The Steel Construction
Institute
S
Design of Structural Steelwork
Lattice Framed Industrial Building (Revised Edition)
/
mit deutscher Zusammenfassung avec résumé français
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con resumen español
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a Construction Métallique
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con sommario italiano Institut de
This document contains 100 pages
Institut für Stahlbau
Istituto di Costruzioni in Acciaio
Instituto de Ia ConstrucciOnMetálica
==
is The Steel ConstructionInstitute. Its aim is to promote the proper and effectiveuse of steel in construction.
Membership is open to all organisations and individuals that are concernedwith the use of steel in construction, and members include designers, contractors, suppliers,
fabricators, academicsandgovernmentdepartmentsin the UnitedKingdom, elsewhere in Europe and in countriesaround the world. SC! is financedby subscriptions from its members, by revenue from research contracts and consultancy servicesandby the sales
of publications.
SCI's work is initiated and guided through the involvement of its members on advisory groups and technical committees. A specialist advisory and consultancy service is available free to members on the use of steel in construction. SC!'s research and development activities cover many aspects of steel construction including multi-storey construction, industrial buildings, use of steel in housing, development of design guidance on the use of stainless steel and cold formed steel, behaviour of steel in fire, fire engineering,use of steel in barrage and tunnel schemes, bridge engineering, offshore engineering, and development of structural analysis systems.
Further information is given in the SC! prospectus available free on request from: The Membership Secretary, The Steel Construction Institute, Silwood Park, Ascot, Berkshire, SL5 7QN. Telephone: (0344) 23345, Fax: (0344) 22944. Although care has been taken to ensure, to the best of our knowledge,
that all data and informationcontainedherein are accurate to the extent that they relate to either matters of fact or accepted practice or matters ofopinionat the time of publication, the Steel ConstructionInstitute, the authors and the reviewersassume no responsibility for any errors in or misinterpretations ofsuch data and/or informationor any loss or damage arising from or related to their use. Publications supplied to the Members ofthe Institute at a discount are notforresale by them.
© The Steel Construction!nstitute 1993
Instituut voor Staalbouwconstructie Staalkonstruktion !nstitut Instituto da Construcao Metálica !voriroisro Yuipó.v iccxraaicevó.v
Institut de la ConstructionMétallique Institut für Stahlbau !stituto di Costruzioni in Acciaio Instituto de Ia ConstrucciónMetálica
SCI PUBLICATION 028
Design of Structural Steelwork Lattice Framed Industrial Building
(Revised Edition)
Entwurf elnes Stah/bau-Gebãudes - G/tterahmen /ndustriegebãude Dimensionnementd'/mmeubles a structure meta/llque - bétiment industriel en cadre et treI/I/s Progettazionedl Ed/f/cl in Accialo: Ed/f/cl Industrial! Inte/alat/ a Tra/lcclo Proyecto de Ed/f/c/os con Estructura de Acero. Ed/f/do /ndustr/al en Ce/os/a
C SOUTHCOMBE ISBN
1
BSc(Eng),
MSc(Eng), CEng, MICE
870004 83 3
British Library
Cataloguing
in Publication
Data
A catalogue record for this book is available from the British Library © The Steel Construction Institute 1993
The Steel Construction Institute Silwood Park Ascot Berkshire SL5 7QN Telephone: 0344 23345 Fax: 0344 22944
FOREWORD This publication is a revised edition of the original text written by Mr W Bates and first published in 1983.
Its purpose is to aid the education of undergraduate students in Engineeringby providing sample calculations for a typical industrial building capable of future extension. The revisionwas made necessary by changes in design Codes and current practice over the past decade.
For their helpful contributions regardingdesign, fabricationand the erectionprocess, the author is indebted to:
Mr. A. Curnow (Blight and White Limited, Plymouth) Mr. R. Fox (F. Parkin and Son Ltd., Exeter) Mr. P. Marozinski(Conder Limited, Winchester)
11
CONTENTS Page FOREWORD
U
SUMMARY
v
1.
INTRODUCTION
1
2.
SCOPE
2
3.
STANDARDS AND CODES OF PRACTICE
4
3.1
3.2 3.3 3.4 3.5 3.6 3.7 3.8 4.
5.
British Standard 5950 - Structural use of steelwork in building BS 5502 - Buildings and structures for agriculture BS 6399: Part 1: 1 984 - Design Loading for Buildings BS 6399: Part 3: 1 988 - Code of practice for imposed roof loads CP3: Chapter V: Part 2: 1972- Wind Loads Statutory regulations National structural steelwork specification for building construction (2nd Edition) Quality assurance
4 4 4 4 4 5 5 5
BUILDING FORM
6
4.1
General
4.2
Low pitch roofs
6 6
LATTICE FRAMED ROOFS 5.1
5.2 5.3 5.4 5.5
8
Simpleforms More complex forms
8
10
Cladding Purlins Side rails
12
6.
CONCEPTUAL DESIGN
16
7.
PRINCIPLESOF DESIGN
19
7.1
19 19 19
7.2 7.3
7.4 7.5
13 14
Purlins and side rails Lattice framed roof girders Stanchions Bracing Connections
21 21
111
CONTENTS - Continued Page
8.
EXAMPLE - DESIGN BRIEF AND APPROACH
24
8.1
24 24
8.2 9.
10.
Brief Cladding
DESIGN OF STEELWORK
27
9.1
Loading
9.2 9.3 9.4 9.5 9.6 9.7 9.8
Assessment of roof load Assessment of wind load on structure
27 27 28
Design of purlins Design of main roof frame Preliminary calculations Loading Cases (for characteristic loads) Analyses
31
36 37
40 40
FINAL DESIGN
49
Top boom 10.2 Bottom boom 10.3 Internal members 10.4 Comparison of member sizes 10.5 Column design - members 1 to 4 and 5 10.6 Gable steelwork
49
10.7 10.8
67 70 73
10.1
10.9
Bracing Column Base (Reference 1. Clause 4.13) Foundation
51
53 54 55 61
11.
ALTERNATIVE FRAME ANALYSIS
75
12.
JOINT DESIGN
78
Application limit check list 12.2 Joint welds
78
FINAL FRAME LAYOUT
84
1 2.1
13.
81
REFERENCES
87
BIBLIOGRAPHY
89
iv
SUMMARY
-
Design of structural steelwork Lattice framed industrial building
Thedesignerof singlestorey buildings for commercial and industrial use will consider a numberof possible solutions. A decisionhas to be made regardingcladding, structural form and material. This publication illustrates for the benefit of students, the many factors which influence the final choiceof a suitable design. Consideration is given to a variety of building forms as well as to the choice of cladding and its supportingelement at the conceptual design stage; other factors influencing the design are related to fabrication, transport and erection. A structural steelwork frame incorporating solid web beams for columns and a latticed structurefor the roof, is chosen and full design details worked out.
The detailed design of a building 30 m wide, 48 m long x 6 m to eaves is providedas an illustration.
The solutionconsiders the main loading calculations and members initially. A detailed analysis is carried out and checks are made of all members, the latticed roof being formed of rectangular hollow section. Typical joints and the foundation are designed.
Entwurf eines Stahlbau-Gebäudes - Gitterahmen Industriegebäude Zusammenfassung Der Konstrukteuer eines eingeschossigen Handels - oder Industrie-Gebaudeswird eine Reihe moglicherLOsungen in Berrachtziehen. Entscheidungen mQssen getroffenwerden hinsichteich Verkleidung, Formgebung und zu verwendender Werkstoffe. Diese Veroffentlichungillustriert zum Nutzen von Studenten die vielen Faktoren, die die endgtlltige Wahi eines geeigneten Entwurfsbeeinflussen. Bei der Konzeptentwickiungwerden verschiedene Gebäude-Formen als auch eine Auswahl von Verkleidungen und ihre Befestigungs - Elemente betrachiet; andere Fakioren, die den Entwuif beeinflussen, betreffen ilerstellung, Transport und Errichtung. Stahlbaurahmen mit soliden Stegträgernfir die Stlitzen und etne GitterstrukiurftJrdas Dach wird gewahit, wozu alle Entwurfs-Einzelheiten ausgearbeiterwurden. Em
Als illustration1st der detaillierteEnlwurfeines Gebäudes mit 30 m Breite und 48 m Lange, sowie 6 m bis zur Unterkante des Daches dargesteilt. Bel der LOsung wurden die wesentlichen Lastberechnungen der Glieder im Ausgangszustand beracksichtigt. Eine deraillicerteAnalyse wurde durchgefilhrr sowie alle Glieder aberprtlft; das Rahmendach wird aus rechteckigen Hohlquerschnirren gebildet. Typische Verbindungen und die Grtindung sind dargesrelir.
V
Dimensionnement d'immeubles a structure métallique - bãtiment industriel en cadre et treitlis Résumé Le projeteur d'immeubles, a un seul niveau, pour usage industrielet commercial peut envisagerde nombreuses solutions constructives, unedecisiondoit être prise concernantla forme structurale, les parios et le matériau. Cette publication discute, a 1'intention des étudiants, les nombreuxfacteursqui influencent le choix d'un bon dimensionnement.
On considèreune grande variétE deformes de bãtiments ainsi que le choix des parois et des éléments qui les supportent, dansle cadre de I'etape de conception du bãtiment. D'autres facteurs qui influencent le dimensionnement et qui sont relatfs a lafabrication, au transport et au montage, sont egalement discutés. Une structure en acier comportantdes colonnes en prof/set une toiture en treillis, est choisie et étudiéeen detail. Le dimensionnement détaillC d'un bâtimentde 30 m de large, 48 m de long et 6 m sous la toiture est donnC comme illustration.
La solutioncomporte une analyse détai!lée et une verfication de tous les Cléments, le treillis de toiture étant rCalisC en profils creux rectangulaires;certains assemblagesainsi que les fondations sontegalementétudiés.
Progettazione di Edilici in Acciaio: Edifici Industriali Intelaiati a Traliccio Sommario Nellaprogettazionedi edfici monopiano ad uso commerciale e industrialedevono essere esaminatedirvesepossibili so!uzioni. E' necessariooperare Ia scelta del rivestimento, della struttura portante e del materiale. Questapubblicazionepresenta, a beneficio degli studenti, tutti queifattori che infiuenzano la sceltafinale in vista di una adeguataprogettazione.
Perlafase preliminare di progettazioneviene presa in considerazione la varieta'delle tipologiestrutturali, !a scelta del rivestimento e dci suoi elementi di collegamento, a/tn fattori che influenzano ii progetto sono que/li relativialla lavorazione, a! trasporto ed al montaggio. Si il/ustra in particolare, sviluppando tutti i dettagli relativi al progetto, un edfici intelaiato in acciaio,formato da colonne ad animapiena e da elementi di copertura realizzati con una struttura a traliccio.
A titolo di esempioviene presentata la progettazione dettagliata di un edficio alto 6 metri con dimensioni in piantadi 30 metri di larghezzae 48 metni di lunghezza. Sonopresentati I principali calcoli relativi ai carichi ed al predimensionamento. L'analisi dettagliata e' seguita dalla verfica di tutti gli elementi portanti. In particolare Ia struttura a traliccio onizzontale e' formatada sezioni rettangolari cave. Vengono inollreprogettati alcuni giunti tipici e le fondazioni.
vi
Proyecto de Edificios con Estructura de Acero. Edificlo Industrialen Celosia Resumen
Elproyectista de edficios de una plantapara usos comerciales o industrialesdisponede dferentes posibles soluciones. Para la selecciOn deben tomarse decisionessobre revestimientos, materialesyfonna de la estructura. Esta publicaciOn aclarapara los estudiantestodos losfactores que infiuyen en Ia eiecciónfinal de un proyecto adecuado. Se analizan dferenresformas de ed/lcios as! como la elecciOn de revestimiento y sus elementos de soporte a nivel de diseflo conceptual. Se tratan ademds otrosfactores influyentes relacionadoscon lafabricaciOn,transportey montaje. Se escoge como modelo una estructura aporticada de aceroformada porperfiles de alma ilena en lospilaresy una celostapara la cubierta, desarroiidndosecompletamente todos los detalles del proyecro. Como ilustraciOn se incluye ci ca/cub dew//ado de un edficiode 30 m. de anchura, 48 m de iongitud y 6 m. de altura. La soluciOn comienza considerandolas cargas principales sobre las barras. A continuaciOn se 1/eva a cabo un análisis dew//ado as( como la comprobaciOn de todas las barras (la celos(a de cubierta estdformada por tubos rectangulares). Tamb(en se proyectan los nudosy zapatas t(picos.
vii
1. INTRODUCTION In general the basic brieffor the design of the majority of single storey buildingsfor industrial and commercial use is to provide, for the client, a structure which has no internalcolumns. If some columnsare essential the numbershould be limited. Thus, in principle,the requirement is for the constructionof four walls and a rooffor a singleor multi bay structure. The walls can be formed of different materials e.g. steel columns with claddingwhich may be of profiled or plain sheet, precast concrete, or masonry load bearing walls etc. The designer will generally considerfor the roof a system of beams or latticed frameworks in structural steel to supportthe roofcladding. Solid web beams will make use of universalbeam sections. The use of lightlatticed frameworks for the roofof an industrial buildingprovides a neat, efficientstructurewhich frequently satisfies architectural requirements. The design of the steelworkis simple. Modern fabrication systems and erectionprocedures make these structural forms economic. This is particularly apparent when it is appreciated how many industrial buildings today employ latticed roof framing and how many makers of standard buildings, as well as suppliers of industrialised buildingsystems, make use of this type of framing in preferenceto solid web beam construction.
The purpose of this publication is to discussthe many factors which can influence the decision making process and can lead to adoptinglatticed framework construction. Alternative design solutions are then illustrated by means of a practical example.
1
2. SCOPE The scope of the publication is mainly restricted to plane frame structures. Other forms, such as space frames, are not considered in detail. Various types of steel sections are used in the construction of the components for this type of structure, viz, hot rolled structural shapes such as universalbeams, universal columns, angles, structural hollow sections and cold formed sections,etc.
Importantfactors which must be considered at the conceptual stageof the design process are the questions of workshopfacilities - including size - and transportationbetweenworkshop and site. Whilst long girders or large sectionsmay appear to be desirable, in order to reduce the number of site connections, this can reduce the numberof fabricatorswho couldtender for a given project.
In the UnitedKingdom, road transport is normally used and loads up to 2.9 m width, 18.3 m
long and 76,200 kg weight may be moved withoutany problems. Above these dimensionsthe Police need to be notified of "Abnormal IndivisibleLoads" and indemnity to Highway and Bridge Authorities is required. Where the dimensions exceed width 6.1 m, length 27.4 m, or weight 152,400 kg a Department of TransportSpecial Order is required. (Reference 'Abnormal IndivisibleLoads', "Aide Memoirefor Requirements as to Notice and Authorisation when not complying with Construction and Use Regulations",Source: Director (Transport), Departments of the Environment and Transport).
It should be noted that the various police authorities have different periods when abnormal
loads are allowed to movethrough their districts. If neighbouring "times" are significantly out of phase and general traffic hold-upscause disruption to the movement of abnormal loads it is possiblefor the latter to be delayed by up to 24 hours. If one or more cranes and associated erection staff are held up by these enforceddelays, the additional costs can be very significant.
Certain towns and cities place length restrictionson materials which can be moved by road e.g. certain areas of London restrict lengthsto 12 m. Girders can be fabricatedand despatched lying flat, the overall height of the load is dependent upon the route travelledand the clear height of any bridges likely to he encountered. Rail transport can accommodate long pieces, but width and height are more restricted. One solutionto limit the length and height of units being transported is to use a system as illustrated in Figure 1. The two external sections are shop welded and the central section is site or shop assembled; the whole being bolted togetheron site. The completed rafter can be craned into position. For export where shipmentis involved, pieces up to the same dimensions as for road transport may be accommodated but it should be appreciated that shipping charges are often based on volume rather than weight. Often there are relatively severe restrictionson the length of a piece that can be carried in the hold of a ship. The ship's engineer may refuse to carry the steelwork as deck cargo. It may be found more economical to despatch the steel piece-small for subsequent assembly on site. Care must then be taken to ensure that the site work is satisfactory.
Other factors of importance which can influence the economics of this type of construction are the facilities available for fabrication and for erectionon site.
2
Many fabricationshops now have equipment which can cut and hole steelworkin a semi-automatic manner thus reducingdirect labour costs. Jigs can also be used for the rapid assembly of components. All these tend to make lattice construction more attractive. On site the lighter overall weight of individual components can result in the use of simple lifting equipment; site costs rise appreciably if heavy cranes have to be installed for erection purposes.
For the design examplein this publication it is assumed that the building is for the home market and that a well equipped fabricatorwill manufactureand erect the steelwork. It follows that the design must be in accordance with the appropriateBritish Standards, codes and regulations. Brief explanatory notes on these publications are given in Section 3. External section
Central
section
7,7,7
Z7rr
Figure 1 Sectionedgirder
3
3. STANDARDS AND CODES OF PRACTICE 3.1 British Standard 5950 - Structural use
of steelwork in building
This document' is in nine parts combining codes of practiceto cover the design, construction and fire protectionof steel structuresand specifications for materials, workmanship and erection.
