Q.
Design the column restrained in position and direction at both ends with length of 3 m
for carrying a factored load of 2300 kN. Use grade 20 concrete and Fe415 steel.
Given: Concrete Grade- 20 N/mm2, Steel grade- 415 N/mm2, Factored Load = 2300 kN
Length- 3 m, Restrained in position and direction at both ends.
Solution:
Step 1: Size of column
Assume 1% of steel.
Asc = 0.01Ag
Ac = Ag – Asc = Ag – 0.01 ×Ag = 0.99Ag
Now, Pu = 0.4 ×fck ×Ac + 0.67 ×fy ×Asc ----------------------------{Cl 39.3, P71, IS456:2000}
2300×103 = 0.4 ×20×0.99Ag + 0.67 ×415 ×0.01 ×Ag
2300×1000
Ag =
(0.4×20×0.99 + 0.67×415×0.01)
Ag = 214943.227 mm2
Provide Square Column, B = Ag = √214943.227 = 463.61 mm ~ 470 mm
Provide column size of 470mm x 470 mm.
Step 2: Check for Short Column
Effective length (Le) = 0.65 × L ---------------------------------{Table 28, P94, IS456:2000}
= 0.65 × 3000 = 1950 mm
Le 1950
= = 4.14 < 12 -------------------------------------{Cl 25.1.2, P41, IS456:2000}
B 470
Hence, it is a short column.
Step 3: Check for Eccentricity
L B
emin = + --------------------------------------{Cl 25.4, P42, IS456:2000}
500 30
1950 470
= + = 19.56 < 20 mm
500 30
eper = 0.05×B = 0.05*470 = 23.5 mm
emin < eper
Hence, design column as axially loaded column.
Step 4: Calculation of steel
Ag = B×B = 470×470 = 220900 mm2
Asc = 0.01×Ag = 0.01× 220900 = 2209 mm2
Assume 20mmϕ bars.
Asc 2209
No of Bars = = = 7.035 ~ 8
Area of 1 bar 314
Actual Steel Provided = No of Bars 1× Area of 1 bar = 8 × 314 = 2512 mm2
Steel Provided
% Pt = ×100
Ag
2512
= ×100 = 1.13 %
220900
0.8 % < % Pt < 6 %
Hence, it is OK.
Step 5: Lateral Ties
Dia of ties = -------------------------------------{Cl 26.5.3.2 c), P49, IS456:2000}
a) ϕ/4 = 20/4 = 5mm
b) 6mm
Maximum of a) and b)
Provide 6mm ϕ lateral ties.
Spacing = -------------------------------------{Cl 26.5.3.2 c), P49, IS456:2000}
a) LLD = 470 mm
b) 16ϕ = 16 × 20 = 320 mm
c) 300mm
Minimum of a), b), c)
Provide stirrup of 6mm ϕ@ 300 mm c/c.
Summary
a) Size of Column = 470mm × 470mm
b) Cover = 40 mm
c) Longitudinal Steel = 20mm ϕ – 8 bars
d) Transverse Steel = 6mm ϕ @ 300 mm c/c.
Detailing:
470-40-40-6-6-20
Distance between two bars = = 179 mm
2
470 mm
6mm ϕ @ 300mm c/c
470 mm
20mm ϕ - 8 bars
470 mm
300mm c/c
Cover = 40 mm
179 mm
