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ABSTRACT

Many analytical softwares do not support design of columns subjected to tension or compression, biaxial bending, shear in both directions and torsion. Tension in columnsoccurs especially in columns of buildings subjected to horizontal loads due to wind or earthquake forces. Hence, a simple method is de eloped using M! excel sheet, fromthe existing "ndian codal procedures. #esults are compared with existing literature andcomparison graphs show that results are matching with the existing literature. Hence,these excel sheets can be used for the design of columns. "mportance of this procedurelies in the fact that modification or alteratation can be made at any step and can beused for a ast number of columns in a short time during the design.

!preadsheets pro ide an easy to use and adaptable alternati e to the traditional charts

and tables used as design aids for reinforced concrete.

$eywords %

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CHAPTER NO. 1

INTRODUCTION

&olumns contain ery low olume of materials compared to the roof it supports.Howe er, extreme care is required to construct columns since failure of columns leadto sudden collapse of the structure, when compared to the failure of the roofingelements it supports. Hence, columns need to design in such a way that failure of

beams occurs prior to failure of columns. "n order to achie e this objecti e, themoment strength of beams or slab shall be less than that of colums at a joint in a

principal plane. This fact shall be checked before the completion of analysis and if found not satisfying, column dimensions shall be increased to satisfy the abo econdition. 'esign of columns are carried o er a group of columns ha ing a low

ariation of design alues. "n this group, design is carried for the highest alues of axial force, moments about ( and ) directions. This practice will reduce safety factor in columns which are critical structural members responsible for the safety of thestructure. Hence, a method is proposed from existing literature for the design of reinforced concrete *#&+ columns considering axial force, moment in ( and ). Thismethod is presented in the excel sheet so that the calculations in each step can be

erified. The results are compared with the literature and erified.

SIGNIFICANCE

&olumn is a critical element in a structure which need to ha e a high moment strengththan beams or slab in the principal plane considered. procedure is formulated usingspreadsheets for the design of columns. !preadsheet facilitates design of number of colums immediately upon entering design data alues. These are significant sincemany analytical softwares do not support design of columns with axial force,moments, shear and torsion.This imposes limitation of using these charts. !- / putslimitation of minimum of 0 bars *for bars places on all faces+ for the charts. Howe er,

the spreadsheet de eloped has no such limitations, since it is de eloped from first principles, for application to rectangular columns. ll the indi idual concepts for design of this problem are a ailable in "! 12/ % 3444 and !- % / 504 . howe er, allthese indi idual concepts are systematically combined in this programme, which takescare of all the rele ant clauses concerning this problem.

!predsheets pro ide an extremely useful,modernalternati e to the traditional chartsand tables.They ha e ad antage which often make them more usableand more

ersatile in many applications.This paper describes spreadsheetwhich is de eloped for design of short rectangular columns.

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METHODOLOGY

!tep by step procedure for the design is as follows. . 6nter axial force, moment about) for top and bottom of column, moment about ( for top and bottom of column.6nter also the breadth of the column and the length of the column. 6nter the grade ofsteel and concrete. This is based on the strains at each row of steel and thecorresponding stresses in steel and the compressi e force in concrete. !tress in steel isfound from the stress strain relations gi en in !-% /. These are gi en for a barconfiguration of four bars per face which is applicable to 7 bars per face and 3 bars per face. This can be extended to any number of bars per face, pro ided, the excel sheet ismodified by simply copying the existing one and changing the bar number from the

present to the required one.The moments shall be compared with the moment due to minimum eccentricity as per

cl. 75.3 of "! 12/ % 3444 and maximum shall be considered. Minimum eccentricitycorresponds to cl. 32.1 of "! 12/ % 3444, which is equal to unsupported length ofcolumn8244 plus lateral dimensions874, subject to minimum of 34 mm. This needs to

be checked about each axis at a time.

VERIFICATION OF SPREADSHEET

"t is required to make sure that the results should match alues as in code. "n case of spreadsheets for concrete design, the results of specific calculations can be compared

with hand calculations. 9urther checks can be made by reworking examples instandard text books, although these do contain errors. Hence, the comparison of resultsobtained from excel sheet is to be made with the results from literature to make surethat each constructs are correct.

&omparison ofsteel stresses fromstrains, for ariousstress le els *fyd :fy8 . 2+, for 9e1 2

!train

fs as per excel*fs as per !- / %

504+

4.44704 fyd *fyd+4.443;/ 4.5;1fyd*4.5;2fyd+4.4431 4.515fyd *4.52fyd+4.44 53 4.5fyd *4.4.54fyd+4.44 /7 4.015fyd

*4.4.02fyd+4.44 11 4.0fyd *4.0fyd+

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CONCLUSIONS

&omputer spreadsheets pro ide a ery con enient aid for the design of column. "t hasgot significant ad antages o er design charts and tables. 9rom the abo e sections, it isclear that, the spreadsheet de eloped for the design of columns subjected tocompression8tension, biaxial bending, can be used effecti ely. !preadsheets also ha ean important ad antage o er con entional computer programs in that they gi e a goodinsight into effect of arious design parameters. The result of any change in the trialdesign is instantly obser ed because of automatic recalculation by the spreadsheet.This encourages the designer to adjust the input data to achie e better design solutions.&hanges can be undertaken ery rapidly and efficiently.