The relevantparts incorporated into this publication are Parts 1 and 5.
3.1.1
BS 5950: Part 1: 1990 Code of practice for design in simple and continuous construction: hot rolled sections
This limit state specification provides limiting values for strength and deformationfor various elements which form part of structures, and for whole systems. The document' covers aspectsrelated to hot rolled sections i.e. UBs, UCs, angles, channels, hollow sections,etc.
3.1.2
BS 5950: Part 5: 1987 Code of practice for design of cold formed
sections
This specification2, using limit state philosophy, provides limiting values for strength and deformation and identifies full design procedures and empirical methods. Within this
publication, it is used in the design of purlins and side sheetingrails.
3.2 BS 5502 - Buildings and structures for agriculture Various parts which cover materials,design, construction and loadings3.
3.3 BS 6399: Part
1:
1984 - Design Loading for Buildings
This is a "Code of practice for dead and imposed loads" for use in designing buildings(4(this is providedas a revisionto CP3 Chapter V Part 1: 1967 which it supercedes).
3.4 BS 6399: Part 3: 1988 - Code of practice for imposed roof loads
This is a "Code of practice for imposed roof loads" and in particular suggests methods of considering snow loads for various buildings5. The loads can be used for permissiblestress design or where factored loads are adopted.
This code recognises the variationin snow loading throughout the United Kingdom and the effect of variable snow loads on a roof due to drifting effects.
Chapter V: Part 2: 1972 - Wind Loads The effect of wind on a buildinghas been found to be very complex and dependentupon
3.5 CP3:
many factors such as the geographical location, the shape of the building and its relationship, to other buildings and natural features. The various rules for calculating the design wind loads on a structureand its cladding are given in this codeof practice6, supplemented by a guide published by the Building Research Establishment7.
4
This code will be replaced by BS 6399: Part 2.
3.6 Statutory regulations In additionto the above the buildings must comply with the requirements of the Building Regulations, which apply in Englandand Wales, and where appropriatewith the special variations or equivalentregulationsapplicable throughout the UK. Particularthermal and sound insulationrequirements of the cladding must also be met. For buildings outside the United Kingdom the local regulations must be observed. Whilstmany places accept structures designedto British Standards care must be taken to considerany unusual features such as typhoonsor earthquakes.
3.7 National structural steelwork specification for building construction (2nd Edition) The object of this publication8 by BCSA and SCI is to achieve greater uniformity in contract specifications issued with tender and contract documents.
3.8 Quality assurance BSI Handbook provides a comprehensive documentof the relevantstandards associated with this topic. Of particular interest to the designer/fabricator/erector is BS 5750 : 1987'° which provides a three level specification of QA requirements in the contractual situation.
5
4. BUILDING FORM 4.1 General Before proceedingto the detailed design of a lattice framed roof it is desirable to consider the alternatives available.
At the outset is must be appreciated that if an industrial building is to be warm during the winter and cool during the summersome form of heating and ventilationis required in addition to the thermal insulation called for by the Thermal Insulation(Industrial Buildings) Regulations. The roof space, which will be heated with the restof the buildingunless cut off completely by a horizontal ceiling, is a constantcharge on running costs withoutcontributing to the work space. There are, therefore, financial advantages in keeping the roof space to a minimum bearing in mind that services can be accommodated in this space. This can be achieved by keeping the roof space as shallow as possible, commensurate with economy of initial cost and efficiency of the cladding.
A flat roof, or a roofwith only a nominal camber, can reduce the roofspace to the minimum but may be expensive to build since the roofcladding will have to be of a more sophisticated nature to ensure adequate weather protection. Again, with a flat roofof any reasonablespan, deflectionof the structureor girders becomes important and extra steelworkmay be required merely to reduce it. A portal frame design helps to reducethe deflection but it does not reduce the cost ofthe cladding and the provisionof the necessary rigid joints is an added cost on the structure. Probably the most economical form of roofconstruction is one of low pitch (say 50 which is the preferred minimum) on which a simple form of cladding can be used with success and which at the same time reduces deflection whilst maintaining reasonable heating costs. However care is required in the selection of the type of sheet, the type of fixing and the sealing of end laps (which should be avoided, if possible). Special care is required where translucent sheets are required (see section 5.3 on cladding). For other than raised seam ° roofing 7½ is the preferredminimum slope.
4.2 Low pitch roofs Such low pitch roofs can be supportedby
either solid web beams, castellated beams or lattice frames. Each has advantages and disadvantages which must he examined before a decision can be made.
4.2.1
Solid web beam
This is the heaviest form though relatively simple and cheap to make. However, the depth of section satisfactoryfor structural purposes may be too shallow for the penetration of service ducting. A monorail or underslung crane can be supported at any positionbut local stiffening of the section may then be required.
4.2.2
Castellated beam
This is a method of increasing the sectional propertiesof a beam without materially increasing the weight. The roof space increases but some servicescan be accommodated in the castellations. Monorails can be located as required but it may be necessary to fill in local castellations and stiffen the flange to carry the load. Castellated beams increase the bending strength and flexural stiffness quite significantly. Enhanced shear capacity at points of high shear can be accommodated by filling the castellations in that region.
6
4.2.3 Lattice frame Figure 2 showsthreedifferent types of rafter and indicates the facilitiesfor services and monorails. It also illustratesthat, notwithstanding its extra depth, the lattice frame has a distinct advantage where serviceshave to be carried in the roof. In addition, the reduction in weight of the girder can result in economy in the supportingstructure and foundations.
This is the lightestform of construction though it requiresmore fabrication. The roof space increases but services can usually be accommodated withinthe depth of the girder. Monorails supported at the panel points cause little problem, but if they are located betweenthem some local stiffeningmay be required. The latticed girder will have a much larger second moment of area and section modulus (about XX axis) than a corresponding solid web beam of a similar weight. Therefore there will be enhanced strength and stiffness.
Solid web beam
Type (a)
Monorail
Type (b)
Type (c)
Figure 2 Typicalroof girders
7
5. LATTICE FRAMED ROOFS 5.1 Simple forms on the overall dimensions of the building,the latticeframed roof can take many forms, some of which are examined below: Depending
5.1.1
Single bay low pitch roof Economicallyspans up to 30m are often fabricated using standard UB, UC section portals. Above this span lighter rafters are providedby latticed girders, as shown in Figures 1 and 3. The advantage of the horizontal boom is that designing for the "kick out" effect, Figure 4, is removed. Columns are then only designed for axial load and moment (due to the eccentricity of the load) from the roof, in addition to wind load on the vertical cladding. A factor to be consideredis the possible lengthening of the bottom boom due to tensile strain.
Figure
3 Single baylow pitch roof
Eaves displacement
— —— /
Figure 4 "Kick out "effect
5.1.2 Multi-bay low pitch roof The single low pitch roofcan be extended into a series of similar bays (Figure 5). Alternate stanchions in the valley can be omitted, the intermediate roof frames being carried on a longitudinal valley girder, spanning two longitudinal bays, as indicated.
5.1.3 Single bay monopitch roof When the slope of the roof is low it is sometimes advantageous to use a monopitch roof (Figure 6). The extra roof space can be compensated for by the saving in drainage since a gutter is requiredonly along one edge and not two. Monopitch roofs are mainly used for relatively small spans.
8
Eaves
gutter
Ridge
Cladding
Side
cladding Figure 5 Multi-bay pitch roof flashing
Side claddi
bolts Figure 6 Single bay monopitch roof
5.1.4 Multi-bay roof In combiningframes to obtain a multi-bay system alternate stanchions can be omitted (Figure 7). The roof is supportedat the apex and the valley by girders spanning two longitudinal bays. Alternatively a multi-bay frame can be provided using a multi-monopitch roofarrangement(Figure 8.) It is preferableto ensure that a valley gutter is wide enough for an erector or maintenance operative to stand in. In the alternativecase using mono pitch roofs (Figure 8) the latticeframes all slope in the same direction. Extra gutters are requiredbut advantage can be taken to introduce lights above the valley gutters. This system is particularlyuseful if direct sunlightinto a building is to be avoided. The glazing can then be provided in the north facing slope of the saw-toothed roof.
Side
Ridge flashing
Longitudinal girders Stanchions at alternate frames
Figure 7 Alternative multi-bay pitch roof
9
North light
Cladding
Figure 8 Alternative form of multi-bay using monopitch roof
5.2 More complex forms Where large internalareas are to be relatively free of stanchions,a doublelatticed system can be adopted. Here, secondary frames in one direction are supportedby primary frames spanning in the other direction between widely spaced stanchions. These notes on latticeframed construction would not be completewithoutsome reference to more complicated forms built up of lattice frames or lattice girders and trusses and of space frames.
5.2.1
Umbrella roof
In thisform of construction light trusses are slung either side of main lattice girders (Figure 9). The pitch of the roof must be sufficient to accommodate the main girders which in turn should be of sufficient depth to avoid excessiveflexibility, bearing in mind the incidental application of imposed and wind loading. Care needs to be taken to ensure adequate provisionfor drainage of rainwater. The trusses act as cantilevers with the bottom chord in compression from imposed loading but
wind loading may cause a reversal of stress. Since these compression members are not laterally restrained (in normal truss construction the rafters are the main compression members and they are restrained by the purlins etc.) a system of inclined or horizontal bracing is required. Eaves
Roof
Ridge
—
— Stanchion
Cantilever trusses Floor, level
cladding Figure 9 Umbrella roof
10
5.2.2 Space frames large areas need to be coveredby a roof, with minimum use of internal columns, a possiblesolution is to use a spaceframe. Generallythese are formed of tetrahedronsas shown in Figure 10. In principle, parallel series of lattice booms (top and bottom) are connected by a system of diagonal members to form a latticed 2-way spanning plate of When
significant stiffness.
Angle section upper ch
Tubular
bars
Secondary
tie bars
Space deck module
Figure 10 Typicalspace frame
11
5.2.3 Butterfly roof Thebutterflyroof (Figure 11) is unlikely to have the drainage problemof the umbrellaroof. Since the latticegirders do not directly govern the slope, the roofcan be flatter. The lattice girders being placed in the valleys do, however, call for increased roof space.
5.2.4
General comment
Thesevarious forms, and indeed many others, are frequentlyadopted to suit the requirements of a particularproject, but it must be remembered that they can increasethe unit cost of a structure compared with the more simple forms. Eaves
Roof ci
H.D. bolts Side
cladding Figure 1 1 Butterfly roof
5.3 Cladding Claddingto a building (roof and walls) has to be provided to satisfy aestheticand functional criteria and to satisfy the economics of the project.
A satisfactoryappearance is accomplished by selecting the appropriatecolour and shape to blend in with the remainderof the building and neighbouring structures. A useful "Product Selector" for "Roofing and Cladding in Steel" has been produced by BSC Strip Mills Products''. This providesdetails of about 70 different products. Functionally,the system has to provideresistanceto atmospheric conditions, sound transmission,and light reflection. It is essential to ensure that both roof and walls are watertight under all conditions, wind causes no damage to either cladding or structure, and adequate insulation is provided against heat and cold. Structurally, cladding has to be of adequate strength and stiffnessto resist induced stresses and excessive deformation. Profiled sheetingis commonly used since it satisfiesthese requirements and is additionally light, durable and easy to erect quickly. Coated steel sheets are extensively used for cladding all types of industrial buildings. They are available in a wide range of profiles(rib depths) and colours. Many proprietary cladding productsprovide integral insulation systems, making use of expanded polystyreneor similar insulationmaterial. Doubleskin metal systems are available and are considered by some designers to be the best type of cladding. Clearly where composite cladding systems are used there is only one operation for the erectors.
12
In general a single skin is used for stores where heat retention is not a significant factor e.g. timber stores etc. In factories and offices where the envelope is dependenton the "U'
value, double skin cladding is a sensiblesolution. However, lining sheets may be a critical factor in the design for wind suction. Sheets, supported by purlins (Figure 12), are available in long lengths. Where possible, sheets are lifted into positionby cranes to providebetter safety conditions for the fixer. Hencethe numberof laps should be minimised in order to reducethe possibility of water ingress, particularlyon shallow slopes. It is possibleto vary the spacingof supportsfor cladding depending upon the thickness and shape of the profile. Three factors generally control the spacing. The first is purlin size and the second is the limitations of lining supports. Often the length of the inverted 'T' sectionsused to support lining panels is limited to about 1.8 m, consequently purlin centres are restrictedto that dimension. Finally purlins are often used to provide lateral restraintto the rafters or frames. All of these factors need to be considered to determinethe most economical solutionto the roofing system. Aluminium sheetingis similar to steel sheeting, although it tends to be lighter. The aluminium coating may provide better resistanceto industrial atmospheres, greater solar heat reflection and brighter appearance. Natural lighting can be provided by the introduction of translucent sheets (which structurally can be very weak), or stretchesof patent glazing. The latter is clearly more expensive and is often limited to slopes greater than 12°. Translucentsheets can be moulded to the profile of the main cladding and would use similar fixings.
Care must be taken in positioningroof lights. It is generallynecessary to have a metal or similar main cladding sheet at the top and bottom of the roof light in order to provide adequatestrengthto the system. When lights are placed near to the eaves and/or ridge there may be inadequate support.
Claddingcan be fixed by the use of selftapping screws or hook bolts. Selftapping screws may have recommended torques. An aspect to be carefully considered is the thicknessof the purlin. It is essential to ensure there is sufficient thickness of metal to accommodate self tapping screws. If there is any doubt it is advisable to check with the cladding and purlin manufacturers of the adequacy and safety of the composite system. Screw sizes vary and their strengths are dependenton their "pull-out" capacity. In checking these the screw manufacturer has to take into account the high "local" wind suction effects. Often gutters are placed inside at eaves level to provide enhanced appearance. However,this advantage needs to be weighed againstthe difficulties which may be encountered in the repair and maintenance of the gutter. With this system the use of overflow weirs should be considered to allow for blocked pipes and freak storms.
5.4 Purlins Purlins are required to support any of the types of cladding available. Cold formed sections have been developed to provideelements of adequate strength and stiffness which also allow maximum speed of erection.
If the design criteria is such that cold formed sections are inappropriatethen use can be made of hot rolled sections.
13
For frame spacings between6.0 m and 10.0 m a propped purlin system can be adopted constructed from either light angle, tee or channel sections or structural hollow sections, as shown in Figure 12. For even wider frame spacingthe use of lattice purlins should be considered. They can be made up in many ways, e.g. using flats with rod lacing or small structural hollow sections. (Cold formed latticepurlins are also available). Castellated beams have been used on occasions.
It should be noted that both propped and lattice purlins can be useful for providingrestraint to the bottom of the main supportingframes.
As indicated in Section 5.3 on cladding it may be necessary to limit the purlin centres to 1.8 m (generally fabricatorsprefer 1.7 m to 1.9 m).
Ofparticular consideration is the locationof the purlins relativeto the node positions of the lattice frame. If they coincidewith the nodes then the top boom would only transmit axial loads. If they are locatedbetweennodes then bendingis induced in the boom member in addition to axial forces. The span of purlins may be controlled by a fixed specification for the main frame centres. Alternatively frame centres can be determined by selecting specific purlins which may have limiting spans. Cold formed sections are normally available in lengthsup to 10 m and depths from 120 mm to 300 mm. Normallyspans are of the order of 4.5 6 m. To enhancethe lateral stiffnessof the purlins it is sometimes necessary to use anti-sag bars - Figure 16. This, however, can increase labour costs and therefore their use should be weighed againstlarger purlins or closer frame centres.
-
An aspect to be considered concernsthe design for snow loads. Cold formedpurlins have generally been developed on the basis of tests carried out using uniformlydistributedloads. Snow loading may be trapezoidal and care is required in the interpretation ofthe manufacturers'literature.
A further design criteria which has implications on purlin size is the incorporation of a dominantopening in the side of a building. This can significantly increasethe uplift due to wind.
Purlins are often used to providelateral restraint to the compression flange of the main supportingframes, and to transmit wind loads to the bracing system. Ifthis is the case combined loading needs to be considered when selecting the appropriatepurlin i.e. it could be subjected to the maximum dead plus superimposed (snow) loads, which induce bending, and additionally axial load from wind effects. Eaves purlins are also available which have a slopingtop flange. Various types of purlins are shown in Figure 12.
5.5 Side rails In general the comments providedin the previous Sectionon purlins are applicable to side rails. The loads acting on these will be different since vertical forces are induced by the self weight of the cladding which acts perpendicularto the wind loads. Sheeting rails are often fixed at about 1.8 m. Generally, a limitof 2 m is placed on their centres. Anti-sag rods are more easily fixed to stiffen these elements.
14
Purlin
Lattice
Purlin
girder
stays
Asbestoscement sheets Hook bolts
Self tapping screws
5°)L
Steel sheets
Hook
i
Sheeting Insulation
Rafter Angle
S%iRaft:):;;:ul:tion Cold formed Z (Anti-sag bars required for spans over 4.5 m)
Rafter
'Structural hollow section (circularor rectangular)
Props to bottom of roof girder
Roof girder Propped angle purlin
Sheeting and insulation
Lattice purlin — Roof girder
Figure 12 Types ofpurl/n
15
6. CONCEPTUAL DESIGN Before considerationis given to the method of analysis and designto be adopted certain decisions have to be taken, which may later be modified as the design progresses. The effect of any modifications clearly can alter the detailed design and alterations to calculations would ensue.
