Worked Examples for the Design of Concrete Buildings - ARUP
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WORKED EXAMPLES
1116
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FOR THE DESIGN
OF CONCRETE
BUILDING
S
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Based on BSI publication DD ENV 1992-1-1:1992. Eurocode 2: Design of concrete structures. Part 1. General rules and rules for buildings.
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This book of worked examples has been prepared by:
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British Cement Association Ove Arup & Partners
and Partners
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S. B. Tietz &
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The work was monitored by the principal authors:
BE(Hons), MSc, DIC, CEng, FIStructE
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Narayanan (p-
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A. W. Beeby BSc, PhD, CEng, MICE, MIStructE, FACI Professor of Structural Design, Dept of Civil Engineering, University of Leeds (formerly Director of Design and Construction, British Cement Association), Partner, S. B. Tietz & Partners, Consulting Engineers, foil
and R. Whittle MA(Cantab), CEng, MICE Associate Director, Ove Arup & Partners,
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and edited by: A. J. Threlfall BEng, DIC Consultant (formerly a Principal Engineer at the British Cement Association).
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This publication was jointly funded by the British Cement Association and the Department of the Environment to promote and assist the use of DD ENV 1992-1-1 Eurocode 2: Part 1.
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S. B. Tietz &
Partners offer consultancy services
in civil,
structural and traffic engineering.
and prices for BCA publications can be obtained from Publication Sales, Centre for Concrete Information, Qom)
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Partners is an international firm offering a wide range of design and specialist services for the construction
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Ove Arup industry.
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The British Cement Association, BCA, is a research and information body dedicated to furthering the efficient and proper design and execution of concrete construction. Membership of BCNs Centre for Concrete Information is open to all involved in the construction process. BCA is funded by subscriptions from cement producers, through joint ventures, sales of publications, information and training courses, and the carrying out of research contracts. Full details are available from the Centre for Concrete Information, British Cement Association, Century House, Telford Avenue, Crowthorne, Berkshire RG11 6YS. Telephone (0344) 725700, Fax (0344) 727202.
at the above address. 43.505
Published by
British Cement Association
°J2
First published 1994 ISBN 0 7210 1446 1
Lma
Century House, Telford Avenue, Crowthorne, Berks RG11 6YS Telephone (0344) 762676
Price group M © British Cement Association 1994
Fax (0344) 761214 From 15 April 1995 the STD Code will be (01344)
50"
(0m
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odd
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0.9
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advice or information from the British Cement Association is intended for those who will evaluate the significance and limitations of its contents and take responsibility for its use and application. No liability (including that for negligence) for any loss resulting from such advice or information is accepted. Readers should note that all BCA publications are subject to revision from time to time and should therefore ensure that they are in possession of the latest version. All
FOR THE DESIGN
OF CONCRETE
BUILD
1
N G S
Based on BSI publication DD ENV 1992-1-1:1992. Eurocode 2: Design of concrete structures. Part 1. General rules and rules for buildings.
Published by the British Cement Association in conjunction with:
Ove Arup
&
Partners
Fitzroy Street London W1 P 6BQ 13
Tel: 071-636 1531
S.B. Tietz & Partners 14 Clerkenwell Close Clerkenwell London ECiR OPQ Tel: 071-490 5050
July 1994
The Department of the Environment 2 Marsham Street London SW1P 3EB Tel: 071-276 3000
FOREWORD
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Eurocode 2: Design of concrete structures, Part 1: General rules and rules for buildings (EC2)(') sets out both the principles for the design of all types of concrete structure, and design rules for buildings. Rules for other types of structure and particular areas of technology, including precast concrete elements and structures, will be covered in other parts of EC2.
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EC2 contains a considerable number of parameters for which only indicative values are given. The appropriate values for use in the UK are set out in the National Application Document (NAD)(') which has been drafted by BSI. The NAD also includes a number of amendments to the rules in EC2 where, in the draft for development stage of EC2, it was decided that 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
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be applied.
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Attention is drawn to Approved Document A (Structure) related to the Building Regulations 1991(3), which states that Eurocode 2, including the National Application Document, is considered to provide appropriate guidance for the design of concrete buildings in the United Kingdom.
Enquiries of a technical nature concerning these worked examples may be addressed to the authors directly, or through the BCA, or to the Building Research Establishment.
CONTENTS 1
INTRODUCTION AND SYMBOLS 5 Introduction 5 Symbols
................. ....................
8.1
SPECIAL DETAILS Corbels
8.2
Nibs
8
...................
8.3
180 ..................... 185 188 Simply supported ends
8.4
Surface reinforcement
2.1
COMPLETE DESIGN EXAMPLE 15 Introduction
2.2
Basic details of structure,
1.1
1.2
2
................
materials and loading
.........
3
BEAMS
3.1
CON-
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2.4 2.5 2.6 2.7 2.8 2.9
9
PRESTRESSED CONCRETE
15
9.1
17
9.2
20 30 34 39 43 49
9.3 9.4
Introduction Design data Serviceability limit state Ultimate limit state
9.5
Mi n
(fl
.................. ................. ..... ................... ................. ................. ..................
Floor slab Main beam Edge beam (interior span) Columns Foundation Shear wall Staircase
2.3
...... ........
............... ............... ...... ..........
.............. .....
10
SERVICEABILITY CHECKS BY CALCULATION
10.1
Deflection
3.4 3.5
Introduction 53 Design methods for shear 53 Shear resistance with concentrated loads close to support 63 70 Design method for torsion Slenderness limits 81
4
SLABS
12
4.1
Solid and ribbed slabs Flat slabs
5.2 L(7
5.3 5.4 5.5 5.6 5.7
COLUMNS Introduction Capacity check of a section by strain compatibility Biaxial bending capacity of a section Braced slender column Slender column with biaxial bending Classification of structure Sway structures
............... .......... .................. (D.
5.1
.................
......
.................. ..... ............
WALLS Introduction
6.2
Example
7
FOUNDATIONS Ground bearing footings Pilecap design
:-I
7.1
7.2
............... .................. .....
.............
82
12.1
109
12.2 12.3
132
12.4 132 12.5 137
............... ..................
LOAD COMBINATIONS Introduction Example 1 frame Example 2 continuous beam 1 Example 3 continuous beam 2 Example 4
-
............... ..................
-
....... .......
222 222
236 237
240 243
- tank ................... 245
141
13
143 147
13.1
151
13.2
154 154
13.3 13.4 13.5
158 172
DESIGN OF BEAM AND COLUMN SECTIONS Concrete grades 246 Singly reinforced rectangular beam sections 246 Compression reinforcement ... 248 Flanged beams 249 Symmetrically reinforced rectangular columns 249 -fl
6.1
6
11.1
208
COY)
5
.......
DEEP BEAMS Introduction 11.2 Example 11
207
.................. 219
10.2 Cracking
WOE
4.2
........ ..... ...........
.................
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3.3
.....
204
i
Reinforcement summary
................
193 193 195
mum and max mum areas of 207 reinforcement i
9.6
3.2
191
REFERENCES
........... .............. ............ ........ .................
256
INTRODUCTION AND SYMBOLS Introduction and symbols
°°6
1.1
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The main objective of this publication is to illustrate through worked examples how EC20) may be used in practice. It has been prepared for engineers who are generally familiar with design practice in the UK, particularly to BS 8110(2).
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The worked examples relate primarily to in-situ concrete building structures. The designs are in accordance with EC2: Part 1 as modified by the UK National Application Document'). Where necessary, the information given in EC2 has been supplemented by guidance taken from other documents.
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The core example, in Section 2, is a re-design of the in-situ concrete office block used in the BCA publication Designed and detailed (SS 8110: 1985), by Higgins & Rogers(). Other design aspects and forms of construction are fully explored by means of further examples in Sections 3 to 12.
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Equations and charts for the design of beam and column sections, taken from the Concise Eurocode for the design of concrete buildings(5), are given in Section 13. Publications used in the preparation of this book, and from which further information may be obtained, are listed in the References. Unless otherwise stated, all references to BS 8110 refer to Part 1.
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'.......
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been adopted in the preparation of this 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. Two conventions have
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The calculations are cross-referenced to the relevant clauses in EC2 and, where appropriate, to other documents; all references in the right-hand margins are to EC2 unless indicated otherwise.
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The symbols used throughout the publication are listed and defined below, and are generally those used in EC2 itself.
E
1.2 Symbols A
Area of cross-section
Ac
Area of concrete cross-section
Act
Area of concrete within tensile zone
Act,ext
Area of concrete tensile zone external to links
Ak
Area enclosed within centre-line of thin-walled section
AP
Area of prestressing tendons
As
Area of tension
A's
Area of compression reinforcement
As,min
Minimum area of tension or, in columns, total longitudinal reinforcement
As, prov
Area of tension reinforcement provided
As,feq
Area of tension reinforcement required
As,surf
Area of surface reinforcement
Ast
Area of transverse reinforcement within flange of beam
As/
Area of tension reinforcement effective at a section of additional longitudinal reinforcement
Asw
Area of shear reinforcement or torsion links
Asw,min
Minimum area of shear reinforcement
E.,eff
Effective modulus of elasticity of concrete
or, in
columns, total longitudinal reinforcement E
E
0
E
__0
or,
for torsion,
area
INTRODUCTION AND SYMBOLS
Er
Secant modulus
E ,,
Secant modulus of elasticity of concrete
ES
Modulus of elasticity of reinforcement or prestressing steel
F
Force due to concrete in compression at ultimate limit state
FS
Force in tension reinforcement or prestressing tendons at ultimate limit state
FSd
Design value of tie force in pilecap
Fsd,sup
Design value of support reaction
Ft
Tie force in corbel or due to accidental action
F
Vertical force applied to corbel or, for sway classification of structures, sum of all vertical loads under service conditions
Gk
Characteristic value of permanent action or dead load
Gkf
Characteristic dead floor load
Gkr
Characteristic dead roof load
H
Overall depth of tank
Hc
Horizontal force applied to corbel
I
Second moment of area of cross-section
It
Second moment
of
area of uncracked concrete section
Iu
Second moment
of
area of cracked concrete section
Ib
Second moment
of
Second moment
of
area of beam section area of concrete section
I..,
Second moment
of
slab
Second moment
of
of elasticity of
concrete at transfer
--E
E
E E E E
E
v J
..,
I
Second moment of Second moment of
area area area area
of column section of slab section of
section
in x
direction
of section in y direction
St Venant torsional stiffness of rectangular section
Jtot
St Venant torsional stiffness of total section
K
Deflection-curvaturefactor dependent upon the shape of the bending moment
diagram Reduction factor for calculation of second order eccentricity
K2
Coefficient taking account of decrease force
M
Bending moment
Mc
Moment of force, FC, about tension reinforcement
Mc,
Moment causing cracking
MCX
Moment of force, Nc, about x axis
MCy
Moment of force, N., about y axis
M.
First order moment
MRd
Design moment of resistance
MRd,c
Moment of force,
M/Rd,c
Moment of force, N'Rd,c' about mid-depth of section
MRd,s
Moment of force,
CD.
K,
NRd,c'
NRd,s'
in
curvature due to increasing axial
about mid-depth of section
about mid-depth of section
6
INTRODUCTION AND SYMBOLS
Design value of applied moment
Msdx
Design moment in x direction
MSdy
Design moment in y direction
MShc
First order moment at end
Msd2
First order moment at end 2
Msd,cs
Design moment in column strip
MSd,ms
Design moment in middle strip
Mspan
Moment in span
M
Moment at support
(1)
SUP
On.
MSd
1
Maximum moment transfer value
MX
Moment about x axis
M
Moment about y axis
E
Mt",,.
y
N
Axial force
N
Axial force due to concrete in compression
NRd
Design resistance to axial force
NRd,c
Design resistance to axial force due to concrete
N'Rd,c
Design resistance to axial force due to concrete of hypothetical section of depth x > h
Design resistance to axial force due to reinforcement Design value of applied axial force
Mean applied axial force
Prestressing force or point load Average prestressing force along tendon profile Initial prestressing force at transfer
Mean effective prestressing force at time t Final prestressing force after all losses
Maximum initial prestressing force at active end of tendon
Required prestressing force Final prestressing force at service
Characteristic value of variable action or imposed load Characteristic value of imposed floor load Characteristic value of imposed roof load Reaction at support A
Reaction at support B First moment of area of reinforcement about centroid of section
First moment of area of reinforcement about centroid of uncracked section First moment of area of reinforcement about centroid of cracked section
Design value of tensile force in longitudinal reinforcement Maximum torsional moment resisted by concrete struts Maximum torsional moment resisted by reinforcement
INTRODUCTION AND SYMBOLS
Design value of applied torsional moment
Tsd,e
Torsional moment applied to flange
Tsd,tot
Total applied torsional moment
Tsd
Torsional moment applied to web
_`L
Tsd
W
Shear force at support A
VB
Shear force at support B
Vod
Design shear resistance provided by concrete
Vex,
Shear force at exterior support
Vnt
Shear force at interior support
VRdl
Design shear resistance of member without shear reinforcement
VRd2
Maximum design shear force to avoid crushing of notional concrete struts
VRd3
Design shear resistance of member with shear reinforcement
Vsd
Design value of applied shear force
Vsdx
Design shear force in x direction
Vsdy
Design shear force in y direction
Vsd,max
Maximum design shear force
V,d
Design shear resistance provided by shear reinforcement
Wb
Section modulus at bottom fibre
Wcp
Section modulus at centroid of tendons
Wk
Characteristic value of wind load
Wt
Section modulus at top fibre
al
COD
0-.
a
COD
VA
Distance or deflection or maximum drape of tendon profile Deflection based on uncracked section
all
Deflection based on cracked section
ac
Distance of load from face of support (corbel) or from centre-line of hanger bars (nib)
a,
Distance from face of support to effective centre of bearing COD
Total deflection
Distance between positions of zero and maximum bending COD
av
Horizontal displacement of the envelope line of tensile force
COD
ato,
0-_
al
COD
Deflection due to concrete shrinkage COD
acs
ax
Deflection at distance x along span
aj,a2
Values of ai at ends of span
b
Width of section or flange width or lateral cover in plane of lap
bav
Average width of trapezoidal compression zone
be
Width of effective moment transfer strip
betf
Effective width of flange
bmin
Minimum width of support beam
br
Width of rib
bsup
Width of support
INTRODUCTION AND SYMBOLS
bt
Mean width of section over the tension zone
bW
Minimum width of section over the effective depth
c
Cover to longitudinal torsion reinforcement
ct,c2
Support widths at ends of beam
d
Effective depth of section
d'
Depth to compression reinforcement
day
Average effective depth for both directions
db
Depth to bar considered Distance of critical section for punching shear from centroid of column
df
Effective depth of flange
dm
Effective depth for punching shear check in column head
dm.
Maximum effective depth for both directions
dmin
Minimum effective depth for both directions
'PL
c"'-
dcnt
dX
Effective depth in x direction
dy
Effective depth in y direction
d,
Effective depth to bars in layer
d2
Effective depth to bars in layer 2
ea
Additional eccentricity due to geometrical imperfections
eay
Additional eccentricity in the y direction
eaz
Additional eccentricity in the z direction
ee
Equivalent eccentricity at critical section
Q1)
eoy
1
First order eccentricity in y direction
eot,eo2
First order eccentricities at ends of column
etot
Total eccentricity
ey
Eccentricity in y direction
eZ
Eccentricity in z direction
e2
Second order eccentricity
e2y
Second order eccentricity in y direction
e2,
Second order eccentricity in z direction
fb
Stress in concrete at bottom fibre
fbd
Design value of ultimate bond stress
cd
Design cylinder strength of concrete
f
Cube strength of concrete at transfer
t
k
ct,eff
f tm 1
Characteristic cylinder strength of concrete
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
fpd
Design tensile strength of prestressing steel
fk
Characteristic tensile strength of prestressing steel
INTRODUCTION AND SYMBOLS
Design value of ultimate bearing stress Stress in reinforcement Stress in concrete at top fibre
Design yield strength of reinforcement
Characteristic yield strength of reinforcement Design yield strength of longitudinal torsion reinforcement
Design yield strength of shear reinforcement or torsion links 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 with biaxial eccentricities
of column section
ha
Active height of deep beam
hC
Overall depth of corbel at face of support
hf
Overall depth of flange
hH
Depth of column head
hmax
Larger dimension of rectangular section
hmin
Smaller dimension of rectangular section
htot
Total height of structure in metres
i
Radius of gyration of section
k
Coefficient or factor
kA
Restraint coefficient at end A
kB
Restraint coefficient at end B
kbottom
Restraint coefficient at bottom
ko
Minimum reinforcement coefficient associated with stress distribution
ktop
Restraint coefficient at top
ki
Crack spacing coefficient associated with bond characteristics
k2
Crack spacing coefficient associated with strain distribution
l
Length or span
if
Length of tendon over which anchorage slip is taken up
Ib
Basic anchorage length
Ib,min
Minimum anchorage length
Ib,net
Required anchorage length
IC
Diameter of circular column
Icon
Height of column between centres of restraints
left
Effective span
leff,slab
Effective span of slab
lH
Distance from column face to edge of column head
In
Clear distance between faces of support
10
INTRODUCTION AND SYMBOLS
to
Distance between positions of zero bending or effective height of column or, for deep beams, clear distance between faces of support
lot
Length of compression flange between lateral supports
is
Required lap length or floor to ceiling height in metres
ls,min
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 COD
u0)
-f6
It
Effective span in x direction
ly
Effective span in y direction
1,,12
Lengths between centres of supports or overall dimensions of rectangular column head ...
c(0
lX
mSd
Minimum design moment per unit width
n
Ultimate design load per unit area or number of tendons or number of
sub-divisions
q
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
rI
Radius of curvature based on uncracked section
rH
Radius of curvature based on cracked section
r,
'23
Radius of curvature due to concrete shrinkage
r.I
Radius of curvature due to concrete shrinkage based on uncracked section
rcsu
Radius of curvature due to concrete shrinkage based on cracked section
riot
Total radius of curvature
s
Spacing of shear reinforcement or torsion links or horizontal length of tendon profile
sf
Spacing
Smax
Maximum spacing of shear reinforcement or torsion links
Srm
Average final crack spacing
t
Thickness of supporting element or wall of thin-walled section
train
Minimum thickness of wall
u
Circumference of concrete section or critical section for punching shear
uk
Circumference of area Ak
VRdl
Design shear resistance per unit length 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 shear reinforcement
VSd
Design value of shear force per unit length of critical perimeter
w
Support width or quasi-permanent load per unit length
wk
Design crack width
(DD
'-.
r-.
