COMPARISON BETWEEN BS 5950: PART 1: 2000 & EUROCODE 3 FOR THE DESIGN OF MULTI-STOREY BRACED STEEL FRAME CHAN CHEE HAN

Transcription

1 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

2 iv ACKNOWLEDGEMENT First of all, I would like to express my appreciation to my thesis supervisor, PM. Dr. Ir. Mahmood Md. Tahir of the Faculty of Civil Engineering, Universiti Teknologi Malaysia, for his generous advice, patience and guidance during the duration of my study. I would also like to express my thankful appreciation to Dr. Mahmood s research students, Mr. Shek and Mr. Tan for their helpful guidance in the process of completing this study. Finally, I am most thankful to my parents and family for their support and encouragement given to me unconditionally in completing this task. Without the contribution of all those mentioned above, this work would not have been possible.

3 v ABSTRACT Reference to standard code is essential in the structural design of steel structures. The contents of the standard code generally cover comprehensive details of a design. These details include the basis and concept of design, specifications to be followed, design methods, safety factors, loading values and etc. 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: This study intends to testify the claim. This paper presents comparisons of findings on a series of two-bay, 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. Design worksheets are created for the design of structural beam and column. The design method by Eurocode 3 has reduced beam shear capacity by up to 4.06% and moment capacity by up to 6.43%. Meanwhile, structural column designed by Eurocode 3 has compression capacity of between 5.27% and 9.34% less than BS 5950: Part 1:2000 design. Eurocode 3 also reduced the deflection value due to unfactored imposed load of up to 3.63% in comparison with BS 5950: Part 1: However, serviceability limit states check governs the design of Eurocode 3 as permanent loads have to be considered in deflection check. Therefore, Eurocode 3 produced braced steel frames which consume 1.60% to 17.96% more steel weight than the ones designed with BS 5950: Part 1: However, with the application of partial strength connections, the percentage of difference had been reduced to the range of 0.11% to 10.95%.

4 vii TABLE OF CONTENTS CHAPTER TITLE PAGE THESIS TITLE DECLARATION DEDICATION ACKNOWLEDGEMENT ABSTRACT ABSTRAK TABLE OF CONTENTS LIST OF TABLES LIST OF FIGURES LIST OF APPENDICES LISTOF NOTATIONS i ii iii iv v vi vii xii xiii xiv xv I INTRODUCTION 1.1 Introduction Background of Project Objectives Scope of Project Report Layout 5

5 viii II LITERATURE REVIEW 2.1 Eurocode 3 (EC3) Background of Eurocode 3 (EC3) Scope of Eurocode 3: Part 1.1 (EC3) Design Concept of EC Application Rules of EC Ultimate Limit State Serviceability Limit State Actions of EC BS Background of BS Scope of BS Design Concept of BS Ultimate Limit States Serviceability Loading Design of Steel Beam According to BS Cross-sectional Classification Shear Capacity, P v Moment Capacity, M c Low Shear Moment Capacity High Shear Moment Capacity Moment Capacity of Web against Shear Buckling Web not Susceptible to Shear Buckling Web Susceptible to Shear Buckling Bearing Capacity of Web Unstiffened Web Stiffened Web Deflection Design of Steel Beam According to EC Cross-sectional Classification Shear Capacity, V pl.rd Moment Capacity, M c.rd 20

6 ix Low Shear Moment Capacity High Shear Moment Capacity Resistance of Web to Transverse Forces Crushing Resistance, R y.rd Crippling Resistance, R a.rd Buckling Resistance, R b.rd Deflection Design of Steel Column According to BS Column Subject to Compression Force Effective Length, L E Slenderness, Compression Resistance, P c Column Subject to Combined Moment and 25 Compression Force Cross-section Capacity Member Buckling Resistance Design of Steel Column According to EC Column Subject to Compression Force Buckling Length, l Slenderness, Compression Resistance, N c.rd Buckling Resistance, N b.rd Column Subject to Combined Moment and 29 Compression Force Cross-section Capacity Member Buckling Resistance Conclusion Structural Beam Structural Column 32 III METHODOLOGY 3.1 Introduction 34

7 x 3.2 Structural Analysis with Microsoft Excel Worksheets Beam and Column Design with Microsoft Excel 36 Worksheets 3.4 Structural Layout & Specifications Structural Layout Specifications Loadings Factor of Safety Categories Structural Analysis of Braced Frame Load Combination Shear Calculation Moment Calculation Structural Beam Design BS EC Structural Column Design BS EC 3 61 IV RESULTS & DISCUSSIONS 4.1 Structural Capacity Structural Beam Structural Column Deflection Economy of Design 75 V CONCLUSIONS 5.1 Structural Capacity Structural Beam 81

8 xi Structural Column Deflection Values Economy Recommendation for Future Studies 84 REFERENCES 85 APPENDIX A1 86 APPENDIX A2 93 APPENDIX B1 100 APPENDIX B2 106 APPENDIX C1 114 APPENDIX C2 120 APPENDIX D 126

9 xii LIST OF TABLES TABLE NO. TITLE PAGE 2.1 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) Resulting moment due to eccentricity of structural columns (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 Weight of steel frame designed by EC Total steel weight for the multi-storey braced frame design Percentage difference of steel weight (ton) between BS design and EC3 design 4.10 Weight of steel frame designed by EC3 (Semi-continuous) Total steel weight of the multi-storey braced frame design 79 (Revised) 4.12 Percentage difference of steel weight (ton) between BS design and EC3 design (Revised)

10 xiii LIST OF FIGURES FIGURE NO. TITLE PAGE 3.1 Schematic diagram of research methodology Floor plan view of the steel frame building Elevation view of the intermediate steel frame (a) Bending moment of beam for rigid construction (b) Bending moment of beam for semi-rigid construction (c) Bending moment of beam for simple construction 80

11 xiv LIST OF APPENDICES APPENDIX TITLE PAGE A1 Frame Analysis Based on BS A2 Frame Analysis Based on EC3 93 B1 Structural Beam Design Based on BS B2 Structural Beam Design Based on EC3 106 C1 Structural Column Design Based on BS C2 Structural Column Design Based on EC3 120 D Structural Beam Design Based on EC3 (Revised) 126

12 xv LIST OF NOTATIONS BS 5950: PART 1: 2000 EUROCODE 3 Axial load F N Sd Shear force F v V Sd Bending moment M M Sd Partial safety factor M0 Radius of gyration M1 - Major axis r x i y - Minor axis r y i z Depth between fillets d d Compressive strength p c f c Flexural strength p b f b Design strength p y f y Slenderness Web crippling resistance P crip R a.rd Web buckling resistance P w R b.rd Web crushing resistance - R y.rd Buckling moment resistance M bx M b.y.rd Moment resistance at major axis M cx M c.y.rd M pl.y.rd Shear resistance P v V pl.y.rd Depth D h Section area A g A Effective section area A eff A eff Shear area A v A v

13 xvi Plastic modulus - Major axis S x W pl.y - Minor axis S y W pl.z Elastic modulus - Major axis Z x W el.y - Minor axis Z y W el.z Flange b/t c/t f Web d/t d/t w Width of section B b Effective length L E l Flange thickness T t f Web thickness t t w

14 CHAPTER I INTRODUCTION 1.1 Introduction Structural design is a process of selecting the material type and conducting indepth calculation of a structure to fulfill its construction requirements. The main purpose of structural design is to produce a safe, economic and functional building. Structural design should also be an integration of art and science. It is a process of converting an architectural perspective into a practical and reasonable entity at construction site. In the structural design of steel structures, reference to standard code is essential. A standard code serves as a reference document with important guidance. The contents of the standard code generally cover comprehensive details of a design. These details include the basis and concept of design, specifications to be followed, design methods, safety factors, loading values and etc. In present days, many countries have published their own standard codes. These codes were a product of constant research and development, and past experiences of experts at respective fields. Meanwhile, countries or nations that do not publish their own standard codes will adopt a set of readily available code as the national reference. Several factors govern the type of code to be adopted, namely suitability of application of the code set in a country with respect to its culture, climate and national preferences; as well as the trading volume and diplomatic ties between these countries.

15 2 Like most of the other structural Eurocodes, Eurocode 3 has developed in stages. The earliest documents seeking to harmonize design rules between European countries were the various recommendations published by the European Convention for Constructional Steelwork, ECCS. From these, the initial draft Eurocode 3, published by the European Commission, were developed. This was followed by the various parts of a pre-standard code, ENV1993 (ENV stands for EuroNorm Vornorm) issued by Comité Européen de Normalisation (CEN) the European standardisation committee. These preliminary standards of ENV will be revised, amended in the light of any comments arising out of its use before being reissued as the EuroNorm standards (EN). As with other Europeans standards, Eurocodes will be used in public procurement specifications and to assess products for CE (Conformité Européen) mark. The establishment of Eurocode 3 will provide a common understanding regarding the structural steel design between owners, operators and users, designers, contractors and manufacturers of construction products among the European member countries. It is believed that Eurocode 3 is more comprehensive and better developed compared to national codes. Standardization of design code for structural steel in Malaysia is primarily based on the practice in Britain. Therefore, 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. Codes of practice provide detailed guidance and recommendations on design of structural elements. Buckling resistance and shear resistance are two major elements of structural steel design. Therefore, provision for these topics is covered in certain sections of the codes. The study on Eurocode 3 in this project will focus on the subject of moment and shear design.

16 3 1.2 Background of Project The arrival of Eurocode 3 calls for reconsideration of the approach to design. Design can be complex, for those who pursue economy of material, but it can be simplified for those pursuing speed and clarity. Many designers feel depressed when new codes are introduced (Charles, 2005). There are new formulae and new complications to master, even though there seems to be no benefit to the designer for the majority of his regular workload. The increasing complexity of codes arises due to several reasons; namely earlier design over-estimated strength in a few particular circumstances, causing safety issues; earlier design practice under-estimated strength in various circumstances affecting economy; and new forms of structure evolve and codes are expanded to include them. However, 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. Besides, this can be achieved if the designer is not too greedy in the pursuit of the least steel weight from the strength calculations. Finally, simple design is possible if the code requirements are presented in an easy-to-use format, such as the tables of buckling stresses in existing BS codes. The Steel Construction Institute (SCI), 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 Lacking analytical and calculative proof, this project is intended to testify the claim.

17 4 1.3 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. 2) To study on the effect of changing the steel grade from S275 to S355 in Eurocode 3. 3) To compare the economy aspect between the designs of both BS 5950: Part 1: 2000 and Eurocode 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. This structure is intended to serve as an office building. All the beam-column connections are to be assumed simple. The standard code used here will be Eurocode 3, hereafter referred to as EC3. A study on the basis and design concept of EC3 will be carried out. Comparison to other steel structural design code is made. The comparison will be made between the EC3 with BS 5950: Part 1: 2000, hereafter referred to as BS The multi-storey steel frame will be first analyzed by using Microsoft Excel worksheets to obtain the shear and moment values. Next, design spreadsheets will be created to calculate and design the structural members.

18 5 1.5 Report Layout The report will be divided into five main chapters. Chapter I presents an introduction to the study. Chapter II presents the literature review that discusses the design procedures and recommendations for steel frame design of the codes EC3 and BS Chapter III will be a summary of research methodology. Results and discussions are presented in Chapter IV. Meanwhile, conclusions and recommendations are presented in Chapter V.

19 CHAPTER II LITERATURE REVIEW 2.1 Eurocode 3 (EC3) Background of Eurocode 3 (EC3) European Code, or better known as Eurocode, was initiated by the Commission of European Communities as a standard structural design guide. It was intended to smooth the trading activities among the European countries. Eurocode is separated by the use of different construction materials. Eurocode 1 covers loading situations; Eurocode covers concrete construction; Eurocode 3 covers steel construction; while Eurocode 4 covers for composite construction Scope of Eurocode 3: Part 1.1 (EC3) EC3, Design of Steel Structures: Part 1.1 General rules and rules for buildings covers the general rules for designing all types of structural steel. It also covers specific rules for building structures. EC3 stresses the need for durability, serviceability and resistance of a structure. It also covers other construction aspects only if they are necessary for design. Principles and application rules are also clearly stated. Principles should be typed in Roman wordings. Application rules must be written in italic style. The use of local application rules are allowed only if they have similar principles as EC3

20 7 and their resistance, durability and serviceability design does not differ too much. EC3 stresses the need for durability, serviceability and resistance of structure (Taylor, 2001). It also covers other construction aspects only if they are necessary for design Design Concept of EC3 All designs are based on limit state design. EC3 covers two limit states, which are ultimate limit state and serviceability limit state. Partial safety factor is applied to loadings and design for durability. Safety factor values are recommended in EC3. Every European country using EC3 has different loading and material standard to accommodate safety limit that is set by respective countries Application Rules of EC3 A structure should be designed and constructed in such a way that: with acceptable probability, it will remain fit for the use for which it is required, having due regard to its intended life and its cost; and with appropriate degrees of reliability, it will sustain all actions and other influences likely to occur during execution and use and have adequate durability in relation to maintenance costs. It should also be designed in such a way that it will not be damaged by events like explosions, impact or consequences of human errors, to an extent disproportionate to the original cause. Potential damage should be limited or avoided by appropriate choice of one or more of the following criteria: Avoiding, eliminating or reducing the hazards which the structure is to sustain; selecting a structural form which has low sensitivity to the hazards considered; selecting a structural form and design that can survive adequately the accidental removal of an individual element; and tying the structure together.

