CIV 3218 STRUCTURAL DESIGN II
WEEK 1
Timber as an engineering material: orthotropic behaviour, engineering
properties, MOE, MOR, tension, compression, shear and bearing strength
parallel, perpendicular and at an angle to the grain.
Design codes and standards. Stress grading, strength grouping, derivation
of design stresses.
INTRODUCTION
01
02
Timber from well-managed
forests is one of the most
sustainable resources
available and it is one of the
oldest known materials used
in construction.
It has a very high strength to
weight ratio, is capable of
transferring both tension and
compression forces, and is
naturally suitable as a flexural
member.
CHARACTERISTICS
There are a number of inherent characteristics
that make timber an ideal construction
material.
• These include its high strength to weight
ratio, its impressive record for durability and
performance and good insulating properties
against heat and sound.
• Timber also benefits from its natural growth
characteristics such as grain patterns, colours
and its availability in many species, sizes and
shapes that make it a remarkably versatile
and an aesthetically pleasing material.
• Timber can easily be shaped and connected
using nails, screws, bolts and dowels or
adhesively bonded together
Wamara Wooden Floor
Scientific Name:Swartzia
Leiocalycina Benth
Wamara is a very dense wood like a typical
rosewood, with dark purple heartwood and
pale yellow sapwood.
Its density makes it hard on your tools, but the
beautiful colours and sheen it takes on finishing
make it very worthwhile. No finishing
difficulties that I have come across, as it is not
an oily wood.
Orthotropic
Nature of
Wood
Wood may be described as an orthotropic material;
that is, it has unique and independent mechanical
properties in the directions of three mutually
perpendicular axes:
I. longitudinal,
II. radial, and
III. tangential.
The longitudinal axis L is parallel to the fiber
(grain); the radial axis R is normal to the growth
rings (perpendicular to the grain in the radial
direction); and the tangential axis T is
perpendicular to the grain but tangent to the
growth rings.
Orthotropic Nature of Wood
Three principal axes of wood with respect to the grain direction and
growth rings
STRENGTH GRADING OF TIMBER
The strength of timber is a function of several parameters
including the species type, density, size and form of
members, moisture content, duration of the applied load
and presence of various strength reducing characteristics
such as slope of grain, knots, fissures and wane.
To overcome this difficulty, the strength grading method
of strength classification has been devised.
Several design properties are associated with a strength
grade; these include modulus of elasticity and bending
strength parallel to the grain, strength properties in
tension and compression parallel and perpendicular to
the grain, shear strength parallel to the grain and density.
Strength grading provides a possibility to assess the strength of a piece of
sawn timber.
This process can be carried out by visual or machine based method.
Visual grading rules and standards define growth characteristics, such as
knots, wane, slope of grain, fissures, distortions, reaction wood, fungal and
insect damage, etc.
Standards provide restrictions on the size, type and number of defects
allowed in each grade. A certified grader visually assesses each piece of
timber.
Grading rules specify acceptable limits for these features. Visual strength
grading produces two strength grades - GS (General Structural) and SS (Special
Structural).
Machine strength grading is based on the relationship between the stiffness
and strength properties of a timber.
SS (Special Structural)
- Used as a basis for strength and modulus of elasticity determination
- Done by subjecting large number of structural sized specimens to short
term load test
- Fifth percentile stresses obtained.
- Fifth percentile is defined as the value at which no more than 5% of the test
results fell.
- The grade stresses were obtained by dividing the fifth percentile stresses by
a reduction factor, which included adjustments for the standard depth of
specimen 300mm, duration of load and factor of safety.
Grade stresses for softwoods graded in accordance with BS 4978: for service classes
1 and 2 (Table 10, BS 5268)
What is the difference between SS and GS?
How do you identify strength graded timber?
Each piece of strength graded timber is clearly marked and provides vital information. The
key elements of these marks are:
• The moisture content condition at the time of grading, i.e. DRY, KD or WET.
• The strength grade or strength class of the piece.
• The timber species or species combination.
• Identification of the grader and company responsible for grading.
• The relevant British/European Standard number.
• The mark of the independent certification body.
• The use of the term 'DRY' or 'KD' confirms that at the time of grading the piece of timber
satisfied the moisture content requirements of the Standard. 'KD' indicates that the timber
has been kiln dried as opposed to air dried. 'DRY' indicates that either air drying or kiln
drying has been undertaken.
Strength classes
The concept of grouping timber into strength classes was introduced into
the United Kingdom with BS 5268-2 in 1984.
Strength classes offer a number of advantages both to the designer and
the supplier of timber.
The designer can undertake the design without the need to check on the
availability and price of a large number of species and grades that might
be used.
Suppliers can supply any of the species/grade combinations that meet the
strength class called for in a specification.
The concept also allows new species to be introduced to the market
without affecting existing specifications for timber.
Strength classes
• In all there are sixteen strength classes, C14, C16, C18, C22, C24,
TR26, C27, C30, C35, C40, D30, D35, D40, D50, D60 and D70, with C14
having the lowest strength characteristics.
• The strength class designations indicate the bending strength of the
timber.
• Strength classes C14 to C40 and TR26 are for softwoods and D30 to
D70 are for hardwoods.
• Strength class TR26 is intended for use in the design of trussed
rafters.
Grade stresses and moduli
of elasticity BS5268
• When the specification specifically
prohibits wane at bearing areas, the
higher value may be used.
The term "wane" is often used to describe
missing areas on lumber. The term typically
refers to defects on the ends or sides of the
lumber, but wane can occur just about
anywhere on milled wood. Wane is often
associated with bark that remains on the
lumber after it's milled.
