SHORT COLUMN DESIGN WITH AXIAL LOAD & BIAXIAL MOMENTS

1. KALINGA INSTITUTE OF INDUSTRIAL TECHNOLOGY
BHUBHANESWAR
Design of Concrete Structure
Presentation on
DESIGN OF SHORT COLUMN WITH AXIAL LOAD & BIAXIAL MOMENT
Date of Presentation :- 22nd April 2020
Presented by :- Guided by :-
Shubham Singh (1801609) Prof. P.C Saha Sir
CV-2, 3RD Year Associate Professor (II)
School of Civil Engineering,
KIIT University, Bhubaneswar
School of Civil Engineering,
KIIT University, Bhubaneswar

2. STEP 1- Given data’s
• Width of Column = b
• Depth of Coluumn = D
• Axial Load = P ,
• Characterstic Compressive strength of Concrete = Fck
• Characterstic strength of Steel = Fy
• Factored moments along given axis
Step 2- Unsuportted Length
• lx =
• lz =

3. Step 3 - Efeective Length
From table 28 of IS 456:2000 we know the Effective Length of Compression Member
• Lex = K x lx
• Lez = K x lz
Step 4 - Check for Short Column
From clause 25.1.2 of IS 456:2000 a compression member may be considered as short when both the
slenderness ratio are less than 12
• Lex / D =
• Lez / b=
Step 5 - Minimum Eccentrity
From clause 25.4, IS 456:2000 we know that all the column shall be designed
for minimum eccentricity, equal to the unsupported length of column /500 + lateral dimension/30
• eminx = l/500 +D/30 =
• eminz = l/500 + b/30 =

4. Step 6- Check for 5% of side with emin
From Clause 25.4 of Is 456:2000 we get that for biaxial bending it is sufficient to ensure that
eccentricity excceds the minimum about one axis at a time
5/100 x D =
5/100 x b =
Let suppose that eminz > 5/100 x b ,
Then moment along minor axis (Muz) = Pu x eminz
Design load ……….(known to us)
Pu, Mux, Muz
Mueq =1.15(Mux2 + Muz2 )1/2 …………………….(i)
Pu /Fck x b x D = …………………......(ii)
Mueq / Fck x b x D2 = ………………………(iii)
d/D = …………………….....(iv)
where, d= spacing + ½ dia of bar
By using value of eqn (i), (ii), (iii) (iv) & fy, From SP – 6 graphs we get Pt / Fck value
Using Pt / Fck we get Pt

5. Assume % of steel (Pt’ )
Pt’ / Fck =
Pu /Fck x b x D =
d’/D =]
Mux1 / Fck x b x D2 =
Mux1 from here we get ( which should be greater than Mux )
d’/b =
Muz1 / Fck x D x b2 =
Muz1 from here we get ( which should be greater than Muz )
From Clause 39.6 of IS Code 456:2000
Puz = 0.45 fck AC + 0.75 Fy ASC
where = AC = Ag – ASC
ASC = Pt’ x Ag /100
Pu / Puz αn
0.2 to 0.8 1.0 to 2.0
< 0.2 1.0
> 0.8 2.0

6. From Clause 39.6 of IS Code 39.6
( Mux / Mux1 ) αn + ( Muy / Muy1 ) αn ≤ 1.0
where Mux, Muy = moments about x and z axes due to design loads
Mux1 Muy1 = maximum uniaxial moment capacity for an axial load of Pu ,
bending about x and z axes respectively

7. THANK YOU