Column design
R/f details
Longitudinal steel
1. Minimum of four bars for rectangular column
2. Minimum of six bars for circular column
100As
3. A ≥ 4.0
col
100𝐴𝑠
100𝐴𝑠
≤ 6.0 &
≤ 8.0𝑓𝑜𝑟 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑙𝑦 & ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑙𝑦 𝑐𝑎𝑠𝑡 𝑐𝑜𝑙𝑢𝑚𝑛𝑠, 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦
𝐴𝑐𝑜𝑙
𝐴𝑐𝑜𝑙
100𝐴𝑠
But at laps
≤ 10.0 𝑓𝑜𝑟 𝑏𝑜𝑡ℎ 𝑡𝑦𝑝𝑒𝑠 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛𝑠
𝐴𝑐𝑜𝑙
As = total area of longitudinal steel
Acol =cross sectional area of the steel
Links
1. Minimum size= ¼ x size of the largest compression bar but not less than 6 mm
2. Maximum spacing = 12 x size of the smallest compression bar
3. The links should be arranged so that every corner bar and alternate bar or
group in an outer layer of longitudinal steel is supported by a link
4. All other bars or groups not restrained by link should be within 150 mm of
restrained bar
5. In circular columns a circular link passing around a circular arrangement of
longitudinal bars is adequate
Design of short columns
Short braced axially loaded columns
Load is perfectly axial.
The ultimate axial resistance, N = 0.4 fcu.Ac + 0.8 fy. Asc
Ac = net area of the concrete
Asc = area of the longitudinal r/f
With small eccentricity, the ultimate resistance, N = 0.4 fcu.Ac + 0.75 fy. Asc
For a rectangular column & to allow for the area of concrete displaced by
the longitudinal r/f, this equation may be modified to,
N = 0.4 fcu.bh + Asc(0.75 fy - 0.4 fcu)
Short braced columns supporting an approximately symmetrical
arrangement of beams
N = 0.35 fcu.bh + Asc(0.7 fy - 0.35 fcu)
Short columns resisting moment & axial forces
The area of longitudinal steel for these columns is determined by,
1. Using design charts or constructing M-N interaction diagrams
2. A solution of the basic design equations
3. An approximate method
Whichever the method used, a column should not be designed for a moment less
than N x emin
emin = lesser value of h/20 or 20 mm
h = overall size of the column cross section in the plane of bending
Design chart & Interaction diagrams
N = Fcc + Fsc + Fs = 0.45fcubs + fscAs’ + fsAs
M = Fcc (h/2 – s/2)
𝑀 = 𝐹𝐶𝐶
ℎ
𝑠
−
2
2
+ 𝐹𝑠𝑐
ℎ
′
−
𝑑
2
ℎ
− 𝐹𝑠 𝑑 − 2
s = 0.9x (depth of the stress block)
use design charts - BS 8110 (Part 3)
Design equations
The symmetrical arrangement of the r/f with As’ = As is suitable for the columns of a
building where the axial loads are the dominant forces. But some members are
required to resist axial forces combined with large bending moments.
N – normal to the section
ℎ
𝑠
𝑁 𝑒 + − 𝑑2 = 0.45𝑓𝑐𝑢 𝑏𝑠 𝑑 − + 𝑓𝑠𝑐 𝐴′𝑠 𝑑 − 𝑑 ′
2
2
This equation can be solved to give a value for 𝐴′𝑠 .
Equilibrium of the axial forces,
𝑁 = 0.45𝑓𝑐𝑢 𝑏𝑠 + 𝑓𝑠𝑐 𝐴′𝑠 + 𝑓𝑠 𝐴𝑠
1. Select a depth of neutral axis x
2. Determine 𝜀𝑠𝑐 & 𝜀𝑠
3. Further values of x may be selected and above steps repeated until a minimum value
for 𝐴′𝑠 + 𝐴𝑠 𝑖𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑.
