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

_

_



PublisheJI

by

o c Structural Engineers Sri Lanka

Colombo Sri Lanka

995

ISBN 955 9347 00 4



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





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





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

1



- 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.

2



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.

3





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

5



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


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.

6



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

7



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 .

8



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

)

9



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 -

1



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.

11



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.


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.

12



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)

13



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

14



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.

15



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

16




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.

7



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



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



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


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.

20



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

21



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.

22



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.

23



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

325



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.

25



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

26



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.

27



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

28



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




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.

3



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

31



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.



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

33



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.



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



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.

6



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.

37



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

38



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.

39



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

4



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 .

41



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

42




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



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


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

44



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



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

46



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

47



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



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.

9



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

il



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


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

51



-

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



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



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).

54



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



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.


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.



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.

57



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.



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.

59



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

60



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.

61




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.

62



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.

63



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



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


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.

6



~

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

66




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.

67



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.

68



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

69



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.

70




~ ~ ~ = 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.

71




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.



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.


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

73




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.



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.


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



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

76



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.


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.



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


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.

78



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



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.

80



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



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



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.

83



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



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


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.

85



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

86



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

87



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 .

88



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



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

90



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



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



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

93



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.

94



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

95



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



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

97



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).

98



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.


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.

99



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

1



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


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).

1 1



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.

102



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)


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.

103



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).


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

104



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

1 5



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.


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



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


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.


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



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



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



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



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



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.

Documents

Prof. Dias Graded Examples in Reinforced Concrete Design Dias