EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 7-15
The lowest value for λ has to be chosen between
λ = 2.5 (1 – 0.874) = 0.315; λ = (1 – 0.3918) / 3 = 0.2027
Then s,cr
0.2027 ⎛ 0.01062 ⎞
σ = 0.6⋅3.086⋅ ⋅ ⎜1+ 15⋅ ⎟=
63.11 MPa
0.01062 ⎝ 0.2027 ⎠
304 63.11 0.2027
wk = ⋅
⎛
1
⎞ ⎛
3.4 50 0.17 26
⎞
0.306 mm
5 ⎜ − ⋅ ⋅ + ⋅ ⋅ =
210 304
⎟ ⎜
0.01062
⎟
⋅ ⎝ ⎠ ⎝ ⎠
k w = 2/3 , ρ s = 69.03 / (50 · 100) = 0.0138
1000 2
− yn + 15⋅6903⋅( 437 − yn) = 0
2
y + 207.1 y − 90498 = 0
2
n n
2
yn 103.5 103.5 90498 214.6 mm
=− + + = , ξ = 0.4292
3
∗ 1000⋅ 214.6
I y = + 15⋅6903⋅( 437 − 214.6) = 8.41⋅ 10 mm
n 3
6 437 − 214.6
σ s = 15⋅600⋅ 10 = 238 MPa
9
8.41⋅10 λ = (1 – 0.4292) / 3 = 0.1903
Table of Content
2 9 4
0.1903 ⎛ 0.0138 ⎞
σ s,cr = 0.6⋅3.086⋅ ⋅ ⎜1+ 15⋅ ⎟=
53.31 MPa
0.0138 ⎝ 0.1903 ⎠
238 ⎛ 53.31⎞ ⎛ 0.1903 ⎞
w k = ⋅ 1 3.4 50 0.17 26 0.213 mm
5 ⎜ − ⎟⋅⎜ ⋅ + ⋅ ⋅ ⎟=
2 ⋅10 ⎝ 238 ⎠ ⎝ 0.0138 ⎠
k w = 1/3, ρ s = 111.51 / (50 · 100) = 0.0223
1000 2
− yn + 15⋅⎡9558⋅( 437 − yn) + 1593⋅( 385 − yn) ⎤=
0
2
⎣ ⎦
y + 334.5 y − 143704 = 0
2
n n
2
yn 167.2 167.2 143704 247.1 mm
=− + + = , ξ = 0.494
3
∗ 1000⋅ 247.1
I y = + 15⋅9558⋅( 437 − 247.1) + 15⋅1593⋅( 385 − 247.1) = 1.06⋅ 10 mm
n 3
6 437 − 247.1
σ s = 15⋅600⋅ 10 = 160 MPa
9
1.06⋅10 λ = (1 – 0.494) / 3 = 0.1687
2 2 9 4