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**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 = 1080000mm^{2}

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

V_{Ed}^{*}= V_{Ed} / 2 + (T_{Ed} ^{.} z) / (2 ^{.} A_{k}) = [1300 ^{.} 10^{3}/2 + (700 ^{. }10^{6} ^{.} 1350) / (2 ^{.} 1.08 ^{.} 10^{6})] ^{. }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 ^{. }10^{3}) / (1350 ^{. }435 ^{.} 2.14) = 0.865mm^{2}/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 ^{.} 10^{6} / (2 ^{.} 1.08 ^{.} 10^{6} ^{.} 435 ^{.} 2.14) = 0.348mm^{2}/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) = 6855mm^{2} 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) = 3198mm^{2} 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

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