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UK Structural recommendations

(TRADA

T E C H N 0 L 0 G

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

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

Contents

Page

Introduction

5

I Loading on structural elements

6

1.1

Structural duties of timber frame elements

6

1.1.1 Timber frame wall panels 6

1.1.2 Horizontal diaphragms 7

1.2

Determination of loads acting on each element

9

1.2.1 Calculation of vertical load 10

1.2.2 Calculation of wind loads 11

1.2.3 Calculation of roof loads 19

1.2.4 Checking strength and stability 20

1.3

Weights of materials

21

1.4

_{Weights of typical constructions}

23

1.4.1 Roofs 23

1.4.2 Floors 23

1.4.3 Walls 24

1.5

General requirements of prefabricated wall, floor and roof elements

24

2 Walls

25

2.1

Wall studs

25

2.1.1 Minimum desirable stud sizes 25

2.1.2 Multiple studs 25

2.1.3 Lateral support 26

2.1.4 Load capacities 26

2.1.5 Design procedure 29

2.1.6 Drilling of studs 30

2.2

_{Framing around openings}

30

2.2.1 Lintel capacities 31

2.3

Racking resistance of timber frame walls

41

2.3.1 Basic design and materials 41

2.3.2 _{Adhesively bonded panels} 42

2.3.3 Overturning effects 43

2.3.4 Deflection 46

3 Floors

48

3.1

Joists and beams

48

3.1.1 Reactions, moments and deflections in a single span, simply supported beam 49

3.1.2 Solid timber beams 50

3.1.3 Structural timber composites 52

3.1.4 Prefabricated engineered timber joists 52

3.1.5 Steel fitch beams 53

4 Roofs

64

4.1

Trussed rafter roofs

64

4.1.1 Bracing 64

4.2

Connections

67

5 Foundations

68

5.1

_Loading

68

6 Multi-storey buildings

69

6.1

General design considerations

69

6.2

Construction

69

6.3

_{Disproportionate collapse}

69

7 Example of calculations for a complete dwelling

70

7.1

Vertical loads

72

7.2

Horizontal loads

73

7.3

Overall stability calculations

75

7.3.1 Wind pressures and self-weight 75

7.3.2 _Overturning 76

7.3.3 _Sliding 77

7.3.4 Resistance to wind uplift 78

7.4

_{Racking calculations}

78

7.4.1 Wind loads 78

7.4.2 Racking forces 79

7.4.3 Design method for racking 79

7.4.4 _{Recommended design procedure} 91

7.5

Wall panel studs

92

7.5.1 Ground floor rear wall — very short term 92 7.5.2 Ground floor rear wall — medium term and long term 94

7.5.3 Ground floor gable wall — very short term 95

7.6

First floor platform

96

7.7

Wall panel lintels

98

7.7.1 2.4 clear span opening at eaves level 98

7.7.2 1.2 m clear span opening at first floor level 99

7.8

Cripple studs

99

7.8.1 First floor rear wall — very short term 99 7.8.2 Check bearing on bottom rail of panel — medium term load 101

7.8.3 Ground floor front wall — very short term 101

7.8.4 Check bearing on bottom rail of panel — medium term load 102

8 References and further reading

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 5268-6.1:1996 Structural use of timber. Code of practice for timber frame walls. Dwellings not exceeding four storeys but it also includes procedures 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.

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.

©

TRADA Technology Ltd 2006 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 wood-based 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

Va

h

adm,1

+

dm,2

Slkedge,1 S2kedge,2

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

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

h = horizontal depth of diaphragm measured in direction of wind (mm)

fadm = permissible load per fastener for very short-term loading (N)

k

= avalueshown 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

1.0

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

1.5

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

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 load-carrying 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

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: 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)

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

,

Shaded areas carry roof or floor loads

Load at each stud point

L

= WrXSX_

2

Load in each stud

4\Nr xsxJ+(Ww xsxh)

Load at each interior joist

support

=(Wf xsx.J+(Ww xsxh)

