rc design i columns

7/21/2019 Rc Design i Columns

1/32

1. DESIGN

of

AXIALLYAND ECCENTRICALLY LOADED COLUMN

1.1. Introduction

Columns are defined as members that carry loads chiefly in compression, even though thebending action may produce tensile forces over part of their cross section.

On the basis of construction and lateral ties, three types of reinforced concretes

compression members are in use.

(i) Members reinforced with longitudinal bars and lateral ties.

(ii) Members reinforced with longitudinal bars and continuous spirals.

(iii) Composite compression members reinforced longitudinally with structural steel

shapes, pipe, or tubing, with or without additional longitudinal bars, and various

types of lateral reinforcement.

Types 1 and are by far the most common.

On the basis of the slenderness ratio columns may be classified as short or long (slender).

! "or isolate columns, the slenderness ratio is defined by

where #e $ is the effective buc%ling length

i ! is the minimum radius of gyration of the concrete section only.

1

i

Le=


2/32

& "or multistory sway frames comprising rectangular sub frames, the following

e'pression may be used to calculate the slenderness ration of the columns in the

same storey

LK

A

i

1=

here *+ the sum of the cross!sectional areas of all the columns of the story.i+the total lateral stiffness of the columns of the story (story rigidity)

ith modulus of elasticity ta%en as unity,#+the story height.

The effective height (length) of a column is the distance between the two

consecutive points of contra fle'ure or -ero bending moments. The figure shownbelow may serve this purpose.

owever in accordance with /0C!, 1223, the effective length #e for an 4CColumn is given as,

a. 5on!sway mode 6.78.7

9.7

++

= mme

L

L

b. way mode( )

13.13.6

:.193.6

1

11 +++++

=

L

Le

Or Conservatively .13.18.71 += me

L

L

"or the theoretical model shown below.

1

1

111

1

1

+=

++=

++

=

m

c

c

kk

kk

kk

kk

here 1and are column stiffness coefficients (/;


3/32

i= is the effective beam stiffness coefficient (/; 5sd$ the design value of the total vertical load.

5cr$ critical vertical load for failure in a sway mode given as

e

ecr

L

EIN

=

/;e+ O./c;c? /s;s (or /;e + ( ) 9.

1Eo

r

M

bal

bal )

/c

( @17311177 =dr

Curvaturefbal

cd

; c+ Moment of inertia of the concrete sections of the substitute column w.r.to centre

;s + Moment of inertia of reinforcement sections of the conc. section

* frame may be classified as braced if its sway resistance is supplied by a bracing system

which is sufficiently stiff to assume that all hori-ontal loads are resisted by the bracing

system. (5ot more than 17A of the hori-ontal loads are attracted by the frame)

Benerally, the slenderness ratio of concrete columns shall not e'ceed 197.

econd!order effects in compressive members need not be ta%en into account in thefollowing cases>

(a). "or sway frames, the greater of the two

d

13

3

(b)."or non $ sway frames

( )( )

1337M

M

@


4/32

here M1and Mare the first!order (calculated) moments at the ends. Mbeing always

positive and greater in magnitude than M1, and M1being positive if member is bent insingle curvature and negative if bent in double curvature.

( )loadaxialdesignNAf

Nsd

ccd

sdd =

1.2. Reinforcement arrangement & Minimum Code Requirements.

"unctions of #ateral 4einforcement.! they hold the longitudinal bars in position in the forms while the concrete is

being placed

! they prevent the slender longitudinal bars from buc%ling out ward bybursting the thin concrete cover.

4ules for the arrangement>

! iameter of ties, t t :mm or 9

! C


5/32

7.778*c*e7.78*c or 7.778 78.7c

e

A

A

! Min. F of bars

=.:

tan9

tarrangemenCircularin

tarrangemengularrecin

! The diameter of longitudinal bars, .1mm

! The minimum lateral dimension of a column shall be at least 137mm and

the minimum diameter of a spiral column is 77mm.

! The Min. cover to reinforcement should never be less than

(a) ormmor n ),97(

(b) ( ) ( ) .@33 mmdifmmOrmm gn >++ dg$ the largest nominal ma'imum aggregate si-e.

/'ample 1.1. (Classification of columns).

