Bending reinforcement

Last modified by Fredrik Lagerström on 2020/05/14 09:41

Bending reinforcement

A keystone in ultimate bending design is the ability to design required reinforcement in a section (whether tensile, compression, or both) with enough flexural capacity for specific section forces. This can be done in several ways, as a trial and error method, a coefficient method, or a direct method. The suitable method depends on section type, moment vector direction, reinforcement rules, and so on. Here we assume bending axis to be parallel to y-axis or z-axis, and then use a direct method to get the most efficient result. In case of biaxial bending, the design problem generally has more than one solution, and we then use an approximate method.

Conditions for the design must be set before the design calculation can begin. These include concrete and reinforcement quality, bar sizes, cover and spacing rules, as well as maximum and minimum rules for bar sizes and number of bars. 

Cover and spacing rules dictate possible positions of reinforcement layers. This applies equally to bottom and top of section as well as possible numbers of bars in each layer at a given bar size. 

From material properties we get maximum compressive strain in concrete, and maximum tensile strain and yield strain in reinforcement. Since we have assumed bending axis to be parallel to the y-axis or z-axis, it follows that the neutral axis is also parallel to the y-axis or z-axis. It is now possible to establish various strain distributions for the section within the limits given by the maximum strain values. A set of internal normal- and moment forces will correspond to each combination of strain distribution and bar positions. An iterative procedure is then mostly necessary to establish a suitable set of bars, which together with an appropriate strain distribution will give the requested capacity. 

Balanced reinforcement and balanced moment

To begin, we choose a strain distribution corresponding to εc = εcmax and εs = fy / Esk. When calculating internal forces in section, we use the method proposed in chapter No reinforcement area is set initially, so the strain distribution will only give the compression force Fc (in the concrete) and its moment Mc with reference to the point R (the point of action for the external section forces). Equilibrium in a direction perpendicular to the section plane will then give the required tensile force in reinforcement, and thus the corresponding reinforcement area. The following expressions thus apply, with forces being positive at tension: 

Fs = Nx - Fc

and the balanced reinforcement area:

As0bal = Fs / σs0 


  • σs0 = fss) is calculated using the stress-strain relation for steel (see chapter 

If Fs < 0 we set Fs = 0 because a negative value does not conform to the chosen strain distribution. 

The moment capacity with balanced reinforcement will be the balanced moment

Mbal = Mcap = Mc + Fs (zR - z0

Thus far we have a moment capacity, which corresponds to a reinforcement area in one lower layer. If current section force My is not equal to Mbal, various steps of action have to be taken depending on whether My < Mbal or not. 

In later iteration steps, additional reinforcement layers (normally filled up) may be present on the tensile side (and compression side) and thus contribute to the internal forces. These layers will be regarded as part of the section, and balanced reinforcement in the lowest unfilled layer can be calculated gradually. The equilibrium equation is modified to

Fs = Nx - Fc - Ft 

and the balanced moment in this case becomes

Mbal = Mcap = Mc + Mt + Fs (zR - z0

Reinforcement with compression bars

Most cases of My > Mbal will give us the greatest economy of reinforcement (= smallest total area), with reinforcement in compression zone and increased tensile reinforcement. 

The additional force in upper layer n and lower layer 0 will then be

1537735242649-279.pngF = (My - Mbal) / (zn - z0

Calculate strain in upper layer n and corresponding stress σsn using the stress-strain relation for steel as above: 

σsn = fsn

Calculate the concrete stress σcn at the position of layer n using the stress-strain relation for concrete (see chapter 

Reinforcement area in compression zone will be

Asn = 1537735242649-279.pngF / (σsn - σcn) and tensile reinforcement area will be

As0 = As0bal + 1537735242649-279.pngF / σs0 

As the reinforcement area corresponds to a number of bars (represented by an integer), we will not get the most economical reinforcement if we put in bars both in compression zone and tensile zone corresponding to respective zone area. Instead we choose to round the compression zone area to the closest integer of bars and calculate the tensile area corresponding to this rounded-off area value. We get 

Asn,adjusted = round (Asn / Abar) Abar 

where round means we round using the common method of rounding (Half Round Up). 

As0,adjusted = As0 - (Asn - Asn,adjusted) (σsn - σcn) / σs0 

Underbalanced reinforcement

When My < Mbal, the balanced reinforcement area in the tensile zone is too great and must be decreased to provide greater economy of reinforcement. 

