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Behavior of a Concrete Section


Behavior of a Concrete Section

Behavior of a Concrete Section

Behavior of a Concrete Section Under Bending and Flexo-Compression Loads at Ultimate Strength

Fig. 1 graphically shows the process that we will be developing in this paper. This is to determine the resistance capacity of any type of concrete section with reinforcement, located at various levels and subjected to different states of loads. Including axial, bending, or the combination of axial loads and flexural loads.


Fig. 1 A process to determine the resistance capacity of a concrete section

As a general case, it is considered a section of any shape with only the condition that it is symmetric in the plane in which the bending moment develops. The reinforcement distribution of the steel bars is equally symmetric in the said plane.

The process described implies the solution of a system of equations composed of:

  1. Compatibility equations of deformations.
  2. Physical equations of the materials.
  3. Equilibrium equations.

The solution of this system of equations permits establishing the necessary ways to obtain the methodology for the design and revision of the sections. Only in simple cases, the process of direct designs is obtained; however, in most problems, only the process of proofs will be applicable. This means that we will propose reinforcements and test that the section can resist the actual loads.

Design hypothesis

The norms establish certain assumptions of design to determine the resistance of axial, flexural, and the combination of axial and flexural loads. Hence, these hypotheses are closely related to the system of equations described previously. In continuation, each one will be studied separately to later integrate them as shown in Fig. 1.


Compatibility of Deformations


The first hypothesis (the plane section remaining plane.)

This means that a linear distribution of the deformations in the concrete and reinforcements are an option. The deformations are proportional to the distance measured from the neutral axis. This hypothesis is very important in the design to determine the deformations and its corresponding tensions in the reinforcement.


Fig. 2 shows a beam subjected to bending. A slice has been chosen in the central section of the longitude ΔI. Under the effect of the bending moments, the upper zone shortens Δc / 2 on both sides and the lower zone lengthens Δs / 2 on both sides.


Note that in this case the slice deforms but remains vertical, that is, the section does not rotate although it is deformed due to being located in the center.


The unit deformations would be εc = Δc / Δb and εs = Δs / Δbd. Giving way to the production of a unitary deformation diagram like the one indicated. This diagram shows the unit deformations of the section so that the diagram is linear as stated in the initial hypothesis. This diagram does not indicate how much the section moves but shows the value of the unit deformations at each point of the cross-section.


Normally, this diagram is associated with the rotation of the section plane; however, as we see in this case, the section does not rotate. The diagram only represents the deformations in each level that allows us to obtain the stresses in the steel at different levels and the compressed zone of the concrete.

Fig. 2 Diagram of unit deformations. The plane sections remain plane.


The diagram in Fig. 2 corresponds to the case of pure bending, if we apply an axial compression load to the diagram of deformations, due to bending, we will be adding the effect of uniform compression deformations. Thus, giving rise to the three possible cases shown in Fig. 3.

Fig. 3 Strain states for different loads states


It is possible to establish equations of compatibility of deformation between the different reinforcement levels and the level at which the maximum compression develops *c = 0.003, giving a particular strain state.

εc/c = εsp/(c-dp) = εs1/(c-d1) = εs2/(c-d2) = εs/(c-ds) = εsi/(c-dsi)


Sometimes it is necessary to relate the deformations of the reinforcements with a deformational state that was established beforehand, that is, for a certain value of k = c/d, then:


εc/k = εsp/(k-dp/ds) = εs1/(k-d1/ds) = εs2/(k-d2/ds) = εs/(k-1) = εsi/(k-di/ds)


For εc = 0.003 there is for reinforcement and reinforcement at the lower edge:

where, for example, for steel at the lower edge, for k = 0.375 εs = -0.005.


All the equations obtained in the deformational state relate the deformations of the most compressed fiber of the concrete with the deformations of the different levels of reinforcement located in the section.


Physical equations of the materials


Steel of ordinary reinforcement

Fig. 4 shows the stress-strain curve of ordinary reinforcing steel with a yield stress fy = 60,000 psi and a breaking stress fr = 85,000 psi. The modulus of elasticity Es = 29000000 psi and the strain in yield is εy = 0.002069 that the norm allows us to approximate εy = 0.002.

Fig. 4 Physical equation of ordinary reinforcing steel


The initial elastic behavior corresponds to the equation:


fs = εs.Es for εs < εy

fs = 60000 psi εs > = εy


while the plastic zone represents fs = fy constant until the breakage of the steel. Furthermore, note that the maximum values ​​of εs are about 0.01 – 0.03 in the limit of influence and 0.15 – 0.20 at the breaking point.


For concrete, the physical equation is more complex to define. The ACI 318 criteria will be used, although there are different approaches that could also be used. The objective that we intend here is to give a guide, and whoever wants to can use the diagram that corresponds to the standard used.


Fig. 5 shows the curve of a concrete subjected to compression, with the maximum deformation of 0.003 according to the ACI code and the maximum value of compression f’c at a strain of εc = 0.002.


The ACI 318 allows to use a simplification of this diagram to an equivalent rectangular diagram with a maximum compression of 0.85f’c and covering a width of β1.εc measured from the maximum deformation.

Fig. 5 Physical equation of concrete. Equivalent rectangular diagram.


This equivalent rectangular diagram is the result of a very original test where it was possible to evaluate for different qualities of concrete the magnitude of the resultant of the stresses in the concrete and the position of the said resultant, which defines the values ​​of β1 for each quality of concrete.