The frame shown in figure below is composed of members with rectangular cross

sections. *ll members are constructed of the same strength concrete (/ is the same forboth beams and columns). Considering bending in the plane of the frame only, classify

column /" as long or short if the frame is (a)braced and (b)unbraced. *ll girders are

@77 ' :77 mm.

olution>

Moments of inertia

Birders> 98@

17391

:77@77mmx

xIg ==

Columns>98

171:1

977@77mmx

xIDE ==

.1761863.171

@37@77 98@

mmxx

IEF ==

3

"

/

;

B

/

C

0

*

M+ 93 5m

M1+ @7 5m

6.3 m2 m

@.87 m

@.87 m

977

:77

:77

@37

@77 ' 977

@77 ' @37

33 5"


6/32

tiffness Coefficients>

( ) ( )

( ) ( )

===

===

=.17.6

6377

1739

.17:2777

1739

>3

8

3

8

ExxE

KK

ExxE

KK

L

EI

KGirdersFIE

cF!E

g

g

g

Columns>

( ) ( )

( )( ) ( )

==

==

=

Ex

x

EK

Exx

xEK

L

EIK

EF

DE

c

c

c

3

@

8

3

@

8

178.

178.@

1761863.17

171.9178.@

171:

The column being considered is column /".

4otational stiffnesses at =oints / and ".

( )

( ) ( )

( )effgf

col

effgf

col

LI

LI

LEI

LEI

( ) ( )@[email protected]

8 xxAI

L

I

L ee ===

.

H::.::93

@7

337

.8.9

s"ortiscolum#"e

ok

=

=

(b) "or unbraced column (sway structure)

( ) ( )designforL

L

FE

FEFEe 13.13.6

:.193.6

+++++

=

+ ( ) ( ) ( )

1@:.1

[email protected]

[email protected]:[email protected]=

++

+++

( ) ( ) mLe @.98.@1@:.1 ==

H::.6

@[email protected]

@783.7

1733

133

6:.97@:.171

9@7

@oknot

xxx

xor =

==

.LongisColumn#"e

1.3. Short Versus ong Co!umns in "#ia! Com$ression.

;n Compression, both the longitudinal steel and concrete contribute to the

resistance of the applied a'ial force. "or the design of short columns in purecompression, /0C! limits the strain in the concrete to 7.77, since generally

this is the strain at which the stress in the concretes is ma'imum. The capacity to

resist compressive force, Dultis appro'imately eEual to>

Dult + fc% (*g!*s)? fy*s , Dd+ (s

s$

c

sgck

AfAAf

+

here + Coefficient, generally ta%en as 7.83.

fc% + Characteristic compressive cylinder strength of concrete

*g+ gross cross!sectional area (bh).

6


8/32

*s + area of longitudinal reinforcement.

fy + yield strength of reinforcement.

cI s + Dartial factors of safety for concrete and steel.

hort columns usually fail by crushing. lender column is liable to fail by buc%ling. The

end moments on a slender column cause it to deflect sideways and thus bring into play anadditional moment. The additional moment causes a further lateral deflection and if the

a'ial load e'ceeds a critical value, this deflection and the additional moment become self

!propagating until the column buc%les.

"or Din ended columns>

Dcr+

EI

* column is classified as short if both #e'


9/32

( ) ss

$

sg

c

cksd A

fAA

fN

+=

83.7

Ta%ing c+ 1.3

s+1.13

5sd+( ) ( ) ( ) ( ) ( )

13.1

.1:789:71:7817777

3.1

@383.7+

+ 221@78 5 + 221.@78 %5

;n tension, the design a'ial load is>

5sd+ ( ) kNNArf

s

$.:9@1:78

13.1

9:7 ==

(b) "or #+6m. #e+ ( ) ( ) .2.966.7 m=

.13@@.1:@.7

27.9

133.19.7

27.9

Longb

L

s"ortn

L

e$

ex

>==

==

The column is slender .

5cr+ ..7

sscce

e

e IEIEEI

L

EI+=

( ) ( )

( )( ) ( ) ( )

..

