 The resonance factor (R2)

R2 

2 .S L z s , n1, x .Rh  h .Rb  b  2.  =0.01

Where

SL = 0.085                     h = 7.66 

Rh= 0.122

b    Rb 

4.6301.893  1.255   208.09 1

b



1 2 b

1 1  e   1.255   2b

2





1 1  e 21.255   0.506 2 21.255 176 

 

R2 

2

2.0.10 

v  0.958

.0.098 . 0.122  . 0.506   0.30

0.30  0.579  0.08 0.52  0.30

k p  2 ln 0.579  . 600  

0.6  3.596  3 2 ln 0.579  . 600 

Cs 

1  7.(0.091). 0.52  0.891 1  7.0.091

Cd 

1  2.3.596  . 0.091. 0.52  0.30  1.091 1  7.0.091 0.52

C.3.9.3.3 External force For windward direction

At 183m height

We=1988.88(0.80) = 1591.10 Pa

For leeward direction

At 183m height

We=1988.88 (-0.65) = -1292.77Pa

Total pressure (by considering non simultaneous action)

At 183m height

We,Total =(0.962)[1591.10-(-1292.77)]=2774.28 Pa

Reference area = 30 x 3.6 = 108 m2 Total force

At 183m height

Fw,e= (0.889)(1.124)(2774.28)108 = 299.39 kN

Internal force

Wi = 1988.88(-0.3) = -596.66 Pa

177   

Fw,i = -596.66 x 108 = -64.44 kN Friction force

Total area parallel to the wind direction = 2 x 30 x183 = 10980 m2 < 0.4 ATotal Hence no need to calculate friction force

C.4 Wind load calculation according to the AS/NZS 1170.2:1989 of 183 m height building

Height of the building = 183m Length of the building = 46 m Width of the building = 30 m

Height 180   6  5 and width 30 A first mode of frequency n 

46 46   0.251  1Hz h 183

Therefore this building is wind sensitive and need dynamic analysis. 50th Floor 

162m 

40th Floor  30th Floor  20th Floor 

126m     90m 

54 m  

10th Floor  18m 

Figure C.5: segment considered in pressure calculation 178   

C.4.1 Design hourly mean wind speed 





From Clause 4.2.2 Vz  V M ( z ,cat ) M s M t M i 

Vz = The design hourly mean wind speed at height z, in meters per second (z = 183m)

V = The basic wind speed = 38ms-1 

M z = An hourly mean wind speed multiplier for a terrain category at height z

Building locates in sub urban area, therefore terrain category is 3. 

M z =0.806

M s = Shielding multiplier Assume Shielding buildings are in more than 12m distance.

M s =1.0 

M t = Topographic multiplier Building locates in flat land. 

Therefore M t =1.0

M i = structure importance multiplier Building has post disaster functions

M i = 1.1 

V z  38  0.806  1 1 1.1  33.69ms 1

C.4.2 Dynamic Wind Pressure at 183 m level 

From Clause 4.3.



q z  0.6 Vz  10 3

179   



q z = The free stream hourly mean dynamic wind pressure at height z, in kilopascals 

Vz = The design hourly mean wind speed at height z, in meters per second 

q z  0.6  33.69 2 10 3  0.68kpa C.4.5 Horizontal force acting on a building or structure at height 183m 



From Clause 4.4.2 Fz  C p ,e q z Az 

Fz = The hourly mean net horizontal force acting on a building or structure at height z C p ,e = The pressure coefficients for both windward and leeward surfaces 



q z = The free stream hourly mean dynamic wind pressure resulting from Vz , in kilopascals Az = The area of a structure or a part of a structure, at height, in square meters

C.4.5.1 For wind normal to 46m wall

d 30   0.65 b 46 C p ,e = 0.8 + 0.5 = 1.3 

Fz  1.3  0.68  46  40.66kN / mheight

C.4.5.2 For wind normal to 30m wall

d 46   1.53 b 30 180   

C p ,e = 0.8 + 0.39 = 1.19 

Fz  1.19  0.68  30  24.28kN / mheight

C.4.6 Gust factor calculation

 SE  2  G  1  r  g v2 B1  w  g 2f    G = a gust factor

r = a roughness factor, twice the longitudinal turbulence intensity at height h

2

v 

2  0.143  0.286 = V  1 . 0 Mt

g v = a peak factor for upwind velocity fluctuation = 3.7 B = a background factor 1

 1

36h

 64b 2 Lh

2



Lh = a measure of the effective turbulence length scale, in meters h = 1000   10 

0.25

 183   1000   10 

0.25

 2068.30

For wind flows normal to 46m wall 1

B= 1

36 183

 64  46 2 2068.30 2



 0.641

181   

For wind normal to 30m wall 1

B= 1

36  180

 64  30 2 2068.30 2



 0.652

w = a factor to account for the second order effects of turbulence intensity gvr B 4

=

For wind normal to 46m wall

w=

3.7  0.286 0.641  0.211 4

For wind normal to 30m wall

w=

3.7  0.286 0.652  0.214 4

g f = a peak factor

=

2 log e 3600na   2 log e 3600  0.251  3.690

S = a size factor to account for the correlation of pressure over a structure =

1        1   3.5na h  1   4na b                  Vh    Vh   

For wind normal to 46m wall =

1    3.5  0.251183    4  0.251 46     1    1      41.67 41.67      

 0.097

182   

For wind normal to 30m wall =

1    3.5  0.251183    4  0.251 30     1    1      41.67 41.67      

 0.119

na= the first mode along-wind frequency of the structure = 0.251 Hz 

V h = the design hourly mean wind speed at height h

N= an effective reduced frequency =

na Lh 

Vh



0.251 2068.30  12.46 41.67

E = a spectrum of turbulence in the approaching wind stream =

0.47 N

2  N 

2 5/6

=

0.47  12.46

2  12.46 

2 5/ 6

 0.087

 = the structural damping capacity = 0.05

 SE  2 G  1  r  g v2 B1  w  g 2f     C.4.6.1 For wind flow normal to 46m wall

 2 3.690 2  0.097  0.087  2    2.114 The gust factor G  1  0.286  3.7  0.641 1  0.211  0 . 05  

