# Shear capacity of the shear reinforcement

The shear capacity of a section with shear reinforcement is considered to be sufficient if

VEd ≤ VRd,s but VEd ≤ VRd,max

where

VRd,s is the shear capacity of a section with shear reinforcement,

VRd,max is the max capacity.

VRd,s = Asw / s z fywd (cot θ + cot α) sin α or s = [Asw z fywd (cot θ + cot α) sin α] / VEd

VRd,max = αcw bw z υ1 fcd (cot θ + cot α) / (1 + cot2θ)

where

fywd is the design yield strength of the shear reinforcement,

Asw is the area of the shear reinforcement,

s is the spacing of the stirrups,

the angle θ should be limited to 1 ≤ cot θ ≤ 2,5 and

υ1 is a strength reduction factor for concrete cracked in shear.

If the design stress of the shear reinforcement is below 80% of the yield stress fyk then

υ1 = 0,6 for fck ≤ 60 MPa

υ1 = 0,91 - fck / 200 > 0,5 for fck > 60 MPa

In this case fywd = 0,8 fywk else

υ1 = 0,6 ( 1 - fck / 250) (fck in MPa)

αcw is a coefficient taking account of the state of the stress in the compression chord,

αcw = 1,0 for non-pre stressed structures.

The maximum effective shear reinforcement Asw,max for cot θ = 1 follows from

Asw,max fywd / (bw s) ≤ 0,5 αcw υ1 fcd / sin α

In regions where there is no discontinuity of VEd (e.g. uniformly distributed loading) the shear reinforcement in any length increment l = z (cot θ + cot α) may be calculated using the smallest value of VEd in the increment.

# National Annex: Denmark

If reinforcement class B or C is used then the inclination θ of the compressive stress evaluates to

tan α / 2 ≤ cot θ ≤ 2.5

where curtailed reinforcement becomes

tan α / 2 ≤ cot θ ≤ 2.0

and where α is the angle between shear reinforcement and the beam axis.

If reinforcement class A is used then cot θ = 1,0.

The strength reduction factor for concrete cracked in shear is calculated as

υ1 = 0,7 (1 - fck / 200) (fck in MPa) (5.103 NA)

# National Annex: Sweden

angle θ should be limited to 1 ≤ cot θ ≤ 2.5 (1 ≤ cot θ ≤ 3.0 for prestressed members)

# National Annex: United Kingdom

If shear co-exists with applied tension then cot θ should be taken as 1,0.

If the design stress of the shear reinforcement is below 80% of the yield stress fyk then

υ1 = 0,54 (1 - 0,5 cos α) for fck ≤ 60 MPa

υ1 = (0,84 - fck / 200) (1 - 0,5 cos α ) > 0,5 for fck > 60 MPa