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 υ₁ f_cd (cot θ + cot α) / (1 + cot²θ)

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

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 f_yk then

υ₁ = 0,6 for f_ck ≤ 60 MPa

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

In this case f_ywd = 0,8 f_ywk else

υ₁ = 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 υ₁ 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

υ₁ = 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

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

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