Floor load W k N/rn2

Load at each stud point

1\NrXSxJ

+(wwXsXh)+\NfXsX_J

Load in each stud

1\NrXSxJ +(wWxsxh)+[wfxs4j

+(W

xsxh)

For

weights of materials and construction, see Section 1.3

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: . L 7.8

Load per stud point at eaves

_{= WrxSx=i.97xO.6x__}

Load at foot of each upper storey

_{= 4.61 +(w}

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

=5.93+IWfxSx

₂ =5.93+11.82 x 0.6 2 = 7.9 +(W xs

x

_h)= 7.9 +(0.35 x0.6x_2.55) =WfXSX =1.82x0.6x7-2 2 = 4.26 +(W x s xh)= 4.26 +(0.25 x0.6x2.4)

For beam designs see Section 3.1

Stud supporting bearn, designed as for lintel support

Typical loading

Self weight of wall W kN/m2

= 4.61 kN 5.93 kN

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

Load at foot of each lower storey

stud

Internal wall:

Load per stud point at 1S floor level Load at foot of each internal wall

stud © TRADA Tecflnology Ltd 2006 10 = 7.9 kN = 8.44 kN = 4.26 kN = 4.62 kN H

Load at each inner roof support point

L

W xdx—

2

Average load in each wall stud (spaced s)

[W1xsx+(Wxsxh)

Careful design required unless d = s

Load at each interior joist

support

(A + B)

= WfXSX 2

Load in each stud

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

+(w xsxh)

End wall

For weights of materials and construction, see Section 1.3.

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

Wr= 1.82 kN/m2 W, = 0.35 kN/m2 for board cladding (lower storey) W = 0.25 kN/m2 for plasterboard internal wall

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

Load per stud point on spandrel panel Load at foot of each upper storey stud Load per stud point at 1st floor level (near end)

Load at foot of each lower storey stud Load at foot of each internal wall stud

=W xdx—=1.97x0.6x——— 2 2 = 2.84 +(W x S x h)= 2.84 +(0.86 xO.6 x 3.15)

=4.47+IWxSx

=4.47+11.82x0.6xi

2)

2

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

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

= 1.82x0.6x—I+(0.25x0.6x2.4)

=2.98kN

2)

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

Typical loading

Roof and ceiling load Wr kN/m2 total

Inner support spacing = d Stud and joist spacing = s Shaded areas carry roof or

loads

Floor load Wf kN/m2

Stud supporting

l

beam, designed

as for lintel support Self weight of wall W kN/m2

= 2.84 kN = 4.47 kN = 5.45 kN = 5.99 kN

1.2.1.3 Other arrangements

Other arrangements are also possible. For example the roof may span front 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 1.3 N _Firstfloor

_N

2.5

Ri

I

R3

5.0 2.5

Ro

N

R4

N 0.3 -- _________________ ____________________ — ___________________ : 0.3

Front elevation End elevation

—

wall I .

I Dimensions in metres

6.8

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

Wall stud

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.

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

Gable walls

Front and rear walls

Diaphragms

Roof diaphragm 6

—---s--- t

--——--- 7.2 m

a1 41.252 +7.22

= 7.31 Ca = 0.968

First floor diaphragm

-= 7.26 Ca = 0.968

First floor diaphragm

6 8 m

a4 42.52 +6.82

= 7.62 Ca = 0.963 - 7 2 m a2 =/2.52 + 7.22 = 7.62 Ca = 0.963

Racking

First floor panels

v

72m

a5 =V1.252 +7.22

= 7.31 Ca = 0.968

Ground floor panels

E

72m

a6 3752 +7.22

= 8.12

Ca = 0.957

First floor panels

- A

y

a7 =2.552 +6.82

= 7.26

Ca = 0.968

Ground floor panels

,,'

6 —7 Ia)

/-a8 5Q52 +6.82

= 8.47 Ca = 0.953

First floor panels

a10 = + 7.22 = 8.77 Ca = 0.950 -d

_{7 m}

a9 = 2.52 + 7.22 = 7.62 Ca = 0.963 = 7.79 = 9.27 Ca = 0.960 Ca = 0.945 6.8 m

a11 43.82 +6.82

Some typical values for the dynamic pressures, q, to be used during 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 -1.17

