Design Example of Timber Column to BS 5268

Example Single column A timber column 200mm  200mm is required to carry a load of 210 kN. The load has been transferred to the column by timber joists such that the end restraint conditions top and bottom may be taken as restrained in position but not in direction. The height of column is 2.8 m and the timber may be taken as strength class C27. The load may be considered as short term. Solution As timber is greater than 100mm thick it would be difficult to dry the section, so use wet stresses. Values found in Table 7 are modified by factor K 2 found in Table 13 From Table 7  c ,= compression parallel to the grain = 8.2 N/mm 2 E min = 8200 N/mm 2 L e =1.0L = 2800 mm

K 2 = 0.6 K 2 = 0.8

I = bh 3 /12 = 200  200 3 /12 = 1.333  10 8 mm 4 A = bh = 200 ´ 200 = 40000 mm 2 i= I/A = 57.7 mm l = L e /i = 2800/57.7 = 48.5 (for both axes) < 180 suitable Ratio E/ c ,= (8200  0.8)/(8.2  0.6) = 1333.3 {modified by factor K 2 } From Table 19: 1300 1400

40 0.809 0.811

50 0.757 0.760

Modification factor K 12 for  = 48.5 and E/ c ,=1333.3

K 12 = 0.767

Alternatively for Table 19 an equivalent slenderness L e /b may be used for rectangular sections, in this example 2800/200 = 14 From Table 19: 1300 1400 as before.

11.6 0.809 0.811

14.5 0.757 0.760

K 3 for short term loading = 1.5 K 8 for non load-sharing member = 1.0 Permissible stress = grade stress parallel to the grain  K 2  K 3  K 8  K 12 = 8.2  0.6  1.5  1  0.767 = 5.66 N/mm 2 Actual compressive stress = Load/Area = 210  10 3 /40000 = 5.25 N/mm 2 As this is less than 5.66 N/mm 2 the column is suitable. Example Column forming part of a partition wall A timber column of 72mm  168mm cross-section supports a medium term axial load of 24 kN. The column forms part of a partition wall that is 3.9 m high and the columns are arranged such that there is no load sharing. The column is restrained in position only top and bottom and is provided with restraining side rails at the third points about the weaker axis. Check the suitability of strength class C22 to carry the load. Solution As there are two differing effective lengths and hence two different slenderness ratios, the critical axis must be identified

L e = 3.9m

L e =1.3m

72  168 12

3

Ix =

Iy =

168  723 12

I = bh 3 /12

i= I/A

28.45  10 6 mm 4

5.23  10 6 mm 4

28.45  10 6 /(168  72)

5.23  10 6 /(168  72)

48.5 mm

20.8 mm

3900 = 80.4 48.5

1300 = 62.5 20.8

l = L e /I Critical axis for buckling is the x–x axis Section is less than 100mm thick so service class 1 or 2 applies (K 2 = 1.0) From Table 7  ,= compression parallel to the grain = 7.5 N/mm 2 c E min

= 6500 N/mm 2

Ratio E/ c ,= 6500/7.5 = 867 From Table 19 for the ratio value of 867 and l = 80.4

K 12 = 0.51

K 3 for medium term loading = 1.25 K 8 for non-load sharing member = 1.0 Permissible stress = grade stress parallel to the grain  K 2  K 3  K 8  K 12 = 7.5  1.0  1.25  1  0.51 = 4.78 N/mm 2 Actual compressive stress

= Load/Area = 24  10 3 /(72  168) = 1.98 N/mm 2

As this is less than 4.78 N/mm 2 the column is suitable.