Worked Examples for Design of Concrete Buildings
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FOR THE DESIGN
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Based on BSI publication DDENV1992-1-1: 1992. Eurocode 2: Design of concrete structures. Part 1. General rules and rules for buildings.
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Thisbook of worked examples has been prepared by: British CementAssociation Ove Arup & Partners and S. B Tietz & Partners The work was monitored by the principal authors:
A. W. Beeby BSc, PhD, CEng, MICE, MlStructE, FACI Professor of Structural Design, Dept of Civil Engineering, University of Leeds (formerly Director of Design and Construction, British CementAssociation), R. S. Narayanan BE(Hons), MSc. DIC, CEng,FiStructE Partner, S. B. Tietz & Partners, Consulting Engineers,
and R. Whittle MA(Cantab), CEng, MICE Associate Director, Ove Arup& Partners, and edited by: A. J. Threlfall BEng, DIC Consultant (formerly a Principal Engineer at the British Cement Association). This publication was jointlyfundedby the BritishCement Associationand theDepartmentofthe Environmentto promote and assist the use of DD ENV 1992-1-1 Eurocode 2: Part 1.
The BritishCement Association,BCA, isaresearchand information bodydedicated tofurtheringthe efficientand proper design and execution of concrete construction. Membership of BCA's Centre for Concrete Information is opento all involved in the construction process. BCA is funded by subscriptions fromcementproducers, through joint ventures, sales of publications, information and training courses, and the carrying out of research contracts. Full details are available fromtheCentre forConcrete Information, British Cement Association, Century House, Telford Avenue, Crowthorne, Berkshire RG11 6YS. Telephone (0344) 725700, Fax (0344) 727202. Ove Arup& Partnersisan internationalfirm offering awide range ofdesignand specialist services for theconstruction industry. S.
B. Tietz & Partnersoffer consultaricy services in civil, structural and traffic engineering.
A catalogue and prices for BCA publications can be obtained from PublicationSales, CentreforConcrete Information, at the aboveaddress. 43.505 First published 1994 10DM A 701A 1AA 1 Price group M © British Cement Association 1994
Published by
British Cement Association Century House, Telford Avenue, Crowthorne, Berks RG11 6YS Telephone (0344) 762676 Fax (0344) 761214 From 15Aprll 1995 the STD Code will be (01344)
All advice or information fromthe British Cement Association is intended forthose who will evaluate the significance and limitations of its contents and take responsibility for its useand application. No liability (including that for negligence) for any loss resulting fromsuchadviceor information is accepted. Readers should note that all BCA publications are subjectto revision fromtime to timeand should therefore ensure that they are in possession of the latest version.
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FOR THE DESIGN
OF CONCRETE
BUILDINGS
Based on BSI publication DDENV1992-1-1: 1992. Eurocode 2: Design of concrete structures. Part 1. General rules and rulesfor buildings.
Published by the British Cement Association in conjunction with: Ove Arup & Partners S.B. Tietz & Partners The Department of the 13 Fitzroy Street 14 Clerkenwell Close Environment LondonW1P 6BQ Clerkenwell 2 Marsham Street London EC1R OPQ Tel: 071-636 1531 London SW1P 3EB Tel: 071-490 5050 Tel: 071-276 3000
July 1994
FOREWORD Eurocode 2: Design of concrete structures, Part 1: General rules and rules for buildings (EC2)1 sets out both the principles for the design of all types of concrete structure, and design rules for buildings. Rulesfor other typesof structureand particular areas oftechnology, including precast concrete elements and structures, will be covered in other parts of EC2. EC2 contains aconsiderable number of parametersforwhichonly indicative valuesare given. The appropriate values for use in the UK are setout in the National Application Document (NAD)1 which has been drafted by BSI. The NAD also includes a number of amendments to the rules in EC2 where, in thedraftfor development stage of EC2, it was decidedthat the EC2 rules either did not apply, or were incomplete. Two such areas are the design for fire resistance and the provision of ties, where the NAD states that the rules in BS 8110(2) should be applied. Attention is drawn to Approved Document A (Structure) related to the Building Regulations 1991, which states that Eurocode 2, including the National Application Document, is considered to provide appropriate guidance forthedesign of concrete buildings in the United Kingdom. Enquiries of a technical nature concerning these worked examplesmay be addressed to the authors directly, or through the BCA, or to the Building Research Establishment.
CoNTENTS 1
8
1.1
INTRODUCTION AND SYMBOLS Introduction 5 5 1.2 Symbols
8.1
180 185
4
188
2 2.1
2.2
23 2.4 2.5 2.6 2.7 2.8 2.9
COMPLETE DESIGN EXAMPLE Introduction Basic details of structure, materials and loading Floor slab Main beam Edge beam (interior span) Columns Foundation Shear wall Staircase
SPECIALDETAILS
Corbels 8.2 Nibs 8.3 Simply supported ends Surface reinforcement
191
15
9 15 17
20 30 34 39 43 49
PRESTRESSED CONCRETE Introduction 193 9.2 Design data 193 93 Serviceability limit state 195 9.4 Ultimate limit state 204 9.5 Minimum and maximum areas of reinforcement 207 9.6 Reinforcement summary 207 9.1
SERVICEABILITY CHECKS BY CALCULATION 10.1 Deflection 208 10.2 Cracking 219 10
3 3.1
3.2
33 3.4 3.5
4
BEAMS 53 Introduction methods for shear 53 Design Shear resistance with concentrated loads close to support 63 70 method for torsion Design Slenderness limits 81
4.1
SLABS Solid and ribbed slabs
4.2
Flat slabs
5
COLUMNS Introduction 132 Capacity check of a section by strain compatibility 132 Biaxial bending capacity of 137 a section Braced slender column 141 Slender column with biaxial 143 bending Classification of structure 147 151 Sway structures
5.1
5.2
53 5.4 5.5 5.6 5.7
WALLS Introduction 6.2 Example
222 222
11.1
12
82 109
LOAD COMBINATIONS Introduction 12.2 Example 1 — frame 123 Example 2 — continuous beam 1 12.4 Example 3 — continuous beam 2 12.5 Example 4 — tank
236
12.1
237 240 243 245
DESIGN OF BEAM AND COLUMN SECTIONS 13.1 Concrete grades 246 13.2 Singly reinforced rectangular beam sections 246 133 Compression reinforcement 248 134 Flanged beams 249 13.5 Symmetrically reinforced 249 rectangular columns 13
...
6
6.1
DEEP BEAMS Introduction 11.2 Example 11
154 154
7
FOUNDATIONS Ground bearing footings 7.2 Pilecap design 7.1
158
REFERENCES
172
fIE
256
I. IHTRODUCTON AND SYMBOLS 1.1
Introduction and symbols
Themainobjective of this publication is to illustrate through worked examples how EC21 may be used in practice. It hasbeen prepared for engineers who are generally familiar with design practice in the UK, particularly to BS 8110(2). Theworked examplesrelateprimarilyto in-situ concrete building structures The designs are in accordance with EC2: Part 1 as modified by the UK National Application Document1. Where necessary,the information given in EC2 has been supplemented by guidance taken from other documents. Thecoreexample, in Section 2, is a re-designofthein-situconcrete office block used inthe BCA publicationDesignedand detailed (BS 8110: 1985), by Higgins & Rogers4. Other design aspects and forms of construction are fully explored by means of further examples in Sections 3 to 12. Equations and charts for the designof beam and column sections, taken from the Concise Eurocode for the design of concrete buildings5, are given in Section 13. Publications used in the preparation of this book, and from which further information may be obtained, are listedirt the References Unless otherwise stated, all references to BS 8110 refer to Part 1. Twoconventionshave been adopted inthe preparation ofthis book. Statements followed by
OK' mark places where the calculated value is shown
to be satisfactory. Green type is used to draw attention to key information such as the reinforcement to be provided. Thecalculationsare cross-referencedtothe relevantclauses in EC2 and, where appropriate, to otherdocuments; all references in the right-hand margins are to EC2 unless indicated otherwise. Thesymbols usedthroughout the publication are listed and defined below,and are generally those used in EC2 itself.
1.2 Symbols A
Ak
A A
A' Amin
Apr Areq
A11
A
A5, As,,,
Aymin
E,
Area of cross-section Area of concrete cross-section Area of concrete within tensile zone Area of concrete tensile zone external to links Area enclosed within centre-line of thin-walled section Area of prestressing tendons Area of tension or, in columns, total longitudinal reinforcement Area of compression reinforcement Minimum area of tension or, in columns, total longitudinal reinforcement Area of tension reinforcement provided
Area of tension reinforcement required Area of surface reinforcement Area of transverse reinforcement within flange of beam Area of tension reinforcement effective at a section or, for torsion, area of additional longitudinal reinforcement Area of shear reinforcement or torsion links Minimum area of shear reinforcement Effective modulus of elasticity of concrete
(tTROOUCTLOL &ND SYMBOLS
E0
E E5
F F5 FSd
Secant modulus of elasticity of concrete at transfer Secant modulus of elasticity of concrete Modulus of elasticity of reinforcement or prestressing steel Force due to concrete in compression at ultimate limit state Force in tension reinforcement or prestressing tendons Design value of tie force in pilecap
at ultimate limit state
Fsdp Design value of supportreaction Tie force in corbel or due to accidental action Fj Vertical force applied to corbel or, for sway classification of structures, sum F of all vertical loads under service conditions Characteristic value of permanent action or dead load Gk Gld
Characteristic dead floor load
Gkr
Characteristic dead roof load
H
Overall depth
H0
I
of tank
Horizontal force applied to corbel Second moment of area of cross-section Second moment of area of uncracked concrete section
1
Second moment of area of cracked concrete section Second moment of area of beam section Second moment of area of concrete section Second moment of area of column section
'slab
Second moment of area of slab section Second moment of area of section in x direction Second moment of area of section in y direction
J
-' K
K1 1
VSd
= 33.9 kN/m
OK
No shear reinforcement required Note: Since shear is rarely a problem for normally loaded solid slabs supported on beams, as the calculation hasshown, it is not usually necessary to check in these instances.
COMPlETE DESiGNEXAMPLE
2.3.7 Deflection Reinforcement ratio provided in span
=
x 149
1000
= 0.0025
Using NADTable 7(1) and interpolating between 48 for 0.15% and 35 for 0.5%, a basic span/effectivedepth ratio of 44 is given. By modifying according to the steel stress, the ratio becomes
NAD Table 7 4.4.3.2(4)
44(400x377) = 422 460 x342 The actual span/effectivedepth ratio is
5000
=
OK
33.6
149
Had EC2 Table 4.14 been used instead of NADTable 7, the basic ratio before modification would have been 35, which would not have been OK.
2.3.8 Cracking For minimum area of reinforcement assume ct.eff
=
4.4.2.2
3 N/mm2
k
= 0.4
k
=
0.8
A0
=
0.5
x 175 x
1000
= 87500 mm2
Hence
A
= kkftffA!o
Eqn 4.78
= 0.4 x 0.8 x 3
x
87500/460 =
183 mm2/m
Area of reinforcement provided = 377 mm/m No further check is necessary as h = 175 Maximum bar spacing
=
OK 200 mm
500 mm
3h
OK
4.4.23(1) NAD Table 3 5.4.a2.1(4)
2.3.9 Tie provisions The NAD requires that ties are provided in accordance with BS 8110. Internal tie in E—W direction, with Tie force
=
F
x
(g
+
7.5
qk)
x—lr
5
=
36 kN/m width, is given by
= 36 x
(4.7
+ 4) x — 5 =
7.5
5
41.8 kN/m
NAD 6.5(g) BS 8110 3.123.4
tONPtETEDESGI4 EXAMPLE
Minimum area
41.8
=
x
i0
=
460
91 mm2/m
Thus this area ofthe bottom reinforcementis the minimum that should be made continuous throughout the slab.
2.3.10 Reinforcement details The reinforcement details are shown in Figure 2.2.
B—B
COVER
o outer
bars 20
Figure 2.2 Slab reinforcement details
2.4 Main beam 2.4.1 Cover for durability and fire resistance Nominal cover for exposure class
1
(internal) is 20 mm.
NAD Table 6
Nominal cover for
1
hour fire resistance is 20 mm.
BS 8110 Table 3.5
Use 20 mm nominal cover to links
COMPLETEQESLG4 EXAMPLE
2.4.2 Loading Permanent load from slab (Section 2.3.3) Self-weight of beam =
(0.5
—
0.175)
Characteristic permanent load (g) Characteristic imposed load (q) Maximum design load = Minimum design load =
=
x 0.3 x
4.7
x
5 =
23.5 kN/m
24 = 2.3 kN/m
= 25.8 kN/m
= 5
x4
= 20 kN/m
l.3Sg + i.5qk = 64.8 kN/m = 34.8 kN/m
2.3.3.1
l35
2.3.22.(4)
2.4.3 Analysis 2.4.3.1 Idealization of structure and load cases
The structure is simplified as a continuous beam attached to columns above and below, which are assumed to be fixedat their upper ends and pinnedat the foundations, as shown in Figure 23.
3500
4000
Figure 23 Idealization of structure 2.4.3.2 Design moments and shears These are summarized in Table 2.2 and Figures 2.4 and 2.5.
2.533
COMPLETE DESIGN EXAMPLE
Table 2.2 Results of frame analysis Loadcase 1
Load case2
Loadcase3
Load per m on 8 m span (kN) Loadper m on 6 m span (kN)
64.8 64.8
64 8
34 8
34.8
64.8
Upper LH column moment (kNm) Lower LH column moment (kNm)
103 68
109 72
50
LH end of 8 m span moment (kNm) LH end of 8 m span shear (kN)
33
—171
—180
233
238
—82 119
242
256
116
RH end of 8 m span moment (kNm) RH end of 8 m span shear (kN)
—382
—345
—242
286
280
159
Upper centre column moment (kNm) Lower centre column moment (kNm)
33 18
55 29
3 2
Middle
of 8 m span moment (kNm)
LH end of 6 m span moment (kNm) LH end of 6 m span shear (kN) Middle
—331
—262
—247
240
146
223
98
20
130
—57
—12 63
—76
7 5
46
of 6 m span moment (kNm)
RH end of 6 m span moment (kNm) RH end of 6 m span shear (kN)
149
34
Upper RH column moment (kNm) Lower RH column moment (kNm)
REDISTRIBUTION Case 1 — Reduce Reduce Case 2—Reduce Reduce Case
3— No
22
AND AT 171 to 126 (see 2) 382 to 268 (—30%) 180 to 126 (_30%) 365 to 268 (see 1)
382
166
30
(1
redistribution
(2) 180
(2) elastic
(R)
(3) eLastic
76
(3)
(1) redistributed (2) elastic (1)
&
325 Envelope
Moments
in
kNm
Figure 2.4 Bending moment envelope
(2) redistributed
COAPLETE
SGI EX&MPLE
(1) (3)
(R) 242
(1) 233 (3)
240 223
119
Case 1 Case 2 Case 3
149 (1) 166 (3)
Redistributed Envelope
Forces in
277 (R)
286 (1)
kN
6000
8000
-J
—I——
Figure 2.5 Shear force envelope
2.4.4 Reinforcement for tiexure 2.4.4.1 Internal support From bending moment envelope
M = 268kNm 8 1iim
=
0.7
=
0.0864 and WIim
and xld
M
—
(8 — 0.44)/1.25
=
= 0.1084 (Section 13, Table 13.2)
268x106
— —
300
bdç<
0.208
=
x 4402 x 32
0.1442
> hm
Therefore compression reinforcement is required = Asfyk =
I'IIm
bdfck
0.87 (1— d'Id)
— = 0.1442 0.0864 —
0.87 (1
50/440)
= 0.0750 (Section 13)
Af = = —-
wurn
bdfck
+ w' = 0.1084 + 0.0750 = 0.1834 (Section 13)
= 0.1834 x 300 Since d'Ix
=
d'/0.208d
Increase w' to 1
—
0.429
1 — 0.546
A' = S
0.0943
x
300
x 440 x
32/460
= 0.546 > 0.075
—
çk'805)
=
= 0.0943
x 440 x
Use 4T25 (1960 mm2) top Use 2T25 (982 mm2) bottom
(1
= 1684 mm2
32/460
= 866 mm2
0.429
2.5.3.4.2 Eqn 2.17
COMPLETE DEStGIL EX&MPLE
2.4.4.2 Near middle of 8 m span From bending moment envelope M = 325kNm 8
>1.0
Effective flange width =
x
325
=
1660
300
+ 0.2 x 0.85 x 8000 =
108
x 4502 x 32
1660 mm
2.5.2.2.1 Eqn 2.13
= 0.030
x/d = 0.068 (Section 13, Table 13.1) Neutral axis is in flange since x =
A8
=
31
< 175 mm
0.035 (Section 13, Table 13.1)
= 0.035 x 1660 x 450 x 32/460 = 1819 mm2
Use 4T25 (1960 mm2)
2.4.4.3 Left-hand end
I
of 8 m span
From bending moment envelope
M
= l26kNm
8
= 0.7 and
= 0.0864 (Section 13, Table 13.2)
126
= 300
x
x
4402
106
=0.0678<
x 32
tim
Therefore no compression reinforcement is required.
= 0.084 (Section 13, Table 13.1) A8
= 0.084 x 300 x 440 x 32/460 = 772 mm2
Using 2T25 bent-up bars, minimum diameter of mandrel
=
13 (Asreq/A,pr,) =
Use 2125 (982 mm2) with
104
r=5
5.2.1.2
NAD Table 8
PETEDESGI4 EXMAPLE 2.4.4.4 Right-hand end of 6 m span From bending moment envelope
M = 76kNm
M
=
76
= 300
bd2fCk
x
106
x 44Q2 x 32
w
= 0.049 (Section 13, Table 13.1)
A
= 450 mm2
Use 2125 (982 mm2) with r
=
4qS
=
0.041
minimum
2.4.4.5 Near middle of 6 m span From bending moment envelope
M = l38kNm Effective flange width
= 300 +
138
= 32
x
x
4502
0.2
108
x 1320
x
0.85
=
x
6000
= 1320 mm
0.0161
= 0.019 (Section 13, Table 13.1) AS
= 0.019
x 1320 x 450 x 32/460 = 785 mm2
Use 2125 (982 mm2) 2.4.4.6 Minimum reinforcement
kkfAIcr
A5
4.4.2.2 Eqn 4.78
where
k
=
0.4
k
=
0.68
=
3 N/mm2
= 300 x 325 mm2 = 460 N/mm2 Therefore
AS 0.6bd
173 mm2
.
0.0015 bd =
OK
203 mm2
OK
5.4.2.1.1(1)
COMPLETE DESIGN EXAMPLE
2.4.5 Shear reinforcement
4.a2
2.4.5.1 Minimum links Here, for comparison with BS 8110 design, grade 250 reinforcement will be used.
5.4.2.2
Interpolation from EC2 Table 5.5 gives Minimum =
= 0.0022 x 300 = 0.66 mm2lmm
A,,js
If VSd
0.0022
refer to Section 2.4.5.3 for VRd2
—
(--)
= 300 mm Sm = lesser of 300 mm or 0.8d Use R12 links @ 300 mm crs. (A/s
2.4.5.2 Capacity of section VAd1
=
TRdk(l .2
Eqn 5.17
= 0.75 mm2/mm)
without shear reinforcement +
40p1)
4.3.2.3
bd
Assume 2125 effective p1
= 982/(300 x
440)
k
=
1.6 —
1.6 — 0.44
=
0.35
TRd
VRd1
d =
=
0.00743
=
1.16
Table 4.8
= 300 x 440 x 0.35 x 1.16 x (1.2 + 40 x 0.00743) x
iO = 80.2 kN
2.4.5.3 Shear reinforcement by standard method
of section
Maximum capacity
= 0.7
4.3.2.4.3
—
k'20° =
0.7
—
32/200
=
0.54
i
0.5
= 0.5 x 0.54 x (32/1.5) x 300 x 0.9 x 440 x iO = 684 kN Design shear force is shear at a distance d from the face of the support. This VRd2
is 590 mm from the supportcentre(ine. A s
VSd
= 0.9
—
x 440 x
80.2 0.87
x 250
=
0.0116 (VSd — 80.2)
Design of shear reinforcement is summarized in Table 2.3.
Eqn 4.21
Eqn 4.25 43.2.2(10)
Eqn 4.23
CO&PL Qc.SG4EX&MPLE Table 23 Design of shear reinforcement Location
8 m span LH end
203 248
1.42
LH end
202
1.41
RH end
128
mm.
RH end
6 m span
s for 12 mm
AJs
VSd
1.95
links
Links
159 116
R12 @ 150 R12 @ 100
160 max.