Average loss of prestressing force per unit length due to friction
transverse reinforcement within flange of beam
0)8
of critical perimeter, for
slab without
of critical perimeter, for CAD
of
0
n""
p'
slab with
INTRODUCTION AND SYMBOLS
E
wmin
Minimum width of support
Neutral axis depth or distance along span from face along tendon or column dimension in x direction
x'
Maximum depth of concrete in compression in direction of minor axis for column section with biaxial eccentricities
xc
Depth of concrete in compression at position of minor axis for column section with biaxial eccentricities
y
Drape of tendon at distance x along profile or column dimension direction
yt
Distance from centroid of uncracked section to extreme tension fibre
z
Lever arm of internal forces
ZCP
Distance from centroid of section to centroid of tendons
a
Reduction factor for concrete compressive stress or modular ratio or deformation parameter
aI
Value of parameter based on uncracked section
au
Value of parameter based on cracked section
as
Effectiveness coefficient for anchorage
ae
'C$
Effective modular ratio
an
Reduction coefficient for assumed inclination of structure due to
of
support or distance
07Q
5'3
x
in y
'-'
0
imperfections aWasy
Moment coefficients in x and y directions
a,
Effectiveness coefficient for lap
a
Coefficient with several applications including shear resistance enhancement, effective height of column, St Venant torsional stiffness, punching shear magnification, design crack width
ared
Reduced value
a,
Coefficient associated with bond characteristics
a2
Coefficient associated with duration of load
-y'
Partial safety factor for concrete material properties
yF
Partial safety factor for actions
TG
Partial safety factor for permanent action or dead load
yG,inf
Partial safety factor for permanent action, in calculating lower design value
yG,sup
Partial safety factor for permanent action, in calculating upper design value
yP
Partial safety factor for actions associated with prestressing force
To
Partial safety factor for variable action or imposed load
'YS
Partial safety factor for steel material properties of reinforcement or
S
Ratio of redistributed moment to moment before redistribution
Eb
Strain in concrete at bottom of section
ECS
Basic concrete shrinkage strain
EeS-
Final concrete shrinkage strain
EP
u°'
Minimum strain in tendons to achieve design tensile strength
EPm
Strain in tendons corresponding to prestressing force
shear resistance enhancement coefficient
o
CC))
of
prestressing tendons
Pm,t
INTRODUCTION AND SYMBOLS
Strain in reinforcement
ES
es(t,t)
Estimated concrete shrinkage strain
esm
Mean strain in reinforcement allowing for tension stiffening effect of concrete
Ultimate compressive strain in concrete
EU
Eyd
Initial yield strain in reinforcement
Distribution coefficient
Moment coefficient
6
Angle of rotation or angle between concrete struts and longitudinal axis
X
Slenderness ratio
Xcrit
Critical slenderness ratio
Xm
Mean slenderness ratio of all columns in storey considered
Xmin
Slenderness ratio beyond which column is considered slender
µ
Coefficient of friction between tendon and duct or applied moment ratio
µlim
Limiting value of applied moment ratio for singly reinforced section
P
Efficiency factor or assumed inclination of structure due to imperfections
vred
Reduced value of assumed inclination
(OD
00..
n
ate)
of
structure
Longitudinal force coefficient
P
Tension reinforcement ratio or density of liquid
P'
Compression reinforcement ratio
PI
Longitudinal tension reinforcement ratio
PIX'PIy
Longitudinal tension reinforcement ratios in x and y directions
Pr
Effective reinforcement ratio
Pw
Shear reinforcement ratio
Pw,min
Minimum shear reinforcement ratio
P1,P2
Principal and secondary reinforcement ratios in solid slabs
(n.
YU
b0)
UC9
Stress in concrete adjacent to tendons due to self-weight and any other permanent actions
app
Average stress in concrete due to axial force
acpo
Initial stress in concrete adjacent to tendons due to prestress
upo
Initial
as
Stress
asr
Value of as under loading conditions causing first cracking
7'
Basic design shear strength
TRd
(On
stress in tendons immediately after stressing (pre-tensioning) or immediately after transfer (post-tensioning) in
tension reinforcement calculated on basis of cracked section
Factor defining representative value of variable action ,yo
Value of
for rare load combination
0t
Value of
for frequent loading
02
Value of 0 for quasi-permanent loading
W
Mechanical ratio of tension reinforcement
W'
Mechanical ratio of compression reinforcement
INTRODUCTION AND SYMBOLS
Limiting value of w for singly reinforced section
E.
Total vertical force
A,/
Anchorage slip or wedge set
AFd
Variation of longitudinal force in section of flange over distance
OHM
Equivalent horizontal force acting on frame at floor imperfections
c,)
wl.m
applied to frame at floor
j
o-°
j
a,
due to assumed
-_p
about mid-depth of section
Moment of force
AMSd
Reduction in design moment at support
ANRd,c
Design resistance to axial force due to concrete in area of hypothetical section lying outside actual section
APC
Average loss of prestressing force due to elastic deformation of concrete
APst
Loss of prestressing force at active end of tendon due to anchorage slip
APt(t)
Loss of prestressing force due to creep, shrinkage and relaxation at time t
ANRd,c
DTT
((DD
cc-
tea)
AMRd,c
APA(x)
Loss of prestressing force due to friction between tendon and duct at distance x from active end of tendon
0o pr
Variation of stress in tendon due to relaxation Bar size or duct diameter or creep coefficient
0(t,to)
Creep coefficient, defining creep between times t and
deformation at 28 days
0(-,td
Final creep coefficient
1
to,
related to elastic
COMPLETE DESIGN EXAMPLE 2.1
Introduction
("O
(CA
-o:(D
CAD
E
E
Note: Reinforcement areas differ somewhat from those given by BS 8110 which permits design for the single load case of maximum load on all spans combined with 20% redistribution. EC2 requires alternate and adjacent spans to be considered. 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 ductility class H (as defined in prEN 10080(8)) reinforcement would need to be specified.
NAD
Table 5
2.3.6 Shear Shear resistance
the slab without shear reinforcement is given by
of
+ 40pl) bwd
=
TRdk (1.2
TRd
=
0.35 N/mm2
k
= 1.6
VRd1
4.3.2.3
Eqn 4.18
where E
Table 4.8
-
d =
1.6
565 Pi
=
0.149 = 1.451 0. 0038
1000 x 149
Hence VRd1
=
102.3 kN/m
> Vsd = 33.9 kN/m
...... .........
OK
No shear reinforcement required
COD
Note: Since shear is rarely a problem for normally loaded solid slabs supported on beams, as the calculation has shown, it is not usually necessary to check in
these instances.
COMPLETE DESIGN EXAMPLE
2.3.7 Deflection 377
=
Reinforcement ratio provided in span
=
0.0025
1000 x 149
NAD
Table 7
(OD
_.a
'2O
Using NAD Table 70) and interpolating between 48 for 0.15% and 35 for 0.5%, a basic span/effective depth ratio of 44 is given. By modifying according to the steel stress, the ratio becomes
4.4.3.2(4)
44 (400 x 377)
=
42.2
460 x 342
The actual spanleffective depth ratio is
5000
=
...........
33.6
149
OK
Had EC2 Table 4.14 been used instead of NAD Table 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
=
3 N/mm2
kC
=
0.4
k
=
0.8
Act
=
0.5
x 175
x
=
1000
87500 mm2 E
fot,ett
4.4.2.2
Hence kckfct ettA ct/as
=
0.4
x 3 x 87500/460
0.8
x
x
x
Eq n 4.78
Area of reinforcement provided
Maximum bar spacing
=
3h
s
..................
=
200 mm
175
500 mm
(1
=
-
1684 mm2
yk/805)
=
0.0943
0.546
0.0943 x 300 x 440 x 32/460 E
Use 4T25 (1960 mm2) top Use 2T25 (982 mm2) bottom
=
866 mm2 E
S
+ w' = 0.1084 + 0.0750 = 0.1834 (Section 13)
0.1834 x 300 x 440 x 32/460
d'lx =
A'
(1
ado
=
0.87
- 0.0864 - 50/440)
0.0750 (Section 13)
bd ,k AS
(1- d'/d)
0.87
0.1442
+
=
w
-
A-N1im
+
-
w
0.429
COMPLETE DESIGN EXAMPLE
2.4.4.2 Near middle of 8 m span From bending moment envelope
M
=
325 kNm
=
300 + 0.2 x 0.85 x 8000
=
1660 mm
E E
Effective flange width
2.5.2.2.1
Eqn 2.13 µ
325 x 106
=
=
0.030
1660 x 4502 x 32
x/d =
0.068 (Section 13, Table 13.1)
Neutral axis is in flange since x
=
31
<
=
0.035 (Section 13, Table 13.1)
AS
=
0.035 x 1660 x 450
(Y)
x
32/460
=
1819 mm2
E
w
175 mm
E
Use 4T25 (1960 mm2)
2.4.4.3 Left-hand end of 8 m span From bending moment envelope
M
=
126 kNm
S
=
0.7 and
It
_
126
300
=
µ,.m
x
x 4402
0.0864 (Section 13, Table 13.2) 106
=
x 32
<
0.0678
µlim
Therefore no compression reinforcement is required.
W
= 0.084 (Section 13, Table 13.1)
AS
=
0.084 x 300 x 440 x 32/460
=
772 mm2
Using 2T25 bent-up bars, minimum diameter of mandrel
5.2.1.2 NAD
=
130 (As,regIA,.prov)
=
100
Use 2T25 (982 mm2) with r = 50
24
Table 8
COMPLETE DESIGN EXAMPLE
2.4.4.4 Right-hand end of 6 m span From bending moment envelope
M
=
A
_
76kNm M
76 x 106
bd2 ck
300 x 4402 x 32
=
0.041
= 0.049 (Section 13, Table 13.1)
W
AS =
450 mm2
Use 2T25 (982 mm2) with r
4¢ minimum
E
=
2.4.4.5 Near middle of 6 m span From bending moment envelope
M
=
138 kNm
300 + 0.2 x
138
u
32
x
x
106
x
4502
0.85
=
x 6000 = 1320 mm E
=
Effective flange width
0.0161
1320
0.019 (Section 13, Table 13.1)
W
=
As
= 0.019
x
1320
x
450
x 32/460 =
785 mm2
Use 2T25 (982 mm 2)
2.4.4.6 Minimum reinforcement 07c
As
kc kfct,eff A
ct
4.4.2.2
/a s
Eq n 4.78
where =
0.4
k
=
0.68
ct.eff
=
3 N/mm2
Act
= 300
or
=
460 N/mm2
>_
173 mm2
4
0.0015 btd
S
x 325 mm2 E E
CY)
kC
AS 0.6b d t
fYk
_o'
Therefore
....................................
OK
.......................
OK
=
203 mm2
5.4.2.1.1(1)
COMPLETE DESIGN EXAMPLE
2.4.5 Shear reinforcement
4.3.2
2.4.5.1 Minimum links Here, for comparison with BS 8110 design, grade 250 reinforcement will be
5.4.2.2
used. Interpolation from EC2 Table 5.5 gives
Minimum =
0.0022
A3W Is
=
0.0022
<
smax
=
(
5)
VRd2
-
=
300
0.66 mm2/mm
refer to Section 2.4.5.3 for
lesser of 300 mm or 0.8d
=
0
If Vsd
x
E
pW
VRd2
300 mm
Use R12 links @ 300 mm crs. (AsW/s
=
Eqn 5.17
0.75 mm2/mm)
2.4.5.2 Capacity of section without shear reinforcement =
7Rdk(1.2
+ 40pl) bald
+
VRd,
4.3.2.3
Assume 2T25 effective pi
= 982/(300
x
440)
k
=
1.6-d
=
1.6-0.44
TRd
k¢¢
=
0.35
VRdl
= 300 x 440 x 0.35 x 1.16 x (1.2 + 40 x 0.00743)
=
0.00743
=
1.16
Table 4.8
x 10-3 = 80.2
kN
.R+
2.4.5.3 Shear reinforcement by standard method
4.3.2.4.3
Maximum capacity of section v
=
0.7
-
ck/200
=
0.7
-
32/200
=
0.54
4
0.5
Eq n 4.21
x 0.9 x 440 x 10-3 = 684 kN
Eqn 4.25
Design shear force is shear at a distance d from the face of the support. This is 590 mm from the support centreline.
4.3.2.2(10)
0.5 x 0.54 x (32/1.5) x 300
x
VRd2
A SW s
Vsd
0.9
x 440
x
80.2 0.87
x 250
= 0.0116 (Vsd
CC)
Design of shear reinforcement is summarized in Table 2.3.
-
Eqn 4.23 80.2)
COMPLETE DESIGN EXAMPLE
Design of shear reinforcement
Table 2.3 Location
m
223
8
Vsa
A.Is
203
1.42
248
1.95
202
1.41
128
min.
s for 12 mm links
Links
159 116
R12 @ 150 R12 @ 100
160
R12 @ 150 R12 @ 300
span
LH end RH end
S23
m span LH end RH end 6
max.
Minimum
R12 @ 300
2.4.6 Deflection Reinforcement percentage at centre of 8 m span
=
x
100
1960/(450
x
1660)
=
4.4.3.2
0.26%
-Ca
0
Interpolating between 0.15 and 0.5%, basic span/effective depth ratio for end span = 40 E
To modify for
NAD
Table 7
steel stress multiply by 400/460
modify for T section multiply by 0.8
To
modify for span > 7 m multiply by
0
To
Q0)
0
0 E
8000/450
=
Actual ratio
E
=
a`)
Therefore permissible ratio =
718
40 x (400/460) x 0.8 x 7/8
=
24.3
...........................
17.8
OK
2.4.7 Cracking For estimate of steel stress under quasi-permanent loads Ultimate load
= 64.8 kN/m
Assuming 02
=
Quasi-permanent load
=
4.4.2.2
NAD
0.3 0.3
x 20 + 25.8 =
(nn
Approx. steel stress at midspan =
460 1.15
x
Table
31.8 kN/m
31.8
=
1
196 N/mm2
64.8
Approx. steel stress at supports allowing for 30% redistribution =
196/0.7
=
280 N/mm2 E
3C.
These are conservative figures since they do not allow for 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. °01
TOO
E
From EC2 Table 4.11, 25 mm bars in spans are satisfactory at any spacing since steel stress < 200 N/mm2 OK
...................................
From EC2 Table 4.12, bar spacing at supports should be
X
=
N-S direction
The slenderness in the E-W direction will be found to be approximately the
same.
0-0
The structure is braced and non-sway (by inspection), hence the Model Column Method may be used with the column designed as an isolated column. =
Xorc
25(2
-
eo1/eo2)
=
50 in both E-W and N-S directions
4.3.5.5.3 Eq n 4.62
Slenderness ratios in both directions are less than Xcr.t, hence it is only necessary to ensure that the column can withstand an end moment of at least =
1801
x 0.3/20 =
27.0 kNm
(Y)
NSdh/20
4.3.5.5.3
Eqn 4.64 This exceeds the first order moments. Hence NSd = 1801 kN and MSd = 27.0 kNm
NSd
0.62
bh ,k Msd
bh 2f
_
27.0
x
3003 x
ck
Assume
=
45 mm E
d'
106
32
= 0.031
COMPLETE DESIGN EXAMPLE
Then
d'lh
=
45/300 = 0.15
=
0.16 (Section 13, Figure 13.2(c))
=
1002 mm2
AS yk
bh ck
Hence AS
E
Use 4T20 (1260 mm2)
Note: In the design by Higgins and Rogers, the slenderness ratio exceeds the equivalent of Xcrc but the design moment is still Nh120. EC2 requires less
reinforcement due to the smaller design load and the assumption of a smaller a1)
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.
Loading and moments for external column 'L7
Roof Main Edge Self-weight
1
2
184
186
55
55
2nd floor Main Edge Self-weight
240
55
55
240
55
55
104
107
9
145 55 9
41
209
209
114
126
126
55 9
55 9
39
41
109
145 55
148
155
399
399
109
114
126
126
55 9
55 9
257
269
589
589
108
113
125
125
55 9
93
93
(t)
Foundations
233
238
55
55
365
382
778
55 9
778
68
1
2
93
98
93
98
103
109
98
98
N..
235
2
1
(f)
1st floor Main Edge Self-weight
235
1
2
c°0
3rd floor Main Edge Self-weight
Bottom
2
1
39
Top
(f)
Load case
Dead
E
Imposed
Total
Column moments (kNm)
(kN)
(D00
Column design loads
Beam loads (kN)
CIO
Table 2.6
72
COMPLETE DESIGN EXAMPLE
2.6.5.2 Design for column between first floor and foundation 675 x 106 x
ktop
0.5
675 x 106
3125 x 106
3500
8000
4000
kbottom
=
00
=
0.85
= 0.71
Hence a
x
Slenderness ratio = l°/i =
4000 = 3400 mm E
Effective height = 0.85
Figure 4.27
3400
12
= 39.3
300
be less than 25
v° will be small so
Hence =
xmin .
> 25, therefore column E
Calculate
L()
X
25
is
slender in N-S direction
Xcrt
bottom moment
e,,,
top moment
eo2
0
_
= 0
85
Hence Xcrit
=
25 (2 + 0) = 50
Slenderness ratios in the E-W and N-S directions are both less than 50, hence it is only necessary to ensure that the end moment is at least NW20. The worst condition occurs with load case 2 at section just above the first floor, where MSd is greatest. =
589 + 0.8
Nh
804 x 0.3
20
20
x 269 = 804 kN 0))
Nsd
=
12.0 kNm
Design end moment = 109 > 12 kNm
Hence
NSd
= 804 kN and MSd = 109 kNm
4.3.5.5.3
COMPLETE DESIGN EXAMPLE
2.6.6 Reinforcement details Maximum spacing of links for internal column
5.4.1.2.2(3)
x 20 = 240 mm 0.67 x 240 = 160 mm
Generally
NAD
12
Table 3
E
Above and below floor
c>)
Maximum spacing of links for external column
x 25
Generally
12
At lap and below floor
0.67
=
5.4.1.2.2(4)
300 mm
x 300 = 200 mm
The reinforcement details are shown in Figure 2.11. LL.
INTERNAL COLUMN
Vertical bars
COLUMN
Vertical bars
Links
Section 1"1
Links
EXTERNAL
F2
F1
Section
f
O R8-6
150
o
350
m
@
4
^
1st.
200
_7 o 7 N
1
4T20-1
n
300
'
1
4
W
_
o
_
4
rn
N F
300
18(0)
o 1
4T25-4
i
a
'r
o
4
E 0 b
°o
o
F
4
Edge beam
2()
300
@
O
1
OOE
225
-°YI
4l
5
m
°
1
links=
.--
Cover to
0 0 @
1
30
4
Fdn'o
s
g
300
2-14R8
s sc
f
St arters, see Fig. 2.13
Cover to links
_
40
+f
0
N T Fdn.
4
m
in
W
N
,
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
E
4.1.3.3(9)
a nominal cover of not less than 40 mm against blinding. EC2 specifies a minimum cover greater than 40 mm. This implies a nominal cover greater than 45 mm, hence the choice of 50 mm. °-.
C?.
te)
BS 8110 specifies
2.7.2 Loading Taken from internal column design.
Ultimate design loads:
Dead
=
1226
Imposed
=
575
Total
=
1801 kN
COMPLETE DESIGN EXAMPLE
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 the dead weight of the base less the weight of soil displaced is 10 kN/m2 over the area of the base.
2.7.3 Size of base Since, at the time of publication, EC7: Geotechnical design(s) and EC2, Part 3: Concrete foundations0o) have not been finalized, the approach used here is based on current UK practice. Use 2.75 m x 2.75 m x 0.6 m deep pad
Bearing pressure under service loads _
1291
=
+ 10
< 200 kN/m2 .................... OK
181
2.752 1801
=
238 kN/m2 E
=
Design pressure at ultimate limit state
2.752
2.7.4 Flexural reinforcement
-
Q))
CAD
-
50
106
x
2.75
1.2252/2
=
491 kNm
25 = 525 mm
= 0.020
2750 x 5252 x 32
x
bd2fck
ASyk
x
491
Msd
x
x
Average effective depth = 600
(0)
238
E
Moment at face of column =
=
0.023 (Section 13, Table 13.1)
=
0.023
bd>ck
Hence
x
2750
x
x
AS
525
x
32/460 = 2310 mm2
E
E E
(Y)
Use 9T20 @ 300 mm crs. each way (2830 mm2)
2.7.5 Shear 2.7.5.1 Shear across base
Shear force may be calculated at a critical section distance d from the face of the column. (VSd)
=
238 x 2.75 x
x
Design shear
(2.75
2
0
0.3)
_
0.525
= 458 kN
4.3.2.2(10)
COMPLETE DESIGN EXAMPLE
calculating VRd1, the influence of the reinforcement will be ignored since, if straight bars are used, they will not extend d + Ib,net beyond the critical section. In
4.3.2.3(1)
(1)
VSd,
x 2750 x 525/1000
Eqn 4.18
652 kN
=
x
1.2
x
x
x
hence no requirement for shear reinforcement c
>
VRd,
0.35 x 1.075
=
VRdl
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
[
4 x 300 +
7r
(2
x
1.5
x 525) ] /1000 =
6.15 m
a))
Figure 2.12 Critical perimeter for punching =
0.35 x 1.075 x 1.2 x 525 x 6.15
((7
VRdl
Area within perimeter = 2.98 m2 Design shear
VSd
(VSd)
_ (7.56
-
=
1458 kN
4.3.4.5.1
Area of base = 7.56 m2
2.98) x 238
1090 kN
=
4.3.4.1(5)
hence no requirement for shear reinforcement C VRdl,
2.7.6 Cracking Approximate steel stress under quasi-permanent loads 460 1.15
X
(908 + 0.3 x 383)
x
2310
= 186 N/mm2
2830
1801
From EC2 Table 4.11 bar size should not exceed 25
> 20 mm used.
