# Reinforced Concrete Column Design per ACI 318-19 in RFEM 6

## Technical Article on the Topic Structural Analysis Using Dlubal Software

### Technical Article

Using the Concrete Design add-on, concrete column design is possible according to the ACI 318-19. The following article will confirm the reinforcement design of the Concrete Design add-on using step-by-step analytical equations per the ACI 318-19 standard including the required longitudinal steel reinforcement, gross cross-sectional area, and tie size/spacing.

#### Concrete Column Analysis

A reinforced square tie concrete column is designed to support an axial dead and live load of 135 and 175 kips respectively using ULS design and factored LRFD load combinations according to ACI 318-19 [1] as presented in Figure 01. The concrete material has a compressive strength f'c of 4 ksi while the reinforcing steel has a yield strength fy of 60 ksi. The steel reinforcement percentage is initially assumed to be 2%.

#### Dimension Design

To begin, the dimensions of the cross-section must be calculated. The square tie column is determined to be compression controlled since all axial loads are strictly in compression. Per Table 21.2.2 [1], the strength reduction factor Φ is equal to 0,65. When determining the maximum axial strength, Table 22.4.2 [1] is referenced which sets the alpha factor (α) equal to 0,80. Now, the design load Pu can be calculated.

Pu = 1.2 (135) + 1.6 (175) = 442 kips

Based on these factors, Pu is equal to 442 kips. Next, the gross cross-section Ag can be calculated utilizing equation 22.4.2.2.

$$Pu = (Φ) (α) [ 0.85 f'c (Ag - Ast) + fy Ast]$$

With:

Φ - Strength reduction factor

α - Alpha factor

f’c - Compressive strength

Ag - Gross cross-section

Ast - Steel reinforcement percentage

442 kips = (0.65) (0.80) [0.85 (4 kips) (Ag - 0.02 Ag) + ((60 ksi) (0.02) Ag)]

Solving for Ag, we receive an area of 188 in2. The square root of Ag is taken and rounded up to set a cross-section of 14” x 14" for the column.

#### Required Steel Reinforcement

Now that Ag is established, the steel reinforcement area Ast can be calculated utilizing Eqn. 22.4.2.2 by substituting the known value of Ag = 196 in2 and solving it.

442 kips = (0.65) (0.80) [0.85 (4 kips) (196 in2 - Ast) + ((60 ksi) (Ast))]

Solving for Ast yields a value of 3.24 in2. From this, the number of bars required for design can be found. According to Section 10.7.3.1 [1], a square tie column is required to have at least four bars. Based on this criteria, and the minimum required area of 3.24 in2, (8) No. 6 bars for the steel reinforcement is used from Appendix B [1]. This provides the reinforcement area below.

Ast = 3.52 in2

#### Tie Selection

Determining the minimum tie size requires Section 25.7.2.2 [1]. In the previous section, we selected No. 6 longitudinal bars which are smaller than No. 10 bars. Based on this information and section, we select No. 3 for the ties.

#### Tie Spacing

To determine the minimum tie spacing (s), we refer to Section 25.7.2.1 [1]. Ties that consist of closed looped deformed bars must have spacing that is in accordance with (a) and (b) from this section.

(a) The clear spacing must be equal to or greater than (4/3) dagg. For this calculation we will be assuming an aggregate diameter (dagg) of 1.00 in.

smin = (4/3) dagg = (4/3) (1.00 in.) = 1.33 in.

(b) The center-to-center spacing should not exceed the minimum of 16 db of the longitudinal bar diameter, 48 db of the tie bar, or the smallest dimension of the member.

sMax = Min (16 db, 48 db, 14 in.)

16 db = 16 (0.75 in.) = 12 in.

48 db = 48 (0.375 in.) = 18 in.

The minimum clear tie spacing calculated is equal to 1.33 in. and the maximum tie spacing calculated is equal to 12 in. For this design, a maximum of 12 in. for the tie spacing will govern.

#### Detailing Check

The detailing check can now be performed to verify the reinforcement percentage. The required steel percentage must be between 1% and 8% based on the ACI 318-19 [1] requirements to be adequate.