R12 @ 150 R12 300
c
R12 @ 300
Minimum
2.4.6 Deflection Reinforcement percentage at centre of 8 m span
=
100
x 1960/(450 x
1660)
4.43.2
= 0.26%
Interpolating between 0.15 and 0.5%, basic span/effective depth ratio for end span = 40
NAD Table 7
To modifyfor steel stress multiply by 400/460 To modifyfor T section multiply by 0.8 To modifyfor span
> 7 m multiply by 7/8
= 40 x (400/460) x 0.8 Actual ratio = 8000/450 = 17.8 Therefore permissible ratio
x 7/8
= 243 OK
2.4.7 Cracking For estimate of steel stress under quasi-permanent loads
4.4.2.2
= 64.8 kN/m = 0.3 Assuming 2 Quasi-permanent load = 03 x 20 +
NAD
Ultimate load
Approx. steel stress at midspan
25.8
= 460 x 1.15
= 31.8 kN/m 31.8
=
Table 1
196 N/mm
64.8
Approx.steel stress at supports allowing for 30% redistribution = 196/0.7 = 280 N/mm2 These are conservativefigures since theydo notallowfor excess reinforcement over what is needed or for moment calculated at centreline of support rather than at face of support. Check limits on either bar size or spacing. From EC2 Table 4.11,25 mm bars in spans are satisfactoryat any spacing since steel stress 200 N/mm2 OK
<
From EC2 Table 4.12, bar spacing at supports should be 150 mm with no limitationon size. As bars are located inside column bars the maximum possible OK spacing is 125 mm
4.4.2.3
COtaPLETEDESIGNEX&MPLE
2.4.8 Curtailment
of reinforcement
Reinforcement must extend for a distance of a, envelope where a,
= 0.9d12
25 1'Qnet=
a,
+
¼net
beyond the moment
5.4.2.13
= 198 mm 460 —
x
1.15
— 1
= 782mm
5.2.a4.1
3.2
net = 980 mm
+
Bars mark 8, whichare located outside theweb, mustextend afurther 150mm — refer to Figure 2.&
2.4.9 Reinforcement details Curtailmentofthemain reinforcementand arrangementofthe linkreinforcement are shown in Figures 2.6 and 2.7. Reinforcement details are shown in Figure 2.8 and given in Table 2.4. 1400 1600
1800
1800
76
Moment envelope Curtailment Line — ——
Figure 2.6 Curtailment diagram of main reinforcement 150 , -
R12
1450'
300
(minimum
242
c/c
300 - ,iOO links) O'(minimum links) 240 - Shear capacity of —
vRdl+ Vwd= 147kN
I 2150,1
286
Figure 2.7 Arrangement of link reinforcement
minimum Links (R12 —300) with
rs
5.4.2.13
tOThPT
5OI4 EXMJIPLE
Figure 2.8 Main beam reinforcement details
COMPLETEDStGN EX&MPLE
Table 2.4 Commentary on bar arrangement Bar marks 1
Notes Tension bars are stopped column bars
20 + 12 = 32 > 25 mm
Nominal cover
2
50 mm from column face to avoid clashing with the
Remaining tension bars stopped off at LH end as shown in Figure 2.6. Bars extended at RH end to provide compression reinforcement (lap = and continuity for internal ties (lap = 1000 mm) Check minimum distance between bars
OK
'iit)
bar size or 20 mm
(300—32x2—4x25)/3=45>25mm 3
Not used
4
Similar to bar mark 1
510
4.13.3(5)
BS 8110 5.2.1.1
OK
Loose U bars are fixed inside the column bars and provide continuity for columnand internal ties
lop legs project from centre line into span,minimum dimensions shown in
5.4.2.13
Bottom legs are lapped 1000 mm to provide continuity for internal ties
BS 8110
Figure 2.6
5
=
lop legs
1800 mm
+ 1000 =
Bottom legs
200
Use r =
both bends
5 for
1200 mm
Notethatthe bottom legs are raised to avoid gap between bars being < 25 mm 10
lop legs
5.2.1.1
1500 mm
Bottom legs = 200 + 1000 = 1200 mm 6,9
216 provided as link hangers are stopped 50 mm fromcolumn face
7,8
Tension bars over the support are stopped Bars mark8 are located outside the web
11
as in Figure 2.6.
Links are arranged in accordance with Figure 2.7 for shear. Links also 150 mm at all laps provide transverse reinforcement with a spacing
5.4.2.1.2(2)
5.2.4.1.2(2)
2.5 Edge beam (interior span) 2.5.1 Cover for durability and fire resistance Nominal cover for exposure class 2b (external) is 35 mm. Nominal cover for 1 hourfire resistance is 20 mm. Use 40 mm nominal cover to links
NAD Table 6 BS 8110 Table 3.5
CQAPLE.TE ES(GI4 XMAPLE
2.5.2 Loading Permanent load from slab
= 4.7 x 5 x 1.25 = 29.4 kN
(assuming 1.25 m strip to be loading on edge beam)
= (035 — 0.175) x 03 x 5 x 24 = 6.3 kN Cladding load @ 5 kN/m = 5 x 5 = 25 kN Characteristic permanent load = 60.7 kN Characteristic imposed load = 4 x 5 x 1.25 = 25 kN Self-weight of beam
=
Total design load
125
x 60.7 + 1.5 x 25 = 119.5 kN
2.5.3 Design moments and shears These are taken from the Concise Eurocode, Appendix, Table 2.5.3.1
Interior support = 0.10 x
Moment Shear
119.5
x 5 = 59.8 kNm
= 0.55 x 119.5 = 65.7 kN
2.5.3.2 Mid-span Moment
=
0.07
x
119.5
x 5 = 41.8 kNm
2.5.4 Reinforcement for flexure 2.5.4.1
Interior support
= 280 mm M = 59.8 x 106 = 0.079 x 300 x 32 2802 bd2ç
Assume effective depth
Alyk
= 0.099 (Section 13, Table 13.1)
bdfCk
xld
=
A
= 579 mm2
0.189
<
OK
0.45
2.53.4.2(5)
Use 2120 (628 mm2)
2.5.4.2 Mid-span Assume effective depth Effective flange width
= 290 mm
= 300 ÷ 0.1 x 0.7 x 5000 =
650 mm
2.5.2.2.1
COMPLETE DESIGN EXAMPLE
M
650
bd2fk
Afsyk
=
x iO
41.8
—
x
= 0.024
2902 x 32
0.028 (Section 13, Table 13.1)
bdfck A5
= 367 mm2
Use 2120 (628 mm2)
The cross-section is shown in Figure 2.9.
650
-j S
.
-I70
2120
. 1
21 20
I
S
350
60
T
300
Figure 2.9 Edge beam cross-section
2.5.5 Shear reinforcement Design shear force may be taken to be at distance d into the span from the face of the support. This can be calculated approximately as VSd
= 65.7 —
V
= 300 x 280
Adi
=
119.5 (0.28
43.2.2(10)
+ 0.15)/5.0 = 55.4 kN
x 0.35 (1.6 — 0.28) x
1.2
+ 40
300
x 628 x 280
4.3.2.3(1)
58.2 kN
This is greater than
Vsd,
hence only minimum links are required.
4.3.2.2(2)
Assuming grade 250 reinforcement for links, EC2 Table 5.5 gives
p
= 0.0022
Hence
A
— S VRd2
= 0.0022 x 300 = 0.66 mm2/mm =
0.5
(
32 32 07 . —— x — x 300 x 200)
1.5
0.9
x 280 = 435 kN
43.2.3(3)
tOJkP.ErEDESP4 EXAMPLE
Since
< (j-) VRd2,
VSd
200 mm spacing gives
A
Smax
= 0.8d = 224 mm
Eqn 5.17
= 132 mm2
Use RiO links at 200 mm crs. (A5
= 2 x 7&5 = 157 mm2)
2.5.6 Deflection = 5000/290 =
Actual span/effective depth ratio
17.2
At mid-span 100 A5
bd
= 100 x 628 = 033% 650
x 290
By interpolation from NAD Table 7, modified for Basic span/effectivedepth ratio
= 460
= 36
Note:
Thiscan be increased allowing for use of a largerthan required steel areato
= 36 x 628/367 = But not greater than 48/1.15
61.6
4.4.3.2(4)
= 41.7
NAD Table 7
Inspection shows this to be unnecessary. Allowable
l/d >
2.5.7 Curtailment
actual
Note 2
lid
OK
of reinforcement
Since the bending moment diagram hasnot been drawn, simplified curtailment rules are needed. These are givenin Section 8 ofthe Appendix tothe Concise Eurocode. Using the rules, the 20 mm bars in the top may be reducedto 12 mm bars at a distance from the face of support
+ 32cb + 0.45d
Concise Eurocode Figure A.12
= 500
+ 32 x 20 + 0.45 x 280
=
0.11
=
1266 mm from the column face
COMPLETE DESIGN EXAMPLE
2.5.8 Reinforcement details The reinforcement details are shown in Figure 2.10.
150
I—
2T20-4 top cover 60 side cover 75
2112—2
side cover 75 ELEVATION
Cover
to links
3443
40
Li
U
1221 A-A
Figure 2.10 Edge beam reinforcement details
2.6 Columns 2.6.1 Idealization of structure The simplification assumed for the design of the main beam is shown in Figure 2.3.
2.6.2 Analysis Moments and column loads at each floor are taken from the ana'ysis for the main beam given in Section 2.43.
2.6.3 Cover for durability and fire resistance Nominal cover for interior columns (exposure class 1) is 20 mm. Nominal cover for exterior columns (exposure class 2b) is 35 mm. Nominal coverfor 1 hour fire resistance is 20 mm. Use 20 mm (interior) and 40 mm (exterior) nominal cover to links
NAD Table 6
COMPLETE DES)G1 EXAMPLE
2.6.4 Internal column 2.6.4.1 Loading and moments at various floor levels These are summarized in Table 2.5. Table 2.5 Loading and moments for internal column Column design loads
Beam loads (kN) Total
Loadcase Roof
8m
6m
Dead
Imposed
1
2
240 202
238 165
1
2
53 43
51
6
278 143
283 236
6m
2
187 159
187 159
9
9
96
57
355
355
131 110
126 17
152 126
152 126
9
9
Self-weight
2nd floor 8 m
283 236
6m
278 143
337
200
642
642
131
126 17
152 126
152 126
9
9
110
Self-weight
1st floor 8 m
286 240
6m
578
343
929
929
280
132 111
126 17
154
146
14
154 129 14
1226
1226
129
Self-weight
Foundations
486
821
Bottom
Top
1
Self-weight
3rd floor Bm
Column moments (kNm)
(kN)
1
2
32
42
33
1
2
30
39
30
55
33
55
49
49
33
29
18
2.6.4.2 Design for column between first floor and foundation Effective height in N—S direction "top
0.5
—
x
675 x
106
5000
+ 675 x
106
3500
3125 x
106
8000 =
0.28 but take not less than 0.4
=
0.8
+
÷
3125
x
6000
106
Eqn 4.60
kbottom Hence Effective height = 0.8 Load case
1
Figure 4.27
x 5000 = 4000 mm
gives worst condition (by inspection).
Imposed load
= 0.7 x 821 =
575 kN
BS 6399: Part 1, Reduction factor
CO&PLETE DESIGN EXAMPLE
Dead load = 1226 kN Nsd
M
=
1801 kN
=
18 kNm (top), 0 (bottom)
x iO x
1801
=
3002
15Ji =
1.5
x 32
= 0.94
43.53.52
< 25
14.5
Hence X
mm
= 25
4000j
=
=
= 46
300
Note:
iji
=
x
(10/h)
X > 25, hence column is slender in N—S direction The slenderness in the E—W direction will be found to be approximately the same.
Thestructure is braced and non-sway(by inspection),hence theModel Column Method may be used with the column designed as an isolated column. =
25(2
—
e1/e2) = 50 in both E—W and N—S directions
Slenderness ratios in both directions are less than hence it is only necessary to ensure thatthe column canwithstand an end moment of at least Nsdh/2O
=
1801
x 03/20 = 27.0 kNm
43.5.53 Eqn 4.64
This exceeds thefirst order moments. Hence Nsd
=
=
1801
kN and Msd
= 27.0 kNm
0.62
bhfck
= bh2fk
x 106 = 0.031 300 x 32 27.0
Assume
d'
= 45mm
43.5.53 Eqn 4.62
I
COMPLETE DES)GT4 EXAMPLE
Then
d'/h
= 45/300 =
Af —--
=
0.15
0.16 (Section 13, Figure 13.2(c))
bhfck
Hence
= 1002 mm2
A
Use 4120 (1260 mm2) Note:
In the design by Higgins and Rogers, the slenderness ratio exceeds the equivalent of >tcrit but the design moment is still Nh/20. EC2 requires less reinforcement due to the smaller design load and the assumption of a smaller cover ratio. If the same cover ratio is used in the Higgins and Rogers design, 4T20 are sufficient in both cases.
2.6.5 External column 2.6.5.1 Loading and moments at various levels These are summarized in Table 2.6. Table 2.6
Loading and moments for external column Beam loads
Column design loads (kN)
(kN) Total
Loadcase
Dead
Imposed
1
2
184 55
186 55
Column moments (kNm) Bottom
Top
1
2
1
2
1
2
39
41
145
145
104
107
55 9
55
1
2
93
98
93
98
103
109
Roof Main Edge Self-weight
9
39
41
209
209
109
114
126 55 9
126 55 9
148
155
399
399
109
114
126 55 9
126 55 9
257
269
589
589
108
113
125 9
125 55 9
778
778
3rd floor Main Edge
235 55
240 55
Self-weight
93
98
2nd floor Main Edge
235 55
240 55
Self-weight
93
98
let floor Main Edge
233 55
238 55
55
Self-weight
Foundations
365
382
68
72
COIAPLETE DESIGN EXMAPLE
2.6.5.2 Design for column between first floor and foundation
x
=
106
(675 4000
x 0.5 + 675 x
3500106)
—
3125 x 106 8000
= 0.71
kbottom Hence
=
0.85
Effective height
=
Slenderness ratio
Figure 4.27
0.85
=
x
i/i =
4000
3400
= 3400 mm
fi
300
= 393
v will be small so isf7 will be less than 25 Hence
X. X
=
25
> 25, therefore column is slender in N—S direction
Calculate
=
bottom moment
top moment
e02
= o = 0 85
Hence
= 25(2 + 0)= 50 Slendernessratios inthe E—W and N—S directions are both lessthan 50, hence it is only necessary to ensure that the end moment is at least Nh120.
Theworst condition occurswith load case2 at section just above the firstfloor, where Msd is greatest. Nsd
Nh —
=
589 + 0.8 x 269
= 804
20
x 03
20
Design end moment Hence Nsd
=
= 804 kN
12.OkNm
= 109 > 12 kNm
= 804 kN and Msd =
109 kNm
43.5.5.3
DSt4 XMJIP.E 2.6.6 Reinforcement details Maximum spacing of links for internal column 12 x 20 = 240 mm Generally
5.4.1.2.2(3)
NAD Table 3
0.67 x 240 = 160 mm
Above and below floor
Maximum spacing of links for external column 12 x 25 = 300 mm Generally
At lap and below floor
0.67
x 300 =
5.4.1.2.2(4)
200 mm
The reinforcement details are shown in Figure 2.11. INTERNAL COLUMN F2
Links
Vertical bars
Section
EXTERNAL COLUMN Fl
Links Vertical bars
0 U,
'0I
.—
.
.-- --. J
0
C
U,
-
6'1
6',
0 0 m
®I
-3 ,
,
I
-
@. to (inks 30 Fdn"
g
Ilu,
I II I I ,
@i '.5 .;s I
—
o: 3E
Fdn.
L.....i.
t
cover
to Links
60
I
US
''?
see Fig 2 13 CM
"-3
'0
®H__ 11 Cover
0 Os
11
@1
IN
II I
E C
U, Cl IN
®6 il.J
III ----- .i ".
U
1st."
Section
I
Figure 2.11
Column reinforcement details
2.7 Foundation Design typical pad footing for internal column.
2.7.1 Cover Use 50 mm nominal cover against blinding _____ _____ BS 8110 specifies a nominal cover of not less than 40 mm against blinding. EC2 specifies a minimum cover greater than 40 mm. Thisimplies a nominal greater than 45 mm, hence the choice of 50 mm.
cer
2.7.2 Loading Taken
from internal column design.
Ultimate design loads:
Dead
=
1226
imposed
=
575
Total
=
LJ
1801 kN
4.1.3.3(9)
COMPLETE DESIGN EXDMPLE
Hence service loads:
Dead
=
908
Imposed
=
383
Total
=
1291 kN
The assumption is made that the base takes no moment. Also it is assumed that thedead weight of the base less the weightof soil displaced is 10 kN/m2 over the area of the base.
of base
2.7.3 Size
Since, at the timeof publication, EC7: Geotechnical design and EC2, Part 3: Concrete foundations10 have not been finalized, the approach used here is based on current UK practice.
Use2.75m x 2.75m x 0.6mdeep pad Bearing pressure under serviceloads
= 2.752
+
=
10
181
VAd1
x
x
x 2750 x 525/1000
=
0.35
Vsd,
hence no requirement for shear reinforcement
1.075
1.2
=
652 kN
43.23(1)
Eqn 4.18
2.7.5.2 Punching shear
The critical perimeter is shown in Figure 2.12.
=
Design load on base
1801 kN
Length of critical perimeter =
u
[4x 300 + R-(2 x 1.5 x 525)]/1000
/—
300
II
I-I
= 6.15m
\\
of_Hift Figure 2.12 Critical perimeter for punching VRdI
=
x
035
1.075
x
1.2
x 525 x 6.15 =
Area within perimeter = 2.98 m2 Design shear (Vsd)
VSd
<
VAd1,
= (7.56
—
2.98)
1458 kN
43.4.5.1
Area of base = 7.56 m2
x 238 =
1090 kN
43.4.1(5)
hence no requirement for shear reinforcement
2.7.6 Cracking Approximate steel stress under quasi-permanent loads =
460
+ 03 x (908
1.15
1801
x 383)
x—
2310 2830
From EC2 Table 4.11 bar size should not exceed 25 Hence cracking
=186N/mm
> 20 mm used.
4.4.23 OK
Table 4.11
COMPLETE DESIGN EXAMPLE
2.7.7 Reinforcement details The reinforcement details are shown in Figure 2.13 and given in Table 2.7.
P
-
-
2
—
—f
9120—1—300 B2
2
22
9120—1—300 Bi
P LA N
4120-2
II
Fdn.
Cover —1.0
_J[_
III •
L1
A- A COVER — B1
= 50, end =75
Figure 2.13 Base reinforcement details Table 2.7 Commentary on bar arrangement Bar marks
Notes
1
Straight barsextend full widthof base less end cover of 75 mm. Bars should extend an anchorage length beyond the column lace
32 x 20 Anchorage length Actual extension 1150 mm 2
3
640 mm
Column starter bars wired to bottom mat Minimum projection above top of base is a compression lap + kicker = 32 x 20
4.13.3(9)
5.23.4.1
5.2.4.13
+
75
= 715 mm
Links are provided to stabilize and locate the starters duringconstruction
COMPLETE DESiGNEXAMPLE
2.8 Shear wall 2.8.1 Structure The structure is shown in Figure 2.14. fLoor —
I—05x Wi nd Load on
building
I
14300
900
Figure 2.14 Shear wall structure
2.8.2 Loading at foundation level Dead load from firstto third floors and roof
= 0.5 (3
x 23.5 + 28.5) = 49.5 kN/m
Self-weight = 0.175 x 24 x 15.5 Characteristic dead load = 49.5
= 65.1 kN/m + 65.1 = 114.6 kN/m
Characteristic imposed load from slabs =
2.5 (1.5
+ 3 x 4) x 0.7 = 236 kN/m
Wind loading is taken as 90% of value obtainedfrom CP3: Ch V: Part 21). Total wind
load on buildingin N—S direction
NAD 4(c)
= 0.9 x 449 = 404 kN
= 404/2 = 202 kN Moment in plane of wall = 202 x 8 = 1616 kNm Wind load on wall
Hence Maximum force per unit length due to wind moment
= +
Mx6 12
x = ± 1616 6 = ±47.4kN/m 14.22
2.8.3 Vertical design load intensities at ultimate limit state Dead load + imposed load = 135 x 114.6 + 1.5 x 23.6 = 190.1 kN/m Dead load
+ wind load = 135 x 114.6 + 1.5 x 47.4 = 225.8 kN/m; or = 1.0 x 114.6 — 1.5 x 47.4 = 43.5 kN/m
Eqn 2.8(a)
Eqn 2.8(a)
COALETE5ESG14 EXAMPLE
Dead load
+ wind load + imposed load = 135 x 114.6 + 1.35 x 23.6 ± 1.35 x 47.4
Eqn 2.8(b)
= 250.6 kN/m or 122.6 kN/m
NAD 4(c)
Therefore maximum design load
= 250.6 kN/m
From analysis of slab (not presented), maximum moment perpendicular 11.65 kNm/m
to plane of wall =
2.8.4 Slenderness ratio
kA
=
kB
=
= 2.05
4a5
Eqn 4.60
Hence
10
1/i 0
=
0.94
=
fl1O
Figure 4.27
= 0.94 x 4
= 3.76 x 1000 x
376 m
=
175
Hence wall is slender
2.8.5 Vertical reinforcement Higginsand Rogers design the shear wallas unreinforced. Plain concrete walls will be covered in EC2 Part 1A which, at the time of publication, has not yet been finalized. The wallwill, therefore, be designed hereas a reinforced wall. As will be seen, the result is the same. Eccentricity due to applied loads e01
e02
= 0 =
11.65
=
0.6
x
1000/250.6
= 46.5 mm
Hence e0
x
46.5
+ 0 = 27.9 mm
Eqn 4.66
Accidental eccentricity
ea
=
— 1
200
x
3760
2
= 9.4 mm
Eqn 4.61
COVIP%TE OESG EXDJJIPLE
Second order eccentricity
=
37602
x
460
2 x 1.15
10
= Assuming '
A further disadvantage of this methodis that with increasing values of cote, i.e., reductions in the concrete strut angle, theforces in thetension reinforcement
BEAMS
increase significantly and may well outweigh any notional savings in shear reinforcement.These forces are, it should be noted, explicitly checked in EC2 but not in BS 8110. Given specialcircumstancesthe '131 method may be required but for most practical situations, the standard method will provide the most economic design.
