design of reinforced concrete structures by krishna raju

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NEWAGE)

Reinforced Concrete Design "

N~ Krishna Raju

R.N. Pranesh

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Copyright 0 2003, New Age 'International (P) Ltd., Publishers P-'Jblished by New Age IntemationaJ (P) Ltd., Publishers First Edition: 2003 Reprint: 2008 All rights reserved. No part ofthis book may be reproduced in any [onn, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the copyright owner. ISllN : 81-224-1460-5 Rs.250.00 C-08-03-2345 5678910 Printed in lndia at Nagari Printers. Delhi.

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The book is dedicated 10 The pioneers and research workers, Isaac Johnson, Thaddeus Hyatt, Koenen, Coignet Whitney, Emperger, Jenson, Chambaud, Hognestad Baker, Evans, Neville, Fintel, Johanssen, Wood, Jones, Regean, Bresler, Park, Paulay, Gerwick, Murashev, Collins, Taylor, Newmark and a host of others W:10 loiled incessantly for the development and widespread use of Reinforced Concrete

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Preface

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The widespread use- of reinforce4 concrete in a variety of structural members in the construction industry has necessitated a proper understanding of the design and detailing procedures adopte,r by civil and structural engineers. The first Indian standard code of practice for plain and reinforced concrete was published in 1953 and revised in 1964 wiih major emphasis on working stress design. The third revision published in 1978 incolporated the Iimit'state design in conjunction with the working stress design. In recent years valuable infonnation regarding the various aspects of reinforced concrete such as ~urability. fire resistance. type of concrete. quaJrty assurance and limit state criteria. has been gathered and hence the recently revised fourth edition of the Indian standard code IS: 2000 incorporates the latest developments and design practices accumulated during the last three decades and also based on the various national codes such as the B.ritish co'de BS" 811 0, the American code ACI: 319, the German code DIN: 1045 the European concrete commiltee (CEB-FIP) model code of 1990, Canadian code CAN-A23.3 and the Australian code AS: 3800. This text book entitled "REINFORCED CONCRETE DESIGN (IS: 456-2000) is a modem comprehensive text meeting the requirements of undergraduate students of civil engineering and as a reference book for civil engineering teachers and practicing structural engineers. The material covered in the book comprises the first ·course forming the foundation for the theory and design of reinforced concrete structures and is class tested over several. years of teaching by the authors. The book is spread over seventeen chapters covering the fundamental topics in reinforced concrete design generally taught in the first course of B.E and B·T.ech (Civil engineering) curriculum in Indian unive~sities to be covered in a time frame of 50 hours. In the first Introductory chapter. the evolution of reinforced COncrete as a structural material for domestic industrial, highway, marine. environmental and storage structures is highlighted and prominent landmarks in the development of the material during the 20th century is examined in the light of continuous research by various investigators about the different properties of structural material. The second and third chapters present a comprehensive description of the various properties of concrete and types of reinforcement used in structural concrete along with the relevant Indian Standard code specifications. ..', i

th:

v:.,':"

t~t~:~

,·;,;)t!t~J~~r>,

x

Introduqtion

y

$ y.-J

x

iOOOIOODI \DDD\DDDI DOq4rDDO

'DDDDDD\

bodOODI 10 B-J 0100 01 DOD,DOD

V"om.."

Section-xx

Fig. 1.4 Flat Slab Floor System

Fig. 1.5 Grid Floor System

1,\'1"-' of roof is generally used for large conference halls and commercial hU\hlings I"cquiring column free space. The grid floor is supported at the \'d~~l'S on solid walls or columns at regular intervals.

\) l\tulllstorey Vertical Framing System

F\~ l.h shows the multistorey vertical frame-comprising columns beams nn\i s.labs forming three-dimensional structure. The gravity loads are transH\itll'd from slab to beams which in turn transfer the loads to columns and l\n..:\lly to the foundations. The rigid _column and beam frame can resist l.\\~'r;\llllads due to wind.

Column

Bam

CO U':'1" Fig. 1.6 Multistorey Vertical Framiilg System

9

10

g)

ImrodllctiOIl

Reinforced Concrete Design ~h'ear

Wall System

b) Design Codes and Handbooks

This system consists of solid concrete walls covering the full height of the building. Generally the shear wall box is located at the lift/staircase regions. Sometimes the shear walls are l-ceated as exterior or interior walls placed along the transverse direction of the tall building to resist lateral loads due to wind. A typical shear wall provided anhe core of a tall structure is shown in Fig 1.7. R.C.C. Wall

Floor slabs

Fig. 1.7 Shear Wall System

1.6 DESIGN CODES AND HAND BOOKS

,a) Objective of Codes Based on extensive research and practical knowledge, various countries haye evolved their national codes, which serve as guidelines for tlie design of structures. The main objectives of the codes are

I) 2) 3)

4)

To provide adequate structural safety by ensuring strength, service· ability and durability. ' To specify simple design procedures, design tables and formulae for' easy computations. To provide legal validity and to prolect structural engineers form any liability due to failures of structures caused by inadequate design and impr~permaterials and lack of proper supervision .during construction, To provide a uniform set of design gnide lines to be followed by various structural designers in the country.

