im Sc, MSc, PhD, FIVVSc ~ a ~ i ~niver~ity, er Edinbur~~
Blackwell Science
0 1999by Blackwell Science Ltd Editorial Offices: Osney Mead, Oxford OX2 OEL 25 John Street, London WClN 2BL 23 Ainslie Place, Edinburgh EH3 6AJ 350 Main Street, Malden MA 02148 5018, USA 54 University Street, Carlton Victoria 3053, Australia 10, rue Casimir Delavigne 75006 Paris, France Other Editorial Offices: Blackwell WissenschaftsVerlagGmbH Kurfurstendamm 57 10707 Berlin, Germany Blackwell Science KK MG Kodenmacho Building 710 Kodenmacho Nihombashi Chuoku, Tokyo 104, Japan The right of the Author to be identified as the Author of this Work has been asserted in accordance with the Copyright, Designs and Patents Act 1988. All rights reserved.No partof thispublication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical,photocopying, recording or otherwise, except as permitted by the UK Copyright, Designs and Patents Act 1988, without prior permission of the publisher. First published 1999
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Preface 1 Timber as a Structural Material 1.1 ~ntroduction 1.2 The structure of timber 1.3Defects in timber 1.3.1 Natural defects 1.3.2Chemicaldefects 1.3.3Conversiondefects 1.3.4Seasoning defects 1.4Typesof timber l .4.1 Softwoods 1.4.2 Hardwoods 1.5Physical properties of timber 1.5. 1 Moisture content 1.5.2 Density 15.3Slopeof grain 1.5.4Timberdefects 1.6 References S 5268 : Part 2 : 11996 2.1 Introduction 2.2Design philosophy 2.3Stress grading of timber 2.3.1Visual grading 2.3.2Machine grading 2.4 Strength classes 2.5Design considerations (factors affecting timber strength) 2.5.1 Loading 2.5.2Serviceclasses 2.5.3 Moisture content 2.5.4 Duration of loading
xi
1 1 3 3 4 5 5 5 5 5 6 6 7 7 8 8 9 9 10 12 12 12 13 14 14 14 16 16 V
vi
Contents
2.5.5Sectionsize 2.5.6 Loadsharingsystems 2.5.7 Additional properties 2.6 Symbols 2.7 References
17 18 18 19 20
3 UsingMathcad@ for Design C ~ c u l a ~ o n s 3.1 Introduction 3.2WhatisMathcad? 3.3WhatdoesMathcaddo? 3.3.1 Asimple calculation 3.3.2Definitions and variables 3.3.3 Entering text 3.3.4 Workingwith units 3.4 Summary 3.5 References
21. 21
4 Design of FlexuralMembers(Beams) 4. l Introduction 4.2 Design considerations 4.3 Bending stress and prevention of lateral buckling 4.3.1EAFective span, L, 4.3.2 Form factor, KG 4.3.3Depth factor, K7 4.3.4 Selection of a suitable section size 4.3.5 Lateral stability 4.3.6An illustrative example 4.4 Deflection 4.4.1Deflectionlimits 4.4.2 Precamber 4.4.3Bendingdeflection 4.4.4Sheardeflection 4.5 Bearing stress 4.5.1Lengthand position of bearings 4.6 Shear stress 4.6.1Shear at notched ends 4.7 Suspended timber flooring 4.8 References 4.9 Design examples Example 4.1 Example 4.2 Example 4.3 Example 4.4
26 26 26 27 28 29 29 30 30 31 32 32 33 33 34 34 35 36 36 37 39 40 40 43 46 50
21 22 22 22 24 24 25 25
Contents
esign of Axially Loaded Members 5. l Introduction 5.2Designofcompressionmembers 5.2. l Design considerations 5.2.2 Slenderness ratio, h 5.2.3 Modification factor for compression members, K12 5.2.4 Members subjected to axial compression only (Clause 2.1 1.5) 5.2.5 Members subjected to axial compression and bending (Clause 2.1l .6) 5.2.6 Design of loadbearing stud walls 5.3 Design of tension members (Clause 2.12) 5.3.1Design considerations 5.3.2Width factor, K14 5.3.3 Members subjected to axial tension only 5.3.4 Combined bending and tensilestresses 5.4Designexamples Example 5.1 Example 5.2 Example 5.3 Example 5.4 Example 5.5
6 Design of GluedLaminatedMembers 6.1 Introduction 6.2 Design considerations 6.3 Grade stresses for horizontally glued laminated members 6.3.1Singlegrademembers 6.3.2 Combinedgrademembers 6.3.3 Permissible stresses for horizontally glued laminated members 6.4 Grade stresses for vertically glued laminated beams 6.4.1Permissiblestresses for verticallyglued laminated members 6.5 Deformation criteria for glued laminated beams 6.6 Curved glued laminated beams 6.7 Bibliography 6.8 Design examples Example 6.1 Example 6.2 Example 6.3 Example 6.4 Example 6.5
vii
56 56 56 56 57 58
60 61 62 63 64 64 64
65 66 66 69 73 75 79
82 82 84 84 84 84 86 87 89 90 90 92 93 93 97 102 113 119
viii
Contents
esign of P l y  w e b b ~Beams 7.1 Introduction 7.2 Transformed (effective) geometrical properties 7.3 Plywood 7 '4 Design considerations 7.4.Bending l 7.4.2 Deflection 7.4.3 Panelshear 7.4.4 Rollingshear 7.4.5 Lateral stability 7.4.6 Webstiffeners 7.5 References 7.6 Design examples Example 7.1 Example 7.2
8 Design of (Spaced) Builtup Columns 8.1 Introduction 8.2 Spaced columns 8.3 Design considerations 8.3.1 Geometricalrequirements 8.3.2 Modes of failure and permissible loads 8.3.3 Shear capacity ofspacerblocks 8.4 Compression members in triangulated frameworks 8.5 Reference 8.6 Design examples Example 8.1 Example 8.2 esign of Timber Connections
9. l Introduction 9.2 Generaldesignconsiderations 9.3 Joint slip 9.4Effectivecrosssection 9.5Spacingrules 9.6Multipleshearlateralloads 9.7Nailed joints 9.7.1 Improvednails 9.7.2 Predrilling 9.7.3Basicsingleshear lateral loads 9.7.4Axiallyloadednails(withdrawalloads) 9.7.5 Permissible load for a nailed joint 9.8Screwed joints 9.8.1Basicsingle shear lateral loads 9.8.2Axiallyloadedscrews(withdrawalloads) 9.8.3 Permissible load for a screwed joint
123
123 l 24 125 127 128 129 130 130 131 131 132 132 132 137 143
143 144 145 145 145 147 148 148 148 148 152 159
159 160 161 162 163 165 166 167 167 169 172 173 175 176 177 178
Contents
9.9 Bolted and dowelled joints 9.9.1Basicsingle shear lateral loads 9.9.2 Permissible load for a bolted or dowelled joint 9.10 Moment capacity of doweltype fastener joints 9.1 1Connectored joints 9.1 l. l Toothedplate connectors 9.1 l .2 Splitring and shearplate connectors 9.1 l .3 Metalplate connectors 9.12 Clued joints 9.12.1 Durability classification 9.12.2 Design considerations for glued joints 9.13 References 9.14 Design examples Example 9.1 Example 9.2 Example 9.3 Example 9.4 Example 9.5 Example 9.6 Example 9.7 Example 9.8 esign to Euroco~e5 10.1 Introduction 10.2 Design philosophy 10.3 Actions 10.4 Material properties 10.4.1Designvalues 10.5 Ultimate limit states 10.5.1 Bending 10.5.2 Shear 10.5.3 Compression perpendicular to grain (bearing) 10.5.4 Compression or tension parallel to grain 10.5.5 Members subjected to combined bending and axial tension 10.5.6 Columns subjected to combined bending and axial compression 10.5.7 Doweltype fastener joints 10.6 Serviceability limit states 10.6.1Deflections 10.6.2 Vibrations 10.6.3 Joint slip 10.7 Reference 10.8 Bibliography
ix
179 180 184 185 188 189 191 193 195 195 196 196 197 197 199 20 1 204 208 210 213 216
219 220 22 1 22 1 223 224 225 227 228 229 230 230 23 1 239 239 240 24 1 242 242
x
Contents
10.9Designexamples Example 10.1 Example 10.2 Example 10.3
242 242 248 25 1
Appendix A: Section Sizes for Softwood Timber Appendix B: Weights of Building ater rials Appendix C: Related BritishStandards for TimberEngineering
257 259 260
Index
263
The increasing recognition of timber as a structural material is reflected in the inclusion oftimber design in many undergraduate courses. majority of design textbooks for undergraduate engineering students neglect, to a large extent, the importance of timber as a structural and building material. As a consequence, relativelyfew textbooks provide information on the designoftimber structures. Structural TimberDesign is intended to address this issue by providing a stepbystep approach to thedesign of all the most commonly used timber elements and joints illustrated by detailed worked examples.This is an approachwhich is recognisedto be beneficial in learning and preferred by most students. The book has been written for undergraduate students on building, civil and structural engineeringand architectural courses and will be an invaluable reference source anddesign aid for practising engineers and postgraduate engineering students. It provides a comprehensive source of information on practical timber design and encourages theuse of computers to carry out design calculations. Chapter 1 introduces the nature and inherent characteristicsof timber such as defects, moisture content and slope of grain, and discusses the types of timber and factors that influence their structural characte includes a comprehensive review of the recently revised t 2: 1996: Structural Use of Timber. The design philosophy of its new approach to the strength classsystem and also the factors affecting timber strength are explained. Chapter 3 gives an overviewofMathcad@,acomputer software programmeused to carry out mathematical calculations, and details its simplicity and the advantages that it provides when used for design calculations. Theaimis to encourage readers to usecomputing as a tool to increasetheir understanding ofhowdesign solutions vary in response to a change in oneof the variables and howalternative design options canbe obtained easily and effortlessly. The design of basic elements is explained and illustrated in Chapters 4 and 5, whilst the design of more specialised elements such as glued laminated straight and curved beams and columns, plywebbed beams and builtup columns isillustrated in Chapters 6, 7 and 8 using numerous worked examples. xi
xii
Preface
InChapter 9thedesignoftimberconnectionsisdetailed.The new approach adoptedby the revised BS 5268:Part 2 in 1996, i.e.the Eurocode 5 approach for the design of timber joints, is described.The chapter includes a comprehensivecoverageofthedesignrequirements for nailed,screwed, bolted and dowelled joints, and the design of connectored joints such as toothedplates, splitrings and shearplates and glued connections is also detailed. Several stepbystep worked examples are provided to illustrate the design methods in this chapter. Chapter 10 provides a comprehensive review of the proposed European code for timber, Eurocode 5: Design of Timber Structures. The limit states design philosophy of EC5 is explainedand the relevant differences with the design methodology of BS 5268 are high~ghtedand discussed. This chapter also provides comprehensive coverage of EC5 requirementsfor the design of flexural and axially 1o.aded members and doweltype connections such as nailed,screwed,bolted and dowelled joints. Again,stepbystepworked examples are provided to illustrate the design methods in the chapter. Alldesignexamplesgiveninthisbook are produced in theform of worksheet files and are available from theauthor on 3 r disks to run under Mathcad computersoftwareVersion6, or higher, in eitherone of its editions: (Student, Standard, Plus or Professional). Details are given at the end of the book. The examples are fully selfexplanatoryand well annotated and the author isconfident that thereaderswhether students, course instructors, or practising design engineerswill find them extremely usefulto produce designsolutions or prepare course handouts. In particular, the worksheets willallowdesignengineers to arrive at themost suitable/ economic solution(s) very quickly. Extracts from British Standards are reproduced with the permission of BSI under licence no. PD\1998 0823. Complete editions of thestandards can be obtained by post from BSI Customer Services, 389 Chiswick High Road, London'W4 4AL. The cover illustration was kindly supplied by MiTek Industries Ltd.
ructural
1.1 Introduction Timber has always been one the of more plentiful natural resources available and consequently is oneof the oldest known materials used in construction. It is a material that is used for a variety of structural forms such as beams, columns, trusses, girders and is also used in building systems such as piles, deck members, railway foundations and for temporary f o m s in concrete. Timber structures can be highly durable when properly treated and built. Examples of this are seen in many historic buildings all around the world. Timber possesses excellent insulating properties, good fire resistance, light weight and aesthetic appeal. A great deal of research carried out since the early part of this century has provided us with comprehensive information on structural properties of timber and timber products'. A knowledge of engineering materials is essential for engineering design. Timber is a traditional building material and over the years considerable knowledge has been gained on its important material properties and their effects on structural design and service behaviour. Many failures in timber buildingsin the pasthaveshown us the safe methods of construction, connection details and design limitations. This chapter provides a brief description of the engineering properties of timber that are of interest to design engineersor architects. But it should be kept in mindthat, unlike some structural materials suchas steel or concrete, the properties of timber are very sensitive to environmental conditions. For example, timber is very sensitive to moisture content, which has a direct effect on the strength and stiffness, swellingor shrinkage of timber. A proper understanding of the physical characteristics of wood aids the building of safe timber structures"
1.2 The structure of timber * Mature trees of whatever type are the source of structural timber and it is important that users of timber should have a knowledge of the nature and growth patterns of trees inorder to understand its behaviour undera variety 1
Structural TimberDesign
of circumstances. Basically, a tree has three subsystems: roots, trunk and ow^^ Each subsystem has a role to play in the growth pattern of the tree. Roots, by spreading through the soil as well as acting as a foundation, enable the growingtree to withstand windforces.They absorb moisture containing minerals from the soiland transfer it via the trunk to the crown. Trunk provides rigidity, mechanical strength and height to maintain the crown, Alsotransports moisture and minerals up to the crown and sap down from the crown. Crown provides as largeas possible a catchment area covered by leaves. These produce chemical reactions that form sugar and cellulose which cause the growth of the tree. A s engineers we are mainly concerned with thetrunk of the tree. Consider a crosssection of a trunk as shown in Fig. 1.1. Wood, in general, is composed of long thin tubular cells. The cell wallsare made up of cellulose and the cellsare bound together by a substance known as lignin. Most cells are oriented in the direction of the axis of the trunk, except for cells known as rays which run radially acrossthe trunk. Rays are present in all trees but are more pronounced in some species, such as oak. In temperate countries, a tree produces a new layer of wood just under the bark in the earlypart of every growing season. This growth ceases at the end of the growing seasonor during winter months. This process results in clearly visible concentric rings knownas annular rings, annual rings or growth rings. In tropical countries where trees grow throughout the year, a tree produces wood cells that are essentially uniform. The age of a tree may be determined by counting its growth rings'.
Fig. 1.l
Crosssection of a trunkof a tree.
Timber as aStructuralMaterial
3
The annular band of crosssection nearest to the bark is called sapwood. The central core of the wood which is inside the sapwood isheartwood. The sapwood is lighter in colour compared to heartwood and is25170mm wide, depending on the species. It acts as a medium oftransportation for sap from the roots to the leaves, while the heartwood functions mainly to give mechanical support or stiffness to the trunk. In general, the moisture content, strength and weights of the two are nearly equal. Sapwood has a lower natural resistance toattacks byfungi and insectsand accepts preservatives more easily than heartwood', In many trees, each annular ring can be subdivided into two layers: an inner layer made up of relatively large cavities called springwood, and an summerwood, Since outer layerofthickwallsandsmallcavitiescalled summerwood is relatively heavy, the amount of summerwood in any section is a measure of the density of the wood.
efects in timber2 Owing to the fact that wood is amaterial which is naturally occurring, there are manydefectswhich are introduced during the growing period and during the conversion and seasoningprocess. Any of these defects can cause trouble in timber in use either byreducing its strength or impairing its appearance. Defects may be classifiedas: natural defects, chemical defects,conversion defects and seasoning defects. t ~ r a defects J
These occur during the growing period. Examplesof natural defects are illustrated in Fig. 1.2(a). These may include: Cracks and3ssures. They may occur in various parts of the tree and may
e
e
e
even indicate the presence of decay or the beginnings of decay. Knots. These are common features of the structure of wood. A knot is a portion of a branch embedded by the natural growth of the tree, normally originating at the centre of the trunk or a branch. Grain defects. Wood grain refers to the general direction of the arrangement of fibres in wood. Grain defects can occur in the form of twistedgrain, crossgrain, flatgrain and spiralgrain, all of which can induce subsequent problems of distortion in use. Fungaldecay. This may occur in growingmaturetimber or even in recently converted timber, and in generalit is goodpractice to reject such timber. \ Annual ring width. This can be critical in respect of strength in that excess width of such rings can reduce the density of the timber.
Structural TimberDesign
Shake
Diagonalgrain
Crossgrain
Flatgrain
(a) Natural and conversion defects
End
Cupping
Springing Twisting
~oneyco~bing
Bowing (b) Seasoning defects
Defects in timber.
These may occur in particular instances when timber is used in unsuitable positions or in association with other materials. Timbers such as oak and western red cedar contain tannic acid and other chemicals which corrode
Timberas a StructuralMaterial
5
metals. Gums and resins can inhibit the working properties of timber and interfere with the ability to take adhesives.
1.
o ~ ~ e r sdefects io~
These are due basically to unsound practice inthe use of milling techniques or to undue economy in attempting to use every possible piece of timber converted from the trunk. A wane is a good example of a conversion defect.
Seasoning defectsare directly related to the movement that occurs in timber due to changes in moisturecontent. Excessive or uneven drying, exposureto wind and rain, and poor stacking and spacing during seasoning can all produce defects or distortions in timber. Examples of seasoning defectsare illustratedinFig. 12(b). Allsuchdefectshave an effect on structural strength as well as on fixing, stability, durability and finished appearance.
Trees and commercial timbers are divided into two groups: softwoods and hardwoods. This terminology hasno direct bearing on the actual softness or hardness of the wood.
Softwoods are generally evergreen with needlelike leaves comprising single cellscalled tracheids, which are likestrawsinplan, and theyfulfil the functions of conduction and support. Rays, present in softwoods, run in a radial direction perpendicular to the growth rings. Their function isto store food and allow the convection of liquids to where they are needed. oft woo^ characteristics 0
Quick growth rate; trees can befelled after 30 years, resulting in low density timber with relatively lowstrength. Generally poor durability qualities, unless treated with preservatives. Due to speed of felling, they are readily available and comparatively cheap.
1.4.2 ~ ~ r d w o o d s
Hardwoods are generallybroadleaved(deciduous)trees that losetheir leaves at the end of each growing season.The cell structure of hardwoods is
6
Structural Timber
Design
more complex than that of softwoods, with thick walled cells, calledJibres, providing the structuralsupportand thin walledcells,called vessels, providing the medium for food conduction. Due to the necessity to grow newleaveseveryyear the demand for sap is high and in some instances larger vessels may be formed inthe springwood these are referred to as rig porous woods. When there is no definite growingperiod the pores tend to be more evenly distributed, resulting in dzfuse porous woods. rdwood characteristics
Hardwoods grow at a slower rate than softwoods. This generally results in a timber of high density and strength which takes time to mature over 100 years in some instances. There is less dependency on preservatives for durability qualities. Due to time taken to mature and the transportation costs of hardwoods, as most are tropical, they tend to be expensive in comparison to softwoods.
ysical properties of timber3 Dueto the fact that timber issucha variable material, its strength is dependent on many factors which can act independently or in conjunction with others, adverselyaffecting the strength and the workability of the timber. Amongmanyphysical properties that influence the strength characteristics of timber, the followingmaybe considered the most important ones.
l' .5.1
isture content
The strength of timber isdependent on its moisture content, as is the resistance to decay. Most timber in the UK is airdried to a moisture content of between 17% and 23% which is generally below fibre saturation point at which the cell wallsare still saturated but moisture is removed fromthe cells. Any further reduction willresult in shrinkage4. Figure 1.3highlights the general relationship between strength andlor stiffness characteristics of timber and its moisture content. The figure shows that there is an almost linear loss in strength and stifkess as moisture increases to about 30%, corresponding to fibre saturation point. Further increasesin moisture content have no influence on either strength or stiffness. It should be noted that, although for mostmechanical properties the pattern of change in strength and stiffness characteristics withrespect to change in moisture content is similar, the magnitude of change is different from oneproperty to another. It is also to be noted thatasthe moisture content decreases shrinkage increases. Timber is described as being hygroscopic which means
Timberas a StructuralMaterial
7
100 h
E
80
U) U)
S U)
60
ti
I
Fibre saturation point
%
0
40
5
P ; 20 0
0
15
30
45
60
75
90
Moisture content(%)
Fig. 1.3 Generalrelationshipbetweenstrengthand/orstiffnessandmoisture content.
thatit attempts toattainan equilibrium moisture content with its surrounding environment, resulting in a variable moisture content. This should alwaysbeconsideredwhenusing timber, particularly softwoods which are more susceptible to shrinkage than hardwoods.
Density is the best single indicator of the properties of a timber and is a major factor determining its strength. Specific gravity or relative density is a measure of timber's solid substance. It is generally expressed as the ratio of the ovendry weight to the weight of an equal volume of water. Sincewatervolume varieswith the moisture content ofthe timber, the specific gravity of timber is expressedat a certain moisture content. specific gravity of commercial timber ranges from 0.29 to 0.8 1, most falling between 0.35 and 0.60. '
Grain is the longitudinal direction of the main elements of timber, these main elements being fibres or tracheids, and vessels in the case of hardwoods. In many instances the angle of the grain in a cut section of timber is not parallel to the longitudinal axis. It is possible that this variation is due to poor cutting of the timber, but more often than not the deviation in grain angle is due to irregular growth of the tree. This effectisoflesser consequence when timber is axially loaded, but leads to a significant drop in bendingresistance.Theangleof the microfibrilswithinthe timber also affects the strength of the timber, as with the effects ofthe grain, if the angle of deviation increases the strength decreases.
8
StructuralTimber Design
15.4 Timber defects As described earlier, defects in timber, whether natural or caused during conversion or seasoning, will have an eEect on structural strength as well as on fixing, stability, durability and finished appearance of timber.
1.6 References Somayaji, S. (1990) StructuralWoodDesign. West Publishing Company, St. Paul, U.S.A. Illston, J.M., Dinwoodie, J.M.and Smith, A.A. (1979) Concrete, Timber and Metals The Natureand Behaviour of StructuralMaterials. Van Nostrand Reinhold International, London. Illston, J.M. (1994) Construction ater rials Their Nature and Behaviour. E.& F.N. Spon, London. Carmichael, E.N. (1984) Timber Engineering. E.& F.N. Spon, London.
hapter
r
Strength capability of timber is difficultto assess as we have no control over its quality and growth. The strength of timber isa function of severalparameters including the moisture content, density, duration of the applied load, size of members and presence of various strengthreducing characteristics such as slope ofgrain, knots, fissures and wane. To overcome this difficulty, the stress grading method of strength classification has been devised'. Guidance onthe use of timber in building and civil engineeringstructures is given inBS 5268: Structural use of timber. This was originally divided into seven parts: Part 1: Limit state design, materials and workmanship. Part 2: Code of practice for permissible stress design, materials and workmanship. Part 3: Code of practice for trussed rafter roofs. Part 4: Fire resistance of timber structures. Part 5: Preservation treatments for constructional timber. Part 6: Code of practice for timber framed walls. Part 7: Recommendations for the calculation basis for span tables. Part l of BS 5268wasnevercompleted and, with the introduction of Eurocode 5: DD ENV 199511: Design of timber structures, the development of this part was completely abandoned. Part 2 of BS 5268, on which the design of structural timber is based, was originally published as CP 112 in 1952 and revised later in 1967 and, with extensive amendment, in 1971. The 'basic stresses' introduced in CP 112 were determined from carryingout shortterm loading tests on small timber specimens free from all defects. The datawas used to estimate the minimum strength which was taken as the value below which not more than 1% of the test results fell. These strengths were multiplied by a reduction factor to givebasicstresses.Thereduction factor made an allowance for the 9
l' 0
StructuralTimber Design
reduction in strength due to durationfor loading, size of specimen andother effects normallyassociated with a safety factor, such as accidental overloading,simplifyingassumptionsmade during designanddesign inaccuracy, together with poor workmanship. Basic stress was defined as the stress that could be permanently sustained by timber free from any strengthreducing characteristics'. Since 1967 there have been continuing and significant changes affecting the structural use of timber. Research studies in the UK and othercountries had shownthe need for a review of the stress values andmodification factors given in the original code. the concept of 'basic stresses' was With theintroduction of BS 5268 in 1984 largely abandonedand the new approach forassessing the strength of timber moved somewhatin line with 'limitstates' design philosophy. In 1996, Part 2 of BS 5268 was revised witha clear aimto bring this code as close as possible to, and to run in parallel with, Eurocode 5: DD ENV 19951 1: Design of timber structures, Part 1.1 General rules and rules for buildings. The overall aim has been to incorporate material specifications and design approaches from Eurocode 5, while maintaining a permissible stress code with which designers, accustomed to BS 5268, will feel familiar and be able to use without difficulty. The first step in this process involvesstrength grading of timber sections. Thereare two Europeanstandards which relate to strength grading:
BS EN 518 :1995 Structural timber. Grading. Requirements for visual strength grading standards. BS EN 519 ;1995 Structurul t i ~ b e rGrading. . Requiremen~s for ~ a c h i n ~ strength graded timber and grading machines. dance for stress grading of the two typ hardwoods, are given in the following
timber, namelysoftwoods
S 4978 :1996 Spec~cationfor softwoods graded for structural use. S 5756 :1997 Spec~cationfor tropical hard woo^ gradedfor str~cturaluse. The current revised versions
of these standards conform with
the
The structural design oftimber members is related to Part2 of BS 5268, and is based on permissible stress design philosophyin which design stressesare derived on a statistical basis and deformations are also limited. Elastic theory is used to analyse structures under various loading conditions to give the worst design case. Thentimber sections are chosen so that the permissible stresses are not exceeded at any point of the structure.
Introduction to BS 5268: Part 2:1996
11
Permissible stressesare calculated by multiplying the ‘grade stresses’, given in Tables7 to 12a ofBS 5268 :Part 2, by the appropriate modification factors, Kfactors, to allow for theeffectsofparameterssuch as load duration, moisture content, loadsharing,sectionsize,etc.Appliedstresseswhich arederived from theserviceloadsshouldbeless than or equal to the permissible stresses. A summary of the Kfactors used for the calculation ofpermissiblestressesisgiveninTable2.1.Owing to changes made to BS 5268 :Part 2 in 1996, some Kfactors which were used in the previous editions, such as IC1,K10, etc., have been withdrawn. The permissible stress design philosophy, asBS in5268 :Part 2, is different from the limit states design philosophy of Eurocode 5 which has two basic requirements. The first is ultimatelimitstates (i.e. safety) which is usually Table 2.1
Summary of Kfactors used for calculation of permissible stresses
Kfactor Description
or application
5268BS
Timber grade stresses and moduli for service class 3 Duration of loading Bearing stress K4 Shear at notched ends K5 Form factor: bending stress for nonrectangular sections K6 Depth factor: bending stress for beams other than K7 300mm deep Load sharing systems K8 To modify Emin for deflection in trimmer beams and lintels K9 Slenderness in compression members K 12 Efktive length of spaced columns K1 3 Width factor for tension members K1 4 Single grade glued laminated members and horizontally K1 520 laminated beams Vertically glued laminated members K2729 Individually designed gluedend joints in horizontally K3032 glued laminated members Curved glued laminated beams K3334 Stress factor in pitched cambered softwood beams K35 Plywood grade stresses for duration of loading and K36 service classes Stress concentration factor for plywebbed beams K37 For tempered hardboards K3841 Fastener slip moduli K,,, Nailed joints K4350 Screwed joints K5254 Bolted and dowelled joints K56, 57 Toothedplate connector joints KS,C,5861 Splitring connector joints KS,C,D,6265 Ks,C,D,6669 Shearplate connector joints K70 Glued joints
K2
K3
:Part 2 :1996
Table 13 Table 14 Table 15 Clause 2.10.4 Clause 2.10.5 Clause 2.10.6 Clause 2.9 Table 17 Table 191Annex B Table 20 Clause 2.12.2 Table 21 Table 22 Table 23 Clause 3.5.3 Clause 3.5.4.2 Table 33 Clause 4.6 Section 5 Table 52 Clause 6.4 Clause 6.5 Clause 6.6 Clause 6.7 Clause 6.8 Clause 6.9 Clause 6.10
12
Structural Timber Design
expressed in terms of loadcarrying capacity and is achieved byfactoringup of load values and factoringdown of material strength properties by partial safety factors that reflect the reliability of the values that they modify. The secondis servi~eabilitylimitstates (i.e. deformationandvibrationlimits) which refersto the ability of a structural system and its elements to perform satisfactorily in normal use. It is important to note that inpermissible stress design philosophy partial safety factors (i.e. modification factors) are applied only to material properties, i.e. for the calculation of permissible stresses, andnot to the loading.
Once timber has beenseasoned it is stress graded; this grading will determine the strength class of the timber to satisfy the design requirements of BS 5268 :Part 2. Strength grading takes into account defects within the timber such as slope ofgrain, existence and extent of knots and fissures, etc. All timber used for structuralwork needs to be strength graded by either visual inspection or by an approved strength grading machine. Clause 2.5 of BS 5268 :Part 2 deals with strength grading of timber.
3.1
ual g r a ~ i n g
Visual grading is a manual process carried out by an approved grader. The grader examines each piece of timber to check the size and frequency of specific physical characteristics or defects, e.g. knots, slope of grains, rate of growth, wane, resin pockets and distortion, etc. The required specifications are given in BS 4978 and BS 5756 to determine if a piece of timber is acceptedinto one of the two visualstressgrades or rejected. These are General Structural (GS) and Special Structural (SS)grades. Table 2 of BS 5268 :Part 2 (reproduced here as Table 2.2) refers to main softwood combinations of species visually graded in accordance with BS 4978,
Machine grading of timber sectionsis carried out on the principle that strength is related to stiffness. The machine exerts pressure and bending is induced at increments along timber length. The resulting deflection is then automatically measured and compared with preprogrammedcriteria, which leads to the grading of timber section. BS 5268 :Part 2, Clause 2.5 specifies that machine graded timber, other than thatcarried out by North American Export Standard for Machine Stressrated Lumber (e.g. 1450f1.3E), should meet the requirements of BS EN 519. To this effect timber is graded directly to the strength class boundaries and marked accordingly.
Introduction to BS 5268: Part 2: 1996
13
In general less material is rejectedif it is machine graded, however timber is also visually inspectedduring machine gradingto ensure major defectsdo not exist.
The conceptof grouping timberinto strength classes was introducedinto the UK with BS 5268:Part 2 in1984~~trength classes offer a number of advantages both to thedesigner and the supplier of timber. The designer can undertake his design without the need to check on the availability and price of a large number of speciesand grades which he might use. Suppliers can supply any of the specieslgradecombinations that meet the strength class calledfor in a specification. The concept also allows new species to be introduced onto the market without affecting existing specificationsfor timber. The latest strength classes used in the current version of BS 5268 :Part 2 : 1996 relate to the European strength classes which are defined in BS EN 338 :1995 Structural timber. Strength classes. There are a total of 16 strength classes, C14 to C40 for softwoods and D30 to D70 for hardwoods asgiven in Table 7 of BS 5268 :Part 2 :1996 (reproduced here as Table 2.3). The number in each strength class refers to its ‘characteristic bending strength’ value,forexample,C40timberhasa characteristic bending strength of 40N/mm2. It is to be noted that characteristic strength values are conBS 5268 :Part 2, as they siderably larger than the grade stress values used in do not include effects of longterm loading and safety factors.
oftw wood grading: Softwoods which satisfy the requirements for strength classes given in BSEN 338 when graded in accordance with BS4978 and American timber standards NLGA and NGRDL are given in Tables 2, 3 , 4 and 5 of BS 5268 :Part 2. The new strength classes for softwoods are C14, C16, C18, C22, C24, TR26, C27, C30, C35 and C40. However it is likely that the old strength class system (i.e. SC1to SC9) may be encountered for some time. A comparison of the lowest of the newstrength class (C classes) against the most common old SC classes can be made: SC3 compares with C16, SC4 with C24, and SC5 with C27. TR26 timber, which is commonly used for axially loaded members (i.e. trussed rafters), is equivalent to the superseded M75 European redwood/whitewood. ~ a r d ~ o grading: od Tropical hardwoods which satisfy the requirements for strength classes given inBS EN 338 when graded to HS grade in accordance with BS 5756 are given in Table 6 of BS 5268 :Part 2 :1996. The strength classes for tropical hardwoods are D30, D35, D40, D50, D60 and D70. Grade stresses: Grade stresses and moduli of elasticity for service classes l and 2 (described in Section 2.5.2)are given in Table7 of BS 5268 :Part 2 for
14
Structural Timber
Design
Table 2.2 Softwood combinations ofspecies
and visual grades whichsatisfy the requirements for various strength classes. Timber gradedin accordance withBS4978 (Table 2, BS 5268 :Part 2) Timber
Strength classes C27 C24 .C22 C18 C16 C14 C30
Imported: Parana pine Caribbean pitch pine Redwood Whitewood Western red cedar
CS
GS CS
Douglas firlarch (Canada and USA) Hemfir (Canada and USA) Sprucepinefir (Canada and USA) Sitka spruce (Canada) Western white woods(USA) Southern pine (USA) British grown: Douglas fir Larch British pine British spruce
GS
ss
GS
ss ss
ss
CS CS
GS GS
GS CS GS
ss ss
ss
ss ss
GS
GS
GS
ss
ss
ss ss
ss
ss
16 strength classes, and in Tables 8 to 12a for individual softwood and hardwood species and grades. Table 7 is reproduced here as Table 2.3.
erations (factorsaffecting tim As mentioned previously, there are several factors which influence timber strength and hence theyshould be considered in the analysisdesign process of all structural timber members, assemblies and frameworks. The main design criteria recommended by BS 5268 :Part 2, Clause l .6 for consideration are listed below.
For the purpose of design, loading should be in accordance with BS 6399 :Parts 1, 2, and 32 and CP 3: Chapter V :Part 23 or other relevant standards, where applicable.
2.5.2 Service classes Due to the effects of moisture content on mechanical properties of timber, the permissible property values should be those corresponding to one of the
introduction to BS 5268: Part 2: 1996
15
16
Structural Timber
Design Table 2.4 Modificationfactor K2 forobtainingstresses and moduli applicable to service class 3 (Table 13,
BS 5268 :Part 2) Property
parallel Bending to grain parallel Tension to grain Compression parallel to grain Compression perpendicular to grain parallel Shear to grain Mean and minimum modulus of elasticity
K2
0.8 0.8 0.6 0.6 0.9 0.8
three serviceclassesdescribedin Clause 1.6.4 and given in Table 1 of BS 5268 :Part 2 :1996. These are summarised below: (1) Service class I refers to timber used internally in a continuously heated building. The average moisture content likely to be attained in service condition is 12%. (2) Service class 2 refers to timber used in a covered building. The average moisture content likely to be attained in service condition if building is generally heated is 15%, and if unheated, 18%. (3) Service class 3 refers to timber used externally and fully exposed. The average moisture content likely to be attained in service condition is over 20%. Grade stressandelasticmodulivaluesgiven in Tables 7 to 12aof BS 5268 :Part 2 apply to various strength classes and timber speciesin service classes1 and 2. For service class3 condition they should be multiplied by the modification factor K2 from Table 13 of the code (reproduced here as Table 2.4).
2.5.
j ~ t ~content re A s moisture content affects the structural properties of timber significantly, BS 5268 :Part 2 :1996 recommends that in order to reduce movement and
creep under load the moisture content of timber and woodbased panels when installed should be close to that likely to be attained in service.
Duration of load affects timber strength and therefore the permissible stresses. The grade stresses (Tables 7 to 12a) and the joint strengths given in BS 5268 :Part 2 are applicable to longterm loading. Because timber and woodbased materials can sustain a much greater load for a short period
Introductionto BS 5268: Part 2: 1996 Table 2.5 Modification factor K3 for duration ofloading(Table
17
14,
BS 5268 :Part 2)
~~~
loading
Duration of
K3
Longterm: dead i.e. +permanent imposeda Mediumterm: i.e. dead +temporary imposed snow Shortterm:i.e.dead +imposed+wind:dead imposed +snow +windb Very shortterm: i.e. dead +imposed +wind (gust)” 1.75
+
+
1.oo 1.25 1S O
___~
For uniformly distributed imposed floor loads K3 = 1except for type2 and type 3 buildings (see Table 5 of BS 6399 :Part 1 :19842)where, for K3 may be assumed to corridors, hallways, landings and stairways only, be 1.5. For wind, shortterm category applies to class (1 C5 S gust) as defined of in CP3 :Chapter V :Part 23or, where the largest diagonal dimension the loaded area a ,as defined in BS 6399 :Part 2,2 exceeds50 m. c For wind, very shortterm category applies to classes A and B (3 S or 5 S gust)asdefinedinCP 3 :Chapter V : Part Z3 or, wherethelargest diagonal dimensionof the loaded area a, as defined inBS 6399 :Part 2: does not exceeds 50m.
a
(a few minutes) than fora long period (several years), the grade stresses and the joint loads may be increased for other conditions of loading by the modification factors given in the appropriate sections of BS 5268 :Part 2. Table 14 of BS 5268: Part 2 (reproduced here as Table 2.5) gives the modificationfactor K3 by whichall grade stresses(excludingmoduliof elasticity and shearmoduli)shouldbemultiplied for various durations of loading. .5.5 Section size
The bending, tension and compression and moduli of elasticity given in Part 2 of BS 5268 are applicable to materials 300 mm deep (or wide, for tension). Because these properties of timber are dependent on section size and sizerelatedgradeeffects,thegradestressesshouldbemodified for section sizesother than 300 mm deep bythe modification factors specifiedin the appropriate sections of the code. In general, it ispossible to designtimberstructuresusing any sizeof timber. However, since the specific use is normally not known at the time of conversion, sawmills tendto produce a range ofstandard sizes known as ‘customary’ sizes. Specifying such customary sizes will often result in greater availability and savings in cost4. The customarylengths and sizesproducedbysawmills in the UK, normally available from stock, are given in Tables NA.1 to NA.4 of the
Structural TimberDesign
National Annex to BS EN 336 :1995 which usestarget sizes as the basis for the standard. Furtherinformation and details of the customarylengths and sizes are given in Appendix A.
The grade stresses given in Part 2 of BS 5268 are applicable to individual pieces of structural timber. Where a number of pieces of timber (in general four or more) at a maximum spacing of 610 mm centre to centre act together to support a common load, then the grade stressescanbemodified (increased) in accordance with the appropriate sections of the code. In a loadsharing systemsuch as rafters, joists, trusses or wall studs spaced at 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 appropriate grade stressescanbemultiplied by the loadsharing modification factor K8 which has avalueof1.1. In addition, BS 5268 :Part 2 recommends that the mean modulus of elasticity should be used to calculate deflections and displacements induced by static loading conditions. Therefore in a loadsharing system: K8 = 1.l ~odificationfactor Modulus of elasticity E = E m e m
It is to be noted that special provisions are provided in BS 5268 :Part 2 for builtup beams, trimmer joists and lintels, and laminated beams; these are given in Clauses 2.10.10,2.10.1 1 and Section 3 of the code. It is also important tonote that the provisions for loadsharing systems do notextend to the calculation of modification factor K I 2for loadsharing columns.
BS 5268 :Part 2 recommends that in the absence of test data, the following grade stress and moduli of elasticity values may be used: tension perpendicular to grain, Ot&l torsional shear, Ttorsjon rolling shear, Tr
5 5 3
= x shear stress parallel to grain, Tg,11 = x shear stress parallel to grain, l"s, // = x shear stress parallel to grain, zg,11 = 1 X Ernean or min
modulus of elasticity Ito grain, E L shear modulus, G = B1 X Ernean or min permissible compressive stress = CFc,ah,//(CFC,ah,//G c , a h , l ) sin a where the load is inclined at an angle a to the grain, C T ~ , ~ ~ , ~
Introduction to BS 5268: Part 2: 1996
1
The following symbols andsubscripts are used to identify section properties of timber elements, applied loading conditions, type of force and induced andpermissiblestresses.Symbols and subscripts are kept as similar as possible to those given in Part 2 of BS 5268 :1996. ~eometrical and mechanical properties n a
A b d E Emean Emjn
G h l
I L Le
m n
h
e Pk Prnean
z
distance angle of grain area breadth of beam, thickness of member diameter modulus of elasticity mean value of modulus of elasticity minimum value of modulus of elasticity modulus of rigidity or shear modulus depth of member radius of gyration second moment of area length, span effective length, effective span mass number slenderness ratio first moment of area characteristic density average density section modulus
bending moment applied bending stress parallel to grain grade bending stress parallel to grain permissible bending stress parallel to grain
applied shear force applied shear stress parallel to grain grade shear stress parallel to grain permissible shear stress parallel to grain applied rolling shear stress pernhssible rolling shear stress
28
Structural Timber Design
A m As
Atotal
A,&
bending deflection shear deflection total deflection due to bending and shear permissible deflection
CF,,,,~/ appliedcompressive stress parallel to grain I/ gradecompressive stress parallel to grain C T , , , ~ ,permissible // compressive stress parallel to grain C F , , , , ~appliedcompressive stress perpendicular to grain O,,g,
o,lg,l
grade compressive stress perpendicular to grain
C F ~ , permissible , ~ , ~ compressive stress perpendicular to grain
Tension
applied tensile stress parallel to grain CJt1g,// grade tensilestress parallel to grain atlU~ permissible , ~ ~ tensile stress parallel to grain
CF~,~,/I
1. Arya, C. (1994) Design of structural elements. E. & F. N. Spon, London. 2. British Standards Institution (1984, 1995, 1988) BS 6399: Loading for buildings. Part 1 :1984 :Code of practice for dead and imposed loa&. Part 2 :1995 :Code of practice for wind loa&. Part 3:1988 :Code of practice for imposed roof loads.BSI, London. :Chapter V :Loading. Part 3. British Standards Institution (1972) CP 3 2:1972 :Wind loads. BSI, London. 4. BritishStandardsInstitution(1995)BS EN 336 :Structural timber. Coniferous and poplar. Sizes. Permissible deviations. BSI, London.
apter
Many academic institutions and designoffices are turning to computer assisted instructions. This is especially true in scienceand engineering where courses are being introduced to teach the use of computers as analysis and design tools. Mathcad’s potential as a powerful and easy to use computational tool has already been recognised by most academicinstitutions and many design offices. The aim of this chapter is to demonstrate how the analysis and design calculations for structural timber can be incorporated into simpletouse electronic notepads or worksheets. Access to a personal computer (PC)and the associated software Mathcad is not a prerequisite for understanding the designcalculationsin the examplesprovidedinthis book. Alldesign examples given are fully selfexplanatory and well annotated. They have been produced in the formof worksheets to run under Mathcad,version 6, or higher,ineitherone of itseditions,i.e. Student, Standard, Plus or Professional. Details are given at the end of this book. The design worksheets given are intended as a source of study, practice and further development by the reader. They should not be seen as complete and comprehensive design worksheets but rather as the foundations of a design systemthat can be developedfurther. The aim isto encourage readers to use computing as a tool to increase their understanding of how design solutions vary in response to a change in oneof the variables and howalternative design options can be obtained easily and effortlessly, allowing the design engineerto arrive at the most suitable/economic solution very quickly. It is important to note that this chapter is not intended to teach Mathcad. It aims only to familiarise the reader with the Mathcad worksheet formats that are used to produce design examples in this book.
Mathcad (developedby MathSoft, Inc.)is an electronic notepad (live worksheet) that allows ath he ma tical calculation to beperformed on a 21
22
Structural Timber Design
computer screen in a format similar to the way it would be done manually with paper and pencil.WhileMathcademploys the usualmathematical , /,=) for algebraic operations, it also uses the convensymbols (i.e. tional symbols of calculus for differentiation and integration to perform these operations. It preserves the conventional symbolic form for subscribing, special mathematical and trigonometrical functions, series operations, and matrix algebra. When expository text is added, Mathcad’s symbolic format leads to reports that are understood easily by others. Data can be presented in both tabular and graphical forms. Mathcad can also be used to answer, amongst many others, the ‘whatif’ questions in engineering problems. With a well structured worksheet, design calculations can be performed whereby parameters can be changed and the results viewed almost immediately on the computer display and/or printed.
+,
3.3 What does Mathcad do?* Mathcad combines the live document interface of a spreadsheet with the WYSIWYG interfaceof a word processor. WithMathcad, equations can be typeset on the screen in exactly the way they are presented intextbooks, with the advantage that it can also do the calculations. Mathcad also comes with multiplefonts and the ability to print what you see on the screen on any Windows supported printer. This, combined with Mathcad’s live document interface, makes it easy to produce uptodate, publicationquality engineering reports and/or design solution sheets. The following subsections demonstrate how some simple operations are carried out in Mathcad. This is to illustrate the formatlmeaning of the operations used to produce the examples in this text. 3.3.1 A simple calculation’ Although Mathcad can perform sophisticated mathematics, it can just as easily be used as a simple calculator. For example, Click anywhere in the worksheet; you will see a small crosshair. Type 15 81104.5 = As soon as the equal result (see Fig. 3.1).
signispressed,
Mathcad computes and shows the
3.3.2 Definitions and variables2 Mathcad’s power and versatility quickly becomes apparent when the variables and functions are being used. By defining variables and functions, equations can be linked together and intermediate results can be used in further calculations.
