Handbook Reinforced Concrete Design in accordance with AS 3600—2009 A joint publication of Cement Concrete & Aggregates Australia and Standards Australia Fifth edition July 2011 Fourth edition February 2002 Third edition May 1995 Second edition July 1991 First published September 1989 CCAA T38 HB71–2011 (Standards Australia) © Cement Concrete & Aggregates Australia 2011 and Standards Australia 2011 Except where the Copyright Act allows otherwise, no part of this publication may be reproduced, stored in a retrieval system in any form or transmitted by any means without prior permission in writing of both Cement Concrete & Aggregates Australia and Standards Australia. The information provided in this publication is intended for general guidance only and in no way replaces the services of professional consultants on particular projects. No liability can therefore be accepted by Cement Concrete & Aggregates Australia or Standards Australia for its use. 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Suite 2, Level 2, 485 Ipswich Road Annerley QLD 4103 TELEPHONE: (61 7) 3227 5200 FACSIMILE: (61 7) 3892 5655 MELBOURNE OFFICE: 2nd Floor, 1 Hobson Street South Yarra VIC 3141 TELEPHONE: (61 3) 9825 0200 FACSIMILE: (61 3) 9825 0222 PERTH OFFICE: 45 Ventnor Avenue West Perth WA 6005 TELEPHONE: (61 8) 9389 4452 FACSIMILE: (61 8) 9389 4451 ADELAIDE OFFICE: PO Box 229 Fullarton SA 5063 TELEPHONE: (61 8) 8274 3758 TASMANIAN OFFICE: PO Box 246 Sheffield TAS 7306 TELEPHONE: (61 3) 6491 1509 FACSIMILE: (61 3) 6491 2529 WEBSITE: www.ccaa.com.au EMAIL: info@ccaa.com.au Cement Concrete & Aggregates Australia ABN 34 000 020 486 Reinforced Concrete Design Handbook iii Preface to Fifth edition This fifth edition is a complete rewrite of the Reinforced Concrete Design Handbook and brings it into line with the 2009 edition of AS 3600 Concrete Structures and Amendment No. 1–2010. It also takes into account changes in other Australian Standards that have occurred since the fourth edition was published, including AS/NZS 1170 Structural Design Actions, Part 0 General principles and Part 4 Earthquake actions in Australia. The 2009 edition of AS 3600 includes significant changes to: n n The maximum concrete strength covered (now includes 100 MPa) Development lengths and lap lengths of reinforcement n Use of Ductility Classes N and L reinforcement n Durability and fire requirements. The opportunity has been taken to review many of the charts and their relevance in the modern design office. In many cases, the previous charts were nomograms from an era when these were a common design tool. These have now been largely replaced by design software or, as in this Handbook, by spreadsheets. The spreadsheets are used to illustrate the design principles of reinforced concrete, the requirements of AS 3600 and the recommendations of this Handbook. They are in keeping with current design technology. The spreadsheets can be downloaded from CCAA website www.ccaa.com.au/publicationextras/. There is a new chapter covering the strut-and-tie design method to reflect the new section in AS 3600. There are also revised rules for crack control in beams and slabs but no charts or tables are provided. However, the Design Example in Chapter 10 includes calculations showing how these requirements can be checked. By-and-large the order in which material is presented follows that of relevant sections in AS 3600, although not all the sections in the standard are covered. The contribution of J Woodside fie aust fasce f i struct e mice, Director, J Woodside Consulting, in reviewing the Handbook and in the preparation of the design charts and spreadsheets is gratefully acknowledged. iv Reinforced Concrete Design Handbook contents Chapter 1 General pages 1.1–1.20 Chapter 2 Design properties for concrete and reinforcement pages 2.1–2.16 Chapter 3 Durability and fire resistance pages 3.1–3.18 Chapter 4 Beams pages 4.1–4.20 Chapter 5 Suspended slabs pages 5.1–5.20 Chapter 6 Columns pages 6.1–6.26 Chapter 7 Walls pages 7.1–7.24 Chapter 8 Footings pages 8.1–8.12 Chapter 9 Strut-and-tie modelling pages 9.1–9.8 Chapter 10 Design examples pages 10.1–10.50 Appendix A The design process pages A.1–A.2 Appendix B Development and use of the spreadsheets pages B.1–B.2 Chapter 1 General How the Performance Requirements are to be satisfied is spelt out in the Building Solutions. There are two broad approaches: [a] Deemed-to-satisfy (DTS) solutions; and [b] Alternative solutions. 1.1 Introduction 1.1.1 Codes and regulations Designers need to understand the framework of regulations and standards within which the design of the building or structure is designed and constructed. For most buildings in Australia, the Building Code of Australia (BCA)1.1 sets out the regulatory requirements for the building *. The Building Code of Australia sets out Objectives, Functional Statements, Performance Requirements and Building Solutions for the various aspects of design, eg structural, and health and amenity. The first two ('Objectives' and 'Functional Statements') are broad descriptors and are used mainly to interpret the latter two ('Performance Requirements' and 'Building Solutions'). The Performance Requirements are qualitative statements, eg that under structural provisions says: A building or structure, to the degree necessary, must: i Remain stable and not collapse; and ii Prevent progressive collapse; and iii Minimise local damage and loss of amenity through excessive deformation, vibration or degradation; and iv Avoid damage to other properties, by resisting the actions to which it may reasonably be subject...' * The terms 'building' and 'structure' are used to signify the same entity by administrations in Australia and in New Zealand. This may lead to some ambiguity where the terms are used interchangeably in some joint AS/ NZS standards. In general, in Australia the term 'building' is used to refer to buildings­—ranging from houses to multi-storey buildings—and 'structures' to refer to structures other than buildings whereas in New Zealand the term 'structure' is used inclusively to cover buildings and other structures. The DTS approach involves designing the members, buildings and structures in accordance with the relevant Australian standards, eg for concrete in accordance with AS/NZS 11701.2 and AS 36001.3. The Alternative Solution approach allows the designer to use other codes or standards, fire test data, etc. (The 2009 edition of AS 3600 omits a number of clauses from previous editions which gave rise to conflicts of interpretation with the BCA in this area, eg those that provided rules on interpretation of test data. Designers should be aware that their omission in AS 3600 does not imply that the approach is invalid but that the rules under which it is done now lie within the BCA under Alternative Solutions, not the DTS approach using the relevant Standard, eg AS 3600.) As will be discussed later, AS 3600 provides for a number of different analysis and strength check approaches. However, the BCA and AS/NZS 1170.0 are written around a linear elastic analysis/limit states approach using characteristic strengths of the materials and factored loads. Designers should be aware that AS 3600 provides minimum solutions, ie compliance is necessary for all buildings but particular buildings may require the application of more-stringent provisions or additional considerations/criteria to meet the client's requirements. However, AS 3600 represents best practice even when it is not called up in the BCA and it cannot be ignored, especially where its requirements are more stringent than those in earlier editions of the standard. 1.1.2 Responsibility of designers and supervisors The division of responsibility between the parties involved in the design and those in the construction of a building should be clearly understood and fully expressed in the terms of engagement between the owner and the designer, and in the contract for construction between the owner and the builder or contractor. 'Design' here includes the architectural and structural design of the building and the preparation of the drawings, specification, and sometimes the conditions of contract and preliminaries. Most projects will involve a number of other disciplines, eg mechanical, electrical and service engineering. Developing and documenting the final design solution will normally involve a design team covering all the required design disciplines. Reinforced Concrete Design Handbook 1.1 ascertaining the appropriate criteria for the particular building and seeing that these are satisfied. The designer responsible for the structural design should be a practising civil or structural engineer eligible for Chartered Status of Engineers Australia or equivalent and experienced in the design of concrete structures of comparable importance. Architects and building graduates should not be expected to have the appropriate skills to undertake, nor should they assume responsibility for, the design of a concrete structure. Graduate engineers while working under guidance can design parts of concrete structures but they should not take responsibly for the overall design of the structure. [b] Safety In service, a structure must be able to safely resist the actions (loads) expected to be imposed on it throughout its intended life. Safety must also be considered during the construction period, particularly while the concrete has not reached its design strength. Loads imposed on it during that period should be analysed as required. The design should also consider unusual load cases arising from any processes to be carried on in the structure. Some thought should also be given to the ultimate demolition of the structure. When designers assign the detail design of any elements to a manufacturer or supplier, they should ensure that this work is fully specified and controlled by way of detailed performance standards, and that these elements are coordinated with the structure as a whole. Examples of this are the detailed design of precast concrete elements and post-tensioned slabs. Designers usually start with a framing plan, which is logical and sensible, and proceed to examine how that structure behaves when subjected to the various actions. In particular, they should review all possible failure modes to ensure that nothing important has been overlooked. This topic is discussed in more detail in Section 1.2.2 The design process. The supervision of construction is the responsibility of the builder. All structures should be supervised by a suitably qualified person. If the structure is complex or incorporates prestressed concrete, a qualified and experienced engineer employed by the builder should be responsible for the supervision of construction. A structure should be robust and possess structural integrity so that it is not unreasonably susceptible to the effects of accidental loads. Damage to a small area of a structure or the failure of a single element should not lead to the collapse of a large part of the structure, eg by progressive collapse. This topic is discussed in more detail in Section 1.4.6 Structural integrity and robustness. Periodic inspection of construction on behalf of the owner or client is often undertaken by the designer, or by an experienced person employed by the owner or client but under the technical direction of the designer. Where the project is complex or unusual, a more‑detailed inspection regime may be required. This arrangement facilitates the resolution, by the designer, of any queries that may arise as to the interpretation of the design documents. Site records should be kept during construction to show the dates of concrete placing, test results, stressing details and any significant departures from the design drawings. These provide the owner with a useful record of the structure as built, should any modification be required in the future. 1.2 Design Process and Procedures 1.2.1 Broad structural design aims [a] General The aim of structural design is to produce safe, serviceable, durable, aesthetic, economical and sustainable structures. Designers should always strive for simplicity, clarity and excellence in their design. Simple design does not mean elementary design but rather well conceived and quality design. As noted above, mere compliance with the appropriate codes and standards will not guarantee a satisfactory design for all buildings as they provide only a set of minimum requirements. The designer is responsible for 1.2 Reinforced Concrete Design Handbook The accidental hazard arising from fire is covered in building regulations, eg the Building Code of Australia. The particular requirements for different structural elements for fire resistance, eg Fire Resistance Levels to guard against structural collapse (structural adequacy), flame penetration (integrity) and heat transmission (insulation), are discussed in Chapter 3 Durability and Fire Resistance. Designers must be alert to prevent gross errors during design, as these, along with those that may arise during construction, are probably the most common cause of failures. An independent check should be made of the design, including a review of the drawings and specification to ensure that the assumptions made in the design are valid. [c] Serviceability Over its design life, during service under normal operating and load conditions, a structure must behave satisfactorily. The structure and its elements should not deflect or deform excessively or vibrate to cause discomfort to the occupants. Any cracking or apparent distress of the concrete should not impair the structure's functionality or spoil its appearance. While some clients consider concrete to be indestructible, some maintenance and repairs of the concrete structure will normally be required during the life of the building, but this should be minimal. [d] Durability A durable structure is one that performs its intended function over its design life in its environment without excessive degradation or unusual maintenance expense. There have been examples of inadequate durability, such as premature rusting reinforcement, spalling concrete, extensive wear and badly weathered concrete surfaces. The procedures necessary to ensure durable concrete structures are discussed in Chapter 3 Durability and Fire Resistance. [e] Aesthetics An integral part of the design of any structure is consideration of its appearance. Buildings and structures such as bridges should be designed and detailed to present an attractive and wellproportioned appearance to suit their surroundings and environment. Architects rather than structural engineers are usually responsible for the appearance of buildings. However, there are many cases where the engineer can provide a significant input by the selection of appropriate framing systems and the proportioning of members to meet functional, load capacity and any aesthetic requirements. [f] Sustainability In recent years, sustainability has become a design consideration for all structures. Sustainable design requires that social, environmental and economic outcomes are balanced. For example, a project is not sustainable if it damages the environment, or if it results in negative social outcomes such as loss of jobs or health problems, or if it results in financial loss. Concrete is an important contributor to sustainable design. Concrete, like all products, has environmental impacts arising from the acquisition of raw materials, processing, transport and recycling at the end of its life. These are, however, significantly outweighed by the benefits that concrete delivers. Designers are referred to the CCAA's Briefings 111.4 and 131.5 and its website, www.ccaa.com.au, for further information on this topic. Sustainable design also requires the designer to design an economical structure. Thus, the adoption of a simplistic, conservative design approach and poor detailing to minimise design costs—but resulting in an overdesigned structure—is not acceptable. [g] Economy An economical structure contributes to limiting the overall cost of the project. This can be measured in terms of the initial cost, the construction time and the life cycle or overall cost. The low cost of concrete and reinforcement alone does not necessarily produce the most economical structure; construction and time-related costs must also be considered. Ease of construction must therefore be taken into account at the design stage. 1.2.2 The design process The design process typically comprises three phases— conceptual design, preliminary design and final design. These are described briefly below and in more detail in Appendix A. Since conceptual design will often be based on limited information, any structural design should be simple, quick and conservative without being heavy-handed. It is not the time for extensive computer modelling. Designers, however, need to carry out sufficient structural design to ensure that concepts are feasible. The preliminary design phase is where the client's requirements for the project are developed in more detail. On major projects, more than one preliminary design may be produced. The final design stage is where the design data is checked and the chosen optimum design is developed and detailed. This will include the preparation of project documentation and specifications. It is important for the designer to remember that the documentation is the means of communicating the design intentions to the contractor/ builder and subcontractors. The documentation should be reviewed from this viewpoint before being issued. There are a number of overseas manuals1.6,1.7,1.8 on the design of reinforced concrete buildings to which the designer can refer for further information and guidance. 1.2.3 Order of design The building should be designed in a logical order for analysis and drafting. For an office building the order of design might be as follows: n n n n n n Establish the design loads (AS/NZS 1170) Confirm the design data such as: survey, geotechnical, environmental, etc The occupancy of the structure, required fire ratings, sound transmissions, etc from the BCA, (normally provided by the architect) Establish exposure classification and durability requirements including concrete strength(s), cover(s) and axis distances, deflection criteria (AS 3600) Establish any other special design requirements Lateral stability for wind and earthquake loadings and general stability in two orthogonal directions n Roof framing including slabs and beams n Plant room slabs and beams n Typical floor slabs and beams n First floor and non-typical slabs and beams n Ground floor slabs and beams Reinforced Concrete Design Handbook 1.3 n Basement floor slabs and retaining walls n Stairs and lift cores including lift motor rooms n Column and wall load rundowns n Column and wall designs n Footings and foundation capacity n Precast or external walling n Robustness check and detailing n n Other architectural elements that may require structural design Checking and review of drawings and specifications. 1.2.4 Structural framing Finding the best structural framing solution for a building is not straightforward and there will typically be alternative solutions. The framing must consider how all the loads find their way through the structure, both horizontally and vertically, to the footings. Framing is a trial-and-error process and adjustment will need to be made as the design proceeds. The process is neither taught nor covered in textbooks and requires a good appreciation of architectural and engineering constraints. Concrete structural frames are commonly used in Australia and have the advantage of good performance in fire. They can be cast in situ, precast or both. The frame for larger projects usually needs to be modelled for input into the computer for analysis. Which members are pinned and which are continuous also need to be established. Certain buildings lend themselves to standard solutions, eg an industrial building or shopping centre. Local conditions will sometimes favour different solutions depending on the local building industry capability, etc. Column/wall locations are often dictated by the intended use of space. For example, for a car parking building the column spacing must suit parking bay sizes; for an office building a column-free space may be required or there may be other spatial requirements developed by the architect from the client's needs. The floor-to-floor height also needs to be considered and the space required for building services, particularly in the space under the floor and above the ceiling. Concrete allows efficient floor solutions, minimising the overall height of a building or maximising the number of floors in a given height. Designers also need to define how lateral loads are resisted, suitable systems can include one or more of shear walls, moment-resisting (space) frames and cantilever columns or walls. 1.4 Reinforced Concrete Design Handbook Assessing, apportioning loads and understanding the load paths can be difficult to appreciate. The assessment of all loads is one of the fundamental design considerations before commencing final analysis and design. If the loads are wrong or apportioned incorrectly, they will affect the design of all members, and extensive rework and extra time will be involved—assuming the errors are found—or, if the errors are not found, possibly an unsafe structure. 1.2.5 Initial estimation of member dimensions The initial estimation of member sizes is generally based on past experience, some quick trial designs or other design information. Design offices may have design guides based on experience of successful designs and recommendations where problems have arisen. The depth of flexural members is usually controlled by deflection considerations. The minimum thickness of walls tends to be governed by construction and cover considerations, and this is also true for column dimensions. The axial load capacity of columns can be significantly increased by increasing the concrete strength and/or increasing the longitudinal reinforcement percentage. Neither of these necessarily increases the column dimensions. However, lateral load bending moments and limiting sway movements may dictate some minimum dimensions. 1.3 Design Checks and Methods of Analysis In a real structure, the behaviour under load of individual elements can be complex, depending on the materials used and many other factors. Generally, idealised models of the frame or structure are developed to analyse how the real structure may behave. The analysis that is carried out to validate a design is generally a two-step process although some computer programs may combine the two steps: n n Structural analysis of the frame or structure Strength check and other design checks at critical cross-sections of members. The first step of analysis is aimed at determining the action effects such as bending moment, shear force, torsion and axial force at critical sections of members necessary for strength design or determining deformations of the structure. The second step is concerned with the strength check of these critical sections along with other design checks such as deflections. AS 3600 makes provision for a variety of methods to be used for strength checks, viz: [a] Procedure for use with linear elastic analysis methods of analysis with simplified analysis methods and for statically determinate structures (see AS 3600 Clause 2.2.2). [b] Procedure for use with linear elastic stress analysis methods (see AS 3600 Clause 2.2.3). [c] Procedure for use with strut-and-tie analysis (see AS 3600 Clause 2.2.4). [d] Procedure for use with non-linear analysis of framed structures (see AS 3600 Clause 2.2.5). [e] Procedure for use with non-linear stress analysis (see AS 3600 Clause 2.2.6). In addition, it is permissible to carry out design checks for strength and serviceability by testing a structure or component member in accordance with the requirements of Appendix B in AS 3600 (see AS 3600 Clause 2.1.1). The first of these procedures, (a), is the one which will be familiar to most designers and was in earlier editions of AS 3600. The other methods have been introduced into the 2009 edition of AS 3600 to permit the use of more-sophisticated computer-based analysis programs, eg Finite Element Analysis. Foster1.9, while Foster et al1.10 give a summary of the other methods, (b) to (e), and indicate where each may be applicable and the provisos associated with their use. Designers should be aware that there is conflict between these latter methods, (b) to (e), and the requirements in the BCA and AS/NZS 1170.0. For example, BCA (Volume 1) BP1.2 mandates use of 5% characteristic material properties and this would preclude the use of some structural check procedures in AS 3600, eg non-linear analysis of framed structures which uses mean values of material properties. AS/NZS 1170.0 called up by the BCA is written around the linear elastic method of analysis and ultimate limit states approach. For example, see Section 2 in that Standard. This may or may not be a problem. However, strut–and–tie analysis may be the only appropriate method of design for non-flexural members. This Handbook is written around the method in (a) which is compatible with both the BCA and AS/NZS 1170.0. No conflict is therefore foreseen with the following discussions, except perhaps for Chapter 9 Strut-and-tie modelling. 1.4 Limit-states Design and Design Checks using Linear Elastic Methods of Analysis 1.4.1 General A limit-states approach to design is assumed by the BCA and AS/NZS 1170.0. The procedure for use with linear elastic analysis methods and for statically determinate structures given in AS 3600 Clause 2.2.2 is compatible with this approach. Limit-states design assumes there will be an acceptable probability that a structure designed and built in accordance with the Standard will not reach any limit state during its design life. That is to say, it will not fail by collapse or instability (ultimate limit states), or become unfit for service by deformation, vibration or cracking (serviceability limit states). In addition, the structure should not deteriorate unduly during its design life and should not be damaged by events such as fire, explosions and impact to an extent disproportionate to the cause. A checklist of design requirements includes: n Stability n Strength n Serviceability — Deflection — Lateral drift (eg under wind or earthquake) — Cracking — Vibration n Durability n Fire resistance n n Structural Integrity/ robustness (prevention of progressive collapse) Other limit states as required. Limit-states design analyses the structure or part for the relevant combination of factored actions (the action effect). It then confirms that the design capacity, ie the nominal capacity multiplied by the capacity factor (capacity reduction factor), f, exceeds the action effect. (The use of a global factor rather than partial safety factors, as adopted in European standards, follows the practice established in ACI 3181.8 and that used in earlier editions of AS 3600.) In essence, following this approach, the steps in design for the ultimate limit state are (the design for serviceability limit states is similar): n Determine the actions on the structure n Determine the appropriate combinations of actions n n Analyse the structure for the applied combinations of actions Design and detail the structure for robustness and earthquake Reinforced Concrete Design Handbook 1.5 n n Determine the design resistance of the structure using AS 3600 Confirm the design resistance exceeds the action effects. AS/NZS 1170.0 Section 5 provides broad guidelines for appropriate methods of analysis. The general nature of the rules allows for the wide variety of structural materials covered by the Standard, while reference is made to the appropriate material standard for specific guidance for that material. 1.4.2 The capacity reduction factor, f, accounts for variations between the basis of the calculation and the likely actual condition, viz: n n n n Variations in the strength of concrete and reinforcement Variations in the dimensions of the member and the location of reinforcement and in the relative position of members, eg eccentricities in columns Inaccuracies in the design equations for calculating internal actions and the strength of the section Mode of failure, eg ductile or brittle, and the resulting warning of failure Importance of the member and its effect on the structure. For example, compare a beam and a column. The column is less ductile and more sensitive to concrete strength variations than the beam; the column usually supports a larger area than the beam, making the consequences of failure likely to be more serious. For these reasons, the capacity reduction factor, f, for pure bending is larger than that for axial compression, eg 0.8 to 0.6. Note that the design aids, spreadsheets and charts prepared to be compatible with one standard such as AS 3600 must be used only with the load factors, load combinations and capacity reduction factors applicable to that standard. 1.4.3 n The total deflection should not adversely affect the appearance or efficiency of the structure. AS 3600 limits this value to span/250. The incremental deflection should not adversely affect other elements such as finishes, services, partitions, glazing and cladding. Where partitions are detailed to minimise the effect of movement, this deflection should be limited to span/500. If not (eg masonry partitions without closely spaced joints), this limit should be reduced to span/1000. In addition, the following requirements as appropriate must be met: n n The deflection for imposed action (live load and dynamic impact) for members subjected to vehicular or pedestrian traffic should not exceed span/800. For transfer members the total deflection should not exceed span/500 where provision is made to minimise the effect of deflection of the transfer member on the supported structure. Otherwise, span/1000. For cantilevers, the deflections are generally half of those noted above when rotation at the support can occur. AS 3600 also states Design limits given or implied in Clauses 2.3.2 and 2.3.3 are based on previous design experience, and reflect requirements for normal structures. In special situations other limits may be appropriate. For further guidance refer to Appendix C of AS/ NZS 1170.0. Design for the serviceability limit states involves reliable predictions of the instantaneous and timedependent deformation of the structure. This is complicated by the non-linear material behaviour of concrete caused mainly by cracking, tension stiffening, creep and shrinkage. Designers can refer to the notes on the CIA Seminar series1.11 for further information. The calculation of deflection comprises two stages – an elastic or immediate component and an inelastic or creep component that occurs over a long period as shown in Figure 1.1. These are considered as Note: Allow for rotation at supports Serviceability – deflection control AS 3600 Clause 2.3.1 requires that Design checks shall be carried out for all appropriate service conditions to ensure the structure will perform in a manner appropriate for its intended function and purpose. 1.6 n Stability and strength The structure as a whole and its parts are designed to prevent instability due to overturning, uplift and sliding. Generally, the design capacity of a member is calculated as the ultimate strength of the section, using a mathematical model to represent the failure condition, multiplied by the capacity reduction factor. n Under working loads, the deflection of slabs and beams must be controlled to meet two general criteria: Reinforced Concrete Design Handbook Elastic Creep Figure 1.1 Elastic and creep deflections short-term and long-term effects with the appropriate combinations of actions (loads) acting in each case. The long-term loading comprises permanent actions (eg self-weight) and the quasi-permanent component of the imposed action (eg live load), while the shortterm loading includes the probable peak value of the imposed action. Typical values for these are given in Section 1.5 Actions and Combinations of Actions. The calculated deflection is measured from a theoretical line diagram representing the member in its as-cast position. The limit on total deflection of span/250 below an assumed horizontal line may not be sufficient to prevent a slab being unsatisfactory for non-structural reasons, eg water ponding on a roof. These problems may be overcome by cambering the formwork (usually not preferred) or by stressing the floor or roof to counter the long‑term deflection. In the case of long spans, these methods are used frequently; the designer should, however, be careful to check that the reduced stiffness of the floor does not result in excessive incremental deflection or vibration under live load, or large rotations and distress at supports. There is also a visual limit of about 25–40 mm even in long-span floors where long-term deflections become noticeable. Building owners will often not accept deflection over these limits and may perceive a large deflection as a failure. In assessing the practical effect of deflections, the designer should allow also for realistic construction tolerances. The limits in AS 3600, discussed in Section 1.7 Material and Construction Requirements, are based on the requirements of structural adequacy and strength. Tighter construction tolerances usually need to be specified or special details developed to meet the serviceability requirement. 1.4.4 Serviceability – cracking The designer and the building owner tend to view cracking differently. Engineers generally regard some cracking as inevitable; owners on the other hand tend to regard any cracking as a major defect. Most cracking in concrete structures is due to shrinkage of concrete. The structure and the steel reinforcement have to be designed and detailed to control the effects of this shrinkage. This will involve first determining whether or not cracks are allowed to occur and, if so, where they can occur in respect of structural integrity and aesthetics. The size of cracks must be limited so as not to cause future durability problems. In buildings, stiff vertical elements such as cores, basement and retaining walls can result in unsightly cracking in slabs unless steps are taken to minimise such cracking by methods such as the provision of construction joints or delayed pour strips. AS 3600 sets out guidelines for the amount of reinforcement to control cracking, eg Clause 2.3.3 states that cracking of beams and slabs under service conditions shall be controlled in accordance with the requirements of Clause 8.6 or 9.4, as appropriate. In a small percentage of cases, cracks are a symptom of structural or durability distress, eg spalling of concrete due to reinforcement corrosion. In these cases, the cause of cracking needs to be diagnosed and appropriate remedial measures taken. For a more detailed discussion on the types of cracking and practices to minimise its occurrence, see Guide to Concrete Construction1.12 and Movement, restraint and cracking in concrete structures1.13. 1.4.5 Serviceability – vibration Design for vibration is outside the scope of this Handbook. Designers should consult a specialist text or seek expert advice if vibration is likely to be a problem. The chapter on vibration in the Precast Concrete Handbook1.14 is recommended. 1.4.6 Structural integrity and robustness There are currently no specific requirements for design for structural integrity (the prevention of progressive collapse) or robustness in the BCA or AS 3600. The BCA mentions progressive collapse, implying that design for it is covered under the requirement of sustaining at an acceptable level of safety and serviceability the most adverse combination of loads. Section 6 in AS/NZS 1170.0 includes minimum structural robustness requirements. However, this still does not fully address the issue. The spectacular 1968 failure in the UK of Ronan Point, a block of flats constructed of large precast concrete panels, focused attention on the susceptibility of this form of construction to accidental or other loading such as gas explosions, as shown in Figure 1.2. Because of this accident, the British Standard for concrete structures1.15 was revised and included specific detailing requirements to provide continuity and ductility. Other high profile cases of progressive failure include the 1995 bomb attack on the Alfred P. Murrah Federal Building in the US and the collapses of the towers at the World Trade Centre in New York in 2001. Other national codes now also include provisions to prevent progressive collapse. A continuously reinforced, cast-in-place, concrete structure is less likely to be at risk of progressive collapse than a precast one because of its inherent ability to redistribute unusual loads and span over possible local failures, assuming the detailing allows for continuity. Normally, only a general review of such a structure would be required to check its possible failure modes. Reinforced Concrete Design Handbook 1.7 Precast beam Precast floor 1 2 1 5 5 1 6 5 Column under 5 5 5 A 3 A 4 2 7 7 3 7 8 Precast column Precast edge beam 1 2 3 4 5 6 7 8 Precast wall Internal floor ties between precast floor units Edge floor ties between precast floor units and beams Internal beam ties Edge beam ties Column ties horizontally to slabs and beams Columns ties vertically Wall ties horizontally to slabs and beams Wall ties vertically Figure 1.2 Ronan Point, UK Tie bar Insitu concrete However, further investigation should be carried out for: n n n precast concrete structures; unusual structural systems or mixed construction using different materials; structures subject to special risks, such as vehicle collision and explosion (eg a chemical factory). If an abnormal load can be identified, then it is possible to design directly for this condition. Usually, however, this is not the case, so other methods must be adopted to control the extent of damage. One method commonly used in Europe from Eurocode 21.16 is to design for specified forces at each level of the structure and to provide a system of horizontal and vertical ties, properly anchored, to resist these forces. Eurocode 2 replaced BS 8110 in the UK in 2010. For precast-panel buildings, this results in longitudinal, transverse and peripheral ties at each floor level interconnected with sufficient continuous vertical ties to restrain the walls at each level as shown in Figure 1.3. Another method is to provide alternative load paths so that the structure can bridge over the gap formed if a part of a floor or wall or column is accidentally removed. For precast-concrete-panel buildings, this method also results in a system of horizontal and vertical ties. By notionally removing a part of each wall in turn, the floor over is designed to act as a catenary, 1.8 Precast beam with projecting ties Reinforced Concrete Design Handbook SECTION A-A Figure 1.3 Integrity and robustness with 3-dimensional tying system for precast concrete which can support a large load although it may sag 300 mm or more. In an insitu-concrete building, floors or beams can act in a similar way provided they are detailed correctly and can span twice their actual span even when sagging excessively in catenary action. Examples of cantilever and beam action for precast buildings are shown in Figure 1.4. Useful references on methods of design for structural integrity are: Mitchell and Cook1.17, FIP1.18, Elliott and Tovey1.19 and the ACI-ASCE1.20. 1.4.7 Durability and fire resistance These aspects are covered in Chapter 3 Durability and fire resistance. Transverse tie to develop cantilever moment Peripheral Vertical ties to tie to anchor suspend panels and transverse tie for shear transfer in horizontal joints Compression capacity in adjacent panel Vertical ties for tension hold-down against cantilever moment Vertical ties connected to footing CANTILEVER ACTION Vertical ties to suspend panels and for shear transfer in horizontal joints Transverse tie to develop beam moment Compression capacity in return wall support Damaged panel BEAM ACTION Actions and Combinations of Actions 1.5.1 General AS/NZS 1170.0 sets out the various actions (loads) and the combinations of actions (load combinations) to be considered in design for ultimate and serviceability limit states. For combinations particular to prestressed concrete refer to AS 3600. 1.5.2 Damaged panel Compression capacity in adjacent panel 1.5 Vertical ties connected to footing Figure 1.4 Structural integrity – alternative load Permanent, imposed and other actions A permanent action (dead load) is defined as 'an action that is likely to act continuously and for which variations in magnitude with time are small compared with the mean value'. Generally, it is taken to comprise the self-weight of the member plus the weight of all materials of permanent construction – walls, floors and ceilings (including finishes), services, permanent partitions and fixed machinery supported by the member. An imposed action (live load) is defined as 'a variable action resulting from the intended use or occupancy of the structure' and is taken to include uniformlydistributed, concentrated, impact and inertia actions as applicable. Wind, snow and earthquake actions are considered separately. Many imposed actions are short-term relative to the life of the structure; however, some may be of long duration, eg storage loads, and thus have an effect similar to a permanent action. AS/NZS 1170.1 specifies minimum values for imposed actions on floors (floor live loads) for various classified occupancies; these are typically in the range of 2 to 5 kPa, except for storage areas where stacked material results in larger values. The specified uniformly distributed loads are blanket values to cover the expected effect of the occupancy for both small and large areas. Surveys of actual loadings in offices indicate that the statutory loads are reasonable for small areas but tend to be conservative for larger areas. Live load reductions are permitted for certain floors and supporting columns, walls and footings. A reduction of up to 50% is allowed according to a formula, which depends on the loaded area (see AS/NZS 1170 Part 1 Clause 3.4.2). If this reduction is used, then the design drawings should state both the nominal live load and that the reduction for area has been applied. During construction, special actions (loading) may occur and may control the design of some members. Staged construction and composite concrete members usually require a specific design check unless fully propped. Precast members such as floor units supporting wet concrete must be designed for construction loads in accordance with AS 36101.21. Construction loads from concrete, formwork, falsework and equipment such as forklifts, cranes and hoists may Reinforced Concrete Design Handbook 1.9 be greater than the imposed actions (live load) and thus may require strengthening of the structure or the provision of temporary supports during construction. Note that a significant proportion of all structural failures occur during construction, often because a critical loading condition is overlooked or the concrete strength at the time of loading is over-estimated. Structures incorporating flat-plate floors are susceptible to progressive collapse during construction for this reason – when failure of an upper floor due to early stripping leads to failure of those below. Other actions such as concrete shrinkage and creep, shortening due to prestress, temperature effects and foundation movements, cause deformations of the structure and, if resisted by the structure, result in internal forces, which are in equilibrium. A ductile structure is able to redistribute these loads so that the capacity of the member to carry the ultimate-strength loads is not affected. However, the deformation may be a significant factor in the serviceability check. 1.5.3 Combinations of actions – strength design Combination factors for actions (load factors) for strength design take into account: n n n n the possibility of unfavourable deviations of the actions from the characteristic values; the possible inaccurate assessment of the action effects and their significance for safety; variations in dimensional accuracy in so far as they affect estimation of the action effects; and the reduced probability of combinations of actions occurring, all at their characteristic values. The value of the load factor depends on the degree of uncertainty, while the combination factor depends on the probability of that combination of loads occurring. The nominal value of a permanent action (dead load), G, is its mean value. The factor 1.2 applied to it assumes that it can be assessed to within 10%. If circumstances arise where this assumption is not warranted, then a conservative estimate of the permanent action should be made, or part of it treated as an imposed action (live load). For the case of load reversal and where the permanent action is beneficial, the factor is taken as 0.9. The nominal value of the imposed action (live load), Q, is intended to be the peak value for a 50-year life with a probability of exceedence of 5%. This is a characteristic value with a probability similar to the definition of characteristic concrete strength, f 'c. Imposed actions vary and usually comprise two components: a sustained relatively constant value for a particular occupancy, and a superimposed extreme value arising from a crowd of people or a 1.10 Reinforced Concrete Design Handbook concentration of objects. The factor of 1.5 reflects this greater variability compared with permanent actions, and the combination factor, y, of 0.4 to 0.6 reflects the probable lower action likely to occur at the same time as another peak-action effect. See Table 1.1. Wind loads are determined in accordance with AS/NZS 1170 Part 2 either by a simplified procedure, applicable only to small buildings, or by a more detailed procedure using either a static or dynamic analysis. For significant structures, consideration should be given to a model tested in a wind tunnel to determine the wind forces more accurately, including effects on surrounding buildings. For earth-retaining structures in accordance with AS 46781.22, the nominal earth load, Feu, should take account of the likely earth and groundwater pressure. The load factor to be applied is 1.0 if the earth pressure is determined using a limit-states method, or 1.5 if earth pressures are determined using working loads. Liquid-retaining structures are usually constructed so that there is an upper limit to the height of the retained liquid and its density is defined. The accuracy of determination is similar to that of a permanent action, so a factor of 1.2 is used for static liquid pressure, Flp. However, if the density is not well defined or the height is not limited, then a value of 1.5 should be used. If dynamic conditions are possible, these should be considered separately and a factor similar to that for imposed actions applied. Earthquake loads and load combinations are specified in AS 1170 Part 41.23 while additional design and detailing requirements for earthquake resistance are given in AS 3600 Appendix C. The prestressing force, P, is limited by the breaking load of the tendons. In checking the ultimate strength of an anchorage or the possibility of a concrete compressive failure at transfer, a factor of 1.15 is specified by AS 3600. In this case, the prestressing force is similar to an external load and is taken as the maximum jacking force at the anchorage. A different situation arises where secondary moments and shears are being calculated in an indeterminate structure. Because these are caused by a prestress force, which is internal to the cross section, a load factor of 1.0 is sufficient. In selecting combinations of actions, the principle adopted is to consider each imposed action at its maximum value taken in turn with other imposed actions at their probable values at any time. The load combinations (combinations of actions) given in AS/NZS 1170.0 and those specified in AS 3600 are shown in Table 1.1. TABLE 1.1 Load combinations Strength 1.35G Permanent action only (does not apply to prestressing forces) 1.2G + 1.5Q Permanent action and imposed actions 1.2G + ycQ + Wu Permanent action and arbitrary-point-in-time imposed and wind actions G + ycQ + Eu Permanent action and arbitrary-point-in-time imposed and earthquake action (given in AS 1170.4) 0.9G + Wu Permanent action and wind action reversal 1.2G + ycQ + Su Permanent action and arbitrary-point-in-time imposed action and the appropriate one of the following actions: snow, liquid pressure, rainwater ponding, ground water and earth pressure 1.15G + 1.15P Permanent action and prestressing force (acting in same direction from AS 3600) 0.9G + 1.15P Permanent action and prestressing force (acting in opposite directions from AS 3600) G + ylQ + thermal action for fire Permanent action and arbitrary point-in-time imposed action and thermal action for fire Stability 0.9G For combinations that produce net stabilising effects 1.35G For combinations that produce net destabilising effects 1.2G + 1.5Q 1.2G + ycQ + Wu G + ycQ + Eu 1.2G + ycQ + Su Serviceability Use appropriate combinations of G, ysQ, ylQ, Ws, Es, P and other actions G + Ws + P eg short-term serviceability G+P eg long-term serviceability G + ysQ + P G + ylQ + P TABLE 1.2 Short-term, long-term and combination factors ys, yl and yc (after AS/NZS 1170.0) Imposed action Short-term factor (ys ) Long-term factor (y l ) Combination factor (yc ) Distributed imposed actions, Q Residential and domestic structures Offices Parking Retail Storage Other 0.7 0.4 0.4 0.7 0.4 0.4 0.7 0.4 0.4 0.7 0.4 0.4 1.0 0.6 0.6 1.0 0.6 0.6 Roof actions Roofs used for floor-type activities Other roofs 0.7 0.4 0.4 0.7 0.0 0.0 Concentrated imposed actions Floors Floors of domestic housing Roofs used for floor-type activities Other roofs 1.0 0.6 as for distributed floor actions 1.0 0.4 as for distributed floor actions 1.0 0.6 as for distributed floor actions 1.0 0.0 0.0 1.0 0.0 0.0 Balustrades Reinforced Concrete Design Handbook 1.11 1.5.4 Combinations of actions – stability In checking for stability, the loads and actions are divided into components tending to cause instability and those tending to stabilise the structure. Where the strength of a member is used to provide stability, then the design strength, f Ru, should be used. 1.5.5 Combinations of actions – serviceability For serviceability checks, both short-term and longterm effects should be considered. For wind loading, only short-term effects need to be considered. Values for yc, yl and ys are shown in Table 1.2. 1.6 Linear Elastic Methods of Analysis 1.6.1 General AS 3600 Clause 2.2.2 sets out the strength limit state for linear elastic methods of analysis with simplified analysis methods, and for statically determinate structures where the design capacity must be greater than or equal to design load effect, ie: Rd ≥ Ed where Rd = f Ru is the design capacity f is a capacity reduction factor given in AS 3600 Table 2.2.2 Ru is the calculated capacity determined in accordance with the relevant sections of AS 3600 Ed is the design action effect or the 'design action' or the ultimate load condition. Where Ed is determined by one of the following methods of analysis: — Linear elastic analysis in accordance with Clause 6.2 — Linear elastic analysis incorporating secondary bending moments due to lateral joint displacement in accordance with Clause 6.3 — One of the simplified methods of analysis in accordance with Clauses 6.9 and 6.10 — Equilibrium analysis of a statically determinate structure. 1.6.2 Linear elastic analysis (AS 3600 Clause 6.2) Concrete structures behave only in a linear elastic manner under small, short-term loads while the sections are uncracked. As loads increase, the sections crack and the behaviour becomes non‑linear and moments are distributed from the peak‑moment regions to less highly stressed sections of the members. Despite this, linear elastic analysis may be used to determine the action effects in structures for both the serviceability and the strength limit states. If the structure is ductile, this procedure has been shown by experience to be safe for strength design. 1.12 Reinforced Concrete Design Handbook ku 0.5 0.4 0.3 0.2 0.1 0 0 5 10 15 20 25 30 35 40 Redistribution (%) Figure 1.5 Moment redistribution versus ku Design moments and shears are calculated by applying factored loads to the structure, which is analysed assuming linear elastic behaviour. The critical sections are then checked for strength, assuming local inelastic action. Some redistribution of moments has to be relied upon when this model is used. This redistribution depends on ductile behaviour, which is ensured by limiting the neutral axis parameter, ku, to 0.36 and placing limits on the amount of redistribution depending on the Ductility Class of the reinforcement. AS 3600 Clause 6.2.7 gives guidance on the amount of moment redistribution that can be assumed in design. If Ductility Class N reinforcement is used, then the amount of redistribution permitted may be calculated using a deemed-to-satisfy approach based on the value of the neutral axis parameter, ku. Up to 30% is permitted provided that static equilibrium is maintained and ku does not exceed 0.2. For values of ku between 0.2 and 0.4, the permissible distribution is obtained by interpolation as shown in Figure 1.5. If Ductility Class L reinforcement is used, no redistribution is permitted unless an analysis is undertaken to show that there is adequate rotation capacity available at the critical cross sections to allow such redistribution to occur. Figure 1.6 illustrates the use of moment redistribution to reduce the maximum values of bending moment to be used in design for Ductility Class N reinforcement. A linear elastic analysis may be used for buildings with floor slabs and for framed structures without slabs. For reasons of equilibrium and static compatibility, the span of flexural members is taken as the distance centre-to-centre of supports. However, the size of these supports is taken into account by the defined critical sections for negative moment and shear force. In prestressed concrete members, the restraint at supports usually induces parasitic reactions and so-called secondary bending moments and shears that are determined by applying the prestress to an assumed unloaded, uncracked structure. As these are internally induced, a load factor of 1.0 is sufficient to obtain the design values that are added to the elastically determined moments and shears for factored dead and live load. These total moments may be redistributed in the same way as for reinforced concrete members. 1.6.3 Linear elastic analysis incorporating secondary bending moments (AS 3600 Clause 6.3) Span a A Span b B Span c C Load case 1: Spans a + b loaded Load case 2: Span b loaded Load case 3: Spans a + c loaded D Modify moments as follows: Load case 1: Reduce negative moment at B and increase positive moments Load cases 2 & 3: Reduce positive moments and increase negative moments Figure 1.6 Examples of moment redistribution This method applies to frames that are not restrained by shear walls or bracing or both and for which the relative displacement at the ends of compression members is less than Lu / 250 under the design load for strength. It is similar to linear elastic analysis except that additional bending moments are calculated to take account of the lateral displacement. 1.6.4 Simplified methods of analysis (AS 3600 Clauses 6.9 and 6.10) [a] Idealised frame method (Clause 6.9) This method may be used for framed structures incorporating reinforced or prestressed two-way slab systems and is not subject to the restrictions on geometry and loading applicable to other simplified methods. The idealised frame is taken as one of a series of idealised full-height frames running longitudinally through the building and a second series taken transversely. A linear elastic analysis is carried out for each frame using one of the standard frame programs or similar and using a number of practical simplifications: n n n n For vertical loads, a simple frame may be taken as comprising one floor and the columns above and below, with these columns fixed at their far ends Figure 1.7. The width of the frame may be taken as the width of the design strip for flat slabs, or the effective width for T-beams and L-beams using the equations in AS 3600 Clause 8.8.2 and as shown in Figure 1.8. Figure 1.7 Idealised frame 0.1a but not greater than 0.5 x clear span of slab bw 0.1a but not greater than 0.5 x clear span of slab 'a ' is distance between points of zero moment along beam (approx. 0.7L) Figure 1.8 Effective width of T-beam The relative stiffness of the members may be calculated using the gross concrete section or the transformed section if the same basis is used throughout. The fully idealised frame must be considered in the analysis for horizontal loads unless it is restrained by shear walls or similar. Reinforced Concrete Design Handbook 1.13 n n Ductility Class L reinforcement must not be used for the main flexural reinforcement. Openings must comply with AS 3600 Clause 6.9.5.5. The arrangement of vertical action (load) to determine the critical moments and shears may be restricted to only a few combinations as set out in AS 3600 Clause 2.4.4. Imposed action (live load) ALL SPANS n If the imposed action (live load) pattern is fixed — the factored imposed action (live load). n Imposed action (live load) ALTERNATE SPANS If the imposed action (live load), Q, does not exceed three-quarters of the permanent action (dead load), G, — the factored imposed action (live load) on all spans. n Imposed action (live load) ADJACENT SPANS — the factored imposed action (live load) on alternate spans Figure 1.9 Examples of pattern actions (loading) [– 141 ] – 1 16 – 1 9 – 1 9 – 1 10 – 1 11 – 1 11 + 1 16 [+ 141 ] + 1 11 as shown in Figure 1.9. [b] Simplified method for reinforced continuous beams and one-way slabs (Clause 6.10.2) Two spans – 1 16 — the factored imposed action (live load) on all spans. Simple support + 1 11 + 1 11 — the factored imposed action (live load) on two adjacent spans Moment = coefficient x Fd Ln2 [– 81 ] [– 81 ] – 1 10 – 1 24 This method provides a simple, approximate and conservative evaluation of the bending moments and shears in continuous reinforced concrete beams and one-way slabs where: n + 1 11 n Spandrel beam More than two spans MOMENTS n Shear = coefficient x Fd Ln 1 2 1.15 2 1 2 1 7 1 2 1 8 1.15 2 1 2 1 7 Spandrel beam End and interior spans SHEARS Figure 1.10 Approximate moments and shears – one-way members. Bracketed blue figures are for Ductility Class L reinforcement 1.14 If the imposed action (live load), Q, exceeds threequarters of the permanent action (dead load), G, Reinforced Concrete Design Handbook the spans are approximately equal, with the longer of two adjacent spans not greater than the shorter by more than 20%; the loads are essentially uniformly distributed and the imposed actions (live load) do not exceed twice the permanent actions (dead load); the members are of uniform cross section and the reinforcement is arranged in a specific way. This method is normally used only on simple structures and slabs, usually supported by loadbearing walls or similar. The coefficients are shown diagrammatically in Figure 1.10. Because these values are not statically compatible, they should not be used for deflection calculations. If moment reversals occur during construction, these should be investigated separately. Note the higher moments (bracketed blue figures) to be used with Class L reinforcement. [c] Simplified method for reinforced two-way slabs supported on four sides (Clause 6.10.3) For reinforced two-way slabs supported on four sides by beams or walls (having corners that are prevented from lifting) and subject to uniformly distributed loads, approximate bending-moment coefficients for a range of edge conditions are tabulated in AS 3600 Tables 6.10.3.2 (A) and (B) depending on whether Class N or Class L reinforcement is used. Detailing must be in accordance with AS 3600 Clause 9.1.3.3 and there must be no openings or penetrations through the thickness of the slab adversely affecting its strength or stiffness. Slabs with Class L reinforcement must be continuously supported on walls. Moments are calculated as: Moment = coefficient b x Fd Lo2 where Lo = L – [0.7 x S (asup at each end of the span)] The coefficients bx and by are given (in decimal values) in AS 3600 Tables 6.10.3.2 (A) and (B). The shear forces in the slab and the reactions of the supporting beams may be determined by allocating the load using 45° lines emanating from the corners of the slabs as shown in AS 3600 Figure 6.10.3.4. Although not stated in AS 3600, the shear force at the face of the first interior support of an end span should be taken as 1.15 times the simple span value, similar to that for one-way slabs. [d] Simplified method for reinforced two-way slab systems having multiple spans (Clause 6.10.4) The simplified method provides a quick and direct method of design for slabs that meet the restrictions of geometry and live loading (imposed actions) as set out in AS 3600 Clause 6.10.4.1. The total static moment is calculated for each span of the design strips taken in two directions at right angles using the effective span, Lo. This is consistent with the idealised-frame method as the critical section for negative bending moment is the same in each case. The design positive and negative moments are then determined by applying the factors tabulated for interior spans and end spans. In the latter case, the distribution is varied to suit the end restraint provided by the exterior support. Where adjacent spans differ, the designer may either use the larger negative moment or distribute the unbalanced moment to the adjoining members to obtain the design negative moment. In addition, a redistribution of the design moments of up to 10% is permitted, if the total static moment is not reduced. Only Class N reinforcement can be used for this design method. Table 1.3 Distribution of design moment to column strip (from AS 3600 Table 6.9.5.3) Column strip moment factor for Location Strength limit state Negative moment interior support exterior support 0.6 to 1.0 0.75 to 1.0 Positive moment 0.5 to 0.7 Serviceability limit state 0.75 0.75 to 1.0* 0.6 * Depending on whether there is a spandrel beam The transverse distribution of these design moments into the column strip and the middle strip is carried out using tabulated factors that are the same for both the simplified method and the idealised-frame method and are shown in Table 1.3. The values for these factors give the designer a considerable range in which to work. Unequal spans or patterned live loads cause unbalanced moments to be transferred from the slab to the column. A minimum value is specified for this unbalanced moment and this is obtained by taking half the design live load as acting on the longer span and no live load on the shorter span. This moment is included in the shear design at the column-slab junction. 1.7 Material and Construction Requirements 1.7.1 General The designer is obliged to set out in the drawings and specification all the requirements for the construction of the structure so that it can be built to meet the intent of the design. AS 3600 Clause 1.4 sets out the design data that should be shown on the drawings. Note that AS 3600 does not contain specification-type clauses relating to construction, it has only general performance-type clauses. The project specification thus needs to spell out the specific requirements for the project's construction. The project specification should include those items of good practice that the designer considers necessary. A useful document is the national building industry specification system, NATSPEC, which is a master specification containing a library of clauses from which designers can select those suitable for their project and which they can supplement with specific clauses as required. Reinforced Concrete Design Handbook 1.15 1.7.2 Concrete materials and manufacture The constituent materials of concrete and its manufacture are covered by a series of standard specifications, most of which are sufficiently comprehensive to require only a citing in the project specification. Where an Australian Standard specification is not available, reference may be made to an overseas standard such as ASTM, or a performance requirement may be set out in the specification. The manufacture of concrete is covered by AS 13791.24, which includes all methods of manufacture: site-mixed, transit-mixed and factory-mixed. It covers the handling, storage and batching of materials, equipment such as mixers and agitators, mixing and delivery of the plastic concrete. 1.7.3 Specification of concrete AS 3600 specifies concrete by referencing AS 1379. This latter standard defines two classes of concrete: normal-class and special-class. Normal-class concrete is satisfactory for the majority of projects, while special‑class concrete is specified only where particular performance criteria are needed or as required by AS 3600, such as for B2, C1 and C2 exposure. The classifications are used to avoid misunderstandings between the builder and the concrete supplier, and the possibility of concrete being ordered only in terms of strength when special requirements are called up in the specification. Normal-class concrete provides for the standard strength grades N20, N25, N32, N40 and N50 with slump of 20, 30, 40, 50, 60, 70, 80, 90, 100, or 120 mm and maximum nominal size of aggregate of 10, 14 or 20 mm. The particular value of each together with the intended method of placement (if project assessment is required) and air entrainment (if required) should be specified. Generally, normal-class concrete with strengths in the range of 25 to 50 MPa is used on low- and medium‑rise building structures. Higher strength concretes are used typically in walls and columns carrying high loads in taller structures, or in special conditions. Designers should be aware that S class concrete with strengths greater than 50 MPa might be difficult to supply to some sites (eg in country areas), while very high strength concrete (> 65 MPa) may not be available in some cities. Before specifying high performance concrete with strengths greater than 50 MPa, designers should check on its availability with their local suppliers. Special-class concrete is specified when there are any different or additional requirements, and only after careful consideration for its need. It will be needed when the concrete is to have: compressive strength, 1.16 Reinforced Concrete Design Handbook slump or aggregate other than those available in normal-class concrete; any limit on ingredients or mix proportions; or any special performance requirement such as a particular limit on shrinkage. Special‑class concrete is designated as S class and can have different prefixes depending on its specific requirement. Where concrete is specified as special class and one of the exposure classifications B1, B2, C1, C2 or U is specified in AS 3600, prefixes to the strength grade shall be SB for concrete in exposure classification B2, SC for concrete in exposure classification C1 or C2 and SU for concrete in exposure classification U. For SB, SC or SU class concrete, the properties specified shall include the aggregate durability class in accordance with AS 2758.1 and the relevant requirements of AS 3600. To avoid misunderstanding when the concrete is specified and ordered, the class must be stated. 1.7.4Quality control of concrete AS 3600 requires that all concrete for structures designed in accordance with it shall be assessed in accordance with AS 1379 for the specified parameters. All concrete must be tested and subjected to production assessment by the supplier to ensure that the appropriate quality is being maintained. Project assessment is optional for normal-class concrete but is mandatory for special-class concrete. If project assessment is specified, then the concrete delivered to that project is subject to additional testing. In this case, the specification should nominate the responsibility for carrying out this extra testing and who will bear the cost. Note that for specified parameters other than strength, the specification has to set out the method of production control, eg test method, the frequency of testing and acceptance criteria. For a large project, project assessment will usually give sufficient samples to obtain a statistically reliable assessment of the concrete supplied to that project and at an acceptable cost. However, for small or medium sized projects, the cost of obtaining sufficient samples for a reliable assessment is usually prohibitive. Production assessment, as specified in AS 1379, will provide a reasonable level of quality assurance for the majority of small structures. 1.7.5 Tolerances for construction Tolerances for structures and members are specified for two reasons. The first is concerned with structural adequacy and strength, ie to ensure that the design assumptions, in particular the f factors used in the strength calculations, are met. The second relates to serviceability and appearance and will normally overrule the requirements of the first. AS 3600 specifies only tolerances that are necessary for the first reason, ie strength and safety. The designer should examine these carefully, and should generally specify tighter tolerances for construction. Limited guidance on this latter type is provided in AS 3610 where tolerances are specified for different classes of surface finish. Experience has shown that concrete structures can typically be built to tolerances of about ±10 mm to ± 20 mm. Where such tolerances are not achieved, the non-structural elements such as walls, windows, etc will often not fit properly, or extra costs will be incurred in achieving flat faces, etc. This topic is also covered in Section 4.5 of the Precast Concrete Handbook. The tolerances specified in AS 3600 reflect design practice and should be easy to achieve. For exposedto-view concrete, the classes of surface finish in AS 3610 cover a range of work from the highest quality, suitable for monumental structures, to average quality, suitable for many structures. The designer should balance the cost of formwork and related construction to produce a given standard of finish against the overall appearance of the project and/or part being considered and the distance from which it will be viewed. The practical difficulty of accurately measuring concrete surfaces and of achieving tight tolerances should not be overlooked. Note that AS 3610 limits the areas where the Class 1 finish can be specified. Its use requires a pragmatic approach. For practical convenience, tolerances in AS 3600 are measured to the surface and not to the centreline of members. Any point on a surface should lie within a tolerance envelope from its theoretical position. For columns and walls in the first 20 storeys of a building, an absolute limit of 40 mm horizontally is specified to control the overall location of the building. For columns and walls, the deviation from plumb, floor-to-floor, must not exceed the greater of the specified dimension divided by 200 or 10 mm. For other members, the deviation from a specified dimension must not exceed the greater of the specified dimension divided by 200 or 5 mm. In checking these tolerances, an allowance must be made for possible movement of members, such as the deflection of floors, axial shortening both vertically and horizontally or thermal movements in slender structures. The acceptable tolerance on location of reinforcement and tendons depends on the effect of any variation on the strength of the member and also on the possible reduction in cover and its effect on durability. Negative tolerances are permitted on cover and have been allowed for in the covers specified for durability and axis distance for fire resistance in AS 3600. Where durability is considered a significant factor, consideration should be given to using a larger cover than the minimum requirement of AS 3600. 1.7.6Formwork The general requirements for formwork are covered in AS 3610. The particular requirements for formwork that affect the safety and serviceability of the concrete structure are specified in AS 3600. These relate essentially to two conditions: n n The removal of formwork and the strength of the recently cast member The loads imposed on the structure by the plastic concrete and its supporting formwork. The builder is usually responsible for the design, erection, stripping and safety of the formwork. This responsibility should be stated clearly in the project specification. In addition, if the designer requires particular constraints on the method of construction or wishes to oversee and review details of the proposed formwork then this should be specified. The removal of formwork from a vertical surface is controlled by the time necessary for the concrete to gain sufficient strength and to avoid damage during stripping the exposed surface. In addition, where the colour of the off-form concrete needs to be consistent, the time of removal should be similar for all elements. Typically such surfaces are stripped at 1–2 days. The removal of soffit formwork from reinforced beams and slabs at an early age is limited by the need for safety, to control cracking in the concrete and to limit the deflection. In the case of a slab with undisturbed shores kept in place, the slab is analysed as a plain concrete member subject to its self-weight plus a construction load of 2 kPa Figure 1.11. The design moment induced by this load must be less than the ultimate strength of the section calculated using characteristic flexural tensile strength of the concrete at the time of form removal. If control samples are taken and the concrete strength is obtained by Design load (slab self-weight + 2 kPa construction load) Slab soffit (crack control by flexural tensile strength) Undisturbed shores Figure 1.11 Form removal from soffit Reinforced Concrete Design Handbook 1.17 testing, then that strength can be used to determine the characteristic flexural strength. As an alternative to testing of concrete, the mean concrete strength at 7 days may be taken as the values specified in AS 1379 Table 2. In lieu of performing the above calculation, the designer may use the minimum stripping times for soffit forms specified in AS 3600 Tables 17.6.2.4 and 17.6.2.5 but formwork supports must stay in place for longer. These tables cover two cases: reinforced slabs continuous over formwork supports and of normalclass concrete cured at various temperatures; and removal of formwork supports from beams and slabs not supporting structure above. In multi-storey structures, the early removal of formwork and its supports is desirable for speed of construction and economical reuse of forms. The floor being cast at any time is supported by the floors below. Depending on the number of supporting floors (usually a minimum of 3 or 4) and the relative stiffness of these floors, the load is shared between them with a series of closely spaced props. The floor directly beneath that being cast is less mature than, and is being supported also by, the floors below it. A simplified elastic analysis of these floors results in quite high values for the maximum load. Depending on the number of sets of forms, this maximum load varies between 2.25 and 2.40 times the self-weight of the floor, with a converged value of 2.0. If the live load is less than the dead load of the floor, clearly there is a danger of overloading the slabs during construction. Further, if the temperature and curing conditions are not as expected there is a danger of a slab failure. Various methods such as prop release and re-shoring have been devised to reduce these apparently very high loads, and measurements have been taken to check if the simplifications of analysis are reasonable. The measured loads generally comply with the simplified analysis; the variation apparently being due to creep and shrinkage warping of the concrete frame. Further guidance on this subject may be obtained from AS 3610 and literature referenced in it. Prestressed floors are usually designed for staged stressing so that the prestress is applied progressively as the concrete gains strength, the floor becomes largely self-supporting, and the forms may be removed in stages, usually at about 7 days when the floor is fully stressed. For multi-storey buildings and several levels of formwork, the sharing of load and analysis is as for reinforced concrete. 