There are four principalcomponents of a light industrial building i.e. the cladding, the cladding supports, the main frame and the foundations. Early decisionsare required on type(s) of cladding and type of purlin and sheetingrails. Since these are all supported by the main frame.
If the frame is considered as a simple portal, Figure 13, it is necessary to decide on the type(s) of fixity to be providedat the base, eaves and ridge. Generally, the columns to the frame will be of I or H section, unless the building incorporates a high capacity overhead travelling crane when a compositecolumn might be required.
If the rafters are to be latticed structural
steelwork it is possibleto use different layouts of the internal members, Figure 14. However, since the diagonals are likely to be subject to stress reversal, dueto wind effect, the warren type truss is generally preferred. In selectingthe layout it is necessary to decide on the positionof purlins. If these are located at node points then local bendingin individual top boom members are avoided. In principle,forces in all of the members are either direct tensile or compressive, with bendingand shear effects being secondary, as a result of deformation of the truss. Analysis of the framework can be carried out by hand calculation,drawing or computer. In the first two methods, it is essential to assume that all joints are pinned and preferably end support conditions to the rafters are such that the truss is statically determinate.
When a software package is used there are a number of options, three of these are: (i)
assume all joints of the truss and the connections to the columns are pinned;
(ii)
assume full rigidity of all joints;
(iii) assume the internal bracing members are pinned to the booms which are considered to be continuous and therefore rigid.
In adopting(i) or (iii) it is necessary to consider the possible effect of secondary stresses caused by: (a)
loads applied betweenthe truss nodes;
(b)
moments resultingfrom the actual rigid joints and truss deflections.
Additionally, in all cases care needs to be taken in member layout, since secondary stresses can be induced by eccentricity at the connections. (Specific reference should be made to BS 5950: Part 1, Clause and Structural SteelDesign'2 by Dowling, Knowles and et al Owens), Dowling suggestsecondary stresses should be calculated for heavy trusses used in industrial buildings(e.g. those supportingoverhead cranes) and bridges. It is traditionally recognised (e.g. in British Steel Publication,Design ofSHS Welded Joints'3)) and Dowl ing et al also suggest that latticed structuresare assumed, for design purposes,to have pinned joints. This may lead to higher defiections than those induced in a rigidjointed truss, but in practice
4.l0'
16
this is unlikely to be significant with the exception of girders supportingcrane beams. The design exampleillustrated uses a package hut initial hand calculations are used to ascertainmember sizes. These are useful for the software data input. Generallya decision will be taken early during the conceptual design process on the type(s) of member(s) to be used for the latticed frame. There are many options: (a) (b) (c)
Hollow sections - circular or rectangular. Traditional sections - angles, tees, channels, UCs. Combination of (a) and (b).
The selected truss should reflect the need not only to producethe lightest frame but also to minimize the cost of fabrication and erection. Rigid
Pinned
Pinned
Fully rigid
Rigid Pinned
Pinned
Rigid
Rigid
Figure 13 Basic arrangementforportal frames
'Pratt' or 'n' truss
'Warren' truss
Figure 14 Typicalarrangementfor latticedgirders 17
An example of compositeform is shown in Figure 15 where the booms are of UC section and the internalmembers RHS. The UCs enable easy connection of servicesto the truss and easy connection to columns. Also bracing in the plane of the roof can be provided using simple in plane membersand simple connections, or by using the relative stiffnessof an I or H section. When hollow sectionsare used with weldedjoints reference should be made to the British Steel Publications, listed in Section 7.5. It is essential to ensure that it is possible to make a full weld. Difficulties can arise where large booms and small internal members are used which may requirejoint stiffeners. These may be expensive and it is likely to be prudentto increasethe member size. The designermust be aware of problems which can arise in the detail design at the joints. The specific advantages of hollow sections(and tubes) when compared with traditional sections(UBs, UCs, Channels,Angles etc.) are the high strengthto weight ratio, maximum efficiency in tension, efficiency as struts, good torsionalproperties, appearance and maintenance. In deciding to use CHS or RHS the designer should remember that some fabricatorsare not fully equipped to use circular hollow section. Their main disadvantages can be the higher cost of connections especially at nodes involving overlapped CHS bracings and chords, the relative difficulties of making on site connections for services (electrical etc.) and higher basic costs than traditionalsectionson a tonnage basis (overall, however, lighter weight frames are produced). Relevantto the design code BS 5950: Part 1(1) is the consideration of section classification (Fable 7 of the code). Tees cut from UBs are generally slender, hence a reduced yield stress has to be used. Tees cut from UCs are not affected in the same manner.
In designing the joint it is necessary to examine whether high local stresses will be induced by the selectedarrangementand member sizes. These high local stressesmay even occur when member axes intersect.
The relative slopes of the internal members are relevantto the detailingfor the fabrication process. If they are parallel to each other then the angle of cut at each end is identical for all members.
The final decisionon the type(s) of member(s) to be used may be influenced by aesthetics and not cost.
CHS
UC
UC
RHS
OHS
RHS
RHS
RHS
CHS
UC
UC
RHS
Figure 15 Alternative lattice girder layouts
18
7. PRINCIPLES OF DESIGN The designof all the steelworkfor low rise lattice framed buildingsshould satisfy the "aims of economical structural design" and "limit state" philosophies outlined in the appropriate Codes of Practice. Basic design assumptions are made as to the behaviourof the various units which make up the structure.
7.1 Purlins and side rails Purlins and side rails can be designed to satisfy the strength and deformation requirements of the appropriatecodes or they can be designed using empirical rules given in Clause 4.12 of BS 5950: Part and Section 9 of BS 5950: Part 5(2)
1'
It is of note that the empirical
rules are based on unfactored loads and also that the tables of section properties (A checklistfor designers'6 published by the SCI) do not list plastic moduli for angles.
Purlins are generallydesigned as continuous members, over two or more spans, supporting uniformly distributed loads. In this case connections have to be made to transmit shear and bending. Cold formed sections can be selected from manufacturers' catalogues where it is guaranteed that the carrying capacity of the various systems is based on the results of extensive research
and development.
Continuity is obtained by the use of sleeves, and the effectivelength of purlins are reduced by the use of anti-sagbars (Figure 16).
When applied loads are not uniformly distributede.g. trapezoidal snow loading or when purlins are used to support ventilation systems etc. then original calculations are required. These will make use of BS 5950: Part 5 and section properties for cold formedpurlins providedin manufacturers'catalogues.
7.2 Lattice framed roof girders As indicatedin Section 6 the design will be based on the assumption that joints are pinned, rigid or a combination of the two.
The girder will support vertically applied dead and superimposed loads plus wind loads. The latter is likely to induce stress reversal in the members. The rafter will also transmit the horizontalwind loads from the vertical cladding and may act to transmit wind loads in the plane of the roof. Typicalload directions are shown in Figure 17.
7.3 Stanchions When pinnedbases are adopted then moment fixity is required at the column head. The column will be designed for axial and shear forces only at the bottom but for axial, shear and bendingin the upper length. Use of fixed bases enables the stanchions to be designed as propped cantilevers, although it should be noted that simply linking the top of the stanchions with the rooftrusses does not providea fully rigid propped system. The column heads and 19
girders can all movetogether. It is of note that the relative stiffnessof the rafter and column are significantly different (possiblyof the order of 4 to 1). Also changes in the overall depth ofthe rafter can significantly increase or decrease the stiffnessof that member. The stanchion size is controlled by its effectivelength, which is likely to differ about orthogonal axes. Care is required in the selection of end and intermediate fixity conditions. Cleat (behind)
rafter
rail
Figure 16 Sleevedpurlln system Dead & imposed
I
\
I Reversible wind loads
Vertical cladding (dead) load
Reversible wind loads
I
Figure 17 Frame loads 20
I Vertical loads
7.4 Bracing Bracing must be provided to accommodate wind loads on the gable columns. This can be used to facilitateplumbingand squaring the building during erection. It can also provide essential stabilityto the steelwork during erection.
Bracing normally consists of diagonal membersbetweencolumns and trusses both in the walls and plane ofthe roof. The bracing can be single diagonal or cross members (Figure 18). If the former system is adopted the members are designedto support compressive and tensile loads. When cross members are used only the members in tension are assumed to be effective, those in compression are designed to satisfy the slenderness criteria, Clause 4.7.3.2 of BS 5950: Part 1: 199O'. When masonry is used as all or part of the vertical cladding, as part of the bracing system.
it is feasible to use that element
/\NN/7NNNN Single diagonal roof bracing
>< ><
x><
Cross member roof bracing
Figure 18 Roof bracing
7.5 Connections A very important aspectof design using any material is the design of connections.
Structural
membersare designed to carry axial loads, shear force, bendingmoment and torsion. Consequently connections must be designed to transmit these forces from one element to another withoutinducingexcessivestressesor deformations.
To producea good design of a complete structural assembly it is essential for the designerto clearly state at an early stage the basic methods by which various members are to be joined. Sophisticated methods of analysis are now available to determineto a good degree of accuracy the forces and deformations throughout both simple and complex structures. This degree of sophistication is not howevergenerally available in connection design. The stresses induced by connections are often indeterminate and their distribution throughout a joint is not always consistent even in identical conditions. Stress is always a function of deformation and the latter can vary with the irregularities of the properties of the members being connected, the type of fasteners,the quality of workmanship in making the connection and "built in" stresses
in the parent members.
21
Most connection design is, at present, only approximate. The essential aim is to provide the type of connection stipulated by the designerwhich is efficient, economical and aesthetically pleasing. The latter is not always essential. Use will be made principally ofthe basic laws of statics i.e.:
EX
=EY =EZ =
EM
=
EM
=0 = 0
i.e. all jointbehaviourwill be considered to be statically determinate. The distributionof internal forces in a connection has to be assumed and either elastic or limit state design may
be appropriate.
The fabricationof connections is particularlylabour intensive and therefore in order to keep overall costs down it is necessary to try to producesimple but efficientmethods ofjoining members, by weldingor bolting.
In general the design of connections will follow the recommendations given in BS 5950: Part 1: 1990, Section Six. Connections. In the case of the following design example using hollowsection the design is carried out using as referencesthe followingpublications produced by British Steel viz: Design of SHS Welded Joints
TD 338'
Jointing Welding
TD
Hot finishedstructural hollow sections; sizes, properties and technical data
325
ID 328' TD 167
Useful reading in the first instance is TD 325 which provides an indication of the wide spectrumof application of RHS.
PublicationTD 338 provides a clear method of DesigningSHS Welded Joints. As indicated in Section Six of BS 5950: Part 1, it is common practice to carry out the analysison the basis of pin-jointed frames with members in direct compression or tension and the centre lines of members intersecting at the nodes, as shown in Figure 19. Often it is necessary to providea gap or overlap as shown in Figure 20. Joints may take a variety of geometric forms as shown in Figure 21. TD 338 details the method of establishing the joint's design capacity in limit state terms, compatible with BS 5950 and Eurocode3.
It should be noted that fillet welds generally providethe most economic method of connecting members in structuressubject to static load. Clearly one exception is the case of end to end connections where butt welds can be provided to develop the full strength of the sections connected. In this case with RHS sections internal backing members are provided, which are formedfrom strips 20-25 mm wide and 3-6 mm thick.
Of note is the recommendation in a paper by N Yeomans,NewDevelopments in the useof StructuralHollowSections17:
"Because ofthe influence of member and joint geometry on the jointbehaviour, it is important that engineers design the joints when determining member sizes; with SHS design this job should not be left to the detailer".
22
a
Figure 19 Nodingjoints
a) Gap joint with positiveeccentricity
b) 100% overlap joint with negative eccentricity
Figure 20 Definition of eccentricity
/
X joints
/
—Y-------k-
N and K joints with gap
/
and Y joints
,es
/ I
I
N and Kjoints with overlap
/
oy[1%e2 -
-*
Figure 21 Joint geometries
23
8. EXAMPLE - DESIGN BRIEF AND APPROACH 8.1 Brief The client requires a single storey, single bay industrial building to be used as a light machine shop. It is to be sited on an industrial estate on the outskirts of Leicester. Main dimensions
-
30 m wide x 48 m long x 6.7 m to eaves
Cladding
-
Colour coatedsteelsheets to roof, sides and ends with 20% natural lighting provided by translucent sheet inserts.
Insulation
-
A lining system to be provided to wall and roof sheeting.
Access
-
A roller shutter door 4 m X 4 m is to be providedin both gable ends with personnel doors 1 m x 2 m adjacentand along the side walls. Note: The possibilitythat the roller shutter doors would be open during a severe storm was discussed with the client. The final decisionwas that the design should be based on the assumption that both doors would be closed during a severe storm.
Services
-
Allowance was to be made to support set-vices from the roof structure. Mechanical handlingwas not required.
General
-
It was agreed that: (a) The roof pitch would be set at 50 (b) The roof to be of hot rolled hollow section latticed framework. (c) Hot rolled I sections would be used for the columns.
The outline of the buildingbased on the above brief is shown in Figure 22. Selection of RHS for the roof structureis based on its enhanced efficiency and the cost effectiveness ofjoints which will, in general, be quite simple. The girders will be shop fabricated in two halves, approximately 15 m in length and 1.2 m deep. Hollow sections can be used in simple, semi-rigid and rigid design and can adequately carry axial (tensile and compressive) loads, bending, shear and torsion.
8.2 Cladding Since the decisionhas already been made to use colour coated steel sheets with insulation
lining and translucent sheet inserts, it is only necessary to settle upon the most suitable thicknessand profile of sheet to be adopted. This need not be the same for both roof and sides and they are therefore considered separately.
24
/
/ I I
0 E
0
(5.; 4,
>. (5
co
/ 30.Om
Plan - Centre lines of frames
5° pitch
Cross section
Roller shutter door End elevation
Figure 22 General layoutof building
25
8.2.1
Roof sheeting The span of the roofis 30 m and with a 5° pitch the length of one slope is marginally over 15 m. Not all manufacturers produce sheeting of such length and it may be necessary to use, say, 2/8 m sheets lapped at the centre. The laps should be bedded in sealant because of the low rise.
A suitablespacingfor the purlins will be 1.85 m, which on a slope length of about 15 m, dividesthe rafter of the roofframe into eight panels. A typical sheetingsystem would be the "Warmclad 1000R", with lining, manufactured as a BSC Profile (Reference 6), this is suitable for roofand walls.
8.2.2
Wall sheeting
The height from the floor to the eaves is 6 m hence sheeting rails can be spaced at 1.5 m c/c. To achievea different architectural effect to the building either a different sheet and/or an alternativecolour could be adopted. Figure 23 shows a cross section of the building.
Lined
roof cladding
and translucent sheets
6.0 m
Lined wall cladding
'Weathering curb
0.7m ,1
Figure
26
23 Cross section
ground level 15.0 m Half span
Ridge tie
9. DESIGN OF STEELWORK 9.1 Loading The loading for which the steelwork must be designed is in four parts: (1)
Dead load
-
from cladding and structure, assessed from the mass/unitof the various items (Reference 4).
(2)
Imposed load
-
(Reference 5).
(3)
Service load
-
allowance for trunking etc.
(4)
Wind load
-
(Reference 6).
required by client.
The first three loading conditions can be calculated as the design proceeds but the assessment of the wind load requires consideration of the complete structure at the outset.
9.1.1
Load factors Table 2 of BS 5950:
Part iW
.yf
Dead load
1.4
Dead load restraininguplift or overturning
1.0
Dead load acting with wind and imposed loads
1.2
Imposed load
1.6
Imposed load acting with wind load
1.2
Wind load
1.4
Wind load acting with imposed load
1.2
Forces due to temperatureeffects
1.2
9.2 Assessment of roof load Part 3 : 1988 provides the detailed method of evaluating minimum imposed roof loads (Clause 4). With no roof access the load is taken as the greater of: BS 6399:
(a)
the uniformlydistributedsnow load.
In this case Figure 1 in the code provides the "basic snow load on the ground", say 0.6 kN/m2 = s0. The roof snow load Sd
=
js
where
0.8 (Figure 4 in the code).
Hence Sd = 0.48 kN/m2
27
(b)
a uniformlydistributed load (u.d.1.) of 0.6 kN/m2
For simplicityin this example drifting has been ignored and the roofsnow load is therefore taken as 0.6 kN/m2.
9.3 Assessment of wind load
on structure
Part 2 : 1972(6) gives a detailed method of calculating the design wind loads on a structure. The code was preparedfollowing extensive investigation at the Building ResearchEstablishment. CP3 : Chapter V :
The dynamic pressure q exerted by the wind is found in the formula: q = kV2 N/rn2 where:
q = =
dynamicpressure constant having a valueof 0.613 design wind speed V x S1 x S2 x S3
S1
= =
basic wind speed for geographical location topography factor, usually taken as unity (except
S3
= =
abnormal) ground roughness, building size and height above ground factor building life factor, usually taken as unity.
k
V where
V
=
where conditions are
S3 is a factor inserted in the formula to enable accountto be taken of any special circumstances which may justify a variation in the design wind speed.
