9
.
1
Chapter 9
:
Columns
9
.1
Introduction
The main vertical load

carrying members in buildings are called
columns
.
•
The ACI Code defines a
column
as a member used primarily to support axial
compressive loads and with a height at least three times its least lat
eral
dimension.
•
The
Code further defines a
pedestal
as an upright compression member having a
ratio of unsupported height to least lateral dimension of 3 or less.
•
The
Code definition for columns include
s
members subjected to combined axial
compression
and bending moment (i.e. eccentrically applied compressive loads).
The three basic types of reinforced concrete columns are shown in Figure 9

1
(p. 302 of the textbook).
•
Tied columns
are reinforced with longitudinal bars enclosed by horizontal, or
lateral, ties placed at specified spacings.

Tied columns are generally square or rectangular.

However, circular tied columns do exist.
•
Spiral columns
are reinforced with longitudinal bars enclosed by a continuous,
closely spaced, steel spiral.

Spir
al columns are normally circular.

The spiral is made of either wire or bar and is formed in the shape of a
helix.
•
Composite columns
are
compression members reinforced longitudinally with
structural steel shapes, pipes, or tubes with or without longitud
inal bars.

Code requirements for composite columns are found in Section 10.13.
•
Columns of
other shapes, such as octagonal and L

shaped columns
, also exist
.
A column is said to be short when its length is such that lateral buckling need not
be consider
ed.
•
The
length of a column is a design consideration for the
ACI Code.
•
As length increases, the useable strength of a given cross section
decrease
s
due to
buckling.
•
C
oncrete columns are more massive and stiffer than their structural steel
counterpart
s.

As a result
, slenderness is less of a problem in reinforced concrete columns.
9
.
2
•
It
is
estimated that more than 90% of typical reinforced concrete columns
existing in braced frame buildings may be classified as short columns
, and
slenderness effects ma
y be neglected.
9

2
Strength of Reinforced Concrete Columns: Small Eccentricity
If a compressive load P is applied
along
the longitudinal axis of a symmetrical
column,
the load
theoretically
induces a uniform compressive stress over the cross

sectional ar
ea.
•
If the compressive load is applied a small distance
“
e
”
away from the
longitudinal axis, there is a tendency for the column to bend due to the moment
M = Pe.

The distance
“
e
”
is called the
eccentricity
.
•
Unlike the zero eccentricity condition, the
compressive stress is not uniformly
distributed over the cross section.

Because of the eccentric load, t
he stress is greater on one side than on the
other.
Earlier Codes defined small eccentricity as follows.
•
For spirally reinforced columns: e/h ≤ 0.
05
•
For tied columns: e/h ≤ 0.10
where
h = the column dimension perpendicular to the axis of bending
The fundamental assumptions for the calculation of column axial strength (
for
small eccentricities) are that at
nominal
strength
•
The concrete is stres
sed to 0.85f
c
’.
•
The steel is stressed to f
y
.
•
The
nominal
axial load strength at small eccentricity is a straightforward sum
of the forces existing in the concrete and the longitudinal steel when each of
the materials is stressed to its maximum.
The fo
llowing ACI notation is used.
A
g
= gross area of the column section (in
2
)
A
st
= total area of the longitudinal reinforcement (in
2
)
P
0
=
nominal
, or
theoretical
, axial load strength at zero eccentricity
P
n
=
nominal
, or
theoretical
, axial load strength at a
given eccentricity
P
u
=
factored
applied axial load at given eccentricity
9
.
3
The following longitudinal steel ratio is used.
ρ
g
= ratio of longitudinal reinforcement area to cross

sectional area of the
column
= A
st
/A
g
The
nominal
, or
theoretical
, axial loa
d strength for the special case of zero
eccentricity is written as
P
0
= 0.85f
c
’(A
g
–
A
st
) + f
y
A
st
This
nominal
or
theoretical
strength
is
further reduced to a maximum usable load
strength using strength reduction factors.
Extensive testing has shown
that spiral columns are tougher than tied columns.
•
Both
types of columns
behave similarly up to the column yield point.

At the column yield point
,
the outer shell spalls off.
•
Above the column yield point, the columns behave differently.

The tied co
lumn fails by crushing, shearing of the concrete, and outward
buckling of the bars between the ties.

The spiral column has an inner core area within the spiral that is laterally
supported and continues to withstand load.
◦
Failure of the spiral column oc
curs when the steel yields following large
deformation of the column.
•
The Code recognizes the greater
strength
of the spiral column (Section 9.3.2)
by the assigned strength reduction factors.

Spiral column:
strength reduction factor = 0.75

Tied colum
n: strength reduction factor = 0.65
The Code
requires
that the basic load

strength relationship is
φP
n
≥ P
u
where
P
n
= the nominal axial load strength at a given eccentricity
φP
n
= the
design axial load strength
The ACI Code recognizes that no practical column can be loaded with zero
eccentricity.
•
If zero eccentricity could exist, then P
n
= P
0
.
9
.
4
•
Because eccentricity does exist, the
ACI Code imposes the following
requirements.

The
strength reduction factor (φ)
is imposed.

T
he
nominal
strengths
are
further reduced by factors of 0.80 and 0.85 for
tied and spiral columns, respectively.
•
This
results in the following expressions for usable axial load strengths.

For spiral columns:
φP
n(max)
= 0.85φ[0.85 f
c
’(A
g
–
A
st
) + f
y
A
st
]
[
ACI Eq. (10

1)]

For tied columns:
φP
n(max)
= 0.80φ[0.85 f
c
’(A
g
–
A
st
) + f
y
A
st
]
[
ACI Eq. (10

2)]
These ex
pressions provide the magnitude of the maximum design axial load
strength that may be realized from any column cross section.
•
This is the design load strength at small eccentricity.
•
Should the eccentricity (and the associated moment) become larger, φP
n
is
reduced
further
(ref. Section 9

9
of the textbook
).
•
The Code equations for φP
n(max)
provide for an extra margin of axial load
strength.

This provide
s
some reserve strength to carry small moments.
9

3
Code Requirements Concerning Column Details
Ste
el area:
Main (longitudinal) reinforcing
should have a cross

sectional area so
that the steel ratio ρ
g
is between 0.01 and 0.08.
Minimum number of bars: T
he minimum number of longitudinal bars
per ACI Code
(Section 10.9)
is
•
Within
rectangular or circul
ar ties
: 4
•
W
ithin triangular ties
: 3
•
Bars enclosed by spirals: 6
•
No minimum bar size is mentioned in the present Code (the 1963 code
recommended a minimum bar size of
#
5).
Clear distance: The clear distance between longitudinal bars must not be less
than
1.5 times the nominal bar diameter nor 1½
”
(ACI Code, Section 7.6.3).
•
This requirement holds true where bars are spliced.
9
.
5
•
Table A

14
(p. 497 of the textbook)
may be used to determine the maximum
number of bars allowed in one row around the periph
ery of circular or square
columns.
Cover:
Cover shall be 1½
”
minimum over primary reinforcement, ties, or spirals (ACI
Code, Section 7.7.1).
Tie
bars
: Tie
bar
requirements are discussed in detail in the ACI Code (Section
7.10.5).
•
Tie bar sizes:

The m
inimum size tie is
#
3 for longitudinal bars
#
10 and smaller.

The minimum size
tie
is
#
4 for longitudinal bars larger than
#
10.

