BS5950 Vs EC3

PSZ 19:16 (Pind.1/97) UNIVERSITI TEKNOLOGI MALAYSIA BORANG PENGESAHAN STATUS TESIS JUDUL: υ COMPARISON BETWEEN BS 5950: PART 1: 2000 & EUROCODE 3 FOR THE DESIGN OF MULTI-STOREY BRACED STEEL FRAME SESI PENGAJIAN: Saya 2006 / 2007 CHAN CHEE HAN (HURUF BESAR) mengaku membenarkan tesis (PSM/ Sarjana/ Doktor Falsafah)* ini disimpan di Perpustakaan Universiti Teknologi Malaysia dengan syarat-syarat kegunaan seperti berikut: 1. 2. 3. 4. Tesis adalah hakmilik Universiti Teknologi Malaysia. Perpustakaan Universiti Teknologi Malaysia dibenarkan membuat salinan untuk tujuan pengajian sahaja. Perpustakaan dibenarkan membuat salinan tesis ini sebagai bahan pertukaran antara institusi pengajian tinggi. **Sila tandakan ( ) SULIT (Mengandungi maklumat yang berdarjah keselamatan atau kepentingan Malaysia seperti yang termaktub di dalam (AKTA RAHSIA RASMI 1972) (Mengandungi maklumat TERHAD yang telah ditentukan oleh organisasi/ badan di mana penyelidikan dijalankan) TERHAD TIDAK TERHAD Disahkan oleh (TANDATANGAN PENULIS) (TANDATANGAN PENYELIA) Alamat Tetap: PETI SURAT 61162, 91021 TAWAU, SABAH. Tarikh CATATAN: PM DR. IR. MAHMOOD MD. TAHIR Nama Penyelia Tarikh: : 01 NOVEMBER 2006 * ** Potong yang tidak berkenaan. : 01 NOVEMBER 2006 υ Jika tesis ini SULIT atau TERHAD, sila lampirkan surat daripada pihak berkuasa/ organisasi berkenaan dengan menyatakan sekali sebab dan tempoh tesis ini perlu dikelaskan sebagai SULIT atau TERHAD. Tesis dimaksudkan sebagai tesis bagi Ijazah Doktor Falsafah dan Sarjana secara penyelidikan, atau disertasi bagi pengajian secara kerja kursus dan penyelidikan, atau Laporan Projek Sarjana Muda (PSM). “I hereby declare that I have read this project report and in my opinion this project report is sufficient in terms of scope and quality for the award of the degree of Master of Engineering (Civil – Structure).” Signature : Name of Supervisor : P.M. Dr. Ir. Mahmood Md. Tahir Date : 01 NOVEMBER 2006 i COMPARISON BETWEEN BS 5950: PART 1: 2000 & EUROCODE 3 FOR THE DESIGN OF MULTI-STOREY BRACED STEEL FRAME CHAN CHEE HAN A project report submitted as partial fulfillment of the requirements for the award of the degree of Master of Engineering (Civil – Structure) Faculty of Civil Engineering Universiti Teknologi Malaysia NOVEMBER, 2006 The report has not been accepted for any degree and is not concurrently submitted in candidature of any other degree. Signature Name Date : : Chan Chee Han : 01 NOVEMBER 2006 .ii I declare that this project report entitled “Comparison Between BS 5950: Part 1: 2000 & Eurocode 3 for The Design of Multi-Storey Braced Steel Frame” is the result of my own research except as cited in the references. iii To my beloved parents and siblings . PM. Mr. I would like to express my appreciation to my thesis supervisor. Tan for their helpful guidance in the process of completing this study. for his generous advice. patience and guidance during the duration of my study. Ir. Shek and Mr. Universiti Teknologi Malaysia. Tahir of the Faculty of Civil Engineering. Without the contribution of all those mentioned above. this work would not have been possible.iv ACKNOWLEDGEMENT First of all. . I would also like to express my thankful appreciation to Dr. Mahmood Md. Finally. I am most thankful to my parents and family for their support and encouragement given to me unconditionally in completing this task. Mahmood’s research students. Dr. Meanwhile. serviceability limit states check governs the design of Eurocode 3 as permanent loads have to be considered in deflection check. Therefore. . However. However.96% more steel weight than the ones designed with BS 5950: Part 1: 2000. These details include the basis and concept of design.60% to 17. Design worksheets are created for the design of structural beam and column.06% and moment capacity by up to 6.v ABSTRACT Reference to standard code is essential in the structural design of steel structures. This paper presents comparisons of findings on a series of two-bay.27% and 9. The design method by Eurocode 3 has reduced beam shear capacity by up to 4. structural column designed by Eurocode 3 has compression capacity of between 5. Eurocode 3 produced braced steel frames which consume 1.95%. with the application of partial strength connections. design methods.34% less than BS 5950: Part 1:2000 design. This study intends to testify the claim. four-storey braced steel frames with spans of 6m and 9m and with steel grade S275 (Fe 460) and S355 (Fe 510) by designed using BS 5950: Part 1: 2000 and Eurocode 3. the percentage of difference had been reduced to the range of 0. Eurocode 3 also reduced the deflection value due to unfactored imposed load of up to 3.11% to 10. safety factors.43%. loading values and etc. specifications to be followed.63% in comparison with BS 5950: Part 1: 2000. The Steel Construction Institute (SCI) claimed that a steel structural design by using Eurocode 3 is 6 – 8% more cost-saving than using BS 5950: Part 1: 2000. The contents of the standard code generally cover comprehensive details of a design. 4 tingkat yang terdiri daripada rentang rasuk 6m dan 9m serta gred keluli S275 (Fe 430) dan S355 (Fe 510).27% – 9. rujukan kepada kod piawai adalah penting. Kandungan dalam kod piawai secara amnya mengandungi butiran rekabentuk yang komprehensif.11% – 10. Eurocode 3 menghasilkan kerangka keluli dirembat yang menggunakan berat besi 1. .95%. Institut Pembinaan Keluli (SCI) berpendapat bahawa rekabentuk struktur keluli menggunakan Eurocode 3 adalah 6 – 8% lebih menjimatkan daripada menggunakan BS 5950: Part 1: 2000. Kertas kerja komputer ditulis untuk merekabentuk rasuk dan tiang keluli. Kertas ini menunjukkan perbandingan keputusan kajian ke atas satu siri kerangka besi terembat 2 bay.60% – 17. penggunaan sambungan kekuatan separa telah berjaya mengurangkan lingkungan berat besi kepada 0. tiang keluli yang direkebentuk oleh Eurocode 3 mempunyai keupayaan mampatan 5. factor keselamatan. didapati bahawa keadaan had kebolehkhidmatan mengawal rekabentuk Eurocode 3 disebabkan beban mati tanpa faktor yang perlu diambilkira dalam pemeriksaan pesongan.63% berbanding BS 5950: Part 1: 2000. nilai beban. Kajian ini bertujuan menguji pendapat ini.06% dan keupayaan momen rasuk sebanyak 6.43%. Namun begitu. dan sebagainya. spesifikasi yang perlu diikuti. Eurocode 3 juga mengurangkan nilai pesongan yang disebabkan oleh beban kenaan tanpa faktor sehingga 3. Namun begitu. Rekebentuk menggunakan Eurocode 3 telah mengurangkan keupayaan ricih rasuk sehingga 4. Justeru. cara rekabentuk. Butiran-butiran ini mengandungi asas dan konsep rekabentuk.vi ABSTRAK Dalam rekabentuk struktur keluli. Selain itu.34% kurang daripada rekabentuk menggunakan BS 5950: Part 1: 2000.96% lebih banyak daripada kerangka yang direkabentuk oleh BS 5950: Part 1: 2000. 1 1.vii TABLE OF CONTENTS CHAPTER TITLE PAGE i ii iii iv v vi vii xii xiii xiv xv THESIS TITLE DECLARATION DEDICATION ACKNOWLEDGEMENT ABSTRACT ABSTRAK TABLE OF CONTENTS LIST OF TABLES LIST OF FIGURES LIST OF APPENDICES LISTOF NOTATIONS I INTRODUCTION 1.3 1.5 Introduction Background of Project Objectives Scope of Project Report Layout 1 3 4 4 5 .4 1.2 1. 1.2 Stiffened Web 2.3.3.Rd 6 6 6 7 7 8 8 8 9 9 9 10 10 10 11 11 11 12 13 13 14 15 15 15 16 16 17 17 18 18 19 20 .1 Cross-sectional Classification 2.viii II LITERATURE REVIEW 2.2 Ultimate Limit State 2.4.1.4 Design of Steel Beam According to EC3 2.4 Loading 2.3.5 Bearing Capacity of Web 2.3.1.1 2.2 Serviceability 2. Mc 2.3.1 Application Rules of EC3 2.1.3.3 Cross-sectional Classification Shear Capacity.2.3.2 BS 5950 2.3.3. Pv Moment Capacity.2.1 Web not Susceptible to Shear Buckling 2.Rd Moment Capacity.2 2.3.1 Eurocode 3 (EC3) 2.2.3.3 Shear Capacity.3.3. Mc.4 Moment Capacity of Web against Shear Buckling 2.4.5.2.3.6 Deflection 2.2 2.4.2 High Shear Moment Capacity 2.1 Ultimate Limit States 2.1 2.1 2.3. Vpl.3.5.4.2.1.2 2.2 Web Susceptible to Shear Buckling 2.1 (EC3) Design Concept of EC3 2.1 Low Shear Moment Capacity 2.1 Unstiffened Web 2.3 Background of BS 5950 Scope of BS 5950 Design Concept of BS 5950 2.2 2.4 Actions of EC3 2.3.1.4.1.3 Design of Steel Beam According to BS 5950 2.3 Serviceability Limit State 2.2.3 Background of Eurocode 3 (EC3) Scope of Eurocode 3: Part 1.3.3. Ry.3 Buckling Resistance.6.1 Column Subject to Compression Force 2.1.6.Rd 2.4.7.3 Compression Resistance.5 Design of Steel Column According to BS 5950 2.2.1. Nb.1 Low Shear Moment Capacity 2.6.7 Conclusion 2.4 Buckling Resistance.Rd 2.Rd 2.2.1 Crushing Resistance.5.1 Buckling Length.4.3 Compression Resistance.4 Resistance of Web to Transverse Forces 2. Rb.1.ix 2.5. Ra.4.Rd 2.1 Introduction 34 .5.5.2. Nc.4.1.1 Cross-section Capacity 2.2 Slenderness.2 Structural Beam Structural Column 31 32 29 30 25 26 26 26 27 27 27 28 29 20 20 21 21 22 22 23 23 23 24 24 24 25 III METHODOLOGY 3. λ 2.1 2.6.5. λ 2.5.6.2 Member Buckling Resistance 2.1 Cross-section Capacity 2.2 Column Subject to Combined Moment and Compression Force 2.5. LE 2.2 Column Subject to Combined Moment and Compression Force 2.6 Design of Steel Column According to EC3 2.4.1.4.1 Column Subject to Compression Force 2.4.1.2 Slenderness.3.7.2 High Shear Moment Capacity 2.4.6.3.2 Member Buckling Resistance 2.4.6.5 Deflection 2.2.Rd 2. l 2. Pc 2.2 Crippling Resistance.4.1 Effective Length.6.1. 5 3.4 Structural Analysis with Microsoft Excel Worksheets Beam and Column Design with Microsoft Excel Worksheets Structural Layout & Specifications 3.10 Structural Column Design 3.1.1 Structural Beam 81 81 .1 4.2 Shear Calculation 3.8.7 3.2 4.9.10.3 Structural Beam Structural Column 66 66 70 73 75 Deflection Economy of Design V CONCLUSIONS 5.x 3.2 3.9 Structural Beam Design 3.4.4.1 Structural Capacity 4.6 3.8.2 4.1 3.8 Structural Layout Specifications 38 38 39 40 41 42 42 42 43 44 46 47 51 57 57 61 35 36 Loadings Factor of Safety Categories Structural Analysis of Braced Frame 3.2 3.1 Load Combination 3.1 Structural Capacity 5.3 Moment Calculation 3.2 BS 5950 EC 3 3.8.3 3.1.10.1 BS 5950 3.2 EC 3 IV RESULTS & DISCUSSIONS 4.1 3.1.9. 2 5.4 Structural Column 82 82 83 84 Deflection Values Economy Recommendation for Future Studies REFERENCES 85 APPENDIX A1 86 93 100 106 114 120 126 APPENDIX A2 APPENDIX B1 APPENDIX B2 APPENDIX C1 APPENDIX C2 APPENDIX D .3 5.1.xi 5.2 5. xii LIST OF TABLES TABLE NO.10 4.2 4.4 4.11 4.4 4.2 3.6 4.3 4.7 4. TITLE PAGE 2.1 2.1 3.9 4.12 Criteria to be considered in structural beam design Criteria to be considered in structural column design Resulting shear values of structural beams (kN) Accumulating axial load on structural columns (kN) Resulting moment values of structural beams (kNm) Shear capacity of structural beam Moment capacity of structural beam Compression resistance and percentage difference Moment resistance and percentage difference Deflection of floor beams due to imposed load Weight of steel frame designed by BS 5950 Weight of steel frame designed by EC3 Total steel weight for the multi-storey braced frame design Percentage difference of steel weight (ton) between BS 5950 design and EC3 design Weight of steel frame designed by EC3 (Semi-continuous) Total steel weight of the multi-storey braced frame design (Revised) Percentage difference of steel weight (ton) between BS 5950 design and EC3 design (Revised) 31 32 43 44 45 67 68 71 71 73 75 76 76 77 78 79 79 Resulting moment due to eccentricity of structural columns (kNm) 46 .1 4.8 4.5 4.2 3.3 3. 3 4.2 3. TITLE PAGE 3.1(c) Schematic diagram of research methodology Floor plan view of the steel frame building Elevation view of the intermediate steel frame Bending moment of beam for rigid construction Bending moment of beam for semi-rigid construction Bending moment of beam for simple construction 37 38 39 80 80 80 .1 3.1(b) 4.xiii LIST OF FIGURES FIGURE NO.1(a) 4. xiv LIST OF APPENDICES APPENDIX TITLE PAGE A1 A2 B1 B2 C1 C2 D Frame Analysis Based on BS 5950 Frame Analysis Based on EC3 Structural Beam Design Based on BS 5950 Structural Beam Design Based on EC3 Structural Column Design Based on BS 5950 Structural Column Design Based on EC3 Structural Beam Design Based on EC3 (Revised) 86 93 100 106 114 120 126 . y.Rd Mpl.y.xv LIST OF NOTATIONS BS 5950: PART 1: 2000 EUROCODE 3 Axial load Shear force Bending moment Partial safety factor Radius of gyration .Major axis .Rd Rb.Rd Mb.y.Rd Vpl.Rd Ry.y.Rd Mc.Rd h A Aeff Av .Minor axis Depth between fillets Compressive strength Flexural strength Design strength Slenderness Web crippling resistance Web buckling resistance Web crushing resistance Buckling moment resistance Moment resistance at major axis Shear resistance Depth Section area Effective section area Shear area F Fv M γ NSd VSd MSd γM0 γM1 rx ry d pc pb py λ Pcrip Pw Mbx Mcx Pv D Ag Aeff Av iy iz d fc fb fy λ Ra. z c/tf d/tw b l tf tw Sx Sy Wpl.Major axis .y Wel.Major axis .z .Minor axis Elastic modulus .Minor axis Flange Web Width of section Effective length Flange thickness Web thickness Zx Zy b/T d/t B LE T t Wel.xvi Plastic modulus .y Wpl. Several factors govern the type of code to be adopted. and past experiences of experts at respective fields. many countries have published their own standard codes. as well as the trading volume and diplomatic ties between these countries. design methods. In the structural design of steel structures. These details include the basis and concept of design. Meanwhile. These codes were a product of constant research and development. countries or nations that do not publish their own standard codes will adopt a set of readily available code as the national reference.1 Introduction Structural design is a process of selecting the material type and conducting indepth calculation of a structure to fulfill its construction requirements. economic and functional building. loading values and etc. climate and national preferences. It is a process of converting an architectural perspective into a practical and reasonable entity at construction site. The main purpose of structural design is to produce a safe. A standard code serves as a reference document with important guidance. The contents of the standard code generally cover comprehensive details of a design.CHAPTER I INTRODUCTION 1. . Structural design should also be an integration of art and science. In present days. safety factors. reference to standard code is essential. specifications to be followed. namely suitability of application of the code set in a country with respect to its culture. The establishment of Eurocode 3 will provide a common understanding regarding the structural steel design between owners. ECCS. Eurocodes will be used in public procurement specifications and to assess products for ‘CE’ (Conformité Européen) mark. The earliest documents seeking to harmonize design rules between European countries were the various recommendations published by the European Convention for Constructional Steelwork. From these. Therefore.2 Like most of the other structural Eurocodes. contractors and manufacturers of construction products among the European member countries. ENV1993 (ENV stands for EuroNorm Vornorm) issued by Comité Européen de Normalisation (CEN) – the European standardisation committee. the move to withdraw BS 5950 and replace with Eurocode 3 will be taking place in the country as soon as all the preparation has completed. designers. These preliminary standards of ENV will be revised. This was followed by the various parts of a pre-standard code. . Eurocode 3 has developed in stages. The study on Eurocode 3 in this project will focus on the subject of moment and shear design. amended in the light of any comments arising out of its use before being reissued as the EuroNorm standards (EN). Therefore. It is believed that Eurocode 3 is more comprehensive and better developed compared to national codes. published by the European Commission. As with other Europeans standards. operators and users. Buckling resistance and shear resistance are two major elements of structural steel design. Standardization of design code for structural steel in Malaysia is primarily based on the practice in Britain. provision for these topics is covered in certain sections of the codes. Codes of practice provide detailed guidance and recommendations on design of structural elements. were developed. the initial draft Eurocode 3. . this project is intended to testify the claim. 2005). Lacking analytical and calculative proof. simple design is possible if a scope of application is defined to avoid the circumstances and the forms of construction in which strength is over-estimated by simple procedures. earlier design practice under-estimated strength in various circumstances affecting economy. even though there seems to be no benefit to the designer for the majority of his regular workload. causing safety issues. for those who pursue economy of material. simple design is possible if the code requirements are presented in an easy-to-use format. The increasing complexity of codes arises due to several reasons. Finally. There are new formulae and new complications to master. but it can be simplified for those pursuing speed and clarity. this can be achieved if the designer is not too greedy in the pursuit of the least steel weight from the strength calculations. namely earlier design over-estimated strength in a few particular circumstances. and new forms of structure evolve and codes are expanded to include them. Design can be complex. in its publication of “eurocodesnews” magazine has claimed that a steel structural design by using Eurocode 3 is 6 – 8% more cost-saving than using BS 5950.2 Background of Project The arrival of Eurocode 3 calls for reconsideration of the approach to design. However.3 1. such as the tables of buckling stresses in existing BS codes. The Steel Construction Institute (SCI). Many designers feel depressed when new codes are introduced (Charles. Besides. hereafter referred to as BS 5950. . 2) To study on the effect of changing the steel grade from S275 to S355 in Eurocode 3. A study on the basis and design concept of EC3 will be carried out. 1.4 Scope of Project The project focuses mainly on the moment and shear design on structural steel members of a series four-storey. 2 bay braced frames. Comparison to other steel structural design code is made. The comparison will be made between the EC3 with BS 5950: Part 1: 2000. The multi-storey steel frame will be first analyzed by using Microsoft Excel worksheets to obtain the shear and moment values. 