For a factory located in the Midlands the basic wind speed V, from the map in the code, is 44 rn/sec. Then, if unity is accepted for S1 and S2 the design wind speed, V V x S2.
2
A wide range of values for are given in Table 3 of the code. These are divided into four groups related to the surroundingsof the building under consideration, and each group is divided into three classes relatingto cladding or structure and its size. The value of also varies with the height above ground level, stepped at intervals.
2
The building in this exampleis located on an industrial estate where it will be surrounded by other buildings. Group 3 is thus applicable, with Class A factors for the cladding and, since none of the main dimensions exceeds 50 m, Class B factors for the structure.
Table 3 gives values of S2 at heightsof 3, 5 and 10 m above ground. As the overall height from ground level to ridge is say 8 m the value of S7 for the structureat that height may be obtained by interpolation between the tabulated values at 5 m and at 10 m and this results in a value of 0.70. The values of S2 are therefore as shown in Figure 24. (For more detailed informationon the values of S, reference should be made to Wind loading handbookpublished by BRE7).
Figure 24 Values of coefficientS2
28
2'
Using these values for thedesignwind speeds and dynamic pressuresover the two height ranges are as tabulated below:
Table 1 Wind speedsand dynamicpressures Height range
Structure
Cladding
V
S2
m
Oto3 (b) 3 to 8 (a)
q
rn/s
N/rn2
0.64
28.16
486
0.75
33.00
668
q
V, rn/s
N/rn2
0.60
26.40
427
0.70
30.80
582
S2
The design wind loads on either the claddingor the structure are found by multiplying the in the case of dynamic pressure, q, by the area applicableand by a further coefficient for the internal effects i.e. internalpressure or suctionwhich externallyapplied forces and dependon the permeability ofthe building shell.
C,
C
Values for the external pressure co-efficient, C , are relatedto the shape and dimensions of the building and are given in Tables 7 and B orc3 : Chapter V: Part 2: 1972(6).
For the building under consideration the relevant information is: Widthofbuilding, w Lengthof building, 1 h Heightto eaves, Then
hiw
and
11w
= 30.0 m = 48.0 m = 6.7 m = 6.7/30 = 0.22 < 0.5 = 48.0/30 = 1.6 i.e. between 1.5 and 4.
From Tables 7 and 8 of the code the values of (a)
wind on side ofbuilding Windward side Leewardside Windward roof slope (5°) Leewardroofslope (50)
Eithergableend (b)
C are:
= +0.7 = -0.25 = -0.9 = -0.4 = -0.6
Wind on end of building Windward gable
Leewardgable Both roofslopes Either side
= +0.7 = -0.1 = -0.8 maximum = —0.5
The followingcalculations are based on the assumption that the doors will be closed during a severe storm, which has validityand also simplifies the design process for this example. Hence is taken as the more onerous of +0.2 and -0.3.
C
29
However, it should be noted, as stated in Appendix E to CP3: Chapter V: Part 2: 1972, that the value taken for the internalpressure coefficient, must be related to the permeabilityof the cladding and the presenceor otherwiseof large openings. It is necessary to ensure the correct decisionis taken concerningthe permeabilityof a buildingwhich is constructed with one or more large doors. It is often, quite incorrectly, assumed that the door(s)will always be closed during a severe storm. In reaching this decisionno account is taken of loadings which will arise during the construction period when it is very likely the doorwill not be in position. Nor does it take into account the use of the buildingwhich, during its lifetime, is likely to be such that the doorwill be unavoidably opened. An argument also put forward is that in the event of the doorbeing open during a severestorm, the sheetingwould probably "blow out" before damageto the structureoccurred.
The various coefficients to be used to determine wind loads on sides and roofof the structure only are shown diagrammatically in Figure 25. As far as general stabilityi.e. overturning, is concerned internalpressure has no effect since the horizontal forces on the sides and horizontal components ofthe forces on the roofall cancel each other. For the design of individual members and for decidingon the minimum anchorage required, the most adverse conditions are taken from (e), (f), (g) and (h) of Figure 25.
0.7f}225 o[_'b}.5
0.5
Wind on side
Values of c1
Wind on end
Pressure
Values of c,,
Suction
4TfO.45
0.
(C,,) ( 0.2)
(C) - ( 0.2)
1.0
TIj..o.os (C) - (— 0.3)
Figure 30
25
Evaluation
0.7
0.
(C) -
C—
0.3)
of totalpressurecoefficients (Cpe and C,,1)
The followingcalculations are made using the values of q applicable to the structure. The cladding and its fixings must be capable of resistingthe higher values shown in Table 1, and Cr1. The areas adjacentto the eaves, ridge and verges, and adjustedby the values of the corners of the building are designed using the local coefficients in Tables 7 and 8 ofthe code, which are aimed at avoidinglocal damage. These local coefficients for the fixing of claddings are significantly different to those shown for the structure, but these are not included within the scope of this publication. This is information which wouldbe requiredby the manufacturers of claddingand fixings.
C
9.4 Design of purlins These may be either cold rolled or hot rolled sections and both a cold rolled zed section and a hot rolled angle are designed for comparison.
The purlins have a span of 6 m and are spaced at 1.85 m centres.
9.4.1
Cold rolled Z purlin
Normally a cold rolled purlin can be selected from a manufacturers' catalogue(e.g. Metsec'8) where safedistributedloads are given for various sizes and shapes. No further design checks are necessary if these have been designed to BS 5950: Part 5: 1987(2). Dead load, sheetingand lining, say 0.19 kNIm2 Imposed load
QQ kN/m
Total dead and imposed loads
0.79 kNIm2
Selfweight, say Total load
kN/m2
0.82 kNIm2
Using a load factor of 1.6 the design load due to dead and imposed is 1.6 x 0.79 = 1.27 kN/m2. Purlin design is based on a normal load acting on the member. The dead and imposed loads are gravitational. Howeverthe difference between gravitational and normal loads for a low pitch roof is not significant.
Froma manufacturer'scatalogue a 202 x65x60x 1.8 mm thick Z purlin (Figure 26) will carry an allowable(unfactored) load of 1.29 kN/m2 (excluding putlin selfweight of 5.01 kg/rn).
The maximum wind uplift on the roof will be 0.582 x 1.1 = 0.64 kN/m2 (see Table 1 and Figure 25(e)).
In selectingan appropriatepurlin it is necessary to considerthe effect of using the purlin as part of the bracing to transmit wind loads in the plane of the roof. See Section 10.7.1.
-
Gross uplift on a purlin spanning 6 m at 1.85 m centres = (0.64 0.22) x 6 x 1.85 = 4.44 kN. With a factor of 1.4 the design wind uplift = 4.44 x 1.4 = 6.3 kN. According to the manufacturer'scatalogue the Z purlin selected is capable of resisting a wind load of 9.215 kN. To support this load no anti-sag bars are required.
Use 202x65x6Ox 1.8
31
H 65b 1
______
—1.8
Figure 26
Z purlin
9.4.2 Hot rolled purlins (I)
Anti sagrods
Purlins(and rails) can be designedin accordance with Clause 4.2' i.e. as beams, or empirically(as in this example) using the rules detailed in Clause4. 12.4'. The elements are designedassuming the cladding provides lateral restraint to the section. Clearly the cladding and fixings have to be capableof providingthe necessarysupport (in particularcare is needed in areas ofhigh local wind effects). Anti-sagrods may not be required when hot rolled sectionsare used for purlins spanningless than 6 m. They may howeverbe used to providestabilityduring erection. This will reduce deflections in the plane of the cladding. (ii)
Anglepurlins (Figure27) wp wp Rafters @ 6 m c/c
Figure
27 Loadingon purlins
Dead + imposed load
= 0.8 kN/ir? (referto Section 9.4.1)
Howeverthe minimum imposed load (Reference 1 Clause 4.12.4.3) should be taken as 0.75 kN/m2. Henceload/purlin = 0.95 x 1.85 = 1.76 kNIm.
32
For selfweightfirst consider: =
6000
B
=
45
45
133
6O
L6000_100
Minimum section to satisfy D and B would be 200xlOOx 10 angle.
This weighs 23 kg/rn.
=
Total load
Wp/m
=
1.76
+ 0.23 =
1.99 kN/m.
Hence W, = 1.99 x 6 = 12.0 kN (note- unfactored loads are used).
Z
Therefore
WL
12x6000
1800
1800
40 cm3
This will be satisfiedby the 200x100x10x23 kg/rn angle (Z = 93.2cm3). (ii!)
RHS Purlins
The limiting factors in this case are: WL
1800 say
40 as before
600040 Minimum sectionto satisfyD and B is a 100x50x6.3 RHS, this weighs 13.4 kg/rn and Z = 40.5 cm3, or consider lOOx6Ox5x 11.7 kg/rn RHS, Z, = 38.5 cm3.
Thus W, = (1.76 + 0.12)6 = 11.3 kN.
zi
11.3
x 6000 1800
37.7 cm3
Section is satisfactory.
Of note is the significant difference in weightsof the cold formedZ section (5.01 kg/rn), the hot rolled angle section (23 kg/rn) and the hot rolled R}IS (11.7 kg/rn).
a
Unless special conditions dictate it is highly unlikely that other than cold rolledsection will
be usedforpurlins.
Ofnote is thepossibilitythat tubes with open ends may be subject to internal corrosion. recom,nended that hollowsections are sealed.
It is
33
9.4.3 Design of side rails Side rails span of 6 m horizontally and are at 1.5 m c/c vertically. From Figures 25(f) and (g) and Table 1 the characteristic wind loads are: These may be cold or hot rolled.
Pressure
1.0 x 0.582 = 0.582 kN/m2
Factored pressure
Suction
0.7 x 0.582 = 0.407 kN/m2
Factored suction
(I)
= 1.4 x 0.582 = 0.815 kN/m2 = 0.570 kN/m2
Cold rolled Z rail
From manufacturer'scataloguea suitablesection is 142 sleeved system, using one row of side rail supports.
x 54 x 49 x
1.8 mm thick Zed
Hot rolled angle rail (BS 5950: Part 1, Clause 4.12.4) Using weight of sheetingand lining of 0.2 kN/m2, the vertical load on a rail = 0.2 x 6 x 1.5 = 1.8 kN. Allow for self weight of 0.2 kN/m. Hence W1 = 1.8 + (6 x 0.2) = 3.0 kN (ii)
(Figure28). W2
= 3.0 kN
1
W1=5.24kN —
—
-Y
Figure 28 Side rail arrangement
(0.582
W1L
x
1.5
1800
x
6)
x 6000 -
1800
—
5.24
x 6000
1800
— —
17 5 cm2
where W1 is the unfactored horizontal load
Z,
17.5 cm3 W2L
3x6000
1200
1200
L —
D
6000
15cm3
134mm
6000100mm
B
A 150x 150x10x23 kg/rn angle satisfies(Z = 56.9, Z, = 56.9 cm3). CheckD since Z1 provided is greater than minimum Z1 required from Table 30 and Clause 4.12.4.4(d)(').
34
17.5 __ x—=41 45 56.9
= 6000
D'.. tmi
ButD B hence check a 120x120x8x14.7 kg/rn angle (Z1 = In this case the controlling factors are Z1 =
=
x
Check D
Use 120
x
(iii)
Hot rolled RHS
Z1
120
17.5 cm2
x8x
17.5 cm3 and Z2
=
Z2
= 29.1
15 cm3,
cm3)
which are satisfied.
80 mm, which is satisfied.
14.7 kg/rn angle.
15 cm3.
Z2
O857
B A 100x60x3.6x8.59 kg/rn RHS would satisfy. Adjustment can be made for D i.e. D1..
6000 70
17.5 x___—52 29.3 —
In this case there is no available lighter section, a thicker walled section might be preferable. Use 142 x54 x49 x 1.8 Zed sleeved system with one row of anti-sag bars.
9.4.4
Eaves detail
At the eaves a possible solutionwhen using a Zed purlin system is to use a cold formed eaves beam. A typical section is shown in Figure 29.
Figure 29 Special eavespurl/n
35
9.4.5 Design of gable rails These will be of different size to the side rails since, althoughthe vertical
spacingwill be set
at 1.5 m c/c, the spans are 7.5 m. Considerationcould be given to maintaining the spans at 6 m. Side rails would then be of the same size throughout.
The loading is the same as the side rails i.e. pressure 0.582 kN/m2 and suction0.407 kN/m2. From manufacturer'scataloguea suitable section is 202 x60x60x 1.6 mm thick Zed single span system, using 2 rows of side rail support. It should be noted the gable rails are single span, whereas the side rails are sleeved. The latter are used providing there are two or more spans. The gable end is broken by the provision of the large roller shutter doors. Use 202 x60 x6Ox 1.6 Z.
9.4.6 Typical system Typical arrangementsfor a building which incorporates Z purlins and Z side rails are shown
in Figure 16.
9.5 Design of main roof frame In orderto design the members it is necessary to decide on the method(s) of analysisto be used to obtain member actions (forces, moments etc) and deformations. In this example it is initially assumed, in order to determinepreliminary member sizes, that the frame is a portal as shown in Figure 30, where: (a)
the columns are pinned at the base;
(b)
the girder is initially treated as a "beam with partial restraints at the end". The bending moment and deflection at the centre can then be evaluated;
(c)
Having deducedthe bendingmoment in the girder the joints of the latticed truss are assumed to be pinned (this complies with Clause 4. lOW).
Figure 30 Lattice girder/pinned portal analogy
9.5.1
Preliminary design - joints of latticed truss pinned
Span 30 m;
spacing6 m.
To proceed with the design it is necessary to estimate the self weight of the girder. Initially assumethat the selfweight will approximate to 50% of the weight of the cladding.
36
The frame adopted is a lattice girder (beam). As such there are two main design criteria to be considered namely strength and deflection.
In the case of strength, assuming use is made of Grade 43 steel, the maximum yield stress allowed is 275 N/mm2(Table 6(1)). Allowance has to be made for slenderness and member classification in order to determinethe design strength (p for struts (Tables 7 and 27(1)). The reduction can vary from 2% to 75% of the full member compressive capacity.
Generally the smaller the member cross section the larger the percentage reduction. With respectto deflection limitations are not specified, in fact Table (1) particularly excludes pitched roofs. Howeverexperience has shown that a deflectionlimitof span/200 is a useful guide for such a shallow pitch.
If the girder is assumed to be subject to a uniformly distributedload (as distinct from 17 No. purlin loads) then, say, deflection limit is
384
El
200
where W (kN) is the unfactored imposed load and L is in metres. (Note: deflection for a simply supportedbeam supportinga udl (14') is 5WL31384 El and for a fixed ended beam it is IVL3/384 El. Hence for partial restraint assume deflection is 214L3/384El). Taking E = 205 kN/mm2 and re-arrangingthe required
I = 0.51 WL2 cm4.
Preliminarycalculations are thus carried out assuming the selfweightof the girder approximates to 50% of the weight ofthe cladding, the compressive stress in the boom is 250 N/mm2, in the diagonal strut 220 N/mm2 and required is 0.51yp2 cm4.
I
9.6 Preliminary 9.6.1
calculations
Loading (excluding wind) - based on full sheeting width say 30.6 m kN
Dead load (characteristic)
Sheets and lining 30.6 x 6 x 0.2 Purlins 18 No. x 6 x 0.04 Selfweight of girder, say Services load, say Total Imposedload (characteristic) 30.6 x 6 xO.6
= = = = =
36.8 4.3 18.4
1L
71.0 110 kN
Design load (Table 2 of BS 5950: Part 1)
F=(1.4x71)+(1.6x 110)=276kN.
37
9.6.2 Initial member size (i)
Main booms
a)
Assumemaximum bendingmoment (BM) in the girder = WL 16
=
x
276
16
30
= 5l8kNm
The assumed BM of WL/16 takes into account the partial restraint producedon the whole girder by the connection to the columnsi.e. BM at the centre lies betweenWLI8 for a simply supportedbeam and WL/24 for a fixed ended beam (both subjected to a udl of WkN). b)
Sincethe girder depth (d) is 1.2 m the approximate force in the top and bottom boom members is 518/1.2 = 432 kN.
c)
Using an assumed compressivestress for the top boom, of say 250 N/mm2 the area required =
432x10 ______ 250
2 = 17.3cm.
The designermay chose to incorporatethe slenderness criteria provided in Clause4.10(1). These limitations have not been brought into this design since it is considered that secondary stresses in this "lightly loaded" truss will not be significant.
d)
(This is checked in Section 11.0).
Use a 120x80x5.0X18.4 kg/rn RHS (A = 18.9 cm2) for top boom. Whilst the bottom boom will be subject to tensionand compression (the latter at the support)select its preliminary size on the basis of a tensile force of 432 kN, at the centre.
For bottom tie area required
= 432
x
275
10
=
15.7 cm2
Use a 120x60x5x13.3 kg/rn. (A = 16.9 cm2) for bottom boom. Deflection
e)
Requiredl=0.51 x 110x302=5.0X Considering
area of top and bottom booms as 16.9 cm2.
