Graded Example to EC2.pdf

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aded .Example i in Concrete Design

E o code 2 (3rd edition)

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Kesawan Sivakumar

Department of Civil Enghnee:ring

University of Mora-tuwa, Sri La~_ka

Society of Structural Engineers - Sri Lank~a

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. 1 Graded Examples in

. r Reinforced Concrete Design : I to Euro code 2 (3rd edition) ::. f .

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Priyan Dias and

Kesawan Sivakumar

Department of Civil Engineering

University of Moratuwa, Sri Lanka

Society -of Structural Engineers - Sri Lanka

PRE-PUBLICATION DRAFT

Published by the Society of Structural Engineers - Sri Lanka

12011 S Vidya Mawatha, off Wijerama Mawatha, Colombo 7 August 2012

This document shall not be reproduced or transmitted in any fonn in whole or par1 without

the express permission of the publisher

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PREFACE

Background and Acknowledgements

This is the third edition of this book. A significant departure from the first two editions is that this one is based on Eurocode 2 for Concrete Structures. Sri Lankan engineers are not so conversant with the new Eurocodes, and it is hoped that this book of examples will reduce the discomfort of the transition from BS8 l l 0 to EC2.

This is also the first time the book has a co-author and the first author wishes to thank the second for his painstaking perusal of a new code. He did this in the period between finishing his undergraduate degree at Moratuwa University and embarking on a PhD at the Queensland University of Technology. He was supported by a grant from the Society of Structural Engineers, Sri Lanka, who are in fact publishing this edition, as they did the first. Dr Premini Hettiarachchi spent a considerable time checking the calculations and also sourcing the relevant literature. Her name does not appear as another co-author only at her insistence! We also thank Dr Kumari Gamage who was involved in this venture at the early stages.

The Transition from BS 8110 to EC2

EC2 is not an easy code to use. It focuses on theoretical principles rather than on practical design. One manifestation of this is the structuring of the code by stress states (e.g. bending, shear, torsion, anchorage, deflection, crack control etc) rather than by structural elements (e.g. beams, slabs, columns, foundations etc) as in BS 8110. Another is the paucity of design equations and design charts. In addition, some of the parameters in the code allow national agencies to supply their own values through "national annexes". We have used relevant values from the draft version of the Sri Lanka National Annexe (SLNA), produced by the Sri Lanka Standards' Institute.

Two approaches have been used to make easier the use of EC2. One is to provide Appendices at the end of this book with some design equations; the second is to refer to charts and other helpful material, primarily in the Institution of Structural Engineers' Manual (IStructE Manual) for the Design of Concrete Structures to EC2. We have also provided some design charts for flexural reinforcement and shear resistance calculations in Appendices C and D respectively; these charts are however not used in these design examples, which are worked out from first principles as per EC2 guidance. In addition to the IStructE Manual that is referred to widely, we have also had at times to refer to the Concrete Society-publications and a few times even to BS 8110 itself - this is to fill any perceived gaps in the EC2 provisions. We should also acknowledge two textbooks that we have been helped by, namely those by Moseley, Bungey & Hulse (whose 6th edition is to EC2), and the other by AH Allen.

Some Differences in Process and Output

We will highlight some key differences between EC2 and BS 8110 with respect to both design process and design output. The partial safety factors for loads are smaller in EC2 than

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in BS 8110, and the "all spans loaded" case in the latter has been replaced by the "adjacent spans loaded" in the new code. EC2 is based on concrete cylinder strength, the lowest structural grade being 20 (from corrosion durability considerations), whereas BS 8110 was

- based on concrete cube strength with a minimum grade of 25 . The depth of the compression stress block in flexure . in EC2 is only 0.8 times the neutral axis depth (x), compared to the depth of 0.9x in BS 8110. In order to ensure ductility, the ratio of neutral axis depth to effective depth (i.e. x/d) is restricted to 0.45 in EC2, whereas 0.50 was permitted in BS 8110.

The design for durability in EC2 is more nuanced than in BS 8110, and hence more complex. It involves (i) the choice of exposure condition from Table 4.1, which gives a much better defined range of conditions than did BS 811 O; (ii) the selection of strength class from Annexe E, .with possible modification through Table 4.3N; and (iii) the determination of the minimum cover value from Table 4.4N. An allowance for construction deviations must be added to the minimum cover value; this is usually taken as 10 mm for conditions employing normal quality assurance. In this book of examples, the grades employed are at the lower end, reflecting Sri Lankan practice. Grade 20/25 is used for indoor exposure conditions (XCl) with a total cover of 25 mm, and grade 25/30 for outdoor exposure conditions (XC3) with a total cover of 35 mm. It should be noted that the indicative strength class has been reduced from C30/37 to C25/30 for outdoor exposure, on the basis that Sri Lankan concretes have fairly low water/cement ratios for even fairly low grades. By this same argument, grade 16/20 concrete could in fact be used for indoor exposure in Sri Lanka, although we have not done any calculations based on it. In addition, the current use of strength class C20/25 for outdoor (non chloride) exposure can also probably be accepted (provided a 35 mm cover is specified). For chloride environments however, the indicative strength class in Annexe Eis C30/37, and while this could be reduced to C25/30 by the above argument, it would not be wise to reduce it below that. Maintaining a minimum grade of 25/30 with a cover of 35 mm for external concrete in buildings along the coast will help to reduce some of the significant corrosion problems experienced by such structures.

The design for shear and torsion are more theoretical and difficult to carry out in EC2. If designed shear reinforcement is not required (due to the inherent resistance of the concrete and the longitudinal steel), minimum reinforcement has still to be provided, as in BS 8110. If shear reinforcement is required however, no advantage is taken of the above inherent resistance, with the entire design shear force having to be carried by the shear reinforcement. In addition, the capacity of notional concrete struts (in a strut and tie analogy) has to be checked. The check is first perfom1ed assuming the minimum strut inclination to the horizontal of 22 (cot 8 = 2.5). If a strut with this inclination is incapable of resisting the shear force, the inclination that can in fact resist the force has to be calculated. If this inclination is greater than 45 (cot 8 = 1 ), the concrete resistance is not sufficient and must be remedied by an increase in beam depth or concrete grade. The design for torsion involves the conversion of the concrete section into an equivalent thin walled closed section.

The design for deflection in EC2 is both different from BS 8110 and can also result in much thinner slabs for lightly reinforced slabs, especially where concrete grades are high too. This is counteracted in the IStructE Manual by specifying that the percentage of reinforcement should not be taken as smaller than 0.35% .when determining the allowable span/depth ratio. In this book, the above percentage has been set at 0.25%; however, in addition, absolute limits to the allowable span/depth ratio have been set (rather arbitrarily, but reflecting current practice) as 45 for two-way slabs, 40 for continuous one-way slabs and flat slabs and 35 for simply supported one-way slabs.

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A Case for Worked Examples (from previous editions by Priyan Dias)

Educational purists may argue that worked examples are detrimental to student learning because there is an element of "spoonfeeding" involved. While acknowledging that there is some truth in this argument, the author would like to contend that worked examples do have a place in the educational process.

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

The set of examples has been developed through the teaching of a course in Reinforced Concrete Design at the University of Moratuwa, Sri Lanka. The examples are graded, leading from an appreciation of reinforced concrete behaviour, through the design of structural elements, to the analysis of a reinforced concrete structure. The student's understanding of the calculations is deepened by the "Notes on Calculations" while the Introductory and Concluding Notes set each example in a wider context. Hence, in this book, design principles are reinforced through practice, with guidance from notes.

However, this book should not be used as a "stand alone" text. It must essentially be complementary to another text or series of lectures that teaches design from a deductive approach - i.e. one which moves students from principles to practice. It can, of course, be readily used by practising engineers, who already have a grasp of reinforced concrete fundan1entals. In order to equip students for real design practice, the book is very much code based, with extensive references given in the calculations to clauses in EC2 and other pertinent literature listed below.

Key References

Allen, A.H. (1988) Reinforced Concrete Design to BS 8110 simply explained. EF & N Spon BS EN 1990: 2002. Eurocode - Basis of Structural Design, British Standards Institution,

London. BS EN 1991-1-1: 2002. Eurocode 1: Design of Concrete Structures: Part 1-1: General actions

- densities, self weight, imposed loads for buildings, British Standards Institution, London.

BS EN 1992-1-1: 2004. Eurocode 2: Design of Concrete Structures: Part f-1: General rules and rules for buildings, British Standards Institution, London.

BS EN 1992-1-2: 2004. Eurocode 2: Design of Concrete Structures: Part 1-2: General rules-structural fire design, British Standards Institution, London.

BS 8110: 1985. Structural use of concrete, British Standards Institution, London.

Concrete Centre (2005-2008) How to Design Concrete Structures using Eurocode 2: Parts 1-11, The Concrete Centre, Camberley.

Institution of Structural Engineers (2006) Manual for the design of building structures to Eurocode 2, IStructE, London.

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Kesawan,S., Dias W.P.S and Hettiarachchi, M.T.P. (2012) Slab design based on Eurocode 2, Modulus, Vol. 22, No. 1, Society of Structural Engineers, Sri Lanka, pp. 8-12.

Mosley, W.H. Bungey, J.H. and Hulse, R. (2007) Reinforced Concrete Design to Eurocode 2, 6th ed., Palgrave Macmillan, Basingstoke.

Sri Lanka Standards Institution (2011) Sri Lanka National Annex to Eurocode 1: Design of Concrete Structures : Part 1-1: General actions - densities, self weight, imposed loads for buildings (Draft), SLSI, Colombo.

Sri Lanka Standards Institution (2011) Sri Lanka Na ti on al Annex to Eurocode 2: Design of Concrete Structures: Part 1-1 : General rules and rules for buildings (Draft), SLSI, Colombo.

Sri Lanka Standards Institution (2011) Sri Lanka National Annex to Eurocode 2: Design of Concrete Structures : Part 1-2: General rules - structural fire design (Draft), SLSI, Colombo.

