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  • 1. dıgest CI/SfB (J4) March 1999 Wind loading on buildings Digest 436 Part 3 BS 6399-2:1997 worked examples – loads on a portal frame building and on an office tower on a podium This is the last of three Parts giving brief guidance on the use of BS6399-2. Part 1 gives advice and guidance on implementing BS6399-2 and suggests certain options. Part 2 demonstrates the recommended options by example calculations for the case of a two-storey timber frame house. This Part gives example calculations for a further two building types: a steel portal frame building, and a 15-storey tower surrounded by a two-storey podium. This Digest is aimed at architects, engineers and professionals who need to know the effect of wind on buildings, and design options that minimise it. The full title of BS 6399-2:1997 is Loading for buildings. Code of practice for wind loads. In this Digest we refer to it as BS 6399-2 or the Standard. It supersedes BS 6399-2:1995 which itself was a technical revision of Loading. Wind loads,CP3:Chapter V:Part 3:1972 (abbreviated here to CP3-V-2 or the Code); references to Code or Code of practice are to this 1972 document. The three stages Part 2 describes how BS 6399-2 is implemented in three distinct stages: 1 the dynamic classification – to ensure that the Standard is suitable for the structural form, 2 the design dynamic pressure for the site which is independent of the structural form of the building, and 3 the pressure coefficients and structural loads on the building. Because Stage 2 is independent of the building, the examples given in Part 2 of each of the options recommended in Part 1 do not need to be repeated. Instead, the worked examples of this Part use only the recommended Option 2(b) – four orthogonal cases using Equation 29 (see Q16 and Q33 of Part 1) – the ‘hybrid’ method permitted by §3.4.2 that gives the optimum balance between conservatism and complexity. Figures and tables in this Digest are denoted by letters (eg Figure A, Table A) to distinguish them from those in BS 6399-2. Except where specifically noted, any numbered reference to a Clause (§), Figure or Table, refers to that in BS 6399-2:1997. The notional site location For the example calculations in this Part, we shall adopt the same site used in Part 2: in the south-west outskirts of Sheffield at grid reference SK320810. This site was chosen to illustrate the options recommended in Part 1 because it is typical of sites in inland towns. The relevant values for this location are given in Table A from which values for the calculations will be taken. The altitude of the site is 120 m above mean sea level and the site is not in the zone of significant topography, defined in Figure 7. See Part 2 for more guidance on these site parameters. Table A Distances-to-sea and in-town for the site at SK320810 Direction (°) 0 30 60 90 120 150 180 210 240 270 300 330 Distance-to-sea (km) 200 128 110 110 200 200 200 200 200 200 112 200 Distance-in-town (km) 7.5 12.5 9.5 3.5 2.5 1.5 1.5 3.5 2.5 2.5 1.5 3.5
  • 2. 2 Portal frame building The principal dimensions of the portal frame building in this example are given in Figure A. Stage 1: Dynamic classification This is a test to determine whether it is appropriate to use BS 6399-2 and to obtain the value of Cr, and is demonstrated in Calculation 1. Note A The value of Cr given by the classification indicates that the building is 3.5% dynamic. α = 10˚ 6m H =7m W = 30 m L = 60 m Figure A Principal dimensions of the portal frame building Calculation 1 Dynamic classification of the portal frame building Clause Action Notes 1.3.3.2 Building height above its base H=7m Use height to eaves 1.6.1 Read value of Kb from Table 1 Kb = 2 When in doubt, take next larger value 1.6.1 Using H and Kb, read Cr from Figure 3 Cr = 0.035 1.6.2 Check Cr < 0.25 If Cr > 0.1, get better value from Annex C If Cr > 0.25, BS 6399-2 is not applicable BS 6399 can be used Yes Stage 2: Design wind speed and dynamic pressure Notional site layout In accordance with §1.7.2 and the distances in Table A, the site is categorised as in-town terrain. We shall assume that the building is in an industrial estate, surrounded by buildings of a similar height, and is aligned with the ridge at 45° east of north. This gives four orthogonal cases – NW, NE, SE and SW (shown in Calculation 2) – that are the same as for the house in Part 2 (Figure C of Part 2). We shall assume that the spacing of the other buildings in the estate is unknown but typical. This gives an obstruction height Ho = 9.7 m and, from Q8 of Part 1, a spacing X = 20 m. Owing to the symmetry of the house, we need to consider only two orthogonal directions: q wind normal to the ridge θ = 0°, and q wind parallel to the ridge θ = 90°. Accordingly, the orthogonal cases SW for θ = 90° and NW for θ = 0° control the design, and the dynamic pressures for the other cases are not required. Stage 3: Pressure coefficients and design loads Pressure coefficients on walls The scaling dimensions for the walls of the portal frame building are determined in Calculation 3 from the dimensions given in Figure A. The pressure coefficients for the walls of the building are shown in Figure B for θ = 0° (Case NW) and in Figure C for θ = 90° (Case SW). 