# Ultimate Limit State (Concrete Beam)

Last modified by Fredrik Lagerström on 2021/08/06 13:18

# Bending reinforcement

Design of required (tensile and/or compression) reinforcement in a section providing enough flexural capacity for specific section forces is done assuming bending axis to be parallel to local y-axis.

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 correspond 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 of constant stress in compressed zone is used with σc = fcd. 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) as Nx = 0 in the beam case:

Fs = - Fc

and the balanced reinforcement area:

As0bal = Fs / σs0

where σs0 = fss) is calculated using the stress-strain relation for steel.
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.

## Under balanced 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  εsy, 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 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.

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

## Stress-strain conditions for concrete

 λ = 0,8 for fck  50 MPa (3.19) λ = 0,8 - (fck - 50) / 400 for 50 < fck  90 MPa (3.20)

and:

 η = 1,0 for fck  50 MPa (3.21) η = 1,0 - (fck - 50) / 200 for 50 < fck  90 MPa (3.22)

Note: If the width of the compression zone decreases in the direction of the extreme compression fiber, the value ηfcd should be reduced by 10%.

The value of the design compressive strength is defined as:
fcd = αcc fck / γc

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

The value of the design tensile strength is defined as:
fctd = αct fctk,0.05 / γc
αct = 1,0
γc is the partial safety factor.

UK Annex
αcc = 0,85

Finnish Annex
αcc = 0,85

## Stress-strain conditions for reinforcement

Ordinary mild reinforcement steel has generally a well-defined yield limit. In most cases following stress-strain relation may be used both at tension and compression.

Strain limit εud = 0,9 εuk

Danish Annex

εud = fyd / Es

Finnish Annex

εud = 0,02

## 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 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.

The total bending reinforcement in a section is calculated considering both positive and negative moments if this occurs as a result of load combinations with active load. Required number of bars in upper and lower layers will in such cases be composed of the largest number of both design calculations.

## Continuity control at supports

When having chosen bar diameters for all spans these will be modified at all inner supports in order to obtain continuity for the reinforcement. The modification has been performed so that the bar diameters and the number of bars at the side of the support containing the largest area of reinforcement will be applied to the other side of the support as well.

## Curtailing

Curtailing is being performed according to EN 1992-1-1 9.2.1.3-4.

The tensile force curve is calculated with respect to the displacement described in EN 1992-1-1 9.2.1.3 (9.2).

# Shear

## Elements not requiring shear reinforcement

The shear capacity of a section without shear reinforcement is considered to be sufficient if:

VEd  VRd,c

where:

• VEd is the shear design force supplied by analysis,
• VRd,c is the shear capacity of the concrete.

however minimum shear reinforcement should nevertheless be provided according to below except for slabs where transverse redistribution of loads is possible.

Shear capacity of the concrete
VRd,c = [CRd,c k (100 ρ1 fck)1/3+ k1 σcp] bw d

with a minimum of:
VRd,c = [νmin+ k1 σcp] bw d

where:

• CRd,c = 0,18 / γc
• k1 = 0,15
• fck is in MPa
• k = 1 + (200/d)0.5  2,0 with d in mm
• ρ1 = Asl / (bw d)
• Asl is the area of the tensile reinforcement, which extends > (lbd + d) beyond the section considered (see figure below),
• bw = the smallest width of the cross-section in the tensile area (mm),
• σcp = NEd / Ac < 0,2 MPa
• NEd is the axial force in the cross-section due to loading or pre stressing (in N) (NEd > 0 for compression). The influence of imposed deformations on NE may be ignored
• Ac is the area of concrete cross section (mm2),
• VRd,c is in (N)
• νmin = 0,035 k3/2 fck1/2

The shear force VEd should always satisfy the condition:

• VEd  0,5 bw d ν fcd

where:

• ν = 0,6 (1 - fck / 250) (fck in MPa).

## Shear capacity with shear reinforcement

The design of members with shear reinforcement is based on a truss model (see figure below).

where:

• α is the angle between shear reinforcement and the beam axis,
• θ is the angle between the concrete compression strut and the beam axis perpendicular to the shear force
• Ftd is the design value of the tensile force in the longitudinal reinforcement
• Fcd is the design value of the compression force in the direction of the longitudinal member axis
• bw is the minimum width between tension and compression chords,
• z is the inner lever arm. In shear analysis without axial force, the approximate value z = 0,9 d may normally be used.

Shear capacity of the shear reinforcement

The shear capacity of a section with shear reinforcement is considered to be sufficient if:

• VEd  VRd,s
and
• VEd,max  VRd,max

where:

• VEd is the shear force at the design section.
• VEd,max is the maximum shear force.
• VRd,s is the shear capacity of a section with shear reinforcement,
• VRd,max is the max capacity,
• VRd,s = Asw/s z fywd (cot θ + cot α) sin α
or
• s = [Asw z fywd (cot θ + cot α) sin α ] γ VEd
• VRd,max = αcw bw z ν1 fcd (cot θ + cot α) / (1 + cot2θ)

where:

• fywd is the design yield strength of the shear reinforcement,
• Asw is the area of the shear reinforcement,
• s is the spacing of the stirrups,
• The angle θ should be limited to 1  cot θ  2,5
• ν1 is a strength reduction factor for concrete cracked in shear.