• For calculation of dead load, the average
density should be used.
https://www.hunker.com/12582774/the-definition-of-wane-inlumber
Permissible
stresses
The grade stresses given in Tables 6.1 and 6.3 were
derived assuming particular conditions of service and
loading.
In order to take account of the actual conditions that
individual members will be subject to during their
design life, the grade stresses are multiplied by
modification factors known as K-factors.
The modified stresses are termed permissible stresses.
Modification
Factors
Modification factors relevant to the design
of simple flexural and compression
members are:
K2: Moisture content factor
K3: Duration of loading factor
K5: Notched ends factor
K7: Depth factor
K8: Load-sharing systems factor
K12: Compression member stress factor.
MOISTURE CONTENT, K2
The strength and stiffness of timber
decreases with increasing moisture
content. This effect is taken into
account by assigning timber used
for structural work to a service
class.
BS 5628 recognizes three service
classes
MOISTURE CONTENT, K2
The grade stresses and moduli of
elasticity shown in Tables 6.1 and
6.3 apply to timber exposed to
service classes 1 and 2.
According to clause 2.6.2 of BS
5268 where service class 3 exists,
the values in Tables 6.1 and 6.3
should be multiplied by a
modification factor K2 given in
the Table.
MOISTURE CONTENT, K2
Because it is difficult to dry thick timber, service class 3 stresses and
moduli should be used for solid timber members more than 100 mm
thick, unless they have been specially dried.
DURATION OF LOADING, K3
The stresses given in Tables 6.1 and 6.3 apply to long-term loading.
Where the applied loads will act for shorter durations e.g. snow and wind, the grade
stresses can be increased. This is because timber and wood based materials can sustain
much greater loads for shorter periods.
NOTCHED ENDS, K5
• Notches at the ends of flexural members will result in high shear
concentrations which may cause structural failure and must, therefore, be
taken into account during design.
• In notched members the grade shear stresses parallel to the grain are
multiplied by a modification factor K5 calculated as follows:
NOTCHED
ENDS, K5
For a notch at the top edge
ℎ ℎ𝑒 − 𝑎 + 𝑎ℎ𝑒
𝐾5 =
ℎ𝑒2
For a notch on the underside
ℎ𝑒
𝐾5 =
ℎ
the effective depth, he, should not be less than 0.5h,
i.e. K5 ≥ 0.5.
DEPTH
FACTOR, K7
The grade bending stresses given in Table 6.3 only
apply to timber sections having a depth h of 300
mm.
For other depths of beams, the grade bending
stresses are multiplied by the depth factor K7,
defined in clause 2.10.6 of BS 5268 as follows:
K7 = 1.17 for solid beams having a depth ≤ 72mm
300 0.11
ℎ
𝐾7 =
300mm
𝐾7 =
mm
for solid beams with 72mm < h <
0.81 (ℎ2 +923000)
(ℎ2 +56800)
for solid beams with h > 300
LOAD-SHARING SYSTEMS, K8
The grade stresses given in Tables 6.1 and 6.3 apply to individual
members, e.g. isolated beams and columns, rather than
assemblies.
When four or more members such as rafters, joists or wall studs,
spaced a maximum of 610 mm centre to centre act together to
resist a common load, the grade stress should be multiplied by a
load-sharing factor K8 which has a value of 1.1 (clause 2.9, BS
5268).
COMPRESSION MEMBERS, K12
The grade compression stresses parallel to the grain given in Tables 6.1 and
6.3 are used to design struts and columns.
These values apply to compression members with slenderness ratios less
than 5 which would fail by crushing.
Where the slenderness ratio of the member is equal to or greater than 5
the grade stresses should be multiplied by the modification factor K12
The factor K12 takes into account the tendency of the member to fail by
buckling and allows for imperfections such as out of straightness and
accidental load eccentricities.
The factor K12 is based on the minimum modulus of elasticity, Emin,
irrespective of whether the compression member acts alone or forms part
of a load-sharing system.
Questions
Discuss
Discuss the factors
which influence the
strength of timber
and explain how the
strength of timber is
assessed in practice.
Give
Give typical
applications of
timber in the
construction industry
and for each case
discuss possible
desirable properties
Distinguish
Distinguish between
softwood and
hardwood and grade
stress and
permissible stress.
Discuss
Discuss the factors
accounted for by the
modification factor
K12 in the design of
timber compression
members.
Beams, rafters and joists are examples of flexural
members.
Flexural
members
All calculations relating to their design are based on
the effective span and principally involves
consideration of the following:
1. bending
2. deflection
3. lateral buckling
4. shear
5. bearing.
Generally, for medium-span beams the design process
follows the sequence indicated above.
Deflection is usually critical for long-span beams and
shear for heavily loaded short-span beams.
Effective span of simply supported beams
According to clause 2.10.3 of BS 5268, for simply supported beams, the
effective span is normally taken as the distance between the centres of
bearings.
BENDING
If flexural members are not to fail in bending, the design moment, M, must not
exceed the moment of resistance, MR
M ≤ MR
The design moment is a function of the applied loads. The moment of resistance
for a beam can be derived from the theory of bending and is given by:
MR = σm,adm,||Zxx
where σm,adm,|| is the permissible bending stress parallel to grain and
Zxx is the section modulus
For rectangular sections Zxx=
𝑏𝑑 2
6
The permissible bending stress is calculated by multiplying the grade bending
stress, σm,g,||, by any relevant K-factors: σm,adm,|| = σm,g,||K2K3K7K8 (as appropriate).
