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REKABENTUK TETULANG KONKRIT(BS8110)

NOTA-NOTA SYARAHAN

Oleh :

Ng Kok SengUnit Struktur,Bahagian Khidmat PakarJabatan Pengairan dan Saliran Malaysia

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CONTENT OF COURSE

1. INTRODUCTION

Objectives of Course Objectives of Design History and Design Methods Properties/Advantages of Reinforced Concrete

2. LIMIT STATE PRINCIPLES (BS8110)

Principles and Requirements (Ultimate and Serviceability Limit States) Code of Practices

Organisation of BS8110 Codes Characteristic Strength of Material Definition for Concrete and Reinforcement Design Strength of Materials/Partial Safety Factors Characteristic Loads Design Loads (Live, dead and Wind loads) Robustness

3. DURABILITY

Reasons for increased emphasis on durability Cover and concrete quality provisions Durability grades Definition of environment Fire Resistance Sulphate attack Alkali-silica reaction Use of PFA, GGBFS and Microsilica

4. ANALYSIS OF FRAMES AND BEAMS

Braced Vs. Unbraced Frame Load Arrangements Redistribution of Moment Continuous Beams

5. REINFORCED CONCRETE BEAMS - ULTIMATE LIMIT

STATE (BS8110)

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General Theory/Assumptions Strain and Stress Distributions Derivation of Design Formula (Simplified Stress Block BS8110 ) Under-reinforced, Over-reinforced and Balanced design Design Charts Rectangular/ Flanged Beams Design Procedures Work Examples

6. REINFORCED CONCRETE COLUMN Column Design End Conditions to determine effective heights of column Design Example

7. SOLID SLABS

General Simplified Loading Patterns 1 -way slabs 2 way slabs Shear/ detailing

8. SHEAR IN BEAMS AND SLABS

9. DESIGN DETAILS

Minimum and maximum steel percentages Anchorage, laps, local bond Anchorage of Bundled Bars Curtailment rules

Bar spacing

10. DESIGN EXAMPLES Framed Buildings - slabs

- beams- columns

Footing Case Study Retaining Wall

11.DEMONSTRATION OF DESIGN SOFTWARE - GTSTRUDL

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1. INTRODUCTION

1.1 Objectives of Design

Design defined simply as a sequence of activitiesperformedto produce a service or utility whichissafe, functional, economic and aestheticallyacceptable.

The objectives are: -

1) Performance of Structure(a) Safety: nodanger against human life;(b) Serviceability: comfortable use of

structure(c)Profitability: profit to user/owner/society(d) Aestheticallycharacteristics :

fine/pleasant view for userand people around structure

2) Cost of Structure

(a) Initial cost: inexpensive material,constructibility, and simple design;

(b) Maintenance cost : maintenance free,easy repair;

(c) Demolition/replacement cost.

1.2 Design Methods

Several design methods for reinforced concretestructures

Working Stress Design (WSD),- established 1910s

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- working stress under service loads

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VARIOUS PROPERTIES OF CONCRETE

Compressive Strength (20 to 60 N/mm2)Flexural Strength (15 % of cube strength)Tensile Strength (10 % of Compressive strength)Modulus of Elasticity,Ec, (varies depending on

concrete strength e.g. fcu=30 N/mm2, mean Ec=26KN/m2 and typical range between 19 31KN/m2)

Poissons Ratio, (typically 0.1 to 0.2)Coefficient of Thermal Expansion( 10 --5per 0C)Shrinkage Strain(depend on constituent materials,

however max. cement ratio specify as 550kg/m3

toprevent excessive shrinkage and thermal effect)

Density( typically 24 KN/m3)Durability

- recent emphasis as in BS8110 where additionalprovisional for Limit States- depends on various factors such as

constituent materials, curing, quality of

workmanship and curing, exposure conditions etc. ImpermeabilityResistance to AbrasionResistance to Sulphate Attackand many others!

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ADVANTAGES AND DIADVANTAGES OF CONCRETEAS A CONSTRUCTION MATERIAL

ADVANTAGES DIADVANTAGES

1. Ability to be casted

2. Economical

3. Durable/Impermeable

4. Fire Resistance

5. Energy Efficient

6. On Site Fabrication

7. Aesthetic Properties

1. Low Tensile Strength

2. Low Ductility

3. Volume Instability

4. Low Strength-weightRatio

Cement Replacement Materials,

e.g. pozzolanas- Pulverised Fuel Ash (PFA)Granulated Ground Blast Furnace Slag

(GGBFS)(BS6699)1992 and BS8110 Pt 1Microsilica

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2. DESIGN PRINCIPLES AND REQUIREMENTS

q Whats Limit State?

A critical state beyond which the structural performance of thestructure, or part of it, tenders the structure unfit for use i.e. itceases to fulfil the function or to satisfy the condition for whichit was designed.

q Whats considered as Failure or unfit for use ?

TABLE 2.2: Failure Criteria

1 Collapse Failure of one or more critical sections, overturning orbuckling.

2 Deflection adversely affects the appearance or efficiency of thestructure.

3 Cracking adversely affect the appearance or efficiency of thestructure.

4 Vibrat ion due to wind or machinery, cause discomfort/alarm,damage the structure or interfere with its properfunction.

5 Durabi l i ty Porosity of concrete.

6 Fat igue cyclic loadings

7 Fire

Resistance

Insufficient resistance to fire leading to 1,2 and 3

above.

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2.0 Limit State Principles (cont)

q Whatre the limit states (BS8110) ?

2 main groups

a) Ultimate Limit StateStructure (or part of it) collapses i.e. it is not sufficient to withstandthe design loads and unstable.- by rupture of one or more critical sections,- transform from elastic or inelastic instability- loss of equilibrium as a rigid body, and so on.

b) Serviceability Limit States- excessive deflection, cracking, vibration, durability, fatigue, fireresistance, lightning, etc

In practice, 3 limit states considered ,

TABLE 2.3: Limit State in BS8110

Serviceability limit statesUltimateLimit State Deflection Cracking

ObjectiveProvision ofadequatesafety

Structure shouldnot deflect so asto impair use ofstructure

Cracking shouldnot be such as todamage finishesor otherwiseimpair usage

Loadingregime

Designultimateloads

Design service load

Performance limit

Structureshould notfail

Deflectionshould notexceed specifiedlimits. Normallymet byspan/effectivedepth ratio

Crack widthshould notexceed 0.3mmgenerally

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q Basic Requirements in design

- consider each limit state and provide suitable margin of safety- Usually design for the Ultimate Limit State, then checks

serviceability limit- Part 2 of BS8110 provided method for calculation of deflections

and crack widths- durability and fire resistance compliance by grade of concrete,

cement content, cover to reinforcements etc. (decided beforecalculation begin)

q Some considerations in design:

- Variations in materials in the structure and in test specimens- Variations in loading- Constructional inaccuracies- Accuracy of design calculations- Safety and serviceability

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CODE OF PRACTICE

q WHY CODE?