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1.. I . CONTENTS

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I PREFACE 111

CHAPTER 1: Analysis of Beam Sections in Flexure Example 1 Analysis of Under-Reinforced Section 01 Example 2 Analysis of Over-Reinforced Section 04 Example 3 Analysis of Doubly-Reinforced Section 07 Example 4 Analysis of Non-Rectangular Section 09

CHAPTER 2: Design of Beam Sections in Flexure ::?~ Example 5 Design of Rectangular Section 11

Example 6 Design of Section with Redistribution 15 Example 7 Structural Analysis of Beam 17

I Example 8 Design of Beam for Flexure 21 1 Example 9 Design of Flanged Section 27

) . CHAPTER 3: Design of Beams for Shear Example 10 Design of Section for Shear 30 Example 11 Design of Beam for Shear 34

CHAPTER 4: Serviceability Checks and Detailing in Beams Example 12 Serviceability Checks and Detailing 40

) CHAPTER 5: Design of Slabs Example 13 One Way Slab (Continuous) 52

l Example 14 One Way Slab (Simply Supported) 58 Example 15 Two Way Slab 64 Example 16 Flat Slab 69

\ . Example 1 7 Ribbed Slab 78 /: ..

' CHAPTER 6: Design of Columns Example 18 Column Classification 83 Example 19 Symmetrically Loaded Short Column 86 Example 20 Short Column with Axial Load and Moment 90

I. Example 21 Slender Column 93 J

CHAPTER 7: Design of Foundations

t. Example 22 Pad Footing 99 Example 23 Combined Footing 106 Example 24 Pile Cap 113

r CHAPTER 8: Design of Staircases - Example 25 Staircase (between Beams) 118 L, Example 26 Staircase (between Landings) 124 -

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CHAPTER 9: Design of wall and Corbel Example 27 Plain Concrete Wall 128 Example 28 Corbel 132

CHAPTER I 0: Design of Beam for Torsion Example 29 Design for Torsion 137

CHAPTER 11: Frame Analysis and moment Redistribution Exmaple 30 Frame Analysis for Vertical Loads 143 Example 31 Frame Analysis for Horizontal Loads 149 Example 32 Redistribution of moments 152

CHAPTER 12: Design for Stability Example 33 Design for Stability 154

CHAPTER 13: Severability limit State Calculations Example 34 Crack Width Calculation 157 Example 35 Deflection Calculation 161

APPENDICES Appendix A Formulae for Design 166 Appendix B Longitudinal Shear in Beams 170 Appendix C Design Charts: Singly Reinforced Section 171 Appendix D Design Charts: Concrete Shear Resistance 172

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CHAPTER 1

Analysis of Beam Sections in Flexure

Example 1 - Analysis of Under-Reinforced Section

Determine the lever ann for the beam section shown in the figure; find also its moment of resistance.

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d = 375

(All dimensions in mm)

Introductory Notes

fek = 20 MPa fyk = 460 MPa

1. This exan1ple is regarding the analysis of an existing beam. The first step in finding the moment of resistance is to find the lever arm.

Note 2

3.2.7(2) Figure 3.8 Note 3 2.4.2.4(1) Table2.1N

Note 4

3.1.6 (1) 2.4.2.4(1) 3.1.6(1)

Area of steel = 942.5 mm2

fyk = 460 MPa fyk

Ys

Ys = 1.15 (for persistent and transient situations) fyd = 0.87fyk

fek = 20 MPa O'.eefek

fed= Yo Ye = 1.5 aee = 0.85 (SLNA)


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Note 5

3.1.7(3)

Eq 3.19 Eq 3.21

Note 6

Note 7

c~1cui~tioils .-. . . , . . - - .

fed = 0.567fck

Assuming that the steel has yielded, T = 0.87 fykAs = (0.87)( 460)(942.5) = 377,189 N

Hence balancing compressive force = 377,189 N (11 fed) (b) (Ii, x) = 377,189 N

'A=0.8and11=I .Oforfck $50MPa

(0.567)(20)(225)(0.8) x = 377,189 N x = 185 mm

x 185 Since d = 375 = 0.493 $ 0.64,

steel has yielded and original assumption is correct. z = d - (0.4)x = 375 - 0.4 x 185 = 301 mm Note: 2/ d = 301h 75 = 0.8 < 0.95; hence OK

Moment of resistance= 377,189 x 301 = 113.53 xl06 Nmm = 113.5 kNm

Notes on Calculations

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T = 377,189 N

x = 185 mm

z = 301 mm

M = 113.5 kNm

2. Clause 3.2.2(3) says that rules for design and detailing in this Eurocode are valid for a specified yield strength range fyk = 400 to 600 MPa. The upper limit of this value varies according to the country's national annex. In Sri Lanka the proposed value is 500 MPa. UK uses a value of 500 MPa for routine design, but in Sri Lanka it is better to consider the steel characteristic yield strength as 460 MPa, as per current practice.

3. We use the horizontal top branch in Figure 3.8, where the steel strain need not be checked.

4. fck is the characteristic cylindrical strength of concrete according to Clause-3.1.2(1).

5. Most singly reinforced sections will be under reinforced in practice. Hence, assuming that the steel has yielded is the most convenient way of starting. (This assumption should be checked later on of course, using the x / d value.)

6. The concrete strain in compression is limited to Ecuz according to Clause 3.1.7(1), where Ecuz is 0.0035 from Table 3.1 - this is the failure criterion for reinforced concrete. The condition that tensile reinforcement has yielded when the concrete strain is 0.0035, is x/ d $ 0.64 (for fyk = 460 MPa). This can be shown by assuming a linear strain distribution. Sri Lanka National Annex recommends that x/d does not exceed 0.6; but EC2 recommends that x/d be less than 0.45. In this example x/d is greater than 0.45. In designing of

Analysis of Beam Sections in Flexure Page 2

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reinforcement it is better to limit x / d to 0.45. This will give enough warning before failure. This limiting x / d value reduces if moment redistribution is carried out (Clause 5.5( 4)). These limitations will ensure ductile failure due to yielding of reinforcement bars and not brittle failure due to crushing of concrete.

7. z is taken as not greater than 0.9 Sd as in BS81l0. This practice is maintained here.

Concluding Notes

8. The lever arm is the distance between the centroids of the tensile and compressive forces. This separation between two opposite forces is what creates the moment of resistance in a flexural element. Because this distance has to be accommodated within the depth of the section, flexural elements have larger cross sections than compressive elements.


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Example 2 - Analysis of Over-Reinforced Section

Detennine the lever ann for the beam section shown in the figure; find also its moment of resistance.

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2-25

d = 300


Introductory Notes

fck = 20 MPa fyk = 460 MPa

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

Reference

Note2

Note 3

Calculations

Area of steel= 981.7 mm2

Assuming that the steel has yielded, T = 0.87fykAs = (0.87)( 460)(981 .7) = 392,876 N Hence, C = (TJ fcct) (b) (!.. x)

= (0.567)fckb(0.8) X = 392,876 (0.567)(20)(150)(0.8) x = 392,876

x = 289 mm .

But x /ct = 289 hoo = 0.963 > 0.64, hence, steel hasll.Q.Lyielded.

We shall try to find a value for x, by trial and error, such that T and C are approximately equal.

Try x = 200 mm C = 0.567fck(b)(0.8)x = (0.567)(20)(150)(0.8)(200)

= 272,160 N - (0.0035) (300 - 200) - -3

Es- 200 -1.75x10

fs = Es Es Es= 200 GPa fs = (1.75 x 10-3 )(200 x 103 ) = 350 N/mm 2


Output

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Reference

Note4

Calculations .

T = (350) (981.7) = 343,595 N

For a better approximation, try x = 210 mm. Then C = 285,768 N and T = 294,510 N.

For a still better approximation, try x = 211 mm Then C = 287,129 N and T = 289,857 N

This approximation is sufficient. Note:- x;d = 211hoo = 0.70 (> 0.64)

z = d - (0.40)x = 300 - (0.40)(211) = 216 mm M = C. z = (287129)(216) = 62 x 106 Nmm

= 62 kNm

Note:- Alternative method of finding x.

Once it is established that the steel has not reached yield point, for any given value of x,

(0.0035)(300 - x) Es =

x

co.0035) (300 - x) 3 I 2 f5 = (200 X 10 ) N mm x

T = (0.00 35)(300 - x) (200 x 103 ) x 981.7 N x

C = (0.567)(20)(150)(0.8)x N

Putting T = C, we have the quadratic equation xz + 505 x - 151497 = 0, giving x = 211.5 or-716.5 mm


qutput

x = 211 mm

M = 62 kNm

x = 211 mm

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

3. Since the steel has not yielded, the stress can no longer be assumed to be 0.87fyk- Rather, the stress is the steel is o&tained 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 3.8 of the code to arrive at the stress (the inclined

part of the design curve is used).

d

Strain diagram

0. 87 x 460 = 400 N/mm2

Strain Stress - Strain diagram

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

Concluding Notes

5. One way of ensuring that the beam failure is ductile is to introduce some compress10n steel, so that the x/d ratio will be reduced to 0.45 (see Example 3) .


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Example 3 - Analysis of Doubly Reinforced Section

Determine the amount of compression steel required, in order to make x/d = 0.4S m Example 2. Find also the moment ofresistance of the resulting beam.

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2-25

d = 300


Introductory Notes

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1. If it is found that a singly reinforced beam is over reinforced and it is desired to make it under reinforced or balanced, tills may be achieved by (i) Increasing the depth of the section, (ii) Increasing the breadth of the section (i ii ) Introducing compression steel.

2. Increasing the breadth of the section will generally be uneconomical. Therefore, ifthe depth of the section cannot be increased due to non-structural reasons, option (iii) above is used.

Note 3

Appendix A.3

Assume a suitable value ford', say SO mm.

For equilibrium of the section, the compression in the top steel plus the concrete must equal the tension in the bottom steel.

Setting x = (0.4S)d = 13S mm (which automatically ensures the yielding of tension steel), we have d'j -x - 0.37 $ 0.43, which means that the compression steel will yield as well.

By balancing the compression and the tension force, (0.87)fykA~ + (O.S67)fckb(0.8)x = (0.87)fykAs (0.87)( 460)A~ + (O.S67)(20)(1S0)(0.8)(13S) = (0.87) ( 460) (981. 7) Hence, A~ = S23 mm 2


d' =SO mm

As'= S23 mm2

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Reference Calculations Output Note 4 Use 3Tl6 (A~ = 603 mm 2) Use 3T16

Note:- (603 mm2) Note 5 lOOA~/ _ (100) (603)/

Ac - (150)(350)

= 1.15 > 0.2; hence OK Appendix A.2 Lever arm for balanced section

= d - (0.4)(0.45d) = (0.82)d = (0.82)(300) = 246 mm Distance between top and bottom steel = 250 mm

Note 6 Hence, taking moments about level of tension steel, moment of resistance = (0.567)(20)(150)(0.8)(135)(246) +

(0.87)(460)(523)(250) = 97,518,318 Nmm M = 97.52 kNm = 97.52 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 depth will be reduced slightly; thi s is desirable, as it will increase the ductility of the section.