National building codes are periodically sevised to reflect the improve· ments in the quality of materials and design procedures evolved as a result of comprehensive research investigations conducted in the various institution.s in the country ·and abroad.

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11

All reinforced concrete structural design in our country should conf~i'm to the recentlv re\'ised Indian Standard Code IS:456-2000 Code of practice for plain u"nd reinforced concrete (Fourth Revision). The corresponding national codes of other countries, which are often referred to, are the American Concrete Institute Code ACI-3I8 and the British Code BS: 8110. . The design examples presented in this book conform to the Indian standard code. The Bureau of Indian standards have released over.the years s~veral handbooks to facilitate reinforced concrete structural designers to d~sign routine structural elements quickly by referring to the various tables and graphs presented in the handbooks. The following handbooks will serve as useful design aids for structural concrete designers.

1) SP: 16-1980"· J:ksign Aids for Reinforced Concrete to IS: 456. 2} SP: 24-1983 ". Explanatory Handbook on IS: 456. , ., 3)', SP: 34-1987 15• Handbook on Concrete Reinforcement and DetailIng. 4):, SP: 23-198~" _ Hand book on Concrete Mixes (B ased on Indian Standards)

1.7', LOADI:"G STA:"DARDS - Reinforced concrete structureS are designed to resi~t the following types of loads: -

a) Dead Loads These are loads that will not change with respect to time. T~e dead loads acting on the structure include the self-weight of the structural. element.s. and finishes which depends upon the type of matenal used 111 Partitions• • 11 'b the structure. The Indian standard code IS: 875(Part-l) - 1987 prescrt es the unit weight of building materials and stored materials to be used in t~e design. Salient dead loads of most common materials used in structural elements are presented in Ta!?le l.l Table I I Dead Loads of Materials

i

,

l\.'.3tetial

Briel4.. Masonry Plain Coocrete

Rejnfor;:;:ed CCClCfete Stone Masonry

T:! gross weight. Slabs Beams

b) Li ve Loads These are loads that change with respect to time. Live or imposed loads include the loads due to people occupying the floor and those due to materials stored or vehicles in garage floors. The imposed floor and roof loads 111 for different occup,mcies are specified in IS 875 (Part-2) - 1987 • Some of the common live loads encountered in the design of buildings are compiled in Table 1.2

loading Class

Types of Floors

Minimum live load k'Nfm 2

2

Floors in dwelling hOuses, tenements, hospital wards, Bedrooms and private siUing rooms in hostels and dormitories.

2

Office Iloors other than entrance hall floors of light work rooms.

2.5·4.0

Floors of banking halls, office entrance halls and reading rooms

3.0

ROOFS Types of Roof

Live load in plan

4.0

ShOp floors used for display and sale of merchandise, floors of work rooms, floors 01 class rooms. restaurants, machinery halls power stations etc, where not occupied . by plant or equipment.

4.0

5.0

Floors of warehouses, workshops, lactories and other buildings or paris of building or similar calegory for light weighlloads, ollice floors for storage and filling purposes. Assembly floor space without fixed seating, public rooms in hotels, dance halls and wailing halls.

3.0

7.5

Staircases Stairs, landings and corridors lor class 2 but not liable to over crowding.

3.0

Flat, Sloping or Curved roof with slopes up to and including to degrees. a) Access provided. b) Access not provided, except lor maintenance. 2 Sloping rool with slope g'reater than 10°:' 0.75 kN/m c) less 0.001 kNfm 2 for every Increase in slope over2 10 degrees up to and Including 20° and 0.002 kN/m for every degree Increase in slope over 20°

,

4.0 2.5

Garages (Hea\!}') Floors used lor garages for vehIcles not exceeding 40 kN gross weight.

Balcony Balconies not liable to over-crowding' for class 2 loading loading for other classes Balconies liable to over crowdIng

Table 1.2 Live or Imposed Loads

2.5

13

"', , Introduction

12;, Rein!orced-C;om;rete Design

3.0 5.0 5.0

-

.

