CE 134- Design of Reinforced Concrete Structures
Chapter 8
Design of Columns
Instructor: Richelle G. Zafra, PhD
Chapter 8 Outline
8.1
Introduction
8.2
Types of Columns
8.3
ACI/NSCP Code Requirements
8.4
Axially Loaded Short Columns
8.5
Short Columns under Combined Axial Load
and Moment
2
Columns
Pier
Column
3
Axially Loaded Columns
• Columns are defined as members that carry
loads chiefly in compression.
• Columns with a ratio of height-to-least lateral
dimension exceeding 3 are used primarily to
support axial compressive load.
• Columns subjected to pure axial load rarely, if
ever, exists.
4
Eccentically Loaded Columns
Columns are subjected to some bending moment
which may be caused by unbalanced floor loads on
both exterior and interior columns.
Giesel Library Building, UC San Diego
5
Eccentically Loaded Columns
Eccentric loads such as crane loads in industrial
buildings also cause moment.
Crane
RC corbel
Industrial Building
6
Eccentically Loaded Columns
Lateral loading due to earthquake
Olive View Hospital,
1971 San Fernando EQ
http://www.smate.wwu.edu/teched
/geology/eq-CA-SanFernd.html
Collapsed parking structure,
California State University,
1994 Northridge EQ
pubs.usgs.gov
7
Eccentically Loaded Columns
Lateral loading due to wind
http://www.wbdg.org/resources/env_wind.php
Office Building Exterior and Curtain Wall,
2000 Forth Worth, USA Tornado
8
Types of RC Columns
ties
spirals
structural
steel
shape
(c) composite column
steel
pipe
pitch
(a) Tied column
(b) Spiral column
(d) Lally column
9
ACI/NSCP Code Requirements
for Tied Columns
1. Min. dimension = 200 mm
2. Min. gross area, Ag = 200 mm x 300 mm col.
= 60, 000 mm2
3. Min. main bars = 4 pcs – 16 mm ø
4. Min. clear bar cover = 40 mm (same as beams)
5. Lateral tie diameter = same method of
determining stirrup diameter
6. Gross steel ratio: ρg = 1% min. to 8% max.
(use only up to 4% when designing)
10
ACI/NSCP Code Requirements
for Tied Columns (Cont.)
7. Main bar spacing, sm
s > 1.5 main bar ø
s > 40 mm
s > 1.5 max size of
coarse aggregate
8. Lateral tie spacing, st
s = 16 main bar ø
s = 48 lateral tie ø
s = least col. dimension
ties
st
sm
main bar
Smallest value
11
ACI/NSCP Code Requirements
for Spiral Columns
1. Min. column diameter = 250 mm
2. Min. main bars = 6 pcs – 16 mm ø
3. Min. clear bar cover = same as tied column
4. Spiral bar diameter = same as ties
5. Gross steel ratio: ρg = same as tied column
6. Main bar spacing = same as tied column
12
ACI/NSCP Code Requirements
for Spiral Columns (Cont.)
7. Min. spiral steel percentage
 Ag
 fc'
Minimum ρs = 0.45
− 1
 Ac
 fy
(1)
where:
Ag = gross cross-sectional area
= π D 2 ; D = column diameter
4
Ac = core-concrete cross-sectional area
= π Dc2 ; Dc = concrete core diameter/
4
outside diameter of spiral
Gross area
Core
concrete
13
ACI/NSCP Code Requirements
for Spiral Columns (Cont.)
8. Actual/ required spiral steel ratio
40mm CC
~Dc
Ac
D
Mean
Dia, Ds
Core Dia
Dc = D - 80
40mm CC
Core concrete
Cover concrete
conc.
core
spiral
spiral pitch
"s"
2
as = (pi/4)ds
14
Rebar cage
ACI/NSCP Code Requirements
for Spiral Columns (Cont.)
vol . of spiral in one (1) turn
vol . of concrete core bounded
asπDc 4as
req ' d ρ s =
=
;
π 2
Dc s Dc s
4
where : as = area of one spiral bar
req ' d ρ s =
s = spiral pitch
Note: To find the spiral pitch, equate req’d ρs to min ρs.
15
ACI/NSCP Code Requirements
for Spiral Columns (Cont.)
9. Spiral pitch limits
s > 40 mm (minimum)
s < 75 mm (maximum)
s < Dc / 6
16
Axially Loaded Short Columns
Capacity Pu = φ Pn (max )
(2)
= φ (0.80Po ) ; φ = 0.65 for tied column
= φ (0.85Po ) ; φ = 0.75 for spiral column
Po = 0.85fc' (Ag − Ast ) + fy Ast
where:
[
(for analysis)
(3)
Ast = ρg Ag
]
Po = Ag 0.85fc' + ρg (fy − 0.85fc′ )
(for design) (4)
17
Axially Loaded Short Columns
where:
φ = strength reduction factor
Ag = gross area of section
Ast = total area of longitudinal reinforcement
fc′ = specified compressive strength of concrete
fy = specified yield strength of reinforcement
18
Example 1
A 500 mm x 500 mm tied column is
reinforced with 8 - 28 mm ø bars. If f’c = 21
MPa and fy = 414 MPa, find the ultimate
axial capacity of the column.
19
Example 2
A 450mm round spiral column is reinforced
with 6-25mm Ф bars having fy=276MPa.
Determine
the
ultimate
strength
if
f’c=34MPa.
20
Example 3
Design a short axially loaded square tied
column for a service dead load of 1080 KN
and a service live load of 990 KN. The
unsupported length is 2.60 m. Use f’c = 34.5
MPa, fy = 414 MPa, ρg = 2 %, 25 mm ø main
bars, 10 mm ø ties, and 40 mm concrete
cover.
21
Example 4
Design a round spiral column to support an
axial dead load of 800kN and an axial live
load of 1350kN. Assume that 2%
longitudinal steel is desired. Diameter of
main bars is 25mm and the diameter of
spiral ties is 10mm. Use f’c=27.6 MPa and
fy=414 MPa.
22
Columns under Combined Axial
Load and Moment
External Forces
Strain Diagram
Internal Forces on Columns
Equilibrium of external and internal axial forces
requires that
Pn = 0.85fc' ab + As' fs' − As fs
(5)
23
Columns under Combined Axial
Load and Moment (Cont.)
External Forces
Strain Diagram
Internal Forces on Columns
Moment about section centerline of internal forces must
be equal and opposite the moment of external force Pn
Mn = Pne =
h
'
0.85fc ab
a
h


