**1.0 Introduction**

Loads from slabs and beams are transferred to the foundations through the columns. In typical cases, columns are usually rectangular or circular in shape. Normally, they are usually classified as short or slender depending on their slenderness ratio, and this in turn influences their mode of failure.

Columns are either subjected to axial, uniaxial, or biaxial loads depending on the location and/or loading condition. Eurocode 2 demands that we include the effects of imperfections in structural design of columns. Column design is covered in section 5.8 of EC2.

**2.0 Column Slenderness in EC2**

Clause 5.8.2 of EN 1992-1-1 deals with members and structures in which the structural behaviour is significantly influenced by second order effects (e.g. columns, walls, piles, arches and shells). Global second order effects are more likely to occur in structures with a flexible bracing system.

Column design in EC2 generally involves determining the slenderness ratio (λ), of the member and checking if it lies below or above a critical value λ_{lim}. If the column slenderness ratio lies below (λ_{lim}), it can simply be designed to resist the axial action and moment obtained from an elastic analysis, but including the effect of geometric imperfections. These are termed first order effects. However, when the column slenderness exceeds the critical value, additional (second order) moments caused by structural deformations can occur and must also be taken into account.

So in general, second order effects may be ignored if the slenderness λ is below a certain value λ_{lim}.

λ_{lim} = (20.A.B.C)/√n —————- (1)

Where:

A = 1/(1 + 0.2ϕ_{ef}) (if ϕ_{ef} is not known, A = 0.7 may be used)

B = 1+ 2ω (if ω is not known, B = 1.1 may be used)

C = 1.7 – r_{m} (if r_{m} is not known, C = 0.7 may be used)

Where; ϕ_{ef} = effective creep ratio (0.7 may be used)

ω = A_{s}f_{yd} / (A_{c}f_{cd}); mechanical reinforcement ratio;

A_{s} is the total area of longitudinal reinforcement

n = N_{Ed} / (A_{c}f_{cd}); relative normal force

r_{m} = M_{01}/M_{02}; moment ratio

M_{01}, M_{02} are the first order end moments, |M_{02}| ≥ |M_{01}|

If the end moments M_{01} and M_{02} give tension on the same side, r_{m} should be taken positive (i.e. C ≤ 1.7), otherwise negative (i.e. C > 1.7). For braced members in which the first order moments arise only from or predominantly due to imperfections or transverse loading rm should be taken as 1.0 (i.e. C = 0.7).

Also, clause 5.8.3.1(2) says that for biaxial bending, the slenderness criterion may be checked separately for each direction. Depending on the outcome of this check, second order effects (a) may be ignored in both directions, (b) should be taken into account in one direction, or (c) should be taken into account in both directions.

**3.0 Solved Example **

Let us consider the structure shown below. The effects of actions on column member BC is as shown below. It is required to design the column using the following data;

f_{ck} = 25 N/mm^{2}, f_{yk} = 460 N/mm^{2}, Concrete cover = 35mm

Second moment of area of beam AB = (0.3 × 0.6^{3})/12 = 0.0054 m^{4}

Stiffness of beam AB (since E is constant) = 4I/L = (4 × 0.0054) / 6 = 0.0036

Second moment of area of column BC = (0.3 × 0.6^{3})/12 = 0.0054 m^{4}

Stiffness of column BC (since E is constant) = 4I/L = (4 × 0.0054)/7.5 = 0.00288

Remember that we will have to reduce the stiffness of the beams by half to account for cracking;

Therefore, k_{1} = 0.00288/0.0018 = 1.6

Since the minimum value of k_{1} and k_{2} is 0.1, adopt k_{1} as 1.6. Let us take k_{2} as 1.0 for base designed to resist moment.

Take the unrestrained clear height of column as 7000mm

l_{o} = 0.5 × 7000√[(1 + (1.6/(0.45 + 1.6))) × (1 + (1.0/(0.45+ 1.0)))] = 6071 mm

Radius of gyration i = h/√12 = 600/√12 = 173.205

Slenderness ratio λ = 6071/173.205 = 35.051

**Critical Slenderness for the x-direction<**/p>

λ_{lim} = (20.A.B.C)/√n

A = 0.7

B = 1.1

C = 1.7 – M_{01}/M_{02} = 1.7 – (-210/371) = 2.266

n = N_{Ed} / (A_{c} f_{cd})