3.2.2 Example 1
—
uniformly distributed loading
The beam shown in Figures 3.2 and 3.3 is to be designed for shear. Ultimate load
385 kN/m
.. Beam span and loading
Figure 3.2
,
(
400 u
Figure 3.3 Typical section
—
—
example
1
643 4mm 2 (8132) Cover to links 50mm A51
example 1
The material strengths are
= 30 N/mm2 (concrete strength class C30/37) cwk
=
250 N/mm2 (characteristic yield strength of links)
Thebeam will be checked for shear reinforcementatthree locations using both the standard and VSI methods for comparison. These are (1)
d from support
(2) Where
VSd
VRd1,
i.e.,
the point beyond which only minimum shear
reinforcement is required (3) An intermediate point between 1 and 2.
3.2.2.1 Standard method
VSd
1155 kM
I
Id
l
4.3.2.2(10) 4.3.2.2(2)
43.2.4.3
The shear force diagram is shown in Figure 3.4.
f
4.3.2.4.3 4.3.2.4.4
i155
I
I I a
J
Figure 3.4 Shearforce diagram
—
example 1
lS
kN
The design shear resistance of the section, VRd1
TRd
VRd1,
is given by
=
[rRdk (1.2 +
+ 0•15J
=
034 N/mm2 for
= 30 N/mm2
k
= 1.6—d41 =
p
=
A
bd w
=
Eqn 4.18 Table 4.8
1
6434
=
bd
4.3.2.3(1)
400x900
=
i
0.018
0.02
(assuming 8T32 throughout span) Nsd
a
VRd1
= 034 x 1 (1.2 + 40 x 0.018) x 400 x 900 = 235 kN
3.2.2.1.1 Position 1
—
=
VSd VSd
= 0
AC
>
at d from support 1155 VAd1,
—
0.9
x 385 =
808.5 kN
shear reinforcement is required
4.3.2.4.3
The shear resistance of a section with shear reinforcement is given by VRd3
+
=
VCd
=
V
=
Eqn 4.22
VRdl =
235kN
— (O.gd)ç
Eqn 4.23
where
A
= area of shear reinforcement
s
=
f
=
For
spacing of shear reinforcement 250/1.15 = 217.4 N/mm2
VSd
V
VSd
—
VCd;
A
— (0•9d
VSd
or —
Therefore
A5 — s
— = (808.5 235) x iO = 3.25 mm2/mm
0.9
x 900 x 217.4
Try R12 links @ 140 mm crs. (4 legs),
A/s
=
3.23 mm2/mm
BEAMS
Check crushing of compression struts
= (j-)vfcbw0.9d(1 + cota)
VRd2
For vertical links, cota = v
=
fCd
=
0.7
— —a
-
200
Eqn 4.25
0
= 0.55
0.5
Eqn 4.21
= 20 N/mm2
1.5
Therefore
= (4) x 0.55 x 20 x 400 = 1782 kN > Vsd max =
VRd2
x 0.9 x 900 x
1
1155 kN
OK
Check maximum spacing of links
=
VSd
—
sbsin a
3V
pbd
4.4.23
452 = _________ 140
x 400
0.0081
Eqn 4.79
— = (808.5 3 x 235) x iO = 35 N/mm2
0.0081
Table 4.13
x 400 x 900
Maximum spacingfor crack control = Since
=
300 mm
(4) VRd2 <
Smax
=
0.6d
5.4.2.2(7) Eqn 5.18
p300mm
140 mm spacing
Check minimum value of
OK
p
Table 5.5
Concrete strength class C30/37 Steel class S250 By interpolation from EC2 Table 5.5 w,mIn
= 0.0022 <
0.0081 proposed
Use R12 links @ 140 mm crs. (4 legs) Note: Using the standard method, the increase in force in the tension reinforcement is best covered by using the shift rule. Itwill, however, becalculated in this exampleto provide a comparison with the values obtained in the subsequent examples using the VSI method.
43.2.1P(6) 5.4.2.13
Force in tension reinforcement Td
=
M — + j- V(cote
—
cota)
Eqn 430
BEAMS
Msd
= 884 kNm,
VSd
= 808.5 kN
cote
=
Therefore Td
a2.2.1.2 Position 2
—
cota = 0 for vertical links
1,
=
z = 0.9d = 810 mm
1091
+ 404 =
where VSd =
43.2.43(5)
1495 kN
= 235 kN
VRd1
From Figure 3.4
x 385
VSd
= 1155 — a
a
= 239 m from support
From Section 3.2.2.1.1, VRd2
=
235 kN
> Vs max
OK
The amountof shear reinforcement provided should be greaterthan w.mIn pw,mun
= 0.0022
Re-arranging EC2 Eqn 5.16 in terms of
-
Table 5.5
—' gives
= pb5ina =
For vertical links sina
1
Hence
— S
= 0.0022 x 400
x1
=
0.88 mm2/mm
Maximum longitudinal spacing (Smax) is given by EC2 Eqns 5.17—5.19. Vsd VRd2
=
235kN
=
1782- kN from Section a2.2.it
Since
(j-) Smax
A
VRd2, EC2
Eqn 5.17 applies
=
0.8d
1300 mm
=
0.88
x 300
= 264 mm2,
Eqn 5.17 4R10
= 314 mm2
Use RiO links @ 300 mm crs. (4 legs) 3.2.2.1.3 Position
3 — at 1.65 m from support
Thisisapointintermediatebetweenthe section atdfromsupportand the point at which shear reinforcement is no longer required. VSd VRd1
= 1155 — 1.65 x 385 = 520 kN = 235 kN
Since
>
VSd
shear reinforcement is required
VRdl,
Re-arranging EC2 Eqn 4.23 —
VSd
=
s
VCd
(520 — 235) x 10 0.9 900 x 217.4
=
x
O.9dfd
=
Try R12 links @ 250 mm crs. (4 legs)
= 1.62 mm2/mm
1.81 mm2/mm
Check maximum spacing of links
4.4.22
A
Eqn 4.79
sbsina For vertical links sina =
1
Hence 452
=
VSd
—
250
x 400
= 0.0045
(520 — 3 x 235) X 10 0.0045 x 400 900
3Vd =
x
pbd
Maximum spacingfor crack control
=
—114 N/mm2
= 300 mm
OK
Table 4.13
Since (i-)VRd2 S
<
(+W'
5.4.2.2(7) Eqn 5.18
= 0.6d p 300 mm
From Section 3.2.2.1.1 VRd2
>
OK
VSd
Provide R12 links @ 250 mm crs (4 legs) To optimizelinkspacing, check the point atwhich shear reinforcementis satisfied by R12 @ 200 mm crs. (4 legs).
A — s
VWd
452 = = — 2.26 mm2/mm 200
=
— (0.9d)y =
2.26
x 0.9 x 900 x 217.4 =
VCd+VWd
VRd3
Equating VRd3 VSd
= =
VSd
and noting that VCd
VAd1
+
=
235
=
VRd1
+ 398 =
633 kN
398 kN
Distance of point from support
1155
=
—
633
= 136 m
The proposed link arrangement is shown in Figure 3.5.
R12 — 200
R12— 140
4
4
4
Legs
136rn
,
I
R12 —300 I R12 —200 U
legs
4
4
Legs
legs
R12 —140
4
Legs
136m
1i
-
2.39m+ 2 39m+ 60m between centres of supports
U
4
Figure 3.5 Link arrangement (standard method)
—
example 1
Note:
In the centre portion of the beam RiO links are required by calculations but R12 (*) are shown to avoid the possible misplacement on site. Distance from the support(+) could be reduced to 1.70 m in this case. 3.2.2.2 Variable strut inclination method
43.2.4.4
This method allows the angle of the concrete compression strut to be varied at the designer's discretion within limits stated in the Code.
It cangive some economy in shear reinforcementbutwill require the provision of additional tension reinforcement. In most cases the standard method
will
suffice.
This reduced shear reinforcementwill only beobtainedat high levelsof design shear and is counter-balanced by increased tension reinforcement. This can be seen by a comparison of EC2 Eqns4.22 and 4.23 in thestandard method and EC2 Eqn 4.27 in the variable strut inclination method.
The standard method gives
=
VRd3
V
=
V+ V
Eqn 4.22
A
—a S
(0.9d)f
Eqn 4.23
Re-arranging gives
= s
VRd3—VCd
(O.9d)fd
The VSI method gives
A = —f' (O.9cf cote
VRd3
S
Re-arranging gives A5,,,
s
—
VRdS
(o.9d)c,,dcote
Eqn 4.27
BEAMS
Note:
In the above equation the contribution of the concrete, resistance of the section is not taken into account.
VCd,
to the shear
Withcote = 1.5 which is the maximum value permitted inthe NAD, reductions in shear reinforcement will only occur when —
VRd3
(O.9d) cWd x VRd3
<
VCd)
VRd3 gives VSd
>
VSd
(O.9d)cWd
l.S(VRd3
Putting VSd If
or 1.5
>
3VCd
then the VSI method will allow a reduction in shear
reinforcement.
Ifthis inequality is not satisfied, useofthe variable strut inclination method will produce an uneconomic amount of shear reinforcement. In this case the standard method should be used. For elements with vertical shear reinforcement, VAd2 is given by
VRd2
b zv f
—
cotO + tane
=
Putting VSd
E
426
Eqn
4.21
and re-arranging gives 1
VSd
cote + tane
bwzvfCd
Figure 3.1 shows cote plotted against 1/(cote + tan8) together with the EC2 and NAD limits for cote. Hence for a given the limits for cote can be found.
V,
Increasing the value of cote will reduce the shear reinforcement required but increase the force in the tension reinforcement.
In this example, cotO will be chosen to minimize the shear reinforcement. 3.2.2.2.1 Position 1
—
at d from support
From above 1
VSd
cote + tanO
bWzPfd
b
=
400mm
z
=
0.9
=
0.7—
x 900 = f 200
=
810mm 0.55
z 0.5
MAS
-- =
fCd
=
VSd
= 808.5 kN
20N/mm
1.5
Therefore
=
1
cote + tane
80&5 400
x
10
=
x 810 x 0.55 x 20
0.22
From Figureai, this liesunderthecurveTherefore, cote = 1.5 canbe chosen which is the maximum value allowed under the NAD limits.
=
VRd3
Eqn 4.27
(?!)zcWdcot8
Now equating VRd3 to VSd and re-arranging
=
VSd
S
808.5 X 10
= 810
zycote
= 3.06 mm2/mm
x 217.4 x 1.5
Check
Afd=
(j-)vç =
1.66
OK
5.5
Try R12 links @ 150 mm crs. (4 legs),
=
A/s
3.01 mm2/mm
Check maximum spacingof links.
=
A
4.4.23
= 0.0075
Eqn 4.79
sbsIna VSd
—
3V
— = (80&5 3
0.0075
bd
x 235) x iO
x 400 x 900 = 300 mm
Maximum spacing for crack control =
0.0075
= 38.3 N/mm2
Table 4.13
= 0.0022
>
OK
Check S VSd
5.4.2.2(7)
=
=
Since S
Table 5.5
(4-)VRd2
808.5 kN
bWzvf cote + tane <
VSd
=
400 x 810 x 0.55 x 20 _______________
=
1644kN
2.167
(+)VRd2
= 0.6d 1 300 mm
E1
Eqn 5.18
Use R12 links @ 150 mm crs. (4 legs)
Check additional force in tension reinforcement. M
+ (2) vSd(cote — cota)
—
Td
This compares with Td =
= 1091
+ 606 =
1697 kN
Eqn 430
1495 kN using the standard method.
Note:
Although not permitted by theNAD, values of cote upto 2.5 are given in EC2. A checkon shear reinforcement usingcote = 2.5 is now given to illustrate the effect of increasing values of e on shear and tension reinforcement. A
VSd
=
zfd cote
s
x iO x 217.4 x 2.5
808.5
= 810
Try R12 @ 225 mm crs. (4 legs), ASW/s =
=
1.84 mm2/mm
2.01 mm2/mm
Check maximum spacingof links
=
0.005
VSd —3V
pbd
= 57.5 N/mm2 = 250 mm
Maximum spacingfor crack control
t 300
= 0.6d
Sm
mm
OK
Table 4.13
OK
Eqn 5.18
Use R12 links @ 225 mm crs. (4 legs)
Check additional force in tension reinforcement.
=
Td
M —
This compares with 3.2.2.2.2 Position2
—
+ (-)Vsd (cote Td
—
cota)
= 1091 + 1011 = 2102 kN
= 1495 kN using the standard method.
where VSd =
VRdl
Since only minimum shear reinforcement is required this case is identical to
that shown in Section 3.2.2.1.2. 3.2.2.2.3 Position VSd
—
3 — at 1.65 m from support = 520 kN =
S
Try R12 links
VSd
zfYWdcote
520 X 10
= 810
x 217.4 x 1.5
2 = 1.96mm/mm
225 mm crs. (4 legs), ASW/s = 2.01 mm2/mm
From Section 3.2.2.2.1 spacing is satisfactory. Use R12 links @ 225 mm crs. (4 Jegs) As in Section a2.2.1.3,check the pointat which the shear requirementis satisfied by R12 @ 200 mm crs. (4 legs).
—
452
=
s
2 2.26mm/mm
=
200
=
VRd3
cote = 2.26 x 810 x
(._)zf
Distance from support
=
1155
597 =
—
217.4
x
1.5
=
597 kN
Eqn 4.27
1.45 m
385
The proposed link arrangement is shown in Figure 3.6.
R12 —150 1
I
R12 —200
R12 —300 S
4 legs
legs
145m "I 239m
4
R12 —200
I
4 legs
4 legs
legs
I
R12 —150
I I I
I
145m
2 39m
ri
60m between centres of supports
Figure 3.6 Link arrangement (VSI method)
—
example 1
5
obtained using the standard Comparing this with the arrangement in Figure method, it can be seen that less reinforcement is required near the support butthis needs to be carried further along the beam. There is little overall saving in this case.
3.3 Shear resistance with concentrated loads close to
support 3.3.1 Introduction Where concentrated loads are located within 2.5d of a support, the value TRd may be modified by afactor when calculating VRd1. Thisenhancement only applies when the section is resisting concentrated loads and the standard method is used. For a uniformly distributed load, an unmodified value of VRd1 should be used.
3.3.2 Example 2
—
concentrated loads only
The beam shown in Figures 3.7 and 3.8 is to be designed for shear.
4.3.2.2(9)
BEAMS
800kN
800kN
Ultimate toads
—
Figure 3.7 Beam span and loading
ioooI.1
A1 4825mm 2 (6T32)
k 400 Figure 3.8 Typical section
—
example 2
Cover to Links
50mm
example 2
The materials strengths are =
30 N/mm2 (concrete strength grade, C30137)
= 250 N/mm2 (characteristic yield strength of links) In the example, VRd1 will be calculated at positions between the support and 2.5d away at intervals of 0.5d. This is done to illustrate the effect even though the critical section will normally be at the position of the concentrated load. 3.3.2.1 Shear reinforcement
The shear force diagram is shown in Figure 3.9.
VRd1
8OOkNf 135m
x
(mod)
\
211kN
1-
2llkN
Figure 3.9 Shear force diagram
—
\ VRd1 (mod)
example 2
h
135m
}0kN
1d1S The basic design shear resistance of the section,
VRdl,
is given by
43.23(1)
VRd1
= [rRd k (1.2 + 40p,) + 0.15 Or]bd
Eqn 4.18
TRd
= 034 N/mm2 for
Table 4.8
= 30 N/mm2
Forconcentrated loads within 2.5d ofthefaceofthe support, an enhancement of shear resistance is permitted. TRd may be multiplied by a factor /3 when determining
VRdl.
= 2.5d/x with 1.0
/3
/3
5.0
Taking values of x between 0.5d and 2.5d gives values Table 3.1. Table 3.1
Eqn 4.17
of /3rRd shown in
Design shear strength fhAd X
TRd
(m)
(N/mm2)
0.45
5
1.7
0.90
2.5
0.85
1.35
1.67
0.57
1.80
1.00*
0.34
2.25
1.00*
0.34
* No enhancement taken, see Figure 3.9 Theequation for VAd1 can be modified to give a range of values corresponding to the distance from the support. VRd1(x)
=
[BTRdk (1.2
+ 40p,) + O.lSç,]bd
= 1.6—d41 =
k
=
p1
— A2,
bd
1
4825 = ________ =
400
x 900
0.013
N _sa0 A C
Values of design shear resistance, VRd1, are given in Table 3.2. Table 3.2 Design shear resistance VRd1 x
VRd1
(m)
(kN)
0.45
1052
0.90
526
125
353
1.80
211
2.25
211
Eqn 4.18 (mod)
BEAMS
Shear reinforcement is required when VSd
>
= 800 kN from x = 0 to x =
From Figure 3.9, VSd
4.3.2.4
VRd1.
1.35 m
Using the standard method
4.3.2.43
= VCd+VWa Putting VRd3 = "Sd
=
VSd VRd1
Eqn 4.22
and VCd =
VRd1 gives
+ VWd
Values of design shear resistanceto be provided by shear reinforcement, are given in Table 3.3. Table 3.3
V,
Design shear resistance VSd
—
=
VRdI
(kN)
VSd (kN)
1052
800
526
800
274
353
800
447
211
0
<
0
211
0
<
0
VAd1
1
VWd
(kN)
<
0
= 353 kN,
Therefore maximum shear reinforcement is required when VRd1 i.e, when x = 135 m.
This should be provided over the entire length from x = 0 to x
2.25 m
4.3.2.2(9)
Note: If a concentrated load is positioned close to a support, it is possible that using to modify VRd1 may lead to only minimum shear reinforcement being provided throughout the beam. In this case, the designer may wish to base the shear resistance on the unmodified VRd1.
4.3.2.2(9)
(0
< x <
2.5d).
Thiscan be illustratedby taking the example above butplacingthe pointload at 0.5d from the support. The modified shear force diagram is shown in Figure 3.10. VRd1 (mod)
1O52kN
—
VSd = BOOk N
VRd1
ZllkN
Note
B = 1 on span side of concentrated load
—
—
0•5
L
I
I
I
I
10
15
2O
25
Position of concentrated load
Figure 3.io Shear force diagram (load at 0.5
E1
—
example 2 modified
In this case it would be prudent tocheckthe shear resistanceonthe unmodified = 211 kN. The required shear reinforcementshould be provided from VAd1 x = Otox = 0.5d Check area of shear reinforcement required in example 2. Re-arranging the equation for
ASW
VWd
=
s
V gives
Eqn 4.23
447x103
= 0.9
0.9df
x 900 x 217.4
Try R12 links @ 175 mm crs. (4 legs),
=
Ajs
=
2.54 mm2/mm
2.58 mm2/mm
Check crushing of compression strut
(4) vfCdbWO.9d (1 + cota) For vertical links, cota = 0 =
VRd2
=
fCd
30 =—= 1.5
0.7
=
— —--
v
Eqn 4.25
0.55
Eqn
200
4.21
20N/mm2
Therefore
x 0.55 x 20 x 400 x 0.9 x 900 x
=
(-j-
=
1782 kN
>
Vsd
1
= 800 kN
OK
Check maximum spacing of links.
4.4.2.3
=
Eqn 4.79
sbsIna For vertical links sina pW
= 175
=
x 400
1
= 0.0064 >
pw,mln
= 0.0022
OK
Table 5.5
Vsd—3V = (800—3x353)x iO bd 0.0064 x 400 x 900 pww Maximum spacingfor crack control =
300 mm
By inspection, EC2 Clause 5.4.2.2(7) is satisfied.
