Design of structural elements

1. Philosophy of design

This chapter is concerned with the philosophy of struc
tural design. The chapter describes the overall aims of
 design and the many inputs into the design process.
 The primary aim of design is seen as the need to ensure
 that at no point in the structure do the design loads
 exceed the design strengths of the materials. This can be
 achieved by using the permissible stress or load factor
 philosophies of design. However, both suffer from draw
backs and it is more common to design according to
 limit state principles which involve considering all the
 mechanisms by which a structure could become unfit
 for its intended purpose during its design life

1.1. Introduction

 The task of the structural engineer is to design a
 structure which satisfies the needs of the client and
 the user. Specifically the structure should be safe,
 economical to build and maintain, and aesthetic
ally pleasing. But what does the design process
 involve?
 Design is a word that means different things to
 different people. In dictionaries the word is de
scribed as a mental plan, preliminary sketch, pat
tern, construction, plot or invention. Even among
 those closely involved with the built environment
 there are considerable differences in interpretation.
 Architects, for example, may interpret design as
 being the production of drawings and models to
 show what a new building will actually look like.
 To civil and structural engineers, however, design is
 taken to mean the entire planning process for a new
 building structure, bridge, tunnel, road, etc., from
 outline concepts and feasibility studies through
 mathematical calculations to working drawings
 which could show every last nut and bolt in the
 project. Together with the drawings there will be
 bills of quantities, a specification and a contract,
 which will form the necessary legal and organiza
tional framework within which a contractor, under the supervision of engineers and architects, can con
struct the scheme.
 There are many inputs into the engineering
 design process as illustrated by Fig. 1.1 including:
 1. client brief
 2. experience
 3. imagination
 4. a site investigation
 5. model and laboratory tests
 6. economic factors
 7. environmental factors.
 1.1 Introduction
 The task of the structural engineer is to design a
 structure which satisfies the needs of the client and
 the user. Specifically the structure should be safe,
 economical to build and maintain, and aesthetic
ally pleasing. But what does the design process
 involve?
 Design is a word that means different things to
 different people. In dictionaries the word is de
scribed as a mental plan, preliminary sketch, pat
tern, construction, plot or invention. Even among
 those closely involved with the built environment
 there are considerable differences in interpretation.
 Architects, for example, may interpret design as
 being the production of drawings and models to
 show what a new building will actually look like.
 To civil and structural engineers, however, design is
 taken to mean the entire planning process for a new
 building structure, bridge, tunnel, road, etc., from
 outline concepts and feasibility studies through
 mathematical calculations to working drawings
 which could show every last nut and bolt in the
 project. Together with the drawings there will be
 bills of quantities, a specification and a contract,
 which will form the necessary legal and organiza
tional framework within which a contractor, under
 The starting-point for the designer is normally
 a conceptual brief from the client, who may be a
 private developer or perhaps a government body.
 The conceptual brief may simply consist of some
 sketches prepared by the client or perhaps a detailed
 set of architect’s drawings. Experience is crucially
 important, and a client will always demand that
 the firm he is employing to do the design has pre
vious experience designing similar structures.
 Although imagination is thought by some to
 be entirely the domain of the architect, this is not
 so. For engineers and technicians an imagination
 of how elements of structure interrelate in three
 dimensions is essential, as is an appreciation of
 the loadings to which structures might be subject
 in certain circumstances. In addition, imaginative
 solutions to engineering problems are often required
 to save money, time, or to improve safety or quality.
 A site investigation is essential to determine the
 strength and other characteristics of the ground
 on which the structure will be founded. If the struc
ture is unusual in any way, or subject to abnormal
 loadings, model or laboratory tests may also be used
 to help determine how the structure will behave.
 In today’s economic climate a structural designer
 must be constantly aware of the cost implications
 of his or her design. On the one hand design should
 aim to achieve economy of materials in the struc
ture, but over-refinement can lead to an excessive
  number of different sizes and components in the
 structure, and labour costs will rise. In addition
 the actual cost of the designer’s time should not be
 excessive, or this will undermine the employer’s
 competitiveness. The idea is to produce a workable
 design achieving reasonable economy of materials,
 while keeping manufacturing and construction costs
 down, and avoiding unnecessary design and research
 expenditure. Attention to detailing and buildability
 of structures cannot be overemphasized in design.
 Most failures are as a result of poor detailing rather
 than incorrect analysis.
 Designers must also understand how the struc
ture will fit into the environment for which it is
 designed. Today many proposals for engineering
 structures stand or fall on this basis, so it is part of
 the designer’s job to try to anticipate and recon
cile the environmental priorities of the public and
 government.
 The engineering design process can often be
 divided into two stages: (1) a feasibility study in
volving a comparison of the alternative forms of
 structure and selection of the most suitable type and
 (2) a detailed design of the chosen structure. The
 success of stage 1, the conceptual design, relies
 to a large extent on engineering judgement and
 instinct, both of which are the outcome of many
 years’ experience of designing structures. Stage 2,
 the detailed design, also requires these attributes
 but is usually more dependent upon a thorough
 understanding of the codes of practice for struc
tural design, e.g. BS 8110 and BS 5950. These many generations of engineers, and the results of
 research. They help to ensure safety and economy
 of construction, and that mistakes are not repeated.
 For instance, after the infamous disaster at the
 Ronan Point block of flats in Newham, London,
 when a gas explosion caused a serious partial col
lapse, research work was carried out, and codes of
 practice were amended so that such structures could
 survive a gas explosion, with damage being con
f
 ined to one level.
 The aim of this book is to look at the procedures
 associated with the detailed design of structural
 elements such as beams, columns and slabs. Chap
ter 2 will help the reader to revise some basic the
ories of structural behaviour. Chapters 3–6 deal with
 design to British Standard (BS) codes of practice
 for the structural use of concrete (BS 8110), struc
tural steelwork (BS 5950), masonry (BS 5628) and
 timber (BS 5268). Chapter 7 introduces the new
 Eurocodes (EC) for structural design and Chapters
 8–11 then describe the layout and design principles
 in EC2, EC3, EC6 and EC5 for concrete, steel
work, masonry and timber respectively
 documents are based on the amassed experience of

