EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 7-20
1⎡ λ −1⎤
and then ξ 1 = ⎢1− ⎥
2 ⎣ λ ⎦ (7.25)
Finally, considering that
The (7.20) is expressed as
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v
I
5 M
48 E I ∗ =
2
max�
(7.26)
c I
2
⎛�⎞ 5 Mmax�
⎧ ⎡ 48⎛ 4 4 3⎞
12 β
⎤⎫
v⎜ ⎟= 1 ( c 1) 1 1 1 ln 2 2 ( 1
∗ ⎨ + − + ξ − ξ − ⎡ −ξ1)
⎤ ⎬
2 48 EcI ⎢ ⎜ ⎟
I
5 3 5 λ
⎣ ⎦⎥
⎝ ⎠ ⎩ ⎣ ⎝ ⎠
⎦⎭
(7.27)
If the value of c previously calculated is inserted in the (7.27) stating β=1 and letting λ
changing in the range 1≤λ≤∞, we obtain the curves reported in Figure 7.15, that show as the
increase of the ratio λ means a decrease for ξ1 and the increase of v(l/2) as a consequence of a
larger cracked part of the beam.
In the same way, a concentrated load Q=200kN, producing the same maximal moment in the
mid-spam section, leads to the following expression for the section displacement
⎛�⎞ M l ⎡ ⎡ 3 3β
⎤⎤
v⎜ ⎟= 1+ ( c−1) 1−8ξ − ( 1−2ξ )
2 12E I
⎢ ⎢ ⎥⎥
⎝ ⎠ ⎣ ⎣ ⎦⎦
in this case
2
max
*
c I
1 2
λ
1
�
2
v1 = Mmax ∗
12EcII (7.29)
(7.28)
ξ1=1/(2λ). (7.30)
The corresponding curves are reported in Figure 7.15. We observe as the displacements in the
two cases of distributed and concentrated load are respectively 0.93 and 0.88 of the
displacement calculated in the stage II. Furthermore, for the same Mmax, the displacement in
case of concentrated load results to be lower because the linear trend of the relative bending
moment is associated to a smaller region of the cracking beam with respect to the case of
distributed load, that is characterized by a parabolic diagram of the bending moments.
Fig. 7.15. Diagrams for v/v 1 , ξ 1 - λ.
The same problems can be solved in a generalized form evaluating numerically the
displacement following the procedure expressed in (11.51). In this way, it is possible the
evaluation the deformation of the whole beam, varying z . The result, for a distributed load