17
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 (A-6)
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 A-A
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 50-cm 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
RB9@0.40
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