21 Ultimate Limit State Ultimate limit states are those associated with collapse, or with other forms of structural failure which may endanger the safety of people. Partial or whole of structure will suffer from failure. This failure may be caused by excessive deformation, rupture, or loss of stability of the structure or any part of it, including supports and foundations, and loss of equilibrium of the structure or any part of it, considered as a rigid body Serviceability Limit State Serviceability limit states correspond to states beyond which specified service criteria are no longer met. It may require certain consideration, 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; and vibration, which causes discomfort to people, damage to the building or its contents, or which limits its functional effectiveness Actions of EC3 An action (F) is a force (load) applied to the structure in direct action, or an imposed deformation in indirect action; for example, temperature effects or settlement. Actions are classified by variation in time and by their spatial variation. In time variation classification, actions can be grouped into permanent actions (G), e.g. self-weight of structures, fittings, ancillaries and fixed equipment; variable actions (Q), e.g. imposed loads, wind loads or snow loads; and accidental loads (A), e.g. explosions or impact from vehicles. Meanwhile, in spatial variation classification, actions are defined as fixed actions, e.g. self-weight; and free actions, which result in different arrangements of actions, e.g. movable imposed loads, wind loads, snow loads.

22 9 2.2 BS Background of BS 5950 BS 5950 was prepared to supersede BS 5950: Part 1: 1990, which was withdrawn. Several clauses were technically updated for topics such as sway stability, avoidance of disproportionate collapse, local buckling, lateral-torsional buckling, shear resistance, members subject to combined axial force and bending moment, etc. Changes were due to structural safety, but offsetting potential reductions in economy was also one of the reasons. BS 5950 comprises of nine parts. Part 1 covers the code of practice for design of rolled and welded sections; Part 2 and 7 deal with specification for materials, fabrication and erected for rolled, welded sections and cold formed sections, sheeting respectively; Part 3 and Part 4 focus mainly on composite design and construction; Part 5 concerns design of cold formed thin gauge sections; Part 6 covers design for light gauge profiled steel sheeting; Part 8 comprises of code of practice for fire resistance design; and Part 9 covers the code of practice for stressed skin design Scope of BS 5950 Part 1 of BS 5950 provides recommendations for the design of structural steelwork using hot rolled steel sections, flats, plates, hot finished structural hollow sections and cold formed structural hollow sections. They are being used in buildings and allied structures not specifically covered by other standards.

23 Design Concept of BS 5950 There are several methods of design, namely simple design, continuous design, semi-continuous design, and experimental verification. The fundamental of the methods are different joints for different methods. Meanwhile, in the design for limiting states, BS 5950 covers two types of states ultimate limit states and serviceability limit states Ultimate Limit States Several elements are considered in ultimate limit states. They are: strength, inclusive of general yielding, rupture, buckling and mechanism formation; stability against overturning and sway sensitivity; fracture due to fatigue; and brittle fracture. Generally, in checking, the specified loads should be multiplied by the relevant partial factors f given in Table 2. The load carrying capacity of each member should be such that the factored loads will not cause failure Serviceability Limit States There are several elements to be considered in serviceability limit states Deflection, vibration, wind induced oscillation, and durability. Generally, serviceability loads should be taken as the unfactored specified values. In the case of combined imposed load and wind load, only 80% of the full specified values need to be considered when checking for serviceability. In the case of combined horizontal crane loads and wind load, only the greater effect needs to be considered when checking for serviceability.

24 Loading BS 5950 had identified and classified several loads that act on the structure. There are dead, imposed and wind loading; overhead traveling cranes; earth and groundwater loading. 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. Loading conditions during erection should be given particular attention. Where necessary, the settlement of supports should be taken into account as well. 2.3 Design of Steel Beam According to BS 5950 The design of simply supported steel beam covers all the elements stated below. 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 Cross-sectional Classification Cross-sections should be classified to determine whether local buckling influences their capacity, without calculating their local buckling resistance. 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. The elements of a cross-section are generally of constant thickness.

25 12 Generally, the complete cross-section should be classified according to the highest (least favourable) class of its compression elements. Alternatively, a crosssection may be classified with its compression flange and its web in different classes. Class 1 is known as plastic section. It is cross-section with plastic hinge rotation capacity. Class 1 section is used for plastic design as the plastic hinge rotation capacity enables moment redistribution within the structure. Class 2 is known as compact section. It enables plastic moment to take place. However, local buckling will bar any rotation at constant moment. Class 3 is known as semi-compact section. When this section is applied, the stress at the extreme compression fiber can reach design strength. However, the plastic moment capacity cannot be reached. Class 4 is known as slender section. Sections that do not meet the limits for class 3 semi-compact sections should be classified as class 4 slender. Cross-sections at this category should be given explicit allowance for the effects of local buckling Shear Capacity, P v The web of a section will sustain the shear in a structure. Shear capacity is normally checked at section part that sustains the maximum shear force, F v. Clause of BS 5950 states the shear force F v should not be greater than the shear capacity P v, given by: P v = 0.6p y A v

26 13 in which A v is the shear area. BS 5950 provides various formulas for different type of sections. p y is the design strength of steel and it depends on the thickness of the web Moment Capacity, M c At sectional parts that suffer from maximum moment, moment capacity of the section needs to be verified. There are two situations to be verified in the checking of moment capacity low shear moment capacity and high shear moment capacity Low Shear Moment Capacity This situation occurs when the maximum shear force F v does not exceed 60% of the shear capacity P v. Clause of BS 5950 states that: M c = p y S for class 1 plastic or class 2 compact cross-sections; M c = p y Z or alternatively M c = p y S eff for class 3 semi-compact sections; and M c = p y Z eff for class 4 slender cross-sections where S is the plastic modulus; S eff is the effective plastic modulus; Z is the section modulus; and Z eff is the effective section modulus.

27 High Shear Moment Capacity This situation occurs when the maximum shear force F v exceeds 60% of the shear capacity P v. Clause of BS 5950 states that: M c = p y (S S v ) < 1.2p y Z for class 1 plastic or class 2 compact cross-sections; M c = p y (Z S v /1.5) or alternatively M c = p y (S eff S v ) for class 3 semi-compact sections; and M c = p y (Z eff S v /1.5) for class 4 slender cross-sections in which S v is obtained from the following: - For sections with unequal flanges: S v = S S f, in which S f is the plastic modulus of the effective section excluding the shear area A v. - Otherwise: S v is the plastic modulus of the shear area A v. and is given by = [2(F v /P v ) 1] 2

28 Moment Capacity of Web against Shear Buckling Web not Susceptible to Shear Buckling Clause of BS 5950 states that, if the web depth-to-thickness d/t 62, it should be assumed not to be susceptible to shear buckling and the moment capacity of the cross-section should be determined using Web Susceptible to Shear Buckling Clause states that, if the web depth-to-thickness ratio d/t > 70 for a rolled section, or 62 for a welded section, it should be assumed to be susceptible to shear buckling. 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 F v buckling resistance, 0.6V w, where V w is the simple shear V w = dtq w where d = depth of the web; q w = shear buckling strength of the web; obtained from Table 21 BS 5950 t = web thickness b) High shear flanges only method If the applied shear F v > 0.6V w, but the web is designed for shear only, provided that the flanges are not class 4 slender, a conservative value M f for

29 16 the moment capacity may be obtained by assuming that the moment is resisted by the flanges alone, with each flange subject to a uniform stress not exceeding p yf, where p yf is the design strength of the compression flange. c) High shear General method If the applied shear F v > 0.6V w, provided that the applied moment does not exceed the low-shear moment capacity given in a), the web should be designed using Annex H.3 for the applied shear combined with any additional moment beyond the flanges-only moment capacity M f given by b) Bearing Capacity of Web Unstiffened Web Clause states that bearing stiffeners should be provided where the local compressive force F x applied through a flange by loads or reactions exceeds the bearing capacity P bw of the unstiffened web at the web-to-flange connection. It is given by: P bw = (b 1 + nk)tp yw in which, - except at the end of a member: n = 5 - at the end of a member: n = b e /k but n 5 and k is obtained as follows: - for a rolled I- or H-section: k = T + r - for a welded I- or H-section: k = T

30 17 where b 1 is the stiff bearing length; b e is the distance to the nearer end of the member from the end of the stiff bearing; p yw is the design strength of the web; r is the root radius; T is the flange thickness; and t is the web thickness Stiffened Web Bearing stiffeners should be designed for the applied force F x minus the bearing capacity P bw of the unstiffened web. The capacity P s of the stiffener should be obtained from: P s = A s.net p y in which A s.net is the net cross-sectional area of the stiffener, allowing for cope holes for welding. If the web and the stiffener have different design strengths, the smaller value should be used to calculate both the web capacity P bw and the stiffener capacity P s 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. Actual deflection is a deflection caused by unfactored live load. Suggested limits for calculated deflections are given in Table 8 of BS 5950.

31 Design of Steel Beam According to EC3 The design of simply supported steel beam covers all the elements stated below. Sectional size chosen should satisfy the criteria as stated below: (i) (ii) (iii) (iv) (v) Cross-sectional classification Shear capacity Moment capacity (Low shear or High shear) Bearing capacity of web a) Crushing resistance b) Crippling resistance c) Buckling resistance Deflection Cross-sectional Classification A beam section should firstly be classified to determine whether the chosen section will possibly suffer from initial local buckling. When the flange of the beam is relatively too thin, the beam will buckle during pre-mature stage. To avoid this, Clause 5.3 of EC3 provided limits on the outstand-to-thickness (c/t f ) for flange and depth-tothickness (d/t w ) in Table Beam sections are classified into 4 classes. Class 1 is known as plastic section. It is applicable for plastic design. This limit allows the formation of a plastic hinge with the rotation capacity required for plastic analysis. Class 2 is also known as compact section. This section can develop plastic moment resistance. However, plastic hinge is disallowed because local buckling will occur first. It has limited rotation capacity. It can also achieve rectangular stress block.

32 19 Class 3 is also known as semi-compact section. The stress block will be of triangle shape. Calculated stress in the extreme compression fibre of the steel member can reach its yield strength, but local buckling is liable to prevent development of the plastic moment resistance. Class 4 is known as slender section. Pre-mature buckling will occur before yield strength is achieved. The member will fail before it reaches design stress. It is necessary to make explicit allowances for the effects of local buckling when determining their moment resistance or compression resistance. Apart from that, the ratios of c/t f and d/t w will be the highest among all four classes Shear Capacity, V pl.rd The web of a section will sustain shear from the structure. Shear capacity will normally be checked at section that takes the maximum shear force, V sd. At each crosssection, the inequality should be satisfied: V sd V pl.rd where V pl.rd = A v (f y / 3) / MO A v is the shear area. f y is the steel yield strength and MO is partial safety factor as stated in Clause Shear buckling resistance should be verified when for an unstiffened web, the ratio of d/t w > 69 or d/t w > 30 k for a stiffened web. k is the buckling factor for shear, and = [235/f y ] 0,5

33 Moment Capacity, M c.rd Moment capacity should be verified at sections sustaining maximum moment. There are two situations to verify when checking moment capacity that is, low shear moment capacity and high shear moment capacity Low Shear Moment Capacity When maximum shear force, V sd is equal or less than the design resistance V pl.rd, the design moment resistance of a cross-section M c.rd may be determined as follows: Class 1 or 2 cross-sections: M c.rd = W pl f y / MO Class 3 cross-sections: M c.rd = W el f y / MO Class 4 cross-sections: M c.rd = W eff f y / M1 where W pl and W el the plastic modulus and elastic modulus respectively. For class 4 cross-sections, W eff is the elastic modulus at effective shear area, as stated in Clause MO and M1 are partial safety factors High Shear Moment Capacity Clause states that, when maximum shear force, V sd exceeds 50% of the design resistance V pl.rd, the design moment resistance of a cross-section should be reduced to M V.Rd, the reduced design plastic resistance moment allowing for the shear

34 21 force. For cross-sections with equal flanges, bending about the major axis, it is obtained as follows: M V.Rd = (W pl A v 2 /4t w ) f y / MO but M V.Rd M c.rd where = (2V sd / V pl.rd 1) Resistance of Web to Transverse Forces The resistance of an unstiffened web to transverse forces applied through a flange, is governed by one of the three modes of failure Crushing of the web close to the flange, accompanied by plastic deformation of the flange; crippling of the web in the form of localized buckling and crushing of the web close to the flange, accompanied by plastic deformation of the flange; and buckling of the web over most of the depth of the member. However, if shear force acts directly at web without acting through flange in the first place, this checking is unnecessary. This checking is intended to prevent the web from buckling under excessive compressive force Crushing Resistance, R y.rd Situation becomes critical when a point load is applied to the web. Thus, checking should be done at section subject to maximum shear force. Clause provides that the design crushing resistance, R y.rd of the web of an I, H or U section should be obtained from: R y.rd = (s s + s ) t w f w / M1 in which s is given by s = 2t f (b f / t w ) 0,5 (f yf / f yw ) 0,5 [1 ( f.ed / f yf ) 2 ] 0,5

35 22 but b f should not be taken as more than 25t f. f.ed is the longitudinal stress in the flange. f yf and f yw are yield strength of steel at flange and web respectively Crippling Resistance, R a.rd by: The design crippling resistance R a.rd of the web of an I, H or U section is given R a.sd = 0.5t w 2 (Ef yw ) 0,5 [(t f / t w ) 0,5 + 3(t w / t f )(s s / d)] / M1 where s s is the length of stiff bearing, and s s / d < 0,2. For member subject to bending moments, the following criteria should be satisfied: F sd R a.rd M sd M c.rd and F sd / R a.rd + M sd / M c.rd 1, Buckling Resistance, R b.rd The design buckling resistance R b.rd of the web of an I, H or U section should be obtained by considering the web as a virtual compression member with an effective b eff, obtained from b eff = [h 2 + s s 2 ] 0,5. R b.rd = ( A f y A) / M1