1.2. Basis of design

Table 1.1 illustrates some risk factors that are asso
ciated with activities in which people engage. It
 can be seen that some degree of risk is associated
 with air and road travel. However, people normally
 accept that the benefits of mobility outweigh the
 risks. Staying in buildings, however, has always beenTable 1.1 Comparative death risk per 108
 persons exposed
 Mountaineering (international)
 Air travel (international)
 Deep water trawling
 2700
 120
 59
 Basis of design
 critical points, as stress due to loading exceeds the
 strength of the material. In order for the structure
 to be safe the overlapping area must be kept to a
 minimum. The degree of overlap between the two
 curves can be minimized by using one of three dis
tinct design philosophies, namely:
 Car travel
 Coal mining
 Construction sites
 Manufacturing
 Accidents at home
 Fire at home
 Structural failures
 56
 21
 8
 2
 2
 0.1
 0.002
 regarded as fairly safe. The risk of death or injury
 due to structural failure is extremely low, but as we
 spend most of our life in buildings this is perhaps
 just as well.
 As far as the design of structures for safety is
 concerned, it is seen as the process of ensuring
 that stresses due to loading at all critical points in a
 structure have a very low chance of exceeding the
 strength of materials used at these critical points.
 Figure 1.2 illustrates this in statistical terms.
 In design there exist within the structure a number
 of critical points (e.g. beam mid-spans) where the
 design process is concentrated. The normal distribu
tion curve on the left of Fig. 1.2 represents the actual
 maximum material stresses at these critical points
 due to the loading. Because loading varies according
 to occupancy and environmental conditions, and
 because design is an imperfect process, the material
 stresses will vary about a modal value – the peak of
 the curve. Similarly the normal distribution curve
 on the right represents material strengths at these
 critical points, which are also not constant due to
 the variability of manufacturing conditions.
 The overlap between the two curves represents a
 possibility that failure may take place at one of the
 Fig. 1.2 Relationship between stress and strength.
 1. permissible stress design
 2. load factor method
 3. limit state design.
 1.2.1 PERMISSIBLE STRESS DESIGN
 In permissible stress design, sometimes referred to
 as modular ratio or elastic design, the stresses in the
 structure at working loads are not allowed to exceed
 a certain proportion of the yield stress of the con
struction material, i.e. the stress levels are limited
 to the elastic range. By assuming that the stress
strain relationship over this range is linear, it is pos
sible to calculate the actual stresses in the material
 concerned. Such an approach formed the basis of the
 design methods used in CP 114 (the forerunner of
 BS 8110) and BS 449 (the forerunner of BS 5950).
 However, although it modelled real building per
formance under actual conditions, this philosophy
 had two major drawbacks. Firstly, permissible design
 methods sometimes tended to overcomplicate the
 design process and also led to conservative solutions.
 Secondly, as the quality of materials increased and
 the safety margins decreased, the assumption that
 stress and strain are directly proportional became
 unjustifiable for materials such as concrete, making
 it impossible to estimate the true factors of safety.
 1.2.2 LOAD FACTOR DESIGN
 Load factor or plastic design was developed to take
 account of the behaviour of the structure once the
 yield point of the construction material had been
 reached. This approach involved calculating the
 collapse load of the structure. The working load was
 derived by dividing the collapse load by a load factor.
 This approach simplified methods of analysis and
 allowed actual factors of safety to be calculated.
 It was in fact permitted in CP 114 and BS 449
 but was slow in gaining acceptance and was even
tually superseded by the more comprehensive limit
 state approach.
 The reader is referred to Appendix A for an ex
ample illustrating the differences between the per
missible stress and load factor approaches to design.
 1.2.3 LIMIT STATE DESIGN
 Originally formulated in the former Soviet Union
 in the 1930s and developed in Europe in the 1960s,
 5
Philosophy of design
 limit state design can perhaps be seen as a com
promise between the permissible and load factor
 methods. It is in fact a more comprehensive ap
proach which takes into account both methods in
 appropriate ways. Most modern structural codes of
 practice are now based on the limit state approach.
 BS 8110 for concrete, BS 5950 for structural
 steelwork, BS 5400 for bridges and BS 5628 for
 masonry are all limit state codes. The principal
 exceptions are the code of practice for design in
 timber, BS 5268, and the old (but still current)
 structural steelwork code, BS 449, both of which
 are permissible stress codes. It should be noted, how
ever, that the Eurocode for timber (EC5), which is
 expected to replace BS 5268 around 2010, is based
 on limit state principles.
 As limit state philosophy forms the basis of the
 design methods in most modern codes of practice
 for structural design, it is essential that the design
 methodology is fully understood. This then is the
 purpose of the following subsections.
 1.2.3.1 Ultimate and serviceability
 limit states
 The aim of limit state design is to achieve accept
able probabilities that a structure will not become
 unfit for its intended use during its design life, that
 is, the structure will not reach a limit state. There
 are many ways in which a structure could become
 unfit for use, including excessive conditions of bend
ing, shear, compression, deflection and cracking
 (Fig. 1.3). Each of these mechanisms is a limit state
 whose effect on the structure must be individually
 assessed.
 Some of the above limit states, e.g. deflection
 and cracking, principally affect the appearance of
 the structure. Others, e.g. bending, shear and com
pression, may lead to partial or complete collapse
 of the structure. Those limit states which can cause
 failure of the structure are termed ultimate limit
