Prof. Dias Graded Examples in Reinforced Concrete Design Dias

REINFORCED CONCRETE DESIGN

W.P.s. Dias BSc{Eng), PhD(Lond),DIC, pEng, MIStructE, MIE(SL) Senior Lecturer Department of Civil Engineering University of Moratuwa Moratuwa Sri Lanka

/

Society of Structural Engineers - Sri Lanka

.,. "'~

.... _._.J

PublisheJI. by

ji

Soc. of Structural Engineers - Sri Lanka, Colombo, Sri Lanka, 1995. ,

ISBN 955-9347-00-4

FOREWORD

The Society of 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 of 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 of books on related topics. As the majority of structures in this country are constructed of reinforced concrete, the selection of GRADED EXAMPLES IN REINFORCED CONCRETE DESIGN as the object of the Society's first book publishing effort constitutes an ideal beginning. Dr 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 of an immediately practical nature, Dr 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.

I

A.C. Visvalingam MA, PhD, DIC, MICE, MIStructE, MIE(SL), CEng PRESIDENT, Society of Structural Engineers - Sri Lanka

2 March 1995

•.

GRADED EXAMPLES IN REINFORCED CONCRETE 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 10 - 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)

104

Serviceability Limit State Calculations (Examples 34 -35)

107

INTRODUCTION

A Case for Worked Examples 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. It can, of course, be used by practising engineers, who already have a grasp of reinforced concrete fundamentals. In 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 BS 8110. In addition, examples are also given for the de3ign for torsion and the calculation of deflection and cracking, dealt with in Part 2 of BS 8110.

Sections of code are referred to by indicating the relevant clause, table Or equation of BS 8110: Part 1. Where clauses, tables, charts or equations from Parts 2 and 3 of BS 8110 are referenced, the relevant Part is also indicated. One very useful feature of BS 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 Using Lower Grades of Concrete Table 3.4 in Part 1 of BS 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.

It must be emphasised here that the index of durability used in BS 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/m3 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. The question arises as to whether Table 3.4 in BS 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 Lanka as a case in point. The 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 are given in TABLE 1. This table is taken from the author's publication "Specifying for Concrete Durability: Part II - The Sri Lankan 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 Sri Lankan practice. In short, these recommendations rationalise satisfactory Sri Lankan practice (especially under mild exposure conditions) with respect to BS 8110, while suggesting improvements to Sri Lankan practice where problem areas (such as concrete exposed to sea spray) are 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 BS 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 Classification

Examples of Exposure

Mild Moderate Severe Very severe Extreme

Indoor Outdoor Driving Rain Sea Spray Abrasive

mm 25

--

Nominal Cover mm mm 20 35

--

20* 30 40 50

mm

mm

20* 25 30 40 60

20* 20 25 30 50

----

--

Maximum free water/cement ratio

. 0.65

0.60

0.55

0.50

0.45

Minimum cement content (kg/m3 )

275 (300)

300 (325)

325 (350)

350 (400)

400 (450)

40

45

Lowest grade of concrete

25

--

30

--

35

.

Note 1

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.

Note 2

A minimum of 25 mm cover to all reinforcement should be maintained in beams and columns.

Note 3

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.

Note 4

The minimum cement content values in parentheses should be maintained if no water-reducing admixtures are used.

Note 5

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.

Note 6

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

EXAMPLE 1 - ANALYSIS OF UNDER-REINFORCED SECTION Determine the lever arm for the beam section shown in the figure; find also its moment of resistance. (

225

)

f eu = 25 N/mm2 f y = 460 N/mm2 3-20

000 (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.

Area of steel Note 2

Note 3

3,4,4. 1(e)

Output

Calculations

Reference

= 942.5 mm2

Assuming that the steel llas yielded, T = (0.87)fy.~ = (0.87)(460)(942.5) = 377189 N Hence, balancing compressive force = 377189 N (0,45)feu .b(O.9)x = 377189 . (0,45)(25)(225)(O.9x) = 377189 x = 166 mm 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 Note :- z/d = 300/375 = 0.8 < 0.95, Hence O.K. Moment of resistance

= (377189)(300) = 113.16 x106 Nmm = 113 kNm

T

= 377189 N

x

=

z

= 300 mm

M

=

166 mm

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 xJd value.) 3.

The condition that tensile reinforcement has ~ielded when the concrete strain is 0.0035, is x/d < = 0.64 (for fy = 460 N/mm ) and x/d < = 0.76 (for fy = 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

o

Id=300

f cu = 25 N/mm2 f y = 460 N/mm 2

0

(All dimensions in mm)


This section is different from that in Example 1, in that it is over-reinforced. The calculation procedure is more complicated here.

6

Reference

Output

Calculations Area of steel

= 981.7 mm2

Assuming that the steel has yielded, T = (0.87)fy ' " = (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 Note 2

But, x/d = 259/300 = 0.86 :>0.64 Hence, steel has nQ1 yielded. We shall try to find a value for x, by trial and error, such that T and C are approximately equal. Try x = 200 mm C = (0.45)fcu .b(0.9)x

Note 3

= (0.45)(25)(150)(0.9)(200) = 303750N

= (0.0035)(300-200)/200 = 1.75 xlO-3 Hence, fs = (1.75 xlO-3)(200 xloJ) = 350 N/mm 2 , and T = (350)(981.7) = 343595 N Es

For a better approximation, try x = 205 mm. Then C = 311344 Nand T = 318454 N.

Note 4

For a still better approximation, try x = 206 mm. Then C = 312863 Nand T = 313572 N. This approximation is sufficient. Note:- x/d = 206/300 = 0.69 (> 0.64)

x

z = d - (0.45)x = 300 - (0.45)(206) = 207 mm M = C.z = (312863)(207) = 64.763 x106 Nmm = 64.8 kNm

M

Note:- Alternative method of finding x. Once it is established that the steel has not reached yield point, for any given value of x, Es = (0.0035)(300-x)/x fs = [(0.0035)(300-x)/x](200 xloJ) N/mm 2 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 x2 + (452.47)x - 135741 = 0, giving x = 206 or -659 mm

x

7

= 206 mm = 64.8 kNm

= 206 mm

Notes on Calculations

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 of concrete. If 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 of the code to arrive at the stress. ~

'"8

8

........

0.87x460=400 N/mm 2

z

I

, , I

I

200 ,

kNAnm 2

Strain Strain diagram

4.

Stress-Strain diagram

It is possible to use this method because the stress-strain curve for steel below the

yield point is a single straight line.

Concluding Notes

5.

One way of ensuring that the beam failure is ductile is to introduce some compression steel, so that x/d will be reduced to 0.5 (See Example 3).

8

EXAMPLE 3 - ANALYSIS OF DOUBLY REINFORCED SECTION

1

Detennine the amount of compression steel required, in order to make Example 2. Find also the moment of resistance of the resulting beam.

~ .150 .) ~ A'd'

s

d=300

f eu fy

x/d

=

0.5 in

= 25 N/mm2 = 460 N/mm2

2-25

o

0



If 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 Note 3

Calculations

Output

Assume a suitable value for d', say 50 mm.

d' = 50 mm

For 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

Note 4

(0. 87)fy.As ' + (0.45)feu ·b(0.9)x = (0.87)f As (0. 87)(460)As ' + (0.45)(25)(150)(0.9)(150 = (0.87)(460)(981.7) Hence, As' = 412 mm 2 , Use 4Tl2 (As' = 452.4 mm 2).

r

9

A s '= 412 mm 2

Use 4T12 (452.4 mm2)

Reference

Output

Calculations

Table 3.27 Note 5

Note:- lOOA s ' / Ac = (100)(452.4) / (150)(350) = 0.86 (> 0.2), Hence O.K. Lever arm for balanced section = d - (0.45~1.) = (0.775)d = (0.775)(300) = 232.5 mm Distance between top and bottom steel = 250 mm 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

Note 6

M = 94.2 kNm


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 mm2).

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'). It

~

150

t - d'=50 I 41~m2 _

d=300

_

1o

150 ""<--~~

>

982 mm2

o:~~~

1

i

~:.-G.....&-'l 232.5 570mm2

0

-

10

I

~

+

It ,

EXAMPLE 4 - ANALYSIS OF NON-RECTANGULAR SECTION Determine the moment carrying capacity of the trapezoidal beam section· shown below.

f

300

I

d=400

h=450

1

1

fcu = 25 N/mm 22fy = 460 N/mm

156



As in previous examples, the moment carrying capacity has to be found by working from first principles. The additional complication in this example is that the section is non-rectangular.

Reference

Output

Calculations Assume values for the neutral axis, x until the compression in concrete is equal to the tension in steel. The area of the section under compression = (0.5)(0.9)x[600 - {(3OO-150)/450}(O.9)x] ~300~ Area of steel = 981.7 mm 2 Assume also that the steel bas yielded. \10.9><

II

W:

Try x = 100 mm 0 Area in compression, Ac ~ = (O.5)('JO){600 - (O.33)(O.9)(IOO)) = 25650 mm2 C = (0.45)fcu .Ac = (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. Note also that x/d = 139/400 = 0.35 < 0.5; hence assumption that steel has yielded is O.K.

11

x

= 139 mm

Note 2

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 Hence, lever arm = 400 - 61.8 = 338.2 mm M

Note 3

Output

Calculations

Reference

= C.z = (392868)(338) = 132.8 x106 Nmm = 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 Ac = (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, x2 - (2000)x + 258684 = 0, giving x = 139 or 1861 mm Since x/d = 139/400 = 0.35 <0.5, steel has in fact yielded, as assumed.

z

M

x

= 338 =

=

mm

133 kNm

139 mm


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 underreinforced (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.

It 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 1 to 4, or by suitably rearranging the form of the equations.

12

EXAMPLE S - 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 , fy = 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.


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.

Calculations

Reference

Output

(a) Singly reinforced section Note 2

Note 3

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 and K = K' = 0.156 Then K = M / (b.d 2.feJ 0.156 = (150 x106) / {(d/2)(<¥)(25)} d3 = {(2)(150 xlQ6)} I {(0.156)(25)} d = 425 mm Choose d = 475 mm, h = 525 mm, b = 225 mm

3.4.4. 1(e)

Now K = M / (b.d2 .feu) = 150 xl06 / {(225)(475t(25) = 0.118 < 0.156 z = d[O.5 + {0.25 - KI(0.9)}o.~ = (475)[0.5 + {0.25 - (0.118)/(0.9)}O.5] = 401 mm < (0.95)(475) = 451 mm; hence O.K.

Table 3.27 Note 4

As = M / (0.87)f Z = (150 xlW) (0.87)(460)(401) = 935 mm 2 Hence, use·21'25 (As = 981.7 mm 2) lOOA/Ae = (982)(100) / (525)(225) = 0.83 > 0.13; hence O.K.

3.4.4.4

r

13

.dmin = 425 mm d =475 mm h = 525 mm b = 225 mm 0(

225 •

~I~I4Th As = 935 mm 2 Use 21'25 (981.7 mm 2)

Calculations

Reference (b)

Note 5

Note 6

Output

Overall depth restricted

If the overall depth is restricted to 400 mm, h = 400 mm, d = 400 - 50 = 350 mm, b = 225 mm (assuming the same breadth as before)

d = 350 mm b = 225 mm

Now K = M 1 {b.d2.fcu> = (150 x106 ) 1 {(225)(350)2(25)} = 0.218 > 0.156 (Le. K') Hence, compression reinforcement is required. Let us assume that d' = 50 mm. As'

= (K-K')fcu .b.d2 1 {(O.87)f (d-d'))

= {(O.218-0.156)(25)(225)(350f} 1

{(O. 87)(460)(350-50)}

= 356 mm2 Use 2Tl6 (~' = 402.1 mm 2) looAs '/Ac = (100)(402.1)1 (400)(225) = 0.45

Table 3.27

z 3.4.4. 1(e)

Use2Tl6 (402.1 mm 2)

< (0.95)(350)

= 333 mm; hence O.K.

Table 3.27 Note 8

Hence, use 3T25 (bottom) and 2Tl6 (top).

10:: Li~ (225 )

= d[O.5 + {0.25 - K'/(O.9)}O.s] = (350)[0.5 + {0.25 - (0. 156)/(O.9)}O.s] = 272 mm

= 356 mm 2

> 0.2; hence O.K.

As = {(K'.fcu .b.d2) 1 (O.87)fy'z} + As' ={(O.156)(25)(225)(350)21 (O.87)(460)(272)) = 1344 mm2 Use 3T25 (As = 1473 mm2) looA/Ac = (100)(1473) 1 (400)(225) = 1.64 > 0.13; hence O.K.

Note 7

A' s

400

000

+ 356

As = 1344 mm 2 Use 3T25 (1473 mm 2)


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. If 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 If 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 N/mm 2 and fy = 460 N/mm 2. Use the methods of formulae and design charts.


If 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 to take place. This capacity is ensured by restricting the x/d ratio to a specified value.

15

Output

Calculations

Reference Cal Using formulae 3.2.2.1(b) 3.4.4.4

I3b

= (1-0.3) 1 1 = 0.7 K' = (0.402)(l3b-0.4) - (0.18)(~-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.feu> = (150 x106) 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 3.4.4. 1(e)

Table 3.27

= 412 mm

- (0.104)/(0.9)}O.5]

< (0.95)(475) = 451 mm; hence O.K.

As' = (K -K')feu ·b.d2 1 {(0.87)fy
As ' = 104 mm 2 Use 2Tl2 (226.2 mm 2)

As = 905 mm 2 Use 2T25 (981.7 mm 2)

Hence, use 2T25 (bottom) and 2Tl2 (top). (225 ~

(bl Using charts

I

0

Chart 3 (part 3)

Appro~riate chart for feu = 25 N/mm 2 , fy = 460 N/mm and d'/d = 50/475 = 0.105 is Chart No.3.

3.2.2. 1(b)

M/bd 2 = (150 x106) 1 (225)(475)2 = 2.95 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 lOOA s'/bd must be read off 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 mm2; Use 2T25 As' = (0.1)(475)(225)1(100) = 107 mm 2; Use 2Tl2 to satisfy minimum steel requirement. 16

525

o

2-1~ 2-25

I

475

0

As = 908 mm 2 Use 2T25 As ' = 107 mm 2 Use 2Tl2


Any combination to the left of the line corresponding to the x/d = 0.3 line will give a feasible combination of lOOAjbd and lOOAs ' /bd. If a point on the line itself is chosen, the solution will generally be 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. It 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 be 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.

17

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/m3 •


This example involves load evaluation and a simple stru,ctural analysis on appropriate loading patterns, in order to find the design ultimate moments.

100

1<:

Beam Section Sectional Elevation _'i' ..