2.8 m +0.6 Windward face Windward4.3) (D/H = 4.3) ( D/H = face -0.4 -0.8 -0.1 -1.3 14 m Side face (isolated) Side face (isolated) Leeward face (Leeward face (D/H = 4.3) D/H = 4.3) Figure B Pressure coefficients for walls of the portal frame building θ = 0° Pressure coefficients on the roof The scaling dimensions for the walls of the portal frame building are determined in Calculation 4 from the dimensions given in Figure A. The pressure coefficients for the roof of the building are shown in Figure D (on page 4) for both orthogonal cases. Calculation 2 Dynamic pressures for the portal 3.9 m +0.6 Windward face Windward face (D/H = 4.3) ( D/H = 6.2) -0.4 Side face (isolated) Side face (isolated) -0.8 19.3 m Figure C Pressure coefficients for walls of the portal frame building θ = 90° -1.3 -0.1 Leeward face Leeward face (D/H = 4.3) ( D/H = 6.2)
  • 3. 3 Note B Ready-reckoner gives values of Sb directly, avoiding the use of Sc, St, Tc and Tt factors. frame building by Option 2(b) Clause Action 2.2.1 Notes From Figure 6, basic wind speed Vb = 22.6 m/s Hourly mean speed 10 m above flat open country From site plan, altitude of site ∆ = 120 m 1.3.3.1 2.2.2.2.1 2.2.2.2 2.1.1.2 2.2.2.3 Site less than halfway up hill From Equation 9, altitude factor Sa = 1.120 Orthogonal direction : NE SE SW 0.85 1.00 Topography not significant Range of NE includes 0°, 30°, 60° and 90° NW 0.99 NE value is biggest of From Table 3, select biggest direction factor in range Sd = 0.78 2.2.2.1 Take from OS 1:50,000 mapping Using Figure 7, check for significant topography = No 0.78, 0.73, 0.73 & 0.74 Ss and Sp taken as unity From Equation 8, site wind speed Vs = 19.74 21.52 25.31 25.06 m/s 1.3.3.4 Surrounded by similar buildings Select lowest obstruction height in range Ho = 9.7 9.7 9.7 9.7 m 20 20 20 m 7.6 7.6 7.6 m Spacing unknown but typical. Select furthest obstruction separation in range Xo = 20 E.2.1 displacement height Hd = 7.6 2.2.3.3 110 200 112 km 1.5 1.5 1.5 km Hr = 9.7 9.7 9.7 9.7 m Larger of He = Hr – Hd = 2.1 2.1 2.1 2.1 m or He = 0.4 Hr = 3.9 3.9 3.9 3.9 m Selected for range from Table A From Table A, shortest distance-in-town in range = 3.5 1.7.3.3 Hd = 1.2Ho – 0.2 Xo Selected for range from Table A From Table A, closest distance-to-sea in range = 110 1.7.3.1 See Q8 of Part 1 Defined in Annex E. From Q10 of Part 1, Safe assumption for whole building Reference height at ridge See Q10 of Part 1 Option 2(b): use Equation 29 or ready-reckoner 3.2.3.2.3 From Table 22, factor Sc = 0.839 0.839 0.839 0.839 Logarithmic interpolation used in From Table 22, factor S t = 0.197 0.197 0.197 0.197 Tables 22 and 23, but linear interpolation From Table 23, factor Tc = 0.702 0.722 0.722 0.722 is adequate From Table 23, factor Tt = 1.687 1.687 1.687 1.687 3.4.2.1 Standard value of factor g t = 3.44 2.2.3.3 From Equation 29, terrain-&- 3.44 3.44 Equivalent to CP3 Class A 3.44 Equation 29 allows for building factor S b = 1.260 1.296 1.296 1.296 2.2.3.1 Now gust speed, equivalent to CP3 Class A effective wind speed Ve = 24.9 2.1.2.1 actual distance-in-town From Equation 12, 27.9 32.8 32.5 m/s Gust dynamic pressure, From Equation 1, dynamic pressure qs = 379.4 476.5 659.5 646.4 Pa equivalent to CP3 Class A Calculation 3 Scaling dimensions for walls of the portal frame building Clause Wind parallel to ridge, θ = 90°, case SW Wind normal to ridge, θ = 0°, case NW 1.3.3.2 Height H = 9.65 m Height of gable Height H = 7 m 1.3.4.3 Breadth B = 30 m Figure A, B = W Breadth B = 60 m 1.3.4.4 Depth D = 60 m Figure.A, D = L Depth D = 30 m 2.2.3.2 H < B, one part, Hr = 9.65 m 2.4.1.2 Span ratio D/H = 6.2 2.4.1.3 Scaling length b = 19.3 m Peak of gable See Table 5 Smaller of B or 2H Height of eaves Figure A, B = L Figure A, D = W H < B, one part, Hr = 7 m Span ratio D/H = 4.3 Scaling length b = 14 m Eaves See Table 5 Smaller of B or 2H Calculation 4 Scaling dimensions for roof of the portal frame building Clause Wind parallel to ridge, θ = 90°, case SW Wind normal to ridge, θ = 0°, case NW 1.3.3.2 Height H = 9.65 m = Hr 2.5.2.2 Scaling length bW = 19.3 m Height of ridge Smaller of W or 2H Height H = 9.65 m = Hr Scaling length bL = 19.3 m Height of ridge Smaller of L or 2H
