Graded Examples in Reinforced Concrete Design Dias

REINFORCED CONCRETE DESIGN

W.P.s. DiasBSc{Eng), PhD(Lond),DIC, pEng, MIStructE, MIE(SL)Senior LecturerDepartment of Civil EngineeringUniversity of MoratuwaMoratuwaSri Lanka

/

Society of Structural Engineers - Sri Lanka

.,. "'~ .... _.._.J

PublisheJI. by

Soc. of Structural Engineers - Sri Lanka,ji 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 learningbecause there is an element of "spoonfeeding" involved. While acknowledging that there issome truth in this argument, the author would like to contend that Worked Examples do havea place in the educational process.

Knowledge can be acquired using two broad approaches - i.e. the deductive approach, havingits roots in Greek rationality, and the inductive approach, having its roots in Renaissanceempiricism. Learning through worked examples is an inductive approach, and both theformat and content of this book reflect that approach.

The book has been developed through the author's teaching of a course in ReinforcedConcrete Design at the University of Moratuwa. The examples are graded, leading from anappreciation of reinforced concrete behaviour, through the design of structural elements, tothe analysis of a reinforced concrete structure. The student's understanding of the calculationsis deepened by the "Notes on Calculations" while the Introductory and Concluding Notes seteach example in a wider context. Hence, in this book, design principles·are reinforcedthrough practice, with guidance from notes.

However, this book caimot and should not be used as a "stand alone" text. It must essentiallybe complementary to another text or series of lectures that teaches design from a deductiveapproach - i.e. one .which moves students from principles to practice. It can, of course, beused 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, withextensive references given in the calculations to clauses in BS 8110 (1985) - "Strueturaluseof concrete". This is another reason for the book's usefulness for Practising engineers. Theexamples cover most of the reinforced concrete elements and stress states dealt with by PartI of BS 8110. In addition, examples are also given for the de3ign for torsion and thecalculation 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 BS8110: Part 1. Where clauses, tables, charts or equations from Parts 2 and 3 of BS 8110 arereferenced, the relevant Part is also indicated. One very useful feature of BS 8110 is thateach table also gives the equation from which its values· are derived. .This is a clearadvantage for computerised design, and even hand calculations. Therefore, although thetables have in fact been referred to in the following calculations, very often it is thecorresponding 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 alsoindicates cement contents and water/cement ratios correspondingro the grade specified. Thebackground 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), whichdescribes how U.K. concrete strengths vary for given cement contents and water/cementratios and shows how the grade specified covers the cement content and water/cement ratiorequirements 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 parameterto measure and control. This is clearly evident in the provisions made in the code forreducing the grade if a checking regime establishes that a lower grade of concrete complieswith the cement content and water/cement ratio limits (Clauses 3.3.5.2 and 3.3.5.3 of Part1). 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 40concrete will have to be used for all but mild and moderate exposure conditions, althoughthe corresponding minimum cement content and maximum water/cement ratio are only325kg/m3 and 0.55 respectively. This seems to be a very stringent condition to be imposedon concreting practice in developing countries, where most concrete specified is still grade20 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. isapplicable in other (especially developing) countries, where materials and practices may bevery 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 ratiovalues 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 theauthor'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 scopeof the specifications, and also conditions under which deviations from the tabulated valuescan be allowed. In particular, Notes 5 and 6 allow reductions in grade and cover values thatbring these recommendations in line with current Sri Lankan practice. In short, theserecommendations rationalise satisfactory Sri Lankan practice (especially under mild exposureconditions) with respect to BS 8110, while suggesting improvements to Sri Lankan practicewhere 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 formild exposure conditions, it was felt that basing the examples on such a low grade wouldhave deviated too much from the provisions of BS 8110, where grade 25 is specified as thelowest grade to be used with normal weight aggregate concrete (Clause 3.1.7.2) and whereall tables and charts have grade 25 as the lowest grade. As such, it is grade 25 concrete thatis used for all the following examples, except in Examples 28 and 29, where the use of grade30 concrete is illustrated.

2

TABLE 1 - NOMINAL COVER TO ALL REINFORCEMENT (INCLUDINGLINKS) TO MEET DURABILITY REQUIREMENTS - ADAPTED FROMBS 8110: 1985 FOR SRI LANKAN PRACTICE

Exposure Examples of Nominal CoverClassification Exposure mm mm mm mm mm

Mild Indoor 25 20 20* 20* 20*Moderate Outdoor -- 35 30 25 20Severe Driving Rain -- -- 40 30 25Very severe Sea Spray -- -- 50 40 30Extreme Abrasive -- -- -- 60 50

Maximum free water/cement ratio . 0.65 0.60 0.55 0.50 0.45

Minimum cement content (kg/m3) 275 300 325 350 400(300) (325) (350) (400) (450)

Lowest grade of concrete 25 30 35 . 40 45

Note 1

Note 2

Note 3

Note 4

Note 5

Note 6

This table applies to normal-weight aggregate OPC concrete of 20 mmnominal maximum aggregate size and river sand fine aggregate. In no caseshould the cover be less than the maximum aggregate size or diameter of mainreinforcement.

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

Cover values marked with asterisks (*) can be reduced to 15 mm, providedthe nominal maximum aggregate size does not exceed 15 mm, subject to theconditions in Notes 1 and 2.

The minimum cement content values in parentheses should be maintained ifno water-reducing admixtures are used.

The grade requirement can be reduced by 5 if a checking regime establishesthat the maximum free water/cement ratio and minimum cement cot\tentrequirements are met.

The above cover values can be reduced by 5 mm, subject to the conditions inNotes 1 and 2 and a minimum of 15 mm, provided a 1:3 cement: sandrendering of 10 mm, 15 mm or 20 mm is applied to concrete made towater/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 ofresistance.

( 225 )

3-20000

f = 25 N/mm2eu

f = 460 N/mm2y

(All dimensions in mm)

Introductory Notes

1. This example is regarding the analysis of an existing beam. The first step in findingthe moment of resistance is to find the lever arm.

Reference Calculations Output

Area of steel = 942.5 mm2

Note 2 Assuming that the steel llas yielded, T = 377189 NT = (0.87)fy.~ = (0.87)(460)(942.5) = 377189 NHence, balancing compressive force = 377189 N

(0,45)feu.b(O.9)x = 377189 .(0,45)(25)(225)(O.9x) = 377189

x = 166 mm x = 166 mm

Note 3 Since x/d = 166/375 = 0.44 < = 0.64,steel has yielded and original assumption is correct.z = d - (0,45)x = 375 - (0,45)(166) = 300 mm z = 300 mm

3,4,4. 1(e) Note :- z/d = 300/375 = 0.8 < 0.95, Hence O.K.

Moment of resistance = (377189)(300)= 113.16 x106 Nmm= 113 kNm M = 113 kNm

Notes on Calculations

2. Most singly reinforced sections will be under-reinforced in practice. Hence, assumingthat 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 is0.0035, is x/d < = 0.64 (for fy = 460 N/mm ) and x/d < = 0.76 (for fy = 250N/mm2). This can be shown by assuming a linear strain distribution. However thecode recommends that x/d < = 0.50, in order to accommodate redistribution up to10% (Clause 3.4.4.4).

Concluding Notes

4. The lever arm is the distance between the centroids of the tensile and compressiveforces. This separation between two opposite forces is what creates the moment ofresistance in a flexural element.

5. Because this distance has to be accomodated within the depth of the section, flexuralelements tend to have larger cross'sections than compressive elements.

EXAMPLE 2 - ANALYSIS OF OVER-REINFORCED SECTION

Determine the moment of resistance of the section shown.

( 150 )

2-25 Id=300o 0


Introductory Notes

f = 25 N/mm2cu

f = 460 N/mm2y

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

6


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

But, x/d = 259/300 = 0.86 :>0.64Note 2 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 mmC = (0.45)fcu.b(0.9)x = (0.45)(25)(150)(0.9)(200)

= 303750N

Note 3 Es = (0.0035)(300-200)/200 = 1.75 xlO-3

Hence, fs = (1.75 xlO-3)(200 xloJ) = 350 N/mm2,

and T = (350)(981.7) = 343595 N

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

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 = 206 mm

z = d - (0.45)x = 300 - (0.45)(206) = 207 mmM = C.z = (312863)(207) = 64.763 x106 Nmm

= 64.8 kNm M = 64.8 kNm

Note 4 Note:- Alternative method of finding x.Once it is established that the steel has not reachedyield point, for any given value of x,Es = (0.0035)(300-x)/xfs = [(0.0035)(300-x)/x](200 xloJ) N/mm2

T = ((0.0035)(300-x)/x](200 xloJ)(981.7) NC = (0.45)(25)(150)(0.9,qNPutting T = C, we have the quadratic equationx2 + (452.47)x - 135741 = 0,giving x = 206 or -659 mm x = 206 mm

7


2. In some rare cases, as in this one, a beam may be over-reinforced, meaning that theyielding of steel will not take place before the crushing of concrete. If such a beamfails, it will do so suddenly, without warning, and hence over-reinforced beams arediscouraged 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.

Strain

II,I,

200 ,kNAnm 2

~

'"88 0.87x460=400 N/mm 2

........z

Strain diagram Stress-Strain diagram

4. It is possible to use this method because the stress-strain curve for steel below theyield point is a single straight line.

Concluding Notes

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

8

f = 25 N/mm2eu

f = 460 N/mm2y

EXAMPLE 3 - ANALYSIS OF DOUBLY REINFORCED SECTION

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

~ .150 .) ~1A'd'

s d=300

2-25o 0


Introductory Notes

x/d = 0.5 in

1. If it is found that a singly reinforced beam is over-reinforced and it is desired to makeit 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 thedepth of the section cannot be increased due to non-structural reasons, option (iii)above is used.


Note 3 Assume a suitable value for d', say 50 mm. d' = 50 mm

For equilibrium of the section, the compression inthe top steel plus the concrete must equal the tensionin the bottom steel.

Setting x = (0.5)d = 150 mm (which automaticallyensures the yielding of tension steel), we haved'/x = 50/150 = 0.33 < = 0.43, which means thatthe compression steel will yield as well.

3.4.4.4 (0. 87)fy.As' + (0.45)feu·b(0.9)x = (0.87)frAs(0.87)(460)As' + (0.45)(25)(150)(0.9)(150 =

(0.87)(460)(981.7) A '= 412 mm2s

Hence, As' = 412 mm2, Use 4T12Note 4 Use 4Tl2 (As' = 452.4 mm2). (452.4 mm2)

9


Table 3.27 Note:- lOOAs' / Ac = (100)(452.4) / (150)(350)Note 5 = 0.86 (> 0.2), Hence O.K.

Lever arm for balanced section = d - (0.45~1.)

= (0.775)d = (0.775)(300) = 232.5 mmDistance between top and bottom steel = 250 mm

Note 6 Hence, taking moments about level of tension steel,moment.of resistance =

(0.45)(25)(150)(0.9)(150)(232.5) +(0.87)(460)(412)(250) = 94187006 Nmm

= 94.2 kNm M = 94.2 kNm


3. The value of d' will depend on the cover, and other requirements (See Example 8).

4. If the compression steel provided is greater than that required, the neutral axis depthwill be reduced slightly; this is desirable, as it will increase the ductility of thesection. When providing four bars within a width of 150 mm, it may be necessary touse the bars as two pairs of bars.

5. When compression steel is provided, a minimum percentage is required. The areaof 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 doublyreinforced section, is to take moments about the level of tension steel. The amountof compression steel to be used in the calculation is the amount required (412 mm2),

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 thesum of the moments of resistance of (i) a balanced section and (ii) a It steel section It ,

consisting of equal amounts of tension and compression steel, separated by (d-d').

150 150~ > ""<--~~

t - d'=50 1I 41~m2 - _ o:~~~ i

d=300 _ ~:.-G.....&-'l 232.5

1982 mm2 570mm2 I

o 0 - ~

10

EXAMPLE 4 - ANALYSIS OF NON-RECTANGULAR SECTION

f = 25 N/mm2cu 2-

fy = 460 N/mmId=400

1f

h=450

1

Determine the moment carrying capacity of the trapezoidal beam section· shown below.

300

156(All dimensions in mm)

Introductory Notes

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


Assume values for the neutral axis, x until thecompression in concrete is equal to the tension insteel.

The area of the section under compression =(0.5)(0.9)x[600 - {(3OO-150)/450}(O.9)x]

Area of steel = 981.7 mm2 ~300~

Assume also that the steel bas yielded. \10.9>< IITry x = 100 mm 0

Area in compression, Ac ~

= (O.5)('JO){600 - (O.33)(O.9)(IOO)) W:= 25650 mm2

C = (0.45)fcu .Ac = (0.45)(25)(25650) = 288563 NT = (0.87)(460)(981.7) = 392876 N

Try x = 139 mmThen, C = 392868 Nand T = 392876 N.This approximation is satisfactory. x = 139 mmNote also that x/d = 139/400 = 0.35 < 0.5; henceassumption that steel has yielded is O.K.

11


The centroid of the compression zone from the top ofthe section will be given byy = {(150)(139)(139/2) + (O.5)(150)(139)(139/3)} 1

{(150)(139) + (0.5)(150)(139)} = 61.8 mmNote 2 Hence, lever arm = 400 - 61.8 = 338.2 mm z = 338 mm

M = C.z = (392868)(338) = 132.8 x106 Nmm= 133 kNm M = 133 kNm

Note:- Alternative method of finding x.Assuming that steel has yielded,T = (0.87)(460)(981.7) = 392876 NFor any x, the area under compression isAc = (O.5)(O.9)x[600 - {(300-150)/450}(O.9)x]C = (O.45)(25)AcPutting T = C, we have the quadratic equation,x2 - (2000)x + 258684 = 0, x = 139 mmgiving x = 139 or 1861 mmSince x/d = 139/400 = 0.35 <0.5, steel has in fact

Note 3 yielded, as assumed.


2. The lever arm cannot be calculated as d - (0.45)x in this case, because thecompression 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 haveto 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. Whendesigning, the amount of steel has to be assumed, and the moment carrying capacitychecked 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 findthe moment of resistance of a given section. This has to be done using straincompatibility concepts from first principles, as illustrated in Examples 1 to 4, or bysuitably 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/mm2, fy = 460 N/mm2, and that thedifference between effective depth and overall depth is 50 mm. Assume also that noredistribution of moments has been carried out.

Introductory Notes

1. This is the first example on the~, as opposed to the analysis of a section.Where beams (as opposed to slabs) are concerned, it will be often found that themoment carrying capacity is more critical than the deflection criterion, and that theformer will govern the selection of cross sectional dimensions.


(a) Singly reinforced section

Note 2 Let us assume that d/b = 2.0In order to find the minimum depth for a singlyreinforced section, we should assume that x/d = 0.5

3.4.4.4 and K = K' = 0.156Then K = M / (b.d2.feJ

0.156 = (150 x106) / {(d/2)(<¥)(25)}d3 = {(2)(150 xlQ6)} I {(0.156)(25)}d = 425 mm .dmin = 425 mm

Note 3 Choose d = 475 mm, h = 525 mm, b = 225 mm d =475 mmh = 525 mm

Now K = M / (b.d2.feu) b = 225 mm= 150 xl06 / {(225)(475t(25) = 0.118

< 0.156 0( 225 •z = d[O.5 + {0.25 - KI(0.9)}o.~

~I~I4Th= (475)[0.5 + {0.25 - (0.118)/(0.9)}O.5]

3.4.4. 1(e) = 401 mm < (0.95)(475) = 451 mm; hence O.K.

As = M / (0.87)frZ

= (150 xlW) (0.87)(460)(401) = 935 mm2

Hence, use·21'25 (As = 981.7 mm2) As = 935 mm2

Table 3.27 lOOA/Ae = (982)(100) / (525)(225) = 0.83 Use 21'25Note 4 > 0.13; hence O.K. (981.7 mm2)

13

Reference Calculations

(b) Overall depth restricted

Output

Note 5

Note 6

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

Table 3.27

3.4.4. 1(e)

Note 7

Table 3.27

Note 8

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 mm2)looAs'/Ac = (100)(402.1)1 (400)(225)

= 0.45 > 0.2; hence O.K.

z = d[O.5 + {0.25 - K'/(O.9)}O.s]= (350)[0.5 + {0.25 - (0. 156)/(O.9)}O.s]= 272 mm < (0.95)(350) = 333 mm; hence O.K.

As = {(K'.fcu.b.d2) 1 (O.87)fy'z} + As'={(O.156)(25)(225)(350)21 (O.87)(460)(272)) + 356= 1344 mm2

Use 3T25 (As = 1473 mm2)looA/Ac = (100)(1473) 1 (400)(225)

= 1.64 > 0.13; hence O.K.

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

A' = 356 mm2s

Use2Tl6(402.1 mm2)

(225 )

40010:: i~000 L

A = 1344 mm2s

Use 3T25(1473 mm2)


2. In practice, the ratio of depth to breadth for a beam will have a value between 1.5and 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 excessof the minimum depth for a singly reinforced section may be chosen, in order toreduce 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 architecturalrequirements. On the other hand, there may be some economy in designing beamswith a marginal amount of compression steel, because longitudinal steel on thecompression 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 depthsis 50 mm.

7. When calculating the are of tension steel, it is sufficient to use the value ofcompression steel required (as opposed to that provided), in this equation.

8. When providing reinforcement, a combination of bar sizes should be adopted, suchthat 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 steelrequired for these sections. The relevant charts are Chart No. 2 for the singlyreinforced 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 stepsof 0.05. The chart with d'/d value closest to the actual value should be used fordesign. If the actual d'/d value lies exactly between the chart values, the chart withthe 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 = 225mm) was carrying an ultimate moment of 150 kNm after a 30% downward redistributionof moment, design the steel reinforcement required. Assume that d' = 50 mm, feu = 25N/mm2 and fy = 460 N/mm2. Use the methods of formulae and design charts.

Introductory Notes

1. If the moment at a section has been reduced by downward redistribution, that sectionmust have adequate rotational capacity at ultimate limit state, in oder for plastic hingeaction to take place. This capacity is ensured by restricting the x/d ratio to a specifiedvalue.

15


Cal Using formulae

3.2.2.1(b) I3b = (1-0.3) 1 1 = 0.73.4.4.4 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.d2.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 - (0.104)/(0.9)}O.5]

3.4.4. 1(e) = 412 mm < (0.95)(475) = 451 mm; hence O.K.

A ' = 104 mm2s

Use 2Tl2(226.2 mm2)

As' = (K -K')feu·b.d2 1 {(0.87)fy<d-d')}= {(O. 118-0. 104)(25)(225)(475t} 1

{(0.87)(46O)(475-50)}= 104 mm2

Use 2Tl2 (As' = 226.2 mm2)looAs'/Ae = 0.19 « 0.2, but acceptable)Table 3.27

As = [(K' .feu.b.d2) 1 {(0.87)fy'z)] + As'= {(0.104)(25)(225)(475)2 1 (0.87)(46O)(412)} + 104= 905 mm2

Use 2T25 (As = ~81.7 mm2)

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

A = 905 mm2s

Use 2T25(981.7 mm2)

Chart 3(part 3)

(bl Using charts

Appro~riate chart for feu = 25 N/mm2, fy = 460N/mm and d'/d = 50/475 = 0.105 is Chart No.3.

