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
gm 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/ gm % less than IS
Ultimate stress in concrete 0.67fcu/ gm
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
else – UNBRACED Cl

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
Contd..
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

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 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
Section Stress Strain
Distribution of stress and strain on a Column-Section
0.67fcu/gm
0.0035 d1 e1 f1
0.5h M
0.9x d x h N e2 f2

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