Prof. Dias Graded Examples in Reinforced Concrete Design Dias
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REINFORCED CONCRETE DESIGN
W s
Dias
BSc{Eng , PhD Lond ,DIC, pEng, MIStructE, MIE SL
Senior Lecturer
Department of Civil Engineering
University of
Moratuwa
Moratuwa
Sri Lanka
ociety
of Structural Engineers Sri Lanka
_
_
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PublisheJI
by
o c Structural Engineers Sri Lanka
Colombo Sri Lanka
995
ISBN 955 9347 00 4
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FOR WOR
The Society o Structural Engineers - Sril;-anka was incorporated in July 1993.
Our membership is very small and our fmancial resources are absolutely
minimal. Nevertheless, the members of our Committee have contributed a great
deal
o
their time and effort to collect funds from various sources to help
advance the knowledge and practice of structural engineering in Sri Lanka
through, inter alia, the publication o books on related topics.
As the majority
o
structures
in
this country are constructed
o
reinforced
concrete, the selection
o
GRADED EXAMPLES REINFORCED
CONCRETE DESIGN
the object o the Society s first book publishing effort
constitutes an ideal beginning.
r Priyan Dias is a brilliant young academic and is highly motivated towards
training engineers to use a thinking approach to solve technical problems.
Whilst this book itself is
o
an immediately practical nature, r Dias and others
will, no doubt, follow up with more publications which will help our engineers
to think laterally so as to come
up
with innovative solutions to any structural
problems they encounter.
A.C. Visvalingam
MA, PhD, DIC, MICE, MIStructE, MIE(SL), CEng
PRESIDENT, Society
o
Structural Engineers - Sri Lanka
2 March 1995
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GRADED EXAMPLES INREINFORCED ON R T DESIGN
with explanatory notes, using
Grade
25 concrete to BS 8110
CONTENTS
Introduction 1
Analysis of Beam Sections in Flexure Examples 1 - 4 5
Design of Beam Sections in Flexure Examples 5 -
9
13
Design
of
Beams for Shear Examples 1 - 11 26
Serviceability Checks and Detailing in Beams Example 12 31
Design
of
Slabs Examples
13
- 17
38
Design of Columns Examples 18 - 21 58
Design
of
Foundations Examples 22 - 24 66
Design of Staircases Examples 25 - 26 76
Design
of
Wall and Corbel Examples
27
- 28 83
Design of Beam for Torsion Example 29 90
Frame Analysis and Moment Redistribution Examples 30 - 32 94
Design for Stability Example 33
1 4
Serviceability Limit State Calculations Examples 34 -35 1 7
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INTRODUCTION
A Case for orkedExamples
Educational purists may argue that Worked Examples are detrimental to student learning
because there is an element
of
spoonfeeding involved. While acknowledging that there is
some truth in this argument, the author would like to contend that Worked Examples do have
a place in the educational process.
Knowledge can be acquired using two broad approaches - i.e. the deductive approach, having
its roots in Greek rationality, and the inductive approach, having its roots in Renaissance
empiricism. Learning through worked examples is an inductive approach, and both the
format and content
of
this book reflect that approach.
The book has been developed through the author s teaching
of
a course in Reinforced
Concrete Design at the University of Moratuwa. The examples are graded, leading from an
appreciation
of
reinforced concrete behaviour, through the design
of
structural elements,
to
the analysis
of
a reinforced concrete structure. The student s understanding
of
the calculations
is deepened by the Notes on Calculations while the Introductory and Concluding Notes set
each example in a wider context. Hence, in this book, design principles·are reinforced
through practice, with guidance from notes.
However, this book caimot and should not be used as a stand alone text. It must essentially
be complementary to another text or series of lectures that teaches design from a deductive
approach - i.e. one .which moves students from principles to practice. tcan, of course, be
used by practising engineers, who already have a grasp
of
reinforced concrete fundamentals.
order to equip students for real design practice, the book is very·much code based, with
extensive references given in the calculations
to
clauses in
BS
8110 (1985) - Strueturaluse
of concrete . This is another reason for the book s usefulness for Practising engineers. The
examples cover most
of
the reinforced concrete elements and stress states dealt
with
by Part
I of S 8110.
addition, examples are also given for the de ign for torsion and the
calculation of deflection and cracking, dealt with in Part 2
of
S 8110.
Sections of code are referred to by indicating the relevant clause, table equation of S
8110:
rt
Where clauses, tables, charts
or
equations from Parts 2 and 3
of
S
8110 are
referenced, the relevant Part is also indicated. One very useful feature of S 8110 is that
each table also gives the equation from which its values· are derived. .This is a clear
advantage for computerised design, and even hand calculations. Therefore, although the
tables have in fact
been
referred to in the following calculations, very often it is the
corresponding equations that have
been
used.
A Case for singLower Grades of Concrete
Table 3.4 in Part 1 of
S
8110 (1985) specifies durability
by
cover and grade, but also
indicates cement contents and water/cement ratios correspondingro the grade specified. The
background
to
this table is given in the paper by Deacon and Dewar ( Concrete durability
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- specifying more simply and surely
by
strength. Concrete,
February
1982,
pp.19-2l ,
which
describes how U.K. concrete strengths vary for given
cement
contents and water/cement
ratios and shows how the grade specified covers the cement content and water/cement ratio
requirements 96 of the time.
must be emphasised here that the index of durability used in
8110 is mix proportions.
However,
it
has related these
mix
proportions to strength, which is a much easier parameter
to measure and control. This is clearly evident in the provisions made in the code for
reducing the grade
if
a checking regime establishes that a
lower
grade
of
concrete complies
with the cement content and water/cement ratio limits (Clauses 3.3.5.2 and
3.3.5.3
of Part
1). Such a relaxation of grade is not allowed, however for concretes using blended cements.
Even a cursory glance at Table 3.4 in BS 8110:
Part
1 will indicate that at least grade 40
concrete will have to be used for all but mild and moderate exposure conditions, although
the corresponding minimum cement content and maximum water/cement ratio are only
325kg/m
3
and 0.55 respectively. This seems to
be
a very stringent condition to
be
imposed
on concreting practice in developing countries, where most concrete specified is still grade
20 to 25. In fact, even in the U.K., the most commonly used grades were grades 20 to 30,
even up to the early 19805.
Th e
question arises as to whether
Table
3.4 in 8110: Part 1, developed for the U.K. is
applicable in other (especially developing) countries, where materials and practices may be
very different. This problem was studied by the author using Sri nk as a case in point.
Th e
strengths that could be achieved for various cement content and water/cement ratio
values were obtained on the basis of a batching plant survey.
Specifications based on the above survey
ar e
given in TABLE 1. This table is taken from the
author s publication Specifying for Concrete Durability: Part
II
- Th e Sri
ankan
Context,
Engineer, Vol. XX, Nos 1-4, 1992, pp. 4-14 . The Notes in
TABLE
1 indicate the scope
of
the specifications, and also conditions under which deviations from the tabulated values
can be allowed. In particular, Notes 5 and 6 allow reductions in grade and cover values that
bring these recommendations
in
line with current
Sr i Lankan
practice. In short, these
recommendations rationalise satisfactory Sri ankan practice (especially under mild exposure
conditions) with respect to 8110, while suggesting improvements to Sri
ankan
practice
where problem areas (such
as
concrete exposed
to
sea spray)
ar e
concerned.
Although the recommendations in TABLE 1 make it possible to use grade 20 concrete for
mild exposure conditions, it was felt that basing the examples on such a low grade would
have deviated too much from the provisions of S 8110,
where
grade 25 is specified as the
lowest grade to be used with normal weight aggregate concrete (Clause 3.1.7.2 and where
all tables and charts have grade 25 as the lowest grade.
As
such, it is grade 25 concrete that
is used for all the following examples, except in Examples 28 and
29 ,
where the use of grade
30 concrete is illustrated.
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TABLE 1 - NOMINAL COVER TO ALL REINFORCEMENT INCLUDING
LINKS TO MEET DURABILITY REQUIREMENTS - ADAPTED FROM
BS
8110: 1985 FOR SRI LANKAN PRACTICE
Exposure
Examples
of
Nominal Cover
Classification Exposure
mm
mm mm
mm mm
Mild Indoor
25
20
20* 20* 20*
Moderate
Outdoor
35 30
25 20
Severe Driving Rain
40
30
25
Very severe
Sea Spray
50
40
30
Extreme Abrasive
60 50
Maximum free water/cement ratio . 0.65
6
0.55 0.50 0.45
Minimum cement content kg/m
3
275
300 325 350
400
300
325
350
400
450
Lowest grade
of
concrete
25
30 35
40
45
Note 1
Note 2
Note 3
Note 4
Note 5
Note 6
This table applies to normal-weight aggregate OPC concrete
of
20 mm
nominal maximum aggregate size and river sand fine aggregate. In no case
should the cover be less than the maximum aggregate size
or
diameter
of
main
reinforcement.
A minimum
of
25 mm cover to all reinforcement should be maintained n
beams and columns.
Cover values marked with asterisks
*
can be reduced to 15 mm, provided
the nominal maximum aggregate size does not exceed 15 mm, subject to the
conditions in Notes 1 and 2.
The minimum cement content values in parentheses should be maintained
if
no water-reducing admixtures are used.
The grade requirement can be reduced
by
5
if
a checking regime establishes
that the maximum free water/cement ratio and minimum cement cot\tent
requirements are met.
The above cover values can be reduced by 5 mm, subject to the conditions in
Notes 1 and 2 and a minimum
of
15 mm, provided a 1:3 cement: sand
rendering
of
10 mm 15 mm or 20 mm
is
applied to concrete made to
water/cement ratios of 0.65, 0.6 and 0.55 respectively.
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EX MPLE
N LYSIS
OF
UNDER REINFORCED SECTION
Determine the lever arm for the beam section shown in the figure; find also its moment
of
resistance.
5
3-20
f
= 25 N/mm
2
u
f
= 460 N/mm
2
y
All dimensions in mm
Introductory Notes
1. This example is regarding the analysis
of
an existing beam. The first step in finding
the moment
of
resistance is to find the lever arm.
Reference
Calculations
Output
Area of steel
=
942.5
2
Note 2 Assuming that the steel llas yielded,
T = 377189 N
T = 0 . 8 7 f y . ~ = 0.87 460 942.5 = 377189 N
Hence, balancing compressive force = 377189 N
0,45 f
u
.b O.9 x = 377189 .
0,45 25 225 O.9x
=
377189
x
=
166
x = 166 mm
Note 3
Since x/d = 166/375 = 0.44 < = 0.64,
steel has yielded and original assumption is correct.
z
=
d - 0,45 x
=
375 - 0,45 166
=
300 mm
z = 300 mm
3 4 4
1 e
Note :- z/d = 300/375 = 0.8 < 0.95, Hence O.K.
Moment
of
resistance = 377189 300
= 113.16 x10
6
Nmm
= 113 kNm
M = 113 kNm
Notes on
Calculations
2. Most singly reinforced sections will be under-reinforced in practice. Hence, assuming
that the steel has yielded is the most convenient way
of
starting. This assumption
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should be checked later on,
of
course, using the
x
value.)
3. The condition that tensile reinforcement has
~ i l
when the concrete strain is
0.0035, is x/d
< =
0.64 for f
y
=
460 N/mm ) and x/d
< =
0.76 for f
y
=
250
N/mm
2
This can be shown
by
assuming a linear strain distribution. However the
code recommends that x/d 0.50, in order to accommodate redistribution
up
to
10
Clause
3.4.4.4).
Concluding Notes
4. The lever arm is the distance between the centroids
of
the tensile and compressive
forces. This separation between two opposite forces is what creates the moment
of
resistance in a flexural element.
5. Because this distance has to be accomodated within the depth
of
the section, flexural
elements tend to have larger cross sections than compressive elements.
EXAMPLE 2 - ANALYSIS OF OVER-REINFORCED SECTION
Determine the moment
of
resistance
of
the section shown.
150 )
2-25 Id=
o 0
All dimensions in mm
Introductory Notes
f = 25
N/mm
2
f
460 N/mm
2
y
1.
This section is different from that in Example
1,
in that it is over-reinforced. The
calculation procedure is more complicated here.
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Reference
Calculations
Output
Area of
steel
=
981.7
mm
2
Assuming that the steel has yielded,
T
= 0.87 f
y
= 0.87 460 981.7 = 392876 N
Hence, C
=
0.45 fcu .b 0.9 x
=
392876
0.45 25 150 0.9 x
=
392876
x = 259
mm
But, x/d = 259/300 = 0.86
:>0.64
Note 2
Hence, steel has
n
yielded.
We shall try
to
find a value for x, by trial and error,
such that
T
and C are approximately equal.
y
x
=
200
mm
C
=
0.45 f
cu
.b 0.9 x
=
0.45 25 150 0.9 200
= 303750N
Note 3
E
s
= 0.0035 300-200 /200 = 1.75 xl
Hence, f
s
= 1.75 xl
3
200 xloJ = 350 N/mm
and
T =
350 981.7
=
343595 N
For
a better approximation,
t y
x
=
205 mm.
Then C
=
311344
Nand
T
=
318454 N.
For a still better approximation, t y x = 206 mm.
Then C
=
312863
Nand T =
313572 N.
This approximation is sufficient.
Note:- x/d
=
206/300
=
0.69
>
0.64
x
=
206 mm
z = d - 0.45 x = 300 - 0.45 206 = 207 mm
M
=
C.z
=
312863 207
=
64.763 x10
6
Nmm
=
64.8 kNm M
=
64.8 kNm
Note 4
Note:- Alternative method
of
finding x.
Once it is established that the steel has not reached
yield point, for any given value
of
x,
E
s
= 0.0035 300-x /x
f
s
=
[ 0.0035 300-x /x] 200 xloJ N/mm
T
= 0.0035 300-x /x] 200 xloJ 981.7 N
C = 0.45 25 150 0.9,qN
Putting
T
=
C, we have the quadratic equation
x
2
452.47 x - 135741
=
0,
giving x = 206 or -659
mm
x
=
206
mm
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otes on alculations
2. In some rare cases, as in this one, a
beam
may be over-reinforced, meaning that the
yielding of steel will not take place before the crushing o
concrete. such a beam
fails, it will do so suddenly, without warning, and hence over-reinforced beams are
discouraged in practice.
3. Since the steel has not yielded, the stress can no longer be assumed to be 0.87fy.
Rather, the stress is the steel is obtained by
i determining the strain in the steel, assuming a linear strain distribution across
the section
and
ii using the stress-strain curve in Figure 2.2 o the code to arrive at the stress.
Strain
I
I
I
200 ,
kNAnm
8
8
0.87x460=400 N mm
2
z
Strain diagram
Stress-Strain diagram
4. t is possible to use this method because the stress-strain curve for steel below the
yield point is a single straight line.
oncluding otes
5. One way of ensuring that the beam failure is ductile is to introduce some compression
steel,
so
that x will be reduced
to
0.5 See Example 3 .
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f
=
25
N/mm
2
eu
f
=
460 N/mm
2
y
EXAMPLE
3 - ANALYSIS
OF
DOUBLY REINFORCED
SECTION
Detennine the amount of compression steel required, in order to make
Example 2. Find also the moment
of
resistance
of
the resulting beam.
.150
d
s d=3
2-25
o 0
All dimensions
in mm
Introductory
Notes
x/d
=
0. 5
in
1 I f
it is found that a singly reinforced beam is over-reinforced and it is desired
to
make
it under-reinforced or balanced, this
may be
achieved by
i) increasing the depth
of
the section,
ii)
increasing the breadth
of
the section
or
iii)
introducing compression steel.
2. Increasing the breadth
of
the section will generally
be
uneconomical. Therefore,
if
the
depth of the section cannot be increased due
to
non-structural reasons, option
iii
above is used.
Reference Calculations Output
Note 3
Assume a suitable value for
d ,
say 50 mm.
d
= 50
mm
Fo r
equilibrium
of
the section, the compression
in
the
top
steel plus the concrete must equal the tension
in the bottom steel.
Setting x = 0.5)d = 150 mm which automatically
ensures the yielding
of
tension steel), we have
d /x =
50/150
=
0.33
< =
0.43, which means that
the compression steel will yield as well.
3.4.4.4
0.87)f
y
.A
s
0.45)f
eu
·b 0.9)x = 0.87)f
r
0.87) 460)A
s
0.45) 25) 150) 0.9) 150
=
0.87) 460) 981.7)
A 412 mm
2
s
Hence,
As
=
412
mm
2
,
Use 4T12
Note 4
Us e 4Tl2 As =
452.4 mm
2
).
452.4 mm
2
)
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Reference Calculations
Output
Table 3.27 Note:- lOOA
s
/A
c
=
100) 452.4) / 150) 350)
Note 5
=
0.86 0.2), Hence O.K.
Lever arm for balanced section
=
d -
0 . 4 5 ~ 1 .
=
0.775)d
=
0.775) 300)
=
232.5 mm
Distance between top and bottom steel = 250 mm
Note 6
Hence, taking moments about level of tension steel,
moment.
of
resistance =
0.45) 25) 150) 0.9) 150) 232.5)
+
0.87) 460) 412) 250)
=
94187006
Nmm
=
94.2 kNm
M = 94.2
kNm
Notes on Calculations
3. The value
of
d will depend on the cover, and other requirements See Example 8).
4.
If
the compression steel provided is greater than that required, the neutral axis depth
will be reduced slightly; this is desirable, as it will increase the ductility of the
section. When providing four bars within a width of 150 mm,
it
may
be
necessary to
use the bars as two pairs of bars.
5. When compression steel is provided, a minimum percentage is required. The area
of concrete is based on the gross section, and the overall depth is taken
as
300
50)
=
350 mm.
6. In general, the most convenient way of fmding the moment of resistance for a doubly
reinforced section, is to take moments about the level of tension steel.
The
amount
of compression steel to be used in the calculation is the amount required 412 mm
2
,
and not the amount provided 452.4 mm
2
.
Concluding Notes
7
The
moment of resistance of a doubly reinforced section
can
be considered to
be
the
sum of the moments of resistance of i) a balanced section and
ii)
a steel section
consisting of equal amounts of tension and compression steel, separated by d-d ).
150 150
~
>
< ~ ~
t -
d =5
1
I
4 ~ m
- _ o : ~ ~ ~
i
d=300 _ ~ : . - G . . . . . - l 232.5
1
982 570mm2 I
o
0 -
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EX MPLE
4 - ANALYSIS
OF
NON-RECTANGULAR
SECTION
f
=
25 N/mm
2
cu
2-
f
y
= 460 N mm
I
=4
f
=45
Determine the moment carrying capacity
of
the trapezoidal e m section·shown below.
3
6
All dimensions in mm)
Introductory
Notes
1. As in previous examples, the moment carrying capacity h s to b e found by working
from first principles. The additional complication in this example is that the section
is non-rectangular.
Reference
Calculations Output
Assume values for the neutral axis, x until the
compression
in
concrete is
equ l
to the tension in
steel.
The area of the section under compression
=
0.5) 0.9)x[600 - { 3OO-150)/450} O.9)x]
Area of steel
=
981.7
mm
2
~ ~
Assume also that the steel s yielded. \10.9><
Try x
=
100 mm 0
Area in compression, A
c
~
=
O.5) JO){600 - O.33 O.9 IOO
25650 mm
2
C
=
0.45)f
cu
.A
c
=
0.45) 25) 25650)
=
288563 N
T
=
0.87) 460) 981.7)
=
392876 N
Try x
=
139 mm
Then, C
=
392868
Nand
T
=
392876 N.
This approximation is satisfactory.
x
=
139 mm
Note also that x/d
=
139/400
=
0.35 < 0.5; hence
assumption that steel has yielded is
O.K.
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Reference
Calculations
Output
The centroid of the compression zone from the top of
the section
will
be given by
y = { 150 139 139/2 O.5 150 139 139/3 } 1
{ 150 139 0.5 150 139 } = 61.8 mm
Note 2 Hence, lever arm
=
400 - 61.8
=
338.2 mm z
=
338 mm
M
=
C z
=
392868 338
=
132.8 x10
6
Nmm
=
133 kNm M
=
133 kNm
Note:- Alternative method of finding
x.
Assuming that steel has yielded,
T
=
0.87 460 981.7
=
392876 N
For any
x
the area under compression is
A
c
= O.5 O.9 x[600 -
{ 300-150 /450} O.9 x]
C
=
O.45 25 A
c
Putting T
=
C, we have the quadratic equation,
x
2
- 2000 x
258684
=
0, x
=
139 mm
giving x = 139 or 1861
mm
Since x/d
=
139 4
=
0.35 <0 5 steel has in fact
Note 3 yielded, as assumed.
Notes on Calculations
2. The lever arm cannot
be
calculated as d - 0.45 x in this
case
because the
compression block is non-rectangular.
3.
This calculation will become a little more complicated if the section is not under
reinforced see Example 2 .
Concluding Notes
4. This approach from first principles, using the idea
of
strain compatibility, will have
to
be
employed even in the desi n
of
beams such as these, which are non-rectangular,
since the design formulae and charts apply only to rectangular sections. When
designing, the amount
of
steel has
to be
assumed, and the moment carrying capacity
checked to ensure that it is greater than the design moment.
5.
should be noted that the form
of
the formulae given in the code is such that,
although they can
be
used to design rectangular sections, they are not meant
to
find
the moment
of
resistance of a given section. This has
to be
done using strain
compatibility concepts from first principles, as illustrated
in
Examples
to 4, or by
suitably rearranging the form
of
the equations.
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EXAMPLE - DESIGN
OF
RECTANGULAR SECTION
Design a rectangular beam to take an ultimate load moment of 150 kNm,
a) as a singly reinforced beam and
b) as a beam whose overall depth is limited to 400 mm.
Use design formulae. Assume that feu =
25
N/mm
2
f
y
= 460 N/mm
2
and that the
difference between effective depth and overall depth is 50 mm. Assume also that no
redistribution of moments has been carried out.
Introductory Notes
1.
This is the first example on the
as
opposed to the analysis of a section.
Where beams as opposed
to
slabs) are concerned, it will be often found that the
moment carrying capacity is more critical than the deflection criterion, and that the
former will govern the selection
of
cross sectional dimensions.
Reference
Calculations
Output
a) Singly reinforced section
Note 2 Let us assume that d/b = 2.0
In order
to
find the minimum depth for a singly
reinforced section, we should assume that x/d = 0.5
3.4.4.4
and K = K = 0.156
Then K = M / b.d
2
f
e
J
0.156 = 150 x10
6
) /
{ d/2) <¥) 25)}
d
3
= { 2) 150 xlQ6 }
I
{ 0.156) 25)}
d = 425 mm
.d
min
= 425 mm
Note 3 Choose d =
475
mm, h = 525 mm, b = 225 mm
d
=475
mm
h = 525 mm
Now K = M / b.d
2
f
eu
)
b = 225 mm
= 150
xl0
6
/
{ 225 475t 25 = 0.118
<
0.156
0
225 •
z = d[O.5 {0.25 - K I 0 . 9 } o . ~
~ ~ T h
= 475)[0.5
{0.25 - 0.118)/ 0.9)}O.5]
3.4.4.1 e)
= 401
mm <
0.95) 475) = 451
mm;
hence O.K.
As = M / 0.87)f
r
Z
= 150
xlW
0.87) 460) 401) = 935
mm
2
Hence, use·21 25
As
= 981.7 mm
2
)
As
= 935 mm
2
Table 3.27 lOOA/A
e
= 982) 100) / 525) 225) = 0.83
Use 21 25
Note 4
>
0.13; hence O.K. 981.7
mm
2)
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eferen e
alculations
b )
Overall depth restricted
Output
Note 5
Note 6
I f
the overall depth
is
restricted
to
400 mm,
h = 400 mm, d = 400 - 50 = 350 mm, d = 350
mm
b
=
225
mm
assuming the same breadth as before) b
=
225
mm
Now K = M
1
{b.d
2
.f
c
= 150 x
10
6
) 1
{ 225) 350)2 25)}
=
0.218 > 0.156 Le. K )
Hence, compression reinforcement is required.
Let
us
assume that
d =
50 mm.
Table 3.27
3.4.4.1 e)
Note 7
Table 3.27
Note 8
As =
K-K )f
cu
.b.d
2
1 { O.87)f d-d ))
=
{ O.218-0.156) 25) 225) 350f}
1
{ O.
87) 460) 350-50)}
=
356
mm
2
Use
2Tl6
~
=
402.1 mm
2
)
looA
s
/A
c
=
100) 402.1)1 400 225
= 0.45 > 0.2; hence O.K.
z = d[O.5
{0.25 - K / O.9)}O.s]
=
350)[0.5 + {0.25 - 0. 156 / O.9 }O.s]
= 272
mm
< 0.95) 350) = 333 mm; hence O.K.
As = { K .f
cu
.b.d
2
)
1
O.87)f
y
z}
As
={ O.156) 25) 225) 350)21 O.87) 460) 272))
356
=
1344
mm
2
Use 3T25
As =
1473 mm
2
)
looA/A
c
=
100) 1473) 1 400) 225)
= 1.64 > 0.13; hence O.K.
Hence, use 3T25 bottom) and 2Tl6 top).
= 356 mm
2
s
Use2Tl6
402.1
mm
2
)
2 2 5
A =
1344 mm
2
s
Use 3T25
1473
mm
2
)
Notes on alculations
2. In practice, the ratio
of
depth to breadth for a beam will have a value between 1.5
and 2.5.
3. Many designers still choose dimensions for beams and columns in steps
of
25 mm,
because 1 inch is approximately
25
mm. Furthermore, depths considerably
in
excess
of
the minimum depth for a singly reinforced section may be chosen,
in
order to
reduce the steel requirement.
4. The check for minimum reinforcement is almost always satisfied for tension steel
in
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beams. A little care should be excercised, however, for compression steel.
5. The overall depth
of
the beam may have to be restricted, due to architectural
requirements. On the other hand, there may be some economy in designing beams
with a marginal amount
of
compression steel, because longitudinal steel on the
compression face will be required anyway,
in
order to support the shear links.
6.
This is keeping with the idea that the difference between overall and effective depths
is 50 mm.
7. When calculating the are of tension steel, it is sufficient to use the value
of
compression steel required (as opposed to that provided), in this equation.
8.
When providing reinforcement, a combination
of
bar sizes should be adopted, such
that the maximum and minimum spacing between bars is kept within specified limits
(see Example 12).
Concluding Notes
9. Design charts (in Part 3 of the code) could also have been used to design the steel
required for these sections. The relevant charts are Chart No. 2 for the singly
reinforced section and Chart
No 4
for the doubly reinforced section, since
d /d
50/350
0.143.
