Timber Framed Buildings.pdf

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YI TECHNO

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TRADA Technology Ltd Chiltern House Stocking Lane Hughenden Valley High Wycombe HP14 4ND

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Timber frame housing: UK Structural recommendations

(TRADA TECHN0L0G

Third edition 2006

TRADA Technology Ltd Chiltern House Stocking Lane Hughenden Valley High Wycombe Buckinghamshire HP14 4ND

t +44 (0)1494 569600 f +44 (0)1494 565487 e [email protected] w 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 the printed book is permitted without the consent of the copyright holder, TRADA Technology, 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, the company cannot accept liability for loss or damage arising from the use of the information 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 Timber frame wall panels Horizontal diaphragms

1.1.1

1.1.2

1.2

Determination of loads acting on each element Calculation of vertical load Calculation of wind loads Calculation of roof loads Checking strength and stability

1.2.1

1.2.2 1.2.3 1.2.4

6 6 7

9 10 11

19

20

1.3

Weights of materials

21

1.4

Weights of typical constructions

23

Roofs Floors

1.4.1

1.4.2 1.4.3

1.5

Walls

General requirements of prefabricated wall, floor and roof elements

2 Walls 2.1

24 25

Wall studs Minimum desirable stud sizes Multiple studs Lateral support Load capacities Design procedure Drilling of studs

2.1.1 2.1.2 2.1.3 2.1.4 2.1.5 2.1.6

23 23 24

25 25 25 26 26 29 30

2.2

Framing around openings

30

2.2.1

Lintel capacities

31

2.3

Racking resistance of timber frame walls Basic design and materials Adhesively bonded panels Overturning effects Deflection

2.3.1 2.3.2 2.3.3 2.3.4

3 Floors 3.1 3.1.1

3.1.2 3.1.3 3.1.4 3.1.5

41 41

42 43 46

48

Joists and beams Reactions, moments and deflections in a single span, simply supported beam Solid timber beams Structural timber composites Prefabricated engineered timber joists Steel fitch beams

3.2 Floor decking

48 49 50 52 52 53

63

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4 Roofs

64

4.1

Trussed rafter roofs

64

4.1.1

Bracing

64

Connections

67

4.2

5 Foundations

68

5.1

68

Loading

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

75

7.3.1 7.3.2 7.3.3 7.3.4

7.4 7.4.1

7.4.2 7.4.3 7.4.4

7.5 7.5.1 7.5.2 7.5.3

Wind pressures and self-weight Overturning Sliding Resistance to wind uplift

Racking calculations Wind loads Racking forces Design method for racking Recommended design procedure

Wall panel studs Ground floor rear wall — very short term Ground floor rear wall — medium term and long term Ground floor gable wall — very short term

75 76 77 78

78 78 79 79 91

92 92 94 95

7.6

First floor platform

96

7.7

Wall panel lintels

98

7.7.1

7.7.2

7.8 7.8.1 7.8.2 7.8.3 7.8.4

2.4 clear span opening at eaves level 1.2 m clear span opening at first floor level

Cripple studs First floor rear wall — very short term Check bearing on bottom rail of panel — medium term load Ground floor front wall — very short term Check bearing on bottom rail of panel — medium term load

8 References and further reading

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98 99

99 99 101 101

102

103

(A


Introduction Timber Frame Housing - Structural Recommendations covers well established principles and methods for the structural design and strength and stability checking of timber frame buildings. The guidance is based on the recommendations in BS 5268-2 Code of practice for the structural use of timber. Permissible stress design, materials and workmanship, BS 52686.1:1996 Structural use of timber. Code of practice for timber frame walls. Dwellings not exceeding four storeys but it also includes procedures now widely used in the design of timber frame houses but which are not specifically covered in these Codes. Worked examples, including calculations for a complete house, are included. The guidance does not exclude other ways of showing that a design satisfies the requirements of Building Regulations and Standards regarding strength and stability. The design of timber frame buildings, for example, can be undertaken using the guidance given in the Eurocodes suite, BS EN 1990:2002 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 structures and currently most buildings are designed or checked using the relevant parts of the British Code, BS 5268.

Materials and components are covered by other standards as listed in the References. Span tables for floor joists, rafters and purlins for roofs, and joists for flat roofs are available in a separate TRADA Technology document Span tables for solid timber members in floors, ceilings and roofs (excluding trussed rafter roofs) for dwellings. This publication deals solely with the engineering aspects of timber frame design; it does not address issues such as fire safety and performance, thermal and acoustic performance, weatherproofing and durability and design detailing. These are covered in a companion publication Timber Frame Construction, last updated in 2001. Timber frame: Standard details for houses and flats, published in 2006, provides typical details for 'open' panel timber frame wall panels for three storey houses requiring 30 minutes fire resistance and for flats requiring 60 minutes fire resistance. The details show floors with solid timber joists which can readily be used for timber frame buildings up to four storeys in height. This is the third edition of Timber Frame Housing - Structural Recommendations. The first edition, published in 1979, was itself an updated version of Section 9 Structural Recommendations of the TRADA Design Guide for Timber Frame Housing first issued in 1967.

TRADA Technology TRADA Technology is the leading independent timber research, consultancy and information provider for the construction industry. 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, design assistance, 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 UK and overseas.

Acknowledgements TRADA Technology wishes to acknowledge the assistance of CCB Evolution in the preparation of this publication. CCB Evolution, Consultants in prefabrication and modular design, is a specialist engineering consultancy with expertise in timber frame structures and cold rolled steel frames. More information is available from www.ccbevolution.co.uk.

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I Loading on structural elements 1.1 Structural duties of timber frame elements 1.1.1 Timber frame wall panels Timber frame panels used in house construction have three major structural duties to perform: support of vertical loading, resistance to deformations caused by horizontal loading in their plane and resistance to wind loading perpendicular to their plane.

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

1ICHN0

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6


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

The illustration shows a floor and roof diaphragm resisting wind load on a gable wall. The ends of the gable wall are supported directly by the front and rear walls of the building, and the bottom edge by the foundation. The load on the rest of the wall is transferred via the horizontal diaphragms into the front and 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 plus half the nett load on the first floor walls. The roof diaphragm takes half the nett load on the first floor walls plus load from the roof. For wind parallel to the ridge it is usual to assume that half the nett horizontal wind load on the spandrels or roof is transferred to the roof diaphragm; for wind perpendicular to the ridge all the nett horizontal roof load is transferred to it. Where adjacent panel edges are fastened with nails or screws to the same timber members such as joists or blocking, the resulting connection between the 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 woodbased panel product is fastened to timber joists, have adequate strength and stiffness as horizontal diaphragms, provided that:

• • •

•

the diaphragm span : depth ratio does not exceed 2:1 in either wind direction (BS 5268-6-2 Clause 6.5) the span does not exceed 12 rn between supporting walls (BS 5268-6-2 Clause 6.5) the fixing around the edges of the panels complies with standard recommendations (eg 3.00 mm diameter ringed shank nails at 150 mm centres for plywood or 3.35 mm ringed shank nails at 300 mm for wood particleboard and OSB, with a length equal to 2.5 times the board thickness) the perimeter of the diaphragm is attached to the walls with fastenings of equivalent strength.

Plasterboard ceilings in roofs that comply with BS 5268-3 Annex A may also be assumed to provide adequate diaphragm action provided that truss clips are used to secure every truss to a head binder and the fixing around the edges complies with standard recommendations (3.5 mm diameter plasterboard nails or screws 40 mm long at 150 centres). It is recommended that in areas of high wind load (eg with a dynamic wind pressure> 1500 N/rn2), and always for horizontal diaphragms outside the range given above, the required fastener spacings should be calculated by ensuring that

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ci:13,


Va

h

adm,1 Slkedge,1

+ dm,2 S2kedge,2

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

h fadm

k

= maximum shear force (normally half the total wind load applied to the diaphragm) (N) = horizontal depth of diaphragm measured in direction of wind (mm) = permissible load per fastener for very short-term loading (N) = a value shown in the following table k

Blocked floors and flat roofs, in which all four edges of the panels are joined to adjacent panels by fastening them to joists or blockings Floors and flat roofs in which adjoining tongued and grooved edges are not fastened to the same timber member but are glued together Unblocked floors and flat roofs in which adjoining tongued and grooved edges are not fastened to the same timber member and are not glued together Plasterboard ceilings in which all four edges of the panels are joined to adjacent panels by fastening them to joists or blockings

1.0

1.5

1.4

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

For wood chipboard and OSB, there is no method in BS 5268-2: 2002 for determining 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 safe to use with suitable structural grade materials. (Group 1 plywoods are listed in a footnote to BS 5268-2 Table 63.) Alternatively a value for fadm in OSB can be calculated using Eurocode 5 (BS EN 1995-1-1) for wind loading in service class 1 and dividing the design value by 1.5. For plasterboard fastened to trussed rafters with plasterboard nails or screws of at least 3.5 mm diameter, the Trussed Rafter Association recommends a value for fadm of 190 N for 12.5mm thick board and 213 N for 15mm thick board.

For unconventional building forms or unconventional types of floor or roof, the structural adequacy of the diaphragm action should be checked, not only for the 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 element Loads 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 the floor, wall and roof systems may be calculated. In each example, a load bearing internal wall is used on the ground floor to bring about economy in design. If a more flexible plan is required the joists may often be designed to span the whole ground floor without interior support. Prefabricated timber I-joists are particularly common for longer spans, while timber-flanged joists with open metal webs, glulam, laminated veneered lumber (LVL) and other structural timber composites can all span significantly further than solid timber joists of the same depth.

Dead load:

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

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

Imposed floor load:

The imposed floor load specified by BS 6399-1 Loading for buildings. Code of practice for dead and imposed loads for a self-contained unit designed for occupation by a single family is a long-term distributed load of 1.5 kNlm2 or a point load of 1.4 kN, whichever is more severe. However, for solid timber floor joists BS 5268-7.1 recommends instead a long-term distributed load of 1.5 kN/m2 for joists with an effective span of 2400 mm or more, or for shorter joists a load of 3.6 kN per metre width of floor (measured perpendicular to the span) uniformly distributed over the entire span. All these loads may also be used 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-term distributed load of 0.25 kN/m2, assuming that the space above the ceiling is used for storage rather than a living room.

Imposed roof load:

The exact value of the imposed roof load depends on the location and size of the building, and BS 6399-3 Loading for buildings. Code of practice for imposed roof loads should be consulted. In general for roofs up to and including 30° pitch where access is limited to maintenance and repair, the imposed load is a medium-term distributed load of 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 distributed load may be obtained by linear interpolation between the value for a 30° pitch and nil. For roof slopes of 60° or more the distributed load is nil. The point load may be ignored for slopes of 30° or more.

Wind loads:

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Wind loads are calculated using BS 6399-2 Loading for buildings. Code of practice 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) Stud and joist spacing = S

Roof and ceiling load Wç kN/rn2 total

Shaded areas carry roof or floor loads

,

Load at each stud point

L 2

= WrXSX_ Load in each stud

4\Nr xsxJ+(Ww xsxh) Load at each interior joist support

=(Wf xsx.J+(Ww xsxh) For beam designs see Section 3.1

Floor load W k N/rn2

Load at each stud point

1\NrXSxJ +(wwXsXh)+\NfXsX_J Load in each stud

Stud supporting bearn, designed as for lintel support

1\NrXSxJ +(wWxsxh)+[wfxs4j +(W xsxh) Self weight of wall W kN/m2

Typical loading 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)

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:

.

7.8

L

Load per stud point at eaves

= WrxSx=i.97xO.6x__

Load at foot of each upper storey

= 4.61 +(w

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

=5.93+IWfxSx 2

h)= 4.61 + (0.86 x 0.6x 2.55)

=5.93+11.82 x 0.6 Load at foot of each lower storey stud

Internal wall:

5.93 kN

= 7.9 kN

2

= 7.9 +(W x s x h) = 7.9 +(0.35 x 0.6 x 2.55) =WfXSX

Load at foot of each internal wall

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

2

= 8.44 kN

=1.82x0.6x72

Load per stud point at 1S floor level stud

= 4.61 kN

= 4.26 kN = 4.62 kN

H

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1.2.1.2 Roof and floor spanning between cross-walls (discontinuous joists supported by load bearing internal wall) Roof and ceiling load Wr kN/m2 total

Typical loading Inner support spacing = d Stud and joist spacing = s Load at each inner roof support point

Shaded areas carry roof or loads

L

W xdx— 2

Average load in each wall stud (spaced s) Floor load Wf kN/m2

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

Load at each interior joist support = WfXSX (A + B)

2

Load in each stud

l

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

Stud supporting beam, designed as for lintel support

Self weight of wall W kN/m2

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. End wall

Load per stud point on spandrel panel

=W xdx—=1.97x0.6x———

= 2.84 kN

Load at foot of each upper storey stud

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

= 4.47 kN

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

=4.47+IWxSx =4.47+11.82x0.6xi 2

= 5.45 kN

Load at foot of each lower storey stud

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

= 5.99 kN

Load at foot of each internal wall stud

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

2

2

2)

= 1.82x0.6x—I+(0.25x0.6x2.4)

2)

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11

=2.98kN


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

rearasin 1.1.1.1.

1.2.2 Calculation of wind loads Wind pressure is the main horizontal force causing racking and bending deformations. If adequate provision is made to resist the wind loadings, all other racking forces can usually be ignored for two-storey domestic buildings. The design wind pressures will depend on the locality, the degree of exposure, 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 a timber frame building are calculated, as for other buildings, according to BS 6399-2. An example of a calculation for a two-storey house can be found in BRE Digest 436 Part 2. Where masonry walls provide an external shell, the resulting wind loads should be based on the overall dimensions of the masonry walls, not those of the load bearing timber frame. However BS 5268-6.1 allows the wind load transferred to the timber frame to be reduced by a factor K100 where masonry walls conforming to Clause 3.2.2 provide a shielding effect, for the calculation of racking forces and overall building stability only (ie not for stud design). This reduction should not be applied to the spandrels of gable walls, because the masonry walls in this area have no returns to make them self-supporting in a direction perpendicular to their plane.

1.2.2.2 Horizontal wind loads The figure below shows the overall dimensions of a semi-detached timber frame house with masonry cladding built on a foundation 0.3 m above ground level. The racking forces on each pair of walls are shown as R1, R2 etc. 1.3

2.5

1.3

Ri

N

Ro

N

Firstfloor I

R3

N

R4

N

5.0 2.5

0.3 -- _________________ ____________________ — ___________________ : 0.3 Front elevation

—

End elevation

wall

I I

6.8

.

Dimensions in metres

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

© TRADA Technology Ltd 2006

12


panels in the storey below or by the foundations. Therefore the total racking load on a timber frame wall is calculated as the racking load transferred from the roof or the storey above it, plus half the wind load applied to the same storey. The next figure shows how the diagonal dimension, a, is calculated for each diaphragm and wall. This dimension is used to determine the corresponding size effect factor, Ca, (BS 6399-2 Clause 2.1.3.4). Since the lowest value is 0.95, it will normally be simpler to assume a value of 1.0 for Ca, and calculate the exact values only if it proves necessary.

Front and rear walls

Gable walls Diaphragms First floor diaphragm

Roof diaphragm 6

—---s----- t

First floor diaphragm

-

--——--- 7.2 m

-72m

68m

a4 42.52 +6.82

a2 =/2.52 + 7.22

a1 41.252 +7.22

= 7.62 Ca = 0.963

= 7.31

Ca = 0.968

= 7.62 Ca = 0.963

= 7.26 Ca = 0.968

Racking First floor panels

Ground floor panels

v

First floor panels

-A y

E

Ground floor panels

/-

,,'

—7

72m

a5 =V1.252 +7.22 = 7.31 Ca = 0.968

72m

a6 3752 +7.22 = 8.12 Ca = 0.957

a7 =2.552 +6.82

a8 5Q52 +6.82

= 7.26 Ca = 0.968

= 8.47 Ca = 0.953

First floor panels

-d

7m a9 = 2.52 + 7.22 = 7.62 Ca = 0.963

6.8 m

+ 7.22 a10 = = 8.77

a11 43.82 +6.82 = 7.79 Ca = 0.960

Ca = 0.950

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

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13

= 9.27 Ca = 0.945

6 Ia)


Some typical values for the dynamic pressures, q, to be used during the normal lifetime of a building, are shown below in Pascals (N/rn2). For buildings with masonry cladding, BS 5268-6.1 Table 1 allows a reduction in the wind loads on the timber frame and allows part of the load on the timber frame to be transferred back into the brickwork through specified wall ties, but the building 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 for shorter periods of construction. Hence there are two design situations in which the building's strength and stability have to be checked: (a) the construction period when there is neither a masonry shielding effect nor a roof load to resist overturning but the normal design wind load is reduced, and (b) the constructed period when the shielding and load-sharing of any masonry may be taken into account, but the full 50 year return wind load is applicable. (a) is particularly important in the construction of multi-storey buildings when overturning forces can be significant. The wind pressures shown in the diagram have been reduced where applicable by the appropriate value of K100, the reduction factor for the masonry shielding effect given in Table 1 of BS 5268-6. Note that this factor is not applied to the spandrel area of the gable walls, for reasons given previously (see Section 1.2.2.1). The pressure coefficients, Cpe and C, for the example building are also shown, obtained from BS 6399-2, Tables 5, 10 and 16. For a semi-detached house the external wind pressure on the party wall may be taken as zero. The overall dimensions of the two dwellings should be used to calculate the pressure coefficients.

