Reinforced Concrete Design To Eurocode 2 4e5ge

then

.f.:1 200

X

10

6X

15

X

10'

3.0N/mm 2

When cracking occurs. the uncracked lengths of concrete try to contract so thnt the embedded steel between cracks is in compression while the steel across the cracks is in tension. This feature is accompanied by localised bond breakdown, adjacent to each crnck. The equilibrium of the concrete and reinforcement is shown in figure 1.8 and calculations may be developed to relate crack widths and spacings to properties of the cross-section; this is examined in more detail in chapter 6, which deals with serviceabi lily requirements. Figure 1.8

Shrinkage forces adjacent toil crack

Thermal movement As the coefficients of thermal expansion of steel and concrete (ar. s and o:,., c) are similar. differential movement between the steel and concrete will only be very small and is unlikely to cause cracking. The dif ferential thermal strain due to a temperature change T may be calculated as

T(a-r.c- Ct,-.,) and should be added to the shrinkage strain Ecs if significant.

.~~_}~ Reinforced concrete design The overall thermal conu·action of concrete is. however, frequen tly effective in producing the first crack in a restrained member. since the required temperature changes could easily occur overnight in a newly cast member. even with good control of the heat generated during the hydration processes.

(

EXAMPLE 1.3 Thermal shrinkage

Find the faJJ in temperature required to cause cracking in a restrained member if ultimate tensile s trength of the concrete .f~t. cn - 2 N/mrn 2• 6'c111 = 16 kN/m m2 and 6 !'l:T, c = OT,, = !0 X 10- per °C. Ultimate tensile strain of concrete Cull

= I 6 X2 .103 =

125

X

10

6

Minimum lcmpcrature tlrop to cause cracking

It should be notctlthal full restraint, as assumed in this example, is unlikely to occur in practice: thus the temperature change required 10 cause cracking is increased. A maximum 'restraint factor' of 0.5 is nflen used. with lower va lue~ where external restrnint is likely to be smal l. The temperature drop required would then be given by the theoretical minimum divided by the ·reS!I'aint factor'. i.e. 12.5/0.5 = 25
1.4

Creep

Creep is the continuous clet'ormntion of a member under sustained loutl. Tt is " phenomenon associated with many materia ls, but it is particularly evident with concrete. The precise behaviour of a pnrticular concrete depends on t·he aggregates and the mix design as wel l as the ambient humidiry, member cross-section, nnd nge at (irst loading, hut the general pattern is illustrated by considering :1 member subjected to axial compression . For such a me1nber, a typical variation of deformation with rjme is shown by the curve in fi gure 1.9. The characteristics of creep are Short·IM>l t•anlc

1. The fi nal deformation of the member can he three to four times the shorHerm elastic deformation. 2. The deformation is roughly proportional to the intensity of loading and to the inverse of the concrete strength.

Figure 1.9 Typical increase of deformation with time for concrete

3. If the load is removed. only the instantaneous elastic deformation will recover- the plastic deformation will nor. 4. There is a redistribution or load between the concrete and any steel present.

Properties of reinforced concrete The redistribution of load is caused by the changes in compressive strains being transferred to the reinforcing steel. Thus th e compressive stresses i n the steel are increased so that the steel takes a larger propo1tion of the load. The effects of creep are particularly important in beams, where the increased deflections may cause tJ1e opening of cracks, damage to finishes. and the non-alignment of mechanical equipment. Redistribution of stress between concrete and steel occurs primarily in the uncracked compressive areas and has little effect on the tension reinforcement other than reducing shrinkage stresses in some instances. T he provision of reinfo rcement in the compressive zone of a nexural member. however. often helps to restrain the de flecti on~ due to creep.

1.5

Durability

Concreto Htructuros, properl y designed and constructed, 11re long lasting und should require little maintcnunt.;o. The durability of the concrete is inlluenced by

1. the exposure conditions; 2. tho cement type; 3. tho concrete quality:

4. the cover to the rei nforcement; 5. the wid th of any cracks. Concrete can be exposed to a w ide range of conditions such as the soil. !.ea water. de-ici ng salts, stored chemicals or the atmosphere. T he severity or the exposure govern s the type of concrete mix required and the minimum cover to the reinforcing steel. Whatever the exposure, the concrete mix should be made from impervious and chemically inert a~grcgates. A dense, well-compacted concrete with a low watercement ratio is all important and for some soi l conditions it is advisable to usc a sulfateresiHti ng cement. A ir entrainment is usually specified where it is necessary to cater for repeated f'reezing and thawing. Ad equate cover is essential to prevent corrosive agents reachi ng the rei nforcement through cracks and pervious concrete. The thickness of cover required depends on the severity of' the exposure and the quality of the concrete (us shown in t.uhlc 6.2). T he cover is also necessary to protect the rein forcemenL against a rapid ri se in temperature and subsequent loss of strength during a fi re. Part 1.2 of EC2 provides guidance on t.his and other nspects of tire design. Durability requirements w ith rclu ted design culeulalions to check nnd control crack w idt hs and depths arc descr ibed in more dctuil in chapter 6.

1.6

Specification of materials

1.6.1

Concrete

T he selection of the type of concrete is frequently governed by the strength required, which in turn depends on tl 1e intensity of loading and the forn1 and size of the structural . For example, in the lower columns of a multi-storey buildi ng a higherS!rength concrete may be chosen in preference to greatl y increasing the size of the column section with a resultant loss in clear floor space.

w.~

··<_:!2

12

~ Reinforced concrete design As indicated in section 1.2.1, the concrete strength is assessed by measuring the crushing strength of cubes or cylinders of concrete made from the mix. These are usually cured, and tested after 28 days according to standard procedures. Concrete of a given strength is identified by its 'class' - a Class 25/30 concrete has a characteristic cylinder crushing strength lfc~:) of 25 N/mm 2 and cube strength of 30 N/mm 2 . Table 1.2 shows a list of commonly used classes and also the lowest class normally appropriate for various types of construction. Exposure conditions and durability can also afTect the choice of the mix design and the class of concrete. A structure suhject to con·osive conditions in a chemical plant, for example, would require a denser and higher class of concrete than, say, the interior of a school or office block. Although Class 42.5 Portland cement would be used in most structures, other cement types can also be used to advantage. Blast-furnace or sulfate-resisting cement may be used to resist chemical ullack, low-heat cements in massive sections to reduce the heat of hydration, or rapid-hardening cement when a high enrly strength is required. ln some circumstances it· mny be usefu l to replace some or the cement by materials such as Pulverised Fuel Ash or Grouud Granu lated Blast Furnace Sing which have slowly developing cernentitious propert'ies. These wi ll reduce the heat of hydration and may also lead to a smnlter pore structure and incn.:ased durability. Generally, naturul aggregates found locnlly <.~re preferred: however, manufactured lightweight material may be used when self-weight is importnnt. or " special dense aggregate when radiation sh ielding is required. The concrete mix may either be ci
Strength classes of concrete

Class

fcl• (N/mm 2)

(16/20 C20/25 C25/30 C28/35

16 20 25 28

(30/37 C32/40 C35/45 (40/50 C45/55 C50/60 C55/67 (60/ 75 C70/85

30 32 35 40 45 50 55 60 70 80 90

CB0/ 95

C90/1 05

Norma/lowest class for use os specified Plain concrete Reinforced concrete Prestressed concrele/Reinforced concrete subject to chlorides Reinforced concrete in foundations

~

Properti es of reinforced concrete

ch loride-induced con·osion). Detailed requirements for mix specification and compliance arc given by BS EN206 ·concrete - Performance. Production, Placing and Compliance Cliteria' and BS8500 'Concrete - Complementary British Srandard to BS E\'206'

1.6.2

Reinforcing steel

Table 1.3 lists the characteri stic design strengths of some of the more common types of 2 reinforcement currently used in the UK. Grade 500 (500N/mm characteri stic strength) has replaced Grade 250 and Grade 460 reinforci ng steel throughout Europe. The a bar is the diameter of an equivalent circular area. nominal si7.e Grade 250 bars arc hot-rolled m.ild-steeJ bars w hich usually have a srnooth surrace so tha t the bond w ith the concrete is by adhesion only. T his type or bar can be more readily bent, so they have in l he past been used where smal l radius bends arc necessary, such as l inks in narrow beams or colunUls, but plain bars arc not now recognised in the l~uropcan U nion and th ey nJe no longer avai lable for general use i n the UK. lligh-yicld bars are manufactured with a ribbed surfnce or in the form of n twisted square. Square twisted bnrs have inferior bond characteristics and have been used in the past, although they are now obsolete. Deformed bars have a mechanical bond w ith the conl:rctc, thus enlumcing ultimate bond ~ tresses ns described in section 5.2. T he bending or high-yield bars through a small radius is liable to cause tension cracking of the steel, and to avoid this t11e radius of the bend should not be less than two times the nominal bar size for small bars (~ 16 mm) or 3\12 times for larger bars (sec figure 5.11 ). The ductility of reinforcing steel is also classified for design purposes. Ribbed high yield bars may be classified as:

or

Class A - which is normally associated with small diameter ( < 12 mm) cold-worked bars used in mesh and fabric. This is the lowest ductil ity category and will include l imit~ on moment redistribution which can be applied (sec section 4.7) and higher quantities for fire resistance. Class B - which is most commonly used for reinforcing bars. Class C- high ducti lity which may be used in eart hquake design or simi lar situations. Floor slabs. walls, shells and roads may be reinforced with a welclecl fa bric of rcin forccmcnL, supplied in rolls and having a square or rectangular mesh. T his can give large economics in the detailing of the reinforcement and also in site labour costs of handling and fi xing. Prefabricated reinforcement bnr assemblies are also becoming increasingly popular for si mi lar reasons. Welded fabric mesh made of ribbed w ire greulcr than 6 mm dinmeter may be of any of the cluct.il ity cl
Table 1.3

Strength of reinforcement

Designation Hot-rolled high yield (854449) Cold-worked high yield (8$4449)

Normal sizes (mm)

Specified characteristic strength fyk (N/mm 1)

All sizes

500

Up to and including 12

500

•Note that BS4449 will be replaced by BS EN1 0080 in due course.

~;~~

q

~

Reinforced concrete design

The cross-sectional areas and perimeters of various sizes of bars. and the crosssectional area per unit width of slabs •are listed in the Appendix. Reinforcing bars in a member should either be straight or bent to standard shapes. These shapes must be fully dimensioned and listed in a schedule of the reinforcement which is used on site for the bending and fixing of the bars. Standard bar shapes and a method of scheduling are specified in BS8666. The bar types as previously described are commonly identified by the following codes: H for high yield steel, irres pective of ductility class or HA, HB, HC where a specific ducti lity class is required: this notation is generally used throug hout this book.

CHAPTER

2

······· ·· ·············•··············

Limit state design CHAPTER INTRODUCTION limit slate design of an engineerin g structure must ensure that (1) under the worst loadings the structure is safe, and (2) during normal working conditions the deformation of the does not detract from the appearance, durability or performance of the structure. Despite the difficulty in assessing the precise loading and variations in the strength of the concrete and steel, these requirements have to be met. Three basic methods using factors of safety to achieve safe, workable structures have been developed over many years; they are 1. The permissible stress method in which ultimate strengths of the materials are divided by a factor of safety to provide design stresses which are usually within the elastic range. 2. The load factor m ethod in which the working loads are multiplied by a factor of sa fety. 3. The limit state method which multiplies the workin g loads by partial factors of safety and also divides the materials' l.lltimate strengths by further partial factors of safety. The permissible stress method has proved to be a simple and usefu l method but it does have som e serious inconsistencies and is generally no longer In use. Because it is based on an elastic stress d istribution, it is not really applicable to a semi-plastic materia l such as concrete, nor is it suitable when the deformations are not proportional to the load, as in slender columns. It has also been found to be unsafe w hen dealing with the stability of structures subject to overturning ~ forces (see example 2.2).

15

r~

16


~

In the load factor method the ultimate strength of the materials should be used in the calculations. As this method does not apply factors of safety to the material stresses, it cannot directly take of the variability of the materials, and also it cannot be used to calculate the deflections or cracking at working loads. Again, this is a design method that has now been effectively superseded by modern limit state design methods. The limit state method of design, now widely adopted across Europe and many other parts of the world, overcom es many of the disadvantages of the previous two methods. It does so by applying parlial factors of safety, both to the loads and to the material strengths, and the magnitude of t he factors may be varied so t hat they may be used either with the plastic conditions in the ultimate state or with the more elastic stress range at working loads. This fl exibility is particularly important if full benefits are to be obtained from development of improved concrete and steel properties.

2.1

limit states

The purpose of design is to achieve acceptable probabilitie1> that a structure will not become unfit for its imendcd use- that is, that it will not reach a limit' state. Thus, ru1y way in wh ich a structure may cease to be fit for use will constitute a l imit stale and the design aim is to avoid any such condition being reached during the expected life of the structure. The two principal types of limit stale are the ultimate limit stntc and the serviceabi lity limit state.

(a) Ultimate limit state This requires that the structure must be able to withstnnd, with an adequnre factor of safety against coll apse. the loads for which it is dc:-;igncd to ensure the safety of the building occupants and/or the safety of' the structure itself. T he possibi li ty of buckling or overturning must also be taken imo uc<.:ount, as mu:-;t the possibility of' ucci.dental damage as c~ u sed , for example, by an intern al explosion.

(b) Serviceability limit states General ly the most important serviceabil ity limit st~tcs arc:

1. Deflection - the appearance or effi ciency of any part of the structure must not be adversely affected by detlection~ nor shou ld the com fort of the building s be adversely affected. 2. Cracking - local damage due to cracking and sp:~ ll ing must not affect the :~ppearancc. efficiency or durability of the struclUrc. 3. Durability- this must be considered in of the proposed life of the structure and its conditions of exposure. Other limit stares that may be reached include:

4. Excessive vibration - which may cau:-;e discomfort or alarm as well as damage. 5. Fatigue- must be considered if' cyclic loading is likely.

Limit state design

n

17

6. Fire resistance - this must be considered in of resistance to collapse. flame penetration and heal transfer. 7. Special ci rcumstances - any special requirements of the structure which are nol covered by any of the more common limit states, such as earthquake resistance, must be taken into . The relative importance of each limit stale will vary according to the nature of the structure. The usual procedure is to decide which is the ciUcial limit state for a particular structure and base the design on this, although durability and fire resistance requirements may well innucnce initial member sizi ng and concrete class selection. Checks must also be made to ensure that all other relevant limit states arc satisfied by the results produced. Except in special cases, such as water-retaining structures, the ultimate limit staLe is generally critical for rei nfo rced concrete although subsequent serviceability checks may affect some of the details of the design. Prestressed concrete design, however, is generally based on serviceabi lity conditions with checks on the ult.imat:c limit stale. In assessing a particular limit state for a structure it is necessary to consider all the possible variable parameters such as· the loads, material strengths and all construct"ional tolerances.

2.2

Characteristic material strengths and characteristic loads

2.2. 1

Characteristic material strengt hs

The strengths of material:. upon which a design is based are, normally, those strengths below which results arc unlikely to fal l. These are called 'characteristic' strengths. ft i~-. assumed that for a given mmerial. the disuibution of sLrcngth will be approximately ·normal ', so that a frequency distribution curve of a large number of sample resu l t~> would be of the form shown in figure 2.1. The characteristic strength is taken as that value helow which it is unlikely that more than 5 per cent of lhe results wi ll fa ll. Th is is given by

Jk. =};,, -

1.64.1'

= characteristic strength,f;, = mean strength and s = standard deviation. The relnrionship between eharHcteristic and mean values aecounts fo r variations in results of test specimens and wi ll, therefore, reliect the method and cont.rol of rnonufncture, qual ity of const ituents, and nature of the materia l.

wherc.f~

Mean strength (f.n)

I

Figure 2.1 Normal frequency distribution

of strengths

Number of

test specimens

Strength

r Reinforced concrete design

]_!._

2.2.2

Characteristic actions

In Eurocode tenninology the set of applied forces (or loads) for which a structure is to be designed are called 'actions· although the ·actions· and 'loads' tend to be used interchangeably in some of the Eurocodes. 'Actions· can also have a wider meaning including the effect of imposed deformations caused by. for example. settlement of foundations. In this text we will standardise on the term 'actions· as much as possible. Ideally it should be possible to assess actions statistically in the same way that material characteristic strengths can be determined statistically, in wltich case characteristic action

= mean action ± 1.64 standard deviations

In most cases it is the maximum value or the actions on a structural member that is critical and the upper, positive value given by this expression is used: but the 10\vcr, minimum value may apply when considering stability or the behaviour or continuous . These characteristic values represent the limits within which at least 90 per cenl of values will lie in practice. It is to be expected that not more than 5 per cent of cases will exceed the upper limit and not more than 5 per cent wi ll fa ll below the lower limit. They arc uesign values that take into the aecmacy with which the structural loading can be predicted. Usually. however. there is insufficient stat istical data to al low actions to be treated in tl1is way. and in this case the standard loadings, such as those given in BS EN 199 1, Eurocode I - Actions on Strut.:tures, should be uscu as representing characteristic values.

2.3

Partial factors of safety

Other possible variations such as constructional tolerances are allowed for by partial factors of safely applied to the strength of the materials and to the actions. ll should theoretically be possible to derive values for these from a mathematical assessment of the probability of reaching each limit state. Lnck of adequate data. however, makes this unrealistic and, in practice, the values adopted arc based on experience anu simpl ified calculations. 2.3.1

Partial factors of safety for materials ('/'m)

. charact:cristic ~ trcn gt.h (f') Destan strenoth = . . . . ' "' "' parttal 'factor oJ· sal'ety b.,) The following factors are considered when selecting n suitable value for ~lm :

1. The strength of the material in an actual member. Th is strength wi ll differ from that measured in a carefu lly prarcu test specimen and it is particularly true l'or concrete where placing, compaction anu curing arc so important to the strength. Steel, on the other hand, is a relatively consistent material requiring a small partial factor of safety. 2. The severity of the limit state being considered. Thus, higher values are taken for the ultimate limit state than for the serviceability limit state. Recommended vallLes for l m are given in table 2. 1 The values in the first two columns should be used when the structure is being designed for persistent design situations (anticipated normal usage) or transient design siruatiom (temporary

Limit state design

~~

Table 2. 1 Partial factors of safety applied to materials (/'m) to ed

Limit state

og

Persistent and transient Concrete

Reinforcing and Prestressing Steel

Concrete

Reinforcing and Prestressing Steel

1.50 1.50 1.50

1 .1 5 1 .15 1.15

1.20 1.20 1. 20

1 .00 1 .00 1.00

1.00

1.00

of

ie. iaJ

Accidental

Ultimate

Flexure Shear Bond

Serviceability IS

!f.

of ill

-;ituations such as may occur during construction) . The values in the last two columns ~ho uld be used when the structure is being designed for exceptional accidental design situations such as tl1e effects of tire or explosion.

e)

lg

m l.