3) To compare the economy aspect between the designs of both BS 5950: Part 1: 2000 and Eurocode 3. All the beam-column connections are to be assumed simple. design spreadsheets will be created to calculate and design the structural members. The standard code used here will be Eurocode 3. Next.3 Objectives The objectives of this project are: 1) To compare the difference in the concept of the design using BS 5950: Part 1: 2000 and Eurocode 3.4 1. hereafter referred to as EC3. This structure is intended to serve as an office building. . Chapter III will be a summary of research methodology.5 1. Results and discussions are presented in Chapter IV. conclusions and recommendations are presented in Chapter V. Chapter II presents the literature review that discusses the design procedures and recommendations for steel frame design of the codes EC3 and BS 5950.5 Report Layout The report will be divided into five main chapters. Meanwhile. Chapter I presents an introduction to the study. 2.CHAPTER II LITERATURE REVIEW 2. It also covers other construction aspects only if they are necessary for design.1 (EC3) EC3. “Design of Steel Structures: Part 1.2 Scope of Eurocode 3: Part 1. The use of local application rules are allowed only if they have similar principles as EC3 . Eurocode 1 covers loading situations. or better known as Eurocode. Eurocode is separated by the use of different construction materials. Application rules must be written in italic style. Eurocode 3 covers steel construction.1 Eurocode 3 (EC3) 2.1 Background of Eurocode 3 (EC3) European Code. Principles should be typed in Roman wordings. Principles and application rules are also clearly stated. serviceability and resistance of a structure.1. Eurocode covers concrete construction. It was intended to smooth the trading activities among the European countries. EC3 stresses the need for durability. It also covers specific rules for building structures. was initiated by the Commission of European Communities as a standard structural design guide. while Eurocode 4 covers for composite construction.1.1 General rules and rules for buildings” covers the general rules for designing all types of structural steel. serviceability and resistance of structure (Taylor. . impact or consequences of human errors. Partial safety factor is applied to loadings and design for durability.7 and their resistance. it will remain fit for the use for which it is required.3 Design Concept of EC3 All designs are based on limit state design. selecting a structural form and design that can survive adequately the accidental removal of an individual element. to an extent disproportionate to the original cause.3. durability and serviceability design does not differ too much. and tying the structure together. It also covers other construction aspects only if they are necessary for design. 2001).1.1. Every European country using EC3 has different loading and material standard to accommodate safety limit that is set by respective countries. It should also be designed in such a way that it will not be damaged by events like explosions. 2.1 Application Rules of EC3 A structure should be designed and constructed in such a way that: with acceptable probability. it will sustain all actions and other influences likely to occur during execution and use and have adequate durability in relation to maintenance costs. 2. selecting a structural form which has low sensitivity to the hazards considered. Potential damage should be limited or avoided by appropriate choice of one or more of the following criteria: Avoiding. having due regard to its intended life and its cost. EC3 stresses the need for durability. and with appropriate degrees of reliability. eliminating or reducing the hazards which the structure is to sustain. Safety factor values are recommended in EC3. which are ultimate limit state and serviceability limit state. EC3 covers two limit states. temperature effects or settlement. 2. It may require certain consideration. Actions are classified by variation in time and by their spatial variation. e. . for example.g. and vibration. which result in different arrangements of actions. rupture.1. damage to the building or its contents. self-weight of structures. This failure may be caused by excessive deformation.8 2. and accidental loads (A).g. In time variation classification. 2. snow loads.3. which causes discomfort to people. imposed loads. e. wind loads or snow loads. e. or loss of stability of the structure or any part of it. including: deformations or deflections which adversely affect the appearance or effective use of the structure (including the proper functioning of machines or services) or cause damage to finishes or non-structural elements. considered as a rigid body. and loss of equilibrium of the structure or any part of it.3 Serviceability Limit State Serviceability limit states correspond to states beyond which specified service criteria are no longer met. Partial or whole of structure will suffer from failure. in spatial variation classification. movable imposed loads. variable actions (Q). actions are defined as fixed actions.2 Ultimate Limit State Ultimate limit states are those associated with collapse. e. or an imposed deformation in indirect action.g. or which limits its functional effectiveness.g.4 Actions of EC3 An action (F) is a force (load) applied to the structure in direct action. including supports and foundations. self-weight. or with other forms of structural failure which may endanger the safety of people.g. e. actions can be grouped into permanent actions (G).1.1.3. explosions or impact from vehicles. Meanwhile. fittings. ancillaries and fixed equipment. and free actions. wind loads. but offsetting potential reductions in economy was also one of the reasons. 2. lateral-torsional buckling.9 2. Part 8 comprises of code of practice for fire resistance design. Part 6 covers design for light gauge profiled steel sheeting. avoidance of disproportionate collapse. which was withdrawn.2 Scope of BS 5950 Part 1 of BS 5950 provides recommendations for the design of structural steelwork using hot rolled steel sections. fabrication and erected for rolled. hot finished structural hollow sections and cold formed structural hollow sections.2. . They are being used in buildings and allied structures not specifically covered by other standards. plates. Part 1 covers the code of practice for design of rolled and welded sections.2.1 Background of BS 5950 BS 5950 was prepared to supersede BS 5950: Part 1: 1990. sheeting respectively. BS 5950 comprises of nine parts. welded sections and cold formed sections. Part 2 and 7 deal with specification for materials. local buckling. flats. Several clauses were technically updated for topics such as sway stability. and Part 9 covers the code of practice for stressed skin design. members subject to combined axial force and bending moment. Part 3 and Part 4 focus mainly on composite design and construction.2 BS 5950 2. Part 5 concerns design of cold formed thin gauge sections. etc. Changes were due to structural safety. shear resistance. The fundamental of the methods are different joints for different methods. stability against overturning and sway sensitivity. . and experimental verification.3 Design Concept of BS 5950 There are several methods of design.1 Ultimate Limit States Several elements are considered in ultimate limit states. in the design for limiting states.3. rupture.2 Serviceability Limit States There are several elements to be considered in serviceability limit states – Deflection. namely simple design. Generally.2.2. only 80% of the full specified values need to be considered when checking for serviceability. Generally. and brittle fracture. 2. Meanwhile. 2.3. the specified loads should be multiplied by the relevant partial factors γf given in Table 2. BS 5950 covers two types of states – ultimate limit states and serviceability limit states. wind induced oscillation. semi-continuous design.2. inclusive of general yielding. The load carrying capacity of each member should be such that the factored loads will not cause failure. In the case of combined imposed load and wind load. In the case of combined horizontal crane loads and wind load. They are: strength. and durability. serviceability loads should be taken as the unfactored specified values. in checking. only the greater effect needs to be considered when checking for serviceability. continuous design. fracture due to fatigue. vibration. buckling and mechanism formation.10 2. the settlement of supports should be taken into account as well. 2. There are dead. imposed and wind loading.1 Cross-sectional Classification Cross-sections should be classified to determine whether local buckling influences their capacity. earth and groundwater loading. without calculating their local buckling resistance. .3 Design of Steel Beam According to BS 5950 The design of simply supported steel beam covers all the elements stated below. The elements of a cross-section are generally of constant thickness. The classification of each element of a cross-section subject to compression (due to a bending moment or an axial force) should be based on its width-to-thickness ratio. Where necessary. All relevant loads should be separately considered and combined realistically as to compromise the most critical effects on the elements and the structure as a whole.11 2.3. overhead traveling cranes.4 Loading BS 5950 had identified and classified several loads that act on the structure. Loading conditions during erection should be given particular attention.2. Sectional size chosen should satisfy the criteria as stated below: (i) (ii) (iii) (iv) (v) (vi) Cross-sectional classification Shear capacity Moment capacity (Low shear or High shear) Moment Capacity of Web against Shear Buckling Bearing capacity of web Deflection 2. the plastic moment capacity cannot be reached. Sections that do not meet the limits for class 3 semi-compact sections should be classified as class 4 slender. Class 4 is known as slender section. However. a crosssection may be classified with its compression flange and its web in different classes. Cross-sections at this category should be given explicit allowance for the effects of local buckling. Class 3 is known as semi-compact section. When this section is applied.2 Shear Capacity. It enables plastic moment to take place. Alternatively. local buckling will bar any rotation at constant moment. the complete cross-section should be classified according to the highest (least favourable) class of its compression elements. Clause 4. Class 2 is known as compact section. given by: Pv = 0. Class 1 is known as plastic section.6pyAv . Shear capacity is normally checked at section part that sustains the maximum shear force. However. Pv The web of a section will sustain the shear in a structure. It is cross-section with plastic hinge rotation capacity.2. Fv. 2. Class 1 section is used for plastic design as the plastic hinge rotation capacity enables moment redistribution within the structure.3 of BS 5950 states the shear force Fv should not be greater than the shear capacity Pv.12 Generally.3. the stress at the extreme compression fiber can reach design strength. 2.3 Moment Capacity.5. 2. and Mc = pyZeff for class 4 slender cross-sections where S is the plastic modulus.1 Low Shear Moment Capacity This situation occurs when the maximum shear force Fv does not exceed 60% of the shear capacity Pv.3. Seff is the effective plastic modulus.3. Mc At sectional parts that suffer from maximum moment.2 of BS 5950 states that: Mc = pyS for class 1 plastic or class 2 compact cross-sections. . and Zeff is the effective section modulus. moment capacity of the section needs to be verified. 2. py is the design strength of steel and it depends on the thickness of the web. Mc = pyZ or alternatively Mc = pySeff for class 3 semi-compact sections.3. Clause 4.13 in which Av is the shear area. There are two situations to be verified in the checking of moment capacity – low shear moment capacity and high shear moment capacity. Z is the section modulus. BS 5950 provides various formulas for different type of sections. in which Sf is the plastic modulus of the effective section excluding the shear area Av.2 High Shear Moment Capacity This situation occurs when the maximum shear force Fv exceeds 60% of the shear capacity Pv.3.2pyZ for class 1 plastic or class 2 compact cross-sections. Mc = py(Z – ρSv/1.5) for class 4 slender cross-sections in which Sv is obtained from the following: - For sections with unequal flanges: Sv = S – Sf. - Otherwise: Sv is the plastic modulus of the shear area Av.3.5) or alternatively Mc = py(Seff – ρSv) for class 3 semi-compact sections. and ρ is given by ρ = [2(Fv/Pv) – 1]2 .5.2. Clause 4. and Mc = py(Zeff – ρSv/1.14 2.3 of BS 5950 states that: Mc = py(S – ρSv) < 1. it should be assumed not to be susceptible to shear buckling and the moment capacity of the cross-section should be determined using 2. if the web depth-to-thickness ratio d/t > 70ε for a rolled section.2 states that.15 2.4. or 62ε for a welded section.3.3.2 Web Susceptible to Shear Buckling Clause 4.4.4. obtained from Table 21 BS 5950 t = web thickness b) High shear – “flanges only” method If the applied shear Fv > 0.1 Web not Susceptible to Shear Buckling Clause 4.6Vw.3.3. but the web is designed for shear only. 2.4. if the web depth-to-thickness d/t ≤ 62ε. Vw = dtqw where d = depth of the web. where Vw is the simple shear buckling resistance.3.4 Moment Capacity of Web against Shear Buckling 2. qw = shear buckling strength of the web.1 of BS 5950 states that. a conservative value Mf for .6Vw. The moment capacity of the cross-section should be determined taking account of the interaction of shear and moment using the following methods: a) Low shear Provided that the applied shear Fv ≤ 0. it should be assumed to be susceptible to shear buckling. provided that the flanges are not class 4 slender.4.4. 3. .1 Unstiffened Web Clause 4.6be/k but n ≤ 5 and k is obtained as follows: .5.3 for the applied shear combined with any additional moment beyond the “flanges-only” moment capacity Mf given by b).or H-section: .for a welded I.or H-section: k=T+r k=T .1 states that bearing stiffeners should be provided where the local compressive force Fx applied through a flange by loads or reactions exceeds the bearing capacity Pbw of the unstiffened web at the web-to-flange connection.5 Bearing Capacity of Web 2. the web should be designed using Annex H. 2.at the end of a member: n = 2 + 0.3.6Vw.except at the end of a member: n = 5 .for a rolled I.2. It is given by: Pbw = (b1 + nk)tpyw in which.5. c) High shear – General method If the applied shear Fv > 0. provided that the applied moment does not exceed the “low-shear” moment capacity given in a). where pyf is the design strength of the compression flange. with each flange subject to a uniform stress not exceeding pyf.16 the moment capacity may be obtained by assuming that the moment is resisted by the flanges alone. 17 where b1 is the stiff bearing length. If the web and the stiffener have different design strengths.netpy in which As. . T is the flange thickness. the smaller value should be used to calculate both the web capacity Pbw and the stiffener capacity Ps.net is the net cross-sectional area of the stiffener. pyw is the design strength of the web. The capacity Ps of the stiffener should be obtained from: Ps = As. allowing for cope holes for welding.3.5. and t is the web thickness.6 Deflection Deflection checking should be conducted to ensure that the actual deflection of the structure does not exceed the limit as allowed in the standard.2 Stiffened Web Bearing stiffeners should be designed for the applied force Fx minus the bearing capacity Pbw of the unstiffened web. Suggested limits for calculated deflections are given in Table 8 of BS 5950. r is the root radius. 2. Actual deflection is a deflection caused by unfactored live load. be is the distance to the nearer end of the member from the end of the stiff bearing. 2.3. 3 of EC3 provided limits on the outstand-to-thickness (c/tf) for flange and depth-tothickness (d/tw) in Table 5. the beam will buckle during pre-mature stage.1 Cross-sectional Classification A beam section should firstly be classified to determine whether the chosen section will possibly suffer from initial local buckling.18 2. This limit allows the formation of a plastic hinge with the rotation capacity required for plastic analysis. . Clause 5. Sectional size chosen should satisfy the criteria as stated below: (i) (ii) (iii) (iv) Cross-sectional classification Shear capacity Moment capacity (Low shear or High shear) Bearing capacity of web a) b) c) Crushing resistance Crippling resistance Buckling resistance (v) Deflection 2. It can also achieve rectangular stress block. To avoid this. It has limited rotation capacity. It is applicable for plastic design.4.4 Design of Steel Beam According to EC3 The design of simply supported steel beam covers all the elements stated below.3. Class 1 is known as plastic section. This section can develop plastic moment resistance. However. Class 2 is also known as compact section.1. plastic hinge is disallowed because local buckling will occur first. When the flange of the beam is relatively too thin. Beam sections are classified into 4 classes. Vpl.1.Rd The web of a section will sustain shear from the structure. At each crosssection. the ratios of c/tf and d/tw will be the highest among all four classes.Rd where Vpl. fy is the steel yield strength and γMO is partial safety factor as stated in Clause 5. and ε = [235/fy]0.4. Pre-mature buckling will occur before yield strength is achieved. The member will fail before it reaches design stress.1. Calculated stress in the extreme compression fibre of the steel member can reach its yield strength. It is necessary to make explicit allowances for the effects of local buckling when determining their moment resistance or compression resistance.5 . the ratio of d/tw > 69ε or d/tw > 30ε √kγ for a stiffened web. but local buckling is liable to prevent development of the plastic moment resistance.Rd = Av (fy / √3) / γMO Av is the shear area. 2. the inequality should be satisfied: Vsd ≤ Vpl. Vsd.19 Class 3 is also known as semi-compact section. Class 4 is known as slender section. The stress block will be of triangle shape. Shear capacity will normally be checked at section that takes the maximum shear force.2 Shear Capacity. Shear buckling resistance should be verified when for an unstiffened web. kγ is the buckling factor for shear. Apart from that. 3 Moment Capacity.Rd = Weff fy / γM1 where Wpl and Wel the plastic modulus and elastic modulus respectively.1 Low Shear Moment Capacity When maximum shear force. when maximum shear force.Rd.7 states that. Vsd exceeds 50% of the design resistance Vpl.4. 