1.5 0.75

0

kNfm 2

c) Wind Loads Wind loads have to be considered in the design of multistorey buildings, towers and poles. Wind loads depend upon the intensity of wind prevailing in the locality of the structure. IS: 875(Part-3) _1987" prescribes basic wind speeds in various zones by dividing the country into 6 zones. The design wind pressure is computed as P x = O.6V:

Where 5.0

(eontd.)

p,,- :;::; design wind pressure in N/ml~2 at a height Z Sd or 450 mm whichever Steel . is smaller 3) Maximum -Diameter of Bars: ::} 1/8 0

4) Cover: { 20 mm nor < diameter of bar whichever is higher Fig.3.1 Reinforcement Speciflcatlolls in R.C. Slabs (IS: 456·2000)

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3.2 REINFORCEMENTS IN BEAMS Generally, beams are provided with main reinforcement on the tension side for flexure and transverse reinforcement for shear and torsion.

a) Tension Reinforcement The minimum area of tension reinforcement shall be not less than that given by the relation,

30

Reinforcement Specificatio~IS For Structural Concrete M('mlv rs

Reinforced Concrete Design

i-b--t

A, = (0.85 bdlf,l

r

Where As = Minimum area of tension reinforcement. b breadth of beam or breadth of web of flanged sections. d effective depth and 2 /y ;:; characteristic strength of reinforcement expressed in N/nuri • D nverall depth of the member

= = =

Effective depth=d

depth=D Stirrups

/

b) Cot:nprcssion Reinforcement The maximum area of compression. reinforcement shall not exceed' 0.04 bD. The compression reinforcement in beams shall be enclosed by stirrups fOr effective lateral restraint as shown in Fig. 3.2~

,--r-.+i.,--l-Tension reinforcement Ast over IS: 456-2000 Specifications

1I Minimum Reinforcement: Ast

0) Side Face Reinforcement

= (~)

or

Ast ~ 0·34 X for mild steel Cf y =250 N/mm )

When the depth of web or rib in a beam exceeds 750 mm, side face reinforcement of cross se,tional area not less than 0.1 percent of the web area is to be provided and distributed equally on two faces and the spacing of the bars not to exceed 300mm or web thickness whichever is smaller..

~ 0·20 X for HYS~ bars

(fy=415 N/mm )

, 2) Maximum Reinforcement: ~ 0'04 bD for both tension and

compression reInforcement i 3} Spacing Between 8ars~ 0( diameter of larger. bar nor lesS than the maximum size of coarse aggregate t 5 mm, wh,lchever l.s greater

d) Transverse or Shear Reinforcement Minimum' or nominal area of shear reinforcement provided in the form of stirrups is computed by the relation,

4) Cover: --t: 25 mm nor less than the 'diameter of bar 51 Curtailment, Refer clause 26·2'3 of IS: 456-2000

0.4 bS, Asv~ 0. 87 / y

Fig.3.2 ReJnforcement Specifications In

R.C. Beams (IS: 456·20011)

circular cross section. Columns are provided with main longitudinul r~iJ~~ forcements and lateral ties to prevent buckling of the main bars. The 11I1111mum and maximum limits of reinforcements, minimum numbcr of bnrs and their size, cover requirements and the diametcr and Spaciilg of Jalcnll ties are illustrated in Fig. 3.3. In R.C.Columns with helical ties, at least six main longitudinal reinforcements have to be provided within the helical .reinforcel)1ellt. The spacing of longitudinal bars measured along the periphery of thc column shall not exceed 300 mm. The pitch of helical reinforcement is Jimilcd 10 a maximum value of 75 mm and a minimum of 25 mm.. Helically reinforced columns have marginally higher load carrying capacity than those. with ordinary lateral ties due to hi,gher degree of confinement of concrete III the core.

=total cross sectional area of stirrup legs in shear =spacing of stin'ups along the length of the member =breadth of beam (or web in a flanged member)

h .;:; characteristic strength of stirrup reinforcement in N/mm 2 which shall not exceed 415 N/mm'.

The maximum spacing of shear reinforcement should not exceed 'O.75d' for vertical 'stirrups and 'd' for inclined stirrups at 450 where'd' is the effective depth. The maximum spacing is restricted to 300mm. The specifications of reinforcements in beams are illustrated in Fig. 3.2. 3.3 REINFORCEMENTS IN COLUMNS Reinforced concrete columns are generally of square; rectangular. or

Hanger bars

'Overall

The maximum area of tension reinforcement shall not exceed 0.04 bD

Where A sv S, b

)'1

I

"".",'>"

·liein!orced COflcr~te. pesign.

CHAPTER 4 Laterat

ties

RectanQular

Elastic Theory of Reinforced Concrete Sections in Flexure

Main reinforcement Cover

15:455-2000 Specifications

4.1 ELASTIC THEORY OF REINFORCED CONCRETE SECTIONS

1) Maximum Reinforcement: :t>o 6·0 X 2) Minimum Reinforcement: ~ 0·8 % 3) Minimum Number of Bars: 4 in rectangular and 6 in circular columns 4) Diameter of Bars:,~ 12 m m 5) Minimum Cover: 40 mm nor diameter of bar whichever is greater 6) Lateral Ties Diameten c4: 1/4 diameter of largest longitudinal/bar nor less than 5mm , Pitch