UsingMathcadforDesignCalculations
15"
23
\
14.923
Fig. 3.1 A simple calculation.
For example, to define a valueofsay 10 to a variable, say t, click anywhere in the worksheet and type t: (the letter t followed by a colon). Mathcad will show the colon as the definition symbol := and will create an empty place holder to its right. Then type 10 in the empty placeholder to complete the definition for t. To enter another definition, press ["I] to move the crosshair below the first equation. For example, to define ace as 9.8, type acc:9.8. Then press [J] again. Now that the variables ace and t are defined, they can be used in other ace
t 2 type acc/ expressions. For example, to calculate the magnitude of 2 2*tA2.The caret symbol represents raising to a power, the asterisk * is multiplication, and the slash / is division, To obtain the result, type =. Mathcad will return the result (as shown in Fig. 3.2). A
acc :=g%
Fig. 3.2 Calculating with variables and functions.
24
Structural Timber
Design
acc :=9.8
Fig. 3.3 Enteringtext.
Mathcad handles text as easily as it does equations. To begin typing text, click in an empty space and choose Create Text %ion from the Text menu or simply click on the icon on the menu bar. Mathcad will then create a text box in which you can type, change font, format andso on asyou would when using a simple Windows based word processor. The text box will grow as the text is entered. Now type, say, ‘Equation of motion’ (see Fig. 3.3). To exit text mode simply click outside the text box.
Units of measurement, while not required in Mathcad equations, can help detect and enhance the display of computed results. Mathcad’s unit capabilities take care of many of the usual chores associated with using units
M = 20 *kN.m
Fig. 3.4 Equations using units.
Using Mathcadfor Design Calculations
25
and dimensions in engineering analysis and design calculations. Once the approp~atedefinitions are entered, Mathcad automatically performs unit conversions and flags up incorrect and inconsistentdimensional calculations. Although Mathcad’s latest edition recognises most common units, you maywish to defineyourown units. For example,N =newton,and kN = IO3N. Toassign units to a number, simply multiplythe number by the name or letter(s) which defines the unit. To illustrate this, calculate the magnitude of the bending moment M at the builtin end of a cantilever of length L =2m induced by a force of P = 10 kN acting at its free end. To do this, click anywhere in a worksheet and type: N :=newton kN := 103*N L :=2*m P := 1O*kN
M
:=P*L
Then type M=.As soon as the = sign is typed, Mathcad will compute the result and also display the units of M (as shown in Fig. 3.4).
Thepreviousexamplesaimed to demonstrate the simplicityofusing Mathcad in producingthe design examples givenin the proceeding chapters of this book. To learn more about Mathcad, refer to the next section in this chapter.
1. Wieder, S. (1992) Introduction to ~ u t h c u for d Scientists and Engineers. McGraw
Hill, Inc., Hightstown.
2. ~ a t ~ c User’s a d Guide, ~ a t h c u d6.0 (1996) Mathsoft, Inc.,
MA.
Flexural members are those subjected to bending. There are several types and forms of flexural timber members that areused in construction. Typical examples are solidsection rectangular beams, floor joists, rafters and purlins. Other examplesinclude glulam beams (verticaland horizontal glued laminated beams), plywebbed beams (Ibeams and boxbeams) and beams of simple composites (Tee and I shaped beams). Although the design principles are essentially the same for all bending members of all materials, the material characteristics are different. Steel for example is ductile, homogeneous, and isotropic. Concrete is brittle and can be assumed homogeneous for most practical purposes. A s for timber, the material properties are differentin the twomain directions: parallel and perpendicular to the grain. Even though the normal stresses due to bending are parallel to grain direction, support conditions may impose stresses that are perpendicular to grain direction. Thesestresses, in addition to the primary stresses, should be checked in the design against the permissible values, which include the effects of environmental conditions, material and geometrical characteristics. This chapter deals in detail with the general considerations necessary for the designofflexuralmembers and describes the design details ofsolid section rectangular timber beams.Designmethods for glued laminated beams and plywebbed beamsare described in Chapters 6 and 7 , respectively.
The main design considerations for flexural members are: (1) bending stress and prevention of lateral buckling (2) deflection (3) shear stress (4) bearing stress. 26
Design of FlexuralMembers(Beams)
27
The crosssectional properties of all flexural members have to satisfy elastic strength andservice load requirements. In general,bendingis the most critical criterion for mediumspan beams, deflection for longspan beams and shear for heavily loaded shortspan beams. In practice, design checks are carried out for all criteria listed above. In Chapter 2 it was mentioned that the design oftimber elements, connections and components is basedon the recommendations ofBS 5268 :Part 2 : 1996 which is still basedon 'permissible stress' design philosophy. The permissible stress value is calculated as the product of the grade stress and the appropriate modification factors for particular service and loading conditions, and is usually compared withthe applied stress in a member or part of a component in structural design calculations. In general: permissible stress (=grade stress x Kfactors )2 applied stress
4.3 Bending stress and prevention of lateral buckling The design of timber beams in flexurerequires the application of the elastic theory of bending as expressed by: 1M.Y ff=I
The term Z/y is referred to as section modulus and is denoted by Z. Using the notations defined in Chapter 2, the applied bending stress about the major (x") axis of the beam (say) (see Fig. 4.1), is calculated from:
I
1
Y Fig. 4.1 Crosssectionof a rectangular beam,
28
Structural Timber Design
where: amla,//= applied bending stress (in N/mm2) M= maximum bending moment (in Nmm) Zxx=sectionmodulus about its major (xx) axis(inmm3). angular sections
For rect
bh3
2 Ixx= second moment of area about xx axis (in mm4) y= distance from the neutralaxis of the section to the extreme fibres
(in mm) h= depth of the section (in mm) b =width of the section (in mm). amla&,// is calculated as the product of The permissible bending stress grade bending stress parallel to grain am,g,// and any relevant modification factors (Kfactors). Theseare K2 for wet exposure condition (if applicable), K3 for loadduration, K6 for solid timber members other than rectangular sections (if applicable), KT for solid timber members other than 300mm deep, and K8 for loadsharing systems (if applicable). Hence:
am,adm,//= Om,g,// x K2
x K3 x K6 x K1 x Kt3
(4.4)
KZ?K3 and K8 are general modificationfactors, which were described in detail in Chapter 2. K6 and K, specifically relate to the calculation of permissible
bending stress, am,a&,// and are described in the following sections.
Clause 2.10,3 of BS 5268 :Part 2 recommends that the span of flexural members should be taken as the distance between the centres of bearings. Required bearing length
Beam orjoist
Clear span Effective span Span to centres of actual bearings I
i
ig. 4.2 Effective span(Baird and Ozelton').
DesignofFlexuralMembers(Beams)
29
Where members extend over bearings, which are longer than is necessary,the spans may be measured between the centres of bearings of a length which should be adequate in accordance withPart 2 of the code (see Fig. 4.2). In determining the effective span, Le, it is usually acceptableto assume an addition of 50 mm to the clear span, between the supports, for solid timber beams and joists and 100mm for builtup beams on spans up to around 12m, but longer spans should be checked.'
Grade bending stress values given the in code applyto solid timber members of rectangular crosssection.For shapes other than rectangular (see Fig. 4.3), the grade bending stress value should bemultipliedby the modification factor K6 where: &j
= 1.18 for solid circular sections, and = 1.41 for solid square sections loaded diagonally (see Fig. 4.3).
The grade bending stressesgiven in Tables 712a of BS 5268 :Part 2 apply to beams having a depth, h, of 300 mm (Clause 2.10.6). For other depths ofbeams,thegradebendingstressshouldbemultiplied by the depth modification factor, K7, where: for h 5 72mm, for 72mm
h <300mm,
K7 = 1.17 K7
= (
for h
300mm,
K:,= 1.0 Fig. 4.3 Form factorKG.
~
~
K7 = 0.81
h2 h2
K4=
1.18
.
l
l
+ 92 300 + 56 800
K4=
1.41
30
Structural Timber Design
4.3.4 Selection of a suitable sectionsize There are two methods commonly used in selecting section:
an appropriate trial
(1) Engineering judgement which is based on experience. (2) By utilising the permissible bending stress criterion in equation (4.2). Thus the expression for calculation of the required section modulus Zxxfor timber members, incorporating all the relevant Kfactors, isas follows:
Thus a suitable section size having a Zxx2 Zxx,required can be selected. The chosen section should then be checkedfor lateral stability, deflection, shear and bearing. The standard (customary) sizes of timber sections in the UK,normally available fromstock, are given in theNational Annex to BS EN 336. A summary of details is given in Appendix A. 4.3.5 Lateraf stability
BS 5268 :Part 2 :1996 recommends that the depth to breadth ratio of solid and laminated rectangular beams should not exceed the valuesgivenin Table 4.1 Maximum depth to breadth ratio for solid and laninated members, (Table 16, BS 5268 :Part 2)
in
Degree of lateral support
Maximum depth to breadth ratio
No lateral support
2
Ends held
3
position
Endsheldinposition and memberheld in line as by purlins or tie rods at centres not more than 30 times breadth of the member
4
Ends heldinposition and compressionedgeheld in line, as by direct connection of sheathing, deck or joists
5
Ends heldinposition and compressionedgeheld in line, as by direct connection of sheathing, deck or joists, together with adequate bridging or blocking spaced at intervals not exceeding 6 times the depth
6
Endsheldinposition
7
andboth edgesheldfirmlyinline
DesignofFlexuralMembers(Beams)
31
(a) Solid blocking
(b) Skewed blocking
(c) Herring bonestrutting (bridging) Fig. 4.4
Examplesof provisions for lateral support.
Table 16 (reproduced here as Table 4.1) of the code corresponding to the appropriate degree of lateral support. Examples of provisions for lateral support are shown in Fig. 4.4.
4.3.6 An iJJustrative exampJe
Determine the value of permissible bending stress parallel to grain andmagnitude of maximumbendingmoment for amainbeam of 50mm x 200 mm deep Canadian Douglas firlarch grade SS under service class 2 and shortduration loading.
32
Structural Timber
Design
BS 5268 :Part Description 2
output
Table 2 Table 7 Clause l .6.4 Table 14 Clause 2.10.5
strength classification grade stress // to grain service class 2 shortduration loading rectangular section
strength class =C24 = 7.5 N/mm2 K2 1 K3 = 1.5 =1
Clause 2.10.6
depth factor
K7 =
2.9 Clause
no loadsharing
Kg = l
gm,g,//
~=ll
(g) 0.11
= 1.045
Permissible bending stress
Allowable maximum bending moment is obtained by rearranging equation (4.5)
M = 3.91 kNm
4.4 Deflection BS 5268 :Part 2, Clause 2.10.7 recommendsthat ‘The dimensions of flexural members should be suchas to restrict deflection within limitsappropriate to the type of structure, having regard to the possibility of damageto surfacing materials, ceilings,partitions and to the functional needs as well as aesthetic requirements.’ 4.4.1 Deflection limits
In most cases, including domestic flooring, the combined deflection due to bending, Am, and shear, As, should not exceed 0.003 of the span to satisfy this r~commendation. Inaddition, for domestic floor joists, the deflection under full load should not exceedthelesserof0.003timesthe span or 14mm. This is to avoid undue vibration under moving or impact loading. In general
Atotal=
and for domestic joists
Atotal
(Am
+ A,) 5 (0.003 X span)
5 lesser of (0.003 x span or 14 mm)
33
DesignofFlexuralMembers(Beams)
4.4.2 ~ r e c a ~ ~ e r Subject to consideration being given to the effect of excessive deformation, timber beams may be precambered to account for the deflection under full dead or permanent load. In this instance, BS 5268 :Part 2 recommends that the deflection due to imposed load only shouldnot exceed 0.003 of the span.
4.4.3 Bending deflection The maximum bending deflection induced by the two most common load cases is given below: (1) For a simply supported beam carrying a uniformly distributed load of wtotal
Am= 5 wtotalL3 384EI (2) For a simply supported beam carrying a concentrated load at midspan of P
Am =
PL3 48EI
(4.7)
where: I/V;otal=
total uniformly distributed load
P =concentrated load acting at midspan
L =effective span I= second moment of area about axis of bending, usually beam’s major (x+ axis E =Emh for a beam acting on its own =Emean for a beam in a loadsharing system.
For a singlespan simply supported beam subjected to a maximum bending moment of Mm,, irrespective of the loading type, the maximum deflection of the beam may be estimated using:
A,
cli
0.1O4MmxL2
EI
34
Structural Timber
Design
Table 4.2 Modification factor Kg used to modify the minimum
modulus ofelasticity for trimmer joists and lintels(Table 17, BS 5268 :Part 2)
Kg
ofNumber Value piecesof
l 2
3
4 or more
Softwoods
Hardwoods
1.oo 1.14 l .21 1.24
1.oo 1.06 1.08 1.10
Note: BS 5268 :Part 2,Clause2.10.11recommends thatfor trimmer joists and lintels which comprise two or more pieces connected together in parallel and acting together to support the loads, the minimum modulus of elasticity modified by the modification factor Kg, given in Table 17 of the code, should be used for calculation of deflections. Table 17 is reproduced here as Table 4.2. 4.4.4 Shear def/ection
Since in timber and wood based structural materials the shear modulus is considerably lower as a proportion of the modulus of elasticity, compared to other structural materials such as steel, the effect of shear deflection can be significant and should be considered in the design calculations. Themaximum shear deflection, A,, inducedinasinglespansimply supported beamof either rectangular or square crosssection,maybe determined from the following equation:
where A is the crosssectional area of the beam, M,, is the maximum bending moment in the beam and E is as defined above.
4.5 Bearing stress Thebearingstresses in timber beams are developeddue to compressive forcesappliedina direction perpendicular to the grain and occur in positions such as points of support or applied concentrated loads. Possible bearing failure positions are shown in Fig. 4.5. Theapplied bearing stress, CT,,~,J iscalculatedfrom the following equation: (4.10)
Design of FlexuralMembers(Beams)
35
Concentrated load
Timber beams
Support wall Fig. 4.5 Possible bearing failures.
where: F = bearing force (usually maximum reaction or concentrated load) Abeuring = bearing area (=bearing length x breadth of the section). In general, the value of appliedbearing stress, C T ~ , should ~ , ~ not exceed the determined ,,~,~ from: permissible bearing stress, C T ~ ~ c , u d m , L= Qc,g,L x K2
x K3 x K4 x K8
1) (4.1 K,, K3 and & are general modification factors, which were described in detail in Chapter 2. K4 relates to the calculation ofpermissible bearing stress, C T ~ , ~ &and , ~ , is described below. 4.5.1 Length and position of bearings
BS 5268 :Part 2, Clause 2.102 recommends that the grade stresses for compression perpendicular to the grain apply to bearings of any length at the endsof a member, and bearings 150mm or morein length at any position. For bearing less than 150 mm long located 75mm or more from the endof a member, as shown in Fig. 4.6, the grade stress should be multipliedby the modification factor K4 given in Table15of the code (reproduced here as Table 4.3).
375 m
Fig. 4.6 Bearing length and position.
Bearing 150 m
36
Structural Timber
Design Table 4.3 Modification factor K4 for bearing stress (Table 15, BS 5268 :Part 2) K4
Length of bearing (mm) 10 15 25 40 50 75
1.74 l .67 1.53 l .33 1.20 1.14 1.10 l .oo
100
150 or more
ear
stress
Thecriticalposition for shear isusually at supports where maximum reactionoccurs.Theappliedshearstress, r, iscalculatedas themaximum (not average) shear stress from the following equation: (4.12)
For a rectangular timber beam, the maximurn applied shear stress parallel to grain, ra,joccurs at the neutral axis and is calculated from: Tal//=
r;,
1.5 
(4.13)
A
where: Fv=maximum vertical shear force (usually maximum reaction) A =crosssectional area Q =first moment of areaabout neutral axis abovethe position where shear stress is required I = second moment ofarea b = breadth of the section at the position where shear stress is required. In general, the value of applied shear stress, Tal//,should not exceed the Permissible shear stress parallel to grain, rahI/, determined from: Tu&,//
= rg,// X
K2 X
K3
X
K5
X
(4.14)
&, K3 and K8 are general modification factors, whichweredescribedin detail in Chapter 2. K5 relates to the calculation of the permissible shear , members with notched ends, and is described below. stress, r a h I /for ear atnotched ends
K5 is a modification factor which allows for stress concentration induced at squarecornered notches at the ends of a flexural member (Clause 2.10.4, l3S 5268 :Part 2) where:
DesignofFlexuralMembers(Beams)
\
37
4
(a) Beam with notch on the underside
I I
(b) Beam with notch on the top edge Fig. 4.7 Notched beams (adapted from BS 5268: Part 2).
(1) For a notch on the underside of a beam, see Fig. 4.7(a),
(2) For a notch on the top edge, see Fig. 4.7(b),
= 1.0
fora >he
uspended timber flooring
A suspended flooring system generally comprises a series of joists closely spaced, beingeither simply supported at their ends or continuous over loadbearing partition walls. The floor boarding or decking is applied on the top of the joists and underneath ceiling linings are fixed. A typical suspended floor arrangement is shown in Fig. 4.8(a) The distance between the centres of the joists is normally governed bythe sizeof the decking and ceiling boards, which are normally available in dimensions of 1200 mm wide x 2400mm long. The size of the decking and ceilingboardsallowsconvenient joist spacingsof 300 mm, 400mm or 600mm centre to centre. In addition, the choice of joist spacing may also be afYectedby the spanning capacity of the flooring material, joist span and other geometrical constraints such as an opening for a stairwell.
38
Structural Timber Design
Header joist Joists
38 or 50 mm
Trimer joists C \
Tongued & grooved boarding
(b)Solid timber tongued&
\
grooved decking
Loadbearing partition wall& spreader beam Masonry wall
Header joist
Trimer joists A
Stairwell
Joists T r i m e r joists Joisthanger nailed together
(a) A typical suspended floor arrangement
trimer (c) A typical joist to joists connection 10 m bolts staggered at 600 m centres
Joists
6 to 1 O m thick steel plate, 10 m less in depth than timber joists
Joists (d) Flitched beam
Solid blocking wall between Masonry joists v
(e) A typical support arrangement
Fig. 4.8 Suspended timber flooringtypical components.
Design of FlexuralMembers(Beams)
39
The most common floor decking in domesticdwellings and timberframed buildingsusessomeformofwoodbased panel products, for example chipboard or plywood. Solid timber decking such as softwood tongued and grooved (t & g) decking is often used in roof constructions, in conjunction with gluedlaminatedmembers, to produce a pleasant, natural timber ceiling with clear spans between the main structural members. The solid timber tongued and grooved boards are normally machined from 150mmwide sections with 3875 mm basic thicknesses [Fig. 4.8(b)]. The supports for joists are provided in various forms depending on the type of construction. Timber wall plates are normally used to support joists on top of masonry walls and foundations, Fig. 4.8(e). In situations where joists are to be supported on loadbearing timberframe walls or internal partitions, headerbeams or spreader members are provided to evenly distribute the vertical loads. Joisthangers are often used to attach and support joists onto the main timber beams, trimmer members or masonry walls [Fig. 4.8(c)]. Timber trimmerjoists are frequently used within timber floors of all types of domestic buildings, seeFig. 4.8(a). There are two main reasons for which trimmer joists may be provided2.First is to trim around anopening such as a stairwell or loft access (Trimmerjoists A), and to supportincoming joists (Trimmer joists B), and second isto reduce the span of floor joists over long open spans (Trimmer joists C), as shown in Fig. 4.8(a). Trimming around openings can usually be achieved by using two or more joists nailed together to form a trimmer beam,Fig. 4,8(c), or by using a single but larger timber section if construction geometry permits. Alternatively, trimmers canbe of hardwood or glued laminated timber, boxed plywebbed beams, or composite timber and steel flitched beams2, Fig. 4.8(d). All flooring systems are required to have fire resistance from the floor below and this is achieved by the ceiling linings, the joists and the floor boarding acting together as a composite construction3.For example, floors in two storey domestic buildings require modified 30 minutes fire resistance (30 minutes loadbearing, 15 minutes integrity and 15 minutes insulation). In generala conventional suspended timber flooring systemcomprising 12.5 mmplasterboard taped and filled, tongued and grooved floor boarding with at least 16mm thicknessdirectlynailed to floor joists, meets the requirements for the modified 30 minutes fire resistance providedthat where joisthangers are used they are formed fromat least 1 mm thick steel ofstrap or shoe type. Further details and specific requirements for fire resistance are given in BS 5268 :Part 4: ‘Fire resistance of timber structures’.
4.8 References 1. Baird and Ozelton (1984) Timber Designer’s Munual, 2nd edn. BSP Professional
Books, Oxford.
40
Structural Timber Design
2. TheSwedishFinnishTimberCouncil (1988) Principles of Timber Framed Construction, Retford. 3. TRADA (1994) Timber Frame Construction, 2ndedn.TimberResearchand Development Association (TRADA), High Wycombe.
4.9 Design examples Example 4 7
Design of a main beam
A main beam of 3 m lengthspans over an opening 2.8 m wide (Fig. 4.9)and supportsa flooring system which exerts a longduration loading of 3.9 kN/m, including its own selfweight, over its span. The beam is supported by 50mm wide walls on either side. Carry out design checks to show that a 75 mm x 225 mm deep sawn section whitewood grade SS under service class 1 is suitable.
Dimensions inmm Fig. 4.9
Beam details (Example4.1 ).
Definitions
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm"2
N :=newton kN:= lo3 .N Direction parallel to grain, // Direction perpendicular to grain, pp
'1. Geometricalproperties Span (clear distance), L Bearing width, bw Effective span, Le
Beam di~ensi5ns: Breadth of the section, b Depth of the section, h
L := 2.8 m bw := 50mm Le := L+bw Le = 2.85 o m b := 75. mm h := 225. mm
Design of FlexuralMembers (Beams)
A:=b*h A = 16 875o mm2 Zxx := b h3 Zxx= 7.12 x lo7 o mm4
Crosssectional area, A
h.
Second moment of area, ,,Z
*
2. Loading Applied uniformly distributed Total load, W
load,
W
:= 3.9 kN m* W:=weLe W = 11.11 okN
W
3. Kfactors
Service class 1 (K2,Table 13) Load duration (&, Table 14) Bearing: 5 0 m , but located <75 mm from end of member (K4, Table 15) Notched end effect (K5, Clause 2.10.4) Form factor (&, Clause 2.10.5)
K2 :=l K3 := 1 for longterm := 1
Depth factor (K7, Clause 2.10.6)
K7 :=
No load sharing (Kg, Clause 2.9) 4.
K5
:= 1 for no notch
K6 := 1
(
300 mm h
) '*l1
K7 = 1.03
K8
:= 1
Gradestresses
BS5268 : Part 2, Tables 2 and 7 Whitewood grade SS Bending parallel to grain Compression perpendicular to grain Shear parallel to grain Minimum modulus of elasticity 5 . Bending stress
Applied bending moment
Section modulus
Applied bending stress
Strength class =C24 am.g.l/:= 7.5 N mm2 :=2.4 N mm2 no wane rg.I:=0.71 N mm"2 Emin :=7200 N mm2

*

+
*
+
*
41
42
Structural Timber Design
Permissible bending stress
0m.adm.// := 0 m . g . j * K2 K3 K6 crrn.,h.l = 7.74 o N mm2
*
K1 ' K8
Bending stress satisfactory 6. Lateralstability
BS S268 :Part 2, Clause 2.20.8 and Table 16 Maximum depth
to breadth ratio, h/b
h =3 b Ends should be held in position
7 . Shearstress
Applied shear force
Fv
W L
Applied shear stress
F,, = 5.56 o kN 3 2 l.,:=Ta = 0.49 o N mm"2 7adm.j :Irz7 g . l K2K3 K5 Tadm.j = 0.71 o N mm2 Shear stress satisfactory
(A) *
Permissible shear stress, no notch
*
*
*
*
8. Bearingstress
Applied load
W 2 Fv= 5.56o kN
Fv:=
Applied bearing stress = 1.48 o N mm2 0c.adm.pp :=ac.g.pp K2 K3 K4 ' K8 (7c.adm.pp = 2.4 o N mm2 Bearing stress satisfactory
,,.,a
Permissible shearing stress

9. Deflection
No load sharing
Deflection due to bending
A, = 6.54 o mm Deflection due to shear
19.2 M (l? h) E A, = 0.63 o mm *
As:=
*
*
Design ofFlexuralMembers
:=A m
(Beams)
43
+A s
Total deflection
AItotal
Permissible deflection
Aadm := 0.003
AtOtal = 7.16 0 m m
Le = 8.55 0 m m Deflection satisfactory Aadm
Therefore a 75 mm x 225 mm sawn section whitewood C24 is satisfactory Example 4.2 Design of floor joists
A timber floor spanning 3.8 m centre to centre is to be designed using timber joists at 400 mm centres. The floor is subjected to a domestic imposed load of 1.S kN/m2 and carries a dead loading, including selfweight of 0.35 kN/m2. Carryout design checks to show that a series of 44mm x 200mm deep sawn section British spruce gradeSS under service class 1 is suitable. Joist han er
Main beam Tongued & grooved boarding
Y X"
"X
~
Y
Crosssection
Section AA Fig. 4.10 Timber floor joists (Example4.2).
Definitions
Force, kN Length, m Crosssectional dimensions,mm Stress, Nmm2
N :=newton kN :==: lo3 * N Direction parallel to grain, // Direction perpendicular to grain, pp
1. Geometricaiproperties Effective span, Le Joist spacing, Js
L e := 3.8 * m
Js :=0.4. m
StructuralTimber Design
Joist d i ~ e n ~ i ~ n s ~ Breadth o f section, b Depth o f section, h Crosssectional area, A Second moment o f area, Zxx
Dead load, DL Imposed load, ZL Total load, W
Service class 1 (KZ, Table 13) Load duration (K3, Table 14) Bearing: assume 50 mm, but located <75 mm from the end o f the member (K& Table 15) Notched end eEect (Ks, Clause 2.10.4) Form factor (K&Clause 2.10.5) Depth factor (K7,Clause 2.10.6) Load sharing applies (Kg,Clause 2.9)
BS 5268 : Part 2, Tables 2 and 7
British spruce grade SS Bending parallel to grain Compression perpendicular to grain Shear parallel to grain Mean modulus o f elasticity, load sharing
Applied bending moment
Section modulus
b := 44 mm h := 200 mm A := b * h A = 8.8 X lo3 0 mm2 Zxx := b h3 Zxx= 2.93 x lo7 o mm4 *
*
DL := 0.35  kN m2
IL := 1.5 kN m2 W:=(DL+IL)*JS*Le W = 2.81 o kN
K2 := 1 K3 :=1 for longterm
& := 1
K5
:=1
K6 :=
1
forno notch
'=(300 hmm
)
'.l1
K7 = 1.05 Kg := 1.1
Strength class = C 18 := 5.8 N mm2 o  ~ . ~ .: ,= ~ 2.2 . N ,mm2 no wane 7.1 := 0.67. N mmm2 Emean := 9 100 N mm"2 o  ~ . ~ . N
*
+
Design of Flexural Members (Beams)
Applied bending stress Permissible bending stress
6. ~ate~at sta~i~i~
BS5268 : Part 2, Clause 2.10.8, and Table 16 Maximum depth to breadth ratio, h/b
h b Ends should be held in position and compression edges held in line
= 4.55
7. ~ ~ estress a p
Applied shear force
W
F, :=yL.
.Fv
= 1.41 o kN
Applied shear stress Permissible shear stress, no notch
Applied load Assume bearing widths, bw, o f 50mm either side Applied bearing stress Permissible bearing stress
W r;, := 2
46
Structural Timber Design
9. ~ e f f e ~ t i o n Load sharing system Deflection due to bending Am = 7.53 o mm 19.2 A4
A,
Deflection due to shear
*
:=
(b h) E A, = 0.32 o mm Atotal := a m As Atotal= 7.85 o mm := 0.003 Le A u h = 11.4 o mm Deflection satisfactory *
*
+
Total deflection Permissible deflection
*
Therefore 44 mm x 200 mm sawn sections in Cl8 timber are satisfactory Example 4.3
Design of floor joists sefection of a suitabfe section and design for notched ends
The crosssection of a suspended timber flooring system is shown in Fig. 4.1 1. It consists of tongued and grooved (t & g) boarding with a selfweight of 0.15 kN/m2 and carries a plasterboard ceiling of 0.2 kN/m2. The floor has an effective span of 4.0 m and is subjectedto a domestic imposedload of 1.5 kN/m2. Designthe timber floor joists using timber in strength class C l 8 under service class 1. If the joists are to be notched at bearings with a 72mm deep notch, check that the notched section is also adequate. Tongued & grooved boarding
Plasterboard
\
Y I
Y
Notched joist Fig.4.11 Timber floor joists (Example 4.3).
Crosssection
Design of FlexuralMembers(Beams)
47
Force, kN Length, m Crosssectional dimensions, mm Stress, Nrnmw2
N :=newton k~ := 103 .N Direction parallel to grain, // Direction perpendicular to grain, pp
Effective span, Le Joist spacing, Js
L e := 4.0 * m
Dead load: t & g boarding (kN/m2), tg Plasterboard ceiling (kN/m2), Pb Selfweight (kN/m2), Swt Imposed load (kN/m2), ZL Total load (kN), W
Service class 1 (KZ,Table 13) Load duration (K3, Table 14) Bearing (.K4, Table 15), assume Notched end effect (Ks, Clause 2.10.4) for no notch Form factor (KG, Clause 2.10.5) Depth factor (K7, Clause 2.10.6) Load sharing applies (K8, Clause 2.9)
135'5268 : Part 2, Table 7 Strength class = C 18 Bending parallel to grain Compression perpendicular to grain Shear parallel to grain Meanmodulus ofelasticity, load sharing
Js :=0.45
m
m
tg := 0.15 kN m2 Pb := 0.2 kN m2 Swt := 0.10 kN mm2 assumed ZL := 1.5 k N m2 W:=(tg+Pb+Swt+ZL).Js.L, W = 3.51 o kN


K2 := 1 K3 := l for longterm & := 1 K5 := 1 := 1
At this stage ignore KT K8 := 1.1
".g.[
:= 5.8 N mm"2
oc.g.pp := 2.2 N
.mm2 no
:= 0.6'7 N mm"2 Ernean :=9100 N mm2 *
wane
Structural Timber Design
Applied bending moment Bemissible bending stress Required section modulus
BS5268: Part 2, Clause 2.10.8 and Table I6 In order to achieve lateral stability by direct fixing of decking to joists, the depth to breadth ratio should be limited to 5, i.e. h 5 5b. Substituting for h = 5b in Zxx= bh2/6 and equating it to Zrezuiredt gives: =
(
Zequired
b (5 b)2 *
6
Thus, minimum breadth of section
b := 6
and depth of section
b = 40.42 o m m h:=5*b h = 202.08 o mm
*
zrequired 52
Selecting a trial section from Table A2, Appendix Try A. 47 mm x 200 mm deep section. Beam dimensions Depth ,h Breadth, b Section modulus Zprovihd
~odificationfactor K7 for section depth
K7
= 3.13
X
(
'=
300 mm h
K7 = 1.05 Actual permissible bending stress
lo5 o mm3
)
' K2 K3 = 6.67 0 N Bending stress satisfactory
Om.adm.1 := gm.g.1 cm.adm./l
Check se~weight BS.5268 : Part 2, Table 7 density Average Total joist selfweight
*
*
*
p := 380 kg. mW3 SWtactual := p g h b *
SWtactual
Assumed total selfweight
'*''
*
*
Le
= 0.14 0 kN
SWtmsumed := SW5 * L e swtms~e= d 0.18 0
JS
kN
Selfweight satisfactory
K6
*
K7
*
K8
Structural Timber Design Atotal :=A m
Total deflection
4A s Atotal= 10.65 o mm Aa& :=0.003 Le nu&= 12omm Deflection satisfactory
Permissible deflection
*
~ere~ore 47 mm x 2 mm sawn ~ t i o n s in 618 tim~erare satisfacto~y esign of a ~ l o o r ~system ng f l o ~boards r an
The ground floor of' a shop is to comprise a series of timber joists at 600 mm centres with tonguedand grooved (t & g) boarding. The joists are simply supportedon 100 mm hangers attached to loadbearing walls 4.2 mapart as shown in Fig. 4.12. Determine a suitable thickness for floor boarding using timber in strength class C18 and a suitable size for joists using timber in strength class C22 under service class2. Assume imposed load is 2.0 kN/m'.
Joist hanger
ued & grooved boarding
A flooring system for Example 4.4.
Force, kN Length, m Crosssectional dimensions, mm Stress, NmmB2
N :=newton kN:= IO3 * N Direction parallel to grain, // Direction perpendicular to grain, pp
~~d~ r o o v~ e~ o ~ r d i ~
To calculate flooring thickness boards may be designed as simply supported beams. Calculations need only be madefor bending strength and deflection as span to thickness ratios and bearing widths are such that the shear and bearing stresses at supports are unlikely to be critical. Assuming tongued and grooved boarding comprises 100mm wide timber beams of thickness (depth) t simply supported on joists.
Design of Flexural Members (Beams)
1. ~ e o ~ e t r i cproperties al Js := 0.6 .m :=100 mm Le := JS
Joist spacing, JS Assume t & g boarding width,bb Effective span, L, 2. ~oading
tg := 0.1 kN m"; IL := 2.0. kN m; W:=(tg+IL)*b.Le W = 0.13 o kN
t & g boarding, tg Imposed load, IL Total load, W
+

3. Kfactors
Service class 2 (K2,Table 13) Table 14) Load duration (K3, Clause 2.10.2) Bearing (K4, Notched end effect (KS, Clause 2.10.4) Form factor (K6, Clause 2.10.5) Depth factor (K7, Clause 2.10.6) for h 5 72mm Load sharing applies (Kg, Clause 2.9) 4.
K; := 1 K3 := 1 for longterm Kb, := 1 assumed K5 := 1 forno notch K4 := 1 K7 := 1.17
Gradestresses
BS.5268 : Part 2, Tables 2 and 7 Strength class = Cl8 Bending parallel to grain Compression perpendicular to grain Shear parallel to grain Mean modulus of elasticity, load sharing

:= 5.8 N mmy2 := 2 .N 2. mmv2 no rg.I := 0.67 N mm"; Ernean := 9 100 N mmw2 +

wane
*
5. Bending stress
W *Le . 8
Applied bending moment
*
Permissible bending stress
Required thickness for t & g can be b t2 obtained by rearranging Z := 6 *
t :=
~
Z y ' r e d,t = 8.72 0 mm
'
51
52
Structural Timber Design
6. ~eflectjon
Load sharing system Permissible deflection Using 5 . WL: * 384 E *A d m rXx= 2.16 X io4 0mm4
I,, :=
b t3 Therefore from I,, = 12 *
{T
t :=3 12 I,, *
t = 13.74 o mm Therefore t 2 (the greater of .13.74mm and 8.72 mm and allowing for wear), thus t := 16 mm *
Adopt 16mm t & g boarding using strength class C18 timber B ~ e s j of ~ floor n joists
7 . ~ e o ~ e t rproperties i~al
Effective span, L, Joist spacing, Js Bearing width, bw
4.1 * m Js := 0.6 m bw := 100. m Le
. Loa Average density (Table 7 ) t & g boarding, tg Selfweight of each joist (kN/m2), Swt Imposed load (kN/m2), IL Total load (kN), W
Service class 2 (K2,Table 13) Load duration (K3, Table 14) Bearing (K4, Table 15), assume Notched end effect (K5, Clause 2.10.4) Clause 2.10.5) Form factor (K6, Depth factor (K7, Clause 2.10.6) Load sharing applies (Ks, Clause 2.9)
p := 410kgm3 tg:=p*g*t tg = 0.06 o kN m2 Swt := 0.10 kN m2assumed

ZL := 2.0 kN 
W:=(tg+SWt+IL)*JS*Le W = 5.32 o kN
K2 := 1 K3 := 1 for longterm & :=1 K5 := 1 forno notch K6 := 1 At this stage ignore K? Kg := 1.1
Design of Flexural Members (Beams)
BS5268 : Part 2, Table 7 Strength class =C22 Bending parallel to grain Compression pe~endicularto grain Shear parallel to grain Mean modulus of elasticity, load sharing
amSg.1 := 6.8 N mm2 oc.g.pp :=2.3 .Nmm2 no wane ~~1:=0.71 N . E m e m := 9700 N mm"2 +
*
*
Applied bending moment Permissible bending stress Required section modulus
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 breadth ratio should be limited to 5, i.e. h 5 5b. Substituting for h = 5b in Zxx = bh2/6 and equating it to Zrequired, gives: b (5 b)2 zrequired = 6 6 zrequired b := 52 )l'? Thus, minimum breadthsection of *
(
and depth of section
*
*
b=44.4omm h:=5*b h = 222.02 o
Selecting atrial section from Table A2, Appendix A. Try 47 xmm 225 mmdeep section. Beam dimensions Depth, h Breadth, b Section modulus
h := 225. mm b := 47 mm b h2 zprovided := 6

*
K7 = 1.03
4
Structural Timber
Design
Actual permissible bending stress
Check S ~ ~  w e i g h t BS5268 : Purt 2, Tuble 7 Average density Total joist selfweight Assumed total selfweight
12. Shearstress
W 2 Fv= 2.66 o kN
Fv := 
Applied shear force
Applied shear stress Permissible shear stress
13. ~
e
astress ~ j
~
~
Applied load Permissible bearing stress Minimum required bearing width
14. ~ e f f e c t i o ~
Second moment o f area, I,. Load sharing system Deflection due to bending
I,, := b h3 I,, = 4.46 x lo7 o mm4 *
Design of Flexural Members (Beams)
Deflection due to shear Total deflection Permissible deflection
m x 225mm sawn sections in C22 ~mberare sa~s~actory
Timbersections are commonlyusedin construction as axially loaded members or members in combined axial force and bending. Members of a truss, posts or columns, vertical wall studs and bracing elements are typical examples, This chapter deals in detail with the general considerations necessary for the design of compression and tension members and describes the design details of solid section rectangular timber members. Design methods for glulamcolumns and columnsofsimple composites are described in Chapters 6 and 8 respectively.
pression mem~ers Compression members include posts or columns, vertical wall studs, and struts in trusses and girders. Permissible stresses for timber compression members are governed by the particular conditions of service and loading defined in Clauses2,6.2,2.8 and 2.9 of BS 5268 :Part 2 :1996 whichrelate to differentserviceclass conditions, duration of loading and loadsharing systems respectively; and also by the additional factors given in Clause 2.11 ofthecodewhich are detailed here. Clause2.1 1 deals withdesignof compression members and divides them into two categories: (l) members subject to axial compression (without bending), and (2) members subject to combined axial compression and bending (this may be due to applied eccentric compressive force).
The main design considerations for compression members are: (1)Slenderness ratio. This relates to positional restraint of ends, lateral restraint along the length and crosssectionaldimensions ofthe member. (2) Axialcompressionandbendingstresses. 56
Design of AxiallyLoadedMembers
57
The loadcarrying capacity of compression members is a function of the slenderness ratio, h, which is calculatedas the effective length,L, divided by the radius of gyration, i: L = yLe (51) l
The radius of gyration, i, is given by
where l i s the second moment of areaand A is the crosssectional area of the member. For rectangular sections, where b is the least lateral dimension (Fig. 5.1), the value of i simplifies to:
Clause 2.1 1.4 ofBS 5268 :Part 2 :1996 recommends that the slenderness ratio should not exceed a value of: X = 180, for compression members carrying dead and imposed loads other than loads resulting from wind, X = 250, for any members subject to reversal of axial stress solely from the effect of windand any compression member carrying selfweight and wind loads only.
The effective length, Le, of a column (given in Clause 2.11.3) should be derived from either: (1) The deflected form of compression memberas affected by any restraint and or fixing moment(s). Then the effective length is consideredas the distance between adjacent points of zero bending moment, or
x
Y
.l Crosssectionof a column.
h
Structural Timber Design
Effective length of compression members
BS 5268 :Part 2)
(Table 18, LC?1L
End conditions Restrained at both ends in position and in direction Restrained at both ends in position and one end in direction Restrained at both ends in position but not in direction Restrained at one end in position and in direction and at the other end in direction but not in position Restrained at one end in position and in direction and free at the other end
0.7 0.85 l .0 1.5
2.0
(2) Table 18 of the code for the particular end conditions at the column ends. Then the effective length is obtained by multiplying a relevant coefficient taken from this table by the actual length, L. Le = coefficient x L
(5.3)
The end conditions defined in Table 18 of BS 5268 :Part 2 are reproduced here as Table 5.1 and are illustrated in Fig. 5.2. For compressionmemberswithslenderness ratios equal to or greater than 5, Clause 2.1 1.5 of the code requires that the permissible compressive stresses are further modified by f y l 2 the modification factor for compression members.
L, = 0.7 L
L, = 0.85 L
L, = 1.0 L
L, = 1.5 L
L, = 2.0 L
Fig. 5.2 Effective lengths and end conditions.
ation factor for c o ~ ~ r e s s i o ~ ne ~ ~ e r s ,
The modification factor for compression members, KI2,can be determined using either Table 19 of BS 5268 or calculated from the equation given in Annex B of the code. For either method, the minimum value of modulus of
Design of Axially Loaded Members
59
Structural TimberDesign
elasticity E (i.e. Emin) should be used in all cases including when load sharing is present. The value of G,,/ for use in either method, should be the grade (given in Tables 712a of the code) modified only compressive stress, oclgI/, for moisture content, duration of loading and size where appropriate. This will necessitate that all relevant load cases be considered, as the KI2value will differ with each load duration. It is to be noted that for memberscomprisingtwo or morepieces connected together in parallel and acting together to support the loads, the minimum modulus of elasticity should be modified by Kg (see Table 17) or 8 (see Table 22) of the code. For horizontally laminated members, the modified mean modulus of elasticity should be used (see Clausesand 3.2 3.6 S 5268 :Part 2 is reproduced here as Table 5.2 and the equation given in Annex B for calculation of K12 is in the following form:
where: o , =permissiblestress for a very short column (h <5), E =minimum modulus of elasticity,Emjn Le h =slenderness ratio, 
x K2K3
i
0.005 h
An axially loaded column hasits line of action of load passing through the centroidal axis of the column, see Fig. 5.3(a). The axial compressive stress, G,,a,//,is given by:
where P is the axial compressive load and A the crosssectional area. is calculated l~, as the product The per~issiblecompressive stress, o c , a ~ of gradecompressivestressparallel to grain, and anyrelevant modification factors (Kfactors).These are K2 for wet exposure condition (if applicable), K3 for loadduration, Kg for loadsharing systems(ifapplicable) and the modification factor for compression members, .Kl2, if appropriate:
Design of Axially Loaded Members
Y
Y
Y
Y
I
I
!
i
i
1
ex !
i
x .eY
! !
l
I
i
i
Y
Y
(a) Concentric load Load = P M=0
Y
(b) Eccentric load Load = P Mxx= P . e,
Y
(c) Eccentric load Load =P MW= P . ex
(d) Eccentric load Load =P Mxx=P ' e , MW=P.ex
Fig. 5.3 Axial concentric and eccentric loads.
The compression member is designedso that the applied compressive stress, oclal//, does not exceed the permissiblecompressive stress parallel to the grain, C f c , a h , / /
ers s#bjected to axial compression and b e ~ d i n ~ e 2.77.6)
This includes compression members subject to eccentric loading, where load acts through a point at a certain distance from the centroidal axis, which can be equated to an axial compression force and bending moment, see Fig. 5,3(b), (c) or (d). Members which are restrained at both ends in position but not direction, whichcoversmostreal situations, should be so proportioned that:
where: C T ~ ,= ~ applied , ~ bending C T ~ , , =permissible ~ , ~
M stress= 
z
bending stress =
X
K2K&K7Kg
62
Structural Timber
Design
P compression stress= 
C S ~ ,= ~ applied , ~
A
=permissible compression stress= CS,,~,/ x &&&K12
K~EI CS, =Euler critical stress= and E = Emin. (LeliI2
Equation (5.7) is the interaction formula used to ensure that lateral instability does not arise in compression members subjectto axial force and bending. Thus if the column is subject to compressive loading only, then C S ~ ,= ~ 0, , ~ and hence the equation simplifies to C F ~ , ~ , ~ / C S ~5, ~1.. LAlter~,~ ~ 0, , ~ then the natively, if the column is subject to bending only, i.e. C S ~ , = equation simplifies to ~ m , ~ , ~ / C F m , u . L5n ,1. i 5.2.6 Design of /oadbearingstud vva//s
Stud walls are often constructed as loadbearing walls in timber framed housing. Details of a typical stud wall are shown in Fig. 5.4. Top raiUplate Studs
Noggings
Diagonal braces Bottom raillplate . ............... ............... ............... ................. ............... ............... ............... ............... ............... ............. . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
. ............... ............. ............. ............ ............. ............ .. .......................