5. When compression steel is provided, a minimum percentage is required. The area of concrete is based on the gross section, and the overall depth is taken as (300 + 50) = 350 mm. The compression reinforcement percentage in beams should be greater than 0.2% as per IStructE Manual (Clause 5.4.4.2) .

6. In general, the most convenient way of finding the moment of resistance for a doubly reinforced section is to take moments about the level of tension steel. The amount of compression steel to be used in the calculation is the amount required (523 mm2), and not the amount provided (603 mm2) .

Concluding Notes

7. The moment of resistance of a doubly reinforced section can be considered to be the sum of the moments of resistance of (i) a balanced section and (ii) a "steel section" , consisting of equal amounts of tension and compression steel, separated by (d-d').

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1 Analysis of Beam Sections in Flexure

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Example 4 - Analysis of Non-Rectangular Section

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

h = 450

300

2-25

I( 150 )I


Introductory Notes

d = 400 fck = 20 MPa fyk = 460 MPa

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

Reference ---

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Calcufations - -

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

The area of the section under compression =(0.5)(0.8)x[600 - {(300 - 150)/450}(0.8)x] Area of steel= 981.7 mm2 t~ 300 ~I

Assume also that the steel has yielded. \ J 0.8 x / 0

Try x = 100 mm )u ~ Area in compression, Ac = (0.5)(0.8)(100){600 - (0.33)(0.8)(100)} -~ = 22,944 mm2 150

C = (0.567)fck.Ac = (0.567)(20)(22,944) = 260,185 N T = (0.87)( 460)(981.7) = 392,876 N

Try x = 155 mm Then, C = 393,078 N and T = 392,876 N. This approximation is satisfactory. Note also that x / d < 155 / 400 = 0.39 < 0.64; hence assumption that steel has yielded is OK


-output

x = 155 mm

Page 9

Reference

Note 2

Note 3

CaklJ.lat~ons .-Depth in compression= (0.8) (155) = 124 mm. The centroid of the compression zone from the top of the section is

{ (150)(124) (1) + (0.5)(150)(124) (-)} y = {(150)(124) + (0.5)(150)(124)}

= 55.11 mm

Hence, lever am1 (z) = 400 - 55.11 = 345 mm M = C.z = (393,078)(345) = 135.6 x106 Nmm

= 136 kNm

Note: - Alternative method of finding x.

Assuming that steel has yielded, T = (0.87)(460)(981.7) = 392,876 N

For any x, the area under compression is Ac= (0.50)(0.8)x[600 - {(300 - 150)/450}(0.8)x] C = (0.567)(20)Ac

Putting T = C, we have the quadratic equation, x 2 - (2273)x + 328079 = 0, giving x = 155 or 2118 mm Since x/d = 155/400 = 0.39 < 0.45, steel has in fact yielded, as assumed.


z = 345 mm

M = 136 kNm

x = 155 mm

2. The lever arm cannot be taken as d - (0.40)x in this case, because the area in compression is non-rectangular.

3. This calculation will become a little more complicated if the section is not under-reinforced (see Example 2).

Concluding Notes

4. This approach from first principles, using the idea of strain compatibility, will have to be employed even in the design of beams such as these, which are non-rectangular, since the design formulae in Appendix A apply only to rectangular sections.


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CHAPTER2

Design of Beam Sections in Flexure

Example 5 - Design of Rectangular Section

Design a rectangular beam to take an ultimate load moment of 15 0 kN m, (i) as a singly reinforced beam and (ii) as a beam whose overall depth is limited to 400 mm. Assume that fck = 20 MPa, fyk = 460 MPa, and that the difference between effective depth and overall depth is 50 mm. Assume also that no redistribution of moments has been carried out.

Introductory Notes

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

Note 2

Appendix A.2

Note 3 Eq A.5

EqA.6

(a) Singly reinforced section

Let us assume that djb = 2.0 In order to find the minimum depth for a singly reinforced section, we should assume that x / d = 0.45

and K = M /bd2fck Then K = K' = 0.167

0.167 = (150 x 106)/{(d/2)(d2)(20)} d3 = {(2)(150 x 106 )/(0.167)(20)} d = 448 mm

Choose d = 475 mm, h = 525 mm, b = 225 mm Now K = M/bd2fck

_ 1so x10 6 I - {(225)(475) 2(20)} = 0.148 < 0.167; hence singly reinforced


dmin=448 mm d = 475 mm h = 525 mm b = 225 mm

Page 11

Reference

Eq A.9

Note 4 9.2.1.1(1) Eq 9.lN

Table 3.1

9.2.1.1(3)

Note 5

Appendix A.2

Eq A.13

. .. . . ,::,_,~ .1\:;,._,.,.!; . 1.-. :.t.,~~-' ;. ~ ... ~ - ~ .Calcu\ations

z = c 475) [ o.5 + j (0.25 - 0148 I i.134)] = 402 mm< (0.95)(475) = 451 mm

So z = 402 mm

As= M/co.87fykz) _ 150 x10 6 I - 0.87 x 460 x 402

= 932 mm2 Hence use 2T25 (As= 981.7 mm2)

Check for minimum reinforcement As.prov has to be greater than 0.26 f;rm btd but not less

yk

than 0.0013btd f

0.26 ;tm btd = 133 mm 2 yk

0. 0013btd = 139 mm 2 [ fctm = 2.2MPa] As.prov= 981.7 mm2 > 139 mm 2 ; hence OK

Check fo r maximum reinforcement 100A5/ _ (982)(100) /

Ac - / (525)(225)

= 0.83 < 4; hence OK

(b) Overall depth restricted

If the overall depth is restricted to 400 mm h = 400 mm, d = 400 - 50 = 350 mm b = 225 mm (assuming the same breadth as before) Now K = M /bd 2 f

ck _ 150 x10 6/ - {(225)(350) 2 (20)} = 0.272 > 0.167(i. e. K')

Hence compression reinforcement is required. Let us assume d' = 50 mm.

CK - K')f bd 2 A' ck s 0.87fyk(d - d')

(0.272 - 0.167)(20)(225)(350) 2 -

0.87(460)(350 - 50) = 482 mm 2

Use 2T16 & 1 Tl2 (A~ = 515 mm2)


..

225 ..

t t 525 475

i 2T25 l As,req = 932 mm2 2T25 (981.7 mmZ)

d = 350 mm b = 225 mm

A~.req = 482 mm 2 Use 2T16 & 1T12 (515 mm2 )

Page12

j.

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l j l.

l.

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Reference Calculations Output ~ IStructE 100A~; _ 100 x sis; Manual Ac - 22S x 400 (5.4.4.2)

= O.S72 > 0.2; hence OK

Eq A.6 z =ct [o.s + j(o.2s - K/i.134)]

z = (3SO) [o.s + J(o.2s- 0167/ 1.134)] = 287 mm< (0.9S)(3SO) = 333 mm

Kbalfckbd 2 Eq A.14 As= +A~

0.87fykZbal As= 1284 mm 2 (0.16 7) (20) (2 2S) (3S0) 2 Use 2T25 & 1 T20

Note 6 = (0.87)(460)(287) + 482 (1296 mm 2 ) = 1284 mm 2

Use 2T2S & 1 T20 (As = 1296 mm 2 )

9.2.1.1(1) Check/or minimum reinforcement As.prov= 1473 mm2 > 139 mm 2 ; hence OK

fctm [ 0.26-f-btd = 133 mm 2 ; yk

0.0013btd = 139 mm 2 ]

Note 7 Check for maximum reinforcement ~ 225 .. 9.2.1.1(3) lOOA~j _ 100 x 1473/ t @) 0 @ t Ac - 225 x 400 2T16& 1TJ2

525 475 = 1.64 < 4.0; hence OK i 2T25 & 1T2C l Note 8 Hence, use 2T2S & 1 T20 (bottom)

and 2T16 & 1 T12 (top)


2. In practice, the ratio of depth to breadth for a beam will have a value between 1.S and 2.S.

3. Many designers still choose dimensions for beams and columns in steps of 25 mm, because an inch is approximately 25 mm. Furthermore, depths considerably in excess of the minimum depth for a singly reinforced section may be chosen, in order to reduce the steel requirement.

4. The check for minimum reinforcement i$ almost always satisfied for tension steel in beams. A little care should be exercised, however, for compression steel. The minimum amount of reinforcement is given to control cracking.

5. The overall depth of the beam may have to be restricted, due to architectural requirements. On

Design of Beam Sections in Flexure Page13

the other hand, there may be some economy in designing beams with a marginal amount of compression steel, because longitudinal steel on the compression face will be required anyway, in order to support the shear links.

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

7. The maximum reinforcement percentage is limited in order to ensure that concrete is properly compacted around the reinforcement.

8. When providing reinforcement, a combination of bar sizes should be adopted such that the maximum and minimum spacing between bars is kept within specified limits (see Example12) .

Concluding Note

9. The Eurocode EC2 does not provide design charts, unlike BS 8110. The IStructE Manual does not contain design charts for beams and slabs, although it does have charts for column design. Moseley et al (2007) give a chart for converting M/bdZfck values to z values. Appendix D of this book provides a table for this same purpose. Nevertheless, all the reinforcement calculations in the examples that follow are done using the equations derived in Appendix A.

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Design of Beam Sections in Flexure Page14

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Example 6 - Design of Section with RedistributioQ ------ -- - -~ ---- --

I If the beam section in Part (a) of Example 5 (h = 525 mm, d = 475 mm and b = 225 mm) was / carrying an ultimate moment of 150 kNm after a 30% downward redistribution of moment, find

the steel reinforcement required. Assume that d' = 50 mm, fck = 20 MPa and fyk = 460 MPa.

)' .. )'"

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Introductory Notes

1. If the moment at a section has been reduced by downward redistribution, that section must have adequate rotational capacity at ultimate limit state, in order for plastic hinge action to take place. This capacity is ensured by restricting the x / d ratio to a value lower than the usually applied limit.

Reference - Calf_ulati?P,s

-

Output - -

..