Hence cracking ...........................................
41
OK
4.4.2.3 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.
c T 9T20-
I
O AA
2
2
2
2
1
-300
B2
3
DA
9T20- -300 1
81
PLAN
rCo
1
Fdn.
I
e-2
Cover =40
I
2R8-3-300
ACOVER
Figure 2.13
-
B1
=
50,
A
end =75
Base reinforcement details
Table 2.7 Commentary on bar arrangement Bar marks
x 20 = 640 mm E
Anchorage length = 32
_-0
Straight bars extend full width of base less end cover of 75 mm. Bars should extend an anchorage length beyond the column face E
1
Notes
4.1.3.3(9) 5.2.3.4.1
°-'
Actual extension = 1150 mm
75 = 715 mm
E
3
5.2.4.1.3 +
+
Column starter bars wired to bottom mat Minimum projection above top of base is a compression lap + kicker = 32 x 20
11o
2
Links are provided to stabilize and locate the starters during construction
COMPLETE DESIGN EXAMPLE
2.8 Shear wall 2.8.1
Structure
-
The structure is shown in Figure 2.14. 1
st floor
0.5 x wind load on building 4000
250
14300
i
T
_,
Figure 2.14 Shear wall structure
2.8.2 Loading at foundation level Dead load from first to third floors and roof x
0.175 x 24 x 15.5 = 65.1 kN/m x
x
Self-weight =
0.5 (3 x 23.5 + 28.5) = 49.5 kN/m CJ)
=
Characteristic dead load =
49.5 + 65.1 = 114.6 kN/m
Characteristic imposed load from slabs
x
4)
x 0.7 = 23.6 kN/m
Wind loading is taken as 90% of value obtained from CP3: Ch
N-S direction
Part
201).
NAD 4(c)
= 0.9 x 449 = 404 kN
z
Total wind load on building in
V:
z
2.5 (1.5 + 3 (J1
=
Wind load on wall = 404/2 = 202 kN
Moment in plane of wall = 202 x 8 = 1616 kNm Hence Maximum force per unit length due to wind moment M x 6
P
-
+
x 6 =
1616
+ 47.4 kN/m
14.22
^O,
2.8.3 Vertical design load intensities at ultimate limit state Dead load + imposed load x
114.6 + 1.5
x 23.6 =
x
1.35
190.1 kN/m
E
=
Eqn 2.8(a)
Dead load + wind load 114.6 + 1.5
x 47.4 = 225.8 kN/m; or
x
x
+
1.35
1.0x114.6-1.5x47.4
=
43.5kN/m
Eqn 2.8(a)
COMPLETE DESIGN EXAMPLE
Dead load + wind load + imposed load =
1.35x
=
250.6 kN/m or 122.6 kN/m
114.6 + 1.35
x 23.6 ± 1.35 x 47.4
Eq n 2.8(b)
NAD 4(c)
Therefore maximum design load = 250.6 kN/m E
E
From analysis of slab (not presented), maximum moment perpendicular to plane of wall = 11.65 kNm/m
2.8.4 Slenderness ratio 0.5 4
kA
+
1
1
3.5
=
2.05
Eqn 4.60
5
=
00
a
=
0.94
10
=
01c0l
kB
Hence a))
_
Figure 4.27 0.94 x 4 = 3.76 m
=
3.76x1000x
12
=
74.4
175
Hence wall is slender
2.8.5 Vertical reinforcement Higgins and Rogers design the shear wall as unreinforced. Plain concrete walls will be covered in EC2 Part 1A which, at the time of publication, has not yet been finalized. The wall will, therefore, be designed here as a reinforced wall. As will be seen, the result is the same. Eccentricity due to applied loads eo1
=
0
eo2
=
11.65
=
0.6
x 1000/250.6 = 46.5 mm
Hence ee
x
46.5+0=27.9
mm
Eqn 4.66
Accidental eccentricity ea
_
1
200
x
3760 2
=
9.4 mm
Eq n 4.61
COMPLETE DESIGN EXAMPLE
Second order eccentricity
3760z
x 2
x
z
10
=
L!7
Assuming K2
=
460
x 1.15
x
x 200000
x
x
_
e
1
0.9
x
122
Kz
Eqns 4.72 & 4.69
51.5K2 1
Design eccentricity
27.9 + 9.4 + 51.5 = 88.8 mm
=
Design ultimate load = 250.6 kN/m
Design ultimate moment = 88.8 M
x
250.6/1000 = 22.3 kNm/m
0.023
bh2fok
N
=
0.045
bhf,k Asfyk
0.01
(Section 13; Figure 13.2(d))
bhfck
Hence 122 mm2/m or 61 mm2/m in each face
E
=
E
As E
Minimum area of reinforcement 0.004 x 1000 x 175 = 700 mm2/m
5.4.7.2
E
=
This exceeds the calculated value. Hence the minimum governs. E
E
E
Use T12 @ 300 mm crs. in each face (754 mm2/m)
2.8.6 Shear (=D
Design horizontal shear = 1.5 x 202 = 303 kN
x
303 x 1000
= 0.12 N/mm2 E
Shear stress =
14300 x 175
..................
OK
Note:
min
not calculated since it must be > 0.12bw d by quick inspection of VRdt is EC2 Eqn 4.18.
2.8.7 Horizontal reinforcement Minimum at 50% of vertical reinforcement provided As
=
5.4.7.3
188 mm2/m (E F)
Minimum for controlled cracking due to restraint of early thermal contraction
45
4.4.2.2
COMPLETE DESIGN EXAMPLE
Eqn 4.78
As
=
kckf
kC
=
1.0
k
=
0.8
ct,eft
=
1.9 N/mm2 (assuming
as
=
AS
= 1.0 x 0.8 x 1.9 x 175 x 1000/360 = 739 mm2/m
t.ettAct/QS
concrete strength to be equivalent to
Table
3.1
C16/20 at time of cracking) 360 N/mm2 (assuming 10 mm bars)
Table 4.11
Use T10 @ 200 mm crs. in each face (785 mm2/m)
2.8.8 Tie provisions at first floor
NAD 65(g)
According to the NAD, these should follow the rules in BS 8110. Ft
2.8.8.1 Peripheral
As
=
BS 8110 3.12.3
36 kN
tie =
36x103_78mm2 460
Use 1T10 (78.5 mm2)
2.8.8.2 Internal tie force Force =
2.5 x 36 (4.7 + 4.0) x
14.3
_
299 kN
5
7.5
Hence As
299 x
103
= 650 mm2
460
Use 5T10 in each face (785 mm2)
i-0
Hence T10 @ 200 mm crs. horizontal reinforcement below slab is adequate.
2.8.8.3 Wall tie
Take the greater of (a) and (b) (a) Lesser of 2.OFt or
lsF 2.5
= 72 or 48 kN
in wall 0.5 m
above and
COMPLETE DESIGN EXAMPLE
(b) 3% of total vertical load = 0.03
x
190.1
= 5.7 kN
Hence Tie force
A
=
48 kN
-
48 x 103
S
= 104 mm2
460
Therefore reinforcement in slab will suffice
2.8.9 Strip footing (1)
EC2, Part 3: Concrete foundations, at the time of publication, has not yet been drafted, hence current UK practice is adopted.
Maximum pressure due to characteristic dead, imposed and wind loads =
For 900 mm wide strip, pressure
=
191
191
=
kN/m
z
114.6 + 23.6 + 47.4/0.9
+
=
212 kN/m2
0.9
z
Allow extra 10 kN/m2 for ground floor loads soil in foundations. This gives 222 kN/m2.
Allowable pressure
=
x 200
1.25
and weight 250 > 222
=
of
concrete displacing
kN/m2........
OK
Use 900 mm wide strip
Calculate reinforcement for flexure
-
(0.9
0.175)2
=
16.5 kNm/m
E
250.6 x
E
=
ion
Moment
8
=
Minimum area
209 mm2/m E E
AS
0.0015bd
=
0.0015
x
5.4.2.1.1
1000
x
Use T12 @ 300 mm crs. (377 mm2/m)
200 = 300 mm2/m E
=
COMPLETE DESIGN EXAMPLE
2.8.10 Reinforcement details The reinforcement details are shown in Figure 2.15 and given
IT-
A
II
I
1T10-7
r 2x1T10-8
1T10
i-
v
o°
cO
P
NO
NN
L
1
bars
-200 links
ci
96T12-3-300
(48N2 -48F2)
-200
-N 200
350
8
i
ooz-
links
a
Y
T B
19F1)
bars
N
M3
Cr Cr
19T10-5
19R8-4
19R8-4 19T10-5
N
0
P P
G N
96T12-1-300
(48 N2+48 F2)
L
Fdn.
i
-7-
la
1
i-
2%4T12-2-300
B2
cover ends
= =
40 75
1
-
2
A-A grid
B-B COVER to
outer bars -
%13R10 -9 EW
- 1000
-T-
t
Table 2.8.
wall tie
1st. SFL
J
B
in
2
EAST N1
=
40, F1=
omitted for clarity) WALL ELEVATION
20
Figure 2.15 Shear wall reinforcement details
COMPLETE DESIGN EXAMPLE
Table 2.8
Commentary on bar arrangement
Bar marks E
Notes
1
Wall starters match vertical reinforcement The projection of the horizontal legs beyond the face of the wall form the tension reinforcement in the footing
This extension must be at least a tension anchorage length _
12
4
460
x 1.15
x 3.2
x
209
= 208
377
mm
..............
....
OK
5.2.2.2 5.2.2.3 5.2.3.4.1
The minimum projection above the top of the base is a compression lap + 75 mm kicker =
32 x 12 + 75 = 459 mm
This is detailed at 525 mm
..................................
Minimum longitudinal reinforcement provided
u')
Minimum horizontal reinforcement provided
4,5,6
OK
5.4.7.3
2
2
4.4.2.2
9
BS 8110 3.12.3.5
Peripheral tie at floor ,r.
7,8
Wall spacers to maintain location of reinforcement
2.9 Staircase 2.9.1 Idealization The idealization of the staircase is shown
in
Figure 2.16.
T
3500
5060
Figure 2.16 Idealization of staircase Design as end span of a continuous beam. Calculations will be given for width.
2.9.2 Durability and fire resistance As for floor slab, Section 2.3, 20 mm nominal cover will be satisfactory.
1
m
COMPLETE DESIGN EXAMPLE
2.9.3 Loading Average slab thickness on plan
=
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
= 0.25 x 24
Self-weight
x
6.5 + 1.5
x 4 =
14.78 kN/m
2.9.4 Analysis Using coefficients in the Concise Eurocode
=
0.11
x
14.78
x 5.062 = 41.6 kNm
Moment near mid-span
=
0.09
x
14.78
x
5.062 = 34.1 kNm
Shear
=
0.6
x
14.78
x
5.06
x
Moment at interior support
2.9.5 Reinforcement for flexure =
Interior support,
175
M
- 20 -
bd2fCk From Section 13, Table
6 = 149 mm
E
Effective depth
41.6
103
x
x
1492
13.1
AS yk
bdf
=
0.072
=
746 mm2/m
(L)
Hence E
E
AS
E
Use T12 @ 150 mm crs. (754 mm2/m)
Span
M
=
0.048
=
0.058
=
601 mm2/m
bd2ck Asf ,k
bd ck Hence
E
E
E
A5
E
L()
E
Use T12 @ 150 mm crs. (754 mm2/m)
10s
x 32
= 0.059
= 44.9 kN
Concise Eurocode Table A.1
COMPLETE DESIGN EXAMPLE
2.9.6 Shear Reinforcement ratio
754
=
=
0.0051
1000 x 149
Near support VRd,
VRd,
0.35 x (1.6
=
- 0.175)
x (1.2 + 40 x 0.0051) x 149 = 104.3 kN
4.3.2.3
Eqn 4.18
>
Vsd
= 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. Since
Yk
Table 4.14
= 460, this should be modified to: =
Actual span/effective depth ratio =
(+'j
32 x 400/460 x 754/601
4.4.3.2(4)
34.9
5060/149 = 34
<
34.9.......
OK
2.9.8 Cracking c
As for floor slab in Section 2.3.8
Thickness of waist
=
175
=
183 mm2/m
E
E
Minimum area of reinforcement
4.4.2.2
< 200 mm
4.4.2.3 (1)
No further check is necessary.
2.9.9 Tie provisions E-W internal
tie,
the minimum area required =
91 mm2/m
(see Section 2.3.9) Total
area for staircase =
91
x 3 = 273 mm2
Provide 2T12 tie bars each side of staircase in adjacent slab
51
BS 8110 3.12.3.2
COMPLETE DESIGN EXAMPLE
2.9.10 Reinforcement details The reinforcement details are shown in Figure 2.17.
0
o
2
750
a
150
3- 2T10-8
6T12-3*6T12-141-125 alternate 3
3Qrd
5T10-I11 -300 ST10-16-300 i
L
16
_
117-.1
7
10T12-12-150
'-1OT12-13-150
{
Cover = 40
I
-
7
10T12-15-150 10
16
117
-9 6
Cover
to
outer
bars
LANDING
0
15
= 20
A-A
I
2nd
V
Aa FLIGHT
Figure 2.17
A
4 10-2-300 L 1OT12-1-150
Staircase reinforcement details
52
3BEAMS 3.1
Introduction C
O
O
O
cO
N
O
C
-
0 0
This Section covers the design of beams for shear and torsion, and supplements the examples given in Section 2. The requirements for adequate safety against lateral buckling are also examined.
N
3.2 Design methods for shear O
3.2.1 Introduction ((]
U
O
O
C
O C
C
U
EC20) 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. Variable Strut Inclination (VSI). C O
Standard
(2)
C
(1)
C
The standard method assumes a concrete strut angle of 450 (coto = 1) and that the direct shear in the concrete, V d, is to be taken into account. This contrasts with the VSI method which permits the designer to choose strut angles between the limits set in the NADO), as shown in Figure 3.1, but ignores the direct shear in the concrete. C
O
C_
C
Z
C
U
O
-C
o
O U
Limits of cotA (VSI method)
O
3.1
O
U
p30
C O
c
U U
O
c
O
C
O
U
C
U
U
c
Figure
(D
C
c
O U C O
O
L
c
C O
O
C
U C
O U O O
C
O
U
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 d). U
U
U
C
O
C
o U
C
C
U
O U E O O O
C
U
°
A further disadvantage of this method is that with increasing values of cote, i.e., reductions in the concrete strut angle, the forces in the tension reinforcement
BEAMS
CD-
°OM
E
COO
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 special circumstances the VSI method may be required but for most practical situations, the standard method will provide the most economic design. AO)
E
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
=
6m
Figure 3.2
-
Beam span and loading
iii
example
1
400 1000
Z
*FJ
Asl
400
Figure 3.3 Typical section
-
=
6434mm
Cover to links
example
2 =
(BT32) 50mm
1
The material strengths are
fck fvWk
=
30 N/mm2 (concrete strength class C30/37)
=
250 N/mm2 (characteristic yield strength of links)
0
The beam will be checked for shear reinforcement at three locations using both the standard and VSI methods for comparison. These are (1)
d from support
(2)
Where Vsd = VRd, i.e., the point beyond which only minimum shear reinforcement is required '
(3)
An intermediate point between
1
4.3.2.2(10) 4.3.2.2(2)
and 2.
3.2.2.1 Standard method
4.3.2.4.3
The shear force diagram is shown in Figure 3.4.
V Sd
1155 k N
4.3.2.4.3 4.3.2.4.4
VRdl
I
I
I
l
1155 kN
a I
Figure 3.4 Shear force diagram
-example
1
BEAMS
The design shear resistance of the section,
=
Table 4.8
I TRdk
T IT
=
0.34 N/mm2 for
=
1.6
-
4.3.2.3(1) Eq n 4.18
=
k
is given by
(1.2 + 40pl) + 0.15acp ] bwd
VRd1
Rd
VRd1,
d 4
1
fok
=
30 N/mm2
1
As,
6434
bwd
400 x 900
0.018
P,
0.02
I>
(assuming BT32 throughout span) NSd
=
0
AC 0.34 x
=
VRd1
3.2.2.1.1 Position
-
1
=
VSd
>
VSd
(1.2 + 40
1
x 0.018) x 400 x 900 = 235 kN
at d from support
-
1155 VRd1'
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 Vcd
VRd3
V V
d
=
d
=
Eq n 4.22
d
235 kN
=
VRd1
A sw
V
+
(0.9d)fywd
Eq n 4.23
s
where
area
A sw
shear reinforcement
of
spacing
VRd3
>
Vwd
?
70C
For
=
217.4 N/mm2
VSd
Al
VSd
-
Vcd; or
A (0.90) ywd Sw
shear reinforcement E E
250/1.15
of
?
VSd
-
Vd
Asw
(808.5
235) x 103
=
3.25 mm2/mm E
-
E
S
-
(Nn
Therefore
0.9 x 900 x 217.4
3.23 mm2/mm
E E
55
=
E
E
Try R12 links @ 140 mm crs. (4 legs), Asw/s
BEAMS
Check crushing of compression struts =
VRd2
For vertical links, cot«
-
0.7
V
30 cd
=
0
ck
=
4
0.55
200 =
Eqn 4.25
+ cots)
bw0.9d(1
(2 )Pf
Eqn 4.21
0.5
20 N/mm2
1.5
Therefore
=
x
x
(2)x0.55x20x400x0.9x900x1
=
VRd2
>
1782 kN
=
Vsd, max
................
1155 kN
OK
Check maximum spacing of links
4.4.2.3
452
ASw
Pw
-
3V
-
d
140
(808.5
-
0.0081
pwbwd
3
x 235) x
Since (5) Smax
VRd2
=
<
(3)
:5
VSd
103
x 400 x 900
Maximum spacing for crack control
=
Eqn 4.79
0.0081
x 400
=
CY)
Vsd
sbwsin a
35 N/mm2
Table 4.13
300 mm 5.4.2.2(7)
VRd2
Eqn 5.18
0.6d > 300 mm
140 mm spacing
..........................................
Check minimum value of
OK
Table 5.5
Pw
Concrete strength class C30/37 Steel class S250
By interpolation from EC2 Table 5.5 Pw,mm
=
<
0.0022
0.0081 proposed
E
Use R12 links @ 140 mm crs. (4 legs)
((1
C.-
O-in
Note: Using the standard method, the increase in force in the tension reinforcement is best covered by using the shift rule.
4.3.2.1P(6) 5.4.2.1.3
'-h
It will, however, be calculated in this example to provide a comparison with the values obtained in the subsequent examples using the VSI method.
Force in tension reinforcement
-
MS
z
+
(
2) VSd(Cote
-
Cota)
Eqn 4.30 G.)
Td
BEAMS
884 kNm,
Vsd
=
808.5 kN
cotO
=
1,
Therefore 3.2.2.1.2 Position 2
Td
-
z
=
0.9d
810 mm
=
3
=
LC)
Msd
cota
=
+ 404 =
1091
where
0 for vertical links
=
=
Vsd
4.3.2.4.3(5)
1495 kN
235 kN
=
VRdl
From Figure 3.4 LC)
-
a x 385
Vsd
=
1155
a
=
2.39 m from support
235 kN
((0
=
E
From Section 3.2.2.1.1,
VRd2
>
...........
Vsd, max
.
......
Pw,min
E
E
The amount of shear reinforcement provided should be greater than =
OK
Pwmin
Table 5.5
0.0022 E
Re-arranging EC2 Eqn 5.16 in terms of
A 5w s
gives
A Pwbwsina S
For vertical links sins
=
1
Hence =
0.0022 x 400 x
Maximum longitudinal spacing
=
1
(smax)
0.88 mm2/mm E
s
E
A aw
is given by EC2 Eqns 5.17-5.19.
