Worked Examples for Design of Concrete Buildings

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Thisbook of worked examples has been prepared by: British CementAssociation Ove Arup & Partners and S. B Tietz & Partners The work was monitored by the principal authors:

A. W. Beeby BSc, PhD, CEng, MICE, MlStructE, FACI Professor of Structural Design, Dept of Civil Engineering, University of Leeds (formerly Director of Design and Construction, British CementAssociation), R. S. Narayanan BE(Hons), MSc. DIC, CEng,FiStructE Partner, S. B. Tietz & Partners, Consulting Engineers,

and R. Whittle MA(Cantab), CEng, MICE Associate Director, Ove Arup& Partners, and edited by: A. J. Threlfall BEng, DIC Consultant (formerly a Principal Engineer at the British Cement Association). This publication was jointlyfundedby the BritishCement Associationand theDepartmentofthe Environmentto promote and assist the use of DD ENV 1992-1-1 Eurocode 2: Part 1.

The BritishCement Association,BCA, isaresearchand information bodydedicated tofurtheringthe efficientand proper design and execution of concrete construction. Membership of BCA's Centre for Concrete Information is opento all involved in the construction process. BCA is funded by subscriptions fromcementproducers, through joint ventures, sales of publications, information and training courses, and the carrying out of research contracts. Full details are available fromtheCentre forConcrete Information, British Cement Association, Century House, Telford Avenue, Crowthorne, Berkshire RG11 6YS. Telephone (0344) 725700, Fax (0344) 727202. Ove Arup& Partnersisan internationalfirm offering awide range ofdesignand specialist services for theconstruction industry. S.

B. Tietz & Partnersoffer consultaricy services in civil, structural and traffic engineering.

A catalogue and prices for BCA publications can be obtained from PublicationSales, CentreforConcrete Information, at the aboveaddress. 43.505 First published 1994 10DM A 701A 1AA 1 Price group M © British Cement Association 1994

Published by

British Cement Association Century House, Telford Avenue, Crowthorne, Berks RG11 6YS Telephone (0344) 762676 Fax (0344) 761214 From 15Aprll 1995 the STD Code will be (01344)

All advice or information fromthe British Cement Association is intended forthose who will evaluate the significance and limitations of its contents and take responsibility for its useand application. No liability (including that for negligence) for any loss resulting fromsuchadviceor information is accepted. Readers should note that all BCA publications are subjectto revision fromtime to timeand should therefore ensure that they are in possession of the latest version.

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FOR THE DESIGN

OF CONCRETE

BUILDINGS

Based on BSI publication DDENV1992-1-1: 1992. Eurocode 2: Design of concrete structures. Part 1. General rules and rulesfor buildings.

Published by the British Cement Association in conjunction with: Ove Arup & Partners S.B. Tietz & Partners The Department of the 13 Fitzroy Street 14 Clerkenwell Close Environment LondonW1P 6BQ Clerkenwell 2 Marsham Street London EC1R OPQ Tel: 071-636 1531 London SW1P 3EB Tel: 071-490 5050 Tel: 071-276 3000

July 1994

FOREWORD Eurocode 2: Design of concrete structures, Part 1: General rules and rules for buildings (EC2)1 sets out both the principles for the design of all types of concrete structure, and design rules for buildings. Rulesfor other typesof structureand particular areas oftechnology, including precast concrete elements and structures, will be covered in other parts of EC2. EC2 contains aconsiderable number of parametersforwhichonly indicative valuesare given. The appropriate values for use in the UK are setout in the National Application Document (NAD)1 which has been drafted by BSI. The NAD also includes a number of amendments to the rules in EC2 where, in thedraftfor development stage of EC2, it was decidedthat the EC2 rules either did not apply, or were incomplete. Two such areas are the design for fire resistance and the provision of ties, where the NAD states that the rules in BS 8110(2) should be applied. Attention is drawn to Approved Document A (Structure) related to the Building Regulations 1991, which states that Eurocode 2, including the National Application Document, is considered to provide appropriate guidance forthedesign of concrete buildings in the United Kingdom. Enquiries of a technical nature concerning these worked examplesmay be addressed to the authors directly, or through the BCA, or to the Building Research Establishment.

CoNTENTS 1

8

1.1

INTRODUCTION AND SYMBOLS Introduction 5 5 1.2 Symbols

8.1

180 185

4

188

2 2.1

2.2

23 2.4 2.5 2.6 2.7 2.8 2.9

COMPLETE DESIGN EXAMPLE Introduction Basic details of structure, materials and loading Floor slab Main beam Edge beam (interior span) Columns Foundation Shear wall Staircase

SPECIALDETAILS

Corbels 8.2 Nibs 8.3 Simply supported ends Surface reinforcement

191

15

9 15 17

20 30 34 39 43 49

PRESTRESSED CONCRETE Introduction 193 9.2 Design data 193 93 Serviceability limit state 195 9.4 Ultimate limit state 204 9.5 Minimum and maximum areas of reinforcement 207 9.6 Reinforcement summary 207 9.1

SERVICEABILITY CHECKS BY CALCULATION 10.1 Deflection 208 10.2 Cracking 219 10

3 3.1

3.2

33 3.4 3.5

4

BEAMS 53 Introduction methods for shear 53 Design Shear resistance with concentrated loads close to support 63 70 method for torsion Design Slenderness limits 81

4.1

SLABS Solid and ribbed slabs

4.2

Flat slabs

5

COLUMNS Introduction 132 Capacity check of a section by strain compatibility 132 Biaxial bending capacity of 137 a section Braced slender column 141 Slender column with biaxial 143 bending Classification of structure 147 151 Sway structures

5.1

5.2

53 5.4 5.5 5.6 5.7

WALLS Introduction 6.2 Example

222 222

11.1

12

82 109

LOAD COMBINATIONS Introduction 12.2 Example 1 — frame 123 Example 2 — continuous beam 1 12.4 Example 3 — continuous beam 2 12.5 Example 4 — tank

236

12.1

237 240 243 245

DESIGN OF BEAM AND COLUMN SECTIONS 13.1 Concrete grades 246 13.2 Singly reinforced rectangular beam sections 246 133 Compression reinforcement 248 134 Flanged beams 249 13.5 Symmetrically reinforced 249 rectangular columns 13

...

6

6.1

DEEP BEAMS Introduction 11.2 Example 11

154 154

7

FOUNDATIONS Ground bearing footings 7.2 Pilecap design 7.1

158

REFERENCES

172

fIE

256

I. IHTRODUCTON AND SYMBOLS 1.1

Introduction and symbols

Themainobjective of this publication is to illustrate through worked examples how EC21 may be used in practice. It hasbeen prepared for engineers who are generally familiar with design practice in the UK, particularly to BS 8110(2). Theworked examplesrelateprimarilyto in-situ concrete building structures The designs are in accordance with EC2: Part 1 as modified by the UK National Application Document1. Where necessary,the information given in EC2 has been supplemented by guidance taken from other documents. Thecoreexample, in Section 2, is a re-designofthein-situconcrete office block used inthe BCA publicationDesignedand detailed (BS 8110: 1985), by Higgins & Rogers4. Other design aspects and forms of construction are fully explored by means of further examples in Sections 3 to 12. Equations and charts for the designof beam and column sections, taken from the Concise Eurocode for the design of concrete buildings5, are given in Section 13. Publications used in the preparation of this book, and from which further information may be obtained, are listedirt the References Unless otherwise stated, all references to BS 8110 refer to Part 1. Twoconventionshave been adopted inthe preparation ofthis book. Statements followed by

OK' mark places where the calculated value is shown

to be satisfactory. Green type is used to draw attention to key information such as the reinforcement to be provided. Thecalculationsare cross-referencedtothe relevantclauses in EC2 and, where appropriate, to otherdocuments; all references in the right-hand margins are to EC2 unless indicated otherwise. Thesymbols usedthroughout the publication are listed and defined below,and are generally those used in EC2 itself.

1.2 Symbols A

Ak

A A

A' Amin

Apr Areq

A11

A

A5, As,,,

Aymin

E,

Area of cross-section Area of concrete cross-section Area of concrete within tensile zone Area of concrete tensile zone external to links Area enclosed within centre-line of thin-walled section Area of prestressing tendons Area of tension or, in columns, total longitudinal reinforcement Area of compression reinforcement Minimum area of tension or, in columns, total longitudinal reinforcement Area of tension reinforcement provided

Area of tension reinforcement required Area of surface reinforcement Area of transverse reinforcement within flange of beam Area of tension reinforcement effective at a section or, for torsion, area of additional longitudinal reinforcement Area of shear reinforcement or torsion links Minimum area of shear reinforcement Effective modulus of elasticity of concrete

(tTROOUCTLOL &ND SYMBOLS

E0

E E5

F F5 FSd

Secant modulus of elasticity of concrete at transfer Secant modulus of elasticity of concrete Modulus of elasticity of reinforcement or prestressing steel Force due to concrete in compression at ultimate limit state Force in tension reinforcement or prestressing tendons Design value of tie force in pilecap

at ultimate limit state

Fsdp Design value of supportreaction Tie force in corbel or due to accidental action Fj Vertical force applied to corbel or, for sway classification of structures, sum F of all vertical loads under service conditions Characteristic value of permanent action or dead load Gk Gld

Characteristic dead floor load

Gkr

Characteristic dead roof load

H

Overall depth

H0

I

of tank

Horizontal force applied to corbel Second moment of area of cross-section Second moment of area of uncracked concrete section

1

Second moment of area of cracked concrete section Second moment of area of beam section Second moment of area of concrete section Second moment of area of column section

'slab

Second moment of area of slab section Second moment of area of section in x direction Second moment of area of section in y direction

J

-' K

K1 1<2

M M0 Mcr

St Venant torsional stiffness of rectangular section St Venant torsional stiffness of total section Deflection-curvaturefactor dependent upontheshapeofthe bending moment diagram Reduction factor for calculation of second order eccentricity Coefficient taking account of decrease in curvature due force Bending moment Moment of force, F0, about tension reinforcement Moment causing cracking Momentof-force, N0, about x axis

M0,,

Moment of force, N, about y axis

M0

First order moment

M'RdC

Design moment of resistance Moment of force, NRdC, about mid-depth of section Moment of force, N'RdC, about mid-depth of section

MAdS

Moment of force, NRdS, about mid-depth

MRd MRdC

of section

to increasing axial

WTRODUCT%OII AHD SYMBOLS

M MsdX

Design value of applied moment Design moment in x direction

Msdl

Design moment in y direction First order moment at end 1

Msd2

First order moment at end 2

Msd/S

Design moment in column strip Design moment in middle strip

Msdy

MSdms

Moment in span Moment at support

Mtm

Maximum moment transfer value

M M

Moment about x axis Moment about y axis

N

Axial force

N

Axial force due to concrete in compression

NRdC

Design resistance to axial force Design resistance to axial force due to concrete

N'RdC

Design resistance to axial force due to concrete

NRd

depthx>h

of hypothetical section of

Nsd

Design resistance to axial force due to reinforcement Design value of applied axial force

NSdm

Mean applied axial force

NRdS

P

P,

Prestressingforce or point load Average prestressing force along tendon profile

m.o

Initial prestressing force at transfer Mean effective prestressing force at time t

Pmoo P0

Final prestressing force after all losses Maximum initial prestressing force at active end

of tendon

req

Requiredprestressing force Final prestressing force at service

k

Characteristic value of variable action or imposed load

kf

Characteristic value of imposed floor load

kr

Characteristic value of imposed roof load

RA

Reaction at support A

RB

Reaction at support B

S

First moment of area of reinforcement about centroid of section

S1

First moment of area of reinforcement about centroidof uncracked section

S

First moment of area of reinforcement about centroidof cracked section

TRd1

Design value of tensile force in longitudinal reinforcement Maximum torsional moment resisted by concrete struts

TRd2

Maximum torsional moment resisted by reinforcement

Td

(ILTRODUCTION AND SYMBOLS

TsdfI

Design value of applied torsional moment Torsional moment applied to flange

Tsd tot

Total applied torsional moment

TsdW

Torsional moment applied to web

VA

Shear force at support A

VB

Shear force at support B

Tsd

VCd Ve,c

t'

VRd1

VRd2 VRd3 VSd VsdX VsdY

Design shear resistance provided by concrete Shear force at exterior support Shear force at interior support Design shear resistance of member without shear reinforcement Maximum design shear force to avoid crushing of notional concrete struts Design shear resistance of member with shear reinforcement Design value of applied shear force Design shear force in x direction Design shear force in y direction

Vmax Maximum design shear force VWd

Wb

Design shear resistance provided by shear reinforcement Section modulus at bottom fibre Section modulus at centroid of tendons

Wk

Characteristic value of wind load

W

Section modulus

a

at top fibre Distance or deflection or maximum drape of tendon profile

a1

Deflection based on uncracked section

a11

a0

a0 a

Deflection based on cracked section Distanceof loadfromfaceof support (corbel) orfromcentre-line of hanger bars (nib) Deflection due to concrete shrinkage

a1

Distance from face of support to effective centre of bearing Horizontal displacement of the envelope line of tensile force

a0

Total deflection

a a

Distance between positions of zero and maximum bending Deflection at distance x along span

a1,;

Values of a

b

Width of section or flange width or lateral cover in plane of lap

at ends of span

be

Average width of trapezoidal compression zone Width of effective moment transfer strip

beff

Effective width of flange

bmin

Minimum width of support beam

br

Width of rib

bay

Width of support

OTOII MiD b

SYMBOLS

Mean width of section over the tension zone

b

Minimum width of section over the effective depth

c

Cover to longitudinal torsion reinforcement

c1,c2

Support widths at ends of beam

d

Effective depth of section

d'

Depth to compression reinforcement Average effective depth for both directions

day

dcrt

Depth to bar considered Distance of critical section for punching shear from centroid of column

dH

Effective depth of flange Effective depth for punching shear check in column head

dm

Maximum effective depth for both directions

dmin

Minimum effective depth for both directions

db

d d d1

d2

Effective depth in

x direction

Effective depth in

y direction

Effective depth to bars in layer 1 Effective depth to bars in layer 2

eay

Additional eccentricity due to geometrical imperfections Additional eccentricity in the y direction

ee

Additional eccentricity in the z direction Equivalent eccentricity at critical section

e,

First order eccentricity in y direction

e01,e02

First order eccentricities at ends of column

ot

Total eccentricity

ey

Eccentricity in y direction Eccentricity in z direction

ea

e

e 1bd

Second order eccentricity Second order eccentricity in

y direction

Second order eccentricity in z direction Stress in concrete at bottom fibre Design value of ultimate bond stress Design cylinderstrength of concrete

Cube strength of concrete at transfer Characteristic cylinder strength of concrete cteff ctm

Effective tensile strength of concrete at time cracking is expected to occur Mean value of axial tensile strength of concrete Characteristic cube strength of concrete Design tensile strength of prestressing steel Characteristic tensile strength of prestressing steel

INTRODUCTION AND SYMBOLS

RdU

f

Design value of ultimate bearing stress Stress in reinforcement Stress in concreteat top fibre Design yield strength of reinforcement Characteristic yield strength of reinforcement

cld

Design yield strength of longitudinal torsion reinforcement

Wd

Design yield strength of shear reinforcement or torsion links

1ywk

h

Characteristic yield strength of shear reinforcement or torsion links Characteristic dead load per unit area Overall depth of section or liquid in tank Reduced value of h for separate check about minor axis of column section with biaxial eccentricities

ha

Active height of deep beam

h0

Overall depth

hf

Overall depth of flange Depth of column head

hH

of corbel at face of support

hmn

Larger dimension of rectangular section Smaller dimension of rectangular section

h0

Total height

hm

of structure in metres

Radius of gyration of section

k

Coefficient or factor

kA

Restraint coefficient at end A

kB

Restraint coefficient at end B

kboflom

Restraint coefficient at bottom

k

Minimum reinforcement coefficient associated with stress distribution Restraint coefficient at top

k1

Crack spacing coefficient associated with bond characteristics

k2

Crack spacing coefficient associated with strain distribution

1

Length or span

1'

Length of tendon over which anchorage slip is taken up Basic anchorage length Minimum anchorage length

1net

Required anchorage length Diameter of circular column

1eff

Height of column between centres of restraints Effective span

1effslab

Effective span of slab Distance from column face to edge of column head

Clear distance between faces of support

ThOD)VT1OI4 MD SYMBOLS

Distance between positions of zero bending or effective height of column or, for deep beams, clear distance between faces of support Length of compression flange between lateral supports

min

Required lap length or floor to ceiling height in metres Minimum lap length Greater of distances in metres between centres of columns, frames or walls supporting any two adjacent floor spans in direction of tie under consideration Effective span in x direction

ç 11,12

mSd

n

Effective span

in y direction

Lengths between centres of supports or overall dimensions

column head

Minimum design moment per unit width Ultimate design load per unit area or number sub-divisions

of rectangular

of tendons or number of

q

Average loss of prestressing force per unit length due to friction Equivalent load per unit length due to prestressing force profile

qk

Characteristic imposed load per unit area

r

Radius of bend or radius of curvature

p'

Radius of curvature based on uncracked section Radius of curvature based on cracked section

rr

Radius of curvature due to concrete shrinkage Radius of curvature due to concrete shrinkage based on uncracked section Radius of curvature due to concrete shrinkage based on cracked section

ot

Total radius of curvature

s

Spacing of shear reinforcement or torsion links or horizontal length of tendon profile Spacing of transverse reinforcement within flange of beam

Sm

Maximum spacingof shear reinforcement or torsion links

t

Average final crack spacing Thickness of supporting element or wall of thin-walled section

tmln

Minimum thickness of wall

u

Circumference of concrete section or critical section for punching shear

Uk

Circumference of areaAk

VAd1

Design shear resistance per unit length of critical perimeter,for slab without shear reinforcement

VRd2

Maximum design shear resistance per unit length of critical perimeter, for slab with shear reinforcement

VRd3

Design shear resistance per unit length of critical perimeter, for slab with shear reinforcement

s

v w

Design value of shear force per unit length of critical perimeter Support width or quasi-permanent load per unit length Design crack width

t4TODUCTION MID SYMBOLS

Wmin

x x'

Minimum width of support Neutral axis depth or distance along span from face of support or distance along tendon or column dimension in x direction Maximum depth of concrete in compression in direction of minor axis for column section with biaxial eccentricities

x

Depth of concrete in compression at position of minor axis for column section with biaxial eccentricities

y

Drape of tendon at distance direction

x along

profile

or column dimension in y

Distance from centroid of uncracked section to extreme tension fibre

z

Lever arm of internal forces Distance from centroid of section to centroidof tendons

a

Reduction factor for concrete compressive stress or modular ratio or deformation parameter

a1

Value of parameter based on uncracked section

a11

Value of parameter based on cracked section

aa

Effectiveness coefficient for anchorage

ae

Effective modularratio

Reduction coefficient for assumed inclination of structure due to imperfections Moment coefficients in x and y directions a1

Effectivenesscoefficient for lap

ared

Coefficient with several applications including shear resistance enhancement, effective height of column, St Venant torsional stiffness, punching shear magnification, design crack width Reduced value of shear resistance enhancement coefficient Coefficient associated with bond characteristics

a2

Coefficient associated with duration of load Partial safety factor for concrete material properties Partial safety factor for actions Partial safety factor for permanent action or dead load

7G.nf

Partial safety factor for permanent action, in calculating lower design value Partial safety factor for permanent action, in calculating upper design value Partial safety factor for actions associated with prestressing force Partial safety factor for variable action or imposed load

of reinforcement or prestressing tendons Ratio of redistributed moment to moment before redistribution Partial safety factor for steel material properties

5

Strain in concreteat bottom of section

ç €pm

Basic concrete shrinkage strain Final concrete shrinkage strain Minimum strain in tendons to achieve design tensile strength Strain in tendons corresponding to prestressing force mt

tVROQUCT(OL &IID SYMBOLS

Strain in reinforcement e(t,t0)

Estimated concreteshrinkage strain

sm

Mean strain in reinforcement allowing for tension stiffening effect of concrete Ultimate compressive strain in concrete Initial yield strain in reinforcement Distribution coefficient Moment coefficient

e

Angle of rotation or angle between concrete struts and longitudinal axis

X

Slenderness ratio

Xcrit

Critical slenderness ratio

Xm Xmin

Mean slenderness ratio of all columns in storey considered Slenderness ratio beyond which column is considered slender Coefficient of friction between tendon and duct or applied moment ratio

Iim v

Limiting value of applied moment ratio for singly reinforced section Efficiency factor or assumed inclination of structure due to imperfections

1'red

Reduced value of assumed inclination of structure

p

p' p1

Longitudinal force coefficient Tension reinforcement ratio or density of liquid Compression reinforcement ratio Longitudinal tension reinforcement ratio Longitudinal tension reinforcement ratios in x and y directions Effective reinforcement ratio

Shear reinforcement ratio w,mIn p1,p2

Minimum shear reinforcement ratio Principal and secondary reinforcement ratios in solid slabs Stress in concrete adjacent to tendons due to self-weight and any other permanent actions Average stress in concrete due to axial force Initial stress in concrete adjacent to tendons due to prestress Initial stress in tendons immediately after stressing (pre-tensioning) or immediately aftertransfer (post-tensioning) Stress in tension reinforcement calculated on basis of cracked section

°r

Value of a8 under loading conditions causing first cracking

TRd

Basic design shear strength Factor defining representative value of variable action Value of

(, for rare

load combination

Value of y' for frequent loading Value of & for quasi-permanent loading Mechanical ratio of tension reinforcement Mechanical ratio of compression reinforcement

(KTRODUCTION AND SYMBOLS

WIim

Limiting value of w for singly reinforced section

A1

Total vertical force applied to frame at floor Anchorage slip or wedge set Variation of longitudinal force in section of flange over distance

AH

Equivalent horizontal force acting on frame at floor imperfections Moment of force about mid-depth of section

j

a

due to assumed

Reduction in design moment at support ANRdC

Design resistanceto axial force due to concrete in areaof hypothetical section lying outside actual section

AP

Average loss of prestressing force due to elastic deformation of concrete Loss of prestressing force at active end of tendon due to anchorage slip

Loss of prestressing force due to creep, shrinkage and relaxation at time t AP(x) Loss of prestressingforce due to friction between tendonand duct at distance x from active end of tendon AP(t)

Variation of stress in tendon due to relaxation

Bar size or duct diameter or creep coefficient (t,t0)

Creep coefficient, defining creep between times deformation at 28 days

(oo,t0) Final creep coefficient

t and t, related to elastic

.

2 COMPLETE DESIGN EXAMPLE 2.1 Introduction Design calculations for the main elements of a simple in-situ concreteoffice block are set out. The structure chosen is the same as that used in Higgins and Rogers' Designed and detailed (BS 8110: 1985)(). Calculations are, wherever possible, given in thesame orderas those in Higgins and Rogersenabling a directcomparison to be made between BS 8110(2) and EC21 designs. For thesame reason,a concrete gradeC32140 is used. This is nota standard grade recognized by EC2 or ENV 206(6), which gives grade C35145in Table NA.1. Some interpolation of the tables in EC2 has, therefore, been necessary.

Theexample was deliberatelychosen to be simple and to cover aconsiderable range of member types Comparisonshowsthat, forthis typeof simplestructure, there is very little difference between BS 8110 and EC2 in the complexity of calculation necessary or the results obtained.

2.2 Basic details of structure, materials and loading These are summarized in Table 2.1 and Figure 2.1. Table 2.1

Design information

Intended use Laboratory and office block

Fire resistance 1 hour for all elements Loading (excludingself-weightof structure) — Roof imposed (kN/m2) — finishes (kNIm2)

Floors

— —

imposed including partition allowance (kN/m2) finishes (kNIm2)

1.5 1.5 4.0 0.5

Stairs

— —

imposed (kN/m2) finishes (kNIm2)

4.0 0.5

External cladding (kNIm)

5.0

Wind load 40

Speed(m/sec) Factors

1.0 0.83 1.0

S1

2 S3

1.1

Cl Exposure class

2b (external) and 1

(internal)

Subsoil conditions Stiff clay — no sulphates

Allowable bearing pressure (kNIm2)

200

Foundation type Reinforced concrete footings to columnsand walls Materials

GradeC32/40 concrete with 20 mm maximum aggregate Characteristic strength of main bars (N/mm2) Characteristic strength of links(NImm2) Self-weight of concrete (kN/m2)

460 250

24

COMPLETE DESG4 EXAMPLE

8

(Th 500O1,0,O00

—175 waIl 1

175 walL

TYPICAL FLOOR PLAN

R

c7 3

Notes

2

I 2

All columns

300 x 300 500 x 300 350 x 300 Internal column bases

Main beams 3. Edge beams

c7

1,

2750 x 2750

0

x 600

TYPICAL CROSS SECTION

Notes 1. N 2

wind direction WIND

C

P

—

— C

.

C

Ft

1

20 + 4n0

— Peripheral tie Int — Internal tie C — Column (external) tie W — Wall (external) tie P

—

—

Ia

—

—

P C

C

TIE PROVISION

C

C

C

l(ey

mt

—

C

CP

C

C

—

W

LateraL bracing in E-.W direction provided by staircase, and infill masonry panels, on grid 1/H-i, and grid 31H—J

RESISTANCE

-

Int

Sendwindshearforcewallsresisted by

C

(at

.,-

C

horizontcit ties

20 + 16

Figure 2.1 Structural details

a 36

60 kN

COMPLETEQES(GM EX&MPLE

2,3 Floor slab 2.3.1 Idealization of structure Consider as a one-waycontinuous slab on knife-edge supportsand design a typical 5 m interior span where

= 32 N/mm2 = 460 N/mm2

2.3.2 Cover for durability and fire resistance Nominal cover for exposure class 1 (internal) is 20 mm. Cover should not be lessthan the bar size when 20 mm maximum aggregate size is used. 175 mm slab with 20 mm cover will give 1.5 hours fire

resistance.... OK

NAD Table 6 4.1.33

NAD 6.1(a)

&BS 8110 Table 3.5 & Figure 3.2

Use 20 mm nominal cover bottom and top

2.3.3 Loading Self-weight of slab = 0.175 Finishes = 0.5 kN/m2

x

24 =

4.2 kN/m2

(g) = 4.7 kN/m2 Characteristic variable load (q) = 4.0 kN/m2 Characteristic permanent load

Design permanent load Design variable load

= 135 x 4.7 = 1.5 x 4.0

= =

Table 2.2

6.35 kN/m2 6.0 kN/m2

2.3.4 Design moments and shears Momentshave been obtained using momentcoefficientsgiven in Reynolds and Steedman's Reinforced concrete designer's handbook7, Table 33.

= 0.079 x 635 x 52 + 0.106 x

Support moment

6.0

x

52

= 2&4 kNm/m Span moment = 0.046 x 635 Design shear force

x 52 + 0.086 x 6.0 x 52

= 0.5 x 6.35 x 5 +

0.6

= 20.2 kNm/m

x 6.0 x 5 =

339 kN/m

2.3.5 Reinforcement 2.3.5.1 Support Assume effective depth = =

0.040

=

0.048,

175

—

20

—

6

= 149 mm

bd2fck

AfS yk bdfck

x/d = 0.092 (Section 13, Table 13.1)

COMPLETE DESIGN EXAMPLE

For zero redistribution, x/d should

A

be less than 0.45

OK

2.5.3.4.2(5)

OK

5.4.2.1.1(1)

= 498 mm2/m

Minimum area of reinforcement 0.6b d 0.0015

bd =

224 mm2/m

Use T12 @ 200 mm crs. (565 mm2/m)

2.3.5.2 Span M

= 0.028

bd2fCk

Afyk

=

0.033,

= 0.063 (Section

x/d

13, Table

13.1)

bdfCk

A

= 342 mm2/m

Use 112 @ 300 mm crs. (377 mm2/m) Note: Reinforcementareasdiffer somewhatfrom those givenby BS 8110 which permits design for the single loadcase of maximum load on all spanscombined with 20% redistribution. EC2 requiresalternateand adjacent spansto beconsidered. In this instance, no redistribution has been carried out but it would have been permissible to carry out 30% redistribution in the EC2 design. This would have resulted in an identical answer to that given by BS 8110 but ductilityclass H (as defined in prEN 10080(8)) reinforcement would need to be specified.

NAD Table 5

2.3.6 Shear Shear resistance of the slab withoutshear reinforcement is given by = VAd1 rRdk(l.2 + 40p,)bd

43.23 Eqn 4.18

where TRd

= 035 N/mm2

k

= =

1.6 —

Table 4.8

d = 1.6

565 1000 x 149

—

0.149

= 1.451

= 0.0038

Hence VRd1

= 1023 kN/m >

VSd

= 33.9 kN/m

OK

No shear reinforcement required Note: Since shear is rarely a problem for normally loaded solid slabs supported on beams, as the calculation hasshown, it is not usually necessary to check in these instances.

COMPlETE DESiGNEXAMPLE

2.3.7 Deflection Reinforcement ratio provided in span

=

x 149

1000

= 0.0025

Using NADTable 7(1) and interpolating between 48 for 0.15% and 35 for 0.5%, a basic span/effectivedepth ratio of 44 is given. By modifying according to the steel stress, the ratio becomes

NAD Table 7 4.4.3.2(4)

44(400x377) = 422 460 x342 The actual span/effectivedepth ratio is

5000

=

OK

33.6

149

Had EC2 Table 4.14 been used instead of NADTable 7, the basic ratio before modification would have been 35, which would not have been OK.

2.3.8 Cracking For minimum area of reinforcement assume ct.eff

=

4.4.2.2

3 N/mm2

k

= 0.4

k

=

0.8

A0

=

0.5

x 175 x

1000

= 87500 mm2

Hence

A

= kkftffA!o

Eqn 4.78

= 0.4 x 0.8 x 3

x

87500/460 =

183 mm2/m

Area of reinforcement provided = 377 mm/m No further check is necessary as h = 175 Maximum bar spacing

=

OK 200 mm

500 mm

3h

OK

4.4.23(1) NAD Table 3 5.4.a2.1(4)

2.3.9 Tie provisions The NAD requires that ties are provided in accordance with BS 8110. Internal tie in E—W direction, with Tie force

=

F

x

(g

+

7.5

qk)

x—lr

5

=

36 kN/m width, is given by

= 36 x

(4.7

+ 4) x — 5 =

7.5

5

41.8 kN/m

NAD 6.5(g) BS 8110 3.123.4

tONPtETEDESGI4 EXAMPLE

Minimum area

41.8

=

x

i0

=

460

91 mm2/m

Thus this area ofthe bottom reinforcementis the minimum that should be made continuous throughout the slab.

2.3.10 Reinforcement details The reinforcement details are shown in Figure 2.2.

B—B

COVER

o outer

bars 20

Figure 2.2 Slab reinforcement details

2.4 Main beam 2.4.1 Cover for durability and fire resistance Nominal cover for exposure class

1

(internal) is 20 mm.

NAD Table 6

Nominal cover for

1

hour fire resistance is 20 mm.

BS 8110 Table 3.5

Use 20 mm nominal cover to links

COMPLETEQESLG4 EXAMPLE

2.4.2 Loading Permanent load from slab (Section 2.3.3) Self-weight of beam =

(0.5

—

0.175)

Characteristic permanent load (g) Characteristic imposed load (q) Maximum design load = Minimum design load =

=

x 0.3 x

4.7

x

5 =

23.5 kN/m

24 = 2.3 kN/m

= 25.8 kN/m

= 5

x4

= 20 kN/m

l.3Sg + i.5qk = 64.8 kN/m = 34.8 kN/m

2.3.3.1

l35

2.3.22.(4)

2.4.3 Analysis 2.4.3.1 Idealization of structure and load cases

The structure is simplified as a continuous beam attached to columns above and below, which are assumed to be fixedat their upper ends and pinnedat the foundations, as shown in Figure 23.

3500

4000

Figure 23 Idealization of structure 2.4.3.2 Design moments and shears These are summarized in Table 2.2 and Figures 2.4 and 2.5.

2.533

COMPLETE DESIGN EXAMPLE

Table 2.2 Results of frame analysis Loadcase 1

Load case2

Loadcase3

Load per m on 8 m span (kN) Loadper m on 6 m span (kN)

64.8 64.8

64 8

34 8

34.8

64.8

Upper LH column moment (kNm) Lower LH column moment (kNm)

103 68

109 72

50

LH end of 8 m span moment (kNm) LH end of 8 m span shear (kN)

33

—171

—180

233

238

—82 119

242

256

116

RH end of 8 m span moment (kNm) RH end of 8 m span shear (kN)

—382

—345

—242

286

280

159

Upper centre column moment (kNm) Lower centre column moment (kNm)

33 18

55 29

3 2

Middle

of 8 m span moment (kNm)

LH end of 6 m span moment (kNm) LH end of 6 m span shear (kN) Middle

—331

—262

—247

240

146

223

98

20

130

—57

—12 63

—76

7 5

46

of 6 m span moment (kNm)

RH end of 6 m span moment (kNm) RH end of 6 m span shear (kN)

149

34

Upper RH column moment (kNm) Lower RH column moment (kNm)

REDISTRIBUTION Case 1 — Reduce Reduce Case 2—Reduce Reduce Case

3— No

22

AND AT 171 to 126 (see 2) 382 to 268 (—30%) 180 to 126 (_30%) 365 to 268 (see 1)

382

166

30

(1

redistribution

(2) 180

(2) elastic

(R)

(3) eLastic

76

(3)

(1) redistributed (2) elastic (1)

&

325 Envelope

Moments

in

kNm

Figure 2.4 Bending moment envelope

(2) redistributed

COAPLETE

SGI EX&MPLE

(1) (3)

(R) 242

(1) 233 (3)

240 223

119

Case 1 Case 2 Case 3

149 (1) 166 (3)

Redistributed Envelope

Forces in

277 (R)

286 (1)

kN

6000

8000

-J

—I——

Figure 2.5 Shear force envelope

2.4.4 Reinforcement for tiexure 2.4.4.1 Internal support From bending moment envelope

M = 268kNm 8 1iim

=

0.7

=

0.0864 and WIim

and xld

M

—

(8 — 0.44)/1.25

=

= 0.1084 (Section 13, Table 13.2)

268x106

— —

300

bdç<

0.208

=

x 4402 x 32

0.1442

> hm

Therefore compression reinforcement is required = Asfyk =

I'IIm

bdfck

0.87 (1— d'Id)

— = 0.1442 0.0864 —

0.87 (1

50/440)

= 0.0750 (Section 13)

Af = = —-

wurn

bdfck

+ w' = 0.1084 + 0.0750 = 0.1834 (Section 13)

= 0.1834 x 300 Since d'Ix

=

d'/0.208d

Increase w' to 1

—

0.429

1 — 0.546

A' = S

0.0943

x

300

x 440 x

32/460

= 0.546 > 0.075

—

çk'805)

=

= 0.0943

x 440 x

Use 4T25 (1960 mm2) top Use 2T25 (982 mm2) bottom

(1

= 1684 mm2

32/460

= 866 mm2

0.429

2.5.3.4.2 Eqn 2.17

COMPLETE DEStGIL EX&MPLE

2.4.4.2 Near middle of 8 m span From bending moment envelope M = 325kNm 8

>1.0

Effective flange width =

x

325

=

1660

300

+ 0.2 x 0.85 x 8000 =

108

x 4502 x 32

1660 mm

2.5.2.2.1 Eqn 2.13

= 0.030

x/d = 0.068 (Section 13, Table 13.1) Neutral axis is in flange since x =

A8

=

31

< 175 mm

0.035 (Section 13, Table 13.1)

= 0.035 x 1660 x 450 x 32/460 = 1819 mm2

Use 4T25 (1960 mm2)

2.4.4.3 Left-hand end

I

of 8 m span

From bending moment envelope

M

= l26kNm

8

= 0.7 and

= 0.0864 (Section 13, Table 13.2)

126

= 300

x

x

4402

106

=0.0678<

x 32

tim

Therefore no compression reinforcement is required.

= 0.084 (Section 13, Table 13.1) A8

= 0.084 x 300 x 440 x 32/460 = 772 mm2

Using 2T25 bent-up bars, minimum diameter of mandrel

=

13 (Asreq/A,pr,) =

Use 2125 (982 mm2) with

104

r=5

5.2.1.2

NAD Table 8

PETEDESGI4 EXMAPLE 2.4.4.4 Right-hand end of 6 m span From bending moment envelope

M = 76kNm

M

=

76

= 300

bd2fCk

x

106

x 44Q2 x 32

w

= 0.049 (Section 13, Table 13.1)

A

= 450 mm2

Use 2125 (982 mm2) with r

=

4qS

=

0.041

minimum

2.4.4.5 Near middle of 6 m span From bending moment envelope

M = l38kNm Effective flange width

= 300 +

138

= 32

x

x

4502

0.2

108

x 1320

x

0.85

=

x

6000

= 1320 mm

0.0161

= 0.019 (Section 13, Table 13.1) AS

= 0.019

x 1320 x 450 x 32/460 = 785 mm2

Use 2125 (982 mm2) 2.4.4.6 Minimum reinforcement

kkfAIcr

A5

4.4.2.2 Eqn 4.78

where

k

=

0.4

k

=

0.68

=

3 N/mm2

= 300 x 325 mm2 = 460 N/mm2 Therefore

AS 0.6bd

173 mm2

.

0.0015 bd =

OK

203 mm2

OK

5.4.2.1.1(1)

COMPLETE DESIGN EXAMPLE

2.4.5 Shear reinforcement

4.a2

2.4.5.1 Minimum links Here, for comparison with BS 8110 design, grade 250 reinforcement will be used.

5.4.2.2

Interpolation from EC2 Table 5.5 gives Minimum =

= 0.0022 x 300 = 0.66 mm2lmm

A,,js

If VSd

0.0022

refer to Section 2.4.5.3 for VRd2

—

(--)

= 300 mm Sm = lesser of 300 mm or 0.8d Use R12 links @ 300 mm crs. (A/s

2.4.5.2 Capacity of section VAd1

=

TRdk(l .2

Eqn 5.17

= 0.75 mm2/mm)

without shear reinforcement +

40p1)

4.3.2.3

bd

Assume 2125 effective p1

= 982/(300 x

440)

k

=

1.6 —

1.6 — 0.44

=

0.35

TRd

VRd1

d =

=

0.00743

=

1.16

Table 4.8

= 300 x 440 x 0.35 x 1.16 x (1.2 + 40 x 0.00743) x

iO = 80.2 kN

2.4.5.3 Shear reinforcement by standard method

of section

Maximum capacity

= 0.7

4.3.2.4.3

—

k'20° =

0.7

—

32/200

=

0.54

i

0.5

= 0.5 x 0.54 x (32/1.5) x 300 x 0.9 x 440 x iO = 684 kN Design shear force is shear at a distance d from the face of the support. This VRd2

is 590 mm from the supportcentre(ine. A s

VSd

= 0.9

—

x 440 x

80.2 0.87

x 250

=

0.0116 (VSd — 80.2)

Design of shear reinforcement is summarized in Table 2.3.

Eqn 4.21

Eqn 4.25 43.2.2(10)

Eqn 4.23

CO&PL Qc.SG4EX&MPLE Table 23 Design of shear reinforcement Location

8 m span LH end

203 248

1.42

LH end

202

1.41

RH end

128

mm.

RH end

6 m span

s for 12 mm

AJs

VSd

1.95

links

Links

159 116

R12 @ 150 R12 @ 100

160 max.

R12 @ 150 R12 300

c

R12 @ 300

Minimum

2.4.6 Deflection Reinforcement percentage at centre of 8 m span

=

100

x 1960/(450 x

1660)

4.43.2

= 0.26%

Interpolating between 0.15 and 0.5%, basic span/effective depth ratio for end span = 40

NAD Table 7

To modifyfor steel stress multiply by 400/460 To modifyfor T section multiply by 0.8 To modifyfor span

> 7 m multiply by 7/8

= 40 x (400/460) x 0.8 Actual ratio = 8000/450 = 17.8 Therefore permissible ratio

x 7/8

= 243 OK

2.4.7 Cracking For estimate of steel stress under quasi-permanent loads

4.4.2.2

= 64.8 kN/m = 0.3 Assuming 2 Quasi-permanent load = 03 x 20 +

NAD

Ultimate load

Approx. steel stress at midspan

25.8

= 460 x 1.15

= 31.8 kN/m 31.8

=

Table 1

196 N/mm

64.8

Approx.steel stress at supports allowing for 30% redistribution = 196/0.7 = 280 N/mm2 These are conservativefigures since theydo notallowfor excess reinforcement over what is needed or for moment calculated at centreline of support rather than at face of support. Check limits on either bar size or spacing. From EC2 Table 4.11,25 mm bars in spans are satisfactoryat any spacing since steel stress 200 N/mm2 OK

<

From EC2 Table 4.12, bar spacing at supports should be 150 mm with no limitationon size. As bars are located inside column bars the maximum possible OK spacing is 125 mm

4.4.2.3

COtaPLETEDESIGNEX&MPLE

2.4.8 Curtailment

of reinforcement

Reinforcement must extend for a distance of a, envelope where a,

= 0.9d12

25 1'Qnet=

a,

+

¼net

beyond the moment

5.4.2.13

= 198 mm 460 —

x

1.15

— 1

= 782mm

5.2.a4.1

3.2

net = 980 mm

+

Bars mark 8, whichare located outside theweb, mustextend afurther 150mm — refer to Figure 2.&

2.4.9 Reinforcement details Curtailmentofthemain reinforcementand arrangementofthe linkreinforcement are shown in Figures 2.6 and 2.7. Reinforcement details are shown in Figure 2.8 and given in Table 2.4. 1400 1600

1800

1800

76

Moment envelope Curtailment Line — ——

Figure 2.6 Curtailment diagram of main reinforcement 150 , -

R12

1450'

300

(minimum

242

c/c

300 - ,iOO links) O'(minimum links) 240 - Shear capacity of —

vRdl+ Vwd= 147kN

I 2150,1

286

Figure 2.7 Arrangement of link reinforcement

minimum Links (R12 —300) with

rs

5.4.2.13

tOThPT

5OI4 EXMJIPLE

Figure 2.8 Main beam reinforcement details

COMPLETEDStGN EX&MPLE

Table 2.4 Commentary on bar arrangement Bar marks 1

Notes Tension bars are stopped column bars

20 + 12 = 32 > 25 mm

Nominal cover

2

50 mm from column face to avoid clashing with the

Remaining tension bars stopped off at LH end as shown in Figure 2.6. Bars extended at RH end to provide compression reinforcement (lap = and continuity for internal ties (lap = 1000 mm) Check minimum distance between bars

OK

'iit)

bar size or 20 mm

(300—32x2—4x25)/3=45>25mm 3

Not used

4

Similar to bar mark 1

510

4.13.3(5)

BS 8110 5.2.1.1

OK

Loose U bars are fixed inside the column bars and provide continuity for columnand internal ties

lop legs project from centre line into span,minimum dimensions shown in

5.4.2.13

Bottom legs are lapped 1000 mm to provide continuity for internal ties

BS 8110

Figure 2.6

5

=

lop legs

1800 mm

+ 1000 =

Bottom legs

200

Use r =

both bends

5 for

1200 mm

Notethatthe bottom legs are raised to avoid gap between bars being < 25 mm 10

lop legs

5.2.1.1

1500 mm

Bottom legs = 200 + 1000 = 1200 mm 6,9

216 provided as link hangers are stopped 50 mm fromcolumn face

7,8

Tension bars over the support are stopped Bars mark8 are located outside the web

11

as in Figure 2.6.

