SSRay - Reinforced Concrete
SSRay - Reinforced Concrete
SSRay - Reinforced Concrete
Ray
Reinforced Concrete
ANALYSIS AND DESIGN
...:
:.:
available
Blackwell
Science
[Best
coEJ
________________________
.
of reinfcrCedconcreteelements in
foundations and superstructures in a
logical, step-by-step fashion. The theory of
reinforced concrete and the derivation of
in a design office
Is
I'
REINFORCED CONCRETE
Analysis and Design
S.S. RAY
BE (Cal), CEng, FICE, MBGS
b
Blackwell
Science
is
document
contains
z pagJ
1995 by
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Library of Congress
Cataloging in PublicationData
Ray, S.S.
design/S.S.
Ray.
cm.
Includesbibliographical referencesand
index.
ISBN 0-632-03724-5
1. Reinforced concrete construction.
p.
I. Title.
TA683.R334 1994
624.1'8341 dc2O
94-13306
CIP
'V
Page blank
in original
Contents
xiii
Preface
References
1.6
1.7
1.8
1.9
1.10
Notation
Introduction
Characteristicstrength of materials
Material factors
Material stressstrain relationship
Design formulaefor reinforcedconcrete sections
1.5.1 Singly reinforcedrectangularsection
1.5.2 The concept of balanced design and redistribution
of moments
1.5.3 Doubly reinforcedrectangularsection
1.5.4 Singly reinforcedflanged beams
Ultimate limit state shear
Serviceability limit state crackwidth
Serviceability limit state deflection
Ultimate limit state torsion
xv
1
1
2
3
3
4
6
6
7
8
9
11
17
18
18
1.11
19
31
1.12
1.13
32
34
2.5
Notation
Analysis of beams
Load combinations
Step-by-step design procedurefor beams
Workedexamples
Example2.1: Simply supported rectangularbeam
Example2.2: Three span continuousbeam
Example2.3: Design of beam with torsion
Figures for Chapter 2
Figure 2.1: Values of f3
Figure2.2: Simplified detailingrules for beams
41
41
43
47
50
65
65
73
85
99
99
100
VII
viii Contents
Notation
Analysis of slabs
Load combinations
Step-by-step design procedurefor slabs
Worked example
Example3.1: Designof a two-way slab panel
Figures and Tables for Chapter 3
Figures 3.1 and 3.2: Elastic and elasto-plasticunit resistances
101
101
103
107
107
120
120
130
151
151
152
154
155
155
161
164
164
164
169
and tension
176
and tension
183
193
193
193
194
197
200
Contents ix
5.5
6.6
6.9
212
212
213
213
213
215
215
218
218
219
222
223
223
223
224
224
225
226
226
226
229
231
232
Sign convention
Essentials of soil mechanics
6.5.1 Ultimate bearingcapacity
6.5.2 Settlementof foundation
6.5.3 Sliding resistance
Bearingpressure calculations
6.6.1 Rectangular Pad uniaxial bending no loss of
contact
232
6.6.2 Rectangular Pad uniaxial bending loss of contact 234
6.6.3 RectangularPad biaxial bending no lossofcontact 234
6.6.4 RectangularPad biaxial bending loss ofcontact
235
6.6.5 Multiplecolumn biaxial bending no lossof contact 238
6.6.6 Circularpad biaxial bending no loss of contact
238
for
239
Step-by-step designprocedure
pads
250
Worked examples
250
6.1:
RC
with
column
Example
pad
single
columns
264
6.2:
RC
with
multiple
Example
pad
side
in
cohesive
6.3:
Mass
concrete
bearing
Example
pad
277
soils
side
6.4:
Mass
concrete
in
Example
pad
bearing
cohesionless soils
283
for
6
289
Figures
Chapter
6.1:
Values
of
and
289
N,
Figure
Nq
N
6.2:
Calculation
of
mean
vertical
stresses
in
soil
290
Figure
6.3:
Plan
on
base
different
zones
291
Figure
showing
biaxial
6.4:
Pressures
under
base
Figure
rectangular
292
bending
6.7
6.8
200
208
x Contents
293
293
355
355
357
8.1
8.2
8.3
Analysis of walls
8.1.1 Walls and propertiesof walls
8.1.2 Modelling for structural analysis
Step-by-stepdesign procedurefor walls
Worked example
Example8.1: Reinforcedconcrete cell
Definitions
296
299
301
302
302
306
309
310
326
326
354
354
354
357
368
370
385
385
403
403
403
404
Analysisof flat slabs
of
flat
slabs
406
Design
412
for
flat
slabs
Step-by-stepdesign procedure
413
Worked example
413
9.1:
Flat
slab
construction
for
a
hall
Example
sports
435
Tables and Graphs for Chapter 9
Tables 9.1 to 9.6: Bending moment coefficients for design
of flat slabs
43843
Table 9.7: Bending moment coefficient for design of columns
in flatslab construction
444
9.1
to
9.18:
Correction
factors
for
moments
Graphs
bending
in flat slabs
44553
9.19
to
9.26:
Correction
factors
for
moments
Graphs
bending
in columns
4547
459
459
459
460
460
Contents
10.4
10.5
v forf
v
v
xi
461
463
464
465
467
467
468
469
470
470
487
488
489
490
491
492
493
494
495
flanged beams
496
Chart 11.4: Modification factor for compression reinforcement
496
Chart 11.5: Modification factor for tension reinforcement
Table 11.6: Nominal cover to all reinforcementincluding links to
497
meet durability requirements
Table 11.7: Nominal cover to all reinforcement including links to
meet specified periodsof fire resistance
497
Tables 11.8 to 11.17: Design tables for rectangularcolumns
498517
Tables 11.18 to 11.27: Design tables for circularcolumns
51837
Index
538
Page blank
in original
Preface
I believe that the contentsof this book will prove to be extremely valuable
concrete design. There are many excellent books available dealingwith the
design of reinforced concrete elements but, in my opinion, they lack
completeness in certain ways. The design of a reinforced concrete
member requires many checks in a systematic structured manner and the
step-by-stepapproach adopted in this book is intended to ensure that the
design process is complete in all respects. It is my view that the member
itself, when fully designed,does not constitute a complete design because
it ignores the connections to other members and to the foundationthat are
needed to provide true completeness of design for the structure. I have
attempted here to elucidate the necessary global analysis. Also, most
books on reinforced concrete design do not deal with the aspects of soil
structure interaction problems and are hence incomplete.
The highly structured step-by-step methodologyI have used makes the
book fully comprehensive and user-friendly. Accordingly, the task ofquality
assurance becomes less arduous and the product or output of a design
office becomes fully standardised if this approach is strictly followed. For
students, the book should prove to be invaluable because the essential
elementsof the theory of reinforcedconcrete are discussed, followed by a
structured approach to the design of all elements in a building, including
foundations and the connections of the reinforced concrete members to
each other to create a complete building. The numerousworked examples
should be very useful to students and practitionersalike. The book also
presents practical advice on designing reinforced concrete elements and
the student should benefit from learning the methods adopted in a design
consultancy.
XIII
xiv
Preface
The opinions expressed in this book are those of the author and the
correctness or otherwiseof the text is the author's responsibility. Taylor
Woodrow and its Group of Companies are in no way associatedwith the
productionofthis book and they have not adoptedthe book or any part ofit
as theirin-house standard.
References
1.
3.
4.
American Concrete Institute (1983) Building code requirements for reinforced concrete. M83. ACI, Detroit, Michigan. USA, ACI 318.
5.
6.
7.
Tomlinson. M.J. (1987) Pile Design and Construction Practice, 3rd edn. E.
9.
Moody, W.T. Moments and Reactions for Rectangular Plates. US Department of the Interior, EngineeringMonograph No. 27, Denver, Colorado,
USA.
14. British Standards Institution (1972) Wind loads. CP3: Chapter V: Part 2.
BSI, London.
xv
xvi
References
General references
Allen, A.H. (1983) Reinforced Concrete Design to BS8IIO Simply
Explained. E. & F.N. Spon, London.
Batchelor & Beeby (1983) Charts for the design of circular columns to
BS81IO. British Cement Association, Slough, UK.
British Standards Institution (1987) Design of concrete structures to retain
aqueous liquids. BSI, London, BS8007.
Park, R. & Paulay, T. (1975) Reinforced Concrete Structures. John Wiley
& Sons, New York.
Pucher, A. (1977) Influence Surfaces of Elastic Plates. Springer Verlag,
Vienna, Austria.
Roark, R.J. & Young, W.C. (1975) Formulaefor Stress and Strain, 5th
edn. McGraw-Hill International, Tokyo.
Chapter
1.0 NOTATION
A.
b
C0
C
d
d'
E
E
fk
frn
f,
ff
h
hf
h,
hmax
hmin
M'
M
N
p
p
q
Q
of elasticity of concrete
in steel
2 Reinforced Concrete
rb
s
S
T
v
v,
v
V
V
x
y
13a
13b
1.1
of a member in bending
Standard deviation
Spacing of shear reinforcement
Internal tensile force in steel reinforcement
Shear stress in concrete (N/mm2)
Design concrete shear stress (N/mm2)
Shear stress in concrete due to torsion (N/mm2)
Shear force in concrete section
Design concrete shear capacity
Design shear capacity of shear reinforcement
Depth of neutral axis from compressionface
Distance from neutral axis
Depth of lever arm
Curvature
INTRODUCTION
Safety
Durability is taken care of by the choice of the right material for the
purpose and also by bearing in mind duringthe design process,the requirements for proper maintenance.
Theory of ReinforcedConcrete 3
Good valueformoneyisperhapsthe most importantcriterion.The designer
should take into account not only the cost of materials but also the
buildability. the time required to build, the cost of temporary structures,
the costofmaintenanceover a period of time and in some casesthe cost of
demolition/decommissioning.
External appearance of structures changes over a period of time. The
designer should be aware of the effects of cracking, leaking, staining,
spalling, flaking, etc. of the materials in use. The designer should make
appropriate allowances to avoid the degradation of appearance.
User comforts are influenced by the vibrationof the structure due to wind,
road/rail traffic or vibratingmachinery. Large deflections under load also
cause alarm to the users. The designer should pay adequate attention to
alleviation of these anticipated discomforts.
Robustness comes with the chosen structural form and is determined by
the additional inherent strength of the structure as a whole to withstand
accidental loadings.Collapse of one key member in the structure must not
initiate global collapse.The design must foresee the dominoeffect' in the
structure and avoid it by careful planning.
1.2 CHARACTERISTIC STRENGTH OF MATERIALS
following formula
fk =
1.64s
f,
MATERIAL FACTORS
To obtain the design strength of materials a further factor called the
material factor 'ye, is applied. The material factor takes into account the
tolerances associated with the geometry, the variability of materials on
ReinforcedConcrete
site, the inconsistency in the manufactureand curingon site and the effects
of long-term degradation.
The values of Ym for theultimate limit state are as follows:
reinforcement
concrete in fiexure
1.15
concrete in shear
bond strength in concrete
bearing stress
1.25
1.40
1.50
0.57S /8m
uJ
u_J
U) U)
(.J_)
u_i
U)
I-_i
0.0035
STRAIN
E = 5.5('kN/mm2
\Im1
N/mm2
Theory
of ReinforcedConcrete 5
= 2.4 x
Note:
10_4)
/8m
yield.
The
but for design purposes it
yield stress of steel reinforcement is
will be taken as f/ym.
The stress after yield remainsconstant and is representedby a constant
stress line.
= 0.87
= 2.0 x i03
forf = 460N/mm2
ReinforcedConcrete
SECTION
STRESS
STRAIN
= O.4O2fbx
()fbo.9x
(LY)A.
= O,87fA.
where
C= T
or
O.4O2fbx = O.87fA
or
x = 2.164k
fb
z = d O.45x =
or
or
/
d1
\fbd
/fA.
\fbd
=10.971-
d
(
z\(fbd
A=i1ii
\ d/\0.97f
M = O.87fAz =
=
M
fbd2
o.9o(1
()(i
=K=0.901-\ dJ\d
)fbdz
Theory
or
[0.5
of ReinforcedConcrete 7
+ (0.25 -
Maximum moment of resistant of concrete section is obtainedfor redistribution not exceeding 10%, when x = d12.
z = d 0.45x
or
0.775d
0.402fb()(0.775d)
= 0.156fbd2
Where redistribution exceeds 10%,
x
(I3b
0.4)d
Similarly,
M'
or
= 0.402fbxz
= 0.402fCUb(13b 0.4)d[d O.45(Pb 0.4)dj
= EO.4O2(I3h 0.4) 0. 1(I3b 0.4)2jfbd2
K' = O.4O2(h
where
ib =
0.4) O.lS(3b
0.4)2
< 0.9
COMPRESSION
FAILURE
COMPARATIVE
STRAIN DIAGRAM
ReinforcedConcrete
From strain diagram,
______
(d x)
0.0035
0.002
or
x = 0.636d
The Code does not allow x to be larger than 0.5d ensuring that the steel
reachesits yield strain before the concretereaches the ultimatestrain. This
is designed to allow sufficient rotational capacity in the section.
The more redistribution of moment is allowed, the more rotational
capacity is needed from the section. The amount of rotation is dependent
on how under-reinforcedthe section is, or in other words, how quickly the
steel in the section reachesthe yield strain before the concrete reaches the
ultimate strain. To make sure that the rotational capacity exists in the
section to allow redistribution, the depth of neutral axis for the design is
SECTION
STRAIN
STRESS
=
=
0.9x(")b +
(K'fcubd2)
+ 0.87fA
Theory of ReinforcedConcrete 9
Equating C = T,
A=l/\K'fbd2\
J+A'
0.87fz /
= (K K')fbd2
d')
In the above formula it is assumed that the compressive steel will attain
yield. This is true provided d' is less than or equal to 0.43x or the strain in
the steel is at least 0.002 for f=460N/mm2. If dlx is greater than 0.43x,
the steel stress f. will be proportionately modified to account for the
reduced strain. Use in the equation for A instead of 0.87f.
0.87f(d
A' = (K
K') fbd2
f(d d')
where
j = \0.57x/
(x__d' E E
fy
=
'(mE's
10 ReinforcedConcrete
/8m
SK 1/7 Stressdiagramfor a
flanged beam section.
C1
(0.67Ju)(b
bW)hf
= 0.45f(b bW)hf
C2
= 0.87fA.
(L)A
M' =
C1d
hf\ +
C2(d 0.45x)
= 0.45f (b
/
bW)hfd
h\/
= fbd2 1/
hf\
b\/
+ 0.201fbd(d 0.225d)
hf\ +
0.157
= 1ffcu
Values of
(Chapter 2).
for different ratios of b/b and dlhf are found in Fig. 2.1
Td
hf\
hf
C20.45x
/ hf\
0.87fAd
M + 0.1fbd(0.45d h)
0.87f(d 0.Shf)
0.1fbd(0.45d
hf)
Theory of ReinforcedConcrete
11
C=
(0.67
hf
lever arni = d
2
Mf =
hf
0.45fcubhf(\d
This is the flange resistance and if the applied moment exceedsthis value,
then the web comes into compression.
The moment to be carried by the web is M, when M is the applied
moment.
= M 0.45f(b
=M-Mf (b
=M
Find Mf, and
Design for
M(1
hf
bW)hfd
b .)
h. Find A1
A1 =
0.45f(b
b)hf
0.87f
= M
0.87fz
when
and z
Note:
1.6
K=
Jcu, ,j2
dLO.5
+/
K\
(\0.25
The design against M may follow Section 1.5.3, which means that the
flanged section may be doubly reinforced, if required.
12
ReinforcedConcrete
where
Shear flow, q = VQ
5TRE55
SHEAR FLOW
T4
1 /dT
= I/
II
M
=
or
b/ \dx
dM
=dT
z
v=
dx
bz
For convenience in the ultimate limit state the Code shear stress index
is
taken as:
V
v=
bd
Effective shear in haunched beams
Veff
= V Csine'
= V C'tanO'
=VM
tanO'
z
VE
A
TENSILE
STEEL
14
ReinforcedConcrete
COLUMN
CONTINUOUS BEAM
1.6.1
Shearreinforcement
V.
truss analogy
Cot
Co
1- p
CRACK
C
d
rUPPORT
SK 1/13 Truss analogyof shear
reinforcement.
FORCE DIAGRAM
From geometry:
S = z(cot
+ cot3)
T'. =
V=
sinc
S
T' =
or
or
V0
z(cot + cot3)
sin c(cotc +
O.8?f,Asb
or
_______
cot13)
V.,
__________________
Zsin c(cot + cot 13)
()
0.87f5A5()
V0
0.87f5A5
==
bz
bS
16 Reinforced Concrete
(v v)
or
Note:
bS(v v)
O.87f
Bond forces are sustained along the length of the beam where shear
reinforcementis
The lever arm zrequired.
is assumed constant over the section with variable
moment producing the shear to be resisted. The diagonalcompressive
stress in concrete struts calculated from the analogy is equal to vI
[sin213(cot + cot 3)1 is sustainable.
T == T + T
M2 + M2
T' is the tensile force in the shear reinforcement.
M2
M1 = Vx = M2 + Vzcofl3 = Tz
or
Note:
T=
M2
z
+ Vcot3
M1 = Vx = M2 +
or
Vzcot = Tz +
M2
(S Tsinc
T=+
Vcot ;
\2Z/
Z
() T'sinc
T'sina = Y
and
S = z(cot + cot3)
we get:
T=
()(cot1
cotc)
cot)
This demonstratesquite clearly that when diagonalcracks form in concrete due to shear exceeding V, additional tensile steel will be required
over and above MIz.
This requirement is not explicitly covered in the Code. The rules of
curtailmentof reinforcementare deemed to satisfy this requirement.
Note:
checked.
The basic assumptions to find crack width for flexure are summarised
below:
to reinforcement.
18
1.8
Reinforced Concrete
SERVICEABILITYLIMIT STATE DEFLECTION
Calculatemomentsat service load without redistribution.
SECTION
________
STRAIN DIAGRAM
________________
section.
f.
Find the depth of neutral axis, x, and find the stresses in concrete,
the stress in steel, by following Step 25 of worked example in
Chapter 2. This method ignores the concrete under the neutral axis.
f.
and
Curvature =
= ft =
Tb
xE
ft
(d x)E.
= K1(
\
hmax hm,n
TORSION
Jhmin/2
hp
o>cJ]hmin/2
SK 1/15 Membrane analogy for
torsion.
Theory of ReinforcedConcrete 19
Applying this analogyto a rectangularsection gives a pyramidaldeflected
membrane.
(hminH'\(
2
hmin(hmax
max
hmin)H
'1mm
By membrane theory it is known that the torsional shear stress is the slope
of the angle of the deflected membrane.
tan0=
H
(hmin/2)
T= 2 x
Substituting
volume
=v
of pyramid =
)m
hmin
hmin
3
2T
v=
h2
mint
1.10
H= Vt(hmmn/2),
(Vthnin\(
or
/
hminHhmax
hmin
max
LOAD
SK 1/16Typical loadstraincurve
of a column.
1.10.1
20 ReinforcedConcrete
yield strain. Hence, the steel reaches its ultimate load-carrying capacity
long before concrete gets there.
The ultimate load-carrying capacity of a short reinforced column may be
written as,
=
where
(O.o7fcu)
(-)A
= O.45fA + O.87fA
A = net concrete cross-sectional area
= area of compressivesteel reinforcement.
N = O.4fA + O.75fA
for a column with nominal eccentricityof load meaning a column with
no design moments and eccentricloads. The eccentricity is allowedfor the
constructionaltolerances.
N = O.35fA + O.67fA
for a column supporting an approximately symmetrical arrangement of
beams. The spans of the beams on either side of the column should not
differ by more than 15%. To allow for a certain eccentricity of loadingdue
to the variations in spans and the location and disposition of live loadings
on spans, the equation has been modified.
1.10.2 Axial load capacity of slender columns
The Code uses the 'Moment Magnifier Method', whereby the effect of
slenderness is transferred into an equivalent deflectionand an additional
momentgiven by the product of this deflection and the applied direct load.
Madd
where
= Na
a=
I3a1(hZ
K(NUZ-N
NbaI
Theory
Nba! = 0.25
fbd
of ReinforcedConcrete
21
To use the charts to find the total area of steel required, the following
NIbh, MIbh2 and the d/h ratio. (See
parameters are required:
Chapter 4 for further explanation.)
f,
1.10.4
Bending Moment
Equivalent System
e = M/N
22 Reinforced Concrete
moment.
Column
Section
Ultimate
Strain
Diagram
Stress
Diagram
Idealised
Stress
Diagram
Balancedfailure
A balancedfailure occurs when the tension steel just reaches yield at the
same time as the extreme compression fibrein concrete reachesthe ultimate
strain of 0.0035.
fy/EsYm
(dxb)
xb
assuming
= 460N/mm2
= 1.15 for steel
Ym
= 200kN/mm2
Theory of ReinforcedConcrete 23
Csf
Strain Diagram
=0.002
..Cc00035
SK 1/19 Strain and force diagram
at balanced failure.
44
Ics
At Balanced
Failure.
Force Diagram.
$N
(dxb)
Xb
OF
0.002
X = 0.636d
Nb
(O.o7fcu)
Yrn
b A,(0.87f)
N = 0.256fbd
The strain in compression steel is governed by the value of d'.
The yield strain in compression steel is fy/EsYm.
E,=200kN/mm2 and Ym= 1.15.
For Grade 460N/mm2 steel, this yield strain =0.002.
At balanced failure' condition:
0.0035
Xb
Assuming
(Xb
d')
d' = 0.273d
or
24 ReinforcedConcrete
= 0.4O2fCUxbb(0.5h
Mb
where
+ Afh(k 0.5)
f. = E.
and r., =
O.4SXb)
+ 0.87Afh(k 0.5)
O.OO3S(xb
d')
Xb
for k0.78
Mb
bh2
= 0.256fk(0.5
0.286k) + 4p(k
for k<0.78
bh2
0.256fk (0.5
+ llp(1.636k 1) (k
Note:
0.5)
0.5)
x).
CcOOO35
mEs
comessureI
(x>
Tensile Steel Does Not Reach Yield
Xb).
When N<Nb and f=f/ym, a tension failure condition will apply. The
column behaves more like a beam in this condition.
N= C + C T
Assuming symmetrical reinforcementand yield strain
compression steel,
N = C = O.402fbx
N
x=
or
0.4O2fb
Check that
= 0.0035(x d')
or
2.331h(1
0002
k)
1N\)
x=
2.4875(
x
h
for
\fbh/
= 0.5[(B2 + 4C) B1
2.331(1
for
k)
< 2.331(1 k)
B = 7.463(-- 2.4875(--
where
\fbh
C =-
17.413(1
k)()
2.331(1k)
M = 0.402fxb(0.5h 0.45x) + 0.87Afh(k
+ 0.87Ajh(k 0.5)
for
<2.33(1
k)
+ Afh(k
for
2.331(1
M
fbh2
0.5)
0.5)
0.5)
k)
0.402((0.5
hi
8p(k
0.45(
\h// +
0.5)
26
Reinforced Concrete
for
<2.331(1 k)
fbh2
llp(k
0.45(1 +
0.5)
h/i
0.402()[0.5
h
7p[(l k)/(x/h)](k
0.5)
Compressionfailure
When N> Nb, a compression failure condition applies.
N = C, + C. T
Assuming symmetrical reinforcement and yield strain in both tension and
compression steel,
N = C = 0.402fbx
or
0.402fbh
IN
2.4875k
fbh
For the tensile steel to be at yield,
0.002
for = 460N/mm2
or
0.0035 (d
or
0.636k
x)
0.002
2.331(1
k)
for 0.636k
<2.331(1 k),
= 0.5[(B2 + 4C) B]
where B =
C=
7.463(
I 2.48751
17.413(1
IN
\fbh
for 0.636k
<
2.331(1
- B]
= 0.5[(B2 + 4C)
B = 27.363("
where
2.4875(--\f11bh
C = 17.413k(1?
\f
for 0.636k
<
<2.331(1 k),
- B]
= 0.5[(B + 4C)
B=
where
fI
34.826(-_"
fbh
2.4875(__
C = l7.413(
Taking moment about the centroid of section.
For both tension and compression steel going into yield,
M = 0.402fxb(0.5h 0.45x) + 0.87A,fh(k
+ 0.87Afh(k
0.5)
0.5)
+ Afh(k 0.5)
0.5) + Afh(k
k)
M
fbh2
for 0.636k
M
fbh2
0.402()[0.5
h
0.45(f)]
h
+ 8p(k
0.5)
<2.331(1 k)
/x\1
= 0.4021/x\1
II 0.5 0.451 II + Fllp(k 0.5)
\h/i
\h!L
f7p[(1
k)/(x/h)1 (k 0.5)
fcu
0.5)
28
ReinforcedConcrete
for 0.636k
fbh2
2.331(1
0.402(10.5
h/L
1('
L
fbh2
Note:
k)
for 0.636k
<<
2.331
0.45(i)]
h
(x/h)f0.5)]
0.5)
(1
= 0.402(10.5
\hJL
k).
7p(2k
0.45(1 +
h/J
1)(k 0.5)
(x/h)f
k = d/h
d1Ih = 1k
0.87f = 0.87x460=400N/mm2
x)/x=700 (d1 x)/xsI400IN/mm2
fi == Eci=200x103x0.0035x(di
700 (d x)/x 14001N/mm2
A == area of concrete in compression=0.9bx = 0.9bh(x/h)
N (0.67fcuIYm)Ac (AI2)(f1 f)
0.402f()
()(fi + f)
Ne = /O.67j\
O.45x)
(1
0.402f()bh2(0.5 0.45(i))
k)h)(f.
bh2()(k
f.)
0.5)(f
fbi)
+
= 0.402f()(0.5 o.45())
0.402f()
x
\h
x
h
or
fi
= 700(1
f=
- ()(f + f)
70(11
fi =
700(k
k
400 N/mm2
_)
h
4001N/mm2
fJ
SK 1/21 Circularcolumn
diagram.
strain
h5
h=2R
30
Reinforced Concrete
20
from compressionface
at d3.
Note:
N=
- ()(fi +
(O.67fcu)A
= 0.45R2f(0 sinOcosO)
N=
fi =
0.45f(0 sinOcosO)
O.OO35ES(d1
R
= 700(_
or
+ fs3)
()fi
+ fs2 + f3)
14001 N/mm2
_x)
- R-)
fs2
14001N/mm2
Id2
R
R
700\__
fs2 =
14001N/mm2
d3
R
14001N/mm2
kcos30
(_-_)(i
cosO)
= 1 + kcos30
R
= 1.11 (1
cosO)
2sin3O
3(8
sinOcosO)
Taking moment about the centre of section assuming that the applied load
N is always at the centre of section,
Ne = O.45R2f(O sinOcos8)(R 1) +
= O.45R3f(e
O.45f(O sin
sin8cosO)(1
()(kRsin6o0)(fsi
(R1)(k
O)(f
fbi)
fbi)
get eIR.
+
OcosO)(l
O.45f(O sin9cos8)
((ksin6o0)(f3
f)
()u + j2 + f3)
= 1.0866k
zd
RRR
Fora range of valuesof k and p. the above equationscanbe solved for
different values of e/R. Tables 11.18 to 11.27 give N/R2 and z/R for
different values of e/R.
Note: The above equations are valid up
to x = 1.111/i = 2.22R.
Vbd
Strari Diagram
32
Reinforced Concrete
F = T + Fcos3 = T +
F = (O.67fcu)b 0.9xcos
= 0.402fbxcosl3
V= FsinI3
= 0.402fbxcos13sin3
a
cos= (a +
sin1=
V
(a + z2)
= = 0.402fx
bd
za
(a + z2)d
= 0.4021 II xa
\dJ \a +
Substitutingx = (d z)I0.45,
0.893k
(a)2(z)2
From
1.12
the above equation the graphs in Fig. 5.1 have been drawn.
Ay
I
Mxy
My
liii A
Mxy
M*x
M == M + M.
M M. +
If M<O, then M=O
and
M=M+
M= M+
M
Top steel
M = M,
M= M
M.
M%.V
M<=M,
M=M
Skew reinforcement
Bottom steel
M = M + 2M cot ct + Mcot2c +
-
M=---+
sin s
+ Mcotc
sincr
+ Mcotc
sincy
34 ReinforcedConcrete
\/
(M Mcot)2
+
M = ()JM
\sin2 cyl \
+
M 2Mcot x + Mcot2
If M< 0, then M = 0
(M + Mcot)2
and M = M1 + 2M cot + M cot2 +
/
and
Top steel
+ Mcot
___________
M = M + 2Mcotc + Mcot2c Mr
sinc
-
M = Sin
+ Mcot
sinc
/
(M1 Mcotc)2
M= /IIlM
\
\sin2J
M + 2Mcota + Mcot2a
1
1.13
1.13.1
M = M + 2Mcotc + Mcot2
(M + Mcota)2
M
Neutrul
Axis
As
SECTION
SIRESS
______
STRAIN
______
Theory of ReinforcedConcrete 35
T=
tensile force
in bars in tension
= strain in compressive
= strain in tensile steel
steel
m = EIE
or
xd'
x
or
f E/
f
f
=-i
E\dd'
E(
C = O.5bxf
C' =
fA fA((x\
d'
= (in
1)fA
d'
x
Id x
T = fA. = mfA(
\
C(d
or
O.5bxf(d
+ C'(d d') = M
+ (m
1)fA
d')(d
- d') = M
C + C' = T
or O.5bxf + (m 1)fA ____ ,l
\ X /
= mfA(fdx
\
+ 2(rn
bd2
(x)2
or
+ 2[(m
- l)p' + mpj() -
2[(m
- 1)'() + mp] =
+
+
+
{[mp (m 1)p'] 2[mp (m
[mp + (m l)p']
i)()p]}2
36 ReinforcedConcrete
0.5bx(d
x/3)
+ (m
1)A(x d')(d
d')/x
M
k2bd2 +
where k2
k3
k3A(d d')
x
\2d/\
= (m
- 1) (i -
Id
ft = tensile stress in steel = mfI
1.13.2
Neutral
Axis
STRESS
STRAIN
C' + C = N. + T and e =
of Reinforced Concrete
Theory
37
The distance of the centroid of the stressed area from the compressive
face of the rectangular section is called g and may be found by taking
moments of all transformed areas of steel about the compressive face of
the section
g
bx+(m1)A+mA
g=
where
_Nc(e+d_g)+C'(d_d')+C(d_)o
or N(e + d g) + (m
+ O.5bxf(d
or (m
l)fA(x
d
x
)(d
=N.l1(e d g) +1
cL
g) +
k1
= [(e
k2
= (-)( 1 x
3d
\2d/ \
= (m
1)(l
\
+ k2bdf = k1N
k3(1
or
k1N
f=
k3(1
+ k2bd
T = C' + C N
or
Af. = (m
l)fA
d')
= k3fA + O.5bxf N
or
f= f(k3A + A,O.5bx)
+ O.5bxf
d')
38 ReinforcedConcrete
f
or
= m(d
x)
X(f)
mf
The procedure is to assume x and then calculate andf. and then check
x. Repeat this process until convergence is reached.
1.13.3
Neufral
Axis
Using the same symbols as in Section 1.13.1 and A1 is the tensile force,
equating the loads:
M
C'+C+NT
and
(m
=
1)(1
+ g)
- i]
d')'-- + 0.5bx(1
Theory of ReinforcedConcrete 39
k1
+ g)
[(e
k2=(-V1\2d/\ 3d
k3
or
= (m
i)(i
k1N
f=
k3(1
or
+ k2bd
Af, = T = C' + C + N
+
+N
f. = f(k3A O.5bx)
d
(1 +f_
mf
asin Section 1.13.2. Checkassumedvalueofx andrepeat until convergence is
reached.
Page blank
in original
Chapter 2
NOTATION
2.0
a'
acr
A
A,
A,
A,
b
bc
Cmin
d
d'
d1
E
E,
f,
ff
F
Fht
G
h
h1
hmax
I
I
42
Reinforced Concrete
4,
Effective span
M
Md
Mr
N
p
p
r
Sb
S,
Applied torsion
Proportion of total torsion carried by each rectangle of an I-, T- or
T
v
v
v1
v,
Vt,mjn
V
Vb
V
V.
Vmax
Vnom
Wmax
x
x1
Yi
13
13b
Em
Emh
L-section
ANALYSIS
OF BEAMS
_______f3
IL
SK 2/2 Continuousbeam.
I=
Cantilever
I = I+
where
'e = smaller of (1 + d) or 10
1.
1 = centre-to-centredistancebetween supports
1e
effective span
12
SECTION A -
b=
b=
+b
b = s-+
7.14
b = k--- +
14.29
b
b
44 Reinforced Concrete
where
Note:
b=
effective
width of
compressionflange
b = average width of web.
Use actual b if it is less than the calculated b using the above formulae.
A typical example may be a precast T-beam.
see
Table 11.2.
I = 0.5 (Ibh3
+
\12
Fbh3)
where
p = 100 bd
where
p' = 100
where
bd
Design
m = modular ratio =
of ReinforcedConcrete Beams
45
b.
E = 200 kN/mm2
Modulusof elasticity of concrete, E, for short-termand long-term loadings
is given in Table 2.1.
Table 2.1 Modulus of elasticity of concrete: short-term and long-term loading.
(N/mm2)
20
25
30
40
50
60
Short-term loading, E
(kN/mm2)
Long-term loading, E
(kN/mm2)
24
25
12
26
28
30
32
13
14
15
16
12.5
Note: Wind load is short-term loading and dead load is long-term loading.
of rectangular section
46 ReinforcedConcrete
h.
1.5
10
0.14
0.20
0.23
0.26
0.29
0.31
hmin
c=
where
I
16L3
k=
/
\
3.36k1 1
k4
12
hmin
T1max
2.1.6
Shear modulus
Shear modulus, G, is given by
G =
where
Note:
(1
+ t) = 0.42E
for concrete
= Poisson's ratio.
Design
of ReinforcedConcrete Beams 47
Note: The shear areaof concrete is entered as input to some computer programs
when the analysis is required to take into account the deformationsdue to
shear.
The coefficients of thermal expansion are given in Table 2.3 for different
typesof aggregate used.
Coefficient(x 10 1/c)
12
It)
1.0 DL
1.4 DL
1.0 DL
1.2 DL
+ 1.4 EP + 1.4 WP
+ 1.4 WL + 1.4 EP + 1.4 WP
+ 1.4 WL + 1.4 EP + 1.4 WP
+ 1.2 LL + 1.2 WL + 1.2 EP +
1.2 WP
Note: Load combinations LC2 and LC4 should be considered when the effects of
dead load and live load are beneficial.
48 Reinforced Concrete
where
DL
= dead load
WL = wind load
WP = water pressure
EP = earth pressure.
The general principleof load combination is to leave out the loads which
have beneficial effect. If the load is of a permanent nature, like dead load,
earth load or water load, use the partial load factor of 1 for that load
which producesa beneficial ratherthan adverse effect. This rule of combination will be used for design as well as for the check of stability of a
structure.
I1
LC1
LC1
LC1
'
I
LC1 ALTERNATEWITH LC2fQR MAXIMUM
MIDSPAN MOMENT
LC1
I
SK 2/7 Continuousbeam loading
sequences.
i2
I'
LC i
LC,
1'
LRedistributed Moment
moment redistribution.
of
moments may
other-
50 Reinforced Concrete
2.3
M(-ve)
M(.ve)
Vmax
a continuousbeam.
Design
Step
Effctive span
Determine effective span
r
flL.4J
V/f/1'//fA
H.
or
250
b2
1<25b or
100
52 ReinforcedConcrete
where
M
or actual d, whichever is lesser
\O.156fb1
\/
b = effective width of the compressionflange
M = design ultimate moment.
d = !(
SK 2/17 Rectangularbeam
stress diagram.
= M + N (0.5h d1)
for N
0.1fbd
Note: For N> 0.1fbd, design as column (see Chapter 4). Md may also be taken
and N may be totally ignored. (Sign
equal to M where
convention: N is +ve for compression.)
N0.1fbd
K=
z
Md
fbd2
d[0.5
= d
0.95d
/(o.25
0.91]
0.45
A=
Md
0.87fz
K' = 0.156
K' =
O.4O2(13b
I3b =
M2
< 0.9
O.87f
0.4) O.l8(13b
exceeds 10%
M'
0.4)2
when redistribution
M2
tribution
When K>K',
x=
A'
d[0.5
d z
0.45
(K
(o.25
O.5d
K')f11bd2
0.87f (d d')
A = K'fbd2 + A'
0.87fz
N
- ___
0.87f
SK 2/18 Straindiagram.
If d'/x>0.43x,
A'
(K K')fbd2
f(dd')
K'f
bd2
O.87fz
+A.
,/xd'\f
0.57x
fS
0.87f
strain
c = (\ 0.57xd'\
because steel
54 Reinforced Concrete
Note: The flanged beam becomes a rectangular beam
produces tension in the flange.