 states. The others are categorized as serviceability
 limit states. The ultimate limit states enable the
 designer to calculate the strength of the structure.
 Serviceability limit states model the behaviour of the
 structure at working loads. In addition, there may
 be other limit states which may adversely affect
 the performance of the structure, e.g. durability
 and fire resistance, and which must therefore also
 be considered in design.
 It is a matter of experience to be able to judge
 which limit states should be considered in the
 design of particular structures. Nevertheless, once
 this has been done, it is normal practice to base
 6
 the design on the most critical limit state and then
 check for the remaining limit states. For example,
 for reinforced concrete beams the ultimate limit
 states of bending and shear are used to size the
 beam. The design is then checked for the remain
ing limit states, e.g. deflection and cracking. On
 the other hand, the serviceability limit state of
 deflection is normally critical in the design of con
crete slabs. Again, once the designer has determined
 a suitable depth of slab, he/she must then make
 sure that the design satisfies the limit states of bend
ing, shear and cracking.
 In assessing the effect of a particular limit state
 on the structure, the designer will need to assume
 certain values for the loading on the structure and
 the strength of the materials composing the struc
ture. This requires an understanding of the con
cepts of characteristic and design values which are
 discussed below.
 1.2.3.2 Characteristic and design values
 As stated at the outset, when checking whether a
 particular member is safe, the designer cannot be
 certain about either the strength of the material
 composing the member or, indeed, the load which
 the member must carry. The material strength may
 be less than intended (a) because of its variable
 composition, and (b) because of the variability of
 manufacturing conditions during construction, and
 other effects such as corrosion. Similarly the load
 in the member may be greater than anticipated (a)
 because of the variability of the occupancy or envir
onmental loading, and (b) because of unforeseen
 circumstances which may lead to an increase in the
 general level of loading, errors in the analysis, errors
 during construction, etc.
 In each case, item (a) is allowed for by using a
 characteristic value. The characteristic strength
 is the value below which the strength lies in only a
 small number of cases. Similarly the characteristic
 load is the value above which the load lies in only
 a small percentage of cases. In the case of strength
 the characteristic value is determined from test re
sults using statistical principles, and is normally
 defined as the value below which not more than
 5% of the test results fall. However, at this stage
 there are insufficient data available to apply statist
ical principles to loads. Therefore the characteristic
 loads are normally taken to be the design loads
 from other codes of practice, e.g. BS 648 and BS
 6399.
 The overall effect of items under (b) is allowed
 for using a partial safety factor: γm
 for strength
Basis of design
 Fig. 1.3 Typical modes of failure for beams and columns.
 7
Philosophy of design
 and γf
 for load. The design strength is obtained by
 dividing the characteristic strength by the partial
 Design strength  
safety factor for strength:
 characteristic strength
 m
 =
 γ
 (1.1)
 The design load is obtained by multiplying the
 characteristic load by the partial safety factor for
 load:
 Design load = characteristic load × γf 
(1.2)
 The value of γm
 will depend upon the properties
 of the actual construction material being used.
 Values for γf
 depend on other factors which will be
 discussed more fully in Chapter 2.
 In general, once a preliminary assessment of the
 design loads has been made it is then possible to
 calculate the maximum bending moments, shear
 forces and deflections in the structure (Chapter 2).
 The construction material must be capable of
 withstanding these forces otherwise failure of the
 structure may occur, i.e.
 Design strength ≥ design load
 (1.3)
 Simplified procedures for calculating the moment,
 shear and axial load capacities of structural ele
ments together with acceptable deflection limits
 are described in the appropriate codes of practice.
 Questions
 1. Explain the difference between conceptual
 design and detailed design.
 2. What is a code of practice and what is its
 purpose in structural design?
 3. List the principal sources of uncertainty in
 structural design and discuss how these
 uncertainties are rationally allowed for in
 design.
 These allow the designer to rapidly assess the suit
ability of the proposed design. However, before
 discussing these procedures in detail, Chapter 2
 describes in general terms how the design loads
 acting on the structure are estimated and used to
 size individual elements of the structure.
 1.3 Summary
 This chapter has examined the bases of three
 philosophies of structural design: permissible stress,
 load factor and limit state. The chapter has con
centrated on limit state design since it forms the
 basis of the design methods given in the codes of
 practice for concrete (BS 8110), structural steel
work (BS 5950) and masonry (BS 5628). The aim
 of limit state design is to ensure that a structure
 will not become unfit for its intended use, that is,
 it will not reach a limit state during its design life.
 Two categories of limit states are examined in
 design: ultimate and serviceability. The former is
 concerned with overall stability and determining
 the collapse load of the structure; the latter exam
ines its behaviour under working loads. Structural
 design principally involves ensuring that the loads
 acting on the structure do not exceed its strength
 and the first step in the design process then is to
 estimate the loads acting on the structure.
 4. The characteristic strengths and design
 strengths are related via the partial safety
 factor for materials. The partial safety
 factor for concrete is higher than for steel
 reinforcement. Discuss why this should be so.
 5. Describe in general terms the ways in
 which a beam and column could become unfit for use.