[email protected]

1:9

1721

@77977

9:

98

@

mmxI

mmxI

s

c

==

==

Ta%e minimum reinforcement Cover + @mm

( ) ( )( ) ( )( )

( ) ( )( )

.:.@919277

178:.68

.178:.68179:.12179.32

[email protected]@@.7

8

888

:8

kNx

N

mmkNxxx

xxEI

cr

e

==

=+=

+=

1.%. esign of Short Members for "#ia! 'orce and nia#ia! ending

2

977mm

@77 mm97mm

7mm

97mm

97mm 7mm 97mm1:


10/32

Beneral>

* column is sub=ected to unia'ial bending when the load applied to a column is eccentric

about one a'is only. The presence of this form of bending in a'ially loaded members can

reduce the a'ial load capacity of the member. ;t is the combined effect of a'ialcompression and bending at the ultimate limit state that tends to govern the design.

esign load for a'ially loaded columns (ideal columns)

!5o moment considered.

( ) K83. cstgcs

$st

s

fOAAF

r

fAF

=

=

;n practice column loads will have eccentricities at least due to imperfect constructions.

esign /ccentricity

ed + ec ? ea ? e

here ee+eEuivalent constant first!order eccentricity of the design a'ial load.

ee + eofor eoeEual at both ends of a column

igher of ee+ 7.: eo?7.9eo1 ee+7.9eo "or moments varying linearly along the length.

eo1and eoare first order eccentricities at the ends with eobeing positive and greater inmagnitude than eo1.

ea+additional eccentricity in account of geometric imperfections.

ea+ mmL

e 7@77

17

fc

fyfy

fc

fy

tress

AA

Pod

ection *!*

7 7.771 7.77 7.77@ train

teel

Concrete

Pod

( )c

cstg

s

$st

csod

fAAfAFF*

K83.7+=+=


11/32

e + econd!order eccentricity .

"or non $ sway frames, e+ ( )17

11 rLK e

[email protected]

@31363.77

1

>=

=

forK

forK

@

17

3

sec1

=

=

dK

tioncriticalt"eatCurvaturer

here d + the Column dimension in the buc%ling plane less the cover to the center of the

longitudinal reinforcement.

+bal

dMM

Md + design moment at the critical section including second!order effects. Mbal + balanced moment capacity of the column.

The sway moments found by a first!order analysis shall be increased by multiplyingthem by the moment magnification factor>

crsd

sNN


12/32

Dlane section before loading remain plane after loading

"ailure of concrete is governed by the ma'imum strain criteria.

The ma'imum compressive strain in the concrete is ta%en to be> 7.77@3 in

bending (simple or compound) 7.77 in a'ial compression.

The ma'imum tensile strain in the reinforcement is ta%en to be 7.71.

The strain diagram shall be assumed to pass through one of the three points *,0

Or C.

The design stress $ strain curve for concrete (fig. a) I steel (fig. b) are as shownbelow.

Consider the rectangular section when sub=ected under an a'ial load Pdwith large

eccentricity e, as shown below. "or the purpose of stress calculation, the actual non!linearstress distribution shown can be replaced with eEuivalent rectangular stress distribution.

*pplying force eEuilibrium.

1

Cross $ section *ctual stress train implified stress

a

s

c fcdPd

Nc

Ns1

Ns2fcd

Pd

Nc

Ns1

Ns2

Pd

d

d

7.71 !7.771 !7.77 !7.77@3

esign iagrams

$k

$d

ff

=

;deali-ed iagram

esign iagram

ckcd

ff

83.7=

( ) 7.7,1371777 = ccdccc forff

;deali-ed iagram

fs

sc

fc

train iagram at L#!7.77

!7.77@3!7.77

7.71

7.71

cssy

7

7

C

0

*

"

6

@

h d

(a) b

e

hd

e

b

*s

*s '

s

a 7.8'


13/32

( ) ( )( )

11

1

K

K

K

1

.

)(

)1(7

sss

sssscdc

sa

cdNs

siscd

fAN

fANAabfN

ddNdNe*OM

NNN*F

=

==

+==

+==

"or very small eccentricity, the stress distribution along the cross!section is as shown

below.

*pplying force eEuilibrium.

"+O Dd+ 5c?5s?5si !!!!!!!!!!!!!!! (@)

= OMN31

Dde+ 5c( ) ( ) ( )913 + ddNd x

ence 5c+ fcd(b x ' $ *st)

53+ *sfs

5si+ *sfs

( ) .


