 # Reinforced Concrete Column Design per ACI 318-14 in RFEM

### Technical Article

001608

05/28/2019

Using RF-CONCRETE Members, concrete column design is possible according to ACI 318-14. Accurately designing concrete column shear and longitudinal reinforcement is important for safety considerations. The following article will confirm the reinforcement design in RF-CONCRETE Members using step-by-step analytical equations per the ACI 318-14 standard including 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 318-14  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 , 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 k) + 1.6 (175 k)

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

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

442k = (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.

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

Solving for Ast yields a value of 3.24 in. From this, the number of bars required for design can be found. According to Sect. 10.7.3.1 , 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 A . This provides a reinforcement area of.

Ast = 3.52 in2

#### Tie Selection

Determining the minimum tie size requires Sect. 25.7.2.2 . 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 Sect. 25.7.2.1 . 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 16db of the longitudinal bar diameter, 48db of the tie bar, or the smallest dimension of the member.

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

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

48db = 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 with 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 318-14  requirements to be adequate.

Steel Percentage = $\frac{{\mathrm A}_{\mathrm{st}}}{{\mathrm A}_{\mathrm g}}\;=\;\frac{3.52\;\mathrm{in}^2}{196\;\mathrm{in}^2}\;=\;0.01795\;\cdot\;100\;\;=\;1.8\%$ O.K.

#### 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 = $\frac{14\;\mathrm{in}.\;-\;2\;(1.5\;\mathrm{in}.)\;-\;2\;(0.375\;\mathrm{in}.)\;-\;3\;(0.75\;\mathrm{in}.)}2\;=\;4.00\;\mathrm{in}.$

4.00 in. is less than 6 in. which is required per 25.7.2.3 (a) . O.K.

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

(a) 1.5 in.

(b) 1.5db = 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 1.33 in.

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

(a) ${\mathrm L}_{\mathrm{dc}}\;=\;\left(\frac{\displaystyle{\mathrm f}_{\mathrm y}\;\cdot\;{\mathrm\psi}_{\mathrm r}}{\displaystyle50\;\cdot\;\mathrm\lambda\;\cdot\;\sqrt{\mathrm f'\;\cdot\;\mathrm c}}\right)\;\cdot\;{\mathrm d}_{\mathrm b}\;=\;\left(\frac{\displaystyle\left(60,000\;\mathrm{psi}\right)\;\cdot\;\left(1.0\right)}{50\;\cdot\;\left(1.0\right)\;\cdot\;\sqrt{4000\;\mathrm{psi}}}\right)\;\cdot\;\left(0.75\;\mathrm{in}.\right)\;=\;14.23\;\mathrm{in}.$

(b) ${\mathrm L}_{\mathrm{dc}}\;=\;0.0003\;\cdot\;{\mathrm f}_{\mathrm y}\;\cdot\;{\mathrm\psi}_{\mathrm r}\;\cdot\;{\mathrm d}_{\mathrm b}\;=\;0.0003\;\cdot\;(60000\;\mathrm{psi})\;\cdot\;(1.0)\;\cdot\;(0.75\;\mathrm{in}.)\;=\;13.5\;\mathrm{in}.$

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

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

${\mathrm L}_{\mathrm{dc}}\;=\;{\mathrm L}_{\mathrm{dc}}\;\begin{pmatrix}{\mathrm A}_{\mathrm s,\;\mathrm{provided}}\\{\mathrm A}_{\mathrm s,\;\mathrm{required}}\end{pmatrix}\;=\;(14.23\;\mathrm{in}.)\;\begin{pmatrix}1\;\mathrm{ft}.\\12\;\mathrm{in}.\end{pmatrix}\;\begin{pmatrix}1.92\;\mathrm{in}.^2\\3.53\;\mathrm{in}.^2\end{pmatrix}\;=\;0.65\;\mathrm{ft}.$

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 add-on module RF-CONCRETE Members and perform the design per the ACI 318-14 . The module will determine the required reinforcement to resist the applied loads on the column. Furthermore, the program will also design the provided reinforcement based on the given axial loads on the column while taking into consideration spacing requirements from the standard. The user can make small adjustments to the provided reinforcement layout in the results table.

Based on the applied loads for this example, RF-CONCRETE Members has determined a required longitudinal bar reinforcement area of 1.92 in2 and a provided area of 3.53 in2. The development length calculated in the add-on module 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 RF-CONCRETE Members is found by multiplying the reciprocal value of gamma by the length determined from 25.4.9.2 . More info on this non-linear calculation can be found in the RF-CONCRETE Members help file linked below. This reinforcement can be previewed in Figure 03.

The provided shear reinforcement for the member within RF-CONCRETE Members was calculated to be (11) No. 3 bars with a spacing s of 12 in. The shear reinforcement layout provided is shown below in Figure 04.

#### Reference

  ACI 318-14, Building Code Requirements for Structural Concrete and Commentary  Manual RF-CONCRETE Members. (2017). Tiefenbach: Dlubal Software. 