EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 6-19
nodes tensioned – compressed by anchor logs in a fixed direction
⎛ fck
⎞
⎜1- ⎟
250 35
σ 2Rd,max = k
⎝ ⎠ ⎛ ⎞
2 f cd = ⎜1-⎟19.83 = 17.05 N / mm
0.85 ⎝ 250 ⎠
nodes tensioned – compressed by anchor logs in different directions
⎛ fck
⎞
⎜1- ⎟
250
35
σ 3Rd,max = k
⎝ ⎠ ⎛ ⎞
3 f cd = 0.88 ⎜1-⎟19.83 = 15 N / mm
0.85 ⎝ 250 ⎠
Actions
F Ed = 700 kN
Load eccentricity with respect to the column side: e = 125 mm (Fig. 6.8)
The beam vertical strut width is evaluated by setting the compressive stress equal to σ1Rd,max:
x1= σ
FEd 700000
= ≅87
mm
b 20.12⋅ 400
1Rd,max
the node 1 is located x1/2 ≅ 44 mm from the outer column side (Fig. 6.9)
We state that the upper reinforcement is located 40 mm from the upper cantilever side; the
distance y1 of the node 1 from the lower border is evaluated setting the internal drive arm z
equal to 0.8⋅d (z = 0,8⋅360 = 288 mm):
y1 = 0.2d = 0.2·360 = 72 mm
rotational equilibrium: FEda Fz c
Table of Content
700000 ⋅ (125 + 44) = F ⋅ 288
= c
700000 ⋅ (125 + 44)
Fc = Ft = = 410763 N ≅ 411 kN
288
node 1verification:
F 411000
c
2 2
σ = = = 7.14 N/mm ≤ σ1Rd,max = 20.12 N/mm
b ( 2 y1 ) 400 ( 2⋅ 72)
Main upper reinforcement design:
F 411000
A 1050 mm
t
2
s = = = we use 8φ14 (As = 1232 mm
fyd 391.3
2 )
Secondary upper reinforcement design:
The beam proposed in EC2 is indeterminate, then it is not possible to evaluate the stresses for
each single bar by equilibrium equations only, but we need to know the stiffness of the two
elementary beams shown in Fig. 6.10 in order to make the partition of the diagonal stress
⎛ Fc
FEd
⎞
⎜Fdiag
= = ⎟ between them;
⎝ cosθ senθ ⎠
2
2