1.18 Reinforced Concrete Design Handbook References 1.1 Building Code of Australia Australian Building Codes Board, 2010. 1.2 AS/NZS 1170 Structural design actions, Standards Australia, Part 0: General principles Part 1: Permanent, imposed and other actions Part 2: Wind actions Part 4: Earthquake actions in Australia. 1.3 AS 3600 Concrete structures Standards Australia, 2009. 1.4 Sustainable Concrete Materials Briefing 11, Cement Concrete & Aggregates Australia, 2010. 1.5 Sustainable Concrete Buildings Briefing 13, Cement Concrete & Aggregates Australia, 2010. 1.6 Manual for the design of reinforced concrete building structures, 2nd Ed, Institution of Structural Engineers (jointly with Institution of Civil Engineers), 2002. 1.7 Manual for the design of concrete building structures to Eurocode 2 Institution of Structural Engineers, 2006. 1.8 ACI 318 Building code requirements for structural concrete American Concrete Institute, 2008. 1.9 Foster SJ 'Design and analysis procedures', paper presented at AS 3600 – 2009 National Seminar Series: What's New? What's Different? Improvements and Developments – what are the implications and what do they mean for you, Concrete Institute of Australia, 2009. 1.10 Foster SJ, Kilpatrick AE and Warner R Reinforced Concrete Basics 2E: Analysis and design of reinforced concrete structures Pearson Education, Australia, 2010. 1.11 Serviceability – Design for Deflection and Crack Control Concrete Institute of Australia Seminar Series, 2010. 1.12 Guide to Concrete Construction (T41/HB64) 2nd Ed, Cement Concrete & Aggregates Australia/Standards Australia, 2002. 1.13 Movement, restraint and cracking in concrete structures, The Concrete Society TR67, 2008. 1.14 Precast Concrete Handbook 2nd Ed, National Precast Concrete Association Australia and Concrete Institute of Australia, 2009. 1.15 BS 8110 Structural use of concrete Part 1: Code of practice for design and construction British Standards Institution, 1997. 1.16 BS EN 1992, Eurocode 2: Design of concrete structures British Standards Institution, 2004. 1.17 Mitchell D and Cook WD 'Progressive collapse of slab structures' Concrete Framed Structures: Stability and Strength Narayanan R (ed), Elsevier Applied Science, 1986. 1.18 FIP Commission on Prefabrication Planning and Design Handbook on Precast Building Structures, Thomas Telford Ltd, 1994. 1.19 Elliott KS and Tovey AK Precast concrete framed buildings: Design guide British Cement Association, 1992. 1.20 ACI-ASCE Committee 550 'Design Recommendations for Precast Concrete Structures', ACI Structural Journal JanuaryFebruary 1993, pp 115–121. 1.21 AS 3610 Formwork for concrete 1995 and AS 3610.1 Formwork for concrete Part 1: Documentation and surface finish Standards Australia, 2010. 1.22 AS 4678 Earth-retaining structures Standards Australia, 2002. 1.23 AS 1170 Structural design actions Part 4: Earthquake actions in Australia Standards Australia, 2007. 1.24 AS1379 Specification and supply of concrete Standards Australia, 2007. Reinforced Concrete Design Handbook 1.19 blank page 1.20 Reinforced Concrete Design Handbook Chapter 2 Design properties for concrete and reinforcement 2.1 Concrete 2.1.1 General The performance of a concrete structure depends on a number of factors ranging from the design to its loading history. Not the least of these factors is the insitu quality of the concrete in it. This in turn is affected by two major factors, the quality of the concrete supplied to the project and the construction process employed. Considering first the quality of the concrete supplied to the project, the major factors influencing this are the type and quality of the constituent materials and the proportions in which they are mixed. However, for design, it is usual to adopt values for the various design properties, eg compressive strength, on the basis of what is a reasonable design value from a memberbehaviour perspective and what can be achieved in the geographical location using local materials. In general, designers should specify the values of only the concrete properties they require and not specify limitations on how the supplier should produce the concrete, except to require that materials and manufacture comply with the relevant Australian standards. Equally important to ensure that the design quality is achieved for the concrete in its final place, is specifying that the concrete is appropriately transported, placed, compacted, finished and cured, not overlooking the fact that to facilitate this, appropriate properties of the fresh concrete need to be specified. However, the importance of actually ensuring that the provisions of the specification are complied with on site must not be overlooked. Ensuring that proper curing is undertaken and that unauthorised addition of water is not allowed are also important 2.1. Design values for concrete are specified in AS 3600 2.2. In general, these are characteristic values, eg f 'c , as they provide appropriate values for strength design accommodating the variation inherent in concrete production and the subsequent construction processes. However, average values are preferred for some properties, eg Ec , as they give a better prediction of the in-service behaviour of the member or structure. Generally, AS 3600 provides for design properties to be taken as either a prescribed value or to be determined from tests carried out on concrete made from the proposed materials. Alternatively, values may be derived from historical records of similar concrete. If records are not available and tests are required, the designer should allow for the time and possible delays to obtain the results. If test results and testing of various properties are to be specified, the designer needs to understand the precision of the test, ie the repeatability and reproducibility. The choice of the design value and the value included in the specification need to take these factors into account. 2.1.2 Strength General AS 3600 specifies concrete with a characteristic 28-day compressive strength in the range 20 to 100 MPa. Although most pre-mixed concrete suppliers can supply concrete in the range 20 to 50 MPa, designers should be aware that concretes above 50 MPa are deemed to be special-class concretes and may not be readily available in some regions. More importantly, while the design methods in AS 3600 cover high strength concretes, additional design and detailing will be required for them than for lower strength concretes. Compressive strength AS 3600 Clause 3.1.1.1 specifies standard strengths of 20, 25, 32, 40, 50, 65, 80 and 100 MPa. In this series, the strength of each grade is about 25% greater than that of the preceding grade. In practice, for members such as slabs and beams, the choice of strength grade will frequently be determined by durability and serviceability considerations rather than the structural requirements. However, for columns and walls it may be determined by load-carrying capacity, ie strength. Non-standard strength grades may be specified but these are deemed to be special-class concretes. Associated properties, eg shrinkage, may need to be specifically determined and project testing is required. Tensile strength The uniaxial tensile strength of concrete is determined from either: n Tests. The flexural tensile strength obtained by testing plain concrete beam specimens and calculating the extreme fibre stress at failure in accordance with AS 1012.11 2.3, or the principal tensile strength obtained using the split-cylinder test method in accordance with AS 1012.10 2.3. In these cases the uniaxial tensile strength, fct, is taken as: fct = 0.6 fct.f or fct = 0.9 fct.sp as appropriate. Reinforced Concrete Design Handbook 2.1 Alternatively, in the absence of more-accurate data the characteristic uniaxial tensile strength may be taken as 0.36√f 'c, and the characteristic flexural tensile strength as 0.6√f 'c. n The uniaxial tensile strength is used in calculations limiting cracking of concrete such as web shear cracking in prestressed beams, while the flexural strength is used in designing plain concrete members such as pavements, and in calculations to control flexural cracking. Stress 0.9 f ' c 0.45 f ' c E 1 2.1.3 Modulus of elasticity c 0 The modulus of elasticity of concrete, Ec, is taken as the secant modulus of the non-linear stress-strain relationship as shown in Figure 2.1 and is used in the calculation of deformations. Figure 2.1 Modulus of elasticity of concrete In most cases, the value of Ec can be taken as the value given by the empirical formulae given in the Standard: the water‑cement ratio. AS 3600 suggests using a density of 2400 kg/m3 for normal-weight unreinforced concrete. Ec = r1.5 x 0.043√fcmi (when fcmi ≤ 40 MPa); and Ec = r1.5 x (0.0243√fcmi + 0.12) (when fcmi > 40 MPa) For normal-weight concrete and up to fcmi = 40 MPa, the formula reduces to Ec = 5055√fcmi . Note that the formula uses fcmi, the mean value of the in situ compressive strength at the age of the concrete being considered, not f 'c. In the absence of more-accurate data fcmi can be taken as 90% of the mean value of the cylinder strength, fcm. The precision of the formula is noted as ± 20%. If a higher precision on the calculation of immediate deflection is required, then values from trial mixes or similar concretes should be used. Properties of standard concrete grades using equations given in the Standard are shown in Table 2.1. 2.1.4 Density To comply with AS 3600 the saturated surface-dry density of the concrete has to be in the range of 1800 to 2800 kg/m3. The density of plain concrete depends on the density of the coarse aggregate and Strain, εc For reinforced concrete, an allowance should be made for the reinforcement. For most structures, a conservative value of 2500 kg/m3 (25 kN/m3) for the unit weight of reinforced or prestressed concrete is satisfactory for design purposes. AS 1170.1 2.4 suggests that the density of reinforced concrete is 24 kN/m3 plus 0.6 kN/m3 for each 1% of reinforcement. 2.1.5 Stress-strain curves The Standard does not prescribe a stress-strain curve for concrete but allows the use of a curvilinear form defined by recognised equations, eg CEB2.5, or determined from test data. For design, the shape of the in situ uniaxial compressive stress-strain curve is taken as that for a maximum stress of 0.9 f 'c for strength and fcmi for serviceability limit states, respectively. Under uniaxial stress, for concrete with characteristic compressive strengths in the range 20 to 100 MPa, the peak stress occurs at a strain of approximately 0.0025, but varies with mix. The shape of the curve changes Table 2.1 Properties of standard concrete grades Grade or characteristic compressive strength, f 'c (MPa) Characteristic flexural tensile strength, f 'ct.f = 0.6 √f 'c (MPa) Uniaxial tensile strength, f 'ct = 0.36 √f 'c (MPa) Modulus of elasticity, Ec.28 (MPa) 20 25 32 40 2.68 3.00 3.39 3.79 1.61 1.80 2.04 2.28 24 000 26 700 30 100 32 800 50 65 80 100 4.24 4.84 5.37 6.00 2.55 2.90 3.22 3.60 34 800 37 400 39 600 42 200 2.2 Reinforced Concrete Design Handbook with strength, with the ascending and descending branches becoming steeper as the strength increases. In the stress-strain curve, the maximum stress is modified from that of the standard cylinder test to account for the differences in the size, environmental/ curing and testing conditions between that of the cylinder test to that of the insitu concrete under local conditions. 2.1.6 Poisson's ratio The Standard provides a value of 0.2 for concrete. This assumes the concrete is uncracked in tension. Table 2.2 Value of final drying basic shrinkage strains for major cities (after AS 3600) City Brisbane Sydney Melbourne Adelaide Perth Hobart Value of final drying basic shrinkage strain (mm/mm x 10–6) 800 800 900 1000 1000 1000 2.1.7 Coefficient of thermal expansion The Standard provides a value of 10 x 10–6/ °C. This should be sufficient for most calculations even though the actual value can vary by ± 20% depending on aggregate type, volume of cement paste and degree of saturation. In other standards, eg Eurocode 2 2.6, the values of the coefficients of thermal expansion for concrete and steel are taken as equivalent, whereas AS 3600 suggests different values for each. 2.1.8 Shrinkage Shrinkage is the decrease in the volume of hardened concrete caused mainly by the loss of moisture as a result of drying, and also by chemical changes in the cement hydration products. It is independent of the load applied to the concrete; it depends chiefly on the relative humidity and temperature of the environment, the size of the member and the constituent materials of the concrete. The basic shrinkage strain is measured by taking standard test specimens wet-cured for 7 days and then stored in air at 23°C at a relative humidity of 50% for 56 days. Tests have shown that the aggregate type has a significant influence on the shrinkage of concrete. The range of shrinkage values for normal concrete in major cities is given in Table 2.2. The figures reflect the best estimate of the value for design purposes of the whole range of normal‑class concretes available in Australia. If designers are concerned about shrinkage, a better estimate for design purposes can be obtained by using measurements on similar local concrete. If shrinkage is a significant design parameter, then special-class concrete should be specified and the desired basic shrinkage strain nominated (remembering that for such concrete extra project testing will be required). Designers should also check that such concrete can be supplied since suitable aggregates may not be available locally to achieve such limits. The Standard gives a method to calculate the design shrinkage strain, ie the sum of the autogenous shrinkage strain and the drying shrinkage strain, of concrete at any time. It also provides a table of typical final design shrinkage strains after 30 years based on a value of 1000 x 10-6 for the final drying basic shrinkage strain. The design shrinkage strain for various environments and size of member may be obtained directly from the series of curves given in the Standard. The four environments covered may be assumed to reflect conditions of increasing humidity. The description 'interior environments' reflects the situation inside non-air-conditioned buildings, while the others reflect the environments defined for durability considerations given in Clause 4.3 of the Standard. For practical design conditions, these general classifications are considered to provide a more useful guide to designers than would an attempt to provide absolute values for the effects of temperature and relative humidity. The suggested accuracy for the calculation of ecs using the nominated figures is ± 30%. The benefit of obtaining more-accurate results should be assessed before embarking on time-consuming and costly methods of data collection and calculation. 2.1.9 Creep of concrete Creep of concrete is defined as the time-dependent increase in strain under sustained loading. The basic creep coefficient is expressed as the ratio of the ultimate creep strain to the elastic strain of a standard specimen initially loaded at 28 days and maintained under a constant stress of 0.4 f 'c. For the practical calculation of the creep of a member, the basic creep coefficient is modified for the effects of member size, exposure environment and the maturity of the concrete at the time of loading. In the absence of specific data for local concrete, the designer may use the average values for the basic creep coefficient and modification factors given in the tables and graphs in the Standard. The suggested accuracy of creep calculations based on this average data is ± 30%. Reinforced Concrete Design Handbook 2.3 The creep under constant stress as determined above is known as pure creep. Practical examples of pure creep include creep due to prestress and sustained or dead load on uncracked concrete, such as axial shortening of concrete columns and loadbearing walls of buildings. However, where stresses are induced by movements such as settlement or shrinkage, the initial stress caused by the induced strain is reduced by creep. This loss of stress is known as relaxation. 2.2 Reinforcement 2.2.1 General Reinforcement (reinforcing steel) is defined by the Standard in Clause 1.6.3.68 as 'steel bar, wire or mesh but not tendons'. This definition precludes the use of fibres (steel and other types), non-metallic reinforcement and non-tensioned prestressing strand, bars and wires if the structure is to comply with the Standard. The Standard also requires (see Clause 17.2.1.1) that reinforcement be deformed bars or mesh (welded wire fabric of either plain or deformed wire) although plain bars or wire may be used for fitments. The following observations on reinforcement relate to the requirements set out in AS/NZS 4671 2.7. 2.2.2 Shape Reinforcing bars can be either plain, deformed ribbed or deformed indented. The shapes are designated by the letters R (Round), D (Deformed ribbed) and I (deformed Indented) respectively. Generally, only deformed ribbed bars will meet the intention of the requirement in AS 3600 that reinforcement be deformed. However, AS/NZS 4671 contains provisions outlining a test method to measure the bond performance of indented bars or ribbed bars with ribs not meeting the specification set out in that standard. 2.2.3 Strength Strength grade is represented by the numerical value of the lower characteristic yield stress, 250, 300, and 500 MPa. Reinforcing steel with a strength grade above 250 MPa is also required to comply with the specification of an upper characteristic yield stress. 2.2.4 Ductility class The three classes of ductility are designated L, N and E for low, normal and earthquake respectively. Ductility Class E has been especially formulated for New Zealand and is not manufactured or available in Australia. 2.4 Reinforced Concrete Design Handbook AS 3600 imposes a number of limitations on the use of Ductility Class L reinforcement. These reinforcing materials may be used as main or secondary reinforcement in the form of welded wire mesh, or as wire, bar and mesh in fitments; but are not permitted 'in any situation where the reinforcement is required to undergo large plastic deformation under strength limit state conditions' (see Clause 1.1.2 (c) (ii)). Importantly it also states (Clause 17.2.1.1) that 'Ductility Class L reinforcement shall not be substituted for Ductility Class N reinforcement unless the structure is redesigned '. The use of Ductility Class L reinforcement is further limited by other clauses in AS 3600. For example, where Ductility Class L reinforcement is used and where the design incorporates moment redistribution, then the designer has to undertake an analysis to show that there is adequate rotation capacity in critical moment regions to allow the assumed redistribution to take place. There is also a different value for the capacity reduction factor f throughout the Standard where Ductility Class L reinforcement is used. For further background as to the reasons for the restrictions on the use of Ductility Class L reinforcement, refer to the national seminars on AS 3600—2009 2.8. 2.2.5 Size The common sizes of bar available in Australia of the various grades and classes are shown in Tables 2.3 and 2.4. 2.2.6 Weldability Reinforcement conforming to AS/NZS 4671 is weldable. Depending on the manufacturing process used and the chemical composition of the steel, the requirements for welding may differ and may be more or less stringent than requirements for other reinforcement complying with that Standard. Designers should consult the steel producer's literature for specific advice. Any structural welding of reinforcing steel should comply with AS/NZS 1554.32.9 and be carried out by qualified operators. More‑detailed information and guidance is provided in the WTIA Technical Note 12.10. Welding of galvanised reinforcement needs care and should be avoided if possible due to possible damage to the coating. Locational tack-welds can be used for pre-assembly of reinforcement cages in lieu of tying at bar intersections. They may be smaller than tack welds as defined in AS/NZS 1554.4 and are (currently) not covered by it. They should be performed by trained personnel and should be executed in a manner that does not cause notching or reduce the cross sectional area of the intersecting bars. Where reinforcement cages are to be lifted care is required to ensure the welds are adequate to support the weight of the cages. Table 2.3 Nominal values for hot-rolled deformed bars of grade D500N Size N10 N12 N16 N20 N24 N28 N32 N36 N40 Cross-sectional area (mm2) 78.5 113 201 314 452 616 804 1020 1260 Mass per metre length (kg/m) 0.617 0.890 1.580 2.470 3.550 4.830 6.310 7.990 9.860 Notes: — These normal-ductility bars are used typically in beams, slabs as flexural reinforcement and in columns and walls as compression reinforcement. — This Table includes sizes outside AS/NZS 4671. — N10 bars are not available in all States and Territories. — N40 bars may be available only on special order for larger quantities. — Larger diameter fitments are made with N12, N16 and N20 bar as required. Table 2.4 Nominal values for high-strength deformed bars of grade D500L Table 2.5 Nominal values for hot-rolled round bars of grade R250N Size Size L4 L5 L6 L7 L8 L9 L10 L11 L12 12.6 17.7 28.3 35.8 45.4 57.4 70.9 89.1 111.2 Mass per metre length (kg/m) 0.099 0.139 0.222 0.281 0.356 0.451 0.556 0.699 0.873 Notes: — These low-ductility bars (sometimes known as wires) are used commonly as fitments in beams and columns generally using L6, L8, and L10 sizes. The other sizes may not be readily available. — Larger size fitments are usually made from N bar as in Table 2.3. Mass per metre length (kg/m) R 6.5 R10 30 80 0.267 0.632 R12 R16 R20 R24 110 200 310 450 0.910 1.619 2.528 3.640 Notes: — The shaded bars, R 6.5 and R10, are used for fitments. — R12 to R24 bars are generally used only for dowel bars. — For dowel bars larger than R24, check with supplier. Note that reinforcement manufactured overseas may not conform to AS/NZS 4671, as it may have a higher carbon equivalent content. Some overseas sources, however, can supply complying reinforcement. Designers should consult the Australian Certification Authority for Reinforcing (ACRS) for details of those suppliers. Generally, welding of reinforcement to comply with AS/NZS 4671 will require the use of: n Cross-sectional area (mm2) Cross-sectional area (mm2) n n hydrogen-controlled electrodes; special precautions in adverse conditions, eg wet weather, temperatures ≤ 0°C; preheating when bars over 25 mm diameter are being welded. Note the limitation on the location of welds in a bar that has been bent and re-straightened specified in AS 3600 (Clause 13.2.1(f)), ie it shall not be welded closer than 3d b to the area that has been bent and re-straightened. 2.2.7 Bending and re-bending reinforcement AS/NZS 4671 specifies for bars of diameter ≤16 mm a 90° bend and rebend test and for bars ≥ 20 mm a 180° bend test. It is thought that these requirements will ensure that bars likely to be restraightened in the field, ie with d ≤16 mm, can be safely re-bent. 2.2.8 Mesh Meshes commonly available in Australia are shown in Table 2.6. Reinforced Concrete Design Handbook 2.5 Table 2.6 Meshes commonly available in Australia Longitudinal bars Cross bars Mesh type and reference No. x dia. Pitch No. x dia. Pitch number (mm) (mm) (mm) (mm) RL 1218 RL 1118 RL 1018 RL 918 RL 818 RL 718 Rectangular 25 x 11.90 25 x 10.65 25 x 9.50 25 x 8.55 25 x 7.60 25 x 6.75 100 100 100 100 100 100 Mass of 6-m x 2.4-m sheet (kg) Cross-sectional area/m width Longitudinal bars (mm2/m) 30 x 7.60 200 157 1112 30 x 7.60 200 131 891 30 x 7.60 200 109 709 30 x 7.60 200 93 574 30 x 7.60 200 79 454 30 x 7.60 200 68 358 Cross bars (mm2/m) 227 227 227 227 227 227 Square, with edge side-lapping bars 10 x 9.5 + 200 30 x 9.5 200 80 354 354 4 x 6.75 100 SL 102 SL 92 10 x 8.6 + 4 x 6.0 200 100 30 x 8.6 200 66 290 290 SL 82 10 x 7.6 + 4 x 6.0 200 100 30 x 7.6 200 52 227 227 SL 72 10 x 6.75 + 4 x 5.0 200 100 30 x 6.75 200 41 179 179 SL 62 10 x 6.0 + 4 x 5.0 200 100 30 x 6.0 200 33 141 141 Square, without edge side-lapping bars 25 x 7.6 100 60 x 7.6 100 105 454 SL 81 454 Trench meshes L12TM L11TM L8TM n x 11.9 n x 10.7 n x 7.6 100 100 100 20 x 5.0 20 x 5.0 20 x 5.0 300 300 300 na na na 1112 899 454 65 65 65 Notes: — The edge bar on SL meshes may be replaced by smaller diameter edge bars of equal or greater total cross-sectional area provided the smaller bars meet the minimum ductility requirements of the bar to be replaced. — Purpose-made mesh can be specified for large projects but designers should first check its availability from reinforcement suppliers. — SL 52 is also usually available along with SL 53 and SL 63 which are available only in WA. — Currently most meshes are made from Ductility Class L wire although normal ductility meshes may be available on special order. 2.3 Stress Development 2.3.1 General The rules for stress development are given in AS 3600 Clause 13.1, the data and tables following are based on that information. Development lengths and lapped splice lengths differ depending on whether the reinforcement is in tension or compression. 2.3.2 Development length for bars in tension AS 3600 gives the option of a two-tier approach for determining the development length in tension. Either it can be taken as the Basic development length or, if desired, that length can be reduced as in the procedure given for the Refined development length. 2.6 Reinforced Concrete Design Handbook For most designs, the basic development length will be used. For bars in tension, the basic development length, Lsy.tb, and the refined development length, is multiplied by: 1.5 for epoxy-coated bars; 1.5 for all plain bars; 1.3 when lightweight concrete is used; 1.3 for all structural elements built with slip forms. Tables 2.7 and 2.8 give development lengths for deformed bars in the various situations as detailed in each table. The lengths are based on the formula provided in AS 3600 Clause 13.1.2.2, ie: Lsy.tb = 0.5 k1k3fsydb / (k2 √f 'c) ≥ 29k1db The values are rounded off to the nearest 10 mm. The values used for the factors k1, k2 and k3 are shown at the top of each table. The factor k1 is to allow for the settlement of wet concrete and accumulation of bleed water under the bar in deep sections, which reduces the value of the bond strength. The factor k2 accounts for the reduction in the average ultimate bond strength as the bar diameter increases. The factor k3 depends on the confining effects of the concrete surrounding the bar. The value of cd used to calculate the factor k3, and to produce Tables 2.7 and 2.8, is a dimension (mm) derived from the clear spacing between adjacent parallel bars (horizontally) and critical covers to the bar under consideration as shown in Figure 2.2. (Note the cover is to the main bar, ie the bar being anchored.) c1 a a c (b) Cogged or hooked bar (c) Looped bars (a) Straight bars cd = c cd = min (a/2,c1,c ) cd = min (a/2,c1) (i) Narrow elements or members (eg beam webs and columns) a a c c (b) Cogged or hooked bar (c) Looped bars (a) Straight bars cd = min (a/2,c1,c ) cd = a/2 cd = c (ii) Wide elements or members (eg flanges, band beam, slabs, walls and blade columns) Development lengths for bars developing a tensile stress, sst, less than fsy can be calculated from the formula: a a Lst = Lsy.t sst / fsy ≥ 12d b Except that for slabs, the minimum lengths given in AS 3600 Clause 9.1.3.1(a) (ii) may be used where appropriate. For wide elements such as band beams, slabs, walls and blade columns, where the bars being lapped are in the plane of the element or member, the tensile lap lengths for either contact or non-contact splices can be determined by multiplying the development lengths by 1.25. A lower value of 1.0 is possible where the area of steel provided is at least twice that required and less than 50% of bars are lapped. Refer to AS 3600 Clause 13.2.2. For narrow elements such as the webs of beams or columns where the bars are in contact or where there is less than 3d b between the bars, the tensile lap length is the development length multiplied by 1.25. Where the bars are further apart than 3d b then additional calculations will be required. Refer to AS 3600 Clause 13.2.2. For splices in tension tie members, only welded or mechanical splices are allowed. 2.3.3 Reducing tensile development length by standard hooks and cogs By definition (AS 3600 Clause 13.1.2.7), the term cog is a 90° bend in a bar while a standard hook can be either a 135° or a 180° bend. The length of bar required to physically make each standard hook (which should be specified) is given in Table 2.9. The overall dimensions of hooks and cogs are given in Table 2.10. c c1 Lsy.t Lsy.t db (iii) Planar view of staggered development lengths of equi-spaced bars Note: For wide elements or members (such as band beams, slabs, walls and blade columns), edge cover , c1, should be ignored. Figure 2.2 Values of c d (after AS 3600) Although hooks and cogs reduce the tensile development length as shown in Figure 2.3, they cause congestion of reinforcement in critical areas such as beam/column joints and ends of simplysupported beams. They can also become a source of corrosion if they are allowed to encroach into the required cover. Straight bars are easier to fix and ensure that the required cover is maintained. Where a short development length is required, an alternative to using standard hooks is to use smaller diameter bars and/or higher strength concrete. Tensile stresses are also generated in the concrete in the plane of hooks because of bearing on the concrete on the inside of the hook when the bar is fully stressed under load. Hooks should not be used in sections thinner than about 12 bar diameters or as top bars in slabs, to avoid splitting or spalling of the concrete cover. Reinforced Concrete Design Handbook 2.7 Table 2.7 Basic development lengths Lsy.t for Grade D500N bars in tension and where there is ≤ 300 mm of concrete under the bar Horizontal bars k1 = 1.0 k2 = (132 – db ) /100 k3 = 1.0 – 0.15 (cd – db ) /db (within limits 0.7 ≤ k3 ≤ 1.0) fsy = 500 MPa ≤ 300 mm Concrete strength f 'c (MPa) cd Bar size Concrete strength f 'c (MPa) cd N12 N16 N20 N24 N28 N32 N36 430 400 390 390 390 390 390 390 390 390 390 390 390 390 390 670 630 600 560 540 540 540 540 540 540 540 540 540 540 540 920 890 850 810 770 740 700 700 700 700 700 700 700 700 700 1200 1160 1120 1080 1040 1000 960 920 890 870 870 870 870 870 870 1490 1450 1410 1370 1330 1290 1250 1210 1170 1130 1090 1050 1050 1050 1050 1790 1760 1720 1680 1640 1600 1550 1510 1470 1430 1390 1340 1300 1260 1250 2100 2100 2060 2020 1970 1930 1890 1840 1800 1760 1710 1670 1620 1580 1540 40 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 390 360 350 350 350 350 350 350 350 350 350 350 350 350 350 600 570 530 500 480 480 480 480 480 480 480 480 480 480 480 830 790 760 730 690 660 630 630 630 630 630 630 630 630 630 1070 1030 1000 970 930 900 860 830 790 780 780 780 780 780 780 1330 1300 1260 1220 1190 1150 1120 1080 1040 1010 970 940 940 940 940 1600 1580 1540 1500 1470 1430 1390 1350 1320 1280 1240 1200 1170 1130 1120 32 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 530 500 470 460 460 460 460 460 460 460 460 460 460 460 460 730 700 670 640 610 580 580 580 580 580 580 580 580 580 580 950 910 880 850 820 790 760 730 700 700 700 700 700 700 700 1180 1150 1110 1080 1050 1020 990 950 920 890 860 830 830 830 830 1410 1390 1360 1330 1290 1260 1230 1200 1160 1130 1100 1060 1030 1000 990 20 30 Bar size N12 N16 N20 N24 N28 N32 N36 35 40 45 50 55 60 65 70 75 80 85 90 95 100 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 470 460 460 460 460 460 460 460 460 460 460 460 460 460 460 650 630 600 580 580 580 580 580 580 580 580 580 580 580 580 850 820 790 760 740 710 700 700 700 700 700 700 700 700 700 1050 1020 1000 970 940 910 880 850 820 810 810 810 810 810 810 1260 1250 1220 1190 1160 1130 1100 1070 1040 1010 980 950 930 930 930 1480 1480 1460 1430 1400 1360 1330 1300 1270 1240 1210 1180 1150 1120 1090 1880 1880 1840 1800 1770 1730 1690 1650 1610 1570 1530 1490 1450 1410 1380 50 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 460 460 460 460 460 460 460 460 460 460 460 460 460 460 460 580 580 580 580 580 580 580 580 580 580 580 580 580 580 580 760 730 710 700 700 700 700 700 700 700 700 700 700 700 700 940 920 890 870 840 810 810 810 810 810 810 810 810 810 810 1130 1120 1090 1060 1040 1010 980 960 930 930 930 930 930 930 930 1330 1330 1300 1280 1250 1220 1190 1170 1140 1110 1080 1060 1040 1040 1040 1660 1660 1630 1600 1560 1530 1490 1460 1420 1390 1350 1320 1280 1250 1220 ≥ 65 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 460 460 460 460 460 460 460 460 460 460 460 460 460 460 460 580 580 580 580 580 580 580 580 580 580 580 580 580 580 580 700 700 700 700 700 700 700 700 700 700 700 700 700 700 700 830 810 810 810 810 810 810 810 810 810 810 810 810 810 810 990 980 960 930 930 930 930 930 930 930 930 930 930 930 930 1160 1160 1140 1120 1090 1070 1050 1040 1040 1040 1040 1040 1040 1040 1040 Notes: — k1 = 1.0 — The basic development lengths have been calculated using the nominal areas as per AS/NZS 4761 and have been rounded (generally to the nearest 10 mm) within the accuracy of normal design limits. — cd = smaller of the cover to the deformed bar or 1/2 clear distance to next parallel bar. — For concrete strength greater than 65 MPa use figures for 65 MPa. 2.8 Reinforced Concrete Design Handbook Table 2.8 Basic development lengths Lsy.t for Grade D500N bars in tension and where there is > 300 mm of concrete under the bar Horizontal bars k1 = 1.3 k2 = (132 – db ) /100 k3 = 1.0 – 0.15 (cd – db ) /db (within limits 0.7 ≤ k3 ≤ 1.0) fsy = 500 MPa > 300 mm Concrete strength f 'c (MPa) cd Bar size Concrete strength f 'c (MPa) cd Bar size N12 N16 N20 N24 N28 N32 N36 560 520 510 510 510 510 510 510 510 510 510 510 510 510 510 870 820 780 730 700 700 700 700 700 700 700 700 700 700 700 1200 1150 1100 1050 1010 960 910 910 910 910 910 910 910 910 910 1550 1500 1450 1400 1350 1300 1250 1200 1150 1130 1130 1130 1130 1130 1130 1940 1880 1830 1780 1730 1670 1620 1570 1520 1460 1410 1370 1370 1370 1370 2330 2290 2240 2180 2130 2070 2020 1970 1910 1860 1800 1750 1690 1640 1630 2730 2730 2680 2620 2570 2510 2450 2400 2340 2280 2230 2170 2110 2060 2000 40 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 500 460 460 460 460 460 460 460 460 460 460 460 460 460 460 780 740 690 650 630 630 630 630 630 630 630 630 630 630 630 1070 1030 990 940 900 860 810 810 810 810 810 810 810 810 810 1390 1350 1300 1250 1210 1160 1120 1070 1030 1010 1010 1010 1010 1010 1010 1730 1680 1640 1590 1540 1500 1450 1400 1360 1310 1260 1230 1230 1230 1230 2080 2050 2000 1950 1900 1860 1810 1760 1710 1660 1610 1560 1510 1470 1460 32 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 450 450 450 450 450 450 450 450 450 450 450 450 450 450 450 690 650 610 600 600 600 600 600 600 600 600 600 600 600 600 950 910 870 830 800 760 750 750 750 750 750 750 750 750 750 1230 1190 1150 1110 1070 1030 990 950 910 900 900 900 900 900 900 1530 1490 1450 1410 1360 1320 1280 1240 1200 1160 1120 1080 1080 1080 1080 1840 1810 1770 1730 1680 1640 1600 1550 1510 1470 1420 1380 1340 1300 1290 20 30 N12 N16 N20 N24 N28 N32 N36 35 40 45 50 55 60 65 70 75 80 85 90 95 100 450 450 450 450 450 450 450 450 450 450 450 450 450 450 450 620 600 600 600 600 600 600 600 600 600 600 600 600 600 600 850 810 780 750 750 750 750 750 750 750 750 750 750 750 750 1100 1060 1030 990 960 920 900 900 900 900 900 900 900 900 900 1370 1330 1290 1260 1220 1180 1150 1110 1070 1060 1060 1060 1060 1060 1060 1640 1620 1580 1540 1510 1470 1430 1390 1350 1310 1270 1240 1210 1210 1210 1930 1930 1890 1850 1810 1770 1730 1690 1650 1610 1570 1530 1490 1450 1410 2440 2440 2400 2350 2300 2240 2190 2140 2090 2040 1990 1940 1890 1840 1790 50 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 450 450 450 450 450 450 450 450 450 450 450 450 450 450 450 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 760 750 750 750 750 750 750 750 750 750 750 750 750 750 750 980 950 920 900 900 900 900 900 900 900 900 900 900 900 900 1220 1190 1160 1120 1090 1060 1060 1060 1060 1060 1060 1060 1060 1060 1060 1470 1450 1420 1380 1350 1310 1280 1240 1210 1210 1210 1210 1210 1210 1210 1720 1720 1690 1660 1620 1590 1550 1520 1480 1440 1410 1370 1360 1360 1360 2150 2150 2120 2070 2030 1980 1940 1890 1850 1800 1760 1710 1670 1620 1580 ≥ 65 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 450 450 450 450 450 450 450 450 450 450 450 450 450 450 450 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 900 900 900 900 900 900 900 900 900 900 900 900 900 900 900 1070 1060 1060 1060 1060 1060 1060 1060 1060 1060 1060 1060 1060 1060 1060 1290 1270 1240 1210 1210 1210 1210 1210 1210 1210 1210 1210 1210 1210 1210 1510 1510 1490 1450 1420 1390 1360 1360 1360 1360 1360 1360 1360 1360 1360 Notes: — k1 = 1.3 — The basic development lengths have been calculated using the nominal areas as per AS/NZS 4761 and have been rounded (generally to the nearest 10 mm) within the accuracy of normal design limits. — cd = smaller of the cover to the deformed bar or 1/2 clear distance to next parallel bar. — For concrete strength greater than 65 MPa use figures for 65 MPa. Reinforced Concrete Design Handbook 2.9 Lsy.t Table 2.9 Overall dimensions (mm) of 180° hooks and 90° cogs db Overall dimension Pin db Pin Bar diameter, d b (mm) diameter (mm) 6 10 12 16 20 Overall dimension Pin 0.5Lsy.t 24 28 32 36 180° hooks 30 50 60 * 3d b 4d b 40 60 70 100 5d b 40 70 80 110 6d b 50 80 100 130 8d b 60 100 120 160 * 120 140 160 200 * 140 170 190 240 * 170 200 220 280 * 190 220 260 320 * 220 250 290 360 90° cogs 3d b 120 4d b 130 130 5d b 6d b 140 8d b 160 * 240 260 290 340 * 280 310 340 400 * 330 360 400 470 * 370 400 450 530 * 420 450 510 600 140 150 160 180 200 160 170 180 200 230 * 200 210 240 280 STRAIGHT BAR * Not to be used Notes: HOOKED OR COGGED BAR Figure 2.3 Reduced development length using hooks or cogs Locate cog within beam cage Top slab bars Standard cog — 5d b pin is the one most commonly used. 150 min Cogs are commonly used with top reinforcement in slabs where the slab sits on beams and the cogged bars sit over the beam bars as shown in Figure 2.4. For fitments with cogs, acting as shear reinforcement, AS 3600 Clause 8.2.12.4 requires that there is 50 mm or more of concrete cover over the cog. Figure 2.4 Cogs with slabs and beams AS 3600 also covers the development lengths of plain bars and headed reinforcement in tension (see Clauses 13.1.3 and 13.1.4 respectively). requirements for fitments around compression lap splices in AS 3600 Clause 13.2.4. 2.3.4 Development length for bars in compression Development lengths for bars in compression are less than those for bars in tension because the detrimental effects of tensile cracking are less and the end bearing of the bar is beneficial. Again, AS 3600 allows a two‑tier approach with a Basic development length, which can be modified as in the Refined development length. For most designs, the basic development length will be used. While no specific comment is made about the effect of cover, bar spacing and confinement by fitments, the general rules for cover and bar spacing (for placing and compacting concrete) given in Sections 4 and 17 of the Standard should be followed. The importance of confinement by fitments is highlighted by the 2.10 Reinforced Concrete Design Handbook The development length, Lsy.c, shall be taken as the basic development length of a deformed bar in compression, Lsy.cb, calculated from: Lsy.cb = 0.22 fsy √f 'c db ≥ 0.0435 fsydb or 200 mm, whichever is the greater. In compression, the basic development length in the above equation is largely independent of the concrete strength as, generally, the minimum length will apply. However, all values of the basic development length for different concrete grades are shown in Table 2.10. A refined development length equal to 0.75 of the basic development length can be used subject to complying with AS 3600 Clause 13.1.5.3 but is not shown in Table 2.10. Table 2.10 Minimum lengths for hooks and cogs (mm) Min. greater of, 4db or 70 mm La on bar centreline Pin dia. di (< 8db) La on bar centreline Pin dia. di db di/2 + db db Dimension in bar schedule di/2 + db Dimension in bar schedule 135° HOOK COG La on bar centreline Pin dia. di db Dimension in bar schedule Min. greater of, 4db or 70 mm di/2 + db 180° HOOK Type of bar Min. pin diameter (mm) Bar diameter, d b (mm) 6 10 12 16 20 24 28 32 36 Fitments: D500L and R250N bars D500N bars 3d b 4d b 100 110 110 130 120 140 * 170 * 200 * 230 * 270 * 300 * 340 Reinforcement other than those below 5d b 120 140 160 180 220 260 300 340 380 Bends designed to be straightened or subsequently rebent 4d b 5d b 6d b 110 * * 130 * * 140 * * 170 * * * 220 * * 260 * * * 330 * * 380 * * 430 Bends in reinforcement epoxy-coated or galvanised either before or after bending 5d b 8d b 120 * 140 * 160 * 180 * * 290 * 340 * 390 * 440 * 500 * Not to be used Notes: — 5d b pin is the one most commonly used. — The overall sizes are nominal. No allowance for spring-back is included, nor is the real oversize diameter of a deformed bar taken into account. — 135° on fitments is the most common hook used, which has the same internal diameter and length as 180° hook. Hooks and cogs cannot be used to reduce the development length in compression. For example, with the use of cogged starter bars in a footing, the overall depth of the footing must allow for the development length of the starter bar in compression, the cog, the bottom reinforcement and the bottom cover as shown in Figure 2.5. 2.3.6 Splicing of reinforcement [a] General As reinforcing bars come in lengths up to about 12 m maximum, splicing is required for most concrete elements during construction, including across construction joints. This is a necessary part of the detailing of the reinforcement for any project. AS 3600 Clause 13.2.1 requires that splices are to be made only as permitted in the drawings or specification. Therefore, the designer has the Lsy.c Refer schedule AS 3600, also has rules for the development lengths of plain bars and bundled bars in compression (see Clauses 13.1.6 and 13.1.7). For column size and reinforcement refer to column schedule Refer schedule Allow for cog, bottom reinforcement and cover Figure 2.5 Development length of column starter bars in compression Reinforced Concrete Design Handbook 2.11 responsibility to detail where and how splices are or can be made. It is important to specify which splices are full strength splices and which are not; otherwise full strength splices may be detailed by the reinforcement scheduler for all locations. The designer, knowing how the structure works, should do this detailing, eg crack control reinforcement may not need full strength splices. Lsy.t.lap = k7 Lsy.t db BARS IN CONTACT OR LESS THAN 3db APART Splices in tension members can be only welded or mechanical splices. Lsy.t.lap = larger of k7 Lsy.t and Lsy.t + 1.5s b db [b] Lapped splices in tension For webs of beams where spliced bars are in contact or spaced less than 3db apart, the lap splice length is the development length multiplied by the factor k7 (determined from AS 3600 Clause 13.2.2) and which is generally 1.25 times the development length Lsy.t.. Bars spliced by non-contact lap splices in flexural members, eg slabs and flanges of beams, spaced transversely further apart than 3d b shall have a splice length not less than the larger of k7 Lsy.t (generally 1.25Lsy.t) and Lsy.t + 1.5s b, where s b is the clear distance between bars of the non-contact lapped splice (mm) as shown in Figure 2.6. The length of the lapped splice can be calculated by multiplying the development lengths in Tables 2.7 and 2.8 by the factor k7 (usually 1.25). However, if s b does not exceed 3d b, then s b may be taken as zero for calculating Lsy.t.lap. Designers should remember that in most designs for bars in tension, bars should not be lapped at the point of maximum tension and good design practice will minimise bars being lapped in high stress areas. An example is top bars in a cantilever beam or slab, which are usually spliced at about the quarter points in the back span, depending on the length of the cantilever span and back span. AS 3600 allows a pro rata reduced development length (and lap splice) where the stress in the bar is less than the yield stress both in tension and compression. For the situation where the stress in the bars is less than 0.5fsy and only half the bars are being spliced at the location, k7 can be taken as 1.0 (see AS 3600 Clause 13.2.2). For tension, there is a minimum development length of 12db or D, whichever is the greater, for slabs as permitted by AS 3600 Clause 9.1.3.1 (a) (ii). A lapped splice for welded mesh in tension shall be made so the two outermost cross-bars (spaced at not less than 100 mm or 50 mm apart for plain or deformed bars respectively) of one sheet of mesh overlap the two outermost cross-bars of the sheet being lapped as shown in AS 3600 Figure 13.2.3. The minimum length of the overlap shall be 100 mm. A lapped splice for welded deformed and plain meshes, with no cross-bars within the splice length shall be determined in accordance with AS 3600 Clause 13.2.2. 2.12 Reinforced Concrete Design Handbook sb BARS MORE THAN 3db APART Figure 2.6 The lap-splice length of adjacent bars in tension in webs of beams and in columns (narrow elements) Table 2.11 Basic development lengths Lsy.cb and lap‑splice lengths (mm) for Grade D500N bars in compression Concrete strength f 'c (MPa) Bar size N12 N16 N20 N24 N28 N32 N36 20 25 32 40 50 65 80 100 300 260 260 260 260 260 260 260 390 350 350 350 350 350 350 350 490 440 440 440 440 440 440 440 590 530 520 520 520 520 520 520 690 620 610 610 610 610 610 610 790 700 700 700 700 700 700 700 890 790 780 780 780 780 780 780 Lap-splice length for bars in contact or spaced at less than 3d b apart – development length or 40d b (see AS 3600 Clause 13.2.4 (a)) 480 640 800 960 1120 1280 1440 0.8 Concessional value* 380 510 640 770 900 1020 1150 * If certain conditions are met (see AS 3600 Clause 13.2.4 (b) and (c) for details). [c] Lapped splices in compression AS 3600 Clause 13.2.4 (a) requires that the lap-splice length for deformed bars in compression be a minimum of 300 mm and not less than 40d b which are independent of the concrete strength. However, there are two conditions in AS 3600 Clause 13.2.4 which allow the lap splice length to be reduced to 0.8 of the 40d b value. This reduced value of 0.8 has also been included in Table 2.11. Primary beam 750 x 750 180 Slab 180 Slab 750 x 750 column under and over Secondary beam 750 x 750 A 2.4 Detailing Structural analysis is only one part of the design process. Good detailing is equally important and requires an understanding of what each bar, fitment or piece of mesh is doing in the structure and what forces it is resisting. Detailing of reinforcement is the interface between the theoretical design and what can be built in practice on site. There is no point in having the most refined analysis and design, if it cannot be constructed. Detailing also has an impact on durability as poor placement of reinforcement leads to insufficient cover and premature failure. Designers need to be practical in terms of what is readily achievable on site and clear in their drawings and their details to allow their concrete structure to be properly built. It is important not to expect reinforcement to end up in precisely the position nominated in the documentation. As with all site work, some tolerance must be allowed. In general, ±10 mm is realistic, while ± 5 mm is the tightest that should ever be specified. Note that AS 3600 Clause 17.5.3 allows –5 mm +10 mm deviation from the specified position controlled by cover for beams, slabs, columns and walls. Generally, reinforcement fixing is a three dimensional problem. Lines, dots, cogs, hooks and laps on drawings have real sizes and location in the formwork. Some adjustments of the reinforcement in formwork will be necessary to make it all fit, especially with larger bars. Reinforcement cannot be shifted quickly from its position without bends and cranks. Figure 2.7 illustrates this point. To gain an understanding of how reinforcement is fixed and the practicalities of work on site, designers should be encouraged to inspect their design work in the field. It should not be assumed that the builder and/or the scheduler will 'work it out', nor that the detailing will just 'happen'. Comprehensive detailing (particularly of complex reinforcement) is vital. Some examples of poor detailing and suggested improvements are shown in Figure 2.8, 2.9 and 2.10. Guidance on detailing of reinforcement is provided in Reinforcement Detailing Handbook 2.11. A 180 Slab Cantilever slab PART PLAN (NTS) Note: Cover to secondary beam reinforcement set by primary beam Ensure enough room to place concrete See DETAIL A Crank bottom reo to miss beam bars or provide a splice bar Move bottom and top bars in primary beam to miss column bars As bottom bars will clash, provide drop-in splice bar Column bars beyond SECTION A-A (NTS) Note : Cover to beam reinforcement is set by cover to slab and size of top reinforcement in slab If corner bar has to move to the right use smaller diameter bar to fit into radius of fitment. Also bar may clash with column bar beyond so may have to move into slab Beam bar N32 spacer bars at 1000 Check that standard radius for both fitment and secondary beam reinforcement will pass between main reinforcement Check that if main bar is displaced it will not clash with any other bar DETAIL A (NTS) Fitment Secondary beam bar Rather than cog top bar, can beam bar run into cantilever slab? Figure 2.7 What was shown on the drawings and how it fits differently on site Reinforced Concrete Design Handbook 2.13 Potential crack 1 Sealant 2 12 000 3 12 000 12 000 280 slab 600 x 400 columns IB1 IB2 IB3 IB2 1200 x 1200 IB3 1200 x 600 A 4 1200 wide band beam PART PLAN Fireproof filler Column over Potential crack POOR DETAILING IB1 1200 x 1200 Joint is likely to fail due to diagonal cracking U bars Additional fitments Provide neoprene bearing pad or bearing as required Sealant ELEVATION 1 1a Column over 9N36 2 18N36 2 layers 3 8N36 3W12@300 Fireproof filler Additional fitments 12N36 in 2 layers IB1 U bars 8N36 IB2 POOR DETAILING GOOD DETAILING 1 Many bars exceed 12 m length in IB1 and IB2. 2 9N36 cogged into column or edge beam and will not fit at grid 1. Consider drop-in bars to match moment into column or reduce number and size of bars. 3 12N36 in 2 layers in bottom of IB1 could be in 1 layer. 4 No side face reinforcement shown. 5 W12 is the wrong designation. Consider N12 fitments at 200 centres in pairs to reduce fixing and to comply with transverse spacing. 6 Starter bars for column at 1a not shown. 7 Cogging 8N36 bottom at grid 3 not required. Suggest 8N20 bars with 4 bars cogged into column. 8 As splice lengths not shown, the scheduler will assume a full splice. Note details are indicative only and are subject to final design Figure 2.8 Detailing of corbel Column Column starter bars − over refer schedule 9N36 1500 300 18N36 in 2 layers 9N20 1000 1000 8N36 5 Consider 6N28 drop-in bars 6 1000 N12 at 200 N12 fitments at 200 in pairs EF 12N36B 1000 6 N12 fitments at 200 in pairs 8N36 8N20 x 3000 long lap 1200 nom each side BETTER DETAILING 8N20 drop-in bars into columns 1000 800 1 All detailing are subject to final design. 2 Splices generally shown so scheduler will not use full strength splices. Figure 2.9 Detailing of beam 2.14 Reinforced Concrete Design Handbook N12 @ 200 Slab References 2N20T Sirivivatnanon V Fitness for Purpose of Residential Slab-on-Ground Proceedings of Concrete 07, 18–20 October 2007, Adelaide, Concrete Institute of Australia. 2.2 AS 3600 Concrete structures Standards Australia, 2009. 2.3 AS 1012 Methods of testing concrete Standards Australia. 2.4 AS/NZS 1170 Structural design actions, Standards Australia, 900 N12 @ 200 2.1 W10 fitments @ 200 cts N12 @ 200 Cantilever slab Beam 150 Part 1: Permanent, imposed and other actions 2N28B N12 @ 200 2.5 Deformability of concrete structures – basic assumptions Bulletin D'Information No. 90, Comité Européen du Béton (CEB), 1973. 2.6 BS EN 1992, Eurocode 2 Design of concrete structures British Standards Institution, 2004. 2.7 AS/NZS 4671 Steel reinforcing materials Standards Australia, 2001. 2.8 Lecture 8, Steel reinforcement National Seminar Series AS 3600—2009, CIA, EA and CCAA. 2.9 AS/NZS 1554 Structural steel welding Standards Australia POOR DETAILING 1 N12 top bars to cantilever slab are not properly anchored. Also hooks in 150 cantilever slab difficult to fit in depth. 2 N12 top bars to slab at top of beam not properly anchored. 3 No side face reinforcement to beam. 4 Designation of fitments is incorrect. 5 Can it be cast in one pour? N12 @ 200 Slab 2 N20T Construction joint if required provide 10 x 10 joint Part 3: Welding of reinforcing steel, 2008. 900 Beam off-form finish as specified CJ L10 fitments @ 200 cts 300 4N12 as sideface reinforcement 1N12 additional 2N28B N12 @ 200 Non slip surface Cantilever slab Part 4: Welding of high strength quenched and tempered steels, 2010. 2.10 Welding Technology Institute of Australia (WTIA), Technical Note 1, 1996. 2.11 Reinforcement Detailing Handbook (Z06), 2nd Ed, Concrete Institute of Australia, 2010. L1O U-bar at 200 centres slope as required to fit in 150 depth GOOD DETAILING 1 Note details are indicative only and are subject to final design. Figure 2.10 Detailing of cantilever slab Reinforced Concrete Design Handbook 2.15 blank page 2.16 Reinforced Concrete Design Handbook Chapter 3 Durability and fire resistance n Durability is a complex topic and compliance with these requirements may not be sufficient to ensure a durable structure. The second point should alert designers to the issues involved in the design and construction of concrete structures for durability and to think about these issues, rather than just meeting the minimum requirements of the Standard. 3.1 DESIGN FOR DURABILITY 3.1.1 General Concrete is one of the most widely used materials in structures and durability is one of its key advantages. Durability can be defined as the ability of a concrete structure to resist during its design life the effects of weathering, chemical attack, abrasion and other deteriorating influences (acting on the structure or its members) arising within the concrete, from the environment or from processes being carried out inside the structure. Although not specifically mentioned in this definition, deterioration due to the corrosion of reinforcement, tendons or other inserts cast into the concrete is an important aspect of durability. In designing for durability the environment in which the structure is to be built, including micro-climates generated by the structure itself, has to be evaluated. The following also need to be taken into account: n n n n n Aggressive agencies and actions of the processes to be carried out in or around the structure The expected wear and deterioration through the intended service life of the structure The amount of periodic inspection and maintenance the structure is likely to receive during its working life (particularly external parts exposed to the environment) The length of time the concrete structure is expected to be operational without repair The difficulty of carrying out repairs and their economic impact. Durability of concrete is covered in Section 4 of AS 3600 Concrete structures 3.1. The Standard applies to plain, reinforced and prestressed concrete structures and members with a design life of 50 years ± 20%, but notes that: n More stringent requirements would be appropriate for structures with a design life in excess of 50 years (eg monumental structures), while some relaxation of the requirements may be acceptable for structures with a design life less than 50 years (eg temporary structures). Nevetheless, structures designed to AS 3600 have generally performed satisfactorily in normal environments and properly designed, proportioned, inspected, placed, finished and cured concrete is capable of providing many years of durable service. The information in this chapter is based on that given in AS 3600, unless noted otherwise. Guidance in AS 3600 is, however, given for only a limited number of aspects of durability – corrosion of reinforcement (based on the concrete strength and cover for various exposure classifications), aggressive soils, freeze-thaw and abrasion – and is restricted to a limited number of exposure conditions. For specific durability-related issues reference to publications relevant to the specific issue in question is recommended (eg abrasion resistance, acid attack, or corrosion of embedded steel in concrete). Documents such as Durable Concrete Structures 3.2, Performance Criteria for Concrete in Marine Environments 3.3 and Guide to Durable Concrete 3.4 provide sound guidance. 3.1.2 AS 3600 requirements The 2009 edition of AS 3600 includes some important changes to exposure classifications. Specifically for surfaces of maritime structures in seawater the exposure classification C has been split into: n C1 (spray zone) and n C2 (tidal/splash zone). It should be noted that there is now a separate standard guide on maritime structures, AS 4997 Guidelines for the design of maritime structures 3.5. Also, CCAA has published a document on Chloride Resistance of Concrete 3.6. AS 3600 now includes specific guidance for sulfate soils and saline soils. There is considerable background information on aggressive soil conditions. CCAA's Technical Note Sulfate-resisting Concrete 3.7 and Guide to Residential Slabs and Footings in Saline Environments 3.8 provide background information in this complex area. Importantly, while durability is a complex topic, for the criteria discussed, AS 3600 essentially manages issues of durability from a compressive strength and cover specification perspective for the particular exposure classification. Concrete mix designs Reinforced Concrete Design Handbook 3.1 Flowchart 3.1 Designing for durability in accordance with AS 3600 Is member subject to abrasion? Is member subject to freeze-thaw cycles? no Is member subject to aggressive soils? no yes yes Determine f 'c from Table 3.4 Determine f 'c and air entrainment requirements from Table 3.5 Input f 'c1 Input f 'c 2 no yes Is member subject to sulfate soils? yes no Is member subject to saline soils? NO yes Determine exposure classification from Table 4.8.1 Determine f 'c from AS 3600 Table 4.4 Determine f 'c from Table 4.8.2 Input f 'c 3 Determine exposure classification for each surface of member from Table 4.3 Determine f 'c and curing period for each surface from Table 4.3 and adopt largest value no (Optional see Clause 4.3.2) Adopt f 'c for next lower concrete grade – and increased cover yes Is member external but with external exposure essentially on one surface only? no A 3.2 Reinforced Concrete Design Handbook Is exposure classification U? yes Obtain advice and recommendations (f 'c , curing, cover etc) from other sources stop A Input f 'c3 Input f 'c 4 (for strength and serviceability) Adopt largest f 'c from f 'c1, f 'c2, f 'c3 and f 'c 4. This should be specified along with associated curing period, and any other additional requirements Is member to be constructed using normal compaction and standard formwork? no Is member to be constructed using intense compaction and rigid formwork? yes yes Determine cover from Table 4.10.3.2 Is concrete cast against the ground? Determine cover from Table 4.10.3.3 no Is member to be constructed using spinning and rolling? no yes Determine cover from appropriate Standard See Clause 4.10.3.6 yes Outside scope of AS 3600 no Adopt cover value determined above Increase cover in accordance with Clause 4.10.3.5 Ensure cover will permit reinforcement to be fixed and the concrete with specified nominal aggregate size to be compacted around reinforcement, tendons and ducts (Clause 4.10.2) Reinforced Concrete Design Handbook 3.3 120° 130° 140° 150° Thursday Is 10° Ashmore Is DARWIN Troughton Is TROPICAL Katherine TEMPERATE Derby B1* Halls Creek Broome Wittenoom Gorge Townsville Camooweal Mt Isa Hughenden Rockhampton A1* Carnarvon Bundaberg Birdsville Meekatharra 30° PERTH Cape Leeuwin Kalgoorlie A2* Taroom Charleville Oodnadatta BRISBANE Laverton Geraldton 20° Mackay Longreach Alice Springs Mundiwindi Cairns Normanton Tennant Creek Port headland North West Cape Cooktown Wyndham ARID 20° 10° Weipa Yirrkala Cook Forest Eucia A2* Marree Tarcoola Bourke Port Agusta Ceduna Dubbo Esperance Cape du Couedic Albany ADELAIDE B1* CANBERRA Portland 1 km Coast Sale Point Hicks Currie Burnie Cape Sorrell A2* B2* 40° Newcastle SYDNEY Wollongong Mildura Echuca Cooma MELBOURNE 50 km A2* 30° Cobar Launceston 40° HOBART 110° 120° 130° 140° 150° * Unless close to industry Figure 3.1 Climatic zones and exposure classifications (after AS 3600) using supplementary cementitious materials (SCM), maximum water-cement ratios, cementitious binder types, etc are not discussed in any detail in AS 3600 except that for exposure classification B2, C1 and C2 special class concrete must be used. Appendix B of AS 1379 Specification and supply of concrete 3.9, lists the various criteria for special class concrete which are additional to or different from those for normal class concrete. As concrete mix design is a specialist area, designers are recommended to seek specialist advice if special mix designs are required for durability. There are number of important issues that designers should discuss with their clients on the durability of concrete at the beginning of any project, including: n 3.4 The design life for the concrete structure and whether Clause 4.1 of AS 3600 with an implied design life of 50 years ± 20% is appropriate Reinforced Concrete Design Handbook (eg an iconic building such as a church might have a required design life of 100 years or more) n n n The need for maintenance and repairs during the life of the building Any specific durability requirements for individual concrete members, as the durability issues may not be immediately apparent The level of inspection during construction to ensure that cover requirements are achieved on site The above should be part of a durability plan and durability report for the building being designed which is accepted by all parties as part of the project risk management. Such durability issues may require a re-assessment of covers and concrete strengths and the need for special concrete mixes. table 3.1 Required concrete properties Exposure Surface and exposure environment classification Concrete properties f 'c (MPa) Curing period (6) (days) External surfaces above ground B2 40 (5) 7 Within 1 km of coastline(1) Within 1 to 50 km of coastline B1 32 7 Further than 50 km from coastline and – within 3 km of industrial polluting area(2) B1 32 7 – in tropical zone(3) B1 32 7 – in temperate zone(3) A2 25 3 – in arid zone(3) A1 20 3 Internal surfaces In industrial building subject to repeated wetting and drying B1 32 7 Non-residential A2 25 3 Residential A1 20 3 Surfaces in contact with the ground In contact with aggressive soils(4) – Sulfate bearing (magnesium content <1g/L) A1 – C2 25–50 3–7 Refer AS 3600 Table 4.8.1 – Sulfate bearing (magnesium content >1g/L) U Designer to assess (7) Protected by a dpm A1 20 in contact with non-aggressive soils – residential buildings A1 20 – other members A2 25 3 3 3 Surfaces in contact with water (8) In soft or running water U Designer to assess (7) In fresh water B1 32 7 In seawater – permanently submerged B2 40(5) 7 – in spray zone C1 50(5) 7 – in tidal/splash zone C2 50(5) 7 Other situations Notes: (1) See Figure 3.1. AS 3600 states that the coastal zone includes locations within 1 km of the shoreline of large expanses of salt water (eg Port Phillip Bay, Sydney Harbour east of the Spit and Harbour Bridges, Swan River west of the Narrows Bridge). Where there are strong prevailing onshore winds or vigorous surf, the distance should be increased beyond 1 km and higher levels of protection should be considered. (2) Industrial polluting areas are defined in AS 3600 as areas where there are industries that discharge atmospheric pollutants. The 3-km distance should be increased if there are strong prevailing winds in one direction. (3) See Figure 3.1. (4) Severity of sulfate attack depends on the type of sulfate, which must be in solution. For example, magnesium and ammonium sulfates are more aggressive than sodium sulfate. The use of sulfate-resisting cement would be adequate for sodium sulfate conditions. For the magnesium and ammonium sulfates conditions, specific consideration should be given to the cement and the concrete that are likely to resist this type of sulfate. (5) Special-class concrete is required for B2, C1 and C2 exposure classifications and this may require items such as the minimum cement content, the cement type, SCM and water-cement ratios to be specified by the designer. Designer to assess (7) (8) U (6) AS 3600 makes provision for accelerated curing regimes to be used by specifying average compressive strengths at the completion of the curing period in column 4 of AS 3600 Table 4.4. Exposure classification f 'c (MPa) A1 A2 B1 B2 C1, C2 20 25 32 40 50 f 'cm at end of accelerated curing (MPa) ≥15 ≥15 ≥20 ≥25 ≥32 (7) Classification U represents an exposure environment not specified in this table but for which a degree of severity of exposure should be appropriately assessed and will involve special class concrete. Protective surface coatings may be taken into account in such an assessment. Further guidance on measures appropriate in exposure classification U may be obtained from AS 3735 Concrete structures for retaining liquids 3.10. (8) For water-retaining structures, designers should consult AS 3735 as its requirements supplement and take precedence over those of AS 3600. It provides more-detailed advice for particular situations and sets out more-stringent requirements for concrete quality and cover to reinforcement and tendons. Reinforced Concrete Design Handbook 3.5 AS 3600 sets out the minimum requirements for the design of concrete structures for durability. As noted earlier these will be adequate in many situations, but in others (particularly when there are unknowns) it will be prudent to exceed these requirements. A sequence of steps in designing for durability in accordance with AS 3600 is provided in Flowchart 3.1. Details of the Standard's requirements for particular concrete members are provided in Tables 3.1 to 3.6, while information for the achievement of appropriate durability in specific circumstances is provided in Sections 3.1.5, 3.1.6 and 3.1.7. In Table 3.1, classifications A1, A2, B1, B2, C1 and C2 represent increasing degrees of severity of exposure. For most capital cities in Australia, external surfaces above ground will be B1 or B2 exposure classification as a minimum. Exposure classification B2 (within 1 km of the coastline) requires the use of special class concrete. 3.1.3 Concrete properties Frequently, the concrete properties (including strength) required to meet the requirements for durability will control that for the design, while the requirements for cover can influence the member size. 3.1.4 Cover to reinforcement AS 3600 sets out the minimum cover required to protect the reinforcement and tendons from fire and long-term corrosion. Since cover has a very significant influence on concrete's durability, greater cover should be specified when there are any durability concerns. Inadequate and inappropriate cover to the reinforcement has been an ongoing durability problem, particularly with concrete exposed to the elements, and especially that in marine or other aggressive environments. Marosszeky and Gamble3.11 have reported that on a number of building sites, where corrosion occurred the cover was as low as 5 mm. In a similar paper, Clarke et al3.12 recognized the difficulty of achieving the right cover on site. Problems with cover were also discussed in a series of papers in Concrete in Australia 3.13. Reinforcement can be complicated and congested. AS 3600 Clause 4.10.2 requires the designer to consider the cover depending on the the size and shape of the member, the size, type and configuration of the reinforcement (and, if present, the tendons or ducts), the aggregate size, the workability of the concrete and the direction of concrete placement. It is important to recognise that reinforcement cannot be expected to end up in precisely the position shown on the drawings. An accuracy of ± 5 mm is the best that can be expected. AS 3600 Clause 17.5.3 allows 3.6 Reinforced Concrete Design Handbook –5 or +10 mm deviation from the specified position for beams, slabs, columns and walls (–10 or +20 mm for slabs on ground and –10 or +40 mm for footings) where cover is critical. These tolerances are, however, sometimes difficult to achieve on site, especially with larger bars. If, for example, 30 mm cover for a wall or beam is the absolute minimum required, then 40 mm should be specified (35-mm bar chairs are not available) as the reinforcement will then end up between 35 to 50 mm from the face of the wall or beam, based on the tolerances in AS 3600 (or between 30 and 50 mm for ±10 mm tolerances). Table 3.2 Required cover (mm) − Standard formwork and standard compaction Concrete strength f 'c (MPa) 20 25 32 40 ≥ 50 Exposure classification A1 A2 20 [50] 20 20 20 20 30 25 20 20 B1 B2 C1 [60] 40 [65] 30 45 [70] 25 35 50 C2 65 Use of figures in brackets to the right of the zigzag line (with the related characteristic strength) is limited to when essentially only one surface of a member is subject to the particular exposure classification. Where concrete is cast on or against ground, then the figures in this table should be increased: Where protected by dpm – add 10 mm Where not protected by dpm – add 20 mm Table 3.3 Required cover (mm) − Rigid formwork with repetitive procedure and intense compaction or self‑compacting concrete Concrete strength f 'c (MPa) 20 25 32 40 ≥ 50 Exposure classification A1 A2 B1 B2 C1 C2 20 20 20 20 20 [45] 30 [45] 20 30 [50] 20 25 35 [60] 20 20 25 45 60 Use of figures in brackets to the right of the zigzag line (with the related characteristic strength) is limited to when essentially only one surface of a member is subject to the particular exposure classification. Table 3.4 Concrete strength for abrasion resistance Member and type of traffic Minimum f 'c (MPa) Footpaths and residential driveways 20 Commercial and industrial floors not subject to vehicular traffic 25 Floors and pavements in public car parks, driveways and parking areas, subject only to light traffic (vehicles ≤ 3 t gross) 25 Floors and pavements in warehouses, factories, driveways and hard standings subject to: – medium or heavy pneumatic-tyred traffic (> 3 t gross) – non-pneumatic-tyred traffic – steel-wheeled traffic 32 40 (to be assessed but ≥ 40) Table 3.5 Freeze-thaw resistance Exposure condition Minimum f 'c (cycles per annum) (MPa) Entrained air for nominal aggregate size (mm) 10–20 40 < 25 32 8–4% 6–3% ≥ 25 40 8–4% 6–3% Table 3.6 Exposure classification for concrete in sulfate soils (after AS 3600) Exposure conditions Exposure classification Sulfates (expressed as SO4 )* In soil (ppm) <5000 5000–10 000 10 000–20 000 >20 000 In groundwater (ppm) <1000 1000–3000 3000–10 000 >10 000 pH > 5.5 4.5–5.5 4–4.5 <4 Soil conditions A** Soil conditions B*** A2 B1 B2 C2 A1 A2 B1 B2 * Approximately 100 ppm SO4 = 80 ppm SO3 ** Soil conditions A – high permeability soils (eg sands and gravels) which are in groundwater *** Soil conditions B – low permeability soils (eg silts and clays) or all soils above groundwater Table 3.7 Strength and cover requirements for saline soils (after AS 3600) Soil electrical conductivity, ECe* 4–8 8–16 >16 Exposure classification Minimum f 'c Minimum cover (mm) A2 B1 B2 25 32 40 45 50 55 * ECe is saturated electrical conductivity in deciSiemens per metre Reinforced Concrete Design Handbook 3.7 Table 3.8 Effect of chemicals on concrete (after ACI 515.1R3.17) Material — Effect Material — Effect Acetone — Liquid loss by penetration. May contain acetic acid as impurity. Castor oil — Disintegrates concrete, especially in presence of air. Acid acetic — Disintegrates concrete slowly. carbonic — Disintegrates concrete slowly. formic — Disintegrates concrete slowly. humic — Disintegrates concrete slowly. hydrochloric — Disintegrates concrete and steel rapidly. hydrofluroic — Disintegrates concrete and steel rapidly. lactic — Disintegrates concrete slowly. nitric — Disintegrates concrete and steel rapidly. oxalic — Not harmful. Protects tanks against acetic acid, carbon dioxide and salt water. Poisonous. Should not be used with food or drinking water. phosphoric — Disintegrates concrete slowly. sulfuric — Disintegrates concrete and steel rapidly. sulfurous — Disintegrates concrete and steel rapidly. tannic — Disintegrates concrete slowly. Cinders — Harmful if wet, when sulfides and sulfates leach out (eg, see sodium sulfate). Acid water (pH of ≤ 6.5)(1) — Disintegrates concrete slowly. Attacks steel in porous or cracked concrete. Alcohol (ethyl, methyl) — Liquid loss by penetration. Coke — Sulfides leaching from damp coke may oxidize to sulfurous or sulfuric acid. Copper sulfate — Disintegrates concrete of inadequate sulfate resistance. Creosote — Phenol present disintegrates concrete slowly. Ethylene glycol(4) — Disintegrates concrete slowly. Fermenting fruit, grains, vegetables or extracts(5) — Industrial fermentation processes produce lactic acid. Disintegrates concrete slowly (see also fruit juices). Ferric sulfate — Disintegrates concrete of inadequate quality. Ferrous sulfate — Disintegrates concrete of inadequate sulfate resistance. Fertilizer — See ammonium sulfate, ammonium superphosphate, manure, potassium nitrate and sodium nitrate. Fish liquor (6) — Disintegrates concrete. Alum (potassium aluminium sulfate) — Disintegrates concrete of inadequate sulfate resistance. Fish oil — Disintegrates concrete slowly. Aluminium chloride — Disintegrates concrete rapidly. Attacks steel in porous or cracked concrete. Flue gases — Hot gases (200–600°C) cause thermal stresses. Cooled, condensed sulfurous and hydrochloric acids disintegrate slowly. Aluminium sulfate — Disintegrates concrete. Attacks steel in porous or cracked concrete. Ammonia, liquid — Harmful only if it contains harmful ammonium salts. Fruit juices — Hydrofluoric, other acids, and sugar cause disintegration (see also fermenting fruits, grains, vegetables or extracts). Ammonia vapours — May slowly disintegrate moist concrete or attack steel in porous or cracked moist concrete. Hydrogen sulfide — Not harmful, but in moist, oxidizing environments converts to sulfurous acid, disintegrates concrete slowly. Ammonium chloride — Disintegrates concrete slowly. Attacks steel in porous or cracked concrete. Kerosene — Liquid loss by penetration. Ammonium hydroxide — Not harmful. Ammonium nitrate — Disintegrates concrete. Attacks steel in porous or cracked concrete. sulfate — as above superphosphate — as above Linseed oils — Liquid disintegrates concrete slowly. Dried or drying films are harmless. Lubricating oil, machine oil — Fatty oils, if present, disintegrate concrete slowly. Magnesium chloride — Disintegrates concrete slowly. Attacks steel in porous or cracked concrete. Automobile and diesel exhaust gases(2) — May disintegrate moist concrete by action of carbonic, nitric or sulfurous acid. Magnesium sulfate — Disintegrates concrete of inadequate sulfate resistance. Beef fat — Solid fat disintegrates concrete slowly, melted fat more rapidly. Margarine — Solid margarine disintegrates concrete slowly, melted margarine more rapidly. Beer — May contain (as fermentation products) acetic, carbonic, lactic or tannic acids. Milk fresh — Not harmful. sour — Disintegrates concrete slowly. Calcium chloride — Attacks steel in porous or cracked concrete. Steel corrosion may cause concrete to spall. Calcium sulfate — Disintegrates concrete of inadequate sulfate resistance. Carbon dioxide(3) — Gas may cause permanent shrinkage (see also carbonic acid). Manure — Disintegrates concrete slowly. Mine water, waste — Sulfides, sulfates, or acids present disintegrate concrete and attack steel in porous or cracked concrete. Ores — Sulfides leaching from damp ores may oxidize to sulfuric acid or ferrous sulfate. continues 3.8 Reinforced Concrete Design Handbook Table 3.8 continued Effect of chemicals on concrete (after ACI 515.1R3.17) Material — Effect Paraffin — Shallow penetration not harmful, but should not be used on highly-porous surfaces like concrete masonry (7) Petroleum oils — Liquid loss by penetration. Fatty oils, if present, disintegrate concrete slowly. Pickling brine — Attacks steel in porous or cracked concrete. Potassium nitrate — Disintegrates concrete slowly. Seawater — Disintegrates concrete of inadequate sulfate resistance. Attacks steel in porous or cracked concrete. Silage — Acetic, lactic acids (and sometimes fermenting agents of hydrochloric or sulfuric acids) disintegrate slowly. 