- - -

~-

- -

- - -

- - -

-

-

-

-

-

- - '-I _

---rr---------------~ I I I

I

'

I I

I I

,

----r~-

3500

I

I

- ---------- - --r

---+~----------------r

I I 3500 I I I I I I ~L----------------L ~~----------------l I

I

I I

____ ___

.Ly-

Plan (All dimensions in mm) 18

Reference Note 2

Calculations

Output

The beam can be idealised as follows.

~-

A~ t---..:.6(XX):.:.:.-----~ _

The critical moments for design will be (i) Hogging moment at B (ii) Sagging moment in span BC Loadin~

on beam (per m

len~th);-

Note 3

From slab = (0.125)(24)(3.5) = 10.5 kN/m From finishes = (1.0)(3.5) = 3.5 kN/m From beam = (0.45-0.125)(0.3)(24) = 2.34 kN/m Total dead load udl = 16.34 kN/m Dead load point load at A = (2.0)(3.5) = 7.0 kN/m Live load udl = (4.0)(3.5) = 14.0 kN/m

3.2.1.2.2

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 kN/m Minimum design ultimate load (udl) = 16.34 kN/m

Hoe;e;ine; moment atB:7.0x1.4

/45.28 kN/rn

~---'ft M B = (7.0)(1.4)(1.95) = 109.7 kNm

+

(45.28)(2.0)2/2

19

M B = 110 kNm (hogging)

Output

Calculations

Reference

SUging moment in BC:7.0 /16.34 kN/rn

/45.28 kN/rn

L:-~ te

A

,t

.

B

x

MB = (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 Mx = (128.1)x - (45.28)x 2/2 dMx/dx = 0 when (45.28)x = 128.1 x = 2.83 m Mmax = (128.1)(2.83) - (45.28)(2.83tl2 = 181.2 kNm

MBC =181 kNm (sagging)


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 can 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 by a proper combination of loading patterns.

20

EXAMPLE 8 - DESIGN OF BEAM FOR FLEXURE Design the reinforcement for hogging and sagging moments in the beam in Example 7. Use feu = 25 N/mm 2 and fy = 460 N/mm 2• 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·

Output

Calculations Effectiye de,pth

Table 3.2 Note 3 Note 4 TABLE 1 Table 3.5

Note 5

Assume moderate exposure conditions, for outdoor exposure. Making use of Notes 5 and 6 of Table 1, we can use a cover of 30 mm. 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 be d = 450 - 30 -10 - 25/2 = 397.5 mm Desi~n

Chart 2 (Part 3)

Table 3.27

"

cover

d

= 30 mm

= 397.5 mm

for ho~~in~ moment

The beam behaves a a rectangular beam. b = 300 mm, d = 397.5 mm, M = 110 kNm M/bd 2 = (110 xl06) 1 (300)(397.5)2 = 2.32 lOOA/bd = 0.67 As = (0.67)(300)(397.5) 1 100 = 799 mm 2 Use 21'20 & ITl6 (As = 829 mm2) looA/bwh = (100)(829) 1 (300)(450) = 0.61 > 0.26; hence O.K.

As = 799 mm 2 Use 21'20 & lT16(829 mm 2) (hogging)

Design for sagging moment

3.4.1.5

The beam behaves as a flanged beam. b = lesser of 3500 mm or 1/5 + bw = {(0.7)(6000)}/5 + 300 Hence, b = 1140 m

21

= 1140 mm bf

= 1140 m

Calculations

Reference

3.4.4.4

Chart 2 (Part 3) Table 3.27 Note 6

Assume that the neutral axis is within the flange. K = M I (b.dz.fcu) = (181 x106)/{(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 the beam can be designed as a rectangular beam with b = 1140 mm. M/bd z = (181 xl06) 1 (1140)(397.5)z = 1.00 l00A/bd = 0.27 As = (0.27)(1140)(397.5) 1 100 = 1224 mm z Use 21'25 & 11'20 (As = 1295 mm2) bwlb = 250/1140 = 0.22 < 0.40 l00A,Ibw h = (100)(1295) 1 (300)(450) = 0.96 > 0.18; hence O.K.

Output

N.A. is in flange

As = 1224 mm 2 Use 21'25 & 11'20 (1295 mm z) (sagging)

TranYerse steel

Table 3.27 Note 7

In ordeJ:l that flanged beam action is ensured, the minimum amount of transverse steel (to be provided in the top of the slab) is given by l00Ast/htl = 0.15 Ast = (0.15)(125)(1000) 1 100 = 187.5 mmz/m Use R6 @ 150 (min.) (Ast = 190 mm2/m)

Transverse steel R6@150 (min) (190 mmz/m)

Slenderness check 3.4.1.6 Continuous portion - clear distance between restraints is 5700 mm (60)bc = (60)(1140) = 68400 mm (250)b/ld = (250)(1140)21 (397.5) = 817358 mm Since these values are > 5700 mm, check is O.K.

Note 8

Cantilever portion - clear distance between restraints is 1850 mm (25)bc = (25)(300) = 7500 mm (l00)b/ld = (100)(300)2 I (397.5) = 22642 mm Since these values are > 1850 mm, check is O.K.

22

Slenderness O.K.


It is sufficient to assume a "moderate" exposure condition for the exteriors of most

structures, which are not subjected to freezing and sheltered from driving rain. 4.

The cover values are obtained from TABLE 1 in the Introduction to this text; this Table is relevant for Sri Lankan concreting practice. The figures in the table can be further modified by Notes 5 and 6 of the table, as has been done here. It 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 15 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.

-

y

h

bars

,Lt

_

b~a

~1

x.....-

shear link

cover

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 beam action takes place and where it does not. Furthermore, such locations will be reversed in systems where upstand beams are used.

10.

If 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 FLAN~ED 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', fcu = 25 N/mm2 ,. fy = 460 N/mm 2.


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. .

Output

Calculations

Reference Note 2

Assume that bw = 225 mm

3.4.1.5

b = lesser of 2000 mm or lilO + bw = {(O.7)(6000)}/lO Hence, b = 645 mm

+ 225

= 645 mm

If the beam is to be singly reinforced, K=K' = 0.156 M 1 (b.d2.fcu) 0.156 (200 x106) 1 {(645)(d)2(25)} = 0.156 d = 2~2 rpm Hence, choose d = 325 mm and h = 375 mm

b = 645 mm

=

Note 3 3.4.4.4

Note 4

equation 2 Note 5 3.4.4.5 Note 6

Then, K = (200 xl06 ) 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 Since this is greater than hf = 100 mm, the neutral axis lies outside the flange. b/b w = 645/225 = 2.87 d/h f = 325/100 = 3.25 {3f = 0.129 {3f.fcu .b.d2 = (0.129)(25)(645)(325)2 = 219.7 xlO6 = 220 kNm > 200 kNm Also, hid = 100/375' = 0.308 < 0.45" Hence, section can be singly reinforced.

24

dmin = 282 mm d = 325 mm h = 375 mm

N.A. is out of flange

~

singly reinforced

Output

Cakulatioas

ReCerence equation 1 Note 7

AI = [M + (0.I)fcu .bw .d{(0.45)d-b,)] I (0.87)f {d-(0.5)hrH

3.12.6.1 Note 8

[(O.87)(460){325-(0.5)(I00))] = 1894 mm Use 21'32 &. 11'20 (As = 1922 mm2) l00A/bwh = (100)(1922) I (225)(375) . = 2.28 < 4.0; hence O.K.

=[(200xl()6)+(0.1X2S~2S){(0.45)(325)-100}f

= 1894 mm2 Use 21'32 &. 11'20' (1922 mm 2)

As


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 Pf • 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 if lapping is done.

Concluding Notes

9.

10.

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. In addition, if the moment is greater than Pf .fcu .b.d2 (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 of cross 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 0.8~. Assume feu = 2S N/mm2 and ~ = 2SO N/mm2• 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.

If possible, mild ~ of fyv = 250 N/mm2 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.

Cakuiatioas

Reference

Note 3 3.4.5.10

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 wfrom the face. Vmax

design .

: ~ '

~d~. VJIWt

~2500 ~

= «('()(5)/2 = 156 leN

"". ""

Table 3.9 Table 3.8

= (156 xloJ) I (225}(400) = 1.67 N/mnJl (0.8)(Wo.s = (0.8)(25)0.5 = 4 N/mm 2 > 1.67 N/mm2 < 5 N/mm2 ; hence O.K. VdeIip = (156)(2500-400) 1(2500) = .126 leN v = V/(bv.d) = (126 xloJ) I (225)(400)' = 1.4 N/mm2 l00AJb"d = 0.8, d =:= 400 nun, feu = 25 N/mm2 ; Hence, ve = 0.58 N/mm2• Since v > ve + 0.4 N/mm2, links have to be

Note 6

designed.

vJIWt

3.4.5.2 Note 4

equation 3 Note 5

Note 7

3.4.5.5 Note 8

A.v > = bv·sv
26

Vmax

= 1.67

v

1.4 N/mm 2

Vc

=;

N/mm2

= 0.58

N/mm2

Links RIO @ 175


This is the simplified method to account for the enhanced shear resistance near supports. The section 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. If 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 N/mm2 is the shear resistance that can be carried by nominal shear links.

7.

When using this inequality for providing links, either the Aav value or Sy value must be chosen. In g~, 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 axis of a section. Gea1etally, it is twice the area of the chosen bar, since in most cases it is links with 2 vertical legs that are used. The resulting Sy value should not exceed (0.75)d, to ensure that at least one link crosses a potential shear crack. The transverse spacing between the legs of a link should be such that it does not exceed -d- and that 00 longitudinal tension bar is greater than 150 mm from a vertical leg.

8.

The link spacing is also often specified in steps of 25 mm, because of the tendency

to think in Imperial units. (1 inch is approximately 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. The mid span. steel consists of 4 Nos. 2S 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 = 2S N/mm2 , fy = 460 N/mm2 and fyv = 250N/mm . Introductory Notes 1.

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.

R.eference

Output

Calculatioos SumutNCl!l

.

= 400 kN v = (400 xl 2.08 N/mm2 < 5 N/mm2 ; V max lDllX

3.4.5.2

Vmax

= 2.08 N/mm2

hence O.K.

equation 4 3.4.5.6 Note 3

Table 3.9 3.4.5.10 Note 4

3.4.5.6

Shear resistance of 2 inclined bars, Vb = A.(O.87)fyb(cosa + sina.eotP)(d-d') I Assumethalll = 67.5° and d' = 50 mm' hence .•stt = (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 ~m2

=

Since 2 bars continue into support, lOOA/bvd = (loo)(981.J 1(350)(550) = 0.51; hence, Vc = 0.50 N/mm Shear force at section "d" from support {(3000-550) I (3000)}(400) = 327 kl '
28

Vb

= 1.44

N/mm 2

Vc

= 0.5 N/mm 2

Reference

Output

Calculations

Asv > = bv.Iv{(1.20)12} 1 (0.87)fyv Putting A.v = 157.1 mm2 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 entire area over which the bent up bars are effective - i.e. for 0.71 m from the face of support. Middle

Table 3.9 Table 3.8

RIO @ 150 mm (support area)

area

l00A/bvd == (100)(1963) 1 (350)(550) == 1.02; hence Vc == 0.63 N/mm2 Shear stress taken by nominal links == 0.63 + 0.4 = 1.03 N/mm2 Shear force taken by nominal links == (1.03)(350)(550)(Hr3) = 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)bv·1v 1 (0.87)fyv Putting Aav = 157.1 mm2 for 10 mm links, Iv < = (157.1)(0.87)(250) 1 (0.4)(350) = 244 mm < (0.75)d; hence O.K. Use RIO links @ 225 mm

Vc

== 0.63 N/mm2

RIO @ 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/mm2 , design shear links. A.v ~= bv·lv(v-vJ 1 (O·87)fw Putting A. v = 157.1 mm2 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

l~

2/1(QjaX)

1~2T25

I.

4T25

2T25'"

1.48m

29

v = 1.58

N/mm2

Vc

= 0.63

N/mm 2

2 RIO@

200 mm (area inbetween)


Since fJ should be taken as > 45° and ~ is restricted to l.S(d..(i'), this assumed value of 67.5° for fJ 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 distance "d" from any point in this section of the beam. Hence, the value of vc will be the same as in the middle area.

6.

If the link spacing is less than around ISO mm, it will be difficult for concreting to be carried out. Hence, as in this case, 2 links can be placed together, spaced wider apart. An alternative would have been to use 12 mm dia. links; however fabrication will be easier if links of the same diameter are used throughout the beam.

a

Concludina Notes 7.

It is not very common practice to use bent up bars as described in this example, although it was in the past.

30

EXAMPLE 12 - 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.


The serviceability checks consist of spanldepth ratio calcu1ations for deflection and bar spacing rule checks for cracking. If 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 Table 3.10 Example 8

Table 3.11 Notes 3&4

3.4.1.4 Table 3.10 Example 8

Table 3.11 Notes 3&4

Consider the man BC; effective span = 6000 mm bwlb = 0.22 < 0.3 .Hence, basic span/depth = 20.8 for continuous, flanged beam. Mlbd2 = 1.00 and f. = (S/8)(460){(1224)/1295)} = 272 N/mm2 Hence, P l -= I.4S (for tension reinforcement) F 2 = 1.0 (as there is no compression reinforcement) Hence, allowablespanldepth ratio = (20.8)(1.45) = 30.16 Actual spanldepth= (6000)/(397.5) = 15.09 < 30.16; hence O.K. Consider fP8D AD: effective span = 2000 mm Basic spanldepth = 7 for cantilever with rectangular beamaetion. Mlbd2 = 2.32 and f. = (5/8)(460){(799)/829)} = 277 N/mm2 Hence F l == 1.07 (for tension reinforcement) F2 = 1.0 (as there is no compression reinforcement) Hence, allowable spanldepth ratio = (7)(1.07) = 7.5 Actual spanldepth = (2000)/(397.5) = 5.03 < 7.5; hence O.K.

Curtailment of reinforcement The bending moment diagram envelope must first be dmwn

31

All. span! depth = 30.2

Act spanI

depth = 15.1 Hence O.K.

All. span! depth = 7.5 Act. span/ depth = 5.03 . Hence O.K.

Cakulatiolls

Reference NoteS

Por 111M Be. the controlling 1oa4.c:ase is when AS has the minimum designultimate1Qed 'aDd Be has the maximum designultimateIoad~ This case bas already been considered in Example 7 .

7.0 /6.34 kN/m

/45.28 kN/m

~•• nn~ 1.95m

Example 7

For span BC,

6.Om

x

Mx = (128.1)x - (45.28)x212

Mx = Oatx - O. Mx is max. at x = 2.83 and equal to 18l.2 kNm Mx .. 0 apiA
lno. We can consider

Example 8

Steel at span BC is 2T25 & curtailing the lno bar.