36 23 where A = 1 and buckling curve c is used at Table and Table 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. Actual deflection is a deflection caused by unfactored live load. Suggested limits for calculated deflections are given in Table 4.1 of EC Design of Steel Column According to BS 5950 The design of structural steel column is relatively easier than the design of structural steel beam. Column is a compressive member and it generally supports compressive point loads. Therefore, checking is normally conducted for capacity of steel column to compression only. This, however, applies only to non-moment sustaining column Column Subject to Compression Force Cross-sectional classification of structural steel column is identical as of the classification of structural steel beam. For a structural steel column subject to compression load only, the following criteria should be checked: (i) (ii) (iii) Effective length Slenderness Compression resistance

37 Effective Length, L E The effective length L E 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. Depending on the conditions of restraint in the relevant plate, column members that carry more than 90% of their reduced plastic moment capacity M r in the presence of axial force is assumed to be incapable of providing directional restraint. For continuous columns in multi-storey buildings of simple design, in accordance of Table 22, depending on the conditions of restraint in the relevant plane, directional restraint is based on connection stiffness and member stiffness Slenderness, The slenderness of a compression member is generally taken as its effective length L E divided by its radius of gyration r about the relevant axis. This concept is not applicable for battened struts, angle, channel, T-section struts, and back-to-back struts. = L E / r Compression Resistance, P c by: According to Clause 4.7.4, the compression resistance P c of a member is given P c = A g p c (for class 1 plastic, class 2 compact and class 3 semi-compact cross-sections)

38 25 P c = A eff p cs (for class 4 slender cross-section) where A eff is the effective cross-sectional area; A g is the gross cross-sectional area; p c the compressive strength obtained from Table 23 and Table 24; and p cs is the value of p c from Table 23 and Table 24 for a reduced slenderness of (A eff /A g ) 0.5, in which is based on the radius of gyration r of the gross cross-section 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 Cross-section Capacity Generally, for class 1 plastic, class 2 compact and class 3 semi-compact cross sections, the checking of cross-section capacity is as follows: Fc A p g y M M x cx M M y cy 1 where F c is the axial compression; A g is the gross cross-sectional area; p y is the design steel strength; M x is the moment about major axis; M cx is the moment capacity about major axis; M y is the moment about minor axis; and M cy is the moment capacity about minor axis.

39 Member Buckling Resistance In simple construction, the following stability check needs to be satisfied: F P c M M x bs p M y y Z y 1.0 where F is the axial force in column; P c the compression resistance of column; M x the maximum end moment on x-axis; M b the buckling resistance moment; p y the steel design strength; and Z y the elastic modulus. 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 Column Subject to Compression Force Cross-sectional classification of structural steel column is identical as of the classification of structural steel beam. For a structural steel column subject to compression load only, the following criteria should be checked: (i) (ii) (iii) (iv) Buckling length Slenderness Compression resistance Buckling resistance

40 Buckling Length, l The buckling length l of a compression member is dependant on the restraint condition at both ends. Clause states that, provided that both ends of a column are effectively held in position laterally, the buckling length l may be conservatively be taken as equal to its system length L. Alternatively, the buckling length l may be determined using informative of Annex E provided in EC Slenderness, The slenderness of a compression member is generally taken as its buckling length l divided by its radius of gyration i about the relevant axis, determined using the properties of the gross cross-section. = l / i For column resisting loads other than wind loads, the value of should not exceed 180, whereas for column resisting self-weight and wind loads only, the value of should not exceed Compression Resistance, N c.rd given by: According to Clause 5.4.4, the compression resistance N c.rd of a member is N c.rd = A f y / M0 (for class 1 plastic, class 2 compact and class 3 semi-compact crosssections)

41 28 N c.rd = A eff f y / M1 (for class 4 slender cross-section) The design value of the compressive force N Sd at each cross-section shall satisfy the following condition: N Sd N c.rd Buckling Resistance, N b.rd For compression members, Clause states that the design buckling resistance of a compression member should be taken as: N c.rd = A A f y / M1 where A = 1 for Class 1, 2 or 3 cross-sections; and A eff / A for Class 4 cross-sections. is the reduction factor for the relevant buckling mode. For hot rolled steel members with the types of cross-section commonly used for compression members, the relevant buckling mode is generally flexural buckling. The design value of the compressive force N Sd at each cross-section shall satisfy the following condition: N Sd N b.rd

42 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 Cross-section Capacity Generally, cross-section capacity depends on the types of cross-section and applied moment. Clause states that, for bi-axial bending the following approximate criterion may be used: M M y. Sd Ny. Rd M M z. Sd Nz. Rd 1 for Class 1 and 2 cross-sections N N Sd M M y. Sd M M z. Sd pl. Rd pl. y. Rd pl. z. Rd 1 for a conservative approximation where, for I and H sections, = 2; = 5n but 1, in which n = N sd / N pl.rd. N Af Sd yd W M el. y y. Sd f yd W M el. z z. Sd f yd 1 for Class 3 cross-sections A N eff Sd f yd M y. Sd W eff. y N f Sd yd e Ny M z. Sd W eff. z N f Sd yd e Nz 1 for Class 4 cross-sections where f yd = f y / M1 ; A eff is the effective area of the cross-section when subject to uniform compression; W eff is the effective section modulus of the cross-section when subject

43 30 only to moment about the relevant axis; and e N is the shift of the relevant centroidal axis when the cross-section is subject to uniform compression. However, for high shear (V Sd 0.5 V pl.rd ), Clause 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 )f y for the shear area, where = (2V Sd / V pl.rd 1) 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 N Sd and major axis moment M y.sd must satisfy the following condition: N N Sd k y. Sd b. y. Rd c. y. Rd y M M 1,0 where N b.y.rd is the design buckling resistance for major axis; M c.y.rd is the design moment resistance for major-axis bending, k y is the conservative value and taken as 1,5; and = M0 / M1 for Class 1, 2 or 3 cross-sections, but 1,0 for Class Conclusion This section summarizes the general steps to be taken when designing a structural member in simple construction.

44 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 CRITERIA EC3 Flange subject to compression Web subject to bending (Neutral axis at mid depth) = (275 / p y ) Cross-sectional Classification Class 1 Plastic Class 2 Compact Class 3 Semi-compact Class 1 Plastic Class 2 Compact Class 3 Semi-compact Flange subject to compression Web subject to bending (Neutral axis at mid depth) = (235 / f y ) 0,5 P v = 0.6p y A v A v = Dt 2.0 Shear Capacity V pl.rd = f y A v / ( 3 x M0 ) M0 = 1,05 A v from section table M c = p y S M c = p y Z M c = p y Z eff 3.0 Moment Capacity Class 1, 2 Class 3 Class 4 M c.rd = W pl f y / M0 M c.rd = W el f y / M0 M c.rd = W eff f y / M1 M0 = 1,05 M1 = 1, Bearing Capacity

45 32 P bw = (b 1 + nk)tp yw Smaller of R y.rd = (s s + s y ) t w f yw / M1 R a.rd = 0,5t 2 w (Ef yw ) 0,5 [(t f /t w ) 0,5 + 3(t w /t f )(s s /d)]/ M1 R b.rd = Af y A / M1 d/t Shear Buckling Resistance Ratio d/t w 69 L / Deflection Limit (Beam carrying plaster or other brittle finish) L / 350 N/A Limit (Total deflection) L / 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 CRITERIA EC3 Flange subject to compression Web (Combined axial load and bending) 80 / 1 + r / r Cross-sectional Classification Class 1 Plastic Class 2 Compact Class 3 Semi-compact Class 1 Plastic Class 2 Compact Flange subject to compression Web (Combined axial load and bending) 396 / (13 1) 456 / (13 1)

46 / 1 + 2r 2 r 1 = F c / dtp yw, -1 < r 1 1 r 2 = F c / A g p yw = (275 / p y ) 0.5 Class 3 Semi-compact 42 / (0,67 + 0,33 ) = 2 M0 a / f y 1 a = N Sd / A = 0,5(1 + M0 w / f y ) w = N Sd / dt w = (235 / f y ) 0,5 P c = A g p c P c = A eff p cs 2.0 Compression Resistance Class 1, 2, 3 Class 4 N c.rd = Af y / M0 M0 = 1,05 N c.rd = A eff f y / M1 M b = p b S x M b = p b Z x M b = p b Z x.eff 3.0 Moment Resistance Class 1, 2 Class 3 Class 4 M c.rd = W pl f y / M0 M c.rd = W el f y / M0 M c.rd = W eff f y / M1 M0 = 1,05 M1 = 1, Stability Check F P c M M x bs p M y y Z y 1.0 N N Sd k y. Sd b. y. Rd c. y. Rd y M M 1,0

47 CHAPTER III METHODOLOGY 3.1 Introduction As EC3 will eventually replace BS 5950 as the new code of practice, 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. The first step is to study and understand the cross-section classification for steel members as given in EC, analyzing the tables provided and the purpose of each clause stated in the code. At the same time, an understanding on the cross-section classification for BS 5950 is also carried out. Analysis, design and comparison works will follow subsequently. Beams and columns are designed for the maximum moment and shear force obtained from computer software analysis. Checking on several elements, such as shear capacity, moment capacity, bearing capacity, buckling capacity and deflection is carried out. Next, analysis on the difference between the results using two codes is done. Eventually, comparison of the results will lead to recognizing the difference in design approach for each code. Please refer to Figure 3.1 for the flowchart of the methodology of this study.

48 Structural Analysis with Microsoft Excel Worksheets The structural analysis of the building frame will be carried out by using Microsoft Excel worksheets. As the scope of this study is limited at simple construction, the use of advanced structural analysis software is not needed. Sections 3.4 to 3.8 discuss in detail all the specifications and necessary data for the analysis of the multi-storey braced frame. Different factors of safety with reference to BS 5950 and EC3 are defined respectively. Simple construction allows the connection of beam-to-column to be pinned jointed. Therefore, only beam shear forces will be transferred to the structural column. End moments are zero. Calculation of bending moment, M and shear force, V are based on simply-supported condition, that is M = wl 2 / 8 V = wl / 2 where w is the uniform distributed load and L the beam span. 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.

49 Beam and Column Design with Microsoft Excel Worksheets The design of beam and column is calculated with Microsoft Excel software. The Microsoft Excel software is used for its features that allow continual and repeated calculations using values calculated in every cell of the worksheet. Several trial and error calculations can be used to cut down on the calculation time needed as well as prevent calculation error. Furthermore, Microsoft Excel worksheets will show the calculation steps in a clear and fair manner. The method of design using BS 5950 will be based on the work example drawn by Heywood (2003). Meanwhile, the method of design using EC3 will be based on the work example drawn by Narayanan et. al. (1995). 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.

50 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=wL 2 /8) Design worksheet development using Microsoft Excel Beams and columns design Fail Phase 3 Checking (Shear, Moment, Combined) Pass Comparison between BS 5950 and EC3 END Figure 3.1: Schematic diagram of research methodology

51 Structural Layout & Specifications Structural Layout In order to make comparisons of the design of braced steel frame between BS : 2000 and Eurocode 3, a parametric study for the design of multi-storey braced frames is carried out. The number of storey of the frame is set at four (4). In plan view, the 4-storey frame consists of four (4) bays; in total, there will be three (3) numbers of 4-storey frames. 4 th storey is roof while the rest will serve as normal floors. Each of the frames longitudinal length is 6m. Two (2) lengths of bay width will be used in the analysis 6m and 9m respectively. The storey height will be 5m from ground floor to first floor; whereas for other floors (1 st to 2 nd, 2 nd to 3 rd, 3 rd to roof), the storey height will be 4m. Please refer to Figure 3.2 and Figure 3.3 for the illustrations of building plan view and elevation view respectively. The intermediate frame will be used as the one to be analysed and designed. 6m 6m 6/9m 6/9m Figure 3.2 : Floor plan view of the steel frame building.

52 39 4m 4m 4m 5m Figure 3.3 : Elevation view of the intermediate steel frame Specifications The designed steel frame structure is meant for office for general use. All the bays will be serving the same function. Meanwhile, flat roof system will be introduced to cater for some activities on roof top. All the roof bays will be used for general purposes. The main steel frame will consist of solely universal beam (UB) and universal column (UC). As this is a simple construction, all the beam-to-column connections are assumed to be pinned. Web cleats will be used as the connection method to create pinned connection. Top flange of beams are effectively restraint against lateral torsional buckling. Meanwhile, all the column-to-column connections are to be rigid.

53 40 Precast concrete flooring system will be introduced to this project. The type of precast flooring system to be used will be solid precast floor panel. Therefore, all floors will be of one-way slab. Consequently, each bay will contribute half of the load intensity to the intermediate frame. The steel frame is assumed to be laterally braced. Therefore, wind load (horizontal load) will not be considered in the design. Only gravitational loads will be considered in this project. 3.5 Loadings Section 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. Therefore, all the values of imposed loads of both BS 5950 and EC3 design will be based on BS For imposed roof load, section 6.2 (Flat roofs) states that, for a flat roof with access available for cleaning, repair and other general purposes, a uniform load intensity of 1.5kN/m 2 is appropriate. In this design, this value will be adopted. Meanwhile, Table 8 (Offices occupancy class) states that the intensity of distributed load of offices for general use will be 2.5kN/m 2. This value will be used as this frame model is meant for a general office usage. 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. For precast floor selfweight, precast solid floor panel of 100mm thick was selected for flat roof. Meanwhile, 125mm think floor panel will be used for other floors. Weight of concrete is given by 24kN/m 3. Multiplying the thickness of the slabs, the intensity of slab selfweight will be 2.4kN/m 2 and 3.0kN/m 2 respectively.