IC

ttl

ld >i

'd

tl

II"

2.3.2 Partial factors of safety for actions (l't) Errors anti inaccuracies may be due to a number of causes:

1. design assumpt ions nnd inaccuracy of calculation: 2. possible unuf'ual increases in the magnirude of the actions; 3. unforeseen stress redistributions: 4. constructional inaccuracies. These cannot he ignore<.!, anti arc taken into by applying a partial factor of ~afcty h f) on the ehaructcristic actions, so that design value of action = characteristic action x partial factor of safety (')r) The value or this factor ~ho u ld al ~o take into the importuncc of the limit srate under considerat ion and reflects to some extent the accuntcy w ith which differclll rypes of actions cnn be predicted, and the probabili ty o r particular com bin ation.~ of actions occurrin g. It should be noted that design errors and constructional inaccuracies have similar effects and are thus sensibly grouped together. T hese factors will udequutely for normal conditions alth ough gross errors in design or construction obviously cannot he catered for. Recommen<.lcd values or partial factors of safety are given in tables 2.2 and 2.3 accordin g to the di fferent categorisations of actions shown in the tnblcs. Actions arc cn tegori setl us either permanent (Od. such as the self-weight of the structure, or Pariab/e (Qk), such as the temporary imposed loading ari sing from the traffic of people, wind and snow lomli ng, anti the li ke. Va riable actions urc also cutegorised as leading (the predomi nant variable action on the structure such as an imposed crowd load - Q~ . 1) and accompanying (secondary vmiable action(s) such as the effect of wind loading. Qk, ;, where the subscript 'i' indicates the i'th action ). The rerms favourable and unfavourable refer to the effect or the action(s) on the design situation under consideration. For example, if a beam. continuous over several spans, is to be designed for the largest sagging bending moment it will have to sustain any action that has rhe effect of increasing the bending moment will be considered unfavourable wh ilst any action that reduces the bending moment will be considcrcu to be favourab le.

19

1·.

20

-

r


Table 2.2

Partial safety factors at the ultimate limit state

Persistent or transient design situation

Permanent actions

Leading variable action

(Gk)

(0...t)

Accompanying variable actions (Qk.i)

Unfavourable

Favourable

Unfavourable

Favourable

Unfavourable

Favourable

(a) For checking the static equilibrium of a building structure

1.10

0.90

1.50

0

1.50

0

(b) For the design of structural (excluding geotechnical actions)

1.35

1.00

1.50

0

1.50

0

(c) As an alternative to (a) and (b) above to design for both situations with one set of calculalions

1.35

1.15

1.50

0

1.50

0

Table 2.3

Partial safety factors at t he serviceabili ty limit state

Design Situation

Permanent actions

Variable actions

All

1.0

1.0

Example 2. I shows how the panial safety factors at the ullirnate limit state from tables 2.1 and 2.2 are used to design the cross-sectional area of a steel cable ing permanent and variable actions.

(

EXAMPLE 2.1

Simple design of a cable at the ultimate limit state Determine the cross-sectional area of steel rcquirecl for a cable which suppons a total charnctcti sti c permanent a<.:tion o f 3.0 kN and a characteristic variabl e action of 2.0 kN as shown in figure 2.2.

Figure 2.2 Cable design

Variable load - 2.0 kN (man ... eqtlipmenl)

Permanent load = 3.0 kN (platform + cable)

Limit state design

21

The characteristic yield stress of the steel is 500 N/mm 2 • Carry out the calculation using limit state design with the following factors of safety: IG = 1.35 for the pennanenl action. 1Q

= 1.5 for the variable action. and

1m

= 1.15 for the steel strength.

Design value = /G x pennanent action + 'YQ x variable action 1.35

X

3.0 + 1.5

X

2.0

= 7.05kN characteristic yield stress Des1gn stress = --------.

'I'm

500

=w

= 434 N/mm2 .

.

desi<>n value

Reglllred cross-sect1onal area=-:---:-."';:....__ _ des1gn stress

7.05

X

103

434

- 16.2 mm~

l

For convenience, the p~utial factors of safety in the example arc the same as those recommended in EC2. Probably. in a practical design. higher factors of safety would he prefcrred for a single ing cable, in view of the consequences of a failure.

___________________________________________)

~·

Example 2.2 shows the design of a foundation to resist uplift at the ultimate limit \late using the partial facto rs or safety from table 2.2. lt demonstrates the benelits or using the limit stare approach instead of the potential ly unsafe overal l factor of safety design used in part (b).

(E XAMPLE 2.2 Design of a foundation to resist uplift

Figure 2.3 shows a beam ed on foundations at A and B. The loads ed by the bC<1m arc its own uniformly distributed permanent weight of 20 kN/m and a 170 kN variable load concentrated at end C. Determine the weight of foundation required at A in order to resist uplift: (a)

hy applying a factor of safety of 2.0 to the reaction calculated for the working loads.

(b) by using an ultimate limit state approach with partial factors of safety of 'Yr. = 1.10 or 0.9 for the permanent action and I'Q = 1.5 for the variable action. Investigate the effect on these designs of a 7 per cent increase in the vatiablc action.

··~

22


Figure 2.3 Uplift calculation example beam

permanenlload 20 kN/m

. 1..

6m

(a)

2m

0.9 x permanent load

(b) Loading arrangement for uplifl at A at the ultimate limit state

(a) Factor of safety on uplift = 2.0 Taking moments about B Uplift

( 17()

R,, =

X

2

20 6.0

X

8 X 2) =:l :lJ kN ....

Weight of foundation required

= 3.33 x safety factor 3.33

X

=6.7 kN

2.0

With a 7 per cent incrc
·up1'lit,.

R _ ( l.07 lA -

X l 70 X 2- 20 X 8 X 2~

fi.O

= 7 .3 ''l\.1~"

1

T hus with a ~l i gh t increase in the vnri nble action there is u significant increase in the

..

upl ifl and the structure becomes unsafe.

(b) Limit state method - ultimate load pattern t\'1> this example includes a cantilever and also involves the requirement for static equi li brium a1 A, partia l factors of safety of 1.10 and 0.9 were chosen for the permanent actions as given in the lina row of values in table 2.2 The arrangement of tJ1e loads for the maximum uplift at A is shown in figure 2.3b. x 20 x 2

= I . I 0 x 20 x 2 = 44 kN

Design permanent action over AR - I<; x 20 x 6

= 0.9 x 20 x 6 = lORkN

Design permanent action over

BC

= IG = ~lo

Design v
x 170 = J .5 x 170

255 kN

Taking moments about B for the ultimate actions

.f R _ (255

UpI l t

A-

X2

+ 44 X 6.0

l - 108 X 3) _ }o kN -

0

T herefore weight of foundation required = 38 kN. A 7 per ccm increase in th e variable action will not endanger the structure, since the acl.u
Limit state design : 23 Parts (a) and (b) of example 2.2 illustrate how the l imit state method or design can ensure a safer result when the stability or strength of a structure is sensitive lo a smal l numerical difference between the effects of two opposing actions of a similar magnitude.

2.4

Combination of actions

Permanent and variable actions wi ll occur in different combina tions, all of which must be taken imo in determi ning the most cri tical design situation for any structure. For example, the seU"-wcight of the structure may be considered in combination with the weight of furnishings and people, w ith or wi thout the effect of wind acting on the building (which may also act in more than one direction) In cal'es where actions are to be combined it is recommended that, in determining suituble design values, each characteri stic action is not only multiplied by the parti al factors of safety, as discussed above, but also by a further f act.or given th e symbol W. T his !"actor is generally taken as 1.0 other than wh ere described below:

(i)

Combination values of variable actions Where more than one variable action is to be considered (i.e a combination) then the variable actions should be multiplied by a value of IJ! (denoted as 'l'o) ns given in table 2.4. This ensures that the probability of a combination or a<.:tions being exceeded is approximately the same a<; that for a single netion. As can be seen in the table thi s is also dependent on the type of structure being designed. Combination values are used for deg for (i) the ultimate limit state and (i i) irreversible servicenbility limit states such as irreversible crucking due to temporary hut excessive overloading of the structure.

(ii)

Frequent values of variable actions Frequent combinations of actions arc used in the consideration of ( i) ultimate limit states involving accidental actions and (i i) reversible l imit states ~uch as the serviceabi lity limi t states or cracki ng and deflection where the m:tions causing these effects ure of a short transitory nature. Tn these cases the varinhle actions are multiplied by a vu lue of 1T1 (denoted as IJ!·1) as given in table 2.4. The values of 1T1 1 give an estim ati on of the proportion of the total var iable acti on that is li k<.:ly to be associrned w ith thi s parti cular combination of actions.

(i ii) Quasi-permanent values of variable action EC2 requires t.hat, in certain situations, th e c fTct:ts of ·quasi-permanent.' a<.:ti ons should be considered. Quasi-permanent (meaning 'almost.' pennnncnl) acti ons are those that may be su~ta i ned over a long period but arc not necessaril y as permanent as, say. the self-weight of the structure. A n example of such a loading wou ld be the effect of snow on the roofs of bui ldings at high altitudes w here the weight of the snow may have to be sustained over weeks or months.

)

Quasi-permanent combinations of actions are used in the consideration of (i) ultimate limil Slates involving accidemal actions and (ii) serviceability limit states attribmable to. for example, the long-term effects of creep and where the actions causing these effects. whilst variable, arc of a more long-tem1. sustained nature. In these cases the variable actions are multiplied by a value of IJ! (denoted as ili 2 ) as given in table 2.4. The values of 'T'2 give an estimation of the proportion of the total variable action that is l ikely to be associated with this particular combi nation of actions.

24

mReinforced concrete design

Table 2.4

Values of \II for different load combinations

Action

Combination

Frequent

Quasi-permanent

'1!,

IJiz

Imposed load in buildings, category (see EN 1991 -1-1) Category A: domestic, residential areas Category B: office areas Category C: congregation areas Category D: shopping areas Category E: storage areas Category F: traffic area, vehicle weigh t < 30kN Category G: traffic area, 30 kN < vehicle weight< 160 kN Category H: roofs

0.7 0.7 0.7 0.7 1.0 0.7 0. 7 0.7

0.5 0.5 0.7 0.7 0.9 0.7 0.5 0

0.3 0.3 0.6 0.6 0.8 0.6 0.3 0

Snow loads on buildings (see EN 1991 -1·3) For sites located at altitude /i > 1000 m above sea level For sites located at altitude H ~ 1000 m above sea level Wind loads on buildings (see EN 1991·1·4)

0. 7 0.5 0.5

0.5 0.2 0.2

0.2 0 0

0.7

Figure 2.4 Wind and imposed load acting on an office building - stability

X

1.5Q,

1.5Q,

HH

Ht

check

t

0.5

X

--

1.5W,

t

0.9G,

t

, 1.1 Gk

I /~

(a)

(b)

Figure 2.4 illustrales how the factors in table 2.2 and 2.4 can be applied when considering the stability of the office building shown for overturning abour point B. Figure 2.4(n) treats lhe wind load (Wk) as the lending variable action ancl lhe live load (Qk) on the roof a.o; the accompanyi ng variable action. Figure 2.4(b) considers the live load as the leading variable action and the wind as the accompanying variable action.

2.4.1

Design values of actions at the ultimate limit state

In general . for p('r.l'istent and transiem design situations the design value can be taken as: Design value (Ed) - (factored permanent actions) combined with (factored si ngle leading variable action) combined with (factored remaining accompanying variable actions) The 'factors' wi ll, in all cases, be the appropriate partial factor of safety hr) taken together with the appropriate value of iii as given in table 2.4.

Limit state design The design value can be expressed formalistically as: £d

= 1:?:"/G.jGk,jl +!'Q,IQk.t + 12:::: lo.;Wo,;Qk,;l J2: 1

(2.1 )

•> I

\'ore that the ..1.. sign in this expression is not algebraic: it simply means ·combined with'. The L: symbol indicates the combined eiiect of all the similar action effects. e.g. 1Lj2:l/'G.jGk.j1 indicates the combined effects of all factored permanent actions, -.ummcd from the first to the 'j'th action. where there are a total of j permanent actions acting on the structure. Two other similar equations are given in EC2, the least tavourable of which can alternatively be used to give the design value. Jlowever. equation (2.1) will normally apply for most standard situations. For accide111al design silltations the design value of actions can be expressed in a -,imi Jar way with the permanent and variable actions being combined with the effect of the accidental design situation such as fire or impact. As previously indicated. such accidentnl design situations wi ll be based on the frequent or quasi-permcment values or actions with the load combinations calculated usiDg the appropriate 111 value(s) from table 2.4

2.4.2 Typical common design values of actions at the ultimate li mit state For the routine design or the within reinforced concrete structures the sl'nndard design loading cases will oflcn consist of combinations of tJ1e permanent action with a -,ingle variable act ion and possibly with wind. If the single variable act ion is considered to he the leading variable action then wind loading will be the accompanying variable .lction. The reverse may, however. be true and both scenarios mu~t be considered. In ' uch cases the factors given in tnble 2.5 can be used to determine the design value of lhe actions. The value of 1.35 for unfavourable permanent actions is conservati ve, and used throughout this book for simplicity. Alternative equations indicated in 2.4.1 may, in some cases, give greater economy. Table 2.5

Combination of actions and load factors at the ultimate limit state

n L d e

Persistent or transient design siluotion

·-

Permanent + Variable Permanent + Wind

fl

25

Permanent actions (Gk)

1.35

{

1.00

Wind

(QI,,l)

Unfavourable Favourable Unfavourable 1.QQ(I) 1.35 1.50 1.35 1.00

Permanent + Variable + Wind Either of these two cases may be critical - both should be considered

Variable action

1.50

Favourable

0 1.50 0

\1! 0 x 1.50 = 0.5(2) X 1 .50

1.35

1.00

IJ, 0 t3l x 1.50

0

-0.75 1.50

(1) For continuous beams with cantilevers, the partial safety factor for the favourable effect of the permanent action should be taken as 1.0 lor the span adjacent to the cantilever (see figure 7.21 ). (2) Based on the 'combination' figure in table 2.4 for wind (3) lflo to be selected from table 2.4 depending on category of building (most typical value = 0.7) (4) The partial safety factor for earth pressures may be taken as 1.30 when unfavourable and 0.0 when favourable

26


2.4.3

Design values of actions at the serviceability limit state

The design values of actions at the serviceability limit state can be expressed in a simi lar way to equation 2. L but taking of the different combinations of actions to be used in the three different situations discussed above. ln the case of serviceability the partial factor of safety, ~1 r will be taken as equal ro 1.0 in all cases. (i)

Combination values of variable actions Ed =

I?:= Gk,jl +Qk. l +I?:= IJ!o.; Qk.il j>l

(2.2)

t> l

(ii) F r equent values of variable actions

Eo~ = IL Gk.jl +l

11

,> 1

t, l

Ql<.l +I?:= \]1 2. 1 Qk,il

(2.3)

t> l

(i ii) Quasi-permanent values of variable ac.tions

(2.4) Note that. as before. the ...1.. signs in these expressions arc not necessnrily algebraic: they simply mean 'combined with' . The in the expressions have the follow ing meanings: the combined effect of all the characteristic permanent actions where the subscript 'j' indicates that there could be between one and 'j' permanent actions on the structure the si ngle leading characteristic variable action mu ltiplied by the factor 1T1. where 111 takes the value of 1, \11 1 or 2 as appropriate from table 2.4. The subscript ·1· inclicmes that thi s relates to the single leading variable action 0 11 th e structure.

'r'

the combined cf'fect or nll the 'accompany ing' characteristi c variable actions each multipliecl by the J.a ' ct.or ~~ . where \]'1 takes the value of 111 0 or 1T1 2 as approprime l'rom table 2.4. The .subscript ' i' indicates that th ere could be up to ' i' variable actjons on t.he structure in ncldil'ion to the single leading varinble action

(

EXA MPLE 2.3

Combination of actions at the serviceability limit state A simply ed reinforced concrete beam forms part of a building within a shopping complex. It is to be designed for a characteristic pcrmanem action of 20 kN/m (its own self-weight and that of the ed structure) together with a characteristic, single leading variable action of 10 kN/m and an accompanying variable action of2 kN/m (both representing the imposed loading on the beam). Calculate each of the serviceability limit state design values as gi ven by equations (2.2) to (2.4).

Limit state design From table 2.4 the building is and 0.6.

w2 =

Combination value

Ed=

IL

GL,,, + Qu +

j~ l

IL Q~;) IL 'Vo.;

= 20 +

10 + (0.7 x 2)

= 0.7, 11! 1 = 0.7

= 31.4 kN/m

i> l

Frequent value

Ed =

classified as category D. Hence. Yio

I~ Gk.jl +\h I QL. I + J? l

qi2,i Qk,tl

= 20 + (0.7 X

10) + (0.6

X

2)

s> l

= 28.2 kN/m Quasi-permanent value

Ed =

~~ Gk,jl + ~~ W2,rQh,il = 20 + (0.6 x 10) + (0.6 x 2) = 27.2 kN/m

2.5 t Global factor of safety The use of partial faclors of safety on materials and aclions offers considerable flexibility. which may he used to allow for special conditions such as very high \tandards of construction and control or, ar tJ1e other extreme, where structural fai lure would be particularly disastrous. T he global factor of safety against a particular type of failure may be obtained by multiplying the appropriate partial factors of safety. For instance, a beam failure cau$Cd by yielding of tensile reinforcement would have a factor of ~(, 11

or

x "fr

1. 15 x 1.35 = L.55 for permanent loads only 1.15 x 1.5 = 1.72 for variable loads on ly

Thus practical cases will have a value between these two figures, depending on the 1.8 which was rclutivc load ing proportions, and this can be compared w ith the value the order of magnitude used by the load !'actor method prior to the introduction of l imit ~ late design.

or

Similarl y, failure by crushing of the concrete in the compression zone hns u f'w.:lor of

1.5 x 1.5 = 2.25 due t.o variable actions only, which reflects !he fact that such failure is generally without warning and may be very serious. Thus the basic values of partial factors chosen arc such !hal under normal circumstances the global factor of sn fety is similar to that used in earlier design methods.