Note 6

M.o.R. of continuing bars (A, .= 981.7 mm2) can be

3.12.9.1

shown to be 148.4kNm. Putting (128.1)x - (22.64)x2 == 148.4 we can obtain x = 1.63 m and 4.03 m. 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.

Calculations

RelereD£e 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)

Ra

+ (16.34)(6~/2

= 167.7kN

My == (7)(1.4){y"o..0S]

My

+ (4S.28)r/2

- (167.7)[y-2.0) - (45.28-16.34)[Y-2.0j2/2 == (9.8){y"o.05] + (22.64)r - (167.7)[y-2.0) - (14.47)[y-2.0r /

A

\

/

A

M.o.R. '\..... M.o.R./2

;B

My = Oaty =OandML = loo.7atB. My == 0 again at y == 4.2J m. Steel at support is 2T2O & ITI6. We can consider curtailing the 1T16 bar. Note 6

3.12.9.1 Note 9 3.12.9.1(e)

M.o.R. of the continuing bars (~ == 628.3 mm2) can be shown to be 90.5 kNm. Putting-(9.8)(y-Q.05) + (22.64)r = 90.5, we can obtainy == 1.80 m for span 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 span Be. These are the theoretical cut-offpoints. To find where the M.o.R. of continuing bars is twice 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

Output

Note 10

Table 3.29 Note 11

Output

Calculatioos

Refereoce

TheGiffe:reoce betweenthe·SIDIlIer y values is (1.80 - 1.22) = 0.58 m or 580 Mm. This is greater than (12)4> (192 mm) or -d- (397.5 mm). 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 y= 3.10 m. Length of 16 mm bar required = 3.10 - 1.22 = 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) Since the distances to either side of B > = (40)41 {Le. (40)(16)= 640 mm}, anchorage is satisfied.

Curtail I T16 top bar 0.78 m (left) and 1.10 MCright)

ofB. I..eIlgth of bar is 1.88 m.

J 'Imine of bars

Note 12

3. 12.9. 1(c) Table 3.29 Note 13

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 Be can be curtailed attbe point of contraflexure 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

3.12.8.13

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 the curtailment of the 21'20 bars, Le. 2.63 0.48 =2.15 m to the right of B. ~:- Min. lap length = gtQterof (15~ (::: ISO mm) or 300 mm is satisfied; also distance between laps will be greater than 75 mm and (6)4> (=72 mm).

Note 15

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. The nl2 bars will start 0.48 - 0.06 = 0.42 m to the right of B.

3.12.8.13 Note 14 3.12.8.11

34

Curtail 2T20 top bars 2.63 m to .right of B. Start 2T12 top bars 2.1~ 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

Output

Calculations lT16 ZIal

mo

I A

>

Zf12

214> <

•

<

t

>'

Zf12

,

),

21Q

2I25 l'IID

B

21"25t c

Crack width check
Cover required = 30 mm Assume link diameter of 10 mm.

Example 8

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 If middle (16 mm) bar is curtailed, clear spacing =

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

ISO rom. The top spacing at the support < 160 mm; hence O.K. (Note:- Since 16120 = 0.8 > 0.45, the 16 mm

1£

(0].-1 ( :m )

bar satisfies-the ·0.45 role·.) However, the spacing role is marginally violated when the middle bar is eurtail~; 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 80 mm; hence O.K. = 6O.7mm < 160/2

Crack Width O.K. at support

Considering the span section (tension on bottom), Clear spacing between bottom bars (2T25 &. 1nO) = {300 - (2)(30) - (2)(10) - (25+25+20)}/2 = 75 mm If 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.

Crack width O.K. in span

Note also that all the above spacings are greater than hagg + 5 mm, if we assume that h = 20 mm. Hence, minimum spacing rules are~so satisfied.

Minimum spacing O.K

=

35

Notes

00

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 3.4.1.4.

3.

The use of eq~on 7 will be more convenient than obtaining F 1 from double interpolation in Table 3.11.

4.

In 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.

S.

If 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.

This calculation is done as in Example 1. The 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 . 3.12.9.1 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 Be to be as far as possible from B.

9.

For sagging momenteurtai1ment, generally condition (e) is the4XNUlOUing one, over (a) and (b) in Clause 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. The 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.

The 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.

36

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. If such redistribution bad been performed at a .support ectioo, the muimum 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.aations of the bar spicing rules can be allowed. In any case, see Note 19.

18.

Both maximum and minimum spacings have to be satisfied. The maximum 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.

If the "deemed to satisfy" serviceabilitycbecks.ae not satisfied, the more aocurate calculations for deflection and crack width in Section 3 of BS 8UO: Part 2 can be resorted to, in order to find out whether the Rlquimnents of Clause 2.2.3 are met.

37

EXAMPLE 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 kNIm2• Taking the density of reinforced concrete to be 24 kN/m3 , feu = 25 N/mm 2 , and fy = 460 N/mm2, design a typical interior panel.


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 ~tions is not accurately known, it is reasonable to U$umeaudl value of I kNlm2• 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 kN/m 2 and 1.5kN1m2 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".

Output

Calculatioas Slab

thickness

Note 3

, to choose a slab thickness, assume tb of 34 .(for a continuous I way slab). Hence, effective depth (5.0 xloJ)/(34) 147 mm

TABLE 1

Note 4

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

NoteS

LoadiOI: (for I m wide strip)

=

=

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

Design load = (1.4)(5.2)

=

h == 175 mm. d -ISO . .

4.2 kN/m

= 1.0 kNlm = 5.2 kN/m (gk)

= 3.0 kN/m = 1.0 kN/m

= 4.0 kNlm ('he)

+ (1.6)(4.0) =

38

13.7 kN/m

design udl = 13.7 kN/m

Calculations

Referenee

Output

Ultimate bendin& moments and shear forces 3.5.2.3 Table 3.13

Since gk > 30 m2 , 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

M~

=

21.6 kNmlm

M rt= 2r6~mlm

v = 34.3 kN/m

Desi&n for bendin& at §l)M Chart 2 (Part 3) Note 6 Note 7 Fig. 3.25 Note 8 3.12.11.2.7

Mlbd 2 = (21.6 x10~ 1 (1000)(150f" = 0.96 100AJbd = 0.26 (> 0.13); hence min. steel O.K. As = (0.26)(1000)(150) 1 (100) = 390 mm2/m Use TI0@175 mm (As = 448 mm2/m) Half theSe be Q1rtailed at (0.2)1 - i.e (0;2)(5) == 1.0 mfrom the centre-line of support. Thenr/f will be TI0 @ 350 mm {«3)(150) "" 450} lOOA/Ac =i (100)(44812) / (1000)(175) = 0.13 Hence crack cootroland minimum steel O.K.

bars_

deflection

Note 9

Check for

Table 3.11 Table 3.10

MJbd2 = 0.96 and f s = (5/8)(460){(390)/(448)} = 250 N/mm2 Hence F 1 = 157 Allowable spanldepth = (26)(1.57) = 40.8 Actualspanldepth =(5000)/(150) = 33.3 < 40.8; ~ O.K. ~n for

Fig. 3.25

Span steel TIO @ 175 mm

Deflection O.K.

bendin& at sUllJlOrt

Since the moment is identical to that in the span, steel provided also can be identical. Half these bars can be curtailed at (0.15)1 = (0.15)(5) = 0.75 m from the face of support (Note: 45 tP = 450 mm< 750mm) and all the steel curtailed at (0.3)1 = (0.3)(5) = 1.5 m from the face of support.

39

Support steel TIO @ 175 mm

Output

calculations

Reference

,

Check for shear

Note 10 Table 3.9 Table 3.17 Note 11

v = (34.3 xl 0.23N/mm2 Hence, no shear reinforcement required.

v

= 0.23

N/mm 2

vc - 0.54 N/mm.2.

Seconda[y reinforcement Table 3.27

3.12.11.2.7

l00As/A c = 0.13 As = (0.13)(1000)(175) I (100) = 227.5 mm 2/m Use TlO @ 350 mm (As = 224 mm2/m) Max. spacing = (3)(150)- 450 mm > 350 mm.

~ o. 7fm J>=7~

~7EmMO.7~ Tl~175

Notel2

cr

nOl3OO~75

TlQfJ350

.. ....

~~Tl~

·--n~·---TlOOI75

, I

0(

I.On

(

'>

SeCondary steel TI0@350mm

,

"

.. ~

Tl............I

'f f

( I.On )

5.0m

)

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 34 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 slab would have an imposed load of around 4 kN/m2 , while a heavily loaded slab would bave one of around 10 kN/m2 •

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/ Ac. However. since the 40

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 As = 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 bar can be curtailed.

9.

The assumption regarding spanldepth ratio must becbecked as early as possible in the design. Hence span moments should be 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.

In general, apart from .some cases in flat slabs, it is sought to avoid shear reinforcement in flat stabs. Hence, if v is greater than vC' the slab thickness is increased. This should always be borne in mind, and perhaps an approximate check for shear made early in the design,especially if the slab 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 approach is given for the design of continuous beams as well, in Table 3.6. The coefficients in this latter table are higher than those for slabs, because the slab 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 fcu .·2S'N/mm2, f y = 460 N/mm2 and density of reinforced concrete 24 kN/m3 •

=


This example has more unknowns than the previous one. It 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 taken regarding the end fixity of the slab.

...i..

The imposed load could be taken as 1.5 kN/m 2, since it is a domestic building. The finishes (onboth top surface and soffit) can be assumed to be a uniformly distributed load of 1 kN/m2 .

3.

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 approach is to idealize this slab as a one way simply supported slab. Any fiXing moments caused by the above partial fixity can be accomodated by taking SO% of the midspan steel into the top face of the slab at the support.

4.

The parapet wall parallel to the span will have to be carried by the slab. It 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/m2• The load from this wall will be distributed only over a limited width of the slab (Clause 3.5.2.2).

Calculations

Reference

Output

Slab thickness Note 5 Note 6

TABLE 1 Example 8 Note 7 3.4.1.2

Approximate span = 3500 mm Assuming Spanldepth ratio of 28 (for a simply supported 1 way slab), effective depth = (3500)/(28) = 125 mm If we take cover = 30 mm (moderate exposure conditions and TABLE 1 values modified by Notes 5 and 6), and bar diameter = 10 mm, we can choose h = 160 mm and d = 160 -30 - 10/2 = 125 mm. Hence, effective· span = lesser of (3500+225) = 3725 mm or (3500+ 125) = 3625 mm

42

h d

= =

160 mm 125 mm

eff. span 3.625 m

=

Reference

Calculations

Output

Loadine (for 1 m wide strip) Self load = (0.16)(1)(24) Finishes (1.0)(1) Total dead load Imposed load = (1.5)(1)

= 3.84 kN/m

=

=

1.00 kN/m 4.84 kN/rn = 1.50 kN/rn

=

Design load = (1.4)(4.84) + (1.6)(1.5) = 9.2 kN/m 3.5.2.2 Note 8

Strip carrying parapet wall = (0.3)(3.615) + 0.12 = 1.21 m Additional dead load in that area = (1.0)(0.12)(23) 1 (1.21) = 2.28 kN/m

design udl = 9.2 kN/rn

Ultimate bending moment and shear force Note 3

Since we assume the slab to be simply supported, Mid span moment = w.l 2/8 = (9.2)(3.625)21 8 = 15.1kNmlm Shear force at support = w.l/2 = (9.2)(3.625) 12 = 16.7 kN/m

Mspan = 15.1 kNmlm V = 16.7 kN/m

Desim for bending Chart 2 (Part 3)

Mlbd2 = (15.1 xl
looA/bd = 0.26 (> 0.13); hence min. steel O.K. = (0.26)(1000)(125) I (100) = 325 mm2/m Use TI0 @ 225 mm (As = 349 mm2/m) Max. spacing allowed = (3)(125) = 375 mm > 225 mm; hence crack width O.K.

A. 3.12.11.2.7

Note 9

Fig. 3.25

However, bar spacing as well as minimum steel requirement will be violated if bars are curtailed. Hence, use TlO @ 187.5 mm (As =·419 mm2/m) Spacing after curtailment = 375 mm. l00A/ Ac 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

span steel TIO @ 187.5 mm

RefeNilee Note 10

3.12.10.3.2

Output

Calculations The rest of the steel could be taken into the support and bent back into the span as top steel to extend a distance from support face of (0.15)1 = (0.15)(3625) = 544 mm {> (45)cP = (45)(10) = 450 mm}, say 0.55 m

support steel TI0 @ 375 mm

Check for deflection

Note II Table 3.11 Table 3.10

Mlbd2 = 0.93 f s = (5/8)(460)(325/419) = 223 N/mm 2 Hence, F 1 = 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.

Check for shear

v = (16.7 xlol) 1(1000)(125) = 0.13 N/mm 2

looA/bvd Table 3.9 Table 3.17

Hence,

Vc

=

0~131

= 0.45 N/mm2 > 0.13 N/mm2 ; hence shear r/f is not required.

Shear O.K.

SecondflO' reinforcement

3.12.11.2.7

Note 12

looA/Ac = 0.13 As = (0.13)(1000)(160) I (100) = 208 mm2/m Use TIO @ 375 mm (i.e. max. spacing allowed - 3d) (As = 209 mm2/m) 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 midspan (and hence TIO @ 350 mm at supports). rt1 110075 TIotm5 ~o.~ II~

r

I :

t

•

I

•

1100187.5

o.ti: ~.

0.:rAJn

44

secondary steel TI0@375 rom

under parapets TI0@ 175 mm (span) TI0@350 (support)

Notes on Calculations 5~

6.

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. 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 taken 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 too 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 be 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. One possible alternative is to use smaller diameter bars, but bars 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 fIgure,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 than required at mid-Span (see Note 9), advantage should be.taken of this by generally calculating the service stress, which will be lower than (5/8)fy and lead to a greater allowable span/depth ratio.

12.

It may be convenient to reinforee the entire slab with TlO @ 175 mm at mid span 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 and support respectively. The small penalty in cost will probably be worth the simpler detailing arrangement.


It 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/m2• The loading fronl finisheaand light partitions can each be taken as 1 kN/m2 • Design a typical interior panel, using feu == 2S N/mm2 , fy == 460 N/mm2 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.

It will be assumed that the comers of this slab are prevented from lifting and that adequate provision is made for torsion. .

