# Bending reinforcement

Design of required (tensile and/or compression) reinforcement in a section providing enough flexural capacity for specific section forces is one of the keystones in ultimate bending design. This can be done in several ways, e.g. as a trial and error method, as a coefficient method or as a direct method. The suitable method depends on section type, moment vector direction, reinforcement rules etc. Here we assume bending axis to be parallel to y-axis or z-axis and use a direct method to get the most efficient result. At biaxial bending the design problem has generally more than one solution and we then use an approximate method, see ch. 3.1.1.6. Before the design calculation can start conditions for the design must be set. 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.

From cover and spacing rules we get possible positions of reinforcement layers both in 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. As we have assumed bending axis to be parallel to y-axis or z-axis it follows that the neutral axis is also parallel to y-axis or z-axis. It is now possible to establish various strain distributions for the section within the limits given by maximum strain values. To each combination of strain distribution and bar positions will corresponds a set of internal normal- and moment forces. An iteration 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 start with a strain distribution corresponding to εc = εcmax and εs = fy / Esk is chosen. When calculating internal forces in section the method proposed in ch 3.2.3.1 is used. From start no reinforcement area has been set, so the strain distribution will only give the compression force Fc in the concrete and its moment Mc with reference to the point R, which is 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. We get (forces are positive at tension):

• Fs = Nx - Fc

and the balanced reinforcement area:

• As0bal = Fs / σs0

where

• σs0 = fss) is calculated using the stress-strain relation for steel (see ch. 3.2.3.3).

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 in 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, not filled layer can be calculated gradually.

Equilibrium equation is modified to:

Fs = Nx - Fc - Ft

and the balanced moment in this case:

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

## Reinforcement with compression bars

When My > Mbal in most cases we get the most economic 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:

ΔF = (My - Mbal) / (zn - z0)

Calculate strain in upper layer n and corresponding stress σsn using 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 ch. 3.2.3.2).

Reinforcement area in compression zone will be:

Asn = ΔF / (σsn - σcn)

and tensile reinforcement area will be:

As0 = As0bal + ΔF / σs0

As the reinforcement area corresponds to an integer number of bars, we will not get the most economic 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 an integer number of bars and calculate the tensile area corresponding to this rounded area. We get:

Asnadjusted = round (Asn / Abar) Abar

where round means round to higher or lower value in usual way.

As0adjusted = As0 - (Asn - Asnadjusted) (σsn - σcn) / σs0

## Underbalanced reinforcement

When My < Mbal the balanced reinforcement area in tensile zone is to large and must be decreased to give better reinforcement economy.

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

ΔF = (My - Mbal) / (zc - z0)

Required compression force will be given by Fc = Fcbal + ΔF (note that ΔF 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 zc will be changed too. Consequently, an iterative procedure is required to calculate ΔF.

Iteration procedure

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

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

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

εs = εc (d - dc) / dc

and we can calculate new forces Fc and Fs0 (which include ΔF).

When a constant εc gives εs > εsmax instead εc must be decreased keeping εs = εsmax 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 Reinforcement with compression bars and Underbalanced reinforcement).

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 is 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 lower layer the tensile reinforcement area will be:

• As0 = Fs0 / σs0

And in upper layer:

• Asn = Fsn / σsn

## Without tensile reinforcement

When My < Mbal the balanced reinforcement area is too large and in some cases not needed at all to establish 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 < Mcbal / Fc and My < Mcbal no tensile reinforcement is needed. As Nx < Fc it is always possible to find a strain distribution that satisfies equilibrium.

When eN > Mcbal / Fc or My > Mcbal 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.

Iteration procedure:

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

• ΔF = (My - Mc) / (zn - zR)

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

σsn = fsn)

Calculate also the concrete stress σcn at the position of layer using the concrete stress-strain relation (see Stress-strain relation for concrete).

Compression reinforcement area will be:

Asn = ΔF / (σsn - σcn)

## Arranging reinforcement bars When calculating required reinforcement area for layers in bottom and top it must be checked that it is possible to arrange the corresponding number of bars in the layer considering rules for cover and spacing. If it is possible the design of reinforcement is finished. If required number of bars is larger than the maximum possible number of bar in a layer we have to settle with this number and increase the number of layers to give space for the missing bars. As the new layer is not so favourably situated as the one used in 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 viewed as bottom and top layer.