(225 ~

I0 2-1~I525 475

2-25o 0

M/bd2 = (150 x106) 1 (225)(475)2 = 2.953.2.2. 1(b) x/d has to be restricted to ({3b-O.4), i.e. 0.3

Note 2

Note 3

Hence, the values for lOOA/bd and lOOAs'/bd mustbe read off the point at which the horizontal lineM/bd2 = 2.95 cuts the x/d = 0.3 line.Thus, looA/bd = 0.85 and looAs'/bd = 0.1As = (0.85)(475)(225)/(100) = 908 mm2; Use 2T25As' = (0.1)(475)(225)1(100) = 107 mm2; Use 2Tl2to satisfy minimum steel requirement. -

A = 908 mm2s

Use 2T25A ' = 107 mm2

sUse 2Tl2

16


2. Any combination to the left of the line corresponding to the x/d = 0.3 line will givea feasible combination of lOOAjbd and lOOAs' /bd. If a point on the line itself ischosen, the solution will generally be the most economical one, in terms of the totalamount 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 forconcrete stress, while the formulae are based on the simplified rectangular one. It isthe 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 plastichinge 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 thefollowing 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/m2.

(iii) Imposed load on slab should be taken as 4.0 kN/m2•

(iv) Density of reinforced concrete = 24 kN/m3•

Introductory Notes

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

100

1<: Beam SectionSectional Elevation

_'i' ..

- - - ~ - - - - - - - - - - - - - - - - '-I _---rr---------------~

I I 'I II II I I, I

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

I I II I II I II I I

____ ~L----------------L___ ~~----------------l

.Ly-

Plan

18

3500

3500


Reference

Note 2

Note 3

3.2.1.2.2

Calculations

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);-

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

The hogging moment at B will be maximum whenthe cantilever portion AB is loaded with themaximum design ultimate load, irrespective of theload on the span BC.The sagging moment in BC will be a maximum whenthe cantilever portion AB has the minimum designultimate load, while the span Be has the maximumdesign ultimate load.Maximum design ultimate load (udl) =

(16.34)(1.4) + (14.0)(1.6) = 45.28 kN/mMinimum design ultimate load (udl) = 16.34 kN/m

Hoe;e;ine; moment atB:-

7.0x1.4 /45.28 kN/rn

~---'ftMB = (7.0)(1.4)(1.95) + (45.28)(2.0)2/2

= 109.7 kNm

19

Output

MB = 110 kNm(hogging)


SUging moment in BC:-

7.0 /16.34 kN/rn /45.28 kN/rn

L:-~te ,t .

A Bx

MB = (7.0)(1.95) + (16.34)(2.0)2/2 = 46.33 kNmTaking moments about B for Be,Rc(6.0) + 46.33 = (45.28)(6.0tl2

Rc = 128.1 kNMx = (128.1)x - (45.28)x2/2dMx/dx = 0 when (45.28)x = 128.1

x = 2.83 mMmax = (128.1)(2.83) - (45.28)(2.83tl2

= 181.2 kNm

Output

MBC =181 kNm(sagging)


2. Idealization is the first step in analysis. Since it is not possible to model the actualstructure with complete accuracy, idealization should be performed such that theresults obtained are conservative. For example, although point C has a certain degreeof restraint, it is impossible to quantify it. However assuming the end C to be simplysupported will give a higher (and hence conservative) moment in the span Be. Therestraint moment at C can be subseqently accounted for by providing a nominalamount of.hogging steel there.

3. Since the beam spacing is 3.5 m, each beam carries the loads acting on a strip 3.5m wide.

Concluding Notes

4. Where dead and imposed loads are combined, as in the case of this example, thedesign moments at critical sections have to be arrived at by a proper combination ofloading patterns.

20

EXAMPLE 8 - DESIGN OF BEAM FOR FLEXURE

Design the reinforcement for hogging and sagging moments in the beam in Example 7. Usefeu = 25 N/mm2 and fy = 460 N/mm2•

Introductory Notes

1. In this example, only the reinforcement for the maximum sagging and hoggingmoments need to be calculated, since the beam section is already specified inExample 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 aspectis considered in Example 12.

Reference· Calculations Output

Effectiye de,pth "

Table 3.2 Assume moderate exposure conditions, for outdoorNote 3 exposure.Note 4 Making use of Notes 5 and 6 of Table 1, we can useTABLE 1 a cover of 30 mm. cover = 30 mmTable 3.5 This will also give a fire resistance of 2 hours.

Assuming a link diameter of 10 mm and areinforcement size of 25 mm, the effective depth will

Note 5 be d = 450 - 30 -10 - 25/2 = 397.5 mm d = 397.5 mm

Desi~n for ho~~in~ moment

The beam behaves a a rectangular beam.b = 300 mm, d = 397.5 mm, M = 110 kNm

Chart 2 M/bd2 = (110 xl06) 1 (300)(397.5)2 = 2.32(Part 3) lOOA/bd = 0.67

As = (0.67)(300)(397.5) 1 100 = 799 mm2 A = 799 mm2s

Use 21'20 & ITl6 (As = 829 mm2) Use 21'20 &Table 3.27 looA/bwh = (100)(829) 1 (300)(450) lT16(829 mm2)

= 0.61 > 0.26; hence O.K. (hogging)

Design for sagging moment

The beam behaves as a flanged beam.3.4.1.5 b = lesser of 3500 mm or

1/5 + bw = {(0.7)(6000)}/5 + 300 = 1140 mmHence, b = 1140 m bf = 1140 m

21


3.4.4.4 Assume that the neutral axis is within the flange.K = M I (b.dz.fcu)

= (181 x106)/{(1140)(397.5t(25)} = 0.040 <0.156z = d[0.5 + {0.25 - K/(0.9)}o.s]

= d[0.5 + {0.25 - (0.04)/(0.9)}o.5]= (0.95)d = (0.95)(397.5) = 377.6 mm

x = (d-z)/(0.45) = (397.5-377.6) 1 (0.45)= 44.2 mm < 125 mm.

Hence, neutral axis is in fact within the flange, and N.A. is inthe beam can be designed as a rectangular beam with flangeb = 1140 mm.

Chart 2 M/bdz = (181 xl06) 1 (1140)(397.5)z = 1.00(Part 3) l00A/bd = 0.27

A = 1224 mm2As = (0.27)(1140)(397.5) 1 100 = 1224 mmzs

Use 21'25 & 11'20 (As = 1295 mm2) Use 21'25 &Table 3.27 bwlb = 250/1140 = 0.22 < 0.40 11'20Note 6 l00A,Ibwh = (100)(1295) 1 (300)(450) (1295 mmz)

= 0.96 > 0.18; hence O.K. (sagging)

TranYerse steel

In ordeJ:l that flanged beam action is ensured, theminimum amount of transverse steel (to be providedin the top of the slab) is given by

Table 3.27 l00Ast/htl = 0.15Ast = (0.15)(125)(1000) 1 100 = 187.5 mmz/m Transverse steel

Note 7 Use R6 @ 150 (min.) (Ast = 190 mm2/m) R6@150 (min)(190 mmz/m)

Slenderness check3.4.1.6

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

Cantilever portion - clear distance between restraintsis 1850 mm(25)bc = (25)(300) = 7500 mm(l00)b/ld = (100)(300)2 I (397.5) = 22642 mm Slenderness

Note 8 Since these values are > 1850 mm, check is O.K. O.K.

22


3. It is sufficient to assume a "moderate" exposure condition for the exteriors of moststructures, 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; thisTable is relevant for Sri Lankan concreting practice. The figures in the table can befurther modified by Notes 5 and 6 of the table, as has been done here. It is assumedin 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 figurebelow.

cover

_ shear link-

y

~1h bars

b~a,Lt x.....-

6. Although the actual steel requirement is calculated using the value of flange width,the minimum steel requirement is based on the web width.

7. This transverse steel will also have to resist the hogging moment in the slab, and agreater 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 oflong, deep cantilevers.

Concluding Notes

9. When designing beam-slab systems, care must be taken to note where flanged beamaction takes place and where it does not. Furthermore, such locations will be reversedin systems where upstand beams are used.

10. If the neutral axis of a flanged beam falls within the flange, the design is identical toa 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 beamsection, 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 midspan.Spacing of beams = 4.0 m; Span of beams = 6.0 m;Thickness of slab= 100 mm', f = 25 N/mm2. f = 460 N/mm2

cu , y .

Introductory Notes

1. An edge beam will have a transverse slab only on one side; hence it is called anL-beam. The beam in the earlier example is called a T-beam, since the slab extendedover both sides of the beam. If the beam is below the slab (as is the case most of thetime), the slab will act as a flange only in the span, when the top of the section is incompression, and not at the supports. .


Note 2 Assume that bw = 225 mm

3.4.1.5 b = lesser of 2000 mmor lilO + bw = {(O.7)(6000)}/lO + 225 = 645 mmHence, b = 645 mm b = 645 mm

If the beam is to be singly reinforced, K=K' = 0.156M 1 (b.d2.fcu) =0.156(200 x106) 1 {(645)(d)2(25)} = 0.156 dmin = 282 mmd = 2~2 rpm d = 325 mm

Note 3 Hence, choose d = 325 mm and h = 375 mm h = 375 mm

3.4.4.4 Then, K = (200 xl06) 1 «645)(325)2(25)} = 0.117z = 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 mmNote 4 Since this is greater than hf = 100 mm, the neutral N.A. is out of

axis lies outside the flange. flangeb/bw = 645/225 = 2.87d/hf = 325/100 = 3.25

equation 2 {3f = 0.129Note 53.4.4.5 {3f.fcu .b.d2 = (0.129)(25)(645)(325)2 = 219.7 xlO6

~

Note 6 = 220 kNm > 200 kNmAlso, hid = 100/375' = 0.308 < 0.45" singlyHence, section can be singly reinforced. reinforced

24

ReCerence Cakulatioas Output

equation 1 AI = [M + (0.I)fcu.bw.d{(0.45)d-b,)] INote 7 (0.87)f {d-(0.5)hrH

=[(200xl()6)+(0.1X2S~2S){(0.45)(325)-100}fAs = 1894 mm2[(O.87)(460){325-(0.5)(I00))] = 1894 mm

Use 21'32 &. 11'20 (As = 1922 mm2) Use 21'32 &.3.12.6.1 l00A/bwh = (100)(1922) I (225)(375) 11'20'Note 8 . = 2.28 < 4.0; hence O.K. (1922 mm2)


2. A web width of 225 mm is around the minimum that is practically desirable, in orderto accommodate the reinforcement. A width of 200 mm can be considered as theabsolute 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 actualcalculation 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 hasto be used (singly reinforced flanged beam design) or whetha" it is sufficient to useClause 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 thanto 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 isrestricted to (0.5)d.

7. This equation for A. is slightly conservative, as it assumes that x = (0.5)<1 , althoughthe actual neutral axis may be somewhere between x = bf and x = (0.5)<1. Since the .width of the web is relatively small, compared to the flange, this discrepency isnegligible 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, carewill have to be exercised if lapping is done.

Concluding Notes

9. This example illustrated the situation where the neutral axis fell below the flange ofa flanged beam. Design charts cannot be used in such a situation, and the equations

.. in Clause 3.4.4.5 have to be employed.

10. In addition, if the moment is greater than Pf.fcu.b.d2 (i.e. compression steel isrequired), or if more than 10% redistribution has been carried out, the beam has tobe designed from strain compatibility fIrst principles as given in Clause 3.4.4.1.

25

EXAMPLE 10 - DESIGN OF SECTION FOB. SIlEAB.

A simply supported beam ofcross section.b = 22S nun and d =400 mmcarries an ultimateload of 60 leN/m over its clear span of 5.0 m. Design the shear reiDforcemeat required nearthe 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 bendingmoment 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 tobend into shape, compared to high yield steel. Links generally have diameters varyingfrom 6 to 12 mm, in steps of 2 mm.

Reference

Note 33.4.5.10

3.4.5.2Note 4

equation 3Note 5

Table 3.9Table 3.8Note 6

Note 7

3.4.5.5Note 8

Cakuiatioas Output

Although the shear force will be maximum .attheface of the support, the deaign &bear force foruniformly distributed loading is at a section wdwfromthe face.

~design

Vmax ' .:

~d~. "".~2500 ~ ""

VJIWt = «('()(5)/2 = 156 leNvJIWt = (156 xloJ) I (225}(400) = 1.67 N/mnJl Vmax = 1.67(0.8)(Wo.s = (0.8)(25)0.5 N/mm2

= 4 N/mm2 > 1.67 N/mm2 < 5 N/mm2;hence O.K.

VdeIip = (156)(2500-400) 1(2500) =.126 leNv = V/(bv.d) = (126 xloJ) I (225)(400)'

= 1.4 N/mm2 v =; 1.4 N/mm2

l00AJb"d = 0.8, d =:= 400 nun, feu = 25 N/mm2;

Hence, ve = 0.58 N/mm2• Vc = 0.58Since v > ve + 0.4 N/mm2, links have to be N/mm2

designed.A.v > = bv·sv<v-vJ I (0.87)fyvAssuming 10 mm links, A.v = 157.1 mm2

Hence, Sy < = (157.1)(0.87)(250) I (1.4-0.58)(225)= 185 mm < (0.75)d = 300 nun; hence O.K. Links

Use R,lO links @ 175 Mm. RIO @ 175

26


3. This is the simplified method to account for the enhanced shear resistance nearsupports. The section considered should be an effective depth away from the face ofthe support. Where support details are not available, it will be comervative tomeasure -d- from the centre-line of support.

4. This is the maximum shear check. If this fails, there is no alternative but to changethe beam dimensions. It is prudent therefore, to make this check fairly early in thedesign procedure.

5. bv for a flanged beam should be taken as the average width of the web below theflange.

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 mustbe chosen. Ing~, the A.., value is assumed and the Sy value c:a1culated. The A.vvalue 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 with2 vertical legs that are used. The resulting Sy value should not exceed (0.75)d, toensure that at least one link crosses a potential shear crack. The transverse spacingbetween 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 tendencyto 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 beencalculated. The requimnent close to mid-span will be much less. This aspect will beconsidered 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 issubject to a triangularly varying shear force diagra,m, with a value of 400 kN at the face ofthe supports. The mid span. steel consists of 4 Nos. 2S mm bars. Design the shearreinforcement required over the entire span, if two of the main bars are bent~ at 45° nearthe 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 nearthe beam supports.

2. The most reasonable way to provide shear reinforcement for the entire span wouldbe 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 Calculatioos Output

SumutNCl!l.

Vmax = 400 kNv

lDllX= (400 xl<P) I (350)(550) ... 2.08 N/nun2 Vmax = 2.08

3.4.5.2 (0.8)(W0.s = .(0.8)(25)0.5 N/mm2

=4 N/mm2 > 2.08 N/mm2 < 5 N/mm2;

hence O.K.

Shear resistance of 2 inclined bars,equation 4 Vb = A.(O.87)fyb(cosa + sina.eotP)(d-d') I3.4.5.6 Assumethalll = 67.5° and d' = 50 mm'Note 3 hence .•stt = (1.41)(d-d') =(1.41)(500) = 705 mm

Vb=(982)(O.87)(460){0.71+(0.71)(0.4 )}(SOO)/(705)= 277890 N

Vb = (277890) I (350)(550) = 1.44 Nir~m2 Vb = 1.44N/mm2

Since 2 bars continue into support,lOOA/bvd = (loo)(981.J 1(350)(550) = 0.51;

Table 3.9 hence, Vc = 0.50 N/mm Vc = 0.53.4.5.10 Shear force at section "d" from support - N/mm2

Note 4 {(3000-550) I (3000)}(400) = 327 kl '<lv = (327 xloJ) I (550)(350) = 1.70 NI mm2

v-ve = 1.70 - 0.50 = 1.20 N/mm2

3.4.5.6 Although this can be resisted by the ben up barsalone, half of this must be resisted by lj;flks.

28


Asv > = bv.Iv{(1.20)12} 1 (0.87)fyvPutting 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 RIO @ 150 mmentire area over which the bent up bars are (support area)effective - i.e. for 0.71 m from the face of support.

Middle area

l00A/bvd == (100)(1963) 1(350)(550) == 1.02;Table 3.9 hence Vc == 0.63 N/mm2 Vc == 0.63Table 3.8 Shear stress taken by nominal links == 0.63 + 0.4 N/mm2

= 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 byA.., > == (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. RIO @ 225 mmUse RIO links @ 225 mm (middle area)

f

Area in-between

Note 5Table 3.8

Note 6

Extent of this area = 3.0 - (2.97)/2 - 0.71 = 0.81 mShear force at distance 0.71 m from support face =

{(3.0-0.71)/(3.0)}(400) = 305 leNv = 1.58 N/mm2

vc = 0.63 N/mm2

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 mmUse 2RIO links @ 200 mm < (0.75)<1; hence O.K.

lCQJlOO 2/1(QjaX) l~

v = 1.58N/mm2

Vc = 0.63N/mm2

2 RIO@200 mm(area inbetween)

1~2T25I. 2T25'"

29

4T25

1.48m


3. Since fJ should be taken as > 45° and ~ is restricted to l.S(d..(i'), this assumed valueof 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 canbe used from the support upto the point where the main bars are cranked up.

S. Although 2 bars are bent up, they also continue for at least adistance "d" from anypoint in this section of the beam. Hence, the value of vc will be the same as in themiddle area.

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

Concludina Notes

7. It is not very common practice to use bent up bars as described in this example,although it was 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 Example8. Also carry out detailing of reinforeement, including curtailment and lapping. Assume thattype 2 defonned bars are used as reinforcement.

Introductory Notes

1. The serviceability checks consist of spanldepth ratio calcu1ations for deflection andbar spacing rule checks for cracking. If these simplified checks are satisfied, the beamis "deemed to satisfy" the serviceability limit state requirements.

Refereace Calculations Output

Check for deflection fSRanIde,pth rules)Note 23.4.1.3 Consider the man BC; effective span = 6000 mm

bwlb = 0.22 < 0.3Table 3.10 .Hence, basic span/depth = 20.8 for continuous,

flanged beam.Example 8 Mlbd2 = 1.00 and

f. = (S/8)(460){(1224)/1295)} = 272 N/mm2

Table 3.11 Hence, P l -= I.4S (for tension reinforcement)Notes F2 = 1.0 (as there is no compression reinforcement)3&4 Hence, allowablespanldepth ratio = (20.8)(1.45) All. span!

= 30.16 depth = 30.2Actual spanldepth= (6000)/(397.5) = 15.09 Act spanI

< 30.16; hence O.K. depth = 15.1Hence O.K.

3.4.1.4 Consider fP8D AD: effective span = 2000 mmTable 3.10 Basic spanldepth = 7 for cantilever with rectangular

beamaetion.Example 8 Mlbd2 =2.32 and

f. = (5/8)(460){(799)/829)} = 277 N/mm2

Table 3.11 Hence F l == 1.07 (for tension reinforcement)Notes F2 = 1.0 (as there is no compression reinforcement) All. span!3&4 Hence, allowable spanldepth ratio = (7)(1.07) = 7.5 depth = 7.5

Actual spanldepth = (2000)/(397.5) = 5.03 Act. span/< 7.5; hence O.K. depth = 5.03 .

Hence O.K.

Curtailment of reinforcement

The bending moment diagram envelope must first bedmwn

31

Reference

NoteS

Cakulatiolls

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

7.0 /6.34 kN/m /45.28 kN/m

~••nn~1.95m 6.Om x

Example 7 For span BC, Mx = (128.1)x - (45.28)x212Mx = Oatx - O.Mx is max. at x = 2.83 and equal to 18l.2 kNmMx .. 0 apiA<al x = 5.66 m

Example 8 Steel at span BC is 2T25 & lno. We can considercurtailing the lno bar.