- gives guidelines on how design to be carried out- compliance is strongly advisable- only a guidelines, designers free to make use of his/her own

knowledge and experience to arrive at a decision

q SOME CODES

- BS 8110 (1985), replacing CP110- ACI (latest rev. 1995)- JSCE etc

q SPECIAL CONSIDERATIONS DIDS PERSPECTIVE

- DID structures located in difficult environment e.g. marine

remote site with poor accesssoft foundation soilaggressive environment exposure conditions

- Consider

higher concrete grade (C30 or greater)generous covermore durable concrete: low w/c ratio, strong dense

aggregate, cement replacement materials, min. cement

content etc. more dense and reduce permeabilitybetter construction practices (3Cs cement content,compaction and curing)

location of construction joints and edges subjected toaggressive hydraulic action of water

detailing of reinforcement for better compaction of concrete

q How is BS 8110: 1985 STRUCTURAL USE OF CONCRETE Organised?

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Table 2.4 Contents of BS8110 Code

Part 1. Code of Practice for design and construction

Section one. General.

Section two. Design objectives and general recommendations.

Section three. Design and detailing: reinforced concrete.

Section four. Design and detailing: prestressed concrete.

Section five. Design and detailing: precast and composite construction.

Section six. Concrete: materials, specification and construction.

Section seven. Specification and workmanship: reinforcement.

Section eight. Specification and workmanship: prestressing tendons.

Part 2. Code of Practice for special circumstances

Section one. General.

Section two. Non-linear methods of analysis forthe ultimate limit state.

Section three. Serviceability calculations.

Section four. Fire resistance.

Section five. Additional considerations in the use of lightweight aggregate concrete.

Section six. Autoclaved aerated concrete.

Section seven. Elastic deformation, creep, drying shrinkage and thermal strains ofconcrete.

Section eight. Movement joints.

Section nine. Appraisal and testing of structures and components duringconstruction.

Appendix A. Bibliography.

Part 3. Design charts for singly reinforced beams, doubt and symmetricallyreinforced rectangular columns.

q Whats Characteristic Strength of Materials (fcuand fY)?

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- value of the cube strength of concrete (f cu) or the yield or proofstress (fY) of reinforcement below which not more than 5% of the testresults fail.

- Statistically, the value is

f k = f m - 1.64s

Where f m= mean strength of actual test resultss = standard deviation (1.64 is the constant complying with5% of test results

FrequencyOf results

strength

Mean strengthCharacteristic

Strength1.64 S

f k = f m - 1.64s f m

5% of

results

to left

of this

line

Strength

Fig 2.1 Characteristic Strength

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- Typical values of f k

Table 2.5: Characteristic strength of reinforcement

(Comply with BS 4449, BS 4461 or BS 4483)

Designation Nominal. fysizes(mm)

(N/mm2)

Hot-rolled mild steelAll sizes 250

High-yield steel All sizes 460

Table 2.6 Characteristic strength of concrete

Conc. grade Char.Compressivestgth

At 28-days (N/mm2)

C10 10.0C12.5 12.5C15 15.0C20 20.0C25 25.0C30 30.0C35 35.0C40 40.0

C45 45.0

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q WHATRE DESIGN STRENGTH OF MATERIALS?

partial safety factor, m

For Concrete, m- allow for possible differences between fkand actual structural

strength

For reinforcement, m- difference in cross-sectional area between assumed and actual

steelproduced, corrosion etc.

Table 2.7: Values of for concrete and steel at different limit states

Values of mLimit State Concrete Steel

Ultimate 1.5. 1.15Deflection 1.0 1.0

Cracking 1.3 1.0

Note:

- m(both materials) at ULS is higher (reduce the probability offailure and failure should also be localised.

- mconcrete is > mreinforcement (failure in concrete issudden)

- mdeflection is related to the whole member, m(bothmaterials) =1.0.

- mcracking bet. 1.0 and 1.5 (concrete), 1.0 for reinforcement (only parts of the member affected)

m

KfstrengthDesign =

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2.0 Limit State Principles (cont)BS8110 design stress-strain curve for concrete, shown in Fig 2

0.67( fcu / m)

stress

2.4 x 10-4 ( fcu / m) 0.0035

FIG. 2.2design stress-strain relation for normal weight concrete (fcu inN/mm2)

Note:

- design strength =characteristic strength/ m- maximum design strength = 0.67 fcu / m.- design strength reduced to 0.67fcu for concrete cube test

strength Vs. actual structural compressive strength- for SLS elastic modulus may be taken from a table in the code,

i.e. a linear stress-strain relationship is assumed with aspecified value of Ec depending on fcu.

For reinforcementthe short-term stress-strain relationship is shown inFig. 2.3 The elastic modulus is 200KN /mm2.

200KN/ mm2.strain

FIG. 2.3 Short-term design stress-strain relation for reinforcement (fy

in N/mm

2

.) 2.0 Limit State Principles (cont)

5.5( fcu / m)KN/mm

2

Paraboliccurve

Compression

Tension

fy/ m

fy / m

Tension

Strain

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q WHATS CHARACTERISTIC LOAD?

For loading, characteristic load (Fk), use- CP 3: Chapter V, Part 1; Dead (GK)and imposed loads (QK)

- CP3 Part 2: Wind loads(WK)- BS 6399 : Part I for dead and imposed loads- Handbooks or manuals e.g. Reinforced Concrete Designers;

Handbook by Reynolds.

Dead loads:

normally permanent and constante.g. self-weight, permanent fixtures and finishesusually calculated on a slightly conservative basis, so that a

member will not need redesigning because of a small change inits dimensions

Impose loads:

transient, variable, more difficult to determined accurately.e.g. wind or human occupants, furniture, or machinery; pressures

of wind, retained earth or water; and the forces caused by thermalexpansion or shrinkage of the concrete.

Wind loadsimposed load but a separate category when its partial factors ofsafety are specified; and when the load combinations of thestructure are being considered.

depends on the locations the building, size and height, openingsin walls, etc.(CP3 Part 2: Wind loads(WK))

Others loads

Dynamic loads by impact or surgeWheel loads are rolling loads, placed in position to give maximummoments and shears.

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q WHATS DESIGN LOADING?

Design load = Characteristic load x f (partial factor of safety

factor);

Why f ?

1. Possible unusual increases in the load beyond those inderiving the characteristic load.

2. Inaccurate assessment of effects of loading.3. Variation in dimensional accuracy achieved in

constructions.

4. The importance of the limit state being considered.

f varies for different limit states combination of loading.

