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Timber frame housing:UK Structural recommendations

(TRADAT E C H N 0 L 0 G

Third edition 2006 TRADA Technology LtdChiltern HouseStocking LaneHughenden ValleyHigh WycombeBuckinghamshire HP14 4ND

t +44 (0)1494 569600f +44 (0)1494 565487e [email protected] www.trada.co.uk

First published in Great Britain by TRADA Technology Ltd. 2006.

Copyright of the contents of this document is owned by TRADA Technology.All rights are reserved. No copying or reproduction of the contents of theprinted book is permitted without the consent of the copyright holder, TRADATechnology, application for which should be addressed to the publisher.

© TRADA Technology 2006

Whilst every effort is made to ensure the accuracy of the advice given, thecompany cannot accept liability for loss or damage arising from the use of theinformation supplied.

ISBN (10 digit): 1-900510-50-2

ISBN (13 digit): 978-1-900510-50-9

Cover illustration courtesy CCB Evolution

Printed in England on paper with a high recycled content

2

TRADA Technology Timber frame housing: UK Structural recommendations

Contents Page

Introduction 5

I Loading on structural elements 6

1.1 Structural duties of timber frame elements 61.1.1 Timber frame wall panels 61.1.2 Horizontal diaphragms 7

1.2 Determination of loads acting on each element 91.2.1 Calculation of vertical load 101.2.2 Calculation of wind loads 111.2.3 Calculation of roof loads 191.2.4 Checking strength and stability 20

1.3 Weights of materials 21

1.4 Weights of typical constructions 231.4.1 Roofs 231.4.2 Floors 231.4.3 Walls 24

1.5 General requirements of prefabricated wall, floor and roof elements 24

2 Walls 25

2.1 Wall studs 252.1.1 Minimum desirable stud sizes 252.1.2 Multiple studs 252.1.3 Lateral support 262.1.4 Load capacities 262.1.5 Design procedure 292.1.6 Drilling of studs 30

2.2 Framing around openings 302.2.1 Lintel capacities 31

2.3 Racking resistance of timber frame walls 412.3.1 Basic design and materials 412.3.2 Adhesively bonded panels 422.3.3 Overturning effects 432.3.4 Deflection 46

3 Floors 48

3.1 Joists and beams 483.1.1 Reactions, moments and deflections in a single span, simply supported beam 493.1.2 Solid timber beams 503.1.3 Structural timber composites 523.1.4 Prefabricated engineered timber joists 523.1.5 Steel fitch beams 53

3.2 Floor decking 63

TRADA© TRADATechnologyLtd2006


4 Roofs 64

4.1 Trussed rafter roofs 644.1.1 Bracing 64

4.2 Connections 67

5 Foundations 68

5.1 Loading 68

6 Multi-storey buildings 69

6.1 General design considerations 69

6.2 Construction 69

6.3 Disproportionate collapse 69

7 Example of calculations for a complete dwelling 70

7.1 Vertical loads 72

7.2 Horizontal loads 73

7.3 Overall stability calculations 757.3.1 Wind pressures and self-weight 757.3.2 Overturning 767.3.3 Sliding 777.3.4 Resistance to wind uplift 78

7.4 Racking calculations 787.4.1 Wind loads 787.4.2 Racking forces 797.4.3 Design method for racking 797.4.4 Recommended design procedure 91

7.5 Wall panel studs 927.5.1 Ground floor rear wall — very short term 927.5.2 Ground floor rear wall — medium term and long term 947.5.3 Ground floor gable wall — very short term 95

7.6 First floor platform 96

7.7 Wall panel lintels 987.7.1 2.4 clear span opening at eaves level 987.7.2 1.2 m clear span opening at first floor level 99

7.8 Cripple studs 997.8.1 First floor rear wall — very short term 997.8.2 Check bearing on bottom rail of panel — medium term load 1017.8.3 Ground floor front wall — very short term 1017.8.4 Check bearing on bottom rail of panel — medium term load 102

8 References and further reading 103

(A© TRADA Technology Ltd 2006


Introduction

Timber Frame Housing - Structural Recommendations covers well establishedprinciples and methods for the structural design and strength and stabilitychecking of timber frame buildings. The guidance is based on therecommendations in BS 5268-2 Code of practice for the structural use oftimber. Permissible stress design, materials and workmanship, BS 5268-6.1:1996 Structural use of timber. Code of practice for timber frame walls.Dwellings not exceeding four storeys but it also includes procedures nowwidely used in the design of timber frame houses but which are notspecifically covered in these Codes. Worked examples, including calculationsfor a complete house, are included.

The guidance does not exclude other ways of showing that a design satisfiesthe requirements of Building Regulations and Standards regarding strengthand stability. The design of timber frame buildings, for example, can beundertaken using the guidance given in the Eurocodes suite, BS EN1990:2002 Eurocode. Basis of structural design, BS EN 1991 Eurocode 1.Actions on structures (in 5 parts) and BS EN 1995-1-1:2004 Eurocode 5.Design of timber structures. General. Common rules and rules for buildings.However, these do not include specific guidance for timber frame structuresand currently most buildings are designed or checked using the relevant partsof the British Code, BS 5268.

Materials and components are covered by other standards as listed in theReferences. Span tables for floor joists, rafters and purlins for roofs, and joistsfor flat roofs are available in a separate TRADA Technology document Spantables for solid timber members in floors, ceilings and roofs (excluding trussedrafter roofs) for dwellings.

This publication deals solely with the engineering aspects of timber framedesign; it does not address issues such as fire safety and performance,thermal and acoustic performance, weatherproofing and durability and designdetailing. These are covered in a companion publication Timber FrameConstruction, last updated in 2001. Timber frame: Standard details for housesand flats, published in 2006, provides typical details for 'open' panel timberframe wall panels for three storey houses requiring 30 minutes fire resistanceand for flats requiring 60 minutes fire resistance. The details show floors withsolid timber joists which can readily be used for timber frame buildings up tofour storeys in height.

This is the third edition of Timber Frame Housing - StructuralRecommendations. The first edition, published in 1979, was itself an updatedversion of Section 9 Structural Recommendations of the TRADA DesignGuide for Timber Frame Housing first issued in 1967.

TRADA TechnologyTRADA Technology is the leading independent timber research, consultancyand information provider for the construction industry. Technical expertise is atthe heart of our business, working with clients to get the most from their timberproducts. Specialist services include frameCHECK (timber frame qualityassessment), structural and condition surveys on site, design assistance,materials and product evaluation. In the event of a dispute, expert witnessservices are also available. Clients cover the whole construction deliverychain, including architects, engineers, designers, specifiers, contractors andbuilders, in the UK and overseas.

AcknowledgementsTRADA Technology wishes to acknowledge the assistance of CCB Evolutionin the preparation of this publication. CCB Evolution, Consultants inprefabrication and modular design, is a specialist engineering consultancywith expertise in timber frame structures and cold rolled steel frames. Moreinformation is available from www.ccbevolution.co.uk.

© TRADA Technology Ltd 2006


I Loading on structural elements

1.1 Structural duties of timber frame elements

1.1.1 Timber frame wall panelsTimber frame panels used in house construction have three major structuralduties to perform: support of vertical loading, resistance to deformationscaused by horizontal loading in their plane and resistance to wind loadingperpendicular to their plane.

Resistance to vertical loads is checked according to normal engineeringprinciples, bearing in mind that in timber design the duration of each load(long-, medium-, short- or very short-term) has to be considered because thestrength of the timber members depends on the duration of the loads.The racking or shear resistance of wall panels to horizontal load is calculatedaccording to procedures set out in BS 5268-6.1 Structural use of timber—Code of practice for timber frame walls — Dwellings not exceeding fourstoreys, which are based on data from tests on typical timber frame wallpanels.

© TRADA Technology Ltd 2006 61 I C H N 0


1.1.2 Horizontal diaphragmsThe horizontal diaphragms formed by floor, ceiling and roof systems areusually required to take loads in their own plane.

The illustration shows a floor and roof diaphragm resisting wind load on agable wall. The ends of the gable wall are supported directly by the front andrear walls of the building, and the bottom edge by the foundation. The load onthe rest of the wall is transferred via the horizontal diaphragms into the frontand rear walls where their shear resistance can transfer it to the foundations.The first floor diaphragm takes half the neff load on the ground floor walls plushalf the nett load on the first floor walls. The roof diaphragm takes half the nettload on the first floor walls plus load from the roof. For wind parallel to theridge it is usual to assume that half the nett horizontal wind load on thespandrels or roof is transferred to the roof diaphragm; for wind perpendicularto the ridge all the nett horizontal roof load is transferred to it.

Where adjacent panel edges are fastened with nails or screws to the sametimber members such as joists or blocking, the resulting connection betweenthe panels enables shear forces to be transferred from one panel to the next.This is an essential part of the diaphragm action.

It may be assumed that conventional floors and flat roofs, in which a wood-based panel product is fastened to timber joists, have adequate strength andstiffness as horizontal diaphragms, provided that:

• the diaphragm span : depth ratio does not exceed 2:1 in either winddirection (BS 5268-6-2 Clause 6.5)

• the span does not exceed 12 rn between supporting walls (BS 5268-6-2Clause 6.5)

• the fixing around the edges of the panels complies with standardrecommendations (eg 3.00 mm diameter ringed shank nails at 150 mmcentres for plywood or 3.35 mm ringed shank nails at 300 mm for woodparticleboard and OSB, with a length equal to 2.5 times the boardthickness)

• the perimeter of the diaphragm is attached to the walls with fastenings ofequivalent strength.

Plasterboard ceilings in roofs that comply with BS 5268-3 Annex A may alsobe assumed to provide adequate diaphragm action provided that truss clipsare used to secure every truss to a head binder and the fixing around theedges complies with standard recommendations (3.5 mm diameterplasterboard nails or screws 40 mm long at 150 centres).

It is recommended that in areas of high wind load (eg with a dynamic windpressure> 1500 N/rn2), and always for horizontal diaphragms outside therange given above, the required fastener spacings should be calculated byensuring that

© TRADA Technology Ltd 2006 ci:13,


Va � h adm,1 + dm,2

Slkedge,1 S2kedge,2

where the suffices 1 and 2 refer to the decking and plasterboard respectively(if present) and

Va = maximum shear force (normally half the total wind load applied tothe diaphragm) (N)

h = horizontal depth of diaphragm measured in direction of wind (mm)fadm = permissible load per fastener for very short-term loading (N)k = a value shown in the following table

k

Blocked floors and flat roofs, in which all four edges of the panelsare joined to adjacent panels by fastening them to joists orblockingsFloors and flat roofs in which adjoining tongued and groovededges are not fastened to the same timber member but are gluedtogether

1.0

Unblocked floors and flat roofs in which adjoining tongued andgrooved edges are not fastened to the same timber member andare not glued together

1.5

Plasterboard ceilings in which all four edges of the panels arejoined to adjacent panels by fastening them to joists or blockings

1.4

For a plywood listed in BS 5268-2 the code provides instructions forcalculating fadm for specified types of hand driven nail. A machine driven nailmade from steel with a tensile strength of 600 N/mm2 or more has a load-carrying capacity at least equal to that of a standard hand-driven nail of thesame length and diameter.

For wood chipboard and OSB, there is no method in BS 5268-2: 2002 fordetermining fadm. The value of fadm for Group I plywoods of a similar thickness,reduced by a factor of 0.9 for OSB or 0.8 for wood chipboard, should be safeto use with suitable structural grade materials. (Group 1 plywoods are listedin a footnote to BS 5268-2 Table 63.) Alternatively a value for fadm in OSB canbe calculated using Eurocode 5 (BS EN 1995-1-1) for wind loading in serviceclass 1 and dividing the design value by 1.5.

For plasterboard fastened to trussed rafters with plasterboard nails or screwsof at least 3.5 mm diameter, the Trussed Rafter Association recommends avalue for fadm of 190 N for 12.5mm thick board and 213 N for 15mm thickboard.

For unconventional building forms or unconventional types of floor or roof, thestructural adequacy of the diaphragm action should be checked, not only forthe fasteners but also for shear, bending and deflection. Where necessary,specialist advice should be sought.

TRADA© TRADAlechnology Ltd 2006

8


1.2 Determination of loads acting on each elementLoads in general: The following figures show various different combinations of loading that occur

in two-storey terraced housing, and how the forces and loads acting on thefloor, wall and roof systems may be calculated.

In each example, a load bearing internal wall is used on the ground floor tobring about economy in design. If a more flexible plan is required the joistsmay often be designed to span the whole ground floor without interior support.

Prefabricated timber I-joists are particularly common for longer spans, whiletimber-flanged joists with open metal webs, glulam, laminated veneeredlumber (LVL) and other structural timber composites can all span significantlyfurther than solid timber joists of the same depth.

Dead load: The dead load is the weight of the construction and should include water tanksplus their contents, and services such as pipes, ducts, heating units andsimilar appliances that are classed as fixtures. Water tanks and heating unitscan be positioned over load bearing walls to avoid subjecting beams, trussesand joists to heavy dead loads. Where tanks are supported by roof trusses theadditional dead load should be considered in the truss design.

The dead loads should be calculated from the unit weights given in BS 648Schedule of weights of building materials or from the actual known weights ofthe materials used. Weights of materials commonly employed in houseconstruction are given in Section 1 .3.

Imposed floor load: The imposed floor load specified by BS 6399-1 Loading for buildings. Code ofpractice for dead and imposed loads for a self-contained unit designed foroccupation by a single family is a long-term distributed load of 1.5 kNlm2 or apoint load of 1.4 kN, whichever is more severe. However, for solid timber floorjoists BS 5268-7.1 recommends instead a long-term distributed load of1.5 kN/m2 for joists with an effective span of 2400 mm or more, or for shorterjoists a load of 3.6 kN per metre width of floor (measured perpendicular to thespan) uniformly distributed over the entire span. All these loads may also beused in self-contained dwellings within multi-storey buildings.

Imposed ceiling load: The imposed ceiling load specified by BS 6399-1 for dwellings is a long-termdistributed load of 0.25 kN/m2, assuming that the space above the ceiling isused for storage rather than a living room.

Imposed roof load: The exact value of the imposed roof load depends on the location and size ofthe building, and BS 6399-3 Loading for buildings. Code of practice forimposed roof loads should be consulted.

In general for roofs up to and including 30° pitch where access is limited tomaintenance and repair, the imposed load is a medium-term distributed loadof 0.75 kN/m2 measured on plan, or a short-term point load of 0.9 kN,whichever is more severe. For roof slopes between 30° and 60° the distributedload may be obtained by linear interpolation between the value for a 30° pitchand nil. For roof slopes of 60° or more the distributed load is nil. The pointload may be ignored for slopes of 30° or more.

Wind loads: Wind loads are calculated using BS 6399-2 Loading for buildings. Code ofpractice for wind loads.



1.2.1 Calculation of vertical load

1.2.1.1 Roof and floor spanning front to rear (load bearing internal wall on ground floor)

Roof and ceiling load Wç kN/rn2 total Stud and joist spacing = S

,Shaded areas carry roof orfloor loads

Load at each stud pointL= WrXSX_2

Load in each stud

4\Nr xsxJ+(Ww xsxh)

Load at each interior joistsupport

=(Wf xsx.J+(Ww xsxh)

Floor load W k N/rn2

Load at each stud point

1\NrXSxJ +(wwXsXh)+\NfXsX_J

Load in each stud

1\NrXSxJ +(wWxsxh)+[wfxs4j+(W xsxh)

For weights of materials and construction, see Section 1.3Wr = 1.97 kN/m2 W = 0.86 kN/m2 for tile hanging (upper storey)

W1 = 1.82 kN/m2 W = 0.35 kN/m2 for board cladding (lower storey)

W = 0.25 kN/m2 for plasterboard internal wall

For L 7.8 rn s 0.6 m, A = 4.2 m, B = 3.6 m, h 2.55 m (2.4 m for internal wall)Front wall: . L 7.8

Load per stud point at eaves = WrxSx=i.97xO.6x__Load at foot of each upper storey = 4.61 +(w h)= 4.61 + (0.86x 0.6x 2.55)

=5.93+IWfxSx2

=5.93+11.82 x 0.62

= 7.9 +(W xsxh) = 7.9 +(0.35 x 0.6 x 2.55)

=WfXSX =1.82x0.6x7-2 2

= 4.26 +(W x s xh) = 4.26 +(0.25 x 0.6 x 2.4)

For beam designssee Section 3.1

Stud supportingbearn, designedas for lintel support

Typical loading

Self weight of wall W kN/m2

= 4.61 kN

5.93 kN

Load per stud point at 1St floor level(near wall)

Load at foot of each lower storeystud

Internal wall:Load per stud point at 1S floor level

Load at foot of each internal wallstud

© TRADA Tecflnology Ltd 2006 10

= 7.9 kN

= 8.44 kN

= 4.26 kN

= 4.62 kN

H

Load at each inner roofsupport point

LW xdx—2

Average load in each wall stud (spaced s)

[W1xsx+(Wxsxh)Careful design required unless d = s

Load at each interior joistsupport

(A + B)= WfXSX2

Load in each stud

\Nr xsx +(w xsxh) +Wf xsx (A+B)+(w xsxh)

End wall

For weights of materials and construction, see Section 1.3.

Wr = 1 .97 kN/m2 W = 0.86 kN/m2 for tile hanging (upper storey)

Wr = 1.82 kN/m2 W, = 0.35 kN/m2 for board cladding (lower storey)

W = 0.25 kN/m2 for plasterboard internal wall

For D = 8.4 m, s = d = 0.6 m, A = 1.8 m, B = 3.0 m, L = 4.8 m, h = 2.55 m for lower storey, 3.15 m (average)for upper storey, 2.4 m for internal wall.

Load per stud point on spandrel panel

Load at foot of each upper storey stud

Load per stud point at 1st floor level(near end)

Load at foot of each lower storey stud

Load at foot of each internal wall stud

=W xdx—=1.97x0.6x———2 2

= 2.84 +(W x S x h)= 2.84 +(0.86 xO.6 x 3.15)

=4.47+IWxSx =4.47+11.82x0.6xi2) 2

= 5.45 +(W x s x h )= 5.45 +(0.35 x 0.6 x 2.55)

=Wf XSX (A+B)J (w xsxh)

= 1.82x0.6x—I+(0.25x0.6x2.4) =2.98kN2)


1.2.1.2 Roof and floor spanning between cross-walls (discontinuous joists supported by load bearinginternal wall)

Typical loadingRoof and ceiling load Wr kN/m2 total

Inner support spacing = d

Stud and joist spacing = s

Shaded areas carry roof orloads

Floor load Wf kN/m2

Stud supportingl beam, designedas for lintel support

Self weight of wall W kN/m2

= 2.84 kN

= 4.47 kN

= 5.45 kN

= 5.99 kN



1.2.1.3 Other arrangements

Other arrangements are also possible. For example the roof may span front torear as in 1.1.1.1 and the floor span between cross-walls as in 1.1.1.2, or theroof may span between cross-walls as in 1.1.1.2 and the floor span front torearasin 1.1.1.1.

1.2.2 Calculation of wind loadsWind pressure is the main horizontal force causing racking and bendingdeformations. If adequate provision is made to resist the wind loadings, allother racking forces can usually be ignored for two-storey domestic buildings.The design wind pressures will depend on the locality, the degree ofexposure, and the overall height and proportions of the building.

1.2.2.1 Wind pressures on the timber frame

The dynamic wind pressures and pressure coefficients for the elements of atimber frame building are calculated, as for other buildings, according to BS6399-2. An example of a calculation for a two-storey house can be found inBRE Digest 436 Part 2.

Where masonry walls provide an external shell, the resulting wind loadsshould be based on the overall dimensions of the masonry walls, not those ofthe load bearing timber frame. However BS 5268-6.1 allows the wind loadtransferred to the timber frame to be reduced by a factor K100 where masonrywalls conforming to Clause 3.2.2 provide a shielding effect, for the calculationof racking forces and overall building stability only (ie not for stud design). Thisreduction should not be applied to the spandrels of gable walls, because themasonry walls in this area have no returns to make them self-supporting in adirection perpendicular to their plane.

1.2.2.2 Horizontal wind loads

The figure below shows the overall dimensions of a semi-detached timberframe house with masonry cladding built on a foundation 0.3 m above groundlevel. The racking forces on each pair of walls are shown as R1, R2 etc.

1.3 1.3

N Firstfloor N2.5 Ri I R3

5.0

2.5 Ro N R4 N

0.3 -- _________________ ____________________ — ___________________ : 0.3Front elevation End elevation

— wall I .I

Dimensions in metres

6.8 PlanWhen calculating the racking loads on the wall panels in a particular storey, itis assumed that the wind load on the upper half of the storey is applied as aracking load to the top of the panels, and the wind on the lower half of thestorey is applied to the bottom of the panels where it is resisted either by the

© TRADA Technology Ltd 2006 12


Wall stud

panels in the storey below or by the foundations. Therefore the total rackingload on a timber frame wall is calculated as the racking load transferred fromthe roof or the storey above it, plus half the wind load applied to the samestorey.

The next figure shows how the diagonal dimension, a, is calculated for eachdiaphragm and wall. This dimension is used to determine the correspondingsize effect factor, Ca, (BS 6399-2 Clause 2.1.3.4). Since the lowest value is0.95, it will normally be simpler to assume a value of 1.0 for Ca, and calculatethe exact values only if it proves necessary.

For the wind load on a wall stud, Ca = 1.0.


Gable walls Front and rear walls

Diaphragms

Roof diaphragm

6—---s----- t

--——--- 7.2 m

a1 41.252 +7.22= 7.31

Ca = 0.968

First floor diaphragm

-

= 7.26Ca = 0.968

First floor diaphragm

6 8 m

a4 42.52 +6.82= 7.62

Ca = 0.963

- 7 2 m

a2 =/2.52 + 7.22= 7.62

Ca = 0.963

Racking

First floor panels

v

72m

a5 =V1.252 +7.22= 7.31

Ca = 0.968

Ground floor panels

E

72m

a6 3752 +7.22= 8.12

Ca = 0.957

First floor panels- Ay

a7 =2.552 +6.82= 7.26

Ca = 0.968

Ground floor panels

,,'6

—7 Ia)/-a8 5Q52 +6.82

= 8.47Ca = 0.953

First floor panels

a10 = + 7.22= 8.77

Ca = 0.950

-d 7 m

a9 = 2.52 + 7.22= 7.62

Ca = 0.963= 7.79 = 9.27

Ca = 0.960 Ca = 0.945

6.8 m

a11 43.82 +6.82


Some typical values for the dynamic pressures, q, to be used during thenormal lifetime of a building, are shown below in Pascals (N/rn2). For buildingswith masonry cladding, BS 5268-6.1 Table 1 allows a reduction in the windloads on the timber frame and allows part of the load on the timber frame tobe transferred back into the brickwork through specified wall ties, but thebuilding must also be stable during construction, when there is no cladding.BS 6399-2 Annex D allows a reduction in the normal 50 year wind load forshorter periods of construction. Hence there are two design situations in whichthe building's strength and stability have to be checked: (a) the constructionperiod when there is neither a masonry shielding effect nor a roof load toresist overturning but the normal design wind load is reduced, and (b) theconstructed period when the shielding and load-sharing of any masonry maybe taken into account, but the full 50 year return wind load is applicable. (a) isparticularly important in the construction of multi-storey buildings whenoverturning forces can be significant.

The wind pressures shown in the diagram have been reduced whereapplicable by the appropriate value of K100, the reduction factor for themasonry shielding effect given in Table 1 of BS 5268-6. Note that this factoris not applied to the spandrel area of the gable walls, for reasons givenpreviously (see Section 1.2.2.1). The pressure coefficients, Cpe and C, for theexample building are also shown, obtained from BS 6399-2, Tables 5, 10 and16. For a semi-detached house the external wind pressure on the party wallmay be taken as zero. The overall dimensions of the two dwellings should beused to calculate the pressure coefficients.

Wind on gable wall — Wind on -1.17________________ rearwall +027 -05

qs = 658 x 1.0 qs = 756 Q.4 qs = 756= 658 +084 _________________ -0.5 ____________________

qs=658x0.5 qs—0 0.11 3.49 0.11 3.49= 329 +084 -0.3 -0.3 -0.5

qs=590x qs=590x0.82=484 0.8=472

_________________ +082 -0.3 -0.3 -0.5

Gable end Front Separating wall Rear Gable end Front

Reference values for BS 6399-2:Table 5: D/H=1.03 Table 10: W=7.2, bw7.2 TableS: D/H=1.36 Table 10:L=13.6 bL=10.6

(A_. TECHNOLOGYflW



1.2.2.3 Loads on diaphragms and walls

In order to calculate the loads on the elements of timber frame buildings, it isconvenient to rewrite BS 6399-2 Equation (7) in Clause 2.1.3.6 as:

Pfinal = O.85x x(1+Cr)xCa(1)

= final load on a diaphragm or the walls in one storey inwhere Pfinal one direction

0 85= factor to allow for the fact that maximum gust speeds

on front and rear walls are not simultaneous

= nett wind load on an element of the building whichP contributes to the final load, unfactored for dynamic

augmentation or size

= proportion of the element which contributes to the finalI load

= dynamic augmentation factor (BS 6399-2 Clause 1.6.1),taken as 0.02 in this example

= relevant size effect factor (see 1.2.2.2)

For detached houses or wind blowing on the front or rear wall

(2)P = (qs,front,iCpe,front,i — qs,rear,iCpe,rear,i)Ai

where A = area of the element on which the wind blows.