Calculatiops

ReferellCe

Note 3 3.5.7 TABLE I

Note 4 Note 5

Assultle.a spanldepth ratio of 40 (for a continuous 2 way slab) effective depth == (5000)/(40) == 125 mm If we take cover == 20 mm (mild exposure conditions and concrete protected by 10 mm·l:3 cement:sand rendering) and bar diameter as 10mm, then we can choose h == 150 mm and dmort == 150·20 - 10/2 == 125 mm and ~oog == 125 . 10 == 115 mm

h == 150 mm dlbort

==

125 mm ~== l1Smrn

Loadine (udl) == 3.6 kN/m2 Self load == (0.15)\:, 1(24) Finishes == (1,0>, == 1.0 kNlm2 Total dead load == 4.6 tN/mt Imposed load == (3.0). == 3.0kN/m2 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 IcN/m 2 Bendine moments This interior panel- has lyIlx == 615 == 1,2

46

n == 12.8 kN/m2

Reference Table 3.15

Chart 2 (Part 3)

3.12.11.2.7 Table 3.27

Note 6

Table 3.11 Table 3.10

Chart 2 (part 3) :l.12.11.2.7 Table 3.27

Calculations

Short way, Short way, Long way, Long way,

edge span edge span

Output

=(0.042)(12.8)(5)2= 13.44 kNm/m =(O.0~)(l2.8)(5)2= 10.24 kNmlm =(0.032)(12.8)(5)2= 10.24 kNmlm =(0.02~)(12.8)(5)2= 7.68 kNm/m

I Desien of reinforcement / Short way, mid-man:Mlbd 2 = (10.24 xld') 1 (lOOO)(l25t = 0.66 100A/bd = 0.17 As = (0.17)(1000)(125) 1 100 = 213 mm2/m Use TlO @ 350 mm (As :: 224 mm2/m) Max. spacing = (3)(125) = 375 mm > 350 mm l00A/Ac = (100)(224) 1 (1()()())(150) = 0.15 >0,13 Hence, bar spacing and min. steel are O.K. but if steel is curtailed, they will be violated.

Short way, span TlO@ 350 mm

(Check for, deflection):Mlbd2 = 0.66 f s = (5/8)(460)(213/224) = 273 N/mm 2 F 1 = 1.65 (for tension steel) 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' 288 mm2/m Use TlO @ 250 mm (A. = 314 mm2/m) Bar spacing and min. steel areIO.K.

Short way, edge TlO@250 mm

Long way. cts. edge:Mlbd2 = (10.24 xlW) 1 (1000)(115)2 = 0.77 l00Aibd = 0.21 As = (0.21)(1000)(115) 1 (100) = 242 mm2 Use TlO @ 325 mm (A. = 242 mm 2/m) Max. spacing = (3)(115) = 345 mm > 325 mm l00AJAc = (100)(242) 1 (1000)(150) = 0.16 >0.13 Hence, bar spacing and min. steel are O.K., but steel cannot be curtailed. Long way. mid-s.pan:= 0.58, 100Aibd =' 0.15, As = 173 mm 2/m Use TlO @ 350 mm (As = 224 mm2/m), since max. clear spacing (345 mm) governs.

"

Long way, edge TlO @ 325 ~m

Mlbd 2

47

Long way, span TlO@ 350 mm

CalcuJatiOllS

Reference

Output

Edge strips:3.5.3.5

Note 7

Table 3.16

Table 3.9

l00A/Ac =0.13 As = (0.13)(150)(1000) I (100) = 195 mm2/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.

Edge strip nO@375 mm (only for short way, cts. edge)

Check for shear Short way sypport:V = (0.39)(12.8)(5) = 25.0 kN/m v = (25.0 xloJ) I (1000)(125)= 0.2 N/mm 2 l00A/bd = (loo~(314) I (1000)~125) = 0.25 vc = 0.53 N/mm > 0.2 N/mm ; hence O.K.

LoDa l)!&Y R>J!Ort:-

Table 3.16

Table 3.9

V == (0.33)(12.8)(5) = 21.1 kN/m v = (21.1 xloJ) I (1000)(115) = 0.18 N/mm2 l00A/bd = (100)(242) I (1000)(115) = 0.21 vc = 0.50 N/mm 2 > 0.18 N/mm2; hence O.K.

o

Fig. 3.25

~

No shear r/f required

< 1800

~ 16T106325'11 ~

600

, "" E-o ...

1.0

l:"-

M

@

0

E-o

N

~

-~lo

1.0 N

, ~l-

0

1.0 M

--@

@

0

0

E-o

E-o

O'l

00

48

-

...1-

E-o

1.0

r-

M

@

0

E-o

N

-

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 be appropriate for heavily loaded slabs. This will of course reduce with the ratio of long to short span, ~hing the value for 1way slabS when the latter ratio becomes 2. The span/depth ratio is calculated with respect to the shorter span, as it is this that controls d~flection.

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. 175 mm required for the slab in Example 13.

5.

In 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.

49

EXAMPLE 16 - FLAT SLAB A. flat~, which _several bays in each direction, has a panel size 0,{5m x 6 m. The ~gn imposed ontbesJab is 3 kN/m2 • Tbe ioa4ingfrom finishes and liaht PlU'titions can each be considered to be 1 kN/m2. Design a typical interior panel, using feu == 2S N/mm2 , fy 460 N/mm 2 and density of reinforced concrete 24 kNlmJ • It may be assUmed that the columns supporting the slab are braced.

=

=


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 be simplified method of analysis described in Clause 3.7:2.7 and Table 3.19 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.

will

Calculations

Reference Slab thickness 3.7.1.4 Note 4 3.7.8 TABLE 1

NoteS

=

Max. value of he =.(114)(5.0) 1.25 m Assuming a trial span/depth of 32, (6000)/(32) 187.5 mm effective depth 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, dshort == 187.5-10 == 177.5 mm, dave == 182.5 mm

=

Loadin~ (for

entire panel)

he

= 1.25 m

=

h == 212.5 mm d y == 187.5 mm dx == 177.5 mm davg==182~5 mm

>"

Note 6 Panel area == (5)(6) == 30 m2 Self load = (0.2125)(30)(24) == 153lcN Finishes == (1.0)(30) = ..1U:.kN Total dead load = 183 leN . Imposed load = (3.0)(30) = 90 leN Partitions = (1.0)(30) = 30 leN Total imposed load = 120 leN Design load = (1.4)(183) + (1.6)(120) = 448 leN

50

.il"

F = 448 leN

Reference Table 3.19 Note 7

Fig. 3.12 Table 3.20

Fig. 3.12

Note 8 Table 3.11 3.7.8

Chart 2 (part 3) Note 9 3.12.11.2.7 Table 3.27 Note 10

Chart 2 (Part 3)

Calculations

Output

Bending mOments Long way:1 = 6.0 - (2/3)(1.25) = 5.17 m Span moment = (0.071)(448)(5.17) = 164 kNm Col. strip (2.5 m) = (0.55)(164) = 90.2 kNm 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 Span moment = (0.071)(441)(4.17) = 133 kNm Col. strip (2.5 m) = (0.55)(133) ::: 73 kNm Mid. strip (3.5 m) = (0.45)(133) = 60 kNm Support moment = (0.055)(448)(4.17) ::: 103 kNm Col. strip (2.5 m) = (0.75)(103) = 77 kNm Mid. strip (3.5 m) = (0.25)(103) = 26 kNm

Design of reinfOrcement Long way. an:(Check for deflection) Total ~ moment = 164 kNm Mlbd2 = (164 xlo6) 1 (5000)(187.5)2 = 0.93 2 If AS,reqd = Aa.(Jrov. fs = 288 N/mm and F 1 = 1.41 for tension steel) Allowable spanldepth = (26)(1.41)(0.9) = 33.0 Actual spanldepth = (6000)/(187.5) = 32 < 33.0; hence O.K.

Deflection O.K.

(Column strip - 2.5 m wide) Mlbd 2 = (90.2 xl
Long way. span (Col. strip) 17 TI0 @ 147 mm

(Middle strip -2.5 m wide) Mlbd2 = (73.8 xl<>6) 1(2500)(187.5)2 = 0.84 looA/bd = 0.23; As = 1078 mm2 Use 14 TI0 @ 179mm (As = 1100 mm 2)

Long way. span (Mid. strip) 14 TI0 @ 179 mm

51

-

Output

Calculations

Reference

Lone

3.7.3,1

Table 3.27

Chart 2 (Part 3)

3.12.11.2.7

Note 11

way, SURPOrt:(Column strip - 2,5 m wide) M/bd2 = 1.08, l00A/bd = 0.29, A. = 1359 mm 2 Use 18 TlO (As = 1414 mm 2) - 12 TIO centred on column @ 104 mm; 6 TI0 @ 208 mm. (Middle strip - 2,5 m wide) M/bd 2 = 0.36; l00A/bd ;., 0.10 Use nominal steell00A/Ac = 0,13; A. = 691 mm 2 Use 9 TI0 @ 278 mm (As = 706,9 mm2)

Long way, sup (Mid, strip) 9 TI0 @ 278

Short way, span:(Column strip - 2,5 m wide) M/bd 2 = (73 xl(6) / (2500)(177.5)2 = 0.93 l00A/bd = 0.26 As = (0.26)(2500)(177.5) / (100) = 1154 mm2 Use 15 TIO @ 167 mm (As = 1178 mm2) Allowable spacing = (3)(177.5) = 532.5 mm

Short way, span (Col. strip) 15 TlO @ 167

(Middle strip - 3.5 m wide) Mlbd2 = (60 xl(6) / (3500)(177.5)2 = 0.54 l00A/bd = 0.15 As = (0.15)(3500)(177.5) / (100) = 932 mm2 Use 12 TIO @ 292 mm (As = 942.5 mm2)

Short way, span (Mid strip) 12 TI0 @ 292

Short way. suPJ)01't:(Column strip - 2.5 m wide) Mlbd2 = 0.98, l00A/bd = 0.27, As 1198 mm 2 2 Use 16 TIO (As = 1257 mm ) - 10 TI0 centred on column @ 125 mm; 6 TlO @ 208 mm.

Short way, sup (Col. strip) 10TlO@ 125 6 TIO@208

(Middle strip - 3.5 m wide) Mlbd 2 = 0.24, lOOA/bd 0.06' hence use l00AJAc 0,13, ~ 967 mmi Use 13 TlO @ 269 mm
Short way, sup (Mid. strip) 13 TlO @ 269

=

3.7.3.1

3.12.11.2.7

=

Check for

3.7.7.4

Long way,sup (Col. strip) 12 TI0 @ 104 6 TI0 @ 208

=

=

=

shear

If square columns are used, size of column head = {(T/4)(1.25)2}O.5 = 1.1 m . 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 m2

S2

a

3.7.6.2

3.7.7.4 Note 12 Table 3.9

Output

Calculations

Reference

Vt = 448 kN Veff = (1.15)V t = (1.15)(448) = 515.2 leN vmax = (515.2 xloJ) I (4.4 xloJ)(182.5) = 0.64 N/mm2 < (0.8)(25)°.5 = 4 N/mm2; 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 N/mm2 (I00A/bd)avg=(112)(I00/182.5){(14l4+ 1257)12500} = 0.29 ve 0.51 N/mm2 > 0.39 N/mm2

=

o~

I

--I

7--:

I

I

Shear r/f not required.

~ ~r, I __ L ----l---l---~ ,1

g ~ ~

I"

I: ,

12Tl00292B

~ :100J

,

: I

13Tl00269T

~--~- -1 --t - -' - : ~

,

~ " S; I ~'

,

,

I

,--+--'" -

I

fJ

,

I

- I - - - t - - ...... I

... - - -1 ~ - - - - '_L Cl'l E-c E-c~

~

~

~

~

~ -

N

E: Cl'l

~ ~

N""

~ ~

-l-l

3Tl0f208T 15Tl00167B lOTlO,125T 3T lO'208T

~~

~ ~

0

0

E: E: E:E: C'?C'-NC'? ........


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

way slab in Example 15; of course. there is the considerable advantage here of not requiring beams. The slab thickness has been 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 too is the opposite of two way slab action. The average value of effective depth is used for punching shear checks. 6.

It is more convenient to determine the loading on an entire 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 each direction. This is because there are no peripheral beams around the slab, the flat-slab acting as both slab and beam.

8.

The deflection check is done early here, even before the steel is designed. This is a conservative approach, but has the advantage that it can detect early any changes that may be required in slab thickness. If this check is made after the steel has been designed, the average of column and middle strip steel can be taken for the As values.

9.

The reinforcement in a flat slab is generally specified in terms of the number of bars in a given strip. As such, the spacing may 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 can easily be satisfied.

11.

Note that the effective depth in the short way direction is 177.Smm (as opposed to 187.5 mm) and that the width of the middle strip is 3.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 also be avetaged. This is because the square sbear perimeters cross both the long way and short way steel.


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 is the column strips that carry most of the moment and are more heavily reinforced.

14.

Where the simplified method used here is not applicable, a frame analysis will have to be carried out according to Clause 3.7.2.

15.

Edge and comer columns of flat slabs will have column strips considerably narrower than those in interior panels (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 kN/m 2 as a one-way spanning slab. Taking the load from light partitions and finishes as 1 kN/m 2 each, the density of reinforced concrete as 24 kN/m3, feu = 25 N/mm 2 and fy = 460 N/mm 2 , design a typical interior panel. Note that a 1 hour fire resistance is required.


This example can be 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).

Reference Choice 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

Output

Calculations

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 10 mm 1:3 cement:sand rendering) and bar size of 20 mm, we can choose h = 225 mm and 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 = (225)~ ~= {(50)(500) + (1I2)(125+250)(175)} 1(500)(225) = 0.51 1e = (225)(0.51) = 115 mm (> 95 mm for 1 he. fire resistance; hence O.K.)

~~ \(

250 500

55

-*-50

if

h d

= 225 mm = 195 mm

te

= 115

mm

Calculations

Reference Note 5

Output

Loading (for 0.5 m strip) Self load = (0.115)(0.5)(24) = 1.38 kN/m Finishes = (1.0)(0.5) = 0.50 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 = 2.0 kNlm Design load =(1.4)(1.88) + (1.6)(2.0) = 5.83 kN/m

design 001 = 5.83 kN/m

Design for bending

3.4.4.4

Table 3.27

Assuming slab is simply sup~rted, moment in span = (5.83)(5) I (8) = 18.2 kNm K = M I (b.d 2.fcu) = (18.2 xl06) 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 « 50 mm) Hence, neutral axis is in flange. As = (18.2 xIW) I (0.87)(460)085) = 246 mm2 Use 1 1'20 (As = 314 mm2) 1ooA/bwh = (100)(314) I {(1/2)(125+250)(225)} = 0.74 > 0.18; hence min. steel O.K.

main r/f 1 1'20 per rib


Table 3.10

Table 3.11

bw'b = (187.5)/(500) = 0.375 Hence, basic span/depth ratio = 16.4 f s = ~5/8)(460)(246/314) = 226 N/mm 2 M1bd = (18.2 x106) / (500)(195t = 0.96 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

Deflection O.K.

Reference Note 7 Table 3.9 3.6.4.7

v = (13.4 xlol) I (187.5)(195) = 0.37 N/mm l00AJbd = (lOO~{314) I (187.5){195) = 0.86 vc = 0.72 N/mm > 0.J.tN/mm2 ; hence shear r/f not required.