________________

rearwall

+027

-05

qs = 658 x 1.0

qs = 756 Q.4

_{qs = 756}

= 658 +084

_{_________________}

-0.5 _{____________________}

qs=658x0.5

qs—0 0.11 3.49

0.11 3.49

= 329 +084 -0.3

-0.3 -0.5

qs=590x

qs=590x

0.82=484

0.8=472

_________________

+082 -0.3

-0.3 -0.5

Gable end Front Separating wall Rear Gable end Front

Reference values for BS 6399-2:

Table 5: D/H=1.03 Table 10: W=7.2, bw7.2 TableS: D/H=1.36 Table 10:L=13.6 bL=10.6

(A

_. TECHNOLOGYflW

1.2.2.3 Loads on diaphragms and walls

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

Pfinal

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

(1)

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

where Pfinal _{one direction}

0 85 = factor to allow for the fact that maximum gust speeds on front and rear walls are not simultaneous

= nett wind load on an element of the building which P 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

(4)

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

(a) Wind on gable end

Equation (3)

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

Each storey: P siorey = 329 X 0.84 x 7.2 x 2.5 = 4974 N Equation (4)

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

Equation (1)

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

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

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

Shear force on first floor panels

Shear force on ground floor panels Nett load on a wall stud, assuming 0.6 centres

(b) Wind on front and rear walls

Equation (2)

Roof diaphragm load

First floor diaphragm load

Racking load on first floor panels

Racking load on ground floor panels

Shear force on first floor panels

Shear force on ground floor panels Neff load on a wall stud, assuming 0.6 m centres ceiling P floor,! P racking, 1 P racking, 0 P shear, 1 P shear, 0 P stud

The results are shown in diagram below

= 6205 N = 10759 N = 9215 N = 9722 N = 8983 N = 9722 N = 18461 N 14119 N = 22714 N = 705 N Pfloor, 1

= 8.98 kN

Pstud = 0.705 kN

©

TRADA Technology Ltd 2006 16 Pshear,i

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

=6313 N Pshear,0

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

= 10324 N Pstud

=

0.85 x 6751 x 0.6 xl .02 xl .0

=488N

Roof: P roof

Each storey: P storey

Walls P wall Equation (4) Equation (1) 7.2 = 756 x(0.4 xO.68 + 0.27 x2.92 —(—)1 .17 xO.68— (—)0.5x 2.92) xtan20°x6.8 = (484 x 0.82 —(—)472 x 0.5)x 6.8 x 2.5 = (484 x 0.82— (—)484 x 0.3)x 6.8 x 2.5 = 0.85 x (6205 x 1 + 10759 x 0.5) x 1.02 x 0.968 = 0.85 x (10759 x 0.5 + 10759 x 0.5) x 1.02 x 0.963 = 0.85 x(6205 xl + 10759 x 0.5)x 1.02 x 0.968

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

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

= 0.85x(6205x1+10759x2)xl.Q2xQ.945 —0.85x 9215 xO.6x 1.02 xl .0 6.8 Pceiling = 3.17 kN

—

Pfloor, 1 = 4.15 Pstud = 0.488

Pceiling = 9.72 kN

_—

Pracking,1 4.26 kN Pshear,1 6.31 kN

—

Pracking,o = 8.34 kN

—s

Pshear,o = 10.32 kN

—

Wind on gable end

Pracking,1 = 9.72 kN

—k

Psyear,i = 14.12 kN

—s

Pracking,o = 18.46 kN

—s

Pshear,0 = 22.71 kN

—k

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 load-carrying 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.

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)

=443N

0.4

(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.

L L 0.7x140x3.464

Slenderness ratio

= e = e

= 113 mm

I b 3

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

Pc 68 N/mm2.

Hence permissible compression load on steel strap:

PCXA

= 68x3x25

5100 N.

5100> 1661, hence permissible load on one strap = 1661 N With straps at 1200 mm centres at eaves level on a 7.2 m wide house the maximum racking load which can be transmitted at eaves level:

1661x 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.