2. Basic structural concepts and material properties

This chapter is concerned with general methods of
 sizing beams and columns in structures. The chapter
 describes how the characteristic and design loads acting
 on structures and on the individual elements are deter
mined. Methods of calculating the bending moments,
 shear forces and deflections in beams are outlined.
 Finally, the chapter describes general approaches to
 sizing beams according to elastic and plastic criteria
 and sizing columns subject to axial loading.

2.1. Introduction

 All structures are composed of a number of inter
connected elements such as slabs, beams, columns,
 walls and foundations. Collectively, they enable the
 internal and external loads acting on the structure
 to be safely transmitted down to the ground. The
 actual way that this is achieved is difficult to model
 and many simplifying, but conservative, assump
tions have to be made. For example, the degree
 of fixity at column and beam ends is usually uncer
tain but, nevertheless, must be estimated as it
 significantly affects the internal forces in the element.
 Furthermore, it is usually assumed that the reaction
 from one element is a load on the next and that
 the sequence of load transfer between elements
 occurs in the order: ceiling/floor loads to beams to
 columns to foundations to ground (Fig. 2.1).
 Chapter 2
 Basic structural
 concepts and
 material properties
 Fig. 2.1 Sequence of load transfer between elements of a
 structure.
 compressive loading. These steps are summarized
 in Fig. 2.2 and the following sections describe the
 procedures associated with each step.
 2.2 Design loads acting on
 structures
 At the outset, the designer must make an assess
ment of the future likely level of loading, including
 self-weight, to which the structure may be subject
 during its design life. Using computer methods or
 hand calculations the design loads acting on indi
vidual elements can then be evaluated. The design
 loads are used to calculate the bending moments,
 shear forces and deflections at critical points along
 the elements. Finally, suitable dimensions for the
 element can be determined. This aspect requires
 an understanding of the elementary theory of
 bending and the behaviour of elements subject to
 The loads acting on a structure are divided into
 three basic types: dead, imposed and wind. For
 each type of loading there will be characteristic and
 design values, as discussed in Chapter 1, which must
 be estimated. In addition, the designer will have to
 determine the particular combination of loading
 which is likely to produce the most adverse effect
 on the structure in terms of bending moments,
 shear forces and deflections.
 2.2.1 DEAD LOADS, Gk
 , gk
 Dead loads are all the permanent loads acting on
 the structure including self-weight, finishes, fixtures
 and partitions. The characteristic dead loads can be
 9
Basic structural concepts and material properties
 Fig. 2.2 Design process.
 Example 2.1 Self-weight of a reinforced concrete beam
 Calculate the self-weight of a reinforced concrete beam of breadth 300 mm, depth 600 mm and length 6000 mm.
 From Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming that the gravitational constant is
 10 m s−2 (strictly 9.807 m s−2), the unit weight of reinforced concrete, ρ, is
 ρ = 2400 × 10 = 24 000 N m−3 = 24 kN m−3
 Hence, the self-weight of beam, SW, is
 SW = volume × unit weight
 = (0.3 × 0.6 × 6)24 = 25.92 kN
 estimated using the schedule of weights of building
 materials given in BS 648 (Table 2.1) or from manu
facturers’ literature. The symbols Gk
 and gk
 are
 normally used to denote the total and uniformly
 distributed characteristic dead loads respectively.
 Estimation of the self-weight of an element tends
 to be a cyclic process since its value can only be
 assessed once the element has been designed which
 requires prior knowledge of the self-weight of the
 element. Generally, the self-weight of the element
 is likely to be small in comparison with other dead
 and live loads and any error in estimation will tend
 to have a minimal effect on the overall design
 (Example 2.1).
 2.2.2 IMPOSED LOADS Qk
 , qk
 Imposed load, sometimes also referred to as live
 load, represents the load due to the proposed oc
cupancy and includes the weights of the occupants,
 furniture and roof loads including snow. Since
 imposed loads tend to be much more variable
 than dead loads they are more difficult to predict.
 10
 BS 6399: Part 1: 1984: Code of Practice for Dead and
 Imposed Loads gives typical characteristic imposed
 f
 loor loads for different classes of structure, e.g.
 residential dwellings, educational institutions,
 hospitals, and parts of the same structure, e.g.
 balconies, corridors and toilet rooms (Table 2.2).
 2.2.3 WIND LOADS
 Wind pressure can either add to the other gravita
tional forces acting on the structure or, equally
 well, exert suction or negative pressures on the
 structure. Under particular situations, the latter may
