Column load transfer from beams and slabs 1) Tributary area method: Half distance to adjacent columns y x Load on column =area ×floor load y x
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Reinforced Concrete Design
Design of Column 1 Column load transfer from beams and slabs Type of Columns Strength of Short Axially Loaded Columns Column Failure by Axial Load Lateral Ties and Spirals Mongkol JIRAVACHARADET
SURANAREE UNIVERSITY OF TECHNOLOGY
INSTITUTE OF ENGINEERING SCHOOL OF CIVIL ENGINEERING
Tributary Area When loads are evenly distributed over a surface, it is often possible to assign portions of the load to the various structural elements supporting that surface by subdividing the total area into tributary areas corresponding to each member. Half the load of the table goes to each lifter.
6m
Half the 100 kg/m2 snow load on the cantilevered roof goes to each column. The tributary area for each column is 3 m x 3 m. So the load on each column is 100 (3 x 3) = 900 kg
3m
100 kg/m2
Column load transfer from beams and slabs 1) Tributary area method: Half distance to adjacent columns Load on column = area × floor load
y y
x
x
6m
6m
4.5 m
9m
12 m
9m
All area must be tributed to columns
C3
C1
C3
C1 6m
C2
C4
C4
C2
6m C4
C2
C4
C2
4.5 m C1
C3
C3
9m
12 m
C1
9m
C1 : Corner column
C3 : Exterior column
C2 : Exterior column
C4 : Interior column
2) Beams reaction method: Collect loads from adjacent beam ends B2
B1
B4 RB1
RB2
RB1 RB2
B1
C1 B3
B2
Load summation on column section for design ROOF
Design section
Load on 2nd floor column = Roof floor + Column wt.
Design section
Load on 1st floor column = load on 2nd floor column + 2nd floor + Column wt.
2nd FLOOR
1st FLOOR
Ground level
Design section Footing
Load on pier column = load on 1st floor column + 1st floor + Column wt.
C1 (A6)
RB2
3.50 m
T1
RB4
0.3 x 0.3 m
RB19
B5
3.50 m
B4
B4
0.3 x 0.3 m
B5
B5 B4
B4 B5
1.50 m
0.4 x 0.4 m
RB2 RB4 RB19 T1 Col.Wt. Floor load
= 5280 kg = 4800 kg = 4416 kg = 960 kg = 756 kg = 16212 kg
2B5 = 10764 kg 2B4 = 14736 kg Col.Wt. = 756 kg Floor load = 26256 kg Cum. load = 42468 kg 2B5 = 10764 kg 2B4 = 14736 kg Col.Wt. = 576 kg Floor load = 26076 kg Cum. load = 68544 kg
Type of Columns
Tie
Spiral
Longitudinal steel s = pitch
Tied column
Spirally reinforced column
Strength of Short Axially Loaded Columns Short columns are typical in most building columns. P0
Steel
fy Stress
∆
f c′ Section AA
A
A
Concrete
.001
.002 Strain
.003
P0
[ ΣFy = 0 ] P0 = Ast fy + fc′( Ag − Ast ) From experiment:
f c′ fy
fy
Fs = Ast fy Fc = (Ag  Ast) f c′
P0 = Ast fy + 0.85fc′( Ag − Ast ) where Ag = Gross area of column section Ast = Longitudinal steel area
Column Failure by Axial Load
Pu ∆
Axial load
Pu
0
Initial failure
Tied column
Axial deformation ∆
Heavy spiral ACI spiral Light spiral
: U = 1.4D + 1.7L
: U = 0.75(1.4D + 1.7L + 1.7W)
U = 0.9D + 1.3W
"# (φ) :
φ = 0.75
φ = 0.70
& '( ) **+& , & '( ACI 1) !"# 2) &' ∅ 9 ".". ) ≤ DB32 * &' ∅ 12 ".". ) DB36 * DB40 3) !! (s) s ≤ 16 ∅
s ≤ 48 ∅ s≤
,"
4) "") * ."!/ 135o * ."!" "!/ 15 0". x
x
x
x ≤ 15 cm x
x > 15 cm