For a given design moment the minimum required section modulus,
Zxx req, can be calculated as follows:
Zxx req ≥
𝑀
σm,adm,||
A suitable timber section can then be selected from Tables NA.2, NA.3 and NA.4
of BS EN 336: Structural timber.
Table NA.2 is reproduced as Table 6.7. Table 6.8 is an expanded version.
DEFLECTION
• Excessive deflection of flexural members may result in damage to
surfacing materials, ceilings, partitions and finishes, and to the functional
needs as well as aesthetic requirements.
• Clause 2.10.7 of BS 5268 recommends that generally such damage can
be avoided if the total deflection, δt ≤ δp
• The permissible deflection is generally given by δp = 0.003 × span
• for longer-span domestic floor joists, i.e. spans over 4.67 m, should not
exceed 14 mm: δp ≤ 14 mm
• The total deflection is the deflection due to bending and the deflection
due to shear.
δt = δm + δ v
LATERAL BUCKLING
• If flexural members are not effectively laterally restrained, it is possible for
the member to twist sideways before developing its full flexural strength
thereby causing it to fail in bending, shear or deflection.
• This phenomenon is called lateral buckling and can be avoided by ensuring
that the depth to breadth ratios are checked
LATERAL BUCKLING
Table 6.10 Maximum depth to breadth ratios Table 19, BS 5268
SHEAR
If flexural members are not to fail in shear, the applied shear stress
parallel to the grain, τa, should not exceed the permissible shear
stress, τadm
τa ≤ τadm
For a beam with a rectangular cross-section, the maximum applied
shear stress occurs at the neutral axis and is given by:
3Fv
τa =
2A
where Fv = applied maximum vertical shear force
A = cross-sectional area
The permissible shear stress is given by τadm = τgK2K3K5K8
where τg is the grade shear stress parallel to the grain
BEARING PERPENDICULAR TO GRAIN
• Bearing failure may arise in flexural members which are supported at their
ends on narrow beams or wall plates.
• Such failures can be avoided by ensuring that the applied bearing stress,
σc,a,⊥, never exceeds the permissible compression stress perpendicular to the
grain, σc,adm,⊥:
σc,a,⊥ ≤ σc,adm,⊥
The applied bearing stress is given by:
𝐹
σc,a,⊥ =
𝑏𝑙𝑏
where F bearing force (usually maximum reaction)
b breadth of section
lb bearing length
σc,adm,⊥ = σc,g,⊥K2K3K8
Moreover, two values for the grade compression stress perpendicular
to the grain are given for each strength class (Table 6.3).
The lower value takes into account the amount of wane which is
permitted within each stress grade.
Solid timber beams: Design for flexure, shear,
bearing, deflection. modification factors,
permissible stresses.
Design of a timber beam (BS 5268)
A timber beam with a clear span of 2.85 m supports a uniformly distributed
load of 10 kN including self-weight of beam. Determine a suitable section for
the beam using timber of strength class C16 under service class 1. Assume
that the bearing length is 150 mm and that the ends of the beam are held in
position and compression edge held in line.
EFFECTIVE SPAN Distance between centres of bearing (l) = 3000 mm
GRADE STRESS AND MODULUS OF ELASTICITY FOR C16
Values in N/mm2 are as follows
MODIFICATION FACTORS
• K2, moisture content factor does not apply since the beam is subject
to service class 1
• K3, duration of loading factor = 1.0
• K8, load sharing factor, does not apply since there is only a single
beam
• K7, depth factor =
300 0.11
h
• Assume h = 250, K7 = 1.020
DEFLECTION
M=
𝑊𝐿
8
=
10 𝑋 3
8
= 3.75 KNm
σm,adm,||(assuming h = 250) = σm,g,||K3K7 = 5.3 × 1.0 × 1.020 = 5.406 N/mm2
Zxx req≥
𝑀
σm,adm,||
=
3.75 x 106 𝑁𝑚
5.046 𝑁/𝑚𝑚2
= 694 x 103 mm3
DEFLECTION
Permissible deflection (δp) = 0.003 × span
The deflection due to shear (δs) is likely to be insignificant in comparison to
the bending deflection (δb) and may be ignored in order to make a first
estimate of the total deflection (δt):
DEFLECTION
δt(ignoring shear deflection) =
5Wl3
384Emin Ixx
Since δp ≥ δt
0.003 x 3000 ≥
5 x 104 x 30003
384 x 5800 x Ixx
Ixxreq ≥ 67.3 × 106 mm4
From Table 6.8, section 75 × 250 provides
Zxx = 781 × 103 mm3 Ixx = 97.7 × 106 mm4
A = 18.8 × 103 mm2
DEFLECTION
Hence total deflection including shear deflection can now be calculated and is
given by
δ t including shear deflection =
5 x104 x 30003
=
384 x 5800 x 97.7 x 106
+
5Wl3
384Emin Ixx
+
12 x104 x 3000
5 x 5800 x 18.8 x 103
12𝑊𝑙
5 𝐸𝑚𝑖𝑛 𝐴
= 6.2mm + 0.7mm = 6.9mm < δp
δp = 0.003 x 3 = 0.009 = 9 mm
Therefore a beam with a 75 × 250 section is adequate for bending and deflection
Lateral Buckling
Permissible d/b = 5
Actual = 250/75 = 3.3 (ok)
Hence the section is adequate for lateral buckling.
Shear
Permissible shear stress is
τadm = τgK2 K3= 0.67 × 1.0 = 0.67 N/mm2
Maximum shear force is =Fv=
𝑊
2
=
10 𝑋 103
2
= 5 X 103 N
Maximum shear stress at neutral axis is τa =
=
3 𝑋 5 𝑋 103
2 𝑋 18.8 𝑋 103
= 0.4 N/mm2 < permissible.