Note 6 M.o.R. of continuing bars (A, .= 981.7 mm2) can beshown to be 148.4kNm.Putting (128.1)x - (22.64)x2 == 148.4we can obtain x = 1.63 m and 4.03 m.

3.12.9.1 These are the theoretical cut-off points.

Note 7 Keep the practical cut-off points an ancboragelength3.12.9.1(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 atx = 1.63 - 0.8 = 0.83 m andx = 4.03 + 0.8 = 4.83 mLength of 20 mm bar required = 4.83 - 0.83

=4.0mDistances to ends from B are 5.17 m and 1.17 m.

32

Curtail 1nobottom bar at1.13 m and5.13 m from B.Length of baris 4.0 m.

RelereD£e Calculations Output

/ \ M.o.R.

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

Note 8 For support B, the controlling loading case is whenspans AB and Behave the max. and min. designultimate loads. respectively.

7.OX1.4 45•28 kNjm /16.34 kNjm

~B"'¥';:'" ... 'leTaking moments about C for ACRs(6.0) == (7.0)(1.4)(7.95) + (45.28)(2.0)(7.0)

+ (16.34)(6~/2

Ra = 167.7kNMy == (7)(1.4){y"o..0S] + (4S.28)r/2

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

- (14.47)[y-2.0r

A

A ;B

My = Oaty =OandML = loo.7atB.My == 0 again at y == 4.2J m.

Steel at support is 2T2O & ITI6. We can considercurtailing the 1T16 bar.

Note 6 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.

3.12.9.1 These are the theoretical cut-offpoints.

Note 9 To find where the M.o.R. of continuing bars is twice3.12.9.1(e) the applied moment, we can put

(9.8)(y-Q.05) + (22.64)y2 = (90.5)/2 for span AB,and (9.8)(y-0.05) + (22.64)r - (167.7)(y-2.0)

- (14.47)(y-2.0r = (90.5)/2 for span Be,giving Y = 1.22 m and 3.10 m

33

Refereoce Calculatioos

TheGiffe:reoce betweenthe·SIDIlIer y values is(1.80 - 1.22) = 0.58 m or 580 Mm. This is greaterthan (12)4> (192 mm) or -d- (397.5 mm). Thedifference 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 arey = 1.22 m and y= 3.10 m.Length of 16 mm bar required = 3.10 - 1.22

Note 10 = 1.88 mDistances to B are (2.0 - 1.22) =O.78m (span AD)

and (3.10 - 2.0) = 1.10 m (span Be)Table 3.29 Since the distances to either side of B > = (40)41Note 11 {Le. (40)(16)= 640 mm}, anchorage is satisfied.

J 'Imine of bars

Output

Curtail I T16top bar 0.78 m(left) and1.10 MCright)ofB.I..eIlgth of bar is1.88 m.

Note 12

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

3.12.8.13Note 14

3.12.8.11

3.12.8.13

Note 15

The continuing 21'20 top bars at B can be curtailed atthe point of contraflexure closer to B in span BC.andlapped with 2T12 bars (which will anchor the shearlinks). Similarly, the continuing 2T25 bottom bats inspan Be can be curtailed attbe point ofcontraflexurecloser to B in span BC and lapped with 2T12 bus.

For top bars, distance of point of contraflexure fromA is 4.23 m. This would be the theoreticalcut-offpoint To find the practical cut-off point, continuethe bars for an effective depth {Le. 397.5 mm (>12cP)}. Heace, cut-off paint is 4.23 + 0.4= 4.63 mfrom A, Le. 4.63 - 2.0 = 2..63 m.to the right of B.The lapped 2T12 bars will start (40)(12) =4S0mmbefore 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~ (::: ISOmm) or 300 mm is satisfied; also distance betweenlaps will be greater than 75 mm and (6)4> (=72 mm).

For bottom bars, distance of point of contraflexurefrom C is 5.66 m, Le. 6.00 - 5.66 = O.34m to theright of B. As before, the practical cut-off pointwould be 397.5 mm beyond this. Hence, it would be0.4 - 0.34 = 0.06 m to the left of B. The nl2 barswill start 0.48 - 0.06 = 0.42 m to the right of B.

34

Curtail 2T20top bars 2.63 mto .right of B.

Start 2T12 topbars 2.1~ m toright of B.

Curtail 2T25bottom bars0.06 m to leftofB.Start 2Tl2bottom bars0.42 m to rightofB.

Relerenee Calculations

lT16mo ZIal 214> Zf12

I > < >' ,• < ),

Zf12 t 21Q 2I25l'IID

A B

Crack width check <Bar mcW!: rules)

21"25tc

Output

Example 8 Cover required = 30 mmAssume link diameter of 10 mm.

Example 8

Table 3.30Note 163.12.11.2.2

Note 17

3.12.11.2.5

Example 8

Table 3.30Note 163.12.11.2.2

Note 17

3.12.11.2.5

3.12.11.1Note 18

Considering the sUI!Wrt section (tension on top),Clear spacing between top bars (21'20 & lT16) ={300 - (2)(30) - (2)(10) - (20+20+ 16)}/2 = 82 mmIf middle (16 mm) bar is curtailed, clear spacing =ISO rom.The top spacing at the support < 160 mm; henceO.K. (Note:- Since 16120 = 0.8 > 0.45, the 16 mmbar satisfies-the ·0.45 role·.)However, the spacing role is marginally violatedwhen the middle bar is eurtail~; this can betolerated, since the service stress in the continuingbars will be small.Comer distance = [{(30+10+2012)2}(2)]O.s - 20/2

= 6O.7mm < 160/2 = 80 mm; hence O.K.

Considering the span section (tension on bottom),Clear spacing between bottom bars (2T25 &. 1nO)={300 - (2)(30) - (2)(10) - (25+25+20)}/2 = 75 mmIf 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, the20 mm bar satisfies the ·0.45 role·.)However, the spacing role is marginally violatedwhen the middle bar is curtailed; as before, this canbe tolerated.Corner distance = [{(3O+ 10+25/2)2}(2)]O'S - 25/2

= 61.7 mm < 160/2 = 80 mm; hence O.K.

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

35

1£ (0].-1( :m )

Crack WidthO.K. at support

Crack widthO.K. in span

Minimumspacing O.K

Notes 00 cakulatioos

2. The span is taken from Example 7. More guidance regarding the calculation ofeffective 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 F1 from doubleinterpolation in Table 3.11.

4. In a practical beam, there will be some bars on the compression face, in order toanchor 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, thecontrolling loading cases for each situation can be considered.Wberethe span BC isconcerned, the controlling case will be that which causes the points of contraflexureto 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 saggingmoment regions and (e) in hogging moment ones.

8. The controlling loading case for the hogging moment steel at support B is that whichproduces the maximum moment at B, while causing the point of contraflexure closerto 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 thetheoretical 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 momentdiagram being convexto the baseline - the lengths of curtailed bars at supports are much smaller than thosein spans.

II. The anchorage length has to be provided on.either side of the critical section fordesign, so that the full strength of the steel can be utilized. The anchorage lengthsvary depending on the surface characteristics of the reinforcement as well as its yieldstrength. The anchorage length check may become critical when curtailing supportsteel.

12. The continuing bars at the top (2T20) and the bottom (2T2S) can be lapped withsmaller bars, when the former are no longer required to carry tensile stresses. Atleast two bars are required at any section for anchoring the shear links. The minimumdiameter for such bars will be around 12 mm, so that the reinforcement cage willhave adequate stiffness during erection.

36

13. In this instance, it is sufficient to satisfy amdilioos (a) and (b) alone in Clause3.12.9.1 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 mayneed to be in some cases.

15. In general, lapping should not be done at supports, Since column or wallreinforcement will add to reinforcement congestion. In this example however, thebottom 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 spacingallowed becomes fairly small.

17. The continuing bars are able to carry twice the moment actually applied, ascurtailment has been done according to condition (e) in Clause 3.12.9.1. As theservice stress will thtn be quite small, margiDal.aationsof the bar spicing rules canbe allowed. In any case, see Note 19.

18. Both maximum and minimum spacings have to be satisfied. The maximum spacingsapply to the tension face and are "deemed to satisfy- rules for crack control. Theminimum spacing roles apply to both faces and eosure that concre.ting can be carriedout 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 aocuratecalculations for deflection and crack width in Section 3 of BS 8UO: Part 2 can beresorted 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/m2as a one way spanning slab. The loading from finishes and lightpartitioos can each beconsidered equivalent to a uniformly distributed load of I kNIm2• Taking the density ofreinforced concrete to be 24 kN/m3, feu = 25 N/mm2, and fy = 460 N/mm2, design a

typical interior panel.

Introductory Notes

1. A slab is similar to a beam in that it is a flexural member. It isdi.ffeRat to a beamin 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 toU$umeaudl value of I kNlm2• Furthermore, partitions whose positions are notknown should be treated as additional imposed load. The imposed load value specifiedin this example corresponds to .that for a school building. Imposed loads assumed foroffice buildings and domestic buildings are 2.5 kN/m2 and 1.5kN1m2 n=spectively.Further guidance can be obtained from "BS 6399: Part I (1984) - Design loading forbuildings: Code of pmctice for dead and imposed loads".

Note 3

TABLE 1

Note 4

NoteS

Calculatioas

Slab thickness

, to choose a slab thickness, assumetb of 34 .(for a continuous I way slab).

Hence, effective depth = (5.0 xloJ)/(34) = 147 mmWe can use a cover of 20 mm (mild exposureconditions; concrete protected by lOmm 1;3cement:sand rendering).Assuming bar diameter to be 10 mm, chooseh = 175 mm and d = 175 - 20 - 10/2 '"" lSOmm

LoadiOI: (for I m wide strip)

Output

h == 175 mm.d -ISO..

Self load = (0.175)(1)(24)Finishes = (1.0)(1)

Total dead loadImposed load = (3.0)(1)Partitions = (1.0)(1)

Total imposed load

= 4.2 kN/m= 1.0 kNlm= 5.2 kN/m (gk)= 3.0 kN/m= 1.0 kN/m= 4.0 kNlm ('he)

design udl =Design load = (1.4)(5.2) + (1.6)(4.0) = 13.7 kN/m 13.7 kN/m

38

Referenee

3.5.2.3

Table 3.13

Calculations

Ultimate bendin& moments and shear forces

Since gk > <JJc and 'be < = 5.0 kN/m2, and if weassume that bay size > 30 m2, for an interior p&!lel,Span moment = (0.063)F.l = (0.063)(13.7)(5.0)2-

= 21.6kNmlmSupport moment = (-Q.063)F.l = (-Q.063)(13.7)(5):l.

== -2L6kNmlmShear at support = (O.S)F = (0.5)(13.7)(5.0)

= 34.3 kN/m

Output

M =~

21.6 kNmlmM rt=

2r6~mlm

v = 34.3 kN/m

Chart 2(Part 3)Note 6

Note 7Fig. 3.25Note 83.12.11.2.7

Desi&n for bendin& at §l)M

Mlbd2 = (21.6 x10~ 1 (1000)(150f" = 0.96100AJbd = 0.26 (> 0.13); hence min. steel O.K.As = (0.26)(1000)(150) 1 (100) = 390 mm2/m Span steelUse TI0@175 mm (As = 448 mm2/m) TIO @ 175 mmHalf theSe bars_ 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) = 0.13Hence crack cootroland minimum steel O.K.

Note 9 Check for deflection

MJbd2 = 0.96 andfs = (5/8)(460){(390)/(448)} = 250 N/mm2

Table 3.11 Hence F 1 = 157Table 3.10 Allowable spanldepth = (26)(1.57) = 40.8

Actualspanldepth =(5000)/(150)= 33.3 < 40.8;~ O.K.

~n for bendin& at sUllJlOrt

Deflection O.K.

Since the moment is identical to that in the span, Support steelsteel provided also can be identical. TIO @ 175 mm

Fig. 3.25 Half these bars can be curtailed at (0.15)1 =(0.15)(5) = 0.75 m from the face of support (Note:45 tP = 450 mm< 750mm) and all the steelcurtailed at (0.3)1 = (0.3)(5) = 1.5 m from the faceof support.

39

Reference calculations Output

Check for shear,

v = (34.3 xl<P) I (1000)(150) = 0.23 N/mm2 v = 0.23Note 10 For looA/bvd = (100)(448) I (1000)(150) = 0.30, N/mm2

d = 150 mm and feu = 25 N/mm2,Table 3.9 vc = 0.54 N/mm2 > 0.23N/mm2 vc - 0.54Table 3.17 Hence, no shear reinforcement required. N/mm.2.Note 11

Seconda[y reinforcement

Table 3.27 l00As/Ac = 0.13As = (0.13)(1000)(175) I (100) = 227.5 mm2/m SeCondaryUse TlO @ 350 mm (As = 224 mm2/m) steel

3.12.11.2.7 Max. spacing = (3)(150)- 450 mm > 350 mm. TI0@350mm

~7EmMO.7~ ~o.7fm J>=7~Tl~175 TlQfJ350 nOl3OO~75

cr ..·--n~·----

, .. ~.... "Notel2

~~Tl~ TlOOI75Tl............ 'f, I I f

0( I.On '>5.0m

( I.On )( )

Notes on Calculatioas

3. Although the bending moment is the controlling factor in the choice of depth forbeams, where slabs are concerned, the controlling factor is the spanldepth ratio,representing the check for deflection. Atrial'V8lue has to be used initially; a valueof around 34 is a reasonable estimate for lightly loaded one way continuous slabs; thisshould be reduced to around 30 for heavily loaded .s1abs. A lightly loaded slab wouldhave an imposed load of around 4 kN/m2, while a heavily loaded slab would bave oneof around 10 kN/m2•

4. Slabs are generally designed such that shear links are not required; hence, noallowance need be made for link diameter.

5. One way and two way slabs are generally designed - Le. loads evaluated andreinforcementcalculated - 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 theminimum steel requirement.

7. Although we can use the sllghtlylarger spacing·of 200 mm (giving As = 392.5mm2/m), we adopt this smaller spacing, as it results in the minimum steelrequirement being satisfied even after half the steel is curtailed.

8. Although 60% of the steel can be curtailed, in practical slabs, curtailing 50% iseasier, because every other bar can be curtailed.

9. The assumption regarding spanldepth ratio must becbecked as early as possible inthe design. Hence span moments should be designed for first and the deflection checkmade 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 shearreinforcement in flat stabs. Hence, if v is greater than vC' the slab thickness isincreased. This should always be borne in mind, and perhaps an approximate checkfor shear made early in the design,especially if the slab is heavily loaded (e.g. witha water load).

12. Where the curtailment of steel is'toncemed, the distances corresponding to top steelare given from the face of the support and those corresponding to bottom steel fromthe centre-line of support.

Concluding Notes

13. Although it is quite easy to satisfy minimum steel requirements and maximum barspacing rules at critical sections (such as midspan and support), care should be takento 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 mostpractical situations. Such an approach is given for the design of continuous beams aswell, 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 allspans 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, theclear distance between walls being 3.5 m. Design the slab,taking fcu .·2S'N/mm2, fy = 460N/mm2 and density of reinforced concrete = 24 kN/m3•

Introductory Notes

1.

...i..

3.

4.

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 loadfrom finishes and parapet wall have to be assumed and a decision taken regarding theend fixity of the slab.

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

The parapet wall which is constructed on the slab perpendicular to its span will givea degree of fixity to the slab. However, the most conservative approach is to idealizethis slab as a one way simply supported slab. Any fiXing moments caused by theabove partial fixity can be accomodated by taking SO% of the midspan steel into thetop face of the slab at the support.

The parapet wall parallel to the span will have to be carried by the slab. It can beassumed 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 alimited width of the slab (Clause 3.5.2.2).


Slab thickness

Note 5 Approximate span = 3500 mmNote 6 Assuming Spanldepth ratio of 28 (for a simply

supported 1 way slab),effective depth = (3500)/(28) = 125 mm

TABLE 1 If we take cover = 30 mm (moderate exposureExample 8 conditions and TABLE 1 values modified by Notes 5

and 6), and bar diameter = 10 mm, we can choose h = 160 mmNote 7 h = 160 mm and d = 160 -30 - 10/2 = 125 mm. d = 125 mm

Hence, effective· span = lesser of3.4.1.2 (3500+225) = 3725 mm eff. span =

or (3500+ 125) = 3625 mm 3.625 m

42


Loadine (for 1 m wide strip)

Output

3.5.2.2

Note 8

Self load = (0.16)(1)(24) = 3.84 kN/mFinishes = (1.0)(1) = 1.00 kN/m

Total dead load = 4.84 kN/rnImposed load = (1.5)(1) = 1.50 kN/rnDesign load = (1.4)(4.84) + (1.6)(1.5) = 9.2 kN/m design udl =

9.2 kN/rnStrip carrying parapet wall = (0.3)(3.615) + 0.12

= 1.21 mAdditional dead load in that area =

(1.0)(0.12)(23) 1 (1.21) = 2.28 kN/m

Ultimate bending moment and shear force

Note 3

Chart 2(Part 3)

3.12.11.2.7

Note 9

Fig. 3.25

Since we assume the slab to be simply supported,Mid span moment = w.l2/8 = (9.2)(3.625)21 8

= 15.1kNmlmShear force at support = w.l/2 = (9.2)(3.625) 12

= 16.7 kN/m

Desim for bending

Mlbd2 = (15.1 xl<f) 1 (1000)(1251 = 0.97looA/bd = 0.26 (> 0.13); hence min. steel O.K.A. = (0.26)(1000)(125) I (100) = 325 mm2/mUse TI0 @ 225 mm (As = 349 mm2/m)Max. spacing allowed = (3)(125)

= 375 mm > 225 mm; hence crack width O.K.

However, bar spacing as well as minimum steelrequirement 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 > 0.13Hence, min. steel andbar spacing are O.K. aftercurtailment.

The steel should be curtailed at (0.1)1 = (0.1)(3625)= 362.5 mm from the point of support, Le. 362.5 (225/2) = 250 mm from the face of support.

43

Mspan = 15.1kNmlm

V = 16.7 kN/m

span steelTIO @ 187.5mm

RefeNilee Calculations Output

Note 10 The rest of the steel could be taken into the support support steeland bent back into the span as top steel to extend a TI0 @ 375 mmdistance from support face of (0.15)1 = (0.15)(3625)

3.12.10.3.2 = 544 mm {> (45)cP = (45)(10) = 450 mm}, say0.55 m

Check for deflection

Note IITable 3.11Table 3.10

Mlbd2 = 0.93fs = (5/8)(460)(325/419) = 223 N/mm2

Hence, F1 = 1.71 (for tension steel)Allowable span/depth = (20)(1.71) = 34.2Actual 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/mm2

looA/bvd = 0~131

Table 3.9 Hence, Vc = 0.45 N/mm2 > 0.13 N/mm2;

Table 3.17 hence shear r/f is not required. Shear O.K.

SecondflO' reinforcement

3.12.11.2.7

looA/Ac = 0.13As = (0.13)(1000)(160) I (100) = 208 mm2/mUse TIO @ 375 mm (i.e. max. spacing allowed - 3d)(As = 209 mm2/m)

secondary steelTI0@375 rom

under parapetsTI0@ 175 mm(span)TI0@350(support)

I

TIotm5 r~o.~

t • •I :

Note:- It can be shown that the spacing of thereinforcement in the edge strips of 1.21 m should beT10 @ 175 mm at midspan (and hence TIO @ 350mm at supports).

rt1 110075II~

Note 12

1100187.5 o.ti:

~.0.:rAJn

44


5~ In order to use Clause 3.4.1.2 tofind the effective span, the clear distance betweensupports is taken as a first approxl.mation of the span.

