# Section properties

## Area properties of concrete gross section

A general integral over an area of arbitrary shape is given as:

• Hmn = ∫ ym zn dA

with m and n as positive integers. Setting m and n equal to 0, or 2, the integral will give values of the area integrals A, Sy, Sz, Iy, Iz and Iyz which are defined as area properties.

At calculation all section types (open as well as closed) are converted to polygon shape and after that the section properties are easily calculated using numeric integration.

## Area properties of reinforcement in cross-section

Section properties for reinforcement composed of bars with cross-section area Asj and positions (yj, zj) are calculated as a summation:

With m and n equal to 0, 1 or 2, the summation will give values of the area integrals A, Sy, Sz, Iy, Iz and Iyz which are defined as area properties.

## Area properties of composite cross-section

Section properties for the composite section are calculated using the section properties of net concrete section (i.e. with area corresponding to reinforcement subtracted) and section properties of reinforcement considering the difference in modulus of elasticity. This section is also referred to as the transformed section.

The difference in modulus of elasticity is expressed by the factor:

• α = Es / Ec

Summation of areas and static moments around coordinate axes give property values of the transformed section:

• A = Acn + α As
• Sy = Scny + α Ssy
• Sz = Scnz + α Ssz

Coordinates for the centroid of the transformed area are then easily found by:

• ycg = Sy / A
• zcg = Sz / A

Moments of inertia are then calculated using Steiner’s theorems:

• Iycg = Icny + α Isy – zcg2 A
• Izcg = Icnz + α Isz – ycg2 A
• Iyzcg = Icnyz + α Isyz – ycg zcg A

# Stresses and strains at linear section analysis

Section forces are assumed to act in the reference point R, which coincide with the centre-of gravity of the uncracked section.

At linear section analysis it is assumed that:

1. Plane cross-sections remain plane at bending (Bernoulli’s hypothesis).
2. Only normal stresses occur in pure bending (Navier’s hypothesis).
3. A linear stress-strain relation applies for the normal stresses in every fiber (Hooke’s law).

Formulated as equations (see also [5] and [6]) we get from a):

• ε = ε0 + ψy (z – zR) + ψz (y – yR)

where:

• ε0 is the strain at the reference point R
• ψy = ∂ε/∂z, curvature in the xz plane
• ψz = ∂ε/∂y, curvature in the xy plane

From conditions b) and c) we get:

• σ = E ε
• N = ∫σ dA
• My = – ∫σz dA
• Mz = ∫σy dA

Substitution of the expression for ε gives:

• N = E (Aε0 + Syψy + Szψz)
• My = – E (Syε0 + Iyψy + Iyzψz)
• Mz = E (Szε0 + Iyzψy + Izψz)

where

• S are static moments for the section area around axis parallel to y and z-axes through reference point R.
• I are moments of inertia for the section area around axis parallel to y and z-axes through reference point R.

In matrix form it can be written:

This equation can be used to determine the stress resultants {N, My, Mz}T when the strain or stress distributions are known.

The inverse can be used to determine for given values of {N, My, Mz}T the axial strain and the curvatures that define the strain distribution:

Multiplication of the strain distribution by E gives the stress distribution with the three stress parameters:

• σ0 = E ε0
• γy = E ψy
• γz = E ψz

and the stress in an arbitrary point as:

• σ = σ0 + γ(z – zR) + γz (y – yR)

## Stresses in reinforcement

When the strain distribution over the section is determined it is it easy to calculate the reinforcement stresses. First calculate the strain in the bar position (y, z) using the expression for the strain distribution. The bar stress is then calculated using Hooke’s law. Thus we get:

• σs = Es0 + ψy (z – zR) + ψz (y – yR)]

In case of initial stresses in the bar we have to add these stresses to the above bending stresses.

## Time-dependent stresses and strains

An accurate determination of stresses and strains due to time-dependent effects is important in the analysis and design of indeterminate structures, particularly in sway frames or other structures where displacements have great influence on the solution. In such structures long-time displacements due to shrink and creep can have a major influence and a rigorous method for calculating these effects are therefore of great importance.

Stress-strain relation for concrete at creep

At serviceability limit state it is assumed that a stress increment Δσc introduced at time t0 and sustained, without change in magnitude, produces at time t a strain given by:

εc(t) = Δσc (1 + ϕ) / Ec(t0)

where:

Ec(t0) is the modulus of elasticity of concrete at the instant t0,

ϕ [= ϕ(t, t0)] is the creep coefficient.

The value of ϕ representing the ratio of creep to the instantaneous strains depends on the properties of the concrete and the environment in which it is kept. Values are proposed in various codes but it must be emphasised that these values are rough and may have weaknesses.