10. The design charts are given for· ,
d /d
values ranging from 0.10 to 0.20, in steps
of
0.05. The chart with
d /d
value closest to the actual value should be used for
design. the actual
d /d
value lies exactly between the chart values, the chart with
the higher
d /d
value should be used in the design, as this is more conservative.
EXAMPLE 6 - DESIGN
OF
SECTION
WITH
REDISTRIBUTION
the beam section in part a
of
Example 5 (Le. h
=
525 mm, d
=
475 mm and b
=
225
mm was carrying an ultimate moment
of
150 kNm after a 30 downward redistribution
of
moment, design the steel reinforcement required. Assume that d 50 mm, feu
25
mm
2
and f
y
460
mm
2
Use the methods of formulae and design charts.
Introductory Notes
1.
the moment at a section has been reduced by downward redistribution, that section
must have adequate rotational capacity at ultimate limit state, in oder for plastic hinge
action
t
take place. This capacity is ensured by restricting the
x/d
ratio to a specified
value.
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Reference
Calculations
Output
Cal
Using formulae
3.2.2.1 b)
b
= 1-0.3) 1 1 = 0.7
3.4.4.4 K = 0.402) l3
b
-0.4) - 0 . 1 8 ~ - 0 . 4 2
=
0.402) 0.7-0.4) - 0,18) 0.7-0.4)2
=
0.104
Now, K = M 1 b.d
2
.f
e
=
150 x10
6
)
1
{ 225) 475)2 25)}
=
0.118
>
0.104
Hence, compression steel is required.
z
=
d[O.5
{0.25 - K / 0.9)}0.5]
= 475)[0.5
{0.25 - 0.104)/ 0.9)}O.5]
3.4.4. 1 e)
=
412
mm
<
0.95) 475)
=
451
mm; hence
O.K.
A = 104 mm
2
s
Use
2Tl2
226.2 mm
2
)
As = K
-K f
eu
·b.d
2
1
{ 0.87)fy<d-d )}
=
{ O. 118-0. 104) 25) 225) 475t} 1
{ 0.87) 46O) 475-50)}
= 104 mm
2
Use 2Tl2 As
=
226.2 mm
2
)
looA
s
A
e
=
0.19 « 0.2, but acceptable)
able 3.27
As = [ K .f
eu
.b.d
2
)
1
{ 0.87)f
y
z ]
As
= { 0.104) 25) 225) 475)2
1
0.87) 46O) 412)}
+
104
= 905 mm
2
Use 2T25
As
=
~ 8 1 . 7 mm
2
)
Hence, use 2T25 bottom) and
2Tl2
top).
A
= 905
mm
s
Use 2T25
981.7
mm
2
)
Chart 3
part
3
bl Using charts
A p p r o ~ r i a t e
chart for
feu
=
25
N/mm
2,
fy
=
460
N/mm and d /d = 5 475 = 0.105
is
Chart
No.3.
225 ~
I
2 ~
I
5 75
2-25
o 0
M/bd
2
= 150 x10
6
)
1 225) 475)2 = 2.95
3.2.2. 1 b) x/d has to be restricted to
{3b-O.4 ,
i.e. 0.3
Note
2
Note 3
Hence, the values for lOOA/bd and
l
s
/bd
must
be read of f the point at which the horizontal line
M/bd
2
= 2.95 cuts the
x/d
= 0.3 line.
Thus,
looA/bd
=
0.85 and looAs /bd
=
0.1
As = 0.85) 475) 225)/ 100) = 908 mm
2
; Use 2T25
As
=
0.1) 475) 225)1 100)
=
107 mm
2
; Use
2Tl2
to satisfy minimum steel requirement. -
A
=
908
mm
2
s
Use 2T25
A = 107
mm
s
Use 2Tl2
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Notes on Calculations
2. Any combination to the left of the line corresponding to the
x
=
0.3 line will give
a feasible combination of lOOAjbd and lOOA
s
/bd. f a point on the line itself is
chosen, the solution will generally e the most economical one,
in
terms of the total
amount of steel required.
3. The differences between the solutions by formulae and charts are very small indeed,
despite the fact that the design charts are based
?n
the parabolic stress block for
concrete stress, while the formulae are based on the simplified rectangular one. is
the design charts that are used for everyday designs.
Concluding Notes
4. Although the applied moment for this section was the same
as
that
in
Example 5,
because
of
the restriction on the neutral axis depth for the purpose
of
ensuring plastic
hinge rotation, this section
had to
be doubly reinforced.
5. Hence, doubly reinforced sections may need to e resorted to when
i) architectural requirements place limits on the beam depth and/or
ii) when a significant degree of redistribution of elastic moments has been
carried out at that section.
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EXAMPLE 7 - STRUCTURAL ANALYSIS OF BEAM
Determine the design ultimate load moments for the beam shown in the figure, using also the
following information.
i) Dead load from the parapet wall can be taken as a line load of 2 0 kN m
ii) Allowance for finishes on the slab can be taken as 1 0 kN/m
2
.
iii) Imposed load
on
slab should
be
taken
as
4 0
kN/m
2•
iv) Density of reinforced concrete = 24 kN m
3
•
Introductory
Notes
1. This example involves load evaluation and a simple stru,ctural analysis on
appropriate loading patterns, in order to find the design ultimate moments.
100
Beam Section
Sectional Elevation
_ i
- - - ~ - - - - - - - - - - - - - - - -
I
_
r r ~
I I
I I
I I
I I I
I
r ~
-
-
~ r
I I I
I I I
I I
I
I I I
____
~
___
~ ~ l
.Ly-
Plan
18
3500
3500
All dimensions
in mm
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Reference
Note 2
Note 3
3 2 1 2 2
Calculations
The beam can be idealised as follows.
_
t - - - . . : . 6 X X : . : . : . - - - - - ~
The critical moments for design w
be
i Hogging moment at B
ii
Sagging moment in span BC
o d i n ~ on beam per m l e n ~ t h ;
From slab
=
0.125 24 3.5
=
10.5
k m
From finishes = 1.0 3.5 = 3.5
k m
From
beam
= 0.45-0.125 0.3 24 = 2.34 k m
Total dead load udl 16.34 k m
Dead load point load at A = 2.0 3.5 = 7 0 k m
Live load udl
4.0 3.5 14.0
k m
The hogging moment at B will be maximum when
the cantilever portion AB is loaded with the
maximum design ultimate load, irrespective
of
the
load on the span BC.
The
sagging moment in BC will
be
a maximum when
the cantilever portion AB has the minimum design
ultimate load, while the span Be has the maximum
design ultimate load.
Maximum design ultimate load udl
=
16.34 1.4 14.0 1.6
=
45.28
k m
Minimum design ultimate load udl 16.34 k m
Hoe;e;ine;
moment
atB:
7 0x1 4 4 5 2 8
k rn
~ f t
M
B
7.0 1.4 1.95 45.28 2.0 2/2
=
109.7 kNm
Output
M
B
110 kNm
hogging
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Reference
Calculations
SUging
moment in BC:-
7 0 1 6 3 4 k rn 4 5 2 8 k rn
L ~
.
A B
x
M
B
=
7.0 1.95 16.34 2.0 2/2
=
46.33 kNm
Taking moments about
B
for
Be
Rc 6.0 46.33
=
45.28 6.0tl2
Rc
=
128.1 kN
M
x
= 128.1 x - 45.28 x
2
/2
dMx/dx = 0 when 45.28 x = 128.1
x
=
2.83 m
M
max
= 128.1 2.83 - 45.28 2.83tl2
= 181.2 kNm
Output
M
BC
=181 kNm
sagging
Notes on Calculations
2. Idealization is the first step in analysis. Since it is not possible to model the actual
structure with complete accuracy, idealization should be performed such that the
results obtained are conservative.
For
example, although point
C
has a certain degree
of restraint, it is impossible to quantify it. However assuming the end C to be simply
supported will give a higher and hence conservative moment in the span
Be
The
restraint moment at
C
c n be subseqently accounted for by providing a nominal
amount of.hogging steel there.
3. Since the beam spacing is 3.5 m, each beam carries the loads acting on a strip 3.5
m wide.
Concluding Notes
4.
Where dead and imposed loads are combined, as in the case
of
this example, the
design moments at critical sections have to be arrived at y a proper combination of
loading patterns.
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EXAMPLE 8 - DESIGN OF BEAM FOR FLEXURE
Design the reinforcement for hogging and sagging moments
in
the
e m
in Example 7. Use
feu = 25 N/mm
and f
y
= 460 N/mm
Introductory Notes
1.
In this example, only the reinforcement for the maximum sagging and hogging
moments need to be calculated, since the
beam
section is already specified
in
Example 7.
2. Furthermore, as the bending. moment diagram for the beam has not been drawn
although it could be), the curtailment
of
reinforcement is not considered. This aspect
is considered
in
Example 12.
Reference·
Calculations
Output
Effectiye de pth
Table 3.2 Assume moderate exposure conditions, for outdoor
Note 3 exposure.
Note 4 Making use of Notes 5 and 6 of Table 1 we
can
use
TABLE 1
a cover
of
30 mm. cover = 30 mm
Table 3.5
This will also give a fire resistance of 2 hours.
Assuming a link diameter of
10
mm
and a
reinforcement size of 25 mm, the effective depth will
Note 5
be d = 450 - 30 -10 - 5 = 397.5
mm
d = 397.5 mm
s i ~ n
for
h o ~ ~ i n ~
moment
The beam behaves a a rectangular beam.
b
=
300 mm, d
=
397.5 mm, M
=
110 kNm
Chart 2
M/bd
= 110 xl
6
) 1 300) 397.5)2 = 2.32
Part 3)
lOOA/bd = 0.67
As
=
0.67) 300) 397.5)1 100
=
799
mm
2 A
=
799 mm2
s
Use 21 20
ITl6 As
= 829 mm
2
) Use 21 20
Table 3.27
looA/bwh = 100) 829) 1 300 450 lT16 829 mm
2
)
= 0.61
>
0.26; hence O.K.
hogging)
Design for sagging moment
The beam behaves as a flanged beam.
3.4.1.5
b = lesser of 3500 mm or
1 5
+
b
w
=
{ 0.7 6000 }/5
+
300
=
1140
mm
Hence, b = 1140 m
b
f
= 1140 m
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Reference
alculations
utput
3.4.4.4
Assume that the neutral
axis is within the flange.
K = M
I
b.dz.f
cu
=
181
x
6
/{ 1140 397.5t 25 }
=
0.040 <0.156
z
=
d[0.5
+
{0.25 -
K/ 0.9 }o.s]
=
d[0.5 + {0.25 - 0.04 / 0.9 }o.5]
= 0.95 d = 0.95 397.5 = 377.6
mm
x =
d-z / 0.45
= 397.5-377.6
1 0.45
= 44.2
mm
<
125 mm.
Hence, neutral axis is in fact
within
the flange, and N.A. is in
the beam can be designed
as
a rectangular beam with flange
b =
1140 m m.
Chart 2
M/bd
z
= 181
xl
6
1 1140 397.5 z = 1.00
Part
3)
l00A/bd = 0.27
A
=
1224
mm
2
As
=
0.27 1140 397.5
1
100
=
1224
mm
z
s
Use 21 25 11 20 As
=
1295
mm
2
Use
21 25
Table 3.27
bwlb
=
250/1140
=
0.22 < 0.40
11 20
Note 6
l00A,Ib
w
h = 100 1295 1 300 450
1295 mm
z
=
0.96
>
0.18; hence O.K.
sagging)
ran erse
steel
In ordeJ:l
that flanged
beam action is
ensured,
the
minimum amount
o
transverse
steel
to
be
provided
in
the top
o
the slab
is
given
by
Table 3.27
l00A
st
/htl = 0.15
A
st
= 0.15 125 1000
1 100
=
187.5 mmz/m
Transverse steel
Note 7 Use R6 @ 150 min. A
st
= 190
2
/m
R6@150
min)
190
mmz/m)
Slenderness check
3.4.1.6
Continuous portion - clear distance
between
restraints
is
5700
mm
60 b
c
=
60 1140
=
68400
mm
250 b/ld
=
250 1140 21
397.5 =
817358 mm
Since these values are
>
5700
mm,
check
is O.K.
Cantilever portion - clear distance between restraints
is
1850
mm
25)b
c
= 25 300
=
7500 mm
l00 b/ld =
100 300 2
I 397.5) = 22642 mm
Slenderness
Note 8
Since these values are > 1850
mm,
check is O.K.
O.K.
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Notes
on
Calculations
3.
is sufficient to assume a moderate exposure condition for the exteriors
of
most
structures, which
re
not subjected to freezing and sheltered from driving rain.
4.
The cover values re obtained from TABLE
1
in the Introduction to this text; this
Table is relevant for Sri nk n concreting practice. The figures in the table can be
further modified by Notes
5
and
6
of
the table, as has been done here.
is assumed
in this example therefore, that the mix proportions correspond to a grade 30 mix
although the strength achieved is only grade 25 and also that a 5 mm min)
cement: sand rendering protects the concrete surface.
5. The calculation
of
effective depth from the overall depth is illustrated by the figure
below.
cover
_
she r link
y
rs
~
t
x
6. Although the actual steel requirement is calculated using the value
of
flange width,
the minimum steel requirement is based on the web width.
7. This transverse steel will also have to resist the hogging moment in the slab, and a
greater amount than this will need to be provided in most cases.
8. This slenderness check is almost always non-critical, except perhaps in the case
of
long, deep cantilevers.
Concluding Notes
9. When designing beam-slab systems, care must be taken
to
note where flanged
e m
action takes place and where it does not. Furthermore, such locations will be reversed
in systems where upstand beams are used.
10. the neutral axis
of
a flanged beam falls within the flange, the design is identical to
a rectangular beam, as seen here.
11. When designing for hogging and sagging moments at support and span respectively,
care must be taken to remember what steel has to be placed at the top
of
the beam
section, and what steel at the bottom.
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EXAMPLE 9 - DESIGN
OF F L N ~ E
SECTION
Design
an
edge beam
of
a beam-slab system
to
take
an
ultimate moment
of
200 kNm at mid
span.
Spacing
of
beams = 4.0 m; Span
of
beams = 6.0 m;
Thickness
of
slab= 100
mm , f
=
25
mm
f
= 460
mm
u
,
y .
Introductory Notes
1. An
edge beam will have a transverse slab only on one side; hence it is called an
L-beam. The beam in the earlier example
is
called a T-beam, since the slab extended
over both sides of the beam.
If
the beam is below the slab as is the case most
of
the
time), the slab will act
as
a flange only
in
the span, when the top
of
the section is
in
compression, and not at the supports. .
Reference
Calculations
Output
Note 2
Assume that b
w
=
225 mm
3.4.1.5
b = lesser of 2000
mm
or
lilO
b
w
=
{ O.7 6000 }/lO
225
= 645
mm
Hence,
b
=
645 mm
b = 645
mm
the
beam
is
to
e
singly reinforced,
K=K
= 0.156
M
1 b.d
2
f
cu
=
0.156
200 x10
6
) 1
{ 645) d)2 25)} = 0.156
d
min
= 282
mm
d =
rpm
d = 325
mm
Note 3
Hence, choose d = 325
mm
and
h
= 375
mm h
= 375 mm
3.4.4.4
Then, K = 200
xl0
6
)
1
«645) 325)2 25)} = 0.117
z = d[O.5
{0.25 - K/ O.9)}O.5]
= d[0.5
{0.25 -
O.117)/ O.9)}O.5]
= O.846)d = 275 mm
x = 325-275)
1
0.45) =
111
mm
Note 4 Since this is greater than
h
f
=
100
mm , the neutral N.A. is out
of
axis lies outside the flange.
flange
b/b
w
=
645/225
= 2.87
d/h
f
=
325/100
= 3.25
equation 2
{
= 0.129
Note 5
3.4.4.5
{3f.f
cu
.b.d
2
= 0.129) 25) 645) 325)2 = 219.7 xlO
6
Note 6 = 220
kNm >
200
kNm
Also,
hid
=
100/375
= 0.308
<
0.45 singly
Hence, section
can be
singly reinforced.
reinforced
24
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ReCerence
Cakulatioas
Output
equation 1
AI
=
+ 0.I f
cu
.b
w
.d{ 0.45)d-b,)] I
Note 7
0.87 f
{d- 0.5)hrH
= [ 2 0 0 x l 6 + 0 . 1 X 2 S ~ 2 S { 0 . 4 5 3 2 5 - 1 0 0 } f
As = 1894 mm
[ O.87) 460){325- 0.5) I00))] =
1894 mm
Use 21 32 .
11 20
As = 1922 mm
Use 21 32 .
3.12.6.1 l00A/bwh = 100) 1922) I 225 375 11 20
Note 8
. = 2.28 < 4.0; hence O.K. 1922mm
2
)
Notes
on Calculations
2. A web width
of
225
mm is
around the minimum that is practically desirable, in order
to accommodate the reinforcement. A width
of
200
mm can be
considered as the
absolute minimum for
all beams
save those which carry very
nominal
loads.
3.
The
difference
between
d and
h
has
been
taken
as
SO
mm,
although the
actual
calculation of cover should be carried out as in Example 8.
4. This trial-and-error
approach
has to be adopted to find
out
wbetbet
Clause
3.4.4.5
has
to be used singly reinforced flanged
beam
design)
or whetha
it is sufficient to use
Clause 3.4.4.4.
rectangular
beam design, since the neutral axis is within the flange).
5. It
is
easier and
more
accurate
to
use equation 2 to obtain
the
value of
P •
rather than
to
resort to double interpolation in Table 3.7.
6. Pf.fcu.b.d i is
the
greatest moment capoci.ty for a singly
reinfcm:ed
section when x is
restricted to 0.5)d.
7. This
equation
for A is slightly conservative, as it assumes that x
=
(0.5)<1 , although
the
actual
neutral axis may be somewhere between x = b
f
and x = (0.5)<1. Since the .
width of the web is relatively small,
compared
to the flange, this discrepency is
negligible
and
conservative.
8. This check for maximum percentage of reinforcement is also almost always satisfIed,
except for very heavily reinforced sections. Although the
check
is
satisfIed
here,
care
will have to
be
exercised
lapping is done.
Concluding Notes
9. This example illustrated the situation where the neutral axis fell
below
the flange of
a flanged beam. Design charts cannot be used in such a situation, and the equations
.. in Clause 3.4.4.5 have to be employed.
10.
In
addition,
if
the moment is
greater
than
P
f
.fcu
.b.d
2
i.e. compression steel is
required), or if more than 10 redistribution
has
been carried out,
the beam
has to
be designed from strain compatibility fIrst principles as given
in
Clause 3.4.4.1.
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EXAMPLE
10
- DESIGN
OF
SECTION
FOB.
SIlEAB.
A simply supported beam ofcross section.b
=
22S
nun
and d =400 mmcarries an ultimate
load
of
60
leN/m
over its clear
span
of 5 0 m
Design the
shear
reiDforcemeat
required
near
the support, assuming
that
the pe.n:entage of teDsionmntOl'CelDalt at tbesupporl is 8 ~
Assume feu
=
S N/mm
2
and
=
SO N/mm
2
•
Introductory Notes
1. The
two
main effects caused by flexure are bending
moment
and shear.
The
bending
moment in
a
concrete
beam is
carried
by
steel
reinforcement parallel
to
the
axis.
The
shear force is carried by steel reinforcement
in
atransvene direction,
generally in the form of:linb
2. possible, mild
of fyv =
250 N/mm
2
is preferred for links,as
it
is easier to
bend into
shape,
compared to high yield steel.
Links
generally
have
diameters varying
from 6
to
12
mm,
in
steps
of
2
mm
Reference
Note 3
3.4.5.10
3.4.5.2
Note
4
equation 3
Note 5
Table 3.9
Table 3.8
Note 6
Note
7
3.4.5.5
Note 8
akuiatioas
Output
Although the shear force will be
maximum .atthe
face
of the
support the
deaign bear
force
for
uniformly distributed
loading
is
at
a section wd
w
from
the face.
s i g n
V
max
.
:
~ d ~
~ 2 5
V
JIWt =
« ) 5)/2
= 156
leN
vJIWt = 156
xloJ
(225}(400) = 1.67
N mnJl
V
max
= 1.67
0.8) Wo.
s
= 0.8) 25)0.5 N/mm
2
= 4 N/mm
2
> 1.67 mm
2
< 5 N/mm
2
;
hence O.K.
VdeIip
=
156) 2500-400)
1 2500
=
.126
leN
v
= V/ bv.d) = 126
xloJ)
I 225) 400)
=
1.4
N/mm
2
v
1.4 mm
l00AJb d
= 0.8, d 400 nun, feu =
25
N/mm
2
;
Hence,
v
e
= 0.58 N/mm
2
• V
c
= 0.58
Since v
>
v
e
0.4 N/mm
2
, links have
to
be N/mm
2
designed.
v
>
=
bv·sv<v-vJ I
(0.87)fyv
Assuming 10
mm
links,
v= 157.1 mm
2
Hence,
Sy
<
=
157.1) 0.87) 250)
I 1.4-0.58 225
= 185 mm
<
0.75)d = 300 nun;
hence
O.K.
Links
Use
R,lO
links @ 175 Mm. RIO
@
175
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Notes on
Calculations
3. This
is
the simplified method
to
account for the enhanced shear resistance near
supports. hesection
considered
should be an effective
depth
away from the
face
of
the support. Where support details are not available, it will be comervative to
measure
d
from the centre-line
of
support.
4.
This
is
the maximum shear check.
f
this
fails, there is
no alternative but
to
change
the
beam
dimensions. It is prudent therefore, to make this
check fairly early
in the
design procedure.
5. b
v
for a flanged
beam
should
be
taken as
the
average width of the
web
below the
flange.
6. 0.4 mm
is the shear
resistance
that can be carried by nominal shear
links.
7. When
using
this
inequality
for providing links, either the
v
value
or
Sy
value must
be
chosen.
In
~ the
A value is
assumed
and
the
Sy value
c:a1culated.
The
A.
v
value refers
to the total
cross section of links at the neutral
xis
of a section.
Gea1etally it is twice the area
of
the chosen
bar,
since in
most
cases it is
links
with
2 vertic l legs that re
used.
he resulting Sy
value should
not
exceed 0.75 d,
to
ensure that
at
least
one
link crosses
a
potential shear crack. he
transverse spacing
between the legs
of
a link should
be
such that it does
not
exceed -d- and that
longitudinal
tension
bar
is
gre ter
than 150 mm from a vertical
leg.
8. he
link
spacing is also often specified
in
steps of 25 mm, because of the tendency
to think in Imperial
units.
1
inch
is pproxim tely
25 mm.
Concluding Notes
9.
In this
example, only the shear reinforcement requirement near
the
support
has
been
calculated. The requimnent close
to mid-span will
be much less. This
aspect will
be
considered
in the next example.
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EXAMPLE 11 - DESIGN
OF
BEAM
FOR SlIEAa
A simply supported beam, with d = 550 mm
and
b ... 350 mm and
clear
span 6.0
m
is
subject to a triangularly varying shear force diagra,m, with a value of 400 kN at
the face
of
the
supports.
Th e
mid
span steel
consists of
4
Nos.
mm bars. Design the
shear
reinforcement required over the entire span,
if
two
of the
main
bars are bent
at
45° near
the supports. Take feu =
N/mm
2
, f
y
= 460
N/mm
2
and
fyv
=
25 N mm .
Introductory Notes
In this example,
two
bent
up
bars are also used to provide shear reinforcement near
the
beam
supports.
2.
The most
reasonable
way to provide shear reinforcement for the entire
span
would
be to consider three areas - Le.
(i) the
support area where bent up
bars are
also effective in addition to links,
(ii)
the middle
of
the
beam,
where only
nominal
links would suffice , and
iii the portion in
between
the above.
eference
Calculatioos
Output
SumutNCl l
.
V
max
=
400 kN
v
lDllX
= (400 xl<P)
I
(350)(550) ... 2.08 N/nun2
V
max
= 2.08
3.4.5.2
0.8) W
0
s
=
.(0.8)(25)0.5
N/mm
2
=
4 N/mm
2
>
2.08 N/mm
2
<
5
N/mm
2
;
hence O.K.
Shear
resistance
of 2 inclined bars,
equation 4
Vb =A.(O.87)fyb(cosa + sina.eotP)(d-d ) I
3.4.5.6
Assumethalll =
67.5° and
d
= 50 mm
Note 3 hence = (1.41)(d-d ) =(1.41)(500)
=
705 mm
Vb
=(982)(O.87)(460){0.7
1
+(0.71)(0.4
} SOO / 705
= 277890 N
Vb
=
(277890)
I
(350)(550) = 1.44 Nir
~
Vb
= 1.44
N mm
2
Since 2
bars
continue into support,
lOOA/bvd
=
loo) 981.J 1 350 550
= 0.51;
Table 3. 9 hence,
V
c
= 0.50
N/mm
V
c
= 0.5
3.4.5.10
Shear force
at
section d from support
N mm
2
Note 4 {(3000-550) I (3000)}(400) = 327
kl
l
v = (327 xloJ) I (550)(350)
=
1.70
NI
mm
2
v-v
e
=
1.70 - 0.50 = 1.20 N/mm
2
3.4.5.6 Although this can
be
resisted by the
ben
up bars
alone, half of this must be resisted by lj;flks
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Reference
Calculations
Output
A
sv
> = b
v
.Iv{ 1.20 12} 1
0.87 fyv
Putting A.
v
= 157.1
mm
2
for 10
mm links,
Iv
< = 157.1 0.87 250 1 350 0.6
=
163
mm < 0.75 <1; hence O.K.
Use
RIO
links @ 150 mm; this can be used over
the RIO
@ 150
mm
entire area over
which
the
bent
up
bars
are support area
effective - i.e.
for 0.71 m
from
the face of
support
Middle r
l00A/bvd
100 1963 1 350 550 1.02;
Table 3.9
hence V
c
0.63 N mm
V
c
0.63
Table
3.8 Shear stress
taken
by
nominal links 0.63 0.4 N/mm
2
=
1.03 N/mm
2
Shear
force
taken
by
nominal
links
1.03 350 550 Hr
3
= 198
kN
Hence, extent of
area covered
by nominal links =
{ 198 / 400 } 6.0 2.97
m
Steel for nominal links is given by
A
>
0.4 b
v
·1v
1 0.87 fyv
Putting
Aav
= 157.1 mm
2
for 10 mm
links,
Iv < = 157.1 0.87 250 1 0.4 350
=
244
mm < 0.75 d; hence O.K.