Wind on gable wall —

Wind on rearwall

________________

+027

-1.17

-05

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

qs = 658 x 1.0

qs—0

qs=658x0.5

= 329 +084 -0.3

0.11

3.49

0.11 3.49

-0.3 -0.5 qs=590x

qs=590x 0.8=472 -0.3 -0.5

0.82=484

+082 -0.3

_________________ Gable end

Front

Separating wall

Reference values for BS 6399-2: Table 5: D/H=1.03 Table 10: W=7.2, bw7.2

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14

Rear

Gable end

TableS: D/H=1.36

Front

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 is convenient to rewrite BS 6399-2 Equation (7) in Clause 2.1.3.6 as:

(1)

Pfinal

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

where Pfinal

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

0 85

= factor to allow for the fact that maximum gust speeds on front and rear walls are not simultaneous

P

= nett wind load on an element of the building which contributes to the final load, unfactored for dynamic augmentation or size = proportion of the element which contributes to the final I

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 the wall panel depends on the difference between the external and internal pressures, so P

(4) = (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

= 329 X 0.84 x 7.2 x 2.5

= 4974 N

= (329 x 0.84 _(_)329 X 0.3)x 7.2 x 2.5

= 6751 N

= 0.85 x (2587 X 0.667 + 4974 X 0.5)x 1.02 x 0.968

= 3535 N

= 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

Each storey: P siorey Equation (4)

Walls: P wail Equation (1)

Roof diaphragm load P ceiling First floor diaphragm load P fioor, 1

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TR.ADA Technology Ltd 2006

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Shear force on first floor panels

Pshear,i

0.85x(2587x1+4974x1)xl.02x0.963

=6313 N

Shear force on ground floor panels

Pshear,0

0.85x(2587x1+4974x2)xl.02x0.950

= 10324 N

Nett load on a wall stud, assuming 0.6 centres

Pstud

= 0.85 x 6751 x 0.6 xl .02 xl .0

=488N

7.2

(b) Wind on front and rear walls Equation (2)

Roof: P roof Each storey: P storey

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

(—)0.5x 2.92) xtan20°x6.8

= 6205 N

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

= 10759 N

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

= 9215 N

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

= 9722 N

Equation (4)

Walls P wall Equation (1) Roof diaphragm load

ceiling

First floor diaphragm load

P floor,!

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

= 8983 N

Racking load on first floor panels

P racking, 1

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

= 9722 N

Racking load on ground floor panels

P racking, 0

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

= 18461 N

Shear force on first floor panels

P shear, 1

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

14119 N

Shear force on ground floor panels

P shear, 0

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

= 22714 N

Neff load on a wall stud, assuming 0.6 m centres

P stud

—

0.85x 9215 xO.6x 1.02 xl .0 6.8

= 705 N

The results are shown in diagram below

Pceiling = 9.72 kN —

Pceiling

= 3.17 kN

— Pracking,1

4.26 kN

Pracking,1 = 9.72 kN

—k

Pfloor, 1

= 4.15

Pshear,1

Pfloor, 1

—

6.31 kN

= 8.98 kN Pstud

Pracking,o = 8.34 kN

Pstud

= 0.705 kN

—s

= 0.488

—

Pracking,o = 18.46 kN

—s

—k

Wind on front and rear walls

Wind on gable end


—s

Pshear,0 = 22.71 kN

Pshear,o = 10.32 kN

©

Psyear,i = 14.12 kN

16


The shear force is used to calculate the required fixing of the wall panels to the floor or sole plate. With a shear force of Va Newtons on one wall, the maximum 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 necessary to check panels against rotation, but this may be necessary where there are low vertical loads, such as in the upper storey of a building with a flat roof, or where particularly narrow or specially stiff panels are installed. This should be carried out for the end panels in each wall and any narrower panels, using an appropriate proportion of the racking load on the complete wall as a moment force at panel height, and the vertical load on the panel (normally a UDL) as a restraining moment, as shown in the diagram in Section 2.3.3. For each load case the minimum vertical load, normally occurring beneath the part of the roof where there is maximum wind uplift, should be used. The vertical load on the panel should include the weight of the panel itself. For details on designing overturning restraint, see Section 2.3.3 (Overturning effects).

1.2.2.4 Asymmetrical buildings In the example shown above, the wind loads are resisted by two equally stiff racking walls so half the load is applied to each. If the resistance of two racking walls differs greatly then the horizontal diaphragms may usually be assumed to redistribute the load so that the stiffer and therefore stronger wall takes a proportionately higher share of the load. All that is necessary is to determine that the total wind loads do not exceed the combined resistance of the walls. For the same reason the resistance of internal load bearing walls may be added to that of the external walls, regardless of their position in the building. However, for buildings with dimensions outside the limits specified in BS 5268-6.2 Clause 6.5, and for unconventional building forms or unconventional types of floor or roof, the assumption about load redistribution may not hold true. In this case, if the building is asymmetrical then the racking and shear resistance 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 racking resistance to wind blowing directly onto the flank wall. In this case part of the load may be transferred to the adjacent house via steel party wall straps. In order to maintain a low level of sound transmission the Building Regulations specify a maximum density of one row of straps at 1200 mm centres installed at each storey height. However to ensure adequate load distribution the spacing needs to be not greater than 1200 mm, so this spacing should always be used. The thickness of the steel is usually 2 to 3 mm and the width of the straps is typically 25 or 30 mm. The cross-sectional dimensions should not exceed 3 mm x 40 mm. It is recommended that each strap be fastened to the underside of the top plate of the party wall panel, with two nails or screws at each end, maintaining the BS 5268-2 minimum edge distances of 5d. (For the design of party walls, see TRADA's Timber Frame Construction.) The loadcarrying capacity should be calculated as the minimum of the permissible load on two fasteners and the buckling strength of the steel.

The following points should be noted. Continuous tiling battens can generally transfer all the horizontal load from the roof of an end terrace house to the roof of the adjoining house. Where the remaining racking load at ceiling level, or racking loads at intermediate floors, are partly transferred via party wall straps to the adjoining house, there will be a corresponding reduction in the racking loads transferred to lower floors in the end terrace house itself. This permits large openings in the front and rear racking walls on the ground floor of an end terrace house.

©


17

(R


Example Consider a Grade 43 (S275) galvanised steel strap 250 mm x 25 mm x 3 mm thick, fastened with a total of four 3.75 mm square twist nails, 50 mm long in C16 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) 0.4

=443N

(As the headside member is steel the specified headside penetration of 46 mm 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.

= Le = Le

Slenderness ratio

I

b

=

0.7x140x3.464 3

113 mm

From BS 449-2: 1969 Table 17a the allowable compression stress: 68 N/mm2.

Pc

Hence permissible compression load on steel strap:

= 68x3x25

PCXA

5100> 1661, hence permissible load on one strap

5100 N. = 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 7200

Fadm

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 x 0.968/1000 = 2.09 kN, so all of this could be transferred to the adjacent property 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 next house, leaving no requirement for racking resistance in the ground floor walls of the end house! Obviously the racking loads have to reach the foundation eventually, so with n houses in a terrace no more than (n-2)/n of the racking load at any one level should be transferred to the adjacent property, starting from the roof. The designer should specify precisely the required materials, dimensions and positions of the party wall straps and fasteners.

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18


1.2.3 Calculation of roof loads The connections between the roof trusses and the head binders should be adequate to resist the nett horizontal wind forces on each truss after allowing for frictional resistance, and the nett uplift forces on them after allowing for the weight of the roof. The uplift forces are also required for stability calculations to determine the overturning moment. The method for calculating these forces is not unique to timber frame buildings, but is illustrated in the example of calculations for a complete dwelling in Chapter 7. As mentioned in Section 1.1.2 the use of truss clips on every truss is recommended to ensure adequate diaphragm action in the plane of the ceiling. Therefore, the truss clips could be called on, if required, to provide any additional resistance to roof uplift or horizontal wind loads. Manufacturers of proprietary truss clips should provide safe working loads or characteristic values for vertical loading and for horizontal loading in the plane of the truss and 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 failure usually occurs in the clip itself, at a lower load than that predicted by calculating the permissible load on the nails. However, if manufacturers' data are not available it may be assumed, based on the tests mentioned above, that a truss clip with at least three 3.25 mm round wire nails in each member can resist a vertical or horizontal wind load of 1.1 kN. (Some kinds of clip can resist significantly more than this in certain directions.) Where, as is normally the case, the clip resists components of load in two directions, it should be verified that 2

Fva 2+1 Fha Fvadm

where

Fhadm

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

with Fva and Fh,3 in kN.


19

1.2


1.2.4 Checking strength and stability Care must be taken to check every relevant loading situation and potential area of instability when designing a timber frame house. During the construction period no allowance should be made for the shielding or load-sharing effect of any masonry walls which may be added later, nor for the weight of the roof in resisting overturning and sliding. Internal walls may not be present initially, and unless the external wall panels are prefabricated with internal linings there may be a period when their plasterboard linings are not attached. Party walls consisting only of two layers of plasterboard and timber framing need particular consideration, but in general their calculated racking resistance may be assumed during the construction period provided that they are adequately braced on erection in accordance with BS 5268-6.1 Clause 4.7.5. To offset the effects of these conditions, BS 6399-2 permits a reduction in the normal design value of the wind load for periods of less than 50 years. For a 1 year maximum construction period, the probability factor described in BS 6399-2 Annex D is 0.749, which means that the dynamic wind pressure for the construction period may be reduced by a factor of 0.7492 or 0.561.

Where wall panels are built in situ there may be a period during which neither the structural sheathing nor the plasterboard linings are attached to the studs, but the studs may nevertheless have to support temporary loads such as packs of OSB or plasterboard. Under these circumstances the studs will not be braced against buckling in their weaker plane. The engineer is therefore advised either to check their load-carrying capacity in these circumstances, or to specify a construction procedure which will ensure that neither heavy construction loads, nor roof weights are applied before the structural sheathing is fully attached, or to specify that the sheathing be attached to wall panels before their erection. During the service life of the building the full value of the design wind loads must of course be used. However, if masonry walls are present, BS 5268-6 allows a reduction in the wind loads on the timber frame racking walls as a result of the shielding effect of the masonry and, provided that the masonry and timber frame walls are tied together in a specified manner, a further contribution from the masonry towards the racking resistance.

©


20


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

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

Aluminium

kg/rn2 1.5—3.4

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

2.0—4.4

Battens, tiling

38mmxl9mm:

Bituminous roofing felt

3 layer felt bonded together, 13 mm granite chippings

Blockwork, concrete

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

Brickwork

3.4

loOmmgauge

34.0—37.0

Hollow, stone aggregate:

100 mm thick

139.0

Solid stone aggregate:

100mm thick

219.0

Aerated:

100 mm thick

82.0

Clay, medium density:

103 mm thick

221.8

Concrete

103 mm thick

237.6

Sand lime:

103 mm thick

206.0

6 mm thick

Calcium silicate and cement

5.5—7.5

fibreboards

Concrete

100 mm thick

Reinforced:

0.6 mm as laid

6.5

Board, semi-compressed:

25 mm thick

4.9

Flooring:

25 mm thick

9.8

Bitumen impregnated insulating board:

13mm thick

3.1 —4.5

Hardboard:

6.3 mm thick

4.2

Insulating board:

12.7mm thick

3.4

Copper roofing

Cork

Fibre building board

240.0

5.0—7.2

Medium board:

9.0 mm thick

Glass

Plate:

6.4 mm cast clear and armoured

Gypsum panels and partitions

Building panels:

75mm to 150mm

43.9—63.5

Dry partitions:

57 mm to 63 mm

20.5—25.9

Glass fibre acoustic insulation for floating floors:

25 mm thick

Glass fibre thermal insulation:

100mm

2.3

Mineral fibre

100mm

1.2—2.4

Insulation

Internal stud partitions

thermal insulation:

2.0

29.6

12 .5mm plasterboard 2 sides

20.4 — 33.6

Proprietary

Lead

16.1

1.7 mm to 3.0 mm

Sheet:

19.5—34.2

OSB (oriented strand board)

9mm to 18mm

Pitch mastic

Flooring:

25 mm thick

Plasterboard

Gypsum:

9.5 mm

6.1

Gypsum:

12.5mm

8.0

Gypsum-

15mm

9.8

Gypsum

19mm

Flexible pvc:

1.6 mm to 3.2 mm

2.4—4.9

Pvc vinyl:

1.6 mm to 4.8 mm

3.4— 10.3

American construction and industrial plywood:

12.5mm to 19mm

7.3—10.9

Canadian Douglas fir and softwood plywood:

12.5 mm to 18.5 mm

7.2— 10.6

Plastic flooring

Plywood


21

5.7— 12.8

14.9—18.1

15.0


Plywood (continued)

Finnish birch:

12mm to 18mm

kg/rn2 8.4— 12.4

Finnish birch faced:

12mm to 18mm

8.1 —11.6

Swedish softwood:

12.5mm to 19mm

Tropical hardwood:

12 mm to 19 mm

PSL (Parallel strand lumber)

5.3—8.0 7.3—11.6 730 - 750

12.5mm

Rendering

Portland cement: sand (1:3):

Roofing felt

3 layers felt and chippings

37.0

20 mm asphalt

46.0 3.2 mm to 9.5 mm

Rubber flooring Sand

Dry per cubic metre

Shingles

Cedar

Slating

Welsh:

5.4— 16.1 1522—1682

7.3

Westmorland:

Cornish:

Tiling, roof

29.3

thin

24.4

thick

48.8

thin

48.8

thick

78.1

thin

29.3

thick

48.8

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

machine made:

100 gauge

63.5

hand made:

100 gauge

70.8

100 gauge

68.4

Concrete, plain:

41.5—56.1

Interlocking, single lap:

Tiling wall

Concrete, plain

68.0

19mm

Weatherboarding Flooring grades, P4 and PS

Wood chipboard

Hardwood: beech, oak

Wood flooring

Softwood: pitch pine

Softwood: redwood

Wood wool

Slabs:

Zinc

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

7.3

15mm thick

10.2—11.2

18mm thick

12.8—14.0

22 mm thick

14.3— 15.7

16mm thick

11.7

19mm thick

13.9

21 mm thick

15.3

28 mm thick

20.3

16mm thick

11.4

19mm thick

13.5

21 mm thick

14.9

28 mm thick

19.8

16mm thick

8.3

19mm thick

9.8

21 mm thick

10.8

28 mm thick

14.4

25 mm thick

14.6

4.4—7.8

To convert kg/rn2 into N/rn2 multiply by 9.81 LSL (laminated strand lumber)*

kg/rn3 610—670

LVL (laminated veneer lumber)*

530— 710

PSL (Parallel strand lumber)*

730 — 750

* The density of structural timber composites depends on the moisture content, species or grade. For more accurate values see 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.


22


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

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

k

Plan load w kg/rn2 =

cosO

where k is the weight of roofing (tiles and battens) in kg/m2 measured on slope 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 25.5

Ceiling dead

0.25 0.25

Ceiling imposed

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 9 mm OSB

35.5 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

22 mm tongued and grooved chipboard

14.3


10.0

140 mm phenolic foamboard Self-weight of joists

4.2

11.0 39.5

©


kN/m2

0.39

Imposed

1.50

Total design load for floor

1.89

23

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


kN

10.0 89.8

0.88

Example (5) WaIl with board cladding 25 mm boards and battens

12.5

9 mm OSB sheathing

5.8

150 mm glass fibre batt insulation

5.0


7.0


10.0

40.3

0.40

Example (6) Interior wall 12.5 mm plasterboard both sides


20.0

5.0 25.0

0.25

1.5 General requirements for prefabricated wall, floor and roof elements BS EN 14372 Timber structures — Prefabricated walls, floor and roof elements, will, when published, specify the product, performance, production and testing requirements for timber frame walls, floor and ceiling elements where these are prefabricated from 'joists" which are fixed to a panel product on one or both sides made of or from timber or gypsum plasterboard. The fixing may be by mechanical fasteners or glue. For glued fixings detailed requirements are given for the permitted types of adhesive, the methods of application and the quality control systems which must be in place.

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24


2 WaIls 2.1 WaIl studs All load bearing studs and rails must be of strength graded timber. Most commonly studs are softwood of Strength Class C16 or C24 in accordance with BS EN 338 Structural timber. Strength classes.

2.1.1 Minimum desirable stud sizes Stud widths should allow for the satisfactory butt jointing of sheathing and plasterboard. BS 5268-6.1 recommends a minimum cross-section of 38 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 walls may appear too flexible to the occupants of the house, although the studs are safe against axial and wind loading. The depth of wall studs is often governed by the thickness of insulation required within the wall panels to meet the requirements of Building Regulations. For example a stud 89 mm deep with 90 mm of insulation material will only meet requirement Li of the 2004 England and Wales Regulations with a reflective breather membrane or the installation of a high efficiency boiler. A 140 mm deep stud with 140 mm of insulation will meet the same requirement without any special measures. Some companies use prefabricated timber I-joists as wall studs, both to obtain the required thickness of insulation and because their thin webs have much lower thermal conductivity than studs made of solid timber.

2.1.2 Multiple studs In domestic-scale buildings studs are normally spaced at 400 mm or 600 mm centres. Most wall panels are factory-produced but for on-site framing studs spaced at 600 mm are preferred to allow workmen room to walk between them. With high axial loads it may be necessary to specify double, triple or even more studs. In order to prevent buckling in their weaker direction, multiple studs should be fastened together in such a way that a total lateral force of 2% of the axial load on each stud can be applied uniformly to each stud. This may be achieved either by nailing through the sheathing material from both sides into the secondary studs with sufficient nails to provide the required force in lateral loading, or by nailing the multiple studs together with sufficient nails to provide the required force in axial (withdrawal) loading. In the latter case ringed shank nails or screws should be specified. The fasteners should be spaced at 300 mm or less. If three studs are fastened together it is preferable to nail or screw the outer sets from the outside of the panel, since the axial fastener load will be greatest in the outer studs. With the intermediate studs the central one in each set should be fastened to the panel.

When checking the combined bending and compression stresses in multiple studs or their deflection, Emin may be increased by the appropriate factor given in 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 the bearing strength of the bottom rail, it may be possible to reduce the number of studs required by using a higher grade of timber in the top and bottom rails, or deeper studs.

©


25


2.1.3 Lateral support According to BS 5268-6.1, solid timber studs covered on one or both sides with a specified board material and fixed as recommended in the standard may be assumed to be fully restrained against buckling in their weaker direction. However there is evidence from North America that for studs braced on one side only by a sheet material, as in a separating wall, the load-carrying capacity is reduced by about a quarter. Caution is therefore advised particularly when it comes to buildings of more than the four storeys covered by BS 5268-6.1. Where necessary nogging pieces may be used to restore full lateral support, provided that they are combined with some form of diagonal bracing to prevent all the studs buckling simultaneously in the same direction. United States recommendations indicate that one row of noggings at midheight is adequate for nominal 50 x 75 mm and 50 x 100 mm studs, and two equally-spaced rows for 50 x 150 mm studs. The sheet material should be nailed to the noggings as well as the studs.

2.1.4 Load capacities Tables 2.1 and 2.2 relate to timber of strength class C16 and C 24 respectively in service class 1. They give the permissible axial load in kN per stud (not per metre) for studs in panels 2400 mm high. It is assumed that there is adequate bracing against buckling in the plane of the wall, as described above. The end loading is assumed to be nominally axial. Where there is significant eccentricity, reduced figures should be calculated.