' 'h
−  + As fs  − d ′  + As fs  d − 
 242 

 2 2
2
(6)
Columns under Combined Axial
Load and Moment (Cont.)
where: Pn
e
b
h
As′
= nominal axial capacity
= eccentricity
= width of section
= height of section
= compression steel area
As = tension steel area
d ′ = location of A’s from the compression face
d = location of As from the compression face
fs′ = stress in the compression steel
fs = stress in the tension steel
25
a = β1c
Columns under Combined Axial
Load and Moment (Cont.)
If we know
Neutral axis
(c, a)
Strain condition
(εs, ε’s)
Stress condition
(fs, f’s)
Can determine
Column Strength
(Mn, Pn)
26
Columns under Combined Axial
Load and Moment (Cont.)
Strain in Tension Steel
d −c
ε s = ε cu
(6) fs = ε s Es ≤ fy (7)
c
where: ε cu = 0.003
Strain Diagram
Strain in Compression Steel
c − d′
ε s′ = ε cu
(8) fs′ = ε s′ Es ≤ fy (9)
c
Concrete Stress Block
a = β1 c ; c ≤ h (10)
Internal Forces on Columns
27
Jiravacharadet (2013)
Columns under Combined Axial
Load and Moment (Cont.)
External
Force on
Column
Stress
Diagram
C
T
28
Interaction Diagram for Combined
Bending and Axial Load
Pn
Mn
e=
Pn
• For any eccentricity
e, there is a unique
pair of Pn and Mn .
(Mn, Pn)
• Plotting a series of
(Mn,
Pn)
pairs
corresponding
to
different e will result
in an interaction
diagram.
emin
Po
e=0
Direct axial
failure
Compression
e range
failure
eb
Tension failure range
e=∞
Mo
Mn
Column Interaction Diagram
29
Interaction Diagram for Combined
Bending and Axial Load
Pn
• Any combination of
loading that falls
inside the curve is
satisfactory
emin
Po
e=0
Direct axial
failure
Compression
failure range
eb
Tension failure range
e=∞
Mo
• However,
any
combination falling
outside the curve
represents failure.
Mn
Column Interaction Diagram
30
Balanced Failure, eb
• Concrete reaches the strain limit εcu at the same
time that the tensile steel reaches the yield strain εy
• Dividing point between compression failure (small
eccentricities)
and tension failure (large
eccentricities)
Strain Diagram
c = cb = d
ε cu
ε cu + ε y
(11)
where: ε cu = 0.003
a = ab = β1 cb
(12)
31
Balanced Failure, eb
Strain Diagram
Mb = eb Pb
(15)
fs′ = ε s′ Es ≤ fy
where:
Pb = 0.85 fc' ab b + As' fs' − As fy
Mb =
h
'
0.85 fc ab b 
cb − d ′
ε s′ = ε cu
cb
(13)
ab 
h