When the stress increment Δσc is gradually introduced from zero at time t0 to its full value at time t the strain at time t is given by:

εc(t) = Δσc (1 + χc ϕ) / Ec(t0)

where:

χc [= χc(t, t0) ≅ 0.8] is the ageing coefficient of concrete.

The strain may also be written as:

εc(t) = Δσc / Eceff

with Eceff [Eceff(t, t0)] is the age-adjusted modulus of elasticity defined by:

Eceff = Ec(t0) / (1 + χc ϕ)

Stress-strain relation for steel at relaxation

At serviceability limit state it is assumed that the stress-strain relation for normal nonprestressed steel is not time dependent, i.e. it will not change during a calculation.

For prestressed steel the stress strain relation is time dependent. The phenomenon, known as relaxation, appears as stress loss at constant strain. The time dependent relation for intrinsic relaxation is usually set to:

• σs (t) = σs (t0) (1 – χs)

Because of changes in steel stress due to loading history (with influences of creep and shrinkage) the intrinsic relaxation have to be reduced in accordance with these stress changes.

Transformed cross sections at creep

For the analysis of stress and strain occurring immediately after load application on reinforced concrete sections, the term transformed section is used to represent a section composed of the area of concrete Ac plus the areas of the reinforcement As multiplied by α = Es / Ec(t0), ), see Area properties of composite cross-section.

When analysis is performed for changes in stress and strain due to creep, shrinkage and relaxation, the term age-adjusted transformed section is used. This is composed of the area of concrete Ac plus the areas of the reinforcement As multiplied by α = Es / Eceff(t0).

Four analyses steps

Four steps can be followed to determine the stress and strain distributions at time t0, immediately after application of section forces, and at time t after occurrence of creep, shrinkage and relaxation.

• Step 1
Apply section forces {N, My, Mz}T on a transformed section to determine {ε0(t0), ψy(t0), ψz(t0)}T which define the instantaneous strain. Multiplication by Ec(t0) gives {σ0(t0), γy(t0), γz(t0)}T, which define the instantaneous concrete stresses. Multiplication of the concrete stress by α(t0) gives the stresses in nonprestressed steel. For the prestressed steel the initial tension must be added.
• Step 2
Determine the hypothetical change, in the period t0 to t, in the strain distribution due to creep and shrinkage if they were free to occur. The change in strain in the reference point R is equal to [ϕ(t, t0) ε0(t0) + εcs] and the changes in curvature are [ϕ(t, t0) ψy(t0) + Δεcsy] and [ϕ(t, t0) ψz(t0) + Δεcsz]. εcs is the free shrinkage of concrete in the period t0 to t, and Δεcsy and Δεcsz are the differential shrinkage in case of not constant shrinkage over the section.
• Step 3
Calculate the artificial stress, which when gradually introduced in the concrete during the period t0 to t, will prevent occurrence of the strain due to free creep and shrinkage calculated in step 2. The restraining stress distribution Δσrestrained is given by:

• Step 4
Determine the section forces, which are the resultants of Δσrestrained using section properties for the net concrete section.

The change in concrete strain due to relaxation of prestressed steel can be artificially prevented by the application of, at the level of the prestressed steel, a restraining force equal to AsΔσrelax, where Δσrelax is the reduced value of the stress relaxation in the period t0 to t.

Summing up gives {ΔN, ΔMy, ΔMz}Trestrained, the restraining forces required to artificially prevent the strain change due to shrinkage, creep and relaxation.

To eliminate the artificial restraint apply {ΔN, ΔMy, ΔMz}Trestrained in reversed direction on an age-adjusted transformed section and calculate the corresponding changes in strains:

and stresses:

Final stresses and strains

The concrete strain distribution at time t is the sum of the elastic strain (step 1) and the change in strain due to shrinkage, creep and relaxation (step 4).

The concrete stress distribution at time t is the sum of the elastic stress (step 1), the restraint stresses (step 3) and the changes in stresses due to the reversed restraint (step 4).

The strain ε in reinforcement at the time is the sum of the elastic strains (step 1) and the change in strain due to the reversed restraint (step 4). The final stress is then simply calculated as Esε. In prestressed reinforcement the initial stresses and the reduced relaxation must be added to get the final.

## Stresses and strain in cracked sections

Stresses and strain in a cracked section can generally not be solved by a direct method because only the compressed part of the section is active and the part in tension is ignored. The properties for the active part are then dependent of the position of the neutral axis and thus this must be determined before calculation of stresses and strains may be completed. On the other hand the position of the neutral axis depends on the stress-strain distribution, so obviously an iterative method is needed.