©


26


Table 21 Permissible axial load per stud in kN — Timber strength class C16 (a) With wind loading: very short term load case Spacing Breadth mm mm 400

Depth mm

Wind pressure in Pa (N/rn2) 0

100

200

300

400

500

600

700

800

900

1000

38

72

11.03

10.03

9.08

8.08

382

1 69

0.00

000

000

0.00

38

89

1

17.60

16.41

1529

5.95 1421

13.18

12.20

11.25

10.29

8.16

38

97

19.24

18.08

16.97

15.91

1488

1388

6.03 11.99

38

114

22.54

1292 21.45

—

--

600

?

120

38

140

38 44

145 72

44_

97

44

120

44

145

47

72

47

97

47

120

A

12.77

11.76

13.64

47

145

38

72

3_

89

12.63

97

38

114

38

- 140

11.67

9.90

10.75

8.23

938

610

397

1.84

0.00

000

20.52

19.42

18.36

17.33

16 33

7.25

5 11

2.98

0.85

0.00

21.19

20.12

19.07

18.06

000

0.00

1.76

0.00

0.00 0.00

38

145

72

44

120

4._

12.77

4.88 13.69

1.69

000

000

1220

10.79

8.16

0.00 4.96

19.24

17.52

15.91

1438

12.92

11.53

9.82

6.62

3.43

21.45

1988

18.36

16.90

15.49

23.12

21.56

2005

18.59

0.00

0.00

0.00

0.00

0.00

16.33

14.89

13.50

12.16

9.01

25.37

23.85

0.00

9.90

7.17

3 97 19.42

13.64

12.15

10.75

8.31

5.11

1.92

0.00

000

0.00

0.00

21.19

19.59

18.06

16.59

15.17

1380

11.81 26.51

1p 145

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 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 Spacing mm

Breadth

400/600

38 38 38

72

38

114

38 38 38 44

120

44

97

44

120

44

145

©

0.78 17.84

145

97

47_ 47 NB

11.28

97__ --

72

—

0.00 1707

8.08 15.29

120

44 44

0.00 15.36

I

9.55 17.00

11.03

38

10.81

20.40 23.66

mm

Depth mm

89 97

140 145 72

47

72

47

97

47

120

47

145

Load duration

Iw I.

i eaiurn

I/ Iw 'iWa JLong

tk I


27


Table 22 Permissible axial load per stud in kN — Timber strength class C24 (a) With wind loading: very short-term load case Wind pressure in Pa (N/rn2)

mm

Depth mm

0

100

200

38

72

13.43

12.55

11.71

38

89

19.28

38 38

97 114

21.01

38

120

38

140

38

145

3O32 31.40

.40

44

72

15.55

14.67

44

97

22.91

44

120

44

145

28.34 34.

.25

47 47

72

16.61

15.73

14.88

14.06

97

23.72

47 47

120

29

145

35.46

38 38

72 89

13.43

12.13

10.91

19.28

8.17 18.19

38 38

97

21.01

.01

114

24

38

120

38 38

140

44 44

72

25.99 30.32 31.40 15.55 22.91

44 44 47

120

34

145

34.25

72

16.61

47

97

47

120

23.72 29.35 35.46

Spacing mm

Breadth

400

600

—

145

97

47

1300

10.91

500

600

700

800

900

1000

9.24

7.11

4.98

0.71

0.00

0.00

18.66

17.73

16.83

2.85 15.96

15.11

14.29

12.24

20.46

19.54

18.66

17.79

31.40 3.52

1.39

24 .32

40 13.00

13.82

12.04

9.91

.40

7.78

1.40

5.65

.25

.

34.25 4.92

.40 22.87

0.00 21.97

.25 2.79

0.66

13.26

11.31

9.18

7.05

4.98 16.83

1.78

0.00

0.00

15.53

14.29

0.00 11.17

7.98

0.00 4.78

20.00

18.66

17.36

16.12

14.66

.46

35.

31.

.46

0.00 1.58 11.47

23.60

1.40 14.24

.25 15.30

.

.32 31.40

31.40

13.00

10.98

7.78

34.

34

14.06

12.38

9.18

1.40 4.58

5.99

.40 1.39

0.00

22.87

21.53

0.00 20.24

2.79

0.00 23.63

0.00 22.32

.40 0.00

0.00

18.99

17.77

25

.25 0.00 19.81

145 .46 . .46 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.

NB

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

Breadth

mrn

rnm

Depth rnm

400/600

38 38

72 89

1114, 13.77

38

97

15.01

38 38 38 38

114

17.64

120 145

i4 21.66 . 22.

44 44

72

12.15

97

16.36

44 44 47

120

20.24

145

2

72

12.58

47 47 47

©

400

140

- 97

I.

Iivieaiurn

Long

,

,.

.

, r4c -vt 9,

,. 7.

rr

.. —

72

.

. t..

09

.

c

19

.

. ,

1

25.33


I

.1891

,

11

,

120 145

I

Load duration

.

, ,. ,, '. , ,

?7 .27

28


The tables show the permissible axial load per stud, checking combined bending and compression, bearing stress in the bottom rail assuming no wane, and deflection. It is assumed that the strength class of the top and bottom rails is the same as that of the studs. The bearing stress in the bottom rail is calculated using the BS 5268-2 bearing factor K.4, as this is applicable to the studs which are not at the ends of each panel, and these studs normally take more load. For bearing stress the load-sharing factor K8 has not been used because there is only one member involved. In the formulae for combined bending and compression and for deflection the effective length, Le, was calculated as 0.85L, as recommended by BS 5268-6.1. For the actual stud length, L, a value of 2324 mm = 2400 — (2 x 38) was used. The formula used to calculate deflection was:

Z = 0.005 Aca + öma X (öemean — öca)

where

A

=

Le 0.85L1if _______ I

h 2

a e,mean

andZ

mean A2

bh2

6 This gives the approximate deflection of a stud subjected to both compressive and bending forces. BS 5268-6.1 states that when calculating defiections the effects of axial loading should be ignored. However for some small studs with high wind loads ignoring the axial deflection can result in excessive deflections, hence axial load has been taken into account in these tables. A maximum deflection of 0.003L has been allowed. Where no value is given in the tables, this deflection is exceeded even without axial load. In particular situations a designer may choose to vary the deflection limit from 0.003L.

2.1.5 Design procedure The 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 nett design wind pressures on the walls. The method of obtaining these is given in Section 1.2.2.3. In determining the design wind pressure, the worst combination of external and internal pressure and suction should be used. When calculating the vertical load on a stud, half the weight of the wall panel supported by the stud should be included. BS 6399-1 Table 2 permits a reduction of 10(n-1) % in the axial load carried by a stud due to imposed floor loads where there are n storeys above it for a maximum value of n = 5. This reduction is particularly important in the lower floors of multi-storey buildings. For the very short-term load case the vertical component of the wind load should be included in the axial load on the wall stud. As shown in grey in the tables, the load per stud is governed by the permissible stress in the bottom rail, except in the case of smaller studs under very short-term loading. Therefore in many cases the vertical load-carrying capacity of a timber frame wall panel can be increased by using higher strength timber or a structural timber composite for the top and bottom rails. Calculations for a stud wall supporting a vertical load and horizontal wind load are given in the example of calculations for a complete dwelling in Section 7.5.

©


29


2.1.6 Drilling of studs Unless otherwise justified by calculation, drilling of studs should conform to the following requirements:

Stud

Location

Load bearing

Non-load bearing partition

Maximum size

Centre line of section. Between 150mm from one end and 0.25 of the length measured from the same end. Holes to be spaced (centre to centre) at a minimum of 4 hole diameters. Centre line of section. Between 150 mm from one end and 0.40 of the length measured from the same end. Holes to be spaced (centre to centre) at a minimum of 4 hole diameters.

0.25 x depth of stud.

0.25 x depth of stud.

Notching anywhere in the stud is prohibited.

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

II II

II II

Stud material from opening shared equally in framing each side

H Lintel: solid timber, structural timber composite or fitch beam.

ii

II

Ii

Lintel below first floor designed to support load from double stud to side of window above, including weight of construction from other studs.

H


30

H

[


2.2.1 Lintel capacities Tables 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 considered below, stronger types of beam may have to be built into the wall above the opening.

The load tables are for strength classes 016, 024, D40 and D50. Where the required span cannot be achieved in a softwood, it is now more common to specify LVL than a hardwood. However there are various makes of LVL which have different mechanical properties, so to obtain these properties the designer must refer to the appropriate manufacturer's certification literature (usually a BBA or BM TRADA Q-mark certificate) and calculate the required size accordingly. Where roof loads from similarly loaded trusses at no more than 600 mm centres are applied, it is generally adequate to convert them to a udl provided that the lintel is fastened to the top rail of the wall panel and the top rail of the wall panel is fastened to the wall plate with a minimum of two 2.85 mm diameter 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 to 10%. However, where larger point loads from cripple studs on the floor above are 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 the lintel (bearing stress). The total load on the lintel must be safely transferred to the foundation by means of cripple studs. For cripple studs, extra stud material may be provided by the principle illustrated in the previous diagram, and normally this will be adequate for bearing. A calculation for a cripple stud is 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 in the load tables. Bending, shear and deflection, have been calculated for the worst value of 'a' within the ranges shown for each case. A maximum initial deflection of 0.003L under dead and imposed load has been allowed. Separate tables are provided for medium term load, ie loads from roofs, and for long term load, ie loads from floors.

©


31


Loads at 600 mm centres

600

k

I

600 1200

1800

600 2400

1200

a 1 mm to 599 mm

Loads at 400 mm centres

_

A400

T

A

1200

00 4

A 1800

1800

4

a 1 mm to 399 mm


_rrrrr

T TA400

400

T

32

2400

2400


Table 2.3 Permissible loads on lintels: Timber strength class C16 — Long-term loading Point loads at 600 mm spacing Span

No of loads

1200

1

2

Span

No of loads

1800

2

3

Span

Noof loads

2400

3

Depth Breadth 72 97 120 72

9!

'1 21

202

2.26 3.04 3.7

1 64

120

145

201

120

1.50

2.7 3.35

Depth Breadth 72 97 120 72 97 120

72

97

120

45

0.33 0.45 0.55 0.29 0.39 0.48 72

0.79

1.47

2.41

1.07 1.32

1 98

0.13 0.18

72 97 120

0.21

10

195

220

245

3.26 4. 5 2

4.23 5.69

5.31 7.1

6.51

245

0.69 0.93

1 27

1.93

1 71

2.60

3.51

1.14 97

2 12 120

3.22

4.35

141)

70

0.33

0.61

0.44 0.54 0.30

0,82

8.77 .84

7.

105

220

245

(

1.4 'I 7j,,, 098i

0.50

0.57 0.76 0.94

72

97

0.72 0.97

189

120

1.21

28

72 97

0.63 0.84

120

1.04

26

Depth Breadth 72

72

97

0.20 0.27

0.48

0.

0.64

1.1 9,

1.91

0.33 0.20 0.27 0.33 72

0.80 0.48 0.64 0.80 97

lÀ 8

2.36

0. 38

1.38 1.86

0.41

13.01

171)

3.25 4.02

1.01

245

17.53 .69

1

1.21

0.22 0.13 0.17

21)

195

8

0.90

Depth Breadth 72 97

170

3

97

120

4

72

15 2.1

32j

I 'f

2.59J

120

145

170

195

220

245

'1

488

170

195

220

245

Point loads at 400 mm_spacing Span

Noof loads

1200

1

2

Span

Noof loads

1800

2

3

No of

Span

loads

2400

Depth Breadth 72 97

97 120 72 97 120 Depth Breadth

3

0.21

4

0.28 0.35 0.20 0.28 0.34

0.15

______

©

deflection governs ______


2

8

4

120

1.

0.39 0.53 0.66 0.38

145

811'ft91a48 a1J35 4.7 ')

1.4

2.30

120

145

2.58 3.19

3.34 4.13

4.20 5.19

1.86 2.51 3.11 175

2.26 3.04 3.77

2.55 3.43 4.25

5.91 2.84 3.83 4.73

1I5

220

24

1.08 1.46 1.80 1.05

0.51

0.68 0.92 1.14 0.66 0.89

0,64

1.10

1.75

1.41

= bending governs

1.41

1.90 2.35 1.39 1.87 2.32

1.77 2.38

2.17 2.92

2 95 1.75

3.61

2.35

2.14 2.88

2.91

3.56

— sh-?r .)c.virft

-


Table 2.4 Permissible loads on lintels: Timber strength class C16 — Medium-term loading Point loads at 600 mm spacing No of 72 Span Depth loads 1200

1

2

Span

loads 1800

2

3

Span

No of loads

2400

72 97

1.21

3

120

2.02 0.90

145

170

220

195

245 13.01

1

.

17.53 1

1.21

1.50

Depth Breadth

72

97

120

145

170

72 97 120 72 97 120

0.33

0,79 1.07

1 47 1 98

1.32

245

2.41 3 4

26

0.45 0.55 0.29

1.27

0.39

0.69 0.93

0.48

1.14

3

Depth Breadth

72

97

2.12 120

72

0.33 0.44 0.54 0.30

0.61

1 05

1

0.82

1.42

2

120 72 97

0.13 0.18 0.22 0.13 0.17

1.01

1.76

2.

3.52

0.98

1.55

2.

1 32

120

0.21

0.50

0.57 0.76 0.94

2.09 2 59

2.7 3.42

97 4

120

1.64

72 97 120

Noof

97

Breadth

195

4

220

245

5.31

6.51

.1

8.77 10.84

1.71

0.41

70

145

1 64

195

220

2.11

3 4.42 57 3.47

246

3.25 4.38 5.42

4.

Point loads at 400 mm spacing Span

No of loads

1200

1

Depth Breadth 72 120

72 97 120

Span

Noof loads

1800

2

3

Span

No. of loads

2400

3

Depth Breadth 72

F

©

145!

170!

195!

220!

245!

97

120

145

17

195

220

245

0.20

1

1

0.48

0.89

97 ______

0.27

______

0.64

1.19

120

0.33

0.80

148

72

0.20

0.48

088

1.

2

1

1.

97

0.27

0.64

1.19

0.33

0.80

1.47

Depth Breadth

72

97

120

145

17C

195

72

0.09

0.21

08

1.41

0.12 0.15 0.08

0.28 0.35 0.20 0.28 0.34

0.39 0.53 0.66 0.38

0.68

97

16

1.90

1.14

80 1.05

2.35

72 97 120

0,11

0.14

= deflection governs


j

I

3.1

.91

120

120

4

120!

0.72 0.97 121 0.63 0.84 1.04 72

97 2

97!

72!

2.

0.51

0.66 0.89

0,64

1.10

= bending governs

34

.11

-

1.41

[1

175

220

2.38 2.95

245

2.17 2.92 3.611

1.39 1.87

1.75

2.141

2.35

2.32

2.91

2.8 3.561

:hear qoverrv,


Table 2.5 Permissible loads on lintels: Timber strength grading class C24 — Long-term loading Point loads at 600 mm spacing No of Span Depth 72! loads 1200

72 97

1

120

2

72 97 120

No of

Span

loads

2

1800

3

No. of load

Span 2400

3

971

120!

4.78 6.44

1.51

3.20

2.03

4.31

2.51 1.12 1 50 1 86

5

7

120

72 _____ 0.41 _____ 0.99 97 _____ 0.55 _____ 1.33 120 0.68 1.64 72 ______ 0,36 ______ 0.85 97 ______ 0.48 ______ 1.15 120 0.59 1.42 72 97 Depth Breadth

1.83 2.46 _____ 3.04 1.58

2

1800

3

No. of

Span

loads

2400

3

©

245

13

4.61

422

6

522

7

5.

145

95

175 I

220 I

97

120

'.44

4.03 4.98

245 I

0.63

2.11

3.

2,84

4

120

1.50

3.51

72 97

0.78

5 06 6.26

4.60 6.20 7.67

93

2.66 3. 3,

21

4.43

145

170

2 59 3. 4 4,50 5.00

5.57

19.

20

245

170

95

220

245

220

45

2.50 3.37 4.17

3.07 4.13

1.05 1.29

Depth Breadth 72 97

7.

97

120

45

0.25 0.33

120

0.41

1.10 1.48 1.83

1 89 2.54 3.14

72 97

0.25 0.33

120

0.41 72

0.59 0.80 0.99 0.59 0.80 0.99 97

145

75

105

0.11

026 U9 255

1..4

98

035

066

14

1,81

' 67

0.44

0;2

2 24

.3.30

047

0.14

0.25 0.34

1.42 0.82

0.64

1.11

0.17

0.42

0.79

1,37

Depth Breadth 72 97 72 97 120

______

2 63 120

1.21

120

4

'20

2.12

0.90

120

loads

170

Breadth

72 97

No. of

I'S

2 78

Point loads at 400 mm spacing No. of 72 Span Depth

Span

ic5

11

2 07[ 2.99 3.76

0.51

120

6 9.21

1.76 2.18 1.22 -. 1.64 2.03

0.26

72 97

2

245

1.31

0.21

120

1

2201

1.02 1.26 0.70 0.95 1.17

0.40 0.54 0.67 0.38

1200

195!

0.76

0.17 0.23 0.28 0.16

loads

170!

1

07

Depth Breadth

72 97

4

145!

Breadth

0.15 0.18 0.10

1.48 1.83 120

= bending governs

5.11

1.30

1 75 '



110

17

= shear governs


Table 2.6 Permissible loads on lintels: Timber strength class C24 — Medium-term loading Point loads at 600 mm spacing

No.of

Span

loads

1200

1

2

72

1.51

97

2.03

120

2.51

72

1.1 2

97

No. of loads 2

Span 1800

4

4.7 6.44

9.21

220

245

220

245

'f.5

220

245

195 11.

7.

12.

16.1

11.

1

97

120

145

170

19

72

0.41

0.99

183

3 13

4.61

1.33

2.46

4 22 5.22

6.22 7

5.98 8

0.55 0.68 0.36 0.48 0.59

1.64

304

0.85

1 58

1,15 1.42

212 263

Depth Breadth

72

97

120

145

' 73

72

0.17

0.40

0.76

1.31

97

0.54

1.02

1.76

2.07 2

120

0.23 0.28

0.67

1.26

2.18

3.

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

97

120

145

170

195

220

245

145

170

195

220

245

170

195

220

245

1 'j8

2.50 3.37 4.17

3.07 4.13

1 90 2 14

2 39 3.22 3.98

97

3

1

170

72

120

2400

145

1.L

72

No. of loads

120

Depth Breadth

97

Span

97

120

120

3

72

Depth Breadth

9.