RIO
@ 225 mm
Use
RIO
links @
225
mm
middle area
f
Area in-between
Note 5
Table
3.8
Note 6
Extent
of this area
= 3.0 -
2.97 /2
- 0.71 = 0.81
m
Shear force at distance 0.71 m from support face =
{ 3.0-0.71 / 3.0 } 400
=
305 leN
v
=
1.58
N/mm
2
vc
=
0.63
N/mm
2
Since v > vc 0.4 N/mm
2
, design shear links.
v
bv·lv v-vJ
1
O·87 fw
Putting A.
v
=
157.1 mm
2
for
{6 mm
links,
Iv
<=
157.1 0.87 250
1 350 1.58-0.63
= 103 mm
Use
2RIO
links
@ 200
mm < 0.75 <1;
hence O.K.
lCQJlOO
2/1 QjaX
v = 1.58
N/mm
2
V
c
=
0.63
N/mm
2
2
R O
200
mm
area in
between
~ T 5
2T25
29
4T25
1 48m
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Notes on Calculations
3. Since
should be
t ken
as
5° and
is restricted
to
l.S(d..(i ), this assumed value
of 67 ° for
is reasonable and easy for calculation purposes.
4. This is the same approach described in Note 3 of Example 10, The links designed can
be used from the support upto the point where the main bars are cranked up.
S Although 2 bars are bent up, they also continue for at least adistance d from any
point in this section of
the beam.
Hence, the value of v
c
will
the
same as in the
middle area.
6.
the l nk spacing is less than around
ISO
mm, it will be difficult for concreting to
be carried out. Hence, as n this case, 2 links can
e
placed together, spaced wider
apart. An alternative would have been to
use 2
mm dia. links; however fabrication
w ll
be easier
if
links
of
the same diameter are
used
throughout the
beam.
Concludina Notes
7. It is not very common practice
to
use bent up bars as
described
in this example,
although
it
was n the past.
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EXAMPLE 2 -
SERVICEABnJTY
CHECKS AND DETAll.JNG
Carry out serviceability checks on the beam analysed in Example 7
and
designed in Example
8. Also carry out detailing of reinforeement, including curtailment and lapping. Assume that
type 2
defonned
bars are used as reinforcement.
Introductory
Notes
1 The serviceability checks consist of spanldepth ratio calcu1ations for deflection and
bar spacing rule
checks
for cracking.
these
simplified checks are satisfied, the beam
is deemed
to
satisfy the serviceability limit state requirements.
Refereace
Calculations
Output
Check for deflection
fSRanIde pth
rules)
Note 2
3.4.1.3 onsiderthe
n
BC; effective span
=
6000 mm
bwlb = 0.22 0.3
Table 3.10 .Hence, basic span/depth = 20.8 for continuous,
flanged beam.
Example 8
Mlbd
2
= 1.00 and
= S/8 460 { 1224 /1295 } = 272 N/mm
2
Table 3.11
Hence
P
l
I.4S (for tension reinforcement)
Notes F
2
= 1.0 (as
there
is no compression reinforcement)
4
Hence allowablespanldepth ratio = (20.8)(1.45)
All
span
=
30.16
depth
=
30.2
Actual
spanldepth= (6000)/(397.5) = 15.09
Act
spanI
< 30.16; hence O.K.
depth = 15.1
Hence O.K.
3.4.1.4
onsiderfP8D AD: effective span
=
2000 mm
Table 3.10
Basic spanldepth = 7 for cantilever with rectangular
beamaetion.
Example 8
Mlbd
2
=
2.32
and
f = (5/8)(460){(799)/829)}
=
277 N/mm
2
Table 3.11
Hence
F
l
1.07 (for tension reinforcement)
Notes
F
2
=
1.0 (as there is no compression reinforcement)
All.
span
3 4
Hence allowable spanldepth ratio = (7)(1.07) = 7.5
depth = 7.5
Actual spanldepth = (2000)/(397.5) = 5.03
Act. span/
< 7.5; hence O.K.
depth = 5.03 .
Hence O.K.
Curtailment of reinforcement
The bending moment
diagram
envelope must first
be
dmwn
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Reference
NoteS
Cakulatiolls
Por
Be. the controlling 1oa4.c:ase is when
AS
has the
minimum designultimate1Qed
aDd Be
has
the
maximum
d e s i g n u l t i m t e o d ~
This case
bas
already been
considered in
Example 7 .
7.0
6 3 4 kN m 4 5 2 8 kN m
••
.95m
Om
x
Example 7 For span BC, x= 128.1 x - 45.28 x212
= Oatx -
O.
is max. at x
= 2.83
and equal
to
18l.2
kNm
Mx .. 0 apiA<al
x
= 5.66
m
Example 8 Steel
at
span
BC
is 2T25
lno.
We can consider
curtailing the lno bar.
Note 6 M.o.R. of continuing bars A, .
=
981.7 mm
2
can
be
shown
to be 148.4kNm.
Putting
128.1 x - 22.64 x
2
148.4
we can
obtain
x =
1.63 m and 4.03 m.
3.12.9.1 These
are the theoretical cut-off points.
Note 7
Keep the
practical cut-off points an ancboragelength
3.12.9.1 c . away from the theoretical ones.
Table 3.29
Anchorage length =
40 20
-=300 mm
This ancborage length is greater
than
12)41 { 12 20 =
240
mm} or
-d-
397.5 mm .
Hence, practical cut-off points are
at
x
=
1.63 - 0.8
=
0.83 m and
x
= 4.03 0.8 =
4.83 m
Length
of
20
mm bar
required
=
4.83 - 0.83
=4.0m
Distances
to
ends from B are 5.17 m and 1.17 m.
32
Curtail 1no
bottom bar at
1.13 m
and
5.13
m from
B.
Length of bar
is 4.0 m.
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RelereD£e
Calculations
Output
/ \ M.o.R.
/ \.....
M.o.R./2
Note 8 For support B, the controlling loading case is when
spans AB and
Behave
the max and min. design
ultimate
loads
respectively.
7 OX1 4
45
•
28
kNjm 16 34 kNjm
~ B ¥ ; :
le
Taking moments about C for
AC
Rs 6.0 (7.0)(1.4)(7.95) (45.28)(2.0)(7.0)
+
1 6 . 3 4 6 ~ / 2
a= 167.7kN
My
(7)(1.4){y o
0S]
4S.28 r/2
- (167.7)[y-2.0) - 45.28-16.34 [Y-2.0j2/2
My
(9.8){y o.05] 22.64 r - (167.7)[y-2.0)
- (14.47)[y-2.0r
A
A ;B
y
=
Oaty =OandM
L =
loo.7atB.
y
0 again at y
4 J m.
Steel at support is 2T2O
ITI6. We can consider
curtailing the 1T16 bar.
Note 6
M.o.R.
of the continuing bars 628.3
mm
2
)
can
be shown to be 90.5 kNm
Putting-(9.8)(y-Q.05) 22.64 r = 90.5,
we
can obtainy 1.80m
for
sp n AB
and
from
(9.8)(y-D.05)
22.64 r
- (167.7)(y-2.0)
- (14.47)(y-2.0r
90.5,
we can obtain
y
2.30
m
for
sp n Be.
3.12.9.1 These are the theoretical cut-offpoints.
Note 9 To find where the M.o.R. of continuing b rs is twice
3.12.9.1(e) the applied moment, we can put
(9.8)(y-Q.05)
(22.64)y2 = (90.5)/2 for span AB
and (9.8)(y-0.05) 22.64 r - (167.7)(y-2.0)
- (14.47)(y-2.0r
=
90.5 /2
for
span
Be,
giving Y = 1.22 m
and
3.10 m
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Refereoce Calculatioos
TheGiffe:reoce betweenthe·SIDIlIer y values is
1 .80 - 1.22) = 0.58 m or 580 Mm. This is greater
than 12)4> 192mm) or -d- 397.5mm). The
difference between the larger
y
values is
3.10-2.30)
=
0.80
m or 800 mm,which
is alsogIeater
than
12)41 or
d
Hence, the
practical cut-off points are
y =
1.22 m
and
3.10
m.
Length of 16 mm bar required = 3.10 - 1.22
Note 10 = 1.88 m
Distances to B are 2.0 - 1.22) =O.78m span
AD)
and 3.10 - 2.0) = 1.10 m
span
Be
Table 3.29 Since the distances
to
either
side
of B
>
=
40)41
Note
11
{Le. 40) 16)=
640
mm}, anchorage
is satisfied.
J
Imine
of bars
Output
Curtail I T16
top bar 0.78 m
left) and
1.10
MCright)
ofB.
I eIlgth
of
bar
is
1.88
m.
Note 12
3.
12.9.
1
c)
Table 3.29
Note 13
3.12.8.13
Note
14
3.12.8.11
3.12.8.13
Note
15
The continuing
21 20
top bars at B
can
be curtailed at
the point of contraflexure
closer
to B in span
BC.and
lapped
with 2T12
bars which will anchor the shear
links).
Similarly
the continuing 2T25 bottom bats in
span
can be curtailed
attbe
point ofcontraflexure
closer
to
B in
span
BC
and
lapped
with
2T12
bus.
For
top bars, distance of point
of
contraflexure from
A is 4.23 m. This
would
be the theoreticalcut-off
point To find the practical cut-off
point,
continue
the bars
for
an effective
depth
{Le. 397.5 mm >
12cP)}. Heace, cut-off paint is 4.23
0.4=
4.63 m
from
A, Le. 4.63 -
2.0
= 2 63 m.to the
right
of B.
The lapped 2T12 bars will start 40) 12)
=4S0mm
before th e
curtailment
of the 21 20 bars Le. 2.63
0.48 =2.15 m to the right of B.
~ Min.
lap
length
= gtQterof
5 ~
::: ISO
mm)
or 300 mm is satisfied;
also
distance between
laps will
be
greater
than 75 mm and 6)4> =72
mm).
For
bottom bars, distance of
point
of contraflexure
from C is 5.66 m, Le. 6.00 - 5.66
=
O.34m
to the
right of
B. As
before,
the
practical cut-off point
would be 397.5 mm beyond this. Hence,
it
would be
0.4 - 0.34
=
0.06
m
to
the
left
of
B.
Th e
n
bars
will start 0.48 - 0.06 = 0.42
m to
the
right
of B.
34
Curtail 2T20
top
bars 2.63
m
to .right
of
B.
Start
2T12 top
bars
2 ~
m to
right of
B.
Curtail 2T25
bottom bars
0.06 m to
left
ofB.
Start 2Tl2
bottom
bars
0.42
m
to
right
ofB.
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Relerenee
Calculations
lT16
mo al
2 4>
Zf 2
I
,
•
Zf 2
t
21Q
2I25
l IID
A
B
Crack width check <Bar mcW : rules)
21 25t
c
Output
Example 8
Cover required =
30 mm
Assume link diameter
of
10 mm
Example
8
Table 3.30
Note 16
3.12.11.2.2
Note 17
3.12.11.2.5
Example 8
Table
3.30
Note 16
3.12.11.2.2
Note 17
3.12.11.2.5
3.12.11.1
Note 18
Considering the sUI Wrt section (tension on top),
Clear
spacing
between
top
bars
(21 20
lT16)
=
{300 - (2)(30) - (2)(10) - 20+20+
16)}/2
= 82 mm
middle 16 mm)
bar is
curtailed, clear spacing =
ISO rom
The top spacing
at
the support
<
160
mm;
hence
O.K.
(Note:- Since
6 2
= 0.8 > 0.45, the 16 mm
bar satisfies-the
·0.45
role·.)
However, the spacing role is marginally violated
when the middle
bar
is e u r t i l ~ this can be
tolerated, since the service
stress
in
the
continuing
bars will be small.
Comer distance = [{(30+10+2012)2}(2)]O.s - 20/2
= 6O.7mm < 160/2 = 80 mm; hence O.K.
Considering the span s tion (tension on bottom),
Clear spacing between bottom bars (2T25 1
nO)
=
{300 - (2)(30) - (2)(10) - 25+25+20)}/2
=
75 mm
middle (20
mm)
bar is curtailed, clear spacing =
170
mm.
The
bottom spacing
near
midspan is
<
160 mm;
hence
O.K.
(Note: - Since 20/25 =
0.8
> 0.45, the
20 mm
bar
satisfies the ·0.45
role·.)
However, the spacing role is marginally violated
when the middle bar is curtailed; as before, this can
be tolerated.
Corner distance
=
[{(3O+
10+25/2)2} 2)]O S
- 25/2
=
61.7 mm
<
160/2
=
80 mm; hence O.K.
Note also that all the
above
spacings
are
greater than
hagg
+
5 mm, i we assume that h
=
20 mm.
Hence, minimum spacing rules are s o satisfied.
35
£
0].-1
(
:m
)
Crack Width
O.K. at support
Crack width
O.K. in span
Minimum
spacing O.K
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Notes
cakulatioos
2. The span is taken from Example 7. More guidance regarding the calculation of
effective
spans
is given in clauses 3.4.1.2
to 3 4 4
3. The use of q ~ o 7 will be more convenient than obtaining F
from double
interpolation in Table 3.11.
4. n a practical
beam,
there will be some
bars on
the compression face, in
order
to
anchor the
shear
links. These may be considered as compression reinforcement;
neglecting them is conservative.
the structure is simple, instead of drawing the entire bending moment envelope,
the
controlling loading cases for each situation can be
considered Wberethe span BC
is
concerned, the controlling case will be that which causes
the
points of contraflexure
to be
as close as possible
to
the supportsB and
C
6. his calculation is done as in Example 1.
he
beam
is under-reinforced.
7. Since the curtailed
bar
will be anchored
in
the tension zone, one
of
the conditions c
to
e in Clause .
3 2 9
must be satisfied. In general c
can be used
in sagging
moment regions
and
e in hogging moment ones.
8.
The
controlling loading case for the hogging moment steel at support B is
that
which
produces the maximum moment at B, while causing the point
of
contraflexure closer
to
B in the span e
to
be as far as possible from B.
9. For sagging momenteurtai1ment, generally condition e is
the4XNUlOUing
one, over
a and b
in lause
3.12.9.1,
in
order
to
determine
the
distance between the
theoretical and practi cal cut-off points. For hogging moment situations, however,
since
the
moment values drop sharply from
the
point of maximum moment,
conditions a and b may govern over e .
10. For the same reason given
in
Note 9 - Le. the bending
moment
diagram
being
convex
to
the baseline -
the
lengths of curtailed
bars
at supports are much smaller
than
those
in
spans.
II
The
anchorage length has
to be
provided on. either side
of the
critical section for
design, so
that
the full strength of the steel can be utilized.
he
anchorage lengths
vary depending on
the
surface characteristics of the reinforcement as well as its yield
strength. The anchorage length check may become critical when curtailing support
steel.
12.
he
continuing
bars
at the top 2T20 and the bottom
2T2S
can
be
lapped with
smaller bars, when the former are no longer required
to
carry
tensile stresses. At
least two bars
are
required at any section for anchoring the shear links.
The
minimum
diameter for such bars will be around 12 mm, so that the reinforcement cage will
have
adequate stiffness
during erection.
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13. In this instance, it is sufficient to satisfy
amdilioos
a) and b) alone in Clause
3.12.9.1 isosed as the
bars will
not
be
anchored
in the
tension zone.
14. All
the references
in
Clause 3.12.8.13 are
to
the
smaller of
the two
lapped bars.
Although the
basic lap
length
does
not need
to be
increased in this example, it
may
need to be in some cases.
15.
In
general, lapping
should not
be
done
at
supports,
Since
column
or
wall
reinforcement will add to
reinforcement congestion. In this example however, the
bottom lap extends into the support.
16. No downward redistribution
of moments has been
carried
out
at this support section.
such
redistribution bad been performed ata support ectioo, themuimum spacing
allowed becomes
fairly small.
17.
The
continuing bars
are
able
to
carry twice the moment actually applied, as
curtailment
has been
done according
to
condition
e)
in
Clause 3.12.9.1.
As
the
service stress will
thtn be
quite small, margiDal.aationsof
the bar spicing rules
can
be
allowed.
In any
case
see Note 19.
18. Both maximum and minimum spacings have
to
be satisfied. hemaximum spacings
apply
to the
tension
face and
are
deemed to
satisfy-
rules
for crack control.
The
minimum spacing roles apply
to
both
faces and
eosure
that
concre.ting can
be
carried
out satisfactorily. The most commonly
used
size
in pmctice is 20
mm
maximum size).
C Netes
19.
the
deemed to
satisfy
serviceabilitycbecks.ae not
satisfied,
the
more aocurate
calculations
for deflection
and
crack width in
ection
of
UO Part
can
be
resorted
to ,
in order
to find out whether the Rlquimnents of
Clause 2.2.3 are
met.
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X MPL 13
- ONE WAY SLAB
A slab which has several continuous spans of 5 m is to carry an imposed load of 3 kN/m2
as a one way spanning
slab.
The loading from
finishes
and lightpartitioos can each be
considered equivalent
to
a uniformly distributed load of
I
kNIm
2
•
Taking the density of
reinforced concrete to be 24 kN/m
3
,
feu = 25
N mm
2
and f
y
= 46 N/mm
2
,
design a
typical interior
panel.
Introductory
Notes
1 A
slab
is similar
to
a beam in
that
it is a flexural member. It isdi.ffeRat
to
a beam
in that it i.s a
two
dimensional element, as opposed
to
being one dimensional.
2. Where
the
loadings
from
light ~ t o s is not accurately known, it is reasonable to
U umeaudl
value of I kNlm
2
• Furthermore, partitions whose positions are not
known should
be treated as
additional
imposed
load. The imposed load value
specified
in this example corresponds
to .that
for a
school
building. Imposed loads assumed for
office buildings
and
domestic buildings are 2.5 k m
and 1.5kN1m
2
n=spectively.
Further guidance can
be
obtained from
BS 6399: Part I 1984) -
Design loading for
buildings: Code
of
pmctice for dead and imposed loads .
Note 3
TABLE 1
Note 4
NoteS
Calculatioas
Slab
thickness
, to
choose a slab
thickness, assume
tb
of
34
. for a continuous
I way slab).
Hence,
effective
depth =
5.0
xloJ)/ 34) = 147
mm
We can use a cover of 20 mm
mild
exposure
conditions; concrete protected by lOmm
1;3
cement:sand rendering).
Assuming
bar diameter
to
be 10 mm, choose
h = 175 mm and d = 175 - 20 - 10/2 lSOmm
LoadiOI: for I m wide strip)
Output
h 175 mm
d
ISO
Self load
=
0.175) 1) 24)
Finishes = 1.0) 1)
Total dead
load
Imposed
load = 3.0) 1)
Partitions =
1.0) 1)
Total imposed load
= 4.2
k m
=
1.0
kNlm
=
5.2
k m
gk
=
3.0 k m
=
1.0 k m
=
4.0
kNlm he)
design
udl
=
Design load =
1.4) 5.2) + 1.6) 4.0)
=
13.7
kN m
13.7
kN/m
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Referenee
3.5.2.3
Table 3.13
Calculations
Ultimate bendin moments
and
shear forces
Since gk > JJc and 'be < =
5.0
kN m
2
, and if we
assume that bay size > 30 m
2
, for an interior p lel,
Span moment
=
(0.063)F.l
=
(0.063)(13.7)(5.0)2-
=
21.6kNmlm
Support moment
=
(-Q.063)F.l
=
(-Q.063)(13.7)(5):l.
-2L6kNmlm
Shear at support = (O.S)F = (0.5)(13.7)(5.0)
= 34.3 kN m
Output
M
=
21.6
kNmlm
M
t
r ~ m l m
v
=
34.3 kN m
Chart 2
Part
3)
Note 6
Note 7
Fig. 3.25
Note 8
3.12.11.2.7
Desi n for bendin
at
§l M
Mlbd
2
=
(21.6
x ~
1
(1000)(150f
=
0.96
100AJbd = 0.26 > 0.13); hence min. steel O.K.
As = (0.26)(1000)(150)1 (100)
=
390 mm
/m
Span steel
Use TI0@175 mm
s
= 448 mm
/m
TIO @ 175 mm
Half
th
bars_ be
Q1rtailed
at (0.2)1 - i.e
(0;2)(5)
1.0 mfrom the centre-line
support.
Thenr/f
will be
TI0
@ 350 mm
{«3 150
450}
lOOA/
i 100 44812 (1000)(175) = 0.13
Hence crack
cootroland
minimum steel O.K.
Note 9 Check for
deflection
MJbd2 = 0.96 and
f
s
= 5/8 460 { 390 / 448 } = 250 N/mm
2
Table 3.11
Hence
F
=
7
Table 3.10 Allowable spanldepth = (26)(1.57) = 40.8
Actualspanldepth
= 5000 / 150
=
33.3
<
40.8;
O.K.
for bendin at
sUllJlOrt
Deflection O.K.
Since the moment is identical to that
in
the span, Support steel
steel provided also can be identical. TIO @ 175 mm
Fig. 3.25
Half
these
bars
can
be
curtailed at (0.15)1
=
(0.15)(5) = 0.75 m from the face
of
support (Note:
= 450 mm<
750mm and all the steel
curtailed at (0.3)1
=
(0.3)(5)
=
1.5 m from the face
of support.
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Reference
calculations
Output
Check for shear
,
v = (34.3 xl<P) (1000)(150)
=
0.23 N mm
v = 0.23
Note 10
For
looA/bvd
=
(100)(448)
(1000)(150)
=
0.30,
N mm
d
=
150 mm and feu
=
25 N mm
Table
3.9
v
c
=
0.54 N mm
>
0.23N/mm
2
v
c
- 0.54
Table 3.17
Hence, no shear reinforcement required.
N mm 2
Note
11
Seconda[y reinforcement
Table 3.27
l00As/A
c
=
0.13
As
=
(0.13)(1000)(175) (100)
=
227.5 mm
2
/m
SeCondary
Use
TlO @ 350 mm
As
=
224 mm
2
/m) steel
3.12.11.2.7
Max. spacing = 3 150 - 450 mm > 350 mm
TI0@350mm
~ 7 E m M O 7 ~
~
o
7fm
J > = 7 ~
T l ~ 7 5 TlQfJ350
n O l O O ~ 7 5
r
· n ~ ·
,
~
Notel2
~ ~ T l ~
TlOOI75
Tl
f
I
I f
0
I.On
5.0m
( I On )
Notes on Calculatioas
3.
Although
the
bending moment is the controlling factor
in the choice of
depth for
beams, where slabs
are
concerned,
the
controlling factor
is
the
spanldepth
ratio,
representing
the
check for deflection. Atrial V8lue
has
to
be
used initially;
a value
of around is a reasonable estimate for lightly loaded one way continuous slabs; this
should be reduced to around 30 for heavily loaded
s1abs
A lightly loaded slabwould
have an imposed load
of
around 4 kN/m
2
,
while a heavily loaded slab would bave one
of around 10 kN/m
2
•
4. Slabs are generally designed such that shear links are not required; hence, no
allowance need
be
made for link diameter.
5. One way and two way slabs are generally designed - Le. loads evaluated and
reinforcementcalculated - on the
basis of
a strip of unit width (e.g. 1 m wide).
6. The minimum steel requirement is in fact based on
looA/
c However. since the
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lOOA/bd is obtained from the design charts, it provides an approximate
check on
the
minimum steel requirement.
7. Although we
can
use the
sllghtlylarger
spacing·of
200
mm (giving
392.5
mm
2
/m), we adopt this smaller spacing, as
it
results in the minimum steel
requirement being satisfied even after half the steel is curtailed.
8. Although 60
of
the steel
can
be
curtailed, in practical slabs, curtailing 50 is
easier, because every other r can be curtailed.
9. The assumption regarding spanldepth ratio must becbecked as early as possible in
the design. Hence span moments should
e
designed for
first
and
the
deflection check
made soon after.
10
The area of steel used here is that of the top (tension) steel at the support.
11.
general, apart from .some c ses in flat slabs, it is sought
to
avoid shear
reinforcement in flat stabs. Hence,
v is greater than vC the
sl
thickness is
increased. This should always be borne in mind, and perhaps an approximate check
for shear m de early in the design,especially if the sl is heavily loaded (e.g. with
a
water load).
12.
Where
the
curtailment
of
steel is toncemed, the distances corresponding
to
top steel
are
given from the face
of
the support and those corresponding
to
bottom steel from
the centre-line of support.
Concluding Notes
13. Although
it
is quite easy
to satisfy
minimum steel requirements and maximum
bar
spacing rules
at
critical sections (such as midspan and support),
care
should
be
taken
to
ensure that the above checks are not violated after curtailment
of
reinforcement.
14.
The
simplified approach to the design
of
slabs, using
Table
3. 13·can be used in most
practical situations. Such an
ppro ch
is given for the design of continuous beams as
well
in Table 3.6. The coefficients in this latter table
are
higher
th n
those for slabs,
because the sl coefficients are based on the less stringent single load case of all
spans loaded, with support moments redistributed downwards
by 20 .
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EXAMPLE 14 - ONE WAY SLAB
A garage
roof
in a domestic building is to function as an accessible platform,
surrounded by a parapet wall; the slab is supported
on
two parallel 225 mm
brick
walls, the
clear distance between walls being 3.5 m. Design
the
slab,taking f
cu
.·2S N/mm
2
, f
y
=
460
N/mm
and density
of
reinforced concrete = 24 kN/m
•
Introductory Notes
1
i
3.
4
This example has more unknowns
than
the previous one. describes a real
situation, where design assumptions will have to be made. The imposed load and load
from finishes and parapet wall have to
be
assumed and a decision t ken regarding
the
end fixity of the slab.
The
imposed load could be taken as 1.5 kN/m
2
, since it is a domestic building.
The
finishes (on
both
top surface and soffit) can be assumed to
be
a uniformly distributed
load
of
1 kN/m
2
.
The parapet wall which is constructed
on
the slab perpendicular
to
its span will give
a degree of fixity to the slab. However, the most conservative
ppro h
is to idealize
this slab as a one way simply supported slab. Any
fiXing
moments caused by the
above
p rti l
fixity can
be
accomodated by taking S of
the
midspan steel
into the
top face of the slab at the support.
The parapet wall parallel to the
sp n
will
have to
be
carried
by
the slab.
t
can
be
assumed that the wall is 1.0 m high and 120 mm thick and that the density
of
the
(brick) wall is 23 kN/m
2
• The load from
this
wall
will
be
distributed
only
over
a
limited
width
of
the slab (Clause 3.5.2.2).
