column design - as per bs

8/13/2019 Column Design - As Per BS

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Column design as Per BS

8110-1:1997

PHK/JSN



Contents :- General Recommendations of the code

Classification of columns

Effective Length of columns & Minimumeccentricity

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√f cu/ gm 10% less than IS

Ultimate stress in concrete 0.67f cu/ 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

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



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



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 = 0.45f cu Ac+0.95fyAsc

Nbal = 0.25f cubd

Value of K found iteratively

Contd..



Contd..

Design Moments in Braced columns :-

Maximum Design Column Moment Greatest of

a) M2

b) Mi+Madd Mi = 0.4M1+0.6M2

c)M1+M

add/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 Columns

Axial Strength of column N = 0.4f cu Ac + 0.8 Ascf y

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/6bhf cu (Check explanatory hand book)

Minimum Pt =0.4% Max Pt = 6%



Shear in Columns

Shear strength v c ’ = v c +0.6NVh/AcM

To avoid shear cracks, v c ’ = v c √(1+N/(Acv c )

If v > vc’, Provide shear reinforcement

If v ≤ 0.8√f cu or 5 N/mm²



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

e1

e2

0.67f cu /gm0.0035



Equilibrium equation from above stress block

N = 0.402f cubx + f 1 A1 +f 2 A2

M =0.402f cubx(0.5h-0.45x)+f 1 A1(0.5h-d1)+f 2 A2(0.5h-d)

f 1 and f 2 in terms of E and f 1 = 700(x-d+h)/x

f 2 = 700(x-d)/x

The solution of above equation requires trial and error method



THANK YOU

column design - as per bs

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