.... ... ... ... ... ..... ... ... ... ... ... .. ... ... ... ... ... .. ... ... ... ... ... .. ... ... ... ... ... ..... ... ... ... ... ..... ... ... ... ... ... .. ... ... ... ... ... .. ... ... ... ... ... .. ... ... ... ... ... ..... ... ... ... ... ..... ... ... ... ... ... .. ... ... ... ... ... .. ... ... ... ... ... .. ... ... ... ... ... ..... ... ... ... ... ..... ... ... ... ... ...
(a) Elevation
(b) Sideelevation
Y
! ! ! I
Studs
! ! L. fi
Y
(c) Crosssection Fig. 5.4 Details of a typical stud wall.
!
610 m maximum for load sharing
Design of Axially Loaded Members
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 ends of the studs. These walls can be designed to resist both vertical and lateral loadings; windloadbeing a typicalexampleof a lateral load.Each stud maybeconsidered to be laterally restrained about its yy axis, fully or partially (for example, at mid height),either by thecladding/sheathing material, such as plasterboard and by internal noggings or diagonalbracing.Thereforein situations where the claddinglsheathingmaterialisproperly attached to the stud along its whole length, the strength of the stud can be calculated about its xx axis;otherwise the greater of theslenderness ratios about the individualstud’s xx and yy axesshouldbeconsideredin the design calculations. The following general considerations may apply in designing stud walls: (1) Studs are considered as a series of posts (columns) carrying concentric (axial compression) or combined bending and axial compressive loads, (2) Loadsharing: If stud spacing is less than 610 mm (which it usually is), the loadsharing factor (Kg = 1.l) applies. (3) Endrestraints: In generaltheindividual studs are assumed to be laterally restrained in position but not in direction (pinned ends), as these walls are normally provided with a top and bottomrail. In other situations, for example where the studs’ bases are cast in concrete, the appropriate end conditions should be used (see Table 5.1).
n of c ens ion members (Clause 2.1 Permissiblestresses for timbertensionmembers are governed by the particular conditions of service and loading defined in Clauses 2.6.2,2.8 and 2.9 of BS 5268 :Part 2 which relate to diEerent service class conditions, duration of loading and loadsharing systems respectively;and also by the additional factors given in Clause 2.12 of the code which are detailed here. With tension members, since there isno tendency to buckle, the ratio of ,) is not critical. In truss length to thickness(i.e.theslenderness ratio, A frameworks sometimes a membermay undergo compression for a short duration dueto wind loading and hence it isrecommended that the slenderness ratio, h, for truss members is limited to 250. The design of tension members involvestrial and error. The capacity of a tension member at itsweakest point, for example at connection points, should be determined. Generally, the crosssectional dimensions are found first by assuming a connector arrangement. After connectors are designed, the strength of the member is again checked basedon its net crosssectional area at the point of connection.
4
Structural Timber
Design
The main design considerations for tension members are: (1) Tensile stress for members subjected to axial loading only. (2) Combined bendingand tensile stresses for members subjected to lateral loading as well as the axial tension. In both cases, as before, the permissible stress value is calculated as the product of the grade stress and the appropriate modification factors for a particular service and loading conditions, and is compared with the applied stress in the member, where: Permissible stress (=grade stress x Kfactors) 2 applied stress
BS 5268 :Part 2 recommends that in the calculationof the permissible tensile stress value the effectof the width of the timber section KI4 should be considered. 5.3.2 W j d t ~ factor, K ,
The grade tension stresses given in Tables 712a of BS 5268 :Part 2 apply to members havinga width (i.e. the greater transverse dimension),h, of 300 mm (see Fig. 5.1). For other widths of members assigned to a strength class, the grade tension stresses should be multiplied by the width modificationfactor, K14, where: for h 5 72mm
K14 = 1.17
and for h >72mm
K14 =
(y )
0.11
5.3.3 M e ~ b e rsubjected s to axial tension only
An axially loaded column hasits line of action of load passing through the centroidal axis of the column. (1) The applied tensile stress, ot,a,/, in an axially loaded timber member is calculated from the following equation:
where:
T = tensile force A,,, =net crosssectionalarea.
Design of AxiallyLoadedMembers
(2)
65
The p e r ~ i s s i ~tensile le stress, G,,,&,// ,is calculatedas the product of the grade tensile stress, and any relevant modification factors (Kfactors) as follows:
where K2,K3 and & are general modificationfactors for service class3, loadduration and loadsharingsystemsrespectively,which were described in detail in Chapter 2. K14 is the width factor as described earlier. o t , a , / / , should
In general, the value of applied tensile stress, ~,I, permissible tensile stress, C T ~ , ~hence:
not exceed the
5.3.4 Combined bendingand tensile stresses
In members that are subjected to lateral loading as well as the axial tension, the position of maximum stress occurs at the point of maximum bending moment (Fig. 5.5). Clause 2.12.3 of the code requires that the sum of the ratios of the applied tensile and bending stresses to those of the permissible ones (i.e. interaction quantity) must not exceed unity: (5.10) w,Mie: Applied tensile stress, permissibletensilestress, applied bending stress,
T Ut,a,ll
=A,,,
cFt,ah,// = "ts,//
=m,a,//
" =
May
xK~K~K~KM
M
z
L Fig. 5.5 Tension member subjected to lateral loading.
Structural Timber Design
xamples
A timber column in strength class C18 is4m in height with a rectangular crosssection of 97mm x 145mm as shown in Fig. 5.6. The column is restrained at both ends in position but not in direction and is subjected to service class 2 conditions. (a) Determine the maximum axial longterm load that the column can support. (b) Check that the column is adequate to resist a longterm axial load of 12kN and a bending momentof 0.8 kNm about its x” axis. Y !
L, = 1.0 x L
Crosssection Fig. 5.6
Column details (Example 5.1).
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm2
N :=newton
k~ := 103 . N Direction parallel to grain, // Direction perpendicular to grain, pp
1. ~ e o ~ e properties t r ~ ~ a ~
133’5268 : Part 2, Table 18 Column length, L Effective length, L, Width of section, b Depth of section, h Crosssectional area, A Second moment of area, I,,
L :=4.0 m Le := 1.0 L Le=40m b := 97mm h := 145 mm A.1”bh A = 14065 omm2 I,, := L. b .h3 12 I,, = 2.46 x l O7 o mm4 *
+
Design of AxiallyLoadedMembers
Secondmomentof
area, Iyy
h
Iyy := h b3 lYy = 1.1 X io7 o
67
*
m 4
For a rectangular section
Radius of
gyration, iyy
Section modulus, Zxx
2. Check slenderness ratio, h
b
ZYY
:= J12
iyy= 28 o mm b h2 zxx:= 6 Zxx= 339 904.17o mm3 *
h : = G? _ lYY
h = 142.85 <180, satisfactory 3. Gradestresses
BS 5268: :Part 2, Table 7 Timber strength class, C18 Bending parallel to grain Compression parallel to grain ~ i n i m u mmodulus of elasticity
:= 5.8 N oc.g.l := 7.1 N Emin := 6000 * N
mm"2  mm2 mm"2 e
4. Kfactors
Service class 2 (K2, Table 13) Load duration (K3, Table 14) Form factor (K6, Clause 2.10.5)
K2 := 1 K3 := 1 for longterm K6 := 1
Depth factor (IC7, Clause 2.10.6)
K7 :=
(
300 mm h
) '*l1
No load sharing (K8, Clauses 2.9 and 2.1 1.5)
The modification factor for compression member, one of the following methods: (l) Using equation in Annex B:
K12,
can be calculated using either
Structural Timber Design
(2) Using Table 19: For
Emin %g.[
*
K2
K3
'
= 845.07 and h = 142.85 thus,
0c.adm.l cTc.adm./
:= 0 c . g . l * K2 * K3
K12
*
= 1.55 o N *mm2
= 0.22
K12
Hence the axial longterm load capacity of the column is calculated by rearranging, CT = P/A. Thus Pcapacity
:= (Jc.ab.//
*
A
2 1.79 0 kN
Pcapacity
P := 12 kN
Axial load Applied moment
*
M := 0.8. kN m
o ~ ~ r e s s jand o n ~ e n d jstresses n~
Applied compressive stress Permissible compressive stress
Applied bending
stress
0c.a.i
*
P 
A = 0.85 o N mm2 0c.adm.i := K2 .K3 Kg 0 C . a h . i == 1.55 o N .mm"2 Compression satisfactory ' "

0c.a.l
Gm.a.1 :=
M &x
= 2.35 o N mm2 0m.adm.l := 0m.g.// K2 K3 cm.a&./ = 6.28 0 Nmm2
om.a,/
~ermissiblebending stress
K12
*
*
*
K6 * K7
*
Check interaction quantity Euler critical stress
n2
0 ,
:= 
0 ,
= 2.9 o N
*
Emh
h2
Interaction quantity is satisfactory Therefore a 9'7 mm x 145 m m timber section in strength class C18 is satisfactory
Design of AxiallyLoadedMembers
69
For the designdata given below, checkthat a 100mm x 250mm rough sawn section,as shown in Fig. 5.7, is adequate as a column if the load is applied 90mm eccentric to its P” axis. The column is 3.75 m in height and has its ends restrainedin position but not in direction.
class
Design data: Timber Service Design load (longterm) 25 Design load (mediumterrn) 30
Whitewood grade SS kN kN Y
90 L = 3.75m
Crosssection
L, = 1.0 x L
Fig. 5.7 Column details (Example5.2).
~efjnjtjons
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm“2
N :=newton k~ :=103. N Direction parallel to grain, // Direction perpendicular to grain, pp
l . Geometrical properties
BS5268 : Part 2, Table 18 Column length, L Effective length, L, Width of section b Depth of section, h
L :=3.75 m L, :=1.0. L L, = 3.75 0 m b := 100 mm h := 250 *
*
90
Structural Timber Design
Crosssectional area, A Second moment of area, I,, Second moment of area, I,,,, For a rectangular section, radius of gyration: i,,,, Section modulus, ZXx
A:=b.h A = 25 000 o mm2 I,, := b h3 I,, = 1.3 X io8 0mm4 Iyy:= & h b3 Iyy= 2.08 x lo7 o mm4 b
h
*
*
*
m
:= iyy = 28.87 o mm iyy
b h2 zx,:= 0
6
Z,, = 1.04 x lo6 o mm3 2. Check slenderness ratio,A Le h := 7
lYY
h = 129.9 <180, satisfactory 3.
Gradestresses
BS5268 : Part 2, Tables 2 and 7 Strength classification Whitewood Bending parallel to grain Compression parallel to grain Minimum modulus of elasticity
grade SS =C24 N mm2 N mmm2 N mm"2
:= 7.5 C T , . ~ .: ~= 7.9 Emin := 7200

 
4. Kfactors
Service class 2 (K2, Table 13) Load duration (K3,Table 14) for long and mediumterms Form factor (KG, Clause 2.10.5) Depth factor (K7, Clause 2.10.6) No load sharing (K8,Clauses 2.9 & 2.115)
K2 :=l
K3.lOng := 1
and
K6 := 1
(
300 mm
K7 := h K7 = 1.02 K8 := 1
K3.med
:= 1.25
) OJ'
Modification factor for compression member,.Kl2, can be calculated using either one of the following methods:
Design of AxiallyLoadedMembers
(1)Using
71
equation in Annex B:
(a) Longterm loading:
K12.med :=
5+ 1
*
( l + q ) * ~ ~* E 3 h2 U, *
:= 0.005 h
U, := u,.~,,// K2 * K3.lOng
+
E := Emin
 x2 E 1.5 h2 *
*
*
U,
(2)UsingTable19: (a) Longterm loading: 0c.g.i
*
K2
= 911.39 and h = 129.9 thus,
K3.long
(b)Mediumterm K2
'
*
= 0.27
K12.med
= 0.23
loading:
Emin Dc.g.//
K12.long
= 729.11 and
h = 129.9 thus,
K3.med
5. Loading
Eccentricity, e,, e,, mm:= 90 (a) Longterm loading: Axial load Plong := 25 kN Moment =axial load X eccentricity M1ong := P1ong e,, Mlong = 2.25 o kN m (b)Mediumterm loading: Axial load Pmed := 30 kN Moment =axial load X eccentricity M m & := Pmed e,, M m e d I== 2.7 o kN m *
*
*
*
6. ~ o ~ p r e s s iand o n ~ e n ~ istress ng
(a) Longterm loading: Applied compressive
K 3 := K3.lOng
stress
Permissiblecompressive stress
K 1 2 := K12.long
P :==PIong M := Mlong
P A u , , ~= . ~1o N mm2
0 c . a . i := 
c~,,,h.i := u,~,J
K2 K 3 +
*
= 2.13 O N.mm2 Compression satisfactory 0c.ah.i
Kg
*
K12
92
Structural Timber
Design
Applied bending stress Permissible bending stress
Check interaction quantity Euler critical stress Oe
= 4.21 o N mmm2 *
Interaction quantity satisfactory
(b) Mediumtermloading: Applied compressive stress Permissible compressive stress
Applied bending stress Permissible bending stress
Check interaction quantity Euler critical stress 0,
= 4.21 o N.mmn2
Interaction quantity satisfactory
Therefore a 1OOmm x 250mm timber section in strength class C24 is satisfactory
Design of AxiallyLoadedMembers
€ x a ~ p /5.3 e
73
Design of a /Qadbearingstud wa//
A stud wall, as shown in Fig. 5.8, has an overall height of 4.0m. the vertical studs are positioned at 600mm centres with noggings at midheight. Carry out design calculations to show that studs of 44 mm x 100mm section inC l 6 timber under service classl are suitable to sustain a longduration load of 10.0kN/m.
Y x
12.5 mm 100.0 m 12.5 mm
.
Plasterboards Y
Crosssectionof a stud(44 m x 100 mm) Fig. 5.8 Stud wall details (Example5.3).
Definitions
Force, kN Length, m Crosssectional dimensions,mm Stress, Nmm"2
1.
N :=newton
k ~ : =1 0 3 . ~
Direction parallel to grain, // Direction perpendicular to grain, pp
Geometrical properties
BS5268 : Part 2, Table 18 For each stud acting as a column, restrained about its yy axis by plasterboards and restrained at both ends in position but not in direction: Stud length, about xx axis, L, Effective length, about XX axis, Le.z
L, :=4.0 m Le.x := 1e0 L, Le.x = 4 o m
Structural TimberDesign
Width of section, b Depth of section, h Crosssectional area A Secondmomentof
area, ,,Z
For a rectangular section, radius of gyration, ixx
b:=44*mm h := 100. mm A :=b*h A = 4400 o mm2 Ixx:= h b h3 ,,Z = 3.67 x lo6 o mm4 h *
lXX =
*
m
ixx= 28.87 o mm
eck s~enderness ratio, h
about xx axis
h,,
L.,
:= 1,
h, = 138.56 < 180, satisfactory 3. Gradestresses
35'5268 : Part 2, Table 7 Timber strength class, Cl 6 Bending parallel to grain Compression parallel to grain Minimum modulus of elasticity 4. ~  f a ~ t o r s
Service class 2 (KZ, Table 13) Load duration (K3, Table 14) Form factor (K6, Clause 2.10.5) Load sharing applies (K8, Clause 2.9) Modification factor for compression member,K12,can be calculated usingeither one of the following methods:
Design of AxiallyLoadedMembers
75
(2) Using Table 19: For
Emin oc.g./l
K2 K3
= 852.94 and h = 138.56 thus, K12 = 0.23
*
5. Loading
Uniformly distributed axial load Stud spacing, Sstud Axial load per stud
6. Compressionstress
Applied compressive stress Permissible compressive stress
Therefore 44mm x 100mm timber sections in strength class C16 are satisfactory Example 5.4 Design of a loadbearing stud wall with lateral load
A stud wall forming part o f an open farm building, as shown in Fig. 5.9, has an overall height of 3.6m. The vertical studs are builtin to concrete at the base, support a roof Roof framework
Concrete foundation' (a) Elevation Fig. 5.9 Stud wall details (Example 5.4).
(b) Sideelevation
76
Structural Timber
Design
structure at the top, and are positioned at 400mm centres with noggingsat midheight. The cladding materials are assumed to make no contribution to the structural stabilityoftheframed structure. Carry out designcalculations to show that studs of 47 mm x 1OOmm section in C22 timber under service class 3 are suitable to sustain a mediumduration load of 5.4 kN/m and a wind loading of 0.975 kN/m2 (3 S gust). efinitions
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmnr2
N :=newton kN := lo3 N Direction parallel to grain, // Direction perpendicular to grain, pp *
1. ~ e o ~ e t r i properties cai BS5268 : Part 2, Table 18 Stud length, L Stud length, about xx axis, L, Stud length, about yy axis, Ly EflEective length, about xx axis, Le,, EEective length, about yy axis, Le.y is the greater of: Thus
L := 3.6 m L, := 3.6 m Ly :=1.8 . m Le., := 0.85 L, Le., = 3.06 o m *
*
Ly or Le,y:=1.0 Ly
Le,y :=0.85
Le.y := 1.O* L y
= 1.8 0 m b:=47*mm h := 100. mm A:=b*h A :=4.7 x io3 o mm2 ,,I := 1. b .h3 12 ,,Z = 3.92 x lo6 o mm4 Iyy:= G h b3 lyy= 8+65x lo5 o mm4 Le.y
Width of section, b Depth of section, h Crosssectional area, A Second moment of area, I,, Second moment of area, Iyy
*
*
For a rectangular section, radius of gyration, iyy
ZYY
:=
b
1/12
iyr = 13.57 o mm
h
radius of gyration, ixx
:= 4 2
i,, = 28.87 o mm
Section modulus, Z,,
b h2 6 Z,, = 7.83 x lo4 o mm3
z,,
:= 
Design of Axially Loaded Members
Le.y
hyy:= 
about yy axis
lYY
hyy = 132.67
h,,
about xx axis
:= Le.,
h,, = 106 Both < 180, satisfactory hyy is the critical one
BS5268 : Part 2, Table 7' Timber strength class, C22 Bending parallel to grain Compression parallel to grain Minimum modulus of elasticity
Service class 3 (K2, Table 13) Load duration (K3,Table 14) Form factor (KG,Clause 2.10.5)
(
Depth factor (K7, Clause 2.10.6)
K7 :=
300 mm h
) '*'I
Load sharing applies (Kg, Clause 2.9) Modification factor for compression member, K12
(l) ~ediumtermloading (no wind): (a) Using equation in Annex B: K 3 := K3.med
h
DC:= Gc.g.1 K2.comp
?l := 0.005 h
:= h y y
*
K3
'
E := Emin K ~ . E
Kl2.med
= 0.26
(b)UsingTable19: For
E o c . g . 1 * Kicornp * K 3
= 924.44 and
h = 132.67 thus, Kl2.med = 0.26
Structural Timber Design
(2) Veryshortterm loading (3s gust): (a) Using equation inAnnex B:
K3 :=K3.v.short
h :"Ih=y y
oc
:=oc.g.// K2.comp ' K3 *
E := Emin K ~ . E
II := 0.005 h
*
(b) UsingTable 19:
5. foadjng
Uniformly distributed axial load Stud spacing, Sstud Axial load per stud, Pstud Applied wind load, W L Wind load per stud, Y s t u d

P := 5.4 kN m1 Sstud := 0.4 pstud := p Sstud Pstud = 2.16 0 k N WL := 0.975 kN m2 *
Wstud :=
WL
Sstud
0.39 0 kN m"
W&d
C o ~ ~ r e s sand j o ~bending stresses
Mediumterm loading (no wind): := 7 Pstud
Appliedcompressivestress per stud
oc.a.1
Permissiblecompressive stress
Gc.a&.//
= 0.46 o N mm2 a
:= oc.g.// K2.comp ' K3.med K8 K12.med *
*
oc.ah,l= 1.63 o N mmw2
Compression satisfactory Veryshort term loading (3 S gust): AppliedcompressivestressAsabove Permissiblecompressive stress Applied bending moment per stud
o c . u= .~ 0.46 o N .mm2 oca&.// :=oc.g.// K2.comp K3.v.short * K8 Kl2.v.short *
*
Gc.udm.l = 1.77 0 N Elm2 Compression satisfactory *
M :=
Wstud
*
Design of Axially LoadedMembers
Check i n t ~ ~ a c tquantity i~n
uler critical stress 0 ,
= 2.92 o N mm2 a
0m.a.f
1.5
*
Gc.a.1 *
Kl2.v.short
%.ah.//
= 0.98 <1
Interaction quantity is satisfactory mber sections in s ~ e n class ~ h C22 are satis
ere~ore
le A trussed rafter tie of 38 mm x 100mm section is2.7 m long and is subjectedto a lateral concentrated load of 0.65 kN at midlength. Determine the maximum mediumterm axial tensile load that the rafter tie can carry. Assume a timber in strength class C22 under service class 2 conditions.
l
T
~__
L
...
/
Position of maximum bending moment
0,325 kN ?
0.65 lsN
L
=
2
.
7
m
t
0.325 kN
2
Fig. 5.10 Trussed rafter tie (Example 5.5).
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm2
N :=newton kN :=lo3 .N Direction parallel to grain, // Direction perpendicular to grain, pp
T
80
StructuralTimber Design
e o ~ e t r i c aproperties l
BS5268 : Part 2, Table 18 Tie length, L EEective length, Le Width of section, b Depth of section, h Crosssectional area, A For a rectangular section, radius of gyration, iyy Section modulus, Zxx
.
~
L :=2.7. m L e := 1.O.L L e = 2.7 o m b := 38 *mm h :=100. mm A := b * h A = 3800 o mm2 b ZYY
:= p j
iyy = 10.97 o mm

b h2 6 Zxx= 63 333.33o mm3
Zxx:=
~slenderness e c ratio, ~ h
Le h. *
iYY
h = 246.13
<250, satisfactory
3. Gradestresses
3S5268 : Part 2, Table 7 Timber strength class, C22 Bending parellel to grain Tension parallel to grain ~ i n i m u mmodulus ofelasticity 4.
:= 6.8 N mmW2 :=4.1 N mm2 Emin:= 6500 N mm2 e
m
+
actors
Serviceclass2 (KZ, Table 13) ~ediumtermloading (K3,Table 14) Clause 2.10.5) Form factor (K6,
K2 := 1
Depth factor (KT,Clause 2.10.6)
300 mm K7 := h K7 = 1.13
Width factor
(K14, Clause 2.12.2)
K3 := 1.25 .&j
:= 1
(
K14 :=
(300~mm~'11
K14 = 1.13
No load sharing (K8, Clause2.9)
)
0.1 1
K8 := 1
Design of Axially LoadedMembers
5. ~ o ~ d i n g
Lateral point load
*
M := 4 M=0.44okNm
moment Applied
6.
P := 0.65 kN P Le
fie^ a n d ~ e r ~ i s s jbending b ~ e stresses
Applied bending stress
om.a.1
:=
M
zxx
mm2 K3 K6 KT o m . a h . 1 = 9.59 o N mmm2 Bending satisfactory Um.a.1 = 6.93 o N
Permissible bending stress
o m a h . / := G m . g . 1 . K2
7 . ~ e r ~ i s stensile i ~ i e stress
8.
interactj~ quantjty ~
where hence
Hence the axial tensile load capacity of the tie is: Tcapacity := 0 t . a . l Tcapacity
A = 61 0 kN *
*
Kg
81
Clued laminated timber, commonlyknown as gZuZarn, is a structural material obtained bygluing together at least four laminations of timber sections of usually 1950 mm basic thickness so that the grain directions of all are parallel.Normally the laminations are dried to around 121 8% moisture content beforebeingmachined and assembled. This leads to a reduction in laminations’ thicknessof around 35mm.Assemblyis commonly carried out by applying carefully controlled adhesive mix to the faces ofthe laminations. They are then placed in mechanical or hydraulicjigs of the appropriate shape and size, and pressure is applied at right angles to the glue lines and is helduntil curing of the adhesive is complete.Clulam is then cut, shaped and any specified preservative and finishingtreatments are applied. In Fig. 6.1 the crosssection of a typical glulam beam is shown.
Y Fig. 6.1 Crosssectionof a glulam beam. 82
Designof GluedLaminatedMembers
83
Since the length of glulam members normally exceedsthe length of commercially available solidtimber, the individual laminations are fingerjointed together to make laminations of the required length. These fingerjoints are then placed randomly throughout the glulam component. Timber sections with thickness a of around 33mm to a maximum of50 mm are used to laminate straight or slightlycurved members, whereas much thinner sections (l2 mm or 19mm, up to about33 mm) are used to laminate curved members. Glued laminated members can also be constructed with variable sections to produce tapering beams, columns, arches and portals. A glulam beam designedto resist loads applied perpendicular to the plane of the laminations is referred to as hurizuntally laminated, and when it is designed to resist loads applied parallel to the plane of the laminations it is vertically laminated, see Fig. 6.2. Construction withglulammembersoffers the advantages ofexcellent strength and stiffnesstoweight ratio, which are usual with timber materials, and for this reason it has been selected for a number of prestigeprojects in the last few decades. Gluedlaminated members can also be designed to have substantial fire resistance.A summary of the advantages of glulam members over solid timber sections is as follows: Glulam sections are built up from thin members and it is therefore possible to manufacture complicated shapes. They can be produced in anysize, length, and shape, and manufacturing and transportation facilities remain as the only practical limiting factors affecting dimensions. Theuseof a numberof laminates enables an even distribution of defectsand the location of better quality timber where it ismost efficiently utilised. It is easily possible to buildin a camber at the time of assembly. Load
i
Load
(a) Horizontally laminated
(b) Vertically laminated
Fig. 6.2 Horizontally and vertically laminated glulam beams.
4
Structural Timber
Design
This chapter details the general considerations necessary for the design of flexural and axially loaded glued laminated members.
Themaindesign considerations for flexural and axiallyloadedglulam members are the same as those described in detail in Chapters 4 and 5 respectively. It is therefore assumed that the reader is familiar with the various modification factors and all other aspects of the design methodology detailed in these previouschapters. They are notrepeated at length here,but when a factor isused,referenceismade to the explanatory itemof the relevant chapter. BS 5268 :Part 2 :1996, Section 3 deals with the design. of glued laminated timber and specifiesthat such timber should be manufacturedin accordance with Clause 3.4 of the code and also with the recommendations of BS EN 386 :1995 Glued laminated timber. Performance requirements and mini~um ~roductionrequirements. It also requires that all timber used for laminated softwood members (similar to solid section timber, see Chapter 2, Section 2.1) should be strength graded visually or mechanically in accordance with S EN 5 18 or BS EN 519, respectively. Glulam strength properties may be determined from the strength class properties given in Table7 of the code (reproduced here as Table 2.3). These are described in detail in the following sections.
If a glulam section is manufactured using the same grade throughout it is referred to as singlegrade glued laminated member. Thegrade stresses given in Tables 712 of BS 5268 :Part 2 apply primarily to solid structural timber. In order to determine the grade stresses for a horizontally glued laminated softwood or hardwood member, Clause 3.2 the of code recommendsthat the strength class stresses given in its Tables 712 for the relevant grade and species are multiplied by the modification factors K15 to K20 according to the nature of stress and the number of laminations. The modification factors $ l l 5 to K20, detailed in Table 21 of the code, are forsinglegrade horizontally glued laminated members (Table 21 is reproduced here as Table 6.1).
For economic reasons there are many instances where members may be horizontally laminated withtwodifferent strength class timber sections.
Design of GluedLaminatedMembers
85
Table 6.1 Modification factors Kls, K17,.Kl8, K19 and K20for singlegradeglued laminated members and horizontally glued laminated beams (Table 21, BS 5268)
Strength classes
Number ofBending // Tension // Compression Compression Shear // Modulus of laminationsa to grain to grain //to grain L to grainb to grain elasticityc (om,g,/)
(ot,g,/)
(oc,g,/)
(oc,g,.d
(zg,/)
K18
K 1 9
Emem
K16
K1 7
1.39
1.39
1 . 1 1 1.49
1.49
1.03
1.07 C24, 2.34 C22, D35, D40
4 1.55 5 7 10 15 20 or more
1.04 1.26 1.34 1.39 1.43 1.48 1.52
1.26 1.34 1.39 l .43 1.48 1.52
C16, C18, D30
4 5 7 10 15 20 or more
l .05 1.16 1.29 1.39 1.49 1.57
1.05 1.16 1.29 1.39 1.49 lS7
1.072.73
1.69
1.17
K1 5
C27,C30, D50, D60, D70
4 or more
K20
aInterpolation is permitted for intermediate numbers of laminations K18should be applied to the lower value givenin Table 7 for compression perpendicular to grain ‘KZ0 should be applied to the mean value of modulus of elasticity
Highergrade laminations are normally placed in the outer portions of the member where they are effectively used, and lowergrade laminations are placed in the inner portion (core) where the lower strength will not greatly affect the overall strength. The combination is permitted provided that the strength classes are not more than three classes apart (as given in Table 7 of the code.) For example C24and C16 can be horizontally laminated but C24 and C14 should not. The code also requires that in horizontally laminated members where combination grades are used, they should be fabricated so that not less than 25% of the depth at both the top and bottom of the memberisof the higher strength class,see Fig. 6.3. For suchmembers the grade stresses should be taken as the product of the strength class stresses (given in Table 7 ) for the higher strength class laminations and the modification factors from Table 21 of the code. In addition, for bending, tensionandcompression parallel to grain, the grade stresses should be multiplied by 0.95.
Structural Timber Design
Highergrade
Lowergrade
onegrade
4 Highergrade
(a) Singlegrade
(b) Combinedgrade
Fig. 6.3 Horizontally glued laminated members.
6.3.3 Permissible stressesfor horizontally glued laminatedmembers
The per~issiblestresses for a horizontally gluedlaminatedmember are 7, by any relevant determined as the product ofgrade stresses, given in Table modification factors, i.e. K2, K3,. ..,KI4and modification factors K15 to K20 appropriate to the number of laminations. These are summarised below: Bending parallel to grain: for asinglegradeglulam, for a combinedgrade glulam,
O m , a h , / / = Om,g,// X O m , a h , / / = Om,g,// X
Tension parallel to grain: for a singlegrade glulam, for a combinedgrade glulam,
G t , a h , f/
Compression parallel to grain: for a singlegrade glulam, for a combinedgrade glulam,
oc,adm,// = Oc,g,// x
Ot,ah,i
K~IY~K~K~K~KI~ K ~ K ~ & & K & ~xs0.95
= O,,g,/x / K2K3&K14K16 = ctts,/X K&3&K14K16 x 0.95
Oc,adm,// =
X
K2K3K8K12K17 &K&K12K17 X 0.95
Compression perpendicular to grain: for both single and combinedgrade glulam members, Oc,ah,l.
= oc,g,l.
X
K2K3K4K8K18
Shear parallel to grain: for both single and combinedgrade glulam members, zah,//
= zg,// x K2K3K5K8K19
Modulus of elasticity: for both single and combinedgrade glulam members,
Design of Glued Laminated Members
It should alsobe noted that for tension perpendicularto grain and torsional shear, permissible stresses should be calculated in accordance with Clause 2.7 of the code, and factors K19 and K27 should be disregarded. For example, For tension perpendicular to grain: for both single and combinedgrade glulam members,
In determining the permissible stresses for glued laminated members, the modification factors used (if relevant) are:
K14
service class 3 (wet exposure condition), see Section 2.5.2, duration of loading, see Section 2.5.4, bearing stress, see Section 4.5.1, shear at notched end, see Section 4.6.1, form factor, see Section 4.3.2, depth factor, see Section 4.3.3, which applies to the full glulam section size, not the individual laminates, load sharing system, see Section 2.5.6, factor for compression member, see Section 5.2.3 (for calculation of K12 for glulam compression members, the value of Egiulam, as defined, should be used, i.e. Egiulam =Ernean x K20), width factor for tension member, see Section 5.3.2, which applies to the full glulam section size, not the individual laminates, are the appropriate grade stress modification factors for singlegrade horizontally laminated members. For combinedgrade members, these factors apply to the highergrade lamination stresses.
Attention is drawn to the definitions for K7, K12 and K14 given above.
ically glued lami A glulambeamdesigned to resist loads applied parallel to the plane of laminations is referredto asa vertically l a m i ~ a tbeam. e ~ This is illustrated in Fig. 6.2. BS 5268 :Part 2, Clause 3.3 specifiesthat permissible stressesfor vertically glued laminated beams are governed by the particular conditions of service and loading defined in its Clauses 2.6.2, 2.8 and 2.9 which relate to service class3(wetexposure) condition, duration of loading and loadsharing systems respectively; and also by the additional modification factors Kz7, K28 and K29 given in Table 22 of the code appropriate to the number of laminations. Table 22 of the code is reproduced here as Table 6.2.
8
Structural Timber Design
Table 6.2 Modification factors (Table 22, BS 5268 :Part 2)
Number of laminations
and
K27,
Bending // to grain, // to grain, @&g,/), and shear I/ to grain, &,l)
K29,
for vertically glued
laminated members,
Modulus of elasticity, Compresison l.to grainb and compression &g,d /I to grain, (Gc;g;/)
( G ~ ~ , /tension ),
K27
K28
K29
Softwoods
Hardwoods
Softwoods Hardwoods Softwoods and hardwoods
2 3 4
1*11 1.16 1.19
1.06 1.os 1.10
1.14 l .21 1.24
1.06 l .os 1.10
5 6 7
1.21 1.23 1.24
1.11 1.12 1.12
1.27 1.29 1.30
1.11 1.12 1.12
8 or more
1.25
1.13
1.13
1.10
1.32
aWhen applied to the value of the modulus of elasticity, E, K28 is applicable to the minimum value of E bIf no wane is present, K29 should have the value 1.33, and, regardless of the grade of timber used, should be applied to the SS or HS grade stress for the species
In general it is not recommended that vertically laminated beamsbe designedusingdifferent grade laminates (i.e.combined strength classes). When a designer is considering a glulam beam with vertical laminations, it is often because a horizontally laminated member with bendingabout the xx axis is also being subjected to bending on its y” axis, for example, due to lateral loading. If the vertically laminated beam is made from different grade laminations it can be assumed that such a member has grade stresses equal to theproduct of weighted strength classstresses, and the modification factors K27, K28 and K29, given in Table 22, appropriate to the total number of laminations in the beam. A weighted strength class stress, G , , may be obtained from:
where CTg,lower and Gg,higher refer to appropriate Stress or modulus of elasticity for the lowergrade and highergrade laminations respectively, as given in Table 7 of the code; and similarly N refers to the appropriate number of laminations in the member. For example, the weighted bending stress parallel to grain for a vertically laminated beam of 12 laminations, where 3 laminations at each end are in
Design of GluedLaminatedMembers
89
C24 and 6 inner ones are in C18 softwood timber, may be calculated as follows: From Table 7 of the code (Table 2.3 in this book): for C l 8 timber, for C24 timber,
om,g,/ om,g,/
= 5.8 N/mm2 = 7.5 N/mm2
Therefore the weighted grade bending stress is: Gw,m,g,/
= 5.8 X
6+ 7.5 X 6= 6.65 N/mm2
Hence the weighted bending stress parallel to grain for this glulam beam is calculated as: From Table 22 of the code (Table 6.2in this book), for 12 laminations, K27 = 1.25. ow,m,/
= 6.65 X 1.25 = 8.31 N/mm2
Thepermissiblestresses for averticallyglued laminated member are determined as the product of grade stresses (or weighted grade stresses inthe case of a member with combinedgradelaminations) by any relevant modification factors, i.e. KZ,&, ...,K14 and modification factors K27,K2g and KZ9 appropriateto thenumberof laminations). These are summarised below: Bending parallel to grain: Tension parallel to grain: Shear parallel to grain: Compression parallel to grain: Compression perpendicular to grain: Modulus of elasticity:
o m , g , / x K2K3K6K7KgK27 = o t , g , / x K2K3KgK~4K27 “a&,/ =“g,/ x K2K3K5Kg~27 ot,adm,/ = Ot,g,/ x K2~3KgK~2~28 o,,,h,~ = oClg,Jx K2K3&K8 K29
om,,&,/
ot,udm,/
Eg1ulum
==I
Emeun
For vertically laminated beams, the E value (for 4 or more laminations) is Erne,, andisonly applicable to Ernin.It should be noted that the size factors K7 and K14 for all glulam members, whetherhorizontally or vertically laminated, should be applied to glulam strengths not to the individual larninate strengths. It should also be noted thatfor tension perpendicular to grain and torsional shear, permissible stresses should be calculated in accordance with Clause 2.7 of the code, and factor K27 should be disregarded. Lateral torsional buckling is unlikely to be a problem as the depthtobreadth ratio related to bending about the yy axis will almost certainly be less than 1.0andanybeam tends to buckleonly at right angles to the direction of bending.
Structural Timber Design
BS 5268 :Part 2 :1996,Clause 3.5 recommends that in addition to the deflection due to bending, the shear deflection may be significant and should be taken into account. Attention is drawn to deflection criteria for flexural members which were covered in detail in Section 4.4, as the same criteria apply to glulam beams. As it is easily possible to buildin a camber at the time of assembly, the code specifies that members can be precambered to offset the deflection under dead or permanent loads. In such cases the deflection under live or intermittent imposed load should not exceed 0.003 of the span.
ed glued l a m i ~ a t ebeams ~ Glued laminated beams are often curved and/or tapered in order to meet architectural requirements, to provide pitched roofs, to obtain maximum interiorclearance and/or to reducewallheightrequirements at the end supports. The most commonly used types are the curved beams with constant rectangular crosssection and the double tapered beams (see Fig. 6.4). The distribution of bending stress,urn,in a curved beam is nonlinearand, in addition, radial stresses, er, perpendicular to the grain are caused by bending moments. If the bending moment tends to increase the radius of curvature, the radial stresses are in tension perpendicular to the grain and if it tends to decrease the radius of curvature, the radial (i.e. 0, = at,L); stresses are in compression perpendicular to the grain (i.e. Gr = u c , ~ ) . Curved glulam beams are formed by bending laminations in purposebuilt jigs during the production process. In order to make allowance for the initial stresses induced inthe laminations, a modification factor .K33 has
(a) Curved beam
(b) Double tapered beam
Fig. 6.4 Curved and double tapered beams.
Designof Glued Laminated Members
been introduced. For curvedglulambeamswith constant crosssection, BS 5268 :Part 2, Clause 3.5.3 specifies the following requirements: (l) The ratio of the radius of curvature, r, to the lamination thickness, t ,
should be greater than the mean modulus of elasticity, in N/mm2, for the strength class divided by 70, for both softwoods and hardwoods (see Fig. 6.5). Therefore check that: r &lean >t
70
r (2) If <240, then the bending, tension and compression grade stresses t
parallel to grain should be multiplied by the modification factor where
K33 = 0.75 + 0.001
t
and K33 5 1.0
/ /
\
Point of tangency
Crosssection AA
Bending stress,
CT,
Radial stress,
G,
Fig. 6.5 Curved glued laminated beam with stress distributions at apex.
Structural TimberDesign
(3) In curvedbeams,where the ratio of the minimummean radius of curvature, rmean,to the depth, h, isless than or equal to 15 (i.e. if rmean 5 15), then the bending stress induced by a mmnentpM , should h be calculated as: (a) Bending stress in the extreme fibre on the concave face, 6M o m = k34 bh2 where: K34 = 1
+
(rmtan) y y) 0.5
for
5 10, and
rmean 5 15 for 10
(1.01
6M = bh2 (4) The radial stress, or,induced by a bendingmoment, M, shouldbe calculated as: Om
where: If the moment tends to increase the radius of curvature of the beam, the radial stress will be tension perpendicular to the grain and the value of or should not be greater than the value derivedin accordance with Clauses 3.2 and 2.7 of the code. Therefore: or 5
ot,adm,l_ = +.g,//
x K2K3K8
(6.5)
~ ,$ T~ ~ , , / . where, from Section 2.5.7, C T ~ , = If the moment tends to reduce the radius of curvature of the beam, the radial stress willbe compression perpendicular to the grain and the value of Or should not be greater than 1.33 times the compression perpendicular to the grain stress for the strength class. Therefore:
or 5 1.33 X
o,,ah,L =
1.33 X
c ~ , , g , lX
K2K3~KaKla
(6.6)
Baird and Qzelton (1984) Timber Designer’s ~ a n u a l 2nd , ed. BlackwellScience, Oxford. Mettem, C.J. (1986) Structural Timber Design and Technology. Longman Scientific and Technical, Harlow.
Design of Glued Laminated Members
TRADA (1995) Gluedlaminatedtimber an introduction, Wood Information, Section 1, Sheet 6. Timber Research and Development Association, High Wycombe, UK.
A series of glulam beams of l15 mm wide x 560 mm deep (14 laminations) with an effective span of 10.5m is to be used in theconstruction of the roofof an exhibition hall. The roof comprises exterior tongued and grooved solid softwood decking exposed on the underside and covered on the top with 3 layer felt and chippings. For the design data given below, (a) check the suitability of the section, (b)determinethecamberrequired. Design data Timber
width Bearing Dead load (longterm): 0.75decking Exterior 3 layer felt and chippings 0.63 0.3 selfweight Beam Imposed load (mediumtem)
Imported whitewood in strength class C24 and service class 1 9Omm end each kN/m kN/m kN/m 2.5 kN/m
.
i
Y Fig. 6.6 Crosssectionof glulam beam (Example6.1).
94
Structural Timber Design
~efinjtions
Force, kN Length, m Crosssectional dimensions, mm Stress, Nrnmw2
N :=newton kN := lo3 N Direction parallel to grain, ,// Direction perpendicular to grain, pp *
1 . Geometrical properties
Effective span, L, Bearing width, bw Beam dimensions: Breadth of section, b Depth of section, h Crosssectional area, A Second moment of area, Zxx
Le := 10.5m bw := 90.mm b := 115.h := 560. mm A:=b*h A = 6.44 X lO4omm2 :=L.b.h3 12 Zxx = 1.68 x lo9 o mm4
2. ~ o a ~ j n ~ Dead load: Selfweight of beam (kN/m), S w t t and g decking, (kN/m), tg 3 layer felt and chippings (kN/m), fchip Total dead load, DL Imposed load: Prescribed imposed load, (kN/m), Pil Total imposed load, IL Total load (kN): Longterm, Wlong Mediumterm, Wmed
Swt := 0.3 kN m] tg := 0.75 kN m]
fchip := 0.63 lcN. m1
+ +
DL :=( S w t tg fihip) Le DL = 17.64 o kN

pil := 2.5 kN m"'
ZL :=pi1 L, IL = 26.25 o kN a
WJong := DL Wlong= 1'7.64o kN w m e d := D L ZL w m e d = 43.89 0 kN
+
3. Kfactors
Service class 1 (KZ, Table 13) Mediumterm loading (K3,Table 14) Table 15) Bearing: 90 mm (K4, (bearing position not defined)
K2 := l
K3 := 1.25 K4 := 1.0
*
Design of GluedLaminatedMembers
Notched end effect (K5,Clause 2.10.4) Form factor (K6, Clause2.10.5) Depth factor (K7, Clause 2.10.6) to provide a dimensionless value for K7 in Mathcad,let: for h >300mm NO load sharing (Kg, clause 2.9)
K5
:= l
K6 := 1
h1
:= h
mm" h; 92 300 0.81 h; + 56 800
K7
*
+
K7 = 0.89 := 1
Glulam m o d ~ c a t ~ factors on (Table 21): Singlegrade, 15 laminates in C24 timber Bending // to grain, K15 Compression perpendicular to grain, K18 Shear // to grain, K19 Modulus of elasticity, K20 4.
Gradestresses
BS5268 : Part 2, Table 7 Timber strength class, C24 Bending // to grain Compression perpendicular to grain Shear // to grain Mean modulus of elasticity
Gm,g.// :=
cF,.g.pp
7.5 N mm2 mm2
:= 1.9N
:= 0.71 N . mm2
Emean :=
mmm2
10800N
5. ~ e ~ d stress j ~ g
Applied bending moment = 57.61 o kN b h2 zxx:= 6 Zxx= 6.01 X lo6 o mm3
Mmed
Section modulus
Applied bending stress Permissible bending stress
*
*
95
96
Structural Timber Design
. ~ateraJ st~bi~ity Clause 2.10.8 and Table 16
Maximum depthtobreadth ratio, h/b
h = 4.87 b Ends should be held in position and compression edges held in line by direct connection of t & g decking to the beams
7. Shearstress
Applied shear force
wmed Fv:=7
Applied shear stress Permissible shear stress
8. Bearingstress
Applied load
Applied bearing stress, (bearing area = b bw) Permissible bearing stress
9. Deflection
Modulus of elasticity of glulam, E Deflection due to bending Am
Deflection due to shear Total deflection
= 34.02 o mm
Design of GluedLaminatedMembers
Permissible deflection
97
Aah := 0.003 Le Aah = 31.5 omm *
Camber required =deflection under permanent (longterm) loading:
Deflection under live load
Along = 13.67 o mm Provide 15 mm camber Alive := A t o t a l  Along Alive 21.83 o mm Deflection satisfactory
Therefore 115 mm x 560 mm glulam sections in singlegrade C24 are satisfactory €xample 6.2 Design of a combinedgrade glued laminated beam selection of a suitable section size
Design of a glued laminated timber beam for the roof of a restaurant is required. The beam is to span 9.8 m centre to centre on 125 mm wide bearings under service class 2 conditions. It is proposed to use a combinedgrade layup using softwood timber in strength classes of C18 and C16 with laminations of36mm finished thickness. The beam is subjectedto a dead load of 0.67 kN/m excluding selfweight,from t & g boarding and roofing, and an imposed mediumterm load of 2.25 kN/m. Definitions
Force, kN Length, m Crosssectional dimensions,mm Stress, Nmmm2 1.