As the balanced reinforcement area is calculated with εs ≥ εy, a decrease in area also means a decrease of tensile force in reinforcement. The decrease of force in compression zone and tensile reinforcement will then be

1537735242649-279.pngF = (My - Mbal) / (zc - z0

Required compression force will be given by Fc = Fc,bal + 1537735242649-279.pngF (note that 1537735242649-279.pngF is negative in this case). With no reinforcement in compression zone, the area of the compression zone then must be decreased to maintain equilibrium, which will influence the position of Fc and thus also change zc. Consequently, an iterative procedure is required to calculate 1537735242649-279.pngF. 

The iteration is performed by estimating a new value of the compression zone depth. A suitable estimation at each step i is

dc (i + 1) = dc (i) My / Mcap (i

Keeping the concrete strain constant will yield a new value of the tensile strain εs by calculating 

εs = εc (d - dc) / dc 

and we may calculate new forces Fc and Fs0 (which include 1537735242649-279.pngF). 

When a constant εc gives εs > εs,max instead εc must be decreased keeping εs = εs,max constant. 

The iteration continues until Mcap (i) = My

When εs0 and Fs0 have been calculated corresponding stress σs0 is calculated using stress-strain relation and required reinforcement area will be

As0 = Fs0 / σs0 

Reinforcement at tensed section

At external tensile normal force (Nx > 0) the eccentricity of the normal force must be checked to be able to decide if tensile reinforcement is needed both in bottom and top of section. 

The eccentricity of the normal force is: 

eN = My / Nx 

When eN > zR - z0 the calculation may be performed as in the usual way as described before (see chapters - 

When eN < zR - z0 equilibrium is not possible without tensile reinforcement in top as well as in bottom. 

In the last case, forces in top and bottom can be calculated directly. Forces in upper layer n and lower layer 0 are calculated as: 

Fs0 = [My + Nx (zR - z0)] / (zn - z0

Fsn = [-My + Nx (zn - zR)] / (zn - z0

We then set the strain distribution equal to εy both in upper and lower layer and use the stress-strain relation to calculate stresses in the layers. 

In the lower layer the tensile reinforcement area will be 

As0 = Fs0 / σs0 

And in the upper layer,

Asn = Fsn / σsn 

Without tensile reinforcement

When My < Mbal the balanced reinforcement area is too large and (in some cases) unnecessary for establishing equilibrium between internal and external forces. 

The eccentricity of the normal force is

eN = My / Nx 

If the balanced reinforcement area is ignored, we can verify that if eN < Mc,bal / Fc and My < Mc,bal , no tensile reinforcement is needed. As Nx < Fc , it is always possible to find a strain distribution that satisfies equilibrium. 

When eN > Mc,bal / Fc or My > Mc,bal we start with assuming that underbalanced reinforcement is a correct solution. If we then at iteration find that for a suitable εs equilibrium requires that εs < 0 no tensile reinforcement is needed and the iteration is changed to find a compression zone depth that satisfies equilibrium. 

The iterative procedure is as follows: the iteration is performed by estimating a new value of the compression zone depth. A suitable estimation at each step i is

dc (i + 1) = dc (i) Nx / Fc (i

Keeping εc constant will give a new value of the tensile strain εs (although no reinforcement is assumed at the lower layer) by calculating

εs = εc (d - dc) / dc and we can calculate a new value of the force Fc

The iteration continues until Fc (i) = Nx .

When My < Mc the capacity is sufficient without reinforcement for the obtained strain distribution. 

When My > Mc , compression reinforcement (and perhaps tensile reinforcement) is needed in order to get sufficient moment capacity. 

The added force in upper layer n required to get My = Mcap will then be

1537735242649-279.pngF = (My - Mc) / (zn - zR

Calculate strain in upper layer n and corresponding stress σsn using the stress-strain relation as above: 

σsn = fsn

Finally, you may calculate the concrete stress σcn at the position of layer n using the concrete stress-strain relation (see chapter 

Compression reinforcement area will be

Asn = 1537735242649-279.pngF / (σsn - σcn

Arranging reinforcement bars

When calculating the required reinforcement area for layers in bottom and top, it must be possible to arrange the corresponding number of bars in the layer according to rules for cover and spacing. If this is possible, the design of reinforcement is finished. If the requisite number of bars is greater than the maximum possible number of bars in a layer, we must accept this number and increase the number of layers to provide space for the missing bars. Since the new layer is not as favourably situated as the one used for calculation, the calculation must be repeated to get a more correct value for the added layer. 

When repeating the calculation, the layers that are filled up will be treated as fixed, and the added layers are treated as bottom and top layer. 

The calculation is repeated as long as reinforcement must be moved to new layers. When there is no space for a new layer on the proper side of the neutral axis, we have exhausted the possibility of applying more reinforcement, and this sets a limit to the capacity of the section.