The maximum size
tie
is usually a
#
5
bar
.
•
Tie bar spacing: The center

to

center spacing of ties should not exceed the
small
er of

16 longitudinal bar diameters, or

48 tie

bar diameters, or

The least column dimension.
•
The ties should be arranged so that

E
very corner
bar
and alternate longitudinal bar
has
lateral support provided
by the corner of a tie having an includ
ed angle of not more than 135°, and

N
o bar
is
farther than 6” clear on each side from such a laterally supported
bar
.
•
Typical tie arrangements are shown in Figure 9

3 (p. 307 of the textbook).
Spiral requirements: Spiral requirements are discussed in
the ACI Code (Sections
7.10.4 and 10.9.3).
•
The minimum spiral size is
#
3 for cast

in

place construction (
#
5 is usually the
maximum).
•
Clear space between spirals must not exceed 3
”
or be less than 1
”
.
•
The spiral steel ratio ρ
s
must not be less than t
he value given by
ρ
s(min)
= 0.45(A
g
/A
ch
–
1)(f
c
’/f
yt
)
[ACI Eq. (10

5)]
where
ρ
s
=
(
volume of spiral steel in one turn
)/(volume of column core in height “
s
”
)
s = center

to

center spacing of spiral (inch) (called the
pitch
)
A
g
= gross cross

sectional
area of the column (in
2
)
9
.
6
A
ch
= cross section area of the core (in
2
) (out

to

out of spiral)
f
yt
= spiral steel yield stress (psi) ≤ 60,000 psi
f
c
’ = compressive strength of concrete (psi)
Th
e minimum
spiral steel ratio results in a spiral that makes up
the strength
lost due to spalling of the outer shell.
An approximate formula for the calculated spiral steel ratio in terms of physical
properties of the column cross section may be determined as follows.
•
From the definition of ρ
s

The volume of spir
al steel in one turn = A
sp
π
D
s

The volume of the column core in height (s) = (π
D
ch
2
/4)(s)
ρ
s
= A
sp
π
D
s
/(π
D
ch
2
/4)(s)
where
A
sp
= the cross

sectional area of the spiral bar
D
s
= the spiral diameter (center

to

center)
D
ch
= the overall core di
ameter (out

to

out of spiral)
If the small difference between D
ch
and D
s
is neglected, then the spiral steel
ratio, expressed in terms of D
ch
, is
C
alculated ρ
s
= 4
A
sp
/D
ch
s
9

4
Analysis of Short Columns: Small Eccentricity
The analysis of short column
s carrying axial loads
will
small eccentricities involves
the following.
•
C
hecking the maximum design load strength and
•
Checking
the various details of the reinforcing.
The procedure is illustrated in the following examples.
9
.
7
Example
–
Analysis of S
hort Columns with Small Eccentricity
Example 9

1 (p. 308 of the textbook)
Given: Column cross section shown.
f
c
’ = 4000 psi
f
y
= 60,000 psi
Find: The maximum design load;
check ties.
Solution
1.
Check the steel ratio for the longitudinal steel.
ρ
g
= A
st
/A
g
= 8.00/(16)(16) = 0.0313
0.01 < 0.0313 < 0.08
OK
2.
Check the maximum number of bars.
From Table A

14
,
13” core (column size less cover on each side)
:
The
maximum number of
#
9 bars is
8
.
Therefore, the number of longitudinal bars is satis
factory.
3.
Calculate the maximum design axial strength.
φP
n(max)
= 0.80φ[0.85 f
c
’(A
g
–
A
st
) + f
y
A
st
]
= 0.80(0.65)[0.85(4.0)(256
–
8.00) + 60.0(8.00)]
φP
n(max)
=
688.1 kips
4.
Check the ties.
#
3
ties are
acceptable for longitudinal bars size up to
#
10
bars.
The spacing of the ties must not exceed the smalle
st
of
the following.
•
48 tie

bar diameters = 48(0.375) = 18”
•
16 longitudinal

bar diameters = 16(1.128) = 18”
•
T
he
least column dimension = 16”
Use 16”
(which matches the tie spacing).
•
Therefore, the tie spacing of 16” is OK.
9
.
8
Check t
he tie arrangement
to ensure that the clear spacing does not exceed 6”.
•
Clear space in excess of 6” would require additional ties in accordance with
ACI Code (Section 7.10.5.3).
T
he clear
space
= ½ [
16
–
2(1.5)
–
2(0.375)
–
3(1.128)] = 4.4” < 6” OK
Therefore, no extra ties are needed.
9
.
9
Example
–
Analysis of Short Columns with Small Eccentricity
Problem
9

4
(p. 3
40
of the textbook)
Given: Column cross section shown.
f
c
’ = 3000 psi
f
y
= 40,00
0 psi
Find: The maximum design load;
check the spiral.
Solution
1.
Check the steel ratio for the longitudinal steel.
A
st
= 8(1.00) = 8.00 in
2
A
g
=
π(18)
2
/4 = 254.5 in
2
ρ
g
= A
st
/A
g
= 8.00/
254.5
= 0.0314
0.01 < 0.031
4
< 0.08
OK
2.
Check the maximum number of bars.
From Table A

14, 15” core (column size less cover on e
ach side)
:
The maximum number of
#
9 bars is
10
.
Therefore, the number of long
itudinal bars is satisfactory.
3.
Calculate the maximum design axial strength.
φP
n(max)
= 0.8
5
φ[0.85 f
c
’(A
g
–
A
st
) + f
y
A
st
]
= 0.8
5
(0.
7
5)[0.85(
3
.0)(25
4.5
–
8.00) +
4
0.0(8.00)]
φP
n(max)
=
604.7
kips
4.
Check the
spiral
.
S
piral s
ize:
3/8 inch spiral
OK
Clear space: 2”
–
2(0.375/2) = 1.625”
1” < 1.625” <
3”
OK
Thus, 3/8” diameter spiral @ 2” spacing
OK
Spiral steel ratio:
Minimum: ρ
s
(min)
=
0.45(A
g
/A
ch
–
1)(f
c
’/f
yt
)
=
0.45(254.5/176.7
–
1)(3.0/40.0) = 0.0149
Actual: ρ
s
= 4A
sp
/
D
ch
s =
4(0.11)/(15)(2) = 0.0147 ≈ 0.0149
The spiral is slightly under

designed.
9
.
10
9

5
Design of Short Columns: Small Eccentricity
The design of reinforced concrete columns involves
•
The proportioning of the steel and concrete areas.
•
The selection of
properly sized and spaced ties or spirals.
Because the ratio of steel to concrete area must fall with a given range (
that is,
0.01 ≤ ρ
g
≤ 0.08), the strength equation is modified to include this term.
ρ
g
= A
st
/A
g
from which
A
st
= ρ
g
A
g
•
For a tied co
lumn,
φP
n(max)
= 0.80
φ
[0.85 f
c
’
(A
g
–
A
st
) + f
y
A
st
]
= 0.80
φ
[0.85 f
c
’
(A
g
–
ρ
g
A
g
) + f
y
ρ
g
A
g
]
φP
n(max)
= 0.80
φ
A
g
[0.85 f
c
’
(1
–
ρ
g
) + f
y
ρ
g
]
Because
P
u
≤ φP
n(max)
, an expression can be written for required A
g
in terms of the
material strengths,
P
u
, and ρ
g
.
•
For tied columns,
R
equired A
g
= P
u
/0.80
φ
[0.85 f
c
’
(1
–
ρ
g
) + f
y
ρ
g
]
•
F
or spiral columns,
R
equired A
g
= P
u
/0.8
5
φ
[0.85 f
c
’
(1
–
ρ
g
) + f
y
ρ
g
]
There
are
many
suitable
choices for the size of column that will provide the
necessary streng
th to carry the load P
u
.
•
A low ρ
g
will result in a larger required A
g
.
•
A high ρ
g
will result in a smaller required A
g
.
•
Other considerations may affect the practical choice of column size.