2. the design moment resistance of a cross-section should be reduced to MV.5.20 2. low shear moment capacity and high shear moment capacity. the reduced design plastic resistance moment allowing for the shear .3.4.4. Vsd is equal or less than the design resistance Vpl.Rd = Wpl fy / γMO Class 3 cross-sections: Mc.Rd = Wel fy / γMO Class 4 cross-sections: Mc.Rd. 2. Mc.3.3. Weff is the elastic modulus at effective shear area.Rd Moment capacity should be verified at sections sustaining maximum moment.Rd. There are two situations to verify when checking moment capacity – that is.4. For class 4 cross-sections. γMO and γM1 are partial safety factors. as stated in Clause 5.Rd may be determined as follows: Class 1 or 2 cross-sections: Mc.2 High Shear Moment Capacity Clause 5. the design moment resistance of a cross-section Mc. accompanied by plastic deformation of the flange. This checking is intended to prevent the web from buckling under excessive compressive force. Ry.Rd of the web of an I. is governed by one of the three modes of failure – Crushing of the web close to the flange. For cross-sections with equal flanges. checking should be done at section subject to maximum shear force.Rd = (Wpl – ρAv2/4tw) fy / γMO but MV.3 provides that the design crushing resistance.21 force. accompanied by plastic deformation of the flange.7.Ed / fyf)2]0. and buckling of the web over most of the depth of the member.Rd ≤ Mc. Clause 5.4 Resistance of Web to Transverse Forces The resistance of an unstiffened web to transverse forces applied through a flange.4. However.4.5 . bending about the major axis. Ry.Rd Situation becomes critical when a point load is applied to the web. Thus. if shear force acts directly at web without acting through flange in the first place. crippling of the web in the form of localized buckling and crushing of the web close to the flange. H or U section should be obtained from: Ry. it is obtained as follows: MV.Rd = (ss + sγ) tw fγw / γM1 in which sγ is given by sγ = 2tf (bf / tw)0.1 Crushing Resistance.Rd where ρ = (2Vsd / Vpl. this checking is unnecessary.Rd – 1)2 2. 2.5 (fyf / fyw)0.4.5 [1 – (σf. For member subject to bending moments.5 2.5tw2(Efyw)0. the following criteria should be satisfied: Fsd ≤ Ra. and ss / d < 0.Rd ≤ 1.Rd The design crippling resistance Ra.5 + 3(tw / tf)(ss / d)] / γM1 where ss is the length of stiff bearing.3 Buckling Resistance.2.5. H or U section should be obtained by considering the web as a virtual compression member with an effective beff.2 Crippling Resistance.5 [(tf / tw)0. Rb. 2.Rd = (χ βA fy A) / γM1 .4. Ra.Rd + Msd / Mc.4.Rd of the web of an I.Sd = 0.Ed is the longitudinal stress in the flange.4.Rd Msd ≤ Mc. obtained from beff = [h2 + ss2]0. H or U section is given by: Ra. fyf and fyw are yield strength of steel at flange and web respectively.Rd The design buckling resistance Rb.Rd of the web of an I. Rb.Rd and Fsd / Ra.4.22 but bf should not be taken as more than 25tf. σf. 1 of EC3.5. 2. For a structural steel column subject to compression load only. Column is a compressive member and it generally supports compressive point loads.2. Therefore. checking is normally conducted for capacity of steel column to compression only.1 and Table 5.5 Deflection Deflection checking should be conducted to ensure that the actual deflection of the structure does not exceed the limit as allowed in the standard.5.1 Column Subject to Compression Force Cross-sectional classification of structural steel column is identical as of the classification of structural steel beam. This. however. 2.5. Actual deflection is a deflection caused by unfactored live load. Suggested limits for calculated deflections are given in Table 4.4. 2.5 Design of Steel Column According to BS 5950 The design of structural steel column is relatively easier than the design of structural steel beam. the following criteria should be checked: (i) (ii) (iii) Effective length Slenderness Compression resistance . applies only to non-moment sustaining column.23 where βA = 1 and buckling curve c is used at Table 5. This concept is not applicable for battened struts.5. λ The slenderness λ of a compression member is generally taken as its effective length LE divided by its radius of gyration r about the relevant axis. and back-to-back struts.2 Slenderness. T-section struts.7.1.1. For continuous columns in multi-storey buildings of simple design. column members that carry more than 90% of their reduced plastic moment capacity Mr in the presence of axial force is assumed to be incapable of providing directional restraint. LE The effective length LE of a compression member is determined from the segment length L centre-to-centre of restraints or intersections with restraining members in the relevant plane. the compression resistance Pc of a member is given by: Pc = Ag pc (for class 1 plastic. directional restraint is based on connection stiffness and member stiffness. Pc According to Clause 4. λ = LE / r 2. 2.5.1. in accordance of Table 22.4. depending on the conditions of restraint in the relevant plane. channel.24 2.5. class 2 compact and class 3 semi-compact cross-sections) . angle. Depending on the conditions of restraint in the relevant plate.3 Compression Resistance.1 Effective Length. 25 Pc = Aeff pcs (for class 4 slender cross-section) where Aeff is the effective cross-sectional area. pc the compressive strength obtained from Table 23 and Table 24.5. Ag is the gross cross-sectional area. class 2 compact and class 3 semi-compact cross sections.5. the checking of cross-section capacity is as follows: My Fc M + x + ≤1 Ag p y M cx M cy where Fc is the axial compression. Mx is the moment about major axis. 2.2.1 Cross-section Capacity Generally. and Mcy is the moment capacity about minor axis. Mcx is the moment capacity about major axis. and pcs is the value of pc from Table 23 and Table 24 for a reduced slenderness of λ(Aeff/Ag)0. .2 Column Subject to Combined Moment and Compression Force For a column subject to combined moment and compression force. the crosssection capacity and the member buckling resistance need to be checked. py is the design steel strength. Ag is the gross cross-sectional area.5. for class 1 plastic. My is the moment about minor axis. 2. in which λ is based on the radius of gyration r of the gross cross-section. py the steel design strength. 2.6. the following criteria should be checked: (i) (ii) (iii) (iv) Buckling length Slenderness Compression resistance Buckling resistance .0 Pc M bs p y Z y where F is the axial force in column. 2.2.6 Design of Steel Column According to EC3 The design of steel column according to EC3 is quite similar to the design of steel column according to BS 5950. and Zy the elastic modulus.5. the following stability check needs to be satisfied: My F Mx + + ≤ 1 .26 2.2 Member Buckling Resistance In simple construction. Mx the maximum end moment on x-axis. Mb the buckling resistance moment.1 Column Subject to Compression Force Cross-sectional classification of structural steel column is identical as of the classification of structural steel beam. Pc the compression resistance of column. For a structural steel column subject to compression load only. 5 states that. provided that both ends of a column are effectively held in position laterally. Nc. Clause 5.1 Buckling Length.27 2.5. the value of λ should not exceed 250.1. Alternatively.Rd According to Clause 5. λ=l/i For column resisting loads other than wind loads. determined using the properties of the gross cross-section.Rd of a member is given by: Nc.1. class 2 compact and class 3 semi-compact crosssections) .3 Compression Resistance. l The buckling length l of a compression member is dependant on the restraint condition at both ends. 2.2 Slenderness. the buckling length l may be determined using informative of Annex E provided in EC3. the buckling length l may be conservatively be taken as equal to its system length L. 2. the value of λ should not exceed 180.6.6. λ The slenderness λ of a compression member is generally taken as its buckling length l divided by its radius of gyration i about the relevant axis.1.Rd = A fy / γM0 (for class 1 plastic.4.6.4. the compression resistance Nc.1. whereas for column resisting self-weight and wind loads only. and Aeff / A for Class 4 cross-sections. Clause 5.Rd 2.28 Nc.1.1 states that the design buckling resistance of a compression member should be taken as: Nc. Nb. The design value of the compressive force NSd at each cross-section shall satisfy the following condition: NSd ≤ Nb. 2 or 3 cross-sections.1.Rd = χ βA A fy / γM1 where βA = 1 for Class 1.5.Rd = Aeff fy / γM1 (for class 4 slender cross-section) The design value of the compressive force NSd at each cross-section shall satisfy the following condition: NSd ≤ Nc.4 Buckling Resistance.Rd .6. For hot rolled steel members with the types of cross-section commonly used for compression members. the relevant buckling mode is generally “flexural” buckling.Rd For compression members. χ is the reduction factor for the relevant buckling mode. Sd + + ≤1 Af yd Wel .6. Clause 5. z f yd for Class 4 cross-sections where fyd = fy/γM1. z . Rd for a conservative approximation where.2.Sd ⎤ ⎢ ⎥ +⎢ ⎥ ≤1 ⎢ M Ny .2 Column Subject to Combined Moment and Compression Force For a column subject to combined moment and compression force.Sd N Sd M z .Sd ⎤ ⎡ M z . for I and H sections. α = 2. y f yd Weff . z f yd for Class 3 cross-sections M y . y . Rd ⎦ ⎣ ⎦ α β for Class 1 and 2 cross-sections M y .1 states that. Weff is the effective section modulus of the cross-section when subject . M y .4.8. Rd M pl .Rd.Sd + N Sd e Nz N Sd + + ≤1 Aeff f yd Weff . 2. for bi-axial bending the following approximate criterion may be used: ⎡ M y . cross-section capacity depends on the types of cross-section and applied moment. the crosssection capacity and the member buckling resistance need to be checked. in which n = Nsd / Npl.Sd + + ≤1 N pl . β = 5n but β ≥ 1. y f yd Wel .29 2.Sd + N Sd e Ny M z .1 Cross-section Capacity Generally.Sd N Sd M z . Aeff is the effective area of the cross-section when subject to uniform compression. Rd M pl . Rd ⎥ ⎣ M Nz .6. 30 only to moment about the relevant axis; and eN is the shift of the relevant centroidal axis when the cross-section is subject to uniform compression. However, for high shear (VSd ≥ 0.5 Vpl.Rd), Clause 5.4.9 states that the design resistance of the cross-section to combinations of moment and axial force should be calculated using a reduced yield strength of (1 – ρ)fy for the shear area, where ρ = (2VSd / Vpl.Rd – 1)2. 2.6.2.2 Member Buckling Resistance A column, subject to buckling moment, may buckle about major axis or minor axis or both. All members subject to axial compression NSd and major axis moment My.Sd must satisfy the following condition: k y M y.Sd N Sd + ≤ 1,0 N b. y . Rd ηM c. y . Rd where Nb.y.Rd is the design buckling resistance for major axis; Mc.y.Rd is the design moment resistance for major-axis bending, ky is the conservative value and taken as 1,5; and η = γM0 / γM1 for Class 1, 2 or 3 cross-sections, but 1,0 for Class 4. 2.7 Conclusion This section summarizes the general steps to be taken when designing a structural member in simple construction. 31 2.7.1 Structural Beam Table 2.1 shown compares the criteria to be considered when designing a structural beam. Table 2.1 : Criteria to be considered in structural beam design BS 5950 Flange subject to compression 9ε 10ε 15ε Web subject to bending (Neutral axis at mid depth) 80ε 100ε 120ε ε = (275 / py)0.5 2.0 Shear Capacity Pv = 0.6pyAv Av = Dt Vpl.Rd = fyAv / (√3 x γM0) γM0 = 1,05 Av from section table 3.0 Moment Capacity Mc = pyS Mc = pyZ Mc = pyZeff Class 1, 2 Class 3 Class 4 Mc.Rd = Wplfy / γM0 Mc.Rd = Welfy / γM0 Mc.Rd = Wefffy / γM1 γM0 = 1,05 γM1 = 1,05 4.0 Bearing Capacity Class 1 Plastic Class 2 Compact Class 3 Semi-compact Class 1 Plastic Class 2 Compact Class 3 Semi-compact CRITERIA 1.0 Cross-sectional Classification Flange subject to compression 10ε 11ε 15ε Web subject to bending (Neutral axis at mid depth) 72ε 83ε 124ε ε = (235 / fy)0,5 EC3 32 Pbw = (b1 + nk)tpyw Smaller of Ry.Rd = (ss + sy) tw fyw / γM1 Ra.Rd = 0,5tw2(Efyw)0,5 [(tf/tw)0,5 + 3(tw/tf)(ss/d)]/γM1 Rb.Rd = χβAfyA / γM1 5.0 Shear Buckling Resistance d/t ≤ 70ε Ratio 6.0 Deflection L / 360 Limit (Beam carrying plaster or other brittle finish) N/A Limit (Total deflection) L / 250 L / 350 d/tw ≤ 69ε 2.7.2 Structural Column Table 2.2 shown compares the criteria to be considered when designing a structural beam. Table 2.2 : Criteria to be considered in structural column design BS 5950 Flange subject to compression 9ε 10ε 15ε Web (Combined axial load and bending) 80ε / 1 + r1 100ε / 1 + 1.5r1 Class 1 Plastic Class 2 Compact Class 1 Plastic Class 2 Compact Class 3 Semi-compact CRITERIA 1.0 Cross-sectional Classification Flange subject to compression 10ε 11ε 15ε Web (Combined axial load and bending) 396ε / (13α – 1) 456ε / (13α – 1) EC3 67 + 0.Sd N Sd + ≤ 1. Rd .0 Stability Check My F Mx + + ≤ 1 .eff Class 1.5(1 + γM0σw / fy) σw = NSd / dtw ε = (235 / fy)0.0 Moment Resistance Mb = pbSx Mb = pbZx Mb = pbZx.33ψ) ψ = 2γM0σa / fy – 1 σa = NSd / A α = 0. 2 Class 3 Class 4 Mc.Rd = Welfy / γM0 Mc.5 2. 3 Class 4 Nc. Rd ηM c. y .0 Compression Resistance Pc = Agpc Pc = Aeffpcs Class 1.Rd = Wefffy / γM1 γM0 = 1.05 Nc.Rd = Wplfy / γM0 Mc.0 Pc M bs p y Z y k y M y.05 γM1 = 1.Rd = Aefffy / γM1 3.5 Class 3 Semi-compact 42ε / (0.05 4. y . -1 < r1 ≤ 1 r2 = Fc / Agpyw ε = (275 / py) 0.Rd = Afy / γM0 γM0 = 1.0 N b. 2.33 120ε / 1 + 2r2 r1 = Fc / dtpyw. Please refer to Figure 3. analyzing the tables provided and the purpose of each clause stated in the code. design and comparison works will follow subsequently. Eventually. Analysis. an understanding on the cross-section classification for BS 5950 is also carried out. .CHAPTER III METHODOLOGY 3. moment capacity. Checking on several elements. comparison of the results will lead to recognizing the difference in design approach for each code. Beams and columns are designed for the maximum moment and shear force obtained from computer software analysis. Next. it is necessary to study and understand the concept of design methods in EC3 and compare the results with the results of BS 5950 design. analysis on the difference between the results using two codes is done. such as shear capacity. buckling capacity and deflection is carried out. bearing capacity.1 Introduction As EC3 will eventually replace BS 5950 as the new code of practice.1 for the flowchart of the methodology of this study. The first step is to study and understand the cross-section classification for steel members as given in EC. At the same time. . Please refer to Appendices A1 and A2 for the analysis worksheets created for the purpose of calculating shear force and bending moment values based on the requirements of different safety factors of both codes. End moments are zero. only beam shear forces will be transferred to the structural column. Different factors of safety with reference to BS 5950 and EC3 are defined respectively. Calculation of bending moment. Therefore. Simple construction allows the connection of beam-to-column to be pinned jointed. As the scope of this study is limited at simple construction. that is M = wL2 / 8 V = wL / 2 where w is the uniform distributed load and L the beam span.8 discuss in detail all the specifications and necessary data for the analysis of the multi-storey braced frame.35 3. M and shear force.2 Structural Analysis with Microsoft Excel Worksheets The structural analysis of the building frame will be carried out by using Microsoft Excel worksheets. Sections 3. the use of advanced structural analysis software is not needed.4 to 3. V are based on simply-supported condition. al. The Microsoft Excel software is used for its features that allow continual and repeated calculations using values calculated in every cell of the worksheet. (1995). The method of design using BS 5950 will be based on the work example drawn by Heywood (2003). Meanwhile.3 Beam and Column Design with Microsoft Excel Worksheets The design of beam and column is calculated with Microsoft Excel software. . Microsoft Excel worksheets will show the calculation steps in a clear and fair manner. Furthermore. the method of design using EC3 will be based on the work example drawn by Narayanan et.36 3. Several trial and error calculations can be used to cut down on the calculation time needed as well as prevent calculation error. Please refer to Appendices B1 to C2 for the calculation worksheets created for the purpose of the design of structural beam and column of both design codes. 1: Schematic diagram of research methodology .37 Determine Research Objective and Scope Phase 1 Literature Review Determination of building and frame dimension Specify loadings & other specifications Phase 2 Frame analysis using Microsoft Excel (V=wL/2. M=wL2/8) Design worksheet development using Microsoft Excel Beams and columns design Fail Checking (Shear. Combined) Pass Comparison between BS 5950 and EC3 Phase 3 END Figure 3. Moment. 1 Structural Layout In order to make comparisons of the design of braced steel frame between BS 5950-1: 2000 and Eurocode 3.4. The storey height will be 5m from ground floor to first floor.2 : Floor plan view of the steel frame building. 2nd to 3rd. In plan view.3 for the illustrations of building plan view and elevation view respectively. Each of the frame’s longitudinal length is 6m. whereas for other floors (1st to 2nd. Please refer to Figure 3. 4th storey is roof while the rest will serve as normal floors. 