' 'h
−  + As fs  − d ′  + As fy  d −  (14)
2  32


2 2 
2
Short Columns Under Combined
Axial Load and Moment
Pn
emin = 0.10h (Tied)
Po
emin = 0.05D (Spiral)
B. Actual Eccentricity
Mu
Mn
e=
or
Pu
Pn
e=0
A. Minimum Eccentricity
(Mn, Pn)
e
e=∞
Mo
Mn
Column Interaction Diagram
33
Behavior at Failure:
Columns under Combined P and M
Pn
Region I: Negligible Moment
e=0
Po
e < emin
Direct axial failure
(Region I)
eb
e=∞
Mo
Capacity: See axially
loaded column
Mn
Column Interaction Diagram
34
Behavior at Failure:
Columns under Combined P and M
Pn
Region II: Proportioned
Axial Load and Moment
Po
e=0
emin < e < eb
Compression
failure range
(Region II)
eb
e=∞
Mo
Capacity: Pn > Pb
Mn
Column Interaction Diagram
35
Approximate Capacity Formulas
(Region II)
1. From straight line relation on interaction curve
Mn
Po
= Po − (Po − Pb )
Pn =
Po  e
M
b
1+  − 1
Pb  eb
(16)
36
Approximate Capacity Formulas
(Region II)
2. Whitney’s Formula
b h fc '
As ' fy
Pn =
+
3he
e
+1
+ 0 .5
2
d − d'
d
Ag fc '
9 .6 D e
(17)
As fy
Pn =
+
(18)
3e
+ 1.18
+1
2
Ds
(0.8D + 0.67D s )
Tied
Column
Spiral
Column
37
Behavior at Failure:
Columns under Combined P and M
Pn
Region III: Proportioned
Axial Load and Moment
Po
e=0
e > eb
Capacity: Pn < Pb
eb
Tension failure range
(Region III)
e=∞
Mo
Mn
Column Interaction Diagram
38
Approximate Capacity Formulas
(Region III)
Approximate Whitney’s Capacity Formulas (Tied Columns)
2

e'  e' 
d '  e' 

Pn = 0.85fc ' bd 1 − ρ − + 1 −  + 2 ρ (m − 1)(1 − ) + 
d
d  d

 d


(19)
As
; As = bars in tension
where: ρ =
bd
fy
m=
'
0.85fc
h
'
e = e + − d'
39
2
Approximate Capacity Formulas
(Region III)
Approximate Whitney’s Capacity Formulas (Spiral Columns)
2 ρ mDs
  0.85e
g