Reference
Calculations
Output
Slab thickness
Note 5
Approximate span
=
3500 mm
Note 6
Assuming Spanldepth ratio of 28 (for a simply
supported 1 way slab),
effective depth = (3500)/(28) = 125 mm
TABLE 1
f we take cover = 30 mm (moderate exposure
Example 8
conditions and TABLE 1 values modified by Notes 5
and 6), and bar diameter = 1 mm, we can choose
h = 160 mm
Note 7
h
=
160
mm
and d
=
160 -30 - 10/2
= 125
mm.
d = 125
mm
Hence, effective· span
=
lesser of
3.4.1.2
(3500+225) = 3725 mm eff. span
=
or 3500 125)
=
3625
mm
3.625 m
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Reference Calculations
Loadine for 1 m wide strip
Output
3.5.2.2
Note 8
Self
load
= 0.16 1 24 = 3.84
kN/m
Finishes =
1.0 1
=
1.00
kN/m
Total dead load
=
4.84 kN/rn
Imposed
load = 1.5 1 = 1.50
kN/rn
Design load = 1.4 4.84
+
1.6 1.5 = 9 2
k m
design udl =
9.2
kN/rn
Strip
carrying
parapet wall
= 0.3 3.615 0.12
=
1.21
m
Additional dead load in
th t area
=
1.0 0.12 23 1 1.21 = 2.28 kN/m
Ultimate bending moment and
sh r
force
Note
3
Chart 2
Par t 3
3.12.11.2.7
Note
9
Fig 3.25
Since we assume the slab to be simply supported,
Mid sp n moment
=
w l
2
= 9.2 3.625 21
8
=
15 1kNmlm
Shear force at support
=
w l 2 = 9.2 3.625
12
=
16.7 kN/m
Desim
for bending
Mlbd
2
= 15.1 xl<f 1 1000 1251 = 0.97
looA/bd =
0.26
>
0.13 ;
hence min. steel O.K.
A.
=
0.26 1000 125
I
100
=
325
mm
2
m
Use TI0 @ 225
mm As = 349
mm
/m
Max.
spacing allowed
= 3 125
= 375 mm > 225 mm;
hence
crack
width
O.K.
However,
bar
spacing as well as minimum steel
requirement will
be
violated
if
bars are curtailed.
Hence, use TlO
@
187.5 mm As =·419 mm
2
/m
Spacing after curtailment = 375 mm.
l00A/ after curtailment
=
100 419/2 1
1000 160 =
0.131 > 0.13
Hence, min. steel andbar spacing are O.K. after
curtailment.
The steel should be curtailed at 0.1 1 = 0.1 3625
=
362.5
mm
from the point
of
support, Le.
362.5 -
225/2 = 250 mm from the face
of
support.
43
Mspan = 15.1
kNmlm
V = 16.7 k m
span steel
TIO @ 187.5
mm
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RefeNilee Calculations
Output
Note 10 The rest of the steel could be t ken into the support support steel
nd bent back into the span as top steel to extend a
TI0
@ 375
mm
distance from support f e of 0.15)1 = 0.15) 3625)
3.12.10.3.2 =
544
mm
{>
45 cP =
45) 10)
=
450
} say
0.55
m
Check for deflection
Note
Table 3.11
Table
3.10
Mlbd
2
= 0.93
f
s
= 5/8 460 325/419 =
223 N mm
Hence, F = 1.71 for tension steel)
Allowable span/depth = 20) 1.71) = 34.2
Actual span/depth = 3625/125 = 29
<
34.2;
hence
O.K.
Deflection
O.K.
Che ck f or s he ar
v = 16.7
xlol
1 1000 125 = 0.13 N mm
2
looA/bvd = ~ 3
Table
3.9 Hence, V
c
=
0.45 N mm
2
> 0.13
N mm
2
;
Table
3.17
hence shear
r f
is not required.
Shear
O.K.
SecondflO reinforcement
3.12.11.2.7
looA/
=
0.13
As = 0.13) 1000) 160)I 100) = 208 mm
2
m
Use TIO @ 375 mm i.e. max. spacing allowed - 3d)
As =
209
mm
m
se ond ry steel
TI0@375 rom
under
parapets
TI0@ 175 mm
span)
TI0@350
support)
I
TIotm
r
~
t
I :
Note:- It
can be
shown that the spacing of the
reinforcement in the edge strips
of
1.21 m should be
T10
@ 175 mm at
midsp n
and hence TIO @
350
mm at supports).
t 110075
Note 12
1100187.5
o.ti:
0.:rAJn
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Notes
on
Calculations
In order to use Clause 3.4.1.2 tofind the effective span, the clear distance between
supports
is
taken as a first approxl.mation
of
the span.
6. For a lightly loaded one-way simply supported slabs, a span/depth ratio of around
26-28 may be assumed. Tbisshould
be
.reduced to around 24 for a heavily loaded
slab.
7. In this instance,
we
have
t ken
a value for h, such that slab thicknesses are assumed
to vary in steps
of
lO
mill. To use steps of 25 mm corresponding to 1 inch) would
be
t Conservative for slabs. Hence either
10
mm steps· or
12.5
mm steps
corresponding to
0.5
inches) should
be
adopted.
8. The edge areas of the slab, Le. the 1.21 m strips carrying the parapet loads, will e
more heavily reinforced than the rest
of
the slab. However, only the central part
of
the slab is actually designed in
this
example.
9 . There may be other alternatives to increasing the mid-Span steel, bot this approach
makes the detailing for curtailment very simple and also helps
to
satisfy the deflection
check, which is very critical
in
slabs. This approach also facilitates the detailing of
steel for support restraint, as shown in the figure.
ne
possible alternative is to use
smaller diameter bars, but rs smaller than
10
mm, if used as main steel, will not
be
very stiff and may deflect significantly during concreting, thus losing their cover.
lO
As shown in the fI gu re,this
is
a very neat method
of
providing top steel at partially
restrained ends
of
slabs
and
beams.
11. Since we have provided more steel th n required at mid Span see Note9 , advantage
should be.taken
of
this by generally calculating the service stress, which will be lower
than
5/8 f
y
and lead to a greater allowable span/depth ratio.
12.
t
may be convenient to reinforee the entire slab with TlO
@
175 mm at mid
sp n
and
TlO@
350 mm at support, since the central part of the slab already has TlO @ 187.5
mm and TlO 375 mm at span an d support respectively. The small penalty in cost
will
probably
be
worth
the
simpler detailing arrangement.
Concluding Notes
13. t is important to keep
in
mind curtailment, bar spacing rules and minimum steel
requirements while designing the reinforcement, because these detailing considerations
may lead to the design being altered, as was the case here.
45
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EXAMPLE
15 - TWO WAY SLAB
A.two way spanning slab which has several bays in each direction.ha a panel.size of S m
x
6
m.
The imposed load
on
the slab is 3 kN/m
2
• The
loading
fronl finisheaand
light
partitions
can
each
be
taken
as
1 kN/m
2
• Design
a
typical interior panel, using
feu
2S
N/mm
2
,
f
y
460
mm
and density of reinforced concrete
24kN1m3.
Introductory Notes
1. The short span length and loading for this example have been anade ideAtica1to those
in Example 13 for
a
one-way spanning slab.
Hence,
results can be
compared.
2.
t
will
be
assumed that the comers
of
this slab are prevented from lifting and that
adequate provision is made for torsion. .
eferell e
Note 3
3.5.7
TABLE I
Note 4
Note 5
Calculatiops
Assultle.a
spanldepth
ratio
of
40 (for a continuous 2
way slab)
effective depth
(5000)/(40)
125
mm
we
take
cover
20 mm (mild exposure conditions
and
concrete protected
y
10 mm·l:3 cement:sand
rendering)
and
bar diameter
as
10mm
then we
can
choose h
150 mm
and
ort 150·20
- 10/2
125
mm and ~ 125 .
10
115 mm
h
150
mm
dlbort
125 mm
l1Smrn
Loadine (udl)
Self load
(0.15)\:, 1 24 3.6 kN/m
2
Finishes 1,0>, 1.0 kNlm
2
Total
dead
load
4 6
tN/m
t
Imposed load (3.0). 3 0kN/m
2
Partitions = (1,0) = 1,0 kNlm
2
Total imposed load
4.0 kN/m
2
Design load= (1,4)(4.6) (1,6)(4.0)
12.8 cN m
n 12.8
kN m
2
Bendine moments
This interior
panel-
has lyIlx 6 1,2
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Reference
Calculations Output
Table 3.15 Short way, edge = 0.042) 12.8) 5)2= 13.44 kNm/m
Short way, span = O . 0 ~ l 2 . 8 5 2 = 10.24 kNmlm
Long way, edge = 0.032) 12.8) 5)2= 10.24 kNmlm
Long way, span
= 0 . 0 2 ~ 1 2 . 8 5 2 =
7.68 kNm/m
Desien
of
reinforcement
I
Short way, mid-man:-
Chart
2 Mlbd
2
= 10.24
xld
1
lOOO) l25t
=
0 66
Part 3)
100A/bd =
0.17
As
=
0.17) 1000) 125)1 100 = 213
2
/m
Use TlO @ 350 mm As
::
224
mm2/m
3.12.11.2.7
Max.
spacing = 3) 125) = 375 mm > 350
mm
Table 3.27
l00A/A
c
= 100) 224)
1
1 ) ) )) 150) = 0.15 >0 13
Hence, bar spacing and min. steel are O.K. but
if
Short way, span
Note 6
steel is curtailed, they will be violated.
TlO@ 35 0 mm
Check for, deflection):-
Mlbd
2
=
0 66
f
s
= 5/8 460 213/224 = 273
N/mm
Table 3.11
F
1
= 1.65 for
tension
steel)
Table
3 10
Allowable span/depth = 26) 1.65) = 42.9
Actual spanJdepth
=
5000)/ 125)
=
40 < 42.9;
hence O.K. Deflection O.K.
Short way. cts. edge:-
MJbd2 = 0.86, l00AJbd = 0.23, As
7
mm
2
/m
Use TlO @ 250 mm A = 314
2
/m
Short way, edge
Bar spacing and min. steel
areIO.K.
TlO@250 mm
Long way. cts. edge:-
Chart
2
Mlbd2 = 10.24
xlW) 1
1000) 115)2 = 0.77
p art 3)
l00Aibd
=
0.21
:l.12.11.2.7 As
=
0.21) 1000) 115)1 100)
=
242
mm 2
Table 3.27
Use
TlO @ 325
mm A
= 242
/m
Max.
spacing
=
3) 115) = 345
mm
> 325 mm
l00AJA
c
= 100) 242)
1
1000) 150) = 0 16 >0 13
Long way, edge
Hence, bar spacing and min. steel are
O K
but steel
TlO
@ 325
cannot
be
curtailed.
Long way. mid-s.pan:-
Mlbd
2
= 0 .5 8, 1 00 Aibd = 0.15, As =
173
/m
Use TlO
@
350 mm
As
= 224
2
/m , since Long way, span
max. clear spacing 345
mm)
governs.
TlO@ 350 mm
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Reference
3.5.3.5
Note 7
Table 3.16
Table
3.9
CalcuJatiOllS
Edge
strips:
l00A/Ac
=0.13
As = (0.13)(150)(1000) I (100) =
195
mm
m
Use TlO @ 375 mm (governed by max. spacing rule
in short way direction)
Use only in short way cts. edge; at other locations,
use middle strip steel for edge steel.
Check for
shear
Short way sypport:-
V = (0.39)(12.8)(5) = 25.0 k m
v = (25.0 xloJ I (1000)(125)= 0.2
mm
l00A/bd = l o o ~ 3 1 4 I
1 0 0 0 ~ 1 2 5 =
0.25
v
c
=
0.53
mm
>
0.2
mm
;
hence O.K.
Output
Edge strip
nO@375 mm
(only for short
way, cts. edge)
Lo a l) Y R>J Ort:-
Table 3.16 V (0.33)(12.8)(5) = 21.1 k m
v = (21.1
xloJ I
(1000)(115) = 0.18 N mm
l00A/bd = (100)(242)
I
(1000)(115) =
0.21
Table 3.9 v
c
= 0.50
mm
> 0.18 mm
hence O.K.
No shear
r f
required
Fig. 3.25
o
<1800 ~
16T1 6325 11
600
-
-
...
~ l -
1
E o
o 0
E o
1.0
1.0
1.0
1.0
l:
N
M
r
M
@
@
M
@
0
0
@
0
0
E o
E o
o O l
00
E o
N
-
N
48
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Notes on
CalcuIatiODS
3. A
trial
value for span/depth ratio
of
40 is reasonable for a lightly loaded continuous
square2-way slab; a ratio of
38 would
appropriate for heavily loaded slabs. This
will
of
course reduce with the ratio of long to short span ~ i n
the
value for 1
way slabS when the latter ratio becomes 2. T he span/depth ratio is calculated with
respect to the shorter span as it is this
that
controls d ~ f l e t i o n
4. It should be noted that the slab thickness required for a two-way slab is less than that
required for a one-way slab of similar span and loading - cf.
7
mm required for the
slab
in
Example 13.
arranging the reinforcement in the slab the short way reinforcement should be
placed outermost
in
order to have the greatest effective depth since the shorter span
controls deflection and since the bending moments and shear forces are greater in the
short way direction as well.
6. Two way spanning slabs are
in
general very lightly reinforced so that curtailing is
often not possible because
of
the minimum steel requirement
or
the maximum spacing
requirement or both.
7 Since the main steel requirements are
also
fairlysmaIi for practical detailing it may
be it may be convenient to use the same reinforcement as the middle strip for the
edge strips except in the case of the short way continuous edge.
Concluding Notes
8. Where an edge
or comer
panel is concerned
in
addition to the main and edge
steel
the
requirements
of
torsional steel reinforcement have
to
bernet at the top and
bottom of the slab according to Clause 3.5.3.5; in many cases the main and edge
steel provided would meet those requirements.
9. Although the loads on a beam supporting a two-way slab will be either triangular or
trapezoidal the code gives coefficients for an equivalent uniformly distributed load
over three quarters
of
its span.
10. In the calculation of moment coefficients from Table 3.15 if there are significantly
differing coefficients on either side
of
a common edge the code suggests a method
of moment distribution to rectify the situation in Clause 3.5.3.6.
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EXAMPLE
16 - FLAT SLAB
A
f l t ~ which _several bays
in
each direction,
has
a panel size 0 {5m x 6
m
The
~
imposed
ontbesJab
is
3
kN/m
2
• Tbe ioa4ingfrom finishes and liaht PlU titions
can
each
be
considered to
be
1 kN/m
2
.
Design
a typical interior panel, using
u
N/mm
2
,
f
y
=
460 N/mm
2
and density of reinforced concrete
=
24
kNlm
J
• It may be assUmed that
the columns supporting the slab are braced.
Introductory Notes
1
This example,
too, can be
compared with Examples 13 and 15.
2. As the columns are braced, and as the I1ab has several bays in each direction, the
simplified
method of
analysis
described
in Clause 3 7:2 7
and
Table
3 19 w ll
be
employed.
3.
It
will
be
assumed that the slab is without
drops and
the maximum
value of
effective
diameter
will be
employed for column.
beads
Reference
3 7 1 4
Note 4
3 7 8
TABLE 1
NoteS
Note 6
Calculations
Slab thickness
Max. value
of
he
=. 114 5.0
=
1.25 m
Assuming a trial span/depth of 32,
effective
depth =
6000)/ 32)
=
187.5 mm
If
we
take
cover =
20
mm mild exposui-e condi\ions
and concrete protected by 10 mm 1:3 et:sand render)
and bar
diameter = 10 mm, we can choose
h = 212.5 mm, d
long
= 212.5-20-10/2
187.5 mm
d
short
187.5-10
177.5 mm,
dave
182.5 mm
L o a d i n ~
for entire panel)
Panel area
5) 6)
30 m
2
Self load = 0.2125) 30) 24) 153lcN
Finishes
1.0) 30) =
1U: kN
Total dead load = 183
le
Imposed load = 3.0) 30) = 90 leN
Partitions = 1.0) 30)
=
30
le
Total imposed load = 120
leN
Design load
=
1.4) 183)
1.6) 120)
=
44 8
le
50
he
=
1.25 m
h
212 5 mm
d
y
187 5 mm
dx 177.5 mm
d a v g = = 8 ~ 5 mm
F =
448
leN
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Reference
Calculations
Output
Table 3.19
Bending
mOments
Note 7 Long way:-
1 =
6.0
- 2/3 1.25
=
5.17 m
Span moment = 0.071 448 5.17 =
164
kNm
Fig. 3.12 Col. strip 2.5
m
= 0.55 164 = 90.2 kNm
Table
3.20
Mid. strip 2.5
m
= 0.45 164 = 73.8 kNm
Support moment = 0.055 448 5.17 = 127 kNm
COL strip 2.5
m =
0.75 127 = 95 kNm
MId. strip· 2.5 m = 0.25 127 = 32 kNm
Short way:-
1 = 5.0
- 2/3 1.25 = 4.17 m
p n
moment
= 0.071 441 4.17 = 133 kNm
Fig. 3.12
Col. strip
2.5
m
= 0.55 133
:::
73 kNm
Mid. strip 3.5
m
= 0.45 133 = 60 kNm
Support moment = 0.055 448 4.17 ::: 103 kNm
Col. strip 2.5
m
= 0.75 103 = 77 kNm
Mid. strip 3.5
m =
0.25 103
= 26
kNm
Design
of
reinfOrcement
Long way. an
Check for deflection
Total moment
=
164 kNm
Mlbd2
=
164 xlo6
1
5000 187.5 2
=
0.93
Note 8
AS,reqd
= Aa Jrov f
s
= 288 N/mm
Table 3.11
and F
1 =
1.
41 for
tension steel
3.7.8
Allowable sp nldepth = 26 1.41 0.9 = 33.0
Actual sp nldepth
= 6000 / 187.5
= 32
< 33.0; hence O.K.
Deflection O.K.
Column strip 2.5 m wide
Chart 2 Mlbd
2
=
90.2 xl<J6
1
2500 187.5 2 = 1.03
part 3
looA/bd
=
0.28
As = 0.28 2500 187.5 1 l00 = 1313
mm
2
Note 9
Use 17 TlO
@
147 mm
As = 1335 mm
2
Long way. span
3.12.11.2.7
Allowable spacing = 3 187.5 = 562.5
mm
Col. strip
Table 3.27
l00A/A
c
= 100 1335 1 212.5 2500
=
0.25
17 TI0
Note 10
Hence
b r
spacing and min. steel are O.K.
@
147 mm
Middle strip -2.5 m wide
Long way. span
Chart 2
Mlbd
2
=
73.8 xl<>6 1 2500 187.5 2
=
0.84
Mid. strip
Part 3
looA/bd = 0.23; As
=
1078
mm
2
14 TI0
Use 14
TI0 @ 179mm
As
=
1100
mm
2
@
179
mm
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-
Reference
Calculations
Output
Lone
way, SURPOrt:-
Long way,sup
Column strip -
2, 5
m wide Col. strip
M/bd
2
= 1.08, l00A/bd =
0.29, A.
= 1359 mm
2
12 TI0 @ 104
3.7.3,1
Use 18 Tl O As = 1414 mm
2
- 12 TIO centred on
6
TI0
@ 208
column @ 104 mm; 6
TI0
@ 208 mm.
Middle strip - 2,5 m wide
M/bd
2
= 0.36; l00A/bd
;.,
0.10 Long way, sup
Table
3.27
Use nominal steell00A/Ac = 0,13; A. = 691 mm
2
Mid, strip
Use 9
TI0
@ 278 mm As = 706,9 mm
2
9 TI0 @ 278
Short way, span:-
Column strip - 2,5 m wide
Chart 2 M/bd
2
= 73
xl 6)
/ 2500 177.5 2 = 0.93
Part 3
l00A/bd
=
0.26
As = 0.26 2500 177.5 / 100 = 1154 mm
2
Short way, span
Use 15 TIO @ 167 mm s= 1178 mm
2
Col. strip
3.12.11.2.7
Allowable spacing
=
3 177.5
=
532.5 mm
15 TlO @ 167
Middle strip -
3. 5
m wide
Note 11
Mlbd
2
= 60 xl 6) / 3500 177.5 2 = 0.54
l00A/bd
= 0.15 Short way, span
As
=
0.15 3500 177.5 / 100
=
932 mm
2
Mi d
strip
Use 12 TIO @ 292 mm s= 942.5 mm
2
12 TI0
@ 292
Short way.
suPJ 01 t:-
Column strip - 2. 5 m wide
Short way, sup
Mlbd2 =
0.98,
l00A/bd = 0.27, As = 1198 mm
2
Col. strip
3.7.3.1
Use 16 TIO s 1257 mm
2
- 10 TI0 centred on
10TlO@
125
column @ 125 mm; 6 Tl O @ 208 mm.
6 TIO@208
Middle strip - 3.5 m wide
Mlbd
2
=
0.24,
lOOA/bd = 0.06 hence use
Short way, sup
3.12.11.2.7
l00AJA
c
=
0,13,
=
967 mm
i
Mid. strip
Use 13 Tl O @ 269 mm s = 1021 mm
2
13
TlO @ 269
Check for sh r
square columns are
used,
size of column head =
a
{ T/4 1.25 2}O.5 = 1.1 m .
3.7.7.4
Perimeter
of
column head = 1.1 4 = 4.4 m
1st
critical perimeter = { 2 1.5 0.1825 1.1} 4
=
1.648 4
=
6.59 m
Area within this perimeter
=
1.648 2
=
2.716 m
2
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Reference
3.7.6.2
3.7.7.4
Note
12
Table
3. 9
Calculations
V
t
= 448 kN
V
ef f
=
1.15 V
t
=
1.15 448
=
515.2
leN
v
max
=
515.2
xloJ)
I
4.4 xloJ 182.5
=
0.64
mm
<
0.8 25 °.5 = 4
mm
Load on 1st crit . perimeter
=
448/30 30-2.716
=
407 kN
v
=
407 xloJ 1.15
I
6.59 xloJ 182.5
= 0.39 mm
2
I00A/bd)avg= 112) I00/182.5){ 14l4+1257 125 }
=
0.29
v
e
= 0.51 N/mm
2
>
0.39
mm
I
7 :
I
L ~
~
I I
12Tl00292B
g
: J
:
I
13Tl00269T
-1
--t
-
- :
3Tl0f208T
I
15Tl00167B
I fJ
t - ......--+--' -
lOTlO,125T
S I I
I
3T lO 208T
... - - -1
_ L
-
-l-l
Cl l E c c ~
~ ~
~ ~ ~ ~ ~ ~
< ill
N
N
< ill
< ill
~ 0 0
~ E: E:
-
Cl l
C ?C N C ?
........
Output
Shear r f
not required.
Notes on
Calculations
4. The
trialspan/depth
ratio should
be
around
0. 9
times
that
used for continuous
one-way slabs See Example 13, Note 3 ; hence a value of around 32 is reasonable.
The deflection is governed by the longer span unlike in two-way slabs; therefore the
slab thicknesses will be greater for flat slabs than for two-way slabs
of
similar
dimensions and.loading.
5. Compare .this much greater overall depth with that
of
150 mm obtained for the two
53
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w y
slab in Example 15; of
course.
there i s
the
considerable advantage here of not
requiring beams. The slab
thickness
has een
chosen
in steps of 12.5 mm
corresponding to 1/2 inch). The greater effective depth should be used for the long
way span - Le. the long way reinforcement should be on the outside - because
deflection is governed by the longer span and the moments in the long way direction
are greater than those in the short way direction; this to o is the opposite of two way
slab action.
The
average
v lue
of
effective depth is used for punching shear checks.
6. is more convenient to determine the loading on an entir e panel for flat slabs, as
opposed to that on a strip
of
unit width.
7.
The flat slab has to be analysed in two mutually perpendicular directions, with the
total load being taken in
e ch
direction. This is ec use there re no peripher l beams
around the slab, the flat-slab acting
as oth
slab and beam.
8. The deflection check is done early here, even before the steel is designed. This is a
conservative approach, but h s the advantage that
it can
detect early any changes that
may be required in slab thickness. this check is made
after
the
steel
h s een
designed, the average of column and middle strip steel
can
be t ken for th e As values.
9.
The reinforcement in a flat
sl
is generally specified in terms of the number
of
bars
in a given strip. As such, the spacing
m y not
be
in preferred dimensions.
10. Curtailment, in this and other instances will not be carried out in this example. In
D0st cases, the minimum steel requirement
will
preclude
such
curtailment, although
the maximum
spacing
requirement n easily
be
satisfied.
11. Note that the effective depth in the
short
way dire tion is 177 Smm as opposed to
187.5 mm) and that the width of the
middle
strip is
S
m as opposed to 2 5 m).
12. Just as the average effective depth is used for punching
shear
calculations, the
lOOA/bd value should lso be avetaged. This is because the square s e r perimeters
cross both the long way and short way steel.
Concluding Notes
13. Unlike in the two-way slab, where the middle strips carry most of
the
moment and
are hence more heavily reinforced, in the flat slab, it i s t he column strips that
carry
most of the moment and are more heavily reinforced.
14. Where the simplified method used here is not applicable, a fr me analysis will have
to be carried out according to Clause
3 7 2
15. Edge and
comer
columns of
fl t
slabs will have column strips considerably narrower
than those in interior
p nels
see Clause
3 7 4 2
Furthermore,
the
enhancement
factors for shear due to moment transfer will be greater
at
these columns see Clause
3.7.6.3).
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EXAMPLE 17· RIBBFD SLAB
A ribbed slab which has several continuous spans of 5 m is to carry an imposed load of 3
k m
s
a one-way spanning slab. Taking the load from light partitions and finishes
s
k m
each, the density of reinforced concrete s 24
kN/m3,
feu = 25
mm
and f
y
=
460
mm
2
, design a typical interior panel. Note that a 1 hour fire resistance is required.
Introductory Notes
1. This example
can e
compared directly with Example 13, where the only difference
is that the slab is solid.
2. Although this slab is continuous, because
of
the difficulty
of
reinforcing the topping
over the supports, it will be treated as a series of simply supported slabs see Clause
3.6.2 .
25
500
\
Reference
Note 3
TABLE 1
Fig. 3.2
Table 3.5
3.6.1.3
Table 3.18
3 6 1 3
Note
4
Fig. 3.2
Calculations
Choice of form
Assuming a trial span/depth ratio of 26,
effective depth
=
5000 / 26
=
192 mm
Assuming cover of 20 mm mild exposure conditions
and concrete protected by 1 mm 1:3 cement:sand
rendering and bar size of 20 mm, we
can
choose
h
=
225 mm nd
d
=
225 - 20 - 20/2
=
195 mm
Min
rib
width for
1
hr. fire resistance
= 125
mm
and min. cover
=
20 mm; hence cover O.K.
