The load-carrying capacity of the column depends upon
●. The grade of concrete.
●. Grade of steel.
●. Area of steel. &
●. The sectional area of the column.
The ultimate load-carrying capacity of the columns should be always greater than the actual load imposed over the column.
Eg:
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Calculate the load-carrying capacity of a column that is carrying an ultimate axial load of 650KN. The size of the column is 230mm х380mm. The column has 4nos. of 16mm. dia. bars. Use M25 & Fe415 steel.
Given data:
Breadth = b = 230mm.
Depth =D = 380mm.
No. of rebars = 4nos.
Dia. of main bar = d = 16mm.
fck = 25N/mm²
fy= 415N/mm².
Calculation:
Load carrying capacity is calculated by the formula [as per IS-456 (2000)]
Pu = 0.4fck*Ac + 0.67fy*Asc
Where,
Pu = Load carrying capacity.
fck = Characteristic compressive strength of concrete.
Ac = Area of concrete.
fy = Grade of steel.
Asc = Area of the main reinforcement.
Now,
◆. Area of concrete
Ac = [Ag - Asc]
Where,
Ag = Gross sectional area of the column.
Ag = [breadth х depth]
= [230mm.х 380mm.]
Ag = 87,400 mm².
◆. The sectional area of steel
Asc = [(area of 1 bar) х No. of bars]
= [(πd² ÷ 4) х 4nos.]
= [(3.142 х 16² ÷ 4) х 4nos.]
= [201.09 х 4nos.]
Asc.= 804.4 mm².
Area of concrete
Ac = [Ag - Asc]
= [ 87,400 - 804.4]
Ac = 86596.6mm².
Pu = [0.4fck*Ac + 0.67fy*Asc]
= [(0.4 х 25 х 86596.6 ) + (0.67 х415 х804.4)]
= [ 865,966 + 2,23,663.4]
Pu = 1,089,629.4N
To get the above result in KN, we have to multiply the value by 10⁻³
Pu = [1,089,629.4 х 10⁻³]
Pu = 1089.6KN > 650KN & hence safe.✔
◆. Check for minimum reinforcement:
Asc% = [Asc ÷ Ag] х 100
= [804.4 ÷ 87400] х 100
Asc% = 0.92% > 0.8% & hence safe ✔
Thank you for going through these calculation steps❤. Have a good day 😄.
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