AXIALLY LOADED COLUMN

INTRODUCTION

COLUMN- if a compression member, effective length > 3* least lateral dimension is called Column.

Normally columns are subjected to axial compressive force, but sometimes it may be subject to to moments on one or both the axes.

STRENGTH OF COLUMNS DEPENDS UPON: Shape, size and cross section of column Length

TYPES OF LOADING

AXIALLY LOADED COLUMNS Columns subjected to loads acting along the

longitudinal axis or centroid of column section.

ECCENTRICALLY LOADED COLUMN (uniaxial loading ) Loads which do not act on the longitudinal axis of

the column section

ECCENTRICALLY LOADED COLUMNS (biaxial loading) If the eccentricity is with respect to both the axis (x

and y axis) the column is said to be under biaxial loading.

REINFORCEMENT REQUIREMENT FOR COLUMNS

REQUIREMENTS AS PER IS :456-2000,PG 48,CL.26.5.3 FOR LONGITUDINAL REINFORCEMENT The cross sectional area of longitudinal reinforcement Shall not be less then 0.8% Shall not be more than 6%

Of the gross sectional area of the column

Minimum number of longitudinal bars shall be 4 in rectangular column and 6 in circular columns

The bars shall not be less than 12mm in diameter

Spacing of longitudinal bars along the periphery should not exceed 300mm

TRANSVERSE REINFORCEMENT

The effective lateral support is given by transverse reinforcement either in form of circular rings capable of taking circular tension or by polygon links (lateral ties ) with internal angle not exceeding 135 degree

If the longitudinal bars are not spaced more than 75mm on either sides, transverse reinforcement need only to go near the corner alternate bars are used for the purpose of lateral supports

TRANSVERSE REINFORCEMENT CONT…

If the longitudinal bars are spaced at a distance not exceeding 48times the diameter of the tie, effectively tied in two directions, additional longitudinal bars in between these bars need to be used in one direction by open ties

Longitudinal bars in compression member placed in more than one row, effective lateral support to the longitudinal bars in the inner rows may be assumed to have been provided if *Transverse reinforcement is provided for the outer most row in accordance with fig 7.6(b).No bars of the inner is closer to the nearest compression face then three times the diameter of the largest bar in inner row fig (7.6)

TRANSVERSE REINFORCEMENT

FOR LATERAL TIES the pitch of the lateral ties shall not be more than the least of the following: The least lateral dimension of the compression member 16 times diameter of the longitudinal bar 300mm Diameter The diameter of the lateral links or lateral ties shall not be less than ¼* diameter of the largest longitudinal bar 6mm Which ever is more

FOR HELICAL REINFORCEMENT

PITCH- the pitch of helical reinforcement shall be: ≥ 75mm 1/6* core diameter of column But ≤ 25mm 3* diameter of helix Diameter :the diameter of helical reinforcement shall be,not less than: ¼* diameter of the largest longitudinal bar 6mm

Which ever is more

MINIMUM ECCENTRICITY IN COLUMNS

IS:456-2000,PG:42,CL.25,4 All the column should be designed for minimum eccentricity Minimum eccentricity(e)= unsupported length of column/500+Lateral dimension/30 emin= L/500+D/30 where L and D are in mm emin≥20mm

ASSUMPTIONS IN DESIGN OF COMPRESSION MEMBER

Plane section normal to the axis remain plane after bending Maximum strain in concrete at the outermost compression

fibre is taken as 0.0035 in bending The maximum compressive strain in concrete in axial is taken

as 0.002 The tensile strength of concrete is ignored

SHORT AXIALLY LOADED COLUMNS

The figure shows rectangular column section subjected to axial load. The column is subjected to uniform strain of 0.002 and uniform stress of 0.446fck.

Axial load where e=0 Pu=Puc+Pus Pu=0.446fck*Ac+fs*Asc Fy = stress in steel corresponding to a strain of

0.002. =0.87fy for Fe 250 =0.79fy for Fe 415 =0.75fy for Fe 500

CONT…

I.S CODE adopts the critical value of 0.75fy for all grades of steel for finding out the pure axial load carrying capacity of columns.