Architectural requirements.

Constructability: The desire t
o maintain column size from floor to floor so
that forms may be reused.
The procedure for the design of short columns for loads at small eccentricities is
illustrated by the following examples.
9
.
11
Example
–
Design of Short Columns with Small Eccentrici
ty
Example 9

3 (p. 311 of the textbook)
Given: A square tied column.
DL = 320 kips
LL = 190 kips
f
c
’ = 4000 psi
f
y
= 60,000 psi
Use ρ
g
= 0.03
Find: Design the column.
Solution
1.
Material strengths and
approximate
ρ
g
are given.
2.
Determine the
factored
axial load.
P
u
= 1.2P
DL
+ 1.6P
LL
= 1.2(320) + 1.6(190) =
688 kips
3.
Calculate the
required
gross column area
A
g
.
R
equire
d A
g
= P
u
/0.80
φ
[0.85 f
c
’
(1
–
ρ
g
) + f
y
ρ
g
]
= 688/{0.80(0.65)[0.85(4.0)
(1
–
0.03) + 60.0
(0.03)]}
= 259.5 in
2
4.
The required size of a square column is
= 16.1”
Use a 16” square column; this
column
size
causes
the actua
l ρ
g
to be slightly
greater than 0.03.
A
ctual A
g
= (16)(16) = 256 in
2
5.
Determine the loads on the concrete and steel areas.
Load on concrete = 0.80
φ
A
g
[0.85 f
c
’
(
1
–
ρ
g
)
]
= 0.80(0.65)(256)[0.85(4.0)(1
–
0.03)]
= 439.0 kips (Approximate since
ρ
g
increase
s
slightly)
Load on steel = 688
–
439 = 249 kips
9
.
12
Because the maximum design axial load strength of the steel is 0.80
φ
(
f
y
A
st
),
the required steel area
is
0.80
φ
(f
y
A
st
) = 249
A
st
= 249/0.80
φ
f
yt
= 249/(0.80)(0.65)(60.0)
= 7.98 in
2
Ba
rs of the same size are distributed evenly around the perimeter of the
column.
•
Bars must be selected in multiples of 4.
•
Use 8

#
9 bars (A
st
= 8.00 in
2
)
•
Table A

14 indicates a maximum of 8

#
9 bars for a 13” core.
6.
Design the ties.
Select a
#
3 t
ie
(ref.
Table A

14
)
.
The spacing
of the ties
must not
exceed the smalle
st
of
the following.
•
48 tie

bar diameters = 48(0.375) = 18”
•
16 longitudinal

bar diameters = 16(1.128)
=
18.0”
•
The l
east column dimension = 16”
Use
#
3 ties spaced at 16” c/c.
Ch
eck the
tie arrangement to ensure that the clear spacing does not exceed 6”.
•
Clear space in excess of 6” would require additional ties in accordance with
ACI Code (Section 7.10.5.3).
The clear space between adjacent bars in the same face is
½ [16
–
2
(1.5)
–
2(3/8)
–
3(1.128)] = 4.43” < 6”
Therefore
,
no additional ties are required by the ACI Code (Section
7.10.5.3).
7.
Sketch the design.
9
.
13
9

6
Summary of Procedure for Analysis and Design of Short Columns with Small
Eccentricities
Analysis
1.
Check the steel ratio ρ
g
for the longitudinal steel.
0.01 ≤ ρ
g
≤ 0.08
2.
Check the
maximum
number of
longitudinal bars to ensure that the bars are
within acceptable limits for clear space (using Table A

14).
•
The minimum number
of bars
is

For ba
rs enclosed with rectangular or circular ties: 4

For bars enclosed by spirals: 6
3.
Calculate the maximum design axial
load strength.
•
For tied columns
:
φ
P
n(max)
= 0.80
φ
[0.85 f
c
’
(A
g
–
A
st
) + f
y
A
st
]
•
For spiral columns
:
φ
P
n(max)
= 0.85
φ
[0.85
f
c
’
(A
g
–
A
st
) + f
y
A
st
]
4.
Check the lateral reinforcing.
•
For ties: check size, spacing
,
and arrangement (clear spacing).
•
For spirals: check size, ρ
s
, and clear distance.
Design
1.
Establish the material strengths.
2.
Determine the
factored
axial
load P
u
.
3.
Calculate
the required gross column area A
g
.
4.
Select the column dimensions. Use full

inch increments.
5.
Determine
the load carried by the concrete and the load carried by the
longitudinal steel.
For a tied column,
Load on concrete = 0.
80
φ
A
g
[0.85 f
c
’
(1
–
ρ
g
)]
Load on steel = P
u
–
load on concrete
Determine the required longitudinal steel area.
A
st
= Load on steel/0.80
φ
f
yt
Select the longitudinal steel.
9
.
14
6.
Design the lateral reinforcing (ties or spiral).
Select a
#
3,
#
4, or
#
5
tie
(ref.
Table A

14
)
.
The spacing of the ties must not exceed the smalle
st
of
the following.
•
48 tie

bar diameters
•
16 longitudinal

bar diameters
•
The
least column dimension
Specify the tie size and spacing.
Check the tie arrangement to ensure th
at the clear spacing does not exceed 6”.
•
Clear space in excess of 6” would require additional ties in accordance with
ACI Code (Section 7.10.5.3).
7.
Sketch the design.
9
.
15
Example
–
Design of Short Columns with Small Eccentricity
Problem 9

10
(p. 34
1
of the textbook)
Given: A short, circular spiral column.
DL = 175 kips
LL = 325 kips
f
c
’ = 4000 psi
f
y
= 60,000 psi
Use ρ
g
= 0.03
Find: Design the column.
Solution
1.
Material strengths and approximate ρ
g
are given.
2.
Determine the
factored
axi
al load.
P
u
= 1.2
P
DL
+ 1.6
P
LL
= 1.2(175) + 1.6(325) =
73
0 kips
3.
Calculate the required gross column area A
g
.
R
equired A
g
= P
u
/0.85
φ
[0.85 f
c
’
(1
–
ρ
g
) + f
y
ρ
g
]
=
730
/{0.85
(0.7
5)[0.85(4.0)(1
–
0.03) + 60.0(0.03)]}
=
224.6
in
2
4.
The requi
red size of a
circular
column is
D = (4A
g
/π)
1/2
= [4(
224.6
)/π]
1/2
= 1
6.91
”
Use a 1
8
”
circular
column; this
column
size
causes
the actual ρ
g
to be
smaller
than 0.03.
A
ctual A
g
=
π
D
2
/4 = π (1
8
)
2
/4 =
254.5
i
n
2
5.
Determine the loads on the concrete and
steel areas.
Load on concrete = 0.8
5
φA
g
[0.85 f
c
’(1
–
ρ
g
)]
= 0.8
5
(0.7
5)(25
4.5
)[0.85(4.0)(1
–
0.03)]
=
535
kips (Approximate since ρ
g
de
crease
s
slightly)
Load on steel =
730

535
=
195
kips
9
.
16
Because the maximum design axial load strength of the s
teel is 0.8
5
φ
(f
y
A
st
),
the required steel area
is
0.8
5
φ
(f
y
A
st
) =
195
A
st
=
195
/0.8
5
φ
f
yt
= 195/(0.85)(0.7
5)(60.0) =
5.10
in
2
Bars of the same size are distributed evenly around the perimeter of the
column.
•
A minimum of 6 bars are required.
•
The
maximum number
s
of bars (depending on the size of the core, the size
of the spiral, and the size of the bar) are listed in Table A

