DESIGN OF COMPRESSION MEMBERS (MS544: Part 2: 2001, Clause 12)
Compression Parallel to Grain – the actual compressive stress shall be calculated as c,act = P / Ae where P = the axial force acting on the member Ae = the effective cross sectional area for the compression member The allowable design compressive stress parallel to grain is c,adm = c,g x Ki Column stability Column stability is the tendency of the member to buckle. The stability becomes most critical for long and thin (slender) columns.
The effective length (Le) of a column is defined as the distance between two points at which the member is assumed to buckle in the shape of a sine wave. It is more simply defined as the minimum distance between inflection points of a buckled compression member. Le = KeL where Ke = buckling length coefficient (see Table 9) L = the unsupported length of the column between points preventing lateral buckling Buckling length coefficient (Ke) is given in Table 9 The thinness of the member is measured by its radius of gyration (i) where i = ( I / A )½ = b / ( 12 )½ The measure of this tendency for columns to buckle is predicted by a unit-less value referred to as the slenderness ratio ( = Le / i ) The maximum value for slenderness ratio as given in Clause 12.4,
Structural Timber Design – Associate Prof Dr David Yeoh
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= Le / i
180
for member with dead and/or imposed loads
= Le / i
250
for member with self weight and wind load only
Compressive stress modification factor, K8 Where 5 without eccentricity of loads, the permissible design compressive stress is calculated as c,adm = c,g x Ki Where 5, the permissible design compressive stress is calculated as c,adm = c,g x Ki x K8 The value K8 can be calculated from the equation given in Appendix 2D or from Table 10 of the MS544: Part 2. The grade compressive stress taken for K8 calculation should be modified for duration of loading and the modulus of elasticity should be the minimum MOE regardless of the member acting alone or in load sharing system.
Example: Design the column size that is axially loaded (DL + IL) per column 60 kN. The actual length of the column is 4.7 m. The column is fixed at one end and pinned at the other. The spacing between columns is 4.5 m. Design the column from SG3 timber, standard grade, green condition.
Solution: Ke = 0.85 (column is restrained at both ends in position and one end in direction) Effective length of column, Le = 0.85 x 4.7 = 4.0 m
(Table 9)
Modification factors: K1 = 1.25
(medium term load, DL + IL)
K2 = 1.0
(no load sharing)
c,g = 12 N/mm2 c,g = 12 x 1.25 = 15 N/mm2 Emin = 9800 N/mm2 (no load sharing) Structural Timber Design – Associate Prof Dr David Yeoh
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Use trial and error method. Try size 125 x 150 (finished size 120 x 145). A = 17400 mm2 Emin/c,g = 653 Le /b = 33 Interpolating from Table 10, K8 = 0.266 c,adm = 12 x 1.25 x 1.00 x 0.266 = 3.99 N/mm2 Permissible axial load = c,adm x A = 3.99 x 17400 = 69 kN > Pact Therefore use column size 125 x 150 Review of a Rectangular Column with equivalent lengths
A column is loaded and supported as indicated. The column is a surfaced 150 x 250, common grade Penaga used at 16 % maximum moisture content. The loading is a combination of DL + FLL + WL. Determine the maximum allowable concentric load, P, that the column can adequately support.
COLUMNS OF OTHER SHAPES Round columns – Design of a column with round cross section can be based on the design calculations for a square column of the same cross sectional area. Acircle = Asquare Review a round column A 150 mm diameter log, 4 m long is used as a compression member in a foundation system. The member is considered to be pinned at the top, but fixed at the bottom. It is made of select grade Kempas, and is not surfaced. It is used in an environment with high moisture content and normal duration loading. Determine the maximum concentric load for the member.
Structural Timber Design – Associate Prof Dr David Yeoh
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