Vsd
=
235 kN
VRd2
=
1782 kN from Section 3.2.2.1.1
Since
=
Eqn 5.17
0.8d > 300 mm
x 300 =
264 mm2,
4R10
= 314 mm2 E
0.88
Asw
EC2 Eqn 5.17 applies
CC)
Smax
VRd2'
U')
(5)
Vsd
E
Use R10 links @ 300 mm crs. (4 legs)
-
at 1.65 m from support E
3.2.2.1.3 Position 3
E
This is a point intermediate between the section at dfrom support and the point at which shear reinforcement is no longer required. aro)
=
1155
VRdl
=
235 kN
LC)
-
Vsd
1.65
x 385 =
520 kN
BEAMS
Since
VSd
>
shear reinforcement is required
VRdt'
Re-arranging EC2 Eqn 4.23
Vd
-
VSd
_
(520
_
0.9df M,d
s
0.9
-
235) x 103
x 900 x 1.81
mm2/mm E
=
1.62 mm 2 /mm
217.4
E
Try R12 links @ 250 mm crs. (4 legs)
=
E
ASw
Check maximum spacing of links
4.4.2.3
A SW
Eqn 4.79
Pw
sb w sina For vertical links sina
=
1
Hence _
452
Pw
0.0045
250 x 400 = (520
-
3
x 235) x
x
3V d
103
0.0045 x 400 x 900
_
-114 N/mm 2
x
pwbwd
E
-
E
Vgd
E
Maximum spacing for crack control = 300 mm
..................
OK
Table 4.13
Since
<
(5)VRd2
smax
=
VSd
5.4.2.2(7)
(3)VRd2
Eqn 5.18 0.6d P 300 mm
From Section 3.2.2.1.1 VRd2
>
VSd,max
.......................................
OK
E
Provide R12 links @ 250 mm crs (4 legs)
optimize link spacing, check the point at which shear reinforcement is satisfied by R12 @ 200 mm crs. (4 legs).
To
ASw
452
S
200
A SW wd
VRd3
=
2.26 mm2/mm
= 2.26 x 0.9 x 900 x 217.4 = 398 kN
(0.9d)fywd
s
=
Vd +
Vwd
Equating
VSd
=
=
VSd
VRdl
and noting that + V ,d +
VRd3
=
Vd
=
235 + 398
VRdl
= 633 kN
BEAMS
-
1155
=
633
=
1.36 m
(Y)
Distance of point from support
385
The proposed link arrangement is shown in Figure 3.5.
R12-200
R12-140 4 legs
4
R12-300 4 legs
legs
R12-200
R12-140
4 legs
4 legs
1-36m
m
+
+
-2.39
2-39m+
a
6.Om between centres of supports
Figure 3.5 Link arrangement (standard method)
-
example
1
Note:
CLL.
r-1.
the centre portion of the beam R10 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. In
3.2.2.2 Variable strut inclination method
4.3.2.4.4
((DD
O-N-0
This method allows the angle of the concrete compression strut to be varied at the designer's discretion within limits stated in the Code.
can give some economy in shear reinforcement but will require the provision of additional tension reinforcement. In most cases the standard method will >"O
It
suffice.
This reduced shear reinforcement will only be obtained at high levels of design shear and is counter-balanced by increased tension reinforcement. This can be seen by a comparison of EC2 Eqns 4.22 and 4.23 in the standard method and EC2 Eqn 4.27 in the variable strut inclination method. The standard method gives VRd3
=
V
=
d
Vd A Sw
+ twd
Eqn 4.22
(0.9d)ywd
Eqn 4.23
Re-arranging gives Asw
s
VRd3
-
Vd
(0.9d)ywd
The VSI method gives A sw VRd3
(0.90)fywd Cot()
s
Re-arranging gives A sw
VRd3
S
(0.9d)ywd cot()
Eqn 4.27
BEAMS
Note:
0
the above equation the contribution of the concrete, resistance of the section is not taken into account. In
V
d,
to the shear
With cot6 = 1.5 which is the maximum value permitted in the NAD, reductions in shear reinforcement will only occur when VRd3
(0.9d) fyWd x 1.5
Putting
Vsd
=
If
>
3V
Vsd
reinforcement.
Vcd Or
(0.9d) f,N,d
1.5(VRd3
G
VRd3
-
VRd3
-
Vd) Vsd
gives
VRd3
>
3V
d
then the VSI method will allow a reduction in shear
d,
of of
the variable strut inclination method will shear reinforcement. In this case the
For elements with vertical shear reinforcement, bWzv VRd2
Putting
cd
+
cote + tan6
Vsd
=
VRd2
and re-arranging gives 1
Vsd
bWzv
cotO + tan6
cd
VRd2
DLO
this inequality is not satisfied, use produce an uneconomic amount standard method should be used. If
is given by
Eqn 4.26
7-O
-'O
Figure 3.1 shows cotO plotted against 1/(cot6 + tan6) together with the EC2 and NAD limits for cot6. Hence for a given Vsd, the limits for cotO can be found.
Increasing the value of cotO will reduce the shear reinforcement required but increase the force in the tension reinforcement. E
(OD
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 V11
1
cotO + tan6
bWzvfcd
bW
=
400 mm
z
=
0.9
x 900
=
810 mm
P
=
0.7
-
=
0.55 K 0.5
fck
200
Eq n 4.21
BEAMS
30 cd
=
20 N/mm2
1.5
808.5 kN
=
Vsd
Therefore 808.5 x 103
_
x
Coo
1
=
400 x 810 x 0.55 x 20
0.22
x
x
cotO + tanO
!z?
From Figure 3.1, this lies under the curve. Therefore, cotO = 1.5 can be chosen which is the maximum value allowed under the NAD limits.
(A s
V Rd3
Now equating
VRd3
Eqn 4.27
) Z ywdCOte
to
Vsd
Vsd
s
zywdcotO
808.5 810
x
103
x 217.4 x
=
3.06 mm2/mm E
Asw
and re-arranging
1.5
Check
_
1.66
-
SIN
Asw ywd
< ('z)ofcd
5.5
=
.........................
E
3.01 mm2/mm
E
=
Try R12 links @ 150 mm crs. (4 legs), Asw/s
OK
Check maximum spacing of links. A sw
=
pw
=
4.4.2.3
Eqn 4.79
0.0075
sbW sin«
sd
pw
cd
(808.5
=
-
3
x 235) x (NA)
V- 3V
Check
>
0.0075
pw.min
= =
38.3 N/mm2
300 mm
Table 4.13
E
Maximum spacing for crack control =
=
0.0075 x 400 x 900
bwd
pw
103
0.0022
...................
OK
5.4.2.2(7)
smax
Vsd
=
808.5 kN bwzv
VRd2
400 x 810 x 0.55 x 20
cd
cotO + tanO
=
VSd
0.6d
=
1644 kN
2.167
+
Since (5) VRd2 smax
Table 5.5
(3) VRd2
D 300 mm
Eqn 5.18
BEAMS
Use R12 links @ 150 mm crs. (4 legs)
Check additional force in tension reinforcement.
2)
1091 + 606
cot«) =
= 1697 kN
Eqn 4.30
1495 kN using the standard method.
=
Td
-
VSd(cote
E
This compares with
(
c..
+
z
+
Msd
=
Td
Note: Although not permitted by the NAD, values of cote up to 2.5 are given in EC2.
check on shear reinforcement using cotO = 2.5 is now given to illustrate the effect of increasing values of 0 on shear and tension reinforcement. A
-
ASW
-
Vsd
z ywd tote
S
808.5 x 103
x
810
217.4
=
1.84 mm2/mm
x 2.5
= 2.01 mm2/mm
Try R12 @ 225 mm crs. (4 legs), A,Wls
Check maximum spacing of links =
pw
0.005
-
3 °d pwbwd
Vsd
=
57.5 N/mm2
................
OK
Table 4.13
.............................
OK
Eqn 5.18
Maximum spacing for crack control
> 300 mm
0.6d
E
=
CEO
Smax
250 mm
=
Use R12 links @ 225 mm crs. (4 legs)
Check additional force in tension reinforcement.
Td
=
Msd
z
This compares with
3.2.2.2.2 Position 2
-
where
-
+ (2)Vsd (tote Td
=
Vsd
cot«) = 1091 +
= 2102 kN
1011
1495 kN using the standard method.
=
VRd1
3.2.2.2.3 Position 3 Vsd
=
at 1.65 m from support
520 kN
zywdCoto
x
103
217.4
x
Try R12 links @ 225 mm crs. (4 legs), ASW/s
=
S
810
2
=
1.96 mm2/mm
1.5
2.01 mm2/mm
E
520 x
VSd
E
AiW
-
E
(D3
Since only minimum shear reinforcement is required this case is identical to that shown in Section 3.2.2.1.2.
BEAMS
From Section 3.2.2.2.1 spacing is satisfactory.
Use R12 links @ 225 mm crs. (4 legs)
E
As in Section 3.2.2.1.3, check the point at which the shear requirement is satisfied by R12 @ 200 mm crs. (4 legs). SW
452
=
2.26 mm2/mm
=
200
s (
Ssw
zfyWd
VRd3
cot6 = 2.26 x 810 x
x
217.4
1.5 =
597 kN
Eq n 4.27
J
Distance from support
=
1155
-
597
=
1.45 m
385 The proposed link arrangement is shown in Figure 3.6.
R12-150
R12-200
4 legs
300_
4 legs
1
R12-200
R12-1SO
4 legs
4 legs
4 legs
1.45m
2-39m
2.
a
39m
6.Om between centres of supports
Figure 3.6
Link arrangement (VSI method)
-
example
1
5
Comparing this with the arrangement in Figure 3.5 obtained using the standard method, it can be seen that less reinforcement is required near the support but this 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 a factor a when calculating VRd1. This enhancement only applies when the section is resisting concentrated loads and the standard method is used. For a uniformly distributed load, an unmodified value of VRdt should be used. -T
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
800 k N
800 kN
Ultimate loads
I
1.35rr
1-35m
6m
1.4
Figure 3.7 Beam span and loading
-
example 2
A
900
1000
A sl =
4825mm
Cover to links
Figure 3.8 Typical section
-
Z
(6T321 50mm
=
example 2
The materials strengths are fck fVwk
=
=
30 N/mm2 (concrete strength grade, C30/37) 250 N/mm2 (characteristic yield strength of links)
the example, VRd, 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. ((DD
In
3.3.2.1 Shear reinforcement
The shear force diagram is shown in Figure 3.9.
Figure 3.9 Shear force diagram
-
example 2
BEAMS
The basic design shear resistance of the section, VRd,
=
TRd
=
[TRd
VRdi,
is given by
4.3.2.3(1)
k (1.2 + 40pl) + 0.15 acp]b,,d
0.34 N/mm2 for Ck
=
Eqn 4.18 Table 4.8
30 N/mm2
E
a-0
For concentrated loads within 2.5d of the face of the support, an enhancement of shear resistance is permitted. -rT Rd may be multiplied by a factor a when E
determining
VRd,.
=
a
2.5d/x with 1.0
Vsd
..........
OK
Check maximum spacing of links
4.4.2.3
By comparison with example 2, requirements are satisfied
Use R12 links @ 125 mm crs. (4 legs) for 0
.........
< x < 2.25
OK
m
For the remainder of the beam beyond x = 2.5d (2.25 m) provide minimum reinforcement as example given in Section 3.2.2.
3.4 Design method for torsion 3.4.1
Introduction The edge beam shown in Figure 3.13 carries the ends of simply supported floor slabs seated on the lower flange. 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: Simply explained(12). Analysis of the structure and the design of the section for flexure is not included. The section will be checked for shear, torsion and the combination of both.
5.4.2.2(7)
BEAMS
Figure 3.13
Beam section
3.4.2 Design data Design torsional moment (Tsd) Design shear (Vsd)
=
120 kNm
=
355 kN
Concrete strength grade is C30/37,
=
3.1.2.4 Table 3.1
30 N/mm2
OWED
Ck
Nominal cover to links is 35 mm.
4.1.3.3 NAD
=
E
d
E
.E-
Assuming 25 mm bars and 10 mm links 1500
-
35
-
-
10
25
Table 6 =
1441.5 say 1440 mm
2
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.
The design shear resistance,
4.3.2.3(1)
VRd1,
+ 40pl)bwd
=
7-
TRd
=
0.34 N/mm2 for
k
=
1.6
-
d =
=
ick
1.6
Eqn 4.18
-
30 N/mm2 E
VRdl
Rdk(1.2
with zero axial load is given by
C')
4.3.3.2.2(4)
.0)
When combined shear and torsion effects are to be considered, shear is to be checked using the variable strut inclination method. The angle 9 of the equivalent concrete struts is to be the same for both torsion and shear design.
1.44 = 0.16 X1.0
Table 4.8
BEAMS
=
VRdl
=
0.0025 > 0.02
0
Assuming 0.25% tensile reinforcement, pi
0.34 x 1(1.2 + 40 x 0.0025) x 250 x 1440 x 10-3
= 159.1 kN < 355 kN ;;z
Therefore shear reinforcement required.
Use the variable strut inclination method. The maximum design shear force, to avoid web crushing is given by VRd2' bWzu VRd2
cd
4.3.2.4.4(2)
(cot6 + tang)
Eqn 4.26
Re-arranging gives VRd2
1
bwzu cd
cot6 + tan6 =
355 kN
bW
=
250 mm
z
=
0.9d
u
=
0.7
ca
=
E
Vsd
°k
=
-
x 1440 =
0.9
°k =
-
0.7
200
=
30
=
1296 mm
30 =
0.55
-9 0.5
4.3.2.4.2(3)
200
20 N/mm2
1.5
-/C
Therefore 355 x 103
VSd
bzu od W
250 x 1296 x 0.55 x 20
1
cot6 + tan6
should be
>_
0.1
=
0.1
CD
By reference to Figure 3.1, it will be seen that the value of cot6 may be taken anywhere between the limits of 0.67 to 1.5.
NAD Table 3 4.3.2.4.4(1)
To minimize link reinforcement, take
VRd3
IA- I
VRd3'
ZYwdcote
=
1.5
for shear reinforcement is given by E
Design shear resistance,
cot6
4.3.2.4.4(2)
Eqn 4.27
BEAMS
Re-arranging gives A SW
VRd3
s
Zfywdcote
Putting
VRd3
equal to
Vsd
Vsd
ASW
Zfywdcote
s
Using high yield reinforcement
-
460
400 N/mm2
=
E E
=
YwK
ywd
1.15
ys
Therefore Sw
0.9
S
ASWfywd
355 x
=
103
=
x 1440 x 400 x
= 0.46 x
bw s
0.46 mm2lmm
1.5
400 = 0.74 < " 250
Ed
= 0.55 x 20 = 5.5 N/mm2.. OK
2
2
4.3.2.4.4(2) Eqn 4.27
Before choosing the reinforcement, the effects of torsion will be considered and the results combined.
The force in the longitudinal reinforcement, NIA
(
SIN
Td
2) Vsd(cotO
For vertical links, cot« Td
-
355
x
= 1.5
-
Td,
ignoring flexure, is given by
Iota)
4.3.2.4.4(5)
Eq n 4.30 0
=
266.3 kN
2
Additional area of longitudinal reinforcement
266.3 x 103
Td
=
666 mm 2
400
ywd
V'.
This area of reinforcement must be combined with the tension reinforcement required for flexure together with the longitudinal reinforcement required for torsion.
3.4.4 Torsional resistance
of
(1)
as the sum
C,)
°.'
Torsional resistance is calculated on the basis of a thin-walled closed section. Solid sections are replaced by an idealized equivalent thin-walled section. Sections of complex shape are divided into sub-sections with each sub-section treated as an equivalent thin-walled section. The torsional resistance is taken
the torsional resistances of the sub-sections. X17
The torsional moment, carried by each sub-section according to elastic theory, may be found on the basis of the St Venant torsional stiffness. Division of the section into sub-sections should be so arranged as to maximize the calculated stiffness.
7
BEAMS
For this example the section will be divided into the sub-sections shown in
Figure 3.14.
310
200
I>
1500
250
x 310
200
Figure 3.14
3.4.4.1 St Venant
I
Dimensions of sub-sections
torsional stiffnesses
J
BS 8110:
ah3minhmax
Part 2
hmax
hmax
h min
=
_
2.4.3 Eq n
flanges (gy
3.4.4.1.1 Top and bottom
-
310 mm,
310
=
hmin
=
200 mm
1.55
200
.
From which R
=
0.203
BS 8110:
Therefore
J
=
0.203 x 2003 x 310
hmax
=
1500 mm,
=
0.5
x
109 mm4
Part 2 2.4.3 Table 2.2
3.4.4.1.2 Web
1500
hmin
250
From which
=
250 mm E
h max
hmin
6
(3
=
0.33
BS 8110: Part 2 2.4.3
Therefore
J
=
0.33
x 2503 x 1500 =
7.7
x
109 mm4
Table 2.2
1
BEAMS
3.4.4.1.3 Total stiffness
[(2 x 0.5) + 7.7] x 109
_
`Jtot
3.4.4.2 Thicknesses
t
=
'4
8.7x
109mm4
of equivalent thin-walled sections
>5
u
=
the actual wall thickness
4.3.3.1(6)
where u
=
outer circumference of the section
A
=
total area within the outer circumference
3.4.4.2.1 Top and bottom flanges
u
=
(310 + 200)2
A
=
310 x 200
_
62x103
=
1020 mm 62 x 103 mm2
=
Therefore x
=
61
mm
E E
t
1020
t may not be less than twice the cover, c, to the longitudinal bars. Hence, with 10 mm links 2(35 + 10)
3.4.4.2.2
u
=
(1500 + 250)2
A
=
1500
+
=
tmin
=
4.3.3.1(6)
90 mm
Web =
x 250 =
3500 mm 375 x 103 mm2
Therefore 375 x 103 CA)
=
3500
t
107 mm
> 2c
..................
OK
Values of t between the limits of A/u and 2c may be chosen provided that the design torsional moment, Tsd, does not exceed the torsional moment that can be resisted by the concrete compression struts. 3.4.4.3 Torsional moments Tsd,tot
=
120 kN m
0
This total moment is shared between the flanges and web in proportion to their torsional stiffness. Therefore 120
TsdJI
6.9 kNm
8.7 120
Tsd,w
Tsd
x 0'5 = x
106 kNm 8.7
must satisfy the following two conditions Tsd
E E
-
net
Table 5.3 E E
1b,
5.2.2.1
400
x
4
calculating
250 mm, good bond may be assumed.
2.7 N/mm2
lb
In
_
20 mm
(3D
(3D
Minimum cover to reinforcement Assume nominal aggregate size Assume maximum bar size
NAD 6.4(a)
25 mm
=
E
Use nominal cover
NAD Table 6
0
0
Note: 20 mm nominal cover is sufficient to meet the NAD requirements
NAD in all
respects.
Table 3 4.1.3.3(8)
Check requirements for fire resistance to BS 8110: Part 2.
NAD 6.1(a)
4.1.3.1.2 Materials
_
vk
vd
_
460 =
l's
yk
=
460 N/mm2 E
Type 2 deformed reinforcement,
400 N/mm2
2.2.3.2P(1) Table 2.3
1.15
C25/30 concrete with 20 mm maximum aggregate size 4.1.3.1.3 Analysis model
Flange depth =
600
_
0
Span 6 m
OK OK
2.5.2.1(5)
E
100 mm
1
x
10
clear spacing between ribs
>_
50 mm
OK
Transverse ribs (at supports only)
=
6 m
>
10
x
slab depth
=
2.75 m E
Spacing
Hence the ribbed slab may not be treated as a solid slab in the analysis under the terms of this clause unless intermediate transverse ribs are incorporated. This is not always desirable.
The model adopted in this example uses gross concrete section properties of the T shape in sagging regions and a rectangular section, based on the rib width, in the hogging region. EC2 Figure 2.3 has been used initially to define the extent of the hogging. This method can clearly be refined.
2.5.2.1(5)
SLABS
4.1.3.1.4 Effective span
Eqn 2.15
+ al + a2
+
In
Jeff
2.5.2.2.2
Assume 300 mm wide supporting beams 5700 mm
at edge beam
-t6
a1
=
=
leff
beam
=
t
Figure 2.4(b)
150 mm
spans between =
0.851,
0.85 x 6000
1
and 1.5 =
5100 mm
E
=
(2)
Figure 2.4(a)
150 mm
=
t
6000 mm
For ratio of adjacent to
_
a.
