Reinforced Concrete Beam Design per ACI 318-14 in RFEM

Technical Article

Using RF-CONCRETE Members, concrete beam design is possible according to ACI 318-14. Accurately designing concrete beam tension, compression, and shear 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 moment strength, shear strength, and required reinforcement. The doubly reinforced concrete beam example analyzed includes shear reinforcement and will be designed under the strength limit state (SLS). 


The doubly reinforced concrete beam section, found in Figure 01, will be designed under SLS according to ACI 318-14 [1] using factored LRFD load combinations. An unfactored uniform dead and live load of 2.0 kip/ft and 3.2 kip/ft are applied to the beam respectively. The selected rectangular beam has an overall cross-section of 25 in ⋅ 11 in. The concrete material has a compression strength f'c of 5,000 psi while the reinforcing steel has a yield strength of 60,000 psi. The compression reinforcement A's consists of two #8 bars with a centroid distance d' of 3.0 in from the top fiber with a total area of 1.57 in². The tensile reinforcement As consists of six #8 bars with a centroid distance d of 20.5 in from the top fiber with a total area of 4.71 in². The shear reinforcement Av includes #4 stirrups for a total area of 0.4 in2. The dimensions and stress/strain diagram of the beam cross-section are shown in Figure 03. 
The dimensions and stress/strain diagram of the beam cross-section are shown in.

Figure 03 - Reinforced Concrete Section: Stress and Strain Diagram

Moment Strength

The required nominal moment, Mu, from the applied loads is found to be 4752.00 kip-in. Deriving the equation to find the nominal moment requires the following assumptions:

Compression steel does not yield: ε's < εy → f's = Es ⋅ ε's
Tensile steel does yield: εs ≥ εy → fs = fy

Analyzing the stress and strain diagram of the beam, the neutral axis can be found with the equation below, derived by setting the compression forces equal to the tension forces in order to satisfy equilibrium:

T= C's + Cc  0 → As ⋅ fy - A's ⋅ f's - 0.85 ⋅ f'c ⋅ a ⋅ b = 0

Utilizing the strain diagram and similar triangles, we can assume:

$\mathrm\varepsilon'_{\mathrm s}\;=\;\frac{{\mathrm\varepsilon}_{\mathrm{cu}}\;\cdot\;({\mathrm c}_{\mathrm{NA}}\;-\;\mathrm d')}{{\mathrm c}_{\mathrm{NA}}}$

We also know:

a = β1 ⋅ CNA

Substituting β1 ⋅ CNA and $\mathrm\varepsilon'_{\mathrm s}\;=\;\frac{{\mathrm\varepsilon}_{\mathrm{cu}}\;\cdot\;({\mathrm c}_{\mathrm{NA}}\;-\;\mathrm d')}{{\mathrm c}_{\mathrm{NA}}}$ for "a" and ε's respectively into the equilibrium equation above, the neutral axis can be calculated as all values are known except CNA.

${\mathrm A}_\mathrm s\;\cdot\;{\mathrm f}_\mathrm y\;-\;\frac{\mathrm A'_\mathrm s\;\cdot\;{\mathrm E}_\mathrm s\;\cdot\;{\mathrm\varepsilon}_\mathrm{cu}\;\cdot\;({\mathrm C}_\mathrm{NA}\;-\;\mathrm d')}{{\mathrm C}_\mathrm{NA}}\;-\;0.85\;\cdot\;\mathrm f'_\mathrm c\;\cdot\;{\mathrm\beta}_1\;\cdot\;{\mathrm C}_\mathrm{NA}\;\cdot\;\mathrm b\;=\;0$

Using Table from the ACI 318 - 14 [1], β1 is equal to 0.80. Solving for CNA, we find it is equal to about 5.83 in. from the top fiber.

The above assumptions (1 and 2) must be verified. Assumption 1 consists of calculating the strain in the compression steel (ε's) and comparing it with the yield strain (εy). If ε's is less than εy, our assumption is correct. Assumption 2 requires calculating the strain of the tensile steel reinforcement (εs) and comparing it to εy. If εs is greater than εy then our assumption is correct. We verify through calculation (not shown) that assumption 1 and 2 are valid. 

Finally, to solve for the nominal moment (Mn), we set the sum of moments about the location of the concrete in compression (Cc) equal to zero. This can be seen in the diagram from Figure 01.