6m 6m 6/9m 6/9m Figure 3. . the 4-storey frame consists of four (4) bays.38 3. a parametric study for the design of multi-storey braced frames is carried out. Two (2) lengths of bay width will be used in the analysis – 6m and 9m respectively. the storey height will be 4m.2 and Figure 3. The number of storey of the frame is set at four (4).4 Structural Layout & Specifications 3. in total. The intermediate frame will be used as the one to be analysed and designed. there will be three (3) numbers of 4-storey frames. 3rd to roof). Top flange of beams are effectively restraint against lateral torsional buckling. all the column-to-column connections are to be rigid.3 : Elevation view of the intermediate steel frame. . 3.2 Specifications The designed steel frame structure is meant for office for general use. flat roof system will be introduced to cater for some activities on roof top.39 4m 4m 4m 5m Figure 3. All the bays will be serving the same function.4. All the roof bays will be used for general purposes. Meanwhile. As this is a simple construction. The main steel frame will consist of solely universal beam (UB) and universal column (UC). Web cleats will be used as the connection method to create pinned connection. Meanwhile. all the beam-to-column connections are assumed to be pinned. Therefore. In this design. precast solid floor panel of 100mm thick was selected for flat roof.2.0kN/m2 respectively. The type of precast flooring system to be used will be solid precast floor panel.4kN/m2 and 3. the intensity of slab selfweight will be 2.40 Precast concrete flooring system will be introduced to this project. this value will be adopted. Only gravitational loads will be considered in this project. all the values of imposed loads of both BS 5950 and EC3 design will be based on BS 6399. Consequently. Table 8 (Offices occupancy class) states that the intensity of distributed load of offices for general use will be 2. Multiplying by 6m (3m apiece from either side of the bay) will result in 9kN/m and 15kN/m of load intensity on roof beam and floor beam respectively. all floors will be of one-way slab. Therefore. Therefore. Weight of concrete is given by 24kN/m3. repair and other general purposes.3 of Concise Eurocode 3 (C-EC3) states that the characteristic values of imposed floor load and imposed roof load must be obtained from Part 1 and Part 3 of BS 6399 respectively. Multiplying the thickness of the slabs. For precast floor selfweight.5 Loadings Section 2. For imposed roof load. each bay will contribute half of the load intensity to the intermediate frame. 3. Meanwhile. 125mm think floor panel will be used for other floors. . wind load (horizontal load) will not be considered in the design. a uniform load intensity of 1. section 6. This value will be used as this frame model is meant for a general office usage. for a flat roof with access available for cleaning.5kN/m2 is appropriate. Meanwhile.5kN/m2. The steel frame is assumed to be laterally braced.2 (Flat roofs) states that. 4 for dead load. depending on the interior designer’s intention. variable actions Q include live loads such as imposed load.0kN/m2 for finishes (superimposed dead load) on all floors will be assumed. for normal design situations. finishes and fittings. the principal combination of loads that should be taken into account will be load combination 1 – Dead load and imposed gravity loads. γM0. γM1.4. is given by 1.41 The finishes on the flat roof will be waterproofing membrane and decorative screed. Partial safety factor for resistance of Class 4 cross-section. Meanwhile. 3.1.4kN/m2 and 4kN/m2 respectively. partial safety factors. γF for dead load. γQ is given by 1.5. the total dead load intensity for roof and floor slabs are 3.05. permanent actions G include dead loads such as self-weight of structure. The factor γM0 is used where the failure mode is plasticity or yielding. is given by 1. Meanwhile.6 for imposed load. Multiplying by 6m (3m apiece from either side of the bay) will result in kN/m and 24kN/m of load intensity on roof beam and floor beam respectively. 2 or 3 cross-section. in the design of buildings not subject to loads from cranes.1. Partial safety factor for resistance of Class 1. Partial safety factors for loads. For other floors. γf should be taken as 1.6 Factor of Safety Section 2. A general load intensity of 1. and 1.05 as well. From Table 2. Combining the superimposed dead load with selfweight.35. a selection of floor carpets and ceramic tiles will be used. γG is given by 1. In EC3.2 “Buildings without cranes” of BS 5950 states that. The . for imposed floor load. 1. in the meantime. namely S 275 (or Fe 430 as identified in EC3) and S 355 (or Fe 510 as identified in EC3). For steel grade S 355.8.42 factor γM1 is used where the failure mode is buckling – including local buckling. For steel grade S 275. Meanwhile. which governs the resistance of a Class 4 (slender) cross-section.1 Load Combination This section describes the structural analysis of the steel frame. fy is 355N/mm2 and 335N/mm2 respectively for the same thickness limits. 3. the load combination will be 1. 3. According to BS 5950.2 “Material properties for hot rolled steel” (C-EC3) limits thickness of flange to less than or equal to 40mm for nominal yield strength fy of 275N/mm2 and larger but less than or equal to 100mm for fy of 255N/mm2.6 times total imposed .4 times total dead load plus 1.8 Structural Analysis of Braced Frame 3.7 Categories In this project. py is 355N/mm2 and 345N/mm2 respectively for the same limits of thickness. in order to justify the effect of design strength of a steel member on the strength of a steel member. py is 275N/mm2 for thickness less than or equal to 16mm and 265N/mm2 for thickness larger but less than or equal to 40mm. for Fe 510. design strength py is decided by the thickness of the thickest element of the cross-section (for rolled sections). two (2) types of steel grade will be used. In BS 5950. 3. 1. the w will be 59.43 load (1.5% between the analyses of both codes.92 Bay Width 9m 216 281. will be 48kN/m. For simple construction. there is a difference of approximately 4.88 6m 137. the resultant load combination.92 From Table 4. the resultant load combination. the w will be 62. will be 45.4DL + 1.9kN/m.6LL). This is done by summating the resultant shear . For all other floors.2 will present the accumulating axial loads acting on the structural columns of the steel frame. BS 5950 results in higher value of shear. Clearly.55 268.1 below: Table 3.7 179. Table 3. According to EC3. For the roof.2 Shear Calculation This steel frame is pinned jointed at all beam-to-column supports.76kN/m. Inputting the resultant load combinations into the formula.28 EC 3 Bay Width 9m 206. the resulting shear values of both bay widths and codes of design can be summarized in Table 3. w.5LL). The next table. For the roof. w.5 times total imposed load (1.35DL + 1. the load combination will be 1.8. V at end connections is given by V = wl/2.64kN/m. For all other floors. where w is the resultant load combination and l is the bay width.1 Resulting shear values of structural beams (kN) BS 5950 Location 6m Roof Other Floors 144 187. the shear.35 times total dead load plus 1. 3. This is solely due to the difference in partial safety factors. Inputting the resultant load combinations into the formula.52 1351. M.78 2026.76 1559. 275. 137. Roof – 3rd 3rd – 2nd 2nd – 1st 1st .68 1415. 216 497.76 1061. Internal columns will sustain axial load two times higher than external columns of same floor level as they are connected to two beams.8.96 992.4 633.7 316.3: .08 Int.39 1013. 432 995. structural beam moment.64 6m Ext.2 Accumulating axial load on structural columns (kN) BS 5950 Floor Int.1 950.54 Int.5%. where w is the resultant load combination and l is the bay width.47 744.28 Int.76 9m Ext.31 Int. 3.52 2123. the resulting moment values of both bay widths and codes of design can be summarized in Table 3. similar with the beam shear.84 707.26 675.94 1488. = External column The accumulating axial loads based on the two codes vary approximately 4.88 779. 144 331. = Internal column Ext.44 force from beam of each floor. 206. can be calculated by using the formula M=wl2/8.62 Ext.55 475.52 EC 3 9m 6m Ext.Ground 288 663.84 1039.3 Moment Calculation For simple construction. 413.98 496.92 519. since all the beam-to-column connections are pinned jointed. Table 3. the depth (D for BS 5950 and h for EC 3) of a structural column is assumed to be 400mm.07 From Table 3. since this is simple construction. there is a difference of approximately 4.3 Resulting moment values of structural beams (kNm) BS 5950 Location 6m Roof Other Floors 216 281.92 EC 3 Bay Width 9m 464. Me. D or h is the depth of column section (m). For the moments of the structural columns. Since this is only preliminary analysis as well. Therefore. However. In this project. can be determined from the following formula: Me = V (e + D/2) = V (e + h/2) where V is resultant shear of structural beam (kN). the eccentricity moment. the higher the load combination of a floor. BS 5950 results in higher value of moment.3.6% between the analyses of both codes.88 Bay Width 9m 486 634. in this case.55 268. there will be a moment due to eccentricity of the resultant shear from the beams. . there will be no end moments being transferred from the structural beams.74 605. the higher the difference percentage will be. the depth of the column has not been decided yet.45 Table 3. This is solely due to the difference in partial safety factors. Regardless of the width of the bay. Subsequently. the eccentricity of the resultant shear from the face of the structural column will be 100mm.23 6m 206. Clearly. e is the eccentricity of resultant shear from the face of column (m).4% to 4. initially. In simple construction. 3.4 84. For EC 3.98 80. V can be expressed as V = (1. . However.78 Int.88 Int.4 below summarizes the moment values due to eccentricity.46 V for external column can be easily obtained from shear calculation. Table 3. V should be obtained by deducting the factored combination of floor dead (DL) and imposed load (LL) with unfactored floor dead load.6 56.5LL) – 1. serviceability check in the form of deflection check will need to be done.6LL) – 1.08 EC 3 9m 6m Ext.84 Ext. for internal column.38 9m Ext.56 6m Ext.5.35DL + 1.66 57.66 53. 32. 20. 30. V can be expressed as V = (1. 30.4 94.9 Structural Beam Design Structural beam design deals with all the relevant checking necessary in the design of a selected structural beam. two major checks that need to be done is shear and moment resistance at ultimate limit state. Roof Other Floors 21. 20. 32. Table 3. Next.68 These values of eccentricity moments will be useful for the estimation of initial size of a column member during structural design in later stage.0DL.6 63.98 86.0DL.4 Resulting moment due to eccentricity of structural columns (kNm) BS 5950 Floor Int.4DL + 1.6 Int. For BS 5950. 21. The moments for floor columns will be evenly distributed as the ratio of EI1/L1 and EI2/L2 is less than 1. Elastic modulus. From the section table. the sections are rearranged in ascending form. This is selected to give a suitable moment capacity. Sx = 1290cm3.88kNm.47 The sub-sections next will show one design example which is the floor beam of length 6m and of steel grade S 275 (Fe 430).88 x 103 / 275 = 1025cm3 From the rearranged table. Zx = 1120cm3.9mm.3. first the mass (kg/m) and then the plastic modulus Sx (cm3). shear capacity.6mm. Depth between fillets. Sx = M / py = 281.1mm. UB section 457x152x60 is chosen. The moment will then be divided by the design strength py to obtain an estimated minimum plastic modulus value necessary in the design.6mm. b/T = 6.3mm.9. Width.92kN and 281. necessary checks for ultimate limit state will be shear buckling. Flange thickness.1 BS 5950 In simple construction. 3. moment capacity and web bearing capacity. ε = √(275/py) = √(275/275) . d = 407. The size will then be checked to ensure suitability in all other aspects. Web thickness. Depth. From the section table for universal beam.99. The shear and moment value for this particular floor beam is 187. T = 13. D = 454.8kg/m. d/t = 50. t = 8. Plastic modulus. the properties of the UB chosen are as follows: Mass = 59. B = 152. which is smaller than 9ε = 9. where neutral axis is at mid-depth. Since both flange and web are plastic.1 x 454. shear buckling needs not be checked. This is the limit for Class 1 plastic section. Mc = 275 x 1290 x 10-3 . For web of I-section. web is Class 1 plastic section.75. section 4. flange is Class 1 plastic section.0. Mc” is checked.0 Sectional classification is based on Table 11 of BS 5950. Pv = 0. Next.6 = 3682. Mc = pySx. actual d/t = 50. shear capacity is adequate. therefore. After clause 4.54kN > Fv Therefore. Since actually d/t < 70.4.6 x 607.4. section 4.92kN Therefore. Meanwhile.26mm2 Pv = 0.6 x 275 x 3682. Shear capacity.26 x 10-3 = 607. Next.2.5 states that if the d/t ratio exceeds 70ε for a rolled section.0 in this design.0. Actual d/t did not exceed 80. Av = 8.3 “Shear capacity” is checked.48 = 1. 0.6pyAv. Therefore. Therefore.57 = 364.5 “Moment capacity. where Av = tD for a rolled I-section.3. this section is Class 1 plastic section.0. clause 4. shear buckling resistance should be checked. For class 1 plastic cross-section.6Pv = 0.5 is checked.57kN > Fv = 187. the limiting value for Class 1 plastic section is 80ε = 80. Actual b/T = 5. it is low shear.2. 49 = 354. n = 2 + 0. bearing capacity of web.3 = 51.02mm Pbw = 98. moment capacity is adequate.75kNm To avoid irreversible deformation under serviceability loads.88kNm from analysis < Mc = 354. section 4.1 x 275 x 10-3 = 218.75kNm Therefore.2pyZx = 1. therefore. M = 281.2pyZx.2mm b1 = t + 1.2 x 275 x 1120 x 10-3 = 369.1 + 1.2 “Bearing capacity of web” is checked.5. If Fv exceeds Pbw. Pbw = (b1 + nk)tpyw r = 10.6r + 2T (Figure 13) = 8.6 x 10. bearing stiffener should be provided.92kN .5mm At support. To prevent crushing of the web due to forces applied through a flange. be = 0. 1.34kN > Fv = 187.6be/k.2 = 23. n = 2 b1 + nk = 98.02mm k=T+r = 13. Mc should be limited to 1.6kNm > Mc.02 x 8.2 + 2 x 13.3 + 10. OK. This is done in the form of deflection check. In this case. L = 6. only unfactored imposed load shall be used to calculate the deflection. Generally. is given by δ = 5wL4 / 384EI = 5 x 15 x 64 x 105 / 384 x 205 x 25500 = 4. δ. . the serviceability load should be taken as the unfactored specified value. The section is adequate.50 Therefore. However.84mm Table 8 (Suggested limits for calculated deflections) suggests that for “beams carrying plaster or other brittle finish). δlim = 6000 / 360 = 16. the vertical deflection limit should be L/360. Therefore. After necessary ultimate limit state checks have been done.5) should be conducted. the deflection is satisfactory.67mm >δ Therefore. it should also satisfy all the required criteria in the ultimate limit state check.0m E = 205kN/mm2 I = 25500cm4 The formula for calculating exact deflection. w = 15kN/m for floors. the serviceability limit state check (Section 2. This calculation is repeated for different sections to determine the suitable section which has the minimal mass per length. the bearing capacity at support is adequate. y = 1051cm3.y = 927cm3. d = 360. necessary checks for ultimate limit state will be shear buckling.92kNm. Shear area. Wel.y = M / py = 268. Area of . Depth between fillets. lateral torsional buckling. Wpl. Depth.6mm.4mm. tw = 7. The moment will then be divided by the design strength py to obtain an estimated minimum plastic modulus value necessary in the design. h = 402. Elastic modulus. Web thickness. Plastic modulus.2 EC 3 In simple construction. the properties of the UB chosen are as follows: Mass = 54kg/m.9cm3 From the rearranged table. resistance of web to crushing.y (cm3).28kN and 268.9cm2. tf = 10.51 This section satisfied all the required criteria in both ultimate and serviceability limit state check.9mm. b = 177.9. the sections are rearranged in ascending form. Therefore. crippling and buckling. UB section 406x178x54 is chosen. The size will then be checked to ensure suitability in all other aspects.6mm. The shear and moment value for this particular floor beam is 179. Width. This is selected to give a suitable moment capacity. From the section table. Wpl. From the section table for universal beam. Flange thickness.92 x 103 / 275 = 977. shear capacity. 3. first the mass (kg/m) and then the plastic modulus Wpl. Av = 32.6mm. moment capacity. it is adequate to be used. Therefore.2.6(a).05) = 497.52 section.5 x 497. limiting c/tf ratio (c is half of b) is 9.6.36cm.Rd = 0. For S275 (Fe 430). c/tf = 8. limiting d/tw ratio is 66.Rd = Av(fy / √3) / γM0.1.48 = 298. fy = 275N/mm2 and ultimate tensile strength. Before checks are done for ultimate limit states.5.28kN γM0 = 1. section classification is a must.2 for Class 1 elements.9 x 100 x 275) / (√3 x 1. tf = 10. neutral axis at mid depth”. iLT = 4.15. d/tw = 47. shear resistance is sufficient.Rd.4 ≤ 66. For “web subject to bending.28kN . VSd from analysis at each cross-section should not exceed the design plastic shear resistance Vpl. Flange is Class 1 element. Next. aLT = 131cm. Based on Table 3.28kN Therefore.48kN > 179. VSd = 179.4. A = 68.9mm. section 5. From Table 5. fu = 430N/mm2. These values must be adopted as characteristic values in calculations. yield strength. that is Vpl. Actual d/tw = 47. Iy = 18670cm4.5Vpl.15 ≤ 9. tf ≤ 40mm. Second moment of area.05 Vpl.49kN > VSd = 179.1 “Shear resistance of cross-section” of beam is checked. Web is Class 1 element. for “outstand element of compression flange. 0. UB section 406x178x54 is Class 1 section.6cm2. flange subject to compression only”.Rd = (32. Actual c/tf = 8.6 for Class 1 elements. The design value of shear force. Actual d/tw = 47.y fy / γM0 for Class 1 or Class 2 cross-section.5[fyf/fyw]0. Ra. crippling resistance. Ry. sy = tf(bf/tw)0.8.6 “Resistance of webs to transverse forces” requires transverse stiffeners to be provided in any case that the design value VSd applied through a flange to a web exceeds the smallest of the following – Crushing resistance.6 “Shear buckling” requires that webs must have transverse stiffeners at the supports if d/tw is greater than 63. the moment capacity is sufficient.5 [1 – (γM0 σf. The beam is fully restrained. not susceptible to lateral torsional buckling. section 5.2 “Moment resistance of cross-section with low shear” the design value of moment MSd must not exceed the design moment resistance of the cross-section Mc.4 < 63. For crushing resistance. Therefore.05 = 275.Rd = 1051 x 275 x 10-3 / 1.Rd = (ss + sy) twfyw / γM1 where at support.Ed/fyf)2]0. low shear.Rd.Rd and buckling resistance.5 “Lateral-torsional buckling” needs not be checked.5.5.53 Therefore. For low shear.Rd. Ry. Section 5.8 and 56.Rd = Wpl. Section 5.5. MSd = 268.92kNm Mc.5 .1 for steel grade Fe 430 and Fe 510 respectively.26kNm > MSd Therefore. section 5. shear buckling check is not required. Rb. Therefore. 