Structural design of timber to BS 5268 –2:1996
The structural design of timber elements is based on permissible stresses and deflections
derived from elastic theory.
Flexural members – members subject to bending
Members subject to bending (i.e. beams) are assumed to behave in accordance with elastic
bending theory provided that the permissible material stresses are not exceeded. The bending
expression can be applied to timber design:
M/I =f/y =E/R
At any point across a section of a beam which is loca ted a distance y from the neutral axis of a
section, a stress f will be developed as a consequence of applying a bending moment M to the
section. The magnitude of the stress developed will vary with the second moment of area of
the section I.
In the timber design code,  is the designation for stress, hence the above equation may be
written as:
M/I = /y
In timber design generally rectangular sections are used, therefore the maximum compressive
and tensile bending stress will occur at the extreme fibres. Thus y is equal to half the depth of
the section.
As both I and y are geometric properties of the section it is convenient to combine the two
terms in a single property which is referred to as the elastic modulus and denoted by the
symbol Z.
Z = I/y
Further, as rectangular sections are being considered, if b is the width of the section and h the
depth then I and y may be expressed as:
I = bh 3 /12 and y =h/2
Hence elastic modulus, Z = (bh 3 /12)/(h/2) = bh 2 /6
Considering, from the bending expression, M/I = /y and combining with the definition of
elastic modulus then:
M = Z
This can be rearranged to determine the maximum bending stress in the beam and then
compared with the permissible stress that the beam may carry.
Other design requirements
The other checks required to timber beams are shear, bearing, deflection and the maximum
depth to breadth ratio.
As the beams are simply supported, and generally carry uniformly distributed loads, shear will
be a maximum at the supports.
Maximum shear stress = –
3
Load

2
Cross-sectional area
The end bearing area is dependent on the contact area with the beam support.
Maximum bearing stress =
Load
Load
=
Contact area
width of beam  bearing length
Both of these values can then be compared with the permissible values obtained from the
design code.
Deflection, as with other structural materials, is a serviceability requirement. Maximum
deflections are determined using the standard deflection formulae and compared with the
deflection limits given in the design code.
Timber as a structural material
Unlike other construction materials, timber cannot be mixed to a pre -determined formula. The
cut wood has to be inspected and graded by visual or mechanical means. The design code
allows for a number of strength classes based on the inspection of the timber, or alternatively,
if the species of timber is known it may be classified as given in Table 8 of the code according
to its standard name.
Appropriate grade stresses are assigned to the graded timber. For flexure the appropriate grade
stresses are:
•
•
•
Bending parallel to the grain
Compression perpendicular to the grain
Shear parallel to the grain
Account must also be taken of the loading and exposure conditions that the timber will be
subject to. The design code lists almost thirty factors that can be applied to the grade stresses.
Only a few will be of concern in this course.
Modification factors
Moisture content of timber related to service class.
It is difficult to artificially dry solid timber more than 100 mm thick, unless it is specially
dried. BS 5268 recognises three services classes that are related to the conditions of end use.
Service classes 1 and 2 generally require the timber to be art ificially dried and the dimensions
and properties of the dried timber can be taken as the grade values. Service class 3 timber is
used when the finished structure is fully exposed or if the timber is more than 100 mm thick. In
this case the grade values must be modified by a factor K 2 found in Table 13 which allows for
the differing load carrying mechanisms of wet and dry timber.
Service
class
Examples of end use of
timber
3
2
2
1
External & fully exposed
Covered and unheated
Covered and heated
Internal use and
continuously heated
Average
moisture
content
> 20 %
18 %
15 %
18%
12%
Moisture content in
each piece at time
each piece at time
24 %
20 %
24%
20%
Service classes 1 and 2 use unmodified stresses and moduli.
Service class 3 timber uses modified stresses and moduli.
i.e. K 2 =1
i.e. K 2 <1
Duration of loading K 3
The grade stresses based on the strength classes of the timber apply to long -term loading on the
structural element. Table 14 gives a modification factor for various load durations and l ist
values of K 3 varying from 1.0 to 1.75. K 3 is applied to the grade stresses only and does not
apply to the modulus of elasticity.
Load-sharing systems K 8
A load-sharing system may be considered as being, for example, a series of four or more floor
joists connected by flooring in such a way that act together – a standard timber floor. Provided
that the joists are no farther apart than 610 mm centres then the grade stresses should be
modified by the modification factor K 8 =1.1.
For all other systems K 8 may be taken as being equal to 1.0.
For load-sharing systems the mean modulus of elasticity should be used to calculate any
deflections except in circumstances where dynamic loads may occur, e.g. gymnasia, where the
minimum value should be used.
Depth factor K 7
The grade stresses based on the strength classes of the timber apply to materials having a depth
(h) of 300 mm. A modification factor K 7 is applied to the grade bending stress of beams
having a depth other than 300 mm.