For semi-detached and terraced houses with wind blowing on the gable wall,the external pressure on the rear wall is zero, so

(3)Pi = qs.front,iCpe,front,iAi

To check the strength and stiffness of wall studs, the nett pressure on thewall panel depends on the difference between the external and internalpressures, so

(4)P = (qs,tront,iCpe,i — qS,fIOI,ICPI,I)A

(a) Wind on gable end

Equation (3)

Gable: P gable = 658 x 0.84 x 7.2 xl .3 x 0.5 2587 N

Each storey: P siorey = 329 X 0.84 x 7.2 x 2.5 = 4974 N

Equation (4)

Walls: P wail = (329 x 0.84 _(_)329 X 0.3)x 7.2 x 2.5 = 6751 N

Equation (1)

Roof diaphragm load P ceiling = 0.85 x (2587 X 0.667 + 4974 X 0.5)x 1.02 x 0.968 = 3535 N

First floor diaphragm load P fioor, 1 = 0.85 x (4974 X 0.5 + 4974 x O.5)x 1.02 x 0.963 = 4153 N

Racking load on first floor panels P racking, 1 = 0.85 (2587 xl + 4974 x O.5)x 1 .02 x 0.968 = 4258 N

Racking load on ground floor panels P racking, 0 = 0.85 x(2587 xl + 4974 xl .5)x 1.02 x 0.957 8337 N

© TR.ADA Technology Ltd 200615


Shear force on first floor panels

Shear force on ground floor panels

Nett load on a wall stud, assuming0.6 centres

(b) Wind on front and rear walls

Equation (2)

Roof diaphragm load

First floor diaphragm load

Racking load on first floor panels

Racking load on ground floor panels

Shear force on first floor panels

Shear force on ground floor panels

Neff load on a wall stud, assuming0.6 m centres

ceiling

P floor,!

P racking, 1

P racking, 0

P shear, 1

P shear, 0

P stud

The results are shown in diagram below

= 6205 N

= 10759 N

= 9215 N

= 9722 N

= 8983 N

= 9722 N

= 18461 N

14119 N

= 22714 N

= 705 N

Pfloor, 1

= 8.98 kNPstud

= 0.705 kN


Pshear,i 0.85x(2587x1+4974x1)xl.02x0.963 =6313 N

Pshear,0 0.85x(2587x1+4974x2)xl.02x0.950 = 10324 N

Pstud0.85 x 6751 x 0.6 xl .02 xl .0= =488N

Roof: P roof

Each storey: P storey

Walls P wall

Equation (4)

Equation (1)

7.2

= 756 x(0.4 xO.68 + 0.27 x2.92 —(—)1 .17 xO.68—

(—)0.5x 2.92) xtan20°x6.8

= (484 x 0.82 —(—)472 x 0.5)x 6.8 x 2.5

= (484 x 0.82— (—)484 x 0.3)x 6.8 x 2.5

= 0.85 x (6205 x 1 + 10759 x 0.5) x 1.02 x 0.968

= 0.85 x (10759 x 0.5 + 10759 x 0.5) x 1.02 x 0.963

= 0.85 x(6205 xl + 10759 x 0.5)x 1.02 x 0.968

= 0.85x(6205x1+10759x1.5)xl.02x0.953

= 0.85x(6205x1+10759x1)xl.02x0.960

= 0.85x(6205x1+10759x2)xl.Q2xQ.945

— 0.85x 9215 xO.6x 1.02 xl .0

6.8

Pceiling

= 3.17 kN—

Pfloor, 1

= 4.15

Pstud= 0.488

Pceiling = 9.72 kN—Pracking,1 4.26 kN

Pshear,1 6.31 kN—Pracking,o = 8.34 kN—s

Pshear,o = 10.32 kN—Wind on gable end

Pracking,1 = 9.72 kN—k

Psyear,i = 14.12 kN—s

Pracking,o = 18.46 kN—s

Pshear,0 = 22.71 kN—k

Wind on front and rear walls


The shear force is used to calculate the required fixing of the wall panels tothe floor or sole plate. With a shear force of Va Newtons on one wall, themaximum fastener spacing is given by:

Smax = Lfadm/Va (mm)

where L = wall length (mm)

fadm = permissible fastener load (N)

With recommended fixing to the foundation or floor it is not usually necessaryto check panels against rotation, but this may be necessary where there arelow vertical loads, such as in the upper storey of a building with a flat roof, orwhere particularly narrow or specially stiff panels are installed. This should becarried out for the end panels in each wall and any narrower panels, using anappropriate proportion of the racking load on the complete wall as a momentforce at panel height, and the vertical load on the panel (normally a UDL) as arestraining moment, as shown in the diagram in Section 2.3.3. For each loadcase the minimum vertical load, normally occurring beneath the part of theroof where there is maximum wind uplift, should be used. The vertical load onthe panel should include the weight of the panel itself.

For details on designing overturning restraint, see Section 2.3.3 (Overturningeffects).

1.2.2.4 Asymmetrical buildings

In the example shown above, the wind loads are resisted by two equally stiffracking walls so half the load is applied to each. If the resistance of tworacking walls differs greatly then the horizontal diaphragms may usually beassumed to redistribute the load so that the stiffer and therefore stronger walltakes a proportionately higher share of the load. All that is necessary is todetermine that the total wind loads do not exceed the combined resistance ofthe walls. For the same reason the resistance of internal load bearing wallsmay be added to that of the external walls, regardless of their position in thebuilding.

However, for buildings with dimensions outside the limits specified in BS5268-6.2 Clause 6.5, and for unconventional building forms or unconventionaltypes of floor or roof, the assumption about load redistribution may not holdtrue. In this case, if the building is asymmetrical then the racking and shearresistance of each wall should be checked separately.

1.2.2.5 Racking resistance in terraced housing

In some end terrace houses it is difficult to provide adequate rackingresistance to wind blowing directly onto the flank wall. In this case part of theload may be transferred to the adjacent house via steel party wall straps. Inorder to maintain a low level of sound transmission the Building Regulationsspecify a maximum density of one row of straps at 1200 mm centres installedat each storey height. However to ensure adequate load distribution thespacing needs to be not greater than 1200 mm, so this spacing should alwaysbe used. The thickness of the steel is usually 2 to 3 mm and the width of thestraps is typically 25 or 30 mm. The cross-sectional dimensions should notexceed 3 mm x 40 mm. It is recommended that each strap be fastened to theunderside of the top plate of the party wall panel, with two nails or screws ateach end, maintaining the BS 5268-2 minimum edge distances of 5d. (For thedesign of party walls, see TRADA's Timber Frame Construction.) The load-carrying capacity should be calculated as the minimum of the permissible loadon two fasteners and the buckling strength of the steel.

The following points should be noted. Continuous tiling battens can generallytransfer all the horizontal load from the roof of an end terrace house to theroof of the adjoining house. Where the remaining racking load at ceiling level,or racking loads at intermediate floors, are partly transferred via party wallstraps to the adjoining house, there will be a corresponding reduction in theracking loads transferred to lower floors in the end terrace house itself. Thispermits large openings in the front and rear racking walls on the ground floorof an end terrace house. (R



Example

Consider a Grade 43 (S275) galvanised steel strap 250 mm x 25 mm x 3 mmthick, fastened with a total of four 3.75 mm square twist nails, 50 mm long inC16 framing members. The free length of the strap between the inner nails, L,is 140 mm.

From BS 5268-2: 2002 Table 61 the basic single shear load F is:

F = 377+O.35x(453—377) =443N0.4

(As the headside member is steel the specified headside penetration of 46mm may be ignored.)

K44 for square twist nails = 1.2

K46 for steel to timber joints = 1.25

K48 for short- and very short-term loads = 1.25

Permissible load on 2 nails = 2 x F x K44 x K46 x K48 = 1661 N

From BS 5268-2: 2002 Table 21 the effective length of the strap Le = 0.7L.

L L 0.7x140x3.464Slenderness ratio = e = e = 113 mmI b 3

From BS 449-2: 1969 Table 17a the allowable compression stress:

Pc 68 N/mm2.

Hence permissible compression load on steel strap:

PCXA = 68x3x25 5100 N.

5100> 1661, hence permissible load on one strap = 1661 N

With straps at 1200 mm centres at eaves level on a 7.2 m wide house the

maximum racking load which can be transmitted at eaves level:

1661x 7200Fadm = 9.97 kN

(1 200 1000)

In the example above, the racking load at ceiling level on the end house,excluding the component from the roof = 0.85 x 4974 x 0.5 x 1 .02 x0.968/1000 = 2.09 kN, so all of this could be transferred to the adjacentproperty if necessary. The remaining racking load on the ground floor panels

4.15 kN, which again could be transferred at first floor level to the nexthouse, leaving no requirement for racking resistance in the ground floor wallsof the end house! Obviously the racking loads have to reach the foundationeventually, so with n houses in a terrace no more than (n-2)/n of the rackingload at any one level should be transferred to the adjacent property, startingfrom the roof.

The designer should specify precisely the required materials, dimensions andpositions of the party wall straps and fasteners.



1.2.3 Calculation of roof loadsThe connections between the roof trusses and the head binders should beadequate to resist the nett horizontal wind forces on each truss after allowingfor frictional resistance, and the nett uplift forces on them after allowing for theweight of the roof. The uplift forces are also required for stability calculationsto determine the overturning moment. The method for calculating these forcesis not unique to timber frame buildings, but is illustrated in the example ofcalculations for a complete dwelling in Chapter 7.

As mentioned in Section 1.1.2 the use of truss clips on every truss isrecommended to ensure adequate diaphragm action in the plane of theceiling. Therefore, the truss clips could be called on, if required, to provide anyadditional resistance to roof uplift or horizontal wind loads. Manufacturers ofproprietary truss clips should provide safe working loads or characteristicvalues for vertical loading and for horizontal loading in the plane of the trussand the wall. It is not safe to use the load-carrying capacity of the nails,because TRADA tests on various makes of truss clip demonstrated that failureusually occurs in the clip itself, at a lower load than that predicted bycalculating the permissible load on the nails.

However, if manufacturers' data are not available it may be assumed, basedon the tests mentioned above, that a truss clip with at least three 3.25 mmround wire nails in each member can resist a vertical or horizontal wind loadof 1.1 kN. (Some kinds of clip can resist significantly more than this in certaindirections.) Where, as is normally the case, the clip resists components ofload in two directions, it should be verified that

Fva 2+1 Fha2

Fvadm Fhadm

where Fv,a and Fh,a = the vertical and horizontal forces applied to each

connection respectively

and Fv,acjm and Fh,adm = the corresponding permissible vertical and

horizontal loads.

When a value of 1.1 kN is used for Fv,adm and Fhadm the formula above

simplifies to

Fva2 + Fha2 � 1.2

with Fva and Fh,3 in kN.



1.2.4 Checking strength and stabilityCare must be taken to check every relevant loading situation and potentialarea of instability when designing a timber frame house.

During the construction period no allowance should be made for the shieldingor load-sharing effect of any masonry walls which may be added later, nor forthe weight of the roof in resisting overturning and sliding. Internal walls maynot be present initially, and unless the external wall panels are prefabricatedwith internal linings there may be a period when their plasterboard linings arenot attached. Party walls consisting only of two layers of plasterboard andtimber framing need particular consideration, but in general their calculatedracking resistance may be assumed during the construction period providedthat they are adequately braced on erection in accordance with BS 5268-6.1Clause 4.7.5. To offset the effects of these conditions, BS 6399-2 permits areduction in the normal design value of the wind load for periods of less than50 years. For a 1 year maximum construction period, the probability factordescribed in BS 6399-2 Annex D is 0.749, which means that the dynamic windpressure for the construction period may be reduced by a factor of 0.7492 or0.561.

Where wall panels are built in situ there may be a period during which neitherthe structural sheathing nor the plasterboard linings are attached to the studs,but the studs may nevertheless have to support temporary loads such aspacks of OSB or plasterboard. Under these circumstances the studs will notbe braced against buckling in their weaker plane. The engineer is thereforeadvised either to check their load-carrying capacity in these circumstances, orto specify a construction procedure which will ensure that neither heavyconstruction loads, nor roof weights are applied before the structuralsheathing is fully attached, or to specify that the sheathing be attached to wallpanels before their erection.

During the service life of the building the full value of the design wind loadsmust of course be used. However, if masonry walls are present, BS 5268-6allows a reduction in the wind loads on the timber frame racking walls as aresult of the shielding effect of the masonry and, provided that the masonryand timber frame walls are tied together in a specified manner, a furthercontribution from the masonry towards the racking resistance.



I 3 Weights of materialsThe weights of materials can be obtained from sample tests, information frommanufacturers or by reference to BS EN 1991-1-1 or BS 648 Schedule ofweights of building materials.

Aluminium

Battens, tiling

Bituminous roofing felt

Blockwork, concrete

Lead

OSB (oriented strand board)

Pitch mastic

Plasterboard

Hollow, stone aggregate:

Solid stone aggregate:

Aerated:

Clay, medium density:

Concrete

Sand lime:

Reinforced:

Board, semi-compressed:

Flooring:

Bitumen impregnated insulating board:

Hardboard:

Insulating board:

Medium board:

Plate:

Building panels:

Glass fibre thermal insulation:

Mineral fibre thermal insulation:

12 .5mm plasterboard 2 sides

Proprietary

Sheet:

Flooring:

Gypsum:

Gypsum:

Gypsum-

Gypsum

Flexible pvc:

Pvc vinyl:

American construction and industrial plywood:

Canadian Douglas fir and softwood plywood:

100 mm thick

100mm thick

100 mm thick

103 mm thick

103 mm thick

103 mm thick

6 mm thick

100 mm thick

0.6 mm as laid

25 mm thick

25 mm thick

13mm thick

6.3 mm thick

12.7mm thick

9.0 mm thick

6.4 mm cast clear andarmoured75mm to 150mm

57 mm to 63 mm

25 mm thick

100mm

100mm

1.7 mm to 3.0 mm

9mm to 18mm

25 mm thick

9.5 mm

12.5mm

15mm

19mm

1.6 mm to 3.2 mm

1.6 mm to 4.8 mm

12.5mm to 19mm

12.5 mm to 18.5 mm

kg/rn21.5—3.4

2.0—4.4

3.4

34.0—37.0

139.0

219.0

82.0

221.8

237.6

206.0

5.5—7.5

240.0

6.5

4.9

9.8

3.1 —4.5

4.2

3.4

5.0—7.2

16.1

43.9—63.5

20.5—25.9

2.0

2.3

1.2—2.4

29.6

20.4 — 33.6

19.5—34.2

5.7— 12.8

14.9—18.1

6.1

8.0

9.8

15.0

2.4—4.9

3.4— 10.3

7.3—10.9

7.2— 10.6

Weights of materials commonly employed in house construction are givenbelow as a convenient reference. These may be taken as a general guide,though the density of individual products can vary considerably. For detailedcalculations the manufacturer's figures should be used.

flat: 0.6 mm to 1.2 mm

corrugated 0.6 mm to 1.2 mm (including 20% for added laps)

38mmxl9mm: loOmmgauge

3 layer felt bonded together, 13 mm granite chippings

Typical values. More accurate figures may be obtained from manufacturers' literature.

Brickwork

Calcium silicate and cement

fibreboards

Concrete

Copper roofing

Cork

Fibre building board

Glass

Gypsum panels and partitions

Insulation

Internal stud partitions

Dry partitions:

Glass fibre acoustic insulation for floating floors:

Plastic flooring

Plywood



12mm to 18mm

12mm to 18mm

12.5mm to 19mm

12 mm to 19 mm

Rubber flooring 3.2 mm to 9.5 mm

PSL (Parallel strand lumber)

Rendering Portland cement: sand (1:3):

Roofing felt 3 layers felt and chippings

20 mm asphalt

Sand Dry per cubic metre

thick 48.8

thin

thick 78.1

thin

thick 48.8

Tiling, roof Typical values. More accurate figures may be obtained from manufacturer's literature

Clay, machine made: 100 gauge

hand made: 100 gauge

Concrete, plain: 100 gauge

Tiling wall

Weatherboarding

Wood chipboard

Interlocking, single lap:

To convert kg/rn2 into N/rn2 multiply by 9.81

kg/rn3610—670

530— 710

730 — 750

* The density of structural timber composites depends on the moisture content, species or grade. For more accurate valuessee the manufacturers' literature.

To convert kg/rn3 into N/mrn multiply by 9.8lbh x io- where the breadth and depth, band h, are in mm.


Plywood (continued) Finnish birch:

Finnish birch faced:

Swedish softwood:

Tropical hardwood:

12.5mm

Shingles

Slating

kg/rn28.4— 12.4

8.1 —11.6

5.3—8.0

7.3—11.6

730 - 750

29.3

37.0

46.0

5.4— 16.1

1522—1682

7.3

24.4

Cedar

Welsh:

Westmorland:

Cornish:

thin

48.8

29.3

Concrete, plain

Flooring grades, P4 and PS

63.5

70.8

68.4

19mm

15mm thick

18mm thick

22 mm thick

16mm thick

19mm thick

21 mm thick

28 mm thick

16mm thick

19mm thick

21 mm thick

28 mm thick

16mm thick

19mm thick

21 mm thick

28 mm thick

25 mm thick

41.5—56.1

68.0

7.3

10.2—11.2

12.8—14.0

14.3— 15.7

11.7

13.9

15.3

20.3

11.4

13.5

14.9

19.8

8.3

9.8

10.8

14.4

14.6

4.4—7.8

Wood flooring Hardwood: beech, oak

Softwood: pitch pine

Softwood: redwood

Wood wool Slabs:

Zinc Sheet 12 to 16 zinc gauge (0.63— 1.04 mm)

LSL (laminated strand lumber)*

LVL (laminated veneer lumber)*

PSL (Parallel strand lumber)*


1.4 Weights of typical constructionsImposed loads from BS 6399-1: 1996

1.4.1 RoofsTo obtain load per square metre on plan of a pitched roof surface, divide theweight of roof surface measured on slope by the cosine of the pitch, ie

2 kPlan load w kg/rn =

cosOwhere k is the weight of roofing (tiles and battens) in kg/m2 measured onslope and B is the pitch angle in degrees.

Example (1) Pitched roof kg/rn2 kN/m2

Medium weight tile and batten (on plan) 59.5 0.57

Allow for self weight of structure 15.0 0.15

Imposed load on roof structure 0.75

Total design load for roof surface: 1.47

Ceiling dead 25.5 0.25

Ceiling imposed 0.25

200 mm mineral wool quilt 2.4 0.02

Total ceiling design load 0.50

Total design load for roof structure 1.99

Example (2) Flat roof (cold deck type)

Bituminous roofing felt with chippings 35.5

9 mm OSB 5.8

12.5 mm plasterboard 10.0

Self-weight of joists 11.5

200 mm mineral wool quilt 2.4

65.2 0.64

Imposed 0.75

Total design load for roof 1.39

1.4.2 Floors

Example (3) Floor kg/rn2 kN/m2

22 mm tongued and grooved chipboard 14.3


140 mm phenolic foamboard 4.2

Self-weight of joists 11.0

39.5 0.39

Imposed 1.50

Total design load for floor 1.89

© TRADA Technology Ltd 200623 ci:13,


1.4.3 Walls

Example (4) WaIl with tile hanging kg

Clay tiles and battens 64.5

9 mm OSB sheathing 5.8

140 mm mineral wool batt 3.5

Self-weight of framework 7.0


Example (5) WaIl with board cladding

25 mm boards and battens

9 mm OSB sheathing

150 mm glass fibre batt insulation

Self-weight of framework

12.5 mm plasterboard

Example (6) Interior wall

12.5 mm plasterboard both sides

Self-weight of framework

89.8 0.88

12.5

5.8

5.0

7.0

10.0

40.3 0.40

20.0

5.0

25.0 0.25

1.5 General requirements for prefabricated wall, floor and roofelements

BS EN 14372 Timber structures — Prefabricated walls, floor and roofelements, will, when published, specify the product, performance, productionand testing requirements for timber frame walls, floor and ceiling elementswhere these are prefabricated from 'joists" which are fixed to a panel producton one or both sides made of or from timber or gypsum plasterboard. Thefixing may be by mechanical fasteners or glue. For glued fixings detailedrequirements are given for the permitted types of adhesive, the methods ofapplication and the quality control systems which must be in place.


kN


2 WaIls

2.1 WaIl studsAll load bearing studs and rails must be of strength graded timber. Mostcommonly studs are softwood of Strength Class C16 or C24 in accordancewith BS EN 338 Structural timber. Strength classes.

2.1.1 Minimum desirable stud sizesStud widths should allow for the satisfactory butt jointing of sheathing andplasterboard. BS 5268-6.1 recommends a minimum cross-section of38 mm x 72 mm for external walls, 38 mm x 63 mm for internal walls.However, when calculations produce small sizes of stud, the resulting wallsmay appear too flexible to the occupants of the house, although the studs aresafe against axial and wind loading.

The depth of wall studs is often governed by the thickness of insulation requiredwithin the wall panels to meet the requirements of Building Regulations. Forexample a stud 89 mm deep with 90 mm of insulation material will only meetrequirement Li of the 2004 England and Wales Regulations with a reflectivebreather membrane or the installation of a high efficiency boiler. A 140 mmdeep stud with 140 mm of insulation will meet the same requirement withoutany special measures. Some companies use prefabricated timber I-joists aswall studs, both to obtain the required thickness of insulation and because theirthin webs have much lower thermal conductivity than studs made of solidtimber.

2.1.2 Multiple studsIn domestic-scale buildings studs are normally spaced at 400 mm or 600 mmcentres. Most wall panels are factory-produced but for on-site framing studsspaced at 600 mm are preferred to allow workmen room to walk betweenthem. With high axial loads it may be necessary to specify double, triple oreven more studs. In order to prevent buckling in their weaker direction,multiple studs should be fastened together in such a way that a total lateralforce of 2% of the axial load on each stud can be applied uniformly to eachstud. This may be achieved either by nailing through the sheathing materialfrom both sides into the secondary studs with sufficient nails to provide therequired force in lateral loading, or by nailing the multiple studs together withsufficient nails to provide the required force in axial (withdrawal) loading. Inthe latter case ringed shank nails or screws should be specified. Thefasteners should be spaced at 300 mm or less.

If three studs are fastened together it is preferable to nail or screw the outersets from the outside of the panel, since the axial fastener load will begreatest in the outer studs. With the intermediate studs the central one in eachset should be fastened to the panel.

When checking the combined bending and compression stresses in multiplestuds or their deflection, Emin may be increased by the appropriate factor givenin BS 5268-2 Table 20 (see Clause 2.11.5). To ensure vertical load sharing,multiple studs should be cut square with a tolerance on their length of+1- 0.5 mm.

Since the vertical load-carrying capacity of studs is often limited by thebearing strength of the bottom rail, it may be possible to reduce the number ofstuds required by using a higher grade of timber in the top and bottom rails, ordeeper studs.



2.1.3 Lateral supportAccording to BS 5268-6.1, solid timber studs covered on one or both sideswith a specified board material and fixed as recommended in the standardmay be assumed to be fully restrained against buckling in their weakerdirection. However there is evidence from North America that for studs bracedon one side only by a sheet material, as in a separating wall, the load-carryingcapacity is reduced by about a quarter. Caution is therefore advisedparticularly when it comes to buildings of more than the four storeys coveredby BS 5268-6.1. Where necessary nogging pieces may be used to restore fulllateral support, provided that they are combined with some form of diagonalbracing to prevent all the studs buckling simultaneously in the same direction.United States recommendations indicate that one row of noggings at mid-height is adequate for nominal 50 x 75 mm and 50 x 100 mm studs, and twoequally-spaced rows for 50 x 150 mm studs. The sheet material should benailed to the noggings as well as the studs.

2.1.4 Load capacitiesTables 2.1 and 2.2 relate to timber of strength class C16 and C 24respectively in service class 1. They give the permissible axial load in kN perstud (not per metre) for studs in panels 2400 mm high.

It is assumed that there is adequate bracing against buckling in the plane ofthe wall, as described above. The end loading is assumed to be nominallyaxial. Where there is significant eccentricity, reduced figures should becalculated.