Links are arranged in accordance with Figure 2.7 for shear. Links also 150 mm at all laps provide transverse reinforcement with a spacing

5.4.2.1.2(2)

5.2.4.1.2(2)

2.5 Edge beam (interior span) 2.5.1 Cover for durability and fire resistance Nominal cover for exposure class 2b (external) is 35 mm. Nominal cover for 1 hourfire resistance is 20 mm. Use 40 mm nominal cover to links

NAD Table 6 BS 8110 Table 3.5

CQAPLE.TE ES(GI4 XMAPLE

2.5.2 Loading Permanent load from slab

= 4.7 x 5 x 1.25 = 29.4 kN

(assuming 1.25 m strip to be loading on edge beam)

= (035 — 0.175) x 03 x 5 x 24 = 6.3 kN Cladding load @ 5 kN/m = 5 x 5 = 25 kN Characteristic permanent load = 60.7 kN Characteristic imposed load = 4 x 5 x 1.25 = 25 kN Self-weight of beam

=

Total design load

125

x 60.7 + 1.5 x 25 = 119.5 kN

2.5.3 Design moments and shears These are taken from the Concise Eurocode, Appendix, Table 2.5.3.1

Interior support = 0.10 x

Moment Shear

119.5

x 5 = 59.8 kNm

= 0.55 x 119.5 = 65.7 kN

2.5.3.2 Mid-span Moment

=

0.07

x

119.5

x 5 = 41.8 kNm

2.5.4 Reinforcement for flexure 2.5.4.1

Interior support

= 280 mm M = 59.8 x 106 = 0.079 x 300 x 32 2802 bd2ç

Assume effective depth

Alyk

= 0.099 (Section 13, Table 13.1)

bdfCk

xld

=

A

= 579 mm2

0.189

<

OK

0.45

2.53.4.2(5)

Use 2120 (628 mm2)

2.5.4.2 Mid-span Assume effective depth Effective flange width

= 290 mm

= 300 ÷ 0.1 x 0.7 x 5000 =

650 mm

2.5.2.2.1

COMPLETE DESIGN EXAMPLE

M

650

bd2fk

Afsyk

=

x iO

41.8

—

x

= 0.024

2902 x 32

0.028 (Section 13, Table 13.1)

bdfck A5

= 367 mm2

Use 2120 (628 mm2)

The cross-section is shown in Figure 2.9.

650

-j S

.

-I70

2120

. 1

21 20

I

S

350

60

T

300

Figure 2.9 Edge beam cross-section

2.5.5 Shear reinforcement Design shear force may be taken to be at distance d into the span from the face of the support. This can be calculated approximately as VSd

= 65.7 —

V

= 300 x 280

Adi

=

119.5 (0.28

43.2.2(10)

+ 0.15)/5.0 = 55.4 kN

x 0.35 (1.6 — 0.28) x

1.2

+ 40

300

x 628 x 280

4.3.2.3(1)

58.2 kN

This is greater than

Vsd,

hence only minimum links are required.

4.3.2.2(2)

Assuming grade 250 reinforcement for links, EC2 Table 5.5 gives

p

= 0.0022

Hence

A

— S VRd2

= 0.0022 x 300 = 0.66 mm2/mm =

0.5

(

32 32 07 . —— x — x 300 x 200)

1.5

0.9

x 280 = 435 kN

43.2.3(3)

tOJkP.ErEDESP4 EXAMPLE

Since

< (j-) VRd2,

VSd

200 mm spacing gives

A

Smax

= 0.8d = 224 mm

Eqn 5.17

= 132 mm2

Use RiO links at 200 mm crs. (A5

= 2 x 7&5 = 157 mm2)

2.5.6 Deflection = 5000/290 =

Actual span/effective depth ratio

17.2

At mid-span 100 A5

bd

= 100 x 628 = 033% 650

x 290

By interpolation from NAD Table 7, modified for Basic span/effectivedepth ratio

= 460

= 36

Note:

Thiscan be increased allowing for use of a largerthan required steel areato

= 36 x 628/367 = But not greater than 48/1.15

61.6

4.4.3.2(4)

= 41.7

NAD Table 7

Inspection shows this to be unnecessary. Allowable

l/d >

2.5.7 Curtailment

actual

Note 2

lid

OK

of reinforcement

Since the bending moment diagram hasnot been drawn, simplified curtailment rules are needed. These are givenin Section 8 ofthe Appendix tothe Concise Eurocode. Using the rules, the 20 mm bars in the top may be reducedto 12 mm bars at a distance from the face of support

+ 32cb + 0.45d

Concise Eurocode Figure A.12

= 500

+ 32 x 20 + 0.45 x 280

=

0.11

=

1266 mm from the column face

COMPLETE DESIGN EXAMPLE

2.5.8 Reinforcement details The reinforcement details are shown in Figure 2.10.

150

I—

2T20-4 top cover 60 side cover 75

2112—2

side cover 75 ELEVATION

Cover

to links

3443

40

Li

U

1221 A-A

Figure 2.10 Edge beam reinforcement details

2.6 Columns 2.6.1 Idealization of structure The simplification assumed for the design of the main beam is shown in Figure 2.3.

2.6.2 Analysis Moments and column loads at each floor are taken from the ana'ysis for the main beam given in Section 2.43.

2.6.3 Cover for durability and fire resistance Nominal cover for interior columns (exposure class 1) is 20 mm. Nominal cover for exterior columns (exposure class 2b) is 35 mm. Nominal coverfor 1 hour fire resistance is 20 mm. Use 20 mm (interior) and 40 mm (exterior) nominal cover to links

NAD Table 6

COMPLETE DES)G1 EXAMPLE

2.6.4 Internal column 2.6.4.1 Loading and moments at various floor levels These are summarized in Table 2.5. Table 2.5 Loading and moments for internal column Column design loads

Beam loads (kN) Total

Loadcase Roof

8m

6m

Dead

Imposed

1

2

240 202

238 165

1

2

53 43

51

6

278 143

283 236

6m

2

187 159

187 159

9

9

96

57

355

355

131 110

126 17

152 126

152 126

9

9

Self-weight

2nd floor 8 m

283 236

6m

278 143

337

200

642

642

131

126 17

152 126

152 126

9

9

110

Self-weight

1st floor 8 m

286 240

6m

578

343

929

929

280

132 111

126 17

154

146

14

154 129 14

1226

1226

129

Self-weight

Foundations

486

821

Bottom

Top

1

Self-weight

3rd floor Bm

Column moments (kNm)

(kN)

1

2

32

42

33

1

2

30

39

30

55

33

55

49

49

33

29

18

2.6.4.2 Design for column between first floor and foundation Effective height in N—S direction "top

0.5

—

x

675 x

106

5000

+ 675 x

106

3500

3125 x

106

8000 =

0.28 but take not less than 0.4

=

0.8

+

÷

3125

x

6000

106

Eqn 4.60

kbottom Hence Effective height = 0.8 Load case

1

Figure 4.27

x 5000 = 4000 mm

gives worst condition (by inspection).

Imposed load

= 0.7 x 821 =

575 kN

BS 6399: Part 1, Reduction factor

CO&PLETE DESIGN EXAMPLE

Dead load = 1226 kN Nsd

M

=

1801 kN

=

18 kNm (top), 0 (bottom)

x iO x

1801

=

3002

15Ji =

1.5

x 32

= 0.94

43.53.52

< 25

14.5

Hence X

mm

= 25

4000j

=

=

= 46

300

Note:

iji

=

x

(10/h)

X > 25, hence column is slender in N—S direction The slenderness in the E—W direction will be found to be approximately the same.

Thestructure is braced and non-sway(by inspection),hence theModel Column Method may be used with the column designed as an isolated column. =

25(2

—

e1/e2) = 50 in both E—W and N—S directions

Slenderness ratios in both directions are less than hence it is only necessary to ensure thatthe column canwithstand an end moment of at least Nsdh/2O

=

1801

x 03/20 = 27.0 kNm

43.5.53 Eqn 4.64

This exceeds thefirst order moments. Hence Nsd

=

=

1801

kN and Msd

= 27.0 kNm

0.62

bhfck

= bh2fk

x 106 = 0.031 300 x 32 27.0

Assume

d'

= 45mm

43.5.53 Eqn 4.62

I

COMPLETE DES)GT4 EXAMPLE

Then

d'/h

= 45/300 =

Af —--

=

0.15

0.16 (Section 13, Figure 13.2(c))

bhfck

Hence

= 1002 mm2

A

Use 4120 (1260 mm2) Note:

In the design by Higgins and Rogers, the slenderness ratio exceeds the equivalent of >tcrit but the design moment is still Nh/20. EC2 requires less reinforcement due to the smaller design load and the assumption of a smaller cover ratio. If the same cover ratio is used in the Higgins and Rogers design, 4T20 are sufficient in both cases.

2.6.5 External column 2.6.5.1 Loading and moments at various levels These are summarized in Table 2.6. Table 2.6

Loading and moments for external column Beam loads

Column design loads (kN)

(kN) Total

Loadcase

Dead

Imposed

1

2

184 55

186 55

Column moments (kNm) Bottom

Top

1

2

1

2

1

2

39

41

145

145

104

107

55 9

55

1

2

93

98

93

98

103

109

Roof Main Edge Self-weight

9

39

41

209

209

109

114

126 55 9

126 55 9

148

155

399

399

109

114

126 55 9

126 55 9

257

269

589

589

108

113

125 9

125 55 9

778

778

3rd floor Main Edge

235 55

240 55

Self-weight

93

98

2nd floor Main Edge

235 55

240 55

Self-weight

93

98

let floor Main Edge

233 55

238 55

55

Self-weight

Foundations

365

382

68

72

COIAPLETE DESIGN EXMAPLE

2.6.5.2 Design for column between first floor and foundation

x

=

106

(675 4000

x 0.5 + 675 x

3500106)

—

3125 x 106 8000

= 0.71

kbottom Hence

=

0.85

Effective height

=

Slenderness ratio

Figure 4.27

0.85

=

x

i/i =

4000

3400

= 3400 mm

fi

300

= 393

v will be small so isf7 will be less than 25 Hence

X. X

=

25

> 25, therefore column is slender in N—S direction

Calculate

=

bottom moment

top moment

e02

= o = 0 85

Hence

= 25(2 + 0)= 50 Slendernessratios inthe E—W and N—S directions are both lessthan 50, hence it is only necessary to ensure that the end moment is at least Nh120.

Theworst condition occurswith load case2 at section just above the firstfloor, where Msd is greatest. Nsd

Nh —

=

589 + 0.8 x 269

= 804

20

x 03

20

Design end moment Hence Nsd

=

= 804 kN

12.OkNm

= 109 > 12 kNm

= 804 kN and Msd =

109 kNm

43.5.5.3

DSt4 XMJIP.E 2.6.6 Reinforcement details Maximum spacing of links for internal column 12 x 20 = 240 mm Generally

5.4.1.2.2(3)

NAD Table 3

0.67 x 240 = 160 mm

Above and below floor

Maximum spacing of links for external column 12 x 25 = 300 mm Generally

At lap and below floor

0.67

x 300 =

5.4.1.2.2(4)

200 mm

The reinforcement details are shown in Figure 2.11. INTERNAL COLUMN F2

Links

Vertical bars

Section

EXTERNAL COLUMN Fl

Links Vertical bars

0 U,

'0I

.—

.

.-- --. J

0

C

U,

-

6'1

6',

0 0 m

®I

-3 ,

,

I

-

@. to (inks 30 Fdn"

g

Ilu,

I II I I ,

@i '.5 .;s I

—

o: 3E

Fdn.

L.....i.

t

cover

to Links

60

I

US

''?

see Fig 2 13 CM

"-3

'0

®H__ 11 Cover

0 Os

11

@1

IN

II I

E C

U, Cl IN

®6 il.J

III ----- .i ".

U

1st."

Section

I

Figure 2.11

Column reinforcement details

2.7 Foundation Design typical pad footing for internal column.

2.7.1 Cover Use 50 mm nominal cover against blinding _____ _____ BS 8110 specifies a nominal cover of not less than 40 mm against blinding. EC2 specifies a minimum cover greater than 40 mm. Thisimplies a nominal greater than 45 mm, hence the choice of 50 mm.

cer

2.7.2 Loading Taken

from internal column design.

Ultimate design loads:

Dead

=

1226

imposed

=

575

Total

=

LJ

1801 kN

4.1.3.3(9)

COMPLETE DESIGN EXDMPLE

Hence service loads:

Dead

=

908

Imposed

=

383

Total

=

1291 kN

The assumption is made that the base takes no moment. Also it is assumed that thedead weight of the base less the weightof soil displaced is 10 kN/m2 over the area of the base.

of base

2.7.3 Size

Since, at the timeof publication, EC7: Geotechnical design and EC2, Part 3: Concrete foundations10 have not been finalized, the approach used here is based on current UK practice.

Use2.75m x 2.75m x 0.6mdeep pad Bearing pressure under serviceloads

= 2.752

+

=

10

181

<200 kN/m2

Design pressure at ultimate limit state

=

OK

= 238 kN/m2 2.752

2.7.4 Flexural reinforcement Moment at face of column = 238 x 2.75 x 1.2252/2 = 491 kNm Average effective depth = 600 — 50 — 25 = 525 mm

M —

2750

bd2fck

Af

491

=

x

106

x 5252 x 32

=0.020

= 0.023 (Section 13, Table 13.1)

bdfck Hence

=

0.023

x

2750

x

525

x 32/460 = 2310 mm2

Use 9T20 @ 300 mm crs. each way (2830 mm2)

2.7.5 Shear 2.7.5.1 Shear across base

Shearforce may be calculated at a critical section distance d from the face of the column. Design shear (V)

= 238

x 2.75 x

0.3)

[2.75_

= 458 kN

—

0.525]

43.2.2(10)

Th?T D5%I4 XMJIPLE the influence of the reinforcement will be ignored since, if VRd1, bars are used, straight they will not extend d + 1bnet beyondthe critical section. In calculating

VRd1

>

VAd1

x

x

x 2750 x 525/1000

=

0.35

Vsd,

hence no requirement for shear reinforcement

1.075

1.2

=

652 kN

43.23(1)

Eqn 4.18

2.7.5.2 Punching shear

The critical perimeter is shown in Figure 2.12.

=

Design load on base

1801 kN

Length of critical perimeter =

u

[4x 300 + R-(2 x 1.5 x 525)]/1000

/—

300

II

I-I

= 6.15m

\\

of_Hift Figure 2.12 Critical perimeter for punching VRdI

=

x

035

1.075

x

1.2

x 525 x 6.15 =

Area within perimeter = 2.98 m2 Design shear (Vsd)

VSd

<

VAd1,

= (7.56

—

2.98)

1458 kN

43.4.5.1

Area of base = 7.56 m2

x 238 =

1090 kN

43.4.1(5)

hence no requirement for shear reinforcement

2.7.6 Cracking Approximate steel stress under quasi-permanent loads =

460

+ 03 x (908

1.15

1801

x 383)

x—

2310 2830

From EC2 Table 4.11 bar size should not exceed 25 Hence cracking

=186N/mm

> 20 mm used.

4.4.23 OK

Table 4.11

COMPLETE DESIGN EXAMPLE

2.7.7 Reinforcement details The reinforcement details are shown in Figure 2.13 and given in Table 2.7.

P

-

-

2

—

—f

9120—1—300 B2

2

22

9120—1—300 Bi

P LA N

4120-2

II

Fdn.

Cover —1.0

_J[_

III •

L1

A- A COVER — B1

= 50, end =75

Figure 2.13 Base reinforcement details Table 2.7 Commentary on bar arrangement Bar marks

Notes

1

Straight barsextend full widthof base less end cover of 75 mm. Bars should extend an anchorage length beyond the column lace

32 x 20 Anchorage length Actual extension 1150 mm 2

3

640 mm

Column starter bars wired to bottom mat Minimum projection above top of base is a compression lap + kicker = 32 x 20

4.13.3(9)

5.23.4.1

5.2.4.13

+

75

= 715 mm

Links are provided to stabilize and locate the starters duringconstruction

COMPLETE DESiGNEXAMPLE

2.8 Shear wall 2.8.1 Structure The structure is shown in Figure 2.14. fLoor —

I—05x Wi nd Load on

building

I

14300

900

Figure 2.14 Shear wall structure

2.8.2 Loading at foundation level Dead load from firstto third floors and roof

= 0.5 (3

x 23.5 + 28.5) = 49.5 kN/m

Self-weight = 0.175 x 24 x 15.5 Characteristic dead load = 49.5

= 65.1 kN/m + 65.1 = 114.6 kN/m

Characteristic imposed load from slabs =

2.5 (1.5

+ 3 x 4) x 0.7 = 236 kN/m

Wind loading is taken as 90% of value obtainedfrom CP3: Ch V: Part 21). Total wind

load on buildingin N—S direction

NAD 4(c)

= 0.9 x 449 = 404 kN

= 404/2 = 202 kN Moment in plane of wall = 202 x 8 = 1616 kNm Wind load on wall

Hence Maximum force per unit length due to wind moment

= +

Mx6 12

x = ± 1616 6 = ±47.4kN/m 14.22

2.8.3 Vertical design load intensities at ultimate limit state Dead load + imposed load = 135 x 114.6 + 1.5 x 23.6 = 190.1 kN/m Dead load

+ wind load = 135 x 114.6 + 1.5 x 47.4 = 225.8 kN/m; or = 1.0 x 114.6 — 1.5 x 47.4 = 43.5 kN/m

Eqn 2.8(a)

Eqn 2.8(a)

COALETE5ESG14 EXAMPLE

Dead load

+ wind load + imposed load = 135 x 114.6 + 1.35 x 23.6 ± 1.35 x 47.4

Eqn 2.8(b)

= 250.6 kN/m or 122.6 kN/m

NAD 4(c)

Therefore maximum design load

= 250.6 kN/m

From analysis of slab (not presented), maximum moment perpendicular 11.65 kNm/m

to plane of wall =

2.8.4 Slenderness ratio

kA

=

kB

=

= 2.05

4a5

Eqn 4.60

Hence

10

1/i 0

=

0.94

=

fl1O

Figure 4.27

= 0.94 x 4

= 3.76 x 1000 x

376 m

=

175

Hence wall is slender

2.8.5 Vertical reinforcement Higginsand Rogers design the shear wallas unreinforced. Plain concrete walls will be covered in EC2 Part 1A which, at the time of publication, has not yet been finalized. The wallwill, therefore, be designed hereas a reinforced wall. As will be seen, the result is the same. Eccentricity due to applied loads e01

e02

= 0 =

11.65

=

0.6

x

1000/250.6

= 46.5 mm

Hence e0

x

46.5

+ 0 = 27.9 mm

Eqn 4.66

Accidental eccentricity

ea

=

— 1

200

x

3760

2

= 9.4 mm

Eqn 4.61

COVIP%TE OESG EXDJJIPLE

Second order eccentricity

=

37602

x

460

2 x 1.15

10

= Assuming '<2

x

200000

1

x 0.9

x

x 122

'<2

Eqns 4.72 &

469

51.51<2

=

1

Design eccentricity

=

27.9

+ 9.4 +

51.5

= 88.8 mm

= 250.6 kN/m Design ultimate moment = 88.8 x 250.6/1000 = 22.3 kNm/m Design ultimate load

_!__ =

0.023

bh2fck

_L

=

0.045

bhck

Al

=

0.01 (Section 13, Figure 13.2(d))

"ck Hence

A

= 122 mm2/m or 61

mm2/m in each face

Minimum area of reinforcement

= 0.004 x 1000 x 175 = 700 mm2/m

5.4.7.2

This exceeds the calculated value. Hence the minimum governs. Use T12 @ 300 mm crs. in each face (754 mm2/m)

2.8.6 Shear Design horizontal shear Shear stress

=

1.5

x 202 = 303 kN

303 x 1000 = = __________ 14300

x 175

0.12 N/mm2

Note:

is not calculated since VRd1 EC2 Eqn 4.18.

it must be

OK

> 0.12bd by quick inspection of

2.8.7 Horizontal reinforcement Minimum at 50% of vertical reinforcement provided = 188 mm2/m (EF)

5.4.73

Minimum for controlled cracking due to restraint of early thermal contraction

4.4.2.2

A

COMPLETE DESIGN EXAMPLE

=

kkfffAIa

kC

=

1.0

k

=

0.8

cte

=

1.9 N/mm2 (assuming concrete strength C16120 at time of cracking)

A9

Eqn 4.78

to be equivalent to

= 360 N/mm2 (assuming 10 mm bars)

= 1.0 x

A9

0.8

x

1.9

x

175

x

1000/360

Table 3.1

Table 4.11

= 739 mm/m

Use T10 @ 200 mm crs. in each face (785 mm2/m)

2.8.8 Tie provisions at first floor

NAD 6.5(g)

According to the NAD, these should follow the rules in BS 8110.

=36kN

F

2.8.8.1 Peripheral tie

=

A

36X10382 460

Use 1110 (7&5 mm2)

2.8.8.2 Internal

tie force

Force

= 2.5 x 36 (4.7 +

4.0)

7.5

x

143 —

5

=

299kN

Hence

AS

= 299 x iO =650 mm2 460

Use 5110 in each face (785 mm2) Hence T10 © 200 mm crs. horizontal reinforcement in wall 0.5 m above and below slab is adequate. 2.8.8.3 WaIl tie Take the greater of (a) and (b)

(a) Lesser of 2.OF or

iF 2.5

= 72 or 48 kN

BS 8110 3.12.3

Gt EXMAPLE (b) 3% of total vertical load = 0.03

x

190.1

=

5.7 kN

Hence

Tie force

A

= 48 kN = 48

x iO = 104mm2

460

Therefore reinforcement in slab will suffice

2.8.9 Strip footing EC2, Part 3: Concrete foundations, at thetime of publication, hasnot yet been drafted, hence current UK practice is adopted. Maximum pressure due to characteristic dead, imposed and wind loads

=

114.6

+ 23.6 + 47.4/0.9 =

For 900 mm wide strip, pressure =

191 —

191

kN/m

= 212 kN/m2

0.9

Allowextra 10 kN/m2for ground floor loadsand weight of concrete displacing soil in foundations. This gives 222 kN/m2. Allowable pressure

= 1.25 x 200 = 250 > 222 kN/m2

OK

Use 900 mm wide strip Calculate reinforcement for flexure Moment A5

'

= 250.6 x '09

—

0175\2 /

=

8

16.5 kNm/m

= 209 mm2/m

Minimum area =

=

0.OOlSbd 0.0015

x 1000 x 200 = 300 mm2/m

Use 112 @ 300 mm crs. (377 mm2/m)

5.4.2.1.1

COMPLETE DESIGN EXAMPLE

2.8.10 Reinforcement details The reinforcement details are shown in Figure 2.15 and given in Table 2.8.

Ep

0 2xITI 0—8

1T1O—7

r_

Walt

1TIO-7—11 lSt.SFL

tie

----: 2

96T12-3—300 C

(48N2+48F23

-

—

a a a350 a

U-

350

U

Os

- 3 x13RiO -9

C..' ¶51

a a a I.-

I._____________

1Os

Os

c%1

(48N2+ 48F2)

Fdn. I -V.

.t0P...

—I

2x 4T12—2-300B2

3

cover ends

AJ (grid 2 omitted for clarity) B—B COVER to outer bars

Figure 2.15

EAST WALL ELEVATION —

—1000 EW

a

+

K K C..'

96T12—i— 300 —

UI —

K

Ni 40 • Fl

20

Shear wall reinforcement details

-

40

75

DEGH EXAMPLE Table 2.8 Commentary on bar arrangement Bar marks

Notes

1

Wall starters match vertical reinforcement Theprojection ofthe horizontallegs beyond theface of thewall formthe tension reinforcement in the footing This extension must be at leasta tension anchorage length

=

—x 4 12

460 1.15 x 3.2

x

5.2.2.2

— =208mm... 209 377

...

OK

5.2.23 5.23.4.1

The minimum projection above the top of the base is a compression lap + 75 mm kicker

=

32x12+75 =459mm

This is detailed

2 4,5,6

.. ..

at 525 mm

OK

Minimum longitudinal reinforcement provided

5.4.73 4.4.2.2

Minimum horizontal reinforcement provided

BS 8110

7,8

Peripheral tie at floor

9

Wall spacers to maintain location

3.12.3.5

of reinforcement

2.9 Staircase 2.9.1 Idealization The idealization of the staircase is shown in Figure 2.16.

/ /, /,

,- ,——————

,-,

3500

Figure 2.16 Idealization of staircase Design as end span of a continuous beam. Calculations will be given for 1 m width.

2.9.2 Durability and fire resistance As for floor slab, Section 2.3, 20 mm nominal cover will

be satisfactory.

COMPLETE DESIGN EXAMPLE

2.9.3 Loading Average slab thickness on plan = = Self-weight = 0.25 x 24

250 mm 6.0 kN/m

Finishes

=

0.5

Characteristic dead load

=

6.5 kN/m

Characteristic imposed load

=

4.0 kN/m

Design ultimate load

= 1.35 x

6.5

+ 1.5 x 4 =

14.78 kN/m

2.9.4 Analysis Using coefficients in the Concise Eurocode = 0.11 x Moment at interior support

14.78

Moment near mid-span

=

0.09

x

14.78

Shear

=

0.6

x

14.78

x x x

5.062

= 41.6 kNm

5.062

=

5.06

= 44.9 kN

2.9.5 Reinforcement for flexure Effective depth = 175 — 20 — 6 = 149 mm Interior support,

M

=

bd2ck

41.6x106

iO x

1492

From Section 13, Table 13.1

Af

=

0.072

=

746 mm2/m

bdfCk

Hence

A

150 mm crs. (754 mm2/m)

Use T12 Span

—

=

0.048

=

0.058

=

601 mm2/m

bd2fck

Af bdfck Hence

A

Use 112 @ 150 mm crs. (754 mm2/m)

x 32

= 0.059

34.1 kNm

Concise Eurocode Table A.1

DSGI4 EXAMPLE 2.9.6 Shear Reinforcement ratio

754

=

1000

x 149

=

0.0051

Near support VRd1

VRd1

= 0.35 x (1.6

>

VSd

—

0.175)

x (1.2 + 40 x 0.0051) x 149 = 1043 kN

43.2.3 Eqn 4.18

= 44.9 kN, hence no shear reinforcement required

2.9.7 Deflection Reinforcement ratio at mid-span =

0.51%

Concrete is lightly stressed, hence basic span/effective depth ratio is 32.

Table 4.14

= 460, this should be modified to: 32 x 400/460 x 754/601 = 34.9

Since fyk

4.43.2(4)

Actual span/effectivedepth ratio = 5060/149 = 34 <

34.9

OK

29.8 Cracking As for floor slab in Section 2.3.8 Minimum area of reinforcement Thickness of waist =

175

= 183 mm2/m

< 200 mm

4.4.2.2

4.4.23 (1)

No further check is necessary.

2.9.9 Tie provisions E—W internal tie, the minimum

(see Section 2.3.9) Total area for staircase =

area required

= 91 mm2/m

BS 8110 3.12.3.2

91

x3=

273 mm2

Provide 2T12 tie bars each side of staircase in adjacent slab

COMPLETE DESIGN EXAMPLE

2.9.10 Reinforcement details The reinforcement details are shown in Figure 2.17.

150

3+ 2T10—8

10112—iS—iSO 10

16

I

FLIGHT

'B

Cover

[1

to outer

bars

r9 6

20

14

1

rI

I

LANDING

JL9 j A-A

Cover= 40

2nd 10112—5—150

FLIGHT

'A'

Figure 2.17 Staircase reinforcement details

3 EMAS U

3,1

Introduction This Sectioncoversthe design of beamsfor shear and torsion,and supplements the examplesgivenin Section 2. The requirementsfor adequate safety against lateral buckling are also examined.

3.2 Design methods for shear 3.2.1 Introduction EC21 differs from BS 8110(2) because the truss assumption used in shear design is explicit. Leading on from this, two alternative methods are given in

the Code. (1)

Standard

(2) Variable Strut Inclination (VSI).

The standard method assumes a concrete strut angleof 450 (cote = 1) and that the direct shear in the concrete, VCd, is to be taken into account. This contrastswith theVSI method which permitsthe designer to choose strut angles between the limits set in the NAD1, as shown in Figure 3.1, but ignores the direct shear in the concrete. NAD Limits

05

0-45

cot e + tan e

04

0•35

0.4

067

1-0

i

1-5

cot

2-0

25

0

EC2 Limits —I

I—

Figure 3.1 Limits of cote (VSI method) Because the direct shear in the concrete is not taken into account in the VSI method, no savings in shear reinforcement can be achieved until the applied shear exceeds three times the concrete shear (Vsd 3V).

>

A further disadvantage of this methodis that with increasing values of cote, i.e., reductions in the concrete strut angle, theforces in thetension reinforcement

BEAMS

increase significantly and may well outweigh any notional savings in shear reinforcement.These forces are, it should be noted, explicitly checked in EC2 but not in BS 8110. Given specialcircumstancesthe '131 method may be required but for most practical situations, the standard method will provide the most economic design.

3.2.2 Example 1

—

uniformly distributed loading

The beam shown in Figures 3.2 and 3.3 is to be designed for shear. Ultimate load

385 kN/m

.. Beam span and loading

Figure 3.2

,

(

400 u

Figure 3.3 Typical section

—

—

example

1

643 4mm 2 (8132) Cover to links 50mm A51

example 1

The material strengths are

= 30 N/mm2 (concrete strength class C30/37) cwk

=

250 N/mm2 (characteristic yield strength of links)

Thebeam will be checked for shear reinforcementatthree locations using both the standard and VSI methods for comparison. These are (1)

d from support

(2) Where

VSd

VRd1,

i.e.,

the point beyond which only minimum shear

reinforcement is required (3) An intermediate point between 1 and 2.

3.2.2.1 Standard method

VSd

1155 kM

I

Id

l

4.3.2.2(10) 4.3.2.2(2)

43.2.4.3

The shear force diagram is shown in Figure 3.4.

f

4.3.2.4.3 4.3.2.4.4

i155

I

I I a

J

Figure 3.4 Shearforce diagram

—

example 1

lS

kN

The design shear resistance of the section, VRd1

TRd

VRd1,

is given by

=

[rRdk (1.2 +

+ 0•15J

=

034 N/mm2 for

= 30 N/mm2

k

= 1.6—d41 =

p

=

A

bd w

=

Eqn 4.18 Table 4.8

1

6434

=

bd

4.3.2.3(1)

400x900

=

i

0.018

0.02

(assuming 8T32 throughout span) Nsd

a

VRd1

= 034 x 1 (1.2 + 40 x 0.018) x 400 x 900 = 235 kN

3.2.2.1.1 Position 1

—

=

VSd VSd

= 0

AC

>

at d from support 1155 VAd1,

—

0.9

x 385 =

808.5 kN

shear reinforcement is required

4.3.2.4.3

The shear resistance of a section with shear reinforcement is given by VRd3

+

=

VCd

=

V

=

Eqn 4.22

VRdl =

235kN

— (O.gd)ç

Eqn 4.23

where

A

= area of shear reinforcement

s

=

f

=

For

spacing of shear reinforcement 250/1.15 = 217.4 N/mm2

VSd

V

VSd

—

VCd;

A

— (0•9d

VSd

or —

Therefore

A5 — s

— = (808.5 235) x iO = 3.25 mm2/mm

0.9

x 900 x 217.4

Try R12 links @ 140 mm crs. (4 legs),

A/s

=

3.23 mm2/mm

BEAMS

Check crushing of compression struts

= (j-)vfcbw0.9d(1 + cota)

VRd2

For vertical links, cota = v

=

fCd

=

0.7

— —a

-

200

Eqn 4.25

0

= 0.55

0.5

Eqn 4.21

= 20 N/mm2

1.5

Therefore

= (4) x 0.55 x 20 x 400 = 1782 kN > Vsd max =

VRd2

x 0.9 x 900 x

1

1155 kN

OK

Check maximum spacing of links

=

VSd

—

sbsin a

3V

pbd

4.4.23

452 = _________ 140

x 400

0.0081

Eqn 4.79

— = (808.5 3 x 235) x iO = 35 N/mm2

0.0081

Table 4.13

x 400 x 900

Maximum spacingfor crack control = Since

=

300 mm

(4) VRd2 <

Smax

=

0.6d

5.4.2.2(7) Eqn 5.18

p300mm

140 mm spacing

Check minimum value of

OK

p

Table 5.5

Concrete strength class C30/37 Steel class S250 By interpolation from EC2 Table 5.5 w,mIn

= 0.0022 <

0.0081 proposed

Use R12 links @ 140 mm crs. (4 legs) Note: Using the standard method, the increase in force in the tension reinforcement is best covered by using the shift rule. Itwill, however, becalculated in this exampleto provide a comparison with the values obtained in the subsequent examples using the VSI method.

43.2.1P(6) 5.4.2.13

Force in tension reinforcement Td

=

M — + j- V(cote

—

cota)

Eqn 430

BEAMS

Msd

= 884 kNm,

VSd

= 808.5 kN

cote

=

Therefore Td

a2.2.1.2 Position 2

—

cota = 0 for vertical links

1,

=

z = 0.9d = 810 mm

1091

+ 404 =

where VSd =

43.2.43(5)

1495 kN

= 235 kN

VRd1

From Figure 3.4

x 385

VSd

= 1155 — a

a

= 239 m from support

From Section 3.2.2.1.1, VRd2

=

235 kN

> Vs max

OK

The amountof shear reinforcement provided should be greaterthan w.mIn pw,mun

= 0.0022

Re-arranging EC2 Eqn 5.16 in terms of

-

Table 5.5

—' gives

= pb5ina =

For vertical links sina

1

Hence

— S

= 0.0022 x 400

x1

=

0.88 mm2/mm

Maximum longitudinal spacing (Smax) is given by EC2 Eqns 5.17—5.19. Vsd VRd2

=

235kN

=

1782- kN from Section a2.2.it

Since

(j-) Smax

A

VRd2, EC2

Eqn 5.17 applies

=

0.8d

1300 mm

=

0.88

x 300

= 264 mm2,

Eqn 5.17 4R10

= 314 mm2

Use RiO links @ 300 mm crs. (4 legs) 3.2.2.1.3 Position

3 — at 1.65 m from support

Thisisapointintermediatebetweenthe section atdfromsupportand the point at which shear reinforcement is no longer required. VSd VRd1

= 1155 — 1.65 x 385 = 520 kN = 235 kN

Since

>

VSd

shear reinforcement is required

VRdl,

Re-arranging EC2 Eqn 4.23 —

VSd

=

s

VCd

(520 — 235) x 10 0.9 900 x 217.4

=

x

O.9dfd

=

Try R12 links @ 250 mm crs. (4 legs)

= 1.62 mm2/mm

1.81 mm2/mm

Check maximum spacing of links

4.4.22

A

Eqn 4.79

sbsina For vertical links sina =

1

Hence 452

=

VSd

—

250

x 400

= 0.0045

(520 — 3 x 235) X 10 0.0045 x 400 900

3Vd =

x

pbd

Maximum spacingfor crack control

=

—114 N/mm2

= 300 mm

OK

Table 4.13

Since (i-)VRd2 S

<

(+W'

5.4.2.2(7) Eqn 5.18

= 0.6d p 300 mm

From Section 3.2.2.1.1 VRd2

>

OK

VSd

Provide R12 links @ 250 mm crs (4 legs) To optimizelinkspacing, check the point atwhich shear reinforcementis satisfied by R12 @ 200 mm crs. (4 legs).

A — s

VWd

452 = = — 2.26 mm2/mm 200

=

— (0.9d)y =

2.26

x 0.9 x 900 x 217.4 =

VCd+VWd

VRd3

Equating VRd3 VSd

= =

VSd

and noting that VCd

VAd1

+

=

235

=

VRd1

+ 398 =

633 kN

398 kN

Distance of point from support

1155

=

—

633

= 136 m

The proposed link arrangement is shown in Figure 3.5.

R12 — 200

R12— 140

4

4

4

Legs

136rn

,

I

R12 —300 I R12 —200 U

legs

4

4

Legs

legs

R12 —140

4

Legs

136m

1i

-

2.39m+ 2 39m+ 60m between centres of supports

U

4

Figure 3.5 Link arrangement (standard method)

—

example 1

Note:

In the centre portion of the beam RiO links are required by calculations but R12 (*) are shown to avoid the possible misplacement on site. Distance from the support(+) could be reduced to 1.70 m in this case. 3.2.2.2 Variable strut inclination method

43.2.4.4

This method allows the angle of the concrete compression strut to be varied at the designer's discretion within limits stated in the Code.

It cangive some economy in shear reinforcementbutwill require the provision of additional tension reinforcement. In most cases the standard method

will

suffice.

This reduced shear reinforcementwill only beobtainedat high levelsof design shear and is counter-balanced by increased tension reinforcement. This can be seen by a comparison of EC2 Eqns4.22 and 4.23 in thestandard method and EC2 Eqn 4.27 in the variable strut inclination method.

The standard method gives

=

VRd3

V

=

V+ V

Eqn 4.22

A

—a S

(0.9d)f

Eqn 4.23

Re-arranging gives

= s

VRd3—VCd

(O.9d)fd

The VSI method gives

A = —f' (O.9cf cote

VRd3

S

Re-arranging gives A5,,,

s

—

VRdS

(o.9d)c,,dcote

Eqn 4.27

BEAMS

Note:

In the above equation the contribution of the concrete, resistance of the section is not taken into account.

VCd,

to the shear

Withcote = 1.5 which is the maximum value permitted inthe NAD, reductions in shear reinforcement will only occur when —

VRd3

(O.9d) cWd x VRd3

<

VCd)

VRd3 gives VSd

>

VSd

(O.9d)cWd

l.S(VRd3

Putting VSd If

or 1.5

>

3VCd

then the VSI method will allow a reduction in shear

reinforcement.

Ifthis inequality is not satisfied, useofthe variable strut inclination method will produce an uneconomic amount of shear reinforcement. In this case the standard method should be used. For elements with vertical shear reinforcement, VAd2 is given by

VRd2

b zv f

—

cotO + tane

=

Putting VSd

E

426

Eqn

4.21

and re-arranging gives 1

VSd

cote + tane

bwzvfCd

Figure 3.1 shows cote plotted against 1/(cote + tan8) together with the EC2 and NAD limits for cote. Hence for a given the limits for cote can be found.

V,

Increasing the value of cote will reduce the shear reinforcement required but increase the force in the tension reinforcement.

In this example, cotO will be chosen to minimize the shear reinforcement. 3.2.2.2.1 Position 1

—

at d from support

From above 1

VSd

cote + tanO

bWzPfd

b

=

400mm

z

=

0.9

=

0.7—

x 900 = f 200

=

810mm 0.55

z 0.5

MAS

-- =

fCd

=

VSd

= 808.5 kN

20N/mm

1.5

Therefore

=

1

cote + tane

80&5 400

x

10

=

x 810 x 0.55 x 20

0.22

From Figureai, this liesunderthecurveTherefore, cote = 1.5 canbe chosen which is the maximum value allowed under the NAD limits.

=

VRd3

Eqn 4.27

(?!)zcWdcot8

Now equating VRd3 to VSd and re-arranging

=

VSd

S

808.5 X 10

= 810

zycote

= 3.06 mm2/mm

x 217.4 x 1.5

Check

Afd=

(j-)vç =

1.66

OK

5.5

Try R12 links @ 150 mm crs. (4 legs),

=

A/s

3.01 mm2/mm

Check maximum spacingof links.

=

A

4.4.23

= 0.0075

Eqn 4.79

sbsIna VSd

—

3V

— = (80&5 3

0.0075

bd

x 235) x iO

x 400 x 900 = 300 mm

Maximum spacing for crack control =

0.0075

= 38.3 N/mm2

Table 4.13

= 0.0022

>

OK

Check S VSd

5.4.2.2(7)

=

=

Since S

Table 5.5

(4-)VRd2

808.5 kN

bWzvf cote + tane <

VSd

=

400 x 810 x 0.55 x 20 _______________

=

1644kN

2.167

(+)VRd2

= 0.6d 1 300 mm

E1

Eqn 5.18

Use R12 links @ 150 mm crs. (4 legs)

Check additional force in tension reinforcement. M

+ (2) vSd(cote — cota)

—

Td

This compares with Td =

= 1091

+ 606 =

1697 kN

Eqn 430

1495 kN using the standard method.

Note:

Although not permitted by theNAD, values of cote upto 2.5 are given in EC2. A checkon shear reinforcement usingcote = 2.5 is now given to illustrate the effect of increasing values of e on shear and tension reinforcement. A

VSd

=

zfd cote

s

x iO x 217.4 x 2.5

808.5

= 810

Try R12 @ 225 mm crs. (4 legs), ASW/s =

=

1.84 mm2/mm

2.01 mm2/mm

Check maximum spacingof links

=

0.005

VSd —3V

pbd

= 57.5 N/mm2 = 250 mm

Maximum spacingfor crack control

t 300

= 0.6d

Sm

mm

OK

Table 4.13

OK

Eqn 5.18

Use R12 links @ 225 mm crs. (4 legs)

Check additional force in tension reinforcement.

=

Td

M —

This compares with 3.2.2.2.2 Position2

—

+ (-)Vsd (cote Td

—

cota)

= 1091 + 1011 = 2102 kN

= 1495 kN using the standard method.

where VSd =

VRdl

Since only minimum shear reinforcement is required this case is identical to

that shown in Section 3.2.2.1.2. 3.2.2.2.3 Position VSd

—

3 — at 1.65 m from support = 520 kN =

S

Try R12 links

VSd

zfYWdcote

520 X 10

= 810

x 217.4 x 1.5

2 = 1.96mm/mm

225 mm crs. (4 legs), ASW/s = 2.01 mm2/mm

From Section 3.2.2.2.1 spacing is satisfactory. Use R12 links @ 225 mm crs. (4 Jegs) As in Section a2.2.1.3,check the pointat which the shear requirementis satisfied by R12 @ 200 mm crs. (4 legs).