The section is adequate in shear.
3Fv
2A
BEARING
Permissible bearing stress is
σc,adm,⊥ = σc,g,⊥K2 K3 = 1.7 × 1.0 = 1.7 N/mm2
End reaction, F, is =
𝑊
2
=
10 𝑋 103
2
= 5 X 103 N
The applied bearing stress is given by:
σc,a,⊥ =
F
blb
=
5 x 103
75 x 150
= 0.44 N/mm2 < permissible
Therefore the section is adequate in bearing.
Since all the checks are satisfactory, use 75 mm × 250 mm sawn C16 beam.
Design of floor joist
A timber floor spanning 3.8 m centre
is to be designed using timber joist at
400 mm centres. The floor is subjected
to domestic imposed load of 1.5kN/m2
and carries dead loading, including self
weight of 0.35 kN/m2.
Carry out design checks to show that a
44 mm x 200 mm deep sawn section
Bruce spruce grade SS under service
class 1 suitable.
Joist Hanger
Joist hangers are designed to
provide support underneath the
joist, rafter or beam to provide a
strong a connection.
Geometrical Properties
Effective span Le = 3.8 m
Joist Spacing Js = 0.4 m
Joist dimensions
Breath of section, b = 44mm
Depth of section, h = 200mm
Area of section = b x h = 8.8 x 103 mm2
1
bh3
12
Second Moment of area Ixx=
Loading
Dead load, DL = 0.35kNm-2
Imposed load, IL = 1.5kNm-2
Total load , W = (DL + IL) .Js . Le
(0.35 + 1.5) x 0.4 x 3.8 = 2.81 kN
= 2.93 x 107 mm4
K-factors
Service Class 1 (K2 table 13)
K2 = 1
Load duration (K3, Table 14)
K3 =1 for long term
Bearing: assume 50mm, but located < 75mm
from the end of the member (K4, Table 15) (K4 = 1)
Notched end effect (K5 clause 2.10.4)
(K5 =1) for no notch
Form factor K6 clause 2.10.5
(K6 =1)
(2.10.5 Form factor Grade bending stresses apply to solid timber members of rectangular cross-section.
For other shapes of cross-section the grade bending stresses should be multiplied by the modification
factor, K6, where K6 = 1.18 for solid circular sections; and K6 = 1.41 for solid square sections loaded on a
diagonal)
Depth factor K7 clause 2.10.6
Load Sharing applies (K8, clause 2.9)
K7 =
K8 =1.1
300 0.11
h
= 1.05
2.9 Load-sharing systems
In a load-sharing system which consists of four or more members such as rafters, joists,
trusses or wall studs, spaced a maximum of 610 mm centre to centre, and which has
adequate provision for the lateral distribution of loads by means of purlins, binders,
boarding, battens, etc., the following permissible stresses and moduli of elasticity
appropriate to the strength class or species and grade should apply.
a) The appropriate grade stresses should be multiplied by the load sharing modification
factor, K8, which has a value of 1.1.
b) The mean modulus of elasticity should be used to calculate deflections and displacements
under both dead and imposed load unless the imposed load is for an area intended for
mechanical plant and equipment, or for storage, or for floors subject to vibrations, e.g.
gymnasia and ballrooms, in which case the minimum modulus of elasticity should be used.
Special provisions for built-up beams, trimmer joists and lintels, and laminated beams, are
given in 2.10.10, 2.10.11 and section 3, respectively.
The provisions of this clause do not extend to the calculation of modification factor, K12,
(given in Table 22 and Annex B) for load-sharing columns
Grade Stresses
BS5268: Part 2, Tables 2 and 7
British Spruce SS
Strength Class = C18
Bending parallel to the grain
σm,g,|| = 5.8N/mm2
Compression perpendicular to the grain σc,g,|| = 2.2N/mm2 (No wane)
Shear parallel to the grain
τg// =0.67N/mm2
Mean modulus of elasticity, load sharing Emean = 9100 N/mm2
Bending Stress
𝑊𝐿𝑒
8
2.82 x 3.8
M=
=
= 1.34 kNm
8
44 x 2002
Zprovided =
= 2.93 x 105 mm3 (For rectangular
6
M
1.34 x 106
2
σm,a,|| =
=
=4.55N/mm
Z provided
2.93 x 105
sections Zxx=
𝑏𝑑 2
)
6
Permissible Bending Stress
σm,adm,|| = σm,g,|| x K2K3K6K7K8 = 5.8N/mm2 x 1.05 x1.1= 6.67N/mm2
Bending stress is satisfactory
Lateral Stability
BS5268: Part 2, Clause 2.10.8 and Table 16
Permissible d/b = 5 (Ends held in position and compression edges held in line)
Actual = 200/44 = 4.55 (ok)
Hence the section is adequate for lateral stability.
Shear Stress
Permissible shear stress is
τadm = τgK2K3K5K8 =
0.67 × 1.1 = 0.74 N/mm2
𝑊 2.81
Applied shear force is =Fv= =
= 1.41kN
2
2
Applied shear stress at neutral axis is τa =
3 𝑋 1.41 𝑋 103
2 𝑋 8.8 𝑋 103
3Fv
2A
=
= 0.24 N/mm2 < permissible.
The section is adequate in shear.