  • 4. 4 1.93 m Note C For the wind angle θ = 0° and roof pitch of α = 10° the pressure coefficients on the roof are always suctions. In determining horizontal components the asymmetric loads provisions of §2.1.3.7 will apply to the beneficial action of the suctions on the upwind roof pitch. (See Q24 of Part 1.) 7.5 m -1.8 -0.4 -0.4 -0.6 -0.35 7.5 m -1.3 -0.8 -0.8 -0.6 -0.6 -0.8 1.4 m -1.3 -0.45 -0.45 -0.6 -0.35 -1.8 -1.55 -1.55 -1.0 -1.0 -1.55 1.4 m 9.65 m 9.65 m Figure D Pressure coefficients for roof of the portal frame building θ = 0° and 90° 6m 6 m -1.3 -0.8 3.9 m Figure E Location and size of roller shutter door Internal pressure coefficients We shall assume that there is a 6 m-high roller shutter door spanning the full width of the end bay on one of the side walls. This will be assumed to be closed at the ultimate limit state, corresponding to Calculation 5. We will treat the door open as a serviceability limit state, corresponding to Calculation 6 (see Q46 of Part 1). Figure E shows the external pressure coefficients over the area of the door at wind angle θ = 90°. We shall take the typical porosity for curtain walling from Q44 of Part 1 as 3.5 × 10-4. We shall assume that all walls are equally permeable and the roof is impermeable. As the internal pressure, door closed, is set by the average flow of wind in and out of the distributed porosity, it is reasonable to take the average height of the walls as the reference height for internal pressure. Because the area of the side walls is twice the area of the gable walls, the average height is close to the eaves height and we shall use the eaves height as the reference height for the internal pressure. This is a pragmatic engineering decision. The reference height for the internal pressure caused by a dominant opening is the reference height for the wall in which the opening occurs, which is also the eaves height. Dynamic pressure Reference heights for the portal frame building are ridge height for the roof and the gable walls and eaves height for the side walls. The values at ridge height were derived in Calculation 2. Corresponding values of dynamic pressure at eaves height and for the serviceability limit are given in Table B. Table B Values of dynamic pressure for the portal frame building Reference height Ultimate Sp = 1 Serviceability Sp = 0.8 Application θ = 0° θ = 90° θ = 0° θ = 90° Hr = 7 m 547 Pa 558 Pa 350 Pa 357 Pa pe side walls, pi (see above) Hr = 9.7 m 646 Pa 660 Pa 414 Pa 422 Pa pe roof and gable Loading of roller shutter door Clearly wind loads occur on the roller shutter door only when it is closed, and values are derived in Calculation 7. We shall treat the door as a single structural element and use the average pressure coefficient over the area of the door. (See Q38 of Part 1.) Note E At higher values of roof pitch α the highestloaded purlin will be in the first (ridge) or second purlin on the downwind slope. Remember that the loaded zones are defined in plan so that the dimensions up the roof slope depend on the pitch angle. Highest-loaded purlin We shall assume that there are eight lines of purlins on each roof slope, spaced equally from eaves to ridge, giving a purlin spacing of 2.17 m in the plane of the roof. Inspection of the pressure coefficients in Figure D reveals that 6m 6m the highest-loaded purlin will be either the first (eaves) or second purlin on the upwind roof slope of the end bay, as 2.17 m 2.17 m shown in 1.42 m Calculations 8a (opposite) and 8b 1.93 m 9.65 m (on page 6). The relevant tributary Wind angleangle θ0˚ 0° Wind angle θθ= 90˚ = = Wind θ Wind angle = 90° areas are shown hatched in Figure F. Figure F Highest-loaded purlin
  • 5. 5 Note D With the door open, the large positive value of internal pressure for wind angle θ = 0° increases the net uplift on the roof, while the large negative value θ = 90° increases the load on the windward wall. Calculation 5 Internal pressure coefficient for the case with the door closed Clause Action Notes Internal pressure coefficient Cpi = –0.3 2.6.1.1 See Q43 of Part 1 From Equation 13, diagonal dimension a = 10 × 3√(volume of storey) = 10 × 3√(60 × 30 × (7 + 2.65/2)) = 248 m Includes roof space From Figure 4, size effect factor Ca = 0.646 2.1.3.4 Site in-town, line C Calculation 6 Internal pressure coefficient for the case with the door open Clause Action From Q46 of Part 1, probability factor Sp = 0.8 2.2.2.5 2.6.2 Notes 2 year return From Equation 15, diagonal dimension is greater of diagonal Opening controls dimension of opening = √(62 + 62) = 8.5 m, the value or 0.2 ×3√(volume of storey) = 5.0 m, a = 8.5 m From Figure 4, size effect factor Ca = 0.952 2.1.3.4 Site in-town, line C Area of opening A1 = 6 × 6 m = 36 m2 Total surface area A2 = 3176 m2 Ratio, opening area to sum of remaining openings = 36 / ( 3125 × 3.5 × 10–4) = 32 Much