27

CHAPTER

3

Analysis of the structure at the ultimate limit state CHAPTER INTRODUCTION

··••·•··•·•·••·• ·••• ·• •·•····· ··· ··· •··•···••·•···•··•···•••·•··· ·•········· ·

A reinforced concrete structure is a combination of beams, columns, slabs and walls, rigidly connected together to form a monolithic frame. Each individual member must be capable of resisting the forces acting on it, so that the determination of these forces is an essential part of the design process. The full analysis of a rigid concrete frame is rarely simple; but simplified calculations of adequate precision can often be made if the basic action of the structure is understood. The analysis must begin with an evaluation of all the loads carried by the structure, including its own weight. Many of the loads are variable in magnitude and position, and all possible critical arrangements of loads must be considered. First the structure itself is rationalised into simplified forms that represent the load-carrying action of the prototype. The forces in each member can then be determined by one of the fo llowing meth ods: 1. applying moment and shear coefficients 2. manual calculations 3. computer methods Tabulated coefficients are suitable for use only with simple, regular structures such as equal-span continuous beams carrying uniform loads. Manual calculations are possible for the vast majority of structures, but may be tedious for large or complicated ones. The computer can be an invaluable help in the analysis of even quite small frames, and for some calculations it is almost indispensable. However, the amount of output from a computer analysis is sometimes almost overwhelming; and then the results are most readily inter~ preted when they are presented diagrammatically.

28

Analysis of the structure

---.... Since the design of a reinforced concrete member is generally based on the ultimate limit state, the analysis is usually performed for loadings corresponding to that state. Prestressed concrete , however, are normally designed for serviceability loadings, as discussed in chapter 11.

3.1 ·

Actions

The actions (loads) on a structure are divided into two types: permanent actions, and variable (or imposed) actions. Permanent actions are those which are normally constant during the structure's life. Variable actions, on the other hand. arc transient and not constant in magnitude, as for example those due to wind or to human occupants. Recommendations f'or the loadings on structures arc given in t·he European Swndarcls. some of which arc EN I 99 1-1- 1General actions, EN I 99 I -1-3 Snovv loads, EN 199 I - 1-4 Wind actions, EN I 99 I - I -7 Accidental actions from impact and explosions, and EN 199 I -2 Tral'lic loads on bridges. A table of values for some useful permanent loads and variable loads is given in the appendix.

3. 1.1

Permanent actions

Permanent actions include the weight of the Stlllcture itself and all architectural components such as exterior cladding, partitions and ceilings. &Juipmcnt and static machinery, when permanent fixwres. arc also often considered as part of the permanent action. Once the sizes of all the structural , and the details or the architectural requiremems and permanent lixtures have been established. the permanent actions cnn be calculated quite accurately; but, first or all , preliminary design calcul ations are generally rcquireu to estimate the probnble sizes and self-weights or the structural concrete clements. For most rcinf'orccd concretes, a typical value for the self-weight is 25 kN per cubic merre. bul a higher density should be taken for heavi ly reinforced or dense concretes. In the case of' a bu ilding, l.he weights of any permanent p
3.1.2

Variable actions

These actions arc more difficult to determine accurately. For many of 1hem. it is on ly possible 10 make conservative estimates based on standard codes of practice or past experience. Examples of variable actions on buildings arc: the weights of its occupants,

29

30

r


furniture, or machinery; the pressures_of wind, the weight of snow, and of retained earth or water: and the forces caused by 'thermal expansion or shrinkage of the concrete. A large building is unlikely to be carrying its full variable action simultaneously on all its floors. For this reason EN 1991-1-1: 2002 (Actions on Structures) clause 6.2.2(2) allows a reduction in the total variable floor actions when the columns, walls or foundations are designed, for a building more than two storeys high. Similarly from the same code. clause 6.3.1.2(10). the variable action may be reduced when deg a beam span which s a large floor area. Although the wind load is a variable action. it is kept in a separate category when its partial factors of safety are specified, and when the load combinations on the structure are being considered.

3.2

Load combinations and patterns

3.2.1

Load combinations and patterns for the ultimate limit state

Various combinations of the characte ri~l.ic va lues of permanent Gk, variable actions Qk, wind actions Wk, and thelr partial factors of safety must he considered for the loading of the structure. The partial factors of safety specified in the code arc discussed in chapter 2. and for the ultimate limit slate the followin g loading combinations from tables 2.2, 2.4 and 2.5 arc commonly used. 1. Permanent and variable nctions 1 .35G~

+ 1 .5Q~

2. Permanent and wind actions

The variable load can usually cover al l or any part or the structure and, therefore, should be ::IITanged to cause the most :;cvcrc stresses. So. for a three-span continuous beam, load combination I would huvc the loading arrangement shown in figure 3.1, in order to cause the maximum sagging moment in the outer spans and the maximum possible hogging moment in the centre span. A stucly of the deflected shape of the beam would confirm this to be the case. Load combination 2. permanent + wind load is used to check the stabi lity of a structure. A load combination of permanent + variable + wi nd load uould have the arrangements shown in figure 2.4 and described in section 2.4 of Chapter 2. Figure 3.1 Three-span beam

1.35G, + 1.50Q,

1.35Gk + 1.50Q,

t

A

l

1.3SG,

j

c

l

(a) Loading arrangement for maximum sagging moment at A and C

(b) Deflected shape

A nalysis of the st ruct ure

=i =i

1.3SC, + l.SOQ,

[

a

s

'·

1.35G, + 1.50Q,

1.35C, + 1.50Q,

F

1.35Ck + l.SOQ,

j"sc, j j'"c'l 1'%

(i) loading arrangements for maximum moments in the spans 1.3SC, + 1.500\

!'"~

1.35C,. + 1.500.

1.3SC, + l.SOQ,

I I l'"c·l

(ii) loading arrangements for maximum moment at A 1.35G, + l .SOQ, 1.3SC,

~~

A

'

t

I

1.35Ck

t ~7

:.?~

t t t at the s according to EC2

1 t (Iii) Load ing for design moments t

\

Note that w hen there Is a cantilever span the minim um load on the span adjacent to the can tilever should be 'I.OG, for loading pattern (i)

Figure 3.2 shows the patterns of vertical loading on a multi-span conl'inuous beam to cause (i) maximum design sngging moments in alternate spans and maximum possible hogging moments in adjacent spans. (ii) maximum clesign hogging moments at ~upport A, and (iii) the design hogging moment at A as 11pccificd by the EC2 code for simplicity. Thus there is a similar loading pattern for the design hogging moment at each internal Mtpport of a continuous beam. It should be nmcd thai the UK :-.rational Annex permits a simpler alternative to load c.:ase (iii) where a <;ingle load ca~c may be considered of all spans loadecl with the maximum loading of

( 1.35(ik ·I 1.50Qd.

s

a

e

3.3

Analysis of beams

To design a Sltl teturc it is necessary ro know the bending moments. tor~ i onn l moments, 'hearing forces and axial forces i n each member. A n elastic analysis is generally used to determine the distribut ion of these forces within the strucmre; bu t hcc:.~ use - to some c,xtent - reinforced concrete is a plastic material, a l imited redistribution of the clasti c.: moments i ~ sometimes allowed. A plastic yield-line theory may be used tn calcu late the moments in concrete slabs. T he properti es of the materials, such as Young's modulus. which arc used i n the structural analysis should be those 11ssociatcd with thei r characteri stic: strengths. T he stiiTncsses of the can be culcu latecl on the basis of any one the fo llowing:

or

the elllire concrete cross-section (ignoring the reinforcement); 2. the concrete cross-section plus the transformed area of reinforcement based on the modular ratio; 3. the compression area only of the concrete cross-section, plus the transformed area of reinforcement based on the modular ratio. The concrete cross-section described in ( 1) is the simpler to calculme and would normally be chosen.

Figure 3.2 Multi-span beam loading patterns

31

32


A structure should be analysed for each of the critical loading conditions which • produce the maximum stresses at any particular section. This procedure will be illustrated in the examples for a continuous beam and a building frame. For these structures it is conventional to draw the bending-moment diagram on the tension side of the .

Sign Conventions 1. For the moment-distrihution analysis anti-clockwise moments arc positive as. for example, in table 3. 1 J'or the fixed end moments (FEM). 2. For subsequently calculating the moments along the span of a member, moments causing sagging nre positive, while moments causing hogging are negative, as illustrated in figure 3.4.

3.3.1

Non -continuous beams

One-span. simply ed beams or slabs are statically determinate and the analysis for bending moments and shearing fo rces is readily performed manually. For the ultimate limit state we need only consider the maximum load of l.35Gk + 1.5Qk on the spun.

(

EXAMPLE 3. 1

Analysis of a non-continuous beam The one-span simply SLlpported hcam shown in rigure 3.3a carries a distributed permanent action including ~elf-weight of 25 kN/m, a permanent concentrated action of 40 kN at micl-spnn, and a distributed variable action of I 0 kN/m. Fig ure 3.3

Analysis of one-span beam (1.35 X 2S + 1.50 X 10)4 = 195 kN

,__

4.0 m

(a) Ultimate l0<1d

124.5 kN

27

27 (b) Shearing Force Dlngmm

'124.5 kN

(c) Bending Moment Diagram

Figure 3.3 shows the values ol' ultimate loacl required in the calculations of the sheru·ing forces and bending moments. Maximum shear force .

.

54

195

= 2 +T = 124.5 kN 54 x 4

Max1mum bend111g moment = -

195 x4

•

- -1 - - - = I:l l.5 kN m 4 8 The analysis is compleLccl by drawing the shearing-force and bcnding-momenL diagrams which would later be used in the design and detailing of the shear and bending reinforcement. )

l~----------------------------------------

Analysis of the structure icb

3.3.2 Continuous beams

be '5e

of

ive nts

as

·::.is the on

The methods of analysis for continuous beams may also be applied to cominuous slabs which span in one direction. A continuous beam is considered to have no fixity with the s so that the beam is free to rotate. This assumplion is not strictly t111e for beams framing into columns and for that type of continuous beam it is more accurate to analyse them as part of a frame. as described in section 3.4. A cominuous beam should be anal ysed for the loading arrangements which give the maximum stresses at each section. as described in section 3.2. 1 and illustrated in figures 3. 1 and 3.2. The analysis to calculate the bending moments can be curri ed out manually by moment distribution or equivalent methods. but tabulated shear and moment coefficients may be adequate fo r continuous beams having approximately equal spans anti uniformly distributed loads. For a beam or slab set monolithically into its s, the design mornenl at th e can be taken as the moment at the face of the .

Continuous beams - the general case Hav ing tletermin cd the moments at the s hy, say, moment distribution, it is necessary to calculate the moments in the spans and also the shear forces on the beam. For a uniformly distributed load, the equations for the shears and th e maximum 1-pan moments can be derived from the rollowing analysis.

Load = w/metre

Figure 3.4

Shears and moments in a beam

S.F

B.tvl

Using th e sign convemion of 11gure 3.4 and taking momems nhout suppon B:

therefore

{3.1 ) and

j

(3.2)

33

34


Maximum span moment

Mmax

occurs at zero shear, and distance to zero shear

V.,n

a3 = -

(3.3)

w

therefore VAB2

MmaJ<

= -2w- + MAB

(3.4)

TI1e points of conlraflexure occur at M

= 0. that is

wxl

VAB X -2+ MAo = 0

where x the distance from A. Taking the roots of this equation gives .Y =

VAs± j(v~,~/ +2wMAo) ------'- -- -- - W

so that CIJ

=

\lAo - j(VAo 2 +2HMi\H)

(3.5)

and

(3.6) A similar analysis can be applied to beams that do not!>upport a uniformly distributed loud. In manual calculations it is usually not considered necessary to calculate the distances a 1• a2 and a3 which locate the points or eontrallexure and maximum moment a sketch of the bending moment is often adequate - hut if a computer is performing the culculations these distances muy us well be determined nlso. /\t the fnce of the . width .v

M~u

(

MAn - ( VAo -

~~r) ~

EXA M PLE 3.2

Analysis of a continuous beam The continuous beam shown in fi gure 3.5 has fl constant cross-section and s a uniformly distributed permanent action including its self-weight of Gk = 25 kN/m and a v;u·iable action Q~ - I0 k.N/m. The critical loading pallerns for the ultimate limit state are shown in figure 3.5 where the ·stars' indicate the region of maximum moments, sagging or possible hogging. Table 3. 1 is the moment distribution carried out for the first loading arrangement: similar calculations would be required for each of the remaining load cases. It should be noted that the reduced stiffness of ~Z has been used for the end spans.

F ...

Analysis of the structure : 35 G, = 25 kN/m Q, = 1 0 kN/m

6.0m

3.3) ( 1.35

X

6.0m

25 + 1.50 X 1 0)

1)

I

X

6 (1.35

= 292.5 kN

3.-l)

I

* ( 1.35

I

(2)

X

25

X

25

X

4,_)- -- - - - ,

I

292 .5 kN

*

I

=2o2 5 kN

(3)

X

135 kN

*

( 1.35 X 25 + 1.50 X 10)

6)

Figure 3.5 Continuous beam loading patterns

195 kN

*

*

292.5kN

195 kN

I

X4

202.5 kN

* 202.5 kN

* 202.5 kN

(4)

195 kN

I

292.5 kN

*

~,,:> ") _

Table 3. 1

Moment distribution for the first loading case A

3.6) uted the

3 I 4 'I 3 1 .,.. 4 ' 6 = 0.125

Stiffness (k)

Distr. factors

phe Load (kN)

0 Balance

1 4 = 0.25

2/ 3

- 292 X 6 8 - 219.4 + 58.1

da

Balance Carry over

+

ere

Balance Carry over

+

Balance

M (kNm)

'T

0

19.4 6.5 2.2 0.7

- 132.5

292

+135 ~ 12

+ 45.0 + 116.3 58. 1

+

be

1/ 3

2/ 3 135

Balance Carry over

Jar

= 0.125

0.25 0.'125 + 0.25

Carry over

~g.

3 I 4'I

I

292 0

D

L

0.125 0.125 + 0.25 - 1/ 3

~m

F.E.M.

c

B

+

-

38.7 19.4

X

45.0 116. 3

292 X 6 +--8 + 219.4 58. 1

19.4

38.7 1- 19.4

12.9 6.5

6.5

-

6.5

I

+

-

4.3 2.2

-

4.3 2.2

-

2.2

1

+

1.5

-

1.5

-

0.7

132.5

0

+ 58. 1

+ 12.9

+

0

- 132.5

+ 132.5

0

36


The sheruing forces. the maximum span bending moments. rutd their positions along the beam, can be calculated using the formulae previously delived. Thus for the first loading a1rangemem and span AB, using the sign convention of figure 3.4: Shear

VAR

load (MAR - MRA) = --..:..._----'--'2

vi3A

Maximum moment, span AB

'-

-292.5

132.5 -

= 292.5 -

124.2

?

- ---6.(}- L4. 2 = load - VAl/

2kN

= 168.3 kN

VAo2

= -2- + M,,R w

where w = 292.5/6.0 = 48.75 kN/m. Therefore: 2

Mm:1x

0 ? =?-X124.2 +0 = I5o._ kNm 48.7.5

['rom A, a,=VAI3 . D!Stance • 111 124.2

= 48.75 -

2.55 m

The bending-moment diagrams for each of the loading arrangements are shown in fi gure 3.6, and tl1e correspondi ng shearing-force diagrams arc shown in figu re 3.7. The individual bending-moment diagrams arc combined in figu re 3.8a to give the bendingmoment design envelope. Similarly. figure 3.8b is the l>hearing-forcc desig11 envelope. Such envelope diagrams arc used in the detailed de!>ign of the beams. as described in chapter 7. ln this example, simple s with no fixity have been assumed for the end s at A and 0. Even so. the sections at I\ and I) should be designed for a hogging momem clue to a partial fixity equal to 25 per ce11l of the maximum moment in the span, lhtll iN 158/4 Figure 3.6 Bending-moment dlagmms (kN 111)

= 39.5 kNm. 133

(1)

'133

Ld

~

"""-7

158

158

108 (2)

108

~

~ 103

~ 103

151

(3)

~

""=/ 151

1~

(4)

~

~

151

A

~

~ 151

Analysis of the structure 124 ( 1)

~

168 67.5

~

r::---... I

83

97.5

C>--.

~

c:::::::::::::J

121

<3> ~

~

119

1 1~ I

"'-J 171

85 (4) ['-:..,

119

"""""J 9 7.5

--=:::::::::::) 8 3

118~

I~

""""J 85

-==z::::J 8 s

171 ~

"""J

11 8

~

I

""'J

151 (a)

Shearing-force diagrams (kN)

.............. I '-..,J 124

~

SSt->,

11 0

""'J 121

151

Figure 3.8

~

Bending-moment and

.-::-----&'---====---->,~-----,., kN.m

\ ·-. ----/,i

11

~

" \: ;::.- - -- :

158

-:l

158

171

~

110

~

(b)

shearing-force envelopes

~

124

~

--- ~

kN

~~ 110 124 171

Continuous beams with approximately equal spans and uniform loading

or

The ultimate bending moments and shearing forces in continuous beams three or more approximately equal spans without canLilevers can be obtained using relevant coefficients provided that the spans c.liff'er by no more than 15 per cent of the longest span, that the loading is uniform, and that the characteristic variable action does not exceed the characteristic permanent action. The values of these coefficients are shown in diagrammatic form in figure 3.9 f'or beams (equi valent si mplified values for slabs are given in chapter R). End Span 0.11FL

_.-/1

(a) Bending Moments

~

Interior Spa11

0.10FL

./1

<:::::::::::7 0.07FL

0.55f

0.45 F

(b)

0.10FL

"'--

0.09H

Shearing Forces

r----__

r---._

-:::::::::]

---=::::::j

0.60F F = Total ultimate load on span = (1.35G 1 + 1.50~) kN L = Effective span

37

Figure 3.7

-=::::::J67 5 .

"'-J168

(2)

~

rnF.::l

0.55f

Figure 3.9 Bending-moment and shearing-force coefficients for beams

38

1

Reinforced concrete design The possibility of hogging moments in any of the spans should not be ignored, even if it is not i ndicated by these coeffici~nts. For example. a beam of three equal spans may have a hogging moment in the centre span if Qk exceeds 0.45Gk.