Simplified design method
As an alternative to the rigorous method, approximate method may be used
with 𝑒 ≥
ℎ
− 𝑑2
2
The moment M and the axial force N are replaced by an increased moment Ma plus a
compressive force N acting through the tensile steel As.
Hence, the member is designed as a doubly r/f section
𝑀𝑎 = 0.156𝑓𝑐𝑢 𝑏𝑑2 + 0.95𝑓𝑦 𝐴′𝑠 𝑑 − 𝑑 ′
0.95𝑓𝑦 𝐴𝑠 = 0.201𝑓𝑐𝑢 𝑏𝑑 + 0.95𝑓𝑦 𝐴′𝑠
The area of As calculated is reduced by amount N/0.95fy
Biaxial bending of short columns (cl. 3.8.4.5)
 For most of the columns, biaxial bending will not govern the design.
 Internal and edge columns will not usually cause large moments in both directions.
 Corner columns may have to resist significant bending about both axes, but axial
loads are usually small and a design similar to the adjacent edge column is
generally adequate.
Simplified method
Column subjected to an ultimate load N & moments Mx and My about the xx and yy
axes, respectively.
If Mx/h'≥ My/b'
increase single axis
design moment
Mx'= Mx+βh'b' My
If Mx/h'< My/b'
increase single axis
design moment
My'= My+βb'h' Mx
Design of slender column
•If the slenderness ratio about either axis
>15 for a braced column
Slender column
>10 for an unbraced column
•There is a general restriction on maximum slenderness, lo ≤ 60b'
•For an unbraced column, lo ≤100b’2/h'
lo = clear distance between end restraints
b'and h'-smaller and larger dimensions of the column section.
A slender column must be designed for an additional moment caused by its
curvature at ultimate condition.
Mt = Mi + Madd = Mi + N*au
Mi = initial moment in the column,
Madd = moments caused by the deflection of the column
au = deflection of the column
𝑎𝑢 = 𝛽𝑎 𝑘ℎ
1
𝑙𝑒 2
𝛽𝑎 =
2000 𝑏′
b’= smaller dimension
K= reduction factor to allow for the fact that the deflection must be less when there is a
large proportion of the column section is compression.
𝑁𝑢𝑧 − 𝑁
𝑘=
≤ 1.0
𝑁𝑢𝑧 − 𝑁𝑏𝑎𝑙
Nuz=ultimate axial load 𝑁𝑢𝑧 = 0.45𝑓𝑐𝑢 𝐴𝑐 + 0.95𝑓𝑦 𝐴𝑠𝑐
Nbal= axial load at balanced failure 𝑁𝑏𝑎𝑙 = 0.25𝑓𝑐𝑢 𝐴𝑐
In order to calculate k, the area Asc of the columns r/f must be known
 Hence, a trial and error approach is necessary
 Taking initial conservative value of k=1.0
 Value of k are also marked on the column design chart
Braced slender column
Maximum additional moment Madd occurs near the mid height of column.
At this location 𝑀𝑖 = 𝑜. 4𝑀1 + 0.6𝑀2 ≥ 0.4𝑀2
M1 = smaller initial end moment due to design ultimate load
M2 = larger initial end moment
For the usual case with double curvature of a braced column, M1 taken as negative &
M2 as positive.
𝑀𝑎𝑑𝑑
Final design moment > 𝑀2 , 𝑀1 + 𝑀𝑎𝑑𝑑 , 𝑀1 +
𝑜𝑟
2
ℎ
𝑁 ∗ 𝑒𝑚𝑖𝑛 𝑤𝑖𝑡ℎ 𝑒𝑚𝑖𝑛 <
𝑜𝑟 20 𝑚𝑚
20
Ratio of the lengths of the sides ≤ 3
Slenderness ratio
𝑙𝑒
ℎ
𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛 𝑏𝑒𝑛𝑡 𝑎𝑏𝑜𝑢𝑡 𝑖𝑡𝑠 𝑚𝑎𝑗𝑜𝑟 𝑎𝑥𝑖𝑠 ≤ 20