2

Top steel 3.6.2

Output

Cakulations

shear r/f not required

over sup,port

This is to control cracking and should be 25 % of midspan steel. As = (114)(246) = 61.5 mm2 Use 1 TIO bar (As = 78.6 mm2), extending (0.15)1 = (0.15){5.00) = 0.75 m into span on each side.

over support 1 TIO per rib

Notes on CalculatioR§ 3.

This trial ratio is reasonable for simply supported one~way slabs - see Note 6 in Example 14.

4.

It 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.

It 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 8.

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 flanged 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 be greater, because of non-planar formwork requirements. 57

EXAMPLE 18 - COLUMN CLASSIFICATION

PI. 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 U1e 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 Notes 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 carry additional moments due to deflection or not).

2.

The effective length of a column will depend on the degree of fixity at its ends.

Caknlations

Reference

Clear height between end restraints, (for ground floor columns) lox = 5.0 m. loy = 5.0 + (0.525-0.175) (for other floor columns) lox = 3.5-0.525 = 2.975 m, loy = 3.5-0.175 ::: 3.~25 m 3.8.1.6.2

Table 3.21 Note 3

3.8.1.3 3.8.1.3

Output

= 5.35 m .

The end conditions for the columns in the direction of beams are all conditionl. H(mce, fJ ::: 0.75 In the other direction. fJ = 0.80 (ground floor columns) fJ = 0.85 (other columns) lex = (0.75)(5000) = 3750 mm (ground floor) ::: (0.15)(2975) ::: 2231 mm (other floors) ley::: (0.80)(5350) ::: 4280 mm (ground floor) = (0~85)(3325) ::: 2826 mm (other floors) 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~jb (2826)/(300) 9.42. < 15; hence short.

=

=

58

All columns are short.


The values of {3 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 (i.e. sum of equations 3 to 6 in section 2.5 of Part 2 of the code. The ratios 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 I) respectively.

"c

Concluding Notes

4.

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. In this case, an approximate value for {3 can be interpolated, based on the actual C¥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.

5.

For a column to be considered short, both le/h and le/b have to be less than 15 (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 {3 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, fcu == 2S N/mm 2, fy = 460 N/mm 2 , and that the imposed loads on the roof and the floors are 1.0 kN/m2 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 part of an internal column of the framea structure described in Example 18. : . " . ".


Since the ground floor .part of 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 part 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 dilDClDsions are 5.0 x 5.0 m.

Hence, area conesponding to column" (5)2 .. 25 m2

Dead loads

Note 3

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

Imposed loads From roof = (1.0)(25) = 25 kNFrom 3 floors = (3)(2.5)(25) = 187.5 kN From partitions- (3)(1.0)(25) =~ Total imposed load = 287.5.kN 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

o.c .. 201 kN

N = (1.4)(601) + (1.6)(201) = 1163 kN

N .. 1163 kN

Desim of main steel 3.8.4.3 equation 38

Note 4 Note 5 Table 3.27

For short columns resisting axial load = (O.4)fcu 'Ac -t (0.75)A,.,.fy Assuming we use 4 T16, Awe =8Q4 mm2 Ac .. (300)2 - 804 III 89196 mm2 (1163 xl03) .. (0.4)(25)(89196) + (0.75)Awe(460) Awe .. 785 mm2 < 804 mm2;hence O.K. ' .' . Hence, use 4 T16 (A.c = 804 mm 2) . Note:- lOOAwc/Ac = (lOO){804) I (300)2 == 0.89 > 0.4; hence min. steel O.K.

N

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 Ac is the net area of concrete. A trial value of Awe C3ft be obtained from equation 38 assuming the gross area of concrete for Ac as a flrst approximation; this 60

area of A~ can then be deducted from the gross area to obtain Ac' The value of Asc obtained from the formula should be less than the original trial value of A.c. 5.

In some cases, a negative value may be obtained for A~; 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.

In 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 A short column of 300 mm x 400 mm cross section carries an ultimate axial load of 800 leN. If 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 N/mm 2 and fy d 460

N/mm2 •

Introductory 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

Calculations

Reference

N = 800 kN, M == 80 kNm Nlbh = (800 xI(3) 1 (400)(300) = 6.67 (for all cases)

Note 2 Chart 23 (part 3)

Chart 23 (Part 3)

(al Major axis bendine b = 300 mm, h = 400 mm Mlbh2 = (80 xl()6) 1 (300)(400)2 = 1.67 lOOA.Jbh = 0.4 A~ = (0.4)(300)(400) I (100) = 480 mm 2 Use 4 T16 (Asc = 804 mm2) (b) Minor axis bendine b '''';' 400 mm,h = 300 mm Mlbh2 = (80 xl()6) 1(400)(300)2 = 2.22 lOOAaclbh = 0.8 A~ = (0.8)(300)(400) 1 (lOOi = 960 mm 2 Use 4 TIO (Ase = 1256 mm ) (<<) ~

3.8.4.5' Note 3

Table 3.24 Chart 23 (Part 3) Note 4 Note 5 3.12.7.1

Note 6

~

0

major axis 4T16

CJ o

0

minor axis 4

no

!;>endin

MKlh' '= (80 xHi~ t(~50)

MyIb'

= (80 x106) 1 (300-50)

hence M,/h'

equation 41

[J

<

M~'

M ' = M +(fl)

(b.Ii.fcu>=

'Ih')~

N1 (800 x1(3) 1 (300)(400)(l.S) = 0.267 hence P = 0.690 ~' = 80 + (0.~(250/300)(8O) = 126 leN Mlbh2 = (126 xl )/ (400)(300)2 = 3.5 . 100~lbh == 1.7 A~ = (1.7)(300)(400) 1(100) = 2040 mm 2 Use 4 TI5 + (2 T16 in each dir.) (A~ = 2366 mm2)

)&silo of links , For major axis~ding, use R6 {> (16/4) = 4 mm} @ 175 mm{ < (12)(16) = 192 mm}. For minor axis bending, use R6{ > (20/4) = 5 mm} @ 225 mm {< (12)(20) = 240 mm}. For biaxial bending, use R 8 {> (25/4) :;: 6.25 mm}! @ 175 mm {< (12)(16) = 192 mm}.

0 • • OeO

biaxial

4 T25 + 4 T16

JiDb

major axis R6@ 175 mm. minor axisR6@225 mm.

biaxial R8@ 175 mm.

Check for shear

= (126)/(800) = 0.158 « (0.75)(0.3) = 0.225 m);

3.8.4.6

MIN

Note 7

hence, shear is not critical.

62

Shear check oot required.

Calculations

Reference 3.8.6

Crack control (0.2)fcu .Ac = (0.2)(25){(3OO)(400) - 804} = 596 kN Axial load = 800 kN > 596 kN. Hence, no check is required.

Output Crack width check not required.


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 bar diameter of 25 mm, then d/h will be (400 - 50.5) 1 (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. It should 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.

If 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 BS 8110 simply explained, E. & F.N. Spon, London, 1988" and in "Rowe, R.E. et al., Handbook to British Standard BS 8ll0: 1985 - Structural use of concrete, Palladian, London, 1987".

5.

Although smaller diameter bars (e.g. TIO) could have been used, the Tl6 bars'are used, so that the link spacing would no~ be too small; furthermore, bars smaller than T12 are not used as column reinforcement, as they would not be stiff enough during erection.

6.

Generally plain mild steel is used for links as it is easier to bend into shape. Furthermore, where bars other than corner bars are used, multiple links may have to be used if (i) there is more than one intermediate bar or (ii) the intermediate bar is greater than 150 mm ~way from a restrained bar (see Clause 3.12.7.2)

7.

Strictly speaking, howe~er, the shear stress should be found in order to check for the limits on v max '

Concluding Notes 8.

In general,sh.ear and crack control are not very critical for columns. 63

EXAMPLE 21 - SLENDER COLUMN

A braced slender column of 300 mm x 400 mm cross section carries an ultimate axial load of 800 kN. It is bent in double curvature about the major axis, carrying ultimate moments of 80 kNm and 40 kNm at its ends. The effective length of the column corresponding to the major axis is 7200 mm. Determine the column reinforcement if feu = 25 N/mm 2 and fy =

460 N/mm 2 •


This example can be compared with Example 20, where the short column was of the same dimensions and carried similar loads.

Reference 3.8.1.3 3.8.3.3 Note2

3.8.3.2 equation 36 3.8.2.4 equation 34 equation 32 equation 35

Output

Calculations Type of column Vh = (7200)/(400) = 18 Also h/b = (400)/(300) = hence, single axis bending.

> IS; hence < 20, 1.33 < 3;

s~nder.

Design moments M 1 = -40 kNm; M 2 = +80 kNm M i = (0.4)(-40) + (0.6)(80) = 32 kNm {= (0.4)(80) = 32 kNm} emin = (0.05)(400) = 20 mm N.emin = (800)(0.020) = 16 kNm (3a = (1I2000)(IJb')2= (112000)('200/300Y= 0.288 au = #a· K.h = (0.288)(1)(0.4) = 0.115 M add == N.au = (800)(0.115) = 92 kNm Hence, critical moment is M j + Madd = 32 + 92 = 124 kNm. However, as K is reduced, if M j + Madd becomes < 80 kNm, M2 will become critical. Design of reinforcement

TABLE 1

Chart 23 (Part 3)

Assuming cover = 30 mol (moderate exposure conditions and TABLE 1 values modified by Notes 5 & 6), link diameter of 8 mm and main bar size of 25 mm, d/h = (400-50.5)/(400) = 0.87. N/bh = (800 x103) I (300)(400) = 6.67 M/bh 2 7 (124 xI06) I (300)(400)2 = 2.58 K = 0.9

64

slender column bent about major axis.

M j = 32 kNm

Madd = 92 kNm

Reference

Output

Calculations

Madd = (0.9)(92) = 83 kNm M = 32 + 83 = 115 kNm (>

Chart 23 (Part 3)

Note 3 Chart 23 (part 3)

80 kNm) Mlbh 2 = (115 xldi) 1 (300)(400)2 = 2..40 K = 0.85 Madd ~ (0.85)(92) = 78 kNm M = 32 + 78 = 110 kNm ·Mlbh2 = 2.29; K = 0.85 (again). Hence, lOOAsc /bh 0.8 Asc (0.8)(300)(400) 1 (100) :;: 960 mm2 Use 4 1'20 (Asc 1256 mm2)

=

=

=

aJnksl 3.12.7.1

Use R 6 {> (20/4) :;: 5mm} @ 225 mm {< (12)(20) :;: 240 mm}

K

= 0.85

main steel -

::I R6@ 225 mm


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 K that is virtually constant. It should be noted that the factor K 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.

65

~LE

22· PAD FOOTING

Design a square pad footing for a 300 mm x 300 mm internal column, which carries an ultimate load of 1100 leN (service load of 760 kN), if the allowable bearing pressure of the soil is 150 kN/m 2 • Use feu = 25 N/mm2 , fy = 460 N/mm2 (deformed type 2) and density of reinforced concrete = 24 kN/m3 •


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 (i) 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.

"

Calculations

Reference

Output

Dimensions of base Note 3 Note 4

=

=

=

=

= = =

NoteS

=

=

=

= =

=

Ultimate bearing pressure

= (1100) 1 <2.4/

= 191 kN/m moment (at face of column) (191)(2.4){(2.4-o.3)/2}2(1/2) = 253 leNm

~ritical bending

TABLE 1 Note 7

footing size ~x2.4m .4m

\

DesilW for ben4in& Note 6

\

Service load 760 leN Expected total load (1.08)(760) 821 leN Required area for base (821) 1 (150) 5.47 m2 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 141 kN/m2 2 < 150 kN/m ;hence O.K. Preliminary check on effective depth:d > 10(N)°'~ 10(1100)°·5 332 mm; hence overall depth of 400 mm is O.K.

=

Assume a cover of 40 mm, for moderate exposure conditions. H bar size of 16 mm is assumed, dmin = 400-40-16-16/2 336 mm and daV2 = 344 mm.

=

66

M

~

= 253 leNm

= 336 mm dan = 344 mm d min

Reference

Calculations

Chart 2 (Part 3)

Mlbd 2 = (253 xIW) I (2400)(336)~ = 0.93

3.11.3.2

= (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 {< (3)(336) + 300 = l308}; Use (3+3) T12 @ 200 mm in two outer bands. (As = 1407 + 678 = 2085 mm2; 1407/2085 > 2/3) Max. spacing = 750 mm; hence O.K. 100AiAe = (100)(2085) I (2400)(400) = 0.22 > 0.13; hence O.K. Anc~ge length :; (40)(16) = 640 mm < (2400-300)/2 = 1050 mm; hence O.K.

3.12.11.2.7

Note 8

l00A/bd = 0.25 As = (0.25)(2400)(336) I (100) = 2016 mm2 (3/4)c

+ (9/4)d

Check for yerticalline 3.4.5.10

Table 3.9

shear

Consider a section at "d" from the column face, .and assume no enhancement to ve. V = (191)(2.4){(2.4-o.3)/2 - 0.336} = 327 kN v = (327 xl 0.41 N/mm2; hence O.K. Check for punchine shear

3.7.7.2

= (1100 xl
<

3.7.7.6

Note 9

Output

(O.8)(25f· S = 4 N/mm2

bottom r/f 3 T12 @ 200 7 T16 @ 200 3 T12 @ 200 (both ways)

[§I

m Shear r/f not required.


Soil bearing pressures are given in terms of service lOads. Hence, service loads have to be 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. Another approach is to first estimate a depth (in this case 0.4 m) and reduce the allow~le bearing pressure by the corresponding weight per unit area (i.e. 0.4 x 24 = 9.6 ~/m2), 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. It is based on a punching shear considerations for commonly used pad footings. If 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.

If 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 Vc should correspond to 100A/bvd < = 0.15 in Table 3.9.

Concluding Notes

10.

I.f the footing carries a bending moment in addition to the axial load, the maximum and minimum pressures under the footing will be given by (l/BL}(N ± 6M1L), with symbols having usual meanings. The maximum pressure should be kept below the allowable bearing pressure.

11.

If 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 axial 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/m2 • Use feu = 25 N/mm 2 and fy = 460 N/mm2 •

o~_""

4_._7m_~_--7>1~~'r

\lOkN C

~D

E


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 pad 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.