14/32

Desig C!i"e!ia

0alanced Condition >!for a given Cross $ section a design a'ial force Dda acts at one

specific eccentricity e+eb(are + eb) to cause failure by simultaneous yielding of tension

steel and crushing of concrete. "or simplification purpose symmetrical reinforcement isconsidered and compressive steel is assumed to be its limiting stress level.

( )38.7

+=== %

EE

Ebdf*nb%dOF

cus

cucdn

ince part of the concrete is under tension,

c+ cn+ 7.77@3 ,and s+ yd + .s

d

Ef$

( )( ) ( )):(

111

11

nb

$dbcd

*

ddfAsabdAsbaf

ebeOMNsi

+

===

in which ab 7.8' +bd

As

bd

As*and

d

cns

cn ==+

.8.7

;n case where compression steel is not at its limiting stress.

( )cus

cucu xbxb

dxbs

+

=

= ,1

1

I fs+/ss1fyd.

henever, fs1N fyd, a value of a force *s(fyd$ fs

1)

hall be subtracted from pub of eEn (s) and fyd in (:) shall be replaced with fs1

Tension failure Controls.

*gain *s+ *s1assumed I both steel are stressed to fyd. The two

eEuilibrum eEuations yield.

Dd + fcdbd

+

++

1

1

11

1

1

1 md

d%

d

e

d

e

heredc

r$ds

ffm

bdAsbdA

===11 I

Compression failure controls (very small eccentricity)

;n this case fsN fyd I it is not %nown whether the steel furthest from the load is under

compression or tension. This situation ma%es the solution procedure more

complicated.Column interaction diagrams can be used to simplify the design.

19


15/32

Column ;nteraction iagram. ( dimensionless ) .

! ;t is a plot of a column a'ial load capacity against the moment it

sustains.

! *ny loading within the curve is a possible fafe loading

combination.

! *ny combination of loading out side the curve represent a faiburecombination.

! *ny radial line from pt. O represents a vonstant ratio of moment to

load Constant eccentricity.

! The full line curve in compression failure range can beconservatively replaced by the dashed line as shown. nowing the

coordinates (O, Ddo) I ( Mnb, Dnb), the design capacity pd for a%nown moment Md, Md + ed pdP can be obtained using the

straight line eEuation as >

( ).811

+

=

ed

ed

*ub

*do

*do%d

ete ed I eb are design eccentricity I eccentricity for balanced conditionrespectu'ly.

hen Dd + O

Md + fcd(7.8:'!*s1) (d!o!9') ?*s1fs

1(d!d1).

;n which [ ]cbb, 9

11

1 ++=

0+ [ ]cdcd$dcuss fbffEA 8.7


16/32

* rectangular column @77'377 (mm ' mm) reinforced with 9 : (*s1+*s+17:7mm

)

one at each corner with .1.71

="d etermine the design strength pd when the design

eccentricity fron centerline of column '! section including all effects isa) eb (balanced ) b) 7mm c) 137mm d) 977 mm e)Rery large approaching .

#o$%"io

"yd + :7.86

18.128.7

[email protected]

77683.7

K ===

==

==

mf

fm

f

fm

bd

A%

cd

$d

cd

$d

s

17

.17:73@71

====

Ast

xAsAs

(a) 0alanced case>!

mmxd

,

xcu$d

cu

b@8937

[email protected]

[email protected].

173

86.:7=

+=

+

=

ab +7.8'b +:mm.

( )[email protected]@.7772::[email protected]@8

681-ds x >=

Compression steel is yielding

1:

377

1+(7!1) (377) + 37 mm+377!37 +937mm

Constants >! fc%+9,fcd+1@.:7


17/32

+

= %bdf*

cu$d

cucdnb

8.7

( ) ( ) ( ) ( )kNN 173:9.173:86

[email protected]@.7

[email protected]@77:7.1@

=

+

=

( )( ) ( )nb

$dab

bcd

*

ddfAsdAsbafeb

11

1

1 +

=

+

( ) ( )( )[ ] ( ) ( ) ( )

mm9@

173:

37.93786.:717:7

:93717:7@77::7.1@

=

+

eb + eb1!(d!d1)


18/32

.)( flexureonl$O*dorDee =

b1 + [ ].8.7< cdcd$dcuss fbffEA ( [ ].:7.1@@778.7


19/32

! The section in Euestion is investigated which load combination it

can sustain. More suitably, for a fi'ed value of ed, determine Ddn(its

capacity) such that .