LIMITATIONS

t present, column with four bars per face is considered, which can be increased toany number of bars per face by a little modification in the spreadsheet. This isapplicable for rectangular columns only. 9or circular columns and odd shapedcolumns, this spreadsheet can be modified and could be adopted. ll notationscorresponds to ref A

Column

&ompression members are structural elements primarily subjected to axialcompressi e forces and hence, their design is guided by considerations of strength and

buckling. 9igures a to c show their examples% pedestal, column, wall and strut. Bhile pedestal, column and wall carry the loads along its length l in ertical direction, thestrut in truss carries loads in any direction. The letters l , b and D represent theunsupported ertical length, horizontal lest lateral dimension, width and the horizontallonger lateral dimension, depth. These compression members may be made of bricksor reinforced concrete.

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9or a compression member, the effecti e length C three times the least lateraldimension.

Short n! Sl"n!"r Com#r"$$%on M"m&"r$

Bhen both slenderness ratios lex8' and ley8b are D 3

E &olumn is a short column

E "f more than 3, then it is long or slender column.

E !lender &olumns are designed for dditional Moments as per &lause 75.; of "!12/

6ffecti e height of column%F

E 9or effecti e column height refer table 30 * nnexure 6+ of "!%12/F3444.

D"$%'n O( Column$ ) Im#ort nt Con$%!"r t%on$

*ii+ @nsupported ?ength G

"n beamFslab construction, it is the clear distance between the floor under side of shallower beam framing into columns in each direction at next higher floor le el.

*iii+ !lenderness limits for columns G

The unsupported length between end restraints shall not exceed /4 times the leastlateral dimension of a column.

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Fig.1 Pedestal, column,

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*i + Minimum 6ccentricity G ll columns shall be designed for "m%n * l+,-- D+/- *0- mm

Bhere l: @nsupported length of column in mm. ': ?ateral dimension of column inthe direction under consideration in mm.

.3 &lassification of &olumns Iased on Types of #einforcement

Iased on the types of reinforcement, the reinforced concrete columns are classifiedinto three groups

*i+ T%"! olumn$ % The main longitudinal reinforcement bars are enclosed withinclosely spaced lateral ties.

*ii+Column$ 2%th h"l% l r"%n(or "m"nt % The main longitudinal reinforcement bars

are enclosed within closely spaced and continuously wound spiral reinforcement.&ircular and octagonal columns are mostly of this type .

*iii+ Com#o$%t" olumn$ % The main longitudinal reinforcement of the compositecolumns consists of structural steel sections or pipes with or without longitudinal bars .

>ut of the three types of columns, the tied columns are mostly common with differentshapes of the crossFsections iz. square, rectangular, TF, ?F, cross etc. Helically boundcolumns are also used for circular or octagonal shapes of crossFsections. rchitects

prefer circular columns in some specific situations for the functional requirement. Thismodule, accordingly takes up these two types *tied and helically bound+ of reinforcedconcrete columns.

.7 &lassification of &olumns Iased on ?oadings

&olumns are classified into the three following types based on the loadings%

*i+ &olumns subjected to axial loads only *concentric+, as shown in 9ig.3a.

*ii+ &olumns subjected to combined axial load and uniaxial bending, as shown in9ig.3b.

*iii+ &olumns subjected to combined axial load and biFaxial bending, as shown in9ig.3c

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.1 &lassification of &olumns Iased on !lenderness #atios

&olumns are classified into the following two types based on the slenderness ratios%

*i+ !hort columns

*ii+ !lender or long columns

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9igure 7 presents the three modes of failure of columns with different slendernessratios when loaded axially. "n the mode , column does not undergo any lateraldeformation and collapses due to material failure. This is known as compressionfailure. 'ue to the combined effects of axial load and moment a short column mayha e material failure of mode 3. >n the other hand, a slender column subjected toaxial load only undergoes deflection due to beamFcolumn effect and may ha e materialfailure under the combined action of direct load and bending moment. !uch failure is

called combined compression and bending failure of mode 3. Mode 7 failure is byelastic instability of ery long column e en under small load much before the materialreaches the yield stresses. This type of failure is known as elastic buckling.

The slenderness ratio of steel column is the ratio of its effecti e length l e to its leastradius of gyration r . "n case of reinforced concrete column, howe er, "! 12/ stipulatesthe slenderness ratio as the ratio of its effecti e length l e to its least lateraldimension.The column may ha e the possibility of buckling depending on the two

alues of slenderness ratios as gi en below%

!lenderness ratio about the major axis : l ex /D

!lenderness ratio about the minor axis : l ey /b

Thus, cl. 32. .3 of "! 12/ stipulates the following%

compression member may be considered as short when both the slenderness ratiosl ex /D and l ey /b are less than 3 where l ex = effecti e length in respect of the major axis,

D : depth in respect of the major axis, l ey : effecti e length in respect of the minor axis, and b : width of the member. "t shall otherwise be considered as a slender compression member.

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9urther, it is essential to a oid the mode 7 type of failure of columns so that allcolumns should ha e material failure *modes and 3+ only. ccordingly, cl. 32.7. of "! 12/ stipulates the maximum unsupported length between two restraints of a columnto sixty times its least lateral dimension.

D"$%'n O( Column$ ) D"$%'n A##ro h

E The design of column is complex as it is subjected to axial loads moments whichmay ery independently.

Column !"$%'n r"3u%r"$ )

GD"t"rm%n t%on o( th" ro$$ $" t%on l !%m"n$%on.

GTh" r" o( lon'%tu!%n l $t""l 4 %t$ !%$tr%&ut%on.

GTr n$5"r$" $t""l.