6)0.62 + 502]0.05 E = 210kN/mm2 Ra.9)(0.5 = 52. OK.5 [(tf/tw)0. A = 227.69) x 7.5 = 405.8kN For buckling resistance.8 x 7.5 x 7.6 x 275 x 10-3 / 1. OK γM1 = 1. fyf = 275N/mm2.14 ≤ 0.14)] / 1.5 + 3(tw/tf) (ss/d)] / γM1 ss/d = 50 / 360.5[h2 + ss2]0.6 = 1731.Rd = βA fc A / γM1 A = beff x tw beff = 0.69mm Ry.Rd = 0.62 (210000 x 275)0.5 [(10.6/10.6 / 7.54 At support.Ed = 0.5 + 0 + 50/2 = 227.4kN For crippling resistance. Rb.5tw2 (Efyw)0. Ra.Rd = (50 + 52. bending moment is zero.4 = 0. sy = 10.6)0.05 = 307.9/7.5 + 3(7.5 [402.Rd = 0.5 + a + ss/2 = 0.8mm beff should be less than [h2 + ss2]0.28mm2 . ss = 50mm at support. σf.2.9 (177.05.7mm. γM0 = 1.05 = 204. the serviceability limit state check (Section 4. From Figure 4.28 x 10-3 / 1. δmax = δ1 + δ2 – δ0 (hogging δ0 = 0 at unloaded state) w1 = 27.6 λ√βA = 118. This is done in the form of deflection check.05 = 197. the serviceability load should be taken as the unfactored specified value.5 d/t = 2. Therefore.8kN Ry. λ = 2.05 For ends restrained against rotation and relative lateral movement (Table 5.13 (rolled I-section).2) should be conducted.4 / 7.Rd = 1 x 119.28kN.Rd = 307.4kN Minimum of the 3 values are 197. OK.6kN/m for floors. (Permanent load) . buckling about y-y axis.8N/mm2 Rb.29). After necessary ultimate limit state checks have been done. fc = 119.6 = 118. curve (a) is used.8 x 1731. Generally.5kN Ra.5 x 360.Rd = 204. deflection should take into account deflection due to both permanent loads and imposed loads. the web of the section can resist transverse forces. fc = 121N/mm2 λ√βA = 120.1.55 βA = 1 γM1 = 1. fc = 117N/mm2 By interpolation. which is larger than VSd = 179. λ√βA = 118.5kN.6 From Table 5. the vertical deflection limit should be L/350 for δ2 and L/250 for δmax. it should also satisfy all the required criteria in the ultimate limit state check. the deflection is satisfactory. max = 6000 / 250 = 24mm > δ1 + δ2 = 18.46mm Table 4. δlim. 2 = 6000 / 350 = 17. This calculation is repeated for different sections to determine the suitable section which has the minimal mass per length.1 (Recommended limiting values for vertical deflections) suggests that for “floors and roofs supporting plaster or other brittle finish or non-flexible partitions”.0m E = 210kN/mm2 Iy = 18670cm4 The formula for calculating exact deflection. . (Imposed load) L = 6. However.88mm δ2 = 5 x 15 x 64 x 105 / 384 x 210 x 18670 = 6.56 w2 = 15kN/m for floors.34mm Therefore.6 x 64 x 105 / 384 x 210 x 18670 = 11. is given by δ = 5wL4 / 384EI δ1 = 5 x 27. The section is adequate. δ. In this case.14mm > δ2 δlim. 57 This section satisfied all the required criteria in both ultimate and serviceability limit state check. Therefore, it is adequate to be used. 3.10 Structural Column Design Structural column design deals with all the relevant checking necessary in the design of a selected structural beam. In simple construction, apart from section classification, two major checks that need to be done is compression and combined axial and bending at ultimate limit state. The sub-sections next will show one design example which is the internal column “ground floor to 1st floor” (length 5m) of the steel frame with bay width 6m and of steel grade S 275 (Fe 430). 3.10.1 BS 5950 In simple construction, apart from section classification, necessary checks for ultimate limit state will be compression resistance and combined axial force and moment. The axial force and eccentricity moment value for this particular internal column are 1415.52kN and 63.08kNm respectively. From the section table for universal column, the sections are rearranged in ascending form, first the mass (kg/m) and then the plastic modulus Sx (cm3). The moment will then be divided by the design strength py to obtain an estimated minimum plastic modulus value necessary in the design. Sx = M / py 58 = 63.08 x 103 / 275 = 229.4cm3 From the rearranged table, UC section 203x203x60 is chosen. This is selected to give a suitable moment capacity. The size will then be checked to ensure suitability in all other aspects. From the section table, the properties of the UC chosen are as follows: Mass = 60kg/m; Depth, D = 209.6mm; Width, B = 205.2mm; Web thickness, t = 9.3mm; Flange thickness, T = 14.2mm; Depth between fillets, d = 160.8mm; Plastic modulus, Sx = 652cm3; Elastic modulus, Zx = 581.1cm3; Radius of gyration, rx = 8.96cm, ry = 5.19cm; Gross area, Ag = 75.8cm2; b/T = 7.23 (b = 0.5B); d/t = 17.3. T < 16mm, therefore, py = 275N/mm2 ε = √(275/py) = √(275/275) = 1.0 Sectional classification is based on Table 11 of BS 5950. Actual b/T = 7.23, which is smaller than 9ε = 9.0. This is the limit for Class 1 plastic section (Outstand element of compression flange). Therefore, flange is Class 1 plastic section. Meanwhile, actual d/t = 17.3. For web of I-section under axial compression and bending, the limiting value for Class 1 plastic section is 80ε / 1 + r1, where r1 is given by r1 = Fc / dtpy. r1 = 1415.52 x 103 / 160.8 x 9.3 x 275 = 3.44 but -1 < r1 ≤ 1, therefore, r1 = 1 Limiting d/t value = 80 x 1 / 1 + 1 = 40 59 > Actual d/t = 17.3 Therefore, the web is Class 1 plastic section. Since both flange and web are plastic, this section is Class 1 plastic section. Next, based on section 4.7.2 “Slenderness” and section 4.7.3 “Effective lengths”, and from Table 22 (Restrained in direction at one end), the effective length, LE = 0.85L = 0.85 x 5000 = 4250mm. λx = LEx / rx = 4250 / 8.96 x 10 = 47.4 Next, based on section 4.7.4 “Compression resistance”, for class 1 plastic section, compression resistance, Pc = Agpc. pc is the compressive strength determined from Table 24. For buckling about x-x axis, T < 40mm, strut curve (b) is used. λx = 46, pc = 242N/mm2 λx = 48, pc = 239N/mm2 From interpolation, λx = 47.4, pc = 239.9N/mm2 Pc = Agpc = 75.8 x 100 x 239.9 x 10-3 = 1818.44kN > Fc = 1415.52kN Therefore, compressive resistance is adequate. M = 31. pb = 233N/mm2 From interpolation. pb = 260. For EI / L1st-2nd : EI / Lground-1st < 1.78 x 652 x 10-3 = 170.03kNm .17 From Table 16 (Bending strength pb for rolled sections). Mi = 63.7 “Columns in simple structures”.19 x 10 = 48. the beam reaction.08kNm. for columns in simple construction.5.78N/mm2 Mb = pbSx = 260. Therefore.7.5L / ry = 0. in proportion to the bending stiffness of each length. is assumed to be acting 100mm from the face of the column. R. The moment is distributed between the column lengths above and below 1st floor. the moment will be equally divided. From frame analysis. when only nominal moments are applied. My / pyZy = 0 Equivalent slenderness λLT of column is given by λLT = 0. the column should satisfy the relationship (Fc / Pc) + (Mx / Mbs) + (My / pyZy) ≤ 1 My = 0.54kNm. therefore.5 x 5000 / 5.17. λLT = 45. λLT = 48.60 Next. Section 4. pb = 250N/mm2 λLT = 50. it is adequate to be used.5cm3 From the rearranged table. the combined resistance against axial force and eccentricity moment is adequate. the sections are rearranged in ascending form. first the mass (kg/m) and then the plastic modulus Wpl. .96 < 1. necessary checks for ultimate limit state will be cross-section resistance (in the form of moment resistance) and in-plane failure about major axis (which is a combination of axial force and eccentricity moment).44 + 31.y = MSd / fy = 57.03 = 0. Therefore. This section satisfied all the required criteria in ultimate limit state check. The size will then be checked to ensure suitability in all other aspects.08kN and 57.88 x 103 / 275 = 210. The axial force and eccentricity moment value for this particular internal column are 1351. This is selected to give a suitable moment capacity. UC section 254x254x73 is chosen. apart from section classification.0 Therefore. From the section table for universal column.10.y (cm3).61 (Fc / Pc) + (Mx / Mbs) = 1415. Wpl.52 / 1818.2 EC 3 In simple construction.88kNm respectively.54 / 170. The moment will then be divided by the design strength fy to obtain an estimated minimum plastic modulus value necessary in the design. 3. Shear area. Beforehand. tf = 14. Flange thickness.86cm. Next.46cm. A = 92. this section is Class 1 section.5cm. fy = 275N/mm2. iy = 11. d/tw = 23. Width. Av = 25.6 “Axially loaded members with moments” will be checked. σw.2 and 10. Depth. c/tf = 8. the web is Class 1. fu = 430N/mm2 Sectional classification is based on Table 5. Wpl. therefore. For web subject to bending and compression. for outstand element of compression flange (flange subject to compression only). Therefore. the classification depends on the mean web stress. Wel. .2mm.6(a) of C-EC3 for Class 1 elements. the limiting values of c/tf for Class 1 and 2 are 9. iz = 6.73N/mm2 Table 5. the properties of the UC chosen are as follows: Mass = 73kg/m.2.3.6 = 784.8.62 From the section table. Iy = 11370cm4. aLT = 98. section 5.2 x 8.08 x 103 / 200. Plastic modulus. Second moment of area.1cm. Radius of gyration. For symmetric I-section of Class 1 or 2.2 respectively.5.5b).2mm < 40mm. σw = NSd / dtw = 1351. Actual c/tf = 8.6mm.6cm2. tw = 8. Elastic modulus. Depth between fillets. d = 200.y = 895cm3.9cm2. Actual c/tf = 8. tf = 14.94.3. iLT = 6. Area of section. b = 254mm.y = 990cm3. Web thickness. from. From this table. Since both flange and web are plastic. flange is Class 1 element.94 (c = 0.2mm.94 < 9.1. with d/tw = 23. section 5.8 gives the limiting values of stress σw for Class 1 and 2 cross-sections. From Table 5. h = 254mm. Rd n ≥ 0.Rd = Av(fy / √3) / γM0 = (25.63 Vpl.08 / 2433. n = NSd / Npl.11 Mpl.1 : MNy.Sd / L = 57.11 Mpl.88 x 103 / 5000 = 11.05 = 2433. the section is subject to a low shear.Rd is such that n < 0.27.05) = 387.Rd = A fy / γM0 = 92.1 = 0.Rd = 1.y.Rd Reduced design plastic moment.58kN 0.555) .Rd = Mpl.555 ≥ 0. MNy.1 Therefore.11 x 259.1kN n = 1351.Sd = My.Rd > Vmax.Rd = 1.3 x (1 – 0.y fy / γM0 = 990 x 10-3 x 275 / 1.y.9 x 102 x 275 x 10-3 / 1.1 : MNy.1kN Maximum applied shear load (at top of column) is Vmax. MN.Sd Therefore.6 x 102 x 275) x 10-3 / (√3 x 1.y.05 = 259.Rd = 1.Rd (1 – n) Npl.5Vpl.Rd = Wpl.y. From Table 5.3kNm MNy.Rd (1 – n) Mpl. allowing for axial force. 13 “Selection of buckling curve for fc”.0 Ly = 0. fc = 249.Rd) + (kyMy.85 x 5000 = 4250mm Slenderness ratio λy = Ly / iy = 4250 / 11. βA = 1 λy√βA = 38.1 x 10 = 38.Sd must satisfy the expression (NSd / Nb. fc = 250N/mm2 λy√βA = 40.3.y.85L = 0.3 Based on Table 5.Rd) ≤ 1.7N/mm2 . buckling curve (b) is used.3 tf ≤ 40mm λy√βA = 38.94kNm Therefore. for buckling about y-y axis. fc = 248N/mm2 From interpolation. Lastly.2 “Axial compression and major axis bending” states that all members subject to axial compression NSd and major axis moment My.y. the moment resistance is sufficient. λy√βA = 38.6. section 5.Sd / ηMc.64 = 128.1kNm > MSd = 28.3. y. it is adequate to be used.95 < 1. This section 254x254x73 UC satisfied all the required criteria in ultimate limit state check.Rd) = (1351.7 x 92.94 / 1 x 128. γM1 = 1.3) + (1. the resistance against in-plane failure against major axis is sufficient. Therefore.5 x 28.05 = 2209.65 Nb.9 x 102 x 10-3 / 1.y.Sd / ηMc.08 / 2209.y.05 = 1 x 249.3kN ky = interaction factor about yy axis = 1. .1) = 0.Rd) + (kyMy.Rd = βA fc A / γM1.0 Therefore.5 (Conservative value) η = γM0 / γM1 =1 (NSd / Nb. structural capacity is sub-divided into beam and column. namely structural capacity. and weight of steel. 4. based on steel grade S275 and S355.1 Structural Capacity Structural capacity deals with shear and moment resistance of a particular section chosen. 4. The results are arranged accordingly. deflection.2 for moment capacity. Shear capacity and moment capacity of each section are being calculated separately. The results are shown in Table 4.1.CHAPTER IV RESULTS & DISCUSSIONS The results of the structural design of the braced steel frame (beam and column) are tabulated and compiled in the next sections. The results based on BS 5950 and EC3 calculation are compiled together to show the difference between each other.1 for shear capacity and Table 5. Here. .1 Structural Beam UB sections ranging from 305x102x25 to 533x210x122 are being tabulated in ascending form. 6 14.09 1012.81 528.02 698.73 -3.2 447.4 -0.14 18.64 0.82 2.78 -25.58 34.99 15.83 0.16 1057.35 730.27 14.55 712.18 8.86 1204.26 2.99 918.35 -1.72 -12.65 0.37 399.13 19.99 660.74 0.68 1007.22 2.29 5.7 -0.74 594.67 644.66 497.08 2.1 493.38 1.32 877.81 -2.5 1.16 551.77 -3.38 20.28 554.35 793.87 -0.6 1.38 542.75 437.47 831.27 819.28 303.21 -24.85 767.2 1102.11 -2 2.47 545.53 356.09 16.44 471.69 4.4 0.94 2.93 334.55 617.27 0.5 461.48 517.65 846.77 1146.83 0.16 4.72 % Diff.06 EC 3 (kN) 284.97 392.86 619.67 Table 4.38 1.13 1091.31 446.42 820.79 11.93 11.21 668.55 3.09 -2.78 942.15 507.14 .66 704.56 -5.44 2.87 433.27 845.33 577.5 642.4 -10.07 942.56 878.09 773.92 2.29 452.82 2.88 876.51 1.85 517.15 3.69 -1.18 358.27 13.23 -9.24 3.45 623.53 564.98 305.15 -16.3 683.47 596.46 2.99 589.65 420.73 -2.92 394.79 2.65 724.51 384.74 -0.15 3.11 1218.26 -8.55 583.1 -2.57 13.7 -0.32 783.55 1.39 511.58 753.21 441.14 583.94 2.8 800.6 10.28 8.61 345.94 559. 2.21 15.78 541.31 2.57 680.4 0.62 1.24 0.91 1011.5 1102.74 2.26 888.85 854.85 405.57 -2.86 -0.7 1.27 0.88 -18.62 515.04 % Diff.52 439.02 6.63 12.33 409.77 6.45 -1.5 1.19 1.3 14.75 -13.51 18.1 Shear capacity of structural beam UB SECTION BS 5950 (kN) 305x102x25 305x102x28 305x102x33 305x127x37 305x127x42 305x127x48 305x165x40 305x165x46 305x165x54 356x127x33 356x127x39 356x171x45 356x171x51 356x171x57 356x171x67 406x140x39 406x140x46 406x178x54 406x178x60 406x178x67 406x178x74 457x152x52 457x152x60 457x152x67 457x152x74 457x152x82 457x191x67 457x191x74 457x191x82 457x191x89 457x191x98 533x210x82 533x210x92 533x210x101 291.74 2.59 460.95 404.41 925.81 1024.91 -19.35 431.05 607.64 0.39 462.34 523.96 6.74 393.79 12.79 2.5 529.5 S355 Difference (kN) 9.65 635.64 5.84 727.03 4.02 12.55 522.37 338.48 759.32 EC 3 (kN) 366.38 811.5 1.2 777.98 1134.89 678.92 588.65 0.81 -3.25 382.17 8.93 1.68 6.02 496.51 -4.81 523.6 1. BS 5950 (kN) 376.2 -2.33 862.78 456.13 705.19 1.37 609.93 11.36 11.56 400.46 478.09 -2.06 1.56 3.21 667.5 -0.32 860.78 -20.7 9.78 15.39 1.96 666.95 2.77 728.32 10.14 784.47 341.58 308.61 340.06 1.19 4.83 938.19 387.55 1.66 24.5 -0.15 343.79 2.34 44.71 429.56 S275 Difference (kN) 7.84 300.81 -3.93 1.53 943.66 5.79 398.52 443.56 15.11 -1.6 405.46 -3. 78 11.68 533x210x109 533x210x122 995. which is approximately 8. 6.6 py Av Av = Dt Vpl.43 160.76 191. Table 4.44 1300.06 % Diff.59 4.64 The difference is based on deduction of shear capacity of EC3 from BS 5950.59 5.43 3.14 8.06 1115.86 125. Also.35 217.13 8.Rd = (Av x fy) / (γM0 x √3) … (EC3) … (BS 5950) Av is obtained from section table. these facts explain the reason why shear capacity of most of the sections designed by EC3 is lower than the one designed by BS 5950.97 EC 3 (kNm) 113.2 Moment capacity of structural beam UB SECTION BS 5950 (kNm) 305x102x25 305x102x28 305x102x33 305x127x37 305x127x42 94.94 162.03 1440.94 -12.07 170.62 182. There are a few explanations to the variations.57 206. For steel grade S275.35 -0. the difference percentage ranges from -3.07 6.51 1007.3% less than 0. This value.05 1099.06%. For steel grade S355.83 132.76 4.55.05 3.01 -16.3 6. however.23 168.8 8.58 -9. the difference percentage ranges from -2.69% to 4.6 as suggested by BS 5950.56 S355 Difference (kNm) 7.41 143. BS 5950 (kNm) 121.59 5.29 S275 Difference (kNm) 6.57% to 4.6 137.45 1431.81 5.77 4.13 -0. Most of the values given are lesser than Dt value.49 1295. meanwhile.91 % Diff.98 141.28 148.58 4.05 110.97 6.43 3.07 . 1 / (γM0 x √3) ≈ 0.85 EC 3 (kNm) 88 106. The shear capacity of a structural beam is given by Pv = 0. Negative value indicates that the shear capacity calculated from EC3 is higher than that from BS 5950. varies with Av = Dt as suggested by BS 5950.43 -1.06%. Therefore.21 -1.57 -4. 65 749.9 163.34 404. .05 0.9 900.44 12.5 654.52 395.49 5.63 7.02 377.73 21.47 955.27 1.16 9.8 799.57 5.5 479.65 5.68 560.08 252.05 11.3 695.95 755.9 11.24 1.75 332.32 10. For steel grade S355.01 182. Positive value indicates that the moment capacity calculated from EC3 is lower than that from BS 5950.31 19.65 590.1 1.28 5.75 431.08 6.02 455.9 619.85 11.33 192.75 199.31 4.45 976.29 2.94 10. For steel grade S275.52 11.25 397.72 9.87 4.42 5.86 4.01 4.17 7.39 682.52 434.29 1.5 34.3 4.45 234.17 24.67 685.07 609.1 5.23 213.35 1104 238.95 385.95 275.2 24.53 5.65 244.28 5.1 285.61 4.24 376.45 521.48 5.53 1.45 18.1 539 619.85 27.6 341.44 14.98 20.55 4.45 769.73 2.95 189.03%.1 244.17 27.33 221.5 302.2 291.11 261.93 740.17 255.01 4.19 370.98 24.69 188.95 848 184.92 13.95 532.58% to 6.33 181.6 300.4 264.36 2.43%.43 4.17 171.8 1082.79 141.29 15.68 12 13.66 2.05 35.08 358.84 13.63 4.65 149.1 5.41% to 6.99 4.96 21.53 171.55 9.25 453.4 277.51 1.24 1.66 5.75 631.35 302.33 198 232.85 5.44 4. meanwhile.37 16. the difference percentage ranges from 1.02 18.81 529.97 14.5 44.06 11.25 517.53 5.21 287.55 433.14 3.43 4.57 355.6 5.95 479.67 20.05 336.98 352.41 19.3 426 479.5 457.08 510.25 5.77 233.75 398.7 211.41 221.71 9.55 21.1 220.5 14.35 731.06 0.93 885.29 202.5 5.85 5.26 312.83 4.05 585.24 17.87 4.95 24.5 354.32 0.5 390.83 5.49 5.12 5.69 305x127x48 305x165x40 305x165x46 305x165x54 356x127x33 356x127x39 356x171x45 356x171x51 356x171x57 356x171x67 406x140x39 406x140x46 406x178x54 406x178x60 406x178x67 406x178x74 457x152x52 457x152x60 457x152x67 457x152x74 457x152x82 457x191x67 457x191x74 457x191x82 457x191x89 457x191x98 533x210x82 533x210x92 533x210x101 533x210x109 533x210x122 195.38 8.26 317.89 1.28 15.55 257.33 471.3 844.91 The difference is based on deduction of moment capacity of EC3 from BS 5950.27 14.05 232.5 15.49 15.13 246.75 484.78 15.04 1.41 5.7 18.16 5.57 5.65 404.67 425.22 13.16 5.58 5.4 838.55 4.86 8. the difference percentage ranges from 0.68 0.85 585.83 275.83 1.5 691.95 566.11 5.96 10.86 4.02 315.73 19.25 497.5 330 371.13 318.46 5.63 4.62 7.11 5.11 242.75 562.55 429.35 624.95 514.08 5.75 300.48 17.14 410.88 10.53 549.35 693.78 487.32 1. are revised. EC3 provides better guidelines to classify a section web. The moment capacity of a structural beam is given by Mc = py Sx Mc.1. plastic modulus based on BS 5950 (Sx) and EC3 (Wpl. for a UB section 406x178x54.4 shows the result and percentage difference of moment resistance. 