− 0.38  +
Pn = 0.85fc ' D 2  
2.5D

  D

e
0
.
85


−
− 0.38 
 D

(20)
40
Bending is about y-axis of
the column.
3-28mmφ
• ultimate axial capacity
at balanced condition;
• load
eccentricity
at
balanced condition;
• ultimate axial capacity if
e = 200 mm.
As
300 mm
For the column shown
with f’c = 28 MPa and fy =
414 MPa, determine:
y
e
As’
3-28mmφ
Example 3
62.5 187.5 187.5
500 mm
Top View
Pn
x
62.5
e
Pn
Elevation View
41
Biaxial Bending
• Axial
compression
is
accompanied by simultaneous
bending about both principal
axes of the section.
• Such is the case of corner
columns of tier buildings
• Beams and girders frame into
the columns in both directions
and
transfer
their
end
moments into the columns in
two perpendicular planes
nees-anchor.ceas.uwm.edu
nisee.berkeley.edu
42
Strength Interaction Diagram for
Biaxial Bending
Uniaxial bending
about Y axis
Uniaxial bending
about X axis
Biaxial bending
Column Interaction Diagram
43
Reciprocal Load Method
• A
simple,
approximate
developed by Bresler.
• Acceptably accurate
provided Pn ≥ 0.10P0
for
design
design
method
purposes
44
Reciprocal Load Method
Bresler’s reciprocal load equation is given by
1
1
1
1
=
+
−
Pn Pnx 0 Pny 0 P0
(21)
where:
Pn = approximate value of nominal load in biaxial
bending with eccentricities ex and ey
Pny 0 = nominal load when only eccentricity ex is
present (ey = 0)
Pnx 0 = nominal load when only eccentricity ey is
present (ex = 0)
45
Po = load for concentrically loaded column
Example 4
y
300 mm
Using Bresler’s equation,
determine the strength of
the column shown given
the biaxial capacities Pnx =
1880 KN, Pny = 1000 KN,
f’c = 21 MPa, fy = 414
MPa.
8-20mmφ
x
200 mm 200 mm
46
Use of Column Interaction
Diagram for Design
• The preceding lectures have clearly shown that
the analysis and design of columns with
eccentricities, using static equations, is very
tedious and complicated.
• Consequently,
designers
resort
almost
completely to tables, computers, or diagrams
(e. g. column interaction diagram) for their
column calculations.
• Interaction diagrams are useful for studying
strength of columns with varying proportions of
loads and moment.
47
ρg
engmechanics.blogspot.com
Use of Column Interaction
Diagram for Design
e/h
φ Pn
Ag
(ksi)
φ Pn e
Ag
×
h
=
φMn
Ag h
(ksi)
48
How to Use Column Interaction
Diagram for Design
In order to correctly use the
column interaction diagram, it
is necessary to compute the
value of γ
γ =
h
γh
γh
h
where:
γh = center to center distance
of bars on each side of
the column
h = column depth
b
49
Use of Column Interaction
Diagram
Note:
• Both γh and h should be taken in the direction of
bending.
• In using the column interaction diagram, be sure
that the column picture at the upper right of the
diagram being used agrees with the column
being considered.
• For example, are there bars on two faces of the
column or on all four faces?
50
Example 5
75
4-25mmφ
4-25mmφ
e
400 mm
Calculate the nominal
axial capacity (in kN) of
the column shown if the
eccentricity of the load
is 200 mm. Use f’c = 21
MPa and fy = 414 MPa.
Use
the
interaction
diagram.
Pn
450 75
600 mm
51
Column Interaction Diagram
(Rectangular Section)
1.2 ksi
52
500 mm
8-22mmφ
375
e
62.5
Calculate the nominal
axial load (in kN) that
can be applied in the
column
at
an
eccentricity of 200
mm. Use f’c = 28 MPa Pn
and fy = 414 MPa.
Use the interaction
diagram.
62.5
Example 6
53
Column Interaction Diagram
(Circular Section)
1.0 ksi
54
Example 7
e
500 mm
Calculate the nominal
axial load (in kN) that can
be applied in the column
at an eccentricity of 400
mm. Use f’c = 21 MPa and
fy = 414 MPa. Use the
interaction diagram.
16-28mmφ
70
Pn
70
360
500 mm
55
Column Interaction Diagram
(Rectangular Section)
0.65 ksi
56
Column Interaction Diagram
(Rectangular Section)
0.82 ksi
57
References
Jiravacharadet, M. Lecture Notes in Reinforced
Concrete Columns. School of Civil Engineering,
Suranaree University of Technology, Thailand.
Accessed June 10, 2013.
Nilson, A. H.
1997.
Design of Concrete
Structures. 12th Edition. McGraw-Hill: Singapore.
McCormac, J. C. and Nelson, J.K. 2005. Design
of Reinforced Concrete. 6th Edition. John Wiley &
Sons, Inc: New Jersey.
58