3.4

Analysis of frames

In situ reinforced concrete structures behave as rigid frames, and should be analysed as such. They can be analysed as a complete space frame or be divided into a series of plane frames. Bridge deck-type structures can be analysed as an equi valent gri llage. whi lst some form of finite-clement analysis can be utilised in solving complicated shear W
3.4.1

Braced frames ing vertical loads only

A bui lding frame can be analysed as a complete frame, or il can be simplified into a series of substitute frames for the vertical loading analysis. The frame shown in figure 3. 10, for example. cw1 be divided i nto any of the sub frames shown in figure 3.11 . T he substi tu te frumc I in figu re 3. 11 consists of one complete lioor beam with its connectin g columns (which arc assumed ri gidly li xcd m th eir remote ends). A n analysis of thi s frame wi ll gi ve th e bending moments and shearin g forces in the beams and colu mns for Lhe lloor level consiclcrccl. Substitute frame 2 is a single span combined with its connectin g columns and two adjacent spans, all li xed at their remote ends. T his frame may be Ul'ed to determine the bending moments and shearing forces in the centrul beam. Provided that the central span is greater than the two adjacent spans, the bendi ng moments in the columns can also be found wi th this frame. Substitute f rame 3 can be used to fi nd the moments in the columns only . It consists of a si ngle j unction, with the remote ends of the fixed. This type of subframe would be used when bean'ts have been analysed as continuous over simple s. I n frames 2 and 3, the assumption of fixed ends to the outer beams over-estimates their stiffnesses. These values are, therefore, halved to allow for the flexibility resulting from continuity. The various critical loading patterns to produce maxjmum stresses have to be considered. In general these loading patterns for the ultimate limit state are as shown in figure 3.2, except when there is also a cantilever span wh ich may have a beneficial minimum loading condition (I.OGk) -sec figure 7.2 1.

Analysis of the structure 0

if

aay

Half stiffness

Half stiffness

l a<;

of (2)

ge. ear

~

my Ha lf sli[fncss

Half sliffn ess

lefS

r or gid

(3)

cal '/

.are

'lltS

the

1-1 1, 11 2=Sto rey Heig hts

1,7,?7;, 1,7,»; 1,7, »? '7i ~ Figure 3.10

Figure 3.11 Substitu te rram es

Building rrame

Ill!>. ~ if

., a

in I I. its ;oiS

lmd •\'0

When considering the critical loading arrangements for a column. it is sometimes necessary to include the case of maximum moment and minimum possible ax ial load, in order to investigate the possibility of tension fa ilure caused by the bending.

( EXA MPL E 3.3

Analysis of a substitute frame The substitute rrnme shown in figure 3.12 is part of the complete frame in fi gure 3.10. The characteristic actions carried by the beams are permanent actions (including selfweight) G~ = 25 kN/m, and variable action, Qk = 10 kN/m, uniformly distributed along the beam . The analysis of the sub frame will be carried out by moment distributiou: thus the member stiffnesses and their relevant distribution factors are fi rs! required.

!he pan

be

K

Figure 3.12

M

L

E ,.,.;

"' A

B

c

D

E

Beam 60 X 300

ues

C! ....