Than: a} Least laleral dimension of member b) 16 times the smaller diameter of longitudinal reil1forcement c) 48 times the diameter of tran sverse reinforcement d} For effective arrangement of lateral ties refer Fig. 6,9, 10 and 11 of

IS:· 455-2000 Fig. 3.3 Reinforcement Specifications in R.C. Columns (IS: 456·2000)

The working stress method of design' of reinforced concrete structures developed during the beginning of 20th century is based on the elastic theory of reinforced concrete sections. The working stress m'ethod is based on the 'assumptions that the structural materials behave in a linear elastic manner and the required safety is ensured by restricting the stresses in the materials under service or' working loads. The permissible stresses in con~ crete and steel are obtained by dividing the characteristic strength of the material by the fac.tor of safety to restrict the working stress in the material under service loads to be well within the linear elastic phase of the materials. 4.2 NEUTRAL AXIS DEPTH AND MOMENT OF RESISTANCE OF SECTIONS Consi,jer a rectangular sec.tion shown in Fig. 4.1 subjected toa moment 'M' under working loads. Let

O'cbc:::= D'SI::::

= =

ASI d = b n = k = m =

C :;:;: T:::: M::::

compressive stress devel9ped in Concrete, tensile stress developed in steel. area of tension reinforcement. effective depth. width of member. neutral axis depth. neutral axis depth factor. modular ratio. = (280/3 a",,) compressive force in concrete. tensile force in Steel. moment of Resistance of the section.

34

Reinforced Concrete Design

For any given section with known values of b, d, OSlO ache' and Asl ' we can evaluate the neutral a~is depth by equating the first moment of areas above and below the neutral axis. " O.Sh.1l 2 =III ,An.(d -11)

(2)

By solving Eq. (2), the value of 'n' and 'k' can be determined In equation (I) substituting, Q = 0.5

Asl

0,,,, k.j

We have M = Q.b:d'.

~

d = VMIQb

..

Fig. 4.1 Stress Distributlon in Rectangular Section

In the cracked section, concrete below the neutral axis is neglected in computations. Below the neutral axis, the steel area is converted into an equivalent area of concrete by multiplying the steel area by modular ratio and this area contributes to the tensile force for equilibrium of th~ section. From the stress distribution diagram shown in Fig. 4.1 we have the

Equation (3) is generally used to check the adequacy of the depth of section assumed to resist the~given moment M. The Moment of resistance of the section c~mputed from the tension side is gi ven by M

=A".n,,(d-i) =A,,' n,,(d - k:) =A".o".d (I-~) = (A;,.o".d.j)

A,,=(~) CJSI").d

relation,

n,,,,)·=(d-II (aJm II

)

(

=

kd )

(3)

(4)

E9uation (4) is generally used to compute the area of tension reinforcement in the section to resist the given moment.

d-kd

From the above relation we get Neutral axis depth factor

Further when the section is subjected to external loading, resisting moment is developed due to compression in concrete and tension in steel.

The neutral axis depth factor 'k' depends only on the permissible stresses in concrete and steel 'ache and OS! and modular ratio 'm', The value of 'k' can be evaluated by the following equations.

Moment of Resistance of the section is given by the relation. Frorn Fig. 4.1,

M = C (d -11/3) =0.5 n,,,,. b.lI(d -11/3) M =0.5 n,,,,. b .kd (d - k:) =0.5 n,,,,. b.k.d'(I- ~)

.,

•

The factor

Solving

(1 - ~)is termed as lever arm factor. and is represented by 'j'

Also

Hence, we have (I)

cbc

_(.l!!!....)

CJ (n.lm)d-kd

k [ 1'] . = 1+ (n.lmn,,,,),

(5)

( mn,,,,)=(~) 1 k O'~l

Substituting (m. 0,,,,) = (280/3) and Solving,

k-(- 280+3(n,,) 280 )

(6)

!lP,;"f.orc'PdCoricrete Design '.

Elastic Theory ofR;!inforced Concrete Sections in Flexure

Equation (5)or (6) can be ·used to evaluate 'k'.· In the analysis of reinforced concrete sections, it is often necessary to evaluate the neutral axis depth factor k using Equation (5) or (6). Equation (3) and (4) is generally used in the design of reinforced concrete sections. The values of the design coefficients k, j, and Q depend only on the permissible stresses ac~' a SI and the modular ratio m. The permissible stresses in steel and concrete according to IS: 456-2000 are shown in Table 4.1 and 4.2 respectively. The values shown

Medium Tensile bars

forcernElint.

Sleel

IS: 432

~.No.

m (ii)

(iii)

" (iv)

(2) Tension (crot or a. v)

I

bars

Bars

18:1766

IS: 432

Grade

(3)

a) Up 10 and inclUding 20mm b) Over 20mm

140 130

Compression in Column 8ars (osc)

130

Compression in bars in beam or slab when the compressIve resJstance of the concrete is laken Into account Compression In bars In a beam or slab where the compressive resistance of the concrete is not laken Into account. a) Up 10 and including 20mm b) Over 20mm

(4) . Half the guaranteed yield sfress subject 10 a maximum

Bending (crcbc )

(5)

DIrect (cr,;o) .

