CE 473– Concrete IILecture 10: 8.8 – 8.108.8) Circular Columns8.9) ACI Code Provisions for Column Design8.10) Design Aids

CE473 – Concrete II Dr. Ammar T. Al-Sayegh

8.8) Circular Columns As mentioned in Section 8.2, when load eccentricities are small, spiral reinforced

columns show greater toughness (greater ductility) than tied columns. That’s why the reduction factor ∅ = 0.75 for spiral columns, compare with ∅ = 0.65 for tied columns. Thus, circular columns permit more economical utilization of materials, particularly for small eccentricities.

CE473 – Concrete II Dr. Ammar T. Al-Sayegh1

Shown is a cross-section of a spirally reinforced column with six rebars. Let the column be loaded with rightward eccentricity so that the strain distribution at the instant when ultimate load is reached can be drawn as shown. The stress block for the same instant is delineated with the shaded area in the cross-section and stress diagram below.

Note that bar groups 2 and 3 are seen to be strained to much smaller values than bar groups 1 and 4.

Any bars with strains in excess of yield strain 𝜖𝜖𝑦𝑦 = 𝑓𝑓𝑦𝑦/𝐸𝐸𝑠𝑠will have stress equal to yield stress 𝑓𝑓𝑦𝑦.

For bars with strains below yield strain, the stress if found from 𝑓𝑓𝑠𝑠 = 𝜀𝜀𝑠𝑠𝐸𝐸𝑠𝑠.

The nominal force and moment capacities can then be calculated as follows:

→ 𝑃𝑃𝑛𝑛 = 0.85𝑓𝑓𝑐𝑐′ℎ2

4𝜃𝜃 − 𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃 + ∑𝑖𝑖=1

# 𝑜𝑜𝑜𝑜 𝑙𝑙𝑙𝑙𝑦𝑦𝑙𝑙𝑙𝑙𝑠𝑠 𝐴𝐴𝑠𝑠𝑖𝑖𝑓𝑓𝑠𝑠𝑖𝑖

→ 𝑀𝑀𝑛𝑛 = 0.85𝑓𝑓𝑐𝑐′ℎ3

12𝑠𝑠𝑠𝑠𝑠𝑠3𝜃𝜃 + ∑𝑖𝑖=1

# 𝑜𝑜𝑜𝑜 𝑙𝑙𝑙𝑙𝑦𝑦𝑙𝑙𝑙𝑙𝑠𝑠 𝐴𝐴𝑠𝑠𝑖𝑖𝑓𝑓𝑠𝑠𝑖𝑖ℎ2− 𝑑𝑑𝑖𝑖

→ 𝑓𝑓𝑠𝑠𝑖𝑖 = 𝑚𝑚𝑠𝑠𝑠𝑠 �𝑐𝑐−𝑑𝑑𝑖𝑖𝑐𝑐

600𝑓𝑓𝑦𝑦

CE473 – Concrete II Dr. Ammar T. Al-Sayegh2

θ

ah/2-a

h

Aconc

ha2

≤

1h 2 a

cosh 2

90

− − θ =

θ ≤ °

φ

a

a-h/2

h

θ

Aconc

ha2

>

1a h 2

180 cosh 2

90 180

− − θ = −

ϕ

° < θ ≤ °

xx

8.9) ACI Code Provisions for Column Design For columns, as all members designed according to the ACI Code, safety margins are

established by magnifying service loads by load factors and reducing nominal strengths by strength reduction factors, such that:

→ ∅𝑀𝑀𝑛𝑛 ≥ 𝑀𝑀𝑢𝑢 (8.6a)

→ ∅𝑃𝑃𝑛𝑛 ≥ 𝑃𝑃𝑢𝑢 (8.6b)

where:

→ ∅ = 0.65 (for tied columns)

→ ∅ = 0.75 (for spirally reinforced columns)

Note that the reduction factors for columns are lower than those for flexure (∅ = 0.90) or shear (∅ = 0.75).