—

452

=

s

2 2.26mm/mm

=

200

=

VRd3

cote = 2.26 x 810 x

(._)zf

Distance from support

=

1155

597 =

—

217.4

x

1.5

=

597 kN

Eqn 4.27

1.45 m

385

The proposed link arrangement is shown in Figure 3.6.

R12 —150 1

I

R12 —200

R12 —300 S

4 legs

legs

145m "I 239m

4

R12 —200

I

4 legs

4 legs

legs

I

R12 —150

I I I

I

145m

2 39m

ri

60m between centres of supports

Figure 3.6 Link arrangement (VSI method)

—

example 1

5

obtained using the standard Comparing this with the arrangement in Figure method, it can be seen that less reinforcement is required near the support butthis needs to be carried further along the beam. There is little overall saving in this case.

3.3 Shear resistance with concentrated loads close to

support 3.3.1 Introduction Where concentrated loads are located within 2.5d of a support, the value TRd may be modified by afactor when calculating VRd1. Thisenhancement only applies when the section is resisting concentrated loads and the standard method is used. For a uniformly distributed load, an unmodified value of VRd1 should be used.

3.3.2 Example 2

—

concentrated loads only

The beam shown in Figures 3.7 and 3.8 is to be designed for shear.

4.3.2.2(9)

BEAMS

800kN

800kN

Ultimate toads

—

Figure 3.7 Beam span and loading

ioooI.1

A1 4825mm 2 (6T32)

k 400 Figure 3.8 Typical section

—

example 2

Cover to Links

50mm

example 2

The materials strengths are =

30 N/mm2 (concrete strength grade, C30137)

= 250 N/mm2 (characteristic yield strength of links) In the example, VRd1 will be calculated at positions between the support and 2.5d away at intervals of 0.5d. This is done to illustrate the effect even though the critical section will normally be at the position of the concentrated load. 3.3.2.1 Shear reinforcement

The shear force diagram is shown in Figure 3.9.

VRd1

8OOkNf 135m

x

(mod)

\

211kN

1-

2llkN

Figure 3.9 Shear force diagram

—

\ VRd1 (mod)

example 2

h

135m

}0kN

1d1S The basic design shear resistance of the section,

VRdl,

is given by

43.23(1)

VRd1

= [rRd k (1.2 + 40p,) + 0.15 Or]bd

Eqn 4.18

TRd

= 034 N/mm2 for

Table 4.8

= 30 N/mm2

Forconcentrated loads within 2.5d ofthefaceofthe support, an enhancement of shear resistance is permitted. TRd may be multiplied by a factor /3 when determining

VRdl.

= 2.5d/x with 1.0

/3

/3

5.0

Taking values of x between 0.5d and 2.5d gives values Table 3.1. Table 3.1

Eqn 4.17

of /3rRd shown in

Design shear strength fhAd X

TRd

(m)

(N/mm2)

0.45

5

1.7

0.90

2.5

0.85

1.35

1.67

0.57

1.80

1.00*

0.34

2.25

1.00*

0.34

* No enhancement taken, see Figure 3.9 Theequation for VAd1 can be modified to give a range of values corresponding to the distance from the support. VRd1(x)

=

[BTRdk (1.2

+ 40p,) + O.lSç,]bd

= 1.6—d41 =

k

=

p1

— A2,

bd

1

4825 = ________ =

400

x 900

0.013

N _sa0 A C

Values of design shear resistance, VRd1, are given in Table 3.2. Table 3.2 Design shear resistance VRd1 x

VRd1

(m)

(kN)

0.45

1052

0.90

526

125

353

1.80

211

2.25

211

Eqn 4.18 (mod)

BEAMS

Shear reinforcement is required when VSd

>

= 800 kN from x = 0 to x =

From Figure 3.9, VSd

4.3.2.4

VRd1.

1.35 m

Using the standard method

4.3.2.43

= VCd+VWa Putting VRd3 = "Sd

=

VSd VRd1

Eqn 4.22

and VCd =

VRd1 gives

+ VWd

Values of design shear resistanceto be provided by shear reinforcement, are given in Table 3.3. Table 3.3

V,

Design shear resistance VSd

—

=

VRdI

(kN)

VSd (kN)

1052

800

526

800

274

353

800

447

211

0

<

0

211

0

<

0

VAd1

1

VWd

(kN)

<

0

= 353 kN,

Therefore maximum shear reinforcement is required when VRd1 i.e, when x = 135 m.

This should be provided over the entire length from x = 0 to x

2.25 m

4.3.2.2(9)

Note: If a concentrated load is positioned close to a support, it is possible that using to modify VRd1 may lead to only minimum shear reinforcement being provided throughout the beam. In this case, the designer may wish to base the shear resistance on the unmodified VRd1.

4.3.2.2(9)

(0

< x <

2.5d).

Thiscan be illustratedby taking the example above butplacingthe pointload at 0.5d from the support. The modified shear force diagram is shown in Figure 3.10. VRd1 (mod)

1O52kN

—

VSd = BOOk N

VRd1

ZllkN

Note

B = 1 on span side of concentrated load

—

—

0•5

L

I

I

I

I

10

15

2O

25

Position of concentrated load

Figure 3.io Shear force diagram (load at 0.5

E1

—

example 2 modified

In this case it would be prudent tocheckthe shear resistanceonthe unmodified = 211 kN. The required shear reinforcementshould be provided from VAd1 x = Otox = 0.5d Check area of shear reinforcement required in example 2. Re-arranging the equation for

ASW

VWd

=

s

V gives

Eqn 4.23

447x103

= 0.9

0.9df

x 900 x 217.4

Try R12 links @ 175 mm crs. (4 legs),

=

Ajs

=

2.54 mm2/mm

2.58 mm2/mm

Check crushing of compression strut

(4) vfCdbWO.9d (1 + cota) For vertical links, cota = 0 =

VRd2

=

fCd

30 =—= 1.5

0.7

=

— —--

v

Eqn 4.25

0.55

Eqn

200

4.21

20N/mm2

Therefore

x 0.55 x 20 x 400 x 0.9 x 900 x

=

(-j-

=

1782 kN

>

Vsd

1

= 800 kN

OK

Check maximum spacing of links.

4.4.2.3

=

Eqn 4.79

sbsIna For vertical links sina pW

= 175

=

x 400

1

= 0.0064 >

pw,mln

= 0.0022

OK

Table 5.5

Vsd—3V = (800—3x353)x iO bd 0.0064 x 400 x 900 pww Maximum spacingfor crack control =

300 mm

By inspection, EC2 Clause 5.4.2.2(7) is satisfied.

Use R12 links @ 175 mm crs. (4 legs) for 0

< x < 2.25 m

Table 4.13 5.4.2.2(7)

—

3.3.3 Example 3

combined loading

The revised loading and shear force diagrams are shown in Figures 3.11 and 3.12 respectively.

Ultimate loads 800 kN

800 kN 100

kN/m I

IAYAYVIW* YAWA

AYAYAAWAVAW*WkWA\ AV*VAWAUAWAW&

1•35m

1.35m

:

6m

liii

Figure au Beam span and loading

—

example 3

1100

135m

kNI I

135m

Ix

l I

KN

F"

jioo Figure 3.12 Shear force diagram

example 3

—

The basic design shear resistance of the section, VRdl, is given by

43.2.3(1)

+ 40p,) + 0.15Or]bd

Eqn 4.18

For concentrated loads within 2.5d of the face of the support, TAd may be increased as in Section aa2. However, no similar enhancement is permitted for uniformly distributed loads.

4.3.2.2(9)

VRd1

=

[rRdk(l.2

must be reduced depending onthe proportion ofconcentrated loads to total design load. 9 can then be written as

red

=

1+(3—1)

Vsd(COflC)

with 1.0

5.0

Vsd(tot)

=

design shear force due to concentrated loads

=

design shear force due to total loads

Vsd(COflC)

Vsd()

Values of the concentrated load ratio and the resulting design shear strength are given in Tables 3.4 and 3.5.

BEAMS

Table 3.4 Concentrated load ratio Vsd(COflC)IVsd(t) x

Vsd(flC)

VSfl)

Vsd

(m)

(kN)

(kN)

(kN)

0.45

800

255

1055

0.76

0.90

800

210

1010

0.79

1.35

800

165

965

0.83

1.80

0

120

120

0

2.25

0

75

75

0

Vsd()IVsd

Table 3.5 Design shear strength flredTRd X

'3redRd

red

(m)

(NImm2)

0.45

5

4.04

1.37

0.90

2.5

2.19

0.75

1.35

1.67

1.56

0.53

1.80

1.0

1.00

0.34

2.25

1.0

1.00

0.34

Theequation for VRd1 can be modified to give a range of valuescorresponding to the distance from the support. VRd1

(x)

= [adrRdk(1.2 + 4Op1) + 0.15

i

As in Section 3.3.2.1

k

= 1,

p1

= 0.013,

a] bd

=

0

Values of design shear resistance, VAd1, and design shear resistance to be are given in Tables 3.6 and 3.7. provided by shear reinforcement,

V,

Table 3.6 Design shear resistance (VRd1) x

VRd1

(m)

(kN)

0.45

848

0.90

464

1.35

328

1.80

211

2.25

211

Table 3.7 Design shear resistance (V) Vsd

=

—

VRd1

VSd

(kN)

(kN)

848

1055

207

464

1010

546

328

965

637

211

120

211

75

<0 <0

(kN)

VWd

Eqn 4.18 (mod)

M4S Therefore maximum shear reinforcement is required when VRd1

= 328 kN, i.e., when x = 135 m

This should

be provided from x = 0 to x =

2.25 m (0

< x < 2.5d)

Checkarea of shear reinforcement required. Re-arranging the equation for VWd

A — s

= ______ = (0.9d)f

Eqn 4.23

x iO = 0.9 x 900 x 217.4 637

Try R12 links @ 125 mm crs. (4 legs), AIs

3.61 mm2/mm

= 3.62 mm2/mm

Checkcrushing of compression strut From example 2,

VRd2

=

1782 kN

>

= 1100 kN

OK

Check maximum spacing of links

4.4.23

By comparison with example 2, requirements are satisfied Use R12 links @ 125 mm crs. (4 legs) for 0

OK

< x < 2.25 m

For the remainder of the beam beyond x = 2.5d (2.25 m) provide minimum reinforcement as example given in Section a2.2.

3.4 Design method

for torsion

3.4.1 Introduction Theedge beam shown in Figure 3.13 carries the endsofsimply supported floor slabs seated on the lowerflange. The beam is fully restrained at its ends. The example chosen is the same as that used in Allen's Reinforced concrete design to BS 8110: SimplyexpIained12. Analysisofthestructure andthe design of thesection forflexure is not included. The section will be checked for shear, torsion and the combination of both.

5.4.2.2(7)

560

T

200

250

FLoor

1500

£LQb

\\

I

Figure 3.13 Beam section

3.4.2 Design data Design torsional moment (T$d) = 120 kNm Design shear (Vsd) = 355 kN

k

Concrete strength grade is 030/37,

=

30 N/mm2

Nominal cover to links is 35 mm.

4.133 NAD Table 6

Assuming 25 mm bars and 10 mm links

d

= 1500

—

35

—

3.1.2.4

Table ai

—

10 —

=

2

1441.5 say 1440 mm

Assume 0.25% tensile reinforcement for flexure

3.4.3 Shear resistance Shear will be taken as acting on the web of the section only. When combined shear and torsion effects are to be considered, shear isto be checked using the variable strut inclination method. The angle e of the equivalent concretestruts isto bethesamefor bothtorsion and shear design.

433.2.2(4

The design shear resistance, VAd1, with zero axial load is given by

43.2.3(1)

VAd1

= rRdk(l.2 + 40p1)bd

Eqn 4.18

TRd

= 034 N/mm2 for

Table 4.8

k

= 1.6

—

d =

= 30 N/mm2

1.6 — 1.44

=

0.16

1.0

Assuming 0.25% tensile reinforcement, p1

= 034 x

VAd1

0.02

+ 40 x 0.0025) x 250 x 1440 x 1O

1(1.2

= 159.1 kN <

= 0.0025

355 kN

Therefore shear reinforcement required. Use the variable strut inclination method. The maximum design shear force, VHd2, to avoid web crushing is given by

=

VRd2

b zvf

43.2.4.4(2) Eqn 4.26

(cote + tane)

Re-arranging gives VRd2

1

bWzucd

cote + tane = 355 kN

VSd

b

= 250mm

z

= 0.9d = 0.9

v

=

0.7

—- = 0.7

-- = 1.5

= 1296 mm =

—

200

=

=

'cd

—

x 1440

0.55

200

.

0.5

43.2.4.2(3)

20N/mm2

Therefore VSd

355

— —

250

bwzufCd 1

cote + tane

x

x iO

1296

x 0.55 x 20

should be

=

0.1

0.1

By referenceto Figure 3.1, it will be seen that the value of cote may be taken anywhere between the limits of 0.67 to 1.5.

To minimize link reinforcement, take cotO =

NAD Table 3 4.3.2.4.4(1)

1.5

Design shear resistance, VRd3, for shear reinforcement is given by

VRd3

=

()

cWdcote

4.3.2.4.4(2) Eqn 4.27

Re-arranging gives

A sw

VRd3

s

zfcote

Putting VRd3 equal to VSd

Asw

VSd

s

zçcote

-

Using high yield reinforcement

=

=

=

400N/mm2

1.15

Therefore

A —

0.9

s

AS

bs

x 1O

355

=

x

x 400 x

1440

= 1.5

0.46 mm2/mm

vi 400 = 20 = 0.46 x— 0.74 — = 0.55 x— = 5.5 N/mm2.. 2

250

W

2

OK

4.3.2.4.4(2)

E"n 427

Before choosing the reinforcement,the effects oftorsion will beconsidered and the results combined.

The force in the longitudinal reinforcement, Td, ignoring flexure, is given by Td

= (4) VSd(cote

=

--2

x

cota)

1.5

0 =

2663kN

Additional area of longitudinal reinforcement Td —

= 266.3

x iO

400

= 666mm

Thisarea of reinforcement must be combined with the tension reinforcement required for flexure together with the longitudinal reinforcement required for torsion.

3A.4 Torsional resistance Torsionalresistance is calculated on the basis of a thin-walled closed section. Solid sections are replaced by an idealized equivalent thin-walled section. Sectionsofcomplex shape are divided into sub-sectionswith each sub-section treated as an equivalent thin-walled section. The torsional resistance is taken as the sum of the torsional resistances of the sub-sections. The torsional moment, carried by eachsub-section according to elastic theory, maybe found on the basis of theSt Venanttorsional stiffness. Division of the section intosub-sectionsshould be so arranged as to maximize the calculated stiffness.

43.2.4.4(5)

Eqn 4.30

=

For vertical links, cota T

—

BEAMS

For this example the section will be divided into the sub-sections shown in Figure 3.14.

1500

200f Figure 3.14 Dimensions of sub-sections

3.4.4.1

St Venant torsional stiffnesses J

=

BS 8110: Part 2

f3h3mh

2.43 Eqn 1

a4.4.1.1 Top and bottom flanges hmax

=

hmax

=

h

200

mm

From which

h.

310 mm,

= 200 mm

1.55

fi = 0.203

BS 8110: Part 2 2.4.3

Therefore

= 0.203 x 200 x

J

310

= 0.5 x

iO mm4

Table 2.2

3.4.4.1.2 Web hmax

= 1500 mm,

hmax

1500

h

250

mm

From which 3 =

hmin

= 250 mm

6 BS 8110:

0.33

Therefore

J

= 033 x 250 x 1500 =

7.7

x iO mm4

Part 2 2.43 Table 2.2

3.4.4.1.3 Total stiffness

ot

= [(2

x 0.5) + 7.7] x iO

=

8.7

x iO mm4

3.4.4.2 Thicknesses of equivalent thin-walled sections

t

=

the actual wall thickness

43.3.1(6)

where

= outer circumference of the section = total area within the outer circumference

u

A

a4.4.2.1 Top and bottom flanges

= (310 + 200)2 = 1020 mm = 310 x 200 = 62 x iO mm2

u A Therefore

= 62 x iO = 61mm 1020

t may not be less than twicethe cover, c, to the longitudinal bars. Hence, with

43.3.1(6)

10 mm links

t

.

mm

= 2(35 +

10)

= 90 mm

3.4.4.2.2 Web

u

A Therefore

t

= (1500 + 250)2 = 3500 mm = 1500 x 250 = 375 x 1O mm2 = 375 x

i0

3500

=

107 mm

>

2c

OK

Values of t between the limits ofA/u and 2c maybe chosen provided that the designtorsional moment, Tsd, does not exceed thetorsional moment that can be resisted by the concrete compression struts. 3.4.4.3 Torsional moments

TSd,tot = l2OkNm This total moment is shared between theflangesand web in proportion totheir torsional stiffness. Therefore

Tsd

TSd,fl

=

120x 8.7

= 6.9kNm

TSd,w

=

l2OxZ

= lO6kNm

8.7

mustsatisfy the following two conditions TRd1

and

TRd2

4.3.3.1(5)

Eqn 4.38 Eqn 439

3.4.4.4 Torsion in flanges

TAd1

=

2vfCdt4k cote + tan8

4.33.1(6)

Eqn 4.40

Re-arranging gives 1

TRd1

cote + tan8 Putting TAd1 equal to Tsd

= 2vfcdtAk TsdfI

1

cote + tane

= 6.9 kNm = 0.7

is

0.7

= 0.7

433.1(6) Eqn 4.41

200 —

=

(0.7

0385

.

0.35

=

20 N/mm2

t

=

90mm

Ak

= area enclosed withinthe centre line of the thin-wall section = (310 — 90) x (200 — 90) = 24.2 x iO mm2

Therefore

T

x 106 2 x 0385 x 20 x 90 x 24.2 x i0 6.9

2ufdtAk

= 0.206

By referenceto Figure 3.1 it maybe seen that thevalue of cote maybe taken anywhere between the limits of 0.67 to 1.5.

NAD Table 3 433.1(6)

To minimize link reinforcement take cote = 1.5. Notethat this valuemust be consistent with the value taken for normal shear.

TAd2

=

A

cot8) 2Ak(f— $

Re-arranging gives

Asw

TRd2

$

2Aicot8

4.33.1(7) Eqn 4.43

Putting TAd2 equal to Tsd

s

T

=

2Akçcote

Using mild steel reinforcement

fywd

=-!= 9=

217N/mm2

1.15

Therefore

A — s

6.9

=

x

106

2x24.2x103x217x1.5

The spacingof torsion links should not exceed

2 = 0.44mm/mm

-

5.4.2.3(3)

where Uk

= the circumference of the areaAk = 2[(310

—

90)

+ (200 — 90)]

=

433.1(7)

660 mm

Therefore

Sm

A

— = 82.5 mm, say 80 mm = 660 = 0.44 x 80 = 35.2 mm2

Use RB links at 80 mm crs. The additional area of longitudinal steel for torsion is given by

Ac,d

= (TAd2

Re-arranging and putting

T

A

=

Eqn 4.44

_!)cote TRd2

equal to Tsd

Uk

Sdk cote Y1d

-

Using high yield reinforcement

fyld

=

= 400N/mm2

1.15

Therefore

A

= 6.9

x 106 x 660 x 1.5

400x2x24.2x 10

Use 4T12 bars

=

353mm2

BEAMS

Reinforcementwill also be required in the bottom flange to cater for flexure of the flange acting as a continuous nib. 3.4.4.5 Torsion in web

= 106 kNm

Tsdw

=

Ak

(1500

—

107)

x (250

—

107)

=

199.2 x

iO mm2

Therefore

TSd

106x106

=

2x

2vfCdtAk

0.385

x 20 x 107 x 199.2 x io

=

032

Again by reference to Figure 3.1, cote should fall within the limits of 0.67 to 1.5. 1.5 Similarly use cotO As the web is subjectto shear and torsion, the combined effects should now be checked to satisfy the condition

T

2

TAd1

TSd

TRdl

4.33.2.2(3)

< —

Eqn 4.47

VRd2

= 106 kNm = =

2ufcdt.4k cote + tane

43.3.1(6) Eqn 4.40

2 x 0385 x 20 x 107 x

199.2

1

1.5 +

x iO =

151.5 kNm

355 kN

VSd

VRd2

2

+

=

bWzufCd

43.2.4.4(2) Eqn 4.26

cote + tane 250

x

1300 xO.55 1.5 +

x

20

= 1650 kN

Therefore

T2 + —V2 TRd1

=

1062

(—)

VRd2

+

()

3552 =

0.54

<

1.0

OK

Wherethe entire section is used to resist normal shear, each sub-sectionshould be checked to satisfy the above interaction condition.

EMAS 3.4.5 Reinforcement in web Link reinforcementfor torsion Using high yield links

A

106x106

=

2x

s

199.2

x iO x 400 x 1.5

=

0.44 mm2/mm

Notethat A5 for torsion relates to a single leg in the wall of the section. Link reinforcement required for shear

—

=

0.46 mm2/mm from Section 3.4.3

S

Notethat A5 for shear relates to the total shear link legs. Assuming single links, total areafor one leg

ASW

0.46s

=

2

+ 0.44s =

0.67s mm

Using 112 links 0.67s

s

113 mm2

= 168 mm, say 160 mm

Maximum link spacing for shear

S

V

<

(-k)VRd2

(+P'

= 0.6d = 864

'

5.4.2.2(7) Eqn 5.18

300 mm

Therefore S

= 300 mm

For cracking VSd —3VCd

50 N/mm2

4.4.2.3(5)

Table 4.13

300 mm

Therefore S For torsion U

k

smax Uk

8

= 2[(1500

107) + (250

—

107)]

=

Therefore Smax

=

3072

8

= 384mm

Maximum spacing to suit all conditions is 300 mm. Use T12 Hnks @ 160 mm crs.

3072 mm

BEAMS

Additional area of longitudinal steel for torsion in web

ASI

x 106 x 3072 x 1.5 400 x 2 x 199.2 x iO

106

= 3065 mm2

Use 16116 bars

The bars in the tension face of the web will need to be increased to provide for theadditional longitudinal steel required for shear and combined with the reinforcement required for flexure. Area required in tension face for combined torsion and shear

=

(3065

x 2) + 666

=

1049 mm2

16

Use 3T25 bars

3.4.6 Summary of reinforcement Top flange 4T12 longitudinai bars R8 iinks @ 80 mm crs Bottom flange 4112 longitudinai bars R8 iinks @ 80 mm crs. Plus reinforcement for flexure of the nib Web 3125 iongitudinai bars in tension face 7116 bars in each side face 112 links 160 mm cr5. Pius reinforcement for flexure

The reinforcement details are shown in Figure 3.15

4T12

R8_S0___f _—T12 —160

Additional bars needed for flexure

in

T16 bars except where shown

otherwise

nib

R8_80__f/ 4T12

3T25

Figure 3.15 Beam reinforcement details

Eqn 4.44

It will be seen from this example that choosing the upper timit value of cote, to minimize the link reinforcement,results in substantial additional longitudinal reinforcement being required. In practice the value of cote should be chosen so as to optimize the total reinforcement in the section.

3.5 Slenderness limits

4.3 5.7

The Code requires that a beam has an adequate factor of safety against buckling. Providing that the following requirements are met, the safety against lateral buckling may be assumed to be adequate

4.3.5.7(2)

'ot

< 50b; and

Eqn 4.77

h

<4b

NAD

b

=

2.5.2.2.1(3)

h

= total depth of the beam

where width of the compression flange, which can be taken as bei for T and L beams

unrestrained length of the compression flange taking lateral bracing into account

1

2.5.2.2.1(4)

For example, consider the beam shown in Figure 3.16. 975x400 beam

B

C

0

Figure 3.16 Beam spans and loading for slenderness check In this example the top of the beam is loaded but unrestrained (for instance, the beam is carrying a wall).

The second requirement is satisfied i.e. h

< 4b

=

1600 mm

In calculating 1, the unrestrained length of the compression flange can be taken as the distance between points of contraflexure.

<

These distances, which need to be EC2 Figure 23. 10(A—.B)

=

(B—C) = = 10(C—D)

0.851(A—B) 0.71(B—C) 21(C—D)

= =

0.85 0.7

= 2x

50b = 20 m, can be obtained from

x 22 =

x 22 11

18.7 m 15.4 m

= 22 m

Spans A—C are satisfactory but span C—D is not, It is too slender and the width will need to be increased, or additional lateral retraintwill need to be provided.

Figure 2.3

I

4 SLABS 4.1

Solid and ribbed slabs 4.1.1 One-way spanning

solid slabs

Example of a one-way spanning slab is given

in Section 2.

4.1.2 Two-wayspanning solid slabs EC21 permits the use of elastic analysis, with or without redistribution, or plastic analysis for ultimate limit state design. Elastic analyses are commonly employed for one-way spanning slabs and for two-way spanning slabs without adequate provision to resist torsion at the corners of the slab and prevent the corners from lifting. Plastic analyses are commonly used in other situations.

2.5.1.1(5)

Tabulatedresultsfor momentsand shearsfrom both typesof analysisare widely

BS 8110 Tables 3.14 & 3.15

available.

Care is necessary in subsequent design to ensure that adequate ductility is present.Where redistributionhasbeen performed,the necessarychecks should be carried out.

2.53.5.1(2)

2.53.2.2(5) 2.5.3.4.2(3) 2.5.3.5.5(2)

4.1.2.1 Design example of a simply-supported two-way spanning solid slab Design a solid slab, spanning in two directions and simply-supported along each edge on brickwork walls as shown in Figure 4.1. The slab is rectangular on plan and measures 5 m by 6 m between the centre of the supports. In addition to self-weight, the slab carries a characteristic dead load of 0.5 kN/m2 and an imposed load of 5.0 kN/m2.

The slab is in an internal environment with no exposure to the weatheror aggressive conditions.

220mm wide

supporting

walk

I

I

I

I

tx

I

I

I

Ii

II

__________

I..

Figure 4.1

Sm

I

I

Ly6m

Layout of slab

4.1.2.1.1 Durability

For a dry environment, exposure class is 1. Minimum concrete strength grade is C25/30.

Table 4.1 ENV 206 Table NA.1

For cement content and wlc ratio, refer to ENV 206 Table 3(6)

= 15 mm = 20 mm

Minimum coyer to reinforcement Assume nominal aggregate size Assume maximum bar size Nominal cover Use nominal cover

NAD Table 6

= 12 mm

20 mm

= 25 mm

NAD 6.4(a) I

Note: respects.

NAD Table 3 4.133(8)

Check requirements for fire resistance to BS 8110: Part 2(2).

NAD 6.1(a)

20 mm nominal cover is sufficient to meet the NAD1 requirements in all

4.1.2.1.2 Materials Type 2 deformed reinforcement

= 460 N/mm2

NAD 63(a)

= 460 = 400 N/mm2

yd

2.23.2 Table 23

1.15

.ys

C25/30 concrete with 20 mm maximum aggregate size 4.1.2.1.3 Loading Assume 200 mm thick slab Gk

= 4.8 + 0.5 = = 5.0 kN/m2 = 135 = 1.5

Ultimate load

=

7GGk

5.3

kN/m

Table 2.2

+ 1O

k

=

14.66 kN/m2

Eqn 2.8(a) NAD 6.2(d)

4.1.2.1.4 Flexural design Bending moment coefficients for simply-supported two-way spanning slabs, without torsionalrestraintat thecorners or provisionto resistupliftatthe corners, based on the Grashof-Rankine Formulae, are widely published and are reproduced in BS 8110.

MSdx = afl12 MSdy = =

For

1.2

lx

=

0.084,

a

=

0.059

—

BS 8110 Table 3.14

Giving

= 30.8 kNrn/rn

MsdX

=

Msdy

21.6 kNm/m

For short span with reinforcement in bottom layer =

d

200

—

25

—

12 — =

169 mm

2

_M

=

0.043

bd2fck

x

= 0.099

—

d

Af

=

< 0.45

OK

2.5.3.4.2(5)

0.052

bdck Therefore

A

=

478 mm2/m

Use T12 @ 200 mm crs. (566 mm2/m) in short span

For longer span

d

= 200—25—12—6 = 157mm

_.

=

M

0.035

bd2fck

At

= 0.042

bdfCk

Therefore A

= 359 mm/m

Use T12 @ 300 mm crs. (377 mm2/m) 4.1.2.1.5 Shear VSdX

VSdy

in long span 4.3.2

=

=

8a (-i) ni 2

8a, (_)

=

=

24.6 kN/m

14.4 kN/m

The shear resistance with no axial load: VRd1

= rRdk(l.2 + 40p,)bd

43.2.3 Eqn 4.18

= 0.3 N/mm2

Table 4.8

Where TRd

Assume p 50% of reinforcement curtailed

k

=

1.6—d =

=

=

at support

lzl

1.431

Assume 0

bd

p

Figure 4.12

0.02

Hence VRd1

=

87.0 kN/m

>

= 24.6 kN/m

VSdX

No shear reinforcement required

4.3.2.1P(2)

4.3.2.2 4.1.2.1.6 Serviceability

—

deflection

Control by limiting span/effectivedepth ratio based on the shorter span for a two-way spanning slab.

As,prov = 566 mm2/m NAD Table based on

p = 0.0033

7 gives basic span/effectivedepth ratios which are assumed to be =

4.43.2

400 N/mm2.

4.43.2(4)

Note 2to NADTable7 statesthatmodificationto thetabulated valuesfor nominal reinforced concrete should not be carried out to take into account service stressesin the steel (referto EC2 Clause 4.43.2(4)).However,it is assumed that the correction ought to be made for concrete with 0.15% p 0.5% but that the resultingvalues should notexceed those tabulated intheNADfor nominally reinforced concreta

<

Basic limiting span/effective depth ratios are: Concrete lightly stressed (p = 0.5%): 25 Concrete nominally reinforced (p = 0.15%): 34 29.4 By interpolation at p = 033%:

NAD 6.4(e)&(f)

Table 7

The actual service steel stress modification factor is 250

— —

s

400

I (A yk

400

— —

IA areq sprov'

460

x 478/566

— —

103

Therefore, permissible span/effective depth ratio

= 1.03 x Since span

i

29.4

= 303

34

7 m, no further adjustment is required.

Actual span/effective depth ratio =

=

29.6

4.43.2(3)

< 303

....

OK

169

Note: No modification to the longer span reinforcement is required in cases where short span reinforcement isincreased to complywith deflection requirements.

BS 8110 3.5.7

SLS 4.1.2.1.7 Serviceability

—

cracking

For a slab with h 200 mm, no further measures to control crackingare necessary if the requirements of EC2 Clause 5.4.3 have been applied.

4.4.23(1)

4.1.2.1.8 Detailing

Detailing requirements for cast in situ solid slabs, including two-way slabs OK Slab thickness, h = 200 > 50 mm

5.43

Forthe short span, usealternatelystaggered bars and anchor50% ofthe midspan reinforcement at the supports. a

5.43.2.2(1)

Anchorage force, F6

NSd

—

=

--

VSd

+ Nsd

5.4.3.1(1)

5.43.2.1(1) 5.4.2.1.4(2)

EqnS.15

0

5.43.2.1(1)

a1

Therefore F6

=

Areq =

=

VSd

24.6 kN/m

24.6 x iO = = ________

F6 —

61.5 mm2/m

400

= 283 mm2/m Net bond length,

net =

OK aalbAsreq

5.23.4.1(1) Eqn 5.4

1b.min

A6pçp,

=

1.0 for straight bars A.

1

b

=

4

f x—

f

All bars in slabs with h 'bd

1

b

=

5.2.3.4.1

5.2.2.3 EqnS.3

250 mm may be assumed to have good bond.

2.7 N/mm2

12 =—

x= 400

444mm

2.7

=

lOqS or

°31b

100 mm

=

133 mm

5.23.4.1(1) Eqn 5.5

NAD 6.5(c)

In calculating net takeA6req as mid-span reinforcement/4 giving

r.et

5.2.2.1 Table 5.3

=

1.0

x

444

x

-

= 222 mm

>

l

5.4.2.1.4(3)

OK

Eqn 5.4

For a direct support, the anchorage length required is (2/3)1b,net

5.4.2.1.4(3)

= 148 mm

Figure 5.12(a)

The reinforcement details are shown in Figure 4.2. 112 —300

/

A

I

(2/3) tb,net

112 — 200 alternateLy staggered

148

220

Figure 4.2 Section through short span support The use of (213)1bn at a direct support is an allowance for the transverse compression due o the support reaction. Minimum area of reinforcement

A

•

0.6b d yk

l

0.0015bd

=

254 mm2/m

5.43.2.1(3) 5.4.2.1.1(1)

Minimum area provided (T12 @ 400 mm crs.) near support

= 283 Maximum bar spacing

=

3h

OK

mm2/m

500 mm

NAD Table 3 5.4.3.2.1(4)

Maximum spacing used =

400 mm near support

OK

4.1.2.2 Design example of a continuous two-way spanning solid slab Design a solid slab spanning between beams, as shown

in Figure 4.3.

In addition to self-weight, the slab carries a characteristic dead load of 1.0 kN/m2 and an imposed load of 5.0 kN/m2.

E1

Supporting beams

6m

6m

6m

I

I

I

72m

72m

72m

j

Figure 43 Layout of slab 4.1.2.2.1 Durability

For a dry environment, exposure class is 1. Minimum concrete strength grade is C25/30. For cement content and wlc ratio, refer to ENV 206 Table 3. Minimum cover to reinforcement Assume nominal aggregate size Assume maximum bar size Nominal cover Use nominal cover

= 15 mm = 20 mm

= 12 mm 20 mm

Table 4.1 ENV 206 Table NA.1 NAD Table 6

NAD 6.4(a)

= 25 mm

Note:

20 mm nominal cover is sufficientto meet the NAD requirementsin all respects.

NAD Table 3 4.1.3.3(8)

Check requirements for fire resistance to BS 8110: Part 2. 4.1.2.2.2 Materials Type 2 deformed reinforcement, ck = 460 N/mm2 C25/30 concrete with 20 mm maximum aggregate size.

1=1

NAD 6.1(a)

4.1.2.2.3 Loading Assume 200 mm thick slab

lfQ

=

4.8

=

5.0 kN/m2

=

1.35

=

1.5

+

1.0

= 5.8 kN/m2 Table 2.2

or 1.0

or 0.0

For non-sensitivestructures, a single designvaluefor permanent actionsmay = 1.35 throughout. be applied throughout the structure, i.e. Maximum ultimate load

= 1.35 x

Minimum ultimate load

=

1.35

5.8

+

1.5

x 5.8 =

x 5.0 =

2.3.2.3

1533 kN/m2

7.83 kN/m2

4.1.2.2.4 Load cases

For continuous beams and slabs in buildings without cantileverssubjected to dominantly uniformly distributed loads, itwill generally be sufficient to consider only thefollowing load cases. Alternate spans carrying the design variable and permanent load (a) + 'yGGk), otherspans carrying only the design permanent

2.5.1.2(4)

(ok load,

7GGk.

(b)

Anytwo adjacentspanscarrying the design variable and permanent load (Q0k + 'YGGk). All other spans carrying only the design permanent load,

7GGk.

4.1.2.2.5 Flexural design Bending moment coefficients for two-way spanning slabs supported on four edges, with provision for torsion at the corners, have been calculated based on both elastic and yield line theory. The coefficients published in BS 8110: Part 1, Table 3.15, are based on yield line analysisand are used in thisexample.

BS 8110

For continuous slabs the effects of rotational restraint from the supports can be ignored.

2.53.3(3)

Yield line methods can only be used for very ductile structural elements. Use high ductility steel Class H to prEN 10080(8).

2.5.3.2.2(5)

No direct checkon rotational capacity is required if high ductility steel is used.

2.53.5.5(3)

The area of steel should not exceed a value corresponding to

2.53.5.5(2)

x

—

d

=

0.25 which is equivalent to

M

=

Table 3.15

NAD Table 5

0.102

bd2ç

For the yield line (kinematic)method, avariety of possible mechanisms should be considered. Thisis assumed in theuse of the published bending moment coefficients.

2.53.5.5(4)

SLABS

The ratio of moments at a continuous edge to the span moment should be between 0.5 and 2.0. This is true for the published coefficients.

2.53.5.5(5)

Consider the design of the corner panel, D, in Figure 4.4.

2.5.1.2

-

-.

C..'

? 0.024

I-

—0•032

.

—0032 —0•037

I

m4

0 (n+

c$

c.$

C..'

0028

C...'

ci

—

A

•

—0037

I

B

ni .0

u ..o

0028

9 —0•037

C4

l n-

1 cY -i•f

ct C

0

Figure 4.4 Bending momentcoefficients cI1 = 1.2 Using the coefficients shown in Figure 4.4 and the method described in BS 8110 to adjust moments for adjacent panels with unequal conditions, the following moments and shears can be calculated for this panel: = 29.7 kNm/m In the 6 m direction,

BS 8110 3.53.6

M5

In the 7.2 m direction,

Mspan

=

28.5 kNm/m

M5

=

21.0 kNm/m

Mspan

=

20.6 kNm/m

Thesupportmoments calculated can be further reduced by an amount IMSd

M$d =

Fsdsup

x b5I8

Eqn 2.16

where Fsd su

=

design supportreaction compatible with the analysismoments.

In the 6 m direction, Fsd

= 81.9 kN/m

In the 7.2 m direction,

=

Fsd SUP

69.9 kN/m

For a 300 mm wide supporting beam: In the 6 m direction,

AMSd

In the 7.2 m direction, Msd

= 3.1 kNm/m = 2.6 kNm/m

Therefore, the design support moments are:

In the 6 m direction, Msup

=

26.6 kNm/m

In the 7.2 m direction, M

=

1&4 kNm/m

sup

2.5.3.3(4)

ri

For the short span, with the reinforcement in the first layer

= 200

d

25

—

—

12 — =

169 mm

2

__

=

0.038

=

0.087

=

0.045

=

414 mm2/m

bd2fIk

Af

_-

<

OK

0.25

2.5.3.5.5(2)

bdck

A

Use T12 @ 250 mm crs. (452 mm2/m)

I

in short span

The span moment is similar to that over the support and the same reinforcement may be used in the bottom For the long span, with the reinforcement in the second layer

d

=

_M

200

—

25

—

12

—

12 — =

157 mm

2

=

0.030

=

0.068

=

0.035

=

297 mm2/m

bd2fCk

x

Af

< 0.45

OK

bdck

A

Use T12 @ 300 mm crs. (377 mm2/m) T in long span

The span moment is again similar to that over the support and the same reinforcement may be used in the bottom Forarrangementsof reinforcementin middle and edge stripsuse BS 8110. The NAD directs the useof BS 8110 where torsion reinforcement is required in the corners of panels 4.1.2.2.6 Shear

BS 8110 3.5.3.5

NAD 6.5(e) 5.43.2.2 4.3.2

Use forces consistent with the analysis moments. In the 6 m direction:

At internal beam, At edge,

V

V.

mt

= 0.47 x 1533 x 6 = = 031 x 15.33 x 6 =

42 kN/m 28.5 kN/m

In the 7.2 m direction:

At internal beam, At edge,

V

V1,

= 0.4 x 1533 x 6 = 36.8 kN/m = 0.26 x 15.33 x 6 = 23.9 kN/m

+ 0.15a] bd

VAd1

= [rRdk(l.2 +

TAd

= 0.3 N/mm2

Table 4.8

j

5.43.2.2

Assume

k

=

4.3.2.3 Eqn 4.18

50% of the bottom reinforcement curtailed at edge support. 1.6 — 0.169

A

p

4Op,)

= —- =

bd

=

1.431

0.00134

0.02

w

Note: Ensuredetailing provides necessary anchorage to A81. See EC2 Figure 4.12 for definition of A81. aCp

=0

Therefore VAd1

=

91.0 kN/m

>

Vs =

It is also clear that VR >

28.5 kN/m at edge support

= 43.2 kN/m at the internal beam.

No shear reinforcement required

43.2.1P(2) 4.3.2.2(2)

4.1.2.2.7 Serviceability — deflection Control by limiting span/effective depth ratio based on the shorter span for a two-way spanning slab. 6000 = Actual span/effectivedepth ratio = 35.5 169

4.43.2 4.43.2(5)

For a corner panel use structural system 2.

It may be normally assumed that slabs are lightly stressed (p

0.5%).

Table 4.14 4.4.3.2(5)

NAD 6.4(e) and (f) allowsthe basic span/effectivedepth ratio to be interpolated, according to the reinforcement provided, for values in the range 0.15% < p

<0.5%.

Basic span/effective depth ratio (p

(p For the span moment A8req Aprov

=

452 mm2/m,

=

= 0.5%) = 32 = 0.15%) = 44

441 mm2/m

p = 0.27%

Basic span/effectivedepth ratio (p

= 0.27%) =

39.9

NAD Table 7

> 400 N/mm2,this value should be multiplied Using reinforcement with to reflect the actual servicesteel stress by the factor

— 250

x 452 = 400 =0.89 460x441

400 _______

=

f

x AsjeqIAsprov

yk

4.4.3.2(4)

Therefore, permissible span/effective depth ratio

= 0.89 x 39.9 =

OK

35.5

Note 2to NADTable 7 is taken to mean that the resultingspan/effective depth ratio, afterthe service stress modification, is limited to the value tabulated for nominally reinforced concrete. In this case the value is 44. 4.1.2.2.8 Serviceability

—

cracking

200 mm no further measures are required to control For a slab with h the cracking, provided requirements of EC2 Clause 5.4.3have been applied. 4.1.2.2.9 Detailing

4.4.2.3(1)

5.43

Slab thickness, h = 200 mm

>

50 mm

OK

For theshortspan, use alternatelystaggered bars and anchor50% ofthe midspan reinforcement at the external support.

5.4.3.1(1)

5.43.2.2(1)

Anchorage force (at external support) F5 Nsd a1

F5

As.req

=

VSd

x

a

d

5.4.2.1.4(2) Eqn 5.15

+ Nsd

= 0 = d =

VSd

5.4.3.2.1(1)

=

28.5 kN/rn 28.5 x 400

F5 =—=

f

=71mm/rn

Aprov = 226 mm2/m

OK

Net bond length

net

=

Clalb

Asjeq

X

1bmiri

AsprcN

1b

=

0.7

=

qSxf

5.2.3.4.1(1) Eqn

for curved bars yd

5.2.2.3

Eqn 53

SLABS

For all bars in slabs with h =

bd

250 mm, good bond may be assumed.