14.
•
Possible selections (required A
st
= 5.10 in
2
, core = 15”):
6

#
9
A
st
= 6.00 in
2
(maximum 10

#
9 bars)
OK
7

#
8
A
st
= 5.53 in
2
(maximum 12

#
8 bars)
OK
9

#
7
A
st
= 5.40 in
2
(maximum 13

#
7 bars)
OK
12

#
6
A
st
= 5.28 in
2
(maximum 14

#
6 bars)
OK
17

#
5
A
st
= 5.27 in
2
(maximum 15

#
5
bars)
NG
Select
7

#
8
bars (A
st
=
5.53
in
2
)
•
A little larger A
st
is sel
ected due to the selected column size.
Check ρ
g
= A
st
/A
g
= 5.53/254.5 = 0.0217
0.01 < 0.0217 < 0.08
OK
6.
Design the lateral reinforcing (spiral).
Select spiral size
and determine the spacing
: Try a
#
3
spiral
A
g
= π
D
2
/4 = π
(18)
2
/4 = 254.5 in
2
D
ch
= 18
–
2(1.5) = 15”
A
ch
= π
D
ch
2
/4 =
π
(15)
2
/4 = 176.7 in
2
Spiral steel ratio:
Minimum
spiral steel ratio
:
ρ
s(min)
= 0.45
(A
g
/A
ch
–
1)(f
c
’/f
yt
)
= 0.45
(254.5/176.7
–
1)(
4
.0/
6
0.0) = 0.0
132
Maximum
spiral
spacing:
s
max
= 4A
sp
/ρ
s
D
ch
= 4(0.11)
/0.0132(15)
=
2.22”
Use 2” spacing.
Check
the
actual spiral steel ratio:
ρ
s
= 4A
sp
/D
ch
s = 4(0.11)/(15)(2) = 0.0147 > 0.0132
OK
9
.
17
Check
the
spiral
c
lear spac
ing
:
2”
–
2(0.375/2) = 1.625”
1” < 1.625” < 3” OK
Thus, 3/8” diameter spiral @ 2” spa
cing OK
7.
Sketch the final design.
•
The final design is shown below.
9
.
18
9

7
The Load

Moment Relationship
If a force P
u
is applied to a cross section at a distance e (eccentricity) from the
centroid, a moment equal to P
u
e is also applied.
•
M
u
(wh
ere M
u
= P
u
e) is defined as the
factored
moment that is applied on a
compression member along with the
factored
axial load of P
u
.
To prevent
the column
from being
overloaded when subjected to a load
with an eccentricity,
•
The load
P
u
must be reduced so
t
hat the column can carry both P
u
and P
u
e.
•
The required reduction
in the load
P
u
depends on the magnitude of
the eccentricity.
9

8
Columns Subjected to Axial Load at Large Eccentricity
As previously seen, u
nder the 2008 ACI Code, the maximum design axial
load
strength φP
n(max)
is given by ACI Equations (10

1) and (10

2).
For spiral columns:
φP
n(max)
= 0.85
φ
[0.85 f
c
’(A
g
–
A
st
) + f
y
A
st
]
[ACI Eq. (10

1)]
For tied columns:
φP
n(max)
= 0.80
φ
[0.85 f
c
’(A
g
–
A
st
) + f
y
A
st
]
[ACI Eq. (10

2)]
These tw
o equations apply for columns with
small
eccentricities, that is,
eccentricities no greater than
•
0.10
h for tied columns and
•
0.05
h for spiral columns.
where
h = the overall dimension of the column
For large eccentricities,
•
ACI Equations (10

1
) and (10

2) no longer apply.
•
φP
n
must be reduced below φP
n(max)
.
The occurrence of columns subjected to eccentricities sufficiently large so that
moment must be a design consideration is common.
9
.
19
•
Interior columns supporting beams of equal spans receive unequal loads from
the beams
due to applied live load patterns (ref. Figure 9

10a, p. 317 of the
textbook).

The unequal loads mean that the column must carry both load and moment.

The resulting eccentricity could be greater than the small eccentricity.
•
In a rigid frame, the ri
gidity of the joint requires the column to rotate along
with the end of the beam that the column supports (ref. Figure 9

10b, p. 317 of
the textbook).
•
The beam reaction
may be
eccentrically applied on the column through a column
bracket (ref. Figure 9

10
c, p. 317 of the textbook).
9

9
φ Factor Considerations
Columns discussed so far have strength

reduction factors applied in a
straightforward manner.
•
φ = 0.65 for tied columns.
•
φ =
0.75 for spiral columns
These φ factors correspond to the
compression

controlled
strain limit or net
tensile strain in the extreme tension reinforcement, ε
t
≤ 0.002.
Eccentrically loaded columns carry both axial load and moment.
For values of ε
t
> 0.002, the φ equations from the ACI Code (Section 9.3.2) give
higher values
than indicated above
(ref. p. 40 of the textbook)
.
For tied columns: φ = 0.65 + (ε
t
–
0.002)(250/3)
0.65 ≤
φ ≤ 0.90
For spiral columns: φ = 0.75 + (ε
t
–
0.002)(200/3)
0.75 ≤
φ ≤ 0.90
9

10
Analysis of Short Columns: Large Eccentricity
The f
irst step in the investigation of short columns carrying loads
with a
large
eccentricity is to determine the strength of the given column cross section that
carries loads at various eccentricities.
•
This may be thought of as an analysis process.
•
For thi
s development, the design axia
l strength φP
n
will be found (where P
n
is
the
nominal
axial load strength at a given eccentricity).
9
.
20
Example
–
Analysis of Short Columns with Large Eccentricity
Example 9

5 (p. 318 of the textbook)
Given: Tied column with cross section shown.
f
c
’ = 400
0 psi
f
y
= 60,000 psi
Bending is about the y

axis.
Find: Column streng
th for the following
conditions.
a)
Small eccentricity (e = 0 to 0.10
h)
b)
e = 5”
c)
The
compression

controlled
strain limit
(balanced condition)
, ε
t
= 0.002
d)
ε
t
= 0.004
e)
The ten
sion

controlled strain limit, ε
t
= 0.005
f)
Pure moment
Solution
A
g
= 20(14) = 280 in
2
A
st
= 6(1.0) = 6.0 in
2
a)
Smal
l eccentricity (e = 0 to 0.10
h)
φP
n
= φP
n(max)
= 0.80
φ
[0.85 f
c
’
(A
g
–
A
st
) + f
y
A
st
]
= 0.80(0.65)[0.85(4.0)(280
–
6.0) + 60.0(6)]
= 671.6 kips
The
maximum small eccentricity is
e
max
= 0.10
h = 0.10(20) = 2.0”
The corresponding maximum moment is
M
u
=
φM
n
= φP
n
e = 671.6(2.0) = 1
343.2 kip

inch (111.9 kip

ft)
b)
e = 5”
In part a), all steel is in compression.
•
As the eccentri
city increases, the steel on the side of the column away from
the load is subjected to less compression.
9
.
21
•
There is
a certain
value of eccentricity at which the stress in the steel
changes from compression to tension.
•
The value of eccentricity where this
change takes place is not known.