If the design stress of the shear reinforcement is below 80% of the yield stress fyk then:

• ν1 = 0,6 for fck  60 MPa
• ν1 = 0,91 - fck / 200 > 0,5 for fck > 60 MPa

In this case fywd = 0,8 fywk, else:

• ν1 = 0,6 (1 - fck / 250) (fck in MPa),
• αcw is a coefficient taking account of the state of the stress in the compression chord,
• αcw = 1,0 for non-pre stressed structures.

The maximum effective shear reinforcement Asw,max for cot θ = 1 follows from:

Asw,max fywd / (bw s)  0,5 αcw ν1 fcd / sin α

In regions where there is no discontinuity of VEd (e.g. uniformly distributed loading) the shear reinforcement in any length increment l = z (cot θ + cot α) may be calculated using the smallest value of VEd in the increment according to EN 1992-1-1 6.3.2 (5).

Rules for shear reinforcement

The shear reinforcement should form an angle α between 45 and 90 degrees to the longitudinal axis of the structural element.

The ratio of shear reinforcement is given by:

• ρw = Asw / (s bw sin α)

where:

• Asw is the area of shear reinforcement within length s1,max

The minimum shear reinforcement ratio is given by:

• ρw,min = (0,08 fck0,5) / fyk

The maximum spacing between shear assemblies should not exceed sl,max

• sl,max = 0,75 d (1 + cot α)

The transverse spacing of the legs in a series of shear links should not exceed st,max:

• St,max = 0,75 d  600 mm

Design of shear reinforcement is performed by the stringer method.

UK Annex

If shear co-exits with applied tension then cot θ should be taken as 1,0.
If the design stress of the shear reinforcement is below 80% of the yield stress fyk then:

• ν1 = 0,54 (1 - 0,5 cos α) for fck  60 MPa
• ν1 = (0,84 - fck / 200) (1 - 0,5 cos α) > 0,5 for fck > 60 MPa

Compression stresses in web

The shear capacity is valid, if it is verified that the compression stress in the diagonal compression field is not exceeded. The compression stress is calculated as:

• σcd = t (1 + cot2θ) / (cot θ + cot α)

With the condition:

• σcd  νv fcd

where:

• νv is the efficiency factor (see below),
• fcd is compression strength of concrete at design.

we get maximum possible shear stress as:

• τmax = νv fcd (cot θ + cot α) / (1 + cot2θ)

If reinforcement class B or C is used then the inclination θ of the compressive stress should be chosen as:

• tan α/2  cot θ  2,5
• α is the angle between shear reinforcement and the beam axis.

where curtailed reinforcement is chosen:

• tan α/2  cot θ  2,0
• α is the angle between shear reinforcement and the beam axis.

If reinforcement class A is used then cot θ = 1,0
The strength reduction factor for concrete cracked in shear is calculated as:

• ν1 = 0,7 (1 - fck / 200) (fck in MPa) (5.10.3NA)

Shear field

At calculation the beam element is divided in zones of length z (cot θ + cot α) where the shear stress is assumed to be constant, thus giving a uniform shear reinforcement area Asv in the zone. However, zones close to support are given the length z cot θ. In order to get constant shear stress, zones influenced by point loads are split at point load position.

The height of the shear field is assumed to be equal to the moment lever z.

Note: Stringer heights are assumed to be small compared with the shear field height, thus indirectly giving a low limit of the shear field height.

Stringer forces

As an effect of the diagonal compression field model the stringer forces will get the following values:

• C = M / z - 0,5 |V| (cot θ - cot α) (compression stringer)
• T = M / z + 0,5 |V| (cot θ - cot α) (tensile stringer)

Compression and tensile capacity of stringers must be checked for these forces, although higher values than corresponding to section with moment maximum is not used.

If the stringer forces exceed stringer capacities the current shear calculation is not valid as exceeding force must be carried by the web. Increase stringer height and recalculate. If stringer heights larger than 0,25 H are required a part of the stringer force must be redistributed to the web, giving the axial web force NSdw in the web.

If NSdw  - |VSd| cot θ shear calculation method for beam with no axial force is still applicable. If NSdw is a compression force it has a positive influence and is ignored as it act as a fictitious reinforcement bar. For section where (|VSd| cot θ + NSdw) > 0 it is required to insert longitudinal bar in shear field corresponding to the force (|VSd| cot θ + NSdw).

The minimum shear reinforcement ratio is given by:

• ρw,min = (0,063 fck0.5) / fyk