2.7 N/mm2

l 12

Table 5.3

400

=-j-x---

1b

In calculating =

= 444mm

take Asq as mid-span reinforcement/4.

0.7

+1

1

x 444 x

Bars to extend into support for

b —

5.2.2.1

= 156 mm >

NAD 6.5(c) 5.4.2.1.4(3) OK

a distance

= 256 mm

Figure 5.12(b)

giving sufficient end cover in 300 mm wide section

at edge beam Design moment = M8/4 = 7.125 kNm/m

OK

4.1.2.2.10 Top reinforcement

M

=

0.01

=

0.012

=

110 mm2/m lz

5.43.2.2(2)

bd2ck

Af bdfCk

A

A1

Minimum area of reinforcement

A8

0.6b d

z

0.001Sbtd

5.43.2.1(3)

= 254 mm2/m

yk

Use T10 @ 250 mm crs. bars extending 0.21 from inner face of supportinto span

The reinforcement details are shown in Figure 4.5.

5.4.3.2.2(2)

T10—500 secondary transverse

reinforcement

112— 300 middle Strip T10—300 edge strip T10 —250

112—250 aiternately staggered

I

Edge

strip

Middle strip

—I——

Figure 4.5 Detail at edge beam 4.1.2.2.11 Secondary transverse reinforcement — top

Principal reinforcement, T10 @ 250 mm crs., A5 Secondary reinforcement,A5 = 0.2 x 314 Maximum spacing

= 314 mm2/m = 63 mm2/m

= 500 mm

5.4.3.2.1(2)

NAD Table 3 5.43.2.1(4)

Use T10 @ 500 mm crs. (157 mm2/m) 4.1.2.2.12 Corner reinforcement

5.4.3.2.3

Use the detailing guidance given in BS 8110.

NAD 6.5(e)

of bottom reinforcement at intermediate supports Retain not less than aquarterof mid-span reinforcementat support and provide

5.4.2.1.5

Providecontinuity bars lapped with bottom reinforcementas shown in Figure4.6.

Figure

Using alternately staggered bars with continuity for 50% of the mid-span reinforcement.

5.13(b)

5.4.3.2.3

4.1.2.2.13 Anchorage

not less than 10 anchorage.

Minimum lap, net

= 1.4 x 444 x

= 310 mm 112— 500

_____

L12250

120 I

C

Figure 4.6 Detail at interior support

4

1

100

120

5.4.2.1.4(1)

SLABS

4.1.2.2.14 Transverse reinforcement at laps No requirement for slabs.

NAD 6.5(b) 5.2.4.1.2

4.1.3 Ribbed slabs EC2 permitsribbedslabsto betreated as solid slabs forthe purposesof analysis, provided that the flangeand transverse ribshave sufficient torsional stiffness. 4.1.3.1 Design example of a ribbed slab Design a ribbed slab spanning between beams as shown in Figure 4.7. In addition to self-weight, the slab carries a characteristic dead load of 1.0 kN/m2 and an imposed load of 5.0 kNIm2.

Supporting beams

I-

6m

6m

Ai PLAN

-

1oo

A- A Figure 4.7

Ribbed slab spanning between beams

2.5.2.1(5)

4.1.aL1 Durability For a dry environment, exposure class is 1. Minimum concrete strength grade is C25130.

Table 4.1 ENV 206 Table NA.1

For cement content and wlc ratio, refer to ENV 206 Table 3. Minimum cover to reinforcement Assume nominal aggregate size Assume maximum bar size Nominal cover

= = =

NAD Table 6

15 mm 20 mm 20 mm 20 mm

NAD 6.4(a)

= 25 mm

Use nominal cover

Note: 20 mm nominal cover is sufficientto meet the NAD requirementsin all respects.

NAD Table 3 4.133(8)

Check requirements for fire resistance to BS 8110: Part 2.

NAD 6.1(a)

4.1.3.1.2 Materials

Type 2 deformed reinforcement, ck

=

=

= 460 N/mm2

= 400N/mm2

2.2.3.2P(1)

Table 23

1.15

C25/30 concrete with 20 mm maximum aggregate size 4.1.3.1.3

Analysismodel

Span 6m

4 x slab depth 4 x 0.275 = 1.lm

Rib spacing Rib depth

= 600 1500mm = 175 4 x rib width = 500 mm

Flange depth

= 100mm 1

Transverse

x clear spacing between ribs

2.5.2.1(3)

OK OK OK

50 mm

...

OK

ribs (at supportsonly)

= 6m

> 10

x slab depth = 2.75 m Hence the ribbed slab maynot betreated as a solid slab inthe analysis under Spacing

2.5.2.1(5)

the terms of this clause unless intermediate transverse ribs are incorporated. This is not always desirable. Themodel adopted in this example uses gross concrete section properties of the shape in sagging regions and a rectangular section, based on the rib

I

width, in the hogging region.

EC2 Figure 23 has been used initiallyto define the extent of the hogging.This method can clearly be refined.

2.5.2.1(5)

SL&BS

4.1.al.4 Effective span 1eff

=

2.5.2.2.2

i, + a1

+

Eqn 2.15

a2

Assume 300 mm wide supporting beams

= 5700 mm a1

at edge beam =

i

= a2 at central beam = 6000 mm

a1

taken as (---) t =

a = (-i-) t

=

Figure 2.4(a)

150 mm

For ratio of adjacent spans between 1 and 10

150 mm

Figure 2.4(b)

1.5

= 0.85i = 0.85 x 6000 = 5100 mm

2.5.2.2.1(4)

Figure 23

4.1.3.1.5 Effective width of flanges

2.5.2.2.1

Effective flange width is assumedconstantacrossthe span for continuousbeams

2.5.2.2.1(2)

in buildings. For a symmetrical T beam

2.5.2.2.1(3)

b

+ (*) 10 be = = 125 + (*) x

b 5100

Eqn 2.13 600

mm

Therefore

beff=

4.1.3.1.6

600mm

Loading

=

3.6+ 1.0

=

5.0 kN/m2

=

1.35

— —

15

= 4.6 kN/m2 Table 2.2 2.3.2.3P(2)

Table 2.2

Maximum ultimate load

= 1.35 x

Minimum ultimate load =

1.35

4.6

x 4.6

+

1.5

x 5.0

=

13.7 kN/m2

6.2 kN/m2

4.tal.7 Flexural design Design for ultimate limit state using linear elastic method, choosing not to redistribute moments.

2.53.2.2

Consider the following load combinations:

2.5.1.2

the design variable and permanent load other spans carrying only the design permanent

(a) Alternate spans carrying

(ok load,

(b)

+

Gk)'

Anytwo adjacent spans carryingthe designvariable and permanent load ('ok + 1GGk). All other spans carrying only the design permanent load, GGk.

SLABS

-305 (—37-0)

26.7 (24 81

267 (248)

BENDING MOMENT ENVELOPE (kNm) Notes

1. Values are per rib 2. Values in brackets are those obtained when I us taken as uniform throughout the span

29-8 (30•8)

210 (20-2)

(— 20-2)

SHEAR FORCE ENVELOPE

(kN)

Figure 4.8 Results of analysis

The following results are taken from the analysis (see Figure 4.8). Mspan

Fsd5UP

26.7 kNm/rib =

—

=

59.6 kN/rib

30.5 kNm/rib

Support moment can be reduced by an amount 1Msd where

Msd =

59.6

x

0.3/8

=

2.2 kNm/rib

Therefore Msup

=

—

d

=

275

b

=

283 kNmlrib —

25

—

10

—

— = 232 mm 2

600 mm (span), 125 mm (support)

2.5.3.3(4)

Eqn 2.16

M

=

0.033

=

0.075

bd2fk

x

d

< 0.45

Neutral axis in flange (x

Afsyk

=

=

0.039

=

295 mm2/rib

OK 17.4

<

100 mm)

2.53.4.2(5)

OK

bdfck AS Use

2T16

M

—E

(403 mm2lrib) bottom in span

=

0.168 >

bd2fck

lLIim

=

0.167 (Section 13, Table 13.2)

Therefore

x d

>

0.45

This section may be analyzed to take account compression zone, as shown in Figure 4.9.

of the varying width of the

+

+ 195

cd

125

Figure 4.9 Analysis of section Consider x

0.45d

= 94 mm as a trial value

-

Using the rectangular stress block diagram with a =

af

=

0.85

x

1.5

0.8x

=

75 mm

bav

=

140mm

=

14.2 N/mm2

0.85 gives

NAD Table 3 4.2.1.3.3(12) Figure 4.4

z

= d—39 = 193mm

F

= a8x(dcd)bav = 75 x

MC

=

AS

= 28.3 x

149.1

x 140 =

(232 — 39)

x

=

io

106

=

400x193

28.8

149.1 kN

> 28.3 kNm

.... OK

367mm/rib

Use 4112 (452 mm2/rib) top at interior support =

Minimum longitudinal reinforcement with b A5

z 0.6 bdIck

1

0.0015 bd =

160 mm

56 mm2/rib <

5.4.2.1.1(1)

A8 ... OK

Maximum longitudinal reinforcement

A 4.1.3.1.8

= 0.04A = 3450

5.4.2.1.1(2)

mm2

>

Asprov

Shear in rib VSd

=

OK 4.3.2

29.8 kN/rib

at interior support

Shear resistance with no axial load TAd k(1.2

VRd1

TAd

k

=

+ 4Op1) bd

4.3.2.3 Eqn 4.18 Table 4.8

0.3 N/mm2

= 1.6—d =

1.368

lzl

Based on top reinforcement:

A

= 452 mm2/rib

b

= 125mm

p,

= —1 =

A

Figure 4.12

0.0155

0.02

bd

Giving VRd1

=

21.6 kN/rib

<

VSd

Therefore shear reinforcement must be provided.

4.a2.2(3)

Use the standard design method for shear:

4.3.2.2(7) 4.3.2.4.3

VRd3

VRd3

Vgci

=

VCd

+

VWd

Eqn 4.22

SLABS

where VCd

=

=

VRd1

21.6 kN/rib

43.2.4.3(1)

Therefore VWd

x 0.9dç

=

29.8

— 21.6

=

2

kN/rib

Eqn 4.23

Check maximum longitudinal spacing of links

5.4.2.2(7)

x 0.9d (1 + cota)

Eqn 4.25

VRd2

= (-i-)

For vertical stirrups, cota =

0

—a

v

= 0.7

VRd2

= 0.5 x 0.575 x

(i-) VRd2 <

—

= 0.575

200

16.7

0.5

Eqn 4.21

x 125 x 0.9 x 232 x iO = 125 kN

(f) VRd2

"Sd

Therefore

Sm

= 0.6d =

1 300 mm

139

Eqn 5.18

Try mild steel links at 125 mm crs. Pw,min

A

= 0.0022

Table 5.5

= 0.0022bs = 35 mm2

-

Use R6 links @ 125 mm cra (A8 =

fywd

=

Vwd

= -?-

57 mm2)

I

= 217N/mm2

1.15

125

x 0.9 x 232

x

iO

=

20.7

> 82 kN/rib

...

OK

Link spacing may be increased where (-k-)

Sm

=

V

= 25 kN/rib

0.8d 1 300

= 185Thm

Eqn 5.17

Use R6 links @ 175 mm crs. apartfrom region within 0.6 m of interior support

V

=

14.7

>

4

kN/rib

OK

4.1.3.1.9

43.2.5

Shear between web and flanges

VSd

a

Eqn 4.33

d

= a

= (j-)

l

= 2550 mm

Figure 4.14

Maximum longitudinal force in the flanges F0

x F

=

acd(O.8x)b

= 0.075 at mid-span = 14.2 x

0.8

x 0.075 x 232 x

= 122 kN

Force to one side of web

iFd

VRd2

VRd3

—

195

122

x

=

41.2 —

= 16.2kN/m

Therefore

vSd

600

=

2x600

= 41.2kN

2.55

= =

O.2çJhf

2.5TRdhf

= +

0.2

x

16.7

x

100 = 334 kN/m

>

VSd

..

OK

Af

Eqn 436 Eqn 434 Eqn 437

Sf

WithA = 0 VRd3

=

2.5

x 03 x 100 =

75 kN/m

>

VSd

OK

Eqn 4.35

No shear reinforcement required

4.12.1.10 Topping reinforcement

No special guidance is given in EC2 regardingthe design ofthe flange spanning between ribs. The Handbookto BS 8110(13) gives the following guidance. Thickness of topping used to contribute to structural strength Although a nominalreinforcementof 0.12%is suggested inthetopping (3.6.6.2), it is not insisted upon, and the topping is therefore expected to transfer load to the adjacentribswithoutthe assistanceof reinforcement.Themode oftransfer involvesarching action and this is the reason forthe insistence that the depth be at least one-tenth of the clear distance between the ribs. 3.6.1.5

Minimum flange depths are the same in EC2 and BS 8110 and the above is therefore equally applicable. Provide minimum reinforcementtransverselyand where top bars in rib, which have been spread over width of flange, are curtailed.

2.5.2.1(5)

SLABS

A

i

c1


Eqn 5.14

o.6bd,ck iz 0.0015bd

Therefore, conservatively 150 mm2/m

A

Use 18

200 mm crs. (251 mm2/m) or consider fabric

4.43.2

4.1.3.1.11 Deflection

Actual span/effective depth ratio

6000

=

232

=

25.9

403

Mid-span reinforcement ratio, p =

600

x

232

= 0.0029

Therefore section is lightly stressed. Basic span/effective depth ratio (interpolating for p) Modification factor for steel stress

Since flange width Since span

=

400 460

x x

403 295

4.43.2.(5) NAD Table 7

= 39.2 =

1.19

> 3 x rib width, a 0.8 modification factor is required.

7 m, no further

modification is required.

Permitted span/effective depth ratio

= 373 >

= 39.2 x

1.19

x 0.8 OK

25.9

4.1.3.1.12 Cracking

For exposure class 1, crackwidth has no influence on durability and the limit of 0.3 mm could be relaxed. However, the limit of 03 mm is adopted for this

4.4.2.1(6)

example.

Satisfythe requirementsforcontrol of cracking without calculation.Check section at mid-span: Minimum reinforcement,

A

=

Note:

can be conservativelytaken as thearea below the neutral axis for theplain concrete section, ignoringthe tension reinforcement,as shown in Figure 4.10.

4.4.23(2) 4.4.2.2(3) Eqn 4.78

92j

-________

Neutru! axis

—______ -

100

175

125

j•_5

Figure 4.10 Tensile zone of plain concrete section Depth to neutral axis =

A= =

as

160

92 mm

x 175 + 600 (100 — 92) =

32800 mm2

= 460 N/mm2

100%fyk

ceff =

recommended value 3 N/mm2

k

=

0.4 for normal bending

k

=

0.8

A

=

0.4

4.4.2.2(3)

x 0.8 x 3 x 32800/460 =

69 mm2 <

Apr,

OK

Check limit on bar size. Quasi-permanent loads

Eqn 4.78 Table 4.11

= Gk + °30k =

6.1 kN/m2

4.4.23(3) 2.3.4

Ratio of quasi-permanent/ultimateloads = —-- =

Eqn 2.9(c) NAD Table 1

0.45

13.7

Estimate of steel stress 0.45

x

As.req

x fyd

=

0.45

= 32

= 132 N/mm2

403

Asprc

Maximum bar size

295 x— x 400

> 16 mm provided

OK

Table 4.11

For cracks caused dominantly by loading, crack widths generally will not be

4.4.23(2)

excessive.

4.1.3.1.13 Detailing

Minimum clear distance between bars Nominal clear distance in rib

=

= 49 mm

20 mm

5.2.1(3)

OK

SLABS

Bond and anchorage lengths:

5.2.2

> 250 mm bottom reinforcement is in good bond conditions. reinforcement is in poor bond conditions. Top

5.2.2.1

For h

Figure 5.1(c)

Therefore, ultimate bond stresses are

Bottom reinforcement, bd = Top reinforcement,

2.7 N/mm2

= 0.7

d

Basic anchorage length, 1b =

x 2.7

=

1.89 N/mm2

5.2.2.2(2)

cbf

5.2.2.3

Eqn 53

qx 400

For top reinforcement, 1b =

5.2.2.2(2) Table 53

4x

=

1.89

x 400 4 x 2.7

For bottom reinforcement, 'b

=

Anchorage of bottom reinforcement at end support.

5.4.2.1.4

Treatas a solid slab and retain not less than halfof the mid-span reinforcement.

5.4.3.2.2(1)

Use 2T12

L bars bottom at end support

Anchorage force for this reinforcement with zero design axial load

F

S

a

= VSd x

5.4.2.1.4(2) Eqn 5.15

d

where

=

21 kN/rib

For vertical shear reinforcement calculated by the standard method a1

a

5.4.2.1.3(1)

= z(1 — cota)/2 1z 0 = 90° and z is taken as 0.9d

Although this ribbedslab falls outside the solid slab classification requirements for analysis, treat as a solid slab for detailing and take a1 = d.

5.4.3.2.1(1)

Therefore F5

=

21 kN/rib

x iO =53mm2
21

Asreq

400

Required anchorage length for bottom reinforcement at support: — — bMeL

aa lbAs.req mIn

A

OK

5.23.4 5.2.3.4.1(1) Eqn 5.4

s.prov

aa

lmIn

= =

0.7 for curved °31b

=

bars in tension

11.14

lz 104 or 100 mm

Eqn 5.5

jI In calculations of net' As req should be taken iz A55/4 =

bnet

5.4.2.1.4(3)

= 226 mm2

As,pr,, 1

NAD 6.5(c)

101 mm2

= 0.7 x 37 x 12 x

=

139 mm

>

OK

1 bmin

Minimum transverse reinforcement (for indirect support):

Eqn 5.4 5.2.a3

ASt = AS/4 = 226/4 = 57 mm2 Use 1T8 bar as transverse reinforcement Minimum top reinforcement at end support:

M5 = (--) 26.7 = M =

5.4.2.1.2(1)

6.7 kNm/rib

0.040

bd2fmk

Therefore nominal reinforcement is sufficient. Use 2112 L bars top as link hangers

The reinforcement details are shown in Figure 4.11.

Figure 5.12 2112 per rib

/

2112

per rib

N18

!L' I

L2116

100 • b13

139 Lb

Figure 4.11

net

=

15mm

=

rib

622

Is

lb.netal

i, for bottom bars: 1smin

For 100% of bars lapped and b > 24,, a1 = 1.4 = 1.0 and = Aspr Hence with Areq

net

per

Detail at edge support

Provide full lap-length,

i

m

=

374,

= 37 x 12 = 444 mm

= 03 aaallb = 187 mm 4 154, or 200 mm

5.2.4.1.3

Eqn 5.7 NAD Table 3 Figure 5.6 Eqn 5.4 Eqn 5.8

SLABS

Therefore S

=

444

x

1.4

= 622 mm >

OK

1 5mm

Transverse reinforcement at lapped splices should be provided as for a beam

<

section. Since 4, 16 mm, nominal shear links provide adequate transverse reinforcement. Anchorage of bottom reinforcement at interior support. Treat as a solid slab and continue 50% of mid-span bars into support.

The reinforcement details are shown in Figure 4.12.

5.2.4.1.2(1) 5.4.2.1.5 5.4.3.2.2(1)

Figure 5.13(b)

4112 per

rib

R6— 125

links

LJ1

h

L116

L L b,

1116

per rib

per rib

I

net

160

41

Figure 4.12 Detail at interior support This detailing prohibitsthe easy use of prefabricated rib cages because of the intersection of the bottom reinforcement with the supporting beam cage. It is suggested that providing suitably lapped continuity bars through the support should obviate the need to continue the main steel into the support.

Thearrangementofthe reinforcementwithinthe sectionincluding theanchorage of the links is shown in Figure 4.13.

5.2.5

NAD Tables

3&8

4Ø4 50 mm

T12

18- 200

R6

T16

InternaL radius

of bend =

Figure 4.13 Arrangement of reinforcement

2

mm

5.4.2.1.2(2) Figure 5.10

SLABS

4.2 Flat slabs 4.2.1 Flat slabs in braced frames The same frame is usedin each of the following examples, but columnheads are introduced in the second case.

4.2.1.1 Design example of a flat slab without column heads Design the slab shown in Figure 4.14 to support an additional dead load of 1.0 kN/m2 and an imposed load of 5.0 kN/m2.

A

N

B

ii

425m

A

I..

I

4•25rn

C

52m

_

I.

I

D

5•2m

_

SQm

5•2rn

4.25m

5E—

+

-

___

Figure 4.14 Plan of structure

The area shown is part of a larger structure whichis laterally restrained in two orthogonal directions by core walls. The slab is 225 mm thick. All columnsare 300 mm square and along grid 5 there is an edge beam 450 mm deep x 300 mm wide.

4.2.1.1.1 Durability

For a dry environment, exposure class is 1. Minimum concrete strength grade is C25130. Since a more humid environment is likely to exist at the edges of the slab, increase concrete strength grade to C30/37. For cement content and w/c ratio, refer to ENV 206 Table 3.

Table 4.1 ENV 206 Table NA.1

Nominal coverto reinforcement = Nominal cover to all bars

Use nominal cover

20 mm

NAD Table 6

iz

bar size

z

nominal aggregate size =

20 mm ..

OK

NAD 6.4(a) 4.1 .3.3(5)

= 20 mm

4.2.1.1.2 Materials

= 460 N/mm2

Type 2 deformed reinforcement,

NAD 63(a)

C30/37 concrete with 20 mm maximum aggregate size 4.2.1.13 Load cases

It is sufficient to consider the following load cases

+

(a)

Alternate spans loaded with spans.

(b)

Any two adjacent spans carrying YGGk + spans carrying 7GGk. Gk

=

'YGGk

7GGk

0.225

x 24 +

= 1.35 x +

ok

=

6.4 8.7

7GGk

1.0 =

2.5.1.2

and 'yGGk on other and all other

6.4 kN/m2

= 8.7 kN/m2

+ 1.5 x 5.0 =

Table 2.2 16.2 kN/m2

Eqn 2.8(a)

4.2.1.1.4 Analysis Analyses are carried out using idealizations of both the geometry and the behaviour of the structure.The idealizationselected shall be appropriate to the problem being considered. No guidance is given in EC2 on the selection of analysis models for flat slabs, or on the division of panels into middleand columnstripsand thedistribution of analysis moments between these strips. This is left to the assessment of individual engineera Therequirementsset down in BS 8110 forthe above points are taken as a means of complying with EC2 Clause 2.5.1.1P(3). EC2 allows analysis of beams and slabs as continuous over pinned supports. It then permits a reduction in the supportmoment given by Sd

= F b /8 Sdsup sup

Theanalysisin this example includes framing intocolumns. Thus the reduction is not taken. Consider two frames from Figure 4.14 as typical: (i) (ii)

Grid 31A—D subframe Grid B.

2.5.1.1.P(3) and P(4)

2.533(3) 2.5.33(4)

Analysis resuttsfor the frames described above are given in Figure 4.15. The results for each frame are practically identical as the analysis for Grid B has an increasedloaded width (5.2 m), sincethis isthe first internalsupportfor frames in the orthogonal direction. Member stiffnesses have been based on a plain concrete section in this analysis. Column moments and reactions are given in Table 4.1.

4250 Id

—

1

5200

5200

1/W

II 'stab 2

3500

3500

'777

77

ANALYSIS MODEL

—198

—199

123

BENDING MOMENT

Figures 4.15 Analysis of frame

ENVELOPE (kNm)

—201.

SLABS

Table 4.1 Column moments and reactions

(kN)

moments (kNm)

Max E column moments (kNm)

End

156.4

37.9

37.9

1st interior

444.7

6.8

21.4

4.2.1.1.5 Flexural

E column

Max. reaction

Suppo

design — Panel A—B/1—2

EC2 doesnot specificallyaddress the problem ofedge column momenttransfer and the provisions of BS 8110 are adopted hera

BS 8110 3.7.4.2

Mt,max = 0.l5bd2f e Cu Column A/2 moment transfer Assuming 20 mm cover and 20 mm bars in the top d1

d2 be

= 225 — 20 — 10 = 195 mm = 195—20 = 175mm = 300 + 300 (say) = 600 mm

I

= 37 N/mm2

M

=

Cu

NAD 4.1 3.3(5)

0.15

x 600 x

1752

x 37 x 106 =

102 kNm

Thisought to be compared with an analysis for a loading of l.4Gk + which would give approximately 5% higher edge moments than the EC2 analysis results above.

M

=

102

>

1.05

x 37.9

=

39.8 kNm

OK

Design reinforcement to sustain edge moment on 600 mm width.

Using;

=

1.5, a

= 0.85 and; =

1.15

Table 2.3

Referring to Section 13, Table 13.1: Msd

=

bd2ck

Afsyk

=

37.9

600

i0

x

x 1752 x 30

= 0.069

0.085

bdfCk

AS

0.085

x 600 x 175 x 30 =

582 mm2

= 970 mm2/m

460

= 0.163 < 0.45 (zero redistribution)

OK

2.53.4.2(5)

SLABS

Use T16 @ 150 mm crs. (1340 mm2/m) Place over width =

top at edge column

900 mm (see Figure 4.16)

Note: This approach gives more reinforcement than is necessary.

:1

I

T 1 •-I--

cII

1

esi

4_i

I

I Sø I—

Figure 4.16 Edge column moment transfer Check above moment against minimum value required for punching shear.

43.4.5.3 Eqn 4.59

mSd

For moments about axis parallel to slab edge

Table 4.9

= ±0.l25perm

V

=

156.4kN

Therefore mSd

= ± 0.125 x

Edge moment

= 0.6

= ± 19.6 kNm/m

156.4

= 632 >

19.6 kNm/m

OK

Design for mSd above in region outside edge column momenttransfer zone. 19.6

= bd2fCk

1000

x 106

x 1752 x 30

Minimum steel sufficient

=

= 0.6b d

f

0.021

z 0.OOlSbd

5.4.2.1.1

yk

= 0.0015 x 1000

x 175

= 263 mm2/m

Use T12 at 300 mm crs. (373 mm2/m) top and bottom (minimum) Maximum spacing

= 3h p 500 = 500 > 300 mm

OK

NAD Table 3 5.43.2.1(4)

SLABS

ColumnA/i moment transfer Assume the design forces for the frame on grid 1 are directlyrelated to those

for grid 3 in proportion to their loaded widths. Load ratio

=

(4.25/2) 5.2

=

0.41

The ratio of the edge column distribution factors for theframes is 2.0. Msd

= 37.9 x

0.41

x 2.0 =

31.1 kNm

Using design approach as for column A12: be

= 300 +

(say)

M

= 0.15 x 450

x 1752 x 37 x

=

450 mm 10-6

= 76 kNm

> 1.05 x 31.1 = 32.7 kNm Design reinforcement to sustain edge moment on 450 mm width MSd

450

bd2ck

Af

31.1

—

—

x

106

x 1752 x 30

=

OK

0.075

0.093

bdck

A

— —

x 450 x 175 x 30 =

0.093

460

Use T16 @ 150 mm crs. (1340 mm2/m)

478 mm2 =

1062 mm2/m

top for a width of 600 mm

Check above moment against minimum value required for punching shear. where

mJ

,

= ± 0.5 per m for corner columns

= ± 0.5 x (0.41 x Edge moment

=

31.1/0.45

156.4) say

=

>

32.1

In region of slab critical for punching shear: MSd

=

321x106 1000

bd2ck

x 1752 x 30

= 0.035

Af

—-- =

0.042

bdfCk

AS

=

0.042

x 1000 x 460

175

x 30

Eqn 4.59 Table 4.9

= ± 32.1 kNm/m

69.1 kNm/m

=

480 mm2/m

43.4.53

OK

Use 116 @ 300 mm crs. (670 mm2/m) top and bottom outside 600 mm wide moment transfer zone and over area determined in punching calculation

The division of panels intocolumn and middle strips is shown in Figure 4.17. Although BS 8110 indicates a 2.36 m wide column strip at column B2, a 2.6 m width has been used in the following calculations. This is considered reasonable as a loaded width of 5.2 m hasbeen taken in the analysis for grid B and grid 2.

A

B

BS 8110 Figure 3.12

C

Figure4.17 Assumed stripwidths(arrangement symmetrical about diagonal

All—C13)

ColumnB/2 support moments Analysis moment = Column strip Msd

b

=

MsdS = bd2fck

198 kNm in both directions

= 0.75 x 198 = 149 kNm

1300x2 = 2600mm 149

x

x

1752

2600

106

x 30

=

0.062

BS 8110 Table 3.20

Af —--

=

0.076

bdfCk

x 2600 x

0.076

AS

175

x 30

460

= 2255mm

Use 13116 (2613 mm2) top in column strip. Provide 9T16 @ 150 mm crs. in central 13 m and 2116 @ 300 mm crs. on either side

BS 8110 a7.3.1

Checkwhetherminimum moment requiredfor punching shear hasbeen met.

43.4.53

,

With

=

Msd

Table 4.9

—0.125

=

=

tJVsd

—0.125

x 444.7 =

—55.6 kNm

Eqn 4.59

Thisis to be carried over a width of 031. Since includes for a loaded width of 5.2 m, it is assumed that the larger panel width may be used. 0.31

=

0.3

x 5.2

=

1.56 m

By inspection reinforcement (9116 in central 13 m) is sufficient Middle strip (using average panel width) MSd

bd2k d

Af —-

0.25

=

(4725 =

0.059

=

0.031

X 198 X

106

2600) x 1752 x 30

—

<

OK

= 0.026

OK

0.45

bdfk A5 —

b

=

0.031

x

1000

x 175 x

30

460

Use 116 @ 300 mm crs. (377 mm2/m)

= 354 mm2/m

top in middle strip

It is noted that EC2 Clause 2.533(5)would allowtheuseofthemoment atthe face of the support (subject to limits in EC2 Clause 2.53.4.2(7)), but this is considered more appropriate to beams or solid slabs and the peakmoment over the support has been used in the above design. Span moments No special provisions are required in EC2. Hence the design basis of BS 8110 is adopted forthedivision of moments.The same pattern of reinforcementwill be provided in all panels. The column strip moments are given in Table 4.2 where

Msd

=

0.55 Msd

2.53.4.2(5)

Table 4.2 Column strip span moments Total

moment

MsdS

b

Msd

(kNm)

(m)

Span

MsdS

b (kN)

(kNm) End

107

589

2.12

278

1st. interior

123

67.7

236

286

Using the greater value:

!M8 b

/

Af

x— =

0.037

A

—

0.071

d

bdfCk

x

0.037

b

175

0.031

x 30

1752

d2fck

— —

X iO =

28.6

1

x 30 x io

<

0.45

—

422 mm2/m

460

OK

2.53.4.2(5)

OK

2.53.4.2(5)

Use 112 @ 250 mm crs. (452 mm2/m) bottom in column strips Using the middle strip moment for the first interior span

b =

4.725

MSd

= 0.45 X 123 X

236 = 2365 m (average panel width)

2365

bd2fk

Afk

—

=

x

1752

X

—

0.031

d

bdfCk

A — b

=

0.031

x

175

106

= 0.026

x 30 =

<

0.059

x 30 x iO

0.45

=

460

354 mm2/m

Use T12 @ 300 mm crs. (377 mm2/m) bottom in middle strips Minimum longitudinal reinforcement, using dmax =

0.6bd

fyk

= 195mm

iz 0.OOl5bd

= 0.0015 x 1000 x

195

= 293 mm2/m

OK

43.4

4.2.1.1.6 Punching

ColumnB12 (300 mm

x 300 mm internal column)

Critical perimeter located at 1.5d from face of column.

d =

4.3.4.1(3) Figure 4.16

185 mm (average)

For a rectangular column/wall check geometry

Perimeter

= 4

length breadth

=

x 300 =

43.4.2.1(1)

11d

1200 mm

= 2035 mm .. OK

2

1

OK

Hence

u VSd

= 2ir x

1.5

x 185 + 1200

= 2944 mm

Figure 4.18

= 444.7 kN

Note: No reduction in this value has been taken. The applied shear per unit length:

VSd

vSd

=

v

—--— u

43.43(4)

internal column =

where

1.15

Eqn 4.50 Figure 4.21

= 444.7

x iO x

1.15

2944

= 174N/mm

Shear resistance withoutlinks VRd1

=

rRdk(l.2

+

43.4.5.1

Eqn 4.56

4Op1)d

TAd

= 034 N/mm2

k

= 1.6

p1

= reinforcement ratio within zone 1.5d from column face (T16 @ 150 mm crs. top each way gives 1340 mm2/m)

P1

= J x i1 p,

d =

—

1340

=

1000

x 185

Table 4.8 1.6 — 0.185

i

=

1.415

1.0

0.015

=

0.0072

Note: The amount oftensile reinforcementin two perpendicular directions Assume p1

+

p

= 2 (0.0072) > 0.005

> 0.5%. OK

4.3.4.1(9)

Therefore

= 0.34 x

VAd1

=

VSd

x (1.2 + 40 x 0.0072) x 185

1.415

>

174 N/mm

133 N/mm

Vf

Therefore shear reinforcement required such that VRd3 Slab depth

=

4.3.4.3(3)

VSd

200 mm

OK

4.3.4.5.2(5)

Check that applied shear does not exceed the maximum section capacity =

VRd2

2.0

VRd1

=

2.0

x

OK

= 4447 x iO = 2.0 N/mm2

Shear stress around column perimeter

o.9[ç =

= 266 > 174 N/mm

133

1200

x

NAD Table 3 4.3.4.5.2(1)

185

4.9 N/mm2

OK

NAD 6.4(d)

Design shear reinforcement using EC2 Eqn 4.58 since VjVRdl =

174/133

EAswyd f

— —

Rd3

NAD 6.4(d)

1.6

Rdl

Eqn 4.58

u

Using type 2 deformed high yield bars as links

= — =

fyd

=

400 N/mm2

Table 2.3

1.15

Therefore

E

A

133)

(174 —

Minimum reinforcement ratio

= 302 mm2

= 100% x value given in EC2 Table 5.5.

NAD Table 3

Pwmin

=

0.0012

by interpolation

5.4.3.3(2) Table 5.5

area within critical perimeter

—

column area

43.4.5.2(4)

Denominator =

+ 3 x 185)2

x 185)2(4 —

= 575000

mm2

Maximum spacing of links is determined by the ratio vsd/vRd2 where assumed that VRd2 is calculated in accordance with EC2 4.3.2.4.3(4).

it is

(300 Thus

A

sw,mun

0.0012

—

(1.5

x 575000

=

—

3002

690 mm2 5.4.3.3(4) 5.4.2.2(7)

VSd

VRd2

= 174 N/mm

= (2) vcdbw x 0.9d f = — —- = 0.7

=

-

0.55

200 =

Eqn 4.25 Eqn

20 N/mm2

1.5

4.21

2.3.3.2 Table 2.3

Therefore VRd2 VsdIVRd2

Smax

=

(j-) x

=

174/916

=

0.8d P 300 mm

0.55

=

x 20 x 0.9 x

185

=

916 N/mm

0.2

0.19

Eqn 5.17

Longitudinal spacing b 0.75d = Transverse spacing p d

NAD 6.5(f)

138 mm

5.4.2.2(9)

Placing shear links on 100 mm grid in 700 mm square gives 48 links with 44 inside the critical perimeter.

4.3.4.5.2(2)

By inspection the minimum preferred bar size will govern and mild steel links could be used.

fyk

=

EA

250 N/mm2 0.0022

x 575000

=

1265 mm2

Table 5.5

Use 44 R8 links (2220 mm2) Where necessary the punching shear resistance outside the shear reinforced area should be checked by considering further critical perimeters.

4.3.4.5.2(3)

Check where VSd

=

VAd1

= 133 N/mm

Hence

u

=——= V5d13

444.7

x iO x

1.15

= 3845mm

133

VRd1

Therefore distance from column face

= (3845

—

1200)/2ir = 420 mm

= 2.27d

This would be approximately at the next critical perimeter taken to be at a distance 0.75d beyond the previous one. No further shear reinforcement required.

Thetensilereinforcement(p16@ 150mm crs.) should extend for afull anchorage length beyond the perimeter at 420 mm from the column face.

BS 8110 Figure 3.17

SLABS

ColumnAll (300 mm

x 300 mm corner column)

Critical perimeter located at 1.5d from face of column (see Figure 4.18).

4.3.4.1(3)

Figure 4.18 Critical perimeter at corner column U VSd

=

600 + 277r/2 = 1035 mm

=

0.41

x

156.4

=

64.1 kN

Applied shear per unit length, with /3 =

vSd

== V5j3

64.1

u

x

x

1.5

1035

1.5

4.3.4.3(4) Figure 4.21

= 93N/mm

Eqn 4.50

Reinforcement within zone 1.5d from column face is T16 @ 150 mm crs. top each way (see Figure 4.19). 300

277

t.u

'1

Figure 4.19 Corner column detail VAd1 VRd1

= 133 N/mm (as for column > VSd

B/2)

Therefore no shear reinforcement required

I

4.3.43(2)

SLABS

ColumnA/2 (300 mm

x 300 mm edge column)

Critical perimeter located at 1.5d from face of column.

u

=

900 + 277T =

V

=

156.4 kN

1770 mm

Appliedshear per unit length, with 3

vSd

VAd1

VAd1

=

=

= 156.4 x

u

=

1.4

iO x 1.4

1770

Figure 4.21

= 124 N/mm

Eqn 4.50

133 N/mm (as for column 6/2)

>

Therefore no shear reinforcement required

43.43(2)

4.2.1.1.7 Deflection Control by limiting span/effective depth ratio using NAD Table 7.

4.43.2

For flatslabs the checkshould be carried out on thebasis ofthelongerspan.

4.43.2(5)(d)

Forspan < 8.5 m, no amendmentto basic span/effectivedepth ratio isrequired.

4.4.3.2(3)

Note 2 to NAD Table 7 states that modifications to the tabulated values for nominally reinforced concrete should not be carried out to take into account service stresses in the steel (refer to EC2: Clause 4.4.3.2(4)). However it is assumed that correction ought to be carried out for 0.15% p < 0.5% but that the resulting values should not exceed those tabulated in the NAD for nominally reinforced concrete.

NADTable 7 gives basic span/effectivedepth ratios which are assumed to be = 400 N/mm2. based on

4.4.a2(4)

when =

460 N/mm2 and Areq

=

A 400 — x s,req

=

Modification factor

=

Aspr 0.87

Aprj Basic span/effectivedepth ratios for flat slabs are

(p = 0.5%) = 30 reinforced nominally (p = 0.15%) = 41 lightly stressed

Span reinforcement is typicallyT12 @ 250 mm crs. (452 mm2/m)

100A = 100 x 452 = 0.26% bd 1000 x 175

NAD Table 7

. (e)

By interpolation (p

= 0.26%), basic span/effectivedepth ratio =

= 5200 =

max span

d

29.7

175

mm

< 37.5 x

0.87

=

37.5

32.6

OK

4.2.1.1.8 Crack control Use method without direct calculation.

4.4.23

Estimateservicestress,

4.4.23(3)

Gk

+ 03K =

=

+

a, under quasi-permanent loads as follows: 6.4

+ 03 x 5 =

Ratio of quasi-permanent to ultimate design loads

=

7.9

kN/m

7.9/16.2 =

23.4

Eqn 2.9(c) 0.49

Therefore

a

=

0.49

xI x

A

A

< 200

A

NAD Table 1

Asprov

prov

Limit bar size using EC2 Table 4.11 or bar spacing using EC2 Table 4.12. The relevant limits are shown in Table 43. Table 4.3

Crack control limits

08

1.0

Steel stress (N/mm2)

Bar size (mm) Bar spacing (mm)

200

160

25

32

Table 4.11

250

300

Table 4.12

Maximum bar size used is less than 25 mm throughout

OK

Check minimum reinforcement requirement

Eqn 4.78

kkfc•iAcua

A2

A

4.4.23(2)

A

=—2

as 1ct,eff

k

x fyk

= 460 N/mm2

=

100%

=

minimum value suggested,

=

0.4,

k=

0.4

x 0.8 x 3 x

0.8

Therefore

A

3 N/mm2

AC

2x460

=

0.001

< 0.0015 bd (minimum flexural steel)

AC OK

SLABS

4.2.1.1.9 Detailing Considercombined requirementsfor flexure/shearand for punching fortop steel over supports.

ColumnB/2 For flexure/shearbars should extend for a distance d point at which they are no longer needed (a1 = diagram). =

-

x —5L

where

f

+ 1net z 2d beyond the d = shift in moment

=

(1)&(2)

Figure 5.11 5.4.3.2.1(1)

= 400 N/mm2

Eqn 5.3

250 mm bond conditions are good and

For h

5.4.2.1.3

Figure 5.1 Table 5.3

3.0 N/mm2

Therefore 400 =—x—

b

Olb

net

=

33

As.req

=

For straight bars

net

=

1.0

5.23.4.1 Eqn 5.4

and if

Asreq

=

1.0

Apr, =

534 mm for 116 bars, say 550 mm.

Curtail alternate bars as shown in Figure 4.20.

Alternate bars curtailed at 1

and

2

N, 2

0•5

0

I571 ,,j..5711 Figure 4.20 Curtailment diagram

SIJBS

Check that bars are anchored past relevant critical punching perimeter. Earlier calculation required column strip reinforcement to extend beyond a perimeter 420 mm from column face i.e. 570 mm from grid. It is assumed sufficient to provide an anchorage 1flet beyond this perimeter. Inspection of Figure 4.20 shows that this is satisfied.

4.2.1.2 Design example of a flat slab with column heads The previousexamplewill be used with column heads introducedatthe internal columnsto avoid the need for shear reinforcement. The rest of the design is unaffected by the change. 4.2.1.2.1 Punching at column B/2 (300 mm

x

300 mm internal column)

In the previous example it was found that VSd = column face where u = 3845 mm. Provide a column head such that

=

1H

VRd1

at 420 mm from the Figure 4.22

l.5hH (see Figure 4.21).

dcrjt

V

.1

I

300

Figure 4.21

IIH

Slab with column head

For a circular column head, assumethat EC2 Equation 4.51 applies to the case where 'H = l.ShH. Note: It is suggested that EC2 Equation 4.55 should read dcrit which reduces to the same as Equation 4.51 when 'H = Assume an effective column diameter, 1 = 300 mm

4.3.4.4(1)

= l.5dH + 05lç l.5hH.