If MrMd,
+ O.1fbd(O.45d
Md
A
h1)
N
_____
O.87f(d O.Shf)
O.87f
If M< Md, follow Section 1.5.4 using the second approach to design of
flanged beams.
vsO.8 Vf5N/mm2
Change beam geometry if this condition is not satisfied.
Check
SK 2/20 Rectangularbeam.
of support
Design
55
VL 'j
jJx
SK 2/22 Critical section for shear. Shear check
based on bottom reinforcementadequately
anchored.
of support).
0.8
\/f
5N/mm2
bOA.
bd
Find
v,,
v = v + 0.6
Note:
0.8 Vf
()
5N/mm2
v=v \I\/7
/(1-1----
Replace v, by
Av
FindA. = 0.4bS
0.87f
f,,,
460N/mm2
for links.
areaof links,
at a spacing of S for the zone where
shear is less or equal to Vnom. From the shear force envelopedetermine
zones where V exceeds Vnom = (v + O.4)bd
Provide minimum
56 Reinforced Concrete
P*1ff+1
. . (4 .
I.,
a.>.
up shear reinforcement.
= bS (v
Find
v)
0.87f
Replace v, by
Provide
consideration.
For a mixture of links and bent-up bars,
Vb + V.,
(v v)bd
where
Vb = A.,b(O.87fV)(cos
V = A.,(0.87f)
+ sin cot)
d')
and 1
Sb
450
0.5(v v0)bd
load.
Step 14 Alternative design for shear
Find Vmax = 0.8
bd, or = Sbd, whichever is less.
Complete the table below:
\/f
0.75d
0.50d
0.25d
Vnom
= Vc,nc+ 0.4bd
Vconc
< Vmax
(v + 0.4)bd
vbd
(1.143v+0.4)bd
1.143v0bd
(1.333v+0.4)bd
1.333vbd
1.fwbd
2.0vbd
(1.6v0+0.4)bd
(2.0v+0.4)bd
(2.67v+0.4)bd
(4 ++ 0.4)bd
(8v
0.4)bd
2.67vbd
4.0vbd
8.0vbd
Design
Distance from face of
support or concentrated
load
of ReinforcedConcrete Beams 57
V
2.OOd
1.75d
1.50d
1.25d
I .OOd
0.75d
0.50d
0.25d
0.4bS
0.87f
(0.S7f)
V Von
A5
If b/b
A,
0.4
0.0013bh
f=
for f =
for
460 N/mm2
460N/mm2
58
Reinforced Concrete
A6
forf = 460N/mm2
0.OOl3bh
A.
0.OO4bhf
0.002bh
0.OO2bh
For flanged beams over full effective flange width near top surface, use
1.5hfmm2/m reinforcement for the whole length of the beam. Normally
this amount of reinforcementis provided in the slab at the top surface over
the beam as part of slab reinforcementwhen the flanged beam forms part
of a beamslab construction.
d= dia. of bar
Sb
I/Sbb
250mm
Design
of ReinforcedConcrete Beams 59
Note: To control cracking on the side faces of beams use small diameter bars at
close spacings. The distributionof these bars should he over two-thirds of
beam's overall depth measured from tension face.
0.00125bh
Step 19 Deflection
b/b
Find
for flanged beams.
Find 1/d.
Find basic span/effectivedepth ratio from Table 11.3.
Note:
If b/b is greater than 0.3, then interpolate between values in Table 11.3
assuming
60
Reinforced Concrete
Find service stress
where
I3b =
\ fA
f()
5
rcd)
prov
M/M'
A.
Step 21
0.04bh
0.04bh
d' is greater of
SK 2/34 Containmentof
U25 Di or 5mm
compression reinforcement.
Ix
Check bearing stress inside bend where it is required to extend the bar for
more than 4 x diameter beyond the bend because the anchorage requirement is not otherwise satisfied.
Satisfy that
bearing stress=
Fb
2f
1+2(i)
diameter of bar
bar
/ 5 \/A
f = fy
i
OIJb/
rcgd
'1s prov
47000
2MSA
300
62 ReinforcedConcrete
1
SK 2/36 Rectangular section torsional shear
stress.
of composite
,2
'tmin
mm
'tmax
(h1 hmax)
vt=
2T,,
:2
min
Tmax
hmin
Design
63
For hollow and other box sections, follow the method in Chapter 8.
If wall thicknessin a rectangularhollow sectionexceedsone quarter of the
dimension in that direction, treat the hollow section as a solid rectangle.
Calculate
Vtmin = 0.067
v1, = 0.8
\/f
Vf <0.4N/mm2
5 N/mm2
ft
-A
(IongitudinaIreinforcement)
ft
of
Spacing
=
Links
SK 2/38 Torsionalreinforcement
in beam.
If
Vtmin
Check
fYi
v < v
(v + vt) <
where v = flexural shear stress.
A.,,
0.8x1y1(0.87f)
Note:
S < x1, or
64
Reinforced Concrete
SECTION OF BEAM
STRAIN
3acr Em
"max =
1+
Em
Note:
= Cl
b1(h
Cmin)
(h x)
x)
(a' x)
3EA(d x)
LC7.
x l.lh.
E,
x=
d{[(rnp
+ (in
(mp + (in
1)p')2 + 2(mp +
(in
1)p)}
bd
A,
bd
M
k2bd2 +
k3A(d d')
x
k2=) 1-
k3 =
f, = rnf
1)
(1
1)
= fs
E,
= (h
Emh
Note:
E,
b(h x)2
3 E,A,(d x)
66
ReinforcedConcrete
UDL
111111111111111111111111
the worked
Design
of ReinforcedConcrete Beams 67
= 500 40
16 = 434mm
10
dia.
= 6.2m
Therefore
1L
= 10 = 6.2m
/ = 6.Om
60b = 60 x
250
Satisfied
250
300 mm
3002
434
= 18.Om
=51.84m
45 18N/mm2
F
tJl
rectangularbeam.
Md = M= 216kNm
=40 N/mm2
1O4
68 Reinforced Concrete
216 )<
K = Md =
fbd2
=
=
d[0.5
40
10
x 300 x 4342
(o.2s
d[O.5 (o.2s
0.88d = 382mm
- 0.0956)]
dz = 434382 = 116 mm
0.45
AS
0.45
Md
0.87fz
= 216 x 106 = 1413mm2
0S7 x 460 X 382
at faceof support
140
x iO
300
434
2d = 870mm
V = 96kN
96 x
300 x 434
= 0.74N/mm2 <0.8 \/fcu = 5N/mm2
100 A
bd
x 1472
300 x 434
100
= 1.13%
Design
Vnom
V<
of ReinforcedConcrete Beams 69
(V + 0.4) bd
at face of support
at
all
in
the
beam.
Vnom
points
Nominal links
Assume
= 0.4b
o.87f
S, = 300mm
= 90mm2
Use 8mm dia. single closed link
300mm centre to centre.
lOOmm
(f = 460N/mm2) at
70 Reinforced Concrete
(-)
(assume Sb = 200mm)
Use 12 dia. Grade 460 bars at approximately 200 centres on the side face
of beam.
= 716mm2
no. 25dia.
A, = 0.00125bh
OK
Note: Strictly speakingthese barson the side face are not required for beams less
than 750mm overall depth but it is good practiceto use them in order to
avoid shrinkage cracks.
Step 19 Check deflection
= 6200 =
I
d
434
14.3
ib = 1,
M
bd2
((A
rcgd = 460
Y\g/ \A, prov/
\8/ \1472/
= 275 N/mm2
216x106
300
x 4342 =3.8
18
>
Design
Step 23 Curtailment of bars
0.081 = 0.08 x 6000
71
= 480mm
The central 25mm dia. bar will be stopped 250mm from the face of the
support.
Step 24 Spacing of bars
______500
SK 2/48 Arrangementof bars at the bottom of
beam.
SK 2/47 Elevationof beam near support.
47000
= 47000
275
where
f=
= 25 mm
275N/mm2
= 170mm
A=1472mm2
A=226mm2
E6 = 200
=
m=
= 144kNm
d'=54mm
d=439mm
10
20
assumed halfway between long and short-term.
72 Reinforced Concrete
0
0
LI
Ax is
=A =
bd
1472
300 x 43
d{[(mp
+ (m 1)p')2 + 2(mp + (m
(mp + (m
k2 =
= 0.0112
1)p'}
= 0.0017
- 1)()p')] -
= 160mm
I Ill
3d
\2d/\
(
160
2x439!
160
3x439
= 0.16
k3
= (m
- 1)(1 -
M
f= k2bd2 + k3A(d
= (10
54 \
- 1)(1 i:)
=
5.96
d')
x 106
0.16 x 300 x 4392 + 5.96 x 226 x (439 54)
144
14.74 N/mm2
f=
= 10 x 14.74 x
\160
= 257N/mm2
f
=
Emh
1O
i0
b(h x)2
______________
3EA(d x)
= 1.566
i0
= 1.425 x
= 1.414 x 60.5
300 x 3402
200 x i03 x 1472 x 279
= 73.0mm
12.5 = 62.9mm
12.5
= V(60.52 + 452)
a1 >
_____________
+
L
of ReinforcedConcrete Beams 73
x\ = /34O = 1.566 x
257
________
= 1.285 x
200 x i03
Design
Cmin)
hx
x 73.0 x
1.425
i0
i+12(73_48)
L
340
OK
SK2/52 Three-spancontinuous
PQQ
_PQ2
For a typical
74 ReinforcedConcrete
All reinforcementto be used will be high yield steel with = 460 N/mm2.
It is expected that the analysis will be carried out using a computer
program with the load combination shown in Section 2.2.
From moment and shear envelope,
MA=0
VAB=300kN
M = 650 kNm
VBC=325kN
Mnc=+37OkNm
where
VAB=250kN
VAB= negligible
VBA= 370kN
VBA = 320kN
V=275kN
or
l5OkNm
exposurecondition= severe
fire resistance required = 2 hours
grade of concrete= C40
maximum cement content = 325 kg/m3
maximum free water cement ratio = 0.55
nominal cover=4Omm
from Tables 11.6 and 11.7
effective depth, d = 550408 16 = 486 mm
Step 7 Effective span
1 = lo = 10.0 m
Step 8 Effective width of compressionflange
Actual b 4.0m
(centre-to-centreof beams)
Calculated
b=
+b
7.14
= 10000 +300
7.14
= 1700mm
1700
Flanged beam
K=
Md
fbd2
600
106
40 x 1700 x 486
= 0.0373
z = d[0.5 +
= d 0.5
/(o.2s
0.9
0.0373
0.9
= 0.95d = 462mm
= dz
0.45
462
486
0.45
=S3mm<hf=lSOmm
Neutral axis in the slab
Md
0.87fz
600 x 106
______________
= 3245 mm2
0.87
x 460 x 462
Use 3 no. 32dia. bars in bottom layer plus 2 no. 25dia. bars in top layer.
40 cover to
links
SK 2/54 Arrangementof
reinforcementat bottom of beam
at midspan.
76 ReinforcedConcrete
= 81.5mm
d = 550 81.5 = 468.5 mm
Recheck reinforcementrequirement with revised effective depth:
K = 0.040
z = 0.95 x468.5=445mm
= 3369mm2 (required)
A provided = 3394 mm2 OK
MB = 650kNm
Rectangularbeam
= 650kNm
Md
650x106
40 x 300 x 4542
A'
(K 0.156)fbd2
(0.263
0.87f(d d')
= 1696mm2
Ii
dLO.SV0.25
0.156
= 0.775d= 352 mm
x = 0.5d = 227mm
and =
= 0.156fbd2 + A
O.87fz
Design
of ReinforcedConcrete Beams 77
x 300 x 4542
+ 1696 = 4435mm
0.87 x 460 x 352
= 0.156
40
to (inks
Links
SK 2/55 Arrangement
of
reinforcementat top of beam over
Support.
3--32
Flanged beam
b=
1700mm
d=55040816=486mm
Md
K=
= 0.023
fbd2
z = 0.95d = 426mm
AS
Md
0.87fz
370 x 106
=
0.87 x 460 x 462
= 2001mm2
= 150kNm
Rectangular beam
b=
300mm
d = 486mm
K=
x 106
=0.053
x
300 4862 x 40
150
140
coverto links
78 ReinforcedConcrete
d[0.5
./(o.25
= 0.94d = 456mm
150 x 10
= 822mm2
0.87 x 460 x 456
Use 2 no. 32dia. bar (1608 mm2) top of beam.
2032
3-032
SK 2/56
Section through
midspan
BC.
flanged beam
Not required.
= V
bd
370x103
300 x 454
\/f
= 2.716N/mm2 <0.8
5N/mm2
OK
Check shear stress at d from face of column.
VAB = 250kN
d = 468.5mm
for span AB
x i03
300 x 468.5
250
= 1.78N/mm2
100A
bd
100
x 3394
300x468.5
= 2.41
v == 0.85 x 1.17
0.99N/mm2
= (v0.4)bd
= 195kN
v > v + 0.4 = 1.39N/mm2
Vnom
bS(v v)
0.87f
(assume
S = 150 mm)
300
150
x (1.78 0.99)
0.87 x 460
= 89mm2
Use 8mm dia. links = 100 mm2 (two legs) at 150 centre-to-centre up to
the point where shear falls to 195 kN. High yield reinforcement
(f = 460N/mm2).
0.4bS
=
Nominal
0.87f
0.4
x 300
=
x 300 = 90mm
0.87 x 460
(f
VBA = 320kN
320 x i03
v= 300
x 454 (d=454mmatB)
= 2.35 N/mm2
100A
100 x 4435
=3.25
p=
300x454
bd
v = 0.91 x
A
bS(v v)
0.87f
300 x 150
x (2.35 1.063)
0.87 x 460
= 144.5 mm2
Use 8mm dia. links = 150 mm2 (3 legs) at 150 centre to centre up to the
point where shear falls to 195kN.
Step 14 Alternative design for shear
Omitted.
Step 15 Minimum tension reinforcement
80 ReinforcedConcrete
Flanged beam
b = 300
b
1700
Reinforcement in the slab over the beam will be a lot more than this
quantity.
Step 18 Reinforcement in side face of beam
Fora 550mm overall depth of beam with 150mm slab, side reinforcement
will not be required.
Step 19 Check deflection
10000
I.
=
=21.3
d
468.5
d = 468.5 mm
for span AB
(5/8)f.
Service stress, f, =
(assumed)
460
= 288N/mm2
A rcgd
prov
M
bd2
3394
600x106
1700
x 468.52
=1.6
=24.75>21.3
OK
OK
=0.25x32=8mm
OK
Maximum spacing of links = 12 x dia. of bar
=12x32mm=384mm
Note: At least one link at
containment.
OK
0.15/= 1500mm
0.101=1000mm
0.25/ = 2500mm
Span AB
Continue3 no. 32dia. + 2 no. 32dia. up to 1000mm from A (end support).
Stop 1 no. 32dia. and 2 no. 32dia. at 1000mm from A.
(See Step 26: reinforcement in span AB increased.)
Oversupport B (top bars)
Continue 6 no. 32dia. bar top up to 1500mm on either side of B.
Stop 2 no. 32dia. bar at 1500 from B.
Stop 2 no. 32dia. bar at 2500 from B.
Continue 2 no. 32dia. through span.
Step 24 Spacing of bars
Minimum clear spacing = MSA + 5 = 25mm
82
ReinforcedConcrete
Clear spacing of bars in tension= 54mm > 25mm
Maximum clear spacing
= 47000
= 47000
288
= 163mm
OK
Note: Under normal circumstancesthis step will deem to satisfy the 0.3 crack
width limitation criteria, but, as Step 26 will prove, when crack width
calculations are actuallycarried out this may not be the case. In span AB
the maximum clear spacing criterion is satisfied but the calculations show
that the crack widths may be exceeded.
Step 25 Check torsional shear stress
Not required.
Step 26 Crack width calculations
Span AR
d = 468.5mm
b=
1700mm
= 3394mm2
= 1608mm2
= 10
m=
[()2
=
<
=
118 mm
z=d
2mAsd]
hf(= 150mm)
= 429mm
f = 400xx 3
f
106
429
= 274N/mm2
274
200x103
=1.37x103
fhd\
7550118\Jxl.37x103
J=l
xl
\468.5
1181
= 1.69 x
io
Emh
b(hx)2
3EA(d x)
= 1.69 x
iO-3
i0
300 x 4322
_____________________
3 x 20(1 x i03 x 350.5 x 3394
= 1.61 x
a1 = V(642 + 642) 16 = 74.5 mm
a2 = \/(642 + 432) 16 = 61,1 mm
acr = 74.5 mm
3a
3x74.5 x
+ 2(acr Cmin)
(h x)
1.61
+ 2(74.5
(550
x i03
48)
118)
= 0.32mm >0.3mm
The calculated crackwidthis greater than allowable. Increasereinforcement
to 5 no. 32dia. bar instead of 3 no. 32dia. plus 2 no. 25dia. No more
checks are necessary.
Over support B
At face of column,
maximum service moment= 390kNm
84 ReinforcedConcrete
Cmin = 48mm
m=
10
See Step 26 of Example 2.1 for explanationof symbols and the equations.
x = 225 mm
= 0.2068
K3 = 6.44
M
f = K2bd2+ K3AL(d
f = 211.6N/mm2
= 20.69N/mm2
d')
= 1.058 x iO3
= 1.502 x iO3
ac1
= 74.5mm
Step 27 Design
OK
Tension
Anchore
2-432
2-432
:IH
Tension
Anchorage
3-+32
2-+32(second tayer)
8 tinks(1800)
at 150c/c
layer)
r_L
nib.
of section
= 168100mm2
500
= 172.5mm
= 3.943 x 10mm4
(gross section)
86
Reinforced Concrete
Assume
Assume p
=0
= 1%
m=
Assume
= 10
F= 6 x
10-2
Shear centre, e =
b2h2r
'*ixx
b=400145=255
h=500105=395
t = 105
Boundary
condition
Loading
spans
l.4DL on AB
1.ODL on BC
LC1 on AB
LC2 on BC
1.OkN/m LL
on AB
Plastic hinge
at A, C fully
restrained
Force Support
A
BM
Shear
270
+135
180
BM
Shear
159.8
+79.9
102.8
BM 311.7
Shear 193.9
BM
Shear
1.OkN/m LL BM
Shear
on AB
Span
AB
8.44
5.06
0
3.86
+155.9
+4.22
+7.23
Support
Span
BC
Support
+135
270
180
125.9
95.2 72.6
+46.0
92.0
186.7
+28.0
70.6
+270
180
180
166.1
3.38
3.94
5.78
5.14
1.0
65.1
+ 1.19
1.45
kNm
kN
kNm
kN
kNm
kN
kNm
kN
+2.89 kNm
kN
88 ReinforcedConcrete
Not required.
Step 5 Determine torsion
11.25 + 1.6
= 34kN/m
x 11.25
110 15(chamfer)
290
+ 78.5
2
(e = 78.5 = shear
centre)
= 318.5 mm
= 34 x 0.3185
= 10.83 kNm/rn
= 5.6kN/m
=
+e
= 172.5 145 + 78.5
= 106mm
Torsion per unit length
Total ultimate torsion in beam
= 5.6 x
0.106
= 0.59kNm/m
= (10.83 + 0.59) X 4.5
= 51.4kNm at the
supports restraining
rotation
Design
of ReinforcedConcrete Beams 89
b=400mm
1
d=l
60b=6Ox40O=24000mm>8500mm
1
280x106 \
=1
I =335mm
x x
\0.156
\0.156fb1
250b =
250
40
4001
x 4002 =
119402mm > 8500mm
Slendernesscheck is satisfied.
Step 10 Design forflexure
As'
steel
K=
fbd2
280 x 106
x
40 400 x 45752
= 0.0836
0.156
No compressivereinforcementrequired.
d[0.5
(o.2s
457.5[0.5
410mm
dz
0.45
457.5 410
0.45
'(o.2s
0.0836)]
90
Reinforced Concrete
= 105mm = hf
Neutralaxisis in the flange.
M
0.87fz
280 x 106
0.87 x 460 x 410
= 1706mm2
K=
fbd2
175x106
40 x 400 x 45752
= 0.052
z = 0.94d
= 430mm
= dz
0.45
= 61mm <
105mm
= hf
0.87fz
175x106
0.87 x 460
x 430
= 1017mm2
Use 2 no. 25mm dia. bars (982 mm2) + 1 no. 12mm dia. bar (113 mm2).
Step 11 Flanged beam
Not required.
Step 12 Check maximum shear stress at support
V
v=
bd
x 1o3
x 457.5
190
290
= 1.43 N/mm2
0.8Vf = 0.8 x \/40 = 5N/mm2
Step 13 Check flexural shear stress
d = 457.5 mm
VA
= 190 40 x 0.457
= 172 kN at effective depth away from support
172
i03
290 x 457.5
= 1.30N/mm2
l00A
bd
x 1964
290 x 457.5
100
= 1.48
v, = 0.72 x 1.7 = 0.84N/mm2
= (v + 0.4)bd
Vaom
= (0.84 + 0.4) x
= 164.5kN
v > v + 0.4
A
290
x 457.5 x
i0
bS(v v)
0.87f
290x
200 x (1.30 0.84)
=
=
S.
0.87 x 460
66.7 mm2
at 200 mm c/c (2 legs)
66.7
200x2
Nominal
A.
0.87f
0.4 x 290
0.87 x 460
= 0.29 (2 legs)
= 0.145 (for each leg)
Area of tension reinforcementrequired to carry weight of slab on the nib
34 kN/m
0.87 x 460
= 85 mm2/m
A=
S
85
1000
d from support.
92
Reinforced Concrete
1964mm2 provided
Not required.
Step 17 Transverse reinforcement in flange
A = 1.5hfmm/2m
Not required.
Step 19 Check deflection
400
M
P4'
20.8)
x (0.725 0.3) = 24
167.2
155.9
(5 '\
(Asrcgd
,jf 's
A
\Ib'
prov
(5\
/155.9\
= II x I
I x 460 x
\8!
\167.21
(1017
I
\1095
= 249 N/mm2
P4
bd2
175x 106
209
400 x 45752
9000
le_
457.5
19.67
< 28.80
OK
f. = 249N/mm2
47000
= 47000 =
249
189 mm
(see Step 5)
[2'
FIRST CHOICE
SK 2/63Calculation of torsional
shear stress.
> 84mm
SECOND CHOICE
SK 2/64 Calculation of torsional
shear stress.
94
Reinforced Concrete
T x 1.22 x iO'
1.245
51.4
x iO'
x 1.22
1.245
= 50.4kNm
Torsion carried by flanges = 0.5(51.4 50.4) = 0.5kNm
2T
Torsional shear stress v in web =
/
2 I,
!tmin 'tmax
ijfl
2x50.4x106
290
2902500
= 2.97N/mm2
Torsional shear stress v in flange =
2x0.5x 106
/
105
1052(\110
=
ttmin
1.21 N/mm2
0.067\/f
= 0.067\/40
= 0.4 N/mm2
= O.8Vf = 5N/mm2
VtuYl
550
5 X 450
550
= 4.1N/mm2
V > ''t. mm' torsional reinforcementrequired.
Torsionalshear stress + flexural shear stress = 2.97 + 1.30
= 4.27 N/mm2
<SN/mm2 OK
Torsional reinforcementin web (vertical)
Design
of ReinforcedConcrete Beams
95
S,,
0.8x1y10.87f
50.4
x 10
(4()(x + y)
(f
5O'14S (FLEXURALSHEAR)
A/sv0OB5 (NIBDIRECT TENSION)
Z
U)
uJ
z
U)
Ui
I.-
zUi
C)
12 at110 I
12at130
=10
SK 2/65
- 660
A/S diagramfor
Example 2.3.
T
0.8x1y10.87f
12at175
087
065
1300
1750
I
I
I
12at200
057
96 Reinforced Concrete
0.5x106
0.8
x (105
= 0.078
48)
= 57mm
or 200mm
(f, = 250N/mm2).
= 0.078
= 0.078 x 1000 x
(Grade 250)
= 144 mm2/m (for 2 legs of mild steel)
= 72mm2/m (for each leg horizontal)
Total requirement= 241 + 72 = 313 mm2/m < 566 mm2/m
Longitudinalreinforcementfor torsion in flange
() (_)xi
= 0.078
= 59mm2
250
+ Yi)
x (57 + 352)
(Step 25)
A or C = 201 kNm
state)
(serviceability limit
Design
of ReinforcedConcrete Beams 97
4-025
dE(M5)
P5Ommc/c
012LINK
408
____
08 (MS)T
2025
5Ommc/c
support.
A. = 982mm2
rn = 10
hf = 105mm
h = 500mm
= 290mm
d' = 42.5mm
x=
dff(mp + (m
+ (m
1)p')2 +
(mp
+ (rn 1)
l)p') }
= 156.3 mm > hf = 105 mm
(mp
+
x= mdA -f0.5bh?
mA, bhf
10 x 455.5 x 1964 + 0.5 x 400 x
it) x 1964 + 400 x 105
= 181 mm
8-012
d hf(3x 2hf)
3(2x
= 410mm
h1)
1052
= 1)0054
()')]'
98 ReinforcedConcrete
M
x 106
1964 x 410
201
= 250N/mm2
250
200x103
==
1.25
x i03
500 181 \
(\455.5
x 1.25 x iO
181)
i0
= 1.45 x
Smh
b(h x)2
3EA(d x)
3a
=
1+
Emh
2(ac.
(h
Cmin)
x)
1+ 2(50.4
(500
32)
181)
-c
100
Reinforced Concrete
010
201.1
0151
_0-15L
65
6OI.7
_010
r1001 60/.7
r20fr
T,f
0151
7ffecti
Span
0151
0101 for
endsupport
Effective Span
LI
'.
50/. '
\-100.,.
0.c
Effective Span
Chapter 3
Design of Reinforced Concrete Slabs
NOTATION
3.0
a'
ab
acr
ASbX
ASb.
b
Cmin
d'
d!
E
E
F
G
h
charts
Moment of inertia using b as unit width for slab
Clear span or span face-to-face of support
101
Reinforced Concrete
102
I.
Effective span
m
Md
M
Mh
Mb
MUN
MHP
MVN
Modular ratio=
Design bending moment per unit width of slab modified to account for
axial load
Moment per unit width about x-axis
Moment per unit width about y-axis
Torsional moment per unit width
WoodArmer design moment for top reinforcementin y-direction
WoodArmer design moment for bottom reinforcement in y-direction
WoodArmer design moment for top reinforcementin x-direction
WoodArmer design moment for bottom reinforcementin x-direction
Ultimate negative moment capacity of slab per unit width about an axis
parallel to H
Ultimate positive moment capacity of slab per unit width about an axis
parallel to H
Ultimate negative moment capacity of slab per unit width about an axis
parallel
Mp
N
p
p
r
rR
R
S
Sb,
ShV
S,
S.
T1
U1
U0
v
v
v,
v,
vc,,
to L
Ultimate positive moment capacity of slab per unit width about an axis
parallel to L
Axial load per unit width of slab in x-direction to be combined with M
Axial load per unit width of slab in y-direction to be combined with M
Percentageof tensile reinforcement
Percentage of compressive reinforcement
Percentage of tensile steel to resist M, about x-axis
Percentage of tensile steel to resist M about y-axis
Loading per unit area used in yield-line analysis (kN/m2)
Ultimate loading per unit area
Restraint factor for computation of early thermal cracking
Ultimate total load on panel of slab
Spacing of vertical links
Spacing of inclined shear reinforcementto resist V per unit width
Spacing of inclined shear reinforcementto resist V per unit width
Spacing of vertical shear reinforcementto resist V per unit width
Spacing of vertical shear reinforcement to resist V per unit width
Differentialtemperature in a concretepourfor calculation of early thermal
cracking
Perimeterof concentratedload on slab at prescribedmultiplesof effective
depth
Design
v1
V
Wmax
x
x
y
103
z
c
13
Eh
Em
Er
Cmii
Ci
Pent
3.1
ANALYSIS
OF SLABS
Continuous
Cantilever
where
1C
I = smaller of (1+ d) or
I=
span
A and A
104
Reinforced Concrete
are for unit width. Convert A and A into equivalent concrete areas
by multiplying by m = E/E. Moment of inertia increment due to
steel = mA(x')2 wherex' is the distance of the steel from the centroidal
axis of the section. The shift of the centroidal axis due to the presence of
reinforcing steel may be neglected.
(I bh3 + Fbh3)
=
where
where
1OOA.
bd
where
Note:
M, M
and
Design
of ReinforcedConcrete Slabs
V, and V,
(2) Shears
Mh,
(3) WoodArmer moments
N, and N
(4) In-plane loads
and
105
Mh
Method I
BS811O: Part 1: 1985, clauses 3.5.2 and 3.5.3, Table 3.15.1']
Method 2
Yield-line method: non-linear use Figs 3.18
Method 3
Finite difference: linear elastic
to 3.33.
Moody's table.191
Method 4
Recommendations
Use Method 2 or Method 3 generally. Use Method 4 (finite element
analysis)only where complicated loadings and geometry render the other
methods unusable. Use elastic analysis charts if boundary conditions and
loadings are appropriate.
106
Reinforced Concrete
Method 1
BS811O: Part 1: 1985, clause 3537['1
Method 2
R and R to
Method 4
Finite element analysis. Use the support reactionsas loadingon the beam.
Recommendations
Method 2 may be used for all applications. Method 3 and Method 4 may
be used when similar methods are used for the analysis of the slab panels.
PLAN OF SLAB
B=
where
1.2x(1
3.2
LOAD COMBINATIONS
Two-way
spanningslabpanels
3.3
Note: One-way spanning slabs should be treated as beams of unit width and
Chapter 2 should be followedexcept for minimum shear reinforcement.
Step 2 Design forces
Draw panel of slab and indicatemaximum design moments, shears and inplane loads, if any, per unit width of slab.
108
Reinforced Concrete
I WIDTH
O67jcu/m
__ 0 9x
Find the following parameters for design moments in Step 2 per unitwidth
of slab.
Md
Note:
=M+
for N
N(
d1)
0.1fbd
For N>0.1fbd, design as wall (see Chapter 8). Md may also be taken
equal to M where N0.1fbd and N may be ignored. (Sign convention:
N is +ve for compression.)
K=
Md
fbd2
+
=
0.45
AS
Md
0.87f
0.87fz
K' = 0.156
K' = O.4O2(13b
10%
=
where
Note:
<0.9
M = moment after redistribution
M' = moment before redistribution.
If K is greater than K', increase depth of slab and start from Step 1 unless
links are provided in the zone where steel in compression is used. The
links are required to provide lateral restraint to bars in compression. Links
in slab should normally be avoided.
When K> K',
z=
d[0.5
(o.2s
x = d z s 0.5d
A' = (K
K')fbd2
f(d d')
N
AS=Kf+A,_
0.87fz
O.87fy
If d'/x>0.43x,
A' = (K
K')fbd2
f.(d d')
____
0.87fz
0.87J
f = Ix
=
d'\
becausesteel strain
e=
(x
d'\
(L)
where
as in Section 1.4.2.
Step 5 Detailing
Convert areas of steel per unit width found in Step 4 to diameter and
spacing of bars.
Step 6 Checkshear
Y
Mx
i)
(for
Mx,V
110
Reinforced Concrete
v=
0.8
\/f
5 N/mm2
vy =
0.8
\/f
5N/mm2
100A
bd
Py=
100A
bd
and
rAsvy(Totat area)
::
If)
(Total area)
UNIT WIDTH
0.4bS
0.87f
>
or ASVY
SK 3/6
links.
0.4bS0.87f
Note: Single vertical bars may be used instead of closed links provided proper
anchorage bond length is available.
If
and
nominal links
Assume
A
S=
0.8bS
0.87f
= S,
Design
1
f_
L_ L_. L_
L........
111
(I)
(
In
In
UNIT WIDTH
=b
tJ'Jj'J']
Provide single vertical bars with proper anchorageover the whole zone at
a grid spacing of S.
If
+ 0.4) <
0.8
5 N/mm2
or
(v +
(v
0.4) <
0.8
\/f
\/f
5 N/mm2
unit width
__j
I =i1--I-=1
==i
:'t::
1i
-:1with
unit
SK 3/8
reinforcement.
bS1(v v1)
O.87f
bS(v v)
0.87f
112
Reinforced Concrete
Asbx
bdSbx(v1
v)
A.sby
v)
+ 0.4) < v
(v
<
and
+
If
(v
0.4)
0.8\/f
v 0.8\/f
5 N/mm2
5 N/mm2
bdSbX(v1
v)
bdSby(vV
and
Note:
v)
Ab and Ab are the areas of bent-up bar required per unit width of slab
equal to b.
Recommendation
Avoid using links
where
V1
0.8
\/f
5N/mm2
Design
of ReinforcedConcrete Slabs
113
where
U1
Calculate
Note:
Take
l0OA/bd.
p=
If v1
1.6
A.S sincr
v,
(v1
sin
v) U1d
0.87f
If 1.6v < v1
A55
0.4U1d
0.87f
2 v,
5(0.7v1
v) U1d
0.87f
0.4U1d
0.87f
114
Reinforced Concrete
FACE OF LOAD
ZO 3=1.5d
PERIMETER U3
d,2
LU
O-Thd
O.75d
>t0.75d
rI
I
- SHEAR
REINFORCEMENT
REIPORCEMENT
COMMON TO
BOTh FAILURE
ZONES 2 AND 3.
COMMON TO
90TH FAILURE
ZONES AND 2.
O.75d
1.5d
FAILURE ZONE2
LPERIMET U2
if v>2v.
where
U2 =
If V2
1.6 v,
Asin
(v2 v)U2d
O.4U2d
OX7f,
O.87f
If 1.6 v, < v2
A sinc
2 v,
5(O.7v2
v)U2d
O.87f,
O.4U2d
O.87f
Similarly check successive failure zones O.75d apart till v v,, is satisfied.
Reinforcementto resist shear will be provided on at least two perimeters
within a failure zone. Spacingof shear reinforcement on the perimeter
should not exceed 1.5d.
Steps to befollowedfor the determination ofpunchingshear reinforcement in
slabs
Design
of ReinforcedConcrete Slabs
115
(1) The first failure zone is from the face of the loaded area to the
perimeter 1.5d away.
The
first perimeter of shear reinforcement should be placed at d12
(2)
from the face of the loaded area.
(3) The second perimeter of shear reinforcement should be placed at
O.75d from the first perimeter of shear reinforcement.
is the sum of areas of all the legs of shear reinforcement in a
(4)
failure zone in the first and second perimeter.
(5) The second failure zone is 1.5d wide and starts at O.75d from the face
of the loaded area.
(6) The successive failure zones are 1.5d wide and are O.75d apart.
(7) The first perimeter reinforcement in the second failure zone is the
same as the second perimeter reinforcement in the first failure zone.
Step 8 Modification due to holes
OPENING
IN SLAB
SHEAR PERIMETER
of shear
perimeterdue to presenceof holes.
SK 3/14 Modification
REDUCTION OF
PERIMETER
WITHIN THE
RADIATING LINE
III
-
CHEcXING
OF PUNCHING
SHE AR
-Third perimeter of
shear reinforcement
-Second perimeter of
shear reinforcement
First perimeter of
shear reinforcement
116
Reinforced Concrete
Step 9 Minimum tension reinforcement
O.OOl3bh
in both directions
At end support of slabs where simple support has been assumed, provide
in the top of slab half the area of bottom steel at midspan or 0.OOl3bh,
whicheveris greater.
Step 10 Torsionalreinforcement
Step 11
rc4d)f
(_)(hls
b
s prov
= MIM'
M = moment after redistribution
M' = moment before redistribution.
13b
Find M/bd2.
Find modification factor for tension reinforcement from Chart 11.5 and
modification factor for compressionreinforcement from Chart 11.4.
Find modified span/depth ratio by multiplying the basic span/depth ratio
with the modification factor for tensile reinforcement and compression
reinforcement, if used.
Check
(%)
1 or over
0.75
0.5
0.3
less than 0.3
A is the area required at the ultimate limit state. The clear spacings as
(1) Thinwall cast on massive base: R = 0.6 to 0.8 at base, R = 0.1 to 0.2 at
top.
(2) Massive pour cast on blinding: R = 0.1 to 0.2.
(3) Massive pour cast on existingmass concrete: R = 0.3 to 0.4 at base,
R=0.1 to 0.2 at top.
(4) Suspended slabs: R = 0.2 to 0.4.
(5) Infillpanels i.e. rigid restraint: R = 0.8 to 1.0.
R = restraint factor
Typical valuesof T1 for Ordinary Portland Cement (OPC) concrete are:
where
3(X)
500
700
10(X)
25C
35C
42C
47C
17C
28C
28C
28C
T1
obtain
2.1.9.