M·10

3.0

2.5

M-15

5.0

4.0

0.6

M·20

7.0

5.0

0.8

230 230

M-25

8.5

6.0

0.9

M·30

10.0

8.0

1.0

130

190

M·35

11.5

9.0

1.1

M·40

13.0

10.0

1.2

li

!i \

I Half the guaranteed yield stress subject to a maximum of190

190 190

Notes: For high yield strength deformed bars of Grade Fe-500, the permissible stress in direct tension and flexural tension shall be 0.55 J,. The permissible stresses for shear and compression reinforcement shall be as for Grade Fe-415. 2) For welded wire fabric conforming to IS:1566, the permissible value in tension is 230 N/mm2. 3) For the purposes of this standard, the yield stress of steels for which there is no clearly defined yield point should be taken to be 0.2 percent proof stress. I)

Permissible stress In bond (Avg.) for plaIn bars In tension, (0.,)

01190

The calculated compressive stress in the sur· rounding concrete mUllJplied by 1.5 limes the modular ratio or 0sc whichever is lower

140 130

Permissible stress In ~ompresslon

HY8D

Fe-415

(1)

Table 4.2 Permissible stresses in concrete (IS: 456·2000) (Tabte-2I orIS: 456-2000) .

Grade of cqncrete

Permissible stresses In N/mnf Mild

When mild steel conforming to Grade II ofIS: 432(Part-l) is used, thO"· permissible stresses in col.3, or if the design details have already been worked out on the basis of mild steel conforming to Grade I ofIS: 432 (Part-I), the area of reinforcement shall be increased by 10% of that required for Grade I steel.

All values in N/mm 2

Table 4.1 Permissible stresses in Steel Reinforcement (IS: 456·2000) (Table-22 of IS: 456-2000)

Type of Stress in Steel Rein-

4)

37

in Table 4.2 are obtained by applying a factor of safety of 3 to characteristic strength of concrete. Accordfngly the permissiole values of stresses in steel are obtained by applying a factor of safety of 1.78. In the design of reinforced concrete members, the most commonly used grades of concrete are M-20 and M-25. The revised Indian standard code IS: 456-2000 prescribes M-20 as the minimum grade of concrete for reinforced concrete while M-15 and M-IO may be used for plain concrete constructions. For Design office use, it is convenient to use the values of design coefficients 'j' and 'Q' to check the depth of the section and to compute the area of reinforcements required to resist the working moment 1M' using equations (3) and (4). The values of design coefficients are compiled in Table 4.3, for the most commonly used gradeS of concrete. 4.3 BALANCED, UNDER REINFORCED AND OVER REINFORCED SECTIONS

III reinforced concrete sections, the depth of neutral axis generally determines the type of section. The analysis of reinforced concrete sections

~

.

I ~;

I

38

Elastic Theory ofReinforced Concrete Sections in Flexure

Reinforced Concrele Design

39

Table 4.3 Design Coefficients

(N/mm

2

(N/mm 2) 140 230 280 140 ! 230 280 140 230 280

)

7

13.33

8.5

11

10

9.33

k

J

Q

OAOO 0.288 0:250

0.87 0.90 0.92

0,400 0.288 0.250

0.87 0.90 0.92 0.87 0.90 0.92

1.22 0.91 0.80 1.48 1.10 0.98 1.74 1.30 1.15

a.

M

ad<

,

Referring to the Fig. 4.3

OAOO 0.288 0.250

_ --,,/L-.:;Actual neutral -~xis

include the deiennination of 'critical neutral axis' which depends only on the pennissible stresses in concrete and steel and modular rat.io and the actual neutral axis, which is inf]uenged by the sectIOnal properlles and the

Fig. 4.3 Depth of Actual Neutral Axis

quantity or'reinforcement used in the section.

Let n. = actual neutral axis depth. By equating the first moment of areas above and below the neutral axis, we have

Referring to the Fig. 4.2 Let

b = width of section d =effective; depth. nc = critical neutral axis depth. As! = Area of tension reinforcement. 0Sl = Permissible Tensile stress in steel. O'cbc = Permissible Compre~sive stress in concrete. 1/! =modular ratio =(280/3",,,)

N

A

rr d

0.5 bli; = mAu (d -

Solving this quadratic equation. the actual neutral axis depth can be determined. Case-! Under reinfol"ced section If na < ne• the section is under reinforced. The moment of resistance is

f-O"cbc-f-

computeq from tension side with steel reaching the maximum permissible stress O'st and the moment of resistance is computed from Fig. 4.4.