54 41 The finishes on the flat roof will be waterproofing membrane and decorative screed. For other floors, a selection of floor carpets and ceramic tiles will be used, depending on the interior designers intention. A general load intensity of 1.0kN/m 2 for finishes (superimposed dead load) on all floors will be assumed. Combining the superimposed dead load with selfweight, the total dead load intensity for roof and floor slabs are 3.4kN/m 2 and 4kN/m 2 respectively. 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. 3.6 Factor of Safety Section Buildings without cranes of BS 5950 states that, in the design of buildings not subject to loads from cranes, the principal combination of loads that should be taken into account will be load combination 1 Dead load and imposed gravity loads. Partial safety factors for loads, f should be taken as 1.4 for dead load, and 1.6 for imposed load. In EC3, permanent actions G include dead loads such as self-weight of structure, finishes and fittings. Meanwhile, variable actions Q include live loads such as imposed load. From Table 2.1, for normal design situations, partial safety factors, F for dead load, G is given by 1,35. Meanwhile, for imposed floor load, Q is given by 1,5. Partial safety factor for resistance of Class 1, 2 or 3 cross-section, M0, is given by 1,05. Partial safety factor for resistance of Class 4 cross-section, M1, is given by 1,05 as well. The factor M0 is used where the failure mode is plasticity or yielding. The

55 42 factor M1 is used where the failure mode is buckling including local buckling, which governs the resistance of a Class 4 (slender) cross-section. 3.7 Categories In this project, in order to justify the effect of design strength of a steel member on the strength of a steel member, two (2) types of steel grade will be used, namely S 275 (or Fe 430 as identified in EC3) and S 355 (or Fe 510 as identified in EC3). In BS 5950, design strength p y is decided by the thickness of the thickest element of the cross-section (for rolled sections). For steel grade S 275, p y is 275N/mm 2 for thickness less than or equal to 16mm and 265N/mm 2 for thickness larger but less than or equal to 40mm. For steel grade S 355, in the meantime, p y is 355N/mm 2 and 345N/mm 2 respectively for the same limits of thickness Material properties for hot rolled steel (C-EC3) limits thickness of flange to less than or equal to 40mm for nominal yield strength f y of 275N/mm 2 and larger but less than or equal to 100mm for f y of 255N/mm 2. Meanwhile, for Fe 510, f y is 355N/mm 2 and 335N/mm 2 respectively for the same thickness limits. 3.8 Structural Analysis of Braced Frame Load Combination This section describes the structural analysis of the steel frame. According to BS 5950, the load combination will be 1.4 times total dead load plus 1.6 times total imposed

56 43 load (1.4DL + 1.6LL). For the roof, the resultant load combination, w, will be 48kN/m. For all other floors, the w will be 62.64kN/m. According to EC3, the load combination will be 1.35 times total dead load plus 1.5 times total imposed load (1,35DL + 1,5LL). For the roof, the resultant load combination, w, will be 45.9kN/m. For all other floors, the w will be 59.76kN/m Shear Calculation This steel frame is pinned jointed at all beam-to-column supports. For simple construction, the shear, V at end connections is given by V = wl/2, where w is the resultant load combination and l is the bay width. Inputting the resultant load combinations into the formula, the resulting shear values of both bay widths and codes of design can be summarized in Table 3.1 below: Table 3.1 Resulting shear values of structural beams (kn) BS 5950 EC 3 Location Bay Width Bay Width 6m 9m 6m 9m Roof Other Floors From Table 4.1, there is a difference of approximately 4.5% between the analyses of both codes. This is solely due to the difference in partial safety factors. Clearly, BS 5950 results in higher value of shear. The next table, Table 3.2 will present the accumulating axial loads acting on the structural columns of the steel frame. This is done by summating the resultant shear

57 44 force from beam of each floor. Internal columns will sustain axial load two times higher than external columns of same floor level as they are connected to two beams. Floor Table 3.2 Accumulating axial load on structural columns (kn) BS 5950 EC 3 6m 9m 6m 9m Int. Ext. Int. Ext. Int. Ext. Int. Ext. Roof 3 rd rd 2 nd nd 1 st st - Ground Int. = Internal column Ext. = External column The accumulating axial loads based on the two codes vary approximately 4.5%, similar with the beam shear Moment Calculation For simple construction, since all the beam-to-column connections are pinned jointed, structural beam moment, M, can be calculated by using the formula M=wl 2 /8, where w is the resultant load combination and l is the bay width. Inputting the resultant load combinations into the formula, the resulting moment values of both bay widths and codes of design can be summarized in Table 3.3:

58 45 Table 3.3 Resulting moment values of structural beams (knm) BS 5950 EC 3 Location Bay Width Bay Width 6m 9m 6m 9m Roof Other Floors From Table 3.3, there is a difference of approximately 4.4% to 4.6% between the analyses of both codes. This is solely due to the difference in partial safety factors. Clearly, BS 5950 results in higher value of moment. Regardless of the width of the bay, the higher the load combination of a floor, the higher the difference percentage will be. For the moments of the structural columns, since this is simple construction, there will be no end moments being transferred from the structural beams. However, there will be a moment due to eccentricity of the resultant shear from the beams. In this project, the eccentricity of the resultant shear from the face of the structural column will be 100mm. Since this is only preliminary analysis as well, the depth of the column has not been decided yet. Therefore, in this case, initially, the depth (D for BS 5950 and h for EC 3) of a structural column is assumed to be 400mm. Subsequently, the eccentricity moment, M e, can be determined from the following formula: M e = V (e + D/2) = V (e + h/2) where V is resultant shear of structural beam (kn), e is the eccentricity of resultant shear from the face of column (m), D or h is the depth of column section (m).

59 46 V for external column can be easily obtained from shear calculation. However, for internal column, V should be obtained by deducting the factored combination of floor dead (DL) and imposed load (LL) with unfactored floor dead load. For BS 5950, V can be expressed as V = (1.4DL + 1.6LL) 1.0DL. For EC 3, V can be expressed as V = (1,35DL + 1,5LL) 1.0DL. Table 3.4 below summarizes the moment values due to eccentricity. The moments for floor columns will be evenly distributed as the ratio of EI 1 /L 1 and EI 2 /L 2 is less than 1.5. Table 3.4 Resulting moment due to eccentricity of structural columns (knm) BS 5950 EC 3 Floor 6m 9m 6m 9m Int. Ext. Int. Ext. Int. Ext. Int. Ext. Roof Other Floors These values of eccentricity moments will be useful for the estimation of initial size of a column member during structural design in later stage. 3.9 Structural Beam Design Structural beam design deals with all the relevant checking necessary in the design of a selected structural beam. In simple construction, two major checks that need to be done is shear and moment resistance at ultimate limit state. Next, serviceability check in the form of deflection check will need to be done.

60 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) BS 5950 In simple construction, necessary checks for ultimate limit state will be shear buckling, shear capacity, moment capacity and web bearing capacity. The shear and moment value for this particular floor beam is kN and kNm. From the section table for universal beam, the sections are rearranged in ascending form, first the mass (kg/m) and then the plastic modulus S x (cm 3 ). The moment will then be divided by the design strength p y to obtain an estimated minimum plastic modulus value necessary in the design. S x = M / p y = x 10 3 / 275 = 1025cm 3 From the rearranged table, UB section 457x152x60 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 UB chosen are as follows: Mass = 59.8kg/m; Depth, D = 454.6mm; Width, B = 152.9mm; Web thickness, t = 8.1mm; Flange thickness, T = 13.3mm; Depth between fillets, d = 407.6mm; Plastic modulus, S x = 1290cm 3 ; Elastic modulus, Z x = 1120cm 3 ; b/t = 6.99; d/t = = (275/p y ) = (275/275)

61 48 = 1.0 Sectional classification is based on Table 11 of BS Actual b/t = 5.75, which is smaller than 9 = 9.0. This is the limit for Class 1 plastic section. Therefore, flange is Class 1 plastic section. Meanwhile, actual d/t = For web of I-section, where neutral axis is at mid-depth, the limiting value for Class 1 plastic section is 80 = Actual d/t did not exceed Therefore, web is Class 1 plastic section. Since both flange and web are plastic, this section is Class 1 plastic section. Next, clause states that if the d/t ratio exceeds 70 for a rolled section, shear buckling resistance should be checked. Since actually d/t < 70.0 in this design, therefore, shear buckling needs not be checked. After clause is checked, section Shear capacity is checked. Shear capacity, P v = 0.6p y A v, where Av = td for a rolled I-section. A v = 8.1 x = mm 2 P v = 0.6 x 275 x x 10-3 = kN > F v = kN Therefore, shear capacity is adequate. Next, section Moment capacity, M c is checked. 0.6P v = 0.6 x = kN > F v Therefore, it is low shear. For class 1 plastic cross-section, M c = p y S x. M c = 275 x 1290 x 10-3

62 49 = kNm To avoid irreversible deformation under serviceability loads, M c should be limited to 1.2p y Z x. 1.2p y Z x = 1.2 x 275 x 1120 x 10-3 = 369.6kNm > M c, therefore, OK. M = kNm from analysis < M c = kNm Therefore, moment capacity is adequate. To prevent crushing of the web due to forces applied through a flange, section Bearing capacity of web is checked. If F v exceeds P bw, bearing capacity of web, bearing stiffener should be provided. P bw = (b 1 + nk)tp yw r = 10.2mm b 1 = t + 1.6r + 2T (Figure 13) = x x 13.3 = 51.02mm k = T + r = = 23.5mm At support, n = b e /k, b e = 0, n = 2 b 1 + nk = 98.02mm P bw = x 8.1 x 275 x 10-3 = kN > F v = kN

63 50 Therefore, the bearing capacity at support is adequate. After necessary ultimate limit state checks have been done, the serviceability limit state check (Section 2.5) should be conducted. This is done in the form of deflection check. Generally, the serviceability load should be taken as the unfactored specified value. Therefore, only unfactored imposed load shall be used to calculate the deflection. w = 15kN/m for floors. L = 6.0m E = 205kN/mm 2 I = 25500cm 4 The formula for calculating exact deflection,, is given by = 5wL 4 / 384EI = 5 x 15 x 6 4 x 10 5 / 384 x 205 x = 4.84mm Table 8 (Suggested limits for calculated deflections) suggests that for beams carrying plaster or other brittle finish), the vertical deflection limit should be L/360. In this case, lim = 6000 / 360 = 16.67mm > Therefore, the deflection is satisfactory. The section is adequate. This calculation is repeated for different sections to determine the suitable section which has the minimal mass per length. However, it should also satisfy all the required criteria in the ultimate limit state check.

64 51 This section satisfied all the required criteria in both ultimate and serviceability limit state check. Therefore, it is adequate to be used EC 3 In simple construction, necessary checks for ultimate limit state will be shear buckling, shear capacity, moment capacity, lateral torsional buckling, resistance of web to crushing, crippling and buckling. The shear and moment value for this particular floor beam is kN and kNm. From the section table for universal beam, the sections are rearranged in ascending form, first the mass (kg/m) and then the plastic modulus W pl.y (cm 3 ). The moment will then be divided by the design strength p y to obtain an estimated minimum plastic modulus value necessary in the design. W pl.y = M / p y = x 10 3 / 275 = 977.9cm 3 From the rearranged table, UB section 406x178x54 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 UB chosen are as follows: Mass = 54kg/m; Depth, h = 402,6mm; Width, b = 177,6mm; Web thickness, t w = 7,6mm; Flange thickness, t f = 10,9mm; Depth between fillets, d = 360,4mm; Plastic modulus, W pl.y = 1051cm 3 ; Elastic modulus, W el.y = 927cm 3 ; Shear area, A v = 32,9cm 2 ; Area of

65 52 section, A = 68,6cm 2 ; Second moment of area, I y = 18670cm 4 ; i LT = 4,36cm; a LT = 131cm; c/t f = 8,15; d/t w = 47,4. Before checks are done for ultimate limit states, section classification is a must. Based on Table 3.1, t f = 10,9mm. t f 40mm. For S275 (Fe 430), yield strength, f y = 275N/mm 2 and ultimate tensile strength, f u = 430N/mm 2. These values must be adopted as characteristic values in calculations. From Table 5.6(a), for outstand element of compression flange, flange subject to compression only, limiting c/t f ratio (c is half of b) is 9,2 for Class 1 elements. For web subject to bending, neutral axis at mid depth, limiting d/t w ratio is 66,6 for Class 1 elements. Actual c/t f = 8,15 Actual d/t w = 47,4 9,2. Flange is Class 1 element. 66,6. Web is Class 1 element. Therefore, UB section 406x178x54 is Class 1 section. Next, section Shear resistance of cross-section of beam is checked. The design value of shear force, V Sd from analysis at each cross-section should not exceed the design plastic shear resistance V pl.rd, that is V pl.rd = A v (f y / 3) / M0. V Sd = 179,28kN M0 = 1,05 V pl.rd = (32,9 x 100 x 275) / ( 3 x 1,05) = 497,48kN > 179,28kN Therefore, shear resistance is sufficient. 0,5V pl.rd = 0,5 x 497,48 = 298,49kN > V Sd = 179,28kN