Table 21 Permissible axial load per stud in kN — Timber strength class C16

(a) With wind loading: very short term load case

Spacing Breadth Depth Wind pressure in Pa (N/rn2)

mm mm mm 0 100 200 300 400 500 600 700 800 900 1000

400 38 72 11.03 10.03 9.08 8.08 5.95 382 1 69 0.00 000 000 0.00

89 1 17.60 16.41 1529 1421 13.18 12.20 11.25 10.29 8.16 6.03

97 19.24 18.08 16.97 15.91 1488 1388 1292 11.99

114 ? 22.54 21.45 20.40120 A 23.66

140

145

72 12.77 11.76 10.81 9.90 8.23 610 397 1.84 0.00 000 0.0097 20.52 19.42 18.36 17.33 16 33 15.36

120

145

72 13.64 12.63 11.67 10.75 938 7.25 5 11 2.98 0.85 0.00 0.0097 21.19 20.12 19.07 18.06 1707

120

145 I

72 11.03 9.55 8.08 4.88 1.69 000 000 0.00 000 0.00 0.00

89 17.00 15.29 13.69 1220 10.79 8.16 4.96 1.76 0.00 0.00

97 19.24 17.52 15.91 1438 12.92 11.53 9.82 6.62 3.43

114 21.45 1988 18.36 16.90 15.49

120 23.12 21.56 2005 18.59

140

145

72 12.77 11.28 9.90 7.17 3 97 0.78 0.00 0.00 0.00 0.00 0.00

97__ -- 19.42 17.84 16.33 14.89 13.50 12.16 9.01

120 25.37 23.85

145

72 13.64 12.15 10.75 8.31 5.11 1.92 0.00 000 0.00 0.00 0.00

38

38

38

—38

38

44

44_44

44

-- 47

47

47

47

600 38

3_38

38

-38

38

44

44

44

4._

97 21.19 19.59 18.06 16.59 15.17 1380 11.81

26.51

—

47_ 1p47 145

NB Values for a wind pressure of zero are given to permit interpolation between 0 and 1 00 Pa.iShaded values are governed by bearing stress in the bottom railValues in italics are governed by a deflection limit of 0.003L. Other values are governed by buckling in the stud.

(b) Without wind loading: medium — and long term load cases

Spacing Breadth Depth Load duration

mm mm mm

400/600 38 72

38 8938 9738 114

38 120

38 140

38 145

44 72

44 97

44 120

44 145

47 72

47 97

47

47

120

145

i eaiurn JLong

I/Iw 'iWaIwt kI


I.


Table 22 Permissible axial load per stud in kN — Timber strength class C24

Spacing Breadth Depth

mm mm mm

400 38 72

38 97

38 114

38 120

38 140

38 145

44 72

44 97

44 120

44 145

47 72

47 97

47 120

47 145

600 38 72

38 89

38 97

38 114

38 120

38 140

38 145

44 72

44 97

44 120

44 145

47 72

47 97—47 120

Wind

0pressure in

100

Pa (N/rn2)

200 1300 400 500 600 700 800 900 1000

13.43 12.55 11.71 10.91 9.24 7.11 4.98 2.85 0.71 0.00 0.00

19.28 18.66 17.73 16.83 15.96 15.11 14.29 12.24

21.01 20.46 19.54 18.66 17.79

24

3O32 .32

31.40 .40 40 .40 1.40 31.40 .40 31.

15.55 14.67 13.82 13.00 12.04 9.91 7.78 5.65 3.52 1.39 0.00

22.91 22.87 21.97

28.34

34. .25 . .25 34.25 .2516.61 15.73 14.88 14.06 13.26 11.31 9.18 7.05 4.92 2.79 0.66

23.72

2935.46 .46 35. .46

13.43 12.13 10.91 8.17 4.98 1.78 0.00 0.00 0.00 0.00 0.00

19.28 18.19 16.83 15.53 14.29 11.17 7.98 4.78 1.58

21.01 .01 20.00 18.66 17.36 16.12 14.66 11.47

24 23.60

25.99

30.32 .32

31.40 1.40 . 31.40 31.40 1.40 .40 .40

15.55 14.24 13.00 10.98 7.78 4.58 1.39 0.00 0.00 0.00 0.00

22.91 22.87 21.53 20.24 18.99 17.77

3434.25 .25 34. 34 25 .25

16.61 15.30 14.06 12.38 9.18 5.99 2.79 0.00 0.00 0.00

23.72 23.63 22.32 19.81

29.35

35.46 .46 . .46NB Values for a wind pressure of zero are given to permit interpolation between 0 and 1 00 Pa.

I Shaded values are governed by bearing stress in the bottom rail

Values in italics are governed by a deflection limit of 0.003L. Other values are governed by buckling in the stud.

(b) Without wind loading: medium — and long term load cases I

Spacing Breadth Depth Load duration

mrn rnm rnm

400/600 38 72

38 89

38 97

38 114

38 120

38 140

38 145

44 72

44 97

44 120

44 145

47 72

47 - 97

47 120

47 145

I.vieaiurn Lon

1114, , .189113.77, . ,.15.01 , r4c -vt17.64 11

i4 9 ,21.66 7.. .. —22., ,12.15 72. .16.36 09. t.. .20.24 19rr . c2

.12.58

, , ,. ,, '. .1 , ,

?725.33 .27


(a) With wind loading: very short-term load case

38 89

47 145

Ii g I


The tables show the permissible axial load per stud, checking combinedbending and compression, bearing stress in the bottom rail assuming nowane, and deflection. It is assumed that the strength class of the top andbottom rails is the same as that of the studs. The bearing stress in the bottomrail is calculated using the BS 5268-2 bearing factor K.4, as this is applicable tothe studs which are not at the ends of each panel, and these studs normallytake more load. For bearing stress the load-sharing factor K8 has not beenused because there is only one member involved. In the formulae forcombined bending and compression and for deflection the effective length, Le,was calculated as 0.85L, as recommended by BS 5268-6.1. For the actualstud length, L, a value of 2324 mm = 2400 — (2 x 38) was used.

The formula used to calculate deflection was:

0.005 Aca + öma Z=(öemean — öca)

X

Le 0.85L1ifwhere A = _______I h

2a mean

e,meanA2

bh2andZ6

This gives the approximate deflection of a stud subjected to both compressiveand bending forces. BS 5268-6.1 states that when calculating defiections theeffects of axial loading should be ignored. However for some small studs withhigh wind loads ignoring the axial deflection can result in excessivedeflections, hence axial load has been taken into account in these tables. Amaximum deflection of 0.003L has been allowed. Where no value is given inthe tables, this deflection is exceeded even without axial load.

In particular situations a designer may choose to vary the deflection limit from0.003L.

2.1.5 Design procedureThe following load cases should be checked using Table 2.1 or 2.2:

1. Dead + imposed + wind: very short-term

2. Dead + imposed: medium term

3. Dead + imposed floor and ceiling loads: long-term

The very short-term axial loads are given in conjunction with a series of nettdesign wind pressures on the walls. The method of obtaining these is given inSection 1.2.2.3. In determining the design wind pressure, the worstcombination of external and internal pressure and suction should be used.

When calculating the vertical load on a stud, half the weight of the wall panelsupported by the stud should be included. BS 6399-1 Table 2 permits areduction of 10(n-1) % in the axial load carried by a stud due to imposed floorloads where there are n storeys above it for a maximum value of n = 5. Thisreduction is particularly important in the lower floors of multi-storey buildings.For the very short-term load case the vertical component of the wind loadshould be included in the axial load on the wall stud.

As shown in grey in the tables, the load per stud is governed by thepermissible stress in the bottom rail, except in the case of smaller studs undervery short-term loading. Therefore in many cases the vertical load-carryingcapacity of a timber frame wall panel can be increased by using higherstrength timber or a structural timber composite for the top and bottom rails.

Calculations for a stud wall supporting a vertical load and horizontal wind loadare given in the example of calculations for a complete dwelling in Section 7.5.



2.1.6 Drilling of studsUnless otherwise justified by calculation, drilling of studs should conform tothe following requirements:

Stud Location Maximum size

Load bearing Centre line of section. Between150mm from one end and 0.25of the length measured from thesame end. Holes to be spaced(centre to centre) at a minimumof 4 hole diameters.

0.25 x depth of stud.

Non-load bearingpartition

Centre line of section. Between150 mm from one end and 0.40of the length measured from thesame end. Holes to be spaced(centre to centre) at a minimumof 4 hole diameters.

0.25 x depth of stud.

Notching anywhere in the stud is prohibited.

2.2 Framing around openingsWhere openings occur, they should be spanned by suitably designed lintelsas indicated below and the load on the lintels should be transmitted to thefoundations by sufficient stud material at all the levels concerned. This studmaterial is introduced in the form of cripple studs of the same cross-sectionand at least the same number as those removed from the opening, as shownin the diagram

Lintel: solid timber,structural timber compositeor fitch beam.


Stud material from openingshared equally in framingeach side

Lintel below first floordesigned to support loadfrom double stud to side ofwindow above, includingweight of construction fromother studs.

II IIII II

H

ii Ii II

H H [


2.2.1 Lintel capacitiesTables 2.3 to 2.10 which follow give the permissible load P (kN) per stud point(NOT per metre run) on solid timber lintels of nominal spans 1200mm,1800 mm and 2400 mm, for studs spaced at 400 mm and 600 mm centres.For larger openings, or where the loading is too great for the lintels consideredbelow, stronger types of beam may have to be built into the wall above theopening.

The load tables are for strength classes 016, 024, D40 and D50. Where therequired span cannot be achieved in a softwood, it is now more common tospecify LVL than a hardwood. However there are various makes of LVL whichhave different mechanical properties, so to obtain these properties thedesigner must refer to the appropriate manufacturer's certification literature(usually a BBA or BM TRADA Q-mark certificate) and calculate the requiredsize accordingly.

Where roof loads from similarly loaded trusses at no more than 600 mmcentres are applied, it is generally adequate to convert them to a udl providedthat the lintel is fastened to the top rail of the wall panel and the top rail of thewall panel is fastened to the wall plate with a minimum of two 2.85 mmdiameter nails at 600 mm centres. In this situation the values in the tables,which assume that the loads act as point loads, may be conservative by up to10%. However, where larger point loads from cripple studs on the floor aboveare supported by a lintel it is unsafe to regard the loads on them as udls.

It should be noted that the load P, for a stud or floor joist bearing on a lintel,could be limited by the compressive stress perpendicular to the grain of thelintel (bearing stress). The total load on the lintel must be safely transferred tothe foundation by means of cripple studs. For cripple studs, extra studmaterial may be provided by the principle illustrated in the previous diagram,and normally this will be adequate for bearing. A calculation for a cripple studis given in the example of calculations for a complete dwelling in Section 7.8.

The following arrangements of load points on lintels have been allowed for inthe load tables. Bending, shear and deflection, have been calculated for theworst value of 'a' within the ranges shown for each case. A maximum initialdeflection of 0.003L under dead and imposed load has been allowed.Separate tables are provided for medium term load, ie loads from roofs, andfor long term load, ie loads from floors.



Loads at 600 mm centres

600 600 I 600

k1200 1800 2400

1200

a 1 mm to 599 mm

Loads at 400 mm centres

_ _rrrrrA400 A 400 A A400T T T T1200 1800 2400

00 4 1800 4 2400

a 1 mm to 399 mm



Table 2.3 Permissible loads on lintels: Timber strength class C16 — Long-term loading

Point loads at 600 mm spacing

Span No ofloads

Depth 72 9! 120 145 170 195 21) 245

Breadth

'1 21 2.26 3 8 10 13.01

1 64 3.04 1 17.53

202 3.7 .69

0.90 2011.21 2.71.50 3.35

72 97 120 45 171) 195 220 245

1200 1 7297

1202 72

97120

Span No ofloads

DepthBreadth

1.47 2.41 3.26 4.23 5.31 6.511 98 3.25 4. 5.69 7.1 8.77245 4.02 5 7. .841 27 1.93 21 71 2.60 3.51

2 12 3.22 4.35120 141) 70 105 220 245

1800 2 72 0.33 0.7997 0.45 1.07

120 0.55 1.323 72 0.29 0.69

97 0.39 0.93120 0.48 1.14

Span Noofloads

Depth 72 97

Breadth2400 3 72 0.13 0.33 0.61 (

97 0.18 0.44 0,82 1.4120 0.22 0.54 1.01 'I 7j,,,

4 72 0.13 0.30 0.57 098i 1 5

97 0.17 0.41 0.76 32j 2.1

120 0.21 0.50 0.94 I 'f 2.59J

189 2'1 488

28 4 8

26

811'ft91a48 a1J359, 1.91 2.58 3.34 4.20 4.78 2.36 3.19 4.13 5.19 5.91

38 1.38 1.86 2.26 2.55 2.84') 1.86 2.51 3.04 3.43 3.83

2.30 3.11 3.77 4.25 4.73145 175 1I5 220 24

1.41 1.77 2.17

1.90 2.38 2.92 -

2.35 2 95 3.61

1.39 1.75 2.141.87 2.35 2.882.32 2.91 3.56

— sh-?r .)c.virft


Point loads at 400 mm_spacing

Span Noofloads

Depth 72 97 120 145 170 195 220 245

Breadth1200 1 72

97120

0.72

0.971.21

97 120 145 170 195 220 245

2 7297

120

0.63

0.841.04

Span Noofloads

Depth 72

Breadth1800 2 72 0.20 0.48 0.

0.64 1.1

0.80 lÀ0.48 0.

0.64 1.

97 0.27

120 0.333 72 0.20

97 0.27120 0.33 0.80 1.4

Span No ofloads

2400 3Breadth

Depth 72 97 120

4

0.21 0.39 0.68 1.08

0.28 0.53 0.92 1.46

0.15 0.35 0.66 1.14 1.80

0.20 0.38 0.66 1.05

0.28 0.51 0.89 1.41

0.34 0,64 1.10 1.75

______ deflection governs ______ = bending governs


Table 2.4 Permissible loads on lintels: Timber strength class C16 — Medium-term loading



Span No ofloads

Depth 72 97 120 145 170 195 220 245Breadth

1.21 13.01

1.64 1 . 17.53

2.02 1

0.901.21

1.50

72 97 120 145 170 195 220 245

1200 1 7297

1202 72

97120

Span Noofloads

DepthBreadth

1800 2 72 0.33 0,79 1 47 2.41 26 4 5.31 6.511 98 3 .1 8.77245 4 10.841.27

1.71

2.12 3120 145 70 195 220 246

0.61 1 05 1 2.11 3.250.82 1.42 2 3 4.381.01 1.76 2. 3.52 4.42 5.420.57 0.98 1.55 2. 570.76 1 32 2.09 2.7 3.470.94 1 64 2 59 3.42 4.

97 0.45 1.07120 0.55 1.32

3 72 0.29 0.6997 0.39 0.93

120 0.48 1.14

Span No ofloads

Depth 72 97Breadth

2400 3 72 0.13 0.3397 0.18 0.44

120 0.22 0.544 72 0.13 0.30

97 0.17 0.41

120 0.21 0.50


Span No ofloads

Depth 72! 97! 120! 145! 170! 195! 220! 245!

1800 2 72 0.20 0.48

Breadth1200 1 72 0.72 1

97 0.97120 121

2 72 0.63 1

97 0.84120 1.04

Span Noof Depth 72 97 120 145 17 195 220 245loads Breadth

0.89 1. 1 2 3.1

______ ______ 1.19 .91

148

088 1.

1.19

1.47 2. .11

120 145 17C 195 220 245

97 0.27 0.64

120 0.33 0.80

3 72 0.20 0.48

97 0.27 0.64

120 0.33 0.80

Depth

2400 3

72 97

72 0.09 0.21

4120

0.39 0.6897 0.12 0.28 0.53

0.15 0.3572 0.08 0.20

Span No. ofloads Breadth

2.172.38 2.922.95 3.6111.75 2.1412.35 2.82.91 3.561

:hear qoverrv,

0816

80-

1.0541

175[1

1.41

1.90

2.351.391.87

2.32

0.66 1.14

97 0,11 0.28 0.51

120

0.38 0.66

0.14 0.34

F = deflection governs j I = bending governs

0.890,64

1.

1.10


Table 2.5 Permissible loads on lintels: Timber strength grading class C24 — Long-term loading



Span No ofloads

1200 1

DepthBreadth

7297

120

72! 971 120! 145! 170! 195! 2201 245

2 7297

120

Span No ofloads

1800 2

DepthBreadth

72 0.41 0.99

1.51 3.20 4.78 6 11

2.03 4.31 6.44 9.212.51 5 1

1.12

1 501 86

7 07 120 I'S 170 ic5 '20 245

_____ _____ 1.83 13 4.61 5.

_____ _____ _____ 422 63.04 522 7

______ ______ 1.58

______ ______ 2.122 63

120 145 175 95 220 245

3

97 0.55 1.33 2.46

120 0.6872 0,36

1.64

Span

0.8597 0.48 1.15

120 0.59No. ofload

DepthBreadth

1.42

72 97

I I I

2400 3 72 0.17 0.40 0.76 1.31 2 07[ 2.99 3.76 4.601.76 2 78 4.03 5 06 6.202.18 '.44 4.98 6.26 7.671.22 -. 93 2.66 3. 3,1.64 2 59 3. 4 4,502.03 21 4.43 5.00 5.57

97 0.23 0.54 1.02120 0.28 0.67 1.26

4 72 0.16 0.38 0.7097 0.21 0.51 0.95

120 0.26 0.63 1.17


Span No. ofloads

Depth 72 97 120 145 170 19. 20 245Breadth

0.90 2.11 3.1.21 2,84 41.50 3.51

0.781.05

1.29

7. 97 120 45 170 95 220 245

1200 1 72

97120

2 7297

120

Span No. ofloads

DepthBreadth

1800 2 72 0.25 0.59 1.10 1 890.33 0.80 1.48 2.54

0.41 0.99 1.83 3.140.25 0.59 1100.33 0.80 1.48

0.41 0.99 1.8372 97 120 145 75 105 220 45

0.11 026 U9 255 1..4 98 2.50 3.070.15 035 066 14 1,81 ' 67 3.37 4.130.18 0.44 0;2 1.42 2 24 .3.30 4.17 5.110.10 0.25 047 0.82 1.30

0.14 0.34 0.64 1.11 1 75

0.17 0.42 0.79 1,37'

17

97120

3 7297

120

Span No. ofloads

DepthBreadth

2400 3 72

97120

4 7297120

______ deflection governs ______ = bending governs = shear governs


4.7 11.

1 6.44 9.21 12. 16.1

7. 11.

2


Span No. ofoads

Depth 72 97 120 145 170 195 220 245Breadth

1200 1 72 090 211

121 2.8

1.50 3.

0.78

11,

72 97 120 145 170 195 220 245

97

120

2 72

97

120

Span No,ofloads

DepthBreadth

1.10 1.891,48 2.541,83 3,141.101.48

1.83

120 145 170 195 220 245

1800 2 72 0.25 0.5997 0.33 0.80

120 0.41 0.993 72 0.25 0.59

97 0.33 0.80120 0.41 0.99

Span No.ofloads

Depth 72 97Breadth

______ ______ ______ 0.49 0.85 1 'j8 2.50 3.07

______ ______ ______ ____ 1.8 2 7 3.37 4.13120 0.18 0.44 0.82 1.42 2 3 30' 4.17 5.11

4 72 0 10 0 25 047 0 82 I 30[ 1 90 2 14 2 39

_____ _____ _____ _____ _____ 1 75 2.56 2.89 3.22

______ _____ 120 0.17 0.42 0.79 1.37 2 ilL 317 3.57 3.98

______ deflection governs ______ = bending governs 1' — shr

_______ 3.

51

35

29

97 0.15 0.35 0.66

97 0.14 0.34 0.64

1.14

1.11


Table 2.6 Permissible loads on lintels: Timber strength class C24 — Medium-term loading


Span No.ofloads

Depth 72 97 120 145 170 195 220 245Breadth

1200 1 72

97

120

1.51

2.03

2.51

2 72

97

120

1.1

11.L

Span No. of Depth 72 97 120 145 170 19 220 245loads Breadth

1800 2 72 0.41 0.99 183 3 13 4.61 5.981.33 2.46 4 22 6.22 81.64 304 5.22 7 9.

0.85 1 58

1,15 2121.42 263

97 120 145 ' 73 'f.5 220 245

97 0.55120 0.68

3 72 0.36

97 0.48

120 0.59

Span No. ofloads

Depth 72

Breadth2400 3 72 0.17 0.40 0.76 1.31 2.07

97 0.23 0.54 1.02 1.76 2

120 0.28 0.67 1.26 2.18 3.4 72 0.16 0.38 0.70 1.22 1

97 0.21 0.51 0.95 1.64 2.59

120 0.26 0.63 1.17 2.03 3.21

2400 3 72 0,11 0.26


Table 2.7 Permissible loads on lintels: Timber strength class D40 — Long-term loading


Span No ofloads

Depth 72 97 120 145 170! 195! 2201 245

I

107215.381 19.941 25.04j 30.69

1444 20.721 26.86j 41.34

786 2Ø4I $,323j 41 7$! 5115829

1 17

3 81145 110 15 220 245

Breadth

1200 1 72 1.57 3.65 6.52

97 2.12 4.92 879120 262 608 1087

2 72 1.16 2.74 496

97 1.57 3.69 6.69

120 1.94 4.56 8.27

Span No ofloads

Depth 72 97 120

Breadth

1800 2 72 0.43 103 1.90 3.26 5!J9 (.40 10.21 13 997 0.58 1,38 2.56 4.39 6.85 99 13.75 18 1Q

120 0.71 1.71 3,17 5.43 8.48 12.33 17.01 22 51

3 72 0.37 0.89 1.64 2.81 4.37 6,34 8.72 11.51

97 0.50 1.20 2.21 3.78 5.89 8.54 11.75 15.50

120 0.62 1.48 2.74 4.68 7.28 10.57 14.54 19.18

Span Noofloads

Depth 72 97 120 145 170 195 220 245

Breadth

2400 3 72 0.17 0.42 0.79 1.36 2.15 3.17 4,44 5.96

97 0.23 0,57 1.06 1.84 2.90 4.28 5.98 8.03120 0.29 0.70 1.31 2.27 3.59 5.29 7.40 9.94

4 72 0.16 0.39 0.73 1.27 2.01 2.96 4.15 5.58

97 0.22 0.53 0.99 1.71 2.70 3.99 5.59 7.52

120 0.27 0.65 1.22 2.12 3.34 4.94 6.92 9.31


Span Noofloads

Depth 72 97 120 145 170 195 220 245

Breadth

6.60 9.97 14.02 18.€

8.90 13.43 18.89 25.11

11.01 1' 2337 31.15.E

.-.. -.