N :=newton kN :=lo3 * N Direction parallel to grain, // Direction perpendicular to grain, pp
Geometrica/properties
Eflt'ective span, L, Bearing width, bw Beam dimensions: Breadth of section, b Depth of section, h Crosssectional area, A Second momentof area, I,,
Le := 9.8 m bw := 125. mm
2. Loading
Dead load: Selfweight of beam (kN/m), Swt

Swt := 0.3 kN .m"
assumed
Structural Timber Design
Prescribed dead load (kN/m), PDL Total dead load, DL Imposed load: Prescribed imposed load (kN/m),Pi1 Total imposed load, IL
PDL := 0.67 kN
+
m'
+
DL := (Swt PDL) Le DL = 9.51 o kN pi1 := 2.25 kN m" +
IL :=pi1 Le IL = 22.05 o kN *
Total load ( k N ) : Longterm,
Mediumterm, Wmed 3. Kfactors
Service class 2 (K2, Table 13) Mediumterm loading (K3, Table 14) Bearing: 125 mm (K4, Table 15) (bearing position not defined) Notched end effect (K5, Clause 2.10.4) Form factor (K6, Clause 2.10.5) Depth factor (KT,Clause 2.10.6)
:=1 K3 := 1.25
K2
K4 := 1.0
K5 := 1 K6 := 1
Since section size is not known, at this stage assume, K7 := 1 Kg := 1
No load sharing (Kg,Clause 2.9)
Glularn ~ o d ~ c a t i factors on (Table ?l): Stress values for the highergrade timber apply BS5268 : Part 2, Clause 3.2 Since number o f laminations is not known, Table 7, for C18 timber ignore at this stage Bending // to grain, K15 Compression perpendicular to grain, &g Shear // to grain, K19 K20 := 1.17 Modulus o f elasticity, K20 L
4.
Gradestresses
BS5268 : Part 2, Table 7 Timber strength class, C18 Bending // to grain Compression perpendicular to grain
Design of GluedLaminatedMembers
Shear // grain Mean modulus of elasticity
99
zg.i := 0.67 N mm2 Ernean := 9100 N mm2 *
*
Selecting a trial section
Using deflection criteria: Modulusofelasticity for glulam, E Permissible deformation under imposed load
E := Emean  K20 E = 1.06 x lo4 o N.mmF2 0.003 L, = 29.4 o mm
&h : =
0
5 I;.t."ed L2 Required second moment of Zxx,rqd := 384 * E . &h area, Zxx using &h underimposed load Zxx.rqd = 1.24 X io9 o mm4 Using lateral stability criteria: In order to achieve lateral stability by direct fixing of deckingto beams, the depthtobreadth ratio should be limited to 5, i.e. h <5b. Substituting for h = 5b in zxx= bh3/i2 andequating it to zxx,rqd: *
Zxx.rgd
=
and depth of
*
(
12
*
*
zxx.rqd 53
'
b :=
section, h
b = 104.36 o mm h:=5*b h = 521.79 o mm
will give 12 b)3
)
114
Try a 110 mm x 540mm deep section with 15laminations of 36 mm in thickness Trial beam dimensions: Depth (mm), h h := 540. mm (mm), Breadth b b : = 11O.mm Modification factor K7 for section depth: to provide a dimensionless value in let: h1 := h mm" for K7 Mathcad, h: for h >300mm K7 :=0.81 h: K7 = 0.89 *
Glulam modtjication factors (Table BS5268 : Part 2, Clause 3.2 Combinedgrade, 8 laminates in Cl8 timber, by interpolation Bending // to grain, K15 Compression perpendicular to grain, KI8 Shear // to grain, K19 Modulus elasticity, of
+ 92 300 + 56 800
21): Stressvalues for the highergraded timber apply
K15
:==1.32
K18 := 1.69
K19 := 2.73 K20 := 1.17
1
Stru~turalTimber Design
Applied bending moment
Section modulus
Applied bending stress Permissible bending stress
Check selfweight: BS5268 : Part 2, Table 7 Density Beam selfweight (kN/m): Actual Assumed
7 . ~ateral sta~iljty
Clause 2.10.8 and Table 16 Maximum depthtobreadth ratio, hlb
h b Ends should be held in position and compression edges held in line by direct connection of t & g decking to beams
= 4.91
8. Shearstress
Applied shear force
Applied shear stress Permissible shear stress
wmed
F,, := 2 F,, = 15.78 o kN
Designof Glued LaminatedMembers
Applied load
Applied bearing stress (bearing area = b bw)
F:=" w m e d 2 F = 15.78 o kN 0c.a.pp :=
(3%)
= 1.15 o N mm"2 K2 K8K3 0 c . a h . p p = 4.65 o N.mm2 Bearing stress satisfactory
0c.a.pp
Permissible bearing stress
0c.adm.pp := 0c.g.pp
Modulus of elasticity for glulam, E
E := Emean K20 E = 1.06 x lo4 o N  mm2 b h3 rxx:=
Second moment of area
*
12
zXx= I .U x io9 o mm4 Deflection due to bending
Deflection due to shear
A,, = 25.16 0 mm 19S2 '
A, :=
(b h) E *
*
A, = 1.17 o mm
1 Cl8 timber
I
Cl6 timber
C18 timber
Fig. 6.7
Crosssectionof glulam beam (Example6.2).
I
K18
101
4 02
StructuralTimberDesign
Total deflection
Atotal := A m +. Atotal = 26.34 o mm
Permissible deflection
A a h := 0.003 L, A a h = 29.4 o mm Deflection satisfactory and no precambering is required
As
*
Therefore adopt a 110 mm x 540 mm glulam section in combinedgrades of C18 and C16 €xamp/e6.3 Design of a f/ooringsystem
Design of aflooring system using gluedlaminated timber beams for the first floor of an art gallery is required. The flooring arrangement comprises main beams with effective span of 9.6 m at 4.5 m centres supporting secondary beams at 0.8 mcentres and 38 mm tongued and grooved(t & g) flooring using hardwoodtimber. The secondary beamsare simply supported on 100mm hangers attached to main beams which are in turn supported on loadbearing walls providing a 1'75 mm bearing width at each end. It is proposed to use softwood timber in strength class of C24 with laminations of 38 mm finished thickness, singlegrade and horizontally glued laminated throughout. The floor is subjected to a dead load of0.4'7 kN/m2, excluding selfweight, and an imposed mediumtern load of 3.25 kN/m2. Definitions
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm"2
N :=newton kN :=lo3 N Direction parallel to grain, // Direction perpendicular to grain, pp

A. Design of tongued and grooved boarding
Assuming t & g boarding comprises 100mm wide timber beams of thickness (depth) 38 mm simply supported on joists.
'1. Loading Secondary beam spacing, Js t & g width, b t & g thickness, t Dead load: BS5.268 : Part 2, Table 7 Average density, D30 hardwood t & g boarding, selfweight ( W m 2 ) , tg
Js := 0.8 m b := 100. mm t := 38. mm
pmean:=640 kg m"3 tg := pman tg v
*
tg = 0.24 o kN m2
"? ! .
4.5 m
l
.
!  _ _ e _
I
4.5 m
1
1
I
! !
I
! !
!
,!
4.5 m F
! ~!.
! !...
b
. v4 v4
.
.
'
Secondary beams Main beams (a) Support structure layout
300 m
t & g hardwood
(b) Section AA: secondary beam details I
4 details (c) Main beam Fig. 6.8
9.60 m
Details of flooring system (Example 6.3).
300 mm
I
Section CC
104
Structural Timber
Design
Prescribed dead load (kN/m2),PDL Total dead load (kW, DL Imposed load (kN/m2), IL IL Total load (kN), W
PDL :=0.47 kN  m2 DL := PDL + tg DL = 0.71 o kN m2 := 3.25 kN .me2 W:=(DL+IL)*b*Js
W = 0.32 o kN
2. Kfactors
Service class 1 (K2, Table 13) Mediumterm loading (K3, Table 14) Bearing (G,Clause 2.10.2) Notched end effect (K5, Clause 2.10.4) Form factor (KG, Clause 2.10.5) Depth factor (K?, Clause 2.10.6) for h 5 72mm Load sharing applies (K8, Clause 2.9)
K2 := 1
K3 := 1.25 K4 := 1 .O
K5 :=1
K6
:= 1
Kg
:= 1.1
assumed
K7 :=1.17
3. Gradestresses
&S135268 : Part 2, Table 7 Strength classification Bending /I to grain Mean modulus of elasticity
Strength class D30 :=9.0 N Emean := 9500 N mme2 *
CT,.~,//
+
4. Bending stress
Applied bending
moment
Permissible bending stress
W *JS
M:=”
8 M = 0.03 o kN m
K3 K6 K7 ’K8 = 14.48 o N mm2
om.adm./l := om.g.1 K2 CT,.,~.,J
Since Z = t2, therefore
6
*
trqd = 1 1.46 0 mm Thickness satisfactory
*
Design of GluedLaminatedMembers
Load sharing system Permissible deflection Second moment of area
Deflection
E := Ehean A a h := 0.003 JS A a h = 2.4 o mm b . t3 zxx:= 12 zXx= 4.57 x io5 o mm4 5. W *Js3 A := 384 E Zxx A = 0.49 o mm Deflection satisfactory *
*
*
Shear and bearing stresses are not critical in decking arrangements. Therefore 38 mm t & g boarding in D30 timber is satisfactory B. Design of secondary beams 1 . Loading
EfTective length, L, Bearing width, bw Beam dimensions: Depth, h Breadth, b Total dead load from flooring (kN/m2), DL Selfweight of beam (kN/m), Swt Imposed load (kN/m2), ZL
Le := 4.35 m bw := 100 mm

*
Not known at this stage DL = 0.71 o kN m2 Swt :=0.3 kN m'assumed ZL :=3.25 kN mW2
Total load (kN): Longterm,
?VIong
Mediumterm, Wmed
2. Kfactors Service class 1 (K2, Table 13) Mediumterm loading (K3, Table 14) Bearing: l00 mm (Ka, Clause 2.10.2)
K2 := 1 K3 :=1.25
K4 := 1
105
1
Structural Timber
Design
Notched end effect (K5, Clause 2.10.4) Form factor (&, Clause 2.10.5) Depth factor (KT, Clause 2.10.6) No load sharing (Ks, Clause 2.9)
K5
:= 1
:= 1 K7 := 1 assumed Kg := 1 K6
Glulam ~ o d ~ c a t ifactors on (Table 21): BS5268 : Part 2, Clause 3.2 Bending I/ to grain, K15 Compression perpendicular Since number of laminations is not known, to grain, KIs ignore at this stage Shear // to grain, K19 Modulus elasticity, of K2@ K20 := 1.07
i
3. Gradestresses
13s5268 :Part 2, Clause 3.2 Table 7, for C24 timber Bending // to grain Compression perpendicular to grain Shear // to grain Mean modulus of elasticity
Em,,,
:= 0.71 N mm"2 :=10 800 N mm"2 *
Selecting a trial section
Using deflection criteria: Modulus of elasticity for glulam, E Permissible deformation
E := Ernean K20 E = 1.16 X 1 0 ~ 0 ~ . m m  ~ &h
A,*
:= 0.003 L e = 13.05 o mm *
Using lateral stability criteria: In order to achieve lateral stability by direct fixing of deckingto beams, the depthtobreadth ratio should be limited to 5, i.e. h <5b. Substituting for h = 5b in Ixx = bh3/i2 and equating it to Ixx.rgd: Ixx.rqd
=
*
(5 b)3 will give *
12
and depth of section,
h
b :=
(
12
*
Ixx.rqd
53
b = 56.64 o mm h :=5.b h = 283.18 o mm
)
1/4
Design of GluedLaminatedMembers
Try a 70mm x 288 mm section with 8 laminations of 36mm in thickness Trial beam dimensions: Depth (mm), h (mm), Breadth b
h mm :=288 b:=70mm
Modification factor K7 for section depth: to provide a dimensionless value for K7 in Mathcad, let: h1 := h mml 300 mm for h <300mm K7 h
( *
+
K7 = 1
Glulam m o ~ ~ c a t ifactors on (Table 21): BS5268 : Part 2, Clause 3.2 Singlegrade, 8 laminates in C24 timber, by interpolation: K15 Bending // to grain, K15 K1g Compression perpendicular to grain, K18 K19 Shear // to grain, KI9 Modulus of elasticity, Kz0 K20
:= 1.40 := 1.55 := 2.34 := 1.07
5. g e n ~ j n g stress
Applied bending moment
M :=wmed Le 8 *
Section modulus, Zxx
Applied bending stress Permissible bending stress
Check selfweight:
BS 5268 : Part 2, Table 7 Average density Beam selfweight, (kN/m) Actual Assumed
p := 420kgm3
)
'*l1
107
Structural TimberDesign
ateraJ sta~iJity
Clause 2.10.8 and Table 16 aximum depthtobreadth ratio, hJb
Q
h = 4.11 b Ends should be held in position and compression edges held in line by direct connection o f decking to the beams
ear stress
Applied shear force
Fv= 7.54 o k N Applied shear stress Permissible shear stress, no notch
Ta,f/ = 0.56 o N
mm?
:= zg.i K2 K3 K5 +
zah.i = 2.08 o N .mmv2
Shear stress satisfactory
Applied load F = 7.54 o kN
Applied bearing stress (bearing area = b bw) Permissible bearing stress
Modulus of elasticity for glulam, E Second moment o f area, Ixx
Deflection due to bending
Kg K19
Design of Glued LaminatedMembers
Deflection due
to shear
19.2 M (b h) E As= 0.68 o mm Atotal P= Am As Atotal= 10.71 o mm :=I= 0.003 L e A a h = 13.05 o mm Deflection okay and no precambering is required *
As:=
*
Total deflection Permissible deflection
10
*
+
8 mm deep sections with 8 l a ~ n a t of i ~36 mm in thickne~
L e := 9.6 m EEective length, L e bw := 175. mm Bearing width, bw Beam dimensions: Depth, h Not known at this stage breadth, b Selfweightofbeam(kN/m), Swt Swt :=0.7. kN .m" assumed Strip of load of width bj = 300 mm, seeFig. 6.8 (b), on the top of a main beamplus its selfweight bj := 300mm Longterm loading Pstrip.long := (DL bj Swt) L e Pstrip.long 8760 kN Mediumterm loading Pstrip.med := (DL IL) bj L e SWt L e Pstrip,med = 18.12 o kN Point (concentrated) loading wlong from secondary beams(kN), PS ps.low := ' 2
}
+
+
~
Ps.long
Total load (kN): Longterm, Piong Mediumterm, Pmed
K1 to KG are as above No load sharing applies (Kg, Clause 2.9)
L==
3.77 o kN
*
+
*
Structural Timber Design
Glularn ~ o d ~ c a t ifactors on (Table 21): BS5268: Part 2, Clause 3.2 Bending // to grain, K15 Compression pe~endicular Since number of lamination^ is not known, to grain, K18 ignore at this stage Shear /I to grain, K19 Modulus of elasticity, KZ, K20 := 1.07 3. Gradestresses
As for the secondary beams (see above) 4. Selecting a trial section
(1) Usingdeflection criteria: For a simply supported beam subjected to a maximum bending momentof Mm,,, irrespective of the loading type, the maximum deflection may be estimated using the equation given in Section 4.4.3, as:
Am =
0.104 M m , *
E
*
*
L:
Ixx
MmaXfor a simply supported beam carrying (n 1) point loads of magnitude P, equally spaced, is given by Mmax =
ne P . L: 8
(reference: Steel Designer's Manual) P/2
P P P P P P P P PP/2
11111111111 t t + n = 10 equal spacings, (nl) = 9 pointloads along
the length 2 @ supports.
Number of pointloads, (n 1)=9 Applied bending moment due to point loads Applied bending moment due to strip of loading including selfweight Total bending moment
n := 10 Le MP:= n Ps.med Iz *
*
U
MP= 180.97 o kN m M S:=
Pstrip.med Le *
0
0
MS=21.740kN.m Mtotal := M S 4" MP
Mtotal= 202.71 0 kN m
Design of Glued Laminated Members
odulus of elasticity for glulam, E ermissible d e f o ~ a t i o n Required second moment of area, I,, using A,h

E := Em,,, K20 E = 1.16 x lo4 o N mm"2 Au(ifn:= 0.003 L, &h = 28.8 o mm 0.104 Mlotal L: *
Ixx.rqd :=
*
0
E
*
= 5.84 X
io9 o mm4 Using lateral stability criteria: In order to achieve lateral stability by direct fixing of deckingto beams, the depthtobreadth ratio should be limited to 5, i.e. h 5 5b. Substituting for h = 5b in Ixx = bh3/i2 andequating it to Ixx,rqd: IxXrqd
Ixx,rqd
=
b (5 b)3 will give 12 *
*
b :=
(
12
*
Ixx.rqd
53
b = 153.86 o mm h:=5*b h = 769.31 o mm ' S,ectionwith 24 laminations of 36mm in thickness Try a 180 mmx 864 mm deep Trial beam dimensions: h := 864mm Depth (mm), h b := 180. mm Breadth (mm), b Modification factor K7 for section depth: To provide a dimensionless value for K7 in Mathcad, let: h1 := h mm" h; 92 300 K7 := 0.81 for h >300mm h: 56 800 K7 = 0.85 and depth of the section, h
+ +
~ l u l ~ao~d ~ ~ a t factors i o n (Table 21): BS5268 : Part 2, Clause 3.2 Singlegrade, 24 laminates in C24 timber K15 := 1.52 Bending // to grain, K15 Compression perpendicular to K18 :=1.55 grain, K18 Shear // to grain, K19 K19 :==: 2.34 K20 := 1.07 Modulus of elasticity, K20
Total applied bendingmoment Section modulus, Zxx
Mtotal= 202.71 o kN .m b h2 zxx:= 6 Zxx= 2.24 x lo7 o mm3
112
StructuralTimber Design
Applied bending stress Permissible bending stress
Check selfweight: BS.5268 : Part 2, Table 7 Average density Beam selfweight, (kNIm) Actual
nilt0taI
0m.a.i
zxx
= 9.05 0 N mm2 Gm.g.1 K2 K3 K6 0m.ah.I 12.05 0 N mm2 Bending stress satisfactory Gm.a,l/
*
om.adm.(
*
*
*
K7
*
K8 K15 *
p := 420 kg m3
Assumed
6 . ~ a t ~ r sat Ja ~ j i j t ~ Clause 2.10.8 and Table 16 Maximum depthtobreadth ratio, h/b
h = 4.8 b Ends should be held in position and compression edges held in line by direct connection o f decking to the beams
Applied shear force
Pmed Fv :=y L
Fv = 84.46 o kN Applied shear stress Permissible shear stress, no notch
Applied load
F"
Pmed 2 F = 84.46 o kN *
Applied bearing stress, (bearing area = b x bw)
0c.a.pp
:=
(A)
= 2.68 o N mmw2
(T,.,.~~
Designof GluedLaminatedMembers
Permissiblebearingstress
:=0c.g.pp K2 K3 ' Kh 'K8 = 3.68 o N mm2 Bearing stress satisfactory 0c.adm.pp
*
*
*
113
K18
0c.ah.pp
9. ~ e f l e c t i o ~ Modulus elasticity of glulam, E Second moment of
for
E := Emean K20 E = 1.16 x lo4 o N mmm2 b h3 I,, :=12 I,, = 9.67 x lo9 o mm4 a
area, I,,
Deflection due to bending
Deflection due to shear
As= 2.17 o mm A m "I= 19.54 o mm
Total deflection
&,tal Atotal
Permissible deflection
Aah := 0.003 * L e A a h = 28.8 o mm
Deflection okay and no precambering is required Therefore adopt 180mm wide x 8 thickness
m m deep sections with 24 la~nationsof 36 mm in
Design ofa curved gluedlaminated timber beam for the roof of a restaurant is required. The beam is to span 9.0m centre to centre supported on 250 mm wide bearings under service class 1 conditions. It is proposed to use a single grade layup using softwood timber in strength class Cl8 with horizontal laminations of 30mm finished thickness. The radius of curvature at midspan bend is to be 6.5m. The beam is subjected to a dead load of 1.65 kN/m, including selfweight, and an imposed mediumterm load of 2.25 kN/m.
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm"2
N :=newton kN :=l o 3 .N Direction parallel to grain, // Direction perpendicular to grain, pp
4
Structural Timber
Design
250 m
**\
\
250 mtn
1 h
Crosssection AA Fig. 6.9 Curved glulam beam (Example 6.4).
Effective length, L, Bearing width, bw Beam dimensions: Depth of section, h Breadth of section, b Dead load including selfweight (kN/m), DL Imposed load (kN/m), IL Total load (kN): Longterm, W&
Le := 9.0 m bw := 250mm
Serviceclass 1 (K2, Table 13) Mediumterm loading (K3, Table 14) Bearing: 250mm wide (K4,Clause 2.10.2) effect Notched end (K5,Clause 2.10.4)
KZ := 1 K3 := 1.25
Not known at this stage DL := 1.65 kN m1 IL := 2.25 kN m’
K4 :=
K5
1.0
:= l
Design of GluedLaminatedMembers
Form factor (K6, Clause 2.10.5) Depth factor (KT, Clause 2.10.6) No load sharing (Ks, Clause 2.9)
l15
K6 := 1 K7 := 1 assumed Kg := 1
Glulam mod~cation factors (Table 21): BS.5268: Part 2, Clause 3.2 Bending // to grain, K15 Compression perpendicular Since number of laminations is not known, to grain, Kls ignore at this stage Shear // to grain, K19 Modulus elasticity, of K20 :==1.17
i
3. Gradestresses
IPS5268 : Part 2, Clause 3.2 Table 7 , for C18 timber Bending // to grain Compression perpendicular to grain Shear // to grain Mean modulus of elasticity
Qm.g.//
:= 5.8
N mmv2
cT,.g.pp := 1.7
N mmv2
Selecting a trial section
Using deflection criteria: Modulus of elasticity for glulam, E Permissible deformation Required second moment of area, Zxx using A a h
E := Ernean K20 E = 1.06 x 1040N.mm"2 &h := 0.003 Le &h = 27 o mm 5 ZL L: Zxx.rqd := 384 E A a h zxx.rqd = 6.69 X ' 01 o mm4 *
*
*
*
*
Using lateral stability criteria: In order to achieve lateral stability by direct fixing of deckingto beams, the depthtobreadth ratio should be limited to 5, i.e. h 5 5b. Substituting for h = 5b in zxx= bh3/12and equating it to Zx.rqd: b (5 12 *
Zxx,rqd
will give
b :=
(
12
*
zxx.rqd
53
)
1/4
b = 89.51 o mm h:=5.b h = 447.55 o mm Try a 105mm x 5 10mm deep section with 17 laminations of 30 mm in thickness Trial beam dimensions: h := 510 mm Depth (mm), h b b := 105 mm Breadth (mm).
and depth of section, h

l’ 16
StructuralTimber Design
Modi~cationfactor K? for section depth: to provide a dimensionless value for K7 in Mathcad, let: h1 := h mml +
for h >300mm
G l u l a ~ ~ o d ~ cfactors at~on (Table Part 2, Clause 3.2 BS5268 : Singlegrade, 17 laminates in C18 timber by interpolation: Bending // to grain Compression perpendicular to grain Shear // to grain Modulus of elasticity
BS5268: Part 2, Clause 3.5.3.1 Radius of curvature, r Lamination thickness, t Ratio r/t should not be less than E,ea,J’70
K7
:= 0.81
h; + 92 300 h? k 56 800
21):
:= 1.52 K18 :=1.69
K15
K19 :=2.73
K20 :=1.17

r := 6500 mm t := 30 mm r = 216.67 is greater than t 27
&mean
130 o N mmm2 70 Satisfactory
” _ .
BS5268 : Part 2, Clause 3.5.3.2 Check if ( r / t )<240 hence
r =216.67 <240 t r K33 := 0.76 0.001 
+
K33 = 0.98
< l .O Satisfactory
Applied bending moment Check if ( r ~ ~ ~
M :=
Wmed * L e
8 M = 39.49 o kN m 0.5. h r m a n := r r m a n = 6.75 o m
+
rman
13.25 < l 5 h Satisfactory
”
*
t
Design of Glued Laminated Members
Since 10 <(rmean/h)5 15
Section modulus
Applied bending stress on concave face Applied bending stress on convex face Permissible bending stress
K33
BS5268 : Part 2, Clause 3.5.3.3
Radial stress, G r , induced by bending moment, M ,
CTr
3." 2 b h rmean = 0.16 o N mm2 :=
*
Or
*
*
*
Since the applied moment, M, tends to increase the radius of curvature, Or is tension perpendicular to grain, and factor K19 is disregarded (Clause 3.2, BS 5268 :Part 2). Permissible radial stress,
Gt.ah.pp =
f
Tah.1
(see Section 2.5.7, Chapter 2)
therefore
.
~ a t ~ rsa ~fa ~ j f j t y
Clause 2J0.8 and Table 16 Maximum depthtobreadth ratio, h/b
.
h
= 4.86 b
Ends should be held in position and compression edges held in line by direct connection of decking to the beams
ear stress
Applied shear force
Fv := Wmed 2 F, = 17.55 o kN ~
118
Structural Timber
Design
Applied shear stress Permissible shear stress, no notch
10. Bearingstress Wmed
Fv := 7
Applied load
L,
Fv = 17.55 o kN
Applied bearing stress (bearing area = b bw)
0c.a.pp :=
(A)
+
Permissible bearing stress
11. Reflection Modulus of elasticity for glulam, E Second moment of area, ,,Z
Precamber =deflection under dead load
Deflection due to bending under imposed load Deflection due to shear
E := Emean K20 E = 1.06 x lo4 o N mm2 b h3 I,, := 12 zxx= 1.16 X io9 omm4 5.DL.L: Camber := 384 E Zxx Camber = 11.41 o mm Provide 12 mm camber 5 IL L: A m := 384 E Zxx A m = 15.55 o mm 19.2 M a, := (b h) E h, = 1.33 o mm *
*
*
*
*
*
*
*
*
4
Total deflection
atotal := a m tAtoral = 16.88 o mm
Permissible deflection
:= 0.003 L e A a h = 27 o mm Deflection satisfactory. *
Therefore adopt a 105 mm wide x 510mm deep section with 7 ' 1 laminations of 30mm in thickness and 12mm camber
Design of Glued aminated ~ e m b e ~ s 1 7
A singlegrade gluedlaminated timber column instrength class Cl8 consists of 5 laminations of 37mm thickness with overall crosssectional dimensions of 9’7 mm x 185 mm. The column is 4m in height and is restrained at both ends in position but not in direction and is subjected to service class 2 conditions. (a) Determine the maximum axial longterm load that the column can support. (b) Check that the section is adequate to resist a longterm axial load of 25 kN and a bending moment of 1.40kN/m due to lateral loading about its major axis. Y
!
x ...
L, = 1.0 x L
Crosssection Fig. 6.10 Glued laminated column (Example 6.5).
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm2
N :=newton kN := lo3 N Direction parallel to grain, // Direction perpendicular to grain, pp
Column length, L Table 18, effective length, L,
L := 4.0 m L, := 1.0 L L,=4om b := 97 .mm d := 185. mm A :=bed A = 1.79 x io4 o mm2 Ixx:= b d3 I,, = 5.12 x lo7 o mm4 lYy := d . b3 rYY= 1.41 X io7 0mm4
Width of section, b Depth of section, d Crosssectional area, A Second moment of area, I,, Second moment of area, Iyy

*
*
120
Structural Timber
Design
For a rectangular section, radius of gyration, i,, and iyy
d := 4 2 ixx = 53.4 o b lYY := 412 lxx
m
Section modulus, Zxx
iyy = 28 o mm b d2 zxx := 6 zXx = 5.53 x io5 o mm3
About xx axis
Le h, := y"
*
lxx
h, = 74.9 About yy axis
L 2 hyy:= 
iYY
hyy= 142.85 <180 satisfactory
BS.5268 : Part 2, Table 7 Strength class, C18 timber Bending // to grain Compression parallel to grain Minimummodulus of elasticity Meanmodulus ofelasticity
N mm"2 7.1 N mm"2 Emin :=6000 .N mm2 Emean := 9 100 N mm2 CF,.~,J := 5.8 C F ~ .: ~ =. ~
*
Serviceclass 2 (K2, Table13) K2 := 1 Longterm loading (K3, Table 14) K3 := 1 K6 := 1 Form factor (&, Clause2.10.5) Depth factor (.K7, Clause 2.10.6)
+
K7 :=
(
300 mm
*
)
K7 = 1.05 No load sharing (K8, Clauses2.9 and 2.1 1.5)
K8 := l
(4 ~ l u i a m m o d ~ c afactors: tio~ Since buckling is critical about column's minor axis, i.e. yy axis, the behaviour under axial loading onlyissimilar to the vertically laminated membersandhence the
Design of Glued LaminatedMembers
421
modification factor K28, given in Table 22 for compression // to grain and modulus of elasticity, applies. Singlegrade 5 laminates in C l 8 softwood timber: Compression // to grain and modulus of elasticity
K2g := 1.27
~odificationfactor for compression member, KI2, can becalculated using either one of the following methods: (1) Using the equation in Annex B: BS5268 : Part 2, Clause 3.3 E := Emin K2g E = 7.62 x lo3 o N mmn2 h :=hyy G , :=cTCsg.I K2 K3 and := 0.005 h *

K12 = 0.26 (2)UsingTable
19:
E %.g./
*
K2 K3
= 1.07 x lo3and
h = 142.85
K12 = 0.26
*
ssi~~ ~ eo ~ ~ r e s Ss i v e
Axial load Applied moment
~lulammod~cationfactors (Table 21): Since bending is about column's major axis, i.e. xx axis, the behaviour is similar to horizontally laminated beams and hence modification factors K15K20 apply, Singlegrade 5 laminates in C18 softwood timber Bending // to grain Compression // to grain Modulus elasticity of
K15 := 1.16 K17 :=1.07 K20 := 1.17
1
Structural Timber Design
~odi~catio factor n for compression member,KI2, can be calculated usingeither one of the following methods:
(l) Using the equation in Annex B: BS.5268 : Part 2, Clause 3.6 E := Ernean K20 E = 1.06 x lo4 o N.mm"2 *
h :=hyy
G, :=
K3 and
:=0.005
h
+ ( l + q ) * 7 t 2 . E  { ( l + ( l + q )  n 2 . E
K12 := 1
2
K2
3*h2.cTc
(2) UsingTable 19: E oc.g.1 * K2
K3
2
3.h2.Uc
= 1.5 x lo3 and h = 142.85
7. ~ o ~ ~ r e s and s i o bendjng n stresses
Applied compressive
stress
oc.a.1
* *
1.5 h2 c~, m
S
K12
= 0.32
K12
= 0.32
P A

= 1.39 o N mmw2
Permissiblecompressive stress

oc.adm.l := cTc.g K2 .l K3 Kg K12 c ~ ~:= . 2~. 4~ 4 o. ~1. m m  ~
K17
Compressive stress satisfactory Applied bending stress Permissible bending stress Check interaction quantity
Euler critical stress
om.a.1
+
oc.a.1 Dc.adm.1
= 0.98 < 1 Interaction quantity is satisfactory
Therefore a97 mm x 185 mm section with 5 laminations of37 mm timber in stren~thclass C18 is satisfactory
Chapter 7
esign of Plywebbed
A plywebbed beam comprisesthree main parts, namely flanges, web(s)and glued or nailed joints between flanges and web(s). The flanges are generally made of solid structural timber sections fingerjointed together to produce the required length.The web(s) are made of plywood, although other woodbased materials such as particleboard or fibreboard may also be used. The flanges are often connected to plywood web(s)by structural timber adhesives in an industrial process,or sometimes by mechanical fasteners suchas nails. Plywebbed beamsare usually of I or boxed shapes. Typical crosssections ofplybox and I beamswhich are symmetrical about both vertical and horizontal axes are shown in Fig. 7.1. Plywebbed beams owe their structural advantage to plywood acting as a shearresistant materialfor the webs, allowingthe solid timber sectionsto be spaced apart. This permits saving in materials and provides extra stiffness and strength for the beams. Plywebbed beams are widely used for spans in excess of those for which solid timber issuitable, i.e. for spans greater than
x
x 
x
a
H
Fig. 7.1 I and boxed plywebbed beams. 123
4
Structural Timber Design
5 m and often up to 101 5 m. Their commonest application is in roof construction, where parallel top and bottom flanges may be used to give a flat roof or the upper profile may be pitched'. Flywebbed beams have also been used as a type of large span (30m or more) builtup portal frame in bridge construction. In addition, the good strength to weight performance of plywebbed beams makes them an attractive option for construction of large crosssectioned beams.For example as a hollow interior boxed bridge construction for a covered pedestrian crossing or as a prefabricated housing unit'. In this chapter, design recommendations for the most common method of construction, i.e.gluedplywebbedbeams, are discussed.Therecommendations may also be applied, with some modification, to all nailed plywebbed beams.
In general a plywebbed beam is built with materials which have different moduli of elasticity, which leadto development of different stress magnitudes at the same depth. In Fig. '7.2 an example of differences in stress values for the two beam profiles subjectedto bending moment is illustrated. For a plywebbed beam comprising materials with different moduli of elasticity, it is common practice to calculate the efective geometrical properties of the crosssection using a technique known as transformed section m e t ~ o dIn . this methodthe whole crosssection is considered as acting as one homogeneous material with the same properties as say the flange material, where the contribution from the web(s) is altered in proportion to the ratio of the moduli of elasticity.
bl2 t bl2 I
i

I (7inzber :
Fig. 7.2 Bending stresses in plywebbed beams.
%nber
Design of PlywebbedBeams
125
Therefore if it isrequired, for example, to transform a section to an equivalent of all timber, i.e. by replacing the plywood web with an equivalent thickness (width) of timber, then:
ttimber = tplywood x fiplywood
(7. la)
Etirnber
Transformed or effective area may be calculated as: Atrawformed
= Atimber(~a~ges) $Ap~ywood(web~) x
Eplywood ~
Etider
(7.1 b)
and transformed or effective second moment of area as: Itramformed
=r
~ ~ b e r ~ ~$~~plywood(webs) g e s ) x
Eplywood Etimber
(7. IC)
Note: In calculation of transformed (effective) geometrical properties of a plywebbed crosssection, extensive researchand experience in practice have for timber is more realistic.' shown that the use of Ernean
Before discussing design considerations for the plywebbed beams it may be appropriate to describe briefly the main properties, particularly the structural behaviour, of plywood. Plywoodis a panel madeup of a number of softwood veneers.The veneers are glued together so that the grain of each layer is perpendicular to the grain of its adjoining layer. Plywoodas a structural material consists of an odd number of layers or plies (at least three) as shown in Fig. 7.3. Plywood for exterior use is generally made with fully waterproof adhesive that is suitable for severe exposure conditions.
h
Grain
Cross ply
Back ply Fig.7.3 The structureof a threeplyplywood.
126
Structural Timber
Design
The structural properties and strength of plywood depend mainly on the number and thickness of each ply, the species and grade, and the arrangement of the individual plies. As with timber, load factors such as type of applied stresses, their direction with respect to grain directionof face ply and duration of loading have great influence on the structural properties of plywood. Plywood may be subjected to bending in two different planes depending on its intended use and the direction of the applied stress and therefore it is important to differentiate between them: (1) Bending about either of the axes (i.e. xx or yy) in the plane of the board, as shown in Fig. 7.4(a), for example in situations where it is used as shelving or as floorboard. (2) Bending about an axis perpendicularto the plane of the panel (i.e.zz), as shown in Fig. 7,4(b), for example when it is acting as a web of a flexural member such as in plywebbed beams.
/
x
(a) Bending about either of the axesin the planeof the board.
axis perpendicular to the plane of the board. (b) Bending about an
Fig. 7.4
Plywood axes of bending.
Design of Plywebbed Beams
Table 7.1 Modification factor (Table 33 of BS 5268 :Part 2)
K36
l' 27
for the grade stresses and moduli of plywood
Duration of loading
Value of K36 Service classes 1
and 2
Service class
3
Stress
Modulus
Stress
Modulus
1'00 1 1.33 1.50
1.oo 0.57
0.83 l .08 1.17
1.06
~"
Long term Medium term Short and very short term
S4 1.43 2.00
BS 5268 :Part 2 :1996, Section 4 deals with plywood for structural use. Dimensions and section properties of plywoods are given in Tables 2532 of the code. They are based on the minimumthicknessespermitted by the relevant product standards and apply to all service classes. The grade stresses and moduli for plywoods are given in Tables 3447 of the code. These apply to longduration loading in service classes 1 and 2, and should be used in conjunction with the corresponding section properties given in Tables 2532. For other durations of load andlor service class 3 condition, the stresses and moduli should be multiplied by the modification factor K36 given in Table 33 of the code. A summary of this table is reproduced here as Table 7.1. It is important to note that thebendingstresses and moduligivenin Tables 3447 of the code apply when the bending is perpendicular to the plane of the plywood panel (i.e. case (1) above). In situations where plywood is subject to bending about an axis perpendicular to the plane of the board (i.e. with the edge loaded, as in a plywebbed beam) the tensile and compressivestressesinducedby the bendingmomentshould be considered individually, and the tension and compression stresses and moduli for the appropriate face grain orientation should be used.
The following criteria should be considered when designing plyweb beams: (1) Bending. (2) Deflection. (3)Panel shear. (4)Rollingshear. (5) Lateral stability. (6) Webstiffeners.
128
Structural Timber
Design
BS 5268 :Part 2,Clause4.6dealswiththeuseofplywoodinflexural members.Thepermissiblestresses for plywoodinflexuralmembers are governed by the particular conditions of service and loading described in Clauses 2.8 and 2.9 of the code (which relate to load duration and load sharingrespectively) and should be taken as the product of the grade stresses given in Tables 3447 and the modification factor from Table 33 of the code (Table 7.1). BS 5268 :Part 2, Clause 2.10.10 specifiesthat the modification factors K27, K28 and K29 (for glulam members) given in Table 22 may also befor used the flanges of glued builtup beams such as I and box plywebbed beams. The number of pieces of timberin each flange should be takenas the number of laminations, irrespective of their orientation, to determine the value of the stressmodification factors K27, and K29 for that flange. As mentioned earlier, the plywood web(s) of a plywebbed beam will be subjected to bending about anaxis perpendicularto the plane of the board; and it is therefore necessary to check that the maximum applied bending stress in the transformed section is not greater than the lesser of: (1) the permissible tensile stress of the timber flange, Ufimber,t,adm,/ (2) the permissible compressive stress of the timber flange, U t i ~ b ~ r , c , u d m , / (3) the permissible transformed tensile stress of the plywood web,

Bplywood =plywood,t,a~,/
" .
Efimber
and (4) the permissible transformed compressive stress of the plywood web, Eplywood uplywood,c,adm,/
'
p
Etimber
Thereforetheappliedbendingstress in thetransformedsection, O m , u , / , should not be greater than the lesser ofthe above permissible stresses, where, the permissible tensile stress for timber flanges is
the permissible compressive stress for timber flanges is =tjmber,c,adm,/
= gradeStress(Table
719
oc,g,/
x K$3K&28
(7.2b)
and for plywood web(s), the permissible tensile or compressive stress is
Eplywood
X"
Etirnber
K8K36
( 7 . 2 ~& d)
Design of PlywebbedBeams
129
The applied bending stress for a beam subjected to a maximum bending moment of M can be calculated from:
Ixx H where: Zxx= y = y Y
BH3 and Ixx=12 12
ba3
a, b, B and H are as indicated in Fig. 7.1.
7.4.2 Deflection Generally, in composite constructions, such as plywebbed beams, the plane crosssections of the beam do notremain plane during bending, as assumed insimple theory. Thisleads to increased deformation due to shear. Therefore deflection of a plywebbed beam, as mentioned in Section 4.4, is determined as the sum of deflections due to bending, A m , and shear, A,: Atotal
=A m
+ As
(7.4)
For deflection calculations Clause12.10.10 of the code also states that the total number of pieces of timber in both flanges should be taken as the number of laminations necessary to determine the value of K28 that is to be applied to the minimum modulus of elasticity, Emin. Bendingdeflection, Am, for a simply supported beamcarrying uniformly distributed load of w/m, may be calculated from:
a
and for asimply supported beam carrying a concentrated load of p at midspan, from:
where:
x K28 (for the total number of flanges in the beam, Table 22) I= second moment of area of the transformed section. Shear deflection, A,. There are many methods available to estimate the sheardeflectioninplywebbedbeams.However, for a simply supported beam, based on Roark’s recommended approximation (1956); the shear deflection may be estimated from:
E = Emin (for timber)
M GwAw
A, = 
(7.7)
1
Structural Timber Design
where: M = the bending moment at midspan Gw= the modulus of rigidity (shear modulus) of plywood web A the area of the plywoodweb (actual, not transformed). 4.3 ~ a ~shear e /
Applied panel shear, z,, is the horizontal shear stress induced ina plywood web and can be calculated from: Z" a 
Ijy Q It
where: E; =maximum shear force (usually maximum reaction) Q =first moment about the neutral axis of the transformed section area above the axis I= second moment of area of the whole transformed section t =total thickness of the plywood (actual, not transformed). Applied panel shear, za, should not be greater than the permissible panel shear for plywood web, zplywood,adm: Ta
5
Zplywood,adm
The term rolling shear is used to describe a failure mode which may result due to overstressing of the glue line between the flange and yeb of a beam by shear force(s) actingat the interface. This is due to the physical construction of plywood (i.e. having ply grains at 90" to each other), which enables the material fibres to roll across each other, creating a shear failure plane. The rolling shear stresswill occur in veneer which has vertical grain. If face grain is vertical, the rolling shear will occur in the veneer that is glued to the flange. If the face grain is parallel to the flange, it will occur in the second veneer or the first glue line.In order to avoid rolling action taking place, the rolling shear must belimitedby providing sufficient glued contact depth between the flange and web. A simplified equation for calculation of applied rolling shear stress, ,,,,z is given by: (7.10)
131
Design of PlywebbedBeams
where: F,, =maximum shear force (usually maximumreaction) Qr =first moment of area of upper flange about neutral axis h =contact depth (depth of flange) I =second moment of area of whole transformed section.
BS 5268 :Part 2, Clause 4.6 recommends
that for rolling shear at the junction of the web and flange of a plywebbed beam,and at the junction of the outmost longitudinal member, the grade stresses given in Tables3447 should be multiplied by the stress concentration modification factor, K37, which has the value 0.5. Applied rolling shear stress, Tr,a, should be less than or equal to permissible rolling shear stress Tr,a
I Tr,adm
where: T r , a h =grade stress, Tr,g
X K8K3&37
and
K37
= 0.5
(7.1 1)
7.45; Lateralstability Lateral stability of I and box beams should be checked with reference to Clause 2.10.10 ofBS 5268 :Part 2. A summary of this clause is given below. The ratio of the second moment of area of the crosssection about the neutral axis, Ixx, to the second moment ofarea about the axis perpendicular to the neutral axis, IYY,should be checked to ensure that there is no risk of buckling under design load. Thisisachievedbyfollowingthecode's recommendations as summarised in Table 7.2. 7.4.6 Webstiffeners
Clause2.10.10 of BS 5268 :Part 2 recommends that builtup members with thin webs, i.e. I and boxed plywebbed beams, should be provided with Table '7.2 Recommendations for lateral support for plywebbedbeams(Clause
2.10.10, BS 5268 :Part 2)
Degree of lateral support
Ratio Ixx]Iy,,
(a) No lateral support necessary is (b) Ends of the beam should beheldinposition at the bottom flange at the supports (c) The beam should be held in line at the ends (d) One edge should between line held inbe (e) The beam should be restrained by bridging or other bracing at intervals of not more than 2.4m (f) The compressionflange should befully restrained
5 or less between 5 and 10 between 10
and 20 20 and 30 between 30 and 40
greater than 40
Structural TimberDesign
Webstiffeners each side
Webstiffeners inan lbeam.
webstiffeners at all points of concentrated load including support positions or elsewhere as may be necessary. Since there is no definite guidance available on the positioning of webstiffeners along the length of the beam, the manufacturers of plywebbed beams often have to revert to tests to prove the chosen centres, or adopt a conservative approach by placing stiffeners closer than probably required. In Fig. 7.5 webstiffeners in an Ibeam are illustrated. S
l . Mettem, C. J. (1986) Structural Timber Design and Technology. Longman
Scientific and Technical, Harlow, 2. Roark, R. J. (1956) Formulas for Stress and Strain. 4th edn. McGrawHill, NewYork.
a ~ ~ 7.1 / eDesign o f a ~ / y  ~ e b box beb ~ e a ~
Show that a glued plywebbed boxbeam with crosssectional details as indicated in Fig. 7.6 is suitable for carrying a longterm uniformly distributed load of l .95 kN/m including selfweight over an effective span of 6.5 m in service class 2.