Applied shear force is =Fv=
𝑊
2
=
2.81
2
= 1.41kN
Assume a bearing length of 50mm on either side 𝑙𝑏 =50mm
The applied bearing stress is given by:
σc,a,⊥ =
𝐹
𝑏𝑙𝑏
=
1.41 x 103
44 x 50
= 0.64 N/mm2
σc,adm,⊥ = σc,g,⊥K2 K3 K4 K8 = 2.2N/mm2 × 1.1 = 2.42 N/mm2
σc,a,⊥ ≤ σc,adm,⊥ (Bearing stress satisfactory)
δt = δm + δv
5Wl3
384Em𝑒𝑎𝑛 Ixx
19.2 𝑀𝑚𝑎𝑥
𝐴𝐸
+
19.2 𝑀𝑚𝑎𝑥
𝐴𝐸𝑚𝑒𝑎𝑛
is the maximum shear deflection induced in a single span simply
supported beam of either rectangular or square cross section.
δm =
Δv =
5Wl3
384Em𝑒𝑎𝑛 Ixx
19.2 𝑀𝑚𝑎𝑥
𝐴𝐸𝑚𝑒𝑎𝑛
=
=
5 x 2810 x 38003
384 x 9100 x 2.93 x 107
19.2x 1.34 x 106
8.8 x 103 x 9100
= 7.53mm
= 0.32mm
δt = 7.53 + 0.32 = 7.85mm
The permissible deflection is generally given by δp = 0.003 × span
=0.003 x 3800
=11.4mm
Therefore the 44mm x 200mm sawn sections in C18 timber is satisfactory
Design of floor joist – selection of suitable and design for notched ends.
The cross-section of a suspended timber
flooring is shown in the figure below. It consist
of tongued and grooved (t & g) boarding with a
self-weight of 0.15kN/m2 and carries a
plasterboard ceiling of 0.2kN/m2. The floor has
an effective span of 4.0 m and is subjected to
domestic imposed load of 1.5 kN/m2. Design
the timber floor joists using timber in strength
class C18 under service class 1.
If the joist are to be notched at bearings with a
72mm deep notch, check that the notched
section is adequate.
Geometrical properties
Effective span Le = 4.0 m
Joist Spacing Js = 0.45 m
Loading
Dead load,
t & g boarding = 0.15kNm-2
Plasterboard ceiling Pb = 0.2kNm-2
Self-weight Swt. = 0.1kNm-2 (assumed)
Imposed load, IL = 1.5kNm-2
Total load , W = (tg + Pb + Swt + IL) .Js . Le
(0.15 + 0.2 + 0.1 + 1.5) x 0.45 x 4= 3.51 kN
K-factors
Service Class 1 (K2 table 16)
K2 = 1
Load duration (K3, Table 17)
K3 =1 for long term
(K4, Table 18) assume (K4 = 1)
Notched end effect (K5 clause 2.10.4)
(K5 =1) for no notch
Form factor K6 clause 2.10.5
(K6 =1)
(2.10.5 Form factor Grade bending stresses apply to solid timber members of rectangular cross-section.
For other shapes of cross-section the grade bending stresses should be multiplied by the modification
factor, K6, where K6 = 1.18 for solid circular sections; and K6 = 1.41 for solid square sections loaded on a
diagonal)
Depth factor K7 clause 2.10.6
Load Sharing applies (K8, clause 2.9)
K7 =
300 0.11
h
K8 =1.1
Grade Stresses
BS5268: Part 2, Table 7
Strength Class = C18
Bending parallel to the grain
σm,g,|| = 5.8N/mm2
Compression perpendicular to the grain σc,g,|| = 2.2N/mm2 (No wane)
Shear parallel to the grain
τg// =0.67N/mm2
Mean modulus of elasticity, load sharing Emean = 9100 N/mm2
Bending Stress
M=
𝑊𝐿𝑒
8
=
3.51 x 4
8
= 1.755 kNm
Permissible bending stress σm,adm,|| = σm,g,|| x K2K3K6K8 = 5.8N/mm2 x 1.1=
6.38N/mm2
Zrequired =
M
σm,a,||
=
1.755 x 106
6.38
= 2.75 x 105 mm3
Lateral Stability
BS5268: Part 2, Clause 2.10.8 and Table 16
In order to achieve lateral stability by direct fixing of decking to joists, the depth
to breath ratio should be limited to 5, i.e. h ≤ 5b. Substituting for h = 5b and Zxx
=bh2/6 and equating Zxx required gives:
Zrequired =
b=
b x (5b)2
6
1
6 x zreq 3
52
=b=
6 x 2.75 x 105 105
52
1
3
= 40.42 mm
h = 5b = 5 x 40.42 = 202.08mm
Selecting a beam section from table A2, Appendix A. Try a 47mm x 200mm
deep section
Beam dimensions
Breath of section, b = 47 mm
Depth of section, h = 200mm
Zprovided =
b x h2
6
= 3.13 x 105
300 0.11
h
Modification factor K7 for section depth =
= 1.05
Permissible Bending Stress
σm,adm,|| = σm,g,|| x K2K3K6K7K8 = 5.8N/mm2 x 1.05 x1.1= 6.67N/mm2
Bending stress is satisfactory.
Self Weight
BS5268: Part 2, Table 7
Average density ρ = 380 kgm-3
Total joist selfweight = Swtactual = ρghbLe (380 kgm-3/1000) (9.81m/s2)(0.047m x 0.2m x
4m)
Swtactual = 0.14kN
Swtassumed =Swt x Le x Js = 0.1kN/m2 x 4m x 0.45m = 0.18kN ( Self weight is
satisfactory)
Shear Stress
Permissible shear stress is
(i) No notch τadm = τgK2K3K5K8 = 0.67 × 1.1 = 0.74 N/mm2
With notch 72mm deep
Effective section depth he = h – notch
he = 128mm
BS5268: Part 2, Clause 2.10.4 K5 =
ℎ𝑒
ℎ
= 128/200 = 0.64
With notch τadm = τgK2K3K5K8 = 0.67 × 1.1 x 0.64 = 0.47 N/mm2
Applied shear force is =Fv=
𝑊
2
=
3.51
2
= 1.75kN
Applied shear stress at neutral axis is τa =
3 X 1.75 X 103
2 X 47 x 200
3Fv
2A
=
= 0.24 N/mm2 < permissible.
