EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 6-23
rotational equilibrium:
⎛ x ⎞
F ⎜a ⎟ F z
⎝ 2 ⎠
1
.
Ed c + = c 500000 (200+31) = Fc 208
500000 ⋅ (200 + 31)
Fc = Ft = = 555288 N ≅ 556 kN
208
node 1 verification
c
2 2
σ = = = 13.37 N / mm ≤ σ 1Rd,max = 20.12 N / mm
b( 2 y1 ) 400 ( 2⋅ 52)
Table of Content
F 556000
Main upper reinforcement design:
F 556000
A 1421 mm
t
2
s = = = we use 8φ16 (As = 1608 mm
fyd 391.3
2 )
Secondary reinforcement design:
(the expression deduced at the beginning of this example is used)
a
2 −1
F z
w = FEd ≅ 204 kN
3
F 204000
w
2
Aw = = = 521 mm
fyd 391.3
EC2 suggests a minimum secondary reinforcement of:
F 500000
A k 0.5 639 mm
Ed
2
w ≥ 2 = = we use 3 stirrups φ 12 (As = 678 mm
fyd 391.3
2 )
node 2 verification, below the load plate:
The node 2 is a compressed-stressed node, in which the main reinforcement is anchored; the
compressive stress below the load plate is:
F 500000
150⋅ 220 33000
Ed
2 2
σ = = = 15.15 N / mm ≤ σ2Rd,max = 17.05 N / mm