I providedis approximately equal to 2A (d/2)2 = 0.5Ad2. = 0.5 x
16.9 x 1202 = 12.2 x deflection will not be critical).
38
cm4, which is satisfactory (and indicates that
-
(ii)
Diagonals
= 276 kN.
Total load
Hence shear in end panel = say 138 kN. The diagonal slopes at an angle approximately tan'(1.2/0.925) = 52.4° and has a length of 1.52 m. Force in diagonal = 138/sin 52.4 = 175 kN (tension and compression). Hence area of strut = 175 x 10/220 = 8.0cm2. Use 60x40x5x6.97 kg/rn RHS. (A = 8.88 cm2)for diagonal struts. Area of tie = 175
x
10/275
= 6.4 cm2.
Use 60x40x4x5.72 kg/rn (A = 7.28 cm2)for diagonal ties. Ciii)
Check SelfWeight
Selfweight
= 30.6 (14.8 + = 1169 kg
13.3) + 16
x
1.52 (5.72 + 6.97)
11.7kN (compared with the assumed value of 18.4 kN). Hence figure used in the final calculation could be reduced. Say selfweight = 13.4 kN (i.e. reduction of 5 kN). (iv)
Final loads, excluding wind
Dead,say Imposed (v)
=66kN
=
110 kN.
Columns (preliminary sizing)
The columnswhich have pinned bases are subjectto axial load, bendingmoments and shear. The combination of axial load and bendingaffects are assessed using the equationsgiven in Clause 4.8 of BS 5950: Part 1. Consequently a simplified approach,based on axial loads and moments, is much more difficult to define. A useful guide to the size of the column is to considerthe second moment of area 'F. The ratio of'girder: 'column normally lies between 4:1 to 1:1. A further guide is to use the relationship 'c"g = 3h/2L. Hence 1c = 'g X 3 X 6.7/(2 i.e. say
x 30)
= Jg/3.
Since 'g = 12.2 x i04 cm4 then assume Ic = 4 x = 4.1 X column would be 457 x 191 x 89 UB (vi)
i0 cm4.
A suitable sectionfor the
i0 cm4).
Preliminarymembersizes
The preliminarymember sizing is assumed to be as calculated and the relevant section properties have been incorporated into the computer analysis.
39
9.7 Loading Cases (for characteristic loads) DEAD (see 9.6.1 (iv)) Dead load on girder = 66.0 kN Load/purlin= 66/16 = 4.13 kN IMPOSED Imposed load on girder = 110 kN Load/purlin = 110/16 = 6.87 kN.
WIND CASE I (refer to Figure 25(e)).
x 15.4 = 92.4 m2. Load/purlin,left hand slope = 1.1 x 0.582 x 92.4/8 = 7.4 kN (uplift). Load/purlin,right hand slope = 0.6 x 0.582 x 92.4/8 = 4.04 kN (uplift). Load on left hand column (rail area = 1.5 x 6 = 9 m2), allowance needs to be made for say 300 mm sheetingoverhang to top and bottomrails i.e. area supported = 6 x 1.05 = 6.3 m2. Lower rails = (0.5 x 0.427 x 9) = 1.92 kN (pressure) Slope area of sheeting = 6
Upper rails = (0.5 x 0.582
x 9)
= 2.62 kN (pressure).
Load on right hand column Lower rails = (0.45
x 0.427 x 9)
Upper rails = (0.45
x 0.582 x 9) = 2.36 kN (suction).
1.73 kN (suction).
WIND CASE II (Refer to Figure 25 (fT))
Load/purlin,both sides Lower rails, both columns Upper rails, both columns
= 6.72 kN (uplift) = 2.69 kN (suction) = 3.67 kN (suction)
WIND CASE Ill (refer to Figure 25 (g))
Load/purlin,left hand slope Load/purlin,right hand slope Lower rails, left hand side Upper rails, left hand side Lower rails, right hand side Upper rails, right hand side
9.8
= 4.04 kN (uplift) = 0.67 kN (uplift) = 3.84 kN (pressure) = 5.24 kN (pressure) = 0.19 kN (pressure) = 0.26 kN (pressure)
Analyses
The general frame layout is shown in Figure 31.
For "hand" analysisthe sizes of respective members do not effect the calculations of pin-jointed frames.
40
vi
C'1
J
2
Figure 31 General layout of frame
Pinned base
(59) - Member number
\
3O.
28 - Joint number
For the computer analysisthe propertiesof the following member sizes are used: Top boom Bottom boom Diagonals - struts
-ties
Columns
120x80x5 120x60x5
RHS RHS 60x40x5 RHS 60x40x4 RHS 457x191x89 UB
For the computer analysisthe columns are assumed to have pinned bases and are continuous from base to eaves e.g. joints 2, 3, 4, and 5 are rigid. The joints of the lattice girder are assumed to be pinned including the connections to the columns at joints 5, 6, 39 and 40. The girder acts as a brace to the two pinned columns. The respectivetension and compression in the top and bottombooms provide an effectivemoment at the top of the column providingan analogous pinned portal, as shown in Figure 30. This is Frame Type No. 1. In Section 11 alternative frame analyses are considered. The positioning of the purlins at node points removes the effect of local bendingbetween joints in the case of the pin-jointed truss. In practice it is possible the numberof panels would have been reduced from 8 to say 6. In this case bending would be induced in the rafters. Designcould then have been in accordance with Clause 4.10(c) of BS 5950: Part 1, incorporating a BM of WLI6 with theaxial forces or using bending moments obtained from
the computer analysis.
Typical loading diagramsare shown in Figures 32 and 33. These were used in the respective computer analyses. The results of the analyses are listed in Tables 2-5.
42
D
In
Imposed
I—
.1
(58)
(5q)
15 x
15 x
kN
kN
30. Om
6.87
4.44
Figure 32 Vertical loading due to dead and imposed loads on girder
kN
kN
(3)
3.fi
2.22
I
3.44
2.22
kN
kN
0
c'J
0
In
U,
It,
c
0.
kN
(60)
(61)
(62)
30.Om
3.7 ,. 2.02
(65)
Figure 33 Loading on columns and girder due to wind case 1
(56)
7 x 1.
(66)
(61)
7
x 4.04
kN
II,
2.02
IcN
kN
0.81 kN
1.13 kN
2.05
2.36
1.13 kN
1.3S kN
kN
Table 2 Axial forces (kN) and bending moments (kNm) in columns MEMBER
DEAD F
IMPOSED
WIND
WIND
I
II
M
F
M
F
M
F
M
1
-33.0
0
-55.0
0
+50.4
0
+54.6
0
2
-33.0
28.5
-55.0
47.4
+50.4
-51.6
+56.6
-38.9
3
-33.0
57.0
-55.0
94.9
+50.4 -100.3
+54.6
-81.8
-33.0
85.5
-55.0
142.3
+50.4 -145.6
+54.6 -129.5
-23.8 104.6
-39.5
173.9
+35.6 -173.3
+39.4
* * * *
-164.3
73
—23.8
104.6
—39.5
173.9
+26.4 -116.0
+39.4 -164.3
74
-33.0
85.5
-55.0
142.3
+37.2
-91.7
+54.6 -129.5
75
-33.0
57.0
-55.0
94.9
+ 37.2
-58.2
76
-33.0
28.5
-55.0
47.4
+ 37.2
-27.8
+ 54.6 + 54.6
77
-33.0
0
-55.0
0
+37.2
0
+54.6
Notes:
—ye
BM indicates tension on inside of column face.
—ye
Force indicates compression.
+ve Force indicates tension.
Wind Case Ill has not been tabulated- examination were not significant.
* *
Members highlighted
—81.8
-38.9 0
of computerprint-out indicatedthat the loads
by two asterisks are critical members for subsequent design.
45
Table 3 Axial forces (kN) in top boom Truss Members MEMBER
DEAD
Pin Jointed IMPOSED
WIND I
WIND II
6
+64.9
+ 107.3
-107.8
-98.7
7
+21.8
+36.2
-37.3
-30.1
8
-14.5
-24.1
9
-44.3
-73.7
+21.6 + 68.8
+27.9 + 75.3
10
-67.6
-112.5
+104.4
+112.2
11
-84.4
-140.4
÷128.2
+138.5
12
-94.8
-157.7
+140.5
+154.2
-98.7
-164.1
+ 141.1
+154.2
14
-98.7
-164.1
+ 135.0
+ 159.3
15
-94.8
-157.7
+124.8
+154.2
16
-84.4
-140.4
+108.3
+138.5
17
-76.6
-112.5
+85.3
+112.3
18
-44.3
-73.3
+56.0
+75.5
19
-14.5
-24.1
+20.4
+28.1
20
+21.8
-36.2
-21.7
-29.9
÷64.9
-107.3
-70.1
-98.4
*
21 Notes:
*
BM indicates tension on inside of column face. Force indicates compression. + ye Force indicates tension. Wind Case III has notbeen tabulated - examination of computer print-out indicatedthat —ye —ye
the loads were not significant.
**
46
Members highlighted by two asterisks are critical members
for subsequentdesign.
Table 4 Axial forces (kN) in internal boom members Truss Members MEMBER
DEAD
Pin Jointed IMPOSED
WIND
I
WIND II
* 22 *
+35.6
+59.3
-58.6
-56.8
*23 *
-38.8
-64.6
+63.9
+61.8
24
÷ 30.6
+ 50.9
-49.6
25
-33.3
-55.5
26
+25.6
+42.6
27
-27.9
28
36
+20.6 -22.4 + 15.6 -16.9 ÷ 10.5 -11.5 +5.5 -6.0 +0.5
-46.4 +34.2 -37.3 + 25.9 -28.2 + 17.5
+54.0 -40.6 +44.2 -31.6 + 34.4 -22.5 + 24.5
-48.6 + 52.9 -40.4 +44.0
37
-0.5
38
41
+ 13.9 -0.5 +0.5 -6.0
42
-32.2
-13.5
+ 35.1 -24.0 + 26.2 -15.8
-19.1
+ 14.7
+ 17.2
-4.5
-7.6
+4.9
+8.3
+4.6 -5.0 -18.9 +5.6 -5.2 + 11.0
+0.6 -0.6
+5.5
+9.2 -10.0 +0.8 -0.9 + 23.1 -0.9 +0.8 -10.0 + 9.2
43
-11.5
-19.1
44
+ 10.5
+ 17.5
45
-16.9
-28.2
46
+ 15.6
+25.9
47
-22.4
48
+20.6
49
-27.9
50
+25.6
51
-33.3 +30.6 -38.8 + 35.6
-37.3 +34.2 -46.4 +42.6 -55.5 + 50.9 -65.5 + 59.3
29 30 31
32 33
34 35
39 40
52 53 54 Notes:
-10.1
+ 16.4 -15.0 +21.7 -20.0 + 27.1 -24.9 + 32.5 -29.8 + 37.9 -34.8 + 43.2 -39.7
—ye BM indicates tensionon inside of column —ye Force indicates compression.
-20.0 -0.7
+0.6 +8.3 -7.6 + 17.2 -15.8 + 26.1 -24.0
+ 35.0 -32.2
+44.0 -40.4
+ 52.9 -48.6 + 61.9 -56.8 face.
+ve Force indicates tension. Wind Case III has not been tabulated - examination of computer print-outindicated that the loads were not significant.
**
Members highlighted by two asterisks are critical members for subsequent design.
47
Table 5 Axial forces (kN) in bottom boom members Truss Members MEMBER
*55*
DEAD
Pin Jointed IMPOSED
WIND I
WIND II
-106.6
-177.2
+170.8
+175.2
* 56 *
60.6
-100.8
+95.2
+ 101.9
57
-21.1
-35.1
+31.1
+39.2
58
+12.0
+19.9
-21.2
-13.0
59
+38.5
+64.0
-62.0
-54.5
60
+58.6
+97.4
-91.0
-85.5
61
+72.2
+120.0
-105.9
*62*
+79.3
+131.8
-108.5 -114.2
+ 79.3
+ 132.9
-108.4
-115.0
+79.9
+ 132.9
-108.4
-115.0
*65*
+79.3
+131.9
-101.7
-115.7
66
+72.2
+120.0
-88.7
-106.0
67
+58.6
+97.4
-69.3
-85.6
68
+38.5
+64.0
-43.6
-54.7
69
+12.0
+19.9
-11.4
-13.1
70
-21.1
-35.1
+27.1
+39.0
71
-60.6
-100.8
+71.9
+101.6
72
106.6
-177.2
+123.2
+174.9
* 63 * * 64 *
Notes:
* *
48
-115.7
BM indicates tension on inside of column face. Force indicates compression. +ve Force indicates tension. Wind Case Ill has not been tabulated- examination of computerprint-out indicatedthat the loads were not significant. —ye —ye
Membershighlighted by two asterisks are critical members for subsequentdesign.
10.
FINAL DESIGN
Use Grade 43C steel throughout.
10.1 Top boom From the tabulationof forces it is noted that members 6 and 13 are the most heavily loaded elements. Maximum axial loads (to 3 significant figures) (Table 3) Member 6
Member 13
65 kN (T) 107 kN (T) 108 kN (C)
Dead Imposed Wind
99 kN (C) 164 kN (C) 159 kN (T)
Maximum compression due to dead and imposed loads (member 13).
F
= (1.4x99)+(1.6x164)=4OlkN(C)
Maximum tension due to dead and Imposed loads (which is significantly greater than the compression due to dead and wind), (member 6). FT
= (1.4
x 65) + (1.6 x
107)
= 263 kN (T)
Effective length (BS 5950: Part 1, Clause 4.10(e)) = 1.85 m
x 80 x 5 RHS
Referringto the preliminary design it is appropriate to try a 120
D= A=
120 mm 18.9 cm2
B = 80 mm t = 5.0 mm = 4.43 cm ry = 3.21 cm
r
Design strength (Table 6, Reference 1), p,
275 N/mm2 since
t < 16.0 mm.
Design as a Compression Member in accordance with Clause 4.7 (Reference 1). Section classification (Clause 3.5)
=
=
1
(Table 7, Reference 1)
py From Figure 3 (Reference 1) The possibleeffect of secondary stressesin the section can be reduced by placing the 80 mm side in the vertical plane, as shown in Figure 34. Hence
x,'
0.85
185
= 40 < 180 < 50 (Clauses 4.7.3.2
and
=J8!
4.10 of Reference 1).
=42<180
4.43
49
This is also a better way for the section when considering connections to the internal bracing members and purlin support cleats.
b = B — 3t therefore —b t
=
d = D — 3t therefore d
=
120-3x5 = 21.0 __________ 5
r
80
—
3x 5
=
13.0
Hence from Table 7 (Reference 1)
<
t
. t
< 39
section is NOT SLENDER
i.e. section capacity will not be reduced by local buckling(Clause3.6, Reference 1).
120
-
—
80
fxs Figure 34 Section through top boom Compressivestrength
= AgPc where, accordingto Table 25 (Reference For
p, = 275 N/mm2
and X
18.9
=
x 10
is obtained from Table 27(a).
= 49 (say)
= 252 N/mm2 (Note this is an 8% Hence
l)
reduction).
252 = 476 kN > 401 kN
SECTION IS SATISFACTORY IN COMPRESSION. Check for tension Clause 4.6.1
Tension capacity
P = A Py
Since Ae = A then
=
18.9
x 10
275 = 520 > 263 kN
SECTION IS SATISFACTORY IN TENSIONAND IN FACT THIS SHOULD BE OBVIOUS SINCEp, > and F1.
p F>
50
As an alternativeto the abovedetailed calculationuse the SC!publication "Steelwork Design Guide to BS 5950 Vol. 1, Section properties membercapacities
For a 120x80x5RI-IS this shows P,
446 kN when Le'
P = 520 k/V.
2.0 i.e. > 1.85 m and
lop boom use 120 x80 x5 RHS, 80 mm size vertical (the reader should check to determine whether a lighter section would be satisfactory).
10.2 Bottom boom Examinationof the force analysis tables and considering member, and therefore effective lengths, it is clear that there are several members in the bottom boom which need to be considered. (a)
Members 62 and 63 Maximum axial loads (Table 5): Member 62 Imposed
80 kN (T) 132 kN (T)
Wind (Case II)
ll6kN(C)
Dead
Member63 80 kN (T) 133 kN (T) 115 kN (C)
Maximum tension due to dead plus imposed loads FT = (1.4
x 80) +
(1.6
x
133) = 325 kN (1').
Compressioneffects are low in comparison with members adjacent to the column. (b)
Members 55 and 56 Maximum axial loads Member 55 Dead Imposed Wind (Case II)
107 kN (C) 177 kN (C) 175 kN (T)
Member 56 61 kN (C) 101 kN (C) 102 kN (I)
Maximum compression in member 55
= (1.4
x
107)
+ (1.6 x
177)
= 433 kN.
Maximum compression in member 56
= (1.4 x 61) + (1.6 x
101)
= 247 kN
From the above the controlling load is 433 kN (C) in member 55. Since the maximum tensile load in the boom (325 kN) is smaller than the maximum compressiveload (433 kN) the bottom boom has only to be designed for compression.
51
Try the assumed size of 120 x60x5 RHS.