Reference

Calculations Service loads

Note 3 External column load External column moment ,Internal column load

= (600)/(1.45) = 414 leN = (60)/(1.45) = 41.4 kNm = (1100)/(1.45) = 759 leN

Dimensions of footine

Note 4

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 under base, use a base of length (2)(3.4) = 6.8 m

69

Output

Note 5

Note 6

Outppt

CalcuJatloDS

Reference

If we assume a thickness of 0.8 m for the base, the allowable bearing pressure is 150 - (0.8)(24) • 131 kN/rr Width of base required = (759+414) I (6.8)(131) = 1.32 m Use base of 6.8 III x 2.0 m x 0.8 m

footing size 6.8 m·x 2.0 m xO.8 m

Analysis of footing Assuming that the C.of G. for ultimate loads is the same as that for service loads, the footing can be

idealised as follows:udl = (1100+600) / (6.8) Note 7

:1T A-&;::':_= ,6E = 250 leN/m

~~

w•

250 kN/m/

x~

Max. moment at C = (250)(1.65Y I 2 = 340 leNm To find max~ moment in AC. Mx == -(250)x212 + llOO(x-1.65) = -(125)x2 + (lloo)x - 1815 putting dMx/dx == -(250}x + 1100 = 0, we have x = 4.4 m Mmax = -(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 leN Design for bending TABLE 1 Note 8 Chart 2 (part 3)

Assume cover of 40 mm, bar size (longitudinal) of 25 mm; hence, d == 800 - 40 - 25/2 = 747.5 mm (Section AC) Mlbd2 == (605 xl06) I (2000)(747.5)2 = 0.54 looAJbd == 0.15 . A. -= (0. 15)(2000}(747.5) I (100) = 2243 mm2 Use 5 ns (As = 2455 mm2) - on top surface; these can be curtailed if required.

70

longitudinal top steel- S n5

Reference

Table 3.27

3.11.3.2

Table 3.27

Table 3.29

Output

Calculations

~fb~~ = (340 x106) 1 (2000)(747.5)2 = 0.30

100A/bd = 0.08; hence use lOOA/Ae = 0.13 As = (0.13)(800)(2000) 1 (100) = 2080 mm2 Use 5 TI5 (As = 2455 mm2) - on bottom surface; these .~ c.ould be curtailed as well. 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.

longitudinal rlf to be evenly spaced

crransverse direction) M = (25012){(2.0-0.3)/2}2(112) = 45 kNm/m d = 747.5 - 25/2 - 20/2 = 725 mm (assuming bar size of 20 mm) M/bd 2 = (45 x106) 1 (1000)(725)2 = 0.09 Use 100A/Ae = 0.13 As = (0.13)(1000)(800) 1 (100) = 1040 mm 2/m Use TIO @ 300 mm (Ag = 1047 mm 2/m) 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 5T25 \. , ·~T20@300 ........

s

..

,

1 '5T25

k

6.&n

,

>I

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 0.31 N/mm2; hence O.K. In transverse direction, a distance "d" from column face is virtually at edge of footing; hence O.K.

71

transverse bottom steel T20@3oo mm, to be evenly spaced.

J

Check for vertical line shear 3.4.5.10

longitudinal bottom steel 5TI5

Reference

Calculations

Output

Check for punCbio& shear 3.7.7.2

Notes

00

vmax (internal column) = (1100 xlol) I (4)(300)(735) = 1.25 N/mm 22 S < (0.8)(25f· = 4 N/mm 2 < 5 N/mm ; hence O.K. As the critical perimeter, (1.5)d from column face, is outside the footing, only vertical line shear need be checked.

Shear r/f is not required

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 a, 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 and the resulting bending moments and shear fol'CeS' at the column faces will ~, somewhat~maller 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.

area

Coocludina Notes 9.

If 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 been 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.

72

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 fy = 460 Nlmm . Introductory Notes

1.

The minimum centre-to-centre distance for piles is twice the least width of piles for end bearing piles and thrice the least width of piles for friction piles.

2.

A pile cap can be considered as a deep beam, and the most appropriate way to analyse forces is to consider truss action in the pile cap.

Reference

Calculations

Output

Pile cap dimensions

Note 3

Use an outstand beyond the piles of half the pile diameter. Hence, pile cap di~ensions are: length = 1250 + 500 + 500 = 2250 mm width = 500 + 500 = 1000 mm Try overall depth of 700 mm; hence, effective depth = 700 - 40 - 25/2 = 647.5 mm > (1250)/2; hence O.K.

dimensions 2250 mm x 1000 mm x 700mm

N

61

Truss action Note 4

The force T is given by T N.l/(2)d = (2500)(625) 1 (2)(647.5) T = 1207 leN N/2 N/2 As = (1207 x1oJ) 1 (0.87)(460) , 21 ~ = 3016 mm2 Use 7 1'25 (As = 3437 mm2) Banding is not required, as pile spacing < (3)

t

3.11.4.2 Note 5

l'

Anchoraee Anchorage length requirec;l beyond centre-line of pile is given by (40)(25)(3016/3437) = 878 mm

73

main steel 71'25

Reference

Calculations

Output

Stress in r/f - (0.87)(460)(3016/3437)= 351 N/mm2 When turning bars upwards, assume that bend starts ' at edge of pile. Stress at start of bend = (351)(878-250)/(878) = 251 N/mm 2 equation 50

Note 6 3.12.8.23

Bend radius, r, should be S.t. Fb!(r.q,) < = (2)fcu/{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 {1 + (25/81)} r > = 159; Use r = 160 mm

r = 160 mm

If we start the bend as close as possible to the edge of pile cap, length from C/L of pile to start of bend = SOO - 40 - 16 - 160 = 284 mm

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

vmax = (2500 xloJ) 1(4)(450)(647.5) = 2.15 N/mm2 < (0.8)(25)°·5_ 4 N/mm2 < 5 N/mm2 ; hence O.K. Since spacing of piles < (3)q" no further check is required. Check for vertical line shear

3.11.4.4 3.11.4.3

Table 3.9 3.4.5.8

av = 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/mm2 625 625 2d/av = (2)(647.5) I (250) = 5.18 looA/bd = (100)(3437) I (1000)(647.5) = 0.53 2 Vc = (0.51)(5.18) - 2.64 N/mm > 1.93 N/mm 2; hence O.K.

74

Shear rlf not required.

Output

Calculations

Reference

Distribution steel Table 3.27

l00A/Ac = 0.13; As = 910 mm2/m Use 16T @ 200 mm (As = 1005 mm 2/m) This steel can also be bent up like the main steel. Horizontal binders

Note 7

Il

7T25 main

Use 25 % of main steel.' As = (0.25)(3016) = 754 mm2 Use 4 T16 (As = 804 mm2) These binders will tie the main and distributiot:l steel.

distribution steel 16T@200 mm

binders 4')16

,

·LJTl~

horizontal binders 4 Tl6


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.

If the width of the column is accounted for, the value of tensile force will 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.

It 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.

If 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 25· 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/mm2 , fy = 460 N/mm 2 (deformed type 2) or 250 N/mm 2 (plain) and density of reinforced concrete = 24 kN/mJ . Assume top finishes (on tread only) as O.S kN/m 2 and the soffit plaster as 0.25 kN/m 2•

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.

Note 3

Then, rise (R) = 2000/12 = 167 mm going (0) = 3000/12 = 250 mm Also 2R + 0 = (2){167) + 250 = '84 mm (approx. 600 mm; hence O.K.) Use nosing of 25 mm, so that tread = 275 mm

rise = 167 mm going = 2SOmm tread

=

275 rom

Waist thickness Note 4 3.10.2.2 TABLE 1

Table 3.5

Assume trial span/depth of (30)(1.15) = 34.5, for a I-way heavily loaded continuous slab, stiffened by stairs. effective depth = (3000)/(34.5) = 87 mm If we assume cover =' 20 mm (mild exposure conditions and concrete protected by 10 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 Note:- 1.5 hr. fire resistance available. L.oadin& (for 1 m wide strip)

Note S

Factor for slope .... (R2 + 0 2)0.5 1 0 = {(167)2 + (250)2}0.5 1 (250)

76

= 1.20

h = 120mm d = 94 mm

Factor for overlap = T/G = 275/250

Note 6

Output

Calculations

Reference

= 1.1

Waist = (0.1~)(1)(24)(1.2) = 3.46 kN/m Steps = (112)(0.167)(1)(24)(1.1) = 2.20 kN/m = 0.55 kN/m Top finishes = (0.5)(1)(1.1) = 0.30 kNlm Soffit plaster = (0.25)(1)(1.20) = 6,51 kNlm Total dead load Imposed load = (5.0)(1) = 5.0 kN/m Design load ... (1.4)(6.51) + (1.6)(5.0) = 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 mm2/m Use Tl2 @ 250 mm (As = 452 mm2/m) Allowable spacing = (3)(94) = 282 mm > 250 mm 100A/Ac = (100)(452) I (1000)(120) = 0.38 > 0,13 Hence,max. spacing and min. steel are O.K.

main steel T12 @ 250 mm


Table 3.11 3.10.2.2

Mfbd2 = 1.75 fa = (5/8)(460)(451/452) = 287 N/mm 2 Hence, F 1 = 1.15 (for tension reinforcement) Allowable spanldepth = (26)(1.15)(1.15) = 34.4 Actual span/depth = (3000)/(94) = 31.9 < 34.4; hence deflection O. K.

Deflection O. K.

Check for shear Note 9

Table 3.9

v = (0.6)F = (0.6){(17.1)(3)) = 30.78 kN/m v = (30.78 x1 0.33 N/mm2 ; Hence, shear r/f not required.

77

shear r/f not

required.

Reference

Output

Calculations Distribution steel

Table 3.27 Note 10

lOOA,IAc = 0.24 (for mild steel)

A.

= (0.24)(1000)(120) I (100) = 288 mm2/m

Use R8 @ 175 mm

(As

distribution steel R8@·175 mm

= 287 mm 2/m)

R8@175

(4)

~

Note 11

Tl2@250 (1)

Tl2@250

3.j


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 be provided so that the tread 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 be greater by a factor of (R} 0 2)0.5 I O. The steps and tread finishes have 2S 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 5.0 kN/m2 - see "BS 6399: Part 1 (1984): Design loading for buildings: Code of practice for dead and imposed loads" .

7.

For a staircase having continuity, we can assume that both 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 face of the waist. Bar types (2) and (3) can be taken 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 sl$ reinforcement, if required. Bar type (4) shows how the upper slab reinforcement can be taken into the beam support.


If the supporting beam for the flight of stairs 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-------/ span

79

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/m2 and the soffit plaster one of 0.25 kN/m 2• The imposed load can betaken as 3 kN/m 2• Using feu = 25 N/mm2 , fy = 460 N/mm 2 (deformed type 2) or 250 N/mm2 (plain) and density of reinforced concrete ;:; ~4 kN/m3 ,design the staircase component of the system.

1.2m

\<:---.....;*

8xO.25

= 2.Om

2.Om

*:------~)'

."

1. 2m

,

,

1.2m I,


In this example of a staircase, the landings span perpendicular to the stairs and support the staircase, unlike in the case described in Note 12 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.

Calculations

Reference

Output

Stair span and waist thickness equation 47 Note 3

3.10.2.2

= la + (O.5)(lb,l + Ib,V = 2.0 + (0.5)(1.2 + 1.8) = 3.5 m length of stairs (%) = {(2.0)/(3.5)}(IOO) % = 57%

Effective span

< 60%; hence span/depth enhancement not possible.

80

eff. span 3.5 m

=

Reference Note 4

CaJcuJatioas

Outpat

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 bar diameter of 12 mm, we can choose h= 140 mm and d= 140-20-1212 = 114 mm

h = 140 mm d = 114 mm

Loa4in& (for 1.2 m wide staiR:ase) Note 5

= (R2 + OZ)0S I G = {(l75f + (250f}o.s I (250) = 1.22 Factor for overlap = TIG = (275)/(250) = 1.1

Factor for slope

Waist = (0.14)(1.2)(24)(1.22) = 4.92 kN/m Steps = (112)(0.175)(1.2)(24)(1.1)= 2.n kN/m Top finishes (0.5)(1.2)(1.1) = 0.66 kN/m Soffit pJasta' (0.25)(1.2)(1.22) 0.37 kNlm Total dead loed = 8.72 kN/m Imposed lOad = (3.0)(1.2) = 3.6 kN/m Design load= (1.4)(8.72) + (1.6)(3.6) = 18.0 kN/m

= =

=


Desi&n for bendine Note 6

The stain:ase can be idealised as follows:

@; RA = (18)(2.0)(1.9) I (3.5) =

19.5 kN

= (18)(2.0) - 19.5 = 16.5 leN Mx = RA.x - w(x - 0.6f12 dMxIdx = 0 when RA - w[x - 0.6] = 0 RB

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 (Part 3)

MIbc¥

= (22.26 xl
c

Use 5 Tl2 (A. = 565 m.r)

81

= 1.43 main steel 5 T12

CaJcu,,-

Wei.,...

Output

Check for cW1ection MJbd2.:: 1.43;

fs

Table 3.11

= (5/8)(460)(534/565) = 2n N/mm2

H~,FJ

== 1.29 (for tension reinforceament)

:: (26)(1.29) == 33.5 Actual spaoIdeptb == (3500)1{U4) == 30.7 < 33.5; hence O.K.

~. spanldepth

Deflection O. K.

Distribution reinfon:ement Table 3.27

lOOAJA.c

= 0.24 (mild steel)

As =(0.24)(140)(1000) I (100) = 336 mm2/m Use R8 @ lSOmm
OD

distribution steel R8@ ISO mm

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

be·cJooe in a IIl3DDel'sinillar to that in Example 25.

7.

Detailing of reinforcement can

8.

Shear can also be checJr£d for, as in Example 25, the maximum.sIlear force being the greater of RA and~ - 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. RA and RB will be uniformly spread over the entire smaller landing and over 1. 8 m of the Jarges landing, respectively.

10.

Where staircase flights surrounding openwe1ls intersect at right angles, the loads from the common landing can be shared between the two perpendicular spans.

82

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 floo~ slab (clear height of wall 4.0 m) if 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 = 8OleN/m; imposed load from 1st floor slab = 7.5 kN/m; imposed load .from above 1st floor slab = 19 kNlm.

=


Given that.even plain concrete walls require horizontal and vertical reinforcement (Clause 3.9.4.19), and if 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 20 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.

It 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 Note 4

Note 5

1.2.4.9 3.9.4.4

Since lateral support is provided by foundation and 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.

Ie = 3.5 m Hence, Ie = (0.875)(4.0) = 3.5 m lJh = 20 Ie/h = (3.5 xloJ)/(17S) = 20 > 15; hence slender. hence, slender < 30; hence max. value not exceeded. braced wall.

83

1./;

l

i (

We sball use serviceability vertic:al Joads to calculate the resultant eccentricity (e) just below the 1st floor

3.9.4.9 Note 6

3.9.4.15 3.9.4.16

slab, assuming the eccentricity of 1st floor slab loading is bl6. and that the eccentricity of loads above this is zero. e== (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 ea == Ie2/(2S00)h == (3500)2 I (2SOO)(175) == 28 mm

.. ex == 8.75 mm e. == 28 mm

Assuming wind acts at mid height of wall, 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:-

Table 2.1

Combination 1. f l == (1.4)(18+ 12+80) + (1.6)(7.5+ 19)(0.7) == 184 kN/m Combination 2. f2 == (1.4)(110) + (1.4)(33.6) == 201 kNlm or f 2 == (1.0)(110) - (1.4)(33.6) == 63 kN/m Combination 3, f3 (1.2){1l0 + (26.5)(0.7) + 33.6} 195 kNlm Note: - no tension arises.