Use R12 links @ 175 mm crs. (4 legs) for 0
< x < 2.25 m
Table 4.13 5.4.2.2(7)
—
3.3.3 Example 3
combined loading
The revised loading and shear force diagrams are shown in Figures 3.11 and 3.12 respectively.
Ultimate loads 800 kN
800 kN 100
kN/m I
IAYAYVIW* YAWA
AYAYAAWAVAW*WkWA\ AV*VAWAUAWAW&
1•35m
1.35m
:
6m
liii
Figure au Beam span and loading
—
example 3
1100
135m
kNI I
135m
Ix
l I
KN
F"
jioo Figure 3.12 Shear force diagram
example 3
—
The basic design shear resistance of the section, VRdl, is given by
43.2.3(1)
+ 40p,) + 0.15Or]bd
Eqn 4.18
For concentrated loads within 2.5d of the face of the support, TAd may be increased as in Section aa2. However, no similar enhancement is permitted for uniformly distributed loads.
4.3.2.2(9)
VRd1
=
[rRdk(l.2
must be reduced depending onthe proportion ofconcentrated loads to total design load. 9 can then be written as
red
=
1+(3—1)
Vsd(COflC)
with 1.0
5.0
Vsd(tot)
=
design shear force due to concentrated loads
=
design shear force due to total loads
Vsd(COflC)
Vsd()
Values of the concentrated load ratio and the resulting design shear strength are given in Tables 3.4 and 3.5.
BEAMS
Table 3.4 Concentrated load ratio Vsd(COflC)IVsd(t) x
Vsd(flC)
VSfl)
Vsd
(m)
(kN)
(kN)
(kN)
0.45
800
255
1055
0.76
0.90
800
210
1010
0.79
1.35
800
165
965
0.83
1.80
0
120
120
0
2.25
0
75
75
0
Vsd()IVsd
Table 3.5 Design shear strength flredTRd X
'3redRd
red
(m)
(NImm2)
0.45
5
4.04
1.37
0.90
2.5
2.19
0.75
1.35
1.67
1.56
0.53
1.80
1.0
1.00
0.34
2.25
1.0
1.00
0.34
Theequation for VRd1 can be modified to give a range of valuescorresponding to the distance from the support. VRd1
(x)
= [adrRdk(1.2 + 4Op1) + 0.15
i
As in Section 3.3.2.1
k
= 1,
p1
= 0.013,
a] bd
=
0
Values of design shear resistance, VAd1, and design shear resistance to be are given in Tables 3.6 and 3.7. provided by shear reinforcement,
V,
Table 3.6 Design shear resistance (VRd1) x
VRd1
(m)
(kN)
0.45
848
0.90
464
1.35
328
1.80
211
2.25
211
Table 3.7 Design shear resistance (V) Vsd
=
—
VRd1
VSd
(kN)
(kN)
848
1055
207
464
1010
546
328
965
637
211
120
211
75
82 kN/rib
...
OK
Link spacing may be increased where (-k-)
Sm
=
V
= 25 kN/rib
0.8d 1 300
= 185Thm
Eqn 5.17
Use R6 links @ 175 mm crs. apartfrom region within 0.6 m of interior support
V
=
14.7
>
4
kN/rib
OK
4.1.3.1.9
43.2.5
Shear between web and flanges
VSd
a
Eqn 4.33
d
= a
= (j-)
l
= 2550 mm
Figure 4.14
Maximum longitudinal force in the flanges F0
x F
=
acd(O.8x)b
= 0.075 at mid-span = 14.2 x
0.8
x 0.075 x 232 x
= 122 kN
Force to one side of web
iFd
VRd2
VRd3
—
195
122
x
=
41.2 —
= 16.2kN/m
Therefore
vSd
600
=
2x600
= 41.2kN
2.55
= =
O.2çJhf
2.5TRdhf
= +
0.2
x
16.7
x
100 = 334 kN/m
>
VSd
..
OK
Af
Eqn 436 Eqn 434 Eqn 437
Sf
WithA = 0 VRd3
=
2.5
x 03 x 100 =
75 kN/m
>
VSd
OK
Eqn 4.35
No shear reinforcement required
4.12.1.10 Topping reinforcement
No special guidance is given in EC2 regardingthe design ofthe flange spanning between ribs. The Handbookto BS 8110(13) gives the following guidance. Thickness of topping used to contribute to structural strength Although a nominalreinforcementof 0.12%is suggested inthetopping (3.6.6.2), it is not insisted upon, and the topping is therefore expected to transfer load to the adjacentribswithoutthe assistanceof reinforcement.Themode oftransfer involvesarching action and this is the reason forthe insistence that the depth be at least one-tenth of the clear distance between the ribs. 3.6.1.5
Minimum flange depths are the same in EC2 and BS 8110 and the above is therefore equally applicable. Provide minimum reinforcementtransverselyand where top bars in rib, which have been spread over width of flange, are curtailed.
2.5.2.1(5)
SLABS
A
i
c1
3 x rib width, a 0.8 modification factor is required.
7 m, no further
modification is required.
Permitted span/effective depth ratio
= 373 >
= 39.2 x
1.19
x 0.8 OK
25.9
4.1.3.1.12 Cracking
For exposure class 1, crackwidth has no influence on durability and the limit of 0.3 mm could be relaxed. However, the limit of 03 mm is adopted for this
4.4.2.1(6)
example.
Satisfythe requirementsforcontrol of cracking without calculation.Check section at mid-span: Minimum reinforcement,
A
=
Note:
can be conservativelytaken as thearea below the neutral axis for theplain concrete section, ignoringthe tension reinforcement,as shown in Figure 4.10.
4.4.23(2) 4.4.2.2(3) Eqn 4.78
92j
-________
Neutru! axis
—______ -
100
175
125
j•_5
Figure 4.10 Tensile zone of plain concrete section Depth to neutral axis =
A= =
as
160
92 mm
x 175 + 600 (100 — 92) =
32800 mm2
= 460 N/mm2
100%fyk
ceff =
recommended value 3 N/mm2
k
=
0.4 for normal bending
k
=
0.8
A
=
0.4
4.4.2.2(3)
x 0.8 x 3 x 32800/460 =
69 mm2 <
Apr,
OK
Check limit on bar size. Quasi-permanent loads
Eqn 4.78 Table 4.11
= Gk + °30k =
6.1 kN/m2
4.4.23(3) 2.3.4
Ratio of quasi-permanent/ultimateloads = —-- =
Eqn 2.9(c) NAD Table 1
0.45
13.7
Estimate of steel stress 0.45
x
As.req
x fyd
=
0.45
= 32
= 132 N/mm2
403
Asprc
Maximum bar size
295 x— x 400
> 16 mm provided
OK
Table 4.11
For cracks caused dominantly by loading, crack widths generally will not be
4.4.23(2)
excessive.
4.1.3.1.13 Detailing
Minimum clear distance between bars Nominal clear distance in rib
=
= 49 mm
20 mm
5.2.1(3)
OK
SLABS
Bond and anchorage lengths:
5.2.2
> 250 mm bottom reinforcement is in good bond conditions. reinforcement is in poor bond conditions. Top
5.2.2.1
For h
Figure 5.1(c)
Therefore, ultimate bond stresses are
Bottom reinforcement, bd = Top reinforcement,
2.7 N/mm2
= 0.7
d
Basic anchorage length, 1b =
x 2.7
=
1.89 N/mm2
5.2.2.2(2)
cbf
5.2.2.3
Eqn 53
qx 400
For top reinforcement, 1b =
5.2.2.2(2) Table 53
4x
=
1.89
x 400 4 x 2.7
For bottom reinforcement, 'b
=
Anchorage of bottom reinforcement at end support.
5.4.2.1.4
Treatas a solid slab and retain not less than halfof the mid-span reinforcement.
5.4.3.2.2(1)
Use 2T12
L bars bottom at end support
Anchorage force for this reinforcement with zero design axial load
F
S
a
= VSd x
5.4.2.1.4(2) Eqn 5.15
d
where
=
21 kN/rib
For vertical shear reinforcement calculated by the standard method a1
a
5.4.2.1.3(1)
= z(1 — cota)/2 1z 0 = 90° and z is taken as 0.9d
Although this ribbedslab falls outside the solid slab classification requirements for analysis, treat as a solid slab for detailing and take a1 = d.
5.4.3.2.1(1)
Therefore F5
=
21 kN/rib
x iO =53mm2 24,, a1 = 1.4 = 1.0 and = Aspr Hence with Areq
net
per
Detail at edge support
Provide full lap-length,
i
m
=
374,
= 37 x 12 = 444 mm
= 03 aaallb = 187 mm 4 154, or 200 mm
5.2.4.1.3
Eqn 5.7 NAD Table 3 Figure 5.6 Eqn 5.4 Eqn 5.8
SLABS
Therefore S
=
444
x
1.4
= 622 mm >
OK
1 5mm
Transverse reinforcement at lapped splices should be provided as for a beam
<
section. Since 4, 16 mm, nominal shear links provide adequate transverse reinforcement. Anchorage of bottom reinforcement at interior support. Treat as a solid slab and continue 50% of mid-span bars into support.
The reinforcement details are shown in Figure 4.12.
5.2.4.1.2(1) 5.4.2.1.5 5.4.3.2.2(1)
Figure 5.13(b)
4112 per
rib
R6— 125
links
LJ1
h
L116
L L b,
1116
per rib
per rib
I
net
160
41
Figure 4.12 Detail at interior support This detailing prohibitsthe easy use of prefabricated rib cages because of the intersection of the bottom reinforcement with the supporting beam cage. It is suggested that providing suitably lapped continuity bars through the support should obviate the need to continue the main steel into the support.
Thearrangementofthe reinforcementwithinthe sectionincluding theanchorage of the links is shown in Figure 4.13.
5.2.5
NAD Tables
3&8
4Ø4 50 mm
T12
18- 200
R6
T16
InternaL radius
of bend =
Figure 4.13 Arrangement of reinforcement
2
mm
5.4.2.1.2(2) Figure 5.10
SLABS
4.2 Flat slabs 4.2.1 Flat slabs in braced frames The same frame is usedin each of the following examples, but columnheads are introduced in the second case.
4.2.1.1 Design example of a flat slab without column heads Design the slab shown in Figure 4.14 to support an additional dead load of 1.0 kN/m2 and an imposed load of 5.0 kN/m2.
A
N
B
ii
425m
A
I..
I
4•25rn
C
52m
_
I.
I
D
5•2m
_
SQm
5•2rn
4.25m
5E—
+
-
___
Figure 4.14 Plan of structure
The area shown is part of a larger structure whichis laterally restrained in two orthogonal directions by core walls. The slab is 225 mm thick. All columnsare 300 mm square and along grid 5 there is an edge beam 450 mm deep x 300 mm wide.
4.2.1.1.1 Durability
For a dry environment, exposure class is 1. Minimum concrete strength grade is C25130. Since a more humid environment is likely to exist at the edges of the slab, increase concrete strength grade to C30/37. For cement content and w/c ratio, refer to ENV 206 Table 3.
Table 4.1 ENV 206 Table NA.1
Nominal coverto reinforcement = Nominal cover to all bars
Use nominal cover
20 mm
NAD Table 6
iz
bar size
z
nominal aggregate size =
20 mm ..
OK
NAD 6.4(a) 4.1 .3.3(5)
= 20 mm
4.2.1.1.2 Materials
= 460 N/mm2
Type 2 deformed reinforcement,
NAD 63(a)
C30/37 concrete with 20 mm maximum aggregate size 4.2.1.13 Load cases
It is sufficient to consider the following load cases
+
(a)
Alternate spans loaded with spans.
(b)
Any two adjacent spans carrying YGGk + spans carrying 7GGk. Gk
=
'YGGk
7GGk
0.225
x 24 +
= 1.35 x +
ok
=
6.4 8.7
7GGk
1.0 =
2.5.1.2
and 'yGGk on other and all other
6.4 kN/m2
= 8.7 kN/m2
+ 1.5 x 5.0 =
Table 2.2 16.2 kN/m2
Eqn 2.8(a)
4.2.1.1.4 Analysis Analyses are carried out using idealizations of both the geometry and the behaviour of the structure.The idealizationselected shall be appropriate to the problem being considered. No guidance is given in EC2 on the selection of analysis models for flat slabs, or on the division of panels into middleand columnstripsand thedistribution of analysis moments between these strips. This is left to the assessment of individual engineera Therequirementsset down in BS 8110 forthe above points are taken as a means of complying with EC2 Clause 2.5.1.1P(3). EC2 allows analysis of beams and slabs as continuous over pinned supports. It then permits a reduction in the supportmoment given by Sd
= F b /8 Sdsup sup
Theanalysisin this example includes framing intocolumns. Thus the reduction is not taken. Consider two frames from Figure 4.14 as typical: (i) (ii)
Grid 31A—D subframe Grid B.
2.5.1.1.P(3) and P(4)
2.533(3) 2.5.33(4)
Analysis resuttsfor the frames described above are given in Figure 4.15. The results for each frame are practically identical as the analysis for Grid B has an increasedloaded width (5.2 m), sincethis isthe first internalsupportfor frames in the orthogonal direction. Member stiffnesses have been based on a plain concrete section in this analysis. Column moments and reactions are given in Table 4.1.
4250 Id
—
1
5200
5200
1/W
II 'stab 2
3500
3500
'777
77
ANALYSIS MODEL
—198
—199
123
BENDING MOMENT
Figures 4.15 Analysis of frame
ENVELOPE (kNm)
—201.
SLABS
Table 4.1 Column moments and reactions
(kN)
moments (kNm)
Max E column moments (kNm)
End
156.4
37.9
37.9
1st interior
444.7
6.8
21.4
4.2.1.1.5 Flexural
E column
Max. reaction
Suppo
design — Panel A—B/1—2
EC2 doesnot specificallyaddress the problem ofedge column momenttransfer and the provisions of BS 8110 are adopted hera
BS 8110 3.7.4.2
Mt,max = 0.l5bd2f e Cu Column A/2 moment transfer Assuming 20 mm cover and 20 mm bars in the top d1
d2 be
= 225 — 20 — 10 = 195 mm = 195—20 = 175mm = 300 + 300 (say) = 600 mm
I
= 37 N/mm2
M
=
Cu
NAD 4.1 3.3(5)
0.15
x 600 x
1752
x 37 x 106 =
102 kNm
Thisought to be compared with an analysis for a loading of l.4Gk + which would give approximately 5% higher edge moments than the EC2 analysis results above.
M
=
102
>
1.05
x 37.9
=
39.8 kNm
OK
Design reinforcement to sustain edge moment on 600 mm width.
Using;
=
1.5, a
= 0.85 and; =
1.15
Table 2.3
Referring to Section 13, Table 13.1: Msd
=
bd2ck
Afsyk
=
37.9
600
i0
x
x 1752 x 30
= 0.069
0.085
bdfCk
AS
0.085
x 600 x 175 x 30 =
582 mm2
= 970 mm2/m
460
= 0.163 < 0.45 (zero redistribution)
OK
2.53.4.2(5)
SLABS
Use T16 @ 150 mm crs. (1340 mm2/m) Place over width =
top at edge column
900 mm (see Figure 4.16)
Note: This approach gives more reinforcement than is necessary.
:1
I
T 1 •-I--
cII
1
esi
4_i
I
I Sø I—
Figure 4.16 Edge column moment transfer Check above moment against minimum value required for punching shear.
43.4.5.3 Eqn 4.59
mSd
For moments about axis parallel to slab edge
Table 4.9
= ±0.l25perm
V
=
156.4kN
Therefore mSd
= ± 0.125 x
Edge moment
= 0.6
= ± 19.6 kNm/m
156.4
= 632 >
19.6 kNm/m
OK
Design for mSd above in region outside edge column momenttransfer zone. 19.6
= bd2fCk
1000
x 106
x 1752 x 30
Minimum steel sufficient
=
= 0.6b d
f
0.021
z 0.OOlSbd
5.4.2.1.1
yk
= 0.0015 x 1000
x 175
= 263 mm2/m
Use T12 at 300 mm crs. (373 mm2/m) top and bottom (minimum) Maximum spacing
= 3h p 500 = 500 > 300 mm
OK
NAD Table 3 5.43.2.1(4)
SLABS
ColumnA/i moment transfer Assume the design forces for the frame on grid 1 are directlyrelated to those
for grid 3 in proportion to their loaded widths. Load ratio
=
(4.25/2) 5.2
=
0.41
The ratio of the edge column distribution factors for theframes is 2.0. Msd
= 37.9 x
0.41
x 2.0 =
31.1 kNm
Using design approach as for column A12: be
= 300 +
(say)
M
= 0.15 x 450
x 1752 x 37 x
=
450 mm 10-6
= 76 kNm
> 1.05 x 31.1 = 32.7 kNm Design reinforcement to sustain edge moment on 450 mm width MSd
450
bd2ck
Af
31.1
—
—
x
106
x 1752 x 30
=
OK
0.075
0.093
bdck
A
— —
x 450 x 175 x 30 =
0.093
460
Use T16 @ 150 mm crs. (1340 mm2/m)
478 mm2 =
1062 mm2/m
top for a width of 600 mm
Check above moment against minimum value required for punching shear. where
mJ
,
= ± 0.5 per m for corner columns
= ± 0.5 x (0.41 x Edge moment
=
31.1/0.45
156.4) say
=
>
32.1
In region of slab critical for punching shear: MSd
=
321x106 1000
bd2ck
x 1752 x 30
= 0.035
Af
—-- =
0.042
bdfCk
AS
=
0.042
x 1000 x 460
175
x 30
Eqn 4.59 Table 4.9
= ± 32.1 kNm/m
69.1 kNm/m
=
480 mm2/m
43.4.53
OK
Use 116 @ 300 mm crs. (670 mm2/m) top and bottom outside 600 mm wide moment transfer zone and over area determined in punching calculation
The division of panels intocolumn and middle strips is shown in Figure 4.17. Although BS 8110 indicates a 2.36 m wide column strip at column B2, a 2.6 m width has been used in the following calculations. This is considered reasonable as a loaded width of 5.2 m hasbeen taken in the analysis for grid B and grid 2.
A
B
BS 8110 Figure 3.12
C
Figure4.17 Assumed stripwidths(arrangement symmetrical about diagonal
All—C13)
ColumnB/2 support moments Analysis moment = Column strip Msd
b
=
MsdS = bd2fck
198 kNm in both directions
= 0.75 x 198 = 149 kNm
1300x2 = 2600mm 149
x
x
1752
2600
106
x 30
=
0.062
BS 8110 Table 3.20
Af —--
=
0.076
bdfCk
x 2600 x
0.076
AS
175
x 30
460
= 2255mm
Use 13116 (2613 mm2) top in column strip. Provide 9T16 @ 150 mm crs. in central 13 m and 2116 @ 300 mm crs. on either side
BS 8110 a7.3.1
Checkwhetherminimum moment requiredfor punching shear hasbeen met.
