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REINFORCED CONCRETE
COLUMN
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Column Design Procedures:
A procedure for carrying out the detailed
design of braced columns (i.e. columns that do
not contribute to resistance of horizontal
actions) is shown in Table 1. This assumesthat the column dimensions have previously
been determined during conceptual design or
by using quick design methods. Column sizes
should not be significantly different from thoseobtained using current practice.
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Column can be classified as:
Braced where the lateral loads are resisted byshear wall or other form of bracing capable oftransmitting all horizontal loading to the foundations;and
Unbraced where horizontal load are resisted bythe frame action of rigidity connected columns,beams and slabs.
With a braced structure, the axial forces and moments in the columns arecaused the vertical permanent and variation action only, whereas with anunbraced structure, the loading arrangement which include the effects of
lateral load must also be considered
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Loading and Moments
For a braced structure, the critical arrangement of the ultimate loadis usually that which causes the largest moment in the column
together with a larger axial load. Figure 2 shows the critical loading
arrangement for design of its centre column at the first floor level
and also the left-hand column at all floor levels.
1.35 Gk + 1.5 Qk
1.35 Gk + 1.5 Qk
1.35 Gk + 1.5 Qk
1.35 Gk
Figure. 2: A critical loading arrangement
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Slenderness ratio of a column
Eurocode 2 states that second order effects may be ignored if theyare less than 10% of the first order effects. As an alternative, if theslenderness () is less than the slenderness limit ( lim), then secondorder effects may be ignored.
The slenderness ratio of a column bent about an axis is given by:
Where:
lo - effective height of the columni - radius of gyration about the axis
I - the second moment of area of the section about the axis
A - the cross section area of the column
A
I
li
l 00
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Effective height lo
of a column lo is the height of a theoretical column of equivalent section but pinned at both ends. This depends on the degree of fixity at each end and of the column.
Depends on the relative stiffness of the column and beams connected to either end of
the column under consideration.
Two formulae for calculating the effective height:
Figure 3: Different
buckling modes and
corresponding effective
height for isolated column
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i) For braced member
ii) For unbraced member the larger of:
And
2
2
1
10
45.0
1
45.0
15.0
k
k
k
kll
kk
xkkll
1
21
0 101
2
2
1
10
11
11
k
k
k
kll
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Where
k1 and k2relative flexibility of the rotational restrains at end 1 and 2 of thecolumn respectively. At each end k1 and k2can be taken as:
k = column stiffness/ beam stiffness
=
=
For a typical column in a symmetrical frame with span approximately equal
length, k1 and k2can be calculated as:
beam
column
lEI
lEI
)/(2
)/(
beam
column
lI
lI
)/(2
)/(
beam
column
lI
lIkkk
)/(
)/(
4
121
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Limiting Slenderness Ratio short or slender columnsEurocode 2 states that second order effects may be ignored if they are less than 10% of the first
order effects. As an alternative, if the slenderness () is less than the slenderness limit (lim), then
second order effects may be ignored. Slenderness,
= lo/i
where i= radius of gyration
Slenderness limit:
Where:
A = 1/(1+0.2ef) (if ef is not known,A = 0.7 may be used)
B =
w= (if w, reinforcement ratio, is not known, B = 1.1 may be used)
C = 1.7 rm (if rm is not known, C = 0.7 may be used see below)
n =
rm =
M01, M02 are the first order end moments, | M02| | M01|
If the end moments M01 and M02 give tension on the same side, rm should be taken positive.
w21
cdc
Ed
fA
N
02
01
M
M
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** Of the three factorsA, B and C, C will have
the largest impact on lim and is the simplest to
calculate. An initial assessment of lim can
therefore be made using the default values
for A and B, but including a calculation for
C. Care should be taken in determining Cbecause the sign of the moments makes a
significant difference. For unbraced
members C should always be taken as0.7.
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Example:
Determine if the column in the braced frame shown in
Figure 4 is short or slender. The concrete strength fck =
25 N/mm2 and the ultimate axial load = 1280 kN
Hcol=3.5m
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Effective column height lo
Icol = 400 x 3003/12 = 900 x 106 mm4
Ibeam = 300 x 5003/12 = 3125 x 106 mm4
k1 = k2= =0.096
= 0.59 x 3.0 = 1.77 m
Slenderness ratio :Radius of gyration, i=
Slenderness ratio
3636
104/1031252(2
103/10900
/2
/
xxx
xx
lI
lI
beambeam
colcol
2
2
1
10
45.01
45.015.0
k
k
k
kll
mmh
bh
bh
A
I
col
col 6.8646.3
12/3
4.206.86
1077.1 30 x
i
l
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For braced column,
> 20.4
)/(/2.26lim cdcED fAN
866.05.1/2585.0300400
101280)/(
3
xxx
xfAN cdcED
25.30866.02.26lim x
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REINFORCEMENT DETAILS
Longitudinal steelA minimum of four bars is required in the
rectangular column (one bar in each corner) and
six bars in circular column. Bar diameter should
not be less than 12 mm.
The minimum area of steel is given by:c
yk
Eds A
f
NA 002.0
87.0
10.0
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Links
The diameter of the transverse reinforcement should not
be less than 6 mm or one quarter of the maximumdiameter of the longitudinal bars.
Spacing requirements
The maximum spacing of transverse reinforcement(i.e.links) in columns (Clause 9.5.3(1)) should not
generally exceed:
20 times the minimum diameter of the longitudinalbars.
the lesser dimension of the column.
400 mm.
