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Design of Short Column

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18.

Design of Short Column

Given Data : Find : Ag & Ast

Compressive strength of concrete f'c := 28 MPa

Yeild strength of reinforcement fy := 350 MPa

Load on Column Pu := 2600 kN

Percentage of rebar ρg := 2 %

Main bar db := 28 mm

Confinement bar dv := 8 mm

Strength reduction facotor ϕ := 0.65 for compression member

Solution Reference

Axial Compression Strength of Column

( )
ϕPn = Pu = ϕ ( 0.8 )  0.85  f'c Ag − Ast + Ast fy

(ACI.10.3.6)

Ast = Ag ρg

( )
ϕPn = Pu = ϕ ( 0.8 )  0.85  f'c Ag − Ag ρg + Ag ρg fy

( )
ϕPn = Pu = ϕ ( 0.8 )  0.85  f'c Ag 1 − ρg + Ag ρg fy

( )
ϕPn = Pu = ϕ ( 0.8 )  Ag 0.85  f'c 1 − ρg + ρg fy

Pu
ϕ 0.8 2
Ag.req := Ag.req = 164885.899  mm
( )
0.85  f'c 1 − ρg + fy ρg
b
λ= λ := 1
h

h := Ag.req h = 406.061  mm Assume h := 400 mm

b := h λ = 400 mm Assume b := 400 mm

2
Ag.pro := b h Ag.pro = 1600 cm

Axial Compression Strength of Column

( )
ϕPn = Pu = ϕ ( 0.8 )  0.85  f'c Ag.pro − Ast + Ast fy (ACI.10.3.6)

(
ϕPn = Pu = ϕ ( 0.8 )  0.85  f'c Ag.pro − 0.85  f'c Ast + Ast fy )
( (
ϕPn = Pu = ϕ ( 0.8 )  0.85  f'c Ag.pro + Ast fy − 0.85  f'c ))
Rebar area require for column

Pu
− 0.85  f'c Ag.pro
ϕ 0.8 2
Ast := Ast = 3654.2  mm
fy − 0.85  f'c

1 1 1
Rebar arranges D :=  1 0 1 
 
1 1 1
Rebar area by layer in rows

π db cols( D)  i
2  1847.256 
As =  1231.504   mm

2
As :=  D
4  
i =1  1847.256 

 As
2
Rebar provides Ast.use := Ast.use = 4926.017  mm

Spacing of Stirrup

(
Scolumn := min 16 db , 48dv , b ) Scolumn = 384 mm (ACI.7.10.5.2)

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