66 53 Therefore, low shear. For low shear, section Moment resistance of cross-section with low shear the design value of moment M Sd must not exceed the design moment resistance of the cross-section M c.rd = W pl.y f y / M0 for Class 1 or Class 2 cross-section. M Sd = 268,92kNm M c.rd = 1051 x 275 x 10-3 / 1,05 = 275,26kNm > M Sd Therefore, the moment capacity is sufficient. The beam is fully restrained, not susceptible to lateral torsional buckling. Therefore, section Lateral-torsional buckling needs not be checked. Section Shear buckling requires that webs must have transverse stiffeners at the supports if d/t w is greater than 63,8 and 56,1 for steel grade Fe 430 and Fe 510 respectively. Actual d/t w = 47,4 < 63,8. Therefore, shear buckling check is not required. Section 5.6 Resistance of webs to transverse forces requires transverse stiffeners to be provided in any case that the design value V Sd applied through a flange to a web exceeds the smallest of the following Crushing resistance, R y.rd, crippling resistance, R a.rd and buckling resistance, R b.rd. For crushing resistance, R y.rd = (s s + s y ) t w f yw / M1 where at support, s y = t f (b f /t w ) 0,5 [f yf /f yw ] 0,5 [1 ( M0 f.ed/f yf ) 2 ] 0,5

67 54 At support, bending moment is zero. f.ed = 0. M0 = 1,05, s s = 50mm at support. f yf = 275N/mm 2. s y = 10,9 (177,6 / 7.6) 0,5 = 52.69mm R y.rd = ( ,69) x 7,6 x 275 x 10-3 / 1,05 = 204,4kN For crippling resistance, R a.rd = 0,5t w 2 (Ef yw ) 0,5 [(t f /t w ) 0,5 + 3(t w /t f ) (s s /d)] / M1 s s /d = 50 / 360,4 = 0,14 0,2. OK M1 = 1,05 E = 210kN/mm 2 R a.rd = 0,5 x 7,6 2 ( x 275) 0,5 [(10,9/7,6) 0,5 + 3(7,6/10,9)(0,14)] / 1,05 = 307,8kN For buckling resistance, R b.rd = A f c A / M1 A = b eff x t w b eff = 0,5[h 2 + s 2 s ] 0,5 + a + s s /2 = 0,5 [402, ] 0, /2 = 227,8mm b eff should be less than [h 2 + s 2 s ] 0,5 = 405,7mm. OK. A = 227,8 x 7,6 = 1731,28mm 2

68 55 A = 1 M1 = 1,05 For ends restrained against rotation and relative lateral movement (Table 5.29), = 2,5 d/t = 2,5 x 360,4 / 7.6 = 118,6 From Table 5.13 (rolled I-section), buckling about y-y axis, curve (a) is used. A = 118,6 A = 118, f c = 121N/mm 2 A = 120, f c = 117N/mm 2 By interpolation, f c = 119,8N/mm 2 R b.rd = 1 x 119,8 x 1731,28 x 10-3 / 1,05 = 197,5kN R a.rd = 307,8kN R y.rd = 204,4kN Minimum of the 3 values are 197,5kN, which is larger than V Sd = 179,28kN. Therefore, the web of the section can resist transverse forces. OK. After necessary ultimate limit state checks have been done, the serviceability limit state check (Section 4.2) should be conducted. This is done in the form of deflection check. Generally, the serviceability load should be taken as the unfactored specified value. From Figure 4.1, deflection should take into account deflection due to both permanent loads and imposed loads. max = (hogging 0 = 0 at unloaded state) w 1 = 27.6kN/m for floors. (Permanent load)

69 56 w 2 = 15kN/m for floors. (Imposed load) L = 6.0m E = 210kN/mm 2 I y = 18670cm 4 The formula for calculating exact deflection,, is given by = 5wL 4 / 384EI 1 = 5 x 27,6 x 6 4 x 10 5 / 384 x 210 x = 11,88mm 2 = 5 x 15 x 6 4 x 10 5 / 384 x 210 x = 6,46mm Table 4.1 (Recommended limiting values for vertical deflections) suggests that for floors and roofs supporting plaster or other brittle finish or non-flexible partitions, the vertical deflection limit should be L/350 for 2 and L/250 for max. In this case, lim. 2 = 6000 / 350 = 17,14mm > 2 lim. max = 6000 / 250 = 24mm > = 18,34mm Therefore, the deflection is satisfactory. The section is adequate. This calculation is repeated for different sections to determine the suitable section which has the minimal mass per length. However, it should also satisfy all the required criteria in the ultimate limit state check.

70 57 This section satisfied all the required criteria in both ultimate and serviceability limit state check. Therefore, it is adequate to be used 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 1 st floor (length 5m) of the steel frame with bay width 6m and of steel grade S 275 (Fe 430) 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 kN 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 S x (cm 3 ). The moment will then be divided by the design strength p y to obtain an estimated minimum plastic modulus value necessary in the design. S x = M / p y

71 58 = x 10 3 / 275 = 229.4cm 3 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, S x = 652cm 3 ; Elastic modulus, Z x = 581.1cm 3 ; Radius of gyration, r x = 8.96cm, r y = 5.19cm; Gross area, A g = 75.8cm 2 ; b/t = 7.23 (b = 0.5B); d/t = T < 16mm, therefore, p y = 275N/mm 2 = (275/p y ) = (275/275) = 1.0 Sectional classification is based on Table 11 of BS Actual b/t = 7.23, which is smaller than 9 = 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 = For web of I-section under axial compression and bending, the limiting value for Class 1 plastic section is 80 / 1 + r 1, where r 1 is given by r 1 = F c / dtp y. r 1 = x 10 3 / x 9.3 x 275 = 3.44 but -1 < r 1 1, therefore, r 1 = 1 Limiting d/t value = 80 x 1 / = 40

72 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 Slenderness and section Effective lengths, and from Table 22 (Restrained in direction at one end), the effective length, L E = 0.85L = 0.85 x 5000 = 4250mm. x = L Ex / r x = 4250 / 8.96 x 10 = 47.4 Next, based on section Compression resistance, for class 1 plastic section, compression resistance, P c = A g p c. p c is the compressive strength determined from Table 24. For buckling about x-x axis, T < 40mm, strut curve (b) is used. x = 46, p c = 242N/mm 2 x = 48, p c = 239N/mm 2 From interpolation, x = 47.4, p c = 239.9N/mm 2 P c = A g p c = 75.8 x 100 x x 10-3 = kN > F c = kN Therefore, compressive resistance is adequate.

73 60 Next, for columns in simple construction, the beam reaction, R, is assumed to be acting 100mm from the face of the column. From frame analysis, M i = 63.08kNm. The moment is distributed between the column lengths above and below 1 st floor, in proportion to the bending stiffness of each length. For EI / L 1st-2nd : EI / L ground-1st < 1.5, the moment will be equally divided. Therefore, M = 31.54kNm. Section Columns in simple structures, when only nominal moments are applied, the column should satisfy the relationship (F c / P c ) + (M x / M bs ) + (M y / p y Z y ) 1 M y = 0, therefore, M y / p y Z y = 0 Equivalent slenderness LT of column is given by LT = 0.5L / r y = 0.5 x 5000 / 5.19 x 10 = From Table 16 (Bending strength p b for rolled sections), LT = 45, p b = 250N/mm 2 LT = 50, p b = 233N/mm 2 From interpolation, LT = 48.17, p b = N/mm 2 M b = p b S x = x 652 x 10-3 = kNm

74 61 (F c / P c ) + (M x / M bs ) = / / = 0.96 < 1.0 Therefore, the combined resistance against axial force and eccentricity moment is adequate. This section satisfied all the required criteria in ultimate limit state check. Therefore, it is adequate to be used EC 3 In simple construction, apart from section classification, 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). The axial force and eccentricity moment value for this particular internal column are 1351,08kN and 57,88kNm 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 W pl.y (cm 3 ). The moment will then be divided by the design strength f y to obtain an estimated minimum plastic modulus value necessary in the design. W pl.y = M Sd / f y = 57,88 x 10 3 / 275 = 210,5cm 3 From the rearranged table, UC section 254x254x73 is chosen. This is selected to give a suitable moment capacity. The size will then be checked to ensure suitability in all other aspects.

75 62 From the section table, the properties of the UC chosen are as follows: Mass = 73kg/m; Depth, h = 254mm; Width, b = 254mm; Web thickness, t w = 8,6mm; Flange thickness, t f = 14,2mm; Depth between fillets, d = 200,2mm; Plastic modulus, W pl.y = 990cm 3 ; Elastic modulus, W el.y = 895cm 3 ; Radius of gyration, i y = 11,1cm, i z = 6,46cm; Area of section, A = 92,9cm 2 ; Shear area, A v = 25,6cm 2 ; Second moment of area, I y = 11370cm 4 ; i LT = 6,86cm; a LT = 98,5cm; c/t f = 8,94 (c = 0,5b); d/t w = 23,3. t f = 14,2mm < 40mm, therefore, f y = 275N/mm 2, f u = 430N/mm 2 Sectional classification is based on Table 5.6(a) of C-EC3 for Class 1 elements. Actual c/t f = 8,94. From this table, for outstand element of compression flange (flange subject to compression only), the limiting values of c/t f for Class 1 and 2 are 9,2 and 10,2 respectively. Actual c/t f = 8,94 < 9,2. Therefore, flange is Class 1 element. For web subject to bending and compression, the classification depends on the mean web stress, w. For symmetric I-section of Class 1 or 2, w = N Sd / dt w = 1351,08 x 10 3 / 200,2 x 8,6 = 784,73N/mm 2 Table 5.8 gives the limiting values of stress w for Class 1 and 2 cross-sections. From Table 5.8, with d/t w = 23,3, the web is Class 1. Since both flange and web are plastic, this section is Class 1 section. Next, section 5.6 Axially loaded members with moments will be checked. Beforehand, from, section 5.5.1,

76 63 V pl.rd = A v (f y / 3) / M0 = (25,6 x 10 2 x 275) x 10-3 / ( 3 x 1,05) = 387,1kN Maximum applied shear load (at top of column) is V max.sd = M y.sd / L = 57,88 x 10 3 / 5000 = 11,58kN 0,5V pl.rd > V max.sd Therefore, the section is subject to a low shear. From Table 5.27, n = N Sd / N pl.rd Reduced design plastic moment, allowing for axial force, M N.Rd is such that n < 0,1 : M Ny.Rd = M pl.y.rd n 0,1 : M Ny.Rd = 1,11 M pl.y.rd (1 n) N pl.rd = A f y / M0 = 92,9 x 10 2 x 275 x 10-3 / 1,05 = 2433,1kN n = 1351,08 / 2433,1 = 0,555 0,1 Therefore, M Ny.Rd = 1,11 M pl.y.rd (1 n) M pl.y.rd = W pl.y f y / M0 = 990 x 10-3 x 275 / 1,05 = 259,3kNm M Ny.Rd = 1,11 x 259,3 x (1 0,555)

77 64 = 128,1kNm > M Sd = 28,94kNm Therefore, the moment resistance is sufficient. Lastly, section Axial compression and major axis bending states that all members subject to axial compression N Sd and major axis moment M y.sd must satisfy the expression (N Sd / N b.y.rd ) + (k y M y.sd / M c.y.rd ) 1,0 L y = 0,85L = 0,85 x 5000 = 4250mm Slenderness ratio y = L y / i y = 4250 / 11,1 x 10 = 38,3 Based on Table 5.13 Selection of buckling curve for f c, for buckling about y-y axis, buckling curve (b) is used. t f A = 1 y A = 38,3 40mm y A = 38, f c = 250N/mm 2 y A = 40, f c = 248N/mm 2 From interpolation, y A = 38,3, f c = 249,7N/mm 2

78 65 N b.y.rd = A f c A / M1, M1 = 1,05 = 1 x 249,7 x 92,9 x 10 2 x 10-3 / 1,05 = 2209,3kN k y = interaction factor about yy axis = 1,5 (Conservative value) = M0 / M1 = 1 (N Sd / N b.y.rd ) + (k y M y.sd / M c.y.rd ) = (1351,08 / 2209,3) + (1,5 x 28,94 / 1 x 128,1) = 0,95 < 1,0 Therefore, the resistance against in-plane failure against major axis is sufficient. This section 254x254x73 UC satisfied all the required criteria in ultimate limit state check. Therefore, it is adequate to be used.

79 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 are arranged accordingly, namely structural capacity, deflection, and weight of steel. 4.1 Structural Capacity Structural capacity deals with shear and moment resistance of a particular section chosen. Here, structural capacity is sub-divided into beam and column Structural Beam UB sections ranging from 305x102x25 to 533x210x122 are being tabulated in ascending form. Shear capacity and moment capacity of each section are being calculated separately, based on steel grade S275 and S355. The results are shown in Table 4.1 for shear capacity and Table 5.2 for moment capacity. The results based on BS 5950 and EC3 calculation are compiled together to show the difference between each other.