To avoid shear reinforcement:

27rdt

3845 mm

dcrit

612mm

'H

612

—

1.5d

—

0.51

= 612

—

1.5(185)

—

150

=

185 mm

Eqn 4.51

SLABS

185 —

hH

=

123 mm say 125 mm

1.5

+

=

21H

670mm

Circular column head 125 mm below slab and 670 mm diameter is sufficient to avoid shear reinforcement

If a square column head is preferred, 11 =

d

12

=

+ 21H

= 1.5d + 0.561(11 12) 1.5d + 0.691 = 1.5d + 0.561 +

Eqn 4.52

To avoid shear reinforcement

d

612mm (612 —

1.12

= —— 1.12

hH

(612

>—= —

149

—

1 .5d —

1.5

x

0.56l)

185

—

0.56

x 300) =

149 mm

100mm

1.5

+

21H

= 600mm

Square column head 100 mm below slab and 600 mm wide is sufficient to avoid shear reinforcement

4.2.2 Flat slabs in laterally loaded frames In the following example, the structure used in Section 4.2.1 is considered to be unbraced in the North—Southdirection. 4.2.2.1 Design example of an unbraced flat slab frame Thisexample considers only theanalysis of the frame on grid B, consisting of three upperstoreys plus a lightweight roofstructure,as shown in Figures 4.22 and 4.23.

SLABS

425m

52m

t.2 52 m 3-

Figure 4.22 Plan of structure

3 Sm

3 Sm

3-5m

3

77

zI,-

Figure 4.23 Frame on grid B 4.2.2.1.1 Design loads

Office floors Dead load = 6.4 kN/m2 Imposed load = 5.0 kN/m2 Roof imparts load to columnsB/i and B/5 Dead load = 20 kN Imposed load = 30 kN

Sm

SLABS

Assume characteristic wind load

=

1.0 kN/m2

This is 90% of the value obtained from CP3: Chapter V: Part 2(11).

NAD 4(c)

Note:

The distribution of horizontal load between eachframe is determined by their relative stiffness. 4.2.2.1.2 Frame classification Determine whether sway frame or non-swayframe.

43.533

Check slenderness ratio of columns in the frame.

A calculation is required for those columns that resist more than 0.7 of the

A.3.2

mean axial load, NSdm at any level. Service loads are used throughout = 1.0).

A.3.2(3)

(i.e.

It is also assumed that these are vertical loads without anylateral loads applied.

Figure A3.4

7FFV

NSd,m

—

F

= E all vertical loads at given level (under service condition)

n

= Numberof columns

n A.3.2(1)

Consider a simple analytical model of the top floor to determine columns concerned as shown in Figure 4.24. 5OkN

SOkN

wkN/m P3

P2

11

P2

R

Figure 4.24 Load arrangement at third floor w

R2 R3

= 11.4 kN/m2 x loaded width = 11.4 x 5.2 (determined in Section 4.2.1.1) = 593 kN/m = 3 x 593 x 4.25/8 + 50 = 145 kN = 5 x 593 x 4.25/8 + (593 x 5.2/2) = 312 kN = 59.3 x 5.2 = 308 kN

100 + (18.9 NSd,m =

x 59.3)

= 244 kN

5

0.7Nsd,m= 171 kN

>R = 1

145 kN

Therefore slenderness of internal columns only needs to be checked. Clearly, this will normally be the case for multi-bay frames unless the edge columns carry large cladding loads. X

=

101i

where

43.53.5(2)

InnI

=

JII =

=

86.6 mm

A

For a horizontally loaded flat slab framedetermine the stiffnesses of the frame and thus the effective lengths of the columns using half the slab stiffness. Consider the centre column from foundation to first floor.

kA

=

El oI d C

C

; Ecm assumed constant

EOJbIleff

I

CCI

12

Eqn 4.60

i0 mm4

= 0.675 x

=

4.3.5.3.5(1)

= 4725 x 225 = 2.24 x iO mm4

2x

=

12

3500 mm

1

= 5200 mm

a

=

1.0

— —

2(0.675

eff

Therefore

kA kB

=

2.5.2.2.2

x 10/3500)

— —

x

05

10/5200) 2(2.24 0° (pinned at foundation)

Assuming that EC2 Figure 4.27(b) is appropriate to determine fi 10

=

f3l

= 2.15 x 3500 = 7500 mm

Figure 4.27

Hence X

= 7500/86.6 = 87

For non-swayframes

A3.2(3)

z 25

=

X

43.5.3.5(2)

= N5 Ultimate design load for centre column, ignoring self-weight of column. Nsd

= 3 =

x 16.2 x 30 = —

5.22

=

= 1314 kN

20 N/mm2

1.5

Therefore v

=

3002

= Since X

1314 x

iO =

x 20 =

17.6

Eqn 4.4 Table 2.3

0.73

.

25

> 25 the structure is classified as è sway frame

4.3.5.3.5(2)

SLABS

Theanalysisand designwould need to follow the requirements of EC2 Clause A3.5 to take into account the sway effects. EC2 Clause 2.53.4.2(4)does not generally allow redistribution in sway frames.

The method above is included to demonstrate its complexity. However, note theomission of guidance in EC2 Clause A3.2(3) on which nomogram to use

in EC2 Figure 4.27.

As an alternativemeans ofdetermining the frame classification, it is suggested that an analysis as detailed in BS 5950(14) is used to demonstrate thatthe EC2 requirements are met for non-sway frames. Assuming in the above examplethat the columnsizes are increased suchthat a non-swayframeresults, the following load cases need to be considered for design.

43.523(3) BS 5950: Part 1 5.13

These same load cases would also be applicable to sway frames where amplified horizontal loads are introduced to take account ofthe sway induced forces, complyingwith EC2 Clause A3.1(7) (b). 4.2.2.1.3 Load cases and combinations

2.5.1.2

With the rigorous approach the design values are given by

+ O,1k.1 + E

E7G Gk

23.2.2 P(2) Eqn 2.7(a)

I>1

where 0k.1

=

primary variable load, 0k2 =

= secondary variable load

0.7 generally

NAD Table 1

values are given in EC2 Table 2.2. Load cases with two variable actions (imposed and wind) are:

The

(a) Imposed load as primary load l35Gk + tSQk + l.O5Wk

(b) Wind load as primary load l3SGk ÷ tOSQk + l.5Wk In addition, load cases with only one variable action are: (c) Dead load plus wind l.OGk (favourable) + l.SWk l3SGk (unfavourable) + l.5Wk (d) Dead load plus imposed l3SGk + tSQk For non-sensitivestructures it is sufficient to consider the load cases (a) and (b) above withoutpatterning the imposed loads.

The NADallows the useof EC2 Equation 2.8(b) to give a single imposed and wind load case: l.3SGk +

135k (all spans) +

l.35Wk

NAD 6.2(e) 2.5.1.2P(1)

Final load combinations for the example given here

+ 1•5k (as Section 4.2.1.13) + l.5Wk (single load case)

1350k

0) (ii) (iii)

l.OGk

l.35Gk l.3SGk

(iv)

+ l.SWk (single load case)

+ 135k + l.3SWk (single load case)

4.2.2.1.4 Imperfections

2.5.13(4)

Consider the structure to be inclined at angle 1

=

0.005 radians

Eqn 2.10

ioop 1

=

frame height =

=

1-1.

=

Vd

NAD

i +1-

10.5 m

where n

n

Table 3

= number of columns = 5

0.78

= cxp = 0.78 x

0.005

= 0.0039 radians

Take accountof imperfections using equivalent horizontal force at each floor.

iH1

2.5.13(6)

= YVVd =

Using l.35Gk

Eti

Eqn 2.11

=

total load on frame on floor

+ (18.9

on each span gives

x 5.2) x

16.2

= 1592 kN

Therefore

= 1592 x 0.0039 = 6.2 kN per floor Assuming the frame by virtue of its relativestiffness picks up 4.725 m width of wind load: Wk

=

(4.725

x 3.5) x

1.0

=

16.5 kN per floor

Therefore the effects of imperfections are smaller than the effects of design horizontal loads and their influence may be ignored in load combinations (ii) to (iv). 4.2.2.1.5 Design

The design of the slab will be as described in Section 4.2.1.1.

2.5.1.3(8)

I

5 COLUMNS 5.1

Introduction Thedesign of column sectionsfrom firstprinciples using the strain compatibility method is covered. Examples of slender column design are also presented to extend the single example given in Section 2.

5.2 Capacity check of a section by strain compatibility 5.2.1 Introduction Two examples are considered: 1.

Where the neutral axis at ultimate limit state lies within the section; and

2.

Where the neutral axis at ultimate limit state lies outside the section.

The firstof theseis very simple whilethe algebra necessary for the second is more complex. For convenience, the same section will be used for both examples. This is shown in Figure 5.1. Assume

fyk

=

460 N/mm2 and fck =

30 N/mm2

350

2Jf

-

2132

_L k 30 tyk 460

500

4125

"j

_k..

Figure 5.1 Column section

5.2.2 Example 1 Calculatethe moment thatthe section can sustainwhencombined with an axial load of 2750 kN. 5.2.2.1 Basic method

If the neutral axis is within the section, thecompressiveforce generated by the concrete at ultimate limit state is given by NRdC

=

O.459fCkbx

4.2.133 Figure 4.2

and the moment by MAdC

=

NRdC(h/2

—

0.416x)

The strain at the more compressed face is taken as 0.0035 The procedure adopted is as follows: (1)

Assume a value for x

(2)

Calculate NRdC

(3) Calculate

the strain at each steel level

(4)

Calculate force generated by reinforcement (NRdS)

(5)

NRd

(6)

If NRd is not close enough to 2750 kN, modify the value of x and return to step (2)

(7)

If NRd is approximately 2750 kN, calculate MRdC and MRdS

(8)

MRd

+ NRdS

=

=

+

MRdC

MRdS

The design yield strain for the reinforcement 460

= 1.15

x 200000

=

0.002

5.2.2.2 First iteration Assumed value for x is 250 mm NRdC

= 0.459

0.0035 250

stOp

Strain NRdS1

cmid €sbot

NRdS3

x 30 x 350 x 250/1000 x 200

= 0.0028

> 0.002; therefore =

2x

804

= 1205 kN

f

=

400 N/mm2

x 400/1000

= 643 kN

= 0 and NRdS2 = 0 =

—

therefore

=

—400 N/mm2

x 491 x 400/1000

—2

=

—393 kN

Hence NRd

= 1205 + 643

—

393

= 1455 kN

This is considerably less than 2750 kN, hence x must be increased. Try new valuefor x

= 250

x 2750 1455

=

473 mm

COLUIA4S

5.2.2.3 Second iteration NRdC NRdSI

x 30 x 350 x 473/1000 =

=

0.459

=

643 kN as before 0.0035

smid

mid NRdS2

473

(473

=

0.00165

=

330

—

=

250)

x 200000

0.00165

= 330 N/mm2

x 2 x 491/1000

=

324 kN

€sbot

= 0.0035 (473

fS

= 0.00017 x 200000 = 34 N/mm2

—

473

NRd = 34 x 2 x

450)

2289 kN

=

491/1000

0.00017

= 33 kN

Hence NRd

= 2289 + 643 + 324 + 33 = 3289

kN

This is too large, hence x should be reduced. Linear interpolation gives

= 250 +

2750

—

(3289

—

1455\ 1455)

(473 —

250)

= 407 mm

5.2.2.4 Third iteration

x 30 x 350 x

NRdC

= 0.459

NAdS1

= 643 kN as before

mid

=

0.0035 407

(407

—

407/1000

250) =

= 1961 kN

0.00135

= 270 N/mm2 and NRd = 265 kN =

0.0035 (407— 450) =

—0.00037

407

f

=

—74

=

1961

N/mm2

and NRd =

—73 kN

Hence NRd

+ 643 + 265 —

73

= 2796

This is within 2% of the given axial load of 2750 kN

kN

OK

5.2.2.5 Moment MRdC

=

x (250 — 0.416 x 407)/1000

1961

MAd

= 643 x

MRdS2

= 0

MRd = 73 x

=

0.2

0.2

=

12&6 kNm

14.6 kNm

= 301.4 kNm

158.2 + 128.6 + 14.6

(

= 158.2 kNm

5.2.3 Example 2 Calculate the moment and axial force that can be sustainedby the sectionwhere the neutral axis depth is 600 mm.

Note:

The example has been given in this way so that repeated iterations are not necessary.These would not provide any new information to the reader. 5.2.3.1 Basic method When the neutral axis is outside the section the ultimate compressive strain is less than 0.0035 and is given by: U

—

0.002x

x

—

x 600 3 x 500/7

0.002

3 h/7

600

—

=

0.0031

(viii) & Figure 4.11

The conditions in the section are shown in Figure 5.2.

E 0•0031 (4

•

I.I

085 30I15 [.4

• • 600

S

/ "I Strain Figure 5.2 Conditions in section for Example 2

4.3.1.2(1)

—/ Stress in concrete

COLUIMtS

Thetechnique adopted for the calculation of NRdC and MRdC is to calculate the effect ofthe stress block on a depth of 600 mm nd then dduct the influence of the part lying outside the section. 5.2.3.2 Concrete forces and moments The equations for the full stress block are: N'RdC

= 05667(1

—

3/3)bxfCk

M'Rd,c = cN'Rd,c where c

= h/2

f3

=

—

4

+ x(j32 — — 12 43

6); and

O.OO2/E

Note: It will be found that, if the first example.

=

0.0035,theseequations give the values used in

The equations for the force and moment produced by the part of the stress block lying outside the section are =

a13)(x — h)bfk

—

0.5667a(1

= where

c' =

a

x — h12

—

—

(x — h) (8 — 3a) 12 — 4a

= = strain at bottom of section

From the strain diagram,

b =

0.00051

Hence

a

= 0.255 and

N'RdC

= 0.5667(1

—

c

= 250—

600(0.6452

=

0.645

0.645/3) x 350

12

x 600 x 30/1000 = 2802 kN

4 x 0.645 + 4 x 0.645

— —

6)

= 5.67mm

Hence M'RdC

=

5.67

x 2802/1000 =

15.9 kNm

= 0.5667 x 0.255(1—0.255/3)(600—500)350 =

—

600 — 250 — (600

— 500)(8 —

12 — 4

x 30/1000 = 139 kN

3 x 0.255) =

x 0.255

—284 mm

COLUMNS

MRdC =

—139

x 284/1000

=

—39.4 kNm

Hence

NRd,c

= 2802

MRd,c =

15.9

—

139

= 2663 kN

+ 39.4

=

553 kNm

5.2.3.3 Steel forces and moments Strain in upper layer of bars = This is

> 0.002; hence

600

x 550

=

0.0028

= 400 N/mm2

NRd,sl = 643kN MRd.sl = 643

x

0.2

= 128 kNm

Strain in middle layer of bars =

0.0031

600

x 350

=

0.00181

Hence f8

= 362 N/mm2

NRd = 355 kN,

MRd = 0

Strain in bottom layer of bars =

0.0031

600

x

150

= 0.000775

Hence

= 155 N/mm2

NRd =

152 kN

MRd =

—30.4 kNm

NRd MRd

= 2663 + 643 + 355 + 152 = 3813 kN = 553 + 128 — 30.4 = 153 kNm

5,3 Biaxial bending capacity of a section 5.3.1

General To carry out a rigorous check of a section for biaxial bending by hand is verytedious but possible if the simplified rectangular stress block is used. It is not suggestedthattheexamplegiven here is a normal design procedure for commonuse but it could be employed in special circumstances.There would be no difficultyin developing an interactivecomputer programto carry out design, in this way, by trial and error.

COLUMNS

5.3.2 Problem Demonstratethat the section shown in Figure 5.3 can carry ultimate design moments of 540 and 320 kNm about the two principal axes in combination with an axial loadof 3000 kN. Thecharacteristic strength of the reinforcement is 460 N/mm2 and the concrete strength is 30 N/mm2. 500

I

H

2038mm2 500

•

•

Ti,I Figure 5.3 Column section

5.3.3 Basic method

4.2.1.33(12)

The conditions in the section are shown in Figure 5.4.

Stress in concrete

Figure 5.4 Conditions in section Note: It is assumed that EC21 Section 4.2.133(12) implies that a should be taken as 0.8 for btaxial bending butthe NAD1 would allow 0.85.

It can be seen from the diagram that the axial force provided by the concrete is given by

NC

= 0.8bxf ccd

NAD Table 3

The moments about the centroidof the concrete section are given by Mcx

=NR c

where

h 1 1! = ———'lx

x

—

2XL

2

-tane2 +

—

btane(x

I

2

btane 6

1

)j

0.8fCdb3tane

M

cy

12

These equations are valid where x' < h. When x' > h, rather simpler equations can be derived. The location of the reinforcement is shown in Figure 5.5.

27

H 14

d

3

— — —— — — —

Figure 5.5 Location of 'reinforcement The stress in a bar is given by

f

x

= (200000

S

0.0035)

.

— —

ZP

yd

where

z

=

I—c -+ I

(b/2

—

d')tanO

0.8

db

=

cose

—

db}

depth from top faceof section to bar considered. This will be d' for top bars and h — d' for bottom bars.

The force in each bar is f5A5 and the moments are obtained by multiplying the forces by the distance of the bars from the centroid of the concrete section. Dimensions to the right or upwards are taken as positive. The total moments and forces carried by the section are the sum of thesteel and concrete contributions. The correct values of x and e have to be found by iteration.

5.3.4 Initial data

f

=—-=-= 1.5

1.5

2ONImm2

COLUIAI4S

Stress over upper 0.8 of the depth of the compression zone

= °8Cd = 16 N/mm2

fyd

f —

=

= 400N/mm2

1.15

As a first estimate of e, assume that the neutral axis is perpendicular to the direction of principal bending. This gives

=

8

= tan1O.59

tan-1

540 Try 8

= 30° which gives

tan8 =

0.58 and cos8

The limiting value of

=

0.87

x is where x'

=h

Hence X

=

h—

(b/2)tane

=

500

—

355

x500

250

x 0.58

= 355 mm

This gives

N

=

x 16

2840 kN

The reinforcement will increase this value significantly, hence x0 will be less = 300 mm. than 355 mm. Try

x

5.3.5 Calculation The simplest way to carry out the calculation is by writing the equations into a spreadsheet and then adjusting the values of and 0 until the correctaxial load and ratio MIM is obtained. The resulting output for the final iteration is given below. It will be seen that the result is satisfactory.

x

Section breadth (b) Overall depth (h) Embedment (ci') Steel area

500 500

50

8152

30

Concrete strength Steel strength

Average stress Design stress

460

Estimateof angle (radians) Estimateof x,

34.2° 05969026 282.5

Tan (angle) Cos (angle)

0.6795993 0.8270806

Neutral axis depth

z

f

F

1329

22378

2 3

363.13 32.29

—21.30

4

400.00 52.26

456.07 815.20 106.50

—192.54

—311.57

—634.97

126.99

742.80 2260.00

356.95 207.29

Steel totals Concrete Design resistances

432.58

Lever arm (x)

1

Barna

16

400

N

3002.80

91.72

M 91.21

1604

M

567.23

—91.21

16304 21.30 12699 220.12

11327

M

333.39

MJM

1.701432

5.4 Braced slender column 5.4.1 General Thecalculation ofthe effectivelength of columns has been adequately covered in Section 2. In the following example, the effective length is assumed.

5,4.2 Problem Calculatethe reinforcementrequired in a400 mm x 400 mm column subjected toa design axial load of 2500 kN combined with thefirst order bending moments shown in Figure 5.6.

The effective length has been calculated as M02

M01

8

m.

43.5.3.5

75 kNm

30 kNm

Figure 5.6 First order moments Assume =

460 N/mm2 and Ck = 30 N/mm2

5.4.3 Slenderness ratio X

= =

101i

=

43.53.2

(l0/h).[.

(8800I400)J =

76.2

5.4.4 Design requirements for slenderness Minimum slenderness ratio = = = Hence

greater of 25 or

15IJ

43.5.3.5(2)

Nsd/(ACfCd)

2500

15/fr

x

10/(400

17.0

<

x 400 x 30/1.5) =

0.78

25

Therefore minimum slenderness ratio =

25

Slenderness ratio>25, therefore column is slender

I

COLUMNS

Critical slenderness ratio Xcrjt e01

=

x

—30

e2 = 75 x

= 25(2

—

x 10)

=

106/(2500

106/(2500

4.3.5.5.3(2)

e01/e02) —12

mm

x

10) = 30 mm

=

60

Hence

= 25(2 +

12/30)

>

Slenderness ratio effects

therefore design is required for second order

5.4.5 Eccentricities Additional eccentricity ea

v ea

=

43.5.4(3)

lI2

= 1/200 =

2.5.12(4) Eqn 2.10

8800/400

= 22 mm

Equivalent first order eccentricity is greater of

x 30 — 0.4 x

0.6e2

+ 0.4e01 =

0.4e02

= 0.4 x 30 = 12.0 mm

0.6

4.3.5.6.2

12

=

13.2 mm

or

Hence ee

=

13.2 mm

Ultimate curvature, hr Assume d

hr =

= 400

=

—

200000

= 340 mm

60

2x

4.3.5.6.3(5) Eqn 4.72

2K2€1j0.9d

460

x K2

=

x 1.15 x 0.9 x 340

13.07K2

x 106 radians

Second order eccentricity e2 1<1

= =

0.1 K1l2(1Ir)

Eqn 4.69

1

Eqn 4.71

Hence e2

= 88002 x

13.07

x iO

1<2

=

101.2 <2 mm

Total eccentricity

=

C

+

ea

+

=

13.2

+ 22 +

101.21<2

mm

43.5.6.2(1) Eqn 4.65

COU)tt4S

to establish K2 and hence A

5.4.6 Iterative calculation

Make initial assumption of 1<2 =

1

This gives

ot

= 136.4 mm

N/bhck

= 2500 x l0I(400 x 400 x 30) =

M/bh2fk= 136.4 d'/h

=

x

60/400

2500

=

x l0/(400 x 30)

0.52

= 0.178

0.15

Using chart in Section 13, Figure 13.2(c) gives Take this modified value of 1<2 to recalculate

1<2

=

0.69

ot

Therefore

e0

=

13.2

+ 22 +

101.2

x 0.69

=

105.0 mm

Hence = 0.137

M/bh2fCk

This reducesK2 to 0.62 and M/bh2fck to 0.128 Try reduction of 1<2 to 0.60

Thisgives M/bh2fck = 0.125 which corresponds to 1<2

=

0.60 in the chart.

Hence

Af —--

=

038

bhfCk

A9

= 3965 mm2

Use 4132 and 2125 (4200 mm2)

5.5 Slender column with biaxial bending 5.5.1 Genera' This example has bending dominantly about one axis and is designed to illustrate the application of EC2 Section 4.3.5.6.4. There is some ambiguityin the drafting of this Section but the interpretation below seems reasonable.

5.5.2 Problem Design a 400 mm square column, having an effective length of 8 m in both directions,to withstandthe design ultimatefirstorder momentsshown in Figures 5.7 and 5.8 combined with adesign axial load of 2000 kN. The concrete strength class is C30/37 and the reinforcement has a characteristic strength of 460 N/mm2.

COLUMNS

X

= (8000/400)

Assume d'Ih

=

,j

=

693 in both directions

0.15

1102

53 kNm

7 Figure 5.7 First order moments in z direction

5.5.3 Assumptions

Figure 5.8 First order moments in

y direction

for design of section

It is assumed that and in EC2 Section 4.3.5.6.4 are the first order eccentricities at the critical section. They will, therefore, be effective values as defined by Eqns 4.66 and 4.67 in EC2 Section 4.3.5.6.2. Since

= 0

e01

Eqn 4.66

=

ee

43.5.6.2

0.6e02

Hence 0.6

e

x 334 x iO = 2000

0.6x53x

—

e

2000 =

(eJb)/(e/h)

=

100mm

10 = 16mm

0.16

<

0.2

4.3.5.6.4

Eqn 4.75

Hence separate checks for the two axes are permissible

eIh

=

100/400

= 0.25 >

0.2

A reducedvalue ofh, therefore, must be used in carrying out acheck for bending in the y direction. The additional eccentricity in the z direction is 0.5l/200 =

20 mm

Hence

ez

+eaz

=

120mm

LJ

43.5.6.4(3)

It is assumed that the intention of EC2 Section 43.5.6.4(3) is that, using the

reduced section, the applied load should just give zero stress at the least stressed face, i.e. as shown in Figure 5.9.

!ez+eaz

z

I

®h

—

ActuaL section

I

-I ELastic stress

distribution

I

Reduced section

ELastic stress distribution on reduced section

Figure 5.9 Assumption for check in the y direction It will be seen that the point of application of the load must lie on the edge of the middle third of the reduced section. Hence

h'

= 3(h12 — e — e) = 3(200 — 120) = 240 mm

5.5.4 Checkfor bending in z direction This check uses the full section dimensions 1

—

r

=

2x460x106K2 1.15

x 0.9 x 340 x

0.2

=

13.07K2

x

10-6

4.a5.63 Eqn 4.72

COLUMNS

Hence = (Since X > 35, =

etot

x 82 x

0.1

106

x 13.07K2 x

K

=

100

+ 20 + 83.7K2mm

10-6

83.7K2 mm

1 in EC2 Eqn 4.69)

As in the previous example, iterate using the design chart in Section 13 Figure 13.2(c) to find the appropriate value for K2 and hence ASç/bhfCk, starting with = 1. This procedure results in 1<2

M/bh2fck

Nbhfck

=

0.8

=

0.194

=

0.417

Hence

Asyk f Ibhfck=

0.55

=

0.55

A

x 4002 x 30/460

= 5739

mm2

Use 12125 (5890 mm2)

5.5.5 Check for bending in

y direction

The assumed section is shown in Figure 5.10.

kI

240 y M0

32 kNm

z_

Figure 5.10 Reduced section for check in y direction e

oy

eay e2y

=

16mm

=

e

=

e2 =

=

36

= 20mm 83.71(2 mm

Hence

ot

N/bhfck

=

Eqn 4.69

+ 83.71(2 mm x 240x400x30 2000

=

0.694

43.5.6.2 Eqn 4.65

COLUMNS

Mlbh2ck =

0.694

x (36 + 83.7K) = 400

0.0625

Using the same design chart as before, iterate

f

Asyk/bhfck. This gives 1(2

M/bh2fCk

=

0.47

=

0.13

=

0.57

=

0.57

+

0.145K2

to obtain

2 and hence

Hence ASck/bhfCk

A

S

x 240 x 400 x 30/460 =

3569 mm2

This is less than required for z direction bending An appropriate arrangement of reinforcement is shown in Figure 5.11.

Figure

5.11

5.6 Classification

v.—

ri

wi

tI

LI

1

OK

Arrangement of reinforcement

of structure

5.6.1 Introduction EC2 provides more detailed rules than BS 8110(2) for decidingwhetheror not a structure is braced or unbraced, or sway or non-sway. While it will normally be obvious by inspection how a structure should be classified (for example, with shear walls it will be braced and non-sway),there may be cases where direct calculationcouldgive an advantage.Thestructurein the examplefollowing is chosen to illustratethe workings of EC2 in this area. It is entirely hypothetical and not necessarily practical.

5.6.2 Problem an appropriate design strategy forthe columns in the structure shown in Figures 5.12 and 5.13. The applied vertical loadsin the lowest storey are set out in Table 5.1. Establish

I.-I

OLM4S

dD.*—300x300

a

750

=-—— 400 x 300

x 450

400 x 300

a

—'.-

d D.*— 300 x 300

Figure 5.12 General arrangement of columns

=:

:

: 35m

=:

:

:

: 35m

=:

: 35m

3-5m

_7_7 -;

7 7

//

7 -, / / 7 / / -,

,?

_7 7

Figure 5.13 Cross-section of structure Table 5.1 Columnsizes and loads column dimension

2nd moment of area

x 1O)

T

(kN)

Ultimate load (kN)

900

1600

1900

2680

900

1600 5695

2100 3300

2960 4660

675

1200

1700

(mm)

(mm

Column type

y

x

a b

300 300

400 400

c

750

450

15820

d

300

300

675

)

InI

Service load

CO%MA%4S

5.6.3 Check if structure can be considered as braced with the

750

x 450 columns forming the bracing elements

To be considered as braced, the bracing elements must be sufficiently stiff to attract90% ofthehorizontal load. Since all columnsare thesame length, this will be so if EIb

5.6.3.1

>

4.3.53.2(1)

0.9

y direction EI,racing

— —

Ei0

4 X 15820

=

6x900+4x15820+2x675

0904

Hence the four 750 x 450 columns can be treated as bracing elements carrying the total horizontal loads and columnstype a, b and d can be designed as braced in the y direction 5.6.3.2

x direction Elbracing

— —

6x

4 X 5695 1600 + 4 x 5695 + 2 x

— —

675

068

Structure cannot be considered as braced in x direction

5.6.4 Check if structure can be considered as non-sway Classificationof structures as sway or non-swayis covered in EC2 Appendix3. 5.6.4.1

y direction For braced structures of four or more storeys,the frame can be classified as non-sway if htotf F/E

I

A3. 2

0.6

EqnA3.2

where

= height of frame in metres

h0

=

4x

= sum of atl vertical loads taking 'y, =

Ecm

I

1

= 4x1900+2x2100+4x3300+2x1200=27400kN = sum of the stiffnesses of the bracing elements.

Taking Ecm as 32000 N/mm2

El

= 14 m

3.5

=

4x

=

2024960

15820

x 32000 x

x io Nmm2

3.1.2.5.2 Table 3.2 106

Nmm2

COLUMNS

Hence

x io 'J 2024960 x iO

F

fEl

/

27400

/mm

=

0.000116

/ mm = 0.116 I m

Note:

Sincethe height of the building is stated to be in metres, it seems reasonable to assume that m unitsshould be usedfor the otherfactors, thoughthis is not stated in EC2. Hence

IF __.!.

E

cmIc

= 14 x

0.116

= 1.62 >

0.6

Therefore the bracing structure is a sway frame in the y direction 5.6.4.2

x direction For frames without bracing elements, if X < greater of 25 or 15IJ for all elements carryingmore than 70% of the mean axial force then the structure may be considered as non-sway. Mean axial force

NSd

=

= sum of ultimate column loads no. of columns

4x2680+2x2960+4x4660+2x 1700 12

= 38680/12 = 3223 kN 7O%NSd

= 2256 kN

Columns type d carry less than this and are therefore ignored. Assume effective length of 400 x 300 columns is 0.8 x 3.5 (i.e. value appropriate to a non-sway condition). X

= 2.8 m

= 24.25<25

Therefore structure is non-sway

5.6.5 Discussion The results obtained in Sections 5.6.4.1 and 5.6.4.2 above are totally illogical as the structure has been shown to be a sway structure in thestiffer direction and non-sway in the less stiff direction. There are two possible areas where the drafting of EC2 is ambiguous and the wrong interpretation may have been mada (1) In Eqn A3.2 it is specifically stated that the height should be in metres. and E. Since the outputfrom Nothing is stated about the unitsfor A.3.2 is the statement of the units is unnecessary Eqn non-dimensional, unless the units for and Ec are different to that for Should Ji,, and be in N and mm units while is in m? If this were so, then the structure would be found to be 'braced' by a large margin.

), F

E

I, F

F

A.3.2(3)

(2)

In A.3.2(3) it doesnot state whetherX should be calculated assuming the columns to be sway or non-sway.In the calculation, the assumption was made thatthe X was a non-swayvalue. If a swayvalue had been adopted, the structure would have proved to be a sway frame by a considerable margin.

Clearly, clarification is required if A.3.2 is to be of any use at all. It is possible to take this question slightly further and make some estimate at what the answer should have been. Considering the y direction, the ultimate curvature of the section of the 750 450 columnsis —

r

0.2

2x460xK2 x 106 x 1.15 x 0.9 x 700

6.351<2

x

x 10-6

Inspection of the design charts and levels of loading suggest 1<2 is likely to be about 0.6. Assumingan effective length under sway conditionsoftwice the actual height gives a deflection of:

(2

x 3.5)2 x 635 x 0.6 10

=

19 mm

Thisisan overestimateofthe actual deflection. It corresponds to an eccentricity of 19/750 of the section depth or 2.5%. Thismustbe negligible, hence, in the y direction, the structure must effectively be non-sway.

5.7 Sway structures 5.7.1 Introduction Although EC2 gives information on how to identify a swaystructure, it does not give anysimple approach to theirdesign. However,Clause A.3.5.(2) statesthat "the simplified methods defined in 4.3.5 may be used instead of a refined analysis, provided that the safety level required is ensured". Clause A.3.5(3) amplifies this slightly, saying that "simplified methods may be used which introduce bending moments whichtake account of second order effects the average slenderness ratio in each storey does not exceed provided 50 or whichever is the greater".

20/ji

EC2 Section 4.3.5 gives the 'model column' method which is developed only for non-swaycases, so it is leftto the user to find a suitable method for sway frames on the basis of the Model Column Method. BS 8110 doesthis, so it is suggestedthat the provisionsof3.&a7 and3.83.8 of BS 8110: Part 1 are adopted, butthat the eccentricities are calculated using the equations in EC2.

5.7.2 Problem Design columns type c in the structure considered in Section 5.6.2 assuming sway in the x direction. The column loads may be taken from Table 5.1. Thedesign ultimatefirstorder moments in the columns are as shown in Figure 5.14.

3 has been assessed from EC2 Figure 4.27(b) as 1.6 for all columns.

COLUMNS

Column

H0

kNm

type ci

80

b c

110 176

d

25

Figure 5.14 First order moments

5.7.3 Average slenderness ratio The slenderness ratios are shown in Table 5.2. Table 5.2 Slenderness ratios Column type

No.

X

a b

4 2 4 2

48.5 48.5

c

d

64.7

=

Mean value

Since >m

43.1

< 50, the simplified method may be used.

1

x 460 x 1.15 x 0.9d

21<2

r

200000

Hence =

x

(1.6 3.5)2 ____________

0.0044K2

p

=

0.00441<2

d

x 106

d

10 1

A.3.5(3)

=

138001<2

d

mm

1

iooI

2.5.1.3 Eqn 2.10

200

This may be multiplied by

a

Eqn

2.11

Where, with 12 columns

=

1(1

+ 1/12)/2 =

0.736

Hence p

ea e01

= 0.00368 = 0.00368

x 1.6 x 3500/2

= 103 mm

Eqn 4.61

=

+ 13800K2/d =

e0 + 10.3

Eqn 4.65

+

ea

The total eccentricities are shown in Table 5.3.

+ 13800K2/d mm

COLUtM4S

Table 5.3 Total eccentricities d

Column type

M (kNm)

(mm)

a b

350 350 400 250

c

d

N

e0

e1

(mm)

80

30

110 176

37 38

25

15

(mm)

40.3 +

0.744 0.822 0.460 0.630

47.3 + 483 + 25.3 +

391<2 391<2 351<2

55K2

As in the previousexamples, the design chartscanbeused iterativelytoestablish and hence 02• This process gives the values shown in Table 5.4. 1<2 Table 5.4 Lateral deflections Column type

K2

a b

0.39 0.41 0.50 0.45

c d

No. of columns

e2

(mm)

4 2 4

15.2 16.0 17.5 24.8

Average deflection =

2

17.7 mm

All columnswill be assumed to deflect by the average value. The resulting designs are shown in Table 5.5 Table 5.5 Summary of designs Column type

a b

c d

M

N

Vyk

(mTh)

bh2ck

bh

bhfCk

58.0 65.0 66.0 43.0

0.108 0.134 0.067 0.090

0.744 0.822 0.460 0.630

e

0.53

075 0.10 0.38

A8 (mm2)

4148 5870 2201 2230

BS 8110 3.8.3.8

6 WALLS U

6.1

Introduction A wall is defined as a vertical load-bearing member with a horizontal length

2.5.2.1(6)

not less than four times its thickness. Thedesign ofwalls is carried out by considering verticalstrips of thewallacting as columns.

6.2 Example Design the lowest level of a 200 mm thick wall in an eight storey building supporting 250 mm thick solid slabs of 6.0 m spanson each side. The storey heights of each floor are 35 m, the heightfrom foundation tothefirstfloor being 4.5 m. The wall is fully restrained at foundation level. The buildingis a braced non-swaystructure.

6.2.1 Design data = 700 kN/m = 5 kNm/m = 2.5 kNm/m

Design axial load (Nsd) Design moment at firstfloor Design moment at foundation

Concrete strength class is C30/37.

3.1.2.4

= 30 N/mm2

Table 3.1

6.2.2 Assessment of slenderness Consider a 1.0 m vertical strip of wall acting as an isolated column. The effective height of a column

=

3l

43.5.3.5(1)

where

= actual height of the column between centres of restraint

9 is a factor depending rigidity kA

=

upon the coefficients of restraint at the column ends.

I

E °'Ii°' E 'slab'1eff,slab

43.53.5(1) Eqn 4.60

Assuming a constant modulus JCCI

ab

kA

=

1

X

02

=

1

x 0.25

6.67

x iO

x iO m4

13 x

=

12

=

of elasticity for the concrete:

6.67

12

=

kA and kB relating to the

+

6.67

i0 x

3.5

m4

iO

I 2x I

1.3

6

x

iO

=0.78

Base of wall is fully restrained. Therefore

kB

=

0.40 which is the minimum value to be used for kA or

k.B

43.5.35(1) Figure 4.27(a)

Hence

=

13

Icol

0.7

= 4500mm

Therefore

= 0.7

10

x 4500

= 3150 mm

The slenderness ratio X = Ic/I

43.5.3.5(2)

where

= radius of gyration

I

=

fY JA

=

1000

x 200

J12x1000x200

=

57.7 mm

Therefore

=3i5o=

X

54.6

57.7

Isolated columnsare considered slender where X exceeds the greater of 25

or 15IJi where =

N 4.3.53.5(2)

= 700kN A0

i0 mm2

= 1000 x 200 = 200 x =

Therefore =

-

=

= 1.5

20N/mm

x io 200 x 10 x 20 700

=

0.175

Hence 15

=

15

=

J0.175 Therefore the wall is slender

6.2.3 Design The wall may now be designed as an isolated column in accordance with EC21 Clause 43.5.6 and as illustrated in the example in Section 5. Although the column or wall has been classifiedas slender, second ordereffects need not be considered if the slenderness ratio X is less than the critical slenderness ratio X•.

WALLS

=

25 (2

—

43.5.5.3(2) Eqn 4.62

e011e02)

where

e1 and e2 are the first order eccentricities at the ends of the member relating to the axial load. =

01

— M

'Sd Msdl

M —

ande02 =

IvSd

and Msd2 are the first order applied moments.

Therefore >crit

= 25 (2

—

Msdl/Msd2)

where

MSdl —
In this example: Xcrit = 25 2 .

—

= 62.5 >

(_-) 5.0

54.6

The column or wall should therefore be designed for the following minimum conditions:

Design axial resistance (NAd)

=

Eqn 4.63

Nsd

=

Design resistance moment (MAd)

43.5.53(2)

Nsd

x

Eqn 4.64

For this example

MRd = 700

x —-- =

7.0

>

5.0 kNm

20

6.2.4 Reinforcement The vertical reinforcement should not be less than 0.004A0 or greater than

5.4.7.2(1)

0.04A0.

Half of this reinforcement should be located at each face.

5.4.7.2(2)

The maximum spacing for the vertical bars should not exceed twice the wall

5.4.7.2(3)

The area of horizontal reinforcement should be at least 50% of the vertical reinforcement.The bar size should not be less than one quarterofthe vertical bar size and the spacing should not exceed 300 mm. The horizontal reinforcementshould be placed betweenthevertical reinforcementand the wall

5.4.7.3

thickness or 300 mm.

face.

=

(1)—(3)

Link reinforcement is required in walls where the design vertical reinforcement exceeds O.O2A.

In normal buildings it is unlikely that walls will be classified as slender. For practical considerations they will generally not be lessthan 175 mm thick and the vertical load intensity will normally be relatively low. Thus the limiting slenderness ratio given by will be high. In cases where the wall is slender,only slenderness about theminor axis need be considered. Even in this case it is likely that only the minimum conditions given in EC2 Clause 4.3.5.5.3(2) Eqns 4.63 and 4.64 will apply.

15I[

5.4.7.4(1)

7 FOUNDATIONS U

7.1

Ground bearing footings 7.1.1 Pad footing Designa square padfooting fora400 mm x 400 mm column carrying aservice load of 1100 kN, 50% of this being imposed load with appropriate live load reduction. The allowable bearing pressure of the soil is 200 kN/m2. 7.1.1.1

Base size With 500 mm deep base, resultant bearing pressure

= 200

x 24 = 188 kN/m2

— 0.5

1100

Area of base required = Use 2.5 m

=

5.85 m2

188

x 2.5 m x 0.5 m deep base

7.1.1.2 Durability For components in non-aggressive soil and/or water, exposure class is 2(a). Minimum concrete strength grade is C30/37.

Table 4.1 ENV 206 Table NA.1

For cement content and w/c ratio refer to ENV 206 Table 3(6) Minimum cover to reinforcement is 30 mm. For concrete cast against blinding layer, minimum cover

> 40 mm.

NAD Table 6 4.1 33(9)

Use 75 mm nominal cover bottom and sides 7.1.1.3 Materials = 460 N/mm2 Type 2 deformed reinforcement with Concrete strength grade C30/37 with maximum aggregate size 20 mm 7.1.1.4 Loading

= l.3SGk + 15k = 1570 kN

Ultimate column load

NAD 6.3(a)

Eqn 2.8(a) Table 2.2

7.1.1.5 Flexural design Critical section taken at face of column Msd

= 1570 (2.5 8

—

0.4)2

x 2.5

2.533(5)

= 346 kNm

Assuming 20 mm bars

d

= 500

—

75

—

20

= 405 mm

Using rectangular concrete stress diagram — —

fck

— —

30

— —

20 N/mm2

C

= 0.85 x 20 = 17 N/mm2

Figure 4.4 Eqn 4.4 Table 23

For reinforcement

=

f

460

=

= 400 N/mm2

2.23.2P(1) Table 23

1.15

7s

For the design of C30/37 concrete members without any redistribution of moments, neutral axis depth factor

d

2.53.4.2(5)

0.45

Using the design tables for singly reinforced beams Msd

2500

bd2fCk

x— d

Af —-

346 X

=

106

x 4052 x 30

= 0.063 <

=

0.028

0.45

OK

= 0.033

bdfCk

Hence

AS

= 0.033 x 2500 x 405

30 = x— 460

2179 mm2

Minimum longitudinal reinforcement

=

0.6btd 460

5.42.2.1 5.4.2.1.1

= 0.OOl3bd iz 0.OOl5bd

= 0.0015 x 2500 x 405 = 1519 mm2 7120 gives 2198 Bar crs.

> 2179 mm2

2500

Maximum spacing

—

2(75)

—

OK

20

6

= 3h

i

500

= 388 mm = 500 > 388 mm

OK

7T20(EW) are sufficientfor flexural design. Additional checks for punching and crack control require 9T20 (EW) — refer to Sections 7.1.1.7 and 7.1.1.8.