;f Dd5 Dd, safe but is it economical

;f Dd5


20/32

! ;f the coordinates (,) lies with in the families of the curve,

the assumed cross $ section is feasible, which otherwise need

to choose new large section.

! The coordinate (,) gives the value of w.

! Obtain or *stusing *st+$d

cdc

$d

cd

f

f+A

f

+b"f= and arrange the

reinforcement. *cbrh

! Chec% min $ and ma', provisions."or such over lap, gma'+7.79.

/'ample 1...esign a column to sustain a design a'ial load of 1177%5 I design bending moment of

1:7%5m which includes all other effects, assume concrete c!@7, steel s! 977 class ; wor% .

*ppro'imate b+7.:h.

olution>

Constants> 8@.@96

:[email protected]

983.7,9

=

===

$d

edck

f

xff

Dda+1177%5, ed + .1931177

171:7@

mmx

*

Md

da

==

Lsing /Euations.

Trial 1> *ssume @77'377mm I :9

( )

1.793737377

.77683.7377@77

3@7J

778.7719.7

1

min

===

===

>=

ndt"atsod

xbdAs

gg

Conditions controlling the design.

Dub+fcdbd .2:68.7

kN$d cu

cu =

+

*b+ .978.7

mmd$d cu

cu =+

7


21/32

( )( ( )mm

*ub

ddf$AsabdAsbafeb

dbcd98

111

1=

+

=

eb +8mm S ed compression controls.

pdo + fcd.692kNAfAA st$dstg =+

Dd+kN%dkN

eb

ed

%ub

%do

%do 1177191

11

=>=

+

afe not economical.

! "or ductility reEuirement, it would be better to go for low steel ratio.Trial procedure>! Change cross $ section fi'ing reinforcements or vice! versa.

! "or offshore structures large tie bars due to corrosion action.

Trial .

( )[email protected]

9393767

93

.713.79993767

=

==

=

x

x

bdAs

+it"x g

Dub+ 1@.:7679737.3@99@8 $ 7.778@P 17!@+68%5.

*b + 7.3@99@8973 + 1:mm

/b1+( ) ( )

@1768

.939738@.@96279

1:9732791:67:.1@

x

+

+ 991mm.

/b+ 991 $ @:7! unia'ial chart 5o P.

*ssume 67 ' 937 d1


22/32

*st + w8@.@96

:.1@[email protected] =

rdf

+Acfed + 1::.61mm

#9 + @.6use 9 9

E'a$e 1*2*+

esign a column to sustain a design a'ial load of 33%5 acting with a design bending

moment of 77%5m including all other effects. Lse the same materials e'ample 1...

olution >

mmx

*M

ed

dd

337

1777 @== + @:9mm.

"ed+ 1@.: Mpa , fyd+ @96. 8@ Mp

T!ia$ 1, 67 ' 937 with 9 :

93767

3@79

=g mmd 97317

937937,716.7 ===

7726.797367

3@7=

==bd

A% s

pnb+ 687%5 , ab + 1:mm, eb+ 9:6mm

eb+ 86 N ed Tension controls.

@[email protected],11.7,38.311

1 ====d

ed

df

fm

cd

$d

Dd+328%5.S Dda+337safe ( ).t"isatsto%to%ossibleisItTrial > #ets use 9 7 ?91:

pd+389 o%Lsing interaction chart > Lnia'ial Chart 5o P

67'937 1.7K

="d d+ 973.

.6.7.@@. ==== b"

f

Mo

b"f

Nd

edcd

w+7.97.

12778@.@96

:[email protected]

f

f+AAst

$d

cdc =

==

79.se I 9 .1:

*+ @19?9'71+7:7mmS1277mm

Circular Columns.


23/32

Consider the cross $ section shown reinforced with : longitudinal bars (the 5o of bars

can vary from : to 17).