6. For a lightly loaded one-way simply supported slabs, a span/depth ratio of around26-28 may be assumed. Tbisshould be .reduced to around 24 for a heavily loadedslab.

7. In this instance, we have taken a value for h, such that slab thicknesses are assumedto vary in steps of lO mill. To use steps of 25 mm (corresponding to 1 inch) wouldbe 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 bemore heavily reinforced than the rest of the slab. However, only the central part ofthe slab is actually designed in this example.

9 . There may be other alternatives to increasing the mid-Span steel, bot this approachmakes the detailing for curtailment very simple and also helps to satisfy the deflectioncheck, which is very critical in slabs. This approach also facilitates the detailing ofsteel for support restraint, as shown in the figure. One possible alternative is to usesmaller diameter bars, but bars smaller than 10 mm, if used as main steel, will notbe 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 partiallyrestrained ends of slabs and beams.

11. Since we have provided more steel than required at mid-Span (see Note 9), advantageshould be.taken of this by generally calculating the service stress, which will be lowerthan (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 andTlO@ 350 mm at support, since the central part of the slab already has TlO @ 187.5mm and TlO @ 375 mm at span and support respectively. The small penalty in costwill probably be worth the simpler detailing arrangement.

Concluding Notes

13. It is important to keep in mind curtailment, bar spacing rules and minimum 'steelrequirements while designing the reinforcement, because these detailing considerationsmay 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 mx 6 m. The imposed load on the slab is 3 kN/m2• The loading fronl finisheaand lightpartitions can each be taken as 1 kN/m2• Design a typical interior panel, using feu == 2SN/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 thosein 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 thatadequate provision is made for torsion. .

ReferellCe

Note 3

3.5.7TABLE I

Note 4Note 5

Calculatiops

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

h == 150 mmdlbort ==125 mm

~==l1Smrn

Loadine (udl)

Self load == (0.15)\:, 1(24) == 3.6 kN/m2

Finishes == (1,0>, == 1.0 kNlm2

Total dead load == 4.6 tN/mtImposed load == (3.0). == 3.0kN/m2Partitions = (1,0)' = 1,0 kNlm2

Total imposed load == 4.0 kN/m2

Design load= (1,4)(4.6) + (1,6)(4.0) == 12.8 IcN/m2 n == 12.8kN/m2

Bendine moments

This interior panel- has lyIlx == 615 == 1,2

46


Table 3.15 Short way, edge =(0.042)(12.8)(5)2= 13.44 kNm/mShort way, span =(O.0~)(l2.8)(5)2= 10.24 kNmlmLong way, edge =(0.032)(12.8)(5)2= 10.24 kNmlmLong way, span =(0.02~)(12.8)(5)2= 7.68 kNm/m

Desien of reinforcement I/

Short way, mid-man:-"

Chart 2 Mlbd2 = (10.24 xld') 1 (lOOO)(l25t = 0.66(Part 3) 100A/bd = 0.17

As = (0.17)(1000)(125) 1 100 = 213 mm2/mUse TlO @ 350 mm (As :: 224 mm2/m)

3.12.11.2.7 Max. spacing = (3)(125) = 375 mm > 350 mmTable 3.27 l00A/Ac = (100)(224) 1 (1()()())(150) = 0.15 >0,13

Hence, bar spacing and min. steel are O.K. but if Short way, spanNote 6 steel is curtailed, they will be violated. TlO@ 350 mm

(Check for, deflection):-Mlbd2 = 0.66fs = (5/8)(460)(213/224) = 273 N/mm2

Table 3.11 F1 = 1.65 (for tension steel)Table 3.10 Allowable span/depth = (26)(1.65) = 42.9

Actual spanJdepth = (5000)/(125) = 40 < 42.9;hence O.K. Deflection O.K.

Short way. cts. edge:-MJbd2 = 0.86, l00AJbd = 0.23, As '7' 288 mm2/mUse TlO @ 250 mm (A. = 314 mm2/m) Short way, edgeBar spacing and min. steel areIO.K. TlO@250 mm

Long way. cts. edge:-Chart 2 Mlbd2 = (10.24 xlW) 1 (1000)(115)2 = 0.77(part 3) l00Aibd = 0.21:l.12.11.2.7 As = (0.21)(1000)(115) 1 (100) = 242 mm2Table 3.27 Use TlO @ 325 mm (A. = 242 mm2/m)

Max. spacing = (3)(115) = 345 mm > 325 mml00AJAc = (100)(242) 1 (1000)(150) = 0.16 >0.13 Long way, edgeHence, bar spacing and min. steel are O.K., but steel TlO @ 325 ~mcannot be curtailed.

Long way. mid-s.pan:-Mlbd2 = 0.58, 100Aibd =' 0.15, As = 173 mm2/mUse TlO @ 350 mm (As = 224 mm2/m), since Long way, spanmax. clear spacing (345 mm) governs. TlO@ 350 mm

47

Reference

3.5.3.5

Note 7

Table 3.16

Table 3.9

CalcuJatiOllS

Edge strips:l00A/Ac =0.13As = (0.13)(150)(1000) I (100) = 195 mm2/mUse TlO @ 375 mm (governed by max. spacing rulein short way direction)Use only in short way cts. edge; at other locations,use middle strip steel for edge steel.

Check for shearShort way sypport:-V = (0.39)(12.8)(5) = 25.0 kN/mv = (25.0 xloJ) I (1000)(125)= 0.2 N/mm2

l00A/bd = (loo~(314) I (1000)~125) = 0.25vc = 0.53 N/mm > 0.2 N/mm ; hence O.K.

Output

Edge stripnO@375 mm(only for shortway, cts. edge)

LoDa l)!&Y R>J!Ort:-Table 3.16 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) =0.21Table 3.9 vc = 0.50 N/mm2 > 0.18 N/mm2; hence O.K.

No shear r/frequired

Fig. 3.25 o

~ <1800 ~

16T106325'11 ~

600

, ~ ,- - - -

..."" ~l- ~l- ...1-

E-o o 0 E-o1.0 1.0 1.0 1.0l:"- N M r-M @ @ M@ 0 0 @0 - - 0- E-o E-o -E-o O'l 00 E-oN - - N

48

Notes on CalcuIatiODS

3. A trial value for span/depth ratio of 40 is reasonable for a lightly loaded, continuoussquare2-way slab; a ratio of 38 would be appropriate for heavily loaded slabs. Thiswill 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 withrespect 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 thatrequired for a one-way slab of similar span and loading - cf. 175 mm required for theslab in Example 13.

5. In arranging the reinforcement in the slab, the short way reinforcement should beplaced outermost, in order to have the greatest effective depth, since the shorter spancontrols deflection and since the bending moments and shear forces are greater in theshort way direction as well.

6. Two way spanning slabs are, in general, very lightly reinforced, so that curtailing isoften not possible because of the minimum steel requirement or the maximum spacingrequirement, or both.

7.. Since the main steel requirements are also fairlysmaIi, for practical detailing it maybe it' may be convenient to use the same reinforcement as the middle strip for theedge 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 edgesteel, the requirements of torsional steel reinforcement have to bernet at the top andbottom of the slab according to Clause 3.5.3.5; in many cases, the main and edgesteel provided would meet those requirements.

9. Although the loads on a beam supporting a two-way slab will be either triangular ortrapezoidal, the code gives coefficients for an equivalent uniformly distributed loadover three quarters of its span.

10. In the calculation of moment coefficients from Table 3.15, if there are significantlydiffering coefficients on either side of a common edge, the code suggests a methodof 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 caneach be considered to be 1 kN/m2. Design a typical interior panel, using feu == 2S N/mm2,

fy =460 N/mm2 and density of reinforced concrete =24 kNlmJ• It may be assUmed thatthe columns supporting the slab are braced.

Introductory Notes

1. This example, too, can be compared with Examples 13 and 15.

2. As the columns are braced, and as the I1ab has several bays in each. direction, thesimplified method of analysis described in Clause 3.7:2.7 and Table 3.19 will beemployed.

3. It will be assumed that the slab is without drops, and the maximum value of effectivediameter will be employed for column.beads.

Reference

3.7.1.4Note 43.7.8TABLE 1

NoteS

Note 6

Calculations

Slab thickness

Max. value of he =.(114)(5.0) = 1.25 mAssuming a trial span/depth of 32,effective depth = (6000)/(32) = 187.5 mmIf we take cover = 20 mm (mild exposui-e condi\ionsand concrete protected by 10 mm 1:3 et:sand render)and bar diameter = 10 mm, we can chooseh = 212.5 mm, dlong= 212.5-20-10/2 == 187.5 mm,dshort == 187.5-10 == 177.5 mm, dave == 182.5 mm

Loadin~ (for entire panel) >"

Panel area == (5)(6) == 30 m2

Self load = (0.2125)(30)(24) == 153lcNFinishes == (1.0)(30) = ..1U:.kN

Total dead load = 183 leN .Imposed load = (3.0)(30) = 90 leNPartitions = (1.0)(30) = 30 leN

Total imposed load = 120 leNDesign load = (1.4)(183) + (1.6)(120) = 448 leN

50

he = 1.25 m

h == 212.5 mmdy == 187.5 mmdx == 177.5 mmdavg==182~5 mm

F = 448 leN

.il"


Table 3.19 Bending mOmentsNote 7 Long way:-

1 = 6.0 - (2/3)(1.25) = 5.17 mSpan moment = (0.071)(448)(5.17) = 164 kNm

Fig. 3.12 Col. strip (2.5 m) = (0.55)(164) = 90.2 kNmTable 3.20 Mid. strip (2.5 m) = (0.45)(164) = 73.8 kNm

Support moment = (0.055)(448)(5.17) = 127 kNmCOL strip (2.5 m)= (0.75)(127) = 95 kNmMId. strip·(2.5 m) = (0.25)(127) = 32 kNm

Short way:-1 = 5.0 - (2/3)(1.25) = 4.17 mSpan moment = (0.071)(441)(4.17) = 133 kNm

Fig. 3.12 Col. strip (2.5 m) = (0.55)(133) ::: 73 kNmMid. strip (3.5 m) = (0.45)(133) = 60 kNm

Support moment = (0.055)(448)(4.17) ::: 103 kNmCol. strip (2.5 m) = (0.75)(103) = 77 kNmMid. strip (3.5 m) = (0.25)(103) = 26 kNm

Design of reinfOrcementLong way. an:-(Check for deflection)Total~ moment = 164 kNmMlbd2 = (164 xlo6) 1 (5000)(187.5)2 = 0.93

Note 8 If AS,reqd = Aa.(Jrov. fs= 288 N/mm2

Table 3.11 and F1 = 1.41 for tension steel)3.7.8 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)Chart 2 Mlbd2 = (90.2 xl<J6) 1 (2500)(187.5)2 = 1.03(part 3) looA/bd = 0.28

As = (0.28)(2500)(187.5) 1 (l00) = 1313 mm2

Note 9 Use 17 TlO @ 147 mm (As = 1335 mm2) Long way. span3.12.11.2.7 Allowable spacing = (3)(187.5) = 562.5 mm (Col. strip)Table 3.27 l00A/Ac = (100)(1335) 1 (212.5)(2500) = 0.25 17 TI0Note 10 Hence bar spacing and min. steel are O.K. @ 147 mm

(Middle strip -2.5 m wide) Long way. spanChart 2 Mlbd2 = (73.8 xl<>6) 1(2500)(187.5)2 = 0.84 (Mid. strip)(Part 3) looA/bd = 0.23; As = 1078 mm2 14 TI0

Use 14 TI0 @ 179mm (As = 1100 mm2) @ 179 mm

51

-Reference Calculations Output

Lone way, SURPOrt:- Long way,sup(Column strip - 2,5 m wide) (Col. strip)M/bd2 = 1.08, l00A/bd = 0.29, A. = 1359 mm2 12 TI0 @ 104

3.7.3,1 Use 18 TlO (As = 1414 mm2) - 12 TIO centred on 6 TI0 @ 208column @ 104 mm; 6 TI0 @ 208 mm.

(Middle strip - 2,5 m wide)M/bd2 = 0.36; l00A/bd ;., 0.10 Long way, sup

Table 3.27 Use nominal steell00A/Ac = 0,13; A. = 691 mm2 (Mid, strip)Use 9 TI0 @ 278 mm (As = 706,9 mm2) 9 TI0 @ 278

Short way, span:-(Column strip - 2,5 m wide)

Chart 2 M/bd2 = (73 xl(6) / (2500)(177.5)2 = 0.93(Part 3) l00A/bd = 0.26

As = (0.26)(2500)(177.5) / (100) = 1154 mm2 Short way, spanUse 15 TIO @ 167 mm (As = 1178 mm2) (Col. strip)

3.12.11.2.7 Allowable spacing = (3)(177.5) = 532.5 mm 15 TlO @ 167

(Middle strip - 3.5 m wide)Note 11 Mlbd2 = (60 xl(6) / (3500)(177.5)2 = 0.54

l00A/bd = 0.15 Short way, spanAs = (0.15)(3500)(177.5) / (100) = 932 mm2 (Mid strip)Use 12 TIO @ 292 mm (As = 942.5 mm2) 12 TI0 @ 292

Short way. suPJ)01't:-(Column strip - 2.5 m wide) Short way, supMlbd2 = 0.98, l00A/bd = 0.27, As = 1198 mm2 (Col. strip)

3.7.3.1 Use 16 TIO (As = 1257 mm2) - 10 TI0 centred on 10TlO@ 125column @ 125 mm; 6 TlO @ 208 mm. 6 TIO@208

(Middle strip - 3.5 m wide)Mlbd2 = 0.24, lOOA/bd = 0.06' hence use Short way, sup

3.12.11.2.7 l00AJAc = 0,13, ~ = 967 mmi (Mid. strip)Use 13 TlO @ 269 mm <As = 1021 mm2) 13 TlO @ 269

Check for shear

If square columns are used, size of column head = a

{(T/4)(1.25)2}O.5 = 1.1 m .

3.7.7.4Perimeter of column head = (1.1)(4) = 4.4 m1st critical perimeter = {(2)(1.5)(0.1825) + 1.1}(4)

= (1.648)(4) = 6.59 mArea within this perimeter = (1.648)2 = 2.716 m2

S2

Reference

3.7.6.2

3.7.7.4

Note 12

Table 3.9

Calculations

Vt = 448 kNVeff = (1.15)V t = (1.15)(448) = 515.2 leNvmax = (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 kNv = (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 __ L ----l---l---~~r, I I

,1 I" 12Tl00292B

g I: ~ :100J :~ , , I 13Tl00269T

~--~- -1 --t - -' - :3Tl0f208T~ , ~ , , I 15Tl00167B~ " I fJ

-I---t- - ......,--+--'" - lOTlO,125TS; I , I

~' I 3T lO'208T... - - -1 ~ -- -- '_L - -l-l

Cl'l E-c E-c~ ~~~ ~ ~ ~ ~ ~<!ill N N"" <!ill <!ill~ ~ ~ ~ 0 0

~ E: E: E: E:E:- Cl'l C'?C'-NC'?........

Output

Shear r/fnot required.


4. The trialspan/depth ratio should be around 0.9 times that used for continuousone-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 theslab thicknesses will be greater for flat slabs than for two-way slabs of similardimensions 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 notrequiring 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 longway span - Le. the long way reinforcement should be on the outside - becausedeflection is governed by the longer span and the moments in the long way directionare greater than those in the short way direction; this too is the opposite of two wayslab 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, asopposed to that on a strip of unit width.

7. The flat slab has to be analysed in two mutually perpendicular directions, with thetotal load being taken in each direction. This is because there are no peripheral beamsaround 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 aconservative approach, but has the advantage that it can detect early any changes thatmay be required in slab thickness. If this check is made after the steel has beendesigned, 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 barsin 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, althoughthe maximum spacing requirement can easily be satisfied.

11. Note that the effective depth in the short way direction is 177.Smm (as opposed to187.5 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, thelOOA/bd value should also be avetaged. This is because the square sbear perimeterscross both the long way and short way steel.

Concluding Notes

13. Unlike in the two-way slab, where the middle strips carry most of the moment andare hence more heavily reinforced, in the flat slab, it is the column strips that carrymost of the moment and are more heavily reinforced.

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

15. Edge and comer columns of flat slabs will have column strips considerably narrowerthan those in interior panels (see Clause 3.7.4.2). Furthermore, the enhancementfactors for shear due to moment transfer will be greater at these columns (see Clause3.7.6.3).

54

EXAMPLE 17· RIBBFD SLAB

A ribbed slab which has several continuous spans of 5 m is to carry an imposed load of 3kN/m2 as a one-way spanning slab. Taking the load from light partitions and finishes as 1kN/m2 each, the density of reinforced concrete as 24 kN/m3, feu = 25 N/mm2 and fy = 460N/mm2, design a typical interior panel. Note that a 1 hour fire resistance is required.

Introductory Notes

1. This example can be compared directly with Example 13, where the only differenceis that the slab is solid.

2. Although this slab is continuous, because of the difficulty of reinforcing the toppingover the supports, it will be treated as a series of simply supported slabs (see Clause

3.6.2).

~~250

500\(

Reference

Note 3

TABLE 1

Fig. 3.2Table 3.5

3.6.1.3Table 3.18

3.6.1.3

Note 4Fig. 3.2

Calculations

Choice of form

Assuming a trial span/depth ratio of 26,effective depth = (5000)/(26) = 192 mmAssuming cover of 20 mm (mild exposure conditionsand concrete protected by 10 mm 1:3 cement:sandrendering) and bar size of 20 mm, we can chooseh = 225 mm and d = 225 - 20 - 20/2 = 195 mm

Min. rib width for 1 hr. fire resistance = 125 mmand min. cover = 20 mm; hence cover O.K.Choose min. rib width of 125 mm, widening-to 250mm and rib spacing of 500 mm

« 1.5 m; hence O.K.)Also use thickness of topping = 50 mm; then ribheight = 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.)

-*-50if

55

Output

h = 225 mmd = 195 mm

te = 115 mm


Note 5 Loading (for 0.5 m strip)

Self load = (0.115)(0.5)(24) = 1.38 kN/mFinishes = (1.0)(0.5) = 0.50 kNlm

Total dead load = 1.88 kN/mImposed load = (3.0)(0.5) = 1.5 kN/mPartitions = (1.0)(0.5) = 0.5 kN/m

Total imposed load = 2.0 kNlmDesign load =(1.4)(1.88) + (1.6)(2.0) = 5.83 kN/m design 001 =

5.83 kN/m

Design for bending

Assuming slab is simply sup~rted,

moment in span = (5.83)(5) I (8) = 18.2 kNm3.4.4.4 K = M I (b.d2.fcu)

= (18.2 xl06) I (500)(195)2(25) = 0.038z = d[0.5 + {0.25 - (0.038)/(0.9)}o.s] = (0.96)dHence use z = (0.95)d = (0.95)(195) = 185 mmx = (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)

Table 3.27 1ooA/bwh = (100)(314) I {(1/2)(125+250)(225)}= 0.74 > 0.18; hence min. steel O.K.


bw'b = (187.5)/(500) = 0.375Table 3.10 Hence, basic span/depth ratio = 16.4

fs = ~5/8)(460)(246/314) = 226 N/mm2

M1bd = (18.2 x106) / (500)(195t = 0.96Table 3.11 Hence, F i = 1.67 (for tension reinforcement)

Allowable spanldepth = (16.4)(1.67) = 27.4Actual span/depth = (5000)/(195) = 25.6

< 27.4; hence O.K.