43.4.53
,
With
=
Msd
Table 4.9
—0.125
=
=
tJVsd
—0.125
x 444.7 =
—55.6 kNm
Eqn 4.59
Thisis to be carried over a width of 031. Since includes for a loaded width of 5.2 m, it is assumed that the larger panel width may be used. 0.31
=
0.3
x 5.2
=
1.56 m
By inspection reinforcement (9116 in central 13 m) is sufficient Middle strip (using average panel width) MSd
bd2k d
Af —-
0.25
=
(4725 =
0.059
=
0.031
X 198 X
106
2600) x 1752 x 30
—
<
OK
= 0.026
OK
0.45
bdfk A5 —
b
=
0.031
x
1000
x 175 x
30
460
Use 116 @ 300 mm crs. (377 mm2/m)
= 354 mm2/m
top in middle strip
It is noted that EC2 Clause 2.533(5)would allowtheuseofthemoment atthe face of the support (subject to limits in EC2 Clause 2.53.4.2(7)), but this is considered more appropriate to beams or solid slabs and the peakmoment over the support has been used in the above design. Span moments No special provisions are required in EC2. Hence the design basis of BS 8110 is adopted forthedivision of moments.The same pattern of reinforcementwill be provided in all panels. The column strip moments are given in Table 4.2 where
Msd
=
0.55 Msd
2.53.4.2(5)
Table 4.2 Column strip span moments Total
moment
MsdS
b
Msd
(kNm)
(m)
Span
MsdS
b (kN)
(kNm) End
107
589
2.12
278
1st. interior
123
67.7
236
286
Using the greater value:
!M8 b
/
Af
x— =
0.037
A
—
0.071
d
bdfCk
x
0.037
b
175
0.031
x 30
1752
d2fck
— —
X iO =
28.6
1
x 30 x io
<
0.45
—
422 mm2/m
460
OK
2.53.4.2(5)
OK
2.53.4.2(5)
Use 112 @ 250 mm crs. (452 mm2/m) bottom in column strips Using the middle strip moment for the first interior span
b =
4.725
MSd
= 0.45 X 123 X
236 = 2365 m (average panel width)
2365
bd2fk
Afk
—
=
x
1752
X
—
0.031
d
bdfCk
A — b
=
0.031
x
175
106
= 0.026
x 30 =
<
0.059
x 30 x iO
0.45
=
460
354 mm2/m
Use T12 @ 300 mm crs. (377 mm2/m) bottom in middle strips Minimum longitudinal reinforcement, using dmax =
0.6bd
fyk
= 195mm
iz 0.OOl5bd
= 0.0015 x 1000 x
195
= 293 mm2/m
OK
43.4
4.2.1.1.6 Punching
ColumnB12 (300 mm
x 300 mm internal column)
Critical perimeter located at 1.5d from face of column.
d =
4.3.4.1(3) Figure 4.16
185 mm (average)
For a rectangular column/wall check geometry
Perimeter
= 4
length breadth
=
x 300 =
43.4.2.1(1)
11d
1200 mm
= 2035 mm .. OK
2
1
OK
Hence
u VSd
= 2ir x
1.5
x 185 + 1200
= 2944 mm
Figure 4.18
= 444.7 kN
Note: No reduction in this value has been taken. The applied shear per unit length:
VSd
vSd
=
v
—--— u
43.43(4)
internal column =
where
1.15
Eqn 4.50 Figure 4.21
= 444.7
x iO x
1.15
2944
= 174N/mm
Shear resistance withoutlinks VRd1
=
rRdk(l.2
+
43.4.5.1
Eqn 4.56
4Op1)d
TAd
= 034 N/mm2
k
= 1.6
p1
= reinforcement ratio within zone 1.5d from column face (T16 @ 150 mm crs. top each way gives 1340 mm2/m)
P1
= J x i1 p,
d =
—
1340
=
1000
x 185
Table 4.8 1.6 — 0.185
i
=
1.415
1.0
0.015
=
0.0072
Note: The amount oftensile reinforcementin two perpendicular directions Assume p1
+
p
= 2 (0.0072) > 0.005
> 0.5%. OK
4.3.4.1(9)
Therefore
= 0.34 x
VAd1
=
VSd
x (1.2 + 40 x 0.0072) x 185
1.415
>
174 N/mm
133 N/mm
Vf
Therefore shear reinforcement required such that VRd3 Slab depth
=
4.3.4.3(3)
VSd
200 mm
OK
4.3.4.5.2(5)
Check that applied shear does not exceed the maximum section capacity =
VRd2
2.0
VRd1
=
2.0
x
OK
= 4447 x iO = 2.0 N/mm2
Shear stress around column perimeter
o.9[ç =
= 266 > 174 N/mm
133
1200
x
NAD Table 3 4.3.4.5.2(1)
185
4.9 N/mm2
OK
NAD 6.4(d)
Design shear reinforcement using EC2 Eqn 4.58 since VjVRdl =
174/133
EAswyd f
— —
Rd3
NAD 6.4(d)
1.6
Rdl
Eqn 4.58
u
Using type 2 deformed high yield bars as links
= — =
fyd
=
400 N/mm2
Table 2.3
1.15
Therefore
E
A
133)
(174 —
Minimum reinforcement ratio
= 302 mm2
= 100% x value given in EC2 Table 5.5.
NAD Table 3
Pwmin
=
0.0012
by interpolation
5.4.3.3(2) Table 5.5
area within critical perimeter
—
column area
43.4.5.2(4)
Denominator =
+ 3 x 185)2
x 185)2(4 —
= 575000
mm2
Maximum spacing of links is determined by the ratio vsd/vRd2 where assumed that VRd2 is calculated in accordance with EC2 4.3.2.4.3(4).
it is
(300 Thus
A
sw,mun
0.0012
—
(1.5
x 575000
=
—
3002
690 mm2 5.4.3.3(4) 5.4.2.2(7)
VSd
VRd2
= 174 N/mm
= (2) vcdbw x 0.9d f = — —- = 0.7
=
-
0.55
200 =
Eqn 4.25 Eqn
20 N/mm2
1.5
4.21
2.3.3.2 Table 2.3
Therefore VRd2 VsdIVRd2
Smax
=
(j-) x
=
174/916
=
0.8d P 300 mm
0.55
=
x 20 x 0.9 x
185
=
916 N/mm
0.2
0.19
Eqn 5.17
Longitudinal spacing b 0.75d = Transverse spacing p d
NAD 6.5(f)
138 mm
5.4.2.2(9)
Placing shear links on 100 mm grid in 700 mm square gives 48 links with 44 inside the critical perimeter.
4.3.4.5.2(2)
By inspection the minimum preferred bar size will govern and mild steel links could be used.
fyk
=
EA
250 N/mm2 0.0022
x 575000
=
1265 mm2
Table 5.5
Use 44 R8 links (2220 mm2) Where necessary the punching shear resistance outside the shear reinforced area should be checked by considering further critical perimeters.
4.3.4.5.2(3)
Check where VSd
=
VAd1
= 133 N/mm
Hence
u
=——= V5d13
444.7
x iO x
1.15
= 3845mm
133
VRd1
Therefore distance from column face
= (3845
—
1200)/2ir = 420 mm
= 2.27d
This would be approximately at the next critical perimeter taken to be at a distance 0.75d beyond the previous one. No further shear reinforcement required.
Thetensilereinforcement(p16@ 150mm crs.) should extend for afull anchorage length beyond the perimeter at 420 mm from the column face.
BS 8110 Figure 3.17
SLABS
ColumnAll (300 mm
x 300 mm corner column)
Critical perimeter located at 1.5d from face of column (see Figure 4.18).
4.3.4.1(3)
Figure 4.18 Critical perimeter at corner column U VSd
=
600 + 277r/2 = 1035 mm
=
0.41
x
156.4
=
64.1 kN
Applied shear per unit length, with /3 =
vSd
== V5j3
64.1
u
x
x
1.5
1035
1.5
4.3.4.3(4) Figure 4.21
= 93N/mm
Eqn 4.50
Reinforcement within zone 1.5d from column face is T16 @ 150 mm crs. top each way (see Figure 4.19). 300
277
t.u
'1
Figure 4.19 Corner column detail VAd1 VRd1
= 133 N/mm (as for column > VSd
B/2)
Therefore no shear reinforcement required
I
4.3.43(2)
SLABS
ColumnA/2 (300 mm
x 300 mm edge column)
Critical perimeter located at 1.5d from face of column.
u
=
900 + 277T =
V
=
156.4 kN
1770 mm
Appliedshear per unit length, with 3
vSd
VAd1
VAd1
=
=
= 156.4 x
u
=
1.4
iO x 1.4
1770
Figure 4.21
= 124 N/mm
Eqn 4.50
133 N/mm (as for column 6/2)
>
Therefore no shear reinforcement required
43.43(2)
4.2.1.1.7 Deflection Control by limiting span/effective depth ratio using NAD Table 7.
4.43.2
For flatslabs the checkshould be carried out on thebasis ofthelongerspan.
4.43.2(5)(d)
Forspan < 8.5 m, no amendmentto basic span/effectivedepth ratio isrequired.
4.4.3.2(3)
Note 2 to NAD Table 7 states that modifications to the tabulated values for nominally reinforced concrete should not be carried out to take into account service stresses in the steel (refer to EC2: Clause 4.4.3.2(4)). However it is assumed that correction ought to be carried out for 0.15% p < 0.5% but that the resulting values should not exceed those tabulated in the NAD for nominally reinforced concrete.
NADTable 7 gives basic span/effectivedepth ratios which are assumed to be = 400 N/mm2. based on
4.4.a2(4)
when =
460 N/mm2 and Areq
=
A 400 — x s,req
=
Modification factor
=
Aspr 0.87
Aprj Basic span/effectivedepth ratios for flat slabs are
(p = 0.5%) = 30 reinforced nominally (p = 0.15%) = 41 lightly stressed
Span reinforcement is typicallyT12 @ 250 mm crs. (452 mm2/m)
100A = 100 x 452 = 0.26% bd 1000 x 175
NAD Table 7
. (e)
By interpolation (p
= 0.26%), basic span/effectivedepth ratio =
= 5200 =
max span
d
29.7
175
mm
< 37.5 x
0.87
=
37.5
32.6
OK
4.2.1.1.8 Crack control Use method without direct calculation.
4.4.23
Estimateservicestress,
4.4.23(3)
Gk
+ 03K =
=
+
a, under quasi-permanent loads as follows: 6.4
+ 03 x 5 =
Ratio of quasi-permanent to ultimate design loads
=
7.9
kN/m
7.9/16.2 =
23.4
Eqn 2.9(c) 0.49
Therefore
a
=
0.49
xI x
A
A
< 200
A
NAD Table 1
Asprov
prov
Limit bar size using EC2 Table 4.11 or bar spacing using EC2 Table 4.12. The relevant limits are shown in Table 43. Table 4.3
Crack control limits
08
1.0
Steel stress (N/mm2)
Bar size (mm) Bar spacing (mm)
200
160
25
32
Table 4.11
250
300
Table 4.12
Maximum bar size used is less than 25 mm throughout
OK
Check minimum reinforcement requirement
Eqn 4.78
kkfc•iAcua
A2
A
4.4.23(2)
A
=—2
as 1ct,eff
k
x fyk
= 460 N/mm2
=
100%
=
minimum value suggested,
=
0.4,
k=
0.4
x 0.8 x 3 x
0.8
Therefore
A
3 N/mm2
AC
2x460
=
0.001
< 0.0015 bd (minimum flexural steel)
AC OK
SLABS
4.2.1.1.9 Detailing Considercombined requirementsfor flexure/shearand for punching fortop steel over supports.
ColumnB/2 For flexure/shearbars should extend for a distance d point at which they are no longer needed (a1 = diagram). =
-
x —5L
where
f
+ 1net z 2d beyond the d = shift in moment
=
(1)&(2)
Figure 5.11 5.4.3.2.1(1)
= 400 N/mm2
Eqn 5.3
250 mm bond conditions are good and
For h
5.4.2.1.3
Figure 5.1 Table 5.3
3.0 N/mm2
Therefore 400 =—x—
b
Olb
net
=
33
As.req
=
For straight bars
net
=
1.0
5.23.4.1 Eqn 5.4
and if
Asreq
=
1.0
Apr, =
534 mm for 116 bars, say 550 mm.
Curtail alternate bars as shown in Figure 4.20.
Alternate bars curtailed at 1
and
2
N, 2
0•5
0
I571 ,,j..5711 Figure 4.20 Curtailment diagram
SIJBS
Check that bars are anchored past relevant critical punching perimeter. Earlier calculation required column strip reinforcement to extend beyond a perimeter 420 mm from column face i.e. 570 mm from grid. It is assumed sufficient to provide an anchorage 1flet beyond this perimeter. Inspection of Figure 4.20 shows that this is satisfied.
4.2.1.2 Design example of a flat slab with column heads The previousexamplewill be used with column heads introducedatthe internal columnsto avoid the need for shear reinforcement. The rest of the design is unaffected by the change. 4.2.1.2.1 Punching at column B/2 (300 mm
x
300 mm internal column)
In the previous example it was found that VSd = column face where u = 3845 mm. Provide a column head such that
=
1H
VRd1
at 420 mm from the Figure 4.22
l.5hH (see Figure 4.21).
dcrjt
V
.1
I
300
Figure 4.21
IIH
Slab with column head
For a circular column head, assumethat EC2 Equation 4.51 applies to the case where 'H = l.ShH. Note: It is suggested that EC2 Equation 4.55 should read dcrit which reduces to the same as Equation 4.51 when 'H = Assume an effective column diameter, 1 = 300 mm
4.3.4.4(1)
= l.5dH + 05lç l.5hH.
To avoid shear reinforcement:
27rdt
3845 mm
dcrit
612mm
'H
612
—
1.5d
—
0.51
= 612
—
1.5(185)
—
150
=
185 mm
Eqn 4.51
SLABS
185 —
hH
=
123 mm say 125 mm
1.5
+
=
21H
670mm
Circular column head 125 mm below slab and 670 mm diameter is sufficient to avoid shear reinforcement
If a square column head is preferred, 11 =
d
12
=
+ 21H
= 1.5d + 0.561(11 12) 1.5d + 0.691 = 1.5d + 0.561 +
Eqn 4.52
To avoid shear reinforcement
d
612mm (612 —
1.12
= —— 1.12
hH
(612
>—= —
149
—
1 .5d —
1.5
x
0.56l)
185
—
0.56
x 300) =
149 mm
100mm
1.5
+
21H
= 600mm
Square column head 100 mm below slab and 600 mm wide is sufficient to avoid shear reinforcement
4.2.2 Flat slabs in laterally loaded frames In the following example, the structure used in Section 4.2.1 is considered to be unbraced in the North—Southdirection. 4.2.2.1 Design example of an unbraced flat slab frame Thisexample considers only theanalysis of the frame on grid B, consisting of three upperstoreys plus a lightweight roofstructure,as shown in Figures 4.22 and 4.23.
SLABS
425m
52m
t.2 52 m 3-
Figure 4.22 Plan of structure
3 Sm
3 Sm
3-5m
3
77
zI,-
Figure 4.23 Frame on grid B 4.2.2.1.1 Design loads
Office floors Dead load = 6.4 kN/m2 Imposed load = 5.0 kN/m2 Roof imparts load to columnsB/i and B/5 Dead load = 20 kN Imposed load = 30 kN
Sm
SLABS
Assume characteristic wind load
=
1.0 kN/m2
This is 90% of the value obtained from CP3: Chapter V: Part 2(11).
NAD 4(c)
Note:
The distribution of horizontal load between eachframe is determined by their relative stiffness. 4.2.2.1.2 Frame classification Determine whether sway frame or non-swayframe.
43.533
Check slenderness ratio of columns in the frame.
A calculation is required for those columns that resist more than 0.7 of the
A.3.2
mean axial load, NSdm at any level. Service loads are used throughout = 1.0).
A.3.2(3)
(i.e.
It is also assumed that these are vertical loads without anylateral loads applied.
Figure A3.4
7FFV
NSd,m
—
F
= E all vertical loads at given level (under service condition)
n
= Numberof columns
n A.3.2(1)
Consider a simple analytical model of the top floor to determine columns concerned as shown in Figure 4.24. 5OkN
SOkN
wkN/m P3
P2
11
P2
R
Figure 4.24 Load arrangement at third floor w
R2 R3
= 11.4 kN/m2 x loaded width = 11.4 x 5.2 (determined in Section 4.2.1.1) = 593 kN/m = 3 x 593 x 4.25/8 + 50 = 145 kN = 5 x 593 x 4.25/8 + (593 x 5.2/2) = 312 kN = 59.3 x 5.2 = 308 kN
100 + (18.9 NSd,m =
x 59.3)
= 244 kN
5
0.7Nsd,m= 171 kN
>R = 1
145 kN
Therefore slenderness of internal columns only needs to be checked. Clearly, this will normally be the case for multi-bay frames unless the edge columns carry large cladding loads. X
=
101i
where
43.53.5(2)
InnI
=
JII =
=
86.6 mm
A
For a horizontally loaded flat slab framedetermine the stiffnesses of the frame and thus the effective lengths of the columns using half the slab stiffness. Consider the centre column from foundation to first floor.
kA
=
El oI d C
C
; Ecm assumed constant
EOJbIleff
I
CCI
12
Eqn 4.60
i0 mm4
= 0.675 x
=
4.3.5.3.5(1)
= 4725 x 225 = 2.24 x iO mm4
2x
=
12
3500 mm
1
= 5200 mm
a
=
1.0
— —
2(0.675
eff
Therefore
kA kB
=
2.5.2.2.2
x 10/3500)
— —
x
05
10/5200) 2(2.24 0° (pinned at foundation)
Assuming that EC2 Figure 4.27(b) is appropriate to determine fi 10
=
f3l
= 2.15 x 3500 = 7500 mm
Figure 4.27
Hence X
= 7500/86.6 = 87
For non-swayframes
A3.2(3)
z 25
=
X
43.5.3.5(2)
= N5 Ultimate design load for centre column, ignoring self-weight of column. Nsd
= 3 =
x 16.2 x 30 = —
5.22
=
= 1314 kN
20 N/mm2
1.5
Therefore v
=
3002
= Since X
1314 x
iO =
x 20 =
17.6
Eqn 4.4 Table 2.3
0.73
.
25
> 25 the structure is classified as è sway frame
4.3.5.3.5(2)
SLABS
Theanalysisand designwould need to follow the requirements of EC2 Clause A3.5 to take into account the sway effects. EC2 Clause 2.53.4.2(4)does not generally allow redistribution in sway frames.
The method above is included to demonstrate its complexity. However, note theomission of guidance in EC2 Clause A3.2(3) on which nomogram to use
in EC2 Figure 4.27.
As an alternativemeans ofdetermining the frame classification, it is suggested that an analysis as detailed in BS 5950(14) is used to demonstrate thatthe EC2 requirements are met for non-sway frames. Assuming in the above examplethat the columnsizes are increased suchthat a non-swayframeresults, the following load cases need to be considered for design.
43.523(3) BS 5950: Part 1 5.13
These same load cases would also be applicable to sway frames where amplified horizontal loads are introduced to take account ofthe sway induced forces, complyingwith EC2 Clause A3.1(7) (b). 4.2.2.1.3 Load cases and combinations
2.5.1.2
With the rigorous approach the design values are given by
+ O,1k.1 + E
E7G Gk
23.2.2 P(2) Eqn 2.7(a)
I>1
where 0k.1
=
primary variable load, 0k2 =
= secondary variable load
0.7 generally
NAD Table 1
values are given in EC2 Table 2.2. Load cases with two variable actions (imposed and wind) are:
The
(a) Imposed load as primary load l35Gk + tSQk + l.O5Wk
(b) Wind load as primary load l3SGk ÷ tOSQk + l.5Wk In addition, load cases with only one variable action are: (c) Dead load plus wind l.OGk (favourable) + l.SWk l3SGk (unfavourable) + l.5Wk (d) Dead load plus imposed l3SGk + tSQk For non-sensitivestructures it is sufficient to consider the load cases (a) and (b) above withoutpatterning the imposed loads.
The NADallows the useof EC2 Equation 2.8(b) to give a single imposed and wind load case: l.3SGk +
135k (all spans) +
l.35Wk
NAD 6.2(e) 2.5.1.2P(1)
Final load combinations for the example given here
+ 1•5k (as Section 4.2.1.13) + l.5Wk (single load case)
1350k
0) (ii) (iii)
l.OGk
l.35Gk l.3SGk
(iv)
+ l.SWk (single load case)
+ 135k + l.3SWk (single load case)
4.2.2.1.4 Imperfections
2.5.13(4)
Consider the structure to be inclined at angle 1
=
0.005 radians
Eqn 2.10
ioop 1
=
frame height =
=
1-1.
=
Vd
NAD
i +1-
10.5 m
where n
n
Table 3
= number of columns = 5
0.78
= cxp = 0.78 x
0.005
= 0.0039 radians
Take accountof imperfections using equivalent horizontal force at each floor.
iH1
2.5.13(6)
= YVVd =
Using l.35Gk
Eti
Eqn 2.11
=
total load on frame on floor
+ (18.9
on each span gives
x 5.2) x
16.2
= 1592 kN
Therefore
= 1592 x 0.0039 = 6.2 kN per floor Assuming the frame by virtue of its relativestiffness picks up 4.725 m width of wind load: Wk
=
(4.725
x 3.5) x
1.0
=
16.5 kN per floor
Therefore the effects of imperfections are smaller than the effects of design horizontal loads and their influence may be ignored in load combinations (ii) to (iv). 4.2.2.1.5 Design
The design of the slab will be as described in Section 4.2.1.1.
2.5.1.3(8)
I
5 COLUMNS 5.1
Introduction Thedesign of column sectionsfrom firstprinciples using the strain compatibility method is covered. Examples of slender column design are also presented to extend the single example given in Section 2.
5.2 Capacity check of a section by strain compatibility 5.2.1 Introduction Two examples are considered: 1.
Where the neutral axis at ultimate limit state lies within the section; and
2.
Where the neutral axis at ultimate limit state lies outside the section.
The firstof theseis very simple whilethe algebra necessary for the second is more complex. For convenience, the same section will be used for both examples. This is shown in Figure 5.1. Assume
fyk
=
460 N/mm2 and fck =
30 N/mm2
350
2Jf
-
2132
_L k 30 tyk 460
500
4125
"j
_k..