5.5(4) 8 2: k 1 +(k2Xu/ d), for fck :::; 50 MPa Eq 5.!0a 8 = (1 - 0.3)/1 = 0.7 .~ \,' 5.5(4) k1 = 0.4 and k 2 = 1 (SLNA)

' ' I

So for 0.7 2: 0.4 + (1)(xu/d) -x

Note 2 ~

Reference .Cakulatioi:ts-- - output : . . . . .

(K - K')f bd 2 Eq A.13 A' _ ck 5

- 0.87f5 k(d - d') (0.148 - 0.12)(20)(225)(475) 2

-

- 0.87(460)(475 - 50) = 167 mm 2 As,req

1 = 167 mmZ

Use 2T12 Note 3 Use 2T12 (A~ = 226.2 mm 2 ) (226.2 mm 2 )

IStructE lOOA~/ _ 100 x 226.2/ 225 Ac - 225 x 475 ~ .. Manual t t (5.4.4.2) = 0.191 < 0.2; but acceptable 2Tl2

< 4.0; hence OK 525 475

K'f bd 2 ~ 2T25 l Eq A.14 A _ ck +A~ s - 0.87fykZbal (0.12) (20) (225) ( 4 75) 2

= (0.87)(460)(418) + 167 As,req = 895 mmz Use 2T25

= 895 mm 2 Use 2T25 (A 5 = 981.7 mm 2 ) (981.7 mm2)

Hence, use 2T25 (bottom) and 2T12 (top). This satisfies the min and max allowable reinforcement.


2. Any value below x/ d = 0.30 will give a feasible combination of 100As/bd and 100As'/bd, but x/ d = 0. 30 will generally minimize the total amount of steel required. Although x/d will vary from 0.45 to 0.6 as the moment redistribution is varies from 15% to 0% according to Eq 5.1 Oa, in no case should the x/ d value be taken as greater than 0.45.

3. Although smaller diameter bars would have satisfied the compression steel requirement, in general bars smaller than 12 mm diameter are not used for main reinforcement in beams and columns, as they will not be stiff enough during fabrication. Also, in this example, smaller diameter bars would not have satisfied the minimum steel requirement.

Concluding Notes

4. Although the applied moment for this section was the same as that in Example 5, this section had to be doubly reinforced because of the restriction on the neutral axis depth for the purpose of ensuring plastic hinge rotation,.

5. Hence, doubly reinforced sections may be required when (i) Architectural reql:lirements place limits on the beam depth and/or

r -, I

(ii) Significant redistribution of elastic moments has been carried out at a section. 1

Design of Beam Sections in Flexure Page 16

. - ; ;

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1 . ) . Example 7 - Structural Analysis of Beam

.... /

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I Determine the design ultimate load moments for the beam shown in the figure, using also the I following information.

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(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 ofreinforced concrete= 25 kN/m3.

Introductory Notes

1.

jo11

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

100

2000 6000

Sectional Elevation

~ : - -----------l--l------------------------------------------L--1 I I

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I I

I

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- -----------i - -~------------------------------------------~--1 I I I I I I

I I I

- -----------+--~------------------------------------------~--' I I

- -----------i;~--------------------~---------------------r I ~ I

Plan

450 _l_

300

Beam Section

3500

3500



.

~

Reference

Note 2

Note 3

Note 4

5. 1.3 (1)

Note 5

Calculations

The beam can be idealised as follows.

t 2000 i fiOOO ~ The critical moments for design will be (i) Hogging moment at B (ii) Sagging moment in span BC

Loading on beam (perm length)

From slab = (0.125)(25)(3.5) = 10.94 kN/m From finishes= (1.0) (3.5) = 3.50 kN/m From beam= (0.45-0.125)(0.3)(25) = 2.44 kN Im Total dead load udl (Gk, 1 ) = 16.88 kN/m Dead load point load at A (Gk,z ) = 2 x 3.5

= 7.0 kN Live load udl = ( 4.0)(3.5) = 14.0 kN/m

The hogging moment at B will be maximum when the cantilever portion AB is loaded with the maximum design ultimate load, irrespective of the load on the span BC. The sagging moment in BC will be a maximum when the cantilever portion AB has the minimum design ultimate load, while the span BC has the maximum design ultimate load.

Maximum design ultimate load (udl) 1.35Gk,1+1.5Gk,2 = (i35)c16.88)+( 1~14.0)

= 4D 9 kN/m ~ Minimum design ultimate load (udl) 1.35 Gk.1 . =

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Reference Calculations Output Hogging moment at B

9.45kN

bvvy4./ '" kN/m A B ct Ms = (1.35)(7.0)(1.95) + ( 43.79) (2.0) 2 /2

= 106 kNm Ms= 106 kNm (hogging)

Sagging moment in BC

7kN .. ~ 43.79kN/m 227~ A B x

M8 = (7)(1.95) + (22.79) (2.0) 2 /2 = 59.23 kNm

Taking moments about B for BC, Rc(6) + 59.23 = (43.79) (6.0) 2 /2

Re = 121.5 kN

Mx = (121.5)x - ( 43.79) (x) 2 /2 Mx = 0 at x = 0 m & 5.55 m

dMx/ dx = 0 gives:

( 43.79)x = 121.5 x = 2.77 m Msc= 168.6 kNm

Mmax = (121.5)(2.77) - (43.79) (2.77) 2/2 (sagging) = 168.6 kNm


2. Idealization is the first step in analysis. Since it is not possible to model the actual structure with complete accuracy, idealization should be performed such that the results obtained are conservative. For example, although point Chas a certain degree of restraint, it is impossible to quantify it. However assuming the end C to be simply supported will give a higher (and hence conservative) moment in the span BC. The restraint moment at C can be subsequently accounted for by providing a nominal amount of hogging steel there.


3. The point load will in fact act at 1.95 m to the left of B. At supports C and B, it is assumed column dimension is similar to that of the depth of the beam. So the effective span becomes the distance between the centres of the supports (Clause 5.3.2.2 (1)).

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

5. All actions originating from the self weight of the structure may be considered as originating from one source and hence there is no requirement to consider different factors on different spans (Clause 3 .2, IStructE Manual). Since the weight of the slab is considered as occurring from one source, the factor of 1.35 is to both spans when calculating the design sagging moment in BC, although the dead load in span AB is 'favourable'. The favourable parapet wall load is of course factored by 1.0.

Concluding Notes

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


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[ ' Example 8 - Design of Beam for Flexure

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Design the reinforcement for hogging and sagging moments in the beam in Example 7. Use fck = 25 MPa and fyk = 460 MPa.

Introductory Notes

1. In this example, only the reinforcement for the maximum sagging and hogging moments needs to be calculated, since the beam section is already specified in Example 7.

2. Furthermore, as the bending moment diagram for the beam has not been drawn (although it could be), the curtailment of reinforcement is not considered. This aspect is considered in Example 12.

Reference

4.4.1.1 (2)

4.4 .1.2(2) Eq 4.2 4.4.1.2(3)

Note 3 Table 4.1 Note 4 4.4.1.2(5)

Table 4.4N 4.4.1.2(6) 4.4.1.2(7) 4.4.1.2(8)

BSEN 1992-1-2 (Table 5.6) IStructE Manual (Table 5.11)

c akuiatibhs .. , . . . .::; : .. ;. - ..... .

Determination of Cover

Cnom = Cmin + llcdev

Cmin = max( Cmin,b; ( Cmin,dur + flCdur,y - flCdur,st -.6Cctur,add); 10 mm}

Cmin,b= 25 mm [It is assumed that reinforcement bars are separated (not bundled together) and T25 bars are used] - -

Assume outdoor exposure without chlorides So class designation is XC 3 Assuming a design working life of 50 years, the structural class is S4.

Hence Cmin dur = 25mm llcdur,y = 0 flCctur st = 0 llcdur add = 0 So Cmin,dur + llcdur,y - llcdur,st - .6 Cdur,add = 25 mm

Provide fire resistance of 2 hours for continuous reinforced beams. One of the possible combination.is bmin = 300 mm and a= 35mm a5 d = 35 + 10 = 45 mm (If link diameter is taken as 10 nun) Then Cmin,f = 35 - 10 - (25/2) = 12.5 mm

Cmin asf = 45 - 10 - (25/2) = 22.5 mm ' .

[22.5 mm is the side cover required to resist fire, and is not needed in the calculation of effective depth]


Output

Page 21

Reference Calculatio-ns :-. . output -~

So Cmin = 25mm 4.4.1.3(1) liCctev = lOmm

Cnom = 25 + 10 = 35mm Cnom = 35 mm

Note 5 Assuming a link diameter of 10 mm and a reinforcement size of 25 mm, the effective depth will be d = 450 - 35 - 10 - 25/2 = 392.5 mm

Design for hogging moment

The beam behaves as a rectangular beam. Example 7 b = 300 mm, d = 392.5 mm, M = 106 kNm

Eq A.5 K = M/bct2f ck

- (106 x 106 ) Ii - ((300)(392 .5) 2(25)} = 0.092

1 . -

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1

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Reference

Eq A.5

Eq A.6

Eq A.9

Cakufations"-. beff,l = (0.2 X 1.6) + (0.1 X 5.1) = 0.83 m 0.21 0 = 0.2 x 5.1 = 1.02 m beffl < b1, 0.210 So beffl = 0.83 m

beff = 0.83 X 2 + 0.3 = 1.96 m

Assume that the neutral axis is within the flange.

K = M/bd2f ck

- (169 x 106)/ - {(129-0)(392.5) 2 (2~)} = 0.022 < 0.167; hence singly reinforced

z = d [o.5 + jco.25- K/1.134)]

z = C392.5) [ o.5 + j ( 0.25 - 0022I1.134)] z = 385 > (0.95)(392.5) = 373 mm; so take z = 373 mm

_ (d - z)/ x - 0.4

- (392.5 - 373)/ x - 0.4

x = 49 mm 0.8 x = 39.2 mm

Output

300

beff = 1:96 m

z = 373 mm

x = 49 mm

Slab thickness is taken as 125 mm, hence, neutral axis is in fact N.A. is inside

within the flange, and the beam can be designed as a rectangular beam with b = 1960 mm. the flange

M A----

s - 0.87fykZ (169 x 106 )

=------(0.87)( 460)(373) = 1132 mm 2

Use 2T25 and 1 T20 (A5 = 1296 mm 2 )

As,req = 113 2 mm2

Use 2T25 & 1T20 (1296 mm2) (sagging)


Reference Note 6

Example 7

Note 7

6.2.4(3)

Appendix B

Example 7

Note 8 6.2.4(4) 6.2.2(6) 6.2.4(4) Note 9

3.1.6(2)

r . ~-, . ~

-Calculations -Transverse reinforcement in the flange

Design longitudinal shear stress= VEct Distance from the point of the zero moment to the point of the maximum moment = 2. 77 m

A

!:ix = 2.77 /2 = 1.39 m

!:iF ct/ VEct = (hr x !ix)

11x = l'/z

!:iM bro fiFct = (d - hr/2) x br

!:iM (bf- bw)/2 fiFct = x ----(d - hr/2) bf

x

' ,-,y I I ..

