Example: longitudinal reinforcement for shear and torsion according to EC2

Given:

For the rectangular section of Figure 1 calculate the longitudinal reinforcement for shear and torsion. The section considered has depth 1500mm, width 1000mm, d = 1450mm. The vertical members have a width of 200mm and the horizontal members are 150mm wide. It is given that f_ck = 30MPa and f_yk = 500MPa, while the results for the actions are V_Ed = 1300kN (force parallel to the larger side) and T_Ed = 700kNm.

Solution:

f_cd = 0.85^. (30/1.5) = 17.0MPa

ν = 0.7 ^. [1-30/250] = 0.616

ν ^. f_cd = 10.5 MPa

f_yd = 500/1.15 = 435 MPa

Geometric elements:

u_k = 2 ^. (1500-150) + 2 ^. (1000-200) = 4300mm

A_k = 1350 ^. 800 = 1080000mm²

The maximum equivalent shear in each member (vertical) is:

V_Ed^*= V_Ed / 2 + (T_Ed ^. z) / (2 ^.  A_k) = [1300 ^. 10³/2 + (700 ^. 10^{6 .} 1350) / (2 ^. 1.08 ^. 10⁶)] ^. 10^-3 = 1087kN

It is noted that z refers to the length of the vertical member.

Verification of compressed concrete with cotθ =1. It results to:

V_Rd,max = t ^.  z ^.  ν ^.  f_cd ^. sinθ ^. cosθ = 200 ^. 1350 ^. 10.5 ^. 0.707 ^. 0.707 = 1417 k N > V_Ed^*=1087kN

Determination of angle θ:

θ=(1/2) ^. arcsin(2 V_Ed^*/( t ^. z ^. ν ^. f_cd)) = (1/2) ^. arcsin(2 ^. 1087000/( 200 ^. 1350 ^. 10.5)) = 25.03^o_. -> cotθ = 2.14.

Reinforcement of vertical members:

(A_sw /s) = V_Ed^*/ (z ^. f_yd ^. cotθ) = (1087 ^. 10³) / (1350 ^. 435 ^. 2.14) = 0.865mm²/mm

which can be carried out with 2-legs 12mm bars, pitch 200mm; pitch is in accordance with [9.2.3(3)-EC2].

Reinforcement of horizontal members, subjected to torsion only:

(A_sw /s) = T_Ed / (2 ^. A_k ^. f_yd ^. cot θ) = 700 ^. 10⁶ / (2 ^. 1.08 ^. 10^{6 .} 435 ^. 2.14) = 0.348mm²/mm which can be carried out with 8mm wide, 2 legs stirrups, pitch 200mm.

Longitudinal reinforcement for torsion:

A_sl = T_Ed ^. u_k ^. cotθ / (2 ^. A_k ^. f_yd) = 700 ^. 10^{6 .} 4300 ^. 2.14 / (2 ^. 1080000 ^. 435) = 6855mm² to be distributed on the section, with particular attention to the corner bars.

Longitudinal reinforcement for shear:

A_sl = V_Ed ^. cotθ / (2 ^. f_yd) = 1300000 ^. 2.14 / (2 ^.  435) = 3198mm² to be placed at the lower end.

List of symbols:

A_k: is the area enclosed by the centre-lines of the connecting walls, including inner hollow areas

A_sw: is the cross-sectional area of the shear reinforcement

f_cd: is the design effective compressive strength

f_yd: is the design yield stress of the longitudinal reinforcement Asl

s: is the spacing of the stirrups

T_Ed: is the design torsional moment

u_k: is the perimeter of the area Ak

V_Ed: is the design transverse force

V_Rd,max: is the maximum design shear resistance

z: is the lever arm of internal forces

θ: is the angle of compression struts

ν: is a strength reduction factor for concrete cracked in shear