NAD Table 3 5.4.a2.1(4)

Use 9120 (EW) 7.1.1.6 Shear Minimum shear reinforcement may be omitted in slabs having adequate provision for the transverse distribullon of loads. Treating the pad as a slab, therefore, no shear reinforcement is required if VSd VRd1.

43.2.IP(2) 4.3.2.2(2)

cu1ooxIoI4s Shear force at critical section, distance d from face of column —

—

(a5

VSd

0.405

=

)

405 kN

4.3.23

Shear resistance, VAd1, with zero axial load VRd1

TAd

k

=

TRdk(l.2

+

4Op1)

Eqn 4.18

bd

Table 4.8

=

034 N/mm2

=

1.6—d = 1.195z1.0

To calculate A1, area of tension reinforcement extending critical section, determine

net

=

d + 1fleI beyond

(a) i

aalb

4.3.2.2(10)

Eqn 5.4

prov

For curved bars with concrete side cover of at least

=

b

— —

3

0.7

Eqn5.3 4

f

For bars in the bottom half of a pour, good bond maybe assumed. Hence for 4,

5.23.4.1(1)

5.2.2.1(2)

32mm

= ao N/mm2 =

4,

x

Table 5.3

400

=

—-

33.34,

For anchorage in tension

min = 03 x 1b z 104, 100 mm = 104, = 200mm

Eqn 5.5

Actual distance from critical section to end of bar =

2500 2


—

400

— — 405 — 75 2

=

570 mm

= 405 + 200 = 605 mm

Therefore

A, VRd1

=

0

= 0.34 x

>

1.2

x

1.195

x

2500

x 405 x

iO-3 = 493 kN

405kN

VSd

No shear reinforcement required Ch,eck that

VSd

=

u

VRd2

VRd2

0.7

to avoid crushing of compression struts. =

—

200

0.55

Eqn 4.20

iz 0.5 N/mm2

= cd'w°9°' = 0.55 x 20

x 2500 x

2

0.9

x 405 x

10

2

= 5012 > 405kN

Eqn 4.19

OK

7.1.1.7 Punching Length of base from face of column

a

= 1050mm

a —

=—>2

h

1050

Figure 4.16

400

By definition the foundation should be considered as a slab. Critical perimeter at 1.5d from faceofcolumn shouldbe checked for punching. U

=

2ir (1.5 x 405) + 4 x

400 =

5417 mm

In foundations the applied shear may be reduced to allowfor the soil reaction within the critical perimeter.

4.3.4.1P(4)

& 43.4.2.2 4.3.4.1(5)

Enclosed area

= (3 x 405) + 400 = 1615 mm = 608 mm Corner radius = 1.5 x 405 = 1.6152 — (4 — ir) 0.6082 = 2.29 m2 Area Total width

= 1570 (1

—

= 995 kN

The applied shear per unit length

43.4.3(4)

v=Y Sd

u

=

1.0 for pads with no eccentricity

of load

FOUNO&TIONS

Therefore 995

=

vSd

i0

x

= 184N/mm

5417

Theamount oftensile reinforcementin two perpendicular directions should be greater than 0.5%. This is assumed to require p1 + p1, > 0.5%. Using 9120 (EW),

=

A

4.3.4.1(9)

2830 mm2 (EW)

For Bi

= 415mm

dx 1OOA

S

= 0.27%

bd For B2

= 395mm

dy 1OOA

S

= 0.28%

bdy 0.27%

÷ 0.28%

=

0.55%

> 0.5%

OK

Punching resistance for a slab withoutshear reinforcement = + TRdk(l.2

VRd1

4.a4.5

40p1)d

The equation produces similar values to the shear check performed above = 034 x 1.195 x 1.2 x 405 = 197 > 184 N/mm VAd1 No shear reinforcement required Checkthe stress at the perimeter of the column

0.90f = 0.90fö

VSd/ud

d

=

405mm

u

=

4x400 = 1570 x i0

Stress =

405

x 1600

=

NAD 6.4(d)

4.9 N/mm2

1600mm =

2.4

<

4.9 N/mm2

....

....

OK

7.1.1.8 Crack control Use method without direct calculation.

4.4.2.3

Estimate service stress in reinforcement under quasi-permanent loads using the following approximation

44 23(3)

Gk

+ '2°k =

+ °•3k = 550 +

0.3

x 550

=

715 kN

22.2 3P(2) 23.4 Eqn 2.9(c)

& NAD Table 1

FOUNDTOHS

=

Hence quasi-permanent load/factored load and estimated service stress

x

= 0.46 x

715/1570

=

0.46

A = 0.46 x 400 x = 142 N/mm2 A prov 2830

Either limit bar size using EC2 Table 4.11(1) or bar spacing using EC2 Table

4.4.23(2)

4.12.

=

4)

20

< 32mm

OK

This has been chosen to comply with Table 4.12 as well. =

Using 9T20 (EW) bar spacing

290

< 300 mm

OK

Check minimum reinforcement requirement

4.4.2.3(2)

kA/a

For

A

4.4.2.2

Eqn 4.78

it is considered conservative to use (h/2)b =

100%

= 460 N/mm2

x

For c.eff use minimum tensile strength suggested by EC2

k

=

=

3 N/mm2

0.4 for bending

For k interpolate a value for h

k

—

0.5

+ 03(80 —

= 50 cm from values given 50)/(80 — 30) = 0.68

Therefore As req

= 0.4

x 0.68 x 3 x 250 x 2500/460

=

1109 mm2

= 2830 > 1109 mm2

OK

7.1.1.9 Reinforcement detailing

Checkthat flexural reinforcement extends beyond critical section for bending for a distance d + net 1b

=

=

3334)

5.43.2.1(1)

& 5.4.2.13

667 mm

Assuming straight bar without end hook 1

=

d + net = Actual distance =

1.0

x 667 x

405

514 mm

Eqn 5.4

+ 514 = 919 mm

2500

2

= 2830

—

400 —

2

—

75

= 975 > 919 mm

The reinforcement details are shown in Figure 7.1.

OK

POUNDATIONS

I

I

lIi.

S

S

U

75 cover

S

U

U)

111500

9120— 300EW 2500

I-.'

Figure

7.1

Detail of reinforcement in pad footing

7.1.2 Combined footing Design a combined footing supporting one exterior and one interior column.

An exterior column, 600 mm x 450 mm, with serviceloadsof 760 kN (dead) and 580 kN (imposed)and an interior column, 600 mm x 600 mm, with service loads of 1110 kN (dead) and 890 kN (imposed) are to be supported on a rectangular footing that cannot protrudebeyond the outerfaceof the exterior column. The columns are spaced at 5.5 m centres and positioned as shown in Figure 7.2.

The allowable bearing pressure is 175 kN/m2, and because of site constraints, the depth of the footing is limited to 750 mm.

--I —

600

Figure 7.2 Plan of combined footing

7.1.2.1 Base size Service loads =

Gk +

Column A: 1340 kN and Co!mn B: 2000 kN Distance to centroid of loads from LH end =

0.3 +

2000

x 5.5

3340

=

3.593m

For uniform distribution of bad under base Length of base

= 2 x 3.593 say 7.2 m

With 750 mm deep base, resultant bearing pressure = 175 — 0.75 x 24 = 157 kN/m2 Width of base required Use 7.2 m

3340

=

7.2

x

157

= 2.96 say 3.0 m

x 3.0 m x 0.75 m deep base

7.1.2.2 Durability

For ground conditions otherthan non-aggressive soils, particular attention is needed to the provisions in ENV 206 and the National Foreword and Annex to that document for thecountry in which the concrete is required. In the UK it should be noted that theuse of ISO 9690 and ENV 206 may not comply with the current British Standard, BS 8110: Part 1: 1985 Table 6.1(2) where sulphates are present. Class 2(a) has been adopted for this design. Minimum concrete strength grade is C30/37. For cement content and wlc ratio refer to ENV 206 Table 3. Minimum coverto reinforcement is 30 mm.

For concrete cast against blinding layer, minimum cover > 40 mm. However, it is suggestedthat nominalcover> 40 mm is asufficientinterpretation of the above clause

Table 4.1 ENV 206 Table NA.1

Tab 6 4.1.33(9)

Use 75 mm nominal cover bottom and sides and 35 mm top 7.1.2.3 Materials

= 460 N/mm2

Type 2 deformed reinforcement with

NAD 6.3(a)

Concrete strength grade C30/37 with maximum aggregate size 20 mm 7.1.2.4 Loading Ultimate column loads

= l.35Gk + 150k

Column A: 1896 kN and Column B: 2834 kN Distance to centroid of loads from LH end

= 03 + 2834 x

4730

5.5

= 3.595m

i.e virtually at centre of 7.2 m long base Assume uniform net pressure =

4730 7.2

= 657 kN/m = 219 kNIm2

See Figures 72, 7.4 and 7.5 for loading, shear force and bending moment diagrams respectively.

Eqn 2.8(a) Table 2.2

1896 kN

2834 kN

kN/m

4723 kN/m

uJT316O

t tt f

'I

t

657 kN/m

49m

kO•6m

ft t ft 06m

+' 1•lm -I

Figure 7.3 Loading diagram

1717 kN

I

957 kN 289m

Figure 7.4 Shear force diagram

2167 kNm

600

2290

2610 I

Figure 7.5 Bending moment diagram

600! 1100_I I

I

7.12.5 Flexural design —

7.1.2.5.1 Longitudinal direction

top steel

Mid-span Msd

d

= 2167 kNm = 750 — 35

—

20

—

32/2

= 679 say 675 mm

Using the design tablesfor singly reinforced beams 2167 x 106 Msd

=

bd2f<

x

—

d

Af bdfck A

=

=

0.053

3000

x 6752 x 30

0.123

< 0.45 limit with zero redistribution

OK

2.5.3.4.2(5)

= 0.064

x 3000 x

= 0.064

675

x

-

460

=

8452 mm2

=

2818 mm2/m

Use 12T32 @ 250 mm crs. (3217 mm2lm) Continue bars to RH end of baseto act as hangers for links. Particular attention is drawn to the clauses for bar sizes larger than 32 mm. These clauses are restrictive about laps and anchorages, such that designers may need to resort to groups of smaller bars instead.

= 3h p 500 = 500 > 250 mm

Maximum spacing

7.1.2.5.2 Longitudinal direction

OK

NAD Table 3 5.4.3.2.1(4)

bottom steel

—

5.2.6.3P(1) & P(2)

At column face = 398 kNm Msd

d Msd

= 750

3000

=

0.012

=

0.012

bdfck AS

75

398

=

bd2ç

Af

—

— 10

x

=

106

x 6652 x 30

665 mm

=

0.010

30 x 3000 x 665 x —

For minimum steel Asmin

= 1561 mm2

= 520 mm2/m

460

= 0.0015bd = 998 mm2/m

Use 12T20 @ 250 mm crs. (1258 mm2/m)

5.4.2.1.1

cou.4oxIo4s 7.1.2.5.3 Transverse direction — bottom steel

=

Msd

1.5

—

0.45

2

219

x

=

178 kNm/m

Minimum steel governs. Use T20 @ 250 mm crs. (1258 mm2/m)

I

7.1.2.6 Shear Critical shear section at distance d from face of column

4.3.2.2(10)

Column B interior side VSd VRd1

TRd

k p,

x 657

=

1717 — 0.675

=

TRd k(1.2

=

0.34 N/mm2

=

1.6—d .z1.0 =

=

0.00476

=

1273 kN

+ 40p,)bd

4.3.23 Eqn 4.18 Table 4.8 1.0

Ensure bars are continued sufficiently. VRd1

VSd

=

957 kN

>

VAd1

Therefore shear reinforcement required. Shear capacity with links

=

VRd3

VCd

+

=

+

VRd1

VWd

4.3.2.43 Eqn 4.22

Therefore 1273 — 957 =

V

=

316 kN

A

— x O.9dfd

Eqn 4.23

S

= 400 N/mm2,

cWd

A s

d

316x103

0.9x675x400

=

675 mm

= 1.30 mm2/mm

Where shear reinforcement is required, the minimum amount is 100% of the EC2 Table 5.5 value.

NAD Table 3 5.4.3.3(2)

With

=

460,

wmIfl

= 0.0012 by interpolation

Table 5.5

For links

pW

=

ASWIsb

=

0.0012 x 3000 =

Eqn 5.16

W

Therefore

A

3.6

>

1.30 mm2/mm

mm

Therefore minimum links govern. Determine link spacing, using EC2 Eqn VRd2

Vsd/VRd2

=

ufcd

5.17—19.

Eqn 4.25

b(0.9 d)12

x

=

0.55

=

1273/10020

20

x 3000 x 0.9 x 675 x 10I2 = =

0.13

10020 kN

< 0.2

Use EC2 Eqn 5.17 to determine link spacing.

Sm

=

1

0.8d (Note: 300 mm limit in Eqn 5.17 does not apply to slabs)

5.4.33(4)

= 506 mm

O.75d

NAD 6.5(f)

Transverse spacing of legs across section

5.4.2.2(9)

dor800mm = 675mm Use 12 legs T10 @ 250 mm crs. in each direction where

A5 — S

=

12

x

78.5

=

3.77

250

>

V>

3.6 mm2/mm

Check diagonal crack control VCd VSd

VSd

= =

VAd1

5.4.2.2(10)

1273 kN (max.)

< 3Vc

Distancesto where

Xb

VSd

= 1502

=

—

657

=

4.4.2.3(5) VRd1

957

from face of columns A and B

= 0.830m

1.157 m

Check shear in areas where bottom steel is in tension and p1

OK

= 957 kN

No further check required.

xa

VRd1

= 0.0015 (mm. steel)

FOUNDATIONS

= 034(1.2 + 0.06)3000 x 665 x 10 = 854> 723 kN .. OK

VRd1

No links required at RH end of base

In orthogonal direction, shear at d from column face

= 219(0

—

0.45

—

0.6

x

2)

=

148 kN/m

2

From above

=

VAdi

= 3.0

284

>

148 kN/m

OK

No links required in orthogonal direction 7.1.2.7 Punching Length of one side of critical perimeter at 1.5d from face of column = 3 690 + 600 = 2670 mm

4.3.4.1P(4) & 4.3.4.2.2

x

= 3000 mm

This extends almost the full width of the base

Hence it is sufficient Just to check line shear as above and shear around perimeter of column face, where VsdIud

0.90

J7

= 0.90 x

= 4.9 N/mm2

NAD 6.4(d)

The shear stress at the column face perimeter with d = 675 mm is less than 4.9 N/mm2 in both cases (see Table 7.1) Table 7.1

OK

Punching shear at column face

column

Perimeter (mm)

Load (kN)

(N/mm2)

A

1650

1896

1.7

B

2400

2834

1.75

Stress

7.1.2.8 Crack control

Use method without direct calculation.

4.4.2.3

Estimateservice stress in reinforcement under quasi-permanent loads, using

the following approximation:

+ '2k

=

Gk

4.4.2.3(3)

+

The relevant loads are shown in Table 7.2. Table 7.2 Column loads for cracking check Load Gk .i-

03k(kN)

l.35Gk Ratio

+ 1•5k

(kN)

columnA

column B

934

1377

1896

2834

0.49

0.48

Estimated steel stress

A

—-

= 0.49 x f x

Aprov

= 0.49 x 400 x Either limit bar size using EC2 Table 4.11

8452 12

x 804

=

172 N/mm2

or bar spacing using Table 4.12.

> 32 mm used.

In Table 4.11 bar size

25 mm

In Table 4.12 spacing

285 mm in pure flexure

> 250 mm used.

. OK

Check minimum reinforcement requirement

A

4.4.2.3(2)

4.4.2.3(2) 4.4.2.2(3) Eqn 4.78

kckfcteffActhYs

it is considered conservative to use (h12)b.

For aS

= 100% x f = 460 N/mm2 yk

For eff use minimum tensile strength suggested in EC2, 3 N/mm2. =

0.4 for bending

For k interpolate a value for h

= 75 cm, which gives k = 0.53.

Therefore A5

0.4

12T32 gives A5

x 0.53 x 3 x >

750

x 3000/(2 x 460) =

1555 mm2

1555 mm2

OK

7.1.2.9 Detailing Check bar achorage detail

at LH end.

The anchorage should be capable of resisting a tensile force F5

=

5.4.2.1.4(2)

VSda,/d

with 5.43.2.1(1)

a1

F5

V

= =

VSd

column reaction =

1896 kN

The bond strength for poor conditions in the top of the pour

x Table 5.3 value = 0.7 x 3 = 2.1 N/mm2 = (4)(ycI'd) = 47'.6q5 = 1524 mm =

bd

0.7

Continuing all T32 bars to end

A prov = 9650 mm2

Asreq

=

VSd/fyd = 1896 x 10/400 = 4740 mm2

5.2.2.1 & 5.2.2.2

Eqn 53

FOUND&TIONS

Hence required anchorage, (+)1net at a direct support =

(--)lb

x 4740/9650 =

500 mm

> °31b

Figure 5.12 OK

of column = 600 — 75 = 525 mm ... OK Theanchorage may be increased to 'bnet' if preferred, by providing a bend at the end of the bar. Anchorage up to face

Therequirementfortransversereinforcementalongthe anchorage length does not apply at a direct support. Secondary reinforcement ratio for top steel p2

d

O.2p1

=

AS

750

—

=

0.2

35

—

x 0.00476

10

=

0.00095

= 705 mm

670 mm2/m

Use 116 @ 250 mm crs. (804 mm2/m) transversely in top

500 mm Spacing The reinforcement details are shown

OK

in Figure 7.6.

12132 — 250 12T20 — 250 T20 —250EW —

LL1

Hill 1—H.J 120—250

Figure 7.6 Detail of reinforcement in combined footing

72 Pilecap design 7.2.1 Pilecap design example using truss analogy

Afour-pile groupsupports a 500 mm squarecolumn which carries a factored load of 2800 kN. The piles are 450 mm in diameter and spaced at 1350 mm centres.

7.2.1.1 Pilecap size Assumea pilecap depth of 800 mm. Allowthe pilecap toextend 150 mm beyond the edge of the piles, to give a base 2.1 m square as shown in Figure 7.7. Use 2.1 m x 2.1 m x 0.8 m deep pilecap

5.2.3.3

5.43.2.1

375

1350

375

Figure 7.7 Pilecap layout 7.2.1.2 Durability

For components in non-aggressive soil and/or water, exposure class is 2(a). • .

.

Minimum concrete strength grade is C30137.

Table 4.1 ENV 206 Table NA.1

For cement content and w/c ratio refer to ENV 206 Table 3. Minimum cover to reinforcement is 30 mm.

NAD Table 6

Use 100 mm nominal bottom cover over piles and 50 mm sides 7.2.1.3 Materials

= 460 N/mm2 Type 2 deformed reinforcement with Concrete strength grade C30/37 with maximum aggregate size 20 mm.

NAD 6.3(a)

7.2.1.4 Element classification

A beam whose span is less than twice its overall depth is considered a deep

2.5.2.1(2)

beam. With the effective span, 1eff —

h

l,

taken to the centre of the piles:

2.5.2.2.2

1350 =—=1.7<2 800

Therefore treat as deep beam for analysis. 7.2.1.5 Loading Ultimate column load Pilecap (self-weight) Ultimate pilecap load

= 2800 kN

x 25 = 20 kN/m2 = 1.35 x 20 = 27 kN/m2

=

0.8

Eqn 2.8(a)

7.2.1.6 Design Deep beams under a concentrated load may be designed using a strut and

tie model.

2.5.3.72

cOU.DLQ&T(OKS

Use a model with a node at the centre of the loaded area and lower nodes over thecentrelines ofthe piles at the level ofthe tension reinforcementtogether with an effectivecolumn load to account forthe pilecap weight of, forexample: Nsd

d

=

2800

=

800

+ 1352 x 27 = —

100

—

25 =

BS 8110 3.11.4.1

2850 kN

675 mm

The total tensile force in each direction FSd

=

Nsd

X

1eff

2850

=

x 1350

4x

4d

=

1425 kN

675

For reinforcement

cd

A

= 460 = 400 N/mm2

=

2.23.2P(1) Table 23

1.15

=

1425

x iO

=

3563 mm2

400 There are no specific requirements within EC2 for the distribution of the calculated reinforcement.Theprovisionsof BS 8110: Part 1: Clause 3.11.4.2 are adopted in this example.

With piles spaced at 3timesthe diameter,the reinforcement may be uniformly distributed.

Use 8T25 at 275 mm crs. (3928 mm2) Maximum spacing = Minimum AS =

0.6bd

3h p 500 = 500

> 275 mm

z a0015 btd = 0.0015 x 2100 x 675 = 2127 mm2..

yk

The reinforcement details are shown in Figure

7.8.

4Ti6 I

800F I•

IJL

6125EW

Figure 7.8 Details of pilecap reinforcement

I

L4J

OK

OK

NAD Table 3 5.43.2.1(4) 5.4.2.1.1(1)

7.2.1.7 Shear

Only in elements such as slabs may shear reinforcement be omitted where

43.2.1P(2)

calculations justify. Despitethe classification for the pilecapgiven above, in line with commonUK practice, it is not intended to provide shear reinforcementwhen VSd VRd1. Takethe critical section for shear to be located at20% ofthe pile diameter into the piles, extending the full width of the pilecap.

43.2.2(2)

BS 8110 Figure 3.23

Distance from centre of loaded area

x

= 1350/2 — 0.3 x 450 = 540 mm

Shear resistance VRd1

=

+ 40p)bd

TAd k(1 .2

TAd

= 034 N/mm2

k

= 1.6—d #zl.0 = =

3928 = 2100 x 675

4.3.2.3 Eqn 4.18 Table 4.8 1.0

0.00277

All of tension steel is to continue sufficiently pastcritical section; check when detailing. VRd1

= 034(1.2 + 40 x 0.00277) 2100 x 675 x

iO = 632 kN

Consider enhanced resistance close to the supports

=

2.5d —

x

1.0

2.5

=

x 675 =

43.2.2(5)

3.125

540

43.2.2(9) OK

5.0

Shear force VSd

2850 =—=1425kN 2

<

=

3.125

x

= 1975 kN

632

No shear reinforcement required Having taken into account the increased shear strength close to the supports,

it is necessary to ensure that the reinforcement is properly anchored.

43.2.2(11)

In this caseall reinforcement will extend to centre line of pile and be anchored OK beyond that position 7.2.1.8 Punching Piles fall within 1.5d perimeter from column face, it is thus only necessary to check shear around column perimeter, where Stress

0.9

=

0.9

x

= 4.9 N/mm2

No enhancement of this value is permitted.

43.4.2.2(1)

NAD 6.4(d) 4.3.2.2(5)

x iO = 4 x 500 x 675 2800

Stress =

2.1

< 4.9 N/mm2

.

OK

7.2.1.9 Crack control Use method without direct calculation.

4.4.2.3

Estimateservice stress in reinforcement under quasi-permanent loads using following method

4.4.2.3(3)

Gk

+ '2°k

=

+

For this example the column loads, Gk

= 785 kN

1200 kN and

Hence the quasi-permanent load/factored load =

1200

+ 03 x 785 = 2800

0.51

Estimated steel stress

A

0.51

x fyd x — = Aprov

0.51

x 400 x

= 3928

185 N/mm2

Either limit bar size to EC2 Table 4.11 value or bar spacing to EC2 Table 4.12

4.4.23(2)

value.

25 mm = 25 mm used

From Table 4.11 bar size From Table 4.12 bar spacing

270 mm

<

OK

275 mm used

Check minimum reinforcement requirement

kC Idct,eff A/a Ct S

As For

4.4.2.3(2) 4.4.2.2(3)

A it is considered conservative to use = 100%

as

x fyk

Eqn4.78 (h/2)b.

= 460 N/mm2

use minimum tensile strength suggested by EC2, 3 N/mm2.

For

k

=

0.4 for bending

k

=

0.5forh

80cm

Therefore A5

1096 mm2

AS,prov =

3928mm2

OK

7.2.1.10 Detailing

The reinforcement corresponding to the ties in the model should be fully anchored beyond the nodes, i.e., past the centres of piles. b

5.4.5(1)

5.2.2.3(2)

— —

bd

For bars in bottom half of a pour, good bond may be assumed.

5.2.2.1

Hence

Ib

= 3.0 N/mm2

(4,

= 25 x 400

=

4x3

b,net

A

Areq > —

Table 5.3

32 mm) 834mm

b.min

prov

Using bobbed reinforcement, aa = 0.7

Ibnet =

0.7

3563 x 834 x — 3928

= 530 mm

Length beyond centre of pile allowing for end cover = 375 — 50 = 325 < 530 mm Bars cannot be anchored in manner shown in EC2 Figure 5.2. Use bent-up bars with large radius bend and anchorage length

x

—

=

756mm

prov

Diameter of bends can be obtained from NAD Table 8(1). Assumethat the limits in the table are equallyapplicableto bar centres given for minimum For 125 @ 275 mm crs., bend diameter = 134,,

cer

bend radius

= (13/2) x 25 = 165 mm

Theuse of NADTable8 is conservative,as it is based on full stress in thebars at the bend. The values given appear to be consistent with BS 8110: Part 1: = 30 N/mm2. Clause 3.12.8.25 using For concrete placed in the UK, it should be possible to demonstrate,compliance with EC2 Clause 5.2.1.2P(1) by usingthe BS 8110Clause above, with the result that smaller diameter bends may be used. For the edge bars, which have a minimum cover > 34, Table 8 gives 200 mm radius bend (see Figure 7.9).

= 75 mm, NAD

Therequirementfortransversereinforcementalong the anchorage length does not apply at a direct support. Provide bars to act as horizontal links, such as 4116 @ 150 mm crs.

NAD Table 8

FOUNDATIONS

3 785

Figure 7.9 Detail of bent-up bars

7.2.2 Pilecap design example using bending theory Take the pilecap from the preceding example but use bending theory to determine the bottom reinforcement. The shear force diagram is shown in Figure 7.10.

1425 kN

1425 kN 425

500

1350

Figure 7.10 Shear force diagram

7.2.2.1 Flexural reinforcement Msd

= 1425

z

= 0.975d = 658 mm

A6

= 2979 mm2

(0.425

+

= 784 kNm

.2)

Because of the difference in modelling, this is less reinforcement than the previous example.

7.2.2.2 Detailing At an end support, the anchorage of bottom reinforcementneeds to becapable

of resisting a force: F6

N

=

VSdal/d

+

5.4.2.1.4(2)

Eqn 5.15

Nsd

= 0 in this case

with 5.43.2.1(1)

a1

F6

A

= =

VSd

= 1425 kN

1425

x

10/400

= 3563 mm2

This is identical to the area of steel required in the previous example.

Use 8T25 as before (3928 mm2) Using the same detail of bobbed bars

net

= 530 mm

EC2 Figure 5.12(a) applies and is taken to require an anchorage length, = 353 mm past the line of contact between the beam and its (213)1flet support. Using a position 20% into the pile to represent the line of contact, the length available for anchorage = 0.3 x pile dia. + 375 — cover =

0.3

x 450 + 375

—

50

= 460 > 353 mm

....

OK

.

B SPECIAL DETAILS 8.1

Corbels 8.1.1 Introduction

2.5.3.7.2

Consider a corbel designed to carry a vertical ultimate designload of 400 kN with the line of action of the load 200 mm from the face of the support (wall, column etc), as shown in Figure 8.1.

Fv

200 300

ac b d

400 kN

=

hc

465 500

1¼

Figure 8.1 Corbel dimensions

8.1.2 Materials =

30 N/mm2 (concrete strength class C30/37)

=

460 N/mm2 (characteristic yield strength)

8.1.3 Design 8.1.3.1 Check overall depth

of corbel

the maximum shear in the corbel should not exceed VRd1. The butthis would give depth of the corbel could be reduced by putting VRd2 an increased tie force and consequent detailing problems. Thevalueof TAd in the expression for VRd1 (EC2 Eqn 4.18(1)) may be modified by the factor defined in EC2 Clause 43.2.2(9). in EC2 Eqn 4.17. Hence By inspection 6 will be a minimum when x = will also be a minimum. VHdl Conservatively,

F

2.53.7.2(5)

43.23

f

Now VRd1

=

[9rRdk(1.2

+ 4Op1) + 0.15a]bd

= 2.5d/x with 1.0 = 2.5 x 465 = 200 TAd

= 0.34 N/mm2

k

= 1.6—d

p1 is assumed

lzl

f3

5.81

5.0

Eqn 4.18 Eqn 4.17

5.0

Table 4.8

= 1.14m

to be 0.006 (4T16)

SPECIALDETAILS

No provision hasbeen made to limit horizontalforces at the support; therefore minimum horizontal force (He) acting at the bearing area should be assumed. This is given by

a

HC

=

= ±8OkN

0.2F

V

=

A/Sd —

2.5.3.7.2(4)

=

where

—80 kN

C

Therefore a

C

=

—80

465

x i0 x 300

=

—0.6 N/mm2

Hence VAd1

=

[5 x 0.34 x 1.23 (1.2 + 40

x 0.006) — 0.15 x 0.6] x 465 x 300 = 407 kN

Therefore VRd1

>

VSd

=

F

= 400 kN

OK

2.5.3.7.2(5)

8.1.3.2 Determine main reinforcement requirement

Now 0.4h

a, therefore a simple strut and tie model may be assumed, as shown in Figure 8.2.

Figure 8.2 Strut and tie model Under the vertical load

Fa = —- ;and z 0.85fck

FC

(0.8x)b cosf3

The determination of x will be an iterative procedure Choose x such that

=

ç

0.002 and

f

= 400 N/mm2

Therefore

x

=

0.0035 0.0035

+ 0.002

x 465 = 296 mm

2.5.3.7.2(1)

SC(&L DrMLs Now z =

d

= 347 mm

— 0.8x12

and

cosf3

=

0.5

Therefore, from above,

F

x 200

= 400

= 231 kNand

347

x 30 x 0.8 x 296 x 300 x

0.85

FC

0.52

= 301

kN

1.5

=

For equilibrium F

F and further refinement gives z =

x = 235 mm,

F

371 mm,

= 216 kN

In addition, EC2 Clause 2.5.3.7.2(4) requires a horizontal force of applied at the bearing area. 0.2FV =

HC

=

0.2

x 400

=

H to be

2.53.7.2(4)

80 kN

296kN

As.req = 296

x io

=

740mm2

460/1.15

Use 4T16 bars

8.1.3.3 Check crushing

of compression strut

This has been checked directlyby the calculation of F above. However, an indirect check may also be made VRd2

= [(j-) vf]b0.9d

p

= 0.7

—

I = —200

Eqn 4.19 0.55

Eqn 4.20

0.5

Therefore VRd2

= (j-) x 0.55 x 20 x 300

x 0.9 x 465

= 690 kN

Hence VRd2

> F, = 400 kN

OK

8.1.3.4 Check link reinforcement requirements

5.4.4(2)

Links are required if: 0.4 ACICd/cd

A

= 500

x 300

fcd

=

=

1.5

Eqn =

150

20 N/mm2

x iO mm2 f

= 1.15

= 400 N/mm2

5.21

5PC1M. DTA1LS

Hence, links are required if 0.4

A5

x

150

x

i0 x 20/400

= 3000 mm2

Now

Apr

= 804 < 3000 mm2

Therefore links are not required Nevertheless,in practice some linksshould be provided to assist in fixing the main reinforcement.

A5

O4Aspr

=

0.4

x 804

322 mm2

5.4.4(2)

Use 4T8 links (8 legs)

8.1.3.5 Check bearing area of corbel Allowable design ultimate bearing stress concrete. Therefore area required

°•8cd for bearing bedded in

= 400x103

Assume transverse bearing = Therefore length of bearing =

0.8

x 20

25000 mm 2

250mm 100mm

8.1.4 Detailing The reinforcement details are shown in Figure 8.3.

30 cover to main bars

4T16 welded to T20 cross—bar 2110

Figure 8.3 Corbel reinforcement details

L

EC2, Part lB

EC(M. DETMLS 8.1.4.1 Anchorage of main bars at front edge of corbel AnchorT16 ties by means of a cross bar running horizontally and welded to the ties. =

296kN

An allowable bearing stress under the cross bar can be obtained from EC2

5.4.81

Eqn 5.22 as Rdu

=

Eqn 5.22 (mod)

Note: Use ofthis stress requires that the concrete be confined by means of links etc. In areas where the cover is small, the designer may wish to use a modified version of Eqn 50 in BS 8110(2).

296x103 = 4485 mm 2

Therefore area of bar required =

3.3

x 20

For a T20 bar, length required is 225 mm. Use T20 cross bar 240 mm long weldedto T16 ties 8.1.4.2 Anchorage of main bars into support Required anchorage length 1 Lnet

= aaib

>

As.req

Eqn 5.4

bmrn

Aprov

Now 1b

fyd

= (I4) (cd bd

Eqn 5.3

= 400 N/mm2

Bond conditions may be considered good as the T16 bars will be anchored into a substantial support(column or wall). =

3 N/mm2

=

(16/4)

5.2.2.1(2)(b)

Table 5.3

x (400/3)

= 533 mm

Now

Asq

= 740 mm2

Asprov

=

804 mm2

Therefore 1

net

=

804

= Provide

1x533xZ9 °31b

=

490mm

10 or 100 mm

= 490 mm (see Figure 8.3)

aa

=

1

5.2.3.4.1(1)

SPECIALDETAILS

The detail at the front edge of the corbel is shown in Figure 8.4 100

I—

Figure 8.4

Detail

20 40

lI4

30

at front edge of corbel

The inside face of the T20 bar is positioned not less than the cover beyond the edge of the bearing area. This is an interpretation of BS 8110 as no guidance is given in EC2

8.2 Nibs 8.2.1 Introduction Considera nib designed to carry a precastconcretefloor slab imposing a vertical ultimate design load of 25 kNIm.

8.2.2 Materials

fck fyk

=

30 N/mm2 (concrete strength class C30137)

=

460 N/mm2 (characteristic yield strength)

8.2.3 Design Provide a 15 mm chamfer to the outside edge of the nib and assume the line of action of the load occurs at the upper edge of the chamfer Permissible design ultimate bearing stress Therefore minimum width of bearing =

= 0•6d for dry bearing 25 x 10 = 0.6

x 20 x 1000

Minimum width of bearing for non-isolated member Allowance for nib spalling

= 40 mm

Nominal bearing width =

+ 20

+ 25 =

52.4

85 mm

Allow an additional 25 mm for chamfer on supported member. Width of nib projection

= 85 + 25 =

ini

BS 8110

Table 5.1 BS 8110

25 mm 40

2.1 mm

5.2.3.2 BS 8110

= 20 mm

Allowance for inaccuracies =

EC2 Part

110 mm

lB

SEC%M. QETMLS

The distance of the line of action of the load from the face of the beam = 110—15 = 95mm Assuming 20 mm coverto the 110 links in the beam

a

= 95 + 20 + 5 = 120 mm

Check minimum depth of nib.

6.

Assuming T8 bars, minimum internal diameter of loop is Therefore minimum depth of nib = 20 + 8 8 + 20 =

x

= 105 mm

Depth of nib

M

= 25 x

Effective depth (d)

M —

=

bd2ck

Afsyk

104 mm

NAD Table 8

= 3 kNm/m

0.12

= 105 — 20

—

3x106 1000 X 812

4 = 81 mm

=

x 30

0.015

=

0.018 (Section 13, Table 13.1)

=

O.OlBbdfCk

bdfCk

A$

fyk

0.018

=

x 1000 x 81 x 30 460

= 95mm2

Check minimum area of reinforcement A5

=

=

0.6

0.6

(r')

Eqn 5.14

0.0015bd

x 1000 x 81

=

460 0.0015

5.4.2.1.1(1)

x 1000 x 81

106mm2

= 122 mm2

Check minimum area of reinforcement for crack control =

4.4.2.2

Eqn 4.78

kckfcteffAciOs

k

=

k

= 0.8forh

Ct5ff

= 3.0 N/mm2

A

= bh = 1000

cr

= 460 N/mm2

0.4 for bending

2

300 mm

x 105 2

= 52.5

x i0 mm2

Therefore

A

= 0.4

x 0.8 x 3.0 x 52.5 x 10/460

=

110 mm2

bPEt'kM.DETAiLS

No further checkfor crack control is necessary as h

= 3h = 315

Maximum bar spacing

=

105

200 mm.

500 mm

4.4.23.(1)

NAD Table 3

Use T8 @ 300 mm crs. (168 mm2/m)

5.43.2.1(4)

The reinforcement details are shown in Figure &5.

2T8 (acer bars

to anchor U—bars

Figure 8.5 Nib reinforcement details Check shear in nib, takingintoaccountthe proximity of the concentrated load to the support. VAd1

=

[!3rAdk(1.2

=

2.5d/x

+

___

= 2.5 x

81

=

120

=

034 N/mm2

k

=

1.6—dlzl

p

=

A —-

TAd

+ o.i&i] bd

'40p1)

=

bd

=

1.52

168

=

1000

x 81

4.3.23

Eqn 4.18

1.69

Eqn 4.17

0.0021

N C

A

Therefore VAd1

= 1.69 x 034 x 1.52(1.2 +40 x 0.0021)x 1000 x 81 = 90.8 kN/m

VSd

= 25 kN/m

Therefore VRdl

>

OK

1"Sd

Check anchorage of T8 bars.

net = aa 1b

(j

1tmin

Eqn 5.4

SPECIAL DETAILS

Now

= (I'I4) (d'bd) = 400 N/mm2

Eqn 53

Bond conditions may be considered good as the bars are anchored at least 300 mm from the top of the member. =

3 N/mm2

Table 5.3

= (8/4) x (400/3) = 267 mm NowA5req = 122 mm2

Apr

= 168 mm2

aa

=

1

Therefore

=

1

b.net

122 1x267x—

= 194mm

168

4l

mm

For bars in tension

=

1bmin

10 or 100 mm

Therefore =

1b.min

1b.net

=

100 mm 194 mm (see Figure 8.5)

For bars in compression = 1b.min °•61b 104; or 100 mm Therefore 1mmn

=

160 mm

= 160 mm (see Figure 8.5)

8.3 Simply supported ends 8.3.1 Directly supported ends Reinforcement anchorage requirements are shown

f

Iw(b)

tb,net

(a) Straight bar

5.2.2.1(2)(b)

Ib)

Figure 8.6 Anchorage at a direct support

in Figure 8.6.

'2

Hook

I

"I

tb,net

SPECM DETMLS

Figure 8.6(a) shows anchorage of tensile reinforcement beingachieved using

a straight bar. It should be noted that EC2 does not permit straight anchorage or bends for smooth bars of morethan 8 mm diameter.

5.4.2.1.4(3) Figure 5.12(a)

5.23.2(2)

Note:

TheCEB—FIP Model Code16 gives a factor of 1.0 for net as opposedto 2/3 in EC2. Designers may wish to consider using the higther value. Typical values for anchorage length and support width, w, can be obtained for (a) and (b) in Figure 8.6. Assume

I

ck

=

30 N/mm2,

=

1.0

f yk

=

460 N/mm2

Note:

NAD 6.5

Areq maybe taken as one quarter of the reinforcement required at mid-span but not less than that required to resistthe tensile force givenby EC2 Eqn 5.15.

5.4.2.1.4(3) 5.4.2.1.4(2)

'bnet

=

()

0alb

Eqn 5.4

iz lmIfl

s.prov

=

(I4 (ci1bd)

Eqn 53

=

3 N/mm2

Table 53

=

400 N/mm2

=

(cb/4)

Therefore 1b

4min

=

x (400/3)

°31b

aa

=

1

aa

=

0.7

= 3334

0 or 100 mm

lz 1

for straight bars; or for curved bars with 3c/ transversecover

Therefore

x 333

1b,net (a)=

1

net (b)=

0.7

x

=

33

3334.

=

23.3q5

Therefore width of support required in Figure 8.6(a), assuming 20 mm cover and 15 mm chamfer

w (a)

=

(j-) x

333çb

+ 35

=

22.2 + 35 mm

and width of support required in Figure 8.6(b), assuming as above

w (b)

=

Eqn 5.5

(+) x

23.3q

+ 35 = 15.5 + 35 mm

5.2.3.4.1(1)

SPECIAL DETAILS

The minimum supportwidth is given by W

= (+) x

10 + 35 = 6.7'qS + 35 mm

where, in Figure 8.6(a), A8req arid,

in Figure

8.6(b), As req

O•3As,pr,

0.43A

= 0.7 can only be used if the concrete cover As noted above, to the This is clearly difficult perpendicular plane of curvature is at least to achieve in beams without end diaphrams for bar sizes in excess of 12 mm.

a

3

The requirements for the various types of hooks, loops and bends are given in EC2 Figure 5.2. The minimum diameters of mandrels are given in NAD' Table a The required support widths are given in Table 8.1. Table 8.i Width of support (mm)

A A

10

12

16

20

25

w(a) wmn

257 102

302

391

479

590

746

116

143

169

203

250

wb)

190

221

283

345

102

116

143

169

423 203

531

wmm

(mm)

1.0

0.3 1.0

0.43

32

250

8.3.2 Indirectly supported ends Reinforcement anchorage requirements are shown

in Figure a7.

{I tb,net

,net

L

J b(b)

I

(b) Hook

(a) Straight bar Figure 8.7 Anchorage at an indirect support

As in Section 83.1 above, anchorage lengths and support widths can be obtained for both straight bars and hooked bars.

Theanchorage lengths are as Section 8.31 but the required supportwidths are increased.

Assuming 20 mm cover

= (333cb + b(b) = (23.3q + b(a)

x 1.5 = 20) x 1.5 = 20)

50q5

+ 30 mm

354

+ 30 mm

SPECL DETMLS The minimum supportbeam width is given by = (lOqS + 20) x 1.5 = lScb + 30 mm bmin where the same conditions apply as in Section 83.1.

3

In these cases, as the beam is indirectly supported, i.e., by another beam, cover perpendicular to the plane of the curvature can be achieved moreeasily = 0.7 can be readily used in EC2 Eqn 5.4. and The required supportbeam widths are given in Table 8.2. Table 8.2 Width of support beam (mm)

'

,

AspraY

mm'/

10

12

16

20

25

32

1.0

b(a)

530 180

630 210

830 270

1030 330

1280 405

1630 510

380 180

450 210

590 270

730 330

905 405

1150 510

0.3

b

mm

1.0

b(b)

0.43

bmm

8.4 Surface reinforcement

5.4.2.4

In certain circumstancesitmaybe necessaryto provide surface reinforcement located outside the links. Surface reinforcementis provided to resistspalling fromfire and where bundled bars or bar sizes greaterthan 32 mm are used.