greater than 3 From Table 17, internal pressure coefficient Cpi = 0.9 × Cpe Wind angle θ = 0°, Cpi = +0.6 × 0.9 = +0.54 See Figure E Wind angle θ = 90°, Cpi = (–1.3 × 3.9 – 0.8 × 2.1) / 6 × 0.9 = –1.01 Calculation 7 Loads on the roller shutter door Clause Action Notes Diagonal dimension of door a = √(62 + 62) = 8.5 m 2.1.3.4 Site in-town, line C From Figure 4, size effect factor Ca = 0.952 θ = 0° Wind angle : 2.4.1 θ = 90° External pressure coefficient averaged over door Cpe = 0.6 2.1.3.1 From Equation 2, pe = qs × Cpe × Ca = 547 × 0.6 558 × –1.125 × 0.952 = 312 Pa 2.1.3.2 × 0.952 = –598 Pa From Equation 3, pi = qs × Cpi × Ca = 547 × –0.3 558 × –0.3 × 0.645 = –106 Pa 2.1.3.3 Cpi and Ca from × 0.645 = –108 Pa earlier calculation From Equation 4, net pressure p = pe – pi = 418 Pa 2.1.3.5 See Figure E –1.125 –490 Pa Load on door P = p × A = 15.0 kN –17.6 kN Ultimate limit Calculation 8a Highest-loaded purlin (ultimate limit state) Clause Action Notes Eaves purlin Diagonal dimension a = 6.1 m 2.1.3.4 6.4 m From Figure 4, size effect factor Ca = 0.982 0.978 Site in-town, line C 13.02 m2 Loaded area A = 6.51 m2 2.1.3.5 Second purlin True area on slope θ = 0° θ = 90° From Table B, External qs = 646 Pa 660 Pa Internal qs = 547 Pa 558 Pa Ultimate limit, door closed 2.1.2 Loaded purlin Eaves purlin Second purlin Second purlin 2.5.2.4 Averaged Cpe = –1.55 –0.62 –0.99 2.1.3.1 pe = qs × Cpe × Ca = –983 Pa –392 Pa –639 Pa pi = qs × Cpi × Ca = –106 Pa –106 Pa –108 Pa Equation 4, net pressure, p = pe – pi = –877 Pa –286 Pa –531 Pa –3.72 kN –6.91 kN 2.1.3.2 2.1.3.3 2.1.3.5 (Ca =0.646) Load on purlin, P = p × A = –5.71 kN
  • 6. 6 Calculation 8b Highest-loaded purlin (serviceability limit state) Clause Action Notes θ = 0° θ = 90° From Table B, External qs = 414 Pa 422 Pa Internal qs = 350 Pa 357 Pa Serviceability limit, door open 2.1.2 Loaded purlin Eaves purlin Second purlin Second purlin Averaged Cpe = –1.55 –0.62 –0.99 (From Calculation 6) Averaged Cpi = +0.54 +0.54 –1.01 pe = qs × Cpe × Ca = –630 Pa –253 Pa –412 Pa –343 Pa 2.5.2.4 2.1.3.1 2.1.3.2 (Ca = 0.952) pi = qs × Cpi × Ca = +180 Pa +180 Pa 2.1.3.3 Equation 4, net pressure p = pe – pi = –810 Pa –431 Pa –66 Pa –5.61 kN –0.86 kN 2.1.3.5 Load on purlin, P = p × A = –5.27 kN Loads on portal frames, wind normal to ridge We will calculate the individual member loads and the overall horizontal base shear for wind normal to the ridge (θ = 0°). The wind loading is transmitted to the portal frames as a series of point loads through the rail and purlin fixings as shown in Figure G. It is possible to sum the individual standard rail and purlin loads, then to apply the appropriate size effect factor for the frame. However, we shall adopt the quicker and commonly used practice of summing Figure G Rail and purlin loads on frame the load over the tributary area of cladding to give an equivalent uniformly distributed load (UDL). (See Q38 of Part 1.) Figure H shows the tributary area (hatched) for the horizontal base Note F 6m (Figure H) shear of a frame viewed in elevation and the corresponding diagonal Because pressure is a dimension for base shear, as recommended in Q20 and Figure C of scalar, we may determine the horizontal Part 1. We shall assume no load sharing between component of load on 9.65 m a frames. The load on the upwind roof slope is a beneficial component the roof by applying the roof pressures to the of the horizontal base shear, so we will need to apply the provisions of horizontally resolved §2.1.3.7. components of the surface area, as shown Inspection of the pressure coefficient zones leads us to expect the in Figure H. This is Figure H Tributary area for highest loading to occur on Frame 2 because the tributary area spans exactly the same as base shear of frame in elevation resolving the pressure the A–C and E–G zone boundaries, but we shall check also the loading load normal to the roof into its horizontal on the central frames (using Frame 4). First we obtain the unfactored component. internal and external pressures in Calculation 9. This and later calculations omit some of the intermediate steps already demonstrated (eg determining the zone areas, to save space and avoid unnecessary duplication). The loads in each member, expressed as a UDL, are found in Calculation 10 by applying the unfactored pressures from Calculation 9 to the relevant tributary areas and size effect factors. Similarly, Calculation 11 develops the horizontal shear at the base of each frame. We shall also calculate the member loads at the serviceability limit with the door open on the windward face. Calculation 12 gives the unfactored pressures and Calculation 13 develops these into member loads expressed as a uniform distributed load. Comparing Calculation 13 with the previous