80 67 Table 4.1 Shear capacity of structural beam UB SECTION S275 S355 BS 5950 EC 3 Difference % Diff. BS 5950 EC 3 Difference % Diff. (kn) (kn) (kn) (kn) (kn) (kn) 305x102x x102x x102x x127x x127x x127x x165x x165x x165x x127x x127x x171x x171x x171x x171x x140x x140x x178x x178x x178x x178x x152x x152x x152x x152x x152x x191x x191x x191x x191x x191x x210x x210x x210x

81 68 533x210x x210x The difference is based on deduction of shear capacity of EC3 from BS For steel grade S275, the difference percentage ranges from -3.57% to 4.06%. For steel grade S355, meanwhile, the difference percentage ranges from -2.69% to 4.06%. Negative value indicates that the shear capacity calculated from EC3 is higher than that from BS There are a few explanations to the variations. The shear capacity of a structural beam is given by P v = 0.6 p y A v (BS 5950) A v = Dt V pl.rd = (A v x f y ) / ( M0 x 3) (EC3) A v is obtained from section table. This value, however, varies with A v = Dt as suggested by BS Most of the values given are lesser than Dt value. Also, 1 / ( M0 x 3) 0.55, which is approximately 8.3% less than 0.6 as suggested by BS Therefore, these facts explain the reason why shear capacity of most of the sections designed by EC3 is lower than the one designed by BS Table 4.2 Moment capacity of structural beam UB SECTION S275 S355 BS 5950 EC 3 Difference % Diff. BS 5950 EC 3 Difference % Diff. (knm) (knm) (knm) (knm) (knm) (knm) 305x102x x102x x102x x127x x127x

82 69 305x127x x165x x165x x165x x127x x127x x171x x171x x171x x171x x140x x140x x178x x178x x178x x178x x152x x152x x152x x152x x152x x191x x191x x191x x191x x191x x210x x210x x210x x210x x210x The difference is based on deduction of moment capacity of EC3 from BS For steel grade S275, the difference percentage ranges from 0.58% to 6.03%. For steel grade S355, meanwhile, the difference percentage ranges from 1.41% to 6.43%. Positive value indicates that the moment capacity calculated from EC3 is lower than that from BS 5950.

83 70 There are a few explanations to the variations. The moment capacity of a structural beam is given by M c = p y S x (BS 5950) M c.rd = W pl.y f y / M0 (EC3) From EC3 equation, 1 / M This is approximately 5% less than 1.0 as suggested by BS Besides that, there are some variations between plastic modulus specified by BS 5950 section table and EC3 section table. For example, for a UB section 406x178x54, plastic modulus based on BS 5950 (S x ) and EC3 (W pl.y ) are 1060cm 3 and 1051cm 3 respectively. There is a variation of approximately 0.85%. Therefore, these facts explain the reason why moment capacity of most of the sections designed by EC3 is lower than the one designed by BS Structural Column In determining the structural capacity of a column, sectional classification tables Table 11 and Table of BS 5950 and EC3 respectively, are revised. For a column web subject to bending and compression, BS 5950 only provides a clearer guideline to the classification of Class 3 semi-compact section. Meanwhile, EC3 provides better guidelines to classify a section web, whether it is Class 1, Class 2 or Class 3 element. A study is conducted to determine independently compression and bending moment capacity of structural column with actual length of 5m. Table 4.3 shows the result and percentage difference of compression resistance while Table 4.4 shows the result and percentage difference of moment resistance.

84 71 Table 4.3 Compression resistance and percentage difference UC SECTION S275 S355 BS 5950 EC 3 Difference % Diff. BS 5950 EC 3 Difference % Diff. (kn) (kn) (kn) (kn) (kn) (kn) 152x152x x203x x203x x203x x203x x203x x254x x254x x254x x254x x254x x305x x305x x305x x305x x305x x305x x305x Table 4.4 Moment resistance and percentage difference UC SECTION S275 S355 BS 5950 EC 3 Difference % Diff. BS 5950 EC 3 Difference % Diff. (knm) (knm) (knm) (knm) (knm) (knm) 152x152x x203x x203x x203x x203x x203x x254x x254x x254x

85 72 254x254x x254x x305x x305x x305x x305x x305x x305x x305x Shear capacity designed by BS 5950 is overall higher than EC3 design by the range of % and % for steel grade S275 (Fe 430) and S355 (Fe 510) respectively. This is mainly due to the partial safety factor M1 of 1,05 imposed by EC3 in the design. Also, the compression strength f c determined from Table 5.14(a) of EC3 is less than the compression strength p c determined from Table 24 of BS Meanwhile, as the size of section increases, the difference percentage changes from % to 8.26% for S275 (Fe 460) and % 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, M b = p b S x p b depends on equivalent slenderness LT, which is also dependant on the member length. The bigger the member size, the higher the radius of gyration, r y is. Therefore, p b increases with the increase in member size. However, moment capacity based on EC3 design, M pl.y.rd = W pl.y f y / M0

86 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 while for EC3, this is symbolized as 2. Table 4.5 Deflection of floor beams due to imposed load UB SECTION L = 6.0m L = 9.0m BS 5950 EC 3 Difference % Diff. BS 5950 EC 3 Difference % Diff. (, mm) ( 2, mm) (mm) (, mm) ( 2, mm) (mm) 305x102x x102x x102x x127x x127x x127x x165x x165x x165x x127x x127x x171x x171x x171x x171x x140x x140x x178x x178x x178x x178x

87 74 457x152x x152x x152x x152x x152x x191x x191x x191x x191x x191x x210x x210x x210x x210x x210x From Table 4.5 above, for a floor beam of 6m long, subject to 15kN/m of unfactored imposed floor load, the difference percentage ranges from -0.22% to 3.61%. Meanwhile, for a floor beam of 9m long, the difference percentage ranges from -0.22% to 3.63%. This is basically same as the range of beam length 6m. It also indicates that deflection value calculated from BS 5950 is normally higher than that from EC3. The first explanation for this difference is the modulus of elasticity value, E. Section Other properties of BS 5950 states that E = 205kN/mm 2. Meanwhile, section Design values of material coefficients of C-EC3 states that E = 210kN/mm 2. Apart from that, there is also slight difference between second moment of area in both codes. For example, for a section 356x171x57, I x = 16000cm 4 from BS 5950 section table. Meanwhile, I y = 16060cm 4 from EC3 section table. The minor differences had created differences between the deflection values. However, the major difference between the deflection designs of these two codes is the total deflection, max, as required by EC3. Different from BS 5950, EC3 requires deflection due to permanent dead load to be included in the final design.

88 Economy of Design After all the roof beams, floor beams, 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 4.6 and Table 4.7 for BS 5950 and EC3 design respectively. To compare the economy of the design, the weight of steel will be used as a gauge. Table 4.6 Weight of steel frame designed by BS 5950 Model Frame Section Designation No Type Universal Beams Universal Columns Total Steel Weight Floor Roof External Internal (tonne) S Bay 457x152x60 406x140x46 To 2nd 203x203x46 203x203x Storey Storey 2nd - 4th 152x152x30 203x203x46 (6m span) Storey 2 2 Bay 533x210x92 533x210x82 To 2nd 203x203x52 203x203x Storey Storey 2nd - 4th 203x203x46 203x203x52 (9m span) Storey S Bay 406x140x46 406x140x39 To 2nd 152x152x37 203x203x Storey Storey 2nd - 4th 152x152x23 152x152x37 (6m span) Storey 4 2 Bay 533x210x82 457x191x67 To 2nd 203x203x46 203x203x Storey Storey 2nd - 4th 152x152x37 203x203x46 (9m span) Storey

89 76 Table 4.7 Weight of steel frame designed by EC3 Model Frame Section Designation No Type Universal Beams Universal Columns Total Steel Weight Floor Roof External Internal (tonne) S Bay 406x178x54 406x140x46 To 2nd 203x203x52 254x254x Storey Storey 2nd - 4th 152x152x37 203x203x46 (6m span) Storey 6 2 Bay 533x210x92 533x210x82 To 2nd 203x203x71 254x254x Storey Storey 2nd - 4th 203x203x46 203x203x71 (9m span) Storey S Bay 406x178x54 356x171x45 To 2nd 203x203x46 203x203x Storey Storey 2nd - 4th 152x152x30 203x203x46 (6m span) Storey 8 2 Bay 533x210x92 533x210x82 To 2nd 203x203x60 254x254x Storey Storey 2nd - 4th 203x203x46 203x203x60 (9m span) Storey Summary of the total steel weight for the multi-storey braced steel frame design is tabulated in Table 4.8. The saving percentage, meanwhile, is tabulated in Table 4.9. Table 4.8 Total steel weight for the multi-storey braced frame design Types of Frame Bay Width Steel Total Steel Weight (ton) (m) Grade BS 5950 EC3 2Bay 4Storey 6 S (Fe 430) S (Fe 510) 2Bay 4Storey 9 S (Fe 430) S355 (Fe 510)

90 77 Table 4.9 Percentage difference of steel weight (ton) between BS 5950 design and EC3 design Frame Bay Steel Total Steel Weight (ton) Width (m) Grade BS 5950 EC3 % 2Bay 4Storey 6 S (Fe 430) S (Fe 510) 2Bay 4Storey 9 S (Fe 430) S355 (Fe 510) As shown in Table 4.9, all frame types, beam spans and steel grade designed by using BS 5950 offer weight savings as compared with EC3. The percentage of saving offered by BS 5950 design ranges from 1.60% to 17.96%, depending on the steel grade. The percentage savings for braced steel frame with 9m span is higher than that one with 6m span. This is because deeper, larger hot-rolled section is required to provide adequate moment capacity and also stiffness against deflection. Regardless of bay width, the percentage savings by using BS 5950 are higher than EC3 for S355 steel grade with respect to S275 steel grade. This is because overall deflection was considered in EC3 design. Meanwhile, unaffected by the effect of imposed load deflection, BS 5950 design allowed lighter section. This resulted in higher percentage difference. Further check on the effect of deflection was done. This time, the connections of beam-to-column were assumed to be partial strength connection. Semi-continuous

91 78 frame is achieved in this condition. For uniformly distributed loading, the deflection value is given as: = wl 4 / 384EI For a span with connections having a partial strength less than 45%, the deflection coefficient, is treated as = 3.5. This is different from pinned joint in simple construction, where zero support stiffness corresponds to a value of = 5.0, which was used in the beam design. Please refer to Appendix D for a redesign work after the value had been revised and the section redesigned to withstand bending moment from analysis process. The renewed beam sections are tabulated in Table 4.10 shown. Columns remained the same as there was no change in the value of eccentricity moment and axial force. Table 4.10 Weight of steel frame designed by EC3 (Semi-continuous) Model Frame Section Designation (Semi-continous) No Type Universal Beams Universal Columns Total Steel Weight Floor Roof External Internal (tonne) S Bay 457x178x52 406x140x46 To 2nd 203x203x52 254x254x Storey Storey 2nd - 4th 152x152x37 203x203x46 (6m span) Storey 6 2 Bay 533x210x92 533x210x82 To 2nd 203x203x71 254x254x Storey Storey 2nd - 4th 203x203x46 203x203x71 (9m span) Storey S Bay 406x140x46 356x127x39 To 2nd 203x203x46 203x203x Storey Storey 2nd - 4th 152x152x30 203x203x46 (6m span) Storey 8 2 Bay 533x210x82 457x151x67 To 2nd 203x203x60 254x254x Storey Storey 2nd - 4th 203x203x46 203x203x60 (9m span) Storey

92 79 Summary of the total revised steel weight for the multi-storey braced steel frame design is tabulated in Table The saving percentage, meanwhile, is tabulated in Table Table 4.11 Total steel weight for the multi-storey braced frame design (Revised) Types of Frame Bay Width Steel Total Steel Weight (ton) (m) Grade BS 5950 EC3 (Semi-Cont) 2Bay 4Storey 6 S (Fe 430) S (Fe 510) 2Bay 4Storey 9 S (Fe 430) S355 (Fe 510) Table 4.12 Percentage difference of steel weight (ton) between BS 5950 design and EC3 design (Revised) Frame Bay Steel Total Steel Weight (ton) Width (m) Grade BS 5950 EC3 (Semi-Cont) % 2Bay 4Storey 6 S (Fe 430) S (Fe 510) 2Bay 4Storey 9 S (Fe 430) S355 (Fe 510)

93 80 From Table 4.12, it can be seen that there is an obvious reduction of steel weight required for the braced steel frame, if it is built semi-continuously. Even though EC3 design still consumed higher steel weight, the percentage of difference had been significantly reduced to the range of 0.11% to 10.95%. The effect of dead load on the deflection of beam had been gradually reduced. The greater difference for steel grade S355 indicated that deflection still plays a deciding role in EC3 design. However, as the connection stiffness becomes higher, the gap reduces. The ability of partial strength connection had enabled moment at mid span to be partially transferred to the supports (Figure 4.1(b)). Therefore, the sagging moment at mid span became less than that of simple construction (Figure 4.1(c)). Eventually, if rigid connection is introduced, with deflection coefficient set as = 1.0, the effect of deflection on the design will be eliminated. The moment capacity will be the deciding factor. Please refer to Figure 4.1(a) for the illustration of rigid connection. wl 2 /8 M R wl 2 /8 M R wl 2 /8 (a) (b) (c) Design moment, M D = wl 2 /8 M R Figure 4.1 Bending moment of beam for: (a) rigid construction; (b) semi-rigid construction; (c) simple construction.