The section is adequate in shear.
Bearing Stress
Applied shear force is =Fv=
𝑊
2
=
3.51
2
= 1.75kN
σc,adm,⊥ = σc,g,⊥K2 K3 K4 K8 = 2.2N/mm2 × 1.1 = 2.42 N/mm2
Minimum bearing width bw required=
Fv
b x σc,adm,⊥
= 15.43mm
δt = δm + δv
5Wl3
384Em𝑒𝑎𝑛 Ixx
Ixx=
𝑏ℎ3
12
δm =
Δv =
+
19.2 𝑀𝑚𝑎𝑥
𝐴𝐸𝑚𝑒𝑎𝑛
= 3.13 x 107 mm4
5Wl3
384Em𝑒𝑎𝑛 Ixx
19.2 𝑀𝑚𝑎𝑥
𝐴𝐸𝑚𝑒𝑎𝑛
=
=
5 x 3510 x 40003
384 x 9100 x 3.13 x 107
19.2x 1.75 x 106
47 x 200 x 9100
= 10.26mm
= 0.39mm
δt = 10.26 + 0.39 = 10.65mm
The permissible deflection is generally given by δp = 0.003 × span
=0.003 x 4000
=12 mm
Therefore the 47mm x 200mm sawn sections in C18 timber is satisfactory
Design of flooring elements and components
in solid timber
Design of a flooring system- floor boards and joist
Design of a flooring system- floor boards and joist
The ground floor of a shop is to comprise a series of timber joist at
600mm centres with tongued and groove boardings (t & g) The
joist are simply supported on 100 mm hangers attached to load
bearing walls 4.2 m apart. Determine the thickness of the floor
boarding using timber in strength class C18 and a suitable size for
joists using timber in C22 under service class 2. Assume imposed
load of 2.0kN/m2
•
•
•
The floor boards may be designed as simply supported beams
Calculations are required for bending strength and deflections
Assume tongued and groove boarding 100mm wide and thickness t simply supported on joist
Geometrical properties
Effective span Le = joist spacing js = 0.6m
Assume a t & g boarding of width b = 100 mm
Loading
t & g boarding = 0.1kNm-2
Imposed load, IL = 2 kNm-2
Total load , W = (tg + IL) . b . Le
(0.1 + 2 ) x 0.1 x 0.6 = 0.126 kN
K-factors
Service Class 1 (K2 table 13)
K2 = 1
Load duration (K3, Table 14)
K3 =1 for long term
Bearing assume (K4, Table 15)
(K4 = 1) assumed
Notched end effect (K5 clause 2.10.4)
(K5 =1) for no notch
Form factor K6 clause 2.10.5
(K6 =1)
Depth factor K7 clause 2.10.6 for h ≤ 72mm
K7 =1.17
Load Shearing applies (K8, clause 2.9)
K8 =1.1
K-factors
Service Class 1 (K2 table 13)
K2 = 1
Load duration (K3, Table 14)
K3 =1 for long term
Bearing assume (K4, Table 15)
(K4 = 1) assumed
Notched end effect (K5 clause 2.10.4)
(K5 =1) for no notch
Form factor K6 clause 2.10.5
(K6 =1)
Depth factor K7 clause 2.10.6 for h<72mm
K7 = 1.17
Load Shearing applies (K8, clause 2.9)
K8 =1.1
Grade Stresses (t&g)
BS5268: Part 2, Tables 8 and 9
Strength Class = 18
Bending parallel to the grain
σm,g,|| = 5.8N/mm2
Compression perpendicular to the grain σc,g,|| = 2.2N/mm2 (No wane)
Shear parallel to the grain
τg// =0.67N/mm2
Mean modulus of elasticity, load sharing Emean = 9100 N/mm2
Bending Stress
𝑊𝐿𝑒 0.126 x 0.6
M=
=
= 9.45 x 10-3 kNm
8
8
Permissible bending stress σm,adm,|| = σm,g,|| x K2K3K6K7K8 = 5.8N/mm2 x 1.1 x 1.17=
7.46N/mm2
M
Zrequired =
σm,a,||
9.45 x 10−3 x 106 𝑁𝑚𝑚
Zrequired =
= 1.27 x 103 mm3
2
7.46 N/mm
Zrequired =
b t2
6
= 1.27 x 103 mm3 where b = 100mm
t = 8.72 mm
Deflection
The permissible deflection is generally given by δp = 0.003 × span
=0.003 x 600
=1.8 mm
δm =
5Wl3
384Em𝑒𝑎𝑛 Ixx
=
5 x 0.126 x 1000 x 6003
=
384 x 9100 x Ixx
Ixx = 2.16 x 104 mm4
Ixx =
b t3
12
t = 13.74 mm
Use t = 16 mm t & g boarding of strength class C18
1.8mm
Design of floor joist
Geometrical properties
Effective span Le = 4.1 m
joist spacing js = 0.6 m
Bearing width bw = 100 mm
Loading
Average density ( Table 7) = 410 kg m-3
t & g boarding = ρgt = 410kg m-3 x 9.81m/s2 x 0.016m = 64Nm-2 = 0.064
kNm-2
Self weight of each joist = 0.1 kNm-2 assumed
Imposed load, IL = 2 kNm-2
Total load , W = (tg + sw+ IL) . js . Le