3.1.5 Abrasion The abrasion resistance of concrete is the ability of a surface to resist being worn away by rubbing or friction. Abrasion can be caused by foot or vehicular traffic or other sources such as mechanical equipment. The resistance is generally proportional to concrete strength. The use of mixes with a low water-cement ratio improve strength and thus the wear resistance of the surface. Table 3.4 sets out requirements for abrasion resistance for foot and vehicular traffic after AS 3600. For other forms of abrasion, specialist advice may be required. 3.1.6Freezing and thawing Sodium chloride — Magnesium chloride, if present attacks, steel in porous or cracked concrete. Steel corrosion may cause concrete to spall. Freezing and thawing of concrete is generally not a major problem in Australia except in sub-alpine and alpine areas and in specialist facilities such as cold stores. Table 3.5 sets out the requirements in AS 3600. Sodium hydroxide 1–10% — Not harmful(8). The entrained air content is determined in accordance with AS 1012.43.14. 20% or over — Disintegrates concrete. Sodium nitrate — Disintegrates concrete slowly. 3.1.7 Soil conditions Sodium sulfate — Disintegrates concrete of inadequate sulfate resistance. Acid sulfate soils The National Strategy for the Management of Coastal Acid Sulfate Soils3.15, indicates that substantial low-lying coastal areas of the Northern Territory, Queensland and New South Wales are affected by acid sulfate soils (ASS). It also notes that there are similar conditions along the northern coastline of Western Australia, and around Perth, Adelaide and Westernport Bay near Melbourne as shown on Figure 3.2. In Australia, the acid sulfate soils of most concern are those which formed within the past 10,000 years, after the last major sea level rise. Sugar — Disintegrates concrete slowly. Turpentine — Mild attack. Liquid loss by penetration. Urea — Not harmful. Urine — Attacks steel in porous or cracked concrete. (1) Waters of pH higher than 6.5 may be aggressive if they also contain bicarbonates. (Natural waters are usually of pH higher than 7.0 and seldom lower than 6.0, though pH values as low as 0.4 have been reported. For pH values below 3, protect as for dilute acid.) (2) Composed of nitrogen, oxygen, carbon dioxide, carbon monoxide, and water vapour. Also contains unburned hydrocarbons, partially burned hydrocarbons, oxides of nitrogen, and oxides of sulfur. (3) Carbon dioxide dissolves in natural waters to form carbonic acid solutions. When it dissolves to extent of 0.9 to 3 parts per million it is destructive to concrete. (4) Used as deicer for airplanes overseas. Heavy spillage on runway pavements containing too-little entrained air may cause surface scaling. (5) In addition to the intentional fermentation of many raw materials, much unwanted fermentation occurs in the spoiling of foods and food wastes, also producing lactic acid. (6) Contains carbonic acid, fish oils, hydrogen sulfide, methyl amine, brine and other potentially reactive materials. (7) Porous concrete which has absorbed considerable molten paraffin and then been immersed in water after the paraffin has solidified has been known to disintegrate from sorptive forces. (8) However, in the limited areas where concrete is made with reactive aggregates, disruptive expansion may be produced. DARWIN BRISBANE PERTH ADELAIDE SYDNEY MELBOURNE Potential pyritic sediments HOBART Figure 3.2 Indicative distribution of coastal acid sulfate soils in Australia. (from National Strategy for the Management of Coastal Acid Sulfate Soil) Reinforced Concrete Design Handbook 3.9 Table 3.9 Protective barrier systems (after ACI 201.2R – 013.18) Severity of chemical environment Total nominal thickness range Typical protective barrier systems Typical but not exclusive uses of protective systems in order of severity Mild Under 1 mm Polyvinyl butyral, polyurethane, epoxy, acrylic, chlorinated rubber, styrene-acrylic copolymer Asphalt, coal tar, chlorinated rubber, epoxy, polyurethane, vinyl, neoprene, coal-tar epoxy, coal-tar urethane – Improve freeze-thaw resistance – Prevent staining of concrete – Protect concrete in contact with chemical – solutions having a pH as low as 4, depending – on the chemical Intermediate 3 to 9 mm Sand-filled epoxy, sand-filled polyester, sand-filled polyurethane, bituminous materials – Protect concrete from abrasion and – intermittent exposure to dilute acids in – chemical, dairy, and food processing plants 0.5 to 6 mm Glass-reinforced epoxy, glass-reinforced polyester, precured neoprene sheet, – Protect concrete tanks and floors during – continuous exposure to dilute mineral, – organic acids (pH is below 3), salt solutions, – strong alkalies Severe 0.5 to 7 mm Composite systems: (a) Sand-filled epoxy system Over 6 mm topcoated with a pigmented but unfilled epoxy (b) Asphalt membrane covered with acid-proof brick using a chemical-resistant mortar Severe plasticised PVC sheet – Protect concrete tanks during continuous or – intermittent immersion, exposure to water, – dilute acids, strong alkalies, and salt solutions – Protect concrete from concentrated acids or – combinations of acids and solvents Table 3.10 Recommended surface finishes (after Guide to Industrial Floors and Pavements 3.19) Typical applications Anticipated traffic Office and administration areas, laboratories Pedestrian or light trolleys Light to medium industrial premises, light engineering workshops, stores, warehouses, garages Light to heavy forklift trucks or other industrial vehicles with pneumatic tyres Exposure / service conditions Pavements to receive carpet, tiles, parquetry, etc Steel float Pavements with skid-resistant requirements Wooden float or broomed tined (light texture) Smooth pavements Steel trowel Dry pavements with skid‑resistant requirements Steel trowel (carborundum dust or silicon carbide incorporated into concrete surface) Wet and external pavements Broomed/tined (hessian drag light to medium texture) or grooved Sloping floors or ramps or high-speed traffic areas Broomed/tined (coarse texture) or grooved Heavy industrial premises, Heavy solid wheel vehicles Pavements subject to severe heavy engineering works, repair workshops, stores and warehouses or steel wheeled trolleys abrasion 3.10 Finish Reinforced Concrete Design Handbook Steel trowel/burnished finish (use of special aggregate monolithic toppings) In themselves, acid sulfate soils are not necessarily a problem. The problem is that the coastal regions are home to the majority of the Australian population and to various industry and production activities including utility supply, agriculture, aquaculture as well as sand and gravel extraction, which can disturb acid sulfate soils. Once disturbed and exposed to oxygen, these soils oxidise and produce sulfuric acid. The economic consequences of exposure to acid sulfate soils have the potential to be significant and can impact on agriculture, fishing, industry, urban development and infrastructure. The deemed-to-satisfy provisions for Design for Fire Resistance are set out in Section 5 of AS 3600, which should be read in conjunction with the requirements given in the Building Code of Australia (BCA)3.21. The BCA sets out the requirements for fire resistance for elements of a building. These depend on the building classification, rise in storeys and exposure to a fire-source feature, taking into account applicable concessions and other general requirements relating to the performance in fires. The rules for design for fire resistance in Section 5 of AS 3600 are based on Eurocode 2 3.22. The requirements for exposure classifications for concrete in sulfate soils provided in AS 3600 Table 4.8.1 are reproduced in Table 3.6. The Standard includes extensive notes to Table 4.8.1. See also CCAA's Technical Note Sulfate-resisting Concrete 3.7. Designing for fire resistance using the deemed-tosatisfy approach largely involves the specification for the various members of the required Fire Resistance Level (FRL). This is a composite number of the grading periods in minutes for each of structural adequacy/ integrity/insulation, as appropriate. The grading periods are specified in Specification A2.3 of the BCA. It is implied that the grading periods are intended to reflect the Fire Resistance Periods (FRP) for the respective criteria when the element is tested in a Standard Fire Test. Saline soils In saline conditions, or where they are likely to develop over time, the requirements for concrete in contact with the ground need to be assessed to ensure its durability and satisfactory performance over the design life of the structure. Table 3.7 sets out AS 3600 requirements for saline soils. Designers can also refer to Building with concrete in saline soils 3.16 and the CCAA Guide to Residential Slabs and Footings in Saline Environments 3.8 for further information. The BCA provides procedures to meet those performance requirements. There are three basic approaches: n 3.1.8 Effect of chemicals The effects of a comprehensive range of chemicals are shown in Table 3.8, while recommended barrier systems are shown in Table 3.9. 3.1.9Floor finishes The surface finish always needs careful consideration; the recommended finishes are set out in Table 3.10. 3.2 DESIGN FOR FIRE RESISTANCE 3.2.1 General Both engineers and regulators consider concrete structures to be inherently fire resistant and that high levels of fire resistance can be achieved by adopting certain axis distances and member dimensions. The reason for this is that concrete has both low thermal conductivity and high heat capacity; concrete elements are therefore naturally resistant to temperature rise due to fire exposure. In addition, experience in real fires has shown that concrete structures generally perform well. CCAA's Fire Safety of Concrete Buildings 3.20 covers these issues including advice on the performance of high-strength concrete. Deemed-to-satisfy constructions specified in terms of required Fire Resistance Levels (FRL) for various elements of construction n Alternative solutions n By fire tests. The data in AS 3600 is provided to complement the first of these three approaches. The Fire Resistance Level (FRL) is the Fire Resistance Periods (FRP) for structural adequacy, integrity and insulation, expressed in that order. Section 5 of AS 3600 sets out deemed-to-satisfy data to determine Fire Resistance Periods for the various member types. While this approach is used for the great majority of buildings, a small but increasing number of buildings are being designed using the other two approaches, involving fire engineering. These approaches involve neither the use of AS 3600 in general nor Section 5 in particular. The BCA clearly states that in the event of conflict between it and clauses in referenced standards, then the rules in the BCA shall take precedence. The fire resistance requirements in AS 3600 nominate 'axis distances' for longitudinal reinforcement (see Figure 3.3), not 'cover' (to any reinforcement, including fitments) as is the case for durability. Reinforced Concrete Design Handbook 3.11 b h≥b a1 a2 a b Figure 3.3 Axis distance, a (after AS 3600) Note: Axis distances are nominal values and no allowance for tolerance need be added Interpolation is permitted between adjacent values in the tables. It should also be noted that in many cases the axis distance will not control the design. For example, in a beam with a single layer of reinforcement with a main bar diameter of 20 mm, fitment size of 10 mm and a cover of 30 mm (for durability), if the axis distance required is less than 50 mm it will not control the design. In cases where small axis distances imply small or negative covers, the minimum cover for durability and concrete placement will determine the axis distance actually provided for the member. Definitions for Fire Resistance Level (FRL), Fire Resistance Period (FRP), structural adequacy, integrity and insulation are given in AS 3600 Clause 5.2. Generally, there are no specific rules for integrity; it is assumed to be satisfied when the requirements for structural adequacy and insulation are met. AS 3600 Clause 5.3.1 states that the FRP for a member shall be established by either one of the following methods: (a) Determined from the tabulated data and figures given in this Section. Unless stated otherwise within this Section, when using the tabulated data or figures no further checks are required concerning shear and torsion capacity or anchorage details. (b) Predicted by methods of calculation. In these cases, checks shall be made for bending, and where appropriate, shear, torsion and anchorage capacities. NOTE: Eurocode 2, Part 1.2 provides a method of calculation to predict the FRP of a member. during the latter part of the heating after cracks have developed in the member and pieces fall off from the corners and arrises of beams and columns. Explosive spalling occurs during the early part of heating when large pieces of concrete, up to 300-mm long, forcibly burst from the member. This is regarded as the most important type from a practical point of view. BS 8110 Part 2 3.24 Clauses 4.1.6 and 4.1.7 suggest that with covers exceeding 40 to 50 mm, rapid rates of heating, large compressive stresses and moisture contents over 2% to 3% by mass of dense concrete there is a high risk of spalling. Forrest3.25, reporting the findings of the committee that formulated the BS 8110 requirements, notes that fitments act to retain concrete around the main bars and that this limitation on cover should be applied to the surface of the fitment and not the main reinforcement. This is the basis of the shaded area on Charts 3.1 and 3.2, suggesting where spalling may need to be considered. When spalling has to be considered, the alternative strategies for its prevention are: n n n n n n the use of polypropylene fibres; the use of an applied finish of plaster or vermiculite, etc or lost formwork; the provision of a false ceiling as a fire barrier; the use of lightweight aggregate or (for columns) limestone aggregate; the use of sacrificial tensile steel reinforcement; if it does not conflict with durability requirements, the use of a supplementary mesh in the cover zone of concrete 20 mm from the concrete face. The use of 500-MPa reinforcement of itself will not tend to influence the spalling tendency, though consequential detailing practices may. However, it has been reported 3.26 that the use of high strength concrete (f 'c > 50 MPa), increased the spalling tendency. Plank3.27 states that for reinforced concrete structures in fire and in particular for slabs, two important phenomena, spalling and diaphragm action are not accounted for in the current simple code approaches. Ignoring spalling in slabs is unconservative but in contrast, tensile membrane action, which is also ignored in simple approaches, can significantly improve the performance of a fire-exposed structure. 3.2.2 Spalling of floors, beams and columns Spalling is defined as the breaking off of pieces of concrete from the surface of a structural element when it is heated in a fire. Malhotra3.23 defines three types of spalling: surface pitting, corner break-off and explosive. Surface pitting is when pieces of aggregate fly off from the surface. This usually occurs during the early part of the heating. Corner break-off occurs 3.12 Reinforced Concrete Design Handbook 3.2.3 Joints AS 3600 Clause 5.3.5 requires that joints between members or adjoining parts be constructed so that the FRL of the whole assembly is not less than that required for the member. Data on the performance of various generic joint and sealant types is limited and information on specific proprietary sealants (including limitations on joint geometry) should be obtained from the manufacturers. The CIA's Design of Joints in Concrete Buildings 3.28 includes charts for calculating the extent of non-combustible fibre blanket needed in a butt joint to walls to provide the required fire‑resistance periods. Clause 5.4.1, and the axis distance to the bottom reinforcement in the slab between the ribs is not less than that given in AS 3600 Table 5.5.2(A). n 3.2.4 Chases and openings AS 3600 Clause 5.3.6 requires that chases in concrete members subject to fire be kept to a minimum. Openings will normally require a fire rated infill to meet the same FRP as the wall, eg fire rated doors and fire rated dampers for services. 3.2.5 Increasing the FRP by addition of insulating material AS 3600 Clause 5.3.7 provides guidance on how to increase the FRP of an element by various techniques. More information is given in Section 3.2.10. 3.2.6 Beams Charts 3.1 and 3.2 reflect the information in AS 3600 Figures 5.4.2(A) and (B). See the discussion in Section 3.2.2 for the basis of the shaded area where spalling may need to be considered. They can also be used for beams exposed on all four sides (see AS 3600 Clause 5.4.6). For two-way ribbed slabs the slabs shall be proportioned so that the width and the average axis distance to the longitudinal bottom reinforcement in the ribs is not less than that given in AS 3600 Tables 5.5.2(C) and (D) as appropriate, and the axis distance to the bottom reinforcement in the slab between the ribs, and the axis distance of the corner bar to the side face of the rib, is not less than that value plus 10 mm. A slab shall be considered continuous if, under imposed actions, it is designed as continuous at one or both ends. Table 3.11 Minimum effective thickness of slabs for insulation (after AS 3600) Fire-resistance period (min) Effective thickness (mm) 30 60 90 60 80 100 120 180 120 150 175 240 3.2.7 Slabs Tables 3.11 to 3.15 reflect the information in AS 3600 Clause 5.5 and Tables 5.5.1 and 5.5.2(A), (B), (C) and (D). AS 3600 Clause 5.5.1 states that for insulation the effective thickness of slab shall be taken as: for solid slabs, the actual thickness; for hollowcore slabs, the net cross-sectional area divided by the width of the cross section; for ribbed slabs, the thickness of the solid slab between the webs of adjacent ribs. For structural adequacy for slabs, AS 3600 requires the following: n n n For solid or hollow-core slabs supported on beams or walls the average axis distance is not less than the value shown in AS 3600 Table 5.5.2(A). For flat slabs, including flat plates, the average axis distance is is not less than that shown in AS 3600 Table 5.5.2(B); and at least 20% of the total top reinforcement in each direction over intermediate supports is continuous over the full span and placed in the column strip. For one-way ribbed slabs, for the appropriate support conditions, the slab is proportioned so that the width of the ribs and the axis distance to the lowest layer of the longitudinal bottom reinforcement in the slab complies with the requirements for beams given in AS 3600 3.2.8 Columns Insulation and integrity of columns is required only where they are part of a wall with a fire-separating function. In this case they must comply with AS 3600 Clause 5.7.1 for walls. The structural adequacy for columns can be determined from Table 3.16 (axis distance and smaller column cross-sectional dimension are not less than the tabulated value for the desired FRP). Where the column is a wall (ie the longer cross section dimension is more than four times the shorter dimension), Table 3.18 can be used. AS 3600 Clause 5.6.2 also provides for an alternative method for columns in a braced structure. Generally, the value of the load level, N *f / Nu, will be taken as 0.7 but designers can calculate the value if they wish. When As is greater than 2% and the required FRP is greater than 90 min, then bars need to be distributed along all faces with a minimum of two bars in any face. The effective length of columns shall not exceed 3 m, and eccentricity shall be limited to 0.15b. Braced columns can be up to 6 m long. For longer columns designers will have to use the Eurocode 2, Part 1.2 method of calculation to predict the FRP of the column. Reinforced Concrete Design Handbook 3.13 chart 3.1 Simply-supported reinforced concrete beams exposed to fire on three or four sides FRP (min) = 30 100 60 90 120 180 240 a 90 am b 80 a 70 a a D≥b 60 50 Average axis distance, am (mm) am b b am 40 30 20 10 0 0 100 200 300 400 500 600 700 800 Width, b (mm) chart 3.2 Continuous reinforced concrete beams exposed to fire on three or four sides FRP (min) = 30 90 60 90 120 180 240 a 80 am b 70 a 60 a Average axis distance, am (mm) a D≥b 50 40 b 30 20 10 0 am b 0 100 200 300 Width, b (mm) 3.14 Reinforced Concrete Design Handbook 400 500 600 700 800 am Table 3.12 Structural adequacy of solid and hollowcore slabs supported on beams or walls and for one-way ribbed slabs (after AS 3600) Axis distance to lower layer of bottom reinforcement (mm) Simply-supported slabs Fire-resistance period for structural adequacy (min) One-way Two-way Continuous slabs l y / l x ≤ 1.5 One- and two-way 1.5 < Ly / l x ≤ 1.5 ≤ 2 30 60 90 10 20 30 10 10 15 10 15 20 10 10 15 120 180 240 40 55 65 20 30 40 25 40 50 20 30 40 Notes: 1 l y is the longer span and l x the short span for two-way slabs. 2 The axis distance assumes slabs are supported on four sides, otherwise they are treated as one-way slabs. Table 3.13 Structural adequacy of flat slabs including flat plates (after AS 3600) Minimum dimensions (mm) Fire-resistance period for structural adequacy (min) Slab thickness Axis distance to lower layer of bottom reinforcement 30 60 90 150 180 200 10 15 25 120 180 240 200 200 200 35 45 50 Table 3.14 Structural adequacy of two-way simply supported ribbed slabs (after AS 3600) Minimum dimensions (mm) Possible combinations of axis distance, as , and width of ribs, b Fire-resistance period for structural adequacy Combination 1 Combination 2 Combination 3 as b as b as b (min) Flange thickness, hs and axis distance, as in flange as hs 30 60 90 15 35 45 80 100 120 — 25 40 — 120 160 — 15 30 — ≥ 200 ≥ 250 10 10 15 80 80 100 120 180 240 60 75 90 160 220 280 55 70 75 190 260 350 40 60 70 ≥ 300 ≥ 410 ≥ 500 20 30 40 120 150 175 Note: 1 The axis distance is measured to the lowest layer of the longitudinal reinforcement. Reinforced Concrete Design Handbook 3.15 Table 3.15 Structural adequacy of two-way continuous supported ribbed slabs (after AS 3600) Minimum dimensions (mm) Possible combinations of axis distance, as , and width of ribs, b Fire-resistance period for structural adequacy Combination 1 Combination 2 Combination 3 (min) as b as b as b Flange thickness, hs and axis distance, as in flange as hs 30 60 90 10 25 35 80 100 120 — 15 25 — 120 160 — 10 15 — ≥ 200 ≥ 250 10 10 15 80 80 100 120 180 240 45 60 70 160 310 450 40 50 60 190 600 700 30 — — ≥ 300 — — 20 30 40 120 150 175 Notes: 1 The axis distance is measured to the lowest layer of the longitudinal reinforcement. 2 For prestressing tendons, the axis distance shall be increased as given in Clause 5.3.3. Table 3.16 Fire resistance periods for structural adequacy of columns Minimum dimensions (mm) Combinations for column exposed on more than one side Fire-resistance period for structural adequacy N*f /N u = 0.2 N*f /N u = 0.5 N*f /N u = 0.7 (min) as b as b as b Column exposed on one side N*f /N u = 0.7 as hs 30 25 200 25 200 32 27 200 300 25 155 60 25 200 36 31 200 300 46 40 250 350 25 155 90 31 25 200 300 45 38 300 400 53 402 350 4502 25 155 120 40 35 250 350 452 402 3502 4502 572 512 3502 4502 35 175 180 452 3502 632 3502 702 4502 55 230 240 612 3502 752 4502 — — 70 295 Notes: 1 as = axis distance b = smaller cross-sectional dimension of a rectangular column or the diameter of a circular column. 2 These combinations for columns with a minimum of 8 bars. Table 3.17 Minimum effective thickness for insulation for walls (after AS 3600) Fire-resistance period (FRP) for insulation (min) Effective thickness (mm) 30 60 90 60 80 100 120 180 240 120 150 175 3.16 Reinforced Concrete Design Handbook Table 3.18 Fire resistance periods (frp) for structural adequacy for walls Minimum dimensions (mm) combinations of as and b N*f /N u = 0.35 N*f /N u = 0.7 Wall exposed Wall exposed Wall exposed Fire-resistance period on one side on two sides on one side for structural adequacy (min) as b as b as b Wall exposed on two sides as b 30 60 90 10 10 20 100 110 120 10 10 10 120 120 140 10 10 25 120 130 140 10 10 25 120 140 170 120 180 240 25 40 55 150 180 230 25 45 55 160 200 250 35 50 60 160 210 270 35 55 60 220 270 350 Legend: as = axis distance b = wall thickness Table 3.19 Thickness of vermiculite/perlite concrete or gypsum-vermiculite/gypsum-perlite plaster to provide increased cover Increased cover required (mm) 5 10 15 20 25 30 Plaster thickness (mm) 4 8 12 15 19 23 Table 3.20 Thickness of plaster to increase the insulation value of slabs Nominal thickness of topping to be added (mm) Increase in thickness required (mm) 10 20 30 40 50 Plain concrete Vermiculite/perlite Gypsum 20 30 40 50 60 18 26 34 42 50 16 22 28 34 40 3.2.9 Walls Tables 3.17 and 3.18 reflect the information in AS 3600 Clause 5.7 and Tables 5.7.1 and 5.7.2. The FRP for insulation depends on the effective thickness as shown in Table 3.17. The effective thickness of the wall to be used in Table 3.17 shall be: for solid walls, the actual thickness; for hollowcore walls (and sandwich walls or similar), the net cross-sectional area divided by the length of the cross-section. AS 3600 requires that for walls that have an FRL, the ratio of the effective height to thickness shall not exceed 40, where the effective height is determined from AS 3600 Clause 11.4. This latter restriction does not apply to walls where the lateral support at the top of the wall is provided by an element not required by the relevant authority to have an FRL. AS 3600 Clause 11.1.(b) (ii) limits the slenderness ratio to 50 assuming the wall is designed as a slab. Reinforced Concrete Design Handbook 3.17 The FRP for structural adequacy for a wall given in Table 3.18 shall be used, provided the axis distance to the vertical reinforcement and the effective thickness of the wall is not less than the corresponding values given in the Table. For walls where the lateral support at the top of the wall is provided on one side only by a member not required by the relevant authority to have an FRL, structural adequacy will be considered to be achieved by satisfying Table 3.17. This would apply to single-storey buildings with precast or tilt-up walls. AS 3600 Clause 5.7.4 covers recesses and chases in walls under various conditions. 3.2.10 Increasing FRPs by use of insulating materials The information given in Tables 3.19 and 3.20 is derived from AS 3600 Clause 5.8 References 3.1 AS 3600 Concrete structures Standards Australia, 2009. 3.2 Durable Concrete Structures (Z07), 2nd Ed, Concrete Institute of Australia, 2001. 3.3 Performance Criteria for Concrete in Marine Environments (Z13), Concrete Institute of Australia, 2001. 3.4 ACI 201.2R-08 Guide to Durable Concrete, ACI Manual of Concrete Practice, 2008. 3.5 AS 4997 Guidelines for the Design of Maritime Structures Standards Australia, 2005. 3.6 Chloride Resistance of Concrete, Technical Report, Cement Concrete & Aggregates Australia, 2009. 3.7 Sulfate-resisting Concrete (TN68), Cement Concrete & Aggregates Australia, 2007. 3.8 Guide to Residential Slabs and Footings in Saline Environments (T57), Cement Concrete & Aggregates Australia, 2005. 3.9 AS 1379 Specification and supply of concrete Standards Australia, 2007. 3.10 AS 3735 Concrete structures for retaining liquids Standards Australia, 2001. 3.11 Marosszeky M and Gamble J Design, Detailing and Construction of Reinforcement for Durable Concrete Building, Research Centre, The University of New South Wales, 1987. 3.12 Clark LA, Shamas-Toma MGK, Seymour DA, Pallet PF and Marsh BK 'How can we get the cover we need', The Structural Engineer, Volume 75, No. 17, September 1997. 3.18 Reinforced Concrete Design Handbook 3.13 'Concrete Cover', Concrete in Australia, Vol 36, No. 1 March 2010. 3.14 AS 1012.4.1 Methods of testing concrete Part 4: Methods for the determination of air content of freshly mix concrete Standards Australia, 1999. 3.15 http://www.ozcoasts.org.au/indicators/econ_ cons_acid_sulfate_soils.jsp. 3.16 Lume E and Sirivivatnanon V Building with Concrete In Saline Soils, Proceedings of UrbanSalt 2007 Conference, 22–23 May 2007. 3.17 ACI 515 A Guide to the Use of Waterproofing, Damp proofing, Protective and Decorative Barrier Systems for Concrete, ACI Manual of Concrete Practice, 2000. 3.18 ACI 201.2R-08 Guide to Durable Concrete, ACI Manual of Concrete Practice, 2009. 3.19 Guide to Industrial Floors and Pavements (T48), Cement Concrete & Aggregates Australia, 2009. 3.20 Fire Safety of Concrete Buildings (T61), Cement Concrete & Aggregates Australia, 2010. 3.21 Building Code of Australia Australian Building Codes Board, 2010. 3.22 European Committee for Standardisation (CEN) (2004), – Eurocode 2: Design of concrete structures Part 1-1: General rules for buildings, The European Standard EN 1992-1-1:2004. 3.23 Malhotra HL Spalling of concrete in fires Technical Note 118, Construction Industry Research and Information Association, 1984. 3.24 BS 8110 Structural use of concrete Part 2: Code of practice for special circumstances British Standards Institution, 1985. 3.25 Forrest JCM 'New Fire-Resistance Data for Concrete', Concrete, UK, Vol. 18, No. 11, November 1984. 3.26 Phan LT Fire performance of high strength concrete: a report of the state-of-the-art, NISTIR 5934, US Department of Commerce, December 1996. 3.27 Plank R The fire resistance of reinforced concrete structures, Concrete Institute of Australia, Biennial Conference, 2007. 3.28 Design of Joints in Concrete Buildings (CPN 24), Concrete Institute of Australia, 2005. Chapter 4 Beams 4.1 Applicability to Ductility Classes of Reinforcing Steel The Charts, Tables and Spreadsheets in this Handbook are dependent on the Ductility Class of the reinforcement. 46714.1 AS/NZS covers three Ductility Classes of reinforcement: N, L and E. Only two of these are available in Australia, ie N and L. Designers should note that they must specify that reinforcement complies with the requirements of AS/NZS 4671 for building projects where AS 36004.2 is used in conjunction with the BCA4.3. AS 3600 imposes limitations on the use of reinforcing steel of Ductility Class L, eg AS 3600 Clause 1.1.2 states that Ductility Class L reinforcement: n n may be used as main or secondary reinforcement in the form of welded wire mesh, or as wire, bar and mesh in fitments; but shall not be used in any situation where the reinforcement is required to undergo large plastic deformation under strength limit state conditions. These limitations on Ductility Class L bar preclude its use as longitudinal tensile reinforcement in beams. As a result, it is not considered in any of the charts and spreadsheets in this Chapter. In addition, the use of Ductility Class L reinforcement is further limited by other clauses in AS 3600. Reinforcing steel of Ductility Class N may be used, without restriction, in all applications referred to in AS 3600 and the Charts and Spreadsheets herein are based on it. It is the designer's responsibility to ensure that the Ductility Class of the reinforcement specified and used on site: reflects the assumptions in the analysis methods, and is appropriate to the situation and the member being designed. The capacity reduction factor, f, for a strength check using a linear elastic analysis for Ductility Class L reinforcement is lower than that for Ductility Class N reinforcement, see AS 3600 Table 2.2.2 to allow for its lower ductility, compared to that of Ductility Class N reinforcement. 4.2 Rectangular Beams in Bending 4.2.1 General To ensure a beam section has adequate ductility (at ultimate strength under bending and/or compression), AS 3600 Clause 8.1.5 states that ko (the ratio, at ultimate strength, without axial force of the depth to the neutral axis from the extreme compression fibre to d ) should not exceed 0.36 and M * should not exceed 0.6 Mu unless specific requirements are met. It should be noted that kuo is not the balanced design condition as balanced sections are not ductile. The basic outline of a rectangular beam in bending is shown in Figure 4.1. The general theory of bending in reinforced concrete members is discussed in more detail in textbooks such as Concrete Structures4.4 and Reinforced Concrete Basics4.5. Cross-sections where kuo is greater than 0.36 are referred to as 'over-reinforced' or 'non-ductile' and have limited ductility. Over-reinforced members may have a number of unfavourable characteristics such as: n n n susceptibility to sudden, brittle failure with little warning; reduced ability to redistribute moments due to unexpected loads or settlement; and limited energy-absorption capacity under seismic or blast loading. When the structural analysis has been carried out in accordance with AS 3600 Clauses 6.2 to 6.6 and an over-reinforced cross-section cannot be avoided, then a minimum amount of compression reinforcement has to be provided, viz 1% of the area of concrete in compression. The design strength in bending of an over-reinforced section is not to be taken as more than the ultimate strength in bending, f Muo , when ku = 0.36, with the force in the tensile reinforcement reduced to balance the reduced compressive force in the concrete. An over-reinforced cross-section is therefore not an economical or preferred design solution. Equivalent stress block b εc dn D α dn/ 2 α 2f 'c α αdn C z d Ast εst Cross-section Strains z M * = Tz f sy Stresses T Forces Figure 4.1 Basic sections, strains, stresses and forces Reinforced Concrete Design Handbook 4.1 The charts, tables and spreadsheets in this chapter for the strength of beams in bending are based on the principles set out in AS 3600 Clauses 8.1.2 and 8.1.3. The rectangular stress block assumes a maximum strain in the extreme compression fibre of the concrete of 0.003 and a uniform compressive stress of a2f 'c acting on an area bounded by the edges of the section and a line parallel to the neutral axis under the loading concerned and located at a distance g kud from the extreme compressive fibre. To calculate the equivalent rectangular stress block the factors a2 and g are taken as: a2 = 1.0 − 0.003 f 'c (within the limits 0.67 ≤ a2 ≤ 0.85), and g = 1.05 − 0.007 f 'c (within the limits 0.67 ≤ γ ≤ 0.85). The values of α2 and γ are shown in Table 4.1 and graphically in Figure 4.2. 25 g 0.85 0.85 a2 0.85 0.85 32 40 50 65 80 0.826 0.77 0.7 0.67 0.67 0.85 0.85 0.805 0.76 0.85 M */bd 2 = f f 'c q (1 - q /1.7) This equation assumes α2 = 0.85, the value given in Table 4.1 for concrete strengths from 20 to 50 MPa. The chart is therefore limited to that range of concrete strengths. This will cover most design situations for beams in bending. For the above equation: f = 0.6 ≤ (1.19 − 13kuo /12) ≤ 0.8 from AS 3600 Table 2.2.2. (When ku ≤ 0.36 as set out in As 3600 Clause 8.1.5, then f = 0.8.) q = Ast fsy / bd f 'c and The maximum design strength in bending, fMuo allowed by AS 3600 occurs at the ductile limit, ie kuo = 0.36. At the ductile limit: pmax = 0.85 g f 'c / fsy kuo = 0.306 g f 'c / fsy and Concrete strength f 'c (MPa) 20 The non-dimensional curves in Chart 4.1 are useful for initial sizing and are derived from the following basic equations: fsy = 500 MPa. Table 4.1 Value of γ and a2 for various concrete strengths, f 'c Factor 4.2.2 Basis of Chart 4.1 100 0.67 0.7 g = 1.05 − 0.007 f 'c (within the limits 0.67 ≤ g ≤ 0.85) and g varies from 0.85 for f 'c = 20 MPa to 0.67 for f 'c ≥ 65 MPa and Mud / bd 2 = 0.306 g f 'c (1 - 0.18g ). 4.2.3 Design of rectangular beams The bending moments in a beam are determined from the structural analysis. Generally they vary along the beam, eg from maximum moment at the middle of a simply-supported beam reducing to nil at the ends. They vary from positive to negative across a span depending on the continuity and spans, as determined by the analysis. It is the responsibility of the designer to establish the critical section(s) for bending and the design ultimate moments. 0.9 α2 0.8 0.7 γ 0.6 0.5 0.4 0.3 0.2 0.1 0 20 25 32 40 50 65 80 100 Concrete strength f 'c Figure 4.2 Relationship of γ and a2 with the concrete strength, f 'c Spreadsheet 4.1 can be used to calculate the reinforcement requirements for a reinforced rectangular concrete beam cross-section in flexure in accordance with Flowchart 4.1. It uses the requirements of AS 3600 and the standard design principles for ultimate strength design. It checks the minimum reinforcement and assumes that ku ≤ kuo but it does not check cover, spacing requirements or detailing requirements, nor does it cover cross-sections with compression reinforcement. As with many design calculations, some initial design parameters are assumed and then checked and adjusted as required. For the design of concrete beams, once the size and concrete strengths are chosen, then an initial area of reinforcing steel is required to be input. 4.2 Reinforced Concrete Design Handbook chart 4.1 Rectangular beam without compression reinforcement f 'c (MPa) = 25 32 40 Ast /bd 50 0.030 0.029 0.028 0.027 d A st b fsy = 500 MPa 0.026 0.025 0.024 0.023 0.022 0.021 0.020 0.019 0.018 0.017 0.016 0.015 0.014 0.013 0.012 0.011 0.010 0.009 0.008 0.007 0.006 0.005 0.004 0.003 0.002 0.001 0.000 0 1 M */[bd 2] (MPa) 2 3 4 5 6 7 8 Reinforced Concrete Design Handbook 4.3 As noted in Reinforced Concrete Basics 4.5 , for an under reinforced beam, the ultimate moment capacity, Mu ≈ 0.85 Ast fsy d (within 10% of a more accurate calculation), ie it is independent of the concrete strength and width of the beam. This approximation can be used to make an initial estimate of the area of reinforcement required. 4.3.2 Design of T- and L-beams The designer then enters an area of reinforcement approximating to this estimate (usually a number of bars of one size, eg 3 N24). The spreadsheet then calculates the actual moment capacity for the chosen area of reinforcement and compares it with the design moment. If the calculated moment is less than or significantly greater than the design moment, it will be necessary to repeat the calculations with a revised area of reinforcement. The spreadsheet assumes the T- or L-beam is not over reinforced and that ku ≤ kuo and calculates an initial approximation of the reinforcement required as noted above for rectangular beams. No eet Sh No Job By e: Dat t shee is re spo nsi ble for its use . read is sp g th sin nu erso ep am Th Be n er: m io t a clai rete ut Dis nc d mp Co force Co lar r rein nt b u o lie e g d C /J ject tan un Pro ec t are ct bje d R tha Su rce eams info r b Re sign fo ut De D n sig De g din en -B d o e Sh mm mm 0 30 mm 0 45 mm 40 0 9 3 N et b Jo No By te: Da b tor ac nF s tio ertie uc ed rop a y R rial P cit MP a pa ate 0.8 MP Ca & M 32 0 φ 50 mm 6 f' c 0.3 f sy 12 k uo r te me dia nt e reo m Fit ed uir req ired the qu of o re on ati of re xim rea pro e a ap t th rst ge fi to a is ired ure qu fig s re ars is a 2b Th rate of 40 0 um Ite Spreadsheet 4.2 can be used to calculate the reinforcement requirements for a singly reinforced rectangular T- or L-beam in flexure in accordance with Flowchart 4.1. It checks minimum reinforcement, but it does not check cover, spacing requirements or detailing requirements. The spreadsheet then checks the flange thickness to see if t ≥ Ast fsy / (α2 f 'c bef ). If t ≥ Ast fsy / (α2 f 'c bef ), the area of concrete in compression is rectangular and within the flange width. The strength of the section in bending can be obtained from the first part of Spreadsheet 4.2 for beams with the stress block within the flange, using the width of the compression flange as bef. u 2 6 φM inim 12 2 m mm *< am 1.4 kN t M 36 0 se the 2.0 0 95 enion .U 2 8 2 17 e to em left ns n 10 .76 23 25 nc uirte ox d ista ) ire ge re.te reoqf 32 4nsio n 1 2.0 0 wb - d mm qu u llo nta .re e x 0 rs e 5 te st 4 y iz st A f 8 .23nsio 0 rce fle stae ts M* al ta the Pe .61% en neare n 2 in r li2m.itla2y8 ers o f2te min to Da y 5 fitm rre 61y6 No 10 etr , b o 3.0 sio it in ts ams fo t - d to en om idth D rs . 41n2 sioninput xou 2 la en de be sizt efl(1 .9y1e ern of 2te Ge W pth, fitm th d fitm un OK 2la te &ein n 2e4 or.r 8% .0v rs . tenand n e. De er to dep r to o, ro 1.6 8% (1 crebeersm 2a r.2 g3o y tio φM u us u 8e ern of ble e 0 v e c 5 re t n e m M 4 v 1 o < e 4 s h φ v 8 n 7 ioon ay la o a s C ti - co sile it < 1.6 e ta M* at ec n r c nuamnbge emine s 3t.9rs (1 m 2 1govyers 10 4% orvn. for M* nt for Eff = D e te OK 4t.0s 8 or ay lam ula baarrrte20 gm vabe n. 0.1 me ed ento ible 0 m 2 ogo ally of th en (1 ng tednecl re rr3ba1en4a2 eminm uir OK mo ns Mu su th s t ta gesscte req o l ate 5.7nagmire en1tn8 ts(1 oarreaayfr over . (U ntroid m po <φ OK enc ugn on ng u kN l A st M* < φM res er L16 ste cti stre ce edmre S cere l aterrrbae6e.0qnu4germemeentnatnsd m ay g vern g iv 6 itia is e n o * d a ate n 1 g t .3 In 8 rs i o e o a 1 0 M e m g rr rc a T ern2c0.9.3 tecereentarr1a8equgem bae pli ltim <0 25 he t afo fored giv .8 Ap U la608ncosenm el t r an nuoiroef m ents may m k u0 ds enin ≤0 kN er 9r0c.0. s9te en arroeseq emre tion forc121r3naguS3iv rea m φM uo em ents rcly ≤Ø sp c in n fo A kN fog 8agguila c1tato 6fo01n80 em el hto r ere 2.0 uir is in 0.6 sin 5 e g3nm 5.8r n 24 .0 ivefo0r0c. stemeCn t req irem its egsly ecetidore 0.8 5 d are th g th tsn1fo .0 Ar Sa 2 c1ta6to m lim n6 n2rc.e en u ser 24 4 f sin e src γ≤ etinore sin fo 8aere 0.8 φM u 0rb Sig3ivefm mute m o00 req tspo ityr oa m 7≤ 2≤ nu 18 m 0.3 2 foirbe ars ecemdiennoi ts n fo φM u 0.6 ≤ α ePnro acfo utorAa8 reSi3nm6mmor0c.em ent 2 frethin f src rso u re m e s o o 7 e ti q m m r c m it u e . k d ptssiz itfgybly ln inni ts fo8m a reo inirfo e bd lim s 0.6 ep n 1 0 0A reSi3n6f 0 2rcem eerirerTesocetaocm o galoNfreth o) uir ceaBnar acin it Th 0 99 quthmula e2 metio imuto9m ku rewqh lim A inmfom req inegm apresa creinittigcbyaly 0.8 er: rs 19 uhqeurireesecmienntsm3uto ma gsaorsnrefgmw nt enqduir gfocr AasTphpeaaoscin re llainby e, am ed m e e i laim ac e ts 8 g s e b vid b r c in ed n rn re 7 d re tu a m th tabgs f sth.peuro ir mien 1u7 vid em nnt t = ndntsangfoclreAastcoppreaaarecoitcflain -0.5 ou Dis ow Aq .pro 26 rireem insim =A st 0.8 5 elb 07 uo (1 forc oemmee ntebmeeenndnt in . tsabndceaAdtopasaccitaybgs re s grce s p l in dwhqeu rxeurem euir emeepnlict ain , M , 20 ma k 3% hA 0.8 ck rein mrc itie t wit s blo 4.9 m an n pendinfoaabngldca stopasc abgs re flee ac en ile atefo omqfoercunirtebam = ob 85 M* . onngsm a0rs.3o6 f'c ga focrhap inenAst.stw es Cli ns ltimrein enmt reinm t/J inti lyemberepnlict in qarrc str blde nlypaseclm )) = 2 = .m L 2 o e= re jec OK s te ensuile imeamtem foeinnlgtyelyambpeenpdnlict aans loe storc m mtiocrein ulm d f' c crete luambing ) A stula , , tPk uo lpha Pro mm ct kN te t).ufoltrcimuamteeanrtre late ss th oauslfo arercnt ampe eaicmaabnindfo Mtobineimsn2d01in0 &Carlces st Micuosa itial A s f sy / (b r con 2) bje u k r /1 m e u c n p te e l t e 2 2 lc 95 o 2.0 ye b imis umteforticinm rrc l in , .m tu an S ltrc ku in 17 24 15 blo (A st d fo rellint)m.uin ed lateemsseth o ulfo a l or Lampeepanr ire aebl innt 1M eusotrigunc teomBe = Phi Ca lcuula anspm tim a fo a re 13 c mm A stm provid ss .5/α2 k u) b P m e e n : te l a − e lc in o y M im in e e c a e te M 0.0 8 in l rc S i r Tarrc sh ppstem ng D st 9 n rme γ u is uateartre str - 0 C re uo la eersmsseth d 0 <A sllnht)emfl.ueltcetitim Ca inlcfouula φM = oicin (1.1 2 daeth 18 tio ncu ck φM ulafondrly aionenudlati rete oxim 9 p te ing d (1 (1-0.5 .require f' c m lc im id a in a s e te = rc s e .3 : re ip 7 u a lt s p s la e t e c a st n m 3 f sy i C 0 ’ A C Ch Ø rsmss ddsllht)em.u etim tu C oinnlcfouuslaperete M m lc n cr city arrearcularessAm le .36 0.0 3 f c n e in ecstiuopm 0. en A st γ k u ing pa a aeth C a lcenfo A st es ptingin csa, ioCno. rced =0 Tit re suidrc la 50 1 i shertim ca ratio C m.r s tim lapte 60om M u = f'c d us CTahoin uo lD ereersmsadasllnht)la 21 1.0 − 0.0 t = ing n hk lceussidrc Sec3k Mheck φ M u = Ølculate C aalc eein forisopf acooraessulkteens einfo = egushm re nd emen ose lapteerersm A era w it C .0 e γ h n fo n φ u C ) 1 b h tr t n a rp h a C l c oinlc sidrcetanadallsig ee ts co s F in , R d k .2 TC = forc el ec k u c d in C tua = nd Ge re s c e e h c n prell &ers ick 2.2 α2 *d) s Cahoenfo Ch eck Ac Rein of ste *a se ble inr residperorst dadseffe9 (Ifor H tr TC .85 tio ab tM a Ch yu re nt * 0 ra (Ta en , em ilpa FTohoeness nspsreth-e20ll0owanm Are rall ns *f sypth om me CDo e re 0 t a ngfor nd K tio ne e.th m / (Øde rce .1.3) ies ula ge utrlaeng M* TIghno36s0no Rbale r a d ert Calc reinfos (Cl 8 = n to 0.5 s as bw ) ADSoe rneur,ita oste .3)form A st pa n rop ate / f sy n tion on .1g f' c o t s p F s s: 8 ti ( f im ti a io u l l l lt ct t f' n 2 r, a s ula ly .6 (Csin u W ota ) ca inp teria No arne tio rm ts u on ten Calc mp .1) = 0 D dn t,/d enlly ed ita reti .6 co f of ata ma info W emca bas an ee( eo to l 8.1 f' ct r, or ed Lim en uiratints ers sh0.2 (C Th ols reoed =ad os uir ove reqmme ns lay ts of s/ mb ch mm tio req try, c * gthutouire on em men 1 pst.mrein de : Sy es en a eq e op lati de lls M r for ly As e uir lcu d Str d R to Co sis siz .1 ce mpis req l ca an late cgo th nt 134 44 er mm w geom ents ete ia u e n g o th it to Ba ll lc ars In in eng : m ig d in erd 0.3 .8 ye tion mom r diam de rc str ca es re rce n b egma 0 d fo d e 1 c tu g fo 1 g n Dre d in um u la in Se din & ba t gra lerts rein ck nc ku n ing gro forc ion re Mincim n a ba ile xis d ck a in Be ber eme sed me ed ns tral a ter ba of re m γ ku tors d sp No fe te u e n d e r Nu forc ll is u e n n m k e g r an an in : d o to para n bloc gre cin nt oth ut Re ere fi se pth is ith spa shea eme or ba De tral ax ressio Inp , d ns sw Wh p u lts tio ell e an φM u inforc Ne com su es dc of Re gg xe r, siz acity um re pth su e De p Bo Inp Spreadsheet 4.1 is available at www.ccaa.com.au n sig t uta mp Co ion De c on dC ce or inf Re am Be e ret 4.3 T-Beams and L-Beams in Bending 4.3.1 General Concrete beams will often be part of the floor or roof structure; in these cases they will be T- or L-beams for part of their span, usually at the centre portion of the span where the flange is in compression. A T- or L-beam is significantly stiffer than a rectangular beam of the same depth and web width. AS 3600 Clause 8.8 sets out the width of flange that can be adopted for such beams. However, over a support, a T- or L-beam will normally be considered as a rectangular beam. s, im mb nt ca on min Nu cti me for A rre Mo cks co CCA e ts, Ch en t the mm ntac o o rc Fo ase c ple t: u utp O ck ba ed Fe : . The determination of the flexural strength of a T- or L-beam depends on whether the depth of the assumed rectangular compressive stress block lies within the flange thickness, t, only or if it lies both in the flange and in part of the lower section of web. If t ≥ Ast fsy / (α2 f 'c bef), the area of concrete in compression is rectangular and is within the flange. The beam may then be designed as a rectangular beam with the web width equal to flange width, bef . If t < Ast fsy / (α2 f 'c bef ), then the area in compression is T- or L-shaped, ie the flange and part of the stem resist compression. The strength in bending can then be calculated from the second part of Spreadsheet 4.2 for a T- or L-beams with the stress block in the flange and part of the web. This uses the outstand of the compression flange, bef – bw, times the full depth of the flange, t f , and the dimensions of the web resisting compression is the width of the web times the effective depth, ie bw α dn. Based on the initial area of tensile reinforcement calculated and depending on whether the flange or the flange and web are in compression, the designer then enters an area of reinforcement approximating to this estimate (usually a number of bars of one size). The spreadsheet then calculates the actual moment capacity for the chosen reinforcement and for the chosen design case and compares it with the design moment. For both design cases, the spreadsheet checks that the actual moment capacity exceeds the design moment capacity. If the actual moment capacity is less than or significantly greater than the design moment, it will be necessary to repeat the calculations with a revised area of reinforcement. No eet Sh No Job By te: Da ctor n Fa s tio uc ertie Red Prop l ity ac eria Cap Mat & φ b ef on pers t is shee read is sp g th usin e Th n er: laim t at io Disc mpu Co t Clien t/Job ec Proj t bjec Su r its le fo onsib resp Lor Tte forced cre rein on der d C e un rce at ar info s th De bw ing d nd Be mm mm 0 40 mm 00 11 mm 0 12 mm 5 mm 72 40 0 65 2 MPa MPa mm o tN . reo er. ed low uir be req bly . the ba ed of pro uir on ill req ati d w reo xim uire of ro a q p ap o re are rst e re t the e a fi th is am to g ure be ed fig or L quir is Th r a T as re 2 Fo rate mm Ite ee Sh ign s am Be D use. 0.8 25 0 50 0.36 12 f' c eter f sy diam k uo ent Fitm tf o bN Jo By te: Da n sio ten sion re. of n n xu fle layers of te nsio in m orre2. layers of te a xu 2 be t fl(1 ers n. n e or.r . lay ver rete ein se ea or 2y go ern. nc geam m nsth(1 su o it v c a e e . n m y go ern for lar arrtea b gm emin ent (1 a nts gu ecl re na ible ov ns rrbae gemme ts m y g tan te on po rec nL sc cerel ate rranquire men s ma res ed givoer oste n l a re nt t is ire ee forc read T ivlaernc0. steeent requ reme sh i rc aggu 360en c.em nt ad rein onfofo u q re n r r e ly ctiin fota ASgiv fo00 m e sp ing ere n n c tor a reSi3n6 r0c.e ent r this a s egsly ecetidore ts fo mA in6fo0 em sig ing etinonmuto for of sthin src us reS3 rc De nts ityr a thinefoirem mA einfo on ec ienni ts uto m r ers me acfo ofre que s m m m ea ep ire anpts itgyly re thheuriree mienni ts um Th eB qu inegmce apascincinyg of re n t w q rireem inim re e er: a it s io ir a e d t c t hqu m im cr en enqu gfodr sappacsclainbg s re cla uta on tsan c topa abg w here em nnt tbre ndnin Dis g mp dC forc oemmee tebmeeendt inabnlde s o ascl in bs w Co t ce rein tefomrc mqeunir embenpnlict a blde st p sla for en ob sile ain mt ore foercnltyeampe pnlict aan le to Cli t/J ein ten suillteimre amteenreinomlfo arrc ampe icab jec R s m u e e l t m n y im e c te Pro in l c i te t).ufoltrcimuateartre la ss th arrc app bje ulfo ne p lcu te Su icin rly is um artre rellint)m.uinltim lateemsseth Ca lcuula ula n e eeintim s upm te aeth a lc la n: C i . rtic late rsms dsllht)m Ca inlcfouurc eete tio a aen eeintim pa a lc sidrc : re C rip ula Caoinnlcfo pereersm dsllhm t is le sc C Tit re ssidrcprersa a shee Cahoenfo De l e d in TC ra re henssid rea ne TCo sp Ge e Th If t < Ast fsy / (α2 f 'c bef), the area in compression is T- or L-shaped and includes the full depth of the flange and extends down into the web of the beam. The strength in bending has then to be calculated using two components: the compression stress in the flange outstands beyond the web and the compression stress in part of the upper section of the web including the portion in the flange. 4.4 Reinforced Concrete Design Handbook Re r beam Fo mm 42 m 67 kN 40 42 rs b ef 67 the 60 90 2 ba to e, 12 4 14 of ng = ce % 5. ed fla t f um uir stan ) 6 M of 4.08 req 36 w inim 11 - di m EA 60 A st. dth nge, , b 20 φM u ze t 5m 75 am LB 10 6 M* met W idth iv e wi of fla ts IN lue *< t si 6. Use en va en ares OR ct tM s CK ft. is 7 fitm fitm ne LO Effe knes AT men 32 4 x le t - d to e th 40 ic D r to SB ire 71 Th th, 80 4 ve h d men unde w bo N AS qu t us co pt n WEB NS RES llo fit IG re ) 8. no D ye Dep ttom e de r to o, ro ctio d d 85 *d ate 28 9 DES E ST TIO AN se the an φM u e do Bo ctiv cove ile re TH it st 36 6 GE ULA AM t at tiv M* sy * 0. *< H into 72 t lim 61 en Effe D - tens M it ga AN LC *f r r .9 t en = WIT BE e ne 10 (Ø pu s fo mom d fo E FLE CA R L S . If ally of th momM* / y th 11 d in tf ire TH ES su size 24 2 O th E ng an 76 t> (U ntroid Onl Web ns & ng TR = requ E TH A T NG 67 ble 45 .9 tio e stre rs ge and ts A st stre tage A st ce E S FLA US R e ta 14 al d ac mat mbe men ate Flan ge rcen 8% ov FO H TH THE 15 Initi ire ultim ab plie Ulti Pe .9 r nu 20 the Flan m 80 49 Ap ba 4 WIT EB IN in requ on 67 fro 31 .5 mm the ted th sed o ea CK ND W t lock in 21 en gges d ar reng ts ba of re A 7 BLO s B ck 22 St an 16 em Su 13 n es Blo 08 rs 1 ate remen atio s 69 ba Str rang 20 .5 im ul of t ar ox equi calc OK the Stres 33 no pr R al en e e ith 34 = NS Ap 12 em os t th w Initi 34 3 TIO ITH E Cho forc am with 68 11 .7 W LA rein CU EAM IN TH L be am 59 ed u AL or L be 60 2 os L B CK EC < φM 80 rT m2 rs op 67 TH T OR B LO mm Pr Fo r T or M* ze bar m of ba d S EB r si mm USE R A RES D W Fo 00 Ba ea of No. quire tal 64 2 FO E ST AN Ar tical rs re m2 To 0 E m re ba m 65 TH ANG mm 2 eo kN .92% Th tual bars ed FL mm 99 vid Ac of 05 pro 5 64 A st. ea 74 25 Ar .83% 64 d re 68 xu ly φM uo fle re on for flexu IL e ) A st FA g r * b ef um A st fo ns lan im * f' c 2 latio Min imum eF OK S (γ cu th N / in kN M Cal u in TIO ITH E * f sy 2 kN φM W LA ign 00 A st ck m *< CU EAM IN TH bd 32 t≥ m M lo on Des o) AL 00 si am ck 36 32 1 es L B CK sB be ma ku EC Che < 0. TH T OR B LO mpr es lar 16 m S k u0 co EB gu ga m Str USE R A RES D W an .5 m e in kN ng ea rect (1-0 FO E ST AN ion E fla t b ef .8 T r B g a kuo ss TH ANG ef 88 the 2 f' c γb ula min ma 14 α .60% of pre f sy FL f' c 99 rt C f = ng assu gam st α2 pa om f'c cta .36 T =A n = 0.25 d with eb eC Re =0 pha2 .70% 83 am s a at k uo al Al ku dW h th be iti d n) s a uo lar an wit IL 5γ on ent M Phi in gu e 0. ti FA m g = la om ctan (dnea lan lcu M φM uo T* a re kN LB OK eF Ca imum heck u= ing r φM m th kN d C 85 su Max To u in yiel 17 as kN φM at 00 ks ck sa m *< is 32 a oc m M lo it bl 15 s g 36 14 6 ss sB on min ) < 0. stre e 19 es su lati - bw k u0 ing flang g as m ef Str lcu in kN t f (b t us in f' c en rce forc ion Ca f 8 2 .1 fo rein ss T-C b w = α st f sy ) 84 Mom on ≤ 0. γ Cf 14 f si e in ck pre f' c ≤Ø T =A T - C 85 Che pres forc α2 0.6 om ) t =( = ed n ≤ 0. 0.85 its vid en 0.30 Com ion γd Cw dn eC pro ≤γ ≤ lim ns .5 om A st. Te 0.67 ≤ α2 (d-0 ku eM h th ing its cw us mat 0.67 wit lim ed t) + at its .5 Ulti OK m lim lcul (d-0 ea late ca Cf ku th cu 80 LB u= eld wid ck 0. Cal r yi φM u Che eb at To φM nd it is to w l p 0.85 sa tsta ing ua in = ed a 85 ou m eq 0. s A stm provid e A st st ng assu nge on MPa < A nly ty fla = in rcing d fla o ci lati 2 e u e an pa = lc forc reinfo web ng ca ratio 3.00 mm Ca Fla nding ent en sion in e in 8 es rce the be cem chos 38 pr forc fo = in Actual einfor steel Com ion sion is R t of ns es ed /12) eb vid en ck ea Te pr uo k pro W Ar lo = om ty 13 A st. d B c Com 7 f' eM ing an paci 9− ss am 00 us (1.1 mat ge ing ca t ratio tre be - 0. ’ c ed n = f S at en f en Ulti o Ø Fla bend cem chos 1.05 003 lcul the es .36 late ca l the tual or ku cu en − 0. typ k uo=0 ck in Ac Reinf of stee d γ = 1.0 Cal wh 0.5 bw oth ith = Che m ea ) k is α2 r b 2) w Ar / f sy ea ( f' c loc 2 f' ctf s fo 2.2. LB 0.6 ) sB on able /d or es (D 1) ctf = lati (T aT f' Str 0.2 1.6. 1.3) lcu ns for mm l 8. the l 8. Ca culatio = (C (C nt n en ts mi ns Cal me 1 A st. wh men ge 161. 8 latio ply m ire mm n cu a a m co Cal requ be 0.24 9 arr to th 6. nt rL 13 ng med me d n To stre Dee ku rce for mm um im nt info l axis γ dn Min me 8 n re utra eter ge 195. 1 mm d o ne ram k an se th to pa bloc 0.30 5 arr 6. ba Dep l axis ession nt 16 tra pr lts me d n su Neu com ku rce Re of fo is th in l ax γ dn Dep n re utra eter d o ne ram k se th to pa bloc ba Dep l axis ession tra pr lts su Neu com Re of th Dep t pu a In Dat ry Geo Spreadshee 4 2 s ava ab e om www ccaa com au .83% 48 16 0.0 5 0.02 00 64 m kN 2 mm m kN 0.3 5 0.02 00 64 φM p A st u 2 mm 4.4 Beams in Shear 4.4.1 General The vertical shear forces in a beam are determined from the structural analysis. They vary from maximum shear at the supports at each end of the beam reducing to nil in the middle of the beam for uniformly loaded beams, to nearly uniform shear for a heavy point load in the middle of a beam. The designer needs to determine which are the critical sections for shear design. (Note that the critical sections for shear will usually differ from those for bending.) Shear reinforcement in the form of fitments (sometimes referred to as ties, ligatures, stirrups or links) is generally provided in beams for shear (and sometimes torsion). They also support the longitudinal reinforcement within the beam. The fitments are usually vertical due to practical fixing considerations. In the past, fitments could be inclined from the vertical, while bent-up bars from the bottom reinforcement to the top reinforcement were sometimes used as shear bars. These methods are not used in modern construction because they are difficult to fix and AS 3600 does not cover bent-up bars. See Figure 4.3 for a typical arrangement of shear reinforcement. Fitments need to anchored at each end, and for practical reasons, are typically closed. For band beams, however, open fitments are used to allow fixing of the beam reinforcement, with the slab reinforcement supporting the top bars. AS 3600 Clause 8.2.12.4 has specific requirements for anchorage with hooks and cogs, and specifies that cogs must not be located within 50 mm of any concrete surface. Top bars Shear bars (fitment) Positive moment reinforcement Figure 4.3 Typical arrangement of beam reinforcement Mesh or bar (or bar made from coil) can be used for fitments. Fitments can be made from either round bar R250N, round bar R500L, deformed bar D500N or deformed bar D500L. Generally, only six sizes of fitments are available, from 6 mm to 20 mm as noted below. Fitments are normally made from either round bar grade 500 low ductility (R500L) or deformed bar grade 500 low ductility (D500L) for the 6-, 8- or 10-mm diameter sizes or from deformed bar grade 500 normal ductility (D500N) for the 12-, 16- and 20-mm diameter sizes. Mesh is generally not used. Designers need to specify the type and spacing of fitment they require for the beam, eg L8 fitments at 200 centres or N16 fitments in threes at 150 centres. Note that an L8 fitment is actually 7.6 mm in diameter and it can be made from either low ductility grade 500 round or deformed bar. The smaller diameters 6, 8 and 10 L bars were previously known as wire. As a result, the designation W6, W8 and W10 is still often used for L6, L8 and L10 fitments indicating grade 500L bar (wire). Grade 250 normal ductility R6N and R10N fitments are only available in some locations. As noted above, fitments generally have two legs and are usually of the closed type with two 135° hooks. A combination of fitments is sometimes used for beams (which might not be rectangular, eg a beam with an upstand section) or for wide shallow beams, especially if the shear is high and a considerable area of shear reinforcement is required, eg one overall fitment and two smaller fitments to form a triple fitment. Detailing of fitments is discussed in more detail in Section 13 of the Reinforcement Detailing Handbook 4.6. Logical models for shear are very complex, so, like most design codes, AS 3600 Clause 8.2 uses an empirical approach to the design for shear. The first check is to ensure the applied shear does not exceed the web-crushing limit, otherwise the beam size and or concrete strength will need to be increased. For the determination of the shear strength of a beam without shear reinforcement, AS 3600 Clause 8.2.7 provides an empirical formula relating the shear strength to the amount of tensile reinforcement and the concrete strength. Modifying factors are included to allow for the effect of the overall beam depth, axial tension or compression in the beam, and for concentrated loads near the face of a support. The determination of the contribution to shear strength by the shear reinforcement is based on the truss analogy method, using an angle of the concrete compression strut to the horizontal that varies from 30° when only the minimum shear reinforcement is required to 45° when the shear approaches the upper Reinforced Concrete Design Handbook 4.5 r % ign t is 16 ee es 38 (D dsh rea b do sp this ng 8.2 usi Cl OK on ar ers m he No ep m h t S T ee n0 er: n mm im Sh 30 io o sig at 0 cla mm 2.8 De 45 ut bN Dis e 8. mm mp ax Jo us 50 am Co Cla u.m 0 Be in nt .2.6 b 39 φV te l8 By ified Clie ct/Jo *< rC cre ec je te: V ea sp the on Pro sh 2 ct Da to area for dC bje ce m e sion um n th Su m es id rc im ista m ) (av) mpr min ew 60 -d th info co kN 18 the ctiv ze st 5m kN ut ial leng Re d) ffe ax t si kN 9 than the nts Inp =e 0 ore ded en eare 74 t to er ter 1.1 bv 20 me ch ta y vi ov 4 fitm to n bjec an st.pro grea 1 ≥ 52 Da ometr idth, D r to fit su d tsion A or lly β rs es W th, d o men unde ve = be Ge (fu l to p 0.8 V* co pth fit , ro mpr ua nt em ≥ e co De om e to n m eq = 1 d o r al tt em for ) is ax on sio β Bo ctive cove sile re n Ag) n (A sv forc u.m ax e diag .7 n n V ff io 14 in te D te .2 E = */ ct io = 8 ided re. u.m e te Re se of ov crea Cl ØV ns n r 1 + (N ile ally of th d o) ) = to t pr 1 xu nt oss ns su o as en ′ b v vd o .6 me Te (U ntroid a cr 33 fle layers of te nsio so √ f c f c′ b ion; 8.2 al at rcem in1. rce ed 10 √ ce tens ion din info CL rce 2.1 yers of te + 0. 0.10 info m o1.rre ial s e orientat itu r re ax 33 ns r fo r re ea 1ea ng Vuc uc + V u.m tio ea s rstnto.ax ten t io sh ea ar n Lo (1xu 2 la ac n sh of ≥ 1.b on mit sh Ø(V g 0)te ed ea nt fle or.r0.8 larsyosuevbjeecrrs o.f suteppnors ut lati g li d o 00 e. ein pli Desi l ar c 8 ho /1re a r.2beg ye rndothfe (1 m Ap us lcu rushin f c′ b v e do ≥ 0. wit 1/3 t iona − n m e a s h e ct y n o em o a 0) .2 = m n c g 6 la v srs it an . ma ea n C eb = 0 s-se 1(r1.c (10sio form r1 2 goloy anb/1e00 emineβs1 nt ≥rs ade rn os for 1. ab Ast sig w max e = la o aypliela − do d gm e to e cr gu arrte of do ed g)nt De lculate V u. ible 1.6 f cv th 2 ov rn. b v re th nβ1 uir ec1(l re aenβ1a em 5Ae =t (1 m ap in do 3.m ns ng eq 2 s er g ea Ca whe cta o=n1. l aluteerrfobr nag−m tre β 3 b v sβ1cte (N po eR rs ir*/e eβ n nt ed thoma1y goMvP rn. rs be yb va β2 res ea rrbaoereq1 ugememeeprnovtid=n(1 d regivoer nL ste crtere y 1 ma ts sh is mem e ve =β 2, 3e eo No intg; r an uir m ed rc ad T ncoInnse cerelndante et For ma3.17 y gokN β late V uc an rR uir d reE he lcu ivlaer 00.oste ea eq e nre bee arrreaqtenrgtheirem en= ts info Rfo a74.7 kN ds Ca Sh WIS a g u rc β1 o R equir n puem el nt grea a 6 m re c u e n g cv e E f re oTHfo fotar n Sg3ivlaecfrot 0tor0c. stemet lnoat rrreq rem nts 52.3 rR oR ly ctiOin d sp e ea ire uibj3n6 en c.e bue nt d e= ing ere ui Sh ar R n c torAnarsgresu is qu kN ive uc e iv fo0r2d0 o/savteme t req iremV ucV uc = Re a s egsly ecetidore be ASgm φV m wa tam Sh g th .9 eo kN β1 cem .5 nots 6asenrc.e en is u2.7 Ø nr mfo aeraeSi3nniv 8.2 74.8 u e sin for oaf sthin efosrc <0 50 m rR to q l re r a ed * u 8. b 0 kN C e ing ti fo m e l m .4 V >7 t ts ityr fo mAatakeg in6fo0 em nt rt C uir rc * 52 Sh on ecemd1,iennoi ts en eq n te kN en D en acfo uto frethinqueires=rc .4 be re 3 c. < V in re r info r e d em = o oR 0 0 75 ers em c S ay ti m m c m o e i fo u rc ts p rc p ire fo e ts 0 f o m d .8 r re r e cn iennonr uto fo mA ei3n6 rcerem r R 25 q u itgyly re thinheβ2 e f V Vu.m ire an 52 ea ea D> Re 9reinfo+ 0.6 b v = = ueirrseceom 1,o im Th ra S foear in 0.5 < φ sh qu inegmce apascincinyglyofre ti 9 q m o Sh mm w r in ts o) e c r: 9 A d P e * a m kN r la n re c e n sh V u .m M d ir c a a itgabgs orefgth u e ir = e ien 1/m rR t 4m mm , 1she Vu 0.6 b v = im Vu ea 7 reiout m in′cni3ts≤uto .2 l in n whqthueβ3 sreem inim en enqu gfodr sapapascin or Sh >φ 26 with cla rneum c+ ir tsan cr topasaccitcaytabgs ore 4d f chev nm f V* um em nnt tbre ndnin am mm omuinim0.6 b vd o Ø(Vu V u.minu.min Dis gfo s re l in f whqeu re=em the c me inim kN whi 8 a be 7 (i) Vu elb forc oemmee tebmeeendt in 33 5 ith + 0 ØV = tsaabnldeceadptopasaccitaybgs re efcvir sof. .5 kN .2 <φ 0 22 , Mm w V2uc 0 26 8.2 e s p l in whqu e mngieth rein tefomrc mqeunir embmenepnnlicdt in V* > 25 /2 t fogbrc .8 Cl anbea b vdoo)n≥, D Bw uc nldca o asc s re er stcreit en 52 ob n in 5 (i) kN 5 eb φV sile ain mt ore foercunirltyeeambpeere > lict aa gblde st nply sclainbegnst. wcShahepara in. ngthmof a0√ fac′ rs Cli 22 cing t/J D ired 8.2. = 0.5 ew a ten suillteimre amteenreinoremqlfo Cl .8 ns o stopaemlab g ) st.m Loeng 0.1Pe arercngt lyampeepnndiin jec a f th requ Sp 00 a 52 s c = eb t o 5 t m u e ro x e in A , tr uc +s, lt w ye im umteenrticinmsinrrc le no P ct o r 00 uc Ma tent).ufoltrc the kN idth late ss th φV a el nt bampepneaicmaabnindforctobesn2d01in0 & rsehesar sn =sVic a a re g is 5D o r 3 .2.5 bje of fo mriocain ulm ew 0.5 min n e ltim l8 = 0.7 lcu te mm rcin imeismupm mia Su lye L pbepnl rt reablein t 1, uo.muclatu idth 52 te V u.B u. rc of 0.5D gC rellint)m.uin lf th lateemsseth info arrc spafortre φV Ca lcuula mm t imoante ew f er 0 ha miocin ulfo olryeam r re in rcin n einhfo te aeth ea lic el n M tr lcu te ss r o 45 lf th a lc la eeinim in kN and ea .m 0 L e sse n: C is uamteartre info sh ppstemeng SCa cre u.m sh late rsmss re ruTlfo arrc 15 mm Vu d ha = sllnt)emfl.ueltcetitim Ca inlcfouurc m eete φV Le en 0 an tio r re n <φ nd y a n ndlati rete adaeth a lc sidrc s oicin e in imis upm 20 50m m ea : e th re = V* C lapte rsm sh 0m ula f2 r a arlssAiomaelcuonc d Co Caoinnlcfo ss ddaesllnht)em.ultetim onartre ativ crip le ereete 25 V uc = mm ro sidrc 0. nts for Vuc eraticuplre a eth hm C neg aalcenufo e. insecistiupsm terers Ø es gin c , Cn. rce Tit r of re 0 s h la ate is = n C mm te la D in 6 u m D re e s C m s 8 5 io ti t s 5φ h re peete adallnt)lareeinim g 3 oainlc .m m 0. 33 TC 22 grea gu lceussidrc Vu s ral foris pf acooraess lkens info 5 e the re AS equireif V* ≤ if fi the uc C 22 ≤φ nlcfoula eter CTahoin kN te ne pereersmngdusllhm V* ceed V ed R eck ign t o rpstr Fainu te , Re V* No m web ce *≤ φ diam C x = d in Ge re ssidrccptarersadaesshffeeec(Itsncporell &ers kN c< k us Cahoenfo Ch s 52.8 te t ex if V 250m f the se Vu ic rio = tio e r no f no t k re d u o in b kN o tr TC s va a 5φ e 0.0 9 s t o a r id fo, Hem pa 0. y of = re 20 .8 kN ts einn if D does Chgecreatere width D doe 0.75 D th ra = 8 k if FTohoeness nspsreth-e20ll0owanm il ing kN rall .m ec 74 ac men th D p k CDo e re 0 t a ngfor nd K eVum 2 1.0 sp the half ne V* ean.d ec s Ch g fit .7 0.5 de 21 >φ mm 2 ed ch ax * rc 74 um uc ge 2 le Vla ce Vfo TIghno36s0no Rbale r a rtie m im to of ms u.m s kN Øu ing min ex m in if e V s ax u. t ea a ≤ n 2 S te r, rm k ac p m o 47 B * c= φV a sp ADoe rneuita os = .4 kN n area at mm Vu nhecre Wide tiodnoes n 75 .0 C t us pro 46 s r, F gs iffoV imum ts ly sp s: tio a io u l l rio 1 t n w m in a s D ta p a a a va W 21 se 47 n t) o ri c Max men p on = in allo in No arne k if tio for en 2 fit m .m cti Sh yideub f te foCrm hec of co ta mate left em mm dn V us et, = Vu us se ita ll reti rc w o a .f n to s in a W o .2 V s d d a e f sy info e = ro tillocw l 8.1.12 d to or Lim ers e an sh b v s / .f r re 243 ire over, .2 ac haa Th y s ols d ov ea e s t u .2 C m / n b 35 a a r ab 8 la f sy sh q m ts c e fo s m do re 0. b v s / .min Cl lar tio rce ukto v tabl 1/3 ee men de : ec Vus Sy s re etry, M* for 1 sp dicu = bb op d r fo A sv ha ee Fit f′ c ll g o a .6 en d S f C 0 √ e e 0 is d o t in rp 2 do C sis h r c + pe th r to As o 4 ing s an 0.06 n s f c′ b v late forc w geom ents ete 31 V uc d (for ne leg 2 sig Ba 8 b v d o) ≥ f cv ing f sy.f rein llo lcu ars Spac ing : De = 0.2 r 2 16 62 mm A sv do ar sig v ye tion mom r diam de rc ca re ard b v s / he bv 1 s fo ′ b ax de c 2 deg fs 20 nd ing b info rea √ fc V u.m β 2 β 3 14 reg ≥0.35 2 mm latu no Se ding & ba t gra lerts = 45 .10 d a 12 40 β1 rou rc an nc +0 n n re cing ibutio V uc ack / f sy.f 3 n t θv a 1 = kg info 11 op uc 13 s V uc ters Be ber eme sed = me 6 Ad = V + Vu rsio pa Contr= V*/Ø - edb f′ c b v s ac me 10 22 m in c No fe 06 √ dia n b g of re nd to nd s θ v) V u.m = Vu 1 t V us er Nu forc ll is u t e 7 = s n 0. g co a 2 re in t θv me ta Vu in 2 le t: 14 co o / s) oth in = Fit 8 of Re ere fi h g pac hear men d / s) pu .f or A sv.m s 45 (f sy area wit h do s e limit of web-crushing failure. Concrete Structures 4.4 and Reinforced Concrete Basics 4.5 provide more information on the concept of the truss analogy. Spreadsheet 4.3 is available at www.ccaa.com.au { Spreadsheet 4.3 is based on Flowchart 4.2 for reinforced concrete beams in accordance with AS 3600 Clause 8.2. n sig ion t uta mp Co 4.4.2 Basis of Spreadsheet for design for shear } The designer has to determine the ultimate shear forces, V *, in a beam from the analysis and determine which sections need to be designed for shear. Analysis for shear requires the input of the material properties of the concrete and fitments, the cover and the section properties of the beam along with the area of tension reinforcement used. For T- and L-beams the flange is ignored for shear design, so they become rectangular beams. De c on dC ce or inf Re am Be e ret 4.5 Beams in Torsion { 4.5.1 General Although torsion occurs in most concrete beams, it becomes important only in a few cases: eg a spandrel beam supporting significant offset loads (especially if it is not cast with an adjoining slab), a curved beam (again if it is not cast with an adjoining slab) or a beam where loads are offset and cause significant torsion in the beam. In W t: tpu Ou ck ba ed Fe s d M u, s on forc an ell sti d c size ity φ rein ge xe c ug Bo ber, apa imum ,s ns m nt c tio min Nu ec A me for rr o A o s M k , c CC ec e nts Ch me ct th om onta rc Fo ase c ple : f /( fsy. = V us sv (A A sv re s= al) Vu efo er min 2 Th s ult s Re ut Inp wed allo ts en ith fitm of tw en ing 91 6 28 57 m (no ia bar m 2 m tD e n one s m Fitm a of 2 leg Are a of Are um fi on ac sp rr ta en im Max m ge an tm ed ter g me dia pacin nt S me m Fit imu x Ma s ba 10 1 13 gs 2 le mm mm . The maximum shear (web crushing),Vu.max = 0.2 f 'c bv do , is then calculated along with the shear strength of the beam without shear reinforcement and the shear strength of the beam with minimum shear reinforcement. If the ultimate shear forces exceed the maximum shear then the concrete strength and/or the section size of the beam need to be increased. AS 3600 Clause 8.3.2 states Where torsional strength is not required for the equilibrium of the structure and the torsion in a member is induced solely by the angular rotation of adjoining members, it shall be permissible to disregard the torsional stiffness in the analysis and torsion in the member, if the torsion reinforcement requirements of Clauses 8.3.7 and the detailing requirements of Clause 8.3.8 are satisfied. The Standard then requires a series of design steps with various outcomes depending on which path has to be taken. The spreadsheet takes the designer through the various design steps. The first is to determine the shear strength of a beam, Vuc , assuming no shear reinforcement, ie: In discussing combined bending, shear and torsion, Warner et al4.4 note that the approach of AS 3600 is to determine if the torsion is large enough to require special reinforcement. If it is, then the reinforcement for a beam is designed separately for flexure, shear and torsion and the results combined. Vuc = β1 β2 β3 bv do fcv Ast 1/3 bv do Then it checks the shear strength of a beam with minimum shear reinforcement where: Vu.min = Vuc + 0.10 √f 'c (bv do ) ≥ Vuc + 0.6 bv do The need for shear reinforcement is then determined depending on whether V * ≤ 0.5fVuc or 0.5fVuc ≤ V * ≤ fVu.min. For shallow beams if V *< fVuc shear reinforcement may not be required. For beams greater than 750 mm in depth (even if V *≤ 0.5fVuc), minimum reinforcement, Asv.min, shall be provided in accordance with AS 3600 Clause 8.2.8. For the case where V * > fVu.min, the spreadsheet also gives fitment sizes and spacing to provide the minimum shear reinforcement of Asv /smin = 0.35 bv /fsy.f . The component of shear Vus= V */ f - Vuc is then determined along with the spacing of fitments and is assumed to be 45°, ie cot qv = 1, which is conservative. 