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

Fva 2+1 Fha

2

Fvadm Fhadm

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

connection respectively

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

horizontal loads.

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

simplifies to

Fva2 + Fha2

1.2

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.

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.

Aluminium

Battens, tiling Bituminous roofing felt Blockwork, concrete

Lead

OSB (oriented strand board) Pitch mastic

Plasterboard

Hollow, stone aggregate: Solid stone aggregate: Aerated:

Clay, medium density: Concrete

Sand lime:

Reinforced:

Board, semi-compressed: Flooring:

Bitumen impregnated insulating board: Hardboard:

Insulating board:

Mediumboard:

Plate:

Building panels:

Glass fibre thermal insulation: Mineralfibre thermal insulation:

12.5mm plasterboard 2 sides Proprietary Sheet: Flooring: Gypsum: Gypsum: Gypsum-Gypsum Flexible pvc: Pvc vinyl:

American construction and industrial plywood: Canadian Douglas fir and softwood plywood:

100 mm thick 100mm thick 100 mm thick 103 mm thick 103 mm thick 103 mm thick 6 mm thick 100 mm thick 0.6 mm as laid 25 mm thick 25 mm thick 13mm thick 6.3 mm thick 12.7mm thick 9.0 mm thick 6.4 mm cast clear and armoured

75mm

to 150mm 57 mm to 63 mm 25 mm thick 100mm 100mm 1.7 mm to 3.0 mm 9mm to 18mm 25 mm thick 9.5 mm 12.5mm 15mm 19mm 1.6 mm to 3.2 mm 1.6 mm to 4.8 mm 12.5mm to 19mm 12.5 mm to 18.5 mm kg/rn2 1.5—3.4 2.0—4.4 3.4 34.0—37.0 139.0 219.0 82.0 221.8 237.6 206.0 5.5—7.5 240.0 6.5 4.9 9.8 3.1 —4.5 4.2 3.4 5.0—7.2 16.1 43.9—63.5 20.5—25.9 2.0 2.3 1.2—2.4 29.6 20.4 — 33.6 19.5—34.2 5.7— 12.8 14.9—18.1 6.1 8.0 9.8 15.0 2.4—4.9 3.4— 10.3 7.3—10.9 7.2— 10.6

Weights of materials commonly employed in house construction are 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.

flat: 0.6 mm to 1.2 mm

corrugated0.6 mmto 1.2 mm (including 20% for added laps)

38mmxl9mm: loOmmgauge

3 layer felt bonded together, 13 mm granite chippings

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

Brickwork

Calcium silicate and cement fibreboards

Concrete Copper roofing

Cork

Fibre building board

Glass

Gypsum panels and partitions

Insulation

Internal stud partitions

Dry partitions:

Glass fibre acoustic insulation for floating floors:

Plastic flooring

12mm to 18mm 12mm to 18mm 12.5mm to 19mm

12 mm to 19 mm

Rubber flooring 3.2 mm to 9.5 mm

PSL (Parallel strand lumber)

Rendering Portland cement: sand (1:3):

Roofing felt 3 layers felt and chippings

20 mm asphalt

Sand Dry per cubic metre

thick 48.8

thin

thick 78.1

thin

thick 48.8

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

Clay, machine made: 100 gauge

hand made: 100 gauge

Concrete, plain: 100 gauge

Tiling wall Weatherboarding Wood chipboard

Interlocking, single lap:

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

kg/rn3

610—670

530— 710 730 — 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.