E The maximum axial load moments acting along the length of column areconsidered for design of the column section.

E The trans erse reinforcement is pro ided to impart effecti e lateral support against buckling to e ery longitudinal bar.

D"$%'n O( Column$ ) R"%n(or "m"nt Pro5%$%on$ $ #"r IS67,89

A. Lon'%tu!%n l r"%n(or "m"nt

E rea of longitudinal reinforcement shall be not l"$$ th n -.:; nor mor" th n 8;o( ro$$ $" t%on l r" of the column.

E Howe er maximum area of steel should not exceed 1J to a oid practical difficultiesin placing compacting concrete.

E "n pedestals, in which the longitudinal reinf. is not taken into account in strengthcalculations, nominal reinforcement should be not be less than 4. 2J of crosssectional area.

E Minimum dia of longitudinal bar should be 3 mm

D"$%'n O( Column$ ) R"%n(or "m"nt Pro5%$%on$ $ #"r IS67,8

A. Lon'%tu!%n l r"%n(or "m"nt

E !pacing between bars D 744mm along periphery of column

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E The minimum number of bars shall be four in rectangular columns six in circular columns.

*I+Trans erse #einforcement %

Trans erse reinforcing bars are pro ided in forms of circular rings, polygonal links*lateral ties+ with internal angles not exceeding 72 o or helical reinforcement. Thetrans erse reinforcing bars are pro ided to ensure that e ery longitudinal bar nearest tothe compression face has effecti e lateral support against buckling. &lause 3/.2.7.3stipulates the guidelines of the arrangement of trans erse reinforcement. The salient

points are%

*a+ Trans erse reinforcement shall only go round corner and alternate bars if thelongitudinal bars are not spaced more than ;2 mm on either side *9ig.1+.

*b+ ?ongitudinal bars spaced at a maximum distance of 10 times the diameter of the tieshall be tied by single tie and additional open ties for in between longitudinal bars*9ig.1*b++.

*c+ 9or longitudinal bars placed in more than one row *9ig.1*c+% *i+ trans ersereinforcement is pro ided for the outerFmost row in accordance with *a+ abo e, and *ii+

no bar of the inner row is closer to the nearest compression face than three times thediameter of the largest bar in the inner row.

*d+ 9or longitudinal bars arranged in a group such that they are not in contact and eachgroup is adequately tied as per *a+, *b+ or *c+ abo e, as appropriate, the trans ersereinforcement for the compression member as a whole may be pro ided assuming thateach group is a single longitudinal bar for determining the pitch and diameter of thetrans erse reinforcement as gi en in sec. 4.3 .5. The diameter of such trans ersereinforcement should not, howe er, exceed 34 mm *9ig.1*d++.

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-itch and 'iameter of ?ateral Ties

*a+ -itch% The maximum pitch of trans erse reinforcement shall be the least of thefollowing%

*i+ the least lateral dimension of the compression membersK

*ii+ sixteen times the smallest diameter of the longitudinal reinforcement bar to be tiedKand

*iii+ 744 mm.

*b+ 'iameter% The diameter of the polygonal links or lateral ties shall be not less thanoneFfourth of the diameter of the largest longitudinal bar, and in no case less than /mm.

*c+ Helical #einforcement

*a+ -itch% Helical reinforcement shall be of regular formation with the turns of thehelix spaced e enly and its ends shall be anchored properly by pro iding one and a

half extra turns

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of the spiral bar. The pitch of helical reinforcement shall be determined as gi en insec. 4.3 .5 for all cases except where an increased load on the column is allowed for on the strength of the helical reinforcement. "n such cases only, the maximum pitchshall be the lesser of ;2 mm and oneFsixth of the core diameter of the column, and theminimum pitch shall be the lesser of 32 mm and three times the diameter of the steel

bar forming the helix.

ssumptions in the 'esign of &ompression Members by ?imit!tate of &ollapse

"t is thus seen that reinforced concrete columns ha e different classificationsdepending on the types of reinforcement, loadings and slenderness ratios. 'etaileddesigns of all the different classes are beyond the scope here. Tied and helicallyreinforced short and slender columns subjected to axial loadings with or without the

combined effects of uniaxial or biaxial bending will be taken up. Howe er, the basicassumptions of the design of any of the columns under different classifications are thesame

*i+ The maximum compressi e strain in concrete in axial compression is taken as4.443.

*ii+ The maximum compressi e strain at the highly compressed extreme fibre inconcrete subjected to axial compression and bending and when there is no tension onthe section shall be 4.4472 minus 4.;2 times the strain at the least compressed extreme

fibre

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9rom the two similar triangles 6L" and H", we ha e

6L8 H : 4.44 284.4472 : 78;, which gi es

6L : 7 D/ ; * 4.3+

The point L, where the two profiles intersect is assumed to act as a fulcrum for thestrain profiles when the neutral axis lies outside the section. nother strain profile N$ drawn on this figure passing through the fulcrum L and whose neutral axis is outsidethe section. The maximum compressi e strain N of this profile is related to theminimum compressi e strain H$ as explained below.

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N : " G "N : " G 4.;2 H$, as we can write "N in term of H$ from two similar triangles NL"

and HL$%

"N8H$ : L68L9 : 4.;2.

The alue of the maximum compressi e strain N for the profile N$ is, therefore,4.4472 minus 4.;2 times the strain H$ on the least compressed edge. This is theassumption *ii+ of this section *cl. 75. b of "! 12/+.