 well lead to critical conditions and must be con
sidered in design. The characteristic wind loads
 acting on a structure can be assessed in accordance
 with the recommendations given in CP 3: Chapter
 V: Part 2: 1972 Wind Loads or Part 2 of BS 6399:
 Code of Practice for Wind Loads.
 Wind loading is important in the design of ma
sonry panel walls (Chapter 5). However beyond that,
 wind loading is not considered further since the em
phasis in this book is on the design of elements rather
11
 Table 2.1 Schedule of unit masses of building materials (based on BS 648)
 Asphalt
 Roofing 2 layers, 19 mm thick 42 kg m−2
 Damp-proofing, 19 mm thick 41 kg m−2
 Roads and footpaths, 19 mm thick 44 kg m−2
 Bitumen roofing felts
 Mineral surfaced bitumen 3.5 kg m−2
 Blockwork
 Solid per 25 mm thick, stone 55 kg m−2
 aggregate
 Aerated per 25 mm thick 15 kg m−2
 Board
 Blockboard per 25 mm thick 12.5 kg m−2
 Brickwork
 Clay, solid per 25 mm thick 55 kg m−2
 medium density
 Concrete, solid per 25 mm thick 59 kg m−2
 Cast stone 2250 kg m−3
 Concrete
 Natural aggregates 2400 kg m−3
 Lightweight aggregates (structural) 1760 + 240/
 −160 kg m−3
 Flagstones
 Concrete, 50 mm thick 120 kg m−2
 Glass fibre
 Slab, per 25 mm thick 2.0–5.0 kg m−2
 Gypsum panels and partitions
 Building panels 75 mm thick 44 kg m−2
 Lead
 Sheet, 2.5 mm thick 30 kg m−2
 Linoleum
 3 mm thick 6 kg m−2
 loads, γf
 (Chapter 1). The value for γf
 depends on
 several factors including the limit state under
 consideration, i.e. ultimate or serviceability, the
 accuracy of predicting the load and the particu
lar combination of loading which will produce the
 worst possible effect on the structure in terms of
 bending moments, shear forces and deflections.
 than structures, which generally involves investigat
ing the effects of dead and imposed loads only.
 2.2.4 LOAD COMBINATIONS AND
 DESIGN LOADS
 The design loads are obtained by multiplying the
 characteristic loads by the partial safety factor for
 Plaster
 Two coats gypsum, 13 mm thick 22 kg m−2
 Plastics sheeting (corrugated) 4.5 kg m−2
 Plywood
 per mm thick 0.7 kg m−2
 Reinforced concrete 2400 kg m−3
 Rendering
 Cement: sand (1:3), 13 mm thick 30 kg m−2
 Screeding
 Cement: sand (1:3), 13 mm thick 30 kg m−2
 Slate tiles
 (depending upon thickness 24–78 kg m−3
 and source)
 Steel
 Solid (mild) 7850 kg m−3
 Corrugated roofing sheets, 10 kg m−2
 per mm thick
 Tarmacadam
 25 mm thick 60 kg m−2
 Terrazzo
 25 mm thick 54 kg m−2
 Tiling, roof
 Clay 70 kg m−2
 Timber
 Softwood 590 kg m−3
 Hardwood 1250 kg m−3
 Water 1000 kg m−3
 Woodwool
 Slabs, 25 mm thick 15 kg m−2
 Design loads acting on structures
Basic structural concepts and material properties
 12
 Table 2.2 Imposed loads for residential occupancy class
 Floor area usage Intensity of distributed load Concentrated load
 kN m−2 kN
 Type 1. Self-contained dwelling units
 All 1.5 1.4
 Type 2. Apartment houses, boarding houses, lodging
 houses, guest houses, hostels, residential clubs and
 communal areas in blocks of flats
 Boiler rooms, motor rooms, fan rooms and the like 7.5 4.5
 including the weight of machinery
 Communal kitchens, laundries 3.0 4.5
 Dining rooms, lounges, billiard rooms 2.0 2.7
 Toilet rooms 2.0
Bedrooms, dormitories 1.5 1.8
 Corridors, hallways, stairs, landings, footbridges, etc. 3.0 4.5
 Balconies Same as rooms to which 1.5 per metre run
 they give access but with concentrated at
 a minimum of 3.0 the outer edge
 Cat walks–1.0 at 1 m centres
 Type 3. Hotels and motels
 Boiler rooms, motor rooms, fan rooms and the like, 7.5 4.5
 including the weight of machinery
 Assembly areas without fixed seating, dance halls 5.0 3.6
 Bars 5.0
Assembly areas with fixed seatinga 4.0
Corridors, hallways, stairs, landings, footbridges, etc. 4.0 4.5
 Kitchens, laundries 3.0 4.5
 Dining rooms, lounges, billiard rooms 2.0 2.7
 Bedrooms 2.0 1.8
 Toilet rooms 2.0
Balconies Same as rooms to which 1.5 per metre run
 they give access but with concentrated at the
 a minimum of 4.0 outer edge
 Cat walks–1.0 at 1 m centres
 Note. a Fixed seating is seating where its removal and the use of the space for other purposes are improbable.
 In most of the simple structures which will be
 considered in this book, the worst possible com
bination will arise due to the maximum dead and
 maximum imposed loads acting on the structure
 together. In such cases, the partial safety factorsfor
 dead and imposed loads are 1.4 and 1.6 respect
ively (Fig. 2.3) and hence the design load is givenby
 Fig. 2.3
 Design load = 1.4Gk
 + 1.6Qk
 However, it should be appreciated that theoret
ically the design dead loads can vary between the
 characteristic and ultimate values, i.e. 1.0Gk
 and
 1.4Gk
 . Similarly, the design imposed loads can
 vary between zero and the ultimate value, i.e. 0.0Qk
 and 1.6Qk
 . Thus for a simply supported beam with
 an overhang (Fig. 2.4(a)) the load cases shown in
 Figs 2.4(b)–(d) will need to be considered in order
 to determine the design bending moments and shear
 forces in the beam