Check for shear

Note 6 Shear force at lid" from support= {(5.83)(5)/2}{1 - (0. 195)/(2.5)) = 13.4 kN

56

main r/f1 1'20 per rib

Deflection O.K.

Reference Cakulations Output

Note 7 v = (13.4 xlol) I (187.5)(195) = 0.37 N/mm2

Table 3.9l00AJbd = (lOO~{314) I (187.5){195) = 0.86

shear r/f notvc = 0.72 N/mm > 0.J.tN/mm2;

3.6.4.7 hence shear r/f not required. required

Top steel over sup,port

3.6.2 This is to control cracking and should be 25 % ofmidspan steel.As = (114)(246) = 61.5 mm2

Use 1 TIO bar (As = 78.6 mm2), extending (0.15)1 over support= (0.15){5.00) = 0.75 m into span on each side. 1 TIO per rib

Notes on CalculatioR§

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

4. It shouldbe noted that the effective thickness of this slab (reflecting the volume ofconcrete that will be used) is much lower than the one-way solid slab of similar spanand loading in Example 13.

5. It is convenient to calculate the loading for a strip of width equal to a repeating crosssectional 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 - Clause3.4.5.10). 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 ofribbed 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 doesnot demand it (Clause 3.6.6.2). It will be quite difficult to place such a mesh in a 50mm topping while maintaining the top and bottom cover requirements.

11. These ribbed slabs probably have a lower material cost than solid slabs, but theirconstruction 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 ofdimension 525 mm x 300 mm in one direetiononly and aone-way $lab of 175 mm thickness.The roof also has a beam-slab ammgement identical to other. floors. The oolumns are ofdimension 300 mm x 300 mm and the soffit to soffit heigbtoffloors is.3.5m; the heightfrom the top of the pad foundation (designed to resist moment) to the soffit of U1e first floorbeams is 5.0 m. If the frame is braced. classify a typical internal column fordifferent storeysas short or slender.

Introductory Notes

1. Columns are classified as unbraced or braced on the one hand (depending on whetherthey' take lateral loads or not) and .as slender or short on the other (depending onwhether they should be designed to carry additional moments due to deflection ornot).

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

Reference Caknlations Output

Clear height between end restraints,(for ground floor columns)lox = 5.0 m. loy = 5.0 + (0.525-0.175) = 5.35 m(for other floor columns) .lox = 3.5-0.525 = 2.975 m,loy = 3.5-0.175 ::: 3.~25 m

3.8.1.6.2 The end conditions for the columns in the directionof beams are all conditionl. H(mce, fJ ::: 0.75In the other direction.

Table 3.21 fJ = 0.80 (ground floor columns)Note 3 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)

3.8.1.3

3.8.1.3

Hence. for ground floor columns,le/h = (3750)/(300)= 12.5 < 15,leylb ::= (4280)/(300) = 14.3 < 15; hence short.for other columns.le/h = (2231)/(300) = 7.44 < 15,l~jb =(2826)/(300) =9.42. < 15; hence short.

58

All columns areshort.


3. The values of {3 in Tables 3.21 and 3.22 have been obtained from the more rigorousmethod for calculating effective column lengths in framed structures, given inequations 3 to 6 in section 2.5 of Part 2 of the code. The ratios "c (i.e. sum ofcolumn 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.

Concluding Notes

4.

5.

Where edge columns are concerned, they will not have beams "on either side" asspecified 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 Tables3.21 and 3.22 (see Note 3 above); otherwise the method in Section 2.5 of Part 2 canbe adopted.

For a column to be considered short, both le/h and le/b have to be less than 15 (forbraced columns) and less than 10 (for unbraced columns), as specified in Clause3.8.1.3. The ratiO"c in section 2.5 of Part 2 or the value {3 in Clause 3.8.1.6 has tobe 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/mm2, fy = 460N/mm2, and that the imposed loads on the roof and the floors are 1.0 kN/m2 and 2.5kN/m2 respectively and that the allowance for partitions and finishes are 1.0 kN/m2 each,design the ground floor part of an internal column of the framea structure described inExample 18. : .". ".

Introductory Notes

1. Since the ground floor .part of an internal column has been found to be short, andsince the arrangement ,of loads symmetrical, this design can be carried out accordingto 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 theappropriate reduction factors for imposed loads specified in "BS 6399: Part 1 (1984):Design loading for buildings: Code of practice for dead and imposed loads". Thepartition loads are taken as imposed loads, since their positions are not fixed.

59


Column grid dilDClDsions are 5.0 x 5.0 m.Hence, area conesponding to column" (5)2.. 25 m2

Dead loadsFrom 4 slabs = (4)(24)(0.175)(25) = 420 kNFrom beams= (4)(0.525-D. 175)(0.3)(24)(5)=50.4 kNFrom columns={(3)(2.975)+5}(0.3Y(24) =30.1 kNFrom finishes = (4)(1.0)(25) = 100 kN

Total dead load = 600.5 kN Ok = 601 kN

= 25 kN= 187.5 kN=~

= 287.5.kNNote 3

3.8.4.3equation 38

Note 4

Note 5

Table 3.27

Imposed loadsFrom roof = (1.0)(25)From 3 floors = (3)(2.5)(25)From partitions- (3)(1.0)(25)

Total imposed loadI.L. reduction due to floor area

= (0.05)(25/50)(287.5)(4) = 28.75 kNI.L. reduction due to 4·floors

= (0.3)(287.5) = 86.25 kNHence, imposed load = 287.5 - 86.25 == 201.25 kN

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

Desim of main steel

For short columns resisting axial loadN = (O.4)fcu 'Ac -t (0.75)A,.,.fyAssuming 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 mm2) .

Note:- lOOAwc/Ac = (lOO){804) I (300)2== 0.89 > 0.4; hence min. steel O.K.

o.c .. 201 kN

N .. 1163 kN

Use 4 T16(at columncomers)

Notes on .Calculations

3. The total imposed load can be reduced either on the basis of the area supported bythe column or the number of floors -supported by the column. In this case, thereduction allowed as a result of the latter is greater and is hence applied - see as6399: 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 fromequation 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 Ascobtained 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 nominalsteel will be sufficient. In any case, bar diameters under 12 mm are generally notused for columns, because they will not be stiff enough for the erection of thereinforcement cage.

Concluding Notes

6. This method of design is applicable for short braced columns, where moments. arenegligible, due to a symmetrical arrangement of loads. Even if this symmetry is onlyapproximate, provided the columns are short and braced, equation 39 can be used inplace of equation 38.

7. In addition to the main reinforcement, columns should be reinforced by links whichsurround 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/mm2 and fy d 460N/mm2•

Introductory Notes

1. This column canies a substantial moment as well as an.axial load. Hence, we shallhave to use the design charts, which will give us a symmetrically reinforced section.

61

,.


N = 800 kN, M == 80 kNmNlbh = (800 xI(3) 1 (400)(300) = 6.67

(for all cases)

(al Major axis bendine [JNote 2 b = 300 mm, h = 400 mmChart 23 Mlbh2 = (80 xl()6) 1 (300)(400)2 = 1.67

~ 0(part 3) lOOA.Jbh = 0.4

A~ = (0.4)(300)(400) I (100) = 480 mm2 major axisUse 4 T16 (Asc = 804 mm2) 4T16

(b) Minor axis bendine

CJb '''';' 400 mm,h = 300 mmChart 23 Mlbh2 = (80 xl()6) 1(400)(300)2 = 2.22(Part 3) lOOAaclbh = 0.8

o 0

A~ = (0.8)(300)(400) 1 (lOOi = 960 mm2 minor axisUse 4 TIO (Ase = 1256 mm ) 4 no

(<<)~ !;>endin3.8.4.5' MKlh' '= (80 xHi~ t(~50)Note 3 MyIb' = (80 x106) 1 (300-50)

hence M,/h' < M~'equation 41 M ' = M +(fl) 'Ih')~

N1 (b.Ii.fcu>= (800 x1(3) 1 (300)(400)(l.S) = 0.267

0Table 3.24 hence P = 0.690~' = 80 + (0.~(250/300)(8O) = 126 leN • •

Chart 23 Mlbh2 = (126 xl )/ (400)(300)2 = 3.5 OeO(Part 3) 100~lbh == 1.7 .

A~ = (1.7)(300)(400) 1(100) = 2040 mm2 biaxialNote 4 Use 4 TI5 + (2 T16 in each dir.) (A~= 2366 mm2) 4 T25 + 4 T16Note 5

)&silo of links JiDb3.12.7.1 , For major axis~ding, use R6 {> (16/4) = 4 mm} major axis -

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

Note 6 @ 175 mm {< (12)(16) = 192 mm}. R8@ 175 mm.

Check for shear3.8.4.6 MIN = (126)/(800) = 0.158

« (0.75)(0.3) = 0.225 m); Shear checkNote 7 hence, shear is not critical. oot required.

62


Crack control3.8.6 (0.2)fcu.Ac = (0.2)(25){(3OO)(400) - 804} = 596 kN Crack width

Axial load = 800 kN > 596 kN. check notHence, no check is required. required.


2. If we assume a cover of around 30 mm (moderate exposure conditions and TABLE1 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 correspondsto a d/h value of 0.85 - can be used. If there is a doubt, the lower d/h value shouldbe used, as this is more conservative. It should be noted that the column design chartshave a lower limit of looAsclbh = 0.4, thus ensuring that the minimum steelrequirement 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, bya similar argument as in Note 2 above.. '

4. If the steel requirement for bi-axial bending is greater than that which can beprovided as corner steel, the additional amount required has to be provided in eachof the, two mutually perpendicular directions, distributed along the faces of thesection. 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 BritishStandard 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'areused, so that the link spacing would no~ be too small; furthermore, bars smaller thanT12 are not used as column reinforcement, as they would not be stiff enough duringerection.

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 tobe used if (i) there is more than one intermediate bar or (ii) the intermediate bar isgreater 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 thelimits on vmax'

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 loadof 800 kN. It is bent in double curvature about the major axis, carrying ultimate momentsof 80 kNm and 40 kNm at its ends. The effective length of the column corresponding to themajor axis is 7200 mm. Determine the column reinforcement if feu = 25 N/mm2 and fy =460 N/mm2•

Introductory Notes

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


Type of column3.8.1.3 Vh = (7200)/(400) = 18 > IS; hence s~nder.

3.8.3.3 < 20, slender columnAlso h/b = (400)/(300) = 1.33 < 3; bent about

Note2 hence, single axis bending. major axis.

Design moments3.8.3.2 M1 = -40 kNm; M2 = +80 kNmequation 36 Mi = (0.4)(-40) + (0.6)(80) = 32 kNm

{= (0.4)(80) = 32 kNm} Mj = 32 kNm3.8.2.4 emin = (0.05)(400) = 20 mm

N.emin = (800)(0.020) = 16 kNmequation 34 (3a = (1I2000)(IJb')2= (112000)('200/300Y= 0.288equation 32 au = #a·K.h = (0.288)(1)(0.4) = 0.115equation 35 Madd == N.au = (800)(0.115) = 92 kNm Madd = 92

Hence, critical moment is M j + Madd = 32 + 92 = kNm124 kNm. However, as K is reduced, if Mj + Maddbecomes < 80 kNm, M2 will become critical.

Design of reinforcement

TABLE 1 Assuming cover = 30 mol (moderate exposureconditions and TABLE 1 values modified by Notes 5

Chart 23& 6), link diameter of 8 mm and main bar size of 25mm, d/h = (400-50.5)/(400) = 0.87.

(Part 3) N/bh = (800 x103) I (300)(400) = 6.67M/bh2 7 (124 xI06) I (300)(400)2 = 2.58K = 0.9

64


Madd = (0.9)(92) = 83 kNmM = 32 + 83 = 115 kNm (> 80 kNm)

Chart 23 Mlbh2 = (115 xldi) 1 (300)(400)2 = 2..40(Part 3) K = 0.85

Madd ~ (0.85)(92) = 78 kNmM = 32 + 78 = 110 kNm

Note 3 ·Mlbh2 = 2.29; K = 0.85 (again). K = 0.85Chart 23 Hence, lOOAsc/bh = 0.8(part 3) Asc = (0.8)(300)(400) 1 (100) :;: 960 mm2 main steel -

Use 4 1'20 (Asc = 1256 mm2)

::IaJnksl3.12.7.1 Use R 6 {> (20/4) :;: 5mm} @ 225 mm

{< (12)(20) :;: 240 mm} R6@ 225 mm


2. When major axis bending takes place, if either the Vh value is greater than 20 or thebib value is not less than 3, in order to account for the deflection due to,'Slendemessabout the minor axis, the column has to be designed asbiaXiaUy Dent, with zeroinitial 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 virtuallyconstant. It should be noted that the factor K should be applied to the original valueof Madd alone. "

Concluding Notes

4. The reinforcement required for this column is the same as for minor axis bending ofthe 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 anultimate load of 1100 leN (service load of 760 kN), if the allowable bearing pressure of thesoil is 150 kN/m2• Use feu = 25 N/mm2, fy = 460 N/mm2 (deformed type 2) and densityof reinforced concrete = 24 kN/m3•

Introductory Notes

1. Square pad footings are the most common foundation type for columns of framedstructures. 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.

Reference Calculations " Output

Dimensions of base

Note 3 Service load = 760 leN \Note 4 Expected total load = (1.08)(760) = 821 leN

Required area for base = (821) 1 (150) = 5.47 m2

Try a base of 2.4 m x 2.4 m x 0.4 mWeight of base = (2.4Y(0.4)(24) = 55 leNActual total load = 760 + 55 = 815 leNBearing pressure = (815) 1 (2.4j = 141 kN/m2

< 150 kN/m2;hence O.K.NoteS Preliminary check on effective depth:-

d > 10(N)°'~ = 10(1100)°·5 =332 mm; footing sizehence overall depth of 400 mm is O.K. ~x2.4m

.4m

DesilW for ben4in& \

~Note 6 Ultimate bearing pressure = (1100) 1 <2.4/= 191 kN/m

~ritical bending moment (at face of column) =(191)(2.4){(2.4-o.3)/2}2(1/2) = 253 leNm M = 253 leNm

TABLE 1 Assume a cover of 40 mm, for moderate exposureNote 7 conditions. H bar size of 16 mm is assumed, dmin = dmin = 336 mm

400-40-16-16/2 = 336 mm and daV2 = 344 mm. dan = 344 mm

66


Chart 2 Mlbd2 = (253 xIW) I (2400)(336)~ = 0.93(Part 3) l00A/bd = 0.25

As = (0.25)(2400)(336) I (100) = 2016 mm2

3.11.3.2 (3/4)c + (9/4)d = (3/4)(300) + (9/4)(336)= 981 mm < Ie = 1200 mm;

hence reinforcement should be banded.Use 7 Tl6 @ 200 mm in band of 1200 mm bottom r/f{< (3)(336) + 300 = l308}; Use (3+3) T12 @ 3 T12 @ 200200 mm in two outer bands. 7 T16 @ 200(As = 1407 + 678 = 2085 mm2; 1407/2085 > 2/3) 3 T12 @ 200

3.12.11.2.7 Max. spacing = 750 mm; hence O.K. (both ways)100AiAe = (100)(2085) I (2400)(400) = 0.22

> 0.13; hence O.K.Note 8 Anc~ge length :; (40)(16) = 640 mm

< (2400-300)/2 = 1050 mm; hence O.K.

Check for yerticalline shear

[§I3.4.5.10 Consider a section at "d" from the column face, .andassume no enhancement to ve.V = (191)(2.4){(2.4-o.3)/2 - 0.336} = 327 kNv = (327 xl<P) I (2400)(336) = 0.41 N/mm2 .l00A/bvd = (100)(2085) I (2400)(336) = 0.26

Table 3.9 ve = 0.42 N/mm2 > 0.41 N/mm2; hence O.K.

Check for punchine shear

m3.7.7.2 Vmax = (1100 xl<P) I (4)(300)(344) = 2.66 N/mm2< (O.8)(25f·S= 4 N/mm2 < 5 N/mm2; hen~ 'O.K.

3.7.7.6 1st critical perimeter = (4){(1.5)(0.344)(2) + 0.3}. = (4).(1.332) = 5.328·m

Area outside perimeter= (2.4)2 - (1.332)2= 4.43 m2

V = (191)(4.43) = 761 kNNote 9 v = (761 xl<P) I (5328)(344) = 0.42 N/mm2 Shear r/f not

= ve (0.42 N/mm2); hence O.K. required.


3. Soil bearing pressures are given in terms of service lOads. Hence, service loads haveto be used to determine the footing area. Service loads can be approximately obtainedfrom 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 factorthe latter by 1.5.

'4. The weight of the footing itself cannot be known until it is sized. An allowance of8% of the column load is generally satisfactory for obtaining a first estimate offooting weight, which should subsequently be calculated accurately. Another approachis to first estimate a depth (in this case 0.4 m) and reduce the allow~le bearingpressure 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 leNand d in mm. It is based on a punching shear considerations for commonly used padfootings. 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 whichis taken directly by the soil reaction, It should not be considered when designing forthe 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 formoderate exposure conditions. However, the cover is increased by a further 5 mm,in case the footing comes into contact with .any contaminated ground water. Theminimum value of "d" should be used in the design for flexure and vertical lineshear, 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 theanchorage 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 maximumand minimum pressures under the footing will be given by (l/BL}(N ± 6M1L), withsymbols having usual meanings. The maximum pressure should be kept below theallowable 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 verticalshear by assuming that the pressure distribution is uniform and equal to the maximumpressure.

12. Where the design for punching shear is concerned, the average pressure can be takenfor calculations, but a factor of 1.15 applied to the shear force, according to theprovisions 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 footingsfor the external and first internal columns have to be combined, as shown. While the internalcolumn carries an ultimate axial load of 1100 kN, the external column carries an ultimatemoment of 60 kNm in addition to an ultimate axial load of 600 leN. The allowable bearingpressure of the soil is 150 kN/m2• Use feu = 25 N/mm2 and fy = 460 N/mm2•

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

\lOkN

C ~D E

Introductory Notes

1. The situation described above is often found in crowded urban areas where buildingsare constructed on very small plots of land.

2. It is difficult to provide an isolated pad footing for the external column, because ofeccentric loading on the footing. Hence, it can be combined with the first internalcolumn footing as shown above.


Service loadsNote 3

External column load = (600)/(1.45) = 414 leNExternal column moment = (60)/(1.45) = 41.4 kNm

,Internal column load = (1100)/(1.45) = 759 leN

Dimensions of footine

Distance of C.of G. from A is given by x, where(759+414)x = (414)(0.15) + 41.4 + (759)(5.15)Hence, x = 3.4 m; therefore, for uniform pressure

Note 4 under base, use a base of length (2)(3.4) = 6.8 m

69

Reference CalcuJatloDS Outppt

Note 5 If we assume a thickness of 0.8 m for the base, theallowable bearing pressure is150 - (0.8)(24) • 131 kN/rrWidth of base required = (759+414) I (6.8)(131) footing size

= 1.32 m 6.8 m·x 2.0 mNote 6 Use base of 6.8 III x 2.0 m x 0.8 m xO.8 m

Analysis of footing

Assuming that the C.of G. for ultimate loads is thesame as that for service loads, the footing can beidealised as follows:-udl = (1100+600) / (6.8) = 250 leN/m

Note 7 ~~ :1TA-&;::':_= ,6Ew • 250 kN/m/ x~

Max. moment at C = (250)(1.65Y I 2 = 340 leNmTo find max~ moment in AC.Mx == -(250)x212 + llOO(x-1.65)

= -(125)x2 + (lloo)x - 1815putting dMx/dx == -(250}x + 1100 =0,we have x = 4.4 mMmax = -(125)(4.4)2 + (1100)(4.4) - 1815

= 605 kNmMax. shear force at C == 1100 - (250)(1.65)

= 688 leNShear force at A = 600 leN

Design for bending

TABLE 1 Assume cover of 40 mm, bar size (longitudinal) ofNote 8 25 mm; hence, d == 800 - 40 - 25/2 = 747.5 mm

(Section AC)Chart 2 Mlbd2 == (605 xl06) I (2000)(747.5)2 = 0.54(part 3) looAJbd == 0.15 .

