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P. 1BS+Wind+Application+BS+6399+P2[1]BS+Wind+Application+BS+6399+P2[1]|Views: 644|Likes: 25Published by taka731220More info:Published by: taka731220 on May 31, 2010Copyright:Attribution Non-commercialAvailability:Read on Scribd mobile: iPhone, iPad and Android.download as PDF, TXT or read online from ScribdFlag for inappropriate content|Add to collectionSee moreSee lesshttps://www.scribd.com/doc/32255506/BS-Wind-Application-BS-6399-P2-104/24/2013pdftextoriginal﻿18Design of Structural Steelwork 1.9 Application of Wind Loads Two methods of analysis for determining the equivalent static wind loads on structures are given in BS 6399:Part 2: 1995, they are: • the standard method • the directional method The standard method is a simplified procedure and is intended to be used by designers undertaking hand-based calculations. This method gives conservative results for a wide range of the most commonly used structures. The directional method, while assessing wind loads more accurately, is more complex and is intended for use with computational analysis; it is not considered in this text. Structures which are susceptible to dynamic excitation by virtue of their structural properties, e.g. mass, stiffness, natural response frequencies or structural fonn such as slender suspended bridge decks or long span cable stayed roofs will generally require more complex mathematical analysis techniques andlor wind tunnel testing. The standard method is illustrated in Examples 1.6 to 1.8. 1.10 Example 1.6 Storage hopper A closed top storage hopper as shown in Figure 1.8 is situated in an industrial development near Edinburgh and adjacent to the sea. Assuming the altitude of the location to be 5.0 m above mean sea level, determine the overall horizontal wind loading on the structure, while considering the wind to be acting in the direction indicated. 8.0 m r 8.0 m Wi nd rl-==""=h direction lio----n I 30 m r::=:> \ i 8.0 m L ~ Wind direction r::=:> Plan Elevation Figure 1.8 Solution: Clause 2.2.3.2 Since the crosswind breadth (8.0 m) is less than the height (30.0 m) a reduction in lateral loading is permitted, Figure ll(e) ofBS 6399:Part 2 H== 30 m, B == 8.0 m :. H> 2B Structural Steelwork 19 Consider the building surface to be divided into a number of pmts A, B, C and D Area = 64 m2 I Area = 56 m2 I Area = 56 m2 I Area = 64 m2 Figure 1.9 Clause 2.1.3.6 The overall load on the building where: P = 0.85(LPfront - r.,Pre.1r) (1 + Cr) LPji'ont is the SUlll of the horizontal components of surface load on the windward facing wall is the SUlll of the horizontal components of surface load on the leeward facing wall is a dynamic augmentation factor sr.; c, Clause 2.1.3.5 Net load on an area of surface where: p is the net pressure across the surface A is the area of surface being considered P=pA Clause 2.2.1 Figure 6 For Edinburgh the basic wind speed Vb = 23.5 m/sec Clause 2.2.2.2.2 Figure 7 Assuming that the topography of the site is not considered significant altitude factor S. = 1 + O.OOlfl. = 1 + (0.001 x 5) S. = 1.005 Clause 2.2.2 site wind speed Vs = Vb X Sa X S« X S, X Sp In many cases the direction factor (Sd), seasonal factor (S.) and probability factor (Sp) can be considered to be equal to 1.0 (see Clauses 2.2.2.3 to 2.2.2.5). Vs = (23.5 x 1.005) = 23.62 m/sec 20 Design of Structural Steelwork Clause 2.2.3 effective wind speed v;, = Vs X Sb where: Sb is