(coefficient
cR
of thermal expansion) from Table 2.3 in Section
2(acr
Assume x = h12
Er
Cmin)
h x
118
Reinforced Concrete
Note:
A= Acpcrit
For suspended slabs and walls,
A=
bh
or 250b
= 0.0035 A
whichever is smaller
Ab =0.35b
= 0.875b
Ab =0.35b
Wmax
Em
2(acr
Cmin)
h x
b(hx)2
_______________
Emh Eh
3EA(d x)
Note:
..
Unit width
zone
1.A(stelinercompression
unit width)
1d'
L
h d
..
.- ft.
SK 3/18 Section of slab with steel
in compression zone.
E,
m=
x=
in tension
per unit width)
+ (m
(mp + (m
1)p')}
()(1 )
k3 = (m - 1) (1 =
Es
M
k2bd2 + k3A(d d')
mfc(
= fs
_x
\d xi
zor'i
A.
k2=
f=
axis
. )D .___
(steel
A,
p=
bd
d{[(mp
iutral
120
Reinforced Concrete
Emh
Note:
b(h x)2
3EA(d x)
//
//
//
//
//
//
//
///
/ _________________ ///
7
o
o
-
//
//
//
____________________________________
I
I
l, = 4.3m
= 6.3m
1II = 0.68
Elastic analysis
Read coefficients from Fig. 3.12:
m11
= 0.035
= 0.021
m2 = 0.075
m,2 = 0.060
Characteristicdead load = 0.15m x 25 kN/m3x 1.4 + 2 x 1.4
= 8.0 kN/m2
Design
of Reinforced ConcreteSlabs
121
= mn1
= 0.035 x 40 x 432
= 25.9kNm/m
M1 = 0.021 x 40 x 432
= 15.5 kNmlm
M3 = 0.075 x 40 x 432
= 55.5 kNm/m
M,2 = 0.060 x 40 x 432
= 44.4 kNm/m
Allowing for 10% redistributionof moments,
Design moments:
M1 = 31.4kNm/m
M1 = 19.9 kNm/m
M3 = 50.0kNm/m
M,2
= 40.0 kNm/m
shears.
Note: These moments do not take into account the WoodArmer effect due to
the presence of
and may be unconservative locally. In ultimate load
design local plastic hinge formation may be tolerated when there is a
possibility of redistribution of loads.
Analysisfollowing BS8110: Part 1: 1985'
Coefficients from Table 3.15.
m1 = 0.039
m1 = 0.024
m3 = 0.052
= 0.032
M1 = 28.8kNm/m
M1 = 17.8 kNm/m
M3 = 38.5kNm/m
M2 = 23.7kNm/m
122
Reinforced Concrete
Note: These momentsare considerably less than the redistributeddesignmoments
found from elastic analysis. Elasticanalysis gives peak values, whereasthe
BS 8110 coefficients tend to smear them across a long stretch of slab.
It is desirable and practical to use the elastic analysis results and allow
10% redistributionwith a view to minimising the appearance of unsightly
cracks in the slab. This is conservative approach.
= 5OkNm/m
(Vertical Negative)
(Vertical Positive)
(HorizontalNegative)
(Horizontal Positive)
Assume that the elastic analysis results will be the maximum plastic
moments in the panel of slab.
MVN
Mp = 31.4kNm/m
MJJN = 4OkNm/m
MHP = 19.9 kNm/m
L (MVN + Mvp'\
H MHN + MHP)
6.3
4.3
(50 + 31.4'\
\40 + 19.9)
= 1.70
= 0.35
x = 0.35 x 6.3 = 2.20m
Unit resistance, r =
5(M + MHP)
x2
Alternatively,
r=
8(MVN
+ Mp)(3L x)
H2(3L 4x)
= 8(50 + 31.4)(3 x
6.3
2.2)
Designed by the resultsof elastic analysis the slab panel has a large reserve
ofstrengthbecausethefailureloadingis58.23 kN/m2 againstdesign ultimate
loading of 4OkN/m2. Similarly, designed by the results of the BS 8110
= 0.44 x 40 x 4.3
= 75.7kN/m
= 0.33 x 40 x 4.3
= 56.8 kN/m
Refer to Table 3.4.
By yield-line principle: assuming
3 r H (1 x/L)
2(3x/L)
3 x 40 x 4.3 x (1
2(3
63.3 kN/m
Vy =
r= 40kN/m2,
0.35)
0.35)
3rx = 3x40><2.2 =
52.8kN/m
5
d=
124
ReinforcedConcrete
Step 4 Design of slab
K=
=
x=
A. =
fbd
d[o.s
0.45
50x106
40
=0.081
x 1000 x 1242
= 0.9d =
/(o.25
111.6mm
= 27.5mm
0.87fz
50x106
0.87
x 460 x
111.6
= 1120mm2/m
K=
fbd
40x106
40 x 1000 x 1122
=0.08
z = 0.9d = 100.8 mm
x = 24.9mm
A = 992 mm2/m
Positive midspan moment in short direction
M = 31.4 kNm/m
K = 0.051
z = 116.5 mm
A = 673 mm2/m
Positive midspan moment in long direction
M = 19.9 kNm/m
K = 0.04
z = 0.95d =
106.4 mm
= 467mm2/m
Step5 Diameterand spacingof bars
Use:
100(1)
iiTii
LLL
Long direction
(565 mm2/m)
at
midspan
It
V
75.7 x io
=
= 0.61 N/mm2
v = ibd 1000 x 124
1000 x 112
bOA,.
100A.
From Fig.
1000
bd
bd
100
1131
x 124
=0.91%
100 x 1131
=1.0%
1000 x 112
11.5,
= 0.OOl3bh
= 0.0013 x 1000 x
= 195mm2/m
150
satisfied
_{
Beam
bd
Beam
1!
126
Reinforced Concrete
Step 11 Check span/effective depth
1
d
4.3
x i03
124
= 34.7
1.21
(As_rcqd\(1
A
'11
prov' \Ib
= 5 x 460 x
8
673
754
1.21
= 212N/mm2
M
bd2
31.4 x 106
1000
x 1242
20
be ignored.
12
200(T)
(1600]
NEL
j
I
Gl2@ 400(T)
(3000]
It)I2 @ 400(T)
I
I
I
BEAM
BEAM
12 300(B)
I2@300(B)
124001
_(BEAM
l2@400(T)
G12@400(B) [3600)
l2@400(T)k2o0J
012@200(T) (2200)
Design
of ReinforcedConcrete Slabs
127
45xdja. ofbars=45x12=540mm
0.15 1=0.15x4.3=645mm
0.30 4,. = 0.30 x 4.3 = 1290mm
0.20
0.20 x 4.3 = 860mm
4=
0.15 1=0.15x6.3=945mm
0.30 1ev = 0.30 x 6.3 = 1890mm
0.20 = 0.20 x 6.3 = 1260mm
Direction
l. top reinforcement
top reinforcement
bottom reinforcement
OK
3d 3 x 112 = 336mm
OK
cracking= 400mm
=
=
= 0.8 T1 R
= 0.8 x 12 x
= 34.56 x
12
10
x 106 x 0.3
slab
128
ReinforcedConcrete
Cmin = 20mm
+2(cr Cmin)
(h x)
3 x 197.6
x 34.56 x 106
1+ 2(197.6
(150
32)
56)
=0.0045mm<0.3mm
OK
A=-2-=
1000x150
2
=75000mm2
By elastic analysis,
maximum bending moment over long support
i0
Design
=d
= 124
f= M =
35.7
=
=
Emh
i06
x 109.7
1131
)
=
288
200x103
b(h
3E.,A.,(d
12
x)2
x)
= 288 N/mm2
43\
1.90
x
1000(150
x 200 x
= 1.69 x iO
3
43)2
Cmin = 20mm
acr = \/(262
+ 502)
3acr
Wcr=
(h
6 = 50mm
1 +
Cmin)
x)
129
109.7mm
1.44
/150
=0.l6mm<0.3mm
OK
130
Reinforced Concrete
1T1
'p
L,2
L12
Elasto-plastic
resistance,
V
4k.
Elastic
resistance,
1MN
L'
L,'
3L
wwvw1
Li
Li
mThcYj
MN
8MN
L
'P
L
,,PI2
L/3
,P/2
L/34L/34
cT
Support reactions,
r, L
L,i
1/2
L. reaction
'P
W
L,'
L/2
L. reaction
5rL
R. reaction
3rL
8
llR,
R. reaction
16
5R,
16
r,L
R
L,i
L/2
.,
r,L
Rr,
L
,P/2
,,P/2
L13L13jLj3
R
2
Fig. 3.2 Support shears for one-way elements (to be read in conjunction
Fig. 3.1).
with
HIL
>
002
003
.005
.007
01
02
03
05
07
0I
02
03
05
0.7
l0
HIL
H/L
H/L
HIL
H/L
000I
0002
0003
0005
00t
0007
002
003
= IJrH2
iiiiii
----7-
0 0.7 0-5
0-3
02
01
11111111
5'O
30 20
H/L
---J-----
iii
IIIEii 1111
XV =
IU 7.0
'007
005
01
.02
03
05
.07
0I
02
03
07
01
10
free.t81
HIL
HIL
H/L
>
supported)1
H/L
H/L
supported)
a
-l
a
a
C.
simply supported.tSl
NIL
NIL
HIL
fixed.151
08
07
06
'
04
08
.0
11/2
L I Mvp
I
[MHN MHDJ
Fig. 3.18 Location of yield lines for two-way element with two adjacent edges
supported and two edges free (values of x).lMJ
J0
09
08
07
06
ylH
0-s
04
0-3
02
0I
L [MvN+Mvp1''2
HL Mm
xIL
138
Reinforced Concrete
Mvp
.LI
H
1/2
-
LMHN 4MHPJ
Fig. 3.20 Locationof unsymmetrical yield lines for two-way element with three
edges supported and one edge free (X2/X1 = 1.0)181
y!H
.L
I_MVN+MVP1V2
H LMHN+MNPJ
Design
of ReinforcedConcrete Slabs
139
05
0-5
0-4
0-4
0-3
0-3
02
0-2
0I
0-I
k(MVN+_Myp\'/2
H kqH.N)
Fig. 3.22 Locationof symmetrical yield lines for two-way element with four edges
supported.t81
L[
Mvp
N LMHNI+MHPJ
Fig. 3.23 Location of unsymmetrical yield lines for two-way element with three
edgessupported and one edge free (X2/X1 = O.1)JsI
140
Reinforced Concrete
'''
1"
H [MHNI+MHPJ
Fig. 3.24 Location of unsymmetrical yield lines for two-way element with three
edgessupported and one edge free (X2/X1 =
Myp 1/2
H LMHNI MHPJ
Fig. 3.25 Locationof unsymmetrical yield linesfor two-way element with three
edges supported and one edge free (X2/X1 =O.5).I1
Design
U.,
of ReinforcedConcrete Slabs
141
ii 4:u27sq/1
---
Mvp
Mvw2
\'\
E"HH!
--
Nooo
01
Q5
05 10
Myp
10
50
8000
]112
H LMHNI+MHPJ
Fig. 3.26 Location of unsymmetrical yield lines for two-way element with three
edges supported and one edge free (X,1X1 = ().75).1
L[
H
Mvp
11/2
[MHNItMHPJ
Fig. 3.27 Location of unsymmetrical yield lines for two-way element with three
edges supported and one edge free (X2/X1 =
142
Reinforced Concrete
.Li
Mvp
H LMHNI+MHP]
Fig. 3.28 Location of unsymmetrical yield lines for two-wayelement with three
edges supported and one edge free (X2/X1 = 1.5).151
LI
H
Mp
1/2
LMHNI+MHPj
Fig. 3.29 Location of unsymmetrical yield lines for two-way element with three
edges supported and one edge free (X2/X1 = 1.75))81
LI
1112
Mvp
H [MHNI+MHPJ
Fig. 3.30 Location of unsymmetrical yield lines for two-way element with three
edges supported and one edge free (X2/X1 =20)181
05
kH [
kM
oe
(Mv+Mvp)&2
+M)V24 (MHHI4MHP)h/2J
Fig. 3.31 Location of unsymmetrical yield lines for two-way element with three
edges supported and one edge free (values of y))81
144
Reinforced Concrete
XI
-I:
J5(MI*I4MHP)
Fig. 3.32 Location of unsymmetrical yield lines for two-way element with four
edges supported (values of X1).151
07
04
05
04
03
02
03
04
05 06
L--
0.7
5JMviMvp
jMHNI+MHP+ J4N2+MHP
Fig. 3.33 Location of unsymmetrical yield lines for two-way element with four
edges supported (values of V1).151
Design
of ReinforcedConcrete Slabs
1/2
TOOI0
Cantilever Slab
Fig. 3.34 Simplifieddetailingrules for slabs.
r5
145
146
Reinforced Concrete
Table 3.1 Graphical summary of two-way elements to be used in conjunction
with Figures 3.3 to 3.17
Fig. 3.3
LL
Fig. 3.4
Fig. 3.5
ii
L
Fig. 3.6
Fig. 3.7
Fig. 3.8
___H
L
Fig. 3.9
Fig.
Fig. 3.12
Fig. 3.14
.ij
LL
III
=1111
Fig. 3.15
Fig. 3.16
Fig. 3.17
Free
3.11
r
Simple
''
Fixed
Four edges
supported
Three edges
supported and
one edge free
Two adjacent
edges supported
and two edges
free
Edge
conditions
"
Y2
y-i
..4
"
t1 t1
I- I
Hf3
h1
i,,,,,J
x-
y <H
yH
Limits
y2
+ Mvp)
5(MVN
H2(3L
2y)
H2(3L
4x)
8(MVN + Mvp)(3L
L2(3H
L2(3H 4y)
M) or 8(MHN + MHP)(3H
or
or
Y)
x)
y)
x Mvp
AIHN)Y
MVN)x
4x)
10
+ MHP)(6H
x) +
H2(3L
L2(3H 2y)
2MVN(3L
4(MHN
2x)
M + (5Mv
H MHN + (5MHP
6L
+ Mvp)
5(MHN + M11)
S(MVN
or
or
or
x2
5(MHN + MHP)
S(MVN
+ MHP)
x2
5(MHN
Table 3.2 Ultimate unit resistance for two-way elements (symmetricalyield-lines) (to be used in
conjunction with Figs 3.18 to 3.23).
-t
Four edges
supported
Three edges
supported and
one edge free
Two adjacent
edges supported
and two edges
free
Edge conditions
l--1
1/2 \3
t
HIj
LL
NIL:::::1
xL2
X L2
Limits
y2
H2 (3L
2X2)
X2)
2Y1
2}'2)
Y2)
or
or
or
or
+ Mvp)
Yi
5(MVNI
5(MHNI + MHP)
2X1
+ MHp)(6H
(3L
X1
Mp)
X)2(3H
Y)2
(L
X)2 (3H
(MHN2 + MHp)(6H
+ Mp)
Y
5(MVN2
MHp)
(3L
2Y1
X1
X2)
Y2)
2Y2)
2X2)
Y)
2Y)
2X1
+ Mp)(6L
5(MHN2 +
(H
(MVN2
(L
MHP)
MVN2L
2X2)
+ MHp)(6H
(MHN2
+ X2) +
5(MHN3
2X1
MVN2)(XI
+ Mvp)(6L
or
5(MVN3
x2 (3H
(MUNI
Y2
(MVNI
(5Mp
or
(MHNI + Mp)(oH Y)
or
x2 (3H 2Y)
or
5(MHNI + MHPI)
Table 3.3 Ultimate unit resistance for two-way elements (unsymmetricalyield-lines) (to be used in conjunction with Figs 3.18 to 3.33).
Design
of ReinforcedConcrete Slabs
149
Table 3.4 Ultimate support shears for two-way elements (symmetricalyield-lines) (to be used in
conjunction with Table 3.2).
Edge conditions
L1
Two adjacent
edges supported
and two edges
free
xL
Three edges
supported and
one edge free
,.2c.1
"
HILJJI
nt,, T1
X L2
y< H
Vertical shear, Vv
3rH(2)
Li
3rx
5
(o-f)
3rL (2
y
HI
(_)
3rH( 1-)
Li
3rx
5
(3f)
-)
2(6)
3rL (2
3ry
5
x-2
L
3rH(
3rx
1I
LI
2(3-f)
Four edges
supported
3ruL(1
2(3_
HI
edges
supported
Four edges
supported and
one edge free
Three
Two adjacent
edges supported
and two edges
free
H36
_<'
Ix"i"
j1 ,,,
HI[iJ
L
h1
fiti
Y2
22
xI-2
x2
xL
limits
y)
Yi
6H
Yi
Y2
3r(L x)(2H Yi
y)
Y2)
Y2)
6H-yl-y2
3rx (2H
3rx2
3rx1
x)(2H
6H y
3rx (L
3rx(2H
3x2r
3x1r
Horizontal shear, VH
V,
x1
3ry
5
x2
x1
3ruY2
3ry1
6Lx1x2
3r(H y)(2L
x2)
3ry(2L x2)
6Lx1x2
3rH(2Lxix2)
Vertical shear,
Ultimate support shears for two-way elements (unsymmetrical yield-lines) (to be used in conjunction with Table 3.3).
Edge
conditions
Table 3.5
CD
CD
-t
CD
CD
CD
-t
CD
Chapter 4
4.0 NOTATION
ay
A
b
b'
c
E
E
f
f,
G
h
h'
h
hmax
hmin
10
'ex
1ev
m
M
of column
Reinforced Concrete
152
M
M
M
M
Madd
Madd
N
Nbaj
p'
Px
Py
T
v.
(N/mm2)
Design concrete shear stress in concrete due to bending about y-axis
(N/mm2)
Shear force in concrete column due to bending about x-axis
Shear force in concrete column due to bending about y-axis
Coefficient to determine effective height of a column
Coefficient to determine modifiedbending moments in biaxial bending
Diameter of reinforcing bar or equivalent diameter of a group of bars
13
4)
Design
153
Braced:
IC
where
3I()
I = effective height
1 = clear height
a
BeamSimpl1
Supported
ip
Monolithic
D1<H2
Connection
D2 H2
D2
Monolithic
Connection
H3
//
______
_______
SK 4/2 Column end conditions.
Q
______________
to Foundation
Moment Connecton
154
Reinforced Concrete
Table 4.1 Values of 13 for braced columns.
End condition
at top
1
2
3
0.75
0.80
0.85
0.95
0.90
0.95
0.80
0.90
1.00
End condition
at top
1
2
3
4
1.2
1.3
1.6
2.2
1.3
1.5
1.8
1.6
1.8
Design
of ReinforcedConcrete Columns
155
Note: Load combinations LC2 and LC4 should be considered only when the
effect of dead and live load are considered to be beneficial.
= dead load
LL = live loador imposed load
where DL
WL
= wind load
WP = water pressure
EP = earth pressure.
The general principle of load combination is to leave out the loads which
have beneficial effect. If the load is of a permanent nature, like dead load,
earth load or water load, use the partial load factor of 1 for that load
which producesa beneficial rather than adverse effect. This rule of combination will be used for design as well as for the check of stability of
structure.
Note:
I.
156
Reinforced Concrete
Note:
For short columns both ratios should be less than 15 for braced and 10 for
unbraced.
60b
lOOb2Ih
60b
N = 0.4fA + 0.75Af
where
Check
(2) Column supporting continuous beams where analysis does not allow for
framing into columns (no moment in column)
Find
Check
N = 0.35fA + 0.67Af
f,
Bending about
major axis only
= Pa171
Braced
Quy
15<2O
h
Icy
Majy = Na
M = + *M
15 <
20
<3
K=
= 0.45
20
<
/lev\2
Maddy =
2000 h
<
M=
+ *Maddv
ay
= ftKb
A. + 0.87fA,
20(X)
Nhal = 0.25fcabd
20
/Ie \2
= Na.
= Myj + *M
Nu
f2
b)
M1(J = Na
<3
h
b
M = M, + *M
= (3Kh
Braced and
unbraced
= [3Kh
=_('i
\/
20<-and/or
h
2000
and/or
M=
auy
h
b
Madth
I ay
2000
Naat
//\2
F
b/
ay
3ayKb
,''2
2000 \ b /
Maddy = Na
M = + *Maddy
Mayldy =
M = Maddy
h/
2000
= Na
M1dCl1
= Madd
+ *Madth
1
(/\2
h/
=
2000
Maddx = Naur
20 <
' =i_-\
ft
Mdd(JX
NhaI
10 < -
= 13Kh
13
N-N
both axes
= ftKb
a=l---I
2000 \ b /
(lex\j2
Unbraced
10
= M1 + *Macldy
a=
13ayKb
1
ay
13
II\2
2000 b
M,dV = Na
M = + *M
* The addition of MacId will be done following sketches SK4/5 and SK4/6 as appropriate. M, is
the initial moment and M is the final moment about x-axis. Madth is the additional moment due
to slenderness.
given by
a
n
where n
= number of columns.
158
ReinforcedConcrete
MdI2.(anger initial
Madd/2Mi
,MaddMu
BRACED COLUMN FREE TO ROTATE
AT EACH END
momentMj
.Madd.Mi
M1
Madd,Mj
Madd ,M
Madd.Mi
Column restrained
Initial moment from analysis
M1
additional moments.
the full additionalmoment may be combined with the initial end moment
of stifferjoint. Madd for the other end may be reduced proportionalto the
joint stiffness.
Determine d/h correspondingto cover found in Step 3.
Find e = MIN and then c/h.
Select appropriate Table from Tables 11.8 to 11.17 correspondingto
and d/h.
f,
Calculate N/bh.
Find from the appropriateTable the value ofp which satisfies the calculated
N/bh against the e/h due to applied moment. From p calculateA.
See note in Step 4.
Step 6 Design ofcolumn
If Mr/h'> My/b',
to biaxial
bending.
M = M + (i)M
If M/b'>M/h,
M = M + (i')M
Find N/fbh.
N/fbh
0
1.00
13
0.1
0.88
0.2
0.77
0.3
0.4
0.5
0.65
0.53
0.42
>0.6
0.30
IfMy/N
O.60h
and Mr/N
O.60b
160
Reinforced Concrete
(2)
If Ms/N> O.60h
Find
v == V/bh'
v,
V/b'h
1OOA
Px
bh'
1OOA
hb'
Note:
+ O.6NVh
O.6NVb
AM
AM
vxvx
Vx
+ Vv
= vvvv
Vx+Vv
v'' = available concrete shear strength for calculation of shear
Vcv
where
reinforcementfor
about x-axis
bending
concrete shear strength for calculation of shear
v = available
reinforcementfor bending about y-axis.
fl sb
Ash
V, =
=
fAh'
0.87
sx
where
characteristic
f.. =
=
of all
Ah
=
S=
Check
t1.87fyv4sib'
S
VA
and
V.,.
162
ReinforcedConcrete
Step 1 Analysis
Note:
For short columns, the ratio 1/h should be less than 15 for braced and 10
for unbraced.
Step 3 Determination ofcover
A=0.257th2A
N = 0.4fA+ 0.75A.f
Check
(2) Column supporting continuous beams orflat slab where analysis does not
allow for distribution of moment to the column
0.35fA+ 0.67Af
Find N
Check
f,
a=
12
2000h
Madd = NaK
Combine
analysis following
conservatism.
= 0.45fA + 0.87
-N
K=-N1
and
N2
Nbal
Step 4.
Step 6 Biaxial moment and direct load
If biaxial moments are present by analysis on the column, combine these
two orthogonal moments by taking the square root of the sum of the
squares and then adding Madd to the combined moment.
Design the column for the combined moment M and the direct load N
+ Mt,)
following Step 4. M =
Step 7 Check shear stress
Find design shear forces V and V. from analysis.
Find MIN, where M=V(M+M).
V=V(V+V)
Find
If M/NO.6Oh
(1)
(2)
v=
0.75A
;0.8Vf5N/mm2
= 50A = 66.7A
0.75A
0.ONVh
AM
f,
2A,
=A
where
V.VV
See note in Step 7 of Section 4.3.1
Check
164
ReinforcedConcrete
at laps of columns
100A
100A
100A
10
4.4
0.2fA
WORKED EXAMPLES
Design
165
00
x
0
SK 4/12 Biaxially loaded column
section.
Assume end condition 2 at bottom and 3 at top for bending about x axis.
Effective height, 'ev = 1.0 X
= 8m for bracedcolumn
Assume end condition 3 at both top and bottom for bendingabout y axis.
1
b
14.4 =
=
24 >
10 for unbraced
0.6
= 20 > 15 for braced
= 8.0
0.4
Exposure= moderate
Fire resistance = 2 hours
MSA=20mm
Minimum nominal cover=3Omm, from Tables 11.6 and 11.7
Diameter of link = 10mm assumed
Diameter of main bars =40mm assumed
166
ReinforcedConcrete
= 540mm
b' = 400 30 10 20
340mm
iO
= 2400kN
NUL
Nbal
8906
2500
8906
2400
a=
(/\2hK
=x(144002x60oxo.98
2000
\600/
= 169.3mm
//\2
' =i----,bK
2000\hJ
1
/8000\2
=xI-I
x400xO.98
2000
\4001
1
= 78.4 mm
Maddx =
NaK
= 2500 x
0.1693
= 423 kNm
Mady = NaK
= 2500 x 0.0784
= 196 kNm
Design
of ReinforcedConcrete Columns
167
A5/2
M=
573
= 1061 kN
h'
0.54
276
= 812kN
b'
0.34
2500 x iO
400 x 600 x 40
bhf
=0.26
(3=0.70
M = M. +
= 573 + 0.70 x
= 880kNm
()
h' 540
k ==---=0.90
h
601)
e = 0.352
h
N
bh
0.600
= 250() x
276
M
880
e==--=0.352m
2500
0.59
i03
400 x 600
10.4 N/mm2
NIbh = 10.95.
By linear interpolation,p = 2.69 for e/h = 0.59, and NIbh = 10.4.
168
Reinforced Concrete
-4-+32
at 350 c/c
Step 7
32
Shear check
M 150
N 2500
= 0.06m < 0.60h
M
80
2500
150
400
i03
x 540
= 80x103 = 0.39N/mm
bh 600 x 340
No shear check is necessary.
satisfied
satisfied
= 8mm
Maximum spacing of links = 12 x smallest bar diameter
= 12x32=384mm
Step 11
Rectangularsection.
h=600mm b=400mrn
= 640 kNm
N = 1280kN
V = 320 kN
= 400kNm
= 20 x 1280 kNmm
= 25.6kNm
4.8
0.6
= 8<
10
l. = 4.8
= 12> 10
b
0.4
a more onerous
170
Reinforced Concrete
Step 3 Determination ofcover
Grade of concrete=40N/mm2
Exposure= severe
Fire resistance= 2 hours
Maximum size of aggregates= 20mm
Minimum nominal cover= 30mm
Diameter of link = 10mm assumed
Diameter of main bars = 25 mm assumed
= 337.5mm
= 1.5 <3
= 8 <20
Additional moment about minor axis can be ignored (see Table 4.3).
12hK
(-)
()
x 600 x 1
(assume K
= 1 for conservatism)
= 43.2mm
Maddx= Na1
= 1280 x 0.0432
= 55.3kNm
M = 640 + 55.3 (see SK 4/6 column restrained at both ends)
= 695.3kNm
M+N
= 695.3 + 1280
= 999.3 kNm
Md
fbd2
(
d1)
0.0625)
999.3 x 106
40 x 400 x 53752
Compression
z = 0.775d
= 444mm
A' = (K O.156)fbd2
0.87f(d d')
(0.216
0.87
0.156)
x 460
= 1459mm2
x 40 x 400 x 53752
x (537.5
+ A,
62.5)
(O.156fcubd2
0.87fz I
O.87f
(0.156 x 40 x 400 x 53752\
0.87 x 460 x
) + 1459
/1280 x i03
O.87 x 460
= 2319 mm2
Use 3 no. 32mm dia. bars each face (2412 mm2)
Design by using Table 11.12.
e = MIN = 0.543
k=
e/h = 0.905
--
h' = 537.5 =
0.90
N = 1280 x
bh
400
i03
x 600
5.33 N/mm2
2 x 400 x
100
600
= 4800mm2
00
(0
172
Reinforced Concrete
Step 6 Bkixial moment and direct load
Not required.
Step
-=
= 0.36m
320 x i03
400 x 536
= V
h'b
= 1.49N/mm2<5 N/mm2
h'=60040816=536mm
bOA.
bh'
x 2412
400 x 536
100
= 1.125
From Fig. 11.5,
v, = 0.77N/mm2
I)' =
V+
0.6NVh
AM
V/i
320x103x600
=
640x106
=0.30<1
V
C
= v /Ii
C\
N\
Av
= 0.771/1 +
\
400x600x0.77
1280 x i03
\
= 2.167N/mm2 > 1.73N/mm2
This modified higher value of design concrete shear strength may not be
used.
Step 8 Minimum reinforcement
Minimum reinforcement= 0.4% satisfied
x 32
= 8mm
Maximum spacingof links = 12 x dia. of bar
= 12 x 32
=384mm>350mm OK
Centre-to-centrespacing of bars= 136mm < 150mm
Central 32mm diameter bar need not be restrained.
Use 2-legged links 8mm diameter at 350 mm centres.
I1 3-32
$
3-32
j2.+20(anticrack)
8 at 350c/c
A = A. = 2412 mm2
d = 536mm
m=
-=
10
d' = 64mm
Servicebending moment,
= 400kNm
The formulae used below are for a triangular concrete stress block (see
Section 1.13.2).
Firsttrial
q1
174 ReinforcedConcrete
SK 4/18 Calculation
width.
of crack
+ mAd + (m 1)Ad'
qi+mA+(m1)A
0.5x 104000x260+(10x2412x536)+(9x2412x64)
0.Sq1x
185.8mm
400
e==
N 800
k1
O.5m = 500mm
= fe _g\
)+
185.8\
(500
1.586
k2 =
(i
_(
260
'2 x 536)
= 0.203
k3
= (m
260
3 x 536
11
=9(1_)
= 6.785
f=
Nk1
/ d'
k3A1 -800 x i03 x 1.586
/ 64
x 400 x 536 + 6.785 x 2412 xli
k2bd +
0.203
536
= 21,90N/mm2
= f(0.5cn + k3A.)
21.90
800
il)3
2412
= 289.1N/mm2
Check:
x=
1
\mf
536
1+i/
289.1
\10 X 21.9
assumed
Second trial
= 96000mm2
g = 182.2 mm
k1 = 1.593
k2 = 0.190
k3 = 6.60
q1
f ==
f.,
23.27 N/mm2
285.0 N/mm2
x = 240.8mm
= f. = 285
200x103
i0
xi
240\
Ixl.425x103
240/
= 1.733 x
=
= 1.425 x
/600
\536
Emh
i0
b(h
x)2
3EA(d x)
= 1.733 X iO
i0
= 1.612 x
= \/(642 + 642)
= 74.5 mm
a2 = \/(642 + 682)
= 77.4mm
16
16
400
200
x (600 240)2
x 2412 x (536 240)
176 ReinforcedConcrete
acr = 77.4mm
3arn.
cr
m
Cmin
\ hx
3 x 77.4 x 1.612 x
/77.4 48
1 + 21
\600 240
i03
ELEY TION
V
400
SECTION
Step 1
Analysis
Not required.
Design
177
h=h
= 600 30 10 20
= 540mm
= 400 30 10 20
=
dia. of bar
340mm
4d'60
60
M = 25OkNm
N = 250kN
M = M N( d1)
/(o.2s
178
ReinforcedConcrete
540[O.5 /(o.25
= 0.95 x 540 = 513mm
M
S
O.87fz
O9]
N
O.87f,
x 10
250 x i03
+
0.87 x 513 x 460 0.87 x 460
190
= 1550mm2
Use 2 no. 32dia. (1608mm2) bars on each short face.
Method 2 Simple steel beam theory
400
___60
_________-
O.8'a
250 x 106
= 1301mm2
0.87 x 460 X 480
Steel required for axial tension on each face =
x 250 x
0.87x460
0.5
= 312mm2
Total steel required on each face = 1301 + 312 = 1613mm2
Again, 2 no. 32dia. (1608 mm2) on each face will be adequate.
2-
32
22O(anticracl)
2-$32
section.
Note:
Both methods produce the same result but Method 2 is very conservative
usually.
V,,
250 x i03
h'b
540
400
bOA.
100 x 1608
540x400
bh'
= 0.74%
From Fig. 11.5,
v, = 0.67N/mm2
V = v. +
0.6NVh
AM
Vh
250x103x600
250x106
V=O.67
0.6
x 0.6 x 250 x
i03
400x600
= 0.295N/mm2 <
Note:
=0.60<1
1.16N/mm2
N is ye in tension.
Shear reinforcementis required.
i0 =
63.7kN
180
Reinforced Concrete
Assume 8mm diameter links (f=460N/mm2) at 100mm centres.
= 0.87fyVAh'
S
i0 x (0.87 x 460 x
100
x 540)
100
=216kN
okay
x 100
400 x 600
3216
= 1.34%
okay
x 32 = 8mm
satisfied
= 12 x dia. of bar
= 12 x 32 = 384mm
satisfied
STRAIN
DLRAM
STRESS
DIAGRAM
Assume eccentricity
compressive fibre.
Firsttrial
M 160 x iO
e==
=1000mm
160
d' = 30 + 8 + 16 = 54mm
d60030816=546mm
x = 300 =
0.55
=A=
q1
1608mm2
E=
E
m=
11)
+
+
g = 0.Sqix mAd (m
1)Ad'
qi+mA+(m1)A
= 183mm
k1
(e
= 1.167
k2 =
= 0.224
k3
= (m
1)(1
= 7.38
f=
k2bd+
Nk1
7
k3A{1
= 3.13N/mm2
f(0.5q1
d'
+ k3A3 + N
= 239.4 N/mm2
Check
x=
\mf
= 62.8 mm < 300mm
assumed
182
ReinforcedConcrete
Second trial
Assume x = 130mm
= 52000mm2
g = 157mm
k1 = 1.119
k2
k3
= 0.11
= 5.26
N/mm2
ff == 5.66
221N/mm2
x=
111mm
x = 115mm
say
221
==
1.105
x io
x\
dx
=1/600 115\J
\546
io
b(h
x)2
3EA(d x)
= 1.016 x
acr = V(542
= 140mm
cr
xl.105x103
1151
= 1.243 x
Emh
200x103
i0
1462) 16
3ac. Em
1 + 2(cr
Cmin
hx
OK
s 47000/f
300mm
f=
221N/mm2
Maximum spacing
47000/221
213mm
Note: Actual clear spacingis 260mm which does not satisfy this condition. Since
crackwidthcalculationsshowthat the crackof0.3mm may not be exceeded,
this spacing of bars need not be changed.
Example 4.4 Design of a member with biaxial moment and tension
Rectangularsection.
Size: 600mm x 400mm
Ultimate direct load in tension = 250 kN
Ultimate bending moment, MA = 250 kNm
Ultimate bending moment, M = 150kNm
Ultimate shear force, V. = 250kN
Ultimate shear force, V = 150kN
Step 1 Analysis
Not required.
Step 2 Check slenderness of member
Not required because the member is in tension.
Step 3 Determination of cover
Grade of concrete=40N/mm2
Exposure= moderate
Fire resistance required = 1 hour
Maximum size
of aggregates = 20mm
Minimumnominal cover=3Omm
from Tables 11.6 and 11.7
Diameter of link = 10mm assumed
Diameter of main bar = 40mm assumed
= 600 30 10 20
= 540mm
b' = 400 30 10 20
= 340mm
Not required.
M = 250kNm
N = 250kN (tension)
M= M
= 250 250(0.3
= l9OkNm
d'
0.06)
M = M N( d')
=
=
150
250(0.2 0.6)
115 kNm
fMf=115kNm
..I
N=Z5OkN
M19OkNm
-l--j 35330
i f
=3421
k43x33O
I
A5t
625rr1
L334
(C)
I
IT)
tlstx
I II
_____
3342J
NIOB7iy
1342
A5
a = 600
2 x 60 = 480 mm
a = 400
2 x 60 = 280 mm
0.87fa,
190x106
0.87
x 460 x 480
= 989 mm2
Assume
A1 =
0.87fa
115x106
0.87 x 460 x 280
= 1026mm2
Assume 3 no. bars of 342mm2 each on each long face.
Area of bar required at a corner of the member due to the transferred
tension
0.87f
= 250 x
0.87 x 460
= 625mm2
= 1297mm2
186
Reinforced Concrete
One no. 40 diameter bar at each corner (1257 mm2) with 1 no. 25 diameter
bar at the centre of eachface (491 mm2 each bar) will be adequate because
491 mm2 is greater than 330mm2 or 342mm2 found before.
Method2 Design as steel beam without transferredtension
My15OkNm
ttI
4464
4446
As
a=
a=
480mm
as before
280mm
as before
=
0.87fa
250x106
0.87
x 460 x 480
= 1301mm2
Assume 3 no. bars of 434 mm2 each on each short face.
M
O.87fa
150x106
0.87 x 460
x 280
= 1338.5 mm2
Assume 3 no. bars of 446mm2 each on each long face.