=
as (2)
E
a2 at central
a. taken
=
Nip
ln
2.5.2.2.1(4)
Figure 2.3 4.1.3.1.5 Effective width of flanges
2.5.2.2.1 ((DD
2.5.2.2.1(2)
buildings.
a symmetrical
T
+ (5)
=
bW
=
125 +
bell
beam
Ell
For
` 0.02
Giving VRdl
=
21.6 kN/rib
<
Vsd
Therefore shear reinforcement must be provided.
4.3.2.2(3)
Use the standard design method for shear:
4.3.2.2(7)
VRd3
VRd3
>
4.3.2.4.3 VSd
Vd
+
VWd
Eqn 4.22
SLABS
where VRdl
=
21.6 kN/rib
x
0.9df,wd
('7
=
Vd
4.3.2.4.3(1)
Therefore
V
A SW d
-
29.8
>_
21.6
= 8.2 kN/rib
Eqn 4.23
s
Check maximum longitudinal spacing of links v
)
x 0.9d
cdbw
For vertical stirrups, cot«
=
P
=
VRd2
(O
VRd2
-
0.7
=
(1
Eqn 4.25
+ COta)
0
ck
=
0.575
>_
Eqn 4.21
0.5
200
0.5
x 0.575 x
<
VSd
<
(
16.7
3)
x 125 x 0.9 x 232 x 10-3=125kN
z
2
+
SIN
(
VRd2
5.4.2.2(7)
VRd2
Therefore smax
=
0.6d =
> 300 mm
139
Eqn 5.18
Try mild steel links at 125 mm crs.
Pw,min
=
0.0022
Table 5.5 =
35 mm2 E
0.00226W s
A
E
Use R6 links @ 125 mm crs. (Asw
f
ywd
V
d
=
250
=
=
57 mm2)
217 N/mm2
1.15
=
57
x
0.9
x 232 x
125
217 =
20.7
> 8.2 kN/rib
...
OK
103
Link spacing may be increased where VSd
smax
300 =
185 mm
Eqn 5.17
Use R6 links @ 175 mm crs. apart from region within 0.6 m of interior support
V,d
=
14.7
> 3.4 kN/rib
.............................
OK
SLABS
,It
4.1.3.1.9
Shear between web and flanges
4.3.2.5
Eqn 4.33
AFd VSd
(2) SIN
a
a
_
V
Figure 4.14
2550 mm
=
to
Maximum longitudinal force in the flanges
_
«fcd(0.8x)b
d
=
0.075 at mid-span
F
=
14.2
FC
x
x
x 0.075 x 232 x600 _
0.8
122kN
103
Force to one side of web =
AFd
122
x
600 - 195 2
=
x 600
41.2 kN
_
Vsd
NIA
Therefore 41.2
=
16.2 kN/m
2.55
VRd2
-
0.2 cdhf = 0.2 x 16.7
V_
=
2.5To,,h,
x 100 = 334 kN/m >
VSd
.. OK
Eqn 4.36 Eq n 4.34 Eqn 4.37
+ s,
With Asf
0 =
2.5 x 0.3
x 100
=
75 kN/m
>
A
VRd3
=
VSd
...........
OK
Eqn 4.35
No shear reinforcement required
4.1.3.1.10 Topping reinforcement .CJ))
car
No special guidance is given in EC2 regarding the design of the flange spanning between ribs. The Handbook to BS 8110(13) gives the following guidance.
361.5
.v)
Thickness of topping used to contribute to structural strength Although a nominal reinforcement of 0.12% is suggested in the topping (3.662), it is not insisted upon, and the topping is therefore expected to transfer load to the adjacent ribs without the assistance of reinforcement. The mode of transfer involves arching action and this is the reason for the insistence that the depth be at least one-tenth of the clear distance between the ribs. E_0
Minimum flange depths are the same in EC2 and BS 8110 and the above is therefore equally applicable. Provide minimum reinforcement transversely and where top bars in rib, which have been spread over width of flange, are curtailed.
2.5.2.1(5)
Asf
4
df
<
7°t
SLABS
0.6btdflfyk hf
=
g
Eqn 5.14
0.0015btdf
100 mm
Therefore, conservatively Asf
4
150 mm2/m
Use T8 @ 200 mm crs. (251 mm2/m) or consider fabric
4.1.3.1.11
4.4.3.2
Deflection 6000 Actual span/effective depth ratio
25.9
232
403 Mid-span reinforcement ratio, p
=
600 x 232
=
0.0029
Therefore section is lightly stressed.
4.4.3.2.(5)
Basic span/effective depth ratio (interpolating for p) = 39.2
Table 7
NAD
400
Modification factor for steel stress
Since flange width
=
x 403
460 x 295
1.19
> 3 x rib width, a 0.8 modification factor
is
required.
Since span > 7 m, no further modification is required. Permitted span/effective depth ratio = 39.2 x 1.19 x 0.8 =
37.3
>
25.9
.................................
OK
4.1.3.1.12 Cracking
For exposure class 1, crack width has no influence on durability and the limit could be relaxed. However, the limit of 0.3 mm is adopted for this
4.4.2.1(6)
of 0.3 mm
example. Satisfy the requirements for control of cracking without calculation. Check section
4.4.2.3(2)
0
-O_
at mid-span: Minimum reinforcement, A.
=
kckct,effAct/as
00^D
Note:
can be conservatively taken as the area below the neutral axis for the plain concrete section, ignoring the tension reinforcement, as shown in Figure 4.10. Act
4.4.2.2(3)
Eq n 4.78
SLABS
92
Neutral axis
1100
175
i
35
Figure 4.10
125
I
35
Tensile zone of plain concrete section
92 mm
=
-
=
160 x 175 + 600 (100
=
100%f ,k
fct,eff
III
Depth to neutral axis
=
recommended value 3 N/mm2
kc
=
0.4 for normal bending
k
=
0.8
AS
=
0.4
Act
460 N/mm2 4.4.2.2(3)
x 3 x 32800/460 = 69 mm2 <
0.8
E E
x
=
E
QS
32800 mm2
=
92)
As,prov
.....
OK
Table 4.11
Check limit on bar size. =
Gk +
0.30 k
Ratio of quasi-permanent/ultimate loads
=
6.1
4.4.2.3(3) 2.3.4
kN/m2
E
Quasi-permanent loads
Eqn 4.78
=136.7 =
Eqn 2.9(c)
0.45
NAD
Table
1
Estimate of steel stress As,req
x fyd
0.45 x 295 x 400 403
=
A S,Prav
Maximum bar size For cracks
=
32 > 16 mm provided
=
(Y)
LC)
0.45 x
132 N/mm2
..................
OK
caused dominantly by loading, crack widths generally will not be
Table 4.11
4.4.2.3(2)
excessive.
4.1.3.1.13 Detailing
Minimum clear distance between bars = 0 r, 20 mm Nominal clear distance in rib
=
49 mm
5
......................
5.2.1(3)
OK
SLABS
Bond and anchorage lengths:
5.2.2
For h > 250 mm bottom reinforcement is in good bond conditions. Top reinforcement is in poor bond conditions.
5.2.2.1
((DD
Figure 5.1(c)
Therefore, ultimate bond stresses are =
fbd
2.7 N/mm2 E
Bottom reinforcement,
5.2.2.2(2)
Table 5.3 Top reinforcement,
=
fbd
0.7
x 2.7
=
1.89 N/mm2
5.2.2.2(2)
= Ofyd
Basic anchorage length, lb
5.2.2.3
4fbd
x 400
=
1b
c4)
For top reinforcement,
Eqn 5.3 =
4 x 1.89
530
0 x 400
For bottom reinforcement, lb
370
4 x 2.7
Anchorage of bottom reinforcement at end support.
5.4.2.1.4
Treat as a solid slab and retain not less than half of the mid-span reinforcement.
5.4.3.2.2(1)
0
0
0
Use 2T12 L bars bottom at end support
0
0
Anchorage force for this reinforcement with zero design axial load a,
F
=
X
VSd
5.4.2.1.4(2)
d
Eqn 5.15
where VSd
=
21
kN/rib
For vertical shear reinforcement calculated by the standard method
-
Iota)/2 4
ar
=
z(1
a
=
90° and z is taken as 0.9d
5.4.2.1.3(1)
0
E
Although this ribbed slab falls outside the solid slab classification requirements for analysis, treat as a solid slab for detailing and take al = d.
5.4.3.2.1(1)
c,5
0
Therefore
Fs
=
As,req
-
21 21
kN/rib
x
103
=
<
53 mm2
As
400
Prov
Required anchorage length for bottom reinforcement at support: 19
A s, Prov
Eqn 5.4 1b,min
as
=
0.7 for curved bars in tension
lb,mn
=
0.31b
=
5.2.3.4 5.2.3.4.1(1)
as 1bAs req 1b,net
OK
11.10
.9 100
or 100 mm
106
Eqn 5.5
SLABS
In calculations of should be taken Ib,net' As,req
A s,prov
NAD 6.5(c) 5.4.2.1.4(3)
101 mm2
226 mm2
x 37 x
0.7
=
=
As,spanA
x
12
101
=
139 mm
E E
lb,net
=
4
226
>
lb,min
.......
OK
=
As14
226/4
=
5.2.3.3
57 mm2
E
=
Ast
(`")
Minimum transverse reinforcement (for indirect support):
Eqn 5.4
Use 1T8 bar as transverse reinforcement
Minimum top reinforcement at end support: MsUp
=
(4)
=
0.040
M
6.7 kNm/rib
=
26.7
5.4.2.1.2(1)
Therefore nominal reinforcement is sufficient. Use 2T12 L bars top as link hangers
The reinforcement details are shown in Figure 4.11.
Figure 5.12 2T12 per
rib
L TO
R6 - 175 links
L
2T16 per rib
2T12
per rib
Lb,
622
b/3
Is
Detail at edge support
Provide full lap length, is
1b,neta1
ls,
for bottom bars:
5.2.4.1.3
m
4.11
Eqn 5.7
-K ls,min
> 20,
For 100% of bars lapped and b
«1
=
1.4
Hence with « a = 1.0 and A s,req = A s,prov lb,net
twin
lb
= =
370
=
37
0.3 «acxllb
=
187 mm
=
X
12
=
NAD
Table 3 Figure 5.6 444 mm
g 150 or 200 mm
m
Figure
net
100
Eq n 5.4 Eqn 5.8
SLABS
Therefore 444 x 1.4
=
is
=
622 mm
>
................
ls,min
OK
(OD
C"6
Transverse reinforcement at lapped splices should be provided as for a beam section. Since 0 < 16 mm, nominal shear links provide adequate transverse
5.2.4.1.2(1)
reinforcement.
The reinforcement details are shown in Figure 4.12.
Figure
OD)
5.4.2.1.5
L()
Anchorage of bottom reinforcement at interior support. Treat as a solid slab and continue 50% of mid-span bars into support.
5.4.3.2.2(1) 5.13(b)
4T12 per rib
R6-
125
links
L 2T16 '-
I
Figure 4.12
b,
net
10
0
=
M6
per rib
per rib
160
Detail at interior support
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. ((D
This detailing prohibits the easy use of prefabricated rib
a(0
-Z-
5.2.5
of the links is shown in Figure 4.13.
NAD --I
The arrangement of the reinforcement within the section including the anchorage
Tables
3&8 --+
5.4.2.1.2(2)
Figure 5.10
Figure 4.13 Arrangement of reinforcement
108
SLABS
4.2 Flat slabs 4.2.1 Flat slabs in braced frames The same frame is used in each of the following examples, but column heads are introduced in the second case. 4.2.1.1 Design example of a
flat slab without column heads
(n-
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.
Figure 4.14
Plan of structure
0
(CD
The area shown is part of a larger structure which is laterally restrained in two orthogonal directions by core walls.
The slab is 225 mm thick. All columns are 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 C25/30. 7-C
v3)
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
SLABS
Nominal cover to reinforcement
=
20 mm
NAD
Table 6 Nominal cover to all bars .9 bar size .9
Use nominal cover
=
NAD 6.4(a)
nominal aggregate size
=
20 mm
.
OK
4.1.3.3(5)
20 mm
4.2.1.1.2 Materials Type 2 deformed reinforcement,
460 N/mm2
=
fvk
NAD 6.3(a)
C30/37 concrete with 20 mm maximum aggregate size
4.2.1.1.3 Load cases It
is sufficient to
consider the following load cases
2.5.1.2
Alternate spans loaded with yGGk + yoQk and yGGk on other
(a)
spans. Any two adjacent spans carrying yGGk + yoQk and all other
(b)
spans carrying
yGGk
0.225 x 24 + 1.0
=
CJ)
Gk
yGGk.
=
1.35
yGGk + yoQk
x 6.4 =
=
=
6.4 kN/m2
8.7 kN/m2
8.7 + 1.5
x 5.0
Table 2.2
=
16.2 kN/m2
Eqn 2.8(a)
4.2.1.1.4 Analysis (On
Analyses are carried out using idealizations of both the geometry and the behaviour of the structure. The idealization selected shall be appropriate to the problem being considered.
=
2.5.1.1.P(3) and P(4)
-00
>-O
cam
(D3
00,
No guidance is given in EC2 on the selection of analysis models for flat slabs, or on the division of panels into middle and column strips and the distribution of analysis moments between these strips. This is left to the assessment of individual engineers. The requirements set down in BS 8110 for the above points are taken as a means of complying with EC2 Clause 2.5.1.1P(3). om'
(C)
EC2 allows analysis of beams and slabs as continuous over pinned supports. It then permits a reduction in the support moment given by AMsa
Fsasup bsup /8 cmn
The analysis in this example includes framing into columns. Thus the reduction AMSd is not taken. Consider two frames from Figure 4.14 as typical: (i)
(ii)
Grid 3/A-D subframe Grid B.
110
2.5.3.3(3) 2.5.3.3(4)
SLABS
Analysis results for 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 increased loaded width (5.2 m), since this is the first internal support for frames in the orthogonal direction. (Dw
O:3
Member stiffnesses have been based on a plain concrete section
in
this analysis.
Column moments and reactions are given in Table 4.1.
4250
5200
5200
Islab 2
3500
3500
/7777
/7777
/777
ANALYSIS MODEL
-199
-198
-204
123
BENDING MOMENT
Figures 4.15
Analysis of frame
ENVELOPE
(kNm)
SLABS
Column moments and reactions
Table 4.1
(kN)
E column moments (kNm)
Max. E column moments (kNm)
End
156.4
37.9
37.9
1st interior
444.7
6.8
21.4
Max reaction .
°°_
Support
4.2.1.1.5 Flexural design
-
Panel A-13/1-2
O.0
EC2 does not specifically address the problem of edge column moment transfer and the provisions of BS 8110 are adopted here. Mt,max
-
BS 8110 3.7.4.2
0.15bed 2fCU
Column A12 moment transfer Assuming 20 mm cover and 20 mm bars in the top E
NAD 4.1.3.3(5)
- 20 - 10 = 195 mm - 20 = 175 mm
225
d2
=
195
be
=
300 + 300 (say) = 600 mm
fCU
=
37 N/mm2
Mt,max
=
0.15
+
E
=
E
d,
E
1752
10-6
x 37 x
x
x 600 x
=
102 kNm
c)'
This ought to be compared with an analysis for a loading of 1 AGk + 1.60k, which would give approximately 5% higher edge moments than the EC2
=
102
>
1.05
x
x
Mt,max
=
37.9
CA)
analysis results above. 39.8 kNm
................
OK
=
1.5,
a
=
0.85 and
-ys
=
E
Using ry
E
E
Design reinforcement to sustain edge moment on 600 mm width.
Table 2.3
1.15
Referring to Section 13, Table 13.1: Msd
37.9
x
106
=
0.069
CA)
600 x 1752 x 30
bd2fck AS Yk
=
0.085
-
0.085 x 600 x 175 x 30
bdtck
S
d
582 mm2
=
970 mmz/m
460
=
CA)
X
=
E
A
0.163
< 0.45 (zero redistribution)
.................
OK
2.5.3.4.2(5)
SLABS
Use T16 @ 150 mm crs. (1340 mm2/m) top at edge column
900 mm (see Figure 4.16)
=
Place over width
Note: This approach gives more reinforcement than is necessary.
Figure 4.16 Edge column moment transfer
Check above moment against minimum value required for punching shear.
?
msd
4.3.4.5.3 Eq n 4.59
Vsd
edge
For moments about axis parallel to slab
Table 4.9
_ ± 0.125 per m
77
=
Vsd
156.4 kN
Therefore
= ± 19.6 kNm/m
_ ± 0.125 x 156.4
msd
Edge moment
=
37.9
=
>
63.2
19.6 kNm/m
0.6
...............
OK
Design for msd above in region outside edge column moment transfer zone.
_
msd bd2f
ck
19.6
x
106
=
0.021
1000 x 1752 x 30
Minimum steel sufficient
=
0t d .6b
f
4 0.0015btd
5.4.2.1.1
vk
=
263 mm2/m E E
= 0.0015 x 1000 x 175
Use T12 at 300 mm crs. (373 mm2/m) top and bottom (minimum)
=
3h
t>
500
=
500
>
300 mm
(Y)
Maximum spacing
.......
OK
NAD Table 3 5.4.3.2.1(4)
SLABS
Column A/1 moment transfer
Assume the design forces for the frame on grid for grid 3 in proportion to their loaded widths. Load ratio =
(4.25/2)
=
1
are directly related to those
0.41
5.2
The ratio of the edge column distribution factors for the frames is 2.0. Msd
=
37.9x0.41 x2.0 =
31.1kNm
Using design approach as for column A/2:
=
be
300 + 320 (say)
Mt max =
0.15
x 450 x
>
1.05
x
450 mm
x 37 x 10-6 = 76 kNm
1752
=
31.1
=
32.7 kNm
.....................
OK
Design reinforcement to sustain edge moment on 450 mm width
Msd
_
rk
106
=
450 x 1752 x 30
bd2lck AS
x
31.1
=
0.075
0.093
bdlck
AS
_
x 450 x
0.093
175
x 30 = 478 mm2 = 1062 mm2/m
460
Use T16 @ 150 mm crs. (1340 mm2/m) top for a width of 600 mm
Check above moment against minimum value required for punching shear.
_
x
=
x 156.4) say =
(0.41
=
31.1/0.45
69.1
kNm/m
0
0
>
32.1
..........
G.)
In region of slab critical for punching shear:
32.1
bdfck
=
0.035
=
0.042
_
0.042 x 1000
x
460
175
x 30 =
480 mm2/m E
As
106
1000 x 1752 x 30
bd2fck
ASfyk
x
(A)
Msd
Eq n 4.59 Table 4.9
± 32.1 kNm/m
(O"
Edge moment
+ 0.5
= ± 0.5 per m for corner columns
where n
"f Msd
0
>_
x
msd
4.3.4.5.3
OK
SLABS
Use T16 @ 300 mm crs. (670 mm2/m) top and bottom outside 600 mm wide moment transfer zone and over area determined in punching calculation LOLL
The division of panels into column and middle strips is shown in Figure 4.17.
BS 8110
Figure 3.12
E
Although BS 8110 indicates a 2.36 m wide column strip at column 132, a 2.6 m width has been used in the following calculations. This is considered reasonable as a loaded width of 5.2 m has been taken in the analysis for grid B and grid 2.