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

Biaxial bending

When the bending axis is not parallel to y- or z-axis an approximate calculation is performed by splitting the calculation of reinforcement in two separate calculations, one for y-axis and one for z-axis. The obtained bar results are then added to one bar result. It may then happen that bars from the two results overlap each other or else lie too close. At present one of the bars is removed in such cases, which may lead to that required capacity is not obtained. At the calculation the compression zone may partly be used twice and consequently insufficient bending capacity may occur because of insufficient compression zone capacity.

Thus, at biaxial bending, a capacity check must always be done after design and possible required addition of bars to get capacity performed manually.

# Shear reinforcement

Design of shear reinforcement is performed by first calculating shear capacity as described in Shear capacity with shear reinforcement.

The shear reinforcement area Asv is converted to vertical 2-cut stirrups using selected diameter and appropriate spacing and shape depending on current section type.

Shear reinforcement at biaxial bending

When shear force appears at both y- and z-axis the calculation is performed separately for each axis. Shear capacity and required shear reinforcement area are shown separately too. An appropriate stirrup type with spacing corresponding to the shear reinforcement area giving the smallest spacing is composed.

At calculation of capacity in the two directions it should be noted that parts of the same web area may be used twice, thus requiring a separate manually check using e.g. an interaction formula.

# Calculation of sections affected by combined axial, bending, shear and torsion forces

## General

In this chapter is described a general calculation method to be used in ultimate state for design of reinforced concrete sections affected to combined axial force, bending, shear and torsion.

The design method is described e. g. in  and .

## Fundamental

In  and  only rectangular sections are handled but as this program has a variety of section shapes available the calculation method has been extended to deal with these section shapes too. The section forces are assumed to act in the center-of gravity of the solid section with positive directions as shown in figure.

At calculation an effective section is used in shape of a thin-walled, closed section with varying thickness ti situated at the perimeter of the section. The cover is excluded except for sides in compression due to bending. Sides are numbered counterclockwise starting at bottom (see figure below). Side numbers are used at result.

## Stress distribution

The section forces acting on the section are, using plastic methods, replaced by stresses on the effective section. At determination of the stresses it is important to secure that equilibrium is satisfied.

Thus, in the individual walls at the sides of the perimeter (side walls) of the effective section the chosen distribution of axial stresses and shear stresses shall satisfy the condition:

• where:

• (x, s) -coordinate system is a local orthogonal coordinate system at side walls, with x-axis parallel to beam axis (i.e. the x-axis).

Furthermore at connections between the side walls shear forces shall be equal and stresses shall not introduce other unforeseen section forces, e. g. that distribution of shear stresses will introduce moments.

Axial stresses by axial force Nx

As forces act in the center-of-gravity of the solid section we calculate axial stresses in the thin-walled section as: where:

• Nx is the axial section force
• N0 is a part of the axial section force working on the entire solid section
• N0 may be set to zero although Nx ≠ 0. Corresponding axial stress is: • Asolid is the area of the solid section remaining when cover has been peeled off,
• Athin is the area of the thin-walled section, Athin = Σti li with ti is thickness and li length of side wall
• and are coordinates of the center-of-gravity of the thin-walled section,
• and are coordinates of the center-of-gravity of the side wall i in the thin-walled section,
• and are moments of inertia of the thin-walled section, calculated excluding the moment of inertia of the side walls around their own axes:
• • Axial stresses by bending moments My and Mz

The axial stresses in the thin-walled section due to bending moments are calculated assuming that stresses are distributed uniformly at side walls inside a compression zone and a tensile zone respectively. The heights of these zones are calculated as required heights in the solid section but not larger than 0,25 Hsolid at bending around y-axis and 0,25 Bsolid at bending around z-axis. A side wall is considered to be inside a zone if its center-of-gravity is inside the zone. At least the topmost and bottommost side walls are always included.