Choose min. rib width
of
125 mm, widening-to 250
mm and rib spacing of 500 mm
«
1.5 m; hence O.K.
Also use thickness
of
topping
=
50 mm; then rib
height
=
225 - 50
= 175 mm
.
{<
4 125
=
500 mm; hence
O.K.}
Now, effective thickness
=
2 2 5 ~
{ 50 500 + 1I2 125+250 175 }
1 500 225
= 0.51
= 225 0.51
=
115 mm
> 95 mm for 1 he fire resistance; hence O.K.
-*-50
55
Output
h = 225 mm
d
=
195 mm
t
e
= 115 mm
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Reference
Calculations
Output
Note 5 Loading for 5 m strip)
Self load = 0.115) 0.5) 24) = 1.38 kN/m
Finishes = 1.0) 0.5) =
5
kNlm
Total dead load = 1.88
kN m
Imposed load = 3.0) 0.5) = 1.5 kN/m
Partitions = 1.0) 0.5) = 0.5 kN/m
Total imposed load = 2 kNlm
Design load = 1.4) 1.88) + 1.6) 2.0) = 5.83
kN/m
design
=
5.83
kN/m
Design for bending
Assuming slab is simply
s u p ~ r t e
moment
in
span = 5.83) 5) I 8) = 18.2 kNm
3 4 4 4 K = M I b. d
2
f
cu
= 18.2
xl
6
) I 500) 195)2 25) = 0.038
z
= d[0.5
+
{0.25 -
0.038 / 0.9 }o.s]
= 0.96)d
Hence use z = 0.95)d = 0.95) 195) = 185
mm
x = 195-185)
I
0.45 = 21.7 mm
«
5 mm)
Hence, neutral axis is in flange.
As
= 18.2 xIW) I 0.87) 460)085)
=
246 mm
2
Use 1 1 20
s
= 314 mm
2
)
Table 3 27 1ooA/bwh
=
100) 314)
I { 1/2 125+250 225 }
= 0.74
>
0.18; hence min. steel O.K.
Check for deflection
bw b
= 187.5 / 500
= 0.375
Table 3.10 Hence, basic span/depth ratio = 16.4
f
s
=
~ 5 / 8 4 6 0 2 4 6 / 3 1 4
= 226 mm
2
M1bd = 18.2 x10
6
) /
500 195t
= 96
Table 3.11 Hence, F
i
= 1.67 for tension reinforcement)
Allowable spanldepth = 16.4) 1.67) = 27.4
Actual span/depth =
5000 / 195
=
25 6
< 27.4; hence
O.K.
Check for shear
Note 6 Shear force at
lid
from support
{ 5.83 5 /2}{1 - 0 195 / 2.5 = 13.4 kN
56
main r/f
1 1 20 per rib
Deflection
O.K.
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Reference Cakulations
Output
Note 7 v = (13.4
xlol)
(187.5)(195) = 0.37 N/mm
2
Table
3.9
l00AJbd
= l O O ~ { 3 1 4 (187.5){195) = 0.86
shear r not
c
= 0.72 N/mm
0.J.tN/mm
2
;
3.6.4.7 hence shear
r/f
not required.
required
Top
steel
over
sup,port
3.6.2
This is to control cracking and should be 25
of
midspan steel.
As
= 114 246 = 61.5 mm
2
Use 1 TIO bar (As = 78.6 mm
2
extending (0.15)1 over support
= (0.15){5.00) = 0.75 m into span on each side. 1 TIO per rib
Notes on CalculatioR§
3. This tri l ratio is reasonable for simply supported o n ~ w y slabs - Note 6 in
Example 14.
4.
should
be noted that the effective thickness
of
this slab (reflecting the volume
of
concrete that will be used) is much lower than the one-way solid slab of similar span
and loading in Example 13.
5. is convenient to calculate the loading for a strip of width equal to a repeating cross
sectional unit.
6. Since support details are not given, the shear force is calculated at a distance d
from the centre-line of support (and not from the
face
of the support - Clause
3.4.5.10 . The approach here is conservative.
7.
The
average width of web below the flange is used for shear stress calculations.
Concluding
Notes
Fire
resistance considerations will, to a large extent, govern the choice
of
form
of
ribbed slabs.
9. The design of these slabs is essentially the same as the design of fl nged beams.
Generally the neutral axis will lie within the flange.
10. Although the code suggests a single layer mesh reinforcement for the topping, it does
not demand it (Clause 3.6.6.2 . It will be quite difficult to place such a mesh in a 50
mm topping while maintaining the top and bottom cover requirements.
11. These
ribbed
slabs probably have a lower material cost than solid slabs, but their
construction costs would e greater, because of non-planar formwork requirements.
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EXAMPLE 18 - COLUMN CLASSIFICATION
P
four
storey building
has
columns on a
grid
9f
5 0 m x
5 0m supporting beams
of
dimension 525 mm x 300 mm in
one direetiononly
and a
one-way
lab of 175
mm thickness.
The roof
also has
a
beam-slab
ammgement
identical to other.
floors.
The
oolumns are
of
dimension 300
mm x 300
mm and the soffit to soffit heigbtoffloors is 3 5m;
the
height
from
the
top of the pad foundation designed to resist moment) to the soffit of
e
first floor
beams is
5.0
m.
If the
frame
is
braced.
classify a
typical
internal
column
for
different storeys
as short or slender.
Introductory
ot s
1. Columns are classified as unbraced or braced on the one hand depending on whether
they take
lateral
loads or
not
and .as slender or short on the
other
depending on
whether they
should be
designed to rry additional moments due to
deflection
or
not).
2. The effective length of a
column
will depend on the degree of fixity at
its
ends.
Reference
Caknlations
Output
Clear height
between
end restraints,
for
ground floor
columns)
lox
=
5 0 m.
loy
=
5 0
0.525-0.175)
=
5.35
m
for
other
floor
columns)
.
lox =
3.5-0.525
=
2.975
m,
loy
= 3.5-0.175 ::: 3 . ~ 2 5 m
3.8.1.6.2
The
end
conditions for
the
columns in the direction
of
beams
are all conditionl. H mce,
0.75
In the
other
direction.
Table
3.21 = 0.80 ground
floor columns)
Note 3
= 0.85
other
columns)
x
= 0.75) 5000) = 3750
mm
ground
floor)
:::
0.15) 2975)
:::
2231
mm
other floors)
ley:::
0.80) 5350)
:::
4280
mm ground floor)
=
0 ~ 8 5 3 3 2 5 ::: 2826
mm other
loors
3.8.1.3
3.8.1.3
Hence. for ground floor columns,
le/h
= 3750)/ 300)= 12.5
<
15,
leylb 4280)/ 300) = 14.3
<
15;
hence short.
for
other columns.
le/h
= 2231)/ 300) = 7.44
<
15,
l ~ j b
=
2826)/ 300)
=
9.42.
<
15;
hence short.
58
All columns are
short.
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Notes
on
Calculations
3. The values of { in Tables 3.21 and 3.22 have been obtained from the more rigorous
method for calculating effective column lengths in framed structures, given in
equations 3
to
6 in section 2.5 of rt 2 of the code.
The
ratios c i.e. sum of
column stiffnessesl sum of beam stiffnesses) have
been
assumed to be
0.5
1.5, 3.0
and
7.0
for conditions 1, 2, 3 and 4 in Clause 3.8.1.6.2
Part
respectively.
Concluding Notes
4
5
Where edge columns are concerned, they will not have beams on either
side
as
specified in the provisions
of
Clause
3.8.1.6.2.
this case, an approximate value for
{ can
be
interpolated, based on the actual ¥c value and the values used in Tables
3.21 and 3.22 see Note 3 above); otherwise the method in Section
2.5 of
Part 2 can
be adopted.
For
a column
to
be
considered short, both
le/h
and
le/b
have
to
be
less than
5
for
braced columns) and less than 10 for unbraced columns), as specified in Clause
3.8.1.3. The
ratiO c in section 2.5
of
Part 2 or the value { in Clause
3.8.1.6
has to
be obtained for beams in one plane
at
a time.
EXAMPLE 19 - SYMMETRICALLY LOADED SHORT
COLUMN
Assuming that the density of reinforced concrete is 24 kN/of u
==
2S
N/mm
2
, f
y
=
460
N/mm
2
, and that the imposed loads on the roof and the floors are
1.0
kN/m
2
and 2.5
kN/m
2
respectively and that the
allowance
for partitions and
finishes are 1.0
kN/m
2
each,
design the ground floor p rt of an internal column
of
the framea structure described in
Example 18. :
Introductory Notes
Since the ground floor partof an internal column has been found to be short, and
since the arrangement ,of loads symmetrical, this design can be carried
out
according
to the provisions
of
Clause
3.8.4.3
using equation 38.
2. The major p rt
of
this exercise consists
of
a load evaluation, taking into account the
appropriate reduction factors for imposed loads specified in BS 6399: Part 1 1984):
Design loading for buildings: Code
of
practice for dead and imposed loads . The
partition loads are taken as imposed loads, since their positions are not fixed.
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Reference
Calculations
Column grid
dilD lDsions are
5 0
x 5 0
m.
Hence, area conesponding to column
5)2
25 m
2
Dead loads
From
4 slabs = (4)(24)(0.175)(25) = 420 kN
From
beams= (4)(0.525 D 175)(0.3)(24)(5)=50 4
kN
From columns={(3)(2.975)+5}(0.3Y(24) =30 1 kN
From finishes = (4)(1.0)(25) = 100 kN
Total
dead
load = 600.5 kN
Ok
= 601 kN
=
25
kN-
=
187.5 kN
=
287.5.kN
Note 3
3 8 4 3
equation 38
Note
4
Note 5
Table 3.27
Imposed loads
From
roof = (1.0)(25)
From 3 floors = (3)(2.5)(25)
From
partitions
(3)(1.0)(25)
Total imposed load
I.L. reduction due
to
floor
area
= (0.05)(25/50)(287.5)(4) = 28.75 kN
I.L. reduction
due
to 4·floors
= (0.3)(287.5) = 86.25 kN
Hence,
imposed
load
= 287.5 - 86.25 201.25
kN
N
= (1.4)(601) + (1.6)(201) = 1163 kN
Desim of main steel
For short columns resisting axial load
N
=
(O.4)f
cu
c t (0.75)A,.,.f
y
Assuming we use 4 T16, Awe =8Q4
mm
2
c
300)2 - 804 89196 mm2
(1163 xl03)
(0.4)(25)(89196) + (0.75)A
we
(460)
Awe
785 mm
2
<
804
mm2;hence O.K. . .
Hence, use 4 T16 A c
=
804 mm
2
) .
Note:-
lOOAwc/
A
c
=
lOO){804)
300)2
0 89 > 0.4; hence
min.
steel
O.K.
o
..
201
kN
N 1163 kN
Use 4 T16
(at
column
comers)
Notes on .Calculations
3.
The
total imposed load can
be
reduced either on the basis of the area supported by
the column or the number of floors-supported by the column. In
this
case, the
reduction allowed as a result of the
latter is
greater and is hence applied - see as
6399:
Part 1 (1984),
referred
to
in
Note 2
above.
4. The term
A
c
is
the net area
of concrete. A trial value of
A
we
ft
be obtained from
equation 38 assuming the
gross
area
of
concrete for c
as
a flrst
approximation;
this
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area of
can
then be deducted from the gross area to obtain A
c
The valueof A
sc
obtained from the formula should be less than the original trial value of
c
5. In some
cases,
a negative value may be obtained
for ~ this
indicates that nominal
steel will be sufficient. In any case, bar diameters under 12 mm
are
generally not
used for columns, because they will not
be
stiff enough for the erection of the
reinforcement cage.
Concluding Notes
6. This method
of
design
is
applicable
for
short braced columns, where moments.are
negligible, due to a symmetrical arrangement
of
loads.
Even
if this symmetry is only
approximate, provided the columns
are
short
and
braced, equation 39 can be used in
place
of
equation 38
7.
addition
to
the main reinforcement, columns should
be
reinforced by
links
which
surround the main reinforcement
as
well. Ttiis will
be shown in the
next example.
EXAMPLE 20 - SHORT COLUMN WITH AXIAL LOAD AND MOMENT
short column
of
3
mm
x
4
mm
cross section carries
an
ultimate
axial
load
of
8
leN.
an ultimate moment
of
80 kNm is applied
a) about the major axis,
b) about the minor axis
c) about both axes
determine the column reinforcement required. Note that
fcu 25
mm
and f
y
d 460
mm
ntrodu tory
Notes
1. This column canies a substantial moment as well as an. axial load. Hence, we shall
have to use the design charts,
which will
give us a symmetrically reinforced section.
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Reference Calculations
Output
N
= 800
kN, M
80
kNm
Nlbh = 800
xI(3) 1
400) 300) = 6.67
forall cases)
al Major axis bendine
[J
ote 2 b
=
300 mm, h
=
400 mm
Chart 23
Mlbh
2
= 80 xl )6)
1
300) 400)2 = 1.67
~ 0
part 3)
lOOA.Jbh = 0.4
A ~
=
0.4) 300) 400) I 100)
=
480 mm
2
major axis
Use 4 T16 Asc
=
804 mm
2
)
4T16
b)
Minor axis bendine
J
;
400
mm,h =
300 mm
Chart 23 Mlbh2
=
80 xl )6) 1 400) 300)2
=
2.22
Part 3)
lOOA
ac
lbh =
0.8
o
0
A ~ =
0.8) 300) 400)
1
lOOi =
960 mm
2
minor axis
Use 4 TIO Ase = 1256 mm )
4
no
<<)
;>endin
3.8.4.5
MKlh
= 80
x ~ t ~ 5 0
Note 3 MyIb = (80 x10
6
)
1 300-50
hence
M,/h
< ~
equation
41
M
=
M
+ fl
I h ) ~
N1
b Ii fcu>= 800 x1 3)
1
300) 400) l.S)
=
0.267
0
able 3.24
hence P
=
0.690
~ =
80
0 . ~ 2 5 0 / 3 0 0 ) 8 O ) = 126 leN
• •
Chart 23
Mlbh
2
= 126xl 400) 300)2 = 3. 5
Part 3)
1 ~ l b h
1.7
.
A ~ = 1.7) 300) 400)
1 100
= 2040
mm
2
biaxial
Note 4
Use 4 TI5
2
T16 in each dir.)
A ~ = 2366 mm
2
) 4
T25
4
T16
Note 5
) silo of links
JiDb
3.12.7.1
,
For
major a x i s ~ d i n g
use
R6 {>
16/4
=
4 mm}
major axis -
@
175 mm{< 12) 16) = 192 mm}.
R6@ 175 mm.
For minor axis bending,
use
R6{
> 20/4
= 5 mm}
minor axis-
@ 225 mm {< 12) 20)
=
240 mm}.
R6@225
mm.
For biaxial bending, use R 8 {>
25/4
; 6.25 mm}
biaxial
-
Note 6
@ 175
mm
{< 12) 16)
=
192 mm}.
R8@ 175 mm.
Check
for shear
3.8.4.6
MIN
=
126)/ 800)
=
0.158
«
0.75) 0.3)
=
0.225
m);
Shear check
Note 7
hence, shear is not critical.
oot required.
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Reference
Calculations Output
Crack
control
3.8.6
0.2)f
cu
.A
c
= (0.2)(25){(3OO)(400) - 804} = 59 6 kN
Crack
width
Axial load = 800 kN > 59 6 kN.
check
not
Hence, no check is required.
required.
Notes on
Calculations
2.
If
we assume a cover of around 30 mm (moderate exposure conditions and
TABLE
1 values modified by Notes 5 and 6), links
of
8 mm and a
ba r
diameter
of
25
mm ,
then
d h
will
be
(400 - 50.5)
(400)
=
0.87
for major axis bending and (300 - 50 5 1
(300)
= 0.83
for minor axis bending. Hence, Chart 23 (Part 3) - which corresponds
to a d/ h value
of 0.85 -
can be used.
If
there is a doubt, the lower d/ h value should
be used, as this is more conservative. tshould be noted that the column design charts
have a lower limit
of
looAsclbh =
0.4,
thus ensuring that the minimum steel
requirement
of
Table 3.27 is met.
3. In this case too, the difference between
hand
h and b and
b
is taken as
50
mm, by
a
similar
argument as in Note 2 above.. '
4. the steel requirement for bi-axial bending
is
greater than that which
can
be
provided as corner steel, the additional amount required has to be provided in each
of the,
two mutually perpendicular directions, distributed along the faces
of
the
section. Other approaches, which are less conservative and more accurate, perhaps,
are given in Allen,
A.H.,
Reinforced concrete design to
8110 simply explained,
E. F.N. Spon, London, 1988 and in Rowe,
R.E. et al., Handbook
to British
Standard
8ll0: 1985 - Structural use
of
concrete, Palladian, London, 1987 .
5. Although smaller diameter bars (e.g. TIO) could have been used, the T l6 bars are
used, so that the link spacing would be too small; furthermore,
bars
smaller than
T12
ar e
not used as column reinforcement, as they would not be
stiff
enough during
erection.
6. Generally plain mild steel is used
fo r
links as it is easier to bend into shape.
Furthermore, where bars other than corner bars
ar e used, multiple links may have to
be
used
if i)
there is more than one intermediate bar
or (ii)
the intermediate
ba r
is
greater than 150 mm ~ y from a restrained bar (see Clause
3.12.7.2)
7. Strictly speaking,
h o w ~ r
the shear stress should be found in
order
to
check
for the
limits on v
m x
Concluding
Notes
8. In general,sh.ear and crack control are not very critical for columns.
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2 - SLENDER COLUMN
ced slender column of 300 mm x 400
mm
cross section carries an ultimate axial load
800 kN.
is bent in double curvature about the major axis, carrying ultimate moments
80 kNm and 40 kNm at its ends. The effective length of the column corresponding to the
jor axis is 7200 mm. Determine the column reinforcement if feu = 25 mm
and f
y
=
mm
ry Notes
This example can be compared with Example 20, where the short column was of the
same dimensions and carried similar loads.
Calculations
Output
Type of column
Vh =
7200)/ 400)
= 8 > IS; hence s ~ n e r
< 20,
slender column
Also
h/b
=
400)/ 300)
= 1.33
<
3;
bent about
hence, single axis bending.
major axis.
Design moments
M
=
-40 kNm; M
2
=
80
kNm
36 M
i
= 0.4) -40) + 0.6) 80) = 32
kNm
0.4) 80)
=
32 kNm} M
j
= 32 kNm
emin = 0.05) 400) = 20 mm
N emin
= 800) 0.020) = 6
kNm
34
3a = 1I2000) IJb )2= 112000) 200/300Y= 0.288
32
u
=
a·
K
.
h
= 0.288) 1) 0.4) = 0.115
35
M
add
au
= 800) 0.115) = 92 kNm
M
add
= 92
Hence, critical moment is M
j
+ M
add
= 32 + 92 =
kNm
124 kNm. However, as K is reduced, if M
j
+ M
add
becomes
<
80 kNm, M
2
will become critical.
Design of reinforcement
ABLE 1
Assuming cover = 30 mol moderate exposure
conditions and TABLE 1 values modified by Notes 5
art 23
6), link diameter of 8 mm and main bar size of 25
mm, d/h = 400-50.5)/ 400) = 0.87.
rt 3)
N/bh = 800 x10
3
) I 300) 400)
=
6.67
M/bh
2
7 124 xI0
6
) I 300) 400)2
=
2.58
K = 0.9
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Reference
Calculations
ut ut
dd
= (0.9)(92) = 83 kNm
M = 32 83 = 115 kNm
80 kNm)
Chart 23
Mlbh
2
= (115
xldi
(300)(400)2 = 2
40
(Part 3)
K =
0.85
dd
(0.85)(92)
=
78 kNm
M = 32 78 = 110 kNm
Note 3
·Mlbh
2
= 2.29; K = 0.85 (again).
K = 0.85
Chart
23 Hence, lOOA
sc
/bh = 0.8
part 3) sc= (0.8)(300)(400) (100) :;: 960 mm
2
main st
Use
4 1 20
sc
= 1256 mm
2
)
Jnksl
3.12.7.1 Use
R 6 {
20/4
5mm} @ 225 mm
{< (12)(20) :;: 240
mm}
R6@
225 mm
Notes on Calculations
2. When major axis bending
takes
place, if either the Vh value is greater
than
20 or the
bib
value is not less than 3, in order to account for the deflection due to, Slendemess
about the minor axis, the column has to be designed asbiaXiaUy Dent with zero
initial moments about the minor axis (see Clauses 3.8.3.4 and
3.8.3.5 .
3. In general; around 2 iterations are sufficient to arrive
at
a
value
of
that
is virtually
constant. It should be noted that the factor should be applied to the original value
of M
add
alone.
Concluding Notes
4. The reinforcement required for this column is the same
as
for minor
axis
bending of
the short column in Example 20.
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~
22·
PAD
FOOTING
Design a square pad footing for a 300 mm x mm internal column,
which
carries an
ultimate load
of
1100
leN
ervice load
of
760
kN), if
the allowable
bearing pressure
of
the
soil is 150 kN/m
• Use feu = 25
N mm
2
, f
y
= 460 N mm
2
deformed type 2) and density
of reinforced concrete = 24
kN/m
•
Introductory Notes
1. Square pad footings
are
the most common foundation type for columns
of framed
structures. Pad footings
are
essentially inverted cantilever flat
slab
elements.
2.
The
design
of
pad footings involves the choice
of
) footing area which is based on soil bearing pressure),
ii) footing depth which is based on shear resistance) and
iii
reinforcement
to
resist bending moment.
Reference
Calculations
Output
Dimensions
of base
Note 3
Service load = 760 leN
\
Note 4
Expected
total load
= 1.08) 760) = 821 leN
Required area
for
base
=
821)
1
150)
=
5 47
m
2
Try a base of 2 4 m x 2 4 m x 0 4 m
Weight
of
base = 2.4Y 0.4) 24) = 55
leN
Actual
total
load = 760
55 = 815
leN
Bearing pressure = 815)
1
2 4j =
kN/m
<
150 kN/m
2
;hence O K
NoteS
Preliminary check on effective depth:-
d > 1 0 N ) ° ~ = 10 1100)°·5 =332 mm;
footing size
hence overall depth
of
400 mm
i s O K
~ x 4 m
4m
DesilW for ben4in
\
ote 6 Ultimate bearing pressure
=
1100)
1 <2 4
= 9 k m
~ r i t i l bending moment at face
of
column) =
191 2.4 { 2.4-o.3 /2}2 1/2
= 253
leNm
M
=
253 leNm
TABLE 1 Assume a cover
of
40
mm, for moderate exposure
Note 7 conditions.
H
bar
size
of
16 mm
is
assumed, d
min
=
d
min
=
336
mm
400-40-16-16/2
=
336 mm and
da
= 344 mm
dan
=
344 mm
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Reference Calculations
Output
Chart 2
Mlbd
2
=
(253 xIW I 2 4 0 0 3 3 6 ~
=
0.93
(Part 3)
l00A/bd
=
0.25
As = (0.25)(2400)(336) I (100) = 2016 mm
2
3.11.3.2
3/4 c
9/4 d
=
3/4 300 9/4 336
= 981
mm
<
Ie
=
1200 mm;
hence reinforcement should be banded.
Use 7 Tl6
@
200 mm in band of 1200 mm
bottom r f
{<
(3)(336) 300
=
l308}; Use 3+3 T12
@
3
T12 @
200
200 mm in two outer bands.
7 T16 @ 200
As
=
1407 678
=
2085 mm
2
;
1407/2085 > 2/3
3
T12 @
200
3.12.11.2.7
Max. spacing
=
750 mm; hence O.K.
(both
ways)
100AiAe
=
(100)(2085)
I
(2400)(400)
=
0.22
>
0.13; hence O.K.
Note 8
n ~ g e length :; (40)(16) = 640 mm
2400-300 /2 =
1050 mm; hence O.K.
Check for yerticalline shear
[§I
.4.5.10
Consider a section at d from the column face, .and
assume no enhancement to v
e
.
V
=
191 2.4 { 2.4-o.3 /2
0.336} = 327
kN
v
=
(327 xl<P)
I
(2400)(336) = 0.41 N/mm
2
.
l00A/bvd
= (100)(2085) I (2400)(336) = 0.26
Table 3.9
v
e
= 0.42 N/mm
2
>
0.41 N/mm
2
; hence O.K.
Check for punchine shear
m
.7.7.2
V
max
=
(1100 xl<P)
I
(4)(300)(344)
=
2.66 N/mm2
O.8 25f·
S
=
4
mm
<
5 N/mm
2
; n ~ O.K.
3.7.7.6
1st
critical
perimeter
=
(4){(1.5)(0.344)(2)
0.3}
. = (4).(1.332) = 5.328·m
Area outside perimeter= (2.4)2 - (1.332)2= 4.43 m
2
V
= (191)(4.43) = 761 kN
Note 9
v = (761 xl<P)
I
(5328)(344)
=
0.42
N/mm
Shear r f not
= v
e
(0.42 N/mm
2
;
hence O.K.
required.
Notes
on
Calculations
3. Soil bearing
pressures are
given
in
terms
of
service
lOads
Hence, service loads have
to
used
to
determine the footing
area.
Service loads
can
be approximately obtained
from ultimate loads by dividing the latter by 1.45
in
reinforced concrete structures.
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In order to estimate ultimate loads from service loads however, it is safer to factor
the latter by 1.5.
'4.
The weight
of the
footing itself
cannot be known
until it is sized.
An
allowance of
8
of
the column load is generally satisfactory for obtaining a first estimate
of
footing weight, which should subsequently be calculated accurately. Anotherapproach
is to first estimate a depth (in
this
case 0.4 m) and reduce
the l l o w ~ l
bearing
pressure
by
the corresponding weight per unit area (i.e. 0.4 x 24
=
9.6
~ / m 2
before fInding the footing area.
5. This formula is not dimensionally homogeneous and can be
used
only if N is in
leN
and
d in mm.
is based on a punching shear considerations for commonly
used
pad
footings.
there is moment transfer to
the
footing as well, use d
>
(11.5)(Nf· 5
6.
As
the weight
of
the footing is considered to be a uniformly distributed load which
is taken directly by the soil reaction,
It
should not be considered
when
designing for
the ultimate limit states
of
flexure
and
shear.
7.
the values
of
TABLE 1 are modified
by
Note 5, a cover
of 35 mm
will suffice for
moderate exposure conditions. However, the cover is increased by a further 5 mm,
in case the footing comes into contact with .any contaminated ground water. The
minimum
value
of
d should
be used in the
design for flexure and vertical line
shear, while the average value of lid can be
used
in checking for punching shear.