the terrain and building factor obtained from Clause 2.2.3.3 and Table 4 Clause 2.2.2.3 Table 4 The effective wind speed for each of the areas A, B, C and D can be detennined assuming an effective height He equal to the reference height H, to the top of each area. Table 4 Area A AreaB AreaC AreaD He == 8.0m He= I5.0m He == 22.0m He=30.0m Sb == 1.73 So = 1.85 Sb == 1.91 Sb = 1.96 Effective wind speeds Area A AreaB AreaC AreaD v;, = 23.62 x 1.73 = 40.9 m/sec v;, == 23.62 x 1.85 = 43.7 rn/sec v;, == 23.62 x 1.91 == 45.1 mlsec Ve = 23.62 x 1.96 == 46.3 m/sec Clause 2.1.2.1 Table2 Dynamic wind pressure q. = 0.613v:,2 Area A q. == 0.613 x 40.92 == 1.03 kNI m2 Area B qs = 0.613 x 43.70 = 1.17 kN/m2 Area C q. = 0.613 x 45.10 = 1.25 kN/m2 Area D qs = 0.613 x 46.30 == 1.31 kN/m2 Clause 2.1.3.1 External surface pressure p; = qsCpeC. where: Cpe is the external pressure coefficients (Clause 2.4) C. size effect factor (Clause 2.1.3.4) Note: In this problem it is not necessary to consider internal pressure coefficients since the overall horizontal loading is being considered. Internal pressure coefficients are considered in Example 1.7. Clause 2.1.3.4 Figures 4 and 5 Diagonal dinlension a == ~h02 + 82 == 31.05 m From Figure 4 use line A in the graph to determine C. ~ 0.89 Structural Steelwork Clause 2.4 Table 5 Figure 12 and Table 5 D = _!. = 0.27 < 1.0 H 30 Cpe windward face = + 0.8 Cpe leeward face = - 0.3 Note: +ve indicates pressure on a surface -ve indicates suction on a surface In this problem the wind loading on the side faces and roof are not being considered. External surface pressure Pfront ::;;; q. x 0.8 x 0.89 == + 0.712qs prear = q. x 0.3 x 0.89 = - 0.267q. Area A pfront = + 0.712 x 1.03 = + 0.73 kN/m2 . prear = - 0.267 x 1.03 = - 0.28 kN/m2 Area B pfront = +0.712 x 1.17 ::: + 0.83 kN/m2 prear ::;;; - 0.267 x 1.17 :;;;; - 0.31 kN/m2 AreaC Pfront = + 0.712 x l.25 = + 0.89 kN/m2 pre:ar = - 0.267 x l.25 = - 0.33 kN/m2 Area A pfront = +0.712·x 1.31 = + 0.93 kN/m2 prear = - 0.267 x 1.31 = - 0.35 kN/m2 Clause 2.1.3.5 Net load on building surface area = p x loaded area Area A Pfront = 0.73 x 64= + 47.72 kN r.: = -0.28 x 64 = - 17.92 F AreaB Pfront = 0.83 x 56= + 46.48 kN r.; = - 0.31 x 56 = _ 17,5C. Area C Prront = 0.89 x 56= + 49.84 kN r.: - 0.33 x 56 = _ ,g. <fR Area A Pltont :;;;; 0.93 x 64= + 59.52 kN r.; = - 0.35 x 64 I::. - '22." "1;b Clause 2.1.3.6 The overall horizontal load on the buildingP = 0.85(2:Pfront -~ fV'e'tV" )( It c.t?) Clause 1.6.1 Table 1 Figure 3 Table 1 Figure 3 Building type factor Dynamic augmentation factor "ZPfront 2:Prear ::: + (47.72 + 46.48 +. = - (17.92 + 17.36 + 18. Overall horizontal load P = 0.85(203.56 - (-76.16>') (\ +0.04) ~ P = 247.3kN 21 22 Design a/Structural Steelwork 1.11 Example 1.7 Industrial warehouse An industrial warehouse is to be designed comprising a series of three-pinned pitched roof portal frames as shown in Figure 1.10. Using the data provided determine the wind loading on the structure. Design Data: Location Altitude of site Closest distance to sea Overall length of building Centres of frames open country near Preston 20.0 m above mean sea level 8.0km 24.0m 4.0111 Frames at 4.0 m centres Pinned base Pinned base L-----20.0 m~---Typical internal frame Figure 1.10 Solution: Clause 2.2.3.2 Figure ll(a) H= 10 m B=24m H<B . '. consider building as one part Clause 2.2.1 Figure 6 Location Preston Basic wind speed Vb = 23.0 misec Clause 2.2.2.2.2 Figure 7 Assuming that the topography is not considered significant 6.s = 20.0 In Altitude factor Sa = 1 + (0.001 X 20.0) = l.02 Clause 2.2.2 Site wind speed V. = Vi, X Sa X Sd X S. X Sp Assuming Sd = S. = Sp = 1. 