1 / γM0 ≈ 0. whether it is Class 1.3. A study is conducted to determine independently compression and bending moment capacity of structural column with actual length of 5m. Besides that. sectional classification tables – Table 11 and Table 5. For example.Rd = Wpl. For a column web subject to bending and compression.70 There are a few explanations to the variations.3 shows the result and percentage difference of compression resistance while Table 4. .1 of BS 5950 and EC3 respectively. BS 5950 only provides a clearer guideline to the classification of Class 3 semi-compact section. there are some variations between plastic modulus specified by BS 5950 section table and EC3 section table.0 as suggested by BS 5950. Meanwhile. Table 4. 4.95.y) are 1060cm3 and 1051cm3 respectively. these facts explain the reason why moment capacity of most of the sections designed by EC3 is lower than the one designed by BS 5950.2 Structural Column In determining the structural capacity of a column.y fy / γM0 … (BS 5950) … (EC3) From EC3 equation. Class 2 or Class 3 element. Therefore. There is a variation of approximately 0.85%. This is approximately 5% less than 1. 71 Table 4.3 Compression resistance and percentage difference UC SECTION BS 5950 (kN) 152x152x37 203x203x46 203x203x52 203x203x60 203x203x71 203x203x86 254x254x73 254x254x89 254x254x107 254x254x132 254x254x167 305x305x97 305x305x118 305x305x137 305x305x158 305x305x198 305x305x240 305x305x283 1027.63 1403.56 1588.95 1818.44 2199.15 2667.72 2341.45 2878.73 3454.34 4291.41 5419.6 3205.31 3901.39 4553.57 5256.95 6612.78 8028.11 9489.33 EC 3 (kN) 956.1 1323.8 1500 1721.2 2067.3 2508.5 2209.3 2715.9 3269.7 4057.6 5117.3 3025.8 3695.7 4292 4965.7 6242.4 7572.7 8958.9 S275 Difference (kN) 71.53 79.76 88.95 97.24 131.85 159.22 132.15 162.83 184.64 233.81 302.3 179.51 205.69 261.57 291.25 370.38 455.41 530.43 S355 Difference (kN) 117.66 142.41 158.24 170.26 213.57 255.76 209.85 256.99 295.49 375.39 486.02 271.11 310.04 385.76 426.68 530.78 641.15 735.89 % Diff. BS 5950 (kN) 1259.66 1773.41 2007.94 2298.26 2780.37 3373.46 2982.65 3668.29 4402.89 5474.39 6918.72 4097.01 4987.14 5821.16 6720.88 8455.58 10267.55 12138.99 EC 3 (kN) 1142 1631 1849.7 2128 2566.8 3117.7 2772.8 3411.3 4107.4 5099 6432.7 % Diff. 6.96 5.68 5.6 5.35 6 5.97 5.64 5.66 5.35 5.45 5.58 5.6 5.27 5.74 5.54 5.6 5.67 5.59 9.34 8.03 7.88 7.41 7.68 7.58 7.04 7.01 6.71 6.86 7.02 6.62 6.22 6.63 6.35 6.28 6.24 6.06 3825.9 4677.1 5435.4 6294.2 7924.8 9626.4 11403.1 Table 4.4 Moment resistance and percentage difference UC SECTION BS 5950 (kNm) 152x152x37 203x203x46 203x203x52 203x203x60 203x203x71 203x203x86 254x254x73 254x254x89 254x254x107 69.47 129.03 146.73 167.96 205.13 249.38 277.94 344.27 413.51 EC 3 (kNm) 80.9 130.2 148.5 171.3 209.8 256.4 259.3 320.8 388.7 S275 Difference (kNm) -11.43 -1.17 -1.77 -3.34 -4.67 -7.02 18.64 23.47 24.81 S355 Difference (kNm) -30.81 -7.67 -9.49 -12.6 -16.45 -21.92 14.12 17.68 16.48 % Diff. BS 5950 (kNm) 73.69 160.33 182.21 208.5 254.35 309.08 348.82 431.88 518.18 EC 3 (kNm) 104.5 168 191.7 221.1 270.8 331 334.7 414.2 501.7 % Diff. -16.45 -0.91 -1.21 -1.99 -2.28 -2.81 6.71 6.82 6 -41.81 -4.78 -5.21 -6.04 -6.47 -7.09 4.05 4.09 3.18 72 254x254x132 254x254x167 305x305x97 305x305x118 305x305x137 305x305x158 305x305x198 305x305x240 305x305x283 521.91 669.51 438.6 538.83 633.77 738.82 946.51 1168.56 1403.39 490.3 633.3 416.2 511.2 600.5 700.6 900.4 1111.3 1287.4 31.61 36.21 22.4 27.63 33.27 38.22 46.11 57.26 115.99 6.06 5.41 5.11 5.13 5.25 5.17 4.87 4.9 8.26 653.96 838.26 575.44 705.68 828.47 964.08 1231.05 1515.42 1815.14 632.9 817.5 537.2 660 775.3 904.4 1162.4 1434.5 1676 21.06 20.76 38.24 45.68 53.17 59.68 68.65 80.92 139.14 3.22 2.48 6.65 6.47 6.42 6.19 5.58 5.34 7.67 Shear capacity designed by BS 5950 is overall higher than EC3 design by the range of 5.27 – 6.96% and 6.22 – 9.34% for steel grade S275 (Fe 430) and S355 (Fe 510) respectively. This is mainly due to the partial safety factor γM1 of 1,05 imposed by EC3 in the design. Also, the compression strength fc determined from Table 5.14(a) of EC3 is less than the compression strength pc determined from Table 24 of BS 5950. Meanwhile, as the size of section increases, the difference percentage changes from -16.45% to 8.26% for S275 (Fe 460) and -41.81% to 7.67% for S355 (Fe 510). This means that smaller sizes designed by EC3 have higher moment capacity than BS 5950 design. From the moment capacity formula of BS 5950, Mb = pbSx pb depends on equivalent slenderness λLT, which is also dependant on the member length. The bigger the member size, the higher the radius of gyration, ry is. Therefore, pb increases with the increase in member size. However, moment capacity based on EC3 design, Mpl.y.Rd = Wpl.y fy / γM0 73 The moment capacity is not dependant on equivalent slenderness. Therefore, when member sizes increase, eventually, the moment capacity based on EC3 is overtaken by BS 5950 design. 4.2 Deflection Table 4.5 shows the deflection values due to floor imposed load. In BS 5950, this is symbolized as δ while for EC3, this is symbolized as δ2. Table 4.5 Deflection of floor beams due to imposed load UB SECTION BS 5950 (δ, mm) 27.56 22.99 19 17.22 15.06 12.89 14.53 12.47 10.55 14.97 12.11 10.2 8.76 7.72 6.33 9.88 7.86 6.6 5.72 5.08 4.52 L = 6.0m EC 3 (δ2, mm) 27.62 22.16 18.54 16.83 14.77 12.68 14.1 12.13 10.31 14.71 11.93 9.98 8.51 7.51 6.17 9.71 7.69 6.46 5.6 4.95 4.39 Difference (mm) -0.06 0.83 0.46 0.39 0.29 0.21 0.43 0.34 0.24 0.26 0.18 0.22 0.25 0.21 0.16 0.17 0.17 0.14 0.12 0.13 0.13 % Diff. -0.22 3.61 2.42 2.26 1.93 1.63 2.96 2.73 2.27 1.74 1.49 2.16 2.85 2.72 2.53 1.72 2.16 2.12 2.1 2.56 2.88 BS 5950 (δ, mm) 139.53 116.41 96.17 87.18 76.23 65.25 73.54 63.14 53.43 75.77 61.28 51.66 44.33 39.07 32.06 50.01 39.81 33.43 28.94 25.72 22.9 L = 9.0m EC 3 (δ2, mm) 139.83 112.19 93.86 85.2 74.79 64.19 71.36 61.42 52.2 74.49 60.42 50.51 43.09 38 31.23 49.17 38.94 32.68 28.33 25.08 22.25 Difference (mm) -0.3 4.22 2.31 1.98 1.44 1.06 2.18 1.72 1.23 1.28 0.86 1.15 1.24 1.07 0.83 0.84 0.87 0.75 0.61 0.64 0.65 % Diff. -0.22 3.63 2.4 2.27 1.89 1.62 2.96 2.72 2.3 1.69 1.4 2.23 2.8 2.74 2.59 1.68 2.19 2.24 2.11 2.49 2.84 305x102x25 305x102x28 305x102x33 305x127x37 305x127x42 305x127x48 305x165x40 305x165x46 305x165x54 356x127x33 356x127x39 356x171x45 356x171x51 356x171x57 356x171x67 406x140x39 406x140x46 406x178x54 406x178x60 406x178x67 406x178x74 56 2.85 15.06 0.66 2.21 2.77 16. Section 3.63 19.33 12.52 0.4 2.84 4.29 0.4 2. Ix = 16000cm4 from BS 5950 section table.61%.1 0.05 0.24 2.53 0.4 0.2 3.7 2. the major difference between the deflection designs of these two codes is the total deflection.6 2.31 2.1.27 3. Apart from that.74 457x152x52 457x152x60 457x152x67 457x152x74 457x152x82 457x191x67 457x191x74 457x191x82 457x191x89 457x191x98 533x210x82 533x210x92 533x210x101 533x210x109 533x210x122 5.1.73 1. It also indicates that deflection value calculated from BS 5950 is normally higher than that from EC3.61 3.07 0.96 1.51 0.7 2.59 2.46 2.63%.21 24.54 2.43 2.8 1.29 0.1 0.42 0.07 1.83 20.06 0.22% to 3.46 2.55 From Table 4.63 2.1 3.3 “Other properties” of BS 5950 states that E = 205kN/mm2.58 0.33 4.25 2.78 3.79 16. Meanwhile.25 16.25 0.16 9.35 0.71 3. the difference percentage ranges from -0. Different from BS 5950. as required by EC3.71 3.83 13.03 9. This is basically same as the range of beam length 6m.05 0.93 2. Iy = 16060cm4 from EC3 section table.07 0.56 2.9 9.49 2.16 11.16 1. for a section 356x171x57. E.26 0.22% to 3.21 3.37 4.85 1.98 21.74 4. The first explanation for this difference is the modulus of elasticity value.45 14. there is also slight difference between second moment of area in both codes.26 2.34 1. for a floor beam of 9m long.7 2.33 3.84 11. For example.56 2.77 4.75 2. section 3.34 18.47 29. . subject to 15kN/m of unfactored imposed floor load.55 23.18 1.38 2. Meanwhile.85 1.12 17.25 13. EC3 requires deflection due to permanent dead load to be included in the final design.68 13.13 0.23 0. The minor differences had created differences between the deflection values.19 2.26 18.01 1.25 2.5 above.08 0.32 10.22 28.62 5. However.36 8. for a floor beam of 6m long.13 8.32 0.06 0.68 2.41 1. Meanwhile.04 2.33 0.08 21.4 “Design values of material coefficients” of C-EC3 states that E = 210kN/mm2.64 4.75 18. δmax.01 2.04 0.51 21.66 0.77 2.37 2.08 0. the difference percentage ranges from -0.01 0.1 0. external columns and internal columns have been designed for the most optimum size. the results of the design (size of structural members) are tabulated in Table 4.744 Roof Section Designation Universal Columns External Internal Total Steel Weight (tonne) .75 4.122 152x152x30 203x203x46 203x203x46 203x203x60 4.4th Storey To 2nd Storey 2nd .889 152x152x23 152x152x37 152x152x37 203x203x52 3.4th Storey 203x203x46 203x203x52 203x203x52 203x203x86 9. the weight of steel will be used as a gauge.6 and Table 4. To compare the economy of the design.3 Economy of Design After all the roof beams.750 533x210x92 533x210x82 457x152x60 406x140x46 To 2nd Storey 2nd .7 for BS 5950 and EC3 design respectively.4th Storey 152x152x37 203x203x46 203x203x46 203x203x71 7.4th Storey To 2nd Storey 2nd . floor beams. Table 4.6 Weight of steel frame designed by BS 5950 Model No Frame Type Universal Beams Floor S275 1 2 Bay 4 Storey (6m span) 2 2 Bay 4 Storey (9m span) S355 3 2 Bay 4 Storey (6m span) 4 2 Bay 4 Storey (9m span) 533x210x82 457x191x67 406x140x46 406x140x39 To 2nd Storey 2nd . 744 EC3 4.571 Total Steel Weight (ton) BS 5950 4. The saving percentage.571 533x210x92 533x210x82 406x178x54 406x140x46 To 2nd Storey 2nd .4th Storey To 2nd Storey 2nd .750 4.313 9.889 9.313 152x152x30 203x203x46 203x203x46 203x203x71 4.645 152x152x37 203x203x46 203x203x52 254x254x73 4. meanwhile.76 Table 4.8. Table 4.821 Roof Section Designation Universal Columns External Internal Total Steel Weight (tonne) Summary of the total steel weight for the multi-storey braced steel frame design is tabulated in Table 4. is tabulated in Table 4.8 Total steel weight for the multi-storey braced frame design Types of Frame 2Bay 4Storey Bay Width (m) 6 Steel Grade S275 (Fe 430) S355 (Fe 510) 2Bay 4Storey 9 S275 (Fe 430) S355 (Fe 510) 7.4th Storey 203x203x46 203x203x71 203x203x71 254x254x107 9.9.7 Weight of steel frame designed by EC3 Model No Frame Type Universal Beams Floor S275 5 2 Bay 4 Storey (6m span) 6 2 Bay 4 Storey (9m span) S355 7 2 Bay 4 Storey (6m span) 8 2 Bay 4 Storey (9m span) 533x210x92 533x210x82 406x178x54 356x171x45 To 2nd Storey 2nd .4th Storey To 2nd Storey 2nd .821 .4th Storey 203x203x46 203x203x60 203x203x60 254x254x89 9.122 9.645 3. The percentage savings for braced steel frame with 9m span is higher than that one with 6m span. Further check on the effect of deflection was done. The percentage of saving offered by BS 5950 design ranges from 1.9 EC3 design Percentage difference of steel weight (ton) between BS 5950 design and Frame 2Bay 4Storey Bay Width (m) 6 Steel Grade S275 (Fe 430) S355 (Fe 510) Total Steel Weight (ton) BS 5950 4. This time.77 Table 4. the connections of beam-to-column were assumed to be “partial strength connection”.60% to 17. Meanwhile.744 3.60 17. This is because deeper.889 EC3 4. the percentage savings by using BS 5950 are higher than EC3 for S355 steel grade with respect to S275 steel grade.29 2Bay 4Storey 9 S275 (Fe 430) S355 (Fe 510) As shown in Table 4.750 9. Semi-continuous . Regardless of bay width.645 9. larger hot-rolled section is required to provide adequate moment capacity and also stiffness against deflection.571 9. This resulted in higher percentage difference.96%. unaffected by the effect of imposed load deflection.821 4.122 7.42 15. all frame types. beam spans and steel grade designed by using BS 5950 offer weight savings as compared with EC3. BS 5950 design allowed lighter section.9.96 5. This is because overall deflection was considered in EC3 design. depending on the steel grade.313 % 1. Columns remained the same as there was no change in the value of eccentricity moment and axial force.10 Weight of steel frame designed by EC3 (Semi-continuous) Model No Frame Type Universal Beams Floor S275 5 2 Bay 4 Storey (6m span) 6 2 Bay 4 Storey (9m span) S355 7 2 Bay 4 Storey (6m span) 8 2 Bay 4 Storey (9m span) 533x210x82 457x151x67 406x140x46 356x127x39 To 2nd Storey 2nd .5. For uniformly distributed loading.10 shown. the deflection value is given as: δ = βwL4 / 384EI For a span with connections having a partial strength less than 45%. Table 4. Please refer to Appendix D for a redesign work after the β value had been revised and the section redesigned to withstand bending moment from analysis process.0. where zero “support” stiffness corresponds to a value of β = 5. which was used in the beam design. The renewed beam sections are tabulated in Table 4.749 Roof Section Designation (Semi-continous) Universal Columns External Internal Total Steel Weight (tonne) .503 152x152x30 203x203x46 203x203x46 203x203x71 4.211 533x210x92 533x210x82 457x178x52 406x140x46 To 2nd Storey 2nd . This is different from pinned joint in simple construction. the deflection coefficient.78 frame is achieved in this condition.4th Storey 203x203x46 203x203x71 203x203x71 254x254x107 9.4th Storey To 2nd Storey 2nd .4th Storey 203x203x46 203x203x60 203x203x60 254x254x89 8. β is treated as β = 3.4th Storey To 2nd Storey 2nd .645 152x152x37 203x203x46 203x203x52 254x254x73 4. 645 3.12.749 % 0.122 9. The saving percentage.749 Table 4.11.503 7.11 .12 Percentage difference of steel weight (ton) between BS 5950 design and EC3 design (Revised) Frame 2Bay 4Storey Bay Width (m) 6 Steel Grade S275 (Fe 430) S355 (Fe 510) 2Bay 4Storey 9 S275 (Fe 430) S355 (Fe 510) 7.645 5.503 9.79 Summary of the total revised steel weight for the multi-storey braced steel frame design is tabulated in Table 4.889 8.11 Total steel weight for the multi-storey braced frame design (Revised) Types of Frame Bay Width (m) 2Bay 4Storey 6 Steel Grade S275 (Fe 430) S355 (Fe 510) 2Bay 4Storey 9 S275 (Fe 430) S355 (Fe 510) 7.42 3.122 9.22 9.744 EC3 (Semi-Cont) 4.889 8. meanwhile.211 Total Steel Weight (ton) BS 5950 4. is tabulated in Table 4. Table 4.744 Total Steel Weight (ton) EC3 (Semi-Cont) 4.95 BS 5950 4.750 4.211 10.750 4. MD = wL /8 – MR 2 (b) (c) Figure 4. with deflection coefficient set as β = 1.1(b)). the percentage of difference had been significantly reduced to the range of 0.12.1(c)). The moment capacity will be the deciding factor. . However. the sagging moment at mid span became less than that of simple construction (Figure 4. The greater difference for steel grade S355 indicated that deflection still plays a deciding role in EC3 design. wL2/8 MR wL2/8 MR wL2/8 (a) Design moment. if it is built semi-continuously. if rigid connection is introduced. it can be seen that there is an obvious reduction of steel weight required for the braced steel frame. The ability of partial strength connection had enabled moment at mid span to be partially transferred to the supports (Figure 4.95%.0. the gap reduces. Please refer to Figure 4. (c) simple construction. the effect of deflection on the design will be eliminated.11% to 10.80 From Table 4. Eventually. as the connection stiffness becomes higher.1 Bending moment of beam for: (a) rigid construction. Therefore. Even though EC3 design still consumed higher steel weight.1(a) for the illustration of rigid connection. The effect of dead load on the deflection of beam had been gradually reduced. (b) semi-rigid construction. calculation based on EC3 had effectively reduced a member’s shear capacity of up to 6. γM0 of 1. Meanwhile.05 in the moment capacity . Av value also caused the difference.06% with regard to BS 5950 due to the variance between constant values of the shear capacity formula specified by both codes. the difference between the approaches to obtain shear area. for the moment capacity of structural beam. 5. The application of different steel grade did not contribute greater percentage of difference between the shear capacities calculated by both codes.1 Structural Capacity 5.43%. In review to the research objectives. Apart from that. This is mainly due to the application of partial safety factor. a summary on the results of the objectives is categorically discussed.1 Structural Beam For the shear capacity of a structural beam.1. calculation based on EC3 had reduced a member’s shear capacity of up to 4. Suggestions of further research work are also included in this chapter.CHAPTER V CONCLUSIONS This chapter presents the summary for the study on the comparison between BS 5950 and EC3 for the design of multi-storey braced frame. EC3 design created majority . there is also a deviation in between the compressive strength. Meanwhile. A reduction in the range of 5. For the same value of unfactored imposed load. 5.82 calculation required by EC3.27% to 9. it was found that for a same value of λ. only compressive resistance comparison of structural column was made.2 Structural Column In simple construction. From interpolation.1.0m long. γM0 of 1. This is due to the implication of partial safety factor. Therefore. only moments due to eccentricity will be transferred to structural column. The design of structural beam proposed by EC3 is concluded to be safer than that by BS 5950. of both codes.24% of column compressive resistance was achieved when designing by EC3. In comparison. This comparison is based on a structural column of 5. a structural beam will be subject to deflection. Therefore. Only gravitational loads will be considered in this project. fc is smaller than pc. The steel frame is assumed to be laterally braced.05 as required by EC3 design. it is obvious that EC3 stresses on the safety of a structural beam. γM of 1.2 Deflection Values When subject to an unfactored imposed load. as compared to the partial safety factor. wind load (horizontal load) will not be considered in the design. 