8. If the distance between the column face and the end
of
the footing is smaller than the
anchorage length, the bars will have to be bent
up
near the end. of the footing;
otherwise, they can be straight. .
9. In most cases, punching shear is more critical than vertical line shear. Furthermore,
if a distance
d
is not available from the critical perimeter to the end of the footing,
the value of V
c
should correspond to 100A/b
v
d = 0.15 in Table 3.9.
on luding Notes
10.
I.f
the footing carries a bending
moment
in
addition
to
the
xi l
load, the maximum
and minimum pressures under the footing will be given by (l/BL}(N ± 6M1L), with
symbols having usual meanings. he maximum pressure should be kept below the
allowable bearing pressure.
11. the difference between maximum and
minimum
pressures is small (say upto 20
of the
maximum
pressure) it may be convenient to design for bending and vertical
shear by assuming that the pressure distribution
is
uniform and equal to the maximum
pressure.
12. Where the design for punching shear is concerned, the average pressure can be taken
for calculations, but a factor of
1.15
applied
to
the
shear force, according to the
provisions of Clause 3.7.6.2.
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EXAMPLE 23 • COMBINED FOOTING
Let
us
assume that
an
external columns
is
flush with the property line and that the footings
for the external and first internal columns have to
be
combined,
as
shown. While the internal
column carries an ultimate xi l
load of 1100
kN the external column carries an ultimate
moment of 60 kNm in addition to
an
ultimate axial load of 600 leN The allowable bearing
pressure of the soil is 150 kN/m
2
•
Use feu 25
N/mm
2
and f
y
460 N/mm
2
•
o ~ _ 4 _ . _ 7 m _ ~ _ 7 > 1 ~ ~ r
\l k
C
~
E
Introductory Notes
1 The situation described above is often found in crowded urban areas where buildings
are constructed
on very small plots
of
land
2. It is difficult to provide
an
isolated
p d
footing for the external column, because of
eccentric loading on the footing. Hence,
it can
be combined with the first internal
column footing
as
shown above.
eference Calculations utput
Service loads
Note 3
External column load
=
600 / 1.45
=
414
leN
External column moment = 60 / 1.45 = 41.4 kNm
,Internal column load
=
1100 / 1.45
=
759 leN
Dimensions of footine
Distance
of
C.of
G
from A
is given by x
where
759+414 x = 414 0.15 41.4 759 5.15
Hence, x
=
3.4
m;
therefore, for uniform pressure
Note 4
under base, use a base of
length
2 3.4 = 6.8 m
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Reference
CalcuJatloDS
Outppt
Note 5
f
we assume a thickness
of
0.8 m
for
the base, the
allowable bearing pressure is
150
- 0.8 24 • 131 k rr
Width
of
base required = 759+414
I 6.8 131
footing size
= 1.32 m
6.8 m·x 2.0 m
Note 6 Use base
of
6.8
x 2.0 m x 0.8 m xO.8 m
Analysis
of
footing
Assuming
that the C.of
G.
for
ultimate
loads is the
s me s
that for service loads, the footing
c n
be
idealised as follows:-
udl =
1100+600 / 6.8
=
250
le m
Note 7
:1T
A- ;:: :_=
E
w • 25 k m
Max. moment at C = 250 1.65Y I 2 = 340 leNm
To find
x ~
moment in
AC.
Mx
- 250 x
2
12
+ llOO x-1.65
= - 125 x
2
+ lloo x - 1815
putting dMx/dx
- 250}x
1100
=
0,
we have
x
= 4.4 m
M
m x
=
- 125 4.4 2 + 1100 4.4 - 1815
= 605 kNm
Max. shear force at C
1100 - 250 1.65
= 688 leN
Shear force at A = 600 le
Design for bending
TABLE 1
Assume cover of 40 mm, bar size longitudinal of
Note 8
25 mm; hence, d 800 -
40
-
25/2
= 747.5 mm
Section AC
Chart 2
Mlbd2
605
xl
6
I 2000 747.5 2 = 0.54
part
3
looAJbd 0.15 .
A. 0.15 2000} 747 .5
I
100 = 2243 mm
2
longitudinal
top
Use 5
s
As
=
2455
mm
2
-
on top
surface; steel-
these can be curtailed i required.
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Reference Calculations
~ ~ ~ = 340 x10
6
) 1 2000) 747.5)2 = 0.30
Table 3.27 100A/bd = 0.08; hence use lOOA/A
e
= 0.13
As =
0.13) 800) 2000)
1
100)
=
2080
mm
2
Use 5
TI5
s
=
2455 mm
2
) - on bottom surface;
these c.ould be curtailed as well.
3.11.3.2 Note:- 3/4 c 9/4 d
=
3/4 300 9/4 747.5
=
1907 > Ie
=
1000 mm;
hence, longituninal steel
can
be evenly distributed.
crransverse direction)
M
= 25012 { 2.0-0.3 /2}2 112 = 45
kNm/m
d
=
747.5 - 25 2 - 2 2
=
725
mm
assuming
r
size of 20 mm)
M/bd2
=
45 x10
6)
1
1000) 725)2
=
0.09
Table 3.27 Use
100A/A
e
=
0.13
As =
0.13) 1000) 800)
1
100)
=
1040 mm
2
/m
Use TIO @ 300 mm
Ag =
1047
mm
2
/m)
Table 3.29 Anchorage length = 20) 40)
=
800
mm
< 2000-300 /2 - 40 cover)
=
810 mm; hence O.K.
This steel
too can
be evenly distributed, as it is
nominal reinforcement; the same nominal steel
can
also be used as distribution steel for the top
longitudinal bars.
t
4.7m
1
T25
\
s
· ~ T @ 3
J
.
5T25
6. n
Check for
vertic l
line shear
ut ut
longitudinal
bottom steel
5TI5
longitudinal r
to
be
evenly
spaced
transverse
bottom steel
T20@3oo
mm
to be
evenly spaced.
3.4.5.10
In the longitudinal direction, check at a distance d
from the internal column face.
V = 688) -
{ 688+600 / 5.0 } 0.15+0.7475
=
457 kN
v =
457
x1<P) 1
2000) 747.5)
=
0.31 N/mm
2
Since nominal steel is used, v
e
= 0.34 N/mm
2
> 0.31 N/mm
2
;
hence O.K.
In transverse direction, a distance d from column
face is virtually at edge
of
footing; hence O.K.
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Reference Calculations
Output
Check
for
punCbio shear
3.7.7.2
v
max
(internal column) (1100 xlol I 4 300 735
=
1.25 mm
2
< 0.8 25f·
S
=
4
mm
2
5
mm
2
; hence
O.K.
As
the critical perimeter, (1.5)d from column face, is
outside the footing, only vertical
line
shear need be
Shear r is
not
checked. required
Notes Calculations
3. Where
service
loads
are
not specified
or
known,
they
can
be
estimated
by
dividing
ultimate
loads
by 1.45 for.reinforced concrete structures. When converting service
to
ultimate
loads,
it is safer
to
multiply the
former
by 1.5.
4. If the footing dimensions are given, as opposed to being designed the pressure
distribution under the
base
may
riot be
uniform.
5. This
base
thickness is
fairly
high, and is governed primarily by shear considerations.
If the distance between columns is large, bending moment considerations
will
also
require fairly deep base.
6. This fairly
large
width has been chosen to reduce
the pressure under the
footing and
satisfy the
shear
criteria. Although increasing the depth is generally more efficient
than
increasing the width,
having a
large depth may
also cause excessive
build up
of
h5t of hydration
temperatures, leading to thermal cracking.
7. Assulning the column
loads
to
be
point loads
is
conservative. In
reality
the load
will
be spread over a (mite r and the
resulting
bending
moments
and shear fol'CeS' at
the column faces
will
s o m w h t ~ m l l r
than those
obtained
from this
analysis.
8.
The argument
used
to choose
the cover is
the same as that
in Note 7 of
Example 22.
Coocludina
Notes
9.
the
perimeter or
section at
which shear should be
checked
falls outside
the footing,
the ,footing can be considered safe for shear.
10.
The analysis
of
the
above footing
has
n
performed assuming
that
both
footing
and
subgrade
are rigid.
If
elastic
foundation
assumptions
had
been used, the soil
pressure
near
the
columns
i.e. loaded
areas)
would increase,
but the midspan bending moment
would decrease.
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EXAMPLE 24 - PILE CAP.
A two-pile group of pile diameter
500
mm
and
spacing 1250
mm
centres supports a 450
mm
square column carrying
an ultimate load
of 2500 leN.
Design
the file
cap,
using concrete of
grade
25 and type
2 deformed
reinforcement
of f
y
= 460
Nlmm
Introductory Notes
1.
The minimum
centre-to-centre
distance
for piles
is twice
the least width of piles
for
end bearing piles and thrice the least width of piles for friction piles.
2.
A pile
cap can be considered
as a
deep beam, and
the
most
appropriate
way to
analyse forces is
to consider
truss
action
in
the pile cap.
Reference Calculations
Output
Pile cap dimensions
Note 3
Use an outstand beyond
the piles of half the pile
diameter.
Hence,
pile
cap i ~ n s i o n s
are:
length = 1250 500 500 = 2250
mm
width = 500
500
=
1000 mm
Try overall depth of
700
mm; hence,
effective
depth = 700
-
4 25/2 =
647.5
mm
>
1250 /2;
hence
O.K.
dimensions
2250
mm
x
1000
mm
x
7 mm
main steel
71 25
N
6
T
l
N 2 N 2
,
21
The
force T is given by
T N.l/ 2 d
= 2500 625 2 647.5
= 1207 leN
As
=
1207 x1oJ 0.87 460
= 3016
mm
2
Use
7 1 25 As
=
3437
mm
2
Banding is
not required, as pile spacing <
3 <p
spacing
of r
::;
1000-80-32-25
I
6
=
144
mm
Truss
action
Note
4
3.11.4.2
Note
5
Anchoraee
Anchorage
length requirec;l beyond
centre-line
of
pile
is
given
by
40 25 3016/3437
=
878
mm
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Reference Calculations Output
Stress
in
r - (0.87)(460)(3016/3437)=
351
N/mm2
When turning bars upwards, assume that bend starts
at edge
o
pile.
Stress at start of bend = 351) 878-250)/ 878)
= 251
N/mm
2
Bend radius, r, should be S.t.
equation
50 Fb (r.q,) <
=
(2)f
cu
/{1
2 q /lljJ}
Critical value for Itl = 40 16 (hor.
loops
25
= 81 mm < 144 mm
251) 491)
I r(25)
<
= (2)(25)
I
{
25/81 }
r
>
= 159;
Use
r = 160 mm r = 160 mm
Note
6
we
start the
bend
as
close
as
possible
to
the edge
of pile cap, length from C L of pile
to
start of bend
=
-
40 - 16 - 160 =
284
mm
3.12.8.23 eff. anchorage of bend =
12 25
=
300 mm
{ (4)r
4 160 = 640 mm}
vertical length
available
= 647.5 -
40
-
160
- 4 25
- 347.5 mm
total anchorage available -
284
300 347.5
. = 931.5 mm > 878 mm required; hence O.K. Anchorage O
K,
Check for punchina shear
3.11.4.5
3.11.4.4
3.11.4.3
Table 3.9
3.4.5.8
v
m x
=
(2500 xloJ) 1 4 450 647.5 = 2.15 N/mm
2
< (0.8)(25)°·5_ 4 N/mm
2
< 5 N/mm
2
;
hence
O.K.
Since spacing of piles < 3)q
no
further check is
required.
Check for
vertical
line shear
v = 625-150-225 = 250 mm
V
(at critical
section
-
2500 I 2 - 1250 kN
v - (1250 xloJ)
I
1000 647;5 k
_ 1.93
N/mm
2
625 625
2d/av = (2)(647.5)
I
250
= 5.18
looA/bd = 100 3437 I (1000)(647.5) = 0 53
V
c
=
(0.51)(5.18) - 2.64
N/mm
2
>
1.93 N/mm
2
;
hence O.K.
74
Shear
rl
not
required.
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Reference
Calculations
Output
Distribution steel
Table 3.27
l A A
c
= 0.13; As = 910 mm
2
/m
Use 16T @200 mm s= 1005 mm
2
/m
distribution steel
This steel can also be bent
up
like the
main
steel.
16T@2
mm
Horizontal binders
Il binders
7T
main
4')16
Note 7
Use
25
of
main
steel
,
As
= (0.25)(3016)
= 754
mm
2
horizontal
Use 4 T16
As
= 804 mm
2
· T l ~
binders
These binders will tie the
4 Tl6
main
and
distributiot:l
steel.
Notes on Calculations
3. The criterion used is that the effective
depth
is equal to at least half the distance
between
pile centres.
When using truss
theory,
this will
result
in
a compressive strut
of
45°
minimum
inclination.
The
cover value has
been
chosen
as
per Note 7 of
Example 22.
4.
the width of the column
is
accounted for, the value of tensile force
w ll
be a little
less. This is given some treatment in nAllen, A.H., Reinforced concrete design to
BS
8110 simply explained, E. F.N. Spon, London, 1988 .
5. Allowance is made here for side cover of 40 mm and a horizontal binder of 16 mm
diameter.
6. We start the bend closer to the edge of the pile cap than assumed in the bend radius
calculation - this is
to
achieve
as
great a length
for
anchorage
as
possible within the
geometry
of the
pile cap.
7. This provision is also given by Allen, referred to in Note 4 above. The main steel
required as opposed to provided) can be used in the calculation.
Concludinl Notes
8. t can be shown
that
less
steel
is
required
if Beam Theory is used. Furthermore, the
anchorage requirement
beyond
the centre line of pile is much
less.
However, Truss
Theory probably describes more accurately the actual behaviour of the pile cap.
9. the spacing of piles exeeds 3 times the pile diameter, an additional check for
punching shear has to be
made,
and the pile
cap
has to be considered
as being
banded for the distribution of tension steel and check
for
vertical line shear.
75
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EXAMPLE · STAIRCASE
A staircase has to
span
between
two beams,
which are 3.0 m apart in plan. The difference
between
the
two
levels is 2.0 m. Assuming that
the
staircase
is
sheltered
and
that it is
subject
to
crowd loading, design the staircase,
using feu
= 25
N mm
f
y
= 46
N mm
deformed type 2) or 250
N mm
plain
and density of reinforced concrete = 24 kN/m
J
.
Assume top finishes on tread only) as Sk m
and the soffit plaster as 0.25
k m
•
IJ;ttroductory Notes
1.
Staircases
are
essentially inclined slabs. The major difference in design approach
is that the loading has to be
obtained
as the loading in plan.
2.
In this
particular example,
the
layout
of the
staircase
has
to be designed
as
well.
Reference
Calculations Output
Choice of layout
Let
us choose
12
stairs.
Then, rise R)
= 2000/12 =
167
mm
rise = 167 mm
going 0
=
3000/12
= 250
mm
going
=
Note
3
Also 2R
0
=
2){167)
250
=
84
mm
2SOmm
approx. 600 mm; hence O.K.
tread
=
Use nosing
of
25
mm,
so that tread = 275 mm
275
rom
Waist
thickness
Note
4
Assume trial span/depth of 30 1.15 = 34.5,
for
a
3.10.2.2 I-way heavily loaded continuous slab, stiffened by
stairs.
effective
depth
=
3000 / 34.5
=
87 mm
TABLE 1
I f
we assume cover
=
20 mm mild exposure
conditions and concrete protected by mm 1:3 ct:
sand rendering and bar diameter = 12 mm, we can
choose h 120 mm and d = 120-20-12/2 =
94
mm
h
=
120mm
Table 3.5 Note:- 1.5 hr. fire resistance available.
d =
94
mm
L.oadin
for 1 m
wide
strip
Note S
Factor for slope
R2
0
2
)0.5 0
=
{ 167)2 250)2}0.5
250) =
1.20
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Reference
Calculations
Factor for overlap
=
T/G
=
275/250 = 1.1
Output
Waist =
0 . 1 ~ 1 2 4 1 . 2
= 3.46 kN/m
Steps
=
112 0.167 1 24 1.1
=
2.20
kN/m
Top finishes = 0.5 1 1.1 = 0.55
kN/m
Soffit plaster = 0.25 1 1.20 = 0.30 kNlm
Total dead load = 6,51 kNlm
Note 6 Imposed lo d = 5.0 1 = 5.0 kN/m design udl =
Design
load ... 1.4 6.51 1.6 5.0 = 17.1 kN/m 17.1 kN/m
PesiKn for
bendine
Note 7
Chart 2
Part 3
3,12,11.2,7
Table 3.27
Note 8
M span and support =
F.l
I 10
= { 17.1 3.0 } 3,0 I 10 = 15,4 kNm/m
M/bd2
= 15.4 x1 6 I
looo 94f =
1.75
l00AJbd = 0.48
A. = 0.48 1000 94 1 100 = 451 mm
/m
Use Tl2
@
250
mm
As
=
452 mm
2
/m main steel
Allowable spacing = 3 94 = 282
mm
> 250 mm T12 @ 250
mm
100A/Ac= 100 452
I
1000 120 =
0.38
> 0,13
Hence,max.
sp ing
and min. steel are O.K.
Check for deflection
Mfbd2 = 1.75
fa = 5/8 460 451/452 = 287 N/mm
Table 3.11 Hence, F
1
= 1.15 for tension reinforcement
3.10.2.2 Allowable spanldepth = 26 1.15 1.15 =
34.4
Actual span/depth = 3000 / 94 = 31.9
<
34.4;
hence deflection O.
K
Check
for shear
Deflection O.
K
Note 9
Table 3.9
v
= 0.6 F = 0.6 { 17.1 3 = 30.78 kN/m
v = 30.78 x1<P I 1000 94 = 0.33 Nfmm
2
l00AJbd = 100 452 I 1000 94 = 0.48
v
c
...
0.66N/mm
2
> 0.33 N/mm
2
;
Hence, shear r not required.
77
shear
r
not
required.
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Reference
Distribution steel
Calculations
Output
Table
3.27
lOOA,IA
c
= 0.24 for mild steel)
Note 10
A.
= 0.24 1000 120 100 = 288
mm
2
/m
Use
R8 @ 175
mm
s
=
287
mm
2
/m)
distribution steel
R8@·175
mm
R8@175
Note
11
Tl2@250 Tl2@250
1
4
Notes on Calculations
3.<In
~ f C > j
3. In
general
the rise
should
vary
from a minimum of 150
mm for
public stairways to
a
maximum
of
175 mm for
private
stairways.
The going
should
vary
from 300
mm
for public
stairways
to 250 mm for private stairways. 2R 0 should be kept as
close
as
possible
to 600 mm.
A nosing
can e
provided
so
that the tre d
is
greater
than the .going, thus making for greater user
comfort.
4. Although the value
used in
Example
13
for
a
continuous one
way slab
was
34,
a ratio
of
30 is
used here, because
the loading is
much heavier
- the
waist
carries the load
of the steps, in addition to
its
own
weight
on an incline, and
also
a high imposed
load.
The
above ratio
is
increased by
15 because
of
the
stiffness contributed by
the stairs.
5. The waist and soffit plaster have thicknesses
that
are measured perpendicular to
the
incline. Hence their
load
in plan
will e
greater by a factor of R} 0
2
0.5
O. The
steps
and tread
finishes
have
mm overlaps for each 250 mm length in plan,
because
of
the provision
of a
nosing. Hence
their
load
will have to
be
factored
by
T/O. This factor
can
be ignored
for the imposed
load, because it
can
be argued that
the
entire
tread will not be
available
for standing.
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6.
The
imposed load corresponding
to
crowd loading is 5 kN/m
2
- see BS 6399: Part
1 1984):
esign
loading for buildings: Code
of
practice for dead and imposed
loads .
7. For a staircase having continuity, we can assume that oth the span and support
moments are approximately F l 1O
8. Staircases
are
generally heavily loaded see Note 4 above), unlike horizontal slabs.
Hence, the check for minimum steel is not very critical.
9 Taking Shear Force as 0.6)F is conservative for staircases such as this, In any case,
as for most slabs, staircases
will
not require shear reinforcement.
10; Mild steel reinforcement is often used for distribution bars, as
in
this case, since the
use
of
high
yield reinforcement may result
in
more steel than that specified by the
minimum
steel
requirement, in order
to
meet the maximum bar spacing rule.
11. When detailing reinforcement, care should
be taken
not to bend tension steel in a
way that an inside comer can get pulled out. Hence
bar type
2) should be continued
from the bottom
face
of the lower slab to the
top f ce of
the waist.
Bar types
2) and
3) can be
t ken
horizontal distances
of
0.3)1 see Clause 3.12.10.3) into the waist
from the faces
of
the beams. Bar types 1), 2) and 3) can be continued into the
lower and upper slabs as
s
reinforcement, if required.
Bar type
4) shows how the
upper slab reinforcement can
be t ken
into the beam support.
Concluding
Notes
12.
If
the supporting
beam
for the flight of
st irs is at
the ends
of
the landings, the entire
system of staircase and two landings can be
taken
as spanning between the supporting
beams see figure below). In this case, the slab system could be considered as simply
supported
if
there is no continuity beyond the landings. The loading
on
the landing
and staircase section would be different in a case such as this.
r
sp n
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EXAMPLE
16
- STAIRCASE
A typical plan area of a stair well is shown· in the
figure,
where the landings span in a
direction perpendicular to
the
flight and span of the stairs. The dimensions of a single stair
are as follows:- rise
=
175 mm; going
=
250 mm;
tread =
275 mm. The top finishes on
tread
only) are equivalent to a distributed load
of
0.5
kN m
and the soffit plaster one
of
25
k m
The imposed load can
betaken
as 3
k m
Using
feu =
25 N/mm
2
,
f
y
=
460
N/mm
2
deformed
type
2)
or
250 N/mm
2
lain) and density
of
reinforced concrete
;:;
kN/m
3
,design
the staircase component
of
the system.
1.2m 8xO.25 = 2 Om
2 Om
\<:---.....;* * : - - - - - - ~
1 2m
1 2m
I
Introductory
Notes
1. In
this
example
of
a staircase, the landings
span perpendicular
to
the
stairs and
support the staircase, unlike in the case described in Note 2
of
Example 2S, where
the landings
also
span in the direction of. the staircase. •
2. In analysing the above system, the
staircase is
assumed
to be
supported along two
edges within the landings. Continuity over the supports can
be
assumed for the
purpose
of
spanldepth Iatio calculations.
Reference
Calculations Output
Stair
span
and
waist
thickness
equation
Effective span
= la
O.5) lb,l
Ib,V
Note 3
=
2.0
0.5) 1.2 1.8)
=
3.5 m
eff. span =
length
of
stairs
)
=
{ 2.0)/ 3.5)} IOO)
=
57 3.5 m
3.10.2.2
<
60 ; hence span/depth enhancement not possible.
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Reference CaJcuJatioas utpat
Note 4 Assume a trial spanldepth ratio of 32, for a
continuous I-way
spanning
sIaiR:ase
Effective depth
=
3500)/ 32)
=
109
mm
Assuming a cover of 20 mm mild exposure
conditions
and
concrete protected by 10 mm 1:3
ct:
sand
rendering),
and
ba r
diameter
of
12
mm, we can h
=
140
mm
choose
h=
140
mm and
d= 140-20-1212 = 114 mm d = 114 mm
Loa in for 1.2 m wide staiR:ase)
Note 5 Factor for slope = R2 + OZ 0S I G
=
{ l75f +
250f}o s I
250)
=
1.22
Factor for overlap = TIG = 275)/ 250) = 1.1
Waist = 0.14) 1.2) 24) 1.22) = 4.92 kN/m
Steps
= 112) 0.175) 1.2) 24) 1.1)= nk m
Top
finishes
=
0.5) 1.2) 1.1) = 0.66
kN/m
Soffit pJasta
=
0.25) 1.2) 1.22)
=
0.37 k lm
Total dead loed
= 8.72
kN/m
Imposed
lOad
= 3.0) 1.2) =
3.6
kN/m design udl =
Design load= 1.4) 8.72) + 1.6) 3.6) = 18.0 k m 18.0 k m
esi n
for
bendine
Note 6 The
stain:ase
can be idealised
as
follows:
@;
R
A
=
18) 2.0) 1.9)
I
3.5)
=
19.5
kN
R
B
= 18) 2.0) - 19.5 = 16.5
leN
M
x
=
RA x
- w x - 0 6f12
d x dx = 0 when R
A
- w[x - 0.6] = 0
Le x= RA/w + 0.6
= 19.5 / 18
+
0.6
= 1.68 m
Muua= 19.5) 1.68) - 18 1.68-G.6t/2
= 22.26 kNm
Chart 2
Pa rt 3)
MIbc¥ =
22.26 xl<n
1
12OO) 114t
=
1.43
l00A/bd
=
0.39; A
c
=
534
mm
2
Use
5 Tl2 A.
=
565
m.r
81
main
steel
5 T12
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Wei. .. .
CaJcu,,-
Output
Check
for
c ection
MJbd2.::
1.43;
f
s
= 5/8 460 534/565 =
N mm
Table 3.11
H ~ J
1.29 for tension reinforceament
spanldepth
: : 26 1.29
33.5
Actual spaoIdeptb 3500 1{U4 30.7
<
33.5; hence O.K. Deflection O.
Distribution reinfon:ement
Table 3.27
lOOAJA.c
=
0.24 mild steel
As = 0.24 140 1000 100 = 336 mm2/m
distribution
steel
Use
R8
@ lSOmm
s 335 m.r/m
R8@
ISO mm
Notes
OD Cakulatious
3.
The
support
Jineforthe stain:ase
is
at
the centre
of
the smaller landing
but
only
0.9
m into the wider Janding, because 1.8 m
js
the maximum distance
,over which the
load
can be assullWldto be spread.
4. This ratio is a liUlep:ater
than that
assumed for the
previous example Example
25 ,
because the-imposed load here is somewhat lower.
5.
These
factors and their use
are deScribed
in Note 5 of Example
25.
6. Although
continuity is assumed oversupports for spanldepdl ratio considerations, it
will be safeeto assume simple suppoIts
when designing
for bending, as tile continuity
extends only
upto the
edge of the Jaadiqg.
The loads
fromd1e landings are.carried by
the
landings
in the direction to
the
flight of the stairs; hence they are
not considered in the analysis.
Concluding Notes
7 Detailing of reinforcement can be·cJooe
in
a IIl3DDel sinillar
to
that
in
Example 25.