0 V. = 23.0 X 1.02 = 3.46 mlsec Structural Steelwork Clause 2.2.3 Table 4 Effective wind speed Clause 2.2.3.3 Table 4 He = H, = actual height closest distance to sea Ve = 23.46 x 1.78 = 10.0 m (see Clause 1.7.3.2) == 8 km ., Sb = 1.78 = 41.76 m/sec Clause 2.1.2.1 Dynamic wind pressure q, = 0.613 Ve2 = (0.613 X 41.762)1103 = 1.07 kN/m2 Clause 2.1.3.1 Extemal surface pressures pe = q.CpeCa Clause 2.1.3.2 Intemal surface pressures Pi = qsCpiCa External Pressure Coefficients Consider the wind acting on the longitudinal face of the building !}_ = 20 ;: 2.0 H 10 20.0 m 11 wind direction Figure l.11 Clause 2.4 Table 5 The values of Cpe in Table 5 can be interpolated in the range 1 < D < 4 H leeward face windward face Cpo = - 0.23 Cpe = + 0.73 t 0.2 D t 0.7 Figure 1.12 In the case of the gable faces, different values of Cpo should be used depending on the gap between adjacent buildings; in this example assume that the building is isolated. 23 24 Design afStructural Steelwork Clause 2.4.1.3 Figure 12 The surfaces on which the wind is blowing are considered as separate zones which are defined in Figure 12 and depend on a variable 'b', where: b ::; crosswind breadth ::: 24.0 m ::; 2 x height = 2 x 10.0 ::: 20.0 m Use the smaller value of b i.e. b ::: 20.0 m Figure 12(b) D = b .', consider the gable surface as two zones A and B TIle width of zone A is equal to 0.2 x b = The width of zone B is equal to 0.2 x 20 = (20.0 - 4.0) = 4.0m = 16.0 m 4m zone zone A B gable faces zone A zone B Cpe = -1.3 Cpe = -0.8 6m Figure l.13 1.3+-D~ 1.3 D.8+- ~ 0.8 Figure 1.14 Consider the wind acting on the gable face of the building c::::J wind direction 20.0 m D 24 - == - == 2.4 H 10 6 bays at 4.0 m :;: 24.0 m Figure 1.15 Structural Steelwork 25 Using Table 5 and interpolating as before windward face leeward face Cpe = +0.71 Cpe = -0.21 0.71 --> 0--> 0.21 Figure 1.16 Clause 2.4.1.3 b S crosswind breadth S 2 x height Use the smaller value of b i.e, "" 20.0m = 2 x 10.0 = 20.0m 20.0m b = Figure 12(b) D = 24.0 > b .. consider the longitudinal surface as three zones A, B and C. The width of zone A is. equal to 0.2 x b The width of zone B is equal to 0.8 x b The width of zone C is equal to D - b = 0.2 x 20 = 4.0 rn = 0.2 x 20 = 16.0m = 24 - 20 = 4.0 m zone A zoneB zoneC Longitudinal elevation Figure 1.17 zone A zoneB zone C 1.3 0.8 0.4 iii c; - 1.3 D Cpe = - 0.8 Cpe = -0.4 ~ ~ ~ 1.3 0.8 0.4 Figure 1.18 Clause 2.5.2.4 Table 10 Figure 20 TIns clause defines the external pressure coefficients for the roof of buildings Figure 20 pitch angle a = tan-I _±_ = 2l.8° 10 Clause 2.5.2.2 tins clause defines the loaded zones which are indicated in Figure 20 and are based on variables bi: and bw where: bL S L (crosswind dimension with the wind on the side of the building) ::;; 2H 26 Design a/Structural Steelwork hw S W (crosswind dimension with the wind on the gable of the building) S 2H L == 24.0 m W= 20.0 m 2H::::;. 