Table 2.8 Values of f at ultimate limit state

Load type

Dead Imposed

Earth &

waterpressur

e

Wind

Load Combination Adverse

Benefi-cial

Adverse Benefi-cial

1 Dead and Imposed(and earth andwater pressure)

1.4 1.0 1.6 0 1.4 -

2 Dead and Wind

(and earth andwater pressure) 1.4 1.0 - - 1.4 1.4

3 Dead and Imposed(and earth andwater pressure)

1.2 1.2 1.2 1.2 1.2 1.2

Note:- arrangement of loads to cause the most severe effects- adverse' partial factor - applied to loads that produce a

more critical design condition.

- 'beneficial factor - applied to loads that produce a lesscritical condition.

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- minimum design load is 1.0 Gk


- Load combination (2) is generally for a stabilitycondition, most critical case may arise when momentsdue to 1.4 Gk on some parts of the structure areadditive to wind moments due to 1.0 Gk on other partsof the structure form the restoring moments.

- Load combination (3), a factor of 1.2 is usedthroughout the structure

For SLS, the load combinations specified by BS8110 are:1. Dead and Imposed Load - 1.0 Gk 1.0 Qk 2. Dead and Wind Load - 1.0 Gk 1.0 Wk3. Dead, Imposed and Wind Load - 1.0 Gk 1.0 Qk 0.8 Wk

q WHATS ROBUSTNESS?

- Related to structural integrity of the structures.- Avoid damage to small areas or failure of singleelements leading to collapse of major parts of the structure.

- Clause 2.2.2.2 / 3.1.4of BS 8110: Part 1 lists theprecautions

All buildings should be capable of resisting a minimumhorizontal force (Notional horizontal force)

All buildings are provided with effective horizontal ties.For buildings >= 5 storeys, check for key elements on

layoutFor buildings >= 5, any vertical load bearing element(other than the key element) should be detailed suchthat its loss will not cause considerable damage.

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3.0 DURABILITY

q WHATS IS DURABLITY OF CONCRETE?

- a durable concrete is one that will withstand, to asatisfactory degree, the effects of serviceconditions to which it will be subjected, such asweathering, chemical action, and wear.

q WHY MORE EMPHASIS ON DURABILITY INBS8110/CP110 ?- Corrosion of rebars in bridges in UK and USA

- Extensive Corrosion of rebars in Middle East- Use of calcium chloride as accelerators, cause

damage in multi-storeys caldding panels in UK- Concern with Alkali-silica reaction (ASR)- Chloride attack on bridges- A lot of problems with durability with post-war

housing in UK- Increase in strength of cement, higher strength with

lower w/c ratio- Use of cement replacement materials, e.g. PFA,

GGBFS and microsilica

q WHATRE THE PROVISIONS FOR DURABILITY IN

BS8110 ?

- dealt in 3 main places in Code

a) Section 2 (2.2.4)- general issues

b) Section 3 (3.3)- choice of suitable cover andappropriate quality of concrete. (most significantset of provisions related to durability to designer)

c) Section 6 (6.2 Durability of structural concrete)

- general treatment of durability with special

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reference to mechanisms leading to degradation ofthe concrete (i.e. sulphate attack, ASR, frost),useof pfa, ggbfs

3.0 durability (cont..)q DEFINITION OF ENVIRONMENT (Provision in

BS8110)

- Environment and the corresponding requirementsfor cover and concrete quality listed in Table 3.1 and3.2

Table 3.1 : Exposure Conditions (From Table 3.2 BS8110)

Environment Definition

Mild Concrete surfaces protected againstweather or aggressive conditions.

Moderate Concrete surfaces sheltered: -

From severe rain or freezing whilst wet.Concrete subject to condensation.Concrete surfaces continuously underwater.Concrete in contact with non-aggressivewater.Concrete in contact with non-aggressivesoil.

Severe Concrete surfaces exposed to: -driving rainalternate wetting and drying andoccasional freezingsevere condensation.

Very severeConcrete surfaces exposed to:

seawater spray

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de-icing salts, directly or indirectlycorrosive fumessevere freezing conditions while wet.

Extreme Concrete surfaces exposed to abrasiveaction e.g.

seawater carrying solidsflowing water with pH < 4.5machinery or vehicles.

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3.0 Durability

Table 3.2: NOMINAL COVER TO ALLREINFORCEMENT (INCLUDING LINKS) TO MEET

DURABILITY REQUIREMENTS (Table 3.4 in BS8110)

Conditions of exposure Nominal coveras definedin Table I

mm mm mm mm mmmm

Mild 25 20 20 20 20

Moderate - 35 30 25 20Severe - - 40 30 25Very severe - - 50 40 30

Extreme - - - 60 50-Maximum freewater/cement ratio 0.65 0.60 0.55 0.500.45-Minimumcement contentkg/m3 275 300 325 350400-Lowest gradeof concrete C30 C35 C40 C45

C50

Notes:

- Normally if strength of concrete meet the requiredgrade, it is deemed to satisfy the w/c ratio andcement content values.

- Actual cover should not be less than nominal coverminus 5mm.

- For precast concrete, where very low w/c ratiosand cement contents were well below the appropriatevalues in Table 3.2, such mixes are also likely to be

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very durable, and the minimum cement contents maybe reduced up to 10% provided the minimum watercement ratio is reduced by the same percentage.

- The addition of pfa or ggbfs together with OPC forthe 'cement' in the Table 3.2 used for checking thew/c ratio and cement content is allowed provided themixes with and without additions would have equaldurability if they were of equal 28-day strength.

3.0 Durability(cont)

q FIRE RESISTANCE

The general arrangement of the material on fire

resistance in the code is as follows:

- Section 3.3 in Part 1 : tables of minimum bar sizeVs. nominal cover for the normally specified periodsof fire resistance. (simple and conservative)

- Part 2 (Section 4): more comprehensive tablesgiven in the BRE guidelines (with slight

modifications in the case of walls).

- Design for fire by direct calculation is permitted,reference to the Institution of StructuralEngineer/Concrete Society recommendations.

q Requirements for concrete exposed to sulphate

attack (Cl.6.2.3.3)

- Table 6.1 BS8110. Specified min. Cementcontent, w/c ratios related to concentration ofsulphate

- Use of pfa or ggbfs in concretes to resist sulphatesis permitted.

q Alkali Silica reaction (ASR) Cl. 6.2.5.4

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- Considered only in cases where there will be ahigh moisture level in the concrete and the mix willhave a high alkali content and a reactive aggregate isbeing used.

- Under such circumstances, Alkali content of theconcrete should be reduced to below 3kg/m

3ofNA20

equivalent by using a low alkali cement, a lowcement content or by using a composite cementincluding at least 30% pfa or 50% ggbfs.

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4.0 ANALYSIS OF FRAMES AND BEAMS

q GENERAL

- Generally, the design procedure can be started by assumingthe ultimate limit state as the critical limit state.

- Computer programs are common nowadays. Thestructure can be analysed as a complete frame. e.g.GTSTRUDL, STAAD III, ATSstructE , Microstran etc.

