Design of Slender Columns According to EC2

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. Slender columns are likely to fail by buckling than by crushing.

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 EN 1992-1-1:2004 (EC2).

Slender Columns Requirements 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_sf_yd / (A_cf_cd); mechanical reinforcement ratio;
A_s is the total area of longitudinal reinforcement
n = N_Ed / (A_cf_cd); relative normal force
r_m = M₀₁/M₀₂; moment ratio
M₀₁, M₀₂ are the first order end moments, |M₀₂| ≥ |M₀₁|

If the end moments M₀₁ and M₀₂ 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 r_m 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.

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/mm², f_yk = 460 N/mm², Concrete cover = 35mm

Second moment of area of beam AB = (0.3 × 0.6³)/12 = 0.0054 m⁴
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 × 0.6³)/12 = 0.0054 m⁴
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, k₁ = 0.00288/0.0018 = 1.6

Since the minimum value of k₁ and k₂ is 0.1, adopt k₁ as 1.6. Let us take k₂ 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
λ_lim = (20.A.B.C)/√n
A = 0.7
B = 1.1
C = 1.7 – M₀₁/M₀₂ = 1.7 – (-210/371) = 2.266

n = N_Ed / (A_c f_cd)
N_Ed = 3500 × 10³ N
A_c = 300 × 600= 180000 mm²
f_cd = (α_cc f_ck)/1.5 = (0.85 × 25)/1.5 = 14.167 N/mm²
n = (3500 × 10³) / (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 l₀/2) = [(1/200) × (6071/2)] = 15.1775 mm
e_iN_Ed = 15.1775 × 10^-3 × 3500 = 53.121 KNm

First order end moment
M₀₁ = M_Bot + e_iN_Ed = -210 + 53.121 = -156.879 KNm
M₀₂ = M_Top + e_iN_Ed = 371 + 53.121 = 424.121 KNm

Equivalent first order moment M_0Ed
M_0Ed = (0.6M₀₂ + 0.4M₀₁) ≥ 0.4M₀₂ = 0.4 × 424.121 = 169.648 KNm
M_0Ed= (0.6 × 424.121 – 0.4 × 156.879) = 191.721 KNm

Nominal second order moment M₂
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/r₀ = ε_yd/(0.45 d)
ε_yd = f_yd/Es = (460 /1.15) / (200 × 10³) = 0.002
1/r₀ = 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 K_r = 0.8
1/r = K_r.Kϕ. 1/r₀ = 0.8 × 1.2099 × 8.2457 × 10^-6 = 7.981 × 10^-6

e₂ is the deflection = (1/r) (l₀²) / 10 = 7.981 × 10^-6 × 6071² / 10 = 29.415 mm

M₂ = N_Ed.e₂ = 3500 × 29.415 × 10^-3 = 102.954 KNm

Design Moment M_Ed
M_Ed = maximum of {M_0Ed + M₂; M₀₂; M₀₁ + 0.5M₂}
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
d₂ = C_nom + ϕ/2 + ϕ_links
d₂ = 35 + 16 + 10 = 61 mm

d₂/h = 61/600 = 0.1016
Reading from chart No 2; d₂/h = 0.10;

M_Ed/(f_ck bh²) = (424.121 × 10⁶) / (25 × 300 × 600²) = 0.1571
N_Ed/(f_ck bh) = (3500 × 10³) / (25 × 300 × 600) = 0.777

From the chart, (A_sF_yk)/(bhf_ck) = 0.53
Area of longitudinal steel required (As) = (0.53 × 25 × 300 × 600)/460 = 5185 mm²

Provide 6Y32 + 2Y20 (A_Sprov = 5452 mm²)

A_s,min = (0.1 N_Ed)/f_yd = (0.1 × 3500 × 1000) / 400 = 875mm², 0.002bh = 0.002 × 300 × 600 = 360 mm²
A_s,max = 0.04bh = 0.04 × 300 × 600 = 7200 mm²

Links
Minimum size = 0.25ϕ = 0.25 × 32 = 8mm < 6mm
We are adopting Y10mm as links
Spacing adopted = 300mm less than {b, h, 20ϕ, 400mm} Provide Y10 @ 300 mm links

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