94 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. In review to the research objectives, a summary on the results of the objectives is categorically discussed. Suggestions of further research work are also included in this chapter. 5.1 Structural Capacity Structural Beam For the shear capacity of a structural beam, calculation based on EC3 had reduced a members shear capacity of up to 4.06% with regard to BS 5950 due to the variance between constant values of the shear capacity formula specified by both codes. Apart from that, the difference between the approaches to obtain shear area, A v value also caused the difference. The application of different steel grade did not contribute greater percentage of difference between the shear capacities calculated by both codes. Meanwhile, for the moment capacity of structural beam, calculation based on EC3 had effectively reduced a members shear capacity of up to 6.43%. This is mainly due to the application of partial safety factor, M0 of 1,05 in the moment capacity

95 82 calculation required by EC3, as compared to the partial safety factor, M of 1.0 as suggested by BS With the inclusion of partial safety factor, it is obvious that EC3 stresses on the safety of a structural beam. The design of structural beam proposed by EC3 is concluded to be safer than that by BS Structural Column In simple construction, only moments due to eccentricity will be transferred to structural column. In comparison, axial compression is much more critical. Therefore, only compressive resistance comparison of structural column was made. A reduction in the range of 5.27% to 9.24% of column compressive resistance was achieved when designing by EC3, compared with BS This comparison is based on a structural column of 5.0m long. This is due to the implication of partial safety factor, M0 of 1,05 as required by EC3 design. Meanwhile, there is also a deviation in between the compressive strength, f c and p c respectively, of both codes. From interpolation, it was found that for a same value of, f c is smaller than p c. The steel frame is assumed to be laterally braced. Therefore, wind load (horizontal load) will not be considered in the design. Only gravitational loads will be considered in this project. 5.2 Deflection Values When subject to an unfactored imposed load, a structural beam will be subject to deflection. For the same value of unfactored imposed load, EC3 design created majority

96 83 lower deflection values with respect to BS 5950 design. The difference ranges from % to 3.63%. The main reason for the deviation is the difference in the specification of modulus of elasticity, E. BS 5950 specifies 205kN/mm 2 while EC3 specifies 210kN/mm 2. Higher E means the elasticity of a member is higher, thus can sustain higher load without deforming too much. However, serviceability limit states check governs the design of EC3 as permanent loads have to be considered in deflection check. Section of EC3 provided proof to this. Therefore, taking into account deflection due to permanent loads, the total deflection was greater. Cross-section with higher second moment of area value, I will have to be chosen, compared with the section chosen for BS 5950 design. 5.3 Economy Economy aspect in this study focused on the minimum steel weight that is needed in the construction of the braced steel frame. The total steel weight of structural beams and columns was accumulated for comparison. In this study, it was found that EC3 design produced braced steel frames that require higher steel weight than the ones designed with BS For a 2-bay, 4-storey, 6m bay width steel frame, the consumption of steel for S275 (Fe 430) and S355 (Fe 510) is tons and tons for BS 5950 design; and tons and tons for EC3 design. For a 2-bay, 4-storey, 9m bay width steel frame, the consumption of steel for S275 (Fe 430) and S355 (Fe 510) is tons and tons for BS 5950 design; and tons and tons for EC3 design.

97 84 The percentages of differences are as follow: (i) 2-bay, 4-storey, 6m bay width, S275 (Fe 430): 1.60% (ii) 2-bay, 4-storey, 6m bay width, S355 (Fe 510): 17.96% (iii) 2-bay, 4-storey, 9m bay width, S275 (Fe 430): 5.42% (iv) 2-bay, 4-storey, 9m bay width, S355 (Fe 510): 15.29% Further study was extended for the application of partial strength connection for beam-to-column connections in EC3 design. The reduction in deflection coefficient from 5.0 to 3.5 had successfully reduced the percentage of difference between the steel weights designed by both codes. The percentages of differences are as follow: (i) 2-bay, 4-storey, 6m bay width, S275 (Fe 430): 0.11% (ii) 2-bay, 4-storey, 6m bay width, S355 (Fe 510): 10.95% (iii) 2-bay, 4-storey, 9m bay width, S275 (Fe 430): 5.42% (iv) 2-bay, 4-storey, 9m bay width, S355 (Fe 510): 7.22% 5.4 Recommendation for Future Studies For future studies, it is suggested that an unbraced steel frame design is conducted to study the behavior, structural design and economic aspect based on both of the design codes. However, since the results of the third objective contradicted with the background of the study (claim by Steel Construction Institute), it is recommended that further studies to be conducted to focus on the economy aspect of EC3 with respect to BS This study showed that steel weight did not contribute to cost saving of EC3 design.

98 85 REFERENCES Charles King (2005). Steel Design Can be Simple Using EC3. New Steel Construction, Vol 13 No 4, Steel Construction Institute (SCI) (2005). EN 1993 Eurocode 3 Steel. Eurocodenews, November 2005, Issue 3, 4. Taylor J.C. (2001). EN1993 Eurocode 3: Design of Steel Structures. ICE Journal, Paper 2658, British Standards Institution (2001). British Standard Structural Use of Steelwork in Building: Part 1: Code of Practice for Design Rolled and Welded Sections. London: British Standards Institution. European Committee for Standardization (1992). Eurocode 3: Design of Steel Structures: Part 1.1 General Rules and Rules for Buildings. London: European Committee for Standardization. Heywood M. D. & Lim J B (2003). Steelwork design guide to BS :2000 Volume 2: Worked examples. Berkshire: Steel Construction Institute. Narayanan R et. al. (1995). Introduction to Concise Eurocode 3 (C-EC3) with Worked Examples. Berkshire: Steel Construction Institute.

99 APPENDIX A1 86

100 87 Job No: 1001 Page 1 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 1.0 DATA No. of Bay = 2 No. of Storey = 4 Frame Longitudinal Length = 6 m Bay Width, l = 6 m Storey Height = 5 m (First Floor) = 4 m (Other Floors) LOADING Roof Dead Load, DL = 4 kn/m 24 kn/m Live Load, LL = 1.5 kn/m 9 kn/m Floors Dead Load, DL = 4.6 kn/m 27.6 kn/m Live Load, LL = 2.5 kn/m 15 kn/m LOAD FACTORS Dead Load, DL = 1.4 Live Load, LL = 1.6 FACTORED LOAD Roof w = 1.4DL + 1.6LL w = 1.4 x x 9 = 48 kn/m Floors w = 1.4 x x 15 = kn/m

101 88 Job No: 1001 Page 2 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 2.0 FRAME LAYOUT 2.1 Selected Intermediate Frame 6m 6m 6 m 6 m 2.2 Precast Slab Panel Load Transfer to Intermediate Frame

102 89 Job No: 1001 Page 3 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 2.3 Cut Section of Intermediate Frame [4] 4m [3] 4m [2] 4m [1] 5m 3.0 LOAD LAYOUT 6 m 6 m 48 kn/m 48 kn/m kn/m kn/m kn/m kn/m kn/m kn/m

103 90 Job No: 1001 Page 4 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Frame Analysis SKUDAI, 4.0 LOAD CALCULATION JOHOR Client: STC, UTM Made by Frame bracing Laterally braced, horizontal load is not taken into account Checked by Beam restraint Top flange effectively restrained against lateral torsional buckling CCH DR. MAHMOOD 4.1 Beam Moment, M = wl 2 / 8 Shear, V = wl / 2 Roof beams, V = 48 x 6 / 2 = 144 kn M = 48 x 6^2 / 8 = 216 knm Floor beams, V = x 6 / 2 = kn M = x 6^2 / 8 = knm 4.2 Column Shear Column Shear (kn) Internal External [4] [3] [2] [1] Moment External column will be subjected to eccentricity moment, contributed by beam shear. Eccentricity = 100 mm from face of column. Universal column of depth 200 mm Internal column - Moments from left and right will cancel out each other.

104 91 Job No: 1001 Page 5 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 5.0 ANALYSIS SUMMARY Moment (knm) Shear (kn) (144) (144) 144 (187.92) 288 (187.92) 144 [1] (187.92) (187.92) [2] (187.92) (187.92) [3] [4]

105 92 Job No: 1001 Page 6 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD Column moment due to eccentricity (knm) [1] [2] [3] [4] Moments are calculated from (1.4DL+1.6LL) - 1.0DL Most critical condition

106 APPENDIX A2 93

107 94 Job No: 1002 Page 1 UTM Job Title: Braced Steel Frame Design (Eurocode 3) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 1.0 DATA No. of Bay = 2 No. of Storey = 4 Frame Longitudinal Length = 6 m Bay Width, l = 6 m Storey Height = 5 m (First Floor) = 4 m (Other Floors) LOADING Roof Dead Load, DL = 4 kn/m 24 kn/m Live Load, LL = 1.5 kn/m 9 kn/m Floors Dead Load, DL = 4.6 kn/m 27.6 kn/m Live Load, LL = 2.5 kn/m 15 kn/m LOAD FACTORS Dead Load, DL = 1.35 Live Load, LL = 1.5 FACTORED LOAD Roof w = 1.35DL + 1.5LL w = 1.35 x x 9 = 45.9 kn/m Floors w = 1.35 x x 15 = kn/m

108 95 Job No: 1002 Page 2 UTM Job Title: Braced Steel Frame Design (Eurocode 3) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 2.0 FRAME LAYOUT 2.1 Selected Intermediate Frame 6m 6m 6 m 6 m 2.2 Precast Slab Panel Load Transfer to Intermediate Frame

109 96 Job No: 1002 Page 3 UTM Job Title: Braced Steel Frame Design (Eurocode 3) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 2.3 Cut Section of Intermediate Frame [4] 4m [3] 4m [2] 4m [1] 5m 3.0 LOAD LAYOUT 6 m 6 m 45.9 kn/m 45.9 kn/m kn/m kn/m kn/m kn/m kn/m kn/m

110 97 Job No: 1002 Page 4 UTM Job Title: Braced Steel Frame Design (Eurocode 3) Subject: Frame Analysis SKUDAI, 4.0 LOAD CALCULATION JOHOR Client: STC, UTM Made by Frame bracing Laterally braced, horizontal load is not taken into account Checked by Beam restraint Top flange effectively restrained against lateral torsional buckling CCH DR. MAHMOOD 4.1 Beam Moment, M = wl 2 / 8 Shear, V = wl / 2 Roof beams, V = 45.9 x 6 / 2 = kn M = 45.9 x 6^2 / 8 = knm Floor beams, V = x 6 / 2 = kn M = x 6^2 / 8 = knm 4.2 Column Shear Column Shear (kn) Internal External [4] [3] [2] [1] Moment External column will be subjected to eccentricity moment, contributed by beam shear. Eccentricity = 100 mm from face of column. Universal column of depth 200 mm Internal column - Moments from left and right will cancel out each other.

111 98 Job No: 1002 Page 5 UTM Job Title: Braced Steel Frame Design (Eurocode 3) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 5.0 ANALYSIS SUMMARY 5.1 Moment (knm) Shear (kn) (137.7) (137.7) (179.28) (179.28) [1] (179.28) (179.28) [2] (179.28) (179.28) [3] [4]

112 99 Job No: 1002 Page 6 UTM Job Title: Braced Steel Frame Design (Eurocode 3) Subject: Frame Analysis SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 5.3 Column moment due to eccentricity (knm) Moments are calculated from (1.35DL+1.5LL) - 1.0DL Most critical condition

113 APPENDIX B1 100

114 101 Job No: 1003 Page 1 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD Grade = S275 Section 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 Mass S x Section Mass S x (kg/m) (cm 3 ) (kg/m) (cm 3 ) x171x x152x x178x x171x x178x x191x x152x x178x x152x x191x x191x x152x x210x x191x x210x x191x x210x x229x x210x x229x x210x x229x x229x x305x x305x x305x M = knm S x = M / f y = x 10^3 / 275 = 1025 cm 3 Try 457x152x60 UB

115 102 Job No: 1003 Page 2 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 1.0 DATA 1.1 Trial Section Initial trial section is selected to give a suitable moment capacity. The size is then checked to ensure suitability in all other aspects. Section chosen = 457x152x60 UB 1.2 Section Properties Mass = 59.8 kg/m Depth D = mm Width B = mm Web thickness t = 8.1 mm Flange thickness T = 13.3 mm Depth between fillets d = mm Plastic modulus S x = 1290 cm 3 Elastic modulus Z x = 1120 cm 3 Local buckling ratios: Flange b/t = 5.75 Web d/t = SECTION CLASSIFICATION Grade of steel = S275 T = 13.3 mm < 16mm Therefore, p y = 275 N/mm 2 = (275/p y ) = SQRT(275/275) = 1 Outstand element of compression flange, Limiting b/t = 9 = 9 Flange is plastic Actual b/t = 5.75 < 9 Class 1 Section is symmetrical, subject to pure bending, neutral axis at mid-depth, Limiting d/t = 80 = 80 Actual d/t = 50.3 < 80 Web is plastic Class 1 Section is : Class 1 plastic section

116 103 Job No: 1003 Page 3 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 3.0 SHEAR BUCKLING If d/t ratio exceeds 70 for rolled section, shear buckling resistance should be checked. d/t = 50.3 < 70 = 70 Therefore, shear buckling needs not be checked 4.0 SHEAR CAPACITY F v = kn P v = 0.6p y A v p y = 275 N/mm 2 A v = td = 8.1 x = mm 2 P v = 0.6 x 275 x x = kn F v < P v Therefore, the shear capacity is adequate 5.0 MOMENT CAPACITY M = knm 0.6P v = 0.6 x = kn F v < 0.6P v Therefore, it is low shear M c = p y S x = 275 x 1290 x = knm 1.2p y Z = 1.2 x 275 x 1120 x = knm M c < 1.2p y Z OK M < M c Moment capacity is adequate

117 104 Job No: 1003 Page 4 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 6.0 WEB BEARING & BUCKLING 6.1 Bearing Capacity P bw = (b 1 + nk) tp yw (Unstiffened web) r = 10.2 mm b 1 = t + 1.6r + 2T = x x 13.3 = mm k = T + r = = 23.5 mm At the end of a member (support), n = b e /k but n 5 b e = 0 = 2 b 1 + nk = x 23.5 = mm P bw = x 8.1 x 275 x = kn F v = kn F v < P bw Bearing capacity at support is ADEQUATE

118 105 Job No: 1003 Page 5 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by 7.0 SERVICEABILITY DEFLECTION CHECK Unfactored imposed loads: Checked by w = 9 kn/m for roofs L = 6 m = 15 kn/m for floors E = 205 kn/mm 2 I = cm 4 5wL 4 = 384EI = 5 x 15 x 6^4 x 10^5 384 x 205 x = 4.84 mm CCH DR. MAHMOOD Beam condition Carrying plaster or other brittle finish Deflection limit = Span / 360 = 6 x 1000 / 360 = mm 4.84mm < 16.67mm The deflection is satisfactory!