Obviously a stress distribution must be determined that satisfies the equilibrium of the cracked section. As before (see Stresses and strains at linear section analysis) we get:

The stress at any fiber is a function of strain in accordance with the stress-strain relations, which generically may be non-linear. However, in serviceability limit state these relations are often considered to be linear in accordance with Hooke’s law.

At determination of the position of the neutral axis position we assume as a first attempt uncracked section:

where the section properties are properties of a transformed uncracked section.

By numerical integration, using first trial values of the strain parameters and the stress-strain relation, determine {N, My, Mz}calculated ignoring concrete in tension.

Determine a vector of residuals:

Use Newton-Raphson iteration to bring {R} close to {0}:

Solution of this equation gives incremental strain parameters leading to improved trial values:

The elements of the partial derivative matrix can be taken equal to Ec(t) multiplied by the area properties of a transformed section in the cracked:

For any trial values of the strain parameters 0, ψy, ψz}T the equation of the neutral axis is:

e0 + yy (z – zR) + yz (y - yR) = 0

This line will indicate the compression zone to be included in calculating the transformed cracked section above. The partial derivative matrix determined by the trial values 0, ψy, ψz}trial 1 may be used in all iterations, but recalculation in every step can speed up the convergence.

Time-dependent stresses in cracked sections

Calculating time-dependent stresses and strains in cracked sections is performed principally as in chapter Time-dependent stresses and strains. However, as a result of the change of stress distribution between concrete and reinforcement the neutral axis changes position. In cracked sections this calls for a minor iteration to determine the final position of the neutral axis after time-dependent changes in stress-strain distributions.

## Cracking

Cracking stress

A section is assumed to be in cracked state when the maximum tensile stress, calculated on basis of an uncracked section, exceeds the tensile strength of the concrete. At bending the crack condition is then expressed as:

• σn + σm ≤ fctm

where:

• σn is stress of normal force
• σm is stress of moment
• fctm is the mean tensile design strength

Cracking moment

Cracking moment is the moment that barely causes cracking. With section forces and with belonging stresses σn and σm we find the factor c that induces cracking at forces and c M by:

• c = (fctd – σn ) / σm

and cracking moment by:

• Mcrack = c M

Crack width calculation

The crack width wk may be calculated from:

• wk = Sr,maxsm - εcm)

where:

• Sr,max is the maximum crack spacing,
• εsm is the mean strain in the reinforcement,
• εcm is the mean strain in the concrete between cracks.

εsm - εcm may be calculated from:

• εsm - εcm = [σs - kt fct,eff / ρp,eff (1 + αe ρp,eff)] / Es ≥ 0,6 σs / Es

where:

• σs is the stress in the tension reinforcement assuming cracked section
• αe is the ratio Es / Ecm
• fct,eff is the mean value of the tensile strength of the concrete when the first crack occur
• fct,eff = fctm
• ρp,eff = (As + ζ12 Ap') / Ac,eff
• Ap' is the area of pre or post-tensioned tendons within Ac,eff
• Ac,eff is the effective area of concrete as calculated below
• ξ1 is the adjusted ratio of bond strength as calculated below
• kt is a factor dependent on the duration of the load,

Effective area of Ac,eff

Ac,eff is the effective area of concrete of depth hc,ef where hc,ef is the lesser of:

2,5 (h - d), (h - x) / 3 or h / 2, see figure below:

Adjusted ratio of bond strength ξ1

ξ1 is the adjusted ratio of bond strength taking into account the different diameters of prestressing and reinforcing steel:

ξ1 = (ξ φs / φp)0,5

where:

• ξ is the ratio of bond strength according to the below table
• φs is the largest bar diameter
• φp is equivalent diameter of prestressing steel

Crack spacing Sr,max

For bonded reinforcement with spacing ≤ 5(c + φ / 2) the crack spacing is calculated as:

• Sr,max = k3 c + k1 k2 k4 φ / ρp,eff

where:

• φ is the bar diameter in mm. If more than one bar size is present an average bar size φeq should be used,
• φeq = (n1 φ12 + n2 φ22) / (n1 φ1 + n2 φ2)
• c is the cover to the longitudinal reinforcement,
• k1 = 0,8 for high bond bars, 1,6 for plain bars (e.g. prestressing tendons),
• k2 = 0,5 for bending, 1,0 for pure tension
• k3 = 3,4
• k4 = 0,425
• ρp,eff as above.

For not bonded reinforcement or reinforcement with spacing > 5 (c + φ / 2) the crack spacing is calculated as:

• Sr,max = 1,3 (h - x)

where:

• x is the neutral axis depth.