D=
B = 60 mm
120 mm
r
t = 5.0 mm A = 16.9 cm2 = 4.24 cm
ry
= 2.43 cm.
Provide a purlin brace from joint 8 as shown in Figure 12. Also place 60 mm side vertically. Check section classification
b
120—(3x5) =
=
-
21
5
= =
.
= 92.5 cm
= =
--4.24 p
Hence
=
x
2.43
5
r r
=
2.43 cm
=
4.24 cm 185 cm
Ley' 0.85
=
1).
33
< 50
= 44
=
257
t
< 2&, hence section is plastic (Table 7, Reference
r' r'
Hence
Lex1
.
<
then
d = 45 =
257 N/mm2 (Table 27a, Reference 1).
x 16.9 = 10
434 kN > 433 kN.
Hence section is satisfactory.
= 1.7 m long. Load in The purlin brace, at an angle of 45° would be approximately 1.2 the brace is 2½% of load in the boom (Clause4.3.2') i.e. 43.3kN. Assuming an effective length of 1.7 m a suitable brace would he a 40x40x3 RHS (Reference 19). Use of sheetingbrace. Occasionally it may he advantageous to brace node 7 of the bottom boom to the sheetingrail. In that case Ley' = 92.5 cm. The reader should check the effect of the bracing. Bottom boom use 120 x60x5 RHS (60 side vertical), braced to nurlin at node 8 using 40x40x3 RHS.
52
10.3 Internal members Diagonal bracing - Members 22 and 23 22 is a tie and Member 23 is a strut.
Note -
Member
Maximum axial loads (Table 4):
Tie (22)
Strut (23)
36 kN (T) 59 kN (T) 59 kN (C)
Dead Imposed Wind
39 kN (C) 65 kN (C) 64 kN (1)
10.3.1 Tie design (22) Maximum tension due to dead + imposed load
=
(1.4x36)+(1.6x59)
=
145kN(T)
Maximum compression due to dead plus wind
(1.0x36)-(1.4x59)=47kN(C). Memberlength = 1.59 m Selecting member size for tensile force, hence area required = 145
x
10
=
275
5.3 cm2.
Try6Ox4Ox3RHS A = 5.60 cm2
=
1.59 cm
Check for compressive strength.
Xy
p C
x 0.85 = 81
=
= =
201 N/mm2 201
(Table27a of Reference 1)
x 5.6 = 10
(See Reference 1, Clause 4.10 allow for some end fixity).
113 kN
> 45 kN
Diagonal ties 60x40x3 RHS
53
10.3.2 Strut design (23) Maximum compressiondue to dead + imposed load
=
(1.4
x 39) + (1.6 x 65) = 159 kN (C).
Maximum tension due to dead
=
-(1.0
+ wind
x 39) + (1.4 x 65) = 52 kN (T)
Clearly compression design will control.
Try 80
x 40 x 5 RHS
r
= 1.55 cm
A
=
X.
=ix0.85=84 1.55
10.9 cm2
=
194 N/mm2
=
194 xlO.9
(Table 27a Reference 4)
= 211 kN >
159kN
Diagonal struts 80 x40 X S RHS
10.4 Comparison
of member sizes
It is perhaps useful to compare the assumed sizes of the girder membersused in the analysis and the calculated values of the RHS sections. Member
Assumed section
Designed section
Top boom Bottom boom
120x80x5
120x80x5
120x60x5
120x60x5
60x40x5 60x40x4
80x40X5
Diagonals: Struts
Ties
60x40x3
i.e. The basis for assumed sectionsizes is basically justified. Note: Using the top and bottom boom with the 120 mm horizontal side is convenient for welding the diagonal strut. The reader could easily check the possibilityof using a 90x50x3.6 RHS (Reference 19) which is lighter and which could be a suitable member.
In Section 9.6.2 (e) it was stated that deflection would not be critical. From the computer analysis the maximum deflections in the girder occurs at joint 23. Dead load deflection
=
35.3 mm
Imposed load deflection
=
58.8 mm
Ratio of imposed load deflection/span
54
= 58.8/30000 =
1/510,
which is satisfactory.
10.5 Column design - members 1 to 4 and 5 Axial loads and moments (Refer to Table 2) Member
to 4
1
Maximum BM
Maximum axial loads
33 kN (C) 55 kN (C) 50 kN (C) 55 kN (1')
Dead Imposed Wind 1
Wind 2
Design load: dead
=
+ imposed
x 33) + (1.6 x 55) =
(1.4
(1.4 X 86) + (1.6
Design load: dead
=
(1.0
(1.0
x
142)
= 348 kNm
+ wind I
x 33) - (1.4 x 50) = 37 kN (T)
Design BM: dead
=
134 kN (C)
dead + imposed
Dead BM:
=
86 kNm 142 kNm -146 kNm -129 kNm
+ wind I
x 86) -(146 x
1.4)
= 118 kN.m.
Hence worst effect in column is due to dead and imposed load effects. Selection of the effective length for the column provides an interesting problem. About the XX axis member 1 to 5 is restrained in positionat the base but not in direction. At the top it is partially restrained in positionand direction. This condition does not relate to any of the standardcases in Table 24 or Appendix D of BS 5950: Part 1. Reference to the computer frame analysisprovides the horizontal and rotational movements. The figures for the separate dead and wind load conditions indicate points of contraflexure in the column occur between joints 3 and 4. Assume contraflexureoccurs at joint 4 which is 4 m above the base. Then the corresponding effective length of an analogous pin-ended strut would be 8 m. As a ratio of the member length, this provides an effective length factor of 8/5.5 = 1.45. Figure 17 of BS 5950: Part 1 indicates Lex = 1 .5L for a fixed base. However, the connection effect of the latticed girder is clearly stiffer than that provided by the roof connection shown, but the (partial)fixity providedat the base is not the same as that used in Figure 19 of BS 5950:
Part 1.
The abovediscussionindicates the careful consideration which must be given to the selection of an appropriateeffective length factor. When in doubt it is essential to use a conservative value i.e. in this case 2.OL = 2.0 x 5.5 = 11 m. Aboutthe YY axis the base is restrained in positionand directionand the head of the column is restrained in position, but not direction(by the eaves rafter and first rafter purlin). Hence as in Figure 17, (Reference 1) assume Ley = 0.85L = 0.85 x 6.7 = 5.7 m. Lex
=
11
m, Ley = 5.7 m. 55
In selectinga column size a useful starting point is item (v) of Section9.6.2 i.e. try the 457x191x89 UB.
D bIT
= 463.6 mm B = 192.0 mm = 5.42 dIr = 38.5 41000 cm4 r = 19.0 cm =114cm2
A
Since 40
> T>
t
=
10.6 mm
T
r = 4.28 cm
= 17.7 mm d = 407.9 mm = 1770 cm3
S,
= 2010 cm3
=28.3
x
= 265 N/mm2 (Table 6(1))
16 mm then
Hence:
=
s/275/265
=
1.02
Section Classification (Table 7 and Clause 3.5.4W).
Examinethe "outstand of the compression flange" and "web generally". For the flange bIT = 5.42 < 8.5E, henceflange is plastic. The web is in combined axial and flexural compression (Figure 35). Determine the positionof the neutral axis and check footnoteto Table 7(1)
192.0 t*17.7 yc
'—I cv,I
(0
x—
(V)
—
-
(0
1
3.854
Figure 35 Stress diagram Check section classification for "web generally".
Length
'<
of web subjected to direct axial load =
47.7 mm
From Figure 35 the plastic neutral axis is (47.7/2) = 23.85 say = 23.9 mm from the XX axis.
The footnote to Table 7(1) indicates that a = In this case
a=2 x
56
d
y = (463.6/2) + 23.9 = 256 mm
256/407.9
= 1.26
79 0.4 + 0.6a
79
=
Hence web and flange
and overall buckling. (a)
x
0.4 + (0.6
1.02 x 1.26)
= 70 >
t
= 38.5
are plastic. Two checks are required for the design: Local capacity
Local capacity (Clause 4.8.3.1')
F __ AgPy
+_Li M M
M +__ M
NowM =
0
For plastic section
= 2010 x 265 x i0 = 1.2 x 1770 x 265 x
= S,p
or
l.2p,,i'
Hence
134 x 10 114 x 265
+
348
532.7
=
0.044
532.7 kNm
i0 = 562.9 kNm +
0.653
=
0.697
<
1.0
Section satisfies local capacity check.
If use is made of the relationship: M
M
+
M
Mrx
= 1.0
thenZ1 = 2.0
and
M=
0.
i.e.
M2
__f_
M
i
Now Mrx = Srx Jy where S,
= K! — K2 (n)2 (Reference 19).
= F/A .p, = 134 x 10/(1 14 x 265) = 0.044 = 201f K1 = 3070 K2 = 2010 - 3070 x ØØ442 = 2004 cm3. Srx Mrx = 2004 x 265/1000 = 531 kNm. n
M2
= 531
This provides
=0.43
a much more conservativeanswer than 0.697 obtained above.
57
Overall buckling (Clause 4.8.3.3.1")
(b)
F ____
+
mM
mM _2
+
AgPc
Compressivestrength,
= =
>x
1
Mb
p
= =
Lex/rx
Lc,/r,,
1100/19.0 = 58 570/4.28 = 133
From Tables 25, 27a and 27b Reference 1,
i' = 234
or 90 N/mm2.
Usep = 90 N/mm2 Bucklingresistancemoment Mb. Use the conservative approach (Clause4•37•7(1))• From Table 10 (Reference 1) on the basis that the base is restrainedlaterally and torsionally and the top of the column has a lateral restraint to the "tension flange", assume:
Le
= 0.7L and m =
Take X
1.0
n = 0.94 (Table 20w).
= 0.7 x 550 x 0.94/4.28 = 85
Whenx = 28.3 and X =
=
184 N/mm2
(Table l9a')
Mb = S,(p = 2010 x The interaction 134
x
10
Therefore,
—
85
184/1000
= 370 kNm.
expression
l.Ox34.8 =
0.131
+ 0.941 > 1.0
lower section of column is not satisfactory
if the simplified approach is used.
There are four options: (i)
use full calculation to obtain Mb (see Clause 4.3.7 Reference 1)
(ii)
increase size of column
(iii) provide a restraint to the compression flange from the sheetingrail, say at joint 4. (iv)
58
use the more exact approach (Clause4.8.3.3.2(1))i.e. check the interactionequation
—1 M mM
where
M is taken as the lesser of: Pcx
or Mb (1
1+ 0.5F
—
F/P)
Pcx
= 265 x 2010/1000 = 533 kNm (Clause4.25').
P=
F
=
Mb
= 370 kNm
134 kN
=
Ag
= Agp
114
x 234/10 = 2668 kN
114
x 90/10 =
1026kN.
Hence
M
=
533
—
x
[1+(0.5
=
(134/2668)]
x
494 kNm
134/2668)]
or
= 370 (1 Hence
—
(134/1026))
=
322 kNm
M = 322 kNm.
To obtain 'm' use Table 18 of Reference is zero. Hence m = 0.57.
= 0, sincethemoment at the base of the column
1
Therefore ?nM = 0.57
x
322
348 = 0.62 <
1
The more exact approachappearsto indicate that the column can adequately support the applied load and bendingmoment. Howeverthe footnoteto Clause 4.8.3.3.2 states "In cases where M or M approaches zero the more exact approach may be more conservativethan the simplified approach. In such situations the values using the simplified approach (should?) be used". Clause 4.8.3.3.2 probablyshould only be used for bi-axial conditions. In this solutionprovidea column/rail restraint.
y
Check member 5. Maximum effects dueto dead and imposed loads (Table 2) Axial load
= (1.4 x 24) + (1.6 x 40) = 98 kN
Bendingmoment = (1.4
x 105) + (1.6 x
174)
= 425 kNm
59
Checkinglocal capacity.
x x
98 114
10 265
=
=
532.7
+ 0.8 =
0.032
0.832
<
1.0.
is also made on the overall bucklingcapacity of member
A check method.
=
425
+
533 kNm 2668 kN
F=
1-5, usingthe more exact
M =
425 kNm i.e. using the value determined for the column section 1-4. 98 kN
Mb is evaluated for X over the full height of column. i.e. X = 0.7 x 670 x 0.94/4.28 = 103.
Hencep = Mb
154 N/mm2
= 2010 x
154/1000
(Table 19a' for x = 30).
= 310 kNm.
is evaluated for = 0.85 L/r (Table 24'). = 0.85 x 670/4.28 = 133. i.e.
p = 90N/mm2 (Table 27b').
Hence
114/10 = 1026 kN (Note it is coincidental that this is the same value as that evaluated for the column length 1-4).
= 90 x
Hence
= 533 x
M
=
(1—(98.2668))
(1 + (0.5
x
504 kNm.
98/2668)
= 310 (1 - (98/1026)) = 280 kNm.
or Hence
= 280 kNm.
M
Therefore
mM
x 087 .
M
280
____
425
0.87
<
1.
Hence overall bucklingof column length 1-5 is satisfied. Check horizontal deflection Deflection at
Joint 4 mm Dead Imposed
Wind 1 Wind 2
-5.0 -8.2 +11.9
+7.3
Deflection at Joint 39 mm
+1.5 +2.5 +3.5 +3.4
Maximum deflection at joint 4, due to dead and imposed loads
= 5.0 + 8.2 = =
1/303
60
13.2 mm
13.2/4000
Maximum deflection at eaves
=
1.5 + 2.5
= 7.5/6700
+ 3.5 = 7.5 mm = 1/893
BS 5950: Part 1 does not provideguidance for portal frame, however Table 8.1 in Reference 20 specifies a limiting deflectionof height/100at the eaves, which is satisfied. The selected section satisfiesall design criteria. The interaction equations, for local and overall buckling, clearly show the main design criteria are bending effects. This justifies the use of a universal beam section. Use 457x 191 x89 UB for column.
10.6 Gable steelwork There are a number of alternative methods of design of the gable steelwork.
Gable columns can be considered as pinned or fixed at the base. Normallythey will be designed as pinned at the top being supported in this positionby the gable rafter and/or "wallbracing" and purlins and/or roof bracing. Consequently gable columns are designed as propped cantilevers or simply supportedbeams. Figure 18 shows typical roof bracing arrangements. It should be noted that the props at the head of propped cantilever gable columns are not fully rigid. The wall and roof bracing can be single diagonal members or cross diagonals. In the former the members need to be designed to transmit direct tensionand compression (normallydue to wind loads). With cross bracing the members in compression are ignored, the members in tensionare assumed to transmit all of the load (to the foundations in the case of wall bracing and into the roof in the case of roofbracing). In the following design singlebracing is used. Purlins need to be checked for the additional axial load since they are providingthe propping reaction.
10.6.1 Gable rafters are simply supportedon a span of 7.5 m betweenstanchions. They are held in place and loaded by the purlins at 1.875 m c/c (Figure 36). The loads will be dead, imposed and wind (uplift) loads on the rooftogether with some end sheeting vertical load. These
W/2
W
W
W/2
4 x 1.875 = 7.5 m Figure
36
Gable rafter loading
61
Loading (see Section9.7) Load/purlin Dead
4.13/2
= 2.07 kN
Imposed
6.87/2
= 3.44 kN
Wind (Uplift)
7.4/2
= 3.70 kN
1.88 x 1.5/2
= 0.3 kN
Vertical Sheeting, say 0.2
x
Maximum downward load (W)
1.4 (2.07 + 0.3) + 1.6 (3.44) = 8.8kN
=
Hence reaction = 8.8 x 2 = 17.6 kN
- 1.4 (3.7) =
-2.8kN.
Maximum bending moment at C due to 'W'
1.5W
Maximum uplift = 1.0 (2.37)
x 3.76 - W X
1.88
= 3.76W.
Max. BM = 3.76 x 8.8 = 33.1 kNm. Allowing for self weight, design say for 34 kNm.
The rafter should be designed as a laterally restrained beam, clearly with low shear. (Clause4.2.5').
Try 203x102x23 UB = 5.46
t
r=
=
t
2.37 cm
= 2090 cm4 Moment capacity M
T=
32.6 S,
=
p, =
9.3 mm
275 N/mm2
= Sp,
232
= 206 cm3
x = 22.6
232 cm3
From Table 7 (Reference 1) section is plastic.
x 275 x
i0
=
63.8 kNm < 1.2
> 34kNm
Bucklingresistance, usingtheconservative approach (Clause4.3.7.lW).
Mb=pbSX m=1.0 As an approximation assume n = X
= Leiry
188/2.37
1.0
= 79
"Roundingup" to assume X = 80, x = 25 then Pb = 201 N/mm2 (Table l9bW). Mb
= 201 x 232 x i0
= 47 kNm > 34 kNm
Note: Reference 19 indicated that Pb satisfactory.
62
46 kNm for n
= 1.0 and Le = 2m.
This is
Z
The reference also states section is "plastic". Clearly wind uplift will not create a critical situation. Deflection check.
Assumetotal imposed load of 6.88 kN acts as a central point load then central deflection is
wi. 3/48E1.
C
=
6.88
x
753 X i05
48x205x2090
=
14.1 mm
This provides a deflection/span ratio of 14.1/7500 = 1/532. Hence showing that deflectionis not significant,even when the loads are considered as a single central point load. Use 203 x 102 x23 UB for gablerafter.