No. of oads

1200

1

72 97

2

No,of

Span

loads

2

1800

3

No.of

Span

loads

3

2400

120

1.50

72

0.78

97

135

120

1,29

3. 51

72

97

120

72

0.25

1.10 1,48 1,83 1.10 1.48 1.83 120

97

0.25 0.33

0.59 0.80 0.99 0.59 0.80

120

0.41

0.99

97

0.33

120 72

0.41

72

Depth Breadth

97

1.89

2.54 3,14

145

0.49

0.85

0.66

1.14 ____

0.18

0.44

0.82

1.42

0 10 97 0.14 _____ _____

0 25

047

0 82

72

______ _____

3.

Depth Breadth

120

©

211 090 _______ 121 2.8

72 ______ 0,11 ______ 0.26 ______ 97 0.15 0.35 ______ ______ ______

4

______

72

Depth Breadth

120

0.17

0.64 _____ 1.11 0.34 _____ _____ 0.42



1.8

2

I 30[ 1 75

2 7 3 30'

2.56

2.89

2 ilL 317 3.57 = bending governs 1' — shr 0.79

1.37

36

5.11


Table 2.7 Permissible loads on lintels: Timber strength class D40 — Long-term loading Point loads at 600 mm_spacing No of loads

Span

1200

1

2

No of

Span

loads

2

1800

3

Noof

Span

loads 2400

3

97

145

120

170!

195!

2201

245

I

72

1.57

97

3.65 4.92

6.52

2.12

879

1072 1444

120

262

608

72

1.16

2.74

1087 496

786 829

19.941 25.04j 30.69 15.381 41.34 20.721 26.86j 2Ø4I $,323j 41 7$! 5115

17

97

1.57

3.69

6.69

1

120

1.94

4.56

8.27

3 81

Depth Breadth

72

97

120

145

110

15

220

245

72

0.43

103

1.90

3.26

5!J9

(.40

10.21

13 9

97

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

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

7.28

10.57

14.54

19.18

120

0.62

1.48

2.74

4.68

Depth Breadth

72

97

120

145

170

195

220

245

72

0.17

0.23 0.29 0.16 0.22 0.27

0.42 0,57 0.70 0.39 0.53

0.79

97

1.06

1.36 1.84

3.17 4.28

4,44 5.98

8.03

1.31

2.27

2.15 2.90 3.59

5.29

7.40

0.73 0.99

1.27

2.01

2.96

4.15

1.71

2.70

3.99

0.65

1.22

2.12

3.34

4.94

5.59 6.92

72

97

120

145

170

195

220

0.94

2.20 2.96

3,97

14.02

18.€

5.35

6.60 8.90

9.97

1.26

13.43

18.89

25.11

120

1.56

3.66

6.61

11.01

1'

2337 .-.. -. 31.1

72

0.81

1.89

3,41

5.E

97

1.09 1.35 72

2.55 3.15 97

4.59 5.68 120

7,61 9.41 14

1 1..,

195

220

245

0.26 0.35 0.43 0.26 0.35 0.43 72

0.62 0.83

1.15 1.54

3.06

4.46

4.13

6.01

1.03 0.62 0.83 1.03 97

1,91

1.97 2.65 3.28 1.96 2.63 3.26 145

5.11

7.43 4.42 5.96 7.37 195

6.15 8.28 10.25 6.09

8.13 10.96 13.56 8.04 10.83 13.40 245

0.11

0.27 0.37 0.46 0.26 0.36 0.44

0.51

120 72 97 120

4

72

Depth Breadth

5.96 9.94 5.58 7.52 9.31

Point loads at 400 mm spacing

Noof

Span

loads 1200

1

2

Depth Breadth 72 97

120

No of

Span

loads

2

1800

Depth Breadth 72 97 120

72 97

3

120

Noof

Span

loads 2400

3

4

Depth Breadth 72 97 120 72 97 120

______

©

0.15 0.19 0.11

0.15 0.18



1.14 1.54 1.90 120

0.69

0.85 0.49 0.67 0.82

0.88 1.19 1.47

0.86 1.15 1.43

= bending governs

3.05 4.10 5.08 170 1.40

1.88

2.33 1.35 1.82 2.26

2.06 2.78 3.44 2.00 2.69 3.33

8.21

10.15 220

2.89 3.89 4.81

2.80 3.77 4.67

245

3.88 5.22 6.46 3.77 5.07 6.28

= shear governs


Table 2.8 Permissible loads on lintels: Timber strength class D40 — Medium-term loading Point toads at 600 mm spacing 72

97

120

157 212 262 116 157 194

365 492

6.52

10.7

879

608 274

1087

1444 1786

496

8.2

3,69

1117

456

669 827

Depth Breadth

72

97

120

145

170

195

220

245

72 97

043 058

103 138

326

509

7.40

171

12.33

17.01

13.50 18.19 22.51

72 97

0.37 0.50 0.62 72

0.89 1,20 1.48 97

1.64

2.81

8.72

11.51

221

3.78

11.75

2.74

4.68

120

145

6.85 8.48 4.37 5.89 7.28 170

9.97

071

4,39 5,43

1021 1375

120

190 256 317

15.50 19.18 245

0.17 0.23

0.42 0.57 0.70 0.39 0.53 0.65

0.79 1.06

1.36 1.84

1.31

2,27

0.73 0.99

1,27

1.22

2.12

Span 1200

1

Breadth 72 97 120

2

72 97 120

Span

Noof loads

1800

2

3

120

Span

Noof loads

2400

3

4

Depth Breadth 72 97 120 72 97

0.29 0.16 0.22

120

0.27

1381

1,71


No. of loads

1200

1

2

No of loads

1800

2

120

145

170

195

72 97 120

0.94

2.20 2.96 3.66

3.97 5.35

6.60

9.97

1.26 1.56

14.02 18.89 23.37

72

0.81

1.89

3.41

5.65

97

1.09 1.35 72

2.55 3.15 97

4,59 5.68 120

7,61 9.41 145

0.26 0.35 0.43 0.26

0.62 0.83 1.03 0.62 0.83 1.03 97

1.15 1.54 1,91

0.27 0.37 0.46 0.26 0.36 0.44

0.51

0.88 1.19 1.47

Depth Breadth 72

72 97

120

2400

Noof

Depth

loads

Breadth

3

0.35 0.43 72

72

0.11

97

0.15 0,19

120

4

72 97 120

0,11

0,15 0,18

1T11 = deflection governs

©

5.29 2,96 3.99 4.94

97

120

Span

428

72

97

3

:

3.17

Depth Breadth

120 Span

2.15 2.90 3.59 2.01 2.70 3.34

6.34 8.54 10.57 195


8.

6.61

11.(

14.54

220 4.44 5.98 7.40

5.96

4.15

5.58

5.59 6.92

7.52

8.03 9.94

9.31

18t 25.1 31.1

170

195

220

245

1.97

3.06

4.46

8.13

2.65

4.13

6.01

6.15 8.28

3.28

5.11

7,43

10.25

1.14

1,96

4.42 5.96 7.37 195

8.04

2.63 3.26 145

3.05 4.10 5.08 170

6.09

1.54 1.90 120

8.21

10,83 13,40 245

1.40 1.88 2.33 1.35 1,82

2.06 2.78 3.44 2.00 2.69 3.33

0.69

0.85 0.49 0.67 0.82

0.86 1.15 1.43

= bending governs

38

2.26

10.15 220 2.89 3.89 4.81

2.80 3.77 4.67

10,96 13.56

3.88 5.22 6.46

3.77 5.07 6.28

= shear governs


Table 2.9 Permissible loads on lintels: Timber strength class D50 — Long-term loading Point loads at 600 mm_spacing Span

No of

loads 1200

1

Depth Breadth

72

97

72 97

2.64 3.56 4.40 2.63 3.26

6 13 8.26 10.22 4.60 6.20 7.66

72

97

120

2

72 97 120

1.95

120

145 14.

13.

1 .64

1123 1389 '20

l4

1800

2

72 0 72 1 72 3 19 97 0,97 2.32 4.30 120 1.20 2.87 5.32 913 4 /2 72 0.62 1.49 2.76 97 0.84 2.01 3.72 6.36 7.87 120 1.04 2.49 4.60 45 72 97 120 Depth Breadth 72 0.29 0.71 1.32 2.20 97 0.39 0.95 1.78 3,09 120 0.49 1.18 2.20 3.82' 2.13 72 0.27 0.66 1.23 97 0.37 0.89 1.66 2.87 120 0.45 1.10 2.05 3.55

loads

2400

3

4

245 39.28 .92

.7

Depth Breadth

Noof

220

16.99 8.34

No of loads

Span

19

9

10.

Span

3

170

/i

"

95

1 73

-— 1243 16 75 20.72 10.21 13.75 17.01

14.24 7.34 9.89 12,2 t

720

16 03 21.59 2671

i7 ' 221 ' 'i

3

$1

/ 18

1 )3 4.54 5.62

8.89 4.98 6.70 8.29

145

170

195

14.73

14.77 19.89 24.61

3/

245

1964 .46 7

245

7.46

9.82

10 05 12 43

13.23 16,37

6.93 9.40

'26'

38

11.63

6—

220

245


Noof loads

1200

1

72 97 120

2

72 97 120

Span

Noof loads

1800

2

Depth Breadth 72 97

loads 2400

3

©

2.62 1.36 1.83 2.26 72

6.15 3.18

4.28 5.29 97

6.(

8.98f 11.11 5.72 7.7 1 120

145

170

195

220

3.30 4.45 5.50 3.29 4.43

5.1E'

7.49 10.09 12.48 7.29 9.82

941

11.531

12.67 15.68

15.53

8.37

9.32

5.48 145

8.512.15 170

195

220

245

1.49

2.35 3.16

3.46 4.67 5.77 3.36 4.53 5.60

4.85 6.53 8.08

6.51

245j .

2.59

1.73

3.21

1.04

1.92

1.40

Depth Breadth

72

97

2.58 3.20 120

72

0.46 0.62

72

0.19 0.26 0.32 0.18

97

0.25

120

0.31

72 97

18.22

19.14 23.23 25.87 25.78 31.30 3486 31.90 3872 43.12

7.67 8.99 10.31 11.83 12.95 10.33 12.11 13.89 15.61 1745 9.0. A 12.78 14.98 17.18 19.38 21.59

1.40

120

______

3.69

4.97

1.93

97 4

1.57

2.12

120

1.04

120

Noof

Span

97

0.43 0.58 0.72 0.43 0.58 0.72

120

3

72

Depth Breadth

1.73

0.76

0.86 1.15 1.43

0.44

0.83

0.60 0.74

1.12 1.38



2.00 2.48 1.44 1.94

2.40

= bending governs

6.9 8. 58 5.1

6.89

3.91

2.28 3.07 3.79

19.21

11.28 12.56

la.95 15.54

4.71

6.34 7.84

8.78 10.86

6.33 8.53

10.55 = shear governs

cC


Table 2.10 Permissible loads on lintels: Timber strength class D50 Medium-term loading Point loads at 600 mm spacing No of 72 Span Depth loads

1200

1

2

Span

Noof loads

1800

2

Span 2400

10.19

72

1.95

460

8.34

97

2.63 3.26 72

6.20 7.66

11.23

120

97

120

145

1.72

3.19 4.30

5.48 7.38 9.13 4.72 6.36 7.87 145

Depth Breadth

0.72 0.97

120

1.20

72

97

0.62 0.84

120

1.04

Noof

Depth

72

loads

Breadth 72 97 120

4

613 826

1db

10.22

72

3

120

2,64 3.56 4.40

97 3

97

Breadth 72 97 120

72 97 120

0.29 0.39 0.49 0.27 0.37 0.45

2.32 2.87 1.49 2.01

170

195

220

245

14

.92

1

1389

5.32 2.76 3.72

2.49 97

4.60

0.71

1.32 1.78 2.20 1.23 1.66

120

95

220

170

195

220

3.62 4.87 6.03 3.37 5.62

5.33 7.18 8.89 4,98 6.70 8.29

9.40 11.63

195

220

170

74c

54

0.95 1.18 0.66 0.89 1.10

2.05

2.29 3.09 3.82 2.13 2.87 3.55

97

120

145

170

1.57

3.69

6.67

10.93

2.12 2.62

4.97 6.15 3.18 4.28 5.29 97

0 98 14.73

14.77 19.89 24.61

4.54

7.

ici

Fl 05l

12.-.13

6.98

2451

16.37j 9.38 12.64

1564

Point loads at 400 mm_spacing Span

Noof loads

1200

1

2

Span

Noof loads

1800

2

3

Span

No of

loads 2400

3

Depth Breadth 72 97 120 72 97 120 Depth Breadth 72 97 120 72

2.26 72

1.04 1.40 1.73

97

11.11

3.21 1,92

1.40

2.58

1.73

97

3.20 120

72

72 97

0.19 0.26 0.32 0.18 0.25

120

0.31

0.46 0.62 0.76 0.44 0.60 0.74

0.86 1.15 1.43 0.83 1.12 1.38

8.99 IO31 11,63 12.9 170

195

220

245

2.35 3.16

3.46 4.67

4.85 6.53

6.51

3.91

577 336

8.08

10.86

4.71

4.53 5.60

6.34 7.84

6.33 8.53

3.30 4.45 5.50 3.29 4.43 5.48 145

1.93

72


145

2.59

1.04

19.14 2323 25.87 25.78 3130 3486 3190 38r2 43.12

12.18 14.98 17.18 19.38 21

9.54 120

0.72

I

767

245

13 1211 13. 15.67 17.4

7.71

120

deflection governs

18.22

5.72

Depth Breadth

120

______

1.36 1.83

0.43 0.58 0.72 0.43 0.58

97 4

72

1.49 2.00 2.48 1.44 1.94

2.40

= bending governs

40

2.28 3.07 3.79

8.78

10.55

= 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 timber frame wall made of standard wall panels which are sheathed with one of a given list of materials nailed to the frame. To comply with the racking values in the Standard, the frame must consist of grade 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 to standards now withdrawn. All wood-based panels used for sheathing must now comply with the relevant technical specification requirements laid down for wall sheathing in BS EN 13986 Wood-based panels for use in construction. Characteristics, evaluation of conformity and marking. Further information is given in the TRADA Wood Information Sheets 0 - 5 Timber frame building: materials specification, 2/3-56 CE marking: Implications for timber products and 2/3 - 57 Specifying wood-based panels for structural use. Materials selected must be appropriate for use in Service Class 2 (BS 52682). OSB/3 is the most commonly used material. If gypsum plasterboard is assumed to make a structural contribution it must be manufactured in accordance with BS 1230-1: 1995 Gypsum plasterboard — Specification for plasterboard excluding materials submitted to secondary operations. BS 5268-6.1 gives a basic racking resistance in kN per metre length for a specified thickness of each sheathing material, fixed with nails of a specified diameter and spacing, in a 2.4 m square panel without openings or vertical load. The additional contribution of a secondary board, nailed independently of the primary board, is allowed for by another, lower, set of figures. The racking resistance thus obtained is modified as necessary by multiplying it by the following factors to obtain a value for the wall which should be similar to results 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, which accounts for interaction between the walls, floors and the stiffening effects of corners in the actual building.

The construction of separating walls, which consist of two or more layers of plasterboard nailed to one side of the timber frame, is laid down in detail in the standard, 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 the racking resistance of a building in a given direction is one third of the total racking resistance in that direction. No contribution to racking resistance may be assumed for glazed areas.

©


41

(A TECHNOLOGY((7


A worked example of the assessment of the racking strength of a house is given 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 in C16 timber framing and, in the case of plywood, on plywoods in group I as defined in Table 63 of BS 5268-2. Examination of this table shows that if C24 framing 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 than the generalised figures provided by the standard: and different materials and forms of construction for which figures are not given may be suitable for sheathing. In such cases manufacturers may produce more appropriate figures by specific tests described in Section 5 of the Standard. The figures thus obtained may be used only for panels which are identical in all respects to 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 from masonry veneer, provided that is connected to the timber frame with specified wall tying and its contribution does not exceed 25% of the racking resistance provided by the timber frame. Special calculations or tests will be required if the designer wishes to take into account any additional contribution to racking resistance 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 fixing sheathing. Eurocode 5 provides a method for calculating the strength of a stapled connection by means of which a designer could determine the equivalent nail diameter and hence obtain an appropriate value for K101 to determine the racking resistance of a stapled panel. However, there are different 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 in Eurocode 5. Hence it is safer to obtain the racking strength of stapled panels by testing rather than by calculation. Eurocode 5 recommends a minimum corrosion protection of Fe/Zn 12c or Z275 for staples used in Service Class 2.

2.3.2 Adhesively bonded panels Within the limits specified in BS 5268-6.1 the racking resistance of a standard wall panel may be increased by increasing the nail diameter, reducing the nail spacing or increasing the board thickness. If the required racking resistance cannot be achieved by these means, shear panels of very high rigidity and strength can be obtained by attaching a structural sheathing material to the framing with a suitable structural adhesive. The manufacture of such panels must be carried out under a fully approved factory production control quality assurance system to ensure consistency of bond strength and structural characteristics. (See TRADA Wood Information Sheet 2/3-31 Adhesively bonded timber connections.) The strength and stiffness of such panels should be 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 maximum applied panel shear stress may be calculated as: = l.SFa N/mm2 tL

where

Fa

= total applied shear load (N)

= thickness of panel product (mm) L

©


= length of panel (mm) 42


This should not exceed the permissible panel shear stress. For plywood the grade stress for panel shear is given in BS 5268-2 Tables 40 to 56. Generic values for the characteristic planar shear strength of OSB, particleboards and fibreboards are given in BS EN 12369-1 Wood-based panels. Characteristic values for structural design. OSB, particleboards and fibreboards. These may be converted to permissible stresses using the method given in BS 5268-2 Clause 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 more appropriate for the response of timber frame wall panels within a building to a gust of wind. 1.33 is the value of K36 given in BS 5268-2 Table 39 for mediumterm loading in Service Classes 1 and 2. For OSB/3 under medium-term loading 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 of the panel, the applied rolling shear stress in the glued joint is:

Ta

where

=

N/mm2 Lh

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 stress concentrations. A further reduction of 0.9 should be applied if the bonding pressure is applied only by nails or staples (Clause 6.10.1.4). These reductions are not required in designs to Eurocode 5 which provides little guidance on the design of adhesively bonded joints, preferring tests to determine their strength. For panel products other than plywood, the permissible planar shear stress maybe 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 testing as described in BS 5268-6. This involves the test method BS EN 594: 1996 Timber structures — Test methods — Racking strength and stiffness of timber frame wall panels. Structural insulated panel systems (SIPS) may be tested in accordance with ETAG 019: 2005 Pre-fabricated wood-based loadbearing stressed skin panels. As previously stated, the racking resistances obtained from tests may not be modified by factors K101, K102 or K103.