Area of steel required for tension= 625 mm2 as before
This area can be divided over the total number of 4 no. corner bars in the
member. Hence, use 4 no. bars of 156mm2 each.
Area of corner bars = 434 + 446 + 156
= 1036mm2 (use 40mm dia. bars = 1257mm2)
The arrangementof reinforcementis exactly the same as before. Use 4 no.
40mm dia. bars in the corners and 1 no. 25 mm dia. bar at the centre of
each face because 1 no. 25mm bar equal to 491 mm2 is bigger than
434 mm2 or 446mm2 found before.
0-40
$-25
M only.
M = 250 kNm
d=
540mm
= 40N/mm2
M
fbd2
=
d[0.5
250x106
40 x 400 x 5402
(o.2s
=0.05
= dIO.5 + 1(0.25
L
0.9
= 0.94d = 508 mm
=
0.87fz
250x106
x 460 x 508
0.87
= 1230mm2
Reinforcement requiredfor M only
M = l5OkNm
d = 340mm
K=
150
106
40 x 600 x 3402
=0.05
z = 0.94d = 320mm
150 x 106
=
0.87
x 460 x
320
= 1171mm2
= 625mm2
188
Reinforced Concrete
Try
= A(0.87f)
= 5180 x 0.87 x 460 x
= 2073kN
i0
A=2no.4)32+ 1no.25
= 2099mm2
0.87fA.
0.402fb
OK
= 481.5 mm
M = 0.87fAz = 0.87 x 460 x 2099 x 481.5
130
= 404kNm
10
150kNm
are zero
= 2099mm2
= 0.87fA. =
0.402fh
x 87 = 301mm
x 460 x 2099 x 301 x 106 = 252.6 kNm
z = d 0.45x = 3400.45
= 0.87fAz = 0.87
Unity equation
r
1)
/ Al \ 1.5
P0
\M/
/ Al \ 1.5
( ivi \ +l-J
1VIj,,,
+1-I
or
250
/250\
OK
unacceptable
Design
of ReinforcedConcrete Columns
189
square sections.
400
3- $32
2-$32
________
Step
3 $32
0.17(1
= 0.17 x
0.3
)vfbd
x
x x
0.3x250x10 Vo.8 40) 400 x 540 x
io
= 142.8kN
Similarly,
= 0.17
0.3x250x
103)
V(32)
x 600 x 340 x
i0
= 134.9kN
p v. = 0.85 x
p
V.
A SVX
S = S, =
p
0.85fd
100 mm
v, = 250
121.4
= 128.6kN
190 ReinforcedConcrete
x i03 x 100
x 400 x 540
128.6
0.85
= 70mm2
V, V, = l5OkN
vsysvy
0.85fd
x i03 x ioo
0.85 x 400 x 340
150
= 130mm2
is the larger of
and
(AIS) = 1.3
= V = 250 x i0 =
400
x 540
1.14N/mm2
bh'
400
540
b'h
=1.18
v, = 0.82 N/mm2
x i03
= 0.74 N/mm2
bd 600x340
to take into account axial tension.
Modify v. and
VCX
= V, =
=v
150
+1f0.6NVh\I
\ AM1
Vh
=0.6<1
M
maximum
bh'
=v
/0.6NVb\
\ AM II
+1
0.6
Vb
=0.4<1
of
x 250 x i03 x
400 x 600 0.4)
= 0.82 (0.6
= 0.57 N/mm2
vv
0.385
x 1.14
1.14+0.74
= 0.23N/mm2
vv
0.57
x 0.74
v + v 1.14 + 0.74
= 0.22 N/mm2
= v''bh' = 0.23 x 400 x 540 x
= 0.22 x 340 x 600 x
V=
vcy
=
i0
iO =
49.7kN
44.9kN
sx
0.87fASh'
0.87
x 460 x
157
150
x 540 x
i0 = 226.2kN
O.87fyvAbb'
vsy
0.87
Check:
V. V
VVV=15044.9=105.1kN<142.4kN
OK
OK
A.
0.4b
0.87f
x
157
= 0.4 600 =0.6<= 1.04 OK
0.87 x 460
150
192 ReinforcedConcrete
long side
Service stress,
f f
=
= x
assumed
460
= 287.5 N/mm2
47000 =
287 5
163 mm
<212mm provided
This means that to reduce the probability of the crack width exceeding
0.3mm, 4 barsshould be used on the long face, i.e. 2 no. 32dia. and 2 no.
25dia. (total 6 no. 32dia. and 4 no. 25dia. in the member).
3-
Chapter 5
Design of Corbels and Nibs
NOTATION
5.0
A
ASh
db
fs
ff
F
F1
Fbt
h
M
p
r
T
v
V
x
5.1
LOAD COMBINATIONS
193
194
Reinforced Concrete
(3) Depth at root of corbel should be such that shear stress V/bd is less
than O.8\/f or 5N/mm2, whichever is the lesser.
Design
195
(4) Depthat outer edge of loaded areashould be at least half the depth at
(5)
the root.
If is greater than d, the corbel should be designed as a cantilever
beam.
Cos
FORCE DIAGRAM
STRAIN DIAGRAM
brWIDTH
STRESS DIAGRAM
OF CORBEL
Draw strut and tie diagram as shown and find the following parameters.
V
v=
bd
with
along
= depth of lever arm x = depth
of neutral axis
O.5V
O.87f
O.87f
T
O.87f
Alternatively,
F = Fcos + T =
Va
+T
\
=
F = (\I'O.67f
15')b O.9xcosl3 O.4O2fbxcos(3
V = Fsin
z = d O.45x
By iteration, findx after assuming x in first trial. With final value of x, find
z and F. From F, find A.
196
Reinforced Concrete
Step 4 Check shear
bOA.
bd
Find
v from Figs 11.2 to 11.5 and multiply by 2dIa to getv for corbel.
effective depth d.
v)
A> bSh(v
O.87f
Total area of all legs of links in a vertical plane should be more than or
equal to O.5A.
AO.O4Obh
Step 7 Check bearingstress inside bend
The following must be satisfied:
bearing stress = Fbr4
See
CU
1+2
ab
ab,
SK 5/5
Bearing
CHOOSE SMALLESTQb
at bend.
________
SECTION 11
f=
5.3
service stress in
300
bar
IJ'
SK 5/6 Typical arrangementof
nibs.
198
Reinforced Concrete
Step 3 Determine nib geometry
(1) Bearing stress under load
0.4f
s0.6f.
40mm
Spatting allowance
for stab
Constructional
inaccuracy
of stab
iiII
I
..i
Spatting allowance
for nib
Slab
of nib
Theoretical bearin
width
_________
Theoreale
Nib
Step 4
Constructional
inaccuracy of wall
Design of nib
M = Va,,
K= M
bd2f
=
b 1 metre
Design
d[0.5
O.87fz
\/(o.25
)]
199
0.95d
per metre
Clear spacing
47000
300 mm
b = 1 metre
100A
bd
Check that
at
0.87f
l=
200
Reinforced Concrete
la
Wall
Le
26+ 2b+s
5.4
WORKED EXAMPLES
Design
201
ff
of corbel = 400mm
= 460N/mm
to reinforcement = 30mm
Assumed diameter of main reinforcement = 32mm
Assumeddiameter of horizontal links = 10 mm
Minimum cover
V
32
<
300
= 800 x
32
><
i0 = 83mm
300
I = length of corbel = a +
dia.
I+
=400+50+5x32+30+10+30
= 680mm say 700 mm
Use h = 750mm at column face.
d=75030 16=704mm
Maximum allowable shear stress at column face = 5 N/mm2
V
d>=
Sb
800x103
5x400 =400mm
Firsttrial
From strut and tie diagram (Step 3 in Section 5.2),
202 ReinforcedConcrete
Va + T
F = FcosI3 + T =
z
Fsin13
z = d 0.45x
Assume x=0.4d=282mm, say.
z = d 0.45x
cotl3 =
= 0.8218
sin
F = sin
x
= 0.6932
cosI3 = 0.5697
= 973.5kN
13
0.402fbcosI3
x i03
0.402 x 40 x 400 x 0.5697
973.5
= 265.7 mm
Secondtrial
x = 265 mm
z = 584.7 mm
cot13 = 0.6841
sin13
0.8254
cos(3 = 0.5646
F = 969.2kN
x=
266.9 mm
OK
Final z=585mm
Va
F=+T
z
= /800
><
i03 x 400\
585
+ 80 x iO = 627 x
103N
f =460
N/mm2
/0.5V\ /
1+1
\0.87f/
O.87f,
1=1200mm2
\O.S7fI
627 x i03
0.87 x 460
= 1567mm2> 1200mm2 OK
= 0.071
= 2.84
40
704
(1.568
= 0.83
100 x 1608
400 x 704
= 0.57
From Fig. 11.5,
v = 0.608N/mm2
=
()vc
a
v) = 400 x
O.8'7f,
x (2.84 2.14) =
140mm2
0.87 x 460
200
Required: 2-legged 10mm diameter links at 200 centres for the upper twothirds of d.
2
3
= 2 x 704 = 470mm
3
204
Reinforced Concrete
100
100
100
100
100
250
SK 5/11 Elevation
corbel.
of designed
32
L 400
SIDE ELEVATION
FRONT ELEVATION
Ft
\(Asreq
no. of bars )
prov
627
1567
2
1608
= 305.5kN
\/f
fbu = 0.5
(for Type 2 deformed bar as obtained from Table 3.28
of BS811O: Part 1)
= 0.5\/40 = 3.16N/mm2
2tfb
x iO
x 32 x 3.16
305.5
= 962 mm
In the column, the straight length of bar before start of bend is taken as
approximately equal to 350mm which is say one-third of the required
anchorage length. Hence
x 305.5 = 203.7kN
= 32mm
= cover + bar diameter for corner bar
= 72 + 32 = 104mm
= 104mm
FbI = 203.7 x io =
2
49.73 N/mm
128x32
2f
1+1I
2 x 40
/ 32
1+21
\104
= 49.52N/mm2 <49.73N/mm2
straight)
(includes 4 diameter
f ==
service stress
226.4 N/mm2
47000
47000
226.4
47000
< 300
fs
(from crack width calculationsin Step 9)
= 208mm
END ELEVATION
OF CORBEL
STRAIN DIAI3RAM
2bd\ 1
mAji' +l
x=Ill
b L\
AmJ
10
x 1608 1/
[1+
400
= 201 mm
2 x 400 x 704\
1608x10
z=d
= 704 201
3
= 637mm
fsb =
M
Az
200 x 106
1608
x 637
= 195.3 N/mm2
= 50
due to flexure
><
1608
load
fs = fsb + fsh
= 195.3 + 31.1
= 226.4 N/mm2
=
226.4
200x103
E.
= 1.132 x iO
(h x'\
\d X/
201\
Ix 1.132x i03
\704 201/
= 1.235 x iO
= /750
b(h
3ESAS
1.235
x 10
= 0.9866 x
x)2
(d x)
3
i0
201)2
400(750
________________________________
3 x 200
x 1608 x (704 201)
= V(882 + 462) 16
x io
83.3 mm
208 ReinforcedConcrete
3a Em
cr
+ 2(Ocr Cmin
\ hx
3
x
105 x 0.9866 x
(10530
1 + 21
201
1
i0
\750
= 0.244mm <0.3mm
Crack width criterion is satisfied.
Example 5.2 Design of concrete nib
Precastconcrete slab
units.
Clear gap between beams= 4.5m
Width of floor units = 400mm
Depth of floor units = 100mm
False floor + finish on units = 2.5kN/m2
Imposed load on floor = 5.0kN/m2
Grade of concrete for beam = C40
Assume dry bearing.
Step 1 Determine cover to reinforcement
Exposure= mild
Fire resistance = 1 hour
Grade of concrete= C40
of aggregate= 20 mm
Minimum thickness of floor =95 mm
Nominal cover =20mm
Maximum size
Design
209
= 13.5kN
= 0.4 x 40 = 16N/mm2
= 13.5 x
300
i03
x 16
40
= 2.8 mm
= 4.48m
=2x20+8x8= lO4mm<300mm
Minimum depth
Note: The depth of the nib can be reduced if 6mm diameter mild steel bars are
used or welded anchor bars are used at straight ends of flexural bars.
Step 4 Design of nib
210
Reinforced Concrete
= 43.5 x 0.5 x
= 4.05kNm/m
K=
15
x 0.12
fbd2
4.05 x 106
40 x 1000 x 812
= 0.0154
z=
d[0.5
= 0.95d =
A=
0.87fz
.V/(o.25
77mm
0.9/i
4.05 x 106
0.87 x 460 x 77
0.95d
= 131mm2/m
= 0.0013 x 1000 x
= 137mm2/m
105
1O Link in beam
V
= =
33.75
x iO
1000x81
bd
= 0.42N/mm2
bOA,
bd
x 201
1000 x 81
100
= 0.25
From Fig. 11.5,
v = 0.62N/mm2
Vc
= v2d
0.62
x 2x
81
120
V
0.87fy
33.75
x i03
0.87 x 460
= 84mm2/m
212 ReinforcedConcrete
5.5
VALUES OF zid
Design
Effective bearing
xJ
213
Effective bearing
V
Member
t4frspoiling
Lin metres
Support spatling
Material of support
Steel
Concrete Grade 30 or over
Brickwork or masonry
Concrete below Grade 30
Reinforced concrete nib less than
300mm deep
Reinforced concrete nib less than
300mm deep with vertical loop
reinforcement exceeding 12mm in
diameter
Distance y (mm)
0
15
25
25
Reinforcement at bearing of
supported member
Straight bars, horizontal loop or
vertical loop reinforcement not
exceeding 12mm diameter
Tendons or straight bars exposed at
end of member
Vertical loop reinforcement of bar
diameter exceeding 12mm
greater
0
Page blank
in original
Chapter 6
NOTATION
6.0
acr
A
b
B
c
Cmin
C
C,
Cka
d
dD
exg
eyg
f
F
h
H
Ha
H
FI
i-direction
Eccentricity of vertical end reaction from ground beams in the
y-direction
Modulusof elasticity of concrete
Modulusof elasticity of steel
Tensile stress in steel reinforcement
Characteristic yield strength of steel
Characteristiccube strength of concrete at 28 days
Frictional resistance to horizontal movementunder pad foundation
Overall depth of concrete section/thickness of pad
Effective depth of soil under foundation for computation of settlement
Active pressure on side of a foundation (kN)
Passive resistance on side of a foundation (kN)
Ultimate factored horizontal load at undersideof a foundation
Unfactored horizontal shear from column on foundation in the
x-direction
215
Reinforced Concrete
216
H1
Ka
Kh
llK
m
M
M
M
M
Mxg
Myg
Mygu
foundation
Sliding resistance
N
Nq
N
N
p0
Px
Py
P.
"Hv
qn
q,,
quit
r or R
R
s
s
T1
U,,
U0
v,
V
V,,
V,,
)Vmax
x
z
Lmax
Cm
Er
Cmii
Pcrit
pressure
Angle of internal friction
218 ReinforcedConcrete
COLUMN
A'
/c:
(
SK 6/1 Typical columnfoundation in reinforced
mm
45
concrete.
plan.
Design
Ne + Hh + M
Ne + Hh + M
In finding the load on the soil at the underside of the pad footing the
directions of the loads, eccentricitiesand moments must be taken into
account. With reversible horizontal loads and moments, all possible combinations should be examined. Eccentric heavy surcharge on part of the
backfill on foundation mayin certain casesproduce higherbearing pressure
and should be investigated.
6.1.2 Single column pads connected by ground beams (bearing pressure calculations)
,- Ground bam
/pad
foundation
SK 6/4 Typicalarrangement
of ground beams
to column foundation.
Assumptions
(1) The pad foundation is assumedrigid and its rotation is very small.
(2) The ground beam may be designed as fixed to the foundation with
zero rotation at the ends.
(3) The horizontalloads in any orthogonaldirection from all columns with
connected foundations will be algebraically added and then divided by
the number of columns.The total horizontal load in any directionwill
be shared equally between connected foundations.
(4) Because of the very high rotational stiffness of the pad foundations
relative to the ground beam, it is assumed that the horizontal loads,
momentsand load eccentricitiesat the top of the foundation will cause
220 ReinforcedConcrete
N = combined
M,
vertical load
unfactored
settlement.
beams.
SK 6/9 Bendingmomentand
eccentricityof load from ground
beams.
exg
eyg
Note: Mxg and Mvr should include the effects of dead load, live load and differential settlements on the ground beam unfactored.
Loads at underside
P=
ofpad on soil
of foundation + weight of
= moment on xx = M1 + Mxg + Ne +
(Vevg) + Hh + M*
= moment on yy = M + Myg + Ne1 + Ve) + H1h + M
H = horizontal shears = H1 and H
In finding the load on the soil at the underside of the pad footing the
directions of the loads, eccentricities and moments must be taken into
account. With reversible horizontal loads and moments,all possible combinations should be examined. Eccentric heavy surcharge on part of the
222
Reinforced Concrete
-ex(
__
- eyg
SK 6/10 Eccentricity of
surchargeon plan of pad
foundation.
xx
unfactored
= algebraic summation of
unfactored
= moment about yy =
+ Ne + Hh + M
+ Ne + H1h + M
Design
H = horizontal shears =
and
Loadfrom column
N = combinedvertical load factored
= combinedmoment about xx
factored
e = eccentricity of N in y-direction
N+
= M111 + Ne + Hh +
weight
of backfill) + 1.6
224 ReinforcedConcrete
+ Ne + H1h
H = H1
and
and
where
surchargeon backfill.
6.2.2 Single column pads connected by ground beams
Note: Use load factorsand combinations
see Section 6.1.2.
Loadfrom columns
and Mygu should include the effects of dead load, live load and
Mxgu
= M1 + Mxgu + Ne +
=
+ Mxgu + Ne +
=
where
(
(
Hh +
Vuexg) + Hh +
Ve5) +
and
and
surchargeon backfill.
6.2.3 Multiple column pads
Note:
LOAD COMBINATIONS
Loads from the columns will be combined using the following principles.
DL = dead load
IL =
imposed load
LC:
LC6:
LC7:
LC5:
LC9:
LC1:
LC11:
LC12:
226
Reinforced Concrete
LC13:
6.4
1.ODL + O.51L
(vertical direct loads only)
SIGN CONVENTION
Bor r
B
of circular
foundation
Design
of Pad Foundations
227
be available:
level
For continuousfoundation
quit = cN, + po(Nq 1) + 0.5yBN7 + p
For squarefoundation
quit = 1.3cN( + po(Nq
1)
+ O.4yBN. + p
For circularfoundation
1)
+ 0.3'1BN, + [3
a = e075
P/2)tanP
N=
q
2
cos2(45 +
= (Nq
4)12)
1)cot4)
K 1
= O.5tan4) 7l
\cos 4)
10.8 12.2
10
14.7
cN + p
35
82.()
40
141.0
45
50
298.() 800.()
228
Reinforced Concrete
Table 6.2 Values of Ne for cohesive soils (as per Reference 6).
Types of footing
DIB or D/2r
Values of N
0
0.5
1.0
6.2
1.5
9.1
2.0
2.5
3.0
3.5
4.0
9.3
9.3
9.3
9.3
9.3
0
0.5
5.2
6.2
7.1
7.7
8.1
8.2
8.2
8.2
8.2
Circular or
square footing
Strip footing
7.3
8.2
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Note: There
which
4
Medium gravel
Sandy gravel
Loose dry sand
Loose saturated sand
Dense dry sand
Dense saturated sand
Loose silty sand
Dense silty sand
Saturated clay
(degrees)
4055
4550
2834
2834
3546
3344
2022
2530
Design
Cohesivestrength, c (kN/m2)
>300
>300
>300
>300
>300
1503(X)
l5030()
150300
75ISo
75150
75150
4075
4075
2040
Note: Presumed allowable bearingcapacities for various types of soil and grades
of chalk and Keuper Marl may be obtained from Tables 2 and 3 of
BS8004: 1986.121
Coefficientof volume
compressibility.m
(m2/MN)
<0.05
<0.05
<0.05
0.050.1
0.050.1
0.100.3
0.100.3
0.100.3
0.301.5
0.301.5
>1.5
1.5B
,/"Average pressure
SK 6/14 Pressure
distributionfor
settlement computation.
0. 10 qn
Consolidationsettlement= m
where
oH
H = 1.5B (metres)
Po =
= Ckd
=C=
Po
effective pressure
(in kN/m2)
/p+o\ metres
-log
p0 Z)
of pad
Sliding resistance
where
F= Ptan
= horizontal movement of foundation into soil (metres)
Amax = maximumallowable horizontal movementon the basis of soil
shear strength (metres)
Lmax =
7K
K/,
+ 2c
For cohesive soil max = '/D
K,,
K11
= n,1/B
= k1/1.5B
232
Reinforced Concrete
Ka
Note:
K, should not
where c
Check:
PF
eO
Pi =
l)
/ P \ + /6M'
P2 =
/ P \ /6M
Friction angle,
(degrees)
____________
233
35
2931
2931
2429
2429
2429
1924
1924
1719
2226
1719
05
_____________
P2i1ii11ii1
PRESSURE DIAGRAM FOR NO LOSS OF CONTACT
P1
234
Reinforced Concrete
Table 6.7 Typical coefficients of horizontal modulus of subgrade reaction.
Values of
h (cohesionless)
(MN/rn3)
Loose
Medium
Dense
2.2
6.6
17.6
1.26
4.4
10.7
Values of k, (cohesive)
(MN/rn3)
Types of clay
Pi
uniaxial bending
Stiff
Very stiff
7.2
14.4
Hard
28.8
loss of contact
2P
(1.5A
3e)B
x = 1.5A 3e
6.6.3 Rectangular pad biaxial bending no loss of contact
Pi =
/ P \ + /6M + 7I
6Mg,
'\AB)
\AB2
/ P ' /6Mb,,, (Mrx
P2)+
/P\
P3
\AB)
/6M
l\AB2
(6MXX
\AB2
/6M\
P
p4= (
AB)
6.6.4 Rectangular pad
(6Mxx\)
\A2B)
e.
6.6.4.1
P1
e = -P
BIB + /B2
=
12
(
e
!v1xx
\
12)
236
Reinforced Concrete
B
s+B'
P2=
P3
= P4 = 0
(s + )tanc
x2
(s
B\
-)tan
AlA
t=I-+
(A2
l2Lex
tanI3=
/
Pt =
)]
12
3(B 2ev)
2(t + e)
12P
A + 2t
\/
Atan) A2 + 12t2
It - Al
21
P4=
AIPt
P2 = P3 = 0
Yi
)tan
Y2 =
A\
( jtan13
237
e e
k=+--A
B
Pi =
()k[12
3.9(6k 1)(1
2k)(2.3
2k)J
V3
s+BP1
=0
A
API
B1
x=S
Note:
To find maximum pressureat a corner the design chart in Fig. 6.4 may be
used. At the initial design stage when the size of the foundation is being
determined, this design chart becomes very useful.
Find eIA and eIB.
Read from Fig. 6.4 the value of K.
Maximum pressure =
PK
This method is valid only when there is no loss of contact between the
foundation and the soil. Use a consistentsign conventionas in Section 6.4.
A = = O.7854D2
x
Design
Z=
tD3
= O.0982D3
M = (M1 +
MD
P
Pmax
PM
+
where
p = 2M/Z.
6.7
ey
Find eccentricities
Assume
e=
--
6e and B
and e
= Mi,,
j
6e
Forbiaxial bending,
find eIA and er/B, and from Fig. 6.4 find the value of K.
PK
Maximum approximatepressure =
Check whether the maximum pressure is lower than presumed allowable
bearing capacity.
Note:
At this stage some of the loads from the self-weight of the foundation,
ground beams, backfill, eccentricitiesof surcharges, etc. have not been
included and hence a margin has to be left in the bearing pressure to
account for these. Moreover, the actual bearing pressure computations
and settlementcomputationsmayfurtherenhancethe sizeofthe foundation.
LIJcy
SK 6/24 Dimensionsof column or pedestal on
pad foundation.
d[
SK 6/25 Effective depth of pad.
Design
= UO/6
=
C2 N/12v
of Pad Foundations
241
Cl
(assume
reinforcement)
of column bars in
/P\
/H'\ 7H
ll+ll+(--------l1
\P/
\PHX/
\P/
where
Section 6.5.3.)
Step 10 Carry out analysis of bearingpressurefor bending moment and shear
Follow Section 6.2.
Step 11
(see
242
ReinforcedConcrete
321
d
d
.4
L Le/2
PLAN
Le/2
PLAN
Class of
exposure
Total SO1
2
3
4
5
(%)
<0.2
0.20.5
0.51.0
1.02.0
>2.0
Minimum cover on
blinding concrete
(mm)
Minimum cover
elsewhere (mm)
35
75
80
90
100
100
40
50
60
60
0.156
K=
JCu
z = d 0.5 +
A=
(o.2s
0.87fz
K\1
0.95d
244
Reinforced Concrete
Note: Increase depth of pad foundation if K is greater than 0.156.
Distribution of tension reinforcement
Case
If
l> 1.5
If l> 1.5 (Cj + 3d) or 1.5(C2 + 3d), whicheveris the lesser, distribute
M
orthogonal
y, or,
other words, the reinforcementin the x-directionis to resist moment about
the y axis and the effective depth is d.
resistance against moment
0.8f
where
V1
or 5 N/mm2
V2
bd
2v
245
Shear at Section 3
Check shear stress:
V3
V3
=
V.
bd
Note: Change the thickness of the pad if the shear stress at any section exceeds
the allowable limit. It is not cost-effective to provide shear reinforcement
in the pad foundation.
Step 16 Check punchingshear
SK 6/30
Punching shear
d = O.5(d + d)
U() = 2(C, + C)
U = (U0 + 12d)
Check:
v0
=
U0d
O.8\/f or 5N/mm2
N pA vc
U1d
Vl=
A1
= (C + 3.Od)(C + 3.Od)
Px_
Py_
1OOA.,
ixux
1OOA,
1_I
V
246
Reinforced Concrete
lx
Aj
ly
Change the thickness of the pad if the punching shear stress exceeds
v, otherwise shear reinforcement will be required as per Step 7 of
Section 3.3.
Note: Apply
a combined
in both directions
(f =
or over
0.75
0.5
0.3
less than 0.3
Note:
The above rules for spacing of bars in tension will in most cases ensure
adequate control of crack widths to 0.3mm where the cover does not
exceed 50 mm.
R = restraint factor.
where
Determine the valueof temperature, T1, for OPC concrete cast on ground
from the table.
Section thickness (mm)
300
17
5(X)
28
28
28
700
1000
Calculate
where
Wmax
T1(C)
e = 0.8T1ctR
c = 12 x 106/C, or values from Table 2.3 may be used
3a. E1
+ 2 (ac. Cmin
\ hx
Note: The design crack width is 0.3 mm. If this is exceeded, closer spacings of
bars may be used.
xh/2T
I
p
DETAIL
(JQ
DETAIL 1,
_________
248
Reinforced Concrete
in Section 6.3.
Find bending moment M across a critical section, as in Step 12.
= 15
m=
I(mA\2 + 2mAsd]
x=HI
b i
L\bJ
I
mA
b
Az
Es
mh =
Wmax
b(h x)2
3EA(d x)
3acr Cmin
Note:
fh-x\
Ehl
Cmin)
+ 2(acrh x
Design
0
3
A-
2D
V
BD
or
AD
O.037f.
2(C + C)D
O.O37f
O.275f
250
Reinforced Concrete
Step 23 Calculate settlement
Follow Method 2 of Section 6.5.2.
Use load combination LC13 of Section 6.3.
Note:
The settlement calculations should be carried out to give a better understanding of the global effects on the structure. It may be necessary to alter
the sizes of some of the pad foundationsin a structure in order to even out
the differential settlements. It is also important in certain cases to feed
back these settlements in the analysis of the structure.
Dead
Imposed
Wind
610
480
42
38
95
105
Frost attack.
Swelling of soil.
Suitable bearing stratum.
Step 2 Select approximate size
Presumed allowablebearing capacity from BS 8004: 198&2] = 150kN/m2
Maximum Vertical load V= 610+ 480 = 1090kN
Maximum eccentricity=
e = MIV=105/1090 = 0.1 m
Design
of Pad Foundations
251
6e=0.6mA
=7.3m2
Assume
B=3m.
Vmax
1622 x i03
VmaxUo
or d
4.38
x 1600 =231mm
[(cf + 4C2)
where C1
C1J
= 430mm
1600
U, =
=
= 267 mm
N
C2==
12v
6
1622
x i03
=300370mm
12 x 0.45
Assumed = 0.45 N/mm2 which corresponds to about 0.3% tension reinforcementfor = 30N/mm2. Chooseoveralldepth of pad equal to 500mm
allowing for adequate cover.
From field and laboratorytests the following soil parametersof the bearing
stratum are known:
ground water table = 2.Om below ground level
252
ReinforcedConcrete
p0 = p
'yh = l8kN/m2
a2
N =
2
cOS2(450
=9.19
+
' tan/K
2 \cos24
quit = 1.3cN
\
/
+ p0(N9
1)
+ 0.4'yBN + p
= 190kN/m2
= 1.ODL + 1.OIL
= 1.ODL + 1.OIL + 1.OWL
LC1:
LC3:
M1=0
foundation.
= 1.4DL + 1.61L
= 1.2DL + 1.21L + 1.2WL
= 1.4DL + 1.4WL
LC5: N = 1.4 x 610 + 1.6 x 480 = 1622kN
LC5
LC6
LC7
H=0 H=O
M1=0
M=0
for a square
Design
LC6:
LC7:
=0
= 1.2 X 105 = 126kNm
N ==1.4 x 610 == 854kN
=0
1.4 x 42 58.8kN
= 1.4 x 105 = 147kNm
1.2
x 42
50.4kN
This step may be ignored since the foundations are not connected by
ground beams and the differential settlements will have little effect on the
design of this foundation.
Floor slab
Backfill
i,,.
400x400 Column
///
3000x 3000
SK 6/35 Section through pad
foundation.
= 5 kN/m2
of surcharge = 0.75 m
254 ReinforcedConcrete
H=0 M1=0
=
P 1090 + 108 + 113.4 + 22.5
J-I=0
LC3:
= 1333.9kN
H1 = 42kN
M11 = 0
= 105 kNm
M
=
105
M=0
H=0
140.8 kNm
LC2:
9x3
190
Note: 25% overstresson allowablebearing capacity may be allowedfor combinations including wind.
P = 1356.4kN
P = 190 kN/m2 x 9m2 = l7lOkN
H1 = 42kN
PHx = 426kN
Design
P
Li 1356.4
+=
1710
'Hx
of Pad Foundations
255
42
426
=0.89<1 OK
Step 10
Carry out analysis of bearing pressure for bending moment and shear
LC5:
N1.
= 1622kN
=0
LC6:
P1.
H,11 = 0
=0
=0
1308 +
1.2
(108 +
113.4
+ 22.5)
= l6OlkN
=0
M1. = M1. + Hh +
= 126 + 50.4 x 0.45 + 1.2 x 22.5 x 0.75
= 168.9 kNm
LC7: P = N1 + 1.4 (foundation + backfill)
= 854 + 1.4 (108 + 113.4)
= 1164kN
=0
+ H14h
M1. =
= 147 + 58.8 x 0.45
= 173.5 kNm
Step 11
fVJ
_____ill
tt t
f ft
3000 x 3000
p = AB = 9 = 222.7 kN/m2
LC6:
P
p,=+
AB
1601
A2B
6x
168.9
27
2227 kN/m2
x400 Column
T.t_4oo
8
U,
3000
3000
2154
167B
kN/m2
kN/m2
JJiJJJJJJJjj142
JJJJJJJi?m2
LC6.
LC7.
P2
P
A2B
= 140.4 kN/m2
LC7:
P
Pi=+
AB
=+
1164
9
129.3
A2B
6x
173.5
27
38.5 = 167.8 kN/m2
=
+
P2 = 129.3 38.5 = 90.8 kN/m2
,jrcharqe
FLoor
Backfill
slab
015
050
-*--
-Concrete of
pad foundation
Pu
:424kN/m2
11111
130O
Pu1
=222-7kNJm2
11.
= 1500 200 =
= M, =
(Pu
1300mm
=/
Pd)Bt
2
180.3 x
3x
1.32
= 457.lkNm
Shearat section 1 =
V1
= (Pu p)Bl
= 180.3 x 3 x 1.3
= 703.2kN
Assume d = 425mm
V2
258 ReinforcedConcrete
215-4
L.
= (182.9 35.5) x 3 x 1 32
x 1.3
= 428.6 kNm
The shears at sections 1, 2 and 3 need not be checked. By inspectionthey
will be less critical than LC5.
LC7 need not be checked. By inspection it will not be critical.
Step 13 Determine cover to reinforcement
From SI report, total SO3 = 0.5%
Class of exposure= 3
(See write-up of Step 13 in Section 6.7.)
75 mm blinding concrete will be used.
Minimum cover on blinding concrete= 50mm
Assume 16mm diameter HT Type 2 deformed bars.
Effective depth of top layer (symmetrical reinforcementin both directions),
d=50050 168=426mm
z = d[0.5 +
I(o.2s
f-)]
0.95d
/(o.25 ft028)]
M 457.1 x 106
d{O.5
0.87fz
0.87
0.95
426
405mm
X 460 X 405
= 2820mm2
Use 15 no. 16 dia. Type 2 HT bars in each direction (3015mm2).
Distribution of tension reinforcement
(See write-up of Step 14 in Section 6.7.)
= C = 400mm
= 500 50 8 = 442mm
= 500 50 16 8 = 426mm
1.5(C + 3d,) = 2517mm < = 3000mm
1.5(C, + 3d) = 2589mm < 1, = 3000mm
ci,
x 2820 =
1880mm2
=
1.678
= 1120mm2/m
11+16 at175
2-16 equa
spaced
SK 6/44 Distribution of
reinforcementin pad foundation.
Use 11 no. 16mm dia. bars at 175mm centres (1149mm2/m) over the
central zone in each direction.
Use 2 no. 16mm dia. bars on each side outside the central zone.
Total number of 16mm bars used = 15(3015mm2)
All bars are HT Type 2.
260
Reinforced Concrete
Step 15 Check shear stress
(See Step 12 LC5.)
Check
v1
0.8\/f
or 5 N/mm2
Check
V2
2v
= 473.3 x
3000
i03
x 426
= 0.37 N/mm2
bd
3000
442
= 30N/mm2,
v = 0.42N/mm2 > v2
No more shear checks are necessary.
d == 442mm
d = =426mm
0.5
+
434mm
d
(442
(I0 = 2
U1
426)
--
v0
or 5N/mm2
N = 1622kN
v0 =
><
= 2.34N/mm2 <0.8Vf
Pi = Pu Pd = 180.3 kN/m2
(see Step 12)
A1 = (C + 3.0d1)(C + 3.0d)
= (400 + 3.0 x 426)(400 + 3.0 x 442) x 106
= 2.90m2
Design
of Pad Foundations
261
N p1A1
(]d
(1622
x 2.9) x i03
6808 x 434
180.3
= 0.37 N/mm2
v = 0.42 N/mm2
OK
x 500
provided
R = 0.15, say
= 28C
= 12 x 106/C
= 0.8T1cvR
= 0.8 x 28 x 12 x 10 x 0.15
= 4.032 x 10
x = h/2 = 250mm (assumed)
acr
V(742 + 137.52) 8 = 148mm
Er
3acrrr
Wmax
Cmin)
+ (2(acrhx
\
3 x 148 x 4.032 x i05
(2(148 66)
1 +
500
250
262
ReinforcedConcrete
Bottom reinforcement = 0.35b
= 0.35 x 3000
= 1050mm2
Note:
Step 21
Pd
296kN/m2
///W// 'H_______
k\w/N\
'18kN/m3
fo
L
11111
AxB
3000 x 3000
x 24 + 5 = 29.6kN/m2
Pd)B1
(150.7 29.6) x
3x
m = 15
A = 3015mm2
bi +
[(42
5 X
2mAd12
= 307kNm
b = 3000mm
mA
2 x 15 X
3015 X
3000
= 99mm
x
99
z=d=426--=393mm
=
f = zA = 307x106
x
M
393
d = 426mm
3015\
[(1
1.32
3015
259N/mm2
426]2
15 x 3015
_____
259
200x103
/hx\
b(h
acr
= 50 +
Wmax
io
10
x)2
3EA(d x)
16
148mm
x 10
= 1.588 x
Cmjn
= 150099'\
= 1.295 x
Emh
Cjn)
h x
3 x 148 x 0.773 x
i0
66)
1+ 2(148
500 99
=0.24mm<0.3mm
Note:
Step 22
OK
The crack width should be checked if the foundation level is below the
water table and the total SO3 is higher than I%.
Design mass concrete foundation
Not required.