\ I / \ - \

\ ( f

\ .;' ~t --.,

'

!:iM = (121.5)(1.39) - ( 43.79) (1.39) 2 /2 = 126.6 kNm

126.6 x 106 (1960 - 300) /2 fiFct = (392.5 - 125/2) x 1960 !if ct = 162.5 kN v - 162.5 x 103/

Ed - (125 X 1390)

= 0.935 N/mm 2

To prevent crushing of the concrete strut in the flange VEct :::; y;fcd sin Sr cos ef

yJ = 0.6 [ 1 - :~~] = 0.54 Take Sf= 26.5 for compression flange vfcd sin ef cos ef = 0.54(0.567)fck sin 26.5 cos 26.5

= 0.54(0.567)(25) sin 26.5 cos 26.5

/ 1 "' :/ -'T '-' .

= 3.06 N/mm 2 > 0.935 N/mm 2 ; hence OK

f _ actfctk,o.os; _ (1) (1.8) / - 1 2 N/mmz ctct - . y c - / (1.5) -


) j'-

- \}

\ .

VEd

Output

r -,

,_

= 0.935 N/mm2

Page 24

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Reference Calculations Output

Table 3.1 [fctk,o.os = 1.8 N/mm 2 for fck = 25 MPa] 6.2.4 (6) kfctd = (0.4)(1.2) = 0.48 N/mm 2 < 0.935 N/mm 2 ;

so extra reinforcement above that for flexure is required .

6.2.4 (4) Asrfyct/ > VEct hr/ Eq 6.21 Sf - cot Sr

A5r(0.87)(460)/ > 0.935 x 125/ Sf - cot 26.5

Asf/sf;:::: 0.146

Note 10 Provide T10@400 mm spacing (196 mm 2 /m) Asf/sf = 78.5 / 400 = 0.196;:::: 0.146 ; hence OK Transverse steel

6.2.4(2) Minimum amount of reinforcement (T10@400 mm) (196 mmZ/m)

fctm 2.6 9.2.1.1(1) 0.26 fyk btd = 0.26 x 460 x 300 x 392.5 = 173 mm 2 /m

0.0013btd = 0.0013 x 300 x 392.5 = 153 mm 2 /m Table 3.1 [ fctm = 2.6 MPa for fck = 25 MPa]

As.prov =196 mm2/m > 173 mm2/m; hence OK


3. Chloride free exposure entails mainly carbonation induced corrosion. Chloride environment are much more damaging to reinforcement.

4. The minimum cover required for the reinforcement is found based on the safe transmission of bond force, the protection of steel against corrosion and adequate fire safety. Then an allowance for deviation of 10 mm is added to get the nominal cover. l'.1Cctev could be reduced further (Clause 4.4.1.3(3)) ifthe fabrication ofreinforcement is subjected to a quality assurance system, e.g. where monitoring and/or rejecting non confirming members take place.

-

5. The calculation of effective depth from the overall depth is illustrated by the figure below.

shear link

h d

....._ ____ _,-=}--cover


6. Transverse reinforcement in flange resists the longitudinal complementary shear stresses that occur in the flanged section along the interface between the web and flange (see Appendix B).

- 7. In Example 7, points of contra flexure will occur in span BC at two points, namely i. At support C ii. At 5.55 m from C along CB The maximum moment occurs at 2.77 m from C (i.e. 3.22 m from B along BC). !J.x is the half the distance between the section where moment is zero and the section where moment is maximum. So from B along BC, !J.x would be 1.39 m {(5.55 - 2.77)/2}. From C along CB too, 6.x would be 2.77 /2 = 1.39 m. To get maximum shear, the location where !J.x is minimum should be used.

8. What follows are checks and designs for shear, which is dealt with more comprehensively in Exan1ples 10 and 11.

9. If we take 8 = 26.5 (the lowest value for compression flanges), we will get the smallest shear resistance and will hence be conservative.

10. The area of transverse bending steel provided should be greater than that given by Eq 6.21 or half that given by Eq 6.21 plus that required for transverse bending (Clause 6.2.4(5)).

Concluding Notes

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

12. If the neutral axis of a flanged beam falls within the flange, the design is identical to a rectangular beam, as seen here.

13. When designing for hogging and sagging moments at support and span respectively, care must be taken to remember what steel has to be placed at the top of the beam section, and what steel at the bottom.

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Example 9 - Design of Flanged Section

Design an edge beam of a beam-slab system to take an ultimate moment of 520 kNm at mid span. Spacing of beams= 4 m; Span of beams= 6.0 m; ..________ Thickness of slab= 100 mm; fck = 25MPa; fyk = 460 MPa.

Introductory Notes

I. An edge beam will have a transverse slab only on one side; hence it is called an L-beam. The beam in the earlier example is called a T-beam, since the slab extended over both sides of the beam. If the beam is below the slab (as is the case most of the time), the slab will act as a flange only in the span, when the top of the section is in compression, and not at the supports

Reference

Note 2 Note 3

5.3.2.1(3)

Note 4

Eq A.5

Eq A.6

. Calcula'tions .. . . .- ._ ..

Assume that bw = 225 mm Choose h = 450 mm and d = 390 mm

beff = I beff,i + bw :S b beff,i = 0.2bi + 0.11 0

b - ( 4 - 0. 2 2 5 )/ - 1 8 9 1 - 2 - . 111

10 = 0.70 XI = 0.70 X 6 = 4.2 m beff,l = 0.2 X 1.89 + 0.1 X 4.2 = 0.8 m 0.21 0 = 0.2 x 4.2 = 0.84 m

beff,l < b11 0.21 0 So beff,l = 0.8 m

beff = 0.8 + 0.225 = 1.025 m ( < 4/2 = 2.0 m)

M (520 x 106 ) I K = /bd 2 fck = / ((1025)(390) 2 (25)}

K = 0.133

z =ct [o.5 + j(o.25- K/i.134)]

z = (390) [o.5 + j(o.25 - 0. 133; 1_134)] = 337.1 mm< (0.95)(390) = 370.5 mm

Hence take z = 337.1 mm

- (390 - 337.1)/ x - . 0.4


Output bw= 225 mm h = 450mm d = 390 mm

Page 27

Reference x = 132.25 mm

Note 5 0.8 x = 105 .6 mm > 100 mm. So the rectangular stress block goes outside the flange region, and the neutral axis goes inside the web region. The above z value is then not valid.

r-Jt o 11 1:!;__ --------------, _ \ r r \ ,,,. \I'll, ) -:~ ~:J!f IStructE (Mu~ 0.567fckCbeff - bw)hf(d - 0.5hr)\ . t.~ / bJ,--- Manual M~ 0.567(25)(1025 - 225)(100)(390 - 0.5(100))

l (5.4.4.l) M =385.6 kNm ~ \ 1 1.1 ' (r.Jl ' j.J ( Note 6 uf \ \ , \L,

M M \- l

Eq A.6

Note 7

9.2.1.1(3) Note 8

,,.--'-. - ur I , Kr=---

._/ fckbwd 2 (520 - 385.6) x 106

Kr= (25)(225)(390) 2 Kr= 0.157 < K' = 0.167; hence no compression reinforcement is required

z = C39o) [ o.5 + j ( 0.25 - 0 157 I i.134)] z = 325.3 mm

385.6 x 106 (520 - 385.6) x 106 As= 0.87(460)(390 - 50) + 0.87(460)(325.3) As = 2834,+ 1032 As= 3866 mm 2 Use 8T25 bars (3928 mm2)

To provide this reinforcement arrangement h has to be increased further to around 475 mm .

Maximum reinforcement A5 max= 0.04Ac = 0.04 X 225 X 475

= 4275 mm2 > 3927 mm2; hence OK


x = 132 mm

As,req = 3866 mm 2

Use 8T25 (3928 mm2) (sagging)

2. A web width of 225 mm is aroW1d the minimum that is practically desirable, in order to accommodate the reinforcement. A width of 200 mm can be considered as the absolute minimum for all beams save those which carry very nominal loads.


1 -

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3. The difference between d and h has been taken as 60 mm (for outdoor conditions), whereas it can be taken as 50 mm for indoor conditions. The actual calculation of cover should be carried out as in Example 8.

4. It is asswned that the beam considered here is continuous at both ends. So according Clause 5.3 .2.1(2), 10 = 0.701.

5. This trial-and-error approach has to be adopted to find out whether the rectangular stress block of compressive stress in concrete at failure is within the flange or whether it extends to the web region.

6. The approach here is to find the reinforcement required to balance (i) The compression in the outstand flange alone, plus (ii) The compression in the web (extending a depth of 0.8x over a width of bw)

7. The difference of 60 mm between h and d is based on asswning a single layer of reinforcement. Since 8T25 has to be provided to resist bending, the beam will contain double layers of reinforcement, and (h - d) can be taken as 60 + 25 = 85 mm.

8. This check for maximwn percentage ofreinforcement is also almost always satisfied, except for very heavily reinforced sections. Although the check is satisfied here, care will have to be exercised iflapping is done. Spacing between bars may also be small.

Concluding Notes

9. This example illustrated the situation where the rectangular compression stress block fell below the flange of a flanged beam.

10. If compression reinforcement is found to be required, it is recommended in the IStructE Manual (Clause 5 .4.4.1) that the section be redesigned, so that compression reinforcement would not be needed.


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Design of Beams for Shear 1~ CHAPTER3

- "" . -,_, __ . -~

1" Example 10 - Design of Section for Shear --------- I / !/" . I -

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A simply supported beam of cross section b = 225 mm and d = 400 mm cifiries.::iail ultimate load of 100 kN/m over its clear span of 5.0 m. Design the shear reinforcement required near the support, assuming that the percentage of tension reinforcement at the support is 0.8%. Assume fck = 25 MPa and fyk = 460 MPa.