5. compared with BS 5950. axial compression is much more critical. With the inclusion of partial safety factor. fc and pc respectively.0 as suggested by BS 5950. 4-storey. it was found that EC3 design produced braced steel frames that require higher steel weight than the ones designed with BS 5950.2. Therefore.889 tons for BS 5950 design.821 tons and 4.1 of EC3 provided proof to this. 6m bay width steel frame. Cross-section with higher second moment of area value.645 tons and 9. The total steel weight of structural beams and columns was accumulated for comparison. Higher E means the elasticity of a member is higher. Section 4. For a 2-bay. the consumption of steel for S275 (Fe 430) and S355 (Fe 510) is 4.3 Economy Economy aspect in this study focused on the minimum steel weight that is needed in the construction of the braced steel frame. 9m bay width steel frame.22% to 3.83 lower deflection values with respect to BS 5950 design. I will have to be chosen.63%. taking into account deflection due to permanent loads. However. and 4. serviceability limit states check governs the design of EC3 as permanent loads have to be considered in deflection check. In this study. 5. The difference ranges from 0.571 tons for EC3 design.750 tons for BS 5950 design. For a 2-bay. . The main reason for the deviation is the difference in the specification of modulus of elasticity. the consumption of steel for S275 (Fe 430) and S355 (Fe 510) is 9.313 tons for EC3 design. thus can sustain higher load without deforming too much. E.122 tons and 7. BS 5950 specifies 205kN/mm2 while EC3 specifies 210kN/mm2. the total deflection was greater. compared with the section chosen for BS 5950 design. 4-storey.744 tons and 3. and 9. S275 (Fe 430): 0. 4-storey. since the results of the third objective contradicted with the background of the study (claim by Steel Construction Institute). 9m bay width. 9m bay width. This study showed that steel weight did not contribute to cost saving of EC3 design.5 had successfully reduced the percentage of difference between the steel weights designed by both codes. 4-storey. However. S355 (Fe 510): 17. 4-storey.42% 2-bay. 4-storey. 6m bay width. it is suggested that an unbraced steel frame design is conducted to study the behavior.95% 2-bay.96% 2-bay. 9m bay width. S355 (Fe 510): 15. 6m bay width. 4-storey.84 The percentages of differences are as follow: (i) (ii) (iii) (iv) 2-bay.60% 2-bay.11% 2-bay.4 Recommendation for Future Studies For future studies.42% 2-bay. The percentages of differences are as follow: (i) (ii) (iii) (iv) 2-bay.29% Further study was extended for the application of partial strength connection for beam-to-column connections in EC3 design. . S355 (Fe 510): 7. 6m bay width. 4-storey. 4-storey. 9m bay width. S275 (Fe 430): 5. S355 (Fe 510): 10.22% 5. it is recommended that further studies to be conducted to focus on the economy aspect of EC3 with respect to BS 5950.0 to 3. S275 (Fe 430): 5. 6m bay width. The reduction in deflection coefficient from 5. S275 (Fe 430): 1. 4-storey. structural design and economic aspect based on both of the design codes. “Eurocode 3: Design of Steel Structures: Part 1. 4.” Berkshire: Steel Construction Institute. .” ICE Journal. 29-32. European Committee for Standardization (1992). Paper 2658.85 REFERENCES Charles King (2005). “Steel Design Can be Simple Using EC3. Heywood M. Taylor J. “Steelwork design guide to BS 5950-1:2000 Volume 2: Worked examples. Issue 3. Steel Construction Institute (SCI) (2005).C. (1995).” New Steel Construction. D.1 General Rules and Rules for Buildings. British Standards Institution (2001). al. “British Standard – Structural Use of Steelwork in Building: Part 1: Code of Practice for Design – Rolled and Welded Sections. Narayanan R et. “EN 1993 Eurocode 3 – Steel.” Berkshire: Steel Construction Institute.” London: European Committee for Standardization.” Eurocodenews. Vol 13 No 4.” London: British Standards Institution. & Lim J B (2003). 24-27. (2001). “EN1993 Eurocode 3: Design of Steel Structures. “Introduction to Concise Eurocode 3 (C-EC3) – with Worked Examples. November 2005. 86 APPENDIX A1 . LL Floors Dead Load. JOHOR Client: Job Title: Subject: Page 1001 1 Braced Steel Frame Design (BS 5950-1 : 2000) Frame Analysis STC.4 x 27.64 kN/m . DL Live Load. of Bay No.87 Job No: UTM 81310 SKUDAI.0 DATA No. of Storey Frame Longitudinal Length Bay Width.6 x 15 = 62.6 2.6 FACTORED LOAD w = 1. UTM Made by Checked by CCH DR.6 + 1.5 kN/m @ 2 kN/m @ 2 27.6LL Roof w = 1. MAHMOOD 1. DL Live Load. DL Live Load.5 kN/m @ 2 kN/m @ 2 24 9 kN/m kN/m = = 4.4 x 24 + 1.6 x 9 = 48 kN/m Floors w = 1.4 1. LL = = 4 1.4DL + 1.6 15 kN/m kN/m = = 1. LL LOAD FACTORS Dead Load. l Storey Height = = = = = = 2 4 6 6 5 4 m m m (First Floor) m (Other Floors) LOADING Roof Dead Load. 1 FRAME LAYOUT Selected Intermediate Frame 6m 6m 6m 6m 2. JOHOR Client: Job Title: Subject: Page 1001 2 Braced Steel Frame Design (BS 5950-1 : 2000) Frame Analysis STC.2 Precast Slab Panel Load Transfer to Intermediate Frame . MAHMOOD 2.0 2. UTM Made by Checked by CCH DR.88 Job No: UTM 81310 SKUDAI. 64 kN/m 62.3 Cut Section of Intermediate Frame 4m [4] 4m [3] 4m [2] [1] 5m 6m 3.64 kN/m 62.0 LOAD LAYOUT 48 kN/m 6m 48 kN/m 62.64 kN/m . UTM Made by Checked by CCH DR. JOHOR Client: Job Title: Subject: Page 1001 3 Braced Steel Frame Design (BS 5950-1 : 2000) Frame Analysis STC.89 Job No: UTM 81310 SKUDAI.64 kN/m 62.64 kN/m 62. MAHMOOD 2.64 kN/m 62. contributed by beam shear. horizontal load is not taken into account Beam restraint Top flange effectively restrained against lateral torsional buckling 4.84 1039.52 144 331. V = 62. .84 707.64 x 6 / 2 = 187. Eccentricity = 100 mm from face of column.90 Job No: UTM 81310 SKUDAI. UTM Made by Checked by CCH DR.0 LOAD CALCULATION Frame bracing Laterally braced.64 x 6^2 / 8 = 281.1 Beam Moment. V = 48 x 6 / 2 = 144 kN M = 48 x 6^2 / 8 = 216 kNm Floor beams. JOHOR Client: Job Title: Subject: Page 1001 4 Braced Steel Frame Design (BS 5950-1 : 2000) Frame Analysis STC.92 kN M = 62. Universal column of depth 200 mm Internal column . Roof beams. MAHMOOD 4.Moments from left and right will cancel out each other.2 Column Shear Column Shear (kN) Internal External 288 663.68 1415.88 kNm 4. Shear.92 519.76 M = wl / 8 V = wl / 2 2 [4] [3] [2] [1] Moment External column will be subjected to eccentricity moment. 92) 663.0 ANALYSIS SUMMARY Moment (kNm) 216 216 281.84 [3] [4] 707. MAHMOOD 5.92 [2] 519.92) 519.92) 288 (187. UTM Made by Checked by CCH DR.76 .88 Shear (kN) (144) (144) 144 (187.88 281.88 281.92) 331.76 1415.84 (187.91 Job No: UTM 81310 SKUDAI.88 281.68 (187.84 (187. JOHOR Client: Job Title: Subject: Page 1001 5 Braced Steel Frame Design (BS 5950-1 : 2000) Frame Analysis STC.52 707.92 (187.88 281.92) 144 [1] 331.88 281.92) 1039. 4DL+1.54 28.6 21.19 28.6 21.19 [4] 28.19 [3] 28.19 28.92 Job No: UTM 81310 SKUDAI.6 31.54 28.6 [2] 28.54 28.19 21.0DL Most critical condition .6LL) .54 28. JOHOR Client: Job Title: Subject: Page 1001 6 Braced Steel Frame Design (BS 5950-1 : 2000) Frame Analysis STC.6 [1] 21.19 Moments are calculated from (1.54 31.19 28.19 21.54 31.19 28.19 31. MAHMOOD Column moment due to eccentricity (kNm) 21.1.6 28. UTM Made by Checked by CCH DR.19 31.19 31. 93 APPENDIX A2 . 5 x 9 = 45.6 15 kN/m kN/m LOAD FACTORS Dead Load. LL = = 1. LL Floors Dead Load.5 x 15 = 59.0 DATA No. JOHOR Client: Job Title: Subject: Page 1002 1 Braced Steel Frame Design (Eurocode 3) Frame Analysis STC.35 x 24 + 1.76 kN/m .94 Job No: UTM 81310 SKUDAI.6 + 1. DL Live Load.5 kN/m @ 2 kN/m @ 2 24 9 kN/m kN/m = = 4.5 kN/m @ 2 kN/m @ 2 27. MAHMOOD 1. DL Live Load.6 2. l Storey Height = = = = = = 2 4 6 6 5 4 m m m (First Floor) m (Other Floors) LOADING Roof Dead Load.35 1. UTM Made by Checked by CCH DR. of Storey Frame Longitudinal Length Bay Width.5LL Roof w = 1.9 kN/m Floors w = 1. DL Live Load.35DL + 1.5 FACTORED LOAD w = 1.35 x 27. LL = = 4 1. of Bay No. 95 Job No: UTM 81310 SKUDAI. MAHMOOD 2. JOHOR Client: Job Title: Subject: Page 1002 2 Braced Steel Frame Design (Eurocode 3) Frame Analysis STC.2 Precast Slab Panel Load Transfer to Intermediate Frame .0 2.1 FRAME LAYOUT Selected Intermediate Frame 6m 6m 6m 6m 2. UTM Made by Checked by CCH DR. 76 kN/m 59.9 kN/m 59.9 kN/m 6m 45.76 kN/m 59. UTM Made by Checked by CCH DR.76 kN/m . MAHMOOD 2.3 Cut Section of Intermediate Frame 4m [4] 4m [3] 4m [2] [1] 5m 6m 3.76 kN/m 59.0 LOAD LAYOUT 45.96 Job No: UTM 81310 SKUDAI.76 kN/m 59. JOHOR Client: Job Title: Subject: Page 1002 3 Braced Steel Frame Design (Eurocode 3) Frame Analysis STC.76 kN/m 59. .76 x 6 / 2 = 179. Shear. Eccentricity = 100 mm from face of column. UTM Made by Checked by CCH DR.08 137.9 x 6 / 2 = 137.97 Job No: UTM 81310 SKUDAI.7 316. Universal column of depth 200 mm Internal column .4 633.7 kN M = 45. Roof beams. MAHMOOD 4.54 M = wl / 8 V = wl / 2 2 [4] [3] [2] [1] Moment External column will be subjected to eccentricity moment.2 Column Shear Column Shear (kN) Internal External 275.96 992.98 496.55 kNm Floor beams.0 LOAD CALCULATION Frame bracing Laterally braced.92 kNm 4. JOHOR Client: Job Title: Subject: Page 1002 4 Braced Steel Frame Design (Eurocode 3) Frame Analysis STC.28 kN M = 59. V = 45.1 Beam Moment. V = 59. horizontal load is not taken into account Beam restraint Top flange effectively restrained against lateral torsional buckling 4.26 675.52 1351.Moments from left and right will cancel out each other.9 x 6^2 / 8 = 206. contributed by beam shear.76 x 6^2 / 8 = 268. 28) 275.1 ANALYSIS SUMMARY Moment (kNm) 206.7 [1] 316.28) 992.54 1351.52 (179.2 Shear (kN) (137.28) 137.92 268.98 [2] 496.92 268.28) 316.92 268. UTM Made by Checked by CCH DR. MAHMOOD 5.55 206.0 5.96 (179.92 5. JOHOR Client: Job Title: Subject: Page 1002 5 Braced Steel Frame Design (Eurocode 3) Frame Analysis STC.92 268.98 Job No: UTM 81310 SKUDAI.98 (179.28) 496.26 [3] [4] 675.26 (179.54 .28) 633.7) 137.92 268.55 268.4 (179.7 (179.7) (137.08 675. 89 28. UTM Made by Checked by CCH DR.94 28.89 Moments are calculated from (1.35DL+1.94 26.0DL Most critical condition .89 26.89 26.5LL) . JOHOR Client: Job Title: Subject: Page 1002 6 Braced Steel Frame Design (Eurocode 3) Frame Analysis STC.1.89 26.3 Column moment due to eccentricity (kNm) 20.66 20.66 26.94 26.66 26.94 26.89 20.71 28. MAHMOOD 5.94 26.89 26.94 28.89 26.66 19.89 19.89 26.89 28.89 28.71 20.99 Job No: UTM 81310 SKUDAI. 100 APPENDIX B1 . 3 82 82.2 28.88 x 10^3 / 275 3 = 1025 cm Try UB 457x152x60 .1 Sx (cm3) 171 259 234 342 258 306 403 353 314 393 481 543 483 539 724 659 623 614 566 775 888 720 711 896 1100 846 1060 Section Mass (kg/m) 57 59. MAHMOOD Grade = S275 Section Mass (kg/m) 19 22 23.1 Sx (cm3) 1010 1290 1200 1210 1350 1470 1450 1500 1630 1650 1830 1810 2060 2010 2380 2230 2610 2880 2830 3280 3200 3680 4140 4590 5550 7490 178x102x19 254x102x22 203x102x23 305x102x25 203x133x25 254x102x25 305x102x28 254x102x28 203x133x30 254x146x31 305x102x33 356x127x33 254x146x37 305x127x37 406x140x39 356x127x39 305x165x40 305x127x42 254x146x43 356x171x45 406x140x46 305x165x46 305x127x48 356x171x51 457x152x52 305x165x54 406x178x54 356x171x57 457x152x60 406x178x60 356x171x67 406x178x67 457x191x67 457x152x67 406x178x74 457x152x74 457x191x74 457x191x82 457x152x82 533x210x82 457x191x89 533x210x92 457x191x98 533x210x101 610x229x101 533x210x109 610x229x113 533x210x122 610x229x125 610x229x140 610x305x149 610x305x179 610x305x238 M = 281.1 40.9 149.2 74.2 109 113 122 125.8 25. JOHOR Client: Job Title: Subject: Page 1003 1 Braced Steel Frame Design (BS 5950-1 : 2000) Beam Design (Floor Beams.1 67.1 51 52.1 82.3 30 31. UTM Made by Checked by CCH DR.1 48.1 32.8 33.1 37 37 39 39.0m) STC.2 74.1 139.1 25.9 43 45 46 46.2 74.3 101 101.2 89.1 67.1 67. L = 6.1 67.8 60.3 92.101 Job No: UTM 81310 SKUDAI.2 179 238.1 24.88 kNm Sx = M / fy = 281.3 54 54.3 41.1 98.2 28. 6 1290 1120 kg/m mm mm mm mm mm 3 cm cm 3 b/T = d/t = 5.9 8. Limiting b/T = 9ε Actual b/T = 5. MAHMOOD 1.3 407. Limiting d/t = 80ε = 80 Actual d/t = 50. JOHOR Client: Job Title: Subject: Page 1003 2 Braced Steel Frame Design (BS 5950-1 : 2000) Beam Design (Floor Beams.3 < 80 Web is plastic Class 1 Section is : Class 1 plastic section . UTM Made by Checked by CCH DR.0 SECTION CLASSIFICATION Grade of steel T= 13. L = 6.1 13. Section chosen = 457x152x60 UB 1.75 50.3 Therefore. The size is then checked to ensure suitability in all other aspects.75 = < 9 9 Flange is plastic Class 1 Section is symmetrical. subject to pure bending.0m) STC.1 DATA Trial Section Initial trial section is selected to give a suitable moment capacity.3 2.102 Job No: UTM 81310 SKUDAI. neutral axis at mid-depth.0 1.8 454. py = = mm 275 S275 < N/mm 2 16mm ε = √ (275/py) = SQRT(275/275) = 1 Outstand element of compression flange.2 Section Properties Mass Depth Width Web thickness Flange thickness Depth between fillets Plastic modulus Elastic modulus Local buckling ratios: Flange Web = D= B= t= T= d= Sx = Zx = 59.6 152. 103 Job No: UTM 81310 SKUDAI, JOHOR Client: Job Title: Subject: Page 1003 3 Braced Steel Frame Design (BS 5950-1 : 2000) Beam Design (Floor Beams, L = 6.0m) STC, UTM Made by Checked by CCH DR. MAHMOOD 3.0 SHEAR BUCKLING If d/t ratio exceeds 70ε for rolled section, shear buckling resistance should be checked. d/t = 50.3 < 70ε = 70 Therefore, shear buckling needs not be checked 4.0 SHEAR CAPACITY Fv = 187.92 kN Pv = 0.6pyAv py = 275 N/mm Av = tD = 8.1 x 454.6 2 = 3682.26 mm 2 Pv = 0.6 x 275 x 3682.26 x 0.001 = 607.57 kN Fv Pv < Therefore, the shear capacity is adequate 5.0 MOMENT CAPACITY M= 281.88 kNm 0.6Pv = 0.6 x 607.57 = 364.542 kN Fv 0.6Pv < Therefore, it is low shear Mc = pySx = 275 x 1290 x 0.001 = 354.75 kNm 1.2pyZ = 1.2 x 275 x 1120 x 0.001 = 369.6 kNm Mc M < < 1.2pyZ Mc OK Moment capacity is adequate 104 Job No: UTM 81310 SKUDAI, JOHOR Client: Job Title: Subject: Page 1003 4 Braced Steel Frame Design (BS 5950-1 : 2000) Beam Design (Floor Beams, L = 6.0m) STC, UTM Made by Checked by CCH DR. MAHMOOD 6.0 6.1 WEB BEARING & BUCKLING Bearing Capacity Pbw = (b1 + nk) tpyw r= 10.2 mm (Unstiffened web) b1 = t + 1.6r + 2T = 8.1 + 1.6 x 10.2 + 2 x 13.3 = 51.02 mm k= T+r = 13.3 + 10.2 = 23.5 mm At the end of a member (support), n = 2 + 0.6be/k = 2 b1 + nk = = = = = < but n ≤ 5 be = 0 51.02 + 2 x 23.5 98.02 mm 98.02 x 8.1 x 275 x 0.001 218.34 kN 187.92 Pbw kN Pbw Fv Fv Bearing capacity at support is ADEQUATE 105 Job No: UTM 81310 SKUDAI, JOHOR Client: Job Title: Subject: Page 1003 5 Braced Steel Frame Design (BS 5950-1 : 2000) Beam Design (Floor Beams, L = 6.0m) STC, UTM Made by Checked by CCH DR. MAHMOOD 7.0 SERVICEABILITY DEFLECTION CHECK Unfactored imposed loads: w= = E= I= δ= 9 15 205 25500 4 kN/m kN/m kN/mm cm 4 2 for roofs for floors L= 6 m 5wL 384EI = 5 x 15 x 6^4 x 10^5 384 x 205 x 25500 = 4.84 mm Beam condition Carrying plaster or other brittle finish Deflection limit = Span / 360 = 6 x 1000 / 360 = 16.67 mm 4.84mm < 16.67mm The deflection is satisfactory! 106 APPENDIX B2 . 107 Job No: UTM 81310 SKUDAI.y (cm3) 1009 1195 1283 1213 1346 1442 1472 1509 1624 1659 1802 1832 2058 2020 2366 2234 2619 2887 2827 3287 3203 3673 4139 4575 5515 7462 178x102x19 254x102x22 203x102x23 203x133x25 254x102x25 305x102x25 254x102x28 305x102x28 203x133x30 254x146x31 305x102x33 356x127x33 254x146x37 305x127x37 356x127x39 406x140x39 305x165x40 305x127x42 254x146x43 356x171x45 305x165x46 406x140x46 305x127x48 356x171x51 457x152x52 305x165x54 406x178x54 356x171x57 406x178x60 457x152x60 356x171x67 406x178x67 457x152x67 457x191x67 406x178x74 457x152x74 457x191x74 457x152x82 457x191x82 533x210x82 457x191x89 533x210x92 457x191x98 533x210x101 610x229x101 533x210x109 610x229x113 533x210x122 610x229x125 610x229x140 610x305x149 610x305x179 610x305x238 M = 268.92 x 10^3 / 275 = 977. UTM Made by Checked by CCH DR.9 cm3 Try 406x178x54 UB . JOHOR Client: Job Title: Subject: Page 1004 1 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.0m) STC.y = M / fy = 268. MAHMOOD Grade = S275 Section Mass (kg/m) 19 22 23 25 25 25 28 28 30 31 33 33 37 37 39 39 40 42 43 45 46 46 48 51 52 54 54 Wpl.y (cm ) 171 260 232 259 307 336 354 408 313 395 481 539 485 540 654 718 626 612 568 773 722 889 706 895 1096 843 1051 3 Section Mass (kg/m) 57 60 60 67 67 67 67 74 74 74 82 82 82 89 92 98 101 101 109 113 122 125 140 149 179 238 Wpl.92 kNm W pl. L = 6. JOHOR Client: Job Title: Subject: Page 1004 2 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.9 360.6 7.108 Job No: UTM 81310 SKUDAI.15 47.6 10.0m) STC. MAHMOOD 1.y = Av = A= Iy = iLT = aLT = c/tf = d/tw = 54 402. = 406x178x54 UB = h= b= tw = tf = d= W pl.6 18670 4.0 SECTION CLASSIFICATION Grade of steel t= 10. UTM Made by Checked by CCH DR.6 177.y = W el.4 1051 927 32.36 131 8. Section chosen 1.2 Section Properties Mass Depth Width Web thickness Flange thickness Depth between fillets Plastic modulus Elastic modulus Shear area.0 1.1 DATA Trial Section L= 6 m Initial trial section is selected to give a suitable moment capacity. cm 4 cm cm cm 2.4 kg/m mm mm mm mm mm 3 cm cm cm 3 2 2 Area of section. The size is then checked to ensure suitability in all other aspects. fy = fu = = mm 275 430 S275 <= N/mm N/mm 2 (Fe 430) 40mm 2 . Second moment of area.9 68.9 Therefore. L = 6. L = 6.2 Flange is Class 1 element Class 1 limit : c/tf = 9.15 <= 9. flange subject to compression only : c/tf = 8.y fy / γMO = 1051 x 275 x 0. JOHOR Client: Job Title: Subject: Page 1004 3 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.001 γMO = 1.2 (b) Web.48 kN VSd < Vpl.109 Job No: UTM 81310 SKUDAI.Rd = W pl.001 / 1.05 √3 = 497. UTM Made by Checked by CCH DR.26 kNm MSd Mc.Rd < Moment capacity is adequate .48 = 298.49 kN VSd 0.Rd Sufficient shear resistance 4. subject to bending (neutral axis at mid depth) : d/tw = 47.05 = 32.0m) STC. it is low shear Mc.0 MOMENT RESISTANCE MSd = 268.Rd = 0.0 SHEAR RESISTANCE VSd = 179. MAHMOOD Classification of Trial Section (a) Outstand element of compression flange.05 = 275.5Vpl.4 > 46. Rd = Av ⎛ f y ⎞ ⎟ ⎜ γ MO ⎜ 3 ⎟ ⎠ ⎝ x 0.5 x 497.28 kN V pl.7 Web is Class 2 element 406x178x54 UB is a Class 2 section Class 1 limit : d/tw = 46.9 x 100 275 1.92 kNm 0.7 3.Rd < Therefore.5Vpl. ss = Stiff bearing at midspan. not susceptible to LTB 6. tw fyw (ss + sy) Ry. Ed × ⎢1 − ⎜ f yf ⎢ ⎜ ⎣ ⎝ ⎞ ⎟ ⎟ ⎠ 2 ⎤ ⎥ ⎥ ⎦ 0. L = 6.8 < 63.05 2 N/mm fyf = 275 sy = 52.110 Job No: UTM 81310 SKUDAI. JOHOR Client: Job Title: Subject: Page 1004 4 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.4 63.001 / 1.28 kN < Ry.0m) STC.6 x 275 x 0.5 Ry.4 kN = VSd = 179.0 RESISTANCE OF WEB TO TRANSVERSE FORCES Stiff bearing at support.69 mm ⎡ ⎛γ σ MO f .5 σf.69) x 7. ss = 50 75 mm mm 7. MAHMOOD 5.Rd = (50 + 52.1 Crushing Resistance Design crushing resistance.0 SHEAR BUCKLING For steel grade S275 (Fe 430). shear buckling must be checked if d/tw d/tw = > 47.Rd Sufficient crushing resistance . ⎛ bf sy = t f ⎜ ⎜t ⎝ w ⎞ ⎟ ⎟ ⎠ 0.0 LATERAL TORSIONAL BUCKLING (LTB) Beam is fully restrained.Rd = γM1 At support.5 ⎛ f yf ⎞ ⎟ ×⎜ ⎜f ⎟ ⎝ yw ⎠ 0.8 Shear buckling check is NOT required 7.05 204. UTM Made by Checked by CCH DR.Ed = Longitudinal stress in flange (My / I) = 0 at support (bending moment is zero) γMO = 1. 