- Hand calculations, carried out by taking parts of thecomplete frame.- Assessment of section sizes will have to be

made. The so-called rule of thumb is available foruse, i.e. the limiting span/depth ratio specified inthe Codes of Practise. Having assumed somesection the structure can now be analysed.

q FRAME ANALYSIS

Classified into 2 categories i.e. braced frame and Unbraced

frame

- Braced Frame

frame not providing lateral stabilityonly vertical loads are concernedseries of sub frames are being considered. Each

sub frame consists of the beams at one leveltogether with the columns above and below,

assumed to be fixed at their ends remote fromthose beams, unless the assumption of pinned-end is more reasonable like for the case offoundation which unable to develop momentrestraint. See Figure 4-1

For continuous beam sub-frame, Table 3.6 (Cl.3.4.2.) M and V can be usedi) Qk < Gkii) Load fairly uniform over 3 or more spans

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iii) Variation in spans < 15 % of the largest

- Unbraced Frame (Cl. 3.2.1.3.2)

frame providing lateral stabilityconsider both lateral and vertical loadsuse of computer program would be very practical since

hand analysis can be very tedious.

will sway even without horizontal loads.Simplified analysis as described in Pg 411, Reinforced and

Prestressed Concrete Design by Kong & Evans (3rd

Edition)

Note: - The Codes thus warns a designer that theeffects of sway should be considered for frames lessthan three bays.

q LOAD ARRANGEMENT (Braced Frame)

See Figure 4-1 for beams

For one-way slabs, (Cl. 3.5.2.3), designed for singleloading case (1.6Qk + 1.4Gk) on all spans M and Vtaken from Table3.13 provided :-

i)area of each bay > 30 m

2

ii) Qk / Gk

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q REDISTRIBUTION OF MOMENTS Clause 3.2.2 (anoptional)

- transferring some of the calculated moment (by elasticanalysis) at one position to another position in a member,thus reducing rebar requirement/ congestion .

- Conditions

Equilibrium between internal forces and externalloads must be maintained i.e. overall height ofbending moments remains the same for anyparticular loading.

Resistance moment at any section should be atleast 70% of moment at that section obtained froman elastic maximum moment.

After redistribution, neutral axis depth, x ( 0.4d) where =ratio of moment after andbefore redistribution.

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Effective Span of Beam

L1 = Clear spacing between supportsL2 = distance between the centres of bearings,

Simply supported: smaller of L2 or ( L1 + d )

Continuous beam: L2

Cantilever : smaller of (L1 + d) or (Up to centre ofSupport)

Slender Beams

The clear distance between lateral restraints

simply supported beam < 60bC, or 250bC2/d

(whichever is the lesser)

Cantilever < 25bC, or 100bC2/d whichever is thelesser.

Shear Resistance of Beams

1. Shear stress v = V / bVd

2. If v > vC the whole of the shearing force should beprovided by shear reinforcement.

vCgiven in Table 4.5 for various % of bendingreinforcement and various effective depths for 30N/mm2

concrete.

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3. v < 0.8fcu or 5 N/mm2 whichever is the lesser.4. If v < vC, prov. nominal shear reinforcement throughout

the span of the beam (Table 4.6)

5. Where v > vC,,shear reinforcement provided throughout the span of the beamas noted in Tables 4.6 Tables 4.7 and 4.8 tabulate values that, when

multiplied by the depth of the beam (mm), give the shear resistance forparticular values of ASv and sv

6. sv

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Local Bond Stress

Local bond stress may be ignored, as force in a bar canbe developed by the appropriate anchorage length

Laps in Bars

Fortension reinforcement: design anchorage tensionlength

For compression reinforcement: design anchoragecompression x 1.25

In both cases lap lengths for bars of unequal size (or wiresin fabric) may be based on upon the smaller bar.

Notes:-1 lap at the top of a section

minimum cover 2 x rebar size,

lap length x 1.4

2. lap at the corner of a section

minimum cover 2 x rebar size, ORclear distance between adjacent laps 75mm

6 x rebarsize (whichever is the

greater)

lap length x 1.4.

3. Both 1 and 2 apply

lap length x 2.0

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(Values for lap lengths are given in Table 4.12as multiples of bar size)

Hooks and Bends

The effective anchorage length of a hook or bend shouldbe measured from the start of the bend to a point fourtimes the bar size beyond the end the bend. This may betaken as the lesser of 24 times the bar size, or

1. For a hook eight times the internal radius of the hookor the active length of the bar in the hook, includingthe straight portion whichever is greater

2. For a bend - four times the internal radius of the bendwith a maximum of 12 times the bar size, or the actuallength of the bar whichever, is greater.

Curtailment of Bars

In any member subject to bending, every curtailed barshould extend (except at end support) beyond thecalculated cut-off point for a distance equal to the effectivedepth of the member or 12 times the bar size, whichever isgreater. In addition, bars should not be stopped off in atension zone, unless one of the following conditions issatisfied:

1. The bars extend an anchorage length appropriate totheir design strength (0.87fy) from the point at which itis no longer required to assist in resisting the bendingmoment.

2. The shear capacity of the section, where thereinforcement stop, provide double the area requiredfor the moment at such points.

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Anchorage of Bars

As a simply supported end of a member one of the following requirements shouldbe fulfilled:

1. Effective anchorage equivalent to 12 bar sizes plus beyond the centre line ofsupport (No hook or bend should begin before the centre of the support)

2. Effective anchorage equivalent to 12 bar sizes plus (d/2from the face of the support. .(No bend should begind/2 from the face of the support.)

3. For slabs if the design ultimate shear stress the face ofthe support is less than half the appropriate value vC,recommended in Table 4.5 a straight length of barbeyond the centre-line of the support equal to eitherone third of the support width or 30 mm, whichever isgreater.

Simplified rules for the curtailment of bars are given in BS8110: Part 1.

Cover

BS 8110: Part 1. Table 4.13 : - conditions of exposure Vs.nominal cover

Minimum Distance Between Bars

The lateral dimension between bars should be themaximum-sized aggregate plus 5 mm or the bar size,whichever is greater. Vertical dimension between barsshould be two-thirds of the maximum-sized aggregate.

Maximum Distance Between Bars in Tension

General rules for establishing the maximum distancebetween bars in tension are given in BS8110 : Part I.

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Table 4.13 Nominal cover to all reinforcement (includingiini

Minimum and Maximum Percentages ofReinforcement in Members

The minimum and maximum percentages of reinforcementappropriate for various conditions of loading and types ofmember are given in BS8110:Part 1

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5. Reinforced Concrete Column

- Ultimate Limit State (BS8110)Notes are in Transparencies Format

6. Solid Slabs

- Ultimate Limit State (BS8110)Notes are in Transparencies Format

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7. SHEAR RESISTANCE OF BEAMS/ SLABS(Clause 3.5.5)

3. Shear stress v = V / bVd

4. If v > vC the whole of the shearing force should beprovided by shear reinforcement.

vCgiven in Table 4.5 for various % of bendingreinforcement and various effective depths for 30N/mm2concrete.

