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Lectur e 33 33 – Columns Concrete Column Column s: Concrete is good at resisting compression but poor in resisting tension. So, it might make sense that concrete would be the material of choice for columns. It is true that concrete IS used for compression members such as columns, piers, bearing walls and pedestals. Members under pure compression could then (theoretically) be unreinforced. These members are often subject to additional forces such as moment that would put some tensile forces into the member and would thus necessitate the addition of tension reinforcement. Most columns have combined compression and bending . They are essentially a “beam-column”. The compression capacity of a reinforced column is reduced by the bending stresses on the column and vice-versa. A graph of the axial axial load capacity capacity of of a column vs. the moment moment capacity capacity of a typical column is shown below:

y t i c a p a c n o i s s e r p m o c l a i x A

Pure compression

Pure bending Bending moment capacity

Lecture 33 - Page 1 of 14

Types of Concrete Column s: There are two types of reinforced concrete columns – tied and spiral – and refer to the type of confining bars used to contain the interior core of concrete. It has been shown that unconfined concrete core will carry MUCH LESS load than a confined core as shown below: Load Load

Concrete & vert. bars “explode” outward

Unconfined core

Insufficient ties


A confined concrete core will carry substantially more load and will NOT explode outward like the unconfined concrete section will. Load

Load

Minor concrete spalling, core remains intact

Confined core Tied Column: A “Tied” concrete column is one in which individual rebar ties are used to wrap completely around the vertical bars to confine the interior core. These ties are usually #3 or #4 bars spaced per ACI requirements: Least column dimension Tie Spacing = smaller of

16 x Vert. Bar Dia. 48 x Tie Bar Dia.

Additionally, the ACI dictates that there must be a minimum of 4 vertical bars having a minimum area of 1% of the column cross-sectional area and a maximum of 8% of the column cross-sectional area. From a constructability standpoint, 4% is the upper maximum that can be readily achieved because of rebar congestion.


A typical tied concrete column looks like the following:

Vertical bars (4 minimum)

Tie bars Tie spacing

Concrete


Spiral Column : A spiral column has a single rebar wrapped around the vertical bars in a spiral and is stronger than a comparable tied column. It is more laborintensive to build than a tied column. The ACI requires a minimum of 6 vertical bars, with the same minimum and maximum areas of steel as a tied column.

Concrete Pitch = 1” → 3” Spiral rebar

Spiral columns have greater ductility than a tied column. This means that a spiral column is capable of sustaining larger strain before failure than a tied column. For this reason, spiral col umns are used in areas with high seismicity to prevent brittle failures. The stress-strain curve below shows the relative yield strength of a tied column and a spiral column are similar, however the strain capacity of the spiral column is much greater than that of a tied column.

Spiral column Yield

Stress

Fracture

Fracture

Tied column

Strain


Column Lo ad Capacity – Small eccentricit y: The ratio of applied moment to the applied axial load is its “eccentricity.” A small eccentricity “e” means small moment, where a large eccentricity means large moment.

Pu = applied factored axial load

Pu

e

e

Side view of column

h

h

Concentric l oad (e = small) Mu = Pue

Eccentric load (e = large) Mu = Pue

= Small

= Large


1) Tied columns with small eccentricity: Small eccentricity if

e h

≤ 0.10

φPn = Usable axial strength of tied column, KIPS = 0.80 c [0.85f’ c(A g -A s )+f y A s] where:

φc = strength reduction factor for column = 0.65 for tied columns Ag = gross cross-sectional area of column, in 2 As = total area of vertical steel bars, in 2 = minimum of 4 vertical bars = 0.01Ag → 0.08Ag f’c = specified concrete compressive strength, KSI f y = yield strength of vertical steel bars, KSI

2) Spiral columns with small eccentricity: Small eccentricity if

e h

≤ 0.10

φPn = Usable axial strength of tied column, KIPS = 0.85 c [0.85f’ c(A g -A s )+f y A s] where:

φc = strength reduction factor for column = 0.70 for spiral columns Ag = gross cross-sectional area of column, in 2 As = total area of vertical steel bars, in 2 = minimum of 6 vertical bars = 0.01Ag → 0.08Ag f’c = specified concrete compressive strength, KSI f y = yield strength of vertical steel bars, KSI


Spiral requirements:

ρ s

s

=

⎛ A ⎞ f ' = 0.45⎜⎜ g − 1⎟⎟ c ⎝ Ac ⎠ f y 4 As ( d c

− d b )

ρ s d c

2

where:

ρs = ratio of spiral bar to gross area s = spacing of spiral bends (pitch), in Ag = gross area of column, in 2 Ac = area of core, in 2 = area measured out-to-out of spiral As = area of spiral bar, in 2 dc = diameter of core measured out-to-out of spiral db = diameter of spiral bar


Example 1 GIVEN: A 16” x 16” square tied concrete column has an applied factored axial load Pu = 300 KIPS and an applied factored moment M u = 35 KIP-FT. In addition, use the following:

• • • •

Concrete f’ c = 4000 PSI 8 - #7 grade 60 vertical bars #3 individual tie bars Concrete cover = 1½”

REQUIRED: 1) Determine the eccentricity “e” and state whether or not it is considered to be “small eccentricity.” 2) Determine the required tie spacing. 3) Determine if the vertical bars are acceptable based on ACI requirements. 4) Determine the usable axial strength of the tied column, φPn, and state whether or not it is acceptable. 16”

16”

8 - #7 vertical bars Step 1 – Determine eccentricity and whether it is “small.” Eccentricity e =

=

M u (12" / ft ) Pu

35KIP − FT (12" / ft ) 300 KIPS

e = 1.4” Small eccentricity if e/h < 0.10 e/h = 1.4”/(16”) e/h = 0.0875 it is small eccentricity


Step 2 – Determine the required tie spacing:

Least column dimension = 16” Tie Spacing = smaller of

16 x Vert. Bar Dia. = 16(⅞”) = 14”

USE

48 x Tie Bar Dia. = 48(⅜”) = 18” Step 3 – Determine if vertical bars are acceptable based on ACI: 8 vertical bars are used > 4 bars OK As = 8 bars(0.60 in 2 per #7 bar) = 4.8 in2 Min. As = 0.01Ag = 0.01(16” x 16”) = 2.56 in 2 < 4.8 in2 OK Max. As = 0.08 Ag = 0.08(16” x 16”) = 20.48 in 2 > 4.8 in2 OK Step 4 - Determine the usable axial strength of the tied column,

φPn:

Since e/h < 0.10 then the formula can be used:

φPn = Usable axial strength of tied column, KIPS = 0.80φc[0.85f’c(Ag-As)+f y As] where:

φc = 0.65 since it is a tied column Ag = 16” x 16” = 256 in2 As = 8 bars(0.60 in 2 per #7 bar) = 4.8 in2

= 0.80(0.65)[0.85(4 KSI)(256 in 2 – 4.8 in 2) + (60 KSI)(4.8 in 2)] Pn = 594 KIPS Since Pn = 594 KIPS > 300 KIPS

colum n is acceptable


Example 2 GIVEN: A 14” diameter spiral column has a factored axial load P u = 400 KIPS and a factored moment = 45 KIP-FT. In addition, use the following:

• • • •

Concrete f’ c = 4000 PSI 6 - #9 grade 60 vertical bars #3 spiral bar Concrete cover = 1½”

REQUIRED: 1) Determine the eccentricity “e” and state whether or not it is considered to be “small eccentricity.” 2) Determine the required spiral pitch. 3) Determine the usable axial strength of the spiral column, φPn, and state whether or not it is acceptable.


Step 1 – Determine eccentricity and whether it is “small.” Eccentricity e =

=

M u (12" / ft ) Pu

45KIP − FT (12" / ft ) 400 KIPS

e = 1.35” e

Small eccentricity if e h

h

=

< 0.10

1.35" 14"

e/h = 0.096

it is small eccentricity

Step 2 – Determine the required spiral pitch “s”:

ρ s

⎛ A ⎞ f ' = 0.45⎜⎜ g − 1⎟⎟ c ⎝ Ac ⎠ f y

where: Ag =

=

π

(Outside _ Dia) 2

4 π

4

Pitch

(14" ) 2

Ag = 153.9 in 2 Ac =

=

π

4 π

4

(Core _ Dia) 2

(11" ) 2

Ac = 95.0 in 2


ρ s

ρ s

⎛ A ⎞ f ' = 0.45⎜⎜ g − 1⎟⎟ c ⎝ Ac ⎠ f y ⎛ 153.9in 2 ⎞ 4 KSI = 0.45⎜⎜ − 1⎟⎟ 2 ⎝ 95.0in ⎠ 60 KSI

ρs = 0.0186 Pitch = s s

=

4 As ( d c

− d b )

ρ s d c

2

⎛ 3 ⎞" ) ⎟ ⎝ 8 ⎠

4(0.11in 2 )(11"−⎜ s

=

(0.0186)(11" ) 2

Pitch s = 2.08” Use pitc h = 2”


Step 3 - Determine the usable axial strength of the spiral column,

φPn:

Since e/h < 0.10 then the formula can be used:

φPn = Usable axial strength of spiral column, KIPS = 0.85φc[0.85f’c(Ag-As)+f y As] where:

φc = 0.70 since it is a spiral column Ag = 153.9 in 2 As = 6 bars(1.00 in 2 per #9 bar) = 6.0 in2

= 0.85(0.70)[0.85(4 KSI)(153.9 in 2 – 6.0 in 2) + (60 KSI)(6.0 in 2)] Pn = 513 KIPS Since Pn = 513 KIPS > 400 KIPS

column i s acceptable


Column 2

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