We calculate axial stresses in the thin-walled section by bending moments as: where:

• My and Mz are the section bending moments,
• is the area of the compression zone or the tensile zone at the thinwalled section at bending around y-axis. Compression zone area is solely used at side walls situated in compression zone and tensile zone area solely at side walls situated in tensile zone
• is the area of the compression zone or the tensile zone at the thinwalled section at bending around z-axis. Compression zone area is solely used at side walls situated in compression zone and tensile zone area solely at side walls situated in tensile zone
• and are coordinates of the center-of-gravity of the compression zone and the tensile zone at the thin-walled section at bending around y-axis
• and are coordinates of the center-of-gravity of the compression zone and the tensile zone at the thin-walled section at bending around z-axis

Shear stresses by torsional moment Tx

Shear stress by torsional moment in a side wall i of the thin-walled section is calculated as

• where:

• Tx is the section torsional moment
• Ak is the area within the center line of the side walls of the thin-walled section
• ti is the thickness of the side wall i of the thin-walled section.

Shear stresses by shear forces Vy and Vz

Shear stresses by shear forces Vy and Vz are related to the variation of the bending moments Mz and My respectively. Considering the condition: we calculate using N = M / z and σx = N / A which gives: Thus we now get: .

Integrating over the length of the side walls will give the desired shear stress distribution τ as long as the start value τ0 is known: The stress distribution τ is applied to the compression zone and the tensile zone. A study of the shear stress distribution shows that the shear stress has a linear variation over the length of a side wall and with more than one side wall in compression zone the same gradient at all the compressed walls. For parts of the thin-walled section between these zones σx = 0, thus giving a constant shear stress distribution here. These constant shear stresses give the start value τstart for the first side wall in compression zone or tensile zone and the end value τend of the last side wall in compression zone or tensile zone by using the condition that at connections between the side walls shear forces shall be equal. If compression zone or tensile zone consists of more than one side wall the condition of equal shear force at connections applies here too.

To get the shear stress values of the constant shear stress zones a calculation of equilibrium around the point of acting of the section forces is made. The calculation is made using iteration with start values of the constant stresses set equal and then successively changed at the iteration steps to get the shear stresses in equilibrium with the shear force.

The shear stresses are calculated at start, center-of-gravity and end of the side walls and assumed to be positive in direction of the side wall.

## Side wall thickness in the effective section

The calculation always start with a minimum thickness equal to bending bar diameter + stirrup bar diameter if cover is peeled off or equal to cover + bending bar diameter if cover is not peeled.

When stress distribution has been calculated it is possible to calculate the required thickness. If any thickness is changed a new stress distribution must be calculated and new checking made in an iterative process. When changing thickness it is not allowed to set smaller thickness than the initial thickness and not larger thickness than the geometry permits i. e. opposite side walls may not overlap. When increase of wall thickness lead to overlapping the wall thicknesses are limited to not overlap. In this case the required wall thickness is not possible and this will show in the result as too large stresses or too low capacity.

## Reinforcement

At design required reinforcement is calculated. When the required side wall thicknesses have been established and a valid stress distribution as well, required reinforcement as stirrup area and longitudinal bar area are calculated at design in accordance with EN 1992-1-1. At side walls with linear variation of the shear stress the decisive shear stress together with other stresses is used on the entire side wall. The longitudinal bar area is split in areas associated with shear stress and axial stress at design result.

## Shear capacity

At control current shear capacity is calculated. Current stirrups and longitudinal bar are distributed to the side walls and transformed to current geometric reinforcement ratios. The shear capacity of every side wall is then calculated considering
reinforcement ratios and concrete capacity. Using the calculated stress distribution current shear forces at side walls are calculated. The capacity ratio ri = Vi,cap / Vi,cur is then calculated for every side wall and the minimum capacity ratio rmin = min (ri) is then used to calculate the section capacity. We get:

• Vy,cap = rmin Vy
• Vz,cap = rmin Vz
• Tx,cap = rmin Tx

It should be noted that as capacity calculation is based on required side wall thicknesses side wall capacity can not be increased above the capacity corresponding to σcd = ν fcd by increasing the reinforcement.