A slender column is a column that is vulnerable to buckling which prevents it from harnessing its maximum strength. Several codes of practice have devised diverse methods to analyse this type of column with focus on stimulating its inherent instability by augmenting the moments from the general firstorder analysis with additional moment. This will enable the column to be designed and sized appropriately for the most onerous case throughout its service life.
The Eurocode 2 also has several methods of analysing a slender column, however the method favored in this article shall be the nominal curvature method as discussed in clause 5.8.8 of the code. Hence, this article presents a worked example of design of slender column to Eurocode 2 using the nominal curvature method.
Worked Example
The column that shall be designed is column BD that has already been analysed here for moment as part of a frame. Since the column has already been evaluated to be subjected to a moment of 42.302KNm, we are left to evaluate the axial force acting on the column.
The axial force of a column in a structural frame may be calculated on the assumption that beams and slabs transmitting force into the particular column are simply supported. By this, the reaction from the column can be calculated which in turn serves as the load on the column. The axial force on the column BD is calculated below.
Computation of Axial load
For easy computation of axial load, the analysed frame is reproduced below:
Since we are computing for the axial load, the maximum ultimate load is used all through as this gives the unfavorable axial load possible.
Third Floor Load take down
Characteristic permanent load on beam (gk) = 23KN/m
Characteristic variable load on beam (qk) = 36KN/m
Maximum ultimate Load acting on beam = 1.35(23) + 1.5(36) = 85KN/m
Load from third floor to column BD = (6/2 + 3.5/2) X 85 = 403.75KN
Second Floor and First Floor Load take down
Characteristic permanent load on beam (gk) = 100KN/m
Characteristic variable load on beam (qk) = 40KN/m
Maximum ultimate Load acting on beam = 1.35(40) + 1.5(100) = 195KN/m
Load from second floor to column BD = (6/2 + 3.5/2) X 195 = 926.25KN
Load from first floor to column BD = (6/2 + 3.5/2) X 195 = 926.25KN
Axial Load on Column BD Total Axial load on column BD = 403.75 + 926.25 + 926.25 = 2256.25KN
Steps in designing the column
Having calculated the axial load on the column, we can now proceed with the design of the column. However, before embarking on the proper design, let us write down the loads and other design parameters necessary for the column design.
Moment on the column (M) = 42.302KNm
Axial load on the column (N) = 2256.25KN
Length (l) = 5m
Breadth (b) = 250mm
Height (h) = 250mm
fck = 25N/mm2
cover = 25mm
1. Check the Slenderness of the column.
The slenderness ratio of the column shall be compared against the limiting slenderness, if the slenderness ratio is greater than the limiting slenderness then the column is declared “slender”. The column is declared “nonslender” if otherwise. This is shown in the below steps
a) Calculate the effective length of the column
Since the column is braced, the Effective length (lo) is calculated thus:
$ _{\,\,l_0=0.5l\sqrt{\left( \text{1}+\,\,\frac{K_1}{\text{0.45}+\,\,K_1} \right) \left( \text{1}+\,\,\frac{K_2}{\text{0.45}+\,\,K_2} \right)}} $
K_{1} and K_{2} are the relative flexibilities of the top and bottom ends respectively
K_{1}= $ \frac{\frac{\varSigma I_c}{l}}{\frac{\varSigma 2EI_b}{l}} $
$ \frac{\frac{\frac{bh^3}{12}}{l}}{2X\frac{\frac{bh^3}{12}}{l}\,\,+\,\,2X\frac{\frac{bh^3}{12}}{l}} $