1093.9kN
264
Reinforced Concrete
AB
121.5 kN/m2
x 3 x 1 x 18 = 162kN
=1
B
162
= 103.5 kN/m2
A = B = 3m
p0 = p
cJ is obtained from Fig. 6.2.
to cause any
2x4 -32$
H.D. BOLTS
265
H
1.0
l.()
1.0
1.0
1.0
1.0
1.0
1.0
ZIB
0.5
1.5
2.5
3.5
4.5
5.5
6.5
0.167
0.5
0.833
1.167
1.5
1.8333
2.167
7.5
2.5
oIq,,
0.8
82.8
0.52
0.35
0.22
0.16
0.10
0.08
0.075
53.82
36.22
22.77
16.56
10.35
8.28
7.76
p,
S (mm)
27
40
48
190
129
7.38
0.5
1.5
2.5
3.5
4.5
5.5
6.5
56
64
72
Ckd
35
35
35
65
65
90
90
120
6.61
5.25
2(X)
1.54
107
170
8()
149
184
165
88
2(X)
0.73
0.60
0.42
DL
IL
CLV
Al
+81)
+40
A2
+80
+50
+40
+20
+51)
+20
B1
B2
+900
CLH
V
WL2
WL1
V
2(X) 12 fl05 9
50
25
9 50 25
200 12 105 9 5() 25
105 9 50 25
105
266 ReinforcedConcrete
1
STIFFENER
Finishedfloor level
000
BASE PLATE
/
O5OO
PEOESTAL
SO
JL\
GROUT
"H.D. BOLTS
becomes 11000 x 6000. This is based on experience and may need some
revision after all the calculations are carried out.
Determine minimum thicknessof pad.
Ultimate vertical load on pad through one pedestal =
N=1.4X(8O+8O+5O+5O)+1.6X(9OO)
= 1850kN including weight of pedestal
U0=2(C+C)=2x(1400+2500)=78OOmm
C1 = (JI6 = 1300mm
C2 = N/v = 1850/0.4 = 4265 mm2
v, =
0.4 N/mm2
assumedfor
30N/mm2
From field and laboratorytests the following soil parametersof the bearing
stratum are known.
+p
p = overburden pressure = yD = 19 x
c = cohesive strength = 75 kN/m2
1.5m
= 28.5 kN/m
D = 1.5m
B = 6.Om
N = 6.7
quit = 75
= 0.25
Allowablebearing capacity =
LC1
(kN)
Al
80
3.0
DL+IL+CLV
A2
3.0
B1
120
1170
3.0
B2
70
3.0
+ CLH
Totals
LC1
Al
CLH
A2
B!
B2
DL + IL + CLV
Totals
320
120
770
3.0
70
3.0
3.0
(kN)
(kN)
+12
+210
+585
+35
270()
+600
+24
0.5
0.5
960
160
60
12
0.5
0.5
2310
+210
+385
+35
2700
+200
24
+240
+360
+40
+0.5
+0.5
3510
3.0
1280
LC2
0.5
0.5
1280
Ve
(kNm)
+360
60
+12
12
LC2
(kN)
Al
A2
225
25
3.0
DL+ CLV
Bi
1255
3.0
B2
155
3.0
+ CLH+ WL,
Totals
LC2
3.0
Al
70
3.0
DL+CLV
A2
+30
3.0
Bi
1200
3.0
B2
3.0
+ CLH WL2
Totals
1160
(kN)
+465
+113
+13
+628
+78
+21
+9
+21
+9
2700
+832
60
0.5
0.5
+210
+90
15
0.5
0.5
3600
+600
3300
+620
0.5
0.5
0.5
0.5
1160
+675
75
3765
+35
25
25
25
25
+12
100
+24
+12
Design
Note: Most of
of Pad Foundations
269
inspection it is clear that load cases LC3 and LC4 will not produce more
onerous design.
Step 5 Calculate approximate settlement
(See Method 1 in Section 6.5.2.)
Soil parameter required from soil investigation report
Assume this is not available.
From Table 6.5 of Section 6.5.2, assume
= m m2IMN
m = 0.15m2/MN
Consolidationsettlement =
qn
maH
assumed
B = width of foundation = 6m
H = 1.5B = 9.Om
0.15
Settlement = mo H =
1000
42kNt
42kMt
6kNm
Step 6
= 1248kN
270
Reinforced Concrete
= 165kN
Eccentricity,e = +1.Om
Backfilland surcharge
over the hatched area
Eccentricity,e = 1.Om
x 11 x 25 = llOOkN
= 224kN
Eccentricity, = 1.15m
Differential settlement = 10mm assumed (see Step 5)
6EI =
126kNm
for each beam
M=
E = 14 x
106kN/m2
12EI = 2M
=
= 42kN
Note: There
Load case
Load type
LC
Combination I
LC1
Vertical,
M,,
H,,
V(kN)
(kNm)
(kNm)
(kN)
(kN)
+600
2700
+24
Column vertical
Column horizontal
1280
Foundation self-weight
1248
956
165
Backfill
271
+28
+ 165
Ground slab
Surcharge on slab
Ground beam
Differential settlement
1100
224
+1100
258
504
Totals
4973
+1635
3204
+24
Vertical,
M,,
M,
V(kN)
(kNm)
(kNm)
+620
+28
3300
115
Load case
Load type
Column vertical
Column horizontal
Foundation self-weight
1160
H,
(kN)
H,,
(kN)
100
+24
1248
LC2
Backfill
956
Combination 2
Ground slab
Surcharge on slab
Ground beam
Differential settlement
165
1100
224
+ 165
+1100
258
504
Totals
4853
+1655
3919
LC2
100
+24
H may be distributedequally among all connectedfoundations. For the sake of conservatismthis has not been done.
272
Reinforced Concrete
Step
___
Jill
A = 11
Ii
r4
p2
LC1:
I p \ /6M\ toM,
p=ll(----ll---
\AB/
\AB2J
\A2B
/ 4973
\6x11/
x 1635\
\11x62!
76
76
3204
\112x6
= 75 + 25 + 26
= 126kN/m2 < 177kN/m2
(see Step 3)
Similarly,
P2 = 75
25 + 26 = 76kN/m2
25 26 = 24kN/m2
=
75
/33
= 75 + 25 26 = 74kN/m2
LC2: Pi = 74 + 25 +32 = 131 kN/m2
P2 = 8lkN/m2
/33 = l7kN/m
/34 = 67kN/m2
=5
tan = 0.09
H = lOOkN
H = 24kN
437
V(H + H)
=4.2 OK
Check horizontal bearing capacity of soil:
Pi
11
= 4950kN
437
V(1002
+ 242)
Design
PHX = PHy
of Pad Foundations
273
3300kN
P = 4853kN H = 100kN H = 24 kN
= allowable bearing capacity x area = 177 x 66 =
++=
v
P
H,
H,
PHx
'7Hv
4853
11682
++
100
24
3300
3300
11682kN
l.4DL
+l.4CLV
+l.4CLH
Al
168
3.0
A2
+112
+1610
+70
3.0
Bl
B2
Totals
3.0
3.0
Ecc.,
Ve,
Ve,
H,
(m)
(kNm)
(kNm) (kN)
0.5
0.5
+504
+336
+84
56
+0.5
+0.5
4830
+16.8
+210
+805
+35
378()
+868
M,
M,
fJ
(kNm)
(kNm) (kN)
+868
+38.6
3780
ey
+1624
(kN)
+ 16.8
+33.6
1.4DL
+l.4CLV
LC9
+1.4CLH
Load type
Column vertical
Column horizontal
Foundation self-weight
1624
Backfill
1338
Ground slab
Ground beam
Differential settlement
Totals
Note:
Vertical,
(kN)
(kN)
33.6
33.6
1747
231
+231
314
361
706
5254
+776.6
4486
274 ReinforcedConcrete
Step 11
ll
Load caseLC9
/ \ /6M\ + I
/6Mw
Pi = \ABJ + II
\AB2J
\A2B
/5254\
/6 x 776.6\ /6 x 4486
1+1
P
=(l+l
\66! \11x621
\112x6
+ 12 + 37 = 129 kN/m2
P2 = 80 12 + 37 = 105 kN/m2
p3 = 80 12 37 = 31 kN/m2
V4 = 80 + 12 37 = 55kN/m2
=
80
129 ______________________55
105
31
31
105
/ P \ /12Mx1
Ptavll+(
\ A3B
\ABI
=
80
12
x 4486 x 4.8
ii3 x 6
= 112kN/m2
Similarly,
Design
y
PA
P1
3000
E
9.0
3000
700
P2
P3
SK 6/60 Bearingpressure
locations on plan and critical
sections for bending moment and
shear.
P2,av =
P3,av =
P4.av =
P5,av =
P.av=
P7,av =
109 kNIm2
105 kN/m2
96kN/m2
8OkN/m2
64kN/m2
55 kNIm2
= 3kN/m2
Reinforced Concrete
276
1800 .1400
2300
2300
TT
H-'
I
70
58
1400 1800
4J
158
3=6x[
= 642kNm
x 1.82 +
(70 58)1.82
Column loads on A1
123 kN
(factored)
Bending
6mx
= OkNm
Shear at section 1
= 6m x
[(70
+ 65) x
= 284kN
Shear at section 2
= 6m x
[(62
1.25
+ 70) )< i--
= 495kN
Shear at section 3
= 6m x
[(70
+ 58) )<
= 691kN
Shear at section 4
= 6m x
[(70
= 423kN
+ 49) x
(1442
+ 123)
x 552
case.
Similarly bending moments and shears should be calculated for all load
cases and all critical sections parallel to the xx and yy axes following
the recommendationsin Step 12 of Section 6.7.
Step 13 Determine cover to reinforcement
Follow Step 13 of Section 6.7.
Step 14 Calculate area of tension reinforcement and distribution
Follow Step 14 of Section 6.7.
Step 15
to Step 23
32m
0E
LI)
E
C.)
,'-
SK 6/62 Roadside
signpost.
Vertical load
= l8kN
-7--
278 ReinforcedConcrete
:':
//
///
/J
//
/
/
/
Ii
H
II
;i
II
II
II
//
//
I
I
I
I
o
g
/
/
///////////
I
600x600
0.6
N = 9.1
1.5
= 460kN/m2
N = l8kN
H = 4.5kN
M = 25kNm
M=0
= 1.4DL + 1.4WL
N = 25.2kN
H = 6.3kN
LC7
M = 35kNm
Step5 Calculateapproximatesettlement
This step can be ignored.
Step 6 Carry out analysis for bearingpressure
Self-weight of foundation = 0.6m
P = 18 + 8.1 = 26.lkN
Step 7 Calculate bearing pressures
Soil is stiff to very stiff clay.
1.5B
14
1.5
x 0.6
= 0.6m
= 15.5 MNIm3
Note: Horizontalload and vertical load are treated separately to find the bearing
pressures.It is assumed that the vertical load will he carried uniformlyon
the base; size 600mm x 600mm. The horizontal load and moment will be
carried by side bearing in the manner shown.
Assumptions:
d(h-x)
155MN/rn3
155 d(h-x)MN/m2
x
PRESSURE
MODULUS OF
SUBORADE REACTION
DISPLACEMENT
if
P = 0.5KhBxd = 4.65dxMN
Q = 0.5KhB(h x)2
= 4.65(h x)2 MN
h = 0.9m
P=H+Q
or
PQ=H
1
(0.9 x)21 = 4.5 x
4.65d1x
I
L
x
or
i0
Ph
x\
Q(hx)
________
x\
M Hh = 0
or 4.65dx(hJ 4.65d(h
\
3x
31
x)3
=25x103+4.5x103x0.9
(0.9
4.65d[x(0.9
x)3]
= 29.05 x
i0
= d = 22mm
Displacement
P = 4.65dx = 47.3kN
= 42.8kN
Q = 4.65(h
x)2()
= 15.5dMN/m2
= 15.5 x 1000 x
= 341 kN/m2
22
kN/m2
= 1.5B
14
______
=
1.5
x 0.9
10.4 MN/m3
P = 0.5KhBxd
Q = 0.5KB(h
x)2
P-Q=H
h = 1.3m
M = 25kNm
H = 4.5kN
B = 0.9m
P = 34.5kN
d = 11.0mm
x = 673 mm
Q = 30.OkN
p = ll4kN/m2
j-- = 2.63
25kNm
C-,
LI _ _
106 kN/m2
FORCE
_____
DIAGRAM
______
SK 6/66 Diagrams
PRESSURE
DIAGRAM
SHEAR
DIAGRAM
MOMENT
DIAGRAM
44.3
0.9x0.9
Not required.
Step 11 to Step 21
Not required.
Z=
M
Bending tensile stress in mass concrete = =
25x106
1.215 x 10
= 0.20N/mm2
Allowablebending tensile stress = 1.85 N/mm2 OK
Shear stress
30
900 x 900
B = im
l8kN/m3
level
= 18 x 1.5m =
27kN/m2
p0 = p
c=0
Ii =
27kN/m2
as h = 0
= 41.4
2cos2(450 +
= 82.0
7K0
1JI =42.4
N=0.5tan4(-
\cos4
quit = 1.3cN + po(Nq 1) + 0.4yBN + p
= 27 (41.4 1) + 0.4 x 18 x 1 x 42.4 + 27
= 1423kN/m2
Allowable vertical bearing capacity =
= 474 kN/m2
284 ReinforcedConcrete
Maximum horizontal bearing capacity is ultimate passive resistance
given by the following equation:
= p0tan2
(450
+ 2ctan(450 +
= yhtan2
(450
= 66hkN/m2
of soil
P = 37.5 +
18
x 25kN/rn3 = 37.5kN
= 55.5kN
H = 4.5kN
M = 25kNm
Step 7 Calculatebearingpressures
Assumptions:
B=
1.Orn
= 6.6MNIm3
assumed
1.65dx2
Q = 6.6hdB (h
x)2
66h
d(
MODULUS OF
DISPLACEMENT
PRESSURE (IDEALISED)
SUBORADE REACTION
P=Q+H
M + Hh +
Q(h
P(h
=0
for idealisedtriangular
distribution
H = 4.5kN
Maximum pressure = p
= 1.65dx
also p =
66x
MN/rn2
2000
(from Step 3)
66x
2000
d = 20mm
Ultimate horizontal passive pressure on side of foundation is reached at a
depth x/2 when horizontal deformationat top reaches 20mm. It is assumed
that moment-carrying capacity of foundation through side bearing will
have a limiting value when deformationreaches 20mm at top.
Find M when
h = 1.5m
d = 20 x 103m
H = 4.5 x 103MN
B = 1.Om
or P = 0.0165x2
and
M = P(1.5
Q = 0.099(1.5
0.5x)
x)2/x
Q(1.5 x)
_________
3
P=Q+H
Hh (MNm)
286 ReinforcedConcrete
P = 19.4 x 103MN
x=
1.092m
Q = 14.9 x 103MN
M = 9.7 kNm
(maximum allowed by side bearing)
Residual moment = 25 9.7 = 15.3 kNm
By rigorous analysis:
Q = 6.6d'BJ
'!"y(X+y)
dy =
[/XY\
6.6d'B[)
Y2
= 1.ld'B(3XY + 2Y2)
x+Y=1.5
/ x\ =0
M+4.5x 1.5+QY_Py1_)
Y=Y- f
y(X+y)dy
f 0 y(X + y)dy
=
Y[1
(X13 + Y/4)
(X/2 + Y/3)
X = 1.0317m
Y = 0.4683m
d' = 8.9 mm pressure = 88.1 kN/m2
d = 19.7mm
Q = 18.57kN P = 23.O7kN
M = 13.25 kNm
= O.314Y
This gives a higher value of M and hence is less conservative.
Vertical load on foundation = P = 55.5kN
M
15.3
e===0.276m>=0.167m
P 55.5
6
(See Section 6.6.2.)
(see Step 6)
P1=
(1.5A 3e)B
2 x 55.5
x
13
(L5
xO.276) xl
= 165kN/m2 < 474 kN/m (see Step 3)
x = 1.5A 3e = 0.672m
Restrainingmoment = 55.5 x
OK
= 55.5 x 0.5
= 27.75 kNm
Overturning moment = 15.3 kNm
OK
4.5kN
25kNm
400
300_
300_
I ii
MASSJ
SK 6/68 Soil pressure diagramon
mass concrete side-bearing
foundationin cohesionless soil.
35.5kN/m
672
L65kN/
CONCRETE
288
Reinforced Concrete
Maximum pressure at x12 from groundlevel = 1.65dx MN/m2
= 1.65 x 20 x
x kN/m2
= 36kN/m2
i0
iO x 1.092
to Step 21
Not required.
f,
38.2 x
00
= 0.025N/mm2
><
negligible OK
Design
6.9
O0
O8
OL
IL
Li 09
op
Ot
Os
(I)
oe
0I
6
B
L
9
cc
0
In
0
in
0
U'
UO!4IJJ
in
1e4 jo san/e,
in
CZ/q
0-1
00
0-1
0-2
0-3
04
05
0-6
----
0-7
I I/
n-n
z/B
1-0
11
:i I
1-4
1-5
/--IIII
is
12
IE
76
17
18
1-9
2-0
2-1
22
,.
2-I
2-5
Fig. 6.2 Calculation ofmean vertical stress (o) at depth z beneath rectangulararea
a X b on surface,loadedat uniform pressure q.
LID
C)
N
C)
C
C)
LID
C
C
C
291
Fig. 6.4 Chart for calculation of maximum pressure under a rectangular base
subject to moments in two directions.
Chapter 7
Design of Piled Foundations
NOTATION
7.0
a
A
A
A.
A
A.
b
b
B
B
c
CH
d
D
Dr
eh
eh
Ef
fE.
f
f
ffd
Reinforced Concrete
294
ftp
f
h
h
h
H
H1
H
H1
H1
H1
K
1
10
11
L
L
Lb
m
m
M
M0
M
M1
M
M
M
M
M
M1
M11
of friction
strand
Effective length of pile for calculation of slendernessratio
Unsupported length of pile
Transmission length of prestressingwires or strands
Design
Madd
Madd ,
N
N
N1,
N11
N.
N,
N11,
NbaI
Pv
P
Pa
P1
q
q
qcs
R
RH
S
T
T11
T11
U
v
v
v
v
v
V
V,
V.
Vcr
of Piled Foundations
295
to slenderness
to slenderness
of pile (kN)
of
Depth lever arm
Weight
13
(3
7.1
Applied load
Ii
I
W:Weight
of pile
t
II
Psil
FSkin friction
F=End bearing
+ Psj w
where
T = P1 + W
P. = ultimate compressiveload on pile
T == ultimate tensile load on pile
skin friction resistance
= end-bearingresistance
W = weight of pile
= A(38N)
where
()
380N (Ar)
= Aq
where
Lib> I dependent on
(see Fig. 7.2).
(See
L = depth of penetration
B = width or diameter of pile
LIB should be greater than L/B as obtained from Fig. 7.2 for the value
of
Note:
Find point resistance by more than one method if soil test data allow and
take the lowest for a conservative estimate.
Determination of skin resistance
=
where A,
ff == 2NkNIm2
NkN/m2
where
= 0.005qkN/m2
where q = cone penetration resistance (kNIm2).
Third method of skin resistance
f, = q,kN/m2
298 ReinforcedConcrete
fs = 1.5q to 2qcs
8, page 603).
K. for low Dr
K0 for high Dr
Pile type
Steel
Concrete
Wood
200
0.5
1.0
0.75
0.67
1.0
1.5
2.0
4.0
Soil condition
DIB
Sands or sandy gravel
overlying stiff to very
stiff cohesive soil
<10
c=50 c=100
1.0
0.75
0.9
1.0
0.9
0.65
0.35
0.75
20
1.0
1.0
>40
>20
10
0.9
>40
1.0
10
c150
c=200 c=250
1.0
1.0
0.4
0.75
0.4
0.75
0.4
0.30
0.70
0.25
0.63
0.2
0.55
0.2
0.7
0.9
0.3
0.3
0.2
0.3
0.2
0.3
0.5
025
a25
HORIZONTAL LOAD
/EB4\i I E
kB=l.3ll
\EfIf! \1
as per Vesic, 1961 (see Reference 6).
where
= 240qkN/m3
where
q = unconfinedcompressionstrength (kN/m2).
Cohesionless soils
k = 80 1C2i/%Iq + C1 (0.5
'
BN)] kN/m3
as per Vesic (see Reference 8, page 631 and page 323, equation 98).
where
300 ReinforcedConcrete
Table 7.3 Values of Nq and N (Reference 8, page 137).
4) (degrees)
0
5
10
15
20
25
30
35
40
45
50
Nq
1.0
1.6
2.5
3.9
6.4
0.4
10.7
18.4
33.3
64.2
134.9
319.0
0.1
1.2
2.9
6.8
15.1
33.9
79.5
200.8
568.5
Rotation about
axis
0-
"-Translations about
x and y axis
Translations
7 .'"-.
Rotationabout
Note:
Translationsx, y
Rotation z
Translationsy
Rotation z
(2) Fixed head pile
Translationsx, y
Rotation z
Translationsy
Rotation z
Free at top
Free at top
Restrained at bottom
Free at bottom
Free at top
Rigid at top
Restrained at bottom
Free at bottom
Materialtype
For sustained horizontal load due to dead load, water pressure, earth
deflection computations.
Software
Use any fully validated softwarewhich has a suite for analysis of 2-D plane
frame with sprung boundaries.
Member type
S
S
where
2B
3B
S = spacing of piles
B = least width or diameter of pile.
Note: Piles carrying horizontal load should not be spaced at less than 3B.
of group.
q == unconfined
compressivestrength
effective stress at bottom of
q
pile group
Nj,,
Note: Total vertical load on a group of piles should not exceed the group
capacity. Individualpile loads inside the group will be limitedby the single
pile capacity. Piles carrying horizontal load and spaced at 3B or more need
not be checked for group effects due to horizontal load.
//
unfactoredin xx direction
unfactoredin yy direction
eccentricityof N from CG of pile group in xx direction
= eccentricity of N from CG of pile group in yy direction
eh = eccentricity of from CG of pile group in xx direction
= eccentricity of from CG of pile group in yy direction
h = depth of pile cap.
H
H
304 ReinforcedConcrete
---
H/R
TrlIz
= Heh + HYeh
Jx = y2 about xx axis passing through CG of pile group
= x2 about yy axis passing through CG of pile group
Iz = jxx + 11vy
R = number of piles in group.
Vertical load on a pile = (P'\ (MY\1 (Mx\1
+
Horizontal load on any pile = resultant of (H
R
Sign convention
Vertical loads:
downwards positive
clockwise positive
clockwise positive
and
T(x2
+ v2)
I,
306 ReinforcedConcrete
M = Hh
R
about x-axis
Design
Idealised gillage
Pile
Pile11 elements
/
7
//
//
7
/
/
/
7
/
/
/
/
--.+.
..
-.
N
N
/ I
/
/7\\\
=
z =i_LL
/l
H!
I
4Pile
4Pile
7/
raft
4 Widthof
idealised as
griIlage element
of raft
idealised
Width
elements grillage
Width of raft
idealised
flexible
analysis.
r
/
tColumn
/
h/2
hI2fiI______
Cap/Raft
TY
I
______
/
A45
45\
Pile modelled
as spring
elements
Pile
nPile
Idealised grillage
element at centre
of raft
M=
Hh
about y-axis
(7) Find horizontal load on each pile by using the following expressions:
=H
and
=H
and
(8) Find bending moments in pile,
correspondingto
to
an
end
to
corresponding
assuming
fixity
pile cap following
method in Section 7.2. Apply these moments to pile cap grillage
model as nodal loads. The pile head to pile cap connection may be
assumed as hinged and then
and
will be zero.
Find
moments
in
(9)
bending
pile cap by grillage analysis. Divide bending
moments by width of hypothetical strips of pile cap representing
grillage beams and obtain M1, M and M1 in pile cap per metre
width. Apply load factors and combinebasic load cases. Modify these
combined moments by WoodArmer method to find design bending
momentsJ''121
(10) Combine basic load cases at serviceability limit state to find reactions
at pile nodes. Compare maximum reaction with pile capacity.
Finite-element model
-Piles
formulae.
Hh
M =
about x-axis
M = H1h
i
about y-axls
Design
of Piled Foundations
309
and
DL = dead load
IL = imposed load
= wind load
EL = earthquake load.
1.2DL +
LC6:
1.4DL +
LC7:
1.ODL +
LC8:
1.4DL +
LC9:
LC1(: 1.4DL +
LC11: 1.4DL +
LC12: 1.2DL +
1.2CLV + 1.2CLH
310
Reinforced Concrete
7.6
Precast
concrete.
Prestressed
concrete.
Steel tube with closed end.
Steel tube filled with concrete.
Small-displacementpiles (driven)
P
cv
RH =
Ct-I
R1
= R1 or
where
vertical load
rated
load
CH
working
capacity of pile horizontal load
H = total horizontal load on pile cap unfactored
= (H +
.
SK 7/14 Determination
of
numberof piles.
approximate
(9)R=
1.1
where
EKE.
Note: The factor 1.1 is introduced to cater for additionalvertical loads from selfweight of pile cap, surchargeon pile caps, backfilling, etc.
Revise
312
Reinforced Concrete
Step 5 Determine size ofpile cap
Allow 1.5B from centre of pile to edge of pile cap.
Depth of pile cap is governedby the following:
Total SO1
percentage
<0.2
0.2toO.5
2
3
4
5
0.5 to 1.0
1.Oto2.0
>2.0
Minimum cover
on blinding (mm)
Minimum cover
elsewhere (mm)
35
40
50
60
75
60
80
90
100
100
fbd2
where
0.156
0.87fz
z=
d[O.5
\/(o.25
0.9/]
0.95d
Column
Enhancement
/5
33.
V=
where
P=
B=
or enhanced v( if applicable
sum of all pile reactions at ultimate loading on left of
section
width of pile cap at critical section
()
O.8\/f, or 5 N/mm2
-Critical section
diameter f
L_1
for shear
'
3$
concrete
vC (design
shear stress)
2d
Vcc
(enhanceddesign
concrete shear
stress)
Column
E-$- .
vc
[1
2d
Pile cap
The value of
can be found from Figs 11.2 to 11.5 depending on
of
tensile
reinforcementand
percentage
Shear capacity of section should be greater than or equal to applied
shear. Ultimate limit state analysis results should be used for checking
f.
shear capacity.
Step 12 Checkpunchingshear stress in pile cap
shear
around
.$15
perimeter
Design
When the spacing of piles is greater than 3 times the diameter of a pile
then the punching shear plane for column should he considered. For
rectangularpiles the plane can be considered at face of pile. The stress on
this punchingshear planeshould not exceed dependingon the percentage
of tensile reinforcement in pile cap.
Check of punching shear stress is also required at perimeter at face of
column or pile. This shear stress should not exceed O.8\/f or 5 N/mm2.
around pile-
v<o.sj5N/mm2
01 loaded area
0
in
Punching shear
perimeter
Pile cap
Punching shear
perimeter
The punching shear planes for piles will depend on location of pile with
respect to edge of pile cap.
Find the perimeter U at punchingshear plane.
v=
v.
Ud
where
i1
= 1.2
= 1.6
Rectangularpiles
N = 0.4fbh + 0.75Af
Check N applied direct load on pile.
(ii) Pile subjected to uniaxial moment
Find e = MIN and then e/h.
Find N/bhand selectappropriatetable from Tables 11.8 to 11.17 depending
on and k = dlh.
From appropriate table find p which satisfies value of N/bh for given e/h.
Find A=pbh/100.
Put A/2 on each face of pile equidistant from axis of moment.
f'
Note:
and b'.
Find Mi/h' and MIb'.
If Mi/h' > Mr/b', then
M = M + 3MV (7)
If Mr/b' > MIh', then
M, = M + 3MI
Ib'
N/fbh
1.00
0.1
0.88
0.2
0.77
0.3
0.65
0.4
0.53
0.5
0.42
0.6
0.30
a =
hK
2000 I-)
h
1 //\2
I bK
a=
2000\b/
I
Select
K= N11-N
ha
NUL = 0.45fA + 0.87fA
NbaI = 0.25fbh
NUL
A = bh
Madd
Madd
= Na
= Na
circularpile.
Use minimum
six bars
318
Reinforced Concrete
= 0.252th2
N = 0.4fA + 0.75Af
Check N applied vertical load on pile.
Madd
(assumeK = 1 conservatively)
= Na
= 4.8N/nim2
Depth (mm)
Upto400
1.0
0.95
0.9
500
600
in concrete =
/P+N\I + /My
/ \
\ A
I
+ /Mx
1P + N\
'Xx
/My
(Mx
\\
m = modular ratio
(10m)%
refer
to strand manufacturer's
brochure.
(3) Loss due to creep of concrete follow clause 4.8.5 of BS 8110: Part
1.111
iJ'
Note: Prestressedpiles designedas fixed to pile cap must extend into pile cap by
320 ReinforcedConcrete
K4
= --(mm)
vi
where
fCu
s 0.6h
and
0.6b
0.8\/f, 5N/mm2
HT,VU/bh'
and Hpxu/hb'
0.8Vf, 5 N/mm2
Shear check is necessary if:
and
and/or
H/bh'
Find Px = 100A/bh'
> 0.6b
=
Vy
Hpxu/hb'
MI,VU/NU
and
and Py = 100A/hb'
Check
vcy
Ac = bh
concrete shear stress
rO.6 NuHxub/MyuAc
Enhancement of desiqn
concrete stress= 06t'(u Hyuh/MxuAc
Hyuh/ Mxu 10
of design
Enhancement
Ac= bh
due
to
Hxub/Myu
10
v1 and
may be enhanced by using the following formulae due to
presence of an axial load N:
v. = v + 0.6NUH
v=v+
O.NUH
A
and
0.8f
5N/mm2
< 0.8Jf
5N/mm
Shear reinforcement
A
where
bS(v v)
0.87f.
Circularpiles
N=
322 ReinforcedConcrete
Asvx bxSv(VxVc'x)
O87j
bySv(VyVy)
O87iyv
Asv(a1Of link)
A5= area of alt bars
longitudinal
M/N
where
O.60h
and HIO.75A
5 N/mm2
A = O.252th2.
MIN> O.60h
p = 1OOA/1.5A
where
in tension
0.bNUH
v, +
AC1 'Pu
0.8Vf
SN/mm2
HpuhlMpu
0.87fA()
where
Find z/R from appropriate table from Tables 11.18 to 11.27 corresponding
to
h/h, p. N/R2 and eIR.
f,
V, + V
Check H11
Prestressing
LAps tofirid Vc
SECTION
ELEVATION
V0 = 0.67bh(f + 0.8fft)
VCF
V=
(i
VCO
0.55f
vbd +
___E_)
or
VCr
resistance
M,V
0.lbd\/f11
design ultimate
shear
d=
= characteristic
b
Prestressing strands
___
Uniformly prestressed
pile section
Uniform Prestress
O8Sip
21
+
Stress due to M0
M0 ZcO8Zjcp
O.4b
O.87f
Design
S
Note:
of Piled Foundations
325
0.87f..d
For biaxial bending and shear, check requirement for shear reinforcement
for each direction of bending separately, but allow for contribution of
concrete shear resistance V in one directionofloadingonly for calculation
of shear reinforcement. (See Step 7 of Section 4.3.1.)
of link.
fJ[pt
L
= Tension Anchorage
Bond Length
Percentage of reinforcement,
100AIbd (%)
1
or over
0.75
0.5
0.3
Less than 0.3
160
210
320
530
3d or 750
Step 27 Connections
See Chapter 10 for connectionof pile to pile cap and column to pile cap.
7.7 WORKED EXAMPLE
Example 7.1 Pile capfor an internal column ofa building
Size of column= 800mm x 800mm
Spacing of column =8m x 8m on plan
Design
H
H
Dead
Imposed
1610
1480
18
28
112
Wind
72
156
112
'148
624
StratumI
Average thickness of layer= 1.5m
Classification: very loose yellow brown to brownish grey sandy silt.
Average N = 3 (SPT)
c = 11.3kN/m2
= 40
= 2fikN/m3
Stratum2
Average thicknessof layer = 9 m
Classification: soft to medium bluish-grey clayey silt.
Average N = 5 (SPT)
c = 20.2 kN/m2
= 50
= 24kN/m3
Ysat = 27 kN/m3
Stratum 3
Average thicknessof layer = 2 m
Classification: stiff to very stiff bluish-grey silty clay.
= 14 (SPT)
c = 6OkN/m2
= 6
Ysat = 26kN/m3
Average N
Stratum4
Average thicknessof layer = 7 m
Classification: dense to very dense mottled brown sandy silt.
Average
N = 24 (SPT)
c = 13.8 kN/m2
= 310
'(sat = 27kN/m3
328 ReinforcedConcrete
I
uJ
Ui
uJ uJ
0><C
c
LU
>- u_
<0
VERY LOOSE YELLOW BROWN
SANDY SILT NS3
I-
-s
DENSE TO VERY DENSE MOTTLED BROWN
SANDY SILT N=24 AVERAGE
0
CN
0
g
N-
g
In
1/)
Design
Stratum5
Average thicknessof layer= 15 m
Classification: very stiff to hard silty clay.
Average
N = 31 (SPT)
c = 71.5 KN/m2
= 8
Vsat = 28kN/m3
Pu = Ppu + Psi
/Lb
A(38N)--
0.62
= 0.283m2
B = 0.fiOm
= 15280kN
Pp = 0.283 x 38 x 31 x --0.6
=
380N(A) = 380 x 31.0 x 0.283 3334kN
Secondmethodofpoint resistance
P, = A(Nc +
= 0.283m2
c = 71.5 kN/m2
= lOkN/m3
= effective vertical stressat pile point
=1.5x26+1.5x24+7.5x27+2x26+7X27+8X27
(27.5 3) x 10
= 489.5 kN/m2
330
Reinforced Concrete
00
STRATUM
526KN/n13
Lfl
c,
t24KN/m3
Water TabLe
0
o
STRATUM
5at=27KN/m'
N-
STRATUM
0
0
tsat=2GKN/m3
0
0
0
N-
tsat2?KN/m3
0
o
tsatr27KN/m3
0
0
U,
N-
STRATUM
STRATUM
L=27.5m
L/B=46
B=O.6Oni
=8
N = 15
=3
and
LIB = 3.5
L
> >
B
Design
of Piled Foundations
0
0
331
-J
li
uJ
=
Used non-displacementpile of 600mm diameter.
First method of skin resistance
f. = NkNIm
Stratum I
Pi = 3 x 2.83 = 8.5kN
Stratum 2
A.2 = it X 0.60
x 9 = 17m2
fs2 = 5kN/m2
= 5 x 17 = 8SkN
Stratum3
A3=tX0.60X2=3.8m2
f3 = l4kN/m2
P3 = 14 x 3.8 = 53.2kN
Stratum 4
A4 =
tx
0.60
x 7 = 13.2m2
f4 = 24kN/m2
P4 = 13.2 x 24 = 316.8kN
Stratum5
A5 =
tx
0.60
x 8 = 15.1m2
fs5 = 3lkN/m2
= 15.1 x 31 x 468.lkN
P1 = 931.6kN
Fourth method of skin resistance
f = cc + 0.5Ktan
= 0.75 c = 20.2kN/m2
PS12 = 0.75 x 20.2 x 17 = 257.2kN
A52 = 17m2
Stratum3
= 0.75 c = 6OkN/m2
Psi3 = 0.75 x 60 x 3.8 = 171 kN
A53 = 3.8m2
Stratum4
A54 = 13.2m2
= effective
vertical stress
at middle of layer
10
=1.5X26+1.5x24+7.5x27+2x26+3.5x27_(16_3)x
=
294 kN/m2
f=
Psi4
+ 0.5 Ktan
cw
2033kN
The fourth method of skin resistance is giving much higher values than the
first method and may be ignored from the point of view of conservatism.
Pu =
Ppu
I'su
= 719 + 932
=
1651kN
= 1651 = 660 kN
----
Designed pile is 600mm diameter bored and cast in-situconcrete pile with
an average length of 27.5m to carry a workingload of 660kN. This is a
conservative theoretical estimate of single pile vertical load capacity and
must be verified by actual pile tests on site.
Step 3 Determine horizontal capacity ofsingle pile
See Section 7.2.
Assume cohesive soil.
Method 1
= 650N
/EsB\1 7
E.
k5B=1.3lJ I
\EfIf/ \1t2
E1
= 28 x
106 kN/m2
= 7t\
I
I
D = \64
\64/
k51 B
1672kN/m2
k52B = 2909 kN/m2
k53B = 8875 kN/m2
k54B = 15914kN/m2
k5B = 20999kN/m2
k51
k52
k53
k54
k55
= 14792 kN/m3
= 26523 kN/m3
= 34998 kN/m3
2787kN/m3
= 4848kN/m3
334 ReinforcedConcrete
Method 2
k == 240qkN/m2
48OckN/m2
k1 = 480 x 11.3 =
5424 kN/m3
= SBkkN/m
00
0)
3
4
____1
5
6
7
8
9
10
I-.