3. v < 0.8fcu or 5 N/mm2 whichever is the lesser.

4. If v < vC, prov. nominal shear reinforcement throughoutthe span of the beam (Table 4.6)

9. Where v > vC,,shear reinforcement provided throughout the span of the beamas noted in Tables 4.6 Tables 4.7 and 4.8 tabulate values that, when

multiplied by the depth of the beam (mm), give the shear resistance for

particular values of ASv and sv

10. sv0.75d

11. Not necessary to provide shear reinforcement in slabs, bases, pile cap andsimilar members if v does not exceed vC.

12. 50 % of shear reinforcement may be in the form of included bars (seeBS8110:Part 1)

Deflection of Rectangular Beams (clause 3.5.7, 3.4.6)

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If the span/effective depth ratio < ratio in Table 3.10

deflection of a beam will not be excessive and 0.3, linear interpolation between the values given inTable 3.10 for rectangular beams and for flanged beamswith bW/b =0.3 may be used.

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DESIGN DETAILS (BS 8110)

Minimum areas of tension reinforcement

(BS8110: Clause 3.12.5) prevent shrinkage/thermalcracks etc.

(a) Rectangular beams:

As(min) 0.13% bh

(b) Flanged beams (web in tension):

bw= web width b = effective flange width

f bw/b< 0.4, As(min) 0.18% bwh

If bw/b > 0.4, As(min) 0.13% bwh

(c) Flanged beams (flange in tension over a continuous

support):

T-beams As(min) 0.26 % bwhL-beams. As(min) 0.2 % bwh

(d) Transverse reinforcement in flanged beams:

Provided near the top surface of the flange over the

full effective width b.

Ast(min) 0.15% hf l where hf=flange thickness l= beam span.

(will prevent longitudinal cracks along web-flangejunction)

Minimum areas of compression reinforcement

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(BS 8110: Clause 3.12.5)

rectangular beam, As' 0.2% bh

flanged-beam (web in compression): As' 0.2%bwh

bars 16 mm

Maximum areas of main reinforcement(BS 8110: Clause 3.12.6)

As' and As 4% of bh (or 4% bwh for flange beam)

Links or stirrups(BS 8110: Clauses 3.4.5 and 3.12.7)

- to resist shear or contain the compression reinforcementagainst outward buckling.

Requirements:

a) bars 8mm

b) Where compression reinforcement is used in abeam

links bars xlargest compression barsspacing 12 x smallest compression bars

c) Min. links are regd along entire spans in beam

Slenderness limits(BS 8110: Clause 3.4.1.6) prevent lateral buckling

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Slenderness limits= clear distance between lateralrestraintsd = effective depthbc= the breadth of the compression face of the beam

midway betweenrestraints.

Table 7-1 Slenderness limits

Type of beam Slenderness limit

Simply supported/ 60bCor 250 bC2/d (whichever is the

lesser)ContinuousCantilever 25 bC

2or 100 bC

2/d (whichever is the

lesser)

Minimum distance between bars(BS 8110: Clause 3.12.11. 1)

(a) Horizontal clear distance between bars

hagg+ 5 mm, or (whichever is greater)(=size of the larger bars if they are unequal)

(b) Vertical clear distance (2 rows)

2/3 haggor (whichever is greater)

(c) Sufficient space should be left between bars toenable the vibrator to be inserted.

Maximum distance between bars (BS 8110: Clause3.12.11.2) - for crack control, only applied to tension bars

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(See Table 3.30 and related Clauses in BS 8110:Part 1)

Minimum lap length (BS 8110: Clause 3.12.8)

For Bars,

lap length 15 x or 300mm(whichever is thegreater)

For fabric reinforcement

lap length250 mm.

Fortension bars: design anchorage tension length

For compression bars: design anchoragecompression x 1.25(In both cases lap lengths for bars of unequal size (orwires in fabric) may be based on upon the smallerbar.)

Notes:-2 lap at the top of a section

min. cover 2 x : lap length x 1.4

b) lap at the corner of a section

minimum cover 2 x , ORclear distance bet. adjacent laps 75mm or

6 x rebar size (whichever is the

greater)lap length x 1.4.

c) Both 1 and 2 apply: lap length x 2.0

Refer Table 3.29 (BS8110) for lap lengths Vs. barsize

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Curtailment and anchorage of bars(BS8110: Clause 3.12.9)

(a) bar extend beyond theoretical cut-off point d or 12

whichever is greater. (except at end support)

Theoret ical cut-of f poin t = location where theresistance moment of the section, considering only thecontinuing bars, is equal to the required moment.

(b) bar stopped in tension zone should satisfy theadditional requirement that it extends a fullanchorage length from the theoretical cut-off point,unless other conditions detailed in Clause 3.12.9.1 ofBS 8110 are satisfied.

(c) Simple end support, tension bar should have aneffective anchorage of 12 bar sizes beyond the centre

line of the support unless other conditions detailed inClause 3.12.9.4 of BS 8110 are satisfied. Theeffective anchorage lengths of hooks and bends areexplained inFig. 6.6- 1.

Anchorage of links(BS 8110, clause 3.12.8.6)

Bars as links

Fully anchored if satisfies

(a) 900 bend : > its own bar size and continuesbeyond for a minimum length of 8 x its own size.

(b) 1800

bend : > its own bar size and continuesbeyond for a minimum length 4 x its own size.

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For Grade 250 bars, r = 2 x bar size

For Grade 460 bars; r = 3 x bar size for bar sizes < 20mm

r = 4 x bar size for bar sizes > 25mm

Fabric as links(BS 8110, Clause 3.12.8.7)

Fully anchored when length of the anchorage either:

2 x welded wires, or

1 x welded wire of size not less than 1.4 timesthe size of the wire being anchored.

Simplified rules for curtailment of bars(BS 8110: Clause 3.12.10.2)

Problems:-In practical design, bending moment diagrams are oftennot drawn for members of secondary importance-and the theoretical cutoff points are not then knownwithout further calculation.

Thus, simplified rules for slab and beam provided

- loading substantially uniformly distributed- in case of continuous slab/beam, spans approx.

equal

See BS 8110 Figure 3.24 and 3.25

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Concrete cover for durability(BS 8110: Clause 3.3)

Exposure conditions,Table 3.2Durability, Table 3.4.

Fire resistance, Table 3.5.

Cover Nominal max. size of aggregate (3.3.1.3).or equiv. size of two or more bars(3.3.1.2).