$ \frac{\frac{\frac{250 X 250^3}{12}}{5000}}{2X\frac{\frac{225 X 500^3}{12}}{6000}\,\,+\,\,2X\frac{\frac{225 X 500^3}{12}}{3500}} $
= 65104.2/(781250 + 1339285.7)
= 0.031
We calculate K_{2}
The bottom end of column BD is assumed fixed in the subframe which implies the flexibility is 0. However, in accordance to Eurocode 2 guidance K2 is taken as 0.1 since fully rigid restraint is rare in actual practice (check 5.8.3.2 (3))
Hence; K2= 0.1
$ _{\,\,l_0=0.5l\sqrt{\left( \text{1}+\,\,\frac{0.031}{\text{0.45}+\,\,0.031} \right) \left( \text{1}+\,\,\frac{0.1}{\text{0.45}+\,\,0.1} \right)}} $ = 2803.2mm
b) Calculate the radius of gyration
$ r\,\,=\,\,\sqrt{\frac{I}{A}}\,\,=\,\,\sqrt{\frac{\frac{bh^3}{12}}{b\,\,X\,\,h}}\,\,=\,\,\sqrt{\frac{\frac{\text{250}X\,\,250^3}{12}}{\text{250}X\,\,250}}\,\,\,\,=\,\,72.17 $
c. Calculate the slenderness ratio of the column
λ = lo/r = 2803.2mm/72.17 = 38.84
d. Calculate the limiting slenderness
λ_{lim }= (20 X A X B X C/√n)
A = 0.7, B = 1.1, C = 0.7 (All these assumptions are based on alternative suggestion by the code, check clause 5:8:3:1)
n = N_{Ed} /A_{c}f_{cd }
_{= n = 2256.25 X 10^3 / (250 X 250 X 0.85 X 25/1.5) = 1.6}
λ_{lim }= (20 X 0.7 X 1.1 X 0.7/√1.6) = 16.4
Since λ_{lim} (16.4) is less than ⅄ (38.84) then the column is slender and secondorder moment is required.
2. Calculate the secondorder moment
Since the column is slender, secondorder moment due to slenderness is necessary. Secondorder moment (M_{2} ) = N_{ED} + e_{2}
where: N_{ED} is the design axial force.
e_{2} is the deflection of the column
Calculate deflection due to slenderness
e_{2}= 1/r l_{o}^{2} / c
c = 10 (see 5:8:8:2 (4) of EC2)
e_{2}= 1/r x 2803.2^{2}/ 10

Calculate the curvature (1/r)
1/r = Kr Ko 1/r_{o}
Axial load correction factor (Kr)
K_{r} = (nu – n)/ (nu n_{bal}) ≤ 1
nu = 1 + ω
ω = A_{s}f_{yd}/A_{c}f_{cd}
Assume As to be 4y16 = 799.7mm2
f_{yd} = 500/1.15 = 434.8
f_{cd} = 0.85 x 25/1.5 = 14.17
ω = (799.7 x 434.8)/ (250 x 250 x 14.17) = 0.393
nu = 1+ 0.393 = 1.393
nbal = 0.4
n = Ned/Ac x fcd = 2256.25/ (250 x 250 x 0.85 x 25/1.5 ) = 0.003
K_{r} = (1.393 – 0.003)/ (1.393 0.4) = 1.4
Since 1.4 > 1, then Kr is taken as 1
Creep Factor (Kϴ)
Kϴ = 1+ B ϕ_{ef} ≥ 1
B = 0.35 + fck/200 – ⅄/150
= 0.35 + 25/200 – 38.84/150 = 0.216
effective creep ratio (ϕ_{ef}) = 2.14 (This conservative value is gotten by making ϕef the subject of formular in Equ 5.13N of Eurocode 2, when A is assumed to be 0.7)
Kϴ = 1+ 0.216 x 2.14 = 1.462
Calculate 1/r_{o}
1/r_{o} = fyd/(Es x 0.45d)
434.8/ (200×10^3 x 0.45 x 250) = 0.00002
curvature (1/r) = 1 x 1.462 x 0.0002 = 0.00003
deflection due to slenderness (e_{2}) = 0.00003 x 2803.2^2 x 10 = 26.69mm
M_{2 }= 2256.25 x 26.69/1000 = 60.2KNm
3. Compute the design moment (M_{Ed})
The design moment is the greatest of:
 max (M_{o2}, M_{o1}) + (N_{ED}) x max (e)
2 M_{oe}+ M_{2}
3. M_{o1}+ M_{2}/2
Where:
M_{o2} = 42.302KN
M_{o1} = 0
M_{oe}= 0.4 M02 + 0.6 M01 = 16.9KNm
M_{2} = 60.2KNm
 max (M_{o2}, M_{o1}) + (N_{ED}) x max (e)
= 42.302 + 2256.25 x 20/1000 = 87.427KNM
2. M_{oe}+ M_{2} = 16.9 + 60.2 = 77.13KNm
3. M1+ M_{2}/2 = 0 + 60.2/2 = 30.1KNM
The design moment is 87.427KNm
4. Use the design Chart to determine the area of reinforcement required
In order to use the design chart, few parameters have to be calculated
Determine N_{ED}/bhfck
N_{ED}/bhfck = 2256.25X 10^3 /250 X 250 X 25 = 1.44
Determine M_{ED}/bh²fck
M_{ED}/bh²fck = 87.427 X 10^6/250 X 250^2 X 25 = 0.22
Calculate d’/h
d’/h determines the exact graph that shall be used to determine the reinforcement area.
