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

Contents :-General Recommendations of the codeClassification of columnsEffective Length of columns & Minimum eccentricityDesign Moments in ColumnsDesign

General Recos of the codegm for concrete 1.5, for steel 1.05Concrete strength CUBE STRENGTHGrades of steel Fe250 & Fe460Primary Load combination 1.4DL+1.6LLE of concrete Ec = 5.5fcu/ gm 10% less than ISUltimate stress in concrete 0.67fcu/ gmSteel Stress-strain curve Bilinear E of steel 200 kN/mm2

Classification of columnsSHORT 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 SLENDERCl.3.8.1.5else UNBRACED

Effective length &minimum eccentricityEffective length le = lo depends on end condition at top and bottom of column.

emin = 0.05 x dimension of column in the plane of bending 20 mm

Deflection induced moments in Slender columns Madd = N auwhere au = aKha = (1/2000)(le/b)2 K = (Nuz N)/(Nuz Nbal) 1Nuz = 0.45fcuAc+0.95fyAsc Nbal = 0.25fcubd

Value of K found iterativelyContd..

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. Contd..Design Moments in Braced columns :-

Braced and 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 = au/n

Design Moments in UnBraced columns :-

Design Moments in ColumnsAxial Strength of column N = 0.4fcuAc + 0.8 Ascfy

Biaxial Bending Increased uniaxial moment about one axis

Mx/h My/b Mx = Mx + 1 h/bMy

Mx/h My/b My = My + 1 b/hMx

Where 1 = 1- N/6bhfcu (Check explanatory hand book)

Minimum Pt =0.4% Max Pt = 6%

Shear in ColumnsShear strength vc = vc+0.6NVh/AcM

To avoid shear cracks, vc = vc(1+N/(Acvc)

If v > vc, Provide shear reinforcement

If v 0.8fcu or 5 N/mm

Design Construction of Interaction Curve

A1

A2 Section Stress Strain Distribution of stress and strain on a Column-Sectiond1dh0.5hf1f2MNx0.9xe1e20.67fcu/gm0.0035

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 andf1 = 700(x-d+h)/x f2 = 700(x-d)/xThe solution of above equation requires trial and error method

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