4.6 Reinforced Concrete Design Handbook For beams subject to torsion combined with bending and shear, AS 3600 Clause 8.3.3 requires the strength of a section to be determined for torsion and shear acting separately and compared to their respective factored web-crushing limits. The combined action must not exceed the following simple interaction equation: T* fTu.max + V* fVu.max ≤1 Torsional reinforcement is not required if: T* V* T * < 0.25 f Tuc or + ≤ 0.5 fVuc fTuc or where the overall depth does not exceed the greater of 250 mm and half the width of the web and T* fTuc + V* fVuc ≤1 Torsional reinforcement consists of both closed fitments and longitudinal bars and is designed using a simple truss analogy equation. } pe rson he :T mer t io n clai a ut Dis mp Co nt ob Clie ct/J je Pro ject Sub us ing d rce this Co sp read re nc te sh ee Be t is ign es resp am De (D sig n- for rs To sh ea ion l rC Cl 8.2. 0 50 0 70 40 0 62 6 er met Dia D mm 3 8.3 the o mm mm mm mm N et e Sh do b 24 o bN Jo 2 % 33 18 % OK By te: Da m OK m mm 74 to th ce 0 wid 70 m ax iv e 08 stan ) OK .m kN ut 18 - di m fect Tu kN m Inp ze t 5m = ef <φ = ts d) 0 t si ares ta ax y kN en T* 38 0 en ore ded v .m ry fitm , b ch Vu fitm to ne 36 rovi met Width , D r to td <φ y an A st.p 50 th cove h d o men unde ull V* Geo pt M* t (f mm 3 Dep tom ing de r to fit o, ro en sh kN V* ru mm em Bot tiv e ve ile re mm C * 0 7 rc T co ec 70 is kN 50 n mm +0 8.8 er or Web nfo Eff D - tens 5E 28 ctio n ct = 4d f chev ter e kN Rei se ctio 2.1 5.77 ally of th 20 .0 whi grea sile t at se s se kN su = 50 = en oss cros the (U ntroid Ten x 15 .0 0 0 85 mom a cr at a ce = inal 5d b Jt= 10 20 t th at ud < 1. en ng rce = git ns ax V* max 3 on tio e stre r fo l mom 58 u.m 3. T L u. 0. = ax + φV d ac mat shea na L 8. e. u.m T* ØT ax us plie Ulti ign torsio ax s s Jt C n T u.m u.m Ap es V φ = n o ax D ig ti ulu m u.m 3 = kN Des ØV ula mod m ing 8.3. ts alc nal kN ac .6 V* u.max CL en sp 86 n C rsio it + at φV .6 2 fitm lim sig te to 60 T* ax ed mm ing De cula = los 4 φT u.m sh mm fc Cal cru 2,32 = so tf’c 8 22 2 2J w eb tail 1,92 do mm = 0. De Tuc late 2.8 ax = ′ bv 8. cu c f 3 u.m 5 T e 2 11 ØTuc Cal us = 8.3. = 0. Cla Cl in ax 20 n m 4 V u.m ea ified = lygo 31 (av) ec ab sion th sp es s a po t ress of of for A leng th leg 16 1 mpr area rest ea co ng the s r2 um op 20 ial er kN = ar fined stre eter ax ov s fo inim dn At de al m to 2 am ea 1 sw an t kN n sion di 5.6 ion the t A ts 3 es 12 d ar bjec lygo 1 en ous en 11 su .9 an tors mpr than 1. po tm √f’c rs 87 fitm vari co ter β1 ≥ 0.8 /s ers be the late al d fi 3J t v ing ts of s= ≥ of em cu on tƟ 0. se grea met 10 Tu s = β1 rm or co eter form men Cal clo T uc = diag dia 71 fo r t t to t m Tu l te u 2A ri ba e fit Ø ts Ag) ua ho of men crea th en f sy.f = pe eq 0 */14 ea r of to Wit Fit 8 d tm sw ut ) is (N = 1. as l ar d fi = A uire (A sv 1+ na of ba so 45 t θv .7 or ose T us ed Req ctio eas ided 8.2 d d co ov 1 ntat ion; h cl 6 s se ar Cl Reo uire 1. 8 t pr g an orie Wit os inous tens al nt en de 28 Req ri ial t cr .m are me sion d 1.07 ax rcem ort r va = 45 rce Reo Inpu fost Tor t to uire pp θv info al info pt low 1.1 subjec on e su r re Req d ea 1 2 r re d th Ado be orsi e be in rs sh uire T Dia mm 1. ea an bl ay be t .m of ≥ 1 ads m en bar = Req em t sh 0) uota ea 0.8 uc a lo See β1 for m ou be φΤ l ar /100 0) ≥ Reo Fitm a of d MP ied 1 ith 1/3 na 5 ay /s al 25 do 0 pl 00 io re w 0. v m ≥ uire /1 A ct ≤ <0. 6− tƟ m kN sion g) β2 = the ap T* Reo ea co Req s-se 1(1. 6 − do 5A V* uc Tor 2.92 5 t A st d o os al 1. 1. ed = kN eo ab */3. + cr V 2A 9. = id f R on β1 1( φ (N o 17 6 si b v the β1 = 1. al f cv f sy.f T* prov β3 = for 1 − c 5. on re Tor gth do 2, sw 12 β1 lue or φTu 0 ired fcv = bv orsi an whe ren =A g; va T m 25 th 3 st in rs β rt nd < r requ V uc = T us kN be ter β2 ea is Inse be ea If D em 1 V uc sh 2 re t gr reo =β rm Ø .2 E pu al .0 < b/ Fo 15 late t no V uc D ≤1 t to WIS bu cu sion ER bjec V* uc Cal o/av Tor β1 TH su 3.7 + Re info Da n sio ten sion re. of xu ers f ten ion s fle in 2. lay ers o f ten ion m a xourre lay rs o. tens n be t fl(1 2 io e n f n e or.r lay ver no. tens rete mein ea nr.2y go yers r f t (1 nc m its co ngeea minensh (1ioom a la ove rs o. for lar arrtea b gm e entot rsts or a2y glayevern le u . g ecl re aena min ent (1 m r 2 go ib ns n arrb gm e emen ts o ay tan scte ern . o c v po r n te l a n i o s re nL ere re bae u m e t (1 m ay g vern is ed givoer oste nc el aterr reqngeiremen nts o et forc read T ivlaernc0. ste creenl tarrraeqnugermeme nts m ay g he ds i rc aggu 360erncoc.enm t rein onfofo rea r n iv fo0r0 stemeenl arrraequ reme nts m ly ctiin sp i ctatorAnaSggeuila n6 n e e t ing ere n fo is qu me r S3ive r0c. te en ign a s egsly ecetidore g th cutato mrAaeagemi3n6fo0enrcs.em nt re quire nots n fo es e sin for oaf sthin efosrc re e u r D b iv 0 S e ti m m ts ityr fo mAa g in6fo0 em nt r on ecediennoi ts n te en acfo uto f thin uiresrc 3 c. am ers mcnticinre tsn formAreS in6fo0r0 eme ylyo re qthineeforireem em pts nim ep Be uir cean acinitg g orefrew hque rseceom Th 99 n metio imuto mAreS3 nforc te req inegm ap s cinitgyly er: s re th 19 i ulaerire ec ienntsmuto tio im nt enqduir gfocr asppaascin cre e, m inni ts m re lainbg onfgwhqthue sreem cla uta me t tbre on tsandcraatopasaccitcaytabgs ore e ir din urn u Dis gfolde sadp rescl iny f re me inim rce mmeenn bmenennt in whqu rireem lbo 07 mp dC e fo . it b s ts t a o e d e c n o e o e s c e a e b M 0 ir embenenlict a grcle st ppa sclaing whqu e m C t rein tefomrc mqeunrc rc n, ,2 racitie mpepnndininfoabndca topa bgs re o a e n t e b e o f en il a re in o o n t. h a li a ir m foqeu lyemberepnlict ing le s ly sclainen w ap . ng C in rs ns ltimre enmt inm t/J o ea re arercngt lyampeendicaabnsd onstopaemlabinsg c) jec Re s te ensuile imeamtem fo mtiocrein ulm Pro lye b pnl t m bnldeforco snd 10 & A es, L s, P ct sainrrc te t).ufoltrcimuamteeanrtre late ss th bje a in le tbe, 20 tur foenltyeampebpenlaict are riocain ulm sic n e ltrc lcu te imeis upmtefom Su rellint)m.uin lateemsseth arrcor Lampeear icaebl innt 1M truc te Ba Ca lcuula t imoanspm artre miocin ulfo n einhfo te aeth a lc la eeinim lyesh ppsl temeng S Tarrc n: C re is uateartre late rsmss re sllnt)emfl.ueltcetitim Ca inlcfouurc } eete oicrin tio ulafo ndrlysaionenudlaticrete Conc idrc adaeth inltimis upm aalcnfo s e te e : re ip s la e t n h { C d r . m r r la rs C oinlcuuspeetemss dasllnt)e u etim tiopa rear ulares Amalc on le . sc a eth hm C a lcenfo e in cis usmhertic p ngin c , Cn. rced Tit re 00 suidrc la C lapte De rsm CTahoin m ti s dasllnt).r seetim s io ereete 36 lceussidrc a e la e in C alc ral foris pf acooraess lkens info re AS C nlcfoula CTahoin ne pereersmngdusllhm ign t o rpstr Fainu te , Re C d in Ge re ssidrccptarersadaesshffeeec(Itsncporell &ers k Cahoenfo se ic os e r t re d u in b tr TC s a e 9 O t o 2d a rs r id fo, Hem pa φV re n ra= ti be as lly T* FTohoeness nspsreth-e20ll0owanm il me for mem taken φTu pth = era CDo e re 0 t a ngfor nd K 0 e. be s 0 rce 1, duce en 70 ≤ 1. may φT ula TIghno36s0no Rbale r a rtie sg or info β2 = T* 0 Pa n0.25to V* rm pe 25 ADSoe rneur,ita oste φTu na = n 1, pa (a) + ≤ 4M g fo n re t pro s r, F s: φV =o tio ti a io u l 3.4 ly s l 7 t T* n ′ in a β3 s ta 8. p a 3. a f W p o = ri c Cl in No arne us φTu tio rm ten = ing com ta mate dn ita lly reti th et, th f of info W = forc da an ca eo Dep he Wid or Lim rein d to ers s f the ati ire over, Th y s ols d al e 0.6 u m n / n m b a al la io 2 m H to s m 0 m re V* tio req try, c * 1 25 sp tors ee mm + au de : Sy op of φV lls for dc d M r for = e T* ter is 5 ts φTu Co sis ce ted ea th 54 0.5 r to en an φTu gr ≤ w geom ents ete 2 em ≤g0.25 ula ne the ing Ba V* llo mm : T* ed rs equir if in T* + alc sig Vuc = ce ard ye tion mom r diam de re 5 de g d c g ba RinChefockrc T*φTu φ not exre c φT 54 eb un if latu in es k Se ding & ba t gra lerts (b) φM do th .6) ce w *< nc n er gro forc ion re Ccheinckg n da 3.4 a e = 8.3 e pth bof k e 8. t M 36 l l c a in B m dth l deeid en rs a p 20 gC .0 gC em se w mb em eral fe 10 53 ≤1 No n b g of re nd to nd s the rcin ) rcin uir * r Nu forc ll is u e ovelf q 0. 0 e .6 fo fo V .3 Th d ha a + re cin in : 1. e re 32 4 ta th rein l8 rein ). φV 20 al 80 ar al T* stat ne ut Re ere fi en a hg gC r oan 10 ion din ile zo 0.680 mit 28 φTu if rcin ). wit d sp u, she rcem Inp so itu tors r li 1. ck 6 Wh ng e tens info zone n φM lls 4 on for .0 61 s fo lo Che fo ti a e re 80 ≤1 th l ts ze 88 l a mm in si 0. 0 e en sion es ina dc & 24 ion ) (in em T* 1. 0 ud mpres gg dit cot θ xe r, siz acity um re 2 ers 6 it T uir if u 23 d b φ g o t 45 a u 61 co n e p B Req ck ,s en num for / s)( l lo the 1.210 im Che ns mb nt ca ts na ) (in gem bar 20 in 2. n n o u = A io e d ti 4 ra te m 4 N m / f )( dit t θ 90 31 t ar es ec A ire ad u co me for en ugg f qu 2.710 A orr for / s)( S em 16 Mo cks Re = (0.5 3. 1 , c CC nts A 2 forc e 20 A 94 me / f )( e nts w ed rein 2.710 ire Ch al t: 12 allo f qu me ct th 3. ion 3 ts Re = (0.5 3 dit tpu 11 en ion 60 om onta A d ad fitm 4.820 ors Ou rc se of 5. rT po Fo ase c m2 rs 5 ing fo m Pro 56 t ac r of ba d size ba sp en ts ple um Bar a of No. quire otal em en 4.5.2 Basis of Spreadsheet for design for torsion Spreadsheet 4.4 is based on Flowchart 4.3 for reinforced concrete beams in accordance with AS 3600 Clause 8.3. The designer has to determine the ultimate torsional moment, T *, along with the associated shear forces, V *, in a beam from the structural analysis and determine which sections need to be designed for torsion. Analysis of the cross-sections for torsion requires the input of the material properties of the concrete and fitments, the cover and the section properties of the beam. For T- and L-beams the flange is ignored for torsion design, so it normally becomes a rectangular beam. In addition, the longitudinal tensile reinforcement is required along with fitment diameter and a notional spacing. The spreadsheet can be used to calculate the torsional modulus, Jt. It then calculates Tu.max = 0.2 Jt f 'c and the maximum shear (web crushing) Vu.max = 0.2 f 'c bv do T* V* Then it checks + ≤1 fTu.max fVu.max If it exceeds unity, the member size and or the concrete strength will need to be increased. The torsional strength of a beam in accordance with Clause 8.3.5 is calculated with: Tuc = 0.3 Jt √f 'c without fitments and Tus = Asw fsy.f 2At cot qt /s for a given fitment size and spacing. The spreadsheet then determines the shear strength of a beam, Vuc, excluding shear reinforcement, ie: Vuc = β1 β2 β3 bv do fcv Ast 1/3 bv do It then checks the requirements for torsional reinforcement in accordance with Clause 8.3.4 (a). T* V* This includes if T * < 0.25 f Tuc or if + ≤ 0.5 fVuc fTuc or where the overall depth does not exceed the greater of 250 mm and half the width of the web and T* V* + ≤1 fTuc fVuc If this equation is not satisfied torsional reinforcement is required. The spreadsheet then calculates the requirements for both additional tensile and compression reinforcement and fitments as required. The total shear reinforcement and tensile and compression reinforcement can then be calculated by adding all the reinforcement required for flexure, shear and torsion. Spreadsheet 4.4 is available at www.ccaa.com.au c s 4.6 uc 1/3 c Deflection of Beams c m kN cv uc T* V* c 4.6.1 General 40 60 12 43 0. 0 1. 60 12 us c u uc n ctio se n ok oss ctio ing ac a cr ss se sp e at cro forc at a ion rce tors r fo 1/3 ign shea Des ign 2Jtf’c b v d o Ast d o ′ Des = 0. 2 f c bv 0. ’c ax f cv T u.m ax = √f do t bv V u.m = 0.3J β3 β2 T uc β1 = V uc nt me Reinforced concrete structures deflect or move in response to loading, changes in the environment and as the concrete dries out. It is important to ensure that the concrete structure, including the beams, performs satisfactorily during its life as discussed in Chapter 1. c 2 us Fit v t sw sy .f 2 sy v t sw lt sy .f sy lt c ba ed Fe im ng fitm Max rra sed ta en h clo mm fitm s w it on 2 leg mm 2 12 ed 0 mm 2 as ed on 23 b r s s mm 5 ete ult Ba 54 Are etical rs re m2 T or l ba m The ctua bars A of a Are k: . Re s m ing dia ac nt Sp me um eo Reo Fit n xim n R Ma sio ssio ten pre al om ion al c it d n Ad ditio Ad 5 54 Assessment of deflections of beams involves the prediction of the time-dependent behaviour of concrete; this is complicated by the fact that concrete is a non-linear material. Deflection is mainly due to cracking, tension stiffening, creep and shrinkage. This is discussed in more detail elsewhere4.4, 4.7, 4.8. AS 3600 Clause 8.5 gives a three-tier approach to the design for deflection as follows: n Refined calculations n Simplified calculations n Deemed-to-comply span-to-depth ratios for reinforced beams. Flowchart 4.4 and Spreadsheet 4.5 set out this approach. Refined calculations The calculation of the deflection of a beam by refined calculation needs to make allowance for cracking and tension stiffening, shrinkage and creep properties of the concrete, the expected load history, the expected construction procedure and the deflection of formwork or settlement of props during construction (particularly when the beam formwork is supported on suspended floors or beams below). This method is too complicated for most designs; specialist advice would be required if it was to be used. Simplified calculations This is commonly known as the Branson formula and involves the calculation of a short-term and long-term component. It was noted 4.8 that for lightly reinforced beams the Branson formula can overestimate the stiffness of a beam after cracking; the Eurocode 2 may be a better design model. The short-term deflections due to external loads, which occur immediately on their application, are calculated using the value of Ecj determined in accordance with AS 3600 Clause 3.1.2 and the value of the effective second moment of area of the member, Ief . Reinforced Concrete Design Handbook 4.7 { } ax T u.m ax V u.m T uc m .0 50 0 kN m kN 0. 36 8 kN 8. m 28 .0 kN 50 15 .6 k 86 5.6 12 V uc = This value of Ief may be determined from the values of Ief at nominated cross-sections as follows: ys + kcsy l for imposed actions (live loads) and both total deflection and deflection that occurs after the addition or attachment of brittle partitions or finishes. Ief = Icr + (I − Icr) (Mcr /M *s)3 ≤ Icf.max For beams, AS 3600 allows a simplified calculation of Ief using equations 8.5.3.1 (2) and (3), ie: Ief = [(5 − 0.04 f 'c ) p + 0.002] bef d 3 ≤ [0.1/ β 2/3] bcf d 3 for p ≥ 0.001 (f 'c)1/3 / β 2/3 or Ief = [0.055(f 'c )1/3 / β 2/3 − 50 p] bcf d 3 ≤ [0.06 / β 2/3] bcf d 3 for p < 0.001 (f 'c)1/3 / β 2/3. In the absence of more-accurate calculations, the long-term deflection due to shrinkage and creep is calculated by multiplying the short-term deflection caused by the sustained loads by a multiplier, kcs, given by: kcs = (2 - 1.2 Asc / Ast) ≥ 0.8 It should be noted that compression reinforcement will reduce kcs and normally some reinforcement is provided in the compression face of a beam to support the fitments, etc that should be included. Deemed-to-comply span-to-depth ratios for reinforced beams This method (see AS 3600 Clause 8.3.4) involves a simple calculation but is limited to beams that are n of uniform section; n fully propped during construction; n Where kcs = (2 - 1.2 Asc / Ast) ≥ 0.8 Again, it should be noted that compression reinforcement will reduce kcs and normally some reinforcement is provided in the compression face of a beam to support the fitments, etc that should be included. The effective design action (load), Fd.ef , is then calculated as follows: or Fd.ef = (1 + kcs) g + (ys + kcsy l) q for total deflection Fd.ef = kcs g + (ys + kcsy l) q for the deflection that occurs after the addition or attachment of brittle partitions or finishes The formulae given for k1 = Ief /bef d 2 in AS 3600 Clause 8.5.4 for rectangular sections assume that β = bef / bw ≥ 1.0 and p = Ast / bef d at midspan, viz: k1 = (5 − 0.04 f 'c) p + 0.002 ≤ 0.1/ β 2/3 or for p ≥ 0.001(f 'c)1/3 / β 2/3 k1 = 0.055 (f 'c)1/3 / β 2/3 − 50 p ≤ 0.06 / β 2/3 for p < 0.001 (f 'c)1/3 / β 2/3 The factor k2 for various end-restraint conditions is: subject to uniformly distributed loads only and where the imposed action (live load), q, does not exceed the permanent action (dead load), g. Deemed-to-comply span-to-depth ratios are likely to result in conservative beam depths and the long-term deflections will be small. The simplified calculations are likely to give a more appropriate depth. Beam deflections are deemed to comply with the requirements of AS 3600 Clause 2.3.2 if the ratio of effective span to effective depth satisfies the following equation: k1(Δ / Lef ) bef Ec 1/3 Lef / d ≤ k2 Fd.ef For simply supported beams k2 = 5/384 For continuous beams where the ratio of the longer to the shorter of the two adjacent spans does not exceed 1.2 and where no end span is longer than an interior span o tN ee Sh o bN Jo By e: Dat k2 = 2.4/384 in an end span ctor n Fa s tio rtie uc pe a Red Pro MP a ity ial 0.8 ac ater MP 32 Cap & M 0 φ 50 f' c f sy D k2 = 1.5/384 in an interior span .4 8.5 Cl ) No d o et o s DH e rati C Sh ly in R o mp n bN co sio t is ee to cus Jo sh ed dis mm read em to y is sp mm B De fer th 0 30 ing mm te: ns e. Re 0 us 30 tio mm Da 0 rson ec ativ m fl 60 pe m e De serv 50 : Th n 0 n - on mer 55 io sig e c clai 2 t at De to b Dis e m mpu mm ng ea kely fla Co B 2 or li nt 60 te ob th 18 Clie t/J mm wid cre are am ojec t on ults 0 be essive Pr 62 lar d C res bjec d gu mpr m Su ide rce ote an co kN m ov pr the rect th of info n A st. to kN of d id ce Re ase put ide . 15 idth e w an ov le .pr 6 = w fectiv dist m ) (P ta In A sc ure e - t 5m = ef ts siz Da ) ex , b r its e fo sibl on resp s an d ψl ctor fa sψ rm -te ctor ng (ψ l) Lo n fa tio na 0.4 mbi or fact d co 0.4 an erm ) 0.4 m ort-t (ψ s -ter Sh 0.4 ng , lo .0 ) 0.6 0.7 erm 70 0.6 ,Q rt-t S 11 n ns 0.7 tio Sho er A actio d ac 0.7 0.4 se tic (aft sed es po 0.7 e. us b Spreadsheet 4.5 calculates the above deemed-tocomply span-to-depth ratios. Im po im dom uted and rib tial Dist siden es Re ctur stru ces Offi ing rk Pa tail Re age pe -ty ns floor Stor er tio for ac Oth of ed Ro s us of s Ro itie ofs tiv ac er ro Oth 1 0 1 0.7 0.7 1/3 1 (f' 00 c) β 2/3 ) 0. p/( of io n Rat sio ten sion of n ers f te sion ry th t en fl ) en ares g ored met W id , b fitm ch fitm ne in 2. lay ers o f ten ion th D ored q Geo r to an t - d to ch 7m W id th, ve h d lly an co men de ourre lay rs o. tens n lly t (fu Dep tom dept to fit roun e0.a0.4 (1 e x b t (fu 2 io men n e f Bot ctiv ver reo, ce y er o ile men te nt fle mor.mr ns Effe D - cotens ce nfor e. = e a 2 la ov rs n. te Rei cψ re mein 80t00h(1 nfor us ally of th ile Rei n e ioonr.ay glayoeverrs o.f su on ψ eam ns its 4.1 (U ntroid sion l Te r c anbge emine s t rs (1 m r 2 g ye ern ble ce for pres Ta dina .5 ts o y ula arrte L gm .0 itu om ento 70 ng ible lC e14n (1 ma 2 lagov n. ng necl re rrbaena eminm 11 Lo ns dina AS /m o l ate nag/dmire entnt3s(1 or ay over . itu cta scte — po ns kN ng an from tio ad /m Lo res ns r sp d reesgivoer nL ste cere l aterrrbaeeLqnugermemee0.n331.tn60ts m ay g vern d acead lo ad kN tio rio is e n /β lie ac lo D o erent a ui m e o inte et App (f'c) an term Live forc forrevaadlu Tpporivlatsernc0. ste c e l arrreqn= gerem 01n13ts m ay g 0 he /β ng or in 10 0 .001 than ds ct ≥0 rc l ofasuggu 360erncoc.enm d lo c) er e t a q= ui m0.e s m 20 nfohtfo an rm fa re np rea 1 (f' oe rig long lyfactorctotithin 80 0 42 Crn iv fo0r0 stemenl arr 00t rm rt teg e 00 is sp Whe 60 L tofo < 0. rt te Shoin term ble an ctatorAnaSggeuila 6 enrc.e teeent r q= uire me1.n 32 n Cn 1(1) 39 3niv s ng Seee taslye re am np Sho d sp this en 65 0 g ofebecetidore tamAraSagm tsn /dfo fo0e0n sem nt re =uire 0.00 or use K Whe sig r a Lof sthin no c n 6 n ing 40 i e re u ct e 3 c s e s fo . torb reSst iv or0 an rc q orets fo 37 e he o u D sp 1 fa cratidtioinLnim dw esm rK nts cfo mArscA/Aa gei3n6f 0 rc.em nt re = teuto on ityr a ffefctreivtheinefo e ir m n he Fo e 2 an e rc rs u pt o re m a 1. e 39 a Eo e 50 0 qthetofodereemcnticnnim m cs r S ed Km A spani3n6fo00 cem 0.04 ce 80 in ep n ire ceanptsacinitLgylyg orefre Be = ir scensomie otsn fouto 34 pa he u t ex 9 u 0 r idre S Th e S q g r l ti 9 q no m m 89 n m m n e tio s i themA info t ua re 9 tio es ab -0.3 er: s wCth ire caap sacinitgAyctly = ualculaela rireesec ienntsdm t re din at uto tio d sl limita vers ,1 s do im cre re 39 an m /binni ts lainbgesoignsnfgre s an ion ntile en enqu gfodr sapapascin whqA u/Ae rireem sp cla rne 0.04 am 40 0 uta nt A late th on tsan cr topasaccitcaDyta = um be eflect for ca 4) 2 em nnt tbre ndnin ou Dis 80 re of jace D ef) gfo s re l inbalgcuof whqeuratiorepe=m me inim ote 00 32 ns mp (N 5 = 0. 4 whe to o ad dC elb 07β forc oemmee ntebmeeenndnt in . (∆/L ir tsabnldeceadptopasaccitCaybgs re e = ctio n tw fle itatio 1/12= 0.00 mad ct of Co t ce grce s p l in orciwnghqeu 1.r0 e m r of ies euir emeepnlict ain , M ,00220≤ 0.1/ 06 /β de rein mrc 0 n is effe ise al lim ans orte >e cit / β or an n0. pendinfoaabngldca stopasc aReibnfgs re rtic ion r sp 2) sh 0. en 1/25 isio the herw ile atefo omqfoercunirtebam ob ve 32 0 ct the ine=nbest.f /wbwchapa1 (f' ) in. ngβ m) a04rsf′c)po+ − 50 p ≤ provimize t, ot 002 Cli inf ed efle ef) fo 1 and 4 re ns lteimrein teenmt reinm t/J lde nly sclm lyetm bepreepnlicdt in or er to 10 qlarrc en 0 rer’s t D e b in /L es 0.00 he ly te ulat a 2; .m o e n o 30 o m u β re jec 00 em a 0. b (∆ o il a lc w t re g = t no pe a R 0 = tu 30 long to s ) 0. st L(f'c) ot β P− ca u foenlgyeambenpnlict ams lde src e s imemum nmtticin mov 1/50 ufac n bu 0 = t (N 250 0.002 ade of 0.01 the Pro for 1/ = m be bn fo toesnl d0inFa1ct0or&= Are( s0.00, 1 icficsient, (5 5(f′c) / ct teearre oamslainrrc t tent).ufoltrc Man catio 1/25 late ss th p e jec of its 4= a a 0 n is effe ise to a n m ifi b u a m 38 im a 2 n e c p/ tio in fo 50 ricin an u me rlyem bpnli t rein le b , 2o.m ctu sef 5/ ec n e ltrc te io ed . Li 1/ 30 isio the herw 0.05 imeis upmtefom of sp e th 004 Su s, e ra 25 0 00 alc ula lect ider ction 25 rellint)m.uin provimize t, ot 001 0.01 0 am re th ; or late msseth 36 YS a rcor Lampeear icaebl innt 1M u truRatio telectBioan co efficient m mor t imoanspm 70 Def ns artre en = 0 rer’s t be fle e DA kN — AS 0.00 miocin ulfo co n einhfo : C aalc 26 min em are lcu laete de ov that n of ted s, whe span ans. 28 eeinim t no = lyesh ppsl temeng S co 00 tu eDef 0= n as ab tal d sp is uateartre C ion ditio mov 1/10 ufac n bu 500 AT 3.2 late rsmsseth ke ppor ruTlfo arrc .04 = m sllnt)emfl.ueltcetitim C inlcfouurc on 1/40 e to flect e ad the a n ndlati rete onckr eflection eete o 1/ , ta y su beam an en rior d 47 ES e 2. d r from kN ti Man catio Th p TI a nt bl m lc th n y im id a de in be c a s e of te in ifi R in te l io rc r e e us ta pl i r a arss m lcu nc : t e re Ta D sula E 20 K K1 ec is u nartre C d sllht)m.ult tim rip ns sim inuo 625 in in lapereersmsseth Th afte men Caoinnlcfo .04 sp than 002 mem OP 0 le 5 k rs ch 4 ns a ul g A ca Co. nt 0.00 391 32 e dC of From n co for 0 sc PR 00 n e eein ecstiuopm at 12 teeadaeth cu 0.factor or rs = C a lcenufo pe 0.00 oc atta partitio n th tion mor Tit 24 re ctio (a) for co 4 = 0.00 TE ssuidrc be la ce 0.00 i he tic pre n C Ty , n .r s tim re lapte De di rsms dsllht)m or fle 60sert ther K RE 0= CTahoin ctio g erete 0= (b) 2.4/38 4 = mem whe to fle e ad the ein forsspf acroomratiss inlkenssio infor NC 1/25 lceussidrc S3 InUse ei K2 k2 = de a aenulahm All C ortin = 2 aalc e ral e of 1/80 CO (MPa) re (i) 1.5/38 e de ter th t of la e pp 00 ns A e e i ad C te u r ve rs n en a) 0. fo n Th su rtitio e af m CThoinlcu speemngdasll ig etts oorpstr Fain a r ct n f'c (MP rs in (ii) n (li rs chm h 0 = n is effe nsfe MP ctor be ry pa tio cu C Fa Ge Ec re oc atta finisd ac mic sidrcctaea es he c ncprell &ers ick, R 1/50ovisio e the e tra ed Cahoenfo 0s Memason or lues n na g se pr imiz of th the e, 10o us inr residperorst dadseffe9 (Ifor H m po d dy ctio tr TC ab r va on t ortin fle min ction ber ructur re es ra30ti pp e im an lly , em Kilpa en above fo ction F ion F su ish Th load ct) de FTohoeness nspsreth-e20ll0owanm fle em d st e rs fin th ra de m m is pa to be ittle p e fle CDo e re 0 t a ngfor nd orte rw 001 im ted n pp othe 0 rceK2 options tal de l deflect lae. Mem r br d=e en su n ies bjec ria 00 othe TIghno36s0no Rbale r a su dest ctio to sg inrtfofactor See ad for to ementa rmu fle rs ert 1/10 de be r or pec an cr ADSoe rneur,ita oste na tal fo : rop Mem cula traffi n reInse gn r in To da olon spy tio hi at s r, F ut lp 28 l ve ns asiti load fo sing sio atly e de W ota ca inp teria No arne ctiv tio rs tic ten mityp sign ffe rm be yu f/d eti dn ita el caso of ata ma et, l Le inEfo e de all mem W or an ua 5 m toof rs he sfer or Effectiv atic e d ver, Lim Act 14.5 he ulus ls e s an d Bea ir T od y s o d Tr e u m n o la em n lues s/ .55 mb rea tio ethm req try, c * Spa ed uto 1 14 r va de : Sy d sp id lu op deeof e fo lls M r for da ov ided e late ov is pr Co sis cu ce ab nndva ea r to Ast prov ed Cal .73 ns late w geom ents ete g th 13 tio ne gc Ma f/d tE Ba Asc f us op llo lcu ars l Le sig rdin ∆/Le .61 ua e: See ye tion mom r diam de rcInpuin ca e r a 15 b Act 14.5 d fo g d c tu PLY un rcing rein g Se ding & ba t gra lerts OM k re .5 cla ro c 14 4 TC n n n in g n a a r O 00 fo d N io 0. ck rein ac Be be eme sed me db late ES cu ba DO m tors d sp No fee Cal .73 ING 13 Nu forc ll is u n en g of nd = S er RN in : LIE WA .61 gre acin ar a ent a oth MP 15 ut Re ere fi CO ith UT p or m he PLY = OK INP Inp OM ns s w and s M u, s force Wh TO TC tio ell lue = NO r va es ES d c size ity φ rein g , fo DO e ht g x c u rig ING S RN Bo ber, apa imum the ,s to s LIE WA c P d n m n M n e an nt CO ctio tio min Nu ov fle = ctio OK ed ab ec A de fle me for us ns de tal A orr be tio To tal Mo cks = op en f to , c CC )] Le See em e F e t ∆/ nts Incr Ch )/(k Inpu t: )] E me ct th F )(b tpu om onta )/(k n (∆/L E n Ou ctio [k rc )(b fle ctio de /L Fo ase c fle ns (∆ io de tal [k To tal lect en )] Def ple em ef o Spreadsheet 4.5 is available at www.ccaa.com.au s l ef ef 2/3 1/3 2/3 1/3 ef ef st sc st ef/ 2/3 2/3 2/3 1/3 c The serviceability load factors in accordance with AS 1170.0 Table 4.1, combined with the long‑term deflection multiplier as defined in AS 3600 Clause 8.5.4, are: 4.8 ) 1(1 ) 1(2 ) 1(2 ) 1(1 f d.e f d.e 1/3 f 2 ef d.e 1/3 c f 2 ef 1 ef d.e c ef 1 ck ba ed Fe Reinforced Concrete Design Handbook 2/3 1/3 1 + kcs for permanent actions (dead loads) for total deflection, and kcs for permanent action (dead loads) for deflection that occurs after the addition or attachment of brittle partitions or finishes. and 2/3 1/3 1/3 ults Res f cr F d.e 2 1/3 In )/(k Ec f)] F d.e )(b ef 2 ef )/(k (∆/L Ec [k 1 )(b ef ef (∆/L [k 1 : . D b L mm mm 0 60 0 mm 2 30 mm 2 00 80 mm 60 18 0 62 0 / 25 =1 4.7 Longitudinal shear in Composite and Monolithic Beams AS 3600 Clause 8.4 applies to the transfer of longitudinal shear forces, across the interface shear planes through webs and flanges of composite and monolithic beams. Generally, for insitu monolithic beams, this is not a critical design case but designers should always satisfy themselves that their designs do comply. The purpose of composite construction is to form a single flexural element. For concrete beams, this requires the transfer of longitudinal shear across the interface between the abutting concrete surfaces. It is important to check the shear at the interface between concrete elements that are cast at different times. Examples of this could be precast concrete beams with insitu concrete slabs on top, or beams constructed in two or more sections for particular design or construction reasons. The latter case could be an upstand beam cast after the bottom section of the slab is cast; the amount of shear to be designed for will depend on whether the lower section is propped or unpropped. The design procedure assumes a degree of roughness of the hardened surface as set out in AS 3600 Table 8.4.3. With such composite members, the shear stress at the interface can be high and vertical reinforcement additional to any shear reinforcement is sometimes required across the interface to increase the longitudinal shear strength. See Figure 4.4 which shows a precast pile cap and roughened surface for a future section to be cast on top. Designers should refer to AS 3600 for specific design requirements when design for longitudinal shear is required. 4.8 Crack control in Beams It is important to understand that all reinforced concrete beams crack when subjected to design loading. Such cracking is inevitable, as it is needed to allow the tensile reinforcement to act. Uncracked beams will occur only in members which cannot flex and/or are limited in span, are significantly overdesigned or are in areas of low stress, eg the ends of simply supported beams. For beams, the crack width should be limited to about 0.3 mm to reduce the risk of long-term corrosion. Unlike some overseas codes, AS 3600 does not provide methods for calculating crack widths but rather provides an approach in Clause 8.6 where it gives minimum requirements to limit the spacing and stress in the bars. AS 3600 Clause 8.6.1 specifies that cracking in reinforced beams subjected to tension, flexure with tension or flexure shall be deemed to be controlled if the appropriate requirements in Items (a) and (b), and either Item (c) for beams primarily in tension or Item (d) for beams primarily in flexure are satisfied. For regions of beams fully enclosed within a building (except for a brief period during construction), and where it is assessed that crack control is not required, only items (a) and (b) need be satisfied. For crack control in the side face of beams, AS 3600 Clause 8.6.3 requires that where the overall depth of the beam exceeds 750 mm, longitudinal reinforcement, consisting of 12-mm bars at 200-mm centres or 16‑mm bars at 300-mm centres, shall be placed in each side face. 4.9 Detailing of Beams For all reinforced concrete beams, appropriate detailing must be shown on the drawings. AS 3600 Clause 8.1.10 sets out general rules for the detailing of flexural reinforcement for beams. It requires the tensile reinforcement to be extended from the theoretical shape of the bending moment by D + Lsy.t past the cut-off point. Clause 8.1.10.6 sets out a deemed-to-comply arrangement for continuous beams that must be used for beams which have been designed using the simplified method of analysis. Detailing for shear and torsion reinforcement is covered in AS 3600 Clauses 8.2.12 and 8.3.8 respectively. Figure 4.4 Precast pile cap with roughened surface Designers should also refer to Chapter 13 of the Reinforcement Detailing Handbook 4.6 for further guidance. Reinforced Concrete Design Handbook 4.9 Areas that designers need to consider in detailing of reinforcement for beams include: n n n Column bars The top cover to the flexural bars in beams will often be controlled by the cover to the top reinforcement in the slab and its size. This often results in covers to top beam bars being of the order of 65−100 mm rather than the 30−50 mm cover to the side or bottom reinforcement. Negative moment reinforcement Beams that intersect will have different effective depths (because of the bars being in layers) and different covers. Where beams are heavily reinforced, bars in layers may be required to allow placing of concrete. It is usual to use an N32 spacer bar at about 600- to 900-mm centres. This of course will reduce the effective depth for flexure which needs to be taken into account. Continuity bars In addition, the following are recommended: n n n n n Avoid lapping bars in tension in high-stress zones and ensure that all laps and splices are adequately detailed on the drawings. Typically, bottom bars are lapped at the points of support and top bars in the middle third of the beam. Allow adequate space (typically 70−100 mm) between top bars in beams to facilitate the placing of concrete and the use of vibrators. AS 3600 Section 17 requires the reinforcement to be placed so as to allow the concrete to closely surround it. Generally, this means a minimum clear spacing between parallel bars of the greater of 25 mm, d b or 1.5 times the maximum nominal aggregate size. Try to use the same size bars in each section of the top and bottom faces as the use of multiple bar-sizes complicates the fixing on site, ie do not specify 2N36 + 1N24 + 1N20 when 3N36 bars would be appropriate. Where bars are lapped it is permissible to change the size but again keep the same size of bars for that section of the beam, eg 2N20 not 1N20 + 1N24. Use the same size fitment and vary the spacing to suit shear requirements (but use only a limited number of different spacings). The spacing must not exceed the limits specified in AS 3600 Clause 8.2.12.2 but may need to be less for earthquake design. Always look at the beam/beam and beam/column junctions especially when the column and beam widths are the same, as the beam bars and column bars will usually clash, assuming the same cover to both. Figure 4.5 illustrates the situation. 4.10 Reinforced Concrete Design Handbook Basic cage Figure 4.5 Intersection of a beam and column n n n n n n n n n Avoid cogging bars into columns because of the congestion it will cause. Top bars can sometimes run into the slab. If cogged bars are required, consider drop-in bars. Provide a minimum of two bars top and bottom to support the fitments. Also provide continuity in longitudinal reinforcement at the supports with the bottom reinforcement and in the middle of the beam for the top reinforcement. The area of the bars should be of the order of 25% of the reinforcement in the other face (where possible) to allow for reversal and robustness. Always curtail the reinforcement where not required, eg excessive top bars in the middle of beams and excessive bottom bars at the ends of beams. Consider spreading top bars into the slabs at the junction of beams with columns to reduce congestion and facilitate concrete placing. Provide seismic detailing as required in AS 3600 Appendix C. Ensure that any compressive reinforcement is adequately restrained by the fitments. For cantilevers, ensure that the top bars are anchored well back in an area of low stress. If beams are shown in a schedule on the drawings then check the schedule to ensure that all detailing fits within the constructed shape. Always provide elevations, sections and details of complicated or unusual beams on the drawings. See Figure 4.6 for an example. Lenton terminators 18N32 in 2 layers 9 10 A B 10N32 A B N12-300 fitments (in pairs) 1500 lap min. midspan (typ) C 4N28 4N32 ELEVATION – 1B1 ELEVATION – 1B2 N12-200 fitments ELEVATION – 1B3 n n Ensure that construction joints in beams are properly considered and specified. Provide side face reinforcement for beams deeper than 750 mm. 4.10 n n n n n General Guidance The following will assist the designer in sizing and designing beams for a particular project. n n AS 3600 Clause 8.9 sets limits on the slenderness of beams The size of beams needs to be considered in the context of the overall building. The depth of beams generally needs to be minimised to reduce the storey height of the building and to allow building services to pass under them, generally above a ceiling. Where beams are exposed, eg around the perimeter of the building or opening, they should have the same depth irrespective of the span, subject to the architectural requirements. Generally, small horizontal penetrations in the middle third of a beam are possible. However, large horizontal penetrations through a beam will require careful design and detailing; in many cases they may not be possible. The reduced section needs to be analysed and the design checks carried out; any additional shear and crack control reinforcement required around the penetration may introduce congestion in the beam. Large services such as sewer, storm water or water pipes should not be built in since they will cause problems if they leak. n 7 E 4N28 N12-200 fitments Figure 4.6 An example of a beam elevation n 7N32 D 4N28 N12-300 fitments (in pairs) 8 D C N12-200 side bars each face (typ) E 1000 lap (typ) ELEVATION – 1B4 Beams should be of a uniform depth in a span. Haunched beams (deepened at the ends) should be avoided if possible because of the formwork costs and extra detailing. If beams are notched in the middle to accommodate ductwork, shear often becomes a design issue. Beams which are deep or have dapped ends (reduced depth to sit on a corbel or accommodate ductwork) are usually designed using strut-and-tie methods (refer to Chapter 9). Linear elastic analysis is recommended for major projects and the simplified methods for small projects and simple elements, subject to the computer analyses available. The following beam design data is divided into: Bending Flowchart 4.1 pages 4.12–4.13 Shear Flowchart 4.2 pages 4.14–4.15 Torsion Flowchart 4.3 pages 4.16–4.17 Deflection Flowchart 4.4 page 4.18 Spreadsheets 4.1, 4.2, 4.3, 4.4 and 4.5 may be downloaded from the Cement Concrete & Aggregates Australia website www.ccaa.com.au Reinforced Concrete Design Handbook 4.11 Flowchart 4.1 Design of reinforced concrete beams for bending AS 3600 Clause 8.1 Input design bending moment M * from structural analysis Input material properties f 'c and fsy AS 3600 Section 3 Input cover for: durability AS 3600 Section 4 and axis distance for fire resistance AS 3600 Section 5 Calculate a 2 and γ for f 'c AS 3600 Clause 8.1.3 Adopt preliminary cross section to suit: — Architectural requirements — Serviceability (Lef /d) — Economical tensile and shear reinforcement no Is beam cross section known? yes Is section a T- or L-beam? no Design from first principles no yes Input flange thickness t f Calculate effective width bef and ku Is it a rectangular beam? yes Calculate the approximate area of tensile reinforcement eg Ast = M u / fsy 0.85d Calculate φ Muo for kuo = 0.36 Increase cross section dimensions (see AS 3600 Clause 8.1.2 if curvilinear stress-strain used) Is t f < Ast fsy / (a 2 f 'c bef )? yes Can cross section dimensions be increased? no no no A 4.12 Reinforced Concrete Design Handbook Is moment M *≤ φ Muo? Use bef as the compression flange width and t f as the depth of compression block yes B C yes A B Calculate tensile reinforcement Asl for moment φ Mu As2 for moment M *– φ Mu Calculate compression capacity of outstanding flange Cf Calculate total tensile reinforcement Ast = Asl + As2 Calculate compressive capacity in web Cw Calculate compression reinforcement Asc = As2 Calculate dn and check ku < kuo Input dsc yes C Calculate the ultimate moment capacity with compression in the flange and web Calculate tensile reinforcement Ast Is compression reinforcement at yield stress? no Calculate increased compression reinforcement Asc = As2 fsy / εsc Es Is compression reinforcement Asc more than minimum? AS 3600 Clause 8.1.5(b) yes no Increase Asc to minimum required Does Ast satisfy requirements of Clause 8.1.6? no Increase tensile reinforcement to minimum Ast yes stop Reinforced Concrete Design Handbook 4.13 Flowchart 4.2 Design of reinforced concrete beams for shear AS 3600 Clause 8.2 Input design shear force V * and area of longitudinal reinforcement from structural analysis Input material properties f 'c and fsy.f AS 3600 Section 3 Input cover for: durability AS 3600 Section 4 and axis distance fire resistance AS 3600 Section 5 Input section dimensions depth D and web width bv Calculate web crushing limit Vu.max (Clause 8.2.6) Calculate shear strength without shear reinforcement Vuc (Clause 8.2.7.1) Calculate shear strength with minimum reinforcement Vu.min (Clause 8.2.9) Is V *≤ φ Vu.max? yes A 4.14 Reinforced Concrete Design Handbook no Increase section size and/or concrete strength A Is V * > 0.5 φ Vuc? no yes no Is V * ≤ φ Vu.min? yes yes Is D >750 mm? no Is V * ≤ φ Vuc and D < 250 or 0.5 bv whichever is greater? Calculate shear strength required from shear reinforcement Vus = (V */φ ) – Vuc yes No shear reinforcement required no Calculate minimum shear reinforcement Asv.min = 0.06 √f 'c bv s /fsy.f ≥ (0.35 bv s) /fsy.f Calculate shear reinforcement Asv and spacing s (Clause 8.2.10) stop (Clause 8.2.8) no Is V * ≤ φ Vuc? yes Is s ≤ smaller of 0.5 D and 300 mm? yes Is s ≤ smaller of 0.75 D and 500 mm? no Reduce s to smaller of 0.5 D and 300 mm no Reduce s to smaller of 0.75 D and 500 mm yes Ensure transverse spacing < smaller of 600 mm and D stop Reinforced Concrete Design Handbook 4.15 Flowchart 4.3 Design of reinforced concrete beams for torsion AS 3600 Clause 8.3 Input design bending moment M * design shear force V * and design torsional moment T * from structural analysis and longitudinal tensile reinforcement Input material properties f 'c , fsy , fsy.f , fitment diameter and spacing and diameter of longitudinal reinforcement AS 3600 Section 3 Input section dimensions x, y Calculate torsional modulus Jt (Clause 8.3.3) Calculate web crushing limits Tu.max (Clause 8.3.3) and Vu.max (Clause 8.2.6) Is (T */φ Tu.max) + (V */φ Vu.max) ≤ 1? yes Calculate torsional strength Tuc (Clause 8.3.5) Calculate shear strength Vuc (Clause 8.2.7) A 4.16 Reinforced Concrete Design Handbook no Increase f 'c and/or section dimensions A no Is T * ≥ 0.25 φ Tuc? yes Is (T * / φ Tuc ) + (V */ φ Vuc ) > 0.5? no No torsional reinforcement required yes stop no Is D > greater of 250 mm and bv /2? yes no Is (T * / φ Tuc ) + (V */ φ Vuc ) > 1? yes Calculate reinforcement polygon area (Clause 8.3.5) At and perimeter ut Calculate area Asw and spacing s of torsional ties so that (T */ φ Tus ) + (V */ φ Vus ) ≤ 1 Minimum torsional reinforcement to be in form of closed fitments (Clauses 8.3.7 and 8.3.8) Calculate longitudinal torsional reinforcement A lt and A lc in flexural tension and compression zones (Clause 8.3.6) Comply with detailing requirements (Clause 8.3.8) Add fitments to those for shear and add flexural reinforcement to that required for bending Reinforced Concrete Design Handbook stop 4.17 Flowchart 4.4 Deemed to comply span-to-depth ratios for reinforced beams AS 3600 Clause 8.5.4 Does beam comply with limitations in Clause 8.5.4 ? yes Input material properties f 'c and ρ no Calculate modulus of elasticity Ec (Clause 3.1.2) stop Can't use method Input section dimensions b, bef and d, area of tensile reinforcement Ast and area of compression reinforcement Asc Calculate Asc /Ast and multiplier kcs AS 3600 Clauses 8.5.3.2 and 8.5.4 Input dead load g and live load q per unit length Input short-term ψs and long-term ψl load factors from AS/NZS 1170.0 Table 4.1 Calculate effective design load for total deflection Fd.ef = (1.0 + kcs) g + (ψs + kcsψl ) q yes Is span-to-depth ratio to be calculated for total deflection? no Calculate effective design load for incremental deflection Fd.ef = kcs g + (ψs + kcsψl ) q Calculate stiffness factor k l = Ief /bef d 3 using formulae given in AS 3600 Clause 8.5.4 Select appropriate deflection limit ∆/Lef AS 3600 Table 2.3.2 and deflection constant k2 for given support conditions Calculate span-to-depth ratio Lef /d = [ k l(∆/Lef) bef Ec / (k2 Fd.ef )]1/3 Increase dimensions (principally depth) and recalculate 4.18 no Reinforced Concrete Design Handbook Are calculated span-to-depth ratios > actual span-to-depth ratios? yes stop References 4.1 AS/NZS 4671 Steel reinforcing materials Standards Australia, 2001. 4.2 AS 3600 Concrete structures Standards Australia, 2009. 4.3 Building Code of Australia Australian Building Codes Board, 2010. 4.4 Warner RF, Rangan BV, Hall AS and Faulkes KA Concrete Structures Longman, 1998. 4.5 Foster SJ, Kilpatrick AE and Warner RF Reinforced Concrete Basics 2nd Ed, Pearson, 2010. 4.6 Reinforcement Detailing Handbook (Z06) 2nd Ed, Concrete Institute of Australia, 2010. 4.7 Lecture 4, Design for Serviceability National Seminar Series on AS 3600—2009, CIA, EA and CCAA, 2009. 4.8 Design for Deflection and Crack Control National Seminar Series on Serviceability, Concrete Institute of Australia, 2010. Reinforced Concrete Design Handbook 4.19 blank page 4.20 Reinforced Concrete Design Handbook Chapter 5 Suspended slabs is obvious, consideration should be given to using an edge beam or similar stiffening member. The span-to-depth ratios shown in Table 5.1 can be used for initial sizing of suspended slabs. 5.1 General Suspended slabs can be both floors and roofs. They are usually relatively thin members acting in flexure supporting their own mass and other vertical loads and transferring lateral loads into walls and columns. In particular, deflections, shear at supported edges, shear at columns and the design of free edges for slabs always require careful consideration. The type of suspended slab chosen will depend on various architectural, structural and construction considerations, which are discussed in the Guide to Long-Span Concrete Floors 5.1. Reinforced Concrete Basics 5.2 also has a chapter on the design of suspended slabs and background on flat slabs. (For slabs-on-ground, reference should be made to CCAA’s Guide to Industrial Floors and Pavements 5.3.) Types of suspended slabs include: n Flat plate n Flat slab (with drop panels at columns) n Ribbed slab (waffle slab or similar where the ribs are at close centres) n Slab and joist n Beam and slab n Band beam and slab n Precast and composite. However, this chapter does not cover precast and composite slabs, only insitu suspended reinforced concrete slabs. 'Flat plate', 'flat slab', 'beam and slab', and 'band beam and slab' are the simplest types of insitu slabs. They generally require only simple formwork and can be used economically for spans up to about 6−8 m, but can span further if required. Deflection will often govern the thickness of slabs (see Guide to LongSpan Concrete Floors for guidance on span ranges for the various slab types). Generally, for spans over 8 m, insitu post-tensioned slabs or pretensioned precast systems should be considered. Because deflections usually govern the design of slabs, they are typically not heavily reinforced for flexure and the values of ku tend to be low. Where edges of slabs are visible and deflection The bending moments in insitu concrete slabs are determined from the chosen method of structural analysis as discussed in Chapter 1. Concrete slabs can have the maximum bending moment at the middle of the span for a simply-supported slabs or varying moments (positive and negative) across a span based on the extent of continuity, spans and any cantilevers as analysed. It is the responsibility of the designer to establish the critical section(s) for bending and shear and the design ultimate moments and to design ultimate shears at these sections. AS 3600 Clause 9.5 requires vibrations in slabs to be considered and appropriate action taken where necessary. Designers should be aware that concrete slabs are not necessarily waterproof even with high levels of reinforcement. Because of the greater control of cracking, post-tensioned slabs perform better than reinforced slabs in this respect. Careful detailing, particularly of joints, is nevertheless always necessary if the risk of water penetration is to be minimised in exposed slabs unprotected by an applied waterproofing membrane or similar. Section F1 in Volume 1 of the BCA covers the requirements for damp-proofing and waterproofing in Class 2–9 buildings. These minimum requirements may sometimes need to be exceeded. Table 5.1 Span-to-depth ratios for reinforced concrete slabs Simply-supported span Type of slab End span Internal span One-way slab 24 26 28 Flat plate 28 30 32 Flat slab 28 32 36 5.2 Applicability to Ductility Classes of Reinforcing Steel Charts and Tables are provided for Ductility Class L and N reinforcing steel. Designers need to remember that AS 36005.4 imposes limitations on the use of Ductility Class L reinforcing steel. Therefore, it is the designer’s responsibility to ensure that the ductility class of the reinforcement specified is appropriate to the situation and member being designed. Other clauses in the Standard impose further restrictions on its use. Reinforced Concrete Design Handbook 5.1 The capacity reduction factor, f, for a strength check using a linear elastic analysis for Ductility Class L reinforcement (ie reinforcing mesh) is in the range of 0.60–0.64 compared to that for Ductility Class N reinforcement (ie bar) of 0.6–0.8 which reflects its lower ductility, see AS 3600 Table 2.2.2. As a result, using Ductility Class L mesh reinforcement may require up to 25% more area of reinforcement for a chosen design moment, than is required when using Ductility Class N bar reinforcement. 5.3Flexural Reinforcement 5.3.1 Spreadsheet 5.1 for flexure Spreadsheet 5.1 can be used to calculate the reinforcement requirements for reinforced rectangular concrete slabs in flexure. The concrete strength and f value are input. The spreadsheet checks the minimum reinforcement to AS 3600 Clause 9.1.1 and assumes that kuo ≤ 0.36, but it does not check cover, spacing requirements, crack control, shrinkage and temperature reinforcement or provide the detailing requirements to AS 3600. As Reinforced Concrete Basics notes, for an underreinforced section, the ultimate moment capacity, Mu is approximately equal to 0.85 Ast fsy d (within about 10% of a more accurate calculation), ie it is independent of the concrete strength and width of the slab. This approximation is used to estimate the area of tensile reinforcement required. Then the spreadsheet calculates the actual moment capacity and compares it with the design moment, which has been input. o tN ee Sh o bN Jo By e: Dat D le sib for . ) ingNo ildet bSuhe o ea bN sid Jo se u its tor Fac ion ties uct per Red Pro a ity ial MP ac ater 0.8 a Cap & M MP 25 0 φ 50 f' c 0.36 f sy k uo If the calculated moment is less or significantly greater than the design moment then a further iteration with the input of a revised reinforcement area will be required. on m n clai tio Dis puta m Co nt ob Clie ct/J je Pro ct bje Su is g th sin nu so per read resp ee sh he T er: t is sp ign es te cre D lab e los nc g (e din en -B d in S n e. Co ed us lar inforc its tangu der re for d Recat are un rce abs th info for sl ign ut Des Inp ta Da ry ible By te: Da reo d ed uire uir q req as re the of erate It on ati tive. a xim ro serv p 10 ap n on 0 19 sio a first to be c quired n 0 re 15 is n ds n re. of te gunresio reo xu it te na of fi 40 flem yers Thoisf telabsnsio are 60 12 s in kN2 la ers Foor fsgetet the 36 lab .1 or 2lay to d 00 b mm mm mm is re Re Spreadsheet 5.1 is available at www.ccaa.com.au 5.3.2 Basis of Charts 5.1 to 5.8 These charts for the strength of slabs in bending are based on the principles set out in AS 3600 Clauses 8.1.2 and 8.1.3 with a maximum strain in the extreme compression fibre of the concrete of 0.003 and the appropriate capacity reduction factor, f, depending on whether Ductility Class N bar reinforcement or Ductility Class L mesh reinforcement is used. The Standard does not allow the use of Ductility Class L bar as main reinforcement. Charts 5.1 to 5.4 are for ck ba ed Fe A S DR HE 1.2 TE T NO : n . rsio Ve 5.2 The stress in the equivalent stress block is f 'c multiplied by a factor α2 = 1.0 − 0.003 f 'c (within the limits 0.67 ≤ α2 ≤ 0.85). From 20 to 50 MPa, α2 = 0.85 but decreases after this for higher strength concrete to α2 = 0.7 for 100-MPa concrete. The charts are limited to 50 MPa with α2 = 0.85, which will cover most design situations. The design bending moment for a given area of reinforcement in a one-metre width is derived from the equation: M * = f f 'c q (1–q /1.7)d 2/1000 kN.m/m where f = 0.8 or 0.64 as appropriate q = p fsy /f 'c p = Ast /(1000d ) and fsy = 500 MPa At the ductile limit, ie kuo = 0.36: pmax = 0.85 g f 'c /fsy kuo = 0.306 g f 'c /fsy and g = 1.05 − 0.007 f 'c g (within the limits 0.67 ≤ g ≤ 0.85) and g varies from 0.85 for f 'c = 20 MPa to 0.7 for f 'c = 50 MPa and is 0.67 for f 'c ≥ 65 MPa. then f Mud /bd 2 = f 0.306 g f 'c (1 - 0.18g ) As reinforcing mesh comes in standard sheet sizes, the moment capacity of slabs reinforced with mesh is limited by the various sizes of mesh commercially available. Slabs with bar reinforcement can have larger moment capacities because of the higher value of the capacity reduction factor, f, and the fact that a greater area of reinforcement can be provided with larger bar sizes and/or closer spacings. In addition, where high levels of reinforcement are required for crack control for flexure, shrinkage and temperature effects, standard meshes may not be adequate in the secondary direction without additional bar reinforcement. ally su eu n ar ow s of y sh slab er htl for ss 21 mm 20 mm is le s ligriate id 10 ,600 u ars 12 80 area rop 0 ntro 4 φM ing ,0 38 of b ce ) 41 Bar t app 32 ac 0 10 *< M* g sp 40 the m M 4 in t 00 8, d no 0 d to t 5m 80 ac = en ire an 10,2 0 ce sp 7,20 0 requ er an ares 16 st. 2D irem 28 ax et 30 st A 8, u 2D 0 m m ne di 6, 0 6 te m or req to inal rdia 040 6,80 9 61 5,60 0 No 18 ba wn ate bar 8, 2 Nom 0m met th, b 5,82 0 n F12 12 r to d do d 5,04 2 30 it st 24 en D 6,43 0 Geo Wid ctio 5,10 3 lim pth - cove unde 11 giv 4,58 0 2 th, th de 18 5,36 4 s-se for for 160 45 4,53 0 os Dep ctive d = D o, ro wid 4,20 0 6, F11 9 8 m) ing 4,59 0 re a cr 4,08 9 tre 89 m2/ 3,60 0 4,92 7 Effe ally nsile forc 20 4,02 3 t at me (m 0 su u 3,70 0 4 18 en rein 3,15 4,10 0 er (U e te 31 3,57 6 bar 4,52 6 of F10 9 < φM th 3,40 4 tp of mom 3,52 0 ns of en 91 70 M* 61 rea 16 3,21 4 r 08 2, A 3, m tio ing ea 0 3 92 3, 8 ar 0 ge 8 2, 1 d fo 0 d ac bend 2,55 3,01 3 20 ire an nal 3,14 2,73 4 F91 1 2,68 7 plie ign 2 arr 2,58 0 ctio requ ft. 58 2,46 0 Ap Des nt se 2,29 0 .8 12 2,51 3 A st x le 2,26 9 me ss 0 8 2,24 3 3 al =0 bo 2,01 2,09 4 11 Ø rce 2,00 8 Cro 2,01 8 Initi F81 4 2,05 0 e llow u ith 1,79 0 info ye 45 1,80 4 φM ntag 1,76 0 1,60 0 rw the 1,57 6 *< ce d re 1,64 7 Ba 0 1,54 se 1,34 9 t M F718 Per N into 1,39 6 en po 1,50 1 1,13 4 8 t it ss 1,14 5 35 1,25 2 Pro 90 irem 00 1,29 0 Cla inpu u 14 3 1, d q m 3 1, 7 75 lity an 1,13 89 e re F81 6 m2/ cti in 4 1,04 7 4 64 bles .m stat Du size bar m 80 u ta 45 5 m st 89 it 1 e m 56 5 ov Bar a of 73 < φM 2 ar r lim 2 78 0 ab 50 M* fb 0 Are 67 s fo F10 4 2 in o 10 5 3 from ze 45 35 .ming 62 ze 1 u 12 h si uoac 3 41 0 h si Sp 50 7 < φM u Mes F92 15 m 37 mes 5 M* 0 kN 3 17 < φM 29 32 r or ar 0 36 M* 3 ba 20 28 < 0. of NB 5 F82 22 ea ss k u0 7 m 0 e ar ty Cla 22 0.8 kN uo 25 os 5 m φM til Ø .64 27 0.8 kN Cho Duc 0 = 0 F72 9 0.6 Ø 30 se Ø U its 0 17 h 2 u 35 0.6 lim wit 0 φM its mm 40 0.85 sh u lim e φM γ mm 0.85 7 LM ku 37 2 ss 2 0.67 0 la C its 15 m mm lim 0.67 lity m2/ its cti m u lim D h reo d = Mes a of ide ov re pr A e = d A st. ide ntag = prov ce d st. A re Per = 2 0.5% xu d As fle 13 = )b ith block s for k uo 36 *w on A st = 2 =0. .5 = M stress OK lati um k uo (1-0 c)) mm /12) im at lcu 8.8% d f' rete k uo 2) k uo uo Min 10 nc s Ca 13 / (b /1 4 t M 2 f' c k uo in 41 9 en lock st f sy 2 for co d ign b K .1 13 m stm ide s a (A O A prov Mo = Ø MP = (1 .19 De st ress 5/ 2 k u) bd Ø = 7.4% um φM uo <A g st - 0. .5 c = 13 2 d = (1 ed im ire 7 f' sin d (1 Ø us -0 qu 3.00 00 ’ in Max Check mm d tu f sy k u (1 A st.re Ø c - 0. OK A stm provide 36 g 3f men = A st 05 0. in o st 9 f'c = u 1. us 0.00 uo 28 <A k M φM = Ø = 2 ed γ = hk ec 0 Ch Check φM u lculat wit in mm d = 1. 2) 2 ck ca A stm provide = *d) 2.2. Che ck k u 7 A st le 85 34 < 0. ab 2 * Che dd s (T *f sy mm on * an (Ø = lati */ tM cu 4 d en =M 27 bw Cal A st 1) mom )0.5 / f sy m th l 8. 2 ( f' c kN f' ctf reng s (C 0.6 m )2 = mm 2 on /d .1 kN 1.3) ate st d = 21 lati (D l 8. bw mm 2 9 f' ctf cu (C ultim 24 / f sy 0.20 ts Cal mm ctf 6 = 3 f' en d on 1) 2 n 50 M* )2 mi em base 0.00 1.6. 2 st. 5 /d t d uir mm Mu 80 l 8. m 85 y A am = w (D en φ 0. b o m pl req ents (C 2 m be re m 7 ts p / f sy 0.24 = co as a gth uirem Mo ity 37 en mm 2 n of ctf 3 = to en 2 f' s = ing pac o n 33 em m atio ) d ed = φ mi Str Req ir ab ti ul m st. m u 5 /d sl lc g ca t ra Ben er 67 yA (D req ca for A st Dee en al th ign endin mpl r corn 97 0.19 = co ng 11 Initi Des al b = rcem sen sen to thei n stre u o o nfo = mi med ns at Act Rei ar Ch l ch um A st. s ee m y im ee = D lu side mpl N B of st co Min co on 4 to lass rea d by ls A yC orte med wal = pp ctilt 3 bD Dee s or Du s su am 1.75 bD= n be slab of 5 tio = 4 for ec 75% of 3. bD d by Dir % orte 6.0 ef 75 pp of ary c = % s su rim bD 75 P n = 75 slab ol tio of 1. 5 bD ntr for irec 3. bD= Co en ck y D 100% of 0 os mm 0% of 6. dar Cra ch 10 on 0% es Sec 10 u siz ol mm ntr nt Co me ck dn rce Cra fo u k is rein d ax u ile al γk er s tr et n neu am ck in n te to par blo o th f u.m is d n d.e .e Dep al ax ssio sfe tr bda pre lts Neu com su uc of uc Re ef th in Dep .m s (1 rs . t r2 ye rn rete men (1 o 2 lagove rn. t r nc co ngoert en (1 omay gove n. r t lar rrapp em gu eal asu abng menents may gove t asrrla ge em tan ste s y t a c a r n l n enbs teete rra qui me ts m d d re la rce ategivs eonnsc.reteel ant re quire men rea r sp info fo r c60n0 seme t re uire cre a lagiv re is n n n rgu S3ive r0c. e ly tioco th eq ing elacr ntafontorAa gein3f6o0 rce. m ent r ing rS o 0 a s thnegsu ectireocnts fo us n m A inf60 em d for ocf ta ign 9 e se metio muto m AreS3 nforc on 99 es nts ityre th forcire ec ienni ts uto ,1 ers rei me aecd ofin que s m m m bD rne ire aprc ityreg oref thheuriree mienni ts um ep ou Sla qu inregincfoapinagcly q m m Th in i w lb e y e 7 c r n re : re e i . 0 e ir d g ac s paacitabgs t er rete ies , M , 20 n en binegnly in whqu recm r d sap ascl in s re c it n im n m a io fo n t e d s c n e ena o ns a g le stop lainbgnt. whapa . gm cla tat rso tieoendt in forc om b d pasecmebsg c) dC on ea for entrabm Dis mpu la in A , L s, P en lict aan lde storc s ly rein cm ity rce fo toesnd010 & res a oommpfoercnet bampepn icaabnin rtie ile aapte ic Co t info ns ltim pe inm arrc lye pepnl rt reablein tb1, 2 uctu Bas c te Cmre en ob Re pro a elen M Str s te eeuar imdato miocin ulfo rlyeam Cli t/J al ete he plictem g sh t).umlt eimuamteartre late ss th jec a rc tinrete oncr teri d lsy aiopnesndla ulafo n eee inltims upmontlyicin t lcu late n te Pro e c th a c r u ma i u la etemss D llnt)em.u etim bje arerearcularessAcmalcCo.n ed C : C aalc nd aeth he in laibssupm 0. lcu rc te Su g i a p 0 n h c la t n , n C . m la r r rs 6 s s C inlcfouu eetems dasllnt) r etimr pa omrati s in essio fo 3 tio gth a e la e in a lc sidrc fo is f c o es lk n in : re AS C rip en ula Caoinnlcfo ing le pereersmngdusllhm ign t o rpstr Fainu te , Re sc str C d in Tit re ssidrcctaea aes heectsncorell & k forc rs te Cahoenfo De se inr residperorst dadseffe9 (Ifor pHm a be atric TC cre rein ral yu re ,e on ilp FTohoeness nspsreth-e20ll0ow anm ne sh rall er CDo e re 0 t a ngfor nd K r, c Ge ne me ign cto ge or es TIghno36s0no Rbale ter a t fa d s n n e ADSoe rneur,ita Fos by na : tio em ut tio at s r, l uc ns W inp ota ca forc No arne tio red dn ita ata M* rein reti ity W an eo ac ed Lim nts sion ap Th ols uir me ten r, c s/ mb req red ve de : mo rs of Sy lls o o o e h c g c C sis c e in s y, w an lay nd etr Ba llo ule lly : be r 1 Ye mod eom re t fu g d ored ign r fo en 4.1 latu All ction es ete ore em nc ble ch anch g d iam Ta Se forc me an V* din r d lly t fully No 0.0 ce e rein en & ba r fu b 17 t n : r en r fo sil S1 ut Fo ber me rcem A ea l ten p s e h m o m In s rc ina Nu fro rati info reinfo ign ud ly re es ngit tors ion mp ile rd lo fac co tens ress Fo a of on p d to inal com L) binati Are L l e. me ud y ee ngit dina and com ula d da L r lo gitu form 28 (D term Fo a of at g lon ns ing ity us Are a of actio d lon lab tic lly an fs las ca Are lied fe ati p term an o tant o p m A s sp lus t uto ort on du tan Sh ctive on c da mo ns e late cti co Eff the fle lcu M* on of de ca cti ed k nd ents ars fle value us b de rou be k an kg mom ing e c to M g ba ing forc /L E ac en ndin rein sp ut e gre of nd Inp ith ign b ing ta c s sw en ell g de spa em d dc xe ndin e an φM inforc o e B iz y re cit r b r, s m Fo mbe capa imu V* nt min Nu rce me for r fo Mo cks ea t: e sh ØV tpu F Ch n nF u d ig O on ectio s an cti /d de V v et, r Ø L defle l defl Ø n n he Fo os u tio ax, l tio tio ta ds V* u.m ec ØV ec rati th ra tota men rea efl efl ØV cks if *> sp ply dep d for re ld ld V c e m is ta ta if in a n Ch cks co To n to n lo g th for 1/3 me a to in e d p ig re )] a s d s c Ch ard .ef me ctual e de ign lo (F d )]1/3 In reg ee v ED s c)/ ck cti sa .ef rd de CK 0E ba (F d Fo ulate s effe tive 00 ed c)/ HE lc f)1 ec r fe 0E /L e Ca ulate s eff YC the 00 lc LL ro k 4(D /L ef)1 o Ca ulate U 3 F ns lc s [k k 4(D tio Ca BE 3 es late gg lcu s [k TO su Ca ulate s, ILL lc on ST cti Ca A rre T ARE co CCA N , e nts ME TS me ct th OM HEE om onta rc R C DS Fo ase c FO EA ple FT PR et e sh ns o sp Ductility Class N bar reinforcement (capacity reduction factor f = 0.8) while Charts 5.5 to 5.8 are for Ductility Class L mesh reinforcement (capacity reduction factor f = 0.64). Reinforced Concrete Design Handbook Note that for each chart the minimum reinforcement for a particular slab needs to be checked along with the minimum reinforcement for crack control for flexure, shrinkage and temperature effects. See AS 3600 Clauses 9.1.1 and 9.4 for when these requirements are applicable. 5.4 Slabs in Shear 5.4.1 Spreadsheet 5.2 for slabs with shear Spreadsheet 5.2 can be used to calculate the shear capacity at supported edges of slabs where shear failure can occur across the width of the slab, without shear reinforcement. Designers should note AS 3600 requirements for the reinforcement to be properly anchored; this may require a hook or cog at simply supported edges for bar reinforcement or the mesh anchored sufficiently beyond the shear plane. o tN ee Sh dR an ts men Com ote this ction s ce en efer o bN Jo is ) ed int t pr no se (N By te: Da o tN ee Sh 00 36 h r AS gt pe stren r as te ea ncre o bN Jo By ctor sh Ø e: t co n Fa s Dat Inpu tio tie ion uc oper 00 36 ns n Red l Pr th AS MPa ity ia bo 0.7 ac ater 4 of re. of te nsio n or Cap & M 8.2. 25 th Cl xu us ng fer φ fle yers of te nsio Re stre its or f' n) s in r 2 la yers of te size b for ctio o se rete la s sla (1 nc os co rn. ible t r 2 layevrs a cr te se e n s o at ea e n (1 r 2 go ern. force cre em incr IL po r nt o y on rt D If FA r c anpgo me (1 ma gov igenrshnea. res y ula arrup ge entnts is e s ma g(Doesv ng teal suse. rrabn emm et a its asla g re cta n sste ay he nt e b ble foter etel rranqui me ts m ds d re la .0% 4 50 8. rce ategivsresponeonsinsc.reteel ant re quire m 1.58 rea e2n 17 Cl p fo r r e c 0 iv s b cre eet isgr 60n s m nt re - Suhieare =1 rein fo d n as e ly tioconn readshrgaulaS3ive r0ce. geq this take ing elacr is spntafontorAa gein3f6o0 rce. m nefonrct inr OK ing =1 rS 0 ei a s thnegsu useingcthtireocnts fo n n as us m A in3f6o0 eacreRm on sed e on uto take for f ta rse 9mm9 sig ti nnim ts) on ts ityreocThe peth rcirem ts m AreS nino fSohr 9 c or e s n 1 i e pp fo r e aecd mer: ofin que s m uto h e su e10, 000 mmmm bD pe ted prc claiityre eem innim ts ignmwitre n em cfo rn r or a th ax tio re ir la 2.92 e ir u e 19 is pp u f ta e c .m ly u D g o m S es su 0 g puin o whqu Vu q inregin apinaCom m ibnDim Th ply sion <φ elb 007 16 te d g ac s paalieccntitaybgobs re hqeuriree e m V* sim v) t re nly pres Sla s. e M (a er: n at n m r r , 2 s th l eg co e w crretecitie 10 c ila ial im leng 8.2. an on, id io foar ndcaptCopPrasojcecint/J t bgs re d ax sim mm ne em enat sbin eonsdin Cl on the ire r or t to qu with er ea cla tat rs ntro t g e s bjeclain nt. wCohapa . gm ok ov bjec o re ce ho 505 forc om kN su dC for trabti endinabnld opSuasecmnaefobrcsedg c) st.minLidoth nfor sh Pea to the cem ) 10 r re rdan g, sion kN rs Dis mpu ea co es 8.2. 0.8 be d sh acco rein tecm ity mpelyn embenpnlict a blde st rc Cl ce est 5m by 0.0 s kN ei ire sl din 0ut A s,e w rce mpr in s, 9.45 t no 30 qu with mem co 10 lef ired rt75ie5 al for o re ce aninfo toRen20In1inpry &tu= refefectiv asic bar - distn anto near anchoredA Co t sile caapa Cmoominfoercnltyeampe pnlict are the requ on g) r re rdan info pe 52 ea 4A n to is diag te re omlarrc apear cabllein tb1,Datueooa.m sh acco met c b Bd r to d dow t (fully V* ten ualtr im b */1 ow r reo te en ato sh (N Re p=ro in t no Jo is ea e eng M GStrWuidth,th, Dete u fo rlyem crea pth - cove unde lef ired Cli 1+ sh K" tes th to al = icin hee .umltim heppslitem ct/ men ed uamte m the requ or "O then t epcrive de = D o, ro ri s as s rc la r If ce n to D n; re je a s V te l is n io so u te a lycu fond y a nendlati re te 2.7 ow r reo nt)ee im s upm If No t 4 tens onEffect usuatellynsile re ile Reinfor sh tesmeth Pro ated i in alc ula a larlssiom lcu nc is ea ma ØV ent Cl 8. ial i s oanrtre Dell t)m.uinltim C la 1.58 jec ient sh (d r K" C s ax e t b or u d im s . ns "O th en e a A d o to n If th t a n eheein laibs upmhe ticupre lc late of e 4 are ca C . alc Su 6 rcem ctio an 00l Te n: C bjec ort If No late 1.58 rsmseth 8.2. info su s m. s timr sparomratinsgin ess,ion forc Ca inlcfouurc pp th eete 3itu6dina ns force at se rs 0.8 CL tio r re e su be adaellnt)lareein a lc sidrc fo is f c o es lk n in ea ng : 0) re tio r = ASng em d th C rip 00g ula Caoinnlcfo rm 1 le pereersmngdusllhm s an in Lo pplied acign shea lations limit Vd stre without sh ign t o rpstr Fainu te , Re sc rcd /1in 0 fo ad C 6 Tit re ssidrcctaea aes heectsncorell & k ) d lo A Des ed rs Cahoenfo kN De cu hing f b te m) plie in1(1.fo .5Ag = 2 re us inr residperorst dadseffe9 (Ifor pHm a be atric Cal crus 0.c TC ap = 1. .0 (bea */3 kN ral 30 (N re ab the 1 OK t } ign web o=n lly h re e for ,e r ilp 1 FTohoeness nspsreth-e20ll0ow anm MPa a sl 5.0 ne { As b d es ided kN or 52 of lu f era ne DCesalculatetor,V c CDo e re 0 t a ngfor nd K th prov Va Ge d ing; n m V* .1 92 ig 2, ng = nd b e r 2. K 76 0 re s kN o h c O be than kN TIgn 36s no Rbale ter a r st re to de sg ter kN ØV pu 8.7 .6 = ea = n fa late shea V = en 10 t to 76 ADSoe rneur,ita Fos by na t gr f : .1 tio ØV bjec 76 t no cu em ut tio su at s r, bu l = uc if ns Cal rs kN ck be V W inp o/av ota = ca forc No arne tio red ØV Che φ V 2d mem kN ta n ØV < dn 8.8 for ita M* rein reti n as ity V* 2.7 ctio 10 5 1, da kN W if an l 8. take 9. eo ac s se ck tC 10 5 2.9 be Lim = kN ire nts sion en 9. cros l 8. ap Che φ V Th ols = may 10 tC qu the rcem > or .6 me f ten en b d r, c s/ d mb V* 76 = re info 1, o e at em e d y e ) 6 re rc rc re r v b 0. d o } = d : S Pa kN lls ea r fo 6b st info V + f b 4M ho = co d g m ers ea t sh d = ce s {A + 0. Co sis kN r re b sh ou b .0 nc y, f ea y V w 0.10 38 ign din with 2f sh d f V etr 30 ab b Ba ya llo Des ule = 0. um kN en 1 la d : bd a sl = 6b inim ØV of Ye mod eom f full = f kN V re n b r for .6 = 10 g nt = 76 drength oredab with m V + 0. 4.1 V + 0. 30 sig latu All ction sl me d oearer st ØV =V de mete nc = ble V* chSh nch of a f b rce V = Se me an lly astrength + 0.10 .0 Ta (ii) ing r dia V* info o y d 2.5 e r ll 8. N l ØV V* en & ba rc e re fu nt fute sheaV = V 170 gC 1 nt : rcin rb e la r fo nsil info ut Fo ber me rcem Calcu AS ea te r re ea m os rce m Inp sh inal sh Nu for fro rati info reinfo ts t ign ud rs men en ly re es ngit ire cto ion rcem mp ile requ ØV rd lo ck info co tens ress V* n fa Fo a of Che ck if o r re p l ti to ea Che V d sh ina l com LL) bina Are >Ø no e. me ud if V* ck y on ee ngit dina and com ula Che da ased L rm rd lo gitu 28 s b (D term Fo a of g fo at esult g lon ns sin ity R Are a of actio d lon lab yu tic all an of s las c Are lied e t ti f n a p rm n so Ap rt te spa sta t tom o on ulu n au Sh ctive on c od sta ted e on cti em ula * Eff fle n c of th alc M de tio d c nts rs ed k 3 eflec alue us un a e v b d e ro b om g k4 an kg ac g m forcin Me /L ef to ing in nb Ec in ac sp ree end f re ut nd hg nb go Inp ta in wit ig en lls des spac ce em g d d rc u n M a fo xe din Bo ben size ity φ rein c r m r, Fo mbe capa imu V* nt ce min Nu me for for in ar Mo cks f t: f V u.m he u e d.e s h Ø tp F d.e C n nF nd Ou on ectio sig a c e u cti /d uc et, rd ØV .min L ef defle l defl Øv he on Fo on os u tio ax, l ta ds cti cti V* u.m ØV rati th ra r tota men fle fle rea > e e ØV cks if ly p d d sp V* e mp to de ad fo incre tal tal this r Ch cks if co en To 1/3 n lo d fo an ing e rem a d to l sp esig rd c f)] e Ch a lo .e a d m n D (F d )]1/3 In reg ctu tive ee sig c)/ ck KE c sa .ef rd de 0E ba (F d Fo ulate s effe tive EC 00 ed c)/ lc f)1 ec CH r fe 0E /L e Ca ulate s eff the 00 LY lc ro k 4(D /L ef)1 Ca ulate UL so F [k 3 n lc (D s o k4 Ca sti BE 3 late ge O lcu s [k ug Ca ulate LT ,s ns lc TIL tio Ca ec A ES A orr NT S AR , c CC E ts e n M T me ct th OM HEE om onta rc R C DS Fo ase c FO EA ple FT PR e. c Spreadsheet 5.2 is available at www.ccaa.com.au o 2 3 2 1 d ide rov v st.p o ax u.m ax u.m o ax u.m 1/3 1 2 1 o o v c ax u.m o v 3 v cv 1 2 1 o 3 uc ax u.m uc cv 1 in uc u.m uc uc in u.m 2 v 3 uc c 1/3 o 1/3 o v c o u.m in u.m ax u.m uc 5.4.2 Basis of Charts 5.9 to 5.12 v c o) uc 2 1 o v c in v cv 3 v o v cv c v v o o uc o) uc uc uc in u.m in u.m in u.m These charts give punching shear strength for slabs at circular columns with no moment transfer and with no shear head, based on the equation in AS 3600 Clause 9.2.3(a): uc in u.m V * = f u dom fcv where f = 0.7 dom = d for uniform slabs fcv = 0.17 (1+2/b h) √f 'c ≤ 0.34 √f 'c A S DR HE 1.2 TE T NO b h = 1.0, for circular columns ck ba ed Fe Figure 5.1 Long-term deflection of unsupported edge of insitu concrete slab : n . 5.5 5.5.1 General Deflection is probably the most important design criterion for slabs. Control of deflection is discussed in general terms in Section 1.4.3. Designers must review all slabs and satisfy themselves that the allowable deflections or span-to-depth ratios to be used are appropriate for the location. Figure 5.1 illustrates long‑term deflection of unsupported edge of slab. As noted earlier for beams, AS 3600 Clause 9.3 has a three-tier approach to deflection of slabs as follows: n Refined calculation n Simplified calculation rsio \ fcv = 0.34 √f 'c ; and the critical shear perimeter Ve u = p (column diameter + d) n and M *v = 0 5.4.3 Basis of Chart 5.13 This chart gives punching shear strength for slabs at rectangular columns with no moment transfer or shear head, based on the equation in AS 3600 Clause 9.2.3(a): V * = f u dom fcv where f = 0.7 fcv = 0.17(1+2 / b h) √f 'c ≤ 0.34 √f 'c b h = (longest dimension of the effective loaded area, Y ) / (shortest dimension of the effective loaded area, X ) and the critical shear perimeter u = 2 (Y + X ) + 4dom and M *v = 0 Deflection of Slabs Deemed-to-comply span-to-depth ratios for reinforced slabs. Refined calculation This method is too complicated for most design. AS 3600 Clause 9.3.2 requires at least six items to be considered; specialist advice is usually required if this method is to be used. Simplified calculation This involves the calculation of a short-term and long-term component using AS 3600 Clause 9.3.3 which in turn refers to AS 3600 Clause 8.5.3 for beams. This method will give thinner slabs than the deemed-to-comply solutions, so it should be used where possible, even though it is more tedious when calculated by hand. A number of the available commercial software programs will carry out these design checks. Deemed-to-comply span-to-depth ratios for reinforced slabs This method is set out in AS 3600 Clause 9.3.4 and involves relatively straightforward calculations but will generally give more conservative results. It is limited to slabs of uniform section and that are: n fully propped during construction; Reinforced Concrete Design Handbook 5.3 subject only to uniformly distributed loads and where the imposed action (live load), q, does not exceed the permanent action (dead load), g. n 5.6 AS 3600 Clause 9.4 has additional requirements for crack control in slabs both for flexure and for shrinkage and temperature effects irrespective of the ductility class of reinforcement. Generally, crack control reinforcement is provided in both faces of the slab except when the slab is very thin. Slab deflections shall be deemed to comply with the requirements of AS 3600 Clause 2.3.2 if the ratio of effective span to effective depth satisfies the following equation: (Δ / Lef )1000 Ec 1/3 Lef / d ≤ k3 k4 Fd.ef For flexure where the slab is internal in a building (except for a brief period during construction), AS 3600 requires the minimum area of reinforcement to be in accordance with Clause 9.1.1 and the centre-to-centre spacing of bars in each direction to not exceed the lesser of 2.0Ds or 300 mm. Bars with a diameter less than half the diameter of the largest bar in the cross‑section shall be ignored. es nc fere ts en mm Co ote d Re an ction d) nte t pri is no s se thi tiv erva e ns (N e co s ar on cti fle rally de ne t No 5.5.2 Spreadsheet 5.3 for deemed-to-comply span-to-depth ratios Ge ee Sh b No Jo reo sile By 00 ten 36 ength inal ud r AS e str te: pe ret longit nc as ctor Ø ut co s of n Fa s es Inp str bs ctio ertie ld sla du op Yie in Re l Pr MPa n reo city ia 0.8 sio pa ater MPa res Ca & M 25 mp 0 co φ 50 no ally f' c rm No f sy Da OK D Spreadsheet 5.3 can be used to calculate the deemedto-comply span-to-depth ratios for reinforced concrete slabs based on the method set out in AS 3600 Clause 9.3.4. o tN ee Sh do e. le for its us this read sp med to co 00 ee us ns ctio e pe Th er: laim tion Disc puta m Co t b Clien ct/Jo Proje ct bje Su -D 10 efle 0 19 0 15 s ctor b o an 0.4 0.8 0.4 0.4 0.4 Sh ng 0.7 0.7 0.4 0.7 1 im dom uted and rib tial Dist siden s re Re uctu str fices Of rking 8 0.6 0.6 0.6 0.7 po Im Lo n fa tio or fact erm ) ort-t (ψ s mbi d co rm -te lo ) m, .0 By ,Q ns t-t:er 1170 n tio orte tio d ac DSha(after AS d ac se tic se es po mm mm mm or fact rm -te ) ng (ψ l ψl na bN Jo 5.4 l 8. y ra mpl ee sh ing rson resp sC tio sib on t is d an ψs 0 1 0.7 Pa n tail Re age snsio or-type Stor r n he actio for flo n io of ed re. oOtfRote us ns cr ofs s xu on itie ofs ion su fle layers oRoacf tivte dC A id 5.75 r ro ns ce he ntro 3 in A Ot ce r it b nfor the m ) to bs or 2 layers of te Rei e to t 5m puf g sla la nc In ) ) of es ta s rs . h ed q dis near ata le ored chor widt Dib rto ry 0.7 te ent (1 or 2 layevern . 