The value for β1 varies with the quality of concrete

For values of fc < 4000 psi, β1 = 0.85


Equations of section equilibrium 


In order to complete the necessary equations that solve the problem, there are missing the equations of equilibrium that ensure the stability of the section. These are:

Equations of equilibrium of forces and equations of equilibrium of moments.

The first of the equations refers to the equilibrium between the resultants of the stresses on component materials, concrete and steel reinforcements and internal actions generated by external loads. Which translates to:

Σ F = 0

The second description refers to the equilibrium of moments of said forces:

Σ M = 0

In both equations, the equilibrium refers between external actions and internal reactions of the component materials. It is therefore important to counteract external forces acting on different areas of concrete and steel bars with counterparts resulting from internal forces.


On the other hand, in pure bending, since there is no axial force, the equilibrium will be between the forces. Resulting from the steel in tension and the resulting compressive forces in the concrete and in the steel that this compressed. Regarding flexo-compression, the balance of forces includes the external axial load; giving different results according to the magnitude from an outside force. This brings as a consequence that in the flexion there will only be a result. While in flexo-compression there will be many possible results that will be present in the so-called interaction diagram. Which is nothing more than the set of last load states that the same section can reach.


In Fig. 6 a rectangular section is shown with reinforcements at different levels. The areas indicate the sum of the areas of all the bars located in the same level and are symmetrical to the vertical axis.

Moreover, suppose the section is subjected to a state of deformation that corresponds to one of the possible failures of the section when the superior fiber reaches its maximum deformation of εc = 0.003.

Fig. 6 Equations of equilibrium for a rectangular section with reinforcement at different levels.


As result, the equations of equilibrium will be:

Up to here, we have obtained the equations that will allow us to obtain the capacity of the nominal axial load Pn that is capable of resisting the section and the bending moment nominal Mn that is capable of resisting independently or jointly with the axial load Pn.


Calculation resistance ΦPn and ΦMn


The standards state that the sections must have a nominal resistance multiplied by a certain resistance reduction factor ϕ given by the norm for each case. In addition, the structural elements must have a resistance of design in all sections ϕSn greater than or equal to the required resistance U calculated for the load and force factors in the combinations required by the Norm of design used.

ϕ Sn >= U

In our case, this equation translates to:

ϕ Pn >= Pu

ϕ Mn >= Mu


The objective pursued in this work is to obtain the values ​​ϕ Pn and ϕ Mn which are the maximum capacities of the section. The factored load values ​​Pu and Mu are obtained by structural analysis for the conditions and type of structure analyzed. The standard establishes the following values ​​of ϕ shown in the table of Fig. 7.

Fig.7 Values for ϕ


Requirements of the ACI 318-19 standard to determinate the ultimate capacity of sections with  moments and axial load


The ACI Standard establishes a series of requirements regarding the determination of the ultimate capacity of the section. Consequently:

  • The deformation of the concrete in the fiber most compressed will reach a value of εc = 0.003.
  • The maximum deformation of the reinforcement closest to the tensile edge cannot exceed the value εs = 0.005 in bending. This is to prevent the section reaches a fragile failure. In standards prior to 2019, the consideration was εc = 0.004.
  • The maximum capacity of a section with axial compression load Po is given by:

where Ag is the total area of the section of concrete and Ast is the total area of reinforcement.


To take into account an accidental eccentricity of the axial load. A limit is established for the nominal capacity of the section between 80 to 85% of the maximum capacity Po. This percentage is equivalent to having the load with eccentricities of the order of 0.10h at 0.15h for rectangular and spiral stirrups respectively. In short, this specification takes into account that it is very unlikely that the axial load is exactly located at the centroid of the section. Fig. 8


The maximum axial nominal load Pn should not exceed:

Pnmax := 0.8 ⋅ Po for rectangular stirrups

Pnmax := 0.85 ⋅ Po for spiral stirrups


Hence, Po’s formula t does not include spaces in the concrete area Ag where Ast steel bars are. As a matter of fact, this discount has not been established in any other situation where the reinforcement is located in the compressed area of ​​the concrete.

Fig. 8 Effect of the axial face with accidental eccentricity


Example of a section in bending with reinforcement to different level




Fig. 9



Fig. 10


Rectangular section 12 X 24 


Concrete Section
Areas od reinforcement and their position. 
.      Asp ∶= 0.93·in²      3 # 5        dp ∶= 2.5·in
.      As1 ∶= 0.62·in²      2 # 5        d1 ∶= 8.5·in
.      As2 ∶= 0.62·in²      2 # 5        d2 ∶= 15.5·in
.      As ∶= 0.93·in²        3 # 5        d ∶= 21.5·in
.   Ast ∶= Asp + As1 + As2 + As = 3.1·in²
.      Asmin ∶= max(((3·psi^½·√fc·b·d)/fy), 0.0033·b·d)           Asmin = 0.851·in²
As a result, trial and error setting c less than

Deformations and tensions in each level of reinforcement (compression +)

Forces in the reinforcement and in the concrete for the value of “c” supposing 

Therefore, the result of all of the forces must be equal to “0”, 

                The nominal and factored bending moment will be: 

Diagram of interactions of rectangular section




Fig. 11



Translation By: Ivet Llerena (Eastern Engineering Group)

Structural Drawings By: Eastern Engineering Group

© 2021 This article was written by Ernesto Valdes and published by Eastern Engineering Group. All rights reserved.

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