Pu=0.446fck*Fck*Ac+0.75fy*Asc Which is approximated as: Pu=0.45fck*Ac+0.75Asc Where Ac = area of concrete in column Asc= area of steel in compression in column If emin≤ 0.05D (IS:456-2000,CL.39.9) Pu=0.4fck*Ac+0.67fy*Asc It should be noted that equation above is the reduction in the capacity of the

column by 10%

COLUMN WITH HELICAL REINFORCEMENT

AS PER IS:456-2000,PG 71,CL.39.4 Stregnth of columns with helical

reinforcement =1.05*strength of column with lateral ties

=factored load/1.05 Let Acr=area of core Ag=gross area asp=area of spiral(helical reinforcement

ρs= asp/Acr ρs= 0.36(Ag/Acr-1)*fck/fy

(IS:456-2000 PG.71

EXAMPLES ON AXIALLY LOADED COLUMNS

QUESTION: Determine the ultimate load carrying capacity of a rectangular column section 400mm*600mm with 6nos , 28mm diameter bars. Consider concrete of grade M25 and steel of Fe415,assume emin is less than 0.05mm times the lateral dimension .M25 and fe 415 There fore fck=20N/mm^2 and fy=415N/mm^2Asc= 6nos – 28 dia = 3694.5mm^2Ac= Ag-Asc (400*600)-3694.5= 236305.5mm^2

EXAMPLE CONT…

AS PER IS:456-2000,PG 71 (emin< 0.05D) Pu=0.4fck*Ac+0.67fy*Asc = 0.4*25*236305.5+0.67*415*3694.5 = 3390.31KN ALLOWABLE SERVICE LOAD = 3390.31/1.5 = 2260.20KN

EXAMPLES

QUESTION:determine the ultimate load capacity of a circular column of diameter reinforced with 6nos-25dia bars with

Lateral ties SpiralsUse M20 concrete and Fe415 steelAlso assume tha emin<0.05D.Fck=20N/mm^2Fy=415N/mm^2Asc= 2945.24mm^2

EXAMPLES CONT….

Ac= Ag-Asc COLUMN WITH LATERAL TIES =(π/4*400^2)-2945.24 pu=0.4fck.Ac +0.67Fy.Asc = 122718.46mm^2 =0.4*25*122718.46+0.67*415*2945.24 =2046108.58 N =2046.11 KN COLUMN WITH SPIRALSPu=1.05*ultimate axial load capacity of the section with lateral ties =1.05*2046.11 (IS:456-2000,P.71) =2148.42 KN

EXAMPLE

Q design a R.C.C short column square in section to carry a design axial load of 300KN. use M20 concrete &steel Fe415.assume 1% oflongitudinal steel.unsupported length of column is 3m. Sketch the reinforcement details.SolutionFck=20N/mm2Fy=415 N/mm2Ag= gross area ofcolumn 1% steelAsc=1/100×Ag=0.01AgAc=Ag-Asc=Ag-0.01Ag=099Ag

Example cont…

Factored load=Pu=1.5×3000=4500KNPu=0.4Fck.Ac+0.67Fy.Asc4500KN=7.92Ag+2.78AgAg=420560.74mm2Side og square column=648.5mmProvide square column=650×650mm.Ag provided =650×650=422500mm2Asc=0.01Ag=0.01×422500=4225mm2Provide 8 nos. -28Ø(Ast=4926mm2)

Dia.of lateral ties (1)1/4×larger dia.=1/4×28=7mm (2)6 mm Taking larger of two values, Provide 8Ø-lateral tiles.

Example cont…

Dia.of lateral ties (1)1/4×larger dia.=1/4×28=7mm (2)6 mm Taking larger of two values, Provide 8Ø-lateral tiles.

Pitch of lateral ties (1)Least lateral dimension =650mm (2)16×smaller dia.=16×28=448mm (3)300mm Provide 8Ø-lateral ties at300mm c/c Distance between corner bars=650-

2×50-28=522mm 48×dia.of ties=48×8=384m As 522mm>384mm,twosets of c losed

ties shall be used.

Example cont……

Check for eccentricityRmin=l/500 +D/30

=3000/500 +650/30 (IS:456-2000,P.42)

=27.66>20mmEmin =27.66mmEmin/D =27.66/650=0.042<0.05 o.k

design as axially loaded short column is satisfactory.