design column

8/13/2019 design column

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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

design column

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