For solid timber beams:
Depth of beam (h) mm
72 mm or less
72 > h < 300
h > 300
K 7 value
1.17
(300/h) 0.11
0.81(h 2 + 92300)
(h 2 + 56200)
Notching of beams K 5
Figure 2
Notching the end of a beam for construction purposes causes stress concentrations that must be
allowed for in the shear calculation. The shear stress should be calculated by using the
effective depth (h e ) shown in Figure 2. The grade shear stress should be multiplied by a
modification factor K 5 to obtain the permissible stress.
For beams notched on the underside K 5 = h e /h
Note: Beams with notches on the top edge are not considered in this unit.
Deflection
The deflection is acceptable if the deflection of the fully loaded beam does not exceed 0.003
times the span of the member or 14mm whichever is the lesser.
Timber flexural members design examples
Boarding
Check the suitability of 20mm tongued and grooved floor boarding spanning between 50mm 
250mm timber joists at 600mm centres. The boards are of strength class C14.
Note that boarding is normally provided in lengths up to 3m long. Each board spans over a
number of joists and for analysis purposes may be treated as a continuous beam.
The maximum moment occurs at an internal support and may be found using M = wL 2 /10. The
maximum shear force (reaction) occurs at the outside support and may be taken as V= 0.4wL.
Additional data:
Dead load inclusive of self-weight of boards
Imposed load
0.15 kN/m 2
1.5 kN/m 2
Solution
Consider a width of boarding (b) = 1m
b=1000 mm
(Actual width of floor is immaterial if width of one metre is assumed)
Length between supports
L = 600mm = 0.6m
Load on boarding
w =dead + imposed = 0.15 + 1.5 = 1.65 kN/m 2
Considering a typical 1m width of board b=1.65  1 = 1.65 kN/m
Bending
Maximum moment M = wL 2 /10 = 1.65  0.6 2 /10 = 0.06 kNm
Elastic modulus of board Z = bh 2 /6 = 1000  20 2 /6 = 66667 mm 3
Actual bending stress, s = M/Z = 0.06  10 6 /66667 = 0.9 N/mm 2
Permissible stress =
grade bending stress parallel to the grain  K 2  K 3  K 7  K 8
Grade stress from Table 7 – C14 = 4.1 N/mm 2
K 2 – wet stresses modification factor – material 20 mm
thick – service class 1
K 3 – duration of loading – on floor this may be taken
as long term
K 7 – depth factor – less than 72 mm
K 8 – load-sharing – boards are load-sharing
K2 = 1
K3 = 1
K 7 = 1.17
K 8 = 1.1
Permissible stress = 4.1  1.0  1.0  1.17  1.1
= 5.28 N/mm 2 > 0.9 N/mm 2 boards suitable in bending
Shear
Maximum shear force V = 0.4wL = 0.4  1.65  0.6 = 0.4 kN
Maximum shear stress v =
3V
3  0.4  103
=
= 0.03 N/mm 2
2bh
2  1000  20
Permissible stress =
grade shear stress parallel to the grain  K 2  K 3  K 8
Grade stress from Table 7 – C14 = 0.6 N/mm 2
The modification factors used for bending are still applicable – except K 7 that is applied to
bending only.
Permissible stress = 0.6  1.0  1.0  1.1
= 0.66 N/mm 2 > 0.03 N/mm 2 boards suitable in shear
Deflection
Considering the beam as continuous,  = wL 4 /384EI
E mean from Table 7 – C14 (one board cannot act on its own) E = 6800 N/mm 2
I =bh 3 /12 = 1000  20 3 /12 = 666667 mm 4
 = wL 4 /384EI = 1.65  600 4 /(384  6800  666667) = 0.13 mm
Permissible deflection (clause 2.10.7) = 0.003  span = 0.003  600 = 1.8 mm
Actual deflection less than permissible – beam is suitable.
Floor joists
The floor joists for the boarding example above also require to be checked. It may be assumed
that the joists are simply supported over a span of 3.6 m and bear on to blockwork supports
100 mm wide. The revised dead load to include for the self -weight of the beam may be taken
as 0.34 kN/m 2 . The joists are strength class C16.
Solution
Centres of joists
Load/joist
600mm = 0.6m
w = (dead + imposed)  centres
= (0.34 + 1.5)  0.6 = 1.1 kN/m
Bending
For a simply supported beam
Maximum moment M = wL 2 /8 = 1.1  3.6 2 /8 = 1.78 kNm
Elastic modulus of board Z = bh 2 /6 = 50  250 2 /6 = 520833 mm 3
Actual bending stress,  = M/Z = 1.78  10 6 /520833 = 3.42 N/mm 2
Permissible stress = grade bending stress parallel to the grain  K 2  K 3  K 7  K 8
Grade stress from Table 7 – C16 = 5.3 N/mm 2
K 2 – wet stresses modification factor – material 20 mm
thick – service class 1
K 3 – duration of loading – on domestic floor this may be
taken as long term
K 7 – depth factor (clause 2.10.5)
K 7 = (300/h) 0.11 = 300/250) 0.1 = 1.02
K 8 – load-sharing – boards are load-sharing
K 2 =1
K 3 =1
K 7 =1.02
K 8 =1.1
The assumption is that the floor boards are of sufficient length to distribute the load over at
least four joists.