20.0 m :. be= bw "" 20.0 m Consider the wind on the longitudinal face of the building G E I FI E C A ! B I A I I 8.0 m 2.0 m 8.0m 2.0 m 10.0 m 4.0 m 10.0 m wind direction e == 00 Plan Figure 1.19 In Table 10 interpolation is required between + 15° and + 30° pitch Zone for e = 0° angle A B C E F G 21.80 - 1.21 - 0.65 - 0.25 -0.92 -0.67 - 0.45 +0.47 + 0.34 + 0.29 Extract from Table 10 BS 6399:Part 2 1995 Both +ve and +ve values are given for zones A, Band C, the most onerous value should be selected when considering combinations with internal pressure coefficients and other load types. A Figure l.20 ---- - Structural Steelwork 27 The zones A, B, E and Fare normally used when designing for local effects where high local suction can occur. When calculating the load on 'entire structural elements such as roofs and walls as a whole, then the values for C and G should be adopted as shown in Figure 1.21 2.0 m 8,0 m 14.0 m Figure 1.22 Figure 1.21 Consider the wind blowing on the gable of the building 5.0m wind 5.0 m ==> direction 5.0 m e = 90° 5.0 m I /J f-E e D l- S l- e D IA As before, using interpolation in Table 10 pitch angle Zone for 8 = 00 A B C D 21.80 -1.42 -1.32 -0.6 -0.25 Extract from Table 10 BS 6399:Part 2 1995 1,32 1,32 1.42 0,6 0.6 0,25 0,25 ~~ Figure 1.23 As before use zones A and B for local effects and zone C which is more onerous than D for entire stmctural elements. Internal Pressure Coefficients: Clause 2.6 Table 16 Table 17 The internal pressure coefficients are given in Clause 2,6 relating to enclosed buildings and buildings with dominant openings, A dominant opening is defined as one in which its area is equal to, or greater than, twice the sum of the openings in other faces which contribute 28 Design of Structural Steelwork porosity to the internal volume containing the opening. In cases when dominant openings occur they will control the internal pressure coefficients and should be determined using Table 17; in other cases Table 16, Clauses 2.6.1. J and 2.6.1.2 should be used. In many cases for external walls Cpi should be taken as either -0.3 or +0.2, whichever gives the larger net pressure coefficient across the walls. i.e, --+ 0.3 0.2 --+ Internal suction I nternal pressure Figure 1.24 A summary of the combined external and internal pressure coefficients is shown in Figure 1.25(a) and (b) and Figure 1.26. 1.3 0.8 1.3 0.8 0.73EBt I 0.23 o.73rnt t 0.23 c::::? ---+ -+ 0.3 ._ ---+ q ---+ ---+ wind direction t wind direction wind angle e = 0° ~ ~ wind angle e = 00 ~ ~ 1.3 0.8 1.3 0.8 Figure I.25(a) 0.45 0.73 --+ --+ 0.3 0.45 0.73 +- --+ --+ +- 0.2 0.45 --+ --+ Case I Case II Figure I.25(b) Structural Steelwork 0.2 i 0.21+- rn -+- 0.4 0.8 +- __., 0.3"- -+- 0.8 1.3 <t- i __.. 1.3 i 0.71 wind direction l'rwind angle e = 900 U 5l2.11 '- ~ O~_3 __ --+-----'1 O~ Case III 29 0.2 t 0.21+- rn-+- 0.4 0.8 +- 4- 0.2 __., -... 0.8 1.3 +- ~ -+- 1.3 t 0.71 wind direction lr wind angle e = 900 U Case IV Figure 1.26 Clause 2.1.3.4 Figure 4 Figure 5 The size effect factor (Ca) is dependent on the diagonal dimension 'a' as defined in Figure 5 for external pressures and in Clause 2.6 for internal pressures. Diagonal dimension for external pressures Figure 5 considering the longitudinal walls considering the gables considering the roof Diagonal dimension for internal pressures a = .J62 +242 a= )62 +202 a = )42 + 102 +242 ;:;J 25 ~ 21 ~ 36 Clause 2.6.1.1 a = 10 x Vintemal volume of storey a= 10 x V(20x24x6)+(0.5x20x4x24) ~ 157 Size factors (ea) Figure 4 Site in country closest distance to sea = 8.0 km Use line A in graph to determine C. He= 10.0 m 30 Design of Structural Steelwork Considering external pressures Size factor for longitudinal walls gables roof Ca ;:::; 0.9 C. ;:::; 0.91 Ca ;:::; 0.88 Considering internal pressures Size factor for all surfaces C, ;:::; 0.79 Consider Case I Clause 2.1. 