(0.064 + 0.1 + 2 ) x 0.6 x 4.1 = 5.32 kN
K-factors
Service Class 2 (K2 table 13)
Load duration (K3, Table 14)
Bearing assume (K4, Table 15)
Notched end effect (K5 clause 2.10.4)
Form factor K6 clause 2.10.5
Depth factor K7 clause 2.10.6
Load Shearing applies (K8, clause 2.9)
K2 = 1
K3 =1 for long term
(K4 = 1) assumed
(K5 =1) for no notch
(K6 =1)
Ignore K7 at this stage
K8 =1.1
Grade Stresses
BS5268: Part 2, Table 8
Strength Class = C22
Bending parallel to the grain
σm,g,|| = 6.8N/mm2
Compression perpendicular to the grain σc,g,|| = 2.3 N/mm2 (No wane)
Shear parallel to the grain
τg// =0.71N/mm2
Mean modulus of elasticity, load sharing Emean = 9700 N/mm2
Bending Stress
𝑊𝐿𝑒 5.32 x 4.1
M=
=
= 2.73 kNm
8
8
Permissible bending stress σm,adm,|| = σm,g,|| x K2K3K6K8 = 6.8N/mm2 x 1.1 =
7.48N/mm2
M
Zrequired =
σm,a,||
2.73 x 106
Zrequired =
= 3.65 x 105 mm3
7.48
Lateral Stability
BS5268: Part 2, Clause 2.10.8 and Table 16
In order to achieve lateral stability by direct fixing of decking to joists, the
depth to breath ratio should be limited to 5, i.e. h ≤ 5b. Substituting for h
= 5b and Zxx =bh2/6 and equating Zxx required gives:
Zrequired =
b=
b x (5b)2
6
1
6 x zreq 3
52
=b=
1
6 x 3.65 x 105 3
52
= 44.4 mm
h = 5b = 5 x 44.4 = 222.02mm
Selecting a beam section from table A2, Appendix A. Try a 47mm x 225
mm deep section
Beam dimension
Depth h = 225mm
Breadth b = 47mm
b x (h)2
6
Section Modulus Zprovided =
Zprovided = 3.97 x 105 mm3
300 0.11
Depth factor K7
K7 =
= 1.03
h
Permissible Bending Stress
σm,adm,|| = σm,g,|| x K2K3K6K7K8 = 6.8N/mm2 x 1.03 x1.1= 7.7N/mm2
Bending stress is satisfactory.
Self Weight
BS5268: Part 2, Table 7
Average density ρ = 410 kgm-3
Total joist self weight = Swtactual = ρghbLe = 410 x 9.81 x 0.047 x 0.225 x 4.1 = 174 N
Swtactual = 0.17kN
Swtassumed =Swt x Le x Js = 0.1 x 0.6 x 4.1 = 0.25kN ( Self weight is satisfactory)
Shear Stress
Permissible shear stress is
(i) No notch τadm = τgK2K3K5K8 = 0.71 × 1.1 = 0.78 N/mm2
Applied shear force is =Fv=
𝑊
2
=
5.32
2
= 2.66 kN
Applied shear stress at neutral axis is τa =
=
3 X 2.66 X 103
2 X 47 x 225
3Fv
2A
= 0.38 N/mm2 < permissible.
The section is adequate in shear.
Bearing Stress
Applied shear force is =Fv=
𝑊
2
=
5.32
2
= 2.66 kN
σc,adm,⊥ = σc,g,⊥K2 K3 K4 K8 = 2.3N/mm2 × 1.1 = 2.53 N/mm2
Fv
Minimum bearing width bw required=
=
b x σc,adm,⊥
mm, OK
2.66 x 1000
47 x 2.53
22.39mm < 100
δt = δm + δv
5Wl3
384Em𝑒𝑎𝑛 Ixx
Ixx=
𝑏ℎ3
12
+
19.2 𝑀𝑚𝑎𝑥
𝐴𝐸𝑚𝑒𝑎𝑛
= 4.46 x 107 mm4
5Wl3
5 x 5320 x 41003
δm =
=
= 11.04mm
7
384Em𝑒𝑎𝑛 Ixx 384 x 9700 x 4.46 x 10
19.2 𝑀𝑚𝑎𝑥 19.2x 2.73 x 106
Δv =
=
= 0.51mm
𝐴𝐸𝑚𝑒𝑎𝑛
47 x 225 x 9700
δt = 11.04+ 0.51 = 11.55mm
The permissible deflection is generally given by δp = 0.003 × span
=0.003 x 4100
=12.3 mm
Therefore the 47mm x 225mm sawn sections in C22 timber is satisfactory
Design of axially loaded members
Compression members include posts or columns, vertical wall stud, and
struts and truss girders.
Permissible stresses are governed by the particle conditions of service
and loading in clause 2.6.2, 2.8 and 2.9 of BS5268: Part 2: 1968.
Clause 2.11 deals with the design of compression members and divides
them into two categories:
• members subjected to axial compression (without bending), and
members subjected to combined axial compression and bending ( due
to eccentric compressive force)
Design considerations
The main design considerations for compression members are:
(i) Slenderness ratio – This relates the positional restraint of ends, lateral restraint
along the length and cross sectional dimensions of the member.
(ii) Axial compression and bending stress.
