# Minimum longitudinal reinforcement

DATA
Member (?)

Select member you want to calculate minimum and maximum reinforcement areas

Dimensions (?)
b(cm)     h(cm)     c(mm)
• b: width section. Valid values from 10 to 150
• h: depth section. Valid values from 10 to 150
• c: Mechanical concrete cover. Valid values from 10 to h/3
Materials
fck(MPa)   (?)   fyk(MPa)
• fck: Characteristic compressive cylinder strength of concrete. Valid values from 20 to 90
• fyk: Characteristic strength of tension reinforcement. Valid values from 400 to 600

INTERNAL FORCES
Bending Mk(Kn·m) (?)

Characteristic bending moment in the section considered (Mk). Valid values from 0 to 3·104

Axial force Nk(KN) (?)

Characteristic axial force in the section considered (Nk). Value positive for compression and negative for tension. Valid values from -fck·Ac to fck·Ac

VALUES FOR USE IN A COUNTRY
Use values recommended
Tensile As,min (?)
·fctm/fyk·b·d

Minimum area of longitudinal tension reinforcement (see clause 9.2.1.1(1))
The recommended value is 0.26·fctm/fyk·b·d. Valid values from (0.13 to 0.50)·fctm/fyk·b·d

As,max (?)
· Ac

Maximum area of longitudinal tension or compression reinforcement (see clause 9.2.1.1(3))
The recommended value is 0.04·Ac. Valid values from (0.02 to 0.08)·Ac

RESULT Minimum tension reinfor. Asmin,T(cm2) Maximum reinforcement Asmax(cm2) 1.2 36

Reinforcement proposed

 Minimum Maximum As1 As2 As1 As2 2Φ10 2Φ6 3Φ32 2Φ25

DETAILS OF CALCULATION

Notation and methodology according to clause 7.3.2 and 9.2 of EC2

Asmin,T (Minimum tensile reinforcement area) = 1.2 cm2

Asmin,T = max (Asmin_cracking ; Asmin_beam), with:

Asmin_cracking (minimum reinforcement to control cracking) = 1.04 cm2
Asmin_cracking = kc · k · fct,eff · Act / σs
Asmin_cracking = 0.4 · 1 · 2.9 · 450 / 500 = 1.04 cm2
where

• σs = fyk = 500 MPa
• fct,eff = fctm = 2.9 MPa
fctm = 0,30 × fck(2/3) = 0,30 × 30(2/3) = 2.9 MPa
• k = 1 (h ≤ 300)
• Act = b · hct = 30 · 15 = 450 cm2
hct (depth tension zone) = 15 cm
hct = h/2 - (h2/12)·(Nk/Mk), (≥ 0 y ≤ h)
hct,aux = 30/2 - (302/12)·(0/7000) = 15 cm
• kc = 0.4·[1 - σc/(k1·h/h*· fct,eff)] ≤ 1
kc,aux = 0.4·[1 - (0)/(1.5·30/30· 2.9)] = 0.4
where
• σc = NEd / (b·h) = 0/(300·300) = 0 MPa
• h* = min (h ; 100) = 30 cm
• k1 = 1,5

Asmin_beam (minimum reinforcement for beam) = 1.2 cm2
Asmin_beam = max (Asmin_beam_1 ; Asmin_beam_2) = max (1.2 ; 1.03) = 1.2 cm2
where

• Asmin_beam_1 = 0.26 · fctm/fyk · bt · d
Asmin_beam_1 = 0.26 · 2.9/500 · 30 · 26.5 = 1.2 cm2
with d (effective depth) = h - c = 30 - 3.5 = 26.5 cm
• Asmin_beam_2 = 0.0013 · bt · d = 0.0013 · 30 · 26.5 = 1.03 cm2

Asmax (Maximum reinforcement in section beam) = 36 cm2
Asmax = 0.04 · Ac = 0.04 · 30 · 30 cm2