7,61

9.4114 1 1.., 195 220 245

1200 1 72 0.94 2.20 3,9797 1.26 2.96 5.35

120 1.56 3.66 6.61

2 72 0.81 1.89 3,41

97 1.09 2.55 4.59120 1.35 3.15 5.68

Span No ofloads

Depth 72 97 120Breadth

1800 2 72 0.26 0.62 1.15 1.97 3.06 4.46 6.15 8.13

97 0.35 0.83 1.54 2.65 4.13 6.01 8.28 10.96

120 0.43 1.03 1,91 3.28 5.11 7.43 10.25 13.56

3 72 0.26 0.62 1.14 1.96 3.05 4.42 6.09 8.04

97 0.35 0.83 1.54 2.63 4.10 5.96 8.21 10.83

120 0.43 1.03 1.90 3.26 5.08 7.37 10.15 13.40

Span Noofloads

Depth 72 97 120 145 170 195 220 245

Breadth

2400 3 72 0.11 0.27 0.51 0.88 1.40 2.06 2.89 3.8897 0.15 0.37 0.69 1.19 1.88 2.78 3.89 5.22

120 0.19 0.46 0.85 1.47 2.33 3.44 4.81 6.46

4 72 0.11 0.26 0.49 0.86 1.35 2.00 2.80 3.7797 0.15 0.36 0.67 1.15 1.82 2.69 3.77 5.07

120 0.18 0.44 0.82 1.43 2.26 3.33 4.67 6.28




Table 2.8 Permissible loads on lintels: Timber strength class D40 — Medium-term loading

Point toads at 600 mm spacing

SpanBreadth

72 97 120

1200 1 72 157 365 6.52 10.7

97 212 492 879 1444120 262 608 1087 1786

2 72 116 274 496 8.2

97 157 3,69 669 1117120 194 456 827 1381


1800 2 72 043 103 190 326 509 7.40 1021 13.50

97 058 138 256 4,39 6.85 9.97 1375 18.19120 071 171 317 5,43 8.48 12.33 17.01 22.51

3 72 0.37 0.89 1.64 2.81 4.37 6.34 8.72 11.51

97 0.50 1,20 221 3.78 5.89 8.54 11.75 15.50120 0.62 1.48 2.74 4.68 7.28 10.57 14.54 19.18


2400 3 72 0.17 0.42 0.79 1.36 2.15 3.17 4.44 5.9697 0.23 0.57 1.06 1.84 2.90 428 5.98 8.03

120 0.29 0.70 1.31 2,27 3.59 5.29 7.40 9.944 72 0.16 0.39 0.73 1,27 2.01 2,96 4.15 5.58

97 0.22 0.53 0.99 1,71 2.70 3.99 5.59 7.52120 0.27 0.65 1.22 2.12 3.34 4.94 6.92 9.31

Point loads at 400 mm spacingSpan No. of Depth 72 97 120 145 170 195

loads Breadth1200 1 72 0.94 2.20 3.97 6.60 9.97 14.02 18t

97 1.26 2.96 5.35 8. 18.89 25.1120 1.56 3.66 6.61 11.( 23.37 31.1

2 72 0.81 1.89 3.41 5.6597 1.09 2.55 4,59 7,61

120 1.35 3.15 5.68 9.41

Span No of Depth 72 97 120 145 170 195 220 245loads Breadth

1800 2 72 0.26 0.62 1.15 1.97 3.06 4.46 6.15 8.1397 0.35 0.83 1.54 2.65 4.13 6.01 8.28 10,96

120 0.43 1.03 1,91 3.28 5.11 7,43 10.25 13.563 72 0.26 0.62 1.14 1,96 3.05 4.42 6.09 8.04

97 0.35 0.83 1.54 2.63 4.10 5.96 8.21 10,83120 0.43 1.03 1.90 3.26 5.08 7.37 10.15 13,40


2400 3 72 0.11 0.27 0.51 0.88 1.40 2.06 2.89 3.8897 0.15 0.37 0.69 1.19 1.88 2.78 3.89 5.22

120 0,19 0.46 0.85 1.47 2.33 3.44 4.81 6.464 72 0,11 0.26 0.49 0.86 1.35 2.00 2.80 3.77

97 0,15 0.36 0.67 1.15 1,82 2.69 3.77 5.07120 0,18 0.44 0.82 1.43 2.26 3.33 4.67 6.28

:

1T11 = deflection governs = bending governs = shear governs



7.49 941 11.531

- 10.09 12.67 15.5358 12.48 15.68 19.21

7.29 8.37 9.3289 9.82 11.28 12.5612.15 la.95 15.54

© TRADA Technology Ltd 2006 cC

Table 2.9 Permissible loads on lintels: Timber strength class D50 — Long-term loading


Span No ofloads

Depth 72 97 120 145 170 19 220 245Breadth

1200 1 72 2.64 6 13 10. 14. 9 39.288.26 13. 1 .64 .92

10.22 16.99 . 74.60 8.34

6.20 11237.66 1389

97 '20 l4 1 73 95 720 245

97 3.56

120 4.402 72 1.95

97 2.63

120 3.26

Span No ofloads

Depth 72

Breadth

-— 1243 16 03 1964/ i 16 75 21.59 .46913 14.24 20.72 2671 74 /2

"7.34 10.21

6.36 9.89 13.757.87 12,2 t 17.01

45 i7 ' 221 245

1800 2 72 0 72 1 72 3 19

97 0,97 2.32 4.30120 1.20 2.87 5.32

3 72 0.62 1.49 2.76

97 0.84 2.01 3.72

120 1.04 2.49 4.60

Span Noofloads

Depth 72 97 120Breadth

2400 3 72 0.29 0.71 1.32 2.20 3 ' 'i 7.46 9.8297 0.39 0.95 1.78 3,09 $1 / 18 10 05 13.23

120 0.49 1.18 2.20 3.82' 1 )3 8.89 12 43 16,374 72 0.27 0.66 1.23 2.13 3/ 4.98 6.93 38

97 0.37 0.89 1.66 2.87 4.54 6.70 9.40 '26'120 0.45 1.10 2.05 3.55 5.62 8.29 11.63 6—


Span Noofloads

Depth 72 97 120 145 170 195 220 245Breadth

1200 1 72 1.57 3.69 6.(

97 2.12 4.97 8.98

120 2.62 6.15 11.11

2 72 1.36 3.18 5.7

97 1.83 4.28 7.7

120 2.26 5.29 9.0.Span Noof

loadsDepth 72 97 120 145 170 195 220 245j

.Breadth1800 2 72 0.43 1.04 1.93 3.30 5.1E'

4.45 6.95.50 8.

3.29 5.1

4.43 6.

5.48 8.5145 170 195 220 245

97 0.58 1.40 2.59

120 0.72 1.73 3.21

3 72 0.43 1.04 1.92

97 0.58 1.40 2.58

120 0.72 1.73 3.20

Span Noofloads

Depth 72 97 120

Breadth

2400 3 72 0.19 0.46 0.86 1.49 2.35 3.46 4.85 6.51

97 0.26 0.62 1.15 2.00 3.16 4.67 6.53 8.78120 0.32 0.76 1.43 2.48 3.91 5.77 8.08 10.86

4 72 0.18 0.44 0.83 1.44 2.28 3.36 4.71 6.3397 0.25 0.60 1.12 1.94 3.07 4.53 6.34 8.53

120 0.31 0.74 1.38 2.40 3.79 5.60 7.84 10.55

14.77 19.14 23.23 25.87f 14.73 19.89 25.78 31.30 3486

18.22 24.61 31.90 3872 43.122 7.67 8.99 10.31 11.83 12.951 10.33 12.11 13.89 15.61 1745

A 12.78 14.98 17.18 19.38 21.59




10.93 14.77 19.14 2323 25.8798 14.73 19.89 25.78 3130 3486

18.22 24.61 3190 38r2 43.12767 8.99 IO31 11,63 12.9

13 1211 13. 15.67 17.412.18 14.98 17.18 19.38 21

Fl05l13 16.37j

Table 2.10 Permissible loads on lintels: Timber strength class D50 Medium-term loading


Span No ofloads

Depth 72 97 120 1db 170 195 220 245Breadth

613 10.19 14

826 1 .9210.22

460 8.346.20 11.23

7.66 138997 120 145 170 95 220 74c

1200 1 72 2,6497 3.56

120 4.402 72 1.95

97 2.63120 3.26

Span Noofloads

Depth 72Breadth

1800 2 72 0.72 1.72 3.19 5.48 54

2202451

97 0.97 2.32 4.30 7.38

195

120 1.20 2.87 5.32 9.133 72 0.62 1.49 2.76 4.72

97 0.84 2.01 3.72 6.36120 1.04 2.49 4.60 7.87

Span Noofloads

Depth 72 97 120 145 170Breadth

2400 3 72 0.29 0.71 1.32 2.29 3.62 5.33 7.97 0.39 0.95 1.78 3.09 4.87 7.18 ici

120 0.49 1.18 2.20 3.82 6.03 8.89 12.-.4 72 0.27 0.66 1.23 2.13 3.37 4,98 6.98 9.38

97 0.37 0.89 1.66 2.87 4.54 6.70 9.40 12.64120 0.45 1.10 2.05 3.55 5.62 8.29 11.63 1564


Span Noofloads

Depth 72 97 120 145 170 195 220 245Breadth

1200 1 72 1.57 3.69 6.670

11.11

5.727.71

9.54120 145 170 195 220 245

97 2.12 4.97120 2.62 6.15

2 72 1.36 3.1897 1.83 4.28

120 2.26 5.29Span Noof

loadsDepth 72 97

Breadth1800 2 72 0.43 1.04 1.93 3.30

97 0.58 1.40 2.59 4.45120 0.72 1.73 3.21 5.50

3 72 0.43 1.04 1,92 3.2997 0.58 1.40 2.58 4.43

120 0.72 1.73 3.20 5.48Span No of

loadsDepth 72 97 120 145

Breadth2400 3 72 0.19 0.46 0.86 1.49 2.35 3.46 4.85 6.51

97 0.26 0.62 1.15 2.00 3.16 4.67 6.53 8.78120 0.32 0.76 1.43 2.48 3.91 577 8.08 10.86

4 7297

120

0.180.25

0.31

0.44

0.60

0.74

0.83

1.12

1.38

1.44

1.94

2.40

2.28

3.07

3.79

3364.53

5.60

4.71

6.34

7.84

6.338.53

10.55

______ deflection governs I = bending governs = shear governs


2.3 Racking resistance of timber frame walls

2.3.1 Basic design and materials

2.3.1.1 Standard wall panels

BS 5268-6 provides a method of assessing the racking resistance of a timberframe wall made of standard wall panels which are sheathed with one of agiven list of materials nailed to the frame.

To comply with the racking values in the Standard, the frame must consist ofgrade 016 softwood or better with a minimum cross-section of 38 mm x 72 mm.For internal walls the section may be reduced to 38 mm x 63 mm with a 15%reduction in racking resistance.

The sheathing materials listed in the 1996 edition of BS 5268-6.1 refer tostandards now withdrawn. All wood-based panels used for sheathing mustnow comply with the relevant technical specification requirements laid downfor wall sheathing in BS EN 13986 Wood-based panels for use inconstruction. Characteristics, evaluation of conformity and marking. Furtherinformation is given in the TRADA Wood Information Sheets 0 - 5 Timberframe building: materials specification, 2/3-56 CE marking: Implications fortimber products and 2/3 - 57 Specifying wood-based panels for structural use.Materials selected must be appropriate for use in Service Class 2 (BS 5268-2). OSB/3 is the most commonly used material.

If gypsum plasterboard is assumed to make a structural contribution it must bemanufactured in accordance with BS 1230-1: 1995 Gypsum plasterboard —Specification for plasterboard excluding materials submitted to secondaryoperations.

BS 5268-6.1 gives a basic racking resistance in kN per metre length for aspecified thickness of each sheathing material, fixed with nails of a specifieddiameter and spacing, in a 2.4 m square panel without openings or verticalload. The additional contribution of a secondary board, nailed independentlyof the primary board, is allowed for by another, lower, set of figures. Theracking resistance thus obtained is modified as necessary by multiplying it bythe following factors to obtain a value for the wall which should be similar toresults from test data.

K101 Variation in nail diameter

K102 Variation in nail spacing

K103 Variation in board thickness

K104 Height of wall panel

K105 Length of wall

K106 Fully framed openings

K107 Vertical load

The final result is multiplied by K108 an interaction factor of 1.1, whichaccounts for interaction between the walls, floors and the stiffening effects ofcorners in the actual building.

The construction of separating walls, which consist of two or more layers ofplasterboard nailed to one side of the timber frame, is laid down in detail in thestandard, and a basic racking resistance of 0.9 kN/m is given for such walls.The maximum allowable total contribution of plasterboard on other walls to theracking resistance of a building in a given direction is one third of the totalracking resistance in that direction.

No contribution to racking resistance may be assumed for glazed areas.

(ATECHNOLOGY((7



A worked example of the assessment of the racking strength of a house isgiven in Section 7.4.

2.3.1.2 Other ways to increase the racking resistance

The figures given in BS 5268-6 are based on the strength of nailed joints inC16 timber framing and, in the case of plywood, on plywoods in group I asdefined in Table 63 of BS 5268-2. Examination of this table shows that if C24framing is used with a group II plywood then an increase in strength of 10%can be obtained.

Another way to obtain higher racking values is to determine them by testing.Particular materials may have a basic racking resistance which is higher thanthe generalised figures provided by the standard: and different materials andforms of construction for which figures are not given may be suitable forsheathing. In such cases manufacturers may produce more appropriatefigures by specific tests described in Section 5 of the Standard. The figuresthus obtained may be used only for panels which are identical in all respectsto the panels used in the test. Hence they may not be modified by factors K101,K102 or K103

BS 5268-6.1 allows for an additional contribution to racking resistance frommasonry veneer, provided that is connected to the timber frame with specifiedwall tying and its contribution does not exceed 25% of the racking resistanceprovided by the timber frame. Special calculations or tests will be required ifthe designer wishes to take into account any additional contribution to rackingresistance which may be provided by cladding material other than masonry.

2.3.1.3 Staples

Some wall panel manufacturers use staples in place of nails for fixingsheathing. Eurocode 5 provides a method for calculating the strength of astapled connection by means of which a designer could determine theequivalent nail diameter and hence obtain an appropriate value for K101 todetermine the racking resistance of a stapled panel. However, there aredifferent kinds of staple, some of which are designed to splay out on insertion,and it may be difficult to meet the minimum edge distance of 20d specified inEurocode 5. Hence it is safer to obtain the racking strength of stapled panelsby testing rather than by calculation.

Eurocode 5 recommends a minimum corrosion protection of Fe/Zn 12c orZ275 for staples used in Service Class 2.

2.3.2 Adhesively bonded panelsWithin the limits specified in BS 5268-6.1 the racking resistance of a standardwall panel may be increased by increasing the nail diameter, reducing the nailspacing or increasing the board thickness. If the required racking resistancecannot be achieved by these means, shear panels of very high rigidity andstrength can be obtained by attaching a structural sheathing material to theframing with a suitable structural adhesive. The manufacture of such panelsmust be carried out under a fully approved factory production control qualityassurance system to ensure consistency of bond strength and structuralcharacteristics. (See TRADA Wood Information Sheet 2/3-31 Adhesivelybonded timber connections.) The strength and stiffness of such panels shouldbe determined by testing in accordance with BS EN 594 (see Section 2.3.2.3).

2.3.2.1 Panel shear

Assuming a single sheet of sheathing material on one side only, the maximumapplied panel shear stress may be calculated as:

= l.SFa N/mm2tL

where Fa = total applied shear load (N)

= thickness of panel product (mm)

L = length of panel (mm)



This should not exceed the permissible panel shear stress. For plywood thegrade stress for panel shear is given in BS 5268-2 Tables 40 to 56. Genericvalues for the characteristic planar shear strength of OSB, particleboards andfibreboards are given in BS EN 12369-1 Wood-based panels. Characteristicvalues for structural design. OSB, particleboards and fibreboards. These maybe converted to permissible stresses using the method given in BS 5268-2Clause 5.3.

Normally the grade stress would be multiplied by K35 = 1.5 for wind loading,but US recommendations suggest that an increase of 1.33 is moreappropriate for the response of timber frame wall panels within a building to agust of wind. 1.33 is the value of K36 given in BS 5268-2 Table 39 for medium-term loading in Service Classes 1 and 2. For OSB/3 under medium-termloading the permissible planar shear stress in Service class 2 is 2.48 N/mm2.

2.3.2.2 RoIling shear

Assuming the worst case, when all the shear load is applied to the top rail ofthe panel, the applied rolling shear stress in the glued joint is:

Ta= N/mm2

Lh

where Fa = total applied shear load (N)

L = length of panel (mm)

h = depth of top rail (N)

Ta should not exceed the grade rolling shear stress for the sheathing material

multiplied by 1.33 x 0.5.

The factor of 0.5 is required by Clause 4.7 of BS 5268-2 to allow for stressconcentrations. A further reduction of 0.9 should be applied if the bondingpressure is applied only by nails or staples (Clause 6.10.1.4). Thesereductions are not required in designs to Eurocode 5 which provides littleguidance on the design of adhesively bonded joints, preferring tests todetermine their strength.

For panel products other than plywood, the permissible planar shear stressmaybe used, calculated by the method given in 2.3.2.1.

2.3.2.3. Values from tests

Alternatively the racking strength of glued panels may be obtained by testingas described in BS 5268-6. This involves the test method BS EN 594: 1996Timber structures — Test methods — Racking strength and stiffness of timberframe wall panels. Structural insulated panel systems (SIPS) may be tested inaccordance with ETAG 019: 2005 Pre-fabricated wood-based loadbearingstressed skin panels. As previously stated, the racking resistances obtainedfrom tests may not be modified by factors K101, K102 or K103.

2.3.3 Overturning effectsWhether adhesive bonding is used or not, a panel must be restrained againstoverturning by means of:

• its attachment to a panel on the windward side• the vertical load applied to it by the weight of the building above it, less

any wind uplift• holding down the end studs to the foundation in some way• holding down the bottom rail to the foundation with sufficient fasteners

into the foundation and the structural sheathing• a combination of these methods.

TECMNOLOGY(17



h

F/////Rv L

R _EflW Fh WVL 2As shown in diagram (a) above, when the racking load is sufficient to causerotation the reactions at the two bottom corners are in opposite directions.This means that if there is an adjacent panel on the windward side itsconnection to that panel can restrain it against uplift with a total restraint forceof Fh/L + W12. For panels of similar length this is greater than the totalrestraint required against uplift, Fh/L - W/2. Hence no further restraint isnecessary provided it can be shown that the connections between the panelsare strong enough to transfer this force between the panels. For the endpanels however, or for panels where further restraint is needed, otherapproaches should be considered.

Diagram (a) shows restraint to overturning provided by a metal holding downstrap at each of the bottom corners. With a total vertical load of W and aracking load of F the restraint strap must resist a force of

R

A similar strap is needed at both ends to cater for a reversal in the winddirection. These are nailed, screwed or bolted to the wide face of the outerstuds and anchored to the concrete foundation. The strength of the strap andits connection at each end should be calculated and exceed R. In addition thefasteners connecting the bottom rail to the binder or sole plate or foundationmust provide a total lateral resistance of F, so their maximum spacing is

where fadm = permissible load for one fastener for short- or very

short-term loading

= basic load x 1.25 N (BS 5268-2 Clause 6.4.9)


(a) Restraint forces applied at corners (b) Restraint forces applied along basew w

FV V

F ..

h

fh=F1L per unit length/// fvimaxU

3'Fh W'f1max= per unit length

— Fh — W

L 2

FL5 =—mm

adm


A second approach is to attach the outer studs securely to the bottom rail bymeans of metal straps which wrap around the outside of the panel at eachcorner. Suitable straps are available in several proprietary forms, either aspierced plates for nailing to the framing or plates with integral teeth forapplication by hydraulic press. In this case all the overturning restraint isprovided by ringed-shank nails or screws or bolts fastened through the bottomrail along its full length. These fasteners must provide a lateral resistance offh = F/L per unit length and a resistance to axial withdrawal of:

— 3(Fh Wfvmax -

per unit length as shown in diagram (b).

BS 5268 does not provide a method to allow for the effects of combined axialand lateral loading, but Eurocode 5 states that for annular ringed shank andhelically threaded nails, and by implication for wood screws, the requirement

+ ----2

1 must be satisfied1\Raxd) R1ad

where Faxd and Flad may be taken as the axial and lateral load respectivelyapplied to each connection, and Rax,d and Rla,d may be taken as thepermissible withdrawal and lateral load respectively for each connection. Fornails and screws, the requirement in BS 5268 terms becomes

2

+ 1

ax,ajm . facim)

where s = fastener spacing (mm)

fv,max = maximum applied vertical load per unit length, as above (N/mm)

faxacim = permissible withdrawal load for one fastener (N)

= basic load x 1.25

fh = applied horizontal load per unit length = F/L (N/mm)

facirn = permissible lateral load for one fastener (N)

= basic load x 1.25

Hence Smax1

v,max 2 +12fax,acim) . acim)

With bolts it is sufficient to ensure that the resistance of each connection tothe lateral loads is adequate and, in the case of anchor bolts to foundations,that the manufacturer's safe working load on each bolt is not exceeded. Thisis because, with bolts of at least 8 mm diameter, failure in two-memberlaterally loaded softwood connections occurs in the timber rather than the bolt.

It may be possible to transmit the overturning moment from the structuralsheathing into the bottom rail without any restraining straps. In this case theshear connection between the sheathing and the framing resists the appliedloads. These loads comprise (i) a horizontal load of fh = F/L per unit length,which produces in an adhesively bonded connection an applied shear stressof ta as previously calculated, and (ii) a maximum vertical force per unit lengthof

3(Fh Wfvmax =

—--——-—jN/mm

which produces an applied shear stress of

T = v,max N/mm2h

where h = depth of the bottom rail (mm).



In mechanically fastened joints the fasteners must provide a total resistance of

fres, a = h2 + v,max2 N/mm

giving a maximum spacing of

S =mmmres,a

In adhesively bonded joints the resultant applied shear stress in the sheathingand the framing is

/2 2Tresa = Ta +T

For the framing the permissible shear stress may be calculated from theformula given in BS 5268-2 Clause 6.10.1.3 as:

Tadma = Tadm,ll(1_0.67510)

where Tadm a = permissible shear stress in the timber

Tadmjl = permissible timber shear stress parallel to the grain

= 1.33 times grade stress as above

a angle between the grain and direction of Tresa

—1 Z=tan —Ta

For OSB the applied shear stress should be compared with the permissibleplanar shear stress. This can be obtained from the characteristic planar shearstrength given in BS EN 12369-1 converted to a permissible stress using themethod given in BS 5268-2. For OSB/3 under very short-term loading thepermissible planar shear stress in service class 2 is 0.496 N/mm2.

For plywood the angle of the shear stress will vary in alternate plies.Assuming that the rolling shear strength is one third of the shear strengthparallel to the fibres, as implied by the above formula, it may be adapted as

Tadma = Tadm,90(3_25ifl0)

where Tadm a = permissible shear stress in the plywood

Tadm 90 permissible rolling shear stress in the plywood

= 1.33 times grade stress as above

a = maximum of tan1 -- and tan1 .-Tv

Where straps are not used the tensile strength of the sheathing should bechecked for an applied tensile stress of fv,max/t where t is the sheathingthickness, and a permissible stress calculated as the grade stress parallel orperpendicular to the grain as appropriate x 1.33 as above.

2.3.4 DeflectionIt is difficult to calculate the horizontal deflection of a timber frame wallbecause:• the connection between panels is not rigid and manufacturing tolerances

permit small gaps between the abutting studs, so the panels actindependently to some extent, but not entirely

• the sheathing tends to bow along the compression diagonal, increasingthe total deflection

• fastener slip between the top and bottom rails of the panel and thesheathing is difficult to calculate accurately

• the framing makes an indeterminate contribution to the racking stiffness• the effect of openings is difficult to quantify.

TRADA© TRADATechnology Ltd 2006

46


In general, however, it may be assumed that if a wall designed to BS 5268-6.1can resist the applied racking loads then its deflection will not be excessive.One reason for this is that the racking resistance of wall panels calculatedaccording to the Standard is based on tests which limit the amount by which apanel may deflect before it is deemed to have failed.

2.3.2.4 Buckling

Timber frame wall panels designed to BS 5268-6.1 within the height range of2.1 — 2.7 m do not require a separate check for the buckling resistance of thesheathing.

(R© TRADA Technology Ltd 2006 ci:13,


3 Floors

3.1 Joists and beamsSpan tables for floor, ceiling and roof joists are provided in TRADA's Spanfables for solid timber members in floors, ceilings and roofs (excluding trussedrafter roofs) for dwellings. This provides important information about thedesign and installation of timber floors using softwood of Strength ClassesC16 and C24. BS 8103-3 provides similar tables but the spans are a littlelower in some cases because they were calculated for the former SC3 andSC4 Strength Classes. Finally suitable member sizes for any weight of floorand imposed load may be obtained quickly using TRADA's Timber Sizersoftware (visit the Software Toolbox at www.trada.co.uk).

When the distance between load bearing walls becomes so great that it is noteconomical to span directly with solid softwood floor joists, it may be possibleto use prefabricated engineered timber joists, hardwood, or a structural timbercomposite such as LVL (laminated veneer lumber), PSL (parallel strandlumber), or LSL (laminated strand lumber).

Typical spans which can be achieved economically are:• Softwoods: up to 5 m• Hardwoods and timber composites: 5 — 7 m• Engineered timber joists: 4—7 m or more.

If even longer spans are required, or if conventional joists in commonlyavailable sizes are preferred, then intermediate floor beams may be used tosupport the joists. These may be deeper sections of hardwoods or structuraltimber composites, doubled engineered joists or steel flitched beams.Doubled engineered joists should be connected together in pairs by fixing webstiffeners between them or by using proprietary fasteners or fixing methods,always in accordance with the manufacturer's instructions.

Strutting between floor joists in the form of solid timber blocking, timberherringbone strutting, or proprietary metal herringbone strutting should bespecified in accordance with the recommendations given in the publicationscited above or see TRADA's Wood Information Sheet 1-41 Strutting in timberfloors. For engineered joists the manufacturers may provide otherrecommendations or strutting methods, such as the strong-backs used withmetal open web joists.

Currently the National House-Building Council (NHBC) recommends adeflection limit of 12 mm for all domestic floors made without strutting,because tests have shown that properly installed strutting significantlyimproves the vibrational performance of a floor, which is an important part ofstructural design.