Introductory Notes

1. The two main effects caused by flexure are bending moment and shear. The bending moment in a concrete beam is carried by steel reinforcement parallel to the beam axis. The shear force is carried by steel reinforcement in a transverse direction, generally in the form of links.

2. Links normally have diameters varying from 6 to 12 mm, in steps of 2 mm.

Reference

6.2.3(3) Eq 6.9

6.2.2(6) Eq 6.6N 6.2.3(1) 6.2.3(3)

Calculations

Shear check at support

Maximum shear force at the face of the support, VEf = (100 x 5)/2 = 250 kN

Crushing strength of the diagonal strut (concrete),

[ fck] v = 0.6 1 - 250 = 0.54 = V1 fed = 0.567fck z = 0.9d acw = 1 for non prestressed members Assume 8 = 22 for uniformly distributed loads

Design of Beams for Shear

Output ~

VEf = 250 kN

Page 30

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Note 3

6.2.1(8)

6.2.2 (1) Eq 6.2.a

Eq 6.2.b

Calculations

VRd,max = (1.0)(225)(0.9 x 400)(0.54)(0.567)(25)

(cot 22 +tan 22) = 215.4 kN < 250 kN; hence not satisfactory.

Put VRd,max = VEf

v -UcwbwZV1fcd Ucwbwzv1fcd sin 28

= Ef - (cot 8 + tan 8) 2 8 = 0.5 sin-1 Ef [ 2V ]

Ucw bwZV1 f cd - . -1 [ 2 x 250 x 103 l

S - 0.5 Sin (225)(0.9 X 400)(0.54)(0.567)(25) 8 = 26.87

22 < 26.87 < 45; hence OK

Check at distance 'd' from the support

VEd = (250)(2500 - 400)

(2500) = 210 kN

Design value of shear resistance of the concrete beam without links

VRd,c = [ CRd,ck(lOOp1fck) 1h + k1 CYcp] bwd

with a minimum of VRd,c = ( Vmin + k1 CYcp)bwd

k = 1 + ~. = 1 +): = 1. 71 :5 2.0 . As1

p1 = bwd = 0.008 :::; 0.02

0.18 0.1'8 c - - =-=0.12 Rd'. c - Ye 1.5

k1 = 0.15

vRd,c;::; [co.12)(1.71)(0.8 x 25) 1h + o] (225)(400) = 50-:-1 kN

Design of Beams for shear

- -

Output

8 = 26.87

VEd = 210 kN

VRd,c = 50.1. kN

Page 31

Reference

Note 4 Eq 6.3N Note 5

Note 6

Eq 6.8 Note 7

Note 8

Note 9

9.2.2(6) Eq 9.6N

9.2.2(5) Eq 9.4

Eq 9.5N

6.2.3(7) Eq 6.18

Note 10

__ - ~ -.. 1 '-' -z":. .:.: - - "'t, " ~ .. ,._ -Cakrifations-

'-:; - -_-_ -: ;-_"'! .- . '- :,-:-

vmin = 0.035k3/2fck

1lz = (0.035)(1.71) 3iz(25) 112 = 0. 39 N/mm 2

bw = 225 mm

VEd = 210kN > 5a.1 kN

So shear reinforcement is necessary Asw

VRct,s = VE ct = - zf ywd cot 8 s

Asw 210 X 103

-s- = {(0.9)( 400)}{(0.87)( 460)} cot(26.87) = 0.7 4

It is asslUTied links are of Tl 0 diameter bars with two upright arms ( 157 mm2)

157 -;:::0.74

s s:::;: 212 mm

So provide T10 at 2ao mm spacing

Maximum longitudinal spacing between links S1,max = 0.75d(1 +cot a:)

a = 9a 0 (vertical shear links provided) S1 max= a.75 X 40a X (1+cot90)

= 300 mm> 2ao mm; hence OK

Shear reiriforcement ratio Asw

Pw=----(s. bw. sin a) 157

Pw = (2aa x 225 x sin 90) = a.oa 35

(a.as~) (a.a8~ Pw,min = f - = 46 a yk

= a.aaa9 < Pw; hence OK

Additional longitudinal tensile force b.Ftd = a.5VEct(cot8 - cot a:)

a = 9a 0 ; vertical links

L'.ffrct = a.5 x 21a cot 26.87 = 2a7 .2 kN


Links Tla@2aa mm

l'iFtct = 207.24 kN

Page 32

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3. The design of shear reinforcement is done based on a truss model. In this model, it is ensured that the compressive stresses that occur in the diagonal compressive struts do not lead to compression failure of the struts. This is why the maximum shear check is done at the support of the beam (Clause 6.2.1(8)). According to Clause 6.2.3(2) 8 should lie between 22and 45. So initially assuming 8 = 22,we check whether VRd,max < VEf. If not, we put VRd,max = VEf, and find 8. If 8 > 45, the diagonal concrete strut is overstressed, so either the beam should be resized or the compressive strength of the concrete increased.

4. This formulae indicates that the concrete has some shear resistance even without the main reinforcement.

5. For a flanged beam, bwwill normally be the web width .

6. IfVEct < VRd,co it does not mean that shear reinforcement is not required . Minimum shear reinforcement should be always provided in beams according to Clause 9.2.2(5).

7. In this truss analysis model, it is taken that all the shear will be resisted by the provision of shear reinforcement without the contribution of the shear carrying capacity of the concrete (Clause 6.2.1 (5)).

8. When using this inequality for providing links, either the Asw value ors value must be chosen. In general. the Asw value is assumed and the s value calculated . The Asw value refers to the total cross section of links at the neutral axis of a section. Generally, it is twice the area of the chosen bar, since in most cases it is links with 2 ve11ical legs that are used. The resulting s value should not exceed 0.75d (Clause 9.2.2(6)), to ensure that at least one link crosses a potential shear crack. The transverse spacing between the legs of a link should be such that it does not exceed St max= 0.75d:::; 600 mm (Clause 9.2.2(8)).

9. The link spacing is also often specified in steps of 25 mm, because of the tendency to think in Imperial units (1 inch is approximately 25 mm). It may be better practice to specify in steps of 10 mm, which is economical too.

I 0. This additional longitudinal tensile force is resisted by extending the c':!rtailment point of th~ longitudinal reinforcement (Cla~se 9.2.1.3(2)). ---- --

Concluding Notes

11. In this example, only the shear reinforcement requirement near the support has been calculated. The requirement close to mid-span will be much less. This aspect will be considered in the next example.

Design of Beams for shear Page 33

Example 11 - Design of Beam for Shear

A simply supported beam, with d = 550 mm and b = 350 mm and clear span 6.0 mis subject to a triangularly varying shear force diagram, with a value of 400 kN at the face of the supports. The mid span steel consists of 4 Nos. 25 mm bars. Design the shear reinforcement required over the entire span, if two of the main bars are bent up at 45 near the supports. Take fck = 25 MPa fyk = 460 MPa.

Introductory Notes

1. In this example, the contribution from two bent up bars are also used to provide shear reinforcement near the beam supports.

2. The most reasonable way to provide shear reinforcement for the entire span would be to consider three areas - i.e. (i) The support area where bent up bars are also effective in addition to links, (ii) The middle of the beam, where only minimum links would suffice, and (iii) The portion in between the above.

Reference

6.2 .3(3) Eq 6.9 Note 3

Eq 6.6N

6.2 .3(2)

6.2.1 (8)

Support area

VEf = 400 kN Crushing Strength of the diagonal strut (concrete),

CT cw hwZV1 f cd v -Rd.max - (cot 8 + tan 8)

[ fck ] V = 0.6 1 - 250 = 0.54 = V1 fed = 0.567fck = (0.567)(25) = 14.2 N/mm 2 z = 0.9d = (0.9)(550) = 495 mm acw = 1 for non prestressed members. 8 = 22 for uniformly distributed loads a = 45

(1.0)(350)(0.9 x 550)(0.54)(14.2) VRct,max = (cot 22 +tan 22)

= 460.6 kN > 400 kN; hence there is no crushing of the

diagonal compressive strut (concrete).

Check at distance 'd' from the support

( 400) (3000 - 550) VEd = . (3000)

= 326.7 kN


VEf = 400 kN

VRd,max = 460.6 kN

VEd = 326.7 kN

Page 34

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Reference

6.2.2(1) Eq 6.2.a

Eq 6.2.b

Note 4

6.2.2(1)

Eq 6.3N

9.2.2(7) Eq 9.7N Note 5

6.2.3(4) Eq 6.13

Calculations Design value of shear resistance of the concrete beam without links

VRd,c = [cRct,ck(100p1fck) 1h -+ k1acp] bwd

with a minimum of VRd,c = ( Vmin + k1 CJcp)bwd O'cp = Nect/ Ac = 0

k = 1 + ~ = 1 + Jffi = 1.6 ,;; 2.0 A51 2 x 491

PI = bwd = 350 X 550 = 0.005 :5 0.02 0.18 0.18

CRd,c = -- = -1 5 = 0.12 Ye k1 = 0.15 vmin = 0.035k3/2fc//z = (0.035)(1.6) 312(25//2

= 0.35 N/mm2 bw = 350 mm

( V111 in + k1 CJcp)bwd = 67.4 kN vRct,c = [co .12)(1.6)(0.2 x 25) 1h + o] (350)(550)

= 62.9 kN < 67.4 kN VEd = 326.7 kN > 67.4 kN So shear reinforcement is necessary.

Shear resistance of two inclined bars (bent up bars)

A5w = 982 mm2 (area of 2T25 bars) Maximum longitudinal spacing of bent-up bars S1 max = 0.6d(1 +cot a)

a = 45 S1 max= (0.6)(550)(1 + cot45)

.= 660 mm ,.

Hence bent up bars has to be provided at spacing not less than 660 mm. Assume bent up bars are provided at a spacing of 660 mm.

Asw . VRcts =-zfywd (cot8+cota)sma

' s z = 0.9d = (0.9)(550) = 495 mm fywd = (0.87) ( 460) = 400 N /mm2

(491 x 2) VRd,s == 660 ( 49 5) ( 400) (cot 22 + cot 45) sin 45

= 724 kN


Output

VRd c = 67.4 kN

Page 35

Reference ".

