Minimum shear reinforcement
EUROCODE-2

Minimum shear reinforcement

DATA
Member (?)

Select member you want to calculate minimum and maximum reinforcement areas

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Dimensions
b(cm)   (?)   h(cm)
  • b: width section. Valid values from 10 to 150
  • h: depth section. Valid values from 10 to 150
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Reinforcement data
c(mm)     (?)     α(ºsex)
  • c: Mechanical concrete cover. Valid values from 10 to h/3
  • α: Angle between shear reinforcement and the longitudinal axis. Valid values from 45 to 90.
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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
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VALUES FOR USE IN A COUNTRY
Use values recommended
Ratio ρw,min (?)
·(√fck)/fyk

Minimum ratio of shear reinforcement (see clause 9.2.2(5))
The recommended value is 0.08·(√fck)/fyk. Valid values from (0.04 to 0.2)·(√fck)/fyk

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Long. spacing sl,max(cm) (?)
·d·(1+cotα)

Maximum longitudinal spacing between shear assemblies (see clause 9.2.2(6))
The recommended value is 0.75·d·(1+cotα). Valid values from (0.3 to 1)·d·(1+cotα)

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Transverse spacing sb,max (?)
·d ≤ mm

Maximum transverse spacing of the legs in a series of shear links (see clause 9.2.2(8))
The recommended value is 0.75·d ≤ 600 mm. Valid values from (0.3 to 1)·d ≤ (300 to 900) mm

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RESULT

Calculated reinforcement
Minimum Area
Asw,min/s (mm2/mm)		 Max. Long. spacing
sl,max (cm)		 Max. Trans. spacing
sb,max (cm)
0.263 19.9 19.9

Proposed reinforcement
3 legged stirrups Φ6mm /19cm




DETAILS OF CALCULATION

Notation and methodology according to clause 9.2.2 of EC2

Asw,min/s (Minimum shear reinforcement area) = 0.263 mm2/mm
Asw,min/s = ρw,min · bw · sinα = 0.00088 · 300 · sin(90º) = 0.2629 mm2/mm
where

  • ρw,min = 0.08·(√fck)/fyk = 0.08·(√30)/500 = 0.00088

sl,max (Maximum longitudinal spacing) = 19.9 cm
sl,max = 0.75·d·(1+cotα) = 0.75·26.5·(1+cot(90º)) = 19.9 cm
where

  • d (effective depth) = h - c = 30 - 3.5 = 26.5 cm

sb,max (Maximum transverse spacing) = 19.9 cm
sb,max = min(0.75·d ; 600) = max (19.9 ; 60) = 19.9 cm