Designof PlywebbedBeams
Fig.7.6 Box plywebbed beam (Example 7.1).
Design data Flanges: Timber in strength class C16 Webs: Finnish birch, 9plies, 12 mm thickness with face grain // to span
Crosssectional details: b= 120mm t = l2mm h=60mm H = 5oomm
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm2
N :=newton kN := lo3 N Direction parallel to grain, // Direction perpendicular to grain, pp
Effective length, L, Longterm load (kN/m),
L, := 6.5 .m W := 1.95 kN m"*
W
K2 := 1 Service class 2 (IC2? Table 13) Longterm loading (IC3? Table 14) K3 := 1 No load sharing (Kg, Clause 2.9) Kg := 1 b := 120mm Width factor, tension member (K14, Clause 2.12.2)
K14
= 1.11
133
4 34
StructuralTimber Design
Clause 2.60.60 Tension l/,(K27, Table 22) Compression //,(K280 Table 22) Modulus of elasticity for deflection (Kz~E, Table 22) sections Clause 4.5 Plywood grade stresses (&as, Table 33) Plywood moduli, ( K 3 6 ~ , Table 33) Plywood rolling shear, (K37, Clause 4.6) 3.
K27
&&
:= 1 for oneglued timber section per flange := 1 for oneglued timber section perflange := 1.l4 for a total of two glued timber
in both flanges K36s
:=1.0
K 3 6: ~= 1.0 K37
:= 0.5
Gradestresses
BS.5268 : Part 2, Table 7 Strength class, C l 6 timber Tension // to grain Compression // to grain Shear // to grain Mean E value Minimum E value
N
CTthbet.t.g./ := 3.2
mm2 mme2 Ttimber,g./ := 0.67 N mm"2 Ethber.mean := 8800 N mm"2 Ethber.min :=5800 N mm"2 CFtimber,c.g./ := 6.8 * N
+
*
*
*
*
*
*
BS.5268 : Part 2, Table 40 Finnish birch, 9plies, 12 mm plywood Tension // to face grain o p ~ y .: t= . g19.16 .~ N mnr2 Compression // to face grain C T ~ ~ ~ .: ~= . 9.8 ~ . / N mm2 Panel shear // and perpendicular zply.g := 4.83 N mm2 to face grain Rolling shear Tr.g := 1.23 N mm"2 Modulus of elasticity in tension Eply := 4250 N mme2 and compression // to face grain Modulus rigidity of G , := 320 N IXIIII~ (shear modulus)

*
6
a
*
4.
*
Geometricalproperties
Transform section to an equivalent of all timber. Etimber := E;imber.mean Using Erneanfor timber Plywood thickness tply := 12 mm *
Transformed web thickness, tply.tfd
tply.tfd := tply
*
EPrv timber
tp[y.ffd= 5.8 0 mm
Design of PlywebbedBeams
Transformed section: Crosssectional dimensions:
135
b := 120. mm B := 2 tply.tfd 4b B = 131.59 o mm H := 500 mm h :=60. mm a:=H2.h a=380omm B  H 3 b.a3 Ixx:=12 12 Zxx= 8.22 x lo8 o mm4 *
+
Secondmomentof
area, Zxx
~
Section modulus, Zxx
zXx = 3.29 x lo6 o mm3 Secondmomentof
area, Zyy
H . B 3 aeb3 zyy ;= 12 12 Iyy= 4.02 x lo7 o mm4
5. g e ~ ~stress i ~ g
The maximumapplied bending stress in transformed sectionshould not be greater than the lesser of:
(4)
Transformedplywood compression // to face grain
otfd.ply.c.adm./ := oply.c.g./
otfd.ply.c.adm./
= 4.73
N
*
G' &'ly Kg
'
Thus, the lowest value is taken as the permissible stress in bending otimber.t.adm./ = 3*54 mm2 Applied load, W W = 1.95 o kN m' W L: Applied bending moment, M 1" := 8 M = 10.3 o kN m *
a
K36s
II 36
Structural Timber Design
Applied bending stress,
Om.a.1 Om.,./ = 3.13 o N mm2 Bending stress satisfactory *
6. Deflection
BS5268: Part 2, Clause 2.10.10 Modulus of elasticity for deflection Deflection due to bending
E :=Etimber.min ' K28E E = 6.61 x lo3 o N mm2 (I
5.W.L: A m := 384 E Ixx A m = 8.34 o mm A , :=2 ( H  tply) A , = 1.2 X lo4 0 m2 *
Web area, A , Deflection due to shear
*
A4
As :=G, A , As = 2.68 o m m &*tal := A m As Atotal = 11.02 o mm &h := 0.003 Le A a h = 19.5o m m Deflection satisfactory *
Total deflection Permissible deflection
TI.
+
*
Panel shear
Applied shear force, F,,
F,, = 6.34 o kN First moment of area, Q
Q = 1.95 x lo6 o mm3 Applied panel shear stress, ,z Permissible panel shear
= 0.63 0 N mm2 :=zp1y.g ' K8 K36s Tab./ = 4.83 ~ N . m m  ~ Panel shear satisfactory
2,
*
zadm./
8. ~ o l l i nshear g Applied shear force, F,,
F,, = 6.34 o kN
First moment o f area o f upper flange, Qr Qr
= 1.58 X lo6 0mm3
ad
Design of PlywebbedBeams
Appliedrollingshearstress,
, z
zu :=
F v
*
137
Qr
2 h I,, ,z = 0.1 o N . mm2 zr.g= 12 3 o N m"2 K37 := 0.5 zudm := %,g ' K8 K368 K37 zU& = 0.61 o N mm2 *
*
BS 5268: Part 2, Clause 30 Permissible panel shear
*
*
Rolling shear satisfactory 9. Lateralstability Clause 2.1OJO &x __ 20.44
Check Ixx/Iyyratio
IYY
Condition (d) applies, i.e. one edge should be held in line Therefore the section is adequate Example 7.2 Design of a ply webbed /beam
An office building has a flat roof which comprises plywebbed Ibeams spacedat 1.1m centreswith an effective span of 7.6rn. The Ibeams support solidsectiontimber secondary beams that are finished flush with their top and bottom flanges and are positioned perpendicular to their longitudinal axes. Carry out design calculations to show that beams of glued construction with crosssectional details indicated below in service class 2 are adequate. Design data including Given loadDead selfweight =0.8 1kN/m2 Imposed (mediumterm) =0.75 kN/m2 Flanges: Timber strength in class Cl 8 Web: Finnish conifer, 9plies, 12 mm thicknessl .4mm veneer sanded with face ply grain perpendicular to span
b=47mm h=97mm t = 12mm H= 580mm
Definitions
Force, kN Length, m Crosssectional dimensions,mm Stress, Nmm2
N :=newton k~ := 103 . N Direction parallel to grain, // Direction perpendicular to grain, pp
1
Structural Timber
Design
4"L.l Fig. 7.7 I plywebbed beam (Example 7.2).
EEective length, Le Beam spacing, Bs Dead load (kN/m2), DL Imposed load (kN/m2), IL ZL Total (kN/m2), TLTL Applieduniformly distributed load (kN/m), W
Service class 2 (K2, Table 13) Mediumterm loading (K3, Table 14) No load sharing (lug, Clause 2.9) Width factor, tension member (Kl4, Clause 2.12.2)
Le := 7.6 m Bs:= 1.1 . m DL := 0.81 kN mW2 *
a
:= 0.75. kN mV2 := DL ZL
+
TL = 1.56 o kN +my2 W := TL .BS W
= 1.72 o kN
ml
K2 := l K3 := 1.25 K8 := 1 h := 97mm K14 :=
(
300  mm h >".l1
K14 = 1.13 Clause 2.10.10 Tension // (&, Table 22)
Compression // (K2gC,Table 22)
K27 := 1.l
for twoglued timber sections per flange := 1.14 for twoglued timber sectionsper flange
Modulus of elasticity for deflection Table 22) K28E := 1.24 for a total of four glued timber sections in both flanges
Design of PlywebbedBeams
Clause 4.5 Plywood grade stresses ( K ~ GTable ~ , 33) Plywood moduli ( K 3 6 ~Table , 33) Plywood rolling shear (K37, Clause 4.6) 3.
K36s
:= 1.33
K36E := 1.54
K37 := 0.5
Gradestresses
BS5268 : Part 2, Table 7 Strength class, C18 timber Tension // to grain Compression // to grain Shear // to grain Mean E value Minimum E value BS5268 : Part 2, Table 41 Finnish conifer, 9plies 12 mm plywood with face grain perpendicular to span Tension perpendicular to Dply.t.g.pp:=7.44 N mm"2 face grain c ~ ~ ~ := ~ 9.8 . ~ N. mmM2 ~ . ~ Compression perpendicular to face grain Panel shear // and perpendicular ?;ply.g := 3.74 N mm2 to face grain Rolling shear := 0.79 N mmm2 Modulus o f elasticity in tension Eply := 2950 N mm2 and compression perpendicular to face grain G, := 270. N mm"2 Modulus of rigidity (shear modulus) +

e
4.
Geometrica/Properties
Transform section to an equivalent o f all timber. Using Erneanfor timber Plywood thickness Transformed web thickness,
Etimber := Etimbermean
tply := l:!
mm
tply.tfd := tply
*
APlY Etimber
tply, tfd
tply.tfd
= 3.89 o mm
~
139
1
Structural Timber
Design
Transformed section: ~rosssectionaldimension:
b := 47 .mm
+
l? := tply.tfd b .2 B = 97.89 o m m H := 580 mm h :=97 mm +
Second moment of area, I,,
a:=H:!.h a = 386omm B H 3 ( 2 . 6 ) . a 3 zx, := 12 12 = 1.14 X io9 0mm4
Section modulus, Z,,
Second moment of area, ZYY
zyy= 1.52 X
io7 o mm4
The ma~imumapplied bending stress in transformed section should not be greater than the lesser of: Timber tension // to grain Timber compression// to grain Transformed plywood tension perpendicular to face grain Transformed plywood compression perpendicular to face grain
~qd.ply.c.adm.pp:= 0ply.c.g.pp *
timber
Thus, the lowest value is taken as the permissible stress in bending G#d.ply.t.adm.pp = 321 N mm2 *
&
K368
Design of PlywebbedBeams
Applied load,
W
W
Applied bending moment, M
CFm.a.1
Applied bending stress,
= 1.72 0 kN m' *
L: M ;= 8 M = 12.39 o kN * m M W
'
om.u.1 :=ZXX
N mm2 Bending stress satisfactory C F ~ .= ,.~ 3.15 o
BS5268 : Part 2, Clause 2.10.10 E Etimber.min ' K28E E = 7 . 4 4 ; ~1 0 3 ~ N * m m " 2 5.W.L: Am := 384 E .IXX A, = 8.78 o
Modulus of elasticity for deflection Deflection due to bending
*
*
A , :=H
Web area, A,
*
tply
A , = 6.96 x lo3o mm2 M A, :=G, A , A, = 6.59 o Atoral := A m a s Atoral = 15.37 o mm A a h :=0.003 L e A,& = 22.8 o Deflection satisfactory
Deflection due to shear
*
+
Total deflection P e ~ i s s i b l edeflection
Applied shear force,
*
Le
F,, := 2 Fv = 6.52 o kN W
F,,
*
First moment of area above neutral axis, Q
Q = 2.37 x
Applied panel shear stress,
za
Fv
, z :=
lo6 o mm3 *
e
(2 tply) , z = 0.56 o N mm2 IXX
*
*
141
142
Structural Timber Design
:=zp1y.g K8 K36s z a h . / = 4.9’7 ~ N . m m  ~ Panel shear satisfactory
Permissible panel shear
Tach./
*
8. Rolling shear
Applied shear force, F,
F, = 6.52 o kN
First momentof area of upper flange above neutral axis, Qr
Qr
:=(2
Qr
= 2.2 X lo6 o mm3
Applied rolling
shear stress,
zu :=
Fv
a
b
*
Qr
BS 5268 :Part 2, Clause 30
2 h Zxx = 0.06 o N K37 := 0.5
permissible panel shear
zuh
*
*
:= %.g K8 K36s K37 z u h = 0.53 o N mm2 *
*
*
Rolling shear satisfactory 9. Lateralstability
Clause 2.10.10
Check Zxx/Zyy ratio
zxx
75.24
”
ZYJJ
Condition (f) applies, i.e. compreSsion flanges should be fully restrained, for example, by direct fixing of roof sheeting Therefore the section is adequate
Timber columns maybe classified as simple or builtup, depending on their method of construction. Simple columns are fabricated from sawn timber or from gluedlaminated sections (glulam)to form one piece, for example,a rectangular section, The design requirements for simple columns are discussed in detail in Chapters 5 and 6 for sawn solid sections and glulam members respectively. ~uiltupcolumns are those composed of two or more sections of timber connected together either by gluing or by means of mechanical fasteners (such as nails, bolts, etc.)Suchcolumnscan provide considerably more strength than the sum of the strength of the sections acting alone. Builtup columns can beconstructed in a great variety of shapes, see Fig. 8.1, to meet special needs or to provide larger crosssections than are ordinarily available, or purely for architectural applications. While glued builtup columns can be assumed to behave as a single member, the full composite strength developmentinmechanically fastened columnsis doubtful.In general mechanically fastened columns may not possess adequate composite action by the connectors to make them act as a unit. In such cases designers are recommended to revert to prototype testing or conservative designsolution.' BS 5268 :Part 2 :1996 providesguidance for design of a special type of builtup column known as a spaced column which is detailed in this chapter.
Fig. 8.1 Typical crosssections for builtup elements. 143
1
Structural Timber
Design
Spaced c o Z ~ ~ nare s thosebuiltup as two or more individualmembers (shafts) with their longitudinal axes parallel, and are separated at the ends and midpoints with spacer bZocks and joined together by gluing, nailing, bolting or other means of timber connectors.A typical makeupof a spaced column is illustrated inFig. 8.2. "he spacer blocks are provided to restrain differentialmovementbetweenthe shafts and to maintain theirinitial spacing under the action of load. The end connectors provide partial fixity to the individual members of the column, thus effectively restraining their buckling tendency. Spaced columns are often used in architectural applications, in trusses as compression chords, and in frame construction. A spaced column, due to the geometry of its crosssection, could haveat least 25% more loadcarrying capacity than a single solid member with a similar volume of timber. Spaced columns, apart from being economical, can providesuitable construction through which other piecessuch as bracing or column to truss connections can be inserted conveniently.
Suitable connections (mechanical or glued)
Spacer blocks
<
i
c
L x ."
Direction A
7 '
Direction B Crosssection
View in direction A
.2 Details of a twoshaft spaced column.
View in direction B
Design of Builtup(Spaced)Columns
1
Design recommendations for spaced columnsare detailed in Clauses2. l 1.8 to 2.1 1.11 of BS 5268 :Part 2 :1996. A summary of the recommendations, which may fall under the following headings, is given below: (1) geometricalrequirements, (2) modes of failure and permissible loads, and (3) shear capacity of spacer blocks. eo~etrical
re~uire~e~ts
With reference to Fig. 8.2, the following criteria must be satisfied: The clear space between the individual shafts, S, should not be greater than three times the thicknessof a single shaft, b (i.e.ensure that S 5 3 x b). The length of end spacer blocks, h,, should be at least six times the thickness of individual shafts (i.e. ensure that h, => 6 x b). The length of intermediate spacer blocks, hi, should be not less than 230 mm (i.e. ensure that hi 2 230mm). A sufficient number of spacer blocks should be provided to ensure that the greater slenderness ratio of each shaft between the spacer blocks (Lend/iww and Lint./iww) is limitedto 70, or 0.7 times the slendernessratio of the whole column, whichever is the lesser. Therefore it is necessary to ensure that: Lend and Lint. 5 the lesser of '70 and the greater of lww
fww
thegreater of 0.7 Le and 0.7 Le ZYY
fXX
where: L =the overall length ofthe spaced column L, =the effective length of the spaced column Lend= the distancebetween the centroids of end and closest intermediate spacer blocks Lint.=the distance between the centroids of intermediate spacer blocks i =radius of gyration with respect to indicated axes, (xx,yy, or w w ) see Fig. 8.2.
.2
des of f a i ~ ~and r e ~ e r ~ i s s loads i~le
Since the following modesof failure are possible, they should be considered and reflected in calculation of the permissible stresses:
1
Structural Timber Design able 8.1 Modi~cationfactor KI3for the effectivelength (Table 20, BS 5268:Part 2)
Method of connection
of
Value
of spaced columns
K13
Ratio of space to thickness of the thinner member
0 Nailed Screwed or bolted Connectored Clued
1.8
1.7
1.4 1.1
1 2.6 2.4 2.4 1.8 1.41.1
2
3
3.1 2.8 2.2 1.3
3.5
3.1
(1)Bucklingof the wholecolumn about its xx axis. In this case the effective length will depend upon full column length and the spacing of the shafts will not improve the performance, hence: Le,Xx
=L x (effective length coefficient, see Table 5.1)
The permissible axial and/or bending stresses should be calculated for the equivalent solid column section. (2) Buckling of the whole columnabout its yy axis. In this casethe type of connection will affect the magnitude of effective lengthof the column. Table 20 of BS 5268 :Part 2 :1996 (reproduced here as Table 8.1) gives the modification factor K13 which should beused to calculate the effective length about the y y axis, hence: Le,yy=L x K I 3x (effective length coefficient, see Table 5.1) (3)
Buckling of an individual shaft aboutits own WW axis. For this purpose, the proposed axial load is shared equally between shafts and the effective length is taken to be the larger distance between the centroids of the connections (spacer blocks), i.e.Lend or L i n t . , whichever is greater, hence:
These modes of failure are reflected in permissible load requirements given in the code. It is to be noted that the greatest slenderness ratio (i.e. the greater of L e / i x x , L e / i y y , Lend/iwwand Lint,/iww) will govern the critical failure mode. For example, if Le/ixxhas the largest value, thenbuckling about the xx axis is the most critical one and will dictate the magnitude of the permissible stresses.
Design of Builtup(Spaced)Columns
A s before, see Section
147
5.2, the applied and permissible stresses may be
calculated from: oc,a,ll =
applied compressive stress permissible compressive stress
o c , a h , / /= om,a,ll =
applied bending stress permissiblebendingstress
3 P
CT,,~,// x K2K3 Kg K12 M 2
om,adm,// = orn,g,/x
K2K3KbK7&
8.3.3 Shear capacity of spacer blocks
The longitudinallength of themechanically and/or glueconnected end spacer blocks shouldnot be lessthan six timesthe thickness ofthe individual shaft and should be sufficient to transmit a shear force, F,,, at the interface between each block and one shaft equal to:
F,, =
1.3Aboc,u,j7 nu
where: A =total crosssectional area of the column b =thickness of the shaft ~ ~ ~ = , ~applied , l l compressive stress n =number of shafts, and a =distance between centres of adjacent shafts Therefore applied shear stress, za,/, is:
where: h =width of the shaft, and h, =length of the end spacer block
Fig. 8.3 Axes in spaced columns.
1
Structural Timber Design
and the permissible shear stress,
~ ~ c tis:. , ~ ,
The code also recommendsthat intermediate spacer blocks shouldbe not less than 230mm long and beable to transmit ashearforce,between theirindividualinterfaces equal to half that of theendblocks,i.e. Fv.int. = ~v
5
In triangulated frameworks suchas trusses and girders (but excluding trussed rafters designed to Part 3 of BS 5268), Clause 2.1 1.11 of BS 5268 :Part 2 : 1996 provides guidance asto effective lengthsfor the compression members and, where these members are considered to be spaced columns, all design should be in accordance with the design requirements for spaced columns except for the design of end connecting joints. In the case of thesejoints, the shear resistance requirements of Clauses 2.1 1.9.1.1 and 2.1 1.9.2do not apply. The requirements for intermediate spacer blocksfor such members are that they should be not less than 200mm long and should be fixed in such a manner as to transmit a tensile force parallel to the xx axis, between the individual members, of not less than 2.5% of the total axial force in the spaced compression member.
1.
Baird and Ozelton (1984) Timber Designer’sManual, 2nd edn. Blackwell Science, Oxford.
a~ple
A glueconnected spaced column is fabricated from two shafts of equal crosssection of 60 mm x 194mm, as shown in Fig. 8.4, and carries a mediumterm axial compression load of 95kN. The columnisfreestandingandisunbraced throughout its length against the tendencyto buckle in either direction. Check the adequacy of the column if it comprises timber sections in strength class C22 under service class 1 conditions.
Force, kN Length, m Crosssectional dimensions, mm Stress, NmmW2
N :=newton kN :=lo3 N Direction parallel to grain, // Direction perpendicular to grain, pp
Design of Builtup(Spaced)Columns
372 mm
”1 882 m
60
60
60 230 m
1
I t 882 m
230 m
1
882 m
W
i
W
230 m
”1 ~
882 m
Crosssection 372 m ~
Fig. 8.4 Spaced column details (Example 8.1).
e~~etrjcaf
re~ujre~ents
Column length, L Number of shafts, n Breadth of each shaft, b Width of each shaft, h Space between shafts, S Distance between centres of shafts, a Length of end block, h, Length of intermediate block, hi Crosssectional area of each shaft, Ashaft Total crosssectional area, AtOtar Second moment of area, Zxx
L := 3.9 m n := 2 b := 60 m m h := 184.mm S := 60 m m

3.9 111
1
150
Structural Timber Design
Second moment of area, Iyy For each shaft, I,,
Iyy: = &  h . ( 2 * b + ~  )12J~  * ~ * s ~ Iyy= 8.61 x lo7 o mm4 I,, := h b3 I,, = 3.31 X lo6 0 mm4
&
0
*
Radius of gyration, ixx ixx = 53.12 o
Radius of gyration, iyy
m
iyy := iyy = 62.45 o mm
Radius of gyration, i,, i,, = 17.32 o mm
Effective length of individual shafts about their own WW axis is the greater of Lend and Lint. Slenderness ratio about WW axis, h,,
Lend := 882 * mm
Effective length of column about XX axis, L e . x x
Le,xx
Slenderness ratio about XX axis
Lint. :=882 * mm
:= Lend
h,,
lww
h,, = 50.92 Le.xx
:=1.0 L = 3.9 0 m
Le.xx h, :=lxx
hxx= 73.42 Effective length of column about yy axis, Le,, depends on the shaft/block connections BS5268 : Part 2, Table 20 For glued connections and for s/b ratio of 1 Effective length of column about y  , ~axis7 L e . y y
K13 := 1.1 Le.yy := 1.O* L
*
L e  y y = 4.29 0 m
K13
L.YY ?Lyy := iYY
hyy= 68.69
= 60 o mm is less
Condition ( I ) Check: S 5 3 x b
S
Condition (2) Check: he 2 6 X b
he
Condition (3) Check: h i 2 230mm
= 372 0 mmis
hi = 230 o mm okay
than
3 . b = 180 o mm okay greater than 6 . b = 360 o mm okay
Design of Builtup (Spaced) Columns
Condition (4) Checkslenderness ratios so thatthe greatest h,, 5 the lesserof(70and the greater of 0.7 h,, and 0.7 hYY)
151
h,, = 50.92
0.7 hxx 5l .4 0.7 hYY= 48.09 Satisfactory *
B. Modes offailure and permissible load checks
From the above calculations it is evident that the greatest slenderness ratio, i.e. h, = 69.64, is the critical one. Therefore buckling about the xx axis should be considered in determination of the permissible compressive stress. 2. Grade stresses
BS.5268 : Part 2, Table 7 Strength class, C22 Compression parallel to grain Shear parallel to grain Minimum modulus of elasticity, Emin
CT,.~,~ :=7.5 N mm2 := 0.71 N mm2 Emin := 6500 N mm"2
"cg.[
*
+
3. Loading and applied compressive stress
Mediumterm axial loading Applied compressive
stress
P :=95 kN P *
Gc.a.1 :=
Atotal
Uc.a.1
= 4.3 o N mm2 *
4. Kfactors
Serviceclass 1(K2, Table 13) Mediumterm loading (K3, Table 14) No load sharing (K& Clause 2.9)
K2 := l
K3
:= 1.25
K8 := 1
Modification factor for the compressionmember, one of the following methods: (1)Using
equation in Annex B:
can be calculated using either
1
Structural Timber
Design
(2)Alternatively,usingTable 4nin %.g.//
*
K2 ' K3
19:
= 693.33 and hxx= 73.42 K2 K3 ' K8 K12.x~ = 4.83 o N mmv2 Satisfactory ~ c . a d m . / / . x x:= Gc.g.//
Permissiblecompressive stress
Uc.ah.i.xx
*
*
*
s ~ e acapacity r of spacer ~ l ~ c k s
BS.5268 : Part 2, Clause 2.11.9
Shear force acting at end blocks, F,,
F,, = 30.88 o kN
Permissible shear stress, z a h , i
K2 K3 Kg = 0.89 o N mmm2
z a h := 1cs.1 z a h
Shear satisfactory The s ~ a c column e~ is a ~ e ~ u a t e
fa an
A nail connected spaced column isfabricated from two shafts of equal crosssection of 63 mm x 225 mm,as shown inFig. 8.5. Assuming that the column is afree standing one and is unbraced throughout its length against the tendency to buckle in either direction, (a) check the adequacy of the column if it comprises timber sections in strength class C24 under service class 2 conditions,
Design of Builtup(Spaced)Columns
115
(b) determine the maximum axial load, under longduration loading, that the column can carry, and (c) check that the section is adequate to resist a very shortterm lateral wind load of 1.25 kN/m of height tending to bend the column about its xx axis together with an axial load of 45 H?.
4.1 m !
w
!
i
b
Crosssection
Spaced column details (Example 8.2).
Force, k N Length, m Crosssectional dimensions, mm Stress, Nmrn2
N :=newton
Column length, L Number of shafts, n Breadth of each shaft, b Width of each shaft, h Space between shafts, S
L := 4.1 + m n:=2 b := 63 mm h :=225 mm S := 100. mm
k~ := 103 . N
Direction parallel to grain, // Direction perpendicular to grain, pp
154.
Structural Timber Design
Distance between centres of shafts, a Length of end block, he Length of intermediate block, hi Crosssectional area of each shaft, Ashaft Total crosssectional area, Atotal Second moment of area, Ixx Second moment of area, Iyy
a = 163 o m he := 380 mm hi := 250. mm Ashaft := b h Ashaft = 1.42 X io4 o mm2 Atotal := 2 Ashaft Atotal = 2.84 X lo4 o mm2 Ixx := (2) b h3 rXx= 1.2 X io8 0mm4 Iyy : = g . h * ( 2 . b + ~ ) 3   i l i . h . s 3 Iyy= 1.98 x 10' o mm4 Iww h *b3 I,, = 4.69 x lo6 o mm4 *
*
*
g
*
:=A*
For each shaft, I,, Radius of gyration, ixx
ixx = 64.95 o mm
Radius of gyration, iyy iyy = 83.5 o mm
7
Radius of gyration, iww i,, = 18.19 omm
Effective length of individual shafts about their own minor WW axis is the greater of Lend and Lint. Slenderness ratio about axis, h,,
WW
Lend := 720
mm Lint. := 760 m m
Lint. h,, := Lww
Effective length of column about xx axis, Le,,,
h,, = 41.79 Le.xx := 1 .O L Le.xx = 4. l0 m
Slenderness ratio about xx axis
h,
*
:=
L x x lxx
h, = 63.12 Effective length of column about p" axis, Le.yy,depends on the shaft/block connections BS5268 : Part 2, Table 20 For nailed connections and for s/b ratio of 1.59
Design of Builtup (Spaced) Columns
Effective length o f column about YY axis, Le.yy Slenderness ratioabout yy axis:
Le.yy Le.yy
155
:= 1.O L K13 = 11.87 o m
Le.YY hyy := ZYY
hyy= 142.14 Condition ( I ) Check:
5 3x b
= 100 o mm is less than 3 b = 189 o mm okay Condition (2) Check: he 2 6 X b he = 380 o mm is greater than 6 . b = 378 o mm okay Condition (3) Check: hi 2 230mm hi = 250 o mm okay Condition (4) Check slenderness h,, = 41.79 ratios so that the greatest 0.7 h , = 44.19 h,, 5 the lesser of (70 and the 0.7 hyy = 99.5 Satisfactory greater of 0.7 h,, and 0.7 hyy) S

S

B. Modes offailure and permissible load checks From the above calculations it is evident that the greatest slenderness ratio, i.e. hyy = 142.14, is the critical one. Therefore buckling about the yy axis should be considered in determination o f the permissible compressive stress.
2. Grade stresses BS 5268 : Part 2, Table 7
Strength class, C24 Compression parallel to grain Shear parallel to grain Bending parallel to grain Minimum modulus o f elasticity, Emin
cT,.g :. = l 7.9 N . mm2 zg.i := 0.71 N mm2 c ~ ~:.=~ 7.5 . j N mm2 Emin := 7200 N mm2 e
3. Kfactors
Service class 2 (K2,Table 13) Longterm loading (K3,Table 14) Very shortterm loading (K3,Table 14) No load sharing (Kg,Clause 2.9)
K2 := 1
K3.iong := 1 K3.v.short
K8
:=175
:= 1
Modification factor for the compression member, one o f the following methods:
K12,
can be calculated using either
Structural Timber Design
Alternatively, using Table 19: = 911.39 and hyy= 142.14 Gc.g.1 K2
*
K3.1ong
Permissiblecompressive stress Allowable axial compressive Padm load,Pa& Appliedcompressive stress
Gc.ah.1
:= Gc.g.// K2 *
~ ~ . ~= h . [
' K3.long
*
K8
*
K12.y~
1.840 N.mm"2 Pa& := G c . a h , / / .Atotal = 52.14 o kN Gc.a,//:=G c . a h . / /
= 1.84 o N mm2 a
BS5268 : Part 2, Clause 2.1l.9
Shear force acting at end blocks, E; Shear force for intermediate blocks, Fv.int. Shear area for end blocks Shear area for intermediate blocks Shear stress at end blocks, Ta,flend Shear stress at intermediate blocks, Ta,//,int. Permissible shear stress, TU&,/
The axial loa~carryin~ capacity of the spaced column is P a h=52.14 o
Design of Builtup(Spaced)Columns
l
In part (c) of Example 8.2 it is required to check the effects of axial load and bending moment due to lateral loading about xx axis:
Axial load, P Very shortterm lateral uniformly distributed load,
Applied compressive
P := 45 kN W

:= l .25 kN m1
W
stress
%a./
P
:= ___ &tal
~ ~ = 1.59 . ~ O N*mm"2 . i
2*b*h2 6 Zxx= 1.06 x lo6 o mm3 W L2 M := 8 M = 2.63 o kN m
Zxx:= 
Section modulus, Zxx
*
Applied bending moment
M
0 m . a . i ;== 
Applied bending stress
Zxx
Gm.a.1
= 2.47 o N mm2
Modification factor for compression member, K12, for a very s~ortdurationloading considering the largest slenderness ratio, i.e. hyy,can be calculated using either one of the following methods: (1)Using
equation in Annex B:
(2) Alternatively, usingTable19: Emin Gc.g.1
K2 K3.v.short
= 520.8 and
hyy= 142.14
K12.yy = 0.15
*
Check interaction quantity: Permissiblecompressive stress Permissiblebendingstress

o c s ~ . i: ,=yoc.g~l y K2 K3.v.short Kg K12.yy +
c~~.~h.1.~~ = 2.05 o N mmp2 om.aa'm.1 := Gm.g.1 K2 .K3.v.short K8 *
om.adm.i
= 13.13 o N mmm2
*
Chapter 9
0
ber Conn
In early timber buildings, many of which are still around, individual members were joined together by contact joints and wooden pins or iron dowels. Techniques suchas halflap, cogging, framed, scarf and tenonjoinery joints (see Fig. 9.1) utilising metal screws hidden by wooden pegs provided the required connection capacity. The connection schemes were simpleprimarily because the building systems were simple, with nearly symmetricalarrangements of structural elements. The present method of building construction has since evolved considerably and requires more complex connection systems and, as such, the designer should be aware of the structural behaviour and material characteristics of the connection components.
(a) Halflap joint
(b) joint Framed
(c) Tenon joint
(d) Cogging joint
Fig. 9.1 Traditional joinery joints. 159
1
Structural Timber
Design
he common connecting systems in str tural timber may be classified as mechanicallyfastened and glued joints. chanicalfastenersmayalso be divided into two groups depending on h hey transfer the forces between ~ o ~ e ~fasteners,  t y ~such ~ as nails, the connectedmembers.Thesebeing screws, boltsand dowels, and connectors, such as toothedplates, splitrings, shear plates and punched metal plates in which the load transmission is arily achieved by a large bearing area at the surface of the members. n this chapter details of the general considerations necessary for the design of doweltype fasteners are described. This is followed by details of the recommendations for the design of connectored and glued timberjoints.
As mentioned in Chapter 2, the most recent revisionof Part 1996 was to bring this code as close as possible and to run E m 199511 Eurocode 5: Design of t i ~ b e structures. r Part l.l ~ e ~ e r a l rules and rulesfor buiZdings (EC5). The overall aim hasbeen to incorporate material specifi~ationsand design approaches fromEC5, while maintaining a issiblestresscodewhichdesigners,accustomed to iar with and be able to use without difficulty. In t gy for doweltype connectionshas been subjected to considerable ge and while it remains a permissible stress design method, it takes the approach to design of nailed, screwed, bolted and dowelled connections. commendations for timber joints are detailed in Section 6 of t 2. This section providesthe basic lateral shear and withdrawal e mostcommondoweltypefastenersinTables 5474. The les have been calculated using the equations and summary of €365 rules and equations is given in ese equations can be used in lieu of the tabulate^ data if preferred, and may be used for other material/fastener applications that are not currently detailed inthe code, for example, the use of bronze or copper and high tensile steel dowels. n this chapter design lations are based on the tabulated values; while the use of e ~ u a t i o is ~s scussed in Chapter 10 which outlines the Eurocode approachto design of timber structures. A summary of the general design considerationsfor timber joints detailed in Section 6 of the code is as follows: (1) Joints should be designed so that the loads induced in each fastener or timber connector unit by the design loads appropriate to the structure shouldnot exceed thepermissiblevaluesdeterminedinaccordance with Section 6 of the code. (2) Whenmore than one nail,screw, bolt, etc.isusedin a joint, the permissible load is the sum of the permissible loads for the individual
Design of TimberConnections
1
F
U .2 Shear stress in the jointed timber.
units providingthat the centroid of the fasteners' group lies on the axis of the member, and that the spacing,edge and enddistances are sufficient to develop the full strength of each fixing unit. If the line of action of a force in a member does not pass through the centroid of the fasteners' group, account should be taken of the stresses and rotation due to the secondary moments induced by the eccentricity. This will impose higher loads on the fasteners furthest from the centroid of the group. If the load on a joint is carried by more than one type of fastener, due account should be taken of the relative stiffnesses. Clue and mechanical fasteners have very different stiffness properties and should not be assumed to act in unison. F& the purpose of designingjoints in glued laminated timber, tabulate values should be taken for the strength class of timber from whichthe glued laminated timber was made. The effective crosssection of a jointed member should be used when calculating its strength. In addition, the shear stress condition shown in Fig. 9.2 should be satisfied in the jointed member. Therefore the design requirement is to ensure that: Y<$.TU&'h,.t (9.1) where: Y =maximum shear force per member at the joint where (Yl Y2 = Fsina) h, =distance from the loaded edge to the furthest fastener =permissible shear stress, and t =member thickness.
+
int sli All mechanically connected timberjoints, under the action of applied load, will undergo slip which occurs between their component parts. Slip can bea
162
StructuralTimber Design Table 9.1 Fastener slip moduli Kserper fastener per shear plane for service classes 1 and 2 (Table 52, BS 5268 :Part 2)
Fastener type
Kser
Timbertotimber or paneltotimber (N/mm)
Steeltotimbera (N/mm)
& p'5d08
h p'.5d0.8
Nails (no predrilling) Nails (predrilled) Screws Bolts and dowels
p'v5d
where p is the joint member characteristics density given in Table 7 (in kg/m3) d is the fastener diameter (in m) a For the values in this column to apply, the holes in the steel plate should be the minimum clearancediameter and thesteel plate should have a thicknessof l .2 mmor 0.3 times the fastener diameter, whichever is the greater
significant factor in the deformationand, in some cases,it can also affectthe strength of some components. Clause 6.2 of BS5268: Part 2 provides a method for calculation of slip in joints, whereby the slip per fastener per shear plane, U, in millimetres, may be obtained from: F U=
&er
where: F =applied load per fastener per shear plane, in Newtons(N) Kser =slip modulus, in Newtons per millimetre per fastener per shear plane, given in Table 52 of the code (reproduced here as Table 9.1). For bolted joints it is recommended that an additional 1mm should be added at each joint to allow for takeup of bolt hole tolerances. It is to be noted that the magnitude of slip is timedependentand the final slipfollowingmediumterm or longterm loading maybe considerably greater than the slipcalculated as above, particularly inconditions of fluctuating moisture content.
ective crosssection Theeffectivecrosssectionof a jointedmembershould beusedwhen calculating its strength. This is because severalconnector units remove part of the crosssectional area of thetimber and henceaffecting the load capacity of thetimbersection. In suchcases the effectivecrosssection should be determined by deducing the net projectedarea of the connectors from the gross area of the crosssection being considered.
Design of Timber Connections
163
The recommendations of BS 5268 :Part 2 for multidoweltype joints are as follow: (1) Nailed and screwed joints. When assessing the effective crosssection of
multinailed or multiscrewed joints, all nails or screws of 5 mmin diameter, or larger, that liewithin a distance offive nail or screw diameters, measured parallel to the grain from a given crosssection, should be considered as occurring at that crosssection. No reduction of crosssection needs to be made for screws of less than 5mm diameter, and for nails of the same diameter but driven without predrilling. (2) Bolted and dowelled~oints.When assessing the effective crosssection of multibolted or multidowelled joints, all boltsor dowels that lie within a distance of two bolt or dowel diameters, measured parallel to the grain from a given crosssection, should be considered as occurring at that crosssection. (3) Connectored joints. When assessing the effective crosssection of multiconnectored joints, such as toothedplate, splitring and shearplate connector joints, all connectors and their bolts that lie within a distance of 0.75 times the nominal size of the connector measured parallel to the grain from a given crosssection should be considered as occurring at that crosssection.
9.5 Spacing rules In order to avoid splittingof timber, all doweltype fasteners must be spaced at suitable distances from eachother and from the endsand edges of timber or woodbased material. The various distancesinvolved are illustrated in Fig. 9.3. Anenddistanceissaid to be loaded when theload on the fastener has a component towards the end of the timber. Otherwise, it is referred to as an unloaded end distance. Loaded end distances need to be greater than unloaded ones. In a similar way the edge distance may be loaded or unloaded. 92
Fig. 9.3 Spacing of doweltype connectors.
l
Structural Timber Design
Thevaluesofspacings and distancesvary from onefastenertype to another, as well as between the various material types. The spacings and distances recommended in Section 6 of BS 5268 :Part 2 are detailed below: Nail spacing. The recommended endand edge distancesand spacing of nails, to avoidundue splitting, are given in Table 53of the code (reproduced here as Table 9.2). Predrilling (with a hole diameter up to 0.8 times the nail diameter) reducesthesplittingtendencyconsiderably and henceallowsmuch closer spacing of the nails. All softwoods, except Douglas fir, may have values of spacing (but not edge distances) for timber to timber joints multiplied by 0.8. For nails driven into a glued laminated section at right angles to the glue surface, the spacings (but not edge distances) can be multiplied by a further factor of 0.9. Screw spacing. The recommended end and edge distances and spacing of screws, to avoid undue splitting, are given in Table 59 of the code (reproduced here as Table 9.3). Bolt and dowel spacings.The recommended end and edge distancesand spacingofbolts and dowels are givenin Table 75 of the code (reproduced here as Table 9.4). able 9.2 Minimum nail spacings (Table 53, BS 5268 :Part 2)
Spacing"
Timbertotimberjoints
Without predrilled holes End distance parallel to10d grain (Q) Sd Edge distance perpendicular to grain (a4) Distance between lines 3d of nails, perpendicular to grain (a2) Distance between adjacent nails in any one line, parallel to grain (al)
20d
With predrilled holes 14d
Without predrilled holes
5d 7d
14d20d 14d
Without predrilled holes
14d
5d 10d
Steel platetoWoodbased timber joints panel productsb totimber joints
c
7d
1Od
where d is the nail diameter aSpacings and distances are illustrated in Fig. 9.3 Plywood, tempered hardboard or particleboard c The loaded edge distance should not be less than: in timber 5d, in plywood 3d, in particleboard and tempered hardboard 6d, and in all other cases the edge distance should not be less than 3d
165
Desian of Timber Connections
Table 9.3 Minimum screw spacings, (Table 59, BS 5268 :Part 2) With predrilled holes
Spacinga
End distance parallel to grain (a3) Edge distance perpendicular to grain (a4) Distance between lines of screws, perpendicular to grain (az) Distance between adjacent screws in any one line, 10d parallel to grain (a,)
10d 5d 3d
where d is the shank diameter of the screw "Spacings and distances are illustrated In Fig. 9.3
Table 9.4 Minimum bolt and dowel spacings (Table 75 and Clause 6.6.7.3, BS 5268 :Part 2) Direction of loading
Loading // to grain Loading L to grain
End distance
Edge distance
Loaded
Unloaded
Loaded
Unloaded
7d 4d
4d 4d
1.5d 4d
4d 1.5d 1.5d
Distance between bolts or rows of bolts Across the Parallel to grain grain 4d 4d
5d a
aWhere the member thickness, t, is less than 3 times the bolt diameter, d, the spacing may be taken as the greater of 3d or (2 t/d)d.
+
(4)
Connectors spacings.Associated with each typeand size oftoothedp~ate, splitring and shearplate connectorsare standardend and edge distances and spacing betweenconnectors which permit the basic load to be used. These standard distances are given in Tables 7779 for toothedplate connectors and inTables 8587 for both splitring and shearplate connectors. If the end distance, edge distance or spacing is less than the standard, then the basic load should be modified. For further details readers are referred to Clauses 6.76.9 of BS 5268 :Part 2 :1996.
shear lateral loads Doweltype fasteners, such as nails, screws, bolts and dowels, are used to hold two, three or more members together to form a joint. In general, they are designed to carry lateral shear loads, but there are occasions where they might be subjected to axial loads (or withdrawal loads in the case of nailed or screwed joints) or a combination of lateral shear and axial loads.
4 66
StructuralTimberDesign
IF
I F/2
l F/2
f
+
4
IF
/F (a)
i" i"
(b)
(c)
Fig. 9.4 Single shear (a) and (b) and double shear (c) joints.
Doweltype fasteners, depending on their type and size and the arrangement ofthe joint components, may be subjectedto single, double ormultiple lateral shear forces (see Fig. 9.4). It is important to note that the dowels in joints (a) and (b), shown in Fig. 9.4, are all in single shear (i.e. one shear plane per dowel) and the dowels in joint (c) are in double shear (i.e. two shear planes per dowel). The basic multiple shear lateral load for each dowel can generallybe obtained by multiplying the value forits basic single shearload (given inthe relevant tables of the code) by the number of shear planes; providedthat the recommended standard thickness for members of the joint and the recommended joint details are met.
9.7 Nailed joints Nails are straight slender fasteners, usually pointed and headed, and are available in a variety of lengths and crosssectional areas. There are also many types and forms of nails; but the most frequently used nails are the bright, smooth, common steel wire nails with circular crosssectional area with a minimumtensile strength of 600N/mm2, which are availablein standard sizesof 2.65 mm to 8mm indiameter.Nails can beplain or enamelled, etched, electroplated, galvanisedor polymer coated, for example, for protection against corrosion. Nailed joints, which should normallycontain at least two nails,are simple to fabricate and are of considerable economic value. They are suitable for lightly loaded structures and for thin members. They are commonly used in framing, walls, decks, floorsand roofs and in nearly everyconstruction that involves light loads and simple elements. For the outer (headside) members larger than 50mm thick, nails are generally replaced by bolts to reduce splitting of the timber.
Designof TimberConnections
167
The performance of anail, both under lateral and withdrawal loading, may be enhanced by mechanically deforming nail shanks to form annular ringed shank or helical threaded shanknails (see Fig. 9.5). Such nails providehigher withdrawal resistancethan plain shank nails ofthe same size. Other forms of improvednails are obtained by grooving or twisting of square crosssectioned nails. The process of twisting not only modifies thenail surface but also workhardens the steel, increasing its yield strength. In order to provide for the enhanced performance of improved nails, I3S 5268 :Part 2, Clause 6.4.4.4 permits a 20% increase in the basic single shear lateral load capacities of the square grooved and square twisted shank nails, and a 50% increase in the basic withdrawal load capacities for the threaded part of annular ringed shank and helical threaded shank nails.