Slenderness Ratio, λ
The load-carrying capacity of compression members is a function of the slenderness
ratio, λ, which is calculated as the effective length Le divided by the radius of gyration,
i:
λ=
𝐿𝑒
𝑖
and the radius of gyration is given by 𝑖 =
𝐼
𝐴
Where I is the second moment of area and A is the cross sectional area of the
member
Slenderness Ratio, λ
For rectangular sections, where b is the least lateral dimension, the
value of i simplifies to:
i=
𝑏
12
Clause 2.11.4 of BS5268: Part 2: 1996 recommends that the
slenderness ratio should not exceed the value of :
λ = 180 for compression members carrying dead and imposed loads
other than loads resulting from wind,
λ = 250 for any member subject to reversal of axial stress solely from
the effect of wind and any compression member carrying self weight
and wind loads only.
Slenderness Ratio, λ
The effective length Le of a column given in clause 2.11.3 should be
derived from either :
1. The deflected form of compression member affected by any
restraint and fixing moment(s). Then the effective length is
considered as the distance between adjacent points of zero bending
moment.
2. Table 21 of the code for particular end conditions at the column
ends. The effective length is obtained by multiplying a relevant
coefficient by the actual length.
Effective lengths and end conditions
Slenderness Ratio, λ
• For compression members with slenderness ratios equal to or greater than
5, Clause 2.11.5 of the code requires that the permissible compressive
stresses be further modified by K12 – the modification factor for compression
members.
• The modification factor for K12 can be determined by using table 22 of
BS5268 or calculated from equation. For either method, the value of the
modulus of elasticity Emin should be used in all cases , including when load
sharing is present.
• The value of σc,|| for use in either method should be the σc,g,|| given in
tables 7 – 12a of the code modified only for moisture content, duration of
loading and size where appropiate.
• Members comprising of two or more pieces connected together in parallel
and acting together to support the loads, the minimum modulus of elasticity
should be modified by K9 ( Table 20) or K28 (Table 25) of the code
Slenderness Ratio, λ
• For horizontally laminated members, the mean modulus of elasticity
should be used ( see clause 3.2 and 3.6 of the code).
K12 may calculated as follows:
K12 =
1
2
+
1+ η π2 𝐸
3λ2 σ𝑐
−
1
2
+
1+ η π2 𝐸 2
3λ2 σ𝑐
−
π2 𝐸 1/2
1.5λ2 σ𝑐
σ𝑐 = permissible stress for very short column λ < 5, σc,g,|| x K2K3
E = minimum modulus of elasticity Emin
λ= 𝑠𝑙𝑒𝑛𝑑𝑒𝑟𝑛𝑒𝑠𝑠
η = 0.005 λ
𝐿𝑒
𝑟𝑎𝑡𝑖𝑜,
𝑖
Members subjected to axial compression only (Clause 1.11.5)
An axially loaded column has its line of action of load passing through
the centroid axis of the column.
The axial compressive stress σc,g,|| is given as
σc,g,|| =
𝑃
𝐴
Where P is the axial compressive load and A is the cross sectional area.
If λ < 5 σc,adm,|| = σc,g,|| x k2k3K8
If λ ≥ 5 σc,adm,|| = σc,g,|| x k2k3K8K12
The Compression member is designed so that σc,a,|| ≤ σc,adm,||
Axial, concentric and eccentric loads
Members subjected to axial compression and bending ( Clause
2.11.6)
This includes compression members subject to eccentric loading,
where the load acts through a point at a certain distance from the
centroidal axis, which can be equated to the axial compression and
bending moment (b, c or d).
Members which are restrained at both ends in position but not
direction should be so proportioned so that:
Interaction
formula
Interaction Formula
The interaction formula is used to ensure lateral instability does not
arise in compression members subject to axial force and bending.
If the column is subjected to compressive loading only σm,a,|| = 0, and
the equation simplifies to σc,a,|| /σc,adm,|| ≤ 1.
If the column is subjected to bending only σc,a,|| = 0, and the equation
simplifies to σm,a,|| /σm,adm,|| ≤ 1.
Design of Load Bearing Stud Walls
Stud walls are often constructed as load bearing walls in timber framed housing.
Details of a typical stud wall
Stud walls
• Stud walls consist of vertical timber members, commonly referred to as studs, which
are held in position by nailing them to timber rails or plates, located along the top
and bottom of the studs.
• The walls can be designed to resist vertical and lateral loadings, wind load being a
typical example of a lateral load.
• Each stud may be considered to be laterally restrained about the y-y axis, fully or
partially restrained, (e.g. at mid-height), either by cladding/sheathing materials, such
as plasterboard and internal noggins or diagonal bracing.
• In situations where the cladding/ sheathing material is properly attached to the stud
along its whole length, the strength of the stud can be calculated about the x-x axis;
otherwise, the greater of the slenderness ratio about the individual stud’s x-x and y-y
axis should be considered in the design calculation.
Stud walls
Design Examples – Load capacity of timber column
A timber column of strength class C18 is 4m in
height with rectangular cross-section of 97mm x
145mm. The column is restrained at both ends in
position but not in direction and is subjected to
service class 2 conditions.
(i) Determine the maximum axial long-term load
that the column could support.
(ii)Check the adequacy of the column to resist a
long-term axial load at 12kN and a bending
moment of 0.8kNm about the x-x axis.
Column Details
Geometrical Properties
K-factors
Service Class 2 (K2 table 16)
Load duration (K3, Table 17)
Form factor K6 clause 2.10.5
Depth factor K7 clause 2.10.6
K7 = 1.08
Load Sharing (K8, clause 2.9)
K2 = 1
K3 =1 for long term
(K6 =1)
K7 =
300 0.11
h
K8 =1.0 ( no load sharing)
K12
Using Table 22