A. -= (0. 15)(2000}(747.5) I (100) = 2243 mm2 longitudinal topUse 5 ns (As = 2455 mm2) - on top surface; steel- S n5these can be curtailed if required.

70


~fb~~ = (340 x106) 1 (2000)(747.5)2 = 0.30Table 3.27 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.

3.11.3.2 Note:- (3/4)c + (9/4)d = (3/4)(300) + (9/4)(747.5)= 1907 > Ie = 1000 mm;

hence, longituninal steel can be evenly distributed.

crransverse direction)M = (25012){(2.0-0.3)/2}2(112) = 45 kNm/md = 747.5 - 25/2 - 20/2 = 725 mm (assuming barsize of 20 mm)M/bd2 = (45 x106) 1 (1000)(725)2 = 0.09

Table 3.27 Use 100A/Ae = 0.13As = (0.13)(1000)(800) 1 (100) = 1040 mm2/mUse TIO @ 300 mm (Ag = 1047 mm2/m)

Table 3.29 Anchorage length = (20)(40) = 800 mm< (2000-300)/2 - 40(cover) = 810 mm; hence O.K.This steel too can be evenly distributed, as it isnominal reinforcement; the same nominal steel canalso be used as distribution steel for the toplongitudinal bars.

t4.7m

15T25 \.s

, ·~T20@300 ........ J. . , ,'5T25

>Ik 6.&nCheck for vertical line shear

Output

longitudinalbottom steel 5TI5

longitudinal rlfto be evenlyspaced

transversebottom steel T20@3oomm, to beevenly spaced.

3.4.5.10 In the longitudinal direction, check at a distance "d"from the internal column face.V = (688) - {(688+600)/(5.0)}(0.15+0.7475)

= 457 kNv = (457 x1<P) 1 (2000)(747.5) = 0.31 N/mm2

Since nominal steel is used, ve = 0.34 N/mm2

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

71


Check for punCbio& shear

3.7.7.2 vmax (internal column) = (1100 xlol) I (4)(300)(735)= 1.25 N/mm2

< (0.8)(25f·S=4 N/mm2 < 5 N/mm2; hence O.K.As the critical perimeter, (1.5)d from column face, isoutside the footing, only vertical line shear need be Shear r/f is notchecked. required

Notes 00 Calculations

3. Where service loads are not specified or known, they can be estimated by dividingultimate loads by 1.45 for.reinforced concrete structures. When converting serviceto 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 pressuredistribution 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 alsorequire a, fairly deep base.

6. This fairly large width has been chosen to reduce the pressure under the footing andsatisfy the shear criteria. Although increasing the depth is generally more efficientthan increasing the width, having a large depth may also cause excessive build up ofh5t of hydration temperatures, leading to thermal cracking.

7. Assulning the column loads to be point loads is conservative. In reality, the load willbe spread over a (mite area and the resulting bending moments and shear fol'CeS' atthe 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.

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 andsubgrade are rigid. If elastic foundation assumptions had been used, the soil pressurenear the columns (i.e. loaded areas) would increase, but the midspan bending momentwould decrease.

72

EXAMPLE 24 - PILE CAP.

A two-pile group of pile diameter 500 mm and spacing 1250 mm centres supports a 450 mmsquare column carrying an ultimate load of 2500 leN. Design the file cap, using concrete ofgrade 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 forend 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 toanalyse forces is to consider truss action in the pile cap.


Pile cap dimensions

Note 3

Use an outstand beyond the piles of half the pilediameter. Hence, pile cap di~ensions are:length = 1250 + 500 + 500 = 2250 mmwidth = 500 + 500 = 1000 mmTry overall depth of 700 mm; hence,effective depth = 700 - 40 - 25/2 = 647.5 mm

> (1250)/2; hence O.K.

dimensions2250 mm x1000 mm x700mm

main steel71'25

N

61t T l'N/2 N/2

, 21 ~

The force T is given byT II: N.l/(2)d

= (2500)(625) 1 (2)(647.5)= 1207 leN

As = (1207 x1oJ) 1 (0.87)(460)= 3016 mm2

Use 7 1'25 (As = 3437 mm2)

Banding is not required, as pile spacing < (3)<pspacing of rlf ::; (1000-80-32-25) I 6 = 144 mm

Truss action

Note 4

3.11.4.2Note 5

Anchoraee

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

73


Stress in r/f - (0.87)(460)(3016/3437)= 351 N/mm2When turning bars upwards, assume that bend startsat edge of pile. 'Stress at start of bend = (351)(878-250)/(878)

= 251 N/mm2

Bend radius, r, should be S.t.equation 50 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

Note 6 If we start the bend as close as possible to the edgeof pile cap, length from C/L of pile to start of bend= SOO - 40 - 16 - 160 = 284 mm

3.12.8.23 eff. anchorage of bend = (12)(25) = 300 mm{< (4)r =' (4)(160) = 640 mm}

vertical length available = '647.5 - 40 - 160 - (4)(25)- 347.5 mm

total anchorage available - 284 + 300 + 347.5. = 931.5 mm > 878 mm required; hence O.K. Anchorage O..K,

Check for punchina shear

3.11.4.5

3.11.4.43.11.4.3

Table 3.93.4.5.8

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 isrequired.

Check for vertical line shear

av = 625-150-225 = 250 mmV (at critical section) -

2500 I 2 - 1250 kNv - (1250 xloJ) I (1000)(647;5) k ~ "

_ 1.93 N/mm2 625 6252d/av = (2)(647.5) I (250) = 5.18looA/bd = (100)(3437) I (1000)(647.5) = 0.53Vc = (0.51)(5.18) - 2.64 N/mm2

> 1.93 N/mm2; hence O.K.

74

Shear rlf notrequired.


Distribution steel

Table 3.27 l00A/Ac = 0.13; As = 910 mm2/mUse 16T @ 200 mm (As = 1005 mm2/m) distribution steelThis steel can also be bent up like the main steel. 16T@200 mm

Horizontal binders Il binders7T25 main 4')16

Note 7 Use 25 % of main steel.',

As = (0.25)(3016)= 754 mm2 horizontal

Use 4 T16 (As = 804 mm2)

·LJTl~binders

These binders will tie the 4 Tl6main and distributiot:l steel.


3. The criterion used is that the effective depth is equal to at least half the distancebetween pile centres. When using truss theory, this will result in a compressive strutof 45° minimum inclination. The cover value has been chosen as per Note 7 ofExample 22.

4. If the width of the column is accounted for, the value of tensile force will be a littleless. This is given some treatment in nAllen, A.H., Reinforced concrete design to BS8110 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 mmdiameter.

6. We start the bend closer to the edge of the pile cap than assumed in the bend radiuscalculation - this is to achieve as great a length for anchorage as possible within thegeometry of the pile cap.

7. This provision is also given by Allen, referred to in Note 4 above. The main steelrequired (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, theanchorage requirement beyond the centre line of pile is much less. However, TrussTheory 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 forpunching 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 differencebetween the two levels is 2.0 m. Assuming that the staircase is sheltered and that it issubject to crowd loading, design the staircase, using feu = 25 N/mm2, fy = 460 N/mm2

(deformed type 2) or 250 N/mm2 (plain) and density of reinforced concrete = 24 kN/mJ .

Assume top finishes (on tread only) as O.S kN/m2 and the soffit plaster as 0.25 kN/m2•

IJ;ttroductory Notes

1. Staircases are essentially inclined slabs. The major difference in design approachis 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.


Choice of layout

Let us choose 12 stairs.Then, rise (R) = 2000/12 = 167 mm rise = 167 mm

going (0) = 3000/12 = 250 mm going =Note 3 Also 2R + 0 = (2){167) + 250 = '84 mm 2SOmm

(approx. 600 mm; hence O.K.) tread =Use nosing of 25 mm, so that tread = 275 mm 275 rom

Waist thickness

Note 4 Assume trial span/depth of (30)(1.15) = 34.5, for a3.10.2.2 I-way heavily loaded continuous slab, stiffened by

stairs.effective depth = (3000)/(34.5) = 87 mm

TABLE 1 If we assume cover =' 20 mm (mild exposureconditions and concrete protected by 10 mm 1:3 ct:sand rendering) and bar diameter = 12 mm, we canchoose h == 120 mm and d = 120-20-12/2 = 94 mm h = 120mm

Table 3.5 Note:- 1.5 hr. fire resistance available. d = 94 mm

L.oadin& (for 1 m wide strip)

Note S Factor for slope .... (R2 + 0 2)0.5 1 0= {(167)2 + (250)2}0.5 1 (250) = 1.20

76


Factor for overlap = T/G = 275/250 = 1.1

Output

Waist = (0.1~)(1)(24)(1.2) = 3.46 kN/mSteps = (112)(0.167)(1)(24)(1.1) = 2.20 kN/mTop finishes = (0.5)(1)(1.1) = 0.55 kN/mSoffit plaster = (0.25)(1)(1.20) = 0.30 kNlm

Total dead load = 6,51 kNlmNote 6 Imposed load = (5.0)(1) = 5.0 kN/m design udl =

Design load ... (1.4)(6.51) + (1.6)(5.0) = 17.1 kN/m 17.1 kN/m

PesiKn for bendine

Note 7

Chart 2(Part 3)

3,12,11.2,7Table 3.27Note 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 = 1.75l00AJbd = 0.48A. = (0.48)(1000)(94) 1(100) = 451 mm2/mUse Tl2 @ 250 mm (As = 452 mm2/m) main steelAllowable spacing = (3)(94) = 282 mm > 250 mm T12 @ 250 mm100A/Ac= (100)(452) I (1000)(120) = 0.38 > 0,13Hence,max. spacing and min. steel are O.K.


Mfbd2 = 1.75fa = (5/8)(460)(451/452) = 287 N/mm2

Table 3.11 Hence, F1 = 1.15 (for tension reinforcement)3.10.2.2 Allowable spanldepth = (26)(1.15)(1.15) = 34.4

Actual span/depth = (3000)/(94) = 31.9 < 34.4;hence deflection O.K.

Check for shear

Deflection O.K.

Note 9

Table 3.9

v = (0.6)F = (0.6){(17.1)(3)) = 30.78 kN/mv = (30.78 x1<P) I (1000)(94) = 0.33 Nfmm2

l00AJbd = (100)(452) I (1000)(94) = 0.48vc ... 0.66N/mm2 > 0.33 N/mm2;Hence, shear r/f not required.

77

shear r/f notrequired.

Reference

Distribution steel

Calculations Output

Table 3.27 lOOA,IAc = 0.24 (for mild steel)Note 10 A. = (0.24)(1000)(120) I (100) = 288 mm2/m

Use R8 @ 175 mm (As = 287 mm2/m)distribution steelR8@·175 mm

R8@175

~Note 11

Tl2@250 Tl2@250 (1)

(4)


3.<In ..~fC----------:>j

3. In general the rise should vary from a minimum of 150 mm for public stairways toa maximum of 175 mm for private stairways. The going should vary from 300 mmfor public stairways to 250 mm for private stairways. (2R + 0) should be kept asclose as possible to 600 mm. A nosing can be provided so that the tread is greaterthan 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 ratioof 30 is used here, because the loading is much heavier - the waist carries the loadof the steps, in addition to its own weight on an incline, and also a high imposedload. The above ratio is increased by 15% because of the stiffness contributed bythe stairs.

5. The waist and soffit plaster have thicknesses that are measured perpendicular to theincline. Hence their load in plan will be greater by a factor of (R} +0 2)0.5 I O. Thesteps 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 byT/O. This factor can be ignored for the imposed load, because it can be argued thatthe 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: Part1 (1984): Design loading for buildings: Code of practice for dead and imposedloads" .

7. For a staircase having continuity, we can assume that both the span and supportmoments 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 theuse of high yield reinforcement may result in more steel than that specified by theminimum 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 away that an inside comer can get pulled out. Hence bar type (2) should be continuedfrom 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 waistfrom the faces of the beams. Bar types (1), (2) and (3) can be continued into thelower and upper slabs as sl$ reinforcement, if required. Bar type (4) shows how theupper slab reinforcement can be taken into the beam support.

Concluding Notes

12. If the supporting beam for the flight of stairs is at the ends of the landings, the entiresystem of staircase and two landings can be taken as spanning between the supportingbeams (see figure below). In this case, the slab system could be considered as simplysupported if there is no continuity beyond the landings. The loading on the landingand 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 adirection perpendicular to the flight and span of the stairs. The dimensions of a single stairare as follows:- rise = 175 mm; going = 250 mm; tread = 275 mm. The top finishes (ontread only) are equivalent to a distributed load of 0.5 kN/m2 and the soffit plaster one of0.25 kN/m2• The imposed load can betaken as 3 kN/m2• Using feu = 25 N/mm2, fy = 460N/mm2 (deformed type 2) or 250 N/mm2 (plain) and density of reinforced concrete ;:; ~4kN/m3,design the staircase component of the system.

1.2m 8xO.25 = 2.Om 2.Om\<:---.....;* *:------~)'

."1. 2m

,,

1.2m

I,

Introductory Notes

1. In this example of a staircase, the landings span perpendicular to the stairs andsupport the staircase, unlike in the case described in Note 12 of Example 2S, wherethe landings also span in the direction of. the staircase. •

2. In analysing the above system, the staircase is assumed to be supported along twoedges within the landings. Continuity over the supports can be assumed for thepurpose of spanldepth Iatio calculations.


Stair span and waist thickness

equation 47 Effective span = la + (O.5)(lb,l + Ib,VNote 3 = 2.0 + (0.5)(1.2 + 1.8)

= 3.5 m eff. span =length of stairs (%) = {(2.0)/(3.5)}(IOO) % = 57% 3.5 m

3.10.2.2 < 60%; hence span/depth enhancement not possible.

80

Reference CaJcuJatioas Outpat

Note 4 Assume a trial spanldepth ratio of 32, for acontinuous I-way spanning sIaiR:ase.Effective depth = (3500)/(32) = 109 mmAssuming a cover of 20 mm (mild exposureconditions and concrete protected by 10 mm 1:3 ct:sand rendering), and bar diameter of 12 mm, we can h = 140 mmchoose h= 140 mm and d= 140-20-1212 = 114 mm d = 114 mm

Loa4in& (for 1.2 m wide staiR:ase)

Note 5 Factor for slope = (R2 + OZ)0S I G= {(l75f + (250f}o.s I (250) = 1.22

Factor for overlap = TIG = (275)/(250) = 1.1

Waist = (0.14)(1.2)(24)(1.22) = 4.92 kN/mSteps = (112)(0.175)(1.2)(24)(1.1)= 2.n kN/mTop finishes = (0.5)(1.2)(1.1) = 0.66 kN/mSoffit pJasta' = (0.25)(1.2)(1.22) = 0.37 kNlm

Total dead loed = 8.72 kN/mImposed lOad = (3.0)(1.2) = 3.6 kN/m design udl =Design load= (1.4)(8.72) + (1.6)(3.6) = 18.0 kN/m 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 kNRB = (18)(2.0) - 19.5 = 16.5 leNMx = RA.x - w(x - 0.6f12dMxIdx = 0 when RA - w[x - 0.6] = 0Le. x= RA/w + 0.6 = (19.5)/(18) + 0.6 = 1.68 mMuua= (19.5)(1.68) - (18)(1.68-G.6t/2

= 22.26 kNm

Chart 2(Part 3)

MIbc¥ = (22.26 xl<n 1 (12OO)(114t = 1.43l00A/bd = 0.39; Ac= 534 mm2

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

81

main steel5 T12

Wei.,... CaJcu,,- Output

Check for cW1ection

MJbd2.:: 1.43;fs = (5/8)(460)(534/565) =2n N/mm2

Table 3.11 H~,FJ == 1.29 (for tension reinforceament)~..spanldepth :: (26)(1.29) == 33.5Actual spaoIdeptb == (3500)1{U4) == 30.7

< 33.5; hence O.K. 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 distribution steelUse R8 @ lSOmm <As == 335 m.r/m) R8@ ISO mm

Notes OD Cakulatious

3. The support Jineforthe stain:ase is at the centre of the smaller landing but only0.9 m into the wider Janding, because 1.8 m js the maximum distance ,over which theload 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, itwill be safeeto assume simple suppoIts when designing for bending, as tile continuityextends only upto the edge of the Jaadiqg. The loads fromd1e landings are.carried bythe landings in the direction.~ to the flight of the stairs; hence they arenot considered in the analysis.

Concluding Notes

7. Detailing of reinforcement can be·cJooe in a IIl3DDel'sinillar to that in Example 25.

8. Shear can also be checJr£d for, as in Example 25, the maximum.sIlear force being thegreater of RA and~ - i.e. 19.5 kN.

9. When designing the landings, in addition to their own dead and imposed loads, theloads from the stain:ase --i.e. RA and RB will be uniformly spread over the entiresmaller landing and over 1.8 m of the Jarges landing, respectively.

10. Where staircase flights surrounding openwe1ls intersect at right angles, the loadsfrom 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 endconcrete shear walls of length 15 m and height 14 m. The servi<;e wind load on one shearwall is 180 leN. Check.whether a plain concrete wall of grade 25 concrete and 175 mmthickness is sufficient for the wall panel between foundation and 1st floo~ slab (clear heightof wall =4.0 m) if it carries the following terVice loads, in addition to the wind load:- selfweight = 18 leN/m; dead load from 1st floor. slab = 12 kN/m; dead load from above 1stfloor slab = 8OleN/m; imposed load from 1st floor slab = 7.5 kN/m; imposed load .fromabove 1st floor slab = 19 kNlm.

Introductory Notes

1. Given that.even plain concrete walls require horizontal and vertical reinforcement(Clause 3.9.4.19), and if this n"inforeement will be ,distributed on two faces (whichis advisable, since crack control reinforcement should be as close to the surface aspossible), then it is very difficult to construct a wall under 175 mm. This isbecause cover requirements will be 20 mm on the inside (mild exposure) and 30 mmon the outside (moderate exposure) - see TABLE 1, including Notes 5 and 6 - andbecause the bar diameter for vertical steel should be at least 12 mm, in order toensure 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): Basicdata for the design of buildings: Loading: Wind loads·, and the method is shown inExample 31.

3. It is assumed that stability for the stn1Cture as a whole has been satisfled. Theoverturning moment due to wind, factored by 1.4 should be less than the resistingmoment due to dead load, factored by 1.0 (see Table 2.1)

Refereoce Cakulatioos Output

Wall CJassi'ierinn

3.9.4.3 Since lateral support is provided by foundation andNote 4 1st floor slab, wall panel can be considered braced.

The slab will give only displacement restraint, whilethe foundation can give displacement as well asrotational .restraint.

Ie = 3.5 mNote 5 Hence, Ie = (0.875)(4.0) = 3.5 m lJh = 201.2.4.9 Ie/h = (3.5 xloJ)/(17S) = 20 > 15; hence slender. hence, slender3.9.4.4 < 30; hence max. value not exceeded. braced wall.