8
A
C
4-25m
5.2
1
I1
4-25m I
=I-I
06m
I
I
I
I
I
I
I
I
I
---
rt ----TI -
I
11.06ml
I
I
I
I
I
1 1.06m
I
I
1-301M
1
I
I
I
I
5-2m
I I
I
I I
I
I
I
-#-----II
I
I
T
-I--,--T
II
1.06m 3
E E
Figure 4.17 Assumed strip widths (arrangement symmetrical about diagonal A/1- C/3) Column 812 support moments
0
Column strip
Msd,cs
bd2tck
=
198 kNm in both directions
=
0.75
1300 x 2
=
x 198 = 2600
149 kNm
mm
E
b
Msd.cs
=
E
Analysis moment
149 x 106
=
2600 x 1752 x 30
11
0.062
BS 8110 Table 3.20
SLABS
Asyk
=
bdyk
x 2600 x 175 x 30
0.076
=
2255 mm2 E
=
s
E
A
0.076
460
E
Use 13T16 (2613 mm2) top in column strip. Provide 9T16 @ 150 mm crs. in central 1.3 m and 2T16 @ 300 mm crs. on either side
rt
-0.125
=
_
Msd
4.3.4.5.3
Table 4.9 =
" Vsd
-0.125 x 444.7
-55.6 kNm
=
CJ)
With
3.7.3.1
C`7
Check whether minimum moment required for punching shear has been met.
BS 8110
Eq n 4.59
This is to be carried over a width of 0.31. Since Vsd includes for a loaded width of 5.2 m, it is assumed that the larger panel width may be used.
0.3x5.2
=
0.31
=
1.56m
By inspection reinforcement (9T16 in central 1.3 m) is sufficient.
.....
OK
Middle strip (using average panel width) 0.25 x 198
_
Msd,ms
bd 2f
ck
x
=
d Asyk
(4725
-
0.059
< 0.45
x
106
=
2600) x 1752 x 30
0.026
...............................
OK
0.031
bdick
As
0.031
x 1000 x 175 x 30
E
354 mm2/m E
=
E
b
460
Use T16 @ 300 mm crs. (377 mm2/m) top in middle strip
_O)
gym-
It is noted that EC2 Clause 2.5.3.3(5) would allow the use of the moment at the face of the support (subject to limits in EC2 Clause 2.5.3.4.2(7)), but this is considered more appropriate to beams or solid slabs and the peak moment over the support has been used in the above design.
Span moments
in all
panels.
The column strip moments are given in Table 4.2 where Msd,cs
=
0.55 Msd
ago
11d =
2035 mm
length
breadth
>
Hence
x
u
=
27r
Vsd
=
444.7 kN
1.5
x 185 + 1200 =
2944 mm
Figure 4.18
Note: No reduction in this value has been taken. The applied shear per unit length:
v,,
=
Vsd R
where
u
Q
internal column
444.7 x 103 x 1.15 Vsd
4.3.4.3(4)
=
=
Eq n 4.50 Figure 4.21
1.15
174 N/mm
2944
CC/)
4.3.4.5.1
+ 40pi)d
VRd1
=
TRd
=
0.34 N/mm2
k
=
1.6
pl
=
7Rdk(1.2
-
`'/
Shear resistance without links
d =
Eqn 4.56 Table 4.8
1.6
-
0.185
=
1.415
>_
1.0
reinforcement ratio within zone 1.5d from column face (T16 (& 150 mm crs. top each way gives 1340 mm2/m)
=
x
),
pl/
Pt
ply
> 0.015
1340
0.0072
1000 x 185
-U)
Note: The amount of tensile reinforcement in two perpendicular directions
Assume plx + ply
=
2 (0.0072)
> 0.005
> 0.5%.
.....................
OK
4.3.4.1(9)
SLABS
Therefore
VSd
x
=
0.34
=
174 N/mm
x
1.415
>
(1.2 + 40
x 0.0072) x 185
>_
200 mm
133 N/mm
VRdl
Therefore shear reinforcement required such that Slab depth
=
E
VRdl
VRd3
>
4.3.4.3(3)
VSd
.....................................
OK
4.3.4.5.2(5)
Check that applied shear does not exceed the maximum section capacity
-
VRd2
2 .0
VRdl
2.0 x
=
1
33
=
266
>
174 N/mm
.....
OK
NAD
Table 3
Shea r stress aro und colu mn perimet er
444.7 x 103
=
4.3.4.5.2(1)
=
2.0 N/mm2
1200 x 185
...
4.9 N/m m2
VSd
Therefore no shear reinforcement required
4.3.4.3(2)
SLABS
Column A12 (300 mm x 300 mm edge column) Critical perimeter located at 1.5d from face of column. U
=
900 + 2777r
Vsd
=
156.4 kN
1770 mm
=
Applied shear per unit length, with a = 1.4 156.4
VSda
u
=
VRd1
x
103
x
1.4
=
m
_ VSd
Figure 4.21 124 N/mm
Eqn 4.50
1770
133 N/mm (as for column B/2)
VSd
VRd1
Therefore no shear reinforcement required
4.3.4.3(2)
4.2.1.1.7 Deflection
Control by limiting span/effective depth ratio using NAD Table
7.
4.4.3.2
For flat slabs the check should be carried out on the basis of the longer span.
4.4.3.2(5)(d)
For span < 8.5 m, no amendment to basic span/effective depth ratio is required.
4.4.3.2(3)
CD'
v0)
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% 25 the structure is classified as a sway frame E
Since
=
=
SLABS
The analysis and design would need to follow the requirements of EC2 Clause A.3.5 to take into account the sway effects.
CD.
fly
-(5
EC2 Clause 2.5.3.4.2(4) does not generally allow redistribution in sway frames.
On.
.s.
(w3
4.3.5.3.3(3) BS 5950: C/)
1a)
0_0
Assuming in the above example that the column sizes are increased such that a non-sway frame results, the following load cases need to be considered for design.
(r)
As an alternative means of determining the frame classification, it is suggested that an analysis as detailed in BS 5950(14) is used to demonstrate that the EC2 requirements are met for non-sway frames.
07?
The method above is included to demonstrate its complexity. However, note the omission of guidance in EC2 Clause A.3.2(3) on which nomogram to use in EC2 Figure 4.27.
Part
1
5.1.3
D))
Via)
owe
These same load cases would also be applicable to sway frames where amplified horizontal loads are introduced to take account of the sway induced forces, complying with EC2 Clause A.3.1(7) (b).
4.2.2.1.3 Load cases and combinations
LTG Gk +
approach the 'dII''esign values are given by
TQ,1
2.3.2.2 P(2) Eq n 2.7(a)
+
With the rigorous
2.5.1.2
Q0 + i>1 E TQi wo,i Qk,i
Qk,1 >Go
= primary variable load, Qk2 = secondary variable load E
sue'
where
=
0.7 generally
NAD Table
1
(g.
The yF values are given in EC2 Table 2.2. Load cases with two variable actions (imposed and wind) are: (a) Imposed load as primary load 1.35Gk + 1.5Qk + 1.05 Wk (b) Wind load as primary load 1.35Gk + 1.050k + 1.5 Wk In
addition, load cases with only one variable action are: (c) Dead load plus wind 1.OGk
(favourable) + 1.5 Wk
1.35Gk (unfavourable) + 1.51Nk
(d) Dead load plus imposed 1.35Gk + 1.50 k
For non-sensitive structures it is sufficient to consider the load (b) above without patterning the imposed loads.
cases
(a)
and
The NAD allows the use of EC2 Equation 2.8(b) to give a single imposed and wind load case: 1.35Gk + 1.35Qk (all spans) + 1.35 Wk
0
NAD 6.2(e) 2.5.1.2P(1)
SLABS
^a)
Final load combinations for the example given here 1.35Gk + 1.50 k (as Section 4.2.1.1.3) 1.OGk + 1.5Wk (single load case) 1.35Gk + 1.5Wk (single load case) 1.35Gk + 1.35Qk + 1.35Wk (single load case)
(i)
L(T
(ii)
(iii) (iv)
4.2.2.1.4 Imperfections
2.5.1.3(4)
Consider the structure to be inclined at angle v
=
1
100
f
NAD
Table 3 =
frame height
an
=
I1 (
,11
l
vred
Eqn 2.10
0.005 radians
>_
l
+1
=
0.78
=
ar,Y
=
10.5 m
where n
=
number of columns
0.78 x 0.005
=
0.0039 radians
)
n
2
=
=
5
Take account of imperfections using equivalent horizontal force at each floor. AHD
=
E V vred
EV
=
total load on frame on floor
Eqn 2.11
2.5.1.3(6)
j
Using 1.35Gk + 1.5Qk on each span gives
EV
x 5.2) x
=
(18.9
=
1592 x 0.0039
16.2
=
1592 kN
Therefore AHi
=
6.2 kN per floor
Assuming the frame by virtue of its relative stiffness 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)
COLUMNS 5.1
Introduction E-0
The design of column sections from first principles 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 5.2.1
compatibility
Introduction Two examples are considered:
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.
D))
1.
The first of these is very simple while the 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. --q
5
Assume
fk =
460 N/mm2 and
Ck
=
30 N/mm2
350
r
2T32 500
=
30
111
f,k
4T25
fyk =460 50
Figure
5.1
5.2.2 Example
Column section
1
Calculate the moment that the section can sustain when combined with an axial load of 2750 kN. 5.2.2.1 Basic method If the neutral axis is within the section, the compressive force generated by the concrete at ultimate limit state is given by
NRd c
=
0.459f kbx
4.2.1.3.3 Figure 4.2
COLUMNS
and the moment by
-
MRd,c
NRd,c(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
(3)
Calculate the strain at each steel level
(4)
Calculate force generated by reinforcement (N Rd)
(5)
NRd
E
(6)
NRd,c
=
NRd,c
If NRd is not
+
NRd,s
close enough to 2750 kN, modify the value of x and return
to step (2)
(8)
MRd
approximately 2750 kN, calculate MRd,c
MRd,c
and
MRd,s
+ MRd,s
(D'
If
+
NRd is
(7)
c
The design yield strain for the reinforcement 460 1.15
=
x 200000
0.002
5.2.2.2 First iteration E
Assumed value for x is 250 mm NRd,c
=
0.459 x 30 x 350 x 250/1000
0.0035 Es
top
x 200
250
0.002; therefore
0.0028
fs
=
400 N/mm2
x 804 x 400/1000 =
NRd,s,
=
2
Es,mld
=
0 and NRd s2
Es,bot
=
-Es'top;
therefore
NRd,s3
=
-2 x
491
=
1205 + 643 - 393
=
1205 kN
E
Strain >
=
=
643 kN
0 fs
-400 N/mm2
=
x 400/1000
=
-393
kN
NRd
(.)
Hence 1455 kN
=
This is considerably less than 2750 kN, hence x must be increased.
250 x 2750 1455
133
=
473 mm
E
=
x
Try new value for x
COLUMNS
5.2.2.3 Second iteration NRd,
=
0.459 x 30 x 350 x 473/1000 =
NRd.s,
=
643 kN as before
EWid
=
0.0035
fs,mid
N
d,s2
(473
473
-
250)
=
=
0.00165 x 200000
=
330 x 2 x 491/1000 0.0035
Es, bot
(473
-
0.00165
330 N/mm2
=
324 kN
=
450)
=
0.00017
473
f
=
0.00017 x 200000
NRd,s3
=
34 x 2 x 491/1000
=
2289 + 643 + 324 + 33
S
2289 kN
=
34 N/mm2
=
33 kN
C))
Hence +
NRd
3289 kN
=
This is too large, hence x should be reduced. Linear interpolation gives
=
250 +
2750
3289
-
1455) (473
-
250)
=
407 mm E E
x
- 1455
5.2.2.4 Third iteration NRd
=
0.459 x 30 x 350 x 407/1000
NRd.si
=
643 kN as before
=
0.0035 (407 407
=
270 N/mm2 and
0.0035 (407 407
Es,bot
s
250)
-
NRd,s2
450)
=
-74 N/mm2 and
=
1961
=
NRd
0.00135
=
265 kN
-0.00037
_
s3
1961 kN
CA)
VAN
fS
-
0
Es,mid
=
=
-73
kN
Hence NRd
+ 643 + 265 - 73
=
2796 kN
This is within 2% of the given axial load of 2750 kN
134
..............
OK
COLUMNS
5.2.2.5 Moment 1961
MRd,c
643 x 0.2
MRd,sl
73 x 0.2
MRd,s3
=
=
0.416
x 407)/1000
158.2 kNm
=
128.6 kNm
0
MRd,s2
MRd
-
x (250
=
14.6 kNm
158.2 + 128.6 + 14.6
=
301.4 kNm
5.2.3 Example 2 Calculate the moment and axial force that can be sustained by the section where 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
c>3
0
When the neutral axis is outside the section the ultimate compressive strain is less than 0.0035 and is given by:
0
0.002x
x
-
3 h17
0.002 x 600
600
-
3
x 500/7
=
0.0031
4.3.1.2(1) (viii) &
Figure
The conditions in the section are shown in Figure 5.2.
Figure 5.2 Conditions in section for Example 2
4.11
COLUMNS
can
The technique adopted for the calculation of NRd,c and MRd,c is to calculate the effect of the stress block on a depth of 600 mm and then deduct 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:
-
0.5667(1
-a1
=
'x2
N'Rd,c
R13)bxf,,k
__
C
Rd,c
Rd,c
where
c
h12-x(02-40
=
-
=
0
40
0
12
+6) ;and
0.002/E
Note: It will be found that, the first example.
0.0035, these equations give the values used in
=
if Eu
The equations for the force and moment produced by the part of the stress block lying outside the section are ANRd,c
A
=
-
0.5667«(1
al3)(x
-
h)b fck
ANRd,cC
Rd,c
where
c'
-
x
-
h12
(x
- h) (8 12-4a
a
=
Eb/0.002
Eb
=
strain at bottom of section
From the strain diagram,
Eb
=
3a)
0.00051
Hence a
=
N'Rd,c
= 0.5667(1
c
=
250
-
=
5.67
x 2802/1000 =
=
0.5667 x 0.255(1-0.255/3)(600-500)350 x 30/1000 = 139 kN
0.255
and a
0.645
0.645/3) x 350 x 600 x 30/1000 = 2802 kN U-)
-
=
x 0.645 +
4
12-4x0.645
6)
=
5.67 mm
E
-
+
600(0.6452
Hence
(A)
600
- 250 - (600 - 500)(8 - 3 x 0.255) 12 - 4 x 0.255 136
_
-284 mm E
C1
CY)
ANRd,c
15.9 kNm
E
M'Rd,c
COLUMNS
-139 x 284/1000
_
III
-39.4 kNm
=
C'7
AMRd,c F-I:
Hence =
2663 kN
15.9 + 39.4
=
55.3 kNm
G.)
=
MRd,c
-
139
= 2802
NRd.c
5.2.3.3 Steel forces and moments 0.0031
x 550
=
0.0028
COD
Strain in upper layer of bars =
600
=
643 kN
MRd,st
=
643 x 0.2 C')
NRd,sl
=
400 N/mm2
=
128 kNm
Strain in middle layer of bars
=
0.0031
x 350
x
This is > 0.002; hence s
=
0.00181
=
0.000775
600
Hence
f
S
NRd,s2
=
362 N/mm2
=
355 kN,
MRd,s2
Strain in bottom layer of bars
=
=
0.0031
0
x 150
600
Hence
f
=
155 N/mm2
NRd,s3
=
152 kN
MRd,s3
=
-30.4 kNm
S
NRd
=
2663 + 643 + 355 + 152
MRd
=
55.3 + 128
-
30.4
=
=
3813 kN
153 kNm
5.3 Biaxial bending capacity of a section General carry out a rigorous check of a section for biaxial bending by hand is very tedious but possible if the simplified rectangular stress block is used. It is not suggested that the example given here is a normal design procedure for common use but it could be employed in special circumstances. There would be no difficulty in developing an interactive computer program to carry out design, in this way, by trial and error.
((D
To
(CD
5.3.1
137
COLUMNS
5.3.2 Problem Demonstrate that 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 load of 3000 kN. The characteristic strength of the reinforcement is 460 N/mm2 and the concrete strength is 30 N/mm2. 500
0
I
500
0
40
Figure 5.3 Column section
5.3.3 Basic method
4.2.1.3.3(12)
The conditions in the section are shown in Figure 5.4.
Figure 5.4
Conditions in section
Note:
NAD
is assumed that EC20) Section 4.2.1.3.3(12) implies that « should as 0.8 for biaxial bending but the NAD(') would allow 0.85.
be taken
It
can be seen from the diagram that the axial force provided by the concrete NC
=
1-w
is given by 0.8bxc tcd
((DD
It
Table 3
COLUMNS
The moments about the centroid of the concrete section are given by
M
Nx
cX
c
where
h 2
M
xc- b tan6 2+
1
2xc
btan6
x_
btan6
c
2
I
6
1
0NCdb3tan6 cy
= 12
c
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.
Location of reinforcement E
Figure 5.5
The stress in a bar is given by
fs
=
(
200000 x 0.0035 ) z x
=
x, 700z
f yd
where xc
z
±
=
(b12
-
d')tan6
-
db
cos8
0.8
=
depth from top face of section to bar considered. This will be d' for top bars and h d' for bottom bars.
-
SCE)
db
-..off
0.-
=ND
The force in each bar is sAs 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. W-0
The total moments and forces carried by the section are the sum of the steel and concrete contributions. have to be found by iteration. poi
9
5.3.4 Initial data fcd
=
ck (f)
1.5
=
30 1.5
=
20 N/mm2
_G)
The correct values of x and
COLUMNS
Stress over upper 0.8 of the depth of the compression zone
ONcd yd
fYk
=
16 N/mm2
=
400 N/mm2
=
1.15
0, assume that the neutral axis is perpendicular to the direction of principal bending. This gives
0
(
0.6
CCD
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 15/ Ffor all elements carrying more than 70% of the mean axial force then the structure may be considered as non-sway.
Mean axial force =
sum of ultimate column loads no. of columns
x
x
4 x 2680 + 2 x 2960 + 4 x 4660 + 2 x 1700
NSd,m
700/oNsd,m
12
=
38680/12
=
2256 kN
=
3223 kN
i6-
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 = 2.8 m (i.e. value appropriate to a non-sway condition). X
=
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 the stiffer direction and non-sway in the less stiff direction.
Eqn A.3.2 it is specifically stated that the height should be in metres. Nothing is stated about the units for C, FV and E.. Since the output from Eqn A.3.2 is non-dimensional, the statement of the units is unnecessary unless the units for F and Ec are different to that for htot. Should IC, FV and Ec be in N and mm units while htot is in m? If this were so, then the structure would be found to be 'braced' by a large margin. In
I
,..°E ,
-CV
c
3`T
(1)
(1)
m
There are two possible areas where the drafting of EC2 is ambiguous and the wrong interpretation may have been made.
E
150
A.3.2(3)
COLUMNS
In A.3.2(3) it does not state whether X should be calculated assuming the columns to be sway or non-sway. In the calculation, the assumption was made that the X was a non-sway value. If a sway value had been adopted, the structure would have proved to be a sway frame by a considerable
(2)
(Q.
margin. Clearly, clarification is required
if
A.3.2 is to be of any use at all.
is possible to take this question slightly further and make some estimate at what the answer should have been. It
E
Considering the y direction, the ultimate curvature of the section of the 750 x 450 columns is 1
2x460xK2
r
0.2x106x1.15x0.9x700
=
6.35K22
x
10_6
0
Inspection of the design charts and levels of loading suggest K2 is likely to be about 0.6. Assuming an effective length under sway conditions of twice the actual height gives a deflection of: x
X
(2
3.5)2
=
19 mm
E E
x6.35x0.6
10
Cam.
OH-
This is an overestimate of the actual deflection. It corresponds to an eccentricity of 19/750 of the section depth or 2.5%. This must be negligible, hence, in the y direction, the structure must effectively be non-sway.
5.7 Sway structures 5.7.1
Introduction
(OD
Z30
5'w
°-'
-^.
((A
:(Q
Although EC2 gives information on how to identify a sway structure, it does not give any simple approach to their design. However, Clause A.3.5.(2) states that "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 which take account of second order effects ...... provided the average slenderness ratio in each storey does not exceed 50 or 201 v, whichever is the greater' .
E
EC2 Section 4.3.5 gives the 'model column' method which is developed only for non-sway cases, so it is left to the user to find a suitable method for sway frames on the basis of the Model Column Method. BS 8110 does this, so it is suggested that the provisions of 3.8.3.7 and 3.8.3.8 of BS 8110: Part 1 are adopted, but that the eccentricities are calculated using the equations in EC2.