Calculation 10 we find, as expected, that the Note G When summing loads on a number of components (eg gable, top and bottom storeys) it is convenient to determine the standard loads on each component by taking Ca = 1, then to apply the value of Ca appropriate to the total loaded area after summation. Calculation 9 Unfactored pressures for θ = 0°, case NW, ultimate limit, door closed Clause Action 2.1.3.2 2.1.3.4 2.1.3.1 Unfactored* internal pressure, pi = qs Cpi = 547 × –0.3 = –164 Pa Size effect factor for internal pressure, Ca = 0.646 Averaged unfactored* external pressures (UDL): Frame 2 Frame 4 on upwind wall, pe front = qs × Cpe front = 328.2 Pa –54.7 Pa *Apply Ca later. See Note G 328.2 Pa on downwind wall, pe rear = qs × Cpe rear = –54.7 Pa Notes on upwind roof slope, pe upw = qs × (CpeA × AA + CpeB × AB + CpeC × AC) / A = –358.0 Pa walls,qs = 646 Pa –324.3 Pa on downwind roof slope, pe dnw = qs × (CpeE × AE + CpeF × AF + CpeG × AG) / A = –282.9 Pa qs = 547 Pa for for roof, from Table B –270.6 Pa
  • 7. 7 Calculation 10 Member loads expressed as UDL for θ = 0°, case NW, ultimate limit, door closed Clause Action Notes Frame 2 Frame 4 Columns: diagonal dimension a = √(72 + 62) = 9.2 m 0.945 Rafters: a = √ ((15/cos(10°))2 + 62) = 16.4 m 16.4 m From Figure 4, size effect factor Ca = 0.892 2.1.3.5 9.2 m From Figure 4, size effect factor Ca = 0.945 2.1.3.4 0.892 Upwind column UDL = (pe front × Ca – pi × Ca) × 6 = 2497 N/m Downwind column UDL = (pe rear × Ca – pi × Ca) × 6 = 326 N/m Upwind rafter UDL = (pe upw × Ca – pi × Ca) × 6 = –1280 N/m Downwind rafter UDL = (pe dnw × Ca – pi × Ca) × 6 = –878 N/m Note H The tributary area of Frame 2 spans the A–C and E–G zone boundaries, so the members of Frame 2 are more highly loaded than Frame 4. But the higher A zone values are beneficial and are reduced to 60% by §2.1.3.7, so the base shear for Frame 4 and the other central frames is marginally higher than for Frame 2. The internal pressure has no effect on the base shear because its effect cancels out over the inside surfaces. between frames would reduce Ca 2497 N/m 326 N/m Expressed as force –1100 N/m per unit length –813 N/m Calculation 11 Horizontal base shear for for θ = 0°, case NW, ultimate limit, door closed Clause Action Notes Frame 2 Diagonal dimension a = √(9.652 + 62) = 11.4 m 2.1.3.4 From Figure 4, size effect factor Ca = 0.925 2.1.3.5 Frame 4 11.4 m 0.925 Site in-town, line C *Apply Ca later Unfactored* loads: Pfront = pe front × Afront = pe front × 6 × 7 = 13 784 N 13 784 N Adverse Prear = pe rear × Arear = pe rear × 6 × 7 = –2297 N –2297 N Adverse Pupw = pe upw × Aupw = pe rear × 6 × 15 × tan(10°) = –5681 N –5146 N Beneficial Pdnw = pe dnw × Adnw = pe rear × 6 × 15 × tan(10°) = –4489 N –4294 N Adverse 2.1.3.6 Note I (Calculation 11) Load sharing Base shear P = 0.85 × (ΣPfront – ΣPrear) × Ca × (1 + Cr) = 2.1.3.7 0.85 × (Pfront + Pupw × 0.6 – Pdnw – Prear) × Ca × 1.035 = 13.97 kN 14.07 kN See Note G See Note H Calculation 12 Unfactored pressures for θ = 0°, case NW, serviceability limit, door open Clause Action Notes Unfactored* internal pressure pi = qs × Cpi = 350 × 0.54 = 189 Pa 2.1.3.2 Size effect factor for internal pressure Ca = 0.955 2.1.3.4 2.1.3.1 Averaged unfactored* external pressures (UDL) : Frame 2 Frame 4 on upwind wall pe front = qs × Cpe front = 210 Pa –35 Pa see Note G 210 Pa on downwind wall pe rear = qs × Cpe rear = –35 Pa *Apply Ca later, on upwind roof slope pe upw qs = 350 Pa for walls, qs = 414 Pa = qs × (Cpe A × AA + Cpe B × AB + Cpe C × AC) / A = –229.4 Pa –207.9 Pa on downwind roof slope pe dnw for roof, from Table B = qs × (Cpe E × AE + Cpe F × AF + Cpe G × AG) / A = –181.3 Pa –173.4 Pa Calculation 13 Member loads expressed as UDL for θ = 0°, case NW, serviceability limit, door open Member loads, expressed as UDL Clause Notes Action Frame 2 Frame 4 Columns: diagonal dimension a = √(72 + 62) = 9.2 m 9.2 m From Figure 4, size effect factor Ca = 0.945 2.1.3.4 0.945 Rafters: a = √((15/cos(10))2 + 62) = 16.4 m From Figure 4, size effect factor Ca = 0.892 2.1.3.5 Upwind column UDL = (pe front × Ca – pi × Ca) × 6 = 111 N/m 16.4 m 0.892 111 N/m Downwind column UDL = (pe rear × Ca – pi × Ca) × 6 = –1278 N/m –1278 N/m Upwind rafter UDL = (pe upw × Ca – pi × Ca) × 6 = –2307 N/m –2192 N/m Downwind rafter UDL = (pe dnw × Ca – pi × Ca) × 6 = –2050 N/m –2008 N/m Load sharing between frames would reduce Ca Expressed as force per unit length dominant opening on the windward face alters the balance of loads between the members. The windward column is less highly loaded, but the leeward column and rafters are more highly loaded. Whether this is significant will depend on the action of the partial factors for ultimate and serviceability loads in combination with the other dead and live loads.
