# Slender bending design

The design is performed according to EN 1992-1-1 5.8. A separate design in each principal direction is made as a first step and if a moment is defined in at least one direction a possible biaxial check is also made according to EN 1992-1-1 5.8.9. The program is adding the defined moment to the moment from initial bow imperfection and if the column is defined as slender also calculates a 2nd order moment depending on the users choice either according the nominal stiffness method EN 1992-1-1 5.8.7 or the curvature method EN 1992-1-1 5.8.8. It is sufficient to check the column according to either of these methods. Finally the total moment is compared to the moment with regard to unintentional load eccentricity according to EN 1992-1-1 6.1 (4). If no 2nd order effects should be calculated e.g. in order to consider EN 1992-1-1 5.8.2 (6) or the criteria in EN 1992-1-1 5.8.3.1 the column should be defined as short as then no 2nd order moment will be added. Finally the required bending reinforcement with regard to this moment is calculated and the utilization is presented in each direction.

## Method based on nominal stiffness

The following model is used to estimate the nominal stiffness:

EI = Kc Ecd Ic + Ks Es Is

where:

• Ecd is the design value of the modulus of elasticity of concrete
• Ic is the moment of inertia of the concrete cross-section
• Es is the design value of the modulus of elasticity of reinforcement
• Is is the second moment of area of reinforcement, about the center of area of the concrete
• Kc is a factor for effects of cracking, creep
• Ks is a factor for contribution of reinforcement

Moment magnification factor

The total design moment, including second order effects, may be expressed as a magnification of the first order moment:

MEd = M0Ed [1 + β / ((NB / NEd) -1)]

where:

• M0Ed is the first order moment,
• β is a factor which depends on distribution of 1st and 2nd order moments,
• NEd is the design value of axial load,
• NB is the buckling load based on nominal stiffness.

In order to calculate according to this method the reinforcement area has to be at least ρ > 0.002. If not, a warning message will be displayed.

## Method based on nominal curvature

Bending moments

The design moment is:

MEd = M0Ed + M2

where:

• M0Ed is the 1st order moment, including the effect of imperfections
• Mis the nominal 2nd order moment.

1st order moment

For statically indeterminate members, M0Ed is determined for the actual boundary conditions, whereas Mwill depend on boundary conditions via the effective length.

2nd order moment

The nominal second order moment Mis:

M= NEd e2

where:

• NEd is the design value of axial force,
• e2 is the deflection = (1/r) lo 2 / c,
• 1/r is the curvature;
• lo is the effective length,
• c is a factor depending on the total curvature. As the first order moment is assumed constant the value c = 8 is used.

Curvature

For members with constant symmetrical cross sections the following may be used.

1/r = Kr Kϕ 1/r0

where:

• Kr is a correction factor depending on axial load as below
• Kϕ is a factor taking account of creep as below
• 1/r0 = fyd / (Es 0,45 d)
• d is the effective depth

## 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. 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.

Stress-strain relation for concrete

A rectangular stress distribution as shown below with height λx may be assumed.

The value of the design compressive strength is defined as:

fcd = αcc fck / γc

where:

• αcc = 1,0
• γc is the partial safety factor as above.

The value of the design tensile strength is defined as:

fctd = αct fctk,0.05 / γc

where:

• αct = 1,0
• γc is the partial safety factor as above.

Finnish annex

αcc = 0,85

Stress-strain relation for reinforcement steel

Strain limit εud = 0,9 εuk

Danish annex

εud = fyd / Es

Finnish annex

εud = 0,02

Balanced reinforcement and balanced moment

To start with a strain distribution corresponding to εc = εcmax and εs = fy / Esk is chosen. 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 above

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 using the stress-strain relation for concrete above.

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 too 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.

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 i:th 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 n and corresponding stress σsn using the stress strain relation as above:

σsn = fsn)

Calculate also the concrete stress σcn at the position of layer n using the concrete stress-strain relation above.

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 required number of bars is larger than the maximum possible number of bars in a layer the number of layers has to be increased to give space for the missing bars. As the new layer is not as favorably 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.

## Capacity with regard to biaxial bending

Biaxial bending is checked according to EN 1992-1-1 5.8.9. The following apply when simplified methods like the stiffness or the curvature method are used. Imperfections need to be taken into account only in the direction where they will have the most unfavorable effect. No further check is necessary if the slenderness ratios satisfy the following two conditions:

λyz < 2 and λzy < 2

and if the relative eccentricities ey/h and ez/b (see figure below) satisfy one of the following conditions:

• (ey/heq) / (ez/beq) < 0,2
or
• (ez/beq) / (ey/heq) < 0,2

where:

• b is the width of the section
• h is the depth of the section
• beq = iy (12)0,5 and heq = iz (12)0,5 for an equivalent rectangular section
• λy, λz are the slenderness ratios with respect to y- and z-axis respectively
• iy, iz are the radii of gyration with respect to y- and z-axis respectively
• ez = MEdy/NEd; eccentricity along z-axis,
• ey = MEdz/NEd; eccentricity along y-axis,

MEdy is the design moment about y-axis, including second order moment,
MEdz is the design moment about z-axis, including second order moment,
NEd is the design value of axial load in the respective load combination.

If the condition above is not fulfilled, biaxial bending should be taken into account including the 2nd order effects in each direction, unless they may be ignored. In the absence of an accurate cross section design for biaxial bending, the following simplified criterion may be used:

(MEdz/MRdz) a + (MEdy/MRdy) a < 1,0

where:

• MEdz/y is the design moment around the respective axis, including a 2nd order moment
• MRdz/y is the moment resistance in the respective direction
• a is an exponent.

If only a moment is defined in one direction a possible biaxial check is made with regard to the moment from unintentional load eccentricity in the other direction.

# Slender compression capacity

When the stiffness method is used the compression capacity is calculated as the buckling load according to EN 1992-1-1 5.8.3.2 (5.17). The stiffness EI is calculated as the nominal stiffness according to EN 1992-1-1 5.8.7.2 (5.21) or the design axial resistance of the section NRd = Ac fcd + As fyd whichever is lesser.