# Shear capacity of the shear reinforcement

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

V_{Ed} ≤ V_{Rd,s} but V_{Ed} ≤ V_{Rd,max}

where

V_{Rd,s} is the shear capacity of a section with shear reinforcement,

V_{Rd,max} is the max capacity.

V_{Rd,s} = A_{sw} / s z f_{ywd} (cot θ + cot α) sin α or s = [A_{sw} z f_{ywd} (cot θ + cot α) sin α] / V_{Ed}

V_{Rd,max} = α_{cw} b_{w} z υ_{1} f_{cd} (cot θ + cot α) / (1 + cot^{2}θ)

where

f_{ywd} is the design yield strength of the shear reinforcement,

A_{sw} 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

υ_{1} 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 f_{yk} then

υ_{1} = 0,6 for f_{ck} ≤ 60 MPa

υ_{1} = 0,91 - f_{ck} / 200 > 0,5 for f_{ck} > 60 MPa

In this case f_{ywd} = 0,8 f_{ywk} else

υ_{1} = 0,6 ( 1 - f_{ck} / 250) (f_{ck} in MPa)

α_{cw} is a coefficient taking account of the state of the stress in the compression chord,

α_{cw} = 1,0 for non-pre stressed structures.

The maximum effective shear reinforcement A_{sw,max} for cot θ = 1 follows from

A_{sw,max} f_{ywd} / (b_{w} s) ≤ 0,5 α_{cw} υ_{1} f_{cd} / sin α

In regions where there is no discontinuity of V_{Ed} (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 V_{Ed} 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

υ_{1} = 0,7 (1 - f_{ck} / 200) (f_{ck} 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 f_{yk} then

υ_{1} = 0,54 (1 - 0,5 cos α) for f_{ck} ≤ 60 MPa

υ_{1} = (0,84 - f_{ck} / 200) (1 - 0,5 cos α ) > 0,5 for f_{ck} > 60 MPa