2.3.3 Overturning effects Whether adhesive bonding is used or not, a panel must be restrained against overturning 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

©



(a) Restraint forces applied at corners

V

F

(b) Restraint forces applied along base

w F

h

..

V

w

h

///// /// F

Rv

fh=F1L per unit length

U fvimax

L

Fh W

R _EflW

VL 2

f1max=

3'Fh W' per unit length

As shown in diagram (a) above, when the racking load is sufficient to cause rotation the reactions at the two bottom corners are in opposite directions. This means that if there is an adjacent panel on the windward side its connection to that panel can restrain it against uplift with a total restraint force of Fh/L + W12. For panels of similar length this is greater than the total restraint required against uplift, Fh/L - W/2. Hence no further restraint is necessary provided it can be shown that the connections between the panels are strong enough to transfer this force between the panels. For the end panels however, or for panels where further restraint is needed, other approaches should be considered. Diagram (a) shows restraint to overturning provided by a metal holding down strap at each of the bottom corners. With a total vertical load of W and a racking load of F the restraint strap must resist a force of

R

— Fh — W

L

2

A similar strap is needed at both ends to cater for a reversal in the wind direction. These are nailed, screwed or bolted to the wide face of the outer studs and anchored to the concrete foundation. The strength of the strap and its connection at each end should be calculated and exceed R. In addition the fasteners connecting the bottom rail to the binder or sole plate or foundation must provide a total lateral resistance of F, so their maximum spacing is FL 5

=—mm adm

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

- 3(Fh W —

per unit length as shown in diagram (b). BS 5268 does not provide a method to allow for the effects of combined axial and lateral loading, but Eurocode 5 states that for annular ringed shank and helically threaded nails, and by implication for wood screws, the requirement +

----

2

1 must be satisfied

1\Raxd) R1ad

where Faxd and Flad may be taken as the axial and lateral load respectively applied to each connection, and Rax,d and Rla,d may be taken as the permissible withdrawal and lateral load respectively for each connection. For nails and screws, the requirement in BS 5268 terms becomes 2

+

ax,ajm where

1

. facim)

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

1

Hence Smax

v,max 2 + fax,acim)

12 . acim)

With bolts it is sufficient to ensure that the resistance of each connection to the 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. This is because, with bolts of at least 8 mm diameter, failure in two-member laterally loaded softwood connections occurs in the timber rather than the bolt. It may be possible to transmit the overturning moment from the structural sheathing into the bottom rail without any restraining straps. In this case the shear connection between the sheathing and the framing resists the applied loads. These loads comprise (i) a horizontal load of fh = F/L per unit length, which produces in an adhesively bonded connection an applied shear stress of ta as previously calculated, and (ii) a maximum vertical force per unit length

of fvmax

=

3(Fh W

N/mm

—--——-—j

which produces an applied shear stress of

T where h

©


= v,max N/mm2 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

=mmm res,a

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

/2

Tresa = Ta +T

2

For the framing the permissible shear stress may be calculated from the formula given in BS 5268-2 Clause 6.10.1.3 as: = Tadm,ll(1_0.67510) Tadma

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 Z =tan —1—

Ta

For OSB the applied shear stress should be compared with the permissible planar shear stress. This can be obtained from the characteristic planar shear strength given in BS EN 12369-1 converted to a permissible stress using the method given in BS 5268-2. For OSB/3 under very short-term loading the permissible 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 strength parallel 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 be checked for an applied tensile stress of fv,max/t where t is the sheathing thickness, and a permissible stress calculated as the grade stress parallel or perpendicular to the grain as appropriate x 1.33 as above.

2.3.4 Deflection It is difficult to calculate the horizontal deflection of a timber frame wall because: • the connection between panels is not rigid and manufacturing tolerances permit small gaps between the abutting studs, so the panels act independently to some extent, but not entirely • the sheathing tends to bow along the compression diagonal, increasing the total deflection • fastener slip between the top and bottom rails of the panel and the sheathing 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 5268-6.1 can resist the applied racking loads then its deflection will not be excessive. One reason for this is that the racking resistance of wall panels calculated according to the Standard is based on tests which limit the amount by which a panel 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 of 2.1 — 2.7 m do not require a separate check for the buckling resistance of the sheathing.

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(R ci:13,


3 Floors 3.1 Joists and beams Span tables for floor, ceiling and roof joists are provided in TRADA's Span fables for solid timber members in floors, ceilings and roofs (excluding trussed rafter roofs) for dwellings. This provides important information about the design and installation of timber floors using softwood of Strength Classes C16 and C24. BS 8103-3 provides similar tables but the spans are a little lower in some cases because they were calculated for the former SC3 and SC4 Strength Classes. Finally suitable member sizes for any weight of floor and imposed load may be obtained quickly using TRADA's Timber Sizer software (visit the Software Toolbox at www.trada.co.uk). When the distance between load bearing walls becomes so great that it is not economical to span directly with solid softwood floor joists, it may be possible to use prefabricated engineered timber joists, hardwood, or a structural timber composite such as LVL (laminated veneer lumber), PSL (parallel strand lumber), 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 commonly available sizes are preferred, then intermediate floor beams may be used to support the joists. These may be deeper sections of hardwoods or structural timber composites, doubled engineered joists or steel flitched beams. Doubled engineered joists should be connected together in pairs by fixing web stiffeners 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, timber herringbone strutting, or proprietary metal herringbone strutting should be specified in accordance with the recommendations given in the publications cited above or see TRADA's Wood Information Sheet 1-41 Strutting in timber floors. For engineered joists the manufacturers may provide other recommendations or strutting methods, such as the strong-backs used with metal open web joists. Currently the National House-Building Council (NHBC) recommends a deflection limit of 12 mm for all domestic floors made without strutting, because tests have shown that properly installed strutting significantly improves the vibrational performance of a floor, which is an important part of structural design.

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48

TICHNOLOGV((5


3.1.1 Reactions, moments and deflections in a single span, simply supported beam 3.1.1.1 Uniformly distributed loads

Rot

t PA

w = load per unit length

Reaction at A

RA=

Reaction at B

w(b—a)(2L —a—b)

R6=

2L

Value of x

Bending moment at a distance x from RA

x
MX=RAX

a
M =RAx—

x> b

= RAX —

\ (RA RA Mmax =RA—+aJatx =—+a

w(b — a)(2x — a — b) 2

Bending deflection at centre

o— < a— < L/2

YbElWb2 L2—— 32 48

o— < a— < L12 b> L12

a> L/2 b> L12

b2

—w

15L

Yb —------———C

2 L2

L2

El

(L — a)2

2 L2

C2

32 48

El

Yb =

—a

32

—

Shear deflection at centre (1)

48

Ys

a2 2 L2—— —a

(L — a)2


3w 1 b2 —a2

a2

32 48 J 32 48

—2

L2

—

c2

32 48

For engineered timber joists, consult the manufacturer

©

2L

Maximum bending moment

w(x—a)2 2

Configuration

b
w(b2 —a2)

Ys

1OGA — —

3w 1 L2—2c2—2a2

2OGA

=

3w {(L — a)2 — c2

1OGA


3.1.1.2 Point loads L

II a

'F

I

C

tP

Pa

Reaction at A

Reaction at B Fa

Fc

RA=—

RB=-[-

Value of x

Bending moment at a distance x from RA

Maximum bending moment

x
M =RAX

M =RAa atx =a

x> a

M = RAx — F(x — a)

M = RAa at x = a

Configuration

Bending deflection at centre

Shear deflection at centre (1)

aL/2

Fa 1L2 a2l

Y=-

Fc IL2 c21

Ys=

3Fa

3Fc

Yb=l-

(1) For engineered timber joists, consult the manufacturer

3.1.2 Solid timber beams Arrangement of loads on effective span Long-term 2.35 kN Medium-term 1.83 kN

4.l8kN

Long-term 0.56 kN/m

Self-weight 0.23 kN/m Long-term 2.62 kN/m Medium-term 2.03 kN/m 4.88 kN/m

3.Om

1.Om

H RB=13.52 kN

RA=11.85 kN

This example outlines a design for a solid hardwood floor beam consisting of 2 No 63 x 250 mm nominal (60 x 245 mm) hardwood members in Strength Class 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 NA


— 0.56x3.O(2x4.Q—3.0) + 4.18x1.0 + 4.88x4.O 2x4.O 4.0 2 —

50

= 0.23 kN/m.

= 11.85 kN


— 0.56x3.02 — _______ + 4.18x3.0+4.88x4.0 4.0 2 2x4.0

RB

— —13.52kN

= 13.52 kN

Rmax

11.85 0.56 + 4.88

Zero shear stress occurs at Mmax

= 2.178 m from RA

= 12.92kNm

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

Stresses

= 12.92x106x6 2x60x245

Gm,a

G,adm

where

= 10.76 N/mm2

= 18.00 N/mm2

16.0X1.25X1.1XJ

1.25

= K3 (duration factor)

1.1

= K8 (load sharing factor) 0.11

= K7 (depth factor)

iSV 1.5x13.52x1000

Ta

=0.690 N/mm2

2x60x245

A

=2.2x1.1

Tadm

where 1.1

=2.42 N/mm2

= K8 (load sharing factor)

Deflection Use

Emin X K9 (Clause 2.10.11)

= 12600 x 1.06 G

= 13356 N/mm2

= E/16 (Clause2.7) = 13356/16

= 834.8 N/mm2

13356260245

El

=

GA

= 834.8 x 2 x 60 x 245

0.56 1.964x1012

=1.964x1012Nmm2

12

5x40004 384

10002

= 24.54 x 106 Nmm2

40002 10002 32

48

3x0.56

+

20x24.54x106

{40002_(2X10002)}

= 0.855mm

= 0.814+0.041

= 4.18x1000x1000 [40002— 10002] 3x4.l8xl000xl000 16 12 1.964x1012 5x24.54x106 =1.951+0.102

=

=2.053mm

4.88 54000 1.964x1012

384

3x4.88

j 20x24.54x106

{40002}

= 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.


51

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Bolting Assuming that the beam is loaded evenly on both sides then only nominal bolting will be required to ensure that both members act together. A single row of M10 bolts at not more than 600 mm centres along the centre line of the beam will be adequate. If the beam is predominantly loaded to one side, it can be reasoned that for a two-member beam, for which K8 = 1.1, the bolting should be capable of transferring approximately 55% of the load.

3.1.3 Structural timber composites Structural timber composites are made by gluing together flakes, strands or veneers of wood under pressure to create new structural materials. They make excellent floor beams which have greater strength, stiffness and dimensional stability than the solid timber from which they are made. The types currently available in the UK are LVL (laminated veneer lumber), PSL (parallel strand lumber) and LSL (laminated strand lumber). Design values for use 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 the literature should be read carefully, as some modification factors are different from those for solid timber and there may be special rules for mechanically fastened joints. Typical grade values for the principal properties are shown in Table 3.1, with values for C24 timber given for comparison. This shows that their main advantage over solid softwood is their superior strength, hence they are more suitable for extra heavy loads than for extra long spans. Further information about these products can be found in TRADA's Structural timber composites: DG1. For information on suppliers, visit the Suppliers Directory at www.trada.co.uk.

Table 3.1 Grade MOE and stresses for C24 softwood and some selected structural timber

composites in N/mm Material

Emean

m

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

c,perp

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

3.1.4 Prefabricated engineered timber joists Prefabricated engineered timber joists manufactured in the UK and other countries are being increasingly used in floor construction because of their very 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 Ijoists comprise a solid timber or structural timber composite flange and a panel product web. Timber composites used for flanges include laminated veneer lumber (LVL) and parallel strand lumber (PSL). Webs are usually oriented strand board (OSB) but may also be hardboard. The web is secured to each flange by an approved weatherproof, structural adhesive within the rout. Metal web joists comprise a solid timber or structural timber composite flange with V-shaped galvanised steel components forming the open web.

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

rECHN0

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52


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

Generally engineered timber joists are used instead of solid timber joists, but they can be used in pairs as beams. In the latter case the manufacturer's advice on fastening them together should be strictly followed. An engineered joist beam is most useful in a floor which is already made with engineered joists: 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 loft conversions and as lintels, or anywhere that an enhanced load-carrying capacity is required. It is recommended that the depth of the steel plate should be 25 mm less than that of the timber members, to allow for shrinkage of the timber or lack of straightness in the timber or steel. This is because the design method assumes that all the applied loads and the reactions at the supports will be applied to the two timber members. A proportion of these loads is transferred to the steel plate via the bolts along the beam and out again via the 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

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3.1.5.2 Dimensions Structural span

L

Length of bearings

Lbearing = 56 mm

Diameterofbolts

d

Breadth

Depth

b mm

h mm

Timber

44

220

Area (total) A mm2 19.36 x i03

Steel

10

195

1950

4 metres

= 10mm Section modulus (total)

Second moment of area (total)

Z mm3

I mm4

0.7099 x 106

78.09 x 106

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 MOE E N/mm2

205000

Shear modulus G N/mm2 78850

Density p kg/m3

7850

Permissible stresses* in N/mm Ym,ads,s

tadm,s

Yc,adm,s

180

110

250

3.1.5.4 Loads, reactions and bending moment For simplicity this example will use the same span, loads and medium-term load 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 additional long-term UDL) = 0.23 kN/m [(2x44x220x420)+(lOxi95x7850)]x9.812x109

As before

RB

=11.85kN =13.52kN

Mmax

= 12.92 kNm

RA

and

3.1.4.5 Load distribution factors The proportions of load taken by the timber, k, and by the steel, k, depend mainly 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

El

= 205000 x 6.179 x 106

= 1.267 x 1012 Nmm2

For steel:

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The proportion of the total load carried by either the two timber members or the steel plate can be represented by one of the following three factors which allow 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

0.8433 Emeanlt = Emeanlt + El 0.8433 + 1.267

= 0.3997

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

= 1.1xEl5 = 1.lxl.267

k5

= 0.7304

EminIt + El 0.6409 + 1.267

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

k 3.1.5.6

=

Eminlt

0.6409

Eminit + EI

0.6409 + 1.267

= 0.3360

Effects of holes Assuming Ml 0 bolts with a hole diameter d' of 11 mm staggered with an offset of h/4 55 mm from the centre line, the bending stress increase factor due to one hole h2(h2 — hd' + 2d'y) 3 3 4

—

kbending —

where

h

h —h d'—hd' +d' —l2hd'y2

= depth of the members = distance between the centre of the hole and the centre line of the beam.

y For the timber:

22o2(2?o2_(22ox11)+(211x55))

kbending,t =

3

3

4

= 1069

2

220 —(220 xll)_220x11 +11 _(12x220x11x55 For the steel plate:

kbendng,s =

1952(1952 _(195x11)+(2x11x55)) 4 195 —(1953 xll)_195x11 +113 _(12x195x11x55

= 1 096 2

Provided that the bolts at the reaction points are above the bearing plates and

there are no other bolts within a distance of h from the reaction points, it should be unnecessary to consider a reduced section when checking the shear strength of the timber. The shear strength of the steel will normally be adequate.

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(R


3.1.5.7 Timber stresses K3

= 1.25

K6

( =I--j 220)

K8

= 1.1 to check bearing strength of stringers

\O.11

=1.035

An investigation into the load distributions between the timber and steel produced by variations in the timber's strength and stiffness has shown that in fitch beams with two timber members the grade bending and shear stresses should be enhanced by only 1.04, not 1.1 as in clause 2.10.11, and only if 0.2Eltotai EsIs where

0.8Eltotai

Eltotai

= Emeanlt + E515

K8A

= 1.04 to check bending and shear strength of stringers

Bending 0m,a

= ktkbending,tMmax

zt = 0.3997x1.069x12.92x106 0.7099 106

= 7.772 N/mm2

Gmadm = amXK3XK6XK8A

= 7.5x1.25x1.035x1.04

= 10.09 N/mm2

Shear a

— —1.5ktRmax ______

At

= 1.5x0.3997x13.52x103

19.36 io tadm

= 0.419 N/mm2

= txK3 xK8A

= O.71x1.25x1.04

= 0.923 N/mm2

= Rmax

= 13.52x i03

= 2.744 N/mm2

Compression Gc9Q,a

2btLbearing

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) 16


56

513N/mm2

_J\__


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

k EminIt

X

O.56[5>4000 _10002 40002 10002 L4i8oxiooo4000 1ooo24885x4ooo4 32

384

48

L

0.3360

=

0.6409 1O12

384

12

L 16

J

J

x21.70x1012

11.37mm

The total shear deflection at the centre =

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

20

5

0.3360

=

x 15.40 x 10

20

6

= 0.52 mm

513x19.36x103

Total deflection

= 11.37 + 0.52

Permissible deflection

= 0.003L = 0.003 x 4000

= 11.89 mm

= 12.00 mm

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

and

ufflG

=

uno

=

(l+kdef) + M12,ikdef)

(1 +W2lkdef) +

i>I where

UinstG

= instantaneous deflection caused by permanent loads (generally self-weight and dead weight) = instantaneous deflection caused by the leading variable load, ie the one which produces the greatest deflection in combination 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. characteristic load combination factor for variable load i taken from Annex K Table Ki.

wo

= quasi-permanent load combination factor for variable load 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 mintin

—

— Lmin

(ujflstG +ujto) (uflflG + UfIQ)

where

= uflStQl + i> I

©


(A


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 done simply. Using Emin as in this example produces an approximate conservative result, giving slightly larger stresses in the steel, loading in the bolts and final deflections than would be obtained using a more exact method.