2.2. Design loads acting on elements

Once the design loads acting on the structure have
 been estimated it is then possible to calculate the
 design loads acting on individual elements. As was
 pointed out at the beginning of this chapter, this
 usually requires the designer to make assumptions
 regarding the support conditions and how the loads
 will eventually be transmitted down to the ground.
 Figures 2.5(a) and (b) illustrate some of the more
 Fig. 2.5 Typical beams and column support conditions.
 In design it is common to assume that all the joints
 in the structure are pinned and that the sequence of
 load transfer occurs in the order: ceiling/floor loads to
 beams to columns to foundations to ground. These
 assumptions will considerably simplify calculations
 and lead to conservative estimates of the design
 loads acting on individual elements of the struc
ture. The actual calculations to determine the forces
 acting on the elements are best illustrated by a
 number of worked examples as follows.
 13
Basic structural concepts and material properties
 Example 2.2 Design loads on a floor beam
 A composite floor consisting of a 150 mm thick reinforced concrete slab supported on steel beams spanning 5 m and
 spaced at 3 m centres is to be designed to carry an imposed load of 3.5 kN m−2. Assuming that the unit mass of the
 steel beams is 50 kg m−1 run, calculate the design loads on a typical internal beam.
 UNIT WEIGHTS OF MATERIALS
 Reinforced concrete
 From Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming the gravitational constant is 10 m s−2, the
 unit weight of reinforced concrete is
 2400 × 10 = 24 000 N m−3 = 24 kN m−3
 Steel beams
 Unit mass of beam = 50 kg m−1 run
 Unit weight of beam = 50 × 10 = 500 N m−1 run = 0.5 kN m−1 run
 LOADING
 Slab
 Slab dead load (gk
 )
 = self-weight = 0.15 × 24 = 3.6 kN m−2
 Slab imposed load (qk
 ) = 3.5 kN m−2
 Slab ultimate load
 = 1.4gk
 + 1.6qk
 = 1.4 × 3.6 + 1.6 × 3.5
 = 10.64 kN m−2
 Beam
 Beam dead load (gk
 ) = self-weight = 0.5 kN m−1 run
 Beam ultimate load = 1.4gk
 DESIGN LOAD
 = 1.4 × 0.5 = 0.7 kN m−1 run
 Each internal beam supports a uniformly distributed load from a 3 m width of slab (hatched \\\\\\\) plus self-weight.
 Hence
 Design load on beam = slab load + self-weight of beam
 = 10.64 × 5 × 3 + 0.7 × 5
 = 159.6 + 3.5 = 163.1 kN
 14
Design loads acting on elements
 Example 2.3 Design loads on floor beams and columns
 The floor shown below with an overall depth of 225 mm is to be designed to carry an imposed load of 3 kN m−2 plus
 f
 loor finishes and ceiling loads of 1 kN m−2. Calculate the design loads acting on beams B1–C1, B2–C2 and B1–B3 and
 columns B1 and C1. Assume that all the column heights are 3 m and that the beam and column weights are 70 and
 60 kg m−1 run respectively.
 UNIT WEIGHTS OF MATERIALS
 Reinforced concrete
 From Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming the gravitational constant is 10 m s−2, the
 unit weight of reinforced concrete is
 2400 × 10 = 24 000 N m−3 = 24 kN m−3
 Steel beams
 Unit mass of beam = 70 kg m−1 run
 Unit weight of beam = 70 × 10 = 700 N m−1 run = 0.7 kN m−1 run
 Steel columns
 Unit mass of column = 60 kg m−1 run
 Unit weight of column = 60 × 10 = 600 N m−1 run = 0.6 kN m−1 run
 LOADING
 Slab
 Slab dead load (gk
 )
 = self-weight + finishes
 = 0.225 × 24 + 1 = 6.4 kN m−2
 Slab imposed load (qk
 ) = 3 kN m−2
 Slab ultimate load
 Beam
 = 1.4gk
 + 1.6qk
 = 1.4 × 6.4 + 1.6 × 3 = 13.76 kN m−2
 Beam dead load (gk
 ) = self-weight = 0.7 kN m−1 run
 Beam ultimate load = 1.4gk
 = 1.4 × 0.7 = 0.98 kN m−1 run
 Column
 Column dead load (gk
 ) = 0.6 kN m−1 run
 Column ultimate load = 1.4gk
 = 1.4 × 0.6 = 0.84 kN m−1 run
 15
Basic structural concepts and material properties
 Example 2.3 continued
 DESIGN LOADS
 Beam B1–C1
 Assuming that the slab is simply supported, beam B1–C1 supports a uniformly distributed load from a 1.5 m width of
 slab (hatched //////) plus self-weight of beam. Hence
 Design load on beam B1–C1 = slab load + self-weight of beam
 = 13.76 × 6 × 1.5 + 0.98 × 6
 = 123.84 + 5.88 = 129.72 kN
 Since the beam is symmetrically loaded,
 RB1
 = RC1
 = 129.72/2 = 64.86 kN
 Beam B2–C2
 Assuming that the slab is simply supported, beam B2–C2 supports a uniformly distributed load from a 3 m width of
 slab (hatched \\\\\) plus its self-weight. Hence
 Design load on beam B2–C2 = slab load + self-weight of beam
 = 13.76 × 6 × 3 + 0.98 × 6
 = 247.68 + 5.88 = 253.56 kN
 Since the beam is symmetrically loaded, RB2
 and RC2
 are the same and equal to 253.56/2 = 126.78 kN.
 Beam B1–B3
 Assuming that the slab is simply supported, beam B1–B3 supports a uniformly distributed load from a 1.5 m width of
 slab (shown cross-hatched) plus the self-weight of the beam and the reaction transmitted from beam B2–C2 which
 acts as a point load at mid-span. Hence
 Design load on beam B1–B3 = uniformly distributed load from slab plus self-weight of beam
 + point load from reaction RB2
 = (13.76 × 1.5 × 6 + 0.98 × 6) + 126.78
 = 129.72 + 126.78 = 256.5 kN
 16
Example 2.3 continued
 Since the beam is symmetrically loaded,
 Column B1
 RB1
 = RB3
 = 256.5/2 = 128.25 kN
 Structural analysis
 Column B1 supports the reactions from beams A1–B1, B1–C1 and B1–B3 and its self-weight. From the above, the
 reaction at B1 due to beam B1–C1 is 64.86 kN and from beam B1–B3 is 128.25 kN. Beam A1–B1 supports only its
 self-weight = 0.98 × 3 = 2.94 kN. Hence reaction at B1 due to A1–B1 is 2.94/2 = 1.47 kN. Since the column height is
 3 m, self-weight of column = 0.84 × 3 = 2.52 kN. Hence
 Design load on column B1 = 64.86 + 128.25 + 1.47 + 2.52
 = 197.1 kN
 Column C1
 Column C1 supports the reactions from beams B1–C1 and C1–C3 and its self-weight. From the above, the reaction at
 C1 due to beam B1–C1 is 64.86 kN. Beam C1–C3 supports the reactions from B2–C2 (= 126.78 kN) and its self
weight (= 0.98 × 6) = 5.88 kN. Hence the reaction at C1 is (126.78 + 5.88)/2 = 66.33 kN. Since the column height
 is 3 m, self-weight = 0.84 × 3 = 2.52 kN. Hence
 Design load on column C1 = 64.86 + 66.33 + 2.52 = 133.71 kN

