# Chapter 9: Columns

Mechanics

Oct 29, 2013 (4 years and 8 months ago)

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9
.
1

Chapter 9
:
Columns

9
.1

Introduction

-
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,

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

-
strength relationship is

φP
n

≥ P
u

where

P
n

= the nominal axial load strength at a given eccentricity

φP
n

= the

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
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

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

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

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.

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)

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

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

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,

80

φ

A
g

[0.85 f
c

(1

ρ
g
)]

u

Determine the required longitudinal steel area.

A
st

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

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.

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)

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

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

-
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

u
.

To prevent
the column
from being

with an eccentricity,

P
u

must be reduced so
t
hat the column can carry both P
u

and P
u
e.

The required reduction
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

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

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.

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

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

-
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

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

-
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

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,

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

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,

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

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,

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.

-

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

-
moment combination.

A
ny point
within

the solid line also represents a
n allowable

-
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

-
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

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

-

-
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
.

-

-
moment
-
strength combinations above this lin
e area
compression
-
controlled

(φ = 0.65 for tied columns; φ = 0.75 for spiral
columns).

-
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
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

n

= P
u

-

T
he
nominal

moment P
n
e = M
u

9
.
34

R
equired K
n

=
P
n
/
f
c
’A
g

=

P
u

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

n

= P
u

-

The
nominal

moment P
n
e = M
u

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

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

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
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.