© TRADA Technology Ltd 2006 22

Plywood (continued) Finnish birch: Finnish birch faced:

Swedish softwood: Tropical hardwood: 12.5mm Shingles Slating kg/rn2 8.4— 12.4 8.1 —11.6 5.3—8.0 7.3—11.6 730 - 750 29.3 37.0 46.0 5.4— 16.1 1522—1682 7.3 24.4 Cedar Welsh: Westmorland: Cornish: thin 48.8 29.3 Concrete, plain

Flooring grades, P4 and PS

63.5 70.8 68.4 19mm 15mm thick 18mm thick 22 mm thick 16mm thick 19mm thick 21 mm thick 28 mm thick 16mm thick 19mm thick 21 mm thick 28 mm thick 16mm thick 19mm thick 21 mm thick 28 mm thick 25 mm thick 41.5—56.1 68.0 7.3 10.2—11.2 12.8—14.0 14.3— 15.7 11.7 13.9 15.3 20.3 11.4 13.5 14.9 19.8 8.3 9.8 10.8 14.4 14.6 4.4—7.8 Wood flooring Hardwood: beech, oak

Softwood: pitch pine

Softwood: redwood

Wood wool Slabs:

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

LSL (laminated strand lumber)*

LVL (laminated veneer lumber)*

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

2

k

Plan load w kg/rn =

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

Ceiling dead 25.5 0.25

Ceiling imposed 0.25

200 mm mineral wool quilt 2.4 0.02

Total ceiling design load 0.50

Total design load for roof structure 1.99

Example (2) Flat roof (cold deck type)

Bituminous roofing felt with chippings 35.5

9 mm OSB 5.8

12.5 mm plasterboard 10.0

Self-weight of joists 11.5

200 mm mineral wool quilt 2.4

65.2 0.64

Imposed 0.75

Total design load for roof _1.39

1.4.2 Floors

Example (3) Floor kg/rn2 kN/m2

22 mm tongued and grooved chipboard 14.3

12.5 mm plasterboard 10.0

140 mm phenolic foamboard 4.2

Self-weight of joists 11.0

39.5 0.39

Imposed 1.50

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

12.5 mm plasterboard 10.0

Example (5) WaIl with board cladding

25 mm boards and battens 9 mm OSB sheathing

150 mm glass fibre batt insulation Self-weight of framework

12.5 mm plasterboard

Example (6) Interior wall

12.5 mm plasterboard both sides Self-weight of framework 89.8 0.88 12.5 5.8 5.0 7.0 10.0 40.3 0.40 20.0 5.0 25.0 0.25

1.5 General requirements for prefabricated wall, floor and 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.

©

TRADA Technology Ltd 2006 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.

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 mid-height 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.

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

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

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

mm mm mm 0 100 200 300 400 500 600 700 800 900 1000 400 38 72 11.03 10.03 9.08 8.08 5.95 382 1 69 0.00 000 000 0.00 89 1 17.60 16.41 1529 1421 13.18 12.20 11.25 10.29 8.16 6.03 97 19.24 18.08 16.97 15.91 1488 1388 1292 11.99 114

_?

22.54 21.45 20.40 120 A 23.66 140 145 72 12.77 11.76 10.81 9.90 8.23 610 397 1.84 0.00 000 0.00 97 20.52 19.42 18.36 17.33 16 33 15.36 120 145 72 13.64 12.63 11.67 10.75 938 7.25 5 11 2.98 0.85 0.00 0.00 97 21.19 20.12 19.07 18.06 1707 120 145 I 72 11.03 9.55 8.08 4.88 1.69 000 000 0.00 000 0.00 0.00 89 17.00 15.29 13.69 1220 10.79 8.16 4.96 1.76 0.00 0.00 97 19.24 17.52 15.91 1438 12.92 11.53 9.82 6.62 3.43 114 21.45 1988 18.36 16.90 15.49 120 23.12 21.56 2005 18.59 140 145 72 12.77 11.28 9.90 7.17 3 97 0.78 0.00 0.00 0.00 0.00 0.00 97__ -- 19.42 17.84 16.33 14.89 13.50 12.16 9.01 120 25.37 23.85 145 72 13.64 12.15 10.75 8.31 5.11 1.92 0.00 000 0.00 0.00 0.00 38 38 38

—

38 38 44 44_ 44 44

--

47 47 47 47 600 38

3_

38 38

-38 38 44 44 44 4._ 97 21.19 19.59 18.06 16.59 15.17 1380 11.81 26.51 — 47_

1p

47 145

NB Values for a wind pressure of zero are given to permit interpolation between 0 and 1 00 Pa. iShaded values are governed by bearing stress in the bottom 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 Breadth Depth Load duration

mm mm mm 400/600 38 72 38 89 38 97 38 114 38 120 38 140 38 145 44 72 44 97 44 120 44 145 47 72 47 97 47 47 120 145 i eaiurn _JLong

I/

Iw 'iWa

Iw

t k

I I.