8. Shear can
also
be checJr£d for, as
in
Example 25, the
maximum.sIlear force
being the
greater
of R
A
n ~ - i.e. 19.5 kN.
9.
When
designing the landings, in addition
to their own
dead
and
imposed loads, the
loads from the stain:ase --i.e. R
A
and R
B
will be uniformly spread
over
the entire
smaller landing and over
1. 8
m of the Jarges landing, respectively.
10. Where staircase flights surrounding
openwe ls
intersect at right angles, the loads
from
the
common landing can
be
shared
between the two perpendicular spans.
8
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t
EXAMPLE 27 - PLAIN CONCRETE
WALL
The lateral loads in
the
short
way direction on a four storey
building
are
taken
by two
end
concrete
shear walls of
length 15 m
and
height 14 m. The servi<;e wind load on one shear
wall is 180
leN.
Check.whether a plain concrete wall
of
grade
25 concrete and
175
mm
thickness
is sufficient for the
wall panel
between foundation
and
1st
o o ~ slab
(clear height
of wall =4 m) it carries
the
following terVice loads, in addition to the wind load: self
weight
=
18
leN m
dead
load
from 1st
floor.
slab
=
12
kN m
dead load from
above
1st
floor
slab
=
OleN m;
imposed load from 1st floor
slab =
7.5 kN m imposed load from
above 1st floor slab = 19 kNlm.
Introductory
Notes
1. Given
that.even
plain concrete walls require horizontal and vertical reinforcement
(Clause
3.9.4.19),
and
this
n inforeement
will
be
,distributed
on two faces
(which
is
advisable,
since
crack control reinforcement
should
be
as close
to
the surface as
possible), then it is very difficult to
construct
a wall under 175
mm.
This
is
because cover requirements will be 2 mm on the inside (mild exposure) and 30 mm
on
the
outside
(moderate exposure -
see TABLE
1 including
Notes 5 and 6 -
and
because
the
bar diameter for
vertical
steel
should
be
at least
12
mm,
in
order
to
ensure sufficient
sti ffJIess for
the rant orcementcage prior
to concreting.
2. Guidance on
calculating
wind loads is
given
in ·CP3: Ch.
V:
Part 2 (1972): Basic
data for the design of buildings: Loading: Wind loads·, and the method is shown in
Example 31
3.
I t is assumed
that stability
for
the stn1Cture
as a whole
has been
satisfled. The
overturning moment due to wind, factored by
1.4
should be less than the resisting
moment
due to dead
load,
factored
by 1.0
(see Table
2.1)
Refereoce Cakulatioos
Output
Wall
CJassi ierinn
3.9.4.3 Since
lateral
support is
provided by
foundation
and
Note
4
1st floor
slab, wall panel
can
be
considered braced.
The
slab will give only displacement restraint, while
the foundation can give displacement as well as
rotational
.restraint.
= 3.5 m
Note
5
Hence
=
(0.875)(4.0)
=
3.5 m
lJh
=
20
1.2.4.9
Ie/h = (3.5 xloJ)/(17S) = 20
>
15; hence
slender.
hence,
slender
3.9.4.4
<
30;
hence
max. value not exceeded. braced
wall.
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3.9.4.9
Note
6
3.9.4.15
3.9.4.16
Table
2.1
equation
43
equation
44
Note 7
3.9.4.18
We
sball
use serviceability vertic:al
Joads
to calculate
the resultant eccentricity e just
below
the 1st floor
slab, assuming the eccentricity
of
1st floor
slab
loading
is
bl6. and
that
the
eccentricity
of loads
above this
is
zero.
12+7.5 175/6 / [ 12+7.5 + {80+ 19 O.8 }]
4.96 mm
Min. ecc hI20 175 / 20 8.75
mm
>
4.96
mm
e
a
Ie
2/ 2S00 h 3500 2
I 2SOO 175
28
mm
Assuming wind
acts at mid
height
ofwall,
wind
moment
180 1412 1260
kNm.
Hence. wind
loading
on wall
± 6 1260 / 15>2
± 33.6kN1m
Hence.
ultimate loads per unit
length of
wall are:
Combination
1.
f
l
1.4 18+12+80 + 1.6 7.5+19 0.7
184
kN/m
Combination
2.
f
2
1.4 110 1.4 33.6
201
kNlm
or
f
2
1.0 110 - 1.4 33.6 63
kN/m
Combination 3
f
3
= 1.2 {1l0
26.5 0.7 33.6} =
195
kNlm
Note: -
no tension arises.
Now,1lw
<
0.3 {h -
2 exlfcu
= 0.3 {175 - 2 8.75 } 25
1182
k m
and
Ow < O.3 {h -
1.
2
l x - 2 eJf
cu
= O.3 {175 - 1.2 8.75 - 2 28 } 25 == 813 k m
These are
satisfied,
since
l w max
201kN/m
Check for shear
Design
horizontal
shear
force
180
kN
Min.
deagn
vertical load 110 15
·1650
kN
{> 4 180 1.4 1008 kN;
hence
O.K.}
84
x 8.75
mm
e. 28 mm
l w ~
~ O I
k m
n ;
W.1IlUI
k m
l w
is O.K.
i
1./;
l
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4
Refereace
Caleulatio
Output
Minimum reinfOfCQDCOt
Min r
=
(0.25)(1000)(175) I 100
= 437.5
mm
/m
(both directions)
Note 8
Use vertical steel
Tl2
@
3
mm in both faces
vertfcal
steel
A
=
753
mm
/m
2
x Tl2
@ 300
Horizontal steel
1'8
@
225
mm in both
faces
horizontal steel
(As
= 446 mm2 m
2xT8@225
Notes on Calculations
4
A
column is considered braced in
a
given plane if it is not required to carry the
lateral forces in
that
plane. A
wall
bowever is considered braced
if
lateral stability is
given to it by other structural elements, when it is carrying in-plane loads.
the
wall
alone
has to
resist transverse loads, it
is
unbraced.
5. Since the end conditions in the given wall are midway· between those specified in
Clause 3 9 4 3
the effective length
factor
is
also
midway between the factors given.
6. The imposed load here
is factored
by
0 8
according
to
BS 6399:
Part
1 (1984):
Design loading for
buildings
Dead and imposed loads, since loads from 3 floors are
involved. Later on, when cbecldng
the
w value for the
wall
.panel, a factor
of
0.7 is
used,
since
loads from
4
floors are involved.
7. Equations
43
and 44 for braCed
walls
correspond to the top (maximum initial
eccentricity) and midway (maximum eccentricity due to deflection) sections.
However is calculated at
thebottom of
the wall,. taking into account the self
weight
of
the
wall
and maximum inoment
due
to wind. This is slightly inconsistent
but conservative.
A
similar approach is used in column design.
8
Since reinforcement to control thermal
and
hydration sbrinkage should be fairly
closely spaced, a spacing
of
300 mm
is
not
exceeded. 12 mm
dia.
bars
are
used
for
vertical steel, in order
to
give stiffness
to the
reinforcement cage prior
to
concreting.
The horizontal reinforcementsbould be placed outside the vertical steel on both faces,
to ensure better crack control, as thenna
and
shrinkage movements will generally be
in the horizontal direction; furthermore it is easier
to fix the
horizontal steel
on
the
outside.
Concluding Notes
9.
The wall reinforcement should
also
be
checked for satisfying tie reinforcements. This
is dealt with in Example 33.
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EXAMPLE 28 - CORBEL
Design a corbel that will
carry
a vertic:alloed of 35 kN into a 300 mm x 300 nun column.
assuming the line of action of the load to be 150 mm from the face
of
the
column.
Take ~ u
= 30 mm
and
f
y
= 460 N mm
defcmned type 2).
Introductory Notes
1. A cos:bcl can be considered to bea -deep
cantilever •
where
truss
Don.
as
opposed
beam
action, predominates and
where
shearing action is
critical
..
2. Compatibility
of strains between the
strut-and-tie system of the
truss
must
be
ensured
at the root of the corbel Clause 5.2.7.2.1 b».
Corbel
dimensions
Calc:ulatioos
Output
5.2.3.4
Note 3
The width of the corbel
can
be the same as that of
the column, Le.
300 mm.
The length of the bearing plate
can
also
be
taken as
300 mm and i f dry bearing
on concrete
is
assumed,
the
width of thebeariag plate, b,
will
be given by
350
xloJ)
I
3OO b <=
).4)f
cu
b 350 xloJ)
I
300) 0.4) 30)= 97
nun
Hence,
choose
bearing
width
of
100 mm
bearing
width
Since the
corbel
has to project out from the bearing 100
mm
area a
distance that wouJd accommodate a.
stressed
bend radius choose corbel projection as
400 mm total
p r o ~ t o n
Corbel
depth bas
lObe
·suclltbat
max.
aIL
shear is 400nun
not exceeded
- i.e.
O;8) 3Of.5 =
4.38 N mm
Hence, d > 35 xl 3 300) 4.38) =
266mm
Choose h = 375
nun
and assuming cover of 2 mm
mild exposure conditions, concrete
protected
by lQ
mm
1:3
et:
sand
rendel)
and
bar
dia.
of
2
mm
d
=
375
2 2 2
=
345
mm h =375
mm
Let
the
depth. vary
from
375 mm to 25 mm
d::;:
345
mm
y
~ l ~ n 35
kN
ultimate
T ~ C l o T
L
:
t
125
Oo9xl[J
0.45f
v cu
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eference
Calculations
Output
5.2.7.1
Now avid 150 / 345 0.43 < 1
Also,
depth
at outer edge of bearing
area >
375/2
mm; hence, definition of corbel
is
satisfied.
Main reinforcement
From strain compatibility
nd
stress block,
C 0.45)fcu 0.9)b.x.CosP .......... 1
Since the
line of
action
of
C must p ss thro the
centroid of stress block, J tan-
1
zlI50),
Le.
{J
tan-I { d - 0.45x)IlSO}
Furthermore, from the triangle of forces for P, T
andC,
C
P
Sin{J
........................
2)
We need
to
find a value
of
x, and hence {J that will
satisfy 1) and
2)
simultaneously.
x 216
mm
will give
{J
SS.SO and C
409
kN
T
350
Tan{J 212
kN
Since x 216 mm, by
str in
compatibility, strain in
steel is { d-x)/x} 0.0035) 2.090 xlO-]
Note 4 Hence, steel
h s
just ~ and f
s
0.87)f
y
Hence,
As
212
d 3
0.87) 460)
530
mm
2
Use 3 T16
s 603
mm2) .
m in
steel
5.2.7.2.1
Min. re required 112) 350
xloJ) /
0.87) 460)
3 T16
437
mm
2
<
603
mm
2
; hence O.K.
NoteS
Also
l00A/bd
100) 603)
300) 345) 0.58
>
0.4 and < 1.3; hence O.K. Detailing O.K.
Shear reinforeeroent
v 350 xloJ) 300) 345)
3.38 N/mm
2
l00AJbd 0.58
Table 3.9
V
c
0.546 30/25 °.J3 2d1ay
0.58) 2/0.43)
3.4.5.8
2.69 mm
< 3.38
mm
Table 3.8
Provide
An
> bv.Sy v-vJ
0.87)£
v
Aav Sy
>
300) 3.38-2.69) 0.87J 460) 0.517
Use lOT
@
300 mm. Since this has to
be
provided
over 213 375 250 mm,
2
bars will suffIce.
5.2.7.2.3
Min. requirement is 603/2
302 mm
2
Use 2
TI0
links
@
175
mm
links
As
314 mm
2
> 302 mm2;
hence
O.K.)
2TlO
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Reference
Note 6
endine m in reinforcement
The bend in
the
m in reinforcement should start a
cover distance (20 mm) from the bearing plate.
It
should
end
a
cover
+
b r
di a
20
10
16
=
46
mm) from the end
of
the corbel. Hence. distance
available for bend radius
200 - 20 - 46 = 134 mm
3.12.8.25.2 Critical value
of
3tJ=
20
10 (link) 16 = 46 mm
Stress
in b rs
= (0.87)(460)(5301603) = 352 mm
equation 50 F
bt
v »
< =
2 ~ {I 2 q J ~ }
352 201
(16)r < = (2)(30)
{I (2)(l6l46)}
r >
=
125mm en radius
Choose r
=
130
mm
<
134
mm; hence O.K.
=
130 mm
100· 100.. 200
r-l30,.,
1
2TIOfH75
Notes
on Calculations
25
3. Varying the depth from a full depth at the root to 2/3
of
the
depth at
the
end ensures
that
one
of
the
conditions for a corbel
in
Clause
5.2.7.1
is
automatically met - i.e.
that the depth at the outer edge
of
bearing is greater
th n half
the depth
at
the root.
Furthermore, it facilitates
the
placing of horizontal she r links in the upper two-thirds
of
the effective depth
of
corbel as specified
in
Clause 5.2.7.2.3.
4.
Using Figure 2.2, the strain
at
yield
is
(0.87)(460)
1 200 xloJ)
= 2.0 x10
3
for steel
of
f
y
460
N/mm
2
,
since the Young s Modulus specified is 200 kN/mm
2
.
5. Although these limits on l00A/bd, where d is the effective depth
at
root
of
the
corbel, are not given in BS 8110, they are specified in Rowe, R.E. et aI., Handbook
to British Standard BS 8110: 1985 : Structural use
of
concrete, Palladian, London,
1987 .
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6 Although the code allows
the end to
st rt
at
the edge of the bearing
pl te
itself the
allowance of a
cover
dist nce from t outer edge of the bearing plate will ensure the
spreading of load from the e ring plate to the level of tie steel before the bend
commences
Concluding Notes
7 Since a fairly large distance is involved in accomodating the bend radius an
alternative way of anchoring ti bars is to weld a transverse
bar
of equal strength
subject to the detailing rules
in
Clause 5 7 In any ease the actual projection
of the corbel beyond the bearing plate can be adjusted right at the end
of
the design
and will not affect preceding calculations
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EXAMPLE 19 - DESIGN FOR TORSION
A cantilever slab
of
clear
span
2.0 m functions as a hood
over
a porch. Its thickness varies
from 200 mm
at
the support to 100 mm at the free end, while
it
carries finishes amounting
to
0.5
kN/m
2
and an imposed load
of
0.5
kN/m
2
• t
is supported by a beam 600
mm
x 300
mm, which spans 4.0 m between columns, which are considered to provide full bending and
torsional restraint. Design the beam for bending and torsion, assuming
feu
30
N/mm
2
,
f
y
= 460 mm
deformed type
2 ,
fyv = 250 N/mm
2
and density of reinforced concrete =
k m
•
IDtroductory Notes
1 t
is instructive
to
classify torsion into two types. Compatibility torsion, which may
arise in statically indeterminate situations, is generally not significant; torsional
moments will be shed back into the elements carrying bending moments a t right
angles to the element carrying torsion , because torsional stiffnesses
are
lower
than
bending stiffnesses.
ny
torsional cracking will
be
controlled by shear links.
However, equilibrium torsion in statically determinate situations, where torsional
resistance is required for static equilibrium, will have significant magnitudes, and has
to
be
designed for.
The
example above
is
such a case see Clause 2.4.1, Part 2 .
2.
Assuming that the
columns
provide full
bending restraint implies
that
they have
infinite stiffness. practice,
of
course this will not
be
the case and the deformation
of
the columns will reduce the
beam
fixed end moments. However, full torsional
restraint has to
be
provided by
the
columns, in order to preserve
static
equilibrium,
where equilibrium torsion is invol lled
Reference
CaleuIatiODS Output
~
~ o L
2000
oad
in
2
on
e m
HoOd
=
{ 0.2+0.1 /2} 2.0 24
= 7.2
kN/m
Finishes
=
0.5 2.3
=
1.15 kN/m
Self
weight
=
0.6 0.3 24
=
4.32
kN/m
Total dead load
=
12.67
k m
Imposed load 0.5 2.3
=
1.15
k m
bending udl
=
Design load={ 1.4 l2.7 + 1.6 1.15 } =19.6
k m
19.6
kN/m
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Reference
Torsional loadio
Calculations
ut ut
assume shear centre is at centroid of beam section
Hood
= 7.2 2/3 0.15+ 1.0
+ 7.2 1/3 {0.15+ 2.0/3 }
= 7.48 kNmlm
Finishes
0.5 2.0 0.15+ 1.0 = 1,15 kNm m
Total dead load torsion 8,63
kNm m
Imposed load torsion = 0.5 2.0 0,15+ 1.0
1,15
k m m
Design load=={ 1.4 8,6 + 1.6 1.15 } = 13.9 kNmlm torsional udl
= 13.9 kNm/m
Desim for bendin
TABLE 1
Example
8
Chart 2
Part
3
Table
3.27
Table 3.9
Note 3
Assume
cover
30
mm
moderate
exposure
conditions, TABLE 1· values modified
by Notes
5
6 , link dia. ,.; 10 mm and main bar dia, 20 Mm.
hence, d = 600 -
30
- 10 - 20/2 = 550 mm
Take M 1I12 w.l
2
for built in beam
1112 19.6 4 2
26.1 IeNm
Mlbd
= 26.1
xlo6
I 300 550f = 0,29
lOOA,lbd 0.08
Use
lOOAJA
c
0,13
As
= 0.13 300 600
I
loo
= 234
mm
2
Same
nominal
steel r f
can
be used at span.
Shear Force = 19.6 4
I
2 = 39.2 leN
max.
v = 39.2
xloJ I
500 300 = 0,24
N/mm
2
<
V
c
Desi D for torsion
y
I =
6 -
2 30 + 10/2 = 530
mm
Xl = 300 - 2 30 + 10/2 = 230 mm
Total torsional moment = 13.9 4 = 55,6 kNm
Torsional restraint at each
end=
55,6/2 = 27,8 kNm
The torsional moment
will
vary as follows:-
27 .8 1.2m
I ~
-27.8
91
d = 550 mm
Y
= 530
mm
Xl
230 mm
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efereace
a a latioos
utput
equation 2
Max. value of v
t
= 2 T I b m i i ~ - huuJ3
Part
2
= 2 27.8 xld I 300 600 - 300/3
=
1.24
N/mm
<
4.38
N/mm
v
tu
Table 2.3
>
0.37
N/mm
vt,min)
Part
2 Thus, beam section is O.K. but requires torsional
r f
Proyision
of
reinforcement
Table 2.4 Since v
<
V
c
for the entire beam, the area where v
t
part 2
< =
vt,min
bas to
be provided with nominal shear r f
and the area where v
t
> vt.min with
designed torsion
r/f.
equation 2 Torque corresponding to edge of nominal
shear
r f
is
Part 2
given by T
= V
t
min hmm
2
tm x
-ohznm 3
I
2 .
Table 2.3 = 0.37 300f 600 -
3 13
I 2 Nm m
Part
2
=
8.33 kNm
Distance from
beam
elL
= 8.33/27.8 2.0 = 0.6 m
Hence, length of beam for nominal shear links
= 2 0.6 = 1.2 m
Nominal shear links given by
Table 3.8 A,;.)Sv
>
=
0.4 300
I
0.87 250 = 0.55 Nominal links
For 10 mm links, sv= 157 mm
2
; Sv
<
= 285 mm RIO@ 5 mm
Use RIO links @ 250 mm {< 0.75 d = 413 mm}
middle 1.2 m)
Designed torsional
links
given
by
2.4.7
A,;.)Sv >
= T
I
0 . 8 ) ~ Y l 0 . 8 7 ) f
0
Part
2
=
27.8
xl
I 0.8 130 530 0.87 250
= 1.31
For 10 mm links, sv= 157
mm
2
;
Sv
<
=
120
mm
Torsion links
2.4.8 Use 2RI0 links @ 200 mm
{<
= 200 m m,
Xl
yl
/2}
2RIO@2oo
Part
2
Length
of beam
at each end for torsional links
mm 1.4 m
= 4.0 - 1.2
I
2 = 1.4 m
from
both
ends
Designed additional longitudinal steel given by
2.4.7
s > A,;.)Sv) fyvlf xl+Yl
Part 2
157/120 2501460 230
+ 530 = 540
mm
2
this is
divided
between 8 bars, each requires 67.5
mm
2
3 at
top and
bottom, 2 in middle .
2.4.9 Since
beam
length is small, assume bending
Part
2
reinforcement is not curtailed; longitudinal
Note 4
reinforcement for torsion
also
cannot be curtailed.
92
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Reference
Calculations
Output
Total steel requirement at top and bottom levels
=
7.5) 3) + 234
=
436.5 mm
2
Use 2Y16 + YIO at top and bottom levels As =
481
mm
2
) and
2 Y
10 at intennediate level As =
157
mm
2
)
This arrangement will satisfy
Table
3.30
) max. spacing for tension
r
= 160 mm top bottom
2.4.9
b) max. spacing for torsional
r =
300 mm
2Y.l6+YIO
part 2)
) torsional r provided in 4 comers
middle 2YlO
Notes
on
Calculations
3. The torsional moment variation in beams, whether for a distributed moment such as
this
or for
a point moment,
is
geometrically identical to the shear force variation
corresponding to distributed
or
point loads respectively.
4. Longitudinal torsion reinforcement has to be extended at least a distance qu l to th
largest dimension of the section beyond the point where it is theoretically not
required. this example, that would extend the reinforcement by 600 mm, exactly
to
the mid point of the beam. Hence, curtailment is not possible
Concluding Notes
5 The links provided for torsion have to be of the closed type as specified in Clause
2.4.8
Part
2), whereas even open links
are
permissible for shear links.
6. If the section carrymg torsion
is
a flanged beam, it has to be divided into component
non-intersecting) rectangles, such that hmin3 hmax is maximized. This can generally
be achieved by making the widest rectangle as long as possible see Clause
2 4 4 2-
Part 2). The torque is divided up amoung the rectangles in the ratios of their
hmin
3
h
max
)
values and each rectangle designed for torsion.
The
torsional links should
be placed such that they do intersect.
division into 2 rectangles
l
J ~ i n t r s t i n g torsional links
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EXAMPLE 3 - FRAME ANALYSIS FOR VERTICAL LOADS
A typical internal braced transverse· frame for a multi-storey office building
is
shown below.
The
frames are located at
5
m
centres
and the length of the building is
40
m.
The
cross
sectional dimensions
of
members are as follows.
i) Slab. thickness roof
and floors) - 150 mm
il) Beams roof and floors) -
6 mm
x
3
mm
iii)
Columns for all
floors) -
3 mm
x
3
mm
The vertical loading is as follows:- .
(i)
Load
corresponding to finishes = 0.5 k m
for roof and floors)
il) Load corresponding to light partitions = 1.0 kN/m
2
for floors only)
iii) Imposed load on roof =
1.5
k m
iv) Imposed load on floors = 2.5 kN/m
2
v) Density
of
reinforced concrete =
24
kN/m
3
Obtain
the design ultimate moments and shear forces from vertical loads for the
beam
ABC
at the first floor level.
Ground Level
Footing Level
oar
oar
Roof
4.0m
2nd FI
4 Om
A B
C
1st FI
\
4.2f.m
T75m
6.0m 6.0m
7
- - - -
Introductory
Notes
.
L The next 4 examples including this one) deal with the entire structure, as opposed
to structural elements.
2. The loading for partitions and imposed loads is the minimum permissible under as
6399:
Part
I 1984):
Design
loading on buildings: Dead and imposed l o a d s ~
3. general, most frames are braced, the lateral load being
taken
by masonry infill or
lift stair wells.
4. Since the frame is braced, it is possible to use either a beam level sub-frame analysis
or
a continuous beam analysis. Since the latter over-estimates moments considerably,
the fanner will be performed.
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efereoce
Stiffn
alculations
ut ut
IlL of
columns
above 1st floor
=
1112 300 4/ 4000 = 0.169 xl 6
mm
3
Ill,tof columns
below
1st
floor
1/12 300 4/ 5000 = 0.135 x10
6
mm
3
Since T-beam action will prevail in the beam , eff.
flange width = 300 + 0.7 6000 /5 = 1140
mm
b
f
= 1140
mm
«
mm .
I
of beam section
1140
9.388
xl
9
mm
4
,
I/L ofbeams = 4
1
so
5
°1¥
9.388 x10
9
6000 =
1.565
~
mm
3
Distribution factors
Only
the
beam
factors
will be considered.
D
AB
=
CB
= 1.565 / 1.565+0. 169+0.
135
= 0.84
DBA
D
BC
1.565/{ 1.565 2 +0.169+0.135}=0.46
Loading
on
beam
Slab = 5 0.15 24 =
18
kN/m
Beam
= 0.45 0.3 24 = 3.24 kN/m
Finishes = 0.5 5 = 2.5
kN/m
Total dead load = 23.74 kN/m
ik
23.7 kN/m
Imposed load floor = 5 2.5 = 12.5 kN/m
Partitions = 5 1.0 = 5.0 kN/m
Total
imposed
load = 17.5 kN/m
BS 6399: Since a beam span carries
30
m
2
of floor area,
Part 1 reduced imposed
load
= 0.97 17.5 = 17.0
kN m
Qk=
17.0 kN/m
Load arrangements
3.2.1.2.2
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Note
5
Arrangement 3 will be
the
mirror image, about B of
Arrangement 2.
Moment distribution kNm)
ut ut
Note
6
Arrangement 1)
0.84 0.46
AB
BA
-181.2 181.2
+
152.2.....,.
+
76,1
- 29,0 257.3
Arrangement 2)
0.46 0.84
Be
CB
-181.2 181.2
- 76,1
<Eo
-152.2
~ 5 7 3 29,0
Arrangement 1
Support moment
at B= 257 kNm
Note 7
0.84 0.46 0.46 0.84
AB BA
BC
CB
-181.2 181.2 - 11.1 71.1
- 25.3 5 -lo0 .6 --_- Q t ~ 6 - 25,3
+
173.5
+
86,8 - 19.3
<t-
- 15.5 - 31.1 -
31.1
-- - 15.5
13.1-
6.5
6,5 t
13 1
- 3.0
t
-
6.0 - 6,0
-
3.0
2. 5 1.3 1.3 2.5
- 0,6 s - 1.2 - 1.2
- 0.6
- 36.5 186.9 -171.5 3.8
Shear forces lkN
The
shear forces R
A
, RBI R
B2
and
can be found
from the following figures:-
Arrangement 2
Support
moments at B
=
187 kNm
. 172 kNm
Arrangement 1 143.2 219.3 219.3 143.2
Arrangement 2
156.1
206.3 99.1 43.2
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Reference
Calculations
Output
Span moments
w
f
0
ree
bending moment is given
Note 8
by
- w.l.x)/2
+
{w.x
2
12
Fixed end moment variation is
1
iven
by
M
1
+
M2-Ml)xll
1
Hence, the points
of
contraflexure an4the points and
values
of max span
moments
can
be obtained.