2.3. Structural analysis

 The design axial loads can be used directly to size col
umns. Column design will be discussed more fully
 in section 2.5. However, before flexural members
 such as beams can be sized, the design bending
 moments and shear forces must be evaluated.
 Such calculations can be performed by a variety of
 methods as noted below, depending upon the com
plexity of the loading and support conditions:
 1. equilibrium equations
 2. formulae
 3. computer methods.
 Hand calculations are suitable for analysing
 statically determinate structures such as simply sup
ported beams and slabs (section 2.4.1). For various
 standard load cases, formulae for calculating the
 maximum bending moments, shear forces and
 deflections are available which can be used to rapidly
 17
Basic structural concepts and material properties
 analyse beams, as will be discussed in section 2.4.2.
 Alternatively, the designer may resort to using vari
ous commercially available computer packages, e.g.
 SAND. Their use is not considered in this book.
 2.4.1 EQUILIBRIUM EQUATIONS
 It can be demonstrated that if a body is in equilib
rium under the action of a system of external forces,
 all parts of the body must also be in equilibrium.
 This principle can be used to determine the bend
ing moments and shear forces along a beam. The
 actual procedure simply involves making fictitious
 ‘cuts’ at intervals along the beam and applying the
 equilibrium equations given below to the cut por
tions of the beam.
 Σ moments (M) = 0
 Σ vertical forces (V ) = 0
 Example 2.4 Design moments and shear forces in beams using
 equilibrium equations
 Calculate the design bending moments and shear forces in beams B2–C2 and B1–B3 of Example 2.3.
 BEAM B2–C2
 Let the longitudinal centroidal axis of the beam be the x axis and x = 0 at support B2.
 x = 0
 By inspection,
 x = 1
 Moment at x = 0 (Mx=0
 ) = 0
 Shear force at x = 0 (Vx=0
 ) = RB2
 = 126.78 kN
 (2.1)
 (2.2)
 Assuming that the beam is cut 1 m from support B2, i.e. x = 1 m, the moments and shear forces acting on the cut
 portion of the beam will be those shown in the free body diagram below:
 From equation 2.1, taking moments about Z gives
 126.78 × 1 − 42.26 × 1 × 0.5 − Mx=1
 = 0
 Hence
 Mx=1
 = 105.65 kN m
 From equation 2.2, summing the vertical forces gives
 126.78 − 42.26 × 1 − Vx=1
 = 0
 18
Example 2.4 continued
 Hence
 x = 2
 Vx=1
 = 84.52 kN
 Structural analysis
 The free body diagram for the beam, assuming that it has been cut 2 m from support B2, is shown below:
 From equation 2.1, taking moments about Z gives
 126.78 × 2 − 42.26 × 2 × 1 − Mx=2
 = 0
 Hence
 Mx=2
 = 169.04 kN m
 From equation 2.2, summing the vertical forces gives
 126.78 − 42.26 × 2 − Vx=2
 = 0
 Hence
 Vx=2
 = 42.26 kN
 If this process is repeated for values of x equal to 3, 4, 5 and 6 m, the following values of the moments and shear
 forces in the beam will result:
 x(
 m) 0 1 2 3 4 5 6
 M(kN m)
 V(kN)
 0
 126.78
 105.65
 84.52
 169.04
 42.26
 190.17
 0
 169.04
 −42.26
 105.65
 −84.52
 0
 −126.78
 This information is better presented diagrammatically as shown below:
 19
Basic structural concepts and material properties
 Example 2.4 continued
 Hence, the design moment (M) is 190.17 kN m. Note that this occurs at mid-span and coincides with the point of
 zero shear force. The design shear force (V ) is 126.78 kN and occurs at the supports.
 BEAM B1–B3
 Again, let the longitudinal centroidal axis of the beam be the x axis of the beam and set x = 0 at support B1.
 The steps outlined earlier can be used to determine the bending moments and shear forces at x = 0, 1 and 2 m and
 since the beam is symmetrically loaded and supported, these values of bending moment and shear forces will apply
 at x = 4, 5 and 6 m respectively.
 The bending moment at x = 3 m can be calculated by considering all the loading immediately to the left of the
 point load as shown below:
 From equation 2.1, taking moments about Z gives
 128.25 × 3 − 64.86 × 3/2 − Mx=3
 = 0
 Hence
 Mx=3
 = 287.46 kN m
 In order to determine the shear force at x = 3 the following two load cases need to be considered:
 In (a) it is assumed that the beam is cut immediately to the left of the point load. In (b) the beam is cut immediately
 to the right of the point load. In (a), from equation 2.2, the shear force to the left of the cut, Vx=3,L
 , is given by
 128.25 − 64.86 − Vx=3,L
 = 0
 Hence
 20
 Vx=3,L
 = 63.39 kN
Example 2.4 continued
 In (b), from equation 2.2, the shear force to the right of the cut, Vx=3,R
 , is given by
 128.25 − 64.86 − 126.78 − Vx=3,R
 = 0
 Hence
 Vx=3,R
 = −63.39 kN
 Summarising the results in tabular and graphical form gives
 Structural analysis
 x(m)
 0
 1
 2
 3
 4
 5
 6
 M(kN m)
 V(kN)
 0
 128.25
 117.44
 106.63
 213.26
 85.01
 287.46
 63.39|−63.39
 213.26
 −85.01
 117.44
 −106.63
 0
 −128.25
 Hence, the design moment for beam B1–B3 is 287.46 kN m and occurs at mid-span and the design shear force is
 128.25 kN and occurs at the supports.
 2.4.2 FORMULAE
 An alternative method of determining the design
 bending moments and shear forces in beams in
volves using the formulae quoted in Table 2.3. The
 table also includes formulae for calculating the
 maximum deflections for each load case. The for
mulae can be derived using a variety of methods
 for analysing statically indeterminate structures, e.g.
 slope deflection and virtual work, and the inter
ested reader is referred to any standard work on
 this subject for background information. There will
 be many instances in practical design where the
 use of standard formulae is not convenient and
 equilibrium methods are preferable.
 21
Basic structural concepts and material properties
 22
 Table 2.3 Bending moments, shear forces and deflections for
 various standard load cases
 Loading Maximum Maximum Maximum
 bending shearing deflection
 moment force
    
WL
 EI
 3
 8    
WL
 2 W
    
W
 2    
WL
 EI
 3
 48    
WL
 4
    
23
 1296
 3 WL
 EI    
WL
 6    
W
 2
    
11
 768
 3 WL
 EI    
W
 2    
WL
 8
    
5
 384
 3 WL
 EI    
W
 2    
WL
 8
 WL W
    
WL
 EI
 3
 3
    
WL
 EI
 3
 60    
W
 2    
WL
 6
    
WL
 EI
 3
 192
 (at supports
 and at midspan)
    
WL
 8    
W
 2
    
WL
 12    
W
 2    
WL
 EI
 3
 384
 at supports
    
WL
 24
 at midspan
23
 Structural analysis
 This load case can be solved using the principle of superposition which can be stated in general terms as follows: ‘The
 effect of several actions taking place simultaneously can be reproduced exactly by adding the effects of each case
 separately.’ Thus, the loading on beam B1–B3 can be considered to be the sum of a uniformly distributed load (Wudl
 )
 of 129.72 kN and a point load (Wpl
 ) at mid-span of 126.78 kN.
 By inspection, maximum moment and maximum shear force occur at beam mid-span and supports respectively. From
 Table 2.3, design moment, M, is given by
    
M Wl    .  = = ×
 8
 25356 6
 8 = 190.17 kN m
 and design shear force, V, is given by
    
V W    .  . = = = 2
 25356
 2 12678kN
 BEAM B1–B3
 By inspection, the maximum bending moment and shear force for both load cases occur at beam mid-span and
 supports respectively. Thus, the design moment, M, is given by
    
M W l Wl      .    .    . = + = × + × = udl pl kN m 8 4
 12972 6
 8
 12678 6
 4 28746
 and the design shear force, V, is given by
    
V W W      .  .  . = + = + = udl pl kN 2 2
 12972
 2
 12678
 2 1282

3. Design in reinforced concrete to BS 8110

This chapter is concerned with the detailed design of
 reinforced concrete elements to British Standard 8110.
 A general discussion of the different types of commonly
 occurring beams, slabs, walls, foundations and col
umns is given together with a number of fully worked
 examples covering the design of the following elements:
 singly and doubly reinforced beams, continuous beams,
 one-way and two-way spanning solid slabs, pad founda
tion, cantilever retaining wall and short braced columns
 supporting axial loads and uni-axial or bi-axial bend
ing. The section which deals with singly reinforced beams
 is, perhaps, the most important since it introduces the
 design procedures and equations which are common to
 the design of the other elements mentioned above, with
 the possible exception of columns