d’ = c + d/2 = 25 + 16/2 = 33
d’/h = 33/250 = 0.13
This is approximated to 0.1 and the design chart peculiar to d’/h = 0.1 is used. Fig 9b of “How to design concrete Structures to Eurocode 2” matches this. The chart is reproduced below.
From the design chart, the intersection of M_{ED}/bh²fck = 0.22 and N_{ED}/bhfck = 1.44 is outside the boundary of the curve. The curve does not even have the value of 1.44 2hich corresponds to N/bhfck as it ends at 1.3. This means the load and moment acting on the column exceeds its capacity. To circumvent this, we shall increase the height of the column (h) to 350mm. By this the column crosssection becomes 250 x 350mm
We redetermine the parameters
 Determine N_{ED}/bhfck
N_{ED}/bhfck = 2256.25 x 10^3 /250 x 350 x 25 = 1.0
2. Determine M_{Ed}/bh2fck
M_{Ed}/bh2fck = 87.427 x 10^6/250 x 350^2 x 25 = 0.11
3. Calculate d’/h
d’/h determines the exact graph that shall be used to determine the reinforcement area.
d’ = c + d/2 = 25 + 16/2
= 33
d’/h = 33/350 = 0.09
d’/h is approximated to 0.1 and the same design chart is used.
M_{Ed}/bh²fck = 0.11 and N_{ED}/bhfck = 1.0 intersects at at 0.8 on the design chart. This is shown by the red mark drawn otn the design chart as displayed below.
Hence from the design chart:
A_{s}f_{yk}/bhf_{ck }= 0.8
A_{s} = 0.8 x bhf_{ck }/f_{yk}
A_{s} = 0.8 x 250 x 350 x 25/ 500
A_{s }= 3500mm2
Provide 8Y25 (3905mm2) as longitudinal bars.
5. Check minimum required reinforcement
A_{smin} = 0.1Ned/fyd ≥ 0.002Ac
= 518.93mm2
Since A_{smin} (518.93mm2) < As (3905mm2), minimum area of reinforcement is satisfied.
6. Check maximum area reinforcement
A_{smax}= 0.4 Ac = 0.4 x 250 x 350 = 3500mm2
Since A_{s} (3905mm2) is greater than A_{smax} (3500), we will increase the dimension of the column to satisfy maximum area requirement. We will increase the breadth (b) of the column to 300mm so that the column crosssection becomes 300 x 350mm.
Recheck maximum area of reinforcement
A_{smax} = 0.4 Ac = 0.4 x 300 x 350 = 4200mm2
Since A_{s} (3905mm2) is less than A_{smax} (4200mm2), maximum area of reinforcement requirement is now satisfied.
7. Determine the link diameter and spacing
1. Link diameter
The link diameter is the greater of;
 ¼ x maximum diameter of longitudinal bar
 6mm
¼ x maximum diameter of longitudinal bar = ¼ x 25 = 6.25mm
Hence use 10mm as the link diameter.
2. Link Spacing
The link spacing should be the least of;
 20 x minimum diameter of longitudinal bar
 lesser of the column dimension
 400mm
 20 x minimum diameter of longitudinal bar = 20 x 25 = 500mm
lesser of the column dimension = 300m
Since 300mm is lesser than 500mm and 400mm, 300mm governs the spacing
Provide 10mm links at 300mm spacing.
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