Column design as Per BS 8110-1:1997 PHK/JSN Contents :     General Recommendations of the code Classification of columns Effective Length of columns & Minimum eccentricity Design Moments in Columns Design 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 10% less than IS Ultimate stress in concrete 0.67fcu/ gm Steel Stress-strain curve – Bilinear E of steel 200 kN/mm2 Classification of columns SHORT – both lex/h and ley/b < 15 for braced columns < 10 for unbraced columns else – SLENDER 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 – UNBRACED Cl.3.8.1.5 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 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 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. Braced and unbraced columns 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 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% 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 Design – Construction of Interaction Curve 0.67fcu/gm A1 d1 e1 f1 d 0.5h 0.9x M h A2 Section 0.0035 N f2 Stress x e2 Strain Distribution of stress and strain on a Column-Section 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 THANK YOU