Stresses and strains at linear section analysis

Last modified by Fredrik Lagerström on 2020/06/09 16:11


Section forces are assumed to act in the reference point R, which coincides with the centre-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 (read more: Makoto Kawakami, Amin Ghali - Time-Dependent Stresses in Prestressed Concrete Sections of General Shape), condition a) means that

ε = ε0 + ψy (z – zR) + ψz (y – yR) where ε0 is the strain at the reference point R, ψy = ∂ε/∂z, and ψz = ∂ε/∂y, represent the curvatures in the xz and xy planes.

From conditions b) and c) we get σ = E * ε

N = ∫σ dA

My = – ∫σz dA Mz = ∫σy dA of-gravity of the uncracked section.

Substitution of the expression for ε gives

N = E (Aε0 + Syψy + Szψz)

My = – E (Syε0 + Iyψy + Iyzψz) My = E (Szε0 + Iyzψy + Izψz)

where S and I are static moments and 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 as


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 the axial strain and the curvatures that define the strain distribution, for given values of {N, My, Mz}t:

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 at an arbitrary point as

σ = σ0 + γy(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. For a more accurate determination see Betonghandbok - Konstruktion utgåva 2, chapter 2.3.

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 + χAdjusted ratio of bond strength c φ) / Ec(t0)


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

The strain may also be written as

εc(t) = Δσc / Ec,eff

with Ec,eff [Ec,eff(t, t0)] being the age-adjusted modulus of elasticity defined by

Ec,eff = Ec(t0) / (1 + χcφ)

Stress-strain relation for steel at relaxation

At serviceability limit state, it is assumed that the stress-strain relation for normal non-prestressed 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 has to be reduced in accordance with these stress changes. For examples, see EN1992-1-1 and Betonghandbok - Konstruktion utgåva 2.

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 / Ec,eff(t0).

Four analysis steps

Step 1
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.

Apply section forces {N, My, Mz}t on a transformed section to determine {ε0(t0), ψy(t0), ψζ(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 non-prestressed steel. The initial tension must be added for prestressed steel.

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) + Δecsy] and [φ(t, t0) ψζ(t0) + Δecsz]. 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 Δσ restrained 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 As Δσ relax, where Δσ relax is the reduced value of the stress relaxation in the period t0 to t.

Summation gives {ΔN, ΔMy, ΔMz}t restrained, 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}t restrained in reversed direction on an age-adjusted transformed section, and calculate the corresponding changes in strains:


as well as 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 t 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, meaning an iterative method is needed.

Thus, 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.

As a first attempt at determination of the position of the neutral axis position we assume 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 {ε0, ψy, ψz}t the equation of the neutral axis is

e0 + ψy(z – zR) + ψz(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 {ε0, ψy, ψz}trial 1 may be used in all iterations, but recalculation at 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 stress

A section is assumed to be in the 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 N and M with belonging stresses σn and σm we find the factor c that induces cracking at forces N 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 occurs,
 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, kt = 0,6 for short term loading and 0,4 for long term loading.

Effective area of concrete

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
  • h / 2


Adjusted ratio of bond strength

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

ξ1 = (ξ øs / øp)0,5

where ξ is the ratio of bond strength according to the below table, øs is the largest bar diameter, øp is equivalent diameter of prestressing steel.

Prestressing steelξ
Pre-tensionedBonded, post-tensioned
≤ C50/60≥ C70/85
Smooth bars and wiresNot applicable0.300.15
Indented wires0.70.600.30
Ribbed bars0.80.700.35
Note: For intermediate values between C50/60 and C70/85, interpolation may be used.

Crack spacing

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

Sr,max = k3 c + k1 k2 k4 ø / ρp,eff


ø is the bar diameter in mm. If more than one bar size is present an average bar size φeq should be used, φeq = (n1 ø12 + n2 ø22) / (n1 ø1 + n2 ø2)

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 + ø / 2) the crack spacing is calculated as

Sr,max = 1,3 (h - x)


x is the neutral axis depth.