EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 6-7
EXAMPLE 6.4b – the same above, with steel S500C f yd = 435 MPa. [EC2 clause 6.2]
The example is developed for three classes of concrete.
a) fck = 30 MPa ; fcd = 17 MPa ; ν = 0.616
Aswfywd 2
= sin θ
obtained for VRd,s = VRd,max
bsf ν
w cd
2 226⋅ 435
it results: sin θ= = 0.417 hence cotθ = 1.18
150 ⋅150⋅0.616⋅17 Asw 226
−3
Then VRd,s = ⋅z⋅fywd ⋅cot θ= ⋅500⋅435⋅1.18 ⋅ 10 = 387 kN
s 150
b) For the same section and reinforcement, with fck = 60 MPa , fcd = 34 MPa;
ν = 0.532, proceeding as above it results:
2 226⋅ 435
sin θ= = 0.242
hence cotθ = 1.77
150⋅150⋅0.532⋅34 Asw 226
−3
VRd,s = ⋅z⋅fywd ⋅cot θ= ⋅500⋅435⋅1.77⋅ 10 = 580 kN
s 150
c) For the same section and reinforcement, with fck = 90 MPa, fcd = 51 MPa; ν = 0.512,
proceeding as above it results:
2 226⋅ 435
sin θ= = 0.167
hence cotθ = 2.23
150 ⋅150 ⋅0.512 ⋅51
Asw 226
−3
VRd,s = ⋅z⋅fywd ⋅cot θ= ⋅500 ⋅435⋅2.23⋅ 10 = 731 kN
s 150
Determination of reinforcement (vertical stirrups) given the beam and shear action VEd
Rectangular beam bw = 200 mm, h = 800 mm, d = 750 mm, z = 675 mm; vertical stirrups
fywd = 391 MPa. Three cases are shown, with varying values of VEd and of fck.
•VEd = 600 kN; fck = 30 MPa ; fcd = 17 MPa ; ν = 0.616 then
1 2VEd
1 2 ⋅600000
θ= arcsin = arcsin = 29.0
2 ( α νf )b z 2 (1⋅0.616⋅17) ⋅200⋅675 Table of Content
cw cd w
Asw VEd 600000
2
It results: = = = 1.135 mm / mm
s z⋅f⋅cotθ 675⋅435⋅1.80 ywd
which is satisfied with 2-leg stirrups φ12/190 mm.
o
hence cotθ = 1.80
The tensile force in the tensioned longitudinal reinforcement necessary for bending must
be increased by ΔFtd = 0.5 VEd cot θ = 0.5·600000·1.80 = 540 kN