12
uJ
E
13
14
15
z2
16
17 _______
18
19
20
21
22
23
24
25
:26
27
28
29
30
31
SK 7/33 Finiteelementmodelof
pile.
= 1003kN/m
Stratum 2
Spring stiffness
Stratum 3
Spring stiffness
= 0.60 x 0.60 x
= 1745kN/m
484S
= 0.60 x 0.60 x
= 5325kN/m
14792
Stratum 4
= 0.283m2
I = 6.36 x 103m4
= 8333kN/m
P
iV
3090
660
= 28 + 18 + 156 = 202 kN = H
H 202
RH = = = 2.4
CH
83
= greater of
R and
RIH
= 4.7
to 10mm
.058
0.21
5
-0.08
-0.28
-0.42
-0.51
-0.56
10
0.57
11
-0.57
12
0.56
13
-0.52
1.
-0.48
15
-0.43
16
-0.37
17
-0.31
18
0.25
19
-0.19
20
-0.14
21
-0.11
22
-0.09
23
-0.05
2t.
-0.02
25
of pile.
Step 5 Determine size ofpile cap
x 0.6 = 0.9m
Design
1800
PLAN
Spacing
= 1.ODL + 1.OJL
N=
1610 + 1480
H1 = 28 + 18
= OkN
M1
= 3090kN
= 46kN
= OkNm
3090kN
Wind in xx direction
H1 = 46 + 156 = 202kN
J1 = OkN
= OkNm
M = 184 + 624 = 808 kNm
M1
Wind in yy direction
H1 = 46kN
H = ll2kN
M1 = 448 kNm
M = 184 kNm
LC4
1800
= 1.ODL + 1.OWL
_900,
338 ReinforcedConcrete
N = l6lOkN
Wind in xx direction
H1 = 28 +
= OkN
156
= 184kN
JJ
= OkNm
M = 112 + 624 = 736kNm
M1
Wind in yy direction
= 28kN
H = ll2kN
M1 = 448kNm
M = 112kNm
H1
Estimation
cap
LC5
= 1.4DL + 1.61L
N = 1.4 x
= 4622kN
H1=1.4x28+1.6X18=68kN
= OkN
M1 = OkNm
M = 1.4 x
112
1.6
N=
1.2
72
= 272kNm
+ 1.2WL
in xx direction
H1 = 1.2 x (28 + 18 + 156) = 242.4kN
= OkN
Wind
M1
= OkNm
= 1.2 x (112 + 72 + 624) = 969.6kNm
Wind in yy direction
+ 156) = 257.6kN
Design
M OkNm
M = 1.4 (112 + 624) = 1030.4kNm
Wind in yy direction
H = 1.4 x 28 = 39.2kN
H = 1.4 x 112 = 156.8kN
M
1.4
x 448 = 627.2kNm
M = 1.4 x
112
= 156.8kNm
This condition may be ignored because it is highly unlikely that wind will
cause any tension in piles.
Step
- i--
1
0
0
U,
overall
dimensions.
B = 1.8m
L = 3.6m
= 35.9kNIm
= 0.75 say (See Section 7.1, Table 7.2)
D = 27.5m
Group friction capacity
340 ReinforcedConcrete
c = 71.5 kN/m2
at bottom of group
= effective stressat bottom of group = 489.5 kN/m2
N'= 31 for
N = 15 J
(see Step 2)
=8
= 16465kN
24461
2.5
x 71.5 + 489.5 x 3)
= 24461 kN
= 7M kN
=6x
660 =
T=
LC6
LC6
LC7
LC7
LC
LC1
LC1
LC1
LC4
LC4
0
112
184
28
68
242.4
55.2
257.6
39.2
736
112
272
969.6
220.8
1030.4
156.8
448
P=N+
627.2
0
0
537.6
0
112
46
184
448
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
e e
MVV=MV+Nek+HVh--M
156.8
134.4
0
0
46
202
184
808
0
0
3090
3090
3090
1610
1610
4622
3708
3708
2254
2254
Heh + He11
Load case
0
0
0
0
0
0
0
0
0
eh
0
0
0
0
0
0
0
0
0
0
e11
0.9
0.9
0.9
0.9
0.9
0.9
0.9
0.9
0.9
0.9
3251
3251
4562
4562
3802
3802
3802
2322
2322
5619
P or P,
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
768.3
658.6
548.8
548.8
0
0
333.2
1187.6
270.5
1262.2
192.1
137.2
225.4
989.8
225.4
901.6
0
0
0
0
0
0
0
0
0
CD
CD
39.2
=
P
768.3
I
+
1262.2
192.1
270.5
1187.6
658.6
548.8
508
711
983
925
920
717
y = 0.9m
17.82m2
x = 1.8m
(see Step 3)
of pile = 0.l2Hmm (see Step 3)
My
12.96m2
156.8
55.2
257.6
134.4
242.4
112
28
68
R = no. of piles = 6
My + Mx
V(H + H)
= 4.86m2
3251
3251
2322
5619
4562
4562
512
767
548.8
112
46
184
901.6
137.2
333.2
665
771
225.4
989.8
225.4
46
202
3802
3802
3802
2322
Qmax
P or P
H=
Qrnax
LC7
LC6
LC6
LC7
LC
LC3
LC3
LC4
LC4
LC1
Load case
Loads on pile.
373
601
367
26.94
42.93
24.21
40.40
595
890
19.24
11.33
30.67
20.18
33.67
7.67
H or
266
262
501
602
496
Qmin
66.8
106.5
100.2
60.0
28.1
47.7
76.1
50.0
19.0
83.5
M or
3.2
5.2
4.8
2.9
2.3
1.4
3.7
2.4
0.9
4.0
(mm)
Li
LJLJ
in pile cap
=487.5kNm
or
x
= 3.6 43.9
x 2.32 =
418.0kNm
= 5.4 x
or =
51.2
2
x 1.42 =
271.0kNm
= 232.3 kNm
= 2264.9 kNm
as found in Step 8.
Q- and Q.
are pile
487.5
418.0
418.0
487.5
487.5
reactions
983
925
676
717
427
Q
271.0
232.3
232.3
271.0
271.0
M2
2264.9
2172.0
1140
907.7
1134.0
M22
199.1
170.7
1816.4 1090.7 170.7
1520.1
542.0 199.1
1105.7 755.0 199.1
M11
1405
M2
1323
2007.6
813
1593.2 1026
2752.4
2590.0
2234.4
M'1'1
298.6
256.0
256.0
298.6
298.6
V44
4315= 120mm
V and V44 are combined bending moments and shears in pile cap
=600mm
M11, M77,
M'1
= 1.4 (Q + Q.)
M'2 = 0.5 (Q + Q2 + Q)
M11 = M1 + M'1
=
=
=
+
+
+
+
V'-3
Q- Q
V' Q1 Q Q V33 V13 V33
and V4 are bending moments and shears in pile cap due to dead load of pile cap + surcharge
M M2,
M1, M2, V3 and V'4 are bending moments and shears in pile cap due to pile reaction
Q, Q,
LC7
LC7
LC
983
925
882 920
542 717
684 711
937
760
890
595
844
367
657
LC5
LC6
Q2
Load case Qi
1626
2052
1138
2646
2810
2280
V44
1434
1966
1850
1596
V33
938.9
1234.9
1766.9
1679.3
1425.3
VII
1753.4
1327.4
2511.4
2024.0
2390.0
V44
346 ReinforcedConcrete
For this load case, pile fixity moment= 19.0 kNm per pile.
Pile fixity moment on pile cap is opposite in sign to moment M11 and may
be ignored.
Assume 20mm diameter reinforcement.
f
d1
b = 3.6m
z=
+
d[0.5
= 0.96d
=
./(o.25
0.95d = 760mm
M11
0.87fz
x 106 =
x 460 x 760 7447 mm2
2264.9
0.87
24 X 314 = 7536mm2
d = 900
Design
K=
M22
fbd2
1134 x
30 x 5400
106
0.011
x 7842
z = 0.95d by inspection
A. =
M22
1134 x 106
0.87 x 460 x 745
0.87fz
Area of 12mm dia. bar =
= 3803mm2
113mm2
IT
0
0.C
0)
Li
Li
Step 11
V33
1220mm
348 ReinforcedConcrete
1.5d = 1.5 x
800
a, > 1.5d1
bd
= 1200mm
x 800
x 800
v = 0.425N/mm2< 0.61 N/mm2
bd
3600
=
2d
1200mm
1.5d = 1200mm
2x800 1.333
1200
= 30N/mm2 to
cap.
v1 = 0.47N/mm2
V2 =
= 40N/mm2in pile
2 x 784
320
= 100A
bd
=4.9
100 x 3482
5400 x 784
= 0.08%
= 0.40N/mm2 for
v2 = 0.40 x 4.9 =
= 40N/mm2
1.96N/mm2
V 2511 x it)3
bd 5400 x 784
U1
Since pile spacing is not greater than 3 times diameter of pile, then
punchingshear stress at critical perimeter for column need not be checked.
U2
350
Reinforced Concrete
N =
U1d
x i03
3200 x 0.5 x (800 + 784)
4622
or
OK
983 x iO
<
><
800
= Q =
OK
983x103
4556
= 0.27N/mm2
Minimum
Vmax
= 42.93kN
420
= 30N/mm2
= 0.029
M = 28.1
= 0.029m
e=
983
= 0.095
x iO = 2.73N/mm2
600 x 600
983
Design
of Piled Foundations
351
Qm,.
x iO = iN/mm2
600x600
h2
e
R
=1
Again use minimum reinforcement.
Step 14 Check stresses in prestressed concrete piles
Not required.
Step 15 Check shear capacity ofRC pile
No shear check is necessary if
M,
N
106.5
x 10
0.60h.
=290mm
x io
= 0.60 x 600 = 360mm
367
0.60h
M/N
0.75A
0.75
x i03
x r x 6002/4
= 0.20N/mm2<0.8\/f
OK
Not required.
Step 17 Check minimum reinforcement in RC pile
100A
A
A. =
0.4
A. x 0.4
100
tx
3002 x 0.4
100
= 1131mm2
Use 6 no. 16mm dia. HT bars (1206mm2).
Step 18 Check minimum prestress in prestressed pile
Not required.
Step 19 Maximum reinforcement in pile
Not required.
352
Reinforced Concrete
Step 20 Containment ofreinforcement in pile
Minimum dia. of links = 0.25 x
Maximum spacing of links
192 mm
12
6mm
12 X 16
x 3600 x 900 =
4212 mm2
6318 mm2
32 x 201 = 6432mm2
in the yy direction.
Step 23 Curtailment
12 x 20
12 x 16
= 240mm
= 192mm
= 40N/mm2
Reinforcementused is Type 2 deformed bars.
From Table 3.29 of BS811O: Part 1: 1985,1'1
tension anchorage length =
32 = 32 x 20 = 640mm
100.4,
bd
= 100
<
7536
= 0.26%
x 800
Maximum allowedclear spacing forp less 0.3% is 3d or 750mm, whichever
3600
is less.
32; 32 x 16 = 512mm
The bars from the pile will project 600mm into pile cap. (See general
recommendationsfor design of connectionsin Chapter 10.)
::
06
efficiency.
.12
o
.4
-o
4..
(.j
20
8, deg
Chapter 8
Design of Walls
NOTATION
8.0
Ah
AT
b
b
B
c
C
d
d1
d0
e
ea
e11
e2
E
Reinforced Concrete
356
G
h
Modulusof rigidity
of wall
of flange of a shear wall section
of web of a shear wall section
hf
Thickness
Thickness
Thickness
Height of wall
Effective height
H0
K
L
m
M
M'
M1
Mo
M0
MUd(,
N
P
p
q
Qi
R
s
Sh
S
T
v,
v0
'oh
vj
V
V1
V
V0
V0
V0
X
z
of wall
Design
of Walls 357
Coefficient
0
w
8.1
ANALYSIS
OF WALLS
ELEVATION
OF WALL
L>4h
SK 8/1 Plan and elevation of
concretewall.
to vertical loads.
Braced wall does not carry any lateral loads (horizontal loads). All horizontalloads are carried by principal structural bracings or lateral supports.
Reinforced wallcontains at least the minimum quantitiesof reinforcement.
Plain wall contains either no reinforcement or less than the minimum
quantity of reinforcement.
addition
358
ReinforcedConcrete
Stocky wall is where the effective height (He) divided by the thickness(h)
does not exceed 15 for a braced wall and 10 for an unbraced wall.
Slender wall is a wall other than
8.1.1.2
8.1.1.2.1
a stocky wall.
Effective heights
=
where
Values of
Values of
0.75
0.80
0.80
0.85
1.2
1.3
1.3
1.5
Design
of Walls
359
D1>h1
H0
_h1
ENO CONDITION
D2>h1
D2< h2
H0
END CONDITION
'1
D3<h2
H1
CONDITION
H0
END
D>h3
WALL FIXED TO
FOUNDATION
FOUNDATION
end conditions.
8.1.1.2.2
H = O.75H0
H = H0
=
He = 2.OH,
______
H0
H = O.75H0
With translation restraint only at any lateral support:
He = H0
Cantilever construction:
SK 8/5 Cantileverwall.
H = 2H0
Design
8.1 .1.2.5
of Walls
361
= 1.5H()
For other walls with lateral restraints:
= 2.0H0
For cantilever plain wall:
= 3.0H0
8.1.1.3 Effective width offlanges for in-plane bending
Theeffective width is width ofwall perpendicularto directionof horizontal
loading which is considered as effective as compression flange, and also
vertical reinforcement provided in this width acts in tension as in tension
flange.Thesefactorsfor effective widtharebased on the recommendationsin
BS5400: Part 5131
7/
H
bs
ACTUAL
WIDTH
bej
2b6
7-
2xACTUAL
WIDTH
be
LJ
ACTUAL
WIDTH
iHy
2
2x
ACTUAL
WIDTH
______
-H
of a channel
362
Reinforced Concrete
be
i1
h
be
fHY
V//I
I
b
b3
ACTUAL WIDTH
ACTUAL WIDTH
b1 = 0.85 i,b1
= Wb2
b3 = 0.85 ipb3
be4 = 0.85 ipb4
be5 = 0.85 ipb5
be6 = Wbo
be2
b/H
0.05
0.10
0.20
0.40
0.60
0.80
1.00
Continuous
Cantilever
Continuous
wall
wall
wall
1.0
1.0
0.91
0.35
0.27
0.21
0.77
0.58
0.41
0.24
0.15
0.12
0.18
0.11
1.0
0.82
0.68
0.52
0.80
0.67
0.49
0.38
0.30
0.24
1.0
0.84
0.67
0.49
0.30
0.19
0.14
0.12
B2
SK 8/13 Type
I shear wall.
for horizontal force H1,
=
+ h3B
area = 0.8B2h2
bi
_________ -i
fI H,
B2
I=
(2h1B)
Design
of Walls
365
B2F_
B3
Note:
8.1.1.5.2
B/h
0.14
1.5
0.20
2
0.23
0.26
0.29
10
0.31
In a global 3-D model each wallof the open cell shear wall maybe modelled
separately as vertical stiffness elements. The property of each wall will
then include the individual torsional stiffness expressed as C= ch3B.
Torsional stiffness,
4A2
(BIh)
4A2
2B1
h1
+ 2B2
h2
The general formula for any single closed cell is given by:
4A2
fds
Jh
where A
h = thickness of wall
J=
T
where
x2
x1
x4
It
SK 8/18 Closed multiple-cell
section subject to torsion.
_________________________________________
Bi=G(xijxil Jxi+,
iI.,
fds
pi = J 7
Pi.i =
f ds
J
x4
3,
pi,i+1
f
J
ds
= 2..41G0
x; = X/2GO
X1p,1,, + X1'p,
X1'+ip,,,
= A,
Xp1
Xp1,2
+ Xp2
X',z2I-n2,n
,ziPn
Xp2,3
= A1
= A2
nP,z1.11
x;,1p,1,,, + X,',p,,
iiI
= A,,
T = 4G0 AX
When T is known, 0 can be found.
to X,.
J=
A,X1'
xl +I =
(i)x;
Modelling as
a combined unit
Design
of Walls 369
to the following:
370 ReinforcedConcrete
Findproperties of wallsystem
Find moment of inertia and shear area (follow Section 8.1.1.4).
Step 2
hE1
hEl
1.
Vi
hEl
01
Find in-plane forces in walls. After analysis the following internal forces
should be availablefor each individual wall section in the system.
M1 = in-plane bending moments
V1
Design
L
_________
Mov
371
MOM/METRE
jlttttttttttt
Mon
of WaIls
Mov
Von/METRE
___________________________
MOM
V0 =
Note: Load combinations LC2 and LC4 should be considered only when dead
and live load have beneficial effects.
where
DL
= dead load
LL = live or superimposedload
WL
H = f3HO
Check slendernessratio HIh.
372 ReinforcedConcrete
a = 3Kh
I
2000
hI
'J/Jj-
BRACED WALL
Madd/2
_____
+ Madd
BRACED WALL
___________________
walL moment
UNBRACED WALL
______________________
1985.111
J7f
'i" /.
_______
BRACED WALL
UNBRACED WALL
CANTILEVER WALL
SK 8/28 Wall additional
moments.
plane.
Step 10 Design stocky braced reinforced wall with approximately symmetrical
arrangementof slabs
Spans of slab on either side of wall within 15% and slab subjected to
uniform load.
fA + O.67Af
n total design ultimate axial load on wall.
0.35
where
d,
Conveniently
a,
SK 8/29 A typicalexampleof
analysis of a shear wall.
etc. and also distances of these groups of bars from comi.e. ati, at2, at3, etc.
face,
pression
(5) Find the following:
x)
AT =
Ast(at x)
S1 = A1 + (m
1)A,
=
where A1
area of concrete in the layer 1 of concrete in compression zone
+ Sa
x=
(x a)aS
(x a)S
mA1 + :s
M
moment
in-plane bending
Nx(e + AT x)
=
(A
fst =
axial compression
A) (x a)S
(d
- x)
x (x
x)A [f
- ae)S -
N]
Design
of Walls
375
+mf
LAYER
AVERAGE STRESS
Neutral Axis
STRESS DIAGRAM
(9) Draw stress diagram for in-plane bending moment and direct axial
load. Divide wall intounitlengths. Over each unit length convert the
average compressive stress in compressionzone into a direct load by
multiplying with the areaof the unit length of wall. This compression
force acts in combination with the out-of-plane bending moment in
that length of wall. Design reinforcement for out-of-plane bending
additional to that already provided in that unit length using Tables
11.8 to 11.17 design tables for rectangular columns.
(10) In the tension zone of the wall subject to in-plane bending moment
and axial load only, assumethe concrete as unstressed. Find reinforcement required for out-of-plane bending moment as in an RC beam
following Step 10 of Section 2.3. Add this reinforcementto reinforcement already provided for in-plane bending moment.
If reinforcementprovidedfor in-plane bending is not fully stressed
to the ultimate limit of 0.87f, then the residual capacity of this
reinforcementmaybe used to withstand out-of-plane bendingmoment.
Average tensile strain in the tensile flange may be found and
converted to an average tensileforce in the flange for computation of
reduced shear stress for out-of-plane bending. Conservatively ignore
concrete shear resistance in tension flange.
SK 8/31 Out-of-planebending of a
panel of a wall.
Step 13 Design
___LF
L/2
L12
12
________________
hf/2
SK 8/32 Analysisof a shear wall
against in-planebending.
Flanged wall
M' = M + N
where
IL
- hf
N = axial load.
K=
fbd2
sO.156
= =+
0.45
'
M'
N
(
A= (\0.87fz!
\0.87f
This reinforcement will be provided in effective width of flange in two
layers as shown. The web of the flanged wall will have the minimum
reinforcement unless dictated by out-of-plane bending moments or
reinforcement requirement as part of tension flange for other directionof
orthogonal load.
If x > hf, then follow Step 11 of Section 2.3.
The out-of-plane bending about a horizontal plane on either the wall
flange or the wall web may be due to the following:
(1) Out-of-planeframingaction with supported slab.
(2) Slenderness of wall.
(3) Eccentric loads from beam or slab or any other structure on wall.
(4) Coacting horizontal loads on wall panel due to wind, earthquake,
water pressure or earth pressure.
(5) Thermal gradient across wall thickness.
COMRES5ION ZONE
DUE TO Mi
Moh
TEN5ON
DUE
ID
ZONE
Mi
5TRE5S
DIAGRAM
and cannot take any more load. Hence, the bending moment will be
resisted by equal amountsof compressive and tensilesteel with a lever arm
equal to the distance between the two layers of steel.
In the tension zone of concrete wall due to in-plane bending moment,
assume that the concrete is unstressed and use the beam theory to find
reinforcementdue to out-of-plane bending moment.
Thereinforcementrequired due to out-of-planebendingmoment will be
added to the reinforcementfound for in-plane bending.
Step 14 Design ofreinforced wall shortand squat cantilever deep-beamapproach
L
Mi
V
/
______________
4Sf-.
Mo
J (5L-2H)/40
PLAN
(5L-2H)/20
This approach may be used for wallswith total height less than or equal to
their length. For in-plane bending consider the wall as a deep-beam and
follow the deep-beam theory of stress distribution.
For horizontal loading to resist in-plane bending moment
oH
H
when70.5
z=--
= 2(H+L)
when0.5<1
2H
20
15L 2H
=L
40
IL
5L2H
M' = M + N I ______
40
IN
A = IM'\
0.87fz) o.87j
Note:
The flexural strain in concrete is very small in a short and squat cantilever
wall and for all practical purposesmay be ignored when designing for the
Design
of Walls
379
d,
hf/2
___________
SK 8/35 Design for in-plane shear.
h/2
In-plane bending
1OOA,
pi =
where
v1
-V.
Lw I
where
v1
O.8\/f,
V1
5N/mm2
380
Reinforced Concrete
11.5)
(see
Out-of-plane bending
IOOAS()
bd,
where
O.8\Jf, s 5 N/mm2
Note:
In Step 12 and Step 13 the wall is designedas flanged beams for in-plane
loading. For out-of-planeshear in the flanges which acts together with the
in-planeloading,the check should be carried out separatelyforcompression
and tension flange. For compression flange the enhancement of design
shear stressdueto axial load may be allowed based on average compressive
stress. For tension flange the concrete may be conservatively ignored and
theshear force will be totallycarried by shear reinforcement.Alternatively,
average tensile strain in concrete may be found and the shear stress
reduction formula may be used.
Design
of Walls
381
Plan
Case 1:
V1
V0h
= /(1
Elevation
Side Elevation
<
oi'\
V,/
V()
O.87AhfYdj
SI
where
V5
of vertical shear
A Ah
sv
Ss
where
et 'i
'oh + V
Reinforced Concrete
382
vco
V0 Voh
+
v = available
V0h
where
V1
V, = hd1
V0 = V() bd0
Provide shear reinforcement in in-plane direction, as
satisfies
V
,
- V)
SK 8/38 Out-of-planeshear
reinforcementin walls.
I
I
I
I
I
S
S
.
..
(V1
--I
.J
JI
44
I
I
.J
41
in Case 1, which
width
S
S
b1Unit
____
S
S
CC-CCU[
________________
Side Elevation
Elevation
Plan
Vso
Note:
0.87fA5d
where
A5
Check
V0 V0
After local analysis of wall panel the bending moments, direct loads and
shears about the vertical plane in the panel are obtained.
Design
Mov
_
C_
of Walls
M0v
Nov
N0-d)
Mov
MovCi_Nov
NovLb
Mov
M0VcNONOV
Mov
Mov
Etevation
of Wall
Panel
NQ
I-4T+Vov
383
V1-+1
1-.INo
1
_____
0.3 (h 2e)f
wall (minimumvalue of
e is h/20).
nn
0.3
0.3
(h
(h
2e)f
1.2e
2ea)fcu
=
2500h
where He
8.1.1.2).
where
e1,2
0.3 (h 2ei)f
0.3 [h 2 (e2 + ea)]fcu
= resultant eccentricity of loads at top of wall
= resultant eccentricity of loads at bottom of wall.
Check
v=
where
v = shear stress
b = unit length(mm).
0.45N/mm2
Note: A plain wall subjected to in-plane shear should satisfy at least one
above checks.
of the
(f
cross-section
Step 21
(f
of gross cross-section
For vertical compressionreinforcementin walls up to 2% of gross crosssectionalarea, use the following minimum horizontal bars:
0.25% of gross concrete area
Horizontal bar diameter should be greater than or equal to size of
vertical bars but not less than 6mm diameter.
For vertical compressionreinforcementin walls greater than 2% of gross
cross-sectional area, use links through the thicknessof wall.
Dia. of linksdia. of vertical bars or 6mm, whichever is greater
in horizontal and vertical direction
Spacing of links 2h
Spacing of links in vertical direction should not be more than 16 times
vertical bar diameter.
Design
e(0
of Walls
385
200
,
'I
.: ::
c'J
VI
r--
L
---
'l
1.
..
Elevation
I, .. -.
SK 8/41 Detailing rules for walls.
Plan
2h
of Wall
_k
_fl_ 1/4
L____Diarneter
oflt
WORKED EXAMPLE
of the
the centroid
x 600 x 112O0
and 4 (ignoringflanges)
70.25m4
x 11200 = 5.38m2
Equivalent beam elements 2 and 3 (ignoring flanges)
Shear area = 0.8 x 600
=
12
in x-direction
-x = x
Shear area
600
0.8
24000 = 691.2m4
Note:
Step 2
of torsional
vertical beam element.
SK 8/43 Location
By taking moments
+ 2 X 0.6 x 24 x 12
___________
6 + 4 + 4 + 6 + 2 x 14.4
11.8m
P1
(lenththicknessof arm
=1(2_x_8.5\
0.6
= 72.5
/10.6\
(10.6
\0.6/
\0.4
1)
Similarly,
P2= (2
x 44\
0.6
= 67.7
x 10.5\
P3= (2
0.6
/2 x 10.6\
\ 04 )
,
/10.6\
\0.61
(10.6\
= 79.2
P1.2 =
P2.3
10.6
= 26.5
= 10.6 = 26.5
Substituting
388
ReinforcedConcrete
Xp1 Xp12 + 0 = A1
V'
Xp12+ A2f32
Jt3P23
A
t12
Xp23 + Xp3 = A3
or 72.5X 26.5X + 0 = 90.1
26.5X + 67.7X 26.5X = 46.6
0 26.5X + 79.2X = 111.3
0
X = 2.38m
X = 2.11m
X3 = 2.2m
J=
(A,X:)
2.38
+ 111.3 x 2.2)
It is always useful to check at this stage the torsional rigidity of the outer
cell, ignoring the internal dividing walls. This gives confidence in the
numerical accuracy of the analysis.
For single outer cell:
4A2
(B/h)
4 x (23.4 x
10.6)2
(2 x 23.4/0.6) + (2 x
= 2171m4
10.6/0.6)
xl
.x2
x3
.4
CELL
x,.2
x23
xl
x3
CELL
CELL
xl
x2
/2T\
Xi =
X2 =
/2T
x3
= 2.18
X 1O3TkN/m
= 2.18 x 1O3TkN/m
diagram.
Design
x3
()x; =
2.02
of Walls 389
x 103TkNlm
Global torsion
+50000kNm (clockwise) for horizontal load in -dirction
= +40000kNm (clockwise)for horizontal load in xdrectIon
Wall no.
Load
case
DL
LL
WL(y)
DL
LL
WL(y)
DL
3
LL
Local
(kN)
(kN/m)
+20000
+1335()
+1200
+13350
+1200
+8
980
615
3180
1325
+20000
+ 1700
101
+35000
+3000
+87
+3u0()
87
LL
WL(x)
DL
LL
WL(x)
7200
3125
7200
3125
+3500()
SK S145Elevation Wall 1.
+97
+1700
LL
DL
V1
(kNm)
3180
1325
980
615
shear flow
M1
WL(y)
DL
WL(y)
5
N
(kN)
///7
10600
390
Reinforced Concrete
Step5 Carry out local analysis
Find out-of-planeinternal forces in wall panels (follow Section 8.1.2.2).
After analysisthe following internal forces are reported:
Wall no.
Line
Load
M011
case
Vo
M0
(kNm/m)
(kN/m)
(kNm/m)
DL
28
26
LL
WL
DL
B
o
0
0
IF
CE
20
Vov
(kN/m)
WL
10
32
30
DL
LL
WL
28
26
DL
10
5
32
30
LL
LL
WL
GJ
(..J
D
H
M
////////////////
///////
///////,'7///////'
8500
- 4400
10500
Design
of Walls
391
N = 4452kN
M1 = 28000 kNm
V1
2380kN
1.4)
rr1
SK 8/47 Section through WaIl
= 30N/mm2
f, = 460N/mm2
= (3 H0
H0 = clear height = 12.Om
Monolithic construction at
He = 15600
= 26 > 10 < 30
3)2 ReinforcedConcrete
Step 8 Find effective width offlanges
2550
J
b=-=4250
H
w
H=12000
= 0.35
= 0.53 for loading at top of wall
a=Kh,H=fH0=0.8x 120009600
Assume
1
K = 1 for conservatism.
/HC\2
256
==0.128
= 0.128h = 76.8mm
Madd = Na (out-of-plane)
= 4452 x 0.0768
= 342kNm
= 342
10.6
= 32.3 kNm/m
Design
of WaIls 393
LL
162
162
slenderness.
Step 12 Design
0
8
0
394 ReinforcedConcrete
OOxl000xO.40 =
2400mm2/m
x = 3000mm assumed.
(3) Divide compression zone into convenient layers of concrete.
(4) Divide tension zone into convenient layers of steel.
(5) Find the following in completingthe table:
S = A + (m 1)A
= (x a)S
C2 = (x a)aS
C3 = (a x)A
C4 = (a1 x)atA
C1
Number
A
(x
1.53
2
3
0.72
0.72
4
5
6
Totals
(x 106)
106)
6120
2880
2880
1.62
2.97
11880
3.14
300
1200
0.76
0.76
A1
3760
3648
5280
3648
3648
6800
3648
8320
3648
9840
6120 10900
24360
2400
C1
C3
C2
4.374
1.312
2.772
1.042
1.368
1.642
1.094
8.317
4.392
13.862
19.407
16.147
24.952
24.553
48.348
52.699
6.198
4.048
117.658
108.259
0.456
AT
= c4 =
x i09
9201mm
+
X = mA1a
Sa = 1735mm
mA1 + S
28000x103
4452
=6289mm
4452x i03
= 3.467 N/mm2
C4
9.426
fstL (atd
F
= /11100
I
iF(f\
IH(xa)S-N
x)AJL\x/
3000
F/3.467\
I x I 1 I x 6.198 x i09 4452 x io
\117.658 x
L\30001
106/
186.6 N/mm2
mf
f == 3.66N/mm2
f.,t
182.4N/mm2
Check
x = 2570mm.
3.25 N/mm2
stress
:83 N/mm2
396 ReinforcedConcrete
15
for m =
15
= 48.75 N/mm2
Over a unit length of wall,
compressive force, N = 3.25
x 600 = 1950kN
= 39 +
(see Step 9)
---
= 55.2 kNm/m
..
d565
35
.-..
. . . .. .
h=600
-f
- _J
Material strengthschosen:
= 30N/mm2
d
565
f=
k===0.95
460N/mm2
M
e==0.028m
= 0.047
Forp = 0.4,
Availabletensile force in bars per metre length of wall per face of wall
= (400 183) x 1200 (area on each face)
= 260.4kN/m
Design
of WaIls 397
fbd2
55.2 x
30 x 1000
= 5.76 x
10
x 5652
i0
dLO.5
Ii'
l0.25
K\1
0.95d
0.95d = 537mm
M
=
Required tensile force in bars =
z
55.2
x io =
102.8 kN/m
537
Not required. The design principle is exactly similar to beam design and
has not been illustrated.
Step 14 Design
= 20640m2
600 x 10900
398
Reinforced Concrete
= 30N/mm2
From Figs 11.2 to
11.5,
v = 0.47N/mm2
= combined in-plane shear
= V1 + Q = 2380 + 1440 = 3820kN
V,
v1
x
= 100A0 = 100
1000
bd0
1200
x 565
= 0.21/o
to 11.5,
v = 0.4N/mm2
= 4.2kN/mm
V0h
V1
on the web
= 0.007N/mm2
l l
v0 \0.471
Shear reinforcement is necessary for in-plane shear.
v0\
= 71 jv
\ V0/
= / 0.007\
---)
)<
0.47
= 0.46 N/mm2
V( =
V1
vhd1 = 3008.4kN
V
Sh
O.87fd,
811.6 x
=
iC)3
= 0.19
If Sh = 300, then Ah = 300x 0.19 = 57mm2 which is 29 mm2 of horizontal
bar on each face at 300mm centres, or, 97mm2 per metre on each face.
=
= 0.19
forf = 460N/mm2
Vertical shear reinforcementadditionalto vertical bars providedfor bending is required if availablevertical bars have no residual capacity.
In the web 2400mm2/m vertical bars are available at a maximum average
stress level of, say, 160N/mm2 (see Step 12). Hence residual capacity
availablein vertical bars in web = 0.87 x 460 160 = 240N/mm2
Modified
= 0.19 x 0.87 x
= 0.32
(modified)
M0 = 1.4 x 28
= 39.2 kNm/m
V0, = 1.4 x 26
= 36.4kN/m
(see Step 5)
550
400 ReinforcedConcrete
K=
fbd2
30
39.2 x 1O
=4.3x103
x 1000 x 5502
d[O.5 \/(o.25
= 0.95d = 522mm
0.87fz
39.2 x 10'
0.87 x 460 x 522
0.95d
= 187.6 mm2
= 187.6 + 97 = 284.6mm2/m
V0 = 36.4 x io = 0.07N/mm2
v0 = --j
1000 x 550
Shear stress is negligible.
(f
Not required.
Step 22 Check containment of wall reinforcement
Vertical reinforcement is less than 2% of gross concrete area.
(20mm)
R = 0.8 at base.
= 32C
= 0.8T1cR
= 0.8 x 32 x
12
i0
50 5 = 65.7)
take x = h/2
max
3acr Er
+ 2(a1
Cmin)
h x
x
3 65.7 x 2.46 x
1+ 2(65.7300
45)
=0.O4mm<0.3mm OK
Check vertical bars for horizontal cracks
J-
25O
125
125
acr =
120 mm
Wmax
25)
=0.O5mm<0.3mm OK
Step 24 Clear spacings of bars in tension
Reinforcementprovided is 20mm diameter at 250mm centres both faces
vertically and 10mm diameter at 100mm centres both faces horizontally.
These spacings satisfy the requirements according to Step 13 of Section
3.3.
Step 25 Connections
Follow Chapter 10.
Chapter 9
9.0 NOTATION
A
b
C.
d
dh
Gk
l
'ho
'h,max
12
M'
M1
Mtmax
Q
V1
Veff
Wk
x
9.1
DEFINITIONS
Flat slab is a reinforced concrete slab supported by columns with, or
without, drops. The columns may be with, or without, column heads.
Drop is a local thickeningof the slab in the region of the column.
403
404
Reinforced Concrete
section. SK
section.
Column head is a local enlargementof the column at the junction with the
slab.
9.2
ANALYSIS
OF FLAT SLABS
I
Lx[
__________
LLY],
Ly
>
Lx
is limited by a 45 dispersionof
h == effective
(4A/t)0.251
diameter of column or column head
h=
A
4,
= l.4Gk + i.Qk
l.4Gk + l.Qk
on all spans
on alternate spans and otherspans loaded with
l.OGk
406 ReinforcedConcrete
M' =
()(i
2h)2
I
SPAN A
Li
SPAN B
Ii
5PAN C
SK 9/7 Negativemoment
limitationfor flat slabs section.
where
12
+ M4) + M7
M'
Increasenegative moments M1, M2, M3, etc. until these conditions are
satisfied.
tOF
COLUMN
STRIP
MIDDLE
5TRIP
OF COL
I_______
CDL
Lx/2
Ly>Lx
COLUMN
STRIP
408
Reinforced Concrete
division of
strips.
Panels are divided into column strips and middle strips as shown.
For slab without drop the column strip is 1/4 wide on either side of the
centreline of column, where l. is the shorter span.
For slab with drop the column strip is the size of the drop. Ignore drop if
the size of the drop is less than 1/3.
9.3.2 Divisionof moments between columns and middle strips
The moments obtained from analysis of frames should be divided as
follows (these percentages are for slabs without drops):
Negative
Positive
Column strip
Middle strip
75%
55%
25%
45%
Note: Where column drops are used and column strips are determined from the
width of the drop, it mayso happen that the middle strip is bigger than the
middle strip in a slab without drop. In that case the momentsin the middle
strip will be proportionately increased and those in the column strip
decreased to keep the total positive and negative moment unchanged.
1/6
At
-116
SK 9/11 Detailingof
reinforcement in flat slabs.
