Rc Design i Columns
Transcript of Rc Design i Columns
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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=
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& "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 (/;
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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
@
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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
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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"
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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]=
++
+++
( ) ( ) mLe @.98.@1@:.1 ==
H::.6
@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
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*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'
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( ) 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+=
( ) ( )
( )( ) ( ) ( )
..
1:9
1721
@77977
9:
98
@
mmxI
mmxI
s
c
==
==
Ta%e minimum reinforcement Cover + @mm
( ) ( )( ) ( )( )
( ) ( )( )
.:.@919277
178:.68
.178:.68179:.12179.32
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:
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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+=+=
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e + econd!order eccentricity .
"or non $ sway frames, e+ ( )17
11 rLK e
@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
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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'
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( ) ( )( )
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
( ) .
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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$ds
ffm
bdAsbdA
===11 I
Compression failure controls (very small eccentricity)
;n this case fsN fyd I it is not %nown whether the steel furthest from the load is under
compression or tension. This situation ma%es the solution procedure more
complicated.Column interaction diagrams can be used to simplify the design.
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Column ;nteraction iagram. ( dimensionless ) .
! ;t is a plot of a column a'ial load capacity against the moment it
sustains.
! *ny loading within the curve is a possible fafe loading
combination.
! *ny combination of loading out side the curve represent a faiburecombination.
! *ny radial line from pt. O represents a vonstant ratio of moment to
load Constant eccentricity.
! The full line curve in compression failure range can beconservatively replaced by the dashed line as shown. nowing the
coordinates (O, Ddo) I ( Mnb, Dnb), the design capacity pd for a%nown moment Md, Md + ed pdP can be obtained using the
straight line eEuation as >
( ).811
+
=
ed
ed
*ub
*do
*do%d
ete ed I eb are design eccentricity I eccentricity for balanced conditionrespectu'ly.
hen Dd + O
Md + fcd(7.8:'!*s1) (d!o!9') ?*s1fs
1(d!d1).
;n which [ ]cbb, 9
11
1 ++=
0+ [ ]cdcd$dcuss fbffEA 8.7
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* 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
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
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
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+
= %bdf*
cu$d
cucdnb
8.7
( ) ( ) ( ) ( )kNN 173:9.173:86
[email protected]@77:7.1@
=
+
=
( )( ) ( )nb
$dab
bcd
*
ddfAsdAsbafeb
11
1
1 +
=
+
( ) ( )( )[ ] ( ) ( ) ( )
mm9@
173:
37.93786.:717:7
:93717:7@77::7.1@
=
+
eb + eb1!(d!d1)
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.)( flexureonl$O*dorDee =
b1 + [ ].8.7< cdcd$dcuss fbffEA ( [ ].:7.1@@778.7
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! 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
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! ;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
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$d cu
cu =+
7
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( )( ( )mm
*ub
ddf$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 .
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
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*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
f
f+AAst
$d
cdc =
==
79.se I 9 .1:
*+ @19?9'71+7:7mmS1277mm
Circular Columns.
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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.
@
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( )[ ].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$%"io*
Dda + 13%5, Md+ 187%5m, c!@7, 3!@77, class ;.
4eEuired> esign a circular column.
Constants> fcd+1@.: ,fyd+:7.86 m+12.18.
*ssume +h+ 377mm I : 9 *s+61
ma'
@
Ok
"A
g
g
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Lsing ;nteraction Chart
!*ssume +377mm *g + 12:.@3 17@ mm"
d1
+ 7.17
9:[email protected]::.1@
1713@
@
=
==xAf
*ged
da
[email protected]@3.12::.1@
17187@
:
=
==
"Af
Md
ged
+ 7.12 from chart
@
1293
86.:7
:.1@[email protected]:12.7mmAs =
=
use : this shows : : is conservative
1.@. hort columns under 0ia'ial 0ending.
Consider an 4C Column section shown when sub=ected to design a'ial force pdacting
with eccentricities ed'I edy, such that ed'+
.,I axiscentroidalfromc*
Me
*
M
d
dxd$
d
d$ =
Computation commences ( begins) with an assumed neutral a'is depth and>
3
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,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.
:
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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
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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$"
ccd
"
MMbAf
M
MM"Af
M
==
=
,
, .
! elect suitable chart which satisfy ratiob
b"
"1
I1
8
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! /nter the chart I pic% w, mechanical steel ratio.
! Compute *tot +$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
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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
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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
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"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.