N_{Ed} = 3500 × 10^{3} N

A_{c} = 300 × 600= 180000 mm^{2}

f_{cd} = (α_{cc} f_{ck})/1.5 = (0.85 × 25)/1.5 = 14.167 N/mm^{2}

n = (3500 × 10^{3}) / (180000 × 14.167) = 1.3725

λ_{lim} = (20 × 0.7 × 1.1 × 2.266 )/√1.3725 = 29.786

Since 29.786 < 35.051, second order effects need to be considered in the design

**Design Moments**

M_{Bot} = 210 KNm, M_{Top} = 371 KNm

e_{i} is the geometric imperfection = (θ_{i} l_{0}/2) = [(1/200) × (6071/2)] = 15.1775 mm

e_{i}N_{Ed} = 15.1775 × 10^{-3} × 3500 = 53.121 KNm

**First order end moment**

M_{01} = M_{Bot} + e_{i}N_{Ed} = -210 + 53.121 = -156.879 KNm

M_{02} = M_{Top} + e_{i}N_{Ed} = 371 + 53.121 = 424.121 KNm

**Equivalent first order moment** **M**_{0Ed}

M_{0Ed} = (0.6M_{02} + 0.4M_{01}) ≥ 0.4M_{02} = 0.4 × 424.121 = 169.648 KNm

M_{0Ed}= (0.6 × 424.121 – 0.4 × 156.879) = 191.721 KNm

**Nominal second order moment M**_{2}

Specified concrete cover = 35mm

Diameter of longitudinal steel = 32 mm

Diameter of links = 10 mm

Thus, the effective depth (d) = h – C_{nom} – ϕ/2 – ϕ_{links}

d = 600 – 35 – 16 – 10 = 539 mm

1/r_{0} = ε_{yd}/(0.45 d)

ε_{yd} = f_{yd}/Es = (460 /1.15) / (200 × 10^{3}) = 0.002

1/r_{0} = 0.002/(0.45 × 539) = 8.2457 × 10^{-6}

^{
}β = 0.35 + f_{ck}/200 – λ/150 (λ is the slenderness ratio)

β = 0.35 + (25/200) – (35.051/150) = 0.2413

Kϕ = 1 + βϕ_{ef} ≥ 1.0 (ϕ_{ef} is the effective creep ratio, assume 0.87)

Kϕ = 1 + (0.2413 × 0.87) = 1.2099 ≥ 1.0

Assume Kr = 0.8

1/r = Kr.Kϕ. 1/r_{0} = 0.8 × 1.2099 × 8.2457 × 10^{-6} = 7.981 × 10^{-6}

^{
}e_{2} is the deflection = (1/r) (l_{0}^{2}) / 10 = 7.981 × 10^{-6} × 6071^{2} / 10 = 29.415 mm

M_{2} = N_{Ed}.e_{2} = 3500 × 29.415 × 10^{-3} = 102.954 KNm

**Design Moment M**_{Ed}

M_{Ed} = maximum of {M_{0Ed} + M_{2}; M_{02}; M_{01} + 0.5M_{2}}

M_{Ed} = maximum of {191.721 + 102.954 = **294.675 kNm;** **424.121 kNm;** -156.879 + (0.5 × -102.954) = **-208.356 kNm**}

**Longitudinal Steel Area**

d_{2} = C_{nom} + ϕ/2 + ϕ_{links}

d_{2} = 35 + 16 + 10 = 61 mm

d_{2}/h = 61/600 = 0.1016

Reading from chart No 2; d_{2}/h = 0.10;

M_{Ed}/(f_{ck} bh^{2}) = (424.121 × 10^{6}) / (25 × 300 × 600^{2}) = 0.1571

N_{Ed}/(f_{ck} bh) = (3500 × 10^{3}) / (25 × 300 × 600) = 0.777

_{s}F

_{yk})/(bhf

_{ck}) = 0.53

Area of longitudinal steel required (As) = (0.53 × 25 × 300 × 600)/460 = 5185 mm

^{2}

Provide 6X32 + 2X20 (A_{Sprov} = 5452 mm^{2})

A_{s,min} = (0.1 N_{Ed})/f_{yd} = (0.1 × 3500 × 1000) / 400 = 875mm^{2}, 0.002bh = 0.002 × 300 × 600 = 360 mm^{2}

A_{s,max} = 0.04bh = 0.04 × 300 × 600 = 7200 mm^{2}

^{
}

**Links**

Minimum size = 0.25ϕ = 0.25 × 32 = 8mm < 6mm

We are adopting X10mm as links

Spacing adopted = 300mm less than {b, h, 20ϕ, 400mm} Provide Y10 @ 300 mm links

Thank you for reading

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best example with best explanation!!!

Would like to ask From chart , why (As.fyk)/bhfck= 0.53 , is it not 0.71 from the reading then As=6945mm2

Provide 8×32 + 2×20 (From As=7062mm2)

Would like to ask From the Chart , why is (Asfyk/bhfck )=53 , is it not 71 then As=6945mm2

Provide 8×32 + 2×20 (From given As=7062mm2) ?