TICHNOLOGV((5© TRADA Technology Ltd 2006 48


3.1.1 Reactions, moments and deflections in a single span, simply supportedbeam

3.1.1.1 Uniformly distributed loads

w = load per unit length

Reaction at A Reaction at B

w(b—a)(2L —a—b)RA= 2Lw(b2 —a2)

R6= 2L

Value of x Bending moment at a distance x from RA Maximum bending moment

x<a MX=RAX

a <x<b w(x—a)2M =RAx— 2

(RA \ RAMmax =RA—+aJatx =—+a

x> b = RAX — w(b — a)(2x — a — b)

2

Configuration Bending deflection at centre Shear deflection at centre (1)

o < a < L/2— —b<L/2

W 2 L2 b2 2 L2 a2Yb b —— —aEl 32 48 32 48 J

3w 1 2 2Ys b —a

1OGA

o < a < L12— —b> L12

— w 15L 2 L2 C2 2 L2 a2Yb —------———C —a ——

El 32 48 32 48— 3w 1 2 2 2

Ys — L —2c —2a2OGA

a> L/2b> L12

Yb = (L — a)2L2

— (L — a)2 — 2 L2—

c2El 32 48 32 48

= 3w {(L— a)2 — c21OGA

For engineered timber joists, consult the manufacturer


t PA Rot


3.1.1.2 Point loads

LII 'F I

a C

t P Pa

Reaction at A Reaction at BFc

RA=—Fa

RB=-[-Value of x Bending moment at a distance x from RA Maximum bending moment

x<a M =RAX M =RAa atx =ax> a M = RAx — F(x — a) M = RAa at x = a

Configuration Bending deflection at centre Shear deflection at centre (1)

a<L/2 Fa 1L2 a2l 3FaY=-a>L/2 Fc IL2 c21Yb=l- 3FcYs=(1) For engineered timber joists, consult the manufacturer

3.1.2 Solid timber beamsArrangement of loads on effective span

RA=11.85 kN

Long-term 2.35 kNMedium-term 1.83 kN

4.l8kN

Self-weight 0.23 kN/mLong-term 2.62 kN/m

Medium-term 2.03 kN/m4.88 kN/m

1.Om

HRB=13.52 kN

This example outlines a design for a solid hardwood floor beam consisting of2 No 63 x 250 mm nominal (60 x 245 mm) hardwood members in StrengthClass D50.

BS 5268-2 Table 8 gives the average density of D50 timber as 780 kg/rn3.

Hence self-weight = 780 x 2 x 60 x 245 x 9.812 x 10

— 0.56x3.O(2x4.Q—3.0) 4.18x1.0 4.88x4.ONA — + +

2x4.O 4.0 2


= 0.23 kN/m.

Long-term 0.56 kN/m

3.Om

= 11.85 kN


— 0.56x3.02 4.18x3.0 4.88x4.0 —RB — _______ + + —13.52kN

2x4.0 4.0 2

Rmax = 13.52 kN

Zero shear stress occurs at 11.85 = 2.178 m from RA0.56 + 4.88

Mmax = (11.85x2.178)_[(o.56+4.88)x2.1782 xO.5j = 12.92kNm

Stresses

12.92x106x6 2Gm,a = = 10.76 N/mm

2x60x245

G,adm16.0X1.25X1.1XJ

= 18.00 N/mm2

where 1.25 = K3 (duration factor)

1.1 = K8 (load sharing factor)

0.11

= K7 (depth factor)

iSV 1.5x13.52x1000 2Ta =0.690 N/mm

A 2x60x245

Tadm =2.2x1.1 =2.42 N/mm2

where 1.1 = K8 (load sharing factor)

Deflection

Use Emin X K9 (Clause 2.10.11)

= 12600 x 1.06 = 13356 N/mm2

G = E/16 (Clause2.7)

= 13356/16 = 834.8 N/mm2

El = 13356260245=1.964x1012Nmm2

12

GA = 834.8 x 2 x 60 x 245 = 24.54 x 106 Nmm2

0.56 5x4000410002

40002 10002+

3x0.56{40002_(2X10002)}1.964x1012 384 32 48 20x24.54x106

= 0.814+0.041 = 0.855mm

4.18x1000x1000 [40002 10002] 3x4.l8xl000xl000=1.964x1012 16

—

12 5x24.54x106

=1.951+0.102 =2.053mm

= 4.88 54000 3x4.88{40002}1.964x1012 384 j 20x24.54x106

= 8.281 + 0.477 = 8.758 mm

Total deflection = 11.67 mm

Permissible deflection = 4000 x 0.003 (Clause 2.10.7) = 12.00 mm

Therefore the beam is satisfactory.

(__© TRADA Technology Ltd 2006 51


Bolting

Assuming that the beam is loaded evenly on both sides then only nominalbolting will be required to ensure that both members act together. A single rowof M10 bolts at not more than 600 mm centres along the centre line of thebeam will be adequate.

If the beam is predominantly loaded to one side, it can be reasoned that for atwo-member beam, for which K8 = 1.1, the bolting should be capable oftransferring approximately 55% of the load.

3.1.3 Structural timber compositesStructural timber composites are made by gluing together flakes, strands orveneers of wood under pressure to create new structural materials. Theymake excellent floor beams which have greater strength, stiffness anddimensional stability than the solid timber from which they are made. Thetypes currently available in the UK are LVL (laminated veneer lumber), PSL(parallel strand lumber) and LSL (laminated strand lumber). Design values foruse with BS 5268-2 are provided in the manufacturer's certification literature,but for designs to Eurocode 5 reference should be made to the manufacturer.In general, structural timber composites are designed like solid timber, but theliterature should be read carefully, as some modification factors are differentfrom those for solid timber and there may be special rules for mechanicallyfastened joints. Typical grade values for the principal properties are shown inTable 3.1, with values for C24 timber given for comparison. This shows thattheir main advantage over solid softwood is their superior strength, hence theyare more suitable for extra heavy loads than for extra long spans. Furtherinformation about these products can be found in TRADA's Structural timbercomposites: DG1. For information on suppliers, visit the Suppliers Directory atwww.trada.co.uk.

Table 3.1 Grade MOE and stresses for C24 softwood and some selected structural timbercomposites in N/mmMaterial Emean m c,perp

C24 softwood 10800 7.5 0.71 2.4 (no wane)

LSL 10300 13.1 2.8 5.3

LVL 11500 17.5 2.0 2.9

PSL 12750 16.5 2.2 3.6NE The values given in the table relate to particular products irom particular manufacturers. Themanufacturers' literature should be consulted for actual design values.

3.1.4 Prefabricated engineered timber joistsPrefabricated engineered timber joists manufactured in the UK and othercountries are being increasingly used in floor construction because of theirvery high strength-to-weight ratio as well as their dimensional stability.

Typically they are of two types; I-joists and open metal web joists. Timber I-joists comprise a solid timber or structural timber composite flange and apanel product web. Timber composites used for flanges include laminatedveneer lumber (LVL) and parallel strand lumber (PSL). Webs are usuallyoriented strand board (OSB) but may also be hardboard. The web is securedto each flange by an approved weatherproof, structural adhesive within therout. Metal web joists comprise a solid timber or structural timber compositeflange with V-shaped galvanised steel components forming the open web.

TRADA Wood Information Sheet 1-42 Timber I-joists: applications anddesigns, provides further information on the properties of prefabricated timberI-joists and their design requirements.

r E C H N 0



Their use is permifted in the UK through third party certification whichestablishes safe design values for use with British and European designcodes. Manufacturers provide span tables for their products, sometimes witha choice of deflection limits depending on the required performance level.

Generally engineered timber joists are used instead of solid timber joists, butthey can be used in pairs as beams. In the latter case the manufacturer'sadvice on fastening them together should be strictly followed. An engineeredjoist beam is most useful in a floor which is already made with engineeredjoists: in other cases it may be simpler to use one of the other beam options.

3.1.5 Steel flitch beams

3.1.5.1 Bolted steel flitch beams

Bolted steel flitch beams like the one shown in the diagram are useful in loftconversions and as lintels, or anywhere that an enhanced load-carryingcapacity is required. It is recommended that the depth of the steel plate shouldbe 25 mm less than that of the timber members, to allow for shrinkage of thetimber or lack of straightness in the timber or steel. This is because the designmethod assumes that all the applied loads and the reactions at the supportswill be applied to the two timber members. A proportion of these loads istransferred to the steel plate via the bolts along the beam and out again viathe bolts at the supports.

The following example illustrates how to design a flitch beam using BS 5268-2.

Arrangement of beam:

_______ 12.5

22O r71: ____Dimensions in mm



3.1.5.2 Dimensions

Structural span L 4 metres

Length of bearings Lbearing = 56 mm

Diameterofbolts d = 10mm

Breadth Depth Area Section modulus Second moment(total) (total) of area (total)

b mm h mm A mm2 Z mm3 I mm4

Timber 44 220 19.36 x i03 0.7099 x 106 78.09 x 106

Steel 10 195 1950 63.38 x103 6.179x 106

3.1.5.3 Properties

Timber stringers C24 from BS 5268-2: 2002 Table 8

Timber bearings 016 from BS 5268-2: 2002 Table 8

Steel properties for grade 43 steel plate from BS 449-2

MOEE N/mm2

Shear modulusG N/mm2

Densityp kg/m3

Permissible

Ym,ads,s

stresses* in N/mm

tadm,s Yc,adm,s

205000 78850 7850 180 110 250

3.1.5.4 Loads, reactions and bending moment

For simplicity this example will use the same span, loads and medium-termload case as were used for the example in Section 3.1.2.

Self-weight of timber and steel (assumed for simplicity to act as an additionallong-term UDL)

[(2x44x220x420)+(lOxi95x7850)]x9.812x109 = 0.23 kN/m

As before

RA =11.85kN

RB =13.52kN

and Mmax = 12.92 kNm

3.1.4.5 Load distribution factors

The proportions of load taken by the timber, k, and by the steel, k, dependmainly on the bending stiffnesses of the timber and steel.

For two-member softwood beams:

Emjn = 1.14 x Emin (BS 5268-2 Table 20) = 1.14 x 7200 = 8208 N/mm2

For timber:

Emjnlt = 8208 x 78.09 x 106 = 0.6409 x 1012 Nmm2

Emean It = 10800 x 78.09 x 106 = 0.8433 x 1012 Nmm2

For steel:

El = 205000 x 6.179 x 106 = 1.267 x 1012 Nmm2



The proportion of the total load carried by either the two timber members orthe steel plate can be represented by one of the following three factors whichallow for the most unfavourable variables in each case:

k for calculating timber stresses

k for calculating bolt loads and steel stresses

k for calculating the deflection of timber members.

To check timber stresses a safe value of k is obtained using Emean

k = Emeanlt = 0.8433 = 0.3997Emeanlt + El 0.8433 + 1.267

To check steel stresses Emin is used as this produces higher stresses in thesteel. An additional factor of 1.1 is recommended to allow for the relativelygreater shear modulus of steel and slip in the fasteners, both of which canincrease its share of the load.

k5 = 1.1xEl5 = 1.lxl.267 = 0.7304EminIt + El 0.6409 + 1.267

Since the timber and steel move together, the deflection of the beam can becalculated by calculating the deflection of the timber members. The modifiedvalue of Emin is used as this produces the least stiff beam.

k = Eminlt 0.6409 = 0.3360Eminit + EI 0.6409 + 1.267

3.1.5.6 Effects of holes

Assuming Ml 0 bolts with a hole diameter d' of 11 mm staggered with an offsetof h/4 55 mm from the centre line, the bending stress increase factor due toone hole

— h2(h2 — hd' +2d'y)kbending —3 3 4 2h —h d'—hd' +d' —l2hd'y

where h = depth of the membersy = distance between the centre of the hole and the centre

line of the beam.

For the timber:

k = 22o2(2?o2_(22ox11)+(211x55)) = 1069bending,t 3 3 4 2220 —(220 xll)_220x11 +11 _(12x220x11x55

For the steel plate:

k = 1952(1952 _(195x11)+(2x11x55)) = 1 096bendng,s 3 3 4 2195 —(195 xll)_195x11 +11 _(12x195x11x55

Provided that the bolts at the reaction points are above the bearing plates andthere are no other bolts within a distance of h from the reaction points, itshould be unnecessary to consider a reduced section when checking theshear strength of the timber.

The shear strength of the steel will normally be adequate.

(R© TRADA Technology Ltd 2006


3.1.5.7 Timber stresses

K3 = 1.25

( \O.11K6 =I--j =1.035

220)

K8 = 1.1 to check bearing strength of stringers

An investigation into the load distributions between the timber and steelproduced by variations in the timber's strength and stiffness has shown that infitch beams with two timber members the grade bending and shear stressesshould be enhanced by only 1.04, not 1.1 as in clause 2.10.11, and only if

0.2Eltotai � EsIs 0.8Eltotai

where Eltotai = Emeanlt + E515

K8A = 1.04 to check bending and shear strength of stringers

Bending

0m,a = ktkbending,tMmax

zt

= 0.3997x1.069x12.92x106 = 7.772 N/mm20.7099 106

Gmadm = amXK3XK6XK8A

= 7.5x1.25x1.035x1.04 = 10.09 N/mm2

Shear

— 1.5ktRmax— ______aAt

= 1.5x0.3997x13.52x103 = 0.419 N/mm219.36 io

tadm = txK3 xK8A = O.71x1.25x1.04 = 0.923 N/mm2

Compression

Gc9Q,a = Rmax = 13.52x i03 = 2.744 N/mm22btLbearing 2 x 44X 56

To check the compression strength of the bearings 1<8 is not used

0c,90,adm = 0c,90 x K3 = 2.2 xl .25 = 2.750 N/mm2

3.1.5.8 Deflection

For convenience the deflection of the timber members is calculated.

To calculate shear deflection

Gmrn = where Emin includes K9 (Clauses 2.7 and 2.10.11)

513N/mm216

© TRADATechnology Ltd 2006 56 _J\__


Using the formulae given in Section 3.1.1, the total bending deflection at thecentre =

k X O.56[5>4000 _10002 40002 10002 L4i8oxiooo4000 1ooo24885x4ooo4EminIt

L384 32 48

JL 16 12

J384

0.3360 12= x21.70x10 11.37mm0.6409 1O12

The total shear deflection at the centre =

ktdef 3x0.56 OOO2 _(2Xl0002)}+ 3x4180x1000 +3x4.88x40002

GminAt 20 5 20

0.3360 6= x 15.40 x 10 = 0.52 mm513x19.36x103

Total deflection = 11.37 + 0.52 = 11.89 mm

Permissible deflection = 0.003L

= 0.003 x 4000 = 12.00 mm

3.1.5.9 Creep

Although BS 5268-2 does not specifically require creep to be taken intoaccount, creep does affect the distribution of forces within a flitch beam.Annex K explains how to allow for creep in solid timber members. For flitchbeams one approach is to calculate a final effective value of Emin for thetimber from the final deflections. According to Annex K, the final deflectionsdue to permanent and variable loads respectively are:

ufflG = (l+kdef)

and uno = (1 +W2lkdef) + + M12,ikdef)

i>I

where UinstG = instantaneous deflection caused by permanent loads(generally self-weight and dead weight)

= instantaneous deflection caused by the leading variableload, ie the one which produces the greatest deflection incombination with the other variable loads

kdef = long term deformation factor taken from Annex K Table K2= 0.6 for solid timber in service class 1.

wo characteristic load combination factor for variable load itaken from Annex K Table Ki.

= quasi-permanent load combination factor for variableload i taken from Annex K Table Ki.

N.B The second formula corrects misprints in the 2002 edition of Annex K.

For a given load case, the final effective value of Emin is

E — (ujflstG +ujto)mintin — Lmin

(uflflG + UfIQ)

where = uflStQl +

i> I

(A© TRADA Technology Ltd 2006


Strictly speaking the values of "u' should be calculated using Emjn,fi, not Emn,but since the purpose of the formula is to calculate Emin,n, this cannot be donesimply. Using Emin as in this example produces an approximate conservativeresult, giving slightly larger stresses in the steel, loading in the bolts and finaldeflections than would be obtained using a more exact method.

The final value of Emn is then used to check the final deflection. For finaldeflections, Annex K recommends a relaxed deflection limit of L/200.

In this example the following results are obtained:

Load 0.56 2.35 1.83 0.23 2.62 2.03_________ kN/m kN kN kN/m kN/m kN/m

Type Storage Permanent Snow Permanent Permanent Residential

1.0 - 0.6 - - 0.7

'V2 0.8 - 0.0 - - 0.3

UjnstG - 1.18 - 0.42 4.79 -

0.88 - 0.92 - - 3.71*

* leading variable load

uIflSG = 1.18 + 0.42 + 4.79 = 6.38mm

= 3.71 + (0.88 x 1) + (0.92 x 0.6) = 5.14mm

ufflG = 6.38(1 + 0.6) = 10.22 mm

ufIQ = 3.71(1 + 0.3 x 0.6) + 0.88(1 + 0.8 x 0.6) + 0.92(0.6 + 0.0 x 0.6)

= 4.378 + 1.302 + 0.552 = 6.23 mm

Hence Ef = 8208 x (6.38 + 5.14) = 5752 N/mm2(10.22+6.23)

Substituting this value for Emin

Emin,rin It = 5752 x 78.09 x 106 = 0.4492 x 1012 Nmm2

— Eminfinlt — 0.4492 —— ___________ — —0.2167Eminnlt + El 0.4492 + 1.267

Final deflection corrected for Emjn fl and k'fl

= ufflG+urflax kufinEmin= (10.22 +6.23)x

0.2617x8208I I

kuEminn 0.3360 x 5752

= 18.28mm

4000The permissible final deflection = = 20 mm

200

3.1 .5.10 Steel stresses

Bending

The steel will now take a larger share of the load so its bending strengthshould be checked using Emjn,fjn

It is assumed that the timber prevents buckling of the steel plate, therefore thefull bending strength of the steel may be utilised. This means that the restraintconditions specified in BS 5268-2 Table 19 must be satisfied for each timbermember.

t44



Therefore the ends must be held in position and the compression edge held inline, as by direct connection of sheathing, deck or joists.

Emjn,fjnlt = 5752 x 78.09 x 106 = 0.4492 x 10 Nmm2

kffl 1.1xEl = 1.lxl.267 =0.8121Emin6n I + El 0.4492 + 1.267

ksfinkbendingsMmax0m,a,s,fin =.7

S

0.8121x1.096x12.92x10 2___________________ = 181.4 N/mm63.38 x i0

Gm,adm,s = 180.0 N/mm2

In practice the simplest solution to the small over-stress would be to increasethe depth of the steel plate by 1 mm.

Bearing and shear strength

The strength of the steel, ie bearing the beneath the bolt and its shearstrength are normally adequate, but in any doubt these should be checkedafter creep has occurred.

Therefore the design is satisfactory.

3.1.5.11 Bolting requirements

To calculate the number of bolts required to transfer k times the UDLs intothe steel, it is assumed that the partial UDL of 0.56 kN/m extends the fulllength of the beam. The bolt spacing can then be constant. Therefore the totalUDL for bolts

FdI = k5ffl(0.56 + 4.88)L = 0.8052 x 5.44 x 4.0 = 17.52 kN

The total bolt load at the larger reaction is

Freaction = k5R6 = 0.8052 x 13.52 = 10.89 kN

The point load

= k5,f X 4.18 = 3.366 kN

From BS 5268-2 Table 77 the permissible medium term single shear loadperpendicular to the grain per shear plane for an M10 bolt by interpolation is

1.42 + 2.63= 2.025 kN

2

Therefore the permissible bolt load per bolt = 4.050 kN

No of UDL bolts required = 17.52 = = 54.050

In order to ensure that the timber and steel remain together a maximumrecommended bolt spacing of 600 mm or 2.5h (whichever is less)

= 550 mm.



3.1.512 Arrangement of bolts

40 mm.

4.Ox 1000Therefore a minimum of —1

550

No of bolts required at each reaction:— 10.89—

4.050

The bearing length of 56 mm is the minimum possible length for each support,giving a maximum clear span of 4000 — 56 = 3944 mm. In practice a standardtimber dimension of, say, 72 mm might be used for each bearing.

Arrangement of bolts40 mm.

7070

3.1.5.13 Summary

The beam should be clearly marked to indicate at which end the additionalbolt for the point load should be positioned. Spacings between bolt holesshould be at least 4d 40 mm.

Minimum end and edge distances in the timber should be 4d and in the steel2d.

Initial

Final

Ym,t tt

0.77 0.45

Applied-to-permissible ratios for medium-term load case.(Values in italics have not been calculated in this example.)

0.63 0.43 0.83 1.00

1 bearing stress beneath bolts at supports.

Clearly the design process for a fitch beam is complicated and manydesigners may prefer to use TRADA's Flitch Beam Designer software whichcan be found in the Software Toolbox area of www.trada.co.uk.


TRADA

40007+1

6.27

= 7 bolts are required spaced at

= 500 mm apart

= 2.69 = 3 required

A minimum of 2 bolts is recommended at the top and bottom of each supportto maintain stability.

No of point load bolts required:3.365= _____ = 0.83 = 1 required4.050

Additional bolt on centre line for point load- 4— - 4-

72 72

Stringers Bearers Steel plate Beam

Yc,9O,t c,9O,bearing Ym,s

0.83 1.00 0.91 0.08

1.00 0.09

c,s 6

0.13 0.99

0.14 0.66


3.1.5.14 Joist support details

There are a number of methods of supporting joists on a flitch beam, some ofwhich are shown below.

Joist hanger

Joist on beam

Timber ledger

Key:

234567

Steel angle hanger

JoistSolid timber beamSteel fitch plateMetal joist hangerMetal angle hanger50 mm x 50 timber ledger nailed to beamFastenings

One form of joist support is a timber ledger nailed to the side of the beam.Table 3.2 gives the permissible loads per metre run for various nail intensities.


Joist on beam — reduced height


Table 3.2 Permissible long-term loads for solid timber ledger with pre-drilled nail holesNail spacing (mm) Permissible load (kNIm)

One row at Two rows at C16 or better C24 or better

150 2.37 2.52

140 2.54 2.70

130 2.74 2.91

120 2.97 3.15

110 3.24 3.44

100 200 3.56 3.78

90 180 3.96 4.20

80 160 4.45 4.72

- 140 5.09 5.40

- 120 5.93 6.30

- 100

80

7.12

8.90

7.56

9.45-

Notesa) Although BS 5268 permits a loaded edge distance of only Sd (5 nail

diameters), Eurocode 5 specifies 7d for nails with predrilled holes and1 Od for nails without predrilled holes. Because of the possibility of splittinga narrow ledger if predrilled holes are not used, these recommendationsare based on a minimum edge distance in both ledger and beam of 7dwith predrilled holes, as specified in Eurocode 5.

b) The loads are for 3 mm diameter round wire nails with a minimumpointside penetration of 35 mm inserted in predrilled holes with adiameter of approximately 2 mm.

c) For one row of nails the minimum size of ledger is 35 mm thick x 47 mmdeep; for two rows the minimum finished size is 35 mm thick x 60 mmdeep.

d) Minimum edge distances of 21 mm must be maintained in the ledger andbeam, and minimum spacing between two rows of nails should be 15 mm.

e) The loads have been calculated from BS 5268-2 Annex G.f) Where joists are notched over the ledge the shear stress in the reduced

section should be checked in accordance with Clause 2.10.4.



3.3 Floor deckingThe following table provides some approximate values for spans achievableby softwood and wood-based panel products certified for use as floor decking.In a final design the permissible span should be calculated using strength andstiffness properties published in BS 5268 or by the manufacturer.

All timber-based panel products used for floor decking must now comply withthe relevant technical specification requirements laid down for flooring in BSEN 13986. Further information is given in TRADA WI Sheets 2/3- 56 and2/3 — 57.

All thicknesses are based on a domestic imposed floor loading of 1 .5 kN/m2.Refer to manufacturers or trade associations for fixing and other information.

Finished board thickness (mm) Max. joist spacing (mm)

Softwood boards, tongued and grooved

16 450

19 600

Chipboard — type PS to BS EN 312

18&19 450

22 600

Plywood — approved grades as listed in BS 5268 — 2

USA constructional and industrial plywood

15or16 450

18or19 600

Canadian exterior softwood and Douglas fir plywood (unsanded grades)

15.5 450

18.5 600

Finnish plywoods (sanded)

Conifer

12 450

15 600

Birch, birch faced

12or15 600

Oriented strandboard — type OSB/3 to BS EN 300

15 450

18 600



4 Roofs

Pitched roofs may be constructed using prefabricated trussed rafters; bolted,connectored or plated trusses with intermediate common rafters, or withtraditional rafters and purlins made on site. Trussed rafters, made of standardcomponents produced by an efficient factory process are widely used as aneconomical means of roof framing. When the roof space is required forstorage or accommodation, athc trusses may be specified. Details of theseand other roof designs, including room in the roof and flat roof construction,can be found in the TRADA publication Timber frame construction.

4.1 Trussed rafter roofsIn timber framed houses, trussed rafters are nearly always used; BS 5268-3Code of practice for trussed rafter roofs defines established methods ofdesign, materials, functional requirements, fabrication, handling, storage anderection.

The European harmonised Standard BS EN 14250 Timber structures —Product requirements for prefabricated structural members assembled withpunched metal plate fasteners: 2004 specifies requirements with which alltrussed rafters must comply in order to satisfy the Construction ProductsRegulations. It includes minimum dimensions for the members and additionallimitations on imperfections in the graded timber. The fasteners themselvesmust comply with BS EN 14545 Timber structures — Connections —Requirements, when this is published.