Fire resistance (BS8110: Clause 3.3.6)

For beam : depends on width and cover, as shown inTables for cover.Fire resistance requirements may dictatethe size of the beam (BS8110 : Figure 3.2)

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KKUURRSSUUSSPPEENNYYEEDDIIAAAANNRREEKKAABBEENNTTUUKK

UUNNTTUUKKTTEEMMUUDDUUGGAAPPRROOFFEESSSSIIOONNAALL

PADA

6 10 MAC 2000

REKABENTUK TETULANG KONKRIT(BS8110)

DESIGN EXAMPLES

Oleh :

Ng Kok SengUnit Struktur,Bahagian Khidmat PakarJabatan Pengairan dan Saliran Malaysia

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Exposure condition-internal mildExternal moderate

Fire resistanceDead loads -partitions and finishes 1.5 kN/m2

external cladding 5 kN/mlitiposed loads-roof 1.5 kN/m2

floors 3 kN/m2

Allowable soil-bearing pressure 200 kN/m2

Characteristic strengthsConcrete: fcu 40 kN/mm2

Reinforcement:fy (main bars) 460 kN/ mm2

fyv(links) 250 kN/ mm2

Example 1 (slab)Design and detail the reinforcement for one panel of thetypical floor shown inFig. 1

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SOLUTION(See Fig. Ex1-1)

St ep 1 Dur ab i l i t y an d f i re res is tance

From Table 3.4,nominal cover for mild exposure condition = 20 mm

From Table 4.4,fire resistance for 1 hr., 20mm cover, continuous slabs

min. thickness > 95 mm O.K.

Nominal cover = 20 mmFire resistance OK

Step 2 Load ing -pe r m e t r e w id t h o f sl ab

Self-weight = (0.180 m) (24 kN/mm3) (5.5 m) = 23.8

Partitions and finishes = (1.5) (5.5) = 8.3Characteristic dead load Gk = 32.1 kN/m

width

Characteristic imposed load Qk= (3) (5.5) = 16.5 kN/mwidth

Design load F = 1.4 Gk+ 1.6Qk

= 44.9 + 26.4

= 71.3 kN/m width

Gk= 32.1 kN/m Qk= 16.5 kN/m F = 71.3 KN/m

Step 3 U lt im a t e mom en t s

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From Table 3.13 (interior support),

M at supports = 0.063FL = (0.063)(71.3)(5.5) = 24.7 kNm/m

M at midspan = 0.063FL = 24.7 kNm/m

St ep 4 M a in re i n fo r cem en t

Effective depth d = 180 - 20 bar = 154 mm, say

Supports : K = M/(fcubd2) = (24.7)(106) =0.026

(40) (1000) (154 2)

z/d = 0.94 xld = 0.13

where z/d = 0.5 + (0.25 K/0.9) and z/d = d 0.45x

(NOTE : z/d 0.95 )

From Cl. 3.4.4.4

As = M/(0.87 fy z) = 24.7 x 106 = 426mm2/m

(0.87) (460) (0.94) (154)

From Cl. 3.10.5.3 Table 3.27

Min. tension steel = 0.13%bh = 234 mm2/m 426 mm2/m

Hence

As= 426 mm2/m , Provide: Top: T10 @ 150 (523 mm2/m)

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Midspan: M/(fcubd2) =0.026 (as at supports)

Bottom: T10 @ 150 (523 mm2/m)

St ep 5 Shear

From Table 3.13,V = 0.5F = (0.5) (71.3 from Step 2) = 35.7 kN/m

v= V/bvd = (35.7)(103) = 0.23 N/m2

(1000) (154)

< 0.8 fcu(= 5.1 N/ mm2)

From Table 3.9,

For As/bvd= 523/(1000)(154) = 0.34%

vc= 0.65 N/mm2

> vShear resistance OK

St ep 6 Def lect ion

From Table 3.10,

basic span/depth ratio = 26 (cont. slab)

M/(bd2) = (24.7)(106) = 1.04(1000) (1542)

From Table 3.11 Mod. Factors for tension bars (cl. 3.4.6.5)

modification factor = 1.38

Hence

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allowable span/depth ratio = (26) (1.38) = 35.9

actual span/depth ratio= 5500/154 = 35.7 < allowableDeflection OK

(Note : if not O.K., change d, As or cal. Actual Defl.)

St ep 7 Crack ing

3d = (3) (154) = 462 mm clear

spacing between bars = 150 - bar < 462 mm OK

For fy =460, h < 200mm , h = 180 mm OK

no further checks are required.

Cracking OK

St ep 8 Second ary re in f o rcem ent

From Cl. 3.12.5.2.,

minimum secondary reinforcement = 0.13% bh

= (0.0013) (1000) (180) = 234 mm/mT10 @ 300 (262 mm2/m)

St ep 9 :Robu stn ess Cl . 3.1 .4 .3 an d3 .12 .3 .4 )

Longitudinal tie force Ft= 20 + 4 times (no. of storeys) or 60(whichever is lesser)

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Ft = 20 + (4)(4) = 36 kN/m < 60

Ft = 36 kN/m

(gk+ qk) (lr/5) Ft= (32.1/5.5) + (16.5/5.5)(5.5)(36) 7.5 7.5 5

= 46.7 kN/m > FtMinimum continuous internal tie = 46.7 x 1000 = 116.7 mm2/m

(0.87)(460)

Bottom T10 at 400 (196 mm2/m)

From Table 3.2,

full anchorage lap length = 32 The required lap length = (32 )[fs/ 0.87fy]

=(32 ) [As req (116.7)]

[As prov.(196) ]= 19 = 190 mm

From 3.12.8,

Min. lap length = 300 mm or 15 (whichever greater)

Min. Lap length = 300mm ( > 19 )

Tie lap length = 300 mm

St ep 10 Rein fo r cemen t de ta i l s

See Figure 1-2 for reinforcemens details. Note the following comments regarding BS 8110's

detailing requirements:

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(a) Main bars at midspan (mark 1)Curtailment: in a continuous slab the main tension bars

at midspan should extend to within 0.21 of the supports,and at least 40% should extend into the support.Bar spacing: the clear spacing should not exceed 3d (462mm) or 750 mm whichever is the less.

(b) Secondary bars (marks 2, 3, 4 and 6)Bar spacing: the clear spacing should not exceed 3d (462mm) or 750 mm whichever is the less.Minimum area:All secondary bars > minimum area of

0.13%bhMinimum lap length: the minimum lap length should not

be less than 15or 300 mm whichever is the less.

(c) Main bars at supports (mark 5)Curtailment: the main bars at supports should extend a

distance of at least 0.15L or 45, from the face of support

whichever is greater, and at least 50% should extend0.3L(d) Transverse reinforcement across main beam (mark 5)Table 3.27 slab & the main beam =flanged beam.Transverse reinforcement > 0. 15 % hfL at the top surfaceand across the full effective width of the T-beam flange. Theeffective width is bw+0.2Lz= 350 + (0.2) (0.7 of 9000) =1610 mm; 1610/2 = 805 mm. That is, the transverse

reinforcement must extend at least 805 mm into the span oneach side of the main beam, and the area of this transversereinforcement should not be less than (0.0015) (5500) (180)= 1485 mm2/5.5 m = 270 mm2. In Fig. 1-2, the bars 'mark5' extend 850 + 175 1025 mm (> 805 mm) into the span,and they have an area of (T10 at 150) 523 mm2/m(> 270mm2/m).