Minimum 6ccentricity

"n practical construction, columns are rarely truly concentric. 6 en a theoreticalcolumn loaded axially will ha e accidental eccentricity due to inaccuracy in

construction or ariation of materials etc. ccordingly, all axially loaded columnsshould be designed considering the minimum eccentricity as stipulated in cl. 32.1 of "!12/ and gi en below

e x min Ogreater of + l/ 244 P D/ 74+ or 34 mm QQ* 4.7+

e y min Ogreater of + l/ 244 P b/ 74+ or 34 mm

where l, D and b are the unsupported length, larger lateral dimension and least lateraldimension, respecti ely.

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&H -T6# 3

!H>#T (" ??) ?> '6' &>?@MA!

xially loaded columns need to be designed keeping the pro ision of resisting somemoments which normally is the situation in most of the practical columns. This isensured by checking the minimum eccentricity of loads applied on these columns asstipulated in "! 12/. Moreo er, the design strengths of concrete and steel are further reduced in the design of such columns. The design can be done by employing thederi ed equation i.e., by direct computation or by using the charts of !-F /.

3. ssumptions #egarding the !trengths of &oncrete and !teel

The stress block of compressi e part of concrete is used in the design of beam by limit

state of collapse. The maximum design strength of concrete is shown as constant at4.11/ f ck when the strain ranges from 4.443 to 4.4472. The maximum design stress of steel is 4.0; f y.

e x min greater of * l/ 244 P D/ 74+ or 34 mm

e y min greater of * l/ 244 P b/ 74+ or 34 mm

The maximum alues of l ex /D and l ey /b should not exceed 3 in a short column.9urther,it is necessary to keep pro ision so that the short columns can resist theaccidental moments due to the allowable minimum eccentricity by lowering the designstrength of concrete by ten per cent from the alue of 4.11/ f ck , used for the design of flexural members. Thus, the design strength of concrete in the design of short columnis *4.5+*4.11/ f ck + : 4.14 1 f ck , say 4.14 f ck .

3.3 o erning 6quation for !hort xially ?oaded Tied &olumns

9actored concentric load applied on short tied columns is resisted by concrete of area Ac and longitudinal steel of areas A sc effecti ely held by lateral ties atinter als. ssuming the design strengths of concrete and steel are 4.1 f ck and 4./; f y,

respecti ely, P u : 4.1 f ck Ac P 4./; f y A sc

where P u : factored axial load on the member,

f ck : characteristic compressi e strength of the concrete,

Ac : area of concrete,

f y : characteristic strength of the compression reinforcement, and

A sc : area of longitudinal reinforcement for columns

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&H -T6# 7

@A" (" ? &>?@MA

"n general, all practical columns are eccentrically loaded. Aormally, the sidecolumns of a grid of beams and columns are subjected to axial load P and uniaxialmoment M x causing bending about the major axis xx, hereafter will be written as M .

These can be isualized by either of the following ways%

+ The load - u on column is at an eccentricity e from the centre line of column,

and then column is subjected to an axial load - u plus the bending momentequal to - ue.

3+ The load is axial but a column has a moment either due to gra ity loads or anyother loads. "t can be said in this case that column is eccentrically loaded withload - u and bending moment M u or the column is eccentricity loaded with aload - u at an eccentricity e: M u 8 - u.

@niaxial bending%'epending on relati e alues of - u and M u, the following two cases occur for design%

i. &ompression o er the whole section where the neutral axis is outsidethe section. This is called compression controls region. This region isfurther subdi ided into two regions*a+ e R e,min*b+ e C e,min

ii. &ompression on one side in the concrete and steel, and tension in thesteel on the other side where the neutral axis lies inside the section.This is called tension controls region.

7. Modes of 9ailure of &olumns

The two distinct categories of the location of neutral axis, clearly indicate the twotypes of failure modes% *i+ compression failure, when the neutral axis is outside thesection, causing compression throughout the section, and *ii+ tension failure, when theneutral axis is within the section de eloping tensile strain on the left of the neutralaxis. Iefore taking up these two failure modes, let us discuss about the third mode of failure i.e., the balanced failure.

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7. . Ialanced failure

@nder this mode of failure, yielding of outer most row of longitudinal steel near theleft edge occurs simultaneously with the attainment of maximum compressi e strain of 4.4472 in concrete at the right edge of the column. s a result, yielding of longitudinalsteel at the outermost row near the left edge and crushing of concrete at the right edgeoccur simultaneously. The different yielding strains of steel are determined from thefollowing%

*i+ 9or mild steel *9e 324+% yS : 4.0; f y /E s . *7. +

*ii+ 9or cold worked deformed bars% yS : 4.0; f y /E s P 4.443 QQQQQ. *7.3+

7. .3 &ompression failure

&ompression failure of the column occurs when the eccentricity of the load P u is lessthan that of balanced eccentricity * e D eb+ and the depth of the neutral axis is more thanthat of balanced failure. Aeither the tensile strain of the outermost row of steel on theleft of the neutral axis reaches yS.

7. .7 Tension failure

Tension failure occurs when the eccentricity of the load is greater than the balancedeccentricity eb. The depth of the neutral axis is less than that of the balanced failure.The longitudinal steel in the outermost row on the left of the neutral axis yields first.

radually, with the increase of tensile strain, longitudinal steel of inner rows, if pro ided, starts yielding till the compressi e strain reach 4.4472 at the right edge. Thedepth of the neutral axis is designated by * k min D+.