The strain situation must be assumed and verified by calculation.
The assumptions at
nominal
strength are
1.
Maximum concrete strain = 0.003
2.
Compression steel:
ε
s
’ > ε
y
, therefore f
s
’ = f
y
3
.
Tension steel:
ε
s
< ε
y
, therefore f
s
< f
y
The unknown quantities are P
n
and c.
Compression force in the concrete:
C
1
= 0.85f
c
’
a
b = 0.85(4.0)(0.85c)(14) = 40.46c (kips)
Compression force in the steel (accounting f
or the area of concrete displaced
by the steel):
C
2
= f
y
A
s
’
–
0.85
f
c
’
A
s
’ = 60
.0(3)(1.0)
–
0.85(4.0)(3)(1.0)
= 169.8 kips
Tension force in the steel:
By similar triangles: 0.003/c = ε
s
/(d
–
c)
S
o ε
s
= (0.003/c)(d
–
c) = (0.003/c)(17
–
c)
T = f
s
A
s
= ε
s
E
s
A
s
=
(0.003/c)(17
–
c)(29,000)(
3)(1.0)
=
261(17
–
c)/c
Summing forces:
P
n
= C
1
+ C
2

T = 40.46c + 169.8
–
261(17
–
c)/c
P
n
= 40.46c + 169.8
–
4,437.0/c + 261.0
P
n
= 430.8 + 40.46c
–
4,437.0/c
Summing moments (∑M
T
= 0):
P
n
(12) = C
1
(d
–
a/2) +
C
2
(14)
= 40.46c(17
–
0.85c/2) + 169.8(14)
P
n
(12) = 687.82c
–
17.20c
2
+ 2,377.2
P
n
= 198.1 + 57.32c
–
1.433c
2
9
.
22
Equate the two equations for P
n
and solve for c.
430.8 + 40.46c
–
4,437.0/c = 198.1 + 57.32c
–
1.433c
2
430.8c + 40.46c
2
–
4,437.0 = 1
98.1c + 57.32c
2
–
1.433c
3
1.433c
3
–
16.86c
2
+ 232.7c

4,437.0 = 0
c
10.0
12.0
14.0
15.0
14.85
14.90
14.86
f(c)

2363.0

1596.2

551.6
+96.4

6.68
+27.43
+ 0.12
Use c = 14.86
”
Solve for P
n
P
n
= 430.8 + 40.46c
–
4,437.0/c
= 430.8 + 40.46(14.86)
–
4,437.0/14.86 = 733.4 kips
Calculate the tensile strain in the extreme tensile reinforcement.
0.003/c = ε
t
/(d
–
c)
0.003/14.86 = ε
t
/(17
–
14.86)
ε
t
= (0.003/14.86)(17
–
14.86) = 0.00043 < 0.002
For ε
t
≤ 0.002, the corresponding tied column streng
th

reduction factor
is
φ = 0.65.
Therefore, φP
n
= 0.65(733.4) = 476.7 kips
Check
the assumptions that were made.
Compression steel: ε
s
’ > ε
y
By similar triangles ε
s
’/(c
–
3) = 0.003/c
ε
s
’/(14.86
–
3) = 0.003/14.86
ε
s
’ = (14.86
–
3)(0.0
03/14.86)
= 0.0024 > ε
y
= 0.00207 (Table A

1)
Since ε
s
’ =
ε
y
,
then
f
s
’ = f
y
OK
Tension steel: ε
t
< ε
y
, therefore f
s
< f
y
.
ε
t
= 0.00043 (as determined above) < ε
y
= 0.00207
f
s
= ε
t
E
s
= 0.00043(29,000) = 12.53 ksi < 60.0 ksi
OK
All assu
mptions are verified.
The design moment strength for an eccentricity of 5”
is
φP
n
e = 476.7 (5) = 2,383.5 kip

inch (198.6 kip

ft)
Therefore, the column has a design load

moment combination strength of 476.7
kips axial load and 198.6 kip

ft moment (applied
about the y

axis).
9
.
23
c)
The
compression

controlled
strain limit
(balanced condition)
.
•
The
compression

controlled
strain limit (balanced condition) exists when the
concrete reaches a strain of
ε
c
=
0.003 at the same time the extreme
tension steel reaches a
strain of
ε
t
=
0.002
(ACI Code, Section 10.3.3).

P
b
is the
nominal
axial load strength at the balance condition.

e
b
is the associated eccentricity.

c
b
is the distance from the compression face to the balanced neutral
axis.
Using the strain diagram
shown
below
, we can calculate the value of c
b
.
•
Using similar triangles,
0.003/c
b
= 0.002/(d
–
c
b
)
0.003/c
b
= 0.002/(17
–
c
b
)
0.003 (17
–
c
b
) = 0.002 c
b
0.051
–
0.003 c
b
= 0.002 c
b
0.005 c
b
= 0.
0
51
c
b
= 10.2
0
”
For ε
t
= 0.002, the tied c
olumn strength reduction factor φ = 0.65.
Using similar triangles, we can determine the strain
and stress
in the
compression steel, ε
s
’.
0.003/c
b
= ε
s
’/(c
b
–
3)
0.003/10.2
0
= ε
s
’/(10.2
0
–
3)
ε
s
’ = (0.003/10.2
0
)(10.2
0
–
3) = 0.0021
2
Because ε
s
’ > ε
y
=
0.00207, the compression steel has yielded and
f
s
’ = f
y
= 60.0 ksi
Determine t
he forces acting on the section.
Compression force in the concrete:
C
1
= 0.85
f
c
’
a
b = 0.85(4.0)[
0.85(10.2
0
)
]
(14) = 412.7 kips
Compression force in the steel (accounting f
or the area of concrete
displaced by the steel):
C
2
= f
y
A
s
’
–
0.85
f
c
’
A
s
’
= 60.0(3)(1.0)
–
0.85(4.0)(3)(1.0
)
= 169.8 kips
Tension force in the steel:
T = f
s
A
s
= ε
s
E
s
A
s
=
(60.0
)(3)(1.0) = 1
80
.0 kips
9
.
24
The
nominal
axial load strength is
P
b
= C
1
+ C
2
–
T = 412.7 + 169.8
–
180.0 = 402.5 kips
The value e
b
may be established by summing moments about T:
P
b
(e
b
+ 7) = C
1
(d
–
a/2) + C
2
(d
–
3)
402.5(e
b
+ 7) = 412.7
[17
–
0.85(10.2
0
)/2] + 169.8(17
–
3)
402.5 e
b
+ 2,817.5 = 5,226.8 + 2,377.2
402.5 e
b
=
4,786.5
e
b
= 11.89”
At the balanced condition,
Axial load strength: φP
b
= 0.65(402.5) = 261.6 kips
Moment strength: φP
b
e
b
= 261.6(11.89) = 3,110.4 kip