4.1.1 BracingIt is important that trussed rafters are correctly braced. Attention is drawn toSections 7 and 11 of BS 5268-3, where it is made clear that the 'buildingdesigner' is responsible for the stability of the roof structure as a whole. Heshould ensure that the details of the bracing are adequate to ensure theoverall stability of the complete roof structure and the associated supportingwalls.

The under-rafter bracing has the combined duty of providing restraint againstlateral buckling due to compression forces in the members and of transmittingthe wind load from the gable end walls to the front and rear walls. BS 5268-3gives a range of spans and pitches for which standard bracing is adequate atdifferent wind speeds. It should be noted that the bracing system (shown inthe diagram), including the longitudinal runners, should be carried throughoutthe length of the entire roof, but not over separating walls. Rafter bracing isnot required in hip ends, nor in the length of roof between hips if this is 1.8 mor less. Specified sarking materials nailed on top of the rafters may besubstituted for standard bracing.

Domestic dwellings with ceilings made of plasterboard or similar rigid materialdo not require bracing in the plane of the ceiling joists, providing that theyconform in every respect with the criteria set out in Appendix A of BS 5268-3.



1 Longitudinal runner at apex2 Longitudinal runners at intermediate nodes, may be omitted if this does not leave more than 4.2 m of

unbraced rafter or more than 3.7 m of unbraced ceiling tie.3 Further bracing is required on these internal members for spans over 8 m in duopitch roofs, 5 m in monopitch

roofs.4 Under-rafter diagonal brace at approximately 45° to the rafters.

The roof and ceiling system provides a horizontal girder which transmits windload from the face of the building to the cross-walls. For houses which do notconform to the criteria in Appendix A (for example, houses built on flat, openland corresponding to ground roughness category 1, or having a combinationof span, pitch and wind which puts them outside the scope of the standardbracing solutions), the bracing requirements for the roof and ceiling must becalculated.

In the plane of the rafters, moisture resistant plywood, P5 particleboard to BSEN 312, or OSB/3 to BS EN 300 may provide adequate bracing; values fortheir strength and rigidity can be obtained from BS 5268-2, or frommanufacturers. In the ceiling plane, well-fixed plasterboard will make acontribution: conservatively 12 mm plasterboard could be taken as equivalentto 6 mm thick plywood. For solid timber bracing the following designprocedure is recommended.

The diagram below shows the ceiling plane only, with bracing nailed to the topof the joists. A similar set of bracing is installed on the underside of the rafters,so that there are two complete sets as shown for a 4.8 m frontage and 9.6 mfront-to-rear dimension. For a building measuring 7.2 m from front to rear, thediagram would be cut off below the long dotted line.

Assume a roof slope of 1 T giving a rise of 1.47 m at the centre for the span of9.6 m.

The total wind load will be taken as 0.746 kN/m2 acting on a height of 2.68 mfrom the middle of the upper storey to ridge level.

The wind load per metre run along the eaves would be 2.68 x 0.746 = 2.0 kNor 1.2 kN per 600 mm interval. This is divided by 2 because there are two setsof bracing, giving the point loads of 600 N in the diagram. Each of these isshared among four braces, so the point loads on each brace will be 150 N asshown below.



'1

605

7 ,525/

300

106

1


150

150

300

600 600 600 600 600 600 600

/\/// ///— — — —

/////

0c'J 583

0(0c3)

1

106

150

106

300

>1525

L

106 106

E—8501

'1 11591< 3400

>1

(A = 2500 mm2; Z = 41670 mm3; i,= 28.87 mm)

M — 180,200Z

—

41670

For 25 x 100 C16 bracing:Max BM = 180,200 Nmm

0m,a=

0c,a=

0m,adm =

Take =

For very short term load

K12 =

Gcadm =

ae =

5832500

5.3 x 1.75 xl100

340028.87

0.1972

6.8 x 1.75 x 0.1972

rr2E — 9.87x5800

[LJ2

—

(118)2

= 4.32 N/mm2

= 0.233 N/mm2

= 10.47 N/mm2

= 118

= 2.35 N/mm2

=4.11 N/mm2


4.32

104711.5x0.233

x0.19724.11

= 0.420 + 0.099 = 0.519

0m,a + Gta =0m,adm 0t,adm

4.32 0.233+

10.47 6.32= 0.413 + 0.037

4.2 Connections

= 0.450 < 1.0 as required.

1 Circular or square toothed plate connectorsused with bolts in lap-jointed members.

3 Proprietary pierced plates attached by tightly-fitting nails.

2 Nailed plywood or OSB gussets.

The first three of the connection methods shown above are suitable for use insite workshops without special equipment. Designs for nailed plywood gussetsmay be obtained from plywood manufacturers associations and designs formethod 4 from system owners who supply proprietary jointing plates to theirlicensees undertaking the work of fabrication. Details of systems owners andtheir licensees can be viewed on the TRADA website (www.trada.co.uk).


Gma 0c.a

Gmadm[1_'K Gcadm

)

0.233+2.35

For brace in tension

Gtadm =(300h1

3.2 xl .75 xl I = 6.32100)

4 Proprietary metal gusset plates with integral teeth,applied by large hydraulic or roller pressures.


5 Foundations

5.1 LoadingTotal loads on foundations for the external walls of two storey timber framehouses vary between 6.3 and 30 kNIm run.

The types of foundation commonly employed are illustrated in the TRADApublication Timber frame construction.

Loads on the foundations of timber frame houses with timber claddings arelighter than those of masonry construction and, therefore, have distinctadvantages where the soil bearing pressures are poor. Table 5.1 shows sometypical foundation weights on the footings, depending on their size and shape.

Table 5.1 Foundation loads in kN/m

LoadExterior finish

Brick veneer Tile hangingTimberboarding

Dead 16-20

19-30

4.5—8.5 3.5—7.5

Dead andimposed

8-18 6.3-17



6 Multi-storey buildings

6.1 General design considerationsThe Timber Frame 2000 project, begun in 1995, was carried out by theBuilding Research Establishment and TRADA Technology Ltd in collaborationwith the British Government and the timber frame industry. It demonstratedbeyond doubt that conventional timber frame construction can be used tobuild economical, safe and serviceable multi-storey dwelling units. The reportMulti-storey timber frame buildings — a design guide concluded that the use ofBS 5268-6 can be extended to the design of platform frame buildings up toeight storeys without excessive deflection.

Particular attention should be given to the following issues:

• each storey should have sufficient strength and stiffness to resist ahorizontal long-term force of 2.5% of the vertical load + live load

• overturning forces should be carefully checked• where party walls separate the structure into separate units, the engineer

should ensure that horizontal forces can be taken by each unitindependently or be transferred across the party walls

• where additional stiffness has to be provided, eg by the introduction ofportal frames, the deflection limit should be appropriate for the structureand finishes, but may be no more than height/500

• resistance to disproportionate collapse should be checked.

6.2 ConstructionThe stability of the building during construction, ie before vertical loads areapplied and plasterboard is fixed, must be considered as part of the designprocess. It is acceptable to reduce the wind load in accordance withBS 6399-2 for the construction process. Particular attention should be given tothe buckling resistance of studs in party walls, which may have neitherplasterboard nor sheathing attached during construction. Significant verticalloads can result from the storage of construction materials such asplasterboard packs, so normally it will be necessary to specify requirementsfor temporary bracing (see BS 5268-6.1).

6.3 Disproportionate collapseWhile the Timber Frame 2000 project demonstrated that timber frameconstruction is remarkably resilient to disproportionate collapse, specificdesign checks against this possibility may be required under the BuildingRegulations. Guidance is expected in a planned revision to BS 5268-2: 2002.Meanwhile, guidance has been published in the UK Timber FrameAssociation Technical Bulletin 3 Design guidance for disproportionatecollapse. This specifies minimum nailing between the lower rails of wall panelsthrough the interfaces to the upper rails of the panels beneath as follows:

3.1 mm diameter nails at 300 mm centres for Class 1 and Class 2A buildings3.1 mm diameter nails at 200 mm centres for Class 2B buildings.

TECIINOLOQV(t7



7 Example of calculations for a complete dwelling

Side elevation

© TRADA Technology Ltd 2006 70\ TECHNOLOGY1

3.9m P

T8T1I!

I I I I I

4 9.5m

Ground floor plan First floor plan

toN- _1L

Front elevation

r

Rear elevation


General description and specification• Site snow load 0.64 kN/m2

• Platform frame construction

• Brick veneer

• Trussed rafters at 600 mm centres

• Concrete roof tiles weighing 49 kg/m2

• Solid concrete ground floor

• All ground floor internal walls are load bearing

• Plasterboard internal linings

• All wall panels 2400 mm high

• Window clear openings 2400 mm or 1200 mm wide, 1200 mm high

Door clear openings 900 x 2000

• Head binder to top of panels.

Note: The plans and elevations do not necessarily represent a typical houselayout but provide a basis for the following calculations.

The calculations rely on the assumptions that:

• The vertical loads described in Section 1.2.1 are all carried by the timberframe wall studs

• The wind loads described in section 1.2.2, are resisted by the horizontaldiaphragms and timber frame walls; however a proportion of these loadscan be transferred back into the brickwork via the wall ties in accordancewith BS 5268-6.1 Clause 4.10

• The wind loads have been previously calculated in accordance with BS6339-2 and section 1.2.2.

YECMNOLOOY(© TRADA Technology Ltd 2006

71


7.1 Vertical loads

Trusses

200 mm insulation quilt


So total dead load

imposed load 2

ceiling imposed load

53

cos35°

18mm OSB deck

47 x 194mm C16 joists at 400 mm centres

12. 5 mm plasterboard

140 mm phenolic foam board

Framing 47 x 97 mm at 600 mm centres


9.0 mm OSB sheathing

Insulation quilt

1 Concrete tiles can weigh from 42 to 58 kg/rn2

2 BS 6399-3: 1988 Clause 4.3.2 (0.64 x 1.25) (60 - 35) / 30

BS6399-1: 1996 Clause 5.2

BS 6399-1: 1996 Clause 5.1.1


Concrete nterlocking tiles 1

Battens

Felt

kg/rn249

3

53 kg/rn2

Roof (350 pitch):

Floor:

External walls:

Partitions:

So dead load

imposed load

65

16

kg/rn2 on plan

kg/rn2

210

94 kg/rn2

0.92

0.67

0.25

1.84 kN/m2

12 kg/rn2

9

10

4

35 kg/rn2

0.34

1.50

1.84 kN/rn2

6 kg/rn2

10

63

25 kg/rn2

0.24 kN/rn2

4 kg/rn2

20

24 kg/rn2

0.24 kN/rn2

So dead load

Framing 38 x 63 rnrn at 600 mm centres

2/12.5 rnrn plasterboard

So dead load

= 0.67)


7.2 Horizontal loadsBuilding dimensions

Overall length (I) 9.5 m

Overall width (w) = 7.0 m

Height to eaves (h) 5.0 m

Height to ridge (H) = 7.5 m

A dynamic wind pressure of q = 0.800 kN/m2 for both directions is assumed tohave been calculated from BS 6399-2.

As stated in Section 1.2.4 this may be reduced for a maximum 1 yearconstruction period to

q = 0.800 x 0.56 1 = 0.449 kN/m2.

When timber frame walls are shielded from the wind by masonry walls, the loadtransferred to them is reduced by a factor K100 which depends on the proportionof the loaded wall occupied by openings. (In other types of construction the fullwind load is taken by the timber frame walls, so K100 = 1.0) Values of K100 aregiven in Table 1 of BS 5268-6.1 and are applied to an entire wall measured tothe eaves. K100 is not applied to the spandrel in a gable wall, which is not self-supporting.

For the dimensions given in the specification a table is made of the values of P,where

= area of openings x 100area of walls including openings

Wall P K100

Front wall 22.0 0.81

Rear wall 27.3 0.82

Gable wall, door end 5.2 0.48

Gable wall, other end 0.0 0.45

Wind pressures on timber frame in kNIm2 in 50 year constructed period, walls shielded by masonry

Wind on;Roof and Gable wall, Gable wall,Front wall Rear wallspandrel door end other end

K100 1.0 0.81 0.82 0.48 0.45

q x K100 0.800 0.648 0.656 0.384 0.360



BS 6399-2 Tables 5, 10 and 16 give the following external and internal pressurecoefficients for the house illustrated.

9.5

-0.4

-0.7 __________ _______ __________________

+05 or -0.1

+0.8 or -0.3

bL= 9.5 Cpe 0.82

Wind

All measurements in metres

For stability checks, the external roof coefficients shown above have beensimplified as follows:

1. Overturning 2. Sliding & racking 3. Racking 4. Roof Uplift

-055 -0 13 -0.5 -0.55 -0.5 -1.15—,.

02 Wind -0 ±05 Wind -0.55 -0.25 Wind -1.15

Wind Wind Wind

= weighted mean of -0.4 and -0.7


= weighted mean of -1.1 -1.2, -0.6 and -0.5


= weighted mean of +0.5 and +0.8

= weighted mean of-il, -1.2, -0.6 and -0.5 forvertical loads in racking strength calculations



mean of -1.1 and -1.2 for roof uplift

For this example a value of 0.02 has been assumed for the dynamicaugmentation factor, Cr, and a value of 1.00 for Ca in all cases. As shown inSection 1.2.2.2 the calculation of Ca values is time-consuming and is unlikely toreduce the design value of the wind load by more than 5%.


c0= -0.5

I'

2.55

0.951

2.55

0.95

3.5 6.0

1.75Cpe 0.83 -

Cpe.nlex 0.8Wind 1.75

7.0 —

1.75

1.75

All Cp values = -0.3

Cpe -0.5-1

.1

2

-0.6

1

-0.5

.1

1

1

-0.62

-0.5

bw= 7.0 Cpe,r,I 0.8

Key1. Overturning -0.55

-0.2

-0.6

2. Sliding and racking -0.5

+0.6

3. Racking -0.55

4. Roof uplift -0.5

-0.25

-1.15


7.3 Overall stability calculationsIt is necessary to check the stability of the building both during and afterconstruction. In the construction period a lower design wind pressure may beused, but there may be no roof weight to add stability nor masonry walls toreduce the nett wind loads. In the constructed period a roof and in some casesmasonry walls contribute to stability, but the design wind pressure is higher.For the purpose of this example, the stability checks will be carried out for theconstructed period only.

7.3.1 Wind pressures and self-weightIn general the wind pressure on a surface is calculated from BS 6399-3 as

p = 0.85 x q x Cp (1+Cr) Ca

= 0.85 qC x 1.02 x 1.0 = 0.867 qC (7.3.1)

Where reversing the wind direction produces different nett pressures, the worsecase occurs when the higher pressure is combined with the higher value of Cpe.Hence the wind on the rear wall and the door end of the gable wall arecalculated.

Wind on rear wall

For nett horizontal pressure expression (7.3.1) becomes

Pnett = 0.867 (qs,windward Cpe,wjndward — S, leeward Cpe, leeward

Nett pressure on walls = 0.867{0656 x 0.82 - 0.648 x -0.5} = 0.747 kN/m2

Effective horizontal pressure on roof (overturning)

= 0.867{0.80 x (-0.2) — 0.80 x (-0.55)}

= 0.243 kN/m2

Effective horizontal pressure on roof (sliding & racking)

= 0.867{0.80 x 0.6 - 0.80 x (-0.5)} = 0.763 kN/m2

For nett vertical pressures instability calculations expression (7.3.1) becomes

Pnett = 0.867 (q Cpe)

The internal pressure coefficients cancel out.

Effective vertical pressure on windward side of roof (overturning)

= 0.867{0.80 x (-0.2)) = -0.139 kN/m2

Effective vertical pressure on leeward side of roof (overturning)

= 0.867{0.80 x (-0.55)) = -0.381 kN/m2

For nett vertical pressures relating to vertical loads on racking panels

expression (7.3.1) becomes

Pnett 0.867 (q Cpe — q C)Effective vertical pressure on windward side of roof (racking vertical load)

= 0.867{0.80 x 0.6 — 0.80 x (-0.3)} = 0.624 kN/m2

Effective vertical pressure on leeward side of roof (racking vertical load)

= 0.867{0.80 x (-0.5)— 0.80 x (-0.3))

= -0.139kN/m2



Wind on gable wall, door end

Nett pressure on walls = 0.867{0.384 x 0.83 - 0.36 x -0.5} = 0.433 kN/m2

Nett pressure on spandrels = O.867{O.80 x (0.83 - (-)O.5)} 0.922 kN/m2

Effective vertical pressure on roof (overturning) (Internal coefficients cancel out)

= O.867{O.80 x (-O.6)} -0.416 kN/m2

Effective vertical pressure on roof (racking vertical load)

= 0.867(0.80 x (-0.55 - (-)0.3)}

Total building self-weight

= -0.173kN/m2

7.3.2 Overturning

Front and rear walls

Roof including overhang

Total

7.3.2.1 Wind on front or rear wall

Overturning moment about xx due to wind load is:

9.5[(5 x 0.747 x 2.5) + (2.5 x 0.243 x 6.25) + (3.5 x 0.139 x 5.25)

+ (3.5 x 0.381 x 1.75)]

The overturning resistance due to building self-weight

= 132.8 x 3.5

Factor of safety against overturning= 464.8/171.2 = 2.71

Required factor of safety against overturning (BS 5268-6.1 Clause 4.4)

1.20

(When checking overturning during the construction period, the roof weight

should be omitted.)


9.5 x 5.0 x 0.24 x 2

Gable walls without spandrels 7.0 x 5.0 x 0.24 x 2

Spandrels 7.0 x 2.5 x 0.5 x 0.24 x 2

First floor 9.5 x 7.0 x 0.34

9.5 x 7.6 x 0.92

= 22.8 kN

= 16.8kN

= 4.2kN

= 22.6 kN

= 66.4 kN

= 132.8 kN

x

= 171.2 kNm

= 464.8 kNm

7.3.3 Sliding

2.5 m

2.5 m

2.5 m

In this example the maximum sliding force results from wind on the rear wall.

Applied sliding force

= 9.5{(5.0 x 0.747) + (2.5 x 0.763)} = 53.60 kN

Therefore required sliding resistance 1.2 x 53.60 = 64.32 kN

Assuming a coefficient of friction of 0.25 the resistance to sliding

= 132.8 x 0.25 = 33.20 kN

Therefore sole plate straps or other fixing methods are needed to provide a totalresistance of

64.32 - 33.20 = 31.l2kN.

For sole plate fixing straps use at least 2 No 3.25 mm diameter nails per strap.Assuming 016 timber and, initially, two members 38 mm thick, BS 5268-2Annex G or TRADA's fastener software (visit www.trada.co.uk) gives a basicsingle shear lateral load of 327 N.

Therefore the permissible load per strap

= 327x2x1.25x1.25x103(K45) (K48)

No of fixing straps needed

= 31.12/1.02

= 31/((7.0 x 2) + (9.5 x 2))

= 1000/0.939

= 30.51 = 31= 0.939 per m

= 1064 mm centres


7.3.2.2 Wind on gable wall

ThL

Overturning moment about xx due to wind load is

7.0{(5 x 0.433 x 2.5) + (2.5 x 0.5 x 0.922 x 5.83) + (9.5 x 0.416 x 4.75)}

= 216.3 kNm

The overturning resistance due to building self-weight

= 132.8 x 4.75 = 630.8 kNm

Factor of safety against overturning

= 630.8/216.3 = 2.92

Required factor of safety against overturning (BS 5268-6.1 Clause 4.4)= 1.20

= 1.O2kN


Since fixing straps or hammered sleeve fixings are normally used at 600 mmcentres to locate the sole plates during setting out, no additional restraint isnecessary to augment the sliding resistance provided normal practice isfollowed.

Adequate sliding resistance at first floor level will normally be provided byrecommendations made by the UK Timber Frame Association to meetrequirements against disproportionate collapse. These specify that the minimumfastening between the bottom rail of the wall panels and the floor deck or rimbeam and between the rim beam and the head binder or top rail of the wallpanel below should be as follows:

3.1 mm diameter nails at 300 mm centres for Class 1 and Class 2A buildings3.1 mm diameter nails at 200 mm centres for Class 2B buildings.

The maximum uplift force occurs with wind on the gable walls.

The maximum roof uplift pressure

0.867x0.80x 1.15x3.5

Roof dead load

= 0.92x7.6/2

Factor of safety against roof uplift

= 3.50/2.79

2.79 kN/m per side

= 3.50 kN/m per side

= 1.25

7.4 Racking calculations

Required factor of safety against uplift (BS 5268-6.1 Clause 4.4)

= 1.20

Section 1.2.3 explains how to calculate the resistance of truss clips to combinedlateral and vertical wind loads should this be necessary.

7.4.1 Wind loads

The method illustrated is in accordance with BS 5268-6.1.

7.4.1.1 Wind on front or rear wall

Horizontal load on roof

F1 = 2.5 x 9.5 x 0.763

With the wind on the rear wall

F2=F3 =2.5x9.5x0.747

= 18.12 kN

= 17.74 kN

© TRADA Technology Ltd 2006 78(fRADAgJOV/


7.3.4 Resistance to wind uplift


7.4.1.2 Wind on gable wall

7.4.2 Racking forces

F5 = F6 = 2.5 x 7.0 x 0.433 = 7.578 kN

The wind loads F are transferred as racking forces Q to the timber frame wallsat right angles to them.

Qa the applied force to be resisted by the first floor front and rear walls and byinternal partitions parallel to them

= F4 + 0.5 F5

= 11.86 kN

Force on ground floor front and rear walls and parallel partitions

= F4 + F5 + 0.5 F6

Force on first floor gable walls and parallel partitions

= F1 ÷ 0.5 F2

Force on ground floor gable walls and parallel partitions

= F1 + F2 + 0.5 F3

= 44.73 kN

(When opposite wind directions produce different racking forces the highervalue is used).

It may be possible to demonstrate that external timber frame walls designed toa minimum specification can support the actual racking forces, without having tocalculate the contribution of other elements such as internal partitions.However, if the calculation shows the strength of the external walls alone to beinsufficient, then as many as necessary of these other elements will have to beconsidered. The complete order to be followed is:

1. Strength of external timber frame walls without vertical load allowance.2. Increase in strength due to vertical loads on top of such walls.3. Contribution of any exterior masonry walls, via wall ties.4. Contribution of internal partitions.


With wind on gable wall, door end

F4 = 2.5 x 7.0 x 0.5 x 0.922 = 8.068 kN

= 19.44 kN

= 26.99 kN

7.4.3 Design method for racking


If the racking resistance thus calculated is still less than the racking force, thena stronger construction for the external timber frame walls must be chosen andthe procedure repeated. The construction can be strengthened by:

5. Decreasing the nail spacing.6. Increasing the nail diameter.7. Increasing the thickness of the sheathing.8. Increasing the thickness of the lining. (This step would be unusual if the

normal 12.5 mm thick plasterboard lining were used).

If there is still insufficient resistance, then

9. Apply a structural sheathing to the internal partitions.

The racking force on the ground floor gable walls will be considered first, since itis likely to be the most onerous condition.

7.4.3.1 Strength of external walls without vertical load allowance

Clause 4.7.2 of BS 5268-6.1 gives a formula for calculating the permissibleracking force on a timber frame wall. This can be rewritten as:-

Qadm =(R+R1)XLXKwhere R and R1 = the basic racking resistance of sheathing and

lining respectively in kN/m, modified by theproduct of the material modification factors for naildiameter, nail spacing and panel thickness

= K101 x K102 x K103 = KM

L = wall length in m

K = product of wall modification factors for panelheight, wall length, openings, vertical load andmasonry contribution

= K104 x K105 x K106 x K107 x K108

This version of the formula allows for the possibility that KM may be different forthe sheathing and lining.

The first step is to determine whether a construction of minimum strength isadequate.

Denoting the ground floor gable wall with a door by the suffix d and theopposite gable wall by o'.

Ld =L0 =7.Om

Note: The most commonly used sheathing material is 9 mm thick OSB Grade 3to BS EN 300. In order to illustrate the full design procedure, however, thisexample uses 7.5 mm thick plywood because this lower specification requiresthe use of various enhancements to meet the design requirements.

Table 2 in the Standard gives basic racking resistances of

Rs,basic = 1.68 kN/m for 9 mm OSB or 9.5 mm plywood sheathing

RI,basic = 0.28 kN/m for 12.5 mm plasterboard lining.

(Higher values of basic racking resistance may be used if figures for theparticular types of material which are specified are available from tests).