Figure 5.1 Column section
5.2.2 Example 1 Calculatethe moment thatthe section can sustainwhencombined with an axial load of 2750 kN. 5.2.2.1 Basic method
If the neutral axis is within the section, thecompressiveforce generated by the concrete at ultimate limit state is given by NRdC
=
O.459fCkbx
4.2.133 Figure 4.2
and the moment by MAdC
=
NRdC(h/2
—
0.416x)
The strain at the more compressed face is taken as 0.0035 The procedure adopted is as follows: (1)
Assume a value for x
(2)
Calculate NRdC
(3) Calculate
the strain at each steel level
(4)
Calculate force generated by reinforcement (NRdS)
(5)
NRd
(6)
If NRd is not close enough to 2750 kN, modify the value of x and return to step (2)
(7)
If NRd is approximately 2750 kN, calculate MRdC and MRdS
(8)
MRd
+ NRdS
=
=
+
MRdC
MRdS
The design yield strain for the reinforcement 460
= 1.15
x 200000
=
0.002
5.2.2.2 First iteration Assumed value for x is 250 mm NRdC
= 0.459
0.0035 250
stOp
Strain NRdS1
cmid €sbot
NRdS3
x 30 x 350 x 250/1000 x 200
= 0.0028
> 0.002; therefore =
2x
804
= 1205 kN
f
=
400 N/mm2
x 400/1000
= 643 kN
= 0 and NRdS2 = 0 =
—
therefore
=
—400 N/mm2
x 491 x 400/1000
—2
=
—393 kN
Hence NRd
= 1205 + 643
—
393
= 1455 kN
This is considerably less than 2750 kN, hence x must be increased. Try new valuefor x
= 250
x 2750 1455
=
473 mm
COLUIA4S
5.2.2.3 Second iteration NRdC NRdSI
x 30 x 350 x 473/1000 =
=
0.459
=
643 kN as before 0.0035
smid
mid NRdS2
473
(473
=
0.00165
=
330
—
=
250)
x 200000
0.00165
= 330 N/mm2
x 2 x 491/1000
=
324 kN
€sbot
= 0.0035 (473
fS
= 0.00017 x 200000 = 34 N/mm2
—
473
NRd = 34 x 2 x
450)
2289 kN
=
491/1000
0.00017
= 33 kN
Hence NRd
= 2289 + 643 + 324 + 33 = 3289
kN
This is too large, hence x should be reduced. Linear interpolation gives
= 250 +
2750
—
(3289
—
1455\ 1455)
(473 —
250)
= 407 mm
5.2.2.4 Third iteration
x 30 x 350 x
NRdC
= 0.459
NAdS1
= 643 kN as before
mid
=
0.0035 407
(407
—
407/1000
250) =
= 1961 kN
0.00135
= 270 N/mm2 and NRd = 265 kN =
0.0035 (407— 450) =
—0.00037
407
f
=
—74
=
1961
N/mm2
and NRd =
—73 kN
Hence NRd
+ 643 + 265 —
73
= 2796
This is within 2% of the given axial load of 2750 kN
kN
OK
5.2.2.5 Moment MRdC
=
x (250 — 0.416 x 407)/1000
1961
MAd
= 643 x
MRdS2
= 0
MRd = 73 x
=
0.2
0.2
=
12&6 kNm
14.6 kNm
= 301.4 kNm
158.2 + 128.6 + 14.6
(
= 158.2 kNm
5.2.3 Example 2 Calculate the moment and axial force that can be sustainedby the sectionwhere the neutral axis depth is 600 mm.
Note:
The example has been given in this way so that repeated iterations are not necessary.These would not provide any new information to the reader. 5.2.3.1 Basic method When the neutral axis is outside the section the ultimate compressive strain is less than 0.0035 and is given by: U
—
0.002x
x
—
x 600 3 x 500/7
0.002
3 h/7
600
—
=
0.0031
(viii) & Figure 4.11
The conditions in the section are shown in Figure 5.2.
E 0•0031 (4
•
I.I
085 30I15 [.4
• • 600
S
/ "I Strain Figure 5.2 Conditions in section for Example 2
4.3.1.2(1)
—/ Stress in concrete
COLUIMtS
Thetechnique adopted for the calculation of NRdC and MRdC is to calculate the effect ofthe stress block on a depth of 600 mm nd then dduct the influence of the part lying outside the section. 5.2.3.2 Concrete forces and moments The equations for the full stress block are: N'RdC
= 05667(1
—
3/3)bxfCk
M'Rd,c = cN'Rd,c where c
= h/2
f3
=
—
4
+ x(j32 — — 12 43
6); and
O.OO2/E
Note: It will be found that, if the first example.
=
0.0035,theseequations give the values used in
The equations for the force and moment produced by the part of the stress block lying outside the section are =
a13)(x — h)bfk
—
0.5667a(1
= where
c' =
a
x — h12
—
—
(x — h) (8 — 3a) 12 — 4a
= = strain at bottom of section
From the strain diagram,
b =
0.00051
Hence
a
= 0.255 and
N'RdC
= 0.5667(1
—
c
= 250—
600(0.6452
=
0.645
0.645/3) x 350
12
x 600 x 30/1000 = 2802 kN
4 x 0.645 + 4 x 0.645
— —
6)
= 5.67mm
Hence M'RdC
=
5.67
x 2802/1000 =
15.9 kNm
= 0.5667 x 0.255(1—0.255/3)(600—500)350 =
—
600 — 250 — (600
— 500)(8 —
12 — 4
x 30/1000 = 139 kN
3 x 0.255) =
x 0.255
—284 mm
COLUMNS
MRdC =
—139
x 284/1000
=
—39.4 kNm
Hence
NRd,c
= 2802
MRd,c =
15.9
—
139
= 2663 kN
+ 39.4
=
553 kNm
5.2.3.3 Steel forces and moments Strain in upper layer of bars = This is
> 0.002; hence
600
x 550
=
0.0028
= 400 N/mm2
NRd,sl = 643kN MRd.sl = 643
x
0.2
= 128 kNm
Strain in middle layer of bars =
0.0031
600
x 350
=
0.00181
Hence f8
= 362 N/mm2
NRd = 355 kN,
MRd = 0
Strain in bottom layer of bars =
0.0031
600
x
150
= 0.000775
Hence
= 155 N/mm2
NRd =
152 kN
MRd =
—30.4 kNm
NRd MRd
= 2663 + 643 + 355 + 152 = 3813 kN = 553 + 128 — 30.4 = 153 kNm
5,3 Biaxial bending capacity of a section 5.3.1
General To carry out a rigorous check of a section for biaxial bending by hand is verytedious but possible if the simplified rectangular stress block is used. It is not suggestedthattheexamplegiven here is a normal design procedure for commonuse but it could be employed in special circumstances.There would be no difficultyin developing an interactivecomputer programto carry out design, in this way, by trial and error.
COLUMNS
5.3.2 Problem Demonstratethat the section shown in Figure 5.3 can carry ultimate design moments of 540 and 320 kNm about the two principal axes in combination with an axial loadof 3000 kN. Thecharacteristic strength of the reinforcement is 460 N/mm2 and the concrete strength is 30 N/mm2. 500
I
H
2038mm2 500
•
•
Ti,I Figure 5.3 Column section
5.3.3 Basic method
4.2.1.33(12)
The conditions in the section are shown in Figure 5.4.
Stress in concrete
Figure 5.4 Conditions in section Note: It is assumed that EC21 Section 4.2.133(12) implies that a should be taken as 0.8 for btaxial bending butthe NAD1 would allow 0.85.
It can be seen from the diagram that the axial force provided by the concrete is given by
NC
= 0.8bxf ccd
NAD Table 3
The moments about the centroidof the concrete section are given by Mcx
=NR c
where
h 1 1! = ———'lx
x
—
2XL
2
-tane2 +
—
btane(x
I
2
btane 6
1
)j
0.8fCdb3tane
M
cy
12
These equations are valid where x' < h. When x' > h, rather simpler equations can be derived. The location of the reinforcement is shown in Figure 5.5.
27
H 14
d
3
— — —— — — —
Figure 5.5 Location of 'reinforcement The stress in a bar is given by
f
x
= (200000
S
0.0035)
.
— —
ZP
yd
where
z
=
I—c -+ I
(b/2
—
d')tanO
0.8
db
=
cose
—
db}
depth from top faceof section to bar considered. This will be d' for top bars and h — d' for bottom bars.
The force in each bar is f5A5 and the moments are obtained by multiplying the forces by the distance of the bars from the centroid of the concrete section. Dimensions to the right or upwards are taken as positive. The total moments and forces carried by the section are the sum of thesteel and concrete contributions. The correct values of x and e have to be found by iteration.
5.3.4 Initial data
f
=—-=-= 1.5
1.5
2ONImm2
COLUIAI4S
Stress over upper 0.8 of the depth of the compression zone
= °8Cd = 16 N/mm2
fyd
f —
=
= 400N/mm2
1.15
As a first estimate of e, assume that the neutral axis is perpendicular to the direction of principal bending. This gives
=
8
= tan1O.59
tan-1
540 Try 8
= 30° which gives
tan8 =
0.58 and cos8
The limiting value of
=
0.87
x is where x'
=h
Hence X
=
h—
(b/2)tane
=
500
—
355
x500
250
x 0.58
= 355 mm
This gives
N
=
x 16
2840 kN
The reinforcement will increase this value significantly, hence x0 will be less = 300 mm. than 355 mm. Try
x
5.3.5 Calculation The simplest way to carry out the calculation is by writing the equations into a spreadsheet and then adjusting the values of and 0 until the correctaxial load and ratio MIM is obtained. The resulting output for the final iteration is given below. It will be seen that the result is satisfactory.
x
Section breadth (b) Overall depth (h) Embedment (ci') Steel area
500 500
50
8152
30
Concrete strength Steel strength
Average stress Design stress
460
Estimateof angle (radians) Estimateof x,
34.2° 05969026 282.5
Tan (angle) Cos (angle)
0.6795993 0.8270806
Neutral axis depth
z
f
F
1329
22378
2 3
363.13 32.29
—21.30
4
400.00 52.26
456.07 815.20 106.50
—192.54
—311.57
—634.97
126.99
742.80 2260.00
356.95 207.29
Steel totals Concrete Design resistances
432.58
Lever arm (x)
1
Barna
16
400
N
3002.80
91.72
M 91.21
1604
M
567.23
—91.21
16304 21.30 12699 220.12
11327
M
333.39
MJM
1.701432
5.4 Braced slender column 5.4.1 General Thecalculation ofthe effectivelength of columns has been adequately covered in Section 2. In the following example, the effective length is assumed.
5,4.2 Problem Calculatethe reinforcementrequired in a400 mm x 400 mm column subjected toa design axial load of 2500 kN combined with thefirst order bending moments shown in Figure 5.6.
The effective length has been calculated as M02
M01
8
m.
43.5.3.5
75 kNm
30 kNm
Figure 5.6 First order moments Assume =
460 N/mm2 and Ck = 30 N/mm2
5.4.3 Slenderness ratio X
= =
101i
=
43.53.2
(l0/h).[.
(8800I400)J =
76.2
5.4.4 Design requirements for slenderness Minimum slenderness ratio = = = Hence
greater of 25 or
15IJ
43.5.3.5(2)
Nsd/(ACfCd)
2500
15/fr
x
10/(400
17.0
<
x 400 x 30/1.5) =
0.78
25
Therefore minimum slenderness ratio =
25
Slenderness ratio>25, therefore column is slender
I
COLUMNS
Critical slenderness ratio Xcrjt e01
=
x
—30
e2 = 75 x
= 25(2
—
x 10)
=
106/(2500
106/(2500
4.3.5.5.3(2)
e01/e02) —12
mm
x
10) = 30 mm
=
60
Hence
= 25(2 +
12/30)
>
Slenderness ratio effects
therefore design is required for second order
5.4.5 Eccentricities Additional eccentricity ea
v ea
=
43.5.4(3)
lI2
= 1/200 =
2.5.12(4) Eqn 2.10
8800/400
= 22 mm
Equivalent first order eccentricity is greater of
x 30 — 0.4 x
0.6e2
+ 0.4e01 =
0.4e02
= 0.4 x 30 = 12.0 mm
0.6
4.3.5.6.2
12
=
13.2 mm
or
Hence ee
=
13.2 mm
Ultimate curvature, hr Assume d
hr =
= 400
=
—
200000
= 340 mm
60
2x
4.3.5.6.3(5) Eqn 4.72
2K2€1j0.9d
460
x K2
=
x 1.15 x 0.9 x 340
13.07K2
x 106 radians
Second order eccentricity e2 1crit
= 25 (2
—
Msdl/Msd2)
where
MSdl —
(_-) 5.0
54.6
The column or wall should therefore be designed for the following minimum conditions:
Design axial resistance (NAd)
=
Eqn 4.63
Nsd
=
Design resistance moment (MAd)
43.5.53(2)
Nsd
x
Eqn 4.64
For this example
MRd = 700
x —-- =
7.0
>
5.0 kNm
20
6.2.4 Reinforcement The vertical reinforcement should not be less than 0.004A0 or greater than
5.4.7.2(1)
0.04A0.
Half of this reinforcement should be located at each face.
5.4.7.2(2)
The maximum spacing for the vertical bars should not exceed twice the wall
5.4.7.2(3)
The area of horizontal reinforcement should be at least 50% of the vertical reinforcement.The bar size should not be less than one quarterofthe vertical bar size and the spacing should not exceed 300 mm. The horizontal reinforcementshould be placed betweenthevertical reinforcementand the wall
5.4.7.3
thickness or 300 mm.
face.
=
(1)—(3)
Link reinforcement is required in walls where the design vertical reinforcement exceeds O.O2A.
In normal buildings it is unlikely that walls will be classified as slender. For practical considerations they will generally not be lessthan 175 mm thick and the vertical load intensity will normally be relatively low. Thus the limiting slenderness ratio given by will be high. In cases where the wall is slender,only slenderness about theminor axis need be considered. Even in this case it is likely that only the minimum conditions given in EC2 Clause 4.3.5.5.3(2) Eqns 4.63 and 4.64 will apply.
15I[
5.4.7.4(1)
7 FOUNDATIONS U
7.1
Ground bearing footings 7.1.1 Pad footing Designa square padfooting fora400 mm x 400 mm column carrying aservice load of 1100 kN, 50% of this being imposed load with appropriate live load reduction. The allowable bearing pressure of the soil is 200 kN/m2. 7.1.1.1
Base size With 500 mm deep base, resultant bearing pressure
= 200
x 24 = 188 kN/m2
— 0.5
1100
Area of base required = Use 2.5 m
=
5.85 m2
188
x 2.5 m x 0.5 m deep base
7.1.1.2 Durability For components in non-aggressive soil and/or water, exposure class is 2(a). Minimum concrete strength grade is C30/37.
Table 4.1 ENV 206 Table NA.1
For cement content and w/c ratio refer to ENV 206 Table 3(6) Minimum cover to reinforcement is 30 mm. For concrete cast against blinding layer, minimum cover
> 40 mm.
NAD Table 6 4.1 33(9)
Use 75 mm nominal cover bottom and sides 7.1.1.3 Materials = 460 N/mm2 Type 2 deformed reinforcement with Concrete strength grade C30/37 with maximum aggregate size 20 mm 7.1.1.4 Loading
= l.3SGk + 15k = 1570 kN
Ultimate column load
NAD 6.3(a)
Eqn 2.8(a) Table 2.2
7.1.1.5 Flexural design Critical section taken at face of column Msd
= 1570 (2.5 8
—
0.4)2
x 2.5
2.533(5)
= 346 kNm
Assuming 20 mm bars
d
= 500
—
75
—
20
= 405 mm
Using rectangular concrete stress diagram — —
fck
— —
30
— —
20 N/mm2
C
= 0.85 x 20 = 17 N/mm2
Figure 4.4 Eqn 4.4 Table 23
For reinforcement
=
f
460
=
= 400 N/mm2
2.23.2P(1) Table 23
1.15
7s
For the design of C30/37 concrete members without any redistribution of moments, neutral axis depth factor
d
2.53.4.2(5)
0.45
Using the design tables for singly reinforced beams Msd
2500
bd2fCk
x— d
Af —-
346 X
=
106
x 4052 x 30
= 0.063 <
=
0.028
0.45
OK
= 0.033
bdfCk
Hence
AS
= 0.033 x 2500 x 405
30 = x— 460
2179 mm2
Minimum longitudinal reinforcement
=
0.6btd 460
5.42.2.1 5.4.2.1.1
= 0.OOl3bd iz 0.OOl5bd
= 0.0015 x 2500 x 405 = 1519 mm2 7120 gives 2198 Bar crs.
> 2179 mm2
2500
Maximum spacing
—
2(75)
—
OK
20
6
= 3h
i
500
= 388 mm = 500 > 388 mm
OK
7T20(EW) are sufficientfor flexural design. Additional checks for punching and crack control require 9T20 (EW) — refer to Sections 7.1.1.7 and 7.1.1.8.
NAD Table 3 5.4.a2.1(4)
Use 9120 (EW) 7.1.1.6 Shear Minimum shear reinforcement may be omitted in slabs having adequate provision for the transverse distribullon of loads. Treating the pad as a slab, therefore, no shear reinforcement is required if VSd VRd1.
43.2.IP(2) 4.3.2.2(2)
cu1ooxIoI4s Shear force at critical section, distance d from face of column —
—
(a5
VSd
0.405
=
)
405 kN
4.3.23
Shear resistance, VAd1, with zero axial load VRd1
TAd
k
=
TRdk(l.2
+
4Op1)
Eqn 4.18
bd
Table 4.8
=
034 N/mm2
=
1.6—d = 1.195z1.0
To calculate A1, area of tension reinforcement extending critical section, determine
net
=
d + 1fleI beyond
(a) i
aalb
4.3.2.2(10)
Eqn 5.4
prov
For curved bars with concrete side cover of at least
=
b
— —
3
0.7
Eqn5.3 4
f
For bars in the bottom half of a pour, good bond maybe assumed. Hence for 4,
5.23.4.1(1)
5.2.2.1(2)
32mm
= ao N/mm2 =
4,
x
Table 5.3
400
=
—-
33.34,
For anchorage in tension
min = 03 x 1b z 104, 100 mm = 104, = 200mm
Eqn 5.5
Actual distance from critical section to end of bar =
2500 2
1.2
x
1.195
x
2500
x 405 x
iO-3 = 493 kN
405kN
VSd
No shear reinforcement required Ch,eck that
VSd
=
u
VRd2
VRd2
0.7
to avoid crushing of compression struts. =
—
200
0.55
Eqn 4.20
iz 0.5 N/mm2
= cd'w°9°' = 0.55 x 20
x 2500 x
2
0.9
x 405 x
10
2
= 5012 > 405kN
Eqn 4.19
OK
7.1.1.7 Punching Length of base from face of column
a
= 1050mm
a —
=—>2
h
1050
Figure 4.16
400
By definition the foundation should be considered as a slab. Critical perimeter at 1.5d from faceofcolumn shouldbe checked for punching. U
=
2ir (1.5 x 405) + 4 x
400 =
5417 mm
In foundations the applied shear may be reduced to allowfor the soil reaction within the critical perimeter.
4.3.4.1P(4)
& 43.4.2.2 4.3.4.1(5)
Enclosed area
= (3 x 405) + 400 = 1615 mm = 608 mm Corner radius = 1.5 x 405 = 1.6152 — (4 — ir) 0.6082 = 2.29 m2 Area Total width
= 1570 (1
—
= 995 kN
The applied shear per unit length
43.4.3(4)
v=Y Sd
u
=
1.0 for pads with no eccentricity
of load
FOUNO&TIONS
Therefore 995
=
vSd
i0
x
= 184N/mm
5417
Theamount oftensile reinforcementin two perpendicular directions should be greater than 0.5%. This is assumed to require p1 + p1, > 0.5%. Using 9120 (EW),
=
A
4.3.4.1(9)
2830 mm2 (EW)
For Bi
= 415mm
dx 1OOA
S
= 0.27%
bd For B2
= 395mm
dy 1OOA
S
= 0.28%
bdy 0.27%
÷ 0.28%
=
0.55%
> 0.5%
OK
Punching resistance for a slab withoutshear reinforcement = + TRdk(l.2
VRd1
4.a4.5
40p1)d
The equation produces similar values to the shear check performed above = 034 x 1.195 x 1.2 x 405 = 197 > 184 N/mm VAd1 No shear reinforcement required Checkthe stress at the perimeter of the column
0.90f = 0.90fö
VSd/ud
d
=
405mm
u
=
4x400 = 1570 x i0
Stress =
405
x 1600
=
NAD 6.4(d)
4.9 N/mm2
1600mm =
2.4
<
4.9 N/mm2
....
....