Span AB
p n
Be
r r n ~ t 1
Points of
contraflexure O.21m, 4.52m 7.48m, 11.79m
Span
moments:-
rom A)
Max.
sagging
141 kNm at 141 kNm at Arrangement 1
moment 2.37 m 9.63 m
141
kNm
both
spans
Arran eJDent 2
Points
Arrangement 2
contraflexure O.25m, 4.92m 8.45m, 11.90m
165 kNm
and
from A
35 kNm
Max. sagging 165 kNm at 35 kNm at
Note 9
moment 2.58 m 10.18 m
257.3
165 141
Bending Moment Diagram kNm
219.3
Note 10
Shear Force Diagram
kN
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Notes on
Calculations
7
8.
9
The distribution factorS have
cc unted· for
the column stiffness,
but
the column
moments have been left out of
the
.calculations for
c o n v e n ~ c e
as we are interested
only in the beam. mOments. Since
th
remOte ejlds of the columns are assumed
be
fixed, there will e no Carty over riidmentsfrom them to the beam-eQlumn joints. The
sign convention adopted is that clockwise moments are positive
and
anticlockwise
moments negative.
In this symmetrical loading arrangement, the calculation is complete with just on e
joint release.
The
difference between the moments MBA and M
BC
arises out
of
the fact that the
columns
take
part
of
the moment
arising out of
asymmetrical loading.
The
sign convention
adopted
in this part of the solution is th t sagging moments are
negative and ho ing moments
p o s i ~ v e
This t w o - ~ frame is typical of most situations, where the maximum support
moments are o t iried when all spans are loaded with
the·
maximum design ultimate
loads 1.4 gk
1.6
qk) nd the maximum
span
moments
are o t ined
when that span
is loaded with
the
maximum design ultimate load 1.4 gk
1.6Q]J while the adjacent
spans
are
loaded with the minimum design ultimate load
1 . 0 ~ .
The
diagrams. for Load Arrangement 3 have not een shown, for
the
sake of clarity,
since they
will
be
mirror images
of
those for
Load
Arrangement 2 about B.
Concluding Notes
II.
The
e m moments could have. been o t ined using a continuous
e m
analysis,
instead
of
a subframe analysis Clause 3.2.1.2.4) as pointed out in Note 4. However,
column moments
will then
have
to be
estimated
as indicated in Clause 3.2.1.2.5.
12. If there are 3
or
more approximately
equ l
ba ys in the frame
nd
the characteristic
imposed load does
not
exceed the characteristic
dead
load,
the
e m moments and
shear forces can
be o t ined
from Table
3.6
for a continuous beam analysis see
l ~ 3.4.3).
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X MPL 31 - FR M ANALYSIS
FOR
HORIZONTAL
W DS
the office building described
in
Bxample 30 was unbraced and located
in
a semi-urban area
where the basic wind speed is 40 mis, determine the moments and shear forces induced in
. a typical internal frame due to the wind load.
Introductory
Notes
1 The wind forces have to be determined using cP 3: Ch. V: Part 2 1972): Basic data
for the design of buildings: Loading: Wind loads:
2. In carrying out the analysis, the entire structure
is
analysed, assuming that only the
wind load acts on it and that points of contraflexure are developed at the centres of
all beams and columns Clause 3.2. 1.3.2). A further assumption is made regarding
the distribution either
of
shear forces or
of
axial loads
in
columns see Note 4 below).
Thus the analysis for the lateral loads is performed on a statically determinate
structure, as opposed an indeterminate one as in the case of vertical load analysis.
Reference Calculations Output
Wind force
Basic wind speed, Vb
40 mls
CP3:Ch.V:
Sl
1.0;
SJ
1.0
Part 2
S2
for ground roughness 3, building class Band
Note 3
H
12.25) is 0.7805
V
s
1) 0.7805) 1) 40)
31.22 mls
Wind pressure, q
0.613) 31.22)2
597.5 N/m
Table 1
I1w
40 / 12 3.33
of
bid
40 / 12 3.33
CP3:Ch.V:
h b
12.25 / 12
1.02
Part 2 Hence, Cf
1.23
Force on one frame
q.Cf.A
e
597.5) 1.23){ 12.25) 5.0))
45014 N
Wind force on a
45
kN
frame = 45 kN
Analysis
The following assumptions are made:-
1 The wind force is applied at floor
and
roof levels,
the force at each level being proportional
to
the
areas shared by them.
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Reference
Calculations
Output
Note 4
2. Points
of
contraflexure are assumed at the centres
of
beams and columns.
3. The vertical column stresses are proportional to
their distances from the centroid
of
the columns.
c ~ 11.03
o
1.225
.J..
1 . 8 4 - . .
c
l
2.llil
14. 7
t - - - 4 ~ - - - + - - ..... ?
2nd
Fl
r
. 2.llil
5.51
t
6.125
The forces at roof, 2nd floor and
1st
floor levels are
2/12.25) 45)
=
7.35
leN
roof)
{ 2+2)/12.25} 45) = 14.7 leN 2nd flopr) and
{ 2+2.S0)/12.25} 45)
=
16.5
,1eN lst
floor)
k 6.llil
M< 6.llil
~
7.35-+
r
.
e
I · lJ
Roof
. r
1.84
-
3.67
1.225
t
if225
1
225
1 . 8 4 3.67 . . . i
Note 5
1 1 . 0 3 ~ )
~ < 4 :
9.64
~
17.83
Note 6
Note 7
6.125
5.51
~ t l
2.Qn
16.5:+...---..--- ---- ---t
Is
t Fl
r
2.fIn
19.28 ,.
9.64
t
t
17.83
Moments and shear forces in ABC
The moments
in·
ABC can be found as those required
to balance the
~ l u m n
moments.
.
3 5 1 . ~ · ~.
35.f
A
J
.
C
B •
.
.
35. f 35.1
Moment at A B
and C is
1
kNm
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Reference
Calculations
Output
The shear forces in the spans are obtained by
Shear force in
dividing the moment by half the span length. AB and BC is
Hence, shear force = 35.1)/ 3.0) = 11.7
l
11 7kN
Notes on Calculations
3. The
S2
factor can be calculated separately fordifferent parts of the stucture or for the
entire structure, using the total height of the structure. Since this is only a 3 storey
structure, it is simple and conservative to work with a single
S2
value
4. If the column sizes
are
uniform, the vertical forces will be proportional to the
distances of the columns from the centroid of the column group. An alternative
assumption to this is to consider that the horizontal shear forces in the columns
are
proportional to the bay sizes.
5.
The analysis is essentially a subframe analysis, but the entire frame has to analysed
step wise, from the top to bottom. At each step, the vertical column reactions
are
obtained first, taking moments for the equilibrium of the entire sub structure, together
with the third assumption referred to in Note above. The horizontal shear forces in
the columns can be found by taking moments about the points of contraflexure in the
beams, for the equilibrium of different parts of the sub-structure.
The
results obtained
from each sub-structure have to be used for analysing the next lower sub structure.
6
the column bases are not designed to resist moments, the point of contraflexure on
lowest column should be moved down to the level of the base as opposed to being
at column mid height).
7. In order to meet stability requirements, the lateral load at each level should be at least
1.5 of the characteristic dead load at each level Clause 3 1 4 2 Since the total
dead load on·a beam Example 30) is 23.7) 12)
=
284.4 kN and 1.5 of this is
4 3
l « 7.35
kN),
the above condition is met.
Concluding
Notes
8. For unbraced frames having three
or
more approximately equal bays, the combined
effect of wind and vertical loads can
be
obtained
by
superposing the results .of an
analysis such as the one above with those of a subframe analysis such as the one in
Example 30, after factoring the loads appropriately Clause 3 2 1 3 2
9. For very slender structures, the overall stability of the structure against overturning
due to lateral wind loads should also
be
checked.
The
appropriate load combination
would be 1.4 W
causing the overturning moment) and 1 0 G
providing the
restoring moment).
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X MPL 32 - REDISTRIBUTION
OF
MOMENTS
Determine the design ultimate moments for the beam ABC in Example 30, after carrying out
moment redistribution.
Introductory Notes
1. Although the design
of
reinforced concrete sections
is
carried out using the plastic
capacity of the section, the analysis of structures is still performed using elastic
methoos. The advantage to the designer arising out of the above plasticity is
incorporated in the analysis by moment redistribution.
2. Moment redistribution has to be performed separately for each load arrangement. In
addition, the redistributed envelope is not allowed
to
fall below
the 70
elastic
moments envelope, to ensure that wide· cracks at the serviceability state will not
develop
see
Clause 3.2.2.1 .
Reference
Example
30
3.2.2.1
Note 3
Note 4
Calculations
Support
moments
The numerically largest elastic moment
is
257.3
kNm at support B Arrangement 1 . This
can be
reduced to
0.7 257.3 180.1
kNm for all load
cases
leaving the support moments at A and C and
also the column moments unchanged.
Hence, the support moments will be given by
Output
support moment
at
B
=
180.1
kNm
AB
BA
BC
CB
Note 5
Arrangement I
-29.0 +180.1 -180.1 +29.2
Arrangement 2 -36.5 +180.1 -180.1 + 3.8
NQt.e:
The shear forces
can
be found by analysing
the sections
AB
and CB, as in Example
30.
Example 30 Span moments
These can be found by superimposing the free
bending moment diagrams on the above fixed end
moment variation.
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Reference
Calculations
Output
Arraneement 1
Points
of
contraflexure
0.19m, 4.97m
7.03m, 11.81m
(from A
span moments:-
Max.
sagging
173
kNm at
173
kNm at
moment
2.58 m
9.42 m
Arrangement 1
173 kNm (both
Arraneement 2
spans)
Points of
contraflexure
0 24m 4 97m
8.57m, 11.91m
Arrange :Dent 2
(from A)
168 kNm
(span
AB
Max.
sagging
168 kNm at
33
kNm at 33 kNm
moment
2.60 m 10.24 m
(span BC)
Notes on Calculations
3. The support moments are reduced as much as possible so that congestion of
reinforcement at beam-column junctions can be minimized. The maxi.mumamount
of
redistribution allowed is
30
- a figure which can
accomodated by rotation at a
section after plastic hinge formation by the appropriate restriction
of
the
x
ratio
see
Clause
3.2.2.1).
4. general, the
x
ratios in columns are larger than those required to permit plastic
hinge formation. Hence, column elastic moments should never
redistributed.
5. The support moments in Arrangement 2 are made equal to 180.1 kNm - the value
obtained after 30 redistribution in Arrangement 1. This requires a much lower
percentage of downward redistribution for the elastic momentBA and an upward
redistribution for the elastic moment BC. Such upward redistribution may help to
reduce
sp
moments.
Concluding Notes
6. Compared with the elastic design moments in Example 30, the redistributed design
moments are such that the support moment at B
is
considerably lower, while the span
moments
are
only slightly higher; hence the advantage in <:arrying out moment
redistribution - the total moment field is considerably reduced.
7.
The
points
of
contraflexure are generally closer
to
the supports for the redistributed
bending moment diagrams than for the elastic bending moment diagrams. In order to
prevent serviceability state cracking on the top surface, the restriction on the
redistributed moment envelope specified in Note 2 above has to be applied.
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EXAMPLE 33 - DESIGN FOR STABILITY
The figure shows the plan o f a 6 storey framed structure, where
the
floor to ceiling height
of
each
storey is 3.5 m The average
dead
and imposed loads per unit area of floor can be
taken
as 5
k m
each. Design the stability
ties
for this structure with steel
of
f
y
= 460
N/mm
2
•
,
-
eans
ChlumS
)
4 x 5 0} = 2 0}
I. In order to ensure the robustness of a structure, it should normally
be
connected
together by a
system
of continuous ties. This example demonstrates the design of
these ties.
2. In addition, .the structure should
be
capable
of
withstanding a notional horizontal
load, which is proportional to its characteristic dead load see Example 31, Note 7).
3 In calculating the amount of reinforcement required, the steel can be assumed to act
at its
h r t e r i s ~
value
Le
m
= 1.0.
Furthermore, reinforcement designed for
other can be used as ties Clause 3.12.3.2).
Reference Calculations
Output
3 12 3 7
V
ertical
ties
TheSe are required, since no. of storeys
>
5.
Area corresponding to
typical column =
15.0 m
2
1 1 2) 6.0) 5.0) =
MaX
design ultiIpate load
=
15.0){ 1.4) 5.0) 1.6) 5.0)} =
225
le
Area o
ties required =
225 xl<Y)/ 460)
=
489
mm
2
vertical ties
This can easily be met by continuous column ri
A
= 489 mm
2
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Reference
Calculations
Output
3.12.3.5 Peripheral ties
3.12.3.4.2
F
t
=
20 + 4 6
=
44
leN «
60 leN
F
t
=
44
leN
Area
of
ties required
=
44
xloJ
I
460
=
96 mm
2
peripheral tie
This can be easily met by peripheral beam r /f that is A
=
96 mm
2
•
continuous.
3.12.3.4
Internal ties - looeitudinal direction
lr
=
5 0 m
Force I unit width
= { 5+5 / 7.5 } 515 44
=
59
kN/m
{>
1.0 44
=
44
leN/m}
Total force
=
59 6.0
=
354 leN
Area of ties required
=
354 x1oJ
I
460 longitudinal
=
770 mm
2
internal tie
carried in the two peripheral beams, area required A
=
770 mm
2
per beam
=
770 /2
=
335 mm
2
•
Note:- spacing of ties
=
6 0 m
<
1.5 5.0
=
7.5 m
3.12.3.4
Internal ties - transverse direction
lr
=
6 0 m
Force I unit width
=
{ 5+5 / 7.5 } 615 44
=
70.4
kN/m
{ > 1.0 44
=
44
le m}
Total force
=
70.4 20
=
1408
kN
Area of ties required
=
1408 xl<P
1 460
transverse
. =
3061 mm
2
internal tie
distributed in the 5 transverse beams, area
A
3061 mm
2
required per beam
=
3061 /5 = 612 mm
2
•
Note:- spacing
of
ties =
5 0
m
<
1.5 6.0 =
9 0
m
Note 4
Peripheral beams
peripheral
Total tie area per
beam
in longitudinal direction =
beams tie rlf
longitudinal -
96 335 = 4 mm
2
431 mm
2
Total tie area per beam in transverse direction
=
transverse -
96 612 = 708
mm
2
708 mm
2
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Reference
Calculations
Output
3
12.3.6
Coltamn ties
Note 5
Force
=
greater of 31100) 225) 6)
=
40.5 kN
and lesser of 2.0) 44) = 88 kN and
and { 3.5)/ 2.5)} 44)
=
61.6
kN
=
61.6 kN
column ties
Area of tie required
=
1.6 xHP)/ 460)
=
134 mm
2
A
=
134 mm
2
5
Since is less than the ties in the beams, part of
the latter can taken into the columns.
Notes on Calculations
4. Although the
beam
reinforcement may be greater than these tie areas required, it must
ensured that continuity of tie reinforcement is provided - this has to be borne in
mind when curtailing
beam
reinforcement.
5.
The 3
load is taken for 6 storeys,since there will
five floor slabs and the
roof
above the level
of
the frrst floor column tie;· using the floor loading for the
roof
as
is a conservativeapproxjmation.
Conduding Notes
6. If a structure has key elements Le. those that carry, say, over 70 m
2
or over 15
of
floor area at a given level), they have to designed to withstand a specified load
Clause
2.6.2
Part 2). Furthermore,
if
it is not possible to tie the structure e.g. in
load bearing masonry construction), bridging elements have to be designed, assuming
that each,vertica110adbearing element is lost
in
tum Clause
2.6.3
Part 2).
}
The overall layout of the structure should also
be
designed to provide robustness and
key elements should preferably be avoided.
6
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EXAMPLE
34 -
CRACK
wmm CALCULA
nON
The
figure shows a cross section
of
a simply supported
beam of
7
mspan,
and supporting
dead and imposed loads of 20
kN/m
each. Determine the crack widths i midway between
the bars,
ii
at the bottom comer and
iii 2S0
mm below the neutral axis.
750
450
T 5 1
9
f =
25 N/mm
2
cu
f
; 460 N/mm
2
y
E
=
200
kN/mm
2
s
All dimensions in mm
Introductory Notes
1.
This crack width calculation can be performed when the bar spacing rules are not
satisfied, to see whether this more accurate method will satisfy the crack width
requirements in Clause 3.2.4
of
Part 2.
It
can also be used
to
estimate the actual
crack width in a flexural element.
Reference Calculations
Output
Sectional data
I
Note 2
M
s
= 20+20 7
2
=
24S kNm M
s
=
245 kNm
equation
E
c
=
20 + 0.2 25
= 25
kN/mm
2
3.8.3
E
eff
=
0.5 E
e
·
=
12.5 kN/mm
2
Part 2
X
e
=
E/Eeff
200 xloJ
1
12.5
xHf)
=
16
Note 3
P =
3 491
1
690 450 = 0.00474
x d = -cxe.p
+
{cxe.p
2
+
cxe.p }o.S
Note 4
= - 16 0.00474
[ 16 0.00474 {2+ 16 0.00474 }]O.S = 0.321
x
=
221
mm
x
=
221 mm
Note 5
~ b d =
1/3 x1d 3 +
a p
{I - x/d }2
= 1/3 0.321 3 + 16 0.00474 1-Q.321Y
=
0.046
e = 6.80 xl0
9
Ie = 6.80 x10
9
mm
4
107
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Reference
3.8.3
Part 2)
Note 6
equation 13
part 2)
equation 13
part 2)
equation 12
part 2)
3.2.4
part 2)
Calculations
Calculation of strains
Strain in steel
=
245
x1
6
) 690-221) I
. 12.5 xl<Y) 6.8 x10
9
)
=
1.35 dO·
3
< 0.8) 460) / 200 xl<Y
=
1.84 dO·
3
At extreme tension fibre bottom of section),
El =
1.35
xlO·
3
) 750-221)/ 690-221) = 1.523 xl0·
3
tension stifferung
=
bt h-x) a -x)
I
3)E
s
.A. d-x)
= 450) 750-221)2
I
3) 200 xI<Y) 3) 491) 690-221)
=
0.304 xlO·
3
Em
=
1 .523 - 0.304) xHr
3
=
1.219 xlO·
At 250 mm below neutral axis,
EI.
=
1.35 xlO
3
) 250)/ 690-221)
=
0.72 xlO·
3
tension stiffening
=
450) 750-221) 250) I
3) 200 xl<Y) 3) 491) 690-221)
=
0.144
xlO·
3
=
0.72 - 0.144) xlO·
=
0.576
dO·
3
Distances to potential crack points
cmin =
750-690- 2512)
= 47.5 mm
<lerl
=
{ 60)2
82.5 2}O.s
- 12.5
=
89.5 mm
3er2 = { 60)2 60)2)}o.s
- 12.5 = 72.4 mm
3er3 = { 60)2
69O-221.250 2}O.S
-12.5
=
214.5 mm
Crack widths
CW
t
=
3) 89.5) 1.219dO·
3
) /
[I { 2) 89.5·47.5)/ 750-221)}]
= 0 .2 82 mm < 0.3 mm; satisfactory.
CW
2
= 3) 72.4) 1.219
dO·
3
) /
[1
{ 2) 72.4-47.5)/ 750-221)}]
=
0.242
mm
<
0.3 mm; satisfactory.
CW
3
=
3) 214.5) 0.576 xlO·
3
) /
[1 { 2 214.5-47.5 / 750-221 }]
=
0.227 mm < 0.3 mm; satisfactory.
108
Output
CW
t
=
0.282
mm
CW
2
= 0.242
mm
CW
3
=
0.227
mm
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Reference lcul tions
Output
Comparison with
bar m cine rul f
3.12.11.2.3
Spacing between bars = {45G- 2) 47.5)- 3) 25)} max. spacing
Table
3 3
=
140 mm
<
160 mm; satisfactOry.
O K
3.12.11.2.5
Comer distance = { 60)2 60 2}O.s 12.5
comer distance
Table 3.30
= 72. 4 mm < 16012 mm; satisfactOry. O K
Notes
on
Calculations
2 The
partial
safety factors for loads in serviceability calculations is unity.
3. The modulus
of
elasticity
of
concrete is halved,
to
account . for creep. This is a
simpler approach compared
to
the one for deflection calculationsc see Example 35).
4.
The
serviceability calculations are
based on
a triangular stress block for concrete
in
the elastic state. There is no restriction on the x1d ratio, as in ultimate limit state
calculations.
5. The
effective second moment
of
area is found by considering only the area of
concrete that is not cracked; the area of steel is converted to an equivalent area of
concrete using the effective modular ratio, Qe
6. The strain at the required level in the concrete
is
found by calculating the strain from
elastic theory
Et),
and reducing from this value·an allowance for tension stiffening
in the concrete; this is because in calculating
e nd El
we assume that the concrete
has no tensile strength, whereas in fact it does.
Concluding Notes
7. All the calculated crack widths are below
3
mm
and hence satisfactory Clause
3.2.4,
Part
2).
This could have been expected, because the maximum spacing and
comer distance rules are satisfied as well. It is these empirical rules that are used in
everyday design, because of their convenience.
8.
For
beams of overall depth exceeding 750 mm, side reinforcement in the form
of
small diameter bars at spacings not exceeding 250 mtn over two thirds of the
be m
depth from the tension
f ce
must also be p r o v i d e d ~ s per Clause 3 12 11 2 6
1 9
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EXAMPLE 35 • DEFLECTION CALCULATION
figure shows a cross section of a simply supported beam of 7 m span.
H
the dead and
imposed loads are both taken as 5 kN/mleach, and if25 of the imposed load is taken as
permanent, calculate the
total
long term deflection
of
the beam at midspan.
The
age of
loading can be taken as 28 days.
225
375 2T25
2 5
o
0 I
ll dimensions in mm)
Introductory Notes
f = 25 N/mm
2
cu
f =
460
N/mm
2
y
= 200 kN mm
1. This deflection calculation can be performed when the span/depth ratio check fails,
to
see whether this more accurate
method will
satisfy the deflection requirements in
Clause 3.2.1.1
of
Part
2. It can also be used to estimate the actual deflection of a
flexural element.
2. Where domestic
and
office space is concerned, 25
of the imposed load can be
considered permanent; where storage areas are
concerned
the above figure should be
increased to 75 .
3.
The
age of loading is when the fonnwork is removed,
at
which point much
of
the
dead load and some imposed construction loads will be acting on the concrete
elements.
Reference CaIculatio.
Output
Initial aSsessment
of
span dq th
ratio
M
ult
= 5) 1.4 + 1.6 7 2/8 = 92 kNm
Chart 2
M1bd
2
= 92 xl<r
I
225) 325)2 = 3.87
Part
3)
l00A/bd = 1. A
a
= 936 mm
2
equation 8
fa
= 0.58) 460) 936/982) = 275 N/mm
2
equation 7 F
I
= 0.55 + { 477-275) / 120 0.9+3.87)} = 0.9
Table
3.10
Allowable span/depth
=
20) 0.9)
=
18
Actual span/depth
=
7000 / 325
=
21.5
>
18;
span/depth
Hence, span/depth check is violated.
check violated
110
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Reference
Calculations
utput
Data for serviceability calcuIiJions
equation
E
e
= 20 + 0.2 25 =
25
k N ~ B \ 2
Part
2
\:=25 kN mm
2
7.3
Eff. section thickness
=
2 375 225 I 2 375 + 225
Part
2
=
141
mm
RH
= 85
assumed for Sri Lanka
Figure 7.1
Long term creep coefficient,
=
1.8
part
2
E
eff
=
25 I 1
+ 1.8
=
8.93 kN/mm
2
Figure 7.2
f
=
120 x10-6
es
part 2
P
=
982 I 225 325
=
0.0134
a
e
=
E
s
I
E
eff
= 200
I
8.93 = 22.4
a
e
=
22.4
x/d
= -
ae.p
+
{a
e
.p 2
+
a
e
p }O 5
= -
22.4 0.0134 +
[ 22.4 0.0134 {2 + 22.4 0.0134 }f.5
Note 4
=
0.53
xld = 0.53
Hence, x
=
0.53 325
= 172
mm x
=
172 mm
Ie/bd
3
= 1I3 x/d 3
+
a
e
p{ - x/d }2
=
113 0.53 3
+ 22.4 0.0134 1 - 0.53 2
=
0.116
Ie
=
0.116 225 325Y
=
896 xH? mm
4
=
896 x1 6
3.6
Ss
=
As d-x
mm
4
Part
2
=
982 325-172
=
150.2
x103
mm3
Determination of serviceability moments
M
tot
=
5+5 7 2
=
61.25 kNm
~
=
{5
+ 0.
25 5 }
7 2
I
8 = 38.28
kNm
Note 5
~ r m r e d ) =
M
{ 1I3 b h-x 3
f
et
I
d-x
=
3 .28 x1oti -
{ 113 225 375-172 3 0.55
I
325-172 }
=
36 x10
6
Nmm
Calculation
of
curvatures
lIrl
p
=
M I Eeff.I
e
=
36 x1 j6 I 8.93 x103 896 x l ~ )
=
4.5 xlO-6 mm-
l
equation 9
1Ir
es
= fes·ae·S
s
I Ie
part
2
=
120 xlO-6 22.4 150.2
xloJ
I
896
x106
=
0.45 xlO-6 mm-
l
111
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Note
6
Table 3.1
(Part 2
3.2.1.1
(Part 2
Note 7
o
find
instantaneous7 · .. ,
Be
=25 k mm
. . .
= 200
I 25 •
.
d
=
0.368
..
Ie
= 459
xl )6
lIrit llri = (Mtot - M.rm> I Ec·l
e
~
(61.25-38.2 )X1()6
I ( 2 ~
xl(3)(459
x l ~
= 2.01 xl<r'
nun-I
lI r = lIrlp + IIr
es
+ (llrit
;
11rin>
1=
(4.5 + 0.45 + 2.01) x 1 0 ~ mm-
I
=
6.96
x l O ~
mm-
I
Estimation
of
det ls;gion
K
=0.104
a =
K . l ~ l I r )
=
0 . 1 0 4 7 ~ 6 . 9 6 x I O ~
35.5 mm
all
=
35.5
I 7000 = 11197
> 11250
··0ulJwt··
lIr=
6.96
x l O ~
mm-
I
a = 35.5 mm
all > 11250
Notes
on
Calculations
4. The serviceability calculations are ~ a triangular stress block for concrete
in
the
elastic
state.
here
\s
n., resttietionon.
the
x1d ratio, as in ultimate limit state
calculations.