! "or bars with strains in e'cess of yield strain ., $ds$d ff =! "or the cross section with tension crac%, [email protected]=cu! "or bars with smaller strains the stress is found using inf sss = which the

strain iss obtained from the strain geometry.! "or analysis or design, the iterative procedure involves the following.

1. *ssume a, eEuivalent stress bloc% depth and compute

.3.7,1

1

= !!a

. /valuates the stress fsor fydassuming ( )cbcucu or == [email protected]@3.7@. etermine Ddand then a (or ') , using the two un%nown eEuilibrium eEuation.

"irst evaluate Ddusing the moment eEuation about e'treme tension steel as.

Dde1+ 5c(d!a1)? 5s1(d!d11)?5s(d!d@1)here a1is the distance from e'treme compression face to the centroid of

the compression force 5c.Lsing the force eEuation recomputed a and of course a +

++== 393@1, NNNNN%asaa sscd is"ere1 to be computed from the

area of the soction 5ote that during computation, if the value of ' obtained is

larger than d, or over all diameter of the cross!section, tension steel can be under

compression very small eccentricity or nearly concentric loading. ;n such

circumstances the eEuation developed should be ree'amined to account for such

effect.

9. 4epeat step @ until pd%?1

and 1 or a%?1

and a%

are nearly eEual or until certainconvergence criteria are satisfied. The subseript % stands for iteration F

To simplify this , on appro'imate empirical formula modified to confirm 13 of

the local standard, for circular column of diameter h failing in compression is givenhitneyP.

@


24/32

( )[ ].18.1

:6.78.7

:.21

@K1

++

++

d""

"e

fA

d"

e

fA*

cdg$ds

d

This appro'imate formula holds, for eeb Ieb for this purpose may be appro'imated aseb +(7.?7.2 pgm

1)h.

"or practical problems interaction charts are available for use .

E'a$e 1*2*-*

esign a circular column for design a'ial load of Dda+13 %5 I design

bending moment ofMd+187%5.m. use c!@7,3!@77, class ; wor%.

#o$%"io*

Dda + 13%5, Md+ 187%5m, c!@7, 3!@77, class ;.

4eEuired> esign a circular column.

Constants> fcd+1@.: ,fyd+:7.86 m+12.18.

*ssume +h+ 377mm I : 9 *s+61

[email protected]

[email protected]:9

ma'

@

Ok

"A

g

g


25/32

Lsing ;nteraction Chart

!*ssume +377mm *g + 12:.@3 17@ mm"

d1

+ 7.17

9:[email protected]::.1@

1713@

@

=

==xAf

*ged

da

[email protected]@3.12::.1@

17187@

:

=

==

"Af

Md

ged

+ 7.12 from chart

@

1293

86.:7

:.1@[email protected]:12.7mmAs =

=

use : this shows : : is conservative

1.@. hort columns under 0ia'ial 0ending.

Consider an 4C Column section shown when sub=ected to design a'ial force pdacting

with eccentricities ed'I edy, such that ed'+

.,I axiscentroidalfromc*

Me

*

M

d

dxd$

d

d$ =

Computation commences ( begins) with an assumed neutral a'is depth and>

3


26/32

,stifstiAstiAsci,cifciAciM-stifstiAsti$fAM

eltensionstetensioninforcetresulfAN

steelcom%inforcetresulfAN

inconcreteforcetresulfAN+"ere

NNN*F

d$

cicicidx

stistist

sciscisc

cicic

stsccd"

++= +=

=

=

=

+==

.tan

tan

tan,

7

The procedure using the e'pressions developed so far is tedious as the determination of

the neutral a'is reEuires several trials. Thus the two commonly used methods proposed

by 0reseles shall be discussed below.

*) ;nverse load us eccentricities gives bowel shaped failure surface.

Consider the dis e.%1

surface in the region of interest at Dt.U where ed'I edyfor the

respective unia'ial eccentricities are appro'imated using pt.*I0. #et pt.c represent thereciprocal of the concentric design load capacity.

The pt. U on the interaction surface is appro'imated by a point of which generallygives a conservative estimate of the strength. On this basis the strength may be obtained

from

+=

+=

d$dxdod$ddx

dod$dx

d

dd$dxd

%%%%%%

%%%%

%%%%

7

7

1111

here Dd + design a'ial force capacity under bia'ial bending edyI ed'.

