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AS1684 SS N2housebracing.Pdf Understanding Residential Timber AS1684 Framed Construction http://www.docstoc.com/docs/139519457/AS16842-SS-Bracing-Example Bracing Example • Win d c lass ifica tion - N2 • Single storey • “L” -shdhaped • Gable & hip roofs • Ce iling he ig ht 2400mm • Eaves 600mm • Rfith225Roof pitch 22.5o Bracing Example • N2 • Single storey • “L”- shaped • Gable & hip roofs • Ceiling height 2400mm Wind Wind direction 2 • Eaves 600mm direct io n 1 • Roof pitch 22.5o 22.5o 7500 600 600 6300 600 10700 8600 7400 600 19200 600 20400 Bracing Design Process (Clause 8.3.1) 1. Determine the wind classification Clauses 1.4.2 & 1.5 & AS4055/AS1170.2 2. Determine the wind pressure Clause 8.3.2 & Tables 8.1 to 8.5 3. Determine the area of elevation Clause 8.3.3 and Figure 8.2(A) or (B) 4. Calculate the racking force Clause 8.3.4 5. Design the bracing systems - Sub-floors Clause 8.3.5 - Walls Clause 8.3.6, Tables 8.18 and 8.19 6. Check even distribution and spacing Clause 8.3.6.6 and 8.3.6.7 7C7. Connec tion o fbf brac ing to Clause 83698.3.6.9 an d83610d 8.3.6.10 roof/ceilings (at walls) and floors AS1684.2 pg112 1. Determine the Wind Classification Refer Clause 1.4.2 [pg 9] and AS 4055 or AS/NZS 1170. 2 N2 (provided by structural engineer, building professional or local building authority) 2. Determine the wind pressure (for both wind directions) See Clause 8.3.2 [pg 112] also Tables81to85[pgs116Tables 8.1 to 8.5 [pgs 116–124] Need: • the roof pitch, • the width of the building, and • whether there are any flat walls, skillion ends, gable or hip ends. Complex designs may require separate pressures within the one wind direction. 2.1 Determine the wind pressure (for Wind direction 1) See Table 8.1 [pg 116] and Table 8.2 [pg 117] Hip End NOTE: For wind direction 1, adopt this elevation as the wind blowing onto the ‘flat’ surface of the gable end will produce a higher force compared to wind blowing Hip roof on the opposite hip end elevation. Gable end 1 Split the house into it’s two components Gable end (with Hip other end) Hip roof (long length of building) 2.1 Determine the wind pressure (for Wind direction 1 – Gable end) See Table 8.1 [pg 116] 1 Pressure (Gable end) = 0. 92kPa (kN/m2) 2.1 Determine the wind pressure (for Wind direction 1 – Hip end – Long length of building) Roof pitch = 22.5° Table 8. 2 is used for determining the ppgressure on single or upper storey elevations where the wind direction is at 90o to a ridge and for wind speeds N1 , N2, N3 & N4. Interpolation permitted Answer rounded up to the nearest 0.05 22.5° 0.65 6.3m 2.1 Determine the wind pressure (for Wind direction 1 – Hip end – Long length of building) Roof pitch = 22.5° Therefore, pressure (Hip end) = 0.65kPa (kN/m2) 2.2 Determine the wind pressure (for Wind direction 2 – Hip end – Long length of building) Table 8.2 is used for Roof pitch = 22.5° determining the pressure on single or upper storey elevations where the wind direction is at 90o to a ridge and for wind speeds N1, N2, N3 & N4. NOTE: Table 8.2 is W = 7.4m applicable as the roof runs the long length of bu ilding an d con ta ins a hip or gable end. Interpolation permitted Answer rounded up to the nearest 0.05 22.5° 0.65 7.4m 2.2 Determine the wind pressure (for Wind direction 2 – Hip end – Long length of building) Roof pitch = 22.5° W = 7.4m Therefore,,p pressure ((pHip end) = 0.65kPa (kN/m2) 3.1 Determine the Area of Elevation - Discussion Whilst the area/s of elevation should be determined relatively accurately, high levels of precision are not really warranted and therefore use of calculation methods (as used in this example), planimeters or by scaling from drawings would all be acceptable. NOTE: The area of elevation of the triangular portion of eaves overhang up to 1000mm wide may be ignored – Figure 8.2(B), Note 3 [pg 114]. The following method has been used in this example to calculate the area of elevation of the triangular roof section: Area of Roof = W/2 x W/2 T’Tan’X’ + 010.15 x W The triangular part of eaves is ignored. ‘X’ 150 mm nominal allowance f or d epth roof frame, battens and roofing. W Increase this if necessary i.e. for exposed rafter roofs. 