:mg t-

be

"lll

=ial

'3so

6.0m

~ 300 Typical column

section

6.0m

~ I

I

Substitute fram e

~ 39

~~~

40 · Reinforced concrete design Stiffnesses, k

Beam I =

OJ x o.63 = 54 .

kAB

= kco =

12 Spans AS and CD

X

lo- 3 m4

5.4x iO J

6

.0

= 0.<) X

-1

)0

-

Span BC koc

=

5.4 x

w-3

4.0

= 1.35 x 10

3

Columns I

= 0.3 ~~· 353 =

1.07

X

10- J m4

Upper ku

=

1.07

X 10-J 3.5

0.31 X 10-J

Lower 3

k1

'

= ].()7 X I () = 0.27 X 10- J 4.0

ku + kL

= (0.3 1 I- 0.27) I0 ·J = 0.58 X

IQ- 3

Distribution foctors

ts A and D

I:k

= 1.4!; O.t)S = 0.61 1.4

0.9+0.58

D.F.Ao

= D.F.oc =

D.F.cols

= I.4S =

0.58

0.39

ts B and C

2:k = 0.9 +

I .35 -l 0.58 = 2.83 0.9 D.F.RA = D.P.co = . = 0.32 2 83

D.F'.lJc

1.35

= D.F.ca = 2.RJ

0.48

' 0.58 ( 0 D. i'.cols = . = ).2 2 83 The critical loading pa!lcrns for the ultimate limit state are identical to those for lhe continuous beam in example 3.2, and they arc illustrated in figure 3.5. The moment distribution for the first loading arrangement is shown in table 3.2. In the table, the distribution for each upper and lower column have been combined, since this simplifies the layout for the ca.lculations.

D'

~·

ia

Table 3.2

~-

Moment distribution tor the first loading case A

D.F.s Load kN

Cols. (L;M)

AB

0.39

0.61

BA

Cols. {LM)

BC

0.32

0.20

0.48

292

Bal.

M (kN m)

0.32

-

44.6

24.2

X

DC

Cols. (l:M)

0.61

0.39

292

-

.,.

48.5

+

-

146 32.3

146

+

+

24.2

44.6

X

I

I

89.1

56.9

16.2

-

-

+

+

+

-

-

-

+

+

6.3

9.9

22.0

13.8

33.0

33.0

13.8

22.0

9.9

6.3

+

+

-

16.5

16.5

5.0 6.9

11 .0

+

-

11.0

5.0

-

-

+

+

-

-

-

4.3

6.7

6.9

4.3

10.3

10.3

4.3

+

3.4

+

+

5.2

5.2 4.1

-

3.4

3.4

c.o. Bal.

0.20

20.2

+ 20.2

c.o. Bal.

0.48

-

+ 32.3

X

CD

48.5

-

89.1

+

Cols. (LM)

45.0

-

16.2

CB

45.0

+ 146

56.9

c.o.

D

135

+ 146

F.E.M. Bal.

c

8

3.4

-

-

+

-

-

1.3

2.1

2.8

1.7

4.1

i"

.;-

-

135.0

40.0

95.0

95.0

-

+

68.8

68.8 ---

I

+

+

6.7

43

-

-

I

+

1.7

2.8

2.1

1.3

-

+

-

-

40.0

135.0

68.8

68.8 )> ::l til

'-< ;:;;· "'

0

:roT ,...,. "' 2 n

,...,.

c ..,

ro

~::;;;;;t

I~

lq

42 . Reinforced concrete design The shearing forces and the maximum span moments can be calculated from the f01mulae of section 3.3.2. For the' first loading arrangement and span AB: Shear VAs

(MAs - MsA ) L

= -load 2

= 292.5 _ 2

( -68.8 + 135.0) 6.0

= 135 kN

l' nA = load - VAB

= 157 kN . moment, span AB = -v"s-1-1 M,,H MaJ<.unum = 292.5-

135

2w

=

1352 2 >< 48.75

68.8

= J 18kNm \1,\B

Distance from A, a3 =-;;;- =

135 _ 48 75

2.8 m

Figure 3.13 shows the bending moments in the beams for each loading pattern; figure 3.14 shows the shearing forces. These diagrams have been combined in figure 3.15 to give design envelopes for bending moments and shearing forces. A comparison of the design envelopes of figure 3.15 and figure 3.8 wiU emphasise the advantages of considering the concrete beam as part of a frame, not as a continuous beam as in example 3.2. Not only is the analysis of a subframe more precise, but many moments and shears in the beam arc smaller in magnitude. The moment in each column is given by ~

krol

Mcnl - LMcol X

"k

L- col

135

Figure 3.13 Beam bending-moment diagrams (kNm)

69

-v ~

( I)

J\{955

135

~~

99551

69

/1~1\

\7

A

118

118

147

(3)

79 114

(4)

~

147

102 80

6

~

A---;-&

67

~

\.1

79

"'---./

A

~ :;·:;

Analysis of the structure ·~ 43 Figure 3.14 Beam shearing-force diagrams (kN)

be (1)

135 97.5~

91 (2)

111

~

I"'-

[">. 133

(3) 92

n: lfl

(4)

133 147

147

Figure 3.15 Bending-moment and shearing-force envelopes

kNm

118

118

:2 Mcol table 3.2 gives

Thus. for the first loading arrangement and raking Column moment MAJ - 68.8 x

0.31 _ 0 58

= 37 kN m

0.27

= 32 kN

MAt:. = 68.8 X _ 0 58

0.31

m

MoK

= 40 x -0.58 = 21 kN

m

MoF

0.27 9 = 40 x - = I kN 0.58

m

This loading arrangement gives the maximum column moments, as plotted in figure 3.16.

44

~~

WReinforced concrete design

Figure 3.16 Column bending moments (kN m)

(

EXAMPLE 3.4

Analysis of a substitute frame for a column n1e substitute frame for this example, shown in fi gure 3.17, is taken from the building frame in figure 3. 10. The loading to cause maximum column moments is shown in the fi gure for Gk = 251cN/m and Qk = IOkN/rn. ~/,

Figure 3.17

___-..-

Substitute frame

E

~ l. lSG, <150Q. ~.35Gk~ ~

~A

=292.5 kN

f

~

=135 kN

B

~

~

C

~""

'"

E

<:!I

'"I

-

'

~ ,~ 6.0

Ill

.....

4.0 ll1

...

1

The stiffnesses of these are identical to those calculated in example 3.3, except that for lhis type of l'rame the beam stillnesses arc halved. Thus kAu

= 2I X 0.9 X 10 J = 0.45 X )()'"3

kBc

= 'I

upper column ku

2

X

1.35

X ) ()- ) =

= 0.3 1 x 10

().675

X

J()- 3

3

= 0.27 x 10 3 I> = (0.45 -1- o.675 -1- 0.3 1 + 0.27) x w-3 = 1.705 x 6 fixed-end moment MsA = 292.5 x ? = .L46 kN m )_ lower column kL

fixed-end moment MHc 18 16 kN m

Figure 3.18 Column moments

10- 3

= 135 x 1~ = 45 kN m

Column moments are 45 ) x -0.31- = l 8. kN m 1.705 0.27 ( 146 - 45) x - - 16 kN m 1.75 0

upper column Mu

= ( 146. -

lower column 1\ll~_

=

=

The column moments are illustrated in figure 3.1 !\. They should be compm·ed with the corresponding moments for the internal column in figure 3.1 6. )

l ---------------------------------------~

Analysis of the structure ln examples 3.3 and 3.4 the second moment of area of the beam was calculated as

bfl 3/ 12 a rectangular section for simplicity, but where ah in situ slab forms a flange to the beam, the second moment of area may be calculated for the T-scction or L-section.

3.4.2 Lateral loads on frames

_)

mg the

Lateral loads on a structure may be caused by wind pressures, by retained catth or by ;;eismic forces. A horizontal force should also be applied at each level of a structure resulting from u notional incl ination oft11e vertical representing imperfections. The value of this depends on building height and number of columns (EC2 clause 5.2). hut w ill typically be less than 1% of the vertical load at that level for a braced structure. This should be added to any wind loads at the ultimate limit state A n unbraced frame subjected to wind forcel:i must be analysed for all the vertical loading combinations described in section 3.2. 1. T he vertical-load ing analysis cun be carried out by the methods described previously . T he anal ysi s fo r the lateral loads ,hould be kept separate. The forces may be calculated by an elastic computer analysis or by a si mplified approx imate method. For preliminary design calculntions. and also only for med iu m-size regular structures, a simplified analysis may well be adequate. A suitable approximate anal ysis is the cantilever method. It assumes that:

1. points of con tra f lexure are located at· the mid-points of all colu mns 11ncl beams; and 2. the direct axial loads in the columns are in proportion to their distances from the centre of gravity of the frame. It is al so usual to assume that ull the column~ in a storey arc of equal cross-sectional area.

It should he emphasised that these approximate methods may give quite inaccurate results for irregular or high-rise structures. Application of thi s method is probably best Illustrated by an example. as follows. 3.

( EXA MPLE 3.5

Simplified analysis for lateral loads - cantilever method Figure 3. 19 shows a building frame subjected to a charucteri stic wind ncti on of 3.0 kN per metre height of the f rame. This action is assumed to be tnmsfcrrecl to the frame as a concentrated loud at each floor level as indicated in the fi gure. By inspection, there is tension in the two columns w the left and compression in the columns to the ri ght; and by assumption 2 the axial forces jn columns arc proporti onal to their distances rrom the centre line of the frame.

--

5.25 kN

E .....

z

.><

0

.,.; II "0

"' !a ~ ;;;

.,

10 .5!!:!._

4th

10.5~

3rd

11 .2~

2nd

12 .OkN

1st

-"

he

_)

6.0 kN ~

~~

6.0

7l.r

"',.;

--I... j

...C!

--r ' C! ....

'?;~

4.0

I

~

--r '

'l/:(//

6.0

'

Figure 3.19 Frame with lateral load

f1 45

46

~


Figure 3.20 Subframes at the roof and 4th floor

I f,= 0.54

F1= O.HkN

F,= 0.675

t .------.:-----r---+.. J--,--.....,.... J t

5.25

-

tJ

J!

---, s' -

H,= 1.70

H;= 0.93

N,=4.0P = 0 .54

N 1= 1.0P

It;= 0 .93 N = l.OP 3

= 0.135

=0.135

(a) Roor

0.54

0.135 1.70

0.93

t

r:s

0.135

1.70

0.54

0.93

l'

5.1

2.70

5.1

0.68

0.68

7.78 2.70

(b) 4 th Floor

Thus Axial force in exterior column : axial force in interior column

= 4.0P : L OP

The analysis of the frame continues by considering a section through the top-storey columns: the removal of the frame below this section gives the remainder shown in figure 3.20a. The forces in this subl'rame arc calculated as follows.

(a) Axial forces in the columns

s, 2: M, - 0. therefore 6.0 - P X 10.0 - 4P X 16.0 ()

Taking moments about point

5.25

X

1.75

+p X

and therefore

P - 0. 135 kN thus

N1

= - N4 = 4.0P = 0.54 kN

N2 = - N3 =: I .OP =: 0. J 35 kN (b) Vertical shearing forces F in the beams For each part of the sub f rame,

L F = 0, therefore

F 1 = N 1 = 0.54kN

F2 = Nr

+ N2 =

0.675 kN

(c) Horizontal shearing forces H in the columns Taking moments about the poi nts of comraflcxurc of each heam,

Hr X 1.75 - Nr x 3.0 H 1 = 0.93kN

=0

L: M = 0, therefore

Analysis of the structure :md (HI

H1

+ H2) 1.75 -

N1 X 8.0- N2 X 2.0

=0

= 1.70kN

The calculations of the equivalent forces for the fourth floor (figure 3.20b) follow a '1milar procedure. as follows. d) Axial forces in the columns

I: M, = 0, therefore 5.25(3 X 1.75) + 10.5 X 1.75 + P X 6.0 - P X 10.0- 4P X P = 0.675 kN

For the frame above section tr',

16.0 = 0

therefore

= 4.0P = 2.70

kN N2 = l.OP = 0.68 kN N1

(e) Beam shears F,

= 2.70 - 0.54 = 2.16kN 2.70 I 0.68 - 0.54 - 0.135

F2

= 2.705 kN (f) Column shears H1 X

l-11

1.75 t 0.93 X 1.75

(2.70 - 0.54)3.0

=0

= 2.78 k.N

1 H2 = 2( 10.5 + 5.25)- 2.7!\

= 5.1kN Values calculated for sections taken below the remaini ng floors are third noor N1 = 7.03 kN F1 = 4.33 kN ll1 - 4.64kN second floor N1 = 14. 14 kN /··1 7. 11 kN

N2 = 1.76kN F2 = 5.41kN H2 = 8.49kN Nz = 3.53 kN

= 6.6 1kN

H2 = 12. 14kN

first floor N 1 = 24.37 kN F, = l0.23kN HI = 8.74kN

Fz = l2.79kN H2 = 16.01 kN

Ht

F2

= 8.88kN

N2 = 6.09k.N

The bending moments in the beams and columns m their connections can be calculated from these results by the following formulae beams

Mu - F x! beam span

columns Me

= H x i storey height

;:;* ~::~

P 47

48

Reinforced concrete design 1.6

Figure 3.21

1.4

1.6

1 .6

3.0

1.6

4.9

8.9

4.9

8.1

14.9

8.1

13.2

24.3

13.2

17.5

32.0

Moments (kN m) and reactions (kN)

17.5 24.4

6. 1

6.1

24.4

External Column

Internal

Beams

Column

so that the roofs external connection Mu

= 0.54 X 2I X 60 = 1.6kN m

Me

= 0.93 X ;;1 X 3.:1- = 1.6 kN 111

As a check at each t. L: MR = 2:: Me. ThL: bL:nding moments due to characteristic wind londs in all the columns ancl beams of this structure are shown in fi gure 3.2 1.

3.5

Shear wall structures resisting horizontal loads

t\ reinforced concrete structure with shear walls is shown in fi gure 3.22. Shear walls are very effective in resisting horizontal loads such as P,. in the figu re which act in the direction of the plane of the walls. As the walls arc relatively thin they of!'er litl'le resistance to loads which arc perpendicular to their plane. The Aoor slabs which arc ed by the walls also act as rigid diaphragms which transfer and distribute the horizontal forces into the shear wa lls. The shear walls act us vertical cantilevers transferring the horizontal loads to the structural rounclations.

3.5.1

Symmetrical arrangement of walls

With a symmetrical an·angemcnt of walls as shown in figure 3.23 the horizontal load is distributed in proportion to the the relati ve stiffness k1 of each wall. The relative


~

i~:~ 49

Figure 3.22 Shear wall structure

-tiffncsscs arc given by the second moment of area of each wall about its major axis -uch that

k, ~ h X b3 here h is the thickness of the wall and b is the length of the wall. The force P; distributed into each wal l is then given by P,

)

f x

i,kk·

( EXA MPLE 3.6

Symmetrical arrangement of shear walls A structure with a symmetrical arrangement of shear walls is shown in figure 3.23. Calculate the proportion of the JOOkN horizontal load carried by each of the walls.

Figure 3.23

lre

Symmetrical arrangement of shear walls

:he

lle

ch

E

0 N

as

E

"'

1-

10m

_._

15m

'

50

~ Reinforced concrete design Relative stiffnesses: Walls A Walls B

kA

= 0.3 X 2()3 = 2400

kR = 0.2 X R3

= 346

I: k = 2(2400 + 346) = 5492 Force in each wall : 2400

kA

P.... = I;k

X

F = 5492

ks

P8 = - x F

l:k

X

100 = 43.7kN

346 =- x 100 = 6 .•3 kN 5492

Check 2(43.7, 6.3) = 100 kN = F

3.5.2 Unsymmetrical arrangement of walls With an unsymmetrical arrangement of shear walls as shown in figure 3.24 there will also be a torsional force on the structure about the centre of rotation in addition to the direct forces caused by the translatory movement. The calculation procedure for this case is:

1. Determine the location of the centre of rotation by taking moments of the wal l stiffnesses k about convenient axes. Such that

where k~ and k,. arc the stiiTncsses or the walls orientated in the x andy directions respectively. 2. Calculate rJ1e torsional moment M, on the group of shear walls as M,

=F x e

where e is the eccentricity of the horizontal force F ahout the centre of rotation. 3. Calcu late the force P, in each wall as the sum of the direct component Pc1 and tl1e torsional rotation component· Pr

P; = Pc~

+ Pr

kx k;r1 = F X I;k., ± M1 X I;(k;r; 2 )

where r; is the perpendicular distance between the axis of each wall and the centre of rotation.

(

EXAM PLE 3.7

Unsymmetrical layout of shear walls Determine the distribution of the I00 kN horizontal force F into the shear walls A, B, C. D and E as shown in figure 3.24. The relative stiffness of each shear wall is shown in the figure in tenus of multiples of k.


yt.

l1= 12.0m

1

Figure 3.24

20m

I

Centre of 1 rotation

A: ZOk

•' -

Unsymmetrical arrangement of shear walls

I

~ I'

'

E o\

'0

....

--- - ~-

E:5k

0: 5k

I

rf

C: 6k

J.'= 6.4m 32m

~Centre of

20m

20m

F = lOOkN

rotation

Taking moments

_

am I

= 20 + 5 + 5 = 30

L:;k,

L:;(kxx)

X =~=

for k.~

2()

X

about YY at wall A

0 +5 X 32 + 5 X 40 30

= 12.0 metres L_k,.= 6 H

10

Taking moments for k, about XX at wall C

l:(k,.y)

6 X 0 - 4 X 16

.)'---·-= l:k,. 10 = 6.4 metres The torsional moment /v/1 is !VI, = F X (20 - x)

100

X

{20 - 12)

= 800 kNm The remainder of these calculations are conveniently set out in wbular form: Wall

kx

A B

D E

20 0 0 5 5

6 0 0

E

30

10

c

ky

12 9.6 6.4 20 28

0 4

kr

J

pd

P,.

PI

240 38.4 38.4 100 140

2880 369 246 2000 3920

66.6 0 0 16.7 16.7

- 20.4 - 3.3 3.3 8.5 11.9

46.2 - 3.3 3.3 25.2 28.6

9415

100

As an example for wall A: PA

-

P,

+ Pr = F X

rkA k

M,

kArA

X

L (k;r; 2)

20 20 X 12 = 100 x 30 - soo x = 66.6 94 15

[f~

20.4

= 46.2 kN

0

100

!1';;

52

~~


Figure 3.25

Shear wall with openings

D

D D (a) Shear Wall

3.5.3

(b) Idealised Plane Figure

Shear walls with openings

Shear walls with openings can be idealised into equivalent plane frames as shown in figure 3.25. I n the plane frame the second moment or area lc of the columns is equivalent to that of the wall on either side of the Clpenings. T he second moment of area lb of the beams is equi valent to that part of the wall between the openings. The lengths of beam that extend beyond the openings as shown shaded in flgu re 3.25 are given a very large stiffnesses so that their second moment of area would be say

100/b. The equi valent plane frame would be analysed by computer with a plane f rame program.

3.5.4

Shear walls combined with structural frames

For simpl icity in the design or low or medium-height structures shear walls or a l ift shaft are usually considered to resist all of the horizontal load. With higher rise :-~ tructu res for reasons of stiffness and economy it often becomes necessaty to inc lude the combined action of' the shear walls and the structural frames in the design. A method of a n a l y~ing a structure with shear walls and structural frames as one equivalent linked-plane frame is illustrated by the example in figure 3.26. I n the actual structure shown in plan there arc fo ur frames of type A and two frames of type B which include shear walls. ln the linked frame shown in elevation the four type A frames arc lumped together into one frame whose member stiiTnesses arc multiplied by four. Similarly the two type B frames are lumped together into one frame whose member sti ffnesses arc doubled. T hese two equivalent frames arc then l inked together by beams pinned
1~ !:·~

Analysis of the struc ture .'.:: 53 A

B

A

B

A

A

Figure 3.26 Idealised link frame lor a structure with shear walls and structural frames

..

shear walls

t

l ateral Load

a) Plan of Structure p ins

s1 - I- s 2 - \..._b_-J

j.., ~-I-

s ~ ...1• sl ....1 very stiff beams

/

in is rca

..

shear walls

.25

say

me

lift

'?.~ 'J. ~

Ide

7?/?7-

4 N o, frames

::?,~::?,~

A~ 1

-

--

-

ll!e

leS

-

L ~nks

"?.~ 2 N o, frames B

7?/~

I

of large cross-sectional:rea pinned at their ends

(b) Elcva tfon or lin k-Frame Model

ur

!l'e

me .ed of

.he

whereas l:his would normally be of a secondary magni!llclc. To overcome th is the cross'>CCtional areas of nil the beams in the model may be increased say to 1000 m2 and this will virtually remove !he effects of axial shortenjng in the beams. In the computer output the member forces for type A frames would need to be divided by a factor or four and those for type 13 fra mes by a factor of two.

be :.0

3.6

Redistribution of moments

i!S.

be ID

~.

Some method of elastic analysis is generally used to calculate the forces in u concrete srmcturc, despite the fact that the strucmre docs not behave elastically near it!> ultimate load. The assumption of elastic behaviour is reasonably true for low stress level!>; but as a section approaches its ultimate moment of resistance, plastic deformation will occur.

54


Figure 3.27

Typical momentcurvature diagram

c

"' E ::E 0

Curvature