At
Column
r:l
cx
L beCxCy J
for transfer of
to find effective width of slab
momentbetweenflat slab and edgecolumn. This moment should be
See sketches above
limited
to
where
= 0.15bd2f
The moment Mt,max should not be less than half the design moment from
an equivalentframe analysis or 70% of the design moment from a grillage
or finite element analysis. The structural arrangement may be changedif
Mt.max does not satisfy the above condition.
M2
MOPIENT FROM
ANALYSIS
EDGE
REDISTRIBUTED
MOMENT TO
ACCOUNT FOR
Mt. max.
to the column.
9.3.5 Shear in flat slabs
v(i + 1.5Mt)
vtx
Design
of Flat Slabs
411
dlMt
nfl
PUNCHING 5HEAR
PERIMETER
where
SK 9/15Definition of dimension x.
by frame analysis
length of side of perimeter considered parallel to axis of
bending.
Alternatively,
= 1.15V
for simplicity
1.5M\
.5+
2
Vt(1
edge
Alternatively,
Vff 1.4V1
The moment M1 may be reduced by 30% where the equivalent frame
analysis is used and both load cases LC1 and LC2 have been considered.
The shear reinforcementwill becalculatedaccording to Step 7ofSection 3.3.
412
ReinforcedConcrete
Find moment connection to edge column as per Section 9.3.4 and redistribute moments if necessary.
Step 3
Step 6 Divide moments between column strips and middle strips as per Section
9.3.2.
Step 7 Determine cover to reinforcement(see Step 3
Step 8 Carry out design for flexure as per Step 4
of Section 3.3).
of Section 3.3.
Staircase and
Lift Block
414
Reinforced Concrete
1200
T
J2OO
[1100
8
"I
q=kV=1kN/m2
= +0.7 and 0.3
C=
C,1
0.3
The above wind pressure coefficients are obtained from CP3: Chapter V
Wind loadsJ141
Live load on roof= 1.5kN/m2
Live load on floor= 5 kN/m
Floor slab has 2000 x 2000 drop at columns
Thickness of roof and floor slab = 200mm
Thicknessat drop of floor slab= 400mm
Continuousperimeter edge beam 400 wide X 800 deep
Centre-to-centreheight of floor = 4.5m
h=
=
l
0.25l =
5000mm
1250mm
()
0.25l
= 0.2 x 25 = 5kN/m2
Qk = 1.5kN/m2
LC1 = l.4Gk
on alternate spans
LC2 = 9.4kN/m2 and 5kNIm2
or LC2 = 56.4kNJm and 3OkN/m
on alternate spans
Floor slab
Gk = 5 kN/m2
at slab without
drop
= lOkN/m2 at slab with drop
(area 2m x 2m)
= 3OkN/m or 30 + 5 x 2 = 4OkN/m
Qk = 5kNIm2
= 3OkN/m
LC1 = l.4Gk + i.6Qk
LC2 = alternate spansloaded with LC and dead load only
Columns
= l.4Gk + l.4W
= 1.2Gk + l.2Qk + l.2Wk
LC3
LC4
14 members
dead
load
5000
5000
5000
12
,- Member
Numbers
0
0
U,
511_
0
0
U,
Joint
10
analysis.
3OkN/m
IIUIIIIUIIIIIIIIIIIII
6
''''''''J.[U'''''1'
25
3
1O9
'ITI\lull!
1T
I1h
4OkN Fm
B6
B5
111111111
11111111]
()
B2
1111111 IllIlIllIlIli
10
B7
()
63
1.
12
i111IIlUh1H111lI
12
3OkN/m
f
-
- -ll!lllIIlIl!1lIIIlllI
IllllIlIIIlII11Ill!
3OkN/m
11
77
64
10
7-
Combinations:
C1=1.4B1+1.6(B2+B3+B4+B5+B6+B7)
C2 = 1.4B1 + 1.6(B2 + B4 + 85 + B7)
C3
C4
of Flat Slabs
Design
417
Outputfrom analysis
Envelope of load cases (vertical loads)
Elastic analysis
Floor slab
Joint
2
5
Midspan
no redistribution
Member 9
Maximum BM
130.80
215.5
Shear
Combination
228.5
258.2
C2
C4
131.2
C2
Shear
239.0
239.0
Combination
Floor slab
Member 10
Joint
5
8
Midspan
Maximum BM
Roof slab
Joint
Member12
3
6
Midspan
Roof slab
Joint
6
9
Midspan
199.9
199.9
112.5
C'4
Maximum BM
67.8
Shear
130.0
Combination
129.0
153.5
C1
82.0
C2
Shear
141.0
141.0
Combination
Member 13
Maximum BM
121.3
121.3
C1
60.4
'\
C3
,4.,
'-.
'
I '
/I 'S
".
SI
SK 9/22 Combination C1
bending momentdiagram.
C4
".
-/
,"
'S
5
".
I
/
'\
,
j
"11
10/
Envelope of load cases (vertical + horizontal loads) (The analysis of horizontal load is carried out with half stiffness of slab)
Elastic analysis no redistribution
Combinations:
C5 = 1.4B1 + L4B
C6 = 1.2B+ 1.2(B1+ B3 + B4 + B5
+ Bh + B7) + 1.2B5
Floor slab
Joint
Member 9
Maximum BM
Shear
Combination
2
5
Midspan
121.9
185.9
215.2
Co
Floor slab
Joint
Member 10
Maximum BM
Shear
Combination
198.6
C6
176.0
176.0
Midspan
71.4
191.4
95.8
198.6
Co
Co
C6
C6
Roof slab
Member 12
Joint
3
6
Midspan
Maximum BM
57.0
Shear
112.2
128.9
Roof slab
Member13
Joint
6
9
Midspan
Maximum BM
Shear
108.1
108.1
119.8
119.8
108
67.1
45.4
Combination
C6
C6
Co
Combination
C6
C6
Co
2155kNm
En'Iope
kN
Design
of Flat Slabs
kNm
419
9kNm
Envelopq
2390kN
SK 9/24 Shear and momentenvelope for member 10.
rPLac Hinges
74
F.
--
'p
10
91
,..7
.,.
,7
10
420
Reinforced Concrete
rPLastic Hinges
,qPtastlc Hinges
_______9_lI
12
,1
1
;r
10
-
2
Ii
()
1
8 $
r7
10
B5
B6
B7
12
17
R2
Qi)
7
Plastic Hinges
Qq)
_9_
f6[
lkNIm
11
10
10
A3
Il11fllHfl iHUHUUfl
2
B
x dead load
or 1.4B1
The method is illustrated for combinations C and C only.
C = 1.6(B + B + B + B,)
B, B, B = lkN/m
B, B,, B, = 0.3kN/m
Combination Frame
type
Memberend bendingmoments(kNm)
Member9
Joint 2 Joint 5
C
C
C
C
C
C
C
C
C
C
F1
F1
62.7
2.0
F2
2.1
3.1
3.1
3.1
F4
F6
F1
F2
F3
F4
F5
2.3
2.3
3.1
3.1
3.1
102.5
3.7
3.6
0
0
0
2.7
2.6
0
0
0
Member 10
Member
12
Member 13
95.5
44.7
97.5
3.4
0.7
0.9
1.1
3.5
2.4
2.2
0
0.8
0.8
0
0
0
1.1
1.1
1.1
0.8
0.9
1.0
1.0
1.0
0
0
0
0
0.8
0
0
0
0
91.6
1.0
91.6
0.5
0.8
0
0
0.2
0.5
0.8
0
0
0.2
0.2
0.2
0
0
0
0
0
0
1.0
Design
of Fiat Slabs
C1 i.e.
421
C = 1kN/m
B-7
is
Each unit of combination C1 produces 1.1 kNm at joint 6 for frame type
F1. Therefore
units of live load in combination C1 required to form first plastic hinges at
joint 6 and joint 9 in members 12 and 14
= 103.2
97.5
1.1
= 5 units, say
Frame type F2 has joints released at joints 6 and 9 for members 12 and 14.
After 5 units of combination C1 the bending moments at joints are as
follows:
Frame typeF1
Member 10
Member 12
Joint 2
Joint 5
Joint 5
Joint 3
Member 13
Joint 6
joint 6
Member
62.7 +
102.5+
95.5 +
44.7 +
97.5 +
91.6 +
5
5
x 2.0 = 72.7kNm
x 3.7 = 121.0 kNm
= 172
121
3.6
= 14 units of combination
C1
Frame type F2
Member 9
Member 10
Member 12
Member 13
Joint 2
Joint 5
Joint 5
Joint 3
Joint 6
Joint 6
72.7
+ 14 ><
2.1
= 102.lkNm
Joint 5 of member 9 and joint 6 of member 13 have gone plastic simultaneously at 19 units of combination C1. Therefore frame type F? is not
considered.
422 ReinforcedConcrete
Frame type F4
Member 9
Joint 2
Joint 5
Joint 5
Joint 3
Joint 6
Joint 6
Member 10
Member 12
Member 13
Frame type F5
After 30 units of combination C1, the bending moments at joints are as
follows:
Member 9
Joint 2
Joint 5
Joint 5
Joint 3
Joint 6
Joint 6
Member 10
Member 12
Member 13
117.6 + 6 x 3.1
66.3
= 136.2 kNm
= 171.4kNm
= 172.5 kNm
+ 6 x 1.1 = 72.9kNm
= 103.0 kNrn
= 103.6 kNm
Similarly
Member 10
111.1
3.7
3.7
4.6
4.6
4.6
2.2
2.2
3.5
3.5
3.5
F2
F7
F4
F7
C
C
C
9
5
4.3
4.3
3.4
3.4
3.4
126.9
Shear
56.3
Fi
Midspan
moment
Member
F1
type
Frame
C7
Combination
2.6
2.8
5.0
1.5
42.7
1.6
Midspan
moment
Member
4.0
4.0
4.0
4.0
4.0
8
4.0
4.0
4.0
4.0
4.0
119.0
Shear
119.0
10
1.0
1.0
1.0
1.1
0.6
61.5
Midspan
moment
Member
Member midspan moments (kNm) and shears (kN) for various frame types under unit loading.
1.4
1.4
1.4
1.4
1.1
1.0
115.6
1.3
1.0
1.0
1.0
Shear
6
94.4
12
1.5
1.5
0.7
1.0
0.5
39.7
Midspan
moment
Member
1.2
1.2
1.2
1.2
1.2
105.0
1.2
1.2
1.2
1.2
1.2
Shear
105.0
13
424
Reinforced Concrete
1036kNm
729kNm
1O36kNm
909
727
1362
729 kNm
I
A
9O
Alfl5kNm
'1362kNm
A411lfIk
I!
SK 9/31 Bendingmomentdiagram
combination C1 (20%
redistribution).
Note: Only one combination C1 has been fully analysed to demonstrate the
procedure for redistribution of moments in a frame structure. In practice
all combinations of loads should be similarlyprocessed to get an envelope
of moments and shears. For all combinations of loads the plastic hinges
will form at the same moment,i.e. 172kNmat firstfloor level and 103.2 kNm
at roof level.
E
E
M, max = 0.15bd2f
b = C1 + = 400 + 400 = 800mm
d = 175 mm
assumed
= 40 N/mm2
M1, max = 0.15 X 800 X
= 147kNm >
1752 X 40
136.2 kNm (member 9, joint 2)
The column slab connection at the edge can transfer the applied moment
and no further redistribution is necessary. It is conservatively assumed in
this analysis that the depth of the slab at the column is 200mm, ignoring
the drop. The moment M1 max is greater than thedesignmoment obtained
from an equivalentframe analysis.
Design
Step 3
1720kNm
1193
.1214 kNrn
22
MEMBER9
MEMBER 10
SK 9/33 Bendingmoments at
critical points combination C1
x400 Column
(20% redistribution).
Member
Joint 5:
= 172
=
Edge of drop =
246 x
0.22- +
(56 + 48)
0.2252
><
119.3 kNm
(top tension)
1000mm from centreline of column
172 246
+ x
x I + (56 48)
22kNm
(bottom tension)
+
232 x 0.225 + (56 48)
Joint 2, similarly:
Bending moment at h/2 = 136.2
= 86.6 kNm
Bending moment at edge of drop
x 0.2252
(top tension)
426
Reinforced Concrete
Joint 5:
Joint 3:
Bending moment at hI2
C/20225
25 kNm
1443kNm
1193 kNm
866 kNi
CoImn
Column
JOINT 5
JOINT 2
\81\
Th2
3/
1 = 5.Om
12
= 6.Om
x
(15
6.0) (s.o
= 264.6kNm
h = 0.225m
2x
(floor slab)
0.225)2
/\
x 1.5 = 9.4kN/m2
= 165.8kNm (roofslab)
Check negative moment limitation
Member 9
237.1)
27.5kNm
Revised negative moments:
148.6)
146 kNm
< M' =
165.8 kNm
Firstfloor slab
Size of drop = 2000mm
l = 5000
= 1667mm < 2000mm
= 2000mm
With drop middle strip = 5000 2000 = 3000mm (yy)
= 6000 2000 = 4000mm (xx)
and
Middle strip in a slab without drop = 6000 1/2 = 6000
Column strip
2500
3500mm
4000
1.14
Design
_____
_____ -
- 2500 Column
Strip
3500 Middle Strip
Th.-I
-
division of
- 2500 Coktnn
Strip
strips.
5000 =
Column strip = 1 =
2500mm
i
Middle strip
(xx)
(yy)
9: negative moments
136.6
X 0.45
X 1.14
115.7
x 0.45 x
115.7
(bottom tension)
In this examplethe reinforcementis found for the flat slab spanningin the
short direction only. Exactly the same method of analysis and design
should be used to find the reinforcementin the long direction.
Detailing of reinforcement
MEMBER9
JOINT 5
MEMBER 10
BENDING MOMENT DIAGRAM
ELASTIC ANALYSIS NO REOISTRIBUTIQN
LOADING CONDITION C4
in '.00 slab
in slab
432
Reinforced Concrete
Use results of elastic analysis of frame before redistribution.
Maximum column moment at
joint 5 = 215.5
199.9
M1 = 15.6 kNni
= 15.6 kNm
d = 400 30 = 370mm
1.5d = 1.5 x 370 = 555 mm
x = 400 + 2 x 555 = 1510mm = 1.510m
/
/
1.5M \
1.5 x 10.9
t)497211+
Vff=Vl( 1+
\ 497.2 x 1.510
Vx I
= 508kN
Maximum shear stress at column perimeter (U. = 4 x 400)
Veff
U0d
508 X 10
4x400x370 =0.86Nfm2<5N/m2
stress
Shear
'=
OK
Veff
Ud
U = 4 x 1510 = 6040
V
= 508x
i0 = 0.23N/mm2 <
v
370
for minimumpercentage of
reinforcement
= 508 4 x 22
= 420kN
Shear stress, v =
420
x iO
10040 x 170
reinforcement
[15d
_.x_1_Cx
J15d
x
15d
P= 2C+ C1 +6d
P2(x+C)+C1+
EDGE COLUMN ON
EDGE OF SLAB
6d
SLAB
IJ
C1f5d
15d
P= +C+3d
15d
P= lesser of :OR
2(x+C) +C1+6d
2(C+ C1)12d
CORNER COLUMN ON
ON EDGE OF SLAB
SLAB
Cl
x and y 15d
P= C+C7+x+y+3d
CORNER COLUMN INSIDE
LA
d = Average effective
depth of slab
LA
SK 9/39 Punchingshearperimeters for flat slab.
434 ReinforcedConcrete
When the column face is more than 1.5d away from a free edge of slab,
then there are two alternative perimeters possible as illustrated.Take the
least of these two alternativesfor the calculation of punchingshear stress.
Punching shear at roofslab
Check joint 3 edge column.
V = l3OkN
Theframe action considered is in thexxdirectionas explainedin Section
9.3.5.
Veff = 1.25V
= 1.25
162.5kN
>( 130
d = 170 mm assumed
= 1.5 x 170 = 255 mm
1.5d
U0d
162.5 X 10
1200 x 170
x 400 = 1200)
OK
Vf
= 2220
162.5
x i03
v2220xi70
= 0.43 N/mm2
Design
of Flat Slabs
435
L10I
Intermediate floor.Typical
sectici n
rE5=267]
77
[ocj
First floor. Typical section
lx
13
10
15
I.,,
----
:----
07
os
03
A-S
lx
03 050-7
ly
1-3 1-5
C4
___J L__
o.o1y O.3ly O.SLy O7Ly
.3L 151y
C3
Points to note:
(1) The flat slab systemshould have at least 3 spans in the direction and
(3) The tables can be used only for uniformly distributed loads with all
spans loaded simultaneously with the maximum load.
(4) To accountfor the possible increase in momentdue to variation of live
loads in different panels of the flat slab, no redistributionof moments
should be carried out.
(5) For horizontal loading a separate frame analysisshould be carried out
and the appropriate moments will be combined with the vertical load
moments. In general, flat slab construction should be fully braced and
the horizontal load should be carried entirely by a shear-wall system.
(6) The coefficients are applicableto a corner panel, an edge panel with a
free edge in the 1, direction, an edge panel with a free edge in the
direction and an internal panel.
(7) The moment triad (M1, M and M) obtained by this method of
analysis should be combined using the WoodArmer method as
described in Section 1.12.
Bending moment
abouty-axis
=M,
shown)
Bending moment
about x-axis
Note: A positive moment denotes hogging. This sign convention is opposite to the WoodArmer
convention. Reverse the signs of the moments before carryingout WoodArmer combination as per
Section 1.12.
1/l
M == nCK11kNm/m
M
nCK1kNm/m
= nCA.VKXV1kNm/m
C
C,
C,,
+0.02633
0.00159
0.03301
0.02496
0.00400
0.01420
0.01954
+0.04690
0.00744
0.01265
0.02620
0.01954
0.00744
+0.15656
+0.15656
0.04291
0.01476
0.03647
0.04247
0.00338
0.07560
+0.01521
0.00159
0.00395
0.05718
0.05718
0.07603
0.01157
0.00562
0.01103
0.07014
+0.02704
0.00619
0.05749
0.03218
0.01214
0.05245
0.06063
0.002 62
0.05423
0.05423
0.00744
0.06515
0.01059
0.01419
0.02030
+0.05482
+0.11571
0.03189
0.01049
0.04132
0.02524
0.00157
0.07603
0.005 62
0.01059
0.06515
0.01419
+0.06370
+0.06370
0.01277
l.O!
0.7!
0.5!
0.3/p.
C1=0.194355 C2=0.45153
01,,
0.02776
0.01048
0.00619
+0.11571
+0.05482
0.03189
0.01214
0.04132
0.01049
0.00262
C1=0.45153
C4=1.15263
0.02030
0.00296
0.01103
+0.02704
0.032 18
0.05749
0.06063
0.05245
0.02524
0.03300
0.05370
0.01667
0.05739
0.001 59
0.00395
0.00113
0.07014
0.00100
0.01322
0.06249
0.(X)338
+0.01521
0.07560
0.02993
0.06357
0.04987
0.01184
0.00541
0.02794
0.05697
0.01476
0.00819
0.01790
0.06357
0.02993
0.03647
+0.04247
0.00380
0.02941
0.02941
0.00819
0.04358
0.00757
0.00757
+0.04690
0.02620
0.01265
0.06249
0.01322
0.05739
0.01667
0.01048
0.02776
l.31,
0.02844
0.00554
0.00113
0.00380
0.00159
0.00400
0.00100
0.00296
0.00541
C,,,,
C,,
C,,
C,,,,
C,,
C,,
C,,
C,,
C,,
C,,,,
C,,
C,
C,,
C,
C,,
0.02230
0.02230
0.00554
0.02844
0.04358
+0.02633
0.05370
0.03300
0.01184
0.04987
0.02496
0.03301
l.5/
C
0.05697
0.02794
Moment
coefficient,
1.5!,,
1.31,,
0.31,,
0!,,
1.0!,,
0.5!,,
(lIl = 1.0).
0.7/,,
Location
of point of
interest
0.08390
0.05595
0.00148
+0.01542
0.00573
0.08230
0.03575
0.01360
0.07664
0.06117
0.00316
0.01399
0.04465
0.02831
C1=0.195075
0/,,
0.08934
0.01134
0.01902
0.01457
+0.09004
+0.06476
0.01871
0.03097
0.05444
0.01106
0.01417
0.04249
0.057 30
0.03994
0.05809
0.005 02
0.02085
0.04080
0.03032
0.00776
C4=!.15344
0.00641
0.10261
0.01169
0.002 99
0.051 11
0.08693
0.01624
0.00337
0.05003
0.03408
0.03764
0.01332
0.00474
0.03473
0.03295
0.71k
(I/! = 1.2).
0.09097
0.01716
0.00108
C2=0.44811 C,=0.453375
0.06907
0.07796
0.05432
0.009 63
0.3l
0.09065
0.03378
0.00421
0.09512
+0.07262
+0.12706
0.04229
0.013 65
+0.00164
0.00158
0.08348
0.02082
0.012 11
0.01414
0.03235
0.022 11
0.01604
0.081 82
0.007 10
0.10176
0.00347
0.00608
0.08586
0.02828
0.07980
0.03557
0.01657
0.05758
0.51k
0.31k
0/
O.Sly
0.7/,,
1.01
1.3i
l.5l
Location
of point of
interest
0.03657
+0.05945
+0.152(X)
0.07482
0.009 35
+0.05316
0.08095
0.002 36
+0.03919
0.03656
0.012 13
+0.07160
+0.19583
+0.18274
0.05087
0.01089
0.07768
0.01930
0.00236
0.04754
+0.05340
1.01k
0.02540
0.03380
0.01081
0.01330
0.02017
0.05576
0.00431
0.01588
0.06046
0.01747
0.01733
0.02569
+0.03907
0.00707
0.02621
0.01604
0.01256
0.00593
0.00466
0.00898
0.02534
1.31k
0.00656
0.06265
0.01176
0.00331
0.05069
0.045 58
0.00115
0.053 11
0.04109
0.00432
0.04690
0.02623
+0.01480
0.00145
0.05675
0.00383
0.04371
0.00543
0.00123
0.03784
0.01780
1.51k
C,,
Cx
Cx
C,,
C,,
C,
Cx
C,
C,
Moment
coefficient,
0.03332
+0.23834
+0.20889
+0.00397
0.00527
0.11078
0.03997
0.014 18
0.01723
0.04721
0.7l
0.00741
0.117 53
0.01189
0.02358
0.01377
+0. 121 67
+0.06593
0.03124
0.133 60
0.01158
0.01071
0.01606
0.07295
0.00382
0.11820
0.04976
0.10580
0.05462
0.00842
0.03596
+0.06438
0.04237
0.02947
0.04028
0.01080
+0. 19234
0.06868
0.01135
0.01473
0.01305
0.01612
0.02520
0.05361
0.07905
0.05479
0.05667
+0.08586
0.00318
0.08560
0.00739
0.07779
0.01135
0.00343
0.06155
0.04740
0.00116
0.05780
0.04970
0.02048
0.05789
0.(X)483
+0.07003
0.05103
0.06281
0.005 75
0.11551
0.05539
0.001 70
0.10476
0.06227
0.003 55
0.02056
0.08747
0.00422
0.01867
0.01662
0.06343
0.02823
0.02227
0.02342
0.02532
0.03513
0.054 10
0.03142
+0.10672
0.057 55
+0.00431
0.00125
0.07210
0.121 86
0.03854
0.004 11
01,,
O.3l,
0.5!
0.03003
0.12354
+0.09384
+0.13729
0.05628
l.Ol,
+0.02551
0.00717
0.06459
0.13173
0.01129
0.001 70
0.01742
+0.03142
0.00649
0.01609
0.01479
0.01615
0.00296
0.01222
l.3l
0.00335
0.06200
0.00687
0.01923
0.00486
+0.11466
0.05114
0.01445
0.00113
0.01037
0.00498
0.02248
0.00369
0.00639
0.05713
0.01520
0.01469
0.02241
0.050 18
0.(X)320
+0.08741
0.11941
0.03062
0.0010()
0.04705
0.03282
1.51k
1.3l
1.0!
0.7l.
(l/l = 1.4).
0.12362
0.02280
0.10961
0.03952
0.02136
0.06491
0.5l
of flat slabs
0.113 18
0.02623
0.3l
0l
design
0.01752
0.03625
1.51
Location
of point of
interest
G,
C,
C
C,
C
C
C
C
C,
C',,
Moment
coefficient,
0.00602
0.06902
0.05576
0.1)1759
0.00211
0.157 13
0.00373
0.15162
0.05566
0.00182
0.00493
0.14273
0.04471
0.01397
0.13655
0.06388
0.00383
0.01988
0.04901
0.03044
0.02479
0.00954
0.7l
0.00110
0.08087
0.04433
0.01871
0.02697
0.05511
0.00504
0.03182
0.05125
0.03231
0.02162
+0.10749
0.08934
0.00408
+0.12493
0.06494
0.06185
C=0.45576 C4=
1.15173
0.00830
0.02810
0.05131
=0.443385
0.04046
0.011 39
0.14962
0.01231
+0.15859
+0.06699
C2
0.08415
0.01145
0.00267
0.01251
0.01189
0.16914
0.15419
0.04868
0.02243
0.03267
0.04757
0.01061
+0.23712
+0.06950
0.04891
0.01559
0.01727
0.08263
C,
0.07773
0.04713
0.025 16
+0.14768
0.06914
0.03274
0.13751
0.05520
0.04392
0.03091
0.00102
0.00799
0.09590
0.01082
0.00346
0.00386
0.08271
0.03077
C,
C,
C.,
C
C
+0.02416
0.00578
0.09017
0.00511
0.041 16
C,
C,,
C
C
C,
+0.28462
+0.23486
0.06314
+0.01704
0.00649
0.08020
0.00261
Moment
coefficient,
0.01489
0.08306
0.00969
0.00091
0.07938
0.01475
1.51k
0.01912
0.01496
0.01737
0.07649
C1=0.199125
0/
0.3l
0.51w
-
0.09266
0.16570
0.02353
0.15531
0.00727
+0.11858
+0.14626
0.07490
l.Ol
0.00236
0.01375
0.01601
0.01167
0.02746
0.00439
+0.15746
0.06652
0.02163
0.16010
0.02981
0.14612
0.03265
1.3!,
0.00478
0.00411
0.02028
0.03924
0.00484
0.00083
0.00367
0.00638
0.02244
0.01984
0.05106
+0.12785
0.06172
0.03274
0.15690
0.03496
0.14272
0.04461
1.51y
0.02596
0.07131
1.31,.
1.01.
0.31,
0.71k
0.51..
(l/I = 1.6).
0!.
Location
of point of
interest
0.00063
0.20370
0.03510
0.00247
0.209 19
0.01121
0.00914
0.00350
0.18209
0.03986
0.01078
0.19028
0.01830
0.00489
0. 17800
0.04986
0.013 13
0.17182
0.06594
0.00403
0.17285
0.05607
0.01458
0.18539
0.01273
0.03292
0.02210
0.04120
0.01976
+0.14687
+0.15402
0.09911
0.021 83
0.05007
0.02965
0.02840
0.09718
0.01047
0.01716
0.07944
0.00908
+0.20072
+0.06785
0.08002
0.00396
0.19464
0.04804
0.19196
0.05675
0.001 84
0.00344
0.04970
0. 19642
0.00171
0.20023
0.03792
0.19796
0.04105
0.17890
0.05063
0.03010
0.07652
0.00608
0.51k
0.3l
01.
1.8).
0.04472
0.10093
0.01143
0.01906
0.08511
0.54680
0.08093
0.06072
0.005 89
0.03425
0.021 59
0.085 82
0.00631
+0.00888
0.09695
0.08364
0.01791
0.01311
0.00442
0.07854
0.03279
0.7l
0!,,
O.3l
0.51,,
0.71,,
1.01,,
1.3!,,
1.5!,,
Location
of point of
interest
Table 9.5 Bending moment coefficients for design of flat slabs (1jL =
+0.28659
+0.07473
0.05597
0.02196
0.03522
0.05594
0.01029
0.01615
0.04005
0.04887
+0.17018
0.08544
0.00503
0.03531
0.05232
0.01792
0.03982
0.025 23
+0.01736
0.00498
0.04997
0.01293
0.03750
0.00447
0.00516
0.09204
+0.15136
0.02708
0.02818
+0.19437
0.06783
+0.33505
+0.26064
0.02388
0.00829
+0.20582
0.00511
0.00431
0.03203
0.01774
+0.17436
0.05001
1.31
1.01k
0.00844
0.11731
0.01027
0.00350
0.10347
0.041 77
0.00099
0.10074
0.045 52
0.00337
10488
0.03368
0.
0.00081
0.11126
0.01359
0.10677
0.01368
0.00181
0.00066
0.10420
0.01633
1.51k
C1,,
C,,
C1
C1
C,,
C1,,
C1
C1
C1
C
C
Cs,,
C1
C1
Moment
coefficient,
0.00326
0.22093
0.04764
0.(X)981
0.00553
0.02266
0.01707
0.00275
0.00537
0. 12976
0.01002
0.057 19
0.01703
0.22465
0.03847
0.081 66
0.00549
+0.24789
+0.06845
0.11841
0.3!,
= 0.204 615
0.25366
0.01109
0.21163
0.01505
0.5/,
C2 = 0.43974
0.04900
0.011 31
0.11907
0.02071
0.05350
+0.34090
+0.08006
0.06341
0.02705
0.08741
+0.22144
0.00635
0.00588
0. 102 99
0.09363
+0.20147
0.03292
0.02278
+0.24672
+0.38991
+0.28634
0.07181
0.02901
0.00105
+0.25960
0.03730
0.06557
0.009 91
0.01663
0.04992
0.04657
0.00488
0.04540
0.04968
0.01656
0.05017
0.02554
0.00436
0.05993
0.01105
0.01064
0.04825
0.00502
0.00368
0.016 14
0.00621
0.04331
0.04700
1.3!
+0.22671
1.0!,
0.09883
0.05954
0.02000
0.10405
0.03591
+0.00116
0.00633
0.11492
0.01098
0.10237
0.02007
0.00386
0.09727
0.03300
0.7/,
C4 = 1.14903
0.00446
0.239 36
0.04790
0.23636
0.05857
0.001 78
C3 = 0.45666
0.013 16
0.21043
0.06837
0.004 18
0.03105
0.10089
0.011 21
01,
0.01179
0.02777
0.00331
0.21636
0.05533
0.23968
0.05573
0.24572
0.04607
0.2283()
0.02909
0.24390
0.04682
0.001 13
0.00044
0.24238
0.04855
0.51,
+0.17869
+0.16068
().()42 10
0.21798
0.05733
0.3/i
0.03352
0.08045
0/,
0.02187
0.05039
0.7!,
1.0/v
1.31
l.S/,
interest
of point of
Location
Table 9.6 Bending moment coefficients for design of flat slabs (lxii, = 2.0).
0.00882
0.14215
0.00979
0.12932
0.03984
0.00361
0.12677
0.04482
0.000 90
0.00283
0.13005
0.03688
0.13551
0.02132
0.001 02
0.00107
0.13312
0.01861
0.00041
0.13147
0.01960
l.5/
C
C,,
C,
Ck
Cr
(iv
C,,
(1,
(iv,
C,
C,
(ivy
C'
C
c
c,
C,
Moment
coefficient,
444 ReinforcedConcrete
____
-.
Lx
Lx
lj1,
1.2
1.6
1.4
1.8
Moment
coefficients
Column
no.
1
0.01175
0.01345
0.01717
0.01916
0.02123
0.01175
0.00244
0.02280
0.01792
0.00233
0.03271
0.02594 0.03603
0.00224 0.00218
0.04525 0.06073
0.02765 0.03301 0.03885
0.04836
0.00216
0.07939
0.06307
0.00218
0.10147
0.00441
0.01040
0.04513
0.01445
0.05184
0.019 19
0.00418 0.00454
0.00506
0.00571
0.02031
0.02667
2
2
3
3
4
4
0.02280
0.00244
0.00376
0.00393
0.00376
0.00659
0.01526
0.00706
0.01028
0.01484
XR
C.
C,
C,
x1xi
0.6
0.7
0.8
0.9
1.1
1.2
1.3
1.4
S= ad L/h
Graph 9.1
10
0.6
07
0.8
0.9
1.1
1.2
1.3
1.4
1.5
ad L/h
10
0.2
0.4
0.6
0.8
1.2
1.4
1.6
1.8
Stzd3L/h4
6
7
Stiffness correction
KA for
coefficients for Zone A, curves for 1./l = 1, 1.2, 1.4,
1.6, 1.8 and 2.
Graph 9.3
10
'C
05
0.7
09
1.1
1.3
1.5
1.7
1.9
Sad3L/h4
10
0.992
0 994
0.996
0.998
1.002
1.004
1.006
1.008
1.01
Sad L/h
1.8
1.6
1.4
10
ttiti
I I II I I I I 2 V
06
0.7
08
0.9
1.1
1.2
1.3
S=ad LIh
Stiffness correction
Graph 9.6 KBX for
coefficients for Zone B, curves for I/IV 1, 1.2, 1.4,
1.6, 1.8 and 2.
10
1.8
1.6
1 .4
1.2
rt
10
ad3 L /
0.9
0.9
0.92
0.92
0.94
0.94
0.96
0.96
0.98
0.98
1.02
1.04
1.04
1.02
1.06
1.06
4
S
ad3L!h4
10
CD
CD
-I
CD
CD
CD
-t
CD
0.5
0.6
0.7
0.8
0.9
1,1
H.
Sad3L/h
H:_
//j7
/\
I I
....
1.6
for
Stiffness correction
coefficients for Zone C, curves for 1.r/1V = 1, 1.4, 1.6
and 2.
Graph 9.9
10
008
0.992
0.994
0.996
0.998
.002
1.004
1.006
1.01
1,012
Sad 3L/h4
Graph 9.10
10
cf
450 ReinforcedConcrete
cD
('1
0
a,
rl
C
I-
-l
r'i
0
0
0
Axa)4
I-
,-.
-j
0
('1
AO
Design
of Flat Slabs
451
C
V
0
-J
H
U,
0
0
I',
0
0)
0)
0)
00!
0
0)
o
0
0)
0
0)
o
0)
U,
C
V
I-
Ii
t
4-
-J
U)
V
C,
.(.)
0
CC)
0
0
CC)
CC)
'-
o0)
o
0)
0)
X3
U)
0
0)
00)
o
U)
00)
452 ReinforcedConcrete
N
0,
N
I-
2
-j
C.,
N
N
0
N
.-
.-
N
0
0,
CC
NJ
(I
-J
0
C.,
N
V
0
Cd,
.0
0
0
Cd,
Cd,
0,
CC
0,
CC
Cd,
a,
U,
I0
IOC
05
0.95
1.1
1.15
1.2
1.25
1.3
ad3 LI h4
10
0.96
0.97
0.98
'0.99
1.01
1.02
ad3LII
Stiffness correction
Graph 9.18 KF for
= 1, 1.2, 1.4,
coefficients for Zone F, curves for
1.6, 1.8 and 2.
10
0.7
0.8
0.9
>.
1.1
1.2
1.3
Sad3LJh4
I
8
10
o-
1.8
1.6
1.4
1.2
IxIly
0.8
0.85
0.9
0.95
N105
1.1
1.15
1.2
1.25
1.3
Sad3L/h4
10
1.4
h- 2
1.8
-- 1.6
1.2
Ix/ly
0.6
0.7
0.8
0.9
1.2
13
1.4
1.5
1.6
Sad 3L/h4
8
10
lxi II,
1.8
1.6
1.4
1.2
1.1
0.7
0.8
09
1.2
1.3
1.4
1.5
1.6
10
S=ad 3L/h4
1.8
-- 2
1.8
1.4
1.2
-.-D
Ix/ly
1.
0.8
0.8
1.2
1.4
1.6
1.8
for Column 3.
8
10
S.ad3L/h4
h-- 2
0.65
0.75
0.85
10
b-- 2
a-- 1.8
---- 1.4
- 1.6
0.95
a-- 1.2
b--- 1.8
1.05
Ix,Iy
0-- 1.6
-- 1.4
0 1.2
-1
Ix/Jy
1.15
1.25
1.35
0.5
10
Sad3L/h4
05
0.7
0.7
1.1
0.9
-- 2
a- 1.8
--c.--- 1.6
-'-- 1.4
S=ad3L/h4
10
- 2
*- 1.8
o- 1.6
'---.--'--- 1.4
1
o- 1.2
1.3
0.9
1.1
1.3
Ix/lY
1.5
1.5
- 1
1.2
1.7
1.7
'XI ly
1.9
1,9
Page blank
in original
Chapter 10
Design of Connections
NOTATION
10.0
a
C
D
lb
f,
fbu
F,
F
H
K
I
of a tie
I,
4
M
n
fl()
0
r
PC
4e
10.1
INTRODUCTION
There are basically two types of connections: rigid and free. The rigid
connectionswill have full moment and other internal force transfer capability. The connectionsclassed as free do not offer resistance to rotation to
membersat the connection.These connectionsshould be capable of transferring shear and axial loads.