4re ns omet h, b = d (d) r to ba down lly anch lly an mm t (1 r 2 go rn n0.c po Ge WDeidt pth,ctiDve dellypth= D - coo,verounded rcementrc(fuement (fu ψ co anpg00oert mmen (1 omay gove n. 4.1 re info Effe usua ble ψ r res ile info Ta Re Re er tens y (d .0 ula arrup 60 ge entnts ile is 70 sion the ns 11 es of e s ma gov ng teal s rrabn .0 emm l Te mpr et AS m dina l Co a L asla 40 g re cta n sste ay itu he nt s fro ng dina ctor Lo itu e b teetel/d arraenqui0 eme ts m ds d re la ns ng n fa /m tio Lo tio r 10 r n L el na rce ategivs eonnsc.rete d ac d kN /m rea loa kN nt e1. q0.80ui eme mbi plie r co Ap Dead load sp lues fore invafo r r c 00 s me giv ir c term e re Liv ng for ocsuonppnorts forgaulaSg3iv6efon0r0ce. = =ent requ his ht d lo ly t ti rig an ctor g e c m n A r fa t r into th term rm suetolaCl of ocnta tor a reSin36 or0ce. ing factos e t te en ort Sh Shor term a e ta r Se fblta us m A ein3f60 rcem thneabgCL ctire nts fouto ng ign Lo fo ocan of sl ercseeedmcetionim rS 99 on t es nts ityivre e sp m A info th 19 fo u/d ir se miennAsts c/As uto ers s e, a L e me aecdfect ofinratioqth n bD ab e m inknim ts m re ls urn of d sl itatio rs ire aprcL Ef ityregpthore ep ne f hqeunsrire m o ch side pa Sla s an n lim ileve 1.00 qu inregincfoapinagcly me imu Th inde wlatio ls op nt am ctio r ca 4) be te d g ac s pSpaaancctoitaynbCgalscu/Are hqeuriree mieikennsas.— withou,tdrdropMpaeneiolbn on eah no0t le0thsse7drops fle t re nly of e er: n De ) fo ote 002 n r ns 2 s or ct pt l g e (N e in 0. w = c ere ctio n ( /L re it, ta ab; ababawinth ch direeroallnde,beyond 5= foar ndcaptopDaesslccigulate Abgs re fle tio 4 wh to em enat sbin eonsdin on 1/12 0.00 ade ct of t. whanepconsatacnte wainywa.sly feltfegslltm laim tatio s ita s sl ea l de ov ess a m g Ca an s n in l rc c in rs t C e an 6 l sp kn r rtica ion limspan 2) s 0 = on is e effe rwise on e ti b c ony L/ haeve a thic io n in b d m a .m nt ve wa ast d 1/25ovisi th othe 2 ct for a tw L for st o-o info m paectmorlaK bin=sdeg0fle=ctc)1.0 A jace Dis mpu le andP slab ed tw yofo pelyntraembeenpdnlict aan le storc pr mizeent, 0.00 ated Defle /L ) 1 an 4 for , s c , re it o ad — ini em r a nd at ne 95 fo 00 s r es lcul the s ( e k 1 d m s 0. 0= c d m 0. tw Fa ie an -li b ot er te te r’s ore ca (N rt mov 1/50 0= ile aapa moom foercnt ampe lict aaninfoe toben20 in &'=tu r05e ex ntreice D is r of r sp ture not m 4 for 2 wh to Co t 1/25 0.00 ade ct of orte terio be info 00 its ufac t ns ltr im pe lye pepnr reabl in t 1, uo.muc '= 1. whicpphBortace3Ds, wher c ate Creinomlarrc to an ion bu = 0. = is m effe ise e e sh an in n Lim M n 0 te b ro 0 . th u rw e on e at o R p 00 to s ctio ered 1/50ovisi th othe 1 ific 1/25 u fo rlyeam eea ed to tesu an 1. than 36 er ea lic el n M tr Cli ec fle id t/J pr mizeent, 0.00 25 er al 2.1 1 kN/m sp than icin AS De cons umte m ction long ng sh t).umltim sh ppstemeng S cre th — 0.00 late ss th fle rs mini em 00 = s; jec a rc .4 r’s ore the is lo 2 of cu de /m artrelyculfo teri 0= n as slab mov 1/10 13 oc of kN tal n ee im saupm ture not m002 an 2.3. ke lcu te d nd y a n ndlati rete on Pro 1/40 ct i in thation or e to , ta orte r ufac but = 0. ratio d sp ble ma = K i s oanrtre Dell t)m.uinltim Th e lateemsseth be r aularerlssAiomaelcuonc d C Ta Ca lcuula en tant pp t ctionaddit of th the Manation 500 bje d im 7.66 . ctor a e su em ns fle no e e u t om b fa n : e 1/ p te n y er a he in lais umhe tic p ng c , Cn. ce m = ific a lc la rt Fr S ere e de r th en ns a wh 5 00 n co pl of 4 Th afte chmrtitio rs specan 12 late rsmseth Inse pe rs sllnt)m.r setimr sparomrati s in essio for Ca inlcfouurc on C 0.00 cu 00 th th eete 3or6K = deflect1.io4 for sim us slab1.s,2 and orwh atta pa Ty be d oc ti 0= g = 0. = a lc S at em c or id a in n s a la e 0 fo rc lk e n th ion ish m : = or ; re g sulaprersm gushmn t isof rpotre au te ein 1/25 uo ed A ct ere C s. e rip an All 1/80 Caoinnlcfo wh to ing ortin ction dit fin an ntin exce d sp le MPa in Fa k e of 2 pp ns fle e ad the e sp str co sc sidrceetaenadaellsig eectsncores & F in k, R t rc su rtitio (liv ct) C a no an en rior d 0.00is madfect fer e for e deter th ent of rs n Tit re s = pa fo h te Th af m es te C hoenforecper rst d ds ffe (I r p ll ers ic be ry tio pa 0 on e ef trans ted D se do 75 or in in re ch d ac im r 1/50ovisi th the ppor e Memason inr sid o a e 9 fo Hm r a b atr TC se mic he atta rein m = 1. 1 fo nc pr mize of e su wis ral po na on yu nF re g ot r 2. o ,e nF mini ctionon th othe 1 e im d dy cti ctio ilp = FTohoeness nspsreth-e20ll0ow anm ne ortines sh 00 Th d an defle fle , fle r rall er ctio = pp de de be ture = 0. fle su finish CDo e re 0 t a ngfor nd K r,Kc loa c Ge tal de rs ne me ign memstru 1000 d to be ittle to e tal r to r 0 s o h 1/ cte a ian c fo br a o em en g or o M bje str TIgn 36s n Rble ter ct em ction de s su de load fle nt pe rs incr gn n fa rt fa de y be r or ADSoe rneur,ita Fos by na tal da : ula traffic tio Inse ctive desi load for eme Mem To 28 hic ut tio at at s r, ve l uc ns ity sign Effe tic W de inp ota ca forc No arne tio rs red ive elas be of dn em ita ata M*Effect rein reti ity ulus rm W an sfe eo ac mod ed Lim Tran nts sion the ap Th ols uir lues of lue me f ten r, c s/ d ove for va mb req o n va e y e ea re s v o d : M m S ll tE ho ns ab co ce s Co sis Inpu ing yers nc e optio y, w la nd etr Ba y a Se llo ule : be r 1 f/d Ye mod eom full re Le g nt d tual ign r fo red 4.1 Ac 40.0 latu All ction me es ete ore cho nc ble ch n rce g d iam .0 Ta Se me an lly a 40 V* info y din ar d ated .0 e = ll No n re c lcul 4 e b 70 fu t fu Ca .9 41 for nsile 11 nt men : r b er & r e t o S a u F m = e A e te .54 50 lues mb os rce forc m Inp sh inal t va Nu = inpu fro in rati info ign ud for ht, es ngit rig ply ile re ion re tors d c the m r lo n fa d to ctio n co tens ress Fo a of e an fle on ctio ov de p fle ab ed tal de us ns d to inal com L) binati To = tio tal be Are to en )] L e op l e. me ud /L Se t y )/(F )] Increm ee ngit dina and com = ula Inpu da 000 E ios L rm rd lo gitu Rat 28 /L )1 0 E )/(F (D term Fo a of ply gnfo 00 at k (D g lon ns Com sfleinctio ction ity [k k (D/L )1 to u Are a of actio d lon lab de c fle y l ti med [k a)] llTota ental de an of s Dee las c Are lied e ts t ti em F ul f p rm n tan m0 aE )/( )] Incr Res so Ap rt te spa to ns lu )100 t u F o o e u n c Sh ctiv d ka( /L )1000 E )/( od on onsta e /L late [k cti em Eff fle n c of th lcu M*[k k ( de tio ca ed k 3 eflec alue nd nts rs us v d rou ome g ba be k4 an kg ac g m forcin Me /L ef to b ing c E in ac en din sp ut gre ben of re nd Inp g n ith ta s w esig pacin en ell d s em d c ing and M u rc e fo x d Bo ben size ity φ rein c r m r, Fo mbe capa imu V* nt ce min Nu me for for in ar Mo cks : f f V u.m he ut e d.e s h Ø tp F d.e C n nF d Ou on ectio sig an c e u c cti /d u et, rd ØV .min L ef defle l defl Øv he on Fo on os u tio ax, l * ta V ds cti cti V u.m rati th ra r tota men fle fle rea >Ø e e ØV cks if ly p * d d sp p l V de ad fo incre e m is tal nta to r Ch cks if co To g th 1/3 me n lo d fo an e a d to l sp esig rdin cre f)] e Ch a lo .e a In g d d m n 1/3 (F ctu tive ee ED k re )] sig c)/ c sa .ef rd ac de CK 0E (F d db Fo ulate s effe tive 00 c)/ HE fee lc f)1 ec 0E er /L e Ca ulate s eff YC 00 lc oth LL k 4(D /L ef)1 or Ca ulate FU [k 3 ns lc (D s o a E 4 ti k C B 3 es late O gg lcu s [k su Ca ulate LT s, lc on TIL cti Ca ES rre AA co C NT S AR E ts, e C n M T me ct th OM HEE om onta rc R C DS Fo ase c FO EA ple FT PR ign lab S ete -D 2 Des 1 41 2 45 . se mm 2 mm 0.7 For flexure with slabs exposed to the weather, in addition to the requirements above, AS 3600 requires compliance with Clause 9.4.1 (c) and (d); this involves limiting the stress in the reinforcing steel in flexure, depending on the size and spacing of the reinforcement. /m kN /m kN d ide rov st.p ed vid pro sc. o s l o ef Spreadsheet 5.3 is available at www.ccaa.com.au 0 ef 65 37 50 34 ef 10 42 80 39 0 20 0 60 0 40 0 80 cs ef 40 st 32 sc 32 3 30 3 25 YS 3 28 S AT 20 0 00 24 DA 26 0 80 0 10 ef ef 0 70 TIE ER OP PR TE RE NC ) CO Pa (M ) f' c Pa (M Ec 4 4 .94 .54 50 f d.e 4 /d L ef tual .0 40 Ac .0 c 40 ed lat lcu Ca .94 41 .54 50 5.5.3 Basis of Chart 5.14 04 0.0 S LIE MP OK CO S LIE MP OK CO This chart gives the proportion of load carried in the shorter span direction, Lx, for slabs supported on four sides as specified in AS 3600 Clause 9.3.3(b) for slab deflections by simplified calculations. The curves are derived from the equation: S LIE MP OK CO For crack control in the primary direction, no additional reinforcement is required to control expansion or contraction cracking if the area of reinforcement in the direction of the span of a one-way slab, or in each direction of a two-way slab, is not less than: 41 0 70 26 f d.e Crack Control of Slabs D b Slab an ed Sp id ov pr ided A st prov ed A sc us /L ef L mm 0 mm 19 mm 00 2 10 00 mm 2 60 1 mm 41 2 45 0 / 25 =1 S LIE MP f ef 1/3 OK f 3 CO d.e c 1/3 d.e ef c 4 ef 3 4 f n 1/3 the area required by Clause 9.1.1; and d.e c f 1/3 d.e ef 3 c 4 ef 3 4 Proportion of load in Lx direction = 1/(a L x4/ L y4 +1) where a = values in AS 3600 Table 9.3.3 for various conditions of edge continuity. 5.5.4 Basis of Chart 5.15 ck ba ed Fe : A S DR HE 1.2 TE T NO The curves in this chart were derived for the slab‑system multiplier k4 (not the deflection coefficient) given in AS 3600 Table 9.3.4.2 , for each of the edge conditions listed. (Note that a value has been interpolated for Ly / Lx = 0.75.) n . n 75% of the area required by one of Clauses 9.4.3.3 to 9.4.3.5, as appropriate. For crack control and shrinkage where the slab is free to expand or contract in the secondary direction, the minimum area of reinforcement in that direction shall be 1.75 D mm2/m width. The requirements for crack control and shrinkage with restrained slabs in the secondary direction inside a building are: rsio Ve n n 5.5.5 Basis of Table 5.2 The moments of inertia of cracked reinforced concrete slabs one metre wide provided in this table were derived from the formula: n Icr = 1000 d 3[k 3/ 3 + n p (1 - k)2] where k = [(n p) 2 + 2n p] 0.5 - n p n = modular ratio = Es / Ec p = Ast /1000d Icr is used in calculating Ief which in turn is used in determining deflections by the simplified calculation method as set out in AS 3600 Clause 9.3.3. 5.4 Reinforced Concrete Design Handbook Where a minor degree of control over cracking is required, Ast must be at least 1.75 D mm2/m width. Where a moderate degree of control over cracking is required and where cracks are inconsequential or hidden from view, Ast must be at least 3.5 D mm2/m width. Where a strong degree of control over cracking is required for appearance or where cracks may reflect through finishes, Ast must be at least 6.0 D mm2/m width. The requirements for crack control and shrinkage with restrained slabs in the secondary direction elsewhere and in exposure classification A1 and A2 are: n Where a moderate degree of control over cracking is required and where cracks are inconsequential or hidden from view, Ast must be at least 3.5 D mm2/m width. TABLE 5.2 Moment of inertia of cracked reinforced concrete sections Icr per metre width (mm4 x 106) np d (mm) d (mm) 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.15 0.20 80 90 4 6 8 11 11 16 14 20 17 24 20 28 22 32 25 35 27 38 29 41 38 55 46 66 80 90 100 110 120 130 8 11 14 18 15 20 27 34 22 29 38 48 28 37 48 61 33 44 57 73 38 51 66 84 43 58 75 95 48 64 83 105 52 70 90 115 57 75 98 124 75 100 130 165 91 121 157 199 100 110 120 130 140 150 160 170 180 190 23 28 34 41 48 57 42 52 63 76 90 106 60 74 89 107 127 150 76 93 113 136 162 190 91 112 136 163 194 228 105 130 157 189 224 263 119 146 177 213 252 297 131 162 196 235 279 328 144 177 214 257 305 359 155 191 231 278 330 388 206 254 308 369 438 515 249 306 372 446 529 623 140 150 160 170 180 190 200 210 220 66 77 88 123 143 164 174 202 232 222 257 295 266 308 354 307 355 409 346 401 461 383 444 510 418 484 557 452 523 602 601 696 800 726 841 966 200 210 220 230 240 101 115 187 213 265 301 337 383 404 459 467 530 526 598 583 662 636 723 688 781 914 1039 1104 1255 230 240 250 260 270 280 290 130 146 163 182 202 240 271 303 338 375 341 383 429 479 532 433 487 545 608 676 519 584 654 729 810 600 674 755 842 936 676 760 851 950 1055 748 842 943 1051 1168 817 919 1029 1148 1276 883 993 1112 1241 1378 1174 1321 1479 1650 1833 1418 1595 1786 1992 2213 250 260 270 280 290 300 310 320 330 340 224 247 272 298 326 416 459 504 553 605 589 649 714 783 857 748 825 908 996 1089 897 989 1088 1193 1305 1036 1143 1257 1379 1508 1168 1289 1417 1555 1700 1293 1427 1569 1721 1882 1412 1558 1714 1879 2056 1526 1684 1852 2031 2221 2029 2239 2463 2701 2954 2450 2704 2974 3262 3567 300 310 320 330 340 350 360 370 380 390 400 356 387 420 455 492 531 660 718 780 845 913 985 935 1017 1104 1196 1293 1395 1188 1293 1403 1520 1643 1773 1424 1549 1682 1822 1970 2125 1645 1790 1944 2106 2276 2456 1855 2018 2191 2374 2566 2768 2053 2234 2426 2628 2841 3065 2242 2440 2649 2870 3102 3347 2423 2637 2862 3101 3352 3617 3222 3506 3807 4124 4458 4810 3891 4234 4597 4980 5384 5809 350 360 370 380 390 400 Table 5.3 Total area of reinforcement, Ast , for crack control and shrinkage (mm2/m width) Depth of slab, D (mm) Ast (mm2/m width) 100 110 120 130 140 150 160 240 250 1.75 D 3.5 D 6.0 D 193 413 660 210 450 720 228 488 780 245 525 840 263 563 900 280 298 315 333 350 368 385 403 420 600 638 675 713 750 788 825 863 900 960 1020 1080 1140 1200 1260 1320 1380 1440 438 938 175 375 600 170 180 190 200 210 220 230 Reinforced Concrete Design Handbook 1500 5.5 n Where a strong degree of control over cracking is required for appearance or where cracks may reflect through finishes, Ast must be at least 6.0 D mm2/m width. n For restrained slabs in the secondary direction in exposure classification B1, B2, C1 and C2, Ast must be at least 6.0 D mm2/m width, which can be a significant amount of reinforcement. n Table 5.3 shows the various areas of reinforcement per metre width required for crack control and shrinkage in slabs up to 250 mm deep. n 5.7 n Detailing of Slabs For all reinforced concrete slabs, appropriate detailing must be shown on the drawings. n AS 3600 Clause 9.1.3 sets out the detailing of flexural reinforcement for slabs. This requires the determination of the shape of the theoretical bending moment and the extending of the tensile reinforcement by the depth of the section D past the cut-off point. The clause also includes other requirements and a deemed-to-comply arrangement for continuous one-way and two-way slabs using the simplified method of analysis. Detailing for shear reinforcement is covered in AS 3600 Clause 9.2.6. Wherever possible, shear reinforcement in slabs should be avoided, because of the difficulty in fixing it in thin members. Allowing for cover and fitment sizes including bends, a slab at least 200−300 mm thick would be required if shear reinforcement is required. It is usually more economical to increase the depth of slab and /or the concrete strength. n n Designers should also refer to Chapter 14 of the Reinforcement Detailing Handbook 5.5 for further guidance. Areas that designers need to consider in detailing of reinforcement for slabs include: n n n 5.6 Minimum reinforcement in accordance with AS 3600 Clause 9.1.1. One-way slabs with bar reinforcement in the direction of the span will also require transverse reinforcement for support of longitudinal reinforcement. AS 3600 Clause 9.4.1 requires the maximum spacing of reinforcement for crack control for slabs in both directions to be the lesser of 2.0 Ds or 300-mm centres. Typically N10 (if available) or N12 bars are used in the transverse direction. For two-way slabs with bar reinforcement, it is important to nominate on the drawings which bar is to be placed first in the bottom layer and in which direction and which bar is to be placed last in the top layer and again in which direction. Reinforced Concrete Design Handbook n n n n For both one-way and two-way slabs with square and rectangular mesh reinforcement, it is important to nominate on the drawings for each layer, where the top and bottom wires are placed. For rectangular meshes, the area of cross wires is limited to 227 mm2/m so additional bar reinforcement may be required in the secondary direction for flexure, shrinkage and temperature control. For two-way slabs with diagonal edges, up to three layers of reinforcement both in the top and bottom of the slab may result. This may cause congestion and will need special consideration. 12-mm and 16-mm Ductility Class N bar can be either stock length bar or, more commonly, cut to length from coil. Stock length bars should be used where possible. Large size bars (20 mm and larger) are usually supplied in stock lengths of 12 m. For example, specifying a bar 4.00 m long instead of 3.75 m long will save cutting and wastage as three bars can be cut from one stock length bar. Staggering will often effectively allow the use of stock length bars. For flat slabs and flat plates with column and middle strips, the use of different size bars in the column and middle strips will make it easier to check the reinforcement layout on site, eg 20-mm in the column strip and 16-mm in the middle strip in the bottom and 24-mm in the column strip and 20-mm in the middle strip in the top. Slabs will often act as horizontal diaphragms carrying lateral actions such as wind and earthquake forces back to the vertical elements such as walls and columns. Designers may need to check the transfer of shear forces between the slab and vertical elements as well as the bending in the slab when acting as a deep beam. Section 5.6 of the Precast Concrete Handbook 5.6 covers this. Lapping of reinforcement in tension in areas of high moment should be avoided, and all laps and splices should be adequately detailed on the drawings. Typically, bottom reinforcement is lapped at the points of support of the slab and top reinforcement in the middle of the slab where laps are nominal. The reinforcement should always be curtailed where not required, eg excessive top reinforcement in the middle of slabs and excessive bottom reinforcement in the ends of slabs. For cantilevers, the top reinforcement should be anchored well back in an area of no or low stress if possible. Small diameter bars provide better crack control than large diameter bars (of the same area). n n n n n Construction joints in the slabs (often at the quarter point of the span) should be properly detailed and shown on the drawings. n n n n n Charts 5.1–5.4 Flexural reinforcement in slabs Ductility Class N reinforcement pages 5.8–5.11 Standard reinforcing meshes are manufactured in sheet sizes of 2.4 m by 6 m and again by proper detailing, cutting and wastage can be minimised. Charts 5.5–5.8 Flexural reinforcement in slabs Ductility Class L mesh reinforcement pages 5.12–5.15 Care should be taken when lapping reinforcing mesh in thin slabs as up to eight layers of wires can occur in one location for one layer of reinforcement. There are techniques to reduce this by offsetting the laps and the sheets of mesh and by the use of tie bars. Charts 5.9–5.12 Punching shear at circular columns where M *v = 0 pages 5.16–5.17 Chart 5.13 Punching shear at rectangular columns where M *v = 0 page 5.18 For slabs exposed to the weather, adequate slopes for drainage (taking into account long‑term deflections) should be provided. Also, where there is no structural reinforcement, crack-control reinforce-ment should be provided in the top of the slab. Chart 5.14 Slabs supported on four sides – proportion of load carried in shorter direction page 5.19 Chart 5.15 Slabs supported on four sides – deflection coefficient k4 page 5.19 Even with simply-supported slabs, concrete edge beams, walls and end supports will provide some restraint to the slab; nominal top reinforcement in the slab will therefore be required in these areas to control cracking. 5.8 n Charts 5.1 to 5.15 Spreadsheets 5.1, 5.2 and 5.3 may be downloaded from the Cement Concrete & Aggregates Australia website www.ccaa.com.au General Guidance The depth of a slab, unless it is a short span, is usually controlled by deflection considerations. The amount of reinforcement and its location are then determined to meet bending and shear requirements and constructability. The strength of concrete chosen must be appropriate for the exposure classification. For B2, C1 or C2 exposure classification, special class concrete needs to be specified as required by AS 3600. It is recommended that designers discuss these special requirements with the concrete suppliers and concrete technologist. In multi-storey construction a slab may be required to support the construction of other slabs over before it has attained its design strength (at 28 days). This may require higher strength concrete and consideration of the construction design loads as a design case5.7. edges always require careful consideration. Generally, vertical penetrations less than, say, D in size are not a problem unless there are many close together. Additional reinforcement may be required around larger penetrations. n n n n Sometimes, services such as electrical cables may need to be cast into the slab as shown in Figure 5.2. Large services such as sewer, storm water or water pipes should not be built in since they will cause problems if they leak. Set downs may be required to the top of the slab, eg for balconies, toilet areas and falls to drains. Consideration must be given to the finish and class of formwork to the formed surface of the slab (ie the underside of the slab) if exposed to view. Linear elastic analysis is recommended for major projects and the simplified methods for small projects and simple elements, subject to the computer analysis software available. As with all concrete members, suspended insitu concrete slabs will shrink. If restrained by stiff elements such as supporting walls or core walls, unsightly cracking can result. Techniques to minimise this cracking include the use of pour strips or slip joints. Slabs are generally not suitable to support heavy line loads (such as masonry walls) or heavy point loads. Vertical penetrations through slabs for services or other openings, embedded items and unsupported DRAF T Figure 5.2 Electrical ducts cast into a slab Reinforced Concrete Design Handbook 5.7 chart 5.1 Flexural reinforcement in slabs Ductility Class N reinforcement 1000 180 d A st 230 160 220 f = 0.8 f 'c = 25 MPa fsy = 500 MPa 210 200 140 Minimum reinforcement for slabs supported by beams or walls on 4 sides 190 180 120 Minimum reinforcement for slabs supported by columns at their corner 170 160 100 For one-way slabs interpolate between the two values shown 150 140 80 130 120 60 110 100 40 90 80 70 20 M * (kN.m/m) Effective depth (mm) Note:To right of dashed line sections are over reinforced 0 100 300 500 700 900 Ast (mm2/m) 5.8 Reinforced Concrete Design Handbook 1100 1300 1500 1700 1900 chart 5.2 Flexural reinforcement in slabs Ductility Class N reinforcement 1000 180 d A st 230 220 160 210 200 140 190 180 170 120 160 150 100 f = 0.8 f 'c = 32 MPa fsy = 500 MPa Minimum reinforcement for slabs supported by beams or walls on 4 sides Minimum reinforcement for slabs supported by columns at their corner For one-way slabs interpolate between the two values shown 140 130 80 120 110 60 100 90 40 80 70 Effective depth (mm) M * (kN.m/m) 20 Note:To right of dashed line sections are over reinforced 0 100 300 500 700 900 1100 1300 1500 1700 1900 Ast (mm2/m) Reinforced Concrete Design Handbook 5.9 chart 5.3 Flexural reinforcement in slabs Ductility Class N reinforcement 1000 180 230 220 160 210 d A st f = 0.8 f 'c = 40 MPa fsy = 500 MPa 200 190 140 Minimum reinforcement for slabs supported by beams or walls on 4 sides 180 170 120 160 150 Minimum reinforcement for slabs supported by columns at their corner For one-way slabs interpolate between the two values shown 140 100 130 120 80 110 100 60 90 80 40 70 Effective depth (mm) M * (kN.m/m) 20 Note:To right of dashed line sections are over reinforced 0 100 300 500 700 900 Ast (mm2/m) 5.10 Reinforced Concrete Design Handbook 1100 1300 1500 1700 1900 chart 5.4 Flexural reinforcement in slabs Ductility Class N reinforcement 1000 180 d 230 220 160 210 A st f = 0.8 f 'c = 50 MPa fsy = 500 MPa 200 190 140 180 170 120 Minimum reinforcement for slabs supported by beams or walls on 4 sides Minimum reinforcement for slabs supported by columns at their corner 160 150 For one-way slabs interpolate between the two values shown 140 100 130 120 80 110 100 90 60 80 40 70 Effective depth (mm) M * (kN.m/m) 20 Note:To right of dashed line sections are over reinforced 0 100 300 500 700 900 1100 1300 1500 1700 1900 Ast (mm2/m) Reinforced Concrete Design Handbook 5.11 chart 5.5 Flexural reinforcement in slabs Ductility Class L mesh reinforcement 1000 90 d 230 80 220 A st f = 0.64 f 'c = 25 MPa fsy = 500 MPa 210 200 70 190 180 170 60 160 Minimum reinforcement for slabs supported by beams or walls on 4 sides Minimum reinforcement for slabs supported by columns at their corner For one-way slabs interpolate between the two values shown 150 50 140 130 120 40 110 100 30 90 80 20 70 Effective depth (mm) Note:To right of dashed line sections are over reinforced M * (kN.m/m) 10 SL82 SL92 SL102 RL818 RL 718 0 100 200 300 400 500 RL 918 600 Ast (mm2/m) 5.12 Reinforced Concrete Design Handbook RL 1018 700 RL 1118 800 900 RL 1218 1000 1100 1200 chart 5.6 Flexural reinforcement in slabs Ductility Class L mesh reinforcement 1000 90 d 230 220 80 A st f = 0.64 f 'c = 32 MPa fsy = 500 MPa 210 200 70 190 180 170 60 160 150 50 Minimum reinforcement for slabs supported by beams or walls on 4 sides Minimum reinforcement for slabs supported by columns at their corner For one-way slabs interpolate between the two values shown 140 130 120 40 110 100 90 30 80 70 20 Effective depth (mm) M * (kN.m/m) 10 SL82 SL92 SL102 RL818 RL 718 0 100 200 300 400 500 RL 918 600 RL 1018 700 RL 1118 800 900 RL 1218 1000 1100 1200 Ast (mm2/m) Reinforced Concrete Design Handbook 5.13 chart 5.7 Flexural reinforcement in slabs Ductility Class L mesh reinforcement 1000 90 d 230 220 80 210 A st f = 0.64 f 'c = 40 MPa fsy = 500 MPa 200 70 190 Minimum reinforcement for slabs supported by beams or walls on 4 sides 180 170 60 160 150 Minimum reinforcement for slabs supported by columns at their corner For one-way slabs interpolate between the two values shown 140 50 130 120 40 110 100 90 30 80 70 20 Effective depth (mm) M * (kN.m/m) 10 SL82 SL92 SL102 RL818 RL 718 0 100 200 300 400 500 RL 918 600 Ast (mm2/m) 5.14 Reinforced Concrete Design Handbook RL 1018 700 RL 1118 800 900 RL 1218 1000 1100 1200 chart 5.8 Flexural reinforcement in slabs Ductility Class L mesh reinforcement 1000 90 d 230 220 80 210 A st f = 0.64 f 'c = 50 MPa fsy = 500 MPa 200 190 70 Minimum reinforcement for slabs supported by beams or walls on 4 sides 180 170 60 Minimum reinforcement for slabs supported by columns at their corner 160 150 For one-way slabs interpolate between the two values shown 140 50 130 120 40 110 100 90 30 80 70 20 Effective depth (mm) M * (kN.m/m) 10 SL82 SL92 SL102 RL818 RL 718 0 100 200 300 400 500 RL 918 600 RL 1018 700 RL 1118 800 900 RL 1218 1000 1100 1200 Ast (mm2/m) Reinforced Concrete Design Handbook 5.15 chart 5.9 Punching shear at circular columns where M *v = 0 Column diameter (mm) = 1000 2500 750 500 250 d/2 Column diameter d/2 Critical shear perimeter 2000 f 'c = 25 MPa 1500 1000 V* (kN) 500 0 100 200 300 400 500 600 700 d (mm) chart 5.10 Punching shear at circular columns where M *v = 0 Column diameter (mm) = 1000 2500 750 500 250 d/2 Column diameter d/2 Critical shear perimeter 2000 f 'c = 32 MPa 1500 1000 V* (kN) 500 0 100 200 300 d (mm) 5.16 Reinforced Concrete Design Handbook 400 500 600 700 chart 5.11 Punching shear at circular columns where M *v = 0 Column diameter (mm) = 1000 3000 750 500 250 d/2 Column diameter d/2 Critical shear perimeter 2500 f 'c = 40 MPa 2000 1500 V* (kN) 1000 500 100 200 300 400 500 600 700 d (mm) chart 5.12 Punching shear at circular columns where M *v = 0 Column diameter (mm) = 1000 3000 750 500 250 d/2 Column diameter d/2 Critical shear perimeter 2500 f 'c = 50 MPa 2000 1500 V* (kN) 1000 500 100 200 300 400 500 600 700 d (mm) Reinforced Concrete Design Handbook 5.17 chart 5.13 Punching shear at rectangular columns where M *v = 0 c1 d/2 c2 d/2 Critical shear perimeter 2500 f 'c (MPa) 2000 50 40 32 25 1 c1 /c2 2 3 4 5 6 V * (kN) = 500 1000 c1 + c2 (mm) = 500 7 1500 1000 450 400 350 300 250 200 150 d (mm) 100 5.18 Reinforced Concrete Design Handbook 1500 2000 chart 5.14 Slabs supported on four sides – proportion of load carried in shorter direction Ly 2.4 Lx 2.2 2.0 1.8 1.6 1.4 Edge condition = 5 8 3 2 7 1, 6, 4 1.2 Ly/Lx α = 5.0 2.5 2.0 1.0 0.5 0.4 0.2 1.0 0.1 0.2 0.3 0.4 PROPORTION OF LOAD IN Lx DIRECTION Slab edge 1 Support 2 3 0.5 0.6 4 0.7 0.8 5 0.9 7 6 1.0 8 9 EDGE CONDITION – Slabs supported on four sides chart 5.15 Slabs supported on four sides – k4 for deflection calculations Ly 4.0 Lx 3.5 3.0 2.5 2.0 k4 1.5 1.0 1.0 Ly /Lx 1.1 1.2 1.3 1 2 3 4 5 1.4 1.5 Four edges continuous One short edge discontinuous One long edge discontinuous Two short edges discontinuous Two long edges discontinuous 1.6 1.7 6 7 8 9 1.8 1.9 2.0 Two adjacent edges discontinuous Three edges discontinuous (one long edge continuous) Three edges discontinuous (one short edge continuous) Four edges discontinuous Reinforced Concrete Design Handbook 5.19 References 5.1 Guide to Long-Span Concrete Floors (T36) 2nd Ed, Cement Concrete & Aggregates Australia, 2003. 5.2 Foster SJ, Kilpatrick AE and Warner RF Reinforced Concrete Basics 2nd Ed, Pearson, 2010. 5.3 Guide to Industrial Floors and Pavements (T48) 3rd Ed, Cement Concrete & Aggregates Australia, 2009. 5.4 AS 3600 Concrete structures Standards Australia, 2009. 5.5 Reinforcement Detailing Handbook (Z06) 2nd Ed, Concrete Institute of Australia, 2010. 5.6 Precast Concrete Handbook 2nd Ed, National Precast Concrete Association Australia and Concrete Institute of Australia, 2009. 5.7 Multi-Storey Formwork Loading (Z35), Concrete Institute of Australia, 1990. 5.20 Reinforced Concrete Design Handbook Chapter 6 Columns concrete columns. Their design is in accordance with AS 3600 6.2 and will depend on the end fixity adopted and whether or not they are braced. Concrete columns can be subject to a combination of actions including: n 6.1 General Concrete columns are small but important structural elements in most buildings. Their design can be more complex than other concrete elements, while, along with walls, they are frequently the most obvious, and sometimes intrusive, parts of a structure. Architectural and engineering judgement is required to determine their position, size, shape, the spans of the horizontal elements they support and their location for the economy of structure. Prior to final design, it is important to ensure that the design team agrees the proposed column sizes and positions, especially where they are located in car parking areas and architecturally sensitive areas. n n Vertical actions. These are calculated by apportioning the actions on each floor to the column by the selected frame-analysis software or on an area basis, sometimes known as column rundowns or similar. Bending moments from slabs and beams. The bending moments are usually assessed from the slab or beam design or by the selected frameanalysis software. Horizontal actions on the structure resulting in shear forces and bending moments in the column when it is used to resist lateral actions as part of the building frame and when there are no shear walls. These actions are usually assessed by the selected frame-analysis software. It is also important to consider the implications of the location of each of the columns on each floor and, if possible, to avoid offsetting columns from one storey to another. Such changes in position usually involve transfer beams, which can be expensive and time‑consuming to build. This might occur for example in a building where there is a basement car park, ground floor retail and upper floor residential areas and optimum column locations will be different for each area. Generally, columns are designed for axial actions (loads), and bending moments about each axis as required, for the various load cases at each floor level. For single-storey buildings and higher buildings where the last lift of columns is supporting a lightweight roof, they may act as vertical cantilever beams carrying only small vertical actions and resisting lateral actions. Insitu concrete columns can require expensive formwork and take time to build. In recent years, there has been a trend to precast concrete columns, particularly in low-rise buildings, for ease of construction and to reduce costs. To facilitate erection and the casting of the floor over, precast columns are usually most economical as single-storey-height elements. Two-storey-height columns are possible, especially if the bracing is kept below the floor and the connection to the floor in the middle of the column is appropriately considered. When precast columns are used, they need to be temporarily braced in two directions until the floor over is cast. The Precast Concrete Handbook 6.1 includes information on precast Off-form columns to a country bank Column bars offset at the top to facilitate joining above the floor level with fitments through the floor level and also closer spacings of the fitments Reinforced Concrete Design Handbook 6.1 Polished round reinforced and post tensioned precast columns These are unusual non-vertical columns, but illustrate how complex columns can be. 6.2 Initial sizing and actions For the design of a column, an initial size, concrete strength, number, size and location of longitudinal bars and fitments, cover for durability and axis distance (for fire) are assumed along with an assessment of whether the column is braced or not and its effective length. The chosen configuration is then checked for adequacy and the various parameters adjusted as required. The quickest way to size a concrete column is usually to design the chosen configuration using appropriate concrete column design software or spreadsheet. It is recommended that designers start with the design of the lower lifts of columns where the actions are highest and the maximum concrete strength will be needed. Initial sizing, concrete strength and reinforcement for columns can also be based on experience, previous designs or chosen from the design Charts 6.1 to 6.6 in this Chapter. Further advice on designing columns is also given by Warner et al 6.3, 6.4. The implications of Clause 10.8 for the transmission of axial forces through the floor systems must be assessed early in the design process. Where possible, the requirement of AS 3600 Clause 10.8 should be met by specifying a concrete strength for the floor system of not less than 0.75 of that specified for the columns (eg 40 MPa for columns with 32 MPa for floors, or 32 MPa and 25 MPa respectively). Where the floor strength is less than 0.75 times that of the column strength. The effective strength of the column, f 'ce, in the column/slab or column/beam area can be significantly less than the strength of the column, f 'c. This reduction is sometimes of the order of 20–30% for very-high-strength columns. Additional reinforcement can be used to compensate for this reduction at the floor level. However, these extra bars in the junction area, in what almost certainly will be heavily reinforced columns, may lead to congestion and difficulties in placing and compacting the concrete. Solutions 6.2 Reinforced Concrete Design Handbook to overcome these difficulties include: the use of mechanical end splices; and, where the reinforcement percentage is high, locating the lap splices outside this area; or, as AS 3600 suggests, confining fitments can be used to increase the effective strength of the concrete in the joint – presumably in accordance with Clause 10.7.3 but the specifics on how precisely this is to be achieved are not spelt out. Alternatives such as 'blobs' of high-strength concrete placed at the column joint when the slab is being cast can result in cold joints in the slab. Further, such processes are difficult to supervise on site, are usually impractical and therefore generally not recommended. For columns there should be a minimum of four bars in a rectangular section and six bars in a circular section. The minimum area of reinforcement, Asc, required by AS 3600 is 1% of the gross concrete area. Generally, the reinforcement ratio should be kept below 2.5% for economy and ease of splicing. AS 3600 permits a maximum area of reinforcement of 4% and this implies a maximum of 8% at laps. For over-sized columns (eg for aesthetic reasons), AS 3600 allows a lower value for Asc to be used provided Asc fsy > 0.15 N *. It is suggested that the minimum area of reinforcement should not be less than about 0.5 %. In the lower levels of high-rise buildings, to keep the columns to a reasonable size, the longitudinal bars are sometimes spliced using end bearing splices, mechanical splices, welding or the bars are bundled and high-strength concrete is used. It should be noted that end bearing splices may not be readily available, mechanical splices will take up more space and welding is an expensive alternative. Assessing the vertical actions carried by columns (and walls) requires a full understanding of the building structure and its behaviour and knowledge of all actions to be carried by the building. These vertical actions from permanent and applied floor actions are calculated by assessing the actions supported by each column (or wall) on a floor-by-floor basis based on the tributary areas to each column (or wall). This can be determined by using a spreadsheet to calculate the actions on the column or using appropriate structural analysis software (or by hand). Nevertheless, judgement is required in assessing the vertical actions regardless of the method used. One of the problems when using a full 3D building model to assess the column actions is that it may not take into account all loading cases and usually does not take into account construction sequences and the redistribution of actions that may occur due to deflections or shortening in some supports, etc. As a result, the 3D building model may be non‑conservative in some cases. On the other hand the use of tributary areas may be conservative in some cases and non‑conservative in others. Column rundowns calculated by spreadsheet (or by hand) are usually based on a simple area or length basis, with the proportion of the actions to each vertical element calculated by taking half the distance in each direction to the adjacent vertical element supporting vertical actions. Further, the column rundown may not include all the loads to the column (or wall) because of continuity and frame action. These additional actions are sometimes known as moment shears and can be up to 15% of the floor actions. For example, an edge column will generally have less actions on it than is implied by a calculation on an area basis while the first column in from the edge of a building will have more. Actions at each level are usually calculated just above the floor below. (For example the heading 'On Level 4' as shown in Table 6.1 on the column rundown means the actions from the floors above, just above the fourth floor and these actions would be used to design the column from Level 4 to Level 5.) designers should always check the architect's and other members of the design team's drawings to see that all loads have been included. Actions to be considered in the design of columns for a multi-storey building can include some or all of the following: In addition, for imposed actions reductions can be deducted from rundowns, where applicable, in accordance with AS 1170.1 Clause 3.4.2 (including a reduction from the moment shears). Structural n Roof (including finishes) n Floors (including finishes) n Internal masonry walls n Internal lightweight walls n External walls including precast walls, curtain walls, etc n Beams framing into the column n Precast and tilt-up concrete walls n Fascias and sun hoods n Lift machinery n Air conditioning and other mechanical plants n Stairs n Heavy load areas, eg storage areas n Special equipment, eg water tank, generator, etc. The bending moments and horizontal shear actions in a column will be determined by the frame-analysis software used or by other structural design methods. Table 6.1 Sample column rundown Load element Unit area length Permanent Imposed actions actions (DL) (LL) Permanent axial actions (DL) Permanent bending moments E/W Permanent bending moments N/S Imposed axial actions (LL) Imposed bending moments E/W Imposed bending moments N/S On Level 4 1 2 3 4 5 Roof Precast edge beam Wall load Column Moment shears 12.96 2.8 0 2.8 0 0.6 0.25 7.8 3.2 7.68 21.5 0.0 1 0.0 0.0 11.52 32.3 0.0 0 0.0 0.0 Total this level 61.5 0 0 3.2 0 0 Total on Level 4 PA (DL) 61.5 IA (LL) 3.2 On Level 3 1 2 3 4 5 Floor Precast edge beam Wall load Column Moment shears 12.96 2.8 4.5 3 12.96 8.4 4 7.68 0 3 0 11.52 0.84 – 0.4 108.9 0 35.2 21.5 12 13.5 34.6 10.9 Total this level 189.3 12.0 35.2 51.8 0.0 3 0.0 0.0 – 5.2 10.3 46.7 10.3 3.0 PA (DL) 189.3 Total on Level 4 IA (LL)46.7 Notes: — Actions (loads) are in kN or kPa. All loads are unfactored. — Moments are in kN.m. Reinforced Concrete Design Handbook 6.3 The bending moments in the columns are the moments at the top or bottom face of the slab or beam at the beam/slab-column joint and not at the centre line of the slab or beam. In column design, it is important to correctly calculate the applied moments, particularly at the top of the building where columns can be relatively slender, vertical actions low but bending moments can still be large. When there are a large number of columns, it is usual to group them into a series of typical columns, in order to rationalise the design and to avoid detailed analysis of every column. Normally, at least one corner column, one edge column, one internal column and all non-typical columns should be designed at each floor level, keeping the size and details as uniform as possible. For example, if the cover (and axis distance indirectly) and concrete strength are determined by durability considerations for the external columns, then the designer should consider using the same size, cover and concrete strength with less reinforcement for internal columns, even though they could be smaller, use a lower strength concrete and could have less cover to the reinforcement. Since the volume of concrete in columns is generally small, the benefit of helping to avoid errors on site will far outweigh the cost of additional concrete. 6.3 Design Process The design process for columns will follow these general steps. Step 1 Assume an initial size of column, concrete strength and reinforcement See Section 6.2. Step 2 Assess the actions on the column These will be determined from the structural analysis of the building in accordance with one of the strength check/analysis procedures given in AS 3600 Section 2. These are usually carried out using a proprietary computer program or a column rundown for vertical actions. Note that the choice of method has a direct influence on the design procedure to be adopted in accordance with AS 3600 Clause 10.2. See Step 4. Columns are usually designed only for strength. Stability and serviceability are considered only for slender and unbraced columns. The strength action effects will be in accordance with Table 1.1 in Chapter 1 of this Handbook. Considering the entire possible axial actions and bending moment combinations for each loading case for a given cross section for each column at each floor, and manually checking the strength of the chosen column size and 6.4 Reinforced Concrete Design Handbook reinforcement configuration can be a tedious process. Frequently the load combination 1.2G + 1.5Q will be the critical design case. Table 6.2 shows an example of various load combinations that may have to be considered for the design of a particular column. Table 6.2 Load combinations for a particular column Load case Load combination 1 2 3 4 5 1.35 G 1.2 G + 1.5 Q 1.2 G + ycQ + Wu 0.9 G + Wu G + ycQ + Eu Axial load (kN) BM top BM bot (kN.m) (kN.m) 1776 1850 1650 1147 1550 10 45 50 45 45 14 55 65 60 60 Step 3 Assess the durability requirements, cover and fire ratings to determine the axis distance and the cover of the longitudinal bars from the column face This is done in accordance with the requirements of AS 3600 Sections 4 and 5. See also Chapter 3 of this Handbook. For example, an external column in a coastal area within 1 km of the sea would have a durability classification of B2 and would require 40-MPa concrete minimum (Special Class Concrete) and 45-mm cover to the fitments, ie 71-mm axis distance assuming 12-mm fitments and 28-mm main reinforcing bars. Suggest adopting a 75-mm axis distance. Next determine the required fire resistance period (FRP) from the BCA. Then determine the required axis distance (cover plus fitment plus half-bar diameter) for the FRP, eg from AS 3600 Table 5.6.3. Assuming the required FRP is 120 minutes, N *f /Nu = 0.7 and that the column is exposed on more than one side, a minimum column size of 350 mm and an axis distance of 57 mm is required. Finally, check whether durability (cover) or fireresistance (axis distance) will govern the axis distance to the main column bars from the outside face of the column. In this example, durability governs and the required axis distance is 75 mm. (However, if the column was inland in an arid area with an A1 Exposure Classification then the axis distance required for fire resistance would govern the distance to the column bars from the outside face of the column.) Step 4 Choose a design procedure based on AS 3600 Clause 10.2 Generally, a linear elastic analysis (Clause 2.2.2) will be used. Design using rigorous analysis, eg a non‑linear analysis (Clause 2.2.5) will be appropriate only in special circumstances, eg for long, slender columns, tapered columns and columns of other special types or shapes. This approach should be undertaken only after careful consideration, because of the specific and detailed requirements for these methods of analysis. Where the axial forces and bending moments are determined by a rigorous analysis in accordance with AS 3600 Clauses 6.5 and 6.6, a column shall be designed in accordance with Clauses 10.6 and 10.7 without further consideration of additional moments due to slenderness. n a short column, in accordance with AS 3600 Clauses 10.3, 10.6 and 10.7; or Clause 10.6.2.4 Balanced point If the bending moment in a column causes significant lateral deflection, the effective eccentricity of the axial load at mid-height is increased, increasing the moment, having an iterative effect. AS 3600 Clause 10.4 defines when a column is sufficiently slender for this to be taken into account. The design procedure applies an amplification factor to the moment acting on the column so that the short column moment-strength interaction design curves can be used. Moment Figure 6.1 Axial load vs moment diagram (after AS 3600 Figure 10.6.2.1) 0.9 0.85 Assuming a linear analysis is to be used, the general design procedure will be: n n Determine the unsupported length of the column, Lu. n Calculate the effective length, Le , in accordance with Clause 10.5.3, in both directions as required and calculate the slenderness ratio, Le /r. (For braced columns Le will be ≤ Lu and for unbraced columns Le will be > Lu .) 20 25 32 40 50 65 80 100 Concrete strength f 'c (MPa) Figure 6.2 Variation of a1 with f 'c for calculating squash load (Clause 10.6.2.2) n n n n Determine the distance of the longitudinal reinforcement from the face of the column based on durability and fire requirements. ·n Determine the minimum moment, 0.05 D N *. 0.7 0.65 Determine if the column is braced or unbraced. Determine the ultimate axial actions and the design moments at each end of the column about each axis, as required, and whether the column is in single or double curvature. 0.8 0.75 Flowchart 6.1 covers the general design of columns in uniaxial or biaxial bending in accordance with AS 3600. n Pure bending point Clause 8.1 a slender column, in accordance with AS 3600 Clauses 10.4 to 10.7. Where the axial actions (forces) and bending moments are determined by an elastic analysis incorporating secondary bending moments due to lateral joint displacements, as provided in AS 3600 Clause 6.3, a column shall be designed in accordance with Clauses 10.6 and 10.7. n Clause 10.6.2.5 Squash load factor, α 1 n Decompression point Clause 10.6.2.3 Axial load Normally, column design will be carried out where the axial actions (forces) and bending moments are determined by a linear elastic analysis. The column is then designed as either: Squash load point Clause 10.6.2.2 n n n Determine if the column is short, or slender. If the column is slender, determine the moment magnifiers including the buckling load depending on whether it is braced or unbraced. Determine the (magnified) moments and choose the larger moment at each end of the column. For each load case chosen, check that the applied axial actions and moment are less than the maximum allowed by the moment-strength interaction design curves calculated in accordance with AS 3600 for the chosen column dimensions, concrete strength and area and configuration of reinforcement. Iterate as required if the column is under-designed or significantly over-designed. Check the design about the other axis, if required, and comply with Clauses 10.6.3 and 10.6.4, if required, for bending about two principal axes. Check minimum and maximum reinforcement ratios along with the spacing of bars and fitments and detail the reinforcement as required in accordance with AS 3600 Clause 10.7. Reinforced Concrete Design Handbook 6.5 Flowchart 6.1 Design of columns in uniaxial or biaxial bending start Increase column size or properties no Is second order analysis to be used? AS 3600 Clause 6.3 Estimate column size and properties eg b, D, L, f 'c , p and cover/axis distance no Is relative dispacement at ends of column < L u / 250? AS 3600 Clause 6.3.1 Is column braced? AS 3600 Clause 10.1.3.1 yes Calculate L e for braced column AS 3600 Clause 10.5 no no Use AS 3600 Clause 6.4, 6.5, 6.7 and 6.8 as appropriate no yes Input Le = L u AS 3600 Clause 10.2.2 Is non-linear stress analysis to be used? AS 3600 Clause 6.6 Is linear analysis to be used? AS 3600 Clause 6.2 yes yes no Increase column size or properties Calculate L e for unbraced column AS 3600 Clause 10.5 Is L e / r ≤ 120? AS 3600 Clause 10.5.1 yes yes Calculate M * direct from analysis AS 3600 Clause 10.2.3 A B 6.6 Reinforced Concrete Design Handbook C D A B C Is column short? AS 3600 Clause 10.3.1 D Is column braced? AS 3600 Clause 10.1.3.1 no yes Is N *< 0.1 f 'c Ag? AS 3600 Clause 10.3.2 no Calculate Nc from AS 3600 Clause 10.4.4 yes Calculate moment magnifiers δ b and δ s AS 3600 Clause 10.4.3 Calculate Nc from AS 3600 Clause 10.4.4 yes no Calculate moment magnifier δ b AS 3600 Clause 10.4.2 no Is N * ≤ 0.75 φ Nuo? yes Are requirements of AS 3600 Clause 10.3.3 met? Calculate M * AS 3600 Clauses 10.1.2 and 10.4.1 no yes Is Is biaxial bending to be considered? AS 3600 Clause 10.6.2 yes ∝ ∝ [(M *x / φ Mux)] n + [(M *y / φ Muy)] n ≤ 1? no AS 3600 Clause 10.6.4 no yes no For each axis is M * / φ Mu ≤ 1? yes no Is design acceptable? yes Check reinforcement details AS 3600 Clause 10.7 and transmission of loads through floor system AS 3600 Clause 10.8 Note: φ Mu , φ Mub , φ Nuo values can be obtained from the design. φ Mu is the design strength in bending under the design axial force N *. Reinforced Concrete Design Handbook 6.7 Because of the many variables involved, design is usually best carried out using appropriate column design software or a suitable spreadsheet developed to AS 3600. 6.4 Basis of Charts 6.1 to 6.6 Load-moment interaction curve In its simplest form, the load-moment interaction curve for a column cross-section is constructed by calculating four points on the boundary of the curve and the connecting straight lines or curves between them. These points plot: the axial strength at zero moment on the vertical axis; the bending capacity at zero axial action on the horizontal axis; the point at which the neutral axis coincides with the outermost layer of tension reinforcement; and the point at which the tension reinforcement just begins to yield. Figure 6.1 shows a typical load-moment interaction curve for a column cross-section using the rectangular stress block. AS 3600 Clause 10.6.1 allows for the strength of column cross-sections to be calculated using stress‑strain relationships as given in Clauses 3.1.4 and 3.2.3. Note that a limit is placed on the maximum value of concrete stress. While the CEB 6.5 curves may be adequate for normal concrete, other stress-strain curves 6.6 may be more appropriate for high strength concrete and such curves can be used, provided they comply with AS 3600. The design booklet 6.7 contains a discussion on the design of columns and the principles of design. The squash load, Nuo, is a theoretical design point because AS 3600 requires a minimum moment of 0.05 D N * for any concrete column. It is calculated in accordance with AS 3600 Clause 10.6.2.2 where using the rectangular stress block given in AS 3600 Clause 10.6.2.5. For the transition from squash load to the decompression point, AS 3600 allows the section strength to be calculated using a linear relationship between the decompression point and the squash load. The transition from decompression point to the balanced point to the bending strength is set out in AS 3600 Clause 10.6.2.5. The design strength in bending, Mub, and the corresponding design strength in compression, Nub, are referred to as the balance point and can be determined as the locus of values for which kuo = 0.545 (with f = 0.6). The value of kuo is determined assuming the balance point is the point at which the steel yields at a strain of 0.0025 and the strain at the extreme compressive fibre of the concrete is 0.003. Because each load-moment strength interaction diagram depends on so many variables, hand calculations are generally not feasible. Either spreadsheets or column design software are used to develop a load-moment strength interaction diagram for each chosen column size and configuration. The calculated load-moment strength interaction diagram as shown in Figure 6.1 is then modified by applying the capacity reduction factor, f, to give a design load-moment strength interaction diagram for design from which the chosen column configuration can then be checked. The capacity reduction factor, f, varies in accordance with AS 3600 Table 2.2.2. 400 x 400 column 4% reo 3% reo 2% reo 1% reo Concrete alone Nuo = α 1 f 'c Ac + Asc fsy and Ac = the net area of concrete and The factor α 1 is to allow for shrinkage and the fact that the concrete will carry less load in the long term than it would when initially loaded. AS 3600 defines α 1 as α 1 = 1.0 − 0.003 f 'c with lower and upper limits of 0.72 and 0.85 respectively. The modification of 0.9 f 'c is included in the calculation of α 1. Figure 6.3 shows graphs of the squash load for a square column with various percentages of reinforcement and various concrete strengths. As can be seen from the graph, the addition of reinforcement has the greatest effect for low concrete strengths. The decompression point is calculated by taking the strain in the extreme compressive fibre as 0.003, the strain in the extreme tensile fibre as zero and 6.8 Reinforced Concrete Design Handbook 16000 14000 12000 10000 Squash load, Nuo (kN) Asc = the area of column reinforcement. 8000 6000 4000 2000 0 20 25 32 40 50 65 Concrete strength f 'c (MPa) Figure 6.3 Squash load vs concrete strength 80 100 Region where the design action effects of combined axial force and bending on a section require confinement to the core Design charts Preliminary design of circular columns can be made using the square colmn charts with some interpolation. The strengths were calculated based on equilibrium and strain-compatibility considerations consistent with the assumptions of AS 3600 Clause 10.6.1. φ Nuo 0.75 φ Nuo φ Nu Design axial force The design charts provided in this Chapter give the design strength of square reinforced concrete cross-sections in combined uniaxial bending and compression in one direction only and are for specific design parameters only. The strength reduction factor, f, is included. The square columns have equal reinforcement only in two faces and have an axis distance of 60 mm. 0.6 φ Mu, φ Nu φ 0.3 Agf 'c φ Mu, φ Nu Design moment φ Muo Figure 6.4 Confinement of the core (after AS 3600) 6.5 Axial shortening All concrete compression members shorten under axial load because of shrinkage and creep. Shortening for columns is typically about 1 mm per m in height but it can be calculated by various methods in accordance with AS 3600. This shortening must be considered in the design of taller buildings including for example: the support of the facades of multi-storey buildings to avoid non-loadbearing walls carrying load, in the detailing of lift guide rails in the services core, and the junction of the high-rise core, of a building with a low‑level podium. Further investigation is also required where a highly stressed column is adjacent to a much-lower-stressed column or wall. For example, an internal column located, say, 2 m from concrete core walls to reduce the floor span may result in significant differential deflections of the slab across the short span. There can also be problems if there is a change in the plan dimensions of floors in a tall building resulting in adjacent columns on the floor below the change carrying significantly different loads, eg one column might be supporting 20 storeys and the next, just a roof. In such a case the designer should try to even out the loads, eg by prestressing the lightly loaded column. Another solution is to provide separation joints between the low-rise section and a high-rise section. an increase in axial capacity but has limited effect on moment capacity. Research in recent years has shown that an increase in strength and ductility of HSC columns can be achieved by well-detailed lateral confinement reinforcement. Robustness and ductility are important for columns and extra attention is required in the detailing of the fitments in HSC columns. AS 3600 Clause 10.7.3 sets out the requirements for confinement reinforcement both in the special confinement areas and outside these areas as shown in Figure 6.4. Where the fitment spacing does not exceed the values set out in the Standard, AS 3600 Clause 10.7.3.4 provides a deemed-to-comply approach for the design of the core confinement for rectangular and circular sections in lieu of detailed calculations of the core confinement by rational methods. For further information on HSC, designers can refer to Foster 6.8, Mendis 6.9, and other papers 6.10. 6.7 Detailing Designers should refer to Chapter 12 of the Reinforcement Detailing Handbook 6.11 for further information on detailing of columns. Designers should: 6.6 High-strength concrete (HSC) For the purposes of this Handbook, high-strength concrete has been defined as a concrete where the concrete strength, f 'c, is greater than 50 MPa. This is consistent with AS 1379, where such concrete is required to be Special Class Concrete. High-strength concrete is becoming increasingly available and is used to increase the load capacity of columns or to reduce their dimensions. HSC is less ductile than normal concrete, while its use results in n n n Provide sufficient information on the drawings to build the columns including elevations, schedules, sections and details along with the column orientations and the required concrete strengths. Provide the north point on the schedules to match the plans and orientation of the column if the column has different reinforcement in different faces or is rectangular. Provide offset dimensions if columns are not concentric on grid lines. Reinforced Concrete Design Handbook 6.9 n n n n n n n n n Show any special detail such as at the footings and floors, column offsets, columns supported by transfer beams, columns terminating or changing size, column heads, cast-in bolts, etc. Always use the same bar size in one lift of a column to avoid errors in fixing on site. Generally provide the longitudinal reinforcement in one lift, ie one storey height as double-storey column bars in larger sizes may be difficult to restrain in the correct position above the formwork. Provide 2 sets of fitments at crank typically Crank bars 1 in 6 max. For circular columns no cranks required Additional fitments if spacing is greater than that specified Note: Starter bars to be in outside face uno ie bars from top to be cranked inside Terminate lower column bars if not required in column over Try to use larger and fewer bars rather than many small bars to reduce the number of fitments. Consider the beam/column intersection as discussed in Chapter 4 of this Handbook and the Reinforcement Detailing Handbook. Consider using the cranked section of the column bars at the top of the column when lapping column bars so the straight bar can go through a beam. An alternative is to crank the bars below the beam so that the straight section goes through the beam. This is not so critical for band beams or flat slabs. Figure 6.5 illustrates this issue. Check that the longitudinal reinforcement provided satisfies the requirements for minimum and maximum reinforcement given in AS 3600 Clause 10.7.1. Provide minimum bar diameter for fitments for rectangular columns and helices for circular columns in accordance with AS 3600 Table 10.7.4.3. Provide appropriate lapped splice lengths for the longitudinal column bars depending on whether the bars are in tension or compression. Construction joint Note: If f 'c of column is greater than 65 provide sets of fitments at 100 cts Sets of fitments as noted in schedule or in sections Figure 6.5 Cranking of longitudinal bars 2 sets of fitments 12 vertical bars 1 set of fitments 12 vertical bars 2 sets of fitments 8 vertical bars Figure 6.6 Restraint of longitudinal bars Clause 10.7.4.2 (iv) Clause 10.7.4.2 (iii) Clause 10.7.4.2 (ii) External fitment Internal fitment Designers should note that: n n n n While end-bearing splices are allowed by AS 3600, the thin sleeve splices to the bars are no longer generally available in Australia. Several types of thicker sleeve systems and a threaded coupler system are available for splicing bars in line. Column cages can often be fabricated away from the forms and lifted into place by a crane. Bundled bars cannot usually be prefabricated. It is usual to terminate the column reinforcement with cogs into the slab or beam at the top of the building for fixity. Consider drop-in bars prior to the column being cast, to allow easier alignment with beam or slab reinforcement. Fitments are provided to prevent the vertical column bars from buckling under compression loads, to resist shear and torsion, for ductility for earthquake actions if required and for HSC to confine the core. A number of design conditions therefore need to be satisfied in choosing the size and spacing of fitments. 6.10 Reinforced Concrete Design Handbook Clause 10.7.4.2 (i) Consecutive fitments alternated end to end along longitudinal axis Figure 6.7 Lateral restraint of longitudinal bars (after AS 3600) n For longitudinal bars in square or rectangular columns, all corner bars, all bars spaced further apart than 150 mm, and every alternate bar where the spacing is less than 150 mm have to be restrained in two directions (see AS 3600 Clause 10.7.4.2 ). Circular columns do not need this restraint provided a circular fitment or helical reinforcement (anchored in accordance with Clause 10.7.4.4) is used and the longitudinal bars are equally spaced around the circumference. See Figure 6.6 for some examples of fitments and restraint of vertical bars. n n n n To facilitate column construction an alternative method of assembly of internal column fitments is permitted by AS 3600 Clause 10.7.4.2. It allows internal fitments to include a 135° hook at one end with a 90° hook at the other, provided consecutive internal fitments are alternated end to end along the longitudinal reinforcement. This allows the internal fitments to be fixed on site. See Figure 6.7. Where bending moments from a floor system have to be transferred to a column and the column is not fully surrounded by a slab or beams of approximately equal depth, then lateral shear reinforcement of area Asv ≥ 0.35 b s /fsy.f in the form of fitments must be provided through the joint (Clause 10.7.4.5). This Clause will usually apply to all external columns. This requirement can sometimes cause difficulty with the fixing of the beam reinforcement. See Figures 6.8 and 6.9. Where there are large changes in column dimensions, column bars cannot usually be offset and separate starter bars will be required to be cast into the column below. See Figure 6.10. Seismic detailing, if required, in accordance with AS 3600 Appendix C. Crank bars 1 in 6 max. For circular columns no crank required Note: Starter bars to be in outside face uno ie bars from top to be cranked inside Note: If f 'c is greater than 65 provide sets of fitments at 100 cts Anchor bars from bottom beam/slab Sets of fitments as noted in schedule or sections Figure 6.10 Change in column size 6.8 General guidance The following points will assist in the design and inspection of columns for a particular project. n Closed fitments n Figure 6.8 Lateral shear reinforcing n n n Figure 6.9 Columns requiring lateral shear reinforcement Additional fitments if spacing is greater than that specified Construction joint n Construction joint Provide 2 sets of fitments at crank typically While AS 3600 permits the use of 200-mm x 200‑mm columns, a minimum dimension (for square, rectangular or circular columns) of 250 mm is recommended. Column dimensions are usually a multiple of 50 mm. The position and shape of columns is often dictated by architectural and spatial requirements, eg circular columns for appearance or rectangular columns in carparks and blade columns in apartment buildings. For circular columns, thin spiral galvanised metal or plastic is sometimes used as permanent formwork. It is important that cleaning out of the bottom of the column is achieved before placing of the formwork. Consider using HSC rather than increasing the reinforcement ratio or changing column sizes, subject to the transmission of axial forces through the floor. AS 3600 allows the use of HSC up to 100 MPa but concrete strengths higher than 50 MPa involve Special Class Concrete with the requirements of project control testing and extra detailing for confinement. Also check if the local concrete suppliers can supply the HSC concrete specified. Before specifying above 65-MPa concrete, an alternative verification path will be required to satisfy the requirements of the BCA until such time as AS 3600—2009 is called up in the BCA. On successively higher floors, reduce the concrete strength and or reinforcement rather than changing the column sizes, as column formwork is expensive. A uniform column size will also often be desirable for visual reasons, particularly with facade columns. Reinforced Concrete Design Handbook 6.11 M20 cast-in ferrule n 2x M20 grade bolts n SHS welded to pfc with packer pl PLAN Note: After installation site-measure position of ferrules etc before any shop details are drawn for approval Figure 6.11 Cast-in ferrules to columns n n n n n n Check that the layout of fitments will allow placing and compaction of the concrete and the forms to be cleaned out. Remember to check the transmission of axial forces through the floor (see AS 3600 Clause 10.8). Edge columns and corner column are often more critical for design than internal columns because of high bending moments. Avoid, wherever possible, casting in large services such as downpipes, sewer stacks, etc in columns because of future maintenance and durability problems. There is also the complication of getting the services in and out of the column and clashes with reinforcement, especially at the bottom of the column. This area of a column is usually highly stressed and there is the problem of assessing what the reduced strength of the column will be with a section of the column removed for a large service duct. The column below a slab is sometimes cast with the floor slab to minimise scaffolding and allow access from the floor formwork to cast the column. In this situation, columns should be poured at least 1–2 hours before the slab over, to allow for settlement of the concrete in the column to take place, before the concrete in the slab is placed. Avoid cast-in bolts and cleat plates projecting from the side of columns because of the complication with formwork (especially in stripping) and the difficulty of placing and compacting the concrete. Consider using cast-in ferrules or a cast-in plate, which allows some tolerance and onto which cleat plates and the like can be welded after the formwork is stripped. See Figure 6.11. 