Nails driven into dense timbers are likely to induce timber splitting. Various research studies have shown that predrilling can prevent timber splitting and also has other advantages of less slip in the joint, increase in lateral loadcarrying capacity and permits a more densely nailedjoint (i.e. reduced nail spacings and distances). The recommendations of BS 5268 :Part 2 for predrilling are as follows: (1) Hardwoods in strength classes of D30 to D70 will usually require predrilling and the code recommendsthat the diameter of predrilledholes should not be greater than 0.8 times the nail diameter. In general it is usual to provide predrilling when the density of the timber is 500 kg/m3 or greater. (2) For nails driven into predrilled holes in softwood strength classes C14 to C40andTR26, the code permits a15%increase in their load capacities. 3
(a) Round wire (b) Annular ringed
Fig. 9.5 Types of nails.
(c) Helical threaded
1 Structural Timber
Design
m
3
0
U
S
5 o!
52
m
3
0
4
E
9 h0 o! 52
Design of Timber Connections
16
1. T i ~ ~ e r  t o  t inailed ~ ~ e rjoints Thebasicsingleshear lateral loads for single round wirenailswith a ~ n i m u mtensile strength of 600N/mm2, driven at right angles to the side grain of timber in service classes l and 2, are given in Table 54 of the code (reproduced here as Table 9.5). For nails driven into the end grain of timber,the values given inTable 54 should be multiplied by the end grain modification factor K 4 3 which has a value of 0.7, i.e. K 4 3 = 0.7. For nails driven into predrilled holes in softwoodstrength classes C14to C40 and TR26, the basic single shear lateral load values given in Table 54 maybemultipliedby1.15,i.e. for convenience introduce a modification factor for predrilling, as Kpredr$/ = 1.15. For square grooved and square twisted shank nails of steel with a yield stress of not less than 375 N/mm2, the basic single shear lateral loads given in Table 54 should be multiplied by the improved nail lateral modification which has a value 1.2, i.e. KM= 1.2. asic loads in Table 54 to apply, the nails should fully penetrate the tabulated standard values in both the headside and pointside members. n addition, for joints where nails are subjected to multiple shear lateral loads, the thickness of the inner member should be at least 0.85 times the standard thickness given in Table 54. Otherwise, the basic shear loadshould be multiplied by the smaller of the following ratios, where relevant:
(1) actual to standard thickness of headside member, or (2) actual penetration to standard pointside thickness, or (3) actual inner member to 0.85 times the standard thickness. is important to note that no loadcarry in^ capacity should be assumed here the ratios given above are less than 0.66 for softwoods and 1.0 for hereimprovednails are used, t ratios maybereduced addition noincrease in to 0.50 for softwoods and 0.75 for hardwood basicload is permitted for thicknesses or pe ations greater than the standard values. erefore, with referenceto Fig. 9.6, it is possible to define a modification factor for thickness of membersand penetration depth (i.e. bearing lengths) for a nailed connection, &ai&bearing, as: for a nail in single shear,
for a nail in multiple shear,
170
F
Structural Timber Design
+ " l a , head fa,
.."F
point
Nails driven into side grain ta, head
F
.
(a) Nails in single shear head
"+=+F/2
inner
poir1t
W
F/2
Nails Fig. 9.6
Member thicknesses in nailed joints.
2.
Steel platetotimber nailed joints
Steelplatetotimbernailed joints canhave a considerablyhigherload capacity than timbertotimber joints withsimilarnailingarrangements. BS 5268 :Part 2, Clause 6.4.5 recommends a 25% higher load capacity for steelplatetotimber joints provided that steelplateshave a minimum thickness equal to the greater of 1.2mm or 0.3 times the nail diameter. Where a predrilled steelplate or component of such thickness is nailed to a timber member, the basic single shear lateral load given in Table 54 should be multiplied by the steeltotimber modification factor, K44, which has a value of 1.25, i.e. K 4 6 = 1.25. It is recommended that the diameter of the hole in the steel plate should not be greater than the diameter of the nail. 3. ~lywoodtotimbernailed joints The basic single shear lateral loads for single nails in a plywoodtotimber joint, where the nails are driven through the plywood at right angles into the side grain of timber in service classes 1 and 2, are given in Table 56 of the code (reproduced here as Table 9.6). For the basic loads inTable 56 of the code to apply, the plywood should be selected from those given in Section 4 of the code,and the nails should be fullyembedded and shouldhave an overalllength not less than those tabulated. 4. Temperedhardboardtotimbernailed joints The basic single shearlateral loads for single nails ina tempered hardboardtotimber joint, where the nails are driven through the tempered hardboard
Table 9.6 Basicsingleshearlateralloadsforroundwirenailsinaplywoodtotimber
joint,
(Table 56, BS 5268 :Part 2) Plywood
Nail __
.~
Strength Nominal Length (mm) (mm) thickness Softwoods (mm)
s
(N)
load, lateralshear single Basic ~
Diameter
C14
C16/18/22
C24
C27/30/35/40 D30/35/40 D50/60/70
groupa Plywood groupa Plywood I
I1
I
I1
I
I1
I
I1
I
I1
I
I1
6
2.65 3 .OO 3.35 3.75 4.00
40 50 50 75 75
207 252 30 1 344 360
216 262 311 373 414
212 258 308 344 360
22 l 268 319 382 424
220 268 318 344 360
230 279 332 399 438
224 273 318 344 360
234 284 339 407 438
267 295 318 344 360
28 1 346 387 419 438
270 295 318 344 360
296 358 387 419 438
9
2.65 3 .OO 3.35 3.75 4.00
40 50 50 75 175
218 26 l 308 366 405
230 280 328 387 427
223 267 315 374 414
24 1 286 335 396 436
23 1 277 327 389 430
250 297 348 412 454
235 28 1 332 395 438
255 302 355 419 463
274 333 397 479 518
301 362 429 514 57 1
286 347 415 495 518
315 379 450 539 599
12
2.65 3.00 3.35 3.75 4.00
45 50 50 75 75
240 28 1 320 383 42 l
256 306 336 410 449
244 287 334 391 430
268 312 353 419 459
253 297 345 406 446
278 324 374 435 477
256 302 351 412 453
282 329 380 443 485
297 353 41 5 492 545
330 389 454 535 590
309 367 432 513 567
345 406 474 559 616
15
2.65 3 .OO 3.35 3.75 4.00
50 65 65 75 75
266 308 353 408 445
285 337 383 440 478
27 1 314 360 416 454
298 343 39 1 449 488
280 324 372 43 1 470
310 356 405 466 506
284 329 378 438 477
314 36 1 412 473 515
327 382 442 518 568
365 424 487 566 619
339 397 459 538 59 1
3 80 441 508 590 645
18
2.65 3.00 3.35 3.75 4.00
50 65 65 75 75
272 339 384 439 476
292 361 417 474 51 1
283 345 391 447 485
30 1 369 425 483 522
305 356 404 462 501
314 386 440 500 540
31 1 361 410 469 509
320 393 446 508 549
361 417 477 552 60 1
388 46 1 525 603 655
374 432 495 572 623
409 479 546 627 68 1
21
2.65 3 .OO 3.35 3.75 4.00
50 65 65 75 75
275 352 41 1 474 510
280 360 432 513 55 1
288 360 426 483 520
2C3 369 443 523 562
306 375 440 499 537
314 385 462 54 1 58 1
31 1 382 446 506 545
320 393 47 1 549 590
372 456 517 59 1 64 1
387 479 570 649 70 1
39 1 472 535 612 664
408 504 592 674 727
29
2.65 3 .OO 3.35 3.75 4.00
50 65 65 75 75
233 352 403 509 552
235 3 56 407 515 572
242 360 423 521 573
245 365 426 527 586
261 375 450 542 603
263 380 456 550 61 1
270 382 458 552 614
273 388 465 560 623
372 46 1 557 675 754
380 470 568 690 770
39 1 483 583 707 789
399 494 597 723 808
aPlywood group I comprises: (a)Americanconstructionandindustrialplywood (b)CanadianDouglas fir plywood (c)Canadiansoftwoodplywood (d)Finnishconiferplywood (e)Swedishsoftwoodplywood Plywood group I1 comprises: (a) Finnishbirchfacedplywood (b)Finnishbirchplywood
72
Structural Timber
Design
at right angles into the side grain of timber in service classes 1 and 2, are given in Table 57 of the code. For thebasic loads in Table 57 of the code to apply, the tempered hardboard should be selected from those given in Section5 of the code, and the nails should be fully embedded and should have an overall length not less than those tabulated.
5. ~arti~leboardtoti~ber nailed joints The basic single shear lateral loads for single nails in a particleboardtotimber joint, where the nails are driven through the particleboard at right angles into the side grain of timber in service classes 1 and 2, are given in Table 58 of the code. For the basic loads in Table 58 to apply, the particleboard should be selected from those given in Section 9 of the code, and the nails should be fully embedded and should have an overall length not less than: (1) 2.5 times the nominal particleboard thickness for particleboards in the range 6 mm to 19 mm thick, and (2) .2.0 times the nominal particleboard thickness for particleboards in the range 20mm to 40mm thick.
Round wire nails are weak when loaded axially and, thus, their strength should not be relied on to any great extent. The best resistance is generally obtained when the nails are driven into side grain (Fig.9.7). N o withdrawal load is permitted by the code for nails driven into the end grain of timber. Some ofthe main factors that influence the withdrawal resistance ofa nail include the densityand moisture content of the timberinto which the nail is
F
ta,point
fa,head
F
Fig. 9.7 Axially loaded nails.
awal Basic
Design of TimberConnections
173
able 9.7 Basic withdrawal loads per millimetre of pointside penetration for smooth round wire nails driven at right angles to the grain (Table55, BS 5268 :Part 2)
Nail diameter (mm.)
2.65 3 .OO 3.35 3.75 4.00 4.50 5 .OO 5 60 6.00 6.70 8 .OO a
C 14
C 161 18/22
1.28 1.45 1.62 1.81 1.93 2.17 2.41 2.70 2.90 3.23 3.86
1SO 1.69 l .B9 2.12 2.26 2.54 2.82 3.16 3.39 3.78 4.5 1
C27/30/35/40 C24 D30/35/40 D50/60/70 1.99 2.25 2.52 2.82 3 .OO 3.38 3.76 4.21 4.5 1 5.03 6.01
2.42 2.74 3.06 3.42 3.65 4.1 1 4.56 5.1 1 5.48 6.12 7.30
5.86 6.64 7.41 8.30 8.85 9.96 1 1.06 12.39 13.28 14.82 17.70
9.80 1 1.09 12.39 13.87 14.79 16.64 18.49 20.71 22.19 24.78 29.58
The pointside penetration of the nail should not be less than 15 mm driven and the nail type. As mentioned earlier, the type and surface condition of nails enhancestheir performance, for example, cement or polymer coated, annular ringed shank and helicalthreaded shank andsquare twisted nails all perform better under withdrawal loading than round wire nails. In addition such improved nailsare less affected by changes in the moisture content of the timber. The basic withdrawal loads for single nails driven at right angles to the side of timber in service classes 1 and 2 are given in Table 55 of the code (reproduced here as Table 9.7). "he values apply to each 1 mm depth of actual pointside penetration achieved; wherethe pointside penetration of the nail should not be less than 15mm. For the threaded part of annular ringed shank andhelical threaded shank nails, BS 5268 :Part 2 recommends that the basic withdrawal loads given in Table 55 should be multipliedby the improved nail withdrawal modification factor, &S, which has a value of 1.5, i.e. &S = 1S .
Clause 6.4.9 of BS 5268 :Part 2 states that the permissible load for a nailed joint should be determined as the sum of the permissible loads for each nail in the joint, where each permissible nail load, FUh, should be calculated from the equation: ~
u
=hF x &8&9KSO
(9.3)
174
Structural Timber Design
where: F isthebasicload for a nailundershearloading(Table 54) or withdrawal loading (Table 55) and is taken from Clauses 6.4.4, 6.4.5.2, 6.4.6.2,6.4.7.2 or 6.4.8.2ofthecodewhichweredescribedearlier. Therefore F is given by: (a) for a nail under single or multiple shear lateral load,
(b) for a nail under axial (withdrawal)load with a pointside penetration depth of tpoint,
where: nshear
=number of shear planes per nail 1.15, the predrilling factor for softwoods in strength classes C14 to C40 and TR26
K&edrilf=
single shear = (the lesserof
ta,head ta,inner tstanhrd
’0.85
*
tstandard
lapint
’tstandard
or 1)
for a nail in multiple shear =0.7, for nailing into the end grain K& = 1.2, improved nail shear lateral modification factor &S = 1.5, improved nail withdrawal modification factor = 1.25, steel platetotimber modification factor. K4, is the modification factor for duration of loading
where: K4, = 1.00, for longterm loads = 1.25, for tempered hardboardtotimber joints for mediumterm loads = 1.4, for particleboardtotimber joints for mediumterm loads = 1.12, for other than tempered hardboardtotimber and particleboardtotimber joints for mediumterm loads = 1.62, for tempered hardboardtotimber joints for short and very shortterm loads =2.10, for particleboardtotimberjoints for short and very shortterm loads = l .25, for other than tempered hardboardtotimber and particleboardtotimber joints for short and very shortterm loads.
Designof TimberConnections 1yQ9
175
is the modification factor for moisture content where: &9 = 1.00, for lateral loads in joints in service classes 1 and 2 =0.70, for lateral loads in timbertotimberjoints in service class3 = 1.00,for lateral loads using annular ringed shank nails and helical threaded shank nails in all service class conditions = 1.00, for withdrawalload in all constant service classconditions =0.25, for withdrawal loads wherecyclicchangesinmoisture content can occur after nailing.
K50
is the modificationfactor for the number of nails of the samediameter in each line, acting in shear, for an axially loaded member, loaded parallel to the line of nails where: K50= 1.OO, for the number of nails in each line <10 =0.9, for the number of nails in each line2 10 = 1.00, for all other loading cases.
9.8 Screwed joints Woodscrews are used in place of nails in applications requiring higher capacities, inparticular in situations where a greater resistanceto withdrawal is required.In general withdrawal resistanceof screws is2 to 3 times that of nails of a similar size. They can be usedfor timbertotimber joints but they are especially suitable for steeltotimber and paneltotimber joints. Screws should always be inserted by threading into predrilled holes in the timber usinga screwdriver and nothammered, and the basic values given in the code are based on this assumption. Otherwisethe loadcarrying capacity, mainly the withdrawal resistance, will decrease significantly. The for holethe shank should havea diameter equal to the shank diameter and should have the same length as the shank. The pilot hole for the threaded portion of the screw should have a diameter of about half the shank diameter. The most common types of wood screws are the countersunk head, round head and coach screws. These are illustrated in Fig. 9.8. The design recommendations given in BS 5268 :Part 2, Clause 6.5 apply to steel screws which conformto BS 1210 :1963. The diameter relates to the smooth shank, which rangesfrom 3.457.72 mm for countersunk and round headedscrews.Coachscrews, in general, range frkm 62Omm in shank diameter. Coach screws, with nutlikehead, require a washer and should always be threaded into predrilled holes. They are available in lengthsof 25300 mm and are most suitable in large connections to hold timber connectors such as dowels in placeor in situations where the use of bolts is not suitable due to onesided access.
Structural Timber Design
(a) Countersunk head
(c) Coach
(b) Round head
Common types of wood screws.
s ~ ~lat~ral a r ~ o a ~ s
1. ~imbertotimberscrewed joints The basic single shearlateral loads for single screws with a minimum tensile strength of 550 N/mm2, threaded at right angles to the side grain of timber in service classes1 and 2, are given inTable 60 of the code (reproduced here as Table 9.8). For screws insertedinto the end grainof timber, the values given inTable 60 should be multiplied by the end grain modificationfactor, K&, which has a value of 0.7, i.e. K 4 3 = 0.7. For the basic loads in Table 60 to apply, the headside member thickness should be equal to the value given in that table and the penetration of the screw in the pointside member should be at least twice the actual headside member thickness. Otherwise, the basic shear load should be multiplied by the ratio of the actual headside thickness to the standard thickness given in Table 60, provided that the pointside screw penetration is at least twice the actual headside thickness. Table 9.8 Basic single shear lateral loads for screws insertedinto predrilled holes in a timbertotimber joint (Table 60, BS 5268 :Part 2) Basic single shear lateral load, (N)
Screw
headside NumberShankdiameterthickness (mm) (mm) 3.45 4.17
6 8 10 12 14 16 18
253 12 390 16 4.88 546 22 5.59 35 6.30 1002 38 44 7.01 1469 47 7.72
Strength class C14 C16/18/22 C24 C27/40/35/40 D30/35/40 D50/60/70
819 1251
271 407 572 863 1056 1319 S549
303 440 622 952 1163 1429 1694
3 S4 456 647 985 1217 1470 1742
396 584 846 1179 1457 1759 2085
455 677 993 1306 1613 1948 2308
Designof TimberConnections
177
F" to, head ta,yornt
into side grain Figg. 9.9 Member thicknessesin screwed joints.
F
Therefore, with referenceto Fig. 9.9, it is possibleto define a modification factor for thicknessoftheheadsidemember(i.e.bearinglength)of a screwed connection, Kscrewbearing, as: &crewbearing
tu,head
tstundard
The m i ~ m u mheadside member thickness should not be less than twice the shank diameter. In addition no increase in basic load is permitted for thicknesses or penetrations greater than the standard values. 2. Steel platetotimber screwed joints BS 5268 :Part 2, Clause 6.5.5 recommends a 25% higher load capacity for steelplatetotimberscrewed joints, provided that steelplateshave a minimum thickness equal to the greater of 1.2 mm or 0.3 times the shank diameter of the screw, Where a predrilled steel plate or component of such thickness is screwed to a timber member, the basic single shear lateral load given in Table 60 should bemultipliedby the steeltotimbermodification factor, which has a value of 1.25, i.e. &6 = 1.25.
3. ~l~woodtotimber screwed joints The basic single shearlateral loads for single screws in a pl~oodtotimber joint, where the screwsofminimumlength are threaded through the plywood at right angles into the side grain of timber in service classesl and 2, are given in Table 62 of the code. For the basic loads in Table 62 to apply, the plywood should be selected from those given in the Section 4 of the code and the diameter of the predrilled hole in the plywood should not be greater than the shank diameter of the screw.
The basic withdrawal loads for single screws inserted at right angles to the side of timber in service classes 1and 2 are given in Table 61 of the code
178
Structural Timber Design
Table 9.9 Basic withdrawal loads per millimetre of pointside penetration for screws turned at right angles to grain (Table 61, BS 5268 :Part 2)
Basic
Screw
withdrawal loada, (N/mm) Strength class
3.45
6 8 10 12 14 16 18 a The
4.17 4.88 5.59 6.30 7.01 7.72
10.96 12.63 14.21 15.73 17.21 18.64 20.04
12.03 13.87 15.61 17.28 18.90 20.48 22.01
14.29 16.47 18.53 20.52 22.44 24.31 26.14
16.06 18.51 20.83 23.06 25.22 27.33 29.38
27.3 1 3 l .49 35.43 39.23 42.9 1 46.49 49.97
37.17 42.85 48.22 53.39 58.40 63.26 68.01
penetration of the screw point should not be less than 15 mm
(reproduced here as Table 9.9). The values apply to each 1mm depth of actual pointside penetration achievedby the threaded part of the screw; where the penetration of the screw point should not be less than 15 mm. No withdrawal load is permitted by the code for screws inserted into the end grain of timber. 9.8.3 Permissible load for a screwed joint
Clause 6.5.7 of I3S 5268 :Part 2 recommends that the permissible load for a screwed joint should be determined as the sum of the permissible loads for each screw in the joint, where each permissible screw load, Fah, should be calculated from the equation:
where: F is thebasicload for a screw undershear(Tables60 or 62) or withdrawal loading (Table61) and is taken from Clauses 6.5.4, 6.5.5.2 and 6.5.6.2 of the code, which were described earlier. Therefore F is given by: for a screw under shear lateral load,
for a screw under axial (withdrawal) load with a pointside threaded penetration depth of tpoint,
Designof TimberConnections
179
where: tathead
Kscrewbearing
="
tstandard
K43 =0.7, for screwing into the end grain = 1.25, steel platetotimber modification factor. K52
is the modification factor for duration of loading where: K52 = 1.00, for longterm loads = 1 12, for mediumterm loads = 1.25, for short and very shortterm loads.
K53
is the modification factor for moisture content where: K53 = 1.00, for a joint in service classes 1 and 2 =0.70, for a joint in service class 3.
K54
isthemodification factor for thenumber ofscrewsof the same diameter in each line, acting in shear, for an axially loaded member, loaded parallel to the line of screws where: K54 = 1.00, for the number of screws in each line 10 =0.9, for the number of screws in each line 2 10 = 1.00, for all other loading cases.
9.9
Bolted and dowelled joints Dowels are cylindrical rods generally made of steel with a smooth surface, and are available in diameters ranging from 630mm. Bolts are threaded dowels with hexagonal or semisphere heads and hexagonal nuts. Dowels and bolts are commonly used in connections requiring higher shear capacity than the nailed or screwed joints, see Fig. 9.10. Threaded dowels with nuts and bolts are also used in axially loaded tension connections. Both dowels and bolts are commonly used to hold two or more members together, but the mostcommonconnectiontypeinvolvesthree or more members in a multiple shear arrangement. The side members can be either timber or steel. When bolts are used, washers are required under the bolt head and under the nut to distribute the loads. The recommended washer diameter is 3 times the bolt diameter ( 3 4 with a thickness of 0.25 times the bolt diameter (0.25d). When tightened,a minimum of one complete thread should protrude from the nut. Dowelled joints are easy to fabricate. They are inserted into predrilled holes with a diameter not greater than the dowel diameter itself. They can produce stiffer joints compared to bolted ones which require 1mm oversize
180
StructuralTimber Design
(a)Dowel
(b) Bolts Fig. 9.10 Typical dowel and bolts.
for clearance; BS 5268 :Part 2 recommends a maximum of 2 mm oversizing for bolt holes. If steel plates are incorporated in a dowelled connection, a 1m m clearance must be provided and due allowance should then be made for the additional deformation that may occur. ask s i ~ ~ shear l e lateral loads
l. Timbertotimber bolted or dowelled joints The basic single shear lateral loads for a single bolt or dowel, with a minimum tensile strength of 400N/mm2, in a twomember timber joint, in which the load acts perpendicular to the axis of the bolt or dowel, and parallel or perpendicular to the grain of the timber, are given in Tables 6368 of BS 5268 :Part 2. Table 64 is reproduced here as Table 9.10. The basic loads appropriate to each shear plane in a t~reememberjoint under the sameconditions are given in Tables6974 of the code. Table 70 is reproducedhere as Table9.1 1. Separate loads are tabulated for long, medium and shortduration loads and therefore there isno need to further modify the basic load values for duration of loading. Unlike nails and screws, bolts and dowels have loading capacities that vary dependingon the direction of the applied load with respectto the grain direction of the timber. This is why the code provides separate basic shear values for loadings parallel and perpendicular to the grain of timber, and recommends the use of Hankinson’s equation for determination of basic loads, F, in situations where the applied load is inclinedat an angle a to the grain of the timber, see Fig. 9. l l, hence:
Design of TimberConnections
1
Basic singleshearloads for one 4.6 grade steel bolt or dowelin a twomember timbertotimber joint: C16/18/22timber (Table 64, BS 5268 :Part 2)
Table 9.10
Minimum duration Load
C 161 18/22
member shear single thickness Basic (m)
(kN) load, Direction of loading Parallel to the grain Perpendicular to Bolt or doweldiameter,(mm)Bolt
M8 M12 M16 M20 M24 M8
the grain or doweldiameter,(mm) M12 M16 M20 M24
0.29 0.40 0.63 0.79 0.85 1.08 1.30 1.39 1.39
0.40 0.49 0.56 0.55 0.67 0.77 0.87 1.06 1.22 1.09 1.34 1.54 1.17 1.43 1.64 1.49 1.83 2.09 1.79 2.19 2.51 2.41 2.95 3.39 3.00 4.47 5.13
0.61 0.84 1.34 1.69 1.80 2.30 2.76 3.72 5.64
Longterm
16 22 35 44 47 60 72 97 147
0.32 0.44 0.70 0.88 0.94 1.20 1.44 1.47 1.47
0.46 0.63 1.00 1.26 1.35 1.72 2.07 2.78 3.23
0.58 0.80 1.28 1.61 1.72 2.19 2.63 3.54 5.37
0.70 0.96 1.52 1.91 2.04 2.61 3.13 4.22 6.39
0.79 1.09 1.73 2.18 2.33 2.97 3.57 4.81 7.29
Mediumterm
16 22 35 44 47 60 72 97 147
0.41 0.57 0.90 1.13 1.21 1.54 1.63 1.63 1.63
0.59 0.81 1.29 1.62 1.73 2.21 2.66 3.58 3.59
0.75 1.03 1.64 2.07 2.21 2.82 3.38 4.55 6.24
0.89 1.23 1.96 2.46 2.63 3.35 4.02 5.42 8.22
1.02 0.37 0.51 1.40 0.51 0.70 2.23 0.81 1.12 2.80 1.02 1.41 2.99 l .09 1.50 3.82 1.39 1.92 4.59 1.55 2.30 6.18 1.55 3.10 9.37 1.55 3.34
0.63 0.86 1.37 1.72 1.84 2.35 2.82 3.79 5.69
0.72 0.99 1.57 1.97 2.1 1 2.69 3.23 4.35 6.60
0.79 1.09 1.73 2.17 2.32 2.96 3.55 4.79 7.26
Shortterm and very shortterm
16 22 35 44 47 60 72 97 147
0.46 0.64 1.01 1.27 1.36 1.73 1.73 1.73 1.73
0.66 0.91 1.45 1.83 1.95 2.49 2.99 3.81 3.81
0.55 1.16 1.85 2.32 2.48 3.17 3.80 5.12 6.62
1.01 1.38 2.20 2.77 2.96 3.77 4.53 6.10 9.24
1.15 1.58 2.51 3.15 3.37 4.30 5.16 6.95 10.54
0.42 0.57 0.91 1.15 1.23 1.56 1.64 1.64 1.64
0.70 0.97 1.54 1.94 2.07 2.64 3.17 4.27 6.04
0.81 1.l1 1.77 2.22 2.37 3.03 3.64 4.90 7.42
0.89 1.22 1.94 2.44 2.61 3.33 4.00 5.39 8.16
0.57 0.79 1.26 1.58 1.69 2.16 2.59 3.49 3.54
where: F 1 =basic load parallel to the grain, obtained from Tables 6374 FL =basic load perpendicular to the grain, obtained from Tables 6374. If a load F acts at anangle p to the axis of the bolt the component of the load perpendicular to the axis of the bolt, (Fsin p), should not be greater
182
Structural Timber Design
Table 9.11 Basic single shear loads for one 4.6 grade steel bolt or dowel in a threemember timbertotimberjoint: C16/18/22timber (Table 70, BS 5268 :Part 2)
C16/18/22
Load duration Minimum outer single Basic member thicknessa
(kN) load, shear
(mm)
Direction of loading Parallel to the grain Perpendicular to the grain Bolt or dowel diameter, (mm)Bolt or dowel diameter, (mm)
M8 M12 M16 M20 M24 M8
M12 M16 M20 M24
Longterm
16 22 35 44 47 60 72 97 147
0.77 1.04 1.10 1.19 1.22 1.38 1.47 1.47 1.47
1.11 1.52 2.29 2.34 2.36 2.50 2.68 3.12 3.23
1.41 1.94 3.09 3.88 3.97 4.06 4.20 4.63 5.60
1.68 2.3 1 3.67 4.62 4.93 6.06 6.14 6.48 7.63
Mediumterm
16 22 35 44 47 60 72 97 147
0.99 1.17 1.28 1.40 1.45 1.63 1.63 1.63 1.63
1.43 1.96 2.58 2.67 2.71 2.94 3.19 3.59 3.59
1.81 2.49 3.97 4.44 4.47 4.63 4.87 5.51 6.24
2.16 2.46 0.90 2.97 3.38 110 4.72 5.38 1.20 5.94 6.77 1.30 6.34 7.23 1.34 6.81 9.23 1.54 6.98 9.53 1.55 7.55 9.97 1.55 9.22 11.55 1.55
Shortterm and very shortterm
16 22 35 44 47 60 72 97 147
1.12 1.24 1.39 1.53 1.59 1.73 1.73 1.73 1.73
1.60 2.20 2.75 2.87 2.92 3.19 3.49 3.81 3.81
2.04 2.43 2.81 3.34 4.46 5.31 4.73 6.68 4.77 7.14 4.98 7.27 5.27 7.49 6.02 8.18 6.62 10.09
1.91 0.70 0.96 2.63 0.96 1.32 4.19 1.03 2.10 5.26 1.10 2.16 5.62 1.13 2.17 7.18 1.27 2.28 8.49 1.39 2.42 8.72 1.39 2.79 9.74 1.39 13.00
2.77 3.81 6.06 7.62 8.13 10.04 10.17 10.73 12.62
1.01 1.17 1.29 1.42 1.47 1.64 1.64 1.64 1.64
aThe corresponding minimum inner member thickness is assumed value
1.18 1.35 1.62 1.85 2.57 2.95 3.23 3.71 3.45 3.96 3.66 5.06 3.76 5.45 4.08 5.66 4.99 16.48
1.48 2.04 3.24 4.08 4.36 5.56 6.67 7.55 8.20
1.23 1.70 2.38 2.45 2.48 2.66 2.87 3.34 3.34
1.51 2.08 3.31 4.04 4.05 4.16 4.32 4.81 5.69
1.73 2.38 3.79 4.77 5.09 6.05 6.15 6.53 7.76
1.91 2.62 4.17 5.24 5.60 7.15 8.33 8.56 9.61
1.39 1.91 2.54 2.63 2.67 2.88 3.13 3.54 3.54
1.70 2.34 3.72 4.29 4.31 4.46 4.66 5.25 6.04
1.95 2.68 2.95 4.27 5.36 5.90 5.73 6.30 6.44 6.58 8.86 7.05
2.15 4.69 8.05
9.18 8.50 10.45
to be double the tabulated
than the basic load given in Tables 6374, modified where appropriate by Hankinson’s equation, see Fig. 9.12. Clause 6.6.4.1of the code specifies that for twomember joints, where parallel members areof unequal thickness, the load for the thinner member
Designof TimberConnections
183
I Load
1'
" J " _
I
J" j _" i
1
.
Applied load is acting perpendicular to the grain of the horizontal member
Load
Applied load is acting an at angle a to the grainof the horizontal member Fig. 9.11
Load at an angle to the grain.
F cos p
F
Fsin p
Fig. 9.1 2 Load at an angle to the of axis the bolt.
should be used. Where members of unequal thickness are joined at an angle, the basic load for each member should be determined, using Hankinson's equation, and the smaller load should be used. For threemember joints, the basic loads given in Tables 6974 apply to joints where the outer members have the tabulated thickness and the inner member is twice as thick. For other thicknesses the load should be taken as
1
Structural Timber
Design
the lesser of the loads obtained by linear interpolation between the two adjacent tabulated thicknesses for each member. The basic load for a joint with more than three members should be taken as the sum of the basic loads for each shear plane, assuming that the joint consists of a series of threememberjoints. 2. Steel platetotimber bolted or dowelled joints BS 5268 :Part 2, Clause 6.6.5.1 recommends a 25% higher load capacityfor steel platetotimber bolted or dowelled joints, provided that steel plates have a minimum thickness equal to the greater of 2.5mm or 0.3 times the bolt or dowel diameter. Where a steel plate or component of such thickness is bolted or dowelled to a timbermember, the basic loads given in Tables 6374 should be multiplied by the steeltotimber modificationfactor, &G, which has a value of l .25, i.e. K46 = 1.25. In dowelled joints where the steel plates are used as the side (outer) members of the joint, the steel plates should be secured in position adequately, i.e. by threading and applying nuts to the dowel ends or by welding dowel ends to the steel plates. No increase should be made to the basic load when the timber is loaded perpendicular to the grain. le load for a ~ o l t e or d do welledj o j ~ t
Clauses 6.6.6 and 6.6.7.4 of BS 5268 :Part 2 state that the permissible load for bolted or dowelled joints should be determined as the sumof the permissible loads for each bolt or dowel in the joint, where each permissible bolt or dowel load, F u h , should be calculated from the equation:
where: F is thebasicload for a bolt or dowel under lateral loading per shear plane (Tables 6374) and is taken from Clauses 6.6.4, 6.5.5 and 6.6.7 of the code which were described earlier. Therefore F is given by: (a) for a bolt or dowel in a twomember joint, F = F b a i c (Tables 6368)
X K46
(9.11)
(b) for a bolt or dowel in a symmetrical threemember joint F=
(c)in
Fbaic
(Tables 6974)
X
2X
K46
(9.12a)
general, for a bolt or dowelin a joint withthree or more members, F=
x
Fbaic
(Tables 6974)
X &6
(9.12b)
Designof TimberConnections
185
where:
Fbmfc =sum
of the modified basic loads platetotimber modification factor.
K46 = l .25,steel
It is to be noted that in all cases,if load is applied at an angle a to the grain of timber, Fbmic is obtained from Fbmic K56
I==
FIIFJFL cos2a
F1 sin2a
+
is the modification factor for moisture content where: K56 = 1.00, for a joint in service classes 1 and 2 =0.70, for a joint made in timber of service class 3 and used in that service class =0.40, for a joint made in timber of service class 3 but used in service classes l and 2.
K57
is the modification factor for the number of bolts or dowels of the samediameterineachlineactinginshear for an axiallyloaded member, loaded parallel to the line of bolts or dowels, where: K57
3(n 1) = 1
for n < 10
= 0.7 = 1.0
for n 2 10 for all other loading cases
100
where n is the number of bolts or dowels in each line.
Structural connections are often required to transmit moments as well as shear andlor axial forces. Figure 9.13 presents some examples of momentresisting timber joints such as (a) a splice joint in a continuous beam, (b) a momentresisting column baseand (c) a knee joint in a frame. Although in this section a bolted connection is used to illustrate a simplified analysis method for determination of connection moment capacity, the method can similarly be applied to nailed, screwed, or, indeed, dowelled joints. Figure 9.14(a) showsa bolted connection which is subjected to a moment of M , and shear and normalforces of Vand H. The effects of the moment M and the forces V and H, all acting at the geometric centre of the bolt group at C, can be considered separately. M. Foran elastic First considerthe effectof theappliedmoment, behaviour of the connection, we can assume that the moment, M, will induce a shearing force, FM, in any of the bolts perpendicular to the line
""
i 
.
" " " " "
M
M
Fig. 9.13 Momentresisting connections.
(a) Connection subjected to M, Vand H
(b) Effect of the applied moment M
, '
(c) Effect of the applied shear force V
\ '
(d) Effect of the applied normal force H
Fig. 9.14 Momentresisting joint subjected to an applied moment and shear and normal force
Designof TimberConnections
187
joining C to the bolt. Therefore the force FM will be perpendicular to the distance r from the bolt to C (see Fig. 9.14(b)). For equilibrium we have: M=
FMer
(9.13)
If we consider FM = k .r, where k is constant for all bolts, then
M = k x r 2
(9.14)
Thus,
k= M r2 By substituting for k, the force acting on a bolt becomes:
c
FM=
M
due to the moment, M,
'
r2 where: c r 2 = r ~ + r ~ + r ~ + r ~ + r ~ +   * .
(9.15)
As an example, the force acting on the bolt a due to the applied momentM is:
Now considering the effects ofshear*and normalforces V and H, assume that each one is distributed equally among the bolts as a shearing force parallel to the line of action of V and/or H (see Fig. 9.14(c), (d)).For a joint with n number of bolts, the force exerted on a bolt due to the applied shear force V is, V F y =n
(9.16)
the force exerted on a bolt due to the applied normal force H is, H
FH=(9.17) n The total force acting on a bolt can then be calculated as the vectorial Fv FH. For example, the total force acting on the summation of FM, and bolt a, Fa, with its angle ofinclination to the grain of timber, a, is obtained as illustrated in Fig. 9.15. a
, /
I
I
I
I
Fig. 9.15 Vectorial summationof fa,^, FV and FH acting on the bolt a.
1
Structural Timber
Design
With forces defined byequations 9.159.17, moment resistingjoints, with rectangular pattern, can be designed so that the load on the furthest fastener (i.e. fastener a) is not greater than the permissible load value. The total force, Fa, acting on the furthest fastener a can be calculated from:'
acting at an angle a to the grain of timber where: (9.19)
A momentresisting connection with a circular pattern, such as that of joint (c) shown in Fig. 9.13, can be designed considering the load on the fastener located on the longitudinal axis. In such cases the total force on the fastener is given by: (9.20) acting at an angle a where: a = arctan
(
+
FM FV F~
)
(9.21)
It is to be noted that for connections with parallel members, Hankinson's formula (equation 9.9) should be usedfor the calculation of the permissible loads for bolts or dowels loaded at an angle a to the grain of timber. For timbertotimber connections with members perpendicularto each other, a modified version of Hankinson's formula may be used. Therefore: for a timbertotimber boltedor dowelled connection with parallel members or steeltotimber, the basic single shear load is givenby:
for bolted or dowelled connections with perpendicular members,' (9.22)
Timber connectors are the most efficient of all mechanical fasteners,as they improve the transfer of loads by increasing the bearing area between the fasteners and the timber. They are generally in the form of rings, discs or plates,partiallyembeddedinthefaces of adjacent timbermembers, to transmit the load from one to the other. There are several types of timber
Design of Timber Connections
connectors ofwhich the mostcommonlyused are toothedplates, splitrings and shearplate connectors which require bolts to draw the members together and metalplate fasteners suchas punched metalplates with integral teeth and nailplates which allow members to remain in the same plane. BS 5268 ::Part 2 includes rules for the design of joints using bolted timber connectors suchas toothedplates, splitrings and shearplates, together with recommendations on joint preparation, spacing of connectors and appropriate edge and end distances.
Toothedplate connectors are madefrom hot dippedgalvanise rolled mild steel. They are commonly available in circularor square shapes with sizes ranging from 38165mm. Toothedplate connectors are manufactured in either double or singlesided forms, see Fig. 9.16.~oublesided toothedplates are suitable for use in timbertotimber connections, whereas singlesided ones can be used in steeltotimber connections or in pairs put backtoback for timbertotimber joints in situations where it maybe necessary to demount the c~nnection.~ The assembly process involves drilling the bolt hole in the members, then the connector is placed between the timber member(s) and the connection is pressed together, by either a hydraulic pressor with the aid of a ~ghstrength bolt and large washers. After pressing, the highstrength bolt is replaced by the permanent blackbolt with associated washers. Toothedplate connectors are not suitable for timber witha density greater than 500 kg/m3. BS 5268 :Part 2, Clause 6.7 deals with toothedplate connector joints. Bolt and washer sizes suitable for different shapes and forms of toothedplate
Singlesided (a) toothedplate Doublesided (b) toothedplate
Fig. 9.16 Toothedplate connectors.
I 9 0
Structural Timber
Design
connectors are detailed in Table 76 of the code, and recommendations for the spacing, end and edge distances are given in Tables 7779. The basic loads parallel and perpendicular to the grain of the timber for toothedplate connector units are given in Table 80 of the code. The basic loads apply to softwoods and hardwood, in strength classes C14 to C40, TR26 and D30 to D40, providedthat the teeth of the connectorscan be fully embedded. Permissible load for a toothedplate connectored joint
Clause 6.7.6 of BS 5268 :Part 2 recommends that the permissible load for a toothedplate connectored joint be determined as the sum of the permissible loads for each unit in the joint, where each permissibleload, Fah, should be calculated from the equation: Fadm
= x KSSK59&0&il
(9.23)
where: I; is the basic load for a toothedplate under single shear loading (Table 80) It is to be noted that if load is applied at an angle M, to the grain of timber, F is obtained from
is the modificationfactor for duration of loading where: K58 = 1.OO, for longterm loads = 1.12, for mediumterm loads = 1.25, for short and very shortterm loads. K59
is the modification factor for moisture content where: = 1.00, for a joint in service classes 1 and 2 =0.70, for a joint in service class 3.
K60
is the modification factor for end distance where: 1y60 = 1.00 if the standard edge and end distances and spacing given in Tables 7779 are used, otherwisethe lesser of Kc and Ksas given in Tables 81, 82 and 79 of the code should be used. is the modification factor for the number of connector units of the same size in each linefor an axially loaded member, loaded parallelto the line of connectors
Designof TimberConnections
191
where: 3(n l ) K(j1 = l100 =0.7 = 1.0
for n <10 for n 2 10 for all other loading cases
where n is the number of connector units in each line.
9.11.2
litring and shearplate connectors
Splitring and shearplate connectors are manufacturedfrom hot rolled carbon steel, pressed steel, aluminium cast alloy or malleable iron. They require precision grooving and boring in timber members for assembly, and can provide a large shear capacity, much greater than that achievable by toothedplate connectors. Splitringconnectors have a split in their ring to provide bearing on both the inner core and the centre surface of the groove and are suitable for timbertotimber connections only, see Fig. 9.17. They are provided in two sizes of 64 mm and 102 mm diameter and require, respectively, M 12 and M20 bolts with associated washers. BS 5268 :Part 2, Clauses 6.8 deals with splitring connector joints. Bolt and washer sizes and dimensions for suitable circular grooves for splitring connectors are detailed in Tables 83 and 84 of the code. The recommendations for the end and edge distances and spacings for splitring connectors for standard dimensions are given in Tables 8587, and for those less than the standard dimensions in Tables 89 and 90. The basic loads parallel and perpendicular to the grain of the timber for splitring connector units are given in Table 88 of the code. The shearplate connectors can be used for timbertotimber joints, in pairs put backtoback, and also as a single unit for steeltotimberconnections because they can remain flush withthe surface of the timber member, see Fig. 9.18. In shearplate connections, after transfer of the load from one
Fig. 9.17 Splitring connector.
192
Structural Timber
Design
Shearplate connectors.
member into its connector, the bolt is loaded via bearing stresses between the shearplate and the bolt, and the load is then transferred into the other member through the second shearplate or directly into a steel member. It is to be noted that in all connectored joints, due to a small tolerance required for insertion of the bolt, some initial slipis expected and therefore an allowance should be made for this in design calculations. S 5268 :Part 2, Clause 6.9 deals with shearplate connector joints. and washer sizes and dimensions for suitable circular grooves for splitring connectors are detailed in Table 91 and Figure 8 of the code. The recommendations for theend and edgedistances and spacings for shearplate connectors for standard dimensions are given in Tables 8587, and for less than the standard dimensions in Tables 89 and 90. The basic loads parallel and perpendicular to the grain of the timber for shearplate connector units for both timbertotimber and steeltotimber joints are given in Table 92 of the code.
Clauses 6.8.5and 6.9.6 of BS 5268 :Part 2 state that the permissible load for such connectoredjoints should be determined as the sum of the permissible loads for each unit in the joint, where each permissibleload, F a h , should be calculated from the equation: for a splitring connector unit,
Designof TimberConnections
1
for a shearplate connector unit,
Fah = F x K66K67K68K69 5 limiting (thevalues Table given (9.25) 93) in is the basic load for a splitring or shearplate connector unit under single shear loading (Tables 88 and 92 respectively). It is to be noted that if load is applied at an angle a to the grain of timber, F is obtained from:
FPJ
F=
F), sin2a
+ FJcos2a
are the modification factors for duration of loading where: K62166 = 1.oo, for longterm loads = 1.25, for mediumterm loads = 1.50, for short and very shortterm loads. are the modification factors for moisture content where:
= 1.00, for a joint in service classes 1 and 2
=0.70, for a joint made in timber of service class 3,
are the modification factors for end distance where: = 1.00,provided that the standard edgedistance, end distance and spacings given in Tables 8587 are used, otherwise K64 and are equal to the lesser of Ks,K c and KD as given in Tables 87, 89 and 90 of the code. ,
are the modification facto& for the number of connector units of the same size in,,each line for an axially loaded member, loaded parallel to the line of connectors where: &55/69
3(n 1) = 11 00
for n < 10
=0.7
for n 2 10 for all other loading cases
= 1.0
where n is the number of connector units in each line.
tal fate connectors These are lightgauge metalplates with integral projections (teeth) punched out in one directionand bent perpendicular to the base ofthe plate, and are
194
StructuralTimber Design
(a) Punched metalplate
(b) Nailplate
Fig. 9.19Metalplateconnectors.
referred to aspunched ~ e t a l  ~ z a t eMetalplates s. with preformed holesthat depend on nails for load transfer may also fall into this category (and are referred to as nail~l~tes), see Fig. 9.19. Both typesare generally usedto join two or more sections of timber of the same thickness in one plane. Punched metalplates and nailplates are generally manufactured from pregalvanisedmildsteel strips or stainlesssteelstrips.Theirthickness ranges from 0.9mm to 2.5 mm for punched metalplatesand aroundl .Omm for nailplates. Manufacturers of nailplates recommendthe use of improved nails, such as square twisted nails, which should be a tight driven fit in the plate holes. Punched metalplate fastenersare suited to factory prefabrication and are able to transfer member forces via small connection areas. They are widely used to form truss connections as well as joints in many other planeframe timber structures. Loads are transferred in a punched metalplate from the timber member into the plate teeth, then from the teeth into the steel plate and across the joint interface, then back down into the teeth in the other member. Joints are designed and constructed with pairs of plates on their opposite sides.173 Many different forms of punched metalplate fastener have been developed, involving a variety of nail patterns, nail sizes and lengths and nail shapes. Most of these are manufactured and used by individual manufacturers intheirprefabricatedtimber products, for example,rooftrusses. As such, their strengthand slip characteristicsare not available in the literature, but they still haveto satisfy the recommendationsof the related British Standards, in particular BS 5268 :Part 3 :1998 Code of practice for trussed rafter roofs.
