0% found this document useful (0 votes)
94 views3 pages

Design of Column

The document summarizes the design of a column that is 500x500mm in size and subjected to an axial load of 2000kN and bending moments of 75kNm and 60kNm. It involves calculating the equivalent moment, determining non-dimensional parameters, finding the required steel reinforcement of 8 bars with 25mm diameter totaling 3750mm2, checking the design is safe, and specifying 8mm diameter lateral ties spaced at 300mm.

Uploaded by

Harini Ramesh
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
0% found this document useful (0 votes)
94 views3 pages

Design of Column

The document summarizes the design of a column that is 500x500mm in size and subjected to an axial load of 2000kN and bending moments of 75kNm and 60kNm. It involves calculating the equivalent moment, determining non-dimensional parameters, finding the required steel reinforcement of 8 bars with 25mm diameter totaling 3750mm2, checking the design is safe, and specifying 8mm diameter lateral ties spaced at 300mm.

Uploaded by

Harini Ramesh
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
You are on page 1/ 3

DESIGN OF COLUMN

DATA:
Size of column = 500x500mm
Pu = 2000KN
M25 Grade of concrete and Fe415 steel
Mux = 75 KNm
Muy = 60 KNm
d’ = 50mm
SOLUTION:
STEP 1: EQUIVALENT MOMENT
CALCULATION
Mu = 1.15 √Mux2 + Muy2
= 110 KNm
STEP 2: NON –DIMENSIONAL PARAMETERS:
Pu / Fck b D = 2000x103/ 25x500x500 =0.32
Mu / Fck b D2 = 110 x 106/ 25x500x5002 = 0.035
d’/ D = 50/500 = 0.10
refer chart SP 16
p/Fck = 0.06
p = 0.06x25 = 1.5
STEP 3 : REINFORCEMENT CALCULATION:
Asc = pxbxD / 100
= 1.5x500x500 / 100 = 3750mm2
Provide 25mm dia bars
No of bars required = 3750/490.87 = 8 nos
Actual % of r/f = 8 x π/4 x 252 = 3927mm2
Pa = Asc x 100 / b D = 1.57
Pa / Fck = 1.57/25 = 0.32
STEP 4 : MOMENT CALCULATION:
Mux1 / Fck b D2 = 0.038
Mux1 = 118 KNm ,Similarly
Muy1 = 118 KNm
STEP 5 : AXIAL LOAD CALCULATION:
Puz = 0.45 Fck Ac + 0.75 Fy Asc
= 3990 KN
Pu / Puz = 2000/3990 = 0.50
α n = 1.6
STEP 6: CHECK FOR SAFETY
[(Mux/Mux1) αn + (Muy/Muy1) αn ] ≤ 1
( 0.48 + 0.33) ≤ 1
0.81 ≤ 1
Hence design is safe.
STEP 7: LATERAL TIES
Dia of ties = 8 mm
Pitch = Minimum of 300
< 16 x 25 = 400mm
< 500mm
Provide 8mm dia lateral ties @ 300 mm c/c spacing.

You might also like