6.12 Reinforced Concrete Design Handbook n Where bolts are cast into the top of the column for the roof structure and the like, use a template to align the bolts correctly. Avoid the use of threaded reinforcement for this purpose because of the difficulty in accurately aligning the reinforcement and achieving the tolerances usually required for holding-down bolts. Consider using concrete spacer blocks when durability is an issue. Care is needed to ensure that voids do not occur under the blocks, so consideration may need to be given to the use of super plasticisers and smaller aggregates. For precast columns, be careful with choice of dowel ducts, which are commonly used to provide sleeves into which the projecting dowel bars from the concrete below are grouted. Generally, dowel ducts should be of thin galvanised steel 2–3 times the size of the dowel bar. Smooth-faced plastic dowel ducts should be used only where the joint is pinned and there is no tension. Charts 6.1 to 6.7 These charts have been developed on the following assumptions: n n n n Equal reinforcement is in only two opposite faces, resisting bending. An axis distance of 60 mm from the face of concrete to the centre of the longitudinal bar, ie a cover of about 30 to 40 mm depending on the fitment and column bar size. Bending moments about the one axis only. Concrete strengths in the range of 25 to 50 MPa. Chart 6.1 300 x 300 Columns pages 6.14–6.15 Chart 6.2 400 x 400 Columns pages 6.16–6.17 Chart 6.3 500 x 500 Columns pages 6.18–6.19 Chart 6.4 600 x 600 Columns pages 6.20–6.21 Chart 6.5 700 x 700 Columns pages 6.22–6.23 Chart 6.6 800 x 800 Columns pages 6.24–6.25 Reinforced Concrete Design Handbook 6.13 chart 6.1 300 x 300 Columns 4000 Cover depends on fitment size 60 mm y 3500 x x M y 3000 Depth = d 60 mm axis distance Width = b SQUARE COLUMNS 2500 f 'c = 25 MPa 2000 300 column 1% 300 column 2% 300 column 3% 300 column 4% Minimum moment Compressive force (kN) 1500 1000 500 0 0 25 50 75 100 125 150 Moment (kN.m) 4000 Cover depends on fitment size 60 mm y 3500 x x M y 3000 Width = b Depth = d 60 mm axis distance SQUARE COLUMNS 2500 f 'c = 32 MPa 2000 300 column 1% 300 column 2% 300 column 3% 300 column 4% Minimum moment Compressive force (kN) 1500 1000 500 0 0 25 50 Moment (kN.m) 6.14 Reinforced Concrete Design Handbook 75 100 125 150 chart 6.1 (continued) 300 x 300 Columns 4000 Cover depends on fitment size 60 mm y 3500 x x M y 3000 Depth = d 60 mm axis distance Width = b SQUARE COLUMNS 2500 f 'c = 40 MPa 2000 300 column 1% 300 column 2% 300 column 3% 300 column 4% Minimum moment Compressive force (kN) 1500 1000 500 0 0 25 50 75 100 125 150 Moment (kN.m) 4000 Cover depends on fitment size 60 mm y 3500 x x M y 3000 Width = b Depth = d 60 mm axis distance SQUARE COLUMNS 2500 f 'c = 50 MPa 2000 300 column 1% 300 column 2% 300 column 3% 300 column 4% Minimum moment Compressive force (kN) 1500 1000 500 0 0 25 50 75 100 125 150 Moment (kN.m) Reinforced Concrete Design Handbook 6.15 chart 6.2 400 x 400 Columns 6000 Cover depends on fitment size 60 mm y x 5000 x M y 60 mm axis distance Width = b 4000 Depth = d SQUARE COLUMNS f 'c = 25 MPa 3000 400 column 1% 400 column 2% 400 column 3% 400 column 4% Minimum moment Compressive force (kN) 2000 1000 0 0 50 100 150 200 250 300 350 400 450 Moment (kN.m) 6000 Cover depends on fitment size 60 mm y x 5000 x M y Width = b 4000 Depth = d 60 mm axis distance SQUARE COLUMNS f 'c = 32 MPa 3000 400 column 1% 400 column 2% 400 column 3% 400 column 4% Minimum moment Compressive force (kN) 2000 1000 0 0 50 100 150 200 Moment (kN.m) 6.16 Reinforced Concrete Design Handbook 250 300 350 400 450 chart 6.2 (continued) 400 x 400 Columns 6000 Cover depends on fitment size 60 mm y x 5000 x M y 60 mm axis distance Width = b 4000 Depth = d SQUARE COLUMNS f 'c = 40 MPa 3000 400 column 1% 400 column 2% 400 column 3% 400 column 4% Minimum moment Compressive force (kN) 2000 1000 0 0 50 100 150 200 250 300 350 400 450 Moment (kN.m) 6000 Cover depends on fitment size 60 mm y x 5000 x M y Width = b 4000 Depth = d 60 mm axis distance SQUARE COLUMNS f 'c = 50 MPa 3000 400 column 1% 400 column 2% 400 column 3% 400 column 4% Minimum moment Compressive force (kN) 2000 1000 0 0 50 100 150 200 250 300 350 400 450 Moment (kN.m) Reinforced Concrete Design Handbook 6.17 chart 6.3 500 x 500 Columns 10000 Cover depends on fitment size 60 mm y 9000 x 8000 x M y 60 mm axis distance Width = b 7000 Depth = d SQUARE COLUMNS 6000 f 'c = 25 MPa 5000 500 column 1% 500 column 2% 500 column 3% 500 column 4% Minimum moment Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 500 600 700 800 900 1000 Moment (kN.m) 10000 Cover depends on fitment size 60 mm y 9000 x 8000 x M y Width = b 7000 Depth = d 60 mm axis distance SQUARE COLUMNS 6000 f 'c = 32 MPa 5000 500 column 1% 500 column 2% 500 column 3% 500 column 4% Minimum moment Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 Moment (kN.m) 6.18 Reinforced Concrete Design Handbook 500 600 700 800 900 1000 chart 6.3 (continued) 500 x 500 Columns 10000 Cover depends on fitment size 60 mm y 9000 x 8000 x M y 60 mm axis distance Width = b 7000 Depth = d SQUARE COLUMNS 6000 f 'c = 40 MPa 5000 500 column 1% 500 column 2% 500 column 3% 500 column 4% Minimum moment Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 500 600 700 800 900 1000 Moment (kN.m) 10000 Cover depends on fitment size 60 mm y 9000 x 8000 x M y Width = b 7000 Depth = d 60 mm axis distance SQUARE COLUMNS 6000 f 'c = 50 MPa 5000 500 column 1% 500 column 2% 500 column 3% 500 column 4% Minimum moment Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 500 600 700 800 900 1000 Moment (kN.m) Reinforced Concrete Design Handbook 6.19 chart 6.4 600 x 600 Columns 14000 Cover depends on fitment size 60 mm y 13000 x 12000 x M 11000 y 60 mm axis distance Width = b 10000 Depth = d SQUARE COLUMNS 9000 f 'c = 25 MPa 8000 7000 600 column 1% 600 column 2% 600 column 3% 600 column 4% Minimum moment 6000 Compressive force (kN) 5000 4000 3000 2000 1000 0 0 200 400 600 800 1000 1200 1400 1600 1800 Moment (kN.m) 14000 Cover depends on fitment size 60 mm y 13000 x 12000 11000 x M y 10000 Width = b Depth = d 60 mm axis distance 9000 SQUARE COLUMNS 8000 f 'c = 32 MPa 7000 600 column 1% 600 column 2% 600 column 3% 600 column 4% Minimum moment 6000 Compressive force (kN) 5000 4000 3000 2000 1000 0 0 200 400 600 800 Moment (kN.m) 6.20 Reinforced Concrete Design Handbook 1000 1200 1400 1600 1800 chart 6.4 (continued) 600 x 600 Columns 14000 Cover depends on fitment size 60 mm y 13000 x 12000 x M 11000 y 60 mm axis distance Width = b 10000 Depth = d SQUARE COLUMNS 9000 f 'c = 40 MPa 8000 7000 600 column 1% 600 column 2% 600 column 3% 600 column 4% Minimum moment 6000 Compressive force (kN) 5000 4000 3000 2000 1000 0 0 200 400 600 800 1000 1200 1400 1600 1800 Moment (kN.m) 14000 Cover depends on fitment size 60 mm y 13000 x 12000 11000 x M y 10000 Width = b Depth = d 60 mm axis distance 9000 SQUARE COLUMNS 8000 f 'c = 50 MPa 7000 600 column 1% 600 column 2% 600 column 3% 600 column 4% Minimum moment 6000 Compressive force (kN) 5000 4000 3000 2000 1000 0 0 200 400 600 800 1000 1200 1400 1600 1800 Moment (kN.m) Reinforced Concrete Design Handbook 6.21 chart 6.5 700 x 700 Columns 19000 Cover depends on fitment size 17000 x 16000 15000 x M y 14000 Depth = d 60 mm axis distance Width = b 13000 SQUARE COLUMNS 12000 f 'c = 25 MPa 11000 10000 9000 700 column 1% 700 column 2% 700 column 3% 700 column 4% Minimum moment 8000 7000 6000 Compressive force (kN) 60 mm y 18000 5000 4000 3000 2000 1000 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 Moment (kN.m) 19000 Cover depends on fitment size 18000 17000 x 16000 15000 Width = b 13000 Depth = d 60 mm axis distance SQUARE COLUMNS 12000 f 'c = 32 MPa 11000 10000 9000 700 column 1% 700 column 2% 700 column 3% 700 column 4% Minimum moment 8000 7000 6000 Compressive force (kN) x M y 14000 5000 4000 3000 2000 1000 0 60 mm y 0 200 400 600 800 1000 1200 Moment (kN.m) 6.22 Reinforced Concrete Design Handbook 1400 1600 1800 2000 2200 2400 2600 2800 chart 6.5 (continued) 700 x 700 Columns 19000 Cover depends on fitment size 17000 x 16000 15000 x M y 14000 Depth = d 60 mm axis distance Width = b 13000 SQUARE COLUMNS 12000 f 'c = 40 MPa 11000 10000 9000 700 column 1% 700 column 2% 700 column 3% 700 column 4% Minimum moment 8000 7000 6000 Compressive force (kN) 60 mm y 18000 5000 4000 3000 2000 1000 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 Moment (kN.m) 19000 Cover depends on fitment size 18000 17000 x 16000 15000 x M y 14000 Width = b 13000 Depth = d 60 mm axis distance SQUARE COLUMNS 12000 f 'c = 50 MPa 11000 10000 9000 700 column 1% 700 column 2% 700 column 3% 700 column 4% Minimum moment 8000 7000 6000 Compressive force (kN) 60 mm y 5000 4000 3000 2000 1000 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 Moment (kN.m) Reinforced Concrete Design Handbook 6.23 chart 6.6 800 x 800 Columns 24000 Cover depends on fitment size 60 mm y 22000 x 20000 x M y 18000 Depth = d 60 mm axis distance Width = b 16000 SQUARE COLUMNS 14000 f 'c = 25 MPa 12000 800 column 1% 800 column 2% 800 column 3% 800 column 4% Minimum moment 10000 Compressive force (kN) 8000 6000 4000 2000 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 Moment (kN.m) 24000 Cover depends on fitment size 60 mm y 22000 x 20000 x M y 18000 Width = b Depth = d 60 mm axis distance 16000 SQUARE COLUMNS 14000 f 'c = 32 MPa 12000 800 column 1% 800 column 2% 800 column 3% 800 column 4% Minimum moment 10000 Compressive force (kN) 8000 6000 4000 2000 0 0 500 1000 1500 2000 Moment (kN.m) 6.24 Reinforced Concrete Design Handbook 2500 3000 3500 4000 4500 chart 6.6 (continued) 800 x 800 Columns 24000 Cover depends on fitment size 60 mm y 22000 x 20000 x M y 18000 Depth = d 60 mm axis distance Width = b 16000 SQUARE COLUMNS 14000 f 'c = 40 MPa 12000 800 column 1% 800 column 2% 800 column 3% 800 column 4% Minimum moment 10000 Compressive force (kN) 8000 6000 4000 2000 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 Moment (kN.m) 24000 Cover depends on fitment size 60 mm y 22000 x 20000 x M y 18000 Width = b Depth = d 60 mm axis distance 16000 SQUARE COLUMNS 14000 f 'c = 50 MPa 12000 800 column 1% 800 column 2% 800 column 3% 800 column 4% Minimum moment 10000 Compressive force (kN) 8000 6000 4000 2000 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 Moment (kN.m) Reinforced Concrete Design Handbook 6.25 References 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.26 Precast Concrete Handbook 2nd Ed, National Precast Concrete Association Australia and Concrete Institute of Australia, 2009. AS 3600 Concrete structures Standards Australia, 2009. Warner RF, Rangan BV, Hall AS and Faulkes KA Concrete structures, Longman, 1998. Warner RF, Foster SJ and Kilpatrick AE Reinforced Concrete Basics 2nd Ed, Pearson, 2010. CEB, Commission IVc Deformability of Concrete Structures – Basic Assumptions Bulletin D'Information No. 90, 1973. Bridge R and Wheeler A 'Advanced Design of Columns' One Steel September 2001. Guide to Reinforced Concrete Design One Steel Design Booklet RCB-3.1(1) Cross-section Strength of Columns – Part 1 AS 3600 Design, August 2000. Foster SJ Design and Detailing of High Strength Concrete Columns, University of NSW, 1999. Mendis P Design of High-Strength Concrete Members, Engineers Australia, 2001. National Seminar Series on AS 3600—2009, Engineers Australia, Lecture 3, 2009. Reinforcement Detailing Handbook (Z06) 2nd Ed, Concrete Institute of Australia, 2010. Reinforced Concrete Design Handbook Chapter 7 Walls 7.1 Wall types Concrete walls can be categorised in terms of their method of construction, ie: n Insitu n Tilt-up n Precast. Insitu concrete walls require extensive formwork and take longer to build than tilt-up or precast. In recent years there has therefore been a considerable shift to the use of tilt-up or precast concrete walls where possible. For further information, designers should refer to Guide to Tilt-up Design and Construction 7.1 for tilt-up walls and to Chapter 2 of the Precast Concrete Handbook 7.2 for precast walls. Tilt-up or precast walls may need to be temporally braced in accordance with AS 3850 Tilt-up concrete construction 7.3, unless they can be fixed in place top and bottom at the time of erection. There are also statutory requirements in some States and Territories and there is a National Code of Practice – for Precast, Tilt-up and Concrete Elements in Building Construction 7.4 for the design and erection of tilt-up and precast concrete walls. This document requires the design for erection of tilt-up and precast concrete walls to be carried out by an Erection Design Engineer. Walls can also be categorised in terms of their form and function within a building, eg: n n n n 7.2 General Walls are thin vertical elements, commonly defined as where the length is four times or greater than their thickness. This definition is implied in AS 36007.5 Clause 5.6.2(b) relating to fire resistance but is not mentioned in AS 3600 Section 11. Depending on their function, walls can act like a column supporting vertical loads, a slab resisting horizontal loads, a cantilever beam in flexure (ie a shear wall), a deep beam, or sometimes a combination of these. The design of walls is set out in AS 3600 Section 11 and is in line with the rules adopted for columns in Section 10. Generally, their design, when carrying large vertical loads does not differ significantly from the design of columns, in that axial loads and moments about each axis need to be assessed and allowed for. Walls resisting lateral loads perpendicular to their face are usually designed as slabs. However, AS 3600 Section 11 gives design rules for only a limited range of walls. Where walls are outside this range, the designer can adopt any appropriate rational design method as permitted under the Building Code of Australia (BCA)7.6. Flowchart 7.1 can be used to determine whether or not Section 11 is applicable. AS 3600 Section 11 applies to the following walls: (a) Braced walls (as defined in AS 3600 Clause 11.3) that are subject to in-plane load effects which have to be designed in accordance with AS 3600 Clauses 11.2 to 11.7. (b) Braced walls that are subject to simultaneous in‑plane and out-of-plane load effects and unbraced walls which have to be designed in accordance with Section 9 for slabs and Section 10 for columns as appropriate. Except that where the stress at the mid-height section of a wall due to factored in-plane bending Basement retaining walls including cantilever retaining walls Core walls (to lift shafts, service ducts and stairs, etc) External and internal walls to single-storey and low‑rise buildings Cladding wall panels to the facades of multi-storey buildings n Internal walls in multi-storey buildings n Walls for housing n Other, eg curved walls. Comments on these various types are given below. Insitu and precast concrete walls Reinforced Concrete Design Handbook 7.1 Flowchart 7.1 Design of walls AS 3600 Section 11 Is wall planar? no Outside AS 3600 Section 11 yes Is wall braced? (Clause 11.3) no Does the stress at mid-height section due to in-plane bending and axial forces exceed the lesser of 0.03 f 'c and 2 MPa? yes Is wall subject to out-of-plane load effects? no yes no yes no Design wall using Sections 9 and 10 as appropriate Is wall subject to only in-plane vertical forces? Is Hwe / tw ≤ 50? alternative May design as slab using Section 9 providing 2nd order deflections are considered in calculation of bending moments yes no Is Hwe / tw ≤ 30? Is any horizontal cross-section of the wall subject to tension over part of the section? no yes 7.2 Reinforced Concrete Design Handbook May design using simplified method in Clause 11.5 no Is wall reinforced on both faces? no Design for in-plane bending in accordance with Section 8 and horizontal shear in accordance with Clause 11.6 or for both in accordance with Section 12 if appropriate yes Outside scope of Section 11 yes Design as column using Section 10 (Note Clause 11.7.4 may override Clause 10.7.4 ) yes and axial forces does not exceed the lesser of 0.03 f 'c and 2 MPa, the wall may be designed as a slab in accordance with Section 9, provided: (i) second-order deflections due to in-plane loads and long-term effects are considered in the calculation of bending moments; and (ii) the ratio of effective height to thickness does not exceed 50. Designers should be aware that the ratio of effective height to thickness may be restricted by design for fire as set out in AS 3600 Section 5. For off-form finishes to walls, designers should refer to the Guide to Off-form Concrete Finishes 7.7. Designers are reminded that AS 1170.4 7.8 Clause 5.2.3 requires that stiff components of a building such as concrete walls need to be considered as part of the seismic-force-resisting system and designed accordingly, or separated from all structural elements such that no interaction takes place as the structure undergoes deflections due to the earthquake effects determined in accordance with AS 1170.4. The use of high-strength concrete in walls is not mentioned in Section 11 except indirectly in Clause 11.7.4 relating to the restraint of vertical reinforcement. Designers should refer to AS 3600 Section 10 and texts such as Design of High-Strength Concrete Members 7.9 for further information where such concrete is used. Concrete walls behave in different ways depending on whether they are subject to in-plane or out-of-plane bending. They can be subject to a combination of actions including: n n Sliding (shear) n Overturning n Bending. The possibility of any of these occurring can be minimised by increasing the vertical load on the wall. For low-rise walls this may be done by increasing the dead load, eg by increasing the thickness of the wall. AS 3600 Appendix C covers specific detailing requirements for earthquake loads for ductile shear walls, including boundary elements. AS 3600 Clause 11.2.1(b) requires that when the horizontal forces in the plane of the wall as shown in Figure 7.1 cause tension over part of the section, then such walls be designed as beams in flexure in accordance AS 3600 Section 8 and that for in‑plane shear they be designed in accordance with Clause 11.6. Alternatively, they can be designed for in‑plane bending and shear in accordance with AS 3600 Section 12, if appropriate. Concrete walls have good strength, fire resistance, acoustic and thermal properties. Typically, two- to four‑hour fire resistance levels can be met by walls from 150–220 and 230–350 mm thick with 25- and 55-mm axis distances respectively. The varying thickness of wall depends on the minimum thickness for insulation. More particularly it depends on the 'structural adequacy', whether the wall is exposed on one or two sides and the ratio of Nf* / Nu as set out in AS 3600 Table 5.7.2. vertical loads when the walls are commonly known as loadbearing walls (walls supporting their own weight only such as facade cladding walls are known as non-loadbearing walls); loads perpendicular to the face of the wall (out-ofplane bending), principally wind and earthquake loads and sometimes other lateral loads such as soil, surcharge and hydrostatic loads; th ng Le Height n n Thickness horizontal loads in the plane of the wall (in-plane bending and shear) when the walls are commonly known as shear walls resisting the lateral loads, although seldom is shear the critical design case. Shear walls that act as vertical cantilever elements, are able to resist large lateral loads such as wind and seismic forces in buildings and usually also support vertical loads. They are an efficient way to resist horizontal loads and generally provide lateral strength much more economically than a framed structure using flexure in columns and beams or slabs. Shear walls have three basic failure modes: Tension Compression Figure 7.1 Loads on a shear wall Reinforced Concrete Design Handbook 7.3 It should be understood that two-sided exposure refers to walls within a single fire compartment which can be surrounded by fire. A wall forming part of the boundary of a fire compartment is deemed to be exposed only from one direction at a time. Exposure from the other direction implies that the fire has breached the compartment boundary and the wall has failed. However, if a wall projects beyond the facade or compartment boundary and is exposed on three or more faces then it is logical to assume that the wall is exposed on two opposite sides. However, all concrete walls have to comply with the requirements for durability and fire resistance as set out in AS 3600. Nevertheless, there are a large number of walls for which the BCA does not specify a Fire Resistance Level (FRL), eg internal walls in Type C construction. 7.3 Basement and retaining walls Concrete basement retaining walls (generally used to retain soil at the base of a building) can be either cantilever walls or, more often, walls spanning vertically between supports such as the basement floor and lower or ground floor. They sometimes span horizontally between columns, walls, abutments or similar. They can be constructed either of insitu or precast concrete. For embedded retaining walls, piling systems or diaphragm walls constructed from the top of the ground can be used to form a retaining wall and then the soil from the basement excavated from within an enclosed area. Designers can refer to the ICE, Specification for Piling and Embedded Retaining Walls 7.10 for further information. Retaining walls resist lateral actions due to soil, water, surcharge loads, etc. When they resist soil loads, their design should be in accordance with AS 46787.11 and should be based on design lateral pressures provided by a geotechnical investigation with appropriate geotechnical advice They may also have to prevent the ingress of water, which will require consideration of hydrostatic pressures, tanking, drainage, sealing of joints, the risk of damage from salts in the ground, etc. The main reinforcement in cantilever walls retaining soil will be located in the tension face. For walls thicker than 175 mm, a layer of mesh or grid of small bars is normally provided in the other face for crack control. Retaining walls spanning horizontally are usually designed as pinned each end. Retaining walls spanning vertically can be designed as 'pinned' or 'fixed' at both the top and bottom, giving a number of design options. Where spans are not too great, it is common to design them as pinned top and bottom. These walls can also be loadbearing, so there may be a number of different design criteria. 7.4 Reinforced Concrete Design Handbook Suspended floor Precast wall units can act as formwork to beam face Wall units to be braced off the basement slab until the suspended slab is cast Precast wall units spanning between footing beam and upper floor beam Temporary braces Tanking to back face of wall Granular backfill with drainage system Basement slab-on-ground Grouted dowel connection to footing Figure 7.2 Precast retaining walls (after the Precast Concrete Handbook) Insitu concrete walls should be constructed with vertical construction joints at 4- to 6-m spacing along the wall. Constructing longer sections of walls may result in thermal cracking/shrinkage cracking at the bottom of the wall with cracks tapering up the wall, due to restraint by the footing or slab under. Retaining walls that span vertically between restraints are typically supported on a strip footing, slab edge, footing beam or similar. Restraint fixing to the top of the wall is usually by cast-in bars to the slab, or similar, as shown in Figure 7.2 for a precast wall. In cantilever retaining walls and those retaining walls spanning vertically, large horizontal forces (reactions) will usually occur at the bottom of the walls, which the horizontal joint may be unable to resist. A shear key or rebate should be provided to give adequate restraint at the base of retaining walls subject to large lateral forces. 7.4 Core walls (to lift shafts, service ducts and stairs) In multi-storey buildings the core walls are usually loadbearing walls and shear walls and are usually insitu concrete because of the need to transmit tension and compression forces down the building and then to the footings. For low-rise buildings, precast walls are sometimes used. Precast walls, however, usually do not have the same structural capacity as insitu walls even when expensive and time-consuming site connections are used. Core walls generally are loadbearing and usually function as shear walls and can be coupled to form a group of walls in accordance with AS 3600 Clause 11.2.2. They are usually designed in accordance with AS 3600 Clause 11.2. When there are number of shear walls, it is common to assume that the concrete floor or roof acts as a rigid element or diaphragm for loads in the plane of the floor, and that the lateral loads on the building are distributed to each shear wall in proportion to its rigidity, based on gross section properties. depth is usually limited for architectural reasons. The wall curvatures are also altered from that of a cantilever wall because of the frame action developed, and the combined walls and coupling beams will require detailed analysis. These matters are discussed in more detail in Chapter 5 of the Precast Concrete Handbook 7.2. Insitu and precast concrete walls A series of insitu concrete core walls including both stair and lift shafts, constructed using climbing formwork before the floors below are cast. 7.5 External walls to single-storey and other low-rise buildings Precast or tilt-up concrete panels are commonly used for external walls of single-storey and low-rise industrial and commercial buildings. They can be either cladding panels or, more commonly, loadbearing panels combined to form a braced box where the roof or floor acts as a diaphragm as shown in Figures 7.3 and 7.4 respectively. They are generally designed as braced walls carrying the vertical and in-plane horizontal loads in accordance with Clause 11.1(b) and as slabs carrying the out-of-plane forces in accordance with AS 3600 Section 9. Each bay braced to create redundancy in structure Precast wall panels The erection of precast concrete core walls for a lift shaft and stair core. Note the bracing of the panels with a minimum of two braces per panel and the precast panels would sit over projecting starter bars which fit into grout ducts in the precast walls, and which are later grouted. When resisting the lateral loads on the building as individual walls, they are often considered as a cantilever beam from the base of the building, which adds flexural stresses to the calculated vertical stresses. This may result in tension in the wall, in which case the wall is designed as a beam in accordance with AS 3600 Section 8. In addition, in-plane shear forces will need to be checked in accordance with Clause 11.6. When two or more walls, or a wall with a number of large openings, are joined together by a coupling or header beam or similar, the effect of combining walls or sections of a wall is to increase the stiffness of the combined wall system – as a result of transfer of shear and flexure through the beam (commonly known as a tie or header beam). This tie beam must be designed for these actions and the shear at the interconnected vertical edges of the walls must be checked. These coupling beams are often heavily reinforced, as their Wind load acting on side wall of building Roof bracing designed to transfer wind load to end walls End walls act as shear walls Figure 7.3 Single-storey concrete panel building Figure 7.4 A two-storey precast concrete panel building (after Hughes7.12) Reinforced Concrete Design Handbook 7.5 A discussion on the design considerations for a single‑storey building can be found in Briefing 087.13. It notes that the roof sheeting and purlins should not be used as part of the bracing system and redundancy in the bracing system is strongly recommended. In general, the effective-height-to-thickness ratio for walls is limited to ≤ 40 (see AS 3600 Clause 5.7.3). However, this restriction is waived where the top of the wall is supported by a member not required to have a FRL, eg concrete panel walls used as cladding for steel-portal-framed buildings or walls for a concrete panel building with a steel-framed roof. For preliminary sizing, an effective-height-to-thickness ratio of up to 50 may be used. Non-loadbearing walls are subject to lateral loads usually perpendicular to the face of the wall, principally wind and earthquake loads and designers should refer to the Precast Concrete Handbook (Chapters 2 and 7) for further information on the design of such precast panels. In addition to lateral loads perpendicular to the face, loadbearing panels need to be designed for vertical and lateral loads in the plane of the wall as applicable. The key design issues for precast concrete facades are appearance, the correct and adequate design and inspection, durability, waterproofing, building movements, and buildability. Coverage of this topic can be found in Woodside 7.14. A minimum practical thickness of external walls is about 125 mm but, more typically, 150 mm or greater thicknesses are used. For walls up to 170–200 mm one layer of reinforcement placed centrally is generally used; above that thickness, two layers are used, one layer in each face. Designers need to exercise caution when considering cantilevering heavy items such as large awnings off concrete walls that have only a central layer of reinforcement because of their limited bending capacity. Single-storey loadbearing and cladding walls are usually supported on a strip footing, a slab edge, beam or similar. Tilt-up or precast walls have to be restrained at the base at the time of erection and usually temporally braced until connected to the structure. It is also important not to provide a moment connection between the steel roof beams and concrete walls for buildings because of the wall's limited bending capacity about its weak axis. Precast cladding wall panels 7.6 Precast external wall panels in multi-storey buildings 7.7 Internal walls in multi-storey buildings Precast external wall panels in multi-storey buildings may be either non-loadbearing, ie serve only as claddings or, more commonly, loadbearing, ie serve as cladding and have a structural role or roles. The design, detailing and layout of panels will involve considerable interaction between the structural engineer, architect and precaster. In multi-storey buildings, internal concrete walls are typically either precast or insitu. They often have a fire‑separation role, carry vertical loads and usually act as shear walls. They are typically supported on a floor‑by-floor basis. Non-loadbearing panels are typically supported on a floor-by-floor basis using corbels and restraint brackets. A suitable horizontal joint (15–20 mm) must be provided at each floor level to accommodate both manufacturing and construction tolerances and for axial shortening of the building if it is a concrete frame. The detailing of this joint is very important to allow for these tolerances and movement. 7.6 Reinforced Concrete Design Handbook These walls are usually designed for vertical loads in accordance with AS 3600 Clause 11.2.1, using either Clause 11.5, the simplified design method, or Section 10 Column Design. However, if they are acting as shear walls they may need to be checked for bending in accordance with AS 3600 Section 8 if the wall is in tension and designed for in-plane shear forces in accordance with AS 3600 Clause 11.6. Tilt-up or precast walls are typically supported on a concrete floor and/or edge beams at ground level, while the concrete slab-on-ground will often provide a suitable element to temporarily brace the tilt-up or precast panels during erection. Designers should note that most temporary brace fixings used to fix the diagonal braces to concrete slab-on-ground are heavy-duty drilled load controlled, torquesetting expansion anchors which commonly require a minimum slab thickness of 125 mm whereas slabs‑on‑ground are commonly only 100 mm thick when designed in accordance with AS 28707.16. Internal precast walls The internal precast walls shown above include a column at one end of each wall panel. They are loadbearing walls supporting precast flooring. They also act as shear walls carrying in-plane horizontal loads from the floors over into the floor below and have vertical starter bars from the floor into dowel ducts in the wall which are subsequently grouted. The columns sections are used to support the precast concrete beams on either side of the walls which in turn support precast floor units. Note how the precast walls are braced with two diagonal braces per panel until the floor over is cast. The main structural design issue is how the concrete walls are laterally restrained by either concrete walls perpendicular to them or by the floor or roof, acting as a diaphragm. Light timber-framed floors and roofs are unlikely to be adequate to act as a diaphragm for the restraint of heavy concrete wall panels under lateral loads such as wind or to provide restraint in the event of fire and/or earthquake. 7.9 Walls Designed as Columns In Section 11 of AS 3600, the floor-to-floor height of the wall is defined as Hw. However, in Section 10 of AS 3600 (used when designing walls as columns), the effective height of a column (wall) is referred to as an effective length, Le. In Table 7.1 and in Charts 7.1–7.5, the effective length, Le , is in fact the effective height of the wall and not the length of the wall in plan. Precast walls for a garage and house extension 7.8 Walls for houses Concrete is increasingly being used for walls in both individual and modular housing. Designers should refer to The Concrete Panel Homes Handbook 7.15 for further information. These walls are usually tilt-up or precast walls and one- or two-storey in height. In traditional houses, the structural engineer would typically design only the concrete footings, concrete slab-on-ground, suspended concrete floors, some wall bracing for lateral loads and long-span beams as required. However, when tilt-up or precast concrete walls are used, full design for all the structure by the structural engineer will be required to ensure adequate restraint of the concrete walls both in the temporary and final conditions. The charts in this chapter are based on the same assumptions as the charts in Chapter 6. They assume the wall is braced and short, ie L /r ≤ 25. For a slender wall, a moment magnifier will need to be applied in accordance with Clause 10.4.2 for slender (columns) walls. As can be seen from Table 7.1, walls 200–300 mm thick are likely to be slender. The charts are for walls reinforced on two faces for f 'c = 25, 32, 40 and 50 MPa with 50-mm axis distance which should provide a FRR of 120 minutes. Because they are relatively thin, such walls cannot resist high bending moments about their smaller dimension. The design procedure for walls when acting as columns will be: n Assess the actions on the wall. n Assume an initial size of wall and concrete strength. n n Assess the durability requirements, cover and FRR to determine the axis distance and the distance of the vertical bars from the wall face. Choose a design procedure based on Clause 11.1 of AS 3600 and the Flowchart 7.1. Reinforced Concrete Design Handbook 7.7 Table 7.1 Slenderness ratios for braced walls when designed as columns Thickness of braced wall (mm) Maximum Effective length* Le height Hw (mm) Factor k = 0.7 Factor k = 0.85 Factor k = 1.0 Radius of gyration short wall Le / r ≤ 25 200 300 400 500 60 90 120 150 1500 2250 3000 3750 2143 3214 4286 5357 1765 2647 3529 4412 1500 2250 3000 3750 Slender wall Le / r ≤ 120 200 300 400 500 60 90 120 150 7200 10 800 14 400 18 000 10 286 15 429 20 571 25 714 8471 12 706 16 941 21 176 7200 10 800 14 400 18 000 * Effective length factor, k, from AS 3600 Figure 10.5.3(A) Table 7.2 Reinforcement area, Ast , for crack control (mm2/m) Thickness of wall (mm) Ast 100 125 150 175 200 225 250 275 300 325 350 375 400 425 450 475 500 1.5 t w 150 188 225 263 300 338 375 413 450 488 525 563 600 638 675 713 750 2.5 t w 250 313 375 438 500 563 625 688 750 813 875 938 1000 1063 1125 1188 1250 3.5 t w 350 438 525 613 700 788 875 963 1050 1138 1225 1313 1400 1488 1575 1663 1750 6.0 t w 600 750 900 1050 1200 1350 1500 1650 1800 1950 2100 2250 2400 2550 2700 2850 3000 n n Determine the effective height, the slenderness ratio and determine the design moments. For the chosen axial load and moments determine the area of reinforcement required. (See Chapter 6 for a more-detailed discussion of these design steps.) Generally, walls are designed for strength only; stability and serviceability are considered only for slender and/ or unbraced walls. The action effects in accordance with Table 1.1 of Chapter 1 of this Handbook that should be considered include: n 1.35G n 1.2G + 1.5Q n 1.2G + ycQ + Wu n n 0.9G + Wu (not generally a design case – may be applicable for a lightweight roof supported by the wall in a cyclonic area and where large uplift forces may have to be resisted) G + ycQ + Eu Considering all the possible axial load and bending moment combinations for each load case for a given cross-section for each wall at each floor using manual determination of the strength of a chosen wall size 7.8 Reinforced Concrete Design Handbook can be a tedious process, although often the load case 1.2G + 1.5Q will be the critical design case. The design of a particular section is a trial-and-error process and is much more easily accomplished with a load-moment interaction diagram calculated for the chosen wall section and reinforcement, using appropriate design software or a spreadsheet that complies with AS 3600. When using Charts 7.1 to 7.5, note that the wall is assumed to have reinforcement in each face. Note that for walls designed as columns the vertical reinforcement needs lateral restraint using fitments when: n n N * > 0.5 f Nu , or the concrete strength is greater than 50 MPa and: — the vertical reinforcement ratio is used as compression reinforcement, or — the vertical reinforcement ratio is greater than 0.02, and a minimum horizontal reinforcement ratio of 0.0025 is not provided. For 200-mm-thick walls, it is difficult to place and fix fitments unless the covers are small. 250 mm is the practical minimum thickness for walls with fitments. The contribution to in-plane shear strength due to reinforcement, f Vus, is given by the equation in AS 3600 Clause 11.6.4, ie: 7.10 Simplified design method for walls subject to vertical compressive forces f Vus = f pw fsy 0.8Lwtw The axial load capacity for a unit length of wall is given in AS 3600 Clause 11.5.1, ie: where N * ≤ f Nu = 0.6 (t w - 1.2e - 2ea) 0.6 f 'c pw is the reinforcement ratio and = p h where ea is taken as = (Hwe)2 / 2500 t w. This requires the effective height to be calculated, a factor k is determined (see AS 3600 Clause 11.4), and the unsupported height Hw is multiplied by k to give the effective height Hwe. or (when Hw / Lw > 1) ote this n is ctio se s ce en efer dR an ts men Com d) inte t pr no (N et No .4 L 11 eC se ight 2.2 he e t ive Tabl en fect ear em r ef forc Fo for Sh i rein Ph of ress d st el Yi pw = lesser of p h and pv (when Hw / Lw ≤ 1). She o bN Jo By s ctor n Fa s tio rtie uc rope Redial P ity a ac ater MP Cap & M a 0.7 MP 40 e: Dat tw The reinforcement ratios p h and pv are for horizontal and vertical directions respectively. S 0 50 φ Lw o tN e he of lan f' c f sy l wal P o N ob J The simplified design approach given in AS 3600 Clause 11.5 is based on BS 81107.17. The equation gives a conservative estimate of the load capacity of the wall and ignores any load capacity due to the vertical reinforcement. this resp r its By te: Da e. us e fo sibl on t is ee sh read sp e us .6 11 Cla mm mm mm mm m2 Hw r Spreadsheet 7.2 based on these equations can be used to calculate the shear capacity for a wall. ing s rce 0 6,00 0 7,60 0 7,60 0 20 1.52 0.79 Floo 00 36 ion AS ns .7) of f te 0.94 11 Cl 1.5 rs o pu .0 with in 30 Com ce ekN1 ye for nt 0.94 us 2 la rdan Clie ct/Job co lls ac H je C8 la or wa in or f 24 Pro h flo 15 ct l =L it t (1 je to rn. ion no wal 0.00 ise or Sub rw flo be ve ns e w men 15 sig n 1.1 n of he ld o all e c tio te ot w D isio ou n e 0.00 25 va a y) sh yg e. of an Ele tio th ut all ht of L rda g Rev (if us 0.00 bo t ra ma . Inp W heig wall orts ers co arran its en or pp ta ry of d em th ac ntsy) tios lay Floor for .5.1 Da met upporte ngth of een su ng forc in teel (ii) me likel reo ra r 2 11 rein Le Geo Uns ible tw stre s (b) od or rall L be all t ns um ire ost un cal o .1 se w po Ove th inim e 11 quis is m0or verti t (1 size rn. eth given s of ng lau us V* res ete ve the m Le knes nm Cla t rebut th3riz6on0tal men is ic ncr go (Note t in C ectdio r a et Th l area /L S ge co 0. meemnent A sfise y e ho al H li 0 fo he u se a os th W n s p ea d io m /t 36 ceinforc erooff rra to ima cr ction Rat io H incr rea rcesrcs e at a se AS fore re .6 less nts sp AIL r fo Fo Rat ) as ea ar the e se t ratio ts to rengithnbystheis1th1e steel If F 00 me this he 36 d sh Sg m re u pw th s en n ire 0.0015 d 30 n lie sigin AS ing of oft ratio rcem t ratiome muear st0.79la ivent ratios qu App Dues us ee n5 tio on all citceymenl reinforcemuenire .6.4 innito shith C t re xc a g emen Sec ers t e 1.00 a w panfor onaltareinforeqenteCrl e11ntribmutcioe= w for l reinforc 600. men ee ep .S no of g caRei Horerizticin g rcemwhno co aHn/L tion rtica 3 orce Th ions s 50 S at V th n r: e f e A o % l to ider ac fos en rdtio c or ve ng din 46 tio im ns ua 27 co sp ar relainb=0 thccAos ra se ontal ts to rein sd eq tre en cla or fire uta es m ls tb n nd Sheo s if p a the horiz lls to Dis than due kn 0.94 t a e t(Note in o1f the eme nimu ed xia en mp OK Wa less 40 us en abl all Hit/Ly er of i thic d ir n a mom Co t of ratio ited ; an ends em plic f a w Wpheancthe less 1 quratiore m p to be the ds sig ss lim en both en 0.62 de ate th ob to k) erne ay be ign tal e forc y ap th o g ca p is in/Lg> re V Cli bo e or 1.0 ratio nd on h t/J m ht for at <φ an riz w on es Sle the ultim /t .4 tions s ement le11ig ncH e ho jec rein ularl eng ndin V* e th D ided ed at s b hea tha 0.63 H K or ro p e .1 rc O op ov e m W m t tr hC r s is l P kN th fo 1 c pr ovid or u rtic r s th be late th . mm ote 0.3 vweall w fo n is is pr ng bje nd po s rein kN (N s) than ion lcu tes nt) inim pa ea ent tatio n Su stre t a e t Input ffigehtctiofee belo 8 t ro rotatio 3 side less or vis Ca lcula eme ll m et is n sh om (s he en abl 9,72 0 0 ains nsst . t on t not size ag ie etivee k Re 6,00 6,81 em plic n : Ca forc ers a she esig te m g int t ag ete it ai or bu des) L th .0 ffec ctor a 30 ncr tio = kN forc y ap ofE Fa 1.6bunckt.linre restarerastpraaincral su.pp co : 4 si H rein nsid read the d ltim rip = se ay ewhe no tio se 1wm c (late.min le t on where rein larl ea sc 1 or Co e sp tes the u nee0.75 here kN >L Tit pp incr ing ling st e ra Clau Orc 3,80 um rticu De H IL 0dw buck& A . h l su la k = 1. V th l re A s n fo ra t) u a im T F = ) te ay in 1 ra n in whe p te If both ke 69 lc kN re ith in ØV -3,2 ne = he e w r re in bTwo uwo.m (H /3L ling (la or Ca lcula eme all m et is Ge 7 n w nc ea el Mk= 1+ay buck1 Ca forc ers dshe 1,30 = ) sio rda sh ste ting ow /L rein nsid rea Tw V (H res acco r and ssion lcula L 1+ k= = kN kN .6.3 mp a s Co e sp l 11 2H kN on l H co ar in she mpre in ca n. kN V % tC .6.2 30 en lati wal 11 k= 8 Th 1 13 kN 24 s in she for f co ess tensio CL lcu ht of a 3,80 1 rcem ll a 1 V fo a n o str in in kN 2,66 n C heig 50 2,66 it V t ) r re s in sig ts rw 8 sig ive /t ea lim .8L OK 63 = kN V* De Effect H Fo wall t de effec r pre bers 9 t sh hing f (0 t ck = ou kN o r fo 2 3,29 L ØV crus = 0. Che with Fo es n s the llow mem 0.8 2 max t all web 91 n f' } ØV Vu. V 8 8L aw V ctio late 63 <φ Do ore ot a for /L of } 0. kN cu ØV V* ØV th : s se ent -1) 1H Cal os /L lue = Ign es n itable reng 1 -0.2 ns va a cr rcem t f' r st 3 rt kN at o su = s f' /(H ea tio info emen /L 66 D 4,71 ce ce Inse * re sh a 0. H t rc .1 or t en it F fo ou ={ >1 late ) for f' +0 ar efer inted) ith 9 V No cu (a rein ngth V t /L .3 .4 dR Lim 05 to She th w Cal V 3,29 0.79 t pr rson er: aim cl Dis us e pe Th ar he es n -pla n tatio fo w w w w 1 Spreadshee 7 2 s ava ab e a www ccaa com au w w w w w w w There are also a number of limitations/restrictions on the use of the simplified method. These include the requirements that: w w w w w ax u.m w we w w 1 w 1 ax w 1 u.m 2 ax u.m w w 1 2 uc (1) 1 w n the wall is assumed to be braced in accordance with AS 3600 Clause 11.3; and t ee Sh n ts men o an n is ote b Jo no Cl ctio C s or 0.96 w 0.96 w w w 1 w 1 2 1 w w we u 2 we 7.11 Design of walls for in-plane shear forces 1 we w a we a we The design shear strength of a wall subject to in‑plane horizontal shear is given by the sum of the concrete shear strength and the strength due to the reinforcement, ie: a fVu = fVuc + fVus with an upper limit of f 0.2 f 'c (0.8Lw tw). ck th E S H or 1.1 TE T NO : Vuc and Vus are determined from AS 3600 Clause 11.6.3 and Clause 11.6.4 respectively. ba ed Fe . L gh e ri u Ø l 11 tC Vuc to th en rt x inse e bo rcem th V uc info /L w in /L w Hw r re Hw io ea sh Rat ratio by the th reng From r st ea sh w t n to 8 L w 0. tio f sy bu ntri = p w co V us late cu Cal V us f' c 17 = 0. .6 us = ign = e re r st reng th du ea Des ar st sh reng gn She ar st si de She red to Fac gth in r ea sh Vu =V uc + Vu =V H we Lw tw uc +V us ØV 0 6,00 .0 00 76 0 20 0.8 u mm mm mm Hw ight he rted po tw up w all Uns th L of w ng ss Le ne ck /L w Thi Hw io Rat AS 3600 C ause 11 7 g ves equ emen s o m n mum e n o cemen n wa s and o c ack con o These equ emen s a e ndependen o he f va ue used lts Fo m n mum ve ca e n o cemen he e n o cemen a o pw sha be he a ge o 0 0015 1 5 w mm2 m o he va ue equ ed o s eng h w w w V uc us (1) tw w su w 1 fo uc w c 0. ={ Re 1 Spreadsheet 7.1 is available at www.ccaa.com.au (b) us uc w en w 2 .6.3 w c W Str c 1 11 uc w w c W W rH (3) w 7 12 Reinforcement requirements for walls Spreadsheet 7.1 can be used to calculate the axial load capacity for a wall using the simplified method. w w c uc .6 l 11 ct Dat 00 n Fa s tio rtie 36 ion uc rope AS ns Redial P ity of f te a ac ater MP Cap & M 1.5 rs o 0.6 e 1 ye 40 us 2 la la r φ C o y f' ith t (1 rn. ion if an t rts ve ns ew n po L go nc geme f te Sup a or y e. o s a L rd an u rs m . co its ye es ts ac el ausrre. for .5.1 en 2 laforc r its 11 ible d in sblte em ossr ion e fo o r e ns i y . n s o si u th (1 p l q e ivspeon if an t re 0 H lau ern wal res t re 60 enmp is ov Supports in C d m r aeetgis re n of . et en AS3 geaml co Pla fie yg 00 he ut pli n refoadsh ds anrtic ma mm 36 cem of to sim ctithois sp rea se AS for .6 arrve nts sp mm s to rein e 11 teeclt to am the e suseing me this r ts s s je ire 4,800 0 d 30 mm2 ing thrson Floo ing en mum Clau ensub qu us y: oThfe pe us m i iv ll mm e 7,60ee e ll r n a n g 0 c i o a citer t ith r a r w mm uir . ex 201.52 ers en fo a w paclaim n q e m w mm 00 ot ep .6 of g caDis omputatioing re her ance on ofods 25 36 cem Th sn w .0 th n ti th C c er: 10 .6 AS for oe ng din l tio im plieant obbs cord secme kN to rein 25 sd wal tre en cla ts kN d sC jesctl/Ja ac theesign s uta es n of ls tb Dis tio f en um an tPoro ctin kN =L kn all va 8 xia en mp ise ic nt ble aubllje y oed d Ele em im rw th 1,95 5 fW n a mom Co t uir e min me plica a wS pacpitlifi 1 58 7 o othe e ig q e y) s th f 1. n an e ate 3,22 .5.2 en er (if to r ob 11 forc ly ap th o g ca Simevision ing re Cli sig e d ltim t/J t walhl ht H orts ht Cl in Floo c g u e r R c in th u je re la en d D pa Inpbsof W d heig n suppeig calculated=0.05*t 1.00 Pro tes the . 1.1 ct t) um icu str t ben d s atasletary rte twee e h pu as l load bje t) ula s im art ar l In ad an Dtoeom uppo L be tivall t ion Su na alc late en min is p she men nt ble G UnsLengtheffeses cof w rtical loof verticacal lo.ad vis ptio ) : C alcu rcem all heet ign mo me lica e ickn ea of ve ity vertiies es (O Re ram C fo ers ds on rce app es ate f thThW1al1l aren.6tricityt.eccentricricitypaofcit .on forc prog in ti d e fo o id a th : y re ns re the ltim n ca esinsi o rip Ecc imeum cent le d by rein larl pr low9 .m sc 0.96 in n ecg rati use eMm d be 0 Co e sp tes the u late Tit ig 30 Acomst ; an ends ns20 um icu De alcu ds tio, the h Cla forc eDnesdin rt& ical l to (C en both Th lcula tes nt). inim part ua ral N* on in ad G th e op in re it eq d ve thrs bo e or ad .4 ne or 1.0 he e w r re in b ulieo.m at Ca lcula eme all m et is on d Lo ad Q e Lo l 11 ea from an lC 0.71 Ge than ided ed at Lo mat n w nc ea el MApp Dea e th lu,eP rm lti wal Ca forc ers dshe prov ovid pe less Live l U mor of e vas sio rda sh ste ting ta io or Nu n is is pr To ig Rat rein nsid rea <Ø 0.3 0ht) prBoparisatic res acco r and ssion lcula tatio tion des) he 0.79 1 p m ro an K e p si ta r o t 0 O th 3 m in ro pe C e s ss ea pre ca . ains nst ec l , tiv2 or k anteap on N* co mm wal as tEff1 FaCcthonoscre e ling straint agint agai pport but not le in hear for sh com ss in nsion Th OK th of en ck re ra lls l su s f l is ng e s) mm 0 Wal tal le dm d Co way buwhere no rest(latera wa s in sign cts o restr in te 80 side L n to 4, 4 r th for 24 /m t on me rce Onek = 0.750 wherecklin.g H Fo wall t de effe r p bers leng kN or re .08 o bu 0 1. r fo pp th rm g A info 46 whe kN k = way 60 1 ) pe leng l su Fo es n s the llow mem tin e >L o 3 tera load r m 1 /3L H Tw S a pe ora k, R ed re Do ore ot a for (H g (la 1,11 0 A MP a ctor load : ic whe orp in k= 1+ bucklin MP 8,44 e fa ed Ign es n itable ns nc ilpatr mat factor ed = 1 ay Ulti l tio K us 9 (I ow ) Do t su 2.12 ua /L Tw ita Act 00 and lly o 5.55 (H 2 N im ra k= 1+ L r e L = 00 oste 0t en = N ns 2H 36 /250 F sg Ø tio l H k= th =H AS rner, ula wal na leng L ity alc ht of ut tio ric rm l er pe ov n C heig 30 cent inp , etc Wa ota ca d sig ive /t nal ec load load ta ll eti dn De Effect H late e. ored ctored da or ditio wa ck an mm fact fa s lcu ula Che an ad he ls ire the a ate mm ate on T o u c = .6 rm e 25 ultim ultimlati al mm s/ as mb req y for 00 um um u actu able g fo im 48 de : im lc Sy tc lls ds .1 ss sin low Max l max ca 46 ce ometr th e l loa Co sis stre ss al ta r yu e w ve To ea Ba llo ge treng rtica h essi stre all H : r Ye tion r s Comprpressivematic s e re ve fo ea f c om tu te o w C to sh la /L Se ncre ity ad au l lo nc for Hw ns H we ic r d ca e o f o tr a rti o C en acti ht m e 50 ve ity late of H c sh tio ntric No ty nd lcu sds Ec lied heig trici height ecce for on ra l en na ulta ca : p slo Ecc ctive tiona tio sio Ree nd g ut Ap ctive s s ffe addi ra u iv p in E d t s re e d ro an In e loa res kg Eff tor k emen pen mp ac e co c mp ive co ress nb Fa forc V uc d for in p for ree 30 Re e of all com hg ed lu r fw ce wit Va ex lls all y o ll fo ot ce icit f wa n w tr d s s d xe the ccen ity o oe loa Bo a of e sd N* ial tric m es ax Are imu ccen load kn for e et, e thic ate t H w y e a he Min ign ion s im h ds it the th ss to De l ult heig ntric rea t: ht leng mpre ta ve e sp u o c ig k c is T tp ti he r m r co of le ec Ou g th ll a in Eff itiona lues ctive d pe d fo w ard d va effe loa loa wall the D Ad ous reg e red red the ri s in ck KE Va cks facto facto s in th stres ba EC e ed Ch ate ate stres sive CH r fe im im s the LY Ult l ult sive pre L ro s s o ta m lt U u s F To pre co ax on res m le .m sti BE Co wab s V u d V u f the ge O o ug All ulate uc an ary o LT ,s lc V ns NT STIL Ca e of mm tio su ec lu ME RE orr CAA Va vides M c A , eC CO TS nts Pro R E me ct th FO HE om onta rc FT DS Fo ase c RA EA ple e D PR 1 uc c W No e: (2) w w No By w u.m (N et w ax By te: Da She uc ax u.m we se this Com bN Jo Hwe / tw ≤ 30. No we 1 1,11 5 42 ØN N* u /m kN /m kN Fo m n mum ho zon a e n o cemen he e n o cemen a o pw sha be no ess han 0 0015 2 5 w mm2 m excep ha o a wa des gned assum ng one way buck ng us ng AS 3600 C ause 11 4 a and whe e he e s no es a n aga ns ho zon a sh nkage o he ma movemen s h s may be educed o ze o he wa s ess han 2 5 m w de o o 0 0015 1 5 w mm2 m eng h o he w se Fo c ack con o and sh nkage whe e he wa s es a ned om expand ng o con ac ng n he ho zon a d ec on he m n mum a ea o ho zon a e n o cemen n ha d ec on sha be n rsio Ve The two equations given in Clause 11.6.3 to determine Vuc together with the lower limit, which is the third equation, are: Fo exposu e c ass fica ons A1 and A2 n n f Vuc = f [0.66 √f 'c – 0.21√f 'c (Hw / Lw)] 0.8Lwtw or for Hw / Lw ≤ 1 n f Vuc = f [0.05 √f 'c + 0.1√f 'c / (Hw / Lw – 1)] 0.8Lwtw for Hw / Lw > 1 with the lower limit of f Vuc ≥ f 0.17 √f 'c (0.8Lwtw). whe e a m no deg ee o con o ove c ack ng s equ ed A mus be a eas 2 5 w mm2 m whe e a mode a e deg ee o con o ove c ack ng s equ ed and whe e c acks a e nconsequen a o h dden om v ew A mus be a eas 3 5 w mm2 m whe e a s ong deg ee o con o ove c ack ng s equ ed o appea ance o whe e c acks may eflec h ough fin shes A mus be a eas 6 0 w mm2 m Reinforced Concrete Design Handbook 7.9 Access Placing order 4 3 2 1 4 Access Access Near face Table 7.2 shows the various areas of horizontal reinforcement per metre of length required for walls up to 500 mm thick for crack control. Note that this reinforcement is the total of all layers. Placing order 2 3 1 AS 3600 requires the spacing of the reinforcement to be adequate to place the concrete but not less than 3d b. Reinforcement must be in two layers, one near each face when: n n n Figure 7.5 Placing order of bars the wall is greater than 200 mm thick; where in any part of the wall structure the tension exceeds the tensile capacity of the concrete under the design ultimate load (this means shear walls may require two layers of reinforcement); n when walls are designed for 2-way buckling. n The maximum spacing of the reinforcement is the lesser of 2.5tw and 350 mm. For walls greater than 200 mm thick, or where the wall is in tension greater than the tensile capacity of the wall, or where the wall is designed for 2-way buckling in accordance with AS 3600 Clause 11.4, the reinforcement has to be in two layers, one in each face. n n 7.13 Detailing All wall elevations and details should be shown on the drawings. n Designers should refer to Chapter 15 of the Reinforcement Detailing Handbook 7.18 for further advice on the detailing of walls. Designers should note that: n n n n Far face For exposure classifications B1, B2, C1 and C2, Ast in the horizontal direction must be at least 6.0 tw mm2/m , which can be a significant amount of reinforcement. Sufficient information should be provided on the drawings (including plans, elevations, sections and details) to enable the walls to be built. n n Standard details, if used, should be appropriate to the walls being designed. The order in which the various layers, ie the vertical and horizontal bars, are to be placed should be specified. For ease of construction for insitu concrete walls, it is often desirable to tie the horizontal bars on the outside of the vertical bars or to place the horizontal bars from one side only such as in a core wall where the inside formwork is placed first and the reinforcement then fixed. This should be taken into consideration when designing the wall, especially considering cover and axis distance, see Figure 7.5. It is important to note that when a wall is designed and reinforced as a column with fitments to restrain the vertical bars, the vertical bars cannot be on the outside. 7.10 Reinforced Concrete Design Handbook n n When header or coupling beams are within an insitu wall, the thickness of the wall must allow for cover/axis distances, heavy reinforcement and allow the concrete to be properly placed. The effect of bend radii at corner and T junctions in plan where walls intersect, especially if they have to resist bending moments around the junctions. Trimmer bars are needed at openings in walls such as doors and windows to minimise cracking at re-entrant corners. Note that trimmer bars at 45° will form a third layer which may make placing of concrete difficult. Minimum reinforcement is to be provided in accordance with AS 3600 Clause 11.7. Trimmer bars usually are one N12 per layer of reinforcement, around the perimeter of precast and tilt-up wall panels. Larger diameter trimmer bars may not fit within thin walls depending in which layer they are in. Appropriate splice lengths for bars and mesh are specified depending on whether the bars and mesh are in tension or compression. For walls the strength of the concrete in the floor slab also needs consideration, as the strength of the floor cannot be less than 0.75 the strength of the wall without specific design as set out in AS 3600 Clause 10.8. This clause allows for the concrete in the walls to be one strength grade higher than that of the slab. For greater differences in strength, additional calculations are required to determine the effective strength of the concrete in the wall for transmission of axial forces through the floor systems Thin insitu concrete walls can be difficult to cast. Designers should consider the use of 10-mm aggregate and super plasticisers to allow adequate compaction when thin walls are proposed. Avoid the use of fitments if possible as they are difficult to fix. 7.14 General Guidance The following will assist the design and inspection of walls for a particular project. n n n n n n n n Stiff walls such as retaining walls, core walls or other walls in a building, can significantly restrain concrete floors and roofs when they are rigidly connected to them. This can result in unsightly diagonal cracking in the concrete floors and roofs (or in the wall) due to shrinkage. Construction techniques to minimise such problems include locating the connecting bars in flat prestressing ducts, providing slip connections and then grouting the ducts after the floor over is cast and some of the shrinkage allowed to occur. If the wall is to have an off-form finish and is exposed to view, are there any specific requirements for class of finish, type of formwork, arrangement of form ties for insitu walls, etc. For precast walls, are any special finishes required including applied finishes such as painted, acid washed, sandblasted, honed or polished? Designers should refer to Chapter 10 of the Precast Concrete Handbook for information on finishes, remembering that not all finishes may be available in the given location. If the construction joints will be exposed to view, consider providing a small vee or rebate to the joints to give a neat appearance, and ensure that this is allowed for in the cover specified. For precast and tilt-up walls, a positive connection must be provided at the bottom of the wall at the time of erection to prevent kick-out prior to unhooking the panel from the crane. When mixing loadbearing and non-loadbearing precast wall panels, differential movements can be a problem. Insitu off form concrete walls Note the attention to the joint layout, construction joint locations (which will occur at the expressed joints) and the locations of the bolt holes or the form ties in this insitu wall to form a uniform pattern. Charts 7.1 to 7.5 These charts have been developed assuming the wall is braced in both directions and is short. If the wall is a slender braced wall then the appropriate moment magnifiers in accordance with AS 3600 Clause 10.4.2 must be used. They are based on moments about the weak axis only. The design of walls as columns should be in accordance with AS 3600 and appropriate design aids or software should be used. Chart 7.1 200 Thick walls pages 7.12 and 7.13 Casting insitu columns and walls as a combined unit should be avoided because of the complications of the formwork. Chart 7.2 250 Thick walls pages 7.14 and 7.15 The rules given in AS 3600 Clause 5.7.4 under Fire resistance for recesses and chases are empirical rules adopted to provide consistency with the rules for masonry. They should therefore be viewed with caution. Chart 7.4 350 Thick walls pages 7.18 and 7.19 Chart 7.3 300 Thick walls pages 7.16 and 7.17 Chart 7.5 400 Thick walls pages 7.20 and 7.21 Spreadsheets 7.1, 7.2 and 7.3 may be downloaded from the Cement Concrete & Aggregates Australia website www.ccaa.com.au Reinforced Concrete Design Handbook 7.11 chart 7.1 200 Thick walls axis distance = 50 mm 8000 Wall reinforced on two faces Note: Transverse reinforcement not shown 7000 Moment 6000 Lw =1000 mm 5000 50 tw (refer chart) f 'c = 25 MPa 200 wall 1% 200 wall 2% 200 wall 3% 200 wall 4% Minimum moment 4000 3000 Compressive force (kN) 50 2000 1000 0 0 50 100 150 200 Moment (kN.m) 8000 Wall reinforced on two faces Note: Transverse reinforcement not shown 7000 Moment 6000 Lw =1000 mm 5000 200 wall 1% 200 wall 2% 200 wall 3% 200 wall 4% Minimum moment 3000 Compressive force (kN) 50 tw (refer chart) f 'c = 32 MPa 4000 2000 1000 0 50 0 50 Moment (kN.m) 7.12 Reinforced Concrete Design Handbook 100 150 200 chart 7.1 (continued) 200 Thick walls axis distance = 50 mm 8000 Wall reinforced on two faces Note: Transverse reinforcement not shown 7000 Moment 6000 Lw =1000 mm 5000 50 tw (refer chart) f 'c = 40 MPa 200 wall 1% 200 wall 2% 200 wall 3% 200 wall 4% Minimum moment 4000 3000 Compressive force (kN) 50 2000 1000 0 0 50 100 150 200 Moment (kN.m) 8000 Wall reinforced on two faces Note: Transverse reinforcement not shown 7000 Moment 6000 Lw =1000 mm 5000 50 tw (refer chart) f 'c = 50 MPa 200 wall 1% 200 wall 2% 200 wall 3% 200 wall 4% Minimum moment 4000 3000 Compressive force (kN) 50 2000 1000 0 0 50 100 150 200 Moment (kN.m) Reinforced Concrete Design Handbook 7.13 chart 7.2 250 Thick walls axis distance = 50 mm 10000 Wall reinforced on two faces Note: Transverse reinforcement not shown 9000 Moment 50 8000 7000 Lw =1000 mm 6000 f 'c = 25 MPa 250 wall 1% 250 wall 2% 250 wall 3% 250 wall 4% Minimum moment 5000 4000 Compressive force (kN) 50 tw (refer chart) 3000 2000 1000 0 0 50 100 150 200 250 300 350 400 Moment (kN.m) 10000 Wall reinforced on two faces Note: Transverse reinforcement not shown 9000 Moment 50 8000 7000 Lw =1000 mm 6000 f 'c = 32 MPa 250 wall 1% 250 wall 2% 250 wall 3% 250 wall 4% Minimum moment 5000 Compressive force (kN) 4000 3000 2000 1000 0 50 tw (refer chart) 0 50 100 150 Moment (kN.m) 7.14 Reinforced Concrete Design Handbook 200 250 300 350 400 chart 7.2 (continued) 250 Thick walls axis distance = 50 mm 10000 Wall reinforced on two faces Note: Transverse reinforcement not shown 9000 Moment 50 8000 7000 Lw =1000 mm 6000 f 'c = 40 MPa 250 wall 1% 250 wall 2% 250 wall 3% 250 wall 4% Minimum moment 5000 4000 Compressive force (kN) 50 tw (refer chart) 3000 2000 1000 0 0 50 100 150 200 250 300 350 400 Moment (kN.m) 10000 Wall reinforced on two faces Note: Transverse reinforcement not shown 9000 Moment 50 8000 7000 Lw =1000 mm 6000 f 'c = 50 MPa 250 wall 1% 250 wall 2% 250 wall 3% 250 wall 4% Minimum moment 5000 4000 Compressive force (kN) 50 tw (refer chart) 3000 2000 1000 0 0 50 100 150 200 250 300 350 400 Moment (kN.m) Reinforced Concrete Design Handbook 7.15 chart 7.3 300 Thick walls axis distance = 50 mm 12000 Wall reinforced on two faces Note: Transverse reinforcement not shown 11000 Moment 10000 9000 Lw =1000 mm 50 50 tw (refer chart) 8000 f 'c = 25 MPa 7000 300 wall 1% 300 wall 2% 300 wall 3% 300 wall 4% Minimum moment 6000 5000 Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 500 600 Moment (kN.m) 12000 Wall reinforced on two faces Note: Transverse reinforcement not shown 11000 Moment 10000 9000 Lw =1000 mm 50 50 tw (refer chart) 8000 f 'c = 32 MPa 7000 300 wall 1% 300 wall 2% 300 wall 3% 300 wall 4% Minimum moment 6000 5000 Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 Moment (kN.m) 7.16 Reinforced Concrete Design Handbook 300 400 500 600 chart 7.3 (continued) 300 Thick walls axis distance = 50 mm 12000 Wall reinforced on two faces Note: Transverse reinforcement not shown 11000 Moment 10000 9000 Lw =1000 mm 50 50 tw (refer chart) 8000 f 'c = 40 MPa 7000 300 wall 1% 300 wall 2% 300 wall 3% 300 wall 4% Minimum moment 6000 5000 Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 500 600 Moment (kN.m) 12000 Wall reinforced on two faces Note: Transverse reinforcement not shown 11000 Moment 10000 9000 Lw =1000 mm 50 50 tw (refer chart) 8000 f 'c = 50 MPa 7000 300 wall 1% 300 wall 2% 300 wall 3% 300 wall 4% Minimum moment 6000 5000 Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 500 600 Moment (kN.m) Reinforced Concrete Design Handbook 7.17 chart 7.4 350 Thick walls axis distance = 50 mm 13000 12000 Wall reinforced on two faces Note: Transverse reinforcement not shown 11000 Moment 10000 Lw =1000 mm 9000 50 50 tw (refer chart) f 'c = 25 MPa 8000 350 wall 1% 350 wall 2% 350 wall 3% 350 wall 4% Minimum moment 7000 6000 5000 Compresive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 500 600 700 800 900 Moment (kN.m) 13000 12000 Wall reinforced on two faces Note: Transverse reinforcement not shown 11000 Moment 10000 Lw =1000 mm 9000 350 wall 1% 350 wall 2% 350 wall 3% 350 wall 4% Minimum moment 7000 6000 5000 4000 Compresive force (kN) 50 tw (refer chart) f 'c = 32 MPa 8000 3000 2000 1000 0 50 0 100 200 300 400 Moment (kN.m) 7.18 Reinforced Concrete Design Handbook 500 600 700 800 900 chart 7.4 (continued) 350 Thick walls axis distance = 50 mm 13000 12000 Wall reinforced on two faces Note: Transverse reinforcement not shown 11000 Moment 10000 Lw =1000 mm 9000 50 50 tw (refer chart) f 'c = 40 MPa 8000 350 wall 1% 350 wall 2% 350 wall 3% 350 wall 4% Minimum moment 7000 6000 5000 Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 500 600 700 800 900 Moment (kN.m) 13000 12000 Wall reinforced on two faces Note: Transverse reinforcement not shown 11000 Moment 10000 Lw =1000 mm 9000 50 50 tw (refer chart) f 'c = 50 MPa 8000 350 wall 1% 350 wall 2% 350 wall 3% 350 wall 4% Minimum moment 7000 6000 5000 Compressive force (kN) 4000 3000 2000 1000 0 0 100 200 300 400 500 600 700 800 900 Moment (kN.m) Reinforced Concrete Design Handbook 7.19 chart 7.5 400 Thick walls axis distance = 50 mm 15000 14000 Wall reinforced on two faces Note: Transverse reinforcement not shown 13000 Moment 50 12000 11000 Lw =1000 mm 50 tw (refer chart) 10000 f 'c = 25 MPa 9000 400 wall 1% 400 wall 2% 400 wall 3% 400 wall 4% Minimum moment 8000 7000 6000 Compressive force (kN) 5000 4000 3000 2000 1000 0 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 Moment (kN.m) 15000 Wall reinforced on two faces Note: Transverse reinforcement not shown 14000 13000 Moment 50 12000 11000 Lw =1000 mm 50 tw (refer chart) 10000 f 'c = 32 MPa 9000 400 wall 1% 400 wall 2% 400 wall 3% 400 wall 4% Minimum moment 8000 7000 6000 Compressive force (kN) 5000 4000 3000 2000 1000 0 0 100 200 300 400 500 Moment (kN.m) 7.20 Reinforced Concrete Design Handbook 600 700 800 900 1000 1100 1200 chart 7.5 400 Thick walls axis distance = 50 mm (continued) 15000 14000 Wall reinforced on two faces Note: Transverse reinforcement not shown 13000 Moment 50 12000 11000 Lw =1000 mm 50 tw (refer chart) 10000 f 'c = 40 MPa 9000 400 wall 1% 400 wall 2% 400 wall 3% 400 wall 4% Minimum moment 8000 7000 6000 Compressive force (kN) 5000 4000 3000 2000 1000 0 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 Moment (kN.m) 15000 Wall reinforced on two faces Note: Transverse reinforcement not shown 14000 13000 Moment 50 12000 11000 Lw =1000 mm 50 tw (refer chart) 10000 f 'c = 50 MPa 9000 400 wall 1% 400 wall 2% 400 wall 3% 400 wall 4% Minimum moment 8000 7000 6000 Compressive force (kN) 5000 4000 3000 2000 1000 0 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 Moment (kN.m) Reinforced Concrete Design Handbook 7.21 7.15 500 N12 700 N16 Some typical details The following figures illustrate some detailing of larger walls. They are for general information and show the sorts of details that may need to be included on the drawings. N32-20 N32-20 N20-200 EF Offset wall Level 1 N28-150 EF N16-300 EF Mezzanine For starter bars refer to wall reinforcement schedule and notes Refer to column schedule for size Figure 7.8 This figure shows the connection between a column and a wall in plan. Usually the column is poured first and either starter bars or cast-in ferrules with screw-in bars which are then used to connect the column to the wall which is poured later. HB 2 500 N12 700 N16 500 N12 700 N16 N32-200 EF N20-300 EF Basement 1A 300 N28-150 EF Ground N20-300 EF N32-200 EF Figure 7.6 This figure shows a heavily reinforced core wall with a header beam under the door openings and heavy vertical reinforcement in each face. 12-N20 (EF) (V) N12-300 fitments and R6 ties Figure 7.7 This figure shows a reinforced wall with straight bars to the vertical reinforcement, together with fitments to all the vertical bars, so the wall is acting as a column. 7.22 Reinforced Concrete Design Handbook Figure 7.9 This figure shows the corner detail between a wall with a single layer of reinforcement and a wall with two layers of reinforcement to achieve a moderate degree of moment capacity at the corner. RE N20 (EF) Fitments type A N16-300 (EF) N32-125 EF C34 C34 N32-125 EF EF Typical header beam reinforcement 10N28 EF EF Typical header beam reinforcement 4N24 top and bottom N16-450 fitments (T) C34 C34 (H) (B) (L) 10N32 EF Fitments type A (T) SW13-LB-02 N12-300 (EF) N16-300 (EF) Figure 7.10 This figure shows the elevation of an internal wall with two layers of reinforcement in each face and a header beam over the door opening. Figure 7.12 This figure shows an elevation of a lift shaft with reinforcement on both faces including the header beam details, etc. 700 Lap typical 2N12 CORE 09 - W03 N20-200-EF (V) N12-300-EF (H) Figure 7.11 This figure shows a section of a lift shaft with reinforcement on both faces including the corner details, etc. 2N12 2N12 CORE 09 - W02 N20-200-EF (V) N12-300-EF (H) CORE 09 - W01 CORE 09 - W04 N20-200-EF (V) N12-300-fitments N20-200-EF (V) N12-300-fitments N20-200-EF (V) N12-300-EF (H) 175 THK SL82 EF Vertical bar to outer faces 40 cover FF, 30 cover NF 40 cover edge D8 2N12 100 2N12 2-W10-150 U-bar 300 110 wide x 300 legs to each vsl duct 110 Figure 7.13 This figure shows an elevation of a precast wall panel with two layers of mesh reinforcement, trimmer bars, grout tubes, etc. Reinforced Concrete Design Handbook 7.23 References 7.1 Guide to Tilt-up Design and construction, Cement Concrete and Aggregates Australia and Concrete Institute of Australia, 2005. 7.2 Precast Concrete Handbook 2nd Ed, National Precast Concrete Association Australia and Concrete Institute of Australia, 2009. 7.3 AS 3850 Tilt-up concrete construction, Standards Australia, 2003. 7.4 The National Code of Practice for Precast, Tilt‑up and Concrete Elements in Building Australian Safety and Compensation Council, 2008. http://www.ascc.gov.au/ascc/AboutUs/ Publications/NationalStandards/National 7.5 AS 3600 Concrete structures, Standards Australia, 2009. 7.6 Building Code of Australia Australian Building Codes Board, 2010. 7.7 Guide to Off-form Concrete Finishes, Cement Concrete and Aggregates Australia, 2006. 7.8 AS 1170.4 Structural Design Actions, Part 4 Earthquake actions in Australia, Standards Australia, 2007. 7.9 Mendis P Design of High-Strength Concrete Members: State-of-the-Art, Engineers Australia, 2001. 7.10 Specification for Piling and Embedded Retaining Walls, 2nd Ed, The Federation of Piling Specialists in Association with BGA and ICE, 2007. 7.11 AS 4678 Earth-retaining structures, Standards Australia, 2002. 7.12 Hughes S R and Crisp B C Structural Precast Concrete in Melbourne Australia Concrete Institute of Australia Biennial Conference 2007. 7.13 Concrete Panel Buildings, Briefing 08, Cement Concrete and Aggregates Australia, 2003. 7.14 Woodside J The Evolution of Architectural Precast Concrete Facades in Australia over the last 50 years, Concrete Institute of Australia Biennial Conference, 2009. 7.15 The Concrete Panel Homes Handbook Cement Concrete and Aggregates Australia, 2001. 7.16 AS 2870 Residential slabs and footings – Construction, Standards Australia, 2011. 7.17 BS 8110 Structural use of concrete Part 1: Code of practice for design and construction British Standard Institution, 1997. 7.18 Reinforcement Detailing Handbook (Z06), 2nd Ed, Concrete Institute of Australia, 2010. 7.24 Reinforced Concrete Design Handbook Chapter 8 Footings 8.1 General Reinforced concrete footings support the columns and walls at the base of a building. As they are usually concealed, typically cannot be inspected, or maintained, and are constructed in variable and sometimes unstable ground, a particular level of care needs to be given to many aspects of their design. These include such matters as the allowable bearing pressures, settlements, durability and cover to reinforcement. In some ground conditions, a blinding layer of 50−75 mm of about 10 MPa unreinforced concrete should be used to seal the bottom of the footing and provide a clean and stable surface for fixing of reinforcement, especially when the excavation is likely to be unstable, wet or muddy. Footings transfer the loads from the structure to its foundation—the natural soil or rock on which it rests. ('Foundation' is sometimes used to describe footings defined above; in this Handbook, it is used to describe the material on which the footing is supported.) Founding conditions vary widely in Australia and more often than not vary across a site. Before any footing design is undertaken, the properties of the foundation material must be determined. This involves an appropriate geotechnical investigation by a suitably qualified person, usually a geotechnical engineer, and testing as required. In addition to assessing the allowable bearing pressure for pad and strip footings, the investigation should include advice on such matters as the expected soil profile across the whole site, any water tables and dewatering requirements, likely short-term and long-term settlements and differential settlements under load between adjacent footings, whether the footing will be cast in aggressive soils, etc. Also, with expansive clay soils, shrink-swell movements may also need to be considered. Local knowledge and contractor's experience will often dictate appropriate footing types and construction techniques. The selection of the appropriate footing system is often a key structural design decision. The design of footings requires a disciplined approach and consideration of many factors such as the site history, geotechnical conditions including the various layers of soils, the site levels and future levels, the site conditions and constraints, environmental conditions, the building and its constraints. Designers should refer to texts such as Craig's Soil Mechanics 8.1 and Structural Foundation Designer's Manual 8.2 for further information on such matters. Using the geotechnical investigation, a decision has to be made on what stratum of soil will support the footing, the allowable bearing pressure, skin friction if it is a pile, the founding level of the footing, possible types of footings and any potential problems with excavation, aggressive soils and durability. This founding level must take account of both the proposed excavated and any future excavated levels, as appropriate. From all these investigations and considerations, various footing options may be considered and the footing system(s) chosen. Footings are usually one or more of the following types: n pad (spread) footings or combined footings n strip footings n piled (or pier) footings n raft footings n balanced or coupled footings. It should be noted that there are a number of variations within each type, including, in some cases, unreinforced footings. Designers should refer to appropriate texts for further information on these various alternatives. This chapter deals only with the detailed design of reinforced concrete pad or spread footings with concentric vertical loads, although strip footings and piled footings are also mentioned. Raft footings, combined and balanced footings and coupled footings are mentioned only briefly. Generally, except for small or light structures, footings should be founded 300−1000 mm into the ground, and 200−600 mm into undisturbed material or engineered fill. The founding layer and allowable bearing pressure should be confirmed on site, by a suitably qualified person or the geotechnical engineer, following the excavation of the footing. Where ground conditions are difficult or variable, changes may be necessary to suit site conditions, eg deeper or larger footings. There are a number of different types of raft footings. Stiffened-raft footings are commonly used in Australia for houses in areas with expansive clay soils and sometimes for slabs-on-ground for houses that have masonry walls. For domestic construction, the stiffened raft footing consists of concrete slab, edge beams and internal beams at close centres, usually all poured at one time. The purpose of the stiffened raft footing in Reinforced Concrete Design Handbook 8.1 domestic construction is generally to provide a stiff element, which will cope with soil moisture movements. Alternatively, strip footings can be used where the ground floor is above the ground level and where lightweight floors are used. Bearing pressures are usually not that critical for such domestic footings. Designers should refer to AS 2870 8.3 for the design of residential footings. For high-rise buildings, raft footings can be thick reinforced concrete plates, sometimes with thickening of 900 to 2000 mm under the columns and loadbearing walls to spread the column and wall loads over as large an area as possible – ie in effect, a very large pad footing. For industrial and other single-storey buildings, strip footings are sometimes combined with the slab-onground as an edge thickening to support external precast concrete walls or the like. 8.2 Spread footings 8.2.1 General For pad or spread footings, the soil contact pressure under axial load produces well-defined conditions of moment, beam or slab shear, and punching shear in the footing. Although it is recognised that the soil pressure distribution is non-linear, for the simplification of the design of concentrically loaded footings linear distribution is usually assumed. The designer may choose a more accurate soil pressure distribution to suit actual conditions, based on geotechnical advice. For a pad footing on deformable soils, it is assumed that the loads are resisted by flexural slab action in which adequate thickness is usually assured by a strength assessment in accordance with AS 3600. However, a footing on rock or other stiff medium may require an assessment as a stiff shear element, in which the ability to assume plate deformation is limited by the stiffness of the supporting medium. In these cases, the designer may choose to assess the structural action from strut-and-tie theory, deep-beam theory or similar. The size of a pad footing is determined from the design actions resulting from the appropriate combination of loads or other actions. The footing may take vertical loads, horizontal loads and moments depending on the assumptions made in the structural analysis and the ability of the footing and foundation to resist such actions. For a particular building it is preferable to maintain similar soil pressures under all pad footings to avoid significant differential settlements. However, large footings will settle more than smaller footings with similar soil pressures, due to a deeper zone of influence. 