This is recognised in EC2, by allowing redistribution of the clastic moments subject to ccnain limitations. Reinforced concrete behaves in a manner midway between that or steel and concrete. The stressstrain curves for the two materials (figures 1.5 and 1.2) show the elastoplastic behaviour of steel and the plastic behaviour of' concrete. The latter will fail at a relatively small compressive strain. The exnct behaviour of a reinforced concrete section depends on the relative quantities nnd t·he iJ1diviclunl properties of the two materials. However, such a section mny be considered vi rtually elastic unti l the steel yields; and then plastic until the concrete fni ls in compres~ ion. Thus the plastic behaviour is limited by the concrete failure; or more specitically, the C()ncrcte fai lure limits the rotation that may tnke place at a section in bending. A typical momentcurvature diagram for a reinforced concrete member is shown in figure 3.27 Thus, in an indeterminate structure, once a beam section develops its ultimate momelll of resistance, M 0 , it then behaves as a plaMic hinge resisting a constant moment of that value. further loading must he taken by other parts of the structure. with the changes in moment elsewhere being just the same as if a real hinge eJtisred. Provided rotation of a hinge docs not cause crushing of the concrete. further hinges will be fonned until a mechanism is pro<.luced. This requirement is considered in more detai l in chapter 4.

(

EXAMPLE 3.8

Moment redistribution -single span fixed-end beam

The beam shown in figure 3.28 is subjected to an increasing uniformly distributed load: 2

Elaslic moment Elastic span moment

wL =12

wL2

In the case where the ultimate bending strengths are equal at the span and at the s, and where adequate rotation is possible, then the additional load w9 , which the member can sustain by plastic behaviour, can be found. At collapse wL2 12

IWu= -

wL2 + additional mid-span moment IIIH 24 where mR = (waL2 )/ 8 as for a simply ed beam with hinges at A and C. = -

1 Analysis of the structure w/unit length load

Elastic BMO MA=Mc = Mu

Additional moments diagram (Hinges at A and C) Collapse mechan ism :: [0

Elastic BMD (Collapse loads} Final Collapse BMD

?le.

~,:,tic

1\1

=3

.-

'' the load to cause IJ1c first pl astic hinge; thu s the beam may cnrry a load of ''ith redistribution.

fn,m the design point of view. the elastic bending-moment diagmm can be obtained • r.:quircd ultimate loading in the ordinary way. Some of theNe moments may then ·e1ruccd; hut thi s will necessitate increasing others to maintain the static equilibrium 'tructurc. Usually it is the maximum moments which arc reduced. so -w..ing in reinforcing steel and also reducing congestion at the columns. The emcnts for applying moment redistribution arc: l:qullihrium between internal and external forces must be mnintaincd, hence it is necessary to n.:calculate the span bending moments and the shear forces for the load ~••'e involved.

2. The cor11inuous beams or slabs are predom inately suhj t~c t to fkxure. 3. The rmio of adjacent sptms be in Lhe range of 0.5 to 2.

lid:

~. The column design moments must not be reduced.

T here arc other restriction s on the :.~mount of moment. redistribution in order to ensure uucti lity of the beams or slabs. T his entails limjtations on the grade of rei nforcing steel ..ncl or th e areas of tensile reinforcement and hence rhe depth or the n eutr:~l axis as described in Chapter Four -'Annlysis of the Section'.

ne he

( EX AMPLE 3.9 Moment redistribution In exnmplc 3.3. ligurc 3. 13, it is required to reduce the maximum upport moment of 147 kN m as much as possible. but without increasing Lhe span moment nbove the present max imum value of ll8 kN m. MRI\

=

Figure 3.28 Moment redistribution, one-span beam

55

56

Reinforced concrete design 147

Figure 3.29 Moments and shears after redistribution

114

(a) Original Moments (kN m) 140

WA1o8

67

~

80 102

4 LG 2 ~ / -;= ~

\

"'---../ 118

79

(b) Redistributed Moments (kN m) 1 34

92 (c) Shears (kN)

Figure 3.29a duplicates the origi nal bending-moment diagram (pan 3 of figure 3.13) of example 3.3 while figure 3.29b shows the redistributed moments, with the span moment set at l18kN m. The moment at B can be calculated. using a rcan·angement of equations 3.4 and 3.1. Thus

V,,B

J[(A'ln~.x - MAu)2wj

and

Mp,,, ~ (v.. .o- ;L)L I MAo 1

= 48.75 kN m, therefore V,., 11 / [(J 18+67) X 2 X 48.75 j = l 34kN 48.75 X 6.()) 6.0 - 67 = 140 kN m MBA= ( 134 2

For span AB,

w

and lloA

= 292.5 -

134

= 158.5 kN Reduction in MoA = 147 - 140

= 7kNm 7 X 100

= - 147 -- = 4.8 per cent

~

Analysis of the structure In order to ensure that the moments in the columns at t B arc not changed by the 'tribution, moment M 8 c must also be reduced by 7·kN m. Therefore

:'!C

'=~•

= 115 - 7 = 108 kN m hogging

Fnr the revised moments in BC:

l BC

= (l 08 -

4

1 co

F

80) + 195

=:

I05 kN

2

= 195 - 105 = 90 kN

r ~pan

BC: 105 2 x . - 108 2 48 75

.

= 5 kN m saggmg

Figure 3.29c shows the revised shearing-force diagram to accord with the moments. This example illustrates how, with redistribution

-~..!Nrihuted

the moments al a section of beam can be reduced without exceeding the maximum c.;,ign moments at other sections; " values of the column moments are not arrcclcd; and 3

3.13) pan ng a

the equilibrium between external loads and internal forces is maimaincd.

.,.! 57

CHAPTER

4

Analysis of the section CHAPTER INTRODUCTION

•·•··· •·· ···• ·· •·······•••·······••• •• •••·•••·•·•··•···· •••• •·••• ··•·• ···• •·•

A satisfactory and econom1c design of a concrete structure rarely depends on a complex theoretical analysis It is achieved more by deciding on a practical overall layout of the structure, careful attention to detail and sound constructional practice. Nevertheless the total des1gn of a structure does depend on the analys1s and design of the individual member sections. Wherever possible the analysis should be kept simple, yet it should be based on the observed and tested behaviour of reinforced concrete . The manipulation and juggling with equations should never be allowed to obscure the fundamental principles that unite the analysis. The three most important principles are

1. The stresses and strains are related by the material properties, Including the stressstrain curves for concrete and steel. 2. The distribution of strains must be compatible with the d1storted shape of the crosssection. 3. The resultant forces developed by the sect1on must balance the applied loads for static equilibrium. These principles are true Irrespective of how the stresses and stra1ns are distributed, or how the member is loaded, or whatever the shape of the cross-section This chapter describes and analyses the action of a member section under load. It derives the basic equations used in design and also those equations required for

--.

58

Analysis of the section

___,. the preparation of design charts. Emphasis has been placed mostly on the analysis a sociated with the ultimate limit state but the behaviour of the section withm the elastic range and the serviceability limit state has also been cons1dered Section 4.7 deals with the redistribution of the moments from an elastic analysis of the structure, and the effect it has on the equat1ons denved and the tiesign procedure. It should be noted that EC2 does not g1ve any explicit equations for the analysis or design of sections. The equations given in this chapter are developed from the principles of EC2 in a form comparable with the (Quations formerly given in BS 8110 .

.1

Stress-strain relations

n-term ~trcs~-strain curvc1-. for concrete and steel are presented in I:C2. These e~ arc in un idealised form wh1ch can be used in the anCIIysh. of memhcr :-.ecuons. .. 1. 1

Concrete

bl:ha"iour of '>tructural concrete (figure 4.1) i~ reprel>ented by a paruholic '>Ire~' m rclauon~hip. up to n strain -,·~ · from which point the 1-.tr:un increa-;e, '' hile the "' rcmam~ con'>tant. The ultimate design :.tress is gl\·cn hy 0.85f.l 1.5

- 0.567l~l rc the factor ol 0.1!5 allow~ for the dillcrence bet\\ecn the bcnd111g \trength and the 1 fer cru1-.h ing stn.:ngth of the concrete. and;'< = 1.5 ;, the w.uul partial \lllety factor 'le strength of concrete. The ultimate strain of feu~ = 0.0035 ~~typical for cla~'c' of CS0/60. Concrete cla:.sc:. < C50/60 will, un less otherwise stated. be

Figure 4.1 Parabolic-recta ngular stress -strain diagram for C011crete m compression

~

E

-

.z

Porabolic

0.851,.

e

,, 0.0020

0.0035 Strain

59

60


cono;idcred throughout this book a~ these are the classes mo~t commonly used in reinforced concrete construction. Also for concrete clas~cs higher than CS0/60 the delining propertie such a-, the ultimate \train . ,111 vary for each of the hjgher classes. Oel>ign equations for the higher cla\),CS of concrete can in general be obtained using similar procedures to those shO\\n in the tC\t with the relative properties and coefficients obtained from the Eurocodes.

4.1.2

Reinforcing steel

The representative !.hort-term design !>tre~s strain curve for reinforcement is given in tigure 4.2. The behaviour of the steel i~ identical in ten,ion and comprc!>sion, being linear in the clastic range up to the de1-ign yteld ~tress of f>kh, where jyk b the characteristic yield ~>tress nnd ), is the partial tnctQr of safely. Figure 4.2 Short-term design wess-stmin CUIVe for relnforcEC>ment

1,.

T. Tens1on and

compression

eE z

200kN/mm1

Stra1n

\\ 1thm the cla:-.tic range. the Stres\

rclauon~hip hct\\~Cn

elaqlc mouuiU\ ,.. stralll

= £, 'iO

that the uel>ign yield E)

Mraut

the Mrel>' and strain is (4.1 )

i1>

= (l~k);r.· ~' ~~,

ulthe ultim:tte limit for.f),~ - 500N/mm 2 ~y = 500/( 1.15

X

200

X

101)

= 0.00217 It !.houlu be noted that EC:! permits lhc u~c of an ulternutive design strc!>s-strain curve to that shown in figure 4.2 with an inclined top branch and the maximum strain limited to a value which is dependent on the clth:-. of reinforcement. However the more commonly used curve shown in figure 4.2 \\ill be U!.Cd 111 th1s chapter and throughout the text.

4.2

Distribution of strains and stresses across a section in bending

The theol) of bending for reinforced concrete assumes that the concrete will crack in the regions of tcnllilc <;train~ and that. after cracking, all the tension is carried by the

Analysis of the section in the

-

as.~

if- P- JFs- 0.8x

E« ' - -

,_•~ ._,_;_d~

•

c

Figure 4.3 Section with strain diagram and stress blocks

n':utral axts

,.._

A,

• • mtoon

Strains

(a)

(b)

(c)

triangular

rectangular paraboloc

cquovalent rectangular

Stress blocks

ment. It i-; also ns~umed thar plane ~ections of n ~tructural member remuin atter ~trnining, so thnt ncros~ the section thl.!re must be a linear distribution of rc l3 .~ hows the cross-section of a member subjected w bending, and the t struin diagram. together with three different types ol stress distribution in the triangular ~tress tlistribution applies when the stre<.sc~> 01rc very nearly ' ntonalto the strain~. which generally (lt;curs ntthc loading h:\cl!<. cncour11crcd e \Hirl,ing condltiOn\ and is, therefore. used ut the scrvicenhJiitv limit ~tate. ed.tngular parabolic \trcs« hlocli. represents the dJstrihution at failure when the 1 essi\c 'trams are\\ ithin the plastic range. and it "associated '~ith the deS~gn I rt ultunatc limit state. C impht1ed altcrnall\e to the rectangulartx_,hc dl'trihuuon - I

:ion

mere ts cornpatihtht) of strain<. bet\\cen the reinforcement and the adjacent e. the ~tecl \truins ~ ·t 111 ten~ion und ...... tn compresston cun he determined from n dragram. The relntlonshtp). het ween the depth of neutrul u\1!-. ( ') and the m concrete \trurn (~
u'

(\ d') .\

61

(4.J)

1' the effective depth of the beam and d' is the depth olthe cumpre~:-.mn ment. n.., determined the \train,, we can evaluate the streS!>es in the reinforcement from ,_,train curve of figure 4.2, together wrth the equations de\ cloped 111

- d

12 '\I\ of u secuon with known steel strains, the depth of the neutral a \i\ can be r d b) rearranging equation 4.2 ac;

tl (4A )

62


At the ultimate limit !.tate the maximum comprel!sive strain in the concrete is taken as feu:!

= 0.0035 for concrete clas-.

~

C50/60

For higher classes of concrete reference should be made to EC2 Table 3.1 - Strength and deformation characteristics for concrete. For \tecl ''ilh.f;t = SOON/mm1 the y1eld strain from section 4.1.2 is f> = 0.002 17. Inserting these values for .:cu~ and :> mto equation 4.4:

\' =

d

0.00217 = 0 ·611d 1

+ 0.0035

Hence. to X~

en~ure

yielding of the tcn!.ion ..,tccl at the ultimate limit state:

0.6J7c/

At the ultimate limit state it JS important that member sections in Aexure should be ducti le and that failure should occur with the gradual yielding of the tension steel and not hy a sudden catastrophic compression failure of the concrete. Also, yielding of the reinforcement enables the formation of p l n~tic hinges \O that rt!distrihution of maximum moments can occur. resulting in a safer and more economical ~trut:ture. To en .. ure rotation of the plastic hinges with ~ufficient yielding of the ten!'ion ... tecl and also to allow for other factors such a~ the strain h:.mlcning of the steel. EC'2 limit~ the depth of neutral axis to 0.4Sd for concrdc da~., CS0/60. Thi' 1:. the hmiting ma\imum \aluc for 1 gl\cn b} I.C2 w1th no redistrihut1on applied to the momcntl> calculated b) an cla\liC anal):-1' ot the c;tructure. a.c; described in Chapter 3. When moment redi,tnbuuon '' apphcd these maxunum values of .t arc reduced a:. described m Section 4 7 The UK Annex 10 EC2 can gl\·e different limiting \alucs for 1. The EC2 value of \ OA5d is within the Aunex·~ rcqu1red limit' and 11 ensures thut a grudual tcn~ion fuilure of the steel occur:, at the ultimate limit \late. and not wdtlcn briulc fuilure of the concrete in compression.

4.3

Bending and the equivalent rectangular stress block

For most reinforced concrete ~tructure~ it is u~ua l to commence the design for the conditiom at the ult i mr~te limit state, fo llowed hy check~> to ensure that the structure i~ adequate for the serv1ceability limit state with(lut excessive dcncction or cracking of the concrete. For this reason the analy\i~ in thi~> chapter will lirst consider the ~implitied rectangular stress block wh1ch can he U\Cd for thc design at the ultimate limit state. The rectangular stress block us shm\n in ligurc 4.4 may be used in preference to the more rigorous rectangular-parabolic strc" block. This \imphfied stress distribution will fac1htate the analysi' and provide more manageable de.,ign equations. in particular \\hen dealing \\ith non-rectangular cro!>s-scctions or when undertaking hand calculation~.

It can be \ecn from figure 4.~ that the stress block doe" not extend to the neutral axis of the section but has a depth~ = 0.8.\. Th1s will result Ill the centroid of the stress block being sf2 OAOx from the top edge of the ~ection, which i~ very ncar!) the same location a' for the more precise rectangular- parabolic stre~., hhx:k. Abo the areas of the


iaten as

0.0035

b

I

'

~uax1

217.

" f ---J

A.

• •

0.85f,Jy, = 0.567("

l..o

_l

-

S=0.8r-.~ ---_:"

Figure 4.4 Singly reinforced section w1th rectangular stre55 block

_L s/2 j L:

l,d

,_ F,

Section

Stress Block

'res of slre~s block arc approximately equal (see 11ection 4.9). Thu~ the momen1 of ,t.tnce of the section will be simi lar using calculations based on either of the two ,, hlods. -he tle11ign cquntions derived in sections 4.4 to 4.6 arc for tcro redistribULion of ncnt!>. When moment redistribution is uppl ied, reference ~hou l d be mudc to 1011 4. 7 which uc,cribe~ how to modify the design equmions.

uld be I and . .,f the mum eo~ure

'o to rpth of

4.4

Singly reinforced rectangular section in bending at the ultimate limit state

4 4.1 Design equations for bending

lfDhcd lkd in I arc

a.:

te of

Bendang of the \ection will mduce a resultant tcn,ilc force F, 1 m thc reinforcmg \lecl, 31ld a resultant compre"t''e force Ill the concrete /\, which act' thmugh the centrotd of the- effective arca nf concrete in compres~ion. a!> 1-hnwn in figure -1...1. f-ur eqtullbnum. the ultimate destgn moment. M. mu't he balanced by the moment of ,t,tance of the 'ection so thai

lerl'\1011

1-.<-:.

H

t the

F,,:

(-1.5)

hcrc ;: the luver arm bet ween the resultant forces F,( anti /· , 1 Mres~

f _.

r the ore is

1lthe "'tfieu .Je. 11he \\ill ... ular hand axic; lock -arne

, the

x areu of uction

o.567J~~ x JJ.1

)lock .md : '>O

1hat

d

s/2

(4.6)

sub~tiluting

in equation 4.5

M Jnd replacing

1

Jrom equation 4.6 gives

M - 1.13-l.fclh(d

;::);::

(4.7}

Rearranging und wbslttuting K = M fbd~f.l: (:/d)'- (:./d)

Soh ing

Lhi~

K 1.1 34

0

quadratic equation:

t~[o.s + j(o.2s- K/ 1.134))

63

(-1.8)*

64

K


M bd1 f..

0.05

0.06

0.07

0.08

0.09

0.10

0.11

0.12

0.13

0.14

0.15

0.16

0.167

0 954

0.945

0.934

0.924

0.913

0.902

0.891

0.880

0.868

0.856

0.843

0.830

0.820

1.00

Figure 4.5 Lever-~rm

maxomum value of z/d accordong to the Concose Code and previous UK practoce

curve

0.95

..,';:; " ~

Compressoon reonforcement required (al Mb•t)

0.90

i1

0.85

0.82 0

010

0.05 K

0 lS 0 167

1

M/bd /l,

The percentage values on the K axos mMk the limits for stngly reonforc.ed sections woth moment redtstrobutoon applted (see Sectoon 4.7 dnd Table 4.2)

tn equation 4.5

F,,

= V~ "' 1 \

"ith ~ ,

I 15

0.87J; ..A, lienee

(4.9) 4

Equations 4.8 <~nd 4.9 can he u~cd to de~ign rhe area of tension reinforcement in ~ singly reinforced concrete section to resi~t an ultimate moment. ;II. Equation 4.R fm the lever arm :can also he used 10 ~ct up a table and drnw o lever· arm curve as shown in ligure 4.5. This curve may h~.: U\CULO determine the lever :trm • • instead nf solving equation 4.8. The lower limi t of 'l. 0.82d in llgurc 4.5 oct:ur~ when the depth of the neutral ax11 equals 0.45d. This is the max imum value allowcc.J hy EC2 foro singly reinforced seclim wirh concrete class less than or equal to ('50/60 in order to provide a ductile section thu will have a gradual ten~ion type failure as already described in section 4.2.

=

4.4.2 The balanced section The concrete ~ection "irh the depth of neutral axi~ at the spectficd m ted reach their ulumare -.rr;un~ at the same time. This occurs a the ma-
= 0.45d


0e depth of the stres:. block is

0.1 67 • 00820

OoR\h,J)

= 008 X 0.45d = 0036d

The force in the concrete stress block is

F ,,

= 00567/.1. x b~ -

00204/c;;bd

For cqUJhbnum Lhe force m the concrete Fccbal mu~t he balanced by the force F,rt>al in

So that

'reel.

Ool:l7.1)~A,".ll -

...,,

F"b.'' = Oo2

1'hc cfore \ o.tl

= 0.234f~k [)(/ /})~

that

23.4/:~ /vk

per cent

u:h is the steel percentage for a balanced section which ~hou l u not be excccucd for a ..:ulc ~ingly reinforced secuono fhuo.,, for example. with f~k 25 N/ mm2 anti f..~ 500 N/mm ~

=

25 J()(JA,"" 1 - '>J • .< bd - -· o"t 500

nt :

"3 04 x ~

~ 500

1.17 1)er cent