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DESIGN MOMENTFor braced slander column, the design bending moment is illustrated
in Figure 5 and defined as:MEd = max {M02, M0e + M2, M01 + 0.5 M2, NEd.e0}
For unbraced slender column:
MEd = max {M02 + M2, NEd.e0}
Where:M01 = min {|Mtop|, |Mbottom|} + ei NEdM02 = max {|Mtop|, |Mbottom|} + ei NEde0 = max {h/30, 20 mm}
ei = lo/400Mtop, Mbottom = Moments at the top and bottom of the
column
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Figure 5: Design bending moment
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M0e = 0.6 M02 + 0.4 M01 0.4 M02
M01 and M02 should be positive if they give tension on thesame side.
M2 = NEd x e2 = The nominal second order moment
Where:
NEd = the design axial loade2 = Deflection due to second order effects =
lo = effective length
c = a factor depending on the curvature distribution,normally
1/r = the curvature = Kr . K . 1/r0
c
l
r
01
102
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Kr = axial load correction factor =
Where, n =
K = creep correction factor =
Where:
ef= effective creep ratio =
= 0, if ( < 2, M/N > h, 1/r0 < 75)
= 0.35 + fck/200 /150
1/r0 =
A non-slender column can be designed ignoring secondorder effects and therefore the ultimate design moment,
MEd = M02.
4.0,1,/ balucdcEd nwnfAN
1/ baluu nnnn
cdcyds fAfAw /
11 ef
EdEqp MjM 00 /
dEfd sydyd 45.0//)45.0/(
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SHORT COLUMN RESISTING MOMENTS AND
AXIAL FORCES
The area of longitudinal reinforcement isdetermined based on:
Using design chart or construction M-Ninteraction diagram.
A solution a basic design equation.
An approximate method
A column should not be designed for a momentless than NEdx emin where emin has a gratervalue of h/300 or 20 mm
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DESIGN CHARTThe basic equation:
NEd design ultimate axial load
MEd design ultimate moment
s the depth of the stress block = 0.8x(Figure 6)
As the area of longitudinal reinforcement in the more highlycompressed face
As the area of reinforcement in the other face
fsc the stress in reinforcementAsfs the stress in reinforcementAs, negative when tensile
sscccEd FFFN
sssscck AfAfbsf '567.0
2'222
h
dFd
h
F
sh
FM sscccEd
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Figure 6: Column section
Figure 7: Example of
column design chart
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Two expressions can be derived for the area of steel required,(based on a rectangular stress block, see Figure 8) one for the axialloads and the other for the moments:
AsN/2 = (NEd fcd b dc) / [(sc st) c]
Where:
AsN/2 = Area of reinforcement required to resist axial load
NEd = Axial load
fcd = Design value of concrete compressive strengthsc (st) = Stress in compression (and tension) reinforcement
b = Breadth of section
c = Partial factor for concrete (1.5)
dc = Effective depth of concrete in compression
= x h = 0.8 for C50/60
x = Depth to neutral axis
h = Height of section
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AsM/2 = Total area of reinforcement required to resist moment
= [M fcd b dc(h/2 dc/2)] / [(h/2d2) (sc+st) c]
Example:
Figure 8 shows a frame of heavily loaded industrial structure for which thecentre column along line PQ are to be designed in this example. The frameat 4m centres are braced against lateral forces and support the followingfloor loads:
Permanent action, gk - 10 kN/m2
Variable action, qk - 15 kN/m2
Characteristic materials strength arefck= 25 N/mm2 and fyk= 500 N/mm2
Maximum ultimate load at each floor:
= 4.0 (1.35gk+ 1.5qk) per meter length of beam
= 4.0 (1.35 x 10 + 1.5 x 15)
= 144 kN/mMinimum ultimate load at each floor:
= 4.0 x 1.35gk
= 4.0 x (1.34 x 10)
= 54 kN per meter length of beam
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Figure 8: Column structure
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Column load:
1st floor = 144 x 6/2 + 54 x 4/2 = 540 kN
2nd and 3rd floor = 2 x 144 x 10/2 = 1440 kN
Column self weight = 2 x 14 = 28 kN
NEd = 2008 kN
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Figure 10: Results summary
Column moments
Member stiffness:
kBC= 1.07 x 10-3kcol= 0.53 x 10-3
k = [0.71 + 1.07 + (2 x 0.53)]10-3 = 2.84 x 10-3
333
1071.0612
7.03.0
2
1
122
1
2
AB
AB
L
bhk
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Distribution factor for column =
Fixed end moments at B are:
F.E.MBA =
F.E.MBC =
Column moment MEd = 0.19 (432 72) = 68.4 kNm
At the 3rd floor
k = (0.71 + 1.07 + 0.53) 10-3 = 2.31 x 10-3
Column moment MEd =
19.084.2
53.0
k
kcol
kNm43212
6144 2
kNm7212
454 2
kNm6.82)72432(31.2
53.0
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400
300
4H25 4H16
H8 at 300
H8 at 300
Ground to 1st
Floor 1st
to 3rd
Floor
Figure 10: Column reinforcement details
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BIAXIAL BENDING
The effects of biaxial bending may be checked using Expression
(5.39), which was first developed by Breslaer.Where:
Medz,y = Design moment in the respective direction including secondorder effects in a slender column
MRdz,y = Moment of resistance in the respective directionA = 2 for circular and elliptical sections; refer to Table 1 for
rectangular sections
NRd =Acfcd +Asfyd
Table 1: Value of a
for a rectangular
section
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Either or
Where ey and ez are the first-order eccentricities in thedirection of the section dimensions b and h respectively.
(a) If then the increased single axis design
moment is
(b) if then the increased single axis design
moment is
2.0/ b
e
h
e yz 2.0/ b
e
b
ezy
,'' b
M
h
M yz
yzz xMbhMM'''
,'' b
M
h
M yz
zyy xMbhMM'''
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The dimension hand bare defined in Figure 11 and the
coefficient is specified as:ck
Ed
bhf
N 1
Figure 11: Section with biaxial
bending