# Stress in reinforcing steel depending on its strain, `σ`_{s}

_{s}

## 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 `ε _{ud}` and a maximum stress of

`kf`/

_{yk}`γ`at

_{S}`ε`, where

_{uk}`k`= (

`f`/

_{t}`f`)

_{y}_{k},

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

where:

`f`_{yk}- is the characteristic yield strength of the reinforcing steel, see § 3.2.2 (3)P.
`γ`_{S}- is the partial factor for reinforcing steel, see § 2.4.2.4 (1).
`k`- is the ductility property, see Table C.1
`ε`_{uk}- is the characteristic strain of the reinforcing steel at maximum load, see Table C.1
`E`_{s}- is the design value of the modulus of elasticity of the reinforcing steel, see § 3.2.7 (4).
`f`_{yd}- is the design yield strength of the reinforcing steel,
`f`=_{yd}`f`/_{yk}`γ`_{S} `ε`_{ud}- is the design strain of the reinforcing steel at maximum load,
`ε`_{ud}=`Coef`(`ε`)⋅_{ud}`ε`_{uk}, see § 3.2.7 (2) for the value of`Coef`(`ε`)._{ud}

If `ε _{s}` ≤

`ε`(with

_{se}`ε`=

_{se}`f`/

_{yd}`E`), the stress in reinforcing steel

_{s}`σ`=

_{s}`E`⋅

_{s}`ε`.

_{s} If `ε _{se}` <

`ε`≤

_{s}`ε`, the stress in reinforcing steel

_{ud}`σ`is equal to:

_{s}-
`f`+ (_{yd}`ε`-_{s}`ε`)⋅(_{se}`k``f`-_{yk}`f`)/(_{yk}`ε`-_{uk}`f`/_{yk}`E`) for inclined top branch,_{s} -
`f`for horizontal top branch._{yd}

This application calculates
the stress in reinforcing steel `σ _{s}`
from your inputs.
Intermediate results will also be given.