7.3 "nteraction 'iagram

"t is now understood that a reinforced concrete column with specified amount of longitudinal steel has different carrying capacities of a pair of P u and M u before itscollapse depending on the eccentricity of the load. 9igure 7. represents one suchinteraction diagram gi ing the carrying capacities ranging from P o with zero

eccentricity on the ertical axis to M o *pure bending+ on the horizontal axis. Theertical axis corresponds to load with zero eccentricity while the horizontal axisrepresents infinite alue of eccentricity. radial line joining the origin > of 9ig.7. to

point 3 represents the load ha ing the minimum eccentricity. "n fact, any radial linerepresents a particular eccentricity of the load. ny point on the interaction diagramgi es a unique pair of P u and M u that causes the state of incipient failure. Theinteraction diagram has three distinct zones of failure% *i+ from point to just before

point 2 is the zone of compression failure, *ii+ point 2 is the balanced failure and *iii+from point 2 to point / is the zone of tension failure. "n the compression failure zone,small eccentricities produce failure of concrete in compression, while largeeccentricities cause failure triggered by yielding of tension steel. "n between, point 2 is

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the critical point at which both the failures of concrete in compression and steel inyielding occur simultaneously.

The interaction diagram further re eals that as the axial force P u becomes larger thesection can carry smaller M u before failing in the compression zone. The re erse is thecase in the tension zone, where the moment carrying capacity M u increases with theincrease of axial load P u. "n the compression failure zone, the failure occurs due too er straining of concrete. The large axial force produces high compressi e strain of concrete keeping smaller margin a ailable for additional compressi e strain line to

bending. >n the other hand, in the tension failure zone, yielding of steel initiatesfailure. This tensile yield stress reduces with the additional compressi e stress due toadditional axial load. s a result, further moment can be applied till the combinedstress of steel due to axial force and increased moment reaches the yield strength.

?et us understand the corresponding compressi e stress blocks of concrete for the twodistinct cases of the depth of the neutral axis% *i+ outside the crossFsection and *ii+within the crossFsection in the following sections.

7.3. &ompressi e !tress Ilock of &oncrete when the Aeutral xis ?ies>utside the !ection

9igure 7.3c presents the stress block for a typical strain profile N$ ha ing neutral axisdepth kD outside the section * k C +. The strain profile N$ in 9ig.7.3b shows that up toa distance of 7 D8; from the right edge *point >+, the compressi e strain is 4.443 and,therefore, the compressi e stress shall remain constant at 4.11/ f Ock . The remaining part

18

9ig.7. Typical interaction diagram

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of the column section of length 1 D8;, i.e., up to the left edge, has reducingcompressi e strains *but not zero+. The stress block is,

therefore, parabolic from > to H which becomes zero at @ *outside the section+. Thearea of the compressi e stress block shall be obtained subtracting the parabolic area

between > to H from the rectangular area between and H. To establish theexpression of this area, it is essential to know the equation of the parabola between >and @, whose origin is at >. The positi e coordinates of X and Y are measured fromthe point > upwards and to the left, respecti ely. ?et us assume that the generalequation of the parabola as

X : aY 3 P bY P c .*7.7+

19

Fig.3.2 Cross section o column, strain !ro"les and stress #loc$or t%e strain !ro"le &' ($) * )+

Fig.3.2(c+ tress #loc$ or t%e strain

Fig.3.2 (#+ train

Fig. 3.2(a+ Cross

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The alues of a, b and c are obtained as follows%

*i+ t Y : 4, X : 4, at the origin% gi es c : 4

*ii+ t Y : 4, dX/dY : 4, at the origin% gi es b : 4

*iii+ t Y : * kD G 7 D8;+, i.e., at point @, X : 4.11/ f ck % gi esa : 4.11/ f ck /D 3*k F78;+3.

Therefore, the equation of the parabola is%

X : 4.11/ f ck /D 3*k G 78;+3UY 3 QQQQQQQ. *7.1+

The alue of X at the point H *left edge of the column+, g is now determined from6q.7.1 when Y : 1 D8;, which gi es

g : 4.11/ f ck 18*;k G 7+U3 QQQQQQQ. *7.2+

Hence, the area of the compressi e stress block : 4.11/ f ck D < G *183 + 18*;k G 7+U3=

: f ck D *7./+

where : 4.11/< G *183 + 18*;k G 7+U3=QQQQQQQQ *7.;+

6quation 7./ is useful to determine the area of the stress block for any alue of k C *neutral axis outside the section+ by substituting the alue of from 6q.7.;. Thesymbol is designated as the coefficient for the area of the stress block.

The position of the centroid of the compressi e stress block is obtained by di iding themoment of the stress block about the right edge by the area of the stress block. Themoment of the stress block is obtained by subtracting the moment of the parabolic part

between > and H about the right edge from the moment of the rectangular stress block of full depth D about the right edge. The expression of the moment of the stress block about the right edge is%

4.11/ f ck D* D83+ G * 87+*1 D8;+ 4.11/ f ck 18*;k G 7+U3 7 D8; P *781+*1 D8;+U

: 4.11/ f ck D3

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Table 7. represents the alues of and 3 for different alues of k greater than , asgi en in Table H of !-F /. 9or a specific depth of the neutral axis, k is known. @singthe corresponding alues of and 3 from Table 4.1, area of the stress block of concrete and the distance of centroid from the right edge are determined from 6qs.7./and 7.5, respecti ely.

