EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 6-25
Top longitudinal reinforcement (A s’): 8φ20 = 2513 mm 2
Truss a) R = R Sdu /2 = 500 kN
Definition of the truss rods position
The compressed longitudinal bar has a width equal to the depth x of the section neutral axis
and then it is x/2 from the top border; the depth of the neutral axis is evaluated from the
section translation equilibrium:
0.8 b x fcd + Es ε’s A’s = fyd As
ε '
s
Table of Content
Fig. 6.15 Truss a.
0,0035
= ⋅ ( x − d')
where d’ = 50 mm is the distance of the upper surface reinforcement
x
x−50 0.8bxf + E 0.0035 A' = f A
x
and then:
x = 99 mm
cd s s yd s
0.0035 f 391.3
( )
'
yd
εs = ⋅ 99 − 50 = 0.00173 ≤ = = 0.00196
99 Es 200000
then the compressed steel strain results lower than the strain in the elastic limit, as stated in
the calculation;
the compressive stress in the concrete is
C = 0.8 b x fcd = 0.8·800·99·14.17
(applied at 0.4⋅x ≅ 40 mm from the upper surface)
while the top reinforcement stress is:
C’ = Es ε’s A’s = 200000·0.00173·2513