(e) Transverse reinforcement across edge beam(mark 7)

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effective flange width of the edge beam =bw+ 0.1L = 710mm; the min. area of the transverse reinforcement is

0.15%Lhf= (0.0015)(5500) (180) = 1485 mm2

= 270 mm2

/m.The area of the bars 'mark 7' (T10 at 150) is 523 mm2/mandthe bars extend the full effective width.

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Example 2 (Beam)

With reference to the typical floor of the building in Fig. 1,design and detail a main floor beam.

SOLUTION


From Table 3.4,

nominal cover for mild exposure condition = 20 mm cover

to main bars = 20 + link = 35 mm, say

From Table 4.4350 mm beam width and 35 mm cover to main bars, thefire resistance > 1 hour

Nominal cover = 20 mm

Fire resistance OK

St ep 2 Load ing

Dead load from 180 slab: = 31.7 kN/m

Self-weight (0.55 - 0.18) x 0.35 x 24 = 3.1 kN/m

Characteristic dead load gk = 34.8 kN/m

Characteristic imposed load qk = 3 x 5.5= 16.5 kN/m

Design load F = 1.4gk + 1.6qk

= 48.7 + 26.4 = 75.1 kN/m

gk=34.8 kN/m qk=16.5 kN/m F =75.1 kN/m

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Step 3 Ul t ima t e m omen t s BS 8110: Clause 3.2.1.2.4

continuous beams in framed structure can be analysed byassuming that the columns provide vertical restraint but norotational restraint. Figure 2-1 bending moment envelope of 3laoding cases with 0 distribution

St ep 4 Shear for ces

Figure 2-2 shows the shear force envelope.

St ep 5 Long i t ud ina l re in fo r cem en t

Internal support B

M = 629 KNm

(No moment redistribution)

K = M/fcubd2 = (629)( 106) = 0.199 > 0.156

(40)(350)(4752)

Compression reinforcement is required.d'= 50 mm

From z/d = 0.5 + (0.25 K/0.9) and z/d = d 0.45x

zld = 0.775 xld = 0.5

From Cl. 3.4.4.4,As = M Mu (where Mu= Kfcubd

2)0.87fy (d -d')

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= (629)(106) - (0.156)(40)(350)(4752)(0.87)(460)(475-50)

= 800 mm2

(From Cl. 3.12.5 when compression steel is required, theminimum area to be provided = 0.2% bh = 385 mm2< 800mm2OK)

As = Mu/(0.87fyz)+ As (where Mu = K fcu bd2)

= (0.156)(40)(350)(4752) + 800

(0.87) (460) (0.775)(475)

= 4145 mm2

Top 6T32 (4825 mm2)

Bottom 2T32 (1608 mm 2)

Span AB Cl. 3.4.1.5,

effective flange width = 350 + 0.2(0.7 of 9000)

= 1610 mm

From bending moment envelope

M = 760 kNm

M/( fcu bd2) = 760 x 106

(40)(1610)(5) where d = 500= 0.047 < 0.156

singly reinforced

From z/d = 0.5 + (0.25 K/0.9) and z/d = d 0.45x

Z/d = 0.94 x/d = 0.13

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Note that: z/d

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From Table 3.29,ultimate anchorage bond length = 32= 800 mm

required anchor length = (1091)(800)(1473)

= 593 mm

Span BC

effective flange width = 350 + (0.2)(0.7 of 7000)

= 1330 mm

From bending moment envelope

M = 460 kNm

K = M/( fcu bd2) = (460)( 106) = 0.035 < 0.156

(40)(1330)(5002)Singly reinf.

z/d = 0. 94 x/d = 0.13 (i.e. 0.9x < hf)N.A in flange

As = M/( 0.87fy z) = (460)( 106)

(0.87)(460)(0.94)(500) = 2446 mm2

5T25 U-bars (2454mm2)

End support C

Design for 40% of the initial fixed-end moment

M = 40% of 307 kNm

= 123 kNm

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K = M/( fcu bd2) = (123)( 106) = 0.035 < 0.156

(40)(350)(5002)

Singly reinf.

z/d = 0.94 x/d =0.13

As = M/( 0.87fy z) = (123)( 106)

(0.87)(460)(0.94)(500) = 654 mm2

2T25 U-bars (982 mm2)From Table 3.29,

ultimate anchorage bond length = 32= 800 mm

required anchor length = (654)(800)(982)

= 533 mm

See Fig. for curtailment diagram

St ep 6 Shear Re in fo rcem ent

Min. tension reinf. = 2T25 i.e. As= 982 mm

2

100As/bvd = (100)(982) = 0.56 (350) (500)

From Table 3.9vc= 0.61 N/mm

2

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Min. Links provided wherev (vc+ 0.4) i.e. V (vc+ 0.4) bvd

(Table 3.8 Cl. 3.4.5.10)

V (0.61 + 0.4)(350)(500) = 177 kN

Fig. shows the 177 kN limits for min. linkssuperimposed on the shear force envelope

Asv(min links) (0.4bvsv)/(0.87fyv)

Asv/sv(min links) (0.4bv)/(0.87fyv)= (0.4)(350) = 0.64 (0.87)(250)

Min. links R12 @ 300mm (Asv/sv=0.75)

Check v

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St ep 7 Def lect ion

Table 3.10 basic span/depth ratio = 20.8

(for bw/b 0.3)

M/bd2= (760)(106) = 1.89(1610) (5002)

Table 3.11Modification Factor =1.1

Allow. Span/depth ratio = 20.8 x 1.1

= 22.9

actual span/depth ratio = 9000/500 = 18.0 < 26.6

span/depth ratio ok

St ep 8 Crack ing

Bar spacing and corner distance

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Ten sion Spacing ab Spacing acbars Moment Actual Allowed Actual Allowed

Redist.Support B 0% 120mm 135mm 60mm 80mmtop

Span AB: 0% 51mm 160 mm 60 mm 80 mmbottom

Span BC 0% 137.5mm 132 mm 60 mm80mm

bottomCrack widths OK

Step 9 Robustness(BS8110:Clauses3.1.4.3,3.12.3.4, and3.12.3.6)

Internal longitudinal ties (BS 81 10: Clause 3.12.3.4. 1).