83

3.9.4.9

Note 6

3.9.4.15

3.9.4.16

Table 2.1

equation 43

equation 44

Note 7

3.9.4.18

We sball use serviceability vertic:al Joads to calculatethe resultant eccentricity (e) just below the 1st floorslab, assuming the eccentricity of 1st floor slabloading is bl6. and that the eccentricity of loadsabove 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

Assuming wind acts at mid height of wall, windmoment == (180)(1412) == 1260 kNm.Hence. wind loading on wall == ± (6)(1260)/(15>2

== ± 33.6kN1mHence. ultimate loads per unit length of wall are:Combination 1.f l == (1.4)(18+ 12+80) + (1.6)(7.5+ 19)(0.7)

== 184 kN/mCombination 2.f2 == (1.4)(110) + (1.4)(33.6) == 201 kNlm orf2 == (1.0)(110) - (1.4)(33.6) == 63 kN/mCombination 3,f3 = (1.2){1l0 + (26.5)(0.7) + 33.6} = 195 kNlmNote: - no tension arises.

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 shear

Design horizontal shear force == 180 kNMin. deagn vertical load == (110)(15) == ·1650 kN

{> (4)(180)(1.4) == 1008 kN; hence O.K.}

84

ex == 8.75 mme. == 28 mm

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

63 kN/m

l1w is O.K.

i (

1./; l

..

4 "

Refereace Caleulatio.. Output

Minimum reinfOfCQDCOt

Min. r/f = (0.25)(1000)(175) I (100)= 437.5 mm2/m (both directions)

Note 8 Use vertical steel Tl2 @ 300 mm in both faces vertfcal steel(A. = 753 mm2/m) 2 x Tl2 @ 300Horizontal steel 1'8 @ 225 mm in both faces horizontal steel(As = 446 mm2/m) 2xT8@225

- - Notes on Calculations

4. A column is considered braced in a given plane if it is not required to carry thelateral forces in that plane. A wall bowever is considered braced if lateral stability isgiven to it by other structural elements, when it is carrying in-plane loads. If the wallalone has to resist transverse loads, it is unbraced.

5. Since the end conditions in the given wall are "midway· between those specified inClause 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 areinvolved. Later on, when cbecldng the Ow value for the wall .panel, a factor of 0.7 isused, since loads from 4 floors are involved.

7. Equations 43 and 44 for braCed walls correspond to the top (maximum initialeccentricity) and midway (maximum eccentricity due to deflection) sections.However OW is calculated at .the bottom of the wall,. taking into account the selfweight of the wall and maximum inoment due to wind. This is slightly inconsistentbut conservative. A similar approach is used in column design.

8. Since reinforcement to control thermal and hydration sbrinkage should be fairlyclosely spaced, a spacing of 300 mm is not exceeded. 12 mm dia. bars are used forvertical 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 bein the horizontal direction; furthermore it is easier to fix the horizontal steel on theoutside.

Concluding Notes

9. The wall reinforcement should also be checked for satisfying tie reinforcements. Thisis 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 ofthe column. Take ~u= 30 N/mm2 and fy = 460 N/mm2 (defcmned type 2).

Introductory Notes

1. A cos:bcl can be considered to bea -deep cantilever"• where truss Don. as opposedbeam action, predominates and where shearing action is critical..

2. Compatibility of strains between the strut-and-tie system of the truss must be ensuredat the root of the corbel (Clause 5.2.7.2.1 (b».

Corbel dimensions

Calc:ulatioos Output

5.2.3.4

Note 3

The width of the corbel can be the same as that ofthe column, Le. 300 mm.The length of the bearing plate can also be taken as300 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)fcub >= (350 xloJ) I (300)(0.4)(30)= 97 nunHence, choose bearing width of 100 mm. bearing widthSince the corbel has to project out from the bearing 100 mmarea a distance that wouJd accommodate a. stressedbend radius choose corbel projection as 400 mm. total proj~tionCorbel depth bas lObe·suclltbat max. aIL shear is 400nunnot exceeded - i.e. (O;8)(3Of.5 = 4.38 N/mm2

Hence, d > (350 xl(3) I (300)(4.38) = 266mmChoose h = 375 nun and assuming cover of 20 mm(mild exposure conditions, concrete protected by lQmm 1:3 et: sand rendel) and bar dia. of 20 mm,d = 375 - 20 - 2012 = 345 mm. h =375 mmLet the depth. vary from 375 mm to 250 mm. d::;: 345 mm

cAy ~l~n: 350 kN (ultimate)

T~C' 'l5o TL1:- t 125 Oo9xl[J

-A 0.45fv cu

86


5.2.7.1 Now avid == (150)/(345) == 0.43 < 1Also, depth at outer edge of bearing area > 375/2mm; 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' thecentroid of stress block, (J == tan-1(zlI50),Le. {J == tan-I {(d - 0.45x)IlSO}Furthermore, from the triangle of forces for P, TandC,C == P 1 Sin{J ........................ (2)We need to find a value of x, and hence {J, that willsatisfy (1) and (2) simultaneously.x == 216 mm will give {J == SS.SO and C == 409 kNT == 350 1 Tan{J == 212 kN

Since x == 216 mm, by strain compatibility, strain insteel is {(d-x)/x}(0.0035) == 2.090 xlO-]

Note 4 Hence, steel has just ~ded and fs == (0.87)fyHence, As == (212 d()3) 1 (0.87)(460) == 530 mm2

Use 3 T16 (As == 603 mm2) . main steel5.2.7.2.1 Min. area required == (112)(350 xloJ) / (0.87)(460) 3 T16

.. == 437 mm2 < 603 mm2; hence O.K.NoteS Also l00A/bd == (100)(603) 1 (300)(345) == 0.58

> 0.4 and < 1.3; hence O.K. Detailing O.K.

Shear reinforeeroent

v == (350 xloJ) 1 (300)(345) == 3.38 N/mm2

l00AJbd == 0.58Table 3.9 Vc == (0.546)(30/25)°.J3(2d1ay) == (0.58)(2/0.43)3.4.5.8 == 2.69 N/mm1 < 3.38 N/mm2

Table 3.8 Provide An > == bv.Sy(v-vJ 1(0.87)£ vAav!Sy > == (300)(3.38-2.69) 1 (0.87J(460) == 0.517Use lOT @ 300 mm. Since this has to be providedover (213)(375) == 250 mm, 2 bars will suffIce.

5.2.7.2.3 Min. requirement is 603/2 == 302 mm2

Use 2 TI0 links @ 175 mm links(As == 314 mm2 > 302 mm2; hence O.K.) 2TlO

87

Reference

Note 6

Bendine main reinforcement

The bend in the main reinforcement should start acover distance (20 mm) from the bearing plate. Itshould end a cover + bar di.a. (20+ 10+16 = 46mm) from the end of the corbel. Hence. distanceavailable for bend radius == 200 - 20 - 46 = 134 mm

3.12.8.25.2 Critical value of 3tJ= 20 + 10 (link) + 16 = 46 mmStress in bars = (0.87)(460)(5301603) = 352 N/mm2

equation 50 Fbt I (vI» < = (2)~ I {I + (2)(qJ~}

(352)(201) I (16)r < = (2)(30) I {I + (2)(l6l46)}r > = 125mmbendradiusChoose r = 130 mm < 134 mm; hence O.K. = 130 mm

100· 100.. 200

r-l30,.,

.1

2TIOfH75(


250

3. Varying the depth from a full depth at the root to 2/3 of the depth at the end ensuresthat 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-thirdsof the effective depth of corbel as specified in Clause 5.2.7.2.3.

4. Using Figure 2.2, the strain at yield is (0.87)(460) 1(200 xloJ) = 2.0 x10-3 for steelof fy '= 460 N/mm2, 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 thecorbel, are not given in BS 8110, they are specified in "Rowe, R.E. et aI., Handbookto 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, theallowance of a cover distance from the outer edge of the bearing plate will ensure thespreading of load from the bearing plate to the level of tie steel before the bendcommences.

Concluding Notes

7. Since a fairly large distance is involved in accomodating the bend radius, analternative 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 projectionof 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 variesfrom 200 mm at the support to 100 mm at the free end, while it carries finishes amountingto 0.5 kN/m2 and an imposed load of 0.5 kN/m2• It is supported by a beam 600 mm x 300mm, which spans 4.0 m between columns, which are considered to provide full bending andtorsional restraint. Design the beam for bending and torsion, assuming feu == 30 N/mm2, fy= 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 mayarise in statically indeterminate situations, is generally not significant; torsionalmoments will be shed back into the elements carrying bending moments (at rightangles to the element carrying torsion), because torsional stiffnesses are lower thanbending stiffnesses. Any torsional cracking will be controlled by shear links.However, equilibrium torsion in statically determinate situations, where torsionalresistance is required for static equilibrium, will have significant magnitudes, and hasto 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 haveinfinite stiffness. In practice, of course this will not be the case and the deformationof the columns will reduce the beam fixed end moments. However, full torsionalrestraint has to be provided by the columns, in order to preserve static equilibrium,where equilibrium torsion is invol"lled.

Reference CaleuIatiODS Output

~$OO

~oILJ 2000)

~.

Loadin2 on beam

HoOd = {(0.2+0.1)/2}(2.0)(24) = 7.2 kN/mFinishes = (0.5)(2.3) = 1.15 kN/mSelf weight = (0.6)(0.3)(24) = 4.32 kN/m

Total dead load = 12.67 kN/mImposed load = (0.5)(2.3) = 1.15 kN/m bending udl =Design load={(1.4)(l2.7)+(1.6)(1.15)} =19.6 kN/m 19.6 kN/m

90

Reference

Torsional loadio&

Calculations Output

(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 kNmlmFinishes == (0.5)(2.0)(0.15+ 1.0) = 1,15 kNm/m

Total dead load torsion == 8,63 kNm/mImposed load torsion = (0.5)(2.0)(0,15+ 1.0)

== 1,15 kNm/mDesign load=={(1.4)(8,6)+(1.6)(1.15)} = 13.9 kNmlm torsional udl

= 13.9 kNm/m

Desim for bendin&

TABLE 1Example 8

Chart 2(Part 3)Table 3.27

Table 3.9

Note 3

Assume cover == 30 mm (moderate exposureconditions, 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 mmTake M == (1I12)w.l2 (for built in beam)

== (1112)(19.6)(4)2 == 26.1 IeNmMlbd2 = (26.1 xlo6) I (300)(550f = 0,29lOOA,lbd == 0.08Use lOOAJAc == 0,13As = (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/mm2 < Vc

Desi&D for torsion

y I = 600 - (2)(30 + 10/2) = 530 mmXl = 300 - (2)(30 + 10/2) = 230 mm

Total torsional moment = (13.9)(4) = 55,6 kNmTorsional restraint at each end= 55,6/2 = 27,8 kNmThe torsional moment will vary as follows:-

27 .8~33 1.2m

I --..~ "

~-27.8

91

d = 550 mm

YI = 530 mmXl == 230 mm

Refereace CaJaJlatioos Output

equation 2 Max. value of vt = (2)T I (bmiJi~ - huuJ3)(Part 2) = (2)(27.8 xld') I (300) (600 - 300/3)

= 1.24 N/mm2 < 4.38 N/mm2 (vtu)

Table 2.3 > 0.37 N/mm2 (vt,min)(Part 2) Thus, beam section is O.K. but requires torsional r/f.

Proyision of reinforcement

Table 2.4 Since v < Vc for the entire beam, the area where vt(part 2) < = vt,min bas to be provided with nominal shear r/f

and the area where vt > vt.min with designed torsionr/f.

equation 2 Torque corresponding to edge of nominal shear r/f is(Part 2) given by T = Vt,min.hmm29tmax -ohznm/3) I 2 .Table 2.3 = (0.37)(300f(600 - 30013) I 2 Nmm(Part 2) = 8.33 kNm

Distance from beam elL = (8.33/27.8)(2.0)= 0.6 mHence, length of beam for nominal shear links

= (2)(0.6) = 1.2 m

Nominal shear links given byTable 3.8 A,;.)Sv > = (0.4)(300) I (0.87)(250) = 0.55 Nominal links

For 10 mm links, Asv = 157 mm2; Sv < = 285 mm RIO@250mmUse RIO links @ 250 mm {< (0.75)d = 413 mm} (middle 1.2 m)

Designed torsional links given by2.4.7 A,;.)Sv > = T I (0.8)~'Yl(0.87)f 0

(Part 2) = (27.8 xl ) I (0.8)(130)(530)(0.87)(250)= 1.31

For 10 mm links, Asv = 157 mm2; Sv < = 120 mm Torsion links2.4.8 Use 2RI0 links @ 200 mm {< = 200 mm, Xl' yl/2} 2RIO@2oo(Part 2) Length of beam at each end for torsional links mm (1.4 m

= (4.0 - 1.2) I 2 = 1.4 m from both ends)

Designed additional longitudinal steel given by2.4.7 As! > (A,;.)Sv)(fyvlf )(xl +Yl)(Part 2) = (157/120)(2501460)(230 + 530) = 540 mm2

If this is divided between 8 bars, each requires 67.5mm2 (3 at top and bottom, 2 in middle).

2.4.9 Since beam length is small, assume bending(Part 2) reinforcement is not curtailed; longitudinalNote 4 reinforcement for torsion also cannot be curtailed.

92


Total steel requirement at top and bottom levels =(67.5)(3) + 234 = 436.5 mm2

Use 2Y16 + YIO at top and bottom levels (As =481 mm2) and 2 Y10 at intennediate level (As = 157mm2)This arrangement will satisfy

Table 3.30 (a) max. spacing for tension rlf < = 160 mm top & bottom2.4.9 (b) max. spacing for torsional rlf < = 300 mm 2Y.l6+YIO(part 2) (c) torsional rlf provided in 4 comers middle 2YlO


3. The torsional moment variation in beams, whether for a distributed moment such asthis or for a point moment, is geometrically identical to the shear force variationcorresponding to distributed or point loads respectively.

4. Longitudinal torsion reinforcement has to be extended at least a distance equal to thelargest dimension of the section beyond the point where it is theoretically notrequired. In this example, that would extend the reinforcement by 600 mm, exactlyto 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 Clause2.4.8 (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 generallybe 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.hmax) values and each rectangle designed for torsion. The torsional links shouldbe placed such that they do intersect.

division into 2 rectanglesl-

IC J ~intersecting torsional links

II

93

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 crosssectional 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/m2 (for roof and floors)(il) Load corresponding to light partitions = 1.0 kN/m2 (for floors only)(iii) Imposed load on roof = 1.5 kN/m2

(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 ABCat the first floor level.

Ground Level

Footing Level

oar

oar

Roofj

4.0m

~ 2nd FI1\

4.Om

A B C 1st FI\

4.2f.m

,T75m. ~". ~ 6.0m 6.0m -.

7- ~ ?!~ - - - -

Introductory Notes '.

L The next 4 examples (including this one) deal with the entire structure, as opposedto structural elements.

2. The loading for partitions and imposed loads is the minimum permissible under"as6399: 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 orlift!stair wells.

4. Since the frame is braced, it is possible to use either a beam level sub-frame analysisor a continuous beam analysis. Since the latter over-estimates moments considerably,the fanner will be performed.

94

Refereoce

Stiffnesses

Calculations Output

(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 bf = 1140 mm« 5000 mm).I of beam section = 11409.388 xl09 mm4 , )

(I/L)ofbeams = 4

1

so5°1¥(9.388 x109) I (6000) =

1.565 xl~ mm3

~Distribution factors

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

Loading on beam

Slab = (5)(0.15)(24) = 18 kN/mBeam = (0.45)(0.3)(24) = 3.24 kN/mFinishes = (0.5)(5) = 2.5 kN/m

Total dead load = 23.74 kN/m ik= 23.7 kN/mImposed load (floor) = (5)(2.5) = 12.5 kN/mPartitions = (5)(1.0) = 5.0 kN/m

Total imposed load = 17.5 kN/mBS 6399: Since a beam span carries 30 m2 of floor area,Part 1 reduced imposed load = (0.97)(17.5) = 17.0 kN/m Qk= 17.0 kN/m

Load arrangements3.2.1.2.2

95

Note 5

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

Moment distribution (kNm)

Output

Note 6

(Arrangement 1)

0.84 0.46AB BA

-181.2 +181.2+ 152.2.....,. + 76,1- 29,0 + 257.3

(Arrangement 2)

0.46 0.84Be CB

-181.2 +181.2- 76,1 <Eo- -152.2~257.3 + 29,0

Arrangement 1Support momentat B= 257 kNm

Note 7

0.84 0.46 0.46 0.84AB BA BC CB

-181.2 +181.2 - 11.1 + 71.1- 25.3~ ......- 5"'-lo0"'"'.6"--_-......50Qt~,6 -+ - 25,3+ 173.5 -+ + 86,8 - 19.3 <t-~- 15.5 +- - 31.1 - 31.1 --+ - 15.5+ 13.1-+ + 6.5 + 6,5 -t- + 13,1- 3.0 -t- - 6.0 - 6,0 -+ - 3.0+ 2.5 -t + 1.3 + 1.3~ + 2.5- 0,6 s- - 1.2 - 1.2 ~ - 0.6- 36.5 +186.9 -171.5 + 3.8

Shear forces lkN)

The shear forces RA, RBI' RB2 and Rc can be foundfrom the following figures:-

Arrangement 2Supportmoments at B= 187 kNm&. 172 kNm

Arrangement 1 143.2 219.3 219.3 143.2

Arrangement 2 156.1 206.3 99.1 43.2

%


Span moments~x /w

f0

~Free bending moment is givenNote 8 by -(w.l.x)/2 + {w.x2)12

Fixed end moment variation is,M1 <Sgiven by M1 + (M2-Ml)xll 1

Hence, the points of contraflexure an4the points andvalues of max. span moments can be obtained.

Span AB Span BeArran~t 1Points ofcontraflexure O.21m, 4.52m 7.48m, 11.79m Span moments:-(from A)Max. sagging 141 kNm at 141 kNm at Arrangement 1moment 2.37 m 9.63 m 141 kNm

(both spans)Arran&eJDent 2Points ~f Arrangement 2contraflexure O.25m, 4.92m 8.45m, 11.90m 165 kNm and(from A) 35 kNmMax. sagging 165 kNm at 35 kNm at

Note 9 moment 2.58 m 10.18 m

257.3

165 141Bending Moment Diagram (kNm)

219.3

Note 10 Shear Force Diagram (kN)

97


s.

6.

7.

8.

9.

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

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

The difference between the moments MBA and MBC arises out of the fact that thecolumns take part of the moment arising out of asymmetrical loading.. .

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

'This two-~ frame is typical of most situations, where the maximum supportmoments are'obtairied when all spans are loaded with the·maximum design ultimateloads (1.4 gk+ 1.6 qk) and the maximum span moments are obtained when that spanis loaded with the maximum design ultimate load (1.4 gk + 1.6Q]J while the adjacentspans 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 characteristicimposed load does not exceed the characteristic dead load, the beam moments andshear forces can be obtained from Table 3.6 for a continuous beam analysis (seeCla~ 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 areawhere the basic wind speed is 40 mis, determine the moments and shear forces induced in

. a typical internal frame due to the wind load.

Introductory Notes

1. The wind forces have to be determined using''cP 3: Ch. V: Part 2 (1972): Basic datafor the design of buildings: Loading: Wind loads:'

2. In carrying out the analysis, the entire structure is analysed, assuming that only thewind load acts on it and that points of contraflexure are developed at the centres ofall beams and columns (Clause 3.2. 1.3.2). A further assumption is made regardingthe 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 determinatestructure, as opposed to an indeterminate one as in the case of vertical load analysis.


