EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 7-8
EXAMPLE 7.3 Evaluation of crack amplitude [EC2 clause 7.3.4]
The crack width can be written as:
σ ⎡ σ s s,cr ⎤ ⎡ φ ⎤
wk = ⎢1− ⎥⋅⎢k3⋅ c+ k1k2k4 λ⎥
Es
⎣ σs ⎦ ⎣ ρs
⎦ (7.1)
with
λ ⎛ ρs
⎞
σ = k ⋅ f ⎜1+α ⎟
ρ ⎝ λ ⎠ (7.2a)
s,cr t ct,eff e
s
⎡ 1−ξ 1⎤
λ= min
⎢
2.5( 1 −δ)
; ;
⎣ 3 2⎥
⎦ (7.3)
Assuming the prescribed values k 3=3.4, k 4=0.425 and considering the bending case (k 2=0.5)
with improved bound reinforcement (k 1=0.8), the (7.2) we get
σ ⎡ σ ⎤ ⎡ φ ⎤
= ⎢ − ⎥⋅⎢ ⋅ + λ⎥
⎣ ⎦ ⎣ ⎦ (7.4)
s s,cr
wk 1 3.4 c 0.17
Es
σs ρs
The (7.4) can be immediately used as verification formula.
As an example let’s consider the section in Figure 7.6
Table of Content
Fig. 7.6. Reinforced concrete section, cracks amplitude evaluation
assuming αe =15, d=548mm, d’=46.0mm, c=40mm, b=400mm, h=600mm, M=300kNm,
As=2712mm 2 (6φ24), As’=452mm 2 (4φ12), fck=30MPa, fct,eff=fctm=2.9MPa Referring to a short time action (kt=0.6). It results then
β=452/2712=0.167, δ=548/600=0.913, δ’=460/600=0.0767,
ρs=2712/(400 ⋅ 600)=0.0113
And the equation for the neutral axis yn is
−400
2
yn + 15⋅2712 ⎡548 − yn + 0.167( 46 − yn) ⎤ = 0
2
⎣ ⎦
and then