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Column design as Per BS 8110-1:1997PowerPoint Presentation

Column design as Per BS 8110-1:1997

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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

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.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

Deflection induced moments in Slender columns

Madd = N au whereau = 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

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.

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

A1

A2

Section Stress Strain

DistributionofstressandstrainonaColumn-Section

0.67fcu/gm

0.0035

e1

d1

f1

0.5h

0.9x

M

d

x

h

N

e2

f2

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

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