5.4.2.4(3)

EC2 also refers to the use of skin reinforcement located inside the links. Skin reinforcement is provided to control cracking in the side facesof beams 1 m or more in depth.

4.4.2.3(4)

8.4.1 Design data A beam section requiring surface reinforcement is shown in Figure 8.8.

ioooL

fll 400

I

A5t=4825mm2 (6132) to Links 50mm

Cover

Figure 8.8 Beam section showing main reinforcement

=1

SPECIAL DETAILS

8.4.2 Area of longitudinal surface reinforcement

Af

=

0.01A

5.4.2.4(5)

From EC2 Figure 5.15

= 2x50x(1000—360)+50 x300 = 79x 103mm2 Therefore

A ssuri = 0.01 Length of

x

79

x iO

=

790 mm2

= 490

internal perimeter

x 2 + 300

= 1280 mm

Hence

AS,SU1/m=

790 = 1.280

617 mm2/m

Use B785 fabric This comprises 10 mm wires @ 100 mm crs. horizontally and 8 mm wires @ 200 mm crs. vertically. Note: EC2 does not directly coverthe use of plain wire fabric.

5.23.43

Surface reinforcementmayalso be used as longitudinal bending reinforcement in the horizontal direction and as shear reinforcement in the vertical direction in some cases.

5.4.2.4(6)

If surface reinforcement is being used to resist shear, EC2 Clause 5.4.2.2(4) should be noted. Itstatesthat a minimum of50% of shear reinforcementshould be in the form of links.

5.4.2.2(4)

The reinforcement detail is shown in Figure 8.9.

r

—

1

—

1

Ifl S 8-132

—

B785 surface reinforcement

5L150

n -l r —.

_'

f

d-x=54O

I

i(

.1.

I.

-'1.-I

Figure 8.9 Beam section showing surface reinforcement.

<600

9 PRESTRESSED CONCRETE 9.1

Introduction Design of a prestressed band beam with bonded post-tensionedtendons, to support a ribbed floor slab, is set out. Thisexample is similar to Examp'e 2 in theConcrete Society's Post tensioned concrete floors: Design handbook.

9.2 Design data Thefloor plan and section for the structure are shown in Figure 9.1. The band beams run alongthe column lines in the longitudinal direction. Thefloor slab contains unbonded tendons, and is not designed here.

7200

j

7200

7200

7200

PLAN B

3900

3900

A—A Figure 9.1 Floor plan and section

=

PRSTRESSEO CONCRETE

9.2.1 Beam section Theeffectiveflangewidth of the beam for calculation of stiffnessesor stresses is taken as

bff

=

b + (*)

10

=

2508 mm


2.5.2.2.1P(2) Eqn 2.13

The beam section is shown in Figure 9.2. 2508 •1 110

60

I 750

750

Figure 9.2 Beam section

9.2.2 Durability For components in dry environment, exposure class is 1.

Table 4.1

Minimum concretestrength grade for post-tensioned members is C25/30.

4.2.a5.2

Minimum coverto—reinforcement is 15 mm.

4.1.3.3

NAD Table 6

Use 25 mm nominal cover to reinforcement Minimum cover to duct is given asthesmaller cross-sectionaldimension of the duct but not less than half the greatercross-sectionaldimension. Use nominal cover to duct

z 50 mm

4.133(11)

BS 8110 4.12.3.2

9.2.3 Materials 9.2.3.1 Reinforcement Type 2 deformed bars, characteristicstrength, ck ductility

= 460 N/mm2 having high

NAD 63(a)

9.2.3.2 Prestressing steel 15.7 mm diameter superstrand, grouped in oval ducts 20

Characteristic strength,

A E8

=

x 75 mm

1770 N/mm2

= 150 mm2 = 190 kN/mm2

9.2.3.3 Concrete

In order that this example can be compared with that given in Example 2 of the Posttensioned concrete floors: Design handbook, a non-standardconcrete strength grade has been chosen of C32/40.

BS 5896

3.3.4.4

psEssc. CGIICRrE

Ecm E1

=

32 N/mm2

=

20 N/mm2 strength at transfer

=

9.5

x (32 + 8)' =

32.4 say 32 kNImm2

=

9.5

x (20 + 8)T =

28.8 say 29 kN/mm2

3.1.2.5.2(3)

9.2.4 Loading

4.2.3.5.4P(2)

Imposed loading

=

5.0 kN/m2

Effective width of slab =

10.22 m 35.60 + 12.60

Self-weight of slab and beam =

= 48.20 kN/m

9.3 Serviceability limit state 9.3.1 Tendon details The tendon profile is shown in Figure 9.3.

4 i4 7200

14—I

7200

Figure 9.3 Tendon profile Initial prestressing force is taken as P0

= 0.75 x

xA

=

199.1

kN per tendon

Losses are assumed to be

15% of P0 at transfer 30% of P0 at service Hence prestressing force At transfer, mQ = 169.3 kN per tendon

At service,

m'

=

139.4

kN per tendon

9.3.2 Maximum drape Span

y

1—2

= kx(s — x) where s is the distance between inflexion points

InI

4.23.5.4P(3)

PRESTRESSED CONCRETE

Using AppendixC of Post tensioned concrete floors: Design handbook

=

k

1.70

x iO- and s =

At mid-span y = 1.70 x

i0 x

28802

5760mm

=

141

mm

Span 2—3 at mid-span

= 350

y

—

62.5

—

45

= 180 mm

62.5

—

9.3.3 Prestress required Take an equivalent balanced load equal to the self-weight. Span

1—2

req

=

x 57602

48.2

8x

141

< 1000

= 1418 kN

Span 2—3 req

=

x 57602 = 8x180x1000 48.2

Number of tendons required

=

1111

kN

1418/139.4

10.17

Use 11 tendons throughout beam

9.3.4 Equivalent loads from prestress The equivalent loads from thelongitudinal tendons, given by q = 8(nPa)aIs2 where n = 11, are calculated in Tables 9.1 and 9.2. Table 9.1 Calculation of equivalent loads fromlongitudinal tendons attransfer, for the full slab width Span

nP (kN)

2

1

1861.6

1861.6

1861 6

2

3

1861.6

1861.6

1861.6

a (mm)

30

—141

40

45

—180

45

s(mm)

1440

5760

1440

1440

5760

1440

q (kN/m)

216.9

289.4

—633

3232

—80.8

3232

Table 9.2 Calculation of equivalent loadsfrom longitudinal tendons after all losses, for the full slab width Span

nP (kN) a (mm) s(mm)

q (kN/m)

2

1

15334

1533.4

15334

30

—141

40

1440

5760

1440

1787

23&4

—521

InI

2

3

15334

45__1440

266.2

1533.4

15334

—180

45

5760

1440

—66.6

266.2

PRE5ThESSEDCOt4CETE

9.3.5 Load cases For continuous beams, the following arrangements of imposed loads should

be considered:

(a) alternate spans loaded; (b)

2.5.1.2(4)

-

any two adjacent spans loaded.

The rare, frequent and quasi-permanent load combinations should be considered where the values of and '2 are taken from NAD Table 1(1) For = 0.6 and imposed loads in offices 2 = 0.3. Rare combination, Gk + P + +P+ Frequent combination,

0•6k

Quasi-permanent combination, Gk

+

k

'

Eqn 2.9(b)

+

Eqn 2.9(c)

As the beam is a T-section, the values of W and Wb are not equal. By calculation it can be shown that = 6.79 x 1O mm4 and that the centroid of the section is at a height of 196 mm from the soffit. A = 635.9 x iO mm2

I

Therefore

Wb

106 mm3 106 mm3

Thecalculation ofthestressesunder eachtoad combination is not shown here. The method follows that given in the Post tensioned concretefloors: Design handbook. Thetop and bottom concrete stressfor transferconditions are given in Table 93 and those after all losses are given in Table 9.4. Table 9.3 Stresses at transfer Zone

1

(support) 1—2

(span)

2 (support)

2 (support) 2—3

(span)

3 (support)

NAD Table 1 Eqn 2.9(a)

9.3.6 Maximum concrete stresses

= 44.1 x = 34.6 x

2.3.4P(2)

Top stresses, (N/mm2)

I

Bottom stresses,

1b

(N/mm2)

max.

mm.

max.

mm.

42

—

—

1.83

15

2.21

40

2.20

&81

—

—

134

4.10

—

—

0.97

3.07

1.23

4.66

2.29

4.45

—

—

0.53

'f

PRESTRESSED CONCRETE

Table 9.4 Stresses after all losses Rare loading zone

Top stresses, (N/mm2)

max.

(N/mm2)

mm.

1

—2.53

(support) 1—2

Bottom stresses, fb

f

max. &75

1.91

305

—4.19

108

—331

9.82

093

4.31

—2.95

9.30

5.64

mm.

—1.73

(span)

2 (support)

2 (support) 2—3

(span)

4.68

3 (support)

—050

Frequent loading Zone

Bottom stresses, b

(N/mm2)

(N/mm2)

max 1

(support) 1—2

(span)

f

Top stresses, 1 mm.

max.

—087

6.62

2.19

269

437

2

—1.96

a02

—1.17

700

(support)

2 (support) 2—3

(span)

344

3 (support)

1.19

398

—0.77

648

mm

—010

1.09

Quasi-permanentloading Zone

lopstresses, 1

Bottom stresses, fb

(N/mm2)

(N/mm2)

max. 1

(support) 1—2

(span)

341

2 (support) 2 2—3

3 (support)

2.81

max.

038

5.02

2.19

269

—029

5.87

0.48

(support) (span)

mm

mm.

1.13

4.89

138

373

—0.87

438

1.89

PESTESSED COCRErE

9.3.7 Allowable compressive stresses

4.4.1

To prevent longitudinal cracks the compressive stress under rare load combinations should not exceed °•6Ck

= 0.6 x 32 =

19.2 N/mm2

The maximum stress from Table 9.4 is 10.8 N/mm2

4.4.1.1(2)

OK

To control creep thecompressivestress under quasi-permanentloading should not exceed

= 0.45 x 32 =

14.4 N/mm2

The maximum stress from Table 9.4 is 5.87 N/mm2

9.3.8 Limit state

4.4.1.1(3)

OK

of cracking

No check is required at transfer since beam is totally in compression. Design crack width for post-tensioned member under frequent load combinations Wk

= 0.2 mm

The method adopted to determine the minimum reinforcement required is to carry out arigorous calculationofthe crackwidth where theflexuraltensilestress under rare loads exceeds 3 N/mm2. If the calculated crack width under frequent loads does not exceed 0.2 mm then furtherbonded reinforcementis not required.

4.4.2.1 Table 4.10 4.4.1.2.(5) 4.4.1.2(7) 4.4.2.4

From Table 9.4 the stress at support 2 under the rare load combination is —4.19 N/mm2 and hence a more detailed calculation is required. As this example is a beam, at leasttwo longitudinal bars at the top and bottom are required to hold the links in place. For this analysis include 2116s in the top of the beam. Wk /3

= =

/3Srmsm

Eqn 4.80

1.7 for load induced cracking

4.4.2.4(2)

The value of Srm can be conservatively calculated as S

= h

—

x

4.4.2.4(8)

The value of sm can be conservatively calculated as

= ._!

Eqn 4.81

The values of a9 and x, the neutral axis depth, for this example were

determined from computer analysis assuming linear stress/strainrelationships and no tension from the concrete. Applied moment

x

=

—377.6 kNm (frequent load case)

=

213mm

=

—95.8 N/mm2

PRESTRESSED CONCRETE

9.3.9 Calculation Pm,t =

P0

=

P

P 9.3.9.1 9.3.9.1.1

95.8

x (350 — 213) x

Hence wk = 1.7

200x103

=

0.11

< 0.2 mm

.

.

OK

of prestress losses per tendon

-

C

- iP (x) -

SI

C

- XP (x) -

SI

—

- P(t)

2.5.4.2 Eqn 2.19 4.2.3.5.4 Eqn 4.8

Short term losses

4.2.3.5.5

Loss due to elastic deformation

(5—8) 4.2.3.5.5(6)

Modular ratio

E5

=

190

=

E.CI

=

6.55

29

Maximum stress in concreteadjacent to tendons at transfer occurs at middle

of span 2—3. From Table 9.3, stress at level of tendons =

4.66

—

(4.66

350

—

= 4.05 N/mm2

1.23)

Average loss of force due to elastic deformation of concrete C

=

0.5

x 4.05 x

6.55

x

150

x

iO

=

1.99 kN

Theloss, which hasbeen conservativelybased on the maximum concrete stress rather than the stress averaged along the length of the tendon, is only 1% of the jacking force and will be neglected. 9.3.9.1.2

Loss due to friction

P(x)

4.2.3.5.5(8)

=

P

=

jacking force

=

0.19 (recommended

k

=

0.0085 (from Posttensioned concrete floors: Design handbook)

0

=

x

=

7.2 m for each span

=

4x30

P0

Span 1—2

0

(1 —

8a

e lL(O+kx)) =

Eqn 4.9 199.1

kN

for strand)

for each span

+

1440

5760

+

4x40 =0.392

199.1(1

=

199.1 — 16.4

Therefore P2

1440

e°19°392 + 0.0085 x 7.2)) =

=

—

8x141

= 182.7 kN

16.4 kN

STSSED C0t4CRErE Span 2—3

8

8x45

=

8x180

+

=

0.500

5760

1440

e°19892 + 00085 x 144)) =

=

199.1(1 —

=

199.1 — 34.9

34.9 kN

Therefore P3

93.9.1.3 Loss due to wedge set

= 164.2 kN

(/, = 6 mm)

= 2p'l' where

199.1 — 164.2

=

p'

I

E isis

=

ii

= 242 kN/m

14.4

Ap

=

J

p'

J

0.006 x 190 x 150 ______________ = 2.4

4m

= 2 x 2.42 x 8.4 = 40.7 kN

si

The resulting force profile is shown in Figure 9.4.

8400

I

158 4 1312

2

Jacking

199

Trn inaI

47'2

—

——

C

ALL forces

in

I352 kN

Figure 9.4 Force profiles 9.3.9.1.4 Percentage losses at transfer At

199.1 — 158.4 1:

x

100

=

20.4%

x

100

=

11.7%

x

100

=

199.1

At2:

199.1

— 175.8 199.1

At3:

199.1

—

164.2

17.5%

199.1

Average loss

=

16.5% (15% assumed)

OK

PRESTRESSED CONCRETE

9.3.9.2 Long term losses 93.9.2.1 Creep and shrinkage data

al.2.5.5

Notional size of section from Figure 9.2,

2A —

2(2508

=

u

x 110 +

x 240) =

1500

+ 240)

2(2508

230mm

For inside conditions and transfer at 7 days,

(oo,t0) e(oo)

=

Table 33

3.0

= 0.00058

Table

9.3.9.2.2 Relaxation data Long term class 2 relaxation loss for initial stress of 0•67pk immediately after transfer

= 9.3.9.2.3

x 0.02a0

1.5

=

4

4.23.4.1(2) Table 4.8 NAD Table 3

O.O3a

Loss due to creep, shrinkage and relaxation "rc+8+r

+ a(t,t0)(o +

ç(t,t0) E8 ÷

= 1

+c

Az2 1 +

A

—-

CP

[1

+ 0.8(t,t0)1

A0

a

At1:z0

E

190

Ecm

32

= 0

Therefore

= 0

0cg

CO

=

11

x 158.4 x iO 635880

= 287.5 —

At 2:

wC 1

cg

196

=

= 6.79 X

=

10

= 2.74 N/mm2 91.5 mm

=

91.5

+ocpa =

74.2

x

106 mm3

11

x

175.8

M P —+—

w0

A

Using moment and force at transfer

a-I-a cg cpo

=

(6.17

+ 3038) x 74.2

=

0.49

x 106

+ 3.04 =

106

+

635880 3.53 N/mm2

x

4.2.3.5.5(9) Eqn 410

=

91.5mm

W,

=

74.2x

0+0 cg CO

=

49.36

At3:z

x

106

+ 2.84

0.67

11

+

x 106

74.2

=

mm3

106

x

164.2

x iO

635880

= 3.51 N/mm2

Losses of prestress 0.00058 x 190 x iO + 31.7 + 5.94 x 3 x 2.74 ____________________________ 1+ 5.94 x 1650 1 + 635880 x 02 (1 +0.8x3) 635880 6.79 x iO

Atl:Aa p.c+s+r =

1

IP(t)

At2:a

+ 31.7 + 48.8 x 10-2(1 + 9.36 x 1O x 02) 110.2

—

+ 5.24

=

190.7 = 1.052

=

181.2

181.2 N/mm2

x 150 x iO

= 27.2 kN

+ 5.94 x 3 x 3.53 x 10_2 (1 + 9.36 x x 91.52)

110.2 + 35.2

= 1

+ 5.24

= 20&3 =

i0

190.6 N/mm2

1.093

1O

= 190.6 x 150 x

P(t)

At3:ap,c+s+r = =

110.2 + 32.8

= 28.6 kN

+ 5.94 x 3

x 3.81

1.093 210.9

= 193.0 N/mm2

1.093

iO

= 193.0 x 150 x

P(t)

= 29.0 kN

Final forces at service (see Figure 9.4) = 15a4 — 27.2 = 131.2 kN At 1:

= 175.8 —

At 2: P

At 3: P

164.2

—

28.6

= 147.2 kN

29.0

=

135.2 kN

9.3.9.2.4 Percentage losses at service At 1:

199.1

—

131.2

x

100

=

34.1%

x

100

=

26.1%

199.1

At2:

199.1

—

147.2

199.1

PRESTRESSED CONCRETE

199.1

At 3:

—

135.2

199.1

Average loss =

9.4 Ultimate

x 100

= 32.1%

30.8% (30% assumed)

OK

limit state of applied moments

9.4.1 Calculation

4.3.1.1 P(2),

P(4) & P(6)

Partial safety factors

= Load cases

2.33.1

=

1.35, —

=

1.5,

Table 2.2

1.0

2.5.4.4.1(2)

as for serviceability

2.5.1.2

of resistance moments

4.3.1.2P(1)

9.4.2 Calculation

The section may be analysed as shown in Figure 9.5. Rectangular stress block for concrete in compression with

a

=

0.85,

fCd =

-

4.3.1.2(4) 4.2.1.3.3(12) Figure 4.4

21.3 N/mm2

1.5

F5

pd A —O•4

xl

d

I

08x1 cd

Figure 9.5 Analysis of section at ultimate limit state Horizontal top branch to stress-strain curve for prestressing steel with

fpd = og

-

0.9

=

x

1770

=

1385 N/mm2

1.15

For stress to reach maximum design value 1385

Minimum strain, E3

190x103

=1

=

0.0073

4.2.3.3.3 Figure 4.6

SSE

COI4CRETE

139.4

Prestrain

pm

150

AE

x 190

=

0.0049

= 0.0073 — 0.0049 = 0.0024

Increment,

Maximum neutral axis/effectivedepth ratio 0.0035 0.0035 + 0.0024

d

For values of x

FS = MRd

=

=

0.593

0.593d

fpdA

= 1385 x

F(d —

0.4x)

p

x 150 x

11

i0

=

2285 kN

= 2.285(d — 0.4x) kNm

where =

F

=

x=

afCdb(O.8x

= 0.85 x 213 x 0.8bx =

14.Sbx

F5 gives

x iO =

2285

157600

b

14.5b

mm

At support 1: b = 1500 mm, d = 196 mm 157600 = 105mm x = 1500

d

=

=

0.536

<

0.593

MRd =

— 0.4 2.285(196

At supports 2 and 3: b

=

x 105)

1500mm,

— MRd= 2.285(287.5 0.4 x In spans:

x

OK

196

b =

=

= MAd

2508 mm,

157600

=

2508

d =

63 mm

2.285(287.5 — 0.4

9.4.3 Comparison

105)

=

351.9 kNm

d =

287.5

mm

= 561.0 kNm

287.5 mm


x 63) =

110 mm

OK

599.4 kNm

of moments

Thecalculation of the moments due to the applied loads (y = 135, vo = 1.5) is notshown here.These moments are combined withthe secondary moments = 1.0) and compared with theresistance moments at due to prestressing each position. The results are summarized in Table 9.5.

(;

PRESTRESSED CONCRETE

Table 9.5 Moments at ultimate limit state Zone

Secondary moments (kNm)

Moments from ultimate loads (kNm)

Applied moments (kNm)

Resistance moments, MRd (kNm)

122.0

—461.1

—339.1

—351.9

350.8

434.4

599.4

45.1

—673.9

—628.8

—561.0

62.6

—628.1

—565.5

—561.0

1

(support) 1—2

836

(span)

2 (support) 2

(support) 2—3 (span) 3

(support)

67.4

309.4

376.8

599.4

72.2

—604.4

—532.2

—561.0

The resistancemoment is inadequate at support 2 and additionalreinforcement is required. Since

M

= F(d — 0.4x) = 14.5bx(d —

x2_2.5dx+2.5M

=

0.4x),

0

14.5b

Hence

x

= 1.25

M 1_Il— 9.06bd

d

4

= 1.25

x = d F

121

—

1

=

287.5

=

j

i

x 106 x 287.52

628.8 9.06 x 1500

—

0.421

=

< 0.593

M

628.8

d — 0.4x

287.5

287.5

—

121

mm

OK

x iO 0.4

x 121

= 2630 kN

Additional area of reinforcement required

AS

=

F

—

A.

= (2630

2116 and 2T20 gives 1030

—

2285)10 =

400

>

863 mm2

863 mm2

OK

Use 2116 top and bottom throughout beam with additional 2120 top at support2

SThESSD cot4cRErE 9.5 Minimum and maximum areas

of reinforcement

5.4.2.1.1

Although it is not clear whatshould be assumed from EC21, the total area of steel has been taken as the sum of the untensioned and tensioned steel.

As+p = AS -i-Ap =

(2

x 201) + (11 x 150) =

2052 mm2

9.5.1 Minimum Minimum area of total tension reinforcement 0.6bd

1

Eqn 5.14 0.0015bd

At support, b = 2508 mm Minimum area

=

0.00.15

0.6

x 2508 x

290

= 948 mm2

460

x 2508 x 290 =

Area provided = 2052

>

1090 mm2

1090 mm2

OK

9.5.2 Maximum Maximum area of total tension and compression reinforcement

=

0.04A

=

0.04

x 635880 =

25435

>

2052 mm2

....

OK

9.6 Reinforcement summary tendons throughout beam 2116stop and bottom throughout beam.Additional 2120stop at support 2 11

These areas are within maximum and minimum limita

j

SERV(CEBIL1TY CHECKS BY CALCULATION

U

10.1

Deflection Calculatethe longtermdeflectionor a 7.0 m span simply supported beam whose section is shown in Figure 10.1. The beam supports the interior floor spans of an office building.

d': 50

1650 I'S

I

I

___ _1_

Figure 10.1 Beam section Deflections will be calculated using the rigorous and simplified methods given in EC21, together with an alternative simplified method. The results will then be compared with the limiting span/effective depth ratios given in EC2.

10.1.1 Design data Span

=

A'9 A9

7.0 m

= 19.7 kN/m = 19.5 kNIm = 402 mm2 = 2410 mm2 = 30 N/mm2 (concrete strength class C30/37)

a-i.2.4 Table 3.-i

10.t2 Calculation method The requirements for the calculation of deflections are given in Section 4.4.3 and Appendix4 of EC2. Two limiting conditions are assumed to exist for the deformation of concrete sections (1)

A4.3(1)

Uncracked

(2) Cracked. Members which are not expected to be loaded above the level which would causethe tensile strength of the concrete to be exceeded, anywhere in the membe will be considered to be uncracked. Members which are expected to crackwill behave in a manner intermediatebetween the uncracked and fully cracked conditions. For memberssubjecteddominantlytoflexure, the Codegives ageneral equation for obtaining the intermediate value of any parameter between the limiting conditions

InI

A4.3(2)

SCE&BUTY CIBCKS BY CALCULATION =

a

+

(1

—

a1

where

A43(2) Eqn A.4.1

a is the parameter being considered

a

are the values of the parameter calculated for the uncracked and fully cracked conditions respectively a1 and

is a distribution coeffient given by

=

1

—

,a2

a

2

Theeffectsof creep are catered forby the useof an effective modulus ofelasticity for the concretegiven by Ecm

E

=

1

+

A43(2) Eqn A.4.2

A43.(2) Eqn A.43

Bond and deterioration of bond undersustained or repeated loading is taken account of by coefficients and in Eqn A.4.2 Curvatures due to shrinkage may be assessed from

r 1 —

=

caS 0 CS

I

A4.3(2)

Eqn A.4.4

Shrinkage curvatures should becalculated forthe uncracked and fully cracked conditions and the final curvature assessed by use of Eqn A.4.1.

In accordance with theCode, the rigorous method of assessing deflections is to calculatethe curvaturesatfrequent sectionsalong the member and calculate the deflections by numerical integration. Thesimplified approach, suggested by theCode, is to calculate the deflection assuming firstly the whole member to be uncracked and secondly the whole member to be cracked. Eqn A.4.1 is used to assess the final deflection.

10.1.3 Rigorous assessment The procedure is, at frequent intervals along the member, to calculate (1)

Moments

(2) Curvatures (3) Deflections.

Here, calculations will be carried out atthe mid-span position only, to illustrate this procedure, with values at other positions alongthe span being tabulated. 10.1.3.1 Calculation of moments

For buildings, it will normally be satisfactory to consider the deflections under thequasi-permanentcombination of loading, assuming this loadto be of long

A4.2(5)

duration.

Thequasi-permanentcombinationof loading is given,forone variableaction, by Gk +

2.3.4 P(2) Eqn 2.9(c)

SWJ1CEABLTY CHECKS BY CALCULATION

1'2

=

NAD Table 1

0.3

Therefore Loading

=

+ (03 x 19.5) =

19.7

=

Mid-span bending moment (M)

25.6 kN/m

25.6

x 72/8

=

156.8 kNm

10.1.3.2 Calculation of curvatures

In order to calculatethe curvaturesit is first necessaryto calculate the properties of the uncracked and cracked sections and determine the moment at which cracking will occur. 10.1.3.2.1 Flexural curvature

Ecm

The effective modulus of elasticity (Eeff) For concrete strength class C30/37, Ecm

= 2 [(1650 U

1

+ çb

= 32 kN/mm2

x 150) + (250 x 300)1 2(1650 + 300)

3.1.2.5.2 Table 3.2

= 165 mm

For internal conditions and age at loading of 7 days

=

A43(2) Eqn A.43

3.1.2.5.5

Table 33 3.1

Therefore

ECff

=

32

1+3.1

=

7.8 kN/mm2

=

Effective modularratio (cx) e

E

—a-

Eeff

Modulus of elasticity of reinforcement (E Therefore

a

==

25.64

7.8

AS

bd

p'

= 200 kN/mm2

=

A' — bd

=

2410 1650

=

x 390

402 1650

x 390

= 3.75x103

=

6.25x

10

For the uncracked section, the depth to the neutral axis is given by x

— bh2/2— (b — b)(h — h ('iLf +hf) + (ae 1) (A'd' +Ad) = 165.2mm = ______________________________ — — 1) bhf + b(h h + (cxe (A's +

3.2.43(1)

bEWV1CEAB1UTI CHECKSBY CALCULATION

The second moment of the area of the uncracked section is given by bh3

=

b(h—h)3 + bh (x — h/2)2 + b(h — 12

+

hf)

2

h+h

—

2

x + (Oe_1) A'8(x—d')2+ (ae1) A(d—x)2 = 7535 x lO mm4

For the cracked section the depth to the neutral axis is given by

÷ (ae — l)p']

—

x

= 0.345d =

— + +j[aeP + (ae 1)p']2 2[aep+(a

—

1)p']

134.6 mm

The second moment of area of the cracked section is given by

r

—

=

bcP

=

x2

3

x—

1

—

+apl—— +(ae e

3d

d

= 5448 x

0.0556bc13

,x d' —1)p--—— d

2

d

106 mm4

The moment which will cause crackingof the section is given by Mcr yt

= h

—

x = 450

—

165.2

= 284.8 mm

ç

= 2.9 N/mm2

For concrete strength grade C30137,

Table 3.1

Therefore

Mcr

x 106 x

= 2.9 x 7535

3.1.2.4

10-6

= 76.7 kNm

284.8

The section is considered to be cracked, since Mcr

< M =

156.8 kNm

Curvature of the uncracked section is given by 1 —

,

M ____ =

EI1

156.8 7.8

x

106

x 7535 x

x

106

= 2.668 x 10- rad./mm

Curvature of the cracked section is given by 1 — =

r.

M

E6In

156.8

= 7.8

x

x 106

x 5448 x

106

= 3.690 x 10-6 rad./mm

Having obtained the values for the two limiting conditions Eqn A.4.1 is used to assess the intermediate va'ue. Hence

3

=

+

(1

—

3I

A43(2) Eqn A.4.1

SE%CEAB%UTY CHECKSBY CALCULATION

=

1

For high bond bars,

= =

?3

For sustained loading,

1.0 0.5

is the stress in the tension steel calculated on the basis of a cracked section Therefore —

aeM (d

— x)

— —

x

x

25.64 156.8 106 (390 — 134.6)

S

= 188.5 N/mm2

5448x106

is the stress in the tension steel calculated on the basis of a cracked section underthe loading which will just cause cracking at the section considered. a

Therefore sr

=

aM.(d — x)

= 25.64 x 76.7 x 5448

106(390 — 134.6)

x

=

92.2 N/mm2

106

Therefore

=

1 —

0.5

92.2 2

=

0.88

188.5

Note:

a—

1 10.1.3.2.2

M may be replaced by .— in the above calculation M

= [(0.88 x 69) + (1—a88) x 2.668] x 10-6 = 567 x 10-6 rad./mm

Shrinkagecurvature The shrinkage curvature is given by

=

A.43.2 Eqn A.4.4

'cs

where

is the free shrinkage strain For internal conditions and 2.401u = 165 mm

=

0.60

i0-

S is the firstmoment of area ofthereinforcementaboutthecentroid ofthe section.

I is the second moment of area of the section. SandIshould becalculatedfor both the uncrackedandfullycracked conditions Curvature of the uncracked section S1

= A(d — x)

—

A'5(x

—

d')

= 495.5 x iO mm3

3.1.2.5.5 Table 3.4

SERVICEABU.VTYChECKS WI CALCULATION

1 —

r1

= 0.60 x

iO x 25.64 x 495.5 x i0

=

x 106

7535

1.0

x

10-6 rad./mm

Curvature of the cracked section S11

1 —

rcsrl

= =

A' (x d') = 581.5 x iO mm3 0.60 x iO x 25.64 x 581.5 x io = 1.64 x

A(d — x)

—

x 106

5448

10 6rad./mm

Therefore 1 —

=

-x—

=

[(0.88

1

+ (1

)x

—

— 1

x 1.64) + (1 — 0.88) x 1.0] x 10_6 = 1.563 x 10_6 rad./mm

Thetotal curvature at mid-span 1. = .1 ÷

r

1 = (3.567 + 1.563) x 10-6 = 5.130 x lO6rad./mm

Theflexural, shrinkage and total curvatures at positions xli along thespan are given in Table 10.1. Table 10.1 Curvatures xli

Moment (kNm)

0

0

x 106 (rad./mm) 1

1

1

-r

7 0

0

i

1

1

0

1.000

1000

—

0.960

1.000

1.960

0.708

2.171

1 453

3.624

0

0.1

56.4

0960

0.2

100.4

1.708

0.3

131.7

2.241

3.100

0830

'2.954

1.531

4485

0.4

150.5

2.561

3.542

0870

3.414

1 557

4.971

0.5

156.8

2.668

3.690

0.880

3.567

1 563

5.130

06

150.5

2.561

3.542

0.870

3.414

1.557

4971

0.7

131.7

2.241

3.100

0.830

2954

1.531

4.485

0.8

100.4

1.708

2.363

0708

2.171

1.453

3624

0.9

56.4

0.960

—

—

0.960

1.000

1.960

1.0

0

0

1.000

1.000

—

0

0

0

10.1.3.3 Calculation of deflections Having calculated the total curvatures, the deflections may be calculated by numerical integration using the trapezoidal rule.

'

Theuncorrected rotation at anypoint may be obtained by thefirstintegralgiven by 1

=

e1 +

1

7+7 2

n

InI

SERVICEABIUTYCHECKSBY CALCULATION

Having calculated the uncorrected rotations,the uncorrected deflections may be obtained by the second integral given by aX

=ax1 +e÷e1 2

I

n

where the subscript x denotes the values of the parameters at the fraction of the span being considered, and the subscript x—1 denotes the values of the parameters at the preceding fraction of the span.

l

is the span

n

is the number of span divisions considered.

Hence the uncorrected rotation at 0.11

e

0.11

=

e0

_1__ r0.11

+

1

r0

2 1.96

= 0+

+

n

+ 1.0 2

10-6

x

10

= 1.036 x

iO rad.

and the uncorrected deflection at 0.11 a0.11

=

a0

+ (e0.11+

= a ÷

e)i

io x (1.036+ 0)

7000 =

0363 mm

Theuncorrecteddeflectionsmaythen becorrectedtocomplywiththe boundary conditions of zero deflection at both supports. Thisis done by subtracting from theuncorrected deflectionsthe valueof the uncorrected deflection at the right hand support multiplied bythe fractionofthe span at thepointbeing considered. Thevalues of the uncorrected rotations, uncorrected and corrected deflections at positions xli along the span are given in Table 10.2. Table 10.2

xli 0

Deflections (mm)

1x106

r

integral

2nd integral

Correction

1.000

0

0

0

1st

x108

Deflection

0

0.1

1.960

1.036

0.363

8.871

— 8.508

0.2

3.624

2.990

1.772

17.742

—15.970

0.3

4.485

5.828

4.858

26.613

—21.755

0.4

4.971

9.138

10.096

35.484

—25.388

0.5

5.130

12.673

17.730

44.356

—26.626

0.6

4.971

16.208

27.838

53.227

—25.388

0.7

4.485

19.518

40.342

62.098

—21.755

0.8

3.624

22.356

54.998

70.969

—15.970

0.9

1.960

24.310

71.331

79.840

— 8.508

1.0

1.000

25.346

88.711

88.711

0

5EWUCIkBfl.%TYCHECKSBY CALCULATiON

Maximum deflection at mid-span atot

. span = 28mm = 26.6 mm = span < .imitof

263

250

10.1.4 Simplified approach The procedure for this approach is to (1) Calculate the maximum bending moment and the moment causing cracking (2)

Calculate the maximum deflections for the uncracked and fully cracked conditions, and use Eqn A.4.1 to assess the final maximum deflection.

From Section 10.1.3.2.1the maximum bending moment M the moment causing cracking Mcr = 76.7 kNm.

=

156.8 kNm, and

The maximum deflection of the uncracked sectiondue to flexure 81

=

5w14

I

384Eeff I

w

=

25.6 kN/m

1

=

7.Om

E

=

7.8 kN/mm2

=

7535

x

106 mm4

Therefore a

=

5x

384

25.6

x 74 x 1012

x 7.8 x i0 x 7535 x

= 13.6mm 106

The maximum deflection of the cracked section due to flexure 5w14

aII

=

I

= 5448

x 106 mm4

=

5x

I

384Eeff II

Therefore a

384

25.6

x 7.8 x

x 74 x 1012 x

5448

x

= 18.8mm 106

Final maximum deflection due to flexure a

= rail +

(1

—

)a1

= M

A4.3(2) Eqn A.4.1

SEVICEABILITYCHECKS BY CALCULATION

=

1.0

=

0.5

=

1 —

Therefore 0.5

(7•

)2

18.8)

+ (1

=

0.88

—

0.88)

Therefore

= (0.88 x

a

x 13.6

18.2 mm

It must be appreciated that the deflection calculated above is due to flexure only. The additional deflection due to shrinkage must also be assessed. The shrinkage curvature at mid-span from Section 10.1.3.2

=

1.563

x

10-6 rad./mm

1 1 2 =——1 8

aCs

= 1.563x106x72x106 = 9.6mm 8

r5

a0 = a + a5 = 182 + 9.6 =

27.8 mm

This figure is close to the rigorously assessed value of 26.6 mm.

10.1.5 Alternative

simplified approach

An alternative simplified approach, which directlytakes account of shrinkage, is given in BS 8110(2). The procedure here is to calculate the total curvature at one point, generally the point of maximum moment. Then, assuming the shape of the curvature diagram to be the same as the shape of the bending momentdiagram, the deflection is given by

=

a

K12

I

BS 8110: Part 2 Section 3

BS 8110:

rtot

Part 2 3.7.2

where

K is a factor dependent upon the shape of the being moment diagram.

Eqn 11

For a simply supported beam with uniformly distributed load

=

K

0.104

BS 8110:

Total curvature at mid-span, 1 —

from Section lOi.3.2

= 5.130 x 10- 6 rad./mm

rtot

Therefore maximum deflection at mid-span =

0.104 x 72

x 5.130 =

26.2 mm

Again this is close to the rigorously assessed value.

Part 2 Table ai

S'JCE.V1UTY C1ECK5 BY CALCULATiON 10.1.6 Comparison with span/effective depth ratio The procedure for limiting deflections by use of span/effectivedepth ratios is set out in EC2 Section 4.4.3. For the example considered A8req

= 2392 =

mm2

=

Aspr

lOOAprov

=

bd

2410 mm2

100 X 2410 = 1650 x 390

Therefore the concrete is lightly stressed, p

027%

0.5%

4.43.2(5)(c)

The NAD1 introduces a category of nominally reinforced concrete corresponding to p = 0.15%

NAD 6.4(e)

Basic span/effectivedepth ratio for a simply supported beam, interpolating for

p = 0.37% 1

=

d

NAD

28

Table7

For flanged beams where b/b > 3.0 the basic span/effective depth ratio should be multiplied by a factor of 0.8

Thespan/effectivedepth ratios givenin NADTable7 are based on a maximum service stress in the reinforcement in a cracked section of 250 N/mm2. The tabulated values should be multiplied by the factor of 250/as for other stress levels, where is the service stress at the cracked section underthe frequent load combination. Asa conservativeassumptionthe Codestates that thefactor may be taken as 250

-

—

fg 400

A

yk A

s,prov

Therefore, for this example, allowable span/effective depth ratio

-

= 28 x 0.8

(allowable)

=

(460 x 2392/2410)

19.6

>

(actual)

=

19.6

= 7000 = 180 390

If the span/effective depth ratio is modified using the service stress in the reinforcementas calculated in Section 10.1.3.2.1 but adjusted for the frequent load combination

= 188.5 x (allowable)

31.4/25.6

=

231 N/mm2

= 28 x 0.8 x 250 = --.-

21.6

>

18.0 (actual)

I

4.43.2(3)

SERVICEABILITYCHECKS BY CALCULATION

It canbe seen from this examplethat whilstthe span/effectivedepth ratio based

on the calculated steel service stress suggests that the deflection should be wellwithin the prescribed limits, the deflection fromthe rigorous and simplified analysis proves to be much nearer to the limit of span/250. Thisis due to the contribution to the deflection from shrinkage, which in this example is approximately a third of the total deflection. Thevalues of shrinkage strain given in EC2 Table 3.4 relateto concrete having a plastic consistence of classes S2 and S3 as specified in ENV 206(6). For concrete of class Si and class S4 the values given in the Table should be multiplied by 0.7 and 1.2 respectively.

3.1.2.5.5(4)

ENV 206 7.2.1

Table 4 of ENV 206 categorises the class in relation to slump as given in Table 10.3. Table 10.3 Slump classes Class

Slump (mm)

Si

10— 40

ENV 206 7.2.1 Table 4

50— 90

S2 S3

100

—

150

160

34

Thus forclasses S2 and S3 the slumpmayvarybetween 50 mm and 150 mm. It is not logical that mixes withthis variationofslump, and hence w/c ratio,should have a standard value of shrinkage strain.

If thevalues in EC2 Table 3.4are assumed to relate to the median slumpfor classes 82 and S3 of 100 mm, then the values for slumps of 40 mm to 100 mm should be multiplied by afactor between 0.7 and 1.0 and values for slumps of 100 mm to 160 mm should be multiplied by a factor between 1.0 and 1.2.

As most normal mixes will have a slump in the order of 50 mm the values of shrinkage strain for the example considered would be: 0.60

x iO 0.7 +

<

(1

= (0.60

10

0.7)

x 10 x 0.75

= 0.45

x iO

Thisfigure relates more closely to thevalue whichwould be given in BS 8110, for the same example, of 0.4 x

BS 8110: Part 2

For the example considered, thecalculated deflection due to shrinkage from the rigorous assessment would be

7.4

iO.

9.1

x 0.75 =

6.8 mm

and the total deflection from the rigorous assessment would be =

26.6

—

9.1

+ 6.8 =

This is well within the limit of span 250

24.3 mm =

28 mm

Figure 7.2

MCfABUTY C1BCIS BY CALCULATION 10.2 Cracking Check by calculation that the longitudinal reinforcement in the reinforced concrete wall section shown in Figure 10.2 is sufficient to control crackingdue to restraint of intrinsic deformation resulting in pure tension.

T16— 200 T12— 125

Figure 10.2 WaIl section

10.2.1 Design data Concrete strength class is C30/37. Cover to reinforcement

= 35 mm

High bond bars with

=

460 N/mm2

NAD Table 6

Exposure class 2(a)

10.2.2 Calculation method Requirementsfor the control ofcrackingare given in EC2 Section 4.4.2.Crack control is normally achieved by the application of simple detailing rules.

Theprocedure forthe calculation of crack widths isfirstto calculate the stress and hence the strain in the reinforcement, taking into account the bond properties ofthe bars and theduration of loading. Next, the averagefinal crack spacingdependent on the type, size and disposition of the reinforcement and the form of strain distribution is established. Thedesign crackwidth maythen be obtained and compared with thelimiting design crack width. In the absence of specific requirements, a limiting crack width of 0.3 mm will generally be satisfactoryfor reinforced concrete members in buildings with respect to both appearance and durability.