=

=

l1w.~ == ~OI kN/m n .:;: -"W.1IlUI

63 kN/m equation 43

equation 44 Note 7

Now,1lw < == (0.3){h - (2)exlfcu = (0.3){175 - (2)(8.75)}(25) == 1182 kN/m and Ow < == (O.3){h - (1. 2lex - (2)eJfcu =(O.3){175 - (1.2)(8.75) - (2)(28)}(25)== 813 kN/m These are satisfied, since l1w.max == 201kN/m Check for

3.9.4.18

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

l1w is O.K.

Caleulatio..

Refereace

Minimum reinfOfCQDCOt

4"

Min. r/f Note 8

= (0.25)(1000)(175) I (100) = 437.5 mm2/m (both directions)

Use vertical steel Tl2 @ 300 mm in both faces

(A. = 753 mm2/m)

Horizontal steel 1'8 @ 225 mm in both faces (As = 446 mm2/m)

- -

Output

vertfcal steel 2 x Tl2 @ 300 horizontal steel 2xT8@225


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. If 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 Ow 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 OW is calculated at .the bottom 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 reinforcement sbould 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 350 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 N/mm 2 and f y = 460 N/mm2 (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».

Calc:ulatioos

Output

Corbel dimensions

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 if dry bearing on concrete is assumed, the width of thebeariag plate, b, will be given by (350 xloJ) I (3OO)b <= (().4)fcu b >= (350 xloJ) I (300)(0.4)(30)= 97 nun Hence, choose bearing width of 100 mm. Since the corbel has to project out from the bearing area a distance that wouJd accommodate a. stressed bend radius choose corbel projection as 400 mm. Corbel depth bas lObe ·suclltbat max. aIL shear is not exceeded - i.e. (O;8)(3Of.5 = 4.38 N/mm2 Hence, d > (350 xl(3) I (300)(4.38) = 266mm Choose h = 375 nun and assuming cover of 20 mm (mild exposure conditions, concrete protected by lQ mm 1:3 et: sand rendel) and bar dia. of 20 mm, d = 375 - 20 - 2012 = 345 mm. Let the depth. vary from 375 mm to 250 mm. cAy

~l~n:

350 kN (ultimate)

'l5o L 1:- t

T~C'

T

125 Oo9xl[J 0.45f

-A

v

86

cu

bearing width 100 mm

total proj~tion 400nun

h =375 mm d::;: 345 mm

Reference

5.2.7.1

Output

Calculations

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 and stress block, C == (0.45)fcu(0.9)b.x.CosP .......... (1) Since the line of action of C must pass 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 1 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 1 Tan{J == 212 kN

Note 4 5.2.7.2.1

.

NoteS

Since x == 216 mm, by strain compatibility, strain in steel is {(d-x)/x}(0.0035) == 2.090 xlO-] Hence, steel has just ~ded and fs == (0.87)fy Hence, As == (212 d()3) 1 (0.87)(460) == 530 mm2 Use 3 T16 (As == 603 mm2) . Min. area required == (112)(350 xloJ) / (0.87)(460) == 437 mm2 < 603 mm 2 ; hence O.K. Also l00A/bd == (100)(603) 1 (300)(345) == 0.58 > 0.4 and < 1.3; hence O.K.

main steel 3 T16

Detailing O.K.

Shear reinforeeroent

Table 3.9 3.4.5.8 Table 3.8

5.2.7.2.3

v == (350 xloJ) 1 (300)(345) == 3.38 N/mm2 l00AJbd == 0.58 Vc == (0.546)(30/25)°.J3(2d1ay) == (0.58)(2/0.43) == 2.69 N/mm1 < 3.38 N/mm 2 Provide An > == bv.Sy(v-vJ 1 (0.87)£ v Aav!Sy > == (300)(3.38-2.69) 1 (0.87J(460) == 0.517 Use lOT @ 300 mm. Since this has to be provided over (213)(375) == 250 mm, 2 bars will suffIce. Min. requirement is 603/2 == 302 mm2 Use 2 TI0 links @ 175 mm (As == 314 mm2 > 302 mm2; hence O.K.)

87

links

2TlO

Reference

Bendine main reinforcement Note 6

3.12.8.25.2

equation 50

The bend in the main reinforcement should start a cover distance (20 mm) from the bearing plate. It should end a cover + bar di.a. (20+ 10+ 16 = 46 mm) from the end of the corbel. Hence. distance available for bend radius == 200 - 20 - 46 = 134 mm Critical value of 3tJ= 20 + 10 (link) + 16 = 46 mm Stress in bars = (0.87)(460)(5301603) = 352 N/mm 2 F bt I (vI» < = (2)~ I {I + (2)(qJ~} (352)(201) I (16)r < = (2)(30) I {I + (2)(l6l46)}

r > = 125 m m b e n d radius Choose r 130 mm < 134 mm; hence O.K. 130 mm

=

=

100· 100.. 200

.1

r-l30,., 250

2TIOfH75(


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 than half the depth at the root. Furthermore, it facilitates the placing of horizontal shear 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 x 10-3 for steel of fy '= 460 N/mm 2 , since the Young's Modulus specified is 200 kN/mm2 .

5.

Although these limits on l00A/bd, where d is the effective depth at th~ 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 bend to start at the edge of the bearing plate itself, the allowance of a cover distance from the outer edge of the bearing plate will ensure the spreading of load from the bearing 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 tie bars is to weld a transverse bar of equal strength, subject to the detailing rules in Clause 5.2.7.2.2. 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.

89

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/m2 and an imposed load of 0.5 kN/m 2 • It 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 N/mm2 (deformed type 2), fyv = 250 N/mm2 and density of reinforced concrete = 24 kN/m3 •

IDtroductory Notes

1.

It 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 (at right angles to the element carrying torsion), because torsional stiffnesses are lower than bending stiffnesses. Any 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. In 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.

CaleuIatiODS

Reference

~oILJ

2000

Output

~$OO )

~.

Loadin 2 on beam

= {(0.2+0.1)/2}(2.0)(24) = 7.2 kN/m HoOd = (0.5)(2.3) Finishes = 1.15 kN/m Self weight = (0.6)(0.3)(24) = 4.32 kN/m Total dead load = 12.67 kN/m Imposed load = (0.5)(2.3) = 1.15 kN/m Design load ={(1.4)(l2. 7)+(1.6)(1.15)} = 19.6 kN/m

90

bending udl 19.6 kN/m

=

Calculations

Reference

Output

Torsional loadio& (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 = 1,15 kNm/m Finishes == (0.5)(2.0)(0.15+ 1.0) Total dead load torsion == 8,63 kNm/m Imposed load torsion = (0.5)(2.0)(0,15+ 1.0) == 1,15 kNm/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

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.l2 (for built in beam) == (1112)(19.6)(4)2 == 26.1 IeNm Mlbd 2 = (26.1 xlo6) I (300)(550f = 0,29 lOOA,lbd == 0.08 Use lOOAJAc == 0,13 As = (0.13)(300)(600) I (loo) = 234 mm2 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 < Vc

d = 550 mm

Desi&D for torsion y I = 600 - (2)(30 + 10/2) = 530 mm

YI

= 530 mm

= 300 - (2)(30 + 10/2) = 230 mm

Xl

== 230 mm

Xl

Note 3

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 ~33

I

1.2m

--..~

"

~

-27.8

91

Refereace

CaJaJlatioos

equation 2 (Part 2)

Max. value of vt = (2)T I (bmiJi~ - huuJ3) = (2)(27.8 xld') I (300) (600 - 300/3) = 1.24 N/mm2 < 4.38 N/mm 2 (vtu ) > 0.37 N/mm2 (vt,min) Thus, beam section is O.K. but requires torsional r/f.

Table 2.3 (Part 2)

Output

Proyision of reinforcement Table 2.4 (part 2)

Since v < Vc for the entire beam, the area where vt < = vt,min bas to be provided with nominal shear r/f and the area where vt > vt.min with designed torsion r/f.

equation 2 (Part 2) Table 2.3 (Part 2)

Torque corresponding to edge of nominal shear r/f is given by T = V t ,min.hmm29tmax - ohznm/3) I 2 . = (0.37)(300f(600 - 30013) I 2 Nmm = 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

Table 3.8

A,;.)Sv > = (0.4)(300) I (0.87)(250) = 0.55 For 10 mm links, Asv = 157 mm2 ; Sv < = 285 mm

Nominal shear links given by

Use RIO links @ 250 mm {< (0.75)d = 413 mm}

Nominal links RIO@250mm (middle 1.2 m)

Designed torsional links given by 2.4.7 (Part 2)

2.4.8 (Part 2)

A,;.)Sv > = T I

(0.8)~'Yl(0.87)f

= (27.8 xl

0

) I (0.8)(130)(530)(0.87)(250)

= 1.31 For 10 mm links, Asv = 157 mm2; Sv < = 120 mm Use 2RI0 links @ 200 mm {< = 200 mm, Xl' yl /2} Length of beam at each end for torsional links = (4.0 - 1.2) I 2 = 1.4 m Designed additional longitudinal steel given by

2.4.7 (Part 2)

2.4.9 (Part 2)

Note 4

As! > (A,;.)Sv)(fyvlf )(xl +Yl) = (157/120)(2501460)(230 +

530) = 540 mm 2 If this is divided between 8 bars, each requires 67.5 mm2 (3 at top and bottom, 2 in middle). Since beam length is small, assume bending reinforcement is not curtailed; longitudinal reinforcement for torsion also cannot be curtailed.

92

Torsion links 2RIO@2oo mm (1.4 m from both ends)

Reference

Table 3.30

2.4.9 (part 2)

Output

Calculations Total steel requirement at top and bottom levels = (67.5)(3) + 234 = 436.5 mm 2 Use 2Y16 + YIO at top and bottom levels (As = 481 mm2) and 2 Y 10 at intennediate level (As = 157 mm 2) This arrangement will satisfy (a) max. spacing for tension rlf < = 160 mm (b) max. spacing for torsional rlf < = 300 mm (c) torsional rlf provided in 4 comers

top & bottom 2Y.l6+YIO middle 2YlO


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 equal to the largest dimension of the section beyond the point where it is theoretically not required. In 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 2.4.4.2Part 2). The torque is divided up amoung the rectangles in the ratios of their (hmin3 .h max) values and each rectangle designed for torsion. The torsional links should be placed such that they do intersect.

division into 2 rectangles

IC

l-

J ~intersecting torsional II

93

links

EXAMPLE 30 - 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) - 600 mm x 300 mm (iii) Columns (for all floors) - 300 mm x 300 mm The vertical loading is as follows:. (i) Load corresponding to finishes = 0.5 kN/m 2 (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 kN/m 2 (iv) Imposed load on floors = 2.5 kN/m2 (v) Density of reinforced concrete = 24 kN/m3 Obtain the design ultimate moments and shear forces from vertical loads for the beam ABC at the first floor level. Roof j 4.0m 1\~

2nd FIoar

4. Om B

A

C

\

1st FIoar

4.2f.m

, T75m. ~". ~

6.0m

7- ~

6.0m

-.

?!~ - - - -

Ground Level Footing Level

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 loads~

3.

In 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

Output

Calculations

Refereoce Stiffnesses

(IlL) of columns above 1st floor

=

(1112)(300)4/ (4000) = 0.169 xl()6 mm3 (Ill,tof columns below 1st floor = (1/12)(300)4/ (5000) = 0.135 x106 mm3 Since T-beam action will prevail in the beam , eff. flange width = 300 + (0.7)(6000)/5 = 1140 mm « 5000 mm). I of beam section = 1140 9.388 xl09 mm4 , (I/L)ofbeams = = (9.388 x109) I (6000) 1.565 xl~ mm3

b f = 1140 mm

°1¥

15 4 so

)

~ Distribution factors Only the beam factors will be considered. DAB = D CB =(1.565)/(1.565 +0. 169+0. 135) = 0.84 DBA =DBC = 1.565/{(1.565)(2)+0.169+0.135} =0.46

Loading on beam = (5)(0.15)(24) = 18 kN/m = (0.45)(0.3)(24) = 3.24 kN/m Finishes = (0.5)(5) = 2.5 kN/m Total dead load = 23.74 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 Since a beam span carries 30 m2 of floor area, reduced imposed load = (0.97)(17.5) = 17.0 kN/m Slab

Beam

BS 6399: Part 1

Load arrangements 3.2.1.2.2

95

ik= 23.7 kN/m

Qk= 17.0 kN/m

Output Arrangement 3 will be the mirror image, about B, of Arrangement 2.

Moment distribution (kNm) Note 5 (Arrangement 1)

0.84

Note 6

0.46

AB -181.2 + 152.2.....,.

- 29,0

0.46

0.84

BA Be CB + 181.2 -181.2 + 181.2 + 76,1 - 76,1
Arrangement 1 Support moment at B= 257 kNm

(Arrangement 2)

0.84

0.46

AB -181.2

0.46

BA BC + 181.2 - 11.1

- 25.3 ~ ...... - 5"'-lo0"'"'.6"--_-......50Qt~,6

+ 173.5 -+ + 86,8 - 15.5 +- - 31.1

+ 13.1-+ + 6.5 - 3.0 -t- - 6.0 + Note 7

+ 1.3 s- - 1.2 + 186.9

2.5 - t

- 0,6 - 36.5

Shear forces

0.84 CB

+ 71.1 -+ - 25,3

- 19.3
Arrangement 2

- 6,0 -+ - 3.0 + 1.3 ~ + 2.5 - 1.2 ~ - 0.6 -171.5 + 3.8

Support moments at B = 187 kNm &. 172 kNm

lkN)

The shear forces RA, RBI' RB2 and from the following figures:-

Rc can be found

Arrangement 1

143.2

219.3

219.3

143.2

Arrangement 2

156.1

206.3

99.1

43.2

%

Output

Calculations

Reference Span moments

~x

/w 0

Note 8

Free bending moment is given by -(w.l.x)/2 + {w.x2)12 Fixed end moment variation is given by M 1 + (M2-Ml)xll

~

f

,M1 1

Hence, the points of contraflexure an4the points and values of max. span moments can be obtained. Span

AB

Span Be

Arran~t 1

Points of contraflexure (from A) Max. sagging moment

Note 9

Arran&eJDent 2 Points ~f contraflexure (from A) Max. sagging moment

O.21m, 4.52m 141 kNm at 2.37 m

O.25m, 4.92m 165 kNm at 2.58 m

7.48m, 11.79m 141 kNm at 9.63 m

8.45m, 11.90m 35 kNm at 10.18 m

257.3

165 141 Bending Moment Diagram (kNm) 219.3

Note 10

Shear Force Diagram (kN)

97

Span moments:Arrangement 1 141 kNm (both spans) Arrangement 2 165 kNm and 35 kNm


s.