The final value of Emn is then used to check the final deflection. For final deflections, Annex K recommends a relaxed deflection limit of L/200. In this example the following results are obtained:

Load

0.56 _________ kN/m Type Storage

'V2 UjnstG

2.35 kN

1.83

0.23

kN

kN/m

2.62 kN/m

2.03 kN/m

Permanent Snow

Permanent

Permanent Residential

1.0

-

0.6

-

-

0.7

0.8 -

-

-

-

1.18

0.0 -

0.42

0.88

-

0.92

-

4.79 -

0.3 3.71*

* leading variable load

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

= 6.38mm = 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)

uIflSG

= 1.18 + 0.42 + 4.79

= 4.378 + 1.302 + 0.552

Hence Ef = 8208 x

= 6.23 mm

(6.38 + 5.14)

= 5752 N/mm2

(10.22+6.23) Substituting this value for Emin = 0.4492 x 1012 Nmm2

Emin,rin It = 5752 x 78.09 x 106 Eminfinlt — ___________ — — —

0.4492

Eminnlt + El 0.4492 + 1.267

—

—0.2167

Final deflection corrected for Emjn fl and k'fl kufinEmin = ufflG+urflax I I

kuEminn

= (10.22 +6.23)x

0.2617x8208 0.3360 x 5752

= 18.28mm The permissible final deflection

= 4000 200

= 20 mm

3.1 .5.10 Steel stresses

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

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

t


44

58


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

= 0.4492 x 10 Nmm2

Emjn,fjnlt = 5752 x 78.09 x 106

1.1xEl

kffl

=

1.lxl.267

Emin6n I + El 0.4492 + 1.267

0m,a,s,fin =

=0.8121

ksfinkbendingsMmax .7

S

0.8121x1.096x12.92x10 ___________________ 63.38 x i0

= 181.4 N/mm2 = 180.0 N/mm2

Gm,adm,s

In practice the simplest solution to the small over-stress would be to increase the 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 shear strength are normally adequate, but in any doubt these should be checked after 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 into the steel, it is assumed that the partial UDL of 0.56 kN/m extends the full length of the beam. The bolt spacing can then be constant. Therefore the total UDL 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 load perpendicular to the grain per shear plane for an M10 bolt by interpolation is 1.42 + 2.63 2

= 2.025 kN

Therefore the permissible bolt load per bolt

No of UDL bolts required =

17.52 4.050

= 4.050 kN =

=5

In order to ensure that the timber and steel remain together a maximum recommended bolt spacing of 600 mm or 2.5h (whichever is less) = 550 mm.

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Therefore a minimum of

4.Ox 1000

550

6.27

—1

= 7 bolts are required spaced at

4000

= 500 mm apart

7+1 No of bolts required at each reaction: — 10.89 —

= 2.69

4.050

= 3 required

A minimum of 2 bolts is recommended at the top and bottom of each support to maintain stability. No of point load bolts required: 3.365 = _____ 4.050

= 0.83

= 1 required

3.1.512 Arrangement of bolts 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 standard timber dimension of, say, 72 mm might be used for each bearing.

Arrangement of bolts 40 mm.

40 mm.

70 70

- 4-

Additional bolt on centre line for point load

- 4—

72

72

The beam should be clearly marked to indicate at which end the additional bolt for the point load should be positioned. Spacings between bolt holes

should be at least 4d 40 mm. Minimum end and edge distances in the timber should be 4d and in the steel 2d.

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

Stringers

Bearers

Steel plate

Ym,t

tt

Yc,9O,t

c,9O,bearing

Ym,s

Initial

0.77

0.45

0.83

1.00

0.91

Final

0.63

0.43

0.83

1.00

1.00

Beam

c,s

6

0.08

0.13

0.99

0.09

0.14

0.66

1 bearing stress beneath bolts at supports.

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

TRADA ©


60


3.1.5.14 Joist support details There are a number of methods of supporting joists on a flitch beam, some of which are shown below.

Joist hanger

Steel angle hanger

Joist on beam — reduced height

Joist on beam

Key: 2 3 4 5 6 7

Joist Solid timber beam Steel fitch plate Metal joist hanger Metal angle hanger 50 mm x 50 timber ledger nailed to beam Fastenings

Timber ledger 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.

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61


Table 3.2 Permissible long-term loads for solid timber ledger with predrilled nail holes Permissible load (kNIm) Nail spacing (mm) 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

7.12

7.56

-

80

8.90

9.45

Notes a) Although BS 5268 permits a loaded edge distance of only Sd (5 nail diameters), Eurocode 5 specifies 7d for nails with predrilled holes and 1 Od for nails without predrilled holes. Because of the possibility of splitting a narrow ledger if predrilled holes are not used, these recommendations are based on a minimum edge distance in both ledger and beam of 7d with predrilled holes, as specified in Eurocode 5. b) The loads are for 3 mm diameter round wire nails with a minimum pointside penetration of 35 mm inserted in predrilled holes with a diameter of approximately 2 mm. c) For one row of nails the minimum size of ledger is 35 mm thick x 47 mm deep; for two rows the minimum finished size is 35 mm thick x 60 mm deep. d) Minimum edge distances of 21 mm must be maintained in the ledger and beam, and minimum spacing between two rows of nails should be 15 mm. e) The loads have been calculated from BS 5268-2 Annex G. Where joists are notched over the ledge the shear stress in the reduced f) section should be checked in accordance with Clause 2.10.4.

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62


3.3 Floor decking The following table provides some approximate values for spans achievable by softwood and wood-based panel products certified for use as floor decking. In a final design the permissible span should be calculated using strength and stiffness properties published in BS 5268 or by the manufacturer. All timber-based panel products used for floor decking must now comply with the relevant technical specification requirements laid down for flooring in BS EN 13986. Further information is given in TRADA WI Sheets 2/3- 56 and 2/3 — 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

63


4 Roofs Pitched roofs may be constructed using prefabricated trussed rafters; bolted, connectored or plated trusses with intermediate common rafters, or with traditional rafters and purlins made on site. Trussed rafters, made of standard components produced by an efficient factory process are widely used as an economical means of roof framing. When the roof space is required for storage or accommodation, athc trusses may be specified. Details of these and 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 roofs In timber framed houses, trussed rafters are nearly always used; BS 5268-3 Code of practice for trussed rafter roofs defines established methods of design, materials, functional requirements, fabrication, handling, storage and erection. The European harmonised Standard BS EN 14250 Timber structures — Product requirements for prefabricated structural members assembled with punched metal plate fasteners: 2004 specifies requirements with which all trussed rafters must comply in order to satisfy the Construction Products Regulations. It includes minimum dimensions for the members and additional limitations on imperfections in the graded timber. The fasteners themselves must comply with BS EN 14545 Timber structures — Connections — Requirements, when this is published.

4.1.1 Bracing It is important that trussed rafters are correctly braced. Attention is drawn to Sections 7 and 11 of BS 5268-3, where it is made clear that the 'building designer' is responsible for the stability of the roof structure as a whole. He should ensure that the details of the bracing are adequate to ensure the overall stability of the complete roof structure and the associated supporting walls. The under-rafter bracing has the combined duty of providing restraint against lateral buckling due to compression forces in the members and of transmitting the wind load from the gable end walls to the front and rear walls. BS 5268-3 gives a range of spans and pitches for which standard bracing is adequate at different wind speeds. It should be noted that the bracing system (shown in the diagram), including the longitudinal runners, should be carried throughout the length of the entire roof, but not over separating walls. Rafter bracing is not required in hip ends, nor in the length of roof between hips if this is 1.8 m or less. Specified sarking materials nailed on top of the rafters may be substituted for standard bracing. Domestic dwellings with ceilings made of plasterboard or similar rigid material do not require bracing in the plane of the ceiling joists, providing that they conform in every respect with the criteria set out in Appendix A of BS 5268-3.

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64


Longitudinal runner at apex 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. 1

2

The roof and ceiling system provides a horizontal girder which transmits wind load from the face of the building to the cross-walls. For houses which do not conform to the criteria in Appendix A (for example, houses built on flat, open land corresponding to ground roughness category 1, or having a combination of span, pitch and wind which puts them outside the scope of the standard bracing solutions), the bracing requirements for the roof and ceiling must be calculated. In the plane of the rafters, moisture resistant plywood, P5 particleboard to BS EN 312, or OSB/3 to BS EN 300 may provide adequate bracing; values for their strength and rigidity can be obtained from BS 5268-2, or from manufacturers. In the ceiling plane, well-fixed plasterboard will make a contribution: conservatively 12 mm plasterboard could be taken as equivalent to 6 mm thick plywood. For solid timber bracing the following design procedure is recommended.

The diagram below shows the ceiling plane only, with bracing nailed to the top of 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 m front-to-rear dimension. For a building measuring 7.2 m from front to rear, the diagram 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 of 9.6 m. The total wind load will be taken as 0.746 kN/m2 acting on a height of 2.68 m from 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 kN or 1.2 kN per 600 mm interval. This is divided by 2 because there are two sets of bracing, giving the point loads of 600 N in the diagram. Each of these is shared among four braces, so the point loads on each brace will be 150 N as shown below.

©


65


600 600 600 600 600 600 600

/ /

/\ /

150

106

300 150

150 106

/ ——

/ /

0 c'J

——

/ /

583

300

1 '1

0 (0 c3)

>1

525

605

/

/

/ /

7

,525

L

300 106

106

106

1

E—8501 '1 3400

1<

1159 >1

For 25 x 100 C16 bracing: (A = 2500 mm2; Z = 41670 mm3; i,= 28.87 mm) Max BM = 180,200 Nmm

0m,a

0c,a

M Z

=

41670

583 2500

=

100

3400 28.87

=

= 4.32 N/mm2

= 0.233 N/mm2

5.3 x 1.75 xl

0m,adm =

Take

— — 180,200

= 10.47 N/mm2

= 118

For very short term load K12

=

0.1972

Gcadm = ae

6.8 x 1.75 x 0.1972 rr2E

=

— 9.87x5800 — (118)2

[LJ2

©


66

= 2.35 N/mm2

=4.11 N/mm2


0c.a

Gma

K Gcadm

Gmadm[1_'

)

4.32

+

1.5x0.233 4.11

10471

0.233

2.35

x0.1972 = 0.519

= 0.420 + 0.099

For brace in tension

(300h1

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

3.2 xl .75 xl

I

= 6.32

100) 4.32

=

10.47

+

0.233

6.32

= 0.413 + 0.037

= 0.450 < 1.0 as required.

4.2 Connections

1

2 Nailed plywood or OSB gussets.

Circular or square toothed plate connectors used with bolts in lap-jointed members.

3 Proprietary pierced plates attached by tightly-

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

fitting nails.

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

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67


5 Foundations 5.1 Loading Total loads on foundations for the external walls of two storey timber frame houses vary between 6.3 and 30 kNIm run. The types of foundation commonly employed are illustrated in the TRADA publication Timber frame construction. Loads on the foundations of timber frame houses with timber claddings are lighter than those of masonry construction and, therefore, have distinct advantages where the soil bearing pressures are poor. Table 5.1 shows some typical foundation weights on the footings, depending on their size and shape.

Table 5.1 Foundation loads in kN/m

Exterior finish Load

©


Brick veneer

Tile hanging

Timber boarding

Dead

16-20

4.5—8.5

3.5—7.5

Dead and imposed

19-30

8-18

6.3-17

68


6 Multi-storey buildings 6.1 General design considerations The Timber Frame 2000 project, begun in 1995, was carried out by the Building Research Establishment and TRADA Technology Ltd in collaboration with the British Government and the timber frame industry. It demonstrated beyond doubt that conventional timber frame construction can be used to build economical, safe and serviceable multi-storey dwelling units. The report Multi-storey timber frame buildings — a design guide concluded that the use of BS 5268-6 can be extended to the design of platform frame buildings up to eight storeys without excessive deflection. Particular attention should be given to the following issues:

• • •

•

•

each storey should have sufficient strength and stiffness to resist a horizontal 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 unit independently or be transferred across the party walls where additional stiffness has to be provided, eg by the introduction of portal frames, the deflection limit should be appropriate for the structure and finishes, but may be no more than height/500 resistance to disproportionate collapse should be checked.

6.2 Construction The stability of the building during construction, ie before vertical loads are applied and plasterboard is fixed, must be considered as part of the design process. It is acceptable to reduce the wind load in accordance with BS 6399-2 for the construction process. Particular attention should be given to the buckling resistance of studs in party walls, which may have neither plasterboard nor sheathing attached during construction. Significant vertical loads can result from the storage of construction materials such as plasterboard packs, so normally it will be necessary to specify requirements for temporary bracing (see BS 5268-6.1).

6.3 Disproportionate collapse While the Timber Frame 2000 project demonstrated that timber frame construction is remarkably resilient to disproportionate collapse, specific design checks against this possibility may be required under the Building Regulations. Guidance is expected in a planned revision to BS 5268-2: 2002. Meanwhile, guidance has been published in the UK Timber Frame Association Technical Bulletin 3 Design guidance for disproportionate collapse. This specifies minimum nailing between the lower rails of wall panels through 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 buildings 3.1 mm diameter nails at 200 mm centres for Class 2B buildings.


69

TECIINOLOQV(t7


7 Example of calculations for a complete dwelling

3.9m

P

T8T1I! I

I

4

I

I

I

9.5m

Ground floor plan

First floor plan

_1L

to N-

r Rear elevation

Front elevation

Side elevation

©


70

\ TECHNOLOGY1


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 house layout 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 timber frame wall studs The wind loads described in section 1.2.2, are resisted by the horizontal diaphragms and timber frame walls; however a proportion of these loads can be transferred back into the brickwork via the wall ties in accordance with BS 5268-6.1 Clause 4.10 The wind loads have been previously calculated in accordance with BS 6339-2 and section 1.2.2.

71

YECMNOLOOY(


7.1 Vertical loads Roof (350 pitch):

Concrete nterlocking tiles 1

49

kg/rn2

3

Battens

Felt 53

cos35° Trusses

53

kg/rn2

65

kg/rn2 on plan

16

kg/rn2

200 mm insulation quilt

2


10

94

So total dead load

0.92

imposed load 2

0.67

ceiling imposed load

0.25

18mm OSB deck

Floor:

47 x 194mm C16 joists at 400 mm centres

1.84

kN/m2

12

kg/rn2

9 10

12. 5 mm plasterboard

140 mm phenolic foam board

4 35

So dead load

1.50

Framing 47 x 97 mm at 600 mm centres 12.5 mm plasterboard

6

kg/rn2

6

Insulation quilt

3

Framing 38 x 63 rnrn at 600 mm centres

So dead load

25

kg/rn2

0.24

kN/rn2

4

kg/rn2

20

24

kg/rn2

0.24

kN/rn2

Concrete tiles can weigh from 42 to 58 kg/rn2

2 BS 6399-3: 1988 Clause 4.3.2 BS6399-1: 1996 Clause 5.2 BS 6399-1: 1996 Clause 5.1.1

©

kN/rn2

9.0 mm OSB sheathing

2/12.5 rnrn plasterboard

1

1.84

10

So dead load Partitions:

kg/rn2

0.34

imposed load

External walls:

kg/rn2


72

(0.64 x 1.25) (60 - 35) / 30

= 0.67)


7.2 Horizontal loads Building dimensions 9.5 m

Overall length (I) =

Overall width (w)

7.0 m 5.0 m

Height to eaves (h)

=

Height to ridge (H)

7.5 m

A dynamic wind pressure of q = 0.800 kN/m2 for both directions is assumed to have been calculated from BS 6399-2. As stated in Section 1.2.4 this may be reduced for a maximum 1 year construction period to

q=

= 0.449 kN/m2.

0.800 x 0.56 1

When timber frame walls are shielded from the wind by masonry walls, the load transferred to them is reduced by a factor K100 which depends on the proportion of the loaded wall occupied by openings. (In other types of construction the full wind load is taken by the timber frame walls, so K100 = 1.0) Values of K100 are given in Table 1 of BS 5268-6.1 and are applied to an entire wall measured to the eaves. K100 is not applied to the spandrel in a gable wall, which is not selfsupporting. For the dimensions given in the specification a table is made of the values of P, where =

area of openings x 100 area 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 spandrel

©

Front wall

Rear wall

Gable wall,

K100

1.0

0.81

0.82

door end 0.48

q x K100

0.800

0.648

0.656

0.384


Gable wall, other end

0.45 0.360

(R


BS 6399-2 Tables 5, 10 and 16 give the following external and internal pressure coefficients for the house illustrated. c0= -0.5 9.5

I' 2.55

-0.4

Cpe 0.83 Cpe.nlex

-0.7

0.951

0.8

Wind

__________

7.0

-1

2

1.75

.1

1

.1

1

-

6.0

-0.6

-0.5

-0.6 0.95

1.75

+0.8 or -0.3 bL= 9.5

-0.5

1 2

bw= 7.0

All Cp values = -0.3

Cpe 0.82

Cpe -0.5

_______ __________________

—

1.75

+05 or -0.1

2.55

1.75

3.5

0.8

Cpe,r,I

Wind All measurements in metres

For stability checks, the external roof coefficients shown above have been simplified as follows: 1. Overturning

-055

02

—,. Wind

4. Roof Uplift

2. Sliding & racking

3. Racking

-0 13

-0.5

-0.55

-0.5

-0

±05

-0.55

-0.25

Wind

Wind

Wind

-1.15 Wind

-1.15

Wind

Key 1.

2.

3.

Overturning

Sliding and racking

Racking

-0.55

= weighted mean of -0.4 and -0.7

-0.2


-0.6

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

-0.5


+0.6

= weighted mean of +0.5 and +0.8

-0.55

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

4.

Roof uplift

-0.5


-0.25


-1.15

mean of -1.1 and -1.2 for roof uplift

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


(R


7.3 Overall stability calculations It is necessary to check the stability of the building both during and after construction. In the construction period a lower design wind pressure may be used, but there may be no roof weight to add stability nor masonry walls to reduce the nett wind loads. In the constructed period a roof and in some cases masonry 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 the constructed period only.

7.3.1 Wind pressures and self-weight In 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 worse case 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 are calculated. 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 = 0.867 (q Cpe)

Pnett

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 0.922 kN/m2

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

Effective vertical pressure on roof (overturning) (Internal coefficients cancel out) -0.416 kN/m2

= O.867{O.80 x (-O.6)}

Effective vertical pressure on roof (racking vertical load)

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

= -0.173kN/m2

9.5 x 5.0 x 0.24 x 2

= 22.8 kN

Total building self-weight Front and rear walls

Gable walls without spandrels 7.0 x 5.0 x 0.24 x 2

= 16.8kN

Spandrels

7.0 x 2.5 x 0.5 x 0.24 x 2

= 4.2kN

First floor

9.5 x 7.0 x 0.34

= 22.6 kN

Roof including overhang

9.5 x 7.6 x 0.92

= 66.4 kN = 132.8 kN

Total

7.3.2 Overturning 7.3.2.1 Wind on front or rear wall

x

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)]

= 171.2 kNm

The overturning resistance due to building self-weight

= 132.8 x 3.5 Factor of safety against overturning = 464.8/171.2

= 464.8 kNm =

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.)