Wind force

Basic wind speed, Vb = 40 mlsCP3:Ch.V: Sl = 1.0; SJ = 1.0Part 2 S2 (for ground roughness 3, building class B andNote 3 H = 12.25) is 0.7805

Vs = (1)(0.7805)(1)(40) = 31.22 mlsWind pressure, q = (0.613)(31.22)2 = 597.5 N/m2

Table 10 I1w = (40)/(12) = 3.33of bid = (40)/(12) = 3.33CP3:Ch.V: h/b = (12.25)/(12) = 1.02Part 2 Hence, Cf = 1.23

Force on one frame = q.Cf.Ae= (597.5)(1.23){(12.25)(5.0)) = 45014 N Wind force on a

= 45 kN frame = 45 kN

Analysis

The following assumptions are made:-1. The wind force is applied at floor and roof levels,

the force at each level being proportional to theareas shared by them.

99


Note 4

2. Points of contraflexure are assumed at the centresof beams and columns.

3. The vertical column stresses are proportional totheir distances from the centroid of the columns.

c!>~ 11.03l'o

1.225.J..

1.84-..c 'l2.llil

14. 7~t--"'-4~---+--.....----1 ?~ 2nd Fl r. 2.llil~ 5.51

t6.125

The forces at roof, 2nd floor and 1st floor levels are(2/12.25)(45) = 7.35 leN (roof){(2+2)/12.25}(45) = 14.7 leN (2nd flopr) and{(2+2.S0)/12.25}(45) = '16.5 ,1eN(lst floor)

k 6.llil M< 6.llil ~

7.35-+r . e I · lJ Roof. r 1.84 +- 3.67 ~:~1.225 t if225

'1-225

1.84 3.67 ... i)

Note 5

11.03~)

~<4:- 9.64'

~17.83

Note 6

Note 7

6.125

5.51 ~tl2.Qn16.5:+...---..---+----+---t Is t Fl r

2.fIn

~ ~ 19.28 ,. '9.64

't t ,17.83

Moments and shear forces in ABCThe moments in· ABC can be found as those requiredto balance the ~lumn moments.

.

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

. .' .. 35.'f 35.1

Moment at A, Band C is ·,~5..1kNm

100


The shear forces in the spans are obtained by Shear force individing the moment by half the span length. AB and BC isHence, shear force = (35.1)/(3.0) = 11.7 leN 11.7kN


3. The S2 factor can be calculated separately fordifferent parts of the stucture or for theentire structure, using the total height of the structure. Since this is only a 3 storeystructure, 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 thedistances of the columns from the centroid of the column group. An alternativeassumption to this is to consider that the horizontal shear forces in the columns areproportional to the bay sizes.

5. The analysis is essentially a subframe analysis, but the entire frame has to analysedstep wise, from the top to bottom. At each step, the vertical column reactions areobtained first, taking moments for the equilibrium of the entire sub structure, togetherwith the third assumption referred to in Note 4 above. The horizontal shear forces inthe columns can be found by taking moments about the points of contraflexure in thebeams, for the equilibrium of different parts of the sub-structure. The results obtainedfrom 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 onlowest column should be moved down to the level of the base (as opposed to beingat column mid height).

7. In order to meet stability requirements, the lateral load at each level should be at least1.5 % of the characteristic dead load at each level (Clause 3.1.4.2). Since the totaldead load on·a beam (Example 30) is (23.7)(12) = 284.4 kN and 1.5% of this is 4.3leN « 7.35 kN), the above condition is met.

Concluding Notes

8. For unbraced frames having three or more approximately equal bays, the combinedeffect of wind and vertical loads can be obtained by superposing the results .of ananalysis such as the one above with those of a subframe analysis such as the one inExample 30, after factoring the loads appropriately (Clause 3.2.1.3.2).

9. For very slender structures, the overall stability of the structure against overturningdue to lateral wind loads should also be checked. The appropriate load combinationwould be 1.4 W k (causing the overturning moment) and 1.0 Gk (providing therestoring moment).

101

EXAMPLE 32 - REDISTRIBUTION OF MOMENTS

Determine the design ultimate moments for the beam ABC in Example 30, after carrying outmoment redistribution.

Introductory Notes

1. Although the design of reinforced concrete sections is carried out using the plasticcapacity of the section, the analysis of structures is still performed using elasticmethoos. The advantage to the designer arising out of the above plasticity isincorporated in the analysis by moment redistribution.

2. Moment redistribution has to be performed separately for each load arrangement. Inaddition, the redistributed envelope is not allowed to fall below the 70% elasticmoments envelope, to ensure that wide· cracks at the serviceability state will notdevelop (see Clause 3.2.2.1).

Reference

Example 30

3.2.2.1Note 3Note 4

Calculations

Support moments

The numerically largest elastic moment is 257.3kNm at support B (Arrangement 1). This can bereduced to (0.7)(257.3) = 180.1 kNm for all loadcases, leaving the support moments at A and C (andalso the column moments) unchanged.

Hence, the support moments will be given by

Output

support momentat B = 180.1

kNm"

AB BA BC CB

Note 5

Arrangement I -29.0 +180.1 -180.1 +29.2

Arrangement 2 -36.5 +180.1 -180.1 + 3.8

NQt.e:- The shear forces can be found by analysingthe sections AB and CB, as in Example 30.

Example 30 Span moments

These can be found by superimposing the freebending moment diagrams on the above fixed endmoment variation.

102


Arraneement 1Points ofcontraflexure 0.19m, 4.97m 7.03m, 11.81m(from A)

span moments:-Max. sagging 173 kNm at 173 kNm atmoment 2.58 m 9.42 m Arrangement 1

173 kNm (bothArraneement 2 spans)Points ofcontraflexure 0.24m,4.97m 8.57m, 11.91m Arrange':Dent 2(from A) 168 kNm

(span AB)Max. sagging 168 kNm at 33 kNm at 33 kNmmoment 2.60 m 10.24 m (span BC)


3. The support moments are reduced as much as possible so that congestion ofreinforcement at beam-column junctions can be minimized. The maxi.mumamount ofredistribution allowed is 30 % - a figure which can ~ accomodated by rotation at asection after plastic hinge formation by the appropriate restriction of the x/d ratio (seeClause 3.2.2.1).

4. In general, the x/d ratios in columns are larger than those required to permit plastichinge formation. Hence, column elastic moments should never ~ redistributed.

5. The support moments in Arrangement 2 are made equal to 180.1 kNm - the valueobtained after 30% redistribution in Arrangement 1. This requires a much lowerpercentage of downward redistribution for the elastic momentBA and an upwardredistribution 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 designmoments are such that the support moment at B is considerably lower, while the spanmoments are only slightly higher; hence the advantage in <:arrying out momentredistribution - the total moment field is considerably reduced.

7. The points of contraflexure are generally closer to the supports for the redistributedbending moment diagrams than for the elastic bending moment diagrams. In order toprevent serviceability state cracking on the top surface, the restriction on theredistributed 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 heightof each storey is 3.5 m. The average dead and imposed loads per unit area of floor can betaken as 5 kN/m2 each. Design the stability ties for this structure with steel of fy = 460N/mm2•..

,

~ -~Beans

ChlumS"(

)4 x 5.(0} = 2(0}

I. In order to ensure the robustness of a structure, it should normally be connectedtogether by a system ofcontinuous ties. This example demonstrates the design ofthese ties.

2. In addition, .the structure should be capable of withstanding a notional horizontalload, 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 actat its characteris~c value -Le. 'Ym = 1.0. Furthermore, reinforcement designed forotherp~ can be used as ties (Clause 3.12.3.2).


3.12.3.7 Vertical ties

TheSe are required, since no. of storeys > 5.Area corresponding to ~ typical column =

15.0 m2'. .1' (1(2)(6.0)(5.0) =MaX. design ultiIpate load =

(15.0){(1.4)(5.0) + (1.6)(5.0)} = 225 leNArea of ties required = (225 xl<Y)/(460)

= 489 mm2 vertical tiesThis can easily be met by continuous column rif. A, = 489 mm2

104


3.12.3.5 Peripheral ties

3.12.3.4.2 Ft = 20 + (4)(6) = 44 leN « 60 leN) Ft = 44 leNArea of ties required = (44 xloJ) I (460) = 96 mm2 peripheral tieThis can be easily met by peripheral beam r/f that is A = 96 mm2

•continuous.

3.12.3.4 Internal ties - looeitudinal direction.

lr = 5.0 mForce I unit width = {(5+5)/(7.5)}(515)(44)

= 59 kN/m {> (1.0)(44) = 44 leN/m}Total force = (59)(6.0) = 354 leNArea of ties required = (354 x1oJ) I (460) longitudinal

= 770 mm2 internal tieIf carried in the two peripheral beams, area required A = 770 mm2

per beam = (770)/2 = 335 mm2 •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 mForce 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 kNArea of ties required = (1408 xl<P) 1 (460) transverse

. = 3061 mm2 internal tieIf distributed in the 5 transverse beams, area A == 3061 mm2

required per beam = (3061)/5 = 612 mm2 •Note:- spacing of ties = 5.0 m < (1.5)(6.0)= 9.0 m

Note 4 Peripheral beams peripheral

Total tie area per beam in longitudinal direction =beams (tie rlf)longitudinal -

96 + 335 = 431 mm2 431 mm2

Total tie area per beam in transverse direction = transverse -96 + 612 = 708 mm2 708 mm2

105


3.. 12.3.6 Coltamn ties

Note 5 Force = greater of (31100)(225)(6) = 40.5 kNand lesser of (2.0)(44) = 88 kN and

and {(3.5)/(2.5)}(44) = 61.6 kN= 61.6 kN column ties

Area of tie required = (61.6 xHP)/(460) = 134 mm2 A = 134 mm25

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


4. Although the beam reinforcement may be greater than these tie areas required, it mustbe ensured that continuity of tie reinforcement is provided - this has to be borne inmind when curtailing beam reinforcement.

5. The,3% load is taken for 6 storeys,since there will be five floor slabs and the roofabove 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. inload bearing masonry construction), bridging elements have to be designed, assumingthat 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 andkey 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 supportingdead and imposed loads of 20 kN/m each. Determine the crack widths (i) midway betweenthe bars, (ii) at the bottom comer and (iii) 2S0 mm below the neutral axis.

750

,

( 450 >-

3T25 1690

000

f = 25 N/mm2cu

f ;"'460 N/mm2y

E = 200 kN/mm2s


Introductory Notes

1. This crack width calculation can be performed when the bar spacing rules are notsatisfied, to see whether this more accurate method will satisfy the crack widthrequirements in Clause 3.2.4 of Part 2. It can also be used to estimate the actualcrack width in a flexural element.


Sectional data I

Note 2 Ms = (20+20)(7)2 18 = 24S kNm Ms = 245 kNm

equation 17 Ec = 20 + (0.2)(25) = 25 kN/mm2

3.8.3 Eeff = (0.5)Ee · = 12.5 kN/mm2

(Part 2) CXe = E/Eeff =' (200 xloJ) 1 (12.5 xHf) = 16Note 3 P = (3)(491) 1 (690)(450) = 0.00474

x/d = -cxe.p + {cxe.p (2 + cxe.p)}o.SNote 4 = -(16)(0.00474)

+ [(16)(0.00474){2+(16)(0.00474)}]O.S = 0.321x = 221 mm x = 221 mm

Note 5 ~/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 xl09

Ie = 6.80 x109 mm4 mm~

107

Reference

3.8.3(Part 2)

Note 6equation 13(part 2)

equation 13(part 2)

equation 12(part 2)3.2.4(part 2)

Calculations

Calculation of strains

Strain in steel = (245 x1(6)(690-221) I. (12.5 xl<Y)(6.8 x109) = 1.35 dO·3

< (0.8)(460) / (200 xl<Y) = 1.84 dO·3

At extreme tension fibre (bottom of section),El = (1.35 xlO·3)(750-221)/(690-221) = 1.523 xl0·3

tension stifferung = bt(h-x)(a'-x) I (3)Es.A.(d-x)= (450)(750-221)2 I (3)(200 xI<Y)(3)(491)(690-221)= 0.304 xlO·3

Em = (1.523 - 0.304) xHr3 = 1.219 xlO·3

At 250 mm below neutral axis,EI. = (1.35 xlO'3)(250)/(690-221) = 0.72 xlO·3

tension stiffening = (450)(750-221)(250) I(3)(200 xl<Y)(3)(491)(690-221) = 0.144 xlO·3

Em = (0.72 - 0.144) xlO·3 = 0.576 dO·3

Distances to potential crack points~

cmin = 750-690-(2512)= 47.5 mm

<lerl = {(60)2 + (82.5)2}O.s- 12.5 = 89.5 mm

3er2 = {(60)2 + (60)2)}o.s- 12.5 = 72.4 mm

3er3 = {(60)2 + (69O-221.250)2}O.S-12.5 = 214.5 mm

Crack widths

CW t = (3)(89.5)(1.219 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)}]

= 0.242 mm < 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.

108

Output

CW t = 0.282mm

CW2 = 0.242,mm

CW3 = 0.227mm


... Comparison with bar macine rulCfS

3.12.11.2.3 Spacing between bars = {45G-(2)(47.5)-(3)(25)} 12 max. spacingTable 3.30 = 140 mm < 160 mm; satisfactOry. O.K.3.12.11.2.5 Comer distance = {(60)2 + (60)2}O.s - 12.5 comer distanceTable 3.30 = 72.4 mm < (16012) mm; satisfactOry. 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 asimpler approach compared to the one for deflection calculationsc(see Example 35).

4. The serviceability calculations are based on a triangular stress block for concrete inthe elastic state. There is no restriction on the x1d ratio, as in ultimate limit statecalculations.

5. The effective second moment of area is found by considering only the area ofconcrete that is not cracked; the area of steel is converted to an equivalent area ofconcrete using the effective modular ratio, Qe'

6. The strain at the required level in the concrete is found by calculating the strain fromelastic theory (Et), and reducing from this value·an allowance for tension stiffeningin the concrete; this is because in calculating Ie and El we assume that the concretehas no tensile strength, whereas in fact it does.

Concluding Notes

7. All the calculated crack widths are below 0.3 mm and hence satisfactory (Clause3.2.4, Part 2). This could have been expected, because the maximum 'spacing andcomer distance rules are satisfied as well. It is these empirical rules that are used ineveryday design, because of their convenience.

8. For beams of overall depth exceeding 750 mm, side reinforcement in the form ofsmall diameter bars at spacings not exceeding 250 mtn over two thirds of the beamdepth 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 andimposed loads are both taken as 5 kN/mleach, and if25 % of the imposed load is taken aspermanent, calculate the total long term deflection of the beam at midspan. The age ofloading can be taken as 28 days.

(225 )-

375 2T25 125

o 0 I(All dimensions in mm)

Introductory Notes

f = 25 N/mm2cu

f = 460 N/mm2y

Bs = 200 kN/mm2

1. This deflection calculation can be performed when the span/depth ratio check fails,to see whether this more accurate method will satisfy the deflection requirements inClause 3.2.1.1 of Part 2. It can also be used to estimate the actual deflection of aflexural element.

2. Where domestic and office space is concerned, 25 % of the imposed load can beconsidered permanent; where storage areas are concerned the above figure should beincreased to 75 %.

3. The age of loading is when the fonnwork is removed, at which point much of thedead load and some imposed construction loads will be acting on the concreteelements.

Reference CaIculatio. Output

Initial aSsessment of span/dq>th ratio

Mult = (5)(1.4 + 1.6)(7)2/8 = 92 kNmChart 2 M1bd2 = (92 xl<r) I (225)(325)2 = 3.87(Part 3) l00A/bd = 1.2~; Aa = 936 mm2

equation 8 fa = (0.58)(460)(936/982) = 275 N/mm2

equation 7 F I = 0.55 + {(477-275) / 120(0.9+3.87)} = 0.9Table 3.10 Allowable span/depth = (20)(0.9) = 18

Actual span/depth = 7000 / 325 = 21.5 > 18; span/depthHence, span/depth check is violated. check violated

110


Data for serviceability calcuIiJionsequation 17

Ee = 20 + (0.2)(25) = 25 kN/~B.\2(Part 2) 1\:=25 kN/mm2

7.3 Eff. section thickness=(2)(375)(225) I (2)(375 + 225)(Part 2) = 141 mm

RH = 85 % (assumed for Sri Lanka)Figure 7.1 Long term creep coefficient, 4> = 1.8(part 2) Eeff = 25 I (1 + 1.8) = 8.93 kN/mm2Figure 7.2 f = 120 x10-6es(part 2) P = 982 I (225)(325) = 0.0134

ae = Es I Eeff = 200 I 8.93 = 22.4 a e = 22.4

x/d = - ae.p + {ae.p(2 + ae.p)}O.5= - (22.4)(0.0134) +

[(22.4)(0.0134){2 + (22.4)(0.0134)}f.5Note 4 = 0.53 xld = 0.53

Hence, x = (0.53)(325) = 172 mm x = 172 mm

Ie/bd3 = (1I3)(x/d)3 + ae.p{l - (x/d)}2= (113)(0.53)3 + (22.4)(0.0134)(1 - 0.53)2= 0.116

Ie = (0.116)(225)(325Y = 896 xH? mm4 Ie = 896 x1()63.6 Ss = As(d-x) mm4

(Part 2) = (982)(325-172) = 150.2 x103 mm3

Determination of serviceability moments

M tot = (5+5)(7)2 18 = 61.25 kNm~rm = {5 + (0.25)(5)}(7)2 I 8 = 38.28 kNm

Note 5 ~rm(red) = Mrrm

- {(1I3)b(h-x)3.fet I (d-x))= (3 .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 1Ires = fes·ae·Ss I Ie(part 2) = (120 xlO-6)(22.4)(150.2 xloJ) I (896 x106)

= 0.45 xlO-6 mm- l

111

Note 6

Table 3.1(Part 2)

3.2.1.1(Part 2)Note 7

To find instantaneous7'·.... .. ,Be =25 kN/mm2 . . .a = 200 I 25 =8 •.... . .'Jd = 0.368 '.."Ie = 459 xl()6lIrit -llri = (Mtot - M.rm> I Ec·le

~ (61.25-38.2$)X1()6 I (2~ xl(3)(459 xl~= 2.01 xl<r' nun-I

lIr = lIrlp + IIres + (llrit .;. 11rin>'1= (4.5 + 0.45 + 2.01) x10~ mm-I

= 6.96 xlO~ mm-I

Estimation of det'ls;gion

K =0.104a = K.l~lIr)

= (0.104)(7~6.96xIO~ == 35.5 mm

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

··0ulJwt··

lIr= 6.96 xlO~mm-I

a = 35.5 mm

all > 11250


4.

5.

6.

7.

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

The Mpermvalue is ~ced because of the tension stiffening contributed by theconcrete, which is asSUfed to have no ~strength in the ~Jlation of Ie. Thelongterm·tensiIe·strencth ofeoncrete is taken as 0.55 N/m,m2 and·the short termstrength as 1 N/mm2.,

The tension stiffening .ffects in Moot and ~rm cancel out here.

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

"

Concluding Note

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

.,: ',~,~" -<~,'~'~'~~"" ' __ ,- '1l:~~."."'$,<.;

-';,'.

~,,:,.(-

;~'\',

i· .... ~

Graded Examples in Reinforced Concrete Design Dias

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