For high eccentricities, the ACI code recognizes that columns behave more like flexural members by providing a linear transition in ∅ from 0.65 and 0.75 to 0.90 as the net tensile strain in the extreme tensile steel 𝜖𝜖𝑡𝑡 increases from 𝜀𝜀𝑡𝑡 = 𝜀𝜀𝑦𝑦 = 𝑓𝑓𝑦𝑦/𝐸𝐸𝑠𝑠(which can be taken as 0.002 for Grade 420 rebars) to 0.005. Thus:

→ ∅ = 0.65 + 2503

(𝜀𝜀𝑡𝑡 − 0.002) (ACI Commentary)

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For zero or very low eccentricities, ACI recognizes eccentricities introduced by construction imperfections and other unforeseen factors by limiting the maximum design strength to 0.80∅𝑃𝑃0for tied columns and 0.85∅𝑃𝑃0 for spirally reinforced columns where 𝑃𝑃0 is the nominal strength of the axially loaded column with 𝑧𝑧𝑧𝑧𝑧𝑧𝑠𝑠 eccentricity. Thus:

→ ∅𝑃𝑃𝑛𝑛(𝑚𝑚𝑙𝑙𝑚𝑚) = 0.85∅[0.85𝑓𝑓𝑐𝑐′(𝐴𝐴𝑔𝑔−𝐴𝐴𝑠𝑠𝑡𝑡) + 𝐴𝐴𝑠𝑠𝑡𝑡𝑓𝑓𝑦𝑦] (8.4a)

For spirally reinforced columns, and for tied columns:→ ∅𝑃𝑃𝑛𝑛(𝑚𝑚𝑙𝑙𝑚𝑚) = 0.80∅[0.85𝑓𝑓𝑐𝑐′(𝐴𝐴𝑔𝑔−𝐴𝐴𝑠𝑠𝑡𝑡) + 𝐴𝐴𝑠𝑠𝑡𝑡𝑓𝑓𝑦𝑦] (8.4b)

CE473 – Concrete II Dr. Ammar T. Al-Sayegh4

8.10) Design Aids The design of eccentrically loaded columns using the strain compatibility method starts

with the selection of a trial column which is then analyzed to determine if it is adequate to carry any combination of 𝑃𝑃𝑢𝑢 and 𝑀𝑀𝑢𝑢. That is to see that 𝑃𝑃𝑢𝑢 and 𝑀𝑀𝑢𝑢 fall within the strength interaction diagram of the selected column. Note that for economical design, 𝑃𝑃𝑢𝑢 and 𝑀𝑀𝑢𝑢 must be close to the limit of the curve.

Computer programs and spreadsheets can be developed to plot the strength interaction diagram for any trial column. Furthermore, design aids are furnished by ACI and CRSI that cover the post frequent practical cases, such as symmetrically reinforced rectangular and square columns and circular spirally reinforced columns. Commercial programs such as PCA Column, are also available for column design and analysis.

Instead of plotting 𝑃𝑃𝑛𝑛 vs. 𝑀𝑀𝑛𝑛, load is plotted as 𝐾𝐾𝑛𝑛 = 𝑃𝑃𝑛𝑛𝑜𝑜𝑐𝑐′𝐴𝐴𝑔𝑔

= 𝑃𝑃𝑢𝑢∅𝑜𝑜𝑐𝑐′𝐴𝐴𝑔𝑔

vs. 𝑅𝑅𝑛𝑛 = 𝑃𝑃𝑛𝑛𝑙𝑙𝑜𝑜𝑐𝑐′𝐴𝐴𝑔𝑔ℎ

= 𝑃𝑃𝑢𝑢𝑙𝑙∅𝑜𝑜𝑐𝑐′𝐴𝐴𝑔𝑔ℎ

.

Families of curves are plotted for various values of 𝜌𝜌𝑔𝑔 = 𝐴𝐴𝑠𝑠𝑠𝑠𝐴𝐴𝑔𝑔

between 0.01 and 0.08.