Ft= 20 + 4 times No. of storeys = 36 kN/m

From Step 2 of the solution to floor design,gk= 32.1-5.5 = 5.84 kN/m

2of floor

qk= 16.5/5.5 = 3.00 kN/m 2 of floor

Check:(gk+ qk) (lr/5) Ft= {(5.84 + 3.00)}(9/5)(36) 7.5 7.5

= 76.4 kN/m > Ft

Hence

Tie force = (76.4) (5.5) = 420 kNMin. cont. internal tie = 420 x 1000 = 1049 mm2

(0.87)(460)

Bottom 2T32 (1608 mm2continuous support B)

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Check lap length (Tension laps) Table 3.29cover to lapped bars= 20 + link = 36 mm < 2(64mm)Cl. 3.12.8.13(a) & (b)

spacing bet. adjacent laps 100 mm< 6(192 mm)

Hence apply a factor of 1.4 to Table 3.29, so that:

full tension lap length = (1.4)(32) = 45

required lap length = (As req.) (45)

(As prov.)=[1049] (45)(32) = 939 mm

[1608]

Extend the tie bars 1500 mm from the column face at thesupport B.

External column tie (BS 8110: Clause 3.12.3.6. 1).

(a) 2Ft = 2(36) = 72 kN.

(Floor height)(Ft)= [4 - 0.18] (36) = 55 kN2.5 2.5

Hence 72 kN need not be considered further

(b) 3% of total design ultimate vertical load carried by column

3% of 1500 kN (45kN)

45 kN < 55 kN in (a)

Hence, tie force 55 kN.

Minimum continuity external tie

={(55)(1000)}/{0.87(46)}= 137 mm2

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The 25 mm U-bars at the external supports provide ample area.

St ep 10 Rein fo r cemen t de ta i l s

See reinf. Details.

Figure shows the reinforcement details.

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Example 3 (Internal Column)

Design and detail an internal column, B9 for to the braced

building frame shown

SOLUTION


From Table 2.5-7,

nominal cover for mild exposure = 20 mm (use 30 mm)

cover to main bars = 30 + link = 40 mm, say

From Table 3.5-1,

for 380 square column with 40 mm cover to main

bars, the fire resistance exceeds 1 hour

Nominal cover = 30 mm

Fire resistance OK

St ep 2 Co lum n and beam s t i f f n esses

Figure for details

Columns: all floorsI = bh3= 380 X 3803 = 1.74 x 109

12 12I = 1.74 x 550 9 = 435 x 103

4000

Floor beamsI = bh3= 350 X 5503 = 4.85 x 109mm4

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12 129m span I = 4.85 x 109 = 539 x 103mm3

9000

7m span I = 4.85 x 109

= 693 x 103

mm3

7000

Roo f beam s

I = bh3= 350 X 5003 = 3.65 x 109mm4

12 129m span I = 3.65 x 109 = 406 x 103mm3

9000

7m span I = 3.65 x 109 = 521 x 103mm3

7000

St ep 3 Mom en t s in co lum n

Floor junctions

(I/l) =(435 + 435 + 270 + 347) x 103

= 1487 x10

3

For Loading Case 1:-Out-of-balance mom. = 507 142 = 365 kNm

M in column = 365 x (435/1487) =107 kNm

Roof junction

(I/l) =(435 + 203 + 261) x 103= 899 x 103

Out-of-balance mom. = 226 kNm

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M at top of 3rdfloor column = 226x (435/899) =109kNm

St ep 4 Ef fec t i ve co lum n h e igh t

Foundation to Ist floorEqn 30 Cl. 3.8.1.6.1

effective col. Ht. l e= lO

From Table 3.21 Cl. 3.8.1.6.2

N-S direction = 0.9 (end conditions: top = 1, bottom =3)

Hencel ex= 0.9 l 0= 0.9 x 4000 = 3600 mm

l ex/ h= 3600/380 = 9.5 < 15

E-W direction: = 1.0 (end conditions: top = 3, bottom =3)

Hencel ey = 1.0 x 4000 = 4000 mm

l ey/ b = 4000/380 = 10.5 < 15short column

lst to 2nd floor; 2nd to 3rd floor; 3rd floor to roof

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From Table 30:

N-S direction = 0.75 (end conditions: top = 1, bottom =1)

Hencel ex= l O = 0.75 x 4000 = 3000 mm

l ex/ h= 3000/380 = 7.9 < 15

E-W direction: = 1.0 (end conditions: top = 3, bottom =3)

Hencel ey= 1.0 x 4000 = 4000 mm

l ey/ b = 4000/380 = 10.5 < 15short column

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St ep 5 Ax ia l loads

Axial loads on column (see Table )

Colum n des ign loads (k N)

BeamLoad

Imposeda DeadaFloorsuppo r t

ed V ( kN) V

=

1 .6qk V= 1 .4 qk

Case 3 1.4gk+1.6q

k

1.4gk+ 1.6qk

Roof 617 b

16

8

168 449 b

Sel f -w eigh t 12 461

3r d

733 25

8

426 475


2nd

733 25

8

684 475


1st 733 25

8

942 475


agk& qk Taken from step 2bFor roof, taken from roof main beam sehich are not shown

Reduced imposed loads (see Table )

3rd floor to roof 100% of 168 = 168 kN

2nd to 3rd floor 90% of 426 = 382 kN

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Ist to 2nd floor 80% of 684 = 547 kN

Foundation to lst floor 70% of 942 = 659 kN

Total axial loads (see Table above)

3rd floor to roof N3r = 168 + 461 = 629 kN2nd to 3rd floor N23 = 382 + 948 = 1330 kNlst to 2nd floor N12 = 547 + 1435 = 1982 kNFoundation to lst flr Nfl = 659 + 1922 = 2581 kN

N3r= 629 kN N23= 1330 kN N1212 = 1982kN: Nfl= 2581 kN

Step 6 Desi gn bend i ng m om en t s

BS 8110's design minimum eccentricity = 0.05h

Minimum design moment = 0.05hN (Cl. 3.8.2.4)

Foundation to lst floor level:

0.05h Nfl = 0.05 x 380 X 10-3X 2581 (see Step 5)

= 49 kNm

Elsewhere, N < Nfl and hence 0.05hN < 49 kNm.Therefore the column design is governed by thecolumn moments in Step 3 as these are larger than0.05hN.

M (roof Junction) = 109 kNm M (floor junction) = 107kNm

St ep 7 Re in fo r cem ent

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fcu = 40 N/mm2; fy= 460 N/mm

d = 380 - 40 - (bar )/2 = 0.85 approx.h 380

Refer to Design chart, and the results are as shown inTable below

location N/bh M/bh2 Ascreqd. A scprov.N/mm2 N/mm2 mm2 mm2

3rdto roof 4.4 2.0 0.4% bh = 578c 4T25(1960)

2ndto 3rdfl. 9.2 2.0 0.4% bh = 578c 4T25(1960)1stto 2ndfl. 13.8 2.0 0.4% bh = 578c 4T25(1960)

Fdn to 1stfl. 17.9 2.0 1.4% bh = 2021 4T25(1960)c

Min. steel ratio = 0.4 %

St ep 8 Rein fo r cem ent de t a i l s(see Fig )

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rc design

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