©


76


7.3.2.2 Wind on gable wall

Th

2.5 m

2.5 m

2.5 m

L

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 = 630.8 kNm

= 132.8 x 4.75 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

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

= 53.60 kN

= 9.5{(5.0 x 0.747) + (2.5 x 0.763)} Therefore required sliding resistance

1.2 x 53.60

= 64.32 kN

Assuming a coefficient of friction of 0.25 the resistance to sliding

= 33.20 kN

= 132.8 x 0.25

Therefore sole plate straps or other fixing methods are needed to provide a total resistance of = 31.l2kN. 64.32 - 33.20 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 5268-2 Annex G or TRADA's fastener software (visit www.trada.co.uk) gives a basic single shear lateral load of 327 N. Therefore the permissible load per strap

= 327x2x1.25x1.25x103

=

1.O2kN

(K45) (K48)

No of fixing straps needed = 31.12/1.02

©


= 30.51

= 31

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

=

= 1000/0.939

= 1064 mm centres

0.939 per m


Since fixing straps or hammered sleeve fixings are normally used at 600 mm centres to locate the sole plates during setting out, no additional restraint is necessary to augment the sliding resistance provided normal practice is followed. Adequate sliding resistance at first floor level will normally be provided by recommendations made by the UK Timber Frame Association to meet requirements against disproportionate collapse. These specify that the minimum fastening between the bottom rail of the wall panels and the floor deck or rim beam and between the rim beam and the head binder or top rail of the wall panel below should be as follows: 3.1 mm diameter nails at 300 mm centres for Class 1 and Class 2A buildings 3.1 mm diameter nails at 200 mm centres for Class 2B buildings.

7.3.4 Resistance to wind uplift The maximum uplift force occurs with wind on the gable walls. The maximum roof uplift pressure

0.867x0.80x 1.15x3.5

2.79 kN/m per side

Roof dead load

= 0.92x7.6/2

= 3.50 kN/m per side

Factor of safety against roof uplift

= 3.50/2.79

= 1.25

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 combined lateral and vertical wind loads should this be necessary.

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

7.4.1 Wind loads 7.4.1.1 Wind on front or rear wall

Horizontal load on roof F1

= 2.5 x 9.5 x 0.763

= 18.12 kN

With the wind on the rear wall

F2=F3 =2.5x9.5x0.747


78

= 17.74 kN

(fRADAgJ OV/


7.4.1.2 Wind on gable wall

With wind on gable wall, door end F4

= 2.5 x 7.0 x 0.5 x 0.922

F5 = F6 = 2.5 x 7.0 x 0.433

= 8.068 kN = 7.578 kN

7.4.2 Racking forces The wind loads F are transferred as racking forces Q to the timber frame walls at right angles to them. Qa the applied force to be resisted by the first floor front and rear walls and by internal 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

= 19.44 kN

Force on first floor gable walls and parallel partitions = F1 ÷ 0.5 F2

= 26.99 kN

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 higher value is used).

7.4.3 Design method for racking It may be possible to demonstrate that external timber frame walls designed to a minimum specification can support the actual racking forces, without having to calculate the contribution of other elements such as internal partitions. However, if the calculation shows the strength of the external walls alone to be insufficient, then as many as necessary of these other elements will have to be considered. 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.



If the racking resistance thus calculated is still less than the racking force, then a stronger construction for the external timber frame walls must be chosen and the procedure repeated. The construction can be strengthened by: 5. 6. 7. 8.

Decreasing the nail spacing. Increasing the nail diameter. Increasing the thickness of the sheathing. 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 it is 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 permissible racking force on a timber frame wall. This can be rewritten as:-

=(R+R1)XLXK

Qadm

where R and R1

= the basic racking resistance of sheathing and lining respectively in kN/m, modified by the product of the material modification factors for nail diameter, nail spacing and panel thickness = K101 x K102 x K103 = KM

L

= wall length in m

K

= product of wall modification factors for panel height, wall length, openings, vertical load and masonry contribution = K104 x K105 x K106 x K107 x K108

This version of the formula allows for the possibility that KM may be different for the sheathing and lining. The first step is to determine whether a construction of minimum strength is adequate. Denoting the ground floor gable wall with a door by the suffix d and the opposite gable wall by o'. =7.Om Ld =L0 Note: The most commonly used sheathing material is 9 mm thick OSB Grade 3 to BS EN 300. In order to illustrate the full design procedure, however, this example uses 7.5 mm thick plywood because this lower specification requires the 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 the particular 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 K101 where d K101

K101 is not applicable to plasterboard, which must be fixed with nails of at least 2.65 mm diameter in order to contribute to calculated racking resistance.

©


80

rHCHN010


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 internal studs of timber frame panels which are permitted when racking calculations are based on values given in the Table. It also specifies standard nail spacings on the perimeter for various materials, s mm. Nail spacings should not be less than the minimum spacings recommended in BS 5268-2, eg 14 diameters for plywood-to-timber or OSB-to-timber joints, without pre-drilled holes. Although the Standard no longer specifies a maximum spacing of 300 mm, for reasons of wall stud stability and in harmony with recommendations for flooring it is recommended that the perimeter nail spacing should not exceed 300 mm, or 200 mm with sheathing less than 10 mm thick.

For a proposed nail spacing of S on the perimeter, the basic racking resistance must be multiplied by K102, where = _____________ S

K102

O.6x—- +0.4 sp

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

In the case of plywood nailed at 200 mm centres on the perimeter 1

=

K102

= 0.833

I0.6x') + 0.4 150)

K103 Sheathing thickness (Clause 4.8.2.3) For 7.5 mm thick plywood the basic resistance of the standard 9.5 mm thick material must be modified by K103 (These are nominal thicknesses). = (2.8B - B2 —0.8)

K103

Where B

= proposed board thickness

Hence K103,5

= 0.789 for plywood

K103,1

standard board thickness

= 0.789 9.5

= 1.0 for plasterboard

Note that B is limited by BS 5268-6 to 0.75 B 1.25 R and R1 Hence if R5 and R1 are the racking resistance of the sheathing and lining respectively, R5

= 1.68 x K101 x K102 x K103

= 1.68 x 0.75 x 0.833 x 0.789 R1

= 0.28 x K103

=0.28 x 1.0

= 0.828 kN/m = 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 framed openings which have a height in excess of half the panel height and are less than 300 mm from a corner of the building, the length of wall from the corner to the farther edge of the opening should be disregarded.

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81

(J


K106 Framed openings (Clause 4.9.3) K106

where

=(1_1.3.8)2

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 less than 300 mm and their heights are each greater than half the panel height, then the area of the opening should be taken as that of the rectangle which encloses both openings. Clause 4.9.4 explains that certain openings not more than 250 mm wide need not be framed. Its intention is that such openings may be discounted when calculating K106.

K107 Vertical loading (Clause 4.9.5) At this stage K107 will be taken as 1.0

K108 Interaction (Clause 4.9.6)

K

K104xK105xK106xK107xK106

1.0

.. K10

K108 = 1.1

:.KWd =

K0 = 1.45 Permissible racking force Qadm Qadm

=(R+Ri)xLxK

Hence 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 alone would have to be considered with an appropriate share of the total racking load. From Section 7.4.2

= 44.73 kN

Qa Qadm

7.4.3.2 K107 Allowance for vertical loading (Clause 4.9.5) Although its calculation is rather tedious, K107 should be considered next, since it increases the maximum allowable contributions to racking of the brick veneer and the internal partitions, which are considered subsequently in steps 3 and 4.

K107

= 1+[(0.09V _0.0015V2)xJ1 where 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 of a negative load on a first floor wall would have to be included in the calculations of the load on the ground floor wall beneath it. Where negative vertical forces are encountered, adequate holding-down straps must be specified. Special provisions are made in this clause to allow for concentrated vertical loads.


82

(R


To calculate the load from the roof, Vr, on each wall, it is necessary first to find the roof area A supported by it, as shown in the diagram. Dimensions for this example are shown in parentheses. Overall length (9.5m)

Plane

of

Ai for rear wall

rafters A3

A4

for

for end wall

end wall

Overall width including overhangs

(7.6m)

A2 for front wall

x=

x

Overall length - 2x

x

(0 3m)

(89m)

(O.3m)

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

Vr

pressure with wind on gable wall) = (33.82 x 0.92)— (33.82 x 0.173) V (total vertical load)

= 25.26 kN

= V1

Load on ground floor front wall (see Section 7 General description and specification and Section 7.1) = 25.26

load on first floor wall (V1)

+ 2.5 x 9.5 x 0.24

first floor front wall

+ 0.5 x 2.7 x 2.8 x 0.34

floor dead load

=32.25kN

= 3.39 kN/m (The wall is 9.5 m long)

Load on ground floor rear wall = 25.26

load on first floor wall

+ 2.5 x 9.5 x 0.24

:. V

©


first floor rear wall

= 3.26 kN/m

30.96kN

83


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)+2.28 V

0.624 ±

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

= 0.72 kN/m

= 5.07 kN

(The walls are 7.0 m long) Load on ground floor gable wall, door end

:. Vd

= 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

= 1.62 kN/m

= 11.31 kN

Load on ground floor gable wall, other end 5.07

V0

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

= 1.99 kN/m

= 13.91 kN

Hence Vd

K107

1.62kN/mand

1.99kN/m

V0

= 1 + [(0.09v — 0.0015V2)]1

Hence K107,d

1.09 and

K107,0

= 1.11

= 1.09*x 1.09

= 1.19

K107d

K,0

= 1.45*x 1.11

= 1.61

K107 0

see Section 7.4.3.1

Hence Qadm,d

= 7(0.828 + 0.28) 1.19

= 9.23 kN

= 7(0.828 + 0.28) 1.61

= 12.49 kN

Hence Qadm,o

Qam = Qadm,d + Qadm,o Qadm

= 21.72 kN

7.4.3.3 Contribution of any exterior masonry walls (Clause 23) Masonry walls constructed and fastened to the timber frame in conformity with Clauses 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 horizontal length of storey-height masonry in the wall, excluding any horizontal lengths of 600 mm or less. From Table 6 of the Standard, with ties at 4.4 per m2 the maximum racking load taken by the masonry is 0.5 kN/m of panel height masonry.

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84

TRADA TECHNOLOGY


Lm,d

= 6.1 m and

Lm,o

= 7.0 m

Hence Qm,rnax,d

= 6.1 X 0.5

Qm,max,o

7.0 X 0.5

= 3.05 kN = 3.50 kN

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

= 0.25 x 9.23

= 2.31 kN

= 0.25 X 12.49

= 3.12 kN

Qm,adm,d

2.31 kN 25 per cent of Qadm.o

= 3.12 kN

Qm,adm,o

Qadm

= 27.15 kN

21.72 + 2.31 + 3.12

Qadm

7.4.3.4 Contribution of internal partitions Clause 4.7.4.1 in the Standard stipulates that the permissible contribution to racking resistance in any one direction from Category 3 plasterboard must not exceed 50 per cent of the resistance provided in that direction by Category 1 or 2 materials (plywood, OSB, bitumen impregnated insulation board, etc). (Note that the 30 mm of plasterboard in two or more layers in a separating wall is classified as a Category 2 material.) In this example the resistance provided by Category 1 or 2 materials in the external wall = Qadm

(R ±R1)

where Qadm and R and R1

= the permissible racking force on the external walls under consideration, after allowing for vertical load (Section 7.4.3.2) = the racking resistance of the sheathing and lining respectively (Section 7.4.3.1).

Similarly the racking resistance provided by the plasterboard in the external walls °adm

(R±R)

Hence, due to the 50 per cent limitation quoted, the maximum permissible contribution to racking resistance from the internal partitions Qi, max

-— 0.5 Qadm

R

- Qadm

(R+R1)

R1

(R+R1)

(o.sR—R)

—

(R+R) For the ground floor gable walls max

= 21.72 (0.5 x 0.828 —0.28) (0.828 + 0.28)

= 2.63 kN

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


85


9.5 m

Ground floor plan

First floor plan

L,8 = 2.9 m

L,i0 = 1.9 m

L1,9 = 1.9 m

L,11 = 3.9 m

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

= 0.90 + 0.45

R

= (0.90x 1 )+(0.45x 1

= 1.35 kN/m.

K103

= 1.35 kN/m.

K103

(Note that K101 and K102 are not applicable to plasterboard) K104

Panel height K105

K10 =1.0

Height of wall panels = 2.4 m

Length of walls

Forwalls0to2.4m K105 =

L

2.4

For walls 2.4 to 4.8 m K105 =

K106

= 0.792

1<105,8

= 1.08

K105,11

= 1.21

K106

=1.0

Framed openings None in this example

K107

K105,9,10

Vertical load

This can reasonably be ignored in the case of internal partitions in a two-storey house K108

Interaction

K

.'. K107

= 1.0

K108

=1.1

1K,11

= 1.33

= K104 x K105 x K106 x K107 x K108

= 1.19

K,g,1o = 0.871

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.

86

= 29.77 kN


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 the perimeter of the panels. = 0.883

K101

= 2.65/ 3

K102

= K103

R

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

1.0 (see Section 7.4.3.1).

= 0.28 kN/m as before.

R1

The wall modification factors K and the vertical load modification factors K107 are 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

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

14.66 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 Section 7.4.3.3). New maximum allowable contribution to Qadm,d

= 3.66 kN

= 0.25 X 14.66

Hence

= 3.05 kN

Qm,adm,d = maximum masonry load

The masonry wall on the other end can contribute 3.50 kN (see Section 7.4.3.3) New maximum allowable contribution to = 0.25 X 19.84

= 4.96 kN

Qm,adm,o maximum masonry load

= 3.50 kN

Qadm,o

Hence

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 the permissible racking load on the external walls allowing for the vertical load but not the masonry contribution (Section 7.4.3.4). Hence Qi,max

= °adm 2 0.5R5 — R1 = 34.50 (0.5 xl .48—0.28)

R5+R1 Qadm,4

= 9.02 kN

(1.48+0.28)

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 wall specifications reached in Section 7.4.3.5 and shown in the table heading. The sequential method shown is intended to obviate any unnecessary calculations. However, if a spreadsheet is used it may be nearly as simple to introduce all the possible factors at once.

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87


The table calculates the racking resistance in four stages as follows: Allowance for: Qadm,1 Qadm,2 Qadm,3 Qadm,4

Vertical load

/' X

1

Masonry contribution

Internal partitions

X X

X

1

K

X

1

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


88

('TRADA\ '\ recnNoLocv(/

Qed,l

Qadm

K

K108

K107

K106

K105

K104

R1

R

K103

K103,

K102

K101

RIb

RS,beSjC

©

82

Permissible racking force


Total permissible racking force on the floor level and in wind direction shown (kN). Must be > Q

=L(R+R1)K(kN)

82

82

81

82

81

Modification factor for wall length Modification factor for framed openings in external walls Modification factor for vertical loading Interaction modification factor K104 x K105 x K106 x K107 x K108

1.00 1.10

1.00

0.28

81

81

0.828

1.00

0.789

1.00

0.883

0.28

1.68

Common factors

81

81

81

81

80

80

80

70

79

Page

Modification factor for panel height

Total racking force to be resisted on floor level and in wind direction shown (kN) Length of each external wall (m) Basic racking resistance of primary sheathing in external walls (kN/m) Basic racking resistance of secondary sheathing in external walls (kN/m) Nail diameter modification factor for primary sheathing Nail spacing modification factor for primary sheathing Board thickness modification factor for primary_sheathing Board thickness modification factor for secondary sheathing (K101 & K102 not applicable to plasterboard) Rebeec x K101 x K102 x K1o3 (kN/m) RI,basjc x K103,1 (kN/m)

Q

Le

Explanation

14.15

0.846

0.583

1.32

9.5

23.06 (> 11.86) Strength adequate

8.91

0.533

0.367

1.32

9.5

11.86

3

17.87

1.45

1.00

1.32

7.0

17.87

1.45

1.00

1.32

7.0

I

I

4

26.99

Other end (o) I

35.74 (>26.99) Strength adequate

I

I I

6

Rear

11.37

0.680

0.468

1.32

9.5

19.44

I

I

..._...

7


10.73

0.642

0.442

1.32

9.5

Front

8.45

1.09

0.75

1.32

7.0

end (d)

44.73

I

I

/(

TRADA

'2

19.70 (<44.73) Strength inadequate

11.25

1.45

1.0

1.32

7.0

end (0)

Wind onto front/rear walls Door I Other I 8 I 9

10 I

1

Door end (d)

2

Front 1

Wind onto front/rear walls

Wind onto gable walls Rear

Ground floor resisting walls (see diagram Section 7.4.3.4) Wind onto gable walls

First floor resisting walls (see diagram Section 7.4.3.4) 5

9 mm OSB/3 nailed at 150 mm centres with 2.25 mm diameter nails + 12.5 mm plasterboard with standard nailing Two boards of 12.5 mm plasterboard Spaced at 4.4 per m2

Symbol

Racking design figures

External walls: Internal partitions: Masonry Ties:


EQadrs.4

Q,sdrs

Oi,rsas

0

K

K108

K105

K104

R1

K103

Ri,basic

L,

Qadn.3

°rmadn

Qm,nax

Q

L

Qd,2

Qd.2

©

R,) (R8+R1)

(0.5Ra Qadm2 x

kN


Q.

Permissible contributions from internal partitions = the lesser of ZO and Qi,max = Qadrs,3 + (kN) Must be >

— —

Maximum permissible contribution from internal partitions

=LRK(kN)

K1 x K105 x K106 x K107 x K108 Possible contribution from internal partitions to racking resistance

Interaction modification factor

Length of each internal partition (m) Basic racking resistance of internal plasterboard partitions (kN/m) Board thickness modification factor for internal partitions R1 basic x K103 (kNIm) Modification factor for panel height Modification factor for wall length

L

Possible contribution from masonry wall to racking resistance = x 0.5 (kN) forties at 4.4 per m2 Maximum permissible contribution from masonry wall = 0.25 Qadn.2 Permissible contribution from masonry wall = the lesser of Q, and Qn,max ZQadro.2 + ZQr,adr (kN). Must be>

(m)

Length of storey height masonry wall

(kN).Mustbe>Qa

Permissible racking force = x K107 (kN) Total permissible racking force on the floor level and in wind direction shown

Uniformly distributed vertical load on external walls (10.5 kN/m maximum) Modification factor for vertical load

V

K107

Explanation

Symbol

Racking design figures Front

Rear

7

Wind on gable walls

62 99

8

87

82

87

9 02

9.02

9 10

3.50

4.96

3.50

7 .0

1.08

2.9


2.23

0.871

0.792

1.9

7.00

1.33

1.21

3.9

11

TRADA'

2.23

0.871

0.792

1.9


19 . 84

1.11

1.