Design of Timber Connections
195
9.12 Glued joints Another formof structural jointing system in timber is gluing, with a history goingbackmanycenturies.Ifgluingisusedcorrectly it can offer the advantages of strong, rigid and durable joints with a neat appearance. Since the beginning of this century modern synthetic adhesives have been used successfullyinmany structural applications,includinggluedlaminated constructions and fingerjointing systems. Nowadays there areseveraltypes and forms ofadhesivesavailable that can be applied to timber and wood based products; but those for use in the assembly of structural components should comply with the requirements of BS EN 301 :1992 Adhesives, phenolic and aminoplastic~for loadbearing timber structures: classijication and performance requirements, and/ or BS1204:1993 Specijication for type MR phenolicandaminoplastic synthetic resin adhesives for wood. In glued joints, the glue should always be stronger than the surrounding timber (i.e. if it is tested to destruction, failure should occur in timber and not in the thickness ofthe glueline). A glued joint has a much lowerstrength in tension than in shear and for this reasonjoints should be designed so that any tension effect on the glueline is prevented. BS 5268 :Part 2, Clause 6.10 recommends that eccentricglued lap joints whichinduce a tensilecomponent of stress perpendicular to the plane of the glueline should not be permitted. Therefore overlapped glued joints should always have symmetrical member and loading arrangements to induce pure shear at gluelines, see Fig. 9.20. 9.1 2.1
Durability classification
The performanceof glues, particularly in relationto durability, is an important criterion for an appropriate selection of a glue type and specification by the designer. BS 5268:Part 2, Clause 6.10.1.2 recommendsthat the adhesive used should be appropriate to environmental conditions in which the joint will be used. Table 94 of the code provides details of four exposure categories, the permitted adhesive typesand the British Standard classifications and references.
Fig. 9.20 An overlapped glued joint.
As mentioned earlier, if a glued timber joint complies with the required recommendations, the glueline(s)will be subjected to pure shear and will be stronger than the surrounding timber. Therefore the design capacity of a glued connection can be calculated as the permissible shear stress on the interfaces of the timber or wood panel products being glued. Service class, loadduration and loadsharing modification factors (K2, K3 and Kg) apply because stresses in the connection are those of timber stresses. Theglued joint mustbeheldinclose contact during curing, and the pressure is best applied by clamps which are removed after curing, or by nails or staples which are leftinthefinalassembly. In theeventwhere bonding pressure is generatedby nails or staples, the code recommends that thepermissibleshearstressbemultiplied by the naillgluemodification factor, K70, which has the value 0.9, i.e. K70 = 0.9. In the event ofa glued joint being designedso that one face is loaded at an angle a to the direction of the grain, the permissible shear stress for the glueline should be calculated using the following equation: T~ = T a b , / ( 1 0.6’7sin a)
where a =ab,/
(9.26)
is the angle between the direction of the load and the grain of the piece is the permissible shear stress parallel to the grain for the timberand may be determined from:
The permissible shear stress for a joint made with different timbers and or wood based panel products (i.e. plywood or particleboard) should be taken as the permissible shear stressfor the timber or the approp~aterolling shear stress for the wood based product, whichever has the lower value.
erences Timber Engineering STEP I , Section C (l 995) Centrum Hout, The Netherlands.
TRADA (1995) Doweltype fasteners for structural timber. Wood I n f o r ~ a t i o ~ ? Section 2/3, Sheet 52. Timber Research and Development Association (TRADA), High Wycombe. TRADA (1996) Connectors and metal plate fasteners for structural timber. Wood Information, Section 2/3, Sheet 53. TRADA, High Wycombe.
Design of TimberConnections
197
S
The nailed joint shown in Fig. 9.21 is subjected to mediumterm tensile loading under e joint comprises six 3.00 mm diameter x 60 mm long round wire nails acting in single shear in C18 timber. Determine the loadcarrying capacity of the connection.
a 4 f 3 4a 2 4 l
I
I
1
I I
4
4
Ill
4 I
4
l l
I
+ " + a3 aI
I
I
97 m
1 Nailed joint details (Example 9.1 ).
Force, k N Length, m Crosssectional dimensions, mm Stress, Nmm"2
Nail diameter, d Nail length, Znaif NO.of nails, nnail Head side penetration, ta,head Pointside penetration, ta,point
BS5268 : Part 2, Table 54 Strength class, C18 timber Standard penetration, tstandard Basic single shear lateral load, Fbasic
N :=newton kN :=lo3 N Direction parallel to grain, // Direction perpendicular to grain, pp +
198
Structural Timber Design
3. ~ o ~ i f i c a t ifactors on
BS S268 : Part 2 , Clause 6.4 Number of shear planes, nshear No predrilling Nail bearing is the lesser of
and
Knail.bearing.2
:=
ta.point ~
tstandard
Knail.bearing.2 = 0.75
Mathcad command to select the minimum value
KnaiLbearing := [KnaiLbearing.1Knail.bearing.21 Pnill(Kmil.bearhg) = 0.75 0.66 (Clause 6.4.4.1)
>
okay Thus KnaiLbearing := min(Knail.bearing) K 4 3 :=1 Side grain nailing, K43 does not apply KM :=1 Ordinary nails, KM does not apply Timbertotimber nailing, K46 does not apply Load duration, mediumterm K48 :=l .l2 Moisture content, K&, service K49 := 1 class 2 No. of nails,
5. Minimum nail spacings
BS5268 : Part 2 , Table 53 For softwoods other than Douglas fir, nail spacing should be multiplied by 0.8 but the edge distance 2 5d. End distance // to grain, a3 Edge distance perpendicular to grain, a4
a3 := 0.8 w20. d a 3 = 48 o m m a 4 := 5 . d a 4 = 150mm
Designof TimberConnections199
Distance between lines of nails perpendicular to grain, a2 Distance between adjacent nails in any one line // to grain, a1

a2 :=0.8 10 d a2 = 24 o mm
a1 :=0.8v20 d a1 = 48 o mm
6. Joint slip
BS5268 : Part 2, Clause 6.2 and Table 52 C l 8 timber p := 320 kgm3 Characteristic density (Table 7) Slip modulus: Since the equation for slip modulus is an empirical one (i.e. with hidden units in the coefficientpart), for Mathcad to produce correct units, multiply it by the units shown inside the parentheses; or alternatively, use the equation without units for p and d. 1 1.5 d0.8 (kg1.5 .m2.7 .N ,105.4) Thus, slip modulus, K,,, &er := B P Kser= 551.42o N mm"' *
Slip per nail, U, under permissible load U
= 0.4'7 o mm
heref fore the load capacity of the joint is 1.55 kN with a joint slipof 0.47 mm € x a ~ ~9.2 l e ~ e s i of ~ ant i ~ b e r  t o  t i ~ bconnection er with nail~n the end ~ r a i ~
Thenailedconnectionshown in Fig. 9.22 issubjected to shortterm loading under service class 2 conditions. The joint comprises two 3.75mm diameter x 65 mm long round wire nails acting in single shear in C22 timber. Determine the load capacity of this connection. P Nails driven into
P Fig.9.22 Nailed joint details (Example 9.2).
2
Structural Timber
Design
Force, kN Length, m Crosssectional dimensions,m m Stress, Nmm2 1
m
N :=newton :=103. N Direction parallel to grain, // Direction perpendicular to grain, p p
~ e o ~ e t r i cdetails al
Nail diameter, d Nail length, lnail NO. of nails, nnail Head side penetration, ta,head Pointside penetration, ta,poht
BS'5268 : Part 2, Table 54 Strength class, C22 timber Standard penetration, tsgandard Basic single shear lateral load, Fbasic a ~ jfactors o ~
BS5268 : Part 2, Clause 4 . 6 Number of shear planes, nshear No predrilling
Nail bearing is the lesser of
and
Knail.bearing.2 Knail.bearing.2
athcad command to select the minimum value Thus End grain nailing, does not apply Ordinary nails, KM does not apply Timbertotimber nailing, K46 does not apply
:=
ta.point p
tstandard
0.71
Knail.bearing := [Knail.bearing.l Knail.bearing.2] min(K,ail.bearing) = 0.71 0.66 okay
>
KnaiLbearing := min(Knail.bearing) K 4 3 := 0.7 K44
:= 1
Design of TimberConnections
201
Load duration, &g7 shortterm &8 := 1.25 &9 := 1 Moisture content, &g, service class 2 No. of nails,
BS5268 : Part 2, Clause 6.2 and Table 52 C22 timber Characteristic density (Table 7) p := 340kg
 m3
Slip modulus: Since the equation for slip modulus is an empirical one (i.e. with hidden units in the coefficient part), for Mathcad to produce correct units, multiply it by the units shown inside the parentheses; or alternatively, use the equation without units for p and d. Thus,slipmodulus,
Kser Kser
:= p1.5 .d0.8 .(kgl.5 .m2.7 .N 105.4) K'er = 721.94 o N mm' Fah U := *
*
Slip per nail, U , under permissible load
&er
U
= 0.38 o mm
le
The nailed joint shown in Fig. 9.23 is subjected to longterm tensile loading under service class 1 conditions. The joint comprises six 3.35 mm diameter x 100 mm long helicalthreaded shank nails acting in double shear in C18 timber. Determine the load capacity of this connection.
Force, kN Length, m Crosssectional dimensions, mm Stress, NmmV2
N :=newton kN :=lo3 N Direction parallel to grain, // Direction perpendicular to grain, pp
202
Structural Timber
Design
P/2
P/2 33 m a4
J
a2 Y f
~,
i
o
.
+
4
l
a3 a1
Fig. 9.23 Naileld joint details (Example9.3).
1 . ~ e o ~ e t details ~ i ~ a l Nail diameter, d Nail length, lnail NO. of nails, nnail Head side penetration, ta,head Inner member, ta,inmr Pointside penetration, t a p k t
135'5268 : Part 2, Table 54 Strength class, CS8 timber Standard penetration, tstandard Basic single shear lateral load, Fbasic dificati~ factors ~
BS5268 : Part 2, Clause 4.6 Number of shear planes, nshear No predrilling Nail bearing is the lesser of
and
72 m m
Design of TimberConnections
and
Knail.bearmg.3
:=
203
ta.point ~
tstandard
Knail.bearing.3 = 0.5 Mathcadcommand to select KnaiLbearing := [Knail.bearing.1 Knail.bearing.2 Knail.bearing.31 min(Knail.bearing) = 0.5 (Clause 6.4.4.1) okay the minimum value KnaiLbearing := min(Knail.bearing) Thus &3 := 1 Side grain nailing, l&3 does not apply Improved nails, KM does KM := 1.2 not apply Timbertotimber nailing, &6 := 1 does not apply Load duration, &S, longterm &8 := 1 Moisture content, K&, service K 4 9 := 1 class 2 No. of nails,
4. Permissibleload
5 . Minimum nail spacings
BS5268 : Part 2, Table 53 For softwoods other than Douglas fir, nail spacing should be multiplied by 0.8 but the edge distance 2 5d.
End distance // to grain, a a3 3 a3
Edge distance perpendicular to grain, a4 Distance between lines of nails perpendicular to grain, a 2 Distance between adjacent nails in any one line // to grain, a1
:= 0.8 20 d = 53.6omm
a 4 :=5 . d
= 16.75 o mm a 2 := 0.8 10 d a 2 = 26.8 o m m a1 :=0.8 a20 d a1 = 53.6 0 mm
a4
6. Joint slip
BS5268 : Part 2, Clause 6.2 and Table 52 C18 timber p := 320 kg m3 Characteristic density (Table 7)
2
Structural Timber
Design
Slip modulus: Since the equation for slip modulus is an empirical one (i.e. with hidden units in the coeflicient part), for Mathcad to produce correct units, multiply it by the units shown inside the parentheses; or alternatively, use the equation without units for p and d. hus, slipmodulus,
Kser:= h. p"5.do8 .(kg"5 .m2.7 .N .105.4) K& = 602.31 0 N mm"
Kser
*
Slip per nail, U, under permissible load
e load c a ~ a cofi ~t
Design of an eave joint of a roof truss is required. The joint is to comprise a Finnish coniferplywood(9 mm thick, 7ply)gussetwith3.00mm diameter x 50 mm long round wire nails acting in single shear. The rafter and ceilingtie are bothin Cl8 timber under service class 2 conditions and are subjected to mediumterm loading, as shown in Fig. 9.24(a).
Plywood gusset : 9 m 7ply Finnish conifer 4.18
47 x 97 m C18timber
2.13 krN
Fig. 9 . 2 ~ ( ~ Nailed ) plywood gussetedjoint (Example 9.4).
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmrn"*
N :=newton k N := lo3.N Direction parallel to grain, // Direction perpendicular to grain, pp
Design of Timber Connections
Nail diameter, d Nail length, lnail NO. of nails, nnail Head side penetration, ta,head Pointside penetration, ta,point
BS5268 : Part 2, Table 56 Plywood group I and C18 timber Basic single shear lateral load, Fbasic
BS.5268 : Part 2, Clause 4.6 Number of shear planes, nshear No predrilling Side grain nailing, &3 does not apply Ordinary nails, KM does not apply Plywoodtotimber, K46 does not apply Load duration, K48, mediumterm Moisture content, service class 2 No. of nails, assume <10, in one line, K50
BS5268 : Part 2, Clause 4.6.9 Permissible shear lateral load nail per
d := 3.00 mm Znail := 50 mm Required ta.head := 9 mm +
*
*
:= lnail
tapoint
= 41 0 mm
Fbmic := 267 N *
K4.4:=l
K50 := 1
Fadm := Fbasic
%hear
K44 = 299.04 0 N *
Fadm
BS5268 : Part 2, Table 53 End distance // to grain, a3 a3
ta.head
ta.point
K43
*
:= 14 d a3 = 4 2 o m m
Kpre.drill K 4K 649 ' K48
' K50
205
206
Structural Timber Design
Edge distance perpendicular to grain, a4 Distance between lines of nails perpendicular to grain, a2 Distance between adjacent nails in any one line // to grain, al
a4
a4
:=5 * d
= 15 omm
a2 := 7 . d
a2 = 21 0 mm
a1 := 14 d a1
= 42 0 mm
6. Rafterconnection
Load in rafter,
Prafter
Number of nails required, nraffer Mathcad command Using 2 rows of 8 nails each will give a rninimum required gusset length parallel rafter of, Lraffer
to
7 . C e i h g tie connection
Load in ceiling tie,
:= 4.18 kN
P tPi et k
Number ofnailsrequired,
*
:=
nrafternrafter
Ptie 
Fah
= 13.98 nrafter :=CeiKnrafter) Adopt: nrafter I= 14
nrafter
Mathcad command
Using 3 rows of nails, on4top, 4 alongthe centre line and 5 at the bottom row, will give a minimum required length for gusset along the bottom nailing row of the ceiling of, tieLtie
L,, := 2 .a3 $5 a1 Ltk = 294 o mm Thus, a suitable gusset length along the bottom edge of the ceiling tie should be Lrafter * COS (27
G) = 336.8
o mm
>294mm
Try a gusset of 350mm in length along the bottom edge of the ceiling tie. 8. Design of the plywood gusset
A freebody diagramof the nonconcurrent forces actingon the joint, with reference to the centre line of the members, is shown in Fig. 9.24(b). Since the three forces are not
Design of TimberConnections
Fig. 9.24( b)
207
Freebody diagram.
coincident the freebody isnot in full equilibrium (i.e. the sum of the moments of the forces (sayabout point 0) is not equal to zero). Thus, the applied nonconcurrent force the systemwillinduce a moment of M, at the joint, whichshouldberesistedby plywood gusset plate. In addition, the plywood gusset should also be able to resist a shearing force of F = 4.7 kN acting along an axis parallel to the rafter and passing through the interface between the ceiling tie and the rafter (i.e. along xx). Therefore, the balancing moment, M , is Plywood thickness, t
M := (2.13 kN) .(106. mm) M = 0.23 o kN m t:=9.mm
Length of the gusset along the rafter, h
Section modulus, z
Stress due to bending moment
h = 392.81 o mm t h2 z := 6 z = 2.31 x lo5 o mm3 M G m :=
z
= 0.98 o N mm2 4.7 kN G, := t*h CY, = 1.33 o N mm2 D t o t a l := G m t Gs G t o t a l = 2.3 0 N mm"2
Gm
Shearing stress along xx Total stress acting along xx
*
~
208
Structural Timber Design
BS5268 : Part 2, Tuble 41 Grade shear panel zg := 3 . 7 4  ~ * m m  ~ Modification factor for plywood K36 := 1.33 grade stresses for mediumterm (K36, Table 33) Permissible panel shear in T&n :=Tg K36 plywood l;,&= 4.97 0 N m m  2 *
*
Plywood size is $ a ~ s f a c t o ~ Rafter member:
Plywood gusset: 9 mm 7ply Finnish conife Ceiling tie: 47 x 9 7 m C18 ti
@ 42 m centres along tie
/ /350m> Fig. 924(e)
Designed joint details.
.5 ~ e s i of ~ ansimple t w o  m e ~ ~ e r
er
~ o l tconnection e ~
The bolted connection shown in Fig. 9.25 is subjected to mediumterm tensile loading under service class 2 conditions. The joint comprises a singleM12 bolt acting in single shear with C18 timber sections. Determine the load capacity of this connection.
Force, kN Length, m Crosssectional dimensions, mm Stress, Nrnm2
Bolt diameter, d bolts, Thinner member thickness
NO. O f
N :=newton kN := lo3 * N
Direction parallel to grain, ,l/ Direction perpendicular to grain, pp
d:= 12.mm nbolt
:= 1
tthinner
:=33 .mm
Design of Timber Connections
P
47 m
33 m
P Fig. 9.25 Bolted connection (Example9.5).
. Basic v a l ~ e ~ BS5268 : Part 2, Table 64 Strength class, C18 timber, load parallel to grain Basicsingleshear lateral load Fbafic :=1.22 k N under mediumterm loading, by interpolation, Fbmic

BS5268 : Part 2, Clause 6.6 Timbertotimber connection K46 does not apply Moisture content, K56, service class 2 No, of bolts one inline, n = 1, K57
BS.5268 : Part 2, Clause 6.6.6 Permissibleshear lateral load per bolt Loadcapacity of the connection, Pcapacity
K46 := 1
K56
:= 1
K57 := 1
Fadm := Fbasic
*
&6
*
K56 ' K51
F a h = l .22 o kN Pcapacity:= Fa& nb& Pcapacity = 1.22 0 kN
209
210
Structural Timber Design
5 . ~ i n i m bolt ~ m endand edge distances
BS5268 : Part 2, Table 75 loaded distance End End distance unloaded Edge distance loaded distance Edge unloaded
7d=84omm 4 d = 48 o m m not applicable 4 d = 48 o m m not applicable 1 . 5l.8d0=m m

+
6. Joint slip
BS5268 : Part 2, Clause 6.2 and Table 52 Strength class, Cl8 timber Characteristicdensity(Table 7 ) p := 320 .kg m"3 Slip modulus: Since the equation for slip modulus is an empirical one (i.e. with hidden units in the coefficient part), for Mathcad to produce correct units, muliply it by the units shown inside the parentheses; or alternatively, use the equation without units for p and d. Thus, slipmodulus, K,,,
h.
K,,, := p1.5 .d . (kg"5 .m2.5 .N . 106) Kser = 3.43 X lo3 o N mm' *
Slipperboltunderpermissible load, U
U :=
&er
= 0.36 o m m := U 1 .OO mm Ujoint = 1.36 o mm
U
Allowing 1 .OOm m additional slip due to oversize bolt hole
r;,,, 
Ujoint
+
Therefore the load capacity of the joint is 1.22 kN with a joint slip of 1.36mm Example 9.6 Design of a threemember timbertotimber bolted connection
The boltedconnectionshown in Fig. 9.26 issubjected to alongterm loading, as indicated, under service class conditions. 2 The joint comprises a singleM16 bolt acting in double shear with C22 timber sections. Checkthe adequacy of the connection. Definitions
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm2 1. Geometrical details
Bolt diameter, d bolts, nbolt
NO. O f
N :=newton
kN :=l o 3 .N Direction parallel to grain, // Direction perpendicular to grain, pp
Design of TimberConnections 5.8 kN
\r,
211
members: Rafter 2 of 35 x 194mm C22timber
,/'k"
5.2 kN
I
72 x 194m C22timber Fig. 9.26
2.63 W
Boltedconnection (Example 9.6).
Side member thickness Inner member thickness Check adequacy o f the section Distance from loaded edge to furthest fastener
touter
:= 35 mm
tinner :=72
*
mm = 72 0 mm >2 194. mm := 2 = 97 o mm *
tinner he he
*
touter
= 70 0 mm okay
2. Basic vafues BS5268 : Part 2, Clause 6.6.4 and Table 70 The support reaction acting on the ceiling tie makes the load in the rafter act parallel to grain, and in the ceiling tie to act at 27" to grain direction. Therefore Hankinson's equation should be used to determine the appropriate basic shear load values. Strength class, C22 timber (Table 70) FraYter.1 :=3.09 kN For 35 mm outer members (rafters), longterm load parallel to the grain for a M16 bolt For 72mm inner member (ceiling tie) corresponding to 72/2= 36 mm outer members for a M16 bolt:
Ftje.l:= 3.18 kN longterm load parallel to grain, by interpolation longterm load perpendicular Ftie.pp := 2.64 kN to grain, by interpolation
212
Structural Timber Design
For load at angle a where a := 27.
Ftie.8
n
180 (3) Grade shearstressforC22 timber (Table 7)
Ftie.8 Ftie.pp
Ftie :=
+
*
Ftie.pp
*
COS(CX)~
= 3.05 o kN 1cg.8 := 0.71 N mm2 Ftie
3. ~ o d i f i c a t i o factors n
BS5268 : Part 2, Clause 6.6 Timbertotimb,er connection, K46 does not apply Moisture content, K56, service class 2 No.of bolts one inline, n = l, K5, Serviceclass 2 (K2, Table 13) Load duration, longterm (K3? Table 14) Clause2.9) Loadsharing (K8,
&6
:= 1
K56 := 1
K57 := 1
K2 := 1
K3
:= 1
K8
:= 1
4. ~ e r ~ i s loads s i ~ l ~
(1) For rafter: Permissible load per shear plane actual load (per shear plane)
(2) For ceilingtie: permissible load per plane shear actual load (per shear plane)
Fadm.rafter := Fra8er.l
k;6
' K56
= 3090 UJ 5.80 kN Prafter := 2 Prafter = 2 9 0 kN Satisfactory Fadm.mafter
*
Fadmetie := Ffie * &6
K56 ' K57
3.05 o kN 5.20 kN Prafter := 2 Prafter = 2.6 o kN Satisfactory
Fah.tie
*
(3) Shear stress in jointed timber BS 5268 : Part 2, Clause 6.1 and Figure 6 Permissible shear stress zadm Zg.// K2 K3 Kg Z a h = 0.71 o N mm2 Permissible shear force vu* . 2. 3 Z a h he tinner V a h =L 3.31 0 kN >2.63kN Satisfactory *
S"
*
*
*
K57
Designof Timber Connections
BS5268 : Part 2, Table 75 For rafter: end distance loaded end distance unloaded edge distance loaded edge distance unloaded For ceiling tie: end distance loaded end distance unloaded edge distance loaded edge distance unloaded Thereforethe connec~onwith ~ be a d ~ u a t e
7 d = 112 o mm not applicable 4d=64omm 4 d = 64 o mm not applicable 1.5.d=24omm 7 . d = 1120mm 4 .d = 64 o mm not applicable 4.d=64omm 1.5.d= 24omm N edgeand ~enddistances u detailed ~ abovewill
esign of a t ~ r e e  m e m ~ e r ~ ~ moment f t e ~ connection
st~eftotim~~r
The bolted connection shown in Fig.9.27 is subjected to a longterm eccentricload of 12.5 kN asindicated. Thejoint comprises two4mm thick steel sideplatesand an inner timbermemberinstrengthclassC22joined together usingeightM12 bolts under service class 2 conditions. Check the adequacy of the connection.
Two member: 4sideplates mm Innersteel (
Fig. 9.27 Eccentric loaded connection (Example 9.7)).
timber
214
Structural Timber
Design
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmm2
N :=newton k~ :=103 N Direction parallel to grain, // Direction perpendicular to grain, pp
1. Geometrical details
Bolt diameter, d NO. of bolts, nbolt Bolting pattern: horizontal spacing, x vertical spacing, y Side member thickness (Clause 6.6.5.1) Inner member thickness Distance from loaded edge to furthest fastener
2.
d:= 12.mm nbolt :=8
x :=90mm y := 85 mm tsteel :=4 mm
>2.5mm or 0.3 d = 3.6 o mm okay
:= 72mm
he:=294.m62*mm he = 232 o mm
lied loading
Applied vertical load, P Eccentricity, ep No applied horizontal load, H Eccentricity will induce a moment of M and a shear force of V at C, the centre of geometry of the connection Distance from C to each bolt, r
Sum of squares of distances fromC to each bolt, J= Force acting on furthest bolt (i.e. bolt a) due to applied moment M Force acting on bolt a due to applied shear force V
r;!
P := 12.5 kN *
ep :=45 mm H:=O*kN M:=P.ep M=0.56okNm V:=P Y = 12.5okN a
ra := ~ x T ra = 123.79 o mm rb :=x rf := x rc :=ra re :=ra rg := ra rh :=y rd := y J:=4.r~+2*r~+2.r~ J = 9.2 X 1040mm2
Design of Timber Connections
21 5
Force acting on bolt a due to applied horizontal load H = 0 For the rectangular fastener pattern, the total force on bolt a is given by
[(F.+
F:=
J ( x 2+3 ) X
Fa.M
'
F = 2.18 o kN This force is acting at an angle o f a to grain of timber, where
a = 76.18 o deg
3. Basic load vaiues
BS5268: Part 2, Clause 6.6.4 and Table 70 72 mm inner member corresponding to 72/2 = 36 mm outer members for a M12 bolt: Longterm load // to grain, by interpolation Longterm load perpendiuclar to grain, by interpolation
Fpp:= 2.105 kN *
F1 FPP F1 sin(a)2 Fpp COS(^)^ F=2.12okN Grade shear stress for C22 timber ~ ~:=0.71 1 N mmm2 (Table 7)
For load acting at angle a
F :=
4. ~odification factors
BS.5268 : Part 2, Clause 6.6 Steeltotimber connection (&fit Clause 6.6.5.1) Moisture content, K56, service class 2 No. o f bolts in one line, n = 3, Service class 2 (KZ, Table 13) Load duration, longterm (K3, Table 14) Load sharing (Kg, Clause 2.9)
&6
:= 1.25
K56 :=
l
K57 := 1 K2 :=
K3
+ *
*
l
:= 1
K8 := I
BS5268 : Part 2, Clause 6.6.6 Permissible load per shear plane

:=F . &6 K56 = 2.64 o kN nshear :=2
Fadm
K57
Fadm
Number of shear planes Permissible load for bolt
Fadm.bolt := Fadm * nshear
= 5290 kN Satisfactory
Fadm.bolt
Shear stress in jointed timber: BS 5268 : Part 2, Clause 6.1 and Figure 6 Permissible shear stress Permissible shear force Hence the timber section can sustain an applied shear force at the connection of Fadm.y 6. ~ i n i ~ bolt u mspacings
BS5268 : Part 2, Table 75 end distance loaded end distance unloaded edge distance loaded edge distance unloaded spacing parallel to grain spacing perpendicular to grain
7 .d = 84 o mm 4 .d = 48 o mm 4 .d = 48 o mm 1.5 .d = 18 o mm 5 .d = 60 o mm 4 .d = 48 o mm
not applicable not applicable 56mm providedokay not applicable <78mm provided okay 73mmprovidedokay
Therefore the connection as detailed above is adequate
Design of an eave joint of a roof truss is required. The joint is to comprise twoFinnish conifer plywood (9mm thick, 7ply)gussets,one on eachside,glued to rafter and ceiling tie membersboth in Cl8 timber under service class2 conditions and is subjected to mediumterm loading, as shown in Fig. 9.28. Determine the contact areas required for the rafter and tie member where bonding pressure is generated by nails.
Force, kN Length, m Crosssectional dimensions, mm Stress, Nmmm2
N :=newton k~ :=103. N Direction parallel to grain, // Direction perpendicular to grain, pp
Designof Timber Connections
2 Plywood gussets:
zone
(9 m 7ply Finnish conife I
l I
8.36
l
/ Ceiling tie: 47 x 97 m C18 timber 4.26 ICN Fig.
Gluedplywoodgussetjoint(Example
9.8).
ear s t r ~ s sval~es
33.7268: Part 2, Clause 6.10.1.4 Because the joint is to beformedbygluingplywoodgusset plates to timber, the permissible shear stress for the joint is the lesser of that for timber and rolling shear stress for the plywood.
C 18 timber (Table 7) zg.i := 0.67 .N mm2 Shear stress parallel to grain Finnish conifer plywood (Table 41) Rolling shear stress z,.g := 0.79 N mmm2 I
Service class 2 (KZ, Table 13) M e d i u m  t e ~loading (K3, Table 14) No load sharing (Kg,Clause 2.9) Plywood grade stresses Table 33) Naillglue modification factor (K70, Clause 6.10.1.4)
a
K2 := 1
K3
:= 1.25
Kg
:= 1 := 1.33mediumterm
K36
loading
K70 :== 0.9
stresses
The support reaction acting on the ceiling tie makesthe load in the rafter act parallel to the grain, and in the ceiling tie to act at 27" to grain direction. Thus Angle of inclination, a
K
a:=27*180
21 8
StructuralTimber Desian
Therefore the shear stress for timber is limiting for both therafter and tie contactfaces.
(1)
For rafter load in rafter, Prafter Prafter required contact area on each side,
(2) For ceiling tie
load in tie, P t i e required contact area on each side, A t i e
:=9.4 k N *
Prafter
Arafter
:=
Arafter
= 6.24
2
*
Tah.rafter
X
l O3 0 mm2
Pi, := 8.36 k N Atie Atie
Ptie
:=
2
*
Tah.tie
= 7.97
X
io3 o mm2
te
Eurocodes are civil and structural engineering design standards, that are published by the European Committee for Standardisation (CEN), for use in all member countries. They are designed to provide a framework for harmonised specifications for use in the construction industry. At present there are nine Eurocodes which are published as European prestandard, or ENV, to serve as an alternative to the existing national codes in each country. They are as follows: EC1 EC2 EC3 EC4 EC5 EC6 EC7 EC8 EC9
Basis of design and actions on structures Design of concrete structures Design of steel structures Design of composite steel and concrete structures Design of timber structures Design of masonry structures Geotechnical design Design of structures in seismic regions Design of aluminium structures.
The draft edition of Eurocode 5 :Part 1.1 for timber structures was published in theUK in late 1994 as DD ENV 199511 :1994, inconjunction with a National Application Document (NAD) for the United Kingdom which contains supplementary information to facilitate its use nationally during the ENV period. The publicationof the definitive EN EC5 is planned for around 2001 and is expected to be used in parallel withBS 5268 until the year 2005 (overlap period). At thetransition, BS 5268 will not be withdrawn but is likely to be termed ‘obsolete’. In thisstate it would not be updated or amended and as such would become less usable with time. In general, DD ENV 199511:1994 (i.e. EC5)contains all the rules necessary for the design of timber structures, but unlike BS 5268 :Part 2 it does not provide material properties and other necessarydesign information. 21
Structural TimberDesign
Such information is available in supporting European standards, for example, BS EN 338 :1995 which contains the material properties, such as bending and shear strength, etc., for the 15 strength classes of timber (i.e. C14 to D70). Therefore to designwithEC5 it isnecessary to have several additional reference documents to hand.
The design oftimber structures is related to Part 1.1 of EC5 and is based on limit states design philosophy which means that the overall performance of structures should satisfy two basic requirements. The first is ultimatelimit states (i.e. safety), usually expressed in terns of loadbearing capacity, and the second is s~rviceabilitylimit states (i.e. deformation and vibration limits), which refersto the ability of a structural system and its elements to perform satisfactorily in normal use.' In general, it is well understood that the violation of ultimate limit states, i.e. safety criteria, may cause substantial damage and risk to human life, whereas exceeding the serviceability limits, i.e. excessive deformation andlor vibration, rarely leads to such severe consequences. However, while this may suggest that serviceability is relativelyunimportant, in practice it is often the key element to ensuring an economical and troublefree structure. Generally, it is necessary to check that design criteria for both ultimate limit states and serviceabilitylimit states are satisfied. or checkingultimate limit states, the characteristic values of both the loads and the material properties are modifiedbyspecified partial safety factors that reflect the reliability of the values that they modify. These factors increase the values of the loads and decrease the values of the material properties. To check serviceability limit states, EC5 requires that both instantaneous and timedependent (creep) deflections are calculated and, in addition, designers are required to demonstrate that vibration in floors is not excessive. The characteristicvalues are generallyfifth percentile valuesderived directly from a statistical analysis of laboratory test results. The characteristic strength is the valuebelowwhichthe strength liesinonly a small percentages of cases (not more than 5%), and the characteristic load is the value above which the load lies in only a small percentage of cases. Throughout EC5, depending on the character of the individual clauses, a distinction is made between princ~lesand application rules. statements, definitions, requirements or analytical models for which no alternative is permitted unless specifically stated (principles are preceded by the letter P),and application rules are generally recognised rules that follow the principles and satisfy their requirements. In EC5, in common withother Eurocodes, decimal points are displayed as commas (so 1.2 is displayed as 1,2).
Design to Eurocode 5
artial safety factors for actions in building structures for persistent and transient design situations (Table 2.3.3.1, EC5)
~ 
Permanent actions
Variable actions
YQ
YG
One with its Others with their characteristic value combination value Normal partial coeficients 1 Favourable effect Unfavourable effect 1.5 Reduced partial coeficients 1 Favourable effect 1.2 Unfavourable effect
.o
0
0 1.5
0
0
1.35
.o
1.35
Actions is the Eurocode terminology used to define (i) direct loads (forces) applied to the structures, and (ii) indirect forces due to imposed deformations, such as temperature effects or settlement. Actions are covered in Chapter 2 of EC5. ermanent actions, denoted by the letter C, are all the dead loads acting on the structure, including the selfweight, finishes and fixtures. Variable actions, denoted by the letter Q, are the imposed, wind and snow loads. ance on obtaining the characteristic values of actions is given in asis of design and actions on structures. The characteristic values for actions are modifiedby partial coefficients, y to take account ofsafety factors, load combinations, etc. Therefore the design values ofactions, Fd, are obtained by multiplyingthe characte~sticactions for permanent, Gk, and/or variable, relevant partial coefficients, yG and yQ, as given in Table 2.3.3.1 of EC5, (reproduced here as Table 10.1). Thus: (10.1)
Chapter 3 of ENV 199511 (EC5 :Part 1.1) deals with the material properties and defines the characteristic strength, stianess and densityvalues ofsolidtimbersections for strength classesC14 to D70. EC5,unlike S 5268 :Part 2 :1996, doesnot contain the material properties values and, as mentioned earlier, this i n f o ~ a t i o ncan be foundin a supporting standard i.e. in Table 1 of S EN 338 :1995 (re oduced here as Table 10.2). A comparison of this table with Table 7 o S 5268 :Part 2 :1996 (see Table 2.3)
222 Structural Timber
Design
?
099
00000d.N . . I
Pm ?c?"
P00
o w
F?? m W 0 0 P0
?"I? md.
00trtoo
P*oo
i : :
Design to Eurocode 5
223
highlightsthedifferencebetween the characteristic values and the grade stress values familiar to users of BS 5268. The characteristic values, which aregenerallyfifthpercentilevalues, are considerablyhigher than the BS 5268 grade values which have been reduced for longterm duration and alreadyincluderelevantsafety factors. In addition, BSEN338 does not include grade TR26 timber. This isan extra grade which has been included in BS 5268 only.
Thecharacteristic strength valuesgivenin Table 1 of BS EN 338 (see Table 10.2) are measured in specimens of certain dimensions conditionedat 20°C and 65% relative humidity, in tests lasting approximately five minutes; and therefore they are applicable for such loading and service conditions. For other loading and service conditions, the characteristic values shouldbe modified using relevant modification factors, which are described later. Clause 2.2.32 of EC5 specifies that the design value, Xd, of a material property is calculated from its characteristic value, x,, using the following equation: for strength properties,
(10.2a) for stiffness properties,
(10.2b)
xk
x d =
Yiu
where: yM is the partial safety factor for material property and its values are given in Table 2.3.3.2 of EC5 (reproduced here as Table 10.3) kmod is the modificationfactor for service classand duration of load. Modification factor kmod takes into account five different loadduration classes: permanent, longterm, mediumterm, shortterm and instantaneous, relating to service classes of 1, 2 and 3 as defined in Clauses 3.1.53.1.7 of EC5. A summary of Clause 3.1.5 and Tables 3.1.6 and 3.1.7 of EC5 is reproduced hereas Table 10.4. For metal components, kmodshould be taken as 1, i.e. kmod= 1.0. The characteristic strength values given in Table 1 of BS EN 338, see Table 10.2, are related to a depth inbending and widthintensionof 15Omm. For depth in bending or width in tension of solid timber, h, less than 150 mm, EC5 in Clause 3.2.2(5) recommends that the characteristic values for fm,k andft,O,k may be increasedby the factor kh where: kh = thelesserof
(kh
=
(yr.’
and kh = 1.3
)
(10.3)
Structural TimberDesign
able 10.3 Partial coefficients for material properties (Table 2.3.3.2, EC5)
Limit states
YM
Ultimate limit states Fundamental combinations Timber and woodbased materials used Steel joints Accidental combinations
in
Serviceability limit
1.3 1.1 1.o l .o
states
Table 10.4 Modification factor kmodfor service classes and duration of load
Values of kmodfor solid and glued laminated timber and plywood (Table 3.1.7, EC5) Load classb duration Service classa 3
2 Permanent Longterm 0.70 Mediumterm Shortterm Instantaneous a
1 0.60 0.70 0.80 0.90
1.10
0.60 0.80
0.90 1.10
0.50 0.55 0.65 0.70 0.90
Load duration classes (Table 3.1.6, EC5): Permanent refers to order of duration of more than 10 years, e.g. selfweight Longterm refers to order of duration of 6 months to 10 years, e.g. storage ~ e ~ i u m  t e rrefers m to order of duration of 1 week to 6 months, e.g. imposed load Shortterm refers to order of duration of less than 1 week, e.g. wind load ~ n s t u n t u n e orefers ~ to sudden loading, e.g. accidental load Service classes (Clause 3.1S , EC5): Service cZuss I relates to a typical servicecondition of 20°C, 650/RH with moisture content 5 12% Service cluss 2 relates to a typical servicecondition of 20°C, 85%RH with moisture content 520% Service cluss 3 relates to a service condition leading to a higher moisture content than 20%
Chapter 5 of EC5 :Part 1.1 sets out the design procedures for members of solid or glued laminated timber with regard to calculation for their strength
225
Design to Eurocode 5
properties(ultimatelimitstates) and treats solid and gluedlaminated members in the same way. Design procedures relevant to flexural and axially loaded members and doweltype joints are described below.
embers should be designed so that the following conditions are satisfied: (10.4a) +
fmlzld fm,yld
kmo m , z l d 5 1
(1 0.4b)
where: km
&
C T , , ~ , ~
ums,d
modification the is factor for combined bending stress, detailed in Clause 5.1.6(2) of EC5 as: for rectangular sections, k m =0.7 for other crosssections, km= 1.O are the design bending stressesabout the principal axes yy and zz, as shown in Fig. 10.1
where: design bending moment about yy axis, and bh2 W '" 6 is the appropriate section modulus
My,d==
where: =design bending momentabout zz axis, and hb2 Cy, = is the appropriate section modulus 6
Y
Fig. 10.1
226
Structural Timber Design
fm,y.d
fm,z,d
are thedesignbending
kmod kh kcrit kls ‘fm,k *
fm,y/z,d
strengths about yy and Zz axes *
*
YM
(10.5)
where: is the modification factor for load duration and service classes as given in Table 3.1.7 of EC5, see Table 10.4, is the modification factor for bending depth (Clause 3.2.2), is the mod~cationfactor for reducing bending strength of a beam where there is a possibility of lateral buckling. For a beam which is laterally restrainedthroughout the lengthof its compression edgeand its ends are prevented from torsional rotation, kcrir = 1.0. For other conditions, Clause5.2.2ofECS and Clause 6.5 of the UK NAD recommend the following: (1) Calculation of the critical bending stress,om,crit ,as given by the following equation for a rectangular section of breadth b and depth h: (10.6) The effective length Lef is governed by the degree of restraint against lateral deflection, rotation in plan and twisting, and may be considered as: Lef= 0.71; for a beam fully restrained against rotation in plan at both ends, Lef= 0.85L for a beam partially restrained against rotation in plan at both ends or fully restrained at one end, Lef= 1.2L for a beam partially restrained against twisting at one or both ends. (2) Calculation of the relative slenderness for bending as: (10.7) (3) The value of kcrir is then determined from: kcrit = 1 for hre1.m 5 0.75 kcrit = 1.56 0.75hrel,m for 0.75
227
Design to Eurocode 5
Membersshouldbedesigned (Clause 5.l .7, EC5):
so that the followingconditionissatisfied
5 fV,d
zd
(10.8)
where: zd isthedesignshearstress, section, is given by:
and for beamswitharectangularcross
(10.9) where Vd is the design shear force (maximum reaction) and A is the crosssectional area, where A =bh. For beams with a notched end A =bh,, see Fig. 10.2. fv,d is the design shear strength, which is given by: fV,d
=
h o d ' kls * k V ' fV,k
(10.10)
YM
where:
f V,k
is the characteristic shear strength (Table 1,BS EN 338 :1995), is the partial coeficient for material properties, (Table 2.3.3.2, YM EC9, kmod is the modificationfactor for load duration and service classes (Table 3.1.7, EC5), is the modification factor for loadsharing systems (kh= 1.1) klS (Clause 5.4.6,EC5), is the modification factor for shear in members with notched kV ends, see Fig. 10.2, where, for beams with no notched ends, kv= 1.0 for beams notched at the unloaded side, kV= 1.O for beams notched at the loaded side,
+l
he/
l (a) Notch on loaded side
!+
l
(b) Notch on unloaded side
Fig. 10.2 Endnotched beams(DD ENV 19951 1:1994).
I
2 2 ~ Structural Timber Design
kV=the lesser of
(10.11) where: kn=5.0 for solid timber = 6.5 for glued laminated timber h =beam depth (mm) x =distance from line of action to the corner 01 =h,/h i =notch inclination.
oc,90,d 5 fc,90,d
(10.12)
where: is the characteristic compressive strength perpendicular to grain (Table 1, BS EN 338 :1995), yM is the partial coefficient for material properties (Table 2.3.3.2, EC5), kmodis the modification factor for load durationand service classes (Table 3.1.7, EC5),
L,go,k
I
l
" " " I
I I
" " " " " " " " " " " " " "
Fig. 10.3 Compression perpendicular to the grain (DD ENV 199511 :1994).
Design to Eurocode 5
kls
is the modification factor for loadsharing systems, i.e. kls= l. 1,(Clause 5.4.6, EC5), is the modification factor for bearing length, which takes into account that the load can be increased if the loaded are shown length, l, is short, see Fig. 10.3. The values of in Table 5.1.5 of EC5 (reproduced here as Table 10.5).
Table 10.5 Values of
(Table 5.1 S , EC5)
a>
I> 150mm
150m>1> 15mm>Z
229
15m
1 1 l
loom 1
1+(l502)
/ 170
1.8
a<
100 mm
l 1 + a (1501) / 17000 1 up25
+
o~pression or tension parallel to grain
For members subjected to axial compression only, the following condition should besatisfiedprovidedthereis no tendency for buckling to occur (Clause 5.1.4, EC5): oc,O,d
5 fc,O,d
(10.14)
where: o,,O,d
is the design compressive stress parallel to grain, and is given by: (10.15)
fc,O,d
where N d is the design axial load and A is the crosssectional area is the design compressive strength parallel to grain, and is given by: (10.16) where: fc,O,k
yM kmod kl,
is the characteristiccompressive strength parallel to grain (Table 1, BS EN 338 :1995), is the partial coefficient for material properties (Table 2.3.3.2, EC5), is the modification factor for load duration and service classes (Table 3.1.7, EC5), and is the modification factor for loadsharing systems, i.e. kts = 1.1,(Clause 5.4.6, ECS).
For members subjected to axial tension a similar condition and procedure apply (Clause 5.1.2, EC5).