ullunnte moment of resil'-tancc of the balanced ~cc1 ion b M ...,1 1

ti

\( 2 - 0o8:!c/

h'tituting for \/1\;11

Fn~,, 1 :hJI where

r ..h..l

anti ::

1

Ool6 7J~~..bd

(4010)

nd 1/J ~hd

(4.9)*

0.167

K NI

.

ment in u

a lever.:r arm. ::

• rral axis eJ ...ection ~uon that

M,t

\\hen the de~1gn moment M11 1~ such thm 0 b , Kbo•l 00167 then the sc~.:ltOn cannot fcl.. (1• 'mgly reinforced and compression reinforcing steel i:-. requino:tf in the compn.:s,ion nl' of the ~ection. Thi~ is the limiting value of K Oo l 67 mar~ctl on the horiz.ontal a.' ' of the lever arm curve l'hown in figure 4.5 .

,. EXA MPLE 4.1

Design of a singly reinforced rectangular section

Jeplh of <;tate the ;ecur' at on with

':'"he ultimate dc~ign moment 10 be resi sred by the sec1ion in figure 4.6 i~ I R5 kN mo Determine the area of ten:>ion remforcement (t\,) required given the charactcnMic 11aterial ~uocngths are fu = 500 :-..tmm1 and f., = 25 N/mm'

"

M

A,

• •

hd'J~ ~

185 :!6()

10"

4401

X

25

00147 < 00 167

tncrefore comprc~~ion :.teel is not required.

Figure 4.6 Design example - singly reinforced section

6

66


Lever arm:

~ = d{ 0.5

T

/

(

0.25-

).~14)}

=-wo{o.5~ /(o.:?5 O.l·H)} 1.134 = 373mm cor allcmalively. the value ol

~

= f.,d

he obtained from the lever-arm diagram,

figure 4.5.)

M

A,== 0.87}~~ :: 185

X

106

= 0.87 X 500 X 373 I 140 mrn 2

Analysis equations for a singly reinforced section 'I he following equations may be used to calculate the moment of resistance of a given

:-.cction \\ ith a kno'' n area of \lee! retnforcement. For equilibrium of the compres-.tve force 111 the concrete and the tensile force in the \tcl'l tn figure 4.4: or

= 0.87J~vl Therefore depth or ~lfC).S block. ·~ 0.567J~~b

~

0.87/yk/\,

(4. 11 )

0. 567~~b

and x = s/0.80 Therefore the moment

or resi\lalll:e of the section

i~

P,1 X::

M

-

0.8~~vl,(d -

.

=- o.87}yiA

(

.1/ 2)

O.!l7{1 ~t\,) d - u3.iJ~~..b

(4.12)

The~e

equations assume the tension reinforcement ha'> >ielded. which will be the case if ca~c. the problem would require solving by trying -.uccessh·e value' of..\ unttl

.\ < 0.617d. If thi., i-. not the

with the Meel Mrains and hence '>tresses being determtncd from equations 4.2 and 4.1. to be u\cd in equation 4.12 instead of 0.87f~k·

Analysis of the section EXA MPLE 4.2

Analysis of a singly reinforced rectangular section in bending Determine the uJlimatc moment of resistance of the cm'>s-~ecuon ~ho\\n in fiuure 4.7 ~I\ en that the chamctcri!>tic strengths are i)l = 500 Nlmm1 for the rcinforcen;ent and :!5 N/mm2 for the concrete. 0.567f,

agram,

Figure 4.7

- ,0

- ---

:;:;

•••

_____)

Analysis example - singly reinforced sect1on neutral axi~

A, .. t470 mm1

For equilibrium of the compressive and

tcn~ite

forces on the

~cct ion

Frc - /·,1 t

a gi\'en

tree m the

0.567/..lb.l' O.X7/;kA 1 0.567 < :!5 3()() X I= 0.87

X

5(X)

X

1470

therefore 150mm .1110 .1

t4.ll)

.1/ 0.8 15CJ; O.X lHH mrn

fhi~ value of .1 is l c~s than the value of 0.6l7d derived froml>Cction 4.2. and therefore the steel has yielded and /,1 = O.X7/y~ a~ a~"umed. Moment ol' re1>istance of the section is

0.~7/yki\,(d

,1'/2)

O.H7 x 500 x 1470{520- 150/2) x lO

6

284 kN m

-l.J2)

case if trying

.ru -l.l, to

4.5

Rectangular section in bending with compression reinforcement at the ultimate limit state

(a) Derivation of basic equations It -.houhJ be noted that the equauons in thb !-.Cc.:tion ha\'e been deri\ed for the case ot zero moment rcdiMrihution. When this is not the ca"e, reference should he made to 'ection 4.7 '' hich deal' with the effect of moment rcdbtrihut1on

68


Figure 4.8 Section w1Lh compress1on remforcemenl

~

0.567(,,

0 0035

b

..... 1

.--- ---,_j_d

1 X~ 0 45d

• A,' •

neutral ax1s

d

.

lbll

A, .

Section

Strains

Stress Block

from the ~ection deal ing with the analysis uf u 1>ingly reinforced section nnd for concrete class not grcall:r than CS0/60 when M

> 0.167fckbd~

the design ultimate moment exceeds the moment of' re~il>turu.:c of' the concrete(Mbal) and therefore compression reinforcement i~ required. For 1his condition the depth of neutral axis, 1 .... 0.45c/. the maximum value allowed hy the code in order to en&ure a tension failure with a ductile ~cc1ion. Therdorc

::t-111 = d- !it-aJ/'2 = d O.Rlh:ll/2

= d - O.t{ = 0.8:!d

0.45cl,2

For equilibrium of the section in ligurc 4.K

~o

that Wtth the reinforcement at yteld O.R?f)~l\~

= 0.567j~~bs

O. K~f~~~~:

or with

0.8

S

X

0.!:\7}~1.A,

0.45d

0.36d

(4. 13)

0.204fdbd 1- 0.8~/ykA~

and taking moments about the centroid of the tension l.lecl.

M - F.c x ;:""' I F\<;(d - d') = 0.204/.kbd X 0.82c/ + 0.87/;kA:(d - d')

= 0.167fckh + 0.87}y~,A' (d- d')

(4.14)

rrom cqunuon 4.14 M- O.l67hbd1 A, = O.R7J;dd- d' ) I

(4. 15)*

Muluplying both ;,ide;, or equation 4.13 by : A

'

=

O.l67fdhd1

0.'07};1.

\\ ith ;:,.,

X

= 0.82d.

.:ba1

.... 1\

I

'

=0.!!2d and rearranging gi vel> (-U6)*

Analysis of the section .. areas of compression steel. A~. and tension l>tecl, A,, can he calculated from 5 and 4.16. 1 1g Arut 0.167 and K = M I bd7cl into these equations would convert

(4.17 )* (4.18)* a }sts it hru, been assumed that the compressiOn steel ha' yielded so that the 'J ... = 0.87}yl.· From the proportions of the strain distribution diagram: 0.0035

and tor

(4. 19)

r ) and neutr.d

a ren ...ion 0.0035

=

\\ith };1. 500 N/mm~. the steel strain l l the compresston o,teel () 002 17 <, 0.38

Therefore for

=0...15tl

() 171

-+.13)

= ~) = 0.00217.

(4.20)•

0.0035 ~

"'

(4.21 )

rnttu or d' / d for the yielding of other grades of steel can be determined by u. . ing tclu ~train in equmion 4.19. but for values of ]yl. les& thun 500N/mrn'. the <. tton or equation 4.2 1 will provide an adequate safe check. II I' cl .. 0. 17 1. then it is necessary to calculate the strain f ,~ from equation 4. 19 and determine J~~ l'rom /:', X • 'c

200000 "' -l.14)

- 15

\,tlue of stress for the compressive 'tccl mu~t then he u~cd in the denominator of ton 4.15 in place of 0.87/yL in order to calcu late the area A: of compression steel. Tbe area of tension steel is calculated from a modified equation 4.16 such that

~

A

- 16

= 0.167/..kbc/~ -A 0.87/yL;J,at '

1

/,..;

0.87Al

ne above equauon' apply for the calo.e \\here the concrete clu"' "' Ic-.-. than or equal ( 50/60. l·or concrete cla!>ses greater than CS0/60 ~imilar equation ... \\llh different 'tam~. can be derived based on the EC2 requtrement for the-.c classes. The constant11 r concretes up to clalo.s CS0/60 arc tabulated in table 4.1.

69

70

Reinforced concrete design Table 4.1

Limiting constant values Concrete class~ CS0/60

Limiting xt...1 d Maximum Zbaf Ktx~l - limiting K Limiting d' d

0.45 0.82d

0.167 0.171 23.41, 0 , fv<

Maximum percentage steel area 1 OOAoa1, bd

(b) Design charts The equations for the design charts arc obtained by taking moments about the neutral axis. Thus

M = 0.567fck0.8x(.x- O.Sx/2) +J~cA~(,I

d') + f,.A,(d- .1)

This equation and 4.13 may be written in the form

A,

.

A~

Y

f.,, bd - 0.45·+fd d t .f-.: hd \-~

M

'

bl l-

= 0.454/,~ -,, (I t-

d') d

0.40)

A,( (~)

f.,, bd I

For ~pccificd ratios of A: / lu/ . .1jd and d' / d. the two non-dimcn:-.ional cquations can be '>olvcd to give \aluc~ for A j btl and M ' /}{/' \o that a ~ct of dc.,1gn charts such as the one 'hm' n in figure 4.9 ma> be plotted. Before 1hc equation' can be ~olved. 1he steel stresse~ and./~ f11U'1,l be calculmed for each value of 1/ d. Thi~ i!> achieved by first determining the rclc\ ant strain:- from 1he strain dwgram (or by applyu1g cquauons 4.2 and 4.3) and 1hen by e\aluating the stresse:-, from the Mres., ·Slram curve of figure 4.2. Values of t:/d below 0.45 will apply when mome111s are rcdi~tnbllled. ltl>hould be noted that EC2 does not give design charts for bending. Hence although 1t 1~ po'~1ble to derive cham; a<. indicated. it may be Simpler to u~e the equations derived earlier 111 thi~ chapter or simple computer program~.

r.

Figure 4.9 Typical de~lgn chart ror doubly relnrorced bec~ms

12

4.0 2.0 15

10

1.0 0.5

4

2

0

0.5

10

1.S

lOOAJbd

2.0

25

3.0

3.5


71

I

( EXA MPLE 4 . 3

Design of a rectangular section with compression reinforcement 'lO moment redistribution) "i'le ~ec lion !thown in figure 4.10 is to resist an ultimate design momcm of 2g5kNm. The c racterislic material strengths are fyk = 500 N/mm 2 and };:~ = 25 N/mm 2. Determine e areas of reinforcement required. M bd~fck

285 ( l 6 260 X 440~ X 25 - ·--

A,

• •

> 0.167

aeutral

~refore compre~sion

d'!r/=50/440

as

111 ~.:quation

Figure 4.10 Design example with compression reinforcemen l,

steel is required

0. 11 <... 0.17 1

no moment redistribution

4.21 and the compres:.ion

!.~eel

will have y1elded.

C mpre 'l>ion :.tecl: (K - K~>•l }j;,bd' L

0.87.1)dd d') (0.226 0.167)25 ) 260 X 4402 =.;.._~~:--0.87 X 500(440 5())

n be one

=438mm2

e~~e~~

0111£ ~

.1nd xld

Kt...tf~kbtf 2

0.87./)l~t~.~l

2Joe~

' a~ nplc

0.167

X

25

I

+-A, X

260

-W()2

= 0.87 500(0.82 4.40) . . . 438 = 1339 + 438 - 1777 mm~ )<

r~-----------------------------------------------------------------------------.,

I

EX AMPLE 4 .4

Analysis of a doubly reinforced rectang ular section Determine the ultimate momem ol resistance of the cross-section shown in figure 4.1I that the characteristiC \trengths are f,k = 500 N/mm 2 for the reinforcement and J, = 25 N/mm 2 for the concrete. For equilibrium of the tcn~ i le and compressive forces on the .o,ection: ,_!IV en

F,1 =

F cc

\~suming

+ F,_

initially that the steel

0.87/ykAs

= 0.567fcL/H

~tresses

O.X7})LA~

fs1 and fs,

arc the design yield values, then

72


Figure 4.11

b~

I•

Analysis example, doubly reinforced section

t

280

0.5671,,

r ..

.. I

•A."= 628•

'

F,.

d =50

s = 0.8~

£ .,..

'

..,

-

.4.~2410

• •

F.,

Section

Stress Block

therefore 0.87/vk(/\,- A~)

.~

0.567f~kb

=

().H7 x 500(2410 628} 0.567 X 25 X 2SO 195 mm

s/0.8 = 244 mm 1/d - 244/510 0.48 < 0.617 1 -

!>O

(~ee ~ection

4.::!)

the tension steel will haw }icldcd. Also d' j.1

= 50/ 225

0.22

<..

0.38 (~cc equation 4.20)

so the compres<;ion steel \\Ill abo have ) tcltled. as Taking moments about the tcn-.iun steel M

= F,,(cl -

~ 2)

r "'( d

d' )

0.567f,kb:.(tl - 1/ 2) ~ O. H1j>~A~ (d ·0.567 x 25 x 280 x 195(510

= 319 -

124

as~umcd.

d')

195/ 2)+0.!!7

>

'i00

:><.

620(510

50)

X

lO

11

443 !...N m

If the depth of neutral axis was such that the comprc~>sive or tensile yielded. it would have been nccc1.sary to try successive vnluc:, of .r unlil

~teel

had not

F" = Fe,· + F.., balances with the &teel 4.1. The btcel

~;tresses

~trains

and

).Lrcs~es

being c.:ulculaled from C4m11ions 4.2, 4J and

at halancc would !hen be used to calculate lhe moment of

resistance.

4.6

Flanged section in bending at the ultimate limit state

T-sections and L-~ections which have their flanges in comprc~~•on can botl1 be designed or analy:-ed in a similar manner. and the equations \\ hich arc derived can be applied to either type or cross-section. As the Range~ gene rail} pro\ ide a large compressiYe area. it j.., usually unnecessary to cons1der the ca~c where compressiOn steel i:-, required; if it .should be required. the design would be ba~cd on the principles derived in ..,ection 4.6.3


73

lbc '1ngly reinforced section it is necessary to consider two conditions: --e'~

block lies within the compression tlonge, and .:'' block extends beiO\\ the flange.

Flanged section - the depth of the stress block lies within the :ge s hr (figure 4.12) ath of stress block. the heam can be considered as an equivalent rectangular 1 llreadth bt equal to the flange width. Thi~ is because the non-rectangular belm the neutral axb i' in tcns10n and b. therefore, con'>idered to be cracked :U\c. rhus K = M /btd~f..~ can be calculated and the lever arm determ1ncd from a m curve of figure 4.5 or equation 4.H. The relation between the lever arm. ;:. \.of the neutral axis is given by d

I

2

- d- ::) O..S671,,

"1 •

, - --n-eu-tr_a_la-xi-5-~ X

Figure 4. 12

1r-1 -~ ·8rx - "F.,

T-~cctlon,

-r· r --S

stress block with1n the rldnge, s h1

s/2

l

, F.,

10 Section

Stress Block

e" than the flange thtd..nes-; Chr). the we~s block ....._.._..___...... 1nd the area of retnforccment is given hy

doc~

lie wtthtn the nange as

M ll S~fvk:.

de,ign of aT-section beam is described further in section 7.2.3 with a worked e.

It

E.

MP LE 4.5

~lysis

of a flanged section

n nc the ultimate moment of resistance of the T-section ~hown in figure 4.l3. dl.tractcri<.tie material ~trength~ arc [ 11. - 5()() N/mm1 and /d -= :!5 N/mm1. '\.5 ume initially that the Mrc~~ block depth lies wi thin the Oang~: and the ~:e mcnl is strained to the yield. ~>O thal /.11 0.87 }yk·

74

Reinforced concrete design 0.567f,,

b, = 800

Figure 4.13 Analysis example or a

T-section, s < llf

~ '

---~~ral~i~- -

- • • •

h1 =150

r-t-=fF-

_j

A,= 1470 mm<

F,.

Section

s/ 2

.

Stress Block

For equilibrium of the \ecuon

Fe,

= F,,

therefore

and

~olving

.\ -

for the depth ol stress

h l o~.;k

0.87 X 5()(} X 1470 0.567 X 25 X 8()()

= 56mm < 111 .\ = 1/ 0.8

150mm

70mm Hence the stress blocl. does lie within the Aangc ami with thi' depth of neutral axi~ the t->lcel wil l have yielued a~ ns:-.umec.J. Lever arm: ~

s/ 2 - 420 - 56/ 2 '392 nun d

Ta~ing moment~

M

about the centmid of the reinforcement the moment of resiMance il>

F" x :.:

;::: 0.567}~kbrs:.: 0.567

X

25

X

8()()

X

56

X

392

X

10

1 '

249 kN m If in the analysio; it had been found that s > h1 • then the procedure would have been :.unilar to that in example 4.7.

4.6.2 Flanged section -the depth of the stress block extends below the flange, s > h, For the design of a nanged section, the procedure dcscribt.:d in section 4.6.1 wil l check if the depth of the stress block extends below the nangc. An alternative procedure is to calculate the moment of resi tance. M1, of the section with s = lt1, the depth of the


75

~e ~h

(see equation 4.22 of example 4.6 following). Hence if the de~ign moment. Md. is that

\/d > Mr '1

the <;tress block must extend

beiO\\o-

the flange. and

ltr

ca\c the design can be carried out b) either:

ht\

w.. ing an exact method to determine the depth of the neutral axis. as in example 4.6

a

or for the conservative condition of x = 0.45d. which ;, the maximum value of r for a ~o. ing l y reinforced section and concrete class ~ C':'i0/60.

tle~igning

r EX AMPLE 4 .6

Design of a flanged section with the depth of the stress block below the flange ~I-tt:

T-section heam )hown in figure 4.14 is required to rcsi~t an ultimate design of IXO kN m. The churacteri~tic material strengths arc .f>l :'iOO N/mm 2 and 2 25 N/mm • Calcu late the area of reinforcement required.

~1 ment

1

Figure 4.14 Design example of a T·sewon

0.5671,,

'the

j

I

h,. 100

'

X

axo}

A

• •

·r-

s I

l

· -~

'""

j

lo

ll

F,.

l~b,. 20'L.

Section

Stress Block

In figure 4. 14 Fe~

is the l'orcc developed in the flange

I c"' is the force developed in the area of web in comprcsl.ion

been

_)

below

. eck if -e

IS

to

f the

\lomcnt of

re~istance.

Mr. of the flange is

Ml

F.~ X ;:1

Jfr

0.567}ckbl ltr(d

ur

0.567

25

= 170 I..N m <

X

4()()

111 / 2) X

100(350- 100/ 2) X

(4.22)"' JO -h

180 k~ m. the design moment

Therefore, the :..tress blod. mu~t extend belo'' the flange . It is now nece~saf) to determine the depth. s,. . of the web in compres,ion. where

h.

76


For equilibrium: Applied moment IHO

/·~r x

= 170

..L

::1 -'- F," x

~!

0.567f.:~ b"s"

- 170-0.567

X

25

;:2

v

200l., (250- 1" / 2)

X

170 + 2835sw(250

~w /2)

X

I()

6

10- h

X

Thi 'i equation can be rearranged into ~ .. 2 - soo~.. t

1.os

< 10

1

=o

Solving this quadratic equauon

v.. - 15 mm ~o

that Lhc depth of neutral axis

(/tr + sw)/0.8

X

= I.Wmm

(100

j

15)/0.8

0.4ld

0.45d compn!~~•on remforcement For the equilibrium of the section

A\ \

i~

not required.

F,1 - F,1 I Few

or 0.87/>kA, - 0.567j;.b,h

+ 0.567 /.:~ b,.. s,.

0.87 < 5(X) X tl ,

X

0 567

25(400

X

100 -r 200

15) = 610

X

101

Then:fore 610

X

10 1

0.87

X

5()()

1402mm~

(

EXAMPLE 4.7

Analysis of a flanged section Detcnnmc the ultimate moment of re~istance of the T-bcam section 'ho\\ n in figure 4.15 g1vcn h~ - 500 N/mm 2 and fck = 25 N/mm 2. rhc compressive force in the nange is F.r

0.567 f.:kb, It,

o 567 x 2s x 450 Then F·,1

tcn~ile

ISO x

Io-3

957 kN

force in the remlorcmg <;teel, a-.suming it has yielded. i'

0.87/ykA, 0.87 x 500

X

2592 x 10- 3 = I 128 kN

Analysis of the section 0.567fc,

/). =450

Figure 4.15 Analysis example of a T-section, s > /If

•ht =150 c

Section

cfore 1-', 1

f "

> F"

Stress Block

so that s > lit and the force in the web is

0.567/dbw{.V - fir) 0.567

X

25

X

3QO(s - 150)

X

lQ

3

4.25(.1 - 150) equi librium

F.

F,1

4 ~5 ( 1

F"

150)..: I 128 - 957

... '-e

= 190 nun 1 10.R

231:1 mm

= 0.43d

\\nh thr\ depth of ncutml a\is the reinforcement has yielded, F, ..

_)

4.25(190

150)

:I\

assumeu. and

170kN

> F,,, the the strc~~> block wou ld not extend beyond the llange and th~: ~cction he analysed Ul. in cxumplc 4.2 for a rectanguiLlr ~>ection of dimension!~ b1 x d.) rng momcms about th~.: centroid of the reinforcement

It f

1

~ld

l

\I

F~ r (d

flt /2) I f· cw(d

[957(550

~/2

hr /2)

150/2) I 170(550- 190/2- 150/2)] x 1() - 1

5.19kN m

!-+.15 ( EXA MPLE 4.8 Design of a flanged section with depth of neutral axis x - 0 4Sd

,Jfc but con:-.crvmive desrgn tor a flanged section with f > Jr1 can be achieved hy ung the depth of neutral axis to .r = 0.45cl, the maximum depth allowed in the code. [l~, rgn equations can IX! derrved for this condition a:. follows. Depth of Mres~ blocl... s

0.8x = 0.8 x OA5d = 0.36d

77

78

Reinforced concrete design O.S67fc~

r-

Figure 4.16 Flanged section with depth of neutral axis x 0 45d

=

x =OASd s" 0.8x

'

_j_

d aXIS

A,

• •

f,,

Section

Stress Block

Divide the flanged ~ection within the depth uf the ~tn:s~ block into areas l nnd 2 ns shuwn in figure 4.16. so that Area I

= bw X s = 0.36b,,,d

Area 2

(br

hw)

ht

X

and the comprc:-.sion forces developed hy these area~ urc

Fd

0.567.f.:~

Fc2 -

0.567.1~~ ~~1 (b, - /J" )

x 0.36b,.d

0.'1f..,b,.d

Taking moment~ about Fr' at the centroid of the flange

M

F,,(d

h1 /2)

Fd (.\j2

h, /2)

= 0.87/ikA,(d- h / '1 )- 0.2/..lb,.d(O. \()(/ 1

11,)1 2

Therefore

!:' + 0.1 /.:lbv. d(0.36d cunr~~ (d -

ht)

(4.23)*

0.511, )

Thi~ equation should not

he u~ed when 11 1 Applying thi::. equation to example 4.6: 180

A,

X

I 06 -

+ 0.1

X

(J.ln

:!5 X

0.36d.

X 200 X 350(0.36 X 15() 500(350 - 100/2)

l~ l 4mm 1 (compared w ith 1407

111111

2

I00)

in cxamrlc 4.6)

