REINFORCED CONCRETE

COLUMN

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 assumes

that the column dimensions have previously

been determined during conceptual design or

by using quick design methods. Column sizes

should not be significantly different from those

obtained using current practice.

Column can be classified as:

Braced – where the lateral loads are resisted by shear wall or other form of bracing capable of transmitting all horizontal loading to the foundations; and

Unbraced – where horizontal load are resisted by the frame action of rigidity connected columns, beams and slabs.

With a braced structure, the axial forces and moments in thecolumns are caused the vertical permanent and variationaction only;

With an unbraced structure, the loading arrangement whichinclude the effects of lateral load must also be considered

Loading and MomentsFor a braced structure, the critical arrangement of the ultimate load

is 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 Qk1.35 Gk

Figure. 2: A critical loading arrangement

Slenderness ratio of a columnEurocode 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 (λ) is less than the slenderness limit (λlim), then second order effects may be ignored.

The slenderness ratio λ of a column bent about an axis is given by:

Where:

lo - effective height of the column

i - 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

l

il 00

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

i) For braced member

ii) For unbraced member the larger of:

And

2

2

1

10

45.01

45.015.0

k

k

k

kll

kk

xkkll

1

210 101

2

2

1

10

11

11

k

k

k

kll

Where

k1 and k2 relative flexibility of the rotational restrains at end ‘1’ and ‘2’ of the

column respectively. At each end k1 and k2 can be taken as:

k = column stiffness/ Σ beam stiffness

=

=

For a typical column in a symmetrical frame with span approximately equal

length, k1 and k2 can be calculated as:

beam

column

lEI

lEI

)/(2

)/(

beam

column

lI

lI

)/(2

)/(

beam

column

lI

lIkkk

)/(

)/(

4

121

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 (λ) is less than the slenderness limit (λlim), then

second order effects may be ignored. Slenderness,

λ = lo/i

where i = radius of gyration

Slenderness limit:

Where:

A = 1/(1+0.2φef) (if φef 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

** Of the three factors A, B and C, C will have

the largest impact on λlim and is the simplest to

calculate. An initial assessment of λlim can

therefore be made using the default values

for A and B, but including a calculation for

C. Care should be taken in determining C

because the sign of the moments makes a

significant difference. For unbraced

members C should always be taken as

0.7.

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

Effective column height lo

Lcol = 3000 – 500 = 2500 mm

Icol = 400 x 3003/12 = 900 x 106 mm4

Ibeam = 300 x 5003/12 = 3125 x 106 mm4

k1 = k2 = = 0.115

= 0.6 x 2.5 = 1.50 m

Slenderness ratio λ:

Radius of gyration, i =

36

36

104/1031252(2

105.2/10900

/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

Slenderness ratio,

For braced column,

> 17.32

)/(/2.26lim cdcED fAN

866.05.1/2585.0300400

101280)/(

3

xxx

xfAN cdcED

25.30866.02.26lim x

32.176.86

105.1 3

0 x

i

l

REINFORCEMENT DETAILS

Longitudinal steel

A 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

Links

The diameter of the transverse reinforcement should not

be less than 6 mm or one quarter of the maximum

diameter 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 longitudinal bars.

■ the lesser dimension of the column.

■ 400 mm.

DESIGN MOMENT

Non Slender Column

MED = max {Mo2, Mmin}

Where:

Mo2 = M + NED . ei

M = max {Mtop, Mbottom}

ei = lo/400

Mmin = NED x e0

eo = max{h/30, 20 mm)

Slender ColumnFor braced slender column, the design bending moment is illustratedin 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 NEd

M02 = max {|Mtop|, |Mbottom|} + ei NEd

e0 = max {h/30, 20 mm}

ei = lo/400

Mtop, Mbottom = Moments at the top and bottom of the column

Figure 5: Design bending moment

M0e = 0.6 M02 + 0.4 M01 ≥ 0.4 M02

M01 and M02 should be positive if they give tension on the same side.

M2 = NEd x e2 = The nominal second order moment

Where:

NEd = the design axial load

e2 = Deflection due to second order effects =

lo = effective length

c = a factor depending on the curvature distribution,

normally

1/r = the curvature = Kr . Kφ . 1/r0

c

l

r

01

102

Kr = axial load correction factor =

Where, n =

Kφ = creep correction factor =

Where:

φef = effective creep ratio =

= 0, if (φ < 2, M/N > h, 1/r0 < 75)

β = 0.35 + fck/200 – λ/150

1/r0 =

A non-slender column can be designed ignoring second order 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/(

SHORT COLUMN RESISTING MOMENTS AND

AXIAL FORCES

The area of longitudinal reinforcement is determined based on:

Using design chart or construction M-N interaction diagram.