3.1 and 2.1. 3.3 windward wall Pe = q,CpeC. = 1.07 x 0.73 x 0.9 ;:= 0.7 kN/m2 Pi = q,CpiCa = 1.07 x - 0.3 x 0.79 ::: - 0.25 kN/m2 net surface pressure P = (p; - Pi) ::: (0.7 + 0.25) = 0.95 kN/m2 --.. windward roof slope Pc = «c,s: = 1.07 x 0.29 x 0.88 ::: 0.27 kN/m2 Pi = q,CpiCa = 1.07 x - 0.3 x 0.79 = - 0.25 kN/m2 net surface pressure P = (p; - Pi) ::: (0.27 + 0.25) = 0.52 kN/m2 <, leeward wall pe = q.CpeC. ::: 1.07 x - 0.23 x 0,9 = - 0.22 kN/m2 Pi = q.CpiCa ::: 1.07 x - 0.3 x 0.79 = - 0.25 kN/m2 net surface pressure P ::: (p, -Pi) = (- 22 + 0.25) ::: 0.03 kN/m2 .._ leeward roof slope P« ::: qsCpeCa = 1.07 x - 0.45 x 0.88 ::: - 0.42 kN/m2 Pi = q,CpiC. = 1.07 x - 0.3 x 0.79 = - 0.25 kN/m2 net surface preSSure P = (Pc - Pi) ::: (- 0.42 + 0.25) = 0.17 kN/m2 .> Gables zone A Pc ::: «c;«; = 1.07 x - 1.3 x 0.91 = -1.27 kN/m2 Pi = »c.c: = 1.07 x - 0.3 x 0.79 = - 0.25 kN/m2 net surface pressure P = (Pe - Pi) ::: (- 1.27 + 0.25) = 1.02 kN/m2 l Gables zone B Pe = qsCpeC. = 1.07 x - 0.8 x 0.91 = - 0.78 kN/m2 Pi = q.CpiCa ::: 1.07 x - 0.3 x 0.79 = - 0.25 kN/m2 net surface pressure P = (p; - Pi) = (- 0.78 + 0.25) = 0.53 kN/m2 1 Structural Steelwork 31 l.02 kN/m2 0.53 kN/m2 i t 0.03 kN/m2 I 0.95 kN/m2 0.03 kN/m2 =::;:::>- I • t I wind direction L. -;--_-;--_---1 wind angle e = 00 ~ ~ 1.02 kN/m2 0.5 3kN/nl 0.95 kN/m2 r zone A zoneB Figure 1.27 Clause 2.1.3.5 The net load on an area of building is given by P = pA A typical internal frame supports surface areas as shown in Figure 1.29 Figure 1.28 Area of wall supportedJframeA =: Area of roof supportedJframeA 4.0 x 6.0 = 24.0 m2 4.0 x 10.77 43.1 m2 Surface loads (as shown in Figure 1.30) windward wall P 0.95 x 24.0 windward roof slope P = 0.52 x 43.1 leeward wall P 0.03 x 24.0 leeward roof slope P 0.17 x 43.1 22.8 kN = 22.41 kN :::: O.72kN 7.33 kN Only Case I is considered here; when designing such a frame all cases must be considered in combination with dead and imposed loads and appropriate partial load factors as given in Table 2 ofBS 5950:Part 2, to determine the critical design load case. 32 Design a/Structural Steelwork 22.41 kN 7.33 kN 22.8 kN ~t'tb-- 0.72 kN Figure 1.29 1.12 Example 1.8 Radar reflector Radar equipment sited on the perimeter of an airfield comprises a revolving reflector mounted on a tripod trestle as shown in Figure 1.30. Using the data provided determine the force exerted by the wind on the reflector. Data: Location Closest distance to sea Altitude above mean sea level r-------10.0 m __ ...., near Aberdeen 5km 15m 1 -'-'-'1'-'-'-'-'- ._._._._._._._._._._._.-::;;;:- ... _._._._._._._._._._.- .. 4.0 m . • I .... ~~ ~ _1_ Solution: Clause 2.1.3.3 (b) When considering free standing canopies and building elements, the net surface pressure is given by Figure 1.30 where Cp is the net pressure coefficient for a canopy surface or element and is defined in Clauses 2.5.9 and 2. 7. In this problem the radar equipment can be treated as indicated in Clause 2. 7.6; the wind force on the members of the trestle will be neglected. :. Cp=1.8 Clause 2.1.3.4 Figure 4 . Size effect factor (Ca) H;= 8.0 ill, Site in country, Distance to sea== 5 km Download and print this documentRead and print without adsDownload to keep your versionEdit, email or read offlineChoose a format:.PDF.TXTDownload
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