.,1 Xu > Dr . the .moment of resistance may be calculated by the equations (6,8), When (D,Ix,) does not exceed 0.43 and when (D,Ix,) exceeds 0.43, the moment of resistance is computed by the equation (6.9) by substituting xu,max by Xu' 6.3.4 Computation of Tension Reinforcement in Tee beam Sections

Case..,! (x, < D,) In this case of Tee-Section the area of tension reinforcement can be computed by treating the section as rectangular and using the equation 6.7 and solving ASI for a given value of M u expressed as

.

[A".f,].

Mu = O.8?J;..A 5I ,d 1.- br.d.fck

Knowing the value of Xu.

x, =Depth of Nentral Axis. Dr = Depth of flange. Yr~Ax.+B Dr

u

Fig. 6.6. Equivalent Stress BI!lcks

6.3.3 Neutral Axis falls outside the Flange (Dr I d > 0.2)

+ 0.45J;k(b,- bw)y~d - 0.5 y,)

TT j ~" _

.(b) Whitney

Note: For Dr I x,> 0.43, Dr to be replaced by y,

M.

-t-0.45Ick-+

0'8x u

7x ... (6.8)

87

(6.10)

The constants A and B are solved by specifying the following two conditions to be satisfied by this equation. I) When Dr = 0.43 x,. Y, = 0.43 x, 2) When Dr =x, Yr =0.80 x, Substituting these conditions in Eq. (6.10) the constants A and B are eval-

xu' Referring to .stress block parameters shown in Fig. 6.5. force equilib-

uated as

~ium yields the following equations:-

For a given value of M u• evaluate Xu by using Eq. 6.8and replacing'xu•mu by

Ultimate Strength of Reinforced Concrete Seelio/ls T, ::: C 1

Case-~

(A"w·O.87 I,) :::: 0.36 fck bw'xu

..

AMW

_r0.36J;,h

-L

[,,>D,

and

89

(j»0.2]

Similar h.l ~sse (3)

w .Xw]

0.871,

6.3.5 l's,' of Design Aids (SP: 16) for Design of Flanged Beams T2 :::: C2

Also

(A",.0.87 1,) = 0.451:, (h, - hw) Dr AMf

=

[0.45J;,(hf -bw)Df ] 0.87 I,

Hence, the total tension reinforcement in the Tee Sections is given by Asl :::: [A.,w + A.lfl =

A"

Case.3

[0.361:,hw.Xw] +[0.45 10k (h, - bw)D,] . 0.871, 0.87 I,

[Xw>D,,(~)}0.2

and

(~»0.43]

In most C';\~ of tee beams used in buildings, the neutral axis falls within the flange- .mJ the computation of steel area can be made as in the design of rectangular t"ams nsing Tables I to 4 of SP: 16. lu the case of tee beams in which th~ neutral axis fnlls in the rib, design Tables provided in I.S. Special publk'llion SP: 24" (Explanatory Hand Book on IS: 456) are very llseful in \..'\Jlilputing the area of reinforcement for a given tee beam to resist a specitit."\t ~nding moment. SP: 16 dt:5igns Tables 57 to 59 are also useful to compute the limiting moment of resistance factor (Mu,lim I bw • d2 • fck) for singly reinforced Tee beams. 111< tables covcr different grades of steel (250,415 and 500) and ratios of to/,') varying (roll] 0.06 to 0.45 and ratios of (blb w ) varying from I to 10. l1lk',,", tables arc preseuted as Tables 6.8, 6.9 covering FcAI5 and 500 grade sh."els in the text.

For a given value of M u• evaluate Xu by using Eq.6.9 and replacing xu,max byReferring to stress block parameters shown in Fig. 6.6 in which the-,

6.3.6 An"I~"is Examples

depth of stress block-is

I)

Xu'

Yr = (0.15

X

w

+ 0.65 Dr) but not greater than Dr·

Force Equilibrium yields the following Equations:Tl

:::

-

C1

(A"w.O.871,) = 0.361:, bw'x" A. IW Also

Tz

= [0.361:khw.xw]

0. 871,

= C2

(A",.0.871,) = 0.45 1:~ (br - bw) y, A

Slf

= [0.45J;,(b,- hw)Y'] 0.87 fy

DClwuiu" the Ultimate flexural strength of the T-beam having the folk'i\\'ing section properties:. Width "I' tlange = 800 mm Depth "I' flange 150 mm Wid,a of rib = 300 mm Eft""l;''' depth 420 ml11 A,,,,ofsteel = 1470 mm'

= =

M-25 Gm

Dr

Hence the assumption that Xu < Dr is not correct. ,Neutral Axis falls outside the flange.

(iH~~~)

P,- [;;i -

Refer Table-3 of SP: 16 and read out. the value of (M,Ibd') corresponding

,};, = 20 N/mm' I, 415 N/mm' A" 4000 mm'

Assuming Neutral Axis to faU within the' flange, compute the depth of Neutral axis.

