EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 6-2
MRd3 must also be known. This results: MRd3 = 6460·(500 – 0,4·950) ·10 -3 = 1655 kNm
Subsequently, for a chosen value of NEd in each interval between two following values of
NRd written above and one smaller than NRd1, the neutral axis x, MRd, and the eccentricity
MRd
e =
N
are calculated. Their values are shown in Table 6.2.
Ed
Table of Content
Table 6.2. Example 1: values of axial force, depth of neutral axis, moment resistance, eccentricity.
NEd (kN) X (m) MRd (kNm) e (m)
600 0,105 2031 3.38
2000 0,294 2524 1.26
5000 0,666 2606 0.52
10000 virtual neutral axis 1000 0.10
As an example the calculation related to NEd = 5000 kN is shown.
The equation of equilibrium to shifting for determination of x is written:
2 ⎛5000000 −5000⋅391−5000⋅0.0035⋅200000 ⋅5000 ⎞ ⎛0.0035⋅200000⋅5000⋅950 ⎞
x −⎜ ⎟x− ⎜ ⎟=
0
⎝ 0.80 ⋅500 ⋅17 ⎠ ⎝ 0.80 ⋅500 ⋅17
⎠
Developing, it results:
x 2 + 66.91x – 488970 = 0
which is satisfied for x = 666 mm
⎛950 ⎞
2
The stress in the lower reinforcement is: σ s = 0.0035⋅200000⋅⎜ − 1⎟= 297N/ mm
⎝666 ⎠
The moment resistance is:
MRd = 5000·391·(500-50) + 5000·297·(500-50) + 0.80·666·500·17·(500 – 0.40 666) =
2606·10 6 Nmm = 2606 kNm
2606
and the eccentricity e = = 0,52m
5000