inch (259
.2 kip

ft)
d)
ε
t
= 0.004
Find the neutr
al axis using similar triangles.
0.003/c = 0.004/(d
–
c)
0.003/c = 0.004/(17
–
c)
0.003(17
–
c) = 0.004c
0.051
–
0.003c = 0.004c
0.007c = 0.051
c = 0.051/0/0.007 = 7.29”
Find the strain in the compression
steel, using similar triangles:
0.003/c = ε
s
’/(c
–
3)
0.003/7.29 = ε
s
’/(7.29
–
3)
ε
s
’ = (0.003/7.29)(7.29
–
3) = 0.00177 < ε
y
= 0.00207
Find the stress in the compression steel:
f
s
’ = ε
s
’ E
s
= 0.00177(29,000) = 51.3 ksi
Determine the forces acting on
the section:
Compression force in the concrete:
C
1
= 0.85
f
c
’
a
b = 0.85(4.0)
[
0.85(7.29)
]
(14) =
295.0
kips
Compression force in the steel (accounting for the area of concrete
displaced by the steel):
C
2
=
f
s
’
A
s
’
–
0.85
f
c
’
A
s
’
=
51.3
(3)(1.0)
–
0.85(4.0
)(3)(1.0
) =
1
43.7
kips
Tension force in the steel:
T = f
s
A
s
= ε
s
E
s
A
s
= (60.0)(3)(1.0) = 180.0 kips
9
.
25
The
nominal
axial load strength is
P
n
= C
1
+ C
2
–
T = 295.0 + 143.7
–
180.0 = 258.7 kips
The value e may be established by summing moments about T:
P
n
(e + 7) = C
1
(d
–
a/2) + C
2
(d
–
3)
258.7(e + 7) = 295.0[17
–
0.85(7.29)/2] + 143.7(17
–
3)
258.7e + 1,810.9 = 4,101.0 + 2,011.8
258.7 e =
4,301.9
e =
16.63
”
For ε
t
= 0.004,
φ = 0.65 + (ε
t
–
0.002)(250/3) = 0.65 + (0.004
–
0.002)(250/3) = 0.8
17
0
.65 < 0.8
17
< 0.90
OK
Therefore, at ε
t
= 0.004,
Axial load strength:
φP
n
= 0.8
17
(258.7) = 21
1.4
kips
Moment strength:
φM
n
= φP
n
e = 21
1.4
(16.63)
= 3,5
15.6
kip

inch (293.
0
kip

ft)
e)
The tension

controlled strain limit, ε
t
= 0.005
Find the ne
utral
axis using similar triangles.
0.003/c = 0.005/(d
–
c)
0.003/c = 0.005/(17
–
c)
0.003(17
–
c) = 0.005c
0.051
–
0.003c = 0.005c
0.008c = 0.051
c = 0.051/
0.008 = 6.38”
Find the strain in the compression steel, using similar triangles:
0.003/c = ε
s
’
/(c
–
3)
0.003/6.38 = ε
s
’/(6.38
–
3)
ε
s
’ = (0.003/6.38)(6.38
–
3) = 0.00159 < ε
y
= 0.00207
Find the stress in the compression steel:
f
s
’ = ε
s
’ E
s
= 0.00159(29,000) = 46.1 ksi
Determine the forces acting on the section:
Compression force in the concret
e:
C
1
= 0.85
f
c
’
a
b = 0.85(4.0)
[0.85
(6.38)
]
(14) = 258.1 kips
9
.
26
Compression force in the steel (accounting for the area of concrete
displaced by the steel):
C
2
=
f
s
’
A
s
’
–
0.85
f
c
’
A
s
’
= 46.1(3)(1.0)
–
0.85(4.0)(3)(1.0) = 128.1 kips
Tension force in the
steel:
T = f
s
A
s
= ε
s
E
s
A
s
= (60.0)(3)(1.0) = 180.0 kips
The
nominal
axial load strength is
P
n
= C
1
+ C
2
–
T = 258.1 + 128.1
–
180.0 = 206.2 kips
The value e may be established by summing moments about T:
P
n
(e + 7) = C
1
(d
–
a/2) + C
2
(d
–
3)
206.2(e +
7) = 258.1[17
–
0.85(6.38)/2] + 128.1(17
–
3)
206.2e + 1,443.4 = 3,687.9 + 1,793.4
206.2 e = 4,037.9
e = 19.58”
For ε
t
= 0.005, φ = 0.90
Therefore, at ε
t
= 0.005,
Axial load strength:
φP
n
= 0.90(206.2) = 185.6 kips
Moment strength:
φM
n
= φP
n
e = 185.6(1
9.58)
= 3,634.0 kip

inch (302.8 kip

ft)
f)
Pure moment
The analysis of pure moment condition is similar to the analysis of the case
where the eccentricity is infinite.
•
We will
determine
the design moment strength φM
n
.
•
P
u
and φP
n
are both zero.
For p
ure moment, the bars on the load side are in compression; the bars on the
side away from the load are in tension.
•
The total tensile and compressive forces must be equal to each other.

The total tensile force T = A
s
f
y

The total compressive force con
sists of the force in the steel and
concrete (i.e. C
1
and C
2
).
◦
C
1
= 0.85 f
c
’ a b
◦
C
2
= A
s
’ f
s
’
•
Since A
s
= A
s
’, A
s
’ must be at a stress less than yield.
9
.
27
Assume
the tensile steel
A
s
is at yield stress.
•
Find the strain in the compression steel using
similar triangles,
0.003/c =
ε
s
’/
(
c
–
3
)
ε
s
’ = (0.003/c)(c
–
3)
f
s
’ = ε
s
’
E
s
= (0.003/c)(c
–
3) 29,000 = (87/c)(c
–
3)
Determine the forces acting on the section:
Compression force in the concrete:
C
1
= 0.85
f
c
’
a
b = 0.85(4.0)(0.85)c(14) = 40.46c (k
ips)
Compression force in the steel (accounting for the area of concrete
displaced by the steel):
C
2
=
f
s
’
A
s
’
–
0.85f
c
’A
s
’
= (87/c)(c
–
3)(3)(1.0)
–
0.85(4.0)(3)(1.0)
= 261.0
–
783.0/c
–
10.2
C
2
=
250.8
–
783.0/c
Tension force in the steel:
T = f
s
A
s
= (60.0)(3)(1.0) = 180.0 kips
For equilibrium,
C
1
+ C
2
= T
40.46c + 250.8
–
783.0/c = 180.0
40.46c
2
+ 250.8c
–
783.0 = 180.0c
40.46c
2
+ 70.8c
–
783.0 = 0
c =
3.61”
Determine the forces acting on the section:
Compression force in the concrete:
C
1
= 40.46c = 146.1 kips
Compression force in the steel (accounting for the area of concrete
displaced by the steel):
C
2
= 250.8
–
783.0/c = 250.8
–
783.0/3.61 =
33.90 kips
Tension force in the steel:
T = f
s
A
s
= ε
s
E
s
A
s
= (60.0)(3)(1.0) = 180.0 kips
Summar
izing the forces (recall P
n
= 0):
∑F =
C
1
+ C
2
–
T =
146.1 + 33.90
–
180.0 =
0
9
.
28
Compute the
internal moment by summing moments about T:
M
n
= C
1
(d
–
a/2) + C
2
(d
–
3)
M
n
= 146.1[17
–
(0.85)(3.61
)
/2] + 33.90(17
–
3)
= 2,259.5 + 474.6 = 2,734.1 kip

inch
(227.8 kip

ft)
For ε
t
≥ 0.005, the strength reduction factor φ = 0.90.
M
oment strength
:
φM
n
= 0.90(227.8) = 205.0 kip

ft
The results of the six parts of this example may be plotted to form an
interaction
diagram
(ref. Figure 9

18, p. 328 of the textbook
).
•
The diagram applies
only
to this example.

Axial load strengths φP
n
are plotted
along
the vertical scale.

Moment strength φM
n
are plotted along the horizontal scale.
•
Any point
on
the solid line represents a
n allowable
load

moment combination.
•
A
ny point
within
the solid line also represents a
n allowable
load

moment
combination (for which the column is
overdesigned
).
•
Any point
outside
the solid line represents an unacceptable load

moment
combination.
Radial lines from the origin represent vario
us eccentricities.
•
The intersection of the e = e
b
line with the solid line represents the balanced
condition.

Any eccentricity less than e
b
results in a
compression

controll
ed
column.

