fundamentals of engineering supplied-reference handbook - Ventech!

SHORT COLUMNS
Limits for main reinforcements:
Ast
ρ g =
A
g
0.01 ≤ ρg ≤ 0.08
Definition of a short column:
KL
r
≤ 34 −
12M
1
M 2
where: KL = Lcol clear height of column
[assume K = 1.0]
r = 0.288h rectangular column, h is side length
perpendicular to buckling axis ( i.e.,
side length in the plane of buckling )
r = 0.25h circular column, h = diameter
M1 = smaller end moment
M2 = larger end moment
M
M
1
2
positive if M1, M2 cause single curvature
negative if M1, M2 cause reverse curvature
LONG COLUMNS − Braced (non-sway) frames
Definition of a long column:
KL 12M
1
> 34 −
r
M 2
Critical load:
2
2
π E I π EI
Pc = = 2
2
( KL ) ( L )
col
where: EI = 0.25 Ec Ig
Concentrically-loaded long columns:
emin = (0.6 + 0.03h) minimum eccentricity
M1 = M2 = Pu emin (positive curvature)
KL
r
> 22
M c =
M 2
Pu
1 −
0.
75Pc
Use Load-moment strength interaction diagram
to design/analyze column for Pu , Mu
118
CIVIL ENGINEERING (continued)
Concentrically-loaded short columns: φPn ≥ Pu
M1 = M2 = 0
KL
≤ 22
r
Design column strength, spiral columns: φ = 0.70
φPn = 0.85φ [ 0.85 fc' ( Ag − Ast ) + Ast fy ]
Design column strength, tied columns: φ = 0.65
φPn = 0.80φ [ 0.85 fc' ( Ag − Ast ) + Ast fy ]
Short columns with end moments:
Mu = M2 or Mu = Pu e
Use Load-moment strength interaction diagram to:
1. Obtain φPn at applied moment Mu
2. Obtain φPn at eccentricity e
3. Select As for Pu , Mu
Long columns with end moments:
M1 = smaller end moment
M2 = larger end moment
M 1
positive if M1 , M2 produce single curvature
M 2
0.
4 M 1
Cm = 0.
6 + ≥ 0.
4
M 2
Cm
M 2
M c =
≥ M 2
Pu
1 −
0.
75Pc
Use Load-moment strength interaction diagram
to design/analyze column for Pu , Mu