x
x
x
x
x
x x ≤ 15 cm
x
x
x > 15 cm
& ' Initial shape
Pu Spiral
Final shape
f2
f2
Increase of compressive strength due to lateral pressure:
ff = fc′ + 4.1f2 Good design: Strength lost by spalling = Strength gain from f2
( Ag − Acore )(0.85fc′) = Acore (4.1f2 )
1
hcore Core
s
s
Ab fy Spiral Ab fy [ ΣFx = 0 ]
hcore s f2 = 2 Ab fy
f2 = 2
2 Ab fy
2
hcore s
1
4.1(2 Ab fy ) Ag − 1 (0.85fc′) = hcore s Acore
3
Define:
ρs
ρs =
Abπ hcore 4 Ab = 2 (π hcore / 4)s hcore s
3
0.42fc′ Ag ρs = − 1 fy Acore Rounding 0.42 to 0.45,
0.45fc′ Ag ACI Code: ρ s = − 1 fy Acore
) **+& , & ' 1) Minimum width or diameter: hmin ≥ 20 cm 2) Reinforcement ratio: 1% ≤ ρg ≤ 8% (usually ≤ 5%) 3) Can use bundled bars in corners (≤ 4) 4) 2.5 cm ≤ Clear stirrup spacing ≤ 8 cm 5) Spiral diameter: db ≥ 9 mm 6) Lap splices: Lsp ≤ min { 48 db , 30 cm }
Minimum Cover for Column Reinforcement Min. Cover
Condition
Reinforcement
Cast against earth
all sizes
7 cm
Exposed to weather or earth
DB20  DB60 DB16 and smaller
5 cm 4 cm
No exposure
main reinforcement, ties, and spirals
4 cm
Limits on percentage of reinforcement
0.01 ≤ ρ g = Ast / Ag ≤ 0.08 Lower limit:
To prevent failure mode of plain concrete
Upper limit:
To maintain proper clearances between bars
ACI Strength Provision: Pu φ Pn Spirally reinforced column:
Pn = 0.85[0.85fc′( Ag − Ast ) + fy Ast ], φ = 0.75 Tied column:
Pn = 0.80[0.85fc′( Ag − Ast ) + fy Ast ], φ = 0.70
Working Stress Design (WSD) of Short Column Spirally reinforced column:
P = Ag (0.25fc′ + fs ρ g ),
ρ g = Ast / Ag
Tied column:
P = 0.85 Ag (0.25fc′ + fs ρ g ),
ρ g = Ast / Ag
where fs = 0.40fy but not exceed 2,100 kg/cm2
Length Effects ACI permits neglect of length effect when
kLu M ≤ 34 − 12 1 r M2
M1
for braced system
M1
where (34  12M1/M2) may not exceed 40 M1 = The smaller bending moment
+

M2 = The larger bending moment M1/M2 is positive for single curvature and negative for double curvature
kLu ≤ 22 r
for unbraced system
M2
M2
Example 11.1 Design for Pure Compression Design a concentrically loaded square column with ties providing lateral reinforcement. Service dead and live loads are 180 and 90 tons, respectively The column has an unsupported height of 3.0 m and is braced against sidesway. Use f’c = 240 kg/cm2 and fy = 4,000 kg/cm2. 1) Determine required strength Pu = 1.4D + 1.7L = 1.4(180) + 1.7(90) = 405 tons 2) Check column slenderness. Assume an 50cm square column k = 1.0 for braced compression member 1 4 2 r = 0.3(50) = 15 cm I/A = h /h = 12
(
)
1/12 h
kLu 1.0 × 3.0 × 100 = = 20 < 34 − 12(M1 / M2 ) = 22 r 15
Neglect length effects
3) Design for column reinforcement Required Pn = Pu/φ = 405/0.70 = 578.6 ton for tied column:
Pn = 0.80[0.85fc′( Ag − Ast ) + fy Ast ]
578.6 × 1,000 = 0.80(0.85×240(50×50  Ast) + 4,000Ast) Ast = 56.2 cm2
50 cm
USE 12DB25 (Ast = 58.9 cm2, ρg = 2.36%) 4) Select lateral reinforcement
12DB25
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USE RB9 ties with DB25 longitudinal bars Spacing not greater than:
16 (2.5) = 40 cm 48 (0.9) = 43.2 cm column size = 50 cm
USE RB9 @ 40 cm
50 cm