Axially loaded columns

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

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)

 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^2
Asc= 6nos – 28 dia
= 3694.5mm^2
Ac= 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
 Spirals
Use M20 concrete and Fe415 steel
Also assume tha emin<0.05D.
Fck=20N/mm^2
Fy=415N/mm^2
Asc= 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 SPIRALS
Pu=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.
Solution
Fck=20N/mm2
Fy=415 N/mm2
Ag= gross area ofcolumn 1% steel
Asc=1/100×Ag=0.01Ag
Ac=Ag-Asc=Ag-0.01Ag=099Ag

Example cont…
Factored load=Pu=1.5×3000=4500KN
Pu=0.4Fck.Ac+0.67Fy.Asc
4500KN=7.92Ag+2.78Ag
Ag=420560.74mm2
Side og square column=648.5mm
Provide square column=650×650mm.
Ag provided =650×650=422500mm2
Asc=0.01Ag=0.01×422500=4225mm2
Provide 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 eccentricity
Rmin=l/500 +D/30
=3000/500 +650/30 (IS:456-2000,P.42)
=27.66>20mm
Emin =27.66mm
Emin/D =27.66/650=0.042<0.05 o.k
design as axially loaded short column is
satisfactory.

Axially loaded columns

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Axially loaded columns