Bending with or without axial force
EUROCODE-2

Bending with or without axial force

DATA
Dimensions (?)
b(cm)     h(cm)     c(mm)
  • b: width section. Valid values from 5 to 150
  • h: depth section. Valid values from 5 to 150
  • c: Concrete cover. Valid values from 10 to h/2
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Concrete (?)
fck (MPa)       γc       αct
  • fck: the characteristic compressive cylinder strength of concrete. Valid values from 20 to 90
  • γc: Partial factor for concrete for ultimate limit state. National choice 2.4.2.4(1). Valid values from 1 to 2
  • αct: National choice 3.1.6 (1)P: Coefficient of fatigue strength of the concrete. Recommended value is 1.0. Valid values from 0.5 to 1
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Reinforcement (?)
Class       fyk(MPa)       γs
  • Class: select the steel class (see Table C.1)
  • fyk: the characteristic strength of reinforcement. Valid values from 400 to 600
  • γs: Partial factor for reinforcement for ultimate limit state. National choice 2.4.2.4(1). Valid values from 1 to 1.8
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detalle datos

INTERNAL FORCES
Bending Md(Kn·m) (?)

Design bending moment in the section considered (Md). Value positive according to figure.
Valid values from -3·104 to 3·104

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Axial force Nd(Kn) (?)

Design axial force in the section considered (Nd).
Valid values from -3·105 to 3·105. Value according to figure:

  • positive for compression,
  • negative for tensile.

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REINFORCEMENTS
Reinfor. inferior
As1 (cm2)  (?)  Φmax
  • As1: Cross-sectional area of inferior reinforcement. Valid values from As,min (2Φ8) to As,max (layer Φ32 with bar spacing s=5 cm).
  • Φmax: max. nominal diameter of a reinforcing bar. Valid values from 5 to 40

You can enter the area directly or through the following data:

typeNum Φ
1
2
Intro
Reinfor. superior
As2 (cm2)  (?)  Φmax
  • As1: Cross-sectional area of inferior reinforcement. Valid values from 0 to As,max (layer Φ32 with bar spacing s=5 cm).
  • Φmax: max. nominal diameter of a reinforcing bar. Valid values from 5 to 40

You can enter the area directly or through the following data:

typeNum Φ
1
2
Intro

VALUES FOR USE IN A COUNTRY
Use values recommended
Max. strain steel (?)
εud = · εuk

εud: strain limit of steel. The recommended value is 0.90·εuk. Valid values from (0.2 to 1.0)·εuk

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RESULT

Design bending Bending at failure Requirement
Md (KN.m) Mu (KN.m) |Md| ≤ |Mu|
250 313.8 OK
 
Design axial force Axial force at failure Requirement
Nd (KN) Nu (KN) |Nd| ≤ |Nu|
10 12.55 OK



DETAILS OF CALCULATION

Notation and methodology according to clause 6.1 of EC2

Internal forces at failure are the point of interaction diagram axial load - bending which Mu/Nu value is equal to Md/Nd

Range check: 3
d · εcu/(εcu+εud) < x(cm)= 4.59 ≤ xlim
d · εcu/(εcu+εud)= 56 · 0.00288/(0.00288+0.045) = 3.4 cm
xlim = εcu·d / (εcu+fyd/Es) = 0.00288·56 / (0.00288+434.78/200000) = 31.9 cm

Nu (Axial force at failure) = 12.55 KN
Nu(x) = η·fcd·λ·x·b + As2·σs2 - As1·σs1
Nu(N) = 0.95·40·0.78·45.87·400 + 628·73.76 - 1257·456.71

Mu (Bending at failure) = 313.8 KN·m
Mu(x) = η·fcd·λ·x·b·(h/2-λ·x/2) + As2·σs2·(h/2-d′) - As1·σs1·(h/2-d)
Mu(N·m) = 0.95·40·0.78·45.87·400·(0.6/2-0.78·0.0459/2) + 628·73.76·(0.6/2-0.04) - 1257·456.71·(0.6/2-0.56)

where:

  • For 50 < fck = 60 ≤ 90 MPa
    η = 1.0 - (60-50)/200 = 0.95
    λ = 0.8 - (60-50)/400 = 0.78
    εc3 = 1,75 + 0,55[(60 - 50)/40] = 1.89(0/00)
    εcu = 2.6 + 35[(90 - 60)/100]4 = 2.88(0/00)
  • For class of steel B:
    k= 1.08; εuk = 0.05; εud = 0.9 · εuk = 0.045
  • x (depth of the neutral axis) = 4.587 cm (from the upper edge)
    Obtained by iteration in the nonlinear system of equations
  • σs2 = Es · εs2 = 200000· 0.00037 = 73.76 MPa
    εs2 = εcu·(x-d′)/x = 0.00288·(4.587-4)/4.587 = 0.00037
    σs1 = fyd + p·(εs1-fyd/Es) = 434.78 + 727.27·(0.03232-434.78/200000) = 456.71 MPa
    εs1 = εcu·(d-x)/x = 0.00288·(56-4.587)/4.587 = 0.03232
    p = (k·fyd-fyd)/(εuk-fyd/Es) = (1.08·434.78-434.78)/(0.05-434.78/200000) = 727.27 MPa
  • d (effective depth) = h – c - Φmax,s1/2 = 60 – 3 – 2/2 = 56 cm
  • d′ = c + Φmax,s2/2 = 3 + 2/2 = 4 cm
  • fcd = αcc · fck / γc = 1 · 60 / 1.5 = 40 N/mm2
  • fyd = fyk / γs = 500 /1.15 = 434.78 N/mm2