T &l" /.1 Str"$$ &lo < # r m"t"r$ C 1 n! C 0 2h"n th" n"utr l =%$ %$ out$%!" th"$" t%on

The area of the stress block is 4.11/ f ck D and the distance of the centroid from the rightedge is 4.2 D, when k is infinite. Lalues of and 3 at k : 1 are ery close to thosewhen k : V. "n fact, for the practical interaction diagrams, it is generally adequate toconsider alues of k up to about .3.

7.3.3 'etermination of &ompressi e !tress nywhere in the !ection when

the Aeutral xis ?ies outside the !ection

The compressi e stress of concrete at any point between and > of 9ig.7.3c isconstant at 4.11/ f ck as the strain in this zone is equal to or greater than 4.443. !o, wecan write

f c : 4.11/ f ck if 4.443 RR !c 4.4472QQ.. *7. +

Howe er, compressi e stress of concrete between > and H is to be determined usingthe equation of parabola. ?et us determine the concrete stress f c at a distance of Y from

the origin >. 9rom 9ig.7.3c, we ha e f c : 4.11/ f ck F g c * 4.31+

where g c is as shown in 9ig.7.3c and obtained from 6q.7.1. Thus, we get

f c : 4.11/ f ck G 4.11/ f ck 8 D 3*k G 78;+3UY 3

or f c : 4.11/ f ck GY "8*kD G 7 D8;+3UQQQ. *7. 3+

'esignating the strain of concrete at this point by !c *9ig.7.3b+, we ha e from similar triangles

!c 84.443 : G Y 8*kD G 7 D8;+

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which gi es

Y : G *S84.443+U*kD G 7 D8;+QQQ.. *7. 7+

!ubstituting the alue of Y from 6q.7. 7 in 6q.7. 3, we ha e

f c : 4.11/ f ck , i.e., up to a distance of 7 kD8; from the right edge, the compressi e strain is4.443 and, therefore, the compressi e O stress shall remain constant at 4.11/ f ck . 9rom

> to @, i.e., for a distance of 1 kD8;, the strain is reducing from 4.443 to zero and thestress in this zone is parabolic as shown in 9ig.7.3c. The area of the stress block shall

be obtained subtracting the parabolic area between > and @ from the total

22

Fig. 3.2 Cross section, strain !ro"le - and stress #loc$ or

Fig. 3.2(c+ tress #loc$ or t%e strain !ro"le -

Fig3.2 (a+ Cross

Fig. 3.2(#+ trainPro"le - ($) * )+

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rectangular area between and @. s in the case when the neutral axis is outside thesection, we ha e to establish the equation of the parabola with > as the origin andthe positi e coordinates X and Y are measured from the point > upwards for X andfrom the point > to the left for Y , as shown in 9ig.7.3c. ssuming the same equationof the parabola as in 6q.7.7, the alues of a, b and c are obtained as%

*i+ t Y : 4, X : 4, at the origin% gi es c : 4

*ii+ t Y : 4, dX/dY : 4, at the origin% gi es b : 4

*iii+ t @, *i.e., at Y : 1 kD8;+, X : 4.11/ f ck % gi esa : 4.11/ f ck 8*1kD8;+3.

Therefore, the equation of the parabola ># is%

X : 4.11/ f ck 8*1kD8;+3UY 3QQQ.. *7. 2+

The area of the stress block : 4.11/ f ck kD G * 87+ 4.11/ f ck *1kD8;+ : 4.7/ f ck kD.!imilarly, the distance of the centroid can be obtained by di iding the moment of areaof stress block about the right edge by the area of the stress block. Therefore, we ha e

rea of the stress block : 4.7/ f ck kD *7. /+

The distance of the centroid of the stress block from the right edge : 4.13 kD*7. ;+

Thus, the alues of and 3 of 6qs.7./ and 7.5, respecti ely, are 4.7/ and 4.13 whenthe neutral axis is within the section. "t is to be noted that the coefficients and 3 aremultiplied by Df ck and D, respecti ely when the neutral axis is outside the section.Howe er, they are to be multiplied here, when the neutral axis is within the section, bykDf ck and kD, respecti ely.

7.3.7 'etermination of &ompressi e !tress nywhere in the &ompressi eWone when the Aeutral xis is within the !ection

The compressi e stress at any point between and > of 9ig.7.3c is constant at4.11/ f ck as the strain in this zone is equal to or greater than 4.443. !o, we can write

f c : 4.11/ f ck if 4.443 RR !c 4.4472 QQ. *7. 0+

Howe er, the compressi e stress between > and @ is to be determined from theequation of the parabola. ?et us determine the compressi e stress f ci at a distance of Y from the origin >. 9rom 9ig.7.3c, we ha e

f c : 4.11/ f ck F g c *7. 5+

where g c as shown in 9ig.7.3c, is obtained from 6q.7. 2. Thus, we get,

f c : 4.11/ f ck G 4.11/ f ck *1kD8;+3UY 3 QQQQQQ.. *7.34+

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'esignating the strain of concrete at this point by !c *9ig.7.3b+, we ha e from similar triangles

!c 84.443 : G Y 8*1kD8;+, which gi es

Y : F !c 84.443U*1kD8;+QQQ.. *7.3 +

!ubstituting the alue of Y from 6q.7.3 in 6q.7.34, we get the same equation, 6q.7. 1of sec.7.3.3, when the neutral axis is outside the section. Therefore,

f c : 4.11/ f ck

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9rom the point @ to the left edge H of the crossFsection of the column, thecompressi e stress is zero. Thus, we ha e

f c : 4 if !c R 4

f c : 4.11/ f ck if !c O 4.443

f c : 4.11/ f ck 3*!c 84.443+ G *!c 84.443+3U, if 4 R !c D 4.443QQQQ *7.31+

7.7 Tensile and &ompressi e !tresses of ?ongitudinal !teel

Lalues of compressi e or tensile stresses of longitudinal steel *9e 324, 9e 1 2 and 9e244+ are as shown in Table of !- /.