OK
7.1.1.8 Crack control Use method without direct calculation.
4.4.2.3
Estimate service stress in reinforcement under quasi-permanent loads using the following approximation
44 23(3)
Gk
+ '2°k =
+ °•3k = 550 +
0.3
x 550
=
715 kN
22.2 3P(2) 23.4 Eqn 2.9(c)
& NAD Table 1
FOUNDTOHS
=
Hence quasi-permanent load/factored load and estimated service stress
x
= 0.46 x
715/1570
=
0.46
A = 0.46 x 400 x = 142 N/mm2 A prov 2830
Either limit bar size using EC2 Table 4.11(1) or bar spacing using EC2 Table
4.4.23(2)
4.12.
=
4)
20
< 32mm
OK
This has been chosen to comply with Table 4.12 as well. =
Using 9T20 (EW) bar spacing
290
< 300 mm
OK
Check minimum reinforcement requirement
4.4.2.3(2)
kA/a
For
A
4.4.2.2
Eqn 4.78
it is considered conservative to use (h/2)b =
100%
= 460 N/mm2
x
For c.eff use minimum tensile strength suggested by EC2
k
=
=
3 N/mm2
0.4 for bending
For k interpolate a value for h
k
—
0.5
+ 03(80 —
= 50 cm from values given 50)/(80 — 30) = 0.68
Therefore As req
= 0.4
x 0.68 x 3 x 250 x 2500/460
=
1109 mm2
= 2830 > 1109 mm2
OK
7.1.1.9 Reinforcement detailing
Checkthat flexural reinforcement extends beyond critical section for bending for a distance d + net 1b
=
=
3334)
5.43.2.1(1)
& 5.4.2.13
667 mm
Assuming straight bar without end hook 1
=
d + net = Actual distance =
1.0
x 667 x
405
514 mm
Eqn 5.4
+ 514 = 919 mm
2500
2
= 2830
—
400 —
2
—
75
= 975 > 919 mm
The reinforcement details are shown in Figure 7.1.
OK
POUNDATIONS
I
I
lIi.
S
S
U
75 cover
S
U
U)
111500
9120— 300EW 2500
I-.'
Figure
7.1
Detail of reinforcement in pad footing
7.1.2 Combined footing Design a combined footing supporting one exterior and one interior column.
An exterior column, 600 mm x 450 mm, with serviceloadsof 760 kN (dead) and 580 kN (imposed)and an interior column, 600 mm x 600 mm, with service loads of 1110 kN (dead) and 890 kN (imposed) are to be supported on a rectangular footing that cannot protrudebeyond the outerfaceof the exterior column. The columns are spaced at 5.5 m centres and positioned as shown in Figure 7.2.
The allowable bearing pressure is 175 kN/m2, and because of site constraints, the depth of the footing is limited to 750 mm.
--I —
600
Figure 7.2 Plan of combined footing
7.1.2.1 Base size Service loads =
Gk +
Column A: 1340 kN and Co!mn B: 2000 kN Distance to centroid of loads from LH end =
0.3 +
2000
x 5.5
3340
=
3.593m
For uniform distribution of bad under base Length of base
= 2 x 3.593 say 7.2 m
With 750 mm deep base, resultant bearing pressure = 175 — 0.75 x 24 = 157 kN/m2 Width of base required Use 7.2 m
3340
=
7.2
x
157
= 2.96 say 3.0 m
x 3.0 m x 0.75 m deep base
7.1.2.2 Durability
For ground conditions otherthan non-aggressive soils, particular attention is needed to the provisions in ENV 206 and the National Foreword and Annex to that document for thecountry in which the concrete is required. In the UK it should be noted that theuse of ISO 9690 and ENV 206 may not comply with the current British Standard, BS 8110: Part 1: 1985 Table 6.1(2) where sulphates are present. Class 2(a) has been adopted for this design. Minimum concrete strength grade is C30/37. For cement content and wlc ratio refer to ENV 206 Table 3. Minimum coverto reinforcement is 30 mm.
For concrete cast against blinding layer, minimum cover > 40 mm. However, it is suggestedthat nominalcover> 40 mm is asufficientinterpretation of the above clause
Table 4.1 ENV 206 Table NA.1
Tab 6 4.1.33(9)
Use 75 mm nominal cover bottom and sides and 35 mm top 7.1.2.3 Materials
= 460 N/mm2
Type 2 deformed reinforcement with
NAD 6.3(a)
Concrete strength grade C30/37 with maximum aggregate size 20 mm 7.1.2.4 Loading Ultimate column loads
= l.35Gk + 150k
Column A: 1896 kN and Column B: 2834 kN Distance to centroid of loads from LH end
= 03 + 2834 x
4730
5.5
= 3.595m
i.e virtually at centre of 7.2 m long base Assume uniform net pressure =
4730 7.2
= 657 kN/m = 219 kNIm2
See Figures 72, 7.4 and 7.5 for loading, shear force and bending moment diagrams respectively.
Eqn 2.8(a) Table 2.2
1896 kN
2834 kN
kN/m
4723 kN/m
uJT316O
t tt f
'I
t
657 kN/m
49m
kO•6m
ft t ft 06m
+' 1•lm -I
Figure 7.3 Loading diagram
1717 kN
I
957 kN 289m
Figure 7.4 Shear force diagram
2167 kNm
600
2290
2610 I
Figure 7.5 Bending moment diagram
600! 1100_I I
I
7.12.5 Flexural design —
7.1.2.5.1 Longitudinal direction
top steel
Mid-span Msd
d
= 2167 kNm = 750 — 35
—
20
—
32/2
= 679 say 675 mm
Using the design tablesfor singly reinforced beams 2167 x 106 Msd
=
bd2f<
x
—
d
Af bdfck A
=
=
0.053
3000
x 6752 x 30
0.123
< 0.45 limit with zero redistribution
OK
2.5.3.4.2(5)
= 0.064
x 3000 x
= 0.064
675
x
-
460
=
8452 mm2
=
2818 mm2/m
Use 12T32 @ 250 mm crs. (3217 mm2lm) Continue bars to RH end of baseto act as hangers for links. Particular attention is drawn to the clauses for bar sizes larger than 32 mm. These clauses are restrictive about laps and anchorages, such that designers may need to resort to groups of smaller bars instead.
= 3h p 500 = 500 > 250 mm
Maximum spacing
7.1.2.5.2 Longitudinal direction
OK
NAD Table 3 5.4.3.2.1(4)
bottom steel
—
5.2.6.3P(1) & P(2)
At column face = 398 kNm Msd
d Msd
= 750
3000
=
0.012
=
0.012
bdfck AS
75
398
=
bd2ç
Af
—
— 10
x
=
106
x 6652 x 30
665 mm
=
0.010
30 x 3000 x 665 x —
For minimum steel Asmin
= 1561 mm2
= 520 mm2/m
460
= 0.0015bd = 998 mm2/m
Use 12T20 @ 250 mm crs. (1258 mm2/m)
5.4.2.1.1
cou.4oxIo4s 7.1.2.5.3 Transverse direction — bottom steel
=
Msd
1.5
—
0.45
2
219
x
=
178 kNm/m
Minimum steel governs. Use T20 @ 250 mm crs. (1258 mm2/m)
I
7.1.2.6 Shear Critical shear section at distance d from face of column
4.3.2.2(10)
Column B interior side VSd VRd1
TRd
k p,
x 657
=
1717 — 0.675
=
TRd k(1.2
=
0.34 N/mm2
=
1.6—d .z1.0 =
=
0.00476
=
1273 kN
+ 40p,)bd
4.3.23 Eqn 4.18 Table 4.8 1.0
Ensure bars are continued sufficiently. VRd1
VSd
=
957 kN
>
VAd1
Therefore shear reinforcement required. Shear capacity with links
=
VRd3
VCd
+
=
+
VRd1
VWd
4.3.2.43 Eqn 4.22
Therefore 1273 — 957 =
V
=
316 kN
A
— x O.9dfd
Eqn 4.23
S
= 400 N/mm2,
cWd
A s
d
316x103
0.9x675x400
=
675 mm
= 1.30 mm2/mm
Where shear reinforcement is required, the minimum amount is 100% of the EC2 Table 5.5 value.
NAD Table 3 5.4.3.3(2)
With
=
460,
wmIfl
= 0.0012 by interpolation
Table 5.5
For links
pW
=
ASWIsb
=
0.0012 x 3000 =
Eqn 5.16
W
Therefore
A
3.6
>
1.30 mm2/mm
mm
Therefore minimum links govern. Determine link spacing, using EC2 Eqn VRd2
Vsd/VRd2
=
ufcd
5.17—19.
Eqn 4.25
b(0.9 d)12
x
=
0.55
=
1273/10020
20
x 3000 x 0.9 x 675 x 10I2 = =
0.13
10020 kN
< 0.2
Use EC2 Eqn 5.17 to determine link spacing.
Sm
=
1
0.8d (Note: 300 mm limit in Eqn 5.17 does not apply to slabs)
5.4.33(4)
= 506 mm
O.75d
NAD 6.5(f)
Transverse spacing of legs across section
5.4.2.2(9)
dor800mm = 675mm Use 12 legs T10 @ 250 mm crs. in each direction where
A5 — S
=
12
x
78.5
=
3.77
250
>
V>
3.6 mm2/mm
Check diagonal crack control VCd VSd
VSd
= =
VAd1
5.4.2.2(10)
1273 kN (max.)
< 3Vc
Distancesto where
Xb
VSd
= 1502
=
—
657
=
4.4.2.3(5) VRd1
957
from face of columns A and B
= 0.830m
1.157 m
Check shear in areas where bottom steel is in tension and p1
OK
= 957 kN
No further check required.
xa
VRd1
= 0.0015 (mm. steel)
FOUNDATIONS
= 034(1.2 + 0.06)3000 x 665 x 10 = 854> 723 kN .. OK
VRd1
No links required at RH end of base
In orthogonal direction, shear at d from column face
= 219(0
—
0.45
—
0.6
x
2)
=
148 kN/m
2
From above
=
VAdi
= 3.0
284
>
148 kN/m
OK
No links required in orthogonal direction 7.1.2.7 Punching Length of one side of critical perimeter at 1.5d from face of column = 3 690 + 600 = 2670 mm
4.3.4.1P(4) & 4.3.4.2.2
x
= 3000 mm
This extends almost the full width of the base
Hence it is sufficient Just to check line shear as above and shear around perimeter of column face, where VsdIud
0.90
J7
= 0.90 x
= 4.9 N/mm2
NAD 6.4(d)
The shear stress at the column face perimeter with d = 675 mm is less than 4.9 N/mm2 in both cases (see Table 7.1) Table 7.1
OK
Punching shear at column face
column
Perimeter (mm)
Load (kN)
(N/mm2)
A
1650
1896
1.7
B
2400
2834
1.75
Stress
7.1.2.8 Crack control
Use method without direct calculation.
4.4.2.3
Estimateservice stress in reinforcement under quasi-permanent loads, using
the following approximation:
+ '2k
=
Gk
4.4.2.3(3)
+
The relevant loads are shown in Table 7.2. Table 7.2 Column loads for cracking check Load Gk .i-
03k(kN)
l.35Gk Ratio
+ 1•5k
(kN)
columnA
column B
934
1377
1896
2834
0.49
0.48
Estimated steel stress
A
—-
= 0.49 x f x
Aprov
= 0.49 x 400 x Either limit bar size using EC2 Table 4.11
8452 12
x 804
=
172 N/mm2
or bar spacing using Table 4.12.
> 32 mm used.
In Table 4.11 bar size
25 mm
In Table 4.12 spacing
285 mm in pure flexure
> 250 mm used.
. OK
Check minimum reinforcement requirement
A
4.4.2.3(2)
4.4.2.3(2) 4.4.2.2(3) Eqn 4.78
kckfcteffActhYs
it is considered conservative to use (h12)b.
For aS
= 100% x f = 460 N/mm2 yk
For eff use minimum tensile strength suggested in EC2, 3 N/mm2. =
0.4 for bending
For k interpolate a value for h
= 75 cm, which gives k = 0.53.
Therefore A5
0.4
12T32 gives A5
x 0.53 x 3 x >
750
x 3000/(2 x 460) =
1555 mm2
1555 mm2
OK
7.1.2.9 Detailing Check bar achorage detail
at LH end.
The anchorage should be capable of resisting a tensile force F5
=
5.4.2.1.4(2)
VSda,/d
with 5.43.2.1(1)
a1
F5
V
= =
VSd
column reaction =
1896 kN
The bond strength for poor conditions in the top of the pour
x Table 5.3 value = 0.7 x 3 = 2.1 N/mm2 = (4)(ycI'd) = 47'.6q5 = 1524 mm =
bd
0.7
Continuing all T32 bars to end
A prov = 9650 mm2
Asreq
=
VSd/fyd = 1896 x 10/400 = 4740 mm2
5.2.2.1 & 5.2.2.2
Eqn 53
FOUND&TIONS
Hence required anchorage, (+)1net at a direct support =
(--)lb
x 4740/9650 =
500 mm
> °31b
Figure 5.12 OK
of column = 600 — 75 = 525 mm ... OK Theanchorage may be increased to 'bnet' if preferred, by providing a bend at the end of the bar. Anchorage up to face
Therequirementfortransversereinforcementalongthe anchorage length does not apply at a direct support. Secondary reinforcement ratio for top steel p2
d
O.2p1
=
AS
750
—
=
0.2
35
—
x 0.00476
10
=
0.00095
= 705 mm
670 mm2/m
Use 116 @ 250 mm crs. (804 mm2/m) transversely in top
500 mm Spacing The reinforcement details are shown
OK
in Figure 7.6.
12132 — 250 12T20 — 250 T20 —250EW —
LL1
Hill 1—H.J 120—250
Figure 7.6 Detail of reinforcement in combined footing
72 Pilecap design 7.2.1 Pilecap design example using truss analogy
Afour-pile groupsupports a 500 mm squarecolumn which carries a factored load of 2800 kN. The piles are 450 mm in diameter and spaced at 1350 mm centres.
7.2.1.1 Pilecap size Assumea pilecap depth of 800 mm. Allowthe pilecap toextend 150 mm beyond the edge of the piles, to give a base 2.1 m square as shown in Figure 7.7. Use 2.1 m x 2.1 m x 0.8 m deep pilecap
5.2.3.3
5.43.2.1
375
1350
375
Figure 7.7 Pilecap layout 7.2.1.2 Durability
For components in non-aggressive soil and/or water, exposure class is 2(a). • .
.
Minimum concrete strength grade is C30137.
Table 4.1 ENV 206 Table NA.1
For cement content and w/c ratio refer to ENV 206 Table 3. Minimum cover to reinforcement is 30 mm.
NAD Table 6
Use 100 mm nominal bottom cover over piles and 50 mm sides 7.2.1.3 Materials
= 460 N/mm2 Type 2 deformed reinforcement with Concrete strength grade C30/37 with maximum aggregate size 20 mm.
NAD 6.3(a)
7.2.1.4 Element classification
A beam whose span is less than twice its overall depth is considered a deep
2.5.2.1(2)
beam. With the effective span, 1eff —
h
l,
taken to the centre of the piles:
2.5.2.2.2
1350 =—=1.7 275 mm
z a0015 btd = 0.0015 x 2100 x 675 = 2127 mm2..
yk
The reinforcement details are shown in Figure
7.8.
4Ti6 I
800F I•
IJL
6125EW
Figure 7.8 Details of pilecap reinforcement
I
L4J
OK
OK
NAD Table 3 5.43.2.1(4) 5.4.2.1.1(1)
7.2.1.7 Shear
Only in elements such as slabs may shear reinforcement be omitted where
43.2.1P(2)
calculations justify. Despitethe classification for the pilecapgiven above, in line with commonUK practice, it is not intended to provide shear reinforcementwhen VSd VRd1. Takethe critical section for shear to be located at20% ofthe pile diameter into the piles, extending the full width of the pilecap.
43.2.2(2)
BS 8110 Figure 3.23
Distance from centre of loaded area
x
= 1350/2 — 0.3 x 450 = 540 mm
Shear resistance VRd1
=
+ 40p)bd
TAd k(1 .2
TAd
= 034 N/mm2
k
= 1.6—d #zl.0 = =
3928 = 2100 x 675
4.3.2.3 Eqn 4.18 Table 4.8 1.0
0.00277
All of tension steel is to continue sufficiently pastcritical section; check when detailing. VRd1
= 034(1.2 + 40 x 0.00277) 2100 x 675 x
iO = 632 kN
Consider enhanced resistance close to the supports
=
2.5d —
x
1.0
2.5
=
x 675 =
43.2.2(5)
3.125
540
43.2.2(9) OK
5.0
Shear force VSd
2850 =—=1425kN 2
<
=
3.125
x
= 1975 kN
632
No shear reinforcement required Having taken into account the increased shear strength close to the supports,
it is necessary to ensure that the reinforcement is properly anchored.
43.2.2(11)
In this caseall reinforcement will extend to centre line of pile and be anchored OK beyond that position 7.2.1.8 Punching Piles fall within 1.5d perimeter from column face, it is thus only necessary to check shear around column perimeter, where Stress
0.9
=
0.9
x
= 4.9 N/mm2
No enhancement of this value is permitted.
43.4.2.2(1)
NAD 6.4(d) 4.3.2.2(5)
x iO = 4 x 500 x 675 2800
Stress =
2.1
< 4.9 N/mm2
.
OK
7.2.1.9 Crack control Use method without direct calculation.
4.4.2.3
Estimateservice stress in reinforcement under quasi-permanent loads using following method
4.4.2.3(3)
Gk
+ '2°k
=
+
For this example the column loads, Gk
= 785 kN
1200 kN and
Hence the quasi-permanent load/factored load =
1200
+ 03 x 785 = 2800
0.51
Estimated steel stress
A
0.51
x fyd x — = Aprov
0.51
x 400 x
= 3928
185 N/mm2
Either limit bar size to EC2 Table 4.11 value or bar spacing to EC2 Table 4.12
4.4.23(2)
value.
25 mm = 25 mm used
From Table 4.11 bar size From Table 4.12 bar spacing
270 mm
<
OK
275 mm used
Check minimum reinforcement requirement
kC Idct,eff A/a Ct S
As For
4.4.2.3(2) 4.4.2.2(3)
A it is considered conservative to use = 100%
as
x fyk
Eqn4.78 (h/2)b.
= 460 N/mm2
use minimum tensile strength suggested by EC2, 3 N/mm2.
For
k
=
0.4 for bending
k
=
0.5forh
80cm
Therefore A5
1096 mm2
AS,prov =
3928mm2
OK
7.2.1.10 Detailing
The reinforcement corresponding to the ties in the model should be fully anchored beyond the nodes, i.e., past the centres of piles. b
5.4.5(1)
5.2.2.3(2)
— —
bd
For bars in bottom half of a pour, good bond may be assumed.
5.2.2.1
Hence
Ib
= 3.0 N/mm2
(4,
= 25 x 400
=
4x3
b,net
A
Areq > —
Table 5.3
32 mm) 834mm
b.min
prov
Using bobbed reinforcement, aa = 0.7
Ibnet =
0.7
3563 x 834 x — 3928
= 530 mm
Length beyond centre of pile allowing for end cover = 375 — 50 = 325 < 530 mm Bars cannot be anchored in manner shown in EC2 Figure 5.2. Use bent-up bars with large radius bend and anchorage length
x
—
=
756mm
prov
Diameter of bends can be obtained from NAD Table 8(1). Assumethat the limits in the table are equallyapplicableto bar centres given for minimum For 125 @ 275 mm crs., bend diameter = 134,,
cer
bend radius
= (13/2) x 25 = 165 mm
Theuse of NADTable8 is conservative,as it is based on full stress in thebars at the bend. The values given appear to be consistent with BS 8110: Part 1: = 30 N/mm2. Clause 3.12.8.25 using For concrete placed in the UK, it should be possible to demonstrate,compliance with EC2 Clause 5.2.1.2P(1) by usingthe BS 8110Clause above, with the result that smaller diameter bends may be used. For the edge bars, which have a minimum cover > 34, Table 8 gives 200 mm radius bend (see Figure 7.9).
= 75 mm, NAD
Therequirementfortransversereinforcementalong the anchorage length does not apply at a direct support. Provide bars to act as horizontal links, such as 4116 @ 150 mm crs.
NAD Table 8
FOUNDATIONS
3 785
Figure 7.9 Detail of bent-up bars
7.2.2 Pilecap design example using bending theory Take the pilecap from the preceding example but use bending theory to determine the bottom reinforcement. The shear force diagram is shown in Figure 7.10.
1425 kN
1425 kN 425
500
1350
Figure 7.10 Shear force diagram
7.2.2.1 Flexural reinforcement Msd
= 1425
z
= 0.975d = 658 mm
A6
= 2979 mm2
(0.425
+
= 784 kNm
.2)
Because of the difference in modelling, this is less reinforcement than the previous example.
