Uploaded by r.corbett2

INDUSTRY GUIDE T38 Reinforced Concrete Design in Accordance with AS3600

advertisement
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.
Helen Rix Design
ISBN 978-1-877023-28-6
CONTACT DETAILS:
For information regarding the development
of Standards contact:
Standards Australia Limited
GPO Box 476 Sydney NSW 2001
phone: 02 9237 6000
fax: 02 9237 6010
email: mail@standards.org.au
internet: www.standards.org.au
Standards Australia develops Australian Standards ®
and other documents of public benefit and national
interest. These Standards are developed through
an open process of consultation and consensus, in
which all interested parties are invited to participate.
Through a Memorandum of Understanding with the
Commonwealth Government, Standards Australia
is recognised as Australia’s peak non-government
national standards body.
For further information visit www.standards.org.au
Australian Standards ®
Committees of experts from industry, governments,
consumers and other relevant sectors prepare
Australian Standards. The requirements or
recommendations contained in published Standards
are a consensus of the views of representative
interests and also take account of comments received
from other sources. They reflect the latest scientific
and industry experience. Australian Standards are
kept under continuous review after publication and
are updated regularly to take account of changing
technology.
International involvement
Standards Australia is responsible for ensuring the
Australian viewpoint is considered in the formulation of
International Standards and that the latest international
experience is incorporated in national Standards. This
role is vital in assisting local industry to compete in
international markets. Standards Australia represents
Australia at both the International Organization
for Standardization (ISO) and the International
Electrotechnical Commission (IEC).
Sales and distribution
Australian Standards®, Handbooks and other
documents developed by Standards Australia are
printed and distributed under license by SAI Global
Limited.
ii
Reinforced Concrete Design Handbook
For information regarding the sale and distribution
of Standards contact:
SAI Global Limited
phone: 13 12 42
fax: 1300 65 49 49
email: sales@saiglobal.com
CCAA OFFICES
SYDNEY OFFICE:
Level 6, 504 Pacific Highway
St Leonards NSW Australia 2065
POSTAL ADDRESS:
Locked Bag 2010
St Leonards NSW 1590
TELEPHONE: (61 2) 9437 9711
FACSIMILE: (61 2) 9437 9470
BRISBANE OFFICE:
Cement Concrete & Aggregates Australia is a not-forprofit organisation established in 1928 and committed
to serving the Australian construction community.
CCAA is acknowledged nationally and internationally
as Australia's foremost cement and concrete
information body – taking a leading role in education
and training, research and development, technical
information and advisory services, and being a
significant contributor to the preparation of Codes and
Standards affecting building and building materials.
CCAA's principal aims are to protect and extend
the uses of cement, concrete and aggregates by
advancing knowledge, skill and professionalism in
Australian concrete construction and by promoting
continual awareness of products, their energy-efficient
properties and their uses, and of the contribution the
industry makes towards a better environment.
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
Download