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 2^nd 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 2^nd 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 = K_c E_cd I_c + K_s E_s I_s

where:

• E_cd is the design value of the modulus of elasticity of concrete
• I_c is the moment of inertia of the concrete cross-section
• E_s is the design value of the modulus of elasticity of reinforcement
• I_s is the second moment of area of reinforcement, about the center of area of the concrete
• K_c is a factor for effects of cracking, creep
• K_s 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:

M_Ed = M_0Ed [1 + β / ((N_B / N_Ed) -1)]

where:

• M_0Ed is the first order moment,
• β is a factor which depends on distribution of 1^st and 2^nd order moments,
• N_Ed is the design value of axial load,
• N_B 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.

This method is also used in Control.

Method based on nominal curvature

Bending moments

The design moment is:

M_Ed = M_0Ed + M₂

where:

• M_0Ed is the 1^st order moment, including the effect of imperfections
• M₂ is the nominal 2^nd order moment.

1^st order moment

For statically indeterminate members, M_0Ed is determined for the actual boundary conditions, whereas M₂ will depend on boundary conditions via the effective length.

2^nd order moment

The nominal second order moment M₂ is:

M₂ = N_Ed e₂

where:

• N_Ed is the design value of axial force,
• e₂ is the deflection = (1/r) l_o 2 / c,
  • 1/r is the curvature;
  • l_o 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 = K_r K_ϕ 1/r₀

where:

• K_r is a correction factor depending on axial load as below
• K_ϕ is a factor taking account of creep as below
• 1/r₀ = f_yd / (E_s 0,45 d)
  • d is the effective depth

This method is also used in Control.

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 λ_x may be assumed.

The value of the design compressive strength is defined as:

f_cd = α_cc f_ck / γ_c

where:

• α_cc = 1,0
• γ_c is the partial safety factor as above.

The value of the design tensile strength is defined as:

f_ctd = α_ct f_ctk,0.05 / γ_c

where:

• α_ct = 1,0
• γ_c is the partial safety factor as above.

 Finnish annex

α_cc = 0,85

Stress-strain relation for reinforcement steel

Strain limit ε_ud = 0,9 ε_uk

 Danish annex

ε_ud = f_yd / E_s

 Finnish annex

ε_ud = 0,02

Balanced reinforcement and balanced moment

To start with a strain distribution corresponding to ε_c = ε_cmax and ε_s = f_y / E_sk is chosen. From start no reinforcement area has been set, so the strain distribution will only give the compression force F_c in the concrete and its moment M_c 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):

F_s = N_x - F_c

and the balanced reinforcement area:

A_s0bal = F_s / σ_s0

where

• σ_s0 = f_s (ε_s) is calculated using the stress-strain relation for steel above

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

The moment capacity with balanced reinforcement will be the balanced moment:

M_bal = M_cap = M_c + F_s (z_R - z₀)

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

F_s = N_x - F_c - F_t

and the balanced moment in this case:

M_bal = M_cap = M_c + M_t + F_s (z_R - z₀)

Reinforcement with compression bars When M_y > M_bal 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 = (M_y - M_bal) / (z_n - z₀)

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

σ_sn = f_s (ε_n)

Calculate the concrete stress σ_cn at the position of layer n using the stress-strain relation for concrete above.

Reinforcement area in compression zone will be:

A_sn = ΔF / (σ_sn - σ_cn)

and tensile reinforcement area will be:

A_s0 = A_s0bal + ΔF / σ_s0

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:

A_snadjusted = round (A_sn / A_bar)

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

A_s0adjusted = A_s0 - (A_sn - A_snadjusted) (σ_sn - σ_cn) / σ_s0

Underbalanced reinforcement

When M_y < M_bal 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 ε_s ≥ ε_y, 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 = (M_y - M_bal) / (z_c - z₀)

Required compression force will be given by F_c = F_cbal + 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 F_c and thus z_c 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:

d_c (i + 1) = d_c ( i) M_y / M_cap ( i)

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

ε_s = ε_c (d - d_c) / d_c

and we can calculate new forces F_c and F_s0 (which include ΔF).

When a constant ε_c gives ε_s > ε_smax instead ε_c must be decreased keeping ε_s = ε_smax constant.

The iteration continues until M_cap ( i) = M_y.

When ε_s0 and F_s0 have been calculated corresponding stress σ_s0 is calculated using stress-strain relation and required reinforcement area will be:

A_s0 = F_s0 / σ_s0

Reinforcement at tensed section

At external tensile normal force (N_x > 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:

e_N = M_y / N_x

When e_N > z_R - z₀ the calculation may be performed as in the usual way as described before.

When e_N < z_R - 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:

F_s0 = [M_y + N_x (z_R- z₀)] / (z_n - z₀)

F_sn = [-M_y + N_x (z_n - z_R)] / (z_n - z₀)

We then set the strain distribution equal to ε_y 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:

A_s0 = F_s0 / σ_s0

And in upper layer:

A_sn = F_sn / σ_sn

Without tensile reinforcement

When M_y < M_bal 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:

e_N = M_y / N_x

If the balanced reinforcement area is ignored we can verify that if e_N < M_cbal / F_c and M_y < M_cbal no tensile reinforcement is needed. As N_x < F_c it is always possible to find a strain distribution that satisfies equilibrium.

When e_N > M_cbal / F_c or M_y > M_cbal we start with assuming that underbalanced reinforcement is a correct solution. If we then at iteration find that for a suitable ε_s equilibrium requires that ε_s < 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:

d_c (i + 1) = d_c ( i) N_x / F_c ( i)

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

ε_s = ε_c (d - d_c) / d_c

and we can calculate a new value of the force F_c.

The iteration continues until F_c ( i) = N_x

When M_y < M_c the capacity is sufficient without reinforcement for the obtained strain distribution. When M_y > M_c 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 M_y = M_cap will then be:

ΔF = (M_y - M_c) / (z_n - z_R)

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

σ_sn = f_s (ε_n)

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

Compression reinforcement area will be:

A_sn = ΔF / (σ_sn - σ_cn)

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:

λ_y/λ_z < 2 and λ_z/λ_y < 2

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

• (e_y/h_eq) / (e_z/b_eq) < 0,2
  or
• (e_z/b_eq) / (e_y/h_eq) < 0,2

where:

• b is the width of the section
• h is the depth of the section
• b_eq = i_y (12)^0,5 and h_eq = i_z (12)^0,5 for an equivalent rectangular section
• λ_y, λ_z are the slenderness ratios with respect to y- and z-axis respectively
• i_y, i_z are the radii of gyration with respect to y- and z-axis respectively
• e_z = M_Edy/N_Ed; eccentricity along z-axis,
• e_y = M_Edz/N_Ed; eccentricity along y-axis,

M_Edy is the design moment about y-axis, including second order moment,
M_Edz is the design moment about z-axis, including second order moment,
N_Ed 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 2^nd 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:

(M_Edz/M_Rdz) a + (M_Edy/M_Rdy) a < 1,0

where:

• M_Edz/y is the design moment around the respective axis, including a 2^nd order moment
• M_Rdz/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.

This method is also used in Control.

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 N_Rd = A_c f_cd + A_s f_yd whichever is lesser.