Method-2 (Using SP: 16 Design Aids) Percentage Reinforcement in the section (Assuming br = b for the rectangular section)

=

1300mm = 100mm = 325mm = 600mm.

b) Depth of Neutral Axis:-

= 206.52 kN.m

=0.166 80 mm.

~ (~.) = 0.80 > 0,43 ( ~) x, 10\.03 x

= 0.48d ~ (0,48 x 565) = 271.2 mm.

u, max .

.

.

M,

~ 0.36 (x,~",) [1 _Q.42 (x,;'')] /".b•.d' + 0.45fo,(b,- bw)y~d - O.sy,) ~ ~ (101.03) _ _ =0.18 (x) d 565·

M,

Assuming neutral axis to fall within the flange, the depth of neutral axis is computed. x.) (0.87 / ,11,,) ( 0.87x4t5x4000 ) ( d = O.36/".b r,d ~ 0.36 x 20 x900 x 650

~ 0.36 (0.18) [1-0.42 X 0.18] (15 X 250 X 565')

~ 0.34

x, = (0.36 x 650) = 221.mm > D,

Hence the assumption that Xu < Dr is not correct. The neutral axis falls outside the flange.

(i) =U~~)

and Y,= [O.l5x, +0. 65DJ

Yr= [(0.15 x 101.03) + (0.65 X 80)] = 67.15mm < Dr

~ 20 Nlmm'

b) Depth of Neutral Axis

'

Hence according to Clause- G.2.3 of I.S.456-2000 the moment of resista~ce of Tee-section is computed by replacing xu,max by Xu in Eq.(6.9).

f"

f, ~ 415 Nlmm'

=0.23>0.2

Neutral axis depth is determined by referring to Fig. 6.5 and by compati. bility of forces as shown below. [C,+CJ =[T,+TJ ~T

+ 0.45 X 15 (950-250) 67.15 (565-0.5 X 67.15) ~ (240.03 X 10')N.mm ~ 240.03 kN.m

C, = 0.36/,.b.""•. ~

..

.

'u.\ti

(0.36x20x300xx.) ~ 2160 x.

C, ~ 0,45/,,(b, ~ b.) Yf

Method 2 (using SP: 16 Design aids) In the present example, x, < X,.m" and hence~P: 16 Design tables 57,58,59 cannot be. used to compute the moment of reslsta.nce. 4) A singly reinforced T-beam has a flange width of 900mm, thickness of flange is 150mm width of rib ~ 300mm, Effective depth ~ 650mm. 2 Area of tensile reinforcement:;;: 4000 mm . M-20 grade concrete and Fe-415 HYSD bars are used. Estimate the ultimate flexural strength of the section using IS: 456-2000 code provisions.

Y,= 0.15x, +0.65 D, ~·._,u

~ 0.15x,

+ (0.65 x 150)

~(O.l5x,+97.5)

C, ~ [(0,45 x 20)(900 - 300)(0. 15x. + 97.5)] ~ [8 lOx, + 526500] T

~ (0.87f,.A,,) ~ (0.87 x415 x 4000) ~ 1444200 N (2160x,+8IOx,+526500)~ 1444200

Ultimate Srrnrgth ofReinfdrced Concrete Sections

101

Reinforced Ca,ncrete Design

100 ..

x") (O,87,r.,~' , _= ' ='( 0,87x415x5520 ,) =0307 0,J6/.,.bJ. 0.36x20 x 1200x750 . (d

xu =309mm>Dr Xu,

"'"~ = 0.48d = 312mm

Hence, according to Clause-G.2,3 of LS: 456-2000, the moment of

resistance of tee~section is computed by repladng XU,I113~ by Xu in Eq. 6.9 M" = 0.36 ( J) [1-0.42(J

)];;k,bw,d'+0.45;;k(b,-bw)Y~d - O.5y,)

(J)=(~~~)=0,475

and

y,=(0,15x"+0,65D,)

I! I.'I

(e, -C,)=(T,+ T,) = T

,

,i

i

: 111 ' C, = 0,45t,(b,-b w)Dr

,1'1

= 796:05 kN,m

r = 0.87fy,A" r = (0,87 x415 x 5520)

erties. Estimate the ultimate moment of resistanrze of the section using I.S code provisions. .

= 1992996 N [:tlM,= 1600kN.m c) Determination of Nentral Axis:-

For the known value of M u• compute moment equation. ,

f

..

(1600x 10') = 0 36 .

Xu

by repla.ei~g x u.max by x u.- in the

(~) [1- 0.42(~)] (20 x 300 x 750') 750 750·"

+ (0.45 x 20)(1500 - 300) [0.15x, + 130] x [750 ~0.5(0.15x, + 130)J Simplifying we have (x~ - 2500 x,..+ 620700) .-

~.

x,= 273 mmand ....