7.2.2.2 Detailing At an end support, the anchorage of bottom reinforcementneeds to becapable
of resisting a force: F6
N
=
VSdal/d
+
5.4.2.1.4(2)
Eqn 5.15
Nsd
= 0 in this case
with 5.43.2.1(1)
a1
F6
A
= =
VSd
= 1425 kN
1425
x
10/400
= 3563 mm2
This is identical to the area of steel required in the previous example.
Use 8T25 as before (3928 mm2) Using the same detail of bobbed bars
net
= 530 mm
EC2 Figure 5.12(a) applies and is taken to require an anchorage length, = 353 mm past the line of contact between the beam and its (213)1flet support. Using a position 20% into the pile to represent the line of contact, the length available for anchorage = 0.3 x pile dia. + 375 — cover =
0.3
x 450 + 375
—
50
= 460 > 353 mm
....
OK
.
B SPECIAL DETAILS 8.1
Corbels 8.1.1 Introduction
2.5.3.7.2
Consider a corbel designed to carry a vertical ultimate designload of 400 kN with the line of action of the load 200 mm from the face of the support (wall, column etc), as shown in Figure 8.1.
Fv
200 300
ac b d
400 kN
=
hc
465 500
1¼
Figure 8.1 Corbel dimensions
8.1.2 Materials =
30 N/mm2 (concrete strength class C30/37)
=
460 N/mm2 (characteristic yield strength)
8.1.3 Design 8.1.3.1 Check overall depth
of corbel
the maximum shear in the corbel should not exceed VRd1. The butthis would give depth of the corbel could be reduced by putting VRd2 an increased tie force and consequent detailing problems. Thevalueof TAd in the expression for VRd1 (EC2 Eqn 4.18(1)) may be modified by the factor defined in EC2 Clause 43.2.2(9). in EC2 Eqn 4.17. Hence By inspection 6 will be a minimum when x = will also be a minimum. VHdl Conservatively,
F
2.53.7.2(5)
43.23
f
Now VRd1
=
[9rRdk(1.2
+ 4Op1) + 0.15a]bd
= 2.5d/x with 1.0 = 2.5 x 465 = 200 TAd
= 0.34 N/mm2
k
= 1.6—d
p1 is assumed
lzl
f3
5.81
5.0
Eqn 4.18 Eqn 4.17
5.0
Table 4.8
= 1.14m
to be 0.006 (4T16)
SPECIALDETAILS
No provision hasbeen made to limit horizontalforces at the support; therefore minimum horizontal force (He) acting at the bearing area should be assumed. This is given by
a
HC
=
= ±8OkN
0.2F
V
=
A/Sd —
2.5.3.7.2(4)
=
where
—80 kN
C
Therefore a
C
=
—80
465
x i0 x 300
=
—0.6 N/mm2
Hence VAd1
=
[5 x 0.34 x 1.23 (1.2 + 40
x 0.006) — 0.15 x 0.6] x 465 x 300 = 407 kN
Therefore VRd1
>
VSd
=
F
= 400 kN
OK
2.5.3.7.2(5)
8.1.3.2 Determine main reinforcement requirement
Now 0.4h
a, therefore a simple strut and tie model may be assumed, as shown in Figure 8.2.
Figure 8.2 Strut and tie model Under the vertical load
Fa = —- ;and z 0.85fck
FC
(0.8x)b cosf3
The determination of x will be an iterative procedure Choose x such that
=
ç
0.002 and
f
= 400 N/mm2
Therefore
x
=
0.0035 0.0035
+ 0.002
x 465 = 296 mm
2.5.3.7.2(1)
SC(&L DrMLs Now z =
d
= 347 mm
— 0.8x12
and
cosf3
=
0.5
Therefore, from above,
F
x 200
= 400
= 231 kNand
347
x 30 x 0.8 x 296 x 300 x
0.85
FC
0.52
= 301
kN
1.5
=
For equilibrium F
F and further refinement gives z =
x = 235 mm,
F
371 mm,
= 216 kN
In addition, EC2 Clause 2.5.3.7.2(4) requires a horizontal force of applied at the bearing area. 0.2FV =
HC
=
0.2
x 400
=
H to be
2.53.7.2(4)
80 kN
296kN
As.req = 296
x io
=
740mm2
460/1.15
Use 4T16 bars
8.1.3.3 Check crushing
of compression strut
This has been checked directlyby the calculation of F above. However, an indirect check may also be made VRd2
= [(j-) vf]b0.9d
p
= 0.7
—
I = —200
Eqn 4.19 0.55
Eqn 4.20
0.5
Therefore VRd2
= (j-) x 0.55 x 20 x 300
x 0.9 x 465
= 690 kN
Hence VRd2
> F, = 400 kN
OK
8.1.3.4 Check link reinforcement requirements
5.4.4(2)
Links are required if: 0.4 ACICd/cd
A
= 500
x 300
fcd
=
=
1.5
Eqn =
150
20 N/mm2
x iO mm2 f
= 1.15
= 400 N/mm2
5.21
5PC1M. DTA1LS
Hence, links are required if 0.4
A5
x
150
x
i0 x 20/400
= 3000 mm2
Now
Apr
= 804 < 3000 mm2
Therefore links are not required Nevertheless,in practice some linksshould be provided to assist in fixing the main reinforcement.
A5
O4Aspr
=
0.4
x 804
322 mm2
5.4.4(2)
Use 4T8 links (8 legs)
8.1.3.5 Check bearing area of corbel Allowable design ultimate bearing stress concrete. Therefore area required
°•8cd for bearing bedded in
= 400x103
Assume transverse bearing = Therefore length of bearing =
0.8
x 20
25000 mm 2
250mm 100mm
8.1.4 Detailing The reinforcement details are shown in Figure 8.3.
30 cover to main bars
4T16 welded to T20 cross—bar 2110
Figure 8.3 Corbel reinforcement details
L
EC2, Part lB
EC(M. DETMLS 8.1.4.1 Anchorage of main bars at front edge of corbel AnchorT16 ties by means of a cross bar running horizontally and welded to the ties. =
296kN
An allowable bearing stress under the cross bar can be obtained from EC2
5.4.81
Eqn 5.22 as Rdu
=
Eqn 5.22 (mod)
Note: Use ofthis stress requires that the concrete be confined by means of links etc. In areas where the cover is small, the designer may wish to use a modified version of Eqn 50 in BS 8110(2).
296x103 = 4485 mm 2
Therefore area of bar required =
3.3
x 20
For a T20 bar, length required is 225 mm. Use T20 cross bar 240 mm long weldedto T16 ties 8.1.4.2 Anchorage of main bars into support Required anchorage length 1 Lnet
= aaib
>
As.req
Eqn 5.4
bmrn
Aprov
Now 1b
fyd
= (I4) (cd bd
Eqn 5.3
= 400 N/mm2
Bond conditions may be considered good as the T16 bars will be anchored into a substantial support(column or wall). =
3 N/mm2
=
(16/4)
5.2.2.1(2)(b)
Table 5.3
x (400/3)
= 533 mm
Now
Asq
= 740 mm2
Asprov
=
804 mm2
Therefore 1
net
=
804
= Provide
1x533xZ9 °31b
=
490mm
10 or 100 mm
= 490 mm (see Figure 8.3)
aa
=
1
5.2.3.4.1(1)
SPECIALDETAILS
The detail at the front edge of the corbel is shown in Figure 8.4 100
I—
Figure 8.4
Detail
20 40
lI4
30
at front edge of corbel
The inside face of the T20 bar is positioned not less than the cover beyond the edge of the bearing area. This is an interpretation of BS 8110 as no guidance is given in EC2
8.2 Nibs 8.2.1 Introduction Considera nib designed to carry a precastconcretefloor slab imposing a vertical ultimate design load of 25 kNIm.
8.2.2 Materials
fck fyk
=
30 N/mm2 (concrete strength class C30137)
=
460 N/mm2 (characteristic yield strength)
8.2.3 Design Provide a 15 mm chamfer to the outside edge of the nib and assume the line of action of the load occurs at the upper edge of the chamfer Permissible design ultimate bearing stress Therefore minimum width of bearing =
= 0•6d for dry bearing 25 x 10 = 0.6
x 20 x 1000
Minimum width of bearing for non-isolated member Allowance for nib spalling
= 40 mm
Nominal bearing width =
+ 20
+ 25 =
52.4
85 mm
Allow an additional 25 mm for chamfer on supported member. Width of nib projection
= 85 + 25 =
ini
BS 8110
Table 5.1 BS 8110
25 mm 40
2.1 mm
5.2.3.2 BS 8110
= 20 mm
Allowance for inaccuracies =
EC2 Part
110 mm
lB
SEC%M. QETMLS
The distance of the line of action of the load from the face of the beam = 110—15 = 95mm Assuming 20 mm coverto the 110 links in the beam
a
= 95 + 20 + 5 = 120 mm
Check minimum depth of nib.
6.
Assuming T8 bars, minimum internal diameter of loop is Therefore minimum depth of nib = 20 + 8 8 + 20 =
x
= 105 mm
Depth of nib
M
= 25 x
Effective depth (d)
M —
=
bd2ck
Afsyk
104 mm
NAD Table 8
= 3 kNm/m
0.12
= 105 — 20
—
3x106 1000 X 812
4 = 81 mm
=
x 30
0.015
=
0.018 (Section 13, Table 13.1)
=
O.OlBbdfCk
bdfCk
A$
fyk
0.018
=
x 1000 x 81 x 30 460
= 95mm2
Check minimum area of reinforcement A5
=
=
0.6
0.6
(r')
Eqn 5.14
0.0015bd
x 1000 x 81
=
460 0.0015
5.4.2.1.1(1)
x 1000 x 81
106mm2
= 122 mm2
Check minimum area of reinforcement for crack control =
4.4.2.2
Eqn 4.78
kckfcteffAciOs
k
=
k
= 0.8forh
Ct5ff
= 3.0 N/mm2
A
= bh = 1000
cr
= 460 N/mm2
0.4 for bending
2
300 mm
x 105 2
= 52.5
x i0 mm2
Therefore
A
= 0.4
x 0.8 x 3.0 x 52.5 x 10/460
=
110 mm2
bPEt'kM.DETAiLS
No further checkfor crack control is necessary as h
= 3h = 315
Maximum bar spacing
=
105
200 mm.
500 mm
4.4.23.(1)
NAD Table 3
Use T8 @ 300 mm crs. (168 mm2/m)
5.43.2.1(4)
The reinforcement details are shown in Figure &5.
2T8 (acer bars
to anchor U—bars
Figure 8.5 Nib reinforcement details Check shear in nib, takingintoaccountthe proximity of the concentrated load to the support. VAd1
=
[!3rAdk(1.2
=
2.5d/x
+
___
= 2.5 x
81
=
120
=
034 N/mm2
k
=
1.6—dlzl
p
=
A —-
TAd
+ o.i&i] bd
'40p1)
=
bd
=
1.52
168
=
1000
x 81
4.3.23
Eqn 4.18
1.69
Eqn 4.17
0.0021
N C
A
Therefore VAd1
= 1.69 x 034 x 1.52(1.2 +40 x 0.0021)x 1000 x 81 = 90.8 kN/m
VSd
= 25 kN/m
Therefore VRdl
>
OK
1"Sd
Check anchorage of T8 bars.
net = aa 1b
(j
1tmin
Eqn 5.4
SPECIAL DETAILS
Now
= (I'I4) (d'bd) = 400 N/mm2
Eqn 53
Bond conditions may be considered good as the bars are anchored at least 300 mm from the top of the member. =
3 N/mm2
Table 5.3
= (8/4) x (400/3) = 267 mm NowA5req = 122 mm2
Apr
= 168 mm2
aa
=
1
Therefore
=
1
b.net
122 1x267x—
= 194mm
168
4l
mm
For bars in tension
=
1bmin
10 or 100 mm
Therefore =
1b.min
1b.net
=
100 mm 194 mm (see Figure 8.5)
For bars in compression = 1b.min °•61b 104; or 100 mm Therefore 1mmn
=
160 mm
= 160 mm (see Figure 8.5)
8.3 Simply supported ends 8.3.1 Directly supported ends Reinforcement anchorage requirements are shown
f
Iw(b)
tb,net
(a) Straight bar
5.2.2.1(2)(b)
Ib)
Figure 8.6 Anchorage at a direct support
in Figure 8.6.
'2
Hook
I
"I
tb,net
SPECM DETMLS
Figure 8.6(a) shows anchorage of tensile reinforcement beingachieved using
a straight bar. It should be noted that EC2 does not permit straight anchorage or bends for smooth bars of morethan 8 mm diameter.
5.4.2.1.4(3) Figure 5.12(a)
5.23.2(2)
Note:
TheCEB—FIP Model Code16 gives a factor of 1.0 for net as opposedto 2/3 in EC2. Designers may wish to consider using the higther value. Typical values for anchorage length and support width, w, can be obtained for (a) and (b) in Figure 8.6. Assume
I
ck
=
30 N/mm2,
=
1.0
f yk
=
460 N/mm2
Note:
NAD 6.5
Areq maybe taken as one quarter of the reinforcement required at mid-span but not less than that required to resistthe tensile force givenby EC2 Eqn 5.15.
5.4.2.1.4(3) 5.4.2.1.4(2)
'bnet
=
()
0alb
Eqn 5.4
iz lmIfl
s.prov
=
(I4 (ci1bd)
Eqn 53
=
3 N/mm2
Table 53
=
400 N/mm2
=
(cb/4)
Therefore 1b
4min
=
x (400/3)
°31b
aa
=
1
aa
=
0.7
= 3334
0 or 100 mm
lz 1
for straight bars; or for curved bars with 3c/ transversecover
Therefore
x 333
1b,net (a)=
1
net (b)=
0.7
x
=
33
3334.
=
23.3q5
Therefore width of support required in Figure 8.6(a), assuming 20 mm cover and 15 mm chamfer
w (a)
=
(j-) x
333çb
+ 35
=
22.2 + 35 mm
and width of support required in Figure 8.6(b), assuming as above
w (b)
=
Eqn 5.5
(+) x
23.3q
+ 35 = 15.5 + 35 mm
5.2.3.4.1(1)
SPECIAL DETAILS
The minimum supportwidth is given by W
= (+) x
10 + 35 = 6.7'qS + 35 mm
where, in Figure 8.6(a), A8req arid,
in Figure
8.6(b), As req
O•3As,pr,
0.43A
= 0.7 can only be used if the concrete cover As noted above, to the This is clearly difficult perpendicular plane of curvature is at least to achieve in beams without end diaphrams for bar sizes in excess of 12 mm.
a
3
The requirements for the various types of hooks, loops and bends are given in EC2 Figure 5.2. The minimum diameters of mandrels are given in NAD' Table a The required support widths are given in Table 8.1. Table 8.i Width of support (mm)
A A
10
12
16
20
25
w(a) wmn
257 102
302
391
479
590
746
116
143
169
203
250
wb)
190
221
283
345
102
116
143
169
423 203
531
wmm
(mm)
1.0
0.3 1.0
0.43
32
250
8.3.2 Indirectly supported ends Reinforcement anchorage requirements are shown
in Figure a7.
{I tb,net
,net
L
J b(b)
I
(b) Hook
(a) Straight bar Figure 8.7 Anchorage at an indirect support
As in Section 83.1 above, anchorage lengths and support widths can be obtained for both straight bars and hooked bars.
Theanchorage lengths are as Section 8.31 but the required supportwidths are increased.
Assuming 20 mm cover
= (333cb + b(b) = (23.3q + b(a)
x 1.5 = 20) x 1.5 = 20)
50q5
+ 30 mm
354
+ 30 mm
SPECL DETMLS The minimum supportbeam width is given by = (lOqS + 20) x 1.5 = lScb + 30 mm bmin where the same conditions apply as in Section 83.1.
3
In these cases, as the beam is indirectly supported, i.e., by another beam, cover perpendicular to the plane of the curvature can be achieved moreeasily = 0.7 can be readily used in EC2 Eqn 5.4. and The required supportbeam widths are given in Table 8.2. Table 8.2 Width of support beam (mm)
'
,
AspraY
mm'/
10
12
16
20
25
32
1.0
b(a)
530 180
630 210
830 270
1030 330
1280 405
1630 510
380 180
450 210
590 270
730 330
905 405
1150 510
0.3
b
mm
1.0
b(b)
0.43
bmm
8.4 Surface reinforcement
5.4.2.4
In certain circumstancesitmaybe necessaryto provide surface reinforcement located outside the links. Surface reinforcementis provided to resistspalling fromfire and where bundled bars or bar sizes greaterthan 32 mm are used.
5.4.2.4(3)
EC2 also refers to the use of skin reinforcement located inside the links. Skin reinforcement is provided to control cracking in the side facesof beams 1 m or more in depth.
4.4.2.3(4)
8.4.1 Design data A beam section requiring surface reinforcement is shown in Figure 8.8.
ioooL
fll 400
I
A5t=4825mm2 (6132) to Links 50mm
Cover
Figure 8.8 Beam section showing main reinforcement
=1
SPECIAL DETAILS
8.4.2 Area of longitudinal surface reinforcement
Af
=
0.01A
5.4.2.4(5)
From EC2 Figure 5.15
= 2x50x(1000—360)+50 x300 = 79x 103mm2 Therefore
A ssuri = 0.01 Length of
x
79
x iO
=
790 mm2
= 490
internal perimeter
x 2 + 300
= 1280 mm
Hence
AS,SU1/m=
790 = 1.280
617 mm2/m
Use B785 fabric This comprises 10 mm wires @ 100 mm crs. horizontally and 8 mm wires @ 200 mm crs. vertically. Note: EC2 does not directly coverthe use of plain wire fabric.
5.23.43
Surface reinforcementmayalso be used as longitudinal bending reinforcement in the horizontal direction and as shear reinforcement in the vertical direction in some cases.
5.4.2.4(6)
If surface reinforcement is being used to resist shear, EC2 Clause 5.4.2.2(4) should be noted. Itstatesthat a minimum of50% of shear reinforcementshould be in the form of links.
5.4.2.2(4)
The reinforcement detail is shown in Figure 8.9.
r
—
1
—
1
Ifl S 8-132
—
B785 surface reinforcement
5L150
n -l r —.
_'
f
d-x=54O
I
i(
.1.
I.
-'1.-I
Figure 8.9 Beam section showing surface reinforcement.
C35145 (x/d).urn =
o
— 0.56
1.25
and ILIim for rectangular sectionsas a function
Equationscan be derived for
of (X1c01 . These are =
0•4533(X/d)im[1
—
O.4(XId)i]
= (X/d).Il .918
Wi.
Table 13.2 gives values of (XId)Jim' 'hum and WIim as a function of the amount of re-distributioncarriedout. EC2 states that plastic design, for exampleyield line analysis,canbe used where xld 0.25.Thelimits corresponding tothisvalue are also included in the table. Table 13.2 Limiting values %
(xId)
ô
redistribution
0 5
1.00
0.95 0.90 0.85 0.80 0.75 0.70
10 15 20 25 30
1im
k>35
k35
0.448 0.408 0.368
0.352
0328
0232
0.1667 0.1548 0.1423 0.1292
0.288 0.248 0.208
0.192 0.152
0.1155
0.112
0.0864
Plastic design
0.312
0.272
0.25
1013
0i,m
>5
Ck>35
137l
0.2336
0.1238
0.2127
0.1835 0.1627
0.1919
0.1418
aio
00954 00803 0.0647 0.0485
0.1020
0.1710
0.1210
0.1502 0.1293 0.1084
o.iooi 00792 0.0584
0.1303
13.3 Compression reinforcement Compression reinforcement is required amount can be calculated from C.,
=
in any section where'h >
'h'hIIm 0.87(1 —
d'/d)
where mechanical ratio of compression steel
=x=
bd
'ck
InnI
1iim The
2.5.3.5.5
DESIGN OP BEAM AND COLUMN SECTiONS
d'
= depth from compression face to centroidof compression reinforcement
A'8
= area of compression reinforcement
The area of tension reinforcement can now be obtained from W
Equations above for cd and
c are valid for d'Ix
1 —
1)805.
13.4 Flanged beams For beams with flanges on the compression side of the section, the formulae for rectangular sections may be applied provided
xld
hf/d
where hf
= thickness of the flange
For beams where the neutral axis lies below the flange, it will normally be sufficientlyaccurate to assume that the centre of compression is located at middepth of the flange. Thus, for singly reinforced beams, approximately
M
=
0.87
ASck(d
—
hI2)
The neutral axis depth is given approximately by
xld =
1.918 (blb)w
—
1:25 (b/b
—
1)h/d
where br is the rib width and the definition of o, is identical rectangular section.
to that for a
13.5 Symmetrically reinforced rectangular columns Figures 13.2(a) to (e) give non-dimensional design charts for symmetrically reinforcedcolumns where the reinforcementcanbeassumedto beconcentrated in the corners. The broken lines give values of '
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