3.1. Introduction

 Reinforced concrete is one of the principal materials
 used in structural design. It is a composite material,
 consisting of steel reinforcing bars embedded in
 concrete. These two materials have complementary
 properties. Concrete, on the one hand, has high
 compressive strength but low tensile strength. Steel
 bars, on the other, can resist high tensile stresses
 but will buckle when subjected to comparatively
 low compressive stresses. Steel is much more
 expensive than concrete. By providing steel bars
 predominantly in those zones within a concrete
 member which will be subjected to tensile stresses,
 an economical structural material can be produced
 which is both strong in compression and strong
 in tension. In addition, the concrete provides cor
rosion protection and fire resistance to the more
 vulnerable embedded steel reinforcing bars.
 Reinforced concrete is used in many civil
 engineering applications such as the construction
 of structural frames, foundations, retaining walls,
 water retaining structures, highways and bridges.
 They are normally designed in accordance withthe recommendations given in various documents
 including BS 5400: Part 4: Code of practice for
 design of concrete bridges, BS 8007: Code of prac
tice for the design of concrete structures for retaining
 aqueous liquids and BS 8110: Structural use of con
crete. Since the primary aim of this book is to give
 guidance on the design of structural elements, this
 is best illustrated by considering the contents of
 BS 8110.
 BS 8110 is divided into the following three parts:
 Part 1: Code of practice for design and construction.
 Part 2: Code of practice for special circumstances.
 Part 3: Design charts for singly reinforced beams, doubly
 reinforced beams and rectangular columns.
 3.1 Introduction
 Reinforced concrete is one of the principal materials
 used in structural design. It is a composite material,
 consisting of steel reinforcing bars embedded in
 concrete. These two materials have complementary
 properties. Concrete, on the one hand, has high
 compressive strength but low tensile strength. Steel
 bars, on the other, can resist high tensile stresses
 but will buckle when subjected to comparatively
 low compressive stresses. Steel is much more
 expensive than concrete. By providing steel bars
 predominantly in those zones within a concrete
 member which will be subjected to tensile stresses,
 an economical structural material can be produced
 which is both strong in compression and strong
 in tension. In addition, the concrete provides cor
rosion protection and fire resistance to the more
 vulnerable embedded steel reinforcing bars.
 Reinforced concrete is used in many civil
 engineering applications such as the construction
 of structural frames, foundations, retaining walls,
 water retaining structures, highways and bridges.
 They are normally designed in accordance with
 Part 1 covers most of the material required for
 everyday design. Since most of this chapter is
 concerned with the contents of Part 1, it should
 be assumed that all references to BS 8110 refer to
 Part 1 exclusively. Part 2 covers subjects such as
 torsional resistance, calculation of deflections and
 estimation of crack widths. These aspects of design
 are beyond the scope of this book and Part 2, there
fore, is not discussed here. Part 3 of BS 8110 con
tains charts for use in the design of singly reinforced
 beams, doubly reinforced beams and rectangular
 columns. A number of design examples illustrating
 the use of these charts are included in the relevant
 sections of this chapter

3.2. Objectives and scope

 All reinforced concrete building structures are
 composed of various categories of elements includ
ing slabs, beams, columns, walls and foundations
 (Fig. 3.1). Within each category is a range of ele
ment types. The aim of this chapter is to describe
 the element types and, for selected elements, to
 give guidance on their design. A great deal of emphasis has been placed in the
 text to highlight the similarities in structural beha
viour and, hence, design of the various categoriesof
 elements. Thus, certain slabs can be regarded for
 design purposes as a series of transversely connected
 beams. Columns may support slabs and beams
 but columns may also be supported by (ground
 bearing) slabs and beams, in which case the latter
 are more commonly referred to as foundations.
 Cantilever retaining walls are usually designed as
 if they consist of three cantilever beams as shown
 in Fig. 3.2. Columns are different in that they are
 primarily compression members rather than beams
 and slabs which predominantly resist bending.
 Therefore columns are dealt with separately at the
 end of the chapter.
 Irrespective of the element being designed, the
 designer will need a basic understanding of the fol
lowing aspects which are discussed next:
 1. symbols
 2. basis of design

3.material properties
 4. loading
 5. stress–strain relationships
 6. durability and fire resistance.
 The detailed design of beams, slabs, foundations,
 retaining walls and columns will be discussed in
 sections 3.9, 3.10, 3.11, 3.12 and 3.13,respectively

3.3. Symbols

 For the purpose of this book, the following sym
bols have been used. These have largely been taken
 from BS 8110. Note that in one or two cases the
 same symbol is differently defined. Where this
 occurs the reader should use the definition most
 appropriate to the element being designed.
 Geometric properties:
 b width of section
 d effective depth of the tensionreinforcement
 h overall depth of section
 x depth to neutral axis
 z lever arm
 d′ depth to the compression reinforcement
 b effective span
 c nominal cover to reinforcement
 Bending:
 Fk
 characteristic load
 gk
 , Gk
 characteristic dead load
 qk
 , Qk
 characteristic imposed load
 wk
 , Wk
 characteristic wind load
 fk
 characteristic strength
 fcu
 characteristic compressive cube strength
 of concrete
 fy
 characteristic tensile strength of
 reinforcement
 γf
 partial safety factor for load
 γm
 partial safety factor for material
 strengths
 K coefficient given by M/fcu
 bd2
 K′ coefficient given by Mu
 /fcu
 bd2=0.156 when
 redistribution does not exceed 10 per cent
 M design ultimate moment
 Mu
 design ultimate moment of resistance
 As
 area of tension reinforcement
 As
 ′ area of compression reinforcement
 Φ diameter of main steel
 Φ′ diameter of links
 Shear:
 fyv
 characteristic strength of links
 sv
 spacing of links along the member

V
 design shear force due to ultimate
 loads

 v
 vc
 Asv
 design shear stress
 design concrete shear stress
 total cross-sectional area of shear
 reinforcement
 Compression:
 b
 h
 bo
 be
 bex
 bey
 N
 Ac
 Asc
 width of column
 depth of column
 clear height between end restraints
 effective height
 effective height in respect of x-x axis
 effective height in respect of y-y axis
 design ultimate axial load
 net cross-sectional area of concrete in
 a column
 area of longitudinal reinforcement