183   

C.4.6.2 For wind normal to 30m wall

 3.690 2  0.119  0.087  2   2.143 The gust factor G  1  0.286  3.7 2  0.652  1  0.214  0.05   C.4.7 Cross - wind response  C fs       M c  0.5 g f q h bh (1.06  0.06k )      M c  =the design peak base overturning moment for a structure in cross-wind direction  



2

g f = a peak factor

2 log e (3600na ) in cross wind direction

= 3.690 

q h =the hourly mean dynamic wind pressure at height h in pascal = 1380 pa B = the breadth of the structure normal to the wind 30 m for X direction and 45 m for y direction h= the height of the structure, in meters = 183 m k= a mode shape power exponent from representation of the fundamental mode shape k=1.0 for building with a central core and moment resisting façade. C fs = the cross- wind force spectrum coefficient generalized for a linear mode

For wind normal to 45m wall h: b: d = 6:1.5:1 

47.88 V   4.15 and nc b 0.251 46

Turbulence intensity at 2h/3 = 0.159 h: b: d= 6:1:1 and Turbulence intensity 0.12 = Cfs= 0.0015 h: b: d= 6:1:1 and Turbulence intensity 0.20 = Cfs= 0.0051 h: b: d= 6:1:1 and Turbulence intensity 0.16 = Cfs= 0.0033

184   

h: b: d= 6:2:1 and Turbulence intensity 0.12 = Cfs= 0.0008 h: b: d= 6:2:1 and Turbulence intensity 0.20 = Cfs= 0.0014 h: b: d= 6:2:1 and Turbulence intensity 0.16 = Cfs= 0.0011 h: b: d= 6:1.5:1 and Turbulence intensity 0.16 = Cfs= 0.0022 When wind flow normal to 30m wall h: b: d = 6:1:1.5 

47.88 V   6.36 and nc b 0.251  30

Turbulence intensity at 2h/3 = 0.159 h: b: d= 6:1:1 and Turbulence intensity 0.12 = Cfs= 0.0075 h: b: d= 6:1:1 and Turbulence intensity 0.20 = Cfs= 0.018 h: b: d= 6:1:1 and Turbulence intensity 0.159 = Cfs= 0.0128 h: b: d= 6:1:2 and Turbulence intensity 0.12 = Cfs= 0.004 h: b: d= 6:1:2 and Turbulence intensity 0.20 = Cfs= 0.006 h: b: d= 6:1:2 and Turbulence intensity 0.16 = Cfs= 0.005 h: b: d= 6:1:1.5 and Turbulence intensity 0.16 = Cfs= 0.0089

 = the fraction of critical damping = 0.05

C.4.7.1 For wind normal to 46m wall    0.0025  M c  0.5  3.690  1380  46  183 2 (1.06  0.061)    1554.51MNm  0.05 

C.4.7.2 For wind normal to 30m wall    0.0089  M c  0.5  3.690  1380  30  183 2 (1.06  0.061)    1912.86 MNm  0.05 

185   

C.5 Wind load calculation according to the AS/NZS 1170.2:2002 of 183 m height building

Location –Building locates in Zone 3 Terrain – Terrain category3, sub urban area Geography – ground slope less than 1 in 20 for greater than 5 kilometers in all direction Dimension – average roof height 183 meters Horizontal dimension – 46 meters * 30 meters (rectangular cross section) Building orientation – major axis is on East - West Reinforced concrete construction. Curtain wall façade on all four faces. Sway frequencies - na  nc  0.251Hertz . Mode shapes are linear (k=1.0) Average building density- 350 kgm-3

C.5.1 Regional wind speed

According to the report on “Technical Assistance to Sri Lanka on Cyclone Resistant Construction” From Table 3.1 Post disaster wind speed on Zone 3 is 85 mph (38m/s) For calculation of acceleration, use 5 year return period Vs  33ms 1

C.5.1.1Wind direction multiplier

According to clause 3.3 Wind direction multiplier for region A M d  1.0 for overturning moment and major structural system for all directions

C.5.1.2 Terrain- height multiplier

According to Clause 4.2.1 Terrain category is category 4 From Table 4.1(B) Z=h=183 m, for terrain category 3, M z ,cat  M 183.cat 3  1.23

186   

C.5.1.3 Shielding

According to clause 4.3.1 There are no other buildings of greater height in any direction. Therefore M s  1.0 for all directions.

C.5.1.4 Topography

According to clause 4.4.1 Topography multiplier M t  M h  1.0

C.5.1.5 Site wind speed

Site wind speed for all directions for overall loads and main structural design Vsit ,  38(1.0)(1.23)(1.0)(1.0)  46.74 ms-1

C.5.1.6 Design wind speed

For all wind directions, the design wind speeds ` Vdes ,  V sit ,  = 47 ms-1 (for overall loads and main structure design) C.5.2 Aerodynamic shape factor C.5.2.1 External pressures

From Table 5.2(A) External pressure coefficient for windward wall C pe  0.8 (wind speed vary with height) From Table 5.2(B) d/b =0.65