The graphs also include radial lines representing different eccentricity ratios 𝑙𝑙ℎ, as well as lines

representing different ratios of stress 𝑜𝑜𝑠𝑠𝑜𝑜𝑦𝑦

or values of strain 𝜀𝜀𝑡𝑡 = 0.002 and 0.005 in the extreme

tension steel.

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Charts permit the design of eccentrically loaded columns in one of two ways as follows.

1. (a) Select trial cross-sectional dimension b and h.(b) Calculate the ratio 𝛾𝛾 based on required cover distances to the bar centroids, and select the

corresponding column design chart. If ⁄𝑙𝑙 ℎ ≥ 0.2 Bending dominates → use only two opposite sides reinforcement. Otherwise, use four sided reinforcement.

(c) Calculate 𝐾𝐾𝑛𝑛 = 𝑃𝑃𝑢𝑢∅𝑜𝑜𝑐𝑐′𝐴𝐴𝑔𝑔

and 𝑅𝑅𝑛𝑛 = 𝑃𝑃𝑢𝑢𝑙𝑙∅𝑜𝑜𝑐𝑐′𝐴𝐴𝑔𝑔ℎ

where 𝐴𝐴𝑔𝑔 = 𝑏𝑏ℎ.

(d) Using appropriate chart, get the required reinforcement ratio 𝜌𝜌𝑔𝑔.

(e) Calculate the total steel area 𝐴𝐴𝑠𝑠𝑡𝑡 = 𝜌𝜌𝑔𝑔𝑏𝑏ℎ.

2. (a) Select the reinforcement ratio 𝜌𝜌𝑔𝑔.

(b) Choose trial value ℎ and calculate 𝑙𝑙ℎ

and 𝛾𝛾.

(c) Read 𝐾𝐾𝑛𝑛 = 𝑃𝑃𝑢𝑢∅𝑜𝑜𝑐𝑐′𝐴𝐴𝑔𝑔

from appropriate chart and calculate required 𝐴𝐴𝑔𝑔.

(d) Calculate 𝑏𝑏 = 𝐴𝐴ℎ.

(e) Revise the trial value of h that if necessary to obtain a well-proportioned section.(f) Calculate the total steel area 𝐴𝐴𝑠𝑠𝑡𝑡 = 𝜌𝜌𝑔𝑔𝑏𝑏ℎ.

CE473 – Concrete II Dr. Ammar T. Al-Sayegh14

Example 1Consider the column cross-section shown with 𝑓𝑓𝑐𝑐′ = 28 𝑀𝑀𝑃𝑃𝑀𝑀 and 𝑓𝑓𝑦𝑦 = 420 𝑀𝑀𝑃𝑃𝑀𝑀 which is bent about its strong axis. This column is subjected to a service load of 988 kN, maximum live load of 1320 kN, dead load moment of 180 kN.m, and live load moment of 260 kN.m. The minimum live load compatible with the full live load

CE473 – Concrete II Dr. Ammar T. Al-Sayegh15

moment is 738 kN, obtained when no live load is placed on the roof but a full live load is placed on the second floor.

(a) Find the required column reinforcement for the condition that the full live load acts.

(b) Check to ensure that the column is adequate for the condition of no live load on that roof.

500 mm

625 mm

62.5mm62.5 mm

Example 2Consider the column cross-section shown with 𝑓𝑓𝑐𝑐′= 28 𝑀𝑀𝑃𝑃𝑀𝑀 and 𝑓𝑓𝑦𝑦 = 420 𝑀𝑀𝑃𝑃𝑀𝑀 which is bent about its strong axis. This column is subjected to factored load and moment 𝑃𝑃𝑢𝑢 = 2140 𝑘𝑘𝑘𝑘 and 𝑀𝑀𝑢𝑢 = 665 𝑘𝑘𝑘𝑘.𝑚𝑚. Studies show that optimum 𝜌𝜌𝑠𝑠= 0.003. Only two layers of rebars are to be used adjacent to the to outer faces and parallel to the axis of bending.

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b

h

62.5mm62.5 mm