5 ⎛t + 3⎜ w ⎜t ⎝ f ⎞⎛ s s ⎟⎜ ⎟⎝ d ⎠ ⎞⎤ 1 ⎟⎥ ⎠⎥ γ M 1 ⎦ ss/d ≤ 50 / 360.3 ≤ = 1.5 ] .14 1. ⎛ bf s y = 2t f ⎜ ⎜t ⎝ w VSd = 0 ⎞ ⎟ ⎟ ⎠ 0.5 OK Buckling Resistance At support. JOHOR Client: Job Title: Subject: Page 1004 5 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams. Ra.4 = γM1 = E= Ra.Rd 268. UTM Made by Checked by CCH DR.5 ⎛ f yf ⎞ ⎟ ×⎜ ⎜f ⎟ ⎝ yw ⎠ 0.05 205 307.2 0.6 0 mm mm beff = 1 2 2 h + ss 2 [ ] 0.111 Job No: UTM 81310 SKUDAI. L = 6.8 VSd = kN/mm kN 179. Ed × ⎢1 − ⎜ ⎜ f yf ⎢ ⎝ ⎣ ⎞ ⎟ ⎟ ⎠ 2 ⎤ ⎥ ⎥ ⎦ 0.5 +a+ ss 2 but beff ≤ h 2 + s s [ 2 0. h= a= 402.5t w (Ef yw ) 2 0.26 7.2 Crippling Resistance Design crippling resistance At support. MSd Mc.5 ⎡⎛ t f ⎢⎜ ⎜ ⎢⎝ t w ⎣ ⎞ ⎟ ⎟ ⎠ 0. MAHMOOD At midspan.28 kN Sufficient crippling resistance 2 At mid span.0m) STC.5 Crushing resistance is OK 7.92 275.Rd = > 0.5 0.5 ⎡ ⎛γ σ MO f . Rd = 0.98 <= 1. Rd = 1 x 119.(118. UTM Made by Checked by CCH DR.8 x 7.118) = 119.0m) STC.5 x SQRT(402.5 d/t = 2. buckling about y-y axis.6 = 118. λ = 2.117) / (120 . Rb.6 l = 0.112 Job No: UTM 81310 SKUDAI.5 kN > At mid span. MAHMOOD beff = 0.6^2 + 50^2) + 0 + 50 / 2 = 227.7 mm Buckling resistance of web.8 x 1731.4 / 7.28 kN Sufficient buckling resistance Sufficient buckling resistance at midspan .6 .8 N/mm 2 Rb.28 x 0. VSd = 0 VSd = 179.28 mm Ends of web restrained against rotation and relative lateral movement.Rd = βA = βAf c A γM1 1 γM1 = 1.5 x 360.6 fc 121 117 fc = 121 .6 2 = 1731.001 / 1.8 mm <= [h + ss ] 2 2 0.5 = 405.05 = 197.75d Rolled I-section. use curve a λ √βA = λ √βA 118 120 118.05 A = beff x tw = 227.118) x (121 . JOHOR Client: Job Title: Subject: Page 1004 6 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams. L = 6. 113 Job No: UTM 81310 SKUDAI. JOHOR Client: Job Title: Subject: Page 1004 7 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.6 15 kN/m kN/m γG = γQ = 1.0m) STC. MAHMOOD 8.34 mm Recommended limiting vertical deflection for δmax is L 250 = δmax < = 6000 250 24 24 mm mm Deflection limit is satisfactory.0 SERVICEABILITY LIMIT (DEFLECTION) Partial factor for dead load Partial factor for imposed floor load Dead Imposed gd = qd = 27.0 1. . UTM Made by Checked by CCH DR.14 mm OK δmax = 11.46 mm mm < L / 350 = 17.88 + 6. L = 6.0 δ2 = Variation of deflection due to variable loading δ1 = Variation of deflection due to permanent loading δ0 = Pre-camber of beam in unloaded state = 0 δmax = δ1 + δ2 .46 = 18.88 6.δ0 Iy = E= δ= δ1 = δ2 = 18670 210 cm 4 2 kN/mm 4 5(gd / qd) x L 384 EI 11. 114 APPENDIX C1 . 6 978. MAHMOOD Grade = S275 Mass (kg/m) 23 30 37 46 52 60 71 73 86 89 97 107 118 129 132 137 153 158 167 177 198 202 235 240 283 287 340 393 467 551 634 Sx (cm3) 184.1 497.08 kNm M= Sx = M / fy = 63.115 Job No: UTM 81310 SKUDAI.4 568. JOHOR Client: Job Title: Subject: Page 1005 1 Braced Steel Frame Design (BS 5950-1 : 2000) Column Design (Internal Column.1 310.08 x 10^3 / 275 3 = 229.4 cm Try 203x203x60 UC .1 652 802. L = 5.8 1228 1589 1485 1953 2482 1875 2298 2964 2680 2417 3457 3436 3977 4689 4245 5101 5818 6994 8229 10009 12078 14247 Section 152x152x23 152x152x30 152x152x37 203x203x46 203x203x52 203x203x60 203x203x71 254x254x73 203x203x86 254x254x89 305x305x97 254x254x107 305x305x118 356x368x129 254x254x132 305x305x137 356x368x153 305x305x158 254x254x167 356x368x177 305x305x198 356x368x202 356x406x235 305x305x240 305x305x283 356x406x287 356x406x340 356x406x393 356x406x467 356x406x551 356x406x634 63.3 247.4 988.0m) STC. UTM Made by Checked by CCH DR. 96 5.2 Section Properties Mass Depth Width Web thickness Flange thickness Depth between fillets Plastic modulus Elastic modulus Radius of gyration.1 8. L = 5.2 = mm S275 < < < N/mm 2 16mm 40mm 63mm Therefore.6 205. Section chosen = 203x203x60 UC 1.52 kN L= 5 m 1. The size is then checked to ensure suitability in all other aspects.116 Job No: UTM 81310 SKUDAI. Gross area. Local buckling ratios: Flange Web = D= B= t= T= d= Sx = Zx = rx = ry = Ag = 60 209.0 DATA Fc = 1415.8 kg/m mm mm mm mm mm 3 cm cm cm cm cm 2 3 b/T = d/t = 7.1 Trial Section Initial trial section is selected to give a suitable moment capacity.3 14. MAHMOOD 1.8 652 581.2 160.23 17.0 SECTION CLASSIFICATION Grade of steel T= 14.2 9.0m) STC.19 75. UTM Made by Checked by CCH DR. py = 275 ε = √ (275/py) = SQRT(275/275) = 1 .3 2. JOHOR Client: Job Title: Subject: Page 1005 2 Braced Steel Frame Design (BS 5950-1 : 2000) Column Design (Internal Column. 3 80ε 1+r1 100ε 1+1. MAHMOOD Outstand element of compression flange.or H-section under axial compression and bending ("generally" case) r1 = Fc dtpy = 1415.23 < < = < 10ε = 15ε = 9 9 10 15 Flange is plastic Class 1 Web of I.8 x 9. L = 5. UTM Made by Checked by CCH DR.1 SLENDERNESS Effective Length About the x-x axis. "Restrained in direction at one end" LEX = 0.117 Job No: UTM 81310 SKUDAI.0 COMPRESSION RESISTANCE Fc = 1415.44 r1 = 1 Actual d/t = < 17.5r1 = 40 All ≥ 40ε < Section is : = 40 Web is plastic Class 1 Class 1 plastic section 3.52 x 1000 / (160.3 x 275) -1 < r1 ≤ 1 = 3.0 3.0m) STC. JOHOR Client: Job Title: Subject: Page 1005 3 Braced Steel Frame Design (BS 5950-1 : 2000) Column Design (Internal Column.85L = 0.4 4.52 kN Pc = pcAg py = Ag = 275 75. Limiting b/T = 9ε Actual b/T = 7.8 N/mm cm 2 2 Buckling about x-x axis .96 x 10) = 47.85 x 5 x 1000 = 4250 mm λx = LEX / rx = 4250 / (8. (47. beam reaction.46) / (48 .0m) STC. MAHMOOD Use strut curve (b) λx = λ 46 48 Interpolation: pcx = 242 .5. Therefore. R is assumed to act 100mm off the face of the column. UTM Made by Checked by CCH DR. the compressive resistance is adequate 5.239) 2 = 239.4 pc 242 239 Therefore. the moment will be equally divided. in proportion to the bending stiffness of each length.118 Job No: UTM 81310 SKUDAI.9 x 75.9 N/mm Pc = pcAg = 239.001 = 1818.54 kNm . L = 5.08 kNm 100 mm Moments are distributed between the column lengths above and below level 2. For EI/L1 : EI/L2 < 1.46) x (242 . JOHOR Client: Job Title: Subject: Page 1005 4 Braced Steel Frame Design (BS 5950-1 : 2000) Column Design (Internal Column.8 x 100 x 0.44 kN Fc < Pc 47.4 .0 NOMINAL MOMENT DUE TO ECCENTRICITY For columns in simple construction. R From frame analysis sheets. Mi = 63. M= 31. 19 x 10) = 48.44 + 31.03 kNm 1415. JOHOR Client: Job Title: Subject: Page 1005 5 Braced Steel Frame Design (BS 5950-1 : 2000) Column Design (Internal Column.78 N/mm Mb = pbSx = 260.17 py = λLT 45 50 275 pb 250 233 N/mm 2 pb = 250 .96 1.17 .78 x 652 x 0.119 Job No: UTM 81310 SKUDAI.45) / (50 .0 4.0m) STC.45) x (233 .03 = < 0. 7.001 = 170. L = 5.0 COMBINED AXIAL FORCE AND MOMENT CHECK The column should satisfy the relationship My Fc Mx + + ≤1 Pc M bs pyZ y λLT = 0.00 The combined resistance against axial force and moment is adequate.(48.5 x 5 x 1000) / (5.0 6. MAHMOOD 6.0 CONCLUSION Compression Resistance = Combined Axial Force and Moment Check = Use of the section is adequate Use : 203x203x60 UC OK OK . UTM Made by Checked by CCH DR.250) 2 = 260.52 1818.5 L/ry = (0.54 170. 120 APPENDIX C2 . 5 cm Try 254x254x73 UC . UTM Made by Checked by CCH DR. MAHMOOD Grade = S275 Mass (kg/m) 23 30 37 46 52 60 71 73 86 89 97 107 118 129 132 137 153 158 167 177 198 202 235 240 283 287 340 393 467 551 634 Wpl. JOHOR Client: Job Title: Subject: Page 1006 1 Braced Steel Frame Design (EC 3) Column Design (Internal Column.88 kNm M= W pl.88 x 10^3 / 275 3 = 210. L = 5.121 Job No: UTM 81310 SKUDAI.y (cm3) 184 248 309 497 567 654 801 990 979 1225 1589 1484 1952 2485 1872 2293 2970 2675 2418 3455 3438 3978 4691 4243 5101 5814 6997 8225 10010 12080 14240 Section 152x152x23 152x152x30 152x152x37 203x203x46 203x203x52 203x203x60 203x203x71 254x254x73 203x203x86 254x254x89 305x305x97 254x254x107 305x305x118 356x368x129 254x254x132 305x305x137 356x368x153 305x305x158 254x254x167 356x368x177 305x305x198 356x368x202 356x406x235 305x305x240 305x305x283 356x406x287 356x406x340 356x406x393 356x406x467 356x406x551 356x406x634 57.0m) STC.y = M / fy = 57. L = 5. Section chosen = 254x254x73 UC 1.y = W el.122 Job No: UTM 81310 SKUDAI. UTM Made by Checked by CCH DR.1 6. JOHOR Client: Job Title: Subject: Page 1006 2 Braced Steel Frame Design (EC 3) Column Design (Internal Column.9 11370 6.5 8.6 14. MAHMOOD 1. = h= b= tw = tf = d= W pl.2 200.2 Section Properties Mass Depth Width Web thickness Flange thickness Depth between fillets Plastic modulus Elastic modulus Radius of gyration.2 990 895 11. Second moment of area.2 Therefore.94 kNm L= 5 m 1.86 98. fy = fu = = mm 275 430 S275 <= N/mm N/mm 2 (Fe 430) 40mm 2 .08 kN Msd = 28.3 kg/m mm mm mm mm mm 3 cm cm cm cm cm 4 cm cm cm 2 3 2.1 Trial Section Initial trial section is selected to give a suitable moment capacity.46 92.94 23.y = iy = iz = A= Iy = iLT = aLT = c/tf = d/tw = 73 254 254 8.0 DATA NSd = 1351. The size is then checked to ensure suitability in all other aspects. Area of section.0m) STC.0 SECTION CLASSIFICATION Grade of steel tf = 14. 2 Flange is Class 1 element Class 1 limit : c/tf = 9.1 Mpl.0m) STC.Rd = Mpl.9 x 100 x 275 x 0.1 Class 3 = 38.Rd Mny.8 3. subject to bending and compression : Classify web as subject to compression and bending d/tw = 23.11 Mpl.05 = 2433.05 Npl.1 kN n = 1351.001 / 1.Rd = 1.5 Web is Class 1 element Therefore.2 Limit c/tf Class 2 = 10.1 n ≥ 0. MAHMOOD Classification of Trial Section (a) Outstand element of compression flange.0 CROSS-SECTION RESISTANCE n= NSd Npl. JOHOR Client: Job Title: Subject: Page 1006 3 Braced Steel Frame Design (EC 3) Column Design (Internal Column.1 28.3 kNm Mny. it is Class 1 section Class 1 limit : d/tw = 30.Rd A fy Npl.Rd = γMO γMO = 1.y.001 / 1.Rd = 92.Rd = 0.Rd(1-n) W pl. flange subject to compression only : c/tf = 8.05 = 259.2 Class 3 = 13.123 Job No: UTM 81310 SKUDAI.1 Mny.94 <= 9. L = 5.3 <= 30.555 >= n < 0.y.y.5 Limit d/tw Class 2 = 35.94 kNm kNm Sufficient moment resistance .9 (b) Web.y fy γMO = 990 x 275 x 0.1 = 0.Rd = > MSd = 128.08 / 2433. UTM Made by Checked by CCH DR. 3 0.1 x 10) = 38. MAHMOOD 4.y.5 x 28. JOHOR Client: Job Title: Subject: Page 1006 4 Braced Steel Frame Design (EC 3) Column Design (Internal Column.y.Rd = = 1351.05 = 2209.0m) STC. UTM Made by Checked by CCH DR.0 IN-PLANE FAILURE ABOUT MAJOR AXIS Members subject to axial compression and major axis bending must satisfy k y M y .y.Rd = 1 x 249.0 N b .(38. sufficient resistance against in-plane failure against major axis .7 x 92.3 <= fc 250 248 40mm fc = 250 .08 2209.85 x 5 x 1000 = 4250 mm Slenderness ratio λy = l y / iy = 4250 / (11.Sd ηMc.3 kN ky = 1.94 1 x 128.248) 2 = 249.001 / 1.95 (Conservative value) + kyMy.1 1 η= = + < γMO / γM1 1 Therefore.y.3 Buckling about y-y axis (Curve b) βA = λy√βA = tf λ√βA 38 40 1 38.38) / (250 . y . Rd η M c .124 Job No: UTM 81310 SKUDAI.Rd 1.5 NSd Nb. L = 5. Rd Nb.Rd = βA f c A γM1 l y = 0.9 x 100 x 0.3 . Sd N Sd + ≤ 1 .38) x (40 .7 N/mm Nb.85 L (Restrained about both axes) = 0. y . Use : 254x254x73 UC OK OK .0 4.0 CONCLUSION Cross Section Resistance In-plane Failure About Major Axis Use of the section is adequate.125 Job No: UTM 81310 SKUDAI.0m) STC.0 3. JOHOR Client: Job Title: Subject: Page 1006 5 Braced Steel Frame Design (EC 3) Column Design (Internal Column. MAHMOOD 5. L = 5. UTM Made by Checked by CCH DR. 126 APPENDIX D . y = M / fy = 268.92 kNm W pl.0m) Rev 1 STC.y (cm3) 1009 1195 1283 1213 1346 1442 1472 1509 1624 1659 1802 1832 2058 2020 2366 2234 2619 2887 2827 3287 3203 3673 4139 4575 5515 7462 178x102x19 254x102x22 203x102x23 203x133x25 254x102x25 305x102x25 254x102x28 305x102x28 203x133x30 254x146x31 305x102x33 356x127x33 254x146x37 305x127x37 356x127x39 406x140x39 305x165x40 305x127x42 254x146x43 356x171x45 305x165x46 406x140x46 305x127x48 356x171x51 457x152x52 305x165x54 406x178x54 356x171x57 406x178x60 457x152x60 356x171x67 406x178x67 457x152x67 457x191x67 406x178x74 457x152x74 457x191x74 457x152x82 457x191x82 533x210x82 457x191x89 533x210x92 457x191x98 533x210x101 610x229x101 533x210x109 610x229x113 533x210x122 610x229x125 610x229x140 610x305x149 610x305x179 610x305x238 M = 268.127 Job No: UTM 81310 SKUDAI. UTM Made by Checked by CCH DR.92 x 10^3 / 275 = 977. L = 6. MAHMOOD Grade = S275 Section Mass (kg/m) 19 22 23 25 25 25 28 28 30 31 33 33 37 37 39 39 40 42 43 45 46 46 48 51 52 54 54 Wpl.y (cm ) 171 260 232 259 307 336 354 408 313 395 481 539 485 540 654 718 626 612 568 773 722 889 706 895 1096 843 1051 3 Section Mass (kg/m) 57 60 60 67 67 67 67 74 74 74 82 82 82 89 92 98 101 101 109 113 122 125 140 149 179 238 Wpl. JOHOR Client: Job Title: Subject: Page 1004 1 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.9 cm3 Try 457x152x52 UB . 8 152.2 Section Properties Mass Depth Width Web thickness Flange thickness Depth between fillets Plastic modulus Elastic modulus Shear area.1 DATA Trial Section L= 6 m Initial trial section is selected to give a suitable moment capacity.6 21370 3. The size is then checked to ensure suitability in all other aspects.6 kg/m mm mm mm mm mm 3 cm cm cm 3 2 2 Area of section.6 10.y = Av = A= Iy = iLT = aLT = c/tf = d/tw = 52 449.0 SECTION CLASSIFICATION Grade of steel t= 10. Second moment of area.9 Therefore. cm 4 cm cm cm 2. UTM Made by Checked by CCH DR.5 66.0 1.128 Job No: UTM 81310 SKUDAI.59 121 6.0m) Rev 1 STC. L = 6.6 1096 950 36.4 7. Section chosen 1.9 407.y = W el. = 457x152x52 UB = h= b= tw = tf = d= W pl. MAHMOOD 1.99 53. fy = fu = = mm 275 430 S275 <= N/mm N/mm 2 (Fe 430) 40mm 2 . JOHOR Client: Job Title: Subject: Page 1004 2 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams. Rd = W pl.0m) Rev 1 STC.0 SHEAR RESISTANCE VSd = 179.2 (b) Web.92 kN VSd < Vpl.7 3.99 <= 9.y fy / γMO = 1096 x 275 x 0. UTM Made by Checked by CCH DR.0 MOMENT RESISTANCE MSd = 268. subject to bending (neutral axis at mid depth) : d/tw = 53.5Vpl.2 Flange is Class 1 element Class 1 limit : c/tf = 9.92 = 331. it is low shear Mc.05 √3 = 551.92 kNm 0.Rd < Moment capacity is adequate .Rd Sufficient shear resistance 4.7 Web is Class 2 element 457x152x52 UB is a Class 2 section Class 1 limit : d/tw = 46.05 = 287.001 / 1.5Vpl.6 > 46.129 Job No: UTM 81310 SKUDAI.05 = 36. MAHMOOD Classification of Trial Section (a) Outstand element of compression flange.001 γMO = 1.5 x 551.28 kN V pl. flange subject to compression only : c/tf = 6.15 kN VSd 0. Rd = Av ⎛ f y ⎞ ⎟ ⎜ γ MO ⎜ 3 ⎟ ⎠ ⎝ x 0.05 kNm MSd Mc. L = 6.Rd = 0.Rd < Therefore. JOHOR Client: Job Title: Subject: Page 1004 3 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.5 x 100 275 1. MAHMOOD 5.05 2 N/mm fyf = 275 sy = 48.0m) Rev 1 STC.8 < 63. tw fyw (ss + sy) Ry.68 kN VSd = 179. UTM Made by Checked by CCH DR.6 x 275 x 0.81 mm ⎡ ⎛γ σ MO f .Rd = γM1 At support.001 / 1. JOHOR Client: Job Title: Subject: Page 1004 4 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.6 63.5 σf.0 RESISTANCE OF WEB TO TRANSVERSE FORCES Stiff bearing at support. L = 6.81) x 7. not susceptible to LTB 6. Ed × ⎢1 − ⎜ f yf ⎢ ⎜ ⎣ ⎝ ⎞ ⎟ ⎟ ⎠ 2 ⎤ ⎥ ⎥ ⎦ 0.5 Ry.Rd Sufficient crushing resistance .5 ⎛ f yf ⎞ ⎟ ×⎜ ⎜f ⎟ ⎝ yw ⎠ 0.Ed = Longitudinal stress in flange (My / I) = 0 at support (bending moment is zero) γMO = 1.8 Shear buckling check is NOT required 7.0 LATERAL TORSIONAL BUCKLING (LTB) Beam is fully restrained.130 Job No: UTM 81310 SKUDAI.0 SHEAR BUCKLING For steel grade S275 (Fe 430). ⎛ bf sy = t f ⎜ ⎜t ⎝ w ⎞ ⎟ ⎟ ⎠ 0.Rd = (50 + 48. ss = 50 75 mm mm 7.05 = 196.1 Crushing Resistance Design crushing resistance. shear buckling must be checked if d/tw d/tw = > 53.28 kN < Ry. ss = Stiff bearing at midspan. 5t w (Ef yw ) 2 0. MSd Mc.05 205 299.2 Crippling Resistance Design crippling resistance At support.3 ≤ = 1.5 0.5 Crushing resistance is OK 7. Rd = 0.5 +a+ ss 2 but beff ≤ h 2 + s s [ 2 0.5 ⎡⎛ t f ⎢⎜ ⎜ ⎢⎝ t w ⎣ ⎞ ⎟ ⎟ ⎠ 0. Ed × ⎢1 − ⎜ ⎜ f yf ⎢ ⎝ ⎣ ⎞ ⎟ ⎟ ⎠ 2 ⎤ ⎥ ⎥ ⎦ 0. L = 6.0m) Rev 1 STC.5 ⎛ f yf ⎞ ⎟ ×⎜ ⎜f ⎟ ⎝ yw ⎠ 0.92 287. UTM Made by Checked by CCH DR.05 7. ⎛ bf s y = 2t f ⎜ ⎜t ⎝ w VSd = 0 ⎞ ⎟ ⎟ ⎠ 0.131 Job No: UTM 81310 SKUDAI.5 OK Buckling Resistance At support.Rd = > 0.6 = γM1 = E= Ra.28 kN Sufficient crippling resistance 2 At mid span.Rd 268.94 <= 1. Ra.5 ⎛t + 3⎜ w ⎜t ⎝ f ⎞⎛ s s ⎟⎜ ⎟⎝ d ⎠ ⎞⎤ 1 ⎟⎥ ⎠⎥ γ M 1 ⎦ ss/d ≤ 50 / 407.8 0 mm mm beff = 1 2 2 h + ss 2 [ ] 0.5 ⎡ ⎛γ σ MO f . MAHMOOD At midspan.16 VSd = kN/mm kN 179.5 ] . JOHOR Client: Job Title: Subject: Page 1004 5 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.2 0.12 1. h= a= 449. 3 mm <= [h + ss ] 2 2 0.1 fc 103 98 fc = 103 . MAHMOOD beff = 0.3 x 7. buckling about y-y axis.130) x (103 .5 x 407. use curve a λ √βA = λ √βA 130 135 134. UTM Made by Checked by CCH DR.1 l = 0.5 d/t = 2.8^2 + 50^2) + 0 + 50 / 2 = 251.88 mm Ends of web restrained against rotation and relative lateral movement.9 x 1909.6 2 = 1909.5 x SQRT(449.0m) Rev 1 STC.6 / 7.132 Job No: UTM 81310 SKUDAI.5 = 452.6 mm Buckling resistance of web.05 A = beff x tw = 251. λ = 2. JOHOR Client: Job Title: Subject: Page 1004 6 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.(134.88 x 0.6 = 134.05 = 179.001 / 1.1 .75d Rolled I-section.9 kN > At mid span. VSd = 0 VSd = 179.Rd = βA = βAf c A γM1 1 γM1 = 1.Rd = 1 x 98. Rb.28 kN Sufficient buckling resistance Sufficient buckling resistance at midspan . L = 6.98) / (135 .130) = 98.9 N/mm 2 Rb. 0 SERVICEABILITY LIMIT (DEFLECTION) Partial factor for dead load Partial factor for imposed floor load Dead Imposed gd = qd = 27.133 Job No: UTM 81310 SKUDAI.21 mm Recommended limiting vertical deflection for δmax is L 250 = δmax < = 6000 250 24 24 mm mm Deflection limit is satisfactory. JOHOR Client: Job Title: Subject: Page 1004 7 Braced Steel Frame Design (EC 3) Beam Design (Floor Beams.0m) Rev 1 STC. UTM Made by Checked by CCH DR.5(gd / qd) x L 384 EI 7.14 mm OK δmax = 7.6 15 kN/m kN/m γG = γQ = 1.95 = 11. .0 1.26 + 3. MAHMOOD 8.26 3.δ0 Iy = E= δ= δ1 = δ2 = 21370 210 cm 4 2 kN/mm 4 3.95 mm mm < L / 350 = 17. L = 6.0 δ2 = Variation of deflection due to variable loading δ1 = Variation of deflection due to permanent loading δ0 = Pre-camber of beam in unloaded state = 0 δmax = δ1 + δ2 .