Before using equation 4.23 for calculating A,, it i\ necessary to confirm thnt compre!>sion reinforcement i~ not required. Thi~ is achil!vcd by using equation 4.24 to chcc.:k that the moment of rcsi~tancc ot the concrete. M 11ul. i~ greater than the design moment, M. )

l---------------------------------------~ 4.6.3

Flanged section wit h compression reinforcement

\V1th x OA5d m figure 4.16 and taking moment\ about A_. the max1mum res1stance moment of the concrete is

Mba!

= Fc1 X ~I + Fe'! X :'2 = 0.167fc..b,.d 2 - 0.567fc~ (bc- bw )(cl -

hr/ 2)

(4.24)

(Note that the value of 0.167 was derived in equation 4.10 for the rectangular section.)


(4.25 )..

.tpplied design moment. M > MbaJ. compression reinforcement i' required. [n

c

,e the area of comprc)o~ion steel can be calculated from M - Mt\Jt 0.87J;dd d') coo~ tdcring

(4.26)

the equilibrium of forces on the section

r the nrco of tens10n steel is
0.87 ;;.~ n. t1

.7

1 /.\

~

(4.27)

< 0.18, otherwi11e the de~ign compre~-o~ive steel ~tre~s i~ less than 0.87 ;~~, .

Moment redistribution and the design equations

pht\tic hcha,·iour of reinforced concrete at the ultimate limn state affects the huuon of moment' 111 a \tructurc. To allow for this. the moment!> denved from an 111••tnal)l>i' may be redi-.trihuted based on the a~sumplion that plastic hinge-. have ed at the section<, \\ tth the largest moment' (see secuon 3.6). 1 he fonnatton ol 1 ~ htnge' require\ relnth ely large rotations with )'ielding of the ten \ion •rcement To ensure large \tram-; in the tension Meel. the code of practice n.:\tncts epth ol the neutral axi~> according to the magnitude of the moment redi,tribution cd out. Tne equation~-. for thi~. gtven by EC2 for concrete ci
i'

2: "I I

k

,\'b,ll

2

d

(4.28) •here

)

moment at section after redistribution ---< 1. 0 moment m section hefore redistribution .nd ~ ~ ore con .. tant~ from the EC2 code and the UK Annex and ,.h•l is the max1mum ah I! of the depth of the neutral axi~ which will take the limiting value of the equality of equation (4.28) but should be less than OA5d for concrete cia<;" < CS0/60. Tht: depth of the 1>trcs' block is

= 0.8.\'t>.l an the le\cl arm ts

(4.29)

79

80

Reinforced concrete design The moment of resistance of the concrete in compression is

Mt>al - Fcc

X

Zt>>l

=- 0.567 fc~bSt->al

X ;:bat

and

Koot = Mbal/bd~fck = 0.567.1hat X :.~>at/d~ This equation for K""1 and the previous equations from 4.28 to 4.29 can be arranged to give

Kt>al - 0.454(/' - k!) j k2 - 0.182"(6- ~l)/k2j~

(4.30)

or alternatively

0.454c~;t') c:')

Kt>ul

From EC2 clause 5.5 the constnnts k, anu k2 arc given as: k 1 0.44 and k2 - 1.25, but from the UK A1u1ex to EC2 k 1 = 0.4 and k2 1.0. The relevant values of Xtll'" Zbal and Kbnl ror varying pcrccntagcs of momem redistribution and concrete c;luss < C50/60 are shown in table 4.2. When the ultimate design moment i~ ~uch that M

> KhaJbd 2f.:l

K > Kbal

or

then compre:.sion !)teel i!> required '>Uch that I

Kto.tift~hd 1 0.87/ydd- d')

(K

A - .;_,..,..--.;_..;;..;.._--.,...

•

(4.31 t•

and

(4.32) where

A

Mto.. t

(4.33)

bd 2./,l

These equations arc iucnlical in form to tho~c derived previou~ly for the design of a ~ection with compre~sion reinforcement untl no moment rcdi:-.tributton. If the value ot

d' /d for the section exceeds that :-.hown in table 4.2, the compression steel will not have yieluetl and the compressive stress will be l cs~ thun 0.87 /y~· In such cases. the compressive stress _he will be E~!~c where the strain 1, i:-. ohtuincd from the proportion) or the strain dingram. This vnlue of ./~c shou ld replace 0.87/Yk in equation 4.31, and equation 4.32 becomes

A, :;:

Kbat/~k bd ~ 0.87 .1) ~ <-t>al

A1

+ '

./~c

X ---

0.87 /vl

It should be noted that for a singly reinforced ~ection (K < Kbatl. the lever arm i~ calculated from equution 4.8. For a l.ection requiring compre~sion 'teel, the lever arm can be calculated lrom equation 4.29 or by U'>ing the equution

-: = d[o.s- )(0.25- Koot/1.13-t)] "hich il> similar to equation 4.8 but with Krut replacing K.

(4.3-t


81

Table 4.2 Moment redistribution design factors x001/d

{o

l=-ed to ~JO )

t5. but mem

Zt>al/d

Kbal

d'/d

0.821 0.853 0.869 0.885 0.900 0.917

0.167 0.142 0.129 0.116 0.101 0.087

0.171 0.140 0.125 0.109 0.094 0.079

According to EC2, UK Annex, k1 - 0.4 and k; - 1.0 0 1.0 0.45 0.82 10 0.9 0.45 0.82 0.82 15 0.85 0.45 0.40 0.84 20" 0.8 0.75 0.35 0.86 25 30b 0.70 0.30 0.88

0.167 0.167 0.167 0.152 0.137 0.120

0.171 0.171 0.171 0.152 0.133 0.114

According to EC2, k, - 0.44 and k2 = 1 25 0 1.0 0.448 10 0.368 0.9 15 0.85 0.328 20. 0.8 0.288 0.248 25 0.75 30b 0.70 0.206

• Maximum perm1tted redistribution for class A normal ductility ste~l b Max1murn perrnltte?d redistribution lor class Band C h1gher ductility ~teel, see section 1.6.2

/EX AMPLE 4 . 9

Design of a section with moment redistribution applied and to

112)*

I 3.3)*

"t n e of IJVC

~

the ons

and

08

The -;cellon 'hown 1n figure 4.17 i::. subject to an ultimate design moment of 230 k.. m niter a 20'/r reduction due to momem redistribution. The characten,tit matcnal Mrcngths Jre ~~~ 'iOO t\/mm 2 and /.;~ = 25 N/mm 2• Determmc the area' of reinforcement required Ul>ing the con,tants ~~ ond ~1 from (a) the EC2 and (b) the UK unncx to EC2.

b e 260

d'• l

so

• • A,'

II

(a) Using EC2

•

(1) From first principles (b - k,)d/k~

Limiting neutral axis depth,xhJI I rum EC2 clau!>e 5.5 therefore StI'C!.S blocI,. depth Lever arm

~1 .thul

~hal :h•l

0.44 and k~ - 1.25, (0.8 - 0.44)490/1.25 = 14 1mm = ox~hJI 11 3 mm = d - .lbal/2 490 l l3/2 = 434mm

'-'1omcnt of rc~btance of the concrete MhJI

F,.,

:t>al

from

0.567 f~~hl'bal

X Zbal

0.567

260

X

25

X

X

113

X

-l34

X

10

II

181kNm <. 230 kN m, the applied moment

therdorc

d'jxt,.al

0

..."'

compre~<..ion

'oteel is required.

= S0/ 141- 0.35 < 0.38

therefore the compression steel has yielded.

( ee equation 4.20 in :.ection 4.5)

A,

"'

·-

Figure 4.17 Dc~lgn example with moment redistnbution, ~ = 0.8

82

Reinforced concrete design Compression steel:

M - MOO!

I

A,

= 0.87[f.:{d -

d')

(230 -181 ) X l(f 50)

= 0.87 X 500(490 256mm~

-

Tension steel: M t>al

A, -

1

0.87 /,LZbJJ J8J

-:-A,

106

X

= 0.87 X 500 X 434 -256

=959 + 256

12 15 mm2

(ii) Alternative solution applying equations developed in Section 4.7 From equations 4.30 to 4J4: Khal

= 0.454(0- k1 )/k1

0.1821(~

- 0.454(0.X 0.44 )/ 1 25 0.131 -0.015 = 0 116

At J/k1':

O.llUI(O.H - 0.44 )/ 1.25/

which agree<. with the value given in tahlc 4.2. M

K

bd2J,L

228

10"

X

260 X 49()1 X 25 0.146 ~ Kbal- 0.116

=

Therefore compression Meel ~~ requtrcd. Comprc~l>ion

steel:

(K - Kbntlfc~bd 1 O.R7.1'ydd- d')

1\ 1 ~

=

{0. 146

0. 116)25 X 260 X 490 2

0.87

X

500(490

50)

Tension steel: Zhul

t\ ,

=d[0.5 ~ J(Ols- Kt>n17TIT4) d[o.s.,.. J(o.2s- o.II6/ Ll34)) = K~oatfckbd~ -

0.87/,L:.t..J 0 116 X 25

0.87

~A X

1

' 260

X 49()~

500 X 0.89

490 = 954 + 244 = 1198 mm' X

X

I 244

o.89d

Analysis of the section ..., Using the UK Annex of EC2 and applying the equations developed in section 4.7

From the UK Annex of EC2 clause 5.5 kt = 0.4 and k1 = 1.0 Ff m equations 4.30 to -U4:

K~.t

= 0.45-l(/i- kt)jk1 -

0 182!(b

kt )/k,f

= 0.454(0.8

0.4)/ 1.0 - 0.182 [(0.8 - 0.4 )/ I.Of 0.182 - 0.029 = 0.153

hich agrees with the value given K

10

table 4.2.

M bd~f~~

230 X 10~ 260 X 4902 X 25 0. 147 < KhJI = 0.153 1bc: r~fon~ <.:om pression steel is not required. '10n '>teel:

,jng equation 4.8 in section 4.4

~~ d(o.s + /(0.25 - Kt""'li.T34)] 490l0.5 I J(0.25 - 0.146/ 1.134 )]

490 x 0.847

415mm

M

O.tnf)L:. 230 x I(~ = 1274mm 2 0.!l7 X 500 X 415

- 8

Bending plus axial load at the ultimate limit state

.1pplled axial force may be ten\tle or compressive. In the analy,i'i that follows. a nn:sstve for<.:e 1:-. considered. For a ten~ile load the ~arne ba~1c principles ot lthrium. compatibility or strain!., and stre),s- strai n rcltttionships would apply, hut it td he necessary to change the 'ign of the upplied load (N) when we consider the 1 hrium of force-. on the cro-.-. ·~ection. (The area of concrete m compression ha~ nut reduced to allow for the concrete di~placed hy the comprc:-.sion steel. This could he en tnto uccount by reducing the stress.f,c in the comrression steel by an :unount cquul 56~~~~.)

cure 4.18 represent-. the cross-~ection of a member with typical struin and strcs' butions for varying position& of the neutral axi~. The cross-~ccuon ~~ subject to a ..,ent M and an nxial compressive force N. and in the figure the direction of rhc -nent i~ such a., to cause compre%ion on the upper part of the section and tcn~ion on •wer part. For ca~e!> ,.,here there is tension in the section (figure 4.18a) the limumg • ete !;train is taken as 0.0035 - the value used in the de,ign and analysts of !>ccuons ~ nding. However for cases where rherc is no tension in the section (figure 4.1 Rh) I miting strain i~ taken as a value of 0.002 at the level of 317 of the depth of the

83

84


Figure 4.18 Bendmg plus axial load w1th varying pos1tion of the neutral axis

r •

A,'

•

d •

d

h

A.

• •

·if--

s/2 neutral aXIS

t,

(a) S= O.Bx< h

r

0.567fc~

.....

0.0035

b

F,

0.5671<,

b

•

H

1 '

A,'

· -~ -

Ltc

d

h

• • A,

Section (b) s

Stral11s

Stress Block

=h: 0.8x > h

Let I·".:~

he the comprcS\IVC force developed in the concrete and acting through the cemroid of the <.tre-;<. block F.... he Lhe compre~~ive force in the reinforcement area and acting through centroid r , be the tensile or compres~i\e force in the rcmforccment area A~ and act • through it!> centroid.

A:

(I) Basic equations and design charts The applied force (N) mu!it be balanced by the forces developed within the cro section, therefore

N

= Fe~ + F + F~ M:

In thi!t equation, /·'_ wi ll be ncgutivc whenever the pn~ition of the neutral axis i~ Stk that the reinforcement/\, i~ in tcn!tion, at-. in ogure 4.18a. Sub1-tituting into thi~ equau the term), for lhe stresses and areas

{4.35

A;

where /.... i~ the comprec;sivc ~tress m reinforcement and .!- is the tensile compressive stress in reinforcement A . The de~ign moment M must he halunccd hy the moment of resi~tance of the fore" developed within the cro<,s-,cction. Hence, t:1king moments about the mid-depth oft" ~>ection


85

(·U6)* he depth of neutral axis is such that 0.8x ~ It, as in part (b) of ligure 4.18, then lc concrete 'cctton 11> 'uhject ton uniform compressive stres<> of 0.567./~... fn thi<> ~ concrete pmvide<; no contribution to the moment or rcsi•aancc and the lirst the right ~ide of the equation 4.36 di:-.appenr1-.. ') mmerricnl ammgement of reinforcement (A: =A, A,.j'2 and d' - It - d). ' -U5 and 4.36 can be re\\ riuen m the folio\\ ing form

0.567.1 f..c A, /~ A, --1·--+· -

!\

It

-l

}~~ bh

1".:1

0.567.1 ( 0.5- ~) It 2/t

1 "Ji.

bit

!-. ~

C'

fckbh II

0.5)

f..

t\,

.fck bh

c0.5) It

(4J7) (4J8)

cquauon' the Meel \tram'>, ami hence the 'tre'se' f, and f,. \ ary "11h the depth m:utral ax1' (\). Thu~ N / bl!f.:• and M j bh 1f,~ can he calculated for 'pccified ratios bh and r/ h so that culumn de))ign charts for a ~ymmcllicul arrangement of nrc ·ment l>Uch a~; the one !>hown 111 hgure 4.19 can he plotted. cltrcct solutiOn of equauonl> -1.37 and 4.38 for the dcs112-n ol column re1nforceml.!nt bl.! very tedious and, therefore, a set of design charts lor the usual case of trical section~ i~ available in publicatiOn!> :.Lu.:h a~ The IJeli[.lnt•r.\ Guide (rd. 20). pie' 'hO\•ing the dc~ign of column 'teel arc gl\:en 111 chapter 9.

\fodes of failure ~·

ltlve magnnude of the moment tM) and the ax1al load CN) governs whether the will fail in tension or in compression. With Iorge effective eccentricity ./ N) a tcn~i lc failure i~ liJ...ely. but\\ ith n '>mall cccentncity a comprc~~ive railure likely. Thi.! magmtude of the ecccntncll) affcch the po~11ion ol the neutral axi:-. hence the stmin~ and !>lrcs,cs in the re1nforceml.!nt.

Figur!! 4.19

TypiCal column design chart

0

01

0.2

0.3

OA

86


Let

A:

be the compressive strain in reinforcement be the tensile or compressi\'e o;train in reinforcement As be Lhe tensile yield strain of steel as -;hown in the stres\-stmin curve

£..., £,

£l

<J

figure 4.2. From the linear strain distribution of figure 4.18(a) = 0.0035

E:-.:

c- d') X

(4.39

and

X)

d e:, = 0.0035 ( -x-

For values of x greater than h, when the neutral axis extends below the section, as shown in figure 4.18b, the steel !.trains are given by the allernmive expressions: 7(x d')

:..c = 0.002 (?x _ j fl)

and 7(x - d) 0.002 {7x 3h)

£, -

The ~lee! ~tres~es and strain'i arc then related according to the stres!'.- Strnin curve 0 1 figure 4.2. Consider the foliO\\ ing mode.., of f:ulure of the \CCtton a~ shown on the interaction dtagram of figure 4.20. (a) Tension failure,

£s

> "'y

Tim l)pc of failure is a-.~ociated with large eccentrictlles ((')and 'imall depth~ of ncutr... ax1~ (.\). l·ailure begms with yteldmg of the teno;Jic retnforcement, followed by cru'ihin,.. of the concrctl! as the tensile stratns raptdly 111crea'ie.

"s - .:y, point b on figure 4.20 When fat lure occurs with yielding or the tcn-.ion !.led und cru'ihing of the concrete at the

(b) Balanced failure,

sume inslant it is described us a ·oalanccd· ft1ilurc. WiLh ~ , d .I = ~"hQI

=

.\hJI

nnd from equation 4.39

t

1

For example,

cy

+ 0.0~35

sub~tiluting

the vn lues ol

£),

= 0.00217 for grade 500 ~tcel

= 0.617d

Equations 4.35 and 4.36 hccome Nto..t

and

M~.J

Fcc- F"" - F, 0.567f..kb X 0.8Xbat ~ }:,A :· 0.87 f> ~A ,

F«G- o.~r~·) -F-.c d')

\\here

.f-.

~

0.87./;k

l·,(d ~)

(-+..tO)


--.-

Figure 4.20 Bending plus axial load chart with modes of failure

e of

b_.cf .. 0

;;;

c

~

::l

"'=

~ , .!:!

M

nt h on the intenmion diagram of figure 4.20. N Nhut. M M"·'' and -0 87 ~~~· When the design load N > Nhul the section wi ll fail in wmpression, there will he ao initial tensile fai lure, wit h yie lding uf

Otroression failure c."e

1

'''"'

and N > N~-o.1 • The t:hange in slope at point r in ligurc 4.20 occurs

rune of uon

fror'l equation 4.39

0.0035d' (0.0035

.:) )

2.63d' for grade 500 l'tccl n

r "ill ot:t:ur tn the tension failure Lone of the imcrm:tton dtagrJm if .11 1 d

< '"·'''

_ O.H7 }~ ... and tcn,ile n

eat the 4.39

d

1

()

'hen

1

f, •

O.R7.f~k

•

d

nnd t:mnpressive

.. n r become:-. very large and the ~ec tion approac hc~ a .pression

0.00217

y

Mate

of uniform axial

lor grade 500 steel

stage, both layer!> of <,tee) will have yielded :md there will he 7ero moment of t-tam:c with a \ymmetrical section, 1.0 that

1 thl\

4.40)

~

0.567}~lbh

87

0.87J;K(A:- A,)

c\t the stage where the neutral a>. is cmncidcs with the bottom of the sectton the :.train ~ram changes from th:ushO\\ n in figure 4.18a to the alternmh c 'itratn diagram -;hown hgure 4.18b. To calculate N and M at this stage, corrcspondtng to potnt s in ,.urc 4.20. equation!> 4.35 and 4.36 should be used, taking the neutral ax.t!'l depth equal the overall section depth. h.

88


Such M-N interaction diagram~ can be constructed for any o;hape of cross-section vv hich has an axis of symmetry h) applying the ba!>ic equilibrium and strain compatibility equations vv ith the :-.tn.:~!>-~train rclauon\, a\ demonstrated in the foUowmg examples. These diagram., can be very u~cful for design purposes.

(

'\

EXA MPLE 4 .1 0

M-N interactio n diagram fo r a no n-symme trical section

Construct the interaction diagram for the ~cction :.hown m figure 4.21 wirh 1:~ = 25 Nlmm 2 and J, . . = 500N/mm2 The bending causes maximum compression on the face udjm:cnt to the steel area tl~. For a <.,ymmetrical cro~s-scction, taking momenb about the centre-line of the concrete section will give M = 0 with N = N11 and both area~ of stt.:cl ut the yield mess. Thi-; ' nn longer true for unsymmetrical steel area., a~ /• " I F~ at yidd therefore, Lheoreticnll~ moments should be calculated about an axi~ rcfcrn.:d to as tht.: 'plastic centroid'. The ultimate axial lond No acting through the pla~>lic ce111rt1id cau~e<. a unifnrm ~train ucru~, the section with compres~ion yielding of all the reinf(lrccment. and thu\ there i~> zcr mwm:nt of resistance. With uniform :.train the neutral-axis depth. x. is at infinity. Figure 4.21 Non ~ymmetr1eal ~l'Ction M ·N lnteraclton examplt'

~

A

lu

I~

.,"' II

II

-'=

...

A

•A,': 1610• •d'= 60

~

0

"'•

b 3SO ---

neutral ax1s A, • 982

' • '--

•

Section

Stratn

Otagr~m

The locution of the plasttc centroid i~ determined by w~ing moment~ of all the stre rewltants about an arhitrary axil. :.uch "' AA in figure 4.21 so that 'LJFccfl /2 I F,cd' I F,d} f' F)

.\p =~( f. L. 'cc

+ ·~r + '

= 0.567J~~A •• X 450/2 + O.S7.{y~A: >< 60

O.X7 J)~A, X 390 0.567 /.:kAcr + 0.87/y~A: 1 0.87(~k/\, 0.567 X ::!5 X 350 X 45t)a /2 I ().87 X 500{ l6JQ X 60 + 982 X 390) 0.567 25 X 350 X 450 + ().1{7" 500( 161() l 982) - -

=::! l::!mm frumAA The fundamental equation!> for calcul..ning points on the interaction diagram varying depth~ of neutral a>.b nre: (I)

'"th

Compatibility of stram~ (used in table 4 3. columns 2 and 3): E"

£,

= 0 0035 (" ~·

d') 0.0035 (d ~ ')

(4.41

Analysis of the section ---ection .. strain • n the

Table 4.3

M-N interaction values for example 4.10 (7)

(2)

(3)

X

tK

.Es

d' 60 263d 158 0617d-241 XbJI d 390 h- 450 X·

. The

1 .JCrDS~

, 1ero

(4)

(5)

'~

fs

(N/m~) (N/mm2)

(mm)

Qcrete Th1s is • cally.

89

0 >0.00217 0 0.00217 >0.00217 0.87 fyt. >0.00217 0.00217 0.87 fyk >0.00217 0 0 87 fy~ >0 00217 0.00047 0 87 fyk 0.00217 0.00217 0.87 fyk

0 87 fy 0.87 fyl 0 .87 fy.. 0 93.3 0 87 fyk

(6) N (kN)

(1) M (kNm)

189 899 1229 2248 2580 3361

121 275 292 192 146 0

or when the neutral axis depth extends below the bottom <>f the section _

., 7{.t

d')

.

,.

c,, - 0.00. (?.I·_ 311) ,tnd ~' t ii 1 Stres~ \train relation~

~ ~ "Y

= 0.00217

=

7(x

0.002 (7.l _

d) 311

(x > II):

)

for the ~tee! (table 4.3. columns 4 and 5):

J - 0.87/yk f = l:: xc

(4.42 )

(Iii) Equil1hrium (t:lble 4.3. column~ 6 and 7):

0.!<1

"

0.8.1

,

N = Fe, ~ r . . + /', N - 0.567 J.~h O.lh f..._ A: t J A N 0.567l~ bh -f.,.,\: ~ f,,\,

'I ul.mg moment~ .tbout the pla,tic centroid O.!ll

h

M

().1:!.\

,

M

= / "( ' p- (Uh/ 2)

t F.,.( lr - d') - 1 (c/ lr ) l._ (.ir- h/ 2) + F,,:(.rp tl') f',(tl .lp)

F, i., negative when{, i\ a tensile

~Ire:.:..

Thc~c equation~ have been applied 10 provide the vo l ue~> in Utblc ·U lor a range of key \ <-. of i. rhcn them n interaction diagmm ha been ploucd in figure from the hlllicl>>>

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Reinforced Concrete Design To Eurocode 2 4e5ge

Overview 26281t

More details 6y5l6z

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