K101 Nail diameter (Clause 4.8.2.1)The Standard permits the use of nails from 2.25 mm to 3.75 mm in diameter.For a diameter of d mm, the basic racking resistance must be multiplied by K101where

dK101

K101 is not applicable to plasterboard, which must be fixed with nails of at least2.65 mm diameter in order to contribute to calculated racking resistance.

r H C H N 0 1 0



Using a 2.25 mm diameter nail for the sheathing

K101=--- =0.75

K102 Nail spacing (Clause 4.8.2.2)Table 2 in the Standard specifies the maximum nail spacings on the internalstuds of timber frame panels which are permitted when racking calculations arebased on values given in the Table. It also specifies standard nail spacings onthe perimeter for various materials, s mm. Nail spacings should not be lessthan the minimum spacings recommended in BS 5268-2, eg 14 diameters forplywood-to-timber or OSB-to-timber joints, without pre-drilled holes. Althoughthe Standard no longer specifies a maximum spacing of 300 mm, for reasons ofwall stud stability and in harmony with recommendations for flooring it isrecommended that the perimeter nail spacing should not exceed 300 mm, or200 mm with sheathing less than 10 mm thick.

For a proposed nail spacing of S on the perimeter, the basic racking resistancemust be multiplied by K102, where

K102 = _____________S

O.6x—- +0.4sp

K102 is not applicable to plasterboard, which must be nailed throughout atcentres not exceeding 150 mm.

In the case of plywood nailed at 200 mm centres on the perimeter

K102 = 1 = 0.833

I0.6x') + 0.4150)

K103 Sheathing thickness (Clause 4.8.2.3)For 7.5 mm thick plywood the basic resistance of the standard 9.5 mm thickmaterial must be modified by K103 (These are nominal thicknesses).

K103 = (2.8B - B2 —0.8)

Where B = proposed board thickness = 0.789standard board thickness 9.5

Hence K103,5 = 0.789 for plywood

K103,1 = 1.0 for plasterboard

Note that B is limited by BS 5268-6 to 0.75 � B 1.25

R and R1Hence if R5 and R1 are the racking resistance of the sheathing and liningrespectively,

R5 = 1.68 x K101 x K102 x K103

= 1.68 x 0.75 x 0.833 x 0.789 = 0.828 kN/m

R1 = 0.28 x K103 =0.28 x 1.0 = 0.28 kN/m

K (Clause 4.9)K104 Height of wall panels (Clause 4.9.1)

Panel height = 2.4 m :. K104 = 1

K105 Length of walls (Clause 4.9.2)

Both walls exceed 4.8 m in length :. K105 = 1.32

Note: Clause 4.9.3 states that in determining K105 for walls with framedopenings which have a height in excess of half the panel height and are lessthan 300 mm from a corner of the building, the length of wall from the corner tothe farther edge of the opening should be disregarded.

(J© TRADA Technology Ltd 200681


K106 Framed openings (Clause 4.9.3)

K106 =(1_1.3.8)2

where Aa = aggregate area of framed openings in a wall

At = total area of wall including openings

For the ground floor gable wall, door end

Aa/At = 0.103, hence K106,d = 0.75.

For the ground floor gable wall, other end

Aa/At = 0.0, hence K1060 = 1.00.

Note: Clause 4.9.3 states that where framed openings are separated by lessthan 300 mm and their heights are each greater than half the panel height, thenthe area of the opening should be taken as that of the rectangle which enclosesboth openings. Clause 4.9.4 explains that certain openings not more than250 mm wide need not be framed. Its intention is that such openings may bediscounted when calculating K106.

K107 Vertical loading (Clause 4.9.5)

At this stage K107 will be taken as 1.0 .. K10 1.0

K108 Interaction (Clause 4.9.6) K108 = 1.1

K K104xK105xK106xK107xK106 :.KWd =K0 = 1.45

Permissible racking force Qadm

Qadm =(R+Ri)xLxKHence Qadm,d = (0.828 + 0.28) x 7 x 1.09 = 8.45 kN

Qadm,o = (0.828 + 0.28)x 7 x 1.45 = 11.25 kN

Qadm, total = Qadm,d + Qadm,o = 19.70 kN

If the two walls were of very dissimilar strength then the weaker wall alonewould have to be considered with an appropriate share of the total racking load.

From Section 7.4.2Qa = 44.73 kN

Qadm <Qa, so proceed to step 2

7.4.3.2 K107 Allowance for vertical loading (Clause 4.9.5)

Although its calculation is rather tedious, K107 should be considered next, sinceit increases the maximum allowable contributions to racking of the brick veneerand the internal partitions, which are considered subsequently in steps 3 and 4.

K107

= 1+[(0.09V _0.0015V2)xJ1where V is the uniformly distributed load per unit run on the top of the wall panel(10.5 kN/m maximum). Any negative values are taken as zero, but the value ofa negative load on a first floor wall would have to be included in the calculationsof the load on the ground floor wall beneath it. Where negative vertical forcesare encountered, adequate holding-down straps must be specified. Specialprovisions are made in this clause to allow for concentrated vertical loads.

(R© TRADA Technology Ltd 2006 82


To calculate the load from the roof, Vr, on each wall, it is necessary first to findthe roof area A supported by it, as shown in the diagram. Dimensions for thisexample are shown in parentheses.

Overall length (9.5m)

x = (the length of any end overhang) + (half the distance from the end

wall to the first rafter or truss or girder truss).

This applies to both gable ends and hipped ends.

For any given wall

Vr = (roof dead weight on A) — (roof uplift on A with wind parallel to wall).

In this example

A1 = A2 = 8.9 x 7.6 x 0.5 = 33.82 m2

A3 = A.4 = 0.3 x 7.6 = 2.28 m2

Hence load on first floor front or rear wall (see Sections 7.1 and 7.3.1)

Vr = ([A1 or A2] x roof dead load) — ([A1 or A2] x effective vertical uplift

pressure with wind on gable wall)

= (33.82 x 0.92)— (33.82 x 0.173)

V (total vertical load) = V1 = 25.26 kN

Load on ground floor front wall (see Section 7 General description and

specification and Section 7.1)

= 25.26

+ 2.5 x 9.5 x 0.24

+ 0.5 x 2.7 x 2.8 x 0.34

=32.25kN

Load on ground floor rear wall

= 25.26

+ 2.5 x 9.5 x 0.24

:. V 30.96kN

load on first floor wall

first floor rear wall

= 3.26 kN/m


Planeof

raftersAi for rear wall

A3 A4forendwall

forendwall

A2 for front wall

Overall widthincludingoverhangs

(7.6m)

x Overall length - 2x x

(0 3m) (89m) (O.3m)

load on first floor wall (V1)

first floor front wall

floor dead load

= 3.39 kN/m

(The wall is 9.5 m long)


Load on each first floor gable wall (see Sections 7.1 and 7.3.1)

= ([A3 or A4] x roof dead load) + ([A3 or A4] x effective vertical

pressure with wind on rear wall) + (weight of spandrel)

= (2.28 x 0.92)+2.280.624 ±

0.139] + (2.5 x 7.0 x 0.5 x 0.24)

V = 5.07 kN = 0.72 kN/m

(The walls are 7.0 m long)

Load on ground floor gable wall, door end

= 5.07 load on first floor

+ 2.5 x 7.0 x 0.24 first floor gable wall

+ 0.5 x 2.8 x 4.3 x 0.34 floor dead load

:. Vd = 11.31 kN = 1.62 kN/m

Load on ground floor gable wall, other end

5.07 load on first floor

+ 2.5 x 7.0 x 0.24 first floor gable wall

+ 0.5 x 3.9 x 7.0 x 0.34 floor dead load

V0 = 13.91 kN = 1.99 kN/m

Hence

Vd 1.62kN/mand V0 1.99kN/m

K107 = 1 + [(0.09v — 0.0015V2)]1

Hence

K107,d 1.09 and K107,0 = 1.11

= 1.09*x 1.09 = 1.19K107d

K,0 = 1.45*x 1.11 = 1.61K107 0

see Section 7.4.3.1

Hence

Qadm,d = 7(0.828 + 0.28) 1.19 = 9.23 kN

Hence

Qadm,o = 7(0.828 + 0.28) 1.61 = 12.49 kN

Qam = Qadm,d + Qadm,o = 21.72 kN

Qadm <Qa (44.73 kN), so proceed to step 3.

7.4.3.3 Contribution of any exterior masonry walls (Clause 23)

Masonry walls constructed and fastened to the timber frame in conformity withClauses 2.7 and 4.10 of BS 5268-6.1 will take a proportion of the racking load,dependent on the density of spacing of the wall ties and on the total horizontallength of storey-height masonry in the wall, excluding any horizontal lengths of600 mm or less. From Table 6 of the Standard, with ties at 4.4 per m2 themaximum racking load taken by the masonry is 0.5 kN/m of panel heightmasonry.

TRADATECHNOLOGY



Lm,d = 6.1 m and Lm,o = 7.0 m

Hence

Qm,rnax,d = 6.1 X 0.5 = 3.05 kN

Qm,max,o 7.0 X 0.5 = 3.50 kN

Qadm, the allowable load taken by the masonry, must not exceed 25 per cent ofthe racking resistance of the adjacent timber frame wall.

25 per cent of Qadm,d = 0.25 x 9.23 = 2.31 kN

Qm,adm,d

2.31 kN

25 per cent of Qadm.o = 0.25 X 12.49 = 3.12 kN

Qm,adm,o = 3.12 kN

Qadm 21.72 + 2.31 + 3.12 = 27.15 kN

Qadm <Qa (44.73 kN), so proceed to step 4.

7.4.3.4 Contribution of internal partitions

Clause 4.7.4.1 in the Standard stipulates that the permissible contribution toracking resistance in any one direction from Category 3 plasterboard must notexceed 50 per cent of the resistance provided in that direction by Category 1 or2 materials (plywood, OSB, bitumen impregnated insulation board, etc). (Notethat the 30 mm of plasterboard in two or more layers in a separating wall isclassified as a Category 2 material.)

In this example the resistance provided by Category 1 or 2 materials in theexternal wall

= Qadm(R ±R1)

where Qadm = the permissible racking force on the external walls underconsideration, after allowing for vertical load (Section 7.4.3.2)

and R and R1 = the racking resistance of the sheathing and liningrespectively (Section 7.4.3.1).

Similarly the racking resistance provided by the plasterboard in the externalwalls

°adm (R±R)Hence, due to the 50 per cent limitation quoted, the maximum permissiblecontribution to racking resistance from the internal partitions

- R R1Qi, max — 0.5 Qadm - Qadm

(R+R1) (R+R1)

— (o.sR—R)(R+R)

For the ground floor gable walls

max = 21.72 (0.5 x 0.828 —0.28) = 2.63 kN(0.828 + 0.28)

Sometimes it will be necessary at this stage to calculate Q, the actual rackingresistance which the internal partitions can contribute, to ensure that they cansupply the above force. Because the strength of a partition is not simplyproportional to its length, the internal partitions must be considered individually.In this example, they are identified in the diagram as 8 to 11.


R basic = 0.90 + 0.45

R = (0.90x 1 )+(0.45x 1K103 K103

(Note that K101 and K102 are not applicable to plasterboard)

K104 Height of wall panels

Panel height = 2.4 m

K105 Length of walls

K106 Framed openings

None in this example

K108 Interaction


K108 =1.1

= 29.77 kN


9.5 m

First floor plan

LForwalls0to2.4m K105 =2.4

For walls 2.4 to 4.8 m K105 =

Ground floor plan

L,8 = 2.9 m L1,9 = 1.9 m L,i0 = 1.9 m L,11 = 3.9 m

With two boards of 12.5 mm thick plasterboard, Table 2 in the Standard givesthe basic racking resistance of the plasterboard in the internal partitions as

= 1.35 kN/m.

= 1.35 kN/m.

K10 =1.0

K105,9,10 = 0.792

1<105,8 = 1.08

K105,11 = 1.21

K106 =1.0

K107 Vertical load

This can reasonably be ignored in the case of internal partitions in a two-storey

house .'. K107 = 1.0

K = K104 x K105 x K106 x K107 x K108

= 1.19 K,g,1o = 0.871 1K,11 = 1.33

Therefore the possible contribution to racking resistance from the internal partitions

= LR1K (See section 7.4.3.1)

= (2.9 x 1.35 x 1.19) + 2 (1.9 x 1.35 x 0.871) + (3.9 x 1.35 x 1.33)

= 4.66 + 2 x 2.23 + 7.00 = 16.13 kN

Since 2.63 < 16.13, the contribution to racking resistance made by the internal

partitions is 2.63 kN.

.'.Qadm =27.14+2.63

Qa = 44.73 kN, so the material specification must be increased.


7.4.3.5 Strengthening the construction

Try 9.0 mm thick OSB/3 with 2.65 mm diameter nails spaced at 150 mm on theperimeter of the panels.

K101 = 2.65/ 3 = 0.883

K102 = K103 1.0 (see Section 7.4.3.1).

R = 1.68 x 0.883 x 1.0 x 1.0 = 1.48 kN/m instead of 0.828 kN/m

R1 = 0.28 kN/m as before.

The wall modification factors K and the vertical load modification factors K107are unchanged, so with the allowances for vertical load given in Section 7.4.3.3

Qadm,d = Ld(Rs + RI)Kwd = 7 X (1.48 + 0.28) X 1.19 14.66 kN

Qadm,o = L0(R + Ri)K,0 = 7 X (1.48 + 0.28) X 1.61 = 19.84 kN

Qadm,2 = 14.66 + 19.84 = 34.50 kN

Qadm,2 < Qa (44.73 kN), so the contribution of the masonry walls must be included.

The masonry wall on the door end can contribute up to 3.05 kN (see Section7.4.3.3).New maximum allowable contribution to

Qadm,d = 0.25 X 14.66 = 3.66 kN

Hence

Qm,adm,d = maximum masonry load = 3.05 kN

The masonry wall on the other end can contribute 3.50 kN (see Section 7.4.3.3)

New maximum allowable contribution to

Qadm,o = 0.25 X 19.84 = 4.96 kN

Hence

Qm,adm,o maximum masonry load = 3.50 kN

Therefore

Qacjm,3 = 34.50 + 3.05 + 3.50 = 41.05 kN

Qadm,3 < Qa (44.73 kN), so the contribution of the internal partitions must be

included.

The maximum permissible contribution of internal partitions depends on thepermissible racking load on the external walls allowing for the vertical load butnot the masonry contribution (Section 7.4.3.4). Hence

Qi,max = °adm 2 0.5R5 — R1 = 34.50 (0.5 xl .48—0.28) = 9.02 kNR5+R1 (1.48+0.28)

Qadm,4 41.05 + 9.02 = 50.07 kN

Qadm,4 > Qa (44.73 kN), so the construction is satisfactory.

7.4.3.6 Tabulation of results

The following table illustrates the design procedure in tabular format for the wallspecifications reached in Section 7.4.3.5 and shown in the table heading. Thesequential method shown is intended to obviate any unnecessary calculations.However, if a spreadsheet is used it may be nearly as simple to introduce all thepossible factors at once.



The table calculates the racking resistance in four stages as follows:

Allowance for: Vertical load Masonry contribution Internal partitionsQadm,1 X X XQadm,2 / X K

Qadm,3 ' XQadm,4 1 1 1

It will be seen that for all the walls, except the ground floor gable walls, theconstruction proposed can be proved adequate without any reference to verticalloads, masonry contribution or internal partitions. In the case of the ground floorgable walls all three additional factors must be taken into account.

('TRADA\'\ recnNoLocv(/© TRADA Technology Ltd 2006 88

TR

AD

A T

echn

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y Tim

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fram

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usin

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K S

truc

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enda

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Ext

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l wal

ls:

Inte

rnal

par

titio

ns:

Mas

onry

Tie

s:

9 m

m O

SB

/3 n

aile

d at

150

mm

cen

tres

with

2.2

5 m

m d

iam

eter

nai

ls +

12.

5 m

m p

last

erbo

ard

with

sta

ndar

d na

iling

T

wo

boar

ds o

f 12.

5 m

m p

last

erbo

ard

Spa

ced

at 4

.4 p

er m

2

Rac

king

des

ign

figur

es

Sym

bol

Exp

lana

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Pag

e C

omm

on

fact

ors

Firs

t flo

or r

esis

ting

wal

ls (s

ee d

iagr

am S

ectio

n 7.

4.3.

4)

Gro

und

floor

res

istin

g w

alls

(see

dia

gram

Sec

tion

7.4.

3.4)

W

ind

onto

gab

le w

alls

W

ind

onto

fron

t/rea

r w

alls

W

ind

onto

gab

le w

alls

W

ind

onto

fro

nt/r

ear w

alls

F

ront

R

ear

1 2

3 D

oor

I O

ther

I

4 I

5 I

6 en

d (d

) en

d (o

) I

I F

ront

I

Rea

r I

7 D

oor

I O

ther

I

8 I

9 I

10

I 1

end

(d)

end

(0)

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Q

Tot

al r

acki

ng fo

rce

to b

e re

sist

ed o

n flo

or le

vel a

nd in

win

d di

rect

ion

show

n (k

N)

79

11.8

6 26

.99

19.4

4 44

.73

Le

Leng

th o

f eac

h ex

tern

al w

all (

m)

70

9.5

9.5

7.0

7.0

9.5

9.5

7.0

7.0

RS

,beS

jC

Bas

ic ra

ckin

g re

sist

ance

of p

rimar

y sh

eath

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in e

xter

nal w

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(kN

/m)

80

1.68

RIb

B

asic

rack

ing

resi

stan

ce o

f sec

onda

ry

shea

thin

g in

ext

erna

l wal

ls (

kN/m

) 80

0.

28

K10

1 N

ail d

iam

eter

mod

ifica

tion

fact

or fo

r pr

imar

y sh

eath

ing

80

0.88

3

K10

2 N

ail s

paci

ng m

odifi

catio

n fa

ctor

for

prim

ary

shea

thin

g 81

1.

00

K10

3,

Boa

rd th

ickn

ess

mod

ifica

tion

fact

or fo

r pr

imar

y_sh

eath

ing

81

0.78

9

K10

3 B

oard

thic

knes

s m

odifi

catio

n fa

ctor

for

seco

ndar

y sh

eath

ing

(K10

1 &

K10

2 no

t ap

plic

able

to p

last

erbo

ard)

81

1.

00

..._.

.. R

R

ebee

c x

K10

1 x

K10

2 x

K1o

3 (k

N/m

) 81

0.

828

R1

RI,b

asjc

x K

103,

1 (k

N/m

) 81

0.

28

K10

4 M

odifi

catio

n fa

ctor

for p

anel

hei

ght

81

1.00

K10

5 M

odifi

catio

n fa

ctor

for w

all

leng

th

81

1.32

1.

32

1.32

1.

32

1.32

1.

32

1.32

1.

32

K10

6 M

odifi

catio

n fa

ctor

for f

ram

ed

open

ings

in e

xter

nal w

alls

82

0.

583

0.36

7 1.

00

1.00

0.

442

0.46

8 0.

75

1.0

K10

7 M

odifi

catio

n fa

ctor

for v

ertic

al lo

adin

g 81

1.

00

K10

8 In

tera

ctio

n m

odifi

catio

n fa

ctor

82

1.

10

K

K10

4 x

K10

5 x

K10

6 x

K10

7 x

K10

8 82

0.

846

0.53

3 1.

45

1.45

0.

642

0.68

0 1.

09

1.45

Qad

m

Per

mis

sibl

e ra

ckin

g fo

rce

=L(

R+

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K(k

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82

14.1

5 8.

91

17.8

7 17

.87

10.7

3 11

.37

8.45

11

.25

Qed

,l T

otal

per

mis

sibl

e ra

ckin

g fo

rce

on th

e flo

or le

vel a

nd in

win

d di

rect

ion

show

n (k

N).

Mus

t be

> Q

23

.06

(> 1

1.86

) S

tren

gth

adeq

uate

35

.74

(>26

.99)

S

tren

gth

adeq

uate

22

.10

(>19

.44)

S

tren

gth

adeq

uate

19

.70

(<44

.73)

S

tren

gth

inad

equa

te

©

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/( T

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TR

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A T

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usin

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K S

truc

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(Ext

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l wal

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nter

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artit

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, Mas

onry

Tie

s; m

easu

rem

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as

bef

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king

des

ign

figur

es

Sym

bol

Exp

lana

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Pag

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ac or

s F

irst f

loor

res

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g w

alls

(see

dia

gram

in

7.4

.3.4

) G

roun

d flo

or r

esis

ting

wal

ls (

see

diag

ram

in 7

.4.3

.4)

Win

d on

2ble

wafls

W

ind

on fr

ont/r

ear

Win

d on

gab

le w

alls

W

ind

on f

ront

/rea

r wal

ls

Fro

nt

Rea

r 1

2 3

Doo

r en

d O

ther

en

d 4

5 6

Fro

nt

Rea

r 7

Doo

r en

d (d

) O

ther

en

d (o

) 8

9 10

11

V

Uni

form

ly d

istr

ibut

ed v

ertic

al lo

ad o

n ex

tern

al w

alls

(10

.5 k

N/m

max

imum

) 82

1

62

. 1

99

.

K10

7 M

odifi

catio

n fa

ctor

for v

ertic

al lo

ad

82

1.09

1.

11

Qd.

2 P

erm

issi

ble

rack

ing

forc

e =

x

K10

7 (k

N)

87

14 6

6 .

19 8

4 .

Qd,

2 T

otal

per

mis

sibl

e ra

ckin

g fo

rce

on th

e flo

or le

vel a

nd in

win

d di

rect

ion

show

n (k

N).

Mus

tbe>

Qa

87

34.5

0 (<

44.7

3) S

tren

gth

inad

equa

te

L Le

ngth

of s

tore

y he

ight

mas

onry

wal

l (m

) 84

6

1 . 7

0 .

Q

Pos

sibl

e co

ntrib

utio

n fr

om m

ason

ry

wal

l to

rack

ing

resi

stan

ce

= L

x 0.5

(kN

) for

ties

at 4.

4 pe

r m

2

85

3.05

3.

50

Qm

,nax

M

axim

um p

erm

issi

ble

cont

ribut

ion

from

mas

onry

wal

l = 0

.25

Qad

n.2

85

3 66

.

4 96

.

°rm

adn

Per

mis

sibl

e co

ntrib

utio

n fr

om m

ason

ry

wal

l = th

e le

sser

of Q

, and

Qn,

max

85

3

05

. 3

50

.

Qad

n.3

ZQ

adro

.2 +

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). M

ust be

>

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41.0

5 (<

44.7

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tren

gth

inad

equa

te

L,

Leng

th o

f eac

h in

tern

al p

artit

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(m)

86

2.9

1.9

1.9

3.9

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asic

B

asic

rac

king

res

ista

nce

of in

tern

al

plas

terb

oard

par

titio

ns (k

N/m

) 86

1

35

.

K10

3 B

oard

thic

knes

s m

odifi

catio

n fa

ctor

for

in

tern

al p

artit

ions

86

1

0 .

R1

R1

basi

c x

K10

3 (k

NIm

) 86

1.

35

K10

4 M

odifi

catio

n fa

ctor

for p

anel

hei

ght

86

1.0

K10

5 M

odifi

catio

n fa

ctor

for w

all l

engt

h 86

1.

08

0.79

2 0.

792

1.21

K10

8 In

tera

ctio

n m

odifi

catio

n fa

ctor

86

1.

1 K

K

1 x

K10

5 x

K10

6 x

K10

7 x

K10

8 86

1.

19

0.87

1 0.

871

1.33

0

Pos

sibl

e co

ntrib

utio

n fr

om i

nter

nal

part

ition

s to

rac

king

res

ista

nce

=LR

K(k

N)

86

4.66

2.

23

2.23

7.

00

Oi,r

sas

Max

imum

per

mis

sibl

e co

ntrib

utio

n fr

om in

tern

al p

artit

ions

—

(0

.5R

a R

,)

— Q

adm

2 x

kN

(R8+

R1)

87

9.02

Q,s

drs

Per

mis

sibl

e co

ntrib

utio

ns fr

om in

tern

al

part

ition

s =

the

less

er o

f ZO

and

Qi,m

ax

82

9 02

EQ

adrs

.4

= Q

adrs

,3 +

Q. (k

N) M

ust

be >

87

50

.07

(>44

.73)

Str

engt

h ad

equa

te

TR

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TR

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06


If plasterboard lined separating walls constructed in accordance with Clause4.7.5 in the Standard were under consideration, their contribution to rackingresistance could be added to the value 0, adrn calculated in the table forinternal partitions.

The table below gives permissible racking loads for selected plywood or OSBwall panels lined with one layer of 12.5 mm thick plasterboard, allowing for thematerial modification factors K101, K102 and K103, but not for any other factors.