7.1 'esign -arameters

The following are the four major design parameters to be determined for any columnso that it has sufficient pairs of strengths * P u and M u+ to resist all critical pairs obtainedfrom the analysis%

*i+ dimensions b and D of the rectangular crossFsection,

*ii+ longitudinal steel reinforcing bars F percentage #, nature of distribution *equally ontwo or four sides+ and d$/D,

*iii+ grades of concrete and steel, and

*i + trans erse reinforcement.

The roles and importance of each of the abo e four parameters are elaborated below%

>%? D%m"n$%on$ & n! D o( th" r" t n'ul r ro$$9$" t%on

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The strength of column depends on the two dimensions b and D. Howe er, preliminary dimensions of b and D are already assumed for the analysis of structure,which are usually indeterminate statically. "n the subsequent redesign, thesedimensions may be re ised, if needed, in iting reFanalysis with the re iseddimensions.

>%%? Lon'%tu!%n l $t""l r"%n(or %n' & r$

"t is a ery important consideration to utilise the total area of steel bars effecti ely. Thetotal area of steel, expressed in percentage # ranges from the minimum 4.0 to themaximum 1 per cent of the gross area of the crossFsection. The bars may be distributedeither equally on two sides or on all four sides judiciously ha ing two or multiple rowsof steel bars. The strain profiles of 9ig. 4.37.3 re eals that the rows of bars may be allin compression or both compression and tension depending on the location of theneutral axis. ccordingly, the total strength of the longitudinal bars is determined byadding all the indi idual strengths of bars of different rows. The effecti e co er d$ ,though depends on the nominal co er, has to be determined from practicalconsiderations of housing all the steel bars.

>%%%? Gr !"$ o( on r"t" n! $t""l

The dimensions b and D of the crossFsection and the amount of longitudinal steel barsdepend on the grades of concrete and steel.

>%5? Tr n$5"r$" r"%n(or "m"nt

The trans erse reinforcement, pro ided in form of lateral ties or spirals, are importantfor the following ad antages in

*a+ pre enting premature 8 local buckling of the longitudinal bars,

*b+ impro ing ductility and strength by the effect of confinement of the core concrete,

*c+ holding the longitudinal bars in position during construction, and

*d+ pro iding resistance against shear and torsion, if present.

Howe er, the trans erse reinforcement does not ha e a major contribution ininfluencing the capacities of the column. Moreo er, the design of trans ersereinforcement in ol es selection of bar diameter and spacing following thestipulations in the design code. The bar diameter of the trans erse reinforcement alsodepends on the bar diameter of longitudinal steel. ccordingly, the trans ersereinforcement is designed after finalizing other parameters mentioned abo e.

"t is, therefore, clear that the design of columns mainly in ol es the determination of percentage of longitudinal reinforcement #, either assuming or knowing the

dimensions b and D, grades of concrete and steel, distribution of longitudinal bars intwo or multiple rows and d$/D ratio from the analysis or elsewhere. Aeedless to

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mention that any designed column should be able to resist se eral critical pairs of P uand M u obtained from the analysis of the structure. "t is also a fact that se eral trialsmay be needed to arri e at the final selection re ising any or all the assumed

parameters. ccordingly, the design charts are prepared to gi e the results for theunknown parameter quickly a oiding lengthy calculations after selecting appropriatenonFdimensional parameters.

Iased on the abo e considerations and making the design simple, quick and fairlyaccurate, the following are the two nonFdimensional parameters%

9or axial load% P u /f ck bD

9or moment% M u /f ck bD 3

The characteristic strength of concrete f ck has been associated with the nonFdimensional

parameters as the grade of concrete does not impro e the strength of the columnsignificantly. The design charts prepared by !-F / are assuming the constant alue of f ck for M 34 to a oid different sets of design charts for different grades of concrete.Howe er, separate design charts are presented in !-F / for three grades of steel *9e324, 9e 1 2 and 9e 244+, four alues of d$/D *4.42, 4. , 4. 2 and 4.3+ and two types of distribution of longitudinal steel *distributed equally on two and four sides+.

ccordingly there are twentyFfour design charts for the design of rectangular columns.Twel e separate design charts are also presented in !-F / for circular sectionsco ering the abo e mentioned three grades of steel and for alues of d$/D ratio.

Howe er, the unknown parameter #, the percentage of longitudinal reinforcement has been modified to #/f ck in all the design charts of !-F /, so that for grades other than M34, the more accurate alue of # can be obtained by multiplying the #/f ck with theactual grade of concrete used in the design of that column.

Howe er, this lesson explains that it is also possible to prepare design chart taking intoconsideration the actual grade of concrete. s mentioned earlier, the design charts are

prepared getting the pairs of alues of P u and M u in nonFdimensional form from theequations of equilibrium for different locations of the neutral axis. Be now take up therespecti e nonFdimensional equations for four different cases as follows%

*a+ Bhen the neutral axis is at infinity, i.e., kD : V, pure axial load is applied on thecolumn.

