Example on Design of Timber Structure part2

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CALCULATION Example 1: Design of Flexural/Bending Member Floor of a building comprises series of timber flooring system with thickness 20mm arranged and nailed on series of timber joist. The spacing between joists is 500 mm. The timber joists supported by main beam with distance 2500mm to each other. The main beam is bolted to column with span 4000mm. Determine the suitable size of beam and joist used to accommodate all the loads if the timber used is from SG3 (standard, dry). Design data given are as below: Live load Dead load Selfweight of floor

= 3.5 kN/m2 = 0.4 kN/m2 = 0.5 kN/m2

Solution: 1.0 Main Beam Try beam size (named size) 100mm×275mm Step 1: Geometrical Properties Beam dimension Named size, b (breadth) h (depth) Dressed size, b (breadth) h (depth)

= 100mm = 275mm = 90mm = 265mm

 (90 × 265 )   Second moment of area,
 = ℎ 12 = 12 = 135.57 × 10 

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CALCULATION Step 2: Loading Loadings acted on the main beam are reaction for the joists, which are points load at 0.5m spacing. The spacing can be consider as closely spaced thus to simplified the design calculation, the points load can equally presented as uniformly distributed load w= beam distance × (dead load+imposed load) = 2.5m×((0.4+0.5)+3.5)kN/m2 = 11 kN/m Design Span, Le= 4000mm (center-to-center of the column and simply supported at the column) Step 3: Check for lateral stability Both beam-ends are held in position using bolts and the top of the beam is hold by the joists. From Table 7 the permissible maximum depth to breadth ratio is 5 depth/breadth = 265/90 = 2.9475mm for length from end member) K3= 1.05 Bearing grade stress, Ctg= 2.97 N/mm2 (compression perpendicular to grain) Permissible bearing stress, Ctp= Ctg× K1× K2× K3 = 2.97 N/mm2×1.0×1.0×1.05 = 3.12 N/mm2 Bearing stress, Cts= 0.75 N/mm2 < permissible bearing stress, Ctp= 3.12 N/mm2. Therefore bearing stress is adequate

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CALCULATION Example 2: Design of Compression Member Check the adequacy of 5.0 m long of timber column for long term loading if a column section of 150 mm x 150 mm (dressed size) is subjected to an axial load of 80 kN (including selfweight of column). The timber used is in SG4 (standard, wet) and the column is not restrained about both axes but restrained at both ends in position

Solution: Step 1: Geometrical Properties Beam dimension No need to taking account reaping effect as the question already stated the dimension is dressed size. Dressed size, b (breadth) = 150mm h (depth) = 150mm Column length, L= 5.0m

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Step 2: Determine effective length and slenderness ratio Effective column length, Le= 1.0L (restrained at both ends in position but not in direction) = 1.0(5m) = 5m Radius of gyration, r= b/√12 = 150/√12 = 43.3 Slenderness ratio, λ= Le/r = 5000/43.3 = 115.5 Permissible compressive stress, Csp= 2.8 N/mm2. Increase column size or provide lateral support to reduce the slenderness of the column.

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Example 3: Design of Compression Member A 2.5 meter compression member with named sized 75mm×150mm is restrained at both end in position. However, the column is restrained about major axis and unrestrained about minor axis at top and column base. Determine the ultimate axial load capacity of the column section for long term loading if the column section in use is SG2 (select, wet). Solution: Step 1: Geometrical Properties Beam dimension Taking account of reaping effect at four side of the timber section Named size, b (breadth) = 75mm h (depth) = 150mm Dressed size, b (breadth) h (depth)

= 65mm = 140mm

Column length, L= 2.5m Step 2: Determine effective length and slenderness ratio Effective column length about x-x axis Lex= 0.7L (restrained at both ends in position and in direction) = 0.7(2.5m) = 1.75m Radius of gyration about x-x axis, rx= h/√12 = 140/√12 = 40.4 Slenderness ratio about x-x axis, λx= Lex/rx = 1750/40.4 = 43.3