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

Spacing Breadth Depth

mm mm mm 400 38 72 38 97 38 114 38 120 38 140 38 145 44 72 44 97 44 120 44 145 47 72 47 97 47 120 47 145 600 38 72 38 89 38 97 38 114 38 120 38 140 38 145 44 72 44 97 44 120 44 145 47 72 47 97 — 47 120 Wind 0 pressure in 100 Pa (N/rn2) 200 ₁₃₀₀ 400 500 600 700 800 900 1000 13.43 12.55 11.71 10.91 9.24 7.11 4.98 2.85 0.71 0.00 0.00 19.28 18.66 17.73 16.83 15.96 15.11 14.29 12.24 21.01 20.46 19.54 18.66 17.79 24 3O32 .32 31.40 .40 40 .40 1.40 31.40 .40 31. 15.55 14.67 13.82 13.00 12.04 9.91 7.78 5.65 3.52 1.39 0.00 22.91 22.87 21.97 28.34 34. .25 . .25 34.25 .25 16.61 15.73 14.88 14.06 13.26 11.31 9.18 7.05 4.92 2.79 0.66 23.72 29 35.46 .46 35. .46 13.43 12.13 10.91 8.17 4.98 1.78 0.00 0.00 0.00 0.00 0.00 19.28 18.19 16.83 15.53 14.29 11.17 7.98 4.78 1.58 21.01 .01 20.00 18.66 17.36 16.12 14.66 11.47 24 23.60 25.99 30.32 .32 31.40 1.40 . 31.40 31.40 1.40 .40 .40 15.55 14.24 13.00 10.98 7.78 4.58 1.39 0.00 0.00 0.00 0.00 22.91 22.87 21.53 20.24 18.99 17.77 34 34.25 .25 34. 34 25 .25 16.61 15.30 14.06 12.38 9.18 5.99 2.79 0.00 0.00 0.00 23.72 23.63 22.32 19.81 29.35 35.46 .46 . .46

NB Values for a wind pressure of zero are given to permit interpolation between 0 and 1 00 Pa. I Shaded values are governed by bearing stress in the bottom rail

Values in italics are governed by a deflection limit of 0.003L. Other values are governed by buckling in the stud. (b) Without wind loading: medium — and long term load cases _I

Spacing Breadth Depth Load duration

mrn rnm rnm 400/600 38 72 38 89 38 97 38 114 38 120 38 140 38 145 44 72 44 97 44 120 44 145 47 72 47

_-

97 47 120 47 145 I. vieaiurn Lon 1114, , .1891 13.77

,

.

,.

15.01

_{, r4c -vt}

17.64 11

i4

9 ,

21.66

.

7. .. — 22.

,

,

12.15

_.

72

.

16.36

_{. t..}

09

_.

20.24

_rr

_.

19

c

2 . 12.58 ,

, ,. ,, '.

.

1 , ,

?7

25.33 .27

©

TRADA Technology Ltd 2006 28

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

38 89

47 145

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:

0.005 Aca + öma Z

=

(öemean — öca) X

Le 0.85L1if

where A =

_______

I h 2 a_e,mean mean A2 bh2

andZ

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.

2.1.6 Drilling of studs

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

Stud Location Maximum size

Load bearing 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.

0.25 x depth of stud.

Non-load bearing partition

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.

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

Lintel: solid timber, structural timber composite

or fitch beam.

© TRADA Technology Ltd 2006 30

Stud material from opening

shared equally in framing each side

Lintel below first floor

designed to support load

from double stud to side of window above, including weight of construction from

other studs.

