IS:456 COLUMN DESIGN SANJIB DAS

BRACED AND UNBRACED COLUMNS

To adjudge braced or unbraced, one is to determine the stability index ‘Q’ of a storey

in a framed multistorey structure as, u

uu

Hsh

PQ

ΣPu = Sum of axial loads in all columns in the storey

hs = height of storey

Δu = elastically found first order lateral deflection of the storey

Hu = Total lateral force acting within the storey

In the absence of bracing elements the flexibility or drift per storey per unit shear is

given by

bL/

bI

beamcE12

2sh

sh/cI01ccE12

2sh

Hu

u

Where Σ Ic = sum of second moment of areas of all columns in the storey in the plane

under consideration.

b

b

L

I = sum of ratios of second moment of area to span of all floor members in the

storey in the plane under consideration.

Ec = modulus of elasticity of concrete.

In the above equation of ‘Q’, it is assumed that the point of contra flexure of all beams

and all columns are at mid span and mid height

If bracing elements such as trusses, shear walls and in fill walls are used, the value of

oHo will reduce considerably.

If Q < 0.04 the columns are treated as no sway or braced columns if Q > 0.04 the column

is to be treated as sway column or unbraced column.

IS: 456 – 2000 Fig.26 and Fig.27 (reproduced here as Fig.2 and Fig.3) are to be used to

get lef

lK , in terms of

1 and

2 .

Where

bL

bI

sh/cI

sh/cI1 for the joint at the top end of the column and

bL

bI

sh/cI

sh/cI2 for the joint at the bottom end of the column.

Fig.2 and Fig.3 indicate the values of K for braced and unbraced columns respectively.

IS:456 COLUMN DESIGN SANJIB DAS

BRACED AND UNBRACED COLUMNS

IS:456 COLUMN DESIGN SANJIB DAS

BRACED AND UNBRACED COLUMNS