9.2.2(4)

6.2.3(3) Eq 6.8 Note 6

9.2.2(6) Eq 9.6N

9.2.2(5) Eq 9.4

Eq 9.5N

VEd = 326.7 kN < VRct,5 (724 kN)

Shear resisted by bent up bars 724.0 . =--> 05 Total shear force 326.7

Although shear resisted by bent up is higher than the design shear force, at least 50% of the shear reinforcement has to be in the form of links.

So shear force carried by the shear links = (326.7)(0.5) = 163.4 kN

Asw VRd,s = VEct/2 =-zfywctCOt8

s

Asw 163.4 X 103 -= = 033

s (0.9)(550)(0.87)(460) cot22

It is assumed links are ofT6 bars with two upright arms (Asw = 56.5 mm2)

56.5 -= 0.33

s s = 171.2 mm

So provide T6@150 mm spacing

Maxjmum longHudJ!1al spadng between }jnks S1,max = 0.75d(1 +cot a)

a = 90 (For vertical links) S1,max = 0.75 x 550 x (1 + cot90)

= 413 mm> 150 mm; hence OK

Shear rejnforcement raNo _Asw/

Pw - (s. bw. sin a) If only the reinforcement ratio of the shear links is

I' considered.

56.5 p = = 0.0011

w (150)(350) (sin 90)

(o.os~) (o.osvlzS) Pw,min = ==

fyk 460 = 0.0009 < Pw; hence OK


T6@150 mm

I I

Page 36

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Reference

6.2.3(7) Note 8

9.2.2(5) Eq 9.5N

Eq 6.13

9.2.2(6) Eq 9.6N

C alcl!.la.tio ns AddUional longitudinal tensile force

From provision of bent up bars LlFrct = 0.5VEct(cot8- cot a)

a= 45 LlFrct,1 = (0.5)(163.4)(cot22 - cot45) = 120.5 kN

From the provision of shear links LlFrct, 2 = (0.5)(163.4)(cot22 - cot90) = 202.2 kN LlFrct,tot = 120.5 + 202.2 = kN

Middle Area

Shear force that could be carried by the minimum links. Asw

Pw,min = 0.0005 = ( b . ) s. w sm a

Asw - = 0.0005 x 350 x sin 90 = 0.175

s

It is assumed links are of T6 bars with two upright arms (A 5 w = 56.5 mm 2) 56.5 - ;:::: 0.175

s 4,.. . s "fP 323 mm

So provide T6@300 mm spacing as minimum links

Shear tJiat could be carried by the minimum shear links Asw

V = -zfywct cot8 s

56.5 V =

300 (0.9)(550)(0.87)(460) (cot22)

V = 92.3 kN Hence the extent of area covered by the minimum links

400!1 v = (3.0)

40011 92 .3 = (3.0) ,

11 = 0.69 m So the extent of area is 0.69 m.

. .

Maxjmum JongHudjnal spadng between links S1,max = 0.75d(1 +cot a)

a = 90 (for vertical links) S1,max-= (0.75)(550)(1+cot90)

= 413 mm> 300 mm; hence OK


Output

LlFrct,tot = 3.227 kN

T6@300 mm

Page 37

Reference

Note 9

9.2.2(6) Eq 9.6N

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Area in-between

Extent of this area = 3.0 - 0.69 - 0.66 = 1.65 m Shear force at distance 660 mm fromth_e support is V1 v - c 400)(3000 - 660)

1 - (3000) = 312 kN

It is asswned links are of T6 bars with two upright arms (Asw = 56.5 mm2)

3 56.5 312 x 10 ::::; -(0.9)(550)(0.87)(460) cot 22

s s ::::; 89 mm

Provide 2/T6@170 mm spacing.

Maxilnum JongHudjnal spadng between Jjnks Si.max = 0.75d(l +cot a)

= (0.75)(550)(1 + cot90) = 413 mm> 170 mm; hence OK

2/IOT6 ~ 1 .. 3T6

4T25 ~

0.66 m I 1.65 m I

0.69 m

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3. Since bent up bars are also used in the support area VRct,max from Equation 6.14 should be considered. However, putting a = 90 in Equation 6.14 gives the minimwn value of VRct,max' in fact the value given in Equation 6.9, which is the one used for checking.

4. Bottom reinforcement comprises 4T25, but near the supports 2 bars have been bent up. So in calculating the shear resisting capacity of the concrete near the support, only the area of 2T25 bars has been considered as giving the dowel action.

5. Since the bars are bent only at one location in this example, the mcp

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7. If small diameter tor steel is not available, mild steel of fyk = 250 MP a will need to be used, with an appropriate change in Equation 6.8.

8. This additional tensile force is not used in any future calculations here, but Example 12 gives one instance of its use.

9. If the link spacing is less than around 150 mm, it will be difficult for concreting to be carried out. Hence, as in this case, 2 links can be placed together, spaced wider apart. An alternative would have been to use 8 mm diameter links; however fabrication will be easier if links of the same dian1eter are used throughout the beam.

Concluding Notes

10. It is not very common practice now to use bent up bars as described in this example, although it was in the past. Hence, when only vertical links are used, we find the distance for which the minimum reinforcement is enough. For the remaining distance, we design links corresponding to the shear force at a distance 'd' from the face of the support.

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Design of Beams for shear Page 39

CHAPTER4

Serviceability Checks and Detailing in Beams

Example 12 - Serviceability Checks and Detailing

Carry out serviceability checks on the beam analysed in Example 7 and designed in Example 8. Also carry out detailing of reinforcement, including curtailment and lapping.

Introductory Notes

1. The serviceability checks consist of span/effective depth ratio calculations for deflection and bar spacing rule checks for cracking. If these simplified checks are satisfied, the beam is "deemed to ~atisfy" the serviceability limit state requirements.

Reference - r--t "I_'

.1:. ou.mut Check for deflection (Span/depth rules)

Note 2 Consider the span BC; 5.3.2.2(1)

t:---Effective span= 6000 mm fck = 25 MPa; fyk = 460 MPa

\i I As,req 1132

7.4.2(2) 1.-d. z.:__(bd = 1960 x 393 = 0.0015 < 0.0025 Note 3 so use p = 0.0025 -

p' = 0 7.4.2(2) Po = .{f';; x 10-3 = 0.005 7.4.2(2) p

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7.4.2(2)

Table 7.4N

Eq 7.16.b

7.4.2(2)

Note 6

Calculations So allowable span/effective depth

(~) = (54.6)(0.8)(1.24) = 54.16 all

(I) 6000 (1) d = 392 5 = 15.3 < d ; hence OK actuJ. all Consider span AB; Effective span= 2000 mm For cantilever with rectangular beam action.

As,req 780 p = bd = 300 x 392 = 00066 p' = 0 Po =ff;; x 10-3 = 0.005 p >Po; So use Eq 7.16.b

K = 0.4

1 r Po 1 ~i d = K 11+1.5fup- p' + 12 -~~p;; l [ 0.005 ] d = 0.4 11 + 1.5ill 0.0066 + 0 1 d = 6.67

310 500A5 prov 500 x 829 - = ---' - = = 1.22 :::; 1.5

f ykAs,req 460 X 7 41

So allowable span/effective depth

(~) = 6.67 x 1.22 = 8.14 all

(1) 2000 (1) -d = --5 = 5.09 < -d ; hence OK act 392. all , Curtailment of bottom reinforcement

The bending moment diagram envelope must first be drawn. For span BC, the controlling load case is when AB has a favourable design ultimate load and BC has the unfavourable design ultimate load. This has been considered in Example 7.

Serviceability Checks and Detailing

Output

All. span/ depth= 54.16 Act. span/ depth= 15.30; hence OK

All. span/ depth= 8.14 Act. span/ depth= 5.09; hence OK

Page 41

Reference

Figure 9.2

Example 7

Note 7

9.2.1.3(2) Note 8

~ ..

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'.~~Icufatj~ns __ . 7kN z: 43.79 kN/m 22.79kN/~ A19~

x

A B c

1 B~--""'--::Z~B-~of 2T25

Bending Moment Diagram . . ~"j ": : -:_ I t, . .... J

BM 1 - Bending moment diagram :.'. BM 2 - Bending moment diagram after "shift rule" is applied. For span BC, Mx = (121.5)x - ( 43.79)(x) 2 /2 Mx = 0 atx = 0 Mxis maximum at x = 2.77 m Mx,max = 168.6 kNm Mx = 0 at again 5.55 m

Steel at span BC is 2T25 & 1 T20. We can consider curtailing the I T20 bar.

M.O.R of continuing bars ( A5 = 982 mm 2 ) can be shown to be 146.5 kNm. Putting (121.5)x- (43.79)(x) 2 /2 = 146.5 x = 1.77 m and 3.78 m. These are the theoretical cut off points when the tension induced in steel due to the bending is considered.

However, tension in longitudinal steel is also induced due to the shear force; this can be incorporated by applying a shift of the cut off points.

z(cot e - cot a) a1 = 2

(0.9)(392.5)(cot22 - cot90) x 10-3 a 1 = 2 = 0.44 m

So theoretical cut off points .are x = (1.77 - 0.44) = 1.33 m

and (3.78 + .44) = 4.22 m


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Page 42

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Reference Note 9 Figure 9.2

8 .4.3(2) Eq 8.3 8.4.2(2) Eq 8.2 8.4.2(2)

8.4.2(2) 3.1.6(2) Table 3.1 3.1.6(2)

8.4.4(1) Eq 8.4 8.4.4(1) Eq 8.6

8.4.4(1)

Table 8.2

Note I 0

Calculation~ The practical cut off points are lbct (anchorage length) away from theoretical cut off points .