-LO
5.7.2 Problem O-0
_0)
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. 0
The design ultimate first order moments in the columns are as shown in Figure 5.14.
a has been assessed from EC2 Figure 4.27(b) as
151
1.6 for all columns.
COLUMNS
Mo
Column
Mo
type
kNm
a
90
b
110
C
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
4
48.5
b
2
48.5
C
4
43.1
d
2
64.7
Mean value
Since
Xm
(a)
=
49.4
< 50, the simplified method may be used. 2K2
1
r
0.0044K2
200000 x 1.15 x 0.9d
Hence _
(1.6
x
3.5)2
d
x 0.0044K2 x
106
_
13800K2
mm
x
e2
x 460
A.3.5(3)
10
of
of
2.5.1.3 Eqn 2.10
200
100 14
This may be multiplied by «n
Eqn 2.11
Where, with 12 columns «n
_
(1
+ 1/12)/2
=
0.736
Hence =
0.00368
ea
=
0.00368 x 1.6 x 3500/2
etot
=
eo + ea
+
V
+ 13800K2/d
=
=
eo + 10.3 + 13800K2/d
The total eccentricities are shown in Table 5.3.
5
10.3 mm
Eqn 4.61
mm
Eqn 4.65
COLUMNS
Table 5.3
Total eccentricities d
Mo (kNm)
N
eV
(mm)
a
350
80
0.744
40.3 + 39K 2
b
350
110
37
0.822
47.3 + 39K 2
c
400
176
38
0.460
48.3 +
d
250
25
15
0.630
25.3 + 55K 2
`''
eo (mm)
3rD
Column type
(mm)
bhfck
30
35K 2
L()
As in the previous examples, the design charts can be used iteratively to establish K2
and hence
Table 5.4
e2. This
process gives the values shown
Lateral deflections K
a b
e2
No. of
(mm)
columns
0.39
15.2
4
0.41
16.0
2
c
0.50
((')
Column type
17.5
4
d
0.45
24.8
2
(D3
in Table 5.4.
z
17.7 mm
---1
=
E
Average deflection
columns will be assumed to deflect by the average value. The resulting designs are shown in Table 5.5 All
Table 5.5
Summary of designs AsfYk
A
bh2ck
N bhfck
bh ck
(mmz)
58.0
0.108
0.744
0.53
4148
65.0
0.134
0.822
0.75
5870
c
66.0
0.067
0.460
0.10
2201
d
43.0
0.090
0.630
0.38
2230
Column type
e (mm)
a b
M
BS 8110 3.8.3.8
6.1
Introduction defined as a vertical load-bearing member with a horizontal length not less than four times its thickness. A wall is
2.5.2.1(6)
3-0(5.
6.2 Example
0)c
The design of walls is carried out by considering vertical strips of the wall acting as columns.
nom
(so
SOS
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 spans on each side. The storey heights of each floor are 3.5 m, the height from foundation to the first floor being 4.5 m. The wall is fully restrained at foundation level. The building is a braced non-sway structure.
6.2.1 Design data Design axial load (NSd)
=
700 kN/m
Design moment at first floor
=
5 kNm/m
Design moment at foundation
=
2.5 kNm/m
Concrete strength class is C30/37.
fCk
=
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
to
=
4.3.5.3.5(1)
01col
where 1COl
=
actual height of the column between centres of restraint
Q.-
0 is a factor depending upon the coefficients
ko
and
relating to the
kR
rigidity of restraint at the column ends. col /lcol
E
4.3.5.3.5(1)
slab/Ieff,slab
Eqn 4.60 C))
=
CA)
kn
E
Assuming a constant modulus of elasticity for the concrete:
I
o
_
1
x
0.23
=
6.67
x 10-4
m4
1.3
x 10-3
m4
12 1
slab =
A
=
12
(6.67x10-4+6.67x10-4)I(2x1.3x 4.5
3.5
c,,
_
k
x 0.253
10-3)
=
0.78
6
Base of wall is fully restrained.
Therefore kB
=
4.3.5.3.5(1)
0.40 which is the minimum value to be used for
kA
or
kB.
Figure 4.27(a)
WALLS
Hence a
0.7
=
4500 mm
=
0.7
x 4500
The slenderness ratio
= l 0li
=
X
3150 mm
E
=
4.3.5.3.5(2)
where
=
i
radius of gyration _
I
1000 x 2003
57.7 mm E
TA
x 200
12 x 1000
Therefore 3150
X
=
54.6
57.7
Isolated columns are considered slender where or 15/ v.
X
exceeds the greater of 25
Nsd
4.3.5.3.5(2)
'4ccd
=
700
M
=
°k
=
30
=
103 mm2
E
cd
200 x
=
20 Nlmm2 E
= 1000 x 200
E
Ac
1.5
yC
Therefore 700 x 103
_
vU
=
0.175
200 x 103 x 20
Hence 15 u
15
=
35.9
0.175
Therefore the wall is slender
6.2.3 Design The wall may now be designed as an isolated column in accordance with EC2(1) Clause 4.3.5.6 and as illustrated in the example in Section 5. Although the column or wall has been classified as slender, second order effects need not be considered if the slenderness ratio X is less than the critical slenderness ratio Xcrlt.
WALLS
25 (2
=
Xcrt
-
4.3.5.5.3(2) Eqn 4.62
eotleoz)
where eoz
are the first order eccentricities at the ends
of
the member E
and
eo1
relating to the axial load. MSd,
=
e of
N
and e o2
MSdz
=
N Sd
Sd
MSdt
and
MSdz
are the first order applied moments.
Therefore 25 (2
-
MSdi/Msdz)
where MShc
`
MSdz
These moments must be given their correct algebraic signs
the equation.
this example: C)7
In
in
-
1-2.5)
=
50
L(7
25 2
Xcrit
62.5
>
54.6
4.3.5.5.3(2)
Design axial resistance (NRd)
Eqn 4.63
E
The column or wall should therefore be designed for the following minimum conditions: Nsd
=
h
Design resistance moment (MRd)
=
Nsd
x
Eqn 4.64
20
For this example
=
700 x 0'2 20
=
7.0
>
5.0 kNm
A
MRd
6.2.4 Reinforcement The vertical reinforcement should not be less than 0.004Ao or greater than
5.4.7.2(1)
0.04A C .
5.4.7.2(2)
The maximum spacing for the vertical bars should not exceed twice the wall thickness or 300 mm.
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 quarter of the vertical bar size and the spacing should not exceed 300 mm. The horizontal reinforcement should be placed between the vertical reinforcement and the wall face.
5.4.7.3
CT0
((DD
te)
Half of this reinforcement should be located at each face.
5
(1)-(3)
WALLS
Link reinforcement is required in walls where the design vertical reinforcement exceeds 0.02AC.
_T_
normal buildings it is unlikely that walls will be classified as slender. For practical considerations they will generally not be less than 175 mm thick and the vertical load intensity will normally be relatively low. Thus the limiting slenderness ratio given by 15/ vU will be high. In
_--.
cases where the wall is slender, only slenderness about the minor 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. ((DD
can
In
15
5.4.7.4(1)
FOUNDATIONS Ground bearing footings 7.1.1 Pad
footing
Design a square pad footing for a 400 mm x 400 mm column carrying a service 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 LO[)
With 500 mm deep base, resultant bearing pressure
=
200
-
0.5
x 24 = 188 kN/m2
Area of base required
1100
=
=
5.85 m2
188
Use 2.5 m x 2.5 m x 0.5 m deep base
Durability For components in non-aggressive soil and/or water, exposure class is 2(a).
Table 4.1
Zoo
7.1.1.2
ENV 206
Minimum concrete strength grade is C30/37.
Table NA.1
For cement content and w/c ratio refer to ENV 206 Table
3(6).
Minimum cover to reinforcement is 30 mm.
NAD
For concrete cast against blinding layer, minimum cover > 40 mm.
Table 6 4.1.3.3(9)
E
E
Use 75 mm nominal cover bottom and sides 7.1.1.3 Materials
=
460 N/mm2 E
Type 2 deformed reinforcement with yk
z
7.1
NAD 6.3(a)
Concrete strength grade C30/37 with maximum aggregate size 20 mm 7.1.1.4 Loading Ultimate column load
=
1.35Gk + 1.50k
=
1570 kN
Eqn 2.8(a) Table 2.2
7.1.1.5 Flexural design
Critical section taken at face of column 1570 (2.5 Msd
-
0.4)2
CC)
8x2.5
=
2.5.3.3(5)
z
7
346 kN m
Assuming 20 mm bars
dave
=
500 - 75 - 20
=
405
Using rectangular concrete stress diagram cd
=
Ck
'Y'
«cd
=
=
30
=
20 N/mm2
=
17 N/mm2
1.5
0.85 x 20
mm Figure 4.4
Eqn 4.4 Table 2.3
FOUNDATIONS
For reinforcement
=
vk
460
=
400 N/mm2
=
2.2.3.2P(1) Table 2.3
1.15
-Y'
0-2
fyd
E
For the design of C30/37 concrete members without any redistribution of
2.5.3.4.2(5)
moments, neutral axis depth factor
x d
2500
2179 mm2
-
2(75)
-
1519 mm2
.............................. 20
=
OK
388 mm
6
Maximum spacing
=
3h
> 500
=
500
> 388 mm
.......
OK
7T20 (EW) are sufficient for flexural design. Additional checks for punching and crack control require 9T20 (EM refer to Sections 7.1.1.7 and 7.1.1.8.
-
NAD Table 3 5.4.3.2.1(4)
Use 9T20 (EW)
7.1.1.6 Shear
Minimum shear reinforcement may be omitted in slabs having adequate provision for the transverse distribution of loads. Treating the pad as a slab, therefore, no shear reinforcement is required if Vsd _
As
4.4.2.3(2) 4.4.2.2
Eq n 4.78
For Act it is considered conservative to use (h/2)b
=
Us
100% x
460 N/mm2
=
fYk
For cteff use minimum tensile strength suggested by EC2 =
kc
3 N/mm2
0.4 for bending
a value for h =
For k interpolate
50 cm from values given
=
0.5 + 0.3(80
-
50)/(80
As,req
=
0.4
x 0.68 x
3
x 250 x 2500/460 =
As,prov
=
2830
k
-
-
30)
=
0.68
Therefore
(V)
CC)
>
1109 mm2
mm2 .......................... OK
1109
7.1.1.9 Reinforcement detailing
+
Check that flexural reinforcement extends beyond critical section for bending for a distance >_ d + lb,net =
lb
33.30 =
5.4.3.2.1(1)
& 5.4.2.1.3
667 mm
Assuming straight bar without end hook
1.0x667x2179
Ib, net
+
+
of
=
Eqn 5.4
514 mm
2830
lb,net
=
Actual distance =
405 + 514 2500 2
-
-
=
400
919 mm
-
75
=
975
2
The reinforcement details are shown in Figure 7.1.
>
919 mm
.....
OK
FOUNDATIONS
500
75 cover
I Figure
7.1
9T20
- 300EW
2500
Detail of reinforcement in pad footing
7.1.2 Combined
footing
Design a combined footing supporting one exterior and one interior column.
q.0
An exterior column, 600 mm x 450 mm, with service loads of 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
c
in
(J)
E
rectangular footing that cannot protrude beyond the outer face of the exterior column. The columns are spaced at 5.5 m centres and positioned as shown Figure 7.2.
Figure 7.2
(AD
_-j
The allowable bearing pressure is 175 kN/m2, and because of site constraints, the depth of the footing is limited to 750 mm.
Plan of combined footing
7.1.2.1 Base size
Service loads
=
Gk
+0
k
Column A: 1340 kN and Column B: 2000 kN Distance to centroid of loads from LH end
2000 x 5.5
=
3.593 m E
+
0.3 +
3340
164
FOUNDATIONS
For uniform distribution of load under base
=
2
x 3.593 say 7.2 m E
Length of base
((DD
With 750 mm deep base, resultant bearing pressure
=
175
-
Width of base required
0.75 x 24
=
157 kN/m2
3340
= 7.2
=
x 157
2.96
say3.0m
Use 7.2 m x 3.0 m x 0.75 m deep base
(n.
7.1.2.2 Durability
c(°
(On
For ground conditions other than non-aggressive soils, particular attention is needed to the provisions in ENV 206 and the National Foreword and Annex to that document for the country in which the concrete is required. In the UK it should be noted that the use of ISO 9690(15) 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.
Table 4.1 ENV 206
Minimum concrete strength grade is C30/37.
Table NA.1
For cement content and w/c ratio refer to ENV 206 Table 3.
NAD
E
Minimum cover to reinforcement is 30 mm.
Table 6
(ti
For concrete cast against blinding layer, minimum cover > 40 mm. it is suggested that nominal cover > 40 mm the above clause.
However, of
4.1.3.3(9)
is a sufficient interpretation
E E
a))
E
E
Use 75 mm nominal cover bottom and sides and 35 mm top
7.1.2.3 Materials Type 2 deformed reinforcement with
=
fYk
NAD 6.3(a)
460 N/mm2
Concrete strength grade C30/37 with maximum aggregate size 20 mm 7.1.2.4 Loading Ultimate column loads
=
1.35Gk + 1.50k
Eqn 2.8(a)
Column A: 1896 kN and Column B: 2834 kN
Table 2.2
Distance to centroid of loads from LH end 0.3 +
2834 x 5.5
=
3.595 m
4730
virtually at centre of 7.2 m long base
((DD
i.e.
Assume uniform net pressure
=
4730
=
657 kN/m
=
219 kN/m2
7.2
See Figures 7.3, 7.4 and diagrams respectively.
7.5 for loading,
shear force and bending moment
FOUNDATIONS
2834 kN
(/3160 kN/m
f t t-" Fib
Figure 7.3
+
4723 kN/m
657 kN/m
4.9
m
Loading diagram
Figure 7.4 Shear force diagram
Figure 7.5
Bending moment diagram
i
6
1.1m
.l
FOUNDATIONS
7.1.2.5 Flexural design 7.1.2.5.1
Longitudinal direction
-
top steel
Mid-span 2167 kNm
d
=
750
-
-
35
20
-
32/2
679 say 675 mm
=
E
=
E
MSd
E
Using the design tables for singly reinforced beams 2167 x 106
Mc14
=
0.123 < 0.45 limit with zero redistribution
=
0.064
d ASfyk
.........
OK
(.J
X
0.053
3000 x 6752 x 30
bd2fCk
2.5.3.4.2(5)
bdf,k E
E
= 8452 mm2 = 2818 mm2/m E
30
0.064 x 3000 x 675 x
AS
460 E
E
E
Use 12T32 @ 250 mm crs. (3217 mm2/m)
awe
Continue bars to RH end of base to act as hangers for links.
c
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 > 500
=
=
500 > 250 mm E
Maximum spacing
.........
OK
5.2.6.3P(1) & P(2)
NAD
Table 3 7.1.2.5.2 Longitudinal direction
-
5.4.3.2.1(4)
bottom steel
_Z)
At column face
398 kNm
of
=
750 - 75 - 10
398 x
mm
106
0.010
3000 x 6652 x 30
bd2fck ASfyk
665
(7N
Mlz
=
E
=
E
MSd
=
0.012
bdck 0.012 x 3000 x 665
x
A
S
30
= 1561 mm2 = 520 mm2/m
460
=
0.0015btd
=
Use 12T20 @ 250 mm crs. (1258 mm2/m)
998 mm2/m E
For minimum steel As,m.n
x
5.4.2.1.1
FOUNDATIONS
-
7.1.2.5.3 Transverse direction
bottom steel
-
2
0.45 Msd
=
-
1.5
219
x
2
=
178 kNm/m
Minimum steel governs. Use T20 @ 250 mm crs. (1258 mm2/m)
7.1.2.6 Shear Critical shear section at distance d from face of column
Column
B
Vsd
VRdl
4.3.2.2(10)
interior side =
1717
=
-T
TRd
-
0.675 x 657
=
1273 kN
k(1.2 + 40pr)bwd
4.3.2.3 Eqn 4.18 Table 4.8 C`7
'Rd
=
0.34 N/mm2
k
=
1.6
P,
=
0.00476
-
d
.9
1.0
=
1.0
Ensure bars are continued sufficiently. VRdi
=
VSd
>
957 kN VRdt
((DD
Therefore shear reinforcement required. ((DD
Shear capacity with links VRd3
=
Vcd
V
+
d
=
4.3.2.4.3
+ Vwd
VRdl
Eqn 4.22 (DD
Therefore V
d
V
d
1273
>-
A sw
x
957
=
316 kN
Eqn 4.23
0.9dfywd
=
d =
400 N/mm2, E
fywd
316 x 103
Asw
675 mm
= 1.30 mm2/mm E
>_
0.9
S
C11
S
x 675 x 400
NAD
EC2 Table 5.5 value.
Table 3
E
Where shear reinforcement is required, the minimum amount is 100% of the
5.4.3.3(2)
With
fyk
= 460,
Pwm
n
=
0.0012 by interpolation
168
Table 5.5
FOUNDATIONS
For links
AIsb SW
PW
Eqn 5.16
W
x 3000 =
0.0012
> 1.30 mm2/mm
3.6
E
Therefore minimum links govern. (1)
Determine link spacing, using EC2 Eqn 5.17-19. VRd2
Vsd/VRd2
Eqn 4.25
=
ofd bW(0.9 d)12
=
0.55 x 20 x 3000 x 0.9 x 675 x 10-3/2 = 10020 kN
=
1273/10020
=
0.13
< 0.2
=
smax
P-
(I1
Use EC2 Eqn 5.17 to determine link spacing. 0.8d (Note: 300 mm limit in Eqn 5.17 does not apply to slabs)
0.75d
=
5.4.3.3(4)
NAD 6.5(f)
506 mm
Transverse spacing of legs across section .
=
675 mm E
d or 800 mm E
3.6 mm2/mm
250
E
3.77
E
=
s
Vsd
957 kN
=
VRdl
Vsd
=
1273 kN (max.)
Vsd
<
3V
d
No further check required. VRdl
xb
=
=
0.830 m E
=
from face of columns A and B E
Vsd
4.4.2.3(5)
1502 - 957 xa
657 1.157 m
Check shear in areas where bottom steel is in tension and P,
=
OK
5.4.2.2(10)
Vd
Distances to where
VRd,
...........
Check diagonal crack control =
>
0.0015 (min. steel)
FOUNDATIONS
=
VRdl
0.34(1.2 + 0.06)3000
x 665 x 10-3 = 854 > 723 kN .. OK
(C]
orthogonal direction, shear at d from column face E
In
0
0
No links required at RH end of base
219(3.0-0.45-0.6x2)
148 kN/m
2
From above
-
854
VRdl
_
=
>
284
148 kN/m ...................
3.0
OK
No links required in orthogonal direction
7.1.2.7 Punching E
Length of one side of critical perimeter at 1.5d from face of column =
3
x 690 + 600 =
4.3.4.1P(4) &
4.3.4.2.2
2670 mm
This extends almost the full width of the base
=
3000 mm
Hence it is sufficient just to check line shear as above and shear around perimeter of column face, where 0.90
0.90 x
=
x
1555 mm2 E
12T32 gives As
x 0.53 x 3 x 750 x 3000/(2 x 460) = 1555 mm2 E
0.4
o')
>_
(Y)
As
OK
7.1.2.9 Detailing
Check bar achorage detail at ((D
LH
end.
The anchorage should be capable of resisting a tensile force
F
=
VSdaild
al
=
d
Fs
=
Vsd
=
column reaction
5.4.2.1.4(2)
Vsd
5.4.3.2.1(1)
=
1896 kN
The bond strength for poor conditions in the top of the pour =
0.7
x Table 5.3 value
fbd
=
0.7
x 3 =
1b
=
(014)(fyd/fbd)
5.2.2.2
2.1 N/mm2
=
47.60
= 1524 mm
Continuing all T32 bars to end A s. prov
=
9650 mm2
As,req
=
Vsdl yd
=
5.2.2.1 &
1896 x 1031400
4740 mm2
CA)
U)-
with
Eqn 5.3
FOUNDATIONS
Hence required anchorage, (3
at a direct support
)lb,net
(3) lb x 4740/9650 Anchorage up to face of column = The anchorage may be increased to the end of the bar.