8.2 Reinforced Concrete Design Handbook P P P θ< B/6 M θ< B/6 B B P/A + qmin M/S B 0 qmax qmax = q Figure 8.1 Eccentric loads For eccentric columns and walls or footings with moments, the bearing pressure will vary under the pad footing which should generally be proportioned so that zero bearing pressure occurs at one edge in the worst case—so that tension does not occur under the footing and q max does not exceed the allowable bearing pressure as shown in Figure 8.1. Square pad footings are used where possible as this simplifies the design, although rectangular ones can be used. Pad footings generally are not reinforced for one-way shear, consequently the thickness of the pad footing may be a function of one-way shear stress in the section. The initial thickness of the footing will often be determined by the development length of the column or wall starter bars, assuming full compression development length is required for the starter bars. The allowable bearing pressure provided by the geotechnical engineer is the maximum bearing pressure that can be applied to the foundation such that there is an adequate factor of safety against instability due to shear failure of the founding material and the maximum tolerable settlement is not exceeded. Settlements can be immediate and/or long term and calculations of settlements if required are generally carried out by the geotechnical engineer based on the loads provided by the structural engineer. Exact estimates of settlements are difficult and often of the order of accuracy of 5−25 mm. Settlements can affect services, finishes, the concrete structure and the building as a whole. Generally, it is differential settlement that is of most concern and whether the superstructure can tolerate such settlements. Examples of where differential settlements may occur are at the junction between a tall section of a building and a low-rise section, or a change in ground conditions under a building, which might require movement joints at appropriate locations. Experience Starter bars same number and size as column bars and engineering judgment is needed in assessing such settlements and how to design for them. The total allowable bearing pressure provided by the geotechnical engineer is normally calculated from the ultimate bearing capacity using a factor of safety (usually of the order of 3). The footing size is then based on the total load divided by the total allowable bearing pressure. While it is possible to deduct the weight of any surcharge loads to get a net allowable bearing capacity for the design of the concrete, designers need to check with the geotechnical engineer that the allowable bearing capacity allows this deduction, for the footing being designed. For the design of both plain concrete pedestals and plain concrete pad footings (unreinforced), designers should refer to AS 3600 Section 15, and for reinforced footings to AS 3600 Section 16, which in turn refers designers to Section 9. 8.2.2 Development of column starter bars The transfer of load from the column to the footing is by a combination of end bearing of the column and the starter bars in the footing. AS 3600 Clause 12.6 requires that, unless special confinement reinforcement is provided, the design bearing stress at a concrete surface shall not exceed 0.9 f f 'c √(A2 / A1) or f 1.8 f 'c whichever is the less. This usually means that the area of the starter bars does not have to match that of the column reinforcement. However, for both columns and walls, it is common to use the same size and number of column or wall starter bars as used for reinforcement in the column or wall above as shown in Figure 8.2. AS 3600 includes no specific requirements for the minimum area of starter bars but ACI318 8.4 requires a minimum area of 0.5% of Ag, which seems to be prudent. This issue is discussed in more detail in the Design Handbook for Reinforced Concrete Elements 8.5 and with design examples. The initial estimate for the footing depth is often based on the ability of the column or wall starter bars to transfer the applied forces into the footing. The usual condition is compression in the column or wall bars, and sufficient depth must be available to develop the expected compressive force (see AS 3600 Clause 13.1.5.1 and Chapter 2). However, if the footing is fixed, ie resisting moment, the development length of the bars on one or more faces of the column or wall may be in tension under certain loads. This may require the development length in tension to be considered, frequently with cogged starter bars being needed. Splice length Scabble surface Development length Two sets of ties for 50 support of starter bars Cogs 50-mm blinding layer (if required) Extend and cog corner bars (typical) Figure 8.2 Typical pad footing Table 8.1 Footing depths (rounded to the nearest 50 mm) assuming full development length of starter bars in compression Starter bar size N12 N16 N20 N24 N28 N32 N36 400 500 600 700 800 900 1000 Table 8.1 can be used for initial sizing of footings. It assumes: a concrete of f 'c ≥ 25 MPa; a bottom reinforcing mat consisting of two layers of N20 bars each way; 50 mm cover to the bottom bar of the mat; the column or wall starter bars have to develop their full development length in compression. Note that in a column the corner bars are usually cogged to support the reinforcement cage off the bottom reinforcement mat and fitments are used at about 300-mm centres to allow the cage to be held in place and fixed. If the column or wall bars do not need to develop their full development length in compression then the depths shown can be reduced. All of these assumptions will need to be checked against the final design requirements and adjusted appropriately. With a minimum embedment length of 200 mm for the column or wall starter bars required by AS 3600, and using the design parameters above, the minimum depth of any reinforced pad or strip footing will be about 300 to 400 mm. The minimum depth of any unreinforced footing allowed by AS 3600 Clause 15.4.1 is 200 mm. Consideration needs to be given to the minimum concrete strength of the footing required to meet the durability requirements of AS 3600 Section 4. Reinforced Concrete Design Handbook 8.3 For large differences in concrete strength between the column and footing, additional calculations may be required to determine the effective strength of the concrete at the column/footing interface; this may require additional starter bars and possible confinement. This is particularly important for high‑strength columns and walls with combined or coupled footings where the column or wall may be at the edge of a footing, eg at a boundary. With this type of footing the moments due to the offset column or wall are resisted by a combined footing or by a tie or coupling beam back to an adjacent footing. Designers can refer to texts such as Reynolds Reinforced Concrete Designer's Handbook 8.6 for the design of combined and coupled footings. Having established an initial pad footing depth, then the shear and bending at the following critical sections are checked as required and the footing depth, concrete strength or both are adjusted as necessary. N* a do a – do do qu qu Critical section for beam shear A 2 = L 2 (a – do ) c2 A2 L2 c1 8.2.3 One-way (beam) shear action The nominal shear stress appropriate to this condition is calculated as for a beam across the critical shear plane located a distance do (the footing depth) from the face of the column. The beam shear perimeter extends across the whole width of the footing, and the load carried is that portion of the total load located between this perimeter and the outer edge of the footing Figure 8.3. Shear is calculated along each axis of the footing using the appropriate value of do for each direction although it is conservative to use the lesser value of do. The value of β1 used in calculating Vuc in AS 3600 Clause 8.2.7.1 will generally be about 0.8, assuming no shear reinforcement is provided. This means that the shear strength, f Vuc , will be up to about 30% less than when using the previous edition of AS 3600 and it may be a critical design case especially for higher bearing pressures. 8.2.4 Two-way or punching shear action Nominal punching shear stress is determined around the critical perimeter at a distance of dom / 2 from the column face, where dom is the mean value of do around the column Figure 8.4. The total shear force to be resisted by the punching-shear perimeter is the total footing reaction minus the load on the zone inside the punching-shear perimeter. At higher bearing pressures, it also may be a critical design case. L1 Figure 8.3 One-way (beam) shear action of spread footing N* dom 2 do qu A2 = L1 L2 dom qu Punching shear perimeter – ∆ A1 ∆ A1 A1 c2 L2 c1 8.2.5Flexural action The design of a footing is similar to that of a cantilevered member. The critical perimeters for bending moment are located, along each axis, at the column face for concrete columns or wall face 8.4 Reinforced Concrete Design Handbook L1 Figure 8.4 Two-way or punching shear action of spread footing N* total allowable bearing pressure) for the design of the concrete based on the actual bearing pressure at the foundation less the weight of the footing. While it checks the minimum reinforcement required, it does not check cover, spacing requirements or detailing requirements. It is limited to footings with a concrete strength of up to 50 MPa. d qu Critical section for bending moment in L2 direction The spreadsheet assumes the footing is not over reinforced and that ku ≤ kuo. Critical section for bending moment in L1 direction c2 L2 c1 L1 Figure 8.5 Flexural action of spread footing for concrete members for walls. The bending moment is calculated as the cantilever moment carrying the design pressure over the full width of the footing Figure 8.5. For simplicity, the smaller value of do is sometimes used, as minimum reinforcement can often control. Generally, pad footings are only lightly reinforced. For an under reinforced section, the ultimate moment capacity, Mu,is approximately equal to 0.85 Ast fsy d within about 10% of a more accurate calculation, ie it is independent of the concrete strength and the width of the footing. This approximation is used to estimate the reinforcing steel required. The spreadsheet checks the actual bearing capacity against the allowable bearing capacity. If it exceeds the allowable bearing capacity, the plan dimensions of the footing will need to be increased. It also makes an initial approximation of the reinforcement required as well as the minimum amount required. The designer then selects and inputs an actual area of reinforcement (provided by a number of bars, usually of one size in both directions) to satisfy the larger of these two amounts. Then the spreadsheet calculates the actual moment capacity and compares it with the design moment, which has been input. It also checks the one-way and two-way shear. If any of these is less than the required or allowable values, then a further iteration will be required. This iteration can involve larger areas of flexural reinforcement, higher concrete strengths or a deeper footing, or a combination of these. o tN ee An appropriate reference should be consulted for the position of critical sections for steel columns. It can, however, conservatively be assumed to be at the edge of the column or, sometimes, halfway between the column face and the edge of the base plate if it is unstiffened. Sh b Jo o bN Jo o tN ee Sh By No te: Da By e: Dat t en pm elo ev ll d N28 0 80 d fu e ne ars rb rte N24 sta 0 ing 70 r ea 2 N3 0 90 s rtie pe 0.8 ro lP ria te h 32 Ma ds 0 an 50 re φ 6 xu um 0.3 fle sion f' c ss g a N20 g in ten tin f sy tin of 0 60 foo foo yers of k uo r . rete la ea depth N16 use nc r 2 h 0 ll its S co (1 o .r 50 d era ernfo . are nt an ry ov 12 osivble qu eme use 1 g s g a L s N it in in or ng ayon L2 0 msp for nd lim lar rra 40 le ntiss re gu l a Be Pre sib meet tan tee on nize epth ireshe rec n s resp rs d quad sig C1 ed give Ba rall t is t rere e De ee . forc r a enis sp sh Ov 00 C2 ng rein n fo 36 cegmth read oti sp gly ctio AS forsin ing Fo n sin e se to reoinn uforc m this ig a m te g ts rs s g re sin en eum for of th pe rein De mm nu nc rcin h bar em inTim nts ity so ng uir e m Co reinfo me ac er: sing m2 per 50 oti req hearim ire ap n he rce 3,3 0 Fo tric bar qu g c n cl n u tio :T ing mm 5 l fo te en ith do ac s wis m ta t re din tio mer 3,3 2 nta sp lab Dcolu pu nc ed w en en mm cre clai n .2 zo uta nd o s te om em nt b Co forc on Dis 11 0 Pla ori N* mm t a e t re C mp 0 rt h forc ome lar rein dC en abl onc D ob s. lie 1,0 0 t on Co t gu nder 999 rein te m em pplic r a c rce eCn ject/Jcitie 7 n n b rc u ile fo a m o r 1 a fo y a fo /J Clie in 5 ns ltim cta re e, ve mo ent.Pro ap .min. 84 rein ularl hear ject Co Re Re that a ourn s te e u no em ing c Abjestct m s c te Pro d h ct i th u u rc t s ge lb it fo S nd & ay bje t). im ar ula s nta rce oting , Me 2007 Su ly w in be .min alc late en min is p 2 w ) % rce on r re in uo m info r fo an on, : C alcu rcem all heet and Pe 100.6 ds ea el M 5m mm C fo ers ds ay Re sign fo o1n.1gm ears on g pa sh ste ng in sid Ln a 1w rest m )e P pti tin : re and sion ulati re a , e o , ce o ri m r a 0 n D e s n c le u 1 n Fo revisi ics Co e sp ecks es sq ear res alc . 20 ctu ista to Tit m2 Re Bas put Pad 50 or r sh omp s in c sion 1, - d ded lD u 4 Th d ch t r f ra In m n tr n c s ten la fo of o o 0 S yo ne rete ta an gu n me tre rou ott n rd 45 25 Ge nd crete onc Da metr th L 1 D ea ze b reo, tan desig ects pres rs in ctio 0 e c , a K e o sh C e .2 rs m g n e S O le 0 b a kP d r si eff for L2 pth r re ot 0. Len g A , Co rced 0G De om b re an - ba tensi Fo es n s the llow mem th kN s tin 36 al tt Do ore ot a for Wid area omin bo flexu bars of the ora aulke einfo AS kN to for 0 in d m R r orp qu Ign es n itable 30 9 tto Pa rall N ver nc ll & F ick, ed d : 6 bo r laye co us tr th a Do t su 9 (I ns kN 2,1 8 Ove oemnt dep over pplaee. lly 00 n, H Kilpa a tt No tio a 89 -2 f uu era kN Beom ive D -c orm ita <B s en 00 Rang r and forc Effect epth trogidfo Bp rtie kN reo Lim 36 sg L1 pe 81 D censin rein o AS rner, Foste ed na 2 to ro kN l u n ) o 8 (= p e ti 4 uir d re l ut l lle L 2 et, a thlly sio rge Wa rner, 3,3 86 ota ca inp teria e toca ara l to mn req ire kP ha ten dn ati B a surc reti olu C 1 p ralle adsh of ata ma 4,2 8 the requ Wa an eo a re fC r, tom ed of ny ers a 1.2 Th ols au y o ength C 2 p sp uir ove the kP ga lay on 8 is s/ L mb etr req try, c * ted din a 29 ati n of r1 de : Sy om mn idth g th kP lls clu ula xim atio Co sis Ge Colu n W rdain ce ome nts M ter fo Bp t in e) alc bars ro u c ty a w re a e m 3 e b p e B g t lu n gA llo arg g aci : nd ap roxim B2p 7 idth m) m) ye tion mom r diam de Co clukmre ap eigh ErcDh re 0 p m) rou forcin lw an th gra gra c ing ao g C g w CyKsu 35 kg ns is n ap latu ac tota l leng Se ding & ba t gra pro pro rogra dbC ac f rein tio earin xcl ft HgEan nc sp n n for ure is a fee the the e p ac nb B o Be ber eme tota nd me .m din er fig by by y th ed able G (E cl YuC ree cing m g 40 ta for kN No is gure ) oth pli d L(ILn tin ted ted d b Nu forc en a hg 60 .m Th fi or Ap Allow Loa U in ula lcula late gram f Foo : g wit d sp 12 Q em kN is u ns tin 32 alc ut Re ad aFd Mu Th tio ells e an φM u inforc oo (C (Ca alc e pro ide o 1,2 2 De eBLEo Inp 2 < φ 36 (C th es d ht fF dc O m iv gg M* ,23 eig Loa d N* d by nders eo xe r, siz acity um re 20 L T u 1 t m o 2 id W s 0 n 10 e p B L , qu rs g rking Loa late at U im 3.4 me ns mb nt ca de tin mm TIL 90 ire 32 tio oo l Wo ate alcu city min Un n Nu 4.0 0 ,2 e S u c F r t 4 a q m e a m g (C p E fo ta 4 lti 4 re 90 AA orr 1 esi Mo cks To l U ctor g Ca 80 city 5,0 83 d NT S AR 4,2 , c CC ta Fa 4.2 tate ire e pa for D n E o ts s a h e qu ri n T 10 C .re e th ty t: 5.0 0 ea ad gC mit 28 MM ET A st quired ct Lo al B rin apaci mm r li tpu ,10 u 16 CO SHE ea co conta fo 4 st.re 5 6 C ct Ou B A r 3 A R es 5.3 D n ng 81 Fo ase 0 n sig siz 4 ari FO EA 6.0 4 3.4 u & 2 De d Be ctio n ple φM FT PR rs 2 et ire 4.0 ,82 Designers should note that if straight bars are used in flexural action in footings, the development length of the bar must be checked at the critical section for the bending moment at the face of the column (see Table 2.7). In addition, the minimum anchorage may be needed to develop the stress in the bar at the outer edge of the footing, as the moment increases. A cog will usually meet that requirement. When using larger diameter bars in the footing, designers should also check the stress at the location where the cog starts; a hook may be required. 8.2.6 Basis of Spreadsheet 8.1 Spreadsheet 8.1 can be used to calculate the reinforcement requirements for a concentrically loaded reinforced rectangular pad footing in flexure and shear in accordance with the Flowchart 8.1. It uses a reduced bearing pressure (which is less than the Spreadsheet 8.1 is available at www.ccaa.com.au 8.2.7 Basis of Charts 8.1 and 8.2 After calculating the flexural reinforcement, designers must check the minimum tensile reinforcement required in accordance with AS 3600 Clause 8.1.6.1. Charts 8.1and 8.2 show these minimum reinforcement requirements assuming 50 mm and 75 mm cover respectively with two layers of N20 bars and various concrete grades. Pad footings are not usually heavily reinforced, so minimum reinforcement may be the critical tensile reinforcement, especially at low bearing pressures. A S DR HE 1.0 TE T NO k: ac db e Fe . n rsio Ve ft. x le ft. bo w ox le llo b d io re ye 6 4 0 be 45 u s w *< 40 L1 ect 6.9 1 the yello um 65 itie r M < φM 5,0 83 the L 2 dir to r n 20 4.2 ac 7.0 2 d fo r M* in 10 ba ap the a in 1 ire nt 5.0 0 the 14 nC are into qu ed fo 9 ted 4,3 2 3 me t in 0 t s ig .4 re o e s g 9 ir n pu area 4 5,1 3 54 gg .0 ion qu De ndin ate M ome 5.3 t d in 81 10 0 Su 16 ect n re OK an inpu 2 Be Ultim te M 6.0 4 dir nt io 1 L1 d 4,5 1 L 1 irect 20 2 me .66 ma OK for 2 an in d 6 13 .0 ge 4,8 0 36 Ulti le L A st in L 2 an 6.9 65 14 6 tab 12 for Mu al arr 9 7.0 2 3 <φ u nt Initi al A st ove table 4,3 0 11 1 M .34 ab M* me 9 21 .0 4,3 2 25 <φ Initi rce the above 9.4 M* 54 22 2 ng .0 e info si 0 2 th u 1 re g 4,4 5 d m 20 .96 rea usin se kN 37 .0 4,5 1 15 .66 da po a 3 m n 6 8 1 u a 3 re 3 6 kN .0 M Pro 94 m2 ng nd a 14 6 ,17 <φ u a 4,2 8 aci 9 ize bar m 13 76 M M* sp cing 8 4,3 0 rs ,1 <φ 6 .34 rs, uo a Ba a of 13 .3 21 .0 ars 25 M* ba sp φM <0 n . of b of ars, 22 2 Are uo d tio 2 no k u0 b m φM ec al No uire otal mm kN se o of 4,4 5 .96 q T Dir rs oo n m 37 .0 15 L 1 oretic ars re m2 f ba Ch ose kN m o 69 e 38 4 o Th ual b bars cing 1,7 9 Ch 2 ,29 6 a f 4 ct 2 A a o l Sp u bd 2 1,7 4 88 φM mm 2 o) d Are mina .0 ars u ku 0 b b m M a φ n . of o) m No 26 d tio mm a ku ku 53 6 ec al No uire otal mm ga mm m q 2 T Dir rs -0.5 gam 53 L 2 oretic ars re m2 f ba 5 m o e o (1 -0.5 84 Th ual b bars cing a ku o (1 ion Act a of l Spa ect n mm a ku dir io ga Are mina L 1 irect f'c mm 2 d ga ed d in No ha f'c de in L 2 rovid lp 6 vi .p 2 st a lA .3 ro ed d =A .p ns =0 itia lph A st rovid As tio t kuo hi_in al A ith block .p A st *w ula uo a = P i_initi = M stress alc ent M ction = Ph s c)) n C om dire ion ck d f' crete n sig m M L1 irect blo / (b ss f sy for co De ximu φMuo L2 d st 2 tre 2 (A ) bd Ma Check φM uo gs / in s ku .5 eck tu - 0 .5 d Ch ire en -0 d (1 qu om f sy k u (1 A st.re st k M on g = A f'c ec ti sin u Ch irec φM = Ø ted u D L 1 Check φM u lcula ca eck Ch ck k u e Ch n tio ec Dir L2 N cto Fa It is common practice to use N12 bars in footings up to 1000 mm wide, N16 bars for those up to 2000 mm Reinforced Concrete Design Handbook 8.5 wide, N20 bars for those up to, say, 4 m wide and N24 bars for those more than 4 m wide. It has been shown in practice that these sizes provide bars at reasonable spacing. Spacing should not exceed about 300 mm. Mesh reinforcement is not normally used for pad footings, as the minimum reinforcement would usually require at least two layers of mesh, unless the footing is very small or has been designed as unreinforced concrete. B L Y Refer to AS 3600 Section 4 and Chapter 3 of this Handbook for minimum concrete cover requirements. 8.3 Strip footings The determination of force resultants in strip footings generally falls into two cases: n Rigid footings n Flexible footings. Strip footings are used to support line loads, which are applied to the footings from a continuous wall or from closely spaced columns. Stiff footings, or strips supporting stiff superstructure elements that will not allow significant differential settlement to occur along the line of the footing, may be designed using rigid body theory with linear distribution of soil pressure. An assessment must be made of the relative stiffness of the combined foundation and superstructure system (including the soil). This will determine the type of design method to be used. Designers should consult relevant texts, or the method proposed by Meyerhof 8.7 may be appropriate. The strength design of a strip footing will follow the same principles as for the spread footing, after the determination of the design forces. Strip footings carrying a stiff longitudinal load, eg a concrete wall, may require only a design check for transverse bending and shear forces. The designer must ensure that the design of the elements containing stiff and strong superstructure components does not result in specific forces that cannot be tolerated by the supporting strip footing. An example of this would be a shear wall with moments parallel to the wall resulting in tension or high bearing pressures under the footing. 8.4 Pile Caps Soils that provide insufficient strength to economically support spread or strip footings often require the installation of piles (or piers) to transfer the applied loads to a stiffer stronger stratum, or by skin friction over the length of the pile, or a combination of both. There are many types of piles and the selection of the type of pile will depend on experience, availability of 8.6 Reinforced Concrete Design Handbook B Z X L L L Figure 8.6 Some simple pile cap layouts piling contractors, ground conditions and other factors. When two or more piles support a column, then a pile cap is normally required to transfer the load from the column to the piles. As for all footings, specialist geotechnical advice should be sought on choice of pile, pile design and pile testing. The action of simple small pile caps can be treated in a similar manner to a beam or slab as described in AS 3600 Clause 8.1. A pile cap transfers the column loads to discrete points on the underside of the footing, instead of the continuous contact area of the footing supported directly by the soil. Figure 8.6. The size of the pile cap (footing) is determined by the number of piles required to carry the vertical and horizontal load and moments with the minimum Flowchart 8.1 Design of spread footings Determine allowable soil pressure, f 'c and cover, taking into account durability requirements Calculate depth of footing based on development length of column starter bars AS 3600 Clause 13.1.5 Calculate plan area of footing based on allowable soil pressure and applied loads AS 1170.0 and Table 1.1 Increase footing depth and/or concrete strength Increase footing depth and/or concrete strength One-way (beam) shear Does footing satisfy requirements Flowchart 4.2? no yes yes Is there a base moment? no no Slab punching shear (M v*> 0) Is V *< φVuo? AS 3600 Clause 9.2.4 Slab punching shear (M v*= 0) Is V *< φVuo? AS 3600 Clause 9.2.3 yes no yes Calculate reinforcement for flexural action Check minimum reinforcement requirements for flexure AS 3600 Clause 16.3.1 stop Reinforced Concrete Design Handbook 8.7 chart 8.1 Minimum reinforcement in pad footings with 50 mm cover 3000 f 'c (MPa) = 100 2500 f 'c (MPa) = 80 f 'c (MPa) = 65 2000 f 'c (MPa) = 50 f 'c (MPa) = 40 f 'c (MPa) = 32 1500 f 'c (MPa) = 25 Area of reinforcement (mm2/m) fsy = 500 MPa 1000 Note: Chart assumes 50 mm cover with 2 layers of N20 bars bottom b 500 d Reinforcement 0 300 400 500 600 Overall depth of footing (mm) 8.8 Reinforced Concrete Design Handbook 700 800 900 1000 1100 D chart 8.2 Minimum reinforcement in pad footings with 75 mm cover 3000 f 'c (MPa) = 100 2500 f 'c (MPa) = 80 f 'c (MPa) = 65 2000 f 'c (MPa) = 50 f 'c (MPa) = 40 f 'c (MPa) = 32 1500 f 'c (MPa) = 25 Area of reinforcement (mm2/m) fsy = 500 MPa 1000 Note: Chart assumes 75 mm cover with 2 layers of N20 bars bottom b 500 d Reinforcement 0 300 400 500 600 700 800 900 1000 1100 Overall depth of footing (mm) Reinforced Concrete Design Handbook 8.9 D practical spacing between the piles. This depends on driving conditions, the driving tolerances and load performance of the pile as specified by the designer or supplier. Typically, the pile spacing is 2½−3 times the pile diameter, but the spacing must be confirmed prior to design. It should be noted that the settlement of a group of piles is greater than a single pile for a similar axial load. Settlement generally increases as the pile spacing decreases. When a single large diameter pile is used, a pile cap may not be required. Design of piles for buildings, including load testing is covered by AS 2159 8.8. Designers can also refer to texts such as Pile Design and Construction Practice 8.9 for the design of piles and pile caps. Piles are typically cast to about 75−150 mm above the bottom of the pile cap and then cut back to about 25 mm above the bottom of the pile cap with the reinforcement projecting into the pile cap as required. The initial pile cap depth, subject to final design, can be taken as the horizontal distance from the centre of the column to the centre of the outermost pile. Pile caps should extend at least 150 mm beyond the face of any pile. The pile cap depth will need to be checked against any required development lengths for wall or column starter bars and for shear. Bending moments and shear (punching and beam) forces are determined as above except for the following: n n Any pile located do / 2 or less from the face of the column is not considered to be adding external shear forces to the cap. For very deep and/or large pile caps, consideration may need to be given to heat of hydration of the concrete and the control of thermal cracking, especially if the pile cap is over, say, 1200 mm deep. 8.5 For all reinforced concrete footings, details of the footings and any special detailing must be shown on the drawings. AS 3600 Clause 9.1.3.1 sets out the detailing of flexural reinforcement for slabs which is used for footings. This requires that where the bending moment envelope has been calculated, the termination and anchorage of flexural reinforcement is based on a hypothetical bending-moment diagram formed by displacing the calculated positive and negative bending-moment envelopes a distance D along the footing from each side of the relevant sections of maximum moment and to which the development length must be added. This development length in pad footings is usually achieved with cogs (or hooks) to the footing bars as shown in Figure 8.2 and discussed above. Designers should also refer to Chapter 11 of the Reinforcement Detailing Handbook 8.10 for further guidance. Areas needing consideration in the documentation of footings include: n n The pile cap must be checked for the punching shear applied by the piles from the underside of the pile cap as well as the punching shear applied from the top of the pile cap by the column. Pile caps can also be designed by the strut-and-tie method as set out in Chapter 9 Strut-and-tie modelling. A minimum cover of 75−100 mm should generally be used for piles and the bottom and sides of pile caps for practical reasons, unless larger covers are required, eg by AS 3600 for durability reasons, as discussed in Chapter 3 Durability and fire resistance. Flexural reinforcement is provided in the top and bottom of the pile cap and the bars usually have cogged ends. Generally, the bars are in two orthogonal directions for rectangular pile caps, although for a pile cap for two piles, fitments are usually used in the short direction. For non-rectangular pile caps there can be three or more layers of flexural reinforcement, top and bottom. Horizontal reinforcement should also be provided to the vertical faces of the pile cap at about 200-mm centres vertically and lapped as required. 8.10 Reinforced Concrete Design Handbook Detailing n n n n A plan of the footing (including pile caps) should be shown on the footing drawings, which may be combined with other schedules. The footing sizes (including pile caps) should be shown on the drawings, which are sometimes combined with the column schedule. It is usual to size pad footings and pile caps in plan dimensions in 100-mm increments and in depth in 50-mm increments, eg 2.0 m x 600 mm deep, or 2.1 m x 3.4 m x 850 mm deep, etc. Footing beams should be shown on a plan or a beam schedule or similar. Details of footings and pile caps should include (as a minimum) the founding level or stratum, the size of the footing or pile cap and the RL to the bottom or top of the footing/pile cap, the reinforcement including starter bars, the concrete strength, the cover and any construction joints. The details should also show if the columns are concentric to the footing or pile cap and if not, offsets should be dimensioned. It is suggested that the minimum concrete strength for reinforced pad and strip footings be 25 MPa unless higher strengths are required by AS 3600, eg for durability reasons as discussed in Chapter 3. The allowable covers in AS 3600 do not often reflect the actual conditions on site when excavating in ground, which can be variable. It is recommended that the minimum cover to reinforcement to the bottom of pad and strip footings be 50 mm when a layer of blinding concrete is used or where the ground is firm, unless larger covers are required by AS 3600, eg for durability reasons as discussed in Chapter 3. Where the ground conditions are difficult, 75 mm or greater cover may be appropriate. n n n n n n n n n n n n AS 3600 Clause 4.10.3.5 specifies that where footings are cast against ground the cover is to be increased by 10 mm when a damp proof membrane (DPM) is provided, otherwise by 20 mm. It is not usual to provide a DPM to pad footings and generally not possible to pile caps. Top reinforcement is generally not needed in pad footings unless the footing has to carry uplift forces (such as in cyclonic conditions), or if it is part of a combined or tied footing system, or is a pile cap. Footing beams that intersect with footings may have different effective depths because of the bars being in layers. The same size bar in flexure should be used in both directions in square pad footings and each direction in rectangular footings, to avoid fixing errors on site. Groove formed by seepage of concrete (overpour) beneath edge of form boards Absorbent compacted sand/rubble fill Fitments in footing beams and strip footings may have to be designed for shear along the member. For footings, the fitments may also act as the flexural reinforcement perpendicular to the span of the footing for the outstand of the footing beyond the wall. Natural soil and vapour barrier extension, sloping towards footing 8.6general guidance On any particular project, other matters that may need consideration include: n n n n n n Complicated or unusual footings, footing beams and pile caps should be shown in plan and elevation. Construction joints in strip footings and footing beams should be properly considered and detailed. If earthquake design is a consideration, compliance with AS 1170.48.11 and AS 3600 Appendix C should be checked. Vapour barrier Figure 8.7 Slab edge dampness Reinforcement at the junction of strip footings, footing beams and spread footings with columns should be reviewed to ensure that it all fits. If footing beams and strip beams are shown on a schedule on the drawings, the schedule should be checked to ensure that all details fit within the constructed shape. Salt attack or damp and efflorescence appears here Paving Side face reinforcement is not normally provided for footings but is usually provided in pile caps or deep strip footings. Footing beams and strip footings will usually have top and bottom reinforcement with fitments. DPM n Advice in the geotechnical investigation whether the footing will be likely to be cast in aggressive soils as defined in AS 3600 Clause 4.8. Tie beams or similar in both horizontal directions may be needed for earthquake design for pad footings or pile caps that are located in or on soils with a maximum vertical ultimate bearing capacity of less than 250 kPa. This is to limit differential horizontal movement during an earthquake as required by AS 1170.4 Clause 5.2. Will blinding concrete be required to the base of the footing or pile cap? Specify who is to confirm the bearing stratum and bearing pressure (assumed for design) in the excavated ground, before placing the blinding concrete or fixing of the reinforcement. Will the ground conditions allow vertical sides of excavations to stand for sufficient time to permit placement of reinforcement and concrete without danger to those carrying out the work? If not, batters may have to be provided and the vertical faces of the footings formed and the excavation later backfilled. This may require advice from the geotechnical engineer. Generally, excavations deeper than 1.5 m will require shoring or battering of the excavated sides, in accordance with the various state and territory Worksafe safety regulations. For strip footings and footings associated with a slab-on-ground with a habitable area above, a 200‑µm DPM must be provided under the entire slab and footing and be returned up the edges of the slab/footing beam. Reinforced Concrete Design Handbook 8.11 n n n n n For strip footings in non-habitable areas consideration should be given to providing a 200‑μm DPM under the footing to assist in preventing any rising damp. For footings and footing beams, projecting concrete outside the theoretical formed face due to poor construction procedures can allow slab edge dampness to occur as shown in Figure 8.7. Slab edge dampness can be a problem in arid and saline type soils and designers should seek specific advice from the geotechnical engineer on this matter as required. Refer to CCAA Data Sheet 8.12 and CPN 30 8.13 for more information. Footing faces and footing beams on property boundaries should be formed or constructed so that the concrete does not encroach on adjoining properties. Excavations of footings and pile caps must not undermine adjacent footings or structure, especially on adjacent sites. Provide any rebates or set downs for walls over as required. References 8.1 Craig R F Craig's Soil Mechanics 7th Ed, Spon Press, 2004. 8.2 Curtin WG, Shaw G, Parkinson G and Golding J (revised by Seward NJ) Structural Foundation Designers' Manual, 2nd Ed, Blackwell Publishing, 2006. 8.3 AS 2870 Residential slabs and footings – Construction Standards Australia, 2011. 8.4 Building Code Requirements for Structural Concrete (ACI 318-08) ACI Manual of Concrete Practice, 2010. 8.5 Beletich AS and Uno PJ Design Handbook for Reinforced Concrete Elements, 2nd Ed, UNSW Press, 2003. 8.6 Reynolds CE, Steedman JC and Threlfall AJ Reynolds Reinforced Concrete Designer's Handbook, 11th Ed, 2008. 8.7 Meyerhof GG 'Some recent foundation research and its application to design' The Structural Engineer (London) Vol. 31, No. 6, June 1953, pp 151–167. 8.8 AS 2159 Piling – Design and installation Standards Australia, 2009. 8.9 Tomlinson M and Woodward J Pile Design and Construction Practice, 5th Ed, Taylor and Francis, 2008. 8.12 Reinforced Concrete Design Handbook 8.10 Reinforcement Detailing Handbook (Z06), 2nd Ed, Concrete Institute of Australia (CIA), 2010. 8.11 AS 1170.4 Minimum design loads on structures Part 4: Earthquake loads Standards Australia, 2007. 8.12 Slab Edge Dampness and Moisture Ingress Data Sheet, Cement Concrete & Aggregates Australia, 2005. 8.13 Slab Edge Dampness (CPN 30) Concrete Institute of Australia, 1998. Chapter 9 Strut-and-tie modelling 9.1 Introduction Strut-and-tie modelling is covered in Section 7 of AS 36009.1, with formal design rules for three types of models. This section should be read in conjunction with Section 12 which also covers the design of non-flexural members, end zones, concrete nibs, corbels, stepped joints and bearing surfaces. AS 3600 Clause 2.2.4 sets out the strength check to be used when using strut-and-tie analysis. Strut-and-tie modelling has been around for many years, has attracted quite a deal of research and is now an established design method allowed by most concrete codes, including the ACI 9.2, since 2002. Chapter 7 of the Precast Concrete Handbook 9.3 has a section on strut-and-tie design based on the ACI which is similar to AS 3600, with two worked examples. This design method is used for non-flexural members such as deep beams, pile caps, corbels and nibs and the like. It can also be used for non-flexural portions of other structural members such as dapped ends of beams, holes in beams, etc. The strut-and-tie approach is based on the designer selecting the load paths and designing according to the chosen design model. This requires the designer to choose realistic load paths within a member in the form of an idealised or notional truss. See Figure 9.1 for a typical example of a simple strut-and-tie model (STM). P Nodal zone Tie R1 Bottle-shaped struts Idealised prismatic struts Figure 9.1 A simple strut-and-tie model R2 Non-flexural members are defined as members where the ratio of the clear span or projection to the overall depth is small. The ratios must not exceed those set out in AS 3600 Clause 12.1.1: n Cantilevers 1.5 n Simply-supported members 3 n Continuous members 4. The geometry of the notional truss is determined by following the flow of forces from the support reaction into the body of the supported element to the points of the applied load(s). The intersection of compressive struts with tension ties or support reactions defines the nodal zones. The axes of the struts and ties are chosen to coincide approximately with the axes of the compression and tension fields in the real element. Struts should be generally parallel to the axes of cracking. The struts, ties, and nodal zones making up the STM all have finite dimensions that must be calculated and must be taken into account in selecting the dimensions of the truss and the member. Some of the key assumptions for strut-and-tie analysis are: n Ties must yield before failure of the struts for ductility. n Forces in the ties and struts are axial only. n Tension in the concrete is ignored. n External forces are applied at nodes. n The ties are fully developed before the node, which requires anchorage outside the node. Traditionally, STMs are developed using the elastic stress distribution and load path methods. These methods involve a trial-and-error process based on the designer's intuition and experience. The STM obtained by such methods is not unique and can vary with the designer's understanding of this method of analysis. While the method is a useful design tool, to achieve the assumed load path, significant redistribution of the internal forces may be required which the structure may not be capable of accommodating due to limitations of the concrete and steel. Care is therefore needed to ensure that the member is capable of accommodating the redistributions required without undue distress to itself or adjacent components. As a general rule, designers should choose an STM that allows the structure to behave as they would expect it to behave and then to design as close to possible to that model, to minimise the demands on the structure. The strut-and-tie method is therefore in many cases a design tool for the experienced engineer and is not a simple standard analysis procedure. Even where the design of an STM is simple, such as for a corbel, the design by inexperienced engineers must be checked Reinforced Concrete Design Handbook 9.1 by others to ensure they have fully understood the design issues involved because of the usual critical nature of the element. STM design involves a trial‑and‑error approach with hand calculations and hand-drawn sketches to scale, etc and usually cannot be undertaken totally on the computer. For further information on the design of STMs refer to Foster et al 9.4, PCA9.5, 9.6 and others9.7. 9.2 C-C-C node C-c-t node C-T-T node Figure 9.2 Various nodes (dashed line indicates compression struts and full lines indicate ties) Truss geometry 9.2.1 General AS 3600 Section 7 requires the following: Force discontinuity (a) Loads shall be applied at nodes, and the struts and ties shall be subjected only to axial force. B-regions D-regions (b) The model shall provide load paths to carry the loads and other actions to the supports or into adjacent regions. h1 (c) The model shall be in equilibrium with the applied loads and the reactions. (d) In determining the geometry of the model, the dimensions of the struts, ties, and nodal zones shall be taken into account. Geometric discontinuity B-regions D-regions h2 h1 h1 (g) For reinforced concrete members, at a node point the angle between the axes of any strut and any tie shall be not less than 30°. The B-regions (B for Bernoulli or Beam) are flexural areas within a member where the assumption that plane sections remain plane can be applied. The D-regions (D for Disturbance) are the areas of discontinuity. For design purposes, D-regions can be idealised as an STM. See Figure 9.3. 9.2.3 Nodes and nodal zones A node is a point in an STM where the axes of the struts, ties, and any applied concentrated forces, if applicable, acting on the joint, all intersect. Around this area is the nodal zone, which is the volume of concrete surrounding the node, and which is the element that transfers the compression and tension through the node. 9.2 Reinforced Concrete Design Handbook h2 h2 C (f) Struts shall cross or intersect only at nodes. 9.2.2 B- and D-regions h1 Figure 9.3 Examples of B- and D-regions in a horizontal member (e) Ties shall be permitted to cross struts. Once the geometry of the truss is chosen, the forces in the struts and ties are determined by statics, with the applied loads and the reactions being in equilibrium. For equilibrium, at least three forces should act on a node. Nodes are classified according to the signs of these forces as C-C-C (all compression, ie only struts entering the node), C-C-T (when there are two or more compression struts and a tension tie) and C-T-T (when there are two or more tension ties entering the node with a compression strut). See Figure 9.2. h1 Strut Extended nodal zone T θ Nodal zone θ ≥ 25° C One layer of steel C Strut Extended nodal zone T θ Nodal zone C Multiple layers of steel Figure 9.4 Extended nodal zone The greater the number of truss members meeting at a node, the less efficient the nodal zone is and this is recognised by the βn factor which varies from 1.0 to 0.6 as set out in AS 3600 Clause 7.4.2. The principal compressive force on any nodal face must not be greater than φst βn 0.9 f 'c. A nodal zone is called a 'hydrostatic' node when its loaded faces are perpendicular to the axis of the struts and ties acting on it, and the loaded faces have equal stresses. The nodal zone often needs to be extended because of the number of ties required. It is then sometimes referred to as an extended nodal zone Figure 9.4. In a hydrostatic C-C-C nodal zone, the ratios of the lengths of the sides of the node are in the same proportion as the forces acting on it. A C-C-T nodal zone at the end of a member can be represented as a hydrostatic node if the tie is assumed to extend through the node and is anchored by a 'notional plate' on the far side of the node as shown in Figure 9.4. The notional plate has bearing stresses that are equal to the stresses in the incoming struts. If the tie is to be anchored beyond the nodal zone, then at least 50% of the anchorage must be made beyond the zone. The use of threaded bar may be appropriate for such ties. 9.2.5 Ties A tie usually consists of either bar reinforcement (sometimes threaded), a prestressing bar or strand. It is usual to include a small portion of the surrounding concrete that is concentric with the axis of the tie to define the zone in which the forces in the struts and ties are to be anchored. However, the concrete in the tie is not used to resist the axial tension in the tie. The effective thickness in elevation of a tie for design can vary with the distribution of reinforcement in it. If the bars are in one layer, the effective thickness can be taken as the diameter of the bars in the tie plus twice the cover to the surface of the bars. Multiple-layers of reinforcement should be distributed approximately uniformly over the thickness and width of the tie. The reinforcement in ties can be anchored by hooks or cogs, or straight bar development beyond the node, or by anchorage plates. Statics must be satisfied at each node. At least 50% of the development length must extend beyond the nodal zone as set out in AS 3600 Clause 7.3.3. Mechanical anchorage, including welding, can also be used outside the node. Nodes can also be non-hydrostatic as shown in Figure 9.5 but the modelling becomes more complex. Foster et al9.4 provide further information on this type of node. 9.2.4 Struts Struts are normally concrete members idealised as either prismatic or uniformly tapered although they can be fan-shaped as shown in Figure 9.6. They can also be thicker at mid-length where the compressed concrete can spread laterally into the adjacent concrete to form a bottle-shaped strut. Prismatic struts can be used only when the stress field cannot diverge. Without proper transverse reinforcement, a bottle‑shaped strut cannot maintain equilibrium after significant cracking. Reinforcement is therefore needed, usually in two orthogonal directions, to maintain the strength of a bottle-shaped strut as well as reduce crack widths under service load. Longitudinal reinforcement may be used, located within the strut to increase its strength. Such reinforcement should be parallel to the axis of the strut and enclosed by ties satisfying AS 3600 Clause 10.7. The longitudinal reinforcement should be properly anchored beyond the nodal zone. The strength of a longitudinally reinforced strut may be calculated as for a prismatic, pin-ended short column of similar geometry. Hydrostatic node Non-hydrostatic node Figure 9.5 Hydrostatic and non-hydrostatic nodes (after Foster et al) Bursting forces Figure 9.6 Prismatic, bottle and fan-shaped struts (after AS 3600) Reinforced Concrete Design Handbook 9.3 9.3 Analysis of strut-and-tie models AS 3600 requires that the analysis of an STM to determine the internal forces in the struts and ties, has to meet the requirements of Clauses 6.1.1 and 6.1.2 and also compliance with Clause 6.8. This may require several STMs to be considered and analysed to compare results and from which a chosen model is then adopted and fully analysed and detailed. AS 3600 Clause 2.2.4 stipulates the following procedure for the strength check for strut-and-tie analysis: (a) The strut-and-tie model shall satisfy the requirements of Section 7 of AS 3600. (b) The forces acting on all struts and ties and nodes shall be determined for the critical combination of factored actions as specified in AS/NZS 1170.0 and Clause 2.4 by an analysis of the strut-and-tie model in accordance with Section 7. (c) The compressive force in any concrete strut shall not exceed the design strength of that strut determined in accordance with Clause 7.2.3. The strength reduction factor (φst ) to be used in determining the design strength shall be in accordance with Table 2.2.4. (d) The tensile force in any tie shall not exceed the design strength of the tie determined in accordance with Clause 7.3.2 where the strength reduction factor (φst) is given in Table 2.2.4. (e) The reinforcement and/or tendons in the ties shall be anchored in accordance with Clause 7.3.3. (f) The design strength of nodes shall be calculated in accordance with Clause 7.4.2 and shall not be exceeded. The strength reduction factor (φst) shall be in accordance with Table 2.2.4. The strength reduction factor, φst, for concrete in compression is taken as 0.6 and for steel in tension is taken as 0.8, reflecting that the tie should yield first. AS 3600 Section 12 defines three types of design models for non-flexural members. Type I is where the load is carried to the supports directly by major struts and ties. It will cover many simple STMs. Type II is when the load is taken to the supports by a combination of primary (major) and secondary (minor) struts. Hanger reinforcement is required to return the vertical components of forces developed in the secondary struts to the top of the member. Type III is similar to a conventional truss, where the load is carried to the supports via a series of minor struts with hanger reinforcement used to return the vertical components of the strut forces to the top of the member. The use of Type II and III models will require careful consideration by the designer. These are discussed in more detail in the paper by Foster and Gilbert 9.8. 9.4 Reinforced Concrete Design Handbook For the best results, it is recommended that a preliminary design be carried out before finalising the design of the chosen strut-and-tie model. The design of a strut-and-tie member will therefore involve the following steps: 1 Check that the member or section of member to be designed complies with the requirements of AS 3600 and define which areas of the member are areas of discontinuity, ie D regions, and which areas are flexural areas, ie B regions. 2 Determine the external actions and where the member is a mixture of B and D regions, determine the boundary actions, including any concentrated and distributed actions such as moments, shear and axial actions that will act on the STM being designed. 3 Sketch to scale one or more suitable strut-and‑tie models to suit the member being designed, including any boundary forces and choose the STM that will best reflect the internal actions and will not cause significant redistribution of internal actions. 4 Analyse the chosen STM to obtain the actions in each of the individual strut-and-tie members of the model. 5 Check the dimensions of the struts, ties and nodes required, including the necessary concrete strength and amend the STM as required. 6 Choose the material for the tie member which is often reinforcement and ensure that the tie capacity and the anchorages beyond the nodes are adequate. If insufficient, then amend the STM as required. 7 Design the struts, including any compression reinforcement, fitments and bursting reinforcement. 8 Complete the design and drafting, including the detailing as set out below. Figures 9.7 to 9.10 illustrate some examples of STMs (including Type I and III models). 9.4 Detailing All strut-and-tie element details must be shown on the drawings. This is a very important aspect for this design procedure. Designers can refer to Chapter 16 of the Reinforcement Detailing Handbook 9.9 for typical detailing of nibs and corbels. Designers should note that: n n Sufficient details including plans, elevations with large scale sections and details of the strut-and-tie elements should be provided. Struts must be properly detailed and with transverse reinforcement to the member in two orthogonal directions as required, so that splitting cannot occur. F1 = F2 B Node A C2 C1 T1 Ax R1 Node B T8 C9 C5 0 C1 T9 C6 1 C1 C1 2 T10 T6 T3 T5 C8 T4 C7 T2 T7 A D C4 C3 C Node C R2 Figure 9.10 A truss model for a deep beam with an opening (Type I and III) Ay Figure 9.7 An STM for a corbel on a column (Type I) Bearing pad Main column reinforcement (fitments not shown) Figure 9.11 Double corbels on precast concrete columns Figure 9.8 An STM for a corbel on a column showing the proposed reinforcement D Vf A C θ Nf Figure 9.12 A reinforcing cage for a precast concrete column showing the reinforcement for a corbel B Figure 9.9 STM for corbel on side of beam Reinforced Concrete Design Handbook 9.5 n n Ties need to be properly anchored beyond the node as set out in AS 3600 Clause 7.3.3. This often requires development with hooks, cogs or anchorage by mechanical anchors such as welding, plates, etc. Hooks and cogs have real dimensions and can take up a considerable space in strut-and-tie members. Smaller bars allow tighter cogs and bends and require smaller development lengths. This detail is suitable when using 16-mm size bar or smaller for main tensile reinforcement Distance between edge of bearing and inside of Two column fitments bar to be a minimum of should be placed the bar size or cover close to corbel top whichever is greater A 7 7 A 2 6 General Guidance The following will assist in both the design and the inspection of a strut-and-tie member for a particular project. n n n n n 9.6 Draw the truss model to scale to see it all fits within the overall member dimensions and is logical. The dimensions of STM elements should allow the strut reinforcement (if required), the bursting reinforcement (as required), the tie reinforcement and any other associated reinforcement all to fit together and within the overall dimensions of the member and to develop the tie anchorages. Strut‑and-tie elements will generally not fit within thin sections. The actual design of the nodes and nodal zones needs to be considered carefully and can involve a considerable amount of hand calculations. Always inspect the reinforcement in place prior to concreting to ensure that the design model used and what was detailed on the drawings is what is being constructed, as reinforcement fixing may not be simple and the reinforcement fixers may have changed the detail to facilitate their task. With stepped joints, corbels, dapped ends and the like, do not ignore horizontal forces at the joints even if they are not evident, as shrinkage, thermal movement, etc may induce horizontal forces due to friction. AS 3600 Clause 12.4 requires a minimum of 20% of the vertical force to be used as a nominal horizontal design force. The actual horizontal design force may be greater than this minimum and this needs to be assessed for each design case. Because of the dimensions of bends to ties and the cover, do not load corbels, nibs, dapped ends and the like, too close to the vertical edge of the concrete as the reinforcement may be inadequate and the edges will spall and may fail. AS 3600 Clause 12.3(a) requires the bearing to be located over the straight portion of the bar where looped (or cogged) bars are used. Always ensure such elements are loaded over the reinforcement and the actual bearing is back from the edge as shown in Figure 9.13. Reinforced Concrete Design Handbook Main tensile reinforcement large radius of bend as required 3 Tension lap 9.5 4 5 1 2 4 1 2 7 7 Secondary horizontal reinforcement. Total area of this should not be less 0.5 times area of main tensile reinforcement Compression anchorage 7 1 Outer compression bars offset or angled to pass inside fitments 7 345 4 6 5 6 3 7 1 7 2 Sectional plan on A-A Figure 9.13 Detailing of a corbel to a column (after IStructE Standard method of detailing structural concrete) Roughen side of beam Overall depth Structural screed Slab depth Beam rebates Neoprene bearing strip Reinforced and prestressed Precast beam depth Figure 9.14 Concrete corbel or nib on a precast beam supporting a precast concrete floor n Always be careful with concrete (and steel) members bearing on the concrete surface in corbels, dapped ends, slip joints and the like, as invariably spalling and cracking will occur if they are not correctly detailed. This can result in both unsightly and sometimes dangerous conditions if the concrete can fall. It is preferable to use a neoprene bearing strip or a slip-type bearing as appropriate, between the two surfaces. As a minimum, provide a chamfer on the corners and set the bearing area back from the edge as discussed above. See Figure 9.14. References 9.1 AS 3600 Concrete structures, Standards Australia, 2009. 9.2 Building Code Requirements for Reinforced Concrete ACI 318–08, American Concrete Institute, 2010. 9.3 Precast Concrete Handbook 2nd Ed, National Precast Concrete Association Australia and Concrete Institute of Australia, 2009. 9.4 Foster SJ, Kilpatrick AE and Warner RF Reinforced Concrete Basics 2E 2nd Ed, Pearson, 2010. 9.5 Strut-and-Tie Model for Structural Concrete Design Professional Development Series, PCA, October 2007. 9.6 Notes on ACI318-08 Building Code – Requirements for Concrete PCA, 2008. 9.7 National Seminar Series on AS 3600—2009, Lecture 2, Engineers Australia, 2009. 9.8 Foster, SJ and Gilbert, RI 'Strut and Tie Modelling of Non-flexural Members', Australian Civil Engineering Transactions Vol. 39, No. 2/3, 1997. 9.9 Reinforcement Detailing Handbook (Z06), 2nd Ed, Concrete Institute of Australia, 2010. Reinforced Concrete Design Handbook 9.7 blank page 9.8 Reinforced Concrete Design Handbook Chapter 10 Design examples The following pages provide examples of the design of the various types of reinforced concrete members in a hypothetical structure, demonstrating the use of the relevant preceding chapters using the design charts, Excel spreadsheets and tables presented in this Handbook and in accordance with AS 3600—2009. Design notes on structural design and computations pages 10.2–10.3 Design philosophy pages 10.4–10.6 Durability, fire and materials considerations pages 10.7–10.12 Design of typical floor pages 10.13–10.25 Spandrel beams – typical internal span pages 10.26–10.30 Beam Level 1 supporting entry facade pages 10.31–10.35 Column and wall design pages 10.36–10.45 Footing design pages 10.46–10.49 Reinforced Concrete Design Handbook 10.1 Design Notes on Structural Design and Computations The structural solutions presented in this Chapter have been chosen for the purposes of illustrating the analysis, design and reinforcement detailing of concrete members to AS 3600 and in the use of this Handbook. No lateral analysis has been undertaken and only selected members of the structure of the building have been chosen for design. Considerably more computations would be required for the complete design of this project. Where possible these computations are based on AS 3600—2009 with limited use of design aids such as spreadsheets, to illustrate concrete design from first principles. While some basic analysis has been carried out using a 2D analysis program and some of these results used for the design computations following, some input values have been chosen to illustrate specific design issues and cannot be validated by an analysis using the load data provided. A typical concrete structure for a building may account for only 25% of the project cost, but the design affects the whole building, including the architecture and services. The choice and details of a building's structure should reflect both buildability and overall building cost. The aim in any design is to provide a functional and economical structure, on time and on budget. A functional structure is one that has sufficient strength to carry the applied actions (loads) and has adequate stiffness to limit deflection and vibrations. It must also be durable against corrosion and deterioration and have adequate resistance against fire, must be aesthetic and sustainable and meet the expectation of those involved in the project. A structure must not be over-designed nor must it be under-designed. Unfortunately, when professional fees are limited, design and detailing is often minimised, resulting in an under designed structure with excess material. For a sustainable structure, a comprehensive design of all structural elements can be one of the most cost-effective of all of the sustainable measures adopted for any building. An economical structure is one that has an optimisation of material and labour costs and contributes to the overall economics of the project. Minimum weight does not always result in minimum cost. A structure that requires a longer time to construct, even though it may result in a lower construction cost, may nevertheless cost more money. Similarly, rationalisation and simplification of reinforcement will normally speed construction, reduce overall construction costs and construction time. However, excessive curtailment and tailoring of reinforcement to save material at the expense of rationalisation can be counter-productive. 10.2 Reinforced Concrete Design Handbook Structures, particularly those exposed to view, should be designed and detailed to be attractive and suitably proportioned and, although the appearance of the structure is often determined by the architect, the structural engineer must be aware of aesthetics and can significantly contribute to this area. The amount of time spent in analysing a structure or member should be related to the value to be gained from the analysis. Rigorous analysis is justified and necessary for a member when the member occurs many times in the structure, as refinement can result in a significant cost saving, whereas the use of simplified methods of design in AS 3600 can be used where the problem is a simple and basic one. The accuracy of structural behaviour theories and analyses, applied loads and material properties is such that determining actions to a high degree of accuracy is meaningless. For example, action assessment to 1% accuracy is all that is justified, eg 260 kN not 259.67 kN. When designing structures, avoid being bogged down in numbers, computer output and paper. Mistakes can be avoided by applying simple checks such as replacing a complicated load pattern by equivalent distributed or concentrated loads. Frequently check your computations and design assumptions to avoid duplication of errors. A few minutes spent checking your work on a regular basis can save hours of corrective work on design and drafting later on. Do not design members in isolation. Recognise that structures and especially reinforced concrete, will want to act structurally how it has been built and reinforced, not necessarily in the manner assumed in the design, including the computer analysis. The structure as a whole may behave differently from individual members. Remember also that when concrete slabs sit on beams or run into beams and beams frame into concrete columns and walls, the reinforcement must pass through these areas and it must all fit together. Aim for a general uniformity of members and details, and do not have a multitude of slab, beam and column sizes, bar sizes, fitment spacing, etc. The drawings and specifications for a building are the means of communicating the structural design to the builder/contractor, for construction. The drawings and specification usually form part of the Building Contract and are therefore important legal documents. Computations are a means of verification of structural behaviour. While they have no legal status in the formal building contract, they still are an important part of the design of the structure of a building. They are the main method of verification used for the design of buildings by structural engineers, to demonstrate compliance with relevant requirements of the BCA. Computations are also the basis of the structural drawings. Therefore, they should be treated with care, respect, and attention to detail. They should be neat, legible, arranged in a suitable order and properly referenced. Always put your conclusions at the end of each section, usually on the right hand side to allow easy identification and checking. CW Column and Wall actions (rundowns) CD Columns RT Retaining walls W Loadbearing Walls including core walls BF Basement Floors MF Mezzanine Floors 1F First Floor Standardisation in approaching the documentation of a project, neat setting out and the order of computations is essential to allow for effective checking and avoiding costly mistakes. TF Typical Floors PF Plant room Floor LM Lift Motor room PR Plant room Roof PC Precast Cladding It is recommended you use a pencil (HB or softer) for hand computations as that allows easy alterations when changes need to be made (and they usually will be) and use squared paper. Make use of sketches in your computations and prepare free-hand drawings of sections and details to illustrate your design, as a picture is much easier to visualise than many lines of computations. A4 photocopies of architectural or structural details can also be used. At all times, sketch neatly and to scale to assist your feel for the structure. Always set out at the beginning of any computations, your design philosophy and assumptions to allow the checkers and others who may have to access your computations, to understand how you have designed the structure. After establishing the design philosophy, actions, lateral analysis etc, consider designing the structure from the top down, even if the final computation may show a different order. R Roof (including lift motor room floor and roof where applicable) ST Stairs CP Construction Procedures M Miscellaneous The design notes scattered through this Chapter discuss various design matters and would not be normally included in any computations. They are provided for background and information for the users of this Handbook. All significant design computations should be sub‑divided into a number of design sections. These sub-divisions will form the basis for page numbering, list of contents, referencing, etc. The particular letter(s) allocated in the list below for a multi-storey building is an example of a lettering system that can be used, but other logical systems can be used. The numbering or lettering system to be used should be set up at the beginning of the computations and the example below is the sort of order in which computations might be made. For small projects, numerical numbering with a suitable index may be sufficient. DP Design Philosophy AC Actions (loads) SK Sketches DFM Durability, Fire and Materials UT Underpinning and Temporary walls WA Wind Analysis EQ Earthquake analysis FT Footings Reinforced Concrete Design Handbook 10.3 DESIGN PHILOSOPHY CLIENT A B Consolidated Pty Ltd ADDRESS 25 Print Street, Coogee, NSW 2034 USE Office building LOCALITY Light industrial area No polluting industry nearby Approximately 1 km to coast. FOUNDATION Dense sands with underlying clays in some local areas. Allowable bearing pressure 300 kPa in dense cemented sands at 1.2 m below NGL with sulfate and saline soil as advised by the geotechnical investigation. GENERAL The project consists of a new office block to an existing factory complex based on an overall master plan for the site. The new office block has been submitted for Development Approval (DA) and has received Council approval. Final structural drawings are being prepared based on the current architectural drawings. The computations along with the structural drawings will be submitted for structural approval by the private certifier or certifying authority. The following computations are part of what would be expected be submitted for review and approval. STRUCTURAL FORM OF THE NEW BUILDING The building will have a reinforced concrete structure with a suspended concrete roof at level 4 for a future floor extension, suspended concrete floors at first, second and third floors and a slab-on-ground at the ground floor. The concrete roof slab will have metal deck roofing over it initially for waterproofing, which will be removed, when the fourth floor is constructed in the future. The future roof and structure over at level 4 will be of lightweight construction with a metal deck roof with a braced steel frame supported on steel columns onto the concrete columns and concrete walls at level 4 and with lightweight walls externally. The floors and roof will be supported on square concrete columns internally, rectangular columns externally or loadbearing concrete walls with all the vertical elements supported on pad footings or strip footings founded on dense cemented sands. The suspended floors will consist of band beams in the east-west direction with one-way slabs in the other direction. A 700 deep x 400 wide edge beam is provided externally around each floor level where there are no concrete walls to support the glass facade and lightweight spandrel walls at each level. Concrete shear walls are provided on the south, west and north elevations. These will carry all lateral loads. In the southwest corner will be a stair and lift core, which will be part of the lateral-loadresisting system. Part of the external columns, footings, the bottom of the slab-on-ground and external faces of the external concrete shear walls are exposed to the environment but the remaining structure is internal except for a brief period during construction. Concrete stairs will be provided to both the core area and internally in a lightweight, fire-rated shaft to access and egress the building and to meet the egress requirements of the BCA as advised by the architect. BUILDING REGULATION Building Code of Australia (BCA). Assume an Importance Level 2 for the building in accordance with Table B.2a, Volume 1 of the BCA. This results in a Deemed-to-Satisfy Provision with an annual probability of exceedence of 500 years for wind and earthquake design. 10.4 Reinforced Concrete Design Handbook DESIGN ACTIONS (loads) Design actions (loads) will be generally in accordance with the following Standards: AS/NZS 1170.1 Structural design actions: Permanent, imposed and other actions AS/NZS 1170.2 Structural design actions: Wind actions AS 1170.4 Structural design actions: Earthquake actions in Australia Permanent actions (dead loads) will include the structure self weight, permanent loads, ceilings and services, partitions and the like. Imposed actions (live loads) of 3.0 kPa for the office areas generally, 5.0 kPa to public areas and 7.5 kPa for storage areas will be adopted at the south-west corner between grids C and D and 1 and 2. All combinations of actions for design will be factored to provide limit state actions as required by AS/NZS 1170.0 and respective design standards. WIND ACTIONS The wind actions applied to the building are determined using a basic regional wind velocity of V500 = 45 m/sec for an A2 region. The site is assumed to be in Terrain Category 3. The lateral actions will be determined from a wind analysis. Although wind actions are likely to be the governing design case for lateral actions on the building as a whole, consideration will be given to earthquake actions on the building and individual elements such as walls, etc. Typically lateral actions are designed using some form of frame program, sometimes three-dimensional. EARTHQUAKE ACTIONS From a consideration of the acceleration coefficient and site factors a lateral analysis for EQ Forces using a static analysis will be required. Also in accordance with the Standard, structural design of parts and detailing is required. FOOTINGS The columns and walls will be supported on pad or strip footings based on the previous satisfactory performance of this type of footing for similar buildings at this industrial complex. The geotechnical investigation has provided the necessary background to decide on the appropriate footing system founded at about 1200 mm below NGL as about 100 mm of top soil is to be removed. The site is known to be variable with moderately reactive soils underlying the dense sands. The site investigation reports prepared by the geotechnical consultant set out the necessary design parameters for footings, slabs-on-ground, any retaining walls, pavements, etc for the project. The slab-on-ground is to be constructed directly on natural ground or engineered fill with appropriate allowance for movement. This approach is based on previous experience on the site. However, detailing will need further consideration. CURTAIN WALLS, WINDOWS & CLADDING To be designed by others to a performance specification prepared by the architect with input from the structural engineer. Separate computations will need to be supplied by the curtain wall designer to the Checking Authority. Standards AS/NZS 1170 Series, Structural design actions AS 3600—2009 Concrete structures Reinforced Concrete Design Handbook 10.5 sketches 2.0 N Office block Covered access Existing factory 13.5 Stair and lift core SITE PLAN 700 x 400 edge beam (typical) 175 thick walls (typical) 1 2 3 4 5 6 A 190 slab (typical) 7.2 450 x 450 columns internally (typical) 23.6 700 x 500 Edge L1 only 2400 x 350 band beams (typical) 8.4 B D Heavy load area 7.2 C 400 overlap 7.2 8.4 8.4 8.4 7.2 800 x 450 columns externally (typical) 40.4 TYPICAL FLOOR PLAN (stair and lift not shown) 3.5 Future extension 3.5 L4 L2 L1 4.8 Void 3.5 3.5 L3 G CROSS SECTION Glass facade and spandrel panels Future extension Future extension Glass 175 insitu concrete wall – off form CS CS CS CS 175 insitu concrete wall – off form CS 200 insitu concrete wall – off form Strip footing SOUTH ELEVATION (North elevation similar) 10.6 Reinforced Concrete Design Handbook CS Beam Combined footing EAST ELEVATION Glass entry DURABILITY, FIRE AND MATERIALS CONSIDERATIONS Durability considerations Reinforced Concrete Determine f 'c and cover options for evaluation with durability before starting Design Handbook (RCDH) structural design. (Follow Flowchart 3.1) AS 3600 Table 4.6 Abrasion resistance – not applicable Clause 4.7 Freezing and thawing – not applicable Clause 4.8 Aggressive soils (as advised by Geotechnical Report) Sulfate soils with 8,000 ppm in soil (Table 4.8.1) Saline soils with a soil electrical conductivity (ECe) = 10 Table 4.3 Table 4.10.3.2 Table 4.10.3.2 External exposure classification B2, internal exposure classification A2 and surface in contact with the ground as per Table 4.8.1 and Table 4.8.2 for saline soils both with a B1 exposure classification Corrosion protection Exposure classification, f 'c and cover Cover (assume standard formwork and compaction) f 'c (MPa) Cover (mm) Member Exposure class Table 4.8.2 and Footings against ground (no dpm) Clause 4.10.3.5 B1 32 40 + 20 = 60 1 (min. cover from Table 4.8.2 is 50 mm \OK) Table 4.8.2 and Clause 4.10.3.5 Ground slab-on-ground Bottom of slab cast on dpm B1 32 50 + 10 = 60 Table 4.10.3.2 Top of floor surface A2 25 30 Table 4.10.3.2 Suspended floors Roof (protected by roof until extension) A2 A 25 25 30 30 Table 4.10.3.2 Beams generally A2 25 30 Table 4.10.3.2 Columns internally Columns externally Walls internal Walls exterior A2 B2 A2 B2 25 40 25 40 30 45 + 15 = 60 2 30 mm 45 + 15 = 60 2 Note 1 Footing cast in ground without DPM. Add 20 mm to cover (Cl 4.10.3.5). Note that Table 4.8.2 requires a minimum cover of 50 mm also \OK Note 2 Both the insitu external columns and walls are to have a 15-mm rebate at the construction joints externally as shown following, so increase the external cover to 60 mm to allow for rebate at a construction joint. 