Dd'I Ddy + Capacities for unia'ial bending with eccentricities edyI ed'respectively.

:


27/32

Ddo+ concentric a'ial force capacity.

0) 0reslers Method ;;

i) *ssume the cross! ectional dimensions, area of steel and its distribution.

ii) Compute concentric load Capacity Ddoanddo

do

**

iii) etermine unia'ial moment capacities Md'o and Mdyoof the section combined

with given a'ial load Ddawith the use of interaction curves for a'ial load and

uni'ial moment.iv) Then the adeEuacy of the column section can be chec%ed either with the

interaction (/U) of interaction curves. "or chec%ing the adeEuacy of column

section with interaction eEuation, determined

dn *

*::6.1::6.7 +=

which shall be 1N nN. then computen

d$o

d$n

dxo

dx

M

M

M

M

+

1,

otherwise the section is unsafe. Then the section is modified and chec%ed

again, for chec%ing the adeEuacy of section with interaction curves, the values

of Md'< Md'oand Mdy


28/32

7.,::6.1::6.7

7.1

+=

=

+

n

do

dan

n

d$o

d$n

dxo

dx

*

*

M

M

M

M

for a circle and ranges b susceplible for buc%ling %eep it as low as possible

1ma'+ 6.79.1

1= for ductility

owever interaction charts prepared for this purpose can be used for actual design using

the following procedure.

Mb

1.711

==b

b

"

"7 range

h1 Mhvalues of 7.73, 7.1, 7.13,..7.3

are available

b1 b b1

! elect * Cross section dimension h b, h1b1

! Compute>

5ormal force ratio + dced

*NAf

N=,

Moment ratio h +

dxb

ced

bb

d$"

ccd

"

MMbAf

M

MM"Af

M

==

=

,

, .

! elect suitable chart which satisfy ratiob

b"

"1

I1

8


29/32

! /nter the chart I pic% w, mechanical steel ratio.

! Compute *tot +$d

cdc

f

fA+

Chec% *tot satisfies the ma'. I min provisions! elect suitable bass

E'a$e 1*+*1

esign a column to sustain a factored design load of 277%5 and bia'ial

moments of Md'+ 187%5m Mdy + 67%5m including all other effects $ *ssume

materials of concrete c!@7, steel s!@77, class ; wor%.

#o$%"io,

Constants fc%+ 9, fcd+ 1@.:, fyd+ :7.86m1+ 12.18

ed'+ mm*

Memm

*

M

da

dxd$

da

d$77@77 ===

teel ratio for top use N 7.7 for comp< tension steel of a raw

;nner columns larger dimension than outer ones.

T!ia$1> *ssume :77 ' 977 with 8 8 arranged as shown.

*st + 8:13 + 927 773.7=g

Dd7 + fed (*g!*st) ? fyd*st + 9987%5

*s + *s1+ @J :13 The two bass on the controidal a'is have negligible moment for both caves of

direction P

./. di!ec"io,

ed'+@77mm, Dda 277%5, d + :77!:7 + 397mm

b + 977mm + 77339.7=d

s

dA

Dnb+ 1:68%5 , e1b + 37mm, eb+ 87mm N ed'+ @77

T!controls

2


30/32

Dd' + bd fed ( ) ( )

+

+ 11

1

1

11

11

ddm

dee

de

+ 13:7.%5;f Dd'is near 277 or less, you should change the Q! section immediately without

further chec% for Ddyb


31/32

1. The strength of Concentrically loaded Columns decreases with increasing

slenderness ratio .rk

. ;n Columns that are braced against sidesway or that are parts of frames braced

against sidesway, the effective length .., eik , the distance b (*ccording to *C;).

Mc+ .Mns

cu

mns

%%

C

63.71= 1.

here Du+"actored #oad.

Dc+Critical #oad ( ) factorlengt"effectivek

EI

u

=

Cm+7.:?7.9 .9.

1 oM

M

@1


32/32

"or members braced against sidesway and without transverse loads between supports.

ere Mis larger of the two end moments, and 1 MM is positive when the end

moments produce single curvature and negative when they Droduce double curvature.Lnbrace frame, cm+1.7.

rc design i columns

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