The distribution' factorS have accOunted· for the column stiffness, but the column moments have been left out of the .calculations for conven~ce, as we are interested only in'the beam. mOments. Since the. remOte ejlds of the columns are assumed to be fixed, there will be no Carty over riidmentsfrom them to the beam-eQlumn joints. The sign convention adopted is that clockwise moments are positive and anticlockwise moments negative.

6.

In this symmetrical loading arrangement, the calculation is complete with just one joint release.

7.

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. .

.

8.

The sign convention adopted in this part of the solution is that sagging moments are negative and ho"ing moments posi~ve.

9.

'This two-~ frame is typical of most situations, where the maximum support moments are' obtairied when all spans are loaded with the· maximum design ultimate loads (1.4 gk + 1.6 qk) and the maximum span moments are obtained 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 been shown, for the sake of clarity, since they will be mirror images of those for Load Arrangement 2 about B.

Concluding Notes II.

The beam moments could have. been obtained using a continuous beam 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 equal bays in the frame and the characteristic imposed load does not exceed the characteristic dead load, the beam moments and shear forces can be obtained from Table 3.6 for a continuous beam analysis (see Cla~ 3.4.3).

98

EXAMPLE 31 - FRAME ANALYSIS FOR HORIZONTAL WADS If 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.


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 to an indeterminate one as in the case of vertical load analysis.

Reference

Calculations

Output

Wind force CP3:Ch.V: Part 2 Note 3

Table 10 of CP3:Ch.V: Part 2

Basic wind speed, Vb = 40 mls Sl = 1.0; SJ = 1.0 S2 (for ground roughness 3, building class B and H = 12.25) is 0.7805 Vs = (1)(0.7805)(1)(40) = 31.22 mls Wind pressure, q = (0.613)(31.22)2 = 597.5 N/m2 I1w = (40)/(12) = 3.33 bid = (40)/(12) = 3.33 h/b = (12.25)/(12) = 1.02 Hence, Cf = 1.23 Force on one frame = q.Cf.A e = (597.5)(1.23){(12.25)(5.0)) = 45014 N = 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

Wind force on a frame = 45 kN

Output

Calculations

Reference

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.

Note 4

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

~

. I · lJ r . ~:~ t e

7.35-+

Note 5

Roof

+- 3.67

1.84

1.225

if225

225

1.225

'11.84

3.67 ...

.J..

1.84-.. c 'l

i)

2.llil 2nd Fl r . 2.llil ~ 5.51

?~

14. 7~ t--"'-4~---+--.....----1 c!>~

11.03

t

l' o

6.125

6.125 11.03~)

Note 6

16.5:+

~tl2.Qn

...---..---+----+---t ~<4:-

9.64'

~ 17.83

Note 7

5.51

~ ~

't

19.28

Is t Fl r 2.fIn '9.64

,.

t

17.83

,

Moments and shear forces in ABC The moments in· ABC can be found as those required to balance the ~lumn moments.

"'35'1'.~·~ .. 35.f . . . . A "J .. • C . '.B. .

. .'

..

35.'f

100

35.1

Moment at A, B and C is ·,~5..1

kNm

Calculations

Reference

The shear forces in the spans are obtained by dividing the moment by half the span length. Hence, shear force = (35.1)/(3.0) = 11.7 leN

Output Shear force in AB and BC is 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 4 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.

If 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 leN « 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 k (causing the overturning moment) and 1.0 G k (providing the restoring moment). 101

EXAMPLE 32 - REDISTRIBUTION OF MOMENTS Determine the design ultimate moments for the beam ABC in Example 30, after carrying out moment redistribution.


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).

Calculations

Reference

Output

Support moments

Example 30

3.2.2.1 Note 3 Note 4

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

Note 5

AB

BA

BC

CB

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

support moment at B = 180.1 kNm "

Output

Calculations

Reference Arraneement 1 Points of contraflexure (from A)

0.19m, 4.97m

7.03m, 11.81m span moments:-

Max. sagging moment Arraneement 2 Points of contraflexure (from A)

Max. sagging moment

173 kNm at 2.58 m

0.24m,4.97m

168 kNm at 2.60 m

173 kNm at 9.42 m

8.57m, 11.91m

33 kNm at 10.24 m

Arrangement 1 173 kNm (both spans) Arrange':Dent 2 168 kNm (span AB) 33 kNm (span BC)


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/d ratio (see Clause 3.2.2.1).

4.

In general, the x/d 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 span 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 'of 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 kN/m 2 each. Design the stability ties for this structure with steel of fy = 460 N/mm 2 •..

ChlumS"( Beans

,

~-

~

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 characteris~c value -Le. 'Ym = 1.0. Furthermore, reinforcement designed for other p~ can be used as ties (Clause 3.12.3.2).

Calculations

Reference 3.12.3.7

Output

V ertical ties

TheSe are required, since no. of storeys > 5. Area corresponding to ~ typical column = '. . 1 ' (1(2)(6.0)(5.0) = 15.0 m2 MaX. design ultiIpate load = (15.0){(1.4)(5.0) + (1.6)(5.0)} = 225 leN Area of ties required = (225 xl
104

vertical ties A, = 489 mm 2

Output

Calculations

Reference 3.12.3.5

Peripheral ties

3.12.3.4.2

F t = 20 + (4)(6) = 44 leN « 60 leN) Area of ties required = (44 xloJ) I (460) = 96 mm2 This can be easily met by peripheral beam r/f that is continuous.

3.12.3.4

F t = 44 leN peripheral tie A = 96 mm 2

•

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) = 770 mm 2 If carried in the two peripheral beams, area required 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 leN/m} Total force = (70.4)(20) = 1408 kN Area of ties required = (1408 xl
Note 4

longitudinal internal tie A = 770 mm 2

Peripheral beams Total tie area per beam in longitudinal direction = 96 + 335 = 431 mm 2 Total tie area per beam in transverse direction = 96 + 612 = 708 mm 2

105

transverse internal tie A == 3061 mm 2

•

peripheral beams (tie rlf) longitudinal 431 mm2 transverse 708 mm 2

Calculations

Reference 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 Area of tie required = (61.6 x HP)/(460) = 134 mm2

Since thi~ is less than the ties in the beams, part of the latter can be taken into the columns.

Output

column ties A5 = 134 mm 2


Although the beam reinforcement may be greater than these tie areas required, it must be 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 be five floor slabs and the roof above the level of the frrst floor column tie;· using the floor loading for the roof as ~ll is a conservativeapproxjmation.

Conduding Notes 6.

If a structure has key elements (Le. those that carry, say, over 70 m2 or over 15% of floor area at a given level), they have to be 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.

106

EXAMPLE 34 - CRACK

wmm CALCULAnON

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. (

450

>f cu = 25 N/mm2

1

f y ;"'460 N/mm 2

690

750

,

3T25 000

Es

= 200 kN/mm 2



This crack width calculation can be performed when the bar spacing rules are not satisfied, to see whether this more accurate method will satisfy the crack width requirements in Clause 3.2.4 of Part 2. It can also be used to estimate the actual crack width in a flexural element.

Reference

Output

Calculations Sectional data

I

= 24S kNm

Note 2

Ms = (20+20)(7) 2 18

equation 17

E c = 20 + (0.2)(25) = 25 kN/mm 2 Eeff = (0.5)Ee · = 12.5 kN/mm 2 CX e = E/Eeff =' (200 xloJ) 1 (12.5 xHf) = 16 P = (3)(491) 1 (690)(450) = 0.00474 x/d = -cxe.p + {cxe.p (2 + cxe.p)}o.S = -(16)(0.00474) + [(16)(0.00474){2+(16)(0.00474)}]O.S = 0.321 x = 221 mm ~/bd3 = (1/3)(x1d)3 + a.,..p {I - (x/d)}2 = (1/3)(0.321)3 + (16)(0.00474)(1-Q.321Y = 0.046 Ie = 6.80 x10 9 mm4

3.8.3 (Part 2) Note 3 Note 4

Note 5

107

M s = 245 kNm

x = 221 mm

Ie

= 6.80 xl09 mm~

Calculations

Reference

Output

Calculation of strains Strain in steel .

equation 13

(part 2) equation 13

(12.5 xl
< (0.8)(460) / (200 xl
3.8.3 (Part 2) Note 6

= (245 x1(6)(690-221) I

= 1.35 dO·3 = 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)Es .A.(d-x) = (450)(750-221)2 I (3)(200 xI
(part 2) At 250 mm below neutral axis, = (1.35 xlO'3)(250)/(690-221) = 0.72 xlO·3 tension stiffening = (450)(750-221)(250) I (3)(200 xl
Distances to potential crack points cmin

~

= 750-690-(2512) = 47.5 mm

+ (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 equation 12

(part 2) 3.2.4 (part 2)

CW t

= (3)(89.5)(1.219 dO·3) / [I

+

{(2)(89.5·47.5)/(750-221)}]

= 0.282 mm < 0.3 mm; satisfactory. CW2 = (3)(72.4)(1.219 dO·3) / [1

+

{(2)(72.4-47.5)/(750-221)}] < 0.3 mm; satisfactory. CW3 = (3)(214.5)(0.576 xlO·3) / [1 + {(2)(214.5-47.5)/(750-221)}] = 0.227 mm < 0.3 mm; satisfactory. = 0.242 mm

108

CW t

= 0.282 mm

CW2

= 0.242, mm

CW3

= 0.227 mm

Calculations

Reference ...

3.12.11.2.3 Table 3.30 3.12.11.2.5 Table 3.30

Comparison with

Output

bar macine rulCfS

Spacing between bars = {45G-(2)(47.5)-(3)(25)} 12 = 140 mm < 160 mm; satisfactOry. Comer distance = {(60)2 + (60)2}O.s - 12.5 = 72.4 mm < (16012) mm; satisfactOry.

max. spacing

O.K. comer distance

O.K.


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 Ie and 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 0.3 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 beam depth from the tension face must also be provided;~as per Clause 3.12.11.2.6.

109

EXAMPLE 35 • DEFLECTION CALCULATION ~e 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

o

0

I

125

fcu

= 25 N/mm2

fy

= 460 N/mm2

Bs = 200 kN/mm2



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

Chart 2 (Part 3) equation 8 equation 7 Table 3.10

Mult = (5)(1.4 + 1.6)(7)2/8 = 92 kNm M1bd2 = (92 xl 18; Hence, span/depth check is violated.

110

span/depth check violated

Output

Calculations

Reference

Data for serviceability calcuIiJions equation 17 (Part 2) 7.3 (Part 2) Figure 7.1 (part 2) Figure 7.2 (part 2)

Ee = 20 + (0.2)(25) = 25 kN/~B.\2 Eff. section thickness =(2)(375)(225) I (2)(375 + 225) = 141 mm RH = 85 % (assumed for Sri Lanka) Long term creep coefficient, 4> = 1.8 Eeff = 25 I (1 + 1.8) = 8.93 kN/mm2 fes = 120 x10-6 P = 982 I (225)(325) = 0.0134 a e = Es I Eeff = 200 I 8.93 = 22.4 x/d

= - ae.p + {ae.p(2 + = - (22.4)(0.0134) +

= 0.53 Hence, x = (0.53)(325) = 172 mm Ie/bd 3

3.6 (Part 2)

= (1I3)(x/d)3

Ie = (0.116)(225)(325Y = 896 xH? mm4 Ss = As(d-x) = (982)(325-172) = 150.2 x103 mm3

M tot = (5+5)(7)2 18 = 61.25 kNm ~rm = {5 + (0.25)(5)}(7)2 I 8 = 38.28 kNm ~rm(red) = M - {(1I3)b(h-x)3.fet I (d-x))

= (3rrm .28 x1oti) -

{(113)(225)(375-172)3(0.55) I (325-172)}

= 36 x106 Nmm Calculation of curvatures lIrlp

= M I Eeff.Ie

= (36 x1(j6) I

(8.93 x103)(896 xl~)"

= 4.5 xlO-6 mm- l equation 9 (part 2)

1Ires

= fes·ae·S s I

= 22.4

xld = 0.53 x = 172 mm

+ a e.p{l - (x/d)}2 (22.4)(0.0134)(1 - 0.53)2

= (113)(0.53)3 + = 0.116

Determination of serviceability moments

Note 5

ae

a e.p)}O.5

[(22.4)(0.0134){2 + (22.4)(0.0134)}f.5

Note 4

1\:=25 kN/mm2

Ie

= (120 xlO-6)(22.4)(150.2 xloJ) I (896 x106) = 0.45 xlO-6 mm- l 111

Ie = 896 x1()6 mm4

7'· .. . ,

··0ulJwt··

To find instantaneous Be =25 kN/mm 2 . . . a = 200 I 25 =8 •.... . .' = 0.368 '.." Ie = 459 xl()6 lIrit -llri = (Mtot - M.rm> I Ec·le ~ (61.25-38.2$)X1()6 I (2~ xl(3)(459 xl~ = 2.01 xl '1= (4.5 + 0.45 + 2.01) x10~ mm- I = 6.96 xlO~ mm-I lIr= 6.96 xlO~

Jd

Note 6

mm- I

Estimation of det'ls;gion Table 3.1

K =0.104 K.l~lIr) = (0.104)(7~6.96xIO~ == 35.5 mm

a = 35.5 mm

~

all > 11250

a

(Part 2)

=

3.2.1.1 (Part 2) Note 7

all = 35.5 I 7000 = 11197 .•. . > 11250


4.

The serviceability calculations are ~on a triangular stress block for concrete in the elastic state. There \s n., resttietionon. the x1d ratio, as in ultimate limit state calculations.

5.

The Mpermvalue is ~ced because of the tension stiffening contributed by the concrete, which is asSUfed to have no ~strength in the ~Jlation of Ie. The longterm·tensiIe·strencth ofeoncrete is taken as 0.55 N/m,m2 and·the short term strength as 1 N/mm 2 . ,

6.

The tension stiffening .ffects in

7.

The final long term d~tlection is greater than spanl250. Hence, it may be visually unacceptable. "

Moot and ~rm cancel out here.

Concluding Note

This caltulation ~ is· tedious, aDd is geuerally adopted only to explore the . possibi1i9' of using a:j,~ orJ~.~even·after the spanldepthcheck fails.

8.

.,:

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'$,<.; -';,'.

~,,:,.(-

;~'\',

"

i· . . ~

' __,-

'1l:~~."."

Prof. Dias Graded Examples in Reinforced Concrete Design Dias

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