230
Structural Timber
Design
10.5.5 Members subjectedto combinedbendingand axia/tension For members subjected to combined bending and axial tension, the following conditions should be satisfied (Clause 5.1.9, EC5): (10.17a) (10.17b) where: Ot,O,d h,O,d (3,,y,d
fm,y,d
& (Tm,,,d & fm,z,d
=0.7 k m = 1.0 km
is the design tensile stress the is design tensile strength parallel to grain are the respective design bending stressesdue to any lateral or eccentric loads (see Section 10.5.1) are thedesignbending strengths for rectangular sections, and for other crosssections (Clause 5.1.6, EC5).
10.5.6 Co/umns subjected to combined bending and axiaf compression
For the design of columns subjected to combined bending and axial compression (Clause 5.2.1, EC5), and also for the design of slender columns in general, the following requirements should be considered: (l) Calculation of the relativeslenderness ratios about the p. and 22 axes of the column as defined by: (10.18a)
(10.18b) where: (10.19a)
(10.19b) where:
is the 5%ile modulus of elasticity parallel to the grain (Table 1, BS EN 338 :1995).
EO,os
Designto Eurocode 5
(2) For both hrel,y 5 0.5 and following conditions:
hrel,z 5 0.5,
( % ) 2 + Z + k m 
fm,z,d
231
the stressesshouldsatisfythe
5 1.0om,z,d
(10.20a)
5 1.0
(1 0.20b)
where: is the design compressive stress is the design compressive strength parallel to grain are the respective design bending stresses due to any oommy,,zd, d lateral or eccentric loads (see Section 10.5.1) are the design bending strengths fm,y.d & fm,z,d km=0.7 for rectangular sections, and for other crosssections km= 1.0 (Clause 5.1.6, EC5). (3) For all other cases, i.e. hre1,y >0.5 and/or Are/,=>0.5, the stresses should satisfy the following conditions: oc,O,d
fc,O,d
om,z,d oc,O,d
+
+km%
5 1.0
(10.21a)
fm,y,d fm,z,d ‘fc,O,d
(10.21b) where kc,, and kc,z are modification factors of compression members and are given by: 1 (10.22a) kc,, = (similarly for kc,z) ky
+ J(G h;el,y)’
ky = 0.5(1
+ Pc(hrel,y 0.5) +
(similarly for k,) (10.22b)
where: pc= 0.2 for solid timber, and pc=0.1 for glued laminated timber. oweltype fastener joints
EC5, Chapter 6 deals with the ultimate limit states design criteria for joints made with doweltype fasteners, suchas nails, screws, bolts and dowels, but does not give any information regarding connectoredtype joints, i.e. toothed plates, splitrings, etc. EC5 provides formulae, based on Johansen’s equations, to determine the loadcarryingcapacity of timbertotimber, paneltotimber and steeltotimber joints, reflecting all possible failure modes. Loadcarrying capacity formulae for timbertotimber and paneltotimber joints, with reference to their failure modes, are compiledTable in 10.6, and for steeltotimberjoints in Table 10.7.
232
Structural Timber Design
Table 10.6 Loadcarrying capacity of timbertotimber and paneltotimber joints (based on Clause 6.2.1, EC5)
Joints in single shear Failure modes
t2
"i t
(10.23a) carrying capacity per dowel, R d , is the lesser of:
Rd,b =fh,l,d * t2
*
d
P
(10.23b)
Rd,c
P(I
+~)]
(10.23~) (10.23d)
(10.230 Joints in double shear Failure modes
(10.23g) (1 0.23h) (10.23j)
(10.23k)
Design to Eurocode5
233
Table 10.7 Loadcarrying capacity of steeltotimber joints (based on Clause 6.2.2, EC5)
Joints in single shear Failure modes
Design loadcarrying capacity per dowel, Rd, is the lesser of:
(i) Thin steel plate (tsteel5 0.5 d) Rd,a = 0.4fh,l,d tl d Rd,b = 11J(2Myld 'fh,l,d d ) *
*
(10.24a) (10.24b)
(ii) Thick steel plate (fsteel >.d) (10.24c) (10.24d)
Joints in double shear
Design loadcarrying capacity per dowel per shear plane Rd, is the lesser of:
(i)Steel plate as the centre member Rd,e = I.Ifh,l,d tl d *
&,f
= llfh,l,d
*
*
(10.24e)
tl * d
(10.24f)
Rd,g = 1.54(2My,d'fh,l,d d ) (ii) Thin steelside plates (tStMl 5 0.5d) &,h = 0*5f,2,d t2 d &,j = 1.1 J(2My1d'fh12,d d ) (iii) Thick steel side plates (tsteel>d) Rd,k = 0.5fhh,2,dt2 ' d Rd,l = 1.5J(2My1d 'fh,2,d d) *
*
*
(10.24g) (10.24h) (10.24j) (10.24k) (10.241)
As an example, for a twomember timbertotimber joint there are six possible modes of failure. One or both of the timber members may fail, or the fastener may fail inone or bothtimber members,see failure modes(a) to (f) in Table 10.6. EC5 therefore provides six different formulae to calculate
234
Structural Timber
Design
the failure load for each mode and recommends that the lowest value is taken as the failure load of the connection. Therefore the design loadcarrying capacity, R d , for a fastener in single shear in a timbertotimber joint isgiven as the lesserof to (reproduced below from Table 10.6): =h,l,d Rd,a =h,l,d
Rd,b
*
tl
*
t2
*
d
(l 0.23a)
d .P
(10.23b)
P ( 1 + ~ ) ]
(10.23~)
(10.23d)
(10.23f) where: tl 432 t 2 fh,i,d 432 fh,2,d
are timber or board thickness or penetration are the designembedding strengths in tl and t2 and are given by:
fh ,i d,=
P
d
kmod,l
h, l ,k
YM
and
h,2,d
=
kmod,2fh,2,k
(10.25)
YM
where, f h , l , k and fh,2,k are the characteristic embedding strengths in tl and t2. The characteristic embedding strength values for different connection types are extracted fromClauses6.36.7of EC5 and are compiled in Table 10.8. As an example, for a nailed timbertotimber joint withpredrilledholes, fh,k = 0.082(1 O.Old)pk in N/mm2 (Clause 6.3.1.2(1)). Values of the modification factor, kmod, are given in Table 3.1.7 andthe values of y M are given in Table 2.3.3.2 of EC5. = h , 2 , d / f h , 1 ,d
is the fastener diameter is the fastener yield moment and is given by: (10.26) where M’,k is the characteristic value for dowel yield moment.
Design to Eurocode 5
235
The characteristic yield moment values for different fastener types are extracted from Clauses 6.36.7 of EC5 andare compiledinTable10.9. As an example, for round wirenails, M y , k = 180d2.6in Nmm (Clause 6.3.1.2). Table 10.8 Values of the characteristic embedding strength, fh,k, in N/mm2 (based on Clauses 6.36.7, EC5)
Fastener type
h,k
Timbertotimber Steeltotimber
Paneltotimber
Nails (d 5 8 mm) and staples
0.082p&0.3 no predrilling 0.082(1 O.Old)pk predrilled
0. l l p,4°3 for plywood 30t0.6d0.3 for hardboard
Bolts and dowels Parallel to grain, fh,)& At an angle a to grain
h,O,k
Screws (d< 8 mm) Screws (d >8 mm)
0.082(1 0.Old)pk 0.11(1 "O.Old)pk /(/cw sin2a cos2a)
+
Rules for nails apply Rules for bolts apply
Rules for nails apply Rules for bolts apply
where pk is the joint member characteristics density, given in Table l (BS EN 338 :1995) (in kg/m3) d is the fastener diameter (in mm); for screws, d is the smooth shank diameter t is the panelthickness kg0 = l .35 0.01 5 d, for softwoods =0.90 0.015 d, for hardwoods
+ +
Table 10.9 Valuesof fastener yield moment, kfy,k, in Nmm (based on Clauses 6.36.7, EC5)
Fastener type Nails round wire Nails square
%,k
180d2.6 270d2*6
Bolts and dowels Screws (d 5 8mm) Screws (d >8 mm)
Rules for nails applya Rules for bolts applya
Staples
Rules for nails apply
where fu,k is the characteristic tensile strength of the bolt (for 4.6 grade steel, fu,k =400 N/mm2) d is the fastener diameter (in mm) a For a screw the effective diameter def = 0.9d, provided that its root diameter 2 0.7d
236
Structural Timber Design
F
1
"i
v2
~
Fig. 10.4 Shear stressinthe jointed timber.
EC5, Clause 6.1 recommends that unless a more detailed calculation is made, for the arrangement shown in Fig. 10.4, it should be shown that the following condition is satisfied, provided that be >0.5 h: vd
5 $'fV,d'be* t
(10.27)
where: vd is the design shear force produced in the member of thickness fasteners at the joint where (Vl + V'= Fsin a) I
l
t
by the
I
(a) Spacing parallel and perpendicular to grain
90' 5 cx 5 90" Unloaded Unloaded Loaded edge Loaded edge end end
90'
a <270"
0" 5 a 5 180'
180' 5 a 5 360"
(b) Edge and end distances Fig. 10.5
Fastener spacings and distances (DD ENV 199511 :1994).
VI
n
2
V AI, % % W
Design to Eurocode 5 237
238
Structural Timber Design
be fV,d
is the distance from the loaded edge to the furthest fastener is the design shear strength of timber member.
For nailedconnections,EC5,Clause6.3.1.2recommends that there should be at least two nails in a joint, that smooth nails should have a pointside penetration of at least 8d, (6d for improved nails) and nails should be driven into predrilledholesintimberwitha characteristic density of 500kg/m3 or more. The recommendations for minimum nail spacings and end distances are given in Table 6.3.1.2 of EC5 (reproduced here as Table 10.10). For bolted connections, EC5, Clause 6.5.1.2 recommends that for more than six bolts in line with the load direction, the loadcarrying capacity of the extra bolts should be reduced by one third, i.e. for n bolts, the effective number, nef,is nef =6 2(n 6)/3. The recommendations for minimum bolt and dowel spacingsand end distancesare given in Tables 6.5.1.2and 6.6a of EC5 respectively (reproduced here as Table 10.1 1). Chapter 6 of EC5 also provides formulae for calculating resistance to axial loading in nailed, screwed and bolted joints, seeClauses6.3.2 and 6.3.3, 6.7.2 and 6.7.3 and 6.5.2 respectively. These cover withdrawal of the fastener from its headside (pullout), pulling through of the fastener head (pullthrough), and also combined axial and lateral loading.
+
Table 10.11 Minimum bolt and dowel spacings and distances (based on Tables 6.5.1.2 and 6.6a, EC5)
Dowels
Spacings and distances (see Fig. 10.5)
Bolts
Spacing parallel (al) Spacing perpendicular (az)
(4 3lcos al)d 4d
(4 3lcos al)d 3d
7d (but 280mm)
7d (but >80mm)
4d (1 lilsin.al)d (but 2 4 4
+
3d qt61sina1 (but 2 3 4
+ 2lsin al)d (but 2 3 4
(2
Loaded end distance (a3J 90" 5 01 5 90" Unloaded end distance (a3,=) 150" 5 a 5 210" 90"
+
(2
3d
where d= bolt or dowel diameter (in mm) a = angle of force to grain direction
+
+ 2/sin al)d(but 2 3 4
3d
Designto Eurocode5
239
ility limit states Chapter 4 of EC5 :Part 1.1 sets out serviceability requirements with regard to limiting values for deflections and vibrations which are often the governing factors in selection of suitable section sizes. Clause 4.2 of EC5 also provides design procedurefor calculation of slipfor joints made with doweltype fasteners, such as nails, screws, bolts and dowels.
In order to prevent the possibilityof damage to surfacing materials, ceilings, partitions and finishes, and to the functional needs as well as any requirements of appearance, Clause 4.3 of EC5 recommends a number of limiting values of deflection for flexural and laterally loaded members. The components of deflection are shown in Fig. 10.6, where the symbols are defined as follows: uo u1
precamber (if applied) deflection due to permanentloads 2.42 deflection due to variable loads unet the net deflection below the straight line joining the supports and is given by: = 4" U2 U0 (10.28) A summary of EC5 recommendations for limiting deflections is given in Table 10.112. Unet
Fig. 10.6 Components of deflection (DD ENV199511 :1994).
Table 10.12 General deflection recommendations (Clause 4.3, EC5)
Maximum allowable deflection Components of deflection CantileversBeams Instantaneous deflection due to variable L/150 actions, u~~~~~ Final deflection due to variable L/100 actions, u2$,, Final deflection due to all applied actions including precamber,
L/300 L/200
where L is the beam span or the length of a cantilever
L/200
L/ 100
240
Structural Timber
Design
Table 10.13 Values of kddfor solid and glued laminated timber and plywood and joints(based on Table 4.1,EC5)
Service
Material/ Load duration classa
2
3
0.80 0.50 0.25 0.00
2.00 1S O 0.75 0.30
1.oo 0.60 0.30 0.00
2.50
1
Solid and glued laminated timber Permanent Longterm Mediumterm Shortterm
0.60 0.50 0.25 0 .oo Plywood
Permanent Longterm Mediumtern Shortterm a Service
0.80 0.50 0.25 0.00
1.80
0.90 0.40
and load duration classes are defined in Table 10.4
The UK NAD and EC5 recommend the following procedurefor calculation of the instantaneous and final deflections: The instantaneous deflection of a solid timber member acting alone should be calculated using the appropriate 5%tile modulus of elasticity, Eo,05 and/or 5% tile shear modulus, Go,o5, where G0,Os =.Eo,O5/16. The instantaneous deflectionofamembermade of two ormore sections fastened together to act as a single member, a member in a loadsharing system, glulam and a compositesection member may be calculated using the mean elastic or shear moduli, i.e.Eo,,,,, and Gme,,. The final deflectionat the end of the design life of acomponent, ufin,is calculated as:
+
u8n = ~inst(1 kdef)
(10.29)
where kdef isthemodification factor which takes into account the increase in deformation with time due to the combined eEect of creep and moisture.Thevalues of k d , , are givenin Table 4.1of EC5 (reproduced here as Table 10.13).
EC5,Clause 4.4.1 states that actionswhich are anticipated to occur frequently should not produce vibrations which might impair thefunctioning of the structure or cause unacceptable discomfort to the users. Clauses
Design to Eurocode 5
241
4.4.2 and 4.4.3of EC5 provide recommendations and detailed rules for limiting vibrations. The UK NAD recommends that for residential IJK timber floors the requirements of EC5 can be met by ensuring that the total instantaneous deflection of the floor joists under load (i.e. U l , j m t ~2,jmt)does not exceed 14mm or L /333, whichever is the lesser, where L is the span in mm.
+
C5, Clause 4.2providesadesignprocedure for the calculation of slip for joints made with doweltype fasteners, such as nails, screws, bolts and dowels. In general, the serviceability requirements are to ensure that any deformation caused by slip does not impair the satisfactory functioning of the structure with regard to strength, attached materials and appearance. The design procedure for calculation of slip for joints made with doweltype fasteners is summarised as follows: (1) The instantaneous slip, uinst,in mm, is calculated as: for serviceability verification,
UiWt
=Fser
(10.30a)
kser
for strength verification,
3 Fd =2 kser *
Uimt
(10.30b)
*
where: Fser& Fd are the service load and design load, in N, per fastener per shear plane kser is the instantaneous slip modulus, in N/mm, and its values are given in Table 4.2 of EC5 (reproduced here as Table 10.14). f the characteristic densitiesof the twojoinedmembers (pk.1 and then pk in formulae given in Table 4.2 of EC5 (see Table 10.14) should be calculated as: ~ k , are ~ ) different,
Pk
= J(Pk,l
*
Pk,2)
(10.31)
(3) The final joint slip, u ~ i should ~, be calculated as:
where kdef is the modification factor for deformation and is given in Table 4.1 of EC5. For a joint made from members with different creep properties (kdexl and kdex2),the final joint slip should be calculated as: Ufln
= UinstJ[(1
+ kdeS,1)(1+ kdef,Z>]
(10.33)
242
Structural Timber
Design
Table 10.14 Values of slip moduli k,,,, per fastener per shear plane (in N/mm) (Table 4.2, EC5)
Fastener type Timbertotimber Paneltotimber Steeltotimber Bolts and dowels Screws Nails (predrilled) Nails (no predrilling)
pk5d
& pk5d0.8
Staples where Pk is the joint member characteristics density, given in Table 1, BS EN 338 :1995 (in kg/m3) d is the fastener diameter (in mm).
(4) For joints made with bolts, EC5 permits 1mm oversize holes which should be added to joint slip values. Thus: Uinst,bolt
&er
=1 
kSW
(10.34)
(10.35)
1. Timber Engineering STEP I (1995) Centrum Hout, The Netherlands.
Timber Engineering STEP 2 (1955) Centrum Hout, The Netherlands. Page, A.V.(1993) The new timber design Code :ECS. The Structural Engineer 71, No. 20, 19 October.
e s i p of floor joists
A timber floor spanning 3.8 m centre to centre is to be designed using timber joists at 600 mm centres. The floor is subjected to an imposed load of 1.5 kN/m2 and carries a
Design to Eurocode 5
243
dead loading, excluding selfweight, of 0.30 kN/m2. Carry out design checks to show that a series of 44mm x 200 mm deep sawn section timber in strength class C22 under service class l is suitable.
Section AA Fig. 10.7 Timber floorjoists (Example 10.1).
Force, kN Length, m Crosssectional dimensions, mm
Stress, Nmm"2 N :=newton kN := lo3 N
1. ~ e o ~ e t r properties i~af
Breadth of beam section, b Depth of beam section, h Span between support centres, L Bearing length, I Joist spacing, Js Crosssectional area, A Second moment of area about yy axis, Iy Section modulus about yy axis, Wy
b:=44+mm h := 200 mm L := 3.8 mm I := 75 mm Js := 600 mm A :=b.h A = 8.8 X lo3 0mm2 b h3 Iy := 12 Iy = 2.93 x lo7 o mm4 b h2 := 6 W y = 2.93 x lo5 o mm3
wy
+
Crosssection
2
Structural Timber
Design
BSEN338 : 1995, Table l Characteristic bending strength Characteristic shear strength Characteristic compression perpendicular to grain Mean modulus of elasticity // to grain 5%tile modulus o f elasticity // to grain Mean shear modulus Average density 3. PartiaJ safety factors
EC:Part 1.1, Table 2."33.1 Permanent actions Variable actions EC:Part 1.1, Table 2.3,3.2 Material factor for timber
Applied dead load, DL Selfweight, Swt Total characteristic permanent load/m length, G k Imposed load, ZL Characteristic variable (imposed) load/m length, e k Design action, F d
A. U l t i ~ a t elimit states
5 . Mo~jfjcatjon factors
Factor for mediumduration loading and service class 1 (kmod, Table 3. l .7) Size factor (kh,Clause 3.2.2) for h > 150mm
&.O'j
:= 6.7 kN mm2
:= 0.63 kN mm2 pmean := 410 kg mm3
Gm,,,
*
*
*
Design to Eurocode 5
Load sharing applies (kls, Clause 5.4.6) Lateral stability (kcrit, Clause 5.2.2): ffective length, Lef for full restraint against rotation
245
kls := 1.1
Critical bending stress (NAD, Clause 6.5) Relative slenderness for bending Lateral instability factor Bearing factor (kc,90,Clause 5.1.5): Bearing length, I No overhang at beam ends, a Distance between bearings, II

I := 75 mm a := 0  m m 21 :=L I zI = 3.73 x 103
mm
In order to make Mathcad accept the following empiricalequation, the indicated units have been added. Bearing factor
kc.go := 1
I ) + a 17(150000 mm mm2 *
Factor for shearinmembers with notched end (kv, Clause 5.1.7.2)
Design bending moment
Design bending stress
Design bending strength
kc.90 = 1 kV :=l for no notch
246
Structural Timber Design
7 . Shearstrength
Design shear stress = 0.53 O N Illm"2
0V.d
Design shear strength
*
fV.d :=
Kmod
*
kV
'fV.k
Kls
YM
fV.d = 1.62 o N mm2 Shear strength satisfactory 8. Bearingstrength
Design bearing load Vd = 3.12 o kN Design bearing stress oc.9o.d
Design bearing strength
= 0.94 O N lllm2
fc.90.d := fc.90.d
*
kmod kc.90 *
kls
'fc.90.k
YM
= 3.45 o N mm2
Bearing strength satisfactory
B. Serviceability limit states Clause 4.3, EC5 For serviceability limit states (Table 2.3.3.2, EG5) Bending moment due to permanent loads, MG Bending moment due to variable loads, MQ
(1) Instantaneous deflection:
For bending and shear due to permanent action, Gk, using equations (4.6) and (4.9)
yM
MG
:= 1.0
= 0.39 o kN m
Qk L2 M Q := 
8
Design to Eurocode 5
247
For bending and shear due to variable action, Q k Maximum allowable deflection due to variable action (Clause 4.3, EC5) %.inst
c U2.inst.adm
Satisfactory (2) Final deflectiondue to variable actions: Modification factor for deformation, for mediumterm under service class 1 (kde$ Table 4.1, EC5)
Maximum allowable deflection (Clause 4,3, EC5) (3) Final deflection due to all actions (permanent and variable): Maximum allowable deflection (Clause 4.3, EC5)
In general, for beams without precamber, it is not necessary to check ~ 2 . f i nas unet.fm is always greater. IO.
~ibration
The UK NAD re: EC5 Clause 4.4.1 Total instantaneous deformations should be less than the lesser of
(&
= 11.41 o m m
and14mm
Therefore 44 mm x 200 mm sawn timber sections in strength class C22 are satisfactory
2
Structural Timber
Design
2 Design of an eccentrica//y/ o a ~c eo ~t ~ ~ n
For the design data given below, check that a 100mm x 250 mm sawn section is adequate as a columnif the load is applied40 mm eccentric about its yy axis. The column is 3.75m high and has its ends restrained in position but not in direction. Design data Timber: C22 Service class: 2
Permanent load: 15kN Variable (mediumterm): load
17 kN
40
~
m
Y
I
Crosssection
Lq= 1.0 x L
Fig.10.8 Column details (Example 10.2).
Force, kN Length, m Crosssectional dimensions, mm
Stress, Nmm” N :=newton kN :=lo3 N
1. G e o ~ e t r j cproperties a~
Column length, L, EEective length, Ley Width of section, b Depth of section, d Crosssectional area, A Second moment of area about yy axis
L = 3.75 rn
L, := 3.75 Lef := 1e 0
m L Ley= 3.75 o m b :=100 mm d :=250 mm A := b  d A = 2.5 X lo4 0 m m 2 I y : = J12” . b . d 3 Iy = 1.3 x lo8 o mm4

a
*
Design to Eurocode 5
Section modulus axis
about yy
b d2 wy:= 6 *
Wy = 1.04 x lo6 o mm3
35 :=
Radius of gyration about yy axis
ly
Slenderness ratio about yy axis
hy :=
Secondmomentof zz axis
Ay = 51.96 I, := h d b3 Iz = 2.08 x lo7 o mm4
iy = 72.17 o mm
% IY
area about
Radius of gyration about 22 axis
a

l, := *
v
%
iz = 28.87 omm
h, := L;,f
Slenderness ratio about 22 axis
iz
h, = 129.9 2. Timber strength properties
BS EN 338 : 1995, Table l Characteristic bendingstrength Characteristic compression parallel to grain 5%tile modulus ofelasticity parallel to grain
fm.k := 22
N mmm2 N
fc.o.k :=20
Eo.05 := 6.7  k N .mm2
3. Partial safety factors
EC5 : Part l .l,Table 2.3.3.1 actions Permanent actions Variable EC5 : Part l .l, Table 2.3.3.2 Material for timber
:= 1.35 Y Q := 1.5
yw := 1.3
4. Actions
Characteristic permanent load, G k Characteristic variable load, Qk Design action, N d Eccentricity, er
G k := 15
kN
Q k := 17 * kN
Nd:=YG'Gk+YQ.Qk N d = 45.75 o kN
ey := 40mm
249
250
Structural Timber
Design
Design moment due to eccentricity about yy axis
5. ~odifjcation factors
Factor formediumduration loading and service class 1 (kmod, Table 3.1.7) Load sharing does not apply (kis, Clause 5.4.6)
kt, :==1.0
6. Bending strength
Design bending moment about yy axis Design bending stress about yy axis
kid
= 1.83 o k N . m
No bending moment about zz axis, thus Design bending strength
7. Compressionstrength
Design compressive stress 0C.O.d
Design compression strength

= 1.83 o N mm2 kmod kls
*
fc.0.k
fc.0.d
:=
fc.0.d
= 12.31 o N.mm2
YM
Buckling resistance (Clause 5.2.1, EC5): Euler critical stresses about yy and zz axes
Design to Eurocode 5
251
fc.0.k
Relative slenderness ratios
0c.crit.y 5el.y
=Z
0.9
Both relative slenderness ratios are >0.5, hence conditions in Clause 5.2.1 (4) apply: For a solid timber section, For a rectangular section (Clause 5.1.6), Thus
pc := 0.2
k m := 0.7
and Hence
kc, := kY
+ Jh:,.
kc., = 0.81
and
l
kc., :=
kc., = 0.18 Check the following conditions:
0c.O.d kc., ‘fc.0.d
+
0m.z.d fm.z.d
+km”
Gmyd fm.y.d
0.91 < 1.0 Satisfactory < 1.0 Satisfactory
Therefore a l00 mm x 250 mm sawn section timber in strength class C22 is satisfactory
Example 70.3 Design of a timbertotimber nailed tension splice joint
A timbertotimber tension splice joint comprises two 47mm x 120mm inner members and two 33 mmx 120 mm side members ofstrength class C22 timber in service class 2. It is proposed to use 3.35 mm diameter, 65mm long round wire nails without predrilling. The joint is subjected to a permanent load of 2 kN anda mediumterm variable load of 3 kN. Determine the required number of nails with a suitable nailing pattern.
Structural TimberDesign
A Nailing zones
(a)
Splice joint (Example 10.3).
Force, kN Length, m CrosssectiQna1dimensions, mm eo~etri~af
Stress, Nmmm2 N :=newton kN := lo3 N +
prope~ies
Thickness of side members, tl Thickness of inner members, t2 Width of timber members, h Crosssectional area of side members, A, Crosssectional area o f inner members, Ai, Nail diameter, d Nail length, Inail Nail pointside penetration, tpoint
:= 33 .mm :=47 mm h := 120 mm A, := h * tl A, = 3.96 x lo3 o mm2 Ai, := h t2 tl
t2
*
*
Ai, = 5.64 X lo3 o mm2 d := 3.35 mm Inail := 65 mm

+
$point
tpoint
Clause 6.3.1.2, EC5 Minimum allowable pointside penetration For overlap nailing without predrilling
. ~
= 32 o mm
tpoint.adm := 8 d tpoint.ah = 26.8 o mm t2 tpoht= 15 o mm >4 d = 13.4 o,mm Both pointside penetration and overlap nailing are satisfactory
is t r e ~ ~properties t h~ ~
BS EN338 : 1995, Table I Characteristic tension parallel to grain Characteristic density
:= haif tl
r
ft.0.k := 13
N mm2
Pk := 340 * kg
*
m3
Design to Eurocode 5
253
~ i asafety l factors
EC5 : Part 1.1, Table 2.3.3.1 Permanent actions Variable actions
YG :=1.35
EC5 : Part I J , Table 2,3,3.2 Material factor for timber Material factor for steel
yjl.l.timber
YQ :=1.5
YM.steel
Characteristic permanent load, G k Characteristic variable load, (& Design action, N d
A. ~ltimatelimit states
:= 3 :=1.1
:= 2 kN Q k :=3 kN
Gk
+
Nd:=YG'GkS'YQ*ek N d = 7.2 o kN
difi~ation factors
Factor for m e d i ~  d u r a t i o n loading and service class 1 (kmod, Table 3.1.7) Size factor ( k h , Clause 3.2.2) for h S50mmis the lesserof:
kmod
kh
:=0.8
:= 1.3 and
k h ==
.
k h :=
1.05
~ e n ~ strength i o ~ of t i ~ b e r
Designtensionstressparallel grain in side members
to
cft.0.d :=
*
cft.0.d
Designtensionstressparallel grain in inner members
Nd 2 As
to
cft.0.d := Dl.0.d
Designtension strength parallel to grain
= 0.91 o N mm2 Nd A in
= 1.28 o N mm2
ft.0.d :=
kmodkh
'ft.0.k
YM.timber
= 8.37 o N .mmw2 Tension strength satisfactory
ft.0.d
7 . ~ ~ ~ e d strength d i n g of timber
Characteristic embedding strength without predrilling Since the following is an empirical equation (with hidden units inthe coefficient part), for Mathcad to produce the correct units, multiply it by the units shown inside the parentheses; or alternatively use the equation without units for Pk and d. Thus:
254
Structural Timber
Design
pk dOv3 (sec2 .m23 1.26 lo5) = 19.42 o N mm2
fh.k := 0.082 fh.k
Design embedding strength for headside timber
8.
+
fh.1.d :=
kmod ' f h . k YM.timber
fh.1.d
for pointside timber
*
= 11.95 o N mm2
fh2.d Z"fh.1.d f h 2 . d = 11.95 0 N
*
Yield momentof nails
Yield moment o f a nail (using a similar treatment for the units as above) Design yield moment

My.k
d2*6.(kg. m*6 s e c 2 n6.31 io4) = 4.17 x lo3 o N mm
1My.d
:=kmod
M y . k := 180
*
My.k
YM.stee1
= 3.03 x lo3 o N mm
My.d
9. Loadcarryingcapacity
EC5, Clause 6.2.1 For a timbertotimber joint with nails in single shear, design resistance per shear plane, R d , is the lesser of to R d f (see Table 10.6),where: is the pointside penetration length and the ratio, P
t2 := tpoint
Failure mode (a)
&.a
t2
fh.1.d
P:=fh.2d p=1
Rd.a
Failure mode (b)
:=fh.l.d
*
= 1.32 X
Rd.b :=fh.l.d t2
d io3 0 N d .P
$1
*
*
Rd.b
= 1.28
Rd.c
= 538.94 o N
X
lo3 0 N
Failure mode (c)
Failure mode (d)
Design to Eurocode5
Failure mode (e)
Failure mode (f) Rd.f
Therefore the design resistance per nail is the ~ n i m u mof:
= 542.17 0 N
R d := [Rd.aRd.bRd.cRd.dRd.e
&.f]
min(Rd) = 538.94 o N
Number o f nails per side required is For a symmetrical nailing pattern adopt 16 nails per side. Thus,
Nnails := 16
l 0. NaiJ spacing EC5, Table 6.3.1.2 Nail diameter Angle to grain Minimum spacing parallel Minimum spacing perpendicular
d := 3.35 a :=0
+
a1 := ( 5 5 lcos(a)l). d a1 = 33.5 o mm +
a2 := 5 . d a2 = 16.75o mm a3.t := (10 5* ( c x ) ~* )d Minimum loaded end distance ~ 3 =, 50.25 ~ o mm . 10 ~ d Minimum unloaded end distance ~ 3 := ~ 3 =. 33.5 ~ o mm not required ~ 4 . 1:= (5 5 lsin(a)l) d Minimum loaded edge distance ~ 4 =. 16.75 ~ o mm not required Minimum unloaded edge distance ~ 4 :=. 5 ~ d ~ 4 =. 16.75 ~ o mm
+ ICOS
+
+
B. Serviceabilitylimitstates 11. Joint slip EC5, Clause 4.2 Nail diameter Timber characteristic density
d :=3.35 mm pk = 340 o kg
+
255
56
Structural Timber
Design
Slip modulus (kser, Table 4.2, EC5) and using a similar treatment for the units of this empirical equation, for nails without predrilling Design load for serviceability limit states Load per nail
~nsta~taneous slip Uimt
= 0.47 0
Assuming allnails to slip by the same amount,eachinner 0.47mmrelative to side members. Therefore the total instantaneous slip is: ~ o d i ~ c a t i ofactor n for deformation due to creep ( k d d , Table 4.1, EC5) for ~ e ~ a n e load, nt for mediumterm variable load, Final slip per nail is calculated as
Final slip of joint
Fig.10.9(b) Nail spacings and distances.
member willmove
by
In the previous versions ofBS 5268:Part 2 the softwood timber section sizes for sizes of sawn and processed were specified to BS4471: 1978 Spec~cation softwood, which gave sizes and tolerances for three types of surface finish: sawn, planedand regularised. BS 5268 :Part 2 :1996 requirements for timber target sizes are those given in BS EN 336 :1995 Structural timber. Coniferous andpoplar.Sizes.Permissibledeviations and in its National Annex. This standard specifies two tolerance classes: tolerance class 1 (Tl) is applicable to sawn surfaces, and tolerance class 2 (T2) applicable to planed timber. Regularised timber can be achieved by specifying T1 for the thickness and T2 for the width. The commonly available lengthsand crosssection sizes are also listed in the National Annex of BS EN 336, and are referred to as target sizes. The targetsize is defined as the desired timber section size (at 20% moisture content) whichcanbeused, without further modification, for design calculations.
A1 Basicsawn softwood timbersectionsizeswhosesizesandtolerances comply with BS4471: 1978, at 20Y0 moisture content
Table
Width
Thickness
(1
300
250 225 200 175 150 125 100 75 36 38 44
47
50 63 75
100
150
200 250 300
X
X
X X X X
X X X X
X X X
X X X
X X X X
X X X X X X
X X X
X X X X X X
X X X X X
X X X X
X X X X X X
x x x
x x x
x x
x x X
x X
257
258
Structural Timber
Design
Table A2 Customary target sizes of sawn structural timber (Table NA.2, BS EN 336 :1995)
Thickness (to tolerlance class 1) 0 225 200 (mm) 175 150 125 100 22 25 38 47
Width (mm) (to tolerance 1) class 300
75 X X X
63
75 100 150 250 300
X
X
X
X
X X X X X X
X X X X X
X X X X X X X
X X X
X X X X X X X
X
X
X
X
x
X
x X X
X
X X
Note I Certain sizes may not be obtainable in the customary range of species and grades which are generally available Note 2 BS EN 336 has a lower limit of 24 mm. However, as thinner material is used in the UK the customary sizes of such material are also listed here
Table A3 Customary lengths of structural timber (Table NA. 1, BS EN 336 :1995) Length (m) 2.10 1.80 2.40 2.70
3.005.10 4.20 3.30 4.50 3.605.70 4.80 3.90
5.40
6.00 6.30 6.60
7.20
6.90
Note Lengths of 5.70m and over may not be readily available without finger jointing
In general, the differences between BS 4471 and BS EN 336 are minor and should not present any problemsto specifiers and suppliers in the UK.'For comparison purposes, in Table A1 the basic sawn softwood timber section BS 4471 are given.The sizeswhosesizes and tolerancescomplywith customary target sizes, whose sizes and tolerances comply with BS EN 336, for sawn strucural timber are given in Table A2. Table A3 gives the range of lengths of sawn softwood structural timber.
Reference 1. Fewell, A.R. (1997) Changes to the requirements and supply of timber structural materials, UKTEG Seminar, I. Struct. Eng., London, February.
Appendix B
Weights of Building Materials
Some typical building material weights,for determination of dead loads, are tabulated in Table B 1. The values are based on BS 648 :1964. Table B1 Weights of building materials (based on BS 648 :1964) Material
Asphalt
Roofing 2 layers, 19mm thick Dampproofing, 19 mm thick
Bitumen roofing felts
Mineral surfaced bitumen per layer
Glass fibre
Slab, per 25 mm thick
Gypsum panels andpartitions Building panels 75 mm thick
Lead
Sheet, 2.5mm thick
Linoleum
3 mm thick
Plaster
Unit mass 42 kg/m2 41 kg/m2 44 kg/m2
25 kg/m2 44 kg/m2
30 kg/m2
6 kg/m2
Two coats gypsum, 13mm thick
22 kg/m2
Corrugated
4.5 kg/m2
Plastic sheeting Plywood
per mm thick Rendering or screeding Cement :sand (1 :3), 13mm thick
0.7 kg/m2
(depending upon thickness and source)
2478 kg/m2
Slate tiles Steel
Solid (mild) Corrugated roofing sheet per mm thick
Tarmacadam
25mm thick
Tiling
Clay, for roof
Timber
Softwood Hardwood
Water Woodwool
Slab, 25mm thick
30 kg/m2
7850 kg/m3 10 kg/m2
60 kg/m2 70 kg/m2 590 kg/m3 1250 kg/m3 1000 kg/m3 15kg/m2
259
itis i
BS 52 Part 2 :1996 Part 3 :1998 Part 4 :1978 Part 5 :1989 Part 6 :S996 Part 7 :1990
EC5
t
Structural use of timber Code of practice for permissible stress design, materials and workmanship. Code of practice for trussed rafter roofs. Fire resistance of timber structures. Preservation treatments for constructional timber. Code of practice for timber frame walls. Recommendations for the calculation basis for span tables.
DD EN77 199511 : 1994 E~rocode5: structures General rules and rules for buildings (together with United Kingdom National Application Document)
BS EN 384 :1995 Structural timber. Determination of characteristic values of mechanical properties and density.
BS EN 385 :1995 Finger jointed structural timber. Performance requirements and minimum production requirements.
BS EN 518 :1995 Structural timber. Grading. Requirements for visual strength grading standards.
BS EN 336 :1995 Structural timber. Coniferous and poplar. Sizes. Permissible deviations.
BS EN 338 :S995 Structural timber. Strength classes. BS EN519 :1995 Structural timber. Grading. Requirements for machine strength graded timber and grading machines. 260
RelatedBritishStandardsfor
Tmber Engineering
261
BS EN 301 :1992 Adhesives, phenolic and aminoplastic, for loadbearing timber structures: classification and performance requirements.
BS EN 386 :1995 Glued laminated timber. Performance requirements and minimum production requirements.
BS EN 390 :1995 Glued laminated timber. Sizes. Permissible deviations.
BS EN 20898Mechanicalproperties
of fasteners.
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actions, 221, 244, 249, 253 adhesives, 82, 195 additional properties, 18 axial compression, 5663, 86, 89, 119, 148, 229 axial compressing and bending, 612, 68, 71, 78,122,157,23031 axial loading, 5681,119,151,153 basic stress, 10 beams axes, 27, 225 circular, 29 curved, 9092, 11318 flitched, 38, 39 glued laminated, 26, 82122, 225 notched, 367,46, 2278 plywebbed, 26, 12342 rectangular, 26, 289 solid section, 26, 225 straight, 83 trimmer, 38, 39 bending deflection 33, 90, 129, 23940 strength, 2256 stress, 278, 612, 65, 86, 128, 157, 2256 bearing length, 356, 2289 bearing stress, 346, 96, 101, 118, 228 bolts, 160, 163, 165, 17988, 231, 235, 238 boxed beams, 123 British grown timber, 14 camber, 33, 83, 90, 239 cantilever, 239 ceiling tie, 204, 21 1 centre of geometry, 185 characteristic values, 220, 222 clear span, 289
columns, see Chapters 5, 6, 8 and 9 builtup, 143 combined stresses, 56, 612, 65, 230 effective length 578, 1456, 230 glued laminated, 83, 119 load sharing, 63 permissible stresses, 6062, 645, 86, 89, 1467 rectangular solid, 5664, 230, 231 slenderness ratio, 5660, 1456, 23031 spaced, 14358 spacer blocks, 1448 combined stresses, 56, 612, 65, 230 computer, 2 1 compression, 5664, 86, 89, 121, 147,22831 concentric load, 6061 connections, see Chapters 9 and 10 bolted, 163, 17988, 231, 235, 238 dowelled, 163, 17988, 231, 235, 238 glued, 1956 nailed, 146, 163, 164, 16675,231,235,238 nailplates, 194 punched metalplates, 160, 1934 screwed, 146, 163, 165, 1759, 238 shearplates, 160,163,165,1913 splitrings, 160,163,165,1913 toothedplates, 160,163,165,18990 conversion, 3, 4, 5, 8 creep, 220, 240, 241 decking, 37,39 defects in timber, 35 deflection bending, 33, 90, 129, 23940 camber, 33, 83, 90, 239 cantilever, 239 creep, 220, 240, 241 final, 24041 263
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Index
deflection ( c o ~ ~ ~ ~ ~ e ~ ) instantaneous, 23942 limits, 32, 239, 241 modification factor, 240 permissible, 32, 239, 241 shear, 34, 90, 129 slip in joints, 1612, 2412 slip modulus, 162, 2412 depth to breadth ratio, 30 depth factor, 29, 87, 224 design philosophy permissible stress, 10 1 1 limit states, 1112, 220 Douglas fir, 14, 164 dowels,160,163,165,17988 dowelled joints, 163, 17988, 231, 235, 238 durability, 195 duration of load, 11, 1617, 127, 174, 179, 18082 eccentric load, 61, 69, 248 effective crosssection, 1623 effective geometrical properties, 1245 effective length, 578, 1456, 230 embedding strength characteristic, 2345 design, 2345 end rotation, 226 European, 2 19 Eurocodes, 2 19 Eurocode 5, see Chapter 10 factor of safety, 221, 223 Finnish plywood, 133, 204, 216 flexural members, see Chapter 4, see also beams floorboards, 379 form factor, 29 geometrical properties, 1245 geometric centre, 185 glued joints durability classification, 195 pressure, 196 strength, 1956 stress, 196 glulam, see glued laminated glued laminated, see also Chapter 6 assembly, 82 bending deflection, 90 bending stress, 86, 89
column, 119 combinedgrade, 845 curved beams, 9092 horizontally laminated members, 847, 121 modification factors, 8587 radial stress, 912 shear deflection, 90 singlegrade, 84 vertically laminated members, 8790, 120 grade stresses, 13, 15 hangers, joist, 39 Hankinson’s equation, 180, 185, 188 hardwoods, 56,13,15,167 holes bolts, 162, 179, 231 dowels, 179, 231 predrilled, 164,167,175 Ibeams, 123,124,131,137 imposed loads, 17 improved nails, 167, 174 instantaneous deflection, 23942 joint slip, 2412 Johansen’s equations, 23 1 joinery joints, 159 joints, see Chapters 9 and 10, see also connections bolted, 163, 17988, 231, 235, 238 dowelled, 163, 17988, 231, 235, 238 glued, 1956 nailed, 146, 163, 164, 16675,231,235,238 nailplates, 194 punched metalplates, 160, 1934 screwed,146,163,165,1759,238 shearplates, 160,163,165,1913 splitrings, 160,163,165,1913 toothedplates, 160,163,165,18990 joists, 18,379 hangers, 39 trimmer, 38, 39 knots, 3 lateral support, restraint, 3031, 131, 226 lateral stability, 303 1, 13 1 length, see effective length limit states, design philosophy, 1112, 220
index
load duration, 11,1617,127,174,179, 18082 load sharing, 18, 63, 226 longterm loading, 17, 127, 174, 179, 18082 224 machine stress grading, 12 Mathcad, see Chapter 3, 215 mechanical fasteners, see connections,joints mediumterm loading, 17,127,174,179, 18082,224 metalplate fasteners, 1934 modulus of elasticity minimum, 15, 58, 60 mean, 15, 87, 222 5th percentile, 222 moisture content, 67, 16, 224 nails, 160, 162, 231, 235, 237 nailed joints, 146, 163, 164, 16675,231, 235, 238 National Application Document (NAD), 219, 226,240,241 noggings, 62,63 notched beams, 367,46,2278 panel shear, 127, 130 panel bending, 126 partial coefficients for actions, 221 for material properties, 221, 223 permissible stresses solid section timber, 27, 28, 35, 36, 60, 61, 62, 65 glued laminated, 867, 89, 92 plywood, 127,128,130,131 plywebbed beams, see Chapter 7 plywood boarding, 39,126 properties, 125,127 permissible stresses, 127, 128, 130, 13 1 punched metalplates, 160,1934 radial stress, 9092 radius of gyration, 57 redwood, 14 rolling shear stress, 13031 scarf joints, 159 screws, 160, 162, 231, 235
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screwed joints, 146, 163, 165, 1759, 238 second moment of area, 33, 57, 131 section size, 17, 30, 257 serviceclasses, 1416, 127, 175, 179, 185, 190, 193, 2234 serviceability limit states, 12, 220, 23942 shear deflection, 32, 34, 90, 129 shear modulus, 18, 12930 shearplate connectors, 160, 163, 165, 19 13 shear stress, 15, 19, 367, 86,89, 13031, 161, 2278, 236 shortterm loading, 1617,127,174,179, 18082, 224 Sitka spruce, 14 slenderness ratio, 5660, 1456, 2303 1 slip, in joints, 1612, 2412 slip modulus, 162, 2412 spaced columns, see Chapter 8 span, clear, design, effective, 289 species, 14 staples, 235 steel plates, gusset plates, 170, 174, 177, 184, 189, 233, 235 stiffeners, 1312 strength classes, 1315, 222 strength properties, 2212 studs, 623 tapered beams, 83, 98 tension members, 635, 229, 230 thickness, plywood, 125 tongued and grooved (t & g) floorboards, 38, 39 toothedplate connectors, 160, 163, 165, 18990 transformed section method, 1245 ultimate limit states, 12, 220, 22438 variable loads, actions, 221 very shortterm loading, 17, 127 wane, 4, 5 webstiffeners, 13 12 whitewood, 14 width factor, 64 wind loading, 17, 63, 224 yield moment characteristic, 2345 design, 2345
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orksheet
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Structural Timber Design
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