3.1 Determine the Area of Elevation (for Wind Direction 1) Work out each area individually Hip End Area Gable End Area 3.75m o 22.5 0.15m 1. 53m CL 1.2m 24m2.4m FL 10.7m 7.4m NOTE: Wind force on the area below half wall height goes straight into floor and does not add to the wall racking (bracing) force. 3.1 Determine the Area of Elevation (for Wind Direction 1 – Gable end) Hip En d + GblEdGable End Area Roof = 0.5x7.4x1.53 + Hip End Area Gable End Area 0.15x7.4 (depth of 3.75m roof frame) 2 1. 53m = 6. 77m (say) 1.2m Area Wall = 747.4x 121.2 (half wall height) = 8.88m2 10.7m 7.4m Total Gable End Area = 6. 77+8. 88 2 = 15.7m (say) 3.1 Determine the Area of Elevation (for Wind Direction 1 – Hip end – Long length of Building) Hip End + Gable End (Eaves 600mm) 3.75m Hip End Area 1.3m 1.2m Area Roof =107x13+= 10.7x1.3 + (10.7-0.6)x0.15 2 = 1553.43m ((pp)Approx) 10.7m 7.4m Area Wall = 10.7x1.2 (half wall Total Hip End Area height) 2 = 12.84m2 = 15.43+12.84 = 28.3m (say) 3.2 Determine the Area of Elevation (for Wind Direction 2) (Eaves 600mm) 4.3m Area Roof 1.53m = (19.2+0.6)x1.53 - (0. 5x (4. 3-06)153)0.6)x1.53) 12m1.2m + (19.2+0.6)x0.15 2 = 30. 43m (say) 0.6m 19.2m Area Wall =19. 2x1. 2 (half wall height) = 23.04m2 Total Area Wind Direction 2 2 = 30.43 + 23.04 = 53.5m (say) 4. Calculate the racking force (for both Wind Directions) UthfUse the formu la: Racking Force = Area of Elevation x Wind Pressure (kN) (m2) (kPa) - (kN/m2) For complex elevations, combine the results of separate calculations to end up with a total racking force in each of the two wind directions. 4. Calculate the racking force (for both Wind Directions) UthfUse the formu la: Racking Force = Area of Elevation x Wind Pressure (kN) (m2)(kN/m2) Total racking force for Wind Direction 1 Gable = 15.7m2 x 0.92 = 14.44kN Hip = 28 . 3m2 x 0650.65 = 1840kN18.40kN = 32.8kN (say) Total racking force for Wind Direction 2 2 = 53.5m x 0.65kPa = 34.8kN (say) 5. Design the wall bracing systems In this example we will use diagonal bracing where possible – usually one diagonal brace per wall and two (opposing) wherever space permits in long walls. Diagram (a) in Table 8.18 AS 1684.2 [pg 141] illustrates the allowable limits for angle braces. NOTE: Bracing should initially be placed in external walls and, where possible, at the corners of the building – Clause 8.3.6.6 [pg 148]. AS1684.2 p141 5. Design the wall bracing systems (cont.) Determine the length of wall that the brace is acting in and multiply this by the bracing capacity. For basic timber and metal angle braces, the bracing capacity is 0. 8kN per metre . Note : The braces should be set up in opposing pairs. Do not allow single braces to be less than 1.8m or exceed 2.7m in wall length. In this example, the angle of the braces is set to give the maximum 2. 7m length for a single brace. 5.1 Design the wall bracing systems (for Wind Direction 1) [Length of each brace indicated on plan below] (2x2.7) + (2x2.7) + 2.7 + 2.7 + 1.8 + 2.7 + 2.7 + 2.2 + 2.7 + 2.2 + 2.7 + 2.2 = 35.4 m NOTE: 1.8m minimum bracinggg length for diag onal 2.7 2.7 bracing. 2.7 2.7 35.4m of bracing x 0.8kN/m (say) 2.7 = 28.3kN 2.7 1.8 2.7 Additional bracing 2.7 2.2 2.7 (racking force) required 222.2 =328= 32.8 – 28. 3 2.2 2.7 = 4.5kN 5.1 Design the wall bracing systems (for Wind Direction 1) Wind Direction 1 (needs an additional 4.5kN) Sheathing in plywood, 1 panel @ 0.9m on the dining room corner and 1 panel @ 0.9m at Bed 2 will add 6.12kN (1.8m x 3.4kN/m) of racking force to wind direction 1, thereby exceeding the 4 .0kN required . Adopt type (g) [pg 143]. NOTE: Nominal bracing (plasterboard) can also be used in certain instances – 0.9 0.9 refer Section 4.3 following. AS1684.2 p143 5.2 Design the wall bracing systems (for Wind Direction 2) 2.4 + (2x1.8) + 2.7 + 2 + (2x1.8) + 2.7 + 2.7 + 2 + 2 + 2. 7 + 2 = 28. 4m NOTE: 1.8m minimum bracing length for diagonal bracing. 1.8 1.8 28.4m of bracing x 0.8kN/m 2 =227kN= 22.7kN (say) 1.8 1.8 2.0 Additional bracing 2.7 2.0 2.7 2.0 (racking force) required 2.4 = 34.
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