Permissible bending stress
= 5.3  1.0  1.0  1.02  1.1
= 5.95 N/mm 2 < 3.42 N/mm 2
Beam satisfactory in bending.
Shear
Maximum shear force V = wL/2 = 1.1  3.6/2 = 1.98 kN
3V
3  1.98  103
Maximum shear stress v =
=
= 0.24 N/mm 2
2bh
2  50  250
Permissible stress = grade shear stress parallel to the grain  K 2  K 3  K 8
Grade stress from Table 7 – C16 = 0.67 N/mm 2
The modification factors used for bending are still applicable
(K 7 is only applicable to bending)
Permissible stress = 0.67  1.0  1.0  1.1
= 0.74 N/mm 2 > 0.24 N/mm 2
joist suitable in shear
Bearing
Value of reaction = wL/2 = 1.1  3.6/2 = 1.98 kN
Joist bears on to a 100mm wide support and width of joist is 50mm
Reaction
1.98  103
Actual bearing stress =
=
= 0.39 N/mm2
Bearing length  width
100  50
Permissible stress = compression perpendicular to the grain  K 2  K 3  K 8
Grade stress from Table 7 – compression perpendicular to the grain – 2.2 N/mm 2
Two values of compression perpendicular to the grain are given in Table 7. Which value
should be used? Reference should be made to Note 1 of the table.
The modification factors used for shear are still applicable
Permissible stress = 2.2  1.0  1.0  1.1
= 2.42 N/mm 2 > 0.39 N/mm 2 bearing length is suitable
Deflection
As the beam is simply supported,  = 5wL 4 /384EI
E mean from Table 7 – C14 (one board cannot act on its own) E = 8800 N/mm 2
I =bh 3 /12 = 50  250 3 /12 = 65.1  10 6 mm4
 = 5wL 4 /384EI = 5  1.1  3600 4 /(384  8800  65.1  10 6 ) = 4.2 mm
Permissible deflection (clause 2.10.7) = 0.003  span £ 14 mm = 0.003  3600 = 10.8 mm
Actual deflection less than permissible – beam suitable.
Notches
If the beam is notched at the support, then the shear cross -sectional area is reduced and the
modification factor K 5 applies (see clause 2.10.4).
Consider the above beam with a 75mm notch on the underside.
Dimension h e = h – 75 = 250 –75 = 175 mm
Maximum shear stress v =
K5 = he/h = 175/250 = 0.7
3V
3  1.98  103
=
= 0.34 N/mm 2
2bh
2  50  174
Permissible stress = grade shear stress parallel to the grain  K 2  K 3  K 8  K 5
Permissible stress = 0.67  1.0  1.0  1.1  0.7
= 0.52 N/mm 2 > 0.34 N/mm 2
Joist is still suitable in shear.
Timber compression members
As with all structural materials, the design of compression members is dependent on the
slenderness ratio.
Where the slenderness ratio,  = L e /i
L e = effective length is found using Table 18, which lists for conditions of end restraint, the
ratio of L e /L, where L is the actual length.
Values given for L e are 0.7L, 0.85L, 1.0L, 1.5L and 2 L.
i is the radius of gyration of the section. As only solid rectangular sections will be dealt with,
there are two possible axes of buckling, x–x and the y–y. Hence there are two values of
slenderness ratio:
 x = L ex /i x
The radius of gyration i x = I x /A
 y = L ey /i y
i y = I y /A
Where I (for a rectangular section) = bh 3 /12 I x =bh 3 /12 and I y =hb 3 /12
Area A = bh
considering x–x axis
considering y–y axis
The critical slenderness ratio is the larger of the two
In no case should the slenderness ratio exceed 180 (see clause 2.11.4)
The permissible stress is based on the comments of clause 2.11.5 which gives two design
procedures:
1.
2.
Compression members with slenderness ratios less than 5 (short columns)
Compression members with slenderness ratios greater than 5 (slender columns)
In both cases the permissible stress is taken as the grade compression stress parallel to the
grain multiplied by the modification factors for moisture content, duration of loading and load
sharing.
Permissible stress = grade stress parallel to the grain  K 2  K 3  K 8
In addition for members with a slenderness greater than 5, the above formula is multiplied by
K 12 given in Table 19.
Factor K 12 varies with slenderness ratio as calculated above and with
E/ c ,
where E = minimum modulus of elasticity of the material, and
 c ,= compression parallel to the grain.
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 mm4
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 i s 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 h ence two different slenderness ratios, the
critical axis must be identified
L e = 3.9m
L e =1.3m
72  168
12
3
I = bh 3 /12
i= I/A
l = L e /I
Ix =
Iy =
168  723
12
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
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
 c ,= compression parallel to the grain = 7.5 N/mm 2
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.