  • 8. 8 Fifteen-storey tower surrounded by a two-storey podium L = 20 m The office tower consists of a 15-storey tower surrounded by a 2-storey podium. The storey height is 3 m and the tower has a 2 m-high parapet. The principal dimensions of the office tower in this example are given in Figure I. W = 15 m H = 47 m H= 6 m W = 40 m L = 40 m Figure I Principal dimensions of the office tower Stage 1: Dynamic classification Calculation 14 determines whether it is appropriate to use BS 6399-2 and obtain the value of Cr. Note J The value of Cr given by the classification indicates that the building is only 5% dynamic. Calculation 14 Dynamic classification of the office building Clause Action Notes 1.3.3.2 Building height above its base H = 47 m Use height to top of parapet 1.6.1 Read value of Kb from Table 1 Kb = 1 When in doubt, take next larger value Using H and Kb, read Cr from Figure 3 Cr = 0.05 If Cr > 0.1, get better value from Annex C. Check Cr < 0.25 Yes If Cr > 0.25, BS 6399-2 is not applicable 1.6.2 Note K The combination of X = 20 m and maximum funnelling is not consistent but, since the spacing is unknown, it is a conservative assumption. Stage 2: Design wind speed and dynamic pressure Notional site layout In accordance with §1.7.2, and the distances in Table A, the site is categorised as in-town terrain. We shall assume that the building is in the commercial area of town and is aligned with the long axis at 45° east of north. This gives four orthogonal cases NW, NE, SE and SW that are the same as for the house in Part 2 (Figure C of Part 2). We shall assume that the surrounding buildings are a mixture of two and three-storeys (2.5 storeys on average), and that their spacing is unknown but typical. From Q8 of Part 1, this gives an obstruction height Ho = 2.5 × 3 = 7.5 m and a spacing X = 20 m. We shall further assume, since the spacing is unknown, that funnelling (see Q40 of Part 1) is at maximum along the sides of the podium. The design dynamic pressure at the top of the tower, Hr = 47 m, is determined in Calculation 15. As the tower and podium will be divided into a number of parts, we shall require dynamic pressures at a number of other reference heights, but their calculation is left as an exercise for the reader. Owing to the symmetry of the office tower, we need to consider only two orthogonal directions: q wind normal to the long face, θ = 0°, and q wind normal to the short face, θ = 90°. 4m -1.3 2.4 m 0.8 -0.8 -0.3 -1.6 0.6 -0.9 Windward face face Windward Side faceSide face -0.1 3m -1.3 2.4 m 0.8 -0.8 -0.3 -0.9 -0.1 -1.6 Windward face face Windward Side faceSide face Accordingly, the orthogonal cases SW for θ = 90° and NW for θ = 0° control the design, and the dynamic pressures for the other cases are not required. Leeward face face Leeward Wind angle θ = 0° 0.6 BS 6399-2 can be used Leeward face face Leeward Wind angle θ = 90° Figure J Pressure coefficients for the walls of the office tower and podium from Table 5 Stage 3: Pressure coefficients and design loads Pressure coefficients on walls and roofs The scaling dimensions for the office tower are determined in Calculation 16 and for the podium in Calculation 17 (on page 10). Figure J shows the pressure coefficients for the walls of the office tower and podium.
  • 9. 9 Note L The ready-reckoner gives the values of Sb directly, avoiding the use of Sc, S t, Tc and Tt factors. Calculation 15 Dynamic pressures for office building by Option 2(b) Clause Action Notes 2.2.1 From Figure 6, basic wind speed Vb = 22.6 m/s Hourly mean speed 10 m above flat open country 1.3.3.1 2.2.2.2.1 2.2.2.2 2.1.1.2 2.2.2.3 From site plan, altitude of site ∆ = 120 m Site less than halfway up hill From Equation 9, altitude factor Sa = 1.120 Orthogonal direction : NE Topography not significant Range of NE includes 0°, 30°, 60° and 90° SE SW NW 0.85 1.00 0.99 NE value is biggest of From Table 3, select biggest direction factor in range Sd = 0.78 2.2.2.1 Take from OS 1:50,000 mapping Using Figure 7, check for significant topography = No 0.78, 0.73, 0.73 and 0.74 Ss and Sp taken as unity From Equation 8, site wind speed Vs = 19.74 21.52 25.31 25.06 m/s 1.3.3.4 Surrounded by similar buildings Select lowest obstruction height in range Ho = 7.5 7.5 7.5 7.5 m 20 20 20 m 5.0 5.0 5.0 m Spacing unknown, but typical. Select furthest obstruction separation in range X o = 20 E.2.1 displacment height Hd = 5.0 2.2.3.3 110 200 112 km 1.5 1.5 1.5 km 47 47 47 m 42 42 42 m Selected for range from Table A From Table A, shortest distance-in-town in range = 3.5 1.7.3.1 1.7.3.3 Other reference heights will be needed Reference height at parapet Hr = 47 See Q10 of Part 1 From Q10 of Part 1, effective height He = Hr – Hd = 42 Hd = 1.2Ho – 0.2 Xo Selected for range from Table A From Table A, closest distance-to-sea in range = 110 See Q8 of Part 1 Defined in Annex E. From Q10 of Part 1, Option 2(b): use Equation 29 or ready-reckoner 3.2.3.2.3 From Table 22, factor Sc = 1.263 1.263 1.263 1.263 Logarithmic interpolation used in From Table 22, factor S t = 0.151 0.151 0.151 0.151 Tables 22 and 23, but linear interpolation From Table 23, factor Tc = 0.911 0.937 0.937 0.937 is adequate From Table 23, factor Tt = 1.286 1.197 1.197 1.197 3.4.2.1 Standard value of factor g t = 3.44 2.2.3.3 From Equation 29, terrain-&- 3.44 3.44 Equivalent to CP3 Class A 3.44 Equation 29 allows for building factor S b = 1.918 1.918 1.918 1.918 2.2.3.1 effective wind speed Ve = 37.9 2.1.2.1 actual distance-in-town Now gust speed, equivalent to CP3 Class A From Equation 12, 41.3 48.5 48.1 m/s Gust dynamic pressure, From Equation 1, dynamic pressure qs = 879 1044 1444 1416 Pa Calculation 16 Scaling dimensions for the office tower Clause Wind angle, θ = 0° equivalent to CP3 Class A 90° Notes From roof of podium 1.3.3.2 Height of tower H = 41 m 41 m 1.3.4.3 Breadth B = 20 m 15 m Depth D = 15 m 20 m 1.3.4.4 2.2.3.2 Aspect ratio H/B = 2.1 2.7 Both H/B > 2. See Figure T 2.4.1.2 Span ratio D/H = 0.37 0.49 D/H < 1. See Table 5 2.4.1.3 Scaling length b = 20 m 15 m Smaller of B or 2H parapet height h = 2 m 2m Assume 2 m high parapet. 0.13 See Table 8 2.5.1.4 parapet height ratio h/b = 0.1 Figure K (on page 10) shows the zones of pressure coefficient on the roof of the tower and the podium for each wind direction. Note that the parapet of the tower reduces values in the A and B zones by a different amount in each wind direction because the ratio h/b differs.