119 APPENDIX B2 106

120 107 Job No: 1004 Page 1 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD Grade = S275 Section 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 Mass W pl.y Section Mass W pl.y (kg/m) (cm 3 ) (kg/m) (cm 3 ) x171x x178x x152x x171x x178x x152x x191x x178x x152x x191x x152x x191x x210x x191x x210x x191x x210x x229x x210x x229x x210x x229x x229x x305x x305x x305x M = knm W pl.y = M / f y = x 10^3 / 275 = cm 3 Try 406x178x54 UB

121 108 Job No: 1004 Page 2 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 1.0 DATA 1.1 Trial Section L = 6 m Initial trial section is selected to give a suitable moment capacity. The size is then checked to ensure suitability in all other aspects. Section chosen = 406x178x54 UB 1.2 Section Properties Mass = 54 kg/m Depth h = mm Width b = mm Web thickness t w = 7.6 mm Flange thickness t f = 10.9 mm Depth between fillets d = mm Plastic modulus W pl.y = 1051 cm 3 Elastic modulus W el.y = 927 cm 3 Shear area, A v = 32.9 cm 2 Area of section, A = 68.6 cm 2 Second moment of area, I y = cm 4 i LT = 4.36 cm a LT = 131 cm c/t f = 8.15 d/t w = SECTION CLASSIFICATION Grade of steel = S275 (Fe 430) t = 10.9 mm <= 40mm Therefore, f y = 275 N/mm 2 f u = 430 N/mm 2

122 109 Job No: 1004 Page 3 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD Classification of Trial Section (a) Outstand element of compression flange, flange subject to compression only : c/t f = 8.15 Class 1 limit : c/t f = 9.2 <= 9.2 Flange is Class 1 element (b) Web, subject to bending (neutral axis at mid depth) : d/t w = 47.4 Class 1 limit : d/t w = 46.7 > 46.7 Web is Class 2 element 406x178x54 UB is a Class 2 section 3.0 SHEAR RESISTANCE V Sd = kn V pl. Rd A v MO f y 3 MO = 1.05 = 32.9 x x = kn V Sd < V pl.rd Sufficient shear resistance 4.0 MOMENT RESISTANCE M Sd = knm 0.5V pl.rd = 0.5 x = kn V Sd < 0.5V pl.rd Therefore, it is low shear M c.rd = W pl.y f y / MO = 1051 x 275 x / 1.05 = knm M Sd < M c.rd Moment capacity is adequate

123 110 Job No: 1004 Page 4 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 5.0 LATERAL TORSIONAL BUCKLING (LTB) Beam is fully restrained, not susceptible to LTB 6.0 SHEAR BUCKLING For steel grade S275 (Fe 430), shear buckling must be checked if d/t w > 63.8 d/t w = 47.4 < 63.8 Shear buckling check is NOT required 7.0 RESISTANCE OF WEB TO TRANSVERSE FORCES Stiff bearing at support, s s = 50 mm Stiff bearing at midspan, s s = 75 mm 7.1 Crushing Resistance Design crushing resistance, t w f yw R y.rd = (s s + s y ) At support, s y t f b t f w 0.5 M1 f.ed = Longitudinal stress in flange (My / I) = 0 at support (bending moment is zero) MO = 1.05 f yf = 275 N/mm 2 f f yf yw MO f yf f. Ed s y = mm R y.rd = ( ) x 7.6 x 275 x / 1.05 = kn V Sd = kn < R y.rd Sufficient crushing resistance

124 111 Job No: 1004 Page 5 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD At midspan, s y 2t f b t f w 0.5 f f yf yw MO f. Ed f yf V Sd = 0 Crushing resistance is OK 7.2 Crippling Resistance Design crippling resistance At support, R a. Rd 0.5t 2 w Ef yw 0.5 t t f w t t w f ss d 1 M 1 s s /d / = 0.14 M1 = 1.05 E = 205 kn/mm 2 Ra.Rd = kn > V Sd = kn Sufficient crippling resistance At mid span, M Sd 1.5 M c.rd = 0.98 <= OK 7.3 Buckling Resistance At support, b eff h = mm a = 0 mm s h ss a 2 s 2 but b 2 2 eff h s s 0.5

125 112 Job No: 1004 Page 6 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD b eff = 0.5 x SQRT(402.6^2 + 50^2) / 2 = mm <= [h 2 + s 2 s ] 0.5 = mm Buckling resistance of web, R b.rd = Af c A M1 A = 1 M1 = 1.05 A = b eff x t w = x 7.6 = mm 2 Ends of web restrained against rotation and relative lateral movement. = 2.5 d/t l = 0.75d = 2.5 x / 7.6 = Rolled I-section, buckling about y-y axis, use curve a A = A f c f c = ( ) x ( ) / ( ) = N/mm 2 R b.rd = 1 x x x / 1.05 = kn At mid span, > V Sd = kn Sufficient buckling resistance V Sd = 0 Sufficient buckling resistance at midspan

126 113 Job No: 1004 Page 7 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Beam Design (Floor Beams, L = 6.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 8.0 SERVICEABILITY LIMIT (DEFLECTION) Partial factor for dead load G = 1.0 Partial factor for imposed floor load Q = 1.0 Dead g d = 27.6 kn/m Imposed q d = 15 kn/m 2 = Variation of deflection due to variable loading 1 = Variation of deflection due to permanent loading 0 = Pre-camber of beam in unloaded state = 0 max = I y = cm 4 E = 210 kn/mm 2 = 5(g d / q d ) x L EI 1 = mm 2 = 6.46 mm < L / 350 = mm OK max = = mm Recommended limiting vertical deflection for max is L 6000 = = 24 mm max < 24 mm Deflection limit is satisfactory.

127 APPENDIX C1 114

128 115 Job No: 1005 Page 1 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD Grade = S275 Section 152x152x23 152x152x30 152x152x37 203x203x46 203x203x52 203x203x60 203x203x71 254x254x73 203x203x86 254x254x89 305x305x97 254x254x x305x x368x x254x x305x x368x x305x x254x x368x x305x x368x x406x x305x x305x x406x x406x x406x x406x x406x x406x634 S x Mass (kg/m) (cm 3 ) M = knm S x = M / fy = x 10^3 / 275 = cm 3 Try 203x203x60 UC

129 116 Job No: 1005 Page 2 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 1.0 DATA F c = kn L = 5 m 1.1 Trial Section Initial trial section is selected to give a suitable moment capacity. The size is then checked to ensure suitability in all other aspects. Section chosen = 203x203x60 UC 1.2 Section Properties Mass = 60 kg/m Depth D = mm Width B = mm Web thickness t = 9.3 mm Flange thickness T = 14.2 mm Depth between fillets d = mm Plastic modulus S x = 652 cm 3 Elastic modulus Z x = cm 3 Radius of gyration, r x = 8.96 cm r y = 5.19 cm Gross area, A g = 75.8 cm 2 Local buckling ratios: Flange b/t = 7.23 Web d/t = SECTION CLASSIFICATION Grade of steel = S275 T = 14.2 mm < 16mm < 40mm < 63mm Therefore, p y = 275 N/mm 2 = (275/p y ) = SQRT(275/275) = 1

130 117 Job No: 1005 Page 3 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD Outstand element of compression flange, Limiting b/t = 9 = 9 Actual b/t = 7.23 < 9 < 10 = 10 Flange is plastic < 15 = 15 Class 1 Web of I- or H-section under axial compression and bending ("generally" case) r 1 = F c dtp y = x 1000 / (160.8 x 9.3 x 275) -1 < r 1 1 = 3.44 r 1 = 1 Actual d/t = 17.3 < 80 All 40 1+r = Web is plastic < 1+1.5r = 40 1 Class 1 Section is : Class 1 plastic section 3.0 SLENDERNESS 3.1 Effective Length About the x-x axis, "Restrained in direction at one end" L EX = 0.85L = 0.85 x 5 x 1000 = 4250 mm x = L EX / r x = 4250 / (8.96 x 10) = COMPRESSION RESISTANCE F c = kn P c = p c A g p y = 275 N/mm 2 A g = 75.8 cm 2 Buckling about x-x axis

131 118 Job No: 1005 Page 4 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD Use strut curve (b) x = 47.4 p c Interpolation: p cx = ( ) / (48-46) x ( ) = N/mm 2 P c = p c A g = x 75.8 x 100 x = kn F c < P c Therefore, the compressive resistance is adequate 5.0 NOMINAL MOMENT DUE TO ECCENTRICITY For columns in simple construction, beam reaction, R is assumed to act 100mm off the face of the column. R From frame analysis sheets, M i = knm 100 mm Moments are distributed between the column lengths above and below level 2, in proportion to the bending stiffness of each length. For EI/L 1 : EI/L 2 < 1.5, the moment will be equally divided. Therefore, M = knm

132 119 Job No: 1005 Page 5 UTM Job Title: Braced Steel Frame Design (BS : 2000) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 6.0 COMBINED AXIAL FORCE AND MOMENT CHECK The column should satisfy the relationship F P c c M M x bs p M y y Z y 1 LT = 0.5 L/r y = (0.5 x 5 x 1000) / (5.19 x 10) = p y = 275 N/mm 2 LT p b p b = ( ) / (50-45) x ( ) = N/mm 2 M b = p b S x = x 652 x = knm = < 1.00 The combined resistance against axial force and moment is adequate. 7.0 CONCLUSION 4.0 Compression Resistance = OK 6.0 Combined Axial Force and Moment Check = OK Use of the section is adequate Use : 203x203x60 UC

133 APPENDIX C2 120

134 121 Job No: 1006 Page 1 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD Grade = S275 Section 152x152x23 152x152x30 152x152x37 203x203x46 203x203x52 203x203x60 203x203x71 254x254x73 203x203x86 254x254x89 305x305x97 254x254x x305x x368x x254x x305x x368x x305x x254x x368x x305x x368x x406x x305x x305x x406x x406x x406x x406x x406x x406x634 W pl.y Mass (kg/m) (cm 3 ) M = knm W pl.y = M / f y = x 10^3 / 275 = cm 3 Try 254x254x73 UC

135 122 Job No: 1006 Page 2 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 1.0 DATA N Sd = kn L = 5 m M sd = knm 1.1 Trial Section Initial trial section is selected to give a suitable moment capacity. The size is then checked to ensure suitability in all other aspects. Section chosen = 254x254x73 UC 1.2 Section Properties Mass = 73 kg/m Depth h = 254 mm Width b = 254 mm Web thickness t w = 8.6 mm Flange thickness t f = 14.2 mm Depth between fillets d = mm Plastic modulus W pl.y = 990 cm 3 Elastic modulus W el.y = 895 cm 3 Radius of gyration, i y = 11.1 cm i z = 6.46 cm Area of section, A = 92.9 cm 2 Second moment of area, I y = cm 4 i LT = 6.86 cm a LT = 98.5 cm c/t f = 8.94 d/t w = SECTION CLASSIFICATION Grade of steel = S275 (Fe 430) t f = 14.2 mm <= 40mm Therefore, f y = 275 N/mm 2 f u = 430 N/mm 2

136 123 Job No: 1006 Page 3 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD Classification of Trial Section (a) Outstand element of compression flange, flange subject to compression only : c/t f = 8.94 Class 1 limit : c/t f = 9.2 <= 9.2 Flange is Class 1 element Limit c/t f Class 2 = 10.2 (b) Web, subject to bending and compression : Class 3 = 13.9 Classify web as subject to compression and bending d/t w = 23.3 Class 1 limit : d/t w = 30.5 <= 30.5 Web is Class 1 element Limit d/t w Class 2 = 35.1 Therefore, it is Class 1 section Class 3 = CROSS-SECTION RESISTANCE n = N Sd N pl.rd N pl.rd = A f y MO MO = 1.05 N pl.rd = 92.9 x 100 x 275 x / 1.05 = kn n = / = >= 0.1 n < 0.1 M ny.rd = M pl.y.rd n 0.1 M ny.rd = 1.11 M pl.y.rd (1-n) W pl.y f y M pl.y.rd = MO = 990 x 275 x / 1.05 = knm M ny.rd = knm > M Sd = knm Sufficient moment resistance

137 124 Job No: 1006 Page 4 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by 4.0 IN-PLANE FAILURE ABOUT MAJOR AXIS Checked by Members subject to axial compression and major axis bending must satisfy N N Sd Slenderness ratio b. y. Rd c. y. Rd N b.y.rd = A f c A M1 l y = 0.85 L (Restrained about both axes) = 0.85 x 5 x 1000 = 4250 mm y = l y / i y = 4250 / (11.1 x 10) = 38.3 Buckling about y-y axis (Curve b) k A = 1 y A = 38.3 t f <= 40mm y M M y. Sd 1.0 CCH DR. MAHMOOD A f c f c = ( ) x (40-38) / ( ) = N/mm 2 N b.y.rd = 1 x x 92.9 x 100 x / 1.05 = kn k y = 1.5 (Conservative value) N Sd N b.y.rd + k y M y.sd M c.y.rd x = MO / M1 = x = 1 = 0.95 < 1 Therefore, sufficient resistance against in-plane failure against major axis

138 125 Job No: 1006 Page 5 UTM Job Title: Braced Steel Frame Design (EC 3) Subject: Column Design (Internal Column, L = 5.0m) SKUDAI, JOHOR Client: STC, UTM Made by Checked by CCH DR. MAHMOOD 5.0 CONCLUSION 3.0 Cross Section Resistance OK 4.0 In-plane Failure About Major Axis OK Use of the section is adequate. Use : 254x254x73 UC