Points above the e = e
b
line and within the solid line represent
c
ompression

controlled
load

moment combinations.
•
Points between the e = e
b
line and
the e = 19.6” line, the column is in the
transition zone (
i.e.
0.002 ≤ ε
t
≤ 0.005)
.
•
For e
ccentricities
greater than 19.6”, the column is
tension

controlled
.
9
.
29
The
calculations involved with column loads at large eccentricities are involved and
tedious.
•
The previous example was an analysis.
•
The d
esign of a cross section using the calculation approach
is
a trial

and
–
error
method and
is
extremely tedious.
Design
and analysis aids have been developed
to
shorten the process.
•
These aids are found in the form of tables and charts
(i.e. ACI interaction
diagrams)
.
•
The design aids are developed in a manner done in Example 9

5.
•
No strength reduction facto
r
s φ are in
corporated into the diagrams.
•
Eight interaction diagrams are included in Appendix
A Diagrams A

15 through
A

22, pp. 498

501 of the textbook).
The diagrams take on the general form of the diagram developed for Example 9

5,
but are generalized to be appli
cable to more situations.
•
The following definitions are useful:
ρ
g
= A
st
/A
g
h = column dimension perpendicular to the bending axis
γ = ratio of distance between centroids of outer rows of bars and column
dimension perpendicular to the axis of bending
Note that the vertical axis and horizontal axis of Diagrams A

15 through A

22 are
in general terms of K
n
and R
n
, where
K
n
= P
n
/f
c
’A
g
R
n
= P
n
e/f
c
’A
g
h
P
n
is the
nominal
axial load strength.
P
n
e is the
nominal
moment strength.
The slope of the radial line fro
m the origin can be represented as
slope = rise/run =
K
n
/R
n
=
(
P
n
/f
c
’A
g
)/(
P
n
e/f
c
’A
g
h) = h/e
Notice the following features of the diagrams.
•
Curves (concentric with the origin) are shown for the range of allowable ρ
g
values from 0.01 to 0.08.
•
A line ne
ar the horizontal axis labeled ε
t
= 0.005 indicates the limit for
tension

controlled
sections.
9
.
30

Columns with load

moment

strength combinations below this line are
tension

controlled
(φ = 0.90).
•
The line labeled f
s
/f
y
= 1.0 indicates the
balanced
conditi
on
.

The
balanced
condition (i.e. compression

controlled strain limit) occurs when
the concrete reaches a strain of 0.003 at the same time the extreme tensile
steel reaches a strain of 0.002
.

Columns with load

moment

strength combinations above this lin
e area
compression

controlled
(φ = 0.65 for tied columns; φ = 0.75 for spiral
columns).
•
Columns with load

moment

strength combinations between these two lines are
in the transition zone.
•
The line labeled K
max
indicates the maximum allowable
nominal
loa
d strength
[φP
n
(max)
] for columns loaded with small eccentricities.

A horizontal line drawn through the intersection of the K
max
line and a ρ
g
curve corresponds to the horizontal line near the top of the interaction
diagram in Figure 9

18 (p. 328 of the
textbook).
The following examples illustrate the use of the ACI interaction diagrams for
analysis and design of short reinforced concrete columns.
9
.
31
Example
–
Analysis of Short Columns with Large Eccentricity (using
the ACI
interaction diagrams)
Exampl
e 9

6 (p.
329 of the textbook)
Note: This example revisits Example 9

5.
Given: Column cross section shown.
f
c
’ = 4000 psi
f
y
= 60,000 psi
e = 5”
Find:
a)
The
axial load strength φP
n
.
b)
T
he
moment strength φM
n
.
c)
Compare the results with Examp
le 9

5
b
.
Solution
First, determine which interaction diagram to use.
•
Select the diagram based on the type of cross section, material strengths, and
the factor γ.
γ =
γh/h =
distance between outer rows of bars
column dimension perpendicu
lar to the bending axis
γ =
γh/h =
14
”
/20
”
= 0.7
Therefore, use interaction diagram A

15
(p. 498 of the textbook)
.
Calculate ρ
g
to establish a curve value.
ρ
g
= A
st
/A
g
= 6(1)/20(14) = 0.0214
0.01 ≤ 0.0214 ≤ 0.08
OK
Calculate the slope of the radi
al line from the origin, which relates
h
and e.
S
lope =
K
n
/R
n
=
h/e = 20
”
/5
”
= 4
Draw a radial line from the origin to an estimated ρ
g
= 0.0214 curve.
•
Select convenient values (e.g. K
n
= 1.0 and R
n
= 0.25
that form a slope of 4 with
the origin
)
.
•
Use
a straight edge to draw the radial line from the origin to intersect with an
estimated ρ
g
= 0.0214 curve.
9
.
32
At the intersection, read the following values.
•
K
n
≈ 0.64
•
R
n
≈ 0.16
Because this combination of load and moment is above the f
s
/f
y
= 1.0 line, th
is is
a
compression

controlled
section and φ = 0.65.
Calculate the axial load strength and the moment strength.
φP
n
=
φK
n
f
c
’
A
g
= 0.65(0.64)(4.0)(20)(14) = 465.9 kips
φM
n
= φR
n
f
c
’
A
g
h = 0.65(0.16)(4.0)(20)(14)(20)
=
2,329.6 kip

inch (194.1 kip

ft
)
or
φM
n
=
φP
n
e = 465.9(5) = 2
,
329.5 kip

inch (194.1 kip

ft)
These results compare reasonably well with the pr
evious results from Example 9

5.
φP
n
= 476.7 kips
φM
n
= 198.6 kip

ft
9
.
33
Example
–
Analysis of Short Columns with Large Eccentricity (using the A
CI
interaction diagrams)
Example 9

7
(p.
3
31
of the textbook)
Given: Circular column cross section.
P
u
= 1100 kips
M
u
= 285 kip

ft
f
c
’ =
4000 psi
f
y
= 60,000 psi
Find: Design a circular spirally
reinforced concrete column.
Solution
Estimate the c
olumn size based on ρ
g
= 0.01 and axial load only.
R
equired A
g
= P
u
/{0.85
φ
[0.85
f
c
’
(1

ρ
g
) + f
y
ρ
g
]}
A
g
= 1100/{0.85(0.75)[0.85(4.0)(1
–
0.01) + 60.0(0.01)]}
= 435.1 in
2
R
equired diameter = (4A
g
/π)
1/2
= [4(435.1)/π]
1/2
= 23.5”
Try a 24”

diamet
er column (A
g
= πD
2
/4 = 452.4 in
2
)
, 3/8” spiral, and
#
9 bars.
D
istance between the outer rows of bars
(γh):
γh
= 24
–
2(1.5)
–
2(3/8)
–
2(1.128/2) = 19.12”
γ =
γh/h =
distance between outer rows of bars
column dimension perpendicular
to the bending axis
γ =
γh/h =
19.12
”
/24
”
= 0.797
Use ACI Interaction Diagram A

21 from Appendix A (p. 501 of the textbook).
Determine the required ρ
g
.
•
Assume that the column is
compression

controlled
(φ = 0.75), subject to a later
check.
•
Calculate
the values of required K
n
and R
n
.

T
he required
nominal
axial load P
n
= P
u
/φ

T
he
nominal
moment P
n
e = M
u
/φ
9
.
34
R
equired K
n
=
P
n
/
f
c
’A
g
=
P
u
/φ
f
c
’A
g
= 1100
/
[
0.75
(4.0)(452.4)
]
= 0.810
R
equired
R
n
=
P
n
e/
f
c
’A
g
h
=
M
u
/φf
c
’A
g
h = 285(12)/
[
0.75(
4.0)(452.4)(24)
]
= 0.
105
•
From Diagram A

21, ρ
g
= 0.024.