10.6.2
Gable columns
These support the gable rafters, which induce vertical
wind load, as shown in Figure 37.
load, and gable rails which transmitthe
Assume gable columns are tixed at the base and pinned at the top therefore consider as a propped cantilever. The central column is say 8 m high. Load from
Sheeting
gable rafters
load
______
Assume uniform wind load
Roof bracing acts as prop
8m
I//I Figure Dead
37
Loads on gable columns
plus imposed load reactions
=2x
17.6 = 35.2 kN.
Sheetingselfweight (Section 9.4.3) = 0.2 kN/m2. Rail weight
= 0.2 kN/m (5 No. at 7.5 m long).
Total weight from sheetingand rails
= (0.2 x 7.5 x 8) + (0.2
Design vertical load = 35.2 + (1.4
x
19.5)
X
5 X 7.5) =
19.5 kN.
= 62.5 kN say 65 kN, allowingfor self weight.
63
Wind load (Section 9.3) and Table 1. On end = (Cpe
—
C) =
1.0.
As a conservativedesign assume wind load over the full heightof the gable is 0.582 kN/m2. Total design load from wind = 1.4 x 1.0
x 0.582 x 8 x 7.5 = 49 kN.
Maximum bending moment in a propped cantileveroccurs at the base and equals WL/8.
Max. BM due to wind = 49
x 8/8 = 49 kNm.
Ignoring wind uplift, design the gable column for vertical load of 65 kN and BM of 49 kNm. It will be noted that bendingis the controlling applied action.
The reaction at the pinned end of a propped cantilever is:
W=x49 = 8
18.4kN
8
This will be the load transmitted to the wind (root) bracing. In consideringbuckling effects: For axial load design the slenderness should be limited to 180 (Clause4.7.3.2') 0.85L = 0.85
Lex Ley
x 8 = 6.8 m (propped cantileverTable 24(1))
= 0.85 x 8 = 6.8 m
(Appendix D').
For lateral torsionalbucklingdesign, the effectivelength is again problematical. Consider the deflected shape for a propped cantilever subject to a uniformly distributedload. Zero slope occurs at a point 0.4215 L from the free end i.e. column head. Allowing for some movement of the prop it is appropriateto suggest Le = 2 x (0.5L) = 1 .OL. HenceLE = L = 1.0 x 8 = 8 m (Table 9(1)) In designingfor buckling use is made of Table 19 and Clause 43•77(1)
To initially assume a section size either use Reference 19 (Axial Load and Bending Tables) or assume a value for Pb For low loads and moments with L = 8 m the use of Reference 19 is notapplicable. Hence assume a value for Pb = 150 N/mm2. Then
S = M/p = 49 x 10/150
Also if X < 180 then
327 cm3
r < 680/180 = 3.78 cm.
Try 305
x
= bIT = = S, =
303.8mm B = 165.1 mm t = 6.1mm 8.09 d/t = 43.6 = 12.9 cm = 3.85 cm 8520 cm4 3 = 624 cm x 31.1 A = 51.5 cm
Note:
dit = 43.6 >
D
64
165
x 40 UB.
r
39€ (Table 7(1))
T = 10.2mm
= 561
cm3
Howeverthe latter is used for webs subject to compression throughout. Whilst this is the situation when considering dead plus imposed loads it does not apply when the bendingdue to wind effects are present. As will be seen the axial effects on the column are small and therefore a reduced value of (Clause3.6(1)) has not been considered.
p
Since T < 16 mm then
p = 275 N/mm2.
Usingthe approach outlined in Section 10.5 it will be deducedthat the section classification is plastic. çFable 7(1))
Local capacity check (Clause4.8.3.1(1)). Whilst it is suggested the relationship: M
M
should be checked for plastic sections, use is made of:
__2.
+
M,.),
F
M
+
AgPy
M
+
M
being a simplified (conservative) approach in this case.
= 0
= 624 x 275
=
F
+
AgPy
=
M
65x10 51.5 x 275
M 0.041
x 1O = +
+ 0.285 = 0.326 <
172 kNm (Check Reference 19)
172 1
Section satisfies local capacity Overall bucklingcheck (Clause 4.8.3.3.1(1)).
p
Compressivestrength,
> = 680/12.9 = 53
k, = 680/3.85 =
177
<
180
From Tables 25, 27a and 27b (Reference 1)
p = 55 N/mm2. Bucklingresistancemoment Mb X
= 800/3.85 = 208.
From Table 19(1)Pb = 75 N/mm2 (say).
Mb=SJb=624 x75 x 103=46.8kNm
65
The critical condition has been induced by the high slenderness value, when considering buckling. The option used in this case is to examine the exact approach Clause 4.8.3.3.2(1). i.e.
mM 1
M where M is the maximum bucklingmoment about the major axis in the presenceof axial load and is taken as the lesser of:
________
or Mb
1
-
0.5F 1+—_
F
Pcx
In this case the moment at the top of the column is zero hence = 172 kNm. m = 0.57,
M
For X = 53 then 248
=
= 248 N/mm2 (Table 27a').
x 51.5 = 10
=
For
= 55
177
then
1278kN.
p,
x 51.5 =
55 N/mm2 (as before) 283 kN.
10
1=
172
0.5F
1+__c_ Mb
1
- -L
Hence take
mM =
M
= 0 (Table 18(1)), therefore
46.8
65 1278
_______________
10.5x65 1278 1
283
= 159 kNm
= 36.1 kNm
M = 36.1 kNm.
0.57 x 46.8 36.1
=
i.e. section is satisfactory. The significant difference between the simplified and exact analysis is dueto the reduction in the applied design bendingmoment i.e. the effect of 'm'. Use 305x 165 x40 UB for gable columns.
66
10.7 Bracing 10.7.1 Roof bracing (at both ends of building) 33.8 kN
33.8
1'
7.7
16.9
16.9
18.4
7.7 kN
Figure 38 Use of cross bracing
The wind load at the top of the central column = 18.4 kN. At the top of the intermediate columns = 18.4 x 7.35/8 = 16.9 kN. (i.e. in proportion to column height).
/N
At the top of the corner columns =
18.4
x 6.71(8 x 2) = 7.7 kN.
If cross bracing, Figure 38, is used the members in tensionare assumed to carry all of the load; compression members need only satisfy the slenderness criteria (Clause4.7.3.2(1)). i.e. 350 where member length is 9.6 m. For ease of construction single bracing is used, Figure 39.
33.8 kN
33.8 1
6m1
¶
¶
16.9
1I, Figure
39
Use
18.4
I
16.9
I
7.7 kN
4x7.5m
of single or tension bracing
Maximum tension occurs in member (a)
= (33.8 — 7.7) x 9.6/6 = 42 kN.
Since single bracing is used the member will be designed for compression forces, noting the load will be smaller since it is only necessary to design for suctionpressure coefficients — = (—0.6 — (+0.2)) —0.8 (see Section 9.3), instead of 1.0. (Cpe Cr1)
67
= 0.8 x 42 = 34 kN and slenderness criteria = 250 (Clause4.7.3.2'). = 34 x 10/275 = 1 .24 cm2. Using circular hollow section, required area of section Design load
Required radius of gyration
Try 114.3 x 5 CHS A = Dit = 114.3/5 23 < 40 Need to satisfy X
P = 17.2
17.2 cm2
r = 3.87 cm I = 257 cm4 S = 59.8 cm3
(Table7(1))hence section is plastic.
P = Ag P
= 960/3.87 =
248
<
250.
1 31 N/mm (Tables
Hencep =
3.84 cm.
960/250
25, 27a''1'').
x 31/10 = 53 kN > 34 kN.
Note a check needs to be made on the selfweight deflection (Clause4.7.3.2(1)).
Selfweight deflection = 5WL31384E1 = 5 x 0.135 x 9.6 x 10/(384 x 205
= 28.3 mm.
x 257)
Since this is greater than L/1000 = 9.6 mmtheeffectof bendingmust be considered. (Clause4.8.3).
= 0.135 x 9.62/8 =
Selfweightbending moment = MCX
= p S (Clause4.2.5) = 275 x 59.8/10 =
1.6 kNm.
16.5 kNm.
In the interaction equation
F
+
M
AgPy
M
= 0.17 <
1
34 x 10 + 17.2 x 275
= 16.5
0.07
+ 0.1
i.e. local capacity is satisfied Checkoverall buckling. BucklingresistanceMb =
=
16.5 kNm.
In the interactionequation
F __ AgPy
+
16 X+....064+010074<1
mM Mb
34 53
16.5
Hence bendingcriteria is satisfied.
68
The disadvantage of havingto use a larger member for the single bracing system (as compared to smaller membersfor cross bracing)is outweighed by the savings in connections and erectioncosts. Use 114.3 x5 CHS for roofbracing in plane oftop chord. Check also the end purlin for the effect of transmitting axial load (gable column reaction) of
18.4 kN.
From Section 9.4.1 it is noted the proposed202x65x60x1.8 mm thick Z purlin will support a factored load of 1.29 kN/m2. The applied total factored load is 1.27 kN/m2. Hence there is little reserve for the axial load effect.
It would be necessary to check with the manufacturer on the effect of axial load plus bending, since theirliterature does not normally considercombined loading. It is also noted that generallythe load tables listed by manufacturers for their cold rolled sectionswere obtained by testing. A possibilitywill be to use a thicker section for the end bays only.
10.7.2
Gable bracing (at both ends of the building)
6.7 m
7.5m Figure 40 Gable bracing Member
(a) is 10.lm long (Figure 40) and has to resist a wind load say,
F = [Yj (Cpe - C q A/2J where (C — Cr1) = 1.0, q = 0.582 kN/m2 (see Section 10.6.2) andA = 3 x
F = (1.4 x
)
6.1= 20.1 m2.
1.0 x 0.582
Forcein member (a) Radius of gyration
x 20.1)/2 = 8.2 kN
8.2
x
1010/250
10.1/7.5 = 11.1 kN.
= 4.04 cm
Use 139.7x5 CHS for gable brace (Reference 19)
10.7.3 Longitudinal side wall bracing (at both ends of the building) is 9 m long (Figure 41). The wind force at the eaves level is 33.8 kN (Figure Hence force in (b) = 9/6 x 33.8 = 50.7 kN. 39) Member (b)
Radius of gyration
900/250 = 3.6 cm.
Use 114.3 x5 CHS (Reference19).
69
6.7 m
71111/
k
—I
7i,i1/i
6m
'I
YiJ/
6m
—I
Figure 41 Longitudinal side wall bracing
10.8 Column Base (Reference 1. Clause 4.13) The basic assumption in the frame analysis was that the connection at the column base would be pinned. The connection has to be capableof transmitting axial compression, shear and
uplift as shown in Figure 42.
Axial compressive force = 135 kN Axial tensile force (uplift) F = 44 kN Shear = 80 kN
F
F
457 x 191
x 89 UB
t = 20mm bleed holes
/ 8
o
a
/
Concrete grade 20 Length of holding down bolts will depend on type used e.g. expansion bolt or bolt set in clearance holes which are grouted (using mortar or epoxy)
Bolt edge distance is satisfied (Cl. 6.2)
Figure 42 Baseplatedetails
A simple slab baseplatewill be used, weldedto the column. To comply with current safety requirements it is necessary to provide four bolts to ensure stabilityof the columnsduring erection. These4 bolts can be positioned closely together, as shown in Figure 42. Keeping the bolts close to the axis of the column will ensure that any moment restraint effect will be kept to a minimum. 70
Unfactored loads (kN)
Loading (Fable 2) Maximum vertical dead load
=
33
Minimum vertical imposed load
=
55
Maximum wind uplift
=
55
Shear in conjunction with the loads (computer analysis) Dead
=19
Imposed
=
32
Wind, with max uplift
=
26
=
135
Vertical design loads
Due to dead + imposed = (1.4 Due to dead + wind = (1.0
x 33) + (1.6 x 55)
x 33) - (1.4 x 55)
= -44 uplift
Shear design loads due to dead and imposed = (1.4 Shear design loads due to dead and wind = (1.4
x
x
19)
19)
+ (1.4 x 26)
+ (1.6 x 33)
=
80
=
63
10.8.1 Baseplate Only thedesignof the main frame baseplate and plate are detailed within this text. These being designed to resist vertical load and shear only.
For the gable post bases referenceshould be made to Reference 21, Steelwork Design Guide to BS 5950: Part I . 1990 Volume 2 "Worked Examples". This details the baseplate design for shear, vertical load and moment.
(f)
Assumeconcreteto base to have a 28 day cube crushing strength of 20 N/mm2. Use an = 270 N/mm2 (Clause4.13.2.2(1)). allowable bearing stress of 0.4f.
135x Hence area required = __________ 0.4 x 20 Minimum size
=
16,875 mm2
This is impracticable for the size of column. Try a baseplate600 x 400 mm, area 240,000 mm2. X 10 Hence actual baseplate pressure = 135 600 x 400
= 0.57 N/mm2
71
Due to a concentric force the minimum baseplate thickness (Clause4.l3.2.2W)
t
=
.w pyp
In this case a = (400 -
t
192)/2
(a
=
—
0.3b2)
104
b = (600 - 463.6)12 = 68 mm
= 7.1mm
x 0.57(1042 — 0.3 x
=
682)]
This is again impracticable. Match the base plate, approximately to the flange thickness (T = 17.7 mm) i.e. use 20 mm base plate. (Clause4.13.2.2'). Baseplate 600x400x2Omm.
10.8.2 Holding down bolts Thesebolts (4 No.) are to be designed to resist the tension due to uplift and accompanying shear (Clause 6.3.6.3) and positioned to satisfy Table
3l'.
Assumebolts are M20 where tensile stress area = 245 mm2,Grade 4.6. Shear capacity (Clause6.3.2).
= =
160 N/mm2
(Table 32W).
(Reference 19, Bolt Capacities).
=4x
160
x 245 x
i0 =
157 kN.
Tensile capacity(Clause6.3.6)
= pA1 = 4 x
195
x 245 x
i0 =
197 kN.
Combined shear and tension (Clause 6.3.6.3).
F
F
PS
+ —
+ 157
1.4
P1
197
= 0.63 < 1.4
Hence use 4 No. M20 Grade 4.6 holdingdown bolts.
72
10.8.2 Welds (Reference 1, Clause 6.6) It is likely that a fillet weld would be providedaround the full perimeterof the UB to avoid
the formationof corrosion pocketsbetween the column and base plate. Weld strength = 215 N/mm2 (Table 36'). It is suggested that for practical purposesthe weld should be at least 6 mm leg length. Total perimeter is approximately (2 xd) + (4x 7) = (2 x407.9) + (4X 192) = 1584 mm. The shear capacity of say a 6 mm fillet weld is 0.7
x 6 x 215 x
(SeeReference19).
i0 = 0.903 kN/mm.
Hence shear strengthof weld
= 0.903 x 1584 = 1430 kN which is far in excess of any forces to be transmitted. Generally it is recommended, web weld
0.5t and flange weld 0.5T.
However, it should be noted the cross-sectional area (and therefore relevant cost) of welds is proportional to the square of the leg length. Hence use 6 mm f.w. for web and 8mm f.w. for flange.
10.9 Foundation The foundationhas to be capable of transmitting the vertical and horizontal loads to the supportingsoil and to resist uplift (Figure 43). Maximum vertical load
and uplift
= =
135 kN
44 kN (both factored)
135 kN 52 kN 450
55O
900 1500
7/
1500
Figure 43 Mass concrete foundation
73
To prevent uplift isolated foundation pad needs to weigh at least 44 kN. Using a concrete weightof 24 kN/m2 then volume of concrete = 1.84 m3. A base 0.9 m thick and 1.5 m square will be adequate. This size needs checking against magnitude and distribution of ground pressure under serviceability load conditions.
Vertical service load Horizontalservice load
= =
(33
+ 55 + self weight)
=
137 kN
52 kN
Taking moments about a bottom corner of the foundation, positionof resultant vertical load is at (137 x 0.75 - 52 x 0.9)/137 = 0.408 m, hence eccentricity = 1.5/2 — 0.408
= 0.342 m.
As this falls outsidethe middle third the linear distribution of ground pressure is triangularon a base of (3 x 0.408) = 1.224 m. Maximum ground pressure = adequate to sustain.
2
x
1.224
137
x
=
1.5
149 kN/m2 for which
the soil must be
Factor of safety against overturning: Overturningmoment = 52 x 0.9 Restoringmoment = 137 x 0.75
= =
46.8 kNm 102.8 kNm
Hence factor of safety againstoverturning 102.8
=
46 8
. . = 2.2 which is satisfactory.
A check should be also required for the factor of safety against sliding. Use foundation 1.5 m square x 0.9 m deep. Grade 20 concrete.
74
11.
ALTERNATIVE FRAME ANALYSIS
As indicatedin Section9.8 two alternative systemswere to be considered for the frame analysis: (a)
assuming base is pinned and all joints of the latticedtruss are pinned (Frame Type 1).
(b)
assuming base is pinned and all joints of the latticed frame are rigid (Frame Type 2).