II

II

II

II

H

ii

Ii

II

H 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.

Loads at 600 mm centres 600 600

I

600

k

1200 1800 2400 1200

a 1 mm to 599 mm

Loads at 400 mm centres

_

_rrrrr

A400

A 400 A

A400

T

T

T T

1200 1800 2400

00 4

1800

4

2400

a 1 mm to 399 mm

© TRADA Technology Ltd 2006 32

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

Span No of loads Depth 72

9!

120 145 170 195

21)

245 Breadth '1 21 2.26 3 8 10 13.01 1 64 3.04 1 17.53

202

3.7 .69 0.90

201

1.21 2.7 1.50 3.35 72 97 120

45

171) 195 220 245 1200 1 72 97 120

2

72 97 120 Span No of

loads Depth_Breadth

1.47 2.41 3.26 4.23 5.31 6.51 1 98 3.25 4. 5.69 7.1 8.77

245

4.02 5 7. .84 1 27 1.93 2 1 71 2.60 3.51 2 12 3.22 4.35 120 141)

70

105 220 245 1800 2 72 0.33 0.79 97 0.45 1.07 120 0.55 1.32 3 72 0.29 0.69 97 0.39 0.93 120 0.48 1.14 Span

Noof

loads Depth 72 97 Breadth 2400 3 72 0.13 0.33 0.61

(

97 0.18 0.44 0,82

1.4

120 0.22 0.54 1.01 'I 7j,,, 4 72 0.13 0.30 0.57

098i

1 5 97 0.17 0.41 0.76 _32j 2.1 120 0.21 0.50 0.94

I 'f

_2.59J

189

2

'1

488

28

4

8

26

811'ft91a48 a1J35

9, 1.91 2.58 3.34 4.20

4.7

8 2.36 3.19 4.13 5.19 5.91 38 1.38 1.86

2.26

2.55

2.84

')

1.86 2.51

3.04

3.43

3.83 2.30 3.11

3.77

4.25

4.73 145 175

1I5

220

24

1.41 1.77 2.17 1.90 2.38 2.92 -2.35 2 95 3.61 1.39 1.75 2.14 1.87 2.35 2.88 2.32 2.91 3.56 — sh-?r .)c.virft Point loads at 400 mm_spacing

Span

Noof

loads Depth 72 97 120 145 170 195 220 245 Breadth 1200 1 72 97 120 0.72 0.97 1.21 97 120 145 170 195 220 245 2 72 97 120 0.63 0.84 1.04 Span

Noof

loads Depth 72 Breadth 1800 2 72 0.20 0.48 0. 0.64 1.1 0.80

lÀ

0.48 0. 0.64 1. 97 0.27 120 0.33 3 72 0.20 97 0.27 120 0.33 0.80

1.4

Span No of loads 2400 3 Breadth Depth 72 97 120 4 0.21 0.39 0.68 1.08 0.28 0.53 0.92 1.46 0.15 0.35 0.66 1.14 1.80 0.20 0.38 0.66 1.05 0.28 0.51 0.89 1.41 0.34 0,64 1.10 1.75

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

©

TRADA Technology Ltd 2006 34

Point loads at 600 mm spacing

Span No of loads Depth 72 97 120 145 170 195 220 245 Breadth 1.21 13.01 1.64 1 . 17.53 2.02 1 0.90 1.21 1.50 72 97 120 145 170 195 220 245 1200 1 72 97 120

2

72 97 120 Span

Noof

loads Depth Breadth 1800 2 72 0.33 0,79 1 47 2.41 26 4 5.31 6.51 1 98 3 .1 8.77

245

4 10.84 1.27 1.71 2.12 3 120 145

70

195 220 246 0.61 1 05 1 2.11 3.25 0.82 1.42 2 3 4.38 1.01 1.76 2. 3.52 4.42 5.42 0.57 0.98 1.55 2. 57 0.76 1 32 2.09 2.7 3.47 0.94 1 64 2 59 3.42 4. 97 0.45 1.07 120 0.55 1.32