  • 10. 10 Calculation 17 Scaling dimensions of the podium Clause Wind angle, θ = 0° = 90° Notes From roof of podium 1.3.3.2 Height of podium H =6m =6m 1.3.4.3 Breadth B = 40 m = 40 m 1.3.4.4 Depth D = 40 m = 40 m 2.2.3.2 Aspect ratio H/B = 0.15 = 0.15 = 10 = 10 D/H > 4. See Table 5 = 12 m = 12 m Smaller of B or 2H 2.4.1.2 Span ratio D/H 2.4.1.3 Scaling length b Clause 2.5.1.7 requires that we consider the effect of the presence of the tower on the pressure coefficients for the podium roof. The zones of additional pressure coefficients from §2.5.1.7 are given in Figure L and must be used with the dynamic pressure for the tower walls. Because the height of the tower gives a large dynamic pressure, these additional zones may control the design pressures on the podium roof. 1.2 m 6m 3m ±0.2 -2.0 -1.65 -1.14 ±0.2 10 m -0.7 2m -0.7 5m 1.2 m ±0.2 5m -1.75 6m ±0.2 -0.7 -1.4 -2.0 -1.2 5m -0.7 7.5 m 1.5 m -2.0 -2.0 Both H/B < 1, one part 3m Dynamic pressure Following §2.2.3.2, the value aspect ratio H/B Wind angle θ = 90° Wind angle θ = 0° for each direction allows the tower to be split Figure K Pressure coefficients for the roofs of the office tower and podium, into the parts shown in Figure M. These parts are used for determining overall lateral loads from Table 8 only and not cladding (see Q34 of Part 1). The Note M middle part, if it exists, may be split into as many slices as desired. We shall align the slices to the 3 m (Dynamic pressure) storey heights. The effective height for each part is taken as the height of the top of the part. Relevant Clause 2.2.3.2 does not values of design dynamic pressure for each effective height are given in Table C. apply to cladding and 3m 3m -1.4 components. See Q34 of Part 1. -0.3 10 m -0.8 -0.8 4m -1.3 7.5 m 0.8 -1.3 -0.8 7.5 m 7.5 m -0.3 3m -1.3 0.8 10 m -1.3 Wind angle θ = 0° -0.8 7.5 m Wind angle θ = 90° Figure L Additional pressure coefficients for the podium roof from §2.5.1.7 Hr = 47 m Hr = 47 m H r = 32 m H r = 26 m Hr = 30 m H r = 27 m 27 m 24 m H r = 21 m Hr = 6 m Hr = 6 m Wind angle θ = 0° Figure M Reference heights for parts of the office tower Wind angle θ = 90°
  • 11. 11 Note N The reader may confirm the values in Table C by substituting each reference height into Calculation 15 and recalculating the results. Table C Values of dynamic pressure for the office building Reference θ = 0° θ = 90° Application height Case NW Hr = 6 m Case SW 503 Pa 513 Pa Podium, structure and cladding Hr = 21 m 1161 Pa Lowest part of tower, θ = 90° Hr = 24 m 1214 Pa 1st storey of middle part of tower, θ = 90° Hr = 26 m Lowest part of tower, θ = 0° 1216 Pa Hr = 27 m 1230 Pa Middle part of tower, θ = 0°. 1255 Pa 2nd storey of middle part of tower, θ = 90° Hr = 30 m 1292 Pa 3rd storey of middle part of tower, θ = 90° Hr = 32 m 1315 Pa Top of middle part of tower, θ = 90° 1444 Pa Upper part of tower, θ = 0° & 90° and cladding of tower Hr = 47 m 1416 Pa Horizontal shear forces on tower for each storey In determining the horizontal shear forces, §2.2.3.2 permits the tower to be divided into the parts defined above. Figure 5(c) of the Standard shows that the diagonal dimension for the horizontal shear at any level is the diagonal of the loaded area above that level in horizontal projection. Table D gives the calculation of horizontal shear PH at the base of each storey (including the height of the parapet) above the level of the podium roof as follows: q qs is the dynamic pressure for each storey from Table C; q Pstorey is the unfactored horizontal load on each storey (see Note G); q a is the diagonal dimension of the loaded area of the loaded area above the base of the storey; q Ca is the size effect factor of the loaded area, using line B of Figure 4 above 10 m and line C below 10 m (see the key to Figure 4); q PH is the overall horizontal shear at the base of each storey from Equation 7 of the Standard (§2.1.3.6), using the dynamic pressure qs for the appropriate part and Cr = 0.05 from the dynamic classification. Note O Loads on the parapets should be determined as if they were freestanding walls (§2.5.1.4.2) using the provisions of §2.8.1. The dynamic augmentation factor Cr does not apply to the parapets. Table D Horizontal shear at the base of each storey of the office tower Storey Wind direction 0° Wind direction 90° qs Pstorey a (Pa) (kN) (m) 15 1416 155.8 20.6 14 1416 93.5 21.5 13 1416 93.5 12 1416 93.5 11 1416 10 9 Ca PH qs Pstorey a (kN) (Pa) (kN) (m) Ca PH 0.893 124 1444 116.8 15.8 0.913 95 0.890 198 1444 70.1 17.0 0.908 151 22.8 0.885 271 1444 70.1 18.6 0.901 207 24.4 0.880 343 1444 70.1 20.5 0.893 261 93.5 26.2 0.875 414 1444 70.1 22.7 0.886 314 1230 81.2 28.3 0.869 474 1292 60.9 25.0 0.878 359 1266 83.6 30.5 0.864 535 1255 62.7 27.5 0.871 405 8 1216 80.3 32.8 0.858 593 1214 60.2 30.0 0.865 449 7 1216 80.3 35.2 0.853 651 1161 60.2 32.6 0.858 491 6 1216 80.3 37.7 0.847 707 1161 60.2 35.3 0.852 533 5 1216 80.3 40.3 0.842 763 1161 60.2 38.1 0.847 576 4 1216 80.3 42.9 0.805 788 1161 60.2 40.9 0.810 594 3 1216 80.3 45.6 0.800 840 1161 60.2 43.7 0.804 633 (kN) Loads on cladding panels of tower We shall assume that the tower is clad with panels that span the 3 m storey height and are either 3 m or 6 m wide. We shall also assume that the porosity of each face is equal and that each storey is open plan, giving Cpi = –0.3. The maximum loads will occur: q inwards, at any position on the windward face, and q outwards, at the upwind edge of the side face.