Plywoods in BS 5268-2 or equivalentplywoods certified for use in the UK

OSB to BS EN 300, Grades 3 and 4

QadmkN/m

dmm

5

mmtmm

QadmkN/m

dmm

Smm

tmm

1.11 2.25 200 7.5 1.33 2.25 200 9.0

1.38 3.00 200 7.5 1.54 2.25 150 9.0

1.51 3.35 200 7.5 1.68 2.50 150 9.0

1.74 2.65 100 7.5 1.93 2.65 125 9.0

1.93 3.00 100 7.5 2.14 2.65 100 9.0

2.13 3.35 100 7.5 2.28 2.50 75 9.0

2.38 3.00 100 9.5 2.40 2.65 75 9.0

2.63 3.35 100 9.5 2.68 3.00 75 9.0

2.91 3.75 100 9.5 2.96 3.35 75 9.0

3.08 3.00 50 9.5 3.14 2.65 50 13.0

3.28 3.75 75 9.5 3.32 3.75 100 13.0

3.51 3.00 50 12.5 3.52 3.00 50 13.0

3.74 3.75 75 12.5 3.75 3.75 75 13.0

3.88 3.35 50 12.5 3.90 3.35 50 13.0

4.32 3.75 50 12.5 4.33 3.75 50 13.0

d = nail diameters = nail spacing on panel perimeter (use 300 mm for spacing on internal studs)

= thickness of sheathing

7.4.4 Recommended design procedure1. First ascertain from the fabricator his preferred sheathing and lining

materials and the range of thicknesses available. If higher figures forbasic racking resistance are available from the board manufacturers thenthese should also be obtained.

2. Nails are normally inserted by nail guns which cannot necessarily use anydiameter of nail. Ascertain from the fabricator his preferred diameters. It issensible to keep to the same diameter throughout a house.

3. Use the same construction for the wall panels throughout each individualfloor.

4. K107, the factor for vertical loading, increases the calculated rackingstrength of a structure by a factor which is generally below 10 per centand almost never more than 15 per cent, and Section 7.4.3.2 shows thatits calculation is rather time-consuming. Therefore, it is considered thatthe calculation of K107, unless it is essential, may be omitted, and its valuetaken as 1.0.

5. In general, start with a minimum practical thickness of sheathing andspecification for nailing, and only if this proves inadequate adopt astronger construction.

6. A table should next be drawn up of the values which will be required todetermine an adequate construction specification. Using these values aseries of values for Qadm can be calculated and tabulated for each floorand wind direction, proceeding as far down the table in each column as isnecessary to obtain values which exceed the corresponding values of Qa•

7. If the initial construction selected is inadequate, increase its strength asnecessary according to the procedure explained in Section 7.4.3.5.



7.5 Wall panel studs

7.5.1 Ground floor rear wall — very short termConsider ground floor studs to rear elevation carrying wind load and also roof,floor and panel loads.

Assume studs at 600 mm centres

Stud height = panel height — (thickness of top and bottomrails)

2400 — (2 x 47) = 2306 mm

Lateral loading due to wind

External pressure coefficient (Section 7.2) = 0.82

Internal pressure coefficient -0.3(BS 6399-2 Table 16)

From Section 7.3.1

neff pressure on studs = 0.867 x 0.48 x (0.82 — (-0.3)) = 0.466 kN/m2

so loading per stud 0.466 x 0.6 = 0.280 kN/m

Maximum bending moment (Mmax) = 0.28 x2.3062= 0.186 kNm

Axial loading (F)

F = 0.6 [(7.6 x 0.5 x 1.84) + (2.7 x 0.5 x 1.84) + (5.0 x 0.24)]roof dining-room floor wall panels

= 6.41 kN

Try 47 x 97 mm (nominal 50 x 100 mm), C16 timber studs

Actual bending stress

M 0.186x106x6 2Gma = — = =2.52 N/mm

Z 47x972

Actual compressive stress

3F 6.41x10 2Gca =— = =1.41N/mm

A 47x97

Permissible bending stress

Gmadm = grade stress x K3 x K7 x K5

where K3 (load duration) = 1.75

300 "° 1K7 (depth factor) =

K8 (load sharing) = 1.1

/ \Q.11= 5.3 x 1.75 x I xl .1 = 11.55 N/mm2

97 )



Permissible compressive stress

Gcadm = grade stress x K3 x K8 x K12

To find K12 from Appendix B, BS 5268-2:

K12 = 1+(1+n)n2E — 1(1+)u2E 2_ u2E2

2 2NA2a 2 2NA2a NA2a

Where A (slenderness ratio) Le = 0.85x2306*70.00

12

N =1.5

grade stress x K3

= 6.8 x 1.75 = 11.9 N/mm2

Emin = 5800 N/mm2

q (eccentricity factor) = 0.005 A

= 0.005 x 70.00 = 0.350* BS 5268—6.1 Clause 4.6.2 states that the effective length of a timber framewall stud should be taken as 0.85 L

1+q = 1+ 0.350 = 0.6752 2

ir2E = 2x58OQ=0.654

NA2a 1.5x(70.O0)2 xll.9

K12 = + (0.675 x 0.654) _[ + (0.675 x 0.654) -

0.6541

= 0.4596

Gcadm = 6.8 x 1.75 x 1.1 x 0.4596 = 6.016 N/mm2

Now check summation of stress ratios

Gma + Gca <1( l.5XG xK12 0c,adm

—

°m,adml 1

rr2E 2x58OOwhere Ge = = ________ = 11.68

A2 7Q2

2.52 1.41therefore

11 55111 .5x 1 .41x0.4596+

6.016 = 0.472< 1

11.68 )

TRADA© TRADATechnology Ltd 2006


7.5.2 Ground floor rear wall - medium term and long termCheck stud for axial load of medium term duration.

= 1.41 N/mm2

Gcadm = grade stress x K3 x K8 x K12

To find K12 using Table 22, BS 5268-2

A = 70.00 (as before)

Emin = 5800 N/mm2

= 6.8 x 1.25 = 8.50 N/mm2

= = 682.48.50

E

Oc

K12 for

2=70600

682.4

700

0.511

0.539

0.545

= 6.8 X 1.25 X 1.1 x 0.539 = 5.04 N/mm2

1.41 <5.04 which is satisfactory

Check stud for axial load of long term duration

F = 0.6 [(7.6 x 0.5 x 0.92) + (2.7 x 0.5 x 1.84) + (5.0 x 0.24)]

= 4.31 kN

3q.JIXIU 2a = ________ = 0.95 N/mmc,a 47x97

K12 using Table 22, BS 5268-2

A = 70.00

= 6.8 x 1.0 = 6.8 N/mm2

= = 852.96.8

--0c

K12 for

2=70800

852.9

900

0.572

0.583

0.593

:. 0cadm = 6.8 x 1.0 x 1.1 x 0.583 = 4.36 N/mm2

0.95 <4.36 which is satisfactory.

The above calculation is unnecessary when the long term load is 80 per centor less of the total design load (imposed roof, imposed floor and dead weightof construction) but has been included to show the complete procedure.

TRADATECHNOLOGY



Check bearing on bottom rail for medium term duration

Actual compressive stress perpendicular to grain

crc±a = Gca = 1.41 N/mm2

Permissible compressive stress perpendicular to grain

0c,±,adm = grade stress x K3 x K4

where grade stress = 2.2 (specification excluding wane)

K4 (bearing) = 1.24

0c,±,adm = 2.2 x 1.25 x 1.24 = 3.41 N/mm2

Therefore 47 x 97 mm (50 x 100 mm nominal) C16 grade studs aresatisfactory provided there is no wane.

Check deflection

An approximate value for the deflection of a loaded stud is given by thefollowing formula:

— (0.OO5AGca +°ma) Z—

(ae—ca) xZ bh2 h

where—A 6bh 6

(0.005 x 70.00 xl .41)+ 2.52 97= x—=4.74mm(11.68—1.4i) 6

3adni = 0.003 x 2306 = 6.92 mm

which is satisfactory.

For particular cases designers may chose to tighten or relax the standarddeflection limit of 0.003L.

7.5.3 Ground floor gable wall - very short termConsider ground floor studs to gable elevation carrying wind load and paneland floor loads plus a proportion of roof load due to overhang.

From Section 7.2

q = 0.384 kN/m2 and Cpe = 0.83.

Following the calculation in Section 7.5.1, nett pressure on studs

qs,nett = 0.867 x 0.384 x (0.83 —(-0.3)) = 0.376 kN/m2.

Maximum bending moment Mmax = 0.376xo:x2.3062= 0.150 kNm

Axial loading (F)

F = 0.6 [(7.5 x 0.24) + (4.0 x 0.5 x 1.84) + (0.3 x 1.84)] = 3.62 kNwall panel living-room floor roof overhang

In this case the bending moment and axial load are both less than those forthe rear elevation studs. Therefore previous specifications will stand, ie allstuds to be 47 x 97 mm (nominal 50 x 100 mm), C16 timber studs at 600 mmcentres.

Note that a more common stud size is 38 x 89 mm. If this proved inadequate itwould be tried at closer centres.



7.6 First floor platformMaximum effective span of floor joists = 4.0 m

Try 47 x 194 mm (nominal 50 x 200 mm), C16 timber joists at 400 mmcentres.

Maximum bending moment

1.84 0.4 x4.02Mmax = ____________ = 1.472 kNm

Actual bending stress

1.472x106x6 2ma = =4.99Nmm47 x 1942


0m,adm = grade stress x K3 x K7 x K8

/ \O.11

=5.3x1.0x1-.I xl.1 6.12N/mm2194)

Check deflection

fL2 5L2 h2öa

1.84x0.4x40002x12 5x40002 1942= =10.1mm880O47194 384 5

adm =0.003x4000 =12mm

Note: Joists at 400 mm centres may be used if a head binder is specified,provided it can be shown that the shear strength and bending strength of thehead binder and top rail of the panel are adequate. (K8 = 1.1 may be used toincrease the strength properties if the two members are nailed together at 600mm maximum centres.)

Consider largest span of remaining floor joists.

Effective span = 2.9 m

Try 47 x 194 mm (nominal 50 x 200 mm) C16 timber joists at 600 mm centres.

1.84 x 0.6 x 2.92Mmax = 1.l6kNm

1.16x106 x6 2Gma

= = 3.93 N/mm47 x 1942

Gmadm 6.12 N/mm2 from previous calculation

Check deflection

1.84x0.6x29002x12 5x29002 1942= 432mm880047194 384 5

6adm = 0.003 x 2900 = 8.7 mm

Therefore 47 x 194 mm (nominal 50 x 200 mm) C16 timber joists at 600 mmcentres are satisfactory for spans up to 2.9 m.



Consider additional joists required to carry full length 2nd storey partition.


Loading — assume joist carries complete internal wall and 0.2m width of floorsince it will be situated at some point between the floor joists.

F = (0.2 x 1.86) + (2.4 x 0.24) = 0.948 kN/m

Check 2 No 47 x 194 mm (nominal 50 x 200 mm), C16 timber joists,connected together.

0.948x 4.02Mmax = _________ = 1.896 kNm

r 6I.OUXIU XU 2a

= __________ = 3.22 N/mm2x47x1942

Gmadm = 6.12 N/mm2 from previous calculations.

Check deflection

Where E=EmnxK9 5800x 1.14 = 6612

N/mm2

0.948x40002x12 5x40002 19426a =8.66mm

6612x2x47x194 384 5

adm = 0.003 x 4000 = 12 mm

Therefore 2 No 47 x 194 mm (nominal 50 x 200 mm), C16 timber joists aresatisfactory.

Consider short span joists 2.1 m clear span to illustrate use of slab loadingrequirements.

When considering joists spanning less than 2.4 m an imposed loading of 3.6kN per metre width of floor (measured perpendicular to the span) is applied.(BS 5268-7 Recommendations for the calculation basis for span tables.)


Loading imposed + dead

= 3.6x0.6 +0.34x0.6 = 1.19 kN/m2.2

1.19x 2.22Mmax = _________ = 0.72 kNm

0.72x106x6 2ama = 2.44N/mm

47 x 1942

Gmadm = 6.12 N/mm2 from previous calculation.

Check deflection

1.19x22002x12 5x22002 1942= +— =1.61mm880047194 384 5

adm = 0.003 x 2200 = 6.6 mm

Therefore 47 x 194 mm (nominal 50 x 200 mm), C16 timber joists at 600 mmcentres are satisfactory.

TRADATECHNOLOGY



7.7 Wall panel lintelsTwo examples have been selected: 2.4 m clear span opening at eaves leveland 1.2 m clear span opening at first floor level.

7.7.1 2.4 m clear span opening at eaves levelEffective span = 2.5 m

Roof loading, dead + imposed (snow) = 1.84 kN/m2

Assume that the truss point loads act as a udl (see Section 2.3.1)

Try 2 No 45 x 200 mm LVL members

2x 510x 45 x 200 x 9.81Self weight = 0.090 kN/m

Total load f = (1.84 x 7.6 x 0.5) + 0.090 = 7.08 kN/m

Bending moment Mmax = 7.08x 2.52= 5.53 kNm

Reaction R = 0.5 x 7.08 x 2.5 = 8.85 kN

Check shear

1.5x8.85x10 2= ____________ = 0.74 N/mm2 x 45 x 200

Tadm = 2.2 x 1.25 x 1.04 = 2.86 N/mm2

where 1.04 = K8, the load sharing factor which may be applied to the strengthproperties of LVL in lintels and trimmer joists made up of two or moremembers connected together.

Check bending

5.53x106x6 2ma = ___________ = 9.22 N/mm2x45x200

m,adm = 19.5 x 1.25 x 2.J x 1.04 = 20.58 N/mm2

Check deflection

Emean = 13500 N/mm2

Gmean = 600 N/mm2

(Mean stiffness properties are used for LVL deflection calculations, but K9 is

not used)

32x45x200 6 4= =60.OxlO mm12

A =2x45x200 =18.OxlO3mm2

5 x 7.08 x (2.5 x iO0O) 3 x 7.08 x (2.5 x 1000)2= + 5.06 mm384x13500x60.0x106 20x600x2x45x200

öadm = 0.003 x 2500 = 7.5 mm

Therefore 2 No 45 x 200 mm LVL members connected together aresatisfactory.2 No 33 x 200 mm members would also be satisfactory.



7.7.2 1.2 m clear span opening at first floor levelThis carries a long-term load.There is no roof load because of the opening above it.


Floor loading = 2.8 x 0.5 x 1.84 = 2.58 kN/m

Wall panel loading 2.4 x 0.24 = 0.58 kN/m

Self dead weight = 0.07 kN/m (assuming 2 No 016 softwoodmembers)

Total udl, f = 2.58 + 0.58 + 0.07 = 3.23 kN/m

RA=RB = 3.23x1.3 =2.l0kN2

3.23 xl 32Mmax = _________ = 0.682 kNm

Try 2 No 47 x 195 mm (nominal 50 x 200 mm), 016 timber members.

Check shear and bending: -

32.1010 2Ta

= ___________ = 0.17 N/mm2x2x47x195

0.682x106x6 2ama = =1.14N/mm

2x47x1952

tadm = 0.67 x 1.1 = 0.74 N/mm2

0.11

Gmadm = 53 xl.1 = 6.11 N/mm2

Check deflection

where E = 6612 N/mm2 from earlier calculation

öa = 3.23x13002x12 5x13002 1952 =0.42mm6612247195 384 5

adm = 1300 x 0.003 = 3.9 mm

Therefore 2 No. 47 x 195 mm (nominal 50 x 200 mm), 016 timber membersare satisfactory.

7.8 Cripple studs

7.8.1 First floor rear wall - very short-termConsider studs supporting 2.4 m clear span lintel in rear wall at eaves level.

Cripple stud height = (full height stud) — (lintel depth)

=2306—200 =2106mm

As in Section 7.5.1, nett pressure on stud qs,nett = 0.466 kN/m2

Load per unit length on one stud

f = pressure x (half overall width of opening)

= 0.466 x 0.598 kN/m2

0.598 x 2.1062Mmax = ____________ = 0.332 kNm



Axial loading to cripple stud reaction from lintel calculation + self-weight ofhalf the lintel

F = 8.85 + (0.5 x 0.090 x 2.6) = 8.97 kN

In general the number of cripple studs required equals the number of studsremoved by the opening, in this case, three, so two cripple studs should beused at each end. But try 1 No 47 x 97 mm (nominal 50 x 100 mm) C16cripple stud at each end.

Applied bending stress

0.332x106 x60m,a = ___________ = 4.50

47 972

N/mm2

Applied compressive stress

8.97aca = ________ = 1.97

47x97

N/mm2


am,adm = 5.3x1.75x.'J = 10.50 N/mm2


acadm = grade stress x K3 x K12

To find K12

Le 0.85x2106A (slenderness ratio) = = __________ = 63.93

N =1.5

I

= 11.9 N/mm2 (from stud calculation in Section 7.5.1

q (eccentricity factor) = 0.005 A = 0.005 x 63.93 = 0.320

1+ri 1+0.320therefore = _______ = 0.660

2 2

rr2xE = u2x5800 = 0.785NA2a 1.5x63.932x11.9

Using formula given in stud calculation section

K12 = [ + (0.660 x 0.785)1 (0.660 x 0.785)1 — 0.785] = 0.545

Gcadm = 6.8 x 1.75 x 0.545 = 6.49 N/mm2

rr2x5800 2ae =

2= 14.01 N/mm

63.93

Check summation of stress ratios using formula in Section 7.5.1

4.50 1.97___________________ = 0.788<110.5011— 1.5x1.97x0.545 6.49

14.01 )

TRADA1 IC H HO LOOt



7.8.2 Check bearing on bottom rail of panel - medium-term loadApplied compressive stress perpendicular to grain

8.85 x103 2Gc±a = =1.94N/mm

47 x 97

Permissible compressive stress perpendicular to grain (Cl 6 timber)

0c,I,adm = 2.2 x 1.25 = 2.75 N/mm2

which is satisfactory where the specification excludes wane.

In some cases it may be necessary to check compressive stressperpendicular to grain in the lintel.

Therefore 1 No 47 x 97 mm (nominal 50 x 100 mm), 016 timber cripple studat each reaction on the lintel is satisfactory.

7.8.3 Ground floor front wall - very short-termConsider cripple studs supporting 1.2 m clear span lintel at first floor level,(Section 7.7.2)

Cripple stud height = 2111 mm

From Section 7.2

q = 0.44 kN/m2 and C0 = 0.82.

Following the calculation in Section 7.5.1, nett pressure on studs

= 0.867 x 0.44 x (0.82 —(-0.3)) = 0.427 kN/m2

Load per unit length on one stud

f = pressure x (half overall width of opening)

=0.427x 1.3/2 =0.278kN/m

0.278x2.1112Mmax = ___________ = 0.155 kNm

Axial loading to cripple stud

F = [Reaction at lintel (Section 7.7.2)] + [Load from cripple stud tosecond storey opening above (ie half the weight of the upper lintel+ half of three truss point loads) (Section 7.5.1)]

= 2.10 + (0.07x1.3)+(3x4.2) = 8.45 kN2

Try 1 No 47 x 97mm (nominal 50 x 100mm), 016 timber cripple stud

Applied bending stress

60.155x10 x6 2Gma = =2.1ON/mm47 x 972

Applied compressive stress

8.4510 2Gca = =1.85N/mm

47x97

(R© TRADA Technology Ltd 2006 101



amadm = 10.50 N/mm2 from previous calculation


Gcadm = 6.49 N/mm2 from previous calculation

(The lengths of the studs are sufficiently close to permit using the samevalues)

Check summation of stress ratios

(K12 and ae as in previous calculation)

2.10 1.85 =0.51<110.5011 1.5 xl .85 x 0.545') 6.49

14.01 )

7.8.4 Check bearing on bottom rail of panel - medium-term loadApplied compressive stress perpendicular to grain

38.45x10 2= 1.85N/mm47x97

Permissible compressive stress perpendicular to grain

ac±adm = 2.75 N/mm2 from previous calculation

Therefore 1 No 47 x 97 mm (nominal 50 x 100 mm), C16 timber cripple studat each reaction of the lintel is satisfactory.

The remaining lintels and cripple studs should then be checked using themethods given above.

TRADATIC I4NOLOCY



8 References and further reading

British StandardsBS 449-2: 1969 The use of structural steel in building — Metric units.

BS 648: 1964 Weights of building materials.

BS 1230-1: 1995 Gypsum plasterboard — Specification for plasterboardexcluding materials submitted to secondary operations.

BS 5268-2: 2002 Structural use of timber. Code of practice for permissiblestress design, materials and workmanship (under revision 2006)BS 5268-3: 1998 Structural use of timber — Code of practice for trussed rafterroofs. (under revision 2006)BS 5268-6.1: 1996 Structural use of timber — Code of practice for timberframe walls — Dwellings not exceeding four storeys. (under revision 2006)B5 5268-6.2: 2001 Structural use of timber — Part 6: Code of practice fortimber frame walls - Buildings other than dwellings not exceeding four storeys.(under revision 2006)

BS 6399-1: 1996 Loading for buildings — Code of practice for dead andimposed loads.BS 6399-2: 1997 Loading for buildings — Code of practice for wind loads.BS 6399-3: 1988 Loading for buildings — Code of practice for imposed roofloads.

BS 8103-3: 1996 Structural design of low-rise buildings - Code of practice fortimber floors and roofs for housing.

BS 8201: 1987 Code of practice for flooring of timber, timber products andwood based panel products.

BS 8212: 1995 Code of practice for drying lining and partitioning usinggypsum plasterboard.

BS EN 338: 2003 Structural timber. Strength classes

BS EN 594: 1996 Timber structures. Test methods. Racking strength andstiffness of timber frame wall panels

BS EN 1995-1-1: 2004 Eurocode 5: Design of timber structures— Part 1.1General rules and rules for buildings.

BS EN 12369-1: 2001 Wood-based panels. Characteristic values for structuraldesign. OSB, particleboards and fibreboards.

BS EN 13986: 2004 Wood-based panels for use in construction.Characteristics, evaluation of conformity and marking

BS EN 14250: 2004 Timber structures — Production requirements forfabricated trusses using punched metal plate fasteners.

BS EN 14545 Timber structures — Connections — Requirements. Inpreparation.

© TRADAlechnology Ltd 2006 103


TRADA PublicationsTimber frame: Standard details for houses and flats. TRADA Technology.ISBN 1 900510 51 0. 2006.

Timber Frame Construction, TRADA Technology Ltd, TBL 58. 2001.

Multi-storey timber frame buildings — a design guide. BRE and TRADATechnology. ISBN 1 86081 605 3. 2003.

Span tables for solid timber members for floors, ceilings and roofs (excludingtrussed rafters) for dwellings. ISBN 1 900510 42 1. 2004, amended 2005.

Structural timber composites. TRADA Technology DG 1. ISBN 1 900510 01 4.1996.

TRADA Wood Information Sheets

0 - 5 Timber frame building: materials specification. 2006.

1 — 41 Strutting in timber floors. 2005.

1-42 Timber I-joists - applications and design. 2003.

2/3-3 1 Adhesively bonded timber connections. 2003.

2/3-56 CE marking: Implications for timber products. 2003.

2/3 - 57 Specifying wood-based panels for structural use. 2005.

Design software available free to TRADA members on the askTRADAwebsite www.trada.co.ukTimber Sizer (calculates solid timber sizes for all structural elements coveredby BS 5268-7).Nail, Screw, Bolt & Dowel Connection Designer (lateral and withdrawalresistance of fasteners calculated to BS 5268-2 and Eurocode 5).Coach Screw Connection Designer (lateral and withdrawal resistance offasteners calculated to BS 5268-2 and Eurocode 5).Flitch Beam Designer (design of bolted steel fitch beams to BS 5268-2 andBS 5950).

Other publicationsBRE Digest 436 Wind loading on buildings. Part 2. 1999. CRC, London.

ETAG 019: 2005 Pre-fabricated wood-based load bearing stressed skinpanels. European Organization for Technical Approvals. Brussels.

UK Timber Frame Association. Design guidance for disproportionate collapse.Technical Bulletin 3.


Timber Frame Housing: UK Structural Recommendations is a time-saving professional manual for structural engineers. Itcovers well-established principles and methods for the structural design, strength and stability checking of timber framebuildings. Worked examples, including calculations for a complete house, are included.

Guidance is based on the recommendations in BS 5268-2 and BS 5268-6 and also includes procedures now widely used in the

design of timber frame houses, but not specifically covered in these Codes. This publication deals solely with the engineeringaspects of timber frame design.

Timber Frame Housing: UK Structural Recommendations saves engineers valuable time and effort because it:

• includes ready-made load tables and calculations relating to a complete house• features worked examples and diagrams throughout• is based on the latest UK British Standards.

This manual has been written and verified by a team of experts at TRADA Technology with assistance from CCB Evolution. Ifwould like to submit feedback on the content please send your comments to [email protected].

TRADA Technology is the leading independent timber research, consultancy and information provider for the constructionindustry. Technical expertise is at the heart of our business, working with clients to get the most from their timber products.

Specialist services include frameCHECK (timber frame quality assessment), structural and condition surveys on site, designassistance, materials and product evaluation. In the event of a dispute, expert witness services are also available. Clients cover

the whole construction delivery chain, including architects, engineers, designers, specifiers, contractors and builders, in the UKand overseas.

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