EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 6-6
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
Table of Content
1 2VEd
1 2 ⋅600000
θ= arcsin = arcsin = 29.0
2 ( α νf )b z 2 (1⋅0.616⋅17) ⋅200⋅675 hence cotθ = 1.80
cw cd w
Asw VEd 600000
2
It results: = = = 1.263 mm / mm
s z⋅f⋅cot θ 675⋅391⋅1.80 ywd
which is satisfied with 2-leg stirrups φ12/170 mm.
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
•VEd = 900 kN; fck = 60 MPa ; fcd = 34 MPa ; ν = 0.532
1 2VEd
1 2 ⋅900000
θ= arcsin = arcsin = 23.74
2 ( α νf )b z 2 (1⋅0.532⋅34) ⋅200⋅675 hence cotθ = 2.27
cw cd w
Asw VEd 900000
2
Then with it results = = = 1.50 mm / mm
s z⋅f⋅cotθ 675⋅391⋅2.27 which is satisfied with 2-leg stirrups φ12/150 mm.
ywd
The tensile force in the tensioned longitudinal reinforcement necessary for bending
must be increased by ΔFtd = 0.5 VEd cot θ = 0.5·900000·2.27= 1021 kN
• VEd = 1200 kN; fck = 90 MPa ; fcd = 51 MPa ; ν = 0.512
1 2VEd
1 2 ⋅1200000
θ= arcsin = arcsin = 21.45
2 ( α νf )b z 2 0.512⋅51⋅200⋅675 cw cd w
As θ is smaller than 21.8 o , cotθ = 2.50
Asw VEd 1200000
2
Hence = = = 1.82 mm / mm
s z⋅f⋅cot θ 675⋅391⋅2.50 ywd
which is satisfied with 2-leg stirrups φ12/120 mm.
The tensile force in the tensioned longitudinal reinforcement necessary for bending
must be increased by ΔFtd = 0.5 VEd cot θ = 0.5·1200000·2.50 = 1500 kN
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