3

72 0.29 0.69 97 0.39 0.93 120 0.48 1.14 Span No of loads Depth 72 97 Breadth 2400

3

72 0.13 0.33 97 0.18 0.44 120 0.22 0.54 4 72 0.13 0.30 97 0.17 0.41 120 0.21 0.50

Point

loads at 400 mm spacing

Span No of loads Depth 72! 97! 120! 145! 170! 195! 220! 245! 1800 2 72 0.20 0.48 Breadth 1200 1 72 0.72 ₁ 97 0.97 120 121 2 72 0.63 1 97 0.84 120 1.04

Span

Noof

Depth 72 97 120 145 17 195 220 245

loads _Breadth 0.89 1. 1

2

3.1

______ ______

1.19 .91 148

088

1. 1.19 1.47 2. .11 120 145 17C 195 220 245 97 0.27 0.64 120 0.33 0.80 3 72 0.20 0.48 97 0.27 0.64 120 0.33 0.80 Depth 2400 3 72 97 72 0.09 0.21

4

120 0.39 0.68 97 0.12 0.28 0.53 0.15 0.35 72 0.08 0.20 Span No. of loads _Breadth 2.17 2.38 2.92 2.95 3.611 1.75 _2.141 2.35

_2.8

2.91 3.561 :hear qoverrv, 08 16 80 -1.05 41

175

[1

1.41 1.90 2.35 1.39 1.87 2.32 0.66 1.14 97 0,11 0.28 0.51 120 0.38 0.66 0.14 0.34

F = deflection governs j I = bending governs 0.89 0,64

1.

Table 2.5 Permissible loads on lintels: Timber strength grading class C24 — Long-term loading Point loads at 600 mm spacing

Span No of loads 1200 1 Depth Breadth 72 97 120 72! 971 120! 145! 170! 195! 2201 245 2 72 97 120 Span No of loads 1800 2 Depth Breadth 72 0.41 0.99 1.51 3.20 4.78 6 11 2.03 4.31 6.44 9.21 2.51

5

1 1.12 1 50 1 86

7

07 120

I'S

170

ic5

'20

245

_____ _____

1.83

13

4.61 5.

_____ _____ _____

422

6

3.04

522

7 ______ ______ 1.58 ______ ______ 2.12 2 63 120 145 175

95

220 245 3 97 0.55 1.33 2.46 120 0.68 72 0,36 1.64 Span 0.85 97 0.48 1.15 120 0.59 No. of

load _BreadthDepth

1.42 72 97 I I I 2400 3 72 0.17 0.40 0.76 1.31

2 07[ 2.99 3.76

4.60 1.76 2 78 4.03 5 06 6.20 2.18

'.44

4.98 6.26 7.67 1.22 _-. 93

2.66 3. 3,

1.64

2 59 3. 4 4,50

2.03 21

4.43

5.00

5.57 97 0.23 0.54 1.02 120 0.28 0.67 1.26 4 72 0.16 0.38 0.70 97 0.21 0.51 0.95 120 0.26 0.63 1.17

Point loads at 400 mm spacing

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

20

245 Breadth 0.90 2.11 3. 1.21 2,84

4

1.50 3.51 0.78 1.05 1.29 7. 97 120 45 170

95

220 245 1200 1 72 97 120 2 72 97 120 Span No. of loads Depth Breadth 1800 2 72 0.25 0.59 1.10 1 89 0.33 0.80 1.48 2.54 0.41 0.99 1.83 3.14 0.25 0.59

110

0.33 0.80 1.48 0.41 0.99 1.83 72 97 120 145 75 105 220

45

0.11

026 U9 255

1..4

98

2.50 3.07 0.15

035

066

14 1,81

' 67

3.37 4.13 0.18 0.44

0;2

1.42 2 24 .3.30 4.17 5.11 0.10 0.25

047

0.82 1.30 0.14 0.34 0.64 1.11 1 75 0.17 0.42 0.79 1,37 ' 17 97 120 3 72 97 120 Span No. of loads Depth Breadth 2400 3 72 97 120 4 72 97 120