This equation becomes:

${\mathrm M}_\mathrm n\;=\;\mathrm A'_\mathrm s\;\cdot\;{\mathrm E}_\mathrm s\;\cdot\;{\mathrm\varepsilon}_\mathrm{cu}\;\cdot\;\frac{({\mathrm C}_\mathrm{NA}\;-\;\mathrm d')}{{\mathrm C}_\mathrm{NA}}\;\cdot\;(\frac{\mathrm a}2\;-\;\mathrm d')\;+\;{\mathrm A}_\mathrm s\;\cdot\;{\mathrm f}_\mathrm y\;\cdot\;(\mathrm d\;-\;\frac{\mathrm a}2)$    We must also calculate "a" by multiplying β1 and CNA together before calculating Mn.

a = 4.66in

By substituting these values into the Mn equation we get the following:

${\mathrm M}_{\mathrm n}\;=\;1.57\;\cdot\;29,000\;\cdot\;\frac{0.003\;\cdot\;(5.83\;-\;2.5)}{5.83}\;\cdot\;\left(\frac{4.66}2\;-\;3.0\right)\;+\;4.71\;\cdot\;60\;\cdot\;\left(20.5\;-\;\frac{4.66}2\right)$ Mn is calculated as 5122.69 kips-in.

Finally, the safety factor (φ) is determined by referencing Table 21.2.2 from the ACI 318 -14 [1]. To determine φ, the tension strain is compared to the ultimate strain of 0.005. εt is equal to 0.00755 and is greater than 0.005. The beam is tension controlled. From Table 21.2.2, φ is equal to 0.90. When multiplying this factor by Mn, φMn is equal to 4610.42 kips-in.
Therefore, the moment capacity of the beam is sufficient to resist the applied bending moment:

φMn > Mu = 4512.00 kips-in o.k.

Shear Strength

Based on Sect. [1], we calculate the nominal shear strength (Vn) of the beam. The following equation is used to calculate nominal shear:

Vn = φ ⋅ (Vc + Vs)

Referencing table [1], concrete shear strength Vc is equal to the minimum of equations a, b, and c calculated in the sections 1, 2, and 3 below: 

1.) Equation "a" is given as: 

Vc-a$=\;\left(1.9\;\cdot\;\mathrm\lambda\;\cdot\;\sqrt{\mathrm f'_{\mathrm c}}\;+\;2,500\;\cdot\;{\mathrm\rho}_{\mathrm w}\;\cdot\;\frac{{\mathrm V}_{\mathrm u}\;\cdot\;\mathrm d}{{\mathrm M}_{\mathrm u}}\right)\;\cdot\;{\mathrm b}_{\mathrm w}\;\cdot\;\mathrm d$,  with λ = 1

Mu occurs at Vu which is a distance "d" away from the support face (section [1]). Therefore, Mu is equal to 1534.98 kips-in. Vu = 61.05 kips.

Vc-a = 44.89 kips

2.) Equation "b" is given as:

Vc-b$=\;\left(1.9\;\cdot\;\mathrm\lambda\;\cdot\;\sqrt{\mathrm f'_{\mathrm c}}\;+\;2,500\;\cdot\;{\mathrm\rho}_{\mathrm w}\right)\;\cdot\;{\mathrm b}_{\mathrm w}\;\cdot\;\mathrm d$ 

Equation "b" requires ρw to be calculated:

${\mathrm\rho}_{\mathrm w}\;=\;\frac{{\mathrm A}_{\mathrm s}}{{\mathrm b}_{\mathrm w}\;\cdot\;\mathrm d}\;=\;0.01992$

Vc-b = 46.26 kips

3.) Equation "c" is given as:

Vc-c$=\;3.5\;\cdot\;\mathrm\lambda\;\cdot\;\sqrt{\mathrm f'_{\mathrm c}}\;\cdot\;{\mathrm b}_{\mathrm w}\;\cdot\;\mathrm d$                                                                    

Vc-c = 61.25 kips

Therefore, selecting the minimum value from the equations above, we find Vc is equal to 44.89 kips. 

Following the nominal shear of the concrete calculation, the minimum shear reinforcement is found utilizing Sect. 9.6.3 [1]. Here, if the required ultimate shear Vu is less than 0.5 ⋅ φ ⋅ Vc then shear reinforcement is required.

Vu < 0.5 ⋅ φ ⋅ Vc


φ = 0.75 (Table 21.2.1[1])

Therefore, Vu = 61.05 kips > 16.83 kips. Stirrups are required.

The theoretical spacing is determined from Sect. [1]:

φ ⋅ Vn > Vu

We substitute (V+ Vs) for Vn.