15 mm Joint is concealed in the shadow of the rebate Second pour Construction joint First pour CONSTRUCTION JOINT Details of wall joint showing rebate in external face Reinforced Concrete Design Handbook 10.7 Fire resistance considerations Design Note: For the structure and each type of member, normally the architect or designer will determine the required FRL from the BCA and then the structural engineer from this information will determine the design requirements (minimum dimensions, axis distance to main reinforcement, etc) to achieve the specified FRPs from AS 3600 Section 5, assuming the deemed-to-comply approach is used. In order to compare axis distance with covers, a notional cover will need to be determined using an estimated fitment size for the column or beam plus an estimate of bar diameter. Generally, covers for durability will be greater than the notional cover determined for fire. BCA reference Clause A 3.2 Determine FRLs Building Classification (office) = Class 5 Clause C1.2 Number of storeys =4 Clause C1.1 Table C1.1 Type of construction =A Spec. C1.1 Table 3 Required FRL Member FRL Floors ignoring concession Spec. C1.1 CC3.3 120/120/120 Roof ignoring concession Spec. C1.1 CC3.3 120/60/30 Columns 120/ - / - External walls (loadbearing) North and South 1-2 North and South 2-6 West East 120/120/120 120/60/30 120/120/120 120/60/30 120/120/120 Internal walls (lift and stair) Design Note: The BCA also contains a number of other requirements regarding compartmentation, distance to egress, separation, etc, which will all have to be complied with. Normally the architect or designer for the project will determine all of these requirements prior to final design and confirm these to the design team. Reinforced Concrete Design Handbook Determine requirements to achieve specified FRPs in accordance with Section 5 AS 3600 AS 3600 Table 5.5.1 Table 5.5.2 (A) FLOORS AND ROOF FRL 120/120/120 SLABS (assume continuous, one way) Insulation min. effective thickness = 120 mm Structural adequacy axis distance = 20 mm ie approximate notional cover = 20–6 = 15 mm Integrity deemed OK if complies with structural adequacy and insulation (AS 3600 Cl 5.3.1) Figure 5.4.1(B) BEAMS (assume continuous) FRL 120/120/120 Structural adequacy bw (mm) Axis distance (mm) (notional cover) 400 35 (15) (Note the notional cover 15 mm assuming with L10 fitment and 20-mm bar) 10.8 Reinforced Concrete Design Handbook Table 5.6.3 COLUMNS FRL 120/ - / Insulation and integrity for columns only required if part of wall Structural adequacy columns for fire resistance (AS 3600 Cl 5.6.3) b (mm) Axis distance (mm) (notional cover) 450 40 (16) (Note the value of axis distance is based N *f / Nu = 0.5 and the notional cover of 16 mm has been calculated assuming N12 fitment and 24-mm column bars. These assumptions will need to be checked, but it is likely that cover for durability will control the design.) WALLS FRL 120/120/120 or FRL 120/ 60/ 30 AS 3600 Clause 5.7.2 Check the structural adequacy of the wall using AS 3600 Cl 5.7.2 Minimum effective thickness for insulation for N *f / Nu = 0.7 is 160 mm and an axis distance 35 mm, assuming wall is exposed to fire on one side only. Design Note: In the design example, the wall is on the boundary of a fire compartment and therefore is subject to fire on one face only. If the wall was not part of the boundary of a fire compartment and could be subjected to fire on both sides, then it would need to be 220 mm thick and have a minimum axis distance of 35 mm. AS 3600 Clause 5.7.3 Limited Hwe / tw ≤ 30 Check if more critical Calculate Hwe from AS 3600 Cl 11.4 using k = 1.0 or 0.75 as noted below Design Note: Clause 5.7.3 of AS 3600 limits the ratio of effective height to thickness to 40. However, as the wall is to be designed in accordance with Clause 11.5 of AS 3600 using the Simplified Design Method, this limits the ratio of effective height to thickness to 30 and that is what has been used in the table below. Wall Hwe = k x Hw Hwe tw = Hwe / 30 Grnd – L1 1.0 x 4800 = 4800 160 L1 – L2 0.75 x 3500 = 2625 88 L2 – L3 0.75 x 3500 = 2625 88 L3 – Roof 1.0 x 3500 = 3500 88 Design Note: Designers can either use one or two layers of reinforcement in walls of say 10- or 12-mm bars (normal ductility) or reinforcing mesh say RF 82 or RF 92. For two layers of reinforcement both ways in each face, this will requires about 80–100 mm between the bars or mesh to place and compact the concrete when placed in vertical formwork. Therefore, the minimum thickness of an insitu reinforced concrete wall using bar reinforcement in two layers, assuming the covers above = 60 + 12 + 12 + 80 + 12 +12 + 30 = 218 mm, say 225 mm minimum thickness. However, if the wall is precast or tilt- up and is poured on flat, then a minimum thickness of only 150 to 180 mm is usually required with two layers of reinforcement, as it is much easier to cast a wall on flat than in a vertical position. AS 3600 requires that walls over 200 thick have two layers of reinforcement. \ Adopt one layer of reinforcement of Ductility Class N bars placed centrally as ductility is important for this wall. \ adopt Grnd – L1 t w = 200 mm > 160 mm \OK L1 – Roof t w = 175 mm > 160 mm \OK Reinforced Concrete Design Handbook 10.9 Table 5.7.2 Required axis distance to vertical reinforcement = 35 mm By inspection, these thicknesses with the reinforcement in the middle of a wall, the axis distance will be adequate for both insulation and integrity and the minimum covers for durability will also be achieved. Design note: Actual axis distance = 95 mm or 80 mm and covers = 85 mm or 75 mm and therefore satisfactory. CHOICE OF CONCRETE STRENGTH From a consideration of durability and fire-resistance requirements adopt the following concrete strengths: — for footings — for slab-on-ground — for suspended slabs including roof — for columns — for walls — rest of the concrete structure f 'c = 32 MPa f 'c = 32 MPa f 'c = 25 MPa f 'c = 40 MPa (SB) f 'c = 40 MPa (SB) f 'c = 25 MPa (SB denotes special class concrete for exposure classification B2) COVER Note the requirements on cover for concrete placement in AS 3600 Clause 4.10.2 — for footings cover = adopt 75 mm 1 — for slab-on-ground (assuming reinforcement mesh in top only) cover = 30 mm (top) — for suspended slabs including roof (cover > axis distance top and bottom) cover = 30 mm — for columns (cover > axis distance and use the same covers both internally and externally to avoid any errors on site) cover = 60 mm — for walls externally (cover > axis distance) cover = 60 mm — for walls internally (cover > axis distance) cover = 30 mm Note 1 Although only 60 mm cover is required by the Standard, a decision was made to adopt 75 mm because of the fact that the footings are likely to be poured during the winter time. If this was also difficult ground, the cover might be increased to say 90 mm. This increased cover of 75 mm is also consistent with the discussion in Chapter 8 of this Handbook. Design Note: A good rule of thumb: n n Internally, cover should be the greater of 30 mm (minimum A2 Exposure classification with 25-MPa concrete is required for all non-residential construction), d b and the maximum nominal size of aggregate Externally, cover should be the greater of 40 mm, 1.5 d b and twice the maximum nominal size of aggregate.) Design Note: To finalise the design of band beams and beams, the effective depths for design for flexure must be determined. These effective depths in turn will depend on which direction the primary reinforcement in slabs will be placed, ie which way the top layer and bottom layers of the slab reinforcement are constructed. As the slabs essentially span one-way, then adopt the primary direction of slab reinforcement as running north/south, ie upper layer of top bars to run N/S and bottom layer of bottom bars to run N/S. 10.10 Reinforced Concrete Design Handbook \ Adopt the following minimum covers: Footings Slab-on-ground Suspended slabs incl roof Band beams east/west External beams east/west External beams north/south Columns Walls Top Bottom External faces Internal faces 75 30 30 40 1 40 1 50 2 – – 75 60 30 30 30 30 – – 75 60 – – – – 60 60 – – – 30 30 30 60 30 Note 1 Top cover to band beams and beams running east west must allow for the cover to slab plus an assumed 16-mm slab top bar, rounded up to nearest 5 mm less 12-mm fitment as slab bars will sit between fitments ie 30 + 16 – 12 = 34 round up to nearest 5 mm. Therefore, adopt 40 mm cover to the fitment of the band beams and beams running east west. Note 2 Top cover to external beams running north south must allow for cover for slab plus 16-mm bar plus 12-mm bar rounded up minus 12-mm fitment = 30 + 16 + 12 – 12 = 46 mm round up to nearest 5 mm say 50 mm to fitment of the beam. LOADS General Reinforced concrete 25.0 kN/m3 Design Note: Some designers use 24.5 kN/m3 for the density of reinforced concrete, but 25.0 kN/m3 is within the required order of accuracy and is slightly conservative. 175 thick walls 200 thick walls 350 deep band beams = 2.4 x 0.35 x 25 = 4.375 kN/m2 = 5.0 kN/m2 = 21.0 kN/m Weight of concrete in band beams below soffit of slab = 2.4 x (0.35 – 0.19) x 25 = 9.6 kN/m 700 x 400 deep beams = 0.7 x 0.4 x 25 190 slabs = 0.19 x 25.0 Lightweight walls 3.5 m high allow 450 x 450 cols = 0.45 x 0.45 x 25 800 x 450 cols = 0.8 x 0.45 x 25 = 7.0 kN/m = 4.75 kN/m = 1.00 kN/m = 5.1 kN/m = 9.0 kN/m Future Roof (L5) Permanent actions (dead loads) Sheeting Purlins Steel beams Ceilings and services AS/NZS 1170.1 Total Imposed actions (live loads) (1.8 / A + 0.12) but not less than 0.05 kPa 0.05 kPa 0.15 kPa 0.25 kPa 0.50 kPa 0.25 kPa Reinforced Concrete Design Handbook 10.11 Roof Slab L4 (Future office) Permanent actions (dead loads) 190 slab Fixed partitions, finishes Ceilings and services 4.75 kPa 1.00 kPa 0.25 kPa Total 6.00 kPa Imposed floor actions (live loads) Generally Office including allowance for moveable partitions, etc 3.0 kPa Imposed floor actions (live loads) Heavy load area Heavy load area including compactus etc 7.5 kPa Floor Slab L1–3 Permanent actions (dead loads) 190 slab Fixed partitions, finishes Ceilings and services 4.75 kPa 1.00 kPa 0.25 kPa Total 6.00 kPa Imposed floor actions (live loads) Generally Office including allowance for moveable partitions, etc 3.0 kPa Imposed floor actions (live loads) Heavy load areas Heavy load area including compactus etc 7.5 kPa Wind actions to walls and facades The wind actions applied to the building have been determined using AS/NZS 1170, Structural design actions Part 2: Wind Actions. Design wind speeds to AS/NZS 1170 Part 2: Wind Actions Region Regional wind speed V500 Wind direction multiplier Md Terrain category Height Mz.cat = 0.94 Ms =1.0 Mt = 1.0 A2 = 45 m/s = 1.0 3 = 18.8 m Vsit Beta = Vr Md (Mzcat Ms Mt) = 42.3 m/sec Cfig = Cp.e + Cp.i = 0.7 + 0.2 = 0.9 p = Qz = 0.5 x 1.2 x Vr2 /1000 x Cfig = 0.6 x 42.3 x 42.3 x 0.9/1000 = 0.97 kPa (ultimate) 10.12 Reinforced Concrete Design Handbook DESIGN OF TYPICAL FLOOR Analysis of Structure Use Linear Elastic Analysis (AS 3600 Cl 6.2) Use a computer program to analyse and to determine bending moments and shears as shown on the following calculations. Assumptions 1 Model of idealised frame – bent on Grid B Level 2 (Level 3 similar) Clause 6.9 of AS 3600 mm 1 2 3 4 5 6 0 7200 15000 24000 32400 39600 1 2 3 4 5 6 0 7200 15000 24000 32400 39600 –2000 0 2000 PLAN VIEW 2000 0 –2000 FULL ELEVATION VIEW 2400 2400 3000 190 350 2 Section properties 7800 General details of bent For band beam, assume T section in middle of span. bef = bw + 0.2a where a = 0.7 L (Clause 8.8.2 of AS 3600) = 2,400 + 0.2 x 0.7 x 7,200 = 3,408 mm I = 10.15 x 10-3 m4 A = 1.032 m2 190 3408 2400 Band beam assume section following at a support 350 350 mm I = 8.575 x 10-3 m A = 0.84 m2 2400 Reinforced Concrete Design Handbook 10.13 450 Internal column assume I = 3.42 x 10-3 m4 A = 0 .203 m2 450 External column assume 800 I = 19.2 x 10-3 m4 A = 0 .36 m2 450 3 Material properties (RCDH Table 2.1) Band beam Column and walls f 'c (MPa) Ec (MPa) 25 40 26,700 32,800 4 Load on band beam Permanent action (dead load) Band beam plus slab Fixed partitions Ceilings and services etc Total = 46.65 kN/m = 7.80 kN/m = 1.95 kN/m = 56.4 kN/m Imposed actions (live load) = 23.4 kN/m Note: Live load ≤ 0.75 x dead load \ Consider live load on all spans (AS 3600 Cl 2.4.4 (c) iii)) Typical Computer Output for Band Beam Grid C Design note: The bent below has been analysed using a computer program; the output from such a program would look something like what is shown following. The output will depend on the model as analysed, which in this case has used the idealised frame method of analysis as set out in Clause 6.9 of AS 3600. Note that this program has calculated the moments and shears at the critical sections, rather than the peak moments and shears at the centre line of supports. Designers are responsible for ensuring that any software used for analysis is appropriate for the analysis being undertaken, as set out Clause 6.1.2 of AS 3600. 10.14 Reinforced Concrete Design Handbook 1 2 3 4 5 6 –800 –446 –496 –541 –532 –541 –532 –446 –496 –600 –400 –267 –267 –200 kNm 0 220 220 303 289 303 200 400 0 7200 24000 39600 32400 15600 mm Moment 1 600 400 200 0 kN –200 –400 –600 2 273 3 366 –331 0 Shear 7200 4 371 –377 15600 5 377 –371 24000 6 331 –366 32400 –273 39600 mm ULTIMATE FLEXURE Design band beam at column C4 between spans 4–5 Band beam properties AS 3600 Cl 8.8.2 190 350 3408 2400 dtop = 350 – 30 – 16 – 14 = 290 mm (depth – cover – dia slab bar – ½ dia of band beam reinforcement assuming N24 top bars) Adopt dtop = 285 mm as actual bar sizes are nominally larger than actual sizes dbot = 350 – 30– 14 = 306 mm (depth – cover – ½ dia of band beam reinforcement assuming N24 as bottom reinforcement) Adopt dbot = 300 mm as actual bar sizes are nominally larger than actual sizes. Flexure Negative BM @ critical section at column C4 AS 3600 Cl 6.9.2 CL 0.7asup asup Critical section for neg. moment Column face asup = 225 mm Design note: The calculations below are for the critical section for negative moment. The bending moment at the centre line of the support is 564.6 kN.m where the shear is equal to 429 kN. These values are not shown above in the results of the computer analysis, but are generally are available in the output data. Normally the analysis program will calculate this figure for the designer at the critical sections, but calculations below by simple statics illustrate the process from first principles. Reinforced Concrete Design Handbook 10.15 429 0.7 asup M* = 564.6 – x [ ]2 2 asup = 564.6 – 23.6 = 541 kN.m D = 350 mm d = 285 mm bef = 2,400 and assumes no T beam action over the supports f 'c = 25 MPa fsy = 500 MPa Calculation of reinforcement required for M * Use RCDH Excel Spreadsheet 4.1 for rectangular beams to calculate reinforcement Ast nominal = 5,583 mm2 (initial estimate) Try 12 – N24 = 5,424 mm2 (spacing = 209 mm nom) Minimum Ast = 5,234 mm2 < 5,424 mm2 \ OK ku = 0.24 < 0.36 \ Well OK f Mu = 561 kN.m > 541 kN.m \ OK f Muo = 859 kN.m > 541 kN.m \ Well OK Ast min = 1,238 mm2 < 5,424 \ Well OK \ Adopt 12 N24 as top reinforcement Positive BM @ mid-span M* = 303 kN.m D = 350 mm d = 300 mm t = 190 mm bef = 3,408 and assumes T beam action at the midspan f 'c = 25 MPa fsy = 500 MPa Use RCDH Excel Spreadsheet 4.2 for T beams with the stress block within flange Ast nominal = 2,971 mm2 (initial estimate) Provide Ast = 3,140 mm2 = 10 N20 or 7 N24 (spacing = 240 or 342 mm nom) Note: Clause 8.6.1(b) requires a maximum spacing between bars in tension to be 300 mm. \ Adopt 10 N20 bottom reinforcement Minimum Ast for flexure = 2,620 mm2 < 3,100 mm2 \ OK ku = 0.17 < 0.36 \ Well OK f Muo = 952 kN.m > 303 kN.m \ Well OK f Mu = 363 kN.m > 303 kN.m \ OK Ast.min = 1,176 mm2 < 3,140 \ Well OK 10.16 Reinforced Concrete Design Handbook Crack Control AS 3600 Clause 8.6.1 Band Beam is fully enclosed within building except for brief period during construction \ only need to satisfy Item (a) and (b) of Cl 8.6.1 ie minimum reinforcement must comply with Cl 8.1.6.1 of AS 3600 See above reinforcement > minimum ie cover to centre of bars ≤ 100 mm and bar spacing ≤ 300 mm. See above. \ OK Bottom reinforcement cover = 30 mm ≤ 100 mm \ OK spacing use 10 N20 9 x 300 = 2,700 mm \ OK Bottom reinforcement 10 N20 Top reinforcement cover = 40 mm \ OK spacing use 12 N24 11 x 300 = 3,300 > 2,400 \ OK Top reinforcement 12 N24 Shear AS 3600 Clause 9.2.2 (a) Shear Design (treat as shallow beam Clause 8.2) (a) Beam shear (follow Flowchart 4.2) CL Column 429 kN 377 kN CL Clear span Asup + Do = 225 + 285 = 510 3690 4200 Clause 8.2.4 (b) At the critical section for shear V* = 429 x 3,690/4,200 = 377 kN Design note: The calculations above are for the critical section for shear. The bending moment at the centre line of the support is 564.6 kN.m and the shear is equal to 429 kN. These values are not shown above in the results of the computer analysis, but are generally are available in the output data. Normally the analysis program will calculate this figure for the designer at the critical sections, but calculations above by simple statics illustrates the process from first principles. Reinforced Concrete Design Handbook 10.17 Beam Properties Ast = 12 – N24 bv = 2400 Ast = 12 N24 = 5,424 mm2 bv = 2,400 mm f 'c = 25 MPa do = 285 mm D = 350 mm V * = 377 kN Clause 8.2.6 Determine f Vu.max = f 0.2 f 'c bvdo = 0.7 x 0.2 x 25 x 2,400 x 285 x 10–3 = 2,394 kN > 377 kN \ well OK RCDH Excel Spreadsheet 4.3 Refer Excel Spreadsheet for beams, which will determine all the following values β1 = 1.447 For members where the cross-sectional area of shear reinforcement provided (Asv) is not equal to or greater than the minimum area specified in Clause 8.2.8 β2 = 1.0 β3 = 1.0 Where fcv = f 'c 1/3 ≤ 4 MPa = 2.92 MPa Vuc = 577 kN Clause 8.2.7 Shear strength of a beam excluding shear reinforcement Determine f Vuc \ f Vuc = 0.7 x 577 = 404 kN \ 0.5 f Vuc = 202 kN Clause 8.2.9 Shear strength of a beam with minimum shear reinforcement Determine f Vu.min = f (Vuc + 0.10 √f 'c bv do) ≥ f Vuc + f 0.6 bv do f Vuc + f 0.6 bv do = 692 kN f (Vuc + 0.10 √f 'c bv d ) = 644 kN f Vu.min = 0.7 x 853 = 692 kN 0.5 f Vuc = 346 kN \ 0.5 f Vuc < V * ≤ f Vu.min Normally A sv.min is required in accordance with Cl 8.2.5(b) but as V * ≤ f Vuc and D < bv / 2 , no shear reinforcement is required in accordance with Cl 8.2.5(1). Design Note: Generally, designers should try to avoid shear reinforcement in band beams as it results in fitments at close centres and multiple fitments which are difficult to fix and are expensive. This is because the transverse spacing of fitments is limited to the lesser of 600 mm or D (Clause 8.2.12.2) and this spacing has to be reconciled with the maximum spacing of bars in tension of 300 mm. It is recommended that the depth or concrete strength of the band beam be increased or width decreased as required, to avoid the need for shear reinforcement where possible. However, designers should always provide nominal fitments to hold the 10.18 Reinforced Concrete Design Handbook bottom bars in place for fixing the reinforcement on site, for crack control to the soffit and sides and to lap with the bottom bars of the adjacent slabs. The sections on page 10.21 clearly illustrate the differences between a band beam with nominal fitments to support the reinforcement and one with the required shear reinforcement (which is significant). \ Provide N12 fitments at 200 cts through the length of the band beam. N12 fitments @ 200 Punching Shear (as a slab) AS 3600 Cl 9.2.2 (b) Case 1 Punching shear at column C3 AS 3600 Cl 9.2.3 (a) Refer to bending moments and shear forces above. V * = 377 + 371 = 748 kN M *v = 541 – 532 = 9 kN.m treat as M *v = 0 Assumed failure planes d/2 d/2 = 145 Beam width = 2,400 > 450 + 285 RCDH Chart 5.13 450 Punching cone \ OK to use Chart 5.13 c1 = 450 mm c2 = 450 mm d = 285 mm f 'c = 25 MPa Calculate c1/c2 = 1 c1 + c2 = 900 mm Read for V *= 748 kN Minimum depth = 240 mm < 350 mm \ OK AS 3600 Cl 9.2.2 (b) Case 2 Punching shear at column C1 AS 3600 Cl 9.2.4 V * = 273 kN 1 M *v = 267 kN.m dom = 285 mm 2400 dom/2 800 x 450 column Edge beam AS 3600 Cl 9.2.4 (a) ie no closed fitments f Vu = f Vuo / [1.0 + (u M *v / 8 V *a dom)] u = [450 + 285 + 2 x (400 + 285/2)] = 1,820 mm Reinforced Concrete Design Handbook 10.19 Design Note: The calculation of u is conservative as dom is assumed the same for band beam and edge beam. f 'c = 25 MPa a = 450 + 285 = 735 mm β h = X / Y = 450/400 = 1.125 fcv = 0.17 (1 + 2 /β h ) √f 'c ≤ 0.34 √f 'c = 0.472 √f 'c > 0.34 √f 'c \ fcv = 0.34 √f 'c = 1.7 MPa Vuo = u dom fcv = 1,820 x 285 x 1.7/1,000 = 881.8 kN f Vu = f Vuo / [1.0 + (u M *v / 8 V *a dom)] (1,820 x 267 x 106 ) \ f Vu = f 882 / [1.0 + ] (8 x 273 x 1,000 x 735 x 285) = 0.7 x 882 (1.0 + 1.06) = 1,273 kN > 273 kN \ Well OK Check depth using Deemed to Comply Span-to-Depth Ratios AS3600 Clause 8.5.4 Determine maximum value of Lef /d Clause 2.3.2 Total deflection limit for ∆/Lef = 1/250 (ie no masonry) f 'c = 25 MPa \ Use RCDH Excel Spreadsheet 4.5 RCDH Spreadsheet 4.5 Determine input values b = 2,400 mm bef = 3,408 mm D = 350 mm c = 30 mm (cover) do = 300 mm Ast = 10 N20 = 3,140 mm2 Asc = 12 N16 = 3,140 mm2 (to lap with 12 N24) g = 56.4 kN/m q = 23.4 kN/m ys = 0.7 y l = 0.4 Lef = 8,400 – 400 + 350 = 8,350 mm (interior span) k1 = 0.0792 (calculated by the program) k2 = 0.00625 Ec = 26,700 MPa Actual Lef /d = 8,350/300 = 27.8 Calculated allowable Lef /d = 31.8 for total deflection and is > 27.8 so the band beam as designed complies. However, if this band beam was supporting a masonry partition it would not comply. Design Note: As discussed in Chapter 4, the depth of the beam determined by the deemed-to-comply method can be conservative, especially for shallow beams. Deflection calculations by the simplified calculations may result in thinner and more 10.20 Reinforced Concrete Design Handbook economical sections. The deemed-to-comply method suggests a deflection of about 25 mm by interpolation. Carrying out a more refined calculation, using a computer analysis program, results in the deflections as shown below. For span/250 long‑term deflection 8,350/250 = 33.4 mm and the results below show a long‑term deflection of 21.9 mm < 33.4 mm so again a 350 band beam is OK using this method of analysis. 1 mm 2 0 –20 –40 –60 3 –8.72 794 0 4 –16.9 –21.9 378 7200 5 –8.72 –21.9 490 15600 6 378 24000 32400 794 TOTAL LONG TERM DEFLECTION 39600 mm Long term deflection as determined by computer analysis for a 350-deep band beam. The following is the proposed reinforcing layout to be shown on the drawings using a beam elevation. 12N24 12N20 12N16 12N24 12N24 12N16 12N16 12N16 12N24 12N16 12N20 600 typical N12 fitments @ 200 600 typical 1 10N16 x 7.000 m 10N20 x 9.000m 2 3 10N20 x 9.000 m 4 10N20 x 9.000m 5 10N16 x 8.000 m 6 ELEVATION Slab reinforcement Use N16 slab reinforcement to support top bars in band beam 500 typical 12N16 or 12N24 N12 fitments at 200 cts 10N20 or 10N16 Dummy bar for support of top bars (bar chair) 350 190 ≤ 300 mm when in tension Clause 8.6.1 (b) SECTION Where only norminal shear reinforcement is required Note: For fixing on site N16 dummy bars at about 1 m centres are used under to support the top reinforcement until the slab reinforcement is placed L8 fitments at 250 cts Slab reinforcement 500 typical 10N20 Dummy bars 12N24 ≤ 350 mm Clause 8.2.12.2 N12 at 250 cts ≤ 300 mm when in tension Clause 8.6.1 (b) SECTION This section illustrates the extent of detailing if shear reinforcement is required REINFORCEMENT LAYOUT Reinforced Concrete Design Handbook 10.21 Flexural Design of Slab in Transverse Direction, eg Grid 4 The following is the design model that was input into the computer using the loads previously calculated. 1 2 3 4 0 7200 15600 22800 mm 2 3 4 7200 15600 22800 mm 1 2 3 4 0 7200 15600 22800 mm 0 mm –200 –400 –600 ELEVATION VIEW –4000 –2000 mm 0 2000 2000 1 0 PLAN VIEW mm 2000 0 –2000 FULL ELEVATION VIEW BMs and shears from computer analysis 1 –800 –600 –400 –200 –197 kNm 0 200 0 2 –622 3 –632 –632 4 –622 –197 169 177 169 7200 15600 22800 mm 2 3 4 Moment 1 600 400 240 200 kN 0 –200 –400 –600 0 383 –379 Shear ULTIMATE FLEXURE 10.22 Reinforced Concrete Design Handbook 379 –383 7200 15600 –240 22800 mm Design for flexure Interior span AS 3600 Clause 6.9.2 (a) Negative BM @ critical section near centre line of the column M * = 632 kN.m M * per metre strip = 632 / 8.4 M * = 75.2 kN.m/m b = 1,000 mm d = 350 – 30 – 10 = 310 mm f 'c = 25 MPa fsy = 500 MPa RCDH Chart 5.1 Try Chart 5.1 M * = 75 kN.m/m Outside the range \ use RCDH Excel Spreadsheet Spreadsheet Clause 9.4 AS 3600 \ Ast nominal required = 713 mm2/m Maximum spacing of bars = 300 mm or 2.0 D = 380 mm Provide Ast = 670 mm2/m (N16 @ 300) Minimum Ast for flexure = 622 mm2 \ Use = N16 @ 300 (670 mm2/m > 622) or = N12 @ 175 (646 mm2/m > 622) ku = 0.06 < 0.36 \ Well OK f Muo = 423 kN.m > 73 kN.m \ Well OK f Mu = 81 kN.m > 73 kN.m \ OK Ast min = 569 mm2 < 670 mm2 \ OK (b) Negative BM @ edge of band beam (from computer analysis) M * = 283 / 8.4 = 33.7 kN.m/m b = 1,000 mm d = 190 – 30 – 10= 150 mm f 'c = 25 MPa RCDH Chart 5.1 fsy = 500 MPa \ Ast nominal required = 661 mm2/m Try Ast = 670 mm2/m = N16 @ 300 (670 mm2/m > 593) or = N12 @ 175 (646 mm2/m > 593) Minimum Ast for flexure = 593 mm2 < 670 mm2 \ OK ku = 0.12 < 0.36 \ Well OK f Muo = 99.1 kN.m > 33.7 kN.m \ Well OK f Mu = 38.1 kN.m > 33.7 kN.m \ OK Ast min = 347 mm2 < 670 mm2 \ OK Adopt N12 @175 top reinforcement Reinforced Concrete Design Handbook 10.23 (c) Positive BM @ midspan (from computer analysis) M * = 177/8.4 = 21.1 kN.m/m b = 1,000 mm d = 190 – 30 – 10 = 150 mm f 'c = 25 MPa fsy = 500 MPa RCDH Chart 5.1 Ast = 370 mm2/m \ use = N12 @ 250 (411 mm2/m > 370) Adopt N12 @ 250 bottom reinforcement Crack Control AS 3600 Clause 9.4.1 Slab is fully enclosed within building except for brief period during construction \ need only to satisfy Items (a) and (b) Item (a) Ast.min = 0.20 (D/d )2 f 'ct.f / fsy bw d = 150 Ast.min = 289 mm2/m < 411 mm2/m \ OK Item (b) Centre to centre spacing ≤ the smaller of 2.0 x 190 = 380 or = 300 \ max spacing 300 mm \ N12 @ 250 mm from above for flexure meets both these requirements. How provide N12 at 500 as top reinforcement in centres of spans for robustness and crack control in the top of the slab. Adopt N12 @ 500 top reinforcement in centre of spans Crack control for shrinkage and temperature in the secondary direction AS 3600 Clause 9.4.3 Slab fully enclosed within building Check area for crack control for shrinkage and temperature effects Where a moderate degree of control over cracking is required in exposure classification in A1 and A2 then As ≤ 3.5 x 1,000 x D x 10–3 mm2/m 3.5 x 190 = 665 mm2/m Where a strong degree of control over cracking is required is required for appearance and for exposure classifications A1, A2, B1, B2, C1 and C2, then As ≤ 6.0 x 100 x D x 10–3 mm2/m = 1,140 mm2/m \ Adopt moderate degree of control over cracking, as concrete is within an enclosed building and carpet will be used as floor coverings, which will hide any cracking. \ use N16 @ 250 mm (801 mm2/m > 665) or use N12 @ 150 mm (673 mm2/m > 665) Adopt N12 @ 300 as top and bottom reinforcement in the transverse direction ie equivalent to one layer of N12 @150. Design note: This transverse reinforcement will serve two purposes as it will be used to support the main reinforcement in the direction of the span of the slab as well as providing crack control. 10.24 Reinforced Concrete Design Handbook Check depth using Span-to-Depth Ratios RCDH Use RCDH Excel Spreadsheet 5.1 Ast = 411 mm2/m Asc = 205 mm2/m g = 5.75 kN/m q = 3.0 kN/m ys = 0.7 y1 = 0.4 Lef = 6,000 mm Ec = 26,700 MPa Calculate Asc / Ast = 0.5 (at mid-span) Calculate kcs = 1.4 Calculate Fd.ef = (1 + kcs) g + (ys + kcs y l)q AS 3600 Clause 9.3.4.1 Clause 9.3.4.1 = 2.4 x 5.75 + 1.26 x 3.0 = 17.58 kPa (b) k3 k3 = 1.0 (one-way slab) (d) D/Lef D/Lef = 1/250 (total deflection) (c) k4 Clause 2.4.2 k4 = 2.1 (interior span) (e) f 'c RCDH Calculated from spreadsheet Lef /d for total deflection = 38.31 f 'c = 25 MPa Calculate actual Lef /d = 6,000 /150 = 40 > 38.31 \ Not OK Design Note: As discussed in Chapter 5, the depths of members determined using the deemed-to-comply method can be conservative, especially for shallow slabs and the deflection calculations by the simplified method will result in thinner and more economical sections. Carrying out a more refined calculation using the chosen analysis program gives the deflections shown below. For span/250 long term > 6,000/250 = 24 mm and the calculation below show a long‑term deflection of 13.4 mm < 24 mm so a 190 slab is well OK. Note: To get the maximum deflection one has to add the deflection of the slab to that of the bonded beam, ie 13.4 + 21.9 = 35.3 mm which is about the maximum visual limit for deflection. 1 mm 0 -10 -20 -30 -40 2 –12.4 3 –13.4 552 0 4 –12.4 618 552 7200 15600 22800 mm TOTAL LONG TERM The following shows the details of the reinforcement for slabs. N12-250 600 typical N12-175 N12-500 500 typical 1 N12-250 N12-250 N12-250 N12-300 top and bottom outside band beams N12-175 N12-200 2 N12-200 N12-250 Edge beam Band beam 3 4 REINFORCEMENT LAYOUT Reinforced Concrete Design Handbook 10.25 SPANDREL BEAMS – TYPICAL INTERNAL SPAN Details of bent to be analysed 1 3000 2000 1000 mm 0 0 2 3 4 7200 15600 24000 2 3 4 7200 15600 24000 PLAN VIEW 1 mm FULL ELEVATION VIEW bef = 990 190 50 30 700 30 30 bw = 400 Beam properties f 'c = 25 MPa fsy = 500 MPa cover c = 30 mm to fitments side and bottom cover c = 50 mm top (see page 10.11) d = 700 – 50 – 12 – 12 = 626 say 620 mm top and = 700 – 30 – 12 – 12 = 646 say 640 mm bottom AS 3600 Clause 8.8.2 31200 mm 5 2000 0 -2000 0 5 bef = bw + 0.1 x 0.7L = 400 + 0.1 x 0.7 x 8,400 = 988 say 990 mm in the middle of the span 10.26 Reinforced Concrete Design Handbook 31200 mm Loads Permanent actions. See pages 10.11 and 10.12 (dead loads) Beam Lightweight walls 190 slab Partitions, ceilings and services = 7.0 kN/m = 1.0 kN/m = 2.9 x 4.75 = 13.8 kN/m = 2.9 x 1.25 = 5.1 kN/m Total = 26.9 kN/m Imposed actions. See page 10.12 (live loads) Floor Total = 3.8 x 3.0 = 11.4 kN/m = 11.4 kN/m Typical Computer Output for Beams along Grid D 1 –350 –280 –210 –140 –70 kNm 0 70 140 210 2 3 –202 –224 –191 93.1 0 –258 4 –257 141 –228 5 –210 99.1 139 7200 15600 24000 31200 mm 2 3 4 5 Moment 1 280 210 140 70 kN 0 –70 –140 –210 –280 137 155 –137 0 –128 163 –164 7200 141 –156 15600 –114 24000 Shear 31200 mm ULTIMATE FLEXURE Design moments Max –ve Max +ve At Column 3 M * = 258 kN.m D = 700 mm d = 620 mm bef = 400 and assume no L beam action over support M * = 258 kN.m M * = 141 kN.m f 'c = 25 MPa fsy = 500 MPa From RCDH Excel Spreadsheet \ Ast nominal required = 1,244 mm2 Provide Ast = 1,232 mm2 \ use = 2 N28 (1,232 mm2 > 1105) use = 3 N24 (1,356 mm2 > 1105) Minimum Ast for flexure = 1,105 mm2 < 1,356 mm2 \ Well OK ku = 0.14 < 0.36 \ Well OK f Muo = 677 kN.m > 258 kN.m \ Well OK f Mu = 288 kN.m > 258 kN.m \ OK Ast min = 379 mm2 < 1,356 mm2 \ OK \ Top reinforcement 3N24 Reinforced Concrete Design Handbook 10.27 At midspan M * = 141 kN.m D = 700 mm d = 640 mm t = 190 mm bef = 990 and assume T-beam action in middle of span f 'c = 25 MPa fsy = 500 MPa From RCDH Excel Spreadsheet 4.2 As a L-beam with the stress block within the flange \ Ast nominal required = 648 mm2 Provide Ast = 628 mm2 (2 N20) Minimum Ast for flexure = 557 mm2 < 628 mm2 \ OK \ use = 2N20 (628 mm2 > 557 mm2 ) ku = 0.03 < 0.36 \ Well OK f Muo = 722 kN.m > 141 kN.m \ Well OK f Mu = 159 kN.m > 141 kN.m \ OK Ast min = 368 mm2 < 628 mm2 \ OK \ Bottom reinforcement 2 N20 Check crack control Beam not exposed to the weather on external surface Therefore need to satisfy Items (a) and (b) only as appropriate of AS 3600 Cl 8.6.1 AS 3600 Clause 8.6.1 (a) As noted above meets the minimum requirements of Cl 8.1.6.1 \ OK (c) Bars less than 100 mm from side and soffit of beam and less than 300 mm spacing \ OK. AS 3600 Clause 8.6.3 Check crack control in side faces of beam. As overall depth < 750 mm it is not required. However, provide 1 N12 each face (EF) at the centre of the beam for crack control, as side face reinforcement (SFR). SFR 1 N12 EF Shear Design AS 3600 Clause 8.2.4 At critical section V * = 164 kN D = 700 mm d = 620 mm bv = 400 Using Spreadsheet 4.3 AS 3600 Clause 8.2.6 f 'c = 25 MPa fsy = 500 MPa Ast = 1,232 mm2 (3 N24) Determine f Vu.max = f 0.2 f 'c bvdo = 0.7 x 0.2 x 25 x 400 x 620 x 10–3 = 868 kN V * < f Vu.max \ OK 10.28 Reinforced Concrete Design Handbook Refer RCDH Excel Spreadsheet 4.3 for beams, which will determine all the following values β1= 1.1 For members where the cross-sectional area of shear reinforcement provided (Asv) is equal to or greater than the minimum area specified in Clause 8.2.8 β2 = 1.0 β3 = 1.0 Where fcv = f 'c 1/3 ≤ 4 MPa = 2.92 MPa Clause 8.2.7 Shear strength of a beam excluding shear reinforcement Determine f Vuc Vuc = β1 β2 β3 bv do fcv Ast 1/3 bv do \ f Vuc = 0.7 x 136.1 = 98.4 kN \ 0.5 f Vuc = 49.2 kN Clause 8.2.9 Shear strength of a beam with minimum shear reinforcement f Vuc + f 0.6 bv do = 202.5 kN Determine f Vu.min = f (Vuc + 0.10 √f 'c bv do) ≥ f Vuc + f 0.6 bv do f (Vuc + 0.10 √f 'c bv d ) = 185.2 kN f Vu.min = 202.5 kN \ 0.5 f Vuc < V* < f Vu.min \ shear reinforcement is required in accordance with Clause 8.2.5. Determine required shear reinforcement RCDH Excel Spreadsheet 4.3 Asv.min = 140 mm2 at 500 spacing \ Adopt fitments L10 @ 300 cts throughout Check Depth using Deemed-to-Comply Span-to-Depth Ratios AS 3600 Clause 8.5.4 Determine maximum value of Lef /d Clause 2.3.2 Total deflection limit for ∆ /Lef = 1/250 f 'c = 25 MPa \ use RCDH Excel Spreadsheet 4.5 RCDH Spreadsheet 4.5 Determine input values b = 400 mm bef = 990 mm D = 700 mm c = 30 mm do = 640 Ast = 2 N20 = 628 mm2 Asc = 2 N20 = 628 mm2 g = 26.9 kN/m q = 11.4 kN/m ys = 0.7 y l = 0.4 Lef = 8,400 mm Reinforced Concrete Design Handbook 10.29 k1 = 0.0383 (calculated by the program) k2 = 0.00391 (internal span) Ec = 26,700 MPa Actual Lef /d = 8,400/640 = 13.1 Calculated allowable Lef /d = 25.8 for total deflection > 13.1 so the beam as designed complies. Design Note: More refined computations suggest the long-term deflection is about 5 mm compared to about 17 mm suggested by the deemed-to-comply approach, again indicating the deemed-to-comply solution can be conservative. 2N24 2N24 W10-300 1 3N24 W10-300 750 2 2N20 (1N12 SFR each face) REINFORCEMENT LAYOUT 10.30 Reinforced Concrete Design Handbook 2N20 900 2N20 W10-300 3 2N20 2N24 2N24 W10-300 4 2N20 5 2N20 BEAM LEVEL 1 SUPPORTING ENTRY FACADE Beam properties 84 100 1000 AS 3600 Clause 8.8.2 500 700 84 112 56 616 588 N32 spacer f 'c = 25 MPa fsy = 500 MPa cover to main reo = 50 + 10 = 60 mm top and bottom bef = bw + 0.1 x 0.7L = 500 + 0.1 x 0.7 x 8,400 = 1,088 mm say 1,000 mm \ assume bef = 1,000 mm Design actions at critical section Max –ve M * = 380 kN.m Max +ve M * = 960 kN.m V * = 360 kN T * = 50 kN.m At column at the critical section for flexure M * = 380 kN.m D = 700 mm d = 615 mm (rationalised from 616 mm above) bef = 500 mm and assume no L beam action over support f 'c = 25 MPa fsy = 500 MPa From RCDH Excel Spreadsheet 4.2 \ Ast nominal required = 1,817 mm2 Provide Ast = 1,808 mm2 (4N24) Minimum Ast for flexure = 1,660 mm2 < 1,808 mm2 \ OK ku = 0.16 < 0.36 \ Well OK f Muo = 833 kN.m > 380 kN.m \ Well OK f Mu = 414 kN.m > 380 kN.m \ OK \ use = 4 N24 (1,808 mm2) 4 N24 top reinforcement Reinforced Concrete Design Handbook 10.31 At centre of span Check if stress block extends into web of beam M * = 960 kN.m D = 700 mm d = 585 mm (rationalised from 588 mm above) t = 100 mm bw = 500 mm bef = 1,000 mm and assume L beam action in middle f 'c = 25 MPa fsy = 500 MPa From RCDH Excel Spreadsheet \ Ast nominal required = 4,827 mm2 t = 116 mm > 100 mm \ stress block extends into web and the compression block is in both the flange and web Provide Ast = 4,928 mm2 (8 N28) Minimum Ast for flexure = 4,563 mm2 < 4,928 mm2 \ OK ku = 0.27 < 0.36 \ OK f Mu = 1,036 kN.m > 960 kN.m \ OK \ use = 8N28 (4,928 mm2) 8 N28 bottom reinforcement Design for Shear and Torsion at critical section near column (follow Flowchart 4.3) Note the results of the calculations for torsion is combined with the shear and flexural design. The calculations below are by hand and validated by the spreadsheets. M * = 380 kN.m V * = 360 kN T * = 50 kN.m Shear Design AS 3600 Cl 8.2.4 At critical section V * = 360 kN D = 700 mm d o = 615 mm bv = 500 f 'c = 25 MPa fsy.t = 500 MPa Ast= 1,808 mm2 (4 – N24) Using the RCDH Excel Spreadsheet 4.3 Determine f Vu.max = f 0.2 f 'c bv do AS 3600 Clause 8.2.6 = 0.7 x 0.2 x 25 x 500 x 615 x 10–3 = 1,076 kN Refer RCDH Excel Spreadsheet 4.3 for beams, which will determine all the following values β1 = 1.11 For members where the cross-sectional area of shear reinforcement provided (Asv) is equal to or greater than the minimum area specified in Clause 8.2.8 β2 = 1.0 β3 = 1.0 Where fcv = f 'c 1/3 ≤ 4 MPa = 2.92 MPa 10.32 Reinforced Concrete Design Handbook Clause 8.2.7 Shear strength of a beam excluding shear reinforcement Determine f Vuc f Vuc = f β1 β2 β3 bv do fcv Ast 1/3 bv do = 0.7 x 1.11 x 1 x 1 x 500 x 615 x 2.92 (1,808/500/615)1/3/1,000 \ f Vuc = 126 kN \ 0.5 f Vuc = 63 kN Clause 8.2.9 Shear strength of a beam with minimum shear reinforcement Determine f Vu.min = f (Vuc + 0.10 √f 'c bv do ) ≥ f Vuc + f 0.6 bv do f Vuc + f 0.6 bv do = 254 kN f (Vuc + 0.10 √f 'c bv d ) = 233 kN \ f Vu.min = 254 kN \ V * < f Vu.max \ OK > f Vuc \ shear reinforcement is required. RCDH Excel Spreadsheet 4.3 Determine required shear reinforcement Calculate Vus = 335.8 kN If Vus < Vus.min then Vus = Vus.min ie Vus = 336 kN Asv = Vus / [(fsy.f do /s) cot θv] = 229 mm2 and a maximum spacing of 300 mm and where θv = 45 deg for vertical fitments Consider N12 fitments where area of fitment (two legs) = 226 mm2 Maximum spacing of N 12 fitment by interpolation = 300 x 226 / 328 = 296 mm \ Fitments N12 @ 225 cts throughout Design note: It is normal for fitment spacing to be a multiple of 25 mm. Do not use spacings such as 296 mm calculated above as it is just not realistic on site. Torsion Design Design using RCDH Excel Spreadsheet 4.4. T *= 50 kN.m Calculate torsion modulus Jt and ignore the flange as it only contributes a very small part to the overall torsional strength. Ast = 1,808 mm2 J t = J web Table 4.1 for x = 500 mm y = 700 mm J web = 57.7 x 106 mm3 AS 3600 Clause 8.3.3 Determine f Tu.max = f 0.2 f 'c Jt Clause 8.2.6 = 0.7 x 0.2 x 25 x 57.7 x 106 x 10-6 = 202 kN.m Determine f Vu.max = f 0.2 f 'c bv do = 0.7 x 0.2 x 25 x 500 x 615 x 10–3 = 1,076 kN Reinforced Concrete Design Handbook 10.33 Clause 8.3.3 Check torsional strength not limited by web crushing T */f Tu.max + V */f Vu.max = 50/202 + 360/1,024 = 0.25 + 0.33 = 0.58 < 1 \ OK Clause 8.3.5 Check if torsional reinforcement is required. Calculate torsional strength of a beam Clause 8.3.5 (a) Calculate f Tuc = f 0.3 Jt √f 'c Clause 8.3.5 (b) = 0.7 x 0.3 x 57.7 x 106 mm3 x √25 /1,000 = 60.6 kN.m Calculate f Tus = f fsy.f (Asw /s) 2 A t cot θv Where A t = area of a polygon with vertices at the centre of longitudinal bars at the corners of the cross-section = (500 – 40 – 40 –12 –12 – 24/2) x (700 – 40 – 40 –12 –12 – 24/2) = 224,256 mm2 ut = perimeter of the polygon defined for A t = 2 [(500 – 40 – 40 –12 –12 – 24/2) + (700 – 40 – 40 –12 –12 – 24/2)] = 1,936 mm Clause 8.2.10 (b) (1) Where θv = angle between the axis of the concrete compression strut and the longitudinal axis of the member and shall be taken as 45° f = 0.7 Asw = 113 mm2 (assumes N12 fitment) fsy.f = 500 MPa (yield strength of fitment) s = 200 mm (spacing of fitments) f Tus = f Asw fsy.f 2 At cot θv /s = 0.7 x 113 x 500 x 2 x 224,256 x 1 / 200 x 106 = 88.7 kN.m Clause 8.2.7 Calculate shear strength of a beam without shear reinforcement Vuc = β1 β2 β3 bv do fcv Ast 1/3 bv do Refer RCDH Excel Spreadsheet 4.3 for beams, which will determine all the following values β1= 1.11 For members where the cross-sectional area of shear reinforcement provided (Asv) is equal to or greater than the minimum area specified in Clause 8.2.8 β2 = 1.0 β3 = 1.0 Clause 8.3.4 (a) (i) Where fcv = f 'c 1/3 ≤ 4 MPa = 2.92 MPa \ f Vuc = 125.0 kN Check if T * ≤ 0.25 φ Tuc = 15.2 kN.m and as T * = 50 kN.m > 15.2 kN.m \ torsional reo is required Clause 8.3.4 (a) (ii) Check if T */ φ Tuc + V */ φVuc ≤ 0.5 = 50/60.6 + 360/125.0 = 3.7 >> 0.5 \ torsional reo is required Clause 8.3.4 (a) (iii) T */ φ Tuc + V */ φVuc > 1.0 \ torsional reo is required 10.34 Reinforced Concrete Design Handbook Clause 8.3.4 (b) Check requirements for torsional reinforcing ie T */ φ Tus ≤ 1.0 = 50/88.7 = 0.56 < 1.0 \ ok Clause 8.3.6(a) Requirements for additional longitudinal tensile reinforcement A lt = (0.5 fsy.f / fsy) (Asw / s)(ut cot2θv ) = 0.5 x 500 / 500 X 113/200 x 1,936 x 1 = 547 mm2 \ provide 2 additional N24 top bars ie 6 total Clause 8.3.6(b) Requirements for additional longitudinal compressive reinforcing A lt = (0.5 fsy.f / fsy) (Asw / s)(ut cot2θv ) = 0.5 x 500 / 500 X 113/200 x 1,936 x 1 = 547 mm 2 \ provide 2 additional N28 bottom bars ie 10 total Requirements for additional torsional fitments = N12 at 225 mm max centres \ reinforcement to spandrel beam is 4 + 2 = 6 N24 top, 8 + 2 = 10 N28 bottom Fitments N12 @ 175 (shear) + N12 @ 175 (torsion) \ say = N12 @ 175 in pairs for extent of torsion and then N12 @ 250 as single fitments for the remainder of the beam. Design note: In the above design example, the extent of shear and torsion along the length of beam has not been fully defined, which would usually be provided by the analysis of the beam. This then would allow the extent of design for shear and torsion to be calculated along the length of the beam and define the extent of fitments for both shear and torsion. In the above example, the N12 @ 175 could possibly be reduced to N12 @ 300 towards the centre of the beam, if the torsion is not along the full length of the beam. AS 3600 Clause 8.3.8 (a) Note all fitments are to be closed. Reinforced Concrete Design Handbook 10.35 column and wall design VERTICAL ACTIONS (LOADS) AND BENDING MOMENTS (COLUMN RUNDOWNS) Design Note: In order to design the columns, both the vertical actions (loads) at each floor and the bending moments need to be determined. As discussed in Chapter 6, assessing the vertical actions (loads) carried by columns and walls requires a full understanding of the building, the behaviour of the structure and all actions (loads) carried by the structure. These vertical actions (loads) are usually calculated by assessing the actions (loads) supported by each column or walls on a floor-by-floor basis based on the tributary areas to each column or wall. This can be calculated by using a spreadsheet or appropriate structural analysis software. Column rundowns calculated by spreadsheet are typically based on a simple area or length basis, with proportioning of the actions (loads) to each vertical element by taking half the distance in each direction to the adjacent vertical element. These rundowns may not include all the actions (loads) to the columns (and walls) because of continuity and frame action. These additional actions (loads) are sometimes known as 'moment shears' or similar and can be up to 15% of the floor actions (loads). For example, an edge column will generally have fewer actions (loads) on it when using an area basis, while the first column in from the edge of a building will have more actions (loads) on it. It is usual not to deduct moment shears from external columns. Usually, actions (loads) at each level are calculated just above the floor below. The heading 'on Level 3' as shown in the spreadsheet on the column rundown means the actions (loads) from the floors above, just above the third floor and would be used to design the column from Level 3 to Level 4. Clause 3.4.2 of AS/NZS 1170.1 also allows a reduction of uniformly distributed imposed actions. However, as slabs are essentially one-way, no reduction is allowed. Column Rundown C4 Design note: The following rundown is for column C4. The column moments have been determined from the 2D analysis of the bents in both directions. Load element Unit area length Permanent Imposed actions actions (DL) (LL) 65.5 3.5 7.8 0.5 0.25 1 0.5 0.25 Permanent axial actions (DL) Permanent bending moments E/W Permanent bending moments N/S Imposed axial actions (LL) Imposed bending moments E/W Imposed bending moments N/S On Level 4 1 Roof 2 Column 3 Moment shears Total this level Total on Level 4 PA (DL) IA (LL) 32.8 3.5 3.9 16.4 0.0 2.0 40.2 18.3 0 0 0 0 40.2 18.3 (continues) 10.36 Reinforced Concrete Design Handbook Column Rundown C4 (continued) Load element Unit area length Permanent Imposed actions actions (DL) (LL) Permanent axial actions (DL) Permanent bending moments E/W Permanent bending moments N/S Imposed axial actions (LL) Imposed bending moments E/W Imposed bending moments N/S On Level 3 1 2 3 4 5 6 Floor 65.5 Band beam 8.4 Other permanent loads 65.5 Column 3 Moment shears 7.8 Live load reduction 65.5 4.75 3 9.6 0 1.25 0 3.15 6.15 3 0 0 Total this level Total on Level 3 PA (DL) IA (LL) 311.1 3 42 80.6 81.9 9.5 48.0 0.0 196.5 1 21 0.0 0.0 0.0 23.4 0.0 531.1 219.9 3.0 42.0 1.0 21.0 571.2 238.2 On Level 2 1 2 3 4 65.5 Floor Band beam 8.4 Other permanent loads 65.5 Column 5 Moment shears 6 Live load reduction 3 7.8 65.5 4.75 9.6 1.25 3 0 0 311.1 3 42 80.6 81.9 3.15 6.15 3 0 0 9.5 48.0 0.0 531.1 Total this level Total on Level 2 3.0 42.0 196.5 0.0 0.0 1 21 0.0 23.4 0.0 219.9 1.0 21.0 PA (DL) 1,102.3 IA (LL) 458.1 On Level 1 1 Floor 2 3 4 5 6 65.5 Band beam 8.4 Other permanent loads 65.5 Column 3 Moment shears 7.8 Live load reduction 65.5 4.75 3 9.6 0 1.25 0 3.15 6.15 3 0 0 Total this level Total on Level 1 311.1 3 42 80.6 81.9 9.5 48.0 0.0 196.5 1 21 0.0 0.0 0.0 23.4 0.0 531.1 219.9 3.0 42.0 1.0 21.0 PA (DL) 1,633.3 IA (LL) 678.0 On footing 1 2 3 4 5 6 Floor 65.5 Band beam 8.4 Other permanent loads 65.5 Column 3 Moment shears 7.8 Live load reduction 65.5 4.75 3 9.6 0 1.25 0 4.8 6.15 3 0 0 Total this level Total on Footing 311.1 3 42 80.6 81.9 14.4 48.0 0.0 196.5 1 21 0.0 0.0 0.0 23.4 0.0 536.0 219.9 3.0 42.0 1.0 21.0 PA (DL) 2,169.3 IA (LL) 897.9 Total PA (DL) Total IA (LL) 2,169.3 kN 897.9 kN Notes: — Actions (loads) are in kN or kPa. All loads are unfactored. — Moments are in kN.m. Reinforced Concrete Design Handbook 10.37 INTERNAL COLUMN C4, LEVEL L1 TO L2 Column properties 88 L1 350 D500L10 88 274 88 450 3500 3150 450 88 274 60 L2 f 'c = 40 MPa fsy = 500 MPa Cover = 60 mm to fitment AS 3600 Clause 10.1.3 Braced column, ie lateral loads resisted by shear walls Design actions (from computer analysis) N * = 1,633 x 1.2 + 678 x 1.5 = 2,977 kN M *nsmax = 42 x 1.2 + 21 x 1.5 = 82 kN.m M *ewmax = 3 x 1.2 + 1 x 1.5 = 5 kN.m Check if column short Calculate Le / r Clause 10.3.1 Assume Le = Lu = 3,150 mm r = 0.3 D = 0.3 x 450 = 135 mm \ Le / r = 3,150/135 = 23.5 Clause 10.3.1(a) < 25 \ column short \ moment magnifiers are not required. Assume column reinforced on four faces f 'c = 40 MPa Design Note: Because of the design decision to use 40 MPa minimum for all columns with 60 cover to the fitment say 70 mm to the main bar or an axis distance of about 85 mm, then the columns in the upper levels are likely to be not heavily reinforced. Therefore try 1% reo as a minimum, ie 4 N28 at 1.2% and see how the column details fit the interaction diagram following. From various design checks, the concrete strength could be reduced to 32 MPa. However, the volume of concrete for all the internal columns is about 5 m3 total per floor (ie a truck load of concrete and all columns on the floor are likely to be cast at the one time) so the saving in changing to a lower strength concrete is minimal and will only result in confusion on site and possibly add cost because of smaller batches of concrete. However, the designer will need to check the transmission of axial forces through the slab because of the column's concrete strength. Refer to Clause 10.8. An alternative is to design the column with 32-MPa concrete but use 40-MPa concrete for practical reasons. 10.38 Reinforced Concrete Design Handbook Input data into columns design software The following interaction diagram has been calculated using an excel spreadsheet for a short column in accordance with AS 3600. As can be seen below the design actions are under the design line for 4 N28 and 40 MPa so well ok. Area of reo = 1.22% < 4% \ OK 9000 8000 X 7000 Strength line Design line Minimum moment Design actions 6000 5000 4000 X Compresive force (kN) 3000 82,2977 2000 1000 0 0 100 200 300 400 500 600 Moment (kN.m) \ Adopt 4 N28 Determine size and spacing of fitments AS 3600 Clause 10.7.4.3 Size use L10 Table 10.7.4.3 Spacing smaller of 15 d b or D 15 d b = 15 x 28 = 420 mm D = 450 mm However, 420 mm too large \adopt 300 mm Fitments L10 @ 300 Reinforced Concrete Design Handbook 10.39 INTERNAL COLUMN C4, FOOTING LEVEL TO LEVEL L1 8N32 bars L10 88 L1 350 Column properties 88 88 274 450 4800 4450 450 88 274 60 Footing f 'c = 40 MPa fsy = 500 MPa AS 3600 Clause 10.1.3 Braced column, ie lateral loads resisted by shear walls Design actions (from computer analysis) N * = 2,169 x 1.2 + 898 x 1.5 = 3,950 kN M *nsmax = 42 x 1.2 + 21 x 1.5 = 82 kN.m M *ewmax = 3 x 1.2 + 1 x 1.5 = 5 kN.m Check if column short Calculate Le / r Clause 10.5.3 Assume Le = 0.85 Lu = 0.85 x 4,450 = 3,783 mm Design Note: Calculate Le from Clause 10.5.3 rather than Clause 10.5.4 as restraint at footing uncertain. Clause 10.5.2 r = 0.3 D = 0.3 x 450 = 135 mm \ Le / r = 3,783/135 = 28 Determine limit for Le / r for the column to be classed as short AS 3600 Clause 10.3.1(a) = 25; or ≤ α c (38 – f 'c /15) (1+ M *1/M *2) whichever is greater Nuo = g 1 f 'c Ac fsy As = 10,257 kN N */0.60 Nuo = 3,950 / 0.6 x 10,257 = 0.64 > 0.15 \αc = √ (2.25 – 2.5 N */ 0.6 Nuo) = √ ( 2.25 -2.5 x 0.64) = 0.81 Check if M *2 > 0.05 D N * 0.05 D N * = 0.05 x 0.45 x 3,950 = 88.9 kN.m > M *2 (82 kN.m) As less than the minimum value and column in single curvature take M *1 /M *2 = –1 \ limit = 25 Le / r = 28 > 25 \ column is slender 10.40 Reinforced Concrete Design Handbook Determine moment magnifier Clause 10.4.2 d b = km/(1.0 – N */Nc) ≥ 1.0 Calculate km = (0.6 – 0.4 M *1 /M *2) as above M *1 /M *2 = –1 \ km = 1.0 N * = 3,950 kN Determine Nc = (π 2/Le2) [182 do(f Mub)/(1 + bd)] Clause 10.4.4 Determine bd = G/(G + Q) = 2,169/ (2,169 + 898) = 0.707 Assume column reinforced on 4 faces with 9 N32, ie 3 bars in each face = 3.57% < 4% \ OK f 'c = 40 MPa From a program for the design of columns Mub = 614 kN.m \ f Mub = 0.6 x 614 = 368.4 kN.m Le = 3,783 mm do = 362 mm \ Nc = (π 2/ 3.7832) x [182 x 0.362 x 368/(1 + 0.71)] = 9,778 kN \d b = 1.0/(1.0 – 3,950 / 9,778) = 1.68 \M * = 1.68 x 82 = 138.1 kN.m See interaction diagram below for bending in one direction so minimum moments in the other direction will need to be considered. By inspection should be OK for bending in the other direction. Design Note: Column software will probably give marginally more accurate results and will probably cover biaxial bending) 12000 10000 X Strength line Design line Minimum moment Design actions 8000 6000 X Compresive force (kN) 4000 138,3950 2000 0 0 100 200 300 400 500 600 700 Moment (kN.m) \ Adopt 9 – N32, 3 in each face Reinforced Concrete Design Handbook 10.41 Determine size and spacing of fitments Clause 10.7.4.3 Use N12 Spacing = 15 d b = 15 x 32 = 480 > D D = 450 mm But adopt closer spacing for fitments better confinement, eg 300 mm. Provide fitments N12 @ 300 restraint to middle bars as per Cl 10.7.4.1 of AS 3600 Note: adopt Option 2 as better arrangement to place concrete into the column. N12 internal fitment at 200 cts alternate N12 at 300 N12 at 300 OPTION 2 OPTION 1 Design Note: For external columns, provide lateral shear reinforcement through the joint as required by Clause 10.7.4.5. WALL D1 – D2 Wall properties 3.5 175th 3.5 Wu 4.8 15.3 3.5 Floor loads 200th 7.6 f 'c = 40 MPa fsy = 500 MPa Clause 11.3 Wall is braced Design loads. Refer to following rundown of actions for this wall G = 1.20 x 1,958 = 2,350 kN Q = 1.5 x 935 = 1,403 kN Ultimate load = 3,752 kN = 493.7 kN/m Check compression stress in wall at bottom of the wall = 3,752 / 0.2 x 7.6 / 1,000 = 2.47 MPa \ wall is not heavily loaded Also Clause C5.3 compression stress < 0.15 f 'c = 6 MPa \ no boundary elements are needed Wind load Wu (from other computations) = 248 kN Table 1.1 RCDH Check stability for overturning parallel to the wall Wu x 15.3 / 2 = 248 x 15.3 / 2 = 1,897 kN.m 0.9 G x 7.6 / 2 = 0.9 x 1,958 x 7.6 / 2 = 6,696 kN.m >1,897 kN.m \ no overturning, ie stable 10.42 Reinforced Concrete Design Handbook Check if any tension in the wall. Tensile stress = M / Z M * = 1,897 kN.m Z = bd2/6 \ ft = 1,897 x 106/200 x 7,6002/6 = 0.98 MPa < 2.47 MPa \ wall is not in tension Clause 11.2.1(a) Wall subject to in plane vertical and horizontal forces. Design for vertical and horizontal forces independently. Use Cl 11.5 for axial forces and Cl 11.6 for shear. Design for vertical forces at change of section from 175 mm to 200 mm thickness at Level 1. Refer to the wall rundown following. Determine load eccentricity, e at level 1 175 wall over N*2+3+4+roof 87.5 CS Level 1 133 CS N*1 190 Expressed joint Slab N*total d e 100 200 wall under Note: Reinforcement not shown e = (d – 100) mm d = (N *2+3+4+ roof x 87.5 + N *1 x 133.3) / N *total = [(14.27 x 1.2 + 790 x 1.5) x 87.5 + (532 x 1.2 + 145 x 1.5) x 133.5] / 3,228 = 98.0 mm \ e = 98.0 – 100 = – 2 mm Clause 11.5.2 < 0.05 t w = 0.05 x 200 = 10 mm \ use e = 10 mm Clause 11.5.1 Calculate the design axial strength of the wall using the simplified design method for walls subject to vertical compression forces Calculate Hwe = 1.0 x 4.8 = 4.8 m ea = Hwe2/ 2,500 t w = 4.8 x 4.8 x 106/ (2,500 x 200) = 46.08 mm f 'c = 40 MPa t w = 200 mm f N u = f (t w – 1.2 e – 2ea) 0.6 f 'c = 0.6 x (200 – 1.2 x 10 – 2 x 46.08) x 0.6 x 40 = 1,381 kN/m > 494 kN/m \ well OK Design note: The RCDH Excel Spreadsheet 7.2 for the design of walls for axial loads using the simplified method can be used for the above calculations. Reinforced Concrete Design Handbook 10.43 Wall actions and bending moments (wall rundown) D1 – D2 Load element Unit area length Permanent Imposed actions actions (DL) (LL) 25.9 7.6 0 0.5 0.25 1.0 0 0 Permanent axial actions (DL) Permanent bending moments E/W Permanent bending moments N/S Imposed axial actions (LL) Imposed bending moments E/W Imposed bending moments N/S On Level 4 1 Roof 2 Wall LW 3 Moment shears 13.0 7.6 0.0 6.5 0.0 0.0 Total this level 20.6 0 0 6.5 0 0 Total on Level 4 DL 20.6 LL 6.5 On Level 3 1 2 3 4 5 Floor Wall 175 thick Edge beam Moment shears Live load reduction 25.9 27.36 7.2 0 25.9 Total this level Total on Level 3 6.0 4.4 26.9 0.0 0.0 DL LL 7.5 0.0 9.3 0.0 0.0 155.4 3 42 119.7 193.7 0.0 0.0 194.3 1 21 0.0 67.0 0.0 0.0 468.8 3.0 42.0 261.2 1.0 21.0 489.3 267.7 On Level 2 1 Floor Wall 175 thick 2 3 4 5 Edge beam Moment shears Live load reduction t25.9 27.36 7.2 0 25.9 Total this level Total on Level 2 7.5 6.0 4.4 26.9 0.0 0.0 DL LL 0.0 9.3 0.0 0.0 155.4 3 42 119.7 193.7 0.0 0.0 1 194.3 21 0.0 67.0 0.0 0.0 468.8 3.0 42.0 261.2 1.0 21.0 958.1 528.9 On Level 1 1 2 3 4 5 Floor Wall 175 thick Edge beam Moment shears Live load reduction 25.9 27.36 7.2 0 25.9 Total this level Total on Level 1 6.0 4.4 26.9 0.0 0.0 DL LL 7.5 0.0 9.3 0.0 0.0 155.4 3 42 119.7 193.7 0.0 0.0 194.3 1 21 0.0 67.0 0.0 0.0 468.8 3.0 42.0 261.2 1.0 21.0 1,426.9 790.1 On footing 1 2 3 4 5 Floor Wall 200 thick Edge beam Moment shears Live load reduction 25.9 36.48 7.2 0 25.9 Total this level Total on Footing 6.0 5.0 26.9 0.0 0.0 DL LL Total DL Total LL 3.0 0.0 9.3 0.0 0.0 155.4 3 42 182.4 193.7 0.0 0.0 77.7 1 21 0.0 67.0 0.0 0.0 531.5 3.0 42.0 144.7 1.0 21.0 1,958.4 934.8 1,958 kN 935 kN Notes: — Actions (loads) are in kN or kPa. All loads are unfactored. — Moments are in kN.m. 10.44 Reinforced Concrete Design Handbook Design wall for in-plane horizontal shear forces Clause 11.6 V * = 248 kN f 'c = 40 MPa Calculate Hw / Lw = 4.8 / 7.6 = 0.63 Calculate Hwe / tw = 4.8 / 0.2 = 24 ≤ 50 \ OK Calculate f Vu.max = 0.2 f 'c (0.8 Lw tw ) = f x 0.2 x 40 x 0.8 x 7,600 x 200/1,000 = 6,810 kN >> 248 kN Clause 11.6.3 Calculate shear strength of a wall without shear reinforcement Hw / Lw = 0.63 ≤ 1 therefore use Clause 11.6.3 (a) Clause 11.6.3 (a) f Vuc = f (0.66 √f 'c – 0.21 Hw / Lw √f 'c ) 0.8 Lw tw = 0.7 x (0.66 x √40 – 0.21 x 0.63 x √40) x 0.8 x 7,600 x 200/1,000 = 2,840 kN >> 248 kN without the contribution of the shear reinforcement which is f Vus = 1192 kN \ shear capacity of wall is well OK Design note: The RCDH Excel Spreadsheet 7.3 can be used for the design of walls for shear for the above calculations. \ As no reinforcement required for shear use minimum reinforcement from Cl 11.7.1. As wall thickness is 200 mm only need to have only 1 layer of reinforcement. Check bending to wall under lateral wind loads Clause 11.1 (b) Note the stress in the wall in the middle is ≈ 2.06 MPa which is about 3% over the 2 MPa limit and as no reduction for moment shears taken, so OK and design as a slab. M* ≈ w l2 /10 = 0.97 x 4.82 /10 = 2.23 kN.m/m For N12 @ 300 Ast = 377 mm2 and f Mu = 14.7 kN.m > 2.23 kN.m \ well OK Design note: Designers do not need to do sophisticated analysis for such simple load cases. They should try to use simple design methods, where possible, as they are in many cases quicker than running a computer program. Clause 6.10.2.3 using the simplified method of analysis was used for the above calculation which would have given a moment of w l2 /11 but was rounded down to w l2 /10 which is slightly conservative. As shown above it does not make a significant difference. AS 3600 Clause 11.7.1 (a) Vertical Reinforcement A s.v = 0.0015 x 200 x 1,000 = 300 mm2/m Adopt N12 @ 200 mm = 565 mm2/m > 300 mm2/m Vertically N12 @ 200 Design note: While the minimum vertical reinforcement is nominally required only for cracking with N12 @ 300, both for crack control and robustness N12 @ 200 has been adopted as vertical reinforcement. (b) Horizontal Table 2.3 RCDH AS 3600 Clause 11.7.2 As.h = 0.0025 x 200 x 1,000 = 500 mm2/m N12 @ 200 mm = 565 mm2/m Check horizontal reinforcement for crack control Exposure classification B2 \ p = 0.006 Table 2.3 RCDH \ As.h = 0.006 x 200 x 1,000 = 1,200 mm2/m N12 @ 90 mm or N16 @ 160 mm Horizontally N16 @150 Reinforced Concrete Design Handbook 10.45 footing Design FOOTING COLUMN C4 Design data Allow bearing pressure (from geotechnical report) qu = 300 kPa Design loads g = 2,169 kN q = 898 kN N * = 3,950 kN M * = 0 kN.m Column dimensions and details c1 = 450 mm c2 = 450 mm longitudinal reo to column = 9 N32 Concrete strength footing f 'c = 32 MPa Concrete cover = 75 mm Estimate footing size. Assume a square footing side length = b initial \ b initial = √[(2,169 + 898) / 300] = 3.2 m Check footing depth for column bar development length in compression Cl 13.1.5 and Table 2.11 of RCDH, the development length for 32 mm bar = 700 mm Initial depth of footing = cover + two layers of 32 mm bar + basic development length. \ Initial depth of footing = 75 + 64 + 700 = 839 say 900 mm \ Effective depth do = 900 – 75 – 32 – 16 = 775 mm Calculate footing weight and confirm initial estimates of sizes wf initial = 3.2 x 3.2 x 0.9 x 25 = 230 kN \ Approx total working load = 2,169 + 898 + 230 = 3,067 + 230 = 3,297 kN Check b final = √3,297/300 = 3.32 m OK \ Adopt footing 3.35 x 3.35 x 0.9m deep wf final = 3.35 x 3.35 x 0.9 x 25 = 252 kN \ Total working load = 2,169 + 898 + 252 = 3,319 kN Actual bearing pressure = 3,319 / 3.35 / 3.35 = 296 kPa < 300 kPa \ OK Actual design bearing pressure on concrete = (3,319 – 252) / 3.35 / 3.35 = 273 kPa Calculate total load \ Total actions Nf * = (2,169 + 252) x 1.2 + 898 x 1.5 =4,252 kN \ Load factor = 4,252 / (2,169 + 252 + 898) = 1.28 Check BM capacity as cantilever about face of column Calculate BM total M * = quL2/2 = 1.28 x 273 x 3.35 x [(3.35 – 0.45) / 2]2 / 2 = 1,230 kN.m 10.46 Reinforced Concrete Design Handbook Use RCDH Excel Spreadsheet 4.1 Calculate Ast = 4,710 mm2 (15 N20) and f Mu = 1,435 kN.m > 1,230 kN.m \ OK AS 3600 Clause 16.3.1 and Table 8.2 of RCDH Min. reo ≈ 1,400 x 3.35 mm2 = 4,690 < 4,710 \ OK Clause 8.2.4 Check one-way beam shear x = [(L1 – c1 ) / 2] – do = [(3,350 – 450) / 2] – 775 = 675 mm \ V * = 0.675 x 3.35 x 273 x 1.28 = 790 kN Use RCDH Excel Spreadsheet 4.1 β1 = 0.91 (no shear reinforcement) β2 = β3 = 1.0 From AS 3600 Clause 8.2,5 (ii) f Vuc = 639 kN < 790 kN, NG \ Increase depth to 1,050 mm and dom to 925 mm f Vuc = 719 kN > 615 kN \ OK \ by inspection one-way shear OK both ways Clause 9.2.3 Check punching shear with M * = 0 with dom = 935 dom = 785 mm Use RCDH Excel Spreadsheet 8.1 V * punching = 3,255 kN f Vu = 6,974 kN > 3,255 kN \OK As depth has increased, so has the minimum reinforcement to = 5,150 mm2, ie 17 N20 Check development length of column bars = 1,050 – 75 – 20 – 20 = 935 > 700 \OK Therefore footing C4 3.35 x 3.35 x 1.05 m deep with 17 N20 both ways bottom with 75 mm cover to bottom and sides FOOTING WALL D1 – D2 Design Data Allow bearing pressure (from geotechnical report) q u = 300 kPa Design loads (see previous wall rundown for actions) g = 1,958 kN q = 935 kN N * = 3,752 kN M * = 0 kN.m Wall dimensions and details c1 = 7,600 mm c2 = 200 mm Longitudinal reo to wall = N12 @ 300 cts Concrete strength footing f 'c = 32 MPa Concrete cover = 75 mm Reinforced Concrete Design Handbook 10.47 Estimate footing size. Area of footing = (1,958 + 935) / 300 = 9.6 m2 Adopt a footing say 8,000 x 1,300 = 10.4 m2 > 9.6 m2 Check footing depth for column bar development length in compression Cl 13.1.5 and Table 2.11 of the RCDH development length for 12 mm bar = 300 mm Depth of footing = 75 + 40 + 300 = 415 say 500 mm \ Effective depth do = 500 – 75 – 20 – 10 = 395 mm Calculate footing weight and confirm initial estimates wf initial = 8 x 1.3 x 0.5 x 25 = 130 kN \ Approx total working load = 1,958 + 935 + 130 = 3,023 kN Check area = 3,023 / 300 = 10.07 m2 > 9.6 m2 \ Adopt footing 8.0 x 1.3 x 0.5 m deep \ Total working load = 1,958 + 935 + 130 = 3,023 kN Actual bearing pressure = 3,023 / 8 / 1.3 = 291 kPa < 300 kPa \ OK Actual design bearing pressure on concrete = (3,023 – 130) / 8 / 1.3 = 278 kPa Calculate total load \ Total load N *f = (1,958 + 130) x 1.2 + 935 x 1.5 = 3,908 kN \ Load factor = 3,908 / (1,958 + 935 + 130) = 1.29 Check BM capacity as cantilever about face of wall Calculate BM total M * = quL2 / 2 = 1.29 x 278 x 8 x [(1.3 – 0.2)/2]2 / 2 = 434 kN.m Use RCDH Excel Spreadsheet 4.1 Calculate Ast = 3,232 mm2 Use N12 at 200 cts nominal = 4,520 mm2 Design note: To calculate the number of fitments take the total length of the footing minus the cover each end and divide by 200 and rationalise up ie (8,000 – 140) divided by 200 = 40 bars. and f M uo = 6,876 kN.m > 434 kN.m \ well OK and f M u = 705 kN.m > 434 kN.m \ well OK AS 3600 Clause 16.3.1 and Table 8.2 Min. reo ≈ 700 x 8 = 5,600 mm2 > 3,232 mm2 \ Number of bars = 5,600 / 113 = 50 \ adopt = say 53 N12 @150 cts = 5,989 mm2 > 5,600 mm2 \ 53 N12 @ 150 mm nom cts as fitments Minimum reo along footing = 1,061 x 1.3 = 1,379 mm2 Provide 7 N16 top and bottom, ie 2,800 mm2 > 1,379 mm2 \ OK. 10.48 Reinforced Concrete Design Handbook Clause 8.2.4 Check one-way beam shear x = (L1 – c1 – 2do) / 2 = (1,300 – 200 – 2 x 395) / 2 = 155 mm \ V * = 0.155 x 8 x 278 x 1.29 (area x bearing pressure x load factor) = 445 kN Use RCDH Excel Spreadsheet 4.1 β1 = 1.32 β2 = β3=1.0 From AS 3600 Clause 8.2,5 (ii) f Vuc = 1,095 kN > 445 kN \ no shear reinforcing required. \ by inspection one-way shear OK both ways Clause 9.2.3 Check punching shear with M * =0. By inspection well OK. Footing D1 4 8.9 x 1.3 x 0.5 deep with 7 N16 long way top and bottom with N12 fitments as closed fitments @ 150 cts and 75 mm cover to top, bottom and sides. See Section below. 200 wall 500 N12 starter bars @ 200 cts CS SECTION 7N16 top and bottom N12 fitments @ 150 cts Reinforced Concrete Design Handbook 10.49 blank page 10.50 Reinforced Concrete Design Handbook Appendix A The design process n n n Purpose This Appendix is to assist the designer in appreciating and understanding some of the issues involved with designing a concrete structure. It should be read in conjunction with Section 1.2 of this Handbook. n Movement and construction joints n n n Basic design information such as site plan, geotechnical information, site constraints, survey, etc n The likely structural form(s) n How the services and structure may be integrated n How it is likely to be constructed n The likely time frames for design and construction n Environmental considerations n n Preliminary budget estimate to confirm that the project appears economically viable The client's approval for the project to proceed to the next phase. Designers need to carry out sufficient structural design to ensure that concepts are feasible and to avoid subsequently finding that final design does not work. The final design stage is where the chosen optimum preliminary design is designed and detailed. This will include the preparation of project documentation and specifications. It is important that the designer remember that the documentation is the means of communicating the design intentions to the contractor/ builder and subcontractors, the documentation should be reviewed from this viewpoint before being issued. This stage will include: n n n — Lateral load-resisting systems in two orthogonal directions including shear and core walls — Vertical load-resisting systems, ie walls and columns — Robustness n Design information such as site plan, site constraints, survey, etc Approximate member sizes for alternative designs so that options can be costed to get the optimum solutions. Final design Following the acceptance of the conceptual design, further development of it, in more detail, in this phase will include: Evaluation of different structural options as required, taking into consideration: Construction sequence and temporary stability if it is an unusual or complicated structure If necessary, a further budget costing is carried out to confirm the project is on budget. Preliminary design n Likely sizes of structural members including footings, columns and walls including sizing of any precast elements to be used, etc (designers should refer to this Handbook for preliminary sizes for footings, columns, walls and Precast Concrete Handbook for preliminary sizes for precast members) Coordination with building services The conceptual design phase will involve considering some or more of the following general issues: The broad principles of the structure (which may include structural sketch plans) to suit the sketches and layout the architect or designer has proposed and to meet the client's needs Structural framing and, for floors, indicative sizes based on span-to-depth ratios, some simple design, experience, etc. Designers should refer to Guide to Long-Span Concrete Floors A.1 for initial sizing of concrete floors) n Conceptual design n Detailed geotechnical and environmental information if possible n A review of all design data to ensure its validity Full analysis of the chosen design for all combinations of lateral and vertical actions. The effect of loads, forces and deformations on the structure and the behaviour of the total structure under the various design load cases are evaluated. Design for durability, fire resistance, deflection and other relevant design loadings should also be carefully considered Coordination of the structural design with the design of other aspects of the building, eg hydraulic services and external cladding, including liaising with other members of the project team (the architect, services engineers, etc) Full design and detailing of the project. The project must be adequately documented including drawings, conditions of contract and specifications as incomplete documents may delay the project and result in extra costs Reinforced Concrete Design Handbook A.1 n n The provision of guidance on how the structure is stabilised during erection of complex or unusual elements such as precast elements. Until lateral stability is achieved by the completed structure, it may be necessary to nominate the sequence for construction to ensure the design concept is not compromised and the structure remains stable during erection Independent design checks and Quality Assurance (QA) procedures. Documentation It is important to detail and document the project sufficiently so that it can be built without undue reference to the designer and to avoid problems with construction. It is well known within the building and construction industry in Australia that poor documentation has led to an inefficient, non‑competitive industry, cost overruns, rework, extensions of time, high stress levels, loss of morale, reduced personal output and adversarial behaviour, and with diminished reputations (see Getting it Right the First Time A.2). References A.1 Guide to Long-Span Concrete Floors (T36) 2nd Ed, Cement Concrete & Aggregates Australia, 2003. A.2 Getting it Right the First Time, Engineers Australia Queensland Division Industry-wide Taskforce on Documentation within the building and construction industry, 2005 (www.qld.engineersaustralia.org.au). A.2 Reinforced Concrete Design Handbook Appendix B Development and use of the spreadsheets General The spreadsheets in this Handbook have been developed to illustrate design to AS 3600—2009 and the formulae used in the Handbook. They are not intended to be all embracing or to replace commercially available or purpose-written software. Some of the following discussion on spreadsheets has been based on the Reinforced Concrete Council's project Spreadsheets for concrete design to BS 8110 and EC2 B.1 and that source is acknowledged. The design of concrete structures has been described as time-consuming and costly. Computer programs are now used extensively but experienced designers are often reluctant to rely on 'black box' technology over which they have little control. Computer spreadsheets, on the other hand, are usually user-friendly, generally transparent, powerful, and are popular in structural engineering. They have good graphical presentation facilities and established links with other software, notably word processing. They are an ideal medium to deal with the intricacies of concrete design in that they can carry out a series of mathematical calculations and, as in manual design, can check whether certain conditions are met. The spreadsheets presented in this Handbook will help students and inexperienced engineers gain an understanding of reinforced concrete design. For the experienced engineer, the spreadsheets will also help in the production of clear and accurate design calculations. In producing the spreadsheets, Microsoft Excel 2003 was adopted as being the de facto standard and the most widely available spreadsheet used. Designers can also refer to texts such as The Engineers Tables B.2 and Engineering with the Spreadsheet B.3 for guidance when preparing such spreadsheets for engineering calculation. Advantages For the design engineer, these spreadsheets will assist in the preparation of clear and accurate design calculations for individual reinforced concrete elements. Spreadsheets allow users to gain experience by studying their own 'what if' scenarios. Should they have queries, users should be able to answer their own questions by chasing through the cells to give them an understanding of the logic used. Cells within each spreadsheet can be interrogated, formulae checked and values traced. Engineers are sometimes criticised for not thoroughly costing their designs. With spreadsheets, it is a very simple matter to multiply the quantities of reinforcement, concrete and formwork required by current rates to give an idea of material costs. Use Spreadsheets are a very powerful tool. Their use is increasingly common in the preparation of design calculations. They can save time, money and effort. They provide the facility to optimise designs and can help gain experience. However, these benefits have to be weighed against the risks involved which must be recognised and managed. In other words, appropriate levels of supervision and checking, including self‑checking must, as always, be exercised when using spreadsheets. In its deliberations, the Standing Committee on Structural Safety (SCOSS) B.4 noted the increasingly widespread availability of computer programs and circumstances in which their misuse could lead to unsafe structures. These circumstances include: n n n n n n People without adequate structural engineering knowledge or training may carry out the structural analysis. There may be communication gaps between the design initiator, the computer program developer and the user. A program may be used out of context. The checking process may not be sufficiently fundamental. The limitations of the program may not be sufficiently apparent to the user. For unusual structures, even experienced engineers may not have the ability to spot weaknesses in programs for analysis and detailing. The committee's report continued: Spreadsheets are, in principle, no different from other software … Reinforced Concrete Design Handbook B.1 Liability A fundamental condition of the use of the spreadsheets in this publication is that the user accepts responsibility for the input data and output of the program / spreadsheet, its interpretation and how they are used. As with all software, users must be satisfied with the answers these spreadsheets give and be confident in their use. These spreadsheets can never be fully validated for every situation but have been through some testing, both formally and informally. However, users must satisfy themselves that the uses to which the spreadsheets are put are appropriate. It is up to the user what use is made of the output. The spreadsheets have been produced to cater for both first-time designers and the more experienced designers without putting off first-time designers. Summary With spreadsheets, long-term advantages and savings come from repeated use but there are risks that need to be managed. Spreadsheets demand an initial investment in time and effort – but the rewards are there for those who make this investment. Good design requires sound judgement based on competence derived from adequate training and experience – not just computer programs. Control Users and managers should be aware that spreadsheets can be changed and must address change control and versions for use. The flexibility and ease of use of spreadsheets, which account for their widespread popularity, also facilitate ad-hoc and unstructured approaches to their subsequent development. Quality Assurance procedures may dictate that spreadsheets are treated as controlled documents and subject to comparison and checks with previous methods prior to adoption. Users' Quality Assurance schemes should address the issue of changes. The possibilities of introducing a company's own password to the spreadsheets and/or extending the revision history contained within the sheet entitled Notes might be considered. Application The spreadsheets have been developed with the goal of producing calculations to show compliance with AS 3600—2009. Whilst this is the primary goal, there is a school of thought that designers are primarily responsible for producing specifications and drawings which work on site and are approved by clients and/ or checking authorities. Producing calculations happens to be a secondary exercise, regarded by many experienced engineers as a hurdle on the way to getting the project approved and completed. From a business process point of view, the emphasis of the spreadsheets might, in future, change to establishing compliance once members, loads and details are known. Certainly, this may be the preferred method of use by experienced engineers. The spreadsheets have been developed with the ability for users to input and use their own preferred material properties, bar sizes, etc. However, user preferences should recognise efficiency through standardisation. B.2 Reinforced Concrete Design Handbook References B.1 Spreadsheets for concrete design to BS 8110 and EC2, Reinforced Concrete Council, Department of Trade and Industry, UK, 2000. B.2 Mote Dr R The Engineer's Tables, Trafford Publishing, 2009. B.3 Christy CT Engineering with the Spreadsheet, ASCE Press, 2006. B.4 SCOSS – the Standing Committee on Structural Safety http://www.scoss.org.uk/. blank page Reinforced Concrete Design Handbook B.3 blank page B.4 Reinforced Concrete Design Handbook