10.2 CONTENTS: TYPE OF CONNECTIONS
zo
V = shear at section
z = lever arm of bending moment
= summationof perimeter of bars in tension.
The local bond stress at ultimate state need not be checked provided there
is adequate anchorage of the bars in tension on both sides of a section.
Theultimate anchoragebond stress is assumedconstant over the anchorage
length of a bar.
where
=
where
---- -
aa
aE
La..
aa
La
La
Es
a..
a..
a..
a..
fla
Vf
Values of 3.
Plain bars
Type 1: deformed
Type 2: deformed
Fabric
In tension
In compression
0.28
0.35
0.40
0.50
0.65
0.50
0.63
0.81
10.3.1
of links
4 x diameter.
Anchorage
in the foundation.
SK 10/3 A
SK 10/2 A 90 bend.
1800 bend.
Laps andjoints
LAP LENGfH
0i
200 IF
AND 02
AND
> 20
C< 1.50,
OR,1502
200mm where both bars at a lap exceed 20mm diameter and the cover is
less than 1.5 times the diameter of the smaller bar.
10.3.2
CI
Ci AND C2<2+
SK 10/6 Case 1
anchorage length.
D <75 OR 6$
SK 10/8 Case 3
anchorage length.
C1
AND C2 <2+
AND/OR
SK 10/9 Case 4
D<75 OR 6+
anchorage length.
464
ReinforcedConcrete
Case I
Bars lapped are at the top of a section and cover is less than 2 times the
size of lapped bars.
Case 2
Bars lapped are at a corner of a section and cover to either face is less than
2 times the size of lapped bar.
Case 3
The distance between adjacent laps is less than 75mm or 6 times bar
diameter, whichever is the greater.
Case 4
Corner bars at the top of a section with less than 2 times diameter of bar
cover to either face.
Lapped bars at the top of a section with distance between them less than
75mm or 6 times bar diameter.
10.3.3
Design
of compression laps
Lap length = 1.25 x compression anchorage length of smaller bar at lap
Effective anchorage length
T.P
240
T.P.
1
20
Design
180 hook
Effective anchorage length
or
of Connections
465
= 8r(244)) + 1 44)
= actual length of bar from tangent point
whichever is larger
90 bend
Effective anchorage length = 4r( 124)) + I 44)
= actual length of bar from tangent point
or
whichever is larger
124)
ora point where theshear capacity of thesection is twice the shear force at
the point, or
a point
where the availablebars continuingbeyond provide a moment
of resistance twice the bending moment at the point.
466 ReinforcedConcrete
L=
(2)
W/3 AND 30
(3) For slabs, if shear stress at face of support is less than half v, then
project a straight length of bar beyond centreline of support equal to
one third of support width or 30mm, whichever is greater.
Anchorage bond lengths in multiples of bar sizes for Type 2 deformed bars
=460N/mm2).
(f
f,
Grades of concrete
(N/mm2)
C25 C30 C35 C40 C45 C50
Tension anchorage and lap 40
32
Compression anchorage
40
Compression lap
37
29
37
34
32
30
27
34
26
24
30
32
28
22
28
Design
of Connections 467
Anchorage bond lengths in multiples of bar sizes for plain grade 250N/mm2
bars
Grades of concrete
(N/mm2)
C25 C30 C35 C40 C45 C50
Tension anchorage and lap 39
32
Compression anchorage
39
Compression lap
36
29
36
33
27
33
31
25
31
29
23
29
27
22
27
Note: The tension anchoragebond lengths will be multipliedby either 1.4 or 2.0,
depending on the location of the bar as described in Section 10.3.2.
10.4 BUILDING TIES
Peripheral ties.
Internal ties.
Horizontal column and wall ties.
Vertical ties.
Ties are continuousfullyanchored and properly lapped welded or mechanically connected tension reinforcement.
The reinforcement required to act as continuous ties is additional to
other designed reinforcement. Available excess design reinforcement if
properly tied and continuous and capable of carrying the prescribed tie
forces may be considered.
10.4.1
Peripheral ties
A continuous tie should be provided at each floor level and roof level
within 1.2m of edge of building or within perimeter wall. This tie should
be capable of resisting a tensile force equal to FkN.
F = 20 + 4n0 or 60 whichever is less
where
n0
F1
This means that the maximum areaof steel for peripheral tie at each floor
and rooflevel is given by:
60x103 =
150mm2
0.87
x 460
Ft
F
Ft
>-
Lx U,
D
>LU
Lx
0 LU
UJ
U,
SK 10/14
forces.
tie
SK 10/15 Typical floor plan showingties
required.
10.4.2
Internal ties
These ties are at floor and roof levels in two orthogonal directions and
anchored to peripheral ties or columns or perimeter walls.
The spacings of these ties will not be greater than 1.51r,where lr is greatest
INTERNAL TIES (xDIRECTION)
INTERNAL TIES (yDIRECTION)
CDL
CDL
CDL
CDL
Sx
5y
1
1
.S
.S
Lx
Ly
Design
distance
of Connections 469
tie.
The ties should be capable of resisting a tensile force equal to the greater
of:
O.O267(g.
+ qk)1FI
or
10.4.3
PERIPHERAL TIE
'1NTERNAL
TIE
TO
PERIPHERAL TIE
ANCHORED
SK 10/17 Anchorage
of ties.
1
CONTINUOUS
PERIPHERAL TIE
CONTINUOUS
PERIPHERAL TIE
,,
U-BAR
EXTERNAL COLUMNS
HORIZONTAL TLE
BACK TO FLOOR
SK 10/18 Externalcolumn
tie back.
elevation showing
U-BAR
HORIZONTAL TIE
as above.
The corner columnwill have horizontalties at each floor level or rooflevel
in each of two directions, capable of developing a tie force equal to either
10.4.4
Each columnand each wall should be tied continuously from the lowest to
the highest level. The tie force in tension will be the maximum design
ultimate dead and live load imposed on column or wall from any one
storey.
10.5
CONNECTIONS
The most commonly occurringstructural connectionsare illustratedin this
section with guidanceon the preferred detailing methods.
Pile-to-pile cap/foundation raft/ground beam
10.5.1
w
L
cD U
Q
-Jw
LU
ui
cC
L LD
>
LU
C
CD
Z
C ZD
C
F
F)
cD
LU
p iii5iiii o
Case 1
Mainly vertical loads.
Small horizontal load.
SK 10/20 Pile-to-foundation
connection.
Design of Connections
471
O.6f
uJ
o U-U-
(.)
H-OZ
U- 0
C.)
C)C)
.) Z
W
C)
C.)
IN-SITU PILE
CASE 2A
CASE 2B
SK 10/22 Pile-to-foundationconnection.
-BARS
SK 10/23 Pile-to-foundation
connection.
BOREDIN-SITU PILE
CASE 2C
472
Reinforced Concrete
Case 2
12
may be considered)
If bearing stress is higher thanallowed, thenembedment may be increased
by adopting solution in Case 2C.
10.5.1.2 Precast reinforced concrete pile
Case 1 (same condition as in Section 10.5.1.1, Case 1)
z
JL
Umz
"flu
L0
L
z
V)
DI.C)
SK 10/24 Pile-to-foundation
connection.
SK 10/25 Pile-to-foundation
connection.
CASE 2B
_r
--
z
Q
uJ
J
o_________
C-)
__
SK 10/26 Pile-to-foundation
connection.
/ = transmission
vfcu
PRECASTPRESTRESSED PILE
CASE 1
length
of prestressing tendons
474
Reinforced Concrete
TOP HAT REINFORCEMENT
TO RESIST PILE
VERTICAL
REACTION IF ALLOWABLE
PUNCHING SHEAR STRESS FOR
DEPTH d' IS EXCEEDED
ELEVATION
PCA$T PRESTRESSED
CASE - 2
PILE
SK 10/27 Pile-to-foundation
connection.
STEEL HPILE OR
CIRCULAR STEEL TUBE
CASE
STEEL H-PILE OR
CIRCULAR STEEL TUBE
SK 10/29 Momentconnection of
steel pile to foundation.
I=
CASE 2
= diameter of bar =
1.21(
\afn
\Df
10.5.2 Column-to-foundation/pilecap/raft
Case 1: Column bars always in compression
CASE
Design of Connections
477
CASE 2
bars.
Wall-to-foundation/pile
cap/raft
Column-to-columnconnection
Case 1: Column bars always in compression
i+2
IL
S2OO IF
AND +2>20
AND C < 15$ OR 1542
SK 10/33 Column-to-column
connection.
Wall-to-wall connection
UBAR
SK 10/35 Type
SK 10/34 Type
1 plan of wall
'
at corner.
px-\ po--o
VERTICAL U-BARS
LAPPED WITH HORIZONTAL
BARS IN THE LAYER UNDER
L
L
SK 10/36 Type 2
SK 10/37 Type 3
plan of wall
at corner.
If the loading causes the corner of the wall to open up then Type 2
connection becomes most efficient. Moreover, with Type 2 detailing the
horizontal bars could be of different diameters at the inside and the
outside faces.
0.15
span
length
Design
of Connections
481
Case 2.
SK 10/40 Connection
column Case 3.
of beam to
OF COLUMN
0.15 span
tension lap
482
Reinforced Concrete
Case 4: Beam assumedfixed to column (ductile connection for reversible
moment)
5O
L2
:r.1 rJ
SK 10/41 Ductile column/beam
connection Case 4.
r = radius of bend
Design
L,
of Connections
483
SO
Hi
J 2
10.5.7
1LL
LMAIN BEAM
SPLICE BARS
FOR MAIN BEAM
J]
Wall-to-beam connection
The same principles apply as in Section 10.5.6.
10.5.8
Wall-to-slabconnection
10.5.8.1
greatest
is the
484
Reinforced Concrete
Case 1: Small diameter bars
1 = same as in Case 1
12
10.5.9
Case 2 detail is used when tension anchorage length cannot be accommodated within bend of U-bar in Case 1.
Column-to-wallconnection
Case 1: No significant tension in column bars
CKER
SK 10/46 Wall-to-column
connection.
12
in column bars
= tension lap length (see Section 10.3.2 for design of lap length)
= tension anchorage length
10.5.10
Slab-to-beam connection
The same principles apply as in Section 10.5.8.1.
Page blank
in original
Chapter
11
487
i:':
mr E5/E
/11/
'0
/,
v'
08
REINFORCEMENTRATIO
04
= Fbd3
100=A/bd
.6
0020_____
0030
0.0. C
0050
0066
v.u,U
Li
LJ
LL
00 10
002C
n.n
//
_____ _____
,b
/
/
/9/
1.2
/bd
16
////
mEs/E
:::_____
t/VbL
0070
Fig. 11.1 Coefficient for moment of inertia of cracked sections with (a) equal
reinforcement on opposite faces and (b) tension reinforcementonly.
"C
"C
C)
I-
ICC
C)
('C
C)
C)
I-
C)
C
C
C)
C
CC
C)
C
C)
C
CC
CL
pq/SyQQ
Reinforced Concrete
490
U,
U,
I-.
C
4)
U,
4)
4)
0
C)
C
U,
0)
1
0
U,
4)
ct
V
V
V
C
C
C
ci,
CC
V
pq/Sy001
492 ReinforcedConcrete
II-
In
62.8
78.5
100.5
25
32
157.1
125.7
804.2
1256.6
1963.5
50.3
20
40
50
490.9
31.4
37.7
113.1
201.1
314.2
78.5
50.3
10
12
16
28.3
(mm2)
Area
18.8
25.1
Perimeter
(mm)
6
8
Diameter
(mm)
9.864
15.413
6.313
2.466
3.854
0.222
0.395
0.616
0.888
1.579
Weight
(kg/rn)
6544
10722
2681
4189
1046
1508
670
377
75mm
12566
4909
8042
503
785
1131
2011
3142
283
100mm
15708
10052
3926
6433
2513
402
628
904
1608
226
125 mm
13090
8377
3272
5361
2094
335
523
754
1340
188
150mm
1571
2454
4021
6283
9817
141
251
392
565
1005
200mm
1963
3216
5026
7854
113
201
314
452
804
1256
250mm
9818
4022
6284
2262
566
1006
1570
50mm
2680
4188
6545
1636
94
167
261
377
670
1047
300mm
CD
CD
-t
Tj
CD
CD
494 ReinforcedConcrete
Table 11.2 Sectionalproperties.
lzO5CBt.O5Cb(h-h1)
C Coefficientin fable 2-2
txj bh3)j(B-b)j
(h/2
.2(Bb)h1
h1 12)2
hB3.(h_2h1 )b3
(I.-2hf) b
izCBt1.O.5Cb3(h_2hj)
CCoeffucient in table 2-2
4A
of wall.
of
side
of
closed
cell on
B1Length each
rihickness of each sideof closedcellon
lxx
Iyy B3h1.(H2hj)e
.2Bh(x-B12)2
.(H2h1 )(x12)2
(H-2h1)
> b,
lz =cet4 .C(H-2h1)b
C Cotfficuent
in table 2-2
b/b
Cantilever
Simply supported
Continuous
7
20
0.3
5.6
16.0
496 ReinforcedConcrete
Chart 11.4 Modification factor for compression reinforcement.
16
14
L)
12
05
15
25
35
45
100 4s/(bd
Cc =
350
04
02
0
2
N/mm2
M /(bd2)
Cf
30
35
40
45
50
25
20
20
20
20
35
30
25
20
40
30
25
50
40
30
60
50
Table 11.7 Nominal cover (mm) to all reinforcement including links to meet
specified periods of fire resistance.
Fire
resistance
(hours)
Beams
Simply
Continuous
supported
0.5
1.0
1.5
2.0
3.0
4.0
20
20
20
40
60
70
Columns
Slabs
Simply
Continuous
supported
20
20
20
30
40
50
20
20
25
35
45
55
20
20
20
20
25
35
45
20
20
25
25
25
Note: cover in excess of 40mm may require additional measure to reduce risk
of spalling
Table11.8 Rectangularcolumns
.I --d
I
= 30N/mm2, k = 0.95.
1.
!!
-'.-dd
!!
!!
cOO cicicicici
ci
!!! n!!! !!
Best available copy)
53+
ci
ci
499
=30N/mm2,k = 0.90.
""
Iu
uI
"" !! !!!
!!
!!
!! 9!
!!
5! n!!!
""
= 30N/mm2,k
0.85.
oo
!! !!
I
OO
dDd
oooo OdQ
ood
nfl!
!!
i.
oo
""
!!
!!!!
!! !!!E !! !!H HH
3!
H!!!
H!U !H!
= 30N/mm2,k = 0.80.
doo od dodd
dOo
!!!
!! !!!
-g.
oooo
!!
""
!!
odo
dod
OdOO
!!
!! !!! !! !!!
!!!U
dOd
!ffl
= 35N/mm2, k = 0.95.
!!
t
q
00
0 0 0 0 0 0 0 0d 00
00 0 0
!!!
0 000 00 0 0 0 00 00 000 00 00 00 0 0 0 0 0 0
"
00
00
00 0 0 0
00 00 00 000 00 00 0 0 00 00 0 00 00 0 0 00 0 0 0 00
!!
00 0 o 00 0 0 0 00 a0 00 000 00 00 0 0 00 00 000 00 00 00 0
U!
M$
-.-N ?8
ggg8J
do-?
= 35 N/mm,k = 0.90.
00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0a 0 0
U1
000 0a a a a a a0 00 00 00 0 0 0 0 0a a0 0 0 0 0 0 0 0
0 0 00 0 0 0 0 0 0
00000 0aa00
'.'- 000
a aa oa a0 a a a0 aa a 0 00 0 a a a 00 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0
0 0a a0 0a a 0 0a 0 0 0 a 0 0 0 0 0 0 0 0 0 0 a 0 0 00 0 0 0 0 00 0 0 0 0 00 0
00 00000
&
ggg
I&
c 00 00 00 00 0
2
i
.E
0 00 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 00 00 0 0 0 0 00 0 0 0 0 00 0
'w- oo
c
2
.-oo
daooa ooooo
--OOOOO
oo.-
.-aOOO
V---
'l
00000
?8 '''' 8888
504 ReinforcedConcrete
uu
"".
I
DO.3
uu
!!
!!
.
2
q
!!
!U
.4
i,N4
,-.-
!.
""
R8 ?$8 88 8888S
dwi
= 35N/mm2, k = 0.80.
!!
'0000
t
I$
0
!!
00 00 00 00 0 0 0 0 00 00 0 0 0 0 0 00 0 0 00 00 00 0 0 0 0 00 0 0 0 0 0 0 0
ao,oo
2
0 000
0000
'iu .-.-
!!
- 00000
0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
u1
-..-
ood
o '"J
00000 00000
0 0 00 00 00 0 0 00 00 00 00 0 0 00 0 0 0 0 0 00 00 0 0 0 0 00 00 00 00 0
i2
!r?
0 0 00
00 00 0 0 00 00 00 00 0 0 00 00 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0
dooo ooooo
OOOO OOOO
E8
28 8888$
506 ReinforcedConcrete
Table 11.12 Rectangular columns
0 0 0 00 00 00 0 00 00 00 a
oa aa 00a a0 a0 0
''-00
000
u1
0 00 00 00 000 00 00 0 0 a0 0 0
-o
-.-.-
!!
""
a 0a aa a a 000 aa aa a aa a a a a aa a a
sg
_aa
!!
"
00 0 00 0 0 0 0 00 00 00a 00 00 0a 0 00 00 0 0 0 0 0 0 0 aa
""
&
0
L
a aa a a
0 000 0
&
f, = 40N/mm2, k = 0.95.
!!
!!!!
!!
!U
f=4ON/mm2, k= 0.90.
00000 00000
o 00 00 00 0 0 00 00 0 0 000 00 0 0 0 0 00 00 00 00 0 0 00 00 00 00 0
tZr;
.-.--.-o
0 0 0 0 0 0 0 0 0 00 00 00 00 0 0 00 00 00 0 0 0 0 0 00 0 0 00 00 00 00 0
--
0OO,
11
!!!!
- .-.--.- -oooo
&
1 .---.-.--
i.
2?
00 00e00
w u' .t fl fl
fl n en Pi ('1
!!
2?2?
odoo ooo 00000
000
0 000 0
('1
gn
0 0 0 00 00 00 0
el
28 ?8 88888
eJeielen enenenen 0eoe..2?
!!
= 4ONIrnm2, k = 0.85.
0 00 00 0 00 00 00 0 00 00 0 0 00 0 0 0 00 00 00 0 0 0 00 o0 0 0 00 0 0
mi
!!
0 00 00 0 00 00 0 0 0 00 00 0 0 000 0 0 00 00 00 0 00 00 00 0 00 00 0
ma
!! !!
H!
!!
!il
!!!
!! ","
""
!!
!!
!!!! !!!!
n!!!
f=4ON/mm2, k=O.80,
00
0 0 0 0 0 0 00 0
0 00 00 00 00 0 0 00 0 0 0 0 0 00 00 0 0 0 0 00 00 00 00 0
1.
4Jr'1.-
2
0 0 00
00 00 0 0 0 0 0 0 0 0 0 0 0
0 00 0 0
0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
U!
&
0 0 0 0 0 0 00 0
!!
0 0 0 00 00 00 0 0 00 00 0 0 00 0 00 0 0 0 0 00 00 00 00 0
&
2
?mwrw u,*te,n
nNr4e4
!!
,,
0000
0 00 0 0 0 00 0 0 0 0 0 0 0 0 00
P4 p4 P4 P4
in en en en in
88888
w
vS
510 ReinforcedConcrete
Table 11.14 Rectangular columns
,-. 0 0
0 000 0
= 45 N/mm, k 0.95.
!!
0 00 00 00 0 00 00 00 0 0 000 0 0 00 00 00 000 00 00 0
!!! !!
!!
0 0 0 00 0 0 0 00 00 00 0 0 0 00 00 0 0 000 0 0 00 00 00 00 0 00 00 0
0 0 0 0 00 0 0 0 0 00 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 00 0 0 00
""
mm
00 0 0 00 00 00 000 00 00 0 0 00 0 0 0 00 00 00 000 00 00 0
!
-000 00000 00000 00000 00000 00000 00000 0
,.1.-
ooaoa
eo d
?8 ?8 88888
f=45N/mm2, k=O.90.
t
000
;2
00 00
ooi-
me).,
0000
ri
"
0
J
uj
0
&
$2
E1
0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
.
z
e,e.
27; S1!
.eie.Iel.,
0000
d dd
*i
-.-.-,-.- ,
S8
= 45N/mm2, k =0.85.
dOO OdOO OO
!!
-OO
oOO
ooo
!!!
!U !! U!
!!
&
.J
!!!
&
&
i
2
22
I.-
""
;
Oo
g8
2gS 000
00000 00000 0000
01
rI N n
8881!
.6.6
p..
= 45 N/mm2, k = 0.80.
0000 0
00 000
0 00 00 00 00 0 0 00 00 00 0 0 0 0 0 00 0
00 00 0 0 00 00
000
.dooo
00 0 0 0 0 0 0 0 0 0 00 0 0 00 00 00 0 0 0 0 0 0 0 0 0 0 0 00 00 00 0 0 00 00
&
!!
!!!
j
,---,- .- odd
r- w in In * en en ci ci N ci N
0d 00 00 0 0 0 0 0 0 0 00 0 0
ci
!!!! !!!
i
2
01 01
(0 0
n!!!
(fl ci ci
!! !!
gn
!U
514 ReinforcedConcrete
Table 11.16 Rectangularcolumns
= 50N/mm2, k = 0.95.
!!
!!
0 0 0 0 0 000 0 0 0 0 0 000 0 0 00000 000 0 0 00 0 00 0 0000 0 0 0
;2
r1 ?
.-.-
a!!!
000 00000 00000 00000 00000 00000 0000
uu
thO4
U;;
0
'5
!U!!
!!
f=5ON/mm2,
k=O.90.
00 0 0 0 0 00 0 00 0 0 00 00 0 0 00 00 00 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0
u
0o0
J4
'.-
..-
0.0
mg
O00
0 00 0
0000
00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 00 00 0 0 00 0 0 0 0 00 0
!!
!! !!
00
zn;2
F.,.w0S,
flnnnN
00 00000
C..
a o ao0 00 00 0 0 00 00 0 0 0
!!!!!
g8 r
e4
ci ci cm
mmmci.
516 ReinforcedConcrete
Table 11.17 Rectangular columns
f,
50N/mm2, k = 0.85.
uu
.- .--e0
!!
!!
""
. -0000
L
Ig
cifl.
,nnr.e4o.
..
dd
;aood
-.d.
-oaod o
88 ?8
888.88
= 50 N/mm2, k = 0.80.
HH
00 00 00 0 0 0 0 0 0 0 0 0 00 00 00 00 0 0 0 0 00 0 00 00 0 0 00 0 0 00 00
&
!!
flN
O)O)
IWflU' 1fl*flfl
flr,NNr4
000
0 0 0 0 0 01 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
!H
0 00 0 0 0 0 00 0 00 0 0 00 00 0 0 00 00 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 00 0
&
O)CON. WM!flflfl
0000 00000
_a
518
ReinforcedConcrete
!!
t
t
= 30N/mm2, k = 0.90.
&
'.-'-
!! !!!! !!!
!!!
t
I
2!
a,
ri
NN
a a a o aa aa d e a00 d0 00
f=3ON/mm2, k=O.80.
0 000
9l
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0000
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&
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00
C
&:
00
I
2
00000 000
0.
0--0
88S
o o0 00 00
&
= 30N/mm2,
0.70.
0--'-.
o 0 0 00 00
&
0 0 0 00 0
00 0 00 0
L
n!!!
!! !!! !! !!
!! !! !! !!! !!!
!!!
,.-000 00000
iz
!!!!!
o00
00000 0000.
i'
000000000
ooeOO 00000
8888 88a8
f3ON/mm2, k=O.6O.
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in in ,. P di di in in 0
p,
en
en
en en en
en
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in
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&
dir.CDinin
0 0 00
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nnrenn t4C4r1
ce
01 01
CD
fl en en en
e.J
ri
0d
0 0 od 00 0d d0 0 0 00 00
0inin10
8$8
522
Reinforced Concrete
Table
11.20
Circular columns
= 35N/mm, k 0.90.
I 00
&
.
.
% I
&
o0 0
&
000
000
!!
!!E !!!
!!!
.
:
00006 d6od
SB8 8888
= 35 N/mm2, k = 0.80.
00 00 0
!!! !!!! !!
o0 0
&
000
&
00
&
00
ci ci ci
0 0a a 0 0 0 0
(N
(N N N ((I (N (N (N (N
a a a o a aa a a a a a a oa aa
4 n 4 in a a
in
in in a, a,
888S
Table
11.21
Circular columns
f=35N/mm2,k=O.70.
&
ooo
l.
'-g
&
!!
CC
CCCCC 0
t
Z
?oi
0 CC
1
;
%
!! !!U
000000000000
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!! !!!!
&
!!!!!
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&
2
%
0
N
22 gg
00000 00
2
%
00 0 0 0
!!
&
00000 dooo
ooo
$88B8
526 ReinforcedConcrete
= 40N/mm2, k = 0.90.
00
&
oo
&
oo
00
!!! !!!
!!!
IL
.-.-.--
-.-.e 00000
p!!!
i
daoao
ooo
Q8 88
8B8.$
f=4ON/mm2, k=O.80.
0000
&
0 00 0
!!!! !!!
!!!! !!!!
!!
0000
!! !!!!
&
00
0
dodo oood.
fl fl
01
olo.o.olo,
88
oioio, oooo
528 ReinforcedConcrete
o0000 0
o 00 0 0
= 40N/mm2, k = 0.70.
-.-'-
&
0'"
!!
!!
!!!
o 00 0
2
.-.-
&
0000'-
,-,-aoa oo000
f=4ON/mm2, k=O.60.
00000 00000 0
a
L
0 00 0 0 0 0 0 0 0
flN- a-r
.nn
W(OIOIflfl
flfl
,-
0 00 0 0 0 0 0 0
gg
0 00 00 00 0
gE
gggg
gg
-.-.-
&
g!
0 00 00 00
00 0 0 0
ggg
0 00 0
-
netnM,w
= 45N/mm2, k = 0.90.
:000
!!!!!
t
I
!!! !!! !! !!
000W-,-.-
00
,l
9l
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.t
00
!!!!
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!! !!U !U
44
f=45N/mm2, k=O.80.
fl
t
C
&
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00 00
o0 0
o0 0 0
N
rJJ -00
&
IN,.-
,-000 00000
oodo o000
532 ReinforcedConcrete
Table
11.25
!!! !!!
;?.
000
!!!
!!!
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0000
o00
I
%
i
z
0000
88
8S8.8
u1
0 0 0 00 aa 00
g;2
0 0 0 00 aa 0
F- U,
00000
U, F- F- tO (9 (9 (0 (0
't
N N N N Ct N N
t.
U,F-tO(DU(
(0(9N
NNN
NNCJN
o a aa0
!! !! !!
QO OOOO OOOO
F.9(NC
0 0 0 00 00 0
94 t4 N
N 94
CU
N N (9
U, (0 (0
(0 F- F- U, U,
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534
Reinforced Concrete
= 50 N/mm2, k = 0.90.
od
.6
%fl1
i ;g
iSfl gs;
t
I
e4
I
%
oa
zgfg
;fl
1u
UU
;
.6oi.6p.t.
nn
sso,o
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.o ini nsn
iei.
nn
wt4riei $rie$NN
!il
!!
!!
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il9
!!!!
U!
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5!! !!! !!!
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oa
n!!!
n!!!
= 50N/mm2, k = 0.80.
!!
!!
&
0
tL
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0
&
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0
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00
&
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-.-o
536
Reinforced Concrete
= 50N/mm2, k = 0.70.
E1
UH
$!$ i.ini
sogo
""
St
oooo c
&
o---,-,-,!!
8E$
4Z1
Ufl1 H
&
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t5;:3
sns
,d.ii en4eici
!! !!
flU
flfl flH
o,-,--,-,-.?
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%
ng
$r.2
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Ne4r4e4e4
e,er4
ireic.3e4
flE OP LE
aoao
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3;22
fl0p.qj rS$!
qj.4j
rZ'dciri
2fl!
2
te.i c'it4e4...
2Sfl
;$flfl
c4nl. S? .-dS ddSth
!!
coo ocooc
Ufl1
thoood 66o66
flfl
; 00000
50N/mm2, k
0.60.
00000
00000 0000
0
&
..oon0
z
0 0 00 0
0 00 0
0 0 00 0
00
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z
0 0 00 0 0
2
00000
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i2
0000
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000
J4
Index
containment,191
crack width, 191
shearreinforcement, 190
shearstress, 189
bond stress, 460
centroidofstressedarea, 37, 38
circularcolumn,
biaxial moment, 162
shearreinforcement, 163
shear stress, 162
short supporting cont. beams
short withmoment, 162
short withno moment, 161
162
slender, 162
slenderness of, 161
circularpad foundation,238
circularpile, 317, 321
shearcapacity,321, 351
Index 539
cohesive strengthofsoil, 229
column,
analysis of, 154
axial load and moment, 21
axial load capacity, 19
balanced failure, 22. 166
braced, 154
charts, 498537
circular section. 29, 161
compression failure, 26
containmentof reinforcement, 164, 168. 172,
180
540
Index
lap length,462
laps,
compression, 464
tension, 463
lever arm, 6, 8, 11
links,anchorageof, 461
loads,
accidental, 49
concentrated on slab, 106
exceptional, 49
on beams from slab, 105
local analysis, 369, 370, 390
local bond stress, 460
losses in prestress, 319
characteristic strength, 3
factors,3
mean strength, 3
stressstrain relationship, 4
mean vertical stressin soil,290
middle strip, 407, 428
modification factor,
compression reinforcement, 496
tension reinforcement, 496
modulus of elasticity, 45
concrete, long-term, 45
concrete, short-term,45
steel reinforcement, 45
modulus of pile material, 299
modulus of rigidity, 367
modulus of soil, 299
modulus ofsubgradereaction,231, 234, 279, 284,
299
Index 541
pad foundation (coin.)
minimum reinforcement, 246. 248, 261
minimum thickness, 240, 251, 265
punchingshear, 245, 249, 260
rectangular,234
shear, 242, 255, 257, 273
shearperimeter, 245, 260
shearstress, 244, 249, 260
spacing ofreinforcement, 246, 261
passive pressure coefficient, 232
passive resistance, 231, 285
peripheral tie, 467
pile,
additional moment, 317
allowable group capacity, 302
allowable load, 298, 333
approximate numberof, 310, 335
bored and cast in-situ, 470
boundary conditions, 300
circular, 317, 350
containmentofreinforcement, 325, 352
degrees offreedom, 300
effective length, 315
fixed head, 300
free head, 300
group capacity, 301, 339
group end-bearing capacity, 302, 339
group frictioncapacity, 301, 339
horizontal stiffness, 335
lateral load, 300, 333
maximum reinforcement, 325
minimum prestress, 325
minimum reinforcement, 325, 351
point resistance, 296, 329
precastprestressed, 473
precast reinforced concrete, 472
prestressed concrete, 318, 323
rectangular, 314
reinforcement. 315, 350
shearreinforcement, 321
skin friction, 296, 331
slenderness, 317
spacing of, 301
steel circular pile, 476
steel H-pile, 474
types of, 310
volume displacement, 297
with biaxial moment, 316
with no moment, 316, 318
withuniaxial moment, 316, 318
pile cap,
bending moments, 305, 343
critical section for bending moment, 305, 343
critical section for shear, 313, 344
critical sections, 305, 313, 343, 344
curtailment of bars, 325, 352
design of, 312, 344
enhancement ofdesign shearstress. 313, 348
flexible, 306
minimum reinforcement, 325. 352
punchingshear, 314, 349
punchingshear perimeter, 315, 349
rigid. 302
shear, 313, 344, 347
spacing ofbars, 325, 353
pile efficiency, 354
pile fixity moments, 306. 346
pile group torsion,304
pile load combinations, 309. 337
plainwall,
braced, 360. 383
in-plane shear. 384
unbraced, 360, 384
plastichinges. 421
Poisson's ratio, 47
prestressed concrete pile, 318. 323
prestressed pile,
links, 325
shearcapacity. 323
punchingshear,
modification due to holes, 15
pad foundation, 245, 249. 260
perimeter, 411, 431, 433
pile cap. 315, 349
reinforcement. 113, 114
slabs, 112
rectangular column,
bending and tension. 176
biaxial bending. 158, 164. 166
modification of shearstrength, 160, 172
shearcracking. 160. 172
shear reinforcement, 160
shear stress, l59, 168. 171
short, supportmg continuous beams. 156
short with no moment. 56
short with uniaxial moment. 156
slender, 157, 164, l65. 169
rectangular pad foundation, 234
rectangular pile. 3l6, 320
shearcapacity. 320
redistribution ofmoments. 7.48,49. 107,418, 420
reinforced wall,
design of, 373, 394
in-plane shear, 379. 397
out-of-plane shear, 380
rigorousmethod, 373. 393
542 Index
reinforcement (cont.)
maximum in wall, 384
minimum compression, 58, 80
minimum in waIl, 384, 400
minimum tension, 57, 79, 116, 125, 196
orthogonalin slab, 33
side face of beams, 58, 69
skew in slab, 33
transverse for torsion, 63, 95
transverse in flange, 58, 80, 92
relative density of sand. 298
residual prestress, 319
restraintfactor, 117,247
sectional properties, 494
service stress, reinforcement, 60, 92
settlement, foundations,225, 228, 230, 249, 263,
269
sheararea, 47, 364, 386
shearcentre, 86
shear envelope, 50, 87, 418
shear flow, 12, 367, 368
shearlag, 363
shear modulus, 46
shear reinforcement, 14
additionaltensile reinforcement,
biaxial bending and tension, 190
circular pile, 322
concrete strut, 16
prestressed pile, 324
rectangular pile, 320
trussanalogy, 14
wall,380, 398
shear stress,
biaxial bendingand tension, 189
design concrete, 14
massconcrete, 249, 281
torsional, 62, 93
shear stress and tension, 179
shear wall, 362
analysis of, 374
cantilever, 363, 378
closed cell, 365
combination of loading, 371, 390
continuous,363
effective breadthratio, 363, 392
globalanalysis, 368, 370, 389
in-planeshear, 379, 397
irregularclosed cell. 366
local analysis, 369, 370, 390
modelling, 368
multiple cell, 367, 388
open cell, 365
out-of-plane shear, 380
shear flow, 367, 368, 388
single closed cell, 366
torsionalstiffness, 365, 386
side frictionresistance, 298
16
momentofinertia, 103
punchingshear, 112
punchingshear perimeter,
115
supportshears, 131
torsionalreinforcement, 116
two-way charts, I326
ultimateunit resistance, 14650
yield-line method, 105, 122
slab panel,
momenttriad, 33
orthogonalreinforcement, 33
skew reinforcement, 33
slenderness ratio, 371
sliding and bearing,combined, 241, 254, 273
sliding resistance, 231, 254, 272
soil cohesion, 227
soil mechanics, essentials of, 226
soil pressure diagrams,233, 282, 288
spacing of bars,
biaxial bending and tension, 192
uniaxial bending and tension, 182
spalling, allowance for, 198,213
span/effective depth ratio, 495
springstiffness, horizontal (soil), 299, 334
SPT,299, 329
standarddeviation, 3
steel beam theory, 178, 184, 186
steel reinforcement,
elastic modulus, 5
stressstrain relationship, 5
yield strength,3
sulphates,
concentration of, 242, 312
surchargeon backfill, 218, 219, 221
tension and shear stress, 179
thermal cracking, 117
Index 543
thermalcracking (cont.)
minimum reinforcement, 118
thermalexpansion, coefficient of, 47
tie force, 467, 469
ties,
horizontal, 469
internal, 468
peripheral, 467
vertical, 469
torsion, 367
membrane analogy, 18
wall,
additional moments, 372, 392
braced, 357, 358
cantilever, 360
containment of reinforcement, 384, 400
crack width,401
deflection, 372, 392
earlythermal cracking, 385, 400
effective heights, 358
effective width of flange, 361. 392
end conditions, 359
in-planeforces. 370
maximum reinforcement, 384
minimum reinforcement, 384. 400
out-of-plane bending, 396, 399
out-of-plane forces. 371, 383
plain, 357, 360, 383
plain slender braced, 383
plain stocky braced, 383
plain unbraced. 384
reinforced, 357, 358
shear reinforcement, 380, 398
slender, 358, 383
slenderness, 371, 391
spacing of bars, 385, 402
stocky, 358, 383
unbraced,357, 358
Wood-Armercombination, 32, 309, 369, 437
yield-line tables, 146-50
yield-lines, location of. 137-44
FramedStructures
Kim S. Elliott
0-632-034 1 5-7
StructuralDetails in Concrete
M. Y. H. Bangash
0-632-02853-X
In preparation
Structural Steelwork
Analysis and Design
S. S. Ray
0-632-03857-8
ISBN 0632037245
II
llI
1110100
9 Oil780632
037247O >
BlackweH
Science
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