The crack spacing should be calculated in the direction of the principle tensile stress as:

• Sr,max = 1 / (cos θ / Sr,max,y + sin θ / Sr,max,z)

where:

• θ is the angle between the reinforcement in the y-direction and the direction of the principal tensile stress
• Sr,max,y is the crack spacings calculated in the y direction
• Sr,max,z is the crack spacings calculated in the z direction

# Stresses and strains at non-linear analysis

Non-linear section analysis is used when stresses or strains are so large that the relation between stress and strain no longer is assumed to be linear. This is generally the case at Ultimate Limit State.

## Internal section forces from stresses and strain

At nonlinear section analysis it is assumed that (compare Stresses and strains at linear section analysis):

1. Plane cross-sections remain plane at bending (Bernoulli’s hypothesis).
2. Only normal stresses occur in pure bending (Navier’s hypothesis).

In order to determine resultants to stresses in the section, generally numerical integration is needed. A simple method to make the integration is to divide the section into a number of concrete layers and reinforcement bars and then analyze each layer and bar separately. Strain and stress are determined in the mid-depth of each component and a numerical integration of the resulting moments and forces is performed:

with stresses according to:

• fci = fcci) where εci is concrete strain at layer position,
• fcj = fccj) where εcj is concrete strain at bar position,
• fsj = fssj) where εsj is steel strain at bar position,

where:

• n, = number of concrete layers and reinforcing bar elements
• bi, hi = width and depth of a concrete layer i
• As = cross-sectional area of reinforcing bar
• zcg = z-coordinate for centroid of section in local coordinate system with y-axis parallel to neutral axis
• fc = stresses in concrete
• fs = stresses in reinforcement
• fc(ε) = stress-strain relation (non-linear) for concrete
• fs(ε) = stress-strain relation (non-linear) for steel

If the section is divided into a sufficient number of layers the accuracy will be satisfactory even if large non-linearity in the stress distribution exists.

## 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

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

UK Annex

αcc = 0,85

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

## Calculation of stress and strains due to section forces

At non-linear section analysis determination of the strain distribution on a section including position of the neutral axis is generally not possible to perform with a direct method. Instead an iterative method must be. The method that is described in Stresses and strain in cracked sections is suitable to use at non-linear calculation too. The calculation of section forces out of assumed strains is then performed using the method described in Internal section forces from stresses and strain.

With this method primarily the strain distribution is determined and after that the stresses are calculated using the non-linear stress-strain relation.

# Elongation and curvature

Elongation and curvature of a section is immediately available when the strain distribution has been calculated (linear or non-linear) as the components of the strain distribution represent the strain e0 in reference point and the curvatures ψy and ψz.

In structural calculations these values may be used directly for an uncracked section. For a cracked section it is unfavourable to use the cracked values directly as influence of uncracked concrete between cracks is not regarded. This influence is taken care of by the distribution coefficient ζ, which approximately regards the distribution between cracked and uncracked part of the element. We get:

The coefficient ζ is generally calculated in connection to cracking.

# Ultimate moment capacity

Moment capacity in ultimate limit state is most easily calculated with a fixed neutral axis inclination, e.g. parallel to either or z-axis. As non-linear material and cracking are assumed iteration is required to establish equilibrium together with external normal force and at the same time calculate current moment capacity.

Start the calculation by setting maximum allowed strain in extreme concrete fiber (εcmax) and extreme reinforcement bar (εsmax) and calculate force resultants in compression zone and tension zone in order to determine if compression or tension is decisive. Iteration continues with adjustment of neutral axis position to get maximum utilization of decisive materials at established equilibrium conditions.

Moment capacity may also be calculated in an optional direction, either as a fixed neutral axis inclination (other than y or z-axis) or as a fixed moment vector inclination. In the latter case an iteration is generally demanded to establish the neutral axis direction that corresponds to the given moment vector inclination.

# 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
• 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
• but VEd ≤ VRd,max

where:

• 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] / 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.

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

Danish Annex

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

where curtailed reinforcement is chosen:

• tan α / 2 ≤ cot θ ≤ 2,0

where:

• 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 = 0,76 (1 - fck / 200) (fck in MPa) (5.103NA)

# Shear capacity without 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 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 is the smallest width of the cross-section in the tensile area (mm)
• σcp = NEd / Ac < 0,2 (MPa)

where:

• NEd is the axial force in the cross-section due to loading or prestressing (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:

• 0,6 (1 - fck / 250) (fck in MPa)

# Rules for shear reinforcement

The shear reinforcement should form an angle a 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 s

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

Danish Annex

The minimum shear reinforcement ratio is given by:

ρw,min = (0,063 fck0,5) / fyk