Anchorage length of bottom reinforcement

11 1 = 1, (good bond conditions are assumed, for bottom reinforcement ) 112 = 1 for cp :::; 32 mm act= 1 fctk,o.os = 1.8 MPa for fck = 25 MPa;

fctd = actfctk,o.os/Yc = 1X1.8/1.5 = 1.2 MPa fbct = 2.25 x 1 x 1.2 = 2.7 MPa

For 20 mm bars CYsct = (fyk/Ys) lb,rqd = (cp/4)(CY5 ct/fbct) = (20/4){(460/1.15) /2.7} lb,rqd = 741 mm

lbct = a1aza3a4aslb,rqd ~ lb,min For anchorage in tension lb,min > max {0.3lb,rqd; 10; 100 mm) lb min >max {0.3 x 741; 10 X 20; 100 mm)

> max {222; 200; 100 mm) lb min = 222 mm

Cct = min{a/2, c1, c} = min{74 mm, 45 mm, 45 mm} = 45 mm

a 1 = 1 because straight bars a 2 = 1- 0.15(cct - cp)/cp = 1- 0.15 (45 - 20)/20

= 0.81 lbct = 1x0.81x1x1x1 x 741 = 600 mm [Confinement by transverse reinforcement is neglected as it is very small. So a 3, a 4, and a 5 are taken as unity] lbct > lb,min; hence OK Hence, the practical cut off points are x = 1.33 - 0.60 = 0.73 m x = 4.22 + 0.60 = 4.82 m Length of20 mm bar required = 4.82 - 0.73

= 4.09 m


Output

fbct = 2.7 MPa

lbct = 600 mm

Length of 20 mm bar= 4.09 m

Page 43

Reference

Note 11

Figure 9.2

Note 7

Calculations Distances to ends from Bare 5.27 m and 1.18 m.

Curtailment of top reinforcement

For support B, the controlling loading case is when spans AB and BC have the unfavourable and favourable design ultimate loads respectively.

7 x 1.35 22.79 kN/m

~ Taking moments about C for AC R8 (6.0) = (7)(1.35)(7.95) + (43.79)(2.0)(7.0)

+(22.79)(6)2/2 Rs= 183.1 kN My= (7)(1.35)[y - 0.05] + 43.79y 2 /2 - (183.l)[y - 2]

-(43.79 - 22.79) [y - 2]2 /2 My= (9.45)[y - 0.05] + 21 .9y2 - (183.l)[y - 2]

-(21) [y - 2]2 /2

~w \~'\. BMI BM2 A Bl

BM 1 - Bending moment diagram BM 2 - Bending moment diagram after "shift rule" is

applied. My= 0 at y = 0 and My = 106 kNm at B My= 0 again aty = 3.55 m Steel at support is 2T20 and 1Tl6. We can consider curtailing the 1Tl6 bar. Moment of Resistance of continuing bars (A 5 = 628.3 mm2) can be shown to be 91.26 kNm. Putting (9.45)[y - 0.05] + 21.9y2 = 91.26 we can obtain y = 1.84 m for span AB. And from (9.45)[y- 0.05] + 21.9y2 - (183.l)[y- 2]

-(21) [y ~ 2]2 /2 = 91.26 we can obtain y = 2.17 m for span BC.


Output

Page 44

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9.2.1 .3(2) Note 8

8.4.3(2) Eq 8.3 8.4.2(2) Eq 8.2 8.4 .2(2) Figure 8.2 Note 12 8.4.2(2) 3.1 .6(2) Table 3.1 3.1.6(2) Note 12

8.4.4(1) Eq 8.4 8.4.4(1)

Table 8.2 Note 10

- . c~l~rifation( . .. '" " These are the theoretical cut off points when the tension induced in steel due to the bending is considered, but tension in longitudinal steel is also induced due to the shear force.

Shift in cut off points considering the tension induced steel due to shear. a1 = z( cot 8 - cot a) /2

(0.9)(392.5)(cot22 - cot90) x 10-3 a1 = 2

= 0.44 m

So theoretical cut off points are y = 1.40 m and 2.61 m The practical cut off points are lbct (anchorage length) away from theoretical cut off points.

Anchorage length for the top reinforcement

T] 1 = 0.7 (poor bond conditions for top reinforcement since h > 250 mm)

T] 2 = 1 for::; 32 mm act= 1 fctk,o.os = 1.8 MPa for fck = 25 MPa; fctd = actfctk,o.os/Y c = 1 X 1.8/1.5 = 1.2 MPa fbct = 2.25 x 0.7 x 1.2 = 1.9 MPa

For 16 mm bars CJ sd = (f yk/Y s) lb,rqd = (/4)(CY5 ct/fbct) = (16/4){(460/1.15) /1 .9} Ib,rqd = 842 mm lb ct = al az a3 a4aslb,rqd ~ lb,min For anchorage in tension Ib,min > max {0.3lb,rqd; 10; 100 mm) Ib,min >max {0.3 X 842; 10 X 16; 100 mm)

> max{253; 160; 100 mm) Ib,min = 253 mm a 1 = 1 because straight bars a2 = 1-Q.15(Cct - cp)/cp =1-0.15(45 - 16)/16

= 0 73


Output

fbct = 1.9 MPa

Page 45

Reference Note 13

Note 14

Note 15 IStructE Manual (Table 5.25)

lbct = 1 x 0.73 x 1 x 1 x 1 x 842 = 615 mm >lb min; hence OK

[Confinement by transverse reinforcement is neglected as it is very small. So a:.3 , a:.4, and a:.5 are taken as unity]

Hence, the practical cut off points y = 1.40 - 0.62 = 0.78 m y = 2.61 + 0.62 = 3.23 m Length of 16 mm bar required = 3.23 - 0.78

= 2.45 m Lapping of bars

The continuing 2T20 top bars at B can be curtailed at the point where tension in steel becomes zero closer to B in span BC and lapped with 2T12 bars (which will anchor the shear links). Similarly, the continuing 2T25 bottom bars in span BC can be curtailed at the point where tension in steel becomes zero closer to B in span AC and lapped with 2T12 bars.

For top bars, distance of point where tension m longitudinal steel becomes zero from A 1s 3.99 m (3.55 + 0.44). This would be the theoretical cut-off point. To find the practical cut-off point, continue the bars for an anchorage length ( 48 x bar diameter). Hence, cut-off point is 3.99 + 0.96 = 4.95 m from A, 1.e. 4.95 - 2.0 = 2.95 m to the right of B.

lbct = 615 mm

Length of 16 mm bar= 2.45 m

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IStructE Manual (Table 5.25) Note 16

The lapped 2T12 bars will start (56)(12) = 672 mmbefore the curtailment of the 2T20 bars, i.e. 2.95 - 0.67 = 2.28 m to the right ofB.

For bottom bars, distance of point where tension m longitudinal steel become zero is (5.55 + 0.44) = 6.0 m. So it is at B. The practical cut-off point would be lbct = 36 x 25 = 900 mm distance beyond this. Hence, it would be (6 + 0.9 - 6) = 0.9 m to the left of B. The lapped 2T12 bars will start 42 x 12 = 504 mm before the curtailment of the 2T20 bars, i.e. 0.9 - 0.5 = 0.4 m to the left of B.

Detailing arrangements at support C and end A The support at C is designed as simple supported, so the sagging moment is zero. So the stress in steel is only caused by shear.

Serviceability Checks and Detailing Page 46

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9.2.1.4

Note 7 Example 7 9.2.1.2(1) Note 17

Example 8 9.2.1(4)

8.4.4(1) Eq 8.6

- .

Calculations . . -,:.:.>-( --:.- .

flFtct = 0.5VEct (cot 8 - cot a) VEct =the maximum reaction at C The reaction is maximwn when span BC carries the

unfavourable load combination and span AB the favourable one.

7 kN k I 22.79 kN/m 437/ N m ~ Moment about B Rc(6.0) = ( 43.79)(6) 2 /2 - (22.79)(2.0)(1.0)

- (7)(1.95) Re = 121.5 kN VEct = 121.5 kN flFtct = 0.5VEct(cot 22 - cot 0) = 150.4 kN A5 = 402 mm 2 (2T16) Considered that 2T16 bars are provided at the simply supported end. Stress in steel = flFtct/ A5 = 150.4 X 103 / 402

= 374N/mm 2

< 0.87fyk; hence OK

Moment carrying capacity of2T16 bars is 60.0 kNm. Maximum span moment= 168.6 kNm 15 % ofmaximwn span moment= 0.15 x 168.6

= 25.3 kNm So 2T16 can resist the moment of 25.3 kNm. Bottom reinforcement area at the span= 1296 mm 2

Required bottom reinforcement at end supports = 1296 /4 = 324 mm 2 < 402 mm 2 ; hence OK

At the end of the cantilever (Point A), sagging moments and shear forces are zero and hogging moment is also zero. The required anchorage length is then the minimum anchorage length, lb.min = 200 mm for 0 = 20 mm . This can easily be provided by bending the bars .


01:1tput

2T16 at simply supported end

2T16 for restraint moment

Page 47

Reference Calculations ..

1T16 2T20 2T20 2T20 2TI 2 2TI 6

1.-----1 . -7 ( ---.... 7-, --,:_ 'U 2Tl2 1'2T25 A B

2T25 1T20

Crack width check (Bar spacing rules)

2Tis1' c

Table 7.IN For external environment, exposure class XC3 and for reinforced concrete, quasi permanent combination of actions should be used. Allowable crack width is 0.3 mm

Note 18 For office, domestic and residential areas, the stress in steel is

Table 7.3 N Note 19

8.2(2) Note 20

fyk Gk+ 0.3Qk f - ----------5 - 1.15 (l.35Gk + 1.5Qk)8

460 16.88 + 0.3 x 14 fs = 1.15 (1.35 X 16.88 + 1.5 X 14)(1) f5 = 193 N/mm 2

Since the stress is close to 200 N/mm2, maximwn allowable clear spacing is 250 mm. Clear cover provided = 35 mm Assume link diameter of 6 mm

Considering the support section (tension on top), Clear spacing between top bars (2T20 & 1 T16) {300 - (2)(35) - (2)(6) - (20 + 20 + 16)}/ 2

= 81mm {If middle (16 nm1) bar is curtailed, clear spacing

= 178 mm.} The top spacing at the support and the spacing when the middle bar is curtailed is less than 250 mm; hence OK

The minimum clear distance = max{k1 x ; dg + k 2 ; 20 mm} = max{l x 20 = 20 mm; 20 + 5 = 25 mm; 20 mm} = 25 mm < 81 mm; hence OK


Output

Maximwn allowable spacing 250mm

Minimwn allowable spacing 25mm

Page 48

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Considering the span section (tension on bottom), Clear spacing between bottom bars (2T25 & 1 T20) = {300 - (2)(35) - (2)(6) - (25 + 25 + 20)}/2 = 74mm If middle (20 mm) bar is curtailed, clear spacing= 168 mm.

The bottom spacing near mid span and the spacing when the middle bar is curtailed is less th

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Graded Example to EC2.pdf