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1. Column design as Per BS
8110-1:1997
PHK/JSN

2. Contents :-
 General Recommendations of the code
 Classification of columns
 Effective Length of columns & Minimum
eccentricity
 Design Moments in Columns
 Design

3. General Reco’s of the code
 γm for concrete 1.5, for steel 1.05
 Concrete strength – CUBE STRENGTH
 Grades of steel Fe250 & Fe460
 Primary Load combination 1.4DL+1.6LL
 E of concrete Ec = 5.5√fcu/ γm 10% less than IS
 Ultimate stress in concrete 0.67fcu/ γm
 Steel Stress-strain curve – Bilinear
 E of steel 200 kN/mm2

4. Classification of columns
SHORT – both lex/h and ley/b < 15 for braced columns
< 10 for unbraced columns
BRACED - If lateral stability to structure as a whole is provided by walls
or bracing designed to resist all lateral forces in that plane.
else – SLENDER
Cl.3.8.1.5else – UNBRACED

5. Effective length &minimum eccentricity
Effective length le = ßlo ß – depends on end condition
at top and bottom of
column.
emin = 0.05 x dimension of column in the plane of bending ≤ 20 mm

6. Deflection induced moments in Slender columns
Madd = N au where au = ßaKh
ßa = (1/2000)(le/b’)2
K = (Nuz – N)/(Nuz – Nbal) ≤ 1
Nuz = 0.45fcuAc+0.95fyAsc
Nbal = 0.25fcubd
Value of K found iteratively
Contd..

7. Contd..
Design Moments in Braced columns :-
 Maximum Design Column Moment Greatest of
a) M2
b) Mi+Madd Mi = 0.4M1+0.6M2
c)M1+Madd/2
d) eminN
Columns where le/h exceeds 20 and only Uniaxially bent Shall be
designed as biaxially bent with zero initial moment along other axis.

8. Braced and unbraced columns

10. Design Moments in UnBraced columns :-
The additional Moment may be assumed to occur at whichever
end of column has stiffer joint. This stiffer joint may be the
critical section for that column.
Deflection of all UnBraced columns in a storey
auav for all stories = Σ au/n

12. Design Moments in Columns
Axial Strength of column N = 0.4fcuAc + 0.8 Ascfy
Biaxial Bending Increased uniaxial moment about one axis
Mx/h’≥ My/b’ Mx’ = Mx + ß1 h’/b’My
Mx/h’≤ My/b’ My’ = My + ß1 b’/h’Mx
 Where ß1 = 1- N/6bhfcu (Check explanatory hand book)
 Minimum Pt =0.4% Max Pt = 6%

13. Shear in Columns
 Shear strength vc’ = vc+0.6NVh/AcM
 To avoid shear cracks, vc’ = vc√(1+N/(Acvc)
 If v > vc’, Provide shear reinforcement
 If v ≤ 0.8√fcu or 5 N/mm²

14. Design – Construction of Interaction Curve
A1
A2
Section Stress Strain
Distribution of stress and strain on a Column-Section
d1
d h
0.5h
f1
f2
M
N x
0.9x
ε1
ε2
0.67fcu/γm
0.0035

15. Equilibrium equation from above stress block
N = 0.402fcubx + f1A1 +f2A2
M =0.402fcubx(0.5h-0.45x)+f1A1(0.5h-d1)+f2A2(0.5h-d)
f1 and f2 in terms of E and f1 = 700(x-d+h)/x
f2 = 700(x-d)/x
The solution of above equation requires trial and error method

16. THANK YOU