Other end (o)


3.05

3.66

3.05

6. 1

14 .66

1.09

1.

Door end (d)

Wind on front/rear walls

4.66

Door end

Ground floor resisting walls (see diagram in 7.4.3.4)

86

2 3

Wind on front/rear Other 4 5 6 end

1.19

1.1

1

wafls on2ble Front Rear Wind

First floor resisting walls (see diagram in 7.4.3.4)

86 86

86

1.0

1.35

86

1.

0

35

86

1.

ac ors

C0rIlmOfl

86

86

86

87

85

85

85

84

87

87

82

82

Page

(Externalwalls, Internal partitions, Masonry Ties; measurements as before)



If plasterboard lined separating walls constructed in accordance with Clause 4.7.5 in the Standard were under consideration, their contribution to racking resistance could be added to the value 0, adrn calculated in the table for internal partitions. The table below gives permissible racking loads for selected plywood or OSB wall panels lined with one layer of 12.5 mm thick plasterboard, allowing for the material modification factors K101, K102 and K103, but not for any other factors. Plywoods in BS 5268-2 or equivalent plywoods certified for use in the UK d 5 t Qadm mm mm mm kN/m 1.11 2.25 200 7.5

OSB to BS EN 300, Grades 3 and 4 d mm 2.25

S

t

kN/m 1.33

mm

200

mm 9.0

Qadm

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 diameter s = nail spacing on panel perimeter (use 300 mm for spacing on internal studs) = thickness of sheathing

7.4.4 Recommended design procedure First ascertain from the fabricator his preferred sheathing and lining materials and the range of thicknesses available. If higher figures for basic racking resistance are available from the board manufacturers then these should also be obtained. 2. Nails are normally inserted by nail guns which cannot necessarily use any diameter of nail. Ascertain from the fabricator his preferred diameters. It is sensible to keep to the same diameter throughout a house. 3. Use the same construction for the wall panels throughout each individual floor. 4. K107, the factor for vertical loading, increases the calculated racking strength of a structure by a factor which is generally below 10 per cent and almost never more than 15 per cent, and Section 7.4.3.2 shows that its calculation is rather time-consuming. Therefore, it is considered that the calculation of K107, unless it is essential, may be omitted, and its value taken as 1.0. 5. In general, start with a minimum practical thickness of sheathing and specification for nailing, and only if this proves inadequate adopt a stronger construction. 6. A table should next be drawn up of the values which will be required to determine an adequate construction specification. Using these values a series of values for Qadm can be calculated and tabulated for each floor and wind direction, proceeding as far down the table in each column as is necessary to obtain values which exceed the corresponding values of Qa• 7. If the initial construction selected is inadequate, increase its strength as necessary according to the procedure explained in Section 7.4.3.5. 1.

©


91


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

Assume studs at 600 mm centres = panel height — (thickness of top and bottom

Stud height rails)

2400 — (2 x 47)

= 2306 mm

Lateral loading due to wind External pressure coefficient (Section 7.2)

= 0.82 -0.3

Internal pressure coefficient (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 x 0.6

so loading per stud

= 0.280 kN/m

=

Maximum bending moment (Mmax)

= 0.466 kN/m2

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 Gma

= M— =

0.186x106x6

Z

47x972

=2.52 N/mm2

Actual compressive stress Gca

F =— = 6.41x10 3

A

=1.41N/mm2

47x97

Permissible bending stress Gmadm = grade stress x K3 x K7 x K5

where K3 (load duration) = 1.75

300 "° 1 K7 (depth factor) = K8 (load sharing) = 1.1

/ = 5.3 x 1.75 x

©


92

\Q.11 I

97 )

xl .1

= 11.55 N/mm2


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 —

2 2NA2a

1(1+)u2E 2 2NA2a

Where A (slenderness ratio) Le

2

2_ u2E

NA2a

= 0.85x2306*

70.00

12

=1.5

N

grade stress x K3

= 6.8 x 1.75

= 11.9 N/mm2

= 5800 N/mm2

Emin

= 0.005 A

q (eccentricity factor)

= 0.005 x 70.00 = 0.350 *

BS 5268—6.1 Clause 4.6.2 states that the effective length of a timber frame wall stud should be taken as 0.85 L 1+q = 1 + 0.350 = 0.675 2 2 ir2E

=0.654

1.5x(70.O0)2 xll.9

NA2a

K12

2x58OQ

=

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

=

0.6541

= 0.4596 Gcadm

= 6.8 x

= 6.016 N/mm2

1.75 x 1.1 x 0.4596

Now check summation of stress ratios Gma

+

Gca <1 —

( l.5XG xK12 0c,adm °m,adml 1 where

Ge

=

2x58OO = ________

rr2E

2.52

therefore

11 55111

= 11.68

7Q2

A2

+

1.41

.5x 1 .41x0.4596 6.016 11.68

= 0.472< 1

)

TRADA ©



7.5.2 Ground floor rear wall - medium term and long term Check 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

=

= 682.4

8.50

E

K12 for

Oc

2=70

600

0.511

682.4

0.539

700

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 3

ac,a

q.JIXIU = ________

47x97

= 0.95 N/mm

2

K12 using Table 22, BS 5268-2

= 70.00

A

= 6.8 x 1.0

= 6.8 N/mm2

=

= 852.9 6.8

--

K12 for

0c

2=70

800

0.572

852.9

0.583

900

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 cent or less of the total design load (imposed roof, imposed floor and dead weight of construction) but has been included to show the complete procedure.

TRADA TECHNOLOGY © TRADA Technology Ltd 2006


Check bearing on bottom rail for medium term duration Actual compressive stress perpendicular to grain

= Gca

crc±a

= 1.41 N/mm2

Permissible compressive stress perpendicular to grain 0c,±,adm

= grade stress x K3 x K4

where grade stress K4 (bearing)

= 2.2 (specification excluding wane) = 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 are satisfactory provided there is no wane.

Check deflection An approximate value for the deflection of a loaded stud is given by the following formula: — —

Z where—

(0.OO5AGca +°ma) Z

(ae—ca) x

bh2

h

6bh 6

A

97 = (0.005 x 70.00 xl .41)+ 2.52x—=4.74mm 6 (11.68—1.4i) = 0.003 x 2306

3adni

= 6.92 mm

which is satisfactory. For particular cases designers may chose to tighten or relax the standard deflection limit of 0.003L.

7.5.3 Ground floor gable wall - very short term Consider ground floor studs to gable elevation carrying wind load and panel and 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 = 0.376 kN/m2.

qs,nett = 0.867 x 0.384 x (0.83 —(-0.3))

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 kN wall panel living-room floor roof overhang

In this case the bending moment and axial load are both less than those for the rear elevation studs. Therefore previous specifications will stand, ie all studs to be 47 x 97 mm (nominal 50 x 100 mm), C16 timber studs at 600 mm centres. Note that a more common stud size is 38 x 89 mm. If this proved inadequate it would be tried at closer centres.

©



7.6 First floor platform Maximum effective span of floor joists

= 4.0 m

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

Maximum bending moment Mmax

0.4 x 4.02 = 1.84 ____________

= 1.472 kNm

Actual bending stress

ma

1.472x106x6

=

=4.99Nmm2

47 x 1942 bending stress

Permissible

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

/

\O.11

=5.3x1.0x1-.I

xl.1

6.12N/mm2

194) Check deflection öa

adm

fL2 5L2

h2

= 1.84x0.4x40002x12 5x40002 1942 384 5 880O47194

=10.1mm

=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 the head binder and top rail of the panel are adequate. (K8 = 1.1 may be used to increase the strength properties if the two members are nailed together at 600 mm maximum centres.) Consider largest span of remaining floor joists. = 2.9 m

Effective span Try 47 x 194 Mmax

Gm a

Gmadm

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

= 1.84 x

0.6 x 2.92

= 1.16x106 x6

47 x

1.l6kNm = 3.93 N/mm

1942

6.12 N/mm2 from previous calculation

Check deflection = 1.84x0.6x29002x12 5x29002 1942 384 5 880047194 6adm

= 0.003 x 2900

432mm = 8.7 mm

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

©


96

2


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

= 4.0 m

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

= 0.948 kN/m

= (0.2 x 1.86) + (2.4 x 0.24)

Check 2 No 47 x 194 mm (nominal 50 x 200 mm), C16 timber joists, connected together. 0.948x 4.02 = _________ = 1.896 kNm Mmax

a

6XU r I.OUXIU = __________ 2x47x1942

= 3.22 N/mm

2

Gmadm = 6.12 N/mm2 from previous calculations. Check deflection

Where E=EmnxK9

= 6612

5800x 1.14

N/mm2 6a

adm

0.948x40002x12 5x40002 1942 384 5 6612x2x47x194 = 0.003 x 4000

=8.66mm = 12 mm

Therefore 2 No 47 x 194 mm (nominal 50 x 200 mm), C16 timber joists are satisfactory. Consider short span joists 2.1 m clear span to illustrate use of slab loading requirements. When considering joists spanning less than 2.4 m an imposed loading of 3.6 kN per metre width of floor (measured perpendicular to the span) is applied. (BS 5268-7 Recommendations for the calculation basis for span tables.) Effective span

= 2.2 m imposed + dead

Loading

= 3.6x0.6 +0.34x0.6

= 1.19 kN/m

1.19x 2.22 = _________

= 0.72 kNm

2.2

Mmax

0.72x106x6

2.44N/mm2

ama

=

Gmadm

= 6.12 N/mm2 from previous calculation.

47 x 1942

Check deflection = 1.19x22002x12 5x22002 1942 384 5 880047194

+— =1.61mm

adm

= 0.003 x 2200

= 6.6 mm

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

TRADA TECHNOLOGY ©



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

7.7.1 2.4 m clear span opening at eaves level Effective 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 9.81

Self 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

1.5x8.85x10 = ____________

= 0.74 N/mm2

Check shear

2 x 45 x 200

= 2.2 x 1.25 x 1.04

Tadm

= 2.86 N/mm2

where 1.04 = K8, the load sharing factor which may be applied to the strength properties of LVL in lintels and trimmer joists made up of two or more members connected together. Check bending

5.53x106x6 = ___________

ma

= 9.22 N/mm

2x45x200

= 19.5 x 1.25 x 2.J

m,adm

x 1.04

2

= 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) =

2x45x200 3 12

A

=2x45x200 = 5 x 7.08 x (2.5 x iO0O) + 3 x 7.08 x (2.5 x 1000)2

384x13500x60.0x106 20x600x2x45x200

öadm

= 0.003 x 2500

=60.OxlO6mm4 =18.OxlO3mm2 5.06 mm = 7.5 mm

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


98


7.7.2 1.2 m clear span opening at first floor level This carries a long-term load. There is no roof load because of the opening above it. = 1.3 m

Effective span = 2.8 x 0.5 x 1.84

Floor loading

= 2.58 kN/m

2.4 x 0.24

Wall panel loading

= 0.58 kN/m

Self dead weight members)

= 0.07 kN/m (assuming 2 No 016 softwood

Total udl, f

= 2.58 + 0.58 + 0.07

= 3.23 kN/m

RA=RB

= 3.23x1.3 2

=2.l0kN

3.23 xl 32 = _________

= 0.682 kNm

Mmax

Try 2 No 47 x 195 mm (nominal 50 x 200 mm), 016 timber members. Check shear and bending: Ta

ama tadm

32.1010 = ___________

= 0.17 N/mm 2

= 0.682x106x6

=1.14N/mm2

2x2x47x195 2x47x1952

= 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

3.23x13002x12 5x13002 1952

öa

=

adm

= 1300 x 0.003

6612247195

384

5

=0.42mm = 3.9 mm

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

7.8 Cripple studs 7.8.1 First floor rear wall - very short-term Consider studs supporting 2.4 m clear span lintel in rear wall at eaves level. Cripple stud height

= (full height stud) — (lintel depth)

=2306—200 As in Section 7.5.1, nett pressure on stud qs,nett

=2106mm = 0.466 kN/m2

Load per unit length on one stud

f

= pressure x (half overall width of opening)

= 0.466 x

Mmax


2

0.598 x 2.1062 = ____________

0.598 kN/m

= 0.332 kNm


Axial loading to cripple stud reaction from lintel calculation + self-weight of half the lintel F

= 8.97 kN

= 8.85 + (0.5 x 0.090 x 2.6)

In general the number of cripple studs required equals the number of studs removed by the opening, in this case, three, so two cripple studs should be used at each end. But try 1 No 47 x 97 mm (nominal 50 x 100 mm) C16 cripple stud at each end. Applied bending stress 0m,a

0.332x106 x6 = ___________

= 4.50

47 972

N/mm2

Applied compressive stress

aca

8.97 = ________ 47x97

= 1.97

N/mm2

Permissible bending stress = 10.50 N/mm2

am,adm = 5.3x1.75x.'J Permissible compressive stress

acadm = grade stress x K3 x K12 To find K12

= Le

A (slenderness ratio)

N

0.85x2106 = __________

= 63.93

I

=1.5 = 11.9 N/mm2 (from stud calculation in Section 7.5.1

= 0.005 A = 0.005 x 63.93

q (eccentricity factor)

1+ri

therefore

=

1+0.320 _______

2

= 0.320 = 0.660

2

rr2xE =

NA2a

u2x5800 1.5x63.932x11.9

= 0.785

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 ae

=

= 6.49 N/mm2

rr2x5800

= 14.01 N/mm

63.932

Check summation of stress ratios using formula in Section 7.5.1

4.50 ___________________

1.97

= 0.788<1

10.5011— 1.5x1.97x0.545 6.49 14.01


100

)

TRADA 1 IC H HO LOOt

2


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

Gc±a

= 8.85 x103 47 x 97

=1.94N/mm2

Permissible compressive stress perpendicular to grain (Cl 6 timber) = 2.75 N/mm2

0c,I,adm = 2.2 x 1.25 which is satisfactory where the specification excludes wane.

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

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

7.8.3 Ground floor front wall - very short-term Consider cripple studs supporting 1.2 m clear span lintel at first floor level, (Section 7.7.2) = 2111 mm

Cripple stud height 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 Mmax

= pressure x (half overall width of opening)

=0.427x 1.3/2

=0.278kN/m

0.278x2.1112 = ___________

= 0.155 kNm

Axial loading to cripple stud F

= [Reaction at lintel (Section 7.7.2)] + [Load from cripple stud to second 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) 2

= 8.45 kN

Try 1 No 47 x 97mm (nominal 50 x 100mm), 016 timber cripple stud Applied bending stress 6 Gma = 0.155x10 x6

=2.1ON/mm2

47 x 972

Applied compressive stress Gca


=

8.4510 47x97

101

=1.85N/mm2

(R


Permissible bending stress

amadm = 10.50 N/mm2 from previous calculation Permissible compressive stress

Gcadm = 6.49 N/mm2 from previous calculation (The lengths of the studs are sufficiently close to permit using the same values) Check summation of stress ratios (K12 and ae as in previous calculation)

2.10

1.85

10.5011 1.5 xl .85 x 0.545') 6.49 14.01

=0.51<1

)

7.8.4 Check bearing on bottom rail of panel - medium-term load Applied compressive stress perpendicular to grain 3 = 8.45x10

1.85N/mm2

47x97

Permissible compressive stress perpendicular to grain = 2.75 N/mm2 from previous calculation

ac±adm

Therefore 1 No 47 x 97 mm (nominal 50 x 100 mm), C16 timber cripple stud at each reaction of the lintel is satisfactory. The remaining lintels and cripple studs should then be checked using the methods given above.

©


102

TRADA TIC I4NOLOCY


8 References and further reading British Standards BS 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 plasterboard excluding materials submitted to secondary operations.

BS 5268-2: 2002 Structural use of timber. Code of practice for permissible stress design, materials and workmanship (under revision 2006) BS 5268-3: 1998 Structural use of timber — Code of practice for trussed rafter roofs. (under revision 2006) BS 5268-6.1: 1996 Structural use of timber — Code of practice for timber frame walls — Dwellings not exceeding four storeys. (under revision 2006) B5 5268-6.2: 2001 Structural use of timber — Part 6: Code of practice for timber 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 and imposed 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 roof loads.

BS 8103-3: 1996 Structural design of low-rise buildings - Code of practice for timber floors and roofs for housing. BS 8201: 1987 Code of practice for flooring of timber, timber products and wood based panel products. BS 8212: 1995 Code of practice for drying lining and partitioning using gypsum plasterboard. BS EN 338: 2003 Structural timber. Strength classes BS EN 594: 1996 Timber structures. Test methods. Racking strength and stiffness of timber frame wall panels BS EN 1995-1-1: 2004 Eurocode 5: Design of timber structures— Part 1.1 General rules and rules for buildings.

BS EN 12369-1: 2001 Wood-based panels. Characteristic values for structural design. 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 for fabricated trusses using punched metal plate fasteners. BS EN 14545 Timber structures — Connections — Requirements. In preparation.

©

TRADAlechnology Ltd 2006

103


TRADA Publications Timber 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 TRADA Technology. ISBN 1 86081 605 3. 2003.

Span tables for solid timber members for floors, ceilings and roofs (excluding trussed 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 askTRADA website www.trada.co.uk Timber Sizer (calculates solid timber sizes for all structural elements covered by BS 5268-7). Nail, Screw, Bolt & Dowel Connection Designer (lateral and withdrawal resistance of fasteners calculated to BS 5268-2 and Eurocode 5). Coach Screw Connection Designer (lateral and withdrawal resistance of fasteners calculated to BS 5268-2 and Eurocode 5). Flitch Beam Designer (design of bolted steel fitch beams to BS 5268-2 and BS 5950).

Other publications BRE Digest 436 Wind loading on buildings. Part 2. 1999. CRC, London. ETAG 019: 2005 Pre-fabricated wood-based load bearing stressed skin panels. European Organization for Technical Approvals. Brussels. UK Timber Frame Association. Design guidance for disproportionate collapse. Technical Bulletin 3.

©


104

Timber Frame Housing: UK Structural Recommendations is a time-saving professional manual for structural engineers. It covers well-established principles and methods for the structural design, strength and stability checking of timber frame buildings. 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 engineering aspects 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. If would 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 construction industry. 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, design assistance, 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 UK and overseas.

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

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

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