EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 7-6
A = 0.00208⋅300⋅ 1800 = 1123mm
s,min
Table of Content
2
this reinforcement has to be put in the web tied area with height over the bottom slab a =
1800 – 809 – 300 = 691 mm.
We use (5+5)φ12 mm equivalent to 1130 mm 2 .
Referring to the bottom slab we get
1−9 8
α f = 0.812 >ξ
and it follows:
2(1−0.4494) −0.1667
ρ s,min = 0.01235⋅ 0,45 = 0.00943
1− 0.4494
A = 0.00943⋅300 ⋅ 1500 = 4243mm
s,min
We use (14+14)φ14 mm equivalent to 4312 mm 2 .
The reinforcement scheme is report in Figure 7.4
2
Fig. 7.4. Minimum reinforcement, case (a).
Case b)
The cracking moment associated to the axial force N = -6000 kN, with eccentricity eN=1800- 809-250 = 741 mm derives from the relation
⎡ N⎛ A ⎞ ⎤
M = ⎢− ⎜1+ e ⎟+
f ⎥W
⎣ ⎝ ⎠ ⎦
cr N ct,eff i
A Wi
and then:
3 6
⎡6000⋅10 ⎛ 741⋅1.825⋅10 ⎞ ⎤
8 −6
Mcr = ⎢ 1+ + 3.8 7.25⋅10 ⋅ 10 = 9585kNm
6 8 ⎥
⎜ ⎟
⎣1.825⋅10 ⎝ 7.25⋅10 ⎠ ⎦
the eccentricity of the normal force in the presence of M cr is then:
3
e =−9585⋅ 10 6000 + 741 =− 856 mm
and the neutral axis position results from the relation