Steel Percentage

$$AstAg = 3.52 in2196 in2 = 0.01795 · 100 = 1.8 %$$

With:

Ast = Total area of nonprestressed longitudinal reinforcement including bars or steel shapes, and excluding prestressing reinforcement

Ag - Gross cross-section

#### Longitudinal Bar Spacing

The maximum longitudinal bar spacing can be calculated based on the clear cover spacing and the diameter of both the tie and longitudinal bars.

Maximum Longitudinal Bar Spacing

$$14 in. - 2 (1.5 in.) - 2 (0.375 in.) - 3 (0.75 in.)2 = 4.00 in.$$

4.00 in. is less than 6 in. which is required per 25.7.2.3 (a) [1].

The minimum longitudinal bar spacing can be calculated by referencing 25.2.3 [1] which states the minimum longitudinal spacing for columns must be at least the greatest of (a) through (c).

(a) 1.5 in.

(b) 1.5 db = 1.5 (0.75 in.) = 1.125 in.

(c) (4/3) db = (4/3) (1.00 in.) = 1.33 in.

Therefore, the minimum longitudinal bar spacing is equal to 1.50 in.

The development length (Ld) must also be calculated with reference to 25.4.9.2 [1]. This will be equal to the greatest of (a) or (b) calculated below.

(a) Development Length

$$Ldc = fy · ψr50 · λ · f'c · db = 60,000 psi · 1.050 · 1.0 · 4000 psi · 0.75 in. = 14.23 in.$$

With:

fy - Specified yield strength for nonprestressed reinfrocement

ψr - Factor used to modify development length based on confining reinforcement

λ - Modification factor to reflect the reduced mechanical properties of lightweight concrete relative to normal weight concrete of the same compressive strength

f'c - Compressive strength

db - Nominal diameter of bar, wire, or prestressing strand

(b) Development Length

$$Ldc = 0.0003 · fy · ψr · db = 0.0003 · (60000 psi) · (1.0) · (0.75 in.) = 13.5 in.$$

With:

fy - Specified yield strength for nonprestressed reinforcement

ψr - Factor used to modify development length based on confining reinforcement

db - Nominal diameter of bar, wire, or prestressing strand

In this example, (a) is the greater value so Ldc = 14.23 in.

Referencing 25.4.10.1 [1], the development length is multiplied by the ratio of required steel reinforcement over provided steel reinforcement.

Development Length

$$Ldc = Ldc As, providedAs, required = (14.23 in.)1 ft.12 in.1.92 in.23.53 in.2 = 0.65 ft$$

The reinforced square tie column is fully designed and its cross-section can be viewed below in Figure 02.

#### Comparison With RFEM

An alternative for designing a square tie column manually is to utilize the Concrete Design add-on in RFEM 6 and perform the design per the ACI 318-19 [1] standard. The add-on will determine the required reinforcement to resist the applied loads on the column. The user is then required to make adjustments manually to the provided reinforcement layout to meet the required reinforcement shown.

Based on the applied loads for this example, RFEM 6 has determined a required longitudinal bar reinforcement area of 3.24 in2. The development length calculated in the Concrete Design add-on is equal to 0.81 ft. The discrepancy in comparison to the development length above calculated with analytical equations is due to the program's non-linear calculations including the partial factor γ. The factor γ is the ratio of ultimate and acting internal forces taken from RFEM. The development length in the Concrete Design add-on is found by multiplying the reciprocal value of gamma by the length determined from 25.4.9.2 [1]. This development length and reinforcement can be previewed in Figure 03 and 04 respectively.

The minimum required shear reinforcement area (Av,min) for the member within the Concrete Design add-on was calculated to be 0.14 in2 bars with a minimum spacing (smax) of 12 in. The required shear reinforcement layout is shown below in Figure 05.

#### Alex Bacon, EIT

Technical Support Engineer

Alex is responsible for customer training, technical support, and continued program development for the North American market.

#### Reference

 [1] Dlubal Software. (2017). Manual RF-CONCRETE Members. Tiefenbach. [2] ACI 318-19, Building Code Requirements for Structural Concrete and Commentary

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• Updated 05/23/2022

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