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Stress in reinforcing steel depending on its strain, σs

Eurocode 2 part 1-1: Design of concrete structures  3.2.7 (2)

For normal design, either of the following assumptions may be made (see Figure 3.8):

a) an inclined top branch with a strain limit of εud and a maximum stress of kfyk/γS at εuk, where k = (ft/fy)k,

b) a horizontal top branch without the need to check the strain limit.

where:

fyk
is the characteristic yield strength of the reinforcing steel, see § 3.2.2 (3)P.
γS
is the partial factor for reinforcing steel, see § 2.4.2.4 (1).
k
is the ductility property, see Table C.1
εuk
is the characteristic strain of the reinforcing steel at maximum load, see Table C.1
Es
is the design value of the modulus of elasticity of the reinforcing steel, see § 3.2.7 (4).
fyd
is the design yield strength of the reinforcing steel, fyd = fyk/γS
εud
is the design strain of the reinforcing steel at maximum load, εud  = Coef(εud)⋅εuk, see § 3.2.7 (2) for the value of Coef(εud).

If εsεse (with εse = fyd/Es), the stress in reinforcing steel σs = Esεs.

If εse < εsεud, the stress in reinforcing steel σs is equal to:

  • fyd + (εs - εse)⋅(kfyk - fyk)/(εuk - fyk/Es) for inclined top branch,
  • fyd for horizontal top branch.

This application calculates the stress in reinforcing steel σs from your inputs. Intermediate results will also be given.

Input
GPa
MPa
%
Output
fyd
MPa
εse
%
k
εuk
%
εud
%
the stress in reinforcing steel σS
MPa Figure 3.8