Note that this is well above the f
s
/f
y
= 1.0 line. Therefore,
◦
The column is
compression

controlled
.
◦
The assumption that φ = 0.75 is OK.
The required area of steel is
R
equired A
s
= ρ
g
A
g
= 0.024(452.4) = 10.86 in
2
Select 11

#
9 bars (A
s
= 11.00 in
2
).
•
Check the maximum number of
#
9 bars from Table A

14
.

B
ased on a
core size of 24
–
2(1.5) = 21”,
the maximum number of bars is
15
.
OK
Design the spiral.
•
Use 3/8

inch diameter spiral.
•
The concrete co
re diameter and area are
D
ch
= 24
–
2(1.5) = 21”
A
ch
= π
D
ch
2
/4 = π
(21)
2
/4 = 346.4 in
2
•
The required steel ratio is
R
equired ρ
s
= 0.45(A
g
/A
ch
–
1)(f
c
’/f
yt
)
= 0.45(452.4/346.4
–
1)
(4.0/60.0) = 0.0092
•
The required spacing is
R
equired s =
4A
sp
/D
ch
ρ
s
= 4(0.11)/21(0.0092) = 2.28”
Use 2

1/4” spacing.
Sketch the final design.
•
The final design is shown
at the
right (ref.
Figure 9

21
,
p. 332 of
the textbook).
9
.
35
Example
–
Analysis of Short Columns with Large Eccentricity (using the ACI
in
teraction diagrams)
Example 9

8
(p.3
32
of the textbook)
Given: Square tied column cross section.
P
u
= 1300 kips
M
u
= 550 kip

ft
f
c
’ = 4000 psi
f
y
= 60,000 psi
Find: Design a square tied reinforced
concrete column.
Solution
Estimate the column si
ze based on ρ
g
= 0.01 and axial load only.
R
equired A
g
= P
u
/{0.80
φ
[0.85
f
c
’
(1

ρ
g
) + f
y
ρ
g
]}
A
g
= 1300/{0.80(0.65)[0.85(4.0)(1
–
0.01) + 60.0(0.01)]}
= 630.4 in
2
R
equired column size = √A
g
= (630.4)
1/2
= 25.1”
Try a 26”

square column (A
g
= 676
.0 in
2
),
#
3 ties, and
#
9 bars.
D
istance between the outer rows of bars
(γh):
γh
= 26
–
2(1.5)
–
2(0.375)
–
2(1.128/2) = 21.12”
γ =
γh/h =
distance between outer rows of bars
column dimension perpendicular to the bending axis
γ =
γh/h
=
21.12
”
/26
”
= 0.812
Use ACI Interaction Diagram
A

1
8
from Appendix A (p. 49
9
of the textbook).
Determine the required ρ
g
.
•
Assume that the column is
compression

controlled
(φ = 0.65), subject to a later
check.
•
Calculate the values of required K
n
and
R
n
.

The required
nominal
axial load P
n
= P
u
/φ

The
nominal
moment P
n
e = M
u
/φ
9
.
36
R
equired K
n
=
P
n
/
f
c
’A
g
=
P
u
/φf
c
’A
g
= 1300/
[(
0.65
)
(4.0)(676.0)
]
= 0.740
R
equired R
n
=
P
n
e/
f
c
’A
g
h
=
M
u
/φf
c
’A
g
h = 550(12)/
[
0.65(4.0)(676.0)(26)
]
= 0.144
•
From Diagram
A

1
8
,
ρ
g
= 0.023.

Note that this is well above the f
s
/f
y
= 1.0 line. Therefore,
◦
The column is
compression

controlled
.
◦
The assumption that φ = 0.65 is OK.
The required area of steel is
R
equired A
s
= ρ
g
A
g
= 0.023(
6
76.0) = 15.55 in
2
Select 16

#
9 bars (A
s
=
16.00 in
2
).
•
Check the maximum number of
#
9 bars from Table A

14.

Based on a core size of 26
–
2(1.5) = 23”,
the maximum number of bars is 20
.
OK
Design the ties.
•
Use
#
3 tie since the longitudinal bars are
#
10
or smaller
.
•
The maximum tie
spacing is the smallest of the following.

16 bar diameters = 16(1.128) = 18.0”

48 tie diameters = 48(0.375) = 18”

The
least column dimension = 26”
Use
#
3 ties at 18” spacing.
Sketch the final design.
•
The final design is shown
at the
right (
ref.
Figure 9

23
,
p. 334 of
the textbook
)
.
9
.
37
9

11
The
Slender Column
Thus far, the design and analysis have been limited to short columns that require
no consideration of strength due to the possibility of buckling.
•
A
l
l compression members experience
buckling as they become longer and more
flexible.
•
A column may be categorized as
slender
if the cross

sectional dimensions are
small in comparison to the unsupported length.
•
The degree of slenderness is expressed in terms of a slenderness
ratio kℓ
u
/
r,
where
k = effective length factor for compression members
ℓ
u
= the unsupported length of a compression member, taken as the clear
distance between floor slabs, beams, or other members capable of
providing lateral support (ACI Code, Section 10.1.1).
r = ra
dius of gyration of the cross section of the compression members (ACI
Code, Section 10.1.2)
= 0.30
h, where h is the overall dimension of a rectangular column
in the
direction of the moment
= 0.25
D, where D is the diameter of a circular column
The numer
ator
k
ℓ
u
is termed the
effective length
.
•
The effective length is a function of

The unsupported length

E
nd conditions

S
idesway
◦
Sidesway
(a.k.a.
lateral drift
) is a deformation that occurs when one end
of a member moves laterally with respect to
another.
For compression members braced against sidesway, the ACI Code (Section 10.6.3)
states that
“
k
”
may be taken as 1.0.
•
Compression members free to buckle in a sidesway mode are weaker than when
braced against sidesway.
Actual structures are rarel
y completely braced (non

sway) or completely unbraced
(sway).
•
Sidesway may be minimized in a number of ways.

The common approach is to use walls or partitions
sufficiently
strong and
rigid in their own planes to prevent horizontal displacement.
9
.
38

Anoth
er method
is
to us
e a rigid
central core that is capable of resisting
lateral loads and lateral displacements due to unsymmetrical loading
conditions.
For cases when it is not readily apparent whether a structure is braced or
unbraced, the ACI Code (Secti
ons 10.10.1 and 10.10.5) provides analytical methods
to aid in the decision.
•
For braced columns, the effect of sidesway can be neglected when
kℓ
u
/
r
≤
34
–
12(M
1
/M
2
)
≤ 40
[ACI Eq. (10

7)]
where
M
1
is the smaller end moment
M
2
is the larger end mo
ment
M
1
/M
2
is positive if the column is bent in single curvature
M
1
/M
2
is negative if the column is bent in double curvature
•
For columns in sway frames (not braced against sidesway), slenderness effects
may be neglected when
kℓ
u
/
r < 22 (ACI Code, Sect
ion 10.10.1).
Fortunately, for ordinary beam and column sizes and typical story heights of
concrete framing systems, effects of slenderness may be neglected in
•
More than 90% of columns in braced (non

sway) frames, and
•
About 40% of columns in unbraced
(sway) frames.
In cases where slenderness must be considered, the ACI Code gives the methods
that can be used.
•
An approximate method (ACI Code, Section 10.10.5).
•
More rigorous methods using computer analysis.
The design of slender reinforced concret
e columns is one of the more complex
aspects of reinforced concrete design and is not within the intended scope of the
textbook.
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