Two membersof the top chord are checked for these two frame systems. Frame Type 1 was used for the design in Section 10. MEMBER 6 FRAME TYPE
Load Type
Number 2
Number I
Max. axial load (kN)
Max. axial load (kN)
Max. BM (kNm)
Dead
65 (1')
64 (T)
0.70
Imposed
107
(1')
107 (T)
1.16
Wind
108 (C)
107 (C)
1.15
It is noted the ("rounded off") axial loads are identical for this member, Frame Type No. 2 includesa small bendingmoment. Considerthe effects of this bendingmoment. Design tensile load (dead
+ imposed)
= =
Design bendingmoment (dead + imposed) = Design compression load (dead + wind) Design bendingmoment
(1.4
x 64) + (1.6 x
=
(1.4 x 0.70) 2.84 kNm
=
(1.0
= = =
107)
261 kN(T)
+ (1.6 x
x 64) - (1.4 x
-86 kN(C) (1.0 x 0.70) + (1.4
1.16)
107)
x 1.16)
2.32 kNm.
75
MEMBER 13
FRAME TYPE Load Type
Number 2
Number I
Max. axial load (kN)
Max. axial load (kN)
Max. BM (kNm)
99 (C)
99 (C)
0.35
Imposed
164 (C)
164 (C)
0.58
Wind
159 (T)
159 (T)
1.55
Dead
Design compressive load
=
(1.4 X 99) + (1.6 X 164)
Design bendingmoment
=
(1.4
x 0.35) + (1.6 X 0.58)
=
401
=
1.42 kNm
kN
Comparingthe forces and moments of the two members a check is made on the compressive capacityof Member 13 and then on the tensile capacityof Member 6. Check Member 13, 120 x80 x5 RHS, for axial compression and moments, considering local capacity and overall buckling. (Clause 4.8.3'). Local CapacityCheck The clause indicates the use, for plastic sections,of the relationship: M Mrx
where Mrx is the reduced moment capacity in the presenceof axial loads obtained from published tables. These values are not generally available for RHS hence use is made of:
F ____ AgPy
+
M _..L
M
whereF = 401 kN MX
= 1.42 kNm
is obtained from Clause 4.2.5 i.e. M, = S but However 5,7 = 1.21 i.e. > 1.2.
Z = 61.7 cm3.
1.2 pZ where S = 75.4 cm3
From footnoteto Clause 4.2.5 it is noted that the value 1.2 is replaced by Factored Load/Unfactored Load ratio = say 1.4. i.e. or i.e.
i0
Use
= 275 x 75.4 x = 1.4 x 275 x 61.7 x = 20.7 kNm (See Reference
i0
= =
20.7 kNm 23.8 kNm
19).
Hence in the interactionequation 401 X 10 18.9 x 275
+
20.7
= 0.77 + 0.07 = 0.84 <
1
Local capacity is satisfied(notice the effect of the moment is relatively small).
76
Overall bucklingcheck (Clause4.8.3.3W). Simplified approach
F
+mMx1 Mb
AgPc
In considering the bucklingresistancemoment Mb reference is made to Appendix The footnoteindicatesthat if X = Le/ry < co (for DIB takenas 1) then lateral torsional bucklingneed not be checked.
PbPy Hence,
andMb=SXpY=20.7kNm
p = 476 kN (as in Section 10.1).
In the interaction equation assume m = 1.0. Clearly for secondary moment design 'm' will be less than unity and a value from Table 18' could be used with advantage. 401
= 0.842+0.069 = 0.911 <
+
1
Satisfactory. Again the effect of the moment is small. Check Member 6 for the axial tensile load and moments (Clause 4.8.2).
F + _____ AePY
x 10 18.9 x 275 263
M
M +
20.7
where F = 263 kN,
MX = 2.84 kNm.
= 0.51 + 0.14
= 0.64 <
1
This is satisfactory. Similar checks can be madeon other members of the framework.
It is apparent that the basic assumption to use pinned joints in the analysisand rigidjoints in the built structure for the latticetruss is justified. The use of pinned joints for the analysis and design process leads to a simpler and therefore quicker solution to the problem. It is of note that relative to the slenderness limitation of 50 stipulated in Clause 10(1) that members55 and 72 are significantly below that value. It is suggested that the reader checks these membersfor the secondary stress effects.
77
JOINT DESIGN
12.
A usefulguidefor the arrangements of connections is the British Steel/CIDECT Publication Construction with HollowSteel Sections22. A check is made of joint 8 (Figure 44) following the method given in design example No 3
(p 52) in TD 338'.
Member 7
Member 6
kN
172 kN
80mm 15
x 3 RHS
80 x 40
Member 23
104
Figure
Member 24
A
44 Joint 8 (K gap) Member
Dead
Imposed
6
65kNcI')
107 kN (T)
262 kN (T)
7
22 kN (T)
36 kN (T)
88 kN (1')
23
39 kN (C)
65 kN (C)
159 kN (C)
24
3lkN(T)
51 kN (T)
125 kN (1')
Total (Factored)
This is a K-Gap Joint with RHS Chordsand bracings,andb0 > h0 hence refer to Table 11 of Reference 13.
12.1 Application limit check list
-i —
H1 b1
h2
=
— —
60
b2
Bracing angles O
b0
to 78
80
— —
=
)
i.e. all satisfactory
=0.5
)
0.5
=0.67
)
2.0
0.67
= 02 = 52.3° 120
5
=
24.0
90°, 35
30°.
to b1
b2
=-
= 1635
=
=
0.67
=
0.50.35
5
120
= 120
b0
=
= 0.33 < 0.35
120
b0
0.35
Hencethis criteria is not satisfiedusing a gap joint. Note: The 90x50x3.6RHS suggestedas an alternative in Section 10.4 would satisfy the criteria for a gapjoint. For the 80x40x5 RHS try an overlap K joint, Table 13(13) (Figure 45) where:
e
0.55 h0
0.25 h0. Member 7
Member 6
58kN
172 kN
0mm x 40 x 3 RHS
80 x 40 x 5 RH mber 23
82 kN
104 kN
4
Figure 45 Joint 8 (overlap) As above
h _, b0
0.5
h1 h2 —, — b1
Bracing angles 900 b1 b0
=
80
120
2.0
b2 02
30°. = 0.67
0.25
79
b2 = 60
=
h=
= 0.5 40
0.25
= 0.33
0.25
Chord
h0 = 80
=
1640
b0
=
2440
—
120
RHS bracing in compression (p = 275) (E = 205 kN/mm2) b1
= 80
=
30
16
30 = 1.1 /E/p1
where
RHS bracing in tension
b2
=
60
1635
=
All limits are satisfied. Use of tabulated allowable loads Pyo
= Pyi = Py2 (i.e. chords and bracingshave same design strength).
b1 BracingsO.8< —
=
80
133> 125
Henceuse followingequation: Allow 30 mm actual overlap. Actual overlap sin 0 Then percentageoverlap = Chord h
=
30 sin 52.3° x ________
100
< 59 >
80%
40
Hence overlap 50
80
= 59%
<
Therefore the allowable joint capacity is given by:
N
=
1 + sin(0 + 0,) — + + ti (2h1 4t1 be be(ov)) pyj 2 sin (0 + 02)
where
b
=
inn
.
.t° .
p
b01t0
p,1 .
10.0
Pyj.tj
b
and
—
jj
h/t
OV
,
Pyi
I
I
Note - suffix 'ov' refers to overlapped member Therefore
be
be(ov)
10
=
x 80 x 275 x 5 x 120x275x5
5
10x275x5x80x5
— —
80
x 275 x
=
333
— —
50
5
N1
= 275x5 [2x40)-4 x 5)+33.3+50)] x [(1+sin 104.6)1(2 sinlO4.6)J x
N1
=200kN.
Since N1
i0
= 200 kN > bracing compression load (104 kN) the joint is satisfactory.
The reader should carry out checks on otherjoints, as considered necessary. It is feasiblethat bottomboom joints might not be satisfactory. These calculations have indicated the effect of the member size and orientation.
12.2 Joint welds Designof welds to latticegirder is carried Out using limit state design and Reference 15, SHS Welding.
Note: use is made of an Amendment to Reference 15 - Revised Text for Pages 26 and 27.
The minimum fillet weld throat size is taken as the larger value of: (i)
a =
Applied factored load
pw
xs
or (ii)
a = ffl) X f(w) x t
81
Where:
pw
=
weld design strength (Table on page steel.
s
=
intersection length
t
=
thickness
24') and equals 215 N/mm2 for Grade 43
(Table 2B'5).
of branch member andf(7) is a functionequal to the higher value of:-
Actual applied factored load Member tension capacity
or Actual applied factored load Joint capacity
The functionf(w) is the ratio of material strength to weld strength. For Grade 43 steel: f(w)
= 275/215 =
1.28.
Design of weld for member 23. This is a single 'T' joint, at 52.3°. When the fusion faces at the toe of the branch member is, at 127.7°, greater than 1200 it is necessary to prepare and makethe bracing toe connection as a butt weld, as shown in Figure 46.
/•. /'\
= 30°to6O°
Figure 46 Preparation
ofbracing member
Bracing memberis a 80x40x5 RHS carrying a factored load of 159 kN (Section 10.3.2).
The factored tensioncapacityof the brace = AePy
10.9
x 275 x 10 = 300 kN.
The chord is a 120x80x5 RHS with an unfactored joint capacity (Section 12.1) of 200 kN. Assumea load factor of 1.5 then design joint capacity = 200 x 1.5 = 300 kN. Total intersection length for 52.3° angle, s = 283mm. (Table 2B)5, b1
82
= 80 mm.
Hence:
(1)
a=
159
215
x 10 = 2.6 mm x 282
or (ii)
a = f(7) X f(w) x t f(7) = 159/300 = 0.53
or
f(l) =
159/300
= 0.53
Hence usef(1) = 0.53 Thus a = 0.53 x 1.28 x 5
= 3.4 mm.
Therefore the minimum throat size = 3.4
> 2.6 mm.
Hencethe minimum leg length = 3.4/0.7 = 4.9 mm. Use a 5 mm fillet weld.
83
13.
FINAL FRAME LAYOUT
Layout of the building steelwork is shown in Figures 47 to 50. Finally a check must be made of the self weight of the girder, as estimated against actual. The designer may need to decide whethera new analysisis required.
4-.
E E a,
I
U 4-. •0.
0 In
U)
x
Eo
gco
U,
.Q X
)<
I-'--
Em
ox
a.O o c'1
0(0
I (I)
(0
1
x
0
'C 'C
I
c'l
IC)
C.,'
In
C
U)
CD
0C'IN
w— 4-.
e0; CO
0
'C
OC
x
.C
0
V C.-
o DC o b o
'4-
w
C .0 C
U)
a, C •0
'C
V
0)
1
0
0
'C
04-
U)
U)
0)
U) 'C
I0) •0
x
'C—
q) (I)
U)
>
a, C U,
U.
wgL'wgL W
8
W
9.
0 V
ci'
N 0, U-
84
rF'
Om
7
3O'x 165 x
/ -7
.—
UB
gable column
0
114.3x 5 CHS
CO
/
/
J-I
457 x
0
191
x 89 UB
Wind girder in plane of top chord
Column
CD
/
.4
Lattice roof truss
0 (0
142x54x49x 1.8
/
Z side rail
0 E co
CD
/
/
0 CO
/
0 CO
\
0 CD
18 lines of 202 x 65 x 60 x 1.8 Z purlins at 1.85 m c/c
/
0 CD
/
/
Wind girder in plane of top chord
Figure 48 P/an ofroof stee/work
85
I
/
.N.
T \
gable column
305 x 165 x 40
N
I
I I
___ — Main column
457 x 191 x 89 UB
7,
142 x 54x 49 side rail
Figure 49 End elevation; general arrangementof steelwork
457 x 191 x 89 UB column
bracing
139.7 x 5
N
114.3 x 5 CHS Wind brace
N
I,7,
Anti-sag bars
gable brace
rail
202 x 60 x 60 x 1.6
nti-sag bars
203 x 102 x 23 UB gable rafter
xl. 8
REFERENCES 1.
BRITISH STANDARDS INSTITUTION BS 5950: Structuraluse of steelwork in building Part 1: Code of practice for design in simple and continuous construction: hot rolled sections BSI, 1990
2.
BRITISH STANDARDS INSTITUTION BS 5950: The structural use of steelwork in building Part 5: Code of practice for design of cold formed sections
BSI, 1987 3.
BRITISH STANDARDS INSTITUTION BS 5502: Buildingand structures for agriculture Various parts
4.
BRITISH STANDARDS INSTITUTION BS 6399: Loadingfor buildings Part 1: Code of practice for dead and imposed loads BSI, 1984
5.
BRITISH STANDARDS INSTITUTION BS 6399: Loadingfor buildings Part 3: Code of practice for imposed roofloads
BSI, 1988 6.
BRITISH STANDARDS INSTITUTION CP3: Chapter V: Loading Part 2: Wind loads (This willbe supersededby BS 6399: Part 2) BSI, 1972
7.
BUILDING RESEARCH ESTABLISHMENT The designer's guideto wind loading ofbuildingstructures Part 1; Butterworths, 1985 Part 2; Butterworths, 1990
8.
THE BRITISH CONSTRUCTIONAL STEELWORK ASSOCIATION / SC! National steelwork specification for buildingconstruction (2nd Edition) BCSA/SCI, 1991
9.
BRITISH STANDARDS INSTITUTION Handbook 22: Quality assurance BSI, 1990
10.
BRITISH STANDARDS INSTITUTION BS 5750: Quallty systems
Parts 0 to 3 BSI, 1987 11.
BRITISH STEEL STRIP PRODUCTS Roofing and cladding in steel - Productselector BS Strip Products, 1987
87
12.
BRITISH STANDARDS INSTITUTION BS 4360: Specificationfor weldable structural steels BSI, 1990
13.
BRITISH STEEL TECHNICAL MANUAL TD 338/5E/91R Design of SHS welded joints BS Welded Tubes, 1991
14.
BRITISH STEEL TECHNICALMANUAL TD 325/1OE/89 Jointing BS Welded Tubes, 1989
15.
BRITISH STEEL TECHNICALMANUAL TD 328/10/90 Welding BS Welded Tubes, 1990
16.
THE STEEL CONSTRUCTION INSTITUTE A check list for designers SC!, 1987
17.
YEOMANS, N. New developments in the use of structural hollow sections SCllBritish Steel Conference, December 1989
18.
METAL SECTIONS LIMITED Zed purlins technical manual to BS 5950: Part5 Metsec, 1988
19.
THE STEEL CONSTRUCTION INSTITUTE Steelwork design guide to BS 5950: Part 1: 1990 Volume 1: Section propertiesand member capacities SCI, 1992
20.
THE STEEL CONSTRUCTION INSTITUTE Steetwork design guide to BS 5950 Volume4: Essential design data for designers SC!, 1991
21.
THE STEEL CONSTRUCTION INSTITUTE Steelwork design guide to BS 5950: Part 1: 1990 (Revised Edition) Volume 2: Worked examples SCI, 1991
22.
BRITISH STEEL GENERAL STEELS WELDED TUBES In conjunction with CIDECT Construction with hollow steel sections First edition 1984
88
BIBLIOGRAPHY In additionto the publications listed in the text, referencehas been made to catalogues producedby: METAL SECTIONS LIMITED Buildingproducts Birmingham Road, Oldbury Warley West Midlands B69 4HE SPACE DECKS LIMiTED Chard Somerset TA2O 2AA Reference has also been made to: BRITISH STEELTECHNICALMANUAL
TD 16716E192Hot finished structural hollow sections; sizes, propertiesand technical data BS Welded Tubes, 1992
89
RECENT SCI PUBLICATIONS: P062
Steel Designers' Manual - 5th Edition Sc, & Blackwell Scientific PublicationsLtd
P079
Steel Construction Yearbook- 1993 Thomas Telford Directories SC!
P083
Design of Composite Trusses Skidmore 0wings Merrill SC!
P101
Curtain Wall Connections to Steel Frames - Interfaces A G Ogden
P114
Steelwork Design Guide to Eurocode 3: Part 1.1 Introducing Eurocode 3 J C Taylor, N R Baddoo, A W Morrow & C Gibbons
P118
Design of Stub Girders
P119
Design of Stainless Steel Fixings and Ancillary Components N R Baddoo
P120
Technical Report - Slim Floor Construction using Deep Decking 0 L Mu/lett & A M Lawson
P123
Concise Guide to the Structural Design of Stainless Steel
/
/
R M Lawson & R McConnel
B A Burgan P124
Technical Report - The Fire Resistance of Web-Infilled Steel Columns G M Newman
P205
Joints in Simple Construction Volume 1: Design Methods (2nd Edition) SC! & BCSA
P206
Joints in Simple Construction Volume 2: Practical Applications BCSA & SC!
For further information please contact: The Publications Department, The Steel Construction Institute, Silwood Park, Ascot, Berkshire, SL5 7QN. Telephone: 0344 23345 Fax: 0344 22944
cto
Typeset and page make-up by The Steel Construction Institute, Ascot, Berks. Printed and boundby Burgess SciencePress, Basmgstoke, Hampshire. 1500/3-1993
o&.