*b+ Bhen the neutral axis is outside the crossFsection of the column, i.e., VO DC kD.

*c+ Bhen the neutral axis is within the crossFsection of the column, i.e., kD D D.

*d+ Bhen the column beha es like a steel beam.

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7.2. AonFdimensional 6quation of 6quilibrium when k : V, *-ure xial?oad+

The total compressi e force due to concrete of constant stress of 4.11/ f ck is%

c : 4.11/ f ck b D *7.32+

Howe er, proper deduction shall be made for the compressi e force of concrete nota ailable due to the replacement by steel bars while computing s.

The force of longitudinal steel bars in compression is now calculated. The steel bars of area #bD/ 44 are subjected to the constant stress of f sc when the strain is 4.443.!ubtracting the compressi e force of concrete of the same area #bD/ 44, we ha e,

s : * #bD/ 44+ * f sc F 4.11/ f ck + *7.3/+

Thus, we ha e from 6q. 7. after substituting the expressions of c and s from6qs.7.32 and 7.3/,

P u : 4.11/ f ck b D P * #bD/ 44+ * f sc F 4.11/ f ck + *7.3;+

'i iding both sides of 6q.7.3; by f ck bD, we ha e

* P u /f ck bD+ : 4.11/ P * #/ 44 f ck + * f sc F 4.11/ f ck + *7.30+

7.2.3 AonFdimensional 6quations of 6quilibrium when Aeutral xis is>utside the !ection *V C k' O '+

9igures 7.3b and c of present the strain profile N$ and the corresponding stress block for this case. The expressions of c, s and appropriate le er arms are determined towrite the two equations of equilibrium. Bhile computing c, the area of parabolicstress block is determined employing the coefficient from Table !- /. !imilarly,the coefficient 3, needed to write the moment equation, is obtained from Table !-

/. The forces and the corresponding le er arms of longitudinal steel bars are to be

considered separately and added for each of the n rows of the longitudinal bars. Thus,we ha e the first equation as,

P u : f ck bD P * 4.1 + + P *X b'p i 8 44fck *fsiF fci++

where : coefficient for the area of stress block

# i : A si /bD where A si is the area of reinforcement in the i th row,

f si : stress in the i th row of reinforcement, taken positi e for compression and negati efor tension,

f ci : stress in concrete at the le el of the i th row of reinforcement, and

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n : number of rows of reinforcement.

Here also, the deduction of the compressi e force of concrete has been made for theconcrete replaced by the longitudinal steel bars.

'i iding both sides of by f ck bD , we ha e

* P u /f ck bD+ : P * 4.13+ P *X pi 8 44fck *fsiF fci+

!imilarly, the moment equation becomes,

M u : f ck bD * D83 F 3 D+ P X b'p i 8 44fck *fsiF fci+yi

Bhere, 3 : coefficient for the distance of the centroid of the compressi e stress block of concrete measured from the highly compressed right edge and

yi : the distance from the centroid of the section to the ith

row of reinforcement, positi e towards the highly compressed right edge and negati e towards the leastcompressed left edge.

'i iding both sides of by f ck bD 3, we ha e

* M u /f ck bD 3+ : *4.2 F 3+ P X pi 8 44fck *fsiF fci+*yi8'+

7.2.7 AonFdimensional 6quations of 6quilibrium when the Aeutral xis iswithin the !ection *k' D '+

The strain profile "A and the corresponding stress block of concrete are presented in9igs.7.7 b and c for this case. 9ollowing the same procedure of computing c, s andthe respecti e le er arms, we ha e the first equation as

P u : 4.7/ f ck kbD P Xb' p i 8 44fck *fsiF fci+

'i iding both sides of by f ck bD , we ha e

P u /f ck bD : 4.7/ k P Xp i 8 44fck *fsiF fci+

and the moment equation as

M u : 4.7/ f ck kbD*4.2 F 4.13 k + D P X b'p i 8 44fck *fsiF fci+yi

'i iding both sides by f ck bD 3, we ha e

* M u /f ck bD 3+ : 4.7/ k *4.2 F 4.13 k + P X p i 8 44fck *fsiF fci+*yi8'+

Bhere, k : 'epth of the neutral axis8'epth of column

7.2.1 AonFdimensional 6quation of 6quilibrium when the &olumnIeha es as a !teel Ieam

This is a specific situation when the column is subjected to pure moment M u = M o only.

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!ince the column has symmetrical longitudinal steel on both sides of the centroidalaxis of the column, the column will resist the pure moment by yielding of both tensileand compressi e steel bars *i.e., f si : 4.0; f y : f yd +. Thus, we ha e only one equationwhich becomes

M u : X b'p i 8 44fck 4.0; f y yi

'i iding both sides of by f ck bD 3, we ha e

* M u /f ck bD 3+ : X p i 8 44fck 4.0; f y yi8'

7./ o erning equation%

7./. A. . outside the section

-u8fckb' : & P X p i 8 44fck *fsiF fci+

Mu8fckb' 3 : & *4.2F&3 + P X p i 8 44fck *fsiF fci+*yi8'+

7./.3 A. . inside the section

-u8fckb' : 4.7/k P X p i 8 44fck *fsiF fci+

Mu8fckb' 3 : 4.7/k *4.2F4.1 /k + P X pi 8 44fck *fsiF fci+*yi8'+

7./.7 !teel beam

-u : 4

* M u /f ck bD 3+ : X p i 8 44fck 4.0; f y yi8'