500 mm >
=
-
600
lb,net'
Figure 5.12
0.31b
75 =
..........
OK
.....
OK
525 mm
preferred, by providing a bend at
if
5.2.3.3
Secondary reinforcement ratio for top steel
5.4.3.2.1
0-0
.--U
The requirement for transverse reinforcement along the anchorage length does not apply at a direct support.
p2
?
0.2p1
d
=
750 - 35 - 10
As
Al
670 mm2/m
>_
0.2 x 0.00476
=
=
705
=
0.00095
mm
Use T16 @ 250 mm crs. (804 mm2/m) transversely in top Spacing
1090 mm2
........................
OK
9.5.2 Maximum Maximum area of total tension and compression reinforcement 0.04AC
= 0.04 x 635880 = 25435 > 2052 mm2
....
OK
9.6 Reinforcement summary 11
tendons throughout beam 0
2T16s top and bottom throughout beam. Additional 2T20s top at support 2
These areas are within maximum and minimum
207
limits.
10 SERVICEABILITY CHECKS BY CALCULATION Deflection C°.
5-0
Calculate the long term deflection of 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.
1650
d'
=
50
ffAss
r
450
d = 390
As
2
250
Figure
Beam section
10.1
10.1.1
CD
N(9
Deflections will be calculated using the rigorous and simplified methods given in EC2(1), together with an alternative simplified method. The results will then be compared with the limiting span/effective depth ratios given in EC2.
Design data =
Span
:-'
10.1
7.0 m
Gk
=
19.7 kN/m
Qk
=
19.5 kN/m
A'
=
402 mm2
AS
=
2410 mm2
ick
=
30 N/mm2 (conc rete stre n gth class C30/37)
S
3.1.2.4 Table 3.1
10.1.2 Calculation method The requirements for the calculation of deflections are given in Section 4.4.3 and Appendix 4 of EC2. Two limiting conditions are assumed to exist for the deformation of concrete
A4.3(1)
sections (1)
Uncracked
(2)
Cracked.
Members which are not expected to be loaded above the level which would cause the tensile strength of the concrete to be exceeded, anywhere in the
member, will be considered to be uncracked. Members which are expected to crack will behave in a manner intermediate between the uncracked and fully cracked conditions. For members subjected dominantly to flexure, the Code gives a general equation for obtaining the intermediate value of any parameter between the limiting
conditions
A4.3(2)
SERVICEABILITY CHECKS BY CALCULATION
a
all +
=
-
(1
Dal
A4.3(2)
Eqn A.4.1
where
0
-(f)
« is the parameter being considered «l and «n are the values of the parameter calculated for the uncracked
and fully cracked conditions respectively is a
-
distribution coeffient given by 1
-
0102 (
a
A4.3(2)
Eqn A.4.2
)2
The effects of creep are catered for by the use of an effective modulus of elasticity for the concrete given by ECM Ec,eft
A4.3.(2) Eqn A.4.3
1+0 p"U
Bond and deterioration of bond under sustained or repeated loading is taken account of by coefficients a1 and a2 in Eqn A.4.2
Curvatures due to shrinkage may be assessed from
as
cs
°(n
E
1
e
A4.3(2) I
r cs
Eqn AAA
Shrinkage curvatures should be calculated for the uncracked and fully cracked conditions and the final curvature assessed by use of Eqn A.4.1. In accordance with the Code, the rigorous method of assessing deflections is to calculate the curvatures at frequent sections along the member and calculate the deflections by numerical integration.
ago
The simplified approach, suggested by the Code, 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
(1)
Moments
(2) (3)
Curvatures Deflections.
0
at frequent intervals along the member, to calculate 0
is,
E
The procedure
_-'
0-0
Here, calculations will be carried out at the mid-span position only, to illustrate this procedure, with values at other positions along the span being tabulated. 10.1.3.1
Calculation of moments For buildings,
it will normally be satisfactory to consider the deflections under the quasi-permanent combination of loading, assuming this load to be of long
A4.2(5)
duration.
The quasi-permanent combination of loading is given, for one variable action, by Gk
+ w2Qk
2.3.4 P(2)
Eqn 2.9(c)
09
SERVICEABILITY CHECKS BY CALCULATION
=
02
NAD
0.3
Table
1
Therefore Loading
=
19.7 + (0.3
x
moment("
CC]
Mid-span bending
25.6 kN/m
=
19.5)
25.6 x 72/8
=
156.8 kNm
=
10.1.3.2 Calculation of curvatures
order to calculate the curvatures it is first necessary to calculate the properties of the uncracked and cracked sections and determine the moment at which CAD
In
cracking
occur.
will
10.1.3.2.1 Flexural curvature
ECM
(Ec,eff) 1
For concrete strength class C30/37, Ecm
2A c
_
=
A4.3(2) Eqn A.4.3
+ ¢
+
The effective modulus of elasticity
2[(1650 x 150) + (250 x 300) ]
165 mm
=
2(1650 + 300)
U
3.1.2.5.2 Table 3.2
32 kN/mm2
For internal conditions and age at loading of 7 days
('')
3.1.2.5.5 Table 3.3
=
4)
3.1
Therefore 32
=
+
+
Ec,eff
1
7.8 kN/mm2
= 3.1
Effective modular ratio ( ae)
E5
_
Ec,eft
Modulus of elasticity of reinforcement
(ES)
=
200 kN/mm2
3.2.4.3(1)
Therefore =
a
200
=
25.64
7.8
e
AS
bd
=
S
bd
=
3.75x10-
-
6.25 x
3
1650 x 390
A' PI
2410 C11
_
=
p
=
402 1650 x 390
10_4
For the uncracked section, the depth to the neutral axis is given by bh2/2
=
= 165.2 mm E
X
- (b - b) (h - h) (h hf + hf) + (ae - 1) (A'Sd' + Asd) -2 bhf + bN,(h - hf) + (ae - 1) (A's + A)
SERVICEABILITY CHECKS BY CALCULATION
The second moment of the area of the uncracked section is given by
bh 3
I'
b W(h
+
-
12
h f )3
+ bhf (x
122
-
-
hil2)2 + bw(h
hf)
12
(h+hf_
Ajd-x)2
+ (ae-1) A'S(x-d')2 + (CYe-1)
X
= 7535 x
106 rnm4
2
For the cracked section the depth to the neutral axis is given by
d =
[Cep + (ae
0.345d
+ [aeP + (ae
- 1)p']
2
+ 2 [«eP+(«e
- 1)P' d
/]
134.6 mm
=
E
=
'r)
x
-
3
1
bd3
3 (
tip
=
In
+
dl
-
cxep(1
2
'
+
-
(Cle
Xl
1)P'
(d
2
d
5448 x 106 mm4
=
0.0556bd3
I>(
The second moment of area of the cracked section is given by
The moment which will cause cracking of the section is given by M cr yt
=
yt
-
h
x =
450
-
=
165.2 fctm
=
2.9 N/mm2
3.1.2.4
E E
For concrete strength grade C30/37,
284.8 mm
Table 3.1
Therefore
2.9x7535x106x10-6
Mcr
=
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
_
rl
M
_
156.8
x
7.8
Ec,effll
103
x
106
= 2.668 x 10-6 rad./mm
x 7535 x
106
Curvature of the cracked section is given by 1
_
M
_
156.8
x
106
= 3.690 x 10-6 rad./mm E
Ec,eff
III
7.8
x
103
x 5448 x
x
tip
ru
106
Having obtained the values for the two limiting conditions Eqn AAA is used to assess the intermediate value. Hence
i
=
(
) "
+
(1
'
A4.3(2) Eqn A.4.1
SERVICEABILITY CHECKS BY CALCULATION
_
-
,
asr
a, a2
)2
as
For high bond bars,
a,
=
1.0
For sustained loading, 62
=
0.5
Q
s
is the
stress
in
the tension steel calculated on the basis of a cracked section
Therefore S
- x)
25.64 x 156.8 x 106 (390
- 134.6)
E
aeM (d
= 188.5 N/mm2
5448 x 106
III
vsr is the stress in the tension steel calculated on the basis of a cracked section under the loading which will just cause cracking at the section considered. Therefore osr
=
aeMcr (d
-
X)
III
25.64 x 76.7 x 106 390
5448 x
( 106
92.2
=
-
134.6 )
92.2 N/mm2
=
Therefore
=
1
-
0.5 (
188.5
0.88
12
Note: osr
may be replaced by
[(0.88 x 3.69) + (1-0.88) x 2.668] x 10-6 = 3.567 x 10-6 rad./mm -ca
=
III
r
°r in the above calculation M
ors
1
M
10.1.3.2.2 Shrinkage curvature
The shrinkage curvature is given by S
1
E
cse of
I
A.4.3.2
Eqn AAA
res
where Ecs
is the free shrinkage strain
For internal conditions and 2AC/u ecs
=
0.60
= 165 mm
3.1.2.5.5
Table 3.4
x 10-3
0
cad
S is the first moment of area of the reinforcement about the centroid of the
section. 1
is the second moment of area of the section.
S and I should be calculated for both the uncracked and fully cracked conditions. Curvature of the uncracked section SI
=
AS(d
-
x)
- A'S(x -
d')
=
495.5
x
103 mm3
SERVICEABILITY CHECKS BY CALCULATION
_
1
0.60
resl
x 10-3 x 25.64 x 495.5 x 7535 x 106
103
x
1.0
10-6
rad.lmm
Curvature of the cracked section
- A'S (x -
AS(d
_
0.60x10-3x
x)
=
581.5
x
103 mm3
25.64 x 581.5 x 103
Therefore
=fix
r CS
1
+(1
-0x
1
rCS1
rCs«
[(0.88 x 1.64) +
_
x 10-brad./mm
1.64
5448 x 106
resn
1
d')
E
1
-
=
L,,
S
(1
- 0.88)
x 1.0] x 10-6 = 1.563 x 10-6 rad./mm
The total curvature at mid-span 1
1
rtot
r
+
1
= (3.567 + 1.563) x 10-6 = 5.130 x 10-brad./mm
rcs
(-)
The flexural, shrinkage and total curvatures at positions x11 along the span are given in Table 10.1.
Moment (kNm)
x/1
0
1
1
1
r.,
rw
0
1.000
1.000
1
r,
1
r
T.
0
0
0.960
-
-
0.960
1.000
1.960
0.2
100.4
1.708
2.363
0.708
2.171
1.453
3.624
0.3
131.7
2.241
3.100
0.830
2.954
1.531
4.485
0.4
150.5
2.561
3.542
0.870
3.414
1.557
4.971
0.5
L()
156.8
2.668
3.690
0.880
3.567
1.563
5.130
0.6
150.5
2.561
3.542
0.870
3.414
1.557
4.971
0.7
131.7
2.241
3.100
0.830
2.954
1.531
4.485
0.8
100.4
1.708
2.363
0.708
2.171
1.453
3.624
0.9
56.4
0.960
-
-
0.960
1.000
1.960
1.0
0
0
0
1.000
1.000
10.1.3.3 Calculation
0
0
())
56.4
0 0
0.1
(J)
0
OW)
0
`I1
Table 10.1 Curvatures x 106 (rad./mm)
of deflections
Having calculated the total curvatures, the deflections may be calculated by numerical integration using the trapezoidal rule.
The uncorrected rotation at any point may be obtained by the first integral given 1
eX
=
6X_ +
x11
by 1
rX + r 2
X_1
x nl
213
SERVICEABILITY CHECKS BY CALCULATION
tea)
(S=
Having calculated the uncorrected rotations, the uncorrected deflections may be obtained by the second integral given by =
ax
ex
ax_, +
+2 ex-1 / n1
1
((D
3-m
-(3
+L+
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.
the span
l
is
n
is the
number of span divisions considered.
Hence the uncorrected rotation at
8+(
r°.,
l+
1 r°
1
2 =
0 +
1.96 + 1.0 2
2
+
+
0 + (1.03
2
0)
x
7000
=
1.036 x
10_3
rad.
10
I
-9
0
l n 10_6
and the uncorrected deflection at all
0.11
0.11
n 10 _3
0 x 7000
=
0.363 mm
(DD
7-0
The uncorrected deflections may then be corrected to comply with the boundary conditions of zero deflection at both supports. This is done by subtracting from the uncorrected deflections the value of the uncorrected deflection at the right hand support multiplied by the fraction of the span at the point being considered. The values of the uncorrected rotations, uncorrected and corrected deflections at positions x/l along the span are given in Table 10.2. Table 10.2
Deflections (mm)
integral riot
1.000
x
2nd integral
Correction
0
0
Deflection
103
0
0
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
-
1.0
1.000
25.346
88.711
88.711
Off)
1.960
((7
t()
0.1
CJ)
0
1st
x 106
1
x/!
8.508 0
SERVICEABILITY CHECKS BY CALCULATION
Maximum deflection at mid-span span = < limit of
28 mm E
263
E
250
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.
E()
c
c
10.1.4
span
=
= 26.6 mm E
atct
From Section 10.1.3.2.1 the maximum bending moment M the moment causing cracking Mc = 76.7 kNm.
=
156.8 kNm, and
The maximum deflection of the uncracked section due to flexure 5w14
at
384Ec,eff II
=
25.6 kN/m
1
=
7.0 m
Ec,eff
=
7.8 kN/mm2
=
7535 x 106 mm4
.4-
w
II
Therefore I
384 x 7.8 x
103
1012
x 7535 x
=
13.6 mm
E
5x25.6x74x
_
a
106
The maximum deflection of the cracked section due to flexure 5 w14 an
384Ec,ettlu
=
5448 x 106 mm4 E
III
Therefore
5x25.6x74x1012 384 x 7.8 x
aII
103
x 5448. x
=
18.8 mm
106
Final maximum deflection due to flexure
a
=
=
aII + (1 1
-
0102
-
a,
M
)2
r
(_M_
A4.3(2)
Eqn A.4.1
SERVICEABILITY CHECKS BY CALCULATION
1.0
=
0.5
=
1
(J)
L'2
=
-
I2
176.7
0.5
=
0.88
-
0.88)
156.8
Therefore
=
a
x
(0.88
18.8) +
(1
x
=
13.6
18.2 mm
must be appreciated that the deflection calculated above is due to flexure due to shrinkage must also be assessed. The shrinkage curvature at mid-span from Section 10.1.3.2 It
only. The additional deflection
=
x 10-6 rad./mm
C))
1
rcs
1.563
l2 acs
=
8 f BSI
a + acs
x
72
x
106
8
=
18.2 + 9.6
=
=
9.6 mm
27.8 mm
E
=
atot
1.563 x 10-6
This figure is close to the rigorously assessed value of 26.6 mm.
10.1.5 Alternative
simplified approach O-o
An alternative simplified approach, which directly takes account of shrinkage, is given in BS 81100.
BS 8110: Part 2 Section 3
n':7
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 moment diagram, the deflection is given by =
a
Kl
2
_
BS 8110: Part 2 3.7.2
1
rtot
(r)
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 Total
0.104
curvature at mid-span, from Section 10.1.3.2 1
=
Table 3.1
5.130 x 10-6 rad./mm
riot
Therefore maximum deflection at mid-span
=
0.104
x
72
x 5.130 =
26.2 mm E
atot
BS 8110: Part 2 3.7.2
Again this is close to the rigorously assessed value.
SERVICEABILITY CHECKS BY CALCULATION
with span/effective depth ratio
10.1.6 Comparison
The procedure for limiting deflections by use of span/effective depth ratios is set out in EC2 Section 4.4.3. For the example considered
A s, req
=
2392 mm2
A s,prov
100As,prov
=
x
100
2410 mm2 2410
P
bd
=
0.37%
3.0 the basic span/effective depth ratio should be multiplied by a factor of 0.8
CAD
.C.
(Nil
m00
The span/effective depth ratios given in NAD Table 7 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 2501as for other stress levels, where as is the service stress at the cracked section under the frequent load combination. As a conservative assumption the Code states that the factor may be taken as 250
_
400 yk(As, reg )
as
A s p rov ,
Therefore, for this example, allowable span/effective depth ratio I
28 x 0 8 .
d I
d
(allowable)
=
19.6
400
460 x 2392/2410) (
>
l
d
(actual)
=
=
7000 390
19 6 .
=
180
the span/effective depth ratio is modified using the service stress in the reinforcement as calculated in Section 10.1.3.2.1 but adjusted for the frequent load combination If
,,a^)
aS
d
=
188.5
(allowable)
=
x 31.4/25.6
28 x 0.8 x
=
2310
231 N/mm2
=
21.6 > 18.0 (actual)
4.4.3.2(3)
SERVICEABILITY CHECKS BY CALCULATION
,--.
It can be seen from this example that whilst the span/effective depth ratio based on the calculated steel service stress suggests that the deflection should be well within the prescribed limits, the deflection from the rigorous and simplified analysis proves to be much nearer to the limit of span/250.
E
30°'=
_.C
000
This is due to the contribution to the deflection from shrinkage, which in this example is approximately a third of the total deflection.
°(O
C'7
The values of shrinkage strain given in EC2 Table 3.4 relate to concrete having a plastic consistence of classes S2 and S3 as specified in ENV 206(6). For concrete of class S1 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
ENV 206
7.2.1 Class
Table 4
Slump (mm)
S1
10
-
40
S2
50 -
90
S3
100 - 150
S4
>_
160
Thus for classes S2 and S3 the slump may vary between 50 mm and 150 mm. It is not logical that mixes with this variation of slump, and hence w/c ratio, should have a standard value of shrinkage strain.
3(D
v0)
If the values in EC2 Table 3.4 are assumed to relate to the median slump for classes S2 and S3 of 100 mm, then the values for slumps of 40 mm to 100 mm should be multiplied by a factor 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:
0.60x
10
0.45 x 10-3
ADD
x 10-) x 0.75 =
This figure relates more closely to the value which would be given in BS 8110, for the same example, of 0.4 x 10-3.
E
For the example considered, the calculated deflection due to shrinkage from the rigorous assessment would be 9.1
x 0.75
=
6.8 mm
and the total deflection from the rigorous assessment would be
atot
=
26.6
-
9.1 + 6.8
This is well within the limit of
=
span
24.3 mm
_
250
218
28 mm
BS 8110: Part 2 7.4 Figure 7.2
SERVICEABILITY CHECKS 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 cracking due to restraint of intrinsic deformation resulting in pure tension.
r
0
T16- 200 0
T12- 125
M
200
a
Figure 10.2 Wall section
10.2.1 Design data Concrete strength class is C30/37. Cover to reinforcement High bond bars with
= =
vk
NAD Table 6
35 mm
460 N/mm2
Exposure class 2(a)
10.2.2 Calculation method (CD
-s;
Requirements for the control of cracking are given in EC2 Section 4.4.2. Crack control is normally achieved by the application of simple detailing rules. (OD
(/)
O=-(D
The procedure for the calculation of crack widths is first to calculate the stress and hence the strain in the reinforcement, taking into account the bond properties of the bars and the duration of loading. Next, the average final crack spacing dependent on the type, size and disposition of the reinforcement and the form of strain distribution is established. 4.4.2.1(6)
(CD
The design crack width may then be obtained and compared with the limiting design crack width. In the absence of specific requirements, a limiting crack width of 0.3 mm will generally be satisfactory for reinforced concrete members in buildings with respect to both appearance and durability.
10.2.3 Check by calculation 10.2.3.1 Calculation
of steel stress and strain
Steel stress: a
=
S
kckct,e0Act 4.4.2.2(3)
AS
Eqn 4.78
where AS
=
area of reinforcement within the tensile zone
=
905 x 2
1810 mm2/m
E
=
SERVICEABILITY CHECKS BY CALCULATION
Act
kc
k
=
area of concrete within tensile zone
=
1000 x 200
=
a coefficient taking account
=
1.0
=
a coefficient allowing for the effect of non-uniform self-equilibrating
200 x 103 mm2
=
of
stress distribution
for pure tension
stresses =
0.8 for tensile stresses due to restraint of intrinsic deformations (h
300 mm)
'gin
=
ct.eff
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