${\mathrm V}_\mathrm s\;>\;\frac{{\mathrm V}_\mathrm u\;-\;\mathrm\phi\;\cdot\;{\mathrm V}_\mathrm c}{\mathrm\phi}$

So, Vs > 36.51 kips.                                                                                                      From Sect. [1], we use the following equation to calculate the required steel shear strength:

${\mathrm V}_\mathrm s\;=\;\frac{{\mathrm A}_\mathrm v\;\cdot\;{\mathrm f}_\mathrm{yt}\;\cdot\;\mathrm d}{\mathrm s}$          

Where, fyt is the yield strength of the steel reinforcement in tension and "d" is the distance from the top fiber to the centroid of the tension reinforcement.

The max spacing (s) is calculated to be 14.79 in. A spacing of 14 in. for the shear reinforcement is used. Using s = 14.00 in. in the above equation for steel shear strength, Vs = 36.51 kips.

Using Table [1], we can determine the maximum shear spacing must be determine. The following equation is calculated to determine which equation in Table is applicable:

$4\;\cdot\;\sqrt{\mathrm f'_\mathrm c}\;\cdot\;{\mathrm b}_\mathrm w\;\cdot\;\mathrm d\;=\;4\;\cdot\;\sqrt{5,000\;\mathrm{psi}}\;\cdot\;11\;\mathrm{in}\;\cdot\;22.5\;\mathrm{in}$

The steel shear strength, Vs =  36.51 kips, is less than the required ultimate shear Vu = 61.05 kips. Referencing Table, the maximum shear spacing can be determined using the minimum value from the following calculations:

${\mathrm s}_\max\;=\;\min\;\left(\frac{\mathrm d}{24},\frac{\mathrm d}2\right)$

The maximum shear spacing is determined to be 11.25 in. The shear spacing determined previously with #4 bars spaced at 14 in. is sufficient. We verify the nominal shear capacity is greater than the required ultimate shear strength to ensure the shear reinforcement and spacing are adequate.                                                                                                                   

Vn = φ ⋅ (Vc + Vs) = 0.75 ⋅ (41.05 + 54.00) > Vu= 71.29 kips  

Vn = 69.95 > 62.33 kips

The final verification includes determining whether the cross-section dimensions are sufficient based off of Sect. [1]. To do this, ultimate shear strength is compared to Eqn. from the ACI 318-14 [1]:

Vu ≤ $({\mathrm V}_{\mathrm c}\;+\;8\;\cdot\;\sqrt{\mathrm f'_{\mathrm c}}\;\cdot\;{\mathrm b}_{\mathrm w}\;\cdot\;d)\;=\;0.75\;\cdot\;\left(44.89\;+8\;\cdot\;\sqrt{5000\;\mathrm{psi}}\;\cdot\;11\;\mathrm{in}\;\cdot\;22.5\;\mathrm{in}\right)$

Vu = 61.05 kip ≤ 105.04 kips

This value of 105.04 is greater than Vu and therefore, the current cross-section dimensions are sufficient.


We compare these results with the RF-CONCRETE Members module in RFEM to verify the accuracy when designing doubly reinforced concrete members. Figure 02 shows the longitudinal reinforcement provided by the program based off of the given cross-section. RF-CONCRETE recommends a top reinforcement A's of 1.57 in². and a bottom reinforcement As of 4.72 in².

Figure 04 - Tension and Compression Reinforcement RFEM Diagram

Figure 02 - Tensions and Compression Reinforcement RFEM Diagram
The shear reinforcement provided from the RF-CONCRETE Members module is shown in Figure 03. The module calculates the recommended bar number and spacing to be #4 bars with 10 in. spacing. 
Figure 03 - Shear Reinforcement RFEM Diagram
It can be determined that RF-CONCRETE Members includes almost identical results to the analytical equations calculated above. Therefore, the module is sufficient for accurate and reliable reinforced concrete member design.

Figure 05 - Shear Reinforcement RFEM Diagram


ACI 318-14 doubly reinforced concrete beam design


[1]   ACI 318-14, Building Code Requirements for Structural Concrete and Commentary


Contact us

Contact Dlubal Software

Do you have any questions or need advice?
Contact us or find various suggested solutions and useful tips on our FAQ page.

(267) 702-2815

RFEM Main Program
RFEM 5.xx

Main Program

Structural engineering software for finite element analysis (FEA) of planar and spatial structural systems consisting of plates, walls, shells, members (beams), solids and contact elements

Price of First License
3,540.00 USD
RFEM Concrete Structures

Add-on Module

Design of reinforced concrete members and surfaces (plates, walls, planar structures, shells)

Price of First License
810.00 USD
RFEM Concrete Structures
ACI 318 for RFEM 5.xx

Module Extension for RFEM

Extension of the modules for reinforced concrete design by the ACI 318 design

Price of First License
360.00 USD