=O.

{~}(~~~)=0.7:l>0.43

016' b .~ ~hl; w

f) ]

0.871,

j

Reinforced concrete ~eams with compression reinforcement will be required in cases where the depth of the beam is restricted and the singly reinforced section is insufficient to resist the moment on the section.The .behaviour of R.C.C. beams 'with compression steel for ultimate load design '(,is sometimes referred ~o as. the Steel beam theory. The ~inal beam section with tension and compression steel is assumed to consist of two separate .beams consisting of a) A singly reinforced . section' which reaches its limiting value of moment of resistance expressed as M lI ,lim' A steel beam without any concrete but reinforced with tension and compression steel. The moment of resistance of the doubly reinforced section will be the ~um of the ,moment of resistance of the two different sections specified in

(a) and (b).

.

'-

."

·6:4.2 Design Equations

\SQnSider t4e doubly reinforced concrete section split -,into two parts as .shown in Fig. 6.7. . . Mill :::;

.x.] + [0.45/o,(b,-b.)(0.15X,+0.65D .. u

I

REINFORCED CONCRETE SECTIONS

MlI :::; moment of resistance of the dou~ly reinforced section. MII,li~::::: the limiting or the maximum moment capacity of the , M 1I2

" [ 0.87 x415

.:

i r I~

6.4 ULTIMATE FLEXURAL STRENGTH OF DOUBLY

J4et d) Area of tensile reinforcement

A =

ii'

':

r

6.4.1 Design principles

= 184mm x"Um = 230 mm

Hence, the section is over r~inforced e) Moment of resistance

L'.M

M. = 0.36fo. x"I'm b (d - 0,42 x",um) +!., A,; (d - d') = [(0.36 x 30 x. 230 x 280)(500 - 0042 x 230)+ 410 x 400 (500 - 50)]10- = 354.37 kN,m

Method-2 (Using SP: 16 Design Tables) a) Compute parameters for use of design tables (d' I d) = ( 50 1500 ) = 0.1 p, = (100 x 400) I (2802< 500) = 0,285 p, = ( 100 x 2450) I (280 x 500 ) = 1.75

b) Refer Table-56 of SP: 16 forf" = 30 N/mm' andf, = 500 N/mm'

Method-1 (Using IS: 456·2000 Code Formulae)

"

I: a) Data b =280mm d = 500 mm d' =50mm A" =2450 mm'

[0.0035(x•.Hm - d')/x•.,,") = 0,0035(230 - 50)/230 = 0.00274

,, ,, ,

10- = 443 kN,m

The moment of resistance computed is nearly equal to that computed

2)

E" =

,,,

A", = if., A,,)/(0.87 f,) = (410 X 400)/(0.87 x 500) = 377 mm' = (A" - A,,,) = (2450 - 377) = 2073 mm' x, (0,87 f, A,,,)/(0.36fo, b)

a) Compute parameters for use of design tables,

a)

x. = X. Jlm ~ 0.46d = (0.46 x 500) = 230 mm

A",

Method.3 (Using SP: 16 Design Tables)

. p, ~ (100A,,)/(bd)

123

fo,

f,

=30 N/mm'

= 500 N/mm' E, = 2 X 10' N/mm' Asc ::: 400 rnm2

For ( = (1.815 x 60) = 108.9 kN.

c) Total Shear Resistance b) Percentage Reinforcement

p,

v. = (Vo< + V,,) = (86.4 + 108.9) = t95.3 kN.

= (100A,,) = (100 x 942) = 0.52 bd 300x600

2)

Refer Table-19 of IS: 456 (Table 6,17 of text) and read out the design shear strength of concrete 'tc corresponding to fck ;; 20 N/mm2, ~, = 0.48 N/mm' c) Shear Resisted by Concrete

I

II Ii I II

IiIi

Ii II

1I j[

r .1 'I

n I, II I

f:

j!,

!

1

v,,, = (~, b d) =

(0.48 x 300 x 600) 10-3 = 86.4 leN

d) Shear resisted by Stirrups:-

V.,

=

i

A reinforced concrete beam of rectangular section has a width of 250mm and an effective depth of 500mm. The beam is reinforced with 4 bars of 25mm diameter on the tension side. Two of the tension bars are bent up at 45° near the support section. In addition the beam is providec\ with two legged stirrups of 8mm diameter at 150mm centres near the supports. If};, = 25 N/mm' andfy = 415 N/mm', estimate the ultimate sheaf strength of the su~port section.

il

Method-l (Using IS: 456·2000 Code Formula)

J

a) Data

[A,,(0.87S, 1,) d] = [100 xO.87200x415 x 600J 10- = 108 .3 kN . 3

e) Total shear resistance of support section:·

V. = [V., + V"J = [86.4 + 108.3] = 194.7 kN

I

' = 25 N/mm' =. 250 mm J