  • 12. 12 From Figure J, we see that on the side face, the 3 m-wide panel will be entirely within the edge A zone, while the 6 m-wide panel will contain the A zone and part of the B zone in proportions A:B = 1:1 on the long face and 3:2 on the short face. We shall calculate this second case (Calculation 18) that occurs for θ = 0°, Case NW, leaving the other case as an exercise for the reader. Note P In this example we find that the largest load on the smaller panel is outwards (suction) while the largest load for the larger panel is inwards (pressure). This is due to the proportion of the panel that falls into the A zone on the side wall. Calculation 18 Cladding panel loads on the office tower, for θ = 0°, Case NW Clause Action From Table B, for θ = 0°, Hr = 47 m, qs = 1416 Pa 2.1.2 Notes See Q34 of Part 1 Internal pressure Internal pressure coefficient Cpi = –0.3 2.6.1.1 See Q43 of Part 1 From Equation 13, diagonal dimension a Above 10 m, line B. = 10 × 3√(volume of storey) = 10 x 3√(20 x 15 x3) = 96.5 m (See key to Figure 4) From Figure 4, size effect factor Ca = 0.777 2.1.3.4 From Equation 3, pi = qs × Cpi × Ca = –330 Pa 2.1.3.2 External pressure Windward Side Panel size = 3 × 3 m 3×6m 3×3m 3×6m Diagonal a = 4.2 m 6.7 m 4.2 m 6.7 m Above 10 m, line B. Size effect Ca = 1.00 0.978 1.00 0.978 (See Figure 4) 2.4.1 Averaged Cpe = 0.8 0.8 –1.3 –1.1 2.1.3.1 pe = qs Cpe Ca = 1133 Pa 1108 Pa –1841 Pa –1523 Pa 2.1.3.4 2.1.3.3 p = pe – pi = 1463 Pa 2.1.3.5 1438 Pa –1511 Pa –1193 Pa Area of panel A = 9 m2 18 m2 9 m2 18 m2 Load P = p × A = 13.2 kN 25.9 kN –13.6 kN See Figure J. Averaged over panel –21.5 kN Concluding comments The worked examples in Parts 2 and 3 of this Digest illustrate much of the guidance given in Part 1 and demonstrate the steps involved in implementing BS 6399-2 for various elements of typical structures. The examples have been chosen to illustrate as many aspects as possible within the space available, but are not complete. Although laid out as manual calculations, the examples were produced using spreadsheets that gave exact solutions to the methods used by BS 6399-2. It is not possible for manual calculations to reproduce these values exactly, for several reasons. q The values of the S-factors in the BS 6399-2 tables are only accurate to 2%. q Values interpolated from these tables will depend on whether effective height is interpolated before or after distancefrom-sea or in-town and whether interpolation is made on a linear or logarithmic scale. q Using values rounded to three significant figures will accumulate round-off errors. Check calculations of the examples made manually should be considered successful if the effective wind speed Ve is within 0.2 m/s and the dynamic pressure, surface pressure or load is within 5% of the example values. Linear interpolation is much easier to apply than logarithmic interpolation. For example, Example 2 in Part 2 gives S b = 1.43 for an effective height of 4.4 m in open country, > 100 km to the sea. Interpolating in Table 4 gives: for H e = 2 m: S b = 1.26 for H e = 5 m: S b = 1.45 for H e = 4.4 m: linear interpolation gives S b = 1.26 + (1.45 – 1.26) x (4.4 –) / (5 – 2) = 1.412 logarithmic interpolation gives S b = 1.26 + (1.45 – 1.26) x (log(4.4) – log(2)) / (log(5) – log(2)) = 1.26 + (1.45 – 1.26) x (0.643 – 0.301) / (0.699 – 0.301) = 1.423 Linear interpolation for effective height underestimates while linear interpolation for distance-to-sea and distance-in-town overestimates. Where values are given in logarithmic steps, it is better to use logarithmic interpolation. This particular choice of examples does not imply that explicit calculations are required for those elements that are permitted to be sized using the prescriptive methods given in Approved Document A to the Building Regulations and in BS 8103. Publications cited in this Part will be found under Other references in Part 1. Technical enquiries to: BRE Enquiries Garston, Watford, WD2 7JR Tel 01923 664664 Fax 01923 664098 Digests Good Building Guides Good Repair Guides Information Papers are available on subscription. For current prices please contact: Construction Research Communications Ltd, 151 Rosebery Avenue, London, EC1R 4GB. E-mail: crc@construct.emap.co.uk Tel 0171 505 6622 Fax 0171 505 6606 Full details of all recent issues of BRE publications are given in Constructing the future, sent free to subscribers. © Copyright BRE 1999 ISBN 1 86081 271 6 Published by Construction Research Communications Ltd by permission of Building Research Establishment Ltd. Applications to copy all or any part of this publication should be made to Construction Research Communications Ltd, PO Box 202, Watford, WD2 7QG