A solution a basic design equation.

An approximate method

A column should not be designed for a moment less than NEd x emin where emin has a grater value of h/30 or 20 mm

DESIGN CHARTThe basic equation:

NEd – design ultimate axial load

MEd – design ultimate moment

s – the depth of the stress block = 0.8x (Figure 6)

A’s – the area of longitudinal reinforcement in the more highly

compressed face

As – the area of reinforcement in the other face

fsc – the stress in reinforcement A’sfs – the stress in reinforcement As, negative when tensile

sscccEd FFFN

sssscck AfAfbsf '567.0

2'

222

hdFd

hF

shFM sscccEd

Figure 6: Column section

Figure 7: Example of

column design chart

Two expressions can be derived for the area of steel required, (based on a rectangular stress block, see Figure 8) one for the axial loads and the other for the moments:

AsN/2 = (NEd – fcd b dc) / [(σsc – σst) γc]

Where:

AsN/2 = Area of reinforcement required to resist axial load

NEd = Axial load

fcd = Design value of concrete compressive strength

σsc (σst) = Stress in compression (and tension) reinforcement

b = Breadth of section

γc = Partial factor for concrete (1.5)

dc = Effective depth of concrete in compression

= λx ≤ h

λ = 0.8 for ≤ C50/60

x = Depth to neutral axis

h = Height of section

AsM/2 = Total area of reinforcement required to resist moment

= [M – fcd b dc(h/2 – dc/2)] / [(h/2–d2) (σsc+σst) γc]

Example:

Figure 8 shows a frame of heavily loaded industrial structure for which the centre column along line PQ are to be designed in this example. The frame at 4m centres are braced against lateral forces and support the following floor loads:

Permanent action, gk - 10 kN/m2

Variable action, qk - 15 kN/m2

Characteristic materials strength are

fck = 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/m

Minimum ultimate load at each floor:

= 4.0 x 1.35gk

= 4.0 x (1.34 x 10)

= 54 kN per meter length of beam

Figure 8: Column structure

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

Figure 10: Results summary

Column moments

Member stiffness:

kBC = 1.07 x 10-3

kcol = 0.53 x 10-3

Σk = [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

Floor NEd (kN)

M (kNm)

𝑁𝐸𝑑 𝑙0400

(kNm)

MEd (kNm)

𝑁𝐸𝑑

𝑏ℎ𝑓𝑐𝑘

𝑀𝐸𝑑

𝑏ℎ2𝑓𝑐𝑘

𝐴𝑠𝑓𝑦𝑘

𝑏ℎ𝑓𝑐𝑘

𝐴𝑠 (mm2)

3rd u.s 540 82.6 3.30 85.90 0.18 0.07 0 240

2nd t.s 743 68.4 3.30 71.70 0.24 0.06 0 240

Column 540

2nd u.s 1274 68.4 6.61 75.01 0.42 0.06 0 240

1st t.s 1468 68.4 6.61 75.01 0.49 0.06 0.10 600

Column 540

1st u.s 2008 68.4 11.75 80.15 0.67 0.07 0.32 1920

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

Design moment allowing for geometric imperfections

MED = M + NEdlo/400

lo = 2.34 ground floor

lo = 1.80m for first and second

19.084.2

53.0

k

kcol

kNm43212

6144 2

kNm7212

454 2

40

0

300

4H25 4H16

H8 at 300

H8 at 300

Ground to 1st Floor 1

st to 3

rd Floor

Figure 10: Column reinforcement details

At the 3rd floor

Σk = (0.71 + 1.07 + 0.53) 10-3 = 2.31 x 10-3

Column moment MEd = kNm6.82)72432(31.2

53.0

BIAXIAL BENDINGThe 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 second order effects in a slender column

MRdz,y = Moment of resistance in the respective direction

a = 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

Either or

Where ey and ez are the first-order eccentricities in the

direction of the section dimensions ‘b’ and ‘h’ respectively.

Where these conditions are not fulfilled, biaxial bending

must be accounted for.

(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/ h

e

b

ezy

,'' b

M

h

M yz

yzz xMb

hMM

'

''

,'' b

M

h

M yz

zyy xMb

hMM

'

''

The dimension h’ and b’ are defined in Figure 11 and the

coefficient β is specified as:

ck

Ed

bhf

N1

Figure 11: Section with biaxial

bending

Design of column for biaxial bending

The column section shown in Figure 4 is to

be designed to resist an ultimate axial load

of 1200 kN plus moment of Mz = 75 kNm

and My = 80 kNm. The characteristic

material strengths are fck = 25 N/mm2 and fyk

= 500 N/mm2.

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