4.4.2.1(6)

10.2.3 Checkby calculation 10.2.3.1 Calculation of steel stress and strain Steel stress: =

kkfcffAt A

S

where

= area of reinforcement within the tensile zone = 905 x 2 = 1810 mm2/m

4.4.2.2(3) Eqn 4.78

SERV(CEABIUTYCHECKS BY CALCULATION

A0 = area of concrete within tensile zone = 1000 x 200 = 200 x iO mm2

k k

=

a coefficient taking account of stress distribution

=

1.0 for pure tension

=

a coefficientallowing fortheeffect of non-uniformself-equilibrating stresses

=

0.8 for tensile stresses due to restraint of intrinsic deformations 300 mm) (h = tensile strength of concrete effective at first cracking ct,eff

=

3.8 N/mm2 (taking

095

but see Section 10.2.3.4)

3.1 .2.4(3)

Table 3.1 Therefore a$

8

1.0 x 0.8 x x 200 x iO _____________________

=

=

336 N/mm2

1810

Mean strain:

sm =

_

1

fl1f2

E

4.4.2.4(2)

(a)2 a

Eqn 4.81

where modulus of elasticity of steel

E

= 200 kN/mm2

asr

3.2.4.3(1)

= a coefficient taking account of bond properties of the bars = 1.0 for high bond bars = a coefficient taking account of load duration = 0.5 = the stressin the reinforcementbased on a cracked section under the load causing first cracking

=

a for intrinsic imposed deformation

Therefore 336 200

x

io

1 —

0.5

= 8.4

x

i0

10.2.3.2 Calculation of crack spacing

The average final crack spacing srm =

50

+ 0.25 k12 k

where Ic

=

a coefficient taking account of the bond properties of the bar

=

0.8 for high bond bars

4.4.2.4(3)

Eqn4.82

MUTYCCKS Y CALCULATION In the caseof imposed deformations,k1 should be multiplied by k, with k being

in accordance with EC2 Section 4.4.2.2.(3). k2

=

a coefficient taking account of the form of strain distribution

=

1.0 for pure tension

=

A the effective reinforcement ratio = —f--

=

the effective tension area.

ceff

The effective tension area is generally the area of concrete surrounding the tension reinforcementto a depth of2.5timesthe distancefromthetension face to thecontroid ofthereinforcement or, for members in tension, halfthe actual member thickness, whichever is the lesser. This is calculated as: 2.5

x (35 +

= 103

12/2)

h/2

= 100 mm

Therefore

Aff= = srm

1000

x 100 1810

2x

100

x

= 100 x

io

iO mm2

= 0.009

= 50+ (0.25 x 0.8 x 0.8 x

1.0

x 12) =

263mm

0.009

10.2.3.3 Calculation of crack width

The design crack width Wk

= 13S€

4.4.2.4 Eqn 4.80

where

= a coefficient relating the average crack width to the design value = 1.3 for restraint cracking in members with a minimum dimension of 300 mm or less. Therefore Wk

=

1.3

x 263 x 8.4 x

iO

= 0.29 < 03 mm (limit)

10.2.3.4 Concluding remark

The Code suggests a minimum value of 3 N/mm2 be taken when the time of cracking cannot be confidently predictedas being less than 28 days. Whilst the values given for seem high, it is difficult at the design stage to assessaccurately the as pladed concrete strength because this often exceeds theclass specified. Consequently,unlessstrictsite control is exercised,it would be prudent to adopt the apparentlyconservativefigures givenin EC2 Table3.1.

j

DEEP BEAMS

U 11,1

Introduction The design of deep beams may be based on analyses applying: (a) linear elastic analysis; (b) an equivalent truss consisting of concrete struts and arches with reinforcement, all preferably following the elastic field;

2.5.1.1(5) 2.5.3.7.3

(c) non-linear analysis.

In EC21 details of theanalysis model and, therefore, much of the design are notgiven and it is leftfor the Engineerto satisfythe principal Code requirements. This can be achieved using CIRIA Guide 2, The design of deep beams in reinforced concrete18, which also provides recommendations on the detailed analysisand design. TheGuide was written for usewith the then current British Standard CP 110(19). Here it has been assumed that a complete design to the CIRIA Guidewould be carried out and then checks made to demonstrate compliance with the specific clauses for deep beams in EC2. To highlight some of the differences between EC2 and design to the CIRIA Guide, the example in Appendix B of the Guide has been used.

A small number of EC2 clauses have been identified as relating specifically to deep beams:

(a) 2.5.2.1(2)

—

definition of deep beams

(b)

—

analysis modelling

(c) 4.4.23(4)

—

skin reinforcement

(d) 5.4.5

—

reinforcement detailing

2.5.3.7.3

11.2 Example

A proposed arrangementof walls and columns is shown in Figure 11.1. Loading details are presented in Figure 11.2. It is intended to justify a design using the Simple Rules of Section 2 of the CIRIA Guide.

Thebeam is aflat vertical plate and thethickness is small compared with other dimensions.

CIRIA

Guide 2 Cl.2.1.1(1)

There are two loads which may be defined as concentrated and no indirect loads or supports.

Cl.2.1.1(4) Cl.2.1.1(5)

In EC2 a beam is classified as a deep beam if the span is less than twice the

2.5.2.1(2)

depth. CIRIA Guide2 classifies deep beams as 'Beams with span/depth ratios of less than2 for single spanbeams or less than 2.5 for multi span beams',thusgiving an extended range of elements to be designed as deep beams in comparison with EC2.

DRIP BEMAS

I 200L

1800

900

k—..1

14220 300

1120

It

+

300—

d-

10 80 300

V

T

1

I I I

I

I

300k

Iv

7140

300

-E

I I I I I I

-0

300

iv

I 3600

30O

iL

I

1 L2SO

25o.....:j1!

5000

1—1

00O 1O0O I

ELEVATION

Figure 11.1 Structural arrangement

=

I

I-

I

DEEP BEAMS

576 kNIm dead load

11e220

10680

÷ 440 kNIm live load

475 kN 1' dead load

kN/m dead load +440 kM/rn I live

I

' I

475 kN

dead load

load

of supports included in bending moment Load bet ween faces

-m

7140

I 57.6 kN/m

dead

load+ 440 kN/rn

live load

4,4,4,

I 786 kN/m self weight + (57.6 kNlm dead toad -F 440 kM/rn I live toad

4600

_+r 4r

4,

4,4,4, 4,4,

4,f

4,4,4,

'N

Figure 11.2 Loading details

11.2.1 Durability For dry environment, exposure class is 1.

Table 4.1

Minimum concretestrength class is C25/30. = 30 N/mm2 The CIRIA Guide example uses

ENV 206 Table NA.1

Use C25130 for design to EC2 to keep examples broadly consistent. Minimum cement content and water cement ratio Minimum coverto reinforcement

= 15 mm

Assume nominal aggregate size

=

20 mm

Assume maximum bar size

20 mm

Nominal cover

20 mm

EU

ENV 206 Table 3 NAD Table 6

NAD 6.4(a)

'DEEP BEAWIS

Use 25 mm nominal cover

Check requirements for fire resistance to BS 8110: Part 2(2).

NAD 6.1(a)

11.2.2 Materials = 460 N/mm2 Type 2 deformed reinforcement with Concrete strength class C25/30, nominal aggregate size 20 mm 11.2.3 Effective dimensions of beam Effective span (1) = 1 + (c112 0.1l) + (c2/2 =

5000 +

2

+

0.1

x

5000

CIRIA

Guide 2 Cl.2.2.1

0.1l)

= 5750 mm

Note that EC2 effective spans typicallycome to the mid-point of the supports.

= lesser of h and I = 10920 > 1 = 5750 mm

Active height (ha) h

Therefore

= 5750 mm Thickness of beam = 300 mm This thickness is used to be consistent with the CIRIA Guide2 example. It will be necessary under EC2 to demonstrate that the required reinforcement can be accommodated within thiswidth. Theeffectivedimensions ofthebeam are h

shown in Figure 113

of beam 10920

Height

h

Effective span

I

5750

Ii .1502j

Clear span

15000

Figure 11.3 Effective dimensions of beam

=

NAD 6.3(a)

Figure 2.4

DEEP BEAMS

11.2.4 Elastic stability

—

slenderness limits

The CIRIA Guide Simple Rules assume no reduction of capacity due to the slenderness of the section or to lack of adequate restraint.Thisis valid if every panel can be defined as braced and not slender. In the examination of this condition, the CIRIA Guide states that the effective height of each panel is taken to be 1.2 x the shortest distance betweencentres of parallel lateral restraints(where there are effective lateral restraintsat all four edges of the panel) or as 1.5 x the distance betweenthe centres of the parallel lateral restraints(where one or two opposite edges of the panel are free). When both rotational and lateral movements are restrained the effective height may be taken as the clear distance between restraints. In EC2 the demonstration is not quite as straightforward. The floors are assumed to be held in position horizontally by an adequate bracing system and are 'braced' in accordance with EC2.

4.3.5.3.2

The floor slabs are monolithic with the wall so the effective height, 10, is calculated fromthe relevant clauses in EC2 referring to columns. However the design example does not give anyinformation on the adjacent structure so kA

4.3.5.a5 Eqn 4.60

1.0

1

ljl

=

cannot be calculated but

x 3540 =

cannot exceed 1.0 and therefore

3540 mm

43.53.5(1)

The wall is considered slender if X exceeds the greater of 25 or

= radius of gyration = X

= =

l/l

=

3540/86

300/f

15/j

= 86 mm

= 41.2

NSIACfCd

Nsd say in lower storey (bottom loads not considered)

=

57.6x3x135+44x3x1.5+2x475x1.35/5.75

= 654 kN/m U

654x

=

15/f =

io

1000

x 300 x 25/1.5

41.6

>X

=

=

Figure 4.27

0.13

41.2

The wall is not slender

11.2.5 Loading Loading details are shown in Figure 11.2 and evaluated in Tables 11.1 and 11.2.

4.3.5.3.5(2)

DEEP BEAMS

Table 11.1 Characteristic loads CIRIA

Total loads

Slab at level 14.220 10.680 7.140 3.600 Self-weight

0.3 x 10.92

(kN)m)

(kNPm)

44.0*

57.6*

44.0* 44.0 44.0

57.6* 57.6 57.6 78.6

x 24

—

176.0

Guide 2 Cl.2.3 CI.2.a1

309.0

at level 14.220; 2 @ 475 kN*, which are considered as dead loads Vertical forces applied above a level of 3.30 + 0.75 x 5.75 = 7.620 are considered as top loading and loads below as hanging loads Point loads

CIRIA Guide 2 01.23.1(1)

considered as top loads

In EC2 differing 'F values produce slightly different design forces to those in CIRIA Guide 2

= 135,

=

1.5

Ultimate distributed top load

=

Gk +

1.5

x 88 +

x 115.2

1.35

=

288 kN/m

Table 11.2 Hanging loads Loads applied within the depth

of the beam

Slab at level 7.140 Self-weight

(kN!m)

(kNPm)

44.0

57.6

—

76.6

44.0

136.2

Slab at level 600

44.0

57.6

Total hanging load

86.0

198

Loads applied to the bottom ofthe beam

The CIRIA Guide Simple Rules apply where the intenstty of any load is less than 0.2f0 and the load is applied over a length which exceeds 0.21. More intense loads and those applied over shorter lengths are considered to be 'concentrated', in which case reference should be made to the Supplementary Rules in the Guide. To allow for design to EC2 where different values and concrete strength classesare used, the check for load intensitymight reasonablybe made against

I

0.2

x ratio of

=

0.2x

1.4

values

25

x (u'k) x

ck

=

0.23fck

= 135

x 475

= 641 kN Allowing for 450 spread of load through thickness of slab Load intensity = 641 x 10/(800 x 300) = 2.67 N/mm2 Ultimate concentrated top load

=

CIRIA

Guide 2 01.23.1(5)

BEM4S

This loading is well below°23ck but, because the length of the loaded area is lessthan 0.21 = 0.2 x 5750 = 1150mm, some additional reinforcement may be required and must be calculated using the Supplementary Rules in

the CIRIA Guide.

11.2.6 Moment and shears

CIRIA Guide 2

11.2.6.1 Reactions due to loads on clear span

23.2

CI.

Thearrangementof loading and supports assumed for calculating the bending moments is shown in Figure 11.4. These loads are for the fully loadedsystem. 1•35x 475 kN

RBZlO3

kN

Figure 11.4 Total loads at ultimate limit state Reactions from total loads are RA

= 1944 kN,

R8

=

2103 kN

Shear forces must be considered for top- and bottom-load cases separately. Consider the bottom-loaded case shown in Figure 11.5. Total bottom load =

135

x 193.8 + 1.5 x 88

= 393.6 kN/m

Reactions from bottom applied loads are = 1027 kN, = 941 kN RAb RBb

15x 86 kN/m I

135x 1938

i

RA1O27 250

I

kN!m

5000

3R8941 I 500 I

kM

Figure11.5 Hanging loads at ultimate limit state Reactions from top loads are thus RAt RBt

= 1944 — 1027 = 917 kN = 2103 — 941 = 1162 kN

11.2.6.2 Additional shear forces due to loads over supports Loads acting over the effective support width apply an additional shear force to the critical section of the beam (i.e., at the support face). In this example, one ofthe pointloads acts atthe centreline ofthe actual support, B, as shown in Figure 11.6. Innnl

CIRIA Guide 2 CI. 23.2

641 kN 2000

Active height

of beam h0= 5750

Support

4

Support

500

5000

.

2000

support 0

Figure 11.6 Additional load at support B Since the effective support width is halfthe actual width, the additional shear force

= 0.5

x 641 x (h

—

= 320.5(5.75 — 0.2 x

0.210)Ih

265 kN

5.0)15.75

11.2.6.3 Total shear forces

At support A Top loading

(VAt)

Bottom loading

(VAb)

Total

(VA)

= 917 kN = 1027 kN = 1944 kN

At support B Top loading

(VBt)

Bottom loading

(VBb)

Total

(VB)

= 1162 + 265 = 1427 kN = 941 kN = 2368 kN

11.2.6.4 Maximum bending moment Position of zero shear (where x is distance from face of supportA) is given by 1944 — (1.5 Therefore x

x 176 + 135 x 309)x

= 0

= 2.85 m

M = 1944 (2.85 + 0.25) — 681

2852 x —--—

2

=

3261 kNm

DEEP REAMS

11.2.7 Strength design

CIRIA

Guide 2 01.2.4 &

Bending capacity check in accordance with CIRIA Guide 2 = 1 < 1.5 i/ha

2.4.1

Hence thereis no need tocheckthecompression in the concrete and the area of steel required may be calculated from a lever arm given as

z

=

0.21

+ 0.4h

=

3450 mm

For the reinforcement area there is no difference in using EC2 equations.

f

yd

I

A8 =

=

-. =

=

400 N/mm2

Table 2.3

1.15

= 3261 x 400

106

x 3450

= 2363 mm2

I

This is 15% less than CIRIA Guide 2, predominantly due to a higher yield strength reinforcementused in this example, but also in part because of lower values used in EC2.

11.2.8 Detailing of principal bending moment reinforcement CIRIA Guide2 states 'Reinforcementis notto becurtailed inthe span and may be distributed over a depth of 0•2ha A minimum steel percentage in relation to the local area of concrete in which it is embedded is given in Table 1'.

=

Minimum steel percentage

CIRIA

Guide 2 01.2.4.1 01.2.6.2

0.71%

A maximum bar spacing for a maximum crack width of 03 mm is given in Table 2 of the CIR(A Guide. Spacing

for b = 300 mm

165 mm,

Reinforcementmay be distributed over depth = i.e. minimum number

of bars in face =

0.2

x 5750

= 1150 mm

= 7 165

Use 14T16 for main tension reinforcement

A

= 2814 mm2 > 2363 mm2 required

OK

EC2 requires for beams a maximum bar size or a maximum bar spacingto limit cracking under quasi-permanent loading. Quasi-permanent loading Ratio to ultimate loading Estimate of steel stress

= =

Gk + 0.56

°3k

= 0.56f x yd

As.prov

4.4.2.3

4.4.2.3(3)

= 188 N/mm2

DEEP

EMS Maximum bar size is 25

> 16 mm

.

Maximum bar spacing is 250 (pure flexure)

>

165 mm

OK

Table 4.11

OK

Table 4.12

Note that only one of these conditions needs to be met. In CIRIA Guide2, the bars must be anchored to develop 80% ofthe maximum ultimateforce beyondthefaceofthe support and 20% ofthe maximum ultimate force at or beyond a point 0.21from theface of thesupport, or at or beyond the farface of the support, whichever is less. The main reinforcement must be anchored so thatthe concrete within theareaof support relied upon for bearing is adequately confined.

CIRIA

EC2 for deep beams requires thatthe reinforcement,corresponding to theties considered in the design model, should be fully anchored beyond the nodes by bending up the bars, by using U-hoops or by anchorage devices, unless a sufficient length is available between the node and the end of the beam permitting an anchorage length of 1flet The EC2 requirements are clearly more onerous.

5.4.5(1)

Guide 2 C.2.4.1

Support A anchorage

net

=

alA ab sjeq

5.23.4.1

min

A

Eqn 5.4

"s.prov

where

= 0.7 for curved bars with side cover

3q

= 2.7 N/mm2 for good bond in bottom half of pour

i

=

A5req

= 2363

b

=

= 37q

mm2,

42.7

As.prov

Table 53

Eqn53

= 2814 mm2

Therefore 1

b.net

= 0.7

x (37 x

16)

x

2363

2814

= 348 mm

There isinsufficientdistanceto accommodatesuch an anchorage length beyond the centre-line of the column.

If U-barsor loopsare provided,the minimum internaldiameterofthe bend needs to satisfy the requirement for curved bars. This is an indirect check on the crushing of the concrete inside the bend and the tabulated value could be multiplied by AreqIAprov• Minimum internal diameter of bend =

13 =

208 mm

Notethat it is necessary to checkthat sufficient space is available in the final detailing.

At support B a straightanchorage will be sufficientto meet both CIRIA Guide 2 and EC2 requirements.

NAD Table 8

QE.P E&AS

11.2.9 Minimum longitudinalsteel CIRIA Guide 2 refers to the British Standard CP 110, and EC2 will be slightly more onerous.

For beams generally

A

i

O.6bd/ck

z

5.4.2.1.1

0.0015bd

Basing the flexural steel on the active height assumed for the beam design

A

=

x 300 x 5750 =

0.0015

2588 mm2

Deep beams should normallybe provided with a distributed reinforcementnear both sides, the effect of each being equivalent to that of an orthogonal mesh with a reinforcement ratio of at least 0.15% in both directions.

A

=

0.003

x 300 x

1000

= 900 mm2/m

Thisreinforcement should also satisfy therequirement that beams with atotal depth of 1.0 m or more, where the main reinforcement is concentrated in only a small proportion of the depth, should be provided with additional skin reinforcement to control cracking on the side faces of the beam. This reinforcementshould be evenlydistributed betweenthe level of thetension steel and the neutral axis, and should be located within the links.

A

5.4.5(2)

4.4.2.3(4)

4.4.2.2(3) Eqn 4.78

= kckfceIas

where

k

= 0.4 assuming value for bending is to be used

k

=

'cteff

= 3 N/mm2 using suggested value = = 460 N/mm2

0.5

4.4.2.3(4)

4.4.23(4)

Hence

A

—s

ACt

= 0.4

x 0.5 x 3/460

=

0.0013

The requirements of either Table 4.11 or Table 4.12 of EC2 should be met. Assume steel stress

= (--) x value for main bars =

94 N/mm2

4.4.23(4) Table 4.11

Maximum bar size 9max

= 32 mm

Maximum bar spacing in 'pure tension' condition = 200 mm S

Use 110 @ 150 mm crs (EF) above level of main reinforcement

Table 4.12 4.4.23(4)

DEEP

MAS A5

= 1048 > 900

mm2/m

.

OK

11.2.10 Shear design CIRIA Guide 2 separates top and bottom loads and deals with the design of these in different ways.

In principle the bottom loads require vertical tension hangers to suspend the loads above the active beam height, ha with horizontal web reinforcement needed in the area of the supports.

CIRIA

Guide 2 Cl.2.4.2

Thetop-load shear calculationsincludetaking intoaccount anyadditional shear force induced by top loads over the supports. Under the simple design rules the top-load shear capacity is not improved by web reinforcement.

A nominal, orthogonal pattern of web reinforcementnot lessthanthe minimum required for walls in BS 8110 is intended. This is augmented for bottom loads and in the area of the supports.

The detail of the CIRIA Guide calculation is not repeated here and reference should be made to the original document. The reinforcementdetailsare shown

in Figures 11.7 and 11.8.

o

1 0

IL

.0

o In

—

0 U'

•

I-

UJ

I

0 -.

. If ILfl -

C.,'.-

10

IL Lu

II

I

—

l00J..

—

! .

I.,

I

I 'Q

.I..t

'.4

Cl

—

I

Guide 2 Figures 93 & 13

.0 I

0

CIRIA

50T16 — 15OEF 36T16-150

— 5000

—

-'1

I

U-bar

10

Figure 11.7 Arrangement of reinforcement

L I

I.

2000

—

Bars continued through from column and tied to verticaL reinforcement

DtEP BEAMS

Hanger bar given a full tension anchorage length above ha

jIJ__ h0

J'

Full tension lap

______ Hanger bar anchored as a link around main bars

Figure 11.8 Detail at bottom of wall Using the standard method in EC2 VSd

=

4.3.2.2(7)

2368 kN maximum at support B

= rRdk(l.2 + 40p1)bd

Eqn 4.18

=

0.30 N/mm2

Table 4.8

A

=

0.15%A0, therefore p1

k

=

lasd>0.6m

d

=

h — 0.2h, say

VAd1

TAd

= 10920

b

—

0.2

=

0.0015

x 5750 =

9770 mm

=

300mm

=

03 (1.2 + 40 x

>

V, therefore shear reinforcement needed

Therefore VAd1

VSd

VRd3

=

+

where

0.0015)300

VCd

=

x

9770

x

i0 =

1108 kN

VRd1

Eqn 4.22

400 N/mm2

Eqn 4.23

Therefore

A

s

2368

—

1260

x

1108

O.9dfd 036 mm2/mm

= 1260 kN with

f

=

Where the load is not acting at the top of a beam, suspension reinforcement should be provided to transfer the load to the top.

4.3.2.4.1P(3)

The bottom load identified previously = 393.6 kN/m Area of hanger steel needed with

— ASh

=

s

393.6

400

f

= 400 N/mm2

2 =0.99mm/mm

Use T16 @ 150 mm crs. (EF) when hangers are needed

A

—a

s

=

2x201 =

2.68

150

> 036 + 0.99 =

I

135 mm2/mm . . . OK

Use T10 @ 150 mm2(EF) elsewhere A5 —

s

2x78 = = _____

1.04

150

> 036 mm2/mm

OK

= 460 N/mm2

Minimum shear reinforcement with

= 0.0012 by interpolation SW

S

= 0.0012b = 036

mm2/mm

Table 5.5 OK

For heavily loaded deep beams it may prove morecomplicated to justifythe shear.

11.2.11

Further guidance CIRIA Guide2 has further guidance for reinforcementin support regions, under concentrated loads and around holes in beams. Supplementary design rules are also provided for deep beams that include arrangements excluded by the simplified method.

Itis because ofthe extentofthis informationthatthe initialsuggestionwas made, that a complete design to the CIRIA Guide is undertaken and then a parallel designto EC2 is performed as appropriate, to demonstrate compliance with the individual clauses.

CIRIA

Guide 2

ULOAD COMBINATIONS 12.1

Introduction EC21 considers all loadsas variables in time and space and applies statistical principles to arrive at the loads for design. There is an underlying assumption that the basicloadsthemselves are described in statisticalterms Thus, when variable loads of different origins, for example superimposed loads on floors and wind loadson thefacesof buildings, haveto be considered acting together in a load case,the probability of both loadsnot beingattheir full characteristic valuesis allowedfor by multiplierscalled factors.Thesefactorstoo are derived statistically and values are given in EC1(20) and the NAD to EC21. Thus when a number of variable loadshave to be considered simultaneously in anyload case, each load is treated in turn as the primaryload and others are consideredsecondaryTheprimary load is applied at its characteristicvalue multiplied by the partial safety factor. All secondary loadsare applied at their characteristicvalues multiplied by the partial safety factor and further multiplied by a f. factor. These factors varydepending upon the limit state and the type of loading being considered. Mathematically the design loadfor ultimate limit state may be represented as:

E 7GJGkJ +

O.1k,1 primary load

+

Ei1 Q1 secondary load

Whiletheabove procedure isthegeneral approach, EC2 also providessimplified rules: (a) where only one variable

= E + 7G.JGkJ

(b)

load occurs the design load

when more than one variable load occurs the design load = E +

It is importantto note thatthis Codepermitsthe useof eitherapproach although in some circumstances the general method may result in higher loading. In practice the simplified procedure will be perfectly satisfactory for most situations and could be used. The following examples are givento illustratethethinking behind the general approach and indicate where the general approach may be required. Usually, when dead loads produce a favourable effect, 1G can be taken as unity. However, if the variation of the magnitude of the dead load is likely to should be taken as 0.9. prove sensitivethen

For the particular case of continuous beams without cantilevers, the Code = 135 for all the spans permits the use of When caFâulating the loads on vertical elements of multi-storey structures the vertical loads may be based on either: (a)

loads from beams multiplied by suitable 4' and 'y values; or

(b)

loads on beams multiplied by 'y values and a global reduction in loading applied using the proceduregiven in BS 6399(21). This is the approach in

the NAD.

12.2 Example 1

—

frame

For the frameshown in Figure 12.1 identifythe various load arrangements, to check the overall stability. Assume office use for this building. Note that the loadarrangements forthe design of elements could be different.

77/7

7-7/7

Figure 12.1 Frame configuration

—

example

1

12.2.1 Notation Characteristic loads/m = dead — roof Gkr Gkf

kr

= dead — floor = imposed — roof =

—

imposed

floor

Characteristic load/frame =

wind

12.2.2 Load cases

—

—

roof or floor

example 1

Fundamental load combination to be used is YGJGk.J

+ Q,1kj +

2.3.2.2P(2) Eqn 2.7(a)

As the stability will be sensitiveto a possible variation of dead loads, it will be necessary to allowfor this as given in EC2 Section 2.3.23(P3). Take 7Ginf

=

0.9,

7G,sup

= 135

= 1.5 = 0.7 for imposed loads (offices)

Table 2.2 NAD Table 1

LO&O COMBIN&TIONS

12.2.2.1 Load case 1

—

example 1

Treat the wind load as the primary load (see Figure 12.2).

l3S6kr ÷ O7(i5Q.kr)

O9Gkr

lSWk

.- .-

1 5Wk

f/'//////j

I

1355kf + 0.7 (15kf

09Gkf

II 135tikf + 0.7 (15akf

09Gkf

I l5Wk

77

7_7 7_7

Figure 12.2 Load case 1 12.2.2.2 Load case 2

—

—

example

7

1

example 1

Treat the imposed load on the roof as the primary load (see Figure 123).

09Gkr

1356kr +

i•sa

l35Gkf ÷

o•7

F3SGkf ÷

07 (l5Qkf)

07 (lSWk)

0.7 (1.SWk)

09Gkf I '/7 /77/7/ —-

0•9tkf

O•7(l5WkJ

(1•5kf)

//////////

7-7Figure 12.3 Load case 2

—

example

1

InI

COWfl4AT1O%4S

12.2.2.3 Load case 3

—

example 1

Treat the imposed load on the floors as the primary load (see Figure 12.4).

135Gkr + 0.7 (l5akr)

O9Gkr

/

0.7 (1•5Wk

VV

'Y

096 kf

135G kf +

096.kf

1356kf ÷ 1.sakf

/ / / / /j f///,///,, I /1/

07(15Wk)

O•l(l•5Wk)

I

___________

l5akf

I!

7_7 /_7 Figure 12.4 Load case 3 12.2.2.4 Load case 4

—

—

example 1

example 1

Consider the casewithout wind loading treating the imposed floor loads as the primary load (see Figure 12.5).

O96kr

/

r

135Gkr

'/

+ 07 (150.kr)

096 kf 135Gkf + I /7/////// II

1•Sakf

l35Gkf ÷

15kf

/ / / / 7/

O•9Gkf I 'A/

r_7 7_7 Figure 12.5 Load case 4

7_7 ?_r —

example 1

LOA.D COMBINATIONS

12.2.2.5 Load case 5

—

example 1

Consider the case without wind loading treating the imposed roof load as the primary load (see Figure 12.6).

0.96 kr

1 35G

_________________________

Y

/

kr 4- 1 .5ak,

•

,//////// /

1356kf ÷ 07 (1•50.)

0.96 kf

I

// / ///

7_7 1_7

7_7

O9G kf

/1/

Figure 12.6 Load case 5

—

135Gkf ÷ 07 c15akf)

rr

example 1

Note: When the wind loading is reversed, another set of arrangements will need to be considered. However, in problems ofthistype, thecritical arrangementsare likely to be found intuitively rather than by directly searching through all the theoretical possibilities.

12.3 Example 2

—

continuous beam 1

Identify the various load arrangements forthe ultimate limit state forthe design of the four-span continuous beam shown in Figure 12.7. Assume that spans 1—2 and 2—3 are subjectto domestic useand spans 3—4 and 4—5 are subjectto parking use.

k1 IUo05) Figure 12.7 Beam configuration

—

example 2

12.3.1 Notation

= characteristic dead load/rn = characteristic irnposed load/rn (domestic use) = characteristic imposed load/rn (parking use) 12.3.2 Load cases

—

example 2

Fundamental load combination

to be used is

+ 'YQlkl +

17Q.uaI0ki For beams withoutcantilevers the same value of self-weight may be applied to all spans, i.e., l.35Gk. The load cases to be considered for the imposed loads are (a) alternate spans loaded; and (b)

adjacent spans loaded.

2.3.2.2P(2)

Eqn 2.7(a) 2.3.2.3(4)

2.5.1.2(4)

LOAD COMBINATIONS

The various load arrangements are shown in Figure 12.8. 135Gk A

Note

f1c\cfhl\rf

N'V

V

Load case A above should be combined with cases B —J below as necessary O7

15ak1

15k2)

{4}

B

Max. -I-ye moment in 1- 2 and max. col. moment

at

1

15ak2

Max. +ve moment in 3—4 and max. coL. moments at 3 and

4

f1

O7 (l-5a2

Max. +ve moment in

2-3 and max. col. moment at 2 and

05 (l.5Q )

v1

Max. -I-ye moment

F

Mox. —ye moment

in

4—5

at

2

15Qk2

and max. col. moment

150k1

at

S

O7 (15Q k2 )

_____

Max. —ye moment

at

3

( see case H also I 0.5 I 1-SQk2

(.5Q

H

3

Max. —ye moment

at 3

Max. —ye moment

at

(see case B also)

F

Figure 12.8 Load cases

4

—

example 2

1

5ak2

LOAD COMBINATIONS

12.4 Example 3

continuous beam 2

—

For the continuous beam shown in Figure 12.9, identify the critical load arrangements for the ultimate limit state. Assume that the beam is subjectto distributeddead and imposed loads, and a point load atthe end ofthecantilever arising from the dead load of the external wall. P

2

Figure 12.9 Beam configuration

—

example 3

12.4.1 Notation

P

=

characteristic dead load/rn

=

characteristic imposed load/rn

=

characteristic point load (dead)

12.4.2 Load cases

—

example 3

The fundamental combinations given in EC2 Section 23.2.2 should be used. Notethat the presence of the cantilever prohibits the use of thesame design values of dead loads in all spans. The various load arrangements are shown in Figures 12.10 to 12.13.

135P

O9 Gk I

\

13SGk +l•SO.k

N

ti

12 Max.—ve moment and anchorage of top bars at 3 Also max. cot. moment at 3 C see F.g. 12.13 also

Figure 12.10 Load case 1

—

example 3

Jr

2.3.23(4)

LO&O COMBIK&TIONS

O9P

l.3SGk+ 50k 2

at

Max.—ve moment

Figure 12.11 Load case 2

3

2

example 3

—

135P

l35Gk+lSO.k

V#VTh(Th

096k

//

YN

1'

35Gk+5Qk

VvVJ

12

2

Max. +ve moment in 1max. cot. moment at 1 cnd max. cot. at 2 (see Fig. 12.13 aLso) moment possibly

Figure 12.12 Load case 3

—

example 3

O9P

O9Gk I

/

135Gk ÷ 1 5Qk

rv'vyN 2

11

IO9G

k

Max.+ve moment in 2-3, max. cot. moment at 2 (see Fig. 12.12 also) and, possibly, max. cot. moment at 3 (see Fig. 12.10 also) Figure 12.13 Load case 4

—

example 3

I

125 Example 4

—

:f

tank

Awatertank, as shown in Figure 12.14, ofdepth H (in metres)hasan operating depth of water h (in metres). Calculate the designlateral loadsfor the ultimate limit state.

Figure 12.14 Tank configuration

—

example 4

According to the draft EC1, earth loads are permanent loads. The same reasoning can be applied to lateral pressurescaused by water. The NAD for EC2 confirms this. Design canthereforebe based onthe pressurediagram shown in Figure 12.15.

density

water

h

1 35ph Figure 12.15 Design load based on operating water depth

—

example 4

Consideration shouldalso be given to the worstcredible water load, which in this casewill correspond to a depth of H, i.e., water up to thetop of the tank. EC2 permits the variation of the partial safety factor depending on the knowledge of the load Gk,.

is not given. The basis adopted in However, the method of establishing BS 8110: Part 2(2) could be used and a factor of 1.15 applied instead of 1.35. In this case the alternative design loading will be as shown in Figure 12.16.

Hj p

density

1 15pH

Figure 12.16 Design loadbased on worstcredible water depth

—

example 4

NAD 6.2(c)

(3DESIGN OF BEAM AND COLUMN SECTIONS 13.1

Concrete grades

3.1.2.4

k'

EC21 uses the cylinderstrength, to define the concrete strength in design the cube equations, although strength may be used for control purposes. The both grade designations specify cylinder and cube strengths in the form C cylinderstrength/cube strength, for example C25/30.

It mayoccasionally be necessary to use cube strengths which do not exactly correspond to one of the specified grades. In such instances a relationship is required betweencylinderandcube strength in orderto obtain an appropriate value for The relationship implicit in EC2 and ENV 206(6) is given in 13.1. Figure

50

40

C'1

E

5

30

10 U 20

C

10

Cube

strength (N/mm2)

Figure i3.i Relationship between cube and cylinder strength of concrete

13.2

Singly reinforced rectangular beam sections The following equations and design tables have been derived from the assumptions given in 43.1 and 4.2.133(b) of the Code combined with the redistribution limitsgiven in 2.53.4.2.They are entirely in accordance with EC2.

4.3.1

4.2.1.3.3 2.5.3.4.2

M4D SOLUIM4

SEcroNs

13.2.1 Equations for singly reinforced rectangular beam sections ASck =

=

0.652 — To.425

bdck xld

—

1.5

= 1.918w

where

=

M bd2fck

Table 13.1 gives w and Table i3.i M

ck

Flexural reinforcement in singly reinforced rectangular sections. ASck I

0.012

0.012

0.014

0.014 0.016

0.016 0.019

0.018

0.021

0.020 0.022 0.024 0.026 0.028 0.030 0.032 0.034 0.036 0.038 0.040 0.042 0.044

0.023 0.026 0.028

0.046

xld

zid

bdfck

0.010

0.048 0.050 0.052 0.054 0.056 0.058 0.060 0.062 0.064 0.066 0.068

xldas a function of i.

0.031

0.033 0.035 0.038 0.040 0.043 0.045 0.048 0.050 0.053 0.055 0.058 0.060 0.063 0.065 0.068

0.022 0.027

M

ABck

'Ck

bdfck

x/d

zid

0.113 0.116

0.217

0913

0223

0.911

0.119

0.228 0.234 0.239 0.245 0.250 0.256

0.909 0.907

0.977 0.975

0.090 0.092 0.094 0.096 0.098 0.100 0.102 0.104 0.106 0.108

0.973

0.110

0.130 0.133 0.136 0.139 0.142

0.073

0.971

0.112

0.145

0.077 0.082

0.969 0.967 0.965 0.963

0.114 0.116

0.148

0.118 0.120

0.154 0.157

0.961

0.122

0.160

0.267 0.272 0.278 0.284 0.289 0.295 0301 0207

0.124

0.163

0.313

0.126 0.128

0.166

0.169

0.130

0.172

0219 0.324 0.330

0.121 0.125

0.960 0.958 0.956 0.954 0.952 0.950

0.132 0.134

0.175 0.179

0.031

0.036 0.040 0.045 0.050 0.054 0.059 0.063 0.068

0.087

0.092 0.096 0.101 0.106 0.111

0.116

0.991

0.989 0.987 0.986 0.984 0.982 0.980 0.978

0.122 0.125 0.127

0.151

0.130

0948

0.136

0.182

0.071

0.135

0.946

0.138

0.185

0.073

0.140 0.145 0.150

0.944 0.942 0.940

0.140 0.142

0.188

0.076 0.078

0.144

0.081

0.155

0.146

0.084 0.086 0.089 0.092 0.094

0.160 0.165

0938 0936

0.148 0.150

0.176 0.181

0230

0.097

0.186

0.080 0.082

0.100 0.102

0.191

0084

0.105 0.108

0.070 0.072 0.074 0.076 0.078

0.086 0.088

0.111

0.170

0.196 0.202 0.207 0.212

0.261

0336 0343 0349

0904 0902 0.900 0.898 0.896 0.893 0.891

0.889 0.887 0.884 0.882 0.880 0.877 0.875 0.873 0.870

0.868 0.865 0.863 0.861

0.355 0361 0267

0.858 0.856 0.853

0.195

0.373

0.851

0.198

0380

0.848

0.201

0.386

0393 0399

0.846 0.843

0.152

0.205 0.208 0.211 0.215

0.405

0.928

0.154 0.156

0.412

0.840 0.838 0.835

0926

0.158

0.218

0.419

0.833

0.924 0921 0919 0917 0915

0.160

0.222 0.225 0.229 0.232

0.425 0.432 0.439 0.446

0.830

0.934 0.932

0.162 0.164 0.166

0.191

0.827

0.824

0822

OtS%OI(

O EMA M4D COLUMN SECTIONS

13.2.2 Limits to use of singly reinforced beam sections Limitsto x/d as afunction ofthe amount of re-distributioncarried outare given in EC2. These can be re-written as For concrete grades

2.5.3.4.2

C35145

— 0.44

(xld). rn =

1.25

For concrete grades > C35145 (x/d).urn =

o

— 0.56

1.25

and ILIim for rectangular sectionsas a function

Equationscan be derived for

of (X1c01 . These are =

0•4533(X/d)im[1

—

O.4(XId)i]

= (X/d).Il .918

Wi.

Table 13.2 gives values of (XId)Jim' 'hum and WIim as a function of the amount of re-distributioncarriedout. EC2 states that plastic design, for exampleyield line analysis,canbe used where xld 0.25.Thelimits corresponding tothisvalue are also included in the table. Table 13.2 Limiting values %

(xId)

ô

redistribution

0 5

1.00

0.95 0.90 0.85 0.80 0.75 0.70

10 15 20 25 30

1im

k>35

k35

0.448 0.408 0.368

0.352

0328

0232

0.1667 0.1548 0.1423 0.1292

0.288 0.248 0.208

0.192 0.152

0.1155

0.112

0.0864

Plastic design

0.312

0.272

0.25

1013

0i,m

>5

Ck>35

137l

0.2336

0.1238

0.2127

0.1835 0.1627

0.1919

0.1418

aio

00954 00803 0.0647 0.0485

0.1020

0.1710

0.1210

0.1502 0.1293 0.1084

o.iooi 00792 0.0584

0.1303

13.3 Compression reinforcement Compression reinforcement is required amount can be calculated from C.,

=

in any section where'h >

'h'hIIm 0.87(1 —

d'/d)

where mechanical ratio of compression steel

=x=

bd

'ck

InnI

1iim The

2.5.3.5.5

DESIGN OP BEAM AND COLUMN SECTiONS

d'

= depth from compression face to centroidof compression reinforcement

A'8

= area of compression reinforcement

The area of tension reinforcement can now be obtained from W

Equations above for cd and

c are valid for d'Ix

1 —

1)805.

13.4 Flanged beams For beams with flanges on the compression side of the section, the formulae for rectangular sections may be applied provided

xld

hf/d

where hf

= thickness of the flange

For beams where the neutral axis lies below the flange, it will normally be sufficientlyaccurate to assume that the centre of compression is located at middepth of the flange. Thus, for singly reinforced beams, approximately

M

=

0.87

ASck(d

—

hI2)

The neutral axis depth is given approximately by

xld =

1.918 (blb)w

—

1:25 (b/b

—

1)h/d

where br is the rib width and the definition of o, is identical rectangular section.

to that for a

13.5 Symmetrically reinforced rectangular columns Figures 13.2(a) to (e) give non-dimensional design charts for symmetrically reinforcedcolumns where the reinforcementcanbeassumedto beconcentrated in the corners. The broken lines give values of '<2 in Eqn 4.73 of EC2. Where the reinforcement is not concentrated in the corners, a conservative approach is to calculate an effective value of d' as illustrated in Figure 13.

Eqn 4.73

DE51O O EMJI kND COLUMNSECTIONS

a In

U)

0 ci

CI)

E

0 0 0)

0 a) C'

C4

-

,-.

a

a

a

C

Q

CM

1

C)

1 0)

C) U-

Figure 13.2(a) Rectangular columns (d'/h =

0.05)

'0

a

II a

a

a

("4

a

a

DE5GN OP BEAM AND COLUMN SECTIONS

(t1

Q

0 0

(I,

C

E

0 0 0) C

0 '9 Q

_ c'J

1 cv,

.0

a) I-

0) Li-

Figure 13.2(b) Rectangular columns (d'Ih =

0.10)

1

a

C'1

a

-; a

DE.S(GM OF BE&M

KD COLUMN SECTIONS

LC)

0

CI,

C

E

0 C) 0) C c'j

C)

ci

C%1

z

0) I-

0)

U-

Figure 13.2(c) Rectangular columns (d'Ih =

0.15)

'I

DESiGN OF BEAM AND COLUMN SECTIONS

UI

6

UI

m

UI

6

a

UI

0 C" o

a

Cl)

E

0 0 I-

UI

8

Ca

C)

0 w c'j

1

C) a) I-

0)

U-

Figure 13.2(d) Rectangular columns (d'Ih

= 0.20) lnnI

U

C4

DESiGN OP BEAM AND COLUMN SECTIONS

'I

4-

U) c'J

d

a,

C

E

0 0 C',

C C',

0

'

a,

'-.

m '—

c'i

—

—

q

u' Q

—

'I

1C) C) LL

Figure 13.2(e) Rectangular columns (d'Ih

=

0.25)

tAI

L

-

Cl

DS1ON OF BEAM AND COLUMN SET1ONS

0

0 h12

7__of

Centroid

Figure 133

.

bars in

half